What is the 1st Order Expansion Problem for a Function with a Logarithm?

[ks_wann]

I'm a bit frustrated at the moment, as this minor problem should be fairly easy. But I seem to go wrong at some point...

So I've got to do a 1st order expansion of the function
\begin{equation}
f=\frac{\cos(\theta)}{\sin(\theta)}\ln(\frac{L\sin(\theta)}{d\cos( \theta)}+1)
\end{equation}

and my steps are:

\begin{equation}

f(0)=0 \\
f^{\prime}(\theta)=-\frac{1}{\sin^2(\theta)}\ln(\frac{L\sin(\theta)}{d\cos(\theta)}+1)+ \frac{\cos(\theta)}{\sin(\theta)}\frac{d\cos(\theta)}{L\sin(\theta)+d \cos(\theta)}(\frac{\cos^2(\theta)+\sin^2(\theta)}{\cos^2(\theta)}\frac{L}{d}).
\end{equation}

When I insert theta=0 I end up divding by 0...
Furthermore, when I make my computer do the expansion, I get the correct result from the assignment I'm working on.

If anyone could help me out, I'd be grateful!

 

[ehild]

f(0) is not defined, you need the limits at theta=0. Better expand the logarithm. You know that ln(1+x) ≈ x-x/2 if |x|<<1.

ehild

 

[LCKurtz]

Or, if you let $x = \tan\theta$ and $k=\frac L d$ you have $\frac{\ln(1+kx)}{x}$. You can find the limit as $x\to 0$ using L'Hospital's rule.

 

[benorin]

If you're not sure about the substitution, you would have:

\lim_{\theta\to 0}f(\theta )=\lim_{\theta\to 0}\frac{\log\left(1+\frac{L}{d}\tan\theta\right)}{\tan\theta} = \lim_{\theta\to 0}\frac{\frac{d}{d\theta}\log\left(1+\frac{L}{d}\tan\theta\right)}{ \frac{d}{d\theta} \tan\theta}= \cdots

 

[ehild]

The function can be written in simpler form as benorin has shown:

f(\theta )=\frac{\log\left(1+\frac{L}{d}\tan\theta\right)}{\tan\theta}

Use the Taylor expansion of log(1+x) = x - x²/2. Let be x=tan(theta).

f(\theta )≈\frac{\frac{L}{d}\tan\theta-\left(\frac{L}{d}\tan\theta\right)^2/2}{\tan\theta}

Simplify by tan(θ): You get an expression linear in tan(θ). You can expand tan(θ) with respect to θ...

ehild

 

[ks_wann]

Thanks for all of your answers, it really helped me out. I've reread the series chapter of my calculus book, and I've come to a much better understanding of that subject in general.

I basically expand the logarithm in the function, and then I expand the function, which gave the correct answer.