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Finding the exact value of a summation.

  • Thread starter NATURE.M
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  • #1
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Homework Statement


The sum we are given is Σ(from x=0->∞) [(x^2)(2^x)]/x!. We are asked to find the exact value of this sum using concepts discussed in class which include poisson random variables, and their expected values.

The Attempt at a Solution


[/B]
So i know the solution to the infinite sum is divergent. I know this since:

e^2 * Σ(from x=0->∞) (x^2) * Σ(from x=0->∞) (2^x)(e^-2)/x! = e^-2 * Σ(from x=0->∞) (x^2) since Σ(from x=0->∞) (2^x)(e^-2)/x! = 1 from poisson distribution. And we know Σ(from x=0->∞) (x^2) is divergent from simple infinite series. So the sum itself is ∞/or diverges.

But I don't know if theres a better way to solve this using only properties of the poisson distributions and their expected values/variances instead of invoking results from infinite series.
 

Answers and Replies

  • #2
Dick
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Homework Statement


The sum we are given is Σ(from x=0->∞) [(x^2)(2^x)]/x!. We are asked to find the exact value of this sum using concepts discussed in class which include poisson random variables, and their expected values.

The Attempt at a Solution


[/B]
So i know the solution to the infinite sum is divergent. I know this since:

e^2 * Σ(from x=0->∞) (x^2) * Σ(from x=0->∞) (2^x)(e^-2)/x! = e^-2 * Σ(from x=0->∞) (x^2) since Σ(from x=0->∞) (2^x)(e^-2)/x! = 1 from poisson distribution. And we know Σ(from x=0->∞) (x^2) is divergent from simple infinite series. So the sum itself is ∞/or diverges.

But I don't know if theres a better way to solve this using only properties of the poisson distributions and their expected values/variances instead of invoking results from infinite series.
The sum is not divergent. A ratio test will tell you that. I don't know how to sum it exactly, but it is convergent.
 
  • #3
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The sum is not divergent. A ratio test will tell you that. I don't know how to sum it exactly, but it is convergent.
oh I think I noticed my error of breaking the sum up since its a property of finite sums only.
 
  • #4
LCKurtz
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Hint: If that sum had an ##e^{-2}## factor, what expected value of what distribution would it be calculating?
 
  • #5
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Hint: If that sum had an ##e^{-2}## factor, what expected value of what distribution would it be calculating?
Yeah I kinda understand that what I'm uncertain about is how does Σ(from x=0->∞) (x^2)(2^x)(e^-2)/x! = 6. The x^2 confuses me ?
 
  • #6
LCKurtz
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Hint: If that sum had an ##e^{-2}## factor, what expected value of what distribution would it be calculating?
Yeah I kinda understand that what I'm uncertain about is how does Σ(from x=0->∞) (x^2)(2^x)(e^-2)/x! = 6. The x^2 confuses me ?
If you would answer my question it might help you.
 
  • #7
lurflurf
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use homogeneous differentiation for these types of exercises

$$\sum_{x=0}^\infty \dfrac{x^22^x}{x!}=\left. \left(r\dfrac{\mathrm{d}\phantom{r}}{\mathrm{d}r}\right)^2\sum_{x=0}^\infty \dfrac{r^x}{x!}\right|_{r=2}$$

or use
$$\sum_{x=0}^\infty \dfrac{r^x}{x!}
=\sum_{x=0}^\infty x\dfrac{r^{x-1}}{x!}
=\sum_{x=0}^\infty x(x-1)\dfrac{r^{x-2}}{x!}=e^x$$
 
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