A limit of a cosine sum using Euler's formula for sin x/x.

[@maxsch.6555]

Because the 2^n is a strictly monotone and divergent sequence, one may also use Stolz-Cesàro theorem to get the answer.

[@ayushmanpaul3931]

Can you please show how to evaluate using the stolz-cesàro theorem

[@caluire24]

at... 9:30... should be cos(1/2^2^n) ?? sum till 2^n of 1/2^k ??? does not change the result actually😂

[@goodplacetostop2973]

13:25

[@birdbeakbeardneck3617]

u popular now

[@robertgerbicz]

The whole video is a big overshoot in many ways. If a(k) converges to c (finite number), then sum(k=1,n,a(k))/n converges to c (in your problem you can take a(k)=cos(1/2^k); c=1 and n=2^m (subsequence)). It is pretty trivial.

[@tomatrix7525]

Prime example of the squeeze thm. Lovely stuff

[@Jaeghead]

An easier way would be to use the fact that (cos(1/k)) converges to 1, which implies that the average (1/k Σ cos(1/m)) (sum from m = 1 to k) also converges to 1.

[@thefranklin6463]

Jaeghead does it matter that it’s sum(cos(1/2^k)) not sum(cos(1/m))?

[@Jaeghead]

@@thefranklin6463 I was just mentioning the simplified idea, in this case you use the fact that (cos(1/2^k)) converges to 1 and therefore the average of 2^n summands converges to 1.

[@thefranklin6463]

Jaeghead i think that’s a great simplification! I wasn’t aware of that identity before, so thanks for exposing me to it. I was curious how to show that the problem in the video could be reduced into the arithmetic mean problem.

[@xulq]

Shouldnt it be cos(1/2^(2^n)) on the right side instead of cos(1/2^n) at 9:10 ?

[@pcbenutzer6651]

Can you pleas make a video about AGM and relation to elltiptic integrals?

[@abdelbarichedad1741]

THANK U Micheal

[@get2113]

I think it follows easily as a Cesaro sum.

[@radhab8761]

Sir please make a video on the proof of l'hospital rule

[@yusufa.ratleh467]

Can you make a video about gamma function

[@avrahambabkoff3776]

To say lim as x->0 of sin(x)/x is 1 by l’hopitals rule is circular logic

[@GhostyOcean]

Isn't the final value cos(1/2^(2^n)), not cos(1/2^n)?

[@krisbrandenberger544]

@ 1:41 The denominator of the argument of cosine should be 2^k, not 2^n.

[@maxjackson6616]

at 9:03 could you use 1 instead of cos(1/2^n) to bound the sum from above? so you'd get 1/2^n sum from 1 to 2^n of 1 which is just 1?

[@AnishSarkarISIDelhi]

Can't you use the fact that if the n-th term converges to a, then the ratio of partial sum and n would also converge to a?

[@tahtouhladeb7671]

There's a much easier answer : using cesaro theorem

[@f5673-t1h]

What's that?

[@ancientwisdom7993]

@@f5673-t1h He's covered this here: kzbin.info/www/bejne/epSWfYt8nr2Wo7s

[@ayushmanpaul3931]

How to solve it using cesaro theroem

[@tahtouhladeb7671]

@@ayushmanpaul3931 cos(1/2^n) converges to 1 as n goes to infinity so 1/n * (sum k from 1 to n of cos(1/2^k) ) converges to 1

[@f5673-t1h]

Using proof by inspection, the answer is 1.

[@dugong369]

You're taking the mean of a lot (infinite number, in limit) of numbers. 100% of those numbers are infinitely close to 1 (within epsilon of 1 for any epsilon). So the mean can't be less than 1, and it can't be more (since all the numbers are less than 1). I'm guessing the theorem referred to in other comments states this more rigorously. I believe this argument can show that limit as n->infinity of 1/n * sum of cos(1/k) for k=1 to n is also 1, even though for finite n this expression is much smaller than the 2^n version.

[@kingkartabyo6206]

That is exactly the outline of proof of Cesaro's theorem, often showed in Real Analysis I class.

[@bmenrigh]

I didn’t understand why the lower bound in the second use of the squeeze theorem was justified. The replacing 2^n with inf part. I thought we were looking at 2^n as n -> inf so that’s inf too. I must have missed something subtle or misunderstood completely?

[@xCorvus7x]

In the second use the function in the middle is essentially a double limit, where once the root goes to infinity and then the product inside the root, too. Since the limit of that product is strictly less than any finite iteration of the product, we can get a lower bound to apply the squeeze theorem by replacing the interior of the root with the result of that limit.

[@Teclis1917]

Hello, mister Penn. Can you please make video about series with f_n(x)=sin(n!*pi*x)?

[@sujalsagtani6868]

All values of x any n for which 2n²+1 is a perfect square

[@sujalsagtani6868]

Please make a video on it this is useful in the concept of balancers on which you made a video For example 1+2+3+4+5=7+8

[@blazedinfernape886]

@Adam Romanov He is probably demanding a video from michael penn

[@cameronspalding9792]

I would have used the formula cos(x)=1+O(x^2) for all real x

[@anupamahalder279]

So nice ❤

[@estebansanches4236]

1

[@giuseppepapari8870]

How about the limit of my patience with getting one ad every 30sec?

[@douglasmagowan2709]

This is overkill. At the limit, there are are infinitely many terms in the series which are arbitrarily close to 1 and finitely many that are not. We can construct a rather simple delta-epsilon argument to show that the limit approaches 1.

[@emanuelorzusa]

First