Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a non-linear ODE that I need to linearise.

I could go about linearising the trig terms using Taylor series...but not sure about a square root of the form:

$$\sqrt{(A+C)^2 + B -2 * \cos (\alpha - \theta)}$$

Thanks.

share|improve this question
2  
Is $\theta$ your variable? –  Mike Spivey Dec 9 '10 at 17:59
    
Could you do a change of variables and make $\sqrt{(A+C)^2+B-2*\cos(\alpha-\theta)}$ a new variable? –  Ryan Budney Dec 9 '10 at 21:59
    
Maybe giving the actual nonlinear ODE might yield answers more useful to you? –  J. M. Dec 10 '10 at 0:55

2 Answers 2

Since you have an ODE, you probably want to be linearizing around an equilbrium point $\theta_0$ (which might not be zero). Note that I'm assuming that $\theta$ is the variable.

Then think about the general form of a Taylor series expansion -- the linearization is given by $$ f(\theta) \approx f(\theta_0) + (\theta-\theta_0) f'(\theta_0) $$ where $$ f(\theta) = \sqrt{(A+C)^2 + B - 2\cos(\alpha - \theta)} $$

share|improve this answer

If $(A+C)^2+B$ is large compared with 1 you can divide it out to get $$\sqrt{(A+C)^2+B}\sqrt{1-\frac{2cos(\alpha-\theta)}{(A+C)^2+B}}$$ and use $\sqrt{1-x}\approx 1-\frac{x}{2}$ for $x<<1$. If you are specifically interested in $\theta$ you can expand the cosine term and still use the approximation to the square root considering $\alpha$ to be fixed. Does that help?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.