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Why the Equilateral Triangle Wins: Maximum Area for a Fixed Perimeter

Suppose you have a fixed length of wire and want to bend it into a triangle. Which triangle encloses the largest possible area?

It is natural to guess that the answer is the equilateral triangle. But why? This is a beautiful example of how a geometric optimization problem can be turned into a multivariable calculus problem and solved using Lagrange multipliers.

Setting up the problem

Let the side lengths of the triangle be
$a, b, c .$
Suppose the perimeter is fixed and equal to $P$ . Thus,
$a + b + c = P .$
We want to determine which values of $a$ , $b$ , and $c$ produce the largest possible area.

Heron’s formula

Let
$s = \frac{P}{2}$
be the semiperimeter. Heron’s formula can be written in squared form as
$A^{2} = s (s - a) (s - b) (s - c) .$
Because the perimeter is fixed, $s$ is also fixed. Moreover, maximizing $A$ is equivalent to maximizing $A^{2}$ . So this form of Heron’s formula is particularly convenient for our problem.

A useful change of variables

Introduce three new variables:
$x = s - a, y = s - b, z = s - c .$
The triangle inequalities imply that $x$ , $y$ , and $z$ are positive.

Adding the three equations gives
$x + y + z = 3 s - (a + b + c) .$
Since
$a + b + c = 2 s,$
we obtain the simple constraint
$x + y + z = s .$
Heron’s formula now becomes
$A^{2} = s x y z .$
Since $s$ is fixed, maximizing the area is equivalent to maximizing
$f (x, y, z) = x y z$
subject to
$x + y + z = s .$
The original geometry problem has therefore become a simple question: among three positive numbers with a fixed sum, when is their product largest?

Using Lagrange multipliers

Define
$f (x, y, z) = x y z$
and let the constraint function be
$g (x, y, z) = x + y + z .$
At a constrained maximum, the gradients of $f$ and $g$ must be parallel:
$\nabla f = λ \nabla g .$
We have
$\nabla f = ⟨ y z, x z, x y ⟩$
and
$\nabla g = ⟨ 1, 1, 1 ⟩ .$
Therefore, the Lagrange multiplier equations are
$y z = λ, x z = λ, x y = λ .$
Thus,
$y z = x z = x y .$
Since $x$ , $y$ , and $z$ are positive, these equations imply
$x = y = z .$
Their sum is $s$ , so
$x = y = z = \frac{s}{3} .$
Returning to the triangle

Recall that
$x = s - a, y = s - b, z = s - c .$
Since $x = y = z$ , we obtain
$a = b = c .$
Because the perimeter is $P$ , each side must therefore have length
$a = b = c = \frac{P}{3} .$
Therefore, the triangle of maximum area is the equilateral triangle.

What is the maximum area?

For an equilateral triangle with side length $\frac{P}{3}$ , the area is
$A_{max} = \frac{\sqrt{3}}{4} {(\frac{P}{3})}^{2} .$
Therefore,
$A_{max} = \frac{\sqrt{3} P^{2}}{36} .$
Why this argument is interesting

We started with a geometric question about triangles. Heron’s formula converted the area problem into an algebraic one. A simple change of variables then transformed it into the problem of maximizing the product of three positive numbers whose sum is fixed.

Lagrange multipliers reveal the symmetry automatically: at the maximum, the three variables must be equal. Translating that condition back into geometry tells us that the three sides of the triangle must also be equal.

This is one of the appealing features of multivariable calculus: a geometric statement that seems intuitively obvious emerges naturally from an optimization calculation.

Conclusion: Among all triangles with a fixed perimeter, the equilateral triangle has the largest area.

For another surprising connection between equilateral-triangle geometry and a classical geometric object, see The Steiner Inellipse and a Surprising Area Characterization .

The equilateral triangle is distinguished by its symmetry. In three dimensions, the regular tetrahedron has similar geometric elegance. Explore its face areas, volume, and a three-dimensional Pythagorean theorem in The Geometry of a Tetrahedron .

September 28, 2026
From Pythagorean Runs to Cubic Runs

The familiar identity
$3^{2} + 4^{2} = 5^{2}$
can be viewed as an equality between two consecutive runs of squares. Surprisingly, this is the first member of a simple infinite family.

Pythagorean runs

This classical family is known as Pythagorean runs. For every positive integer $K$ , consider
$\sum_{j = 0}^{K} {(A + j)}^{2} = \sum_{j = K + 1}^{2 K} {(A + j)}^{2} .$
Thus the first $K + 1$ consecutive squares have the same sum as the next $K$ consecutive squares.

There is exactly one positive value of $A$ for each $K$ :
$A = K (2 K + 1) .$
To see this, move the right-hand side to the left and simplify. The difference factors as
$\sum_{j = 0}^{K} {(A + j)}^{2} - \sum_{j = K + 1}^{2 K} {(A + j)}^{2}$ $= (A + K) (A - K (2 K + 1)) .$
Since $A > 0$ , the factor $A + K$ cannot vanish. Therefore
$A = K (2 K + 1) .$
So there is a Pythagorean run for every $K$ .

The first few are
$3^{2} + 4^{2} = 5^{2},$ $10^{2} + 11^{2} + 12^{2} = 13^{2} + 14^{2},$
and
$21^{2} + 22^{2} + 23^{2} + 24^{2}$ $= 25^{2} + 26^{2} + 27^{2} .$
This naturally raises another question:

What happens if squares are replaced by cubes?

Looking for cubic runs

The most direct analogue is to look for two runs of consecutive cubes having the same sum. In other words, this is the case $d = 1$ , where the terms on the right also differ by 1.

A computer search found no nontrivial examples. This does not prove that none exist, but it suggests looking at the next possibility.

For $d = 2$ , we compare $2 m$ consecutive cubes with $m$ later cubes whose indices differ by 2:
$\sum_{j = 0}^{2 m - 1} {(A + j)}^{3} = \sum_{j = 0}^{m - 1} {(B + 2 j)}^{3},$ $B > A + 2 m - 1 .$
A computer search produced the remarkable example
$\sum_{j = 0}^{175} {(705 + j)}^{3} = \sum_{j = 0}^{87} {(913 + 2 j)}^{3} .$
Written out, this is
$705^{3} + 706^{3} + \dots + 880^{3}$ $= 913^{3} + 915^{3} + \dots + 1087^{3} .$
Both sides equal
$88681384000 .$
Here
$(m, A, B) = (88, 705, 913) .$
At first, such a large numerical identity looks like something that might have occurred by accident. But it is actually the beginning of a much deeper pattern.

A Pell equation appears

Introduce the centered variables
$X = 2 A + 2 m - 1,$ $Y = B + m - 1 .$
For the example above,
$X = 1585, Y = 1000,$
so
$X : Y = 317 : 200 .$
We therefore set
$X = 317 z, Y = 200 z .$
Substitution into the cubic-run equation and simplification lead to
$48329 z^{2} - 156 m^{2} = 161 .$
This is a generalized Pell equation.

Our first cubic run corresponds to
$(z, m) = (5, 88),$
since
$48329 {(5)}^{2} - 156 {(88)}^{2} = 161 .$
The importance of the Pell equation is that one solution need not stand alone. Pell equations can generate further integer solutions, and in this case they produce infinitely many cubic runs.

Thus the huge identity found by computer search is not an isolated numerical curiosity. It belongs to an infinite Diophantine family.

The next cubic run

The next ordered solution is already enormous:
$m = 12005615252944088,$
with
$A = 96105931914993680$
and
$B = 124412740794926913 .$
Therefore the next run can be written as
$S = \sum_{j = 0}^{24011230505888175} {(96105931914993680 + j)}^{3},$ $S = \sum_{j = 0}^{12005615252944087} {(124412740794926913 + 2 j)}^{3} .$
The final cube on the right has index
$148423971300815087 .$
The enormous jump from $m = 88$ to
$m = 12005615252944088$
helps explain why these identities are so difficult to discover by direct search.

The computer found the first example.

The Pell equation explains why there are infinitely many more.

Reference

Michael Boardman, “Proof Without Words: Pythagorean Runs,” Mathematics Magazine, Vol. 73, No. 1 (2000), p. 59.

See also the corresponding sequence and additional historical references in OEIS A059255.

September 27, 2026
The Gaussian Integral and Beyond: From e^(-x²) to a Family of Integrals

Some integrals are easy to write down but surprisingly difficult to evaluate. One of the most famous examples is
$\int_{- \infty}^{\infty} e^{- x^{2}} d x .$
The function $e^{- x^{2}}$ has no elementary antiderivative. Yet the improper integral has the remarkably simple value
$\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π} .$
Even more interesting is the method used to obtain this result. By turning a one-dimensional integral into a two-dimensional one, we can exploit geometry. Once we understand that idea, it leads naturally to integrals involving $e^{- x^{4}}$ , $e^{- x^{p}}$ , and even integrals in which the Gaussian is multiplied by a sine or cosine.

Example 1: The Gaussian integral

Let
$I = \int_{0}^{\infty} e^{- x^{2}} d x .$
Instead of trying to find an antiderivative, square the integral:
$I^{2} = (\int_{0}^{\infty} e^{- x^{2}} d x) (\int_{0}^{\infty} e^{- y^{2}} d y) .$
Thus,
$I^{2} = \int_{0}^{\infty} \int_{0}^{\infty} e^{- (x^{2} + y^{2})} d x d y .$
Now something important has happened. The expression
$x^{2} + y^{2}$
suggests polar coordinates. In the first quadrant,
$x = r \cos (θ), y = r \sin (θ) .$
Here
$0 \leq r < \infty, 0 \leq θ \leq \frac{π}{2} .$
Since
$d x d y = r d r d θ,$
we obtain
$I^{2} = \int_{0}^{\frac{π}{2}} \int_{0}^{\infty} e^{- r^{2}} r d r d θ .$
The radial integral is elementary:
$\int_{0}^{\infty} e^{- r^{2}} r d r = \frac{1}{2} .$
Therefore,
$I^{2} = \frac{π}{2} \cdot \frac{1}{2} = \frac{π}{4},$
and hence
$\int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2} .$
By symmetry,
$\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π} .$
The essential idea was not integration by parts or an ingenious substitution. It was to increase the dimension.

Example 2: Changing the scale

Consider
$\int_{0}^{\infty} e^{- a x^{2}} d x, a > 0 .$
Let
$u = \sqrt{a} x .$
Then
$d x = \frac{d u}{\sqrt{a}},$
so
$\int_{0}^{\infty} e^{- a x^{2}} d x = \frac{1}{\sqrt{a}} \int_{0}^{\infty} e^{- u^{2}} d u = \frac{\sqrt{π}}{2 \sqrt{a}} .$
One geometric calculation has already produced an entire family of integrals.

Example 3: An integral involving $e^{- x^{4}}$

First consider
$\int_{0}^{\infty} x^{3} e^{- x^{4}} d x .$
The substitution $u = x^{4}$ works immediately because
$d u = 4 x^{3} d x .$
Hence,
$\int_{0}^{\infty} x^{3} e^{- x^{4}} d x = \frac{1}{4} \int_{0}^{\infty} e^{- u} d u = \frac{1}{4} .$
Now remove the factor $x^{3}$ .

Example 4: What about $e^{- x^{4}}$ itself?

Consider
$J = \int_{0}^{\infty} e^{- x^{4}} d x .$
Let $u = x^{4}$ . Then
$d x = \frac{1}{4} u^{- \frac{3}{4}} d u,$
and therefore
$J = \frac{1}{4} \int_{0}^{\infty} u^{- \frac{3}{4}} e^{- u} d u .$
This integral is not elementary, but it is a standard special function. The Gamma function is defined, for $s > 0$ , by
$Γ (s) = \int_{0}^{\infty} u^{s - 1} e^{- u} d u .$
Therefore,
$\int_{0}^{\infty} e^{- x^{4}} d x = \frac{1}{4} Γ (\frac{1}{4}) .$
The Gaussian was already hiding the Gamma function

Apply the same substitution to the Gaussian integral. With $u = x^{2}$ ,
$\int_{0}^{\infty} e^{- x^{2}} d x = \frac{1}{2} Γ (\frac{1}{2}) .$
But our two-dimensional calculation showed that the same integral equals $\frac{\sqrt{π}}{2}$ . Consequently,
$Γ (\frac{1}{2}) = \sqrt{π} .$
From $e^{- x^{2}}$ to $e^{- x^{p}}$

Now consider the general integral
$I_{p} = \int_{0}^{\infty} e^{- x^{p}} d x, p > 0 .$
Set $u = x^{p}$ . Then
$d x = \frac{1}{p} u^{\frac{1}{p} - 1} d u .$
Therefore,
$I_{p} = \frac{1}{p} \int_{0}^{\infty} u^{\frac{1}{p} - 1} e^{- u} d u .$
By the definition of the Gamma function,
$\int_{0}^{\infty} e^{- x^{p}} d x = \frac{1}{p} Γ (\frac{1}{p}) .$
Using the identity $Γ (s + 1) = s Γ (s)$ , we can also write
$\int_{0}^{\infty} e^{- x^{p}} d x = Γ (1 + \frac{1}{p}) .$
For example,
$\begin{array}{l} \int_{0}^{\infty} e^{- x} d x = 1, \\ \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2}, \\ \int_{0}^{\infty} e^{- x^{3}} d x = \frac{1}{3} Γ (\frac{1}{3}), \\ \int_{0}^{\infty} e^{- x^{4}} d x = \frac{1}{4} Γ (\frac{1}{4}) . \end{array}$
An even larger family

We can include a power of $x$ . Consider
$\int_{0}^{\infty} x^{q} e^{- x^{p}} d x,$
where $p > 0$ and $q > - 1$ . Again let $u = x^{p}$ . The same substitution gives
$\int_{0}^{\infty} x^{q} e^{- x^{p}} d x = \frac{1}{p} Γ (\frac{q + 1}{p}) .$
For example,
$\int_{0}^{\infty} x e^{- x^{4}} d x = \frac{1}{4} Γ (\frac{1}{2}) = \frac{\sqrt{π}}{4},$
while
$\int_{0}^{\infty} x^{3} e^{- x^{4}} d x = \frac{1}{4} Γ (1) = \frac{1}{4} .$
Thus three very similar-looking integrals can have rather different-looking answers:
$\begin{array}{l} \int_{0}^{\infty} e^{- x^{4}} d x = \frac{1}{4} Γ (\frac{1}{4}), \\ \int_{0}^{\infty} x e^{- x^{4}} d x = \frac{\sqrt{π}}{4}, \\ \int_{0}^{\infty} x^{3} e^{- x^{4}} d x = \frac{1}{4} . \end{array}$
Why did circles appear?

There is a geometric reason the Gaussian calculation worked so beautifully. When we squared the Gaussian integral, the exponent became
$x^{2} + y^{2} .$
The level curves
$x^{2} + y^{2} = c$
are circles. Polar coordinates are therefore perfectly adapted to the problem.

If instead we square $\int_{0}^{\infty} e^{- x^{4}} d x$ , we obtain
$\int_{0}^{\infty} \int_{0}^{\infty} e^{- (x^{4} + y^{4})} d x d y .$
The corresponding level curves are
$x^{4} + y^{4} = c .$
They are not circles. More generally, $e^{- x^{p}}$ is naturally connected with regions of the form
${| x |}^{p} + {| y |}^{p} \leq r^{p} .$
For $p = 2$ , these are ordinary disks. For larger values of $p$ , their boundaries become increasingly square-like. The Gaussian is the particularly beautiful case in which the geometry becomes ordinary Euclidean geometry.

A surprising turn: add a cosine

Consider
$C (b) = \int_{0}^{\infty} e^{- x^{2}} \cos (b x) d x .$
The answer is
$C (b) = \frac{\sqrt{π}}{2} e^{- \frac{b^{2}}{4}} .$
This is remarkable: multiplying a Gaussian by an oscillating cosine produces another Gaussian, now as a function of the parameter $b$ .

Here is a calculus derivation. Differentiate with respect to $b$ :
$C^{'} (b) = - \int_{0}^{\infty} x e^{- x^{2}} \sin (b x) d x .$
Since
$x e^{- x^{2}} = - \frac{1}{2} \frac{d}{d x} (e^{- x^{2}}),$
integration by parts gives
$C^{'} (b) = - \frac{b}{2} C (b) .$
Therefore,
$\frac{C^{'} (b)}{C (b)} = - \frac{b}{2} .$
Integrating gives
$C (b) = A e^{- \frac{b^{2}}{4}} .$
At $b = 0$ ,
$C (0) = \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2} .$
Thus,
$\int_{0}^{\infty} e^{- x^{2}} \cos (b x) d x = \frac{\sqrt{π}}{2} e^{- \frac{b^{2}}{4}} .$
For example, taking $b = 2$ gives
$\int_{0}^{\infty} e^{- x^{2}} \cos (2 x) d x = \frac{\sqrt{π}}{2 e} .$
What happens with sine?

Now consider
$S (b) = \int_{0}^{\infty} e^{- x^{2}} \sin (b x) d x .$
Unlike the cosine integral, this does not reduce to an elementary expression involving only exponentials and $π$ . It can be written using a special function called Dawson’s integral,
$F (z) = e^{- z^{2}} \int_{0}^{z} e^{t^{2}} d t .$
The result is
$\int_{0}^{\infty} e^{- x^{2}} \sin (b x) d x = F (\frac{b}{2}) .$
The difference between sine and cosine has a simple symmetry explanation. The function
$e^{- x^{2}} \cos (b x)$
is even, while
$e^{- x^{2}} \sin (b x)$
is odd. Therefore, over the entire real line,
$\int_{- \infty}^{\infty} e^{- x^{2}} \sin (b x) d x = 0,$
whereas
$\int_{- \infty}^{\infty} e^{- x^{2}} \cos (b x) d x = \sqrt{π} e^{- \frac{b^{2}}{4}} .$
One function keeps returning

We started with $e^{- x^{2}}$ . It has no elementary antiderivative, so at first it seems difficult to work with. But instead of disappearing, the Gaussian keeps returning.

Geometry gives
$\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π} .$
The Gamma function places it inside the larger family
$\int_{0}^{\infty} e^{- x^{p}} d x = Γ (1 + \frac{1}{p}) .$
Adding a power of $x$ produces
$\int_{0}^{\infty} x^{q} e^{- x^{p}} d x = \frac{1}{p} Γ (\frac{q + 1}{p}) .$
And adding an oscillating cosine gives another Gaussian:
$\int_{- \infty}^{\infty} e^{- x^{2}} \cos (b x) d x = \sqrt{π} e^{- \frac{b^{2}}{4}} .$
This last identity is a glimpse of a much deeper fact: under the Fourier transform, the Gaussian essentially transforms into itself.

So a single integral that cannot be evaluated by ordinary antiderivatives opens the door to geometry, the Gamma function, differential equations, generalized $L^{p}$ geometry, and Fourier analysis.

Sometimes an integral becomes easier not by finding a better antiderivative, but by finding a larger mathematical structure around it.

From one dimension to two: squaring the Gaussian integral reveals circular level curves, making polar coordinates the natural choice.

Another Famous Improper Integral

The Gaussian integral shows how an integral over an infinite interval can be evaluated by introducing an extra dimension and exploiting symmetry. Another celebrated improper integral has a very different appearance:

$\int_{0}^{\infty} \frac{\sin (a x)}{x} d x = \frac{π}{2}, a > 0 .$

What is especially surprising is that the answer does not depend on the positive parameter $a$ .

Continue exploring: A Surprising Improper Integral: Why the Integral of sin(ax)/x Is Always π/2

Another Problem Where Two Dimensions Help

The Gaussian integral becomes manageable after a surprising change of viewpoint: instead of attacking a one-dimensional integral directly, we square it, create a double integral, and use two-dimensional geometry.

The same general idea appears in a completely different problem. The famous series $1 + \frac{1}{4} + \frac{1}{9} + \dots$ can also be approached through a double integral.

Continue exploring: Proving 1 + 1/4 + 1/9 + ⋯ = π²/6 with a Double Integral

From the Gaussian Integral to a Real Signal

The Gaussian function is much more than an elegant calculus example. Gaussian distributions arise naturally when many small independent effects are added together, which makes them fundamental in probability, statistics, physics, and engineering.

A particularly interesting example appears in OFDM communication signals. The in-phase and quadrature components become approximately Gaussian, but the signal magnitude follows a different distribution: the Rayleigh distribution.

Continue exploring: Why Does an OFDM Signal Have a Rayleigh Distribution?

September 27, 2026
How High Should the Compression Ratio of a Gasoline Engine Be?

A gasoline engine becomes more efficient when its compression ratio is increased. So why not simply make the compression ratio as large as possible?

There is a physical obstacle: increasing the compression ratio also increases the pressure inside the cylinder. An engine can withstand only a limited pressure.

This gives us a natural optimization problem:

For a fixed amount of heat released during combustion and a fixed maximum allowable cylinder pressure, what compression ratio gives the greatest possible efficiency?

The answer comes from combining a simple model of a gasoline engine with calculus.

The ideal Otto cycle

We use the ideal Otto cycle, the standard simplified model for a spark-ignition gasoline engine.

Let
$r = \frac{V_{1}}{V_{2}}$
be the compression ratio, where $V_{1}$ is the cylinder volume before compression and $V_{2}$ is the volume after compression.

Let $P_{1}$ and $T_{1}$ be the initial pressure and temperature.

For an ideal gas undergoing adiabatic compression,
$T_{2} = T_{1} r^{γ - 1}$
and
$P_{2} = P_{1} r^{γ} .$
Here
$γ = \frac{c_{p}}{c_{v}},$
and for air we use the familiar approximation
$γ \approx 1.4 .$
Efficiency increases with compression

The thermal efficiency of the ideal Otto cycle is
$η (r) = 1 - \frac{1}{r^{γ - 1}} .$
Differentiate:
$η^{'} (r) = (γ - 1) r^{- γ} .$
Since $r > 1$ and $γ > 1$ , we have
$η^{'} (r) > 0 .$
Thus, according to the ideal model, efficiency always increases as the compression ratio increases.

So there is no unconstrained maximum. Mathematics would simply tell us to keep increasing $r$ .

A real engine, however, cannot withstand unlimited pressure. This is where the optimization problem becomes interesting.

Adding a pressure constraint

Suppose combustion adds a fixed amount of heat $q$ per unit mass of air.

In the ideal Otto model, heat is added at constant volume. Therefore,
$q = c_{v} (T_{3} - T_{2}) .$
Hence
$T_{3} = T_{2} + \frac{q}{c_{v}} .$
Because the volume does not change during combustion, the ideal-gas law gives
$\frac{P_{3}}{P_{2}} = \frac{T_{3}}{T_{2}} .$
Therefore,
$P_{3} = P_{2} (1 + \frac{q}{c_{v} T_{2}}) .$
Now substitute
$P_{2} = P_{1} r^{γ}$
and
$T_{2} = T_{1} r^{γ - 1} .$
We obtain
$P_{3} = P_{1} (r^{γ} + \frac{q}{c_{v} T_{1}} r) .$
The key equation

Define
$B = \frac{q}{c_{v} T_{1}} .$
Then the maximum pressure reached during the idealized cycle is
$P_{3} = P_{1} (r^{γ} + B r) .$
Suppose the engine can safely withstand a maximum cylinder pressure $P_{max}$ . Then
$P_{3} \leq P_{max} .$
Define
$A = \frac{P_{max}}{P_{1}} .$
The pressure constraint becomes
$r^{γ} + B r \leq A .$
Where does the maximum occur?

We already proved that the efficiency $η (r)$ is increasing.

Therefore, the most efficient engine uses the largest compression ratio permitted by the pressure constraint.

The optimum must occur when the pressure reaches its allowable maximum:
$r^{γ} + B r = A .$
This is an interesting kind of optimization problem. We do not find the optimum by solving $η^{'} (r) = 0$ . There is no critical point.

Instead, calculus tells us that efficiency is increasing, and the physical constraint tells us where we must stop.

A numerical example

Take
$P_{1} = 100 kPa, T_{1} = 300 K .$
Use
$c_{v} = 0.718 kJ/(kg K)$
and suppose combustion supplies
$q = 1800 kJ/kg .$
Then
$B = \frac{1800}{(0.718) (300)} \approx 8.36 .$
Suppose the maximum allowable cylinder pressure is
$P_{max} = 10 MPa = 10000 kPa .$
Therefore,
$A = \frac{10000}{100} = 100 .$
Using $γ = 1.4$ , the optimal compression ratio satisfies
$r^{1.4} + 8.36 r = 100 .$
Solving this equation numerically gives
$r \approx 9.27 .$
Thus, in this simplified model, the greatest possible efficiency under the pressure restriction occurs at a compression ratio of approximately 9.27:1.

What efficiency does this give?

For $γ = 1.4$ , the ideal Otto-cycle efficiency is
$η = 1 - \frac{1}{r^{0.4}} .$
Using $r \approx 9.27$ ,
$η \approx 1 - \frac{1}{{9.27}^{0.4}} \approx 0.590 .$
So the theoretical efficiency is approximately
$η \approx 59.0 % .$
This is the efficiency of the idealized mathematical model, not the efficiency we should expect from a real gasoline engine. Real engines have friction, heat loss, pumping losses, finite combustion time, changing specific heats, and other effects that the ideal Otto cycle does not include.

An unexpected seventh-degree polynomial

There is one more mathematical surprise.

We used
$γ = 1.4 = \frac{7}{5} .$
Therefore the equation determining the optimal compression ratio has the form
$r^{\frac{7}{5}} + B r = A .$
Let
$x = r^{\frac{1}{5}} .$
Then
$r = x^{5}$
and
$r^{\frac{7}{5}} = x^{7} .$
So our engine-design equation becomes
$x^{7} + B x^{5} - A = 0 .$
For our numerical example,
$x^{7} + 8.36 x^{5} - 100 = 0 .$
A practical question about the design of a gasoline engine has led us to a seventh-degree polynomial.

We do not need to solve this polynomial symbolically. A numerical method gives the physically relevant positive root and therefore the optimal compression ratio.

The mathematical lesson

Without a pressure restriction, the ideal Otto model says

larger compression ratio → greater efficiency.

There is no finite optimum.

But an engine has to withstand the pressure produced inside its cylinder. Once we impose the constraint
$P_{3} \leq P_{max},$
the optimization problem has a finite solution.

The optimum occurs precisely when increasing the compression ratio any further would violate the pressure constraint:
$P_{3} = P_{max} .$
This illustrates an important idea in applied calculus: sometimes the optimum is not created by a critical point of the function—it is created by the constraint.

Graph showing thermal efficiency and peak cylinder pressure versus compression ratio, with the optimal compression ratio of 9.27 determined by the 10 MPa pressure constraint.

September 27, 2026
When Does x cos(x) Take the Same Value Twice?

Consider the function
$f (x) = x \cos (x)$
on the interval
$0 \leq x \leq \frac{π}{2} .$
At both ends of the interval the function is zero:
$f (0) = f (\frac{π}{2}) = 0 .$
Between these endpoints, the function rises to a single maximum and then falls back to zero. This means that every value strictly between zero and the maximum is attained at exactly two points.

Why is there only one maximum?

Differentiate:
$f^{'} (x) = \cos (x) - x \sin (x) .$
At an interior critical point,
$\cos (x) = x \sin (x),$
or equivalently,
$x \tan (x) = 1 .$
The function x tan(x) is strictly increasing on $(0, \frac{π}{2})$ because
$\frac{d}{d x} [x \tan (x)] = \tan (x) + x \sec^{2} (x) > 0 .$
Therefore the equation $x \tan (x) = 1$ has exactly one solution. Numerically,
$x \approx 0.8603335890,$
and the maximum value is approximately
$f_{\max} \approx 0.5610963382 .$
The main idea

Usually, finding the two points at which $x \cos (x)$ has the same value leads to a transcendental equation. But something interesting happens if the second point is an integer multiple of the first.

Suppose the two points are $x$ and $n x$ , where $n > 1$ is an integer. To keep both points inside the interval, we require
$0 < x < \frac{π}{2 n} .$
We want
$f (x) = f (n x) .$
Thus
$x \cos (x) = n x \cos (n x) .$
Since $x > 0$ , we can divide by x:
$\cos (x) = n \cos (n x) .$
This is where Chebyshev polynomials enter the problem.

What is a Chebyshev polynomial?

The Chebyshev polynomial of the first kind, denoted by $T_{n}$ , is defined by the identity
$T_{n} (\cos θ) = \cos (n θ) .$
For example,
$T_{2} (c) = 2 c^{2} - 1,$ $T_{3} (c) = 4 c^{3} - 3 c,$
and
$T_{5} (c) = 16 c^{5} - 20 c^{3} + 5 c .$
Now set
$c = \cos (x) .$
Then
$\cos (n x) = T_{n} (c),$
so our transcendental equation becomes the algebraic equation
$c = n T_{n} (c) .$
This is the key observation: a question about two equal values of a transcendental function has turned into a polynomial equation.

Case 1: n = 2

We have
$c = 2 (2 c^{2} - 1) .$
Therefore
$4 c^{2} - c - 2 = 0 .$
The root satisfying the required interval condition is
$c = \frac{1 + \sqrt{33}}{8} .$
Hence
$x = \arccos (\frac{1 + \sqrt{33}}{8}) .$
The two different inputs $x$ and $2 x$ therefore give exactly the same value of $x \cos (x)$ .

Case 2: n = 3

Now
$c = 3 (4 c^{3} - 3 c) .$
Since c is positive, division by c gives
$1 = 12 c^{2} - 9 .$
Thus
$c^{2} = \frac{5}{6},$
and therefore
$x = \arccos (\sqrt{\frac{5}{6}}) .$
So x and $3 x$ give another exact pair with the same value of the function.

Case 3: n = 5

Using
$T_{5} (c) = 16 c^{5} - 20 c^{3} + 5 c,$
the equation $c = 5 T_{5} (c)$ becomes, after dividing by c,
$20 c^{4} - 25 c^{2} + 6 = 0 .$
Setting $u = c^{2}$ gives
$20 u^{2} - 25 u + 6 = 0 .$
Therefore
$u = \frac{25 \pm \sqrt{145}}{40} .$
However, not both algebraic roots correspond to our original problem. We need
$0 < x < \frac{π}{10},$
so $c = \cos (x) > \cos (\frac{π}{10}) .$ Only the larger root satisfies this condition. Hence
$c = \sqrt{\frac{25 + \sqrt{145}}{40}},$
and
$x = \arccos (\sqrt{\frac{25 + \sqrt{145}}{40}}) .$
Thus x and $5 x$ form a third exact pair.

What makes this interesting?

The graph tells us immediately that $x \cos (x)$ takes most of its values twice. But the locations of those two points are usually not available in closed form.

Requiring the two inputs to have the special form x and nx changes the problem completely. The multiple-angle identity
$\cos (n x) = T_{n} (\cos (x))$
turns the transcendental equation into a polynomial equation. For $n = 2, 3, 5$ , that polynomial equation gives particularly clean exact answers.

A Related Same-Value Problem

The function $x \cos (x)$ leads to one kind of same-value problem. A surprisingly different example appears when we ask the same question about $x^{1 / x}$ .

Continue exploring: When Does x^(1/x) Take the Same Value Twice?

September 27, 2026
A Surprising Integral on the Sphere: Why Every Direction Is the Same
Consider the integral
I ( u ) = ∫ S n − 1 | u · x | d S ( x )
where $S^{n - 1}$ is the unit sphere in $R^{n}$ and $u$ is a fixed vector. At first glance, this looks like a difficult high-dimensional integral. The absolute value creates a nonsmooth integrand, but the symmetry of the sphere makes the calculation surprisingly simple.

The key observation

Write
u = ∥ u ∥ e
where $e$ is a unit vector. Then
| u · x | = ∥ u ∥ | e · x |
Therefore,
I ( u ) = ∥ u ∥ ∫ S n − 1 | e · x | d S ( x )
The remaining integral does not depend on the direction of $e$ . The sphere is rotationally symmetric, so we may rotate the coordinate system and assume that
e = ( 1 , 0 , … , 0 )
Then
e · x = x 1
and therefore
I ( u ) = ∥ u ∥ ∫ S n − 1 | x 1 | d S ( x )
A geometric interpretation

For a point $x$ on the sphere, let $θ$ be the angle between $x$ and the chosen direction $e$ . Then the projection of $x$ onto this direction is
e · x = cos ( θ )
so
| u · x | = ∥ u ∥ | cos ( θ ) |
The integral is therefore the total absolute projection of all points on the sphere onto a fixed direction.

Measuring spheres

The notation $| S^{n} |$ means the surface area of the unit sphere $S^{n}$ .

For example,
- $S^{0}$ consists of two points, so $| S^{0} | = 2$ .
- $S^{1}$ is the unit circle, so $| S^{1} | = 2 π$ .
- $S^{2}$ is the ordinary unit sphere, so $| S^{2} | = 4 π$ .
Now consider the sphere $S^{n - 1}$ . Fix the angle $θ$ between a point $x$ on this sphere and a fixed direction $e$ .

All points with the same angle $θ$ form a lower-dimensional sphere $S^{n - 2}$ . Therefore, $| S^{n - 2} |$ is the surface area of this slice of the sphere.

This is the geometric reason that $| S^{n - 2} |$ appears when we compute the integral using spherical coordinates.

The lower-dimensional sphere

To compute the integral, we slice the sphere by fixing the angle $θ$ . Each slice is itself a sphere of one lower dimension.

The notation
| S n − 2 |
means the surface area of the unit sphere $S^{n - 2}$ one dimension lower. For example,
- $| S^{0} | = 2$ , because it consists of two points;
- $| S^{1} | = 2 π$ , because it is the unit circle;
- $| S^{2} | = 4 π$ , because it is the usual sphere.
Using spherical coordinates, the surface element becomes
dS = sin ( θ ) n − 2 dθ d S n − 2
Therefore,
∫ S n − 1 | x 1 | dS = 2 | S n − 2 | ∫ 0 π/2 cos ( θ ) sin ( θ ) n − 2 dθ
Finishing the computation

The remaining one-dimensional integral is elementary. Let
y = sin ( θ )
so that
dy = cos ( θ ) dθ
Therefore,
∫ 0 π/2 cos ( θ ) sin ( θ ) n − 2 dθ = ∫ 0 1 y n − 2 dy = 1 n − 1
Substituting this result gives
∫ S n − 1 | x 1 | dS = 2 | S n − 2 | n − 1
Finally,
∫ S n − 1 | u · x | dS = 2 | S n − 2 | n − 1 ‖ u ‖
Examples

The formula becomes especially simple in low dimensions.

The circle $S^{1}$

For the unit circle we have $n = 2$ . The lower-dimensional sphere is
S 0
which consists of two points, so
| S 0 | = 2
Therefore,
∫ S 1 | u · x | d S = 2 · 2 1 ‖ u ‖ = 4 ‖ u ‖
The sphere $S^{2}$

For the ordinary unit sphere in $R^{3}$ we have
n = 3
and the lower-dimensional sphere is the unit circle:
| S 1 | = 2 π
Hence,
∫ S 2 | u · x | d S = 2 · 2 π 2 ‖ u ‖ = 2 π ‖ u ‖
In both examples, the direction of $u$ does not matter. Only its length remains. This is a direct consequence of the rotational symmetry of the sphere.

The main idea

The calculation is simple because the sphere has no preferred direction. A rotation can move any vector $u$ to a coordinate axis without changing the geometry of the sphere.

Therefore, the integral depends only on the length of the vector:
∫ S n − 1 | u · x | d S = C ( n ) ‖ u ‖
where $C (n)$ is a constant that depends only on the dimension.

The important lesson is not the integration itself, but the symmetry behind it: whenever a problem on a sphere involves a single fixed vector, the first question should be whether a rotation can remove the direction completely.
September 26, 2026

Gabriel’s Horn: When Can an Infinite Horn Be Painted?

Can a shape stretch forever, hold a finite amount of liquid, and still have an infinite surface to paint? Gabriel’s horn does exactly that. The more interesting question is what can happen for other horns. Can we classify every possibility?

Illustration of Gabriel’s horn, formed by rotating y equals one over x for x at least one about the x-axis. Its circular opening is wide at x equals one; the radius narrows as the horn continues indefinitely to the right. — Gabriel’s horn is generated by rotating the graph of y = 1/x for x ≥ 1 about the x-axis. The illustration shows only a finite portion.

The familiar paradox

At position x, the horn has radius 1/x. The disk method gives its volume:

$V=\pi\int_1^\infty\frac{dx}{x^2}=\pi$

But the lateral surface area satisfies

$S = 2 π \int_{1}^{\infty} \frac{1}{x} \sqrt{1 + \frac{1}{x^{4}}} d x$ $S \geq 2 π \int_{1}^{\infty} \frac{1}{x} d x = \infty$

So the horn can be filled with π cubic units, although painting its entire outside would require infinite area. This is the usual painter’s paradox. [1, 2]

A general horn

Now let f be a nonnegative continuously differentiable function on [a, ∞), and rotate the region under its graph about the x-axis. “Painting” means covering the curved lateral surface; including the single disk at x = a changes no finite-versus-infinite result. The familiar formulas are [1]

$V = π \int_{a}^{\infty} {f (x)}^{2} d x$ $S = 2 π \int_{a}^{\infty} f (x) \sqrt{1 + {f^{'} (x)}^{2}} d x$

To determine whether the second integral converges, use the elementary inequality

$\max\{1,|v|\}\le\sqrt{1+v^2}\le1+|v|$

Apply it with v = f′(x), multiply by f(x) ≥ 0, and integrate. We obtain a useful if and only if statement:

$S < \infty \Leftrightarrow \int_{a}^{\infty} f (x) d x < \infty$ $and \int_{a}^{\infty} f (x) | f^{'} (x) | d x < \infty$

The first integral measures the radii accumulated along the axis. The second detects rapid changes in the radius. Both must be finite to paint the horn.

The four possibilities

Volume	Lateral area	Condition
Finite	Finite	Both integrals in the painting test converge.
Finite	Infinite	The integral of f² converges, but at least one integral in the painting test diverges.
Infinite	Finite	Impossible.
Infinite	Infinite	The integral of f² diverges.

Why is the third row impossible? Since the derivative of f² is 2ff′, the painting test gives the bound

${f (x)}^{2} \leq {f (a)}^{2} + 2 \int_{a}^{\infty} f (x) | f^{'} (x) | d x$

Thus a paintable horn has a bounded radius. If M is an upper bound for f, then f² ≤ Mf. The painting test also says the integral of f is finite, so the integral of f² is finite. In short: every paintable horn is fillable. The converse fails, as Gabriel’s horn shows. The implication is also noted in a Calculus II laboratory abstract by Royer. [3]

An entire family of examples

Consider

$f(x)=x^{-p},\qquad x\ge1,\quad p>0$

Here f(x) = x^−p decreases from f(1) = 1 toward 0. Its derivative is negative, so |f′(x)| = −f′(x). Thus the second integral in the painting test is

\int_{1}^{\infty} f (x) | f^{'} (x) | d x = - \int_{1}^{\infty} f (x) f^{'} (x) d x

Now (f(x)²)′ = 2f(x)f′(x). Integrating the expression on the right gives

- \int_{1}^{\infty} f (x) f^{'} (x) d x = \frac{{f (1)}^{2} - \lim_{x \to \infty} {f (x)}^{2}}{2} = \frac{1}{2} .

Therefore the second painting integral is finite for every p > 0. The volume is finite when 2p > 1, while the surface area is finite when p > 1.

Exponent	What happens?
p > 1	Both fillable and paintable.
1/2 < p ≤ 1	Fillable but not paintable; p = 1 is Gabriel’s horn.
0 < p ≤ 1/2	Neither fillable nor paintable.

What if the horn wiggles?

For a decreasing graph, the integral involving |f′| needs no separate test. For an oscillating graph, it really matters. Consider the smooth positive function

$f(x)=\frac{2+\sin(x^4)}{(1+x)^2}$

The function is bounded above by 3/(1+x)², so both the integral of f and the volume integral of f² converge. Yet the oscillations become increasingly rapid. The term from differentiating sin(x⁴) makes f|f′| comparable, up to an integrable error, to |cos(x⁴)|/x. Substituting u = x⁴ shows that its integral diverges like the integral of |cos u|/u. Therefore this horn is fillable but not paintable even though the integral of its radii is finite.

Gabriel’s horn fails the painting test because it shrinks too slowly. The oscillating horn fails because its surface becomes too corrugated. The two-integral criterion catches both.

References

Gilbert Strang and Edwin “Jed” Herman, Calculus Volume 1, OpenStax (2016), §6.2, “Determining Volumes by Slicing” and §6.4, “Arc Length of a Curve and Surface Area.”
APEX Calculus, §7.4, “Arc Length and Surface Area,” Example 7.4.18 (Gabriel’s horn).
Melvin G. Royer, “Gabriel’s Other Equipment,” abstract, Joint Mathematics Meetings (2010).

Another Surprise from an Improper Integral

Gabriel’s Horn shows how an improper integral can produce a result that seems geometrically impossible: a solid can have finite volume while its surface area is infinite.

Improper integrals contain other surprises as well. One of the most famous is an oscillating integral involving $\frac{\sin (a x)}{x}$ , whose value turns out to be remarkably independent of $a$ when $a > 0$ .

Continue exploring: A Surprising Improper Integral: Why the Integral of sin(ax)/x Is Always π/2

September 24, 2026

Can a Circle Cut a Triangle in Half?

Orange is inside the circle; blue is outside. Each occupies half the triangle.

Take a triangle $ABC$ . Place a compass point at $A$ , draw a circle, and adjust its radius until half the triangle lies inside the circle. What does the answer look like?

For an equilateral triangle of side $1$ , the calculation is pleasantly short. The portion inside the circle is a sector with angle $\pi/3$ . Since the triangle has area $\sqrt3/4$ , we want

$\frac{\pi r^2}{6}=\frac{\sqrt3}{8}, \qquad\text{hence}\qquad r=\sqrt{\frac{3\sqrt3}{4\pi}}.$

The circle has not yet reached the opposite side $BC$ , so the sector calculation really does describe the part of the triangle inside the circle. A vertex-centered circular arc bisecting an equilateral triangle also appears in Sanjoy Mahajan’s 2008 MIT course materials.

Now change the triangle. The circle might meet $BC$ before it captures half the area. Or it might grow past one of the other vertices. The figure above shows the three possible pictures.

A radius always exists

As the radius increases from $0$ , the area of the triangle inside the circle increases continuously from $0$ to the area of the whole triangle. It increases strictly until the farther vertex is reached. Therefore exactly one radius divides the triangle into equal areas.

We can scale the triangle without changing which of the three pictures occurs. To simplify the calculations, label the vertices so that $AC\le AB$ , and set $AC=1$ . Write $A,B,C$ for the angles, measured in radians. By the sine rule,

$AB=L=\frac{\sin C}{\sin B}, \qquad T=\operatorname{area}(ABC)=\frac{L\sin A}{2}.$

The desired area inside the circle is $T/2$ .

Case 1: A sector is enough

While the circle remains inside the triangle’s angle at $A$ , its area inside the triangle is $Ar^2/2$ . The candidate radius is thus

$\boxed{r_0=\sqrt{\frac{T}{A}}=\sqrt{\frac{L\sin A}{2A}}.}$

We still need to check whether the circle reaches $BC$ . If both base angles are acute, the distance from $A$ to $BC$ is $h=\sin C$ , so this is the answer precisely when $r_0\le h$ . The test can be written entirely in angles:

$\boxed{\cot B+\cot C\le 2A.}$

If $C$ is right or obtuse, the nearest point of the segment $BC$ to $A$ is $C$ . In that situation the sector answer works when $r_0\le AC=1$ .

Case 2: The circle crosses $BC$ twice

Suppose both base angles are acute, $h<r<1$ , and the circle cuts $BC$ at two points. Start with a sector and subtract the circular segment lying beyond $BC$ :

$S(r)=\frac{Ar^2}{2} -r^2\arccos\!\left(\frac{h}{r}\right) +h\sqrt{r^2-h^2}, \qquad h=\sin C.$

The equal-area radius is the unique solution of $S(r)=T/2$ in $(h,1)$ . We can determine before solving whether it lies there. At $r=1$ ,

$S(1)=\frac{C-B}{2}+\sin C\cos C.$

If the sector candidate already exceeds $h$ , and $S(1)\ge T/2$ , we are in this case. Equality gives the boundary case $r=1$ .

For example, a triangle with $A=150^\circ$ and $B=C=15^\circ$ has an equal-area radius of approximately $0.335593$ when $AC=1$ . Its circle crosses $BC$ twice.

Case 3: The circle passes a vertex

If half the area has still not been captured at $r=1$ , the circle passes the nearer vertex $C$ . The remaining part outside the circle sits near $B$ , and $1<r<L$ .

Let $D$ be the circle’s intersection with $BC$ , and set $\phi=\angle BAD$ . The sine rule in triangle $ABD$ gives

$\phi=\arcsin\!\left(\frac{\sin C}{r}\right)-B.$

The outside area equals the area of triangle $ABD$ minus a sector centered at $A$ :

$U(r)=\frac{Lr\sin\phi-r^2\phi}{2}.$

Set $U(r)=T/2$ , or equivalently $Lr\sin\phi-r^2\phi=T$ . This determines the unique radius between $1$ and $L$ . As an example, $A=80^\circ, B=20^\circ, C=80^\circ$ gives $r\approx1.014031$ when $AC=1$ .

The equilateral triangle leads to a square-root formula. Other triangles lead to a circular segment, and some require us to measure the outside area instead. The same question changes character as the angles change.

Another Equal-Area Problem

Here the challenge was to use a circle to divide the area of a triangle exactly in half. What happens if we reverse the roles and ask a circle to divide the area of another circle in half? Surprisingly, we can even require the center of the cutting circle to lie outside the original circle.

Continue exploring: Can a Circle Centered Outside Another Circle Cut Its Area Exactly in Half?

September 23, 2026

When Does x^(1/x) Take the Same Value Twice?

We begin with a simple equality:

2^{\frac{1}{2}} = 4^{\frac{1}{4}} .

The problem

Consider the function $f (x) = x^{\frac{1}{x}}, x > 0 .$ The equality above shows that this function can take the same value at two distinct positive real numbers.

x^{\frac{1}{x}} = y^{\frac{1}{y}}, x \neq y, x > 0, y > 0 .

Prove that infinitely many such pairs exist, and describe all of them.

Solution

Consider the function $f (x) = x^{\frac{1}{x}}$ on $[1, \infty) .$ It increases until $x = e$ and then decreases toward $1 .$ Thus, every horizontal line strictly between $1$ and the maximum intersects the graph twice.

A horizontal line meets the graph twice. Here it gives the approximate pair

(1.471, 7.857) .

A parametrization of all pairs

We now describe all the solutions. Assume first that $x < y$ and set

t = \frac{y}{x} > 1 .

Then

y = t x .

Taking logarithms of $x^{\frac{1}{x}} = y^{\frac{1}{y}}$ gives

\frac{\ln x}{x} = \frac{\ln (t x)}{t x} .

Multiplying by tx, we obtain

t \ln x = \ln x + \ln t .

Therefore,

(t - 1) \ln x = \ln t,

and hence

x = t^{\frac{1}{t - 1}} .

Since y = tx, it follows that

y = t^{\frac{t}{t - 1}} .

Thus, all solutions with $x < y$ are given by

(x, y) = (t^{\frac{1}{t - 1}}, t^{\frac{t}{t - 1}}), t > 1 .

Conversely, direct substitution shows that every $t > 1$ produces a solution. Therefore, this parametrization gives every distinct pair with the smaller number written first. Reversing the two entries gives the solutions with $x > y .$ Since there are infinitely many choices of $t > 1,$ there are infinitely many such pairs.

A Few Examples

Choosing $t = \frac{n + 1}{n}, n = 1, 2, 3, \dots,$ gives the infinite family of rational solutions

(x, y) = ({(\frac{n + 1}{n})}^{n}, {(\frac{n + 1}{n})}^{n + 1}) .

Examples from the rational family
n	t	x	y
1	2	2	4
2	3/2	9/4	27/8
3	4/3	64/27	256/81
4	5/4	625/256	3125/1024

Approaching e

These rational solutions approach $e$ from opposite sides. Indeed, if

x_{n} = {(1 + \frac{1}{n})}^{n}, y_{n} = {(1 + \frac{1}{n})}^{n + 1},

then

x_{n} ↑ e and y_{n} ↓ e .

Thus, the smaller number in each pair approaches $e$ from below, while the larger number approaches $e$ from above.

Another Same-Value Problem

The function $x^{1 / x}$ is not the only familiar-looking function that can take the same value at different inputs. A related question leads to a very different kind of equation.

Continue exploring: When Does x cos(x) Take the Same Value Twice?

September 21, 2026

Blog

Setting up the problem

Heron’s formula

A useful change of variables

Using Lagrange multipliers

Returning to the triangle

What is the maximum area?

Why this argument is interesting

Pythagorean runs

Looking for cubic runs

A Pell equation appears

The next cubic run

Reference

Example 1: The Gaussian integral

Example 2: Changing the scale

Example 3: An integral involving e−x4

Example 4: What about e−x4 itself?

The Gaussian was already hiding the Gamma function

From e−x2 to e−xp

An even larger family

Why did circles appear?

A surprising turn: add a cosine

What happens with sine?

One function keeps returning

Another Famous Improper Integral

Another Problem Where Two Dimensions Help

From the Gaussian Integral to a Real Signal

The ideal Otto cycle

Efficiency increases with compression

Adding a pressure constraint

The key equation

Where does the maximum occur?

A numerical example

What efficiency does this give?

An unexpected seventh-degree polynomial

The mathematical lesson

Why is there only one maximum?

The main idea

What is a Chebyshev polynomial?

Case 1: n = 2

Case 2: n = 3

Case 3: n = 5

What makes this interesting?

A Related Same-Value Problem

The key observation

A geometric interpretation

Measuring spheres

The lower-dimensional sphere

Finishing the computation

Examples

The circle S 1

The sphere S 2

The main idea

The familiar paradox

A general horn

The four possibilities

An entire family of examples

What if the horn wiggles?

References

Another Surprise from an Improper Integral

A radius always exists

Case 1: A sector is enough

Case 2: The circle crosses BCBC twice

Case 3: The circle passes a vertex

Another Equal-Area Problem

The problem

Solution

A parametrization of all pairs

A Few Examples

Approaching e

Another Same-Value Problem

Example 3: An integral involving $e^{- x^{4}}$

Example 4: What about $e^{- x^{4}}$ itself?

From $e^{- x^{2}}$ to $e^{- x^{p}}$

The circle $S^{1}$

The sphere $S^{2}$

Case 2: The circle crosses $BC$ twice