The most underappreciated formula | Exploring high-dimensional spheres
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Two Warm-Up Puzzles 0:07
The talk opens with a promise to explain one of the most underappreciated formulas in mathematics, a formula for the volume of high-dimensional spheres. Before getting there, two puzzles are used to set the stage. The first asks: if you pick two random numbers X and Y between -1 and 1, what is the probability that the sum of their squares is less than or equal to one. The audience immediately recognizes this as a question about a circle inside a square, since X squared plus Y squared less than one describes the inside of a unit circle, and the answer becomes the area of that circle divided by the area of the surrounding square, which works out to pi over 4.
Extending Into Higher Dimensions 3:32
The same idea is pushed further. With three random numbers, the question becomes about the volume of a unit sphere sitting inside a cube, using the formula four thirds pi r cubed, giving a probability a little above fifty percent. With four numbers, you would need the volume of a four-dimensional ball, and with one hundred numbers you would need the volume of a one-hundred-dimensional unit ball. Even though these higher-dimensional balls have no literal physical existence, the geometry is treated as genuinely useful, since any list of numbers can be interpreted as a point in a high-dimensional space, which is exactly what happens inside large language models like ChatGPT, Claude, or Gemini, where chunks of text become long lists of numbers understood as points in a high-dimensional space.
A Warning About Intuition 6:30
The second puzzle serves as a caution against trusting geometric intuition once dimensions increase. Start with a 2 by 2 square with a unit circle at each of its four corners, then ask for the radius of a circle centered at the origin that grows until it touches all four corner circles. Using the diagonal of the square, which is the square root of 2, and subtracting the corner circle's radius of 1, the inner circle's radius comes out to about 0.4, which looks perfectly reasonable.
The Same Trick In 3d And 4d 9:01
Moving to three dimensions, the same setup uses eight unit spheres sitting at the corners of a cube, and the inner sphere tangent to all of them has radius root 3 minus 1, about 0.7. The distance rule keeps building on the ordinary two-dimensional Pythagorean theorem rather than introducing any new law of space, since higher-dimensional distance is inherited from all of its two-dimensional subspaces. In four dimensions the corner distance becomes exactly the square root of 4, or 2, meaning the inner ball fits snugly among the sixteen unit balls at the corners of a four-dimensional cube.
The Sphere That Escapes The Box 11:01
Generalizing to n dimensions, the distance to a cube's corner is the square root of n, so the inner sphere's radius is the square root of n minus 1. In ten dimensions that comes to about 2.16, and comparing it to a 4 by 4 by 4 bounding box, whose distance from center to edge is only 2, reveals something strange: the inner sphere is bigger than the bounding box itself and pokes outside it, even though it started out as the small sphere squeezed between the corner spheres.
Spheres Are Round Cubes Are Weird 12:30
This result has led some to say that high-dimensional spheres must be spiky, but that idea is rejected outright, since a sphere is simply defined as every point at a fixed distance from the center, so it is always perfectly round. The real culprit is the cube, whose corners get farther and farther from the center as dimensions increase, because reaching a corner means walking along every one of the many axes in turn, and even taking the diagonal shortcut only buys a square root improvement, not enough to keep pace. The loose mental picture offered is a cube whose corners stretch far out from its center, with unit spheres sitting so far out at those corners that the sphere trying to touch all of them, and there are two to the n corners in n dimensions, has no choice but to swell up enormously.
Building The Volume Chart 15:02
With the warnings delivered, the real subject is introduced: a formula for the volume of a unit ball in any number of dimensions. The familiar building blocks are laid out in a chart, with circumference 2 pi r and area pi r squared in two dimensions, and surface area 4 pi r squared and volume four thirds pi r cubed in three dimensions. The chart labels the bottom row as the ball itself in n dimensions and the top row as its boundary, and the key relationship between the two rows is that the boundary formula is the derivative of the volume formula, reflecting the idea that a tiny increase in radius adds a thin shell of volume equal to the boundary's measure times that tiny thickness, an idea that works precisely as that thickness shrinks toward zero, and that runs in reverse as integration, which is why the r in a boundary formula becomes r squared over two, or r cubed over three, when climbing back down to the volume.
Filling In One Dimension 19:30
To warm up before tackling the more interesting higher dimensions, the chart is extended down to one dimension. A one-dimensional ball is a line segment of all points within distance r of a center, so its volume, using volume in the general sense, is simply 2r. Taking the derivative gives 2, which matches the idea that the boundary of a one-dimensional ball is just two points, so its zero-dimensional volume is naturally understood as a count of two.
Archimedes And The Sphere Surface 21:01
Attention turns to why the sphere's surface area, 4 pi r squared, is exactly four times the circle's area, a coincidence traced back to Archimedes. His method involved projecting small patches of a sphere's surface outward onto an enclosing cylinder, and although this projection would seem to distort area, Archimedes showed that a patch gets stretched in one direction while simultaneously getting squeezed in another, and the two effects cancel exactly. The proof begins by naming the sphere's radius R and the distance from the z-axis to a chosen patch d, then comparing two similar triangles, one from the original patch and one from its projection, whose sides must share the same ratios because the triangles are similar.
Stretch and shrink cancel out 23:30
When you project the sphere's surface onto a surrounding cylinder, each small rectangle gets stretched in one direction by a factor of capital R over little d, and squished in the perpendicular direction by that exact same factor. Because these two effects are inverses of each other, the projection preserves area, which means the sphere's surface area equals the cylinder's surface area.
Unwrapping the cylinder 25:00
A cylinder is easy to measure because you can unwrap it flat into a rectangle. One edge of that rectangle is the circle's circumference, 2 pi R, and the other edge is the sphere's height, 2R. Multiplying them gives 4 pi R squared, the familiar sphere surface area formula, and from there you can integrate or differentiate to move between surface area and volume.
Archimedes as a knight's move 25:30
The real question is how to generalize Archimedes' trick to higher dimensions. The key insight is to see his method as a kind of knight's move: the boundary of a three-dimensional ball can be thought of as the interior of a one-dimensional ball multiplied by the boundary of a two-dimensional ball, a circle. Choosing a coordinate like z first seems to create a dependency, since the size of the remaining circle depends on z, but Archimedes showed you can treat it as if there were no dependency and just multiply the two pieces, because the stretching and shrinking from projection cancel out.
Extending the trick to four dimensions 29:00
The same idea applies one dimension higher: the boundary of a four-dimensional ball behaves like the interior of a two-dimensional ball multiplied by a circle. Even though choosing coordinates z and w first seems to constrain the remaining x and y options, the projection's stretching and squishing again cancel, so you can multiply the disk's area by the circle's circumference, essentially treating the four-dimensional sphere as if it projects onto a solid donut shape.
Volume of the four dimensional ball 32:00
Multiplying pi r squared by 2 pi r gives 2 pi squared r cubed for the four-dimensional sphere's boundary measure, picking up an extra factor of pi. Integrating that expression turns r cubed into r to the fourth divided by four, giving the volume of a four-dimensional ball as pi squared over two times r to the fourth. You can check this isn't just asserted but empirically testable: simulate choosing four random numbers between negative one and one and check how often the sum of their squares is under one, and the result converges to this formula.
Building the general recurrence 33:32
The same knight's move keeps working going up in dimension: the boundary of a five-dimensional ball behaves like the interior of a three-dimensional ball times a circle, and so on. Each step up multiplies by 2 pi r, then integrating divides by the new dimension number n, giving a recurrence relation where the volume constant in dimension n equals the constant two dimensions earlier, multiplied by 2 pi and divided by n. This lets you compute the volume of, say, an eight-dimensional ball by repeatedly stepping back two dimensions at a time to four, then two.
Pushing back to zero dimensions 38:30
Following the recurrence relation backward naturally raises the question of a base case at zero dimensions rather than two. A zero-dimensional unit ball turns out to have volume one, since in zero dimensions there is nothing to move along, just a single point, and its boundary is empty, giving zero. Donald Knuth, responding to this idea at a talk, pointed out you can even verify it by choosing zero random numbers and asking the probability their sum of squares is less than one, which is trivially one.
Reconciling odd and even dimensions 40:00
Writing out the general formula, the volume constant is pi raised to n over two, divided by the factorial of half of n, times r to the n. For odd dimensions this means taking factorials of half-integers like one-half or three-halves, which seems strange, but the logic behind the knight's move and integration never distinguished even from odd, so the same rule should hold for all dimensions. Working out what one-half factorial would need to be, given the known one-dimensional volume of two, gives the value square root of pi divided by two, and from that every other half-integer factorial follows recursively, matching the values given by the gamma function, which is the standard way mathematicians extend the factorial beyond whole numbers.
Volumes rise then shrink 44:31
Plugging in actual numbers reveals something surprising: the volume of a unit ball increases through the low dimensions, reaching about 4.93 in four dimensions, but growth slows by five dimensions and then reverses. By six dimensions the volume is smaller, and it keeps shrinking through seven, eight, nine, and beyond, so that the ten-dimensional unit ball is actually smaller than the two-dimensional one. This unexpected turnaround, rooted entirely in the same recurrence of multiplying by 2 pi and dividing by n, raises the question of why five dimensions marks the peak and what happens as dimensions climb even higher.
Where the growth turns around 47:00
Stepping through the dimensions two at a time, each jump multiplies the previous volume by two pi divided by the new dimension number. Since two pi is bigger than three, four, and five, the volume keeps growing through those steps, but by the time you reach six and seven the denominator starts winning, and growth turns into shrinkage. Five turns out to be the peak because that is the point where the numerator and denominator of this ratio are as evenly matched as possible.
Balls become infinitesimally small 48:01
Push the dimension count up to one hundred and the ball's volume, pi to the fiftieth divided by fifty factorial, works out to about 2.37 times ten to the negative fortieth, which is essentially nothing. This connects to the opening puzzle: pick one hundred random numbers between negative one and one, square them, and add them up, and the chance the sum is under one is tiny. It is tempting to blame the two to the hundredth in the denominator, but the numerator is just as puny, so both sides of the fraction are working against you. This shrinking-volume effect shows up constantly in machine learning, cryptography, and quantum mechanics.
Why the integration factor grows 50:01
The one third that appears in a sphere's volume formula comes from integrating shells proportional to r squared, the same way a pyramid's one third times base times height comes from stacking squares whose sides grow linearly. In higher dimensions this integration factor keeps getting larger, and that is the underlying reason the overall volume keeps shrinking as dimension count rises.
Volume hugs the boundary 51:00
In a two-dimensional circle, shrinking the radius by one percent only removes about two percent of the area, so boundary effects are mild. In ten thousand dimensions, shrinking every one of those directions by one percent multiplies the volume by 0.99 raised to the ten-thousandth power, which comes out to about two times ten to the negative forty-fourth, meaning virtually all the volume sits right at the outer boundary. A related effect concerns the surface itself: in high dimensions almost all of a sphere's surface area clusters near the equator, because moving even slightly away from it involves shrinking across so many directions that the volume there effectively vanishes.
Comparing spheres across dimensions 54:30
Comparing a one-dimensional ball's length to a three-dimensional ball's volume seems like comparing incomparable units, but the numbers pi, four thirds pi, and so on are really just ratios comparing a sphere's volume to that of a unit cube of the same dimension. So the shrinking pattern really says that as dimension count rises, cubes become enormous compared to spheres. Revisiting the earlier corner-sphere puzzle, the eight corner spheres inside a two-by-two-by-two cube exactly cancel out to the plain sphere-to-cube ratio, and in one hundred dimensions those corner spheres take up essentially none of the cube's volume, matching the image of a high-dimensional cube as something spiky, where corners sit far further out than edges.
Closing thoughts and extra material 57:01
The mathematical beauty connecting factorials and circle areas only becomes visible once you step back to view them within a more general formula. After the lecture, an added note points viewers to a virtual career fair at 3b1b.co slash talent, describing the people involved as technically talented, curious, and high agency, and mentioning that past mentions of this fair led to actual hires. Finally, the opening animation showing a circle morphing into lines of latitude on a sphere is explained as the same process extended one dimension further, showing spheres of latitude on a four-dimensional hypersphere projected down into three dimensions, and a related puzzle using this sphere-volume formula appears in an older Numberphile video.
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