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I'm currently relearning Taylor series and yersterday I thought about something that left me puzzled. As far as I understand, whenever you take the Taylor series of any function $f(x)$ around a point $x = a$, the function is exactly equal to its Taylor series, that is:

$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n $$

For example, if we take $f(x) = e^x$ and $x = 0$, we obtain: $ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $

My doubt is: the only variables in the Tayor series formula are $f(a), f'(a), f''(a),$ etc., that is, the successive derivatives of the function $f$ evaluated in one point $x = a$. But the Taylor series of $f(x)$ determine the whole function! How is it possible that the successive derivatives of the function evaluated in a single point determine the whole function? Does this mean that if we know the values of $f^{(n)}(a)$, then $f$ is uniquely determined? Is there an intuition as to why the succesive derivatives of $f$ on a single point encode the necessary information to determine $f$ uniquely?

Maybe I'm missing a key insight and all my reasoning is wrong, if so please tell where is my mistake.

Thanks!

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    $\begingroup$ Ignoring convergence issues, intuitively think about a person walking on a path. In order to know where the person is immediately going next, you need to know their position ($f(a)$) and which direction they are headed ($f'(a)$). Now, if you knew which way the person was planning to turn, or curve their path ($f''(a)$), then you could predict their path a little farther. Each increment of knowledge on the persons future trajectory gives you more accuracy on where they will be arbitrarily far into the future. $\endgroup$ – Ninad Munshi Nov 14 '19 at 9:24
  • $\begingroup$ @NinadMunshi. Very good explanation ! May I reuse it ? Cheers :-) $\endgroup$ – Claude Leibovici Nov 14 '19 at 9:40
  • $\begingroup$ Thanks for the intuitive explanation @NinadMunshi, it really helped me in understanding how is is possible that, if the function is analytic, then it is uniquely determined by the derivatives on a point. $\endgroup$ – David O. Nov 14 '19 at 9:54
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    $\begingroup$ @ClaudeLeibovici Absolutely! My explanation would not do much good if it just collected dust on stackexchange. $\endgroup$ – Ninad Munshi Nov 14 '19 at 10:29
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You're right, in general $f$ is not determined by its derivatives at one single point. Functions satisfying this condition are called analytic. But not all smooth functions are analytic, for example

$$x\mapsto\left\{\begin{array}{c}e^{-\frac{1}{x^2}}, x>0\\0, x\leq 0\end{array}\right.$$ is a smooth function and the derivatives at zero are all zero, hence the Taylor series developed at zero does not determine the function.

Furthermore the exact statement of Taylor's theorem is quite different from what you said. It is as follows:

If $f\in C^{k+1}(\mathbb{R})$, then $$f(x)=\sum_{n=0}^k f^{(n)}(a)(x-a)^n\frac{1}{n!} + f^{(k+1)}(\xi)\frac{1}{(k+1)!}(x-a)^{k+1}$$

If you now take $k\rightarrow\infty$ it is in general not clear, that this error term converges to zero.

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    $\begingroup$ As an additional note: The existence of non-analytic smooth functions is incredibly important in advanced analysis and differential geometry, where you need test functions ($C^{\infty}_c$-functions are necessarily non-analytic unless they are $0$) to develop weak derivatives by duality, and you need partitions of unity to glue local properties. $\endgroup$ – WoolierThanThou Nov 14 '19 at 9:33
  • $\begingroup$ Thanks for the explanation, I was not familiar with the concept of analytical functions. Is there any theorem that characterises such functions? $\endgroup$ – David O. Nov 14 '19 at 9:56
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    $\begingroup$ @DavidO. That is quite a big topic and maybe worth a new question. Note that the examples here are all or nothing: your example converges for all $x$ and the example here converges for no positive $x$. However, it is possible that the series converges up to a certain distance from $a$ called the radius of converge. Are you familiar with complex numbers? They help the understanding of this topic. $\endgroup$ – badjohn Nov 14 '19 at 10:33
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    $\begingroup$ @DavidO. In some ways, differentiation is similar with complex functions but one big difference is that in the complex world a once differentiable function is analytic. $\endgroup$ – badjohn Nov 14 '19 at 10:45
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    $\begingroup$ @David O: The easiest characterization is done by using complex analysis. In a sense, complex analysis is the "most fully developed" theory of analytic functions. A real function is real analytic if and only if there exists a complex function which is complex-differentiable (a more elegant condition) in a neighborhood of the real line and equals the given real function at that point (up to the obligatory "type coercion" of course :) ). $\endgroup$ – The_Sympathizer Nov 15 '19 at 4:06
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Functions which are the sum of their Taylor series within the interval (or disk for functions of a complex variable) of convergence are known as analytic functions. Many basic elementary functions are analytic: $\;\exp, \sin,\cos,\sinh,\cosh $ and of course polynomials are analytic on $\mathbf R$ (or $\mathbf C$).

It is not true that, in general, an infinitely differentiable function of a real variable is analytic on the interval of convergence of its Taylor series, as @humanStampedist's example shows.

However, for a function of a complex variable, simply being differentiable suffices to ensure the function is analytic (one usually says holomorphic in this case). This is due to the very strong constraints of the Cauchy-Riemann equations.

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  • $\begingroup$ Thanks for these insights @Bernard. I'll take a look on those topics in order to learn more about analytical functions $\endgroup$ – David O. Nov 14 '19 at 10:43
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    $\begingroup$ @DavidO.- just FYI, the term is "analytic", not "analytical". If you search for "analytical", you will have to rely on the search engine's ability to correctly determine what you are really after. $\endgroup$ – Paul Sinclair Nov 14 '19 at 17:56
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HumanStampedist has adequately answered the question. I'd like to mention that, just as there are continuous functions that are nowhere differentiable, such as the Weierstrass function, there are smooth functions (all $n$th derivatives exist at every point) that are nowhere analytic, i.e. at no point does the Taylor series converge to the original function. An example is the Fabius function.

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One doesn't have to try very hard to find a function that does not agree with its Taylor series everywhere. The absolute value function is a familiar enough function. The Taylor series of $|x|$ at $x = 1$ is $x$. (The constant term is zero and all the higher degree terms are zero. If you ignore the left half of the graph of $|x|$, you should see that this function is "trying" to be a straight line in any reasonably small and/or bounded-above-zero-on-the-left open neighborhood of the expansion point, $1$.)

This Taylor expansion is identical to the function on $x \geq 0$, and is hilariously wrong for $x < 0$. However, on expanding a Taylor series at any point on the left half of the real line gives $-x$. This is identical to the function on $x \leq 0$ and hilariously wrong on the right half of the real line.

Why did the Taylor series not "work" everywhere? In any little neighborhood of an $x$ that does not include $0$, the function $|x|$ looks like either a line with slope $1$ or a line with slope $-1$, so this is all the derivatives can see. The sudden change in behaviour at $x = 0$ is not signalled in the derivatives anywhere (except that none of the derivatives exist at $x = 0$). It's almost as if the undefinedness at $x=0$ acts as a barrier -- the Taylor series on one side of that barrier does not replicate behaviour from the other side. (... except in carefully contrived accidents, like $\frac{x^2}{x}$ which is undefined at $0$ so has no derivatives there, but agrees with any of its Taylor series.)

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  • $\begingroup$ Wow, thank you for this example! It's a very simple function yet it reflects the ideas on why the Taylor series doesn't equal the function. One question though: in your example it is clear that the derivatives do not exist at $x=0$, and I can see how this acts as a "barrier". However, in the example proposed by @HumanStampedist, the derivatives do exist on $x = 0$. How would we explain then that the Taylor series does not equal the function in an intuitive way? $\endgroup$ – David O. Nov 15 '19 at 10:56
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    $\begingroup$ @DavidO. : HumanStampedists example does have a barrier, infinitesimally close to $x = 0$ ... in the complex plane. See [math.stackexchange.com/questions/717676/… for more on this. $\endgroup$ – Eric Towers Nov 15 '19 at 11:13
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    $\begingroup$ @DavidO. : I've added some plots to the cited answer to perhaps mke it easier to see that there is a barrier pressed right up against the origin along the imaginary axis. $\endgroup$ – Eric Towers Nov 17 '19 at 21:10

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