# Taylor Series Clarification

For $\sin(x)$, $e^x$, $\cos(x)$... When we are building the $n$-th taylor polynomial, why is it that we always evaluate the functions first $k$ derivatives at $x=0$? In my textbook when they were deriving the formula for each of those 3 functions, they would always evaluate all the derivatives at $x=0$ but why not at some other value of $x$ that is more general?

• Because one is Maclaurin ($x=0$) and the other is Taylor (general) but the later contains the former. – Chinny84 Mar 21 '15 at 0:21
• You can find it at a different point, for instance $e^{x}$ has the Taylor series at $x=a$ $$e^x = e^a \sum_{k=0}^{\infty}\frac{(x-a)^k}{k!} .$$ – science Mar 21 '15 at 0:27
• For a lot of series (like $e^{x}$) the $a=0$ choice is convenient because you actually know what that value is. For example, in science's comment above, if you pick a value other than $a=0$ how do you evaluate the $e^{a}$ term? – TravisJ Mar 21 '15 at 0:43

1. For the functions you list, you don't gain a lot by expanding about another point, because you have identities like $$e^x = e^{a+(x-a)} = e^a e^{x-a},$$ so the Taylor series of $e^x$ about the point $x=a$ is $$e^x = \sum_{k=0}^{\infty} e^a \frac{(x-a)^k}{k!},$$ which has much the same properties as the Taylor series of $e^x$. You can do the same thing with $\sin{x}$ and $\cos{x}$ using the addition formulae like $$\sin{(a+(x-a))} = \sin{a}\cos{(x-a)} + \cos{a} \sin{(x-a)},$$ and you know how to expand those about $x=a$ from your results at $0$.
2. More generally, it is often useful to expand about a point other than $x=0$. A classic example is the logarithm: $\log{x}$ is not differentiable at $x=0$, but at $x=1$ we do have an expansion, which you may not have noticed because we often write it a different way: $$\log{(1+x)} = x-\frac{x^2}{2}+\frac{x^3}{3}+\dotsb$$
3. The most general answer is that the Maclaurin series of the function is the Taylor series of the translate of a function ($f(x-a)$, basically). Often, when looking at a finite number of terms, it's just easier to deal with a polynomial in $x$
And sometimes, expansions about different points can look quite different. Consider the following series: $$1-x^2+x^4-\dotsb = \frac{1}{1+x^2} \quad (|x|<1) \tag{1}$$ Now consider this series: $$\tfrac{1}{2}-\tfrac{1}{2}x+\tfrac{1}{4}x^2-\tfrac{1}{8}x^4+\tfrac{1}{8}x^5-\dotsb. \tag{2}$$ What you don't know about this series is that it's actually the Taylor series of $$\frac{1}{1+(1+x)^2},$$ about $x=0$, so it's also the Taylor series of $(1+x^2)^{-1}$ about $x=1$. (2) looks a lot worse than (1) (you and I would struggle to guess what (2)'s $n$th term is, for one thing), but (using techniques you learn in Analysis courses at university), you can show that (2) actually converges for $|x-1|<\sqrt{2}$, so in a sense it's more useful, because it covers a larger range of possible $x$ values.
Therefore I'd say the normal reason people would give when asked is that there are just some points for which the series is easier to calculate, and this series works for the values of $x$ we're interested in, which is good enough. (In the case of the exponential and sine and cosine, all values of $x$, in fact.)