With a view to better understanding real Taylor series, I have examined some books on basic Calculus, with an eye for the proofs of the Taylor series theorem and the possible authors' comments on its derivation. (My reaction when I first saw a proof of it, many years ago, was a mixture of great surprise and anxiety. And still, while I understand the individual steps, the way they all combine to produce e.g. the series for sinx strikes me as little short of miraculous.)

Up to now, from the books I have seen, I get the same impression: that this theorem is a technical exercise in repeated applications of the mean-value theorem. And we are lucky that some useful functions happen to have all derivatives bounded, so the remainder tends to zero and a nice series occurs, with nothing else to be said. But some authors do place some comments close to what I feel, albeit not very encouraging, e.g:

from Calculus, by Karl Menger: "Taylor's formula (...) is one of the great marvels of mathematics. (...) This is something like a mathematical action at a distance (...)"

from Real Analysis, by Laczkovich & Sós: "The statement of Theorem (...) is actually quite surprising (...) the derivatives of f at a alone determine the values of the function at every other point (...)"

from Introduction to the Calculus, by Osgood: "(...) Since it took the race two centuries to develop this formula after the Calculus was invented, the student will not be surprised that the reasons which underlie it cannot be given him in a few words. Let him accept it as a deus ex machina."

Now all this inquiry may be overly romantic and obsessive on my part, and Taylor series be a perfect example of the "cold and austere beauty of mathematics" as Russell has expressed. But I think that sharing mental experiences helps the mind to improve its turns and horizons, so may I ask:
What was your reaction when you first saw this theorem? And has your general understanding of it changed ever since, by some other way of looking at it and proving it?

  • $\begingroup$ I think the most intuitive "derivation" of Taylor series expansion is to first prove it for polynomials - that is, show that you can find the coefficients by repeatedly differentiating, setting $x = 0$, and normalizing (dividing by the appropriate factorial). Then it's not too surprising that it sometimes works for non-polynomials too, as $n$ tends to infinity. (But admittedly, I was also rather amazed when I first saw Taylor series expansions.) $\endgroup$ Aug 6, 2019 at 20:13
  • $\begingroup$ Related threads (highly recommended): math.stackexchange.com/questions/481661/… ; math.stackexchange.com/questions/2025122/… $\endgroup$
    – littleO
    Aug 6, 2019 at 20:40
  • $\begingroup$ I also highly recommend reading Alon Amit's intuitive explanation of Taylor series here: quora.com/… $\endgroup$
    – littleO
    Aug 6, 2019 at 20:41

2 Answers 2


It's simple to discover Taylor series. Let's start with $$ \tag{1}f(x) = f(a) + \int_a^x f'(s) \, ds, $$ which of course is just the fundamental theorem of calculus. Now if we are feeling playful we might note (again by FTC) that $f'(s) = f'(a) + \int_a^s f''(t) \, dt$. Plugging this into (1), we find that \begin{align} f(x) &= f(a) + \int_a^x f'(a) + \int_a^s f''(t) \, dt \,ds \\ \tag{2}&= f(a) + f'(a)(x - a) + \underbrace{\int_a^x \int_a^s f''(t) \, dt \, ds}_{\text{remainder}}. \end{align} We can keep going like this for as long as we want. The next step is to note that $f''(t) = f''(a) + \int_a^t f'''(u) \, du$. Plugging this into (2), we find that \begin{align} f(x) &= f(a) + f'(a) (x - a) + \int_a^x \int_a^s f''(a) + \int_a^t f'''(u) \, du \, dt \, ds \\ &= f(a) + f'(a)(x - a) + \int_a^x f''(a)(s - a) + \int_a^s \int_a^t f'''(u) \, du \, dt \, ds \\ &= f(a) + f'(a)(x - a) + f''(a) \frac{(x-a)^2}{2} + \underbrace{\int_a^x \int_a^s \int_a^t f'''(u) \, du \, dt \, ds}_{\text{remainder}}. \end{align} You see the pattern. So we have discovered the Taylor polynomial approximation to $f(x)$, and we have a formula for the remainder.

By the way, if $| f'''(u) | \leq M$ for all $u \in [a,x]$, then the remainder $R(x)$ satisfies \begin{align} | R(x) | &\leq \int_a^x \int_a^s \int_a^t | f'''(u) | \, du \, dt \, ds \\ &\leq \int_a^x \int_a^s \int_a^t M \, du \, dt \, ds \\ &= M \frac{(x-a)^3}{3!}. \end{align} You see what the bound on the remainder will be for higher order Taylor series approximations. So we see that the remainder will be small if $x$ is close to $a$.

(If $f$ is sine or cosine, we can take $M = 1$. If $f$ is the exponential function, we can take $M = e^x$.)

  • 2
    $\begingroup$ @ littleO This argument is very beautiful. The integral gathers the error and ultimately zeroes it out, together with my anxiety about this subject. Thank you. $\endgroup$
    – exp8j
    Aug 7, 2019 at 6:51

Consider the function $p(x)$=$\sum_{i=0}^\infty c_i(x-a)^i$=$c_0+c_1(x-a)+c_2(x-a)^2...$

Here $c_i$ represents some scalar.

Consider an arbitrary function $f(x$). We want $p(x)$=$f(x)$ for all x $\epsilon$ $R$. In other words, we want p(x) to "match" f(x). We will assume that this function $f(x)$ is "nice enough" in that it is defined on all $R$ and is of class $C^\infty$ (meaning that it is continuously differentiable).

We start by picking a point a $\epsilon$ $R$. Note that then $p(a)$=$c_0$. Now if $p(x)$ is to equal $f(x)$ on $R$, we must have $p(a)$=$f(a)$, so it follows that $c_0$=$f(a)$.

Now consider $p ' (x)$=$\sum_{i=1}^\infty c_ii(x-a)^{i-1} $=$c_1+2c_2(x-a)+3c_3(x-a)^2...$

If $p(x)$ is to equal $f(x)$ on R, we need $p'(a)$=$f'(a)$. The slope of the tangent lines to both functions should have the same value!

Since $p'(a)$=$c_1$, it is clear that $c_1$=$f'(a)$.

Now we look at $p ' '(x)$= $\sum_{i=2}^\infty c_i(i)(i-1)(x-a)^{i-2} $=$2(1)c_2+(3)(2)c_3(x-a)+c_4(4)(3)(x-a)^2...$

If $p(x)$ is to equal $f(x)$ on R, we need $p ' '(a)$=$f ''(a)$ (The instantaneous rates of change of the first derivatives of each function should be equal). Since $p''(a)$=$2(1)c_2=2!c_2$, it follows that $c_2$=$f''(a)\over2!$. The reason why I wrote 2(1) as 2! should also become immediately clear in the next paragraph.

Let's look at the general nth derivative of $p(x)$, denoted $p^{(n)}$(x).

$p^{(n)}$(x)= $\sum_{i=n}^\infty c_i(i)(i-1)...(i-(n-1))(x-a)^{i-n} $=$n!c_n+c_{n+1}(n+1)(n)...(2)(x-a)+...$

We set $p^{(n)}$(a)=$f^{(n)}$(a). Since $p^{(n)}$(a)=$n! c_n$, we have that $c_n$= $f^{(n)}(a)\over n!$.

Using the above facts, we can rewrite $p(x)$ as:

$p(x)$=$\sum_{i=0}^\infty {f^{(i)}(a)\over i!}(x-a)^{i}$= $f(a)$+$f'(a)(x-a)$+ $f''(a)\over2!$$(x-a)^2$ +.... +$f^{(n)}(a)\over n!$ $(x-a)^n$+...

(Note that $ f^{(0)} $(a)=$f(a)$)

$p(x)$ is the Taylor series of $f(x)$ at x=a. The formula should be completely intuitive. If two functions p(x) and f(x) are to "match" one another, then all of their derivatives at a single point x=a must equal.

On a final note, in general, $p(x)$ will not equal a function for all x $\epsilon$ $R$ that does not satisfy the requirements stated in the second paragraph.

  • $\begingroup$ @ JG123 If you assume the existence of power-series expansion, then your reasoning is valid. The problem is how to ensure this existence. $C^\infty$ smoothness is not enough. We must examine the behavior of the sequence of derivatives at the points between $a$ and $x$. Check this out in calculus books. (A well-known example is f(x)=$e^{-1/x^2}$ if $x≠0$ and $f(0)=0$. Here $f^{(n)}(0)=0$ for every $n$ , so the expansion fails around $x=0$). $\endgroup$
    – exp8j
    Aug 8, 2019 at 5:48
  • $\begingroup$ That is a good point. The Taylor series in that example does not equal f(x) for all x. I probably should have also made it clear in my answer that a power series, (assuming it exists) will only equal a function within its radius of convergence. Regardless, I hope that I conveyed some intuition on the Taylor series in my original answer. $\endgroup$
    – JG123
    Aug 8, 2019 at 14:09
  • $\begingroup$ Furthermore, I assume that if I made it clear that f(x) was analytic in my answer, there would be no issues. $\endgroup$
    – JG123
    Aug 8, 2019 at 14:30
  • $\begingroup$ @ JG123 Perhaps I didn't make it clear enough that my question concerned the sufficient conditions for the Taylor expansion to exist, not the necessary ones. As in the above accepted answer, where the boundedness of the derivatives is crucial. $\endgroup$
    – exp8j
    Aug 8, 2019 at 15:29

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