Tag: Pi

Compute π using the Taylor polynomial of arctan x

by RoRi

December 21, 2015

For this exercise define

$\alpha = \arctan \frac{1}{5}, \qquad \beta = 4 \alpha - \frac{1}{4} \pi.$

Using the trig identity
$\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

twice, once with $A = B = \alpha$ , and then the second time with $A = B = 2 \alpha$ , show that

$\tan (2 \alpha) = \frac{5}{12}, \qquad \tan (4 \alpha) = \frac{120}{119}.$

Then use the same identity again with $A = 4 \alpha$ and $B = -\frac{1}{4} \pi$ to show

$\tan \beta = \frac{1}{239}.$

This establishes the identity

$\pi = 16 \arctan \frac{1}{5} - 4 \arctan \frac{1}{239}.$
Using the Taylor polynomial approximation $T_{11} (\arctan x)$ at $x =\frac{1}{2}$ prove that
$3.158328934 < 16 \arctan \frac{1}{5} < 3.158328972.$
Using the Taylor polynomial approximation $T_3 (\arctan x)$ at $x = \frac{1}{239}$ prove that
$-0.016736309 < -4\arctan \frac{1}{239} < -0.016736300.$
Using the above parts show that the value of $\pi$ to seven decimal places is 3.1415926.

Proof. Letting $A = B = \alpha = \arctan \frac{1}{5}$ we have
$\tan (A+B) = \tan (2 \alpha) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{(2/5)}{1 - (1/25)} = \frac{5}{12}.$

Letting $A = B = 2 \alpha = 2 \arctan \frac{1}{5}$ we have

$\tan (A+B) = \tan (4 \alpha) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{(10/12)}{1 - (25/144)} = \frac{120}{119}.$

Letting $A = 4 \alpha$ and $B = -\frac{\pi}{4}$ we have (recalling that $\beta = 4 \alpha - \frac{\pi}{4}$ )

$\tan (\beta) = \frac{\tan (4\alpha) + \tan\left(-\frac{\pi}{4}\right)}{1 - \tan (4\alpha) \tan \left( -\frac{\pi}{4} \right)} = \frac{(120/119) - 1}{1 - (120/119)(-1)} = \frac{1}{239}.$

But then

$\begin{align*} && \tan \left( 4 \alpha - \frac{\pi}{4} \right) &= \frac{1}{239}\\[9pt] \implies && 4 \alpha - \frac{\pi}{4} &= \arctan \left( \frac{1}{239} \right) \\[9pt] \implies && \frac{\pi}{4} &= 4 \arctan \frac{1}{5} - \arctan \frac{1}{239} \\[9pt] \implies && \pi &= 16 \arctan \frac{1}{5} - 4 \arctan \frac{1}{239}. \qquad \blacksquare \end{align*}$
Proof. We know the Taylor polynomial approximation for $\arctan x$ from this exercise (Section 7.8, Exercise #3):
$T_{11} (\arctan x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11}.$

Therefore, we can compute an approximation to $\arctan x$ ,

$\begin{align*} &&\arctan x &= \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} \right) + E_{10}(x) \\[9pt] \implies && 16 \arctan \left(\frac{1}{5} \right) &= 16 \left( \frac{1}{5} - \frac{1}{375} + \frac{1}{15625} - \frac{1}{546875} + \frac{1}{17578125} - \frac{1}{537109375} \right) + E_{10}\left( \frac{1}{5} \right)\\[9pt] \implies && 16 \arctan \left( \frac{1}{5} \right) &= 3.158328957 + E_{10} \left( \frac{1}{5} \right), \end{align*}$

where

$\left|E_{10}\left(\frac{1}{5} \right)\right| \leq \frac{(1/5)^{11}}{11} = .000000018.$

Therefore,

$3.158328934 < 16 \arctan \left( \frac{1}{5} \right) < 3.158328972. \qquad \blacksquare$
Proof. Again using the Taylor polynomial approximation to $\arctan x$ we have
$\begin{align*} &&T_3 (\arctan x) &= x - \frac{x^3}{3} \\[9pt] \implies && \arctan x &= x - \frac{x^3}{3} + E_2 (x), \qquad |E_2 (x)| \leq \frac{x^3}{3} \\[9pt] \implies && \arctan \left( \frac{1}{239} \right) &= \left( \frac{1}{239} \right) - \frac{(1/239)^3}{3} + E_2 \left( \frac{1}{239} \right) \\[9pt] \implies && -4 \arctan \left( \frac{1}{239} \right) &= -0.01673630401 + E_2 \left( \frac{1}{239} \right) \\[9pt] \implies && -0.016736309 &< -4 \arctan \left( \frac{1}{239} \right) < -0.016736300. \qquad \blacksquare \end{align*}$
Finally,
$\begin{align*} \pi &= 16 \arctan \frac{1}{5} - 4 \arctan \frac{1}{239} \\ &\approx 3.158328957 - 0.01673630401 \\ &\approx 3.1415926. \end{align*}$

Prove formulas for shifting sine and cosine by π

by RoRi

July 31, 2015

Prove

$\sin (x + \pi) = - \sin x \qquad \text{and} \qquad \cos (x+\pi) = -\cos x$

for all $x \in \mathbb{R}$ .

Proof. First,

$\sin (x+ \pi) = \sin \left( x + \frac{\pi}{2} + \frac{\pi}{2} \right) = \cos \left( x + \frac{\pi}{2} \right) = - \sin x.$

And then,

$\cos (x+ \pi) = \cos \left( x + \frac{\pi}{2} + \frac{\pi}{2}\right) = - \sin \left( x + \frac{\pi}{2} \right) = - \cos x. \qquad \blacksquare$

Use integrals to establish the formula for the area of an ellipse

by RoRi

July 30, 2015

Denoting the unit circle, $x^2 + y^2 = 1$ , by $C$ we define an ellipse $E$ to be the set of points

$E = \{ (ax, by) \mid (x,y) \in C, a > 0, b > 0 \}.$

Show that the points on $E$ satisfy the equation:
$\left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1.$
Prove that the area of $E$ is measurable and that
$a(E) = \pi a b.$

Proof. If $(x,y)$ is a point on $E$ then $\left( \frac{x}{a}, \frac{y}{b} \right)$ is a point on $C$ (since all points of $E$ are obtained by taking a point of $C$ and multiplying the $x$ -coordinate by $a$ and the $y$ -coordinate by $b$ ). By definition of $C$ , we must then have,
$\left( \frac{x}{a} \right)^2 + \left(\frac{y}{b} \right)^2 = 1. \qquad \blacksquare$
Proof. From part (a) we know $E$ is the set of points $(x,y)$ such that $\left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1$ . But, this implies,
$y = b \cdot \sqrt{ 1 - \left( \frac{x}{a} \right)^2}, \qquad \text{or} \qquad y = -b \cdot \sqrt{1 - \left( \frac{x}{a} \right)^2}.$

Hence, the area of $E$ is the area enclosed from $-a$ to $a$ by the graphs of

$g(x) = b \cdot \sqrt{1 - \left( \frac{x}{a} \right)^2} \qquad \text{and} \qquad f(x) = -b \cdot \sqrt{1 - \left( \frac{x}{a} \right)^2}.$

To show this region is measurable and has area $\pi a b$ we begin with this identity (from Apostol, 2.4 Exercise 17),

$\begin{align*} \pi = 2 \int_{-1}^1 \sqrt{1-x^2} && \implies && \pi b &= 2 \int_{-1}^1 b \sqrt{1-x^2} \, dx \\ && \implies && \pi a b &= 2a \int_{-1}^1 b \sqrt{1-x^2} \, dx \\ && \implies && \pi a b &= 2 \int_{-a}^a b \sqrt{1 - \left( \frac{x}{a} \right)^2} \, dx \\ && \implies && \pi ab &= \int_{-a}^a \left( b \sqrt{1 - \left( \frac{x}{a} \right)} - \left(-b \sqrt{1 - \left( \frac{x}{a} \right)^2} \right)\right) \, dx. \end{align*}$

In the second to last line we have used the expansion/contraction of the interval of integration. Hence, we know the integral from $-a$ to $a$ of $g(x) - f(x)$ exists and has value $\pi a b$ . Thus, $E$ is measurable and $a(E) = \pi a b. \qquad \blacksquare$

Use dodecagons to deduce an inequality about π

by RoRi

July 30, 2015

By considering dodecagons inscribed and circumscribed about a unit disk, establish the inequalities

$3 < \pi < 12(2-\sqrt{3}).$

First, we draw some pictures of the situation for reference.

Rendered by QuickLaTeX.com

( Note: I don’t know a way to do this without using trig functions, which haven’t been introduced in the text yet. If you have an alternative approach without them, please leave a comment and let us know about it. )

Since these are dodecagons, the angle at the origin of the circle of each triangular sector is $2 \pi / 12 = \pi/6$ , and the angle of the right triangles formed by splitting each of these sectors in half (shown in the diagrams) is then $\pi/12$ . Then we use the fact that

$\tan \left( \frac{\pi}{12} \right) = 2 - \sqrt{3},$

$\sin \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} - 1}{2 \sqrt{2}},$

$\cos \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} + 1}{2 \sqrt{2}}.$

Now, for the circumscribed dodecagon we have the area of the right triangle $T$ with base 1 in the diagram on the left given by

$a(T) = \frac{1}{2} bh = \frac{1}{2} \cdot 1 \cdot (2 - \sqrt{3}) = 1 - \frac{\sqrt{3}}{2}.$

Since there are 24 such triangles in the dodecahedron, we then have the area of the circumscribed dodecahedron $D_c$ given by

$a(D_c) = 24 \left(1 - \frac{\sqrt{3}}{2} \right) = 12 (2 - \sqrt{3}).$

For the inscribed dodecagon we consider the right triangle $T$ with hypotenuse 1 in the diagram. The length of one of the legs is then given by $\sin \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} - 1}{2}$ and the other is given by $\cos \left( \frac{\pi}{12} \right)$ . So the area of the triangle is

$a(T) = \frac{1}{2} bh = \frac{1}{2} \cdot \frac{\sqrt{3}-1}{2 \sqrt{2}} \cdot \frac{\sqrt{3}+1}{2 \sqrt{2}} = \frac{2}{16} = \frac{1}{8}.$

Since there are 24 such triangles in the inscribed dodecahedron, $D_{i}$ we then have,

$a(D_i) = 24 \cdot \frac{1}{8} = 3.$

Since the area of the unit circle is, by definition, $\pi$ , and it lies in between these two dodecahedrons, we have,

$3 < \pi < 12(2 - \sqrt{3}). \qquad \blacksquare$

Compute some integrals in terms of π

by RoRi

July 29, 2015

Recall we have defined $\pi$ as the area of the unit disk. We have proved (Apostol, Section 2.3, Example 3) that

$\pi = 2 \int_{-1}^1 \sqrt{1-x^2} \, dx.$

Use this and the theorems on the property of integrals to compute the following in terms of $\pi$ :

$\displaystyle{\int_{-3}^3 \sqrt{9-x^2} \, dx}$ .
$\displaystyle{\int_0^2 \sqrt{1-\frac{1}{4}x^2} \, dx}$ .
$\displaystyle{\int_{-2}^2 (x-3) \sqrt{4-x^2} \, dx}$ .

Using the expansion/contraction property of the integral (Apostol, Theorem 1.19) we compute:
$\begin{align*} \int_{-3}^3 \sqrt{9-x^2} \, dx &= 3 \int_{-1}^1 \sqrt{9 - (3x)^2} \, dx & (\text{Exp/Cont property}) \\ &= 9 \int_{-1}^1 \sqrt{1-x^2} \, dx & (\text{Factoring out a constant}) \\ &= \frac{9}{2} \pi &(\text{Expression for } \pi \text{ given}). \end{align*}$
Using the expansion/contraction property of the integeral (Apostol, Theorem 1.19) we compute:
$\begin{align*} \int_0^2 \sqrt{1 - \frac{1}{4}x^2} \, dx &= 2 \int_0^1 \sqrt{1 - x^2} \, dx \\ &= \int_{-1}^1 \sqrt{1-x^2} \, dx \\ &= \frac{\pi}{2} \end{align*}$
Using the linearity properties of the integral, as well as, expansion/contraction and the formula for $\pi$ given above. We also use this exercise (Apostol, Section 1.25, Exercise #26 part (b)) which establishes that the integral from $-b$ to $b$ of an odd function is 0.
$\begin{align*} \int_{-2}^2 (x-3) \sqrt{4 - x^2} \, dx &= \int_{-2}^2 x \sqrt{4-x^2} \, dx - 3 \int_{-2}^2 \sqrt{4-x^2} \, dx \\ &= - 6 \int_{-1}^1 \sqrt{4-4x^2} \, dx & (x \sqrt{4-x^2} \text{ is an odd function})\\ &= -12 \int_{-1}^1 \sqrt{1-x^2} \, dx \\ &= -6 \pi. \end{align*}$