Group Abstract Group Abstract

Message Boards Message Boards

0
|
7.1K Views
|
1 Reply
|
3 Total Likes
View groups...
Share
Share this post:

Compute a Taylor series of a function near zero?

Posted 6 years ago
POSTED BY: Claude Mante

Include an assumption that ? is positive:

Series[Cas2a[?], {?, 0, 1}, Assumptions -> ? > 0] //TeXForm

$$\frac{\sqrt{\frac{-p+p \left(-\log \left(\frac{1}{p+1}\right)\right)-\log \left(\frac{1}{p+1}\right)}{p (p+1)^2}}}{\phi }-\frac{(p+1) \log ^2\left(\frac{1}{p+1}\right) \sqrt{\frac{-p+p \left(-\log \left(\frac{1}{p+1}\right)\right)-\log \left(\frac{1}{p+1}\right)}{p (p+1)^2}}}{4 \left(-p-(p+1) \log \left(\frac{1}{p+1}\right)\right)}+\frac{1}{2} \phi \sqrt{\frac{-p+p \left(-\log \left(\frac{1}{p+1}\right)\right)-\log \left(\frac{1}{p+1}\right)}{p (p+1)^2}} \left(-\frac{(p+1)^2 \log ^4\left(\frac{1}{p+1}\right)}{16 \left(-p-(p+1) \log \left(\frac{1}{p+1}\right)\right)^2}-\frac{(p+1) \left(\log ^3\left(\frac{1}{p+1}\right)+\pi ^2 \log \left(\frac{1}{p+1}\right)\right)}{6 \left(-p-(p+1) \log \left(\frac{1}{p+1}\right)\right)}\right)+O\left(\phi ^2\right)$$

POSTED BY: Carl Woll
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard