HSC 2013 MX2 Marathon (archive) (2 Viewers)

SpiralFlex · Jan 12, 2013

Re: HSC 2013 4U Marathon

1.

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

$J=\frac{1}{2}\int (\frac{x^2+1}{x^4+1}-\frac{x^2-1}{x^4+1})$

$=\frac{1}{2}\int (\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}-\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}})\;dx$

$=\frac{1}{2}\int [(\frac{1}{(x-\frac{1}{x})^2+2}\;d(x-\frac{1}{x})}-\frac{1}{2}\int \frac{1}{(x+\frac{1}{x})^2-2}\;d(x+\frac{1}{x})]$

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

$1. \int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C$

$2. \int \frac{dx}{x^2+a^2}=\frac{1}{2a}\ln|\frac{a+x}{a-x}|+C$

vbzxwgy · Jan 12, 2013

Re: HSC 2013 4U Marathon

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

Omed62 · Jan 12, 2013

Re: HSC 2013 4U Marathon

Immortalp00n said:
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 · Jan 13, 2013

Re: HSC 2013 4U Marathon

$Show using integration that$

$\frac{\pi}{4} < \sum_{k=1}^{\infty} \frac{1}{1+k^2} < \frac{\pi}{2}$

Sy123 · Jan 13, 2013

Re: HSC 2013 4U Marathon

$Question Prime: 11 marks$

$For the context of this question, a prime number is a natural number such that it is only divisible by itself and 1. 1 itself is not a prime number$

$$The first few prime numbers are$ \ \ \ \ 2, 3, 5, 7, 11, 17 ....$

$You may assume that there are an infinite number of prime numbers$

$\prod_{p } f(p) \ \ \ \ \ \ $refers to the infinite product of primes of the function f$$

$\sum_{p} f(p) \ \ \ \ \ \ \ $refers to the infinite sum of of primes of the function f$$

$$i) By considering the geometric series$ \ \ \ \ 1+\alpha + \alpha^2+...+\alpha^{n-1} \ \ \ |\alpha| < 1$

$$Show that$ \ \ \ z+\frac{z^2}{2}+ \frac{z^3}{3} + ... + \frac{z^n}{n} = \int_{0}^{z} \frac{1-\alpha^n}{1-\alpha} \ d\alpha \ \ \ \ \fbox{1}$

$$ii) Show that if n increases without bound that$ \ \ \ \ \ -\ln(1-z) = \sum_{k=1}^{\infty} \frac{z^k}{k} \ \ \ \ \ \fbox{2}$

$iii) Now consider the sequence$

$S=\sum_{k=1}^{\infty} \frac{1}{k} \equiv 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$

$$By considering$ \ \ \ \ \lim_{n \to \infty} \int_{1}^{n} \frac{1}{x} \ dx$

$$Show that$ \ \ \ S \to \infty \ \ \ $as n increases without bound$ \ \ \ \ \ \fbox{2}$

$iv) It can be shown that$

$\sum_{n=1}^{\infty} \frac{1}{n} = \prod_{p} \frac{1}{1-p^{-1}}$

$$Using part (ii) show that$ \\ \\ \ln \left( \sum_{n=1}^{\infty} \frac{1}{n} \right) < \left( \sum_{p} \frac{1}{p} \right) + \sum_{p } \frac{1}{p^2} \left (1+\frac{1}{p}+\frac{1}{p^2}+ \frac{1}{p^3}+ \dots \right) \ \ \ \ \ \fbox{3}$

$v) It can be shown that$ \ \ \ \sum_{p } \frac{1}{p-1} - \frac{1}{p} = C \ \ \ \ $for some positive constant C less than 1.$

$$Hence show that$ \ \ \ \ \ \ \ln (S) < \sum_{p } \frac{1}{p} + C \ \ \ \ \ \fbox{1}$

$$vi) Using parts (v) and (iii) deduce the following$ \\ \\ \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \dots + \frac{1}{p} + \dots = + \infty \ \ \ \ \ \ \fbox{2}$

vbzxwgy · Jan 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Question Prime: 11 marks$

$For the context of this question, a prime number is a natural number such that it is only divisible by itself and 1. 1 itself is not a prime number$

$$The first few prime numbers are$ \ \ \ \ 2, 3, 5, 7, 11, 17 ....$

$You may assume that there are an infinite number of prime numbers$

$\prod_{p } f(p) \ \ \ \ \ \ $refers to the infinite product of primes of the function f$$

$\sum_{p} f(p) \ \ \ \ \ \ \ $refers to the infinite sum of of primes of the function f$$

$$i) By considering the geometric series$ \ \ \ \ 1+\alpha + \alpha^2+...+\alpha^{n-1} \ \ \ |\alpha| < 1$

$$Show that$ \ \ \ z+\frac{z^2}{2}+ \frac{z^3}{3} + ... + \frac{z^n}{n} = \int_{0}^{z} \frac{1-\alpha^n}{1-\alpha} \ d\alpha \ \ \ \ \fbox{1}$

$$ii) Show that if n increases without bound that$ \ \ \ \ \ -\ln(1-z) = \sum_{k=1}^{\infty} \frac{z^k}{k} \ \ \ \ \ \fbox{2}$

$iii) Now consider the sequence$

$S=\sum_{k=1}^{\infty} \frac{1}{k} \equiv 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$

$$By considering$ \ \ \ \ \lim_{n \to \infty} \int_{1}^{n} \frac{1}{x} \ dx$

$$Show that$ \ \ \ S \to \infty \ \ \ $as n increases without bound$ \ \ \ \ \ \fbox{2}$

$iv) It can be shown that$

$\sum_{n=1}^{\infty} \frac{1}{n} = \prod_{p} \frac{1}{1-p^{-1}}$

$$Using part (ii) show that$ \\ \\ \ln \left( \sum_{n=1}^{\infty} \frac{1}{n} \right) < \left( \sum_{p} \frac{1}{p} \right) + \sum_{p } \frac{1}{p^2} \left (1+\frac{1}{p}+\frac{1}{p^2}+ \frac{1}{p^3}+ \dots \right) \ \ \ \ \ \fbox{3}$

$v) It can be shown that$ \ \ \ \sum_{p } \frac{1}{p-1} - \frac{1}{p} = C \ \ \ \ $for some positive constant C less than 1.$

$$Hence show that$ \ \ \ \ \ \ \ln (S) < \sum_{p } \frac{1}{p} + C \ \ \ \ \ \fbox{1}$

$$vi) Using parts (v) and (iii) deduce the following$ \\ \\ \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \dots + \frac{1}{p} + \dots = + \infty \ \ \ \ \ \ \fbox{2}$

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 · Jan 13, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
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?

vbzxwgy · Jan 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
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 · Jan 13, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
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 · Jan 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
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.

asianese · Jan 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Show using integration that$

$\frac{\pi}{4} < \sum_{k=1}^{\infty} \frac{1}{1+k^2} < \frac{\pi}{2}$

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 · Jan 13, 2013

Re: HSC 2013 4U Marathon

asianese said:
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

$\lim_{n \to \infty}\int_{0}^{n} \frac{1}{1+x^2} \ dx$

asianese · Jan 13, 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?)

Sy123 · Jan 13, 2013

Re: HSC 2013 4U Marathon

asianese said:
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 · Jan 14, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Show using integration that$

$\frac{\pi}{4} < \sum _{k=1} ^{\infty} \frac{1}{1+k^2} < \frac{\pi}{2}$

Consider $\lim_{k \rightarrow \infty} \int ^k_0 \frac{1}{1+x^2}dx$ .

If we draw out the graph of $y=\frac{1}{1+x^2}$ and then draw out the lower rectangles of breadth 1 unit, we can determine that $\sum _{k=1} ^{\infty} \frac{1}{1+k^2} < \lim_{k \rightarrow \infty} \int ^k _0 \frac{1}{1+x^2}dx = \frac{\pi}{2}$

Now we consider the upper rectangles, which leads us to $\sum _{k=0} ^{\infty} \frac{1}{1+k^2} > \lim_{k \rightarrow \infty} \int ^k _0 \frac{1}{1+x^2}dx = \frac{\pi}{2}$

$\sum _{k=1} ^{\infty} \frac{1}{1+k^2} > \frac{\pi -1}{2} > \frac{\pi}{4}$

$\therefore \frac{\pi}{4} < \sum _{k=1} ^{\infty} \frac{1}{1+k^2} < \frac{\pi}{2}$

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

Sy123 · Jan 14, 2013

Re: HSC 2013 4U Marathon

nightweaver066 said:
Consider $\lim_{k \rightarrow \infty} \int ^k_0 \frac{1}{1+x^2}dx$ .

If we draw out the graph of $y=\frac{1}{1+x^2}$ and then draw out the lower rectangles of breadth 1 unit, we can determine that $\sum _{k=1} ^{\infty} \frac{1}{1+k^2} < \lim_{k \rightarrow \infty} \int ^k _0 \frac{1}{1+x^2}dx = \frac{\pi}{2}$

Now we consider the upper rectangles, which leads us to $\sum _{k=0} ^{\infty} \frac{1}{1+k^2} > \lim_{k \rightarrow \infty} \int ^k _0 \frac{1}{1+x^2}dx = \frac{\pi}{2}$

$\sum _{k=1} ^{\infty} \frac{1}{1+k^2} > \frac{\pi -1}{2} > \frac{\pi}{4}$

$\therefore \frac{\pi}{4} < \sum _{k=1} ^{\infty} \frac{1}{1+k^2} < \frac{\pi}{2}$

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

Correct - good job.

Alternatively consider:

$\frac{1}{2} \int_{0}^{\infty} \frac{1}{1+x^2} \ dx < \sum_{k=1}^{\infty} \frac{1}{1+k^2} < \int_{0}^{\infty}\frac{1}{1+x^2} \ dx$

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 · Jan 14, 2013

Re: HSC 2013 4U Marathon

nightweaver066 said:
Consider $\lim_{k \rightarrow \infty} \int ^k_0 \frac{1}{1+x^2}dx$ .

If we draw out the graph of $y=\frac{1}{1+x^2}$ and then draw out the lower rectangles of breadth 1 unit, we can determine that $\sum _{k=1} ^{\infty} \frac{1}{1+k^2} < \lim_{k \rightarrow \infty} \int ^k _0 \frac{1}{1+x^2}dx = \frac{\pi}{2}$

Now we consider the upper rectangles, which leads us to $\sum _{k=0} ^{\infty} \frac{1}{1+k^2} > \lim_{k \rightarrow \infty} \int ^k _0 \frac{1}{1+x^2}dx = \frac{\pi}{2}$

$\sum _{k=1} ^{\infty} \frac{1}{1+k^2} > \frac{\pi -1}{2} > \frac{\pi}{4}$

$\therefore \frac{\pi}{4} < \sum _{k=1} ^{\infty} \frac{1}{1+k^2} < \frac{\pi}{2}$

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

There is a contradiction in your working.

Sy123 · Jan 14, 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:

$i) Show that$

$\binom{n}{j} \frac{1}{n^j} = \frac{1}{j!} \left ( 1 \cdot (1-\frac{1}{n}) \cdot (1-\frac{2}{n}) \cdot \dots \cdot (1-\frac{j-1}{n}))\right )$

$$ii) You are given that the definition of e is$ \ \ \ \ \left (1+ \frac{1}{n} \right)^n \rightarrow e \ \ \ $if$ \ \ \ n \to \infty$

$$Prove that$ \ \ \ \ \sum_{j=0}^{\infty} \frac{1}{j!} \rightarrow e \ \ \ \ $if$ \ \ \ \ n \to \infty$

asianese · Jan 14, 2013

Re: HSC 2013 4U Marathon

Isn't it $\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e$ not the other way around? idk O_O

The 'proof' that

$e \rightarrow \sum_{j=0}^{\infty} \frac{1}{j!} \text{ if } n\rightarrow \infty$ 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)

twinklegal19 · Jan 14, 2013

Re: HSC 2013 4U Marathon

^yeah this

HSC 2013 MX2 Marathon (archive) (2 Viewers)

Well-Known Member

New Member

Omed62

Guest

This too shall pass

This too shall pass

New Member

This too shall pass

New Member

This too shall pass

New Member

Σ

This too shall pass

Σ

This too shall pass

Well-Known Member

This too shall pass

Administrator

This too shall pass

Σ

.

Users Who Are Viewing This Thread (Users: 0, Guests: 2)