Activity for Mithrandir24601
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #281586 |
Whether or not it's off-topic is up to the community - for reference, we created a category on physics for 'problems', where people essentially ask for help with their attempts to whatever problem (not 'the solution', just help), which is similar-ish to the idea of homework and exercises (more) |
— | almost 3 years ago |
| Comment | Post #278558 |
Worth mentioning that the paper has now been taken down as they were detecting "dead time in the NuSTAR detectors". This is apparently "further evidence for why standard timing methods should not be used with NuSTAR data" (more) |
— | over 3 years ago |
| Comment | Post #278638 |
I'd added an additional (yet more convoluted) edit that should explain in more detail what's going on but honestly, 'it's required because $10nx - x$ is an integer' is by far the best answer because this whole 'left concatenation' process follows immediately and directly from that. Call it a geometri... (more) |
— | over 3 years ago |
| Comment | Post #278638 |
"with change ... the change being what needs to be added to digits further to the left." Is this carrying over process. What more do you want me to say? All you're doing is multiplying a digit by n and shifting it to the left one place. Repeat, summing everything up, this is *precisely* what's going... (more) |
— | over 3 years ago |
| Comment | Post #278638 |
I've updated my answer (although honestly, I feel that the first bit is about 100 times clearer). I've included the bit about the powers of $n$ and what you're referring to as 'backwards concatenation' is what I'm referring to as 'multiplying a digit by $n$ to get the digit to the left'. It works bec... (more) |
— | over 3 years ago |
| Comment | Post #278553 |
Except that's not what I'm doing here, so why is that relevant to the question? (more) |
— | over 3 years ago |
| Comment | Post #278553 |
I'm not sure who told you that, but we [really don't](https://en.m.wikipedia.org/wiki/Degree_of_coherence#Degree_of_first-order_coherence) - is useful *because* it doesn't average over time (or frequency if that's your thing). In classical physics, $g^{(2)}$ is intensity correlations. Anyway, this qu... (more) |
— | over 3 years ago |
| Comment | Post #278553 |
See my comment [to this answer](https://math.codidact.com/a/278553/278562) - we're generally not time averaging anything - the expectation values are taken at specific spatial and temporal locations (more) |
— | over 3 years ago |