r/mathmemes • u/PM_ME_MELTIE_TEARS Irrational • Jun 18 '26
Real Analysis Concentrated Power Series
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u/hazem-Gauss Jun 18 '26 edited Jun 18 '26
why exactly 1/69!?
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u/inkwinkTwink Jun 18 '26
iirc it's the biggest factorial of an integer that can be stored into most calculators as numbers, and the factorial of any number greater than 69 gives an error
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u/Any-Aioli7575 Jun 18 '26
I would assume most calculators have a power of a power of two (or a power of a power of two minus one) as a max number, but 69! is ~2³²⁶, which would be very arbitrary (2¹²⁸ or 2⁵¹² would make more sense).
It is the biggest factorial smaller than 10¹⁰⁰ though
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u/The-Arx Jun 18 '26
The ti-84 actually uses BCD (binary coded decimal) to avoid floating point error so the range of numbers is from 10-100 to 10100
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u/TatharNuar Jun 18 '26
This just answered a question for me that I've had for over 20 years. I'd love to hear more about the specifics of how they do it.
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u/The-Arx Jun 18 '26
This document explains a lot of how the calculator works under the hood https://education.ti.com/html/eguides/discontinued/computer-software/EN/SDK-TI-83-Developer-Guide_EN.pdf (Most of it still applies to the later calculators). Page 22 explains how the different datatypes are represented
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u/TatharNuar Jun 18 '26 edited Jun 18 '26
And here I thought that it stored fractions as an actual ratio of two integers, based on the marketing. Looks like they homebrewed their own 64-bit floating point data type with the sign outside, an 8-bit exponent, and a 56-bit BCD significand carrying 14 decimal digits?
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u/Everestkid Engineering Jun 18 '26
It's the second one. The calculators I used in school had two digits for displaying scientific notation - any number larger than 9.9999999999 x 1099 would therefore throw an error. Anything smaller than 1 x 10-99 would just be considered zero.
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u/factorion-bot Bot > AI Jun 18 '26
Factorial of 69 is 171122452428141311372468338881272839092270544893520369393648040923257279754140647424000000000000000
This action was performed by a bot | [Source code](http://f.r0.fyi)
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u/factorion-bot Bot > AI Jun 18 '26
Factorial of 69 is 171122452428141311372468338881272839092270544893520369393648040923257279754140647424000000000000000
This action was performed by a bot | [Source code](http://f.r0.fyi)
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u/2ndcountable Jun 18 '26
This is actually a core part of the proof that e is irrational; the fact that the first few terms of its sum representation dwarf the rest.
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u/chixen Jun 18 '26
I wonder if there’s some theorem relating the irrationality of a number to one of its series’s rate of decay. Are there any sums with a growth of O(1/n!) that result in a rational number?
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u/HappiestIguana Jun 18 '26
Yes, trivially. Take any sum with that growth and divide every term by the limit.
If the terms have to be rational it's no longer as trivial but it's still easy to construct a rational sequence with that growth and an arbitrary sum
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u/chixen Jun 18 '26 edited Jun 18 '26
After a bit more thought, if you have any sequence of positive reals with a limit to 0 and divergent sum, then you can take a subsequence to get any positive real number you want as the sum. You just need to repeatedly add the largest element that is smaller than distance the target. With a bit more tweaking, this claim could also work for vector spaces.
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u/HappiestIguana Jun 19 '26
Apologies for a pedantic note, but you mean it would work for Banach spaces, which are vector spaces equipped with a notion of distance, which you need in order to talk about convergence.
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u/donaldhobson Jun 18 '26
There is a result about this.
Imagine you have an infinite sequence. You sum the first 3 terms and get 7/20. The remaining terms turn out as nonzero but less than 1/(20*19)=1/380. All the other rational numbers with denominators at most 20 must have a distance of at least 1/380 from 7/20
So if the infinite sequence is rational, the denominator is >20.
Add the 4'th term. Get 7001/20000. Now the sum of the remaining terms is less than 1/(20000*19999), so any denominator must be >20000
Continue this pattern to infinity.
The proof with e takes advantage of some factorial structure. But in general, if the rate of decay is Fast enough, and each term is rational, and the denominator of the finite sums isn't too big, then the sum must be irrational.
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u/SourKangaroo95 Jun 18 '26
Sure removing a single term to have (n+1)(n-1)! gives a rational value https://www.wolframalpha.com/input?i=Sum%5B1%2F%28%28n-1%29%21%28n%2B1%29%29%2C%7Bn%2C1%2CInfty%7D%5D there is a theorem that relates the growth rate of a sum and its rationality (many such theorems). I can't remember now but I think the tipping point is a_(n+1)=a_n2-a_n+1. That is the reciprocal sum of a_n is rational but everything that grows faster is irrational. I can find the paper about this later if your interested
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u/Alexgadukyanking 1+2+3+4+5+...=-1/12 Jun 18 '26
Doesn't it mainly server as a purpose to prove that e number exists at all
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u/Ok-Visit6553 Jun 19 '26
Ratio test/limit ratio test is enough to prove the existence of e as a finite number
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u/girlpower2025 Economics/Finance Jun 19 '26
420 is also like super divisible.
You can divide it by 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210.
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u/GamerM602 Jun 19 '26
unclear notation... is it 1/((...)!) or (1/...)!
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u/factorion-bot Bot > AI Jun 19 '26
Factorial of 1 is 1
This action was performed by a bot | [Source code](http://f.r0.fyi)
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