I understand that when using standard Maclaurin series (e.g. e^x=1+x+x^(2)/(2!)+... , we can substitute x with other functions such as z=x^2 to transform the standard Maclaurin series into e^(x^2)=1+x^2+x^(4)/(2!)+...Why doesn't this require a (dz)/(dx)=2x component in the expansion since the standard Maclaurin series are derived from differentiation of the original term.

Bergsteinj0

Bergsteinj0

Answered question

2022-09-30

using standard Maclaurin series (e.g. e x = 1 + x + x 2 2 ! + . . . , we can substitute x with other functions such as z = x 2 to transform the standard Maclaurin series into e x 2 = 1 + x 2 + x 4 2 ! + . . . Why doesn't this require a d z d x = 2 x component in the expansion since the standard Maclaurin series are derived from differentiation of the original term.

Answer & Explanation

Abigayle Lynn

Abigayle Lynn

Beginner2022-10-01Added 12 answers

f ( x ) = d d x e x 2 = 2 x e x 2
and f ( 0 ) = 0 ! [Not a factorial]

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?