Math Help - Maclaurin series question

1. 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!

$

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

$

2. and the rest...

$

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

$

F''(0) = 0.5 again...

3. Originally Posted by MexicanGringo
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?

$\ln \frac{1}
{{\sqrt {1 - x} }}$
Since $f(x)=\ln \frac{1}
{{\sqrt {1 - x} }}=-\ln\sqrt{1-x}$
, 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

4. That makes sense up to there, but the third derivative gives a different interval?

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

Is my method of differentiating wrong?

5. Originally Posted by MexicanGringo
That makes sense up to there, but the third derivative gives a different interval?

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

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

6. Shot thanks a lot!

7. Originally Posted by MexicanGringo
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!

$

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

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

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

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

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

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

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

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

$=\ln(1-x)$

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

$\therefore\quad\boxed{\ln\left(\frac{1}{\sqrt{1-x}}\right)=\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{ n+1}}{n+1}\quad\forall{x}\backepsilon|x|<1}$

8. Originally Posted by Mathstud28
$\ln\left(\frac{1}{\sqrt{1-x}}\right)$

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

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

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

9. Originally Posted by Chris L T521
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

10. Originally Posted by Mathstud28
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