The function \(\f\) satisfies \(\f(x+1)= \f(x)\) and \(\f(x)>0\) for all \(x\).
Give an example of such a function.
The function \(\F\) satisfies
\[
\frac{\d \F}{\d x} =\f(x)
\]
and \(\F(0)=0\).
Show that \(\F(n) = n\F(1)\), for any positive integer \(n\).
Let \(y\) be the solution of the differential equation
\[
\frac{\d y}{\d x} +\f(x) y=0
\]
that satisfies \(y=1\) when \(x=0\). Show that \(y(n) \to 0\) as \(n\to\infty\), where
\(n= 1,\,2,\, 3,\, \ldots\)
Solution:
\(f(x) = \lfloor x \rfloor+1\)
Clearly \(\displaystyle F(x) = \int_0^x f(t) \d t\), in particular:
\begin{align*}
&& F(n) &= \int_0^n f(t) \d t \\
&&&= \sum_{i=1}^n \int_{i-1}^i f(t) \d t \\
&&&= \sum_{i=1}^n \int_{0}^1 f(t-i+1) \d t \\
&&&= \sum_{i=1}^n \int_{0}^1 f(t) \d t \\
&&&= n \int_{0}^1 f(t) \d t\\
&&&= n F(1)
\end{align*}
\(\,\)
\begin{align*}
&& 0 &= \frac{\d y}{\d x} +f(x) y \\
\Rightarrow && \int -f(x) \d x &= \int \frac1y \d y\\
\Rightarrow && -F(x) & = \ln y + C \\
x=0,y=1: && C &= -F(0) \\
\Rightarrow && y &= \exp(F(0)-F(x))
\end{align*}
Well this \(F(0)-F(x)\) is equivalent to \(-F(x)\) where \(F(0) = 0\), in particular \(F(n) = nF(1)\), so
\(y(n) = e^{-nF(1)}\) which tends to zero as long as \(F(1) > 0\), but since \(f(x) > 0\) for all \(x\) this must be true.