thing was you know different things for different places and and the facebook thing was always going to be the videos were more going to be on youtube and the facebook was going to be more for longer form written content it's just that i haven't got around to writing the long form content i do intend to use um uh javis to do some writing and stuff like that but i haven't i haven't signed up for it because i know that i haven't got the time to devote to it yet and i know that you don't need a lot of time for it but there's all of these tools that you hear about and uh yeah it's it's one of those things isn't it there's there's loads of tools and everyone will say oh you've got to get on this you've got to do this you've got to use start using the script and all these sorts of things and it's it's great you have got to you know but it's there's a time and a place and i just don't have time to sort of get into the more yet but uh the hdmi plugs yes uh those i've been there that has been quite a popular video actually that one that i did about the little uh i've usually got one to him actually the little hdmi dongles yeah so um that is that is one of those one of those videos that just seems to keep going i always noticed like in my amazon stuff uh there's a lot of those that have been gone through my amazon links for those little hdmi dongles uh i could write a the the twitch processing i certainly could write a little keyboard my stro thing to do that you're right rich but i'm almost thinking what's the point do i need to be growing do i want to be focused on growing a video based audience on multiple platforms i know that you've then got the thing that like all the things changed on one prop platform you've got the other one but um yeah i'm just i'm just not sure that i need to even be there to be honest is is the the thing that i'm thinking about is that yeah do i actually need even need to be on twitch what's the reason for it is it just uh uh was it just like a vanity thing in the beginning like oh i can be on all these different platforms i don't know i think it was more just a case of i found out about restream like as i was setting up the channel and thought oh this is amazing you can stream to multiple platforms um and so then i just added in all these other things that i was never intended to be on um so yeah i think it's for me it's just a case of what's what's the point in being there i suppose um so yeah i probably will just actually switch that off i just won't bother scheduling it to go there next week i might keep the twitter one for the moment because um i do intend to do more stuff on twitter so i actually intend i actually intend to be active on twitter a little bit more active than i am but yeah i tend to be a little bit more active there so that's probably the route i'll go down uh and very well thank you very much steve thank you for asking um oh no that's terrible a burst pipe uh things involving water in the home are never fun are they i'm really sorry to hear that uh orbri oh no well i hope it's uh i hope it's not not done too much damage and uh you can get that all sorted out swiftly yeah it's never never never fun so uh steve so you're streaming to uh you do you know what i've just just realized this is ridiculous i've got my teleprompter right there and i've got my comments just down here so i can look down here and i can glance at a comment and i can put it up on the screen but for some reason i'm still reading them off the econ window over there what an idiot what's the point of having a teleprompter right up here with my output and everything and the comments on it and then looking over here to read the comments i don't know silly billy this is much better isn't it i contact guys i contact i stream to youtube linkedin it's even easier to read here than there so i don't know what i'm looking over there for i stream to youtube linkedin facebook and twitter i've decided to make content specific for linkedin and twitter but it's all about workflow exactly my thing it is yeah uh kill youtube you might not be the guy but i would like to see someone who has a perfect way or reason to use everything yeah very sorry to hear about that orbrie i'm now i am looking down at the comments but as soon as i pop them up there i'll flick through them uh i'm glad i'm glad it's not just me steve i'm glad it's not just me i've got no idea i i should probably just actually cover this window up here go i'll if i move because i've also got is i've got my uh i've got my youtube comment pop out as well just in case i want to drop a comment into the chat then i've got that open so if i cover my program window now i actually can't see my e-con window that's what i should have done from the beginning it also gives me some some more space doesn't it so that's that's solved that problem right well that was a bit of a rambling livestream i don't feel i kept much on topic throughout that but there you go it was supposed to be about my my productivity of the week my new tools that i've been using and then also obviously the the course and stuff that i've got uh got planned my my zoom course uh but it was a little bit of a rambling one that was so i'll definitely have a bit more of a think though miles about these uh sort of q and a sessions where people can come and just bring specific problems and stuff like that related to the stuff that i talk about on my channel uh that might be a good idea for a different sort of date with a q and a uh i'll tell you another thing that i've um i've not been uh not been watching which i've got to get back into is um uh greg's zealous app so that's another thing that i've been meaning to uh meaning to spend more time on i mean i went and settled my account before it was uh was it when it was in beta or beta i can't remember but um anyway yeah that would be a great thing for actually doing a sort of q and a where people could come on and actually just sort of share their screen walkthrough stuff and things like that so uh yeah uh i think greg might have dropped out by now but definitely zealous is still on my uh on my list there as well of things to get to get used to it's a great great platform that is so he says having not really got into using it properly but i do see the uh yeah the benefit for doing q and a is that way as well uh wait so what do you feel is uh your most valuable hardware and software well that is uh that is very um specific to different things i'll tell you first of all the hardware um stream deck is just a total game changer so that is something that i use in all of my workflow through all of my work now i got it for live streaming or for video uh but i use it for everything so stream deck without a quite doubt i'm still evaluating loop deck and when i say i'm still evaluating it what i mean is i've had time in the past two and a half weeks to finish setting it up to actually begin to evaluate it but uh yeah certainly stream deck is then in terms of uh software and if you talk about hardware in in terms of um yeah minus minus ecamp yeah um but in terms of the um like the the hardware in my studio the first thing i bought was this mic the short mv7 and i do really like that because i like the fact that it's got a usb and the um xlr so i can plug it straight into my computer it's semi-mobile so i could throw it in a bag and take it somewhere as well without needing to have any sort of external interface or something like that and it just made a biggest difference in terms of you know the quality of the uh the videos the course material that i was producing outside of this channel but then obviously for the channel and then on zoom calls and stuff like that i'm still using my 10-year-old canon eos 60d when i get my whole studio setup i will be getting a sony zv e10 um uh but for the moment i'm still using all camera so uh but yeah for me hardware sitting on the desk is the stream deck now that i'm using my ipad mini uh that is going to the new mini because i've been on the sort of ipad pro but i just wasn't using it for um for years because uh it's not didn't fit into my well first of all it didn't fit into this bag which i take everywhere my little peak design tech pouch but the ipad mini fits in that i can take everything in this bag with me throw it in my the car whatever when i'm going out carry it around with me and it's basically got my little mini office in it now as far as i'm concerned so um uh but in terms of software so there's there's there's a different apps really so obviously apart from ecamm which are useful of this there's there's sort of like core power apps for productivity uh and there's there's some that are kind of i think of like the things that allow me to get my stuff done so that would be things like keyboard maestro i don't know what i'd do without keyboard maestro for all the stuff that i have to do text expander don't know what i do without text expander it saves me a load of time and then i do still have some things that are left over in automator and stuff like that which is the mac automation app so those are kind of like you know they form the backbone of everything i do use dropbox as well uh that i know that um you know there's pros and cons to it but i still do use that so these are kind of like support utility apps if you like that i find uh to do lots of stuff but then there's that like my um the the sort of the actual tasks of planning and stuff like that so i use um omni focus for my task planner i would be lost without that that is what i use to uh use the uh david allen's getting things done system to get all my work done and stuff like that um so and then there is my notes taking system so i use nv alt to manage all of my text notes which i mentioned uh talked about all this at the beginning actually but yeah so omni focus for task planning nv alt to manage all my tasks uh not all my text notes rather text files i've got about 40 50 000 text files and then now i've just started adding in notion as of last week i've just started looking into using notion um but then for the rest of the thing it just totally depends on what i'm doing uh most valuable hardware and software excel i i've been using excel for you know you could measure it in decades now so uh it's probably like what's nearly it's coming up to 30 years i suppose because i'm 43 i started using excel when i was like 15 or whatever it was at that time so yeah 30 years into it now uh and i do use you know there's some people who uh not disparaging anybody here at all uh there's some people who use excel like for just some you know creating basic spreadsheets and stuff like that i would definitely put myself in the the sort of power user bracket of excel do lots of really heavy sort of maths and stuff like that in it mathematics um uh use things like power query and stuff like that which is a bit of a game changer in excel terms uh to be able to you know reference sequel databases and all this sort of stuff and bring it all into excel uh yeah do some quite heavy sort of computational stuff with uh with excel so excel's a real real big one for me as well i spend a lot of my time in excel uh although not obviously for the channel a couple of people had asked about excel and if i could do some excel videos the fact is uh that it would be pointless to do excel videos on my channel because uh i would need to create a whole new channel for excel excel's just such a deep subject and to just throw in a couple of random little examples of the sort of stuff i do with it would be uh probably out of place on my channel so yeah won't be doing any excel tutorials on my on my channel um but uh yeah there will be some notion videos i didn't know how it was going to fit into my workflow but then once i started getting into it i could totally see how it would so i will be doing some uh uh notion videos probably not this week i need to get a bit more into it and a bit more into my workflow uh but yeah as soon as i have i'll i'll get those uh those videos out but yeah so then and then there's other bits of software so for example uh i don't really do this anymore as such but so for 10 15 10 years 12 years uh we've run an architectural design and construction company here in thailand and so we've built uh like i think 40 houses something like that um and designed countless others and designed restaurants hotels all this sort of stuff uh and so that was what i actually was doing you know day to day the most for you know best part of a decade and so for that i was using architectural software i had for and for that my hardware was my 3d mouse that could manipulate models and all this sort of stuff in you know in all three dimensions and stuff like that so yeah different hardware and software for different tasks really but um yeah certainly for the the planning and stuff notion is is one of those that's up there one of the things that i'm going to do for vlogmas by the way while we've been talking about vlogmas um is the idea is for my idea for my vlogmas videos they are all going to be on my youtube channel um however they're going to be a um a bit different to my regular stuff and the idea is that it's going to be a bit more about like my uh i suppose a bit more about my background and the way that i use these things so i will be giving some examples of like the way that i've used the technology in my architectural design stuff uh in the trading stuff that we do and then also in the social media uh marketing and advertising company stuff as well so how i'm using these tools and like using the video to the ecam uh live to create videos for those different things uh is is the sort of things i'll be covering and then just some sort of general background to give an indication as to how i became the sort of i describe it as happily broken person i am uh and anybody in the uh in docs group will understand what i mean by that we're all a unique bunch of people uh united by our uh our common flaws i hope that's uh that's not offensive to anybody but uh yeah so um i suppose that now would be a good time to plug actually by the way if you are not aware of doc rocks uh office hours group then head over to buymeacoffee.com slash not take one tech but doc rock and there you can sign up to uh to office hours in fact let me just quickly get that up shall i because uh it's such a great bunch of people and uh the same time it's about the same time as my uh my live stream we all uh all meet up for office hours and let me just get up here one second uh here we go the drop squad i can take this one down actually now as well while i'm on here there we go so yeah buymeacoffee.com slash doc rock and then look at this you can sign up to the the membership uh getting quick before the uh the the price changes because this is seriously underpriced at the moment and uh yeah sign up there and then it's every week we have a weekly zoom call um it's scheduled for two hours they're regularly three hours though and uh yeah it's just a awesome group of uh people all creating different things all from different uh different backgrounds different professions different walks of life and everything but all there with the same goal to actually get stuff done and so yeah i've got a lot of respect for doc for creating that community it's not a community it's a family let's let's get clear about this about terminology here it is a family and we're all there looking out for each other then helping each other so i know that most of the people are in this these comments are in it anyway uh but just for those that might be dropping by and not then they're definitely go and check that out well guys and girls i think that this is around about the time for me to wrap it up because uh i have heard my little kids are up and about upstairs which means it's time to uh go up and do my uh my daddy duties but thank you all for for stopping by it's always uh always great and thanks for all the great questions and comments and stuff like that it's uh it's an absolute pleasure to chat with you all and uh yeah i will be back i'll be back obviously in the week with videos but i'll be back this time next week to give you an update on my course i am going to for those who've joined late i'm creating a zoom course and i'm going to sort of go through the weekly update of where i'm at in terms of creating the content getting it online and all these sorts of things as well so i'll be sharing more of that progress with you next week but for now though for those watching on the replay i will stick up some uh some other related videos over on the right hand side and to all of those watching live i'll see you all later oh yeah i'll probably see you actually thanks rich i'll see some of you in office hours tomorrow have a great day everyone bye bye


Hi everyone, Sean with jaspyscasebreaks.com here doing a 2020 mosaic football hobby one-pack break random team number 12 Where we're giving away eight spots in NT one box first off the line break itself So 32 spots when gets a random team in the NFL one-pack break 2020 mosaic hobby No vet common ship at the end of the break will take all 32 customers names randomize them top eight names When a spot in the 2019 first off the line one box break So here is our list of customers from Ryan down to Mike and The teams from the Cardinals down to the Washington football team this dice roll and randomizer is for the pack break itself There will be separate dice roll and randomizer at the end of the break to give away the eight spots So three to five eight times on both spots eighth and final time from Ryan down to Brett So you can see eight times there eight times on the dice now eight times on the teams eighth and final time From the Lions down to the Denver Broncos Eight times you can see there eight times So we're not gonna read through every single one of these since it is a filler But we'll stop for some of the highlights Chargers Brandon Hall chances at Justin Herbert bangles Joe Burrell Braden Hall Hackers Jordan Love Sarah Bernard Eagles Justin or sorry Jalen Hertz Damien Kovaleski Dolphins to a Brian Heyman Steelers Chase Claypool Jeff Shea Vikings Justin Jefferson Alec Harrison There you go. So let's print this out rip and then give away those thoughts So here you go. Here is our hobby pack There's the final list Nice Joey Joe mixing silver for the bangles Braden Hall Charles Woodson red for the Raiders Damien Kovaleski Jalen Hertz NFL debut for the Eagles Damien Kovaleski Jonathan Taylor for the Colts Zachary Goldberg Mahomes MVPs for the Chiefs James Connell and Jordan Love base for the Packers Sarah Bernard nice pack Pretty pretty good So that was the pack now. Let's give away those spots member top eight Get spots in the one box and T first off the line break. So here is the customer list again from Ryan back down to Mike dice roll Top eight names after a two and a five seven times Seventh and final time so seven times on the randomizer seven times on the dice Top eight so some spots dying down in 32. Thank you guys very much for getting in on this and hopefully I can do better for you Next time there is another filler in the store now go get your spots so we can do that for you later But special congratulations to Robert Stowell Brandon Hall Alec Harrison Sarah Bernard James Connell James Connell Tony Barkie Brett Gibbs You eight now have spots in the first off the line and T one box break Seven times top eight names seven times on the dice and there you go. That was 2020 mosaic football hobby one pack break random team number 12 from jaspercasebricks.com. I'm Sean jaspercasebricks.com Thanks for hanging out. I'll see you next time


Hey America, I'm heading your way in September. I'm gonna be in two places primarily, in Washington DC and in Dallas. In both those places, I'll be bringing my new live event called the Seven Levels of Authentic Communication. I recently toured this event in Australia and Indonesia and I've sold our events in our seven city tour. And so I figured, you know what, why not bring this event to America? A country I love so much and want to invest in you, emerging and experienced communicators in that country. Hope you can join me in September, link in bio for more information. I hope to see you guys then. All right, cheers.


Welcome to the Sheikot Africa screen shot updates project meeting, it's 25 April topics that I had on the list while we're waiting for our participants to arrive. DT and I were going to discuss automation ideas for screenshot updates and talk through different concepts and things. This is not the focus of the Sheikot Africa project at all, but it might align with it. We're going to identify it aligns with it because we're identifying screenshots in this project and aligns because we are updating the screenshots interactively. So we'll have to go through steps to reach the point where we can take the screenshot. So Aditya here's here's some conceptual model for me which was okay the documentation includes a screenshot right that is preceded by some some words explaining the screenshot or its contact and that you'd expect that. So that's that's how we do screenshots. However, what it that those words are not precise enough to tell a computer how to do it to how to how to duplicate that they and they can't be they naturally cannot be because we write for human beings. Not for we write for people not for machines right the documentation's purpose is not to satisfy a machine is to satisfy a person. So then the question is how do we represent. How would we represent those steps, the machine executable steps to reach the screenshot. There are some additional questions like how would we expressed that a portion of the screen is needed, not the whole screen, because in many of the examples in the screenshot sheet, we can see let's let's jump to one and take a look at it so here. It says that the health icons have changed so when I opened that page. We're going to see okay this one is the whole screen. So that one mechanically I could see how we could grab it. These individual pictures of the health icons are not the whole page. They are they are something different and in fact they might be better. Let's let's phrase this is another way. Some screenshots aren't screenshots at all some screenshots. A few of the screenshots are images in the Jenkins source code. So maybe those we just say hey that's not a screenshot at all it should just extract the image from the Jenkins source code that is used for partially sunny that's used for cloudy whatever that image is should be should be embedded here. If these may be up to date I'll have to look to be sure but so so that's the that's the idea then is okay how do we how do we get that information into an out into a machine into an automation system. Now back to the first question how would we represent machine executable steps. So Python's PI doc facility invites developers to write tests in the documentation and and they do that by by expressing the steps in a machine readable format. So that's one alternative another might be behavioral behavior driven development uses a encourages the creation of creation of a domain specific language, dedicated to the specific tasks, the specific product being tested, and the tasks in that product. So it might be things like let's see let's let's grab an example. And we could probably since Jenkins is Java we could use. Here we go. Let's let's take a look at an example right here we go. This one is a lot of fun because it shows exactly how to do. Here is the. Here is the description that they use to describe precisely. Oh, thank you very much for the ad describe precisely how to how to prepare how to prepare for and execute this. So that it's written in English English language adding an item to order I want to be able to add an item to a current order scenario describes what we're going to do before we start the test. I have not yet ordered anything as a pre pre condition then I go to the burgers category, select a cheese burger, have a new order and the order has one item in it so this is described and then it gives data that's used to describe it so there's a category in the category sandwiches I have the item a chicken sandwich price $9. And now we see as they take us through it how that how that gets implemented. And they really are creating a testing description language, which I think could also be a screenshot description language in this case is one idea of how this kind of thing has been done. Now I did you I've been talking a lot. Does any of this resonate for you or give you a sense of ways we might approach it. Yes, definitely. So I was not aware of the pie dock thing. Look into it. I have done some behavioral development development we did in the past and familiar with cucumber what exactly what you showed me. So good. Yeah, so that's something I'm familiar with and I never thought that it could be used for something like this. So that's some a new way to look at it and yeah that's something again I'll definitely look into when you talked about screenshot automation the first thing that came to my mind was actually those end to end testing libraries in the front end that we have. So, in front end like if you want to test your login page for example, it would just pin up an instance it will actually type in the login it will, you know, automatically type in the login credentials hit the login button. It will actually happen that way. So I thought if we have a part to actually go to the page where the screenshot is in the Jenkins deployment we can actually write an end to end test like an interactive test as you are mentioned that actually does that and there we can have that in and this is something hypothetical. I'm not sure how will you capture a particular section of the screen taking the screen is an easy task as you mentioned taking screenshot of the whole thing. But yeah we'll have to see how we can do a section of the screen and then that can be basically will be the updated screenshot and we can just update it. So that's one thing that I thought of. I've used Cypress and all the open source tools that are there Cypress.io in the past and I think we can give it a try as well how easy or difficult to write something. And Cypress.io is an example of end to end test automation right with a web browser is that correct I'm not experienced with Cypress. I'm I had used Selenium once in the past Selenium. And Cypress is conceptually something similar right where it's, it's actually driving a web browser and while driving the word the web browser. I know at least Selenium has the facility to take a picture at the end to show people what something failed as so I think it's that kind of thing. Yes, Mark you're right. Now, there was another one that I had heard of called Zikuli. That is, it is, is, if I remember correctly isn't a test framework that actually compares screenshots. So it's it clearly must be taking screenshots because it's comparing them. It's, it's a yet another way of considering that same thing I think. Let me look to be sure is it did I spell it correctly. Test automation. Web test automation tooling. Nope. Oh, I spelled it wrong. Okay. Spelled with an S not an S is the Zikuli. Here it is. Okay. Zikuli is a is the same kind of thing only it. It uses very much if I understand correctly it absolutely uses pictures, and it compares images so it's asserting the context contents of images. So it's quite different from Cypress where they're, I think they're inserting a certain things inside the browser. Right there they're asserting is the DOM correct and is the is the is the the object model inside the browser correct. And that has all sorts of strengths and power to it. Good. Okay. And then, yeah, certainly selenium. And, oh, there was a behavior driven thing around that. Oh, Ruby. What is it called the Ruby version there's a Ruby thing that does Brian Merrick did it just a minute and a long time. Brian Merrick Ruby web test automation. I don't remember the name of it. I hadn't heard of that one. Ruby Ruby web testing framework. I should remember this. I apologize that I don't that's really kind of sad. Not a copy of water. That's the one that I had seen before. Okay, so water. And you said Sahi or Sati. It's Sahi. But I haven't used it. It's just something that came up in a website sometime back. Ah, okay. All right. So, so it's Sahi is another is test automation for for desktop apps. Yeah. Okay, all right. Okay. Okay, I think probably not something that we end up using. Well, again, it does play back on browsers. And so that that could let us capture the screenshots. Good. So, so those were the things that were on my mind as possible. I'm sure that Gavin Morgan has other ideas and his, he's certainly got a lot more experience in this area than I do, but, but for me the, the, the question of approach how do we, how do we represent the machine executable steps is the is the bigger challenge, right? It, if we represented them as, as you said, as interactive test automation, that gives us interactive tests that are now regular being run and could be used in more places than just documentation generation. Right. So the benefit of that technique usable in other locations, not just test automation. There's a screenshot automation. There's a, there's this, the acceptance test harness for Jenkins that does something like this, but I don't think it's using any of those frameworks. Just a minute, let's take a look at it, because it might be that we could conceivably. So it does it using Selenium, right? Let's see, where is it? End-to-end test suite for Jenkins Automation Server and Plugins, tests controlling Jenkins under test through the UI and REST APIs. So it has the benefit that it knows about Jenkins REST APIs and some of those REST APIs could be much more helpful to do initial configuration than trying to drive through a web browser. So if we need, if we need certain objects created, it's probably much faster and easier and more reliable to do them through a REST API than it is through a web browser. So, so let me put a link to this one. And where is it that it says it uses Selenium? I just, oh yes, here it is. So this might be another, another approach where we say, hey, we want to get two gains from this is we'll get a few more tests for the acceptance test harness. And these things happen also as a side effect to generate screenshots that we can use. And now, I don't see any mention here of, oh, debugging tests in a container. Does it, I think it may. No, it doesn't show us how to grab. Oh, it shows how to run the debugger. So you don't have to have to be and it assumes you're looking at Firefox locally. So it's not attempting to grab you a screenshot from somewhere else. Let me make a note of that. Okay, so Jenkins acceptance harness uses Selenium for end to end testing. So, so for me that might lobby that we should consider Selenium, if only because we've got people who have created a significant body of tests and Jenkins acceptance test harness. And if you look at the source code, the tests, there's quite a collection of tests. All right, so let's see for instance, well, what have we got here maybe the best acceptance. You know, there aren't a lot here. I thought we had a bunch that were being run by, by, by others maybe I'm there on a different branch or I need to have a conversation with people to be sure I understand. But it's for me acceptance test harness might be a place that we would consider. Hey, might be an effective way to do it because it already exists and we've got people with skills, and it's using Selenium as a known technology. I agree with you on that part. Selenium has a community support right. Right. All right, so those were, those were the kinds of ideas I had anything else you wanted to add before we close our meeting Aditya? Nothing else that comes to mind right now, but I'll go through these and probably you can have another discussion next time. Great. Thank you. Thanks very much. Thanks for being here today. I'll save a copy of this recording so that it's available and get all the other recordings uploaded that I'm supposed to have uploaded. Thanks again, Aditya. Thank you. Bye bye.


A picture is worth a thousand words. A picture is worth a thousand words. When you look at this image, what are the first words that come to mind? Victory. Victory. World War II. Returning. Returning. How about this one? The eyes. The eyes. How about this one?   That's a good one. How about this one? Nice one. Ooh, it went from here to here. A picture is worth a thousand words. Why should this matter to your business? Do you use photos on your website and social media? Are those photos helping or hurting your business? Which of these is $1,000 a night? Their words. The Buckhead Luxury or the Luxury Home. Show of hands. This one or this one. They're both $1,000 a night. I'm thinking to change that. So now let's choose between these two restaurants. Which one are we going to go to? This restaurant. Oh, y'all are pointing this way. Okay, this restaurant. Your photos can turn people away, which is loss of customers, loss of referrals, trust, credibility, loss of revenue. Or your photos can inspire people to do business with you. Say you need to have surgery using these actual photos from LinkedIn and hospital websites. Let's choose a surgeon. Are we going to go with the recent graduate? The fisherman. How about the lady having drinks at TGI Fridays? All too often we choose photos that are good enough. We need a photo so we take a quick snapshot or grab any old photo that's available. Not realizing the message that it's sending. People just aren't searching for hotels and restaurants and doctors. We're searching for our businesses too. 80% of people will research you and your company online before making a decision.  So while it said we shouldn't judge a book by its cover, the reality is we all do it. It's just how our brains function. Imagine searching online for a recipe without the photos. Imagine a cook for dinner tonight. Hmm. It's the photos that make us curious enough to want to click to learn more to read more. Imagine Amazon without any photos. What are we going to buy? It's the photos that make us curious to click, drive that connection to learn more to read more about what is being offered. We're just the perfect words and just the perfect font. I spent two days picking out a single font for my business card. But we can't spend all our time on our words that we completely forget our images.  Visual information like those photos get to your brain 60,000 times faster than text. Here's an example. The deepest part of our brain, the survival part, it's hard wire to form an impression the instant we see someone. Your subconscious brain makes a conclusion about a person in less than one half of one-tenth of a second. So how can we win someone over in less than a second? What's going to make somebody searching on the internet stop in all of the sea of people that are available and say, I want to do business with this person? The simple answer? A connection. People want to do business with people that they like, know, and trust, not faceless websites, especially when it comes to a service-based business. How are we going to make that connection? We can do it through our headshots, our profile photos, those all-important business portraits, which are commonly referred to as a profile photo or a headshot. That headshot is the most important photo on your website and social media because it's the photo that represents you. It's the photo. It's the first thing that people see when they browse your profile and it's the mental image they are going to recall. Remember, we are hard-wired to form an impression of the instant we see someone. We can't help it. So let's look at some common missed opportunities that people do with those all-important profile photos. From a psychology standpoint, arms crossed say, I'm closed off. Leaning away says, doesn't care, makes us less approachable. Too far away hints at insecurity. It's hard to establish trust with this person. Too close. Even in a photograph we don't like it when somebody is in our personal space. Too high asserts dominance. While too low shows weakness and lacks confidence. And the selfies often have too many distractions, especially the bathroom ones. Think about all the information that is in front of our eyes, colors, shapes, lights, movement, words, texture, all vying for our attention. Remember, you have less than a second and anything that distracts the viewer is going to take away from establishing that connection. The opposite of the selfie, the everybody photo, which one are you? There's no personal connection. And my personal favorite, the disembodied friend. There's no connection because we're all wondering who that person was. And covering our eyes, there's no connection. What are we hiding? And looking away, we tend to follow the gaze of others. We break that connection. And wide eyes makes us uncertain and eager. And tired eyes, there's no confidence. It's hard to trust this person. And clothing says so much. This is, I would rather be anywhere else than doing business today. These people, actual LinkedIn photos would rather be anywhere else than doing business. Poor lighting creates wrinkles. It ages us, makes us less trustworthy. Seriously, nobody wants that. And a fake smile, it doesn't generate trust. Most people can spot it. But we have an opportunity to make a connection in less than a second. How can we do that with a single photo? We do it through that all-important headshot. We narrow it down to just our head and shoulders because it reduces unnecessary distractions, giving us a better chance of establishing that instant connection.  So for this type of photo, look directly into the camera because it communicates credibility. And a warm expression says, I'm open to connection. And a genuine smile makes us more approachable and has been said to generate trust instantly. And good lighting makes us more trustworthy, more approachable, and more radiant. And a good photographer can take it a step further. Because while this is reality, that is marketing. Reality, marketing. Reality, marketing. We live in a highly competitive world. And what you market with matters.  And marketing is about standing out from your competition. Connecting with and engaging your target audience. Building a brand reputation and boosting your word of mouth.  The right photos can actually give you a competitive edge. But you have to make the effort when it comes to that all important headshot and profile photos and all those photos on your social media. To stand out from the competition and connect with your target audience. Think about this.   Why? Because we care about the impression that we're giving to the person sitting across from us. Your headshot is that first impression. And online it represents you 24-7-365 so you have to choose it carefully. How do we do this? Here's just one way. This is something that I do with my clients during brainstorming sessions. Those all important strategy sessions. This is the time we're going to narrow this down to just two. I'll put that back up at the end. Fill in the blanks with at least three positives about yourself and your business. What is it that you're wanting people to know about you? To tell others like oh my goodness you need to do business with this web designer or this person because they are fill in the blanks. The words that I most often hear from my clients are I want to look approachable, friendly, and professional. These words are going to vary based on your profession, your unique personality, and what it is that sets you apart. For example, a defense attorney might want their words to be confident, no-nonsense, and top of their game. Whereas a veterinarian might want their words to be friendly, respectable, and knowledgeable. What are your words? Write them down. Then show your headshot to five to ten people preferably clients, your target audience, even random strangers. What is it that they say when they look at that photo? Write down their words and make a list. Once you have a long list it's time to pull back out your words and tally up all of the words. Do your words match their words? Or is there a disconnect? Are you showing the world what it is that you think you're showing the world? You might like your photo and it might be well lit. But is it connecting with and engaging your target audience? Even the slightest variation in facial expressions can turn someone away or inspire them to connect with you. Let's choose a headshot. By a show of hands which one do you prefer? There's no wrong answer here. A or B? What if I told you choose a real estate agent? A or B? What if I told you she was a book editor? A or B? How about an attorney? A or B? How many of you changed your mind based on her profession and her target audience? With every photo that you post on your website and social media you have to keep your target audience in mind. What is going to connect with and engage them? Then once you have chosen the right photo it's time to update it in all the places that your headshot is on the internet. It's in a lot of places. For the sake of time I'm going to go over just a few of these. Do you have a headshot on your about page? When somebody lands there to learn about you how is the facial expression and the body language and your distractions what does your photo look like on your about page? Have you looked at social media and put a headshot there? Because think about this our clients come to us from all different channels across the internet. Are you grabbing their attention and connecting with them and engaging them no matter where it is that they're finding you? Specialty websites. Do you have your services listed on specialty websites? What does your headshot look like on there? Do you compare to the competition that is offering the same thing that you are? My favorite have you thought about those job proposals? By putting a really good headshot in your job proposals you are putting yourself in the room connecting with and engaging the person that is making a decision about who to award that job to and reviews. By placing a headshot with reviews think about Amazon earlier did you want to click on anything without the photos? By placing a photo with your reviews you are increasing the likelihood that people are going to read them which increases your credibility and conversion rates. People come to us from many different channels your product or service is just one of many the consumer has to choose from. Many of you in this room offer similar services these are essentially all sodas they all market themselves differently one of them you are sharing with a friend another you are jumping off a mountains how are you marketing yourself are your photos connecting with and engaging your target audience those words that you wrote down are you conveying that are you expressing that it's important that you use images that are relevant and cohesive to what it is that you do for example these are both clothing companies clothing company A clothing company A is doing a really good job of using relevant and cohesive images clothing company B it's a little confusing about what it is that they are selling and offering you shouldn't confuse your target audience as a Tennessee we do this on social media we like a photo so we go ahead and post it but is it going to connect with and engage your target audience it's important that we use images that show what it is that we offer and the problems that we solve for example this is two architects if you are in need of an architect is it architect A or B that is going to inspire you that you are going to go oh my goodness what is it that they can do for me A is doing a pretty good job where B doesn't really show anything to do with what they are offering or the problems that they would be able to solve it's important that you use images that inspire connection and engagement so using this heat map tool which is a tool used to show user behavior red is where the viewer's eyes linger the most what is it that we can learn as humans we are attracted to the human face and we tend to follow the gaze of others so when you have a headshot and you are wanting the viewer to connect with you you want to look directly into the camera have good body language and a good facial expression when you are wanting the viewer to interact with something like a product or something else you want to turn your body or your head or your eyes towards that product like the all important call to action button the buy more sign up for this newsletter call us don't look into the camera for this you want to look over at what it is you are wanting the viewer to actually interact with your photos should do their job what is that as a business your photos are to entertain to educate to inform to inspire, provoke, engage how are we going to go about getting those images we can hire a professional we can purchase stock photos we can do it ourselves or we can do a combination of any of these above so let's talk a little bit about each of them stock photography this is the one earlier that got the ooo there's dos and don'ts of stock photography don't purchase cheesy images they just don't connect with most people don't select images for yourself make sure you keep your viewer in mind if you are posting what is happening in your office this is not what people do in an office environment day to day this is more like something that you would see in an office this is more likely to connect with your target audience don't select random images so this is on hundreds of companies websites while it might work well for the travel company not so much for the oil change company that it's advertising for don't choose an image before you've done your homework and research google has where you can right click on that image and go where else is this on the internet for example the girl in the car is on thousands of companies websites over 100 car rental companies over 100 car insurance companies remember the whole point of having a website is to show the world I'm different look at me do business with me it's hard to do that when you're using the exact same image as another company and especially your competition it's hard to do that when you're using the exact same spokes model as thousands of other companies so you have to do your homework before purchasing those images and I would be dropping the ball if I didn't tell you there's study after study showing that when you actually replace the stock photo with real life images of you and your company your conversion rates are going to increase here's two examples the top one is a moving company they swapped out the stock photo with the real life movers these are the people coming into your home and carrying away your worldly possessions when they swapped that out their conversions went up by 45% think about this who in here asked for the password from their clients they want to like know and trust you they're giving you their livelihood their password, their business they want to trust that person the bottom image was actually confusing while this laptop's a pretty good stock photo this was not an online conference that you would think of when you actually see a laptop it was an in-person conference so when they swapped out the image from the previous year's participants their sign up rate went up by over 40% and if you still choose to use stock photos at least use the ones with Vince Vaughn and Dave Franco just saying DIY do-it-yourself photography how many of you like doing photography oh yeah so there are so many resources out there where you can learn good lighting and good competition composition free you can learn online photography not like I was 20 something years ago where you're just scouring to try to find stuff don't go out and buy expensive equipment you have a really good cellphone in your pocket so if you learn to know what you're doing for those all important social media posts you can take really good compelling images with that cellphone because the best camera is the one that you have on you when you take images pay attention to the details right, is there a palm tree sticking out the top of somebody's head are there empty water bottles what is their body language what is their cough and leaning away what is their facial expressions really pay attention to the details of those photos and then stockpile your own images those words that you wrote down earlier you know what it is you're wanting to convey the world so as you just go about your day to day think about those images that you need and start taking and capturing those images so when you need them down the road you already have your own stockpile of images for my clients when you use your cellphone to take images go ahead and hit the favorites button so it goes into a separate folder so a month from now, a year from now when you need that image you're not scrolling through thousands of images to find that one for that blog post you already have it in its own folder and then as with anything just practice practice practice and have fun photography is fun and hiring a professional it's pretty easy to take a general list of what it is that you need first you need a really good compelling head shot how many pages do you have on your website four? now you're up to five images that you need take a couple of hours and knock that out research photographers ask around if you see somebody that has good images ask them you know who did you use find the right photographer for you look at their website facial expressions you know if a photographer ever asks you to smile smile pretend you have an emergency gather your stuff and leave because it's not your job it's the photographer's job to entertain you engage you and capture something genuine that people are going to connect with and engage with once you find the right photographer go ahead and make an appointment get it on your calendar it doesn't exist unless it's on the calendar and a good head shot should last you three to five years so go ahead and take an hour of your time and invest in your


Now, publisher of Aviation magazine Deli Mamoru has joined many cretis of the federal government action in banning Twitter's operation in Nigeria. According to him, NBC's order to media houses to deactivate their Twitter handles is a draconian, or is draconian rather, and against the ethical demand of journalists. He asks journalists to remain undaunted in the face of the neo-regulation. The choice of the president to join Twitter was also personal. It had nothing to do with the people of Nigeria, and he used it effectively, and I used it with him effectively, to promote him in 2015. And I still use it almost on daily basis to promote the activities of the president of the federal republic of Nigeria. I'm sure as of the time he ran into problems with Twitter, he must have acquired maybe over four million followers. I have acquired 1.6 million followers, which took me about 12 years to acquire. Nobody can ask me to wipe it out in one day. It's the tool of my job. The president has the freedom to delete himself voluntarily from Twitter, but he cannot delete 14 million others from Twitter because we don't work for the presidency, and he does not own us, and he cannot curtail our freedom from doing our jobs to the best of our abilities.


Thanks very much. It's a real honor to just speak on this occasion of Don's birthday. Happy birthday, Don. I missed doing math with you, and I hope to see you in person soon. It's been too long. But nice to see you in a little corner on Zoom. I thought I would talk today about a problem that goes back a long time, maybe five years, to a Vander Warden. It's a problem I really like because it's very elementary, elementary to state. And yet the techniques that have gone into it have been really varied, ranging from sieve techniques to algebraic ideas of resolvance, the determinant method. And what I hope to do today is to introduce this problem and then introduce a couple new techniques that can be into this area that can be used to attack this problem. So the problem concerns Gallup groups of random integer polynomials. Let me just state the problem for you again. I think many people are familiar with it. So the question concerns taking random monic integer polynomials with small Gallup group. So let E of H denote the number of monic integer polynomials, f of x of degree n such that each coefficient is bounded by h in absolute value. So it's like a big box of side h of monic integer polynomials. And we want to count the number of them for which the Gallup group of the polynomial is not sn. So we expect 100% of these polynomials to have Gallup group sn. That's celebrate irreducibility. But can we make that quantitative? That's the question. So what is, how many of these, so the number of sets polynomials is about 2h plus 1 to the end. And we're asking how many of them have Gallup group not sn. So they're clearly at least on the order of h to the n minus 1 such polynomials for which the Gallup group is not sn. The total is about on the order of h to the end, but we, but clearly they're at least h to the n minus 1 polynomials that don't have Gallup group sn, namely just set a n equals 0. So if this last coefficient is 0, then f of x is reducible. And the other coefficients can be anything they want. Their n minus 1 remaining coefficients, they can be anything they want. And that gives you a polynomial that doesn't have Gallup group sn. So they're at least on the order of h to the n minus 1 such polynomials that don't have Gallup group sn in this big box of size h, size length h. And in 1936, van der Rodden made this tantalizing and lasting conjecture that big O of h to the n minus 1 should in fact also be the correct upper bound for the count of such polynomials. That we've produced h to the n minus 1 of them, and really that's sort of producing the right order of magnitude. No other Gallup group will occur more than just these intransitive Gallup groups that come from reducible polynomials. The conjecture has been open for general degree, but it has been proven for degrees n less than or equal to 4. In van der Rodden's original work, he gave a heuristic argument for why it should be true for n equals 3. And this was made rigorous by Tiao and Dietman, not just for n equals 3, but also for n equals 4. So the conjecture is known up to degrees n equals 4. So there's been a lot of work on this problem since 1936. There's no way I can give the full history of the literature. But of course, Hilbert's irreducibility, as I said, implies that e n of h is little o of h to the n. So in other words, 100% harmonic polynomial is degree n are going to be irreducible and half Gallup group s n. And the question is, can we make this quantitative? The ideal would be big o of h to the n minus 1, but can we even do better than this exponent n here? And van der Rodden himself gave the first such sort of power saving bound on e n of h. So in 1936, van der Rodden showed that e n of h is big o of h to the n minus some function of n, but it's positive. So it's a power saving. And successive improvements to van der Rodden's bound were then given over the years since 1936 every, every few years, there was a new world record. So 1956, no block proved that e n of h is big o of h to the n minus 1 over 18 n times n factorial cubed. And then in 1973, this was a real breakthrough, Gallup introduced his large sieve, which has been so useful throughout number three, but one of his original motivations to introduce his large sieve was this problem. And 1973, he proved using his large sieve that e n of h is at most big o of h to the n minus 1 half, that's epsilon. So in other words, halfway there from n to n minus 1, he achieved using his large sieve. And as a comment in these problems, it always takes a lot of work to, I mean, this plus epsilon, you know, just getting up to n minus half the epsilon and then removing the epsilon, sometimes the epsilon removal is very difficult. And this was done many years later by Zamina, who refined this to e n of h equals big o of h to the n minus 1 half, that was using the larger sieve, so the finement of the large sieve. After these uses of sieve methods, DeepMind introduced a new technique into the subject. In 2010, he proved using algebraic resolvance and the determinant method that e n of h can be bounded by big o of h to the n minus 2 plus square root of 2. So if you think about it, this is a little bit less than n minus half. So that's improvement in 2010 by different methods. And then just very recently, I think just this week, there's a paper of Anderson, Gaffney, Lemke, Oliver, Lowry, Douda, Shekhan, and Zhang. Just recently, using a selberg sieve, they proved that e n of h is the most big o of h to the n minus 2 thirds, approximately, plus a little bit. So almost two thirds of the way to the conjectured answer from n to n minus 1. So these are these are the successive world records, but just this is, I was a fan of this problem watching from outside for a long time, just seeing the different techniques that would get interested in this, and then used for other problems. I made this a very attractive problem to me. So this is something I started to think about as well. And so what I want to do today is I hope to prove the following week van der Waarden conjecture. And by that, so that's what it's referred to. We've van der Waarden just refers to that it's the conjecture of van der Waarden with an epsilon attached. So I hope to prove today that e n of h is a big o of h to the n minus 1 plus epsilon and positive epsilon. So in other words, the exponent of van der Waarden's conjecture is essentially correct. So I haven't talked about this before or shared the preprint with anyone yet. So I'm talking about this in front of a lot of people I realize. So please, please feel free to interject anytime with questions or point out mistakes because there could be some that hopefully not. So yeah, please feel free to interrupt anytime. Okay, so that's the that's the goal for today. I'm trying to give a full proof of this conjecture, this weak conjecture with the epsilon, weak formula conjecture with the epsilon, showing that the exponent n minus 1 is essentially correct. There's a more general question that one can ask here, which is if you take any permutation group G in SN on n letters, then you can define n sub n of g comma h to be the number of monic integer polynomials with coefficients bounded by h and absolute value says that the Galois group of f is exactly g. So it's the counting function for those polynomials that have Galois group G. And one can ask, well, what are the asymptotics of that and how did that depend on G? So in particular, the above theorem, this weak underword conjecture amounts to proving that n sub n of gh is big O of n minus 1 plus epsilon, big O of h to the n minus 1 plus epsilon for all permutation groups G in SN. But of course, one can ask the more general question, well, for certain groups can we do even better than this and the answer is yes. So I just want to point out the methods that I'll be talking about today can in fact be used to give the best known balance for n and gh for various individual groups G. But for today, I just want to concentrate on proving this theorem. So basically proving that n n of gh is big O of h to the n minus 1 plus epsilon for all permutation groups G in SN. Okay, so the fact that n sub n of g comma h is big O of h to the n minus 1, the fact that that holds for intransitive groups, as I said, was already shown by van der Worden, using the fact that polynomials having intransitive Galois groups are exactly those that are reducible over Q. So for that, you just have to count the polynomials that factor over Q. And for that, van der Worden had already shown that big O of h to the n minus 1 holds for reducible polynomials. And in fact, an asymptotic of the form cn times h to the n minus 1 plus a smaller order term for an explicit constant cn was obtained by Tela. So we know the exact order of magnitude for intransitive groups, reducible polynomials, it really is h to the n minus 1 times a constant. And that constant does depend on n. Meanwhile, there's another case that can be handled by using the fact that there are subfields of the algebra cut out by f. So in the case that the permutation group is transitive, so the polynomials are reducible, there's still the possibility that the field cut out by f has subfields. And so you can use those subfields to give an estimate for how many such polynomials there are. So this is the case of impermative Galois groups G. So an impermative permutation group is one that fixes a non-trivial partition of 1 through n. And so G fixes a non-trivial partition of 1 through n. Then in terms of Galois theory, that means that if a polynomial has a Galois group that's impermative, that means that q bracket x modulo f of x as a non-trivial subfield. And so Widmer is given excellent bounds on the number of such polynomials having impermative Galois groups using the fact that such polynomials have Galois groups that correspond to number fields having a non-trivial subfield. And subfield could be a smallest degree too. And so specifically, Widmer gets, proves the result that if G in SN is transitive but impermative, then the total number of polynomials that have such an impermative Galois group, degree n with coefficients bounded by h is the most big O of h to the n over 2 plus 2. So way better than the expect, no, than the total bound on the number of polynomials having Galois with not SN. So this is basically h to the n over 2 here as compared with the h to the n minus 1 that we wanted for all groups. So that's the case of impermative groups. So in transitive groups where the polynomials are reducible and impermative groups where there's a subfield, those are both handleable by thinking about those subfields. And so in order to prove the theorem that for all Galois groups we have h to the n minus 1 as the order of magnitude, it's suffice to just prove the theorem n of gh is O of h to the n minus 1 plus epsilon just for primitive permutation groups G. So these are these correspond to polynomials that cut out a field that doesn't have a subfield. Okay, so for now we'll we'll we'll restrict them to primitive permutation groups. These are the building blocks of all polynomials and suffices to understand those building blocks right here. Can I ask a quick question? Yeah. I may be misremembering the introduction slides, but I saw earlier that this was known for n equals 3 and 4. The the case of transitive but impermative is only lower than n minus 1 if n is 6 or larger. Is there something for n equals 5 that I'm missing here? There are no transitive but impermative groups when n is 5. Oh, okay. Well then that's great. Yeah, so you can only have a transitive impermative group when n is composite. Okay, okay, awesome. Thank you. Okay, so primitive groups, as I said, they're the building groups of all permutations groups G, but then they when you once you restrict a primitive, you can exploit the fact that you can exploit the primitivity and primitivity implies various things. So for example, if you have a primitive group that's not SN and it has a transposition, well that can't happen. So the proposition is if G is an SN is a primitive permutation group on n letters and contains a transposition, then it must actually be SN. And that's that's why n equals 5 cannot have any error. When n is prime, you can't have any primitive transitive permutation groups because you'll have an n cycle and a transposition and that would mean it's all of SN. But in general, suppose you have a group G and SN that's primitive and it contains a transposition. Well you can define an equivalence relation on all the elements from one through n that G is acting on by saying that i is equivalent to j if the transposition i, j is in the group, the transposition that interchanges i and j is in the group. And that's an equivalence relation on one through n that's preserved by the group. And so if that equivalence relation doesn't contain everything, then you'll have a non-trivial partition that's preserved by G contradiction since G is primitive. So once G is primitive and it has a transposition, then it must actually be SN. This is a standard proposition here, permutation groups. And so a consequence of this is that if you have a primitive Galois group that's not SN, well the Galois group then cannot have a transposition and therefore all ramification in the number field, it can't be simple ramification.  If the Galois group of f is not SN and it's primitive, then it doesn't contain a transposition but a previous proposition and therefore the discriminant of k, the discriminant of kf field cut out by kf must be square full, square full meaning that every prime dividing it must occur to order at least two. Everyone see why that's true? Because the group cannot contain a transposition, you can't just have a simple ramification of a prime, there's always going to be extra ramification and so the discriminant will be square full. So primitivity is sort of the building block of all groups but on the other hand, once your Galois group is primitive, then the discriminant has to be square full, all primes dividing it have to occur to order at least two. And so that's what we're going to try to exploit. The fact that if you have square full discriminant, then that shouldn't happen that often. Okay, so in other words, kf, if f has Galois group not equal to SN and it's primitive, then kf cannot have simple ramification. And so the discriminant has to be square full. Okay, so that's all we need to actually prove the theorem. So I'll jump into the proof of the theorem now. So again, just reminding we want to prove that en of h is big O of h to the n minus one. And to do that, we're going to divide the set of all irreducible monoclonal integer polynomials with coefficients bounded in absolute value by h, such that the Galois group is strictly less smaller than SN and it's primitive. And we're going to divide these sets of polynomials that we want to show our at most h to the n minus one and number into three subsets. So they're going to be three subsets. So again, that kf be on the field cut out by f. So the case one is going to be where the product C of the ramified primes in kf is at most h, but the discriminant is at least h squared. So the product the ramified primes is less than h, but the discriminant is at least h squared. That's the first case. Second case is where the product C of the ramified primes is at most h, but the discriminant is less than or equal to h squared. That's where using this h squared is a cut off point for the discriminant. And then finally, the last case is where the product C of the ramified primes in kf is greater than h. So remember h is the side length of the box. So in the product, so the two main cases where the product of the ramified primes is less than the side of the box and where the product of the ramified primes is greater than the side of the box and within that case where the product of the ramified primes is less than eight to the h, there are two cases namely on whether the discriminant is bigger than eight squared less than eight squared. So these are clearly three different cases and we're going to estimate them in turn. So I hope those three cases are clear and I'll write down what each case means on each slide so you remember. So we'll start with case one and that's the case for the product of the ramified primes c is the most h but the discriminant is at least eight squared. It's the product of the ramified primes is at most h and the discriminant is at least eight squared. So in this case so considering those f was the product c ramified primes is at most h the discriminant is more than eight squared in absolute value. So given such a k the polynomial is f such that kf is isomorphic to k must satisfy congruence conditions modulo c because c is the product of the ramified primes and if there's a certain kind of ramification in kf at a prime dividing c then that can be detected in f modulo p by just looking at the factorization and seeing whether or not much ramification is occurring in the factorization of f. So given such a k the polynomial is f for which kf is isomorphic to k will satisfy congruence conditions modulo c that have density about one over d because d is the discriminant that's how much ramification is occurring above c and those conditions about being ramified that much so that the discriminant is d can be detected so let's see. So the density is about o of one over d it's actually o of two to the omega d over d that's one reason why this epsilon will pop up in the final in the final estimate the reason is that for every prime dividing d the number of polynomials that have square full discriminant at p is not one over p squared it's actually two over p squared because there are two ways that the discriminant can become a multiple p squared that polynomial namely you can have a triple root or you can have two double roots and each of them leads to one over p squared density and if you multiply the two over p squared over all primes dividing c you get two to the omega c or omega c or omega d denotes the number okay so the point is that when we restrict to the product the ramified prime is being less than h then you can then you can detect the discriminant just by congruence conditions modulo c and the size of c is less than the side of the box and so you have this box and you have congruence conditions that are less than the size of the box and so you can really get a good estimate so the number of such f can be counted directly within the box because c is less than h so the number of f with this discriminant d right corresponding to whatever the ramification was in k can be just detected by c you can just count that directly because congruence conditions are imposing are less than the side of the box and so we immediately have the estimate a big O of h to the n two to the omega d over d for the number of such f though for each d where we're imposing some congruence conditions modulo c sees the square free part of d and c is less than h so we can so we get the estimate for each d that it's big O of h to the n divided by d more or less for the number of such f for any given d and then we just we just need to sum over all such d and that's where we use the fact that d is at least h squared so summing the estimate big O of h to the n over d which we got by just imposing congruence conditions module of the square free part of d which by assumption was less than h and so congruence conditions are smaller in size than the box so we get this estimate and then we just need to sum this over all square full d remember we can assume d is always square full because we're in the primitive case all square full d bigger than eight squared so we just want to sum h to the n over d over all square full d bigger than eight squared now so you can think of this as being summing h to the n over x squared over all x bigger than h and the x squared becomes an x so you get h to the n divided by by h is the estimate and this omega d of course causes an x amount to be so so this gives the desired estimate big O of h to the n minus one plus x one in this case because we did a summation of our estimate for each individual d over all d bigger than a squared better square full and that gives us the order of magnitude so the point was that we restricted to congruence conditions that are less than the side of the box we get good estimates and we restricted the discriminant being big because then when we sum over all d bigger than that better square full converges and it converges not to an estimate that it's good and so next we have to to go to the case where c is less than equal to h and d is less than eight squared because if we did the sum with d less than eight squared then this exponent when we summed would converge but it would be bigger than that so we have to remember that case separately so that will do on the next page where we replace this condition d greater than eight squared by d less than equal to eight squared and for that we have to apply a different technique okay so here we're we're considering those f for us the product c ram of fact times is at most h and the absolute discriminant d of kf is at least eight squared well in this situation the kf's that you can get right k equals kf is the number field of degree n having absolute discriminant at most eight squared that's the condition they're not that many number fields having small discriminant the number of such number fields k by a result of schmitt so schmitt has a general bound on how many number fields you have degree n namely if you counting all number fields up to discriminant x they're at most x to the n plus two over four the number fields having discriminant absolute discriminant most x and since we're only we've now replaced this condition with d less than equal to eight squared they're not that many number fields kf that can be cut out by such f d is less than eight squared it can be at most eight squared to the n plus two of the four which is at most you know h to the n plus two over two that's that's the most that's the most number of kf that could possibly arise for a polynomial f that has discriminant d less than equal to eight squared so this is not very many number fields this is way less than n minus one h to the n minus one so as long as they're not too many polynomials that occur for each number field it will be in good shape and it's not too hard to see that there can't be that many polynomials that occur for each number field if you fix the number field k how many polynomials can monic polynomials can there be integer polynomials that cut out that field where the coefficients are all bounded by h well there can't be that many if you just look at even the constant coefficient we're saying that the norm of some algebraic integer in k has to be at most h in size up to units they're at most h to the one plus epsilon such numbers this result has been made in progress by a result of empty algebra and thorn once that they prove that for each number field k each number field k arises for at most big o of h basically polynomials f the big o of h to the one plus epsilon polynomial f and you actually get to divide by a little bit of small power of the discriminant here so it's even better so they're not that so the point is they're not that many number fields that arise and the number fields that do arise they can't give rise to two means as polynomials so it's just we're just using the fact that the discriminant of the number field is so small that there can't be many number fields and each number field can give rise to only a few polynomials are you using somewhere that you've got the small product of ramified primes no we're not yeah we're not even so the total number of f in this case is at most we go from h to the n plus two over two that's the total number of number fields that can arise if the discriminant is at most a squared and then after that uh linking all over and thorn group that eats this number field arrives in the most h times log in log to the n minus one h ways and so when you multiply these two things together uh well you can see you're basically at h to the n over two plus a little bit and so that's big o of h to the n minus one when n is less than or equal to seven and even n of plus two over two is still a little bit bigger than n over two so until n hits seven this isn't a big o of h to the n minus one but when you plug in n equals six you're just on the border it's basically h to the n minus one plus epsilon and if you just throw in this extra factor that lumpy all over and thorn actually had it's the denominator then also takes care of n equals six and then finally what about n less than or equal to five well if n is less than or equal to five then we actually know exact results on the number of number fields of discriminant at most x and if we use those instead of the cruder schmitt bound we can get big o of h to the n minus one and n less than or equal to five as well that's fine in this case and just to just to note schmitt bound for large n has been improved in some spectacular ways first by ellenberg and benkatesh in the mecovanias and most recently of mp all over and thorn so this latest result says that we can actually from the large n we can actually replace this exponent n equals two over four by just log squared n so for large n all these this estimate actually gets very good but anyway we have a quick question um well what happens if you replace like the h square here with like h cubed what happens to the estimates oh um well i mean so instead of if we put h cubed here we would have got um and then right and for the case one oh in case one yes case one would have we would have done better yeah so i didn't do i didn't do use the optimal trade-off you're absolutely right h cubed would do better than eight squared i see but it doesn't work would that get rid of the epsilon it does yeah it will get rid of the epsilon so there's no epsilon for the case one in case two if you just choose the cutoff eight squared to be a little bit larger i chose eight square just to make sure that we handle the small n be right but once n gets a little bit big you could have chosen a much better and the epsilon would not be there three questions any other any other questions on this case so the reason this case was was handleable was because we were now in the case where the discriminant is small of the number field and not too many polynomials can arise per assessment okay so uh so we're left then with case three and that's the case where where c is bigger than equal to h so the product of the ram of five primes is bigger than the side of the box so you can't just impose kind of its conditions out there okay so so that's the last case a case where the product of the ram of five primes in case of f is bigger than h so let's consider such f uh fix such an f uh so what i want to observe in this case is that for every prime p dividing c you know there's every prime that's ramified in kf well the polynomial f has at least a triple root or at least a pair of double roots my zelo p because the group is primitive has square full discriminant this is the only way you can have square full discriminant is that you at least have a triple root or at least a pair of double roots monzolo p and as i said before that condition is detectable monzolo p maybe when you look monzolo p you can see whether it has triple root or double roots monzolo p it's in my p condition now so the point is that if you change f by a multiple of p it doesn't change the fact that p squared divides the discriminant right so in other words just the discriminant of f is a multiple of p squared but it's for mod p reasons you can change f by any multiple of p and it still has discriminant multiple p squared usually when a polynomial is multiple p squared it's for mod p squared reasons if you change it by just multiple p it won't necessarily still have value multiple p squared but in this case it's at this f the discriminant function is becoming a multiple p squared for mod p reasons because it's this condition of having a triple root or pair of double roots right now so so here's a proposition just as a general proposition about integer polynomials that take values that are multiples of p squared so if h of x1 through xn is an integer polynomial such that if you evaluate it at some integer c1 through cn and you get a multiple of p squared and suppose moreover that when you change these ci by multiples of p the value of h remains a multiple of p squared no matter what multiple of p you change the argument spine so in other words if h of c1 through cn is a multiple of p squared for mod p reasons you can change the argument spine multiples of p and it doesn't change the fact that it's multiple p squared if that's true then then the partial derivative of h with respect to any one of the variables i'll pick the last one xn partial derivative of h with respect to any one of the variables evaluated at that same point will be a multiple of p so if h is a multiple of p squared in this really strong way that forces any partial derivative to be multiple and so why is that proof is simple we'll just we'll just do the Taylor expansion around so we'll fix c1 through cn minus one that we're plugging in we'll keep the last variable and do the Taylor expansion about xn the last variable and so h of c1 through cn minus one comma xn is equal to oh so we're expanding around cn xn equal cn so the constant term is h of c1 through cn and the next term is the partial with respect to the last variable of h evaluated at c1 through cn times xn minus cn and then the rest is multiple of xn minus cn squared and now if you look at this well the first term is multiple of p squared by hypothesis and if you plug in so if you plug in xn any value of xn that's congruent my p to cn well this doesn't change this is a multiple p squared but this is also multiple p squared if xn is congruent to cn my p squared and so if the hypothesis is that no matter when you plug in an xn that's congruent to cn my p squared gives you multiple p squared that means this always has to be multiple p squared when you plug in xn that's cn plus a multiple p but this is only guaranteed to be then a multiple p not a multiple p squared so that means this guy has to be multiple so the partial with respect to xn evaluated at at our point has to be multiple so so that's the proof of that proposition and we can apply this proposition to this discriminant situation that we have here so here we have a discriminant which is a function polynomial integer polynomial in the coefficients of f and they're taking it's taking a value that's a multiple p squared for and it's if you change it by multiple p it's still going to take that right a value that's a multiple p squared so it satisfies hypotheses of this proposition and what that means is that if you take the derivative of the discriminant function with respect to any of the variables a1 through an for such an f it will be multiple so let's do it with respect to the last variable and what that's saying is that the derivative of the discriminant of f with respect to say the constant coefficient a n is going to be a multiple of c in this scenario where where the Galois group is it's primitive it's not xn okay if that makes sense so just this imprimity just this primitivity condition and the fact that the Galois group's not xn means that the that the discriminant of f is a multiple of p squared for a few reasons and that implies that we have this auxiliary polynomial namely the derivative of the discriminant that has to be a multiple of p okay but the fact that this derivative of the discriminant is a multiple of p and the discriminant is a multiple of p those two facts together imply that the resultant of the discriminant and its derivative is a multiple of p but the resultant of a polynomial and its derivative is the discriminant so if the derivative of discriminant is a multiple of c and the discriminant to the multiple of c that means that the discriminant of the discriminant is a multiple of p, right, because the discriminant is just the resultant of a polynomial and its derivative.  So here we're taking the discriminant of f as a polynomial and a1 through an, and we're viewing that discriminant as a polynomial and a1 through an, and taking the discriminant of that polynomial as a polynomial and an. So this is an iterated discriminant. First, we took the discriminant with respect to x for f of x, and then we take the discriminant again with respect to an. And that has to be multiple of p for such f that have galore groups that are primitive and not s. So why did we do this? The purpose of producing this number that's going to be a multiple of c is that when you take the discriminant with respect to an, an is not there anymore. This is a polynomial in a1 through an, and so on. So we've produced the polynomial just an a1 through an minus one that has to be a multiple of c. It doesn't involve an. So if f of x, which has coefficients a1 through an has galore group that's primitive and not sn, then, and if p ramifies in that extension, then p has to extra ramify p square has to divide the discriminant for my reasons. And that implies that the discriminant of the discriminant is multiple p, which does not involve an. So we found a, if this, so f, f satisfies these conditions, then they're, and p ramifies and p actually has to divide a polynomial that only involves a1 through an. And that gives us a way then to approach this problem. We only have to fix the values of a1 through an minus one evaluate the discriminant of the discriminant on those first n minus one variables. And that's the number and the ramified primes have to divide that. And so c is determined by just the first n minus one variables. And once c is determined, then you can say that an, an is determined. So let me, let me explain that. So, okay, so recall, we're considering those f for which the product c ramified primes and kf is greater than h and we're in a primitive non-sn galore group. So for such f, the polynomial dd of a1 through a minus one, I'm calling a dd because it's the double discriminant or it's the discriminant of the discriminant of f of x is a multiple of c. So if you have such a polynomial f, then just this polynomial in a1 through an minus one has to be multiple of c. We're c is the product of ramified primes and kf. So we can just fix a1 through a minus one and evaluate dd on it and see what it is. And then c is determined as it's the factor of dd. But that doesn't work if dd of a1 through a minus one is zero. But the number of a1 through a and minus one in this box such that dd this polynomial applied to it is zero is big O of h to the n minus two. You can save a variable. In fact, using the large sieve, you can actually decrease this to n minus two to two and a half. And you can actually do better. In that case, you can always save one variable if you're on a variety in a box. The total number of points in the box is hn minus one, but it's satisfying a non-trivial polynomial condition. So you save it there. So the number of a1 through a n minus one is O of h to the n minus two. And then a1 can be anything, any of the h values. And so the number of f with such a1 through a n minus one is big O of h to the n minus one. Okay, so this case where dd vanishes on a1 through a n minus one is taken care of. So we can assume that we're only looking at a1 through a n minus one says that we plug it into dd and we get something non-zero. So let's fix a1 through a n minus one says that dd of a1 through a n minus one is non-zero. Then that means that once it's non-zero, we know that c has to be a factor of this. But number of factors of this is at most O of h to the epsilon. Because all these, this is a fixed polynomial and a1 through a minus one are all bounded by h. So number, so this is at most h to the power. So the number of factors is at most O of h to the epsilon. So once you fix a1 through a minus one, this c is determined as a factor of dd of a1 through a n minus one. And so the number of possibilities for c is at most O of h to the epsilon. And once c is determined by a1 through a n minus one, then I claim we can just solve for a n, the number of solutions for a n modulo c to the discriminant of f being zero mod c, right? Discriminant of f is a multiple of c and it's a polynomial in a n. So you're just solving this polynomial equation a n. It's a one variable polynomial equation a n for a n, namely discriminant of f to zero mod c. And the number of such solutions to a polynomial equation by the fundamental theorem of algebra is at most the degree of discriminant of f in a n. But you have to do that modulo every prime divided c so that you have to multiply that over all prime factors of c. So that's why you have this omega c factor there. That's another reason for the epsilon. But anyway, it's just a constant. The degree is just a constant. In fact, the degrees n minus one raised to the number of factors of c. And again, that's a big O of h to the epsilon. Okay, so what happened here? We fixed a1 through a n minus one. We assumed that when you apply dd, you don't have zero. You don't get zero because that case was already handled. And if it's nonzero, then c has to divide it. And that there are most O of h to the epsilon possibilities for c. And then once you have c, that determines a n by just solving the polynomial equation. So basically, upshot is when you fix a1 through a n minus one, you can actually determine a n up to a bounded number. And that's why you get even when c is bigger than a step, the number of, oh, I should also say, we only in this, in this paragraph here, we only determined a n modulo c, because we're solving the polynomial equation modulo c. But since c is bigger than h, the side of the box, if you determine a n modulo c, you've actually determined a n in that box. It could exist. You can't have multiple values of a n in the box that are congruent to the, what you want modulo c if c is bigger than h. So since c is bigger than h, the number of a n in minus h comma h is also legal. And so the total number of f in this case is, again, big O, h to the n minus one, s x one. Yeah, very good. So, yeah, so that basically proves, that proves all three cases. And so, so we've proven, so we've proven that during the 1, 2, namely if the n of h is the number of monocentric polynomials, the degree n with each ai, most h and f subvalue, such that the Gallo group is not s n, then e n of h is the goal of h to the n minus one to s x one. So that confirms Van der Waart, uniform Van der Waart's conjecture. Confirms that that exponent is correct essentially. So thank you very much. That's what I wanted to prove today. Happy birthday again, Don. I have to take some questions, but otherwise we should hear stories about Don. And here's a drawing for Don due to Kate. Thanks, Kate, for this. Happy birthday again, and thanks so much for your attention. I have a very quick question, apart from thank you for this unbelievable talk and theorem. The question is simply at the very beginning, you said that one can't do better than O of h to the n minus one, because you could just consider the reducible polynomials. But can't you just assume from its irreducible right from the beginning and then ask the question, then there's, so the lower bound in principle could be smaller for. Yeah, definitely. We're all confused. No, no, you're absolutely right. If you were stick to the irreducible polynomials, then we'd expect a much smaller order of magnitude for sure way smaller than n minus one is probably expected to be true. And if you assume that the Gallo group is also not and then actually one can do even a lot better. So I didn't mean I knew I wouldn't have that much time in this talk. So but there are a number of other directions to go in. When is that first of all, I just want to mention that van der Waarden originally conjectured the non monocase and somehow authors went into the monocase over the years and that became van der Waarden's conjecture instead. But he actually considered the non monocase and the techniques I described today work also for the non monocase gives you the same savings and save an H. All right, but to your question, Don, yeah, these techniques actually do a lot better once you start making assumptions about irreducibility and not being certain Gallo groups. So for example, if you assume that polynomials irreducible, and the Gallo group is not SN or AM, so I already made your assumption, I didn't even write it down because I knew you'd ask this. So if you assume polynomials irreducible and Gallo group is not SN or AM, and N is at least nine, then these techniques actually yield an N minus two, it's an N minus one question. And if you assume that N is at least 16 with the same assumptions, Gallo groups not SN or AM and it's irreducible, then you actually get an N minus three using these arguments. And we're generally for sufficiently large N, you can actually save an unbounded amount. So you can save a function that grows within unbounded. Is the idea here to use the other vanishing of the partial derivative? That's right. If P divides the discriminant to a much higher order, then it's not just the first partial that vanishes, but a number of partials will vanish and second partials will vanish. And so you can use that to get additional savings. How many of the variables, can you use those? The other variables vanishing, the first partial derivative. Oh, you can. You can use the other first partials too, but usually it doesn't give too much. It just cuts down the variety, but it doesn't cut down the code dimension. But it helps actually to use the other partials as well. But it's the higher partials that really start cutting down the dimension of the variety. Thanks. Yeah. So the group theory behind this is that if you have a primitive group and it's not SN or AN, and it contains a three-cycle, then it can't contain it. So if you have a primitive group, permutation group that's not SN or AN, then it can't contain a three-cycle. And it can't contain a double transposition once N is at least nine. And so suddenly your ramification has to be even bigger than just a triple root or a pair of double roots. It has to be even more than that. And so suddenly higher partial start to come in. And we can assume that those are also multiples of three. So you can get additional savings. Great talk, Marjul, from your neighbor. Yeah. Thanks, Peter. Yeah, Marjul. Yeah, thanks a lot for this great talk. I'm sorry that to be a little bit passive because the bar froze here, but our computer expert helped. And yeah, so now we can start after your great talk with your fantastic results. We can start with the happenings for Don. And the first thing we wanted to do was to play a birthday song because, of course, that's the first thing to do. And the birthday song is by Noam Altis.


And now for your strong Europe spotlight. General David Perkins, commander of the U.S. Army Training and Doctrine Command, describes the role U.S. Army Europe plays in the Army operating concept. When you take a look at the Army operating concept, one of the purposes of it is to describe the environment that we think the Army will operate in the future. When we describe the environment, we describe it as a complex environment that is very unpredictable and is constantly changing. And when you look at United States Army Europe, it really is exactly what the Army operating concept was written to deal with. And so the role that United States Army Europe plays in this is actually twofold. One is actually helped us visualize what the future was going to be like by looking at the operating environment here. And then two, it gives us an idea of what role the Army plays as part of the joint force. So it helps us describe the future, but also shows us how you operate in that future. For more on the Army in Europe, go to eur.army.mil.