Answer to Question #114643 in Calculus for ANJU JAYACHANDRAN

Question

Expand e^2x in powers of (x-1), up to four terms

Answer 1

$By\, Taylor's\, theorem,\, we\, know\, that \, f(x) =$$f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+......$

$So, here f(x) =e^{2x}$ and $a=1.$

$e^{2x} =f(1)+(x-1)f'(1) +\frac{(x-1)^2}{2!}f''(1)+\frac{(x-1)^3}{3!}f'''(1)+.......$

$f(1)=$ $e^2 , f'(1)=2e^2, f''(1)=4e^2, f'''(1)=8e^2$

Putting the above values in the expansion of $e^{2x}$ , we get

$e^{2x} = e^2+2(x-1)e^2+2(x-1)^2e^2+\frac{4}{3}(x-1)^3e^2$ (expansion is done up to 4 terms)

Therefore, expansion of $e^{2x}$ in powers of $(x-1)$ up to 4 terms is

$e^{2x} = e^2+2(x-1)e^2+2(x-1)^2e^2+\frac{4}{3}(x-1)^3e^2 .$