Hard Questions (2 Viewers)

[leehuan]

Wow.

 

[Paradoxica]

Wow.

I have discovered a truly marvellous proof of this, which this post is too small to contain.

.

 

[leehuan]

Yeh lol i was talking about the integral, the second question made my eyes glaze

The brutal way to do it is just to let u^2=tan(x), followed by another substitution and then the partial fractions is tedious.

But it's been asked quite a few times on the 2015 integration marathon, with a variety of answers.

 

[leehuan]

.

I saw lol, I'm just saying wow

 

[InteGrand]

I think the wording of the original post was the main reason for lots of posting of "joke" questions.

(Original post:

Tbh, i am just bored. Could someone post some very hard math problems.
thanks

)

 

[Nailgun]

The brutal way to do it is just to let u^2=tan(x), followed by another substitution and then the partial fractions is tedious.

But it's been asked quite a few times on the 2015 integration marathon, with a variety of answers.

hmm
why can't you use reverse chain rule btw? (I know you can't but I don't understand why)

 

[InteGrand]

hmm
why can't you use reverse chain rule btw? (I know you can't but I don't understand why)

 

[Nailgun]

Nah like y=(tanx)^1/2, so the integral is [2((tanx)^3/2)] /[ 3(secx)^2]

 

[Paradoxica]

Nah like y=(tanx)^1/2, so the integral is [2((tanx)^3/2)] /[ 3(secx)^2]

Where is the necessary factor of sec^2{x} ? Without the presence of sec^2{x}, an arbitrary function of tanx is most likely impossible to integrate by elementary means.

 

[InteGrand]

Nah like y=(tanx)^1/2, so the integral is [2((tanx)^3/2)] /[ 3(secx)^2]

The dividing by the derivative of tan(x) won't work because tan(x) isn't a linear function. We can only generally do this (divide by the derivative of the function) if that function is linear (ax+b (a non-zero)).

 

[leehuan]

Nah like y=(tanx)^1/2, so the integral is [2((tanx)^3/2)] /[ 3(secx)^2]

You completely lost me here. I guess yeah refer to what InteGrand said.

 

[Nailgun]

The dividing by the derivative of tan(x) won't work because tan(x) isn't a linear function. We can only generally do this (divide by the derivative of the function) if that function is linear (ax+b (a non-zero)).

oh right that makes sense
i just realised it doesn't work (by trying) that it doesn't work with non-linear functions
i thought you could use it with something like (x^2+3)^3

thankies

 

[Nailgun]

You completely lost me here. I guess yeah refer to what InteGrand said.

I was trying to apply this - I thought it worked for all functions

 

[InteGrand]

Yeah. Well that's hard. But jokes aside, do you know that the Riemann Zeta function WAS actually examined in the HSC in 1975. So here is a not-so-hard question - HSC 1975 4 unit Q9i:



See if you can do it.

The Riemann zeta function was in the HSC syllabus before, right?

 

[leehuan]

I was trying to apply this - I thought it worked for all functions

Oh. Try integrating (ax^2+bx+c)^n then and you realise oh wait.

 

[Nailgun]

Oh. Try integrating (ax^2+bx+c)^n then and you realise oh wait.

yeah thats what i tried immediately after Integrands post lele

 

[InteGrand]

Here is a somewhat harder question.

Suppose d_n=sum of positive divisors of n, h_n=1/1+1/2+...+1/n and f_n=h_n+e^h_nlnh_n.

Prove that for n>1, f_n>d_n.

Lol.

 

[InteGrand]

I think the point is that convergence was in the syllabus before.

My Riemann zeta query was in reference to your "do you know that the Riemann Zeta function WAS actually examined in the HSC in 1975" comment. Anyway, I had a look back at buchanan's old post (http://community.boredofstudies.org/238/extracurricular-topics/102519/arc-length.html#post2220688) about old topics that have been removed, and it says that the Riemann zeta function was in the Level 1 (1966-1980) course.

 

[Paradoxica]

Yeah. Well that's hard. But jokes aside, do you know that the Riemann Zeta function WAS actually examined in the HSC in 1975? So here is a not-so-hard question - HSC 1975 4 unit Q9i:



See if you can do it.

The infinite sum is a first-order linear approximation of the following integral:



Two approximations are constructed such that they are definitively greater and lesser than the true value of the integral. Since the integral only converges for s>1, it follows that the sum only converges for s>1. QED.

This isn't a formal proof, strictly speaking, but I cbbs rn. I'll come back to patch it up later.

 

[Paradoxica]

Indeed you are right (but also wrong).

What do you mean?

Formality = Wrong or Argument = Wrong ?