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).

  • $\begingroup$ 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 $9$? $\endgroup$ – Galc127 Feb 16 '15 at 13:12
  • $\begingroup$ 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. $\endgroup$ – Martigan Feb 16 '15 at 13:20
  • $\begingroup$ @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. $\endgroup$ – Jay Feb 16 '15 at 13:28
  • $\begingroup$ @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. $\endgroup$ – Travis Willse Feb 16 '15 at 13:32
  • $\begingroup$ @Travis You are right, this is not "obvious" as in "true in most cases". In fact I was pointing more toward $\sin x$. $\endgroup$ – 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.

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