Equivalence of $\pi$ is the first positive zero of the taylor series for $\sin(x)$ and $\pi/4 = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$

Question

For $x\in\mathbb{R}$, define $\sin (x) = x - x^3/3!+x^5/5!-\cdots$ and $\pi = 4(1-\frac{1}{3}+\frac{1}{5} -\frac{1}{7}+\cdots)$. Then show that $\sin(\pi/2) = 1$

In the prologue of Real and Complex Analysis by Walter Rudin, pi is defined as the first positive zero of the series defined as $\sin(x)$; I want to check that pi is same as above defined Pi.

Answer 1

first attempt: i don't have an answer but here are some thoughts on this problem. Let $$S = x - x^3/3! + x^5/5!-\cdots \mbox{ and }C = 1 - x^2/2! + x^4/4! - \cdots $$

(a) derive differential equations $\frac{dS}{dx} = C, \frac{dC}{dx} = -S$

(b) show $S^2 + C^2 = 1$

(c) establish that $C > 0, S > 0 \mbox{ for some $\delta > 0$ and } 0 < x < \delta $

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i don't see how to use the series for $\pi.$

here is my second attempt at using the series for $\pi.$ define a function $t$ by $$ t = \int_0^m {du \over 1 + u^2}, -\infty < m < \infty$$ we will first establish:

(a) $\ t(\infty) = 2t(1)$

(b) $\ t(1) = \pi/4, t(\infty) = \pi/2$

(c) $\ t$ is an odd function and is increasing on $(-\infty, \infty)$

(d) $\displaystyle {dt \over dm} = {1 \over 1 + m^2}$

proof of (a):

$\begin{eqnarray} t(\infty) &=&\int_0^\infty {du \over 1 + u^2}= t(1) + \int_1^\infty{du \over 1 + u^2} = f(1) + \int_1^0 {d(1/u) \over 1 + (1/u)^2}\\ &=& t(1) + \int_0^1 {du \over 1 + u^2} = 2t(1). \end{eqnarray}$

$$t(1) = \int_0^1 {du \over 1 + u^2} = \int_0^1 (1 - u^2 + u^4 + \cdots) du = \{u - u^3/3 + u^5/5 + \cdots \}|_0^1 = \pi/4$$

Now we define $m$ on $-\pi/2, \pi/2)$ as the inverse function of $t.$ then $m$ has the following properties:

(a) $m$ is odd and increasing on $(-\pi/2, \pi/2)$

(b) $m(0) = 0, m(\pi/4) = 1, m(\pi/2 -) = \infty$

(c) $\displaystyle {dm \over dt} = 1 + m^2$

now define $$c = {1 \over \sqrt{ 1 + m^2}} \mbox{ on the interval } 0 \le t < \pi/2 \mbox{ set } c(\pi/2) = 0 \mbox{ to make it continuous at } t = \pi/2. $$

My aim is to show the generic properties of $c = \cos$ and to find the Mclaurin series expansion about $x = 0$

I will get back to it later, but if someone wants to continue they are welcome to do so.

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The properties of $m$ makes $$c(0) = 1, c(\pi/4) = 1/\sqrt 2, c(\pi/2) = 0, c \mbox{ is decreasing } 0 \le x \le \pi/2.$$

We will establish the Mclaurin series of $c$ by finding second order differential equation satisfied by $c.$

Differentiating $ c^2(1+m^2) = 1$ and using ${dm \over dt} = 1 + m^2,$ we find that $c$ satisfies the differential equation $${dc \over dt} = -mc$$ We also have $\frac{dc}{dt} = 0$ at $t = 0.$ Differentiating once more

$$ {d^2c \over dt^2 } = -c, \ c(0) = 1,\ c^\prime(0) = 0 $$ which gives the series $$c = 1 - t^2/2! + t^4/4! + \cdots $$ and we have shown that $$c(\pi/2) = 0$$ where $\pi$ is defined by $$\pi/4 = 1 - 1/3 + 1/5 + \cdots. $$

If one needs things about $\sin$ define $\sin = \sqrt{1-c^2}$ and derive a similar second order differential equation for $s$.

Answer 2

Here's a detailed outline for how to do this in a fairly standard line of development of trig functions from their power series definitions.

(A "plug in" combinatorial proof would be neat to see as well, of course.)

Step 1. Show the basic trig identities (double angle formulas and $\sin(x)^2 + \cos(x)^2 = 1$) for the taylor series definitions of $\sin(x)$ and $\cos(x)$.

[The double angle formulas are nice to prove using the power series for $e^x$. The second is immediate upon differentiation.]

Step 2. Let $pi$ (not the greek symbol for now) be the first root of $\sin(x)$. From the derivative of $\sin(x)$ at zero and the intermediate value theorem, we get that $\sin(x) > 0$ for $0 < x < pi$.

Step 3. Use trig identities to show that $\sin(2x) = 0$ if and only if $\sin(x) = 0$ or $\cos(x) = 0$. Conclude that $x = pi/2$ is the first time $\cos(x) = 0$.

Step 4. Let $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Show that $\frac{d}{dx} \tan(x) = \frac{1}{\cos(x)^2}$. Conclude that $\tan(x)$ is well-defined, increasing, and unbounded on $(-pi/2,pi/2)$.

Step 5. Define $\arctan(x)$ to be the inverse of $\tan(x): (-pi/2,pi/2) \rightarrow \mathbb{R}$. Show that $\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$ from trig identities.

Step 6. Show that $\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$ from the series expansion for $\frac{1}{1+x^2}$ and differentiation of power series theorems and the fact that $\tan(0) = 0$. This converges and hence is valid on $(-1,1)$.

Step 7. The series also converges for $x = 1$ and hence (by a theorem on power series convergence), we get $\arctan(1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots$. Thus it remains to show $2 \arctan(1) = pi/2$.

Step 8. Show from trig identities that $\tan(x) = \pm 1$ if and only if $\sin(x) = \pm \cos(x)$ if and only if $\cos(2x) = 0$. Conclude that $2 \arctan(1) = pi/2$, as desired (because $2 \arctan(1)$ must be the first zero of $\cos(x)$).