Science Forums 2

[modest]

Originally Posted by Don Blazys

I can't get your graph no matter how many times I click on it.

It's just an image. It's nothing interactive. I graphed it using this website:
Graphing Calculator

EDIT: in case you mean you can't see them at all, I've changed the images to jpg format which any browser can view so long as "show images" is selected in your internet browser's settings. Private message me if you have continued trouble.

Originally Posted by Don Blazys

Also I don't calculate .8818 as a "zero" but as -6501.971... , so it's probably an asymptote, and not a zero.

I believe you're correct. The site I used obviously misinterpreted that part of the function - it does look like an asymptote.

Originally Posted by Don Blazys

By the way, your LaTex rendition is almost correct, exept for the subtraction of unity at the end.

Fixed.

Originally Posted by CraigD

Here’s a way to generate a starting value for Don’s generating formula...

That is really clever thinking. I also checked it up to 500 primes successfully.

Originally Posted by CraigD

More importantly, we haven’t proven that the as given by

Plugging your 500-digit into Don's gives a value no closer to zero than Don's 25 digit approximation of x.

Originally Posted by CraigD

If modest or someone else with a working program like the one in post #28 can modify it to use the new equation (it shouldn’t be hard to differentiate, needing only the product and chain rules, and some carefulness), it would be interesting to compare A few hundred digits of the estimated zero of to my big rational number above. To 500 decimals, it’s...

Using an online derivative finder, I get,
=

((sin(x^(1/2)))^(-1)-1)^(-1) / ((π)^2 + (ln(ln(2*((2* (x^(-1) +1))^(-1) +1))))^(-1))-1

=

-(((-3*π^2*cos(√(x))*x^2-5*π^2*cos(√(x))*x-2*π^2*cos(√(x)))*ln(3*x+2)
+(3*π^2*cos(√(x))*x^2+5*π^2*cos(√(x))*x+2*π^2*cos( √(x)))*ln(x+1))*ln(ln(3*x+2)
-ln(x+1))^2+((-3*cos(√(x))*x^2-5*cos(√(x))*x-2*cos(√(x)))*ln(3*x+2)+(3*cos(√(x))*x^2
+5*cos(√(x))*x+2*cos(√(x)))*ln(x+1))*ln(ln(3*x+2)-ln(x+1))+(2*sin(√(x))^2
-2*sin(√(x)))*√(x))/(√(x)*((((6*π^4*sin(√(x))^2-12*π^4*sin(√(x))+6*π^4)*x^2
+(10*π^4*sin(√(x))^2-20*π^4*sin(√(x))+10*π^4)*x+4*π^4*sin(√(x))^2
-8*π^4*sin(√(x))+4*π^4)*ln(3*x+2)+((-6*π^4*sin(√(x))^2+12*π^4*sin(√(x))-6*π^4)*x^2
+(-10*π^4*sin(√(x))^2+20*π^4*sin(√(x))-10*π^4)*x-4*π^4*sin(√(x))^2+8*π^4*sin(√(x))
-4*π^4)*ln(x+1))*ln(ln(3*x+2)-ln(x+1))^2+(((12*π^2*sin(√(x))^2-24*π^2*sin(√(x))
+12*π^2)*x^2+(20*π^2*sin(√(x))^2-40*π^2*sin(√(x))+20*π^2)*x+8*π^2*sin(√(x))^2
-16*π^2*sin(√(x))+8*π^2)*ln(3*x+2)+((-12*π^2*sin(√(x))^2+24*π^2*sin(√(x))-12*π^2)*x^2
+(-20*π^2*sin(√(x))^2+40*π^2*sin(√(x))-20*π^2)*x-8*π^2*sin(√(x))^2+16*π^2*sin(√(x))
-8*π^2)*ln(x+1))*ln(ln(3*x+2)-ln(x+1))+((6*sin(√(x))^2-12*sin(√(x))+6)*x^2+(10*sin(√(x))^2
-20*sin(√(x))+10)*x+4*sin(√(x))^2-8*sin(√(x))+4)*ln(3*x+2)+((-6*sin(√(x))^2+12*sin(√(x))-6)*x^2
+(-10*sin(√(x))^2+20*sin(√(x))-10)*x-4*sin(√(x))^2+8*sin(√(x))-4)*ln(x+1)))

Graphing (green) and (red) gives:



Red does look like green's derivative to me. Unless it can be simplified algebraically I don't have much hope for Newton's method, but without any better ideas, I could give it a try.

Originally Posted by CraigD

From a practical computational perspective, like so many candidate prime number generators, this one’s not very useful, because in order to generate many primes, its must be very precise, and arithmetic on very precise numbers take a lot of computing time and CPU. From a math insight perspective, however, it’s wonderful if true, and a nifty (and maybe not easy) exercise to prove true or not.

I agree with both points.

~modest