Let \(f(x)\) be defined and positive for \(x > 0\).
Let \(a\) and \(b\) be real numbers with \(0 < a < b\) and define the points \(A = (a, f(a))\) and \(B = (b, -f(b))\).
Let \(X = (m,0)\) be the point of intersection of line \(AB\) with the \(x\)-axis.
Find an expression for \(m\) in terms of \(a\), \(b\), \(f(a)\) and \(f(b)\).
Show that, if \(f(x) = \sqrt{x}\), then \(m = \sqrt{ab}\).
Find, in terms of \(n\), \(a\) function \(f(x)\) such that \(m = \frac{a^{n+1} + b^{n+1}}{a^n + b^n}\).
Let \(g_1(x)\) and \(g_2(x)\) be defined and positive for \(x > 0\). Let \(m = M_1\) when \(f(x) = g_1(x)\) and let \(m = M_2\) when \(f(x) = g_2(x)\).
Show that if \(\frac{g_1(x)}{g_2(x)}\) is a decreasing function then \(M_1 > M_2\).
Hence show that
$$\frac{a+b}{2} > \sqrt{ab} > \frac{2ab}{a+b}$$
Let \(p\) and \(c\) be chosen so that the curve \(y = p(c-x)^3\) passes through both \(A\) and \(B\). Show that
$$\frac{c-a}{b-c} = \left(\frac{f(a)}{f(b)}\right)^{1/3}$$
and hence determine \(c\) in terms of \(a\), \(b\), \(f(a)\) and \(f(b)\).
Show that if \(f\) is a decreasing function, then \(c < m\).
Solution:
The line \(AB\) has equation:
\begin{align*}
&& \frac{y+f(b)}{x-b} &= \frac{f(a)+f(b)}{a-b} \\
\Rightarrow && \frac{f(b)}{m-b} &= \frac{f(a)+f(b)}{a-b} \\
\Rightarrow && m &= \frac{a-b}{f(a)+f(b)}f(b) + b \\
&&&= \frac{af(b)+bf(a)}{f(a)+f(b)}
\end{align*}
Suppose \(f(x) = \sqrt{x}\) then
\begin{align*}
m &= \frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}} \\
&= \frac{\sqrt{ab}(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}} \\
&= \sqrt{ab}
\end{align*}
Suppose \(f(x) = x^{-n}\) then
\begin{align*}
m &= \frac{a b^{-n}+ba^{-n}}{a^{-n}+b^{-n}} \\
&= \frac{a^{n+1}+b^{n+1}}{b^n + a^n} \\
\end{align*}
Without loss of generality, we can scale \(g_1(x)\) and \(g_2(x)\) so that \(g_1(a) = g_2(a)\) and \(m\) won't change for either of them. Then since \(\frac{g_1(b)}{g_2(b)} < 1\) (this function is decreasing) our line connecting \((a,g_i(a))\) and \((b,-g_i(b))\) must interect the axis first for \(g_2\), in particular \(M_1 > M_2\).
Suppose \(g_1(x) =1, g_2(x) = \sqrt{x}, g_3(x) = x^{-1}\), the notice that \(\frac{g_1(x)}{g_2(x)} =\frac{g_2(x)}{g_3(x)}= x^{-1/2}\) are decreasing, therefore:
\begin{align*}
\frac{a+b}{1+1} &> \sqrt{ab} > \frac{1+1}{a^{-1}+b^{-1}} \\
\frac{a+b}{2} &> \sqrt{ab} > \frac{2ab}{a+b} \\
\end{align*}
We must have:
\begin{align*}
&& p(c-a)^3 &= f(a) \\
&& p(c-b)^3 &= -f(b) \\
\Rightarrow &&\left ( \frac{c-a}{c-b} \right)^3 &= -\frac{f(a)}{f(b)} \\
\Rightarrow && \frac{c-a}{b-c} &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \\
\Rightarrow && c-a &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}(b-c)\\
\Rightarrow && c \left (1 + \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \right) &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}b + a \\
\Rightarrow && c &= \frac{\left (\frac{f(a)}{f(b)} \right)^{\tfrac13}b + a}{1 + \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}} \\
&&&= \frac{b[f(a)]^\tfrac13+a[f(b)]^\tfrac13}{[f(a)]^\tfrac13+[f(b)]^\tfrac13}
\end{align*}
We have that \(\frac{c-a}{b-c} = \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \) and \(\frac{m-a}{b-c} = \frac{f(a)}{f(b)}\). Since \(f\) is decreasing, \(\frac{f(a)}{f(b)} > 1\) and so \(\left (\frac{f(a)}{f(b)} \right)^{\tfrac13} < \frac{f(a)}{f(b)}\), therefore \(m > c\).
The numbers \(f(r)\) satisfy \(f(r)>f(r+1)\) for $r=1, 2,
\dots\(. Show that, for any non-negative integer \)n$,
\[
k^n(k-1) \, f(k^{n+1}) \le \sum_{r=k^n}^{k^{n+1}-1}f(r) \le k^n(k-1)\,
f(k^n)\,
\]
where \(k\) is an integer greater than 1.
By taking \(f(r) = 1/r\), show that
\[
\frac{N+1}2 \le \sum_{r=1}^{2^{N+1}-1} \frac1r \le N+1 \,.
\]
Deduce that the sum \(\displaystyle \sum_{r=1}^\infty \frac1r\) does not converge.
By taking \(f(r)= 1/r^3\), show that
\[
\sum_{r=1}^\infty \frac1 {r^3} \le 1 \tfrac 13 \,.
\]
Let \(S(n)\) be the set of positive integers less than \(n\) which do not have a \(2\) in their decimal representation and let \(\sigma(n)\) be the sum of the reciprocals of the numbers in \(S(n)\), so for example \(\sigma(5) = 1+\frac13+\frac14\). Show that \(S(1000)\) contains \(9^3-1\) distinct numbers.
Show that \(\sigma (n) < 80\) for all \(n\).
Notice that if \(f(r) = 1/r\) then \(f(r) > f(r+1)\) so we can apply our lemma, ie
\begin{align*}
&&&2^N(2-1) \frac{1}{2^{N+1}} &\leq & \sum_{r=2^N}^{2^{N+1}-1} \frac1r &\leq&\quad 2^N(2-1) \frac{1}{2^{N}} \\
\Leftrightarrow &&& \frac12 &\leq &
\sum_{r=2^N}^{2^{N+1}-1} \frac1r &\leq&\quad 1 \\
\Rightarrow &&& \frac12+\frac12+\cdots+\frac12 &\leq &
\underbrace{\sum_{r=2^0}^{2^{0+1}-1} \frac1r+\sum_{r=2^1}^{2^{1+1}-1} \frac1r+\cdots+\sum_{r=2^N}^{2^{N+1}-1} \frac1r}_{N+1 \text{ terms}} &\leq&\quad 1 +1+\cdots+1\\
\Rightarrow &&& \frac{N+1}{2} &\leq &
\underbrace{\sum_{r=1}^{2^{N+1}-1} \frac1r}_{N+1 \text{ terms}} &\leq&\quad N+1
\end{align*}
Therefore the sum \(\displaystyle \sum_{r=1}^{2^{N+1}-1} \frac1r\) is always greater than \(N+1\) and in particular we can find an upper limit such that it is always bigger than any value, ie it diverges.
To count the number of numbers less than \(1000\) without a \(2\) in their decimal representation we can count the number of \(3\) digit numbers (where \(0\) is an acceptable leading digit) which don't contain a \(2\) and remove \(0\). There are \(9\) choices for each digit, so \(9^3-1\). Notice this is true for \(10^N\) for any \(N\), ie \(S(10^N) = 9^N-1\).
Notice also that we can now write:
\begin{align*}
&& \sum_{r=10^N }^{10^{N+1}-1} \frac{1}{r} \mathbb{1}_{r \in S} & < \frac{1}{10^{N+1}}\#\{\text{number not containing a }2\} \\
&&&= \frac{1}{10^{N+1}}((9^{N+1}-1)-(9^N-1)) \\
&&&= \frac{9^N}{10^N}(9-1) \\
&&&= 8 \cdot \left (\frac9{10} \right)^N \\
\\
\Rightarrow && \sum_{r=1}^{\infty} \frac{1}{r} \mathbb{1}_{r \in S} &< 8\left ( 1 + \frac9{10} + \cdots \right) \\
&&&= 8 \frac{1}{1-\frac{9}{10}} = 80
\end{align*}