Problem Solving Answers to Problem Set 5

08 July 2012

1. Let f be a real-valued function with n + 1 derivatives at each point of ℝ. Show that for each pair of real numbers a, b, a < b, such that

ln (\frac{f(b) + f′(b) + ⋯ + f^{(n}(b)}{f(a) + f′(a) + ⋯ + f^{(n}(a)}) = b − a

there is a number c in the open interval (a, b) for which

f^{(n+1)}(c) = f(c).

Answer: Let

F (x) = f(x) + f′(x) + ⋯ + f^{(n)}(x).

Then

F′(x) − F (x) = f^{(n+1)}(x) − f(x).

We are told that

\frac{ln F (b) − ln F (a)}{b − a} = 1.

By the Mean Value Theorem,

\frac{ln F (b) − ln F (a)}{b − a} = \frac{F′(c)}{F (c)}

for some c ∈ (a, b).

Thus

\frac{F′(c)}{F (c)} = 1,

ie

F′(c) = F (c),

ie

f^{(n+1)}(c) − f(c).

2. Given one million non-zero digits a_{1}, a_{2}, …, a_{1000000} (ie each a_{i} ∈ {1, 2, …, 9}) show that at most 100 of the million numbers

a_{1}, a_{1}a_{2}, a_{1}a_{2}a_{3}, …, a_{1}a_{2}a_{3} … a_{1000000}

are perfect squares Answer: This is based on an idea that comes up occasionally — squares are quite widely distributed, since there is no square between n^{2} and n^{2} + n, Consider the numbers in the subsequence with an even number of digits, omitting the first, ie

a_{1}a_{2}a_{3}a_{4}, a_{1}a_{2}a_{3}a_{4}a_{5}a_{6}, ….

We claim that there is at most one square in this sequence. For suppose the first square in the sequence is

a_{1}a_{2} … a_{2r} = b^{2};

and suppose there is a later square in the sequence

a_{1}a_{2} … a_{2r} … a_{2r+2s} = c^{2}.

Then

(10^{s}b)^{2} = 10^{2s}b^{2} = a_{1}a_{2} … a_{2r} \overbrace{0 … 0}^{2s 0’s}

Now

(10^{s}b + 1)^{2} = 10^{2s}b^{2} + 2b · 10^{s} + 1

> a_{1}a_{2} … (a_{2r} + 1) \overbrace{0 … 0}^{2s 0’s},

which is not in the sequence. Similarly there is at most one square in the subsequence with an odd number of digits, omitting the first, ie

a_{1}a_{2}a_{3}, a_{1}a_{2}a_{3}a_{4}a_{5}, ….

It follows that there are at most 2 + e squares in the sequence, where e is the number of squares with 1 or 2 digits. Evidently e = 9; so there are at most 11 squares in the sequence.