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I have read an article about Bernoulli Polynomial. I found that Bernoulli Polynomial has explicit formula like this:
$$B_n(x)=\sum_{k=0}^{\infty}\binom nk B_kx^{n-k}$$
And the article said that the explicit formula is obtained from Bernoulli Polynomial's generating function:

$$\frac{te^{xt}}{e^t-1}=\sum_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}$$

My question is:

How to get that explicit formula of Bernoulli Polynomial by using its generating function?

I just thought about that its explicit formula was containt binomial theorem and application of Pascal's triangle.
Thank you so much for your helps.

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Use $\quad t/(e^t-1) = \sum_{n=0}^\infty B_nt^n/n!\quad$ and $\quad e^{xt}=\sum_{n=0}^\infty x^n t^n/n!\quad$ and multiply the two exponential generating function power series in $t$. The resulting generating function power series coefficients is the binomial convolution which gives the formula for $B_n(x)$ in your first equation. Read the section on convolution in Wikipedia article about generating function for this.

By the way, the binomial theorem is one consequence of $\quad e^{(x+y)t}=e^{xt}e^{yt},\quad$ by using the binomial convolution of the coefficients to get $\quad (x+y)^n=\sum_{k=0}^n \binom nk x^k y^{n-k}.$

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  • $\begingroup$ I have tried your advance then $$\frac{te^{xt}}{e^t-1}=\frac{t}{e^t-1}e^{xt}$$ $$=\sum_{n=0}^{\infty}B_n\frac{t^n}{n!}\cdot\sum_{n=0}^{\infty}x^n\frac{t^n}{n!}$$ $$\sum_{n=0}^{\infty}\Big(\frac{t}{n!}\Big)^2\cdot\sum_{n=0}^{\infty}B_nx^n$$ then, by comparing coefficient, we get $$\sum_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}=\sum_{n=0}^{\infty}\Big(\frac{t}{n!}\Big)^2\cdot\sum_{n=0}^{\infty}B_nx^n$$ $$B_n(x)=\frac{t}{n!}\cdot\sum_{n=0}^{\infty}B_nx^n$$ then by using convolution (Cauchy products), we get $$B_n(x)=\sum_{k=0}^{n}\binom nk B_nx^{n-k}$$ is it right? thanks @somos $\endgroup$ – Boo230196 Nov 18 '17 at 12:46
  • $\begingroup$ @Boo230196 Okay, you made a good start, but your first misake was thinking that $(a+b+c+...)\cdot(d+e+f+...)=(a\cdot d)+(b\cdot e)+(c\cdot f)+...$ which is wrong. You want $=(a\cdot d)+(a\cdot e+b\cdot d)+(a\cdot f+b\cdot e+c\cdot d)+...$ $\endgroup$ – Somos Nov 18 '17 at 13:42
  • $\begingroup$ oh, i see.. i am not careful when do the multiplication of two summation there. Okay, thanks for your correction :) @somos $\endgroup$ – Boo230196 Nov 19 '17 at 3:29

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