“A bird does not sing because it has an answer. It sings because it has a song.” ~ Chinese proverb?

I am an applied mathematician. I earn my living doing computer models of power plants. In some of my spare time, I do whatever mathematics interests me.

I have a B.S. from Caltech, and an M.S. from Carnegie-Mellon, both in mathematics. As an undergraduate I took a fair amount of physics, enough that as a grad student, I TA’d freshman physics. Of course, I TA’d calculus; but best of all was designing and teaching a summer course in mathematical modeling, for each of the 3 years I was at CMU.

It might be too conservative to say that my goal is to understand all the mathematics used through second-year graduate school in mathematics, physics, chemical engineering, electrical engineering, and economics. But i’ll go with it for now.

That’s a tall order – for me at least – and I’m not there. Frankly, I don’t really expect to ever get there.

“Ah, but a man’s reach should exceed his grasp, or what’s a heaven for?” ~ Robert Browning (via ThinkExist.com).

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December 7, 2007 at 1:46 pm

It’s about time – Math Guy!

April 5, 2008 at 10:04 am

Hi Rip,

You just answered a question I had on Mathematica to Tex. I was just browsing through your blog. Thanks so much for your reply! And as you note, I want to do what you are doing. And I gather then that you just edit the mathematica output as needed…guess I’ll have to do that too and get used to it! However, wolfram’s mathworld is generated from Mathematica notebooks directly with something called transmogrify, I will look into how they did it.

My wiki is not public yet, otherwise I would show it to you. It is right now just a “Lab notebook” to share with my advisor and thesis committee.

Thanks again,

Ramiro

April 5, 2008 at 12:51 pm

hi ramiro,

glad to have helped. but i learned something, too, from Jean-Marc Gulliet’s post. i had never seen //TeXForm (ok, he wrote TeXForm[stuff], but i prefer the postfix form.)

i’ve just confirmed, for a matrix in MatrixForm, that %//TeXForm gives me what i need to put into a text-only editor. this means i can limit my translations to what i need, instead of saving an entire file as tex.

you asked a great question on mathgroup. good luck, and please let me know about your wiki.

vale,

rip

October 20, 2010 at 10:20 pm

Hi Rip,

I’ve read your articel about the matrix expontial, that helps me to solve a problem with LAPACK. Thanks a lot.

I’ve added a link on my page to your blog, because your work is great

Phil

October 21, 2010 at 2:38 pm

Hi Phil,

Thanks for letting me know you found something useful here. I appreciate it, and the link.

Rip

May 27, 2011 at 7:59 am

dear rip, is there any chance that I could ask you a couple of color math questions? its apparent you are one of a handful of individuals that totally understand how color spectrographic data may be corelated against the human visual response and then derive CIELab values. I have a couple of questions about this topic, and would even be willing to pay you for your time spend answernig my questions. anyways, if you might consider this, please send me an email at my hotmail account.

heres my first such question, so that you know my want here is sincere…

it seems that most spectrophotometers operate in a ‘relative intensity’ (eg: count) mode. the typical spectrographic plots have x axis = wavelength (nm) and the y axis = relative counts. ive been told that its suggested that when one uses a spectrometer, they adjust the ‘integration time’ (eg; like the shutter on a camera) to cause the maximum count value to be at 80% of full scale.

in addition, one may have their spectrometer ‘radiometrically calibrated’, in a manner where the y axis = micro watts / cm squared / nm

so there are two different numerical values that may be obtained in the y axis. either a ‘relative count’, or an ‘absolute micro watt / cm squared / nm’ value.

since you know how to do the calcs from spectral data to CIELab – i was curious how you feel the calculated CIELab values would be ‘different’ if, for example, the set of spectral data was acquired from a ‘relatively’ measured set of data where max value was at 80% full scale, as opposed to spectral data was acquired from a ‘relatively’ measured set of data whare max value was 70%, or 90% of full scale. or if the spectral data was obtained from an ‘absolutely’ measured micro watts / cm squared / nm.

so one will, in practice, get a great variety of spectrographic y value sets. when the same color sample is being read by different instruments.

im unclear, since i dont yet understand all the calcs involved in going from spectral to CIELab – if in the end, these spectral Y values are irrelevant to the CIELab values, or if different y values in the spectral data will give different CIELab values.

May 29, 2011 at 8:28 am

Hi Greg,

First, I appreciate the complement about “totally understanding”, but no, far from it.

Second, I appreciate the offer of payment for my time — but it’s time for this blog rather than money that I need more of.

Third, this question is easy to answer mathematically. Computing CIELab coordinates requires two sets of tristimulus values: X, Y, Z for the color stimulus of interest, and Xn, Yn, Zn for the “reference white”. The equations for L, a, b then use the three ratios X/Xn, Y/Yn, and Z/Zn.

In practice, then, the only question is: are each of X/Xn, Y/Yn, and Z/Zn independent of whether the measurements are relative or absolute? I would hope so. If you measure both X and Xn in relative, or in absolute, mode, I would think that the equipment was designed to give the same answer for the ratio X/Xn.

Try it.

If it doesn’t work, try talking to the manufacturer. Now you’re outside of math and into engineering.

Finally, feel free to ask another question. Bur please put it into one of the color posts rather than here. Perhaps this post.

June 2, 2011 at 3:27 pm

im a little overwhelmed here (like you with time demands) and mentally a bit challenged by all the math you present here. i will try to see if i can understand the formulas to better understand your answer. as i embark on this journey, iwill definitely post my newbie questions to your color blog. thanks very much!

June 3, 2011 at 12:27 pm

There’s certainly a lot of material on my blog… but for the specific question you asked, all you need is the 3 equations for converting from XYZ and (Xn,Yn,Zn) to CIELab, i.e. L*, a*,b*.

And those equations are nowhere on my blog, as of this date. If you don’t have them in your own books, you can find them here:

http://en.wikipedia.org/wiki/Lab_color_space

August 29, 2012 at 11:19 am

Very cool site, stumbled upon it looking for an elementary result in my review HW for Grad Astro. I knew that I knew the way there but…brain fart… oh well, found it on your post of an into to orbital mechanics.

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ABSTRACT

In considering the future of teacher education at the present time, I believe that it is relevant to consider the wider social and political context in which schools and institutions of teacher education are placed at this time. In particular I wish to draw attention to what Prime Minister Tony Blair had to say in his speech to the 1998 Labour Party Conference, where he argued that

1 INTRODUCTION

Since α’’ is always normal to M, that means that the dot product of α’’ and a vector in in TpM is always zero. In particular, α’’ • α’ = 0. Let s(t) be the speed of α at t; . Differentiating s with respect to t,

(1)

This shows that s(t) is a constant function, and so a geodesic must be a constant-speed curve. In fact, we can parameterize any geodesic so that it is unit-speed.

Given an orthogonal coordinate patch x in a geometric surface M, geodesics can be defined by differential equations called, appropriately, the geodesic equations. Consider a curve α in M. Express α(t) = x(u(t), v(t)). Then, , and so

(2)

If α is a geodesic, then it is normal to the surface, and hence

(3) α•xu = 0 and α•xv = 0

So, using (2) and (3) and the fact that xu•xv = 0 yields the differential system

(4) |xu|2 u + u2 xuu•xu + 2uvxuv•xu + v2xvv•xu = 0

(5) |xv|2 v + v2 xvv•xv + 2uvxuv•xv + u2xuu•xv = 0

which a curve must satisfy be a geodesic

to1 INTRODUCTION

These two equations are called the geodesic equations, because solving them gives geodesics. In fact, these equations allow us to generalize the definition of a geodesic to a surface that is not in R3. For such a surface, we merely define a geodesic to be a solution of the geodesic equations. An immediate result of this system of differential equations is the following theorem:

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