How to Find a Taylor Series for an Integral with x=-1 as Center Point?

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Homework Help Overview

The discussion revolves around finding a Taylor series for the integral of the function [sin(x+1)]/(x^2+2x+1) centered at x=-1. Participants are exploring the process of integrating a function after expanding it into a Taylor series.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial steps of finding the Taylor series for sin(x+1) and the denominator (x^2 + 2x + 1). There is a question about the relevance of the Taylor series expansion and how to proceed after finding the series. Some participants suggest integrating the series term by term.

Discussion Status

There is an ongoing exploration of the integration process after expanding the function into a Taylor series. Some participants have provided guidance on the form of the series and the implications of integrating it. Multiple interpretations of the problem are being discussed, particularly regarding the nature of the series (Taylor vs. Laurent).

Contextual Notes

Participants are questioning how many terms of the series should be evaluated before integration, indicating a lack of consensus on this aspect. There is also a mention of the potential for the resulting expression after integration to lack negative powers in x.

Jamie K
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Homework Statement


Hi everyone, determine a Taylor Series about x=-1 for the integral of:
[sin(x+1)]/(x^2+2x+1).dx


Homework Equations


As far as I know the only relevant equation is the Taylor Series expansion formula. I've just started to tackle Taylor Series questions and I've been able to do things like find Taylor series but the integration part seems to get me lost.


The Attempt at a Solution


Basically I've found a Taylor series for sin(x+1), being (x+1)+[1/6(x+1)^3]+[1/120(x+1)^5]+[1/5070(x+1)^7] and for (x^2)+2x+1, being (x-1)^2. Note: I presumed because it's about x=-1 this means a=-1. Did I do this right? Is this even relevant?
Basically I don't quite know where to go from here. Any type of advice of help would be great. Thank you.
 
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Is there anyone who could help, I would be very grateful for any assistance.
 
sin(x+1)/(x2+2x+1) = sin(x+1)/(x+1)2

what you have is a function of the form sin(u)/u2, where u = x+1.

sin(u) = u - u3/3! + u5/5! - u7/7! + ...

since u = x+1, the above series is already centered at x = -1.

sin(u)/u2 = 1/u - u/3! + u3/5! - u5/7! + ...

integrate the above series term for term.
 
are you hoping to integrate this function? what I am getting at is are you trying to do this integration by first expanding the integrand in Taylor series and then try to integrate term by term? (by the way, taylor series for x^2 +2x +1 about x=-1 is (x+1)^2 instead

Note: I presumed because it's about x=-1 this means a=-1
yes, i think so if your a is this a
f(x) = f(a) + f'(a)(x-a) + ...
 
I think I'm beginning to get the picture. Is there a way of knowing how many terms to evaluate before beginning to integrate or is there just a standard number?
Yes mjsd I am trying to expand to a Taylor series and then integrating. Thank you for the help guys.
 
Jamie K said:
I think I'm beginning to get the picture. Is there a way of knowing how many terms to evaluate before beginning to integrate or is there just a standard number?
What do you mean by "how many terms". A Taylor's series requires ALL terms!

Yes mjsd I am trying to expand to a Taylor series and then integrating. Thank you for the help guys.

Since skeeter has already told you that the entire series is
[itex]sin(u)/u^2 = 1/u - u/3! + u^3/5! - u^5/7! + ...[/itex]
[itex]= (x+1)^{-1}- (x+1)/3!+ (x+1)^3/5!+\cdot\cdot\cdot + (-1)^n (x+1)^{2n-1}/(2n+1)!+ \cdot\cdot\cdot[/itex]
You should have no troulbe integrating every term of that.
(Strictly speaking, that's not a "Taylor's:" series; it is a "Laurent" series because of the initial negative power.)
 
Just out of curiosity could this question have a Taylor series also? Or does every question have only either a Taylor or Laurent series applicable to it?
Thanks HallsofIvy for clearing up the confusion.
 
after integration, you may end up with somthing that has no negative powers in x.
 

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