• Support PF! Buy your school textbooks, materials and every day products Here!

Taylor series of an integral function

  • Thread starter Hernaner28
  • Start date
  • #1
263
0

Homework Statement



$$ \displaystyle f\left( x \right)=\int\limits_{0}^{x}{\frac{\sin t}{t}dt} $$

Calculate the Maclaurin series of third order.

Homework Equations





The Attempt at a Solution



What I do is:

$$ \displaystyle f'\left( x \right)=\frac{\sin x}{x} $$

$$ \displaystyle f''\left( x \right)=\frac{x\cos x-\sin x}{{{x}^{2}}}$$

and so on...

but if I try to evaluate theese two on x=0 it doesn't make sense. I've seen I have to transform the integrand to the taylor polynomial but why is my "method" wrong?

Thanks
 
Last edited:

Answers and Replies

  • #2
20,145
4,215

Homework Statement



$$ \displaystyle f\left( x \right)=\int\limits_{0}^{x}{\frac{\sin t}{t}dt} $$

Calculate the Maclaurin series of third order.

Homework Equations





The Attempt at a Solution



What I do is:

$$ \displaystyle f'\left( x \right)=\frac{\sin x}{x} $$

$$ \displaystyle f''\left( x \right)=\frac{x\cos x-\sin x}{{{x}^{2}}}$$

and so on...

but if I try to evaluate theese two on x=0 it doesn't make sense. I've seen I have to transform the integrand to the taylor polynomial but why is my "method" wrong?

Thanks
f' and f'' both have specific limiting values at x = 0. For example, in the limit of x approaches zero, f' = 1.
 
  • #3
Dick
Science Advisor
Homework Helper
26,258
618
It's a lot easier if you expand sin(t) in a power series. Then divide by t. You are simply trying to do it the hard way.
 
  • #4
Zondrina
Homework Helper
2,065
136

Homework Statement



$$ \displaystyle f\left( x \right)=\int\limits_{0}^{x}{\frac{\sin t}{t}dt} $$

Calculate the Maclaurin series of third order.

Homework Equations





The Attempt at a Solution



What I do is:

$$ \displaystyle f'\left( x \right)=\frac{\sin x}{x} $$

$$ \displaystyle f''\left( x \right)=\frac{x\cos x-\sin x}{{{x}^{2}}}$$

and so on...

but if I try to evaluate theese two on x=0 it doesn't make sense. I've seen I have to transform the integrand to the taylor polynomial but why is my "method" wrong?

Thanks
The beauty of a Taylor series is that it allows you to express functions which are complicated to integrate in terms of an infinite sum of polynomials which are easy to integrate.

The Maclaurin series is simply the Taylor series centered at 0.

Recall that the Maclaurin series for [itex]sinx = \sum_{n=0}^{∞}\frac{x^(2n+1)}{(2n+1)!}(-1)^n[/itex]

Using this fact you should be able to solve your integral.
 
  • #5
20,145
4,215
It's a lot easier if you expand sin(t) in a power series. Then divide by t. You are simply trying to do it the hard way.
I agree. That's how I would have done the problem if it had been posed to me. But I wanted to show how it could also have been done using the more difficult way that the OP had pursued.
 
  • #6
263
0
Thank you guys, now I understand ;)
 

Related Threads on Taylor series of an integral function

  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
10
Views
938
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
10
Views
421
  • Last Post
Replies
6
Views
674
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
Top