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

 

[tywebb]

kappa

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.

 

[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

 

[tywebb]

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.

 

[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.

 

[tywebb]

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

I think the point is that convergence was in the syllabus before. And I mean more formal methods of proof thereof - not the Mickey Mouse hand wavy waffle that we see now in the current syllabus.

The non-Mickey Mouse version present before allows for such questions to be easily dealt with, but unfortunately with the current dumbed down syllabus, most students won't have a clue how to deal with it.