# Calculating a double summation

1. Jul 10, 2011

### sabbagh80

Hi,

Can every body solve this problem in terms of $a$ and $b$.

$e^{-(a+b)}\sum _{n=0}^{\infty} \frac{a^n}{n!} \sum_{m=0}^{n}\frac{b^m}{m!}$

2. Jul 10, 2011

### tiny-tim

hi sabbagh80!

hmm … looks a bit like a binomial expansion, doesn't it?

it's over all pairs (m,n) with m ≤ n

try rearranging the ∑∑ into a ∑∑ over different variables

3. Jul 10, 2011

### sabbagh80

I mean that is it possible to further simplify this expression or maybe rewrite it by a more known functions. Because computing this expression in a long loop is time consuming process. isn't it?

4. Jul 10, 2011

### tiny-tim

yes … try rearranging the ∑∑ into a ∑∑ over different variables

5. Jul 15, 2011

### nickalh

Are you meaning
*. e^−(a+b) * ( ∑a^n/n!) (∑b^m/m!)
or
**. e^−(a+b) * ∑ (a^n/n! (∑b^m/m!) )

For *, we can evaluate each sum separately, then multiply the result. Hint: try a power series formula sheet. * simplifies beautifully.

I suspect you mean **, where the b_m/m! sum gets evaluated first. Then multiplied by a single "a" term.

Arggh, almost had it until I realized the sum b^m/m! is not an infinite number of terms, but n terms.

6. Jul 15, 2011

### sabbagh80

Can everybody show me how could I do the above suggestion?

7. Jul 15, 2011

### sabbagh80

What is your meaning, I am confused!

8. Jul 15, 2011

### spamiam

Let me first say that even with tinytim's suggestion, I still haven't figured this out!

Anyway, the idea is straightforward enough. If we sketch out the area over which you are summing with m on the horizontal axis and n on the vertical, then we see the sum is taken over all pairs (m,n) on or above the line m=n (or the line y=x if that is any clearer). Now at the moment, your sum is calculated by picking a value for n and then summing the first n terms of bm. So, using our picture, you're summing horizontally. However, you could also sum vertically, or by diagonals parallel to the line m=n or by diagonals parallel to m=-n.

That said, I'm not sure which method will make things easier, though tinytim's comment makes me think it might be the last one I mentioned...

9. Jul 15, 2011

### tiny-tim

hi spamiam!
yup …

instead of summing over all values of m and n,

sum over all values of m and m+n

10. Jul 17, 2011

### matphysik

ok. didnt see the n.

Last edited: Jul 17, 2011
11. Jul 17, 2011

### sabbagh80

It is a cute joke!
m is from 0 to n not to infinity.
by the way, thanks for your participation.

12. Jul 19, 2011

### matphysik

The outline of the derivation is as follows:
First, check that the double sum converges!

Next, the summation from m=0 to m=n. Rewrite this as the difference of the summation from m=0 to ∞ and the summation from m=n+1 to ∞. The first member of this difference is eᵇ. For the 2nd member set p:=m-(n+1), where p runs from zero to ∞.

We may differentiate this 2nd member (n+1) times (= eᵇ). Then (indefinite) integrate this (n+1) times to get,

(*)- eᵇ + c₁bⁿ/n! + c₂bⁿ⁻¹/(n-1)! +...+cⁿ⁺¹,

where the c` are integration constants which may all be set equal to unity.

Last, multiply Σaⁿ/n! with each term of (*) and sum each resulting term to get the result.

NOTE: Nobody cares about making computations such as this by hand anymore. They do this by computer.

Addendum: See, for example, the work of Doron Zeilberger.

Last edited: Jul 19, 2011