HSC 2013 MX2 Marathon (archive) (2 Viewers)

Riproot · Jan 20, 2013

Re: HSC 2013 4U Marathon

C the answer is always C

Sy123 · Jan 21, 2013

Re: HSC 2013 4U Marathon

$Consider the set of polynomials that are defined by the 3 recurrence relations where$

$1) \ \ B_0(x)=1$

$2) \ \ B_n '(x) = nB_{n-1}(x)$

$3) \ \ \int_{0}^{1} B_n (x) \ dx = 0$

$$i) Find$ \ \ \ B_1 \ \ $in terms of$ \ \ x \ \ \ \ \fbox{2}$

$$ii) If$ \ \ B_n (x+1)-B_n(x) = nx^{n-1}$

$$Prove that if$ \ \ g(x) = B_{n+1}(x+1) - B_{n+1}(x) \ \ \ $then$ \\ g'(x)= n(n+1)x^{n-1} \ \ \ \ $hence deduce an expression for$ \ \ g(x) \ \ \ \ \fbox{3}$

$$iii) Hence prove via Mathematical Induction or otherwise that$ \\ B_p(x+1) - B_p(x) = px^{p-1} \ \ \ \ \fbox{2}$

$iv) By using the result proven in part (iii), prove that$

$\sum_{k=1}^{m} k^j = \frac{B_{j+1}(m+1) - B_{j+1}(0)}{j+1} \ \ \ \fbox{4}$

===========

Still 2 questions (pretty good questions imo) still unanswered on the page before this one.
Here and here

OMGITzJustin · Jan 22, 2013

Re: HSC 2013 4U Marathon

Hey Sy,
Bit off topic, just wondering how you went in the hsc maths papers last year?

Sy123 · Jan 22, 2013

Re: HSC 2013 4U Marathon

OMGITzJustin said:
Hey Sy,
Bit off topic, just wondering how you went in the hsc maths papers last year?

I PMed you

RealiseNothing · Jan 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Consider the set of polynomials that are defined by the 3 recurrence relations where$

$1) \ \ B_0(x)=1$

$2) \ \ B_n '(x) = nB_{n-1}(x)$

$3) \ \ \int_{0}^{1} B_n (x) \ dx = 0$

$$i) Find$ \ \ \ B_1 \ \ $in terms of$ \ \ x \ \ \ \ \fbox{2}$

$B_1'(x) = B_0(x) = 1$

Integrating both sides gives:

$B_1(x) = x + C$

Is this what you mean?

RealiseNothing · Jan 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Consider the set of polynomials that are defined by the 3 recurrence relations where$

$1) \ \ B_0(x)=1$

$2) \ \ B_n '(x) = nB_{n-1}(x)$

$3) \ \ \int_{0}^{1} B_n (x) \ dx = 0$

$$ii) If$ \ \ B_n (x+1)-B_n(x) = nx^{n-1}$

$$Prove that if$ \ \ g(x) = B_{n+1}(x+1) - B_{n+1}(x) \ \ \ $then$ \\ g'(x)= n(n+1)x^{n-1} \ \ \ \ $hence deduce an expression for$ \ \ g(x) \ \ \ \ \fbox{3}$ ]

Since it is given that:

$B_n(x+1) - B_n(x) = nx^{n-1}$

Then by just translating $n \to n+1$ we deduce that:

$g(x) = B_{n+1}(x+1) - B_{n+1}(x) = (n+1)x^n$

Then by differentiating both sides:

$g'(x) = n(n+1)x^{n-1}$

Now by integrating this expression we get:

$g(x) = (n+1)x^n + C$

Which is what we originally had anyway, so idk if that's the right expression or not.

RealiseNothing · Jan 22, 2013

Re: HSC 2013 4U Marathon

Also I probably have to use the 3rd relation to find a value for C.

RealiseNothing · Jan 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Consider the set of polynomials that are defined by the 3 recurrence relations where$

$1) \ \ B_0(x)=1$

$2) \ \ B_n '(x) = nB_{n-1}(x)$

$3) \ \ \int_{0}^{1} B_n (x) \ dx = 0$

$$ii) If$ \ \ B_n (x+1)-B_n(x) = nx^{n-1}$

$$Prove that if$ \ \ g(x) = B_{n+1}(x+1) - B_{n+1}(x) \ \ \ $then$ \\ g'(x)= n(n+1)x^{n-1} \ \ \ \ $hence deduce an expression for$ \ \ g(x) \ \ \ \ \fbox{3}$

$$iii) Hence prove via Mathematical Induction or otherwise that$ \\ B_p(x+1) - B_p(x) = px^{p-1} \ \ \ \ \fbox{2}$

I'm not sure if I'm massively misinterpreting your notation or what the question is asking, but:

We have to use the result in part (ii) to prove by induction the result in part (iii) yes? (I assume so because it starts off with "hence"). But what we are trying to prove in part (iii) was assumed in part (ii) to deduce the result which we use in the induction. So if I'm not misinterpreting this, we are proving the result in part (iii) by using a result which we proved by assuming the result from part (iii), which we can't do.

If I'm completely wrong then can you clarify a bit of the notation and concept of the question? It's a really good question.

RealiseNothing · Jan 22, 2013

Re: HSC 2013 4U Marathon

BOS isn't letting me edit my above post, but basically we are proving the result in part (iii) by first assuming that the result in true in the first place.

Sy123 · Jan 22, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I'm not sure if I'm massively misinterpreting your notation or what the question is asking, but:

We have to use the result in part (ii) to prove by induction the result in part (iii) yes? (I assume so because it starts off with "hence"). But what we are trying to prove in part (iii) was assumed in part (ii) to deduce the result which we use in the induction. So if I'm not misinterpreting this, we are proving the result in part (iii) by using a result which we proved by assuming the result from part (iii), which we can't do.

If I'm completely wrong then can you clarify a bit of the notation and concept of the question? It's a really good question.

The first 3 parts of the question is ripped off of Moriah 2001, but the last part is my own which needed this, so I just copy pasted it in.
For induction for 'Step 2' we are basically applying an assumption for n=k (which we can justify to be true for 1 value which we proved in Step 1), at the start of ii, I am basically repeating hte assumption where IF (blah blah blah) so I think it is legitimate

Also these polynomials are called the Bernoulli Polynomials if you are interested.

vbzxwgy · Jan 23, 2013

Re: HSC 2013 4U Marathon

something i proved

let r be a real number

i) prove that the sequence: cis(r*pi), cis(2r*pi), cis(3r*pi),...
takes on finitely many values if and only if r is rational.

ii) prove that for any complex z with |z|=1 and for any irrational r and for any real c > 0, there are infinitely many integers n > 0 with |cis(nr*pi)-z| < c.

vbzxwgy · Jan 23, 2013

Re: HSC 2013 4U Marathon

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

.

Sy123 · Jan 23, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
something i proved

let r be a real number

i) prove that the sequence: cis(r), cis(2r), cis(3r),...
takes on finitely many values if and only if r is rational.

ii) prove that for any complex z with |z|=1 and for any irrational r and for any real c > 0, there are infinitely many integers n > 0 with |cis(nr)-z| < c.

For (i)

If r = pi then

$cis r = -1 \ \ cis 2r = 1 \ \ cis 3r = -1 ...$

Which is finitely many values (only 2 values), yet pi is irrational?

Moreover if we let r=p/q, since there is a loop of 2pi, the only way I see if there are finitely many values is if r IS irrational, I think if its rational then there are infinitely man values. Although if its r is in degrees then yes the result is probably true.

For example look at r= 1 radian

the next value that is the same is 2pi + 1, 4pi + 1, 6pi + 1 ....
However none of these values are 1, 2, 3, 4, 5, .... since pi is irrational

And (ii) is a direct result from (i) so....

vbzxwgy · Jan 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
For (i)

If r = pi then

$cis r = -1 \ \ cis 2r = 1 \ \ cis 3r = -1 ...$

Which is finitely many values (only 2 values), yet pi is irrational?

Moreover if we let r=p/q, since there is a loop of 2pi, the only way I see if there are finitely many values is if r IS irrational, I think if its rational then there are infinitely man values. Although if its r is in degrees then yes the result is probably true.

For example look at r= 1 radian

the next value that is the same is 2pi + 1, 4pi + 1, 6pi + 1 ....
However none of these values are 1, 2, 3, 4, 5, .... since pi is irrational

And (ii) is a direct result from (i) so....

fixed, forgot to type pi in.

Sy123 · Jan 23, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
.

Let:

$P(x) = (x-a)(x-(a+d))(x-(a+2d))(x-(a+3d))$

$$Let$ \ \ \alpha(x) = (x-a)(x-(a+d)) \ \ \ \ \beta (x) = (x-(a+2d))(x-(a+3d))$

$\therefore P'(x) = \alpha '(x) \beta (x) + \alpha (x) \beta '(x)$

$\alpha '(x) = 2x - (2a+d)$

$\beta '(x) = 2x - (2a+5d)$

$\therefore \ P'(x) = (2x - (2a+d)) ( x- ( a+ 2d)) (x-(a+3d)) + (2x-(2a+5d)) ( x - a)(x-(a+d))$

Now if 3 numbers are in arithmetic progression, that is

$u-v, u , u+v$

Then their sum:

$= 3u$

For any 3 numbers in arithmetic sequence, therefore for a polynomial of degree 3 to have all roots in arithmetic progression, one third of the SUM of the roots, is a ROOT.

$\alpha = \frac{-1}{3} \sum \alpha \div \ \ ($co-efficient of leading term$)$ Where alpha is a root, and sum alpha is sum of roots.

EDIT made above

So we need to prove the same thing for P'(x), we first take the sum of roots of P'(x), which happens to be the co-efficient of x^2.

Take co-efficient of x^2 in P'(x) by partially expanding it:

$P'(x) = 4x^3 - (12a+18d) x^2 +...$

Note that I do not want the other co-efficient as it is too much and not needed

Now, if we take one third of the sum of roots:

$1/3 ((12a+18d)/4) = a+3d/2$

In order for P'(x) to be a polynomial in arithmetic progression, then a + 3/2 d must be a root:

$P'(a+3/2 d) = (2d)(-d /2 )(-3d / 2 ) + (-2d)(3d/2 ) (d/2) = 0$

$\therefore P'(x) \ \ $ has roots in arithmetic progression$$

$\square$

Sy123 · Jan 23, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
something i proved

let r be a real number

i) prove that the sequence: cis(r*pi), cis(2r*pi), cis(3r*pi),...
takes on finitely many values if and only if r is rational.

ii) prove that for any complex z with |z|=1 and for any irrational r and for any real c > 0, there are infinitely many integers n > 0 with |cis(nr*pi)-z| < c.

vbzxwgy said:
fixed, forgot to type pi in.

Alright then:

(i)

Now, since the arguments of complex numbers are periodic by 2pi, in order for the sequence:

$cis(r\pi) , \ cis(2r\pi), \ cis(3r\pi), \ ...$

To have finitely many values, then we must observe that:

$2\pi n + mr\pi = kr \pi$

That is, the kth term in the sequence must have the same value as the nth period of the mth term, due to the periodicity by 2pi and all modulus being 1.

In order for the sequence to have finitely many values the equality must be true:

$2n + r = kr \ \ 2n = r(k-m) \\ \therefore r = \frac{2n}{k-m}$

$n \in N \ \ k \in N \ \ m=1,2,3.... \\ \therefore \ r \in \ \ $rational$$

$\square$

EDIT: I made a change to the wording in order for it to make sense.

(ii)

I think there is something wrong with the 'any' conditions, the way the question is structured, if we let c=0.5 and z=0 the inequality isn't true.
So is there something wrong with the wording?

EDIT: Is it maybe supposed to be c > 1 ?

RealiseNothing · Jan 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I think there is something wrong with the 'any' conditions, the way the question is structured, if we let c=0.5 and z=0 the inequality isn't true.
So is there something wrong with the wording?

but then $|z| \neq 1$

Sy123 · Jan 23, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
but then $|z| \neq 1$

Hahaha biggest failure ever on my part :/

vbzxwgy · Jan 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Alright then:

(i)

Now, since the arguments of complex numbers are periodic by 2pi, in order for the sequence:

$cis(r\pi) , \ cis(2r\pi), \ cis(3r\pi), \ ...$

To have finitely many values, then we must observe that:

$2\pi n + mr\pi = kr \pi$

That is, the kth term in the sequence must have the same modulus as the nth period of the mth term:

In order for the sequence to have finitely many values the equality must be true:

$2n + r = kr \ \ 2n = r(k-m) \\ \therefore r = \frac{2n}{k-m}$

$n \in N \ \ k \in N \ \ m=1,2,3.... \\ \therefore \ r \in \ \ $rational$$

$\square$

(ii)

I think there is something wrong with the 'any' conditions, the way the question is structured, if we let c=0.5 and z=0 the inequality isn't true.
So is there something wrong with the wording?

EDIT: Is it maybe supposed to be c > 1 ? [/tex]

"That is, the kth term in the sequence must have the same modulus as the nth period of the mth term:" doesn't make sense but the rest is pretty much right.

i am quite sure that ii is correct as is.

Sy123 · Jan 23, 2013

Re: HSC 2013 4U Marathon

vbzxwgy said:
something i proved

let r be a real number

i) prove that the sequence: cis(r*pi), cis(2r*pi), cis(3r*pi),...
takes on finitely many values if and only if r is rational.

ii) prove that for any complex z with |z|=1 and for any irrational r and for any real c > 0, there are infinitely many integers n > 0 with |cis(nr*pi)-z| < c.

vbzxwgy said:
"That is, the kth term in the sequence must have the same modulus as the nth period of the mth term:" doesn't make sense but the rest is pretty much right.

i am quite sure that ii is correct as is.

I fixed the wording for my answer to part (i)

(ii)

$$for$ \ \ c>0$

Establish the fact that there will always exist some complex u |u|=1 , such that:

$|u-z| < c \ \ \ $for$ \ \ c>0$

There are an infinitely many u such that this condition is true, for ANY z.

Now, if we look at the complex number:

$cis nr\pi$

For $n=1,2,3,4 ...$ That complex number

$cis nr\pi$ is a 'region' of the WHOLE unit circle, that is, every and any complex number will satisfy $cis nr\pi$ if and only if r is irrational, this is the converse from the result proven in part (i)

Hence there must exist infinitely many cis nrpi such that the inequality holds true for any z.

Q.E.D

EDIT: Fixed the proof.

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