1. ## Maclaurin Polynomial

Good morning,

I am working on Taylor Polynomials this morning, and am not entirely sure I understand what I am supposed to be doing. Would someone mind telling me if the following makes sense?

Compute the Maclaurin Polynomials of degree one, two, and three for $\displaystyle f(x) = e^{-2x}$

As I understand this the fact that this is a Maclaurin Polynomial simply means that the term $\displaystyle (x-a)^n$ becomes simply $\displaystyle x^n$.

I get the following:

$\displaystyle f'(x)=-2e^{-2x}$

$\displaystyle f''(x)=4e^{-2x}$

$\displaystyle f'''(x)=-8e^{-2x}$

$\displaystyle f'(0)=-2$

$\displaystyle f''(0)=4$

$\displaystyle f'''(0)=-8$

$\displaystyle T_1(x)=1-2x$

$\displaystyle T_2(x)=(2x-1)^2$

$\displaystyle T_3(x)=-8x^3+4x^2-2x+1$

Am I way off target here?

2. Don't forget to divide by the factorial.

3. Ah, indeed...

$\displaystyle T_1(0)=-2x+1$

$\displaystyle T_2(0)=2x^2-2x+1$

$\displaystyle T_3(0)=-\frac{4}{3}x^3+2x^2-2x+1$

I had to look up the notation $\displaystyle n!$. Apparently it means $\displaystyle n(n-1)(n-2)...(2)(1))$. Is that correct?

Is that all there is to Taylor Polynomials?

4. Looking good there, now. That is what the factorial means. By convention, 0!=1. The generalization of the factorial is the gamma function, which is the reason for that convention. It also makes the MacLaurin expansion easier to write:

$\displaystyle f(x)=\displaystyle{\sum_{k=0}^{\infty}\frac{f^{(k) }(0)}{k!}\,x^{k}},$

rather than separating out the earlier terms.

Another convention is that the zeroth derivative of a function is simply the function itself.

5. As much as I hate asking "is this right"...

Is there an easy to verify a Taylor Polynomial so that I don't need to ask if my solution is correct?

6. You can always go to WolframAlpha and use the Series command.

Series[Exp[-2x],{x,0,3}]

will give you the 3rd order MacLaurin expansion of the function in this thread.