SOLUTIONS

No problem is ever permanently closed. Any comments, new solutions, or new insights on old problems are always welcomed by the editor.

10. Proposed by Curtis Cooper and Robert E. Kennedy, Central Missouri State University, Warrensburg, Missouri.

Show

∑_{{k=1 \atop (k,45)=1}}^{45} sin^{8} \frac{kπ}{45} = \frac{51}{8} .

Solution by the proposers.

Let w = exp \frac{πi}{n}. Then

∑_{k=1}^{n} sin^{4m} \frac{kπ}{n} = ∑_{k=1}^{n}(\frac{w^{k} − w^{−k}}{2i} )^{4m}

= \frac{1}{2^{4m}} ∑_{k=1}^{n} ∑_{j=0}^{4m}(−1)^{j}\binom{4m}{j}(w^{k})^{4m−j}(w^{−k})^{j}

= \frac{1}{2^{4m}} ∑_{k=1}^{n} ∑_{j=0}^{4m}(−1)^{j}\binom{4m}{j}(w^{4m−2j})^{k}

= \frac{1}{2^{4m}} ∑_{j=0}^{4m}(−1)^{j}\binom{4m}{j} ∑_{k=1}^{n}(w^{4m−2j})^{k} .

Now

∑_{k=1}^{n}(w^{4m−2j})^{k} = \{ \begin{matrix}n, & if w^{4m−2j} = 1, \\ \frac{w^{4m−2j}(1−(w^{n})^{4m−2j})}{1−w^{4m−2j}} = 0, & if w^{4m−2j} ≠ 1. \end{matrix}

But w^{4m−2j} = 1 iff \frac{2m−j}{n} is an integer. Also, if n | 2m − j, then there exists an integer k such that 2m − j = kn. Therefore,

0 ≤ j = 2m − kn ≤ 4m

so

−\frac{2m}{n} ≤ k ≤ \frac{2m}{n} .

Using the above,

∑_{k=1}^{n} sin^{4m} \frac{kπ}{n} = \frac{1}{2^{4m}} \begin{matrix}⌊ & \frac{2m}{n} & ⌋ \\ ∑ \\ k=⌈− \frac{2m}{n}⌉ \end{matrix}(−1)^{2m−kn}\binom{4m}{2m − kn}n

= \frac{n}{2^{4m}} \begin{matrix}⌊ & \frac{2m}{n} & ⌋ \\ ∑ \\ k=⌈− \frac{2m}{n}⌉ \end{matrix}(−1)^{−kn}\binom{4m}{2m − kn}

= \frac{n}{2^{4m}} ∑_{k≥0}(−1)^{2m−n⌊\frac{2m}{n}⌋+kn}\binom{4m}{2m − n⌊\frac{2m}{n}⌋ + kn} .

The last equality follows from examining limits, signs, and coefficients of the previous sum. Next, using Problem 14(a) from Chapter 2 [1] and the above results

∑_{{k=1 \atop (k,n)=1}}^{n} sin^{4m} \frac{kπ}{n} = ∑_{d|n} μ(\frac{n}{d})(∑_{k=1}^{d} sin^{4m} \frac{kπ}{d} )

= ∑_{d|n} μ(\frac{n}{d})\frac{d}{2^{4m}} ∑_{k≥0}(−1)^{2m−d⌊\frac{2m}{d}⌋+kd}\binom{4m}{2m − d⌊\frac{2m}{d}⌋ + kd}

= \frac{1}{2^{4m}} ∑_{d|n} d · μ(\frac{n}{d}) ∑_{k≥0}(−1)^{2m−d⌊\frac{2m}{d}⌋+kd}\binom{4m}{2m − d⌊\frac{2m}{d}⌋ + kd} .

Finally, applying this result when n = 45 and m = 2 we have

∑_{{k=1 \atop (k,45)=1}}^{45} sin^{8} \frac{kπ}{45} = \frac{51}{8} .

\underline{Reference}

1. T. M. Apostol, Introduction to Analytic Number Theory, Springer–Verlag, 1984, p. 48.

11. Proposed by Stanley Rabinowitz, Westford, Massachusetts. A \underline{strip} is the closed region bounded between two parallel lines

in the plane. Prove that a finite number of strips cannot cover the entire plane.

Solution by the proposer.

Since there are only a finite number of strips, there must be some line, L, that is not parallel to the bounding lines of any of the strips. Each strip intersects L in a finite line segment. These line segments cannot completely cover L, so there must be some point of L that is not in any of these line segments. This point is not covered by any of the strips.

The proposer also submitted a second solution.

13. Proposed by James H. Taylor, Central Missouri State University, Warrensburg, Missouri.

(a) Show

∑_{n=0}^{l} ∑_{m=0}^{n} \frac{(−1)^{m}}{m!(l − n)!} = 1

for any non–negative integer l.

(b) For any positive integer n, define

(2n − 1)!! = 1 · 3 · · · · · (2n − 1)

and

(2n)!! = 2 · 4 · · · · · (2n) .

Show

∑_{k=1}^{n} \frac{(2k − 1)!!}{(2k)!!} = \frac{(2n + 1)!!}{(2n)!!} − 1 .

Solution I to part (a) by Joseph E. Chance, Pan American University, Edinburg, Texas.

The result follows from successive Cauchy Products of the series

(1)  e^{−x} = ∑_{l=0}^{∞} \frac{(−1)^{l}}{l!} x^{l} ,

(2)  \frac{1}{1 − x} = ∑_{l=0}^{∞} 1 · x^{l} , for |x| < 1 , and

(3)  e^{x} = ∑_{l=0}^{∞} \frac{1}{l!}x^{l} .

The Cauchy Product of series (1) and (2) is

(4)  \frac{e^{−x}}{1 − x} = ∑_{l=0}^{∞}( ∑_{m=0}^{l} \frac{(−1)^{m}}{m!} )x^{l} for |x| < 1 .

The Cauchy Product of series (3) and (4) is

\begin{aligned}\frac{1}{1 − x} & = e^{x} · \frac{e^{−x}}{1 − x} \\ & = ∑_{l=0}^{∞}(∑_{n=0}^{l} \frac{1}{(l − n)!}( ∑_{m=0}^{n} \frac{(−1)^{m}}{m!} ))x^{l} , for |x| < 1 . \end{aligned}

The last series and series (2) represent the same function, thus must have identical coefficients. Therefore

∑_{n=0}^{l} ∑_{m=0}^{n} \frac{(−1)^{m}}{m!(l − n)!} = 1 .

Solution II to part (a) by the proposer.

We prove this result by induction on l. The result is true for l = 0. Suppose the result is true for some l = k ≥ 0. Then by rearranging terms and using the induction hypothesis

∑_{n=0}^{k+1} ∑_{m=0}^{n} \frac{(−1)^{m}}{m!(k + 1 − n)!} = ∑_{n=1}^{k+1} ∑_{m=0}^{n−1} \frac{(−1)^{m}}{m!(k + 1 − n)!}

+ ∑_{n=0}^{k+1} \frac{(−1)^{n}}{n!(k + 1 − n)!}

= ∑_{n=0}^{k} ∑_{m=0}^{n} \frac{(−1)^{m}}{m!(k − n)!}

+ ∑_{n=0}^{k+1} \frac{(−1)^{n}}{n!(k + 1 − n)!}

= 1 + ∑_{n=0}^{k+1} \frac{(−1)^{n}}{n!(k + 1 − n)!} .

But

∑_{n=0}^{k+1} \frac{(−1)^{n}}{n!(k + 1 − n)!} = \frac{1}{(k + 1)!} ∑_{n=0}^{k+1} \frac{(−1)^{n}(k + 1)!}{n!(k + 1 − n)!}

= \frac{1}{(k + 1)!} ∑_{n=0}^{k+1}(−1)^{n}\binom{k + 1}{n}

= \frac{1}{(k + 1)!}(1 − 1)^{k+1}

= 0 .

Thus

∑_{n=0}^{k+1} ∑_{m=0}^{n} \frac{(−1)^{m}}{m!(k + 1 − n)!} = 1

so the result is true for l = k + 1. Hence by induction,

∑_{n=0}^{l} ∑_{m=0}^{n} \frac{(−1)^{m}}{m!(l − n)!} = 1

for any non–negative integer l.

Solution I to part (b) by Joseph E. Chance, Pan American University, Edinburg, Texas.

From the binomial theorem

(5)  (1 − x)^{− \frac{1}{2}} = 1 + ∑_{k=1}^{∞} \frac{1 · 3 · · · · · (2k − 1)}{2 · 4 · · · · · (2k)}x^{k}

= 1 + ∑_{k=1}^{∞} \frac{(2k − 1)!!}{(2k)!!}x^{k} , for |x| < 1

The Cauchy Product of series (2) and (5) is

\frac{1}{1 − x}(1 − x)^{− \frac{1}{2}} = 1 + ∑_{n=1}^{∞}(1 + ∑_{k=1}^{n} \frac{(2k − 1)!!}{(2k)!!})x^{n} .

But

\frac{1}{1 − x}(1 − x)^{− \frac{1}{2}} = (1 − x)^{− \frac{3}{2}} ,

and from the binomial theorem

\begin{aligned}(1 − x)^{− \frac{3}{2}} & = 1 + ∑_{n=1}^{∞} \frac{1 · 3 · · · · · (2n + 1)}{2 · 4 · · · · · (2n)}x^{n} \\ & = 1 + ∑_{n=1}^{∞} \frac{(2n + 1)!!}{(2n)!!}x^{n} \end{aligned}

Equating coefficients of the respective series representations for (1− x)^{− \frac{3}{2}} gives

\frac{(2n + 1)!!}{(2n)!!} = 1 + ∑_{k=1}^{n} \frac{(2k − 1)!!}{(2k)!!} .

Composite Solution II to part (b) by Russell Euler, Northwest Missouri State University, Maryville, Missouri and the proposer.

The result will be established by using mathematical induction on n. For n = 1, both sides of the alleged identity equal \frac{1}{2}.

∑_{k=1}^{n+1} \frac{(2k − 1)!!}{(2k)!!} = ∑_{k=1}^{n} \frac{(2k − 1)!!}{(2k)!!} + \frac{(2n + 1)!!}{(2n + 2)!!}

= \frac{(2n + 1)!!}{(2n)!!} − 1 + \frac{(2n + 1)!!}{(2n + 2)!!}

= \frac{(2n + 1)!!(2n + 2) + (2n + 1)!!}{(2n + 2)!!} − 1

= \frac{(2n + 1)!!(2n + 3)}{(2n + 2)!!} − 1

= \frac{(2n + 3)!!}{(2n + 2)!!} − 1 ,

as desired.