# Thread: taylor series problem

1. ## taylor series problem

Hi.
Evaluate.

$\displaystyle \lim_{\ x\to 0}\frac{\log{cosx}-xsinx}{2e^{x^2}-cos2x-1}$

2. Originally Posted by Shock
Hi.
Evaluate.

$\displaystyle \lim_{\ x\to 0}\frac{\log{cosx}-xsinx}{2e^{x^2}-cos2x-1}$
use L'Hopital ... twice.

3. Originally Posted by skeeter
use L'Hopital ... twice.

The prof said we have to solve it with taylor series.

4. Originally Posted by Shock

The prof said we have to solve it with taylor series.
hmm ... I think you'll still have to use L'Hopital with the polynomials.

maybe someone knows a shortcut. I apologize that I do not.

5. Look, this is kinda' news to me but it seems to fall out right.. So if we can obtain the Taylor series:

$\displaystyle \log(\cos(x))-x\sin(x)=-3/2 x^2+x^4/10+\cdots$

$\displaystyle 2e^{x^2}-\cos(2x)-1=4x^2+x^4/3+\cdots$

and then use long division:

$\displaystyle \begin{array}{rc@{}c} & \multicolumn{2}{l}{\, \, \, } \vspace*{0.12cm} \\ \cline{2-3} \multicolumn{1}{r}{4x^2+x^4/3+\cdots \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} -3/2 x^2+x^4/12+\cdots} \\ & \multicolumn{2}{l}{\phantom{xxxxxxxxxxxxxxxxxxxxxx }} \end{array}$

note the first term is -3/8 and I assume the remainder will have powers of x so that as x->0, the limit is -3/8 which is the limit given by Mathematica.

I'm not entirely sure about the validity of this method with the other one since $\displaystyle \sqrt{x}$ has a branch-point at the origin.

6. Originally Posted by shawsend
Look, this is kinda' news to me but it seems to fall out right.. So if we can obtain the Taylor series:

$\displaystyle \log(\cos(x))-x\sin(x)=-3/2 x^2+x^4/10+\cdots$

$\displaystyle 2e^{x^2}-\cos(2x)-1=4x^2+x^4/3+\cdots$
not by "hand" I reckon ... ?

7. Originally Posted by skeeter
not by "hand" I reckon ... ?
Why not skeeter? They're both analytic at zero so it's easy to compute the first few terms of the Taylor series for each just be taking a few derivatives at x=0. I guess all this is ok. Never solved a limit this way before. Maybe some issues of uniform convergence is required for it to be valid though.

8. Originally Posted by shawsend
Why not skeeter? They're both analytic at zero so it's easy to compute the first few terms of the Taylor series for each just be taking a few derivatives at x=0. I guess all this is ok. Never solved a limit this way before. Maybe some issues of uniform convergence is required for it to be valid though.
... still seems like a pain-in-the-@ way to do it, IMHO.