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Math Help - [SOLVED] Manipulation of power series

  1. #1
    Senior Member mollymcf2009's Avatar
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    [SOLVED] Manipulation of power series

    Find a power series representation of the function:

    f(x) = ln(5-x)

    Here is my work:

    \int ln(5-x) = \frac{1}{5-x}  =  5(\frac{1}{1-\frac{x}{5}})

    \sum^{\infty}_{n=0} (\frac{x}{5})^n

    Integrate.

    = \frac{(\frac{x}{5})^{n+1}}{n+1}

    Multiply by -1 then the series is:

    \sum^{\infty}_{n=0} \frac{(-1)(\frac{x}{5})^{n+1}}{n+1}

    = \sum^{\infty}_{n=1} \frac{(-1)^{n-1} (\frac{x}{5})^n}{n}

    Here is where I fall apart...Do I need to integrate again? No idea....
    Thanks!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Find a power series representation of the function:

    f(x) = ln(5-x)

    Here is my work:

    \int ln(5-x) = \frac{1}{5-x}  =  5(\frac{1}{1-\frac{x}{5}})

    \sum^{\infty}_{n=0} (\frac{x}{5})^n

    Integrate.

    = \frac{(\frac{x}{5})^{n+1}}{n+1}

    Multiply by -1 then the series is:

    \sum^{\infty}_{n=0} \frac{(-1)(\frac{x}{5})^{n+1}}{n+1}

    = \sum^{\infty}_{n=1} \frac{(-1)^{n-1} (\frac{x}{5})^n}{n}

    Here is where I fall apart...Do I need to integrate again? No idea....
    Thanks!!
    first mistake. \int \ln (5 - x)~dx \ne \frac 1{5 - x}. that is closer to finding the derivative, and even that is wrong.

    also, \frac 1{5 - x} \ne 5 \cdot \frac 1{1 - \frac x5}

    note that \ln (5 - x) = - \int \frac 1{5 - x}~dx

    now, find the power series for - \frac 1{5 - x} and integrate term by term

    you also took the integral of the power series incorrectly (if you can't do it in your head, use substitution)
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  3. #3
    MHF Contributor matheagle's Avatar
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    The DERIVATIVE of \ln x is 1/x not the INTEGRAL.
    And do not forget the chain rule.
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Jhevon View Post
    first mistake. \int \ln (5 - x)~dx \ne \frac 1{5 - x}. that is closer to finding the derivative, and even that is wrong.

    also, \frac 1{5 - x} \ne 5 \cdot \frac 1{1 - \frac x5}

    note that \ln (5 - x) = - \int \frac 1{5 - x}~dx

    now, find the power series for - \frac 1{5 - x} and integrate term by term

    you also took the integral of the power series incorrectly (if you can't do it in your head, use substitution)

    Ok, I got the right answer, but it is entirely possible that it was pure luck LOL Can y'all check my work?

    f(x) = ln(5-x)

    -\int\frac{1}{5-x}

    My power series is:

    -\frac{1}{5} \sum^{\infty}_{n=0} (\frac{x}{5})^n

    -\int \frac{1}{5} + \sum^{\infty}_{n=0} \int (\frac{x}{5})^n

    <br />
= - ln(5) + \sum^{\infty}_{n=0} \frac{(x)^{n+1}}{n+1} \cdot \frac{(\frac{1}{5})^{n+1}}{n+1} ***IS THIS LEGIT FOR THIS INTEGRATION?

    = - ln(5) + \sum^{\infty}_{n=1} \frac{x^n (\frac{1}{5})^n}{n}

    = ln(5) - \sum^{\infty}_{n=1} x^n \cdot \frac{1}{5^nn}

    <br />
= ln(5) - \sum^{\infty}_{n=1}\frac{x^n}{5^nn}
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Ok, I got the right answer, but it is entirely possible that it was pure luck LOL Can y'all check my work?

    f(x) = ln(5-x)

    -\int\frac{1}{5-x}

    My power series is:

    -\frac{1}{5} \sum^{\infty}_{n=0} (\frac{x}{5})^n

    -\int \frac{1}{5} {\color{red}+} \sum^{\infty}_{n=0} \int (\frac{x}{5})^n
    this is wrong. where did that + sign come from??

    <br />
= {\color{red}- ln(5)} + \sum^{\infty}_{n=0} \frac{(x)^{n+1}}{n+1} \cdot {\color{blue}\frac{(\frac{1}{5})^{n+1}}{n+1}} ***IS THIS LEGIT FOR THIS INTEGRATION?
    huh?... \int k~dx = kx + C, for any constant k

    and the thing in blue is just wrong. where you got that from, i don't know. i think you again thought that you could integate a constant the same way you do a variable. this is wrong!

    but your integral was wrong to begin with, because you put that plus sign in. and review your integration rules.
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  6. #6
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Jhevon View Post
    this is wrong. where did that + sign come from??

    huh?... \int k~dx = kx + C, for any constant k

    and the thing in blue is just wrong. where you got that from, i don't know

    but your integral was wrong to begin with, because you put that plus sign in. and review your integration rules.

    Ok, I officially don't understand this. The correct answer is:

    = ln(5) - \sum^{\infty}_{n=1}\frac{x^n}{5^nn}

    So, how did the +/- sign after the log get there? I am really tired, I know that \frac{1}{5} is just a constant, and I'm actually pretty good at my integrations, I'm just foggy brained and basically going through the motions at this point. Do you mind just helping me get through this problem so I can at least understand what I am doing? I really am understanding this better than yesterday, thanks to you Jhevon, but like I said yesterday, Molly & series are not seeing eye to eye.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Ok, I got the right answer, but it is entirely possible that it was pure luck LOL Can y'all check my work?
    so it was luck...and a little bit of cheating on your part

    = {\color{red}-} ln(5) + \sum^{\infty}_{n=1} \frac{x^n (\frac{1}{5})^n}{n}

    = ln(5) {\color{red}-} \sum^{\infty}_{n=1} x^n \cdot \frac{1}{5^nn}
    how did you change the signs here? you seem to just be doing things to make your work look like the answer. naughty!


    anyway, here's the solution:

    \ln (5 - x) = \ln 5 \left( 1 - \frac x5 \right)

    = \ln 5 + \ln \left( 1 - \frac x5 \right)

    = \ln 5 - \frac 15 \int \frac 1{1 - \frac x5}~dx

    = \ln 5 - \frac 15 \int \sum_{n = 0}^\infty \left( \frac x5 \right)^n~dx

    = \ln 5 - \frac 15 \sum_{n = 0}^\infty \int \left( \frac x5 \right)^n~dx

    = \ln 5 - \frac 15 \sum_{n = 0}^\infty \frac {5\left( \frac x5 \right)^{n+1}}{n + 1}

    = \ln 5 - \sum_{n = 1}^\infty \frac {\left( \frac x5 \right)^n}n

    = \ln 5 - \sum_{n = 1}^\infty \frac {x^n}{5^nn}

    Now you can go to bed
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  8. #8
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Jhevon View Post
    so it was luck...and a little bit of cheating on your part

    Hey, you gotta love Webassign and multiple choice


    how did you change the signs here? you seem to just be doing things to make your work look like the answer. naughty!
    BUSTED!!! LOL At this point, I guess I am doing that just because I can't figure out any other way to get this to sink in. I figure if I work enough of these, and have someone (ha ha...um...YOU) blast my every move, that eventually I'll get it?!?!? or NOT?!?!? LOL


    anyway, here's the solution:

    \ln (5 - x) = \ln 5 \left( 1 - \frac x5 \right)

    = \ln 5 + \ln \left( 1 - \frac x5 \right)

    = \ln 5 - \frac 15 \int \frac 1{1 - \frac x5}~dx

    = \ln 5 - \frac 15 \int \sum_{n = 0}^\infty \left( \frac x5 \right)^n~dx

    = \ln 5 - \frac 15 \sum_{n = 0}^\infty \int \left( \frac x5 \right)^n~dx

    = \ln 5 - \frac 15 \sum_{n = 0}^\infty \frac {5\left( \frac x5 \right)^{n+1}}{n + 1}

    = \ln 5 - \sum_{n = 1}^\infty \frac {\left( \frac x5 \right)^n}n

    = \ln 5 - \sum_{n = 1}^\infty \frac {x^n}{5^nn}

    Now you can go to bed
    You are THE BEST!!!!!!! Again, I can not tell you how much I appreciate your help on all of this! Now I AM going to bed! You get some rest too, as tomorrow and next week I'll be harassing you with Taylor & MacLaurin
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