Waste less time on Facebook — follow Brilliant.
×

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

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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

Jon Haussmann - 2 years, 10 months ago

Log in to reply

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

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

Thanks.

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

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, 10 months ago

Log in to reply

Oh, thank you so much!!!

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

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

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

Jake Lai - 2 years, 10 months ago

Log in to reply

@Jake Lai Right !!!, see this .

Azhaghu Roopesh M - 2 years, 10 months ago

Log in to reply

@Azhaghu Roopesh M @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, 10 months ago

Log in to reply

@Kishlaya Jaiswal \(\ddot\smile\) \(\ddot\smile\) \(\ddot\smile\) \(\ddot\smile\) \(\ddot\smile\) \(\ddot\smile\)

Azhaghu Roopesh M - 2 years, 10 months ago

Log in to reply

@Kishlaya Jaiswal Haha thanks .

Azhaghu Roopesh M - 2 years, 10 months ago

Log in to reply

@Jake Lai @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, 10 months ago

Log in to reply

@Kishlaya Jaiswal You flatter me :)

Jake Lai - 2 years, 10 months ago

Log in to reply

What is the geometric meaning of double summation?

U Z - 2 years, 10 months ago

Log in to reply

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 .

Azhaghu Roopesh M - 2 years, 10 months ago

Log in to reply

Try studying a bit of discrete calculus.

Jake Lai - 2 years, 10 months ago

Log in to reply

@Jake Lai Can you suggest some sources ?

Azhaghu Roopesh M - 2 years, 10 months ago

Log in to reply

@Azhaghu Roopesh M 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, 10 months ago

Log in to reply

@Jake Lai 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 ?

Azhaghu Roopesh M - 2 years, 10 months ago

Log in to reply

I am too a learner

U Z - 2 years, 10 months ago

Log in to reply

I wish I had as much brains as you do!

Chirag Trasikar - 2 years, 10 months ago

Log in to reply

Indeed, you have much more brain than me.

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

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

can u plse tell all of us the proof

Navdeep Nainwal - 2 years, 10 months ago

Log in to reply

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

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

This identity only seems to hold for \(x>-1\)?

Michael Mendrin - 2 years, 10 months ago

Log in to reply

Yes Sir, I mentioned that in the wiki.

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

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.

Michael Mendrin - 2 years, 10 months ago

Log in to reply

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

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

@Kishlaya Jaiswal 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?

Michael Mendrin - 2 years, 10 months ago

Log in to reply

@Michael Mendrin 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.

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

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

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, 10 months ago

Log in to reply

@Michael Mendrin 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.

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

@Kishlaya Jaiswal 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.

U Z - 2 years, 10 months ago

Log in to reply

@Michael Mendrin I'll follow your instructions.

Thanks.

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

@Kishlaya Jaiswal Best of Luck @Kishlaya Jaiswal

Azhaghu Roopesh M - 2 years, 10 months ago

Log in to reply

@Michael Mendrin Thank You!

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

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

there r already so many names

Incredible Mind - 2 years, 10 months ago

Log in to reply

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)?

U Z - 2 years, 10 months ago

Log in to reply

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

Kishlaya Jaiswal - 2 years, 10 months ago

Log in to reply

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.

Ronak Agarwal - 2 years, 10 months ago

Log in to reply

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.

U Z - 2 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...