I need a mapping from the unit Hypercube [0,1]^n to a given simplex

[benorin]

I need a mapping from the unit Hypercube $C^n:= \left[ 0,1\right] ^n$ to a given simplex, namely $S^n:=\left\{ \vec{x} \in\mathbb{R}^n |0\leq x_1\leq 1, 0\leq x_{k}\leq x_{k-1}\text{ for } k=2,3,\ldots , n\right\}$. Anybody know one? I have other requirements I need satisfy, so if you know more than one such mapping (transformation equations) please do post them.

Moderator: I’m not 100% certain that this is the appropriate sub-forum in which to post this question, so please feel free to move it. Thanks!

 

[FactChecker]

You should probably indicate any requirements like continuity, onto, 1-1, etc.
Without those requirements, one mapping that comes to mind is $f(x_1,x_2,...,x_n) = [x_1/2, x_2/4+1/2, x_3/8+3/4, ... , x_n/(2^n)+(2^{n-1}-1)/2^{n-1}]$
It is continuous but not onto or 1-1. I suspect that this is not what you want.

 

[benorin]

I need it for change of variables in an integral, so continuous, differentiable, non-zero Jacobian. Also I require that the integrand times the Jacobian is expressible as a function of one variable, to wit

$$\frac{dx_n dx_{n-1}\dots dx_1}{1-\prod_{k=1}^{n}x_k}\cdot \left|\frac{\partial (x_1,x_2,\ldots , x_n)}{\partial (u_1, u_2, \ldots,u_n)}\right|:=g(u_n)$$

Laundry list I know, but this should allow me to represent the integral over the hypercube as an iterated integral so I can generalize the integral to complex values of n. I hoping to get some ideas and maybe tweak them to work. This is just a practice problem. I may equivalently solve the fractional-integral equation

$$I_{\alpha}g(z)=\zeta (z)$$

where $\zeta (\cdot )$ is the Riemann Zeta function--which may well be easier but I wanted to try it this way to make use of my earlier work.

 

[FactChecker]

CORRECTION: THIS POST AND POST #6 ARE WRONG. I THINK THAT POST #7 IS CORRECT.
How about this:
$f(x_1,x_2, ... , x_n) = [y_1, y_2, ... , y_n],$
where
$y_1=x_1;$
$y_2=y_1+x_2*y_1+(1-x_2);$
$y_3=y_2+x_3*y_2+(1-x_3);$
$y_4=y_3+x_4*y_3+(1-x_4);$
$...$
$y_n=y_{n-1}+x_n*y_{n-1}+(1-x_n);$

The idea here is that each $y_i$ has the value between $y_{i-1}$ and 1, which is the weighted average of those two points with weights $x_i$ and $(1-x_i)$.

 

[benorin]

Is * multiplication? Or some other operator? Thanks for your help!

 

[FactChecker]

Is * multiplication? Or some other operator? Thanks for your help!

CORRECTION: THIS POST IS WRONG. I THINK THAT POST #7 IS CORRECT.
Yes, it is multiplication. I think that I made an error on the equations. A corrected version would be:
$f(x_1,x_2,...,x_n)=[y_1,y_2,...,y_n],$
where
$y_1=x_1;$
$y_2=(1-x_2)y_1 +x_2;$
$y_3=(1-x_3)y_2 +x_3;$
$y_4=(1-x_4)y_3 +x_4;$
...
$y_n=(1-x_n)y_{n-1} +x_n;$

Sorry for the confusion.

 

[FactChecker]

I misread your definition of $S^n$ and made $y_i \ge y_{i-1}$ where it should have been $y_i \le y_{i-1}$.

To get that corrected:
$f(x_1,x_2,...,x_n)=[y_1,y_2,...,y_n],$
where
$y_1=x_1;$
$y_2=x_2y_1;$
$y_3=x_3y_2;$
$y_4=x_4y_3;$
...
$y_n=x_ny_{n-1};$

Sorry again for the confusion.

 

[sysprog]

@FactChecker It's really nice of you to, in the process of self-correcting a slightly incorrect explanation, go loud on yourself having earlier slightly erred; tip of the hat to you on that Sir.

 

[FactChecker]

@FactChecker It's really nice of you to, in the process of self-correcting a slightly incorrect explanation, go loud on yourself having earlier slightly erred; tip of the hat to you on that Sir.

Thanks. I just wish I could be more accurate the first time. It's a problem I have even on subjects that I should be much sharper at.

 

[sysprog]

Moi aussi -- I think that I'm ok at fixing bugs -- and I think that it would better if I didn't write bug-inclusive code in the first place, but whenever I wake out of laziness, I got to code, it's just a thing.

 

[benorin]

I misread your definition of $S^n$ and made $y_i \ge y_{i-1}$ where it should have been $y_i \le y_{i-1}$.

To get that corrected:
$f(x_1,x_2,...,x_n)=[y_1,y_2,...,y_n],$
where
$y_1=x_1;$
$y_2=x_2y_1;$
$y_3=x_3y_2;$
$y_4=x_4y_3;$
...
$y_n=x_ny_{n-1};$

Sorry again for the confusion.

Thanks, that’s much easier to work with.

 

[benorin]

I get that $y_1=x_1$ and for $y_{k-1}\neq 0$ we have $x_k=\frac{y_k}{y_{k-1}}$ for $k=2,3,\ldots , n$. So we have that

$$\frac{\partial x_i}{\partial y_j}=
\begin{cases} 1, & i=j=1 \\ \frac{1}{y_{i-1}}, & i=j\neq 1 \\ -\frac{y_i}{y_{i-1}^2}, & i=j+1 \\ 0, & \text{otherwise} \end{cases}$$

So the Jacobian determinant is just the product along the diagonal, correct? So it's $J=\frac{1}{y_1 y_2\cdots y_{n-1}}$?

 

[FactChecker]

So you are looking at the inverse function of $f$, right? Each $y_i$ is a function of all $y_j, j\lt i$, so I think that makes it more complicated. Since you are talking about the inverse of the function $f$, I don't think that you can consider the $y_i$s as being independent variables. I think that this is getting beyond my comfort level and I will leave it to others to help you further.

 

[benorin]

Now I see that I made a mistake and the integral equation that I could alternatively solve should have been

$$\lim_{x\to a} I_{\alpha}f(x)=\zeta (\alpha )$$

Whereas I had not included the limit and also had $\zeta (z)$ on the right hand side back in post #3.

 

[benorin]

So I solved the fractional-Integral equation but with the Hadamard Left Fractional Integral Operator instead of the R-L fractional integral operator, and it turns out that post #12 has the Jacobian determinant correct. I arrived at this conclusion in a round about way. Beginning with the well known integral

$$\Gamma (z) \zeta (z)=\int_{u=0}^{\infty} \frac{u^{z-1}}{e^u-1}\, du, \, \, \Re\left[ z\right] >1$$

Make the substitution $u= \log \frac{x}{t}\Rightarrow du=-\frac{dt}{t}$ and use the negative from du to flip the bounds of integration to get

$$\zeta (z)=\frac{1}{\Gamma (z)}\int_{t=0}^{x} \log ^{z-1}\left( \frac{x}{t}\right) \frac{1}{\frac{x}{t}-1}\, \frac{dt}{t}, \, \, \Re\left[ z\right] >1$$

which, after setting $x=1$ in the factor of the integrand following the log factor, is exactly the Hadamard fractional integral of $\frac{t}{1-t}$ of order z and, get this, the Hadamard FI operator interpolates the n-fold integral

$$\int_{0}^{x_0}\int_{0}^{x_1}\cdots\int_{0}^{x_{n-1}}f(x_n)\frac{dx_n\ldots dx_1}{x_n \cdots x_1} = \frac{1}{\Gamma (n) }\int_0^{x_0}\log ^{n-1}\left( \frac{x_0}{t}\right) f(t) \frac{dt}{t}$$

Hence the integral we were looking at in this thread should have been

$$\int_{0}^{y_0}\int_{0}^{y_1}\cdots\int_{0}^{y_{n-1}}\frac{y_n}{1-y_n}\frac{dy_n\ldots dy_1}{y_{n}\cdots y_1} $$

so the Jacobian determinant of post #12 was correct after multiplying by a factor of $\frac{y_n}{y_n}$.

 

[zinq]

One way to conceive of such a mapping is: First map one vertex v of the cube [0,1]ⁿ to some vertex f(v) of the simplex ∆_n. Then extend this to the n edges of [0,1]ⁿ emanating from v ... to the n edges of ∆_n emanating from f(p) — just map each of them linearly.

Now consider the far vertices of those n edges of the cube — call them v₁, ..., v_n. Now each line from the original vertex v that heads into the cube will pass through a unique convex combination of the v_j's. (That is, a linear combination of the v_j's whose coefficients are non-negative real numbers whose sum equals 1.) For each such convex combination (call it w), send the line in the cube from v through w until it exits the cube to the corresponding line from f(v) through the corresponding convex combination of the f(v_j)'s until it exits the simplex. Again, just map the first interval linearly onto the second for each convex combination.

It shouldn't be hard to derive a formula from this. But don't expect it to be a diffeomorphism (differentiable homeomorphism with differentiable inverse) from the entire closed cube to the entire closed simplex, because that will not exist.

 

[microprediction]

Very interesting problem. I'm having trouble piecing together the solution suggested but would like to.

Meanwhile I can share a draft of an article I'm writing on precisely this problem here.

 

[WWGD]

If you want to include the interior, you may use the idea of a space-filling curve. There may also be conditions so that the map is a simplicial map.