So, we continue with our discussion of analytic continuation, so you see last time I told you that there is 2, so to recall the various notions of analytic continuation, so that is this notion of direct analytic continuation which means direct analytic continuation or extension or extension, so this just involves a pair of data namely an open set which is connected open set namely a domain and a function defined on that and another such pair, another open set with a function defined on that, so this is the open set U is open set V, in fact U and V are supposed to be domains they are not just open they are supposed to be connected and then the requirement is that we say that f is a direct analytic g is a direct analytic continuation of f or f is a direct analytic continuation of g if on this intersection f restricted to this intersection is equal to g restricted to this intersection okay, so f restricted to U intersection V is equal to g restricted to U intersection V. So we say that the pair we say that the analytic function f has been continued to the analytic function g directly and we can also say symmetrically that the analytic function g has been continued to the analytic function f directly and the point about direct analytic continuation is that on this union they both functions can be glued together to define a single analytic function, you can define a function which is equal to f on U and which is equal to g on V and that defines a function properly on the union U into U union V that is because on the intersection f and g coincide, so it defines a well defined function which is also analytic okay. So but then what we are interested in is the notion of indirect analytic continuation, so this indirect analytic continuation or you can also say indirect analytic extension. So usually in the literature people do not use this adjective indirect they just say the analytic continuation and even the word extension is not often used okay but I am stressing on using this word because an indirect analytic continuation can be defined in two ways basically and what we were doing in the last couple of lectures was trying to show that you know both definitions are equivalent alright. So there is definition 1, definition 1 is we say v, g we say analytic function we say g is an indirect analytic continuation of f if g is obtained from f by a finite chain of successive direct analytic continuation. So in other words what is happening is that you have not two pairs but you have finite number of pairs such that each successive pair consists of a direct analytic continuation okay and the first pair is f the function corresponding to the first pair is f and the function corresponding to the last pair is g okay, so it is like this, so the picture is something like this so you have u1, f1 which is defined on this domain which is u1 that intersects with this domain u2 on which there is an analytic function f2 and this u2, f2 is direct analytic continuation of u1, f1 which means that f2 and f1 coincide on this intersection okay and then this goes on there is one more pair u3, f3 and u3, f3 is direct analytic continuation of u2, f2 which means f3 and f2 coincide on this intersection which is u2 intersection u3 and then it goes on like this and finally you end up with a last pair un, fn this is un which is a of course all the ui's are domains the way I am drawing them I am drawing them as if they are bounded domains but they not be bounded okay they just need to be open connected sets I am drawing a picture like this so that it is easy for you to imagine so you know so this will be the intersection with u3 and u4 and this will be the intersection with un of un with un-1 okay and so u2, f2 is a direct analytic continuation of u1, f1, u3, f3 is direct analytic continuation of u2, f2 mind you this does not imply that u3, f3 is a direct analytic continuation of u1, f1 unless and until u1, u2 and u3 all intersect okay it may happen that u3 never intersects u1 alright. So all you can say is u3 is an indirect analytic continuation of u1 because it is a 2 step chain okay so in this way if you have finitely many steps and you have f1 is the function f you started with and fn is the function g then we say this fn which is g is an indirect analytic continuation of f1 which is f okay. So an indirect analytic continuation according to this definition is just given by a finite chain of successive direct analytic continuation okay and of course you may wonder why worry about indirect analytic continuation so this is so whenever we do something you should ask the question why do we do it at all okay. So the answer is the following the answer is that it is rather funny and it is rather very involved on the one hand and it is rather funny on the other hand that you know if you start like this and you have a chain and assume that the last domain is the same as the first domain okay assume that un is the same as u1 okay then what could happen is still you could get a different function okay. So you can even think of it like this you can start with the function on a disk and then you can start moving the disk to get a successive direct analytic continuation and then you come back to the disk you might end up with a different function that is the whole point and then the question is what is this different function so even if u1 is the same as u1 fn need not be the same as f1 fn may be different from f1 that is the whole point and then the question is what is the relationship between this f1 and fn the answer is that they are branches of the same function. So in other words the whole point of studying analytic continuation is to try to get hold of all possible branches of an analytic function okay that is the whole aim alright so well you know this is one definition which is given like this there is another definition and this definition is what is called this definition also gives a definition of indirect analytic continuation but it is called analytic continuation along a path okay using a power series. So this definition is called indirect analytic continuation so I will use I will simply call it analytic continuation along a path or arc and what is this this is the second definition of analytic continuation well so you see so what I am having is that I have this open interval I mean I have this closed interval ab on the real line and then I have this map gamma continuous function which maps this closed interval onto a path on the complex plane so I have the complex plane here and you know I have a path gamma so for any point t I get the corresponding point gamma of t and of course this is this is z0 which is gamma of a is initial point of the path and this is z1 which is gamma of b so I have path like this and then what and suppose suppose so what does it mean to have an indirect analytic continuation along this path that is an analytic continuation along this path so we are given a power series ft of z given by sigma n equal to 0 to infinity an of t z minus gamma of t to the power of n convergent with radius of convergence radius of convergence r of t greater than 0 and and disc of convergence mod z minus gamma of t is less than r of t ok so which means for every point if you give me any point gamma of if you give me any point gamma t so let me draw the gamma t here if you give me a point gamma of t then you know that is this radius r of r of t so I have this disc centered at gamma t radius r of t which is the disc of convergence of this power series ft of z ok and for every such t I am given such a power series so the point about this power series is as t changes this power series changes here how does it change the coefficients change ok and the centers of the power series also change so the centers of power series of the power series are moving along the path and they are parameterized and since the path is parameterized by t the centers are also parameterized by t which are just the points on the path and the coefficients of the corresponding power series are also parameterized by t ok so if you are given a power series like this and assume that the radius of convergence is always positive alright and you know such that you know for every for such that for t prime close to t ft is equal to ft prime as as functions ok so in other words see this ft is a power series which varies as t varies ok but the way we want it to vary is that we want it to locally represent the same function ok so that means if you give me a t then for all t prime in a small neighborhood of t the corresponding ft prime should be the same as ft ok by that we mean that the power series you see when you say ft prime is equal to ft these are same as functions ok but it does not mean that a and t prime is the same as a and t it does not mean that that is because you see a and t prime will be different from a and t because the centers have changed gamma t prime is different from gamma t ok if t prime is different from t then gamma t prime and gamma t are different points on the path and therefore the if you therefore these power series at t and t prime are certainly different power series ok but our requirement is that the functions they represent are the same ok so you must not make the mistake of thinking that if t prime is close to t ft is equal to ft prime means actually that you know this a and t is equal to a and t prime that is not correct because for ft and ft prime for t and t prime different the gamma t and gamma t prime are different ok they are usually different ok and therefore the power series are different right so the coefficients are also different but what we want is that the functions to which they converge they are the same ok that is the requirement so this is a continuity requirement so what so if you have something like this then we then say we then say that f a f b is the result is the result of analytic continuation of f a by a on on gamma on the path so you are saying the starting function here the starting function the function here is the analytic function here is ft ok when you start here it is f a at that end it is f b so you are saying the analytic function f a is analytically continued along this path to give the analytic function f b ok so this is another definition when you say when one function is analytic continuation of the other ok along a path and what I was trying to explain in the previous lecture was that these two definitions are one in the same these two definitions are no different ok so the what we saw we saw last time we saw last time last time that definition 2 implies definition 1 namely I proved that if you give me an analytic continuation along a path then you can then the final function the function at f b is an indirect analytic continuation of f a in the sense of the first definition namely it is obtained by a finite chain of successive direct analytic continuation this is what I proved last time ok. And therefore the in other words I am saying that this definition is this implies this definition ok and the converse is also true it is also easy to see that definition 1 implies definition 2 ok it is also easy to see definition 1 implies definition 2 because it is very easy to do it see because you know let me draw that diagram again so here is my so you know here is my u1 this f1 and then I have something like this this is u2 f2 is u3 f3 and so on and at the end I have some un fn ok and u2 f2 is a direct analytic continuation of well u1 f1 u3 f3 is a direct analytic continuation of u2 f2 and so on and the final one un fn is a direct analytic continuation of un minus 1 fn minus 1 and then we say that according to the first definition fn is an indirect analytic continuation of f1 ok. Now I want to say that fn is also an analytic continuation of f1 along a path in the sense of definition 2 how do you prove that it is very very simple what you do is you simply start with the point z0 in u1 ok then you choose a point here z1 in u1 intersection u2 ok and since z0 and z1 belong to the same set u1 which is connected and open mind you an open connected set is path connected ok for a path connected set topologically is always a connected set but usually what happens is a connected set need not be path connected but if the connected set is an open set then it will be path connected. So since this is a open connected set it is path connected and therefore I can join z0 and z1 by a path ok and I can call that path as gamma1 ok and then I can choose a point here call it z2 and then again z1 and z2 belong to u2 which is again a domain it is connected so it is path connected so I can join z1 and z2 by path gamma2 ok and then I can go on like this. So you know I choose a point z3 here which is in the intersection of u3 and u4 and you know I connect z2 to z3 because both of them belong to u3 which is path connected by a path gamma3 and I go on like this I end up with a point so I end up with a point zn here so this is a point zn and it come it is joined to zn-1 which is in the intersection of un-1 and un-2 by a path gamma n ok yeah so this has to be zn-1 I guess ok. So each gamma i lies in ui so and it connects zi-1 to zi ok so gamma n-1 has to lie in un-1 and it has to connect zn-2 to zn-1 ok so this is how it is going to be and then finally I choose this point well if you want let me call it z prime ok and I join it by another path which is gamma n so this is the path gamma n and finally what I will do is I put gamma equal to union of all the of the gamma is you take the union of all these paths ok you take the union of all these paths that is of course a path alright it is just concatenation of paths if you have 2 paths you join them you again get a path alright so you put them all together you get a single path alright and then what you do is define ft of z to be equal to so you know this is the definition the definition is you define it to be power series expansion of ft of z centered at gamma of t if gamma t belongs to ui ok. So if so I mean by that I mean gamma t is if gamma t is the portion of the path if gamma t is gamma i ok so what I am doing is I am doing the following thing see for every point from z0 to z1 ok on this path I am simply taking the power series of f1 centered at that point which make sense ok then from z1 to z2 for every point along this path gamma 2 the corresponding power series is a power series expansion of the function f2 which is defined on this analytic function centered at that point. So in this way I get this ft ok and it is very clear that I have given you I have given you a power series ok for each t of course the t is of course going to vary in such a way that when t for the initial value of t it is going to be z0 and the final value of t is going to be z prime alright and what is going to happen is to check that this is a this is a analytic continuation along this path I have to only check that look that for any t for t prime close to t the ft prime and ft represent the same function but that is obvious because you know if you take any if you take any t if you take any point it is going point on gamma t it is going to be on some gamma i of t ok on that gamma i of t in a small neighbourhood of t it is going to just represent fi it is this all the ft are just power series expansion of the same function fi. So the condition this condition that this condition for that for t prime close to t ft is equal to ft prime is automatically satisfied alright because let me again repeat if you take any particular value of t and look at particular value of gamma t it is going to be it is going to be one of the gamma it is going to be a point of one of the gamma is and if it is going to be on gamma i if it is going to be a point of gamma i then all the ft is there all the ft primes nearby for t prime nearby to t they are all just expansion of the function fi because fi is a function on ui inside which you have taken the portion of the path gamma i ok. So you get this condition that for t prime close to t ft is equal to ft prime automatically you get this condition therefore it is clear that fn is analytic continuation of f1 along this along the path gamma ok. So let me write that down clearly ft is equal to ft prime is equal to fi for t with gamma of t belonging to gamma of i gamma i ok. So hence fn is the analytic continuation of f1 along gamma ok. So what this is telling you is that definition one is also the same as definition two. So definition one is the same as definition two alright. So either you see as analytic indirect analytic continuation as a chain of success finite chain of successive direct analytic continuation either you see it like that or you see it as analytic continuation along the path using parametrized by power series both are the same ok. So this is one thing that was the aim of the previous lecture and now I need to make a few statements. So the first statement is of course that so looking at it from the point of view of the analytic continuation along the path there are several observations ok. So the first observation is that if you think of analytic continuation along the path then the radii of convergence rt this function r of t is a continuous function of t and this a n of t is also continuous function of t so that is the first observation. So here is a lemma if given an analytic continuation ft of z is equal to sigma n equal to 0 to infinity a n of t into z minus gamma t to the power of n to the power of n mod z minus gamma t less than rt a n of t and r of t are continuous functions functions of t ok. So the first thing is that this is more or less intuitively clear ok this is more or less intuitively clear but then a little bit of justification is required and it will use a lemma that we proved a result that we proved in an earlier lecture ok maybe it was 2 or 3 lectures ago ok but I will recall it. So what is the proof? The proof is you see see so you know so the diagram so let me again draw a diagram. So gamma is from ab to c this is your path where ab is an close interval ab is a close interval on the real line ok. Of course by gamma I also think of the geometric path that it traces and also the function gamma ok I mean it is abuse of language I am using both at the same time but this gamma but when I am thinking of a path this gamma is fixed this function is fixed that which means that I am dealing not just with a geometric path but I am dealing with a fixed parameterization of that path that is what I want you to remember ok. So a path can be parameterized in many ways so when I say I am looking at a path gamma I mean I am fixing my not only my geometric path that is the path traced as t moves but I am also fixing the function gamma ok that is something that you have to remember right. Now you see so you know you also have these functions a so you know an of an will be a function again from ab to c and r will also be a function from ab to c an is going to give you the coefficient of for every t an of t is going to be the coefficient of it is going to be the nth n plus month coefficient of ft ok and it is n plus 1 because I am starting with n equal to 0 alright and r for every t r of t is going to be the radius of convergence of the power series ft ok that is the notation alright. So an and r are both functions on this closed interval ab on which gamma is defined ok and the claim is that an and r are continuous functions ok they are continuous complex functions. So of course I have to say I have to say more this is does not actually not complex valued it takes values on the real line and positive real values ok. So here in fact I can put I can in fact change it to r r greater than or equal to r greater than 0 which is subset of c if you want ok because I am giving you for each t I am always assuming that the radius of convergence is positive ok it is an analytic function each ft is a convergent power series ok and it is not by convergent power series I mean something that has positive radius of convergence ok it is not 0 radius of convergence. So all r t's are always positive ok and of course at each I mean if you have an analytic function at a point then if you expand it in a power series about that point you will get a positive radius of convergence because if it is analytic at a point then it is analytic in a disc surrounding that point and that disc is certainly contained in the disc of convergence of the power series for that expansion of that function centered at that point and therefore the and the radius of convergence therefore is certainly positive ok. So these r t's are always positive right. Now what I want to tell you is that see to check that a you know a function is continuous it is enough to check it locally ok to check that a function is continuous it is enough to check that condition locally. So to check that an and r are continuous it is enough to check them locally ok. Now you look at this so you look at this so let me write that down to check an and r are continuous it is enough to check they are continuous local ok because continuity is a local property right and now here is where I am going to use this fact I am going to use this fact in this definition of analytic continuation along the path which says that you know for nearby points for all t prime near to t f t and f t prime should represent the same function ok that is the that is the key that is the key hypothesis that helps you to check that they are continuous locally. So you know so start with start with a t in a t in a b ok then you know there exists an epsilon of t positive such that so this is something that we repeatedly used in the previous lecture where we proved that definition 2 imprised definition 1. So for every t there exists a positive epsilon of t such that for every t prime in t minus epsilon t t plus epsilon t intersection a b we have f t for every t prime we have f t prime is equal to f ok. So this is just writing in terms of epsilons this continuity condition ok so the continuity condition is that you know if t prime is close to t then the power series f t prime and the power series f t represent the same function the same analytic function I am just putting in more precise language the condition that t prime is close to t I am just saying that t prime belongs to an epsilon t neighbourhood of t inside a b ok and for every t prime in an epsilon t neighbourhood of t inside a b I am saying that f t prime is equal to f t it is just a translation of this condition ok the condition that the power series the power series very continuously ok. Now you see now what I want to do is I want to so you know so I will draw a diagram so here is my t so here is my point gamma t this is my path this is gamma of a and this is gamma of b and well here I have I have this disc this is the disc of convergence with radius of convergence r t ok and what is going to happen is and I am looking at all t prime close to t so which means I am looking at you know a portion of the arc like this ok I am looking at a portion of so when t prime is lying in this neighbourhood so I will get a neighbourhood like this ok so this point will be assuming that gamma t is an interior point otherwise it could I will get only one side of it so if it is in if gamma t is an interior point of this path namely it lies between gamma and a and gamma b I will get this whole interval so this point will correspond to gamma of t minus epsilon t and this point will correspond to gamma of t plus epsilon t ok and well for all t prime inside this open arc f t prime and f t are the same ok so you see on this whole disc the function f t lives f t is defined on this whole disc because it is the after all I have taken f t is the power series centred at gamma t radius r t and within the disc of convergence you know the power series represents an analytic function it is infinitely differentiable and all its derivatives will have the same disc of convergence ok so this f t is the analytic function that lives here and you know this f the other power series f t prime close to f t they are the same they all represent the same function f t because of this choice of epsilon t ok now watch you see you know we have seen we have seen a lemma a lemma so here is another lemma so this lemma is using another lemma which I proved a couple of lectures ago and the lemma is if f is analytic on a domain D and if for each z belonging to D r z is the radius of convergence of the power series the power series f z of f centred at z then r of modulus of r z1 minus r z2 is less than modulus of z1 minus z2 for all z1 and z2 in mean this was essentially the statement that the radius of convergence is a continuous function of z ok I proved this I proved this lemma a couple of lectures ago and now what this lemma will if I apply this lemma then you see that in then you know if you then if you see then for all t prime that correspond to points here ok if I apply this lemma you can see that the radius of convergence r t will be a continuous function of t because r is a continuous function of z and on gamma z is a continuous function of t so r t which is a composition of continuous functions is continuous ok so what I want to say is that here r r t is actually r of gamma of t r t prime is r of gamma of t prime ok it is r of z with respect to the function f t ok throughout all the for all t prime line in this range ok all the functions all the power series represent the same function namely f t ok and therefore if you look at the radius of convergence they are the radius of convergence of f t at these various points ok and that is what r t prime is for t prime in that in this small neighbourhood of t ok and you know t prime is a continuous function of gamma I mean sorry gamma is a continuous function of t prime and r sub f is a continuous function of the variable inside so r t prime is a continuous function of t prime so this implies that r t prime so here you know if you want some notation I can put here if you want to make things clear I instead of just saying r z I will call let me call it r f z this is the radius of convergence of the power series expansion of f at z ok if you want I will put that f to make it clear so in this case I will get this subscript f here alright that might help so what I am looking at is r f t r t prime is just r f t it is the radius of convergence of the power series expansion of f t at the point gamma t prime because f t is the same as f t prime ok and you know if you give me a point and you give me an analytic function the power series representation the power series expansion around that point is unique it is an identity theorem for power series ok the power series expansion of an analytic function about a point is unique ok therefore that is why you get this ok and so r t prime is a composition of gamma and r f t ok and it is a composition of continuous function so r t prime is continuous in t prime for t prime close to t. So what I managed to show I managed to show that r t prime r is a locally continuous function of t ok for every t you should take t prime sufficiently close to t ok I have proved that r is a continuous function of t prime ok a function which is a globally defined function which is locally continuous is globally continuous because continuous is a local property. So I have just verified it locally and how am I able to verify it locally is because locally the point is all the power series they are all representing the same function that is what I am using ok that is what I am using and so long as if you have a fixed analytic function then you know this lemma tells you that the radius if you expand it as power series at different points then the radii of convergence are continuous functions of the points that is what this lemma tells you ok. So this tells you that r t prime is a continuous function of t prime for t prime close to t and this means that r t is locally continuous which means r t is continuous because a function is locally continuous is continuous that is all ok. So this establishes the fact that r is continuous function of t ok now I have to show that a n is a continuous function of t that is also pretty easy that is also pretty easy why is that so I can try to attain a line it is simply again the fact that in that for all t prime in that neighbourhood of epsilon t neighbourhood of t inside a b all the f t primes all the functions given by the power series f t primes they are the same as f t it is the same analytic function. So you see a n of t prime is what this a n of t prime is the this is the n plus 1 coefficient ok in the power series expansion of f t prime in the power series f t prime ok but what is that if you take a power series of an analytic function what is the n plus 1 coefficient it is just the n plus 1 derivative evaluated at the centre divided by factorial n plus 1 ok. So this will be just so it is a n t prime is the n plus 1 coefficient of f t prime so it has to be given by n plus 1 derivative of f t prime centre at gamma t prime by factorial n plus 1 right you know if you have if you have if you have a domain D and if you have a point z0 and if you have an analytic function f on that then you know if you write if you write the Taylor expansion of f at z0 you will get f of z is equal to sigma n equal to 0 to infinity a n z minus z0 to the power of n and what is this a n this a n is the nth derivative of f so I think probably I missed yeah this should just be n not n plus 1 this should be factorial n so I will get nth derivative of f at the centre by factorial n you know this this is just Taylor's formula ok. So I am just applying it after all a n t prime is the a n corresponding to f t prime ok corresponding to f t prime for f t prime the corresponding a n is a n t prime and therefore a n t prime has to be given by the nth derivative of f t prime evaluated at the centre which is gamma t prime divided by factorial n that is all I am writing but then you know but you see the point I am going I am using is that for t prime close to t f t prime is same as f t you see all the for all the t primes close to t the f t primes are all one and the same f t so this is equal to f t n at gamma t prime by factorial n for t prime in that neighbourhood namely t minus epsilon t t plus epsilon t intersection a b that is because in that for all t prime in that neighbourhood f t and f t prime represent the same analytic function ok if f t and f t prime represent the same analytic function then the nth derivative of f t should be the same as nth derivative of f t prime because they represent the same function the derivative depends only on the function ok. So I will get this but now you see you see you know but now you are done why because you see f you know if you have a analytic function then you know all its derivatives are also analytic ok. So each nth derivative of f t is also analytic so it is continuous and gamma is continuous function of t prime therefore f t n of gamma of t prime is a composition of continuous function so it is a continuous function of t prime therefore you get a n t prime is a continuous function of t prime for all t prime close to t ok. Since so that is it so we have proved that a n is locally a continuous function of t and therefore it is globally a continuous function of t ok so since f t n z is a continuous function of z and since of course gamma is a continuous function we see that f t prime sorry a n t prime is a continuous function of t prime for t prime close to t as above close to t means t prime lying in this epsilon t neighbourhood of t inside a b. So this implies that it tells you that it tells you therefore that a n is locally continuous and therefore it is continuous so that finishes the proof of this lemma ok. So this lemma tells you something that is intuitively already clear when you define analytic continuation along a path it is given by a family of power series along the path ok with centres along the path and you want that family of power series locally to represent the same function along the path. This condition actually that is what the lemma explains it tells you that this condition ensures that the coefficients a n are continuous it also ensures that the radio of convergence of each f t namely r that is also a continuous function ok and which is which we intuitively feel should be clear ok. So I will stop here and then in the next lecture what I am going to do is I am going to tell you that I am going to tell you because of this very important fact the important fact is if you give me a fixed parameterization gamma of a path ok and you give me an analytic continuation like this then everything is completely fixed by the initial function ok. f t for every t greater than or equal to every t greater than a including b is completely determined by f a so I am saying in other words if you give me a path if I start if you give me a path with a parameterization I start with an analytic function at the starting point then if you do an any analytic continuation along the path the each of those functions that you get in between they are all unique ok. So the everything depends only on the starting function ok and that will crucially use this continuity right so I will explain that in the next lecture.


Freedom. Welcome to the Anthony show unedited. Now I do not plan on having the Anthony show unedited become a thing. However, right now I'm doing a windows reinstall and I just don't have the software for it. But this is important and I wanted to get it out there because I've been contacted by a member of the Freedom recruiting staff. They wanted me to let you, the Freedom partners, know that we are in fact looking for recruiters. We're looking for very sociable types who love to get out into the community and talk to people and for lack of a better word evangelize about who we are and what we do. If you think that you could make a couple extra dollars a month doing this, if you think you could make it a job, if you think that you would like to work with Freedom as a recruiter full-time, fill out the application in the description below and you may be contacted by somebody about moving your career forward with Freedom. If you have any other questions, comments, matters of concern, let us know in the comment section down below. As you all know, I'm active in the comments and that's not changing anytime soon and hopefully I'll have my editing software back up and running to 200% of the previous capability. I can't edit this the normally this is where the troll would pop out of my shoulder. Till next time, like, subscribe, comments, please.


the IoT button BFF. This little BFF attaches to the back of a cutie pie and it basically adds two things to your cutie pie. A nice big 12 millimeter tactile momentary button and a neopixel. Those are on pins A2 and A3 respectively. I've got my BFF, got some socket headers. Here's a cutie pie.  I'm using whippersnapper with this. I have this paired with my Wi-Fi router. When I press a button I'm gonna get that IoT button BFF indicator on my IoT page to blink. I have this color widget. Oh and pick a different color. Let's go to green. Hit save there and now that updates. That's my product pick of the week this week. It is the IoT button BFF with neopixel for cutie pie and jow board.


So, David, you, a couple weeks ago during the race to deliver or the deliverance service, you were wrecked and touched by God in such a powerful way. But before then, can you let the audience know what were you before then? What was your life like before then? Basically, I've been battling with ADHD depression. I was 15, 16, I started dabbling in dark arts, satanism, tarot cards, Ouija board, and I believe that when I did that, I brought something through. And ever since then, I've been just battling my life, depression, like I said, suicide. I can't focus, I couldn't hardly, you know, I get the point where I couldn't breathe because it was, the anxiety was so bad. And so, it was just, it was just a completely uphill battle for me every day just to get out of bed. Wow. That's crazy. You mentioned that, you mentioned that you had ADHD and what were some things that you just couldn't do or that were hard for you to do before the prayer line? Well, I dropped out of school at a young age and I battled trying to go back and recently I actually tried to go back into school and I had just a hard time focusing. I'd given up all hope of completing my high school diploma. And at the prayer line though, I've been able to start doing my courses and actually focusing on being able to get them done and I've already passed one of them. Come on. That's so awesome. Praise the Lord. When you were going through the prayer line, what were you experiencing? What were you going through then? That's kind of a great question. Well, I was out there and the pastor was like, the pastor was like, you know, saying let it go, let it go. And I was screaming at the top of my lungs and I looked over to my friend Heather and I said, Heather, I don't feel well. And I was like, something's wrong and I'm like, okay, so and I really don't remember a lot after that. I just remember just this, just feeling I was something wanting to come out of me. Just this feeling like being pulled out of me and it actually hurt like it physically hurt me. Like I was and I just remember waking up over here on the floor and I just felt this awesome relief.  That's so good. That's so good. So my question for you is, do you feel different now? Is there a change in your life now? Completely actually.  And so when I really needed it, God delivered to me and I thought it was so unbelievable. I was like, if people don't believe, believe that's all I'm going to say.  So no matter what, even if you turn your back on him, he's still waiting for you to come back. Come on. That's so good. Thank you, David. Awesome. Let's give Jesus all of the glory. This is amazing. This is so good. Thank you, Jesus. That no matter how far you have gone, the other way, he has never left you, never left you in Jesus' name.


Apparently we match today. So what?  I'm Rick and you can follow us on Instagram Twitter And once again, we are filming via Skype because we weren't planning on filming today We yet we got some other stuff to do, but I knew trailer dropped It's called on it. We're on it Roddy Roddy It's starring Salomon the Khan Disha Patini Roddy Hoda Jackie Shroft, and I believe it's directed by Prabhudeva. Prabhudeva. I think he directed it I don't I don't know if he's in it, but we know he's directed like Gandhi Bada I know he directed with with Jade and I know he's directed a bunch of other stuff, but It's the next one. I so I'm expecting obviously we've only seen one Salomon Khan thing, but I know he's the action guy So that's what I'm expecting me to from this a big action flick So Let's just get into this ready. Let's drive drive in dive into the Salomon Khan Three red two one I That's the worst thing you can do, Salman. Don't take it out. Oh, damn. Yea. That's exactly what I expected from a Salman Khan movie. That's exactly what a huge action. You're going to see Salman Khan be Salman, I believe they call him Salman Bai. That's his affectionate name from his fans. all the all the cheesy lines you're gonna get you're gonna get him kicking some ass but it did look like it's gonna have some great action in it and all that but it definitely it was it was it's definitely one of those star vehicles you can tell yeah it's no different than back in the day when Stallone and Schwarzenegger were at the top of the action game and nobody went thinking they were gonna watch anything that had to do with great acting great story they just wanted to see those guys that they love kicking butt and doing great snout work and action and this I mean it looks like it's gonna deliver and the visuals look great this work looks great you want just pop popcorn and watch things get blown up and people kicking each other's ass this looks like it's a lot of fun that I mean that's always that's why I that's why I and I was harping back to John Wick is I'm not a fan of Keanu Reeves I I don't think he's a good actor I've never liked him but those films do really well because one they don't make him act they just make him shoot people and kick their ass and it's wonderful to watch exactly great comparison but it did look ready ready puto actually looked really good in this who was Jackie yeah Jackie and I want to talk about Jackie he's super interesting we haven't seen a lot of him and I'd like to see you guys can recommend old new whatever but he's doing a lot of quirky roles it seems like I bet it seems like he's a very like he he has a good sense of humor about him but because when we saw him in the okay computer trailer he's like this guy who's like fully nude and just as quirky weird dude and I also think of him also as kind of a almost a thespian even though we haven't seen him I just know his old reputation and so it's super super interesting that he's just going for these weird quirky roles now yeah I mean I why wouldn't you want to be in an action film just to be able to be in an action film you know yeah and so yeah I mean if this is one of those things where as long you know you go into it prepared and you're not expecting the acting to be good you're not expecting writing to be good you're just looking for nonstop action great stunt work visual effects a lot of funny one-liners and just an escape kind of and I'm guessing this would be considered like a masala film my bet is most salmon con films are gonna be masala films outside of the one we watched I guess and I guess I don't know I don't know if that would be considered a one and we said in that even though it was over a year ago salmon wasn't really the problem in that film and but we enjoyed that film like he did not like hurt that film at all but I do know that a lot of people like that's like his best acting performance he's ever given yeah I don't know is he is he comparable to the examples I gave in terms of I'd say like an actor like Schwarzenegger in Stallone Vin Diesel kind of Dwayne Johnson yeah yeah people of that ilk that are just there to be like big action stars good with action they can hold their own they don't make blaring mistakes but they never do anything where you're like okay yeah you're just you know then there's always Jean-Claude Van Damme and Steven Seagal who that's just to watch the martial arts expertise but then also you get these cheesy lines that I'm going to clean this city I know which that's I'm sure it started before but here in America as you all know I'm sure that that's that's all that's freaking Arnold Schwarzenegger it's the I'll be back it's yeah and the and the you know throwing knife hits him in the eye gets my point wait what film does he say that in I just made it up oh it's like what in the world also does he usually because I know he's Muslim does he usually release his stuff on Eid kind of similar to others that released their stuff on like other Hindu holidays yeah Diwali and stuff like that I wonder if that's like I don't know if it's on per I mean obviously it's on purpose but I don't know if like if you're like a Muslim actor you released on Eid or I don't know I was just wondering because I know a lot of people release on big holidays like that and I know I know he's big and I remember I think it was over he did he's done a couple of his own songs like he's sang and and music videos I was told when one came out I was like should we react to it and the team was like no so I don't they might be fun for like cringe I don't know watched it if we watched it not going in like when we watch a big B thing when we know he's gonna sing we go win I mean it actually he turned out to be a better singer than expected yeah it's big B's actually quite a quite a decent singer but let us know what we think and also Prabhu Deva we haven't seen anything he's directed outside of just music trailer reactions yeah and we've seen we've seen it I know remember we saw one trailer reaction where we were shocked about the fact that we didn't know he was a director yeah but I that I would the biggest poll for me and sincerity would be the two-fold sense of if you just want a good action movie it be a great escape to enjoy that and I'd also love to see the way Prabhu Helms an action film yeah I would like to see him directing as well so let us know down below what if his best film is and is there another Salman Khan film we would enjoy outside of debong three let us know down below you


Item number SCP-3006 Object Class, Euclid Special Containment Procedures Online video distribution sites impacted by SCP-3006 are to be monitored by Foundation webcrawlers. Manifestations of videos containing SCP-3006 are to be removed upon posting and or detection. Foundation personnel embedded in the Alphabet Corporation are to ensure that no members of the public are able to view SCP-3006. It is estimated that, without proper implementation of the containment procedures, that most videos on the global public internet would be affected within 128 hours. The Foundation has not found any ways to prevent additional SCP-3006 videos from being posted. No other SCP objects are to be referred to in the documentation. Description SCP-3006 is a video titled, We Are Number One except every time you play it, there are twice as many Robbie Rotten's, but the room is the same size, not safe for work. Which was posted to the YouTube video hosting service on October 12, 2016. It features a musical clip from the show LazyTown that, on repeated playbacks, caused a number of humanoids identical to Stefan Karl Steffensen to appear. Videos impacted by SCP-3006 are capable of spreading their effects, primarily by being posted in threads where multiple non-effective videos are present. Within one to five minutes of being linked to, all other videos linked in the same thread will begin manifesting SCP-3006's effect. Humans or living humanoids featured in the video will double with each playback, including individuals who are off-screen such as camera men and sound personnel. The effect is exponential in that, upon each playback, the number of subjects in each video doubles. For example, a video containing two subjects will instead have four upon the second playback. The third would have eight, and the fifth thirty-two. None of these entities appear to have knowledge of SCP-3006's effect and find it highly distressing. Although affecting linked videos, SCP-3006's effect is unique to each viewer. That is, SCP-3006's effect is dependent on how many times the individual has watched the video. If more than one subject is viewing it, this effect will occur based on who physically activated the video. With the subject views a series of videos, the effect will also carry over to the next one. SCP-3006 primarily appears to affect videos taking place in enclosed environments, although this cannot be confirmed as a deliberate effect. Once affected by SCP-3006, it is not possible for recorded subjects to exit the area in which the video is taking place. At a certain point, affected subjects will begin perishing due to overheating, exposure, or crushing force from exponential humanoids instantaneously appearing at the beginning of video playback. This does not prevent SCP-3006's effect from persisting. Eventually, playback reaches a point where the camera and sound equipment are destroyed due to intense viscera pressure in the enclosed space. At this time, the appearance of the video is variable, with non-anomalous graphics spliced with corrupted digital images. SCP-3006 was initially discovered after social media reports of unusual videos came to foundation attention. 11,346 individuals who viewed the first anomalous instance of SCP-3006 were tracked down and given Class-C amnestics. After interrogation, several subjects featured in the initial video were determined to be non-anomalous and were given Class-B amnestics. As of January 18, 2017, SCP-3006 is classified as Euclid. Addendum, Examples of Effective Videos, Title, Video Description, SCP-3006 Effect Colin Furze, Title, So What If Loads of Fireworks Exploded in My Van?  Notes, During the second playback, it was initially unclear where additional humanoid entities were, however, analysis showed that human hands were visible against the windows of the vehicle as the fireworks were going off. None of the other subjects featured in the video appeared unnoticed, and playback continues as normal. Mystery Science Theater 3000, Title, Mystery Science Theater, Mano's Hands of Fate, Free, Mano's to Hands of Fate. Video Description, An illegal upload of the January 30, 1993 episode of the television show Mystery Science Theater 3000, which features low-budget films and humorous commentary. Film attached to the episode is the 1966 film Mano's to Hands of Fate. SCP-3006 Effect Due to the metallic nature of the puppets, after 12 playbacks, there were several hundred copies of the host being impaled by the metallic implements created when the puppets fell apart. In addition, the film being rift on had the actors appearing exponentially, with leg braces worn by one character causing fatal amounts of puncture wounds in the other actors. Game Grumps, Title, Dudes with Attitude in a Not Too Distant Future, Part 1 Video Description, Two adult males play a video game with footage from the game and only sound produced in the recording space being audible. SCP-3006 Effect, Neither of the hosts appear to be aware of the imminent SCP-3006. Episode Preceeds Normally. Title, Dudes with Attitude, That's No Moon, Part 2 Number of Entities Present, Four Note, subjects display confusion and argue for the entirety of the episode. Several attempts to exit the recording area can be heard, all of which are unsuccessful. Title, Dudes with Attitude, Scotty Doesn't Know, Part 3 Number of Entities Present, Six Note, subjects become aware of the anomalous effect. The sounds of violence can be heard, with at least one of the entities appearing to be terminated. This does not appear to have affected the number of entities present in subsequent videos. Title, Dudes with Attitude, Wait, Is This Real? Part 4 Number of Entities Present, 24 Notes, mass confusion can be heard. The game is completely ignored from this point on, with graphics not advancing past any title screen. In the background, subjects appear to have organized and only speak when holding a magic stick. During the course of a debate, it is decided to store soiled clothing in a basket and rationing a food of water while they attempt to exit the recording area. Several outbreaks of violence are suppressed by the other entities. Slightly before the end of the video, the sound of a sofa being used of the battering ram can be heard. Title, Dune Buggy, Hot Sand Massage, Part 11 Number of Entities Present, 576 Notes, Continuous incomprehensible screaming Title, Please Oh Please Why This Always Happened To Me, Part 77 Number of Entities Present, Estimated 500,000 Notes, Sound of Gentle Sloshing with Scrambled Reddish Digitized Video Graphics Addendum, Transcript of Interview, 3006-J Video Redacted Interviewed ██████ Interviewer, Dr. Krang Forward, On February 14, 2017, another video producer ██████ discovered several of their videos affected by SCP-3006. This transcription is for the iteration of the interview. The original has been lost or may never have existed. Begin log. Can you state your name for the record and spell it? Sure. Kind of like a newspaper interview, I guess. I'm ██████, but you already knew that, right? When did you first become aware of the videos? Someone linked it to me on Twitter, I think.  You feel bad, but they're just little computer people, right? So not really a big deal. That's what I thought, at least. How long did it take for you to discover the videos you were featured in? Yeah, well, you know, it was one of the suggested videos. God. Dude, I… I just… It's not something I wanted to describe. So I tried reporting it, and now I'm here with you guys. How many times did you watch the video? I don't know. At a certain point you lose track because it's funny, right? It's weird seeing a ton of yourself. I didn't know what else to do there, to be perfectly honest. What do you mean? Well, it's just… I don't know why I went. Felt like… You ever do something, and then when you're done you don't even know why you did it? That's what it felt like. It felt like it was a dream or something. What were you expecting to happen? From what they told me, well, they said it'd be like they were in there with each other forever, but it's not them, it's me. There were thousands of me, and they're dead now, crushed, viscera peri. One of the eyes floated by, and I saw myself in it. Dude, that's fucked. At this point both the interviewer and interviewee become aware of several humanoids identical to themselves present in a room, standing around the edges. Jesus! Several iterations of Dr. Crang immediately attempt to exit the room, but are unsuccessful. All instances begin fighting one another, while the copies of ██████ congregate in the opposite side of the room around the original interviewee, covering their faces with their hands. Get away! Stop! Help! Copies of ██████ begin violently ripping flesh and skin off of them, while copies of Dr. Crang do the same. This violence continues until there are only one iteration of each left. It is unknown whether or not these are the originals. Video playback ends, with both attempting unsuccessfully to exit the interview area. And log. Closing statement. Subject with issued Class-C amnestics at the interview. The SCP-3006 anomaly continues to persist, and several other instances of SCP-3006 have been subsequently documented in Foundation Video Archives. An upgraded caterer is under consideration.


Everybody forced here up in a box of leaf decadence pop century edition for drew B Leave pop century decadence here we go drew Three premium autograph frame cards per box All right, and it doesn't feel tonight. We'll rip tomorrow night You can only cut with the late knife by the way only works delay Hey legal Empty box just just so you know Legit box. All right. Is this gonna spoil everything or just a little covered. All right, here we go I don't know which way to go. I'll just take this one. All right. Here we go first one First one is a Carmen Electra Carmen Electra that is a they're all framed. I guess two of three. They're all in case two of three Carmen Electra auto Yeah, I bet Right next up next up is a dual auto of Liv Tyler and Alicia Silverstone How about that one of one on that one Liv Tyler Alicia Silverstone? That's what's up Yeah, that is awesome. I didn't seen either of them in the other than the regular pop century. That's what's up that's awesome and The final hit for you drew is Kristen Bell Wow Kristen Bell four of seven all ladies in this one Kristen Bell Forgetting sir Marshall here you go Obviously a lot of other stuff It's very prominent actor actress actor actress There you go great box there. I think it's awesome man. Appreciate you drew. We'll get those out to you