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3

Check out the following FullSimplify[(x^2 + y^2) Cos[ 4 ArcTan[y/x]] - ((8 x^4)/(x^2 + y^2) - 7 x^2 + y^2)] (* 0 *) The two expressions are identical so it is not suprising that they produce identical plots.

18

There are several important things about the way computer systems represent real numbers, which most of the time can be blithely ignored, just like the safety of bridges in the United States. One important thing is that numbers are discrete. With regular machine precision (double precision), the mantissa has 53 bits, which provides a lot of resolution. ...

2

Since your number is quite large, you can use Stirling's approximation to do this. It's also very common to use this approximation in statistical mechanics: For large number $n$ $$\log(n!)\approx n\log(n)-n$$ So in your case $$n=\frac{Nn}{3}$$ then  \log\left[\frac{(3n)!}{n!\times n!\times n!}\right]\\ =\log[(3n)!]-3\log[n!]\\ \approx 3n \log(3n)-3n ...

0

It looks like there isn't a good way to bypass Overflow[]. However, for very large factorial calculations, it's useful and incredibly accurate to use Stirling or Nemes approximations, depending on the size of the factorial. @Mathematica devs, an idea- maybe catch overflow errors, tell Factorial to substitute the Stirling or Nemes approximation, then try to ...

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