Polynomials (2 Viewers)

aa180 · Sep 13, 2018

Just a fun question I made up.

fan96 · Sep 13, 2018

a) Consider the graph of $y = P(z)$ on the cartesian plane, for real $z$ .

$P(0) = -k$

$P'(z) = 3(z^2-2kz+1)$

$P'(0) = 3$

$P'(z) = 0,\, z = k \pm \sqrt{k^2-1}$

(Note that we require $|k| \geq 1$ for a turning point to exist - otherwise $P(z)$ has exactly one real root)

We will prove that both values of $z$ are either all positive or all negative.

Expanding $(a-b)^2 \geq 0$ gives

$\frac{a^2+b^2}{2} \geq ab$ (equality occurs only when $a = b$ .)

Setting $a = \sqrt{k+1},\, b = \sqrt{k-1}$ gives $k - \sqrt{k^2-1} >0$ for $k \geq 1$ .

Setting $a = \sqrt{-(k+1)},\, b = \sqrt{-(k-1)}$ gives $k + \sqrt{k^2-1} <0$ for $k \leq -1$ .

Since $\sqrt{k^2-1} \geq 0$ , both possible values of $z$ are positive for positive $k$ , and negative for negative $k$ .

Suppose $k \geq 1$ . Then the $y$ intercept is negative, and the gradient at that point is positive. Because all turning points of $P(z)$ are for positive $z$ , $P(z)$ is negative definite for $z \leq 0$ and thus must have only positive roots.

Now suppose $k \leq -1$ . Then the $y$ intercept is positive, and the gradient at that point is positive. Because all turning points of $P(z)$ are for negative $z$ , $P(z)$ is positive definite for $z \geq 0$ and thus must have only negative roots.

b) Expanding $(a-b)^2 \geq 0$ gives the required answer.

c) Expanding the RHS and simplifying gives the required answer.

d) From b),

$\begin{cases} a^2+b^2 \geq 2ab \\ a^2+c^2 \geq 2ac \\ c^2+b^2 \geq 2bc\end{cases}$

Adding gives $2(a^2+b^2+c^2)\geq 2(ab+ac+bc) \implies a^2+b^2+c^2 - ab-ac-bc\geq 0$

Applying this to c) gives $a^3+b^3+c^3 = 3abc + \mathrm{(a\, positive\, number)} \implies a^3+b^3+c^3 \geq 3abc$

Then by replacing each variable with its cube root, we get the desired identity.

e) Let the three (real) roots be $\alpha,\,\beta,\,\gamma$ .

Now, taking the sum of roots two at a time gives $\alpha \beta + \alpha\gamma + \beta \gamma = 3$ .

From a) we know that $|k| \geq 1$ is required.

From d), since the roots are real we have another requirement:

$\frac{\alpha \beta + \alpha\gamma + \beta \gamma}{3} \geq \sqrt[3]{(\alpha \beta) (\alpha\gamma )(\beta \gamma)}$ .

(Note that because the roots are either all positive or all negative from a), multiplying any two together always gives a positive number, so this inequality can be used.)

Simplifying gives $1 \geq k^2 \implies |k| \leq 1$ .

If all three roots are real then both $|k| \geq 1$ and $|k| \leq 1$ are required, i.e. $|k| = 1 \implies k = \pm 1$ .

It follows that if $k \neq \pm 1$ , then not all of three roots could be real (the contrapositive).

Because $P(z)$ has real coefficients, the complex conjugate root theorem applies - any non-real roots must come in pairs. And by the fundamental theorem of algebra, $P(z)$ has exactly three roots. Therefore if not all of the three roots of $P(z)$ are real, then it has exactly one real root.

aa180 · Sep 14, 2018

It looks good, BUT... when you were showing that $k\pm\sqrt{k^{2}-1}$ both have the same sign for any $k$ (in fact, the same sign as $k$ ), there should be a total of four cases to consider. These are:

1) $k-\sqrt{k^{2}-1} > 0 \text{ for } k\geq1$
2) $k+\sqrt{k^{2}-1} < 0 \text{ for } k\leq-1$
3) $k+\sqrt{k^{2}-1} > 0 \text{ for } k\geq1$
4) $k-\sqrt{k^{2}-1} < 0 \text{ for } k\leq-1$

You only addressed the first two of these cases, and then stated that $\sqrt{k^{2}-1} \geq 0$ from which you concluded what you were trying to prove. I just don't see how it follows from that, without proving cases 3 and 4 directly as you did with 1 & 2.

fan96 · Sep 14, 2018

aa180 said:
It looks good, BUT... when you were showing that $k\pm\sqrt{k^{2}-1}$ both have the same sign for any $k$ (in fact, the same sign as $k$ ), there should be a total of four cases to consider. These are:

1) $k-\sqrt{k^{2}-1} > 0 \text{ for } k\geq1$
2) $k+\sqrt{k^{2}-1} < 0 \text{ for } k\leq-1$
3) $k+\sqrt{k^{2}-1} > 0 \text{ for } k\geq1$
4) $k-\sqrt{k^{2}-1} < 0 \text{ for } k\leq-1$

You only addressed the first two of these cases, and then stated that $\sqrt{k^{2}-1} \geq 0$ from which you concluded what you were trying to prove. I just don't see how it follows from that, without proving cases 3 and 4 directly as you did with 1 & 2.

I didn't address those because they seemed trivial to me - my bad.

If $k$ is positive, then $k + \sqrt{k^2-1}$ must be positive since two positive numbers are being added.
If $k$ is negative, then $k - \sqrt{k^2-1}$ must be negative too since a positive number is being subtracted from a negative number.

Another way to think about it is that if $k-\sqrt{k^{2}-1} > 0$ for positive $k$ then it follows that $|k| > |\sqrt{k^2-1}|$ i.e. the magnitude of $k$ is always greater than the magnitude of $\sqrt{k^2-1}$ . Therefore, adding or subtracting $\sqrt{k^2-1}$ can never change the sign of $k$ .

aa180 · Sep 14, 2018

fan96 said:
I didn't address those because they seemed trivial to me - my bad.

If $k$ is positive, then $k + \sqrt{k^2-1}$ must be positive since two positive numbers are being added.
If $k$ is negative, then $k - \sqrt{k^2-1}$ must be negative too since a positive number is being subtracted from a negative number.

Lol, I didn't even see that - yes, they are trivial..whoops.

You can also do part (a) by contradiction if you wanted to.

aa180 · Sep 14, 2018

fan96 said:
Another way to think about it is that if $k-\sqrt{k^{2}-1} > 0$ for positive $k$ then it follows that $|k| > |\sqrt{k^2-1}|$ i.e. the magnitude of $k$ is always greater than the magnitude of $\sqrt{k^2-1}$ . Therefore, adding or subtracting $\sqrt{k^2-1}$ can never change the sign of $k$ .

Yep, you can also use that same line of reasoning for the proof by contradiction approach.

fan96 · Sep 14, 2018

I'm having some trouble with part (h).

If one of the non-real roots had real part $k$ , then by the complex conjugate root theorem, the sum of the two non-real roots should be $2k$ .

Then, using the sum of roots, the last root should be $k$ .

But $(k,\,P(k))$ is the point of inflection of $P(z)$ .

So therefore the point of inflection lies on a root, meaning that $P(z)$ should have a triple root at $x= k$ , which contradicts the question.

To double-check, I substituted $k = \sqrt{1/2}$ into $P(z)$ and put it into Wolfram Alpha to check the roots. The non-real roots did not have real part $\sqrt {1/2}$ .

Have I made a mistake with this reasoning?

aa180 · Sep 14, 2018

fan96 said:
I'm having some trouble with part (h).

If one of the non-real roots had real part $k$ , then by the complex conjugate root theorem, the sum of the two non-real roots should be $2k$ .

Then, using the sum of roots, the last root should be $k$ .

But $(k,\,P(k))$ is the point of inflection of $P(z)$ .

So therefore the point of inflection lies on a root, meaning that $P(z)$ should have a triple root at $x= k$ , which contradicts the question.

To double-check, I substituted $k = \sqrt{1/2}$ into $P(z)$ and put it into Wolfram Alpha to check the roots. The non-real roots did not have real part $\sqrt {1/2}$ .

Have I made a mistake with this reasoning?

You are actually correct. I just remembered that a while ago when I was trying to come up with polynomials questions, I tried to set one up that had a monic cubic equation with the remaining coefficients being all arbitrary, such that the cubic had a non-real root corresponding to its point of inflection in the real plane, and I derived pretty much the same contradiction you found - that the point of inflection of the cubic will forcibly coincide with the real root of the polynomial, which then gives three real roots instead of one. I must've completely forgotten about this when I was busy writing this question.

I tried to amend (i.e change) the question so that the non-real root of P now corresponds to the point of inflection reflected about the y-axis, then tried to suppose that the centre of the circle through the three roots of P is the origin, but it turns out that that's also impossible lol. Seems like you can't get any nice geometric properties out of playing with roots of polynomial equations. I'll try to find some sort of way to reconcile this issue and will to get back to you shortly.

aa180 · Sep 14, 2018

Okay so I just realized I made yet another mistake and my amendment actually is possible, but you end up getting $k = \pm\sqrt{1+\sqrt{6}/2}$ which isn't very 'nice-looking' so I think I'll still try to change it further.

jyu · Sep 15, 2018

If k=1, z^3-3z^2+3z-1=(z-1)^3, if k=-1, z^3+3z^2+3z+1=(z+1)^3, .: exactly one real root each.
I think you need to review your question.

fan96 · Sep 15, 2018

jyu said:
If k=1, z^3-3z^2+3z-1=(z-1)^3, if k=-1, z^3+3z^2+3z+1=(z+1)^3, .: exactly one real root each.
I think you need to review your question.

Are those not triple roots?

jyu · Sep 15, 2018

Exactly one real root and no complex roots when k=+/-1.
For other real values of k, there are 3 roots according to the Fundamental Theorem of Algebra, one must be real according to the Conjugate Roots Theorem.

aa180 · Sep 15, 2018

jyu said:
Exactly one real root and no complex roots when k=+/-1.
For other real values of k, there are 3 roots according to the Fundamental Theorem of Algebra, one must be real according to the Conjugate Roots Theorem.

There's one real root of multiplicity 3 when $k = \pm 1$ , i.e there are three real roots - they just all happen to have the same value.

jyu · Sep 15, 2018

There is only one real x such that P(x)=0. Not three real xs. Meaning of root.

aa180 · Sep 15, 2018

Okay, I will change the question to this:

Suppose that $\alpha$ is a non-real root of $P$ occurring when $k\neq\pm1$ such that its position in the Argand plane corresponds to the midpoint between the origin and the point of inflection of $P$ in the Cartesian plane.

(f) Show that $\text{Re}(\alpha) = k/2$ and find an expression for $\text{Im}(\alpha)$ .

(g) If $\alpha$ , the real root of $P$ , and its y-intercept are all collinear, show that $k = \pm\frac{\sqrt{7}}{2}.$

(sorry for changing it so many times)

jyu · Sep 15, 2018

I tried k=sqrt(3.5), plotted x and y intercepts and alpha, they are not collinear.

fan96 · Sep 15, 2018

Also, I noticed that when you find the third (real) root using sum of roots and then sub these three roots into the product of roots, you seem to get an equation with certain solutions for $k$ that contradict the question.

aa180 · Sep 15, 2018

jyu said:
I tried k=sqrt(3.5), plotted x and y intercepts and alpha, they are not collinear.

Try $k = \pm\frac{\sqrt{7}}{2}$ .

aa180 · Sep 15, 2018

fan96 said:
Also, I noticed that when you find the third (real) root using sum of roots and then sub these three roots into the product of roots, you seem to get an equation with certain solutions for $k$ that contradict the question.

Yeah I just tried that and you're right. I think that means it's impossible to have an imaginary root that lies halfway between the origin and the point of inflection. I'll try a few other ideas and see where it gets me..

fan96 · Sep 15, 2018

aa180 said:
Yeah I just tried that and you're right. I think that means it's impossible to have an imaginary root that lies halfway between the origin and the point of inflection. I'll try a few other ideas and see where it gets me..

Well since there are still solutions that exist then there should be at least one complex number that works.

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