Funny sum
For ω = e 1 5 2 π i find:
k = 1 ∑ 1 4 ( ω k − 1 ) 3 1
The answer is 21.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
Stretch of my solution. I will find the sum: S ( n ) = k = 1 ∑ 2 n ( w k − 1 ) 3 1 when w = e 2 n + 1 2 π i and our problem comes putting n = 7 .
Note that w k = w k are the ( 2 n + 1 ) t h complex roots of unity. Consider for x = 1 P ( x ) = k = 0 ∑ 2 n x k = x − 1 x 2 n + 1 − 1 Now put x = y y + 1 (with y = 0 ) and our polynomial P ( x ( y ) ) becomes after same simple simplification: P ( y ) = y 2 n 1 j = 0 ∑ 2 n ( j 2 n + 1 ) y j Note that the root of P ( y ) are exactly y k = w k − 1 1 and so our sum S ( n ) is just S ( n ) = k = 1 ∑ 2 n y k 3 . Now using Vieta or Newton formulae it comes that the sum is equal to S = 2 n ( n − 1 ) So S ( 7 ) = 2 1 .
There is no need to restrict attention to 2 n + 1 to do this. For any integer n ≥ 2 we note that f ( X ) = j = 0 ∑ n − 1 X j has roots ω j = e n 2 π i j 1 ≤ j ≤ n − 1 and that ( X − 1 ) f ( X ) = X n − 1 so that X f ( X + 1 ) = ( X + 1 ) n − 1 = j = 1 ∑ n ( j n ) X j and hence g ( X ) = f ( X + 1 ) = j = 0 ∑ n − 1 ( j + 1 n ) X j has zeros ω j − 1 for 1 ≤ j ≤ n − 1 . Thus h ( X ) = X n − 1 g ( X − 1 ) = j = 0 ∑ n − 1 ( j n ) X n − 1 − j = j = 0 ∑ n − 1 ( j n ) X j has zeros ( ω j − 1 ) − 1 for 1 ≤ j ≤ n − 1 Using Vieta's formulae we obtain e 1 = − 2 1 ( n − 1 ) e 2 = 6 1 ( n − 1 ) ( n − 2 ) e 3 = − 2 4 1 ( n − 1 ) ( n − 2 ) ( n − 3 ) and hence j = 1 ∑ n − 1 ( ω j − 1 ) − 3 = e 1 3 − 3 e 1 e 2 + 3 e 3 = 8 1 ( n − 1 ) ( n − 3 ) Obviously, replacing n by 2 n + 1 retrieves your formula.
please use Euler's identity
\(\begin{align} f(x)&= x^{14} + x^{13} + x^{12} + ... + x + 1 \\ & w^k \text{ Roots of } f(x)\\ f(x)&= \prod_{k=1}^{14}(x-w^k)\\ \ln\left (f(x) \right )&= \sum_{k=1}^{14}\ln(x-w^k)\\ \frac{d^3}{dx^3}\ln\left (f(x) \right )&= \frac{d^3}{dx^3} \sum_{k=1}^{14}\ln(x-w^k)\\ \frac{f'''(x)}{f(x)} -3\frac{f'(x) \cdot f''(x)}{f^2(x)} + 2 \left ( \frac{f'(x)}{f(x)} \right )^3 &= \sum_{k=1}^{14}\frac{2}{(x-w^k)^3}\\
\frac{f'''(1)}{f(1)} -3\frac{f'(1) \cdot f''(1)}{f^2(1)} + 2 \left ( \frac{f'(1)}{f(1)} \right )^3 &= \sum_{k=1}^{14}\frac{2}{(1-w^k)^3}\\
-42 &= \sum_{k=1}^{14}\frac{-2}{(w^k - 1)^3}\\ &\Rightarrow \sum_{k=1}^{14}\frac{1}{(w^k - 1)^3}= 21 \end{align} \\ \color{red} \begin{matrix} f(x)=\sum_{k=0}^{14} x^k & | & f'(x)=\sum_{k=1}^{14} k\cdot x^{k-1} & | & f''(x)=\sum_{k=2}^{14} k(k-1)\cdot x^{k-2} & | & f''(x)=\sum_{k=3}^{14} k(k-1)(k-2)\cdot x^{k-3} \\ \\ f(1)=\sum_{k=0}^{14} 1 = 15 & | & f'(1)=\sum_{k=1}^{14} k = 105 & | & f''(1)=\sum_{k=2}^{14} k(k-1) = 910 & | & f''(1)=\sum_{k=3}^{14} k(k-1)(k-2) = 8190 \end{matrix}\)