My solution to Purdue's no2 Problem of the week

1. Eduardo Enrique Escamilla Saldaña
Monterrey, Nuevo Leon, Mexico
February 2, 2014
Purdue Problem of the Week No.2
It is known that, for any positive integer m,
𝑛
/
𝑚𝑘
lim
!→!
!!!:!"!!
!
!!!
1
𝑛
𝑗 = 𝑚.
Prove this for m=2.
I will prove the general result. The denominator:
𝑛
𝑛
is equivalent to adding the binomial coefficients
such that
!!!:!"!!
𝑚𝑘
𝑘
𝑛
𝑘 ≡ 0 (𝑚𝑜𝑑 𝑚), namely !≡ ! (!"# !)
.
𝑘
Consider the roots of unity 𝜁 ! = cos 2𝜋𝑖𝑘 𝑚 + 𝑖 sin 2𝜋𝑖𝑘 𝑚 of 𝜁 ! − 1 = 0.
𝜁 ! − 1 = 𝜁 − 1 1 + 𝜁 + 𝜁 ! + ⋯ + 𝜁 !!!
Therefore:
!!!
𝑚 𝑖𝑓 𝑘 ≡ 0(𝑚𝑜𝑑 𝑚)
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
𝜁 ! =
!!!
In other words:
1
𝑚
!!!
𝜁 ! =
!!!
1 𝑖𝑓 𝑘 ≡ 0(𝑚𝑜𝑑 𝑚)
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
However by the binomial theorem:
1 + 𝑥 ! 𝜁!
1 + 𝜁!
!
!
!
=
!
=
!!!
!!!
!
!
𝑥 !! 𝜁!" Making 𝑥 = 1
𝑛 !"
𝜁
𝑘
Therefore
!≡ ! (!"# !)
𝑛
=
𝑘
𝑛
𝑘
!
!!!
1
𝑚
!!!
𝜁!
!!!
Note that
𝑒 !! + 𝑒 !!!
cos Φ =
2
2 cos Φ𝑗 = 𝑒 !!! + 𝑒 !!!!
If Φ = 2𝜋𝑗 𝑚 2 cos 2π𝑗/𝑚 = 𝑒 !
1
=
𝑚
!"!/!
!!!
1 + 𝜁!
!
!!!
+ 𝑒 !!
!"!/!
= 𝜁! + 𝜁 !!
Taking advantage of this equality
1 + 𝜁!
!
=
1
𝜁
+ 𝜁!/!
!/!
!
𝜁!/!
!
= 𝜁 !!/! + 𝜁!/!
!
𝜁!"/!
= 2 cos π𝑗/𝑚
! !"/!
𝜁

2. Since we are interested in real numbers we can take the real part of 𝜁!"/! , and using
De Moivre's theorem we get
𝑛𝑗𝜋
𝑅𝑒 𝜁!"/! = cos 2𝜋𝑗𝑛/(2𝑛) = cos
𝑚
!≡ ! (!"# !)
!!!
=
1
𝑚
𝑛
=
𝑘
2 cos π𝑗/𝑚
!!!
1+1
!
!
=
!!!
!
!!!
!
𝑛
𝑘
1
𝑚
cos
!!!
𝜁
1
=
𝑚
!
!!!
𝑛𝑗𝜋
2!
=
𝑚
𝑚
!!!
!
1 + 𝜁!
!!!
!!!
!
cos π𝑗/𝑚
cos
!!!
𝑛𝑗𝜋
𝑚
𝑛 ! !!!
1 1
= 2!
𝑗
Therefore:
𝑛
/
𝑚𝑘
lim
!→!
!!!:!"!!
= lim
!→!
2!
= lim 𝑚
!≡ ! (!"# !)
!→!
= lim
!→!
1
𝑚
= 1/𝑚
!!!
!!!
!
𝑛
𝑗
!!!
!
𝑛
/
𝑘
𝑛
𝑗 .
!!!
!
cos π𝑗/𝑚
𝑛𝑗𝜋
𝑚
cos
2!
!!!
!
cos π𝑗/𝑚
𝑛𝑗𝜋
𝑚
cos
!!!
In the case where m = 2, using the above methodology, we get
lim
!→!
= lim
!!!:!"!!
!!!
!→!
1
𝑚
1
= lim
!→! 2
𝑛
/
𝑚𝑘
cos
!!!
!
!!!
π𝑗
𝑚
𝜋𝑗
cos
2
!
𝑛
𝑗 =
!!!
!
cos
𝑛𝑗𝜋
𝑚
cos
𝑛𝑗𝜋
2
!
1
𝜋𝑗
= lim ( 1 + cos
!→! 2
2
1
= lim ( 1 + 0)
!→! 2
=1/2
!!!
)