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\int_0^{\frac{1}{2}} \frac{\tan^{-1}\left ( x^3 + \frac{3x}{4}\right ) + 2 \tan^{-1}\left (e^{\frac{1}{3} \sinh^{-1}(4x)} \right )}{1 + x^2} \ \mathrm{d}x = \cot^{-1}(2)^2 + \frac{\pi}{2}\cot^{-1}(2)

If I remember correctly, the proof of Fermat's Last Theorem does utilise some mod p results which 2^n + 3^n = 5^n would almost certainly satisfy.

I mean same, but x(x+10^-7)/(0.20-x), and the resulting quadratic has coeff for x 4.3E-7 x as opposed to 3.3E-7 x

There was a question which you had to determine the pH of a solution given the Keq was something of the order 10^-7, and the pH was like 10.smth. To do that q accurately, you had to have the initial concentration of [OH-] in pure water to be 10^-7, otherwise your quadratic is a bit off. I don't...

\displaystyle \int_0^{\frac{\pi}{2}} \dfrac{\ln \left ( a^2 \sin^2 (x) + b^2 \cos^2 (x) \right )}{a^2 \sin^2 (x) + b^2 \cos^2 (x)} \; \mathrm{d}x = \dfrac{\pi}{ab} \ln \left ( \dfrac{2ab}{a + b} \right )

no, not quite, but close also, if i did maths all the time, i'd probably bomb out in my atar, so we'll see

Everyone who's got experience in oly maths can confirm international maths olympiad is much harder than 4u maths. 4u maths - requires maybe a month or so to get good at oly maths - years of training required, thousands of hours doing maths

That last section which results in a dilogarithm is entirely 4U-able, and you may see that if you try the Feynman in a slightly different way.

Oh right, that makes sense. I'll edit that.

No because that swapping is taken into account in 2^4 - 2

No, what you're saying is you're either choosing any two of the 4 possible spots for one digit (e.g. you're choosing the last two digits to be 1 if your initial 2 digits are 2, 1) and iterating over all possible ordered pairs of numbers, meaning you also count the time you choose the first two...

That counts codes like 2211 twice?

Thanks! I've replaced the images with the pdf.

Of course, I can make mistakes, so any errors found would be great. EDIT: Newest version now a file on google drive as my last edit caused it to become too large: https://drive.google.com/open?id=1U_buMWY-R6i_ti3HunPg9TbSj_o74zFR

We're actually identical twins uh what?

yo, if anyone could chuck us the paper, that'd be great

Considering I had contributed to the paper, it obviously couldn’t be me.

For 1984, the question asked about re-understanding loneliness, so I did a very conceptual essay on how there are different ways in which Winston is lonely: - he is living in a society where the only expected loyalties are to the Party, so he is lonely with his beliefs - he is living in a...

See if it can :), you might have another approach to it.

After a partial evaluation of the integral, you should get the original integral, which you can use for symmetry. See how that's possible. (note that this is quite cancerous to do, so you need to have some persistence)