E an EasyMethod!
1 + 2 − α 1 1 + 2 − α 2 1 + 2 − α 3 1 + ⋯ + 2 − α n − 1 1 = ?
If 1 , α 1 , α 2 , … , α n − 1 are n t h roots of unity, find the value of the above expression, taking n = 9 .
The answer is 4.508806.
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3 solutions
WoW. Cool method. This was the method my Dad was referring to.
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This is the (conceptually) "easy" solution... every step is easy to justify...just algebra
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And that is exactly what I did to solve this problem. Would be my favorite approach.
Nice approach...
Given that 1 , α 1 , α 2 , ⋯ , α n − 1 are n t h roots of unity.
⇒ ( z − 1 ) ( z − α 1 ) ( z − α 2 ) ⋯ ( z − α n − 1 ) = z n − 1 Taking lo g ⇒ lo g ( z − 1 ) + lo g ( z − α 1 ) + lo g ( z − α 2 ) + ⋯ + lo g ( z − α n − 1 ) Differentiation w.r.t z ⇒ z − 1 1 + z − α 1 1 + z − α 2 1 + ⋯ + z − α n − 1 1 Putting x = 2 , n = 9 ⇒ 1 + 2 − α 1 1 + 2 − α 2 1 + 2 − α 3 1 + ⋯ + 2 − α n − 1 1 = lo g ( z n − 1 ) = z n − 1 n x n − 1 = 2 9 − 1 9 ⋅ 2 8
∴ 1 + 2 − α 1 1 + 2 − α 2 1 + 2 − α 3 1 + ⋯ + 2 − α n − 1 1 = 5 1 1 9 ⋅ 2 5 6 = 4 . 5 0 8 8 0 6
This method is perhaps not as "very easy" as it looks, although it does produce the right result: Complex logarithms and complex differentiation are subtle concepts. After "Taking log", you use the rule lo g ( a b ) = lo g ( a ) + lo g ( b ) ... but what does that mean for complex logarithms?
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It should be true,
Let a = r 1 e i θ 1 and b = r 2 e i θ 2 . Then
ln ( a b ) = ln r 1 + ln r 2 + i θ 1 + i θ 2 = ln ( r 1 e i θ 1 ) + ln ( r 2 e i θ 2 ) = ln a + ln b
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Since the complex exponential function is not one-to-one (we have e z = e z + 2 π i ) , we cannot simply define the complex logarithm as "the inverse of the exponential function", as we do in real analysis. One way to avoid this ambiguity is to pick a principal value of the logarithm, insisting that its imaginary part be on the interval ( − π , π ] , for example. But then the rule lo g ( z w ) = lo g z + lo g w no longer holds.
I agree that I should prove that it's differentiable (which, here, I assumed)
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To fully understand your solution, one would have to delve deeply into complex analysis: What is complex differentiation? Is lo g ( z ) differentiable, and what is its derivative? Which rules of real analysis generalize to the complex case, and why (you are freely using the sum rule, the chain rule, and the power rule)?
Done by the same method.
From fundamentals,
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Let z = 2 − α 1 . Since α n = 1 , we have ( z 2 z − 1 ) n = 1 or ( 2 z − 1 ) n = z n or ( 2 n − 1 ) z n − n 2 n − 1 z n − 1 . . . = 0 . By Viete, the sum we seek is 2 n − 1 n 2 n − 1 , or 2 9 − 1 9 ∗ 2 8 ≈ 4 . 5 0 9 for n = 9 .