Skip to main content

A personal annotations of Veach's thesis (15) p.168

p.168 D3 and Equation 5.30 updated

I got a comment about Equation 5.30 from my friend Daniel. However, Equation (1) has absolute value operator at |f'(x_0)|, I don't understand this yet. To see this Equation, we could think about this is a chain rule. If we write this in an integral form (as the substitution rule) as in the Equation (2). This is substantially equivalent with the chain rule. But still, the domain doesn't match with Equation 5.30.


On the other hand, if we read the thesis until p.170, Equation 5.35 is a definition and this looks like showing a linearity. It is Equation (3). Maybe the question is, what is the linear coefficient a_{\beta}. That's my guess. What coefficient makes this equation consistent through the Dirac's \delta, that could be the definition of Equation 5.35.

This has an advantage, if this can be, we don't need integrate every time, this \delta looks like just an ordinary function. This is convenient. (as the Veach said in the text, this kind of stuff is the great part of this paper.)

The following is a slightly changed from the paper, but, basically the same with p.170's Equations. Notice, \beta is a bijective function and \beta(x) = x', x = \beta^{-1}(x'),
Therefore, comparing the first equation and the last one, we have
Yes, finally we got it in this way. In the paper, this x and x' replacement is not at one point, maybe some reason, but, for my understanding, I replace them at once.

In the end, the domain of measure linearly changed. This is interesting. But, integration operator is a linear operator, maybe it is natural. Still, this result is interesting.

The last question, why |f'(x_0)| has the absolute operator? This Equation 5.30's  question still remains for me. But it is much better, I think I almost see it. (or I see an illusion as usual...)


Acknowledgements:
I thank Daniel S. to give me hints of this question. I also thank his great patience on my stupid basic questions.

Comments

Popular posts from this blog

Why A^{T}A is invertible? (2) Linear Algebra

Why A^{T}A has the inverse Let me explain why A^{T}A has the inverse, if the columns of A are independent. First, if a matrix is n by n, and all the columns are independent, then this is a square full rank matrix. Therefore, there is the inverse. So, the problem is when A is a m by n, rectangle matrix.  Strang's explanation is based on null space. Null space and column space are the fundamental of the linear algebra. This explanation is simple and clear. However, when I was a University student, I did not recall the explanation of the null space in my linear algebra class. Maybe I was careless. I regret that... Explanation based on null space This explanation is based on Strang's book. Column space and null space are the main characters. Let's start with this explanation. Assume  x  where x is in the null space of A .  The matrices ( A^{T} A ) and A share the null space as the following: This means, if x is in the null space of A , x is also in the n...

Gauss's quote for positive, negative, and imaginary number

Recently I watched the following great videos about imaginary numbers by Welch Labs. https://youtu.be/T647CGsuOVU?list=PLiaHhY2iBX9g6KIvZ_703G3KJXapKkNaF I like this article about naming of math by Kalid Azad. https://betterexplained.com/articles/learning-tip-idea-name/ Both articles mentioned about Gauss, who suggested to use other names of positive, negative, and imaginary numbers. Gauss wrote these names are wrong and that is one of the reason people didn't get why negative times negative is positive, or, pure positive imaginary times pure positive imaginary is negative real number. I made a few videos about explaining why -1 * -1 = +1, too. Explanation: why -1 * -1 = +1 by pattern https://youtu.be/uD7JRdAzKP8 Explanation: why -1 * -1 = +1 by climbing a mountain https://youtu.be/uD7JRdAzKP8 But actually Gauss's insight is much powerful. The original is in the Gauß, Werke, Bd. 2, S. 178 . Hätte man +1, -1, √-1) nicht positiv, negative, imaginäre (oder gar um...

Why parallelogram area is |ad-bc|?

Here is my question. The area of parallelogram is the difference of these two rectangles (red rectangle - blue rectangle). This is not intuitive for me. If you also think it is not so intuitive, you might interested in my slides. I try to explain this for hight school students. Slides:  A bit intuitive (for me) explanation of area of parallelogram  (to my site, external link) .