Tag: Approximation

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,
$\begin{align*} \pi &= 16 \arctan \frac{1}{5} - 4 \arctan \frac{1}{239} \\ &\approx 3.158328957 - 0.01673630401 \\ &\approx 3.1415926. \end{align*}$

Use Taylor polynomials to approximate the integral of sin x / x

by RoRi

December 21, 2015

Using the Taylor polynomial approximation of $\sin x$ find an approximation for the integral

$\int_0^1 \frac{\sin x}{x} \, dx.$

Give an estimate for the error of the approximation. [Define $\frac{\sin x}{x} = 1$ when $x = 0$ .]

We know (from this exercise, Section 7.8, Exercise #1) that the Taylor polynomial approximation of $\sin x$ is given by

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + E_{2n}(x), \qquad \text{where} \quad |E_{2n}(x)| \leq \frac{|x|^{2n+1}}{(2n+1)!}.$

This implies

$\begin{align*} &&\frac{\sin x}{x} &= 1 - \frac{x^2}{3!} + \frac{x^4}{5!} + \frac{E_{2n}(x)}{x} \\[9pt] \implies && \int_0^1 \frac{\sin x}{x} \, dx &= \int_0^1 \left( 1 - \frac{x^2}{3!} + \frac{x^4}{5!} + \frac{E_{2n}(x)}{x} \right) \, dx \\[9pt] &&&= \left( x - \frac{x^3}{3\cdot 3!} + \frac{x^5}{5 \cdot 5!} \right) \Bigr \rvert_0^1 + \int_0^1 \frac{E_{2n}(x)}{x} \, dx \\[9pt] &&&= 0.9461 + R \end{align*}$

where

$|R| \leq \int_0^1 \frac{x^6}{7!} \, dx = \frac{1}{7\cdot 7!} \approx 0.00002.$

Use Taylor polynomials to approximate an integral of sin (x²)

by RoRi

December 21, 2015

Show that
$\sin x = x - \frac{x^3}{3!} + r(x), \qquad \text{where} \quad |r(x)| \leq \frac{1}{2^5 \cdot 5!}$

when $0 \leq x \leq \frac{1}{2}$ .
Using part (a) find an approximation for the integral
$\int_0^{\frac{\sqrt{2}}{2}} \sin (x^2) \, dx.$

Proof. We know from this exercise that
$\sin x = \sum_{k=1}^n \frac{(-1)^{k-1}x^{2k-1}}{(2k-1)!} + E_{2n}(x) \qquad \text{where} \quad |E_{2n}(x)| \leq \frac{|x|^{2n+1}}{(2n+1)!}.$

Therefore, when $0 \leq x \leq \frac{1}{2}$ we have

$\sin x = x - \frac{x^3}{3!} + r(x) \qquad \text{where} \quad |r(x)| \leq \frac{1}{2^5 \cdot 5!}. \qquad \blacksquare$
From part (a) we know
$\begin{align*} \sin x = x - \frac{x^3}{3!} + r(x) && \implies && \sin (x^2) &= x^2 - \frac{x^6}{3!} + r(x^2)\\[9pt] && \implies && \int_0^{\frac{\sqrt{2}}{2}} \sin (x^2) \, dx &= \int_0^{\frac{\sqrt{2}}{2}} \left( x^2 - \frac{x^6}{3!} + r(x^2) \right) \, dx \\[9pt] &&&&&= \left( \frac{x^3}{3} - \frac{x^7}{7 \cdot 3!} \right) \Bigr \rvert_0^{\frac{\sqrt{2}}{2}} + \int_0^{\frac{\sqrt{2}}{2}} r(x^2) \, dx \\[9pt] &&&&&= \frac{55\sqrt{2}}{672} + R \end{align*}$

where

$|R| \leq \int_0^{\frac{\sqrt{2}}{2}} \frac{1}{2^5 \cdot 5!} \, dx = \frac{1}{2^5 \cdot 5!} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{7680}.$

Use Taylor polynomials to approximate the nonzero root of arctan x = x²

by RoRi

December 21, 2015

Show that $r = \frac{\sqrt{21}-3}{2}$ is an approximation of the nonzero root of the equation
$\arctan x = x^2$

using the cubic Taylor polynomial approximation to $\arctan x$ .
Given that
$\sqrt{21} < 4.6 \qquad \text{and} \qquad 2^{16} = 65536$

prove that the number $r$ from part (a) satisfies

$|r^2 - \arctan x| < \frac{7}{100}.$

Determine if $(r^2 - \arctan r)$ is positive or negative and prove the result.

Proof. From a previous exercise (Section 7.8, Exercise #3) we know
$\arctan x = x - \frac{x^3}{3} + E_{2n}(x).$

So, to approximate the nonzero root of $x^2 - \arctan x$ we have

$\begin{align*} x^2 - x + \frac{x^3}{3} \approx 0 && \implies && x^2 + 3x - 3 &\approx 0 \\ && \implies && x &\approx \frac{\sqrt{21}-3}{2}. \qquad \blacksquare \end{align*}$
We know from the same previous exercise we used in part (a) that the error term $E_{2n}(x)$ for $\arctan x$ satisfies the inequality
$|E_{2n}(x)| \leq \frac{x^{2n+1}}{2n+1}.$

Using the values for $\sqrt{21}$ and $2^{16}$ given we have

$\begin{align*} \left|E_5 \left( \frac{\sqrt{21}-3}{2} \right)\right| &\leq \left| \frac{\left( \frac{\sqrt{21}-3}{2} \right)^5}{5} \right| \leq \left| \frac{0.8^5}{5} \right| = \frac{(4/5)^5}{5} \\[9pt] &= \frac{2^{10}}{5^6} = \frac{2^{16}}{10^6} = \frac{65536}{1000000} < \frac{7}{100}. \qquad \blaacksquare \end{align*}$

Use Taylor polynomials to approximate the nonzero root of x²=sin x

by RoRi

December 21, 2015

Using the cubic Taylor polynomial approximation of $\sin x$ , show that the nonzero root of the equation
$x^2 = \sin x$

is approximated by $r = \sqrt{15} - 3$ .
Using part (a) show that
$| \sin r - r^2 | < \frac{1}{200},$

given that $\sqrt{15} - 3 < 0.9$ . Determine whether $(\sin r - r^2)$ is positive or negative and prove the result.

Proof. The cubic Taylor polynomial approximation of $\sin x$ is
$\sin x = x - \frac{x^3}{3!} + E_{2n} (x).$

This implies

$x^2 - \sin x \approx x^2 - x + \frac{x^3}{6}.$

Therefore, we can approximate the nonzero root by

$\begin{align*} x^2 - \sin x = 0 && \implies && x^2 - x + \frac{x^3}{6} &\approx 0 \\ && \implies && 6x^2 + x - 1 &\approx 0 \\ && \implies && x &\approx \sqrt{15} - 3. \qquad \blacksquare \end{align*}$
Proof. We know from this exercise (Section 7.8, Exercise #1) that for $\sin x$ we have
$|E_{2n} (x)| \leq \frac{|x|^{2n+1}}{(2n+1)!}.$

So, for $n = 2$ , and using the given inequality $\sqrt{15} - 3 < 0.9$ , we have

$|E_{2n} (x)| \leq \frac{|0.9|^5}{120} = \frac{9^5}{10^5 (120)} < \frac{1}{200}.$

Furthermore, $(\sin r - r^2) > 0$ since

$\sin r - r^2 = E_{2n}(x) = \frac{x^5}{5!} - \frac{x^7}{7!}+ \cdots$

with the absolute value of each term in the sum strictly less than the absolute value of the previous term (since $x < 1$ and $(n+2)! > n!$ ). Thus, each pair is positive, so the whole series is positive $. \qquad \blacksquare$

Use the weighted mean value theorem to approximate an integral

by RoRi

September 8, 2015

Note that

$1+x^6 = (1+x^2)(1-x^2 + x^4),$

and use the weighted mean value theorem (Theorem 3.16 in Apostol) to prove that for positive $a$ :

$\frac{1}{1+a^6} \left( a - \frac{a^3}{3} + \frac{a^5}{5} \right) \leq \int_0^a \frac{dx}{1+x^2} \leq a - \frac{a^3}{3} + \frac{a^5}{5}.$

Using $a = \frac{1}{10}$ compute the integral to six decimal places.

Recall the weighted mean value theorem:

For functions $f$ and $g$ continuous on $[a,b]$ , if $g$ never changes sign in $[a,b]$ then there exists $c \in [a,b]$ such that

$\int_a^b f(x) g(x) \, dx = f(c) \int_a^b g(x) \, dx.$

Proof. Let

$f(x) = \frac{1}{1+x^6}, \qquad g(x) = 1 - x^2 + x^4.$

Then,

$\begin{align*} \int_0^a \frac{dx}{1+x^2} &= \int_0^a \frac{1-x^2+x^4}{1+x^6} \, dx \\ &= \int_0^a f(x) g(x) \, dx \\ &= \frac{1}{c} \int_0^a g(x) \, dx \end{align*}$

for some $c \in [0,a]$ . Since $f$ is strictly decreasing on $[0,a]$ , we know $f(a) \leq f(c) \leq f(0)$ ; thus,

$\frac{1}{1+a^6} \leq f(c) \leq 1.$

Furthermore,

$\int_0^a g(x) \, dx = \int_0^a (1-x^2 + x^4) \, dx = a - \frac{a^3}{3} + \frac{a^5}{5}.$

Thus,

$\left( \frac{1}{1+a^6} \right) \left( a - \frac{a^3}{3} + \frac{a^5}{5} \right) \leq \int_0^a \frac{dx}{1+x^2} \leq \left( a - \frac{a^3}{3} + \frac{a^5}{5} \right). \qquad \blacksquare$

Now, taking $a = \frac{1}{10}$ we compute,

$\begin{align*} && (.9999999)(.0996687) \leq \int_0^a \frac{dx}{1+x^2} \leq .0996687 \\ \implies && .0996686 \leq \int_0^a \frac{dx}{1+x^2} \leq .0996687 \\ \implies && \int_0^a \frac{dx}{1+x^2} \approx .099669. \end{align*}$