an aesthetic double sum

[@SanketGarg]

Such a great problem with elegant solution.

[@tomatrix7525]

Another beauty Michael. Hello from Ireland

[@keinKlarname]

I really like such problems with elegant solutions.

[@winteringgoose]

I wrote out this expression for a coworker yesterday, and to my astonishment, the instant I finish writing, she blurts out, "It's 6!" Shocked, I ask her how. She proceeds to explain: "Because m and n are both 0, so 2 and 2 and 2 has to be 6!" I told her she was right but that wasn't why...we didn't go any further than that, but it was a funny little moment to share.

[@toamatau8785]

Why would you write it out for someone who thinks 2^0 is 2 or more worryingly that 2(2+2)=6

[@winteringgoose]

@@toamatau8785 Because I don't know what someone will think about a thing before I tell them about it

[@JM-us3fr]

Dang it, I didn't notice the common denominator things leads to a cancellation. I left the fractions un-combined as (1/2^m+1/2^n) and had to do a lot more work.

[@JL-vh8ne]

This equation are called "Macintosh Plus".

[@goodplacetostop2973]

12:10 Can’t believe Ajax Amsterdam put 13 goals yesterday. Anyway, daily homework... For notation reasons, let’s introduce surd(x,n) being the n-th real root of x. So, surd(x,2) is the square root of x, surd(x,3) is the cube root of x, and so on. In the development of the binomal (a • surd(a/3,5) - b/surd(a^3,7))^n, determine the terms that contains a to the power of three, if the sum of the binomial coefficients that occupy uneven places in the development of the binomial is equal to 2048.

[@alphacharlie1979]

-264 a^3 b^7

[@padraiggluck2980]

Wow! Thank you for your videos, Prof Penn.

[@andcivitarese]

let S=sum[m=0 to inft] (m+1)/2^m. Split into sum[m=0 to inft] m/2^m + sum[m=0 to inft] 1/2^m. The second term converges to 2. In the first term let m=k+1 and the sum becomes sum[k=-1 to inft] (k+1)/2^(k+1). Bring a 2 outside the denominator and notice that we can neglect the term for k=-1 which is 0, and so the first term becomes (1/2) sum[k=0 to inft] (k+1)/2^k, i.e. (1/2)S. Putting all toghether we get an equation for S: S=S/2+2 from which S=4

[@toamatau8785]

Was very pleasantly surprised i was able to follow along with this, lol.

[@shanmugasundaram9688]

Separating 'x' outside the summation and writing it as differential of product functions is indeed genius.The double summation converges interestingly to a natural number and that too to a perfect number' 6'.

[@VaradMahashabde]

I am just sad he didn't use an arithmetico-geometric series

[@TheMauror22]

That's what I call supersymmetry

[@shahinjahanlu2199]

You are great and I enjoy your solution

[@skwbusaidi]

But wolframalpha gives that 1/2 (m+n+mn)/(2^(m+n)) coverages to 6 which is correct

[@zecareca42]

Extremely elegant

[@PlayerMathinson]

Can someone please explain me how did covergence lead to him exchanging the terms of the summation and having no effect?

[@Milan_Openfeint]

If a sum converges, you can reorder the terms. If the sum (result) is infinite, you can't. It's... some rule that everybody knows ;-)

[@Dhanush-zj7mf]

@@Milan_Openfeint I was wondering why is that rule true....😊

[@PlayerMathinson]

@@Milan_Openfeint Yes, but any intuition or proof you can give? Thanks.

[@PlayerMathinson]

@ゴゴ Joji Joestar ゴゴ Thank you for your reply. Never knew of this theorem, will find more about it.

[@buxeessingh2571]

The series converges absolutely, so you can exchange sums.

[@BrainsOverGains]

Isn't assuming that the series converges too show that the series converges circular reasoning?

[@Qsdd0]

He's assuming the series converges to find it's value. The argument for the convergence is that the exponential beats the polynomial, but that of course isn't quite rigorous.

[@TheMauror22]

Maybe now, with an epsilon-N argument, one could show that the limit of the partial sums is 6.

[@buxeessingh2571]

It is absolutely convergent, so you can exchange the limit operations.

[@BrainsOverGains]

@@buxeessingh2571 and why does it absolutely converge? 🤔

[@skwbusaidi]

Wolframalpha can not solve this. It gives that series is diverges

[@JivanPal]

It seems that WolframAlpha fails to apply the limit test properly to nested series. For example, it even incorrectly states that Σ_m Σ_n [1/(mn)] fails to converge by virtue of the limit test, which is incorrect, as the limit test gives an inconclusive result on that series.

[@EebstertheGreat]

These videos are great to watch at 2x speed.

[@k3dr1]

Very aesthetic indeed

[@jkid1134]

One blackboard ❤

[@JorgeGomez-li9td]

Very nice!

[@Schaex1]

This is really beautiful! The only thing I don't really understand is why do you evaluate the sum at x=0.5? Did I miss something or what's the reason behind that?

[@easymathematik]

Evaluating x^n at x = 0.5 gives you 2^n in the denominator. :)

[@Schaex1]

@@easymathematik Oooh, I see. Thanks a lot ^^

[@easymathematik]

@@Schaex1 Your welcome. :)

[@MarcoMate87]

Before commuting the order of the sums we should prove that the double series is convergent (absolutely, but it has positive terms so simple and absolute convergence are coincident). This calculus you showed doesn't prove that the series converges, it only proves that if the series converges, the value of the double sum is 6.

[@user-A168]

Good

[@guill3978]

6 is already 110

[@tasnimmahfuznafis8892]

At around 1:40, he exchanged m and n and it changed the form of the problem yet did not affect the total sum somehow. I am in high school now and do not understand how that could have happened. Can someone please explain to me how that happened or what topic I need to study to understand that part?

[@sensei9767]

𝓐 𝓔 𝓢 𝓣 𝓗 𝓔 𝓣 𝓘 𝓒

[@kshitijsharma3170]

😭it's difficult

[@l1mbo69]

I'd say it's easier than most Olympiad problems.. although the first step would be quite unnatural for someone's first time (it was for me at least)

[@lucatanganelli5849]

This is a good place to stop saying this is a good place to stop. Whoever thought of editing that "good place to stop" compilation just made this guy's endings so awkward. Anyway great video Mike.