HSC 2016 MX2 Marathon (archive) (4 Viewers)

Paradoxica · Sep 30, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
$\noindent Let $P(z) = az^{2} + bz + 1$ ($a\neq 0$) be a quadratic with real coefficients.$

$\noindent a) Find necessary and sufficient conditions on the coefficients $a$ and $b$ in order for $P(z)$ to have a root on the unit circle (i.e. a root $\alpha \in \mathbb{C}$ with $|\alpha| = 1$).$

$\noindent b) Find necessary and sufficient conditions on $a$ and $b$ in order for both roots of $P(z)$ to lie strictly outside the unit disk (i.e. both roots have modulus greater than $1$).$

From the modulus of the roots, we have:

$$\noindent$ |\alpha|^2 = \alpha \bar{\alpha} = 1 \Rightarrow a = 1$

From completion of the square, we require it to be strictly a sum of two positive squares.

$$\noindent$ 1-\frac{b^2}{4} > 0 \Leftrightarrow -2 < b < 2$

From the modulus of the roots, we have:

$$\noindent$ |\alpha|^2 = \alpha \bar{\alpha} = \frac{1}{a} > 1 \Rightarrow 0<a<1$

Completing the square is more cumbersome so we resort to the discriminant.

$$\noindent b^2 < 4a \Leftrightarrow -2\sqrt{a}<b<2\sqrt{a}$

seanieg89 · Sep 30, 2016

Re: HSC 2016 4U Marathon

You are ignoring the possibility that both roots of the quadratic are real and do not coincide, in which case they are not complex conjugates of each other.

(Also, you are allowing for the discriminant to equal 0 in the first part but not the second, is there a reason for this?)

Paradoxica · Sep 30, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
You are ignoring the possibility that both roots of the quadratic are real and do not coincide, in which case they are not complex conjugates of each other.

How so? the discriminant is a sufficient condition to determine that the roots are complex. and the completion of the square is equivalent.

Discriminant equal to zero was a slip-up.

seanieg89 · Sep 30, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
How so? the discriminant is a sufficient condition to determine that the roots are complex. and the completion of the square is equivalent.

Discriminant equal to zero was a slip-up.

I know, but the question asks you to find N&C conditions on coefficients for:

a) single root on unit circle
b) both roots in unit disk.

You found this condition by assuming that the roots are of the form r,conj(r), and that the coefficients of the polynomial were such that the polynomial had non-real roots. However it is of course possible for a polynomial to have real roots and satisfy either of a) or b). Your stated conditions do not pick up this possibility.

For examples: (z-1)(z-69)/69 does not have coefficients satisfying the conditions you responded with for the first part, yet it most certainly has a root on the unit circle.

Paradoxica · Sep 30, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
I know, but the question asks you to find N&C conditions on coefficients for:

a) single root on unit circle
b) both roots in unit disk.

You found this condition by assuming that the roots are of the form r,conj(r), and that the coefficients of the polynomial were such that the polynomial had non-real roots. However it is of course possible for a polynomial to have real roots and satisfy either of a) or b). Your stated conditions do not pick up this possibility.

For instance. (z-1)(z-69)/69 does not have coefficients satisfying the conditions you responded with for the first part, yet it most certainly has a root on the unit circle.

Ah.

As I discovered once again, I cannot read.

KingOfActing · Oct 4, 2016

Re: HSC 2016 4U Marathon

$$Define $ S(x) = \sum_{n = 0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ and $C(x) = \sum_{n = 0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\\\\$Show that $S^2(x) + C^2(x) = 1$ for all values of $x.$

leehuan · Oct 4, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
$$Define $ S(x) = \sum_{n = 0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ and $C(x) = \sum_{n = 0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\\\\$Show that $S^2(x) + C^2(x) = 1$ for all values of $x.$

For once in my life I am tempted to say by inspection...

KingOfActing · Oct 4, 2016

Re: HSC 2016 4U Marathon

leehuan said:
For once in my life I am tempted to say by inspection...

You cannot

If you want to assert S(x) another popular function, you must do so rigorously

KingOfActing · Oct 4, 2016

Re: HSC 2016 4U Marathon

$By considering the function $f(x) = e^{-ix}(\cos{x} + i\sin{x})$, prove that $e^{ix} = \cos{x} + i\sin{x}$. Hence find expressions for $\cos{x}$ and $\sin{x}$ in terms of $ e^{ix}$.$

seanieg89 · Oct 5, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
$By considering the function $f(x) = e^{-ix}(\cos{x} + i\sin{x})$, prove that $e^{ix} = \cos{x} + i\sin{x}$. Hence find expressions for $\cos{x}$ and $\sin{x}$ in terms of $ e^{ix}$.$

For this question to be well-posed, you very much need to specify what you mean by the quantity "e^(ix)". I.e. what is your definition of the quantity you are asserting to be equal to cis(x)?

You either need to define what it means to raise a real number to a complex power, or use another method of defining f(z)="e^z" for complex z.

Also the differentiation of functions from R to C is not quite discussed in the high school syllabus, but I am sure that students are comfortable with the notion of just differentiation the real and complex components separately.

(This question has come up before in these forums / elsewhere, am not the biggest fan of the sloppy logic it can lead students to adopt.)

Paradoxica · Oct 5, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
$$Define $ S(x) = \sum_{n = 0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ and $C(x) = \sum_{n = 0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\\\\$Show that $S^2(x) + C^2(x) = 1$ for all values of $x.$

By the ratio test, both series converge for all real x. We assume the function, whatever it happens to be, is continuous and differentiable at every point in the domain.

We begin with the left hand side.

Differentiating both sides with respect to x using the chain rule, we obtain the following:

$2\left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} \right)\left( \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n} \right) + 2\left( \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n} \right)\left( \sum_{n=1}^\infty \frac{(-1)^n x^{2n-1}}{(2n-1)!} \right)$

The second bracket of the second expression is near-identical to the first bracket of the first expression, with shifted summation index. Unshifting the index, it is clear that the two expressions are exactly the same, but opposite in sign, and cancel completely. Thus, the derivative of the expression is 0 for all values of x.

Based on the assumptions and results, the left hand side is equivalently constant for all real x. It suffices to find a single value of the expression. It is clear that by letting x=0, the left hand side is equal to 1, and the desired result is obtained.

Paradoxica · Oct 5, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
For this question to be well-posed, you very much need to specify what you mean by the quantity "e^(ix)". I.e. what is your definition of the quantity you are asserting to be equal to cis(x)?

You either need to define what it means to raise a real number to a complex power, or use another method of defining f(z)="e^z" for complex z.

Also the differentiation of functions from R to C is not quite discussed in the high school syllabus, but I am sure that students are comfortable with the notion of just differentiation the real and complex components separately.

(This question has come up before in these forums / elsewhere, am not the biggest fan of the sloppy logic it can lead students to adopt.)

Yeah, I advocate for either a geometric proof or a series expansion proof.

KingOfActing · Oct 5, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
For this question to be well-posed, you very much need to specify what you mean by the quantity "e^(ix)". I.e. what is your definition of the quantity you are asserting to be equal to cis(x)?

You either need to define what it means to raise a real number to a complex power, or use another method of defining f(z)="e^z" for complex z.

Also the differentiation of functions from R to C is not quite discussed in the high school syllabus, but I am sure that students are comfortable with the notion of just differentiation the real and complex components separately.

(This question has come up before in these forums / elsewhere, am not the biggest fan of the sloppy logic it can lead students to adopt.)

e^z = e^(x+iy) for complex z is to be defined as the unique holomorphic function that agrees with e^x for real x? (I think that's a good definition for this)

Since it's holomorphic that means it's complex differentiable, and so the chain and product rules apply in the same way.

Is what I'm saying even close to correct?

Paradoxica · Oct 5, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
e^z = e^(x+iy) for complex z is to be defined as the unique holomorphic function that agrees with e^x for real x? (I think that's a good definition for this)

Since it's holomorphic that means it's complex differentiable, and so the chain and product rules apply in the same way.

Is what I'm saying even close to correct?

Now do the same for Log(z)

DatAtarLyfe · Oct 5, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
By the ratio test, both series converge for all real x. We assume the function, whatever it happens to be, is continuous and differentiable at every point in the domain.

We begin with the left hand side.

Differentiating both sides with respect to x using the chain rule, we obtain the following:

$2\left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} \right)\left( \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n} \right) + 2\left( \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n} \right)\left( \sum_{n=1}^\infty \frac{(-1)^n x^{2n-1}}{(2n-1)!} \right)$

The second bracket of the second expression is near-identical to the first bracket of the first expression, with shifted summation index. Unshifting the index, it is clear that the two expressions are exactly the same, but opposite in sign, and cancel completely. Thus, the derivative of the expression is 0 for all values of x.

Based on the assumptions and results, the left hand side is equivalently constant for all real x. It suffices to find a single value of the expression. It is clear that by letting x=0, the left hand side is equal to 1, and the desired result is obtained.

For the first time in my life, i actually understood one of your proofs

penapplepineapplepen · Oct 5, 2016

Re: HSC 2016 4U Marathon

I kanot understand nglish wary mutch so i was onedering how youse or i can get two zi funthementals off nglish thred pliez

zanking yous wary mutch

PPAP

InteGrand · Oct 5, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
e^z = e^(x+iy) for complex z is to be defined as the unique holomorphic function that agrees with e^x for real x? (I think that's a good definition for this)

Since it's holomorphic that means it's complex differentiable, and so the chain and product rules apply in the same way.

Is what I'm saying even close to correct?

Yeah but most 4U students wouldn't understand the meaning of this of course. (Nor are they expected to.)

KingOfActing · Oct 5, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
Yeah but most 4U students wouldn't understand the meaning of this of course.

Exactly, which is why I would've just expected a 4u student to differentiate as if i was any other constant with the property i^2 = -1, and the rest of the proof follows

penapplepineapplepen · Oct 5, 2016

Re: HSC 2016 4U Marathon

help me pliez, youse are just ignoreing me

seanieg89 · Oct 5, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
e^z = e^(x+iy) for complex z is to be defined as the unique holomorphic function that agrees with e^x for real x? (I think that's a good definition for this)

Since it's holomorphic that means it's complex differentiable, and so the chain and product rules apply in the same way.

Is what I'm saying even close to correct?

It is a true statement, but then that sweeps the issue under the rug, as it is not obvious to someone who has not studied some analysis that a) the real exponential would have an analytic extension or b) there is a unique such extension. In other words, there is mathematical content in your definition. (Students would also have to know what analyticity even meant in order to understand why a) and b) are true.)

This content is also higher level than the theory required to just define and understand the trigonometric/exponential functions in terms of power series, which is pretty much the cleanest way to rigorously work with such functions.

(Of course in this framework Euler's is not so remarkable, and the proof is rather boring.)

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