May 18th, 2019 by Aziz Lokhandwala
Let α1,α2 ,...,αn be the roots of the equation: $$f(x) = A(n)X(n) + A(n-1)X(n-1) + ... + A(1) x + A(0) = 0$$ We want to find: $$S(k) = {α1}^k +{α2}^k +...+{αn}^k$$
Here, k = 1,2,3,4,5,6,...
Now, as α1,α2 ,...,αn are roots of equation, thus we can write: $$f(x) = An(x - α1)(x - α2)...(x - αn)$$ $$ln |f(x)| = ln|An| + ln | x – α1 | + ln | x – α2 | + ... + ln | x – αn |$$ Now, differentiating above equation we have, $${f'(x) \over f(x)} = {1\over x – α1} + {1\over x – α2}+... + {1\over x – αn}$$ Now, corelating above equation with term, $${1 \over (1-x)} = 1+x+x^2 +x^3 +...$$ $${f'(x) \over f(x)} = n({1 \over x}) + (α1+α2+...+αn)({1 \over x^2}) + ({α1^2}+{α2^2}+...+{αn^2})({1 \over x^3}) + ...$$
Result: Coefficient of \(1 \over x^{k+1}\) in expansion of \(f'(x) \over f(x)\) is \(S(k) = {α1}^k +{α2}^k +...+{αn}^k \)
Question for you: Let, $$f(x) = (x-1)^3$$ Find S(1), S(2), S(3), S(4), S(5),S(6)
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