Examples of problems that are easier in the infinite case than in the finite case.

Question

I am looking for examples of problems that are easier in the infinite case than in the finite case. I really can't think of any good ones for now, but I'll be sure to add some when I do.

Answer 1

One can compute the value of $$\int_{0}^{\infty}e^{-t^{2}}\,\mathrm{d}t$$ exactly. This is known as the Gaussian integral, and it has its own Wikipedia page. The answer turns out to be $\frac{1}{2}\sqrt{\pi}.$

But one cannot do the same with $$\int_{0}^{x}e^{-t^{2}}\,\mathrm{d}t$$ because the antiderivative of the integrand is not an elementary function. This is why we gave a name to the error function $$\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^{2}}\,\mathrm{d}t,$$ which also has its own Wikipedia page.

In that sense, the infinite case is easier than the finite case.

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Addendum: The same phenomenon occurs for variants of this integral, in particular we can transform the integrand to evaluate $$\int_{-\infty}^{\infty}ae^{-(t-b)^{2}/(2c^{2})}\,\mathrm{d}t = \sqrt{2}a\lvert c\rvert \sqrt{\pi}$$ as detailed here on Wikipedia.

Answer 2

First thought would be series vs. partial sums (or improper vs. definite integrals) e.g.

$$ \sum_{k=0}^{\infty} \frac{1}{k!} \;=\; e $$

$$ \sum_{k=0}^{n} \frac{1}{k!} \;=\; \frac{e \cdot \Gamma(n+1,1)}{n!} $$

Answer 3

$\min c^Tx$

subject to $x \in P$

where $P$ is a bounded polyhedral can be solved easily by linear programming method. Ellipsoid method shows that the problem is in $P$ class.

$\min c^Tx$

subject to $x \in P \cap \mathbb{Z}^d$

that is imposing integer constraint increase the difficulty of the question.

Here $P$ is an infinite set in the sense that it contains infinitely many points but $P \cap \mathbb{Z}^d$ contains only finitely many points. Linear programming problem is easier than integer programming problem.

Answer 4

Circle packing. Packing in a plane (infinite case) vs. packing in arbitrarily given bounded area (finite case).

Answer 5

Proving that a polynomial is zero

Suppose you have an unknown polynomial $f$ of degree $d$, but you are able to say things about the values of $f$.

If you are working over an infinite field, an easy test is to check if $f(a) = 0$ for $d+1$ distinct values of $a$.

If you are working over a finite field of $d$ elements or less, you can have the unenviable problem that $f(a) = 0$ for every $a$, even when the polynomial $f(x)$ is nonzero, so this approach won't work in general.

(example: $f(x) = x^q - x$ when working over the field of $q$ elements)

Answer 6

In digital sample analysis (signal, image processing, data science), data is generally discrete and finite. "Discrete" generally means "regularly sampled", and indexed by integer indices. "Finite" here means having a compact support (not talking about the field in which the values are taken, except as a side note). This answer is about the difference in analysing continuous data $x(t)$, $t\in \mathbb{R}$, or discrete data $x[n]$, $n\in [1,\ldots,N]$. Similar problems arise in higher dimensions with $t\in \mathbb{R}^d$ and $n\in [1,\ldots,N]^d$, I will consider only $d\in\{1,2,3\}$.

Tools from harmonic analysis have been used for a while in signal (and image) processing : all sorts of integral transformations, such as Fourier, Radon, Hilbert transforms, and wavelet analysis. Such representations are generally invertible when considered on functions of continuous parameters (like a time $t$). However, the discretization of harmonic analysis transformations with a finite number of samples, which is mandatory for computer operations, while retaining nice properties (orthogonality, invertibility, symmetry) is often complicated.

A standard example is the theory of continuous wavelet transforms. A 1D signal $x(t)$ can be transformed to a 2D domain with a location ($b$) and scale ($a$) variable:

$$ X_{w}(a,b)={\frac {1}{|a|^{1/2}}}\int_{-\infty }^{\infty }x(t)\overline {\psi}\left({\frac {t-b}{a}}\right)\,dt$$ which admits an inversion $$x(t)=C_{\psi }^{{-1}}\int _{{-\infty }}^{{\infty }}\int _{{-\infty }}^{{\infty }}X_{w}(a,b){\frac {1}{|a|^{{1/2}}}}{\tilde \psi }\left({\frac {t-b}{a}}\right)\,db\ {\frac {da}{a^{2}}}$$ with a quite mild admissibility condition (SE.math) on $ \psi$ and $\tilde{\psi}$, which I do not detail here. In plain words, any couple of such functions (avoid pathological ones, still) vanishing in the Fourier domain at $0$, and decaying fast enough at $\infty$, will do the job. So one can use many smooth wiggling functions, with analytical formulas, like Gaussian derivatives, for instance the Mexican hat:

However, the building of discrete wavelets was difficult, apart from trivial examples in early works by A. Haar in 1910 and P. Franklin in 1928. It took some time before J.-O. Strömberg, Y. Meyer and I. Daubechies produced discrete orthogonal wavelet bases in the eighties. Daubechies wavelets of low order are notoriously not so regular, without analytical formulas, and asymmetric:

This is for the transformation. But the signal is quantized on bits. It often takes values in $[0,\ldots,2^B-1]$ If we restrict the output to by quantified too, the problem is even more difficult.

Aside, dealing with (and defining) discrete objects on a digital grid is often more complicated than with their continuous counterparts. Think for instance about discrete arcs of circles (in 2D), drawn in "pixels", that do not scale as well as standard circles:

Those topics are addressed in digital geometry, for which algorithms are developed to draw discrete lines (Bresenham algorithm), discrete planes (in 3D, see below, from 3D Noisy Discrete Objects: Segmentation and Application to Smoothing, 2009, Provot et al.), with less or more thickness, or conics for instance.

Answer 7

Geometric series finite vs. infinite case for $|q|<1$ \begin{align*} \sum_{j=0}^Nq^j=\frac{1-q^{N+1}}{1-q}\qquad\text{vs.}\qquad\sum_{j=0}^\infty q^j=\frac{1}{1-q} \end{align*}

Answer 8

The linear-quadratic control problem, that is, finding a function $u(t)$ that minimizes $\int_0^{t_1} x^TQx + u^TRu\,\mathrm{d} t$ under the constraint $\dot{x} = Ax + Bu$.

In the infinite-horizon case $t_1=\infty$, the solution is obtained by solving an algebraic Riccati equation (ARE). In the finite-horizon case $0<t_1 \in \mathbb{R}$, it is obtained by solving a differential Riccati equation (DRE), which is more difficult to solve; for instance, some numerical methods for DREs involve solving an ARE at each time step.

Answer 9

Let's take a look at Goodstein sequences $G_k(n)$ and their interesting behaviour. We associate to each natural $n$ a sequence $(G_k(n))_{k\geq 0}$.

• Step $0$: We start by representing $n$ in base $2$ notation and also all powers and powers of powers, so that $n$ is written without using any number greater than $2$.

  Taking $n=266$ we obtain \begin{align*} G_0(266)=2^8+2^3+2^1=2^{2^{2+1}}+2^{2+1}+2^1 \end{align*}

• Step $k\geq 1$: Replace each occurrence of base $k+1$ in $G_{k-1}(n)$ by $k+2$, represent it in base $k+2$ notation without using any number greater than $k+2$ and subtract $1$

  We obtain \begin{align*} G_1(266)&=3^{3^{3+1}}+3^{3+1}+3^1-1=3^{3^{3+1}}+3^{3+1}+2\simeq 10^{38}\\ G_2(266)&=4^{4^{4+1}}+4^{4+1}+1\simeq 10^{616}\\ G_3(266)&=5^{5^{5+1}}+5^{5+1}\simeq 10^{10921}\\ G_4(266)&=6^{6^{6+1}}+6^{6+1}-1\\ &=6^{6^{6+1}} +5\cdot6^{6}+5\cdot6^{5}+5\cdot6^{4}+5\cdot6^{3}+5\cdot6^{2}+5\cdot6^{1}+5\simeq 10^{217832}\\ G_5(266)&=7^{7^{7+1}} +5\cdot7^{7}+5\cdot7^{5}+5\cdot7^{4}+5\cdot7^{3}+5\cdot7^{2}+5\cdot7^{1}+4\simeq 10^{4871822}\\ \end{align*}

Although this sequence is enormously increasing at the beginning, subtracting $1$ in each step is sufficient that this sequence and in fact each Goodstein sequence will eventually terminate with $0$ (admittedly with $k$ often rather large).

The interesting thing is to prove this behaviour can be done by going from finite to infinite. It is done by using transfinite ordinals and arguing with the well-ordering property. See this nice introductory presentation for more information.

Answer 10

Kolmogorov's Zero-One Law

Informally, in certain probabilities spaces, events which do not depend on any finite amount of information must have probability zero or one.

For example, suppose you flip a coin over and over again infinitely many times. Any event $E$ which is independent from the outcome of any finite number of coin flips must have probability $0$ or $1$. Examples include

• $E =$ the event that 100 consecutive heads are flipped infinitely times (prob $=1$)
• $E =$ the event that $\limsup_{n \to \infty} H_n - T_n = \infty$, where $H_n$ ($T_n$ resp.) is the number of heads (tails resp.) in first $n$ tosses. (prob $=1$)
• $E =$ the event that $\lim_{n \to \infty} H_n - T_n = \infty$. (prob $=0$)
• $E =$ the event that $\sum_{k=1}^{\infty} \frac{\sigma(k)}{k}$ converges, where $\sigma(k) = 1$ if $k$'th toss is heads and $\sigma(k) = -1$ if $k$'th toss is tails. (prob $=1$)

Answer 11

The Game of Chomp:

Although a winning strategy is guaranteed to exist, finding a winning strategy for a $a \times b$ is an unsolved problem. However, finding a winning strategy for a $a \times \omega$ ($a\geq3$) board is not that hard.

First note that you lose for a $2 \times \omega$ board, because the only losing position is a board of the form of $a+1$ chocolates and then $a$ chocolates. Your opponent can always make sure that you get such board. Now note that you can always reduce a $a \times \omega$ ($a\geq3$) board to a $2 \times \omega$ board.

Answer 12

The asymptotic distribution of primes: n^th prime ≈ n ln n as n → ∞

Try finding a formula for the n^th prime in the non-asymptotic case!

Answer 13

Almost the entirety of calculus and all applications thereof

This needs a bit of explaining: The notion of infinity is an artificial one. There is nothing in reality that we know to be truly infinite¹. So why do we bother with infinity? Because it’s easier to deal with than “very big numbers”! And working with infinity is what calculus (née infinitesimal calculus) is about. Take any word problem in a high-school calculus text book and try to formulate and then solve it without any notion of infinity, continuity, and so on – it will be a pain in the arse.

To be even more concrete, let’s look at the classical setup of exponential growth (e.g., for populations, interest, and so on): You have a quantity $x$ whose growth is proportionally to its current amount and want to describe its long term behaviour. Employing the notion of infinity, you can describe and solve this with the differential equation $\dot{x} = cx$. Without calculus you have to find some finite time $\mathrm{d}t$ and then can state that $x(t+\mathrm{d}t) = (1+c)·x(t)$. (In some cases, e.g., interest applied in daily steps, this may even be more realistic.) So far, there is no big difference in terms of easiness.

But now try to quickly calculate a good approximation of $x(1000 · \mathrm{d}t)$. Without employing the concept of infinity, you probably end up with performing thousands of multiplications. With infinity, you can approximate the behaviour with the exponential function, whose value you can approximate in more efficient ways. Of course, you can formulate some equivalent or almost equivalent statements for obtaining this value without infinity. But neither would the statement itself nor deriving it would be what I would consider easy.

Some of the existing answers provide further examples for this: 1, 2, 3, 4.

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^{¹ or truly infinitesimally small, continuous, etc.}

Answer 14

Another example of an improper integral being "easier" (in this case, by having an elementary result) than a proper one would be integrating along the entire real axis by the residue theorem and Jordan's lemma or estimation lemma

For example, consider: $$f = \frac{e^{i x}}{x^2 + 1}$$ $$\int_{-a}^a f \:\mathrm{d}x$$

For the infinite case ($a \to \infty$), we consider a semicircle with its base along the real axis and curved section in the upper positive imaginary axis, and then take the limit of its radius approaching infinity.

With the residue theorem and Jordan's lemma, we have: $$ \begin{align*} \int_{-\infty}^\infty f \:\mathrm{d}x &= 2 \pi i \operatorname{Res}(f, i) \\ &= 2 \pi i \frac{e^{-1}}{2 i} \\ &= \frac{\pi}{e} \end{align*} $$

For the finite case ($a \in \mathbb{R}^+$), there is no elementary solution, so we have to solve in terms of the exponential integral: $$ \begin{align*} \int f \:\mathrm{d}x &= \frac{i}{2} \int \frac{e^{i x}}{x + i} - \frac{e^{i x}}{x + i} \:\mathrm{d}x \\ &= \frac{i}{2} \int e \frac{e^{i (x + i)}}{x + i} - e^{-1} \frac{e^{i (x - i)}}{x} \:\mathrm{d}x \\ &= \frac{i}{2 e} \left(e^2 \operatorname{Ei}(i x - 1) - \operatorname{Ei}(i x + 1)\right) \end{align*} $$

Answer 15

Notation: $$(i). \text { $|A|$ is the cardinal of the set $A.$ }$$ $$(ii). [A]^2 =\{ \;\{x,y\}\subset A: x\ne y\}.$$ $$(iii). f''A=\{f(x):x\in A\}$$ for any function $f$ and any set $A\subset dom(f).$ $$(iv). \text { $\omega$ is the first infinite cardinal.}$$

For finite or infinite cardinals $a, b$ the arrow notation $$(a)\to (b)^2_2$$ means that if $|A|=a$ then for any function $f:[A]^2\to \{0,1\}$ there exists $B\subset A$ with $|B|=b$ and $|f''[B]^2|=1.$.... Observe that if $a'\geq a$ and $b'\leq b,$ then $(a)\to (b)^2_2$ implies $(a')\to (b')^2_2.$

$$\text {Theorem 1. For any $m\in \mathbb N$ there exists $n\in \mathbb N$ such that $(n)\to (m)^2_2.$}$$ Theorem 1 implies that for all sufficiently large $n ,$ any two-coloring of the edges of the complete graph on $n$ points will have a complete sub-graph on $m $ points with edges all the same color.

$$\text {Theorem 2. $(\omega)\to (\omega)^2_2.$}$$ This is fairly easy to prove. Now there is a result in Model Theory: If a theory has arbitrarily large finite models then it has an infinite model. We can use this, and Theorem 2, to prove Theorem 1 by contradiction: If for some $m\in \mathbb N$ we have $\neg [(n)\to (m)^2_2]$ for all $n \in \mathbb N$ then there exists infinite $a$ such that $\neg [(a)\to (m)^2_2],$ contradicting Theorem 2.

Prof. William Weiss once told me that there is a long finitistic proof of Theorem 1, but that the infinite method "uses some technology."

Answer 16

Without the axiom of infinity, it's very hard (impossible, I think) to prove that the TREE function is well-defined, even though it is a theorem about finite objects. This is taking a fairly liberal view of what "the finite case" means!

Answer 17

I am not sure if this qualifies, but I just want to share it here in case you're interested. I got it from a blog post by Mr. Greg Muller : The Axiom of Choice is Wrong

Let me write it down here the main idea of the post.

$100$ prisoners are placed in a line, facing forward so they can see everyone in front of them in line. The warden will place either a black or white hat on each prisoner’s head, and then starting from the back of the line, he will ask each prisoner what the color of his own hat is (ie, he first asks the person who can see all other prisoners). Any prisoner who is correct may go free. Every prisoner can hear everyone else’s guesses and whether or not they were right. If all the prisoners can agree on a strategy beforehand, what is the best strategy?

There is a strategy that ensures at least $99$ prisoners to go free. Let's consider a variance to that problem:

A countable infinite number of prisoners are placed on the natural numbers, facing in the positive direction (ie, everyone can see an infinite number of prisoners). Hats will be placed and each prisoner will be asked what his hat color is. However, to complicate things, prisoners cannot hear previous guesses or whether they were correct. In this new situation, what is the best strategy?

This sounds hopeless, right? However, there is a strategy that guarantee all but finitely many prisoners to go free! The paradoxical result stems from the fact that there're non-measurable sets (but some would also attribute it to the sigma-additive nature of probability).

Follows the link if you are interested. In the comment section there are also explanation/discussion of this perplexing phenomenon by many mathematician, Prof. Tao being one of them.

Answer 18

Every single Machine Learning problem ever.

Optimal statistical inference is generally much easier in the case of infinite data than finite data.

Answer 19

I know this is trivial, but here's one: $x$ is finite and unknown. Determine whether $x < n$.

Easy when $n$ is infinite!

Answer 20

Stability analysis of finite difference stencils

GKS stability requires that the two semi-infinite problems defined on $[a,\infty]$ and $[-\infty,b]$ must be stable in order for the finite problem defined on $[a,b]$ to be stable. Thus, you can't do the finite case without at least doing the infinite ones first.

Answer 21

A Markov chain is characterized by a discrete probability distribution of the initial state of a system, 𝜋₀, and a transition matrix P such that p_ij is the probability of going from state i to state j after any one transition. Under certain assumptions, the probability distribution of states after the first transition is 𝜋₁ = 𝜋₀P, after two transitions is 𝜋₂ = 𝜋₁P = 𝜋₀P², after n transitions, 𝜋_n = 𝜋₀Pⁿ.

By constrast, the steady-state (as n -> ∞) probability distribution of states, 𝜋, must satisfy 𝜋 = 𝜋P, and 𝜋𝛽 = 1, where 𝛽 is a column vector of 1's. This is an over-determined (by one) system of equations, but the upshot is that the steady-state probability distribution of states is obtained by inverting a matrix.

See, for example, the Markov chain Wiki for details.

Answer 22

Burnside problem

The solution for Burnside problem in the following formulation:

For which positive integers $m$, $n$ is the free Burnside group $B(m, n)$ finite?

is not known for any $m$, $n$, but there are some results for infinite cases:

• $B(m,n)$ is infinite for all odd $n > 665$
• $B(m,n)$ is infinite for even $n > 2^{48}$ and $n$ divisible by $2^9$
• $B(m,3)$, $B(m,4)$ and $B(m,6)$ are finite for all $m$

And the example for the question is $B(2,5)$, for which it is not known whether it is finite. $B(2,5)$ also has the smallest $m$ and $n$ for which the problem remains open.

Answer 23

Finding Nash equilibria for $N$ players vs $N \to \infty$ players; see mean field games.

Answer 24

The Strategy-stealing argument in game theory would seem to fit here. Thus we know some games (e.g hex) are a first player win, but that knowledge doesn't help determine the necessary strategy.

Answer 25

In the spirit of D Poole's comment on the Zero-One Law, long-term probabilistic behavior is often a lot easier to deal with than short-term. A lot of elementary theorems in probability tell us about the asymptotic behavior of a sequence $(X_k)_{k \in \mathbb{N}}$ of random variables under suitable conditions. For instance, suppose I have a six-sided die, and I want to know the probability that in the first $600$ rolls, $1/3$ of the rolls were $4$. The formula for this value might be quite ugly when we want to look at this event after some specified number of rolls, but the strong law of large numbers tells us the probability of this happening over the long run (i.e. that $1/3$ of the rolls will be $4$) is $0$ (and further we can fairly easily provide more specific descriptions of the "size" of this outcome, e.g. Hausdorff dimension). So in general, though descriptions of the behavior of random variables might be ugly after finite time, we're often able to provide nicer descriptions of the long-term ("infinite time") behavior.

Answer 26

I don't believe nobody mentioned applications of the Central Limit Theorem (CLT).

For example calculating this: \begin{align} \int^{n+\sqrt[]{n}}_0 \frac{x^{n-1}}{(n-1)!}e^{-t}\,dt \end{align} is very easy when $n\to \infty$ using the CLT.

And yet another application of CLT itself is Wilk's Theorem very strong statement about the infinity case.