# Burden Numerical Analysis Lagrange Interpolation Question

I have been trying to solve a problem on Lagrange Interpolation from the book Numerical Analysis 10th Edition by Richard Burden. I have been stuck on the first question it for hours and cannot figure it out.

The question is: For the given functions $$f(x)$$ let $$x_0 = 0, x_1= 0.6, x_2 = 0.9$$ Construct interpolation polynomials of degree at most one and at most two to approximate $$f(0.45)$$

$$(a) f(x) = \cos x$$

The answer I get for two approximate is: $$6.79012 x^2 - 7.40741x + 1.5$$

The answer in the book is: $$-0.452592x^2 - 0.0131009x +1$$

I tried this in python using scipy interpolate Library's Lagrange function and the answer it gave was: $$-0.4311x^2 - 0.03246x + 1$$

May I ask if anyone can provide correct working of this question.

My steps:

Step 1 Calculate $$L_0, L_1, L_2$$

$$L_0 = (1/0.36)*(x-0.6)*(x-0.9), L_1 = (1/0.36)*x*(x-0.9), L_2 =(1/0.81)*x*(x-0.6)$$

Step 2: $$L_0+L_1+L_2$$

Solving this I get = $$6.79012 x^2 - 7.40741x + 1.5$$

It is difficult to say what's the problem without seeing what you did exactly, but here are some tips

• Make sure you're using radians, for the looks of it, that may be the problem

• Your solution must pass for all the input points, at least $$f(0) = \cos(0) = 1$$

• Here's just a plot showing your results, just to help you confirm the solution in the book is actually correct

This is the full procedure

$$\begin{eqnarray} L(x) &=& \cos(0) \frac{(x - 0.6)(x - 0.9)}{(0 - 0.6)(0 - 0.9)} + \cos(0.6)\frac{(x - 0)(x - 0.9)}{(0.6 - 0)(0.6 - 0.9)} + \cos(0.9) \frac{(x - 0)(x - 6)}{(0.9 - 0)(0.9 - 0.6)} \\ &=& \frac{\cos(0)}{0.54}(x - 0.6)(x - 0.9) - \frac{\cos(0.6)}{0.18}x(x - 0.9) + \frac{\cos(0.9)}{0.27} x (x - 0.6) \\ &=& 1.85185 (x - 0.6)(x - 0.9) - 4.5852x(x - 0.9) + 2.30226 x(x - 0.6) \\ &=& -0.431087 x^2 - 0.0324552 x + 1 \end{eqnarray}$$

• added my steps to the question above. May 17 '19 at 0:45
• @Varun Included the full procedure in an update May 17 '19 at 0:58