Finding Power Series and Evaluating as a Power Series

Printable View

April 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)
April 18th 2008, 12:33 PM
flyingsquirrel

Hello

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

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

3. 4. If you know the power series representation of $\arctan x$ , you only have to do a substitution. Otherwise, $\frac{\mathrm{d}}{\mathrm{d}x}\left(\arctan x\right)=?$
April 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.
April 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 $(5(1-\frac{x^2}{5})^2$ and using the fact taht $(ab)^2=a^2b^2$ wee se that this is equal to $25(1-\frac{x^2}{5})^2$
April 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 $arctan\bigg(\frac{x}{11}\bigg)=\frac{1}{11}\int_0^ {x}\frac{1}{1+\bigg(\frac{1}{11}t\bigg)^2}dt$ ...and so applying ur knowledge of the fact taht $\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$ you can obtain the power series for the integrand...substitute...then integrate
April 18th 2008, 06:16 PM
Mathstud28

Quote:

Originally Posted by Mathstud28

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

...o and by the way the answer is $\sum_{n=0}^{\infty}\frac{(-1)^{n}\bigg(\frac{1}{11}x\bigg)^{2n+1}}{2n+1}$
April 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?
April 18th 2008, 06:18 PM
Mathstud28

Quote:

Originally Posted by Mathstud28

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

Or as flying squirrel said just imput $\frac{1}{11}x$ into where the x goes in the normal $arctan(x)$ power series
April 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)
April 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 $\lim_{n \to 0}\bigg|\frac{\bigg(\frac{1}{11}x\bigg)^{2n+3}}{2n +3}\cdot\frac{2n+1}{\bigg(\frac{1}{11}x\bigg)^{2n+ 1}}\bigg|<1$ solve it for x...then when you have finished you should have something to the effect of $a<x<b$ and to check for the inclusion/exclusion of those endpoints put them in the power series and check for convergence
April 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 $X=\frac{x}{11}$
Then, $\arctan X=\sum_{k=0}^{\infty}(-1)^k\frac{X^{2k+1}}{2k+1}$ where $|X|\leq 1$ , that is to say $\arctan \frac{x}{11}=\sum_{k=0}^{\infty}(-1)^k\frac{\left( \frac{x}{11}\right)^{2k+1}}{2k+1}$ where $\left|\frac{x}{11}\right| \leq 1 \Leftrightarrow |x| \leq ?$ and here is the radius of convergence