# 2015.11.6 Notes

## 2217 days ago by krsteph

Kylie Stephens

MATH 3600

C10726950

What is the series for $tan(x)$ for small $x$?

The MacLaurin expansion for $f(x)$ about $x = 0$ is

$f(0) + (x-0)f'(0) + (x-0)^2\frac{f''(0)}{2!} + ...$

$f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + ...$

appr = taylor(tan(x),x,0,6) print appr
 2/15*x^5 + 1/3*x^3 + x 2/15*x^5 + 1/3*x^3 + x
A = plot(appr,x,0,2)
B = plot(tan(x),x,0,1.56,color='red')
show(A + B)  Note $tan(x)$ is bigger than our appr.

Note that it is really small until about 0.4.

plot(tan(x)-appr,x,0,1) $tan(x) = \frac{p(x)}{x - \frac{\pi}{2}} ?$

Then $p(x) = (x - \frac{\pi}{2})tan(x)$

We should acutally make this $p(x) = (x - \frac{\pi}{2})(x + \frac{\pi}{2})tan(x)$

What is the MacLaurin seires for this?

appr2 = taylor((x^2-pi^2/4)*tan(x),x,0,20)/(x^2-pi^2/4) show(appr2)
 \newcommand{\Bold}{\mathbf{#1}}\frac{2 \, {\left(221930581 \, \pi^{2} - 2190367044\right)} x^{19} + 342 \, {\left(3202291 \, \pi^{2} - 31605346\right)} x^{17} + 2907 \, {\left(929569 \, \pi^{2} - 9174480\right)} x^{15} + 1220940 \, {\left(5461 \, \pi^{2} - 53898\right)} x^{13} + 23808330 \, {\left(691 \, \pi^{2} - 6820\right)} x^{11} + 1309458150 \, {\left(31 \, \pi^{2} - 306\right)} x^{9} + 5892561675 \, {\left(17 \, \pi^{2} - 168\right)} x^{7} + 247487590350 \, {\left(\pi^{2} - 10\right)} x^{5} + 618718975875 \, {\left(\pi^{2} - 12\right)} x^{3} + 1856156927625 \, \pi^{2} x}{1856156927625 \, {\left(\pi^{2} - 4 \, x^{2}\right)}}

$appr2 = \frac{series for tan(x)(x^2 - \frac{\pi^2}{4})}{(x^2 - \frac{\pi^2}{4})}$

plot(appr2-tan(x),x,-0,1.56) How does $\frac{tan(x)}{(x^2 - \frac{\pi^2}{4})}$ behave near $x = \pi/2$?

plot(tan(x)*(x^2-pi^2/4),x,-pi,pi) $tan(x) = \frac{sin(x)}{cos(x)}$

$cos(x) = sin(x + \frac{\pi}{2})$

How does $\frac{sin(t)}{t}$ behave as t approaches 0?

plot(sin(x)/x,x,-1,1) $\frac{sin(t)}{t} \simeq 1$ as t approaches 0.

$(x - \frac{\pi}{2})*(\frac{sin(x)}{cos(x)})*(x + \frac{\pi}{2})$ near $x = \frac{\pi}{2}$

Note: $x + \frac{\pi}{2} = 1$ at $x = \frac{\pi}{2}$

$\frac{x - \frac{\pi}{2}}{cos(x)} = \frac{x - \frac{\pi}{2}}{-sin(x - \frac{\pi}{2})} = \frac{t}{sin(t)}$     where $t = x - \frac{\pi}{2}$

- approaches -1 as $t$ approaches 0 OR $x$ approaches $\frac{\pi}{2}$

f(x) = taylor((x^2-pi^2/4)*tan(x),x,0,20)
f(pi/2).n()
 -3.14159265449082 -3.14159265449082

Attempting a rational function with rational coefficients to approximate $tan(x)$.

Try series for $sin(x)$, $cos(x)$ to some number of terms?

$tan(x) = \frac{sin(x)}{cos(x)}$

$cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} +$ ...

As we use better approximations to $cos(x)$, we'll get bettter approximations to the root at $\frac{\pi}{2}$.

A = plot(cos(x),0,2*pi) B = plot(1-x^2/2+x^4/factorial(4),x,0,pi,color='red') show(A + B)  find_root(1-x^2/2.+x^4/24.-x^6/factorial(6)+x^8/factorial(8),1,2)
 1.5708210679533907 1.5708210679533907
(pi/2).n()
 1.57079632679490 1.57079632679490