# For $f(x) = \tan(\pi \cdot x)$, find the Interpolation $Q(x) = b_0 + b_1 x + b_2 \frac{1}{x-\frac{1}{2}}$

Progress so far:

In a previous task, I determined a polynomial interpolation using a system of linear equations.

The data points to be used were $$(0, f(0)), (\frac{1}{6}, f(\frac{1}{6})), (\frac{1}{4}, f(\frac{1}{4}))$$

The linear equation used was of the form:

$$\begin{pmatrix}0^2&0^1&0^0\\ \:\frac{1}{6}^2&\frac{1}{6}^1&\frac{1}{6}^0\\ \:\frac{1}{4}^2&\frac{1}{4}^1&\frac{1}{4}^0\end{pmatrix}$$$$\begin{pmatrix}a_2\\ \:a_1\\ \:a_0\end{pmatrix}$$=$$\begin{pmatrix}0\\ \frac{1}{\sqrt{3}}\\ 1\end{pmatrix}$$

and the polynomial of the form

$$p_2(x) = a_0 + a_1 x + a_2 x^2$$ was determined to have the coefficients

$$a_0 = 0, a_1 = -8 + \frac{18}{\sqrt{3}}, a_2 = 48 - \frac{72}{\sqrt{3}}$$

I am not aware of how I would use a system of linear equations to determine $$Q(x)$$. Is it perhaps possible to derive the coefficients from $$p_2$$? Or is there some other method of interpolation I should pursue?

• In my answer, I assumed that you would also use $x=\frac 12$. If this is not the case, for $Q(x)$ you still have three linear equations. – Claude Leibovici Dec 3 '18 at 11:26

## 1 Answer

Instead of working with $$Q(x)$$ $$Q(x) = b_0 + b_1 x + \frac{b_2}{x-\frac{1}{2}}$$ work with $$\Phi(x)$$ an $$\Psi(x)$$ $$\Phi(x)=\left(x-\frac 12\right) Q(x)=\left(b_2-\frac{b_0}{2}\right)+x \left(b_0-\frac{b_1}{2}\right)+b_1 x^2=\alpha+\beta x+\gamma x^2$$

$$\Psi(x)=\left(x-\frac 12\right) F(x)=\left(x-\frac 12\right) \tan(\pi x)$$ $$\Psi(x)$$ is a very nice function for the range of interest.

Compute $$\alpha,\beta ,\gamma$$ and go back to $$b_0,b_1,b_2$$.