Optimization is one of the most useful tools in all of math. Many of the visualizations here based in finding the optimum solution to something (and indeed, it drives many of the solution methods of differential equations). Here's an optimization puzzler brought to you by FiveThirtyEight. What are the optimal thicknesses of a 3-layer cake (each layer is cylindrical) that can fit into a given (right, circular) cone, that maximizes the volume? My solution is complicated, but here's a picture of the result... (it's red velvet)

## Saturday, August 27, 2016

## Wednesday, May 18, 2016

### A Calculation in Statistical Mechanics

We've been harping about phase and state space for a long while, now, and have mentioned at more than one point that statistical mechanics is a subject dealing with, in some sense, the ultimate state space, of many gazillions of dimensions. It is amazing that we can even calculate anything about such spaces. So we take a break from tori, and return to the (music of the) spheres, today, to give an example. We consider $N$-dimensional spheres, for $N = 0, 1, 2, 3, \dots$ and their "volumes". Note, of course, by $N$-sphere, we always mean the curved surface of points in $(N+1)$-dimensional space with unit distance from the orgin, and not the whole ball. [The $3$-sphere, as we know, is foliated by nested tori. We aren't going TOO far afield].

To talk about this right, we need a little physics. Most of you probably have heard of the principle of conservation of energy... If you haven't, or even if you have, make sure to read about how Hamiltonian Mechanics is Awesome. Anyway, so if you imagine little billiard balls of uniform mass $m$ freely zipping about, banging into stuff, each particle having position $\mathbf{q}_i$ and momentum $\mathbf{p}_i$, then since we know Hamiltonian Mechanics is Awesome, this says that the two quantities evolve in such a way that the energy of the $i$th particle \[

E_i = \frac{|\mathbf{p}_i|^2}{2m}

\]

stays constant. But if you imagine having a very, very large number of such billiard balls (several billiards of such (1 billiard = $10^{15}$ in most dialects of English, or such equivalent cognates) HAR HAR!), and compute the

H = \sum_i E_i = \sum_i \frac{|\mathbf{p}_i|^2}{2m}.

\] To say that the

\sum_i \frac{|\mathbf{p}_i|^2}{2m} = E

\] for some fixed number $E$. But what does this say? If we write out each \[

|\mathbf{p}_i|^2 = p_{ix}^2 + p_{iy}^2 + p_{iz}^2,

\] then \[

2mE=\sum_i p_{ix}^2 + p_{iy}^2 + p_{iz}^2,

\] meaning that if we sum over all the particles, say $N$ of them, where $N \approx 10^{24}$ (a milliard billiard of billiard balls), this means

this is the

Well, what's one of the first things you think about when talking geometry? Areas and volumes, of course. There are situations in which you need to

Suffice to say, however, let's talk as if calculating the volume of the $3N$-ball does indeed count the total number of momentum states with energy less than or equal to $E$, i.e., let's calculate \[

\frac{1}{(2\pi \hbar)^{3N}}\int_{|\mathbf{p}|^2 \leq 2mE} d\mathbf{q}d\mathbf{p} =\frac{V^N}{(2\pi \hbar)^{3N}}\int_{|\mathbf{p}|^2 \leq 2mE} d\mathbf{p} .

\]

where $V$ is the ordinary 3D volume of the spatial region where all our particles live. For the momentum integral, we need the formula (whose derivation is fodder for another post!) for the volume of an $K$-ball \[

V_K(R) = \frac{\pi^{K/2}}{\Gamma\left(1+\frac{K}{2}\right)}R^K

\]

where $\Gamma(1+ K/2)$ is $(K/2)!$, the factorial of $K/2$, if $K$ is even, and $\pi^{1/2} K!!/2^{K/2}$ where $K!!$ is the double factorial $1\cdot 3 \cdot 5 \cdots k$, for $k$ odd.

Plugging this in with $R = \sqrt{2mE}$ and $K=3N$ we have the total volume of a $3N$-ball in momentum space: \[

V_{3N}(\sqrt{2mE}) = \frac{\pi^{3N/2}}{\Gamma\left(1+\frac{3N}{2}\right)} (2mE)^{3N/2}

\]

To make headway here, we can use Stirling's Approximation for the Gamma function:\[

\Gamma\left(1+\frac{3N}{2}\right) \approx \sqrt{3N\pi} \left(\frac{3N}{2e}\right)^{3N/2}

\] This gives \[

V_{3N}(\sqrt{2mE}) \approx \frac{1}{\sqrt{3\pi N}}\left(\frac{4\pi meE}{3N} \right)^{3N/2}

\] or, the total number of states is \[

\#\text{states} = \frac{V^N V_{3N}(\sqrt{2mE})}{(2\pi \hbar)^{3N}} \approx \frac{1}{\sqrt{3\pi N}}\left(\frac{4\pi meEV^{2/3}}{3N (2\pi \hbar)^2} \right)^{3N/2}

\]

Considering the smallness of $\hbar$ ($10^{-34}$ in everyday units) and the exponent $3N/2$ being on the order of $10^{24}$, this could be a staggeringly huge number (but that's what you expect: for lots of particles, they really can occupy even more states. For example, two particles behaving independently is the

k\ln(\#\text{states}) = \frac{3Nk}{2} \left( \ln \left(\frac{4\pi mEV^{2/3}}{3N(2\pi \hbar)^2}\right) +1\right) -\frac{1}{2}\ln(3\pi N)k.

\]

Since those last terms are small potatoes (not being multiplied by $N$), we usually just drop it to arrive at the

\[

S = \frac{3Nk}{2} \left( \ln \left(\frac{4\pi mEV^{2/3}}{3N(2\pi \hbar)^2}\right)\right)+\frac{3Nk}{2}.

\] And for you chemists, $Nk = nR$, where $n$ is the number of moles and $R$ is the gas constant:

\[

S = \frac{3nR}{2}\left( \ln \left(\frac{4\pi mEV^{2/3}}{3N(2\pi \hbar)^2}\right)\right)+\frac{3nR}{2}.

\]

That's really nasty. How about we try to plug numbers into this and see what it is? We'll do that more concretely next time. But suffice to say, let's get back to what's interesting: we calculated the volume of some high-dimensional stuff!

To talk about this right, we need a little physics. Most of you probably have heard of the principle of conservation of energy... If you haven't, or even if you have, make sure to read about how Hamiltonian Mechanics is Awesome. Anyway, so if you imagine little billiard balls of uniform mass $m$ freely zipping about, banging into stuff, each particle having position $\mathbf{q}_i$ and momentum $\mathbf{p}_i$, then since we know Hamiltonian Mechanics is Awesome, this says that the two quantities evolve in such a way that the energy of the $i$th particle \[

E_i = \frac{|\mathbf{p}_i|^2}{2m}

\]

stays constant. But if you imagine having a very, very large number of such billiard balls (several billiards of such (1 billiard = $10^{15}$ in most dialects of English, or such equivalent cognates) HAR HAR!), and compute the

*total energy*, summing them all up,\[H = \sum_i E_i = \sum_i \frac{|\mathbf{p}_i|^2}{2m}.

\] To say that the

*total*energy is constant, then, is to say \[\sum_i \frac{|\mathbf{p}_i|^2}{2m} = E

\] for some fixed number $E$. But what does this say? If we write out each \[

|\mathbf{p}_i|^2 = p_{ix}^2 + p_{iy}^2 + p_{iz}^2,

\] then \[

2mE=\sum_i p_{ix}^2 + p_{iy}^2 + p_{iz}^2,

\] meaning that if we sum over all the particles, say $N$ of them, where $N \approx 10^{24}$ (a milliard billiard of billiard balls), this means

*we're summing the squares of a $3N$-dimensional vector and equating it to a constant*. So if we ask ourselves:*What is the collection of all possible momenta of $N$ particles that have one constant energy level $E$,*this is the

*same*as saying,

*What are all the points on the $(3N-1)$-dimensional sphere of radius $\sqrt{2mE}$?**(the -1 from the fact that you're expressing the surface as the solution set to one equation: the boundary of the $k$-ball is the $(k-1)$-sphere)*

*Interesting*, you might say, but is there any occasion which actually requires you to*truly*do calculations treating it precisely as a geometrical $(3N-1)$-dimensional object?Well, what's one of the first things you think about when talking geometry? Areas and volumes, of course. There are situations in which you need to

*integrate*over such energy "surface", or within the $N$-ball contained within, for $N \sim 10^{24}$. Basically, to find quantities like entropy, one needs to "count some number of cells" inside the $3N$-ball . This yields the total "number" of energy states with level*less than or equal to*$E$. "Number" in quotations because classically, of course, the actual number of states is the number of points on a sphere, which is of course uncountably infinite. We instead mean, per unit $3N$-dimensional volume, which actually leads to the*density*of states. The distinction is a little blurry, because, with a little help of quantum mechanics, we can think of each state as a cell of volume $(2\pi \hbar)^{3N}$, and each distinct state of momentum, which must be quantized, ultimately arises from boundary conditions in solving Schrödinger's equation, or eigenstuff.Suffice to say, however, let's talk as if calculating the volume of the $3N$-ball does indeed count the total number of momentum states with energy less than or equal to $E$, i.e., let's calculate \[

\frac{1}{(2\pi \hbar)^{3N}}\int_{|\mathbf{p}|^2 \leq 2mE} d\mathbf{q}d\mathbf{p} =\frac{V^N}{(2\pi \hbar)^{3N}}\int_{|\mathbf{p}|^2 \leq 2mE} d\mathbf{p} .

\]

where $V$ is the ordinary 3D volume of the spatial region where all our particles live. For the momentum integral, we need the formula (whose derivation is fodder for another post!) for the volume of an $K$-ball \[

V_K(R) = \frac{\pi^{K/2}}{\Gamma\left(1+\frac{K}{2}\right)}R^K

\]

where $\Gamma(1+ K/2)$ is $(K/2)!$, the factorial of $K/2$, if $K$ is even, and $\pi^{1/2} K!!/2^{K/2}$ where $K!!$ is the double factorial $1\cdot 3 \cdot 5 \cdots k$, for $k$ odd.

Plugging this in with $R = \sqrt{2mE}$ and $K=3N$ we have the total volume of a $3N$-ball in momentum space: \[

V_{3N}(\sqrt{2mE}) = \frac{\pi^{3N/2}}{\Gamma\left(1+\frac{3N}{2}\right)} (2mE)^{3N/2}

\]

To make headway here, we can use Stirling's Approximation for the Gamma function:\[

\Gamma\left(1+\frac{3N}{2}\right) \approx \sqrt{3N\pi} \left(\frac{3N}{2e}\right)^{3N/2}

\] This gives \[

V_{3N}(\sqrt{2mE}) \approx \frac{1}{\sqrt{3\pi N}}\left(\frac{4\pi meE}{3N} \right)^{3N/2}

\] or, the total number of states is \[

\#\text{states} = \frac{V^N V_{3N}(\sqrt{2mE})}{(2\pi \hbar)^{3N}} \approx \frac{1}{\sqrt{3\pi N}}\left(\frac{4\pi meEV^{2/3}}{3N (2\pi \hbar)^2} \right)^{3N/2}

\]

Considering the smallness of $\hbar$ ($10^{-34}$ in everyday units) and the exponent $3N/2$ being on the order of $10^{24}$, this could be a staggeringly huge number (but that's what you expect: for lots of particles, they really can occupy even more states. For example, two particles behaving independently is the

*square*the total number of states of for one particle). Thus to measure entropy, we take the natural log of all of this (and add a factor of the Boltzmann constant, $k$ which is on the order of $10^{-23}$ in ordinary units), we get \[k\ln(\#\text{states}) = \frac{3Nk}{2} \left( \ln \left(\frac{4\pi mEV^{2/3}}{3N(2\pi \hbar)^2}\right) +1\right) -\frac{1}{2}\ln(3\pi N)k.

\]

Since those last terms are small potatoes (not being multiplied by $N$), we usually just drop it to arrive at the

**entropy**of our system of particles as\[

S = \frac{3Nk}{2} \left( \ln \left(\frac{4\pi mEV^{2/3}}{3N(2\pi \hbar)^2}\right)\right)+\frac{3Nk}{2}.

\] And for you chemists, $Nk = nR$, where $n$ is the number of moles and $R$ is the gas constant:

\[

S = \frac{3nR}{2}\left( \ln \left(\frac{4\pi mEV^{2/3}}{3N(2\pi \hbar)^2}\right)\right)+\frac{3nR}{2}.

\]

That's really nasty. How about we try to plug numbers into this and see what it is? We'll do that more concretely next time. But suffice to say, let's get back to what's interesting: we calculated the volume of some high-dimensional stuff!

## Friday, May 13, 2016

### The Volume of Intersecting Cylinders

Now for another classic, although not as "kewel" as the spiral example. This time, it was

*I*who asked for the help: once upon a time in a calculus class, we were asked to compute the volume of the solid that is the intersection of two cylinders at right angles. This is a surprisingly subtle calculation, and one of the more difficult-to-visualize things. Fortunately, I've managed to reproduce it above, with some transparency so the intersection is a little bit more visible. Now, here is the full intersection itself without the extra cylinder parts (called the Steinmetz Solid):
It is said that the Chinese mathematician 祖沖之 (Zu Chongzhi) computed the volume of this solid in the 400s ... I don't know how he did it without calculus, which is the way we're going to do it today. To do this, we write the equations of the two cylinders, say of radius $R$, which we will assume to have their axes in the $xy$-plane:

\[
x^2 + z^2 = R^2
\] \[
y^2 + z^2 = R^2. \]Let's see what the actual intersection is. We set the two equations equal to each other:

\[

x^2 + z^2 = y^2 +z^2

\]

which means $x^2 = y^2$. This in turn means $y = \pm x$. In the $xy$-plane, this looks like two intersecting lines in an "X" shape, which you can see by looking at the solid above from the very tippy top. Of course, this is not exactly what the intersection actually looks like: we still have that there is a $z$-coordinate, given by $z = \pm \sqrt{R^2-x^2} =\pm \sqrt{R^2-y^2}$, so that the complete parametrization of the curves (four of them corresponding to each choice of $+$ or $-$) are given by, for example

\[

x = t

\]\[

y = \pm t

\]\[

z = \pm\sqrt{R^2-t^2}

\]

for $t$ varying from $-R$ to $R$ (the maximum allowable range given that it has to lie on a cylinder). This gives the "corners" of the solid given above. The solid itself is given by stacking squares with edges running from one curve to another (you can also see that in the solid above). To calculate the volume, we integrate the area of cross-sections perpendicular to the $z$-axis:

\[

V=\int_{-R}^R A(z)\,dz = \int_{-R}^R s^2\,dz.

\]

Therefore, we must find the side length $s$, which requires us to solve things in terms of $z$: $z =\pm\sqrt{R^2-x^2}$ means $z^2 = R^2 - x^2$, or $x^2 = R^2 - z^2$, so finally $x = \pm\sqrt{R^2 - z^2}$. Since the sides of the square are the same on either side, we can choose the positive square root. Then, since one side of the square goes between the two different values of $y$ for a given $x$, we find $s = \sqrt{R^2 - z^2}-(-\sqrt{R^2 - z^2}) = 2\sqrt{R^2 - z^2}$. Finally, we can find the integral:

\[

V = \int_{-R}^R s^2\, dz = \int_{-R}^{R} 4(R^2-z^2)\,dz = \left.4R^2 z - \frac{4}{3}z^3\right|_{-R}^R = \left(8-\frac{8}{3}\right)R^3 = \frac{16}{3}R^3.

\]

## Wednesday, May 11, 2016

### Yitz..'s Kewel Spiral

Every once in a while, I get an "assignment" to calculate some interesting parametrization (because parametrizations are awesome like that). For example, the bowtie pasta was originally a request from a student. Today's goes way back, to someone asking a question on a forum in math way back when. He needed a spiral that started out and twisted its way through space as it got larger. In essence, a spiral that

To discover what its equation might be, we first consider an Archimedean spiral ($r = a\varphi$) in the $r\varphi$-plane (or $yz$-plane), or possibly a logarithmic one, like $e^{\varphi/3}$, which grows as it goes around (like some spirals last time) the "pole" which is now the $x$-axis. This ends up growing as you move clockwise, rather than counterclockwise, since $\varphi$ increases as it goes from north toward east.

This is most efficiently represented in a parametric manner: first, in the $yz$-plane:\[

x=0,

\] \[

y = at \sin t,

\] \[

z = at \cos t.

\]

gives the parametric equation of the Archimedean spiral in the $yz$-plane going clockwise. This gives $\varphi = t$ and $r = at$, which recovers the original polar equation $r = a\varphi$.

How do we give it a twist? We reason that, after each revolution, we want it twisted out of the plane by some small amount (say, rotated about the $z$-axis by $b$ times a full revolution): we apply the rotation to the $x$ and $y$, giving:\[

x=at\sin t \cos (bt),

\] \[

y = at \sin t \sin(bt),

\] \[

z = at \cos t.

\]

Here's the original one I delivered to Yitz.. (it is actually exponential):

*doesn't*stay in the same plane. Obviously, polar curves are out of the question, since they are, well, in the plane. But what's one dimension up from polar coordinates?*Spherical*coordinates $(r,\varphi,\theta)$, of course (or if you're a physicist, $(r,\theta,\phi)$).^{1 }We fall back on a good ol' tool, parametrization.To discover what its equation might be, we first consider an Archimedean spiral ($r = a\varphi$) in the $r\varphi$-plane (or $yz$-plane), or possibly a logarithmic one, like $e^{\varphi/3}$, which grows as it goes around (like some spirals last time) the "pole" which is now the $x$-axis. This ends up growing as you move clockwise, rather than counterclockwise, since $\varphi$ increases as it goes from north toward east.

This is most efficiently represented in a parametric manner: first, in the $yz$-plane:\[

x=0,

\] \[

y = at \sin t,

\] \[

z = at \cos t.

\]

gives the parametric equation of the Archimedean spiral in the $yz$-plane going clockwise. This gives $\varphi = t$ and $r = at$, which recovers the original polar equation $r = a\varphi$.

Archimedean Spiral in the $yz$-plane, with $a = 1/20$. |

How do we give it a twist? We reason that, after each revolution, we want it twisted out of the plane by some small amount (say, rotated about the $z$-axis by $b$ times a full revolution): we apply the rotation to the $x$ and $y$, giving:\[

x=at\sin t \cos (bt),

\] \[

y = at \sin t \sin(bt),

\] \[

z = at \cos t.

\]

Twisting out of the plane with parameter $b = 1/20$, also. |

Here's the original one I delivered to Yitz.. (it is actually exponential):

Pretty

*kewel*, right? (that was his exact word to describe it).^{1 }I always use the convention that if I indeed need to refer to a physics text, do calculations to be consistent... but I always write my $\phi$'s as non-curly in that context... I haven't yet figured how to draw my $\theta$s different though. Yes, I know there is $\vartheta$, but it hasn't caught on in my notation. It just doesn't feel enough like a $\theta$.## Saturday, April 30, 2016

### More Polar Classics: Spirals

Some more from the classic polar curves collection... today we specialize in

(here the angle is in radians and the equation is $r = \theta$). Of course, some people thought this was a bit boring, so Fermat spiced it up and thought what if we make the spirals wind tighter as you go around? Basically, make it a concave-down, increasing function, $r = \sqrt{\theta}$. Actually, it's more fun to also consider $r = -\sqrt{\theta}$:

*spirals*. The basic concept of what a spiral should be, in polar coordinates, is any plot $r$ varying with $\theta$, $r=f(\theta)$, where $f$ is always increasing, and $\theta$ is allowed to go beyond the usual range of $0\leq \theta < 2\pi$ (or $-\pi < \theta \leq \pi$). The most fundamental one, of course, is $r = \theta$, which says the distance from the origin is proportional to the amount you've wound around the origin:(here the angle is in radians and the equation is $r = \theta$). Of course, some people thought this was a bit boring, so Fermat spiced it up and thought what if we make the spirals wind tighter as you go around? Basically, make it a concave-down, increasing function, $r = \sqrt{\theta}$. Actually, it's more fun to also consider $r = -\sqrt{\theta}$:

($r=\sqrt{\theta}$ is in blue, and $r = -\sqrt{\theta} is in yellow$)

This of course could more succinctly be written $\theta = r^2$. But perhaps the second-most famous, due to its ubiquity in nature (because stuff tends to grow exponentially) is the famous logarithmic spiral $r = e^{c \theta}$ or $\theta = \frac{1}{c} \log r$, which magnifies every time you spin it.

($r=e^{\theta/5}$)

And finally, where would we be in polar graphs if we didn't have the analogue of stuff with asymptotes? Consider $r^{-2} = \theta$ (instead of $r^+2$):

This is called the

*lituus of Cotes.*There's all sorts of curves from back in the day with very funny names; I used to look for a lot of them on the MacTutor History of Math Page (more specifically, its Curves Index).
There's one other spiral that's my favorite, but (1) it isn't polar, and (2) there's a lot of interesting differential geometry going on in it, so I'm saving it for another post.

## Thursday, April 28, 2016

### Variance Solids: Volumes and $\mathscr{L}^2$, Part 2

Last time, we talked about how the classical $\mathscr{L}^2$ norm of a (one-variable) function could be visualized as a solid consisting of squares. Of course, there's another formula that is defined by almost the same formula, with an extra factor of $\pi$: the

*solid of revolution*of the function about the $x$-axis: \[

V = \int_a^b \pi f(x)^2\,dx

\]

It looks pretty cool, although not quite as cool as twisted square cross-sections.

I wanted to mention another common application of $\mathscr{L}^2$ norms that probably is more familiar than any other: the statistical concept of

*variance*. For functions $f$ defined on a continuous domain, we can define an average value:\[

\overline{f} = \frac1{b-a}\int_a^b f(x)\,dx.

\] which basically is the height of a rectangle over the same domain $[a,b]$

*that has the same area as the area under*$f$. For example, for \[

f(x) = \sin \left( x \right)\sin \left( t \right)+\frac{1}{3}\sin \left( 5x \right)\sin \left( 5t \right)+\frac{1}{3}\sin \left( 10x \right)\sin \left( 10t \right)+\frac{1}{5}\sin \left( 15x \right)\sin \left( 15t \right)

\]

(choosing $t=3.030303$), it looks like:

*which is precisely the same quantity as the solid obtained by revolving $f(x)$ around its mean $y=\overline{f}$.*The resulting solid is the title picture.

## Sunday, April 24, 2016

### Volumes of Solids and (squared) $\mathscr{L}^2$ Norms

The volume of this solid is the squared $\mathscr{L}^2$ distance between $y=x^2$ and $y=x^3$ on the interval $[0,\frac{3}{2}]$. |

The good ol' $\mathscr{L}^2$ norm has an outsized importance in many applications of math, principally, because Hilbert spaces (spaces with inner product) are useful for, well, just about anything. Much of a fan of Hilbert spaces I might have been for much of my recent life, it is only when teaching how to find the volume of solids (and a little help from Math Homie William) when it hit me what the best visualization for them is, at least for functions defined on $\mathbb{R}$. (For the longest time, I just said "It's sort of like the area between two curves, but, you know, squared, and stuff.") But we learn in integration theory that we can base a solid by putting squares (or semidisks or triangles, or whatever) in a perpendicular, 3rd dimension, over the area between curves; the volume of the solid is \[

V=\int_a^b A(x) \, dx

\] where $A(x)$ is the area of the cross section. But of course there's an obvious candidate: if you choose

*squares*to have side length being the height of your function $f(x)$, then $A(x)$ is just $f(x)^2$. Thus,\[

V = \int_a^b f(x)^2 dx.

\] But that's none other than the squared $\mathscr{L}^2$ norm of $f$! Similarly, for the volume of a solid consisting of squares between two curves $f(x)$ and $g(x)$, this is just \[

V=\int_a^b (f(x) - g(x))^2dx,

\] the (squared) $\mathscr{L}^2$ distance between the two functions. Now if we treat each of these squares like a deck of cards, and rotate them, it shouldn't change the volume. So we apply this to the functions $f(x)=x^2$ and $g(x)=x^3$ on $[0,\frac{3}{2}]$:

Now making vertical segments between the curve, and making a square with that side length pop half out of the plane, and half into the plane, gives us a nice solid. Rotating it gives the picture at the top.

## Friday, April 15, 2016

### Polar Coordinate Classic

In some sense, my first "Nested Tori" was a notebook which contained really cool plots and graphs. There's lots of old gems there worth posting and updating. Today, I bring you a nice polar plot. I was really intrigued by plots in polar coordinates, because it represented the first real departure from simply graphing $y=f(x)$. We instead (most commonly) write $r = f(\theta)$ where $r$ is distance from the origin and $\theta$ is counterclockwise angle, and let things get farther or closer as we go around. Without further ado, here is the famed butterfly (it does have a little post-warping afterward, though), which shows you that you don't need fancy-schmancy 3D stuff to still find great things:

## Monday, April 11, 2016

### Pasta Parametrization, High Resolution

Integrating (ha ha) Nested Tori back into life (of course I'm doing a lot of integrating, if I'm teaching, anyway). Lots of change was (and is) afoot.

Anyway, instead of continually getting peeved with Apple for

*not*ever fixing any bugs in Grapher for over 10 years (a job that I would*gladly*take the mantle on, by the way: anyone at Apple listening??), I took matters into my own hands and started experimenting with some open source plotting libraries. In particular, I've taken somewhat of a liking for matplotlib, a famous Python package. The boon is, it can generate pdf files properly (unlike Grapher, which basically puts the bitmap into a pdf file and adds the extension .pdf, but when you zoom in, it's blocky and clearly still a bitmap). Honestly, though, it*is*in fact a lot more work (which was mitigated somewhat by defining some functions).
Here are the results for plotting the Pasta Parametrization in matplotlib (specifically, the mplot3d toolkit):

Of course, I should probably figure out how to control the lighting better: it doesn't look as good as Grapher's (at a scale for which the resolution issue is not a problem). And here are the matching formulas (at a larger size than the old version):

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