# Is this correct reasoning about Taylor series?

Is the following correct reasoning about the Taylor series? I'm just trying to build some intuition but just want to make sure it's correct.

If a function $f(x)$ has a power series representation it will be of the form: $$f(x)=\sum_{n=0}^{\infty}a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+ \text{ ... } \;\;\;\; \text{(1)}$$ Now we have written the coefficients as $a_k$, but we could also just as well write them as: $$a_k=\frac{f^{(k)}(0)}{k!}$$ Of course to get this we need to know that $f(x)$ is infinitely differentiable. Since we found $a_0$ by letting $x=0$ therefore giving $f(0)=a_0$; similarly for $a_2$, we let $x=0$ for the derivative of $f(x)$, or $f'(x)=a_1$; and so on. Now according to this, if $f(x)$ has a power series representation then it must be of the form (I've just centred it around $0$): $$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n \;\;\;\;\; \text{(2)}$$ Since effectively - assuming the function is infinitely differentiable - we've just re-written the general coefficients $a_k$ into an equally general form, but just in terms of the functions derivative. Thus if a function is infinitely differentiable then (2) is completely equivalent to (1). It's just written in this way so we have a mechanism for finding the coefficients (though it could also be done by setting $x=0$ and considering the various derivatives of $f(x)$ to find the coefficients manually).

The reason I'm asking whether this is correct (though not necessarily rigorous, I'm trying to build more of an intuition), is because I was curious why $f(x)=\sin(x)$ could have a Taylor series representation that is entirely defined by the functions derivative at a single point. From the above reasoning it seems clear(er) that the derivatives at a point are just a mechanism for finding the coefficients of the polynomial. And since if there's a power series expansion for $\sin(x)$ it must be of this form; and I would expect there to exist such a power series that could equal the sine function (given enough terms). It just happens to be in the form of Taylor series (which we've defined in terms of derivatives at a point).

• It seems like you got some intuition about the method, but your reasoning is lacking. Look on the sentence "the derivatives at a point are just a mechanism for finding the coefficients of the polynomial" - It is not true that the expansion at some point $x_0$ will be the same as the expansion at some point $x_1$. Look on the function $\ln(1+x)$. What its expansion at $0$? What its expansion at $9$? – Galc127 Feb 16 '15 at 13:12
• I agree with @Galc127. You could do the expansion of $\sin x$ at $x=\dfrac{\pi}{2}$. It would be different. However this expansion still defines your function everywhere, and the value at one point are, obviously, the same whatever the expansion. – Martigan Feb 16 '15 at 13:20
• @Galc127 If you considered the power series as being the sum of $a_n(x-a)^n$ then you would require $a_n=f^{(n)}(x)/n!$. So the two forms: $f(x)=\sum_{n=0}^{\infty}a_n(x-a)^n$ and $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x)}{n!}(x-a)^n$ would surely be completely equivalent. We've just changed where it's centred. – Jay Feb 16 '15 at 13:28
• @Martigan I'm not sure that this is "obvious"; in fact, this sort of statement is true for general function. For example, the power series for $\frac{1}{1 - x}$ has power series $\sum_{k = 0}^{\infty} x^k$ based at $x = 0$ and $\sum_{k = 0}^{\infty} \frac{1}{2^{k + 1}} (x + 1)^k$ based at $x = -1$. The latter converges at $x = -1$ but the former does not. In general, changing where a power series is based can change its interval of convergence. – Travis Feb 16 '15 at 13:32
• @Travis You are right, this is not "obvious" as in "true in most cases". In fact I was pointing more toward $\sin x$. – Martigan Feb 16 '15 at 14:41

This is essentially correct. Some specific comments:

(1) It is sometimes misleading to use the equality symbol $=$ to denote a function has a given power series expansion (say, around zero), as in $$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots,$$ for two reasons, which illuminate other aspect of your question:

(a) First, the power series may not actually converge everywhere the function $f$ is defined. A standard example is $a(x) := \frac{1}{1 - x}$, which has power series $$\sum_{k = 0}^{\infty} x^k;$$ this series converges only on $(-1, 1)$. (Where it does converge, it converges to $a(x)$.)

(b) Second, the power series may converge but not converge to the given function on any open interval containing $0$ (or whatever the base point is): Again, a standard example is the function $$b(x) := \left\{ \begin{array}{cl} 0, & x = 0 \\ e^{-1/x^2}, & x \neq 0 \end{array} \right. .$$ One can show via an induction argument that $b^{(k)}(0) = 0$ for all $k = 0, 1, 2, \ldots$, so the power series for $b$ at $0$ is just the zero series; in particular, it agrees with $b$ only at $0$.

On the other hand, if for every $x_0$ in the (say, open) domain of an infinitely differentiable function $f$ the power series of $f$ based at a point $x_0$ converges to $f$ on some open interval containing $x_0$, we say that $f$ is (real-)analytic. Most of the usual smooth functions one encounters in practice are real-analytic.

For these reasons, unless a function's power series really does converge to that functioneverywhere on its domain, I prefer to write something like $$f(x) \sim a_0 + a_1 x + a_2 x^2 + \cdots.$$ (You'll still see $=$ sometimes, which is an abuse, but not a terrible one.)

(2) Roughly speaking, the only functions given by a finite power series are polynomials, more or less by definition. In particular, one cannot expect any truncation of the power series for $\sin x$ at finite order to be equal to $\sin x$, which (because it is bounded but nonconstant) is not a polynomial.