Question #eb1ff

Question

1667 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-08T12:17:29

$(1+x^2)|x|$

Taylor expansion is

$f(x)=f(0)+xf'(0)+x^2/(2!)f''(0)+x^3/(3!)f'''(0)+x^4/(4!)f''''(0) +...$ ..

$f=xe^(x^2)$ ; f(0)=0.

$f'=2x^2e^(x^2)+e^(x^2)=2xf+e^(x^2); f'(0)=1$ .

$f''=2xf'+4f; f''(0)=0$

$f'''=6f'+2xf''. f'''(0)=6$

$f''''=8f''+2xf'''; f''''(0)=0$ .

The Taylor biquadratic polynomial, about x = 0 is

x+x^3 and the upper bound is given by

$(x+x^3|<=|x||1+x^2| =(1+x^2)|x|$