Question

Taylor Series

Answer 1

of \[e^{-x^2}\]

Answer 2

\[e^x=1+x+\frac{ x^2 }{ 2! }+\frac{ x^3 }{ 3! }...\]

Answer 3

How about if we substitute x in the e^x expansion by -x^2,

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

Answer 4

so youre saying...

Answer 5

Say I had
\[e^{-4x^3}=1-4x^3-\frac{ (4x^2)^3 }{ 2! }-\frac{ (4x^3)^3 }{ 3! }...\]

Answer 6

Yes, except for the sign. If the exponent is negative, then the sign would be alternating, i.e. negative for odd powers/terms.

Answer 7

ah yup i see. thanks!

Answer 8

\(\large e^{-4x^3}=1-4x^3\color{red}{+}\frac{ (4x^\color{red}{3})^3 }{ 2! }-\frac{ (4x^3)^3 }{ 3! }\color{red}{+}...\)