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
\[f(0) + (x-0)f'(0) + (x-0)^2\frac{f''(0)}{2!} + ...\]
\[f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + ...\]
2/15*x^5 + 1/3*x^3 + x 2/15*x^5 + 1/3*x^3 + x |
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Note $tan(x)$ is bigger than our appr.
Note that it is really small until about 0.4.
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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?
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$appr2 = \frac{series for tan(x)(x^2 - \frac{\pi^2}{4})}{(x^2 - \frac{\pi^2}{4})}$
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How does $\frac{tan(x)}{(x^2 - \frac{\pi^2}{4})}$ behave near $x = \pi/2$?
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$tan(x) = \frac{sin(x)}{cos(x)}$
$cos(x) = sin(x + \frac{\pi}{2})$
How does $\frac{sin(t)}{t}$ behave as t approaches 0?
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$\frac{sin(t)}{t} \simeq 1$ 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 $\frac{\pi}{2}$
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-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 $\frac{\pi}{2}$.
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1.5708210679533907 1.5708210679533907 |
1.57079632679490 1.57079632679490 |
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