# Finding Power Series and Evaluating as a Power Series

• Apr 18th 2008, 10:52 AM
thegame189
Finding Power Series and Evaluating as a Power Series
I'm really bad at this stuff and I can't figure it out.

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

1. f(z) = ln (7-z)

2. f(x) = x^3 / (5-x)^2

3. f(z) = arctan(z/11)
• Apr 18th 2008, 12:33 PM
flyingsquirrel
Hello

1. $\displaystyle \ln(7-z)=\ln \left(7\left(1-\frac{z}{7}\right) \right) =\ln7+\ln\left(1-\frac{z}{7}\right)$ then, if you find it useful, you can substitute $\displaystyle Z=\frac{z}{7}$ and finally give the power series representation. (I guess you know the power series representation of $\displaystyle \ln(1-z)$ ? If you don't, it comes from $\displaystyle \int_0^z\frac{1}{1-x}\,\mathrm{d}x$)

2. $\displaystyle \frac{x^3}{(5-x^2)}= \frac{x^3}{25}\frac{1}{\left(1-\frac{x}{5}\right)^2}$ aknowledged that $\displaystyle \frac{1}{(1-x)^2} =\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{1-x}\right)$

3. 4. If you know the power series representation of $\displaystyle \arctan x$, you only have to do a substitution. Otherwise, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left(\arctan x\right)=?$
• Apr 18th 2008, 02:19 PM
thegame189
2. I don't see what you did to get x^3/25.

3. Can somebody walk me through the arctan(z/11) problem? I know the power series of arctan(x) but I don't know what to do. I'm completely lost. This is like a foreign language to me honestly.
• Apr 18th 2008, 06:07 PM
Mathstud28
Quote:

Originally Posted by thegame189
2. I don't see what you did to get x^3/25.

3. Can somebody walk me through the arctan(z/11) problem? I know the power series of arctan(x) but I don't know what to do. I'm completely lost. This is like a foreign language to me honestly.

He pulled out a 5 to get $\displaystyle (5(1-\frac{x^2}{5})^2$ and using the fact taht $\displaystyle (ab)^2=a^2b^2$ wee se that this is equal to $\displaystyle 25(1-\frac{x^2}{5})^2$
• Apr 18th 2008, 06:11 PM
Mathstud28
Quote:

Originally Posted by thegame189
2. I don't see what you did to get x^3/25.

3. Can somebody walk me through the arctan(z/11) problem? I know the power series of arctan(x) but I don't know what to do. I'm completely lost. This is like a foreign language to me honestly.

I will give you a hint $\displaystyle arctan\bigg(\frac{x}{11}\bigg)=\frac{1}{11}\int_0^ {x}\frac{1}{1+\bigg(\frac{1}{11}t\bigg)^2}dt$...and so applying ur knowledge of the fact taht $\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$ you can obtain the power series for the integrand...substitute...then integrate
• Apr 18th 2008, 06:16 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
I will give you a hint $\displaystyle arctan\bigg(\frac{x}{11}\bigg)=\frac{1}{11}\int_0^ {x}\frac{1}{1+\bigg(\frac{1}{11}t\bigg)^2}dt$...and so applying ur knowledge of the fact taht $\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$ you can obtain the power series for the integrand...substitute...then integrate

...o and by the way the answer is $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}\bigg(\frac{1}{11}x\bigg)^{2n+1}}{2n+1}$
• Apr 18th 2008, 06:18 PM
thegame189
Edit: That's what I got as my answer but the computer isn't taking it. I'll have to show my professor the work. Our homework is online and you only get 4 chances and it won't take my answer.

2nd Edit: How do I find the radius of convergence?
• Apr 18th 2008, 06:18 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
...o and by the way the answer is $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}\bigg(\frac{1}{11}x\bigg)^{2n+1}}{2n+1}$

Or as flying squirrel said just imput $\displaystyle \frac{1}{11}x$ into where the x goes in the normal $\displaystyle arctan(x)$ power series
• Apr 18th 2008, 06:20 PM
Mathstud28
Quote:

Originally Posted by thegame189
I don't get the substitute part. What am I substituting?

I answered that above...and you know you should feel free to use the thanks button (Cool)
• Apr 18th 2008, 06:30 PM
Mathstud28
Quote:

Originally Posted by thegame189
Edit: That's what I got as my answer but the computer isn't taking it. I'll have to show my professor the work. Our homework is online and you only get 4 chances and it won't take my answer.

2nd Edit: How do I find the radius of convergence?

Use the ratio test...solve this $\displaystyle \lim_{n \to 0}\bigg|\frac{\bigg(\frac{1}{11}x\bigg)^{2n+3}}{2n +3}\cdot\frac{2n+1}{\bigg(\frac{1}{11}x\bigg)^{2n+ 1}}\bigg|<1$ solve it for x...then when you have finished you should have something to the effect of $\displaystyle a<x<b$ and to check for the inclusion/exclusion of those endpoints put them in the power series and check for convergence
• Apr 18th 2008, 10:35 PM
flyingsquirrel
Quote:

Originally Posted by thegame189
2nd Edit: How do I find the radius of convergence?

You can also do this :
Let's denote $\displaystyle X=\frac{x}{11}$
Then, $\displaystyle \arctan X=\sum_{k=0}^{\infty}(-1)^k\frac{X^{2k+1}}{2k+1}$ where $\displaystyle |X|\leq 1$, that is to say $\displaystyle \arctan \frac{x}{11}=\sum_{k=0}^{\infty}(-1)^k\frac{\left( \frac{x}{11}\right)^{2k+1}}{2k+1}$ where $\displaystyle \left|\frac{x}{11}\right| \leq 1 \Leftrightarrow |x| \leq ?$ and here is the radius of convergence