 My Math Forum Question about proving the availability of double real root of a polynomial
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 October 31st, 2018, 01:50 AM #1 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 Question about proving the availability of double real root of a polynomial 23A5BCBB-CA79-4DA7-B726-23E5A84070E1.jpg 2BC82CF6-7D69-4698-87BF-11388DD9C0A7.jpg E2282EA8-2E2A-45F4-B13F-D9D0ACAFB0B4.jpg For part b, it is assuming that n is an odd number, so r-a = r+a Since the exponent of n-1 would be an even number for n being an odd number, is the equation of (r-a)^(n-1) = -(r+a)^(n-1) undefined? It is because any real number to a power of an even number cannot be negative. Please rectify me if I have made some mistakes. October 31st, 2018, 08:07 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 I can't read your attachments so I cannot actually check your work. $f(x) \text { is a polynomial such that } f(x) = (x - r)^2 * g(x) \text { and } g(r) \ne 0.$ $f'(x) = (x - r)^2 * g'(x) + 2(x - r) * g(x) \implies f'(r) = 0 * g'(r) + 2 * 0 * g(r) = 0.$ $p(x) = (x - a)^n + (x + a)^n = \displaystyle = \left ( \sum_{k=0}^n \dbinom{n}{k} * (-\ 1)^k * x^{(n-k)}a^k \right ) + \left ( \sum_{k=0}^n \dbinom{n}{k} * x^{(n-k)}a^k \right ).$ $n = 2m \implies p(x) = \displaystyle \left ( \sum_{j=0}^m 2 * \dbinom{2m}{2j} * (x^{(m - j)})^2 * (a^j)^2 \right ) > 0 \ \because a \ne 0 \text { by hypothesis.}$ Every term in that sum is non-negative except the last, which is positive, so the sum is positive. There are no real roots, let alone a real double root. We skip the trivial case where n = 1. $p'(x) = \displaystyle \left ( \sum_{k=0}^{n-1} (n - k) \dbinom{n}{k} * (-\ 1)^k * x^{(n - 1 - k)}a^k \right ) + \left ( \sum_{k=0}^{n - 1} (n - k) * \dbinom{n}{k} * x^{(n-1-k)}a^k \right ).$ $n = 2m + 1 \implies p'(x) = \displaystyle \left ( \sum_{j=0}^m 2 * (2m + 1 - 2j) \dbinom{2m + 1}{2j} * (x^{(m - j)})^2(a^j)^2 \right ).$ $\therefore n = 2m + 1 \text { and } x \ne 0 \implies p'(x) > 0.$ $n = 2m + 1 \text { and } p'(x) = 0 \implies x = 0.$ $p(0) = a \ne 0.$ If n is odd, the derivative is not zero at any real root, and therefore no real root is a double root. October 31st, 2018, 07:40 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,020 Thanks: 2255 Using latex: Let $f(x)$ be a polynomial. If $f(x) = (x - r)^2g(x)$ where $g(r) \ne 0$, we say that $r$ is a double root of $f(x) = 0$. (a) $\$ If $f(x) = 0$ has a double root, prove that $f(r) = f'(r) = 0$. (b) $\$ Show that $p(x) = (x - a)^n + (x + a)^n$, where $a$ is a non-zero real number, does not have a double real root. 2. $\ \$ (a) $\$ If $f(x) = 0$ has a double root $r$, \displaystyle \begin{align*} f(x) &= (x - r)^2g(x) \\ \hspace{80px}f(r) &= (r - r)^2g(r) \\ &= 0 \\ f'(x) &= \frac{d}{dx}[(x - r)^2g(x)] \\ &= (x - r)^2\frac{d}{dx}[g(x)] + g(x)\frac{d}{dx}(x - r)^2 \\ &= (x - r)^2g'(x) + g(x)(2)(x - r)\frac{d}{dx}(x - r) \\ &= (x - r)^2g'(x) + 2(x - r)g(x) \\ f'(r) &= (r - r)^2g'(r) + 2(r - r)g(r) \\ &= 0\end{align*} $\hspace{80px}\therefore$ If $f(x) = 0$ has a double root $r$, then $\hspace{100px}f(r) = f'(r) = 0$ $\hspace{28px}$(b) $\$ Assume $p(x)$ has a double real root $r$, \displaystyle \hspace{66px}\therefore\ p(r) = p'(r) = 0 \\ \begin{align*}\hspace{66px}p'(x) &= \frac{d}{dx}[(x - a)^n + (x + a)^n] \\ &= n(x - a)^{n-1} + n(x + a)^{n-1} \\ \end{align*} \displaystyle \begin{align*}\hspace{100px}p'(r) &= n(r - a)^{n-1} + n(r + a)^{n-1} = 0 \\ (r - a)^{n-1} &= -(r + a)^{n-1}\end{align*} Thanks from justusphung November 1st, 2018, 01:49 AM #4 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 I am quite confused with the binomial expression, though I think, for the p(x), you are proving that the graph does not cut the x-axis given that a is not equal to zero. Therefore, the graph has no any real roots at all. Anyway, thx for your help Jeff. I checked up the answer of the question again. The derivative of p(x) is n(x+a)^(n-1) + n(x-a)^(n-1) = 0 by assuming that there is a double real root. We can re-write the equation to the form of (x-a)^(n-1) = -(x+a)^(n-1). Before, I was wondering why if n-1 was an even number, then x-a = x+a. Someone just told me that only when x-a = x+a = 0, the above equation would be valid. And now, I am convinced that x-a = x+a when n-1 is an even number. This leads to the result that a=0 which contradicts the fact of the constant a being an non-zero real number to disprove the assumption for n-1 being an even integer. November 1st, 2018, 05:38 AM   #5
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 Originally Posted by justusphung I am quite confused with the binomial expression, though I think, for the p(x), you are proving that the graph does not cut the x-axis given that a is not equal to zero. Therefore, the graph has no any real roots at all. Anyway, thx for your help Jeff. I checked up the answer of the question again. The derivative of p(x) is n(x+a)^(n-1) + n(x-a)^(n-1) = 0 by assuming that there is a double real root. We can re-write the equation to the form of (x-a)^(n-1) = -(x+a)^(n-1). Before, I was wondering why if n-1 was an even number, then x-a = x+a. Someone just told me that only when x-a = x+a = 0, the above equation would be valid. And now, I am convinced that x-a = x+a when n-1 is an even number. This leads to the result that a=0 which contradicts the fact of the constant a being an non-zero real number to disprove the assumption for n-1 being an even integer.
First, skipjack has provided a far more elegant proof than mine, and I think you have grasped it.

Second, I think you misunderstood mine. My proof is split into 2 parts.

If n is even, p(x) has no real root and therefore no real double root. For example

$p(x) = (x - a)^2 + (x + a)^2 = x^2 - 2ax + a^2 + x^2 - 2ax + a^2 = 2x^2 + 2a^2.$

And $2x^2 + 2a^2 > 0 \text { unless } x = 0 = a.$

$\text {But } a \ne 0 \text { by hypothesis.}$

So no real root and thus no double real root.

It is a bit more complex if n is odd. Then p(x) certainly has at least one real root. But p'(x) has no real root. But that is a requirement for a double real root. For example

$p(x) = (x - a)^3 + (x + a)^3 =$

$x^3 - 3ax^2 + 3a^2x - a^3 + x^3 + 3ax^2 + 3a^2x + a^3 = 2x^3 + 6a^2x \implies$

$p'(x) = 6x^2 + 6a^2 > 0 \text { unless } x = 0 = a.$

$\text {But } a \ne 0 \text { by hypothesis.}$

So no value of x can be a double root because p'(x) would have to equal 0 there.

Last edited by JeffM1; November 1st, 2018 at 05:41 AM. November 2nd, 2018, 10:41 PM #6 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 Really appreciate for your work, Jeff. It seems that it is more convenient for me to understand the steps by showing some counter-examples to disprove the assumption rather than by showing in the form of algebraic expressions. Really thanks for your time And also jack for your suggestion. Last edited by justusphung; November 2nd, 2018 at 11:04 PM. November 3rd, 2018, 07:14 AM #7 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 p(x)=(x-a)^n+(x+a)^n=0 If n even, impossible. If n odd (a-x)^n=(a+x)^n $\displaystyle \rightarrow$ a-x=a+x $\displaystyle \rightarrow$ x=-x x=0 is only real root Tags availability, double, polynomial, proving, question, real, root Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post wrightarya Algebra 1 July 18th, 2014 04:42 AM honzik Real Analysis 4 June 2nd, 2014 01:16 PM wuzhe Algebra 8 October 14th, 2012 06:43 AM oasi Calculus 1 March 15th, 2012 04:41 AM johnny Calculus 4 June 14th, 2011 10:29 PM

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