Newton's Divided Difference: The Ultimate Interpolation Masterclass
Break free from the rigid constraints of evenly spaced data and master the universal, "upgradable" algorithm for practical, real-world interpolation.
Breaking Free From Equal Spacing
In our previous guides, we built a powerful toolkit with Newton's forward and backward formulas. They were efficient and intuitive, but they had a strict, non-negotiable requirement: all your data points had to be equally spaced. But the real world is messy. Data is almost never this neat.
Imagine you are taking measurements from a scientific experiment. You take a reading at t=0s, t=1s, and t=2s, but then the system changes rapidly, so you take more readings at t=2.1s, t=2.2s, and t=2.5s. Your step size, $h$, is no longer constant. The simple machinery of the finite difference table, which relies on constant spacing, breaks down completely. Does this mean we have to retreat to the less efficient Lagrange method?
Thankfully, no. The true genius of Newton's approach is a more general formulation that works for any set of distinct data points, equally spaced or not. This universal tool is Newton's Divided Difference Interpolation Formula. It replaces the simple "finite differences" with a more robust concept of "divided differences" to create the ultimate, practical interpolation polynomial.
The Core Tool: Divided Differences
The problem with a simple difference like $y_{i+1} - y_i$ for unequally spaced data is that it's not a fair comparison. A change of 10 units over a distance of 0.1 is much more dramatic than the same change over a distance of 100. To fix this, we need to "normalize" our differences by the distance between our x-coordinates. This is the core idea of a divided difference.
Defining Divided Differences
We define them recursivelyA recursive process is one that is defined in terms of itself. To find a 2nd-order difference, you first need to find the 1st-order differences., starting with the simplest case and building up.
- Zeroth-Order: This is just the value of the function at a point. $$ f[x_i] = y_i $$
- First-Order: This is the slope of the secant lineA straight line that connects two distinct points on a curve. Its slope represents the average rate of change between those two points. between two points. $$ f[x_i, x_{i+1}] = \frac{f[x_{i+1}] - f[x_i]}{x_{i+1} - x_i} $$
- Second-Order: This is the difference of the first-order differences, divided by the "total" distance between the outermost points. $$ f[x_i, x_{i+1}, x_{i+2}] = \frac{f[x_{i+1}, x_{i+2}] - f[x_i, x_{i+1}]}{x_{i+2} - x_i} $$
This pattern continues for all higher orders. While the formula looks more complex than a simple finite difference, it's doing the crucial job of accounting for the spacing of our data points at every level.
The Divided Difference Table Generator
Enter your own, arbitrarily-spaced data points below (one point per line, e.g., "0, 1"). The lab will automatically compute the full divided difference table. The crucial diagonal coefficients for the formula are highlighted.
The Newton's Divided Difference Formula
Once we have our divided difference table, the diagonal of coefficients ($f[x_0], f[x_0, x_1], f[x_0, x_1, x_2]$, etc.) gives us everything we need to build the polynomial. The structure is beautifully "nested" and incremental.
The full, general formula is:
This formula represents the pinnacle of practical interpolation. It is computationally efficient and, most importantly, it is upgradable. If you decide you need more accuracy and add a new data point, you don't have to throw away your work. You simply calculate a new row in the divided difference table and add one more term to the end of your polynomial. All your previous coefficients remain unchanged.
The Final Showdown: Newton vs. Lagrange
At this point, you might be wondering: we now have two methods, Lagrange and Newton's Divided Difference, that can both create the unique polynomial for any set of data points. Which one is better?
The answer is that while they produce the exact same final polynomial, they have different strengths and weaknesses that make them suited for different tasks.
| Feature | Lagrange's Method | Newton's Divided Difference |
|---|---|---|
| Data Spacing | Works for any spacing | Works for any spacing |
| Conceptual Basis | Builds the whole curve at once from "spotlight" basis polynomials | Builds the curve piece by piece, adding more detail with each term |
| Adding a New Point | Must recalculate everything from scratch | Highly efficient; just add a new term to the existing polynomial |
| Computational Cost | More expensive, especially for multiple evaluations | Less expensive, especially when using efficient evaluation schemes like Horner's method |
The Verdict: For theoretical work and proofs, the elegance of Lagrange's formula is often preferred. But for practical, computational applications—especially when your dataset might grow or you need to evaluate the polynomial at many points—Newton's Divided Difference method is the clear winner due to its efficiency and "upgradable" nature.
Solving Numerical Problems: A Step-by-Step Guide
Problem 1: Newton's Divided Difference Interpolation
Question: Given the data points (1, -3), (3, 2), and (4, 5), find the Newton's divided difference interpolating polynomial and use it to estimate the value at $x=2$.
Step 1: Construct the Divided Difference Table
| x | f[] | f[ , ] | f[ , , ] |
|---|---|---|---|
| 1 | -3 | 2.5 | 0.1667 |
| 3 | 2 | 3 | |
| 4 | 5 |
Let's verify the calculations:
$f[x_0, x_1] = \frac{2 - (-3)}{3 - 1} = \frac{5}{2} = 2.5$
$f[x_1, x_2] = \frac{5 - 2}{4 - 3} = \frac{3}{1} = 3$
$f[x_0, x_1, x_2] = \frac{3 - 2.5}{4 - 1} = \frac{0.5}{3} \approx 0.1667$
Step 2: Construct the Polynomial
The coefficients from the top diagonal are $f[x_0]=-3$, $f[x_0, x_1]=2.5$, and $f[x_0, x_1, x_2]=0.1667$. Our points are $x_0=1, x_1=3$.
$$ P(x) = f[x_0] + f[x_0, x_1](x-x_0) + f[x_0, x_1, x_2](x-x_0)(x-x_1) $$
$$ P(x) = -3 + 2.5(x-1) + 0.1667(x-1)(x-3) $$
Step 3: Estimate the value at x=2
Now, we just plug $x=2$ into our polynomial:
$$ P(2) = -3 + 2.5(2-1) + 0.1667(2-1)(2-3) $$
$$ P(2) = -3 + 2.5(1) + 0.1667(1)(-1) $$
$$ P(2) = -3 + 2.5 - 0.1667 = -0.6667 $$
Final Answer: The estimated value at $x=2$ is approximately -0.6667.
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