Generalizing a problem to make it easier

Question

One of the many articles on the Tricki that was planned but has never been written was about making it easier to solve a problem by generalizing it (which initially seems paradoxical because if you generalize something then you are trying to prove a stronger statement). I know that I've run into this phenomenon many times, and sometimes it has been extremely striking just how much simpler the generalized problem is. But now that I try to remember any of those examples I find that I can't. It has recently occurred to me that MO could be an ideal help to the Tricki: if you want to write a Tricki article but lack a supply of good examples, then you can ask for them on MO.

I want to see whether this works by actually doing it, and this article is one that I'd particularly like to write. So if you have a good example up your sleeve (ideally, "good" means both that it illustrates the phenomenon well and that it is reasonably easy for others to understand) and are happy to share it, then I'd be grateful to hear it. I will then base an article on those examples, and I will also put a link from that article to this MO page so that if you think of a superb example then you will get the credit for it there as well as here.

Incidentally, here is the page on this idea as it is so far. It is divided into subpages, which may help you to think of examples.

Added later: In the light of Jonas's comment below (I looked, but not hard enough), perhaps the appropriate thing to do if you come up with a good example is to add it as an answer to the earlier question rather than this one. But I'd also like to leave this question here because I'm interested in the general idea of some kind of symbiosis between the Tricki and MO (even if it's mainly the Tricki benefiting from MO rather than the other way round).

Answer 1

Great question. Maybe the phenomenon is less surprising if one thinks that there are $\infty$ ways to generalize a question, but just a few of them make some progress possible. I think it is reasonable to say that successful generalizations must embed, consciously or not, a very deep understanding of the problem at hand. They operate through the same mechanism at work in good abstraction, by helping you forget insignificant details and focus on the heart of the matter.

An example, which is probably too grandiose to qualify as an answer, since your question seems very specific, is Fredholm theory. At the beginning of last century integral equations were a hot topic and an ubiquitous tool to solve many concrete PDE problems. The theory of linear operators on Banach and Hilbert spaces is an outgrowth of this circle of problems. Now, generalizing from $$ u(x) + \lambda \int _a ^b k(x,s) u(s) ds = f(x) $$ to $$ (I+ \lambda K) u = f $$ makes the problem trivial, and we do it without thinking. But it must have been quite a shock in 1900.

Answer 2

Weyl's equidistribution theorem states that for $\alpha \notin \mathbb{Q}$, the sequence of fractional parts $x_n = \{ n \alpha \}, n=1,2,\dots$ is equidistributed in the unit interval, i.e. $\lim_{N \to \infty}\frac{|\{x_n\}_{n=1}^{N} \cap (a,b)|}{N} = b-a$ for every subinterval $(a,b) \subset [0,1]$.

The fact that this sequence is dense in $[0,1]$ is a simple application of the pigeonhole principle, but the fact that it is equidistributed seems a lot harder to prove at first. However, this is very easy to establish if one generalizes the definition of equidistribution (or more precisely, the objects appearing in it). Namely, by very easy arguments one can show that $x_n$ is equidistributed in $[0,1]$ if and only if $\frac{1}{N} \sum_{n=1}^{N} f(x_n) \to \int_{0}^{1}f(x)dx$ for every continuous (or Riemann integrable) function $f$ on $[0,1]$. Now one can't resist the temptation to look for functions $f$ for which this is easy to show. The complex exponentials $f_m (x) = e^{2 \pi i m x}$, where $m \in \mathbb{Z}$, are such functions and by classical Fourier analysis (or the Stone-Weierstrass theorem) it is enough to show the above convergence for such functions. One is thus led to the Weyl criterion, which reduces the problem of showing that the sequence $\{ n \alpha \}$ is equidistributed in $[0,1]$ to a simple computation involving the sum of a geometric sequence. (Of course, Weyl's criterion is useful in studying other, more involved sequences as well)

In essence, we started with a problem which concerns indicator functions of subintervals of $[0,1]$, generalized it to a problem involving a much bigger class of functions $\mathcal{F}$ and then found a nice, special subclass with which we can "capture" the whole class $\mathcal{F}$ (and in particular the functions we started with). This procedure of "generalize and then specialize" seems quite common in analysis. In this case it also has some relation to the Tricki article "Turn sets into functions".

Answer 3

The canonical example to introduce this idea early to students is the Fundamental Theorem of Calculus. In order to figure out one area $\int_{a}^b f(t)dt$, you must come to grips with the generalized problem $x \mapsto \int_{a}^x f(t)dt$

Answer 4

There are many examples where introducing one or more extra parameters into an integral that you want to evaluate or an identity that you want to prove makes things easier. For example, Feynman was fond of evaluating integrals by differentiating under the integral sign, and in his advanced determinant calculus, Christian Krattenthaler explicitly urges you to introduce more parameters into any determinant you are having trouble evaluating.

For the Tricki, maybe one of the simplest examples would be the evaluation of $\int_0^\infty {\sin x \over x}dx$ by considering $\int_0^\infty e^{-xt} \bigl({\sin x \over x}\bigr) dx$ and differentiating with respect to $t$.

Answer 5

I can't resist mentioning the following problem (and requesting that nobody gives away the solution any more than it is already given away by my mentioning it here).

Call a real number repetitive if for every k you can find a string of k digits that appears more than once in its decimal expansion. The problem is to prove that if a real number is repetitive then so is its square.

Answer 6

From Noam Elkies' AMS article Lattices, Linear Codes, and Invariants, Part I: Elkies has been discussing how difficult it is to find the minimal nonzero length of an element of a lattice $C$.

Sometimes an appropriate response to a difficult mathematical problem is to pose a much harder problem. Here we find the minimal nonzero length intractable, and thus ask for *all* the lengths of vectors of $C$ and their multiplicities. Equivalently, we ask for the following generating function of all the squared lengths, called the *theta function* (or *theta series*) of $C$: $$\Theta_C(z) = \sum_{x \in C} z^{\langle x, x \rangle} = 1 + \sum_{m >0}^{\infty} N_m(C) z^m$$ where $N_m(C)$ is the number of lattice vectors of length $\sqrt{m}$.

It is hard to consider particular lengths but easier to consider the entire theta function because you give the problem more structure, and then you have access to stronger tools like Poisson summation.

Answer 7

The following was a conjecture for several years. Put a light bulb and a switch at every vertex of an $m\times n$ grid ($mn$ vertices in all). Each bulb can be on or off. Each switch changes the state of the bulb at its vertex and all its neighbors. (A neighbor is a vertically or horizontally adjacent vertex.) Then whatever the initial set of bulbs that are lit, it is possible to turn all the lights off by some set of switches. Sutner showed in 1989 that the corresponding result is true for any graph. Caro gave a simpler proof in 1996. The proof is an elegant application of linear algebra mod 2. Looking at grids adds an extra layer of complexity that obscures the underlying theory. One reference is http://mathworld.wolfram.com/LightsOutPuzzle.html.

Answer 8

I like the matroid intersection theorem. Basically, this hammer often renders a problem trivial once you generalize to matroids. Also, it's nice that the hammer itself is not hard to prove. First the statement of the theorem.

Matroid Intersection Theorem. Let $M_1$ and $M_2$ be matroids on the same ground set $E$, with rank functions $r_1$ and $r_2$ respectively. Then the size of a maximum common independent of $M_1$ and $M_2$ is

\[ \min_{A \subseteq E} \ r_1(A)+r_2(E-A). \]

I won't include a proof, but you can find one in say Volume B of Combinatorial Optimization by Schrijver. It turns out that this theorem simultaneously proves many theorems such as König's theorem and Nash-Williams' theorem on covering graphs with $k$ edge-disjoint spanning trees. Here's another application.

Rainbow Spanning Trees. Let $G=(V,E)$ be a graph with a (not necessarily proper) colouring of $E$. Suppose you are trying to decide if $G$ contains a rainbow spanning tree. This is a spanning tree such that no two edges of the tree are the same colour. One obvious necessary condition is as follows. If I choose $t$ colours and remove all the edges with these colours, then the resulting graph better have at most $t+1$ components. Is this also sufficient? This seems that it could be rather tricky to prove. However, there is an easy proof using matroids.

Let $M_1$ be the matroid whose independent sets are those subsets of edges which contain at most one edge of each colour. Let $M_2$ be the cycle matroid of $G$. Then, $G$ has a rainbow spanning tree if and only if $M_1$ and $M_2$ have a common independent set of size $|V|-1$.

By the matroid intersection theorem it suffices that

\[ \min_{A \subseteq E} \ r_1(A)+r_2(E-A) \geq |V|-1. \]

Note that $r_1(A)$ is the number of colours among $A$. On the other hand, $r_2(E-A)$ is $|V|-c(E-A)$, where $c(E-A)$ is the number of components of the subgraph $(V,E-A)$. Substituting and simplifying, we conclude that $G$ has a rainbow spanning tree if and only if for all $A \subseteq E$,

\[ c(E-A) \leq r_1(A)+1, \]

as claimed.

Answer 9

Sometime around 25 years ago, Dr. Jeffrey Vaaler at UT Austin gave me the following problem.. He needed the result as a lemma for a paper he was working on.

Let $n$ be a square-free integer with $k$ distinct prime factors and thus $\sigma(n) = 2^k$ divisors. Split the divisors into two sets of equal size: the small divisors $S$ and the large divisors $T$. The statement he was trying to prove was: $$\text{There exists a bijection}\ f: S \rightarrow T \hspace{2mm} \text{such that} \hspace{2mm} d \hspace{1mm} | \hspace{1mm} f(d) \hspace{2mm} \text{for every} \hspace{2mm} d \in S.$$ I was an undergraduate and highly motivated to demonstrate my usefulness, but I didn't really have many ideas about how to go about it. The obvious approach is by induction on $k$, but I never really got anywhere despite spending many hours on the problem.

A year later I ran into Dr. Vaaler in the hall and asked if he ever solved it. Of course, he had, by induction on $k$. He went on to explain the "trick" to making the induction work. He proved a more general result. Introduce a parameter $0 \le r \le \frac{1}{2}$ and consider $S_r$ and $T_r$, the smallest $\lfloor r \cdot 2^k \rfloor$ divisors and the largest $\lfloor r \cdot 2^k \rfloor$ divisors respectively, and instead prove the above statement with $S_r$ and $T_r$ in place of $S$ and $T$.

The lemma is then the special case with $r = \frac{1}{2}$.

This example stuck with me. How could it be easier to prove something more general? Though I understand the concept better today, it still surprises me.

Answer 10

This may be a little trivial, but there are a number of identities for the Fibonacci numbers that are most easily proved by generalizing them. For example, proving

$f_{2n-1}=f_{n+1}f_{n}-f_{n-1}f_{n-2}$

requires a rather convoluted process involving a couple lemmas unless one realizes that it is far easier to prove

$f_{m+n}=f_{m}f_{n+2}-f_{m-2}f_{n}$

and then substitude $m=n,n=n-1$.

I think a classic example of generalizing in order to prove a simple result is Galois Theory. Ruffini's attempted proofs of the unsolvability of the quintic were enormously long and tremendously complicated. However, once the machinery of Galois Theory is developed, which is rather easy, it is almost trivial to demonstrate that there exist quintic equations that are not solvable by radicals.

Answer 11

A nice question !

An extreme case of the phenomenon you are asking about happens frequently enough to be worth mentioning (and I think that it even partially "explains" it); sometimes you can generalize a problem and then realize that the more general problem is already solved ! The point is that there are usually many ways to look at or formulate the same problem, and when the problem is looked at from some points of view it seems novel, but when correctly reformulated you can recognize it as a special case of a solved problem, perhaps even a classic one. Here is an example that I admit is a bit forced, but it illustrates the idea. Show that if a triangle C can be divided by a line from a vertex to an opposite side into two triangles A and B that are similar to the original triangle, and that if corresponding sides of the three triangles are c,a, and b, then $a^2 + b^2 = c^2$. This is of course completely trivial, since the areas of similar triangles are obviously proportional to the squares of corresponding sides while the areas of A and B clearly add up to the area of C. On the other hand, a special case of this: dropping the perpendicular from the right angle vertex of a right triangle onto the hypotenuse, gives a proof of the Pythagorean as a special case. (As you may know, this is how Einstein was supposed to have found a proof of the Pythagorean Theorem for himself.)

Answer 12

How about the Bohr-Mollerup theorem:

There is a unique log-convex $f:[0,\infty)\to(0,\infty)$ that satisfies $f(x+1) = (x+1)f(x)$ and $f(0)=1$ (namely $f(x) = x! = \Gamma(x+1)$).

Here's a generalization that I think is easier to prove:

If $\delta:[0,\infty)\to\mathbb{R}$ is such that $\delta^{(n)}$ decreases monotonically to zero, then there is a unique $f$ such that $f(x+1) = f(x) + \delta(x)$, $f(0) = 0$, and $f^{(n)}$ is monotonically increasing.

(To recover the original statement, take $n=1$ and $\delta(x) = \log(x+1)$.)

The benefit of this formulation is that it can be proved by induction on $n$. The inductive step is pretty routine and involves applying the $(n-1)$-case to $f'$ and $\delta'$. The basis case for $n=0$ still requires a nontrivial argument, but to me it feels much simpler and more intuitive than the case $n=1$ (for the original theorem), and it's actually easy to remember (in contrast to the direct proof of the Bohr-Mollerup theorem, which I find hard to remember). All you have to do is apply the functional equation $k$ times to get $f(x+k)-f(k) = f(x) + \sum_{j=0}^{k-1}[\delta(x+j)-\delta(j)]$ and take the limit as $k$ approaches infinity; you end up with an increasing function of $x$ that is zero whenever $x$ is an integer and is thus zero for all $x$. Thus $f(x) = \sum_{j=0}^{\infty}[\delta(j) - \delta(x+j)]$ is the unique solution. (Since $\delta$ is decreasing, we do have an increasing function, and the sum converges since $\delta(x)\to 0$ as $x\to +\infty$.)

Answer 13

Doron Zeilberger wrote a very nice expository article entitled, "The method of undetermined generalization and specialization illustrated with Fred Galvin's amazing proof of the Dinitz conjecture," in which he discusses how repeated generalization and specialization can lead one to the solution of a difficult problem.

Answer 14

I remember there was an exercise in linear algebra:

Find the determinant of the following matrix: \begin{pmatrix} 0 & a & a & \cdots & a\\ b & 0 & a & \cdots & a\\ \cdots & \cdots & \cdots & \cdots & \cdots\\ b & b & \cdots & 0 & a\\ b & b & \cdots & b & 0 \end{pmatrix}

A simple solution is to generalize this to a function: $$f(x) = \det\begin{pmatrix} x & a + x & a + x & \cdots & a + x\\ b + x & x & a + x & \cdots & a + x\\ \cdots & \cdots & \cdots & \cdots & \cdots\\ b + x & b + x & \cdots & x & a + x\\ b + x & b + x & \cdots & b + x & x\end{pmatrix} $$

Namely, add $x$ to every entry of the matrix.

It is then obvious that:

• $f$ is a linear function of $x$;
• $f(-a)$ and $f(-b)$ are easily computed;
• the original determinant is just $f(0)$.

Answer 15

A very simple example of this phenomenon is also given by the following problem: prove that the maximum determinant of $n\times n$ matrices that have entries in the set $\{-1,+1\}$ is divisible by $2^{n-1}$. In fact, it is much easer proving by induction that all matrices with coefficients in $\{-1,+1\}$ have determinant divisible by $2^{n-1}$, because you can use induction and just say that changing a $-1$ to a $+1$ will change the determinant by the amount $2 * 2^{n-2}k$, by row expansion, and you can do so till when you get a matrix will all entries $1$, which has determinant $0$ when $n>1$. Note that it is not clear how you could use induction to prove the weaker statement about the maximum determinant.

Actually when a statement can be proved by induction it happens quite often that the correct statement that "makes induction work" is a somewhat generalized version of the result to be proved.

Answer 16

One common example of the "generalize the problem" strategy is found in parameter differentiation.

$$\int^\infty_0 \frac{\sin^2(x)}{x^2} \, dx$$

We then create a more general case of $I(a)=\int^\infty_0 \frac{\sin^2(ax)}{x^2} \, dx$ where $a>0$.

\begin{align} I'(a) & =\int^\infty_0 \frac{2\sin(ax)\cdot\cos(ax)\cdot x}{x^2}\,dx \\[10pt] & =\int^\infty_0 \frac{\sin(2ax)}{x} \, dx \end{align}

So $I(a)=1/2\pi a$ and $I(1)=1/2 \pi$

Another similar cute problem is evaluate $\sum_{k=1}^n\frac{k^2}{2^k}$, we instead evaluate $S(x)=\sum_{k=1}^nk^2x^k.$

Edit: Just to finish the second example:

We know that $S(x)=\sum_{k=1}^nx^k=\frac{1-x^{n+1}}{1-x}.$ We take the derivative of this, then multiply that whole mess by x, and take the derivative again, and multiply by x again.

We get $S(x)=\sum_{k=1}^nk^2x^k= \frac{x(1+x)-x^{n+2}-x^{n+1}(nx-n-1)^2}{(1-x)^3}$, so $S(\frac{1}{2})=6-(\frac{n^2+4n+6}{2^n})$.

Answer 17

Perhaps the example of the calculation of fundamental groups works in this situation. Calculating, for instance, the fundamental group of the circle uses hard ideas from complex analysis (at least in many of the traditional approaches) but if one generalises to calculating the fundamental groupoid of the circle based at two points and one uses the groupoid van Kampen theorem, (not the group vKT) then the proof reduces to quite simple algebra. (The importance of that is really that it links the original complex analytic machinery into the geometric machinery in a neat way helping one to see the 'unity' or close symbiotic relationship of the two areas, which of course have a common heritage in Poincaré.)

Answer 18

Here's an example in planar euclidean geometry. Consider an equilateral triangle of side $a$ and a general point in the plane distant $b$, $c$, and $d$ from the respective vertices. Then

$3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2$.

This is an an awful slog to get by planar trigonometry. Even harder to do by trig in three dimensions is the corresponding result for the regular tetrahedron. However, it's easy to get the $(n - 1)$-dimensional result for a regular $(n - 1)$-dimensional simplex of side $d_0$, with vertex distances $d_1$ ,..., $d_n$ :

$n(d_0^4 + ... + d_n^4) = (d_0^2 + ... + d_n^2)^2$.

You can do this by embedding the euclidean $(n - 1)$-dimensional space as the hyperplane of points $(x_1 ,..., x_n)$ in euclidean $n$-space such that $x_1 + ... +x_n = d_0/\sqrt2$. The vertices of the simplex can then be represented as the points $(d_0/\sqrt2)(1, 0 ,..., 0)$, ... , $(d_0/\sqrt2)(0 ,..., 0, 1)$ in the hyperplane, and the result drops out in a few lines.

Answer 19

The free cocompletion. A lot of adjoint couples of functors are just particular case of this construction, and often it is easier to use the general theorem than working out a particular case by hand (for example for $i_{!}$ and $i^{!}$).

Answer 20

Both methods are obviously useful, and I may have under appreciated one of them, but on the relative value of generalization versus specialization in problem solving, I offer the following opinion from Hilbert:

"If we do not succeed in solving a mathematical problem, the reason frequently consists in our failure to recognize the more general standpoint from which the problem before us appears only as a single link in a chain of related problems. After finding this standpoint, not only is this problem frequently more accessible to our investigation, but at the same time we come into possession of a method which is applicable also to related problems. The introduction of complex paths of integration by Cauchy and of the notion of the ideals in number theory by Kummer may serve as examples. This way for finding general methods is certainly the most practicable and the most certain; for he who seeks for methods without having a definite problem in mind seeks for the most part in vain.

In dealing with mathematical problems, specialization plays, as I believe, a still more important part than generalization. Perhaps in most cases where we seek in vain the answer to a question, the cause of the failure lies in the fact that problems simpler and easier than the one in hand have been either not at all or incompletely solved. All depends, then, on finding out these easier problems, and on solving them by means of devices as perfect as possible and of concepts capable of generalization. This rule is one of the most important levers for overcoming mathematical difficulties and it seems to me that it is used almost always, though perhaps unconsciously."

Answer 21

It may be matter of taste or being used to certain ways of thinking, but I find that this proof that the opposite of a category of finitary algebras isn't itself a category of finitary algebras again, is somewhat involved, compared to the proof of the statement that the opposite of a locally presentable category is not again locally presentable (if it's not a poset) - this has both weaker hypothesis and stronger conclusion.

(the original statement follows because a category of finitary algebras is locally finitely presentable, see here)

Answer 22

It seems people have kept on posting answers here rather than in the older thread, so I'll put this one here as well.

I remember thinking about the phenomenon you described when I first came across the tensor power trick. I can't summarise the idea any better than the quick description given in that link; the point relevant here is that, even though one might want to prove something for a single object or a single family of objects, if one can prove it for a family that includes the one of interest and that is also closed under taking 'tensor products', then that might be easier.

Here is a quick example from the book by Tao and Vu. If $A$ and $B$ are finite non-empty subsets of an abelian group $G$, then a natural argument gives the sumset inequality

$$ |2B-2B| \leq 16 \frac{|A+B|^4 |A-A|}{|A|^4}.$$

Now, it is possible to get rid of the factor of 16, but if we had only proved the inequality for $G = \mathbf{Z}/p\mathbf{Z}$, say, then we might very well have trouble doing so. Given the more general statement, one can get rid of the factor by applying the inequality to the group $G\oplus G \oplus \cdots \oplus G$ (say with $M$ copies of $G$) with sets $A \oplus \cdots \oplus A$ and $B \oplus \cdots \oplus B$ and taking $M$th roots.

Another scenario that springs to mind is when one wants to prove a statement involving several instances of a single object $X$: it can sometimes be easier to prove the statement if one replaces some of the instances by possibly distinct objects $X_i$. For example, it seems to be easier to prove the Cauchy-Davenport inequality $|A+B| \geq \min(|A|+|B|-1,p)$ for sets $A,B \subset \mathbf{Z}/p\mathbf{Z}$ rather than its corollary $|A+A| \geq \min(2|A|-1,p)$, since one can induct on the size of $B$ (say) separately from the size of $A$. (For the particular induction proof I have in mind I guess this can be seen as an example of the 'strengthening the induction hypothesis' idea as well.)

Answer 23

Here is a blog post I wrote about this. I contains three elementary examples of problems from the IMC that you can solve by generalizing.

Answer 24

The family of examples that leaps to my mind is the sub-trick of "strengthening the inductive hypothesis," which I thought I wrote a Tricki page on but now see I abandoned after doing epsilon of editing. (I may still have a rough draft, or at least notes, somewhere; I'll try to dig it up in the next week or so.) My all-time favorite example of this is Carsten Thomassen's proof that planar graphs are 5-choosable, which in fact proves the following:

Let $G$ be a planar graph whose interior is triangulated; let $v_1, v_2$ be adjacent vertices lying on the infinite face of $G$; and let $\{L_v\}$ be a family of lists associated to the vertices $v \in V(G)$, $v \neq v_1, v_2$ such that $|L_v| = 3$ if $v$ lies on the infinite face, and $|L_v| = 5$ otherwise. Furthermore, fix the colors of $v_1, v_2$ (ensuring that they are not colored the same). Then $G$ is $L$-choosable.

The above is proved by a fairly straightforward induction; the 5-choosability theorem follows immediately as a corollary.

Answer 25

In the case of proofs by induction, the reason it may be easier to prove a stronger result can be simply that one can use a stronger induction hypothesis.

(I think of the example of proving Łos's theorem in model theory. It says something about formulas that may have free variables. It's proved by induction on the formation of first-order formulas. Imagine trying to prove it only in the case of sentences without free variables. A weaker statement. The proof by induction doesn't work for that case.)

Answer 26

The Amitsur-Levitski Theorem says that the standard non-commutative polynomial $S_{2n}$ vanishes identically over the matrix algebra ${\bf M}_n(k)$. Recall that $$S_r(X_1,\ldots,X_r)=\sum_{\sigma\in{\frak S}_r}\epsilon(\sigma)X_{\sigma(1)}\cdots X_{\sigma(r)}.$$ For instance, $S_2(X,Y)=XY-YX$ is the commutator.

The original proof was about unreadable. When S. Rosset had the idea to embed the problem into ${\bf M}_n(k)\otimes\Lambda_2(k^{2n})$, the theorem became an easy consequence of the fact that if $N,N^2,\ldots,N^n$ are traceless ($N$ an $n\times n$ matrix), then $N^n=0$.

Answer 27

Thomas Royen has recently proven the Gaussian correlation inequality by generalizing from Gaussian distributions to gamma distributions.

Says Royen: "...it occurs frequently that a seemingly difficult special problem can be solved by answering a more general question."

Quanta Magazine has a profile of Royen, and a (very brief) discussion of the proof.

https://www.quantamagazine.org/20170328-statistician-proves-gaussian-correlation-inequality/

Royen sounds like another exception that proves some rules? (Not knowing LaTeX, almost self-publishing, etc.)

Answer 28

I see several answers to this question that rely on the fact that it is often easier to prove a stronger result in the case of proofs by induction. This phenomenon is explained in this answer : https://mathoverflow.net/a/69157/120363

A generalisation of this remark is the following: when you want to prove that an object has a property $(A)$, it is sometimes easier to prove that it has a stronger property $(B)$, because $(B)$ has more nice preservation properties than $(A)$. This is exactly what happens in the case of inductions, since sometimes a stronger induction hypothesis passes more easily from the case $n$ to the case $n+1$.

Another wonderful (in my opinion!) example of this phenomenon is the following. Say that a quasi-ordering (i.e. a reflexive, transitive binary relation) is a well quasi-ordering (wqo) if it has no infinite strictly decreasing sequences and no infinite antichains. This is an important notion in combinatorics, complexity, etc., but it has little preservation properties. In the middle of the 20th century, there were several open and seemingly very difficult problem about this notion. For instance:

1. Is the class of countable linear orderings well-quasi-ordered by the embeddability relation? (Known as Fraïssé's conjecture.)
2. Is the class of countable trees well-quasi-ordered by the minor relation?

In the late 60's - 70's, a stronger notion than wqo, the notion of better quasi-ordering (bqo), was defined and studied by Nash-Williams. The definition of a bqo is quite scary, to say so. And, for applications of the theory (to combinatorics, complexity, etc.), knowing that a particular quasi-ordering is bqo is not more useful that knowing that it is wqo. What's the point of this notion, then? The point is that the bqo property is closed under much more interesting operations than the wqo property. So in general, it is much easier to prove that a particular ordering is bqo rather than wqo. This notion was successfuly used by Nash-Williams to prove conjecture 2., and a few years later by Laver to prove conjecture 1.

To give you an idea, this is the definition of bqo: a quasi-ordering $(Q, \leqslant)$ is bqo if for every borel mapping $f\colon [\mathbb{N}]^\infty \to Q$ with countable image (where $[\mathbb{N}]^\infty$ is the set of infinite subsets of $\mathbb{N}$ with its usual Polish topology), there exists $M \in [\mathbb{N}]^\infty$ such that $f(M) \leqslant f(M \setminus \min(M))$. It's easy to prove that every bqo is wqo, however this definition is quite awful, isn't it?

Answer 29

Here is another of my favourite examples:

Prove that $({\mathbb R},+)$ and $({\mathbb R}[x],+)$ are isomorphic as abelian groups.

It is fairly easy to prove that they are actually isomorphic as ${\mathbb Q}$-vector spaces, which is a stronger result; other than that I don't know any way of proving this.

Answer 30

I've already posted an answer on this thread, but I found another example I'd like to describe separately. Let $r > 0$ and consider the following problem, coming from compound interest or as one definition of $e^r$:

Show that $f(n) = (1 + \frac{r}{n})^n$ increases with $n$.

One generalize the problem strategy is to allow $n$ to be a continuous variable (probably this trick could have its own article). Now, see if you can prove that $f(n)$ still increases. If you take this mindset, it's natural to use the definition of $n$th power for $n \in {\mathbb R}$ and write

$f(n) = e^{n \log(1 + \frac{r}{n})}$

And the problem has reduced to showing that $x \log(1 + \frac{r}{x}) = \int_0^1 \frac{r}{(1 + \frac{sr}{x})} ds$ increases with $x$, which it clearly does. (Here we've used the integral definition of the logarithm, but written in a way typically helpful for analyzing such products.)

Another problem that can be solved through allowing a discrete parameter to be continuous is to prove Stirling's approximation for $n!$ (although to make that proof very clean you can also use other labor saving tricks like Taylor expansion by integration by parts and the dominated convergence theorem).

If you ran into this problem from compound interest, or you were hoping for something more elementary which did not use such a heavy understanding of the exponential function, then you probably want to find a different proof. But finding a different proof still seems to require "generalizing the problem", but in a different way.

Another proof, goes as follows. Imagine that interest at a rate $r$ works as follows: once an amount of money is invested, the value of each unit after a time $t$ is given by $(1 + tr)$. That is, the value of the money grows linearly. Now imagine you had the opportunity to withdraw and immediately reinvest your money at a time of your choice. Having this ability would allow you to raise more money, because it would allow you to accrue interest on the interest you've already earned (hence the name "compound interest"). With this interpretation, the number $(1 + \frac{r}{n})^n$ is the value of each unit of money after time $1$ and $n$ regularly spaced compoundings.

The proof now goes as follows: if you had a choice of when these compoundings would occur, then the more compoundings the better, and the best way to allocate $n$ compoundings is to have them occur at $n$ regularly spaced time intervals. That is, we interpret $(1 + \frac{r}{n})^n = \max \prod_{i=1}^n (1 + a_i r)$ under the constraint that $0 \leq a_i \leq 1$ with $\sum a_i = 1$.

For example, it is better to have one compounding than to have none at all, because after withdrawing and reinvesting the money, now not only does the initial investment grow linearly, but also the interest you earned before the withdrawal grows linearly. For the same reason, given $a_1, \ldots, a_n$, the opportunity to compound once more during, say, $0 < t < a_1$, would allow you to increase the amount of money at all later times.

The fact that the best choice of $(a_1, \ldots, a_n)$ is to have $a_1 = a_2 = \ldots = a_n = \frac{1}{n}$ is the principle that the largest product you can obtain when the sum of positive numbers is fixed is to have all the terms equal. This is easy to check with two variables: you can either find the largest rectangle to fit inside an isosceles triangle, or otherwise just note that if $a_1 \neq a_2$, then changing to $a_1' = \frac{(a_1 + a_2)}{2} = a_2'$ gives an improvement for $(1 + a_1 r)(1+ a_2 r) < (1 + a_1' r) (1 + a_2' r)$. The case of $n$ variables actually follows from this observation.

So if you really wanted some elementary solution to the problem, this one would do. It's an interesting example because you can see that either solution involves some kind of generalization, but the two generalizations are unrelated to each other. The first one does not need to / is unable to consider these non-even partitions. The second does not need to / is unable to consider fractional $n$.

By the way, does anyone know how to prove in an elementary way (i.e. expanding) that $\prod_{i=1}^n (1 + a_ir)$ tends to $e^r = \sum \frac{r^k}{k!}$ as $\max |a_i| \to 0$ with $0 \leq a_i \leq 1$ and $\sum a_i = 1$? An easy solution goes by writing the product with the exponential function so that you get the exponential of $\sum \log(1 + a_i r) = \sum \int_0^1 \frac{a_i r}{(1 + s a_i r)} ds$.

You can then integrate by parts (i.e. Taylor expand) to obtain $\sum a_i r - \sum \int_0^1 (1-s) \frac{(a_i r)^2}{(1 + s a_i r)^2} ds$. Now, $\sum a_i r = r$ is the main term. After you take $\max |a_i|$ to be less than $.5 / |r|$, the error term is bounded in absolute value by $C \sum (a_i r)^2 \leq \max \{ |a_i| \} \cdot \sum a_i |r|^2$. I can, of course, move this question to a different thread.

EDIT: I realized later on that there is a completely elementary proof, and it is also completely obvious even though I didn't think of it. Namely, you expand $(1 + \frac{r}{n})^n$ into powers of $r$, and it is easy to see after a little algebra that each coefficient increases with $n$. I still find the other solutions interesting, but this turns out not to be a good demonstration of how generalizing can make a problem easier. By the way, the last question I had asked was answered in this thread:

A limiting product formula for the exponential function