/sci/ - Science & Math

(a) 0.

>>9388887
You gotta supply a proof. Also, that has to be wrong.

>>9388745
The solutions are
[eqn] \frac{1}{1985} \sqrt{\pi} e^{-3970} \sqrt{1985} \\
- \frac{8}{15} \pi^8 \\
1 - \gamma [/eqn]

Proof:
Think

>>9388901
I know the first and third one are wrong. The second one is probably also wrong.

>>9388926
They are all correct.
For the first one you can do the substitution [math]u = \frac{n}{x} [/math]. Then you're left with
[eqn] \lim_{n \to \infty} \left( \int_1^n \frac{1}{u^2} f(u) du \right) [/eqn]
This integral can easily be solved by Maple like the other ones.

>>9388969
>u-floor(u) is the distance to the nearest integer.

The nearest integer for 5.9 is 6 so the distance is 0.1. But according to you it would be 5.9 - 5 = 0.9

This really gets that nogging jogging, you know. It gives that really deep insight into how retarded someone who just plugs and chugs is.

Anyways, I do trust the first two solutions because maple did them, not you. It is just that a real cool solution would imply using advanced techniques like relating the integrals to the gamma function and the riemann zeta function.

So, those are the hints. Hopefully someone smarter takes them and shares something cool. But for now, we can safely point and laugh at the retard who thinks that the distance to the nearest integer of 5.9 is 0.9

>>9388976
>advanced techniques like relating the integrals to the gamma function and the riemann zeta function.
well there goes my integration by parts attempt

>>9388994
That's fine. First, I'm sure there are other techniques that could yield solution. And second, the third one does not need this stuff. Typical integration is enough, even simple stuff like finding the antiderivative. It is just that you need to be very careful with what you do. So even if you do not know these techniques you can still attempt the third one. And after you have a complete explicit expression for the integral then for the final step try looking at the different known product representations for [math] \pi [/math] and see if you can spot a pattern or similarity with what you have.

[eqn]\lim_{n \to \infty} \left( \frac{1}{n} \int_1^n f \left( \frac{n}{x} \right)dx \right) = 2 \ln(2) - \ln(\pi) [/eqn]

[math]\int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}}\lnb{\dfrac{2x^{2}+2x+1}{2x^2-2x+1}}dx [/math]

>>9388969
>needing software to solve the last integral
[eqn]\int_k^{k+1} \frac{x - \left \lfloor x \right \rfloor}{x^2} \ dx = \int_0^1 \frac{x}{(x+k)^2} \ dx = \int_0^1 \frac{x+k-k}{(x+k)^2} \ dx = \left.
\ln(x+k) + \frac{k}{x+k}
\right|_{0}^1 = \ln\frac{k+1}{k} - \frac{1}{k+1}[/eqn]
Then just sum for k from 1 to n-1 to obtain
[eqn]\int_1^n \frac{x - \left \lfloor x \right \rfloor}{x^2} \ dx = \ln n - H_n + 1[/eqn]
and plug in the definition of the Euler-Mascheroni constant.

This doesn't give the closest integer anyway, but still, step it up brainlet.

>>9388745
[eqn]\int_0^\infty \frac{x^7}{1-e^x}dx = \int_0^\infty \frac{x^7 e^{-x}}{e^{-x}-1}dx = \int_0^\infty - x^7 e^{-x} \left ( \sum_{n=0}^\infty e^{-nx}\right ) dx = - \sum_{n=0}^\infty \int_0^\infty x^7 e^{-(n+1)x} dx = - \sum_{n=0}^\infty \frac{1}{(n+1)^8} \int_0^\infty u^7 e^{-u} du = -\zeta(8) \Gamma(8) = - \frac{\pi^8}{9450} 5040 = -\frac{8}{15}\pi^8 [/eqn]
The integral and infinite sum can be exchanged because the Taylor series for [math]\frac{1}{1-x}[/math] converges absolutely uniformly on compacta, and [math]e^{-x}[/math] is bounded (above by 1) on [math][0,\infty)[/math], so the series converges uniformly. And since x = 0 is a removable singularity of the integrand, we can represent the integral as the limit of the integral between [math]\epsilon[/math] to [math]R[/math] as [math]\epsilon \rightarrow 0[/math] and [math]R \rightarrow \infty[/math] and guarantee that the series converges along the whole interval the integral is taken. And you can calculate [math]\zeta(8)[/math] using contour integrals if you want to go all the way.

the distance to nearest integer function is just going to be a triangle wave right
for 1<=x<=2 it will be 1-1/x because f(x) is closer to 1 than zero, but after 2 it's closer to zero so it's just going to be 1/x
you could do the same thing for n/x but it would be easier to just make a substitution and then separate the integral into two integrals

will do it when i get home