We know that we can generate the Bernoulli numbers using the relation $(1+B)^n=B^{[n]}$ where $B_n$ is $n$th Bernoulli number. But how we can prove this works? Thanks to all.
Edit 2: is there a website or book that can give me good information?
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If $B_n$ is a Bernoulli number then what is $B$? What is $B^{[n]}$? – Soham Chowdhury May 23 '13 at 18:34
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@SohamChowdhury for example :$(1+B)^2=B^{[2]}$ so $1+2B^{[1]}+B^{[2]}=B^{[2]}$ so $1+2B_1+B_2=B_2$ so $B_1=-1/2$ – mnsh May 23 '13 at 18:46
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that way is only for Bernoulli number – mnsh May 23 '13 at 18:47
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is there any website or book can give me good information ? – mnsh Jun 21 '13 at 01:08
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As a possible "good information": you might look at/like http://go.helms-net.de/math/pascal/bernoulli_en.pdf , very elementary treatize which was just motivated by that expression with the Bernoulli-numbers. It was my first deeper encounter with number theory so please don't mind that it is much amateurish. – Gottfried Helms Jun 21 '13 at 05:06
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Hmm, while $(1+B)^2 = B^{[2]} \to B_1=-1/2$ - what does that make for $(1+B)^1 = B^{[1]} \to B_1$ ? We've a contradiction ... – Gottfried Helms Jun 21 '13 at 05:14
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The relation simply doesn't hold for $n=1$, and it isn't expected to. – anon Jun 21 '13 at 08:24
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@anon: in the OP's question there is no such reservation. The same generosity: saying simply "for n is [some formula]" valid and not mentioning, that for $n=1$ this is not true, and that it is even not expected, came across me when I read in Eric Weissstein's mathworld-page for Bernoulli-numbers. I think such missing information at a formula intended for generality is not tolerable. – Gottfried Helms Jun 21 '13 at 12:55
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1 Umbral calculus - Wikipedia 2 Umbral calculus - Encyclopedia of Mathematics 3 What's umbral calculus about? - Mathematics Stack Exchange – Calum Gilhooley Feb 23 '19 at 23:55
1 Answers
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The identity can be proven using generating functions. We have
$$1=\frac{t}{e^t-1}\frac{e^t-1}{t}=\left(\sum_{k=0}^\infty B_k\frac{t^k}{k!}\right)\left(\sum_{m=0}^\infty\frac{t^m}{(m+1)!}\right)=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}\right)t^n. $$
Comparing coefficients of both sides yields for $n\ge1$:
$$\sum_{k=0}^n\frac{1}{k!(n-k+1)!}B_k=0\iff \sum_{k=0}^n\binom{n+1}{k}B_k=0\iff \sum_{k=0}^{n+1}\binom{n+1}{k}B_k=B_{n+1}.$$
Formally this is the relation $(B+1)^{n+1}=B^{n+1}$ expanded via binomial theorem then with the powers taken from superscript to subscript. The identity sometimes takes the recursive form
$$B_n=-\frac{1}{n+1}\sum_{k=0}^{n-1}\binom{n+1}{k}B_k.$$
This proof is present in these notes on Bernoulli numbers in the section on basic properties.
There are many resources available on GF techniques, notably generatingfunctionology.
