# Finding coefficients

## Homework Statement

Find coefficients ##b_n##
$$\sum^{\infty}_{n=0}\frac{t^n}{n!}b_n=e^{\frac{4at}{3}}e^{-\frac{t^2}{3}}$$
a is constant.

## Homework Equations

$$e^x=\sum^{\infty}_{n=0}\frac{x^n}{n!}$$

## The Attempt at a Solution

Here
$$e^{\frac{t}{3}(4a-t)}=\sum^{\infty}_{n=0}\frac{( \frac{t}{3}(4a-t))^n}{n!}$$
I tried to use this. But I did not succeed to find given coefficients.

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BiGyElLoWhAt
Gold Member
This has to be true for all t, no? So that means you can let t be anything you want. If the coefficients are constants (not functions of t), there should obviously be no dependence on t. 0 comes to mind, but upon inspection, doesn't seem that useful. 1, however... This is assuming b_n = constant. If that's not the case, this doesn't work so well. Considering the language they use is "coefficient", my guess is that it's only a function of n.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Find coefficients ##b_n##
$$\sum^{\infty}_{n=0}\frac{t^n}{n!}b_n=e^{\frac{4at}{3}}e^{-\frac{t^2}{3}}$$
a is constant.

## Homework Equations

$$e^x=\sum^{\infty}_{n=0}\frac{x^n}{n!}$$

## The Attempt at a Solution

Here
$$e^{\frac{t}{3}(4a-t)}=\sum^{\infty}_{n=0}\frac{( \frac{t}{3}(4a-t))^n}{n!}$$
I tried to use this. But I did not succeed to find given coefficients.

It will be messy, but in principle it is do-able: just expand ##(4a-t)^n## using the binomial expansion, so that
$$\frac{1}{n!} \left( \frac{1}{3}t (4a-t) \right)^n = \sum_{k=0}^n {n \choose k} (-1)^k \frac{(4a)^{n-k}}{n!\: 3^n} t^{n+k} = \sum_{k=0}^n c_{n,k} x^{n+k} .$$
For given ##N## the coefficient of ##x^N## is given by summing all the ##c_{n,k}## for which ##n+k=N##.

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