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