# The Unapologetic Mathematician

## The Geometric Interpretation of the Jacobian Determinant

We first defined the Jacobian determinant as measuring the factor by which a transformation scales infinitesimal pieces of $n$-dimensional volume. Now, with the change of variables formula and the mean value theorem in hand, we can pull out a macroscopic result.

If $g:S\rightarrow\mathbb{R}^n$ is injective and continuously differentiable on an open region $S\subseteq\mathbb{R}^n$, and $X$ is a compact, connected, Jordan measurable subset of $S$, then we have

$\displaystyle c(Y)=\lvert J_h(x_0)\rvert c(X)$

for some $x_0\in X$, where $Y=g(X)$.

Simply take $f(y)=1$ in the change of variables formula

$\displaystyle c(Y)=\int\limits_Y\,dy=\int\limits_X\lvert J_g(x)\rvert\,dx$

The mean value theorem now tells us that

$\displaystyle c(Y)=\int\limits_X\lvert J_g(x)\rvert\,dx=\lambda\int\limits_X\,dx=\lambda c(X)$

for some $\lambda$ between the maximum and minimum of $\lvert J_g(x)\rvert$ on $X$. But then since $X$ is connected, we know that there is some $x_0\in X$ so that $\lambda=\lvert J_g(x_0)\rvert$, as we asserted.

January 8, 2010 - Posted by | Analysis, Calculus

1. It depends on me, but I can learn here a lot, things I am not familiar with, such as $c(X)$, the Jordan content of $X$.

Comment by Américo Tavares | January 8, 2010 | Reply

2. Well like I said when I introduced it, Jordan content is the first naïve step on the road to Lebesgue measure, and is to the Riemann integral as Lebesgue measure is to the Lebesgue integral. The biggest innovation Lebesgue had was to turn the order around and define the integral in terms of a notion of “volume” rather than the other way around.

Comment by John Armstrong | January 8, 2010 | Reply

3. Is the Lebesgue integral important for the applications outside Mathematics?

At least I know that the Stieltjes integral is used in Probabilities to unify the discrete and continuous cases.

Long time ago I found it in the 2nd year of Engineering without having a previous knowledge of it. I studied only the minimum just to understand that application.

Later I learned more in a post of yours and looking at Taylor’s book on Advanced Cauculus.

Comment by Américo Tavares | January 8, 2010 | Reply

4. The Lebesgue integral and measure theory further generalize the Riemann-Stieltjes integral. Modern probability theory is largely rooted in a part of that theory, although it asks different questions and proceeds in different directions.

Comment by John Armstrong | January 8, 2010 | Reply

5. An wonderful exercise. Evaluate

$\zeta (2)=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}\displaystyle\frac{1}{1-xy}dx\,dy$

by making a change of coordinates

$u=\displaystyle\frac{y+x}{2}$

$v=\displaystyle\frac{y-x}{2}.$

Hence $\displaystyle\frac{1}{1-xy}=\displaystyle\frac{1}{1-u^{2}+v^{2}}$ and the Jacobian determinant
of the transformation is $\left\vert J\right\vert =2.$ Computing twice the double integral, one for $v\in \left[ 0,1/2\right]$ and the other for $v\in\left[ -1/2,0\right]$, and observing that, due to symmetric reasons with respect to $v$, both integrals are equal, we have:

$\zeta (2)=2\left( \displaystyle\int_{0}^{1/2}\int_{0}^{u}\displaystyle\frac{1}{1-u^{2}+v^{2}}2\,du\,dv+\displaystyle\int_{1/2}^{1}\displaystyle\int_{0}^{1-u}\frac{1}{1-u^{2}+v^{2}}2\,du\,dv\right) =$

$\cdots =\displaystyle\frac{\pi ^{2}}{6}$

Source: A paper by Tom Apostol and Proofs from the book by Martin Aigner and Günter M. Ziegler.

Comment by Américo Tavares | January 9, 2010 | Reply

6. There’s something dreadfully wrong with your iterated integrals when you calculate $\zeta(2)$. Clearly you didn’t read my post at the end of last week closely enough!

Comment by John Armstrong | January 9, 2010 | Reply

7. If you say that “there’s something dreadfully wrong with your iterated integrals when you calculate $\zeta(2)$”, I have to respect your opinion and try to find out.

It can be errors with the above formulas, the notation, the names (the Jacobian determinant, in particular), the arguments, the exposition, or something else I do not see.

My sentence “Computing twice the double integral, one for (…) both integrals are equal, we have” is not clear! I just wanted to explain briefly the second integral formula by symmetry and the areas shrinking (second factor 2) due to the change of variables. The first factor 2 appears because the areas above and under $u$-axe are equal, as well as the associate integrals (page 37 of Aigner and Ziegler’s book).

I will read your “post at the end of last week” again and again till I see where my error(s) is(are).

More details on the references I used: the Apostol’s paper is A Proof that Euler Missed: Evaluating $\zeta(2)$ the Easy Way, and the page numbers of Proofs from the book are 36-37 (Three Times $\pi^2/6$).

Many thanks for your reply, because it helped me. I try not to make errors, but I know I cannot guarantee there is none.

I apologize my poor English and promise not bothering you further on this matter.

Comment by Américo Tavares | January 9, 2010 | Reply

8. How can the integrals with respect to $u$ be taken over intervals that depend on $u$?

Comment by John Armstrong | January 9, 2010 | Reply

9. Only now I understand your question, at least partially. Some parentheses can and/or must be added to avoid any misunderstanding. By the integral

$2\left( \displaystyle\int_{0}^{1/2}\displaystyle\int_{0}^{u}\dfrac{1}{1-u^{2}+v^{2}}2\,du\,dv+\displaystyle\int_{1/2}^{1}\displaystyle\int_{0}^{1-u}\dfrac{1}{1-u^{2}+v^{2}}2\,du\,dv\right)$

I meant

$2\left( \displaystyle\int_{0}^{1/2}du\displaystyle\int_{0}^{u}\dfrac{1}{1-u^{2}+v^{2}}2\,\,dv+\displaystyle\int_{1/2}^{1}du\displaystyle\int_{0}^{1-u}\dfrac{1}{1-u^{2}+v^{2}}2\,\,dv\right)$

or

$2\left( \displaystyle\int_{0}^{1/2}\left( \int_{0}^{u}\dfrac{1}{1-u^{2}+v^{2}}2dv\right) du+\int_{1/2}^{1}\left( \displaystyle\int_{0}^{1-u}\dfrac{1}{1-u^{2}+v^{2}}2dv\right) du\right)$

$=4\displaystyle\int_{0}^{1/2}\dfrac{1}{\sqrt{1-u^{2}}}\tan ^{-1}\left( \dfrac{u}{\sqrt{1-u^{2}}}\right) du+4\displaystyle\int_{1/2}^{1}\dfrac{1}{\sqrt{1-u^{2}}}\tan ^{-1}\left( \dfrac{1-u}{\sqrt{1-u^{2}}}\right) du$

$=4\displaystyle\int_{0}^{1/2}\left( \int_{0}^{u}\dfrac{1}{1-u^{2}+v^{2}}dv\right) du+4\displaystyle\int_{1/2}^{1}\left( \int_{0}^{1-u}\dfrac{1}{1-u^{2}+v^{2}}dv\right) du$

Comment by Américo Tavares | January 9, 2010 | Reply

10. Correction:

Only now I understand your question, at least partially. Some parentheses can and/or must be added to avoid any misunderstanding. By the integral

$2\left( \displaystyle\int_{0}^{1/2}\displaystyle\int_{0}^{u}\dfrac{1}{1-u^{2}+v^{2}}2\,du\,dv+\displaystyle\int_{1/2}^{1}\displaystyle\int_{0}^{1-u}\dfrac{1}{1-u^{2}+v^{2}}2\,du\,dv\right)$

I meant

$2\left( \displaystyle\int_{0}^{1/2}du\displaystyle\int_{0}^{u}\dfrac{1}{1-u^{2}+v^{2}}2\,\,dv+\displaystyle\int_{1/2}^{1}du\displaystyle\int_{0}^{1-u}\dfrac{1}{1-u^{2}+v^{2}}2\,\,dv\right)$

or

$2\left( \displaystyle\int_{0}^{1/2}\left( \int_{0}^{u}\dfrac{1}{1-u^{2}+v^{2}}2dv\right) du+\int_{1/2}^{1}\left( \displaystyle\int_{0}^{1-u}\dfrac{1}{1-u^{2}+v^{2}}2dv\right) du\right)$

$=4\displaystyle\int_{0}^{1/2}\dfrac{1}{\sqrt{1-u^{2}}}\tan ^{-1}\left( \dfrac{u}{\sqrt{1-u^{2}}}\right) du+4\displaystyle\int_{1/2}^{1}\dfrac{1}{\sqrt{1-u^{2}}}\tan ^{-1}\left( \dfrac{1-u}{\sqrt{1-u^{2}}}\right) du$.

Comment by Américo Tavares | January 9, 2010 | Reply