Power Series Representations of Functions

Apr 2013
22
0
United States
I'm working on a few problems and I'm stuck on two problems. Would you mind giving me a clue where to go?

1.) Find a power series in x that has the given sum, and specify the radius of convergence:
\(\displaystyle \frac{x}{1-x^4}\)
This is what I have so far:
\(\displaystyle a=x\)
\(\displaystyle r=x^4\)
\(\displaystyle ar^n \rightarrow xx^{4x} \rightarrow \sum_{n=0}^{\infty }x^{4n+1}; r=1\)
I'm a bit confused on how to find the radius; I read the section but I think I'm missing a small detail for it to make sense.

2.) Use a power series representation obtained in this section to find a power series representation for f(x).
\(\displaystyle f(x)=x^4arctan(x^4), |x|<1\)
I know \(\displaystyle a=x^4\) but that's all I got. Is \(\displaystyle r=x^4\)?

Thank you!
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
In part 1, you have \(\displaystyle r=x^4\), and then later, \(\displaystyle r=1\). Which is correct?

For part 2, how do you find the power series for \(\displaystyle \arctan(x)\)? You use implicit differentiation. If \(\displaystyle y=\arctan(x)\), then \(\displaystyle \tan(y) = x\). Take the derivative of both sides: \(\displaystyle \sec^2(y)\dfrac{dy}{dx} = 1\). Solve for the derivative: \(\displaystyle \dfrac{dy}{dx} = \dfrac{1}{\sec^2(y)} = \cos^2(y)\). If \(\displaystyle \tan(y) = \dfrac{x}{1} = \dfrac{\mbox{opp}}{\mbox{adj}}\), then \(\displaystyle \cos(y) = \dfrac{\mbox{adj}}{\mbox{hyp}} = \dfrac{1}{\sqrt{1+x^2}}\). Hence, \(\displaystyle \cos^2(y) = \left( \dfrac{1}{\sqrt{1+x^2}} \right)^2 = \dfrac{1}{1+x^2}\). This lets you find a power series for the derivative of arctan. Then, you integrate term-by-term to get the power series for arctan itself.

So, now you have a different function. You can use a similar process to find the power series.

\(\displaystyle y = \arctan(x^4)\)
\(\displaystyle \tan(y) = x^4\)
\(\displaystyle \sec^2(y) \dfrac{dy}{dx} = 4x^3\)
\(\displaystyle \dfrac{dy}{dx} = 4x^3\dfrac{1}{1+(x^4)^2} = \dfrac{4x^3}{1+x^8}\)
\(\displaystyle \dfrac{dy}{dx} = 4x^3\sum_{n\ge 0}(-1)^n x^{8n} = 4\sum_{n\ge 0}(-1)^n x^{8n+3}\)
Now, integrate term-by-term:
\(\displaystyle \begin{align*}\arctan(x^4) & = 4\sum_{n\ge 0}(-1)^n \dfrac{x^{8n+4}}{8n+4} \\ & = \sum_{n\ge 0}(-1)^n \dfrac{x^{8n+4}}{2n+1} \\ & = \sum_{n\ge 0}(-1)^n \dfrac{(x^4)^{2n+1}}{2n+1}\end{align*}\)

This last result is exactly what you would have gotten if you had plugged in \(\displaystyle x^4\) wherever you saw \(\displaystyle x\) in the original power series for \(\displaystyle \arctan(x)\). So, what is the radius of convergence? Go back to the derivative. It is the same as the radius of convergence of \(\displaystyle \dfrac{1}{1+x^8}\) since that was the original power series you used to find this new power series.

Note that I omitted the initial \(\displaystyle x^4\). The power series I gave is not the complete answer to your problem. It just illustrates where the radius of convergence comes from.
 
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Prove It

MHF Helper
Aug 2008
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I'm working on a few problems and I'm stuck on two problems. Would you mind giving me a clue where to go?

1.) Find a power series in x that has the given sum, and specify the radius of convergence:
\(\displaystyle \frac{x}{1-x^4}\)
This is what I have so far:
\(\displaystyle a=x\)
\(\displaystyle r=x^4\)
\(\displaystyle ar^n \rightarrow xx^{4x} \rightarrow \sum_{n=0}^{\infty }x^{4n+1}; r=1\)
I'm a bit confused on how to find the radius; I read the section but I think I'm missing a small detail for it to make sense.

2.) Use a power series representation obtained in this section to find a power series representation for f(x).
\(\displaystyle f(x)=x^4arctan(x^4), |x|<1\)
I know \(\displaystyle a=x^4\) but that's all I got. Is \(\displaystyle r=x^4\)?

Thank you!
Surely you know that an infinite geometric series with ratio r is only convergent if \(\displaystyle \displaystyle \begin{align*} |r| < 1 \end{align*}\). If \(\displaystyle \displaystyle \begin{align*} r = x^4 \end{align*}\) then what can you determine about \(\displaystyle \displaystyle \begin{align*} |x| \end{align*}\)?