Hi, everyone!

Today, I found an amazing new identity, and I wish to share with you all. Well, it states

\[\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m{n+m \choose m}} = \log \left(\frac{1}{1-x}\right)\]

I have added a wiki on it - Kishlaya's Identity

Since, I was very excited that I didn't thought about having a new name for it. So, do you have some suggestions for name?

All kind of feedbacks will be appreciated.

Thanks,

Kishlaya Jaiswal

## Comments

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TopNewestHere is a simple proof of the identity. First, we write \[\sum_{n = 1}^\infty \sum_{m = 1}^\infty \frac{x^n}{m \binom{n + m}{m}} = \sum_{n = 1}^\infty x^n \sum_{m = 1}^\infty \frac{m! n!}{m (m + n)!}.\]

Then \[ \begin{align*} \sum_{m = 1}^\infty \frac{m! n!}{m (m + n)!} &= \sum_{m = 1}^\infty \frac{(m - 1)! n!}{(m + n)!} \\ &= \frac{1}{n} \sum_{m = 1}^\infty \frac{(m - 1)! n! \cdot n}{(m + n)!} \\ &= \frac{1}{n} \sum_{m = 1}^\infty \frac{(m - 1)! n! \cdot [(m + n) - m]}{(m + n)!} \\ &= \frac{1}{n} \sum_{m = 1}^\infty \left( \frac{(m - 1)! n! \cdot (m + n)}{(m + n)!} - \frac{(m - 1)! n! \cdot m}{(m + n)!} \right) \\ &= \frac{1}{n} \sum_{m = 1}^\infty \left( \frac{(m - 1)! n!}{(m + n - 1)!} - \frac{m! n!}{(m + n)!} \right). \end{align*} \] This sum telescopes to \(1/n\), so \[\sum_{n = 1}^\infty x^n \sum_{m = 1}^\infty \frac{m! n!}{m (m + n)!} = \sum_{n = 1}^\infty \frac{x^n}{n} = \log \frac{1}{1 - x}.\] – Jon Haussmann · 2 years, 1 month ago

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I'll definitely add this proof (by your name) in my article and wiki.

Thanks. – Kishlaya Jaiswal · 2 years, 1 month ago

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Kishlaya , the name's fine. Not everyone is lucky enough to have an identity with their name attached to it .

Congrats \(\ddot\smile\) – Azhaghu Roopesh M · 2 years, 1 month ago

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Actually, I wanted to have some appealing and striking name but I'd no idea for the same. – Kishlaya Jaiswal · 2 years, 1 month ago

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– Jake Lai · 2 years, 1 month ago

Kishlaya's a beautiful name! Suiting for a beautiful identity.Log in to reply

this . – Azhaghu Roopesh M · 2 years, 1 month ago

Right !!!, seeLog in to reply

@Azhaghu Roopesh M You're truly dynamic, restless, independent, ready to accept challenges, and outspoken. You enjoy change, travel, and new experiences.

And this is far more better. – Kishlaya Jaiswal · 2 years, 1 month ago

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– Azhaghu Roopesh M · 2 years, 1 month ago

\(\ddot\smile\) \(\ddot\smile\) \(\ddot\smile\) \(\ddot\smile\) \(\ddot\smile\) \(\ddot\smile\)Log in to reply

– Azhaghu Roopesh M · 2 years, 1 month ago

Haha thanks .Log in to reply

@Jake Lai I am sure that soon I'll be seeing another beautiful identity named after a beautiful name Jake Lai, the great mathematician . – Kishlaya Jaiswal · 2 years, 1 month ago

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– Jake Lai · 2 years, 1 month ago

You flatter me :)Log in to reply

What is the geometric meaning of double summation? – Megh Choksi · 2 years, 1 month ago

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I've searched a lot of places but got no satisfactory info .

Thanks for the same . – Azhaghu Roopesh M · 2 years, 1 month ago

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– Jake Lai · 2 years, 1 month ago

Try studying a bit of discrete calculus.Log in to reply

– Azhaghu Roopesh M · 2 years, 1 month ago

Can you suggest some sources ?Log in to reply

This is a pretty good source, and covers the difference, sum, and rules pretty thoroughly. Kishlaya's own identity uses the Beta function, though, rather than discrete methods. – Jake Lai · 2 years, 1 month ago

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– Azhaghu Roopesh M · 2 years, 1 month ago

Thanks . Can you tell me from where do you study topics on Calculus and other topics so that I can solve good problems here on Brilliant ?Log in to reply

– Megh Choksi · 2 years, 1 month ago

I am too a learnerLog in to reply

I wish I had as much brains as you do! – Chirag Trasikar · 2 years, 1 month ago

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Getting a new identity doesn't makes me more intelligent than you. :) – Kishlaya Jaiswal · 2 years, 1 month ago

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can u plse tell all of us the proof – Navdeep Nainwal · 2 years, 1 month ago

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Kishlaya's Identity – Kishlaya Jaiswal · 2 years, 1 month ago

Yep, you can read the complete article/wiki (proof) here -Log in to reply

This identity only seems to hold for \(x>-1\)? – Michael Mendrin · 2 years, 1 month ago

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– Kishlaya Jaiswal · 2 years, 1 month ago

Yes Sir, I mentioned that in the wiki.Log in to reply

– Michael Mendrin · 2 years, 1 month ago

Oh there it is, I see it. I have to admit, that's quite an awesome identity. And it's one of those fascinating cases where 2 different functions can be identical over a certain range, and suddenly go separate ways.Log in to reply

generalized version of this identity may amaze you more. – Kishlaya Jaiswal · 2 years, 1 month ago

Sir, then theLog in to reply

– Michael Mendrin · 2 years, 1 month ago

I've checked into Wolfram's Special Functions site to see if it already has something similar to what you have. It does have something , but it's still too far different from what you have. Maybe you should notify Wolfram about your identity?Log in to reply

Sir, that's just too great of you and thank you so much. – Kishlaya Jaiswal · 2 years, 1 month ago

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MathWorld

Wolfram Research, Inc.

100 Trade Center Drive

Champaign, IL 61820-7237

USA

fax: +1-217-398-0747

and then there's Wolfram Special Functions site, which is more tightly limited, but will still consider contributions of interest, and may be contacted at:

comments@functions.wolfram.com

Do not contact "WolframAlpha", you really should start with www.wolfram.com and go from there. It' s a pretty big organization, so you'll need patience in finding the right contact interested in your work. Understand that even if Wolfram decides not to publish your identity, they still may want to use it as part of their "knowledge base". Wolfram Mathematica relies on voluntary contributions from a large base of mathematicians. – Michael Mendrin · 2 years, 1 month ago

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Can you suggest what to do next because I am not sure if a demonstration can still be made out of that identity?

[P.S. - In the meantime, I've found two other interesting identities/series.]

Thanks. – Kishlaya Jaiswal · 2 years, 1 month ago

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– Megh Choksi · 2 years, 1 month ago

Hi Kishlaya Jaiswal , instead of posting your 2 newly founded indentities with proofs and examples( in the form of wiki) , post a note containing this - To prove - 1)(identitiy 1) and 2 and so on , and in the end write I will post a wiki on it , the suitable time you think so that we too can think on it and our skills may increase and there could be more than one proof to a problem. Thanks.Log in to reply

Thanks. – Kishlaya Jaiswal · 2 years, 1 month ago

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@Kishlaya Jaiswal – Azhaghu Roopesh M · 2 years, 1 month ago

Best of LuckLog in to reply

Well, I'm researching over it yet, because I still think, there may exist some other interesting variant. – Kishlaya Jaiswal · 2 years, 1 month ago

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there r already so many names – Incredible Mind · 2 years, 1 month ago

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Let's start from basics(I am confused right now)

What's the definition of summation and integral according to you?

How can you simultaneously works with 2 variables?

What does the upper limit as infinity means?

i am unable to understand the logic behind interchanging integral and sum ?

How can you consider the summation part as a slope for the integral?

I am totally confused seeing 2 sigma right now(what do they really mean , why are you taking it independent?) Can I consider infinite sigma with infinite integral such as( a reverse picture of what you were having i.e summand upper limit infinity and integgral upper limit as n) \( \displaystyle \sum_{a}^{\infty} \sum_{b}^{\infty} \cdots \int_{0}^{n} da \int_{0}^{n} db \cdots = \int_{0}^{n} \sum_{a}^{\infty} da \int_{0}^{n} \sum_{b}^{\infty}db \cdots\) ?

At last what is calculus according to you(I think I understood calculus in a wrong manner)? – Megh Choksi · 2 years, 1 month ago

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Actually, it's just a special case of Fubini/Tonelli theorems.

Whereas I'll try my best to update that wiki to completely clear your doubts but in the meantime you can look up here – Kishlaya Jaiswal · 2 years, 1 month ago

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\(\displaystyle \int { f(x)dx } +\int { g(x)dx } = \int { f(x)+g(x)dx } \)

What does this mean. This means that you have evaluating an expression in two ways, you first integrate the functions seperately and than add it, or rather you add the two functions to get the new funciton and then integrate it. You could do it with 3 functions as follows :

\(\displaystyle \int { f(x)dx } +\int { g(x)dx } +\int { h(x)dx } =\int { f(x)+g(x)+h(x)dx } \)

Or with \(n\) functions :

\(\displaystyle \int { { f }_{ 1 }(x)dx } +\int { { f }_{ 2 }(x)dx } +.....+\int { { f }_{ n }(x)dx } =\int { { f }_{ 1 }(x)+{ f }_{ 2 }(x)+.....{ f }_{ n }(x)dx } \)

So if you write that in terms of sigma notation you have :

\(\displaystyle \sum _{ r=1 }^{ n }{ \int { { f }_{ r }(x)dx } } =\int { \sum _{ r=1 }^{ n }{ { f }_{ r }(x) } dx } \)

This is called interchange of summation and integral.

I hope this clear things a bit. – Ronak Agarwal · 2 years, 1 month ago

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– Megh Choksi · 2 years, 1 month ago

Yes , yes thank you at that time I was confused to see double summation with integral , the after sometime I understood it. OK nice explanation thank you.Log in to reply