Multiple root polynomial

Let $c$ be a fixed real number.Show that a root of the equation \[x(x+1)(x+2).....(x+2014) = c\] can have multiplicity at most 2.Also determine the number of values of $c$ for which the equation has a root of multiplicity $2$.

#Polynomials #Root(Equation) #CosinesGroup #Goldbach'sConjurersGroup #TorqueGroup

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

We can clearly assume $c\neq 0$ - otherwise, there is nothing to prove, since all the roots are simple. If we suppose that $x$ is a root of $f(x)=x(x+1)\cdot\ldots\cdot(x+2014)-c$ with multiplicity three or more, we have that $f(x)=0,f'(x)=0$ and $f''(x)=0$ . Since: $f'(x) = \left(\frac{1}{x}+\frac{1}{x+1}+\ldots+\frac{1}{x+2014}\right)\cdot x(x+1)\cdot\ldots(x+2014),$ $f''(x) = -\left(\frac{1}{x^2}+\frac{1}{(x+1)^2}+\ldots+\frac{1}{(x+2014)^2}\right)\cdot x(x+1)\cdot\ldots(x+2014)+\left(\frac{1}{x}+\frac{1}{x+1}+\ldots+\frac{1}{x+2014}\right)^2\cdot x(x+1)\cdot\ldots(x+2014),$ the three conditions give: $\sum_{j=0}^{2014}\frac{1}{(x+j)^2} = 0,$ that is impossible since all the terms are positive numbers.

The possible values of $c$ for which a double root exists are given by $y(y+1)\cdot\ldots\cdot(y+2014)$ where $y$ is a root of $\sum_{j=0}^{2014}\frac{1}{y+j}=0.$

Jack D'Aurizio - 7 years, 1 month ago

It will take a lots of time to type the solution so I am writing the keypoints In order to have a multiplicity of 3 or more than that, the second derivative of the polynomial P(x) = x(x+1)....(x+2014) - c should be equal to zero. We can find the second derivative and apply Cauchy schwarz to know that it cannot be equal to zero.

Siddharth Kumar - 7 years, 1 month ago

Let the left side be $f(x)$ . First note that $c = 0$ doesn't work, as all the roots have multiplicity 1. Next, look at $\frac{f'(x)}{f(x)} = \frac1{x} + \frac1{x+1} + \cdots + \frac1{x+2014}$ . The right side is a decreasing function with $2015$ vertical asymptotes; considering the graph, you can see that it crosses the $x$ -axis once in every interval $(-N-1,-N)$ , where $N$ is a nonnegative integer $\le 2013$ . Since $f'(x)$ is a polynomial of degree $2014$ , this shows that it has $2014$ distinct roots.

Since $f'(x)$ has no roots of multiplicity $\ge 2$ , $f(x)$ cannot have any roots of multiplicity $\ge 3$ . So we're done with the first half.

This also shows that there are $2014$ values $a$ such that $f(x) = f(a)$ has a root $a$ of multiplicity $2$ . This might lead to fewer than $2014$ values of $c$ , though. I don't think it does (but only because $2014$ is even!). But my reasoning is probably better confined to a second post, mostly because I haven't fully worked it out yet.

Patrick Corn - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$