Estimating f(2) using Cubic Interpolation

In summary, for the given cubic function that satisfies f(0) = -32, f(1) = 0, f(3) = 10 and f(4) = 0, the use of cubic interpolation can estimate f(2) to be 12. This can be done by using the Lagrange interpolating polynomial in the form of g(x) = a + bx + cx^2 + dx^3 and solving for the constants given the points. Alternatively, one can use the polynomial P(x) = Σ(i=1 to n) π(x) / π'(x_k) * f(x_i) / (x-x_i), where n = 4 and the points x1 =
  • #1
escobar147
31
0
If a cubic function satisfies f(0) = -32, f(1) = 0, f(3) = 10 and f(4) = 0,
use cubic interpolation to estimate f(2)



I'm not sure how to approach this since I have only ever done quadratic interpolation and linear interpolation, is it just an extension of the lagrange interpolating polynomial?

If so could someone please show me what form to put it in?

The correct answer is 12, however my attempt at extending lagrange give me nowhere near that?!

any help would be hugely appreciated!
 
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  • #2
could you just start with g(x) = a+bx+cx^2+dx^3 and solve for the constants given the points you have?
 
  • #3
There are many ways one of which being
[tex]P(x)=\sum_{i=1}^n \frac{\pi(x)}{\pi\prime(x_k)}\frac{f(x_i)}{x-x_i}[/tex]
where in this case n=4 and
π(x)=(x-0)(x-1)(x-3)(x-4)
x1=0;x2=1;x3=3;x4=4
 

Related to Estimating f(2) using Cubic Interpolation

What is cubic interpolation?

Cubic interpolation is a method used to estimate values between two known data points. It involves fitting a smooth curve between the data points and using that curve to predict the value at a specific point within the data set.

How is cubic interpolation different from linear interpolation?

Cubic interpolation uses a more complex mathematical formula to determine the curve between data points, while linear interpolation simply connects the points with a straight line. This results in a smoother and more accurate estimation with cubic interpolation.

When is cubic interpolation typically used?

Cubic interpolation is commonly used when the data is continuous and the values between data points are expected to be smooth and gradually changing. It is often used in computer graphics, signal processing, and numerical analysis.

What are the advantages of using cubic interpolation?

Cubic interpolation produces more accurate results compared to linear interpolation, especially when the data points are not evenly spaced. It also allows for more flexibility in fitting the curve between data points, which can lead to more precise estimations.

Are there any limitations or drawbacks to using cubic interpolation?

One limitation of cubic interpolation is that it can produce unrealistic values if the data is not continuous or has large gaps between data points. Additionally, the accuracy of the estimation can be affected by the number and spacing of data points, as well as the choice of the interpolation method.

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