More fun in 2015, Part 21

Geometry Level 4

Find ∏ k = 0 2014 2 sin ⁡ ( π 186 + k π 2015 ) \prod_{k=0}^{2014}2\sin\left(\frac{\pi}{186}+\frac{k\pi}{2015}\right)

Bonus question : More generally, what is ∏ k = 0 n − 1 2 sin ⁡ ( a + k π n ) \displaystyle\prod_{k=0}^{n-1}2\sin\left(a+\frac{k\pi}{n}\right) ?


The answer is 1.

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Otto Bretscher by Otto Bretscher , 65, USA

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2 solutions

Otto Bretscher
Oct 15, 2015

Let us prove that ∏ k = 0 n − 1 2 sin ⁡ ( a + k π n ) = 2 sin ⁡ ( n a ) \prod_{k=0}^{n-1}2\sin\left(a+\frac{k\pi}{n}\right)=2\sin(na)

We start with z n − 1 = ∏ k = 0 n − 1 ( z − e − 2 k π i / n ) z^n-1=\prod_{k=0}^{n-1}(z-e^{-2k\pi i/n}) We plug in z = e 2 a i z=e^{2ai} to find that e 2 n a i − 1 = ∏ k = 0 n − 1 ( e 2 a i − e − 2 k π i / n ) e^{2nai}-1=\prod_{k=0}^{n-1}(e^{2ai}-e^{-2k\pi i/n}) Now we divide both sides by e n a i e^{nai} and also by i i . On the right hand side, we will multiply numerator and denominator by ∏ k = 0 n − 1 e k π i / n = e ( n − 1 ) π i / 2 = i n − 1 \prod_{k=0}^{n-1}e^{k\pi i/n}=e^{(n-1)\pi i/2}=i^{n-1} e n a i − e − n a i i = ∏ k = 0 n − 1 1 i ( e a i e k π i / n − e − a i e − k π i / n ) , \frac{e^{nai}-e^{-nai}}{i}=\prod_{k=0}^{n-1}\frac{1}{i}(e^{ai}e^{k\pi i/n}-e^{-ai}e^{-k\pi i/n}), which is the equation we seek to prove.

5 Helpful 0 Interesting 1 Brilliant 0 Confused

Brilliant pro(o)f!

Kenny Lau - 5 years, 8 months ago

I can solve this by Chebyshev polynomials as well, but yours is clearly way much better and shorter! Thanks for the solution!

Pi Han Goh - 5 years, 4 months ago
Kenny Lau
Oct 15, 2015

∏ k = 0 2014 2 sin ⁡ ( π 186 + k π 2015 ) \displaystyle\quad\prod_{k=0}^{2014}2\sin\left(\frac\pi{186}+\frac{k\pi}{2015}\right)

= i 2015 ∏ k = 0 2014 ( exp ⁡ ( − i π 186 − k i π 2015 ) − exp ⁡ ( i π 186 + k i π 2015 ) ) =\displaystyle i^{2015}\prod_{k=0}^{2014}\left(\exp{\left(-\frac{i\pi}{186}-\frac{ki\pi}{2015}\right)}-\exp{\left(\frac{i\pi}{186}+\frac{ki\pi}{2015}\right)}\right)

= i 2015 exp ⁡ ( − 2015 i π 186 − ( 2014 × 2015 / 2 ) × i π 2015 ) ∏ k = 0 2014 ( 1 − exp ⁡ ( i π 93 + 2 k i π 2015 ) ) =\displaystyle i^{2015}\exp{\left(-\frac{2015i\pi}{186}-\frac{(2014\times2015/2)\times i\pi}{2015}\right)}\prod_{k=0}^{2014}\left(1-\exp{\left(\frac{i\pi}{93}+\frac{2ki\pi}{2015}\right)}\right)

= exp ⁡ ( − i π 3 ) ∏ k = 0 2014 ( 1 − exp ⁡ ( i π 93 + 2 k i π 2015 ) ) =\displaystyle \exp{\left(-\frac{i\pi}3\right)}\prod_{k=0}^{2014}\left(1-\exp{\left(\frac{i\pi}{93}+\frac{2ki\pi}{2015}\right)}\right)

z 2015 = 1 z^{2015}=1 : z = exp ⁡ 2 k i π 2015 z=\exp{\dfrac{2ki\pi}{2015}} , k k from 0 to 2014.

( exp ⁡ ( − i π 93 ) z ) 2015 = 1 \left(\exp{\left(-\dfrac{i\pi}{93}\right)}z\right)^{2015}=1 : z = exp ⁡ ( i π 93 + 2 k i π 2015 ) z=\exp{\left(\frac{i\pi}{93}+\frac{2ki\pi}{2015}\right)} , k k from 0 to 2014.

( exp ⁡ ( − i π 93 ) ( 1 − z ) ) 2015 = 1 \left(\exp{\left(-\dfrac{i\pi}{93}\right)}(1-z)\right)^{2015}=1 : z = 1 − exp ⁡ ( i π 93 + 2 k i π 2015 ) z=1-\exp{\left(\frac{i\pi}{93}+\frac{2ki\pi}{2015}\right)} , k k from 0 to 2014.

exp ⁡ ( i π 3 ) ( 1 − z ) 2015 = 1 \exp{\left(\dfrac{i\pi}3\right)}(1-z)^{2015}=1

Product of roots by Vieta's Formulas = exp ⁡ ( i π 3 ) − 1 exp ⁡ ( i π 3 ) = exp ⁡ ( i π 3 ) \displaystyle\frac{\exp{\left(\frac{i\pi}3\right)}-1}{\exp{\left(\frac{i\pi}3\right)}}=\exp{\left(\frac{i\pi}3\right)}

∏ k = 0 2014 2 sin ⁡ ( π 186 + k π 2015 ) \displaystyle\quad\prod_{k=0}^{2014}2\sin\left(\frac\pi{186}+\frac{k\pi}{2015}\right)

= exp ⁡ ( − i π 3 ) ∏ k = 0 2014 ( 1 − exp ⁡ ( i π 93 + 2 k i π 2015 ) ) =\displaystyle \exp{\left(-\frac{i\pi}3\right)}\prod_{k=0}^{2014}\left(1-\exp{\left(\frac{i\pi}{93}+\frac{2ki\pi}{2015}\right)}\right)

= exp ⁡ ( − i π 3 ) exp ⁡ ( i π 3 ) =\displaystyle \exp{\left(-\frac{i\pi}3\right)}\exp{\left(\frac{i\pi}3\right)}

= 1 =\displaystyle 1

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Looks good( upvote)! Thanks! (I will check the details later.) All the sums, ∑ \sum should be products, ∏ \prod .

For future reference, can you generalize: What is ∏ k = 0 n − 1 2 sin ⁡ ( a + k π n ) \prod_{k=0}^{n-1}2\sin\left(a+\frac{k\pi}{n}\right) ?

Otto Bretscher - 5 years, 8 months ago

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∏ k = 0 n − 1 2 sin ⁡ ( a + k π n ) \displaystyle\quad\prod_{k=0}^{n-1}2\sin\left(a+\frac{k\pi}n\right)

= i n ∏ k = 0 n − 1 ( exp ⁡ ( − a i − k i π n ) − exp ⁡ ( a i + k i π n ) ) =\displaystyle i^n\prod_{k=0}^{n-1}\left(\exp{\left(-ai-\frac{ki\pi}n\right)}-\exp{\left(ai+\frac{ki\pi}n\right)}\right)

= i n exp ⁡ ( − a n i − ( n − 1 ) i π 2 ) ∏ k = 0 n − 1 ( 1 − exp ⁡ ( 2 a i + 2 k i π n ) ) =\displaystyle i^n\exp{\left(-ani-\frac{(n-1)i\pi}{2}\right)}\prod_{k=0}^{n-1}\left(1-\exp{\left(2ai+\frac{2ki\pi}n\right)}\right)

= i n exp ⁡ ( − a n i − ( n − 1 ) i π 2 ) exp ⁡ ( 2 a i ) n ∏ k = 0 n − 1 ( exp ⁡ ( − 2 a i ) − exp ⁡ ( 2 k i π n ) ) =\displaystyle i^n\exp{\left(-ani-\frac{(n-1)i\pi}{2}\right)}\exp{\left(2ai\right)}^n\prod_{k=0}^{n-1}\left(\exp{\left(-2ai\right)}-\exp{\left(\frac{2ki\pi}n\right)}\right)

= i n exp ⁡ ( − a n i − ( n − 1 ) i π 2 ) exp ⁡ ( 2 a i ) n ( exp ⁡ ( − 2 a i ) n − 1 ) =\displaystyle i^n\exp{\left(-ani-\frac{(n-1)i\pi}{2}\right)}\exp{\left(2ai\right)}^n\left(\exp{\left(-2ai\right)}^n-1\right)

= i n exp ⁡ ( − a n i ) ( − i ) ( n − 1 ) exp ⁡ ( 2 a i ) n ( exp ⁡ ( − 2 a i ) n − 1 ) =\displaystyle i^n\exp{\left(-ani\right)} (-i)^{(n-1)} \exp{\left(2ai\right)}^n\left(\exp{\left(-2ai\right)}^n-1\right)

= i n exp ⁡ ( − a n i ) ( − i ) ( n − 1 ) exp ⁡ ( 2 a n i ) ( exp ⁡ ( − 2 a n i ) − 1 ) =\displaystyle i^n\exp{\left(-ani\right)} (-i)^{(n-1)} \exp{\left(2ani\right)}\left(\exp{\left(-2ani\right)}-1\right)

= i n ( − i ) ( n − 1 ) ( exp ⁡ ( − a n i ) − exp ⁡ ( a n i ) ) =\displaystyle i^n (-i)^{(n-1)} \left(\exp{(-ani)}-\exp{(ani)}\right)

= i n ( − i ) ( n − 1 ) ( − 2 i sin ⁡ ( a n ) ) =\displaystyle i^n (-i)^{(n-1)} (-2i\sin(an))

= 2 sin ⁡ ( a n ) =\displaystyle 2\sin(an)

This is probably the most beautiful thing I have seen in my life.

Kenny Lau - 5 years, 8 months ago

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I agree that these trigonometric product formulas are quite beautiful and fascinating... but it seems like few people share our passion ;)

Otto Bretscher - 5 years, 8 months ago

Alternatively, for the middle part:

exp ⁡ ( − i π 3 ) ∏ k = 0 2014 ( 1 − exp ⁡ ( i π 93 + 2 k i π 2015 ) ) \quad\displaystyle \exp{\left(-\frac{i\pi}3\right)}\prod_{k=0}^{2014}\left(1-\exp{\left(\frac{i\pi}{93}+\frac{2ki\pi}{2015}\right)}\right)

= exp ⁡ ( − 2 i π 3 ) ∏ k = 0 2014 ( exp ⁡ ( − i π 93 ) − exp ⁡ 2 k i π 2015 ) =\displaystyle\exp{\left(-\frac{2i\pi}3\right)}\prod_{k=0}^{2014}\left(\exp{\left(-\frac{i\pi}{93}\right)}-\exp{\frac{2ki\pi}{2015}}\right)

= exp ⁡ ( − 2 i π 3 ) ( ( exp ⁡ ( − i π 93 ) ) 2015 − 1 ) =\displaystyle\exp{\left(-\frac{2i\pi}3\right)}\left(\left(\exp{\left(-\frac{i\pi}{93}\right)}\right)^{2015}-1\right)

= exp ⁡ ( − 2 i π 3 ) ( exp ⁡ ( i π 3 ) − 1 ) =\displaystyle\exp{\left(-\frac{2i\pi}3\right)}\left(\exp{\left(\frac{i\pi}3\right)}-1\right)

= exp ⁡ ( − 2 i π 3 ) exp ⁡ ( 2 i π 3 ) =\displaystyle\exp{\left(-\frac{2i\pi}3\right)}\exp{\left(\frac{2i\pi}3\right)}

= 1 =1

Kenny Lau - 5 years, 8 months ago

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