Computing the sum of a particular series.

In summary, the problem involves finding the solution to a series that is similar to e^x, but the comparison to other functions/series does not lead to a clear solution. The solution is found by shifting the start of the series and using the Poisson distribution, which results in the answer of 1/(e^-x). However, the denominators for (x-1) and 1 are incorrect and the solution should be written as e^(x-1)-x/(x-1)^2.
  • #1
ShizukaSm
85
0

Homework Statement


Capture.PNG


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

[itex]\sum_{n->0}^{\infty}\frac{x^n}{n!}[/itex]

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!
 
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  • #2
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.
 
  • #3
ShizukaSm said:

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

[itex]\sum_{n->0}^{\infty}\frac{x^n}{n!}[/itex]

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).
 
  • #4
mfb said:
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?

number0 said:
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.
 
  • #5
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).
 
  • #6
mfb said:
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?
 
  • #7
Well, that is just a matter of taste (and you have less e hanging around now), but I would write it as $$\frac{e^{x-1}-x}{(x-1)^2}$$
 
  • #8
Hmm indeed.

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

Related to Computing the sum of a particular series.

What is a series in computing?

A series in computing refers to a sequence of numbers that are added together. The sum of the numbers in the series is known as the total or the result of the computation.

What is the purpose of computing the sum of a series?

The purpose of computing the sum of a series is to find the total value of a sequence of numbers, which can be useful in a variety of mathematical and scientific calculations.

What is the formula for computing the sum of a series?

The formula for computing the sum of a series is Sn = a + (a + d) + (a + 2d) + ... + (a + (n-1)d), where n is the number of terms in the series, a is the first term, and d is the common difference between each term.

What are some common types of series that are computed?

Some common types of series that are computed include arithmetic series, geometric series, and harmonic series. These series differ in the way the terms are added together, which affects the formula used to compute the sum.

What are some methods for computing the sum of a series?

Some methods for computing the sum of a series include using the formula, manually adding each term, using a calculator or computer program, and using mathematical properties such as the sum of an arithmetic series formula or the geometric series formula.

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