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Geometry Level 5

Evaluate − lim ⁡ n → ∞ ∑ m = 1 n ∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) . -\lim_{n\to\infty}\sum_{m=1}^{n}\prod_{k=1}^{m}\cos\left(\frac{2\pi{k}}{2m+1}\right) .


The answer is 0.6.

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Otto Bretscher by Otto Bretscher , 65, USA

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1 solution

Isaac Buckley
Sep 2, 2015

This question is beautiful.

First thing we do is notice: ( z − e i θ ) ( z − e − i θ ) = z 2 − 2 z cos ⁡ ( θ ) + 1 (z-e^{iθ})(z-e^{-iθ})=z^2-2z\cos(θ)+1

Therefore we can see: z 2 m + 1 − 1 = ( z − 1 ) ∏ k = 1 m ( z 2 − 2 z cos ⁡ ( 2 π k 2 m + 1 ) + 1 ) z^{2m+1}-1=(z-1)\prod\limits_{k=1}^{m}\left(z^2-2z\cos\left(\frac{2πk}{2m+1}\right)+1\right)

If we substitute z = i z=i we can see a wonderful cancelation and rearrange to get:

∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) = i 2 m + 1 − 1 ( i − 1 ) ( − 2 i ) m \prod\limits_{k=1}^{m}\cos\left(\frac{2πk}{2m+1}\right)=\frac{i^{2m+1}-1}{(i-1)(-2i)^m}

So when m = 2 r − 1 m=2r-1 we have: ∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) = ( − 1 ) r 1 2 2 r − 1 \prod\limits_{k=1}^{m}\cos\left(\frac{2πk}{2m+1}\right)=(-1)^r\frac{1}{2^{2r-1}}

When m = 2 r m=2r we have: ∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) = ( − 1 ) r 1 2 2 r \prod\limits_{k=1}^{m}\cos\left(\frac{2πk}{2m+1}\right)=(-1)^r\frac{1}{2^{2r}}

A better way to write this would be: ∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) = ( − 1 ) m ( m + 1 ) / 2 1 2 m \prod_{k=1}^{m}\cos\left(\frac{2\pi{k}}{2m+1}\right)=(-1)^{m(m+1)/2}\frac{1}{2^m}

Now we can write out the sum:

S = − 1 2 − 1 4 + 1 8 + 1 16 − . . . S=-\frac{1}{2}-\frac{1}{4}+\frac{1}{8}+\frac{1}{16}-...

When the signs change we can cancel to get:

S = − 1 2 − 1 8 + 1 32 − 1 128 + . . . S=-\frac{1}{2}-\frac{1}{8}+\frac{1}{32}-\frac{1}{128}+...

It makes a geometric series which we sum up:

S = − 1 2 − 1 8 1 1 − − 1 4 = − 0.6 S=-\frac{1}{2}-\frac{1}{8} \frac{1}{1--\frac{1}{4}}=-0.6

∴ − lim ⁡ n → ∞ ∑ m = 1 n ∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) = 0.6 \large \therefore -\lim_{n\to\infty}\sum_{m=1}^{n}\prod_{k=1}^{m}\cos\left(\frac{2\pi{k}}{2m+1}\right)=\boxed{0.6}

12 Helpful 0 Interesting 0 Brilliant 0 Confused

Another beautiful problem by Otto sir and a beautiful solution too.

Knowing the identity -

∏ k = 1 2 m cos ⁡ ( 2 π k 2 m + 1 ) = 1 4 m \displaystyle \prod_{k=1}^{2m}{\cos\left(\frac{2\pi k}{2m+1}\right)} = \frac{1}{4^m}

(which can be proved much easily using roots of unity method)

Now, observe that the terms after n n th one(in short, the terms we don't want for this problem)

cos ⁡ ( 2 ( m + 1 ) π 2 m + 1 ) , cos ⁡ ( 2 ( m + 2 ) π 2 n + 1 ) , ⋯ , cos ⁡ ( 2 ( 2 m ) π 2 n + 1 ) \displaystyle \cos\left(\frac{2(m+1)\pi}{2m+1}\right), \cos\left(\frac{2(m+2)\pi}{2n+1}\right), \cdots, \cos\left(\frac{2(2m)\pi}{2n+1}\right)

We can write these as -

cos ⁡ ( 2 π − 2 m π 2 n + 1 ) , cos ⁡ ( 2 π − 2 ( m − 1 ) π 2 n + 1 ) , ⋯ , cos ⁡ ( 2 π − 2 π 2 m + 1 ) \displaystyle \cos\left(2\pi - \frac{2m\pi}{2n+1}\right), \cos\left(2\pi - \frac{2(m-1)\pi}{2n+1}\right), \cdots, \cos\left(2\pi - \frac{2\pi}{2m+1}\right)

This series is just the reverse of the series of the terms we have not considered now( cos ⁡ ( 2 π − x ) = cos ⁡ ( x ) \cos(2\pi -x) = \cos(x) ). Hence,

∏ k = 1 2 m cos ⁡ ( 2 π k 2 m + 1 ) = ∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) 2 = 1 4 m \displaystyle \prod_{k=1}^{2m}{\cos\left(\frac{2\pi k}{2m+1}\right)} = {\prod_{k=1}^{m}{\cos\left(\frac{2\pi k}{2m+1}\right)}}^{2} = \frac{1}{4^m}

∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) = ± 1 2 m \displaystyle \prod_{k=1}^{m}{\cos\left(\frac{2\pi k}{2m+1}\right)} = \pm \frac{1}{2^m}

Now, we can check the positive and negative case by observing few terms(we've to just check in which quadrant cos ⁡ \cos lies and then deduce whether the whole product is > 0 >0 or < 0 <0 . Observation gives Positive ∀ ( m = 4 i + 3 , 4 i + 4 ) ; Negative ∀ ( m = 4 i + 1 , 4 i + 2 ) ; ∀ i ∈ Z + + 0 \text{Positive} \forall (m = 4i + 3, 4i + 4) ; \text{Negative} \forall (m = 4i + 1, 4i + 2) ; \forall i \in {Z}^{+} + {0}

Kartik Sharma - 5 years, 9 months ago

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@Otto Bretscher Is this approach correct?

Kartik Sharma - 5 years, 9 months ago

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Yes, exactly, that's essentially how I did I (upvote). It boils down to saying that ∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) = ( − 1 ) q 1 2 m , \prod_{k=1}^{m}\cos\left(\frac{2\pi{k}}{2m+1}\right)=(-1)^q\frac{1}{2^m}, where q q is the number of angles 2 π k 2 m + 1 \frac{2\pi{k}}{2m+1} in the second quadrant. A straightforward count shows that q = ⌊ m + 1 2 ⌋ q=\lfloor{\frac{m+1}{2}}\rfloor so that ( − 1 ) q = ( − 1 ) m ( m + 1 ) / 2 (-1)^q=(-1)^{m(m+1)/2} , leading to the formula I stated, ∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) = ( − 1 ) m ( m + 1 ) / 2 1 2 m \prod_{k=1}^{m}\cos\left(\frac{2\pi{k}}{2m+1}\right)=(-1)^{m(m+1)/2}\frac{1}{2^m} .

Alternatively, you can use Isaac's formula, ( − 1 ) q = i 2 m + 1 − 1 ( i − 1 ) ( − i ) m , (-1)^q=\frac{i^{2m+1}-1}{(i-1)(-i)^m},

PS: All the n ′ n' s in your post should be m ′ m' s.

Otto Bretscher - 5 years, 9 months ago

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@Otto Bretscher – Oops, sorry. Edited! Nicely done!

Kartik Sharma - 5 years, 9 months ago

Beautiful solution (upvote)! Thanks! You can write the formula ∏ k = 1 m cos ⁡ ( 2 π k 2 m + 1 ) = ( − 1 ) m ( m + 1 ) / 2 1 2 m \prod_{k=1}^{m}\cos\left(\frac{2\pi{k}}{2m+1}\right)=(-1)^{m(m+1)/2}\frac{1}{2^m} that works for even as well as odd m m .

Otto Bretscher - 5 years, 9 months ago

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So much more elegant. I'll add it into the solution.

I'm also using similar techniques to get a few other cool results. I think I'll post a similar problem tomorrow for my 100 day streak.

Thanks for inspiring me yet again. I've missed your problems.

Isaac Buckley - 5 years, 9 months ago

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Maybe you can apply your powerful approach to a closely related problem that I have posted earlier: If w 1 , . . , w n w_1,..,w_n are the n n th roots of unity, for odd n n , find ∏ i < j ( w i + w j ) \prod_{i<j}(w_i+w_j) . Give a closed-form solution.

Otto Bretscher - 5 years, 9 months ago

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