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Math Help - pls help

  1. #1
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    pls help

    S_1 = \sum_{k = 1}^n k cos k

     S_2 = \sum_{k = 1}^n k sin k

     S_1, S_2 = ? using: S_1 + iS_2
    pls help
    thanx
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  2. #2
    Newbie hurrem's Avatar
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    Quote Originally Posted by ely_en View Post
    S_1 = \sum_{k = 1}^n k cos k

     S_2 = \sum_{k = 1}^n k sin k

     S_1, S_2 = ? using: S_1 + iS_2
    pls help
    thanx
    What I understand from S1 is that is is S_1 = \sum_{k = 1}^n cos k = cos1+cos2+cos3+cos4+ .... +cos(n-1)+cosn which is kx{[cos(n)]x[cos(n+1)]/2}

    and the same thing for S2 (Im assuming) and then adding up as in x+iy form.

    Hope that helped...
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  3. #3
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    S_1 = \sum_{k = 1}^n  k*cos  k, not  \sum_{k = 1}^n cos k
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  4. #4
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    Hello, ely_en!

    I haven't solved it yet . . . but I ran across something interesting.


    S_1 \:= \:\sum_{k=1}^n k\cos k

     S_2 \:=\: \sum_{k=1}^n k\sin k

    Find S_1,\:S_2 . . . using: S_1 + iS_2

    . . . S_1 \;=\;1\cos1 + 2\cos2 + 3\cos3 + 4\cos4 + \hdots
    . . . iS_2 \:=\:i\sin1 + 2i\sin2 + 3i\sin3 + 4i\sin4 + \hdots

    Add: . S_1+iS_2 \;=\;(\cos1+i\sin1) +2(\cos2 + i\sin2) + 3(\cos3 + i\sin3) + 4(\cos4+i\sin4) + \hdots


    And we have: . . . S_1+iS_2 \;=\;e^i + 2e^{2i} + 3e^{3i} + 4e^{4i} + \hdots\;\;{\color{blue}[1]}

    Multiply by e^i\!:\;\;e^i(S_1 + iS_2) \;=\;\qquad \;e^{2i} + 2e^{3i} + 3e^{4i} + \hdots\;\;{\color{blue}[2]}


    \text{Subtract }{\color{blue}[1] \text{ - } [2]}\!:\;\;(1-e^i)(S_1+iS_2) \;=\;\underbrace{e^i+ e^{2i} + e^{3i} + e^{4i} + \hdots}_{\text{geometric series}}

    \text{Hence: }\;(1-e^i)(S_1+iS_2) \;=\;\frac{e^i}{1-e^i}

    \text{Therefore: }\;S_1+iS_2 \;=\;\frac{e^i}{(1-e^i)^2}


    A fascinating result . . .
    . . absolutely useless, but still fascinating.

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  5. #5
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    Wink

    thanx
    I've tried this too, but I used the moivre's formula and then the formula for the sum of a geometric progression.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Lemma: \sum_{n=1}^{N}nx^n=\frac{x^{N+1}\left(xN-N-1\right)+x}{(x-1)^2}

    \therefore\sum_{n=1}^N{n\cos(n)}


    =\text{Re}\bigg[\sum_{n=1}^N{n\left(e^i\right)^n}\bigg]


    =\text{Re}\bigg[\frac{\left(e^i\right)^{N+1}\left(e^iN-N-1\right)+e^i}{(e^i-1)^2}\bigg]

    =\frac{\left((\cos(1)-1)N-1\right)\cos\left(N\right)-\sin(1)N\sin\left(N\right)+1}{2\left(\cos(1)-1\right)}

    Do the same except its

    \sum{n\sin(n)}=\text{Im}\bigg[\sum{n\left(e^i\right)^n}\bigg]
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Soroban View Post

    \text{Subtract }{\color{blue}[1] \text{ - } [2]}\!:\;\;(1-e^i)(S_1+iS_2) \;=\;\underbrace{e^i+ e^{2i} + e^{3i} + e^{4i} + \hdots}_{\text{geometric series}}

    \text{Hence: }\;(1-e^i)(S_1+iS_2) \;=\;\frac{e^i}{1-e^i}
    Hmm, you seemed to apply the rule \sum_{n=1}^{{\color{red}{\infty}}}x^n=\frac{x}{1-x}


    But this series is clearly divergent as N\to\infty

    Besides...the OP did not ask for the infinite series.

    Hope I have made sense,

    Mathstud.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Since it doesnt appear the OP isn't going to post the answer to \sum_{n=1}^{N}n\sin(n) I will present it

    \begin{aligned}<br />
\sum_{n=1}^{N}n\sin(n)&=\text{Im}\bigg[\sum_{n=1}^{N}n\left(e^i\right)^n\bigg]\\<br />
&=\text{Im}\bigg[\frac{\left(e^i\right)^{N+2}N-\left(e^i\right)^{N+1}N-\left(e^i\right)^{N+1}+e^i}{\left(e^i-1\right)^2}\bigg]\\<br />
&=\frac{\sin(1)N\cos\left(N\right)+\cos(1)N\sin\le  ft(N\right)-N\sin\left(N\right)-\sin\left(N\right)}{2\left(\cos(1)-1\right)}<br />
\end{aligned}
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