Challenge complex numbers problem for current students (1 Viewer)

[Heinz]

Originally posted by Faera
for (b)

z + z^2 + z^3 + z^4 + .... + z^n = [z(z^n - 1)]/(z-1) ... sum of n terms in a geometric sequence.

is series and sequences assumed knowledge? I thought it was part of the 2u HSC course.

 

[CM_Tutor]

Originally posted by Heinz
is series and sequences assumed knowledge? I thought it was part of the 2u HSC course.

Everything in the 2u course must be covered by 4u students. Just like there is a Harder 3u part for the 4u course, there is a Harder 2u part for the 3u course. Thus, anything in 2u is fair game for a 3u paper.

 

[Faera]

ok.. i think i have the solution for (d)-

|cos@ + cos2@ + cos3@ +...+ cosn@| = |[sin(n@/2)]/[sin(@/2)] * [cos(n+1)@/2]| .... (from (c))

The Inequality becomes:
|[sin(n@/2)]/[sin(@/2)] * [cos(n+1)@/2]| <= |cosec(@/2)|

multiplying both sides by |sin(@/2)|

|sin(n@/2) * cos((n+1)@/2)| <= 1

But since [-1 <= sine of anything <=1] and the [-1 <= cosine of anything <=1], when two values of sine and cosine are multiplied, their value will be equal to or less than one, as required.

Is that right... ?

 

[:: ck ::]

looks rite to me !

 

[CM_Tutor]

Faera, your reasoning is correct, but HSC markers don't like you working backwards from the answer in inequalities questions, so you need to either construct the answer forwards, or surround your reasoning by the structure of a proof by contradiction.

 

[:: ck ::]

so how would you write the "model" answer?

 

[CM_Tutor]

Writing the answer forwards:

Since -1 <= sin x <= 1 for all x, ie. |sin x| <= 1, we know that |sin (n@ / 2)| <= 1 _______ (1)
Similarly, as |cos x| <= 1 for all x, we know that |cos[(n + 1)@ / 2]| <= 1 _______ (2)
Multiplying (1) by (2), we get |sin (n@ / 2)| * |cos[(n + 1)@ / 2]| <= 1
And so |sin (n@ / 2) * cos[(n + 1)@ / 2]| <= 1
Dividing both sides by |sin(@ / 2)| (provided this is non-zero) gives
|sin (n@ / 2) * cos[(n + 1)@ / 2]| / |sin(@ / 2)| <= 1 / |sin(@ / 2)|
|{sin (n@ / 2) * cos[(n + 1)@ / 2]} / sin(@ / 2)| <= |1 / sin(@ / 2)|
|{sin (n@ / 2) * cos[(n + 1)@ / 2]} / sin(@ / 2)| <= |cosec(@ / 2)|
But, from (c), {sin (n@ / 2) * cos[(n + 1)@ / 2]} / sin(@ / 2) = cos@ + cos(2@) + ... + cos(n@)
Thus, |cos@ + cos(2@) + ... + cos(n@)| <= |cosec(@ / 2)|, as required.

Using Proof by Contradiction:

Theorem: |cos@ + cos(2@) + ... + cos(n@)| <= |cosec(@ / 2)|

Proof: Assume the theorem is false.
It follows that |cos@ + cos(2@) + ... + cos(n@)| > |cosec(@ / 2)|
Now, from (c), we know that cos@ + cos(2@) + ... + cos(n@) = [sin(n@ / 2) / sin(@ / 2)] * cos[(n + 1)@ / 2]
and it then follows that |[sin(n@ / 2) / sin(@ / 2)] * cos[(n + 1)@ / 2]| > |cosec(@ / 2)|
and thus, by multiplying by |sin(@ / 2)|, we get |sin(n@ / 2) * cos[(n + 1)@ / 2]| > 1 _______ (A)

Now, since -1 <= sin x <= 1 for all x, ie. |sin x| <= 1, we know that |sin (n@ / 2)| <= 1 _______ (1)
Similarly, as |cos x| <= 1 for all x, we know that |cos[(n + 1)@ / 2]| <= 1 _______ (2)
Multiplying (1) by (2), we get |sin (n@ / 2)| * |cos[(n + 1)@ / 2]| <= 1
And so |sin(n@ / 2) * cos[(n + 1)@ / 2]| <= 1 _______ (B)

Statements (A) and (B) are contradictions. It follows that are assumption (that the theorem is false) must, itself, be false.

Thus, the theorem is true.

 

[CM_Tutor]

Still waiting on (e) ... (Just making sure the thread doesn't get lost )

 

[Faera]

ooh, i see how you have to do it now.. yeah... that makes much more logical sense than mine.

for (e), i'm also not quite sure about my answer but here goes:

0 =cos@ + cos2@ + cos3@ + cos4@ = [sin(n@/2)/sin(@/2)][cis(n+1)@/2] ......... where n = 4.

0 = [ sin2@ / sin(@/2) ] * cos(5@ / 2)
note: sin@/2 cannot = 0
therefore, @ cannot = 0, 2pi

0 = [ 2sin@cos@/sin(@/2) ] * cos(5@/2)

0 = [ 4sin(@/2)cos(@/2) * cos@ / sin(@/2) ] * cos(5@/2)

0 = cos(@/2) * cos@ * cos(5@/2)

We have:
1. cos(@/2) = 0
@ = pi
2. cos@ = 0
@ = pi/2, 3pi/2
3. cos(5@/2) = 0
@ = pi/5, 3pi/5, pi, 7pi/5, 9pi/5

therefore @ = pi, pi/2, 3pi/2, pi/5, 3pi/5, 7pi/5, 9pi/5.


...?

 

[CM_Tutor]

Originally posted by Faera
note: sin@/2 cannot = 0
therefore, @ cannot = 0, 2pi

Excellent! That was the trick to this one.

therefore @ = pi, pi/2, 3pi/2, pi/5, 3pi/5, 7pi/5, 9pi/5.

Absolutely correct - big tick.

Note - you could have just solved sin2@ = 0 and cos(5@ / 2) = 0, and saved yourself the effort of rewriting the expressions.

 

[OLDMAN]

Originally posted by CM_Tutor
[
2. Bill and Ted play a game with a fair coin, where each toss the coin n times. Show that the probability that they get the same number of heads is (2n)! / [2^(2n) * (n!)^2] [/B]

Must not allow a nice problem go to waste. A hard problem even by 4Unit standard but offers some useful insight to binomial coefficients.

P=BoTo+B1*T1+B2*T2+...+BnTn where Bi & Ti mean probability of Bill and Ted each getting i heads

Bi and Ti = [nCi * (1/2)^i * (1/2)^(n-1)] = nCi * (1/2)^n

Thus P =SUM(i:0--->n) [(1/2)^2n * (nCi)^2]

= (1/2)^2n * SUM(i:0--->n)[ (nCi)^2]

But SUM(i:0--->n)[ (nCi)^2] = (2n)Cn .....why?

Recall a 3Unit question : expand (1+x)^n * (1+x)^n in two ways, hence prove that (nC0)^2+(nC1)^2+...+(nCn)^2 = 2n Cn

The rest is straightforward.

btw, very nice and relevant problems you've been posting. I am sure all the students in this forum appreciate your effort.

 

[Faera]

lol... yeah, i could've... -.-' oops, oh well.
i got it in the end... yay!

 

[CM_Tutor]

Originally posted by OLDMAN
Recall a 3Unit question : expand (1+x)^n * (1+x)^n in two ways, hence prove that (nC0)^2+(nC1)^2+...+(nCn)^2 = 2n Cn
The rest is straightforward.

In some ways this is easier done with the identity (1 + x)²ⁿ = xⁿ(1 + x)ⁿ(1 + 1 / x)ⁿ, but a stuent would need to establish the proof of this identity first.

btw, very nice and relevant problems you've been posting. I am sure all the students in this forum appreciate your effort.

Thanks - hopefully they'll be learning something, too!

 

[Grey Council]

hehe, we sure are. ^__^ and yes, I, for one, really do appreciate CM_Tutor's problems. I'd post for the questions in this thread, but i hate trig. I just don't get it. hrmph

anyway:

1. We know that x^0 = 1, and 0^x = 0, so what is 0^0?

2. Bill and Ted play a game with a fair coin, where each toss the coin n times. Show that the probability that they get the same number of heads is (2n)! / [2^(2n) * (n!)^2]

second question's been answered by OLDMAN, and it was outside my range anyway.

but for the first question, isn't the first question just BS? 0^0 doesn't exist.
how do you get x^0 = 1?
x=x^1
divide by x on both sides:
x/x = x^0
= 1

but if its:
x=0^0
you can't divide by zero. hrm, David kinda explained this to me, although i've forgotten already. lol

 

[CM_Tutor]

Originally posted by Grey Council
second question's been answered by OLDMAN, and it was outside my range anyway.

but for the first question, isn't the first question just BS? 0^0 doesn't exist.
how do you get x^0 = 1?

Remember that I originally posted these in response to Spice Girl's comment. The second question is a perfectly legitimate Q, but would have more lead in on an exam - I'm sre I've seen a varient of it on a 3u paper from JRAHS.

As for the first question - which is really not an HSC type question - it's more a topic for discussion than one that has a definitive answer.

As for where I got x⁰ = 1 from, what is the value of 2⁰, or 5⁰, or 10⁰, or even (-200)⁰? All are 1. Can you point to evidence that 0⁰ does not exist?

 

[Grey Council]

well, not really. Cause if i graph x^x I get this kinda parabola on the positive side that touches the 0 point.

but 0^0 is kinda dodgy.

ah never mind. I'll let finer minds than mine (spice girls and Turtle's) ponder over that question. ^_^

PS I didn't ask where you got 0^0 from, i was just posing a question that I wanted to prove. lol

 

[Faera]

we had to draw x^x once...

perhaps that has something to do with it...

but then again... the graph was undefined at x = 0... so maybe not.

just a thought.

 

[CM_Tutor]

If you maintain that 0⁰ is undefined would it be reasonable to suggest that it might be lim (x --> 0) x^x?

Originally posted by Grey Council
PS I didn't ask where you got 0^0 from, i was just posing a question that I wanted to prove. lol

Really?

Originally posted by Grey Council
but for the first question, isn't the first question just BS? 0^0 doesn't exist.
how do you get x^0 = 1?

 

[Faera]

and lim(x --->0) x^x = 1

but...

Isnt that... as x approaches... ? not when x = 0?

So it wouldnt be the actual answer to x^x, would it?

0.o

 

[CM_Tutor]

Originally posted by Faera
and lim(x --->0) x^x = 1

Does it?
For lim(x --> 0) x^x = 1, we'd need lim(x --> 0+) x^x = 1 and lim(x --> 0-) x^x = 1. The fist of these is straight forward, but given the lack of continuity over the reals of y = x^x for x < 0, the second limit is problematic.

but...

Isnt that... as x approaches... ? not when x = 0?

So it wouldnt be the actual answer to x^x, would it?

True, but it's a good argument that 0^x <> 0 in the case of x = 0. Isn't it?

Remember that this is ultimately a question that can only be discussed, not definitively answered. But it is (IMO) interesting.