University doesnt matte-

Post the hardest then.

Definitely, thats more of a hard A level problem

this is how you know the OP is a roleplaying amerifat, amerifats don't distinguish between college/university in speech

I used these papers as practice for the final year of high school in the higher mathematics courses here in Australia, they are pretty fun.
pic related is one of the hard questions in high school mathematics in Australia

>tfw cant do any of them
am i fucked?

That's pretty cool, quite similar to STEP except you get some more hand-holding.

I wish A-Level Maths had been more like that, so people would have to understand what's going on rather than being able to pass by blindly memorising techniques.

To be fair I posted a rather exceptional example, there is usually only 1 of these kinds of questions per year in the final paper (highest maths course in NSW). Even there you can sort of get by with just memorization, but it won't get you as high of a score as you could by actually understanding the content and having problem solving skills.

Lol. There is no way I would've been able to solve this after school. I'm not a maths student though. Also brainlet question: How are these two equal?

f(x) is a set of functions
I couldn't do this to save my life either

It is certainly possible, because although [math]f((x - x^{-1})^2) > x^{-2} f((x-x^{-1})^2) [/math] for ONLY [math]x > 1 [/math], but also [math] f((x - x^{-1})^2) < x^{-2} f((x-x^{-1})^2) [/math] for [math] 0 < x < 1 [/math], meaning it is possible that by integrating over the entire domain (0, infinity) we "cancel the difference out", leading to the same value for the integral.

The proof anyway is by setting [math]u = \frac{1}{x} [/math], this is a valid substitution as the function is monotone, we change the limits from (0, infinity) to (infinity, 0) and we substitute in du, to obtain:

[math]J = \int_{\infty}^0 f((u^{-1} - u)^2) (-u^{-2}) \mathrm{d}u = \int_0^{\infty} u^{-2} f((u - u^{-1})^2) \mathrm{d} u [/math]

Of course, since it is a definite integral we can change the u for an x and it remains the same.

Was there a typo?

proof?

[math]I_{n+1}[/math] is very ugly written that way; it's easier to prove that
[eqn]I_{n+1}=\frac{(2n-1)!!}{(2n)!!}\cdot\frac\pi2[/eqn]

>not

[math]\displaystyle \frac{4^n I_{n+1}}{\binom{2n}{n}} = \frac{\pi}{2} [/math]

Thanks m8

wtf how do i do the first problem

This is what happened.

>cherrypicking

[eqn] I_n - I_{n+1}= \int_0^\infty \frac{u^2}{(1+u^2)^{n+1}}du \\
= \frac{1}{2n} \int_0^\infty \frac{2nu}{(1+u^2)^{n+1}} \, u \; du \\
= \frac{1}{2n} \left( \left[\frac{-1}{(1+u^2)^n} u \right]_0^\infty + \int_0^\infty \frac{1}{(1+u^2)^n} du \right) \\
= \frac{1}{2n} I_n [/eqn]

No, it really isn't newfag. Look at the archive for yourself:
>archive.is/Txkqc
It is evident that Veeky Forums has decreased in quality. In fact, the occurrence of pointless and easily defused contrarians, like you, is evidence of that.

teach me your ways, sensei

[math]I_1=arctan(u)|_0^ \infty = \frac{\pi}{2}\\
I_{n+1} = I_{n} \left(1-\frac{1}{2n}\right) = I_n \left(\frac{2n-1}{2n}\right) = I_n \frac{1}{2} \frac{2n-1}{n}
\\ \prod_{k=1}^{n} 2k-1 = \frac{(2n)!}{2^n n!}
\\ I_{n+1} = \frac{\pi}{2}\frac{1}{2^n}\frac{(2n)!}{2^n (n!)^2} = \frac{(2n)!\pi}{2^{2n+1}(n!)^2}
[/math]
QED

let u=(x-1/x), then du = 1+1/x^2 dx

what happened to the factor of 1/2?

nvm, the bounds become (-∞,∞) and it gets reduced to 2x (0,∞) by being even.

>I know amerifat speech
No, you do not -- for if you did, you would know
Amerifatica is a large country with many dialects,
and many different idioms -- but you know nothing about them.

I mean, I can do these fairly easily now, but I am at the last year of my math studies.
I don't think I could do them before I entered uni.
We weren't taught improper integrals, we never tried solving integrals using inverse trigonometric functions, and we hadn't even defined factorials as far as I remember.
After like a couple of days of training I probably could solve them, but it would be really hard cause no mathematical maturity.

(iii)
[math]
\int_{0}^\infty dx \frac{x^{2n-2}}{(x^4-x^2+1)^n} = \int_{0}^\infty dx \frac{x^{-2}}{(x^2 - 1 + 1/x^2)^n} = \int_{0}^\infty dx \frac{x^{-2}}{(x^2 - 2 + 1/x^2 + 1)^n} = \int_{0}^\infty dx \frac{x^{-2}}{((x - 1/x)^2 +1)^n} = \int_{0}^\infty du \frac{1}{(u^2 +1)^n} = \frac{(2n-2)!\pi}{2^{2n-1}(n-1!)^2}
[/math]

Even though I have a math degree, I always hated this kind of problems and avoided any area that involved a lot of computations.
Also, I noticed that the ones who were good at solving these usually weren't interested in generalized abstract nonsense...
I guess solving integrals doesn't prove much more than that you're well trained in solving integrals.

Just because it's hard to do mentally doesn't mean not everyone who graduated from high school can do it with a draft sheet at disposal, you stupid memester.

CS major here from some no name uni. Even I can do this, lmaoing at you mathfags

>retarded black hole post
>qc thread
>clockbro (generic mainstreamers think X thread)
>transhumanist
>atheist
>diet
>is it viable to X (hint: no)
>explain X
>philosophy questions
>google this for me
>why do we X (even through we have 14 years of general education)
>explain this meme
>more transhumist shit
>strawpoll tell me what to do because I'm a brainlet
>frogpost
>areufuckingkiddingme
>womyn in stem
any one of these threads could have been posted last week. Veeky Forums looks very much the same to me.

Wait, dont you go to uni at 19? in norway you go to uni after 13th grade, in which you're 19 years old at the end of.

Post your original discordo please mang. I need to learn your ways!

>doing integrals in high school meme
i dislike this retarded curriculum which focuses only on calculus and analysis
if you're ahead of your class and want to learn more math, why not learn something that is actually fun

Did you have to study math beforehand or is this stuff obvious to non-brainlets?

h-how early did you start the prep? is it too late to start in feb?

Spent ~3 hours a week for about a year doing STEP/ Olympiad / other interesting maths.

February isn't too late, but I wouldn't leave it any longer. Consistently doing this stuff allows you to gain more mathematical maturity. Cramming isn't really feasible.

This sort of maths is quite fun though, so if you really enjoy maths it won't feel like hard work.
If you hate the prep then you will realise maths isn't for you. The people who were lukewarm about maths before coming to Cambridge now fucking hate it so be warned!

Done

I graduated with a degree in Physics but this sounds fun. How can I get started on this?