# How to determine the number of real solutions of th equation f(x)=a, f(x)=1/(x-1) + 1/(x-2)+....+1/(x-2009)?

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### 1 Answer

a=1/(x-1)+1/(x-2)+..1(x-2009)

Multiply by the LCM both sides and rewrite as:

P(x)=a(x-1)(x-2)(x-b)... (x-2009)

- (x-2)(x-3)--(x-2009)

-(x-1)(x-3)--(x-2009)

-(x-1)(x-2)(x-4)----(x-2009)

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-(x-1(x-2)(x-3)---(2-2008)

Put x=1, x=2,x=3 suceesively and see that

p(1) =0 -(1-2)(1-3)+..+(1-2009)+0+0++0 =-(2008 -ve factors)=**-ve.**

p(2)=0-(2-1)((2-3)(2-4)+...+(2-2009)+0+..+0=-(one positive)(2007 -ve factor) = **+ve,**

Thus we can see that P(3) **negative** and P(4)** +ve** , P(5) **-ve** and so on and P(n) and P(n+1) have different signs.

This indicates there is real roots between 1 and 2, between 2and 3, between 3 and 4 and so on and between 2007 and 2008.

So there are 2007 real solutions and the other one has to be real as imaginary roots should occur in pairs for any polynomial with rational coefficients.