Computing the sum of a particular series.

[ShizukaSm]

Homework Statement

The Attempt at a Solution

Alright, so, I'm clueless about doing this one. I do know that it's extremely similar to e^x

\sum_{n->0}^{\infty}\frac{x^n}{n!}

But really, that means nothing! Usually there's another function/series I can compare and then integrate, in this case I don't have any idea on how to procceed.

Can someone tip me in the right direction, please?

Thanks in advance!

 

[mfb]

Those quotation marks are "n"?

$\sum_{n=0}^\infty \frac{(x-1)^n}{(n+2)!} = \sum_{n=2}^\infty \frac{(x-1)^{n-2}}{n!} = \frac{1}{(x-1)^2} \sum_{n=2}^\infty \frac{(x-1)^n}{n!}$
Just 2 steps left to get an exponential function.

 

[number0]

Homework Statement

View attachment 55181

The Attempt at a Solution

Alright, so, I'm clueless about doing this one. I do know that it's extremely similar to e^x

\sum_{n->0}^{\infty}\frac{x^n}{n!}

But really, that means nothing! Usually there's another function/series I can compare and then integrate, in this case I don't have any idea on how to procceed.

Woops. I think I answered the wrong problem. But you can use the above method to solve your problem.

Woops I think I answered the wrong question. But you can use this example to probably solve your problem.

Can someone tip me in the right direction, please?

Thanks in advance!

You can try to use the Poisson distribution find the solution from there.

In other words,

multiply your summation with the e^(-x)/(e^(-x)). Move the denominator out. We know that the summation(e^(-x) * x^n/ n!) must be 1 since it is a distribution (Poisson). So the answer is 1/(e^-x).

 

[ShizukaSm]

Those quotation marks are "n"?

$\sum_{n=0}^\infty \frac{(x-1)^n}{(n+2)!} = \sum_{n=2}^\infty \frac{(x-1)^{n-2}}{n!} = \frac{1}{(x-1)^2} \sum_{n=2}^\infty \frac{(x-1)^n}{n!}$
Just 2 steps left to get an exponential function.

I think they're "n", I'm not absolutely sure, but since the ebook was scanned with auto recognition, it's very possible that when scanned the "n" was with a poor print, and it recognized as ", it's the only explanation I could come up with.

Regarding the answer, whoa, I never heard of shifting the start of the sum to do that, it was very clever, so the answer would be:
$\frac{1}{(x-1)^2} \sum_{n=2}^\infty \frac{(x-1)^n}{n!} = \frac{1}{(x-1)^2}*(e^{(x-1)} - \frac{x-1}{6} - \frac{1}{2}) = \frac{-ex+6e^x-2e}{6e(x-1)^2}$

Is this correct?

You can try to use the Poisson distribution find the solution from there.

In other words,

multiply your summation with the e^(-x)/(e^(-x)). Move the denominator out. We know that the summation(e^(-x) * x^n/ n!) must be 1 since it is a distribution (Poisson). So the answer is 1/(e^-x).

I haven't learned Poisson distribution yet, I'm going to see if I find this further in the book.

 

[mfb]

I think the denominators for (x-1) and 1 are wrong.
And I would not split that factor of e at e^(x-1), that looks weird (at least to me).

 

[ShizukaSm]

I think the denominators for (x-1) and 1 are wrong.
And I would not split that factor of e at e^(x-1), that looks weird (at least to me).

Ohh, I think I get it, I used the wrong term: (N+2)! , I should have used N!

So in fact:
$\frac{1}{(x-1)^2} \sum_{n=2}^\infty \frac{(x-1)^n}{n!} = \frac{1}{(x-1)^2}*(e^{(x-1)} - \frac{x-1}{1!} - \frac{1}{0!}) = \frac{e^x - ex}{e(x-1)^2}$

Perfect!Thanks!

Why do you say it looks weird? You have a more elegant solution in mind?

 

[mfb]

Well, that is just a matter of taste (and you have less e hanging around now), but I would write it as $$\frac{e^{x-1}-x}{(x-1)^2}$$

 

[ShizukaSm]

Hmm indeed.

Well, thanks you for all your help so far mfb! That was great.