How do you find the sum $$S(\theta)=\displaystyle\sum_{r=1}^{n} r^2\cos {(r\theta)}$$?
I observed that if $f(\theta)$ is $\sum\cos {(r\theta)}$, then $S(\theta)=-f''(\theta)$.
Help will be appreciated.
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http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression – lab bhattacharjee Jul 19 '14 at 15:01
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do you need an exact value? It looks it can be easily approximated with an integral. – Alex Jul 19 '14 at 15:07
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@Alex What integral? – user1001001 Jul 19 '14 at 15:09
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$\int_{1}^{n}x^2 \cos (x \theta) dx$. Partial integration twice with $u= x^2, dv = \cos (x \theta) $ – Alex Jul 19 '14 at 15:11
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$\cos x$ is the real part of $e^{ix}$. So $\sum_{k=1}^n\cos(k\theta)$ is the real part of $\sum_{k=1}^n \left(e^{i\theta}\right)^k$. Letting $z=e^{i\theta}$, this is:
$$z\frac{z^n-1}{z-1} = \frac{z^{n+1}-z}{z-1}\frac{z^{-1} - 1}{z^{-1}-1} = \frac{z^{n}-1-z^{n+1}+z}{2-z-z^{-1}}$$
Now $z^{-1}=\cos\theta -i\sin\theta = \bar z$, and $2-z-z^{-1}=2-2\cos\theta$. So the real part is:
$$\frac{\cos n\theta -\cos (n+1)\theta +\cos\theta-1 }{2-2\cos\theta}=-\frac12 +\frac{\cos n\theta-\cos(n+1)\theta}{2-2\cos\theta}$$
So that is the value of $\sum_{k=1}^n \cos k\theta$, and you'll have to take the double derivative to get $-\sum k^2\cos k\theta$. That's gonna be ugly.
Thomas Andrews
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I think some trig identities will simplify the answer a bit. Wolfram|alpha should handle it nicely. If I recall correctly it can be expressed without the sums in the denominator and numerator. – zibadawa timmy Jul 19 '14 at 15:25
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1Yes, $$\cos x - \cos y = 2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{y-x}{2}\right)$$ and $$2-2\cos\theta =4\sin^2\frac{\theta}{2}$$ – Thomas Andrews Jul 19 '14 at 15:29
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Wolfram says: http://www4b.wolframalpha.com/Calculate/MSP/MSP1960221b4a4d8913b95e00000hh4d00c56f4aa73?MSPStoreType=image/gif&s=34&w=409.&h=49. – Jul 19 '14 at 15:35
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Let us find $\displaystyle S=\sum_{r=1}^nr^2z^r$ where $z=e^{i\theta}$
We need to find the real part of $S$
$z\cdot S=n^2z^{n+1}+\sum_{r=2}^n(r-1)^2z^r$
$\displaystyle\implies(z-1)S=n^2z^{n+1}-\sum_{r=1}^n(2r-1)z^r$
Let $\displaystyle T=\sum_{r=1}^n(2r-1)z^r$
$\displaystyle\implies z\cdot T=\sum_{r=1}^n(2r-1)z^{r+1}=(2n+1)z^{n+1}+\sum_{r=2}^n(2r-3)z^r$
$\displaystyle\implies (z-1)T=(2n+1)z^{n+1}-2\sum_{r=2}^nz^r,$ the last part is clearly a finite Geometric Series
Can you take it home from here?
lab bhattacharjee
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