Newton's interpolating polynomial

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SUMMARY

The discussion focuses on interpolating the function ln(x+1) using a Newton's polynomial of degree 2, specifically at points x_0=0, x_1=0.6, and x_2=0.9. The polynomial is defined as p_2(x)=c_0+c_1(x-x_0)+c_2(x-x_0)(x-x_1). The coefficients calculated using divided differences yielded c_0=0, c_1≈0.78, and c_2≈-0.68. However, the polynomial fails to accurately interpolate the function at x=0.9, leading to confusion about the calculations, particularly regarding the unequal spacing of data points.

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Homework Statement


Interpolate the function ln(x+1) using a Newton's polynomial of degree 2. The points to be interpolated are x_0=0, x_1=0.6 and x_2=0.9.


Homework Equations



p_2(x)=c_0+c_1(x-x_0)+c_2(x-x_0)(x-x_1).

The Attempt at a Solution


So I used divided differences in order to calculate the coefficients of the polynomial.
f[x_1, x_2]=\frac{f(x_2)-f(x_1)}{x_2-x_1} \approx 0.1718502569.
f[x_0,x_1]\approx 0.7833393821.
f[x_0,x_1,x_2]=\approx -0.6794.
This gives c_0=0, c_1\approx 0.78 and c_2 \approx -0.68.
I've plotted both ln(x+1) and the polynomial and... it doesn't fit as it should. In fact it interpolates ln(x+1) in 0 and 0.6 but not in 0.9 as it should.
Where did I go wrong?! I've redone the algebra twice, even with a calculator and I'm still getting this wrong result.
Thanks for any help.
 

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fludistic - I get a different value for c2 of approx -.2344... Don't forget, your data is spaced unequally.
 
TheoMcCloskey said:
fludistic - I get a different value for c2 of approx -.2344... Don't forget, your data is spaced unequally.

Thanks. I've redone all from scratch and now get your result. :)
 

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