91Ó°ÊÓ

Divided differences are a method used in numerical analysis to compute the coefficients of the Newton polynomial. This method is particularly helpful for creating interpolation polynomials from a given set of data points.
To understand divided differences, let's start with the first divided difference. It is calculated between two points and represents the slope of the line connecting them. The formula for the first divided difference is:
\( f[x_{0}, x_{1}] = \frac{f(x_{1}) - f(x_{0})}{x_{1} - x_{0}} \)
From our example:
Given:
* \(x_{0} = 0.1\), \(f(x_{0}) = 0.2\)
* \(x_{1} = 0.2\), \(f(x_{1}) = 0.24\)
Substituting these into the formula, we have:
\( f[x_{0}, x_{1}] = \frac{0.24 - 0.2}{0.2 - 0.1} = 0.4 \)

The second divided difference involves three points and indicates the curvature of the polynomial that passes through them. Its formula is:
\( f[x_{0}, x_{1}, x_{2}] = \frac{f[x_{1}, x_{2}] - f[x_{0}, x_{1}]}{x_{2} - x_{0}} \)
First, calculate \( f[x_{1}, x_{2}] \):
Given:
* \(x_{1} = 0.2\), \(f(x_{1}) = 0.24\)
* \(x_{2} = 0.3\), \(f(x_{2}) = 0.30\)
Substituting these values:
\( f[x_{1}, x_{2}] = \frac{0.30 - 0.24}{0.3 - 0.2} = 0.6 \)

Now, substitute back to find \( f[x_{0}, x_{1}, x_{2}] \):
\( f[x_{0}, x_{1}, x_{2}] = \frac{0.6 - 0.4}{0.3 - 0.1} = 1 \)
Understanding these differences helps us build interpolation polynomials effectively.

Linear interpolation is the simplest form of interpolation that estimates a value within two known values on a straight line. The formula for the linear interpolation polynomial is:
\( P_{1}(x) = f(x_{0}) + f[x_{0}, x_{1}] (x - x_{0}) \)
Using our example, we substitute the values:
* \(f(x_{0}) = 0.2\)
* \(f[x_{0}, x_{1}] = 0.4\)
* \(x = 0.15\)
The polynomial becomes:
\( P_{1}(0.15) = 0.2 + 0.4 (0.15 - 0.1) \)
Simplify the expression:
\( P_{1}(0.15) = 0.2 + 0.4 \times 0.05 \)
\( P_{1}(0.15) = 0.2 + 0.02 = 0.22 \)
Thus, the linear interpolate at \( x = 0.15 \) is 0.22. This method is straightforward but can become inaccurate for more complex data sets. Still, it's an excellent starting point for understanding interpolation.

Quadratic interpolation uses a polynomial of degree 2 to estimate a value within three known points. The formula for the quadratic interpolation polynomial is:
\( P_{2}(x) = f(x_{0}) + f[x_{0}, x_{1}] (x - x_{0}) + f[x_{0}, x_{1}, x_{2}] (x - x_{0})(x - x_{1}) \)
Substituting from our example:
* \(f(x_{0}) = 0.2\)
* \(f[x_{0}, x_{1}] = 0.4\)
* \(f[x_{0}, x_{1}, x_{2}] = 1\)
* \(x = 0.15\)
The polynomial becomes:
\( P_{2}(0.15) = 0.2 + 0.4 (0.15 - 0.1) + 1 (0.15 - 0.1)(0.15 - 0.2) \)
Break it down step by step:
* First term: \(0.2\)
* Second term: \(0.4 \times 0.05 = 0.02\)
* Third term: \(1 \times 0.05 \times (-0.05) = -0.0025\)
Now add them up:
\( P_{2}(0.15) = 0.2 + 0.02 - 0.0025 = 0.2175 \)
Quadratic interpolation generally gives a better approximation than linear interpolation, especially when the data points exhibit curvature.