let c be a fixed real number. show that a root of the equation x(x+1)(x+2).....(x+2009)=c can

ankit jhunjhunwala

let c be a fixed real number. show that a root of the equation x(x+1)(x+2).....(x+2009)=ccan have multiplicity at most 2.Determine the number of values of c for which the equation has a root of multiplicity 2.

Deepak Aggarwal

If an equation f(x) = c has a root x = a of multiplicity 2 then f'(a) = 0. If the root has multiplicity greater than 2, then also f"(a) = 0.

As, x(x + 1)(x + 2) .............. (x + 2009) = c

Taking log on both sides,

$sum_{n=0}^{2009}ln(x+n) = ln c$

Let $f(x) = sum_{n=0}^{2009}ln(x+n)$ .

or, $f'(x) = sum_{n=0}^{2009} rac1{x+n}$

Also, $f''(x) = -sum_{n=0}^{2009} rac1{(x+n)^2}$

Since all the terms in the sum have the same sign, it's clear that their sum cannot be zero.

So $f(x) = ln c$ has no roots of multiplicity greater than 2.

To find the number of roots of multiplicity 2, note that the derivative of a function has a root between each pair of consecutive roots of the function. Since x(x+1)(x+2)...(x+2009) = 0 has 2010 roots, its derivative will have 2009 roots.

Hari Shankar

Umm, that reasoning is unfortunately not correct. The defined function is not continuous at the roots of the given equation. So differentiability even once is out of the question!

Solution:

Consider

and

Both f and g are of degree 2010

Its obvious that

Now if f(x) has a root of multiplicity greater than 2, then f'(x) has a root of multiplicity 2. In other words f'(x) has at most 2008 real roots.

But we know that g(x) has 2010 real and distinct roots and so g'(x), by Rolle's Theorem has 2009 real and distinct roots.

But and LHS has at most 2008 roots while g'(x) has exactly 2009 roots and we have a contradiction!

Hari Shankar

The answer to the second part is also fairly simple.

Let be the roots of f'(x).

Then if you let for , then for any of these ,

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