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Old 2008-07-17, 04:27   #34
Visu's Avatar
Nov 2006

3×52 Posts

Originally Posted by R.D. Silverman View Post
You admit you know nothing about the application, but offer "advice"???
You talk about "trying a different exponent" in complete ignorance????

This entire thread is STUPID. The OP was trying to curve fit a function
that counts twin primes... But if the OP had bothered to do any READING
he would have found out that there is a well established THEORY
that tells us what the curve should be. The number of twin primes
less than N is uniformly asymptotic to C N/(log^2 N) where C is the
twin prime constant. [given by an infinite product]. Trying to fit
ANY kind of "power law" curve to this data is MORONIC. It can not
possibly fit a curve of that type......

If you want to know more (i.e. the derivation of C) then ask.

But for someone at the level of the OP to even attempt work in this
area is crazy.
Seems to me that the curve that the OP is referring to is different from the one that you are referring to.
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Old 2008-07-17, 07:36   #35
davieddy's Avatar
Dec 2006

194A16 Posts

Originally Posted by roger View Post
Example: if I have the equation y=0.3165*x1.947, by what method do I find the equivilant equation (with a different k-value) with an n-value of 2?
You can find the Taylor expansion about x=a to obtain
an approximation to the curve near x=a in the form
y(x-a) = c0 + c1*(x-a) + c2*(x-a)^2 + c3*(x-a)^3 + ...

Since 1.947 is close to 2, I would guess that c3,c4... are
smaller than usual.

Last fiddled with by davieddy on 2008-07-17 at 07:44
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Old 2008-07-17, 08:15   #36
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Oct 2006

26010 Posts

I was trying to change the 'messy' equations into the form y=A*n^2, as you yourself, Mr Silverman, suggested earlier (as shown in the post below):
Originally Posted by robert44444uk
Originally Posted by roger
I stitched the posts above into one excel chart and graphed them, with and without my data. I added a trendline, and the equations turned out to be (averages):
k=0.2973*n^1.9555 [for data of n=1->5000, no gaps]
k=0.3823*n^1.9191 [for data of n=1->10000, with gaps]
[EDIT]: 91.5% done data of first 10,000 n-values suggests a curve of 0.3002*n^1.9545.[/EDIT]

Attached is the excel file.
This accords with my findings, but Bob Silverman guided me to retrofit to theory, which deal with A*n^2, with A a variable or a constant, and it was also concluded that we should look at medians rather than averages - try to place the values in percentiles over a block of n and you will see that the rogue values have a large influence over the average.

When you look at A constant, the data fits quite well to A=0.24.

You might want to try to plot Norman's formula that he suggested a couple of days ago.
I got my excel charts to find the A-value for n=2 though. Unfortunately, my data differs a bit. The last viable data point (n=8000) has an A-value of 0.1998. The last time the A-value was 0.24 was n=143. Here's my graph:
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Last fiddled with by roger on 2008-07-17 at 08:16
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