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The Proof is Trivial!

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Original post by bananarama2
Since when have the compscis become such realists?


When they had to work within the confines of real hardware? (seriously) :tongue:
Original post by ukdragon37
When they had to work within the confines of real hardware? (seriously) :tongue:


Tell bloody games designers that, designing games that run on computers that cost 32 billions pounds :nothing: :rant:
Original post by bananarama2
Tell bloody games designers that, designing games that run on computers that cost 32 billions pounds :nothing: :rant:


If you fancy a girl who only dates handsome guys, and you aren't up to the standard, who do you blame exactly? :tongue:

Clearly those people make enough money catering to those who do have the adequate hardware than to be bothered with you. :tongue:
Original post by ukdragon37
If you fancy a girl who only dates handsome guys, and you aren't up to the standard, who do you blame exactly? :tongue:

Clearly those people make enough money catering to those who do have the adequate hardware than to be bothered with you. :tongue:


Well I think there's a market for a game which runs on just not quite the best computer.

(I have an i7 and some games still run like it's a DOS machine.)
Reply 1064
Avatar for Slumpy
Slumpy
16
Original post by bananarama2
Well I think there's a market for a game which runs on just not quite the best computer.

(I have an i7 and some games still run like it's a DOS machine.)


I love DOS games thank you very much...
Original post by Slumpy
I love DOS games thank you very much...


I don't disagree :smile: Lemmings and Legacy FTW.
Original post by bananarama2
Well I think there's a market for a game which runs on just not quite the best computer.

(I have an i7 and some games still run like it's a DOS machine.)


There are plenty! And many of them are great too! E.g. Minecraft, many of the games from the Humble Indie Bundles....

Also for something most definitely new you can try looking at Kickstarter and invest in the start-up games at an amount that promises an eventual copy when it is released. Most of them are not hardware-intensive so as to appeal to the broadest market. Plus you get a nice surprise a few months down the line when the game you invested but forgot about sends you something.
(edited 10 years ago)
Original post by ukdragon37
There are plenty! And many of them are great too! E.g. Minecraft, many of the games from the Humble Indie Bundles....

Also for something most definitely new you can try looking at Kickstarter and invest in the start-up games at an amount that promises an eventual copy when it is released. Most of them are not hardware-intensive so as to appeal to the broadest market.


I guess, but they are all of a certain style. Minecraft is good, although surprisingly memory intensive :biggrin:

I'll have a look :smile:
Reply 1068
Avatar for MW24595
MW24595
And here I am, thinking, if I didn't have an Internet connection that flickered 20 times a minute, I'd be blessed.
Original post by MW24595
And here I am, thinking, if I didn't have an Internet connection that flickered 20 times a minute, I'd be blessed.


Admittedly it's a spoiled kid problem I have.:smile:
Cool stuff!

Solution 163

Define ti=a+i(b−a)nt_i=a+\dfrac{i(b-a)}{n}

n{∫abf(x) dx−b−an∑i=1nf(ti)}=n∑i=1n{∫ti−1tif(x) dx−b−anf(ti)}=n∑i=1n{∫ti−1tif(x)−f(ti) dx}=n∑i=1n{ti−1[f(ti)−f(ti−1)]−∫ti−1tixf′(x) dx}=n∑i=1n{ti−1∫ti−1tif′(x) dx−∫ti−1tixf′(x) dx}=∑i=1n∫ti−1tin(ti−1−x)f′(x) dx=∫0b−a−x∑i=1n[1n f′(xn+ti−1)] dx    (n(x−ti−1)→x)=∫0b−a−xb−a∫abf′(z) dz dx    (n→∞)=b−a2(f(a)−f(b))\displaystyle\begin{aligned} n\left\{\int_a^b f(x)\,dx-\frac{b-a}{n}\sum_{i=1}^n f(t_i) \right\} &=n\sum_{i=1}^n \left \{ \int_{t_{i-1}}^{t_i} f(x)\,dx-\frac{b-a}{n} f(t_i) \right\} \\&=n\sum_{i=1}^n \left \{ \int_{t_{i-1}}^{t_i} f(x)-f(t_i)\,dx \right \} \\ &=n\sum_{i=1}^n \left \{ t_{i-1}\Big[ f(t_i) -f(t_{i-1}) \Big]-\int_{t_{i-1}}^{t_i} xf'(x)\,dx\right\} \\ &=n\sum_{i=1}^n \left \{ t_{i-1}\int_{t_{i-1}}^{t_i}f'(x)\,dx-\int_{t_{i-1}}^{t_i} xf'(x)\,dx\right\} \\&=\sum_{i=1}^n\int_{t_{i-1}}^{t_i} n(t_{i-1}-x)f'(x)\,dx \\& = \int_{0}^{b-a}-x\sum_{i=1}^n\left[\frac{1}{n}\, f'\left(\frac{x}{n}+t_{i-1}\right)\right]\,dx\;\;(n(x-t_{i-1})\to x) \\&=\int_0^{b-a} -\frac{x}{b-a} \int_a^b f'(z)\,dz\,dx\;\;(n\to\infty) \\&= \frac{b-a}{2}\Big( f(a)-f(b)\Big) \end{aligned}

Solution 164

Unparseable latex formula:

\begin{aligned} \displaystyle\int_0^{ \infty} \frac{\arctan x}{x^{\alpha}} \,dx=\frac{1}{\alpha-1}\int_0^{ \infty} \frac{dx}{x^{\alpha-1}(1+x^2)}



Set t=(1+x2)−1:\displaystyle t=(1+x^2)^{-1}:

∫0∞dxxα−1(1+x2) dx=12∫011t(1t−1)−α2 dt=12∫01tα2−1(1−t)−α2 dt=12 B(α2, 1−α2)=12 Γ(α2)Γ(1−α2)=π2csc⁡(πα2)    (r.f.)\displaystyle \begin{aligned} \int_0^{ \infty} \frac{dx}{x^{\alpha-1}(1+x^2)} \,dx &= \frac{1}{2} \int_0^{1} \frac{1}{t} \left( \frac{1}{t}-1 \right)^{-\frac{\alpha}{2}} \,dt \\ &= \frac{1}{2} \int_0^{1} t^{\frac{ \alpha}{2}-1} \left(1-t \right)^{-\frac{\alpha}{2}} \,dt\\ &=\frac{1}{2}\,\text{B}\left( \frac{ \alpha}{2},\, 1-\frac{\alpha}{2} \right) \\& = \frac{1}{2}\, \Gamma \left(\frac{\alpha}{2}\right) \Gamma \left(1-\frac{\alpha}{2}\right) \\&= \frac{\pi}{2} \csc \left(\frac{\pi \alpha}{2} \right)\;\;(\text{r.f.})\end{aligned}

Hence ∫0∞arctan⁡xxα dx=π2(α−1)csc⁡(πα2)\displaystyle\int_0^{ \infty} \frac{\arctan x}{x^{\alpha}} \,dx= \frac{\pi}{2(\alpha-1)} \csc\left(\frac{ \pi \alpha}{2}\right)

Time for bed. :biggrin: Here is a nice twist on a classic integral:

Problem 168*

∫0π2ln⁡sin⁡xln⁡cos⁡x dx\displaystyle\int_0^{\frac{\pi}{2}} \ln \sin x\ln \cos x\,dx
(edited 10 years ago)
Original post by Lord of the Flies

Spoiler



Well done.

Original post by jack.hadamard

Spoiler



By the way, the second question is much more difficult. I can't think of anything simpler than L-functions.


Solution 168

Let I=∫0π2ln⁡sin⁡xln⁡cos⁡x dx\displaystyle I = \int_{0}^{\frac{\pi}{2}} \ln \sin x \ln \cos x \, dx; J=∫0π2ln⁡2tan⁡x dx\displaystyle J = \int_{0}^{\frac{\pi}{2}} \ln^{2} \tan x \, dx; K=∫0π2ln⁡2sin⁡x dx\displaystyle K = \int_{0}^{\frac{\pi}{2}} \ln^{2} \sin x \, dx

It is clear that K=∫0π2ln⁡2sin⁡x dx=x↦π2−x∫0π2ln⁡2cos⁡x dx\displaystyle K = \int_{0}^{\frac{\pi}{2}} \ln^{2} \sin x \, dx \mathop =^{x \mapsto \frac{\pi}{2}-x} \int_{0}^{\frac{\pi}{2}} \ln^{2} \cos x \, dx
Furthermore, ln⁡2sin⁡2x dx=x↦x2∫0π2ln⁡2sin⁡x dx=K\displaystyle \ln^{2} \sin 2x \, dx \mathop =^{x \mapsto \frac{x}{2}} \int_{0}^{\frac{\pi}{2}} \ln^{2} \sin x \, dx = K

Now, 2I+2K=∫0π2(ln⁡cos⁡x+ln⁡sin⁡x)2 dx=K−3π2ln⁡22\displaystyle 2I+2K = \int_{0}^{\frac{\pi}{2}} \left( \ln \cos x + \ln \sin x \right)^{2} \, dx = K - \frac{3\pi}{2} \ln^{2} 2.

We also have 2K−2I=∫0π2(ln⁡sin⁡x−ln⁡cos⁡x)2 dx=J\displaystyle 2K - 2I = \int_{0}^{\frac{\pi}{2}} \left( \ln \sin x - \ln \cos x \right)^{2} \, dx = J. Thus I=π2ln⁡22−16J\displaystyle I = \frac{\pi}{2} \ln^{2} 2 - \frac{1}{6}J

Here we shall be done if we manage to evaluate JJ.

We first observe that J=tan⁡x↦x∫0∞11+x2ln⁡2x dx\displaystyle J \mathop =^{\tan x \mapsto x} \int_{0}^{\infty} \frac{1}{1+x^{2}} \ln^{2} x \, dx.

Secondly, we consider f(z)=ln⁡3z1+z2\displaystyle f(z) = \frac{\ln^{3} z}{1+z^{2}}. This function has simple poles at z=±iz = \pm i.


∫γR,ϵln⁡3z1+z2 dz=∫ϵRln⁡3x1+x2 dx−∫ϵR(ln⁡x+2iπ)31+x2 dx+∫CR+∫Cϵ=2iπ(Res(f(z),i)+Res(f(z),−i))\begin{aligned} \displaystyle \int_{\gamma_{R, \epsilon}} \frac{\ln^{3} z}{1+z^{2}} \, dz &= \int_{\epsilon}^{R} \frac{\ln^{3} x}{1+x^{2}} \, dx - \int_{\epsilon}^{R} \frac{( \ln x + 2i \pi)^{3}}{1+x^{2}} \, dx + \int_{C_{R}} + \int_{C_{\epsilon}} \\& = 2i\pi \left(Res(f(z),i)+Res(f(z),-i) \right) \end{aligned}


Clearly, as R→∞R \to \infty and ϵ→0\epsilon \to 0, we have ∫CR→0\displaystyle \int_{C_{R}} \to 0 and ∫Cϵ→0\displaystyle \int_{C_{\epsilon}} \to 0


Separating the real and imaginary parts, we get
Unparseable latex formula:

\begin{aligned} \displaystyle -6\pi i J + 8\pi^{3} i \int_{0}^{\infty} \frac{1}{1+x^{2}} \, dx = 2\pi i \left(Res(f(z),i)+Res(f(z),-i)) \end{aligned}




We easily find Res(f(z),i)=−π316\displaystyle Res(f(z),i) = -\frac{\pi^{3}}{16} and Res(f(z),−i)=27π316\displaystyle Res(f(z),-i) = \frac{27\pi^{3}}{16}.

Thereupon, −3J+2π3=13π38\displaystyle -3J + 2\pi^{3} = \frac{13\pi^{3}}{8}, which gives J=18π3\displaystyle J = \frac{1}{8}\pi^{3}.

Therefore, I=π2ln⁡22−148π3\displaystyle I = \frac{\pi}{2} \ln^{2} 2 - \frac{1}{48} \pi^{3}.
Reply 1072
Avatar for Zakee
Zakee
Original post by bananarama2
Admittedly it's a spoiled kid problem I have.:smile:



A potato generated internet connection would be more stable than his connection.
I think we need an amendment to the first post. **** now means LOTF and Mladenov difficulty.
Original post by bananarama2
I think we need an amendment to the first post. **** now means LOTF and Mladenov difficulty.

:lol:
Or rather, * isn't really A-Level knowledge :cool:
Reply 1075
Avatar for Jkn
Jkn
11
Original post by Lord of the Flies

Unparseable latex formula:

\begin{aligned} \displaystyle\int_0^{ \infty} \frac{\arctan x}{x^{\alpha}} \,dx=\frac{1}{\alpha-1}\int_0^{ \infty} \frac{dx}{x^{\alpha-1}(1+x^2)} \,dx


How did you do this step? :eek:
Original post by Jkn
How did you do this step? :eek:


By parts...?
Original post by Jkn
How did you do this step? :eek:

As above, by parts and evaluating the limits around uv. :smile:
Original post by Lord of the Flies
Solution 160

f(t)=∫01ln⁡(1+tx)1+x2 dx\displaystyle f(t)=\int_0^1 \frac{\ln (1+tx)}{1+x^2}\,dx


Is this differentiation under the integral sign? Or if not what is this technique, is it useful?
Original post by bananarama2
I think we need an amendment to the first post. **** now means LOTF and Mladenov difficulty.


I'm feeling increasingly disturbed that they are just about to (and most certainly will) encroach beyond techniques I understand soon. :laugh:

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