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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|#