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

Power Series ArcTan?

  • Thread starter sammiekurr
  • Start date
  • #1
Power Series ArcTan???

Homework Statement


Let f be the function given by f(t) = 4/(1+t^2) and G be the function given by G(x)= Integral from 0 to x of f(t)dt.
A) Find the first four nonzero terms and the general term for the power series expansion of f(t) about x=0.
B) Find the first four nonzero terms and the general term for the power series expansion of G(t) about x=0.
C) Find the interval of convergence of the power series in part (B). Show the analysis that leads to your conclusion.


Homework Equations


d/dtArctan(t)=1/(1+t^2)


The Attempt at a Solution


A) a=4, R=-t^2. f(t)=Sum from n=1 to infinity of 4 * (-1)^n * t^2n
First four terms: -4t^2 + 4t^4 - 4t^6 + 4t^8

B) Integral from 0 to x of 4/(1+t^2)dt = 4arctan(t) from 0 to x = 4arctan(x)

Now I don't know where to go from here. I don't know how to write the power series for the antiderivative of the original power series, since it is not in the standard form of a power series. Can anybody help?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
955


Homework Statement


Let f be the function given by f(t) = 4/(1+t^2) and G be the function given by G(x)= Integral from 0 to x of f(t)dt.
A) Find the first four nonzero terms and the general term for the power series expansion of f(t) about x=0.
B) Find the first four nonzero terms and the general term for the power series expansion of G(t) about x=0.
C) Find the interval of convergence of the power series in part (B). Show the analysis that leads to your conclusion.


Homework Equations


d/dtArctan(t)=1/(1+t^2)


The Attempt at a Solution


A) a=4, R=-t^2. f(t)=Sum from n=1 to infinity of 4 * (-1)^n * t^2n
First four terms: -4t^2 + 4t^4 - 4t^6 + 4t^8
No. You have the formula right but when n= 0, 4(-1)^n t^2n is 4. The first four terms are 4- 4t^2+ 4t^4- 4t^6.

B) Integral from 0 to x of 4/(1+t^2)dt = 4arctan(t) from 0 to x = 4arctan(x)

Now I don't know where to go from here. I don't know how to write the power series for the antiderivative of the original power series, since it is not in the standard form of a power series. Can anybody help?
Why is it "not in the standard form of a power series"?

You have that [itex]4/(1+ t^2)= 4\sum_{n=0}^\infty (-1)^n t^{2n}[/itex] and can integrate "term by term".
 

Related Threads on Power Series ArcTan?

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
6
Views
12K
  • Last Post
Replies
5
Views
39K
Replies
1
Views
427
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
2
Views
643
  • Last Post
Replies
6
Views
2K
Top