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# A New Wiki And A New Identity

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

Note by Kishlaya Jaiswal
2 years, 10 months ago

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Here 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 $\sum_{n = 1}^\infty x^n \sum_{m = 1}^\infty \frac{m! n!}{m (m + n)!} = \sum_{n = 1}^\infty \frac{x^n}{n} = \log \frac{1}{1 - x}.$

- 2 years, 10 months ago

This is definitely amazing.......Genius!!!

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

Thanks.

- 2 years, 10 months ago

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

Congrats $$\ddot\smile$$

- 2 years, 10 months ago

Oh, thank you so much!!!

Actually, I wanted to have some appealing and striking name but I'd no idea for the same.

- 2 years, 10 months ago

Kishlaya's a beautiful name! Suiting for a beautiful identity.

- 2 years, 10 months ago

Right !!!, see this .

- 2 years, 10 months ago

@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.

- 2 years, 10 months ago

$$\ddot\smile$$ $$\ddot\smile$$ $$\ddot\smile$$ $$\ddot\smile$$ $$\ddot\smile$$ $$\ddot\smile$$

- 2 years, 10 months ago

Haha thanks .

- 2 years, 10 months ago

@Jake Lai I am sure that soon I'll be seeing another beautiful identity named after a beautiful name Jake Lai, the great mathematician .

- 2 years, 10 months ago

You flatter me :)

- 2 years, 10 months ago

What is the geometric meaning of double summation?

- 2 years, 10 months ago

Hey, can you please tell me where can I learn double and triple summation from ? Or can you please write a wiki on it here on Brilliant ?

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

Thanks for the same .

- 2 years, 10 months ago

Try studying a bit of discrete calculus.

- 2 years, 10 months ago

Can you suggest some sources ?

- 2 years, 10 months ago

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.

- 2 years, 10 months 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 ?

- 2 years, 10 months ago

I am too a learner

- 2 years, 10 months ago

I wish I had as much brains as you do!

- 2 years, 10 months ago

Indeed, you have much more brain than me.

Getting a new identity doesn't makes me more intelligent than you. :)

- 2 years, 10 months ago

can u plse tell all of us the proof

- 2 years, 10 months ago

Yep, you can read the complete article/wiki (proof) here - Kishlaya's Identity

- 2 years, 10 months ago

This identity only seems to hold for $$x>-1$$?

- 2 years, 10 months ago

Yes Sir, I mentioned that in the wiki.

- 2 years, 10 months 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.

- 2 years, 10 months ago

Sir, then the generalized version of this identity may amaze you more.

- 2 years, 10 months 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?

- 2 years, 10 months ago

Can you help me with that? (I have no idea about how to notify Wolfram|Alpha about it. I googled it out but found no clues for the same.)

Sir, that's just too great of you and thank you so much.

- 2 years, 10 months ago

Okay, there's Wolfram MathWorld, which actively seek contributions, and may be contacted at:

MathWorld
Wolfram Research, Inc.
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:

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.

- 2 years, 10 months ago

Sir, I wrote up an email to Wolfram, a few days ago. In reply, I was suggested, to create a demonstrations project using Wolfram Mathematica and submit it at their site. But I doubt because it isn't some kind of demonstration and actually a simple identity.

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.

- 2 years, 10 months 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.

- 2 years, 10 months ago

Thanks.

- 2 years, 10 months ago

Best of Luck @Kishlaya Jaiswal

- 2 years, 10 months ago

Thank You!

Well, I'm researching over it yet, because I still think, there may exist some other interesting variant.

- 2 years, 10 months ago

there r already so many names

- 2 years, 10 months ago

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)?

- 2 years, 10 months ago

Well, I got what you are confused with.

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

- 2 years, 10 months ago

Consider this :

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

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.

- 2 years, 10 months 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.

- 2 years, 10 months ago