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# Advanced Math/Speed of Light Computation.

Question
Dear Prof Scott

en.wikipedia.org/wiki/Speed_of_light
www.speed-light.info
galileoandeinstein.physics.virginia.edu/lectures/spedlite.html

The Speed of Light = 299792458 m/s - Universal Physical constant.

1 What is the method used for computing the speed of Light and how this value for the speed of light given above is computed and derived ?.

2 Is there a single method or there can be several methods to calculate
the Speed of Light and derive the above value?.

Thanks & Regards,
Prashant S Akerkar

Numbers from 1 to 100
In http://www.speed-light.info/, the speed of light has been found to be  299792.458 km/s.
Depending on the motion and which direction of movement there is, it varies ever so slightly.

In http://galileoandeinstein.physics.virginia.edu/lectures/spedlite.html, the year 1728 is mentioned.  What I find it interesting about 1728 is that I recognize that number as 12*12*12,
which reminds me that 1729 is the smallest sum of two different cubes.  1729 = 1^3 + 9^3 and 1729 = 10^3 + 12^3.  Not only is it the smallest number with this property, it is the only number I know off the top of my head.  In my dozens and dozens of papers on numbers, I think I found another, but can't recall what it is.  By the way, these papers have several lines on each the number from 1 to several hundred, and numbers here and there out to 1,073m741,824.
They have that because it is 2^30, and they also have every power of 2 between 1 and 30.

That paper also has other information, like the difference of squares, the number of factors, the numbers in different bases, etc.  More on that later.

The calculate the speed of light, take the known time between sightings of one of the planets, the observed time, and the distance to that planet.  The difference between when it happens in observation and when it should occur give two different values.  Taking the distance over the difference in time gives the speed of light.

This can be done in many different planets and even different objects in space, so there are many ways.

Now back to what I was talking about ...

First I give what I have, then I explain it.  Also, there are even more examples given.

As an example, it has, for 105, that it is
105:=3*5*7; f:3/3/8; =sum(1:14); =prod_odd(1:7); =126B9; (1+2+6=9) =151B8; =210B7; =253B6{B6B10}; =410B5;{4+1=5}; =1,221B4;{ref}; =10,220B3; =1,101,001B2; =11^2-4^2; =13^2-8^2; =19^2-16^2; =53^2-52^2;

What this means is
f:3/3/8: 3 unique prime factors, 3 prime factors, and 8 total factors;
=sum(1:14) means it is the sum of 1+2+3+4+5+6+7+8+9+10+11+12+13+14;
=prod_odd(1:7) means it is 1*3*5*7;
=126B9 means it is 126 in base 9; it also gives it in base 8, base 7, base 6, base 5, base 5, base 3, and base 2.

For the difference in squares, this is known because of the many factors.  For example,
105 = 7*15.  The difference in squares can be found by taking the difference and sum each over 2.  15-7=8, and 8/2 = 4, and 15+7=22 and 22/2 = 11.  Since this is true, it is 11^2 = 4^2 =
121 - 16 = 105.

As another example is 6.  Since 6 = 2*3, and (3-2)/2 = .5, and (3+2)/2 = 2.5, it is known that 6 = 2.5^2 - 0.5^2.  This gives 6 = 6.25 - .25, which is correct.

From this, if there are many factors, there are many difference in squares.

For the number of factors, the total can be found by taking the number of each unique factor, adding to one to that, and multiplying together.  For example, 2^5 = 32, so there are 5+1=6 factors.  288=*2^5)(3^2), so there are (5+1)(2+1) = 6*3 = 18 factors.  I'll give a smaller number to demonstrate this.

On the factors, consider 18.  The factors are 2/3/6.  This is because it is divisible by 2 and 3 (thus the 2), it is equal to 2*3*3 (thus the 3), and the factors are 1,2,3,6,9, and 18 (6 factors).

For larger numbers, I only have the info down on a few, not every number.  In my last file (of which there are almost 30 files), I have 11 pages on numbers between 105,000 and 12,345,678,987,654,321.  For the last one, it is = [3^3]*[37^3].  It is also 111,111,111^2.
The number of factors are 2/6/16.  That is, two factors, 3 and 37; there are 6 total, as in =3*3*3*37*37*37, and there are 16 factors.

For a number to have an odd number of factors, it has to have an even number of each factor.
For example, 2^2=4, factors = 1,2,4, so there are 3; (2^4)(3^2) = 144, and there are
(4+1)(2+1) factors, which is 5*3 = 15.

One of the other things of interest is difference in higher order power.  Such as
26 = 3^3 - 1^3, 61 = 5^3 - 4^3, 665 = 3^6 - 2^6, etc.  I know the way to find these is very complicated, but it is there.  Squares are simple because if c = a^2 - b^2, then
c = (b+a)(b-a).  This method was used to do what I did earlier on finding the difference of squares.

I'm not sure if this is of interest, but there is the first file attached.
I haven't used this for a long time and don't remember how files are to be sent,
but this might be right. I believe it chops of in the 80's.

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#### Scott A Wilson

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I can answer any question in general math, arithetic, discret math, algebra, box problems, geometry, filling a tank with water, trigonometry, pre-calculus, linear algebra, complex mathematics, probability, statistics, and most of anything else that relates to math. I can also say that I broke 5 minutes for a mile, which is over 12 mph, but is that relevant?

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My master's paper was published in the OSU journal. The subject of it was Numerical Analysis used in shock waves and rarefaction fans. It dealt with discontinuities that arose over time. They were solved using the Leap Frog method. That method was used and improvements of it were shown. The improvements were by Enquist-Osher, Godunov, and Lax-Wendroff.

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Master of Science at OSU with high honors in mathematics. Bachelor of Science at OSU with high honors in mathematical sciences. This degree involved mathematics, statistics, and computer science. I also took sophmore level physics and chemistry while I was attending college. On the side I took raquetball, but that's still not relevant.

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I earned high honors in both my BS degree and MS degree from Oregon State. I was in near the top in most of my classes. In several classes in mathematics, I was first. In a class of over 100 students, I was always one of the first ones to complete the test. I graduated with well over 50 credits in upper division mathematics.

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My clients have been students at OSU, people who live nearby, friends with math questions, and several people every day on the PC. I would guess that you are probably going to be one more.