Determination of three fitting parameters to approximate $\cos\theta \left( 1-\cos\theta \right) \Theta \left( \frac{\pi}{2} - \theta \right)$

Question

I would like to determine the coefficients $\alpha$, $\beta$, and $\lambda \in \mathbb{R}$ such that $$ f(\theta) = \sin\theta \left( 1+\lambda\cos\theta \right) \left( 1 + \alpha \cos\theta + \beta\left( 5\cos^2\theta - 1 \right) \right) $$ is the best fit for $$ g(\theta) = \cos\theta \left( 1-\cos\theta \right) \Theta \left( \frac{\pi}{2} - \theta \right) \, , $$ in the interval $[0, \pi]$. Here, $\Theta$ denotes the Heaviside step function. Any hints/ ideas are highly appreciated. Thank you!

Answer 1

The most rigorous way is to consider the norm $$\Phi(\alpha,\beta,\lambda)=\int_0^\pi \Big[f(\theta)-g(\theta)\Big]^2\,d\theta$$ which is the same as a curve fit based on an infinite number of data points.

Then, as usual, you need to solve the three equations $$\frac{\partial \Phi(\alpha,\beta,\lambda)}{\partial \alpha}=0\qquad \frac{\partial \Phi(\alpha,\beta,\lambda)}{\partial \beta}=0\qquad \frac{\partial \Phi(\alpha,\beta,\lambda)}{\partial \lambda}=0$$

The calculation of $\Phi(\alpha,\beta,\lambda)$ is not very difficult and I shall let you doing it.

Using the first and second equations, we easily obtain $\alpha(\lambda)$ and $\beta(\lambda)$ but, plugging in the third equation gives a nonic polynomial in $\lambda$ which has three real roots; so, at this point, only numerical methods would do the job.