SOLUTIONS

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7. Proposed by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

Evaluate

L = lim_{x→0}[\frac{sin (tan x) − tan (sin x)}{sin^{−1}(tan^{−1} x) − tan^{−1}(sin^{−1} x)}] .

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

It is well–known that as x → 0

(1)

sin x = x − \frac{1}{6}x^{3} + \frac{1}{120}x^{5} − \frac{1}{5040}x^{7} + O(x^{9}) ,
tan x = x + \frac{1}{3}x^{3} + \frac{2}{15}x^{5} + \frac{17}{315}x^{7} + O(x^{9}) ,
sin^{−1} x = x + \frac{1}{6}x^{3} + \frac{3}{40}x^{5} + \frac{5}{112}x^{7} + O(x^{9}) ,
tan^{−1} x = x − \frac{1}{3}x^{3} + \frac{1}{5}x^{5} − \frac{1}{7}x^{7} + O(x^{9}) .

Using (1), as x → 0

sin (tan x) = tan x − \frac{1}{6} tan^{3} x + \frac{1}{120} tan^{5} x

− \frac{1}{5040} tan^{7} x + O(tan^{9} x)

= x + \frac{1}{3}x^{3} + \frac{2}{15}x^{5} + \frac{17}{315}x^{7} + O(x^{9})

− \frac{1}{6}(x + \frac{1}{3}x^{3} + \frac{2}{15}x^{5} + \frac{17}{315}x^{7} + O(x^{9}))^{3}

+ \frac{1}{120}(x + \frac{1}{3}x^{3} + \frac{2}{15}x^{5} + \frac{17}{315}x^{7} + O(x^{9}))^{5}

− \frac{1}{5040}(x + \frac{1}{3}x^{3} + \frac{2}{15}x^{5} + \frac{17}{315}x^{7} + O(x^{9}))^{7}

+ O(x^{9})

= x + \frac{1}{3}x^{3} + \frac{2}{15}x^{5} + \frac{17}{315}x^{7} + O(x^{9})

− \frac{1}{6}(x^{3} + x^{5} + \frac{11}{15}x^{7} + O(x^{9}))

+ \frac{1}{120}(x^{5} + \frac{5}{3}x^{7} + O(x^{9}))

− \frac{1}{5040}(x^{7} + O(x^{9})) + O(x^{9})

= x + \frac{1}{6}x^{3} − \frac{1}{40}x^{5} − \frac{55}{1008}x^{7} + O(x^{9}) .

Repeating this result and deriving three others similarly we have that as x → 0

(2)

sin (tan x) = x + \frac{1}{6}x^{3} − \frac{1}{40}x^{5} − \frac{55}{1008}x^{7} + O(x^{9}) ,
tan (sin x) = x + \frac{1}{6}x^{3} − \frac{1}{40}x^{5} − \frac{107}{5040}x^{7} + O(x^{9}) ,
sin^{−1}(tan^{−1} x) = x − \frac{1}{6}x^{3} + \frac{13}{120}x^{5} − \frac{341}{5040}x^{7} + O(x^{9}) ,
tan^{−1}(sin^{−1} x) = x − \frac{1}{6}x^{3} + \frac{13}{120}x^{5} − \frac{173}{5040}x^{7} + O(x^{9}) .

Therefore, from (2) we have that as x → 0

sin (tan x) − tan (sin x) = −\frac{1}{30}x^{7} + O(x^{9})

and

sin^{−1}(tan^{−1} x) − tan^{−1}(sin^{−1} x) = −\frac{1}{30}x^{7} + O(x^{9}) .

Thus, as x → 0

\frac{sin (tan x) − tan (sin x)}{sin^{−1}(tan^{−1} x) − tan^{−1}(sin^{−1} x)} = \frac{−\frac{1}{30}x^{7} + O(x^{9})}{−\frac{1}{30}x^{7} + O(x^{9})} .

Therefore,

L = lim_{x→0}[\frac{sin (tan x) − tan (sin x)}{sin^{−1}(tan^{−1} x) − tan^{−1}(sin^{−1} x)}] = 1 .

Also solved by the proposer.

8. Proposed by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

The Fibonacci numbers F_{n} satisfy F_{1} = 1, F_{2} = 1, and F_{n+2} = F_{n+1} + F_{n} for n = 1, 2, 3, …. Find two solutions of x^{n} = F_{n}x + F_{n−1} for all integers n ≥ 2.

I. Solution by Bob Prielipp, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

We shall show that α = \frac{1+\sqrt{5}}{2} and β = \frac{1−\sqrt{5}}{2} are the required solutions. (In fact when n = 2, the given equation becomes x^{2} − x − 1 = 0 which has α and β as its only solutions.) Since F_{k} = \frac{α^{k}−β^{k}}{α−β},

α^{n} − F_{n}α = α^{n} − \frac{α^{n} − β^{n}}{α − β} α

= \frac{α^{n+1} − α^{n}β − α^{n+1} + αβ^{n}}{α − β}

= \frac{(−αβ)(α^{n−1} − β^{n−1})}{α − β}

= F_{n−1} (because αβ = −1) .

Also

β^{n} − F_{n}β = β^{n} − \frac{α^{n} − β^{n}}{α − β} β

= \frac{αβ^{n} − β^{n+1} − α^{n}β + β^{n+1}}{α − β}

= \frac{(−αβ)(α^{n−1} − β^{n−1})}{α − β}

= F_{n−1} .

This completes our solution.

II. Composite solution by Charles J. Allard, Polo R–VII Schools, Polo, Missouri and Enis Alpakin (student), Central Missouri State University, Warrensburg, Missouri (independently).

Suppose x is a solution of x^{n} = F_{n}x + F_{n−1} for all integers n ≥ 2. Then for all integers n ≥ 1,

x^{n+1} = F_{n+1}x + F_{n} ,
x · x^{n} = F_{n+1}x + F_{n} ,
x(F_{n}x + F_{n−1}) = F_{n+1}x + F_{n} ,
F_{n}x^{2} + (F_{n−1} − F_{n+1})x − F_{n} = 0 .

Now since F_{n} ≠ 0 and F_{n−1} − F_{n+1} = −F_{n} for n ≥ 1,

x^{2} − x − 1 = 0.

Therefore,

x = \frac{1 ± \sqrt{5}}{2} .

III. Composite solution by Joseph E. Chance, Pan American University, Edinburg, Texas; Alejandro Necochea, Pan American University, Edinburg, Texas; Leonard L. Palmer, Southeast Missouri State University, Cape Girardeau, Missouri; W. F. Wheatley III (student), Central Missouri State University, Warrensburg, Missouri; and the proposer (independently).

We will show by induction on n that x = \frac{1±\sqrt{5}}{2} are two solutions of x^{n} = F_{n}x + F_{n−1} for all integers n ≥ 2. This statement is true for n = 2. Assume the result is true for some n ≥ 2. Then

x^{n+1} = x^{n} · x
= (F_{n}x + F_{n−1}) · x
= F_{n}x^{2} + F_{n−1}x
= F_{n}(x + 1) + F_{n−1}x
= (F_{n} + F_{n−1})x + F_{n}
= F_{n+1}x + F_{n} ,

so the result is true for n + 1. Thus, by induction on n, \frac{1±\sqrt{5}}{2} are solutions of x^{n} = F_{n}x + F_{n−1} for all integers n ≥ 2.