Series representation of a function

Once you do that, you can modify the equation to get the one with x2 tan-1(x3) in it.In summary, the conversation discusses finding a power series representation for the function f(x)=x^{2}tan^{-1}(x^{3}) and determining its radius of convergence. The process involves using the Taylor Series Expansion formula and applying it to the problem. After some trial and error, the correct solution is found and the conversation ends with a discussion on evaluating the integration constant and modifying the equation to get the series representation for x^{2}tan^{-1}(x^{3}). The radius of convergence is found using the ratio test.
  • #1
iFargle
7
0

Homework Statement


Find a power series representation for the function and determine the radius of convergence.

Homework Equations


[tex]f(x)=x^{2}tan^{-1}(x^{3})[/tex]

The Attempt at a Solution


I don't have any idea on how to even start this. First I differentiated [tex]\frac{d}{dx} arctan(x)=\frac{1}{x^2+1}[/tex], thinking I could try and manipulate it to fit a geometric series, but that quickly came to a dead end. I've tried thinking of any relate-able series representations, but all I know so far is the geometric series and natural logarithms.
 
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  • #2
Google for Taylor Series Expansion formula, and try to apply it to your problem. If you have any questions about it, you can ask :)
 
  • #3
I think I have it. I am overjoyed! Tell me if this is correct:

[tex]f(x)=x^2tan^{-1}(x^3)[/tex]
First,
[tex]\frac{d}{dx}tan^{-1}(x) = \frac{1}{x^2+1}[/tex]
[tex]\int(tan^{-1}(x))dx=\int\frac{1}{x^2+1}dx[/tex]
Rearranging it so it resembles the geometric series:
[tex]\int\frac{1}{1-(-x^2)}dx=\int\sum((-x^2)^n)dx[/tex]
Integrated and re-written:
[tex]\int\sum((-x^2)^n)dx=\sum\frac{(-1)^nx^{2n+1}}{n+1}[/tex]
So now we put x^3 where x is.
[tex]\sum\frac{(-1)^n(x^3)^{2n+1}}{n+1}=\sum\frac{(-1)^n(x)^{6n+3}}{n+1}[/tex]
So we finally have:
[tex]f(x)=x^2\sum\frac{(-1)^n(x)^{6n+3}}{n+1}=\sum\frac{(-1)^n(x)^{6n+5}}{n+1}[/tex]
For the radius of convergence, I just use the ratio test. I get that part pretty easily.
 
  • #4
Almost right. Check your integration.
 
  • #5
iFargle said:
[tex]\int{(tan^{-1}(x))dx} = \int{\frac{1}{x^2+1}dx}[/tex]

I didn't understand how did you make this equality. Am i missing a point?
 
  • #6
Sorry, it was supposed to read:
[tex]tan^{-1}(x)=\int\frac{1}{x^2+1}dx[/tex]
Got carried away with formatting, haha
 
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  • #7
iFargle said:
[tex]\int\sum((-x^2)^n)dx=\sum\frac{(-1)^nx^{2n+1}}{n+1}[/tex]

You have a problem here. The denominator is n+1, but the integral of x2n isn't x2n+1/(n+1).
 
  • #8
[tex] \int\frac{1}{1-(-x^2)}dx=\int\sum(-x^{2})^{n}dx[/tex]
[tex]=\sum(-1)^{n}\frac{x^{2n+1}}{2n+1}dx[/tex]
Correct? I still feel like my answer is wrong, though.
 
  • #9
iFargle said:
[tex] \int\frac{1}{1-(-x^2)}dx=\int\sum(-x^{2})^{n}dx[/tex]
[tex]=\sum(-1)^{n}\frac{x^{2n+1}}{2n+1}dx[/tex]
Correct? I still feel like my answer is wrong, though.
Drop the dx. Where's the integration constant?
What you have so far is:[tex]\tan^{-1}(x)=C+\sum^{\infty}_{n=0}(-1)^{n}\frac{x^{2n+1}}{2n+1}\,.[/tex]

Evaluate the integration constant. Modify the above to get a series representation for x2 tan-1(x3). Find the radius of convergence.
 
  • #10
SammyS said:
Drop the dx. Where's the integration constant?
What you have so far is:[tex]\tan^{-1}(x)=C+\sum^{\infty}_{n=0}(-1)^{n}\frac{x^{2n+1}}{2n+1}\,.[/tex]

Evaluate the integration constant. Modify the above to get a series representation for x2 tan-1(x3). Find the radius of convergence.

So then that would be...
[tex]C+x^{2}\sum^{\infty}_{n=0}(-1)^{n}\frac{x^{6n+3}}{2n+1}=C+\sum^{\infty}_{n=0}(-1)^{n}\frac{x^{6n+5}}{2n+1}[/tex]
or am I missing something huge? I don't get what you mean by evaluate the integration constant.
 
  • #11
The equation

[tex]\tan^{-1}(x)=C+\sum^{\infty}_{n=0}(-1)^{n}\frac{x^{2n+1}}{2n+1}[/tex]

holds for only one value of C. You need to figure out what that is.
 

FAQ: Series representation of a function

What is a series representation of a function?

A series representation of a function is a way of expressing a function as an infinite sum of simpler functions. It is commonly used in mathematics to approximate a function or to understand its behavior.

How is a series representation of a function different from a Taylor series?

A Taylor series is a specific type of series representation that is used to represent a function as a sum of its derivatives evaluated at a specific point. On the other hand, a series representation of a function can use a variety of simpler functions in the sum, not just derivatives.

Why do we use series representations of functions?

Series representations of functions are useful because they allow us to approximate complex functions with simpler ones. This can make it easier to understand the behavior of a function or to solve problems involving the function.

Can any function be represented as a series?

In theory, yes. However, some functions may require an infinite number of terms in the series to accurately represent them. Also, some functions may not have a closed form series representation, meaning the terms in the series cannot be written in a simple formula.

How do we know if a series representation of a function is accurate?

The accuracy of a series representation of a function depends on the number of terms used in the series. The more terms included, the closer the approximation will be to the actual function. In some cases, we can also use mathematical techniques to determine the error in the series representation.

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