4 ms·

Missing one that still makes no sense to me: The Ramanujan Summation: 1 + 2 + 3 + ⋯ + ∞ = -1/12 https://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_%E2%8B%AF https:/

by niccl 5y ago

Missing one that still makes no sense to me: The Ramanujan Summation: 1 + 2 + 3 + ⋯ + ∞ = -1/12 https://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_%E2%8B%AF https://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_%E2%8B%AF

t8e56vd4ih 5y ago
what? that's incredible ...
- kadoban 5y ago
  Literally incredible, because it's not. It's just really awful notation. It's really more like F(1+2+3+...) = -1/12 for a certain F (pedantically it's not a function I don't think, but whatever).
bpodgursky 5y ago
> where the left-hand side has to be interpreted as being the value obtained by using one of the aforementioned summation methods and not as the sum of an infinite series in its usual meaning
charcircuit 5y ago
If you don't use the regular definitions of how things are defined, you can get surprising results. If I define + as -, it might be surprising if I said 1 + 1 = 0.
barbs 5y ago
This is a good video explaining it: https://www.youtube.com/watch?v=w-I6XTVZXww https://www.youtube.com/watch?v=w-I6XTVZXww Although it relies on a particular interpretation of the left side of the equation, it's used in some areas of physics.
- crvdgc 5y ago
  I find a response video to it more enlightening. It's often found as the number one related video. https://youtube.com/watch?v=YuIIjLr6vUA https://youtube.com/watch?v=YuIIjLr6vUA
Kranar 5y ago
I love that one, what this taught me is that infinite sums are fundamentally different from finite sums, and that when operations performed on the finite are generalized to work with the infinite, there are often ambiguities and subtleties about how to perform that generalization. We wish to use the same familiar notation with the infinite that we do with the finite, but we must keep in mind that while they do share similarities, they are not the same operation. The way certain ambiguities are resolved in order to extend the finite to the infinite can lead to counterintuitive results and absurdities that may not even be apparent at first. For me, seeing how one approach to generalizing infinite sums yields the -1/12 result, and how that result actually has some relevance in physics is quite profound and insightful and I am happy that you reminded me of it.
macrolocal 5y ago
Ok, here's my attempt at an elementary explanation. Consider the function f(N) = floor(N) = 1 + 1 + 1 + ... + 1, whenever N's an integer = 1^0 + 2^0 + 3^0 + ... + N^0. Then f(N) ~ N - 1/2 is f's simplest unbiased polynomial estimator. It's a succinct description of f and its asymptotic behavior. Next, let's do something like integration, and increase those zeroth powers to one, ie. consider the function g(N) = 1^1 + 2^1 + 3^1 + ... + (N - 1)^1 = 1 + 2 + 3 + ... + (N - 1). In this case, the estimator ends up being g(N) ~ N^2/2 - N/2 - 1/12 ~ integral(N - 1/2) - 1/12, where I've written this estimate as the difference of a divergent term and a constant term. Here's the point: the divergent term is really just left over (integrated up or 'induced') from our asymptotic description of the degree 0 sum. In a sense, it's degenerate and doesn't provide any sort of 'new' information. So in the same sense, the interesting semantic content of 1 + 2 + 3 + ... is just the constant term: -1/12. Ps. This way of thinking is related to trace formula truncation.

by niccl 5y ago

t8e56vd4ih 5y ago

what? that's incredible ...

kadoban 5y ago
Literally incredible, because it's not. It's just really awful notation. It's really more like F(1+2+3+...) = -1/12 for a certain F (pedantically it's not a function I don't think, but whatever).

bpodgursky 5y ago

> where the left-hand side has to be interpreted as being the value obtained by using one of the aforementioned summation methods and not as the sum of an infinite series in its usual meaning

charcircuit 5y ago

If you don't use the regular definitions of how things are defined, you can get surprising results. If I define + as -, it might be surprising if I said 1 + 1 = 0.

barbs 5y ago

This is a good video explaining it: https://www.youtube.com/watch?v=w-I6XTVZXww https://www.youtube.com/watch?v=w-I6XTVZXww Although it relies on a particular interpretation of the left side of the equation, it's used in some areas of physics.

crvdgc 5y ago
I find a response video to it more enlightening. It's often found as the number one related video. https://youtube.com/watch?v=YuIIjLr6vUA https://youtube.com/watch?v=YuIIjLr6vUA

Kranar 5y ago

I love that one, what this taught me is that infinite sums are fundamentally different from finite sums, and that when operations performed on the finite are generalized to work with the infinite, there are often ambiguities and subtleties about how to perform that generalization. We wish to use the same familiar notation with the infinite that we do with the finite, but we must keep in mind that while they do share similarities, they are not the same operation. The way certain ambiguities are resolved in order to extend the finite to the infinite can lead to counterintuitive results and absurdities that may not even be apparent at first. For me, seeing how one approach to generalizing infinite sums yields the -1/12 result, and how that result actually has some relevance in physics is quite profound and insightful and I am happy that you reminded me of it.

macrolocal 5y ago

Ok, here's my attempt at an elementary explanation. Consider the function f(N) = floor(N) = 1 + 1 + 1 + ... + 1, whenever N's an integer = 1^0 + 2^0 + 3^0 + ... + N^0. Then f(N) ~ N - 1/2 is f's simplest unbiased polynomial estimator. It's a succinct description of f and its asymptotic behavior. Next, let's do something like integration, and increase those zeroth powers to one, ie. consider the function g(N) = 1^1 + 2^1 + 3^1 + ... + (N - 1)^1 = 1 + 2 + 3 + ... + (N - 1). In this case, the estimator ends up being g(N) ~ N^2/2 - N/2 - 1/12 ~ integral(N - 1/2) - 1/12, where I've written this estimate as the difference of a divergent term and a constant term. Here's the point: the divergent term is really just left over (integrated up or 'induced') from our asymptotic description of the degree 0 sum. In a sense, it's degenerate and doesn't provide any sort of 'new' information. So in the same sense, the interesting semantic content of 1 + 2 + 3 + ... is just the constant term: -1/12. Ps. This way of thinking is related to trace formula truncation.