$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$.
Why should analytically continuing to $\zeta(-1)$ give the right answer?
They attach a value to divergent series, which is, as far as I can tell, an implicit mistake.
Numberhile's James Grime (and others) did the same, and used the same "trick" to "prove" $$1-1+1-1+\ldots=\frac{1}{2},$$ which is true only if we consider Cesaro summability. And more importantly, this very same method can be easily shown to yield contradictions.
If you mean that you cannot consistently attach a single value to any divergent series without other context, then yes they're "wrong"
But whenever such sums emerge in physics (and some context is math) you can show rigorously that taking the analytic continuation of the zeta function is the right thing to do
Without that context, you lose that justification.
@VincenzoOliva @MikeMiller Well, Lubos (who I agree with apparently the majority of Math.SE is incredibly abrasive most of the time) begins his answer by stating exactly what he means by "right"
@VincenzoOliva Well without context, sure it yields contradictory results. But in the right context its totally fine
I would never teach at a class that $\displaystyle 1+2+3+\cdots =-1/12$, but I'd try to explain that Riemann zeta function continue to exist outside the values $Re(s)>1$.
Let me try to explain. Consider a pretty simple integral $\int_{-\infty}^{\infty} \text{sech}( \frac{\pi}{2} x) dx$
Clearly the integral converges. The answer happens to be $2$
Now, suppose that we go to the complex plane and note that the function decreases everywhere exponentially in the UHP, except along the imaginary axis
So strictly speaking the circular part of the contour does not necessarily give 0 contribution and you can't close this way and naively apply the residue theorem.
But suppose you do it anyway! What answer do you get?
Well, you get $2 \pi i (\frac{2 i}{\pi})(-1 +1 -1 +1.....)$
And if you use Cesaro summation of these residues, you get exactly the correct result in the end. $2$
@VincenzoOliva Now, of course there's a very rigourous way to compute this integral and show that doing the Cesaro sum is the 'right thing' in this case. But it's tedious and a longer process than just doing what I did. So we skip it!
All this hemming and hawing about the series doesn't converge and this and that just gets in the way of getting to the correct answer for most practitioners
Arguing about this is like saying someone should always use the real numbers as constructed from Dedekind cuts or Cauchy sequences. You do it once to show that it works and its consistent and then you just skip it from then on, treating the real numbers like you always have.
@VincenzoOliva I think it is. It is not know that the soundness and consistency of first order logic with identity (AKA arithmetic) is undecidable within that same system.
But no one goes around worrying that addition is inconsistent and maybe there's no way to assign a definite meaning to the expression that we write down
@KevinDriscoll I have yet to study integrals, so I can't say much about that. But remaining on divergent series, the fact that one wants to show $\zeta(1)=- 1/12$ does not allow to use that method, because accepting it gives contradictory results in the same context. (and the same (divergent) series used in the Numperhile's "proof")
And I think you're optimistic about no one teaching it. If you watch those videos, you'll see that there are actual mathematicians that seriously think: "Your calculator will always give bigger and bigger numbers, but that's because you can't sum up all the integers! If you could, you'd find $\displaystyle -\frac{1}{12}$".
We can assign a value to the regularized sum, but it's pointless to say that summing all the integers gives that.
@Chris'ssis I do :) (you did warn me about how hard it is to quit ... :P every 3rd day I think of quitting it and on every second day I buy a new pack :P lol)
@r9m Have you ever experience very tired eyes? This is what I've experienced for a couple of day, I cannot focus on things anymore ... :-(. I worked too much in the last period of time, without breaks.
@Gato It would be useful, especially in a room with lots of conversations, to link your comments to the ones to which they reply.
@Gato well, you've shown one $e^{ip}$ near $1$, but you need to show density. You have been looking at this for a while, so you may see that $\{e^{ipn}\}_n$ partition the circle, but it would be nice to show that for people who have not been looking at the problem for a long time.
@DanielFischer Hello!!! :) I had a test today in Data Structures.. It was given an AVL-tree and I had to find the height of the node with key k.
@DanielFischer I tried the following:
Height(T,K){ if (T==NULL) return; pointer R; R=LookUp(T,K); int k=height(R); return k; }
height(pointer R){ static j=0,l=0; if (R==NULL) return 0; if (R->lc!=NULL) l=1+height(R->lc); if (R->rc!=NULL) j=l-height(R->lc)+height(R->rc); if (R->rc==NULL and R->lc==NULL) return 1; return max(l,j); }
Could you tell me if it is right or completely wrong? :/
@Hippalectryon Related to my question with radicals I posted yesterday, do you have any idea how to do it? I mean an idea, not a solution. I'm curious to see the perception of people on it.
@TedShifrin Taking a break by doing mathematics here, you are courageous! Yes I work on it. I have shown one $e^{ip}$ near $1$ so to prove the density I need to prove that $\{e^{ipn}\}_n$ partition the circle.
$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$.
Why should analytically continuing to $\zeta(-1)$ give the right answer?
