More questions from the ends of papers (1 Viewer)

[nike33]

Originally posted by CM_Tutor
Nike33, your answer to 3(a) is fine, but with 3(b), how do you know that f(x) = x⁴ + 2x³ + 3x² + 2x + 1?

I know it is true, but in an exam, you would need to provide evidence to support such an assertion...

i origionalyl typed all 3 and it didnt work, so when retyped qn3 i left a few things out..also after reading the entire post i found the much better way of doing it i think you mentioned..of equating the sum / product roots etc to get alpha cubed = 1

 

[Xayma]

Originally posted by CM_Tutor
Do people want me to post the answers to those questions that have not yet been answered?

Have any not been answered if they havent can you do that.

 

[CM_Tutor]

Originally posted by nike33
i origionalyl typed all 3 and it didnt work, so when retyped qn3 i left a few things out..also after reading the entire post i found the much better way of doing it i think you mentioned..of equating the sum / product roots etc to get alpha cubed = 1

I understand. My point was that you have used b = 2 and e = 1, which lead to d² = 4 and c = d + 1, and you have jumped straight to d = 2 and c = 3. This neglects the possibility of d = -2, c = -1. All I'm saying is that in an exam, such a jump would need an explanation.

Originally posted by Xayma
Have any not been answered if they havent can you do that.

I'll make sure there are answers to all the questions in this thread, and the Conics thread. I'm planning to leave it until about the middle of the holidays, as I know there are people still working on them.

 

[CM_Tutor]

1. (For this question, I'm going to use a in place of alpha, for the sake of clarity.)

(a) Provided sin(a / 2) <> 0, show that cos(a / 2) = sin a / 2sin(a / 2)
Similarly, show that if sin(a / 2) and sin(a / 4) are both <> 0, cos(a / 2) * cos(a / 4) = sin a / 4sin(a / 4)

sin a = sin[2 * (a / 2)] = 2sin(a / 2)cos(a / 2)
So, sin a / 2sin(a / 2) = cos(a / 2), on division by 2sin(a / 2), as sin(a / 2) <> 0 (given).
Then, noting that sin(a / 2) = 2sin(a / 4)cos(a / 4), we have sin a / 4sin(a / 4)cos(a / 4) = cos(a / 2)
and so sin a / 4sin(a / 4) = cos(a / 2)cos(a / 4), on multiplication by cos(a / 4)

(b) Prove by mathematical induction that if n is a positive integer, and sin(a / 2ⁿ) <> 0, then
cos(a / 2) * cos(a / 4) * ... * cos(a / 2ⁿ) = sin a / 2ⁿsin(a / 2ⁿ)

Theorem: cos(a / 2) * cos(a / 4) * ... * cos(a / 2ⁿ) = sin a / 2ⁿsin(a / 2ⁿ), for n a positive integer, provided that sin(a / 2ⁿ) <> 0

Proof: By induction on n:

A Put n = 1: LHS = cos(a / 2) = sin a / 2sin(a / 2), from part (a)
= sin a / 2¹sin(a / 2¹) = RHS

So, the result is true for n = 1:

B Let k be a value of n for which the result is true. That is,
cos(a / 2) * cos(a / 4) * ... * cos(a / 2^k) = sin a / 2^ksin(a / 2^k) ______(**)
We must now prove the result for n = k + 1. That is, we must prove that
cos(a / 2) * cos(a / 4) * ... * cos(a / 2^k+1) = sin a / 2^k+1sin(a / 2^k+1)

Now, LHS = cos(a / 2) * cos(a / 4) * ... * cos(a / 2^k+1)
= [cos(a / 2) * cos(a / 4) * ... * cos(a / 2^k)] * cos(a / 2^k+1)
= [sin a / 2^ksin(a / 2^k)] * cos(a / 2^k+1), using the induction hypothesis (**)
= [sin a / 2^ksin(a / 2^k)] * sin(a / 2^k) / 2sin(a / 2^k+1), using @ = a / 2^k in cos(@/2) = sin@ / 2sin(@/2), from (a)
= sin a / [2^k * 2sin(a / 2^k+1)]
= sin a / 2^k+1sin(a / 2^k+1)
= RHS

So, if the result is true for n = k, then it follows that it is also true for n = k + 1

C It follows from A and B by mathematical induction that the result is true for all positive integers n

(c) Hence, deduce that a = sin a / {[cos(a / 2)] * [cos(a / 4)] * [cos(a / 8)] * ... } and that
pi = 1 / {(1/2) * sqrt(1/2) * sqrt[(1/2) + (1/2)sqrt(1/2)] * sqrt{(1/2) + (1/2)sqrt[(1/2) + (1/2)sqrt(1/2)]} * ...}

Rearranging the result from (b), we have 2ⁿsin(a / 2ⁿ) = sin a / [cos(a / 2) * cos(a / 4) * ... * cos(a / 2ⁿ)]
Now, as n --> +inf, a / 2ⁿ --> 0⁺ and so sin(a / 2ⁿ) --> a / 2ⁿ, as lim (x --> 0) (sin x) / x = 1
Thus, 2ⁿ * (a / 2ⁿ) = sin a / [cos(a / 2) * cos(a / 4) * cos(a / 8) * ...]
and so a = sin a / [cos(a / 2) * cos(a / 4) * cos(a / 8) * ...], as required

Now, taking a = pi / 2, we get pi / 2 = sin(pi / 2) / [cos(pi / 4) * cos(pi / 8) * cos(pi / 16) * ...]
and dividing both sides by 1 / 2 gives pi = 1 / [(1 / 2) * cos(pi / 4) * cos(pi / 8) * cos(pi / 16) * ...] _____ (*)

Now, we know that cos(pi / 4) = 1 / sqrt(2) = sqrt(1 / 2), and also that cos²@ = (1 / 2) + (1 / 2) * cos2@.
So, putting @ = pi / 8, cos²(pi / 8) = (1 / 2) + (1 / 2) * cos(pi / 4) = (1 / 2) + (1 / 2) * sqrt(1 / 2)
Thus, cos(pi / 8) = sqrt[(1 / 2) + (1 / 2) * sqrt(1 / 2)]
And, putting @ = pi / 16, cos²(pi / 16) = (1 / 2) + (1 / 2) * cos(pi / 8)
= (1 / 2) + (1 / 2) * sqrt[(1 / 2) + (1 / 2) * sqrt(1 / 2)]
Thus, cos(pi / 16) = sqrt{(1 / 2) + (1 / 2) * sqrt[(1 / 2) + (1 / 2) * sqrt(1 / 2)]}

Substituting these into (*), we get:
pi = 1 / {(1/2) * sqrt(1/2) * sqrt[(1/2) + (1/2)sqrt(1/2)] * sqrt{(1/2) + (1/2)sqrt[(1/2) + (1/2)sqrt(1/2)]} * ...}, as required

 

[CM_Tutor]

2. (a) Let @ = tan^-1x + tan^-1y. Show that tan@ = (x + y) / (1 - xy)

tan@ = tan(tan^-1x + tan^-1y) = [tan(tan^-1x) + tan(tan^-1y)] / [1 - tan(tan^-1x)tan(tan^-1y)] = (x + y) / (1 - xy), as required

(b) If tan^-1x + tan^-1y + tan^-1z = pi / 2, show that xy + yz + zx = 1

tan(tan^-1x + tan^-1y + tan^-1z) = [tan(tan^-1x + tan^-1y) + tan(tan^-1z)] / [1 - tan(tan^-1x + tan^-1y)tan(tan^-1z)]
= [(x + y) / (1 - xy) + z] / [1 - (x + y)z / (1 - xy)]
= [x + y + z(1 - xy)] / [(1 - xy) - (x + y)z], on multiplication by (1 - xy) / (1 - xy)
= (x + y + z - xyz) / (1 - xy - yz - zx)

Now, if tan^-1x + tan^-1y + tan^-1z = pi / 2, then tan(tan^-1x + tan^-1y + tan^-1z) is undefined.
Hence, 1 - xy - yz - zx = 0, and so xy + yz + zx = 1, as required

(c) Let W_n = tan^-1x₁ + tan^-1x₂ + ... + tan^-1x_n, where n is a positive integer.
Show, by mathematical induction or otherwise, that tan W_n = - Im(w_n) / Re(w_n)
where w_n = (1 - ix₁)(1 - ix₂)...(1 - ix_n)

Induction is a much longer way of solving this problem, so I'm simply going to post the 'otherwise' approach, with a minimum of working (IMO, this is a good example of a case where elegance is helpful ):

Notice that, 1 - ix_k is in quadrants 1 or 4 for k = 1, 2, 3, ..., n, and so arg(1 - ix_k) = tan^-1(-x_k)
Then arg w_n = sum (k = 1 to n) arg(1 - ix_k) = sum (k = 1 to n) tan^-1(-x_k) = sum(k = 1 to n) -tan^-1x_k = -W_n
So, tan W_n = -tan(-W_n) = -tan(arg w_n) = - Im(w_n) / Re(w), as required

 

[nike33]

i used induction

 

[CM_Tutor]

Again, I'm going to use a in place of alpha, for the sake of clarity, and also b for beta

3. The quartic polynomial f(x) = x⁴ + bx³ + cx² + dx + e has two double zeroes at alpha and beta, where the coefficients b, c, d, and e are real, but the zeroes alpha and beta may be complex.

(a) Express b, c, d and e in terms of the double zeroes alpha and beta, and hence show that:
(i) d² = b²e
(ii) b³ + 8d = 4bc

Usual root theory gives:
b = -2(a + b) _____ (1)
c = (a + b)² + 2ab _____ (2)
d = -2ab(a + b) _____ (3)
e = (ab)² _____ (4)

Putting (1) into (3) and dividing by b gives d / b = ab _____ (5)
(5) can be put into (4) to give e = (d / b)², and thus, d² = b²e, as required.

Rearranging (1) to a + b = -b / 2, and putting this and (5) into (2) gives:
c = (-b / 2)² + 2(d / b), which on multiplication by 4b gives 4bc = b³ + 8d as required.

(b) Now, suppose that b = 2 and e = 1, and that the double zeroes alpha and beta are both non-real complex numbers.
(i) Show that alpha and beta are cube roots of 1.
(ii) Write f(x) as the product of polynomials irreducible over the real numbers.

Using (1) and (4), we know that the roots a and b satisfy a + b = -1 and (ab)² = 1. Thus, ab = +/- 1.

Taking ab = -1 and eliminating b, we get a² + a - 1 = 0, which is a quadratic equation with two real roots - it has a positive discriminant.
This contradicts the question as a is a non-real complex number, and so ab <> -1

So, we know that ab = 1, and eliminating b we get a² + a + 1 = 0, which is a quadratic equation with a negative discriminant as expected.
Multiplying by a - 1 transforms this quadratic equation to a³ - 1 = 0, and hence a is a non-real cube root of unity.
Our original equation has real coefficients, and so the other root is b = a_bar, and thus both a and b are cube roots of 1, as required.

It follows that f(x) = (x² + x + 1)²