Finding Coefficients b_n: e^{\frac{4at}{3}}e^{-\frac{t^2}{3}}

[LagrangeEuler]

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.

 

[BiGyElLoWhAt]

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 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 $n+k=N$.