Antiderive complex function f(z) and express as power series

In summary, we are trying to find the anti-derivative F(z) of the function f(z) = cos(z^3) with F(0) = 0. To do this, we will use the Taylor series expansion for cosine, replacing "z" with "z^3". Then, we will integrate the resulting series term by term to find the general (nth) term of F(z). Another approach is to use the formula F(z) = F(0) + z F'(0) + z^2 F''(0)/2! + ..., where F(0) is known and F' etc. can be replaced by the corresponding derivatives of f(z).
  • #1
Jd303
35
0
Let F(z) be the anti-derivative of the function f(z) = cos(z^3) with F(0) = 0. Express F(z) as a power series around z=0, giving both the first 3 non-zero terms and the general (nth) term.

Hey guys really struggling with this integration and how to then express this as a power series. Any help at all would be greatly appreciated because I am getting nowhere!

Thanks in advance!
 
Physics news on Phys.org
  • #2
Jd303 said:
Let F(z) be the anti-derivative of the function f(z) = cos(z^3) with F(0) = 0. Express F(z) as a power series around z=0, giving both the first 3 non-zero terms and the general (nth) term.

Hey guys really struggling with this integration and how to then express this as a power series. Any help at all would be greatly appreciated because I am getting nowhere!

Thanks in advance!

Start with the Taylor series of cosine. Is that enough to get you going?
 
  • Like
Likes 1 person
  • #3
Jd303 said:
really struggling with this integration
You don't need to integrate anything. Just write out the general Taylor series for this unknown function F in the vicinity of 0.
 
  • #4
haruspex said:
You don't need to integrate anything. Just write out the general Taylor series for this unknown function F in the vicinity of 0.
How is he going to find the anti-derivative without integrating?

Jd303, Goa'uld's suggestion is best- write out the Taylor series for cos(z), replace "z" by z3, then integrate "term by term".
 
  • Like
Likes 1 person
  • #5
Thanks guys I think I got it:
Let u = z^3
find Taylor series for f(u)
sub back in z^3
integrate term by term to find F(z)
 
  • #6
Jd303 said:
Thanks guys I think I got it:
Let u = z^3
find Taylor series for f(u)
You mean cos(u), right?
sub back in z^3
integrate term by term to find F(z)
Another way: F(z) = F(0) + z F'(0) + z2F"(0)/2! + ...
You are given F(0), and you can substitute for F' etc. using f.
 

1. What is the definition of the antiderivative of a complex function f(z)?

The antiderivative of a complex function f(z) is a function F(z) whose derivative is equal to f(z) for all values of z in the complex plane. In other words, F(z) is the reverse process of differentiation and is also known as the indefinite integral of f(z).

2. How is the antiderivative of a complex function f(z) expressed as a power series?

The antiderivative of a complex function f(z) can be expressed as a power series by using the Cauchy integral formula. This formula allows us to write the antiderivative as a sum of terms with coefficients that are determined by the values of f(z) at the points along the contour of integration.

3. Is the antiderivative of a complex function f(z) unique?

No, the antiderivative of a complex function f(z) is not unique. Since adding a constant term to an antiderivative does not change its derivative, there can be infinitely many antiderivatives for a given complex function f(z).

4. Can the antiderivative of a complex function f(z) be used to find the definite integral of f(z)?

Yes, the antiderivative of a complex function f(z) can be used to find the definite integral of f(z). This is because the definite integral is equal to the difference between the antiderivative evaluated at the upper and lower limits of integration.

5. Are there any special cases where the antiderivative of a complex function f(z) cannot be expressed as a power series?

Yes, there are some special cases where the antiderivative of a complex function f(z) cannot be expressed as a power series. These cases include functions with discontinuities or singularities, as well as functions whose antiderivatives involve non-elementary functions such as logarithms or trigonometric functions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
165
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Replies
1
Views
556
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Topology and Analysis
Replies
2
Views
576
Back
Top