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I recently learned an interesting fact about what the value of a Lagrange multiplier represents: suppose the maximum of some real-valued function $f(\vec{x})$ subject to a constraint $g(\vec{x})=c$ is $M$ (of course, $M$ depends on $c$), which you obtained via Lagrange multipliers (solving $\nabla f = \lambda \nabla g$). Then, it's easy to show (using the chain rule) that the multiplier $\lambda$ can be interpreted as the change of the maximum with respect to perturbations of the level set $g=c$. That is, $$ \lambda = \frac{d M}{dc} $$ I think this is a pretty cool result and I never it heard about during my time as an undergraduate.

What are some interesting calculus (or undergraduate mathematics) results nobody told you about during your calculus (or undergraduate) education?

eeeeeeeeee
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6 Answers6

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When I first learned Taylor's theorem it didn't really click with me; I thought it clunky and difficult. Then I saw the simplest, most intuitive proof of Taylor's theorem there's got to be. If $f \in C^n$ then

$$f(x) = f(x_0) + \int_{x_0}^x f'(t)$$

Iterate this relation and we get

$$f(x) = f(x_0) + \int_{x_0}^x (f'(x_0) + \int_{x_0}^x f''(t))$$

$$f(x) = f(x_0) + \int_{x_0}^x f'(x_0)\,dt + \int_{x_0}^x\int_{x_0}^u f''(t)\,dtdu$$

and then

$$f(x) = f(x_0) + \int_{x_0}^x (f'(x_0) + \int_{x_0}^x (f''(x_0) + \int_{x_0}^x f'''(t)))$$ $$ f(x)= f(x_0) + \int_{x_0}^x f'(x_0)\,dt + \int_{x_0}^x\int_{x_0}^u f''(x_0)\,dtdu + \int_{x_0}^x\int_{x_0}^u\int_{x_0}^wf'''(t)\,dtdwdu$$

so on and so forth, which reduces to

$$f(x) = f(x_0) + f'(x_0)(x-x_0) + f''(x_0)\frac{(x-x_0)^2}{2} + ... + R_n(x)$$

where

$$R_n(x) = \int_{x_0}^x\int_{x_0}^x...\int_{x_0}^x f^{(n)}(x) = \frac{1}{n-1!}\int_{x_0}^x f^{(n)}(t)(x-t)^{n-1}\,dt$$

So that Taylor's theorem is really just iterated integration and derivation all derived from the seemingly obvious formula

$$f(x) = f(x_0) + \int_{x_0}^x f'(t)$$

  • Very neat. I'm used to think of Taylor as a generalized version of the Mean Value Theorem, but this is definitely clearer. – Martin Argerami Feb 24 '17 at 01:41
  • I love this derivation and it deserves to be more well known. In your final expression for $R_n(x)$, how did you get the upper limits to all be $x$ rather than $u,w,...$? – littleO Feb 24 '17 at 01:41
  • @littleO that was just abuse of notation, it should be $u$'s and $w$'s; the equality follows from Cauchy's iterated integral formula https://en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration. –  Feb 24 '17 at 01:42
  • @littleO you say this derivation deserves to be more well known, but it is already one of the standard proofs of the integral form of the remainder. Find a book that proves the integral form of the remainder and I bet the argument will be this one. In fact it is even in Stewart's widely used calculus book. – KCd Feb 24 '17 at 02:50
  • @KCd Hmm, you might be right, maybe it just wasn't emphasized in my math classes for some reason. There is a different approach based on integration by parts that is presented a lot. – littleO Feb 24 '17 at 02:52
  • By the way, here is a math.stackexchange post that gives more details about the final form of the remainder: http://math.stackexchange.com/a/492165/40119 – littleO Feb 24 '17 at 02:53
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    @KCd My prof proved the integral form of the remainder using something like the mean value theorem. I had never seen this proof in a book or a class, only on the internet. It's nice to hear that it's more common than I thought; a great proof. –  Feb 24 '17 at 03:01
  • @littleO whoops, I see I did not really follow what was going on. The standard proof is by repeated integration by parts, not this iterated integration. I think it would be better if the integrals did not have the $dt$'s ignored everywhere, and make the bounds right: in the iterated relation, it's not true that $f'(t) = f'(x_0) + \int_{x_0}^x f''(t),dt$; the letter $t$ is being used to mean two different things there. – KCd Feb 24 '17 at 03:51
  • @kcd yeah abusing notation; I do that often. It still sparks the same semantic network though :) I didnt have the patience to add all those $dt$'s and dummy variables. –  Feb 24 '17 at 03:56
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    It's not just abusing notation: if you claim $f'(t) = f'(x_0) + \int_{x_0}^x f''(t),dt$ then $f'(t) - f'(x_0) = \int_{x_0}^x f''(t),dt$, and that makes no sense since the right side depends on $x$ and the left side does not. Can you make the derivation accurate by writing $f'(t) = f'(x_0) = \int_{x_0}^t f''(t),dt$ (to abuse notation) or $f'(t) = f'(x_0) + \int_{x_0}^t f'(u),du$ (not to abuse notation)? – KCd Feb 24 '17 at 04:00
  • This is a standard proof available on Wikipedia. Integral form of remainder is quite well known, for example, Apostol and G H Hardy both discuss it in their books. However I agree that proof via repeated integration is the easiest route to Taylor's theorem. – Paramanand Singh Feb 24 '17 at 04:46
  • @ParamanandSingh The proof on wikipedia uses integration by parts; it's different than the derivation given here. – littleO Feb 24 '17 at 05:06
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    I agree with @KCd that this is beyond merely abusing notation; I think it should be fixed because this answer is getting some popularity. It's easy to fix. See the link I gave above for a correct version. – littleO Feb 24 '17 at 05:12
  • @littleO: the formula for the remainder at the end comes from integration by parts and that is the key for obtaining remainder in a useful form (like for obtaining error estimates). – Paramanand Singh Feb 24 '17 at 05:49
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Every basic fact about computing derivatives is easiest to understand if you allow yourself to use infinitesimals, the use of infinitesimals can be rigorously justified, and it doesn't require using nonstandard analysis. Moreover, in practice the fastest way to compute derivatives of complicated functions by hand is to use infinitesimals.

Here's a random example. Suppose I wanted to compute the derivative of $\frac{e^t \cos t}{1 - \tan \log (1 + t)}$ at $t = 0$. You might think I would have to use the quotient rule, then the product rule in the numerator, then the chain rule twice in the denominator. But actually I can do something much faster, which is to pretend that $t^2 = 0$, or in other words to repeatedly take Taylor series and cut them off after the linear term, and then just use ordinary algebra. This produces

$$\frac{(1 + t)(1)}{1 - \tan t} = \frac{1 + t}{1 - t} = 1 + 2t$$

(because $(1 + t)(1 - t) = 1$, so $\frac{1}{1 - t} = 1 + t$). So the derivative at $t = 0$ is $2$. A similar trick where you pretend that $t^3 = 0$ can be used to compute second derivatives, which I'll demonstrate on the above example because maybe only getting the first derivative looked too easy: this produces

$$\frac{\left( 1 + t + \frac{t^2}{2} \right) \left( 1 - \frac{t^2}{2} \right)}{1 - \tan \left( t - \frac{t^2}{2} \right)} = \frac{1 + t}{1 - t + \frac{t^2}{2}} = (1 + t) \left( 1 + t + \frac{t^2}{2} \right) = 1 + 2t + \frac{3t^2}{2}.$$

So the second derivative is $3$.

Here's a harder and less elementary example. Consider the function $X \mapsto \det X$ where $X$ is a square matrix. What is its derivative at the identity $X = I$? Well, a standard fact about determinants tells you that

$$\det (1 + t X) = 1 + \text{tr}(X) t + O(t^2)$$

(where $O(t^2)$ - Big O notation - is one way of justifying more rigorously what I'm doing when I ignore second-order and higher terms in the Taylor series). The coefficient of the linear term is $\text{tr}(X)$, so that's the derivative. It is really annoyingly difficult to try to do this computation by writing out the formula for the determinant in full and then differentiating it; I've never tried and I don't plan to.

This method of computing derivatives is so easy you can teach a computer how to do it.

Qiaochu Yuan
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    +1. You probably know this, but Knuth has actually suggested teaching calculus via O-notation. (It's there in a few places on the internet e.g. I've posted it here.) Infinitesimals have a bad reputation, but this isn't even infinitesimals per se, just formal manipulation of "there exists" expressions, with good notation. It's a nice letter and demonstrates that this way of thinking can be successful. – ShreevatsaR Feb 24 '17 at 04:58
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    Isn't this basically computing first couple of terms of Taylor series? If so, it is completely rigorous in standard analysis. What I'm wondering is if there is easy to see justification that you can calculate Taylor polynomial of order $n$ by calculating in $\Bbb R[t]/(t^{n+1})$? – Ennar Feb 24 '17 at 05:57
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    @Ennar: there are a couple of ways to say it. One is that in the ring of functions which are $(n+1)$ times continuously differentiable at $0$, the subset of functions which are $O(t^{n+1})$ (as $t \to 0$) is an ideal, and you can quotient by it. This takes care of products and quotients when they make sense but not necessarily compositions; for compositions maybe a conceptually cleaner way to go forward is to embed $C^{\infty}$ functions into formal power series. – Qiaochu Yuan Feb 24 '17 at 08:08
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Nonelementary antiderivatives. While you may have heard that the antiderivative $$f(x)=\int_{0}^{x} e^{-\frac{t^2}{2}}dt$$ is not expressible as an elementary function you definitely haven't seen the proof.

freakish
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During the 1st semester of calculus I teach my students about differentials. When I was a student nobody told me what is its geometric interpretation.

szw1710
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  • I too always teach the geometric interpretation of differentials. I find it alarming that someone would teach differentials without it. – eeeeeeeeee Feb 23 '17 at 22:21
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    I was taught about the secant line and how you can take a limit to turn it into the tangent line, but they didn't actually do anything with differentials or infinitesimals. I fear this may have been lost in the rush to stamp out the latter from (standard) analysis. – Kevin Feb 24 '17 at 00:07
  • @eeeeeeeeee I must admit I hate thinking of differentials geometrically without first justifying this with techniques far past basic Calculus courses. – Brevan Ellefsen Feb 24 '17 at 03:43
  • @BrevanEllefsen I completely understand and appreciate the desire for rigor but I feel like motivating the topic with basic geometry first can only help. – eeeeeeeeee Feb 24 '17 at 14:41
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Misconception about L'Hospital's Rule

I learned about L'Hospital's Rule (LHR) in my first calculus course as a high school student. For one of the indeterminate forms, the teacher presented the conditions as

Suppose $f$ and $g$ are differentiable in a deleted neighborhood of $x_0$, and $\lim_{x\to x_0}f(x)=\infty$, $\lim_{x\to x_0}g(x)=\infty$, and $g'(x) \ne 0$ in a deleted neighborhood of $x_0$. If $\lim_{x\to x_0}\frac{f'(x)}{g'(x)}$ exists, then $\lim_{x\to x_0}\frac{f(x)}{g(x)}$ also exists and

$$\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f'(x)}{g'x)}$$

The misconception is that the limit of $f$ is required to be $\infty$. This is not the case. In fact, the limit $\displaystyle \lim_{x\to x_0}f(x)$ need not even exist for LHR to be valid provided that $\displaystyle g\to \infty$

EXAMPLE:

There are many applicable examples that might be considered trivial. Here is one that might be less obvious.

Let $f(x)=\log(1/x)+\int_0^1\frac{1-t^x}{1-t}\,dt$ and $g(x)=x$.

Now, without even knowing whether the limit, $\lim_{x\to \infty}f(x)$ exists or not (it is actually equal to $\gamma$, the Euler-Mascheroni constant), we know that $f$ is differentiable for $x>0$ with $f'(x)=-\frac1x-\int_0^1\frac{\log(t)t^x}{1-t}\,dt$.

Inasmuch as $g$ is differentiable with $\lim_{x\to \infty}g(x)=\infty$, L'Hospital's Rule asserts that

$$\begin{align} \lim_{x\to \infty}\frac{f(x)}{g(x)}&=\lim_{x\to \infty}\frac{\log(1/x)+\int_0^1\frac{1-t^x}{1-t}\,dt}{x}\\\\ &=\lim_{x\to \infty}\frac{-\frac1x-\int_0^1\frac{\log(t)t^x}{1-t}\,dt}{1}\\\\ &=0 \end{align}$$

Mark Viola
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Integration by parts - visualization

Not that shocking, but my teacher never showed me this. For many people this will be familiar, but for those who never saw it: it can be an eye opener.

enter image description here

Image source: Wikipedia

The area of the blue / red region is:

$$A_1=\int_{y_1}^{y_2}x(y)dy$$

$$A_2=\int_{x_1}^{x_2}y(x)dx$$

So we have:

$$\overbrace{\int_{y_1}^{y_2}x(y)dy}^{A_1}+\overbrace{\int_{x_1}^{x_2}y(x)dx}^{A_2}=\biggl.x > . y(x)\biggl|_{x1}^{x2} = \biggl.y . x(y)\biggl|_{y1}^{y2}$$

Assuming the curve is smooth within a neighborhood, this generalizes to indefinite integrals:

$$\int xdy + \int y dx = xy$$

Rearranging yields the well-known formula:

$\int xdy = xy - \int y dx$

Job Bouwman
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