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 \(2\).

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 \(\sum_{j=0}^{2014}\frac{1}{y+j}=0.\)

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.

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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWe 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 \(\sum_{j=0}^{2014}\frac{1}{y+j}=0.\)

Log in to reply

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.

Log in to reply

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.

Log in to reply