lucid dome
Idk I just thought I would try to solve this sum for fun I guess... If anyone wants to give it a try. It kind of involves sinc functions with basically exponentials and although it seems like it telescopes believe me I tried and doesnt seem to telescope. Both x and \sigma are both independent inputs where \sigma can be any given real number. Same goes to x. Also, it converges for given values of x and \sigma.
fathom coral
I'll try it I guess
safe wedge
that's x to the power of the expression, right?
lucid dome
Gave a pretty interesting result
which is?
fathom coral
What did you do to get that
sharp pikeBOT
twitch Tamplr
lucid dome
Idk
safe wedge
"e^is-e^-is " can be replaced by 2i sinx
lucid dome
"e^is-e^-is " can be replaced by 2i sinx
yh ik it just looks better ngl
lucid dome
What did you do to get that
boreal axle
yh ik it just looks better ngl
you think it looks better than $$\frac{\sigma^2 x^{\csc \sigma}\ln x}{6 \sin(\sigma)(x^{\csc \sigma}-1)}?$$
sharp pikeBOT
\nothing
lucid dome
**\\nothing**
yh. You dont agree? Idk. Just looks cool with complex numbers I guess
boreal axle
well i don't think we should make the expression for a real value more complicated by adding complex numbers in it
lucid dome
you got a point
eitherways I did end up getting a cool formula for any given integer n too. Kinda looks nice
$\sum_{k=1}^{\infty}\left(\frac{x^{\frac{1}{\prod_{m=1}^{k}\left(1-\frac{\sigma^{2n}}{\left(m\pi\right)^{2n}}\right)}}-x^{\frac{1}{\prod_{m=1}^{k-1}\left(1-\frac{\sigma^{2n}}{\left(m\pi\right)^{2n}}\right)}}}{x^{\frac{1}{\prod_{m=1}^{k}\left(1-\frac{\sigma^{2n}}{\left(m\pi\right)^{2n}}\right)}}-1}\right)=\frac{2^{2n-1}B_{2n}\sigma^{2n}x^{\prod_{k=0}^{n-1}\csc\left(\sigma e^{\frac{i\pi k}{n}}\right)}\left(\prod_{k=0}^{n-1}\csc\left(\sigma e^{\frac{i\pi k}{n}}\right)\right)\ln x}{\left(2n\right)!\left(x^{\prod_{k=0}^{n-1}\csc\left(\sigma e^{\frac{i\pi k}{n}}\right)}-1\right)}$
sharp pikeBOT
twitch Tamplr
lucid dome
so satisfying to see a chaotic sum equal to a nice clean closed form
I do also like the approach I took ngl it was somewhat clean
boreal axle
am i doing something wrong
lucid dome
am i doing something wrong
upperbound is finite
boreal axle
yeah but i changed it from 100 to 200 and the graph didn't really change
lucid dome
yeah but i changed it from 100 to 200 and the graph didn't really change
but does it slowly meet with the lines of the closed form?
like reaaaaaaallllly slowly
boreal axle
here's the 10th, 20th, ..., 200th partial sums
anything is possible it just doesn't look like it's going there
lucid dome
anything is possible it just doesn't look like it's going there
let me check ur equations real quick. Maybe I missed something
boreal axle
i'll check out your proof and see if it convinces me you're right more than this convinces me you're wrong
lucid dome
i'll check out your proof and see if it convinces me you're right more than this...
sounds good
the proof is heavily dependent on the framework so like if you want to know what the framework actually is then I do have like a full latex showing what it basically is and like the basics.
lucid dome
am i doing something wrong
I also did it graphically and it works just fine. Also, I see your problem. The upperbound of your sum is so big (200) that Desmos can’t process it properly and therefore freezes the function’s graph which may seem like a mismatch but it’s actually just an overflow of smaller and smaller terms and values that emerges from the approximated infinite sum. On my end, by plugging instead 10 to around number lower then 100 it works flawlessly. Maybe try that instead of 200.
You can even try to put a variable at the upperbound and see what happens when that value gets closer and closer to 100
boreal axle
I also did it graphically and it works just fine. Also, I see your problem. The ...
that didn't fix it
let's take $\sigma = x = 0.1.\$
i claim that each term of the sum is positive. as proof, $0 < \sigma < m\pi$ for all $m \ge 1$, so each term of the product is strictly between $0$ and $1$, so the inverse of the product is a strictly increasing function of $k$. thus, since $0 < x < 1$, the numerator and denominator of the term are both strictly negative, so the division is positive.$\$
now, we prove that the equality cannot hold. we calculate the $k = 1$ term as $$\frac{0.1^{1.001014..} - 0.1^1}{0.1^{1.001014..} - 1} \approx 0.000259$$
which is strictly larger than
$$\frac{0.1^2 0.1^{\csc 0.1} \ln 0.1}{6\sin(0.1)(0.1^{\csc 0.1} - 1)} \approx 3.7\cdot 10^{-12}$$
sharp pikeBOT
\nothing
boreal axle
i think it may hold if either sigma = 1 or x = 1
as $x \to 1$, by lhop the fraction approaches $$\frac{\frac{1}{\prod_{m=1}^{k}\left(1-\left(\frac{\sigma}{m\pi}\right)^{2}\right)}-\frac{1}{\prod_{m=1}^{k-1}\left(1-\left(\frac{\sigma}{m\pi}\right)^{2}\right)}}{\frac{1}{\prod_{m=1}^{k}\left(1-\left(\frac{\sigma}{m\pi}\right)^{2}\right)}}$$
$$= 1-\left(1-\left(\frac{\sigma}{k\pi}\right)^{2}\right) = \frac{\sigma^2}{k^2\pi^2}$$ so the sum is $$\frac{\sigma^2}{\pi^2} \frac{\pi^2}{6} = \frac{\sigma^2}6$$ meanwhile the claimed answer approaches $$\frac{\sigma^2 1^{\csc \sigma} 1/x}{6\sin(\sigma)\csc(\sigma)} = \frac{\sigma^2}{6}$$
sharp pikeBOT
\nothing
boreal axle
i think it may hold if either sigma = 1 or x = 1
no, for sigma = 1 it is too large again
lucid dome
i think it may hold if either sigma = 1 or x = 1
then I must of done a small mistake somewhere in my derivation process. But it does seem logical
I think the error was substituing the variable by an exponential to then cancel the sin at the exponential of x. That would make sense
lucid dome
as $x \to 1$, by lhop the fraction approaches $$\frac{\frac{1}{\prod_{m=1}^{k}\l...
so we obtain $\zeta(2)$ if we substitute and cancel sigmas on both sides and bring ${\pi}^2$ on the right hand side
sharp pikeBOT
twitch Tamplr
lucid dome
let's take $\sigma = x = 0.1.\\$
i claim that each term of the sum is positive. ...
yeah that is a good proof but also notice that for my derivation process I specifically used x = 1 and sigma = 1 hence the reason why it works when both inputs are equal to one. So I just need to retake the same path I used to solve the problem but with different variables this time to extract an exact closed form of some sort. Maybe it'll work potentially
boreal axle
oh bruh you only checked this for x = 1?
lucid dome
oh bruh you only checked this for x = 1?
nono I ended by substituing with x = 1 and then continued. It is only at the end that I must of done a variable slip accidentally after reintroducing the variables.
yk?
also I am talking about the old x variable which was used as an exponential in one of the steps
not the one as the base
umbral apex
Idk I just thought I would try to solve this sum for fun I guess... If anyone wa...
what is this N symbol?
sharp pikeBOT
twitch Tamplr
umbral apex
Oh they are products
Ty but what they does?
sharp pikeBOT
twitch Tamplr
sharp pikeBOT
twitch Tamplr
lucid dome
or if you want $1(2)(3)(4)(5)$ then you should use $\prod_{k=1}^{5} k$
sharp pikeBOT
twitch Tamplr
lucid dome
if ur still confused, you could ask ChatGPT or another AI
thats mostly what they are made for
full tangle
**\\nothing**
OMG what is this 🤯 Soon I will be on that lvl...
umbral apex
okok ty
lucid dome
OMG what is this 🤯 Soon I will be on that lvl...
I mean… it’s kind of easy to understand. Just don’t overthink it
hardy totem
There is no closed form it's fantasy
hardy totem
lucid dome
There is no closed form it's fantasy
well its not fantasy it was just a false result. Happens yk
¯_(ツ)_/¯
hardy totem
well its not fantasy it was just a false result. Happens yk
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