Series expansion of a function of a funtion

Write down the Taylor expansion around x = 0, up to the order x^3 included, of the expression sin(e^x - 1)

So sinx = x-x^2/2!+x^4/4!+...

and (e^x-1)=x+x^2/2!+x^3/3!+...

how do I combine them?

Do I just sub in x+x^2/2!+x^3/3! for all the xs in the sin expansion?

So sin(e^x - 1)=x+x^2/2!+x^3/3! - (x+x^2/2!+x^3/3!)^2/2!

and the squared term can be reduced to x^2/2!+x^3/8

But this is wrong.

Reply 1

brianeverit

9

Original post by complexnumberij

Write down the Taylor expansion around x = 0, up to the order x^3 included, of the expression sin(e^x - 1)

So sinx = x-x^2/2!+x^4/4!+...

and (e^x-1)=x+x^2/2!+x^3/3!+...

how do I combine them?

Do I just sub in x+x^2/2!+x^3/3! for all the xs in the sin expansion?

So sin(e^x - 1)=x+x^2/2!+x^3/3! - (x+x^2/2!+x^3/3!)^2/2!

and the squared term can be reduced to x^2/2!+x^3/8

But this is wrong.

The squared term is