Complex numbers with the geometric series doesn't work...?

Question

kainui · Answer

Specifically I'm thinking of this:

$$S = 1+x+x^2+x^4$$
$$xS=x+x^2+x^3+x^5$$
$$xS+1-x^5 = S$$
$$S=\frac{x^5-1}{x-1}$$

Plug in $i^k = x$. Doesn't work. Why?

kainui · Answer

If you don't do the simplification you can check:

$$S(0)=1$$
$$S(1)=S(2)=S(3)=0$$

I guess I'm dividing by zero or something haha.

kainui · Answer

err where $S(k) = 1+i^k+i^{2k}+i^{3k}$

kainui · Answer

S(0)=4 sorry I had normalized it when I was playing around earlier and forgot hehe

kainui · Answer

If this works, I have almost a nice closed form of the divisor counting function, $\tau(n)$. So it would be really great if this worked haha.

parthkohli · Answer

$$S_k = \frac{i^{4k}-1}{i^k-1}$$I mean of course that formula doesn't work for common ratio = 1, haha.

kainui · Answer

$$S(k)=1+i^k+i^{2k}+i^{3k}$$
$$S(0)=1+i^0+i^{0}+i^{0} = 1$$
$$iS(k)=i+i^{k+1}+i^{2k+1}+i^{3k+1}$$
$$iS(k)+1-i^{3k+1}=S(k)$$
$$S(k)= \frac{i^{3k+1}-1}{i-1}$$
Did that fix it, that seems different than what I was thinking it should be...?

kainui · Answer

That new equation gives S(0)=1... ??? I don't think I even ever divided by zero in this entire derivation did I?

kainui · Answer

errm somehow I wrote that 1+1+1+1=1 when really it equals 4, so there's definitely a discrepancy here... and not just in my brain haha.

parthkohli · Answer

Yeah when the common ratio is $1$ then you're subtracting something from itself so you're dividing by zero but this sounds alright idk.

ganeshie8 · Answer

that derivation is wrong  $iS(k)+1-i^{3k+1}=1+i+i^{k+1}+i^{2k+1} \ne S(k) $

kainui · Answer

Ah ok I see, that makes sense why that step is wrong.

ganeshie8 · Answer

$S(k)=1+i^k+i^{2k}+i^{3k} = \dfrac{i^{4k}-1}{i^k-1}$

i think this should work fine for all $k$ such that $k \not \equiv 0\pmod{4}$

kainui · Answer

Alright this makes sense, it's 0 for all the cases it should be with the exception of being 4 when k=0 since I am dividing by zero. I wonder if I can get around this by using L'Hopital's rule and seeing if it's valid for $k \ne 0$ as well?

kainui · Answer

Meh, not too interesting. What a shame.

kainui · Answer

For anyone who's wondering:

$$\lim_{k \to 0} \frac{i^{4k}-1}{i^k-1} = \lim_{k \to 0} \frac{4 i^k \ln i}{i^k \ln i} =\lim_{k \to 0} 4 i^{3k} = 4$$

ganeshie8 · Answer

$$\sum _{k=1}^n\:\left(z+z^2+\cdots+z^k\right)=\frac{nz}{1-z}-\frac{z^2}{\left(1-z\right)^2}\left(1-z^n\right),\:z\ne 1 $$

ganeshie8 · Answer

http://math.stackexchange.com/questions/1012704/complex-numbers-and-geometric-series

kainui · Answer

Ohhh interesting! This is just the kinda thing I was looking for I think.

ganeshie8 · Answer

how does that lhop work ?

kainui · Answer

I fixed a typo in the differentiation step, so since at k=0 you have an indeterminante form, 
$$\frac{i^{4k}-1}{i^k-1} = \frac{0}{0}$$

$$\lim_{k \to 0} \frac{i^{4k}-1}{i^k-1} = \lim_{k \to 0} \frac{\tfrac{d}{dk}(i^{4k}-1)}{\tfrac{d}{dk}(i^k-1)}=\lim_{k \to 0} \frac{4 i^{4k} \ln i}{i^k \ln i} =\lim_{k \to 0} 4 i^{3k} = 4$$

kainui · Answer

Well ok I realize that this is still semi ambiguous because somehow I've left $i^4 \ne 1$ unevaluated and perhaps there are infinitely many periodic things we could have done here so this is quite weird maybe I'm naive in this derivation that ends up giving the right answer.

kainui · Answer

instead of the limit approaching 0 maybe I should have let $k \to 4n$ with n an integer while taking the limit? That would be the true limit in mind at least, however I am quite uncertain about this limit now because it's very uncomfortable to say:

$$\frac{d}{dk} i^{4k} = 4 i^{4k} \ln i$$
Since this looks suspiciously like:
$$\frac{d}{dk} 1 = 0$$

kainui · Answer

On second thought, no.

$$i^{4k} \ne 1$$

proof: let k=1/2.
$$i^{4/2} = -1 \ne 1$$

No issues after all.

kainui · Answer

Alright now I guess I'll try to find the closed form for $\tau (n)$ now that I have this geometric thing after all. I'll show my progress so far:

The geometric series part that is 1 every time it hits a divisor:
$$ \large \nu_k(n) =\frac{1}{k} \sum_{a=0}^{n-1} e^{\frac{i 2 \pi n}{k}a} $$

$$\large \tau(n) = \sum_{k=1}^n \nu_k(n)$$
or you could take the upper bound on that to be infinity I guess, not sure if there's any consequence, but this would be the preferred thing anyways.
$$\large \tau(n) = \sum_{k=1}^\infty \nu_k(n)$$

Thanks, I guess if anyone's interested in playing with this here it is haha.