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

In summary: If that's the case, then you can write it as a Taylor's series around x= 0, but for some other function of x. In summary, the conversation discussed finding a Taylor series for the integral of a given function and how to integrate it term by term. The Taylor series for sin(x+1)/x^2+2x+1 was found and the importance of knowing the number of terms to evaluate before beginning to integrate was mentioned. The difference between a Taylor series and a Laurent series was also briefly discussed.
  • #1
Jamie K
4
0

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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  • #2
Is there anyone who could help, I would be very grateful for any assistance.
 
  • #3
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.
 
  • #4
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) + ...
 
  • #5
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.
 
  • #6
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.)
 
  • #7
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.
 
  • #8
after integration, you may end up with somthing that has no negative powers in x.
 

1. What is a Taylor series for integrals?

A Taylor series for integrals is a mathematical tool used to approximate the value of an integral using a polynomial function. It is based on the Taylor series, which is a representation of a function as an infinite sum of terms.

2. How is a Taylor series for integrals calculated?

To calculate a Taylor series for integrals, we first need to find the derivatives of the function at a given point. Using these derivatives, we can then construct a polynomial function that closely approximates the original function. The more terms we include in the polynomial, the more accurate the approximation will be.

3. What is the purpose of using a Taylor series for integrals?

The main purpose of using a Taylor series for integrals is to simplify the calculation of integrals. It allows us to approximate the value of an integral without having to perform complex integration techniques. It is also useful for evaluating integrals at points where the function may be difficult to integrate directly.

4. Are there any limitations to using a Taylor series for integrals?

Yes, there are a few limitations to using a Taylor series for integrals. The most important one is that it can only be used for smooth functions, meaning that the function and its derivatives must exist and be continuous at the point of approximation. Additionally, the accuracy of the approximation depends on the number of terms included in the polynomial function.

5. Are there any real-world applications for Taylor series for integrals?

Yes, Taylor series for integrals have many real-world applications in fields such as physics, engineering, and economics. They can be used to approximate values in situations where the exact value cannot be calculated easily, such as in differential equations or optimization problems. They are also used in numerical methods for solving differential equations and in financial modeling.

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