62from the symmetry of the coefficients, you can say that if xi is a root then 1/xi is also a root
the expression given by you is hence equal to \frac{1}{x_1 -1}+\frac{1}{x_2 -1}+\frac{1}{x_3 -1}...+\frac{1}{x_n -1}=\frac{x_1}{1-x_1}+\frac{x_2}{1-x_2}+\frac{x_3}{1-x_3}...+\frac{x_n}{1-x_n}
Hence you can say that \\2S=\frac{1}{x_1 -1}+\frac{1}{x_2 -1}+\frac{1}{x_3 -1}...+\frac{1}{x_n -1}+\frac{x_1}{1-x_1}+\frac{x_2}{1-x_2}+\frac{x_3}{1-x_3}...+\frac{x_n}{1-x_n} \\2S=\frac{1-x_1}{x_1 -1}+\frac{1-x_2}{x_2 -1}+...+\frac{1-x_n}{x_n-1}=-n \\\\S=-n/2
Have to still see the relevance of the second part to your question !!
3im gettin the second exp to be zero ...[dat cot thing] can u confirm sir?????
341Nishant Sir's method is beautiful. But, another way to arrive at this answer may prove useful.
If x1, x2,..., xn are the n roots of P(x) = 0, then
\frac{P'(x)}{P(x)} = \frac{1}{x-x_1} + \frac{1}{x-x_2} + ...+ \frac{1}{x-x_n}
Here by putting x =1, we obtain \frac{1}{1-x_1} + ...+\frac{1}{1-x_n} = \frac{P'(1)}{P(1)} = \frac{n(n+1)}{2(n+1)} = \frac{n}{2}
Now, we must remember that the xi are nothing but the complex roots of zn+1=1, which are given by z_k = \cos \frac{2k \pi}{n+1} + i \sin \frac{2k \pi}{n+1}; 1 \le k \le n
Substituting and simplifying we get LHS = \frac{n}{2} - i \sum_{k=1}^{n} \cot \frac{k \pi}{n+1} which immediately leads to \sum_{k=1}^{n} \cot \frac{k \pi}{n+1}=0
62
Lokesh Verma
·2009-09-27 03:07:21
A slight comment on what prophet sir has done for basic users...
How to see that the expression given is actually f'(x)/f(x)
see the expression here is the derivative of Σln(xi-x)
Hence the derivative if ln(f(x)
hence f'(x)/f(x)
I know that for some of you this might be very obvious.. but for the others keep this in mind too :)
