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Math Help - Maclaurin series question

  1. #1
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    Maclaurin series question

    Hi All, help much appreciated!

    These are the steps to determine the first two terms in a maclaurin series,
    but they are the same so my working must be out? Its spread over 2 posts cos of laTex limit!

    <br /> <br />
\begin{gathered}<br />
  \ln \frac{1}<br />
{{\sqrt {1 - x} }} \hfill \\<br />
  f(x) = \ln (1 - x)^{ - 1/2}  \hfill \\<br />
  f(0) = 0 \hfill \\<br />
  f'(x) = \frac{1}<br />
{{(1 - x)^{ - 1/2} }}.\frac{{ - 1(1 - x)^{ - 3/2} .( - 1)}}<br />
{2} \hfill \\<br />
  f'(x) = \frac{{(1 - x)^{ - 1} }}<br />
{2} \hfill \\ <br />
\end{gathered} <br /> <br /> <br />
    Last edited by MexicanGringo; July 29th 2008 at 11:57 AM.
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  2. #2
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    and the rest...

    <br /> <br />
\begin{gathered}<br />
  {\text{f'(0) = 0}}{\text{.5}} \hfill \\<br />
  {\text{f''(x) = }}\frac{{{\text{\{ 2( - 1(1 - x)}}^{{\text{ - 2}}} {\text{.( - 1)\}  - \{ (1 - x)}}^{{\text{ - 1}}} {\text{.0\} }}}}<br />
{{{\text{2}}^{\text{2}} }} \hfill \\<br />
  {\text{f''(x) = }}\frac{{{\text{2(1 - x)}}^{{\text{ - 2}}} }}<br />
{{\text{4}}} \hfill \\<br />
  {\text{f''(x) = }}\frac{{{\text{(1 - x)}}^{{\text{ - 2}}} }}<br />
{{\text{2}}} \hfill \\ <br />
\end{gathered} <br /> <br /> <br />

    F''(0) = 0.5 again...
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by MexicanGringo View Post
    Hi All, help much appreciated!

    These are the steps to determine the first two terms in a maclaurin series, but they are the same so my working must be out?

    \ln \frac{1}<br />
{{\sqrt {1 - x} }}
    Since f(x)=\ln \frac{1}<br />
{{\sqrt {1 - x} }}=-\ln\sqrt{1-x}, we need to find consequent derivatives:

    f'(x)=-\frac{1}{2(x-1)}

    f''(x)=\frac{1}{2(x-1)^2}

    So, at x=0, f'(0)=f''(0)=\tfrac{1}{2}

    Thus, the first two terms are \frac{1}{2}x+\frac{1}{2}\frac{x^2}{2!}=\frac{1}{2}  x+\frac{1}{4}x^2...

    Does this make sense? I got the values of f'(0)=f''(0)...
    --Chris
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  4. #4
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    That makes sense up to there, but the third derivative gives a different interval?

    <br />
\begin{gathered}<br />
  f'''(x) = \frac{{(1 - x)^{ - 2} }}<br />
{2} \hfill \\<br />
  f'''(x) = \frac{{\{ 2( - 2)(1 - x)^{ - 3} .( - 1)\} }}<br />
{4} \hfill \\<br />
  f'''(x) = (1 - x)^{ - 3}  \hfill \\<br />
  f'''(0) = 1 \hfill \\ <br />
\end{gathered} <br />

    Is my method of differentiating wrong?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by MexicanGringo View Post
    That makes sense up to there, but the third derivative gives a different interval?

    <br />
\begin{gathered}<br />
  f'''(x) = \frac{{(1 - x)^{ - 2} }}<br />
{2} \hfill \\<br />
  f'''(x) = \frac{{\{ 2( - 2)(1 - x)^{ - 3} .( - 1)\} }}<br />
{4} \hfill \\<br />
  f'''(x) = (1 - x)^{ - 3}  \hfill \\<br />
  f'''(0) = 1 \hfill \\ <br />
\end{gathered} <br />

    Is my method of differentiating wrong?
    That's correct! Its not wrong...however, you're doing it the long way...you don't have to use quotient rule here. Just use power rule and chain rule.

    \frac{1}{2}(1-x)^{-2}=\frac{1}{2}(-2)(1-x)^{-3}(-1)=\frac{1}{(1-x)^3}

    --Chris
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  6. #6
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    Shot thanks a lot!
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by MexicanGringo View Post
    Hi All, help much appreciated!

    These are the steps to determine the first two terms in a maclaurin series,
    but they are the same so my working must be out? Its spread over 2 posts cos of laTex limit!

    <br /> <br />
\begin{gathered}<br />
\ln \frac{1}<br />
{{\sqrt {1 - x} }} \hfill \\<br />
f(x) = \ln (1 - x)^{ - 1/2} \hfill \\<br />
f(0) = 0 \hfill \\<br />
f'(x) = \frac{1}<br />
{{(1 - x)^{ - 1/2} }}.\frac{{ - 1(1 - x)^{ - 3/2} .( - 1)}}<br />
{2} \hfill \\<br />
f'(x) = \frac{{(1 - x)^{ - 1} }}<br />
{2} \hfill \\ <br />
\end{gathered} <br /> <br /> <br />
    \ln\left(\frac{1}{\sqrt{1-x}}\right)

    =-\ln\left(\sqrt{1-x}\right)

    \frac{-1}{2}\ln(1-x)

    Now consider

    -\int_0^x\frac{dx}{1-t}=\ln(1-t)\quad{|t|<1}

    Now we have that

    \frac{1}{1-t}=\sum_{n=0}^{\infty}x^n

    So now consider that |x|<1 is the interval of convergence of the geometric series, therefore it is uniformly convergent on that interval. So we can see that

    -\int_0^x\sum_{n=0}^{\infty}t^n~dt\quad{|x|<1}

    =-\sum_{n=0}^{\infty}\int_0^xt^n~dt

    =-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}\quad|x|<1

    =\ln(1-x)

    \therefore\frac{-1}{2}\ln(1-x)=\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+  1}\quad|x|<1

    \therefore\quad\boxed{\ln\left(\frac{1}{\sqrt{1-x}}\right)=\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{  n+1}}{n+1}\quad\forall{x}\backepsilon|x|<1}
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \ln\left(\frac{1}{\sqrt{1-x}}\right)

    =-\ln\left(\sqrt{1-x}\right)

    \frac{-1}{2}\ln(1-x)

    Now consider

    -\int_0^x\frac{dx}{1-t}=\ln(1-t)\quad{|t|<1}

    Now we have that

    \frac{1}{1-t}=\sum_{n=0}^{\infty}x^n

    So now consider that |x|<1 is the interval of convergence of the geometric series, therefore it is uniformly convergent on that interval. So we can see that

    -\int_0^x\sum_{n=0}^{\infty}t^n~dt\quad{|x|<1}

    =-\sum_{n=0}^{\infty}\int_0^xt^n~dt

    =-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}\quad|x|<1

    =\ln(1-x)

    \therefore\frac{-1}{2}\ln(1-x)=\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+  1}\quad|x|<1

    \therefore\quad\boxed{\ln\left(\frac{1}{\sqrt{1-x}}\right)=\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{  n+1}}{n+1}\quad\forall{x}\backepsilon|x|<1}
    You and you're series...pfft..

    I totally knew how to do that ...

    I'm reteaching myself this stuff [infinte series, and power series], and it makes perfect sense...for now...

    --Chris
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    You and you're series...pfft..

    I totally knew how to do that ...

    I'm reteaching myself this stuff [infinte series, and power series], and it makes perfect sense...for now...

    --Chris
    Yeah, but its not as fun with uniform convergence
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  10. #10
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Yeah, but its not as fun with uniform convergence
    I saw that stuff...and I was "what?"

    Abel's test for uniform convergence [I belive that's what its called] looks interesting...did Weirestrass make contributions to uniform convergence?

    --Chris
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