Which of the following statements are true
and which are false? Justify your answers.
\(a^{\ln b}=b^{\ln a}\) for all \(a,b>0\).
\(\cos(\sin\theta)=\sin(\cos\theta)\) for all real \(\theta\).
There exists a polynomial \(\mathrm{P}\) such that
\(|\mathrm{P}(\theta)-\cos\theta|\leqslant 10^{-6}\)
for all real \(\theta\).
\(x^{4}+3+x^{-4}\geqslant 5\) for all \(x>0\).
Solution:
True. \begin{align*}
&& \ln a \cdot \ln b &= \ln b \cdot \ln a \\
\Leftrightarrow && \exp ( \ln a \cdot \ln b) &= \exp ( \ln b \cdot \ln a) \\
\Leftrightarrow && \exp ( \ln a )^{\ln b} &= \exp ( \ln b )^{\ln a} \\
\Leftrightarrow && a^{\ln b} &= b^{\ln a} \\
\end{align*}
False. Consider \(\theta = 0\). We'd need \(\cos 0 = 1 = \sin 1\), but \(0 < 1 < \frac{\pi}{2}\) so \(\sin 1 \neq 1\)
False. If the polynomial has positive degree, then as \(n \to \infty\), \(\P(x) \to \pm \infty\), in particular it must be well outside the interval \([-1,1]\). Therefore it can't be within \(10^{-6}\) of \(\cos \theta\) which is confined to that interval. The only polynomial which is restricted to that range are constants, but then \(|\cos 0 - c| \leq 10^{-6}\) and \(|\cos \pi - c| \leq 10^{-6}\) \(2 = |1-(-1)| \leq |1-c| + |-1-c| \leq 2\cdot 10^{-6}\) contradiction.