I'm working on the following problem from Spivak, and would like to clarify a few points:
For $\text{(b)}$, what I've done is to show that $\dfrac{\ln(\pi)}{\pi}$ is less than $\dfrac{1}{e}$, as in $\text{(a)}$, it has been shown that the maximum value of $\dfrac{\ln(x)}{x}$ is at $(e, 1/e)$. Rearranging this, we have that $\ln(\pi^e) < \pi$. Since $e^x$ is an increasing function, $\ln(\pi^e) < \pi$ implies $\pi^e < e^\pi$. The problem with this though, is that it rests on the assumption that $e$ is not equal to $\pi$. Spivak defines $e$ as the value such that the integral of $\dfrac{1}{x}$ equals $1$ on $[1, e]$, and $\pi$ as the two times the value of the integral of $\sqrt{1 -x^2}$ on $[1, -1]$. It's not immediately apparent that they are different values, so am I allowed to assume they are different values? And if I'm not allowed to assume this, could you avoid giving a hint on how to show $e$ is not equal to $\pi$ (I would still like to attempt this myself)?
For $\text{(c)}$, am I allowed to assume that both $x$ and $y$ are positive? Spivak has only defined the meaning of $a^x$ for $a>0$.
Thank you so much in advance.
I think that you can assume that $e\approx2.71818$ and that $\pi\approx3.1415$. Anyway, note that\begin{align}\log3&=\int_1^3\frac{\mathrm dt}t\\&>\sum_{k=1}^8\frac14\times\frac1{1+k/8}\\&=\frac{28271}{27720}\\&>1\end{align}and that therefore $3>e$. On the other hand,\begin{align}\pi&=6\arcsin\left(\frac12\right)\\&=6\left(\frac12+\frac16\left(\frac12\right)^3+\frac3{40}\left(\frac12\right)^5+\cdots\right)\\&>6\times\frac12\\&=3.\end{align}So, $\pi>e$.
Yes, you can assume that $x$ and $y$ are positive.