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HSC 2015 MX2 Marathon ADVANCED (archive) (3 Viewers)

Thread starter Trebla
Start date Dec 6, 2014

Status: Not open for further replies.

Trebla

Administrator

Administrator

#1

Post any questions within the scope and level of Mathematics Extension 2 mainly targeting Q16 difficulty in the HSC.

Any questions beyond the scope of the HSC syllabus should be posted in the Extracurricular Topics forum:
http://community.boredofstudies.org/238/extracurricular-topics/

Once a question is posted, it needs to be answered before the next question is raised.

I encourage all current students in particular to participate in this marathon.

I will get the ball rolling:

Prove that

$\dfrac{3}{8}+\dfrac{4}{16}+\dfrac{5}{32}+\dfrac{6}{64}+....= 1$

Last edited: Dec 7, 2014

SilentWaters

Member

#2

Re: HSC 2015 4U Marathon - Advanced Level

Denote the summation on the LHS as $S$ so that:
$2S = \frac{3}{4} +\frac{4}{8} + \frac{5}{16} + \dots$

By subtraction:
$S = \frac{3}{4} + \left(\frac{4}{8}-\frac{3}{8}\right) + \left(\frac{5}{16}-\frac{4}{16}\right)+\dots\\=\frac{3}{4}+\frac{1}{8}\sum\limits_{k=0}^{\infty}{\frac{1}{2^n}}$
which by evaluating the limiting geometric sum equates to $\frac{3}{4} + \frac{1}{8}\cdot 2 = 1.$

Sy123

This too shall pass

#3

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Let$ \ a_0 > 0 \ $and$ \ a_{n+1} = a_n + \frac{1}{\sqrt[k]{a_n}} \\ \\ $Show that$ \ a_n \ $does not converge as$ \ n \ $increases without bound$$

-----------------------

$\\ $HINT: Consider$ \ \lim_{ n\to \infty} (a_{n+1} - a_n) \ $. Then make an interesting logical argument from there!$$

Last edited: Dec 8, 2014

Sy123

This too shall pass

#4

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let$ \ a_0 > 0 \ $and$ \ a_{n+1} = a_n + \frac{1}{\sqrt[k]{a_n}} \\ \\ $Show that$ \ a_n \ $does not converge as$ \ n \ $increases without bound$$

-----------------------

$\\ $HINT: Consider$ \ \lim_{ n\to \infty} (a_{n+1} - a_n) \ $. Then make an interesting logical argument from there!$$

Here is my solution:

$\\ $Take the proposition that$ \ a_n \ $converges$ \\ \\ $1. If$ \ a_n \ $converges, then$ \ \lim_{n \to \infty} (a_{n+1} - a_n) = 0 \\ \\ $2. If$ \ \lim_{n \to \infty}(a_{n+1} - a_n) = 0 \ $then$ \ \left(\lim_{n \to \infty} \frac{1}{a_n} \right)^k = 0, \ $(by the recurrence relation)$ \\ \\ $3. If$ \ \left(\lim_{ n \to \infty} \frac{1}{a_n} \right)^k = 0 \ $then$ \ a_n $is divergent!$ \\ $For if it were convergent, it would converge to a value, $ \ C \ $but then$ \ 0 = \left(\lim_{n \to \infty} \frac{1}{a_n} \right)^k = C^{-k} \neq 0 \\ \\ $4. Therefore, if$ \ a_n \ $is convergent, then$ \ a_n \ $is divergent! (1,2,3)$ \\ \\ $5. Hence$ \ a_n \ $is not convergent as$ \ a_n \ $cannot both be convergent and divergent$$

F

FrankXie

Active Member

#5

Re: HSC 2015 4U Marathon - Advanced Level

Chlee1998 said:
Can someone help with Question 8 part b ii gamma?

the link to the paper is https://onedrive.live.com/view.aspx...33&app=WordPdf&authkey=!AIZ8Uo1ng7VFwpg&wdo=1

thanks to whoever can help

It is just the limiting sum formula for a GP with common ratio absolute value less than 1.

Sy123

This too shall pass

#6

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $NEW QUESTION - required knowledge: Polynomials and Complex Numbers$$

$\\$Let the polynomial$ \ E(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} \\ \\ $i) Prove that$ \ E(x) \ $has no double root$ \\ \\ $ii) Prove that if$ \ n \ $is even, then$ \ E(x) \ $has no real roots$$

seanieg89

Well-Known Member

#7

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
Here is my solution:

$\\ $Take the proposition that$ \ a_n \ $converges$ \\ \\ $1. If$ \ a_n \ $converges, then$ \ \lim_{n \to \infty} (a_{n+1} - a_n) = 0 \\ \\ $2. If$ \ \lim_{n \to \infty}(a_{n+1} - a_n) = 0 \ $then$ \ \left(\lim_{n \to \infty} \frac{1}{a_n} \right)^k = 0, \ $(by the recurrence relation)$ \\ \\ $3. If$ \ \left(\lim_{ n \to \infty} \frac{1}{a_n} \right)^k = 0 \ $then$ \ a_n $is divergent!$ \\ $For if it were convergent, it would converge to a value, $ \ C \ $but then$ \ 0 = \left(\lim_{n \to \infty} \frac{1}{a_n} \right)^k = C^{-k} \neq 0 \\ \\ $4. Therefore, if$ \ a_n \ $is convergent, then$ \ a_n \ $is divergent! (1,2,3)$ \\ \\ $5. Hence$ \ a_n \ $is not convergent as$ \ a_n \ $cannot both be convergent and divergent$$

The "taking the limit" step is the real key. We can pass to the limit immediately from the question statement to get:

$C=C+C^{-1/k}$

Which trivially has no real (or complex) solutions. (ie, we have the desired contradiction.)

This is all just streamlining though of course.

dan964

what

#8

Re: HSC 2015 4U Marathon - Advanced Level

** If f(nx) and g(nx) are to such that in terms of a polynomial; whereby
the sum of the roots in any amount (individually, in pairs, triples etc.) or the product of the roots is the same value - an integer
==> the coefficients of the polynomial are identical and integer values.

And also that g(x) = sqrt(1 - f(x)^2)
Show that n=5 is the first integer solution, apart from any trivial cases.

This question requires a bit of intuition.
edit: identical is wrong word maybe.

Last edited: Jan 5, 2016

Axio

=o

#9

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $NEW QUESTION - required knowledge: Polynomials and Complex Numbers$$

$\\$Let the polynomial$ \ E(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} \\ \\ $i) Prove that$ \ E(x) \ $has no double root$ \\ \\ $ii) Prove that if$ \ n \ $is even, then$ \ E(x) \ $has no real roots$$

I. E(x)-E'(x)=(x^n)/n! Which cannot equal zero. But for a double root, this must equal zero when the root is subbed in. Therefore a double root does not exist.

II. I'll try it tomorrow

F

FrankXie

Active Member

#10

Re: HSC 2015 4U Marathon - Advanced Level

Chlee1998 said:
No it's not. It's asking for the value of the coefficient not the sum of everything.
Can you post your full working?

$$ It follows from the limiting sum formula that $ F(z)=\frac{A}{1-\frac{z}{\alpha}}+\frac{B}{1-\frac{z}{\beta}}=A\sum_{k=0}^{\infty}\left(\frac{1}{\alpha}\right)^kz^k+B\sum_{k=0}^{\infty}\left( \frac {1}{\beta} \right)^kz^k=\sum_{k=0}^{\infty}\left\{A \left( \frac{1}{\alpha} \right)^k+B \left( \frac{1}{\beta} \right)^k\right\}z^k,$

$$ on the other hand, $ F(z)=\sum_{k=0}^{\infty}t_k(x)z^k$

Now equating the coefficients of like powers of these two series, we have

$t_k(x)=A \left( \frac{1}{\alpha} \right)^k+B \left( \frac{1}{\beta} \right)^k$

Last edited: Dec 9, 2014

F

FrankXie

Active Member

#11

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
If f(nx) and g(nx) are identical polynomial of the form
A1^n + A2^(n-1) + ... + An
Ax are constants

And g(x) = sqrt(1 - f(x)^2)
Show that n=5 is the first integer solution.

I still don't understand, if g(x) = sqrt(1 + f(x)^2) and f(nx)=g(nx), isn't it that f(nx)^2=1/2 identically?

Last edited: Dec 9, 2014

Axio

=o

#12

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $NEW QUESTION - required knowledge: Polynomials and Complex Numbers$$

$\\$Let the polynomial$ \ E(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} \\ \\ $i) Prove that$ \ E(x) \ $has no double root$ \\ \\ $ii) Prove that if$ \ n \ $is even, then$ \ E(x) \ $has no real roots$$

Axio said:
I. E(x)-E'(x)=(x^n)/n! Which cannot equal zero. But for a double root, this must equal zero when the root is subbed in. Therefore a double root does not exist.

II. I'll try it tomorrow

II. E'(x)=1+x+x^2/2!...+x^(n-1)/(n-1)!
At stationary point, E'(x)=0, -1=x+x^2/2!...+x^(n-1)/(n-1)!
Therefore at stationary point E(x)=1-1+x^n/n!=x^n/n!, and as n is even, E(x) is positive at any stationary point and as the polynomial is of even degree and has a positive leading coefficient, this means that the graph is always above the x-axis and therefore has no real roots.

Last edited: Dec 9, 2014

Sy123

This too shall pass

#13

Re: HSC 2015 4U Marathon - Advanced Level

Axio said:
II. E'(x)=1+x+x^2/2!...+x^(n-1)/(n-1)!
At stationary point, E'(x)=0, -1=x+x^2/2!...+x^(n-1)/(n-1)!
Therefore at stationary point E(x)=1-1+x^n/n!=x^n/n!, and as n is even, E(x) is positive at any stationary point and as the polynomial is of even degree and has a positive leading coefficient, this means that the graph is always above the x-axis and therefore has no real roots.

Yep that's the right idea

A more rigorous way of doing so would be

"if E has no double roots, then each root is a single root, and so then there is some value 'c' So that E(c) < 0, but E(x) -> +inf as x grows large and E(x) -> +inf as x becomes negatively large, so, there must be another root, lets call these pair of roots A and B, since E(A) = E(B) = 0, then there is a point C where E'(C) = 0 And E(C) < 0, but as you have shown, E(C) > 0, so there is no root"

Sy123

This too shall pass

#14

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Let$ \ p(x) \ $be a polynomial with degree$ \ n \ $with real roots$ \ \alpha_1, \alpha_2, \dots \alpha_n \\ \\ $Prove that:$ \ p'(x) = \frac{p(x)}{x - \alpha_1} + \frac{p(x)}{x - \alpha_2} + \dots + \frac{p(x)}{x - \alpha_n}$

C

Chlee1998

Member

#15

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let$ \ p(x) \ $be a polynomial with degree$ \ n \ $with real roots$ \ \alpha_1, \alpha_2, \dots \alpha_n \\ \\ $Prove that:$ \ p'(x) = \frac{p(x)}{x - \alpha_1} + \frac{p(x)}{x - \alpha_2} + \dots + \frac{p(x)}{x - \alpha_n}$

Rhs is (x-a2)(x-a3).....(x-an)+(x-a1)(x-a3)(x-a4)....(x-an).+...+ (x-a1)(x-a2).... (xn-1)
And its easy to see that this is p'(x) through extended product rule

Last edited: Dec 10, 2014

Sy123

This too shall pass

#16

Re: HSC 2015 4U Marathon - Advanced Level

Chlee1998 said:
Rhs is (x-a2)(x-a3).....(x-an)+(x-a1)(x-a3)(x-a4)....(x-an).+...+ (x-a1)(x-a2).... (xn-1)
And its easy to see that this is p'(x) through extended product rule

In HSC exams I would think that you'd need to justify the use of the "extended product rule", so how would you do it?

seanieg89

Well-Known Member

#17

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
In HSC exams I would think that you'd need to justify the use of the "extended product rule", so how would you do it?

Where p is nonzero we have

$\frac{p'}{p}=(\log |p|)'=(\sum_i \log |x-\alpha_i|)'=\sum_i \frac{1}{x-\alpha_i}.$

The result extends to all x by continuity.

Sy123

This too shall pass

#18

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $A polynomial$ \ p(x) \ $has integer coefficients,$ \ p(x) \ $also has a root of the form$ \\ p + \sqrt{q} \ $where$ \ p, q \ $are rational and$ \ q \ $is not a square of a rational$ \\ \\ $Prove that$ \ p(x) \ $also has a root of the form$ \ p -\sqrt{q}$

SilentWaters

Member

#19

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $A polynomial$ \ p(x) \ $has integer coefficients,$ \ p(x) \ $also has a root of the form$ \\ p + \sqrt{q} \ $where$ \ p, q \ $are rational and$ \ q \ $is not a square of a rational$ \\ \\ $Prove that$ \ p(x) \ $also has a root of the form$ \ p -\sqrt{q}$

For non-zero $p$ :

If $p(x)$ has integer coefficients, then the product of its roots must be of a rational form (the quotient of two integers, the leading coefficient and the constant). To rationalize such a product with some irrational $p+\sqrt{q}$ we must have its conjugate as a necessary root. This comes by way of the difference of two squares identity:

$(p+\sqrt{q})(p-\sqrt{q}) = p^{2}-q,$

which is rational.

For $p = 0$ :

Express $\sqrt{q}$ as a mixed surd $Q\sqrt{Q'}$ where $Q\in\mathbb{Z}\neq 0$ and $Q'$ is coprime with the square numbers. For $Q=1$ it is obvious that we need $-\sqrt{q}$ as a root for a rational sum of roots one at a time. For $Q\neq 1$ we could negate the irrational $\sqrt{q}$ with the sum of some number of smaller surds greater than 1. We are then dealing with at least 3 potential roots. Any summation of roots an odd number at a time (from 3 onwards) would produce an irrational product retaining $\sqrt{Q'}$ . The only way we can eliminate the irrational part in every case of summation would be to refrain from decomposing that root in the first place, and qualifying $-\sqrt{q}$ as a root. This is because taking cyclic sums will always either directly multiply the two simple surds to give the rational $-q$ or multiply both by the same factors (comprising other roots), allowing one to factorise out $\sqrt{q}-\sqrt{q} = 0$ and thus eliminate the irrational.

Last edited: Dec 10, 2014

seanieg89

Well-Known Member

#20

Re: HSC 2015 4U Marathon - Advanced Level

SilentWaters said:
To rationalize such a product with some irrational $p+\sqrt{q}$ we must have its conjugate as a necessary root.

You should probably provide more justification for this statement.

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