# Power Series Representations of Functions

#### DrKittenPaws

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
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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1 person

#### Prove It

MHF Helper
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*}?