Taylor Polynomial

**akhayoon** · April 21st 2008, 09:49 AM

Question was

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

my answer was

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

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???

**Opalg** · April 21st 2008, 10:41 AM

Originally Posted by akhayoon

Question was

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

my answer was

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

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.

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