Polynomials question (1 Viewer)

integral95 · Jul 3, 2018

Let a,b,c,d be the roots of P(z)

Consider sum of each of the square of the roots i.e

$a^2 + b^2+ c^2 + d^2 = (a+b+c+d)^2 - 2($sum of roots 2 at a time$)$

You'll see that this is negative, that proves that there is at least one complex root, since P(z) has real coefficients, the 2nd root must a complex conjugate root of the first.

anddddddd I'm lost with proving the other 2 roots atm.

aa180 · Jul 3, 2018

integral95 said:
Let a,b,c,d be the roots of P(z)

Consider sum of each of the square of the roots i.e

$a^2 + b^2+ c^2 + d^2 = (a+b+c+d)^2 - 2($sum of roots 2 at a time$)$

You'll see that this is negative, that proves that there is at least one complex root, since P(z) has real coefficients, the 2nd root must a complex conjugate root of the first.

anddddddd I'm lost with proving the other 2 roots atm.

Lol for a second there I thought your original solution was correct (though I think I spotted the error), but either way it still differs from my proof.

fan96 · Jul 3, 2018

The coefficients of the polynomial are suspiciously symmetrical.

Consider:

$\begin{aligned} P\left(\frac1z\right) &= \frac{2}{z^4} + \frac{5}{z^3} + \frac{7}{z^2} + \frac{5}{z} + 2 \\ &= \left(\frac{1}{z^4}\right)P(z)\end{aligned}$

Therefore, if $\alpha$ is a (nonzero) root of $P(z)$ then so is its reciprocal $\alpha^{-1}$ .

By the Fundamental Theorem of Algebra and the conjugate root theorem, the four roots of $P(z)$ are $\alpha,\,\overline{\alpha},\,\alpha^{-1},\,\overline{\alpha^{-1}}$ .

This means that the roots of $P(z)$ are either all real or all non-real.

integral95 has already proved that $P(z)$ has non-real roots, so therefore $P(z)$ has no real roots.

aa180 · Jul 3, 2018

fan96 said:
The coefficients of the polynomial are suspiciously symmetrical.

Consider:

$\begin{aligned} P\left(\frac1z\right) &= \frac{2}{z^4} + \frac{5}{z^3} + \frac{7}{z^2} + \frac{5}{z} + 2 \\ &= \left(\frac{1}{z^4}\right)P(z)\end{aligned}$

Therefore, if $\alpha$ is a (nonzero) root of $P(z)$ then so is its reciprocal $\alpha^{-1}$ .

By the Fundamental Theorem of Algebra and the conjugate root theorem, the four roots of $P(z)$ are $\alpha,\,\overline{\alpha},\,\alpha^{-1},\,\overline{\alpha^{-1}}$ .

This means that the roots of $P(z)$ are either all real or all non-real.

integral95 has already proved that $P(z)$ has non-real roots, so therefore $P(z)$ has no real roots.

Are you sure?

You have not accounted for the possibility that $\alpha$ lies on the unit circle which would give $\overline{\alpha} = \alpha^{-1}$ and $\alpha = \overline{\alpha^{-1}}$ , meaning only two roots are recovered and the possibility that the remaining two being purely real is still completely valid at this point.

fan96 · Jul 3, 2018

aa180 said:
Are you sure?

You have not accounted for the possibility that $\alpha$ lies on the unit circle which would give $\overline{\alpha} = \alpha^{-1}$ and $\alpha = \overline{\alpha^{-1}}$ , meaning only two roots are recovered and the possibility of the remaining two being purely real is still completely valid at this point.

whoops... my bad.

aa180 · Jul 3, 2018

fan96 said:
whoops... my bad.

Lol but you were definitely right in observing the implication that the roots occur in reciprocal pairs due to the symmetry of the coefficients, as this is integral to the proof (unless there's some shortcut way I haven't picked up on).

mrbunton · Jul 4, 2018

aa180 said:
Are you sure?

You have not accounted for the possibility that $\alpha$ lies on the unit circle which would give $\overline{\alpha} = \alpha^{-1}$ and $\alpha = \overline{\alpha^{-1}}$ , meaning only two roots are recovered and the possibility that the remaining two being purely real is still completely valid at this point.

if two roots lie on unit circle and are reciprocal to each other:
(using fan96's and integral95's progress)
let x+iy denote one of these imaginary roots.
2x^2-2y^2+a^2+b^2=-3/4 if i didnt make any mistakes, from integral's working.
anyway if the root is on unit circle than x^2+y^2=1; x^2=1-y^2
a and b are also reciprocal to each other from product of roots where ab=1; a=1/b
hence:
2-4y^2+a^2+1/a^2=-3/4

the min value of a^2+1/a^2 is 2 (using calc.) and the max value for y is 1 (unit circle)
putting all this in the min value for the equation should be 2-4+2=0; but it's negative so hence a (and b) must be imaginary as well.

is this a correct addition to the solution?

aa180 · Jul 4, 2018

mrbunton said:
if two roots lie on unit circle and are reciprocal to each other:
(using fan96's and integral95's progress)
let x+iy denote one of these imaginary roots.
2x^2-2y^2+a^2+b^2=-3/4 if i didnt make any mistakes, from integral's working.
anyway if the root is on unit circle than x^2+y^2=1; x^2=1-y^2
a and b are also reciprocal to each other from product of roots where ab=1; a=1/b
hence:
2-4y^2+a^2+1/a^2=-3/4

the min value of a^2+1/a^2 is 2 (using calc.) and the max value for y is 1 (unit circle)
putting all this in the min value for the equation should be 2-4+2=0; but it's negative so hence a (and b) must be imaginary as well.

is this a correct addition to the solution?

I believe you are correct which thus completes the proof, though my solution is still different from what you, fan96, and integral95 have proposed, but that which still utilizes some of the ideas you guys have put forth. Well done for still managing to solve it otherwise.

KAIO7 · Jul 4, 2018

My gut feeling tells me that re-arranging/expressing the equation differently and restricting the range of the possible roots to yield a contradiction could work

Also you could express it as the product of two quadratic functions and show that the discriminant is less than zero.

aa180 · Jul 5, 2018

mrbunton said:
what is your method?

$\text{It is given that } P(z) = 2z^{4} + 5z^{3} + 7z^{2} + 5z + 2. \\ \text{To carry out the proof, we will need some preliminary results.}$

$\text{Lemma 1: If } \alpha \text{ is a root of } P, \text{ then } \frac{1}{\alpha} \text{ is also a root}. \\ \text{Proof: See fan96's reply.}$

$\text{Lemma 2: } a\in\mathbb{R}\backslash\{0\} \Longrightarrow |a+\frac{1}{a}| \geq 2. \\ \text{Proof: Fix } a\in\mathbb{R}\backslash\{0\}. \text{ By the triangle inequality, we have:}$

$|a+\frac{1}{a}| \leq |a|+|\frac{1}{a}| \text{, with equality IFF } a \text{ and } \frac{1}{a} \text{ have the same sign.} \\ \text{But the sign of } a \text{ equals the sign of } \frac{1}{a}, \space \forall{a}\in\mathbb{R}\backslash\{0\}, \text{ therefore:}$

$\begin{align*} |a+\frac{1}{a}| &= |a| + |\frac{1}{a}| \\ &= |a| + \frac{1}{|a|} \end{align*}$

$\text{Now consider the inequality } (\sqrt{|a|} - \frac{1}{\sqrt{|a|}})^{2} \geq 0, \text{ which is true } \\ \forall{a}\in\mathbb{R}\backslash\{0\}. \text{ Expanding and rearranging gives:}$

$|a| + \frac{1}{|a|} \geq 2, \text{ i.e. } |a+\frac{1}{a}| \geq 2, \text{ as required.}$

$\text{Now suppose that two of the roots are real and are equal to } a_{1} \text{ and } a_{2}, \text{ respectively.} \\ \text{ By Lemma 1, } \frac{1}{a_{1}} \text{ and } \frac{1}{a_{2}} \text{ are the remaining two roots (since neither root is zero).} \\ \text{Now consider the sum of roots one at a time:}$

$a_{1}+\frac{1}{a_{1}}+a_{2}+\frac{1}{a_{2}} = -\frac{5}{2} = -2.5$

$\Longrightarrow |(a_{1}+\frac{1}{a_{1}})+(a_{2}+\frac{1}{a_{2}})| = |-2.5| = 2.5$

$\Longrightarrow |a_{1}+\frac{1}{a_{1}}| + |a_{2}+\frac{1}{a_{2}}| \geq 2.5 \hfill \text{(by the triangle inequality)}$

$\text{Now, equality holds precisely when } a_{1} \text{ and } a_{2} \text{ have the same sign. But}$

$\begin{align*} |a_{1}+\frac{1}{a_{1}}| + |a_{2}+\frac{1}{a_{2}}| &\geq 2+2 \space \text{(by Lemma 2)} \\ &= 4 \\ &> 2.5 \end{align*}$

$\therefore \text{ Equality cannot take place which forces the conclusion that } a_{1} \text{ and } a_{2} \text{ have opposing signs.} \\ \text{Now consider the sum of roots two at a time:}$

$\begin{align*} a_{1}\frac{1}{a_{1}}+a_{1}{a_{2}}+a_{1}\frac{1}{a_{2}}+\frac{1}{a_{1}}{a_{2}}+\frac{1}{a_{1}}\frac{1}{a_{2}}+a_{2}\frac{1}{a_{2}} &= \frac{7}{2} \\ 1+a_{1}a_{2}+\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{1}}+\frac{1}{a_{1}a_{2}}+1 &= 3.5 \\ a_{1}a_{2}+\frac{1}{a_{1}a_{2}}+\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{1}} &= 1.5 \end{align*}$

$\text{By inspection, the LHS must be negative while the RHS is positive - contradiction!} \\ \text{We therefore conclude that the roots of } P \text{ cannot all be real.} \\ \text{The proof that } P \text{ cannot have a pair of real and a pair of complex roots is similar and left as an exercise to the reader.}$

$\text{QED}$

aa180 · Jul 5, 2018

KAIO7 said:
Also you could express it as the product of two quadratic functions and show that the discriminant is less than zero.

I originally thought of expressing it as a sum of two positive definite quadratics (as opposed to your suggestion of writing it as a product) but was unable to find two that were suitable.

By quadratics I mean a quadratic and a biquadratic.

InteGrand · Jul 6, 2018

aa180 said:
$\text{Let } P(z) = 2z^{4} + 5z^{3} + 7z^{2} + 5z + 2. \\ \text{ Without solving } P(z) = 0, \text{ prove that } P \text{ has no real roots.}$

$\noindent Well essentially we have $P(z) = z^2 \left( 2z^2 + 5z + 7 + 5z^{-1} + 2 z^{-2}\right)$. Hence $P(z) = z^2 \left( 2 \left( z + z^{-1}\right)^{2} + 5\left(z + z^{-1}\right) + 3\right)$.$

$$\noindent So if $x$ is a real root of $P$, then $x \neq 0$ and $q(u):= 2u^2 + 5u + 3 = 0$, where $u \equiv x + x^{-1}$. Since $x$ is real, we have $|u| \geq 2$ (using AM-GM or whatever). The leading coefficient of $q$ is positive. Also $q(2) > 0$ and $q(-2) = 8 -10+ 3 = 1 > 0$. So $q(t) > 0$ for all real $t$ with $|t| \geq 2$ Hence $q$ has no real root with absolute value at least $2$. Therefore, $P$ has no real root.$

aa180 · Jul 6, 2018

InteGrand said:
$\noindent Well essentially we have $P(z) = z^2 \left( 2z^2 + 5z + 7 + 5z^{-1} + 2 z^{-2}\right)$. Hence $P(z) = z^2 \left( 2 \left( z + z^{-1}\right)^{2} + 5\left(z + z^{-1}\right) + 3\right)$.$

$$\noindent So if $x$ is a real root of $P$, then $x \neq 0$ and $q(u):= 2u^2 + 5u + 3 = 0$, where $u \equiv x + x^{-1}$. Since $x$ is real, we have $|u| \geq 2$ (using AM-GM or whatever). The leading coefficient of $q$ is positive. Also $q(2) > 0$ and $q(-2) = 8 -10+ 3 = 1 > 0$. So $q(t) > 0$ for all real $t$ with $|t| \geq 2$ Hence $q$ has no real root with absolute value at least $2$. Therefore, $P$ has no real root.$

Are you assuming the points (-2, q(-2)) and (2, q(2)) lie on opposite sides of the axis of symmetry of the parabola y = q(t)? Because if they happen to lie on the same side then the minimum value of q(t) will occur away from these endpoints (but still within the domain of q), and so you needed to justify this assumption.

InteGrand · Jul 6, 2018

aa180 said:
Are you assuming the points (-2, q(-2)) and (2, q(2)) lie on opposite sides of the axis of symmetry of the parabola y = q(t)? Because if they happen to lie on the same side then the minimum value of q(t) will occur away from these endpoints (but within the domain of q), and so you needed to justify this assumption.

$\noindent Oh yes, I was assuming that. (The axis of symmetry for $q(t)$ is at $t = -5/4$.)$

aa180 · Jul 6, 2018

InteGrand said:
$\noindent Oh yes, I was assuming that. (The axis of symmetry for $q(t)$ is at $t = -5/4$.)$

Yeah I thought so, but otherwise your proof is acceptable.

InteGrand · Jul 6, 2018

aa180 said:
Yeah I thought so, but otherwise your proof is acceptable.

Thanks.

The main reason I wrote it was to show students the idea of writing the polynomial as a polynomial in (z+ 1/z). In fact, the polynomial P(z) is an even degree palindromic polynomial, and any even degree palindromic polynomial of degree 2N (in z) can be written as z^N times a polynomial of degree N in (z +1/z). This meant the given quadratic polynomial could effectively be written as a quadratic polynomial, which students should find easier to analyse.

aa180 · Jul 6, 2018

InteGrand said:
Thanks.

The main reason I wrote it was to show students the idea of writing the polynomial as a polynomial in (z+ 1/z). In fact, the polynomial P(z) is an even degree palindromic polynomial, and any even degree palindromic polynomial of degree 2N (in z) can be written as a polynomial of degree N in (z +1/z). This meant the given quadratic polynomial could effectively be written as a quadratic polynomial, which students should find easier to analyse.

Yeah when I originally wrote this question, I knew the roots were easily attainable through division of the equation $P(z) = 0$ by $z^{2}$ , and since I had my own alternative method for proving the roots aren't real, I wanted to deny the reader the option of solving P(z) = 0 so as to make the question more of a challenge. Your method is kinda sneaky in that it circumvents the restriction imposed by the question while simultaneously utilizing the property of reducing P(z) to a quadratic in (z+1/z).

Also, thanks for giving me the correct terminology for a polynomial with symmetric coefficients. I know what a palindromic word and a palindromic number is, but for some reason I didn't realize the same name applies for a polynomial haha

aa180 · Jul 9, 2018

Polynomials question (1 Viewer)

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