I need some help understanding the maclaurin solution tothis

[Riazy]

Hi guys, here is my problem

determine Maclaurin development of order 3 with residual term with o notation e^cosx

My way of trying to solve it using a hint

1. e^cosx is the same as e^(1-1/2 * x^2 + O((x^4))
2. e * e^-x^2/2=
3. e* (1-1/2 *x^2) O(x^4) = e-e/2 * x^2 + O(x^4)

My question is how, did the values and numbers in the exponent "jump" down to the ground, if i would put it that way,
in other words how did they become multiplied by e instead of being part of the exponent

Some rule? I am not aware of?

 

[galactus]

Re: I need some help understanding the maclaurin solution to

Taking the first derivative of \(\displaystyle e^{cos(x)}\), we get

\(\displaystyle f'(x)=-sin(x)e^{cos(x)}\). MacLaurin series are centered at x=0, so we have 0. Along with all odd terms being 0.

\(\displaystyle f''(x)=(sin^{2}(x)-cos(x))e^{cos(x)}\), when x=0, we get \(\displaystyle -e\), resulting in

\(\displaystyle \frac{-e\cdot x^{2}}{2}\).

\(\displaystyle f^{(4)}(0)=4e\), resulting in \(\displaystyle \frac{4e\cdot x^{4}}{24}=\frac{e\cdot x^{4}}{6}\)

and so on.

So, the MacLaurin series ends up being:

\(\displaystyle e-\frac{e\cdot x^{2}}{2}+\frac{e\cdot x^{4}}{6}-\frac{31e\cdot x^{6}}{720}+.............\)