Question

Let $f\left(x\right)=e^{x+x^2}$for real x .From among the following choose the Taylor series approximation of f(x) around x=0, which includes all powers of x less than or equal to 3.

• A. $1+x+x^2+x^3$
• B. $1+x+\frac{3}{2}{x}^2+x^3$
• C. $1+x+\frac{3}{2}{x}^2+\frac{7}{6}{x}^3$
• D. $1+x+3x^2+7x^3$

Answer 1

The correct option is C $1+x+\frac{3}{2}{x}^2+\frac{7}{6}{x}^3$

$f\left(x\right)=e^{x+x^2}$
$f^{\prime }\left(x\right)=e^{x+x^2}(1+2x)$
$f^{\prime \prime }\left(x\right)=(1+2x{)}^2{e}^{x+x^2}+e^{x+x^2}\times 2$
$=e^{x+x^2}(3+4x^2+4x$)
$f^{\prime \prime \prime }\left(x\right)=(3+4x+4x^2)(1+2x)e^{x+x^2}+e^{x+x^2}(4+8x)$
Now, $f^{\prime }\left(0\right)=1,f^{\prime \prime }\left(0\right)=3,f^{\prime \prime \prime }\left(0\right)=3+4=7$
Taylor series of f(x) around x=0 is
$f\left(x\right)=f\left(0\right)+\frac{f^{\prime }\left(0\right)}{1!}x+\frac{f^{\prime \prime }\left(0\right)}{2!}{x}^2+\frac{f^{\prime \prime \prime }\left(0\right)}{3!}{x}^3+......$
$\because f\left(x\right)=1+x+\frac{3}{2}{x}^2+\frac{7}{6}{x}^3$