4 ms·

No need to invoke the Riemann Zeta function and analytic continuation. In high school they summed this series(valid for x < 1): 1 + x + x^2 ... = 1/(1-x) Plug

No need to invoke the Riemann Zeta function and analytic continuation. In high school they summed this series(valid for x < 1): 1 + x + x^2 ... = 1/(1-x) Plug in x = 2 to get: 1 + 2 + 4 + 8 ... = -1 There's a million of these series in your dusty old Calculus textbook. Or you could look in 'generatingfunctionology' to find others. Complex analysis makes it more interesting, because the additional Cauchy-Riemann constraints make the solution unique. And so people are more willing to say that the unique solution is the "true" answer. Really, I say this is just another example of the complex numbers being weird. I took a couple different courses in it(one in college, 2 online) because it was clear something interesting was happening with them, but there was never a unifying theme. I can kind of spot a rule like "analytic functions preserve 90-degree angles" in the Cauchy-Riemann equations, but it hardly explains all the crazy theorems. That series is just another example of someone taking a well-behaved series, analytically continuing it into the complex numbers, and now it's clear something interesting is happening but it's not clear what.

thaumasiotes 10y ago
Who says it's only valid for |x| < 1? That series isn't just a cute bit of trivia, it is the basis for signed integer mathematics in your CPU. "True" enough for you?
- messe 10y ago
  > Who says it's only valid for |x| < 1? Anybody who understands the definition of convergent series. > That series isn't just a cute bit of trivia, it is the basis for signed integer mathematics in your CPU. "True" enough for you? Eh. That's only half true. It's related, certainly, by means of a similarity between 2-adic numbers and two's complement representation, but I'd hardly call it the basis for signed integer mathematics.
- MichaelBurge 10y ago
  Are you doing something like this: * Define the series over the 2-adic numbers rather than the reals * Associate every 2-adic integer with its corresponding sequence in the inverse limit Z/2^nZ for n = 1..infinity * Truncate the sequence to a certain precision(an index i), and make an equivalence class a ~ b if a and b agree on the first i elements. You get a field Z_(2^n), and identities in 2-adic analysis should carry down to the equivalence classes. Here 1 + 2 + 4 ... = -1 in the 2-adics. That isn't the way I would've thought about signed integers, but it seems like it would work.
  - thaumasiotes 10y ago
    It's not necessary. Just interpret the bits as coefficients of powers of 2, in a purely conventional manner. Say that leading digits match the high bit rather than being 0 regardless of the high bit. And you're done, you have working signed arithmetic that displays all the behavior you expect and conforms to the geometric series equality. Here 1 + 2 + 4 + ... = -1 in the integers.

thaumasiotes 10y ago

Who says it's only valid for |x| < 1? That series isn't just a cute bit of trivia, it is the basis for signed integer mathematics in your CPU. "True" enough for you?

messe 10y ago
> Who says it's only valid for |x| < 1? Anybody who understands the definition of convergent series. > That series isn't just a cute bit of trivia, it is the basis for signed integer mathematics in your CPU. "True" enough for you? Eh. That's only half true. It's related, certainly, by means of a similarity between 2-adic numbers and two's complement representation, but I'd hardly call it the basis for signed integer mathematics.
MichaelBurge 10y ago
Are you doing something like this: * Define the series over the 2-adic numbers rather than the reals * Associate every 2-adic integer with its corresponding sequence in the inverse limit Z/2^nZ for n = 1..infinity * Truncate the sequence to a certain precision(an index i), and make an equivalence class a ~ b if a and b agree on the first i elements. You get a field Z_(2^n), and identities in 2-adic analysis should carry down to the equivalence classes. Here 1 + 2 + 4 ... = -1 in the 2-adics. That isn't the way I would've thought about signed integers, but it seems like it would work.
- thaumasiotes 10y ago
  It's not necessary. Just interpret the bits as coefficients of powers of 2, in a purely conventional manner. Say that leading digits match the high bit rather than being 0 regardless of the high bit. And you're done, you have working signed arithmetic that displays all the behavior you expect and conforms to the geometric series equality. Here 1 + 2 + 4 + ... = -1 in the integers.