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I will soon have to teach Taylor's formulas (with Lagrange and Peano¹ remainder terms) to first-year students in math-related curricula (computer science, physics, engineering), and I am not terribly satisfied with the proofs I have seen. I do not care so much about having optimal hypotheses, so I am happy to derive Taylor-Peano for $C^k$ functions from Taylor-Lagrange. But the only proof I have found (in numerous places) of Taylor-Lagrange looks like a clever trick: one introduce the number $A$ such that the function defined from $f$ by $$g(x) := f(b)-\Big(f(x) + (b-x) f'(x) + \dots + (b-x)^k \frac{f^{(k)}}{k!} +(b-x)^{k+1} \frac A{(k+1)!}\Big)$$ satisfies $g(a)=0$, and then applies Rolle' theorem to $g$.

Do you know of a proof that carries more intuition about why the result is true? Alternatively, how would you explain the intuition, if any, behind this proof?

What bothers me is that the mean value theorem is deduced from Rolle's theorem in a very clear way; maybe this is linked with the statement itself being more intuitively explained.

Added in edit 1. Let me clarify what bothers me here: the mean value theorem is pretty clear in its statement, as it basically says that a car that drove 60 miles in 2 hours must have had instant speed 30 mph at some point. Moreover a simple drawing explains both the statement and the proof. Somehow, I feel there should be some sort of equivalent explanations, only slightly more involved, for the Taylor-Lagrange formula and its proof. An example of what bothers me is that I found no proof that explicitly involved acceleration for the $k=1$ case.

Added in edit 2. Let me precise the background of my students: they are assumed to have the background of French scientific section of high school, plus one semester course on reals, sequences and functions (including derivatives, Rolle's theorem and the mean value theorem). My goal would be to show them proofs that they feel they could have discovered, if given time to think about them.

Too many proofs look (to students) like aliens gave them to us; this can be either because we choose to teach shorter but less conceptual proofs, or because they where discovered after long and hard thinking by mathematicians with much more training and experience and intuition than they currently have. Independently of why it is so, I think this matter of fact tends to reinforce their common belief that math is a collection of magical formulas that have no true meaning and can only be learned by heart (and beware what happens if you use the wrong formula!)

¹ This is called after Young in France.

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    $\begingroup$ Could you perhaps elaborate slightly on the background of your students? What they find intuitive will be determined partially by what they already know. Do they know the MVT? Why does the clarity of the connection between MVT and Rolle bother you? $\endgroup$ – Shai Dec 31 '14 at 2:13
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    $\begingroup$ As I recall, the proof is really the same as the MVT. The difference being we need an $n$-th order approximating polynomial as opposed to the secant line. So, geometrically, the idea is clear enough. I will admit, I'm not sure I have a direct understanding of the remainder term. Anyway, while this may just be the same as you've already seen, check out: page 202 of supermath.info/OldschoolCalculusII.pdf $\endgroup$ – James S. Cook Dec 31 '14 at 4:03
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    $\begingroup$ @MichaelE2: you should post this as an answer. It will probably not help me too much, as I plan for other reasons to treat integration after Taylor formulas, but it is nonetheless a good answer to my question. $\endgroup$ – Benoît Kloeckner Dec 31 '14 at 10:44
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    $\begingroup$ Tim Gowers on this exact point: google.co.uk/amp/s/gowers.wordpress.com/2014/02/11/… $\endgroup$ – Will R Feb 8 '18 at 19:57
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If you do not wish to push for optimal hypothesis (on the function), or for optimal conclusion (i.e., you can live with the "error bound"), I think what I'm giving below would be a perfectly fine approach.

Start with the Fundamental Theorem of Calculus

$$ f(x) = f(a) + \int_a^x f'(x_1)\,dx_1 $$

and substitute the integrand by

$$ f'(x_1) = f'(a) + \int_a^{x_1} f''(x_2)\,dx_2 $$

so that

$$ \begin{align} f(x) &= f(a) + \int_a^x \left( f'(a) + \int_a^{x_1} f''(x_2)\,dx_2\right)\,dx_1 \\ &= f(a) + f'(a)(x-a) + \int_a^x\int_a^{x_1} f''(x_2)\,dx_2\,dx_1. \end{align} $$

Now, if $|f''(x_2)|\leq M$ for all $a<x_2<x$, we have a simple bound for the remainder: $$ \left|\int_a^x\int_a^{x_1} f''(x_2)\,dx_2\,dx_1 \right| \leq \int_a^x\int_a^{x_1} M\,dx_2\,dx_1 = \frac{M}{2}(x-a)^2. $$

Similarly we can readily deduce (the "Taylor polynomial" and) the remainder $R_n$ as an iterated integral, which then yields the Lagrange error bound $$ |R_n| \leq \frac{M}{n!}(x-a)^n $$ ($M$ being a bound for $f^{(n)}$ on the interval).

With a little bit of multi-variable thinking, we can even get the Lagrange remainder: $$ R_n = \frac{f^{(n)}(\xi)}{n!}(x-a)^n \quad\text{for some}\quad \xi\in(a,x)$$ (a simple application of extreme value theorem and intermediate value theorem). The hypothesis for all this to work is that $f\in C^n$.

I probably should ask a separate question that I am puzzled by: Where do we truly need the actual remainder, in any of the forms — exact but with an unknowable $\xi$)? It seems to me that the error bound suffices for all applications, such as showing convergence of Taylor series). I understand the Mean Value Theorem is used in a crucial step in the Fundamental Theorem of Calculus, but where does the Lagrange remainder (for $n\geq 2$) show up other than to deduce the error bound (which as I just showed is a much simpler result)?

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  • $\begingroup$ For what it's worth, I have more details of the various remainders at brilliant.org/wiki/taylors-theorem-with-lagrange-remainder $\endgroup$ – liuyao Dec 21 '17 at 20:37
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    $\begingroup$ Nice and simple, thank you. The Lagrange remainder is handy to prove inequalities such as $x-x^2/2 < \ln(1+x) < x$ for $x>0$ because you can use the sign of the derivative. Similarly, you can use it to compute approximations of many numbers ($\sqrt{2}, e, \ln(2)$ which are better than with other remainder, for the same reason. $\endgroup$ – Benoît Kloeckner Dec 22 '17 at 8:18
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I think that teaching for understanding is a noble cause and that teaching for procedural proficiency is a necessary part of learning mathematics - I am still making my mind up about where and when it is best to stress the motivation or the procedure in my own teaching and, certainly, there comes a point where it is better to be able to do something first and then understand it later. Taylor polynomials are one of those grey areas for me, so I will share my thoughts.

Suppose we want to think about the Taylor polynomial of $f(x)$ about $0$ and we want to know how good the approximation is at some point $x > 0.$ In order to avoid using integration I think it is necessary to consider taking the Taylor polynomial at different points along the curve $(x,f(x))$ whilst keeping $x$ fixed and considering what happens to the remainder at $x.$ I will talk about the Taylor polynomial of degree $n=1$ first.

In the picture below, $A$ is the point where the Taylor polynomial has been taken, $B$ has $x$ coordinate $x$ and is the point where we are interested in the size of the remainder. The size of the remainder is represented by the vertical segment $BC$ and, clearly, it changes as the position of $A$ changes and is $0$ when $A=B$ because the value of a function always coincides with the value of the Taylor polynomial at the point where the polynomial is taken.

A first degree Taylor polynomial of a typical curve

You can see that a 'more curved curve' - i.e. one whose second derivative is a large number for $t \in [0,x]$- will impose a larger remainder than a less curved curve: this is the relationship between the case $n=1$ and 'acceleration' (in this case, curvature) you were thinking of.

Now for the details. Let $$ f(x) = f(t) + f^{\prime}(t)(x-t) + R(t) $$ so that we have taken the degree $1$ Taylor polynomial of $f$ at $t$ and the remainder is $R(t).$ Think of $t$ as the $x$-coordinate of $A$ and $x$ as the $x$-coordinate of $B$ in the image. $R(t)$ is the length $BC.$ If we rearrange we get $$ R(t) = f(x) - f(t) - f^{\prime}(t)(x-t) $$ and we are not surprised to find that $R(x) = 0$ because the remainder when $t=x$ should be $0$ since when $t = x, A= B$ in our picture. The thing we are interested to find is $R(0),$ and for a very curvy curve, $R(0)$ will be larger than for a less curvy curve. It is natural to ask: what is the rate of change of $R(t)$ as $t$ changes? Well it's pretty clear that... $$R^{\prime}(t) = -f^{\prime\prime}(t)(x-t) $$ for each $t.$ And so we see, the remainder will certainly be effected by the curvature of $f.$ Now, to find $R(0),$ we could try to get a handle on $$ \frac{R(x) - R(0)}{x-0} = -\frac{R(0)}{x} $$ The mean value theorem tells us that this is equal to $R^{\prime}(t)$ for some fixed $t \in (0,x),$ so we get $$ -\frac{R(0)}{x} = R^{\prime}(t) = -f^{\prime\prime}(t)(x-t)$$ or $$ R(0) = xf^{\prime\prime}(t)(x-t) $$ where $t$ is now a fixed number, which is the Cauchy form of the remainder term when $n=1.$

In summary, when you want to know that value of something ($R(0),$ here) it is natural to find its value at a later point (for us, $R(x)$) and to try to work out information about how it has changed from the initial to the final position. Not surprisingly, this is the essence of the fundamental theorem of calculus: $$ \begin{align*} \text{Value at end} & = \text{Value at beginning} + \text{Amount of change} \\ g(b) & = g(a) + \int_{a}^{b} g^{\prime}(t) \, dt \end{align*} $$

As for higher values of $n,$ I think it is normal to lose some amount of intuition: 'the rate of change of the rate of change of the rate of change of...' just gets intractable, but we are used to taking such leaps into the abstract (as we start thinking about $\mathbb{R}^{n},$ for example, as $n$ increases our intuition is based more on algebra and comparisons to smaller $n$s than direct diagrams and tangible examples).

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  • $\begingroup$ Nice answer, thanks. I am still puzzled that one gets the Cauchy remainder in a more direct way than Lagrange one, something I noticed when trying to find another proof for the Lagrange remainder by myself. Do you now a simple way to go from Cauchy remainder to Lagrange remainder? $\endgroup$ – Benoît Kloeckner Dec 31 '14 at 15:14
  • $\begingroup$ Apparently, you can apply the Cauchy Mean Value Theorem to the functions $R(t)$ and $g(t)=(x-t)^{2}$ to get the Lagrange form. I say apparently because I looked this up in a textbook. That, to me, does seem a little plucked out of thin air. $$ \frac{R(x) - R(0)}{g(x) - g(0)} = \frac{R^{\prime}(t)}{g^{\prime}(t)} $$ for some $t \in [0,x].$ Generally, you choose $g(x) = (x-t)^{n+1}.$ $\endgroup$ – Shai Dec 31 '14 at 17:07
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    $\begingroup$ I have only now realised how delightfully backwards that is. The Cauchy form is found by applying the Lagrange Mean Value Theorem and the Lagrange form is found by applying the Cauchy Mean Value Theorem. $\endgroup$ – Shai Dec 31 '14 at 17:17
  • $\begingroup$ How was that animation made? It's great. $\endgroup$ – Noah Jan 6 '15 at 3:40
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    $\begingroup$ @Noah I used GeoGebra software which I use a lot in the classroom, it's just wonderful. You can download here geogebra.org/download There are tutorials to get started here wiki.geogebra.org/en/Tutorials The animation is provided by creating a parametrised image using 'sliders' and then exporting the image as a .gif - it's all done automatically. Another highly desirable feature is that drawn images can be converted automatically into TiKz code for your LaTeX documents; all you need to do is tweak the generated code to suit your preferences. And it's free :) $\endgroup$ – Shai Jan 6 '15 at 19:43
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You can start with powers.

f (x) = $x^0$ -> f (x) = $x^0 / 0! * f (0)$.

f (x) = $x^1$ -> f (x) = $x^1 / 1! * f' (0)$

...

f (x) = $x^4$ -> f (x) = $x^4 / 4! * f^{iv} (0)$

I'd write it down without the x^0, ^1, 0! and 1! and fill them out later so that all the formulas are identical. Then we observe for example when f (x) = $x^3$, then the formula $x^4 / 4! * f^{iv} (0)$ gives 0. So we can add in all the other formulas like

f (x) = $x^4$ -> f (x) = $x^0 / 0! * f (0) + x^1 / 1! * f' (0) + ... + x^4 / 4! * f^{iv} (0) + ...$

and the same formula applies for all powers. It also works for all polynomials. So now the obvious question is what happens for other functions?

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    $\begingroup$ The problem is that, now that we don't assume that all functions are analytic, this does not provide a proof. $\endgroup$ – Benoît Kloeckner Jan 3 '15 at 21:29
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    $\begingroup$ Re paragraphs 2 and 3 of the Motivation, That is exactly the viewpoint in Dieudonné, Infinitesimal Calculus, Chapter 3, Asymptotic Expansions. $\endgroup$ – schremmer Apr 5 '15 at 16:42

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