Let's minimise sin!

Geometry Level 5

P n = ∏ k = 0 n − 1 ( 1 − 2 sin ⁡ ( 2 k π n ) ) P_n=\prod_{k=0}^{n-1}\left(1-2\sin\left(\frac{2k\pi}{n}\right)\right)

Find the minimal value of P n P_n , where n n is a positive integer.

If you come to the conclusion that no minimum is attained, enter 666 as your answer.


The answer is -3.

Otto Bretscher by Otto Bretscher , 65, USA

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1 solution

Otto Bretscher
Oct 16, 2015

1 − 2 sin ⁡ ( 2 k π n ) = 2 ( cos ⁡ ( 2 k π n + π 2 ) − cos ⁡ ( 2 π 3 ) ) = − 4 sin ⁡ ( k π n + 7 π 12 ) sin ⁡ ( k π n − π 12 ) 1-2\sin\left(\frac{2k\pi}{n}\right)=2\left(\cos\left(\frac{2k\pi}{n}+\frac{\pi}{2}\right)-\cos\left(\frac{2\pi}{3}\right)\right)=-4\sin\left(\frac{k\pi}{n}+\frac{7\pi}{12}\right)\sin\left(\frac{k\pi}{n}-\frac{\pi}{12}\right) According to the formula discussed here , the product is P n = 4 ( − 1 ) n + 1 sin ⁡ ( 7 n π 12 ) sin ⁡ ( n π 12 ) P_n=4(-1)^{n+1}\sin\left(\frac{7n\pi}{12}\right)\sin\left(\frac{n\pi}{12}\right)

P n P_n attains a minimum of − 3 \boxed{-3} when n n is divisible by 4 but not by 12.

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P n \displaystyle\quad P_n

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

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

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

= i n exp ⁡ ( − ( n − 1 ) i π ) ∏ k = 0 n − 1 ( − i exp ⁡ ( 2 k π n ) − 1 + exp ⁡ ( 4 k π n ) ) =\displaystyle i^n\exp\left(-(n-1)i\pi\right)\prod_{k=0}^{n-1}\left(-i\exp\left(\frac{2k\pi}{n}\right)-1+\exp\left(\frac{4k\pi}{n}\right)\right)

= i n exp ⁡ ( − ( n − 1 ) i π ) ∏ k = 0 n − 1 ( exp ⁡ ( 2 k π n ) − exp ⁡ ( i π 6 ) ) ( exp ⁡ ( 2 k π n ) − exp ⁡ ( 5 i π 6 ) ) =\displaystyle i^n\exp\left(-(n-1)i\pi\right)\prod_{k=0}^{n-1}\left(\exp\left(\frac{2k\pi}{n}\right)-\exp\left(\frac{i\pi}{6}\right)\right)\left(\exp\left(\frac{2k\pi}{n}\right)-\exp\left(\frac{5i\pi}{6}\right)\right)

= i n exp ⁡ ( − ( n − 1 ) i π ) ∏ k = 0 n − 1 ( exp ⁡ ( i π 6 ) − exp ⁡ ( 2 k π n ) ) ( exp ⁡ ( 5 i π 6 ) − exp ⁡ ( 2 k π n ) ) =\displaystyle i^n\exp\left(-(n-1)i\pi\right)\prod_{k=0}^{n-1}\left(\exp\left(\frac{i\pi}{6}\right)-\exp\left(\frac{2k\pi}{n}\right)\right)\left(\exp\left(\frac{5i\pi}{6}\right)-\exp\left(\frac{2k\pi}{n}\right)\right)

= i n ( − 1 ) n − 1 ( exp ⁡ ( n i π 6 ) − 1 ) ( exp ⁡ ( 5 n i π 6 ) − 1 ) =\displaystyle i^n(-1)^{n-1}\left(\exp\left(\frac{ni\pi}{6}\right)-1\right)\left(\exp\left(\frac{5ni\pi}{6}\right)-1\right)

= − ( − i ) n ( exp ⁡ ( n i π ) − exp ⁡ ( n i π 6 ) − exp ⁡ ( 5 n i π 6 ) + 1 ) =\displaystyle -(-i)^n\left(\exp\left(ni\pi\right)-\exp\left(\frac{ni\pi}{6}\right)-\exp\left(\frac{5ni\pi}{6}\right)+1\right)

= − ( − i ) n ( ( − 1 ) n − exp ⁡ ( n i π 6 ) − exp ⁡ ( 5 n i π 6 ) + 1 ) =\displaystyle -(-i)^n\left((-1)^n-\exp\left(\frac{ni\pi}{6}\right)-\exp\left(\frac{5ni\pi}{6}\right)+1\right)

= − i n ( 1 − exp ⁡ ( 7 n i π 6 ) − exp ⁡ ( 11 n i π 6 ) + ( − 1 ) n ) =\displaystyle -i^n\left(1-\exp\left(\frac{7ni\pi}{6}\right)-\exp\left(\frac{11ni\pi}{6}\right)+(-1)^n\right)

= − i n ( 1 − 2 cos ⁡ ( 2 n π 3 ) exp ⁡ ( 9 n i π 6 ) + ( − 1 ) n ) =\displaystyle -i^n\left(1-2\cos\left(\frac{2n\pi}3\right)\exp\left(\frac{9ni\pi}{6}\right)+(-1)^n\right)

= − i n ( 1 − 2 ( − i ) n cos ⁡ ( 2 n π 3 ) + ( − 1 ) n ) =\displaystyle -i^n\left(1-2(-i)^n\cos\left(\frac{2n\pi}3\right)+(-1)^n\right)

Thus the function is shown to be periodic, with a period of 12 12 .

Trying all the 12 cases gives minimum − 3 -3 .

Formulas used:

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

exp ⁡ ( a i ) + exp ⁡ ( b i ) = 2 cos ⁡ ( a − b 2 ) exp ⁡ ( a i + b i 2 ) \displaystyle \exp(ai)+\exp(bi)=2\cos\left(\frac{a-b}2\right)\exp\left(\frac{ai+bi}2\right)

Kenny Lau - 5 years, 8 months ago

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Wow, this looks quite complex (upvote)! Thank you! I get -4 when I make n n a multiple of 12 in your formula...is there a sign error hidden somewhere?

Otto Bretscher - 5 years, 8 months ago

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Yes, and I fixed it.

Kenny Lau - 5 years, 8 months ago

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@Kenny Lau – There is still a discrepancy between our formulas for odd n n . I believe the answer for n = 3 n=3 should be -2, for example.

Otto Bretscher - 5 years, 7 months ago

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@Otto Bretscher – I hope it is alright now. I always make stupid careless mistakes.

Kenny Lau - 5 years, 7 months ago

+1 Did the same way. The product can also be written as P n 2 = ∏ k = 0 n − 1 ( cos ⁡ ( 6 k π n ) cos ⁡ ( 2 k π n ) ) \displaystyle {P_{n}}^2 = \prod_{k=0}^{n-1} \left(\dfrac{\cos \left(\frac{6k\pi}{n}\right)}{\cos \left(\frac{2k\pi}{n}\right)}\right) Can we find a way to evaluate it through this route?

Ishan Singh - 5 years, 8 months ago

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The two products are not equal! (Square root can only be positive)

Kenny Lau - 5 years, 8 months ago

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@Kenny Lau – Oh yeah, sorry. I mean,

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

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

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

⟹ P n 2 = ∏ k = 0 n − 1 ( 1 − 4 sin ⁡ 2 ( 2 k π n ) ) \displaystyle \implies {P_{n}}^2 = \prod_{k=0}^{n-1} \left( 1 - 4\sin^2\left(\frac{2k\pi}{n}\right) \right)

= ∏ k = 0 n − 1 ( 4 cos ⁡ 2 ( 2 k π n ) − 3 ) \displaystyle = \prod_{k=0}^{n-1} \left( 4\cos^2\left(\frac{2k\pi}{n}\right) - 3 \right)

⟹ P n 2 = ∏ k = 0 n − 1 ( cos ⁡ ( 6 k π n ) cos ⁡ ( 2 k π n ) ) \displaystyle \implies {P_{n}}^2 = \prod_{k=0}^{n-1} \left(\dfrac{\cos \left(\frac{6k\pi}{n}\right)}{\cos \left(\frac{2k\pi}{n}\right)}\right)

Ishan Singh - 5 years, 7 months ago

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@Ishan Singh – Do we need to worry about the denominator being zero?

Otto Bretscher - 5 years, 7 months ago

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@Otto Bretscher – We can take limit since it will be an indeterminate form (For m m zeroes in the denominator, there will be m m zeroes in the numerator)

Ishan Singh - 5 years, 7 months ago

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@Ishan Singh – Limit with respect to what variable?

Otto Bretscher - 5 years, 7 months ago

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@Otto Bretscher – Sorry, I guess the above transformation is valid only if n n is not a multiple of 4 4

Ishan Singh - 5 years, 7 months ago

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@Ishan Singh – But multiples of 4 give us the solution... so, you may have to look for a different approach.

Otto Bretscher - 5 years, 7 months ago

@Ishan Singh – You can just use the formula 2 cos ⁡ θ = exp ⁡ ( i θ ) + exp ⁡ ( − i θ ) 2\cos\theta=\exp(i\theta)+\exp(-i\theta) .

Kenny Lau - 5 years, 7 months ago

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@Kenny Lau – I know that, but I was hoping to calculate the more general product ∏ k = 0 n − 1 cos ⁡ ( r k π n ) \displaystyle \prod_{k=0}^{n-1} \cos \left(\frac{rk\pi}{n}\right) , where r r is a positive integer and then put values for r r . If we simplify it using complex numbers, in the end we'd have to solve ∏ k = 0 n − 1 ( w 2 r + 1 ) \displaystyle \prod_{k=0}^{n-1} (w^{2r} + 1) where w 2 n = 1 w^{2n} = 1 Any insights?

Ishan Singh - 5 years, 7 months ago

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@Ishan Singh – Still the same method as is shown in my solution. ( ω 2 k + 1 ) = ( ω k + i ) ( ω k − i ) = ( − i − ω k ) ( i − ω k ) = ( ( − i ) k − 1 ) ( i k − 1 ) = 1 − ( − i ) k − i k + 1 = 2 − 2 ( − 1 ) k cos ⁡ ( k π / 2 ) (\omega^{2k}+1)=(\omega^k+i)(\omega^k-i)=(-i-\omega^k)(i-\omega^k)=((-i)^k-1)(i^k-1)=1-(-i)^k-i^k+1=2-2(-1)^k\cos(k\pi/2)

Kenny Lau - 5 years, 7 months ago

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@Kenny Lau – How is ( − i − ω k ) ( i − ω k ) = ( ( − i ) k − 1 ) ( i k − 1 ) (-i-\omega^k)(i-\omega^k) = ((-i)^k - 1)(i^k - 1) ? Also note that r ≠ n r \neq n

Ishan Singh - 5 years, 7 months ago

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@Ishan Singh – Oops, I meant that ( − i − ω k ) ( i − ω k ) = ( ( − i ) n − 1 ) ( i n − 1 ) (-i-\omega^k)(i-\omega^k)=((-i)^n-1)(i^n-1) . This is the first formula that I gave in my solution.

Kenny Lau - 5 years, 7 months ago

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@Kenny Lau – I think you misread my question. The product is ∏ k = 0 n − 1 ( exp ⁡ ( i r k π n ) + 1 ) \displaystyle \prod_{k=0}^{n-1} \left(\exp\left(i\dfrac{rk\pi}{n}\right) +1 \right)

Ishan Singh - 5 years, 7 months ago

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@Ishan Singh – ∏ 2 cos ⁡ ( k π / n ) \prod2\cos(k\pi/n)

= ∏ ( exp ⁡ ( k i π / n ) + exp ⁡ ( − k i π / n ) ) =\displaystyle\prod(\exp(ki\pi/n)+\exp(-ki\pi/n))

= ∏ exp ⁡ ( − k i π / n ) ( exp ⁡ ( 2 k i π / n ) + 1 ) =\displaystyle\prod\exp(-ki\pi/n)(\exp(2ki\pi/n)+1)

= ( ∏ exp ⁡ ( − k i π / n ) ) ( − 1 ) n ( ( − 1 ) n − 1 ) =\displaystyle\left(\prod\exp(-ki\pi/n)\right)(-1)^n((-1)^n-1)

= exp ⁡ ( − ( n + 1 ) i π / 2 ) ( − 1 ) n ( ( − 1 ) n − 1 ) =\displaystyle\exp(-(n+1)i\pi/2)(-1)^n((-1)^n-1)

= ( − i ) n + 1 ( 1 − ( − 1 ) n ) =\displaystyle(-i)^{n+1}(1-(-1)^n)

I am writing the general solution now. This is just an example of how I would do it.

Kenny Lau - 5 years, 7 months ago

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@Kenny Lau – Yeah, that's my point, it doesn't seem to work in the general case. Anyway, eagerly waiting for your general solution.

Ishan Singh - 5 years, 7 months ago

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@Ishan Singh – So:

∏ ( 2 cos ⁡ ( r k π n ) ) \quad\displaystyle\prod\left(2\cos\left(\frac{rk\pi}n\right)\right)

= ∏ ( exp ⁡ ( r k i π n ) + exp ⁡ ( − r k i π n ) ) =\displaystyle\prod\left(\exp\left(\frac{rki\pi}n\right)+\exp\left(\frac{-rki\pi}n\right)\right)

= ∏ exp ⁡ ( − r k i π n ) ( exp ⁡ ( 2 r k i π n ) + 1 ) =\displaystyle\prod\exp\left(\frac{-rki\pi}n\right)\left(\exp\left(\frac{2rki\pi}n\right)+1\right)

= exp ⁡ ( − r ( n + 1 ) i π 2 ) ∏ ( exp ⁡ ( 2 r k i π n ) + 1 ) =\displaystyle\exp\left(\frac{-r(n+1)i\pi}2\right)\prod\left(\exp\left(\frac{2rki\pi}n\right)+1\right)

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

= exp ⁡ ( − r ( n + 1 ) i π 2 ) ( − 1 ) n r ∏ ∏ m = 0 r − 1 ( exp ⁡ ( ( 2 m + 1 ) i π 2 ) − exp ⁡ ( 2 k i π n ) ) =\displaystyle\exp\left(\frac{-r(n+1)i\pi}2\right)(-1)^{nr}\prod\prod_{m=0}^{r-1}\left(\exp\left(\frac{(2m+1)i\pi}2\right)-\exp\left(\frac{2ki\pi}n\right)\right)

= exp ⁡ ( − r ( n + 1 ) i π 2 ) ( − 1 ) n r ∏ m = 0 r − 1 ( exp ⁡ ( ( 2 m + 1 ) n i π 2 ) − 1 ) =\displaystyle\exp\left(\frac{-r(n+1)i\pi}2\right)(-1)^{nr}\prod_{m=0}^{r-1}\left(\exp\left(\frac{(2m+1)ni\pi}2\right)-1\right)

Looks like I have disappointed you, haven't I. At least I have eliminated k k .

Kenny Lau - 5 years, 7 months ago

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@Kenny Lau – Looks good. I'll inform if I get anything. Nice talking to you. Night.

Ishan Singh - 5 years, 7 months ago

@Kenny Lau – Btw, there's a typo. It should be exp ⁡ ( ( 2 m + 1 ) n i π r ) \displaystyle\exp\left(\frac{(2m+1)ni\pi}{r}\right) instead of exp ⁡ ( ( 2 m + 1 ) n i π 2 ) \displaystyle\exp\left(\frac{(2m+1)ni\pi}{2}\right) in the last and second last line.

Ishan Singh - 5 years, 7 months ago

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