Question

Find the Taylor polynomial of order 4 (at x = 0) for the function : f(x) = e^{x^2}

Answer 1

In general,

the Taylor polynomial of order n for f(x) centered at x=a is given by

$T_n(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2+\frac{f'''(a)}{6}(x-a)^3...+\frac{f^{(n-1)}(a)}{(n-1)!}(x-a)^{n-1}+\frac{f^{(n)}(a)}{n!}(x-a)^n$

Therefore,

Find the respective derivatives of f(x)

f(x)=$e^{x^2}$ ;f(0)=1

f'(x)=$2xe^{x^2}$ ;f'(0)=0

f''(x)=$(4x^2+2)e^{x^2}$ ;f''(0)=2

f'''(x)=$(8x^3+12x)e^{x^2}$ ;f'''(0)=0

f''''(x)=$(16x^4+48x^2+12)e^{x^2}$ ;f''''(0)=12

Hence,

T₄(x)=$1+0+x^2+0+\frac12x^4$

$T_4(x)=1+x^2+\frac12x^4$