- #1

benorin

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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!

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- #1

benorin

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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!

- #2

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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.

- #3

benorin

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

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.

- #4

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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)##.

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)##.

Last edited:

- #5

benorin

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

- #6

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CORRECTION: THIS POST IS WRONG. I THINK THAT POST #7 IS CORRECT.Is * multiplication? Or some other operator? Thanks for your help!

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.

Last edited:

- #7

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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.

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.

Last edited:

- #8

sysprog

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- #9

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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.

- #10

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- #11

benorin

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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.

- #12

benorin

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$$\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}}##?

- #13

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- #14

benorin

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

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

- #15

benorin

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

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}##.

- #16

zinq

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Now consider the

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.

- #17

microprediction

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

- #18

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