• Want to help us with this year's BoS Trials?
    Let us know before 30 June. See this thread for details
  • Looking for HSC notes and resources?
    Check out our Notes & Resources page

HSC 2013 MX2 Marathon (archive) (1 Viewer)

Status
Not open for further replies.

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
Re: HSC 2013 4U Marathon

1.

There are more elegant ways to do 1st integral Sy.







You can either do it partial fracts or using the standard integrals not listed.



 
Last edited:

vbzxwgy

New Member
Joined
Jan 12, 2013
Messages
17
Gender
Undisclosed
HSC
N/A
Re: HSC 2013 4U Marathon

if a quartic has four real roots in AP prove that its derivative has three real roots in AP.
 
O

Omed62

Guest
Re: HSC 2013 4U Marathon

thanks so much bro
ur a good guy
yeah ill try my best answering them
try to answer the questions lol, what are you doing... you say something but you dont do it...you just wait for people to post you some questions....
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon







































 
Last edited:

vbzxwgy

New Member
Joined
Jan 12, 2013
Messages
17
Gender
Undisclosed
HSC
N/A
Re: HSC 2013 4U Marathon

a couple of questions

what does a complex integral mean?

if the sum of 1/n is infinity then how can we take its log and use it in inequalities?
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

a couple of questions

what does a complex integral mean?

if the sum of 1/n is infinity then how can we take its log and use it in inequalities?
I have not defined alpha nor z in the complex plane.

Would changing the order of iv and iii change the rigour issue here?
EDIT: Or rather how about changing it so we have only a partial sum and a partial product and consider the inequality of the limit instead?
 
Last edited:

vbzxwgy

New Member
Joined
Jan 12, 2013
Messages
17
Gender
Undisclosed
HSC
N/A
Re: HSC 2013 4U Marathon

I have not defined alpha nor z in the complex plane.

Would changing the order of iv and iii change the rigour issue here?
wouldn't you still be logging something that might be infinity? (and is)
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

wouldn't you still be logging something that might be infinity? (and is)
Alright what about taking the limit to infinity of a partial sum/product?

If not, what can I do to fix this?
 

vbzxwgy

New Member
Joined
Jan 12, 2013
Messages
17
Gender
Undisclosed
HSC
N/A
Re: HSC 2013 4U Marathon

Alright what about taking the limit to infinity of a partial sum/product?

If not, what can I do to fix this?
i dont know man its your question, my ugess is that it works but makes the connection to primes more annoying.
 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
Re: HSC 2013 4U Marathon

I think this involves converting the infinite sum to a limiting argument of the function 1/(1+x^2)...then letting that be an integral which turns into arctangent...which is between pi/4 and pi/2 between x=1 and x-> infinity but why is it strictly less than...hmm I'm missing something..
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

I think this involves converting the infinite sum to a limiting argument of the function 1/(1+x^2)...then letting that be an integral which turns into arctangent...which is between pi/4 and pi/2 between x=1 and x-> infinity but why is it strictly less than...hmm I'm missing something..
Not sure what you mean by the first sentence. You are correct in that it involves the integral of 1/(1+x^2)
HINT: Notice how the upper bound is

 
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
Re: HSC 2013 4U Marathon

Lol that's what I meant, ie go back to the limit definition of an integral...

Wait how come your lower bound is now 0? (Typo?)
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

Lol that's what I meant, ie go back to the limit definition of an integral...

Wait how come your lower bound is now 0? (Typo?)
Lower bound is pi/4 from the question, upper bound is pi/2. The lower limit of the integral is 0 however, and that integral that I just posted = pi/2
 

nightweaver066

Well-Known Member
Joined
Jul 7, 2010
Messages
1,585
Gender
Male
HSC
2012
Re: HSC 2013 4U Marathon

Consider .

If we draw out the graph of and then draw out the lower rectangles of breadth 1 unit, we can determine that

Now we consider the upper rectangles, which leads us to





Edit: not sure if there is enough integration in this for you haha
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

Consider .

If we draw out the graph of and then draw out the lower rectangles of breadth 1 unit, we can determine that

Now we consider the upper rectangles, which leads us to





Edit: not sure if there is enough integration in this for you haha
Correct - good job.

Alternatively consider:



Where we can justify the lower bound because the height of the biggest rectangle when we make the lower rectangles of 1/(1+x^2) is 1/2

Hence the area underneath 1/2 ( 1/(1+x^2) ) is always lower than the rectangles.
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,179
Gender
Male
HSC
2006
Re: HSC 2013 4U Marathon

Consider .

If we draw out the graph of and then draw out the lower rectangles of breadth 1 unit, we can determine that

Now we consider the upper rectangles, which leads us to





Edit: not sure if there is enough integration in this for you haha
There is a contradiction in your working.
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon

This may or may not bypass the rigour issues that happened last time I posted this idea as a question:







 
Last edited:
Joined
Sep 20, 2010
Messages
2,225
Gender
Undisclosed
HSC
2012
Re: HSC 2013 4U Marathon

Isn't it not the other way around? idk O_O

The 'proof' that

doesn't make sense either - e is just a constant, it shouldn't matter what n is doing! (I think it's a directional issue here)
 
Last edited:
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top