21the answer is 2n(n+1)?
Well, you can remove one sigma here...
Proof:
Help please!
\sum_{1\leq i\leq j\leq n}^{}{}\sum{\frac{9}{(2-\alpha _{i})(2-\alpha_{j})}}
Where αi , αj are the roots of unity.
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21
341Vivek can you clarify whether you mean say for n=3,
\frac{1}{(2-1)(2-\omega)}+\frac{1}{(2-1)(2-\omega^2)}+\frac{1}{(2-\omega)(2-\omega^2)}
i mean in the summation is it i≤j or as i suspect i<j ?
21srry my above post is wrong
ignore
I have another proof, w/o calculus
but i feel i≤j is ok...
can't we separate i<j and i=j
and when i= j
it is simply Σ(3/(2-a_i))^2 , which is = (Σ 3/(2-a_i))^2 - 2Σ(1≤i<j≤n)9(2-a_i)(2-a_j) ??
21\sum_{1\leq i<j\leq n}^{{}}\frac{1}{(2-a_i)(2-a_j)}
= \left [ x^{n-2} \right ]\prod_{i=1}^{n}\left ( x-\frac{1}{(2-a_i)} \right )
= \left [ x^{n-2} \right ]\; \; x^n\prod_{i=1}^{n}\left ( 1-\frac{1}{x(2-a_i)}\right )
= \left [ x^{n-2} \right ]\; \; \frac{x^n}{\prod_{i=1}^{n}(2-a_i)}\prod_{i=1}^{n}\left ( 2-a_i - \frac{1}{x} \right )
= \left [ x^{n-2} \right ]\; \; \frac{x^n}{2^n-1}\prod_{i=1}^{n}\left ( \left ( 2-\frac 1 x \right ) -a_i\right )
=\left [ x^{n-2} \right ]\; \; \frac{x^n}{2^n-1}\left ( \left ( 2-\frac 1 x \right )^n-1 \right )
= \left [ x^{n-2} \right ]\; \; \frac{\left ( 2x-1 \right )^n - x^n}{2^n-1}
= \frac{(2^{n-2})\cdot\binom{n}{2}}{2^n-1}
71Sorry, I don't know much. Can you be abit more clear on what you did!
21[x^(n-2)] f(x) means the coefficient of x^{n-2} in f(x)
sorry
first step u may see as vietas relations...
or u can realize that the summation sought is 9C(n,2) f''(x)/f(x)
341If r_1, r_2,..,r_n are the n roots of a nth degree polynomial f(x), then The expression
\sum_{1\le i<j \le n} \frac{1}{(x-r_i)(x-r_j)}
is equivalent to (when x is not one of the roots)
\frac{f"(x)}{2 f(x)}
You can easily prove this in atleast two ways and I suggest you try it out yourself.
In the given case, f(x)=x^n-1.
Hence
\sum_{1\le i<j \le n} \frac{1}{(x-r_i)(x-r_j)} = n(n-1)\frac{x^{n-2}}{2(x^n-1)}
Put x=2 and finish
341One way to prove it:
We assume WLOG that the leading coefficient is 1.
So f(x) = (x-r_1)(x-r_2)...(x-r_n)
Now if you differentiate, you will get
\frac{df(x)}{dx} = (x-r_2)...(x-r_n)+(x-r_1)(x-r_3)...(x-r_n)+...+(x-r_1)(x-r_2)...(x-r_{n-1})
I guess you already know that if you now divide by f(x), the expression becomes
\frac{1}{x-r_1}+\frac{1}{x-r_2}+...+\frac{1}{x-r_n}
Differentiate f'(x) again and divide by f(x) and you will get \frac{f"(x)}{f(x)}=2 \sum_{1\lei<j\len} \frac{1}{(x-r_i)(x-r_j)}