# Taylor expansion

• Jun 10th 2009, 12:12 AM
dhiab
Taylor expansion
Give the Taylor expansion order 1 for the function :
http://www.mathramz.com/xyz/latexren...2a7211c1bf.png
and dudect that graph of f accept asymtot of which they ask
his psition and graph of f .
• Jun 10th 2009, 07:05 AM
TheEmptySet
Quote:

Originally Posted by dhiab
Give the Taylor expansion order 1 for the function :
http://www.mathramz.com/xyz/latexren...2a7211c1bf.png
and dudect that graph of f accept asymtot of which they ask
his psition and graph of f .

$\displaystyle f(a)=a^2\arctan\left( \frac{1}{a+1}\right)$

$\displaystyle f'(x)=2x\arctan\left( \frac{1}{x+1}\right)+2x\left(\frac{1}{(\frac{1}{1+ x})^2+1} \right)\left( \frac{-1}{(x+1)^2}\right)=$

$\displaystyle f'(x)=2x\arctan\left( \frac{1}{x+1}\right)-\left(\frac{2x}{(1+x)^2+1} \right)$

$\displaystyle f'(a)=2a\arctan\left( \frac{1}{a+1}\right)-\left(\frac{2a}{(1+a)^2+1} \right)$

Since you didn't say where it was centered I centered it at "a"

$\displaystyle T_1(x)=f(a)+f'(a)(x-a)$

$\displaystyle T_1(x)=a^2\arctan\left( \frac{1}{a+1}\right)+\left[2a\arctan\left( \frac{1}{a+1}\right)-\left(\frac{2a}{(1+a)^2+1} \right) \right](x-a)$