Question

What is the value of `log_2(Pi_(m=1)^2017Pi_(n=1)^2017(1+e^((2 pi i n m)/2017)))` ?

Answer 1

`4033`

Explanation:

We will show, more generally, that for and odd prime `p`,
`log_2(prod_(m=1)^p prod_(n=1)^p(1+e^((2pi i n m)/p)))=2p-1`

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Lemma: `prod_(k=1)^((p-1)/2)cos^2((kpim)/p) = 1/2^(p-1), m in ZZ^+`

Proof: Using the identity `sin(2theta) = 2sin(theta)cos(theta)`, we have `cos((kpim)/p) = sin((2kpim)/p)/(2sin((kpim)/p))`. Substituting, we get

`prod_(k=1)^((p-1)/2)cos^2((kpim)/p) = prod_(k=1)^((p-1)/2)(sin((2kpim)/p)/(2sin((kpim)/p)))^2`

`=1/2^(p-1)(prod_(k=1)^((p-1)/2)sin((2kpim)/p)/sin((kpim)/p))^2`

Note that `sin((kpim)/p) = (-1)^(m-1)sin(((p-k)pim)/p)`. Performing this substitution in the denominator of the above product for each odd `k`, we get, for some `l in ZZ`,

`1/2^(p-1)(prod_(k=1)^((p-1)/2)sin((2kpim)/p)/sin((kpim)/p))^2 = 1/2^(p-1)((-1)^lprod_(k=1)^((p-1)/2)sin((2kpim)/p)/sin((2kpim)/p))^2`

`=1/2^(p-1)`

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Proceeding, we examine the product `prod_(m=1)^p prod_(n=1)^p(1+e^((2pi i nm)/p))`. Note that `e^((2pi i nm)/p) = 1` if `n = p` or `m = p`. Thus, for

`(m, n) in {(1, p), (2, p), ..., (p-1, p), (p, 1), (p, 2), ..., (p, p)}`

we have `1+e^((2pi i nm)/p) = 2`

As there are `2p-1` elements of the above set, we may rewrite the product as

`prod_(m=1)^p prod_(n=1)^p(1+e^((2pi i nm)/p)) = 2^(2p-1)prod_(m=1)^(p-1)prod_(n=1)^(p-1)(1+e^((2pi i nm)/p))`

Making the substitution `e^((2pi i mn)/p) = e^(-2pi i m)e^((2pi i mn)/p) = e^((2pi i m(n-p))/p)` for `n>=(p+1)/2`, we can rewrite the second product as

`prod_(n=1)^(p-1)(1+e^((2pi i nm)/p)) = prod_(n=-(p-1)/2)^((p-1)/2)(1+e^((2pi i nm)/p))`

Pairing factors with corresponding positive and negative values of `n`, we find

`(1+e^((2pi i mn)/p))(1+e^(-(2pi i mn)/p)) = 2 + e^((2pi i mn)/p) + e^(-(2pi i mn)/p)`

`=2 + (cos((2pimn)/p)+isin((2pimn)/p))+(cos(-(2pimn)/p)+isin(-(2pimn)/p))`

`=2(1+cos((2pimn)/p))`

`=>prod_(n=-(p-1)/2)^((p-1)/2)(1+e^((2pi i nm)/p)) = prod_(n=1)^((p-1)/2)2(1+cos((2pimn)/p))`

`=2^((p-1)/2)prod_(n=1)^((p-1)/2)(1+cos((2pimn)/p))`

Using the identities `cos^2(theta)+sin^2(theta) = 1` and `cos(2theta) = cos^2(theta)-sin^2(theta)`, we find

`1+cos(2(pimn)/p) = 2cos^2((pimn)/p)`

`=> 2^((p-1)/2)prod_(n=1)^((p-1)/2)(1+cos((2pimn)/p)) = 2^(p-1)prod_(n=1)^((p-1)/2)cos^2((pimn)/p)`

`=2^(p-1)*1/2^(p-1)" "` (by the lemma)

`=1`

So `prod_(n=1)^(p-1)(1+e^((2pi i nm)/p))=1`, meaning

`prod_(m=1)^p prod_(n=1)^p(1+e^((2pi i nm)/p)) = 2^(2p-1)prod_(m=1)^(p-1)prod_(n=1)^(p-1)(1+e^((2pi i nm)/p))`

`=2^(2p-1)prod_(m=1)^(p-1)1`

`=2^(2p-1)`.

Thus `log_2(prod_(m=1)^p prod_(n=1)^p(1+e^((2pi i nm)/p))) = log_2(2^(2p-1)) = 2p-1`
∎

The answer to the question is a direct application of the above, substituting `p=2017` to arrive at the answer of `2(2017)-1 = 4033`

Answer 2

If `N` is odd and prime then
`log_2(Pi_(m=1)^N(Pi_(n=1)^N(1+e^(2pi i (mn)/N)))) = 2N-1`

Explanation:

Analysing the unit roots

`z^N-1 = Pi_(n=1)^N(z-e^(2pi i n/N))`

if `1 le m_0 le N`

`z^N-1 = Pi_(n=1)^N(z-e^(2pi i (m_0n)/N))`

because

`{e^(2pi i m_0/N),e^(2pi i (2m_0)/N),e^(2pi i (3m_0)/N),cdots,e^(2pi i (Nm_0)/N)}` is a rotation of
`{e^(2pi i 1/N),e^(2pi i 2/N),e^(2pi i 3/N),cdots,e^(2pi i )}`

If `N` is odd and prime, making `z = -1` and considering `1 le m_0 < N`

`-1-1=Pi_(n=1)^N(-1-e^(2pi i (m_0n)/N))` then

`2 = Pi_(n=1)^N(1+e^(2pi i (m_0n)/N))` and for `m_0=N`

`Pi_(n=1)^N(1+e^(2pi i (Nn)/N)) = 2^N` so

`Pi_(m=1)^N(Pi_(n=1)^N(1+e^(2pi i (mn)/N))) = 2^(N-1)2^N = 2^(2N-1)`

Finally for `N` prime

`log_2(Pi_(m=1)^N(Pi_(n=1)^N(1+e^(2pi i (mn)/N)))) = 2N-1`

Answer 3

`3025/log 2`, after self-corrections, in three editions

Explanation:

Heralding the new year 2017, the design of the product is indeed

splendid.

Consider the variation in n, from 1 to 2017, in the inner product.

A typical factor is

`1+e^(i((2pimn)/2017))`

`=1+cos((2pimn)/2017)+i sin((2pimn)/2017)`

`=2 cos^2((pimn)/2017)+i 2 sin((pimn)/2017)cos((pimn)/2017)`

`=2 cos((pimn)/2017)(cos((pimn)/2017)+i sin((pimn)/2017))`

`=2 cos((pimn)/2017)e^(i(pimn)/2017)`

So the whole inner product is

`2^2017 Picos((pimn)/2017)e^(i(pimn)/2017)`

`=2^2017 Picos((pimn)/2017)Pie^(i(pimn)/2017)`

`=2^2017 (Picos((pimn)/2017))e^(i(pim(1+2+...2016+2017))/2017)`

`=2^2017e^(impi((2017)((2018)/2))/2017)Picos((pimn)/2017)`

`=2^2017e^(i1009mpi)Picos((pimn)/2017)`

`=(-1)^m2^2017Picos((pimn)/2017)`, n varying from 1 to 2017

So, the given expression is

`log_2(2^2017Pi((-1)^m(Picos((pimn)/2017)))`

`=(2017+sum(-1)^m sumcos((pimn)/2017))/log 2`
using `log_2=log a/log 2`

`=(2017+sum(-1)^m sumcos((pimn)/2017))/log 2`

`=(2017+ sum(-1)^mcos((pimn)/2017))/log 2`

For now, I break here.I hope that I could report my brief answer, a little later.

Using Lagrange's identity,

the inner `sumcos((mnpi)/2017)`

`= 1/2(1+sin((2017+1/2)mpi/2017)/sin(((mpi/2)/2017)))`

`=1/2(1+sin(mpi+((mpi/2)/2017))/sin(((mpi/2)/2017)))`

`=1/2(1+(-1)^m)`

= 1, when m is even, and 0, when m is odd.

So, the answer simplifies to

(2017 +( count of even numbers less than 2017) )/log 2

`=3025/log 2`