Interpolation | Sahithyan's Notes

Interpolation estimates values between known data points.

Given $n+1$ points $(x_i,y_i)$ with distinct nodes

x_0 < x_1 < \cdots < x_n,

find a function $\phi$ satisfying

\phi(x_i)=y_i, \qquad i=0,1,\ldots,n

Use $\phi(x)$ for $x\in[x_0,x_n]$ . This is interpolation. Values outside this interval are extrapolation. They are less reliable.

Choosing an Interpolant

Choose the form from the data.

Data	Interpolant	Key Property
Few points	Polynomial	One formula through every point
Many non-periodic points	Piecewise polynomial	Low-degree formula on each interval
Periodic data	Trigonometric	Repeating sine and cosine terms
Poles or sharp changes	Rational	Ratio of polynomials
Values and slopes known	Hermite	Matches values and derivatives

Polynomial Interpolation

Theorem

For distinct nodes $x_0,\ldots,x_n$ and values $y_0,\ldots,y_n$ , exactly 1 polynomial $p_n$ of degree at most $n$ satisfies

p_n(x_i)=y_i

The Lagrange form is

p_n(x)=\sum_{i=0}^{n}y_iL_i(x), \qquad L_i(x)=\prod_{\substack{j=0\\j\ne i}}^{n}\frac{x-x_j}{x_i-x_j}

Each basis polynomial has 1 job.

L_i(x_j)= \begin{cases} 1, & i=j,\\ 0, & i\ne j. \end{cases}

Therefore $p_n(x_j)=y_j$ . At node $x_j$ , every term disappears except $y_jL_j(x_j)$ .

f the function, 1/(1 + 25x²), drawn dashed. p₁₀ the degree 10 polynomial through the nodes, which oscillates near both ends. piecewise linear straight segments through the same nodes, which stay close to f.

Piecewise Polynomial Interpolation

Split $[x_0,x_n]$ into intervals. Use a low-degree polynomial on each interval.

Piecewise linear
Straight segments. Continuous with corners at the nodes.
Piecewise quadratic
Parabolic segments. Slopes can jump at the nodes.
Cubic spline
Cubic segments with continuous value, slope and curvature.

Other Forms

For periodic data,

\phi(x)=\frac{a_0}{2}+\sum_{k=1}^{m}\left(a_k\cos(kx)+b_k\sin(kx)\right)

For rational interpolation,

\phi(x)=\frac{p(x)}{q(x)}

For Hermite interpolation, $\phi$ matches both $y_i$ and $y_i'$ at each node.