Math 499, Problem Set #6 Solutions. 1. Find all solutions to f(x)+2f(1 ...

Math 499, Problem Set #6 Solutions.
1. Find all solutions to f(x) + 2f( 1 x ) = 3x2 .
Solution: Given f(x) + 2f( 1 x ) = 3x2 , replacing x with 1 x yields f( 1 x ) + 2f(x) = 3 x 2 .
Multiply the second equation by 2 and subtract from the first equation to obtain the
unique solution f(x) = 2 x 2 − x 2 .
2. Suppose that {u n }, {v n }, and {w n } are real-valued sequences which satisfy the two
relations u n + v n + w n = 3 and u 2 n + vn 2 + wn 2 = 3 for all n.
If the limits exist, find lim u n , lim v n , and lim w n .
n→∞ n→∞ n→∞
Solution: Multiplying the first relation by −2 and adding it to the second relation
yields u 2 n − 2u n + vn 2 − 2v n + wn 2 − 2w 2 = −3. Now, we complete the square to obtain
(u n − 1) 2 + (v n − 1) 2 + (w n − 1) 2 = 0, which immediately implies that
u n = v n = w n = 1 for all n ∈ N. Therefore, lim u n = lim v n = lim w n = 1.
n→∞ n→∞ n→∞
3. Recall the Fibonacci Sequence {F n } is defined by F 1 = F 2 = 1 and F n+2 = F n+1 + F n
for all n ≥ 0. (We[ will also ] find it useful [ to define F 0 ] = 0.)
0 1
Show that if A = , then A
1 1
n Fn−1 F
=
n
for all n ∈ N.
F n F n+1
Note: This relation can often be used to prove Fibonacci identities.
Solution: This is easy to prove by induction. First, it is readily checked that this
true for n = 0 and[ n = 1. Assuming ] [ that]
the[ claim is true for n ] = k, [ we have ]
A k+1 = A k Fk−1 F · A =
k 0 1 Fk F
=
k−1 + F k Fk F
=
k+1
,
F k F k+1 1 1 F k+1 F k + F k+1 F k+1 F k+2
thereby completing the induction.
4. Compute
∞∑
n=1
(−1) n+1 (2n + 1)
.
n 2 + n
Solution: First of all, let’s pair off every positive and negative term to rewrite the
∞∑ [
2(2k − 1) + 1
series as
(2k − 1) 2 + (2k − 1) − 2(2k) + 1 ] ∞∑ 2
=
(2k) 2 + 2k (2k − 1)(2k + 1) .
k=1
k=1
r∑ [
1
This is a telescoping sum, upon rewriting it as lim
r→∞ 2k − 1 − 1 ]
, which
2k + 1
k=1
[
telescopes to lim 1 − 1 ]
= 1.
r→∞ 2r + 1

5. Consider y = f(x), where f(0) = f( 1 ) = 0 for all n ∈ N. If the graph of f(x) on [0, 1]
n
consists of the congruent sides of an isosceles triangle of height 1 for each n ∈ N,
compute ∫ 1
f(x) dx.
0
Solution: Since this integral represents the sums of areas of triangles with base
length 1 − 1 and height 1 (for each n), we have
n n+1
∫ 1
∞∑ 1
( 1
f(x) dx =
2 n − 1 )
= 1 (via telescoping sums).
n + 1 2
0
n=1
6. Compute lim
x→1
sin( √ x + 3 − 2)
tan( √ x 2 + 5x + 3 − 3) .
Solution: Here is a solution without L’Hôpital’s Rule.
sin θ
tan θ
Using lim = 1, we also have lim = 1.
θ→0 θ
θ→0 θ
Letting θ = √ x + 3 − 2 as well as θ = √ √ x 2 + 5x + 3 − 3, the limit in question is
x + 3 − 2
equivalent to lim √ . Now, we rationalise both the numerator and
x→1 x2 + 5x + 3 − 3
denominator by their respective conjugates to rewrite this as
((x + 3) − 2 2 )( √ x
lim
2 + 5x + 3 + 3)
x→1 ((x 2 + 5x + 3) − 3 2 )( √ x + 3 + 2) = lim ( √ x 2 + 5x + 3 + 3)
x→1 (x + 6)( √ x + 3 + 2) = 3
14 .
7. Let P (x) be a polynomial of degree 2012 such that P (k) = 1 k
What is P (0)
for all k = 1, 2, ..., 2013.
Solution: The key idea is to consider Q(x) = xP (x) − 1. Then, Q(k) = 0 for all
k = 1, 2, ..., 2013. So, the Factor Theorem yields Q(x) = A(x − 1)(x − 2)...(x − 2013)
for some constant A. However, since Q(0) = −1, we conclude that A = 1
2013! .
Having determined Q(x), now we can compute P (0). Since P (x) is a polynomial, it is
continuous at x = 0 in particular. So, we can write that
1 + Q(x) 1 + 1 (x − 1)(x − 2)...(x − 2013)
2013!
P (0) = lim = lim
.
x→0 x x→0 x
1
By L’Hôpital’s Rule, we have P (0) = lim
x→0 2013! · d
[(x − 1)(x − 2)...(x − 2013)].
dx
To compute this fairly painlessly, using logarithmic differentiation yields
1
[
P (0) = lim
x→0 2013! · (x − 1)(x − 2)...(x − 2013) · 1
x − 1 + 1
x − 2 + ... + 1
]
.
x − 2013
Finally, we conclude that P (0) = 1 + 1 2 + ... + 1
2013 .

8. For any n > 0, evaluate
∫ ∞
0
⌊x⌋e −nx dx.
Solution: Due to the definition of the floor function, split this integral at every
∫ ∞
∞∑
∫ k+1
integer value: ⌊x⌋e −nx dx = k · e −nx dx.
This equals
0
k=1
∞∑
k(e −nk − e −n(k+1) ) = (1 − e −n ) ·
k=1
k
∞∑
k=1
k
e nk =
e n
(e n − 1) 2 .
To evaluate this latter sum, differentiate the geometric series and then multiply both
x
sides by x:
(1 − x) = ∑ ∞
kx k . Finally, let x = 1 2 e . n
k=1
9. Show that 14x 2 + 15y 2 = 7 2012 has no integer solutions.
Solution: Show that such an integer solution exists. Since 14x 2 and 7 2012 are both
divisible by 7, we see that y is divisible by 7. Writing y = 7m for some integer m, the
equation simplifies to 2x 2 + 15 · 7m 2 = 7 2011 . As before, now we see that x is divisible
by 7. So writing x = 7n for some integer n, we obtain 14n 2 + 15m 2 = 7 2010 , which is
just like the original equation, but with the power of 7 decreased by 2. So, we can
repeat the above argument until we reduce the equation to 14r 2 + 15s 2 = 1 for some
integers r and s. However, this equation clearly has no integer solutions, and so we
have reached a contradiction.