Taylor's Polynomial Evaluation Hi guys now started off doing some taylor expansion wanted some help with the following question:
$$ f:[0,\infty)\to \mathbb{R}$$
$$ f(x) = 2e^{-x/2} + e^{-x} $$
Determine the nth degree taylor's polynomila $T_n$ of f about c = 0
My attempt to this question can be seen below;
$$P(x) = f(x) + f'(x)(x-c) + \frac{f''(x)(x-c)^2}{2!} + \frac{f'''(x)(x-c)^3}{3!} + \frac{f''''(x)(x-c)^4}{4!}$$
Evaluation of the function up to the fourth derivative 
$$f'(x) = -e^{-x/2} - e^{-x}$$
$$f''(x) = \frac{1}{2}e^{-x/2} + e^{-x}$$
$$f'''(x) = -\frac{1}{4}e^{-x/2} - e^{-x}$$
$$f''''(x) = \frac{1}{8}e^{-x/2} + e^{-x}$$
substituting x = 0 and we obtain the following
$$f'(x) = -2 $$
$$f''(x) = \frac{3}{2} $$
$$f'''(x) = \frac{-5}{4}$$
$$f''''(x) = \frac{9}{8} $$
substituting c = 0 and the derivatives into the taylor's polynomial we obtain the following:-
$$P(x) = 3 - 2x + \frac{\frac{3}{2}x^2}{2!} - \frac{\frac{5}{4}x^3}{3!} + \frac{\frac{9}{8}x^3}{4!} $$
I am not sure if my evaluation is correct and if so I am having difficulty putting into the form as required below hoping someone can help.
$$T_n (x) = \sum_{k=o}^\infty \frac{f^{k}(c)(x-c)^k}{k!}$$
 A: You got all the derivatives right, with $f(0) = 3, $$f'(0) = -2$, $f''(0) = \dfrac{3}{2}$, $f'''(0) = -\dfrac{5}{4}$, and so on. As a side note, keep in mind you're evaluating $f^{(n)}$ at $x = 0$, so you saying $f'(x) = -2$, $f''(x) = \dfrac{3}{2}$, and so on isn't correct. (Note that by "derivatives," I'm referring to their values at $x = 0$.)
So the $n^{\text{th}}$-degree Taylor polynomial at $x = 0$ is given by $T_n(x) = \displaystyle\sum_{k = 0}^{n} \dfrac{f^{(k)}(0)}{k!}x^k$. Now, you need to find some way to represent $f^{(n)}(0)$:


*

*First, notice that the signs are alternating. Since the even-order derivatives are positive, you can introduce a factor of $(-1)^k$. This means even $k$'s give $+1$ while odd $k$'s give $-1$.

*Notice how the numerator is always one more than the denominator. Also, notice how the denominators are just powers of two, with the power always being one less than the order. For instance, you have $4 = 2^{3-1}$ for the third derivative. This would indicate a factor of $\dfrac{2^{k-1}+1}{2^{k-1}}$. Does this work for $f(0)$ (which you can think of as the zero-order derivative)? Testing it shows that it does:


$$\frac{2^{0-1}+1}{2^{0-1}} = \frac{\dfrac{1}{2}+1}{\dfrac{1}{2}} = 3 = f(0)$$
Combining the two points gives $f^{(k)} (0) = (-1)^k \dfrac{2^{k-1}+1}{2^{k-1}}$. This means 
$$\boxed{T_n(x) = \displaystyle\sum_{k = 0}^{n} (-1)^k \dfrac{2^{k-1}+1}{\left(2^{k-1}\right)k!}x^k}$$
A: If you start with the usual
$$e^{-t}=\sum_{n=0}^\infty \frac{(-1)^n}{n!} t^n$$ and apply it twice
$$f(x)=2 e^{-x/2}+e^{-x}=2\sum_{n=0}^\infty \frac{(-1)^n}{2^n\,n!}x^n+\sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\left(1+2^{-n+1} \right)x^n$$
