pi: the Equivalent Characterizations: Examples and Counterexamples

At $r=1$, [L1] gives semicircle length $\pi$.

At $r=1$, [L2] gives circumference $2\pi$ and diameter $2$, so their ratio is $\pi$.

At $r=1$, [L3] gives disc area $\pi$.

These three substitutions give every entry and normalization in the table.

Multiplying [L1] by $4$ gives $\mathrm{\mid }\pi -4S_N\mathrm{\mid }\le 4\mathrm{/}(2N+3)$.

If $N$ is even, the remainder in [L1] has negative sign, so $\pi \le 4S_N$; if $N$ is odd, it has positive sign, so $4S_N\le \pi$.

Direct summation gives $S_0=1,S_1=\frac{2}{3},S_2=\frac{13}{15},S_3=\frac{76}{105}\mathrm{.}$

Hence the first certified brackets include $\frac{8}{3}<\frac{304}{105}\le \pi \le \frac{52}{15}<4,$ with the weak inequalities adjacent to $\pi$ supplied by step 1.2.

The error bound decreases only on the scale $1\mathrm{/}N$, so the certification also records the slow convergence of these partial sums.

Dividing $I_{2m+1}\le I_{2m}$ in [L1] by the positive odd-over-even product gives $W_m\le \pi \mathrm{/}2$.

Using the formula for $I_{2m-1}$ obtained from [L1] with $m-1$, the other inequality $I_{2m}\le I_{2m-1}$ gives $\frac{\pi }{2}\frac{2m}{2m+1}\le W_m\mathrm{.}$

Thus $\frac{\pi }{2}\frac{2m}{2m+1}\le W_m\le \frac{\pi }{2},$ which is the exact finite trap coming from the adjacent integrals.

Multiplication gives the three displayed values; for $m=1,2,3$, step 2.1 respectively places them in $\frac{\pi }{3}\le \frac{4}{3}\le \frac{\pi }{2},\frac{2\pi }{5}\le \frac{64}{45}\le \frac{\pi }{2},\frac{3\pi }{7}\le \frac{256}{175}\le \frac{\pi }{2}\mathrm{.}$

The exact bounds in step 2.1 certify every finite product without decimal approximations, while [L2] supplies their limit $\pi \mathrm{/}2$.

Starting with $a_1=\sqrt{2}$, the recurrence gives $a_2=\sqrt{2+\sqrt{2}},a_3=\sqrt{2+\sqrt{2+\sqrt{2}}},a_4=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}\mathrm{.}$

Since every $a_k$ is positive, each $P_n>0$, and ${\mathrm{\Pi }}_n:=\frac{2}{P_n}=\frac{2^{n+1}}{a_1{a}_2\cdots a_n}$ is defined.

Substitution into ${\prod }_{k=1}^n(a_k\mathrm{/}2)$ gives exactly the four products displayed in the Example, as also identified by [L1].

By [L2] and [L3], the nonzero-limit quotient law gives ${\mathrm{\Pi }}_n\to 2\mathrm{/}(2\mathrm{/}\pi )=\pi$. Thus doubling the reciprocals of the displayed finite products gives certified approximants to $\pi$, with convergence supplied by the finite identity rather than a numerical pattern.

Put $p=2\pi$. By [L1], $p>0$, and by [L3], $\sin p=0$.

Yet $2\pi \mathrm{\ne }\pi$ because $\pi >0$.

Thus a positive zero need not equal $\pi$. The correct characterization in [L2] requires the least positive zero.

Since $r>0$, division in [L1] gives $C(r)\mathrm{/}r=2\pi$.

By [L2], $2\pi \mathrm{\ne }\pi$, so the asserted ratio is false.

The valid normalization is circumference divided by diameter: [L1] gives $C(r)\mathrm{/}(2r)=\pi$. The false statement omits this factor of $2$.

Each point $\gamma (s)$, $0\le s\le 2\pi$, equals $\beta (s\mathrm{/}2)$, while periodicity in [L3] reduces every $\beta (t)=\gamma (2t)$ to a parameter in $[0,2\pi ]$. Thus $\gamma$ and $\beta$ have the same unit-circle trace.

By [L2], $\parallel {\gamma }^{\mathrm{\prime }}(t){\parallel }_2=1$ and $\parallel {\beta }^{\mathrm{\prime }}(t){\parallel }_2=2$ throughout the interval.

By [L4] and [L5], $L(\gamma )={\int }_0^{2\pi }1dt=2\pi ,L(\beta )={\int }_0^{2\pi }2dt=4\pi \mathrm{.}$

The traces coincide by step 1.1 but the lengths differ by step 2.1, so the statement is false. The once-around qualification in the definition of circumference prevents this multiplicity ambiguity.