2015.11.6 Notes

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

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

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

Note $tan(x)$ is bigger than our appr.

Note that it is really small until about 0.4.

How about trying

\[tan(x) = \frac{p(x)}{x - \frac{\pi}{2}} ?\]

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?

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

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

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

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

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

$\frac{sin(t)}{t} \simeq 1$ 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 $\frac{\pi}{2}$

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 $\frac{\pi}{2}$.