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Math Help - Root of unity sum

  1. #1
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    Root of unity sum

    Let X be one of the nth roots of unity.

    Find the sum: 1 + 2x + 3x^2 + ... + nx^{n-1}


    I've been trying to figure this out but nothing...
    I know that sum of roots of unity is 0. But I have no clue about this one.
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  2. #2
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    Re: Root of unity sum

    Let f(x)=x+x^2+...+x^n
    This is geometric progression with first member x and q=x.
    Therefore, its sum is f(x)=\frac {x(x^n-1)}{x-1}
    We obtained equality: x+x^2+...+x^n=\frac {x(x^n-1)}{x-1}
    Differentiating with respect to x yields:
    1+2x+...+nx^{n-1}=\frac {nx^{n+1}-(n+1)x^n+1}{(x-1)^2}
    Since x is n-th root of unity then x^n=1 and we can write above equality as
    1+2x+...+nx^{n-1}=\frac {nx^{n+1}-(n+1)*1+1}{(x-1)^2}
    Or
    1+2x+...+nx^{n-1}=n \frac {x^{n+1}-1}{(x-1)^2}
    Again, since x^n=1 then n \frac {x^{n+1}-1}{(x-1)^2}=n \frac {x x^n-1}{(x-1)^2}=n \frac {x-1}{(x-1)^2}=\frac {n}{x-1}
    Finally,
    1+2x+...+nx^{n-1}=\frac {n}{x-1}

    As can be seen from formula x=1 can't be applied to the formula.
    When x=1 1+2x+...+nx^{n-1}=1+2+3+...+n=\frac{n(n+1)}{2}

    Therefore, if x=1 1+2x+...+nx^{n-1}=\frac{n(n+1)}{2}, otherwise 1+2x+...+nx^{n-1}=\frac {n}{x-1}
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  3. #3
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    Re: Root of unity sum

    Hello, gregx22!

    \text{Let }x\text{ be one of the }n^{th}\text{ roots of unity.}

    \text{Find the sum: }\:S \;=\;1 + 2x + 3x^2 + \hdots  + nx^{n-1}

    \begin{array}{ccccccc}\text{We have:} & S \;=\; 1 + 2x + 3x^2 + 4x^3 + \hdots + nx^{n-1}  \qquad\qquad\quad \\ \text{Multiply by }x\!: & xS \;=\; \quad\; x + 2x^2 + 3x^3 + \hdots + (n-1)x^{n-1} + nx^n  \end{array}

    \text{Subtract: }\:S - xS \;=\;\underbrace{1 + x + x^2 + x^3 + \hdots + x^{n-1}}_{\text{geometric series}} - nx^n

    \text{The geometric series has: first term }a = 1\text{, common ratio }r = x\text{, and }n\text{ terms.}
    . . \text{Its sum is: }\:\frac{1 - x^n}{1-x}
    \text{But }x\text{ is an }n^{th}\text{ root of unity. }\;\text{Hence, }1 - x^n \:=\:0

    \text{So we have: }\:S(1-x) \;=\;-nx^n \quad\Rightarrow\quad S \;=\;\frac{nx^n}{x-1} \;=\;\frac{n}{x-1}
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