Use Fourier Analysis to solve

[corey2014]

Homework Statement

let f[k]=1/k!, then let fsub2[k]=f[k] convoluted with f[k]
what is a simple formula for fsubm[k]?

Homework Equations

f[k] convoluted with f[k] = summation from negative infinity to infinity of 1/m! * 1/(n-m)!

The Attempt at a Solution

I tried a base case, but now I feel like this should be e^x, because the rest of the terms will drop out? Please Help I don't really know where to start...

 

[LCKurtz]

Homework Statement

let f[k]=1/k!, then let fsub2[k]=f[k] convoluted with f[k]
what is a simple formula for fsubm[k]?

Homework Equations

f[k] convoluted with f[k] = summation from negative infinity to infinity of 1/m! * 1/(n-m)!

The Attempt at a Solution

I tried a base case, but now I feel like this should be e^x, because the rest of the terms will drop out? Please Help I don't really know where to start...

You don't have any x in your sequence. Do you know the relation between convolution of sequences and power series? This problem can be done using that, if you know it, or it can be worked directly. But I think you want to write the convolution more carefully. Your sequence f is only defined for k = 0,1,2... and the convolution of f with itself would be$$
f\star f = g \hbox{ where }g(n)=\sum_{k=0}^n f(k)f(n-k)$$If you write out $g(n)$ for your particular function, you might be surprised and recognize it from somewhere.

 

[corey2014]

no I don't know it, or maybe I just don't recognize it... When i broke it down I got 1/n!+1/(n+1)!+...+1/(n+1)!+1/n!

 

[LCKurtz]

You don't have any x in your sequence. Do you know the relation between convolution of sequences and power series? This problem can be done using that, if you know it, or it can be worked directly. But I think you want to write the convolution more carefully. Your sequence f is only defined for k = 0,1,2... and the convolution of f with itself would be$$
f\star f = g \hbox{ where }g(n)=\sum_{k=0}^n f(k)f(n-k)$$If you write out $g(n)$ for your particular function, you might be surprised and recognize it from somewhere.

no I don't know it, or maybe I just don't recognize it... When i broke it down I got 1/n!+1/(n+1)!+...+1/(n+1)!+1/n!

Each term in the sum for $g(n)$ would have two factorials, one for $f(k)$ and one for $f(n-k)$ wouldn't it? Substitute your formulas for $f$ directly in the sum for $g(n)$ and leave the summation sign in. What do you get? See if you can find any connection to binomial expansions.