/sci/ - Science & Math

[Anonymous No.9485524]

>University doesnt matte-

This is from a test for 17-18 year olds who want to do math tripos at Cambridge. You should be able to solve it then.

[Anonymous No.9485529]

>>9485524
That's trivial.

[Anonymous No.9485532]

>>9485524
Do your own homework.

[Anonymous No.9485555]

>>9485532
>>9485529
t. brainlets who are insecure about their shit college

[Anonymous No.9485570]

>>9485555
>t. brainlets who are insecure about their shit college
I went to a university, not a college.

[Anonymous No.9485571]

Something is going on, on /sci/, it may be that the board is better when there are no school days, I noticed this board was less insane just 48 hours ago. Not great, but not as bad.

[Anonymous No.9485588]

Rape

[Anonymous No.9485603]

>>9485524
(i) Use partial integration twice to get the recursion then use induction to solve it.
(ii) Use u-substitution.
(iii) Divide numerator and denominator by x^(2n) then apply the formulas from (ii) and (i) with f(x) = 1/(1+x)^n.

[Anonymous No.9485620]

Mathmo at Cambridge here, got 100+ on both STEP II and III.
Feels good to not be a brainlet.

[Anonymous No.9485635]

Also that question is on the very easy end of STEP problems...

[Anonymous No.9485644]

>>9485635
Post the hardest then.

[Anonymous No.9485646]

>>9485635
Definitely, thats more of a hard A level problem

[Anonymous No.9485662]

>>9485570
this is how you know the OP is a roleplaying amerifat, amerifats don't distinguish between college/university in speech

[Anonymous No.9485677]

>>9485524
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

[Anonymous No.9485679]

>tfw cant do any of them
am i fucked?

[Anonymous No.9485687]

>>9485677
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.

[Anonymous No.9485717]

>>9485687
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.

[Anonymous No.9485774]

>>9485524
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?

[Anonymous No.9485782]

>>9485774
f(x) is a set of functions
I couldn't do this to save my life either

[Anonymous No.9485793]

>>9485774
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.

[Anonymous No.9485804]

>>9485793
>>9485774
Was there a typo?

[Anonymous No.9485828]

>>9485620

proof?

[Anonymous No.9485899]

>>9485524
[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]

[Anonymous No.9485909]

>>9485899
>not

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

[Anonymous No.9485921]

>>9485793
Thanks m8

[Anonymous No.9485959]

wtf how do i do the first problem

[Anonymous No.9486141]

>>9485571
This is what happened.

[Anonymous No.9486166]

>>9486141
>cherrypicking

[Anonymous No.9486168]

>>9485959
[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]

[Anonymous No.9486180]

>>9486166
No, it really isn't newfag. Look at the archive for yourself:
>http://archive.is/Txkqc
It is evident that /sci/ has decreased in quality. In fact, the occurrence of pointless and easily defused contrarians, like you, is evidence of that.

[Anonymous No.9486962]

>>9485620
teach me your ways, sensei

[Anonymous No.9487005]

>>9486168
[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

[Anonymous No.9487014]

>>9485774
let u=(x-1/x), then du = 1+1/x^2 dx

what happened to the factor of 1/2?

[Anonymous No.9487018]

>>9487014
nvm, the bounds become (-∞,∞) and it gets reduced to 2x (0,∞) by being even.

[Anonymous No.9487031]

>>9485662
>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.

[Anonymous No.9487039]

>>9485524
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.

[Anonymous No.9487064]

>>9485524
(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]

[Anonymous No.9487175]

>>9485524

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.

[Anonymous No.9487177]

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.

[Anonymous No.9487191]

>>9485524
CS major here from some no name uni. Even I can do this, lmaoing at you mathfags

[Anonymous No.9487192]

>>9486180
>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. /sci/ looks very much the same to me.

[Anonymous No.9487366]

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.

[Anonymous No.9487403]

>>9485620
Post your original discordo please mang. I need to learn your ways!

[Anonymous No.9487468]

>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

[Anonymous No.9487477]

>>9485620
Did you have to study math beforehand or is this stuff obvious to non-brainlets?

[Anonymous No.9487486]

>>9485620
h-how early did you start the prep? is it too late to start in feb?

[Anonymous No.9487521]

>>9487486
>>9487477
>>9486962

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!

[Anonymous No.9487567]

Done

[Anonymous No.9487719]

>>9487521
I graduated with a degree in Physics but this sounds fun. How can I get started on this?