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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 = $6.79012 x^2 - 7.40741x + 1.5$

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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

enter image description here

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}

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

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