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Week 7 Topic 3c Interpolation
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Key takeaways
Linear and Polynomial Interpolation
Interpolation estimates values between known data points. By contrast, extrapolation estimates values beyond the range of known data. This lesson focuses on estimating intermediate values using linear and polynomial interpolation.
Polynomial Interpolation
For (n+1) data points, there is exactly one polynomial of order (n) that passes through all of them:
- Two points determine one straight line, or first-order polynomial.
- Three points determine one parabola, or second-order (quadratic) polynomial.
Once constructed, the polynomial can be used to estimate values between the data points. The lesson highlights Newton’s divided-difference polynomial as a useful way to express an interpolating polynomial.
Linear Interpolation
Linear interpolation estimates a value by connecting two data points with a straight line. Its formula is derived from similar triangles. The slope between the points is also a finite divided-difference approximation of the first derivative.
General procedure:
- Select the two data points surrounding the value to estimate.
- Substitute their (x)- and (f(x))-values, along with the target (x), into the linear interpolation formula.
- Evaluate the estimate and, if the true value is known, compare the result with it.
Example: Estimating (\ln 2)
- Interpolating between (\ln 1 = 0) and (\ln 6 \approx 1.791759) gives an estimate of 0.3583519, with a reported error of 48.3%.
- Using the narrower interval from (\ln 1 = 0) to (\ln 4 \approx 1.386294) gives an estimate of 0.4620981, with a reported error of 33.3%.
This example illustrates that a smaller interval can improve a linear estimate, although it does not eliminate the error caused by approximating a curved function with a straight line.
Quadratic Interpolation
When three data points are available, a second-order polynomial can introduce curvature and may provide a better fit than a straight line.
Coefficient-finding approach:
- Write the polynomial form with unknown coefficients.
- Set (x = x_0) to determine the constant term from (f(x_0)).
- Use (x = x_1) to determine the next coefficient, based on the slope between the first two points.
- Substitute the first two coefficient results and the third data point into the polynomial to determine the remaining coefficient.
- Evaluate the resulting polynomial at the desired (x).
Using three data points in the lesson’s (\ln 2) example produces an estimate of 0.56584.
Newton’s Divided Differences
Newton’s method builds higher-order interpolating polynomials from divided differences.
- A first divided difference is the change in function values divided by the change in the corresponding (x)-values.
- A second divided difference is formed from two first divided differences.
- Higher-order divided differences are calculated in the same way, using differences of the preceding order.
- These divided differences provide the coefficients for the Newton interpolating polynomial.
Four-point example procedure:
- Calculate the function values at the four data points.
- Compute first divided differences for each neighboring pair.
- Compute second divided differences from neighboring first divided differences.
- Compute the third divided difference from the second divided differences.
- Substitute the values into the Newton polynomial and evaluate it at the target point.
The lesson reports a four-point estimate for (\ln 2) of 0.6287686, with a relative error of about 9.3%. In this example, using more data points and a higher-order polynomial improves the estimate.
Closing Activity
The instructor assigns an exercise to reinforce linear and polynomial interpolation and says that the actual results are provided for students to check their work. The exercise’s specific questions are not included in the subtitles.
Speakers or Sources Featured
- One instructor/narrator presents the Numerical Analysis and Finite Element Method lesson. The subtitles do not identify the speaker by name.
- No other speakers or external sources are featured.
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