How Do You Calculate This Double Summation?

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In summary, the conversation discusses a problem involving exponential and summation expressions in terms of variables a and b. The main suggestion is to rearrange the double summation and use power series formulas to simplify the expression. The conversation also mentions the importance of checking convergence and the use of computers to perform such calculations.
  • #1
sabbagh80
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Hi,

Can every body solve this problem in terms of [itex] a [/itex] and [itex] b [/itex].

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

Thanks in advance for your participation.
 
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  • #2
hi sabbagh80! :smile:

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

it's over all pairs (m,n) with m ≤ n

try rearranging the ∑∑ into a ∑∑ over different variables :wink:
 
  • #3
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
sabbagh80 said:
I mean that is it possible to further simplify this expression …

yes … try rearranging the ∑∑ into a ∑∑ over different variables :smile:
 
  • #5
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
tiny-tim said:
yes … try rearranging the ∑∑ into a ∑∑ over different variables :smile:

Can everybody show me how could I do the above suggestion?
 
  • #7
nickalh said:
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.

What is your meaning, I am confused!
 
  • #8
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
hi spamiam! :smile:
spamiam said:
… though tinytim's comment makes me think it might be the last one I mentioned...

yup …

instead of summing over all values of m and n,

sum over all values of m and m+n :wink:
 
  • #10
sabbagh80 said:
Hi,

Can every body solve this problem in terms of [itex] a [/itex] and [itex] b [/itex].

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

Thanks in advance for your participation.

ok. didn`t see the `n`.
 
Last edited:
  • #11
matphysik said:
Answer =1.

It is a cute joke!
m is from 0 to n not to infinity.
by the way, thanks for your participation.
 
  • #12
sabbagh80 said:
Hi,

Can every body solve this problem in terms of [itex] a [/itex] and [itex] b [/itex].

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

Thanks in advance for your participation.
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:

FAQ: How Do You Calculate This Double Summation?

1. How do you calculate a double summation?

To calculate a double summation, you first need to identify the two variables involved in the summation and the range of values for each variable. Then, you can use a nested loop to iterate through each combination of values and add them together to get the final summation.

2. What is the formula for a double summation?

The formula for a double summation is ΣΣf(x,y), where f(x,y) represents the function or expression being summed and the two sigma symbols indicate the double summation.

3. How do you represent a double summation in mathematical notation?

A double summation is represented in mathematical notation using two sigma symbols, with the inner sigma indicating the variable being summed and the outer sigma indicating the range of values for the variable.

4. What is the purpose of calculating a double summation?

Calculating a double summation is often used in statistics and mathematics to find the total sum of a function or expression that depends on two variables. It can also be used to evaluate complex series or to find the area under a curve.

5. Are there any shortcuts or tricks for calculating a double summation?

Yes, there are some shortcuts and tricks that can be used to simplify a double summation. One approach is to use known summation formulas, such as the arithmetic or geometric series formulas, to reduce the number of calculations needed. Another approach is to use symmetry or patterns in the summation to reduce the number of terms to be summed. Additionally, using a computer program or calculator can help automate the calculation process and reduce the chance of errors.

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