How Do You Calculate the Taylor Expansion of e^(2-x)?

[zzmanzz]

Homework Statement

Find e^{2-x}

using taylor/mclaurin expansion

Homework Equations

e^1 = \sum_{n=0}^\inf \frac{1}{n!}

e^x = \sum_{n=0}^\inf \frac{x^n}{n!}

The Attempt at a Solution

Can I just do:

e^{1+1-x}

[\sum_{n=0}^\inf \frac{1}{n!}*\sum_{n=0}^\inf \frac{1}{n!} * \sum_{n=0}^\inf \frac {(\frac{-x}{2})^n}{n!}]

 

[vela]

I'm not sure why you'd want to. About which point are you supposed to expand the function? If it's about x=0, try using the fact that $e^{2-x}=e^2e^{-x}$.

 

[zzmanzz]

I'm not sure about which point to expand. I think that a = 0 is a requirement.

Ok so:

e^2*e^{-x} =

f(x) = C*e^{-x}

C = e^2 = \sum_{n=0}^\inf \frac{2}{n!}

e^{-x} = \sum_{n=0}^\inf \frac{(-x)^n}{n!}]

\sum_{i=0}^\inf \frac{2}{i!} * \sum_{j=0}^\inf \frac {(-x)^j}{j!}]

\sum_{i=0}^\inf \sum_{j=0}^\inf [ \frac{2}{i!} ]* [ \frac {(-x)^j}{j!}]

 

[vela]

Looks good.

Remember that $e^2$ is just a constant. It doesn't need to be expressed in series form, so you have
$$e^{2-x} = e^2\sum_{n=0}^\infty \frac{(-x)^n}{n!}.$$ Typically you pull the negative sign out and write $(-x)^n = (-1)^n x^n$.

 

[zzmanzz]

ah. yes, I totally forgot so much fundamental calculus stuff it's embarrassing.

Please bare me with me for one more question.

In order to take

\frac{de^{2-x}}{dx}

It's

h'(x)*e^{h(x)}

h(x) = 2-x

h'(x) = -1

f'(x) = -1*e^{2-x}

f''(x) = 1*e^{2-x}

f'''(x) = -1*e^{2-x}

can I now do
f(x) = f(0) + f'(0)x + f''(0)\frac{x^2}{2!} + f'''(0)\frac{x^3}{3!} + ...+ f^n(0)\frac{x^n}{n!}

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

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

so that equals the above sum it seems. Much clearer with the constant that you took out. The only thing I don't get is where the factorial n! comes from in the expansion? Sorry I didn't factor out the negative.. just wnated to post quick. I will do it from now on ! I think I might've screwed up in taking the derivatives.

 

[vela]

The series should be in powers of x, not powers of -x, when you substitute in for the derivatives evaluated at 0, but otherwise your work is fine.

To see where the factorial comes from, look up the derivation of the Taylor series in your textbook.

 

[zzmanzz]

Usually when I expand using the taylor/maclaurin series correctly the factorial is present inside the derivatives (i.e. I can factor it out or pinpoint it in the formula)

Like when I expanded ln(1+x) for example, the factorial came from the derivative n times
f'(x) = 1/(1+x) ,
f''(x) = -2*(1+x)^(-2)
f'''(x) = 2*3*(1+x)^(-3) .. etc. ?

In this one, I can't.

I will take your word for it and review it.

 

[vela]

The log series, $\ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$, actually doesn't end up with a factorial in it. The denominator is simply n.

For $\ln (1+x)$, the nth derivative evaluated at x=0 is, as you noted, $f^{(n)}(0) = (-1)^{n+1}(n-1)!$. When you plug this into the formula for the Taylor series, the coefficient of xⁿ simplifies to
$$\frac{f^{(n)}(0)}{n!} = \frac{(-1)^{n+1}(n-1)!}{n!} = \frac{(-1)^n}{n},$$ as expected. So the derivatives generate a factorial, but there's also the factorial already present in the Taylor series formula.