Riproot
Addiction Psychiatrist
- Joined
- Nov 10, 2009
- Messages
- 8,228
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- HSC
- 2011
- Uni Grad
- 2017
Re: HSC 2013 4U Marathon
C the answer is always C
C the answer is always C
I PMed youHey Sy,
Bit off topic, just wondering how you went in the hsc maths papers last year?
Since it is given that:
I'm not sure if I'm massively misinterpreting your notation or what the question is asking, but:
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.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.
.if a quartic has four real roots in AP prove that its derivative has three real roots in AP.
For (i)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.
fixed, forgot to type pi in.For (i)
If r = pi then
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....
Let:
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.
Alright then:fixed, forgot to type pi in.
but thenI 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?
Hahaha biggest failure ever on my part :/but then
"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.Alright then:
(i)
Now, since the arguments of complex numbers are periodic by 2pi, in order for the sequence:
To have finitely many values, then we must observe that:
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:
(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]
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.
I fixed the wording for my answer to part (i)"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.