# Taylor Polynomial

• April 21st 2008, 10:49 AM
akhayoon
Taylor Polynomial
Question was

evaluate the 4th taylor polynomial of $\sqrt{3+x^{2}}$ around a=-1

$f4(x)=2-\frac{-(x+1)}{2}-\frac{(x+2)^{2}}{(8)2!}-\frac{3(x+1)^{3}}{(32)3!}-\frac{15(x+1)^{4}}{(128)4!}$

the marker notes however that there was something missing in the 3rd term

that the 3 in the fourth term was wrong, and that the 15 and 128 in the last term is wrong as well...

but after looking at my derivatives I can't figure out why this is???
• April 21st 2008, 11:41 AM
Opalg
Quote:

Originally Posted by akhayoon
Question was

evaluate the 4th taylor polynomial of $\sqrt{3+x^{2}}$ around a=-1

$f4(x)=2-\frac{-(x+1)}{2}-\frac{(x+2)^{2}}{(8)2!}-\frac{3(x+1)^{3}}{(32)3!}-\frac{15(x+1)^{4}}{(128)4!}$

the marker notes however that there was something missing in the 3rd term

that the 3 in the fourth term was wrong, and that the 15 and 128 in the last term is wrong as well...

but after looking at my derivatives I can't figure out why this is???

$f'(x) = x(3+x^2)^{-1/2},\ \ f'(-1) = {\textstyle-\frac12}$.

$f''(x) = (3+x^2)^{-1/2} - x^2(3+x^2)^{-3/2},\ \ f''(-1) = {\textstyle\frac12 - \frac18 =\frac38}$.

$f'''(x) = -3x(3+x^2)^{-1/2} + 3x^3(3+x^2)^{-5/2},\ \ f'''(-1) = {\textstyle\frac38 - \frac3{32} =\frac9{32}}$.

That gives $f_3(x) = 2 - \frac{(x+1)}{2} + \frac{3(x+1)^{2}}{(8)2!} + \frac{9(x+1)^{3}}{(32)3!}$. I'll leave you to sort out the (x+1)^4 term.