1945 the war was over came back to the ruins of Europe the war was over but there was no peace despair crouched over the continent hopelessness circled Europe like a bird of prey why what were the forces what were the issues in a war that turned nations into rubble heaps and populations into beggars the people wanted the answers they wanted to know what happened and why in the Palace of Justice in Nuremberg the people of the world came together for there sat the international military tribunal to judge the chief Nazi war criminals the opening statement for the prosecution the privilege of opening the first trial in history for crimes against the peace of the world imposes a grave responsibility the wrongs which we seek to condemn and punish have been so calculated so malignant and so devastating that civilization cannot tolerate their being ignored because it cannot survive their being repeated that four great nations flushed with victory and stung with injury stay the hand of vengeance and voluntarily submit their captive enemies to the judgment of the law is one of the most significant tributes that power has ever paid to reason this inquest represents the practical effort of four of the most mighty of nations with the support of 15 more to utilize international law to meet the greatest menace of our time aggressive war the common sense of mankind demands that law shall not stop with the punishment of petty crimes by little people it must also reach men who possess themselves of great power and who make deliberate and concerted use of it to set in motion evils which leave no home in the world untouched in the prisoners dock sit 20 odd broken men reproached by the humiliation of those they have led almost as bitterly as by the desolation of those they have attacked their personal capacity for evil is forever past merely as individuals their fate is of little consequence to the world what makes this inquest significant is that these prisoners represent sinister influences that will lurk in the world long after their bodies have returned to dust they are living symbols of the arrogance and cruelty of power of racial hatred of terrorism and violence they are symbols of fierce nationalism and of militarism of intrigue and war making which have embroiled Europe generation after generation crushing its manhood destroying its home and impoverishing its life they have so identified themselves with the philosophies they conceived and with the forces they directed that any tenderness to them is a victory and an encouragement to all the evils which are attached to their names what these men stand for we will patiently and temperately disclose we will give you undeniable proof of incredible event the catalogue of crimes will omit nothing it may be that these men of troubled conscience do not regard a trial as a favor but they do have a fair opportunity to defend themselves a favor which they rarely extended their fellow country we will not ask you to convict these men on the testimony of their foes there is no count of the indictment that cannot be proved by books and records and we will show you the defendants own film you will see their own conduct and hear their own voices as they reenact for you from the screen some of the events in the course of the conspiracy the acts of the defendants have bathed the world in blood and set civilization back a century they have subjected their European neighbors to every spoliation and deprivation they have brought the German people to the lowest of the wretched they have stirred hatred and incited domestic violence and every continent these are the things that stand in the dark shoulder to shoulder with these prisoners the real complaining party at your bar is civilization


Bitcoin is forming a double bottom pattern. What does this mean exactly? It means that if this pattern plays out, we would see a complete reversal of the price that could take us all the way back up to $36,000. So into this video, we will break down exactly what a double bottom pattern is and how it works. We will also talk about why $36,000 is our target for Bitcoin with this pattern and exactly how accurate this pattern is in a study done with 10 years of data and over 200,000 patterns tracked. Also stay to the end for a special bonus that you won't wanna miss. Let's go ahead and dive right into today's video. Remember guys that this content is for educational purposes only and this is not financial advice. Always do your own research before risking any of your money. And as a disclaimer, I've been invested in Bitcoin since 2016. So what exactly is a double bottom pattern? The double bottom is one of the most common reversal price patterns. It is also one of the most accurate price patterns. So as you can see in this example here, we have a bottom. This would be considered the first bottom. Then we have somewhat of relief rally. This creates the neckline. Then we have the second bottom here. Now in order for the double bottom pattern to play out, it needs to break above this neckline as you guys see here. Once the break happened, as you see in this example, the target of this pattern is the same distance from the low to the neckline. Once the breakout occurs from the breakout point to the top, as you guys can see here. Here's another example of what the double bottom looks like. You see we have the first bottom. We create the resistance line or the neckline, second bottom and then we have a breakout with of course a test of the breakout which happens very common as well before it continues to the upside. So now if we come back here to our chart, you can see that we have the clear trend to the bottom. We have a clear bottom here. Then we have a rally back up which sets up a resistance and a neckline and another drop back down into the same range as the first bottom creating the second bottom. If we measure the low to the top of the neckline, it gives us our potential target here once the breakout occurs which gives us a price target of about $36,000 if this were to play out. We'd be looking for something like this where we retest this neckline here, then get a breakout. I'd imagine we probably retest this breakout and kind of continue up all the way to reach our target. This is just an idea of what it could look like but you see we have the break of the neckline here. We test this Fibonacci level which is a previous support back here. Retest the breakout here, breakout here and retest another previous resistance up here around the $31,000, $32,000 range. And then finally we get to our target based on the low to the neckline which also happens to be a Fibonacci level. That if we zoom out a bit, you can see that same exact level has been important many, many times over. You can see we actually, last time we dropped this low when we broke above that level, it ended up sending us to our necks and our current all-time high. So you can see here the current neckline is sitting just about around $25,000. That was the high that we set last time after our first bottom. Now we still haven't seen a rally back up after our second bottom so that's currently what we're waiting for, what we're expecting. And this is of course all based on this chart pattern. Things of course can change depending on events outside of technical analysis. So rate hikes, inflation, wars, all things outside of technical analysis could of course make this double bottom invalid. But as we currently stand, it is a valid structure and for it to come into play we must break above that neckline which is sitting at $25,000. So now I've told you exactly what the double bottom pattern is and how it works, where the neckline is where we must break out in order for this pattern to be valid, exactly how we found our price target on a breakout above the neckline. And now I wanna go over what the odds are exactly of this playing out, the success rate and basically the probability of success with these double bottom patterns. So if we look at a study that was done with 10 years of data in over 200,000 price patterns tracked, it shows that a double bottom pattern has a 78.55% chance of success. So although it wasn't the most successful price chart pattern in this study, it is one of the highest with that 78% success rate. And if we do test that neckline and fall right back down to support, there is something called the triple bottom pattern as you guys can see here. And in that instance, the percentage for success actually goes up, moving us from 78% to 79%. So if we don't break out this time and fall back to support, as long as we hold, then the triple bottom pattern would be in play at that point. So with this information, what can you do to profit based on this? So there's a few things you can do. The first one is buying here at support. This is the current bottom area. Anywhere between 18,000 to $20,000 is a good time to add to your long-term position. Anything you add here should be probably locked up until the next bull run, which I expect around 2024. So any extra cash that you have lying around that you're doing nothing with that you can lock up for the next two to three years, this is the range where you wanna be adding those funds. Now the next thing that I'm also doing is opening long positions for the short term with leverage in this range. So now you don't just wanna open one entry here around 19, you wanna have multiple entries all the way down to about $18,000 and probably have your stop loss somewhere below $18,000 range. Then you also wanna have take profit areas set up on the way up. So you can find those take profit areas by looking for previous resistant zones here. So we can see here, this was a key resistant zone. That's around 21,500. You can see up here another resistant zone. We can see some support areas here. That's between 22,500 up to around 24,000. Of course, we have the neckline here that is going to be a resistance, that's around 25,000. Another area of interest right here, that's around 21,500 again. And right here, this is around 22,000. So you can use all of those basically to set up different take profit areas as the price bounces back up. And if you just wanna open a leverage position for the long term, you can also do that and use these long term resistances like the neckline, the Fibonacci level up here, which is around $29,000, $30,000. And then our main target up here, which is around $36,000. You do have to hold a bit longer because we don't have a time frame on how long this will all take, but it definitely would pay off, especially if it hit all our targets here. And the final way that you can use this is remember that Bitcoin is the king of all of crypto, right? Of all the markets. So when Bitcoin goes down, the rest of the crypto market goes down. When Bitcoin goes up, the rest of the crypto market also goes up. So if this double bottom pattern were to play out, that means that the rest of the crypto market would also go up. And you know that outcoins always over exaggerate Bitcoin's move. So if Bitcoin were to go up to $36,000, which is pretty much a 100% return from current prices, that means that outcoins will probably do a 2, 3, 4, 5X, maybe even more in that same period of time.  And you can even start opening up some leverage trades as well, some small positions and letting them ride up as well if this double bottom pattern plays out. Where you see we have the break of the neckline here. We test this Fibonacci level, which is a previous support back here. We test the breakout here, breakout here and retest another previous resistance up here around the $31,000, $32,000 range. And then finally we get to our target based on the low to the neckline, which also happens to be a Fibonacci level. That if we zoom out a bit, you can see that same exact level has been important. Many, many times over you can see we actually last time we dropped this low when we broke above that level, it ended up sending us to our necks and our current all-time high. So you can see here the current neckline is sitting just about around $25,000. That was the high that we set last time after our first bottom. Now we still haven't seen a rally back up after our second bottom. So that's currently what we're waiting for and look out for as you load up your bags here for 2023, 2024, 2025. There's a lot of potential out there once Bitcoin does break above and start running back up. Now if you guys are interested in opening any of these leverage type positions, I personally use Bybit to open these positions. I have a link in the bottom in the description where you can earn up to a $4,000 bonus on your deposit. If you wanna know exactly what trades I'm taking, what I'm buying, when I'm selling, in real time, then you need to get into my mentorship program. In this group I share every single trade that I take. I break it completely down for you as you guys can see in this example with the entries, stop loss, and exactly where I'm taking profits. You can see I gave this trade on the third of this month and we were able to hit on our first take profit area and our second one as well as you can see the trade in total took about three days. We also break down exactly what's going on every day in the market and what might affect the market. Like for example, this week we have the CPI reports which come out today. I teach you guys how to trade the right way, how to use leverage, how to position size correctly, everything that you will need to know so that you never blow up your account again. For those who watch till the end today, you can get 25% off on your first month by using this code 25off. So just type the numbers 25 off all capital letters. You'll get 25% off on your first month in the group so you can try it out for yourself, see if it's a fit for you. So if you guys are interested, go to the link in the description below. Crypto is a once in a generation opportunity and the way that it is right now, the easy opportunities to make money will not last forever. So if you guys are interested, sign up in the link in the description below. If not, I recommend you to watch this video on the screen right now that will show you exactly what I'm buying during this bear market to 10X your portfolio in the upcoming bull run. Thank you guys so much for watching. I'll see you on the next one. As always, peace and love.


Welcome to NPTEL NOC course on point set biology part 2, module 13, we shall continue our study of para-compactness. For subspecies of RN, there is a result which says that every open cover has a subordinate partition of unity consisting of smooth function. The word smooth will not have any meaning when you are studying arbitrary topological spaces and para-compactness of that or partition of unity of that. Inside RN such a thing is possible and what is the additional thing that we have to do that namely on every disk you can have what is called as a bump function. For more you can see this reference here. The second thing is the local compactness in a very special way for RN namely the closed disks themselves are compact. So, if you carefully study the proof that we have given earlier, then we can get a similar result for any metric space with the help of decomposition of any open set into a countable union of increasing disk like open set. What I mean to say that is compactness is not all that necessary here the local compactness. So, something else will come to help namely the metric property and that is what we are going to do for more general results you may consult again Kelly's book. So, I will do the bare minimum here to expose you to the ideas behind this course is not something you can say concise or comprehensive and so on. Every pseudo metric space is para-compact. So, this is a big theorem now. So, slowly we have to have some patience slowly we will develop it and the development itself is quite educative. So, first of all let us have some notation here for every subset of this pseudo metric space for each integer n let us have a notation a n a is a subset and n is an integer a n is set of all x inside x such that distance of x from a complement is bigger than 1 by 2 power n. a may be any set you take the complement and from there a distance must be at least 1 by 2 power n of automatically these are subsets of a subsets of a. So, one thing very easy to see is that using triangle inequality distance between a n and a n plus 1 complement will be bigger than equal to 1 by 2 power n minus 1 by 2 power n plus 1 which is 1 by 2 power n plus 1. So, I again leave this element I think as an exercise clue. Note that for each a and each n what we have is this one this also easy a n is contained in a n bar that is obvious always, but a n bar is contained inside a n plus 1 and again a n plus 1 is contained inside a. So, already we have this is what we want it right only thing is these a i's are not compact or anything, but we have this increasing even increasing the phenomenon a n contained in a n bar contained a n plus 1 contained inside a and each a n is open. So, that is because this is this condition bigger than 1 by 2 power n ok. So, closure will be what bigger than or equal to that is all. So, that is the reason why you have these are open subsets and a n contained inside a n bar etcetera. Moreover if a itself is open then only a will be the union of a n in what we have proved earlier was when you have locally compact Lindelof space every open subset can be written like this with each a n bar compact and so on that is not that is not what we have here, but something which you have saved here namely with the use of matrix space we were able to write every open such that union of countable union union of a n with this property a n contained a n bar contained a n plus 1 ok. So, that is what I meant by you know writing every open set as a union of disk like open sets since a n minus 1 that is a n and the whole thing is that a slowly you will approach this a n ok. They are nested very strongly in the sense that the the previous thing is contained in a n plus 1 that is just nested, but the closure itself is contained in a n plus ok. So, these are just notations now you have to remember this one for the rest of the proof. Let now you be an open cover for x we want to extract a locally finite open refinement that is our purpose finally choose a well order on you. So, again here you have used a Ximov's choice or this one every set can be well order ok. For each n belonging to n and for each u inside u let us set up another set of notation now u n be the be defined as in one u n there is no there is no change ok u n is just instead of a here put u ok u n is except all x long time such that distance between x and u complement is bigger than 1 by 2 power n.  So, it is an ordering you take all the initial elements to u and take all there u v n plus 1 part only not the full v then take the union of all of them ok n is fixed here v is varying v is varying only this part initial segment ok take the interior of all this union and throw it away ok u n star is that so here is a picture of u and u 1 and v 1 ok u is the given one v occurring this v occurring before u ok. So, if you look at v 1 star there is nothing suppose v 1 was v was the first one and u was the second one there is no throwing because there is there is nothing below before that.  So, that is u 1 star only this part ok because of u n and v n definition if you take any element of u element of v 2 here the distance will be at least one fourth in particular distance between any element here and any element here will be at least one fourth ok. So, this is the picture that is showing. So, why we are making this kind of you know adding or subtracting some portions of earlier elements only that is what we are doing here ok u n star is this one. So, only u 1 v 1 picture I have shown.  So, for all v before u which occur before you in this well order well order would be any order does not matter just one well order you have to fix once for all ok.  So, why is a close subset what I have just deleted some open subset that is interior of something whatever it is if you delete an open subset it is a close subset of whatever wherever you have deleted close subset of u n ok.   So, that is the definition therefore, in either case what happens is from this general remark here distance between a n and a n plus 1 complement is bigger than 1 by 2 power n plus 1 which I showed you in this picture what happens is distance between u n star and v n star is always bigger than equal to 1 by 2 power n plus 1. So, in this picture it was n equal to 1. So, it is one fourth ok.  After that you have to just use whatever you have proved before all right. So, if you use property 2 this will be obvious. The next thing is each x belongs to u n for some n and some u right first of all it belongs to some u because u is a cover, but once it belongs to some u from the complement of u its distance will be positive. So, there will be someone by n for which it is be smaller than that bigger than that. So, that is all you have to show then it will be inside u n ok. So, first you chose choose u first u such that x is inside u and then since u is open and we have union of u n is equal to u ok. So, one of x must be inside one of the u n ok. We put now u n hat is see you can get both. So, first you have u n star here now u n hat is set of all x such that distance between x and u n star is less than 1 by 2 power n plus 3 ok. And u n griddle is all those x set distance x and u n star is 1 by 2 power 1 more 2 power n plus 4 ok. So, both of them are here this is less than equal to that you have to pay attention each u n is open once again and u n griddle is closed because there is equality here and because you see 2 power n plus 4 u n griddle will be contained inside u n hat. Once again the same property 2 will tell you the distance between u n hat and v n hat is bigger than or equal to 1 divided by 2 power n plus 2 ok for every u for every v inside u. So, this time you can directly use the fifth property here that distance with u n star union star u n divided by 2 n plus 4 ok. So, u n hat v n hat similarly you can talk about u n griddle this u n griddle really will not be needed in the final proof, but it will play some auxiliary role. So, I have kept it ok. So, distance between u n griddle and v n griddle is 1 by 2 power sorry u n hat 1 by 2 power n plus 2 and they are open subsets no matter whether u occurs first or v occurs first. Next for each n inside natural number put v n equal to this collection u n hat where u range is over all of u ok. So, each member is a open subsets. So, this is an open family of open sets all right.  So, we have written v as a countable union of these families. What is the property of these v n's and v? v is an open cover for x not v n's, but when you take all of them v that is an open cover for x and v is a refinement of u ok. So, how to check x this 10th one this can be checked as follows as in the case of 6 what we have what we have done here that each x belongs to some u n ok. So, each x will belong to some u n star also is what you have to think ok in as in 6 choose first u. So, that x is inside u then x will be inside some u n first ok, but then it is also in u n star because it does not belong to any v all those v's which you have subtracted it will not be inside v. So, that will be in u n star all right, but why it is in u n hat since u n star is contained inside u n hat we are done ok. Indeed it is also proved that x itself is in u n twiddle also because u n stars are contained in the u n twiddle also ok. So, remember these are less than less than. So, u n hat and u n twiddle actually fatten the u n star right all those which are of distance smaller than that one. So, you have to understand this inequality he has in the beginning we are bigger than bigger than right. So, both of them are used left and right you are cutting down things all right.  So, that much we have done ok. And it is a refinement of u because all these elements are for each u n is subset of u corresponding u right. So, they are inside all right. Now, look at u n star is contained as a u n. So, u n hat which is actually all those x belong to x. So, the distance between x and u n is 1 by 2 power n plus 3 I am just recalling this definition is contained inside u n plus 2 ok. And that is contained inside u n was contained in u n plus 1, but u n hat is not contained in u n plus 1, but in contained in u n plus 2 ok. So, that is contained inside u ok. So, all these all these u n hats are also it is also an open cover for it is also a refinement of u ok. So, finally, I will have one more notation here let u n check equal to u n hat minus union of all vk twiddles v belonging to u and k is less than n. So, this is where the twiddles were used ok this is only case where we have to do what we are doing now u n check case I do not want even u n hat I am I am checking away some portion of that namely these are closed subsets now take the union of all these vk hats where v is inside u, but integer k is smaller than n now all v I am taking inside u, but k must be less than n. So, you you throw away that part note that this vk twiddles where k is less than n there is a bound for under k vk twiddles, but v serviced in the entire of you this is a locally finite family of closed sets ok. Therefore, it follows that each u n check is open when you have locally if I remember this thing about locally finiteness locally finite family of closed sets when you take arbitrary union it is still a closed set ok. So, this whole thing is closed set the complement will be open subset now u n hats were open subset this u n twiddles were not they were closed subset. So, u n check these are open subset ok. I have a question why this family is locally finite because you have taken only finitely many of them ok. So, what is vk if you take any two of them different from same v the same k they are actually disjoint.  Sir, I have one more question here. So, starting with the open cover first we made u n stars the collection of which is also a cover for x, but that was not open that is why you consider u n hat. So, they are not they are closed subset. Yeah and so, this u n hat collection was open refinement, but that may not be locally finite. So, that is why you are coming to u n check right. Yes that this precise is subtracting these things you know vk hat that will make it u n check locally finite we will see that ok yeah. So, first of all u n check is open now you see we we did not even stop at u n hats also ok. So, first of all these are open itself you you have to look for that these these things are locally finite that is fine vk hat, but they are closed things right. So, now these are open subset first thing. So, what we want to do is that this is now take w to be the collection of all subsets of the form u n hat u range over u n range over n that is like all u n's first n fixed and then taking the union over n you can say double that that is a cover of x it is an open refinement this is locally finite itself all right there is no sigma locally finiteness here this w is actually locally finite. So, we have to prove this 13, 14 and 15 all right. So, say how do you prove 13? 13 is that that w covers the whole thing ok given any x again let n be the first inch is as x belongs to u n twiddle for some u ok once it is in some u there will be some n for which it belongs to ok. So, let us take x belong to some the first u n twiddle for some u it follows that it is not occurring here at all right it is always before that I have k less than n I have taken, but n is the first one to which it belongs to it will not be inside some other u k k less than n. So, this part it is not there so it must be inside u n check once it is here it will be inside u n check. So, that is the trick here. So, these things cover ok 14th is what it is an open refinement we have we have told this open and these are refinements they are all subsets of you that is clear 15 is why it is locally finite ok to see the 15th one notice that u n twiddle chosen as above is a neighborhood of x and does not intersect any v m check for m bigger than n because these n twiddle u n twiddle would have been subtracted from v m right. So, it does not intersect v m twiddle at all therefore, if we choose 0 less than r less than 1 divided by 2 power n plus 3 ok then this ball v r x it will be contained inside u n twiddle ok I sorry what I want to say r must be less than this it may be even further smaller so that v r x is contained u n twiddle because u n twiddle are open subsets neighborhoods ok they are open subsets also then v r x will intersect at most one member of v m ok because m less than equal to n n less than equal to m both of them it cannot be as soon as m m it will not intersect anything smaller or bigger 161 of them it will intersect ok so that is why this 1 by 2 power n plus 3 I have to show once there is an r such that v r of x is in u n twiddle you can make it smaller than any further also you can take it smaller than 1 by 2 power n plus 3 also ok. So, if such a choice is possible then it happens that it will intersect only one of them ok. So, this completes the proof of a theorem there is a remark here which which is a bit deep it takes you a little deeper. So, I do not mind even if you do not understand in the first reading ok so, but I will make this remark a family a of subsets of the topological space is called sigma discrete family if each x belong to x has a neighborhood which means at most one member of a. So, one of you ask this question that is why this remark will be even more relevant here now. So, why is sigma locally what happens is you see this condition right in the beginning distance between these two is bigger than 1 by 2 power n plus 1. So, this u n star and v n star same n, but u and v are different elements of the same cover what happens to these sets they are disjoint right not only that this is stronger than being disjoint namely I can take small open subsets around them right for all of them around all of them simultaneously. So, that all these neighborhoods are disjoint. So, because of this metric property we are able to do that one such a thing is you can make that as an axiom in the general case then it is called sigma discrete that is what I am trying to say here family is discrete family if each point x belongs to x has a neighborhood which means at most one member of a. You see if the distance between as something positive each point x I can take the ball of radius half of that distance whatever positive I have say half the half that radius then what I get is that that open ball cannot intersect both of them that is all. So, that is what we have achieved here it is called sigma discrete if it is union of countable union of a n's where each a n is discrete. So, that is why that n has come when you fix n it is a discrete family you take the union it becomes a cover and so on. Only countable unionality such a thing is called sigma discrete this family a itself is not discrete that is one thing you have to understand is sigma discrete. Properties is 8 to 9 say with that this v which we have constructed is sigma discrete open refinement of view. So, in general what one does it without assuming metric space you would like to prove this one out of some other properties by making this definition sigma discrete. We have not used this concept anywhere in the core and so that is what I want to say if you prefer you can simply ignore it for the time being if you are interested more then you can look into Kelly's book ok. So, coming back to R n we have remarked earlier that due to the existence of some smooth functions we get politicians affinity subordinate to an open cover. Moreover, since step one of the proof of 3.6 is valid for all open subsets of R n it follows that every open subset of R n is paracompact indeed it is also true that every subspace of R n is normal because it is a metric space, but what is important here is that given an open cover for any subspace there is a smooth partition affinity subordinate to that open cover but the functions are all defined on the entire of R n ok not on just that open subset thinking a little further along this line ok you will be able to prove the following theorem ok this is this I will get it as a corollary to whatever we have done so far. All this remark is for getting such a motivation from R n let x be a second countable locally compact half dot space then every closed subset of x is the precise zero set of a continuous real valued function you can choose the the codement to be 0 1 the closed interval 0 1 once you have this it follows that such a thing is a g delta set also because the precise set it is a zero set of a continuous function is a g delta set ok you can just write it as intersection of inverse image alpha inverse of 0 closed 1 by n o 1 ok so how do we prove this one it is not difficult start with any f contained in the x closed subset for each x in the complement choose a function alpha x from x to 0 1 such that alpha x at x is 1 and alpha x of f is 0 ok consider the open cover u is alpha x inverse of open 0 1 closed x belonging to this fc I have such function as alpha x so take u to be this open set this will be an open cover for fc ok fc being a closed subset of a second countable say is second countable locally compact and t2 ok therefore it is para compact so any open cover so this open cover u there is a locally finite open refinement further by second countability any open cover you can get a countable sub cover it will be again a locally finite refinement ok so we get a countable locally finite open refinement u n n belonging to n I can write like this of u ok these u n's are not necessarily subsets of u I mean they are not members of u but they are refinements they are contained in some members of u since each u n is contained in some alpha x inverse of 0 1 that is the way it of course right for some x we can select one alpha x such that and relabel it as alpha n instead of alpha x alpha x n I am cutting it short to alpha n that is all ok so define alpha now from x to 0 infinity no index here right equal to sum of all alpha n's but divided by 2 power n each of them after dividing 2 power n take the sum these 2 power n's are there obviously to make the whole sum convergent these alpha n's are bounded by you know they are 0 to 1 so any number 0 to 1 divided by 2 power n summation is convergent ok not only convergent uniformly convergent so alpha will be automatically continuous function now it is elementary thing to check that this alpha x is 0 exactly on f that is all we wanted to prove ok so go through this this prove carefully again and again maybe three times does not matter ok the each step each way why you are doing all this somewhat circus kind of thing they have meaning there next time we will do some general rate result which comes from nowhere but the motivation is here if you know this one you know where it is coming ok thank you


True or false, the Dallas Cowboys are gonna win the NFC East, let's dive into it. So the Cowboys have actually made their way into first place for the division, but whether or not they stay there remains to be seen. Their arch rivals in the Philadelphia Eagles actually got their easiest remaining schedule moving forward, and the Cowboys, I ain't gonna lie, they kind of got a little bit of a gauntlet. I already know this one's about to come down to the wire, this two-man race is about to have serious critical playoff implications. I can't wait to see who wins it.


The work involves the synthesis of 1, 2, 3 triazole derivatives through a series of reactions, characterization by FTR and NMR spectroscopy, and evaluation of their antibacterial activity against two multi-drug resistant pathogens. This article was authored by Rua Waseem Adam, Hutham Mahmoud Yusuf Al-Laban, Ahmed Abdul-Jabbar Jalub Al-Janabi, and others.