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

Answer 1

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

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#
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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#

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#
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Answer 3

#4033#

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#
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)#
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#
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Answer 4

The value of ( \log_2 \left( \prod_{m=1}^{2017} \prod_{n=1}^{2017} \left(1 + e^{\frac{2 \pi i n m}{2017}}\right) \right) ) is zero.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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