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Avoid overflow for pandigital Sabey formula for calculation of e?

Posted 7 years ago

I want Mathematica to help computationally validate Richard Sabey’s pan-digital equation for e. The equation is:

(1+9^-4^6*7)^3^2^85

Just the part outside the parens makes the system overflow (i.e. 3^2^85).

Is it possible to modify the precision or use arbitrary precision to compute this equation which is supposedly a correct approximation of e to a trillion digits.

I would even be happy to validate it to Mathematica's maximum approximation, which I have apparently exceeded so far. Is this machine precision?

Thanks in advance for any assistance.

Ref: Incredible Formula - Numberphile

enter image description here

POSTED BY: Aeyoss Antelope
4 Replies

Thank you Mariusz and Daniel.

I do understand the derivation of the formula.

However, since the Mma documentation talks about “arbitrary precision”, I guess I thought that the system would, upon being given something as ill-behaved as this, emit a partial result to machine precision (i.e. 10^308) or just run for a “long time” and then emit an approximation to much less than 10^24 digits (the magnitude of the “near miss”).

Mma doumentation says 2*$MaxMachineNumber converts to “arbitrary precision”.

I guess this is actually (1+”too small number” raised to large power) raised to a “too large number”) because 9^-4^42 underflows.

N[E,10^5] result shown immediately

N[E,10^10] computation result $Aborted after 1+ hour

I guess I do not understand the implementation limits of “arbitrary precision” in Mma.

Ref:

http://www.johndcook.com/blog/2014/03/30/amazing-approximation-to-e/

http://mathworld.wolfram.com/eApproximations.html

POSTED BY: Aeyoss Antelope

3^(2^85) > 2^(2>85), and that latter has 2^85 bits. Any computation attempting to deal with a number that big is going to need substantially more atoms than are available on this planet. So yes, of course this will overflow.

POSTED BY: Daniel Lichtblau

The point of the formula is to give something correct to more places than you can compute. To understand it rewrite 9^(-(4^(67))) as (3^2)^(-(4^(67))) = 3^(-(24^(67))). This is the same as 3^(-(2^85)). So you have an expression of the form

(1+1/n)^n

where n=3^(-(2^85)). To estimate the error between this and E, take logs and see how close we get to 1, then use that to get the multiplicative factor by which we might be off. So the log is of the form n*Log[1+1/n] with n again as above. Expand as a series, truncate at second order, this is n*(1/n-1/(2*n^2)+...) or 1-1/(2*n)+error where error<<1/n. Exponentiating shows our result is E*Exp[-1/(2*n)+error], now use what we know about n to show how close this comes to E.

POSTED BY: Daniel Lichtblau

Maybe this this helps why is overflow, but not solve your questions.

POSTED BY: Mariusz Iwaniuk
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