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Math Help - Finding Power Series and Evaluating as a Power Series

  1. #1
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    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)
    Last edited by thegame189; April 18th 2008 at 02:10 PM.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    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)=?
    Last edited by flyingsquirrel; April 18th 2008 at 01:34 PM.
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  3. #3
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    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.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by thegame189 View Post
    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
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by thegame189 View Post
    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
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    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}
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    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?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    ...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
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by thegame189 View Post
    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
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by thegame189 View Post
    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
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  11. #11
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by thegame189 View Post
    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
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