What is math?

What is math?


Most people’s conception of math was drilled into them during grade school.

In my experience, grade school math goes something like this: The teacher says that we need to calculate a thing. He then shows how to calculate that thing, with seven slight variations. Your homework is to calculate six of each of the variations. The test will have five of those seven.


After a decade of this, most walk away thinking that math is calculation. And because of the rote way the material was introduced, many get the impression that math is set in stone. If you perform a particular set of arcane, incomprehensible steps, you will be led to the mythical “right answer.” No other steps are allowed, and heaven help you if you don’t happen to remember the right steps for a particular problem. In that case there is nothing to be done but despair.

And, of course, they believe that all math has been handed down to them from on high, as wisdom from the ancients. It is imperturbable, impenetrable, impeccable.


But that is not what math is.

So, what is math?

Calculation is a useful tool, but it is definitely not what math is.

Math is a quest for understanding. And like any good epic fantasy series, it seems to never quite be finished.1

And the understanding we mathematicians seek is an odd sort of understanding. The goal of science is to understand what is, to describe and understand the universe around them.

Mathematicians, on the other hand, seek to understand what must be.

After all, the questions a mathematician asks are not generally about things that could even exist. Have you ever seen a perfectly straight, infinitely thin line? Or an angle of precisely 90 degrees? But, if I have a perfectly flat triangle with a 90 degree angle, I know the side lengths have a certain relationship, a^2 + b^2 = c^2.

And sure, we can count 37 cows, but do the cows care that there are a prime number of them? But 37 is prime, and so the 37 cows cannot be evenly split between more than one person.


I sometimes like to describe this by saying that I, as a mathematician, try to figure out what even God cannot do. Even an all-powerful God cannot create a perfect flat triangle with a 90 degree angle, whose side lengths do not obey the Pythagorean relationship. Neither could He evenly divide 37 cows between more than one person.2

The basis for deciding what must be are the definitions and axioms of mathematics.

Definitions and axioms are different, but very closely related.

Definitions describe the things we talk about. For instance, a straight line (versus a curved one) might be defined as “a line which lies evenly with the points on itself,” as in Euclid.


Axioms describe what we can do with the things we’ve defined. These tend to be very basic, “obvious” things. For example, the axiom of symmetry says that “If A=B, then B=A.” In this example, you could see the axiom as something you can do (“You can switch the sides of an equation.”) or you could see the axiom as defining what two things being equal really means.

On top of this foundation, mathematics is built with logic. Given the definitions and axioms, certain conclusions follow as inescapable consequences. These conclusions we call theorems or lemmas or propositions.3

Because mathematics is taught in such an authoritative way, it can appear that the definitions and axioms of mathematics are in someway intrinsic, that they have existence outside of the creation of man. It can feel like the axioms and definitions are part of the “what must be” that mathematicians are searching for.

To some extent that may be true, but I don’t think this is completely true, and it’s certainly not how math is done.

When you read a textbook, the most recent thinking of the definitions and axioms that are thought to be important are presented. But that hides to some extent the fact that it took hundreds, or even thousands, or years to decide that those axioms should be the ones to form the foundation of the rest of mathematics.

Math evolves. Math changes. The definitions and axioms we use today are not the same ones that were used by Newton.


Referencing Newton actually brings up a good example of how math changes.

Newton (and Leibniz) invented calculus around 1670. It immediately proved its use in solving any number of important questions in physics and mathematics.

But Newton’s calculus was not built on what we would today consider a rigorous foundation.


In order to explain their ideas, both Newton and Leibniz used some idea of “infinitesimals,” quantities that were infinitely small.


Infinitesimals can be very useful in an intuitive explanation of calculus. (I often use them informally when I teach calculus myself.) And so Newton and Leibniz’s proofs of their results were accepted, even though some were uncomfortable with the idea of an infinitely small quantity.

But as mathematicians delved deeper into the ideas of calculus, it became clear that the infinitesimal arguments weren’t quite complete. There were important theorems that could not be carefully proven because the foundations of calculus were not proven with sufficient rigor.4


Thus, one of the major mathematical projects of the 1800’s was to prove the “soundness” of calculus, and make sure the foundations were correct.

This involved inventing new definitions. For instance, one of the key ideas of calculus is the limit. Informally, the limit asks, “As the inputs get close to a number, what do the outputs get close to?”

The intuition for limits is not difficult; you plug in numbers closer and closer to the one you want, and see if the outputs get close to some other number. But the careful definition for limit that we use today, the \epsilon-\delta (epsilon-delta) definition, was not introduced till the 1820’s by Augustin-Louis Cauchy.


Mathematics is not static, and the axioms and definitions we use are not necessarily natural, sitting there for us to find. As we seek deeper understanding, we often come to a point where we realize our earlier understanding was incomplete, or even incorrect, and we seek to fix the foundations. This has occurred over and over and over again to get to our “fixed” modern ideas of mathematics.5


To summarize, mathematics is a quest for understanding what must be. But the very concepts we try to understand are not set in stone. The objects of mathematics are defined by people, and as we understand them better, the definitions and axioms we base our understanding on change.

In the next post I want to talk more in depth about why these definitions change, and how and why mathematicians come up with new definitions.

This post was mostly about the philosophy of math, which is quite a bit different than my normal post. But as we’ll see in a few weeks, Gödel’s incompleteness theorem is so weird that it is impossible to talk about it without discussing the philosophy of math. Gödel’s theorem puts a fundamental limit on mathematicians’ quest for understanding.


  1. I’m looking at you, George R. R. Martin… 
  2. Well, at least without taking the King Solomon approach and cutting the 37th cow in half! 
  3. Usually “theorems” are bigger, more important conclusions, while “lemmas” are littler conclusions that are needed along the way to show the theorems are true. Propositions can go either way. On the other hand, sometimes lemmas end up being more important than the theorems. 
  4. More recently, mathematicians have come up with rigorous methods to talk about infinitesimals, for instance the hyperreal numbers. However, infinitesimal methods are no longer considered standard. 
  5. Even the work on calculus done in the early 1800’s was not final. The “Riemann” integral, which was the formalization of the integral by Riemann, is what is taught in high schools and early college math. But at the graduate level, we use the “Lebesgue” (Luh-bayg) integral instead, which was introduced in the early 1900’s. Both are rigorous approaches to the integral, but the Lebesgue integral makes a few key lemmas and theorems much easier to prove. The basis of the Lebesgue integral is less intuitive at first, but easier and more powerful in the end. 

Can You Hear the Shape of a Drum? (Part 2)

In the last post, we explained how vibrating strings work.complicatedTo summarize, the string’s position, given as a function f(x,t), is controlled by a differential equation, \dfrac{d^2 f}{dt^2} = \dfrac{d^2 f}{dx^2}.1

5The left hand side of this equation, \dfrac{d^2 f}{dt^2}, is the vertical acceleration of the string. Meanwhile, \dfrac{d^2 f}{dx^2} measures how much the string is curving at one instant of time, as you move from left to right. The more the string is curving, the bigger this is.6The punchline of the last post was that any vibration of the string, including the complicated one above, can be represented by a sum of special “self-similar” solutions to the differential equation. These solutions keep their overall shape, and simply vibrate by scaling up and down.In order to keep their shape, these solutions had to satisfy a special equation, \dfrac{d^2 f}{dx^2}(x) = -\nu^2 f(x), which says that the second derivative of the function (i.e., how much it is bending) must be equal to a constant -\nu^2 times the function itself. If we want the string to have length \pi, the \nu had to be natural numbers, \nu 1, 2, 3, \cdots. The video above are the solutions when \nu is 1, 2, 3, or 7.

The solutions f(x) of \dfrac{d^2 f}{dx^2}(x) = -\nu^2 f(x), which represent the initial states of the string, are eigenfunctions of the “transformation” \dfrac{d^2}{dx^2}, with eigenvalues \lambda = -\nu^2.

The key observation is that the eigenvalue of the self-similar solutions controls the speed of the vibration. As you can see above, the \nu=7 solution vibrates seven times as fast as the \nu=1 solution.

The tone you hear from an instrument is related to how quickly the air (and thus the string) is vibrating. The self-similar solutions are the vibrations of the string that produce pure tones — a single note.

Since any vibration is a combination of these basic pure tone vibrations, when the string vibrates, you will hear all the tones of all the pure tone vibrations you combined together, each at different volumes depending on how much of that pure tone vibration you used. For instance, this vibration would have the \nu=1 and \nu=2 tones in equal amounts:sin1and2However, these pure tones are the only tones the string can produce.

In other words, the eigenvalues (of \dfrac{d^2}{dx^2}) control the tones a string can produce, while the tones a string can produce tell you the eigenvalues. The set of all eigenvalues (of \dfrac{d^2}{dx^2} for a particular string) is called the spectrum.

We finally get to the first question: Can you hear the length of a string?hearLengthHow could we hear the length?

If we vibrate a string at random, our vibration will probably2 be a combination of small amounts of all the pure tone vibrations. If we had perfect pitch (and could hear tones both very low, and infinitely high), we could hear all of those notes, and, from them, deduce the spectrum of the string.

37octavesSince the spectrum and the possible tones are equivalent, a more mathematical way to ask the question is: If you know the spectrum of a string, can you somehow calculate its length?3

To answer that, we need to know how the eigenvalues of \dfrac{d^2}{dx^2} change when the length of the string changes.

So far, we’ve assumed the length of the string was \pi, for simplicity. In that case, the eigenfuctions associated to the eigenvalue \lambda = -\nu^2 was f_\nu(x) = \sin(\nu x). While f_\nu(x) is an eigenfunction for any \nu, it could only represent a vibration on our string of length \pi if \nu was 1, 2, 3, etc. That was because we needed both sides of the string to be fixed.sinesIf our string is instead length L, we will still need both sides of the string to be fixed. For that to happen, we will need f_\nu(x) to be zero at the two ends, x=0 and x=L. That will happen when \nu = \dfrac{\pi}{L} n, where n is some natural number. (Notice that if L=\pi, this is exactly what we had before.) In other words, our eigenvalues are -\dfrac{\pi^2}{L^2} n^2.

That means you can hear the length of the string.doubtedMeFor example, the spectrum for a string of length \pi would be \{-1, -4, -9, \cdots\}, and you would hear the associated tones.

The spectrum for a string of length 2\pi would be \left\{-\dfrac{1}{4}, -1, -\dfrac{9}{4}, \cdots \right\}. Those two spectra (plural of spectrum) are quite different. In fact, you can even tell them apart even by just looking at the first eigenvalue.

You can hear the length of a string just by hearing its lowest tone.4

We’re finally prepared to talk about the original question we had: Can you hear the shape of a drum?yippeeAs with strings, the vibration of a drum is controlled by a differential equation. Unsurprisingly, it’s very similar to the one for a vibrating string. The function f(x, y, t) representing the position of the drum head at any particular time is more complicated now, though. It depends on time and both x and y, since the drum head is two dimensional.

The differential equation is \dfrac{d^2 f}{dt^2} = \Delta f. The left hand side, \dfrac{d^2 f}{dt^2} is the vertical acceleration of the drum head, as before.

The right hand side looks different, but mostly just because I’m using notation you may not be familiar with. The symbol \Delta (capital Delta) is called the Laplacian, and is involved in many of the most important differential equations. Since we have two space dimensions for our drum head, \Delta f = \dfrac{d^2 f}{dx^2} + \dfrac{d^2 f}{dy^2}, the sum of the second derivatives in the two space directions.5

That means the Laplacian is really just the same thing we had on the right hand side of the equation for a vibrating string. The quantity \Delta f adds up how much the drum head is bending in both the x and y directions.

So, as before, where the drum head is bending the most, it accelerates the most as well, bringing it back toward the resting position.restEven on a square drum, solutions to this differential equation can be quite complicated.3Dcomplicated.gifBut, as before, every solution is a combination of small amounts of special self similar solutions.

To find self similar solutions, we again need \dfrac{d^2 f}{dt^2} = -\nu^2 f — the acceleration of the drum head must be equal to a (negative) multiple of the current position.

Since this is the same equation as before, it has the same solution. If we had an initial position f(x, y, 0), the position at time t would be f(x, y, t) = f(x, y, 0) \cos(\nu t). The initial position simply vibrates up and down sinusoidally.3DBasic.gifThe frequency of the vibration, and thus the tone produced by the drum, is controlled by the value \lambda = -\nu^2. The larger \nu is, the higher the frequency, and thus the higher the tone produced.

That leaves the hard part of the problem.hardpartIn order to find initial positions of the drum head that will lead to these simple, self-similar solutions, we need to find solutions to the equation \Delta f = -\nu^2 f.

This is again an eigenvalue problem. The Laplacian transforms our function f into a new function. Usually this new function, \Delta f would be unrelated to the original function f, but we are looking for those special functions for which this transformation simply scales the original function by some constant -\nu^2.

This may not seem that much harder than it was for a string, but it turns out to be very hard to solve explicitly.

For a rectangular drum, it’s not too bad. It turns out the eigenfunctions are just products of sine waves–one in the x direction and one in the y direction. For instance, on a square of side length \pi, f(x,y,0) = \sin(2x)\sin(3y) is an eigenfunction with eigenvalue \lambda = -2^2-3^2 = -13.3D23.gifYou can also find explicit solutions for a circular drum, and for some triangles (equilateral, isosceles right, and 30-60-90), but for just about anything more complicated than those examples, we have no idea how to find explicit eigenfunctions and eigenvalues.6

But, fortunately, finding eigenvalues and eigenfunctions was not the question.

The question was “Can you hear the shape of a drum?” In other words, if you already know the eigenvalues (the spectrum), can you figure out the shape of the drum?

It’s been known for a long time that there are some things we can tell about the drum from the spectrum.

The most “obvious” one is the area of the drum head. A bigger drum head makes a lower tone.

Unlike the string, though, the lowest tone (equivalently, the eigenvalue closest to zero) is not sufficient to tell you the area. It’d be a good start, but it’s not enough.inadequateInstead, we have to look at how the eigenvalues are spread out.

For the string, the spectrum for length \pi was \{-1, -4, -9, \cdots\}. For length 2\pi, the spectrum was \left\{-\dfrac{1}{4}, -1, -\dfrac{9}{4}, \cdots \right\}.

The thing to notice is that, for the longer string, the spectrum is packed closer together. Another way to say this is that, for a given number -R, the spectrum for the string of length 2\pi has more eigenvalues closer to zero than -R than the spectrum for the string of length \pi. Though the formula may not be obvious, there is a way to calculate the length of the string based on this way of interpreting the spectrum.lengthMoreEigensFor a drum, a similar idea works. Suppose we knew the spectrum for a drum completely. We then could count how many eigenvalues are closer to zero than -R for any -R. We’ll call that number of eigenvalues N(R). (N for number of eigenvalues.)N10As R grows, the number of eigenvalues closer to zero than -R also grows. In fact, they grow in a very specific way.

Hermann Weyl (said “vile”) showed in 1911 that, for a drum head, N(R) will grow roughly linearly in R. Not only that, the slope of that line predicts the area of the drum head!HearAreaPrecisely, the slope can be measured as \displaystyle\lim_{R\to \infty} \dfrac{N(R)}{R}. (This is saying something like, “See how many times bigger N(R) is than R for large R. If N(R) were precisely a line, this would give the slope. It works just as well for N(R) that are almost a line.)

Once you have that slope, it’s easy to find the area. In fact, A = 4\pi\displaystyle\lim_{R\to \infty} \dfrac{N(R)}{R}!AreaWeyl’s law tells us that knowing the spectrum determines the area.

But the area is not the entire shape.

Besides the area, what other information can we figure out from the spectrum?

Weyl’s law says that N(R) \approx \dfrac{A}{4\pi}R. But we can be more precise than that. If we look at N(R) - \dfrac{A}{4\pi}R (i.e., see how far off our approximation was), that new thing looks about like a multiple of the square root function \sqrt{R}! And, again, the multiple tells us something about our shape.

Weyl conjectured (and it was proven by Victor Ivrii in 1980) that N(R) - \dfrac{A}{4\pi}R \approx \dfrac{P}{4\pi}\sqrt{R}, where P is the perimeter of the drum head. In other words, if we know the spectrum, a more careful analysis would give us the area and the perimeter of the drum.PerimeterThat would be enough to tell you the shape of a drum, if you knew that it was rectangular. You could hear the difference between a square (lots of area, not so much perimeter) and a stretched out rectangle (no so much area, lots of perimeter.)

This is hopeful. This \sqrt{R} level approximation was good, but there’s still more information lurking about in the spectrum. Though no better approximation results have been proven (that I know of), you might hope that you can hear the shape of a drum.

Unfortunately, you can’t.cursesIn 1991, Gordon, Webb and Wolpert found some shapes that are obviously different, but have the same spectrum. There are lots of examples now, but here are two of them:isospectralNotice that they both have the same area and perimeter. They have to, thanks to Weyl’s and Ivrii’s work. But, despite being different, they have the exact same spectrum.

Of course, these drums heads are not exactly normal shapes for drum heads. Most drums that I’ve seen have had drum heads that were at least convex, meaning they don’t cut in.ExoticDrumsIf you assume a few reasonable things about the shape of the drum head, it turns out that you can hear the shape of a drum. Our earlier example of being able to hear the shape of a rectangle is a strong example.


Within the last 10 years, Steve Zelditch proved a much better result. We assume the drum head is convex, has no holes, has very smooth boundary7, has at least one mirror symmetry, and a few other smaller technical assumptions.

DifferentDrumsIf we assume the shape follows those rules, Zelditch proved you can hear the shape of a drum. If you know the spectrum, you can reconstruct what shape the drum had.SoundsLikeThat’s it for drums. Next time, we’ll start talking about the limits of proof in mathematics–Gödel’s incompleteness theorem! It may be a meandering path, passing through ideas about how math is done, what axioms are, famous paradoxes in math, and more.

Short Bibliography

  1. This equation is often called the wave equation, since it controls the waves of the string. 
  2. Mathematically, a 100% chance. The probability of choosing a pure tone vibration, or even some finite combination of them, has a 0% chance of happening. 
  3. We are assuming that all strings are the same material, thickness, tautness, etc., so that the only question is what the length of the string is. 
  4. The lowest tone since the lowest eigenvalue \lambda represents the lowest frequency vibration possible, and thus the lowest sounding tone. Though, it should be mentioned one more time that this is assuming all the strings we’re comparing have the same material, thickness, tautness, etc. 
  5. In higher dimensions, we just continue adding second derivatives in each of the directions. 
  6. Though there are very good algorithms for estimating both eigenfunctions and eigenvalues for any shape of drum (and any dimension) on a computer. 
  7. The boundary needs to be analytic, which means you need to be able to take its derivative infinitely many times, and, in addition, the Taylor series for the boundary has to converge. This is a strong condition. For instance, this means there can’t be any corners on the shape. 

Can You Hear the Shape of a Drum? (Part 1)

It’s pretty easy to hear the difference between a big bass drum and a small snare drum. Big drums make low sounds and small drums make high sounds.

Why do bigger drums make lower sounds? Can you tell how big a drum is by what sounds it makes?

Could you possibly even tell the shape of the drum by listening? Can you hear the shape of a drum?


In order to understand vibrating drums, we first need to understand the simpler case of vibrating strings.

Vibrating strings vibrate the air, which is what you perceive as sound. So, in order to understand the sound, we need to understand how strings vibrate.

Let’s pluck a guitar string.


When you pluck a string, it stretches the string out, like this:


When you release, it moves. After a fraction of a second, it might look like this:


The vibrations are controlled by a differential equation, i.e., an equation with derivatives in it. If the height of the string (compared to the base, still position), x away from one end, at time t is given by a function f(x,t), then this differential equation is \dfrac{d^2 f}{dt^2} = \dfrac{d^2 f}{dx^2}.1


At each t, f(x,t) gives the position of the string at each x value. For instance, f(x,0) gives the position of the string at the start, t=0. It represents the initial position of the string.

To explain the equation, \dfrac{d^2 f}{dt^2} is the vertical acceleration of the string. Meanwhile, \dfrac{d^2 f}{dx^2} measures how much the string is curving at one instant of time, as you move from left to right. The more the string is curving, the bigger this is.


In other words, the differential equation tells us that the curvature of the string controls the strings acceleration. In the above example, the string would accelerate downward quickly near the top, where \dfrac{d^2 f}{dx^2} is large (and negative), but wouldn’t accelerate much at all where \dfrac{d^2 f}{dx^2} is small. That seems reasonable.

There are infinitely many solutions to this equation. For instance, a more complicated vibrating string might look like this:2


How could we come up with particularly simple examples?

The simplest kind of vibration would be one where the string keeps its shape, but simply scales that shape up and down. In the simplest case, that would look like this:


For the string to keep its shape, its acceleration needs to be (proportional to) its current height. That way, the string accelerates back to equilibrium faster when it’s higher, and more slowly when it’s already low. In other words, we need \dfrac{d^2 f}{dt^2} = -\nu^2 f, where \nu (Greek letter nu) is some constant.3

Mathematically, if that holds, then it’s easy to check that f(x,t) = f(x,0)\cdot cos(\nu t), where f(x,0) is the position of the string when you first let go. In other words, if our string were in this special self-similar vibration, then the original shape would simply vibrate up and down sinusoidally (i.e., like a (co)sine wave), just like in the video above.

What can these special self-similar vibrations look like?

If we plug \dfrac{d^2 f}{dt^2} = -\nu^2 f back into the equation for a vibrating string, we find that we also need \dfrac{d^2 f}{dx^2}(x,0) = -\nu^2 f(x,0). We need the second derivative of the function to be a multiple of the function itself.

In other words, we need f(x,0) to be an eigenfunction of \dfrac{d^2}{dx^2}!


The first place most students see eigen-stuff, like eigenvectors and eigenvalues, is in a college linear algebra course, but the basic idea is pretty simple.

When you multiply a vector v by a matrix A you end up with a new vector A\cdot v. For instance, \begin{bmatrix}2&1\\0&2\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} =  \begin{bmatrix}2\\0\end{bmatrix}.

A good way to think about this is that the matrix A transforms vectors (like v)  into a new vector, that we call A \cdot v. For most vectors, this transformed vector will have nothing much to do with the original.

But eigenvectors are special; they are the vectors that A transforms in the simplest possible way, by simply scaling them.

In terms of an equation, if you have a matrix A and a vector v so that A \cdot v happens to equal \lambda \cdot v, for some number \lambda, then v is an eigenvector of A, with eigenvalue \lambda. The eigenvalue v is the amount that A scales the eigenvector v.

For example, \begin{bmatrix}2&1\\0&2\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} = 2 \begin{bmatrix}1\\0\end{bmatrix}, and so \begin{bmatrix}1\\0\end{bmatrix} is an eigenvector of the matrix \begin{bmatrix}2&1\\0&2\end{bmatrix}, with eigenvalue 2.


When/if you take linear algebra in college,  you may not realize how important eigen-stuff is. I know that when I took it, the equation Ax=\lambda x seems like some sort of useless, gimmicky math problem of the sort some teachers love.


But it turns out that eigen-stuff is a key step in understanding any number of different ideas in mathematics and physics.

An eigenfunction is similar to an eigenvector.5 We can take a function (instead of a vector) and transform it in some way. In our string example, the transformation is taking two derivatives of the function. For most functions, this gives some function that doesn’t seem to have much to do with the original, like these two functions:


But eigenfunctions are special. They are the functions that, when you take two derivatives, you end up with the same function, perhaps scaled by a constant:


The scaling constant is the eigenvalue for that eigenfunction.

To bring it back to vibrating strings, if we want a self-similar vibrating string, we need to find eigenfunctions and eigenvalues of the transformation \dfrac{d^2}{dx^2}. Equivalently, we need to find solution functions f(x) for the equation \dfrac{d^2 f}{dx^2}(x) = -\nu^2 f(x). (So, the eigenvalue is \lambda = -\nu^2.)

It turns out the only solutions of this equation are pretty simple to write down. If \lambda is positive, the eigenfunctions are the exponential functions, f(x) = e^{\sqrt{\lambda} x} and f(x) = e^{-\sqrt{\lambda} x}.4 (For these, the \lambda =-\nu^2 we’re using isn’t convenient, but, as you’ll see, we don’t need to worry about these.)

If \lambda is negative, so we can use \lambda = -\nu^2, the eigenfunctions are the sine and cosine functions, f(x) = \sin(\nu x) and f(x) = \cos(\nu x).5

Of course, if we have a vibrating string, we have one more restriction.

Any solution should have the two endpoints fixed. Our vibrating string is clamped down at either end, and so shouldn’t be moving at either end.


That means we can easily throw out the exponential eigenfunctions. They’re never zero, so, while they may be eigenfunctions, they can’t represent a vibrating string.

The boundary condition also restricts what sinusoidal functions we could have.

If we call the left end of our string x=0, that means \cos(\nu x) can’t be a vibrating string. So we’re left with sine.

If the two ends of the string were, say, \pi apart, that means our sine wave has to be zero at x=\pi as well.6 For that to happen, we need \nu to be a natural number, like 1 or 2 or 3.


In our example, \nu is conveniently the number of humps, though that’s because we chose the length of our string to be \pi. More generally, though, the larger the \nu, the more oscillations the solution will have.

These eigenfunctions, like \sin(5x), are supposed to represent initial positions for the string. What do they look like as they begin to move?

We already said that for these special self-similar solutions, if we start at f(x,0), then the position over time is f(x,t) = f(x,0) \cos(\nu t). Thus, since our eigenfunctions \sin(\nu x) are supposed to be the initial position of the string (i..e, f(x,0)), the function f(x,t) = \sin(\nu x) \cos(\nu t) is a solution to the differential equation!

What do these solutions look like?

Here are a couple of them.

Clockwise from upper left: \nu = 1, 2, 3 and 7.

We already talked about how larger \nu mean more bumps. But you might also note that they vibrate faster as well. That’s because the \nu shows up in the \cos(\nu t) factor of the solution as well!

In fact, if t was measured in seconds, the string would vibrate once (so, go from its original position back to its original position) in 2\pi/\nu seconds. Equivalently, the frequency of the oscillation is \dfrac{\nu}{2\pi} hertz.

Remember that the vibrating string vibrates the air, and that’s what you hear. So, if the string is vibrating at, say, 261.6 hertz, you would hear the same frequency–a middle C!

These self-similar solutions are the special solutions that give pure tones.

Most solutions, though, aren’t these special, self similar ones. For instance, we could add together the \nu =1 and \nu=2 ones, and we’d get this vibration:


In this case, we’d hear two tones: the one for \nu =1 and the one for \nu =2.

The complicated example I gave at the beginning of the post was some combination of the solutions for \nu =1 through \nu = 6.


In no way is this obvious, but it turns out that every vibration of the string is really just a (perhaps infinite) summation of bits of the pure tone solutions. For instance, if you wanted the vibration for a string that started out in a triangle, you could add together a bit of each of the solutions up to \nu=20 in the right proportions, like this:


When you listen to one of these combined vibrations, there’s usually a main tone that you hear the loudest. But the secondary, usually higher frequency, tones can also be heard. They are the overtones.

Since any vibration is a combination of bits of the pure tone solutions, when you pluck the string, you hear a bit of each of the pure tones your vibration solution included.

In fact, if you just pluck the string at random, you’d probably get a bit of all the tones the string can produce.

Let’s return to our original question: can you hear the shape of a drum?

Strings, though, don’t really have shapes. As long as the string is made of the same material and is tightened to the same tension, the “shape” of a string is just its length.

So, the equivalent question for a string is: can you hear the length of a string?7


The key is the eigenvalues \lambda = -\nu^2 we found earlier. They control the frequency of the sounds the string can produce.

It’s important “all the tones the string can produce” is not the same as “all the tones.” Since every solution to our differential equation (and thus every vibration) is a combination of our pure tone solutions, the string can only produce tones corresponding to the \nu that we found, or, in other words, corresponding to a given eigenvalue \lambda.8

The spectrum (of \dfrac{d^2}{dx^2}) of the string is the set of all the eigenvalues of \dfrac{d^2}{dx^2}. In this case, that would be \{-1^2, -2^2, -3^2, \cdots\}. To summarize, if you know the spectrum of \dfrac{d^2}{dx^2} (i.e., know its eigenvalues), you know the tones the string can make. And similarly, if you know the tones, you know the spectrum.

So, can you hear the length of a string? More technically, if you can hear all the tones, and thus know the spectrum, can you deduce the length of the string?

The answer is yes, but this post is already so long that an explanation will have to wait till next time!


  1. This is a math blog more than a physics blog, so I won’t derive this equation from the physics. The equation assumes that the string isn’t stretched too much and that there’s no friction/air resistance/etc. Also, usually there would be a constant multiplying one of the sides, but let’s say that the strings we’re using are all the same thickness, material, tightness, etc., so that that constant just happens to be 1. The math isn’t any harder, just a bit more ugly. 
  2. The amount the string vibrates up and down is greatly exaggerated from reality, so that it’s easier to see what’s going on. It also vibrates much, much slower. 
  3. It seems weird I use -\nu^2 instead of just a negative number \lambda. But while it looks confusing here, it will avoid a \sqrt{-\lambda} later, so I think it’s worth it. 
  4. Technically, an eigenfunction is an eigenvector. In fact, it’s often just called an eigenvector. This is because you can think of the space of all functions as a vector space; like vectors, you can add and subtract functions, and scale them by a constant.
  5. I promise, these are the simple ones. Most eigenfunctions are so complicated you can’t write down a formula for them. 
  6. Really, these are the same thing, believe it or not. If you allow imaginary numbers and know how to take e to a complex number, then the eigenfunctions are f(x) = e^{\sqrt{\lambda} x} and f(x) = e^{-\sqrt{\lambda} x} for any complex number \lambda. It’s weird, but exponentials of imaginary things give sine waves. But that’s not important to understand for vibrating strings. 
  7. The distance of \pi was chosen just for simplicity of the coefficients. 
  8.  We aren’t playing a guitar, where we use the fret board to momentarily change the length of the string. We are only plucking it. The length of the string is fixed. 
  9. Sound is quantized! In other words, for a given string with given length, you can only produce a discrete set of tones, and not any tone you want. The quantization of energy in quantum mechanics actually comes about in the same way. The energy levels represent eigenvalues for that more complicated system.