Question 1137179
Find the Maclaurin series for e^(kx), k is a real number.

Calculate the derivatives:

{{{f}}}′{{{(x)=(e^(kx))}}}′={{{ke^(kx)}}}, {{{f}}}′′{{{(x)=(ke^(kx))}}}′={{{k^2*e^(kx)}}},…....,{{{f^(n)(x)=k^n*e^(kx)}}}


Then, at {{{x=0}}} we have

{{{f(0)=e^0=1}}},  {{{f}}}′{{{(0)=ke^0=k}}}, {{{f}}}′′{{{(0)=k^2*e^0=k^2}}},…{{{f^(n)(0)=k^n*e^0=k^n}}}


Hence, the Maclaurin expansion for the given function is

{{{e^(kx)=sum(f^(n)(0)(x^n/n!),n=0,infinity)}}}={{{1+kx+(k^2x^2)/2!+(k^3x^3)/3!}}}+…={{{sum((k^n*x^n)/n!,n=0,infinity)}}}