1. Maclaurin Series

Use an appropriate local quadratic approximation to approximate tan61

http://img212.imageshack.us/img212/1256/52475955jo9.png

this is the answer from the solution mannual

Can anybody explain this more clearly?
Thank you

2. Originally Posted by soleilion
Use an appropriate local quadratic approximation to approximate tan61

http://img212.imageshack.us/img212/1256/52475955jo9.png

this is the answer from the solution mannual

Can anybody explain this more clearly?
Thank you
Work in radians and expand as a Taylor series about $60^{\circ}=\pi/3 {\rm{rad}}$

$
f(x+\varepsilon)=f(x)+\varepsilon~\left. \frac{df}{dx}\right|_{x}+\frac{\varepsilon^2}{2}~\ left. \frac{d^2f}{dx^2}\right|_{x}+..
$

Now:

$\frac{d}{dx}\tan(x)=(\sec(x))^2=1+(\tan(x))^2$,

and

$\frac{d^2}{dx^2}\tan(x)=2 (\sec(x))^2 \tan(x)=2\left[\frac{d}{dx}\tan(x)\right] \tan(x)$.

Now all you need is that $\tan(\pi/3)=\sqrt{3}$, and the $\varepsilon$ corresponding to $1^{\circ}$ is $\varepsilon=\pi/180$

RonL

3. Originally Posted by CaptainBlack
Work in radians and expand as a Taylor series about $60^{\circ}=\pi/3 {\rm{rad}}$

$
f(x+\varepsilon)=f(x)+\varepsilon~\left. \frac{df}{dx}\right|_{x}+\frac{\varepsilon^2}{2}~\ left. \frac{d^2f}{dx^2}\right|_{x}+..
$

Now:

$\frac{d}{dx}\tan(x)=(\sec(x))^2=1+(\tan(x))^2$,

and

$\frac{d^2}{dx^2}\tan(x)=2 (\sec(x))^2 \tan(x)=2\left[\frac{d}{dx}\tan(x)\right] \tan(x)$.

Now all you need is that $\tan(\pi/3)=\sqrt{3}$, and the $\varepsilon$ corresponding to $1^{\circ}$ is $\varepsilon=\pi/180$

RonL
As the answer from my website link, why it just shows f'', second prime? why not third or more or less than second?

4. Originally Posted by soleilion
As the answer from my website link, why it just shows f'', second prime? why not third or more or less than second?
Because it asks for a locally quadratic approximation which means that the highest order term in $\varepsilon$ should be $\varepsilon^2$ which means we only go as far as the second derivative.

RonL