Forum ► Technical Corner ► Approximating Irrational Numbers w/ Other Irrational Numbers

sorry for the double post, but ^{5}√306 is also quite close

edit:^{9}√29809

edit:

126.411 days ago

Apr 12, 2022 - 6:34 PM

126.408 days ago

Apr 12, 2022 - 6:40 PM

its not elegant/simple but it is accurate to 12 digits: ^{29}√261424513280000

126.355 days ago

Apr 12, 2022 - 7:55 PM

ln(-1)/i = π(2k+1) ∀ k ∈ Z (restrict the natural log's complex extrapolation to [0, 2π)*i for best results)

Alternatively use infinite sums to find both!

6*√(Σ1/k²) = π (k all positive integers)

Use this formula for pi later

e^(ln(π²/6))=π²/6

ln(π²/6) = Σ(1/k*1/(p^{k})) (p is all primes, k is all positive integers)

So e = (π²/6)^{1/ln(π²/6)}

If you want to approximate, stop at any prime/integer, but be warned that the way to approximate the natural log converges super slowly and is incredibly intensive on your processor

I get that at no point except in the limit do you ever get an irrational number, but I still like this one

Alternatively use infinite sums to find both!

6*√(Σ1/k²) = π (k all positive integers)

Use this formula for pi later

e^(ln(π²/6))=π²/6

ln(π²/6) = Σ(1/k*1/(p

So e = (π²/6)

If you want to approximate, stop at any prime/integer, but be warned that the way to approximate the natural log converges super slowly and is incredibly intensive on your processor

I get that at no point except in the limit do you ever get an irrational number, but I still like this one

121.939 days ago

Apr 17, 2022 - 5:55 AM

Forum > Technical Corner > Approximating Irrational Numbers w/ Other Irrational Numbers

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"html": "2\u221a3 ~= pi",
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"id": "1209393",
"time": "1649553788",
"html": "even 2\u221a2 is closer to \u03c0 than 2\u221a3",
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"html": "shhh",
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"html": "sqrt2+sqrt3 is pretty close",
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"html": "sorry for the double post, but <sup>5</sup>\u221a306 is also quite close<br /><br /><br />edit: <sup>9</sup>\u221a29809",
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"html": "<sup>5</sup>\u221a306 is very good. Not quite as elegant as <sup>3</sup>\u221a31 but it is a lot closer",
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"time": "1649793334",
"html": "its not elegant/simple but it is accurate to 12 digits: <sup>29</sup>\u221a261424513280000",
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"html": "ln(-1)/i = \u03c0(2k+1) \u2200 k \u2208 <span style=\"font-weight:bold;\">Z</span> (restrict the natural log's complex extrapolation to [0, 2\u03c0)*i for best results)<br />Alternatively use infinite sums to find both!<br />6*\u221a(\u03a31/k\u00b2) = \u03c0 (k all positive integers)<br />Use this formula for pi later<br />e^(ln(\u03c0\u00b2/6))=\u03c0\u00b2/6<br />ln(\u03c0\u00b2/6) = \u03a3(1/k*1/(p<sup>k</sup>)) (p is all primes, k is all positive integers)<br />So e = (\u03c0\u00b2/6)<sup>1/ln(\u03c0\u00b2/6)</sup><br />If you want to approximate, stop at any prime/integer, but be warned that the way to approximate the natural log converges super slowly and is incredibly intensive on your processor<br /><br />I get that at no point except in the limit do you ever get an irrational number, but I still like this one",
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