8 ms·

Weird Constants in Math Problems

96 points by weinzierl 6y ago

seth_tr 6y ago
Seems the first constant is wrong >>> 1/(2 * (1 - math.exp(-2))) 0.5782588213748329 Instead it should be >>> 1/2 * (1 - math.exp(-2)) 0.43233235838169365 Which handily appears in OEIS (a great resource for random constants and sequences) as http://oeis.org/A247847 http://oeis.org/A247847
- madcaptenor 6y ago
  Michael Lugo (author of the original post that Mark was responding to) here. I had written $1/2 (1-e^{-2})$ in my post, meaning the 0.4323 constant, but with the usual rules of precedence you do in fact get 0.5782; he just propagated my error.
  - mjd 6y ago
    I fixed it on my side also. Thanks to both of you for the correction.
jonathanstrange 6y ago
I find series even more fascinating than unexpected constants. Series also show up in the weirdest places and it's often very non-trivial to explain (or not yet explained at all) why the some series shows up in two different areas/applications. Check out the comments in OEIS[1] entries, it's fascinating. [1] https://oeis.org https://oeis.org
tikej 6y ago
This is nice. I think OEIS and in general libraries of constants will be of great help in tackling problems with no analytical solutions. Actually I do a lot of integrals this way – symbolically but checking whether the transformation was legit evaluating definite integral on few random intervals. For me is very helpful and I believe it can be extended to problems beyond integrals.
- throwlogon 6y ago
  > It's easy to use Monte Carlo methods and find that when n is large, the average cluster size is approximately 2.15013 It never occurred to me before, but running simulations and looking up the result in a dictionary of constants might be a helpful learning approach for people who are more adept at going from practice to theory.
  - madcaptenor 6y ago
    The problem with this approach is that to get enough decimal places to be able to look things up in a dictionary of constants can require a very long simulation.
  - nullc 6y ago
    I solve a lot of problems this way, thanks to OEIS. How many fromblaz of size N are there? lets count them for N=1 ... N=8 or whatever is tractable to count using naive methods then hope a useful result shows up in OEIS. OEIS will also return equations for simple linear integer expressions without an OEIS entry. Unfortunately, the tools for single constants are significantly less good than OEIS.
hoseja 6y ago
https://www.wolframalpha.com/input/?i=2.15013 https://www.wolframalpha.com/input/?i=2.15013 It's there as one of "Possible closed forms". it's there even for 0.4323 but further down, you need to request more closed forms.
rocqua 6y ago
I recall doing a numerical approximation to something, and it seemed to converge to about 2.5 Enough for me to, in my overconfidence, assume the constant was 2.5. Later, when I found an exact derivation, it turned out the actual constant was pi^2/4 or 2.46740110027. Which was apparently close enough to fool me.
- madcaptenor 6y ago
  My PhD advisor wrote a paper titled "when is 0.999... equal to 1": https://www.math.upenn.edu/~pemantle/papers/nearly.pdf https://www.math.upenn.edu/~pemantle/papers/nearly.pdf (It wasn't, in the case he was interested in - he had a sum that he suspected might be 1, from numerical evidence, but that actually turned out to be a combination of zeta values equal to 0.999222...)
- deleted 6y ago
  [deleted]
- mjd 6y ago
  A while back I looked at a problem where Monte Carlo simulation said that the answer was probably 4. But analysis showed that the actual answer was 9304682830147 ÷ 2329089562800 = 3.994987…. https://blog.plover.com/math/breaking-pills.html https://blog.plover.com/math/breaking-pills.html
- SamReidHughes 6y ago
  Damn, I was expecting sqrt(2pi) which is 2.5066.
bo1024 6y ago
I think these kinds of constants are very normal (and very cool) in these kinds of problems. "e" shows up a lot. For example, the probability that a given bin is empty is the probability that every ball misses it, or (1 - 1/n)^n --> 1/e.
- SamReidHughes 6y ago
  Another interesting way to formulate it is as a differential equation in the number of empty buckets as a function of time. Given k empty buckets, the probability of picking an empty bucket is k/n. For large values of n and k, we can pretend it's a continuous and not-particularly-stochastic differential equation dk/dt = -k/n, with k(0) = n. That's classic exponential decay, the solution being k = ne^(-t/n). Thus k(n) = ne^(-1).
lifeformed 6y ago
One I found as a kid was 0.73908513321. It appears when you just press the Cos button on a calculator repeatedly (for any starting value). I can't find it used anywhere else.
- qubex 6y ago
  I suppose that’s the limit value of repeatedly taking the cosine of any value and then the cosine of that iterating up to infinity. Sounds easy enough to check, and easy enough to prove, if the hunch proves right.
- esnard 6y ago
  Its name is the "Dottie Number": [0] https://en.wikipedia.org/wiki/Dottie_number https://en.wikipedia.org/wiki/Dottie_number [1] https://mathworld.wolfram.com/DottieNumber.html https://mathworld.wolfram.com/DottieNumber.html
- deleted 6y ago
  [deleted]
- sgdpk 6y ago
  You can see that your constant is the solution to x=cos(x) (if the equality is true, you can replace the x on the right side with cos(x), and repeat this process). The solution apparently has a name, Dottie constant [1]. [1] https://mathworld.wolfram.com/DottieNumber.html https://mathworld.wolfram.com/DottieNumber.html
  - ja3k 6y ago
    Beautifully, iterating any continuous function will yield a fixed point of the function if it converges. https://en.wikipedia.org/wiki/Fixed-point_iteration https://en.wikipedia.org/wiki/Fixed-point_iteration
    - Sharlin 6y ago
      And in case it neither converges to a single value nor diverges to infinity, a whole new worldof wondrous math emerges [1][2][3] [1] https://en.wikipedia.org/wiki/Logistic_map https://en.wikipedia.org/wiki/Logistic_map [2] https://en.wikipedia.org/wiki/Mandelbrot_set https://en.wikipedia.org/wiki/Mandelbrot_set [3] https://en.wikipedia.org/wiki/Attractor#Strange_attractor https://en.wikipedia.org/wiki/Attractor#Strange_attractor
deleted 6y ago
[deleted]
godelzilla 6y ago
A probability problem has Euler's constant in its answer? And numeric approximations are close to that number? Who cares? That's how probability works. Nothing weird.
nullc 6y ago
Too bad that Plouffe's inverter can't seem to stay reliably online, except in intermittently available cut down forms. The database behind it grew huge, making it expensive to operate, and it seems any copy of it online goes down whenever someone graduates.
SamReidHughes 6y ago
One bit of the analysis is left out (if you follow the link to the Math Stack Exchange post) -- the probability of a bin being empty is not independent of the probability of other bins being empty. From the perspective that the number of groupings Y(x) is contingent on the total number of bins filled, X, you might say they're assuming E[Y(X)] = E[Y(E[X])]. But as n -> infinity, the effect of this becomes negligible.

96 points by weinzierl 6y ago

Seems the first constant is wrong >>> 1/(2 * (1 - math.exp(-2))) 0.5782588213748329 Instead it should be >>> 1/2 * (1 - math.exp(-2)) 0.43233235838169365 Which handily appears in OEIS (a great resource for random constants and sequences) as http://oeis.org/A247847 http://oeis.org/A247847

madcaptenor 6y ago
Michael Lugo (author of the original post that Mark was responding to) here. I had written $1/2 (1-e^{-2})$ in my post, meaning the 0.4323 constant, but with the usual rules of precedence you do in fact get 0.5782; he just propagated my error.
- mjd 6y ago
  I fixed it on my side also. Thanks to both of you for the correction.

jonathanstrange 6y ago

I find series even more fascinating than unexpected constants. Series also show up in the weirdest places and it's often very non-trivial to explain (or not yet explained at all) why the some series shows up in two different areas/applications. Check out the comments in OEIS[1] entries, it's fascinating. [1] https://oeis.org https://oeis.org

tikej 6y ago

This is nice. I think OEIS and in general libraries of constants will be of great help in tackling problems with no analytical solutions. Actually I do a lot of integrals this way – symbolically but checking whether the transformation was legit evaluating definite integral on few random intervals. For me is very helpful and I believe it can be extended to problems beyond integrals.

throwlogon 6y ago
> It's easy to use Monte Carlo methods and find that when n is large, the average cluster size is approximately 2.15013 It never occurred to me before, but running simulations and looking up the result in a dictionary of constants might be a helpful learning approach for people who are more adept at going from practice to theory.
- madcaptenor 6y ago
  The problem with this approach is that to get enough decimal places to be able to look things up in a dictionary of constants can require a very long simulation.
- nullc 6y ago
  I solve a lot of problems this way, thanks to OEIS. How many fromblaz of size N are there? lets count them for N=1 ... N=8 or whatever is tractable to count using naive methods then hope a useful result shows up in OEIS. OEIS will also return equations for simple linear integer expressions without an OEIS entry. Unfortunately, the tools for single constants are significantly less good than OEIS.

hoseja 6y ago

https://www.wolframalpha.com/input/?i=2.15013 https://www.wolframalpha.com/input/?i=2.15013 It's there as one of "Possible closed forms". it's there even for 0.4323 but further down, you need to request more closed forms.

rocqua 6y ago

I recall doing a numerical approximation to something, and it seemed to converge to about 2.5 Enough for me to, in my overconfidence, assume the constant was 2.5. Later, when I found an exact derivation, it turned out the actual constant was pi^2/4 or 2.46740110027. Which was apparently close enough to fool me.

madcaptenor 6y ago
My PhD advisor wrote a paper titled "when is 0.999... equal to 1": https://www.math.upenn.edu/~pemantle/papers/nearly.pdf https://www.math.upenn.edu/~pemantle/papers/nearly.pdf (It wasn't, in the case he was interested in - he had a sum that he suspected might be 1, from numerical evidence, but that actually turned out to be a combination of zeta values equal to 0.999222...)
deleted 6y ago
[deleted]
mjd 6y ago
A while back I looked at a problem where Monte Carlo simulation said that the answer was probably 4. But analysis showed that the actual answer was 9304682830147 ÷ 2329089562800 = 3.994987…. https://blog.plover.com/math/breaking-pills.html https://blog.plover.com/math/breaking-pills.html
SamReidHughes 6y ago
Damn, I was expecting sqrt(2pi) which is 2.5066.

bo1024 6y ago

I think these kinds of constants are very normal (and very cool) in these kinds of problems. "e" shows up a lot. For example, the probability that a given bin is empty is the probability that every ball misses it, or (1 - 1/n)^n --> 1/e.

SamReidHughes 6y ago
Another interesting way to formulate it is as a differential equation in the number of empty buckets as a function of time. Given k empty buckets, the probability of picking an empty bucket is k/n. For large values of n and k, we can pretend it's a continuous and not-particularly-stochastic differential equation dk/dt = -k/n, with k(0) = n. That's classic exponential decay, the solution being k = ne^(-t/n). Thus k(n) = ne^(-1).

lifeformed 6y ago

One I found as a kid was 0.73908513321. It appears when you just press the Cos button on a calculator repeatedly (for any starting value). I can't find it used anywhere else.

qubex 6y ago
I suppose that’s the limit value of repeatedly taking the cosine of any value and then the cosine of that iterating up to infinity. Sounds easy enough to check, and easy enough to prove, if the hunch proves right.
esnard 6y ago
Its name is the "Dottie Number": [0] https://en.wikipedia.org/wiki/Dottie_number https://en.wikipedia.org/wiki/Dottie_number [1] https://mathworld.wolfram.com/DottieNumber.html https://mathworld.wolfram.com/DottieNumber.html
deleted 6y ago
[deleted]
sgdpk 6y ago
You can see that your constant is the solution to x=cos(x) (if the equality is true, you can replace the x on the right side with cos(x), and repeat this process). The solution apparently has a name, Dottie constant [1]. [1] https://mathworld.wolfram.com/DottieNumber.html https://mathworld.wolfram.com/DottieNumber.html
- ja3k 6y ago
  Beautifully, iterating any continuous function will yield a fixed point of the function if it converges. https://en.wikipedia.org/wiki/Fixed-point_iteration https://en.wikipedia.org/wiki/Fixed-point_iteration
  - Sharlin 6y ago
    And in case it neither converges to a single value nor diverges to infinity, a whole new worldof wondrous math emerges [1][2][3] [1] https://en.wikipedia.org/wiki/Logistic_map https://en.wikipedia.org/wiki/Logistic_map [2] https://en.wikipedia.org/wiki/Mandelbrot_set https://en.wikipedia.org/wiki/Mandelbrot_set [3] https://en.wikipedia.org/wiki/Attractor#Strange_attractor https://en.wikipedia.org/wiki/Attractor#Strange_attractor

deleted 6y ago

[deleted]

godelzilla 6y ago

A probability problem has Euler's constant in its answer? And numeric approximations are close to that number? Who cares? That's how probability works. Nothing weird.

nullc 6y ago

Too bad that Plouffe's inverter can't seem to stay reliably online, except in intermittently available cut down forms. The database behind it grew huge, making it expensive to operate, and it seems any copy of it online goes down whenever someone graduates.

SamReidHughes 6y ago

One bit of the analysis is left out (if you follow the link to the Math Stack Exchange post) -- the probability of a bin being empty is not independent of the probability of other bins being empty. From the perspective that the number of groupings Y(x) is contingent on the total number of bins filled, X, you might say they're assuming E[Y(X)] = E[Y(E[X])]. But as n -> infinity, the effect of this becomes negligible.