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How to crunch numbers in Mathematica. How to properly use it to test if a given method solves faster

Posted 1 month ago

I have been trying to factor semiPrimes for years now. I have a simple algebraic equation that gives one factor knowing only N. The only problem I have to find the value of x where y equals 1. I thought this was a simple thing, but with 200 digit numbers finding where y equals 1 on the graph takes some computation. I believe the computation is much smaller than recursive division, but factoring it is the only way to be sure.

My question is how is Mathematica on computation? If I programmed the math algorithm in Mathematica will it be as efficient as a Xeon processor crunching numbers. I have tried computations in Mathematica before and it let me know it was processing but I had no information on the processes it was going through.

And what functions in Mathematica would be used to find y equals 1 on a graph that is f(x} equals x?

Seems like it should be easy. If I were doing the calculation by hand, I would choose test values and find the derivative or slope. But to find the x value with a 200 digit number there would be many close options for x. I know there are equations to solve this but I have never used them.

So I need to know if Mathematica can handle the computational problem of 200+ digits, and if I think my method is less computational will Trurl’s method will be less computational to work in Mathematica.

POSTED BY: Bobby Joe Snyder
4 Replies
Posted 10 days ago

If I programmed the math algorithm in Mathematica will it be as > efficient as a Xeon processor crunching numbers?

Mathematica has a Benchmark command. You can run it on your Xeon machine and compare its results with those of other processors. If you have MMA installed on machines with other processors, you can run the Benchmark on them and compare results with those running on the Xeon machine.

POSTED BY: Mark Harder
POSTED BY: Michael Rogers

The symbolic solution is found easily:

Solve[1 == (((((pnp^2/x) + x^2))/x)/pnp), x]
Solve[1 == (((((pnp^2/x) + x^2))/x)/pnp), x, Reals]

I have no idea how fast it would be for number crunching.

POSTED BY: Gianluca Gorni
POSTED BY: Bobby Joe Snyder
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