4.2 Undetermined Coefficients

According to our interpolating error, the difference between $f$ and an interpolating polynomial is a multiple of $f^{(n + 1)} (ξ_{_{x}})$ thus if $f$ has a degree less than $n + 1$ our error $(P - f) (x) = 0$ . So if $f$ has degree less than $n + 1$ and $P_{n}$ is computed for $n + 1$ nodes then we have an exact approximation:

P_{n} = f \forall x

Our interpolating polynomials have all had the form:

\sum_{i = 0}^{n} a_{i} f (x_{i})

where $x_{0}, x_{1}, \dots, x_{n}$ are our nodes of our interpolating polynomials. Knowing our approximation must be exact for polynomials of degree $n$ we can create $n + 1$ equations with $n + 1$ unknowns, $a_{0}, a_{1}, \dots, a_{n}$

Derivatives

We want to approximated the $k^{t h}$ derivative of $f$ with our values:

f (x_{0} + θ_{0} h), f (θ_{1} + θ_{1} h), \dots, f (x_{n} + θ_{n} h)

Such that:

f^{(k)} (x_{0} + θ h) \approx \sum_{i = 0}^{n} a_{i} f (x_{0} + θ_{i} h)

Where we know $p_{j} (x) = (x - x_{0})^{j}, j = 0, 1, \dots, n$ will be exact by the interpolating error.

p_{j}^{(k)} (x_{0} + θ h) = \sum_{i = 0}^{n} a_{i} p_{j} (x_{0} + θ_{i} h)

Noting that $p_{j} (x_{0} - θ_{i} h) = ((x_{0} + θ_{i} h) - x_{0})^{j} = (θ_{i} h)^{j}$

p_{j}^{(k)} (x_{0} + θ h) = a_{0} + \sum_{i = 1}^{n} (θ_{i} h)^{j} a_{i}

for $j = 0, 1, \dots, n$

Example

so for:

Pasted image 20250316175748.webp

We have $θ = 0, θ_{0} = - 1, θ_{1} = 0,$ and $θ_{2} = 1$
We want:

\begin{array}{r} f^{'} (x_{0}) \approx a_{0} f (x_{0} - h) + a_{1} f (x_{0}) + a_{2} f (x_{0} + h) \\ f^{″} (x_{0}) = b_{0} f (x_{0} - h) + b_{1} f (x_{0}) + b_{2} f (x_{0} + h) \end{array}

and we want each of the formulas to be exact when $f = p_{0}, f = p_{1}, f = p_{2}$

\begin{array}{lcl} p_{0}^{'} (x_{0}) & = & a_{0} p_{0} (x_{0} - h) + a_{1} p_{0} (x_{0}) + a_{2} p_{0} (x_{0} + h) \\ p_{1}^{'} (x_{0}) & = & a_{0} p_{1} (x_{0} - h) + a_{1} p_{1} (x_{0}) + a_{2} p_{1} (x_{0} + h) \\ p_{2}^{'} (x_{0}) & = & a_{0} p_{2} (x_{0} - h) + a_{1} p_{2} (x_{0}) + a_{2} p_{2} (x_{0} + h) \end{array}

\begin{aligned} p_{0} (x) & = (x - x_{0})^{0} ⟹ p_{0}^{'} (x_{0}) = 0 \\ p_{1} (x) & = (x - x_{0})^{1} ⟹ p_{1}^{'} (x_{0}) = 1 \\ p_{2} (x) & = (x - x_{0})^{2} ⟹ p_{2}^{'} (x_{0}) = 2 (x_{0} - x_{0}) = 0 \end{aligned}

\begin{array}{rcl} 0 & = & a_{0} + a_{1} + a_{2} \\ 1 & = & - h a_{0} + h a_{2} \\ 0 & = & h^{2} a_{0} + h^{2} a_{2} . \end{array}

We get:

f^{'} (x_{0}) \approx \frac{f (x_{0} + h) - f (x_{0} - h)}{2 h}

which we see here

Next we can do:

\begin{array}{rcl} p_{0}^{''} (x_{0}) & = & b_{0} p_{0} (x_{0} - h) + b_{1} p_{0} (x_{0}) + b_{2} p_{0} (x_{0} + h) \\ p_{1}^{''} (x_{0}) & = & b_{0} p_{1} (x_{0} - h) + b_{1} p_{1} (x_{0}) + b_{2} p_{1} (x_{0} + h) \\ p_{2}^{''} (x_{0}) & = & b_{0} p_{2} (x_{0} - h) + b_{1} p_{2} (x_{0}) + b_{2} p_{2} (x_{0} + h), \end{array}

Now $p_{0}^{″} (x_{0}) = 0, p_{1}^{″} (x_{0}) = 0, p_{2}^{″} (x_{0}) = 2$

\begin{array}{lll} 0 & = & b_{0} + b_{1} + b_{2} \\ 0 & = & - h b_{0} + h b_{2} \\ 2 & = & h^{2} b_{0} + h^{2} b_{2} . \end{array}

where the final solution is:

f^{''} (x_{0}) \approx \frac{f (x_{0} + h) - 2 f (x_{0}) + f (x_{0} - h)}{h^{2}}

Integrals

Same procedure as above. We want a linear combination to approximate our integral using the values

f (x_{0} + θ_{0} h), f (x_{0} + θ_{1} h), \dots, f (x_{0} + θ_{n} h)

\int_{a}^{b} f (x) d x \approx \sum_{i = 0}^{n} a_{i} f (x_{0} + θ_{i} h)

And in theory this will be exact for all polynomials of degree $n$ or less.

\int_{a}^{b} p_{j} (x) d x = a_{0} + \sum_{i = 1}^{n} (θ_{i} h)^{j} a_{i} j = 0, 1, \dots, n