--- title: "Polynomial Interpolation and Its Error Term" description: "Fit the unique polynomial of degree at most n through n + 1 points with distinct x values, build its Lagrange and Newton forms, and bound its error." canonical: https://lightmysky.com/learn/mathematics/polynomial-interpolation-and-its-error-term-mt_NSnzY9Xodd source: https://lightmysky.com/learn/mathematics/polynomial-interpolation-and-its-error-term-mt_NSnzY9Xodd.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Polynomial Interpolation and Its Error Term Fit the unique polynomial of degree at most n through n + 1 points with distinct x values, build its Lagrange and Newton forms, and bound its error. Subject: Mathematics · Area: Calculus & Analysis · Ages 20 to 21 Page: https://lightmysky.com/learn/mathematics/polynomial-interpolation-and-its-error-term-mt_NSnzY9Xodd ## Ready when they can - Construct an interpolating polynomial in both Lagrange and divided-difference form - State the error term and use it to bound the interpolation error on an interval - Show with a worked example why more equally spaced points can make the fit worse ## Lesson: One polynomial through the points For n+1 points with distinct x values, exactly one polynomial of degree at most n matches every value. Its degree may be smaller. Higher-degree polynomials can also match. To prove uniqueness within the limit, subtract two candidates. Their difference has n+1 distinct roots, more than a nonzero polynomial of degree at most n can have. The difference must be zero. Repeated inputs with different outputs cannot be matched by any function. For data (a,u) and (b,v), where a differs from b, the interpolating line is p(x)=u+m(x-a), with slope m=(v-u)/(b-a). Equivalently, p(x)=u(b-x)/(b-a)+v(x-a)/(b-a). To check any proposed interpolant, substitute each node and compare with its recorded value. The residual is the polynomial's value minus the recorded value; it must be zero at every matched node. *(drawing: The line matches both recorded values. Its slope gives the value 3 at x=1.)* For nodes x_0,...,x_n and values y_0,...,y_n, form L_i(x) by multiplying (x-x_j)/(x_i-x_j) over all j other than i. Distinct nodes make the denominators nonzero. This basis polynomial is 1 at its own node and 0 at every other node. Lagrange form is p(x)=sum of y_i L_i(x). Evaluate each basis term, weight it by its data value and add. To find a coefficient, expand the weighted terms and collect the required power. **Example.** For (-1,2), (0,1) and (1,2), the basis polynomials are x(x-1)/2, 1-x^2 and x(x+1)/2. Multiply them by 2, 1 and 2. Their sum is x(x-1)+(1-x^2)+x(x+1)=x^2+1. The x^2 coefficients add to 1-1+1=1. Substitution checks the data: p(-1)=2, p(0)=1 and p(1)=2. *(drawing: A basis polynomial selects its own node. It need not be zero away from the nodes.)* Write f[a] for the value at node a. Divided differences are f[a,b]=(f[b]-f[a])/(b-a) and f[a,b,c]=(f[b,c]-f[a,b])/(c-a). Newton's quadratic form is p(x)=f[a]+f[a,b](x-a)+f[a,b,c](x-a)(x-b). At each higher level, subtract adjacent differences and divide by the spacing of the outer nodes. Each new coefficient multiplies the product of factors for the preceding nodes. **Example.** For (0,2), (2,6) and (4,18), the adjacent first differences are (6-2)/2=2 and (18-6)/2=6. The second difference is (6-2)/(4-0)=1. Newton form gives p(x)=2+2x+x(x-2)=x^2+2. Both Newton and Lagrange form give this polynomial because each matches the same data within the same degree limit. To add a point (t,v), keep the old interpolant p_n and add K times product(x-x_i) over the old nodes. This product vanishes at every old node, preserving its value. Choose K=(v-p_n(t))/product(t-x_i) to match the new value. The new node t must differ from all old nodes. If the old polynomial already matches the new value, K is zero. *(drawing: Add x(x-2)(x-4)/3. It is zero at the old nodes and equals the missing value 5 at x=5.)* Let p_n be the polynomial of degree at most n matching f at n+1 distinct nodes x_i. Assume f has n+1 continuous derivatives on a closed interval containing the nodes and x. For a non-node x, f(x)-p_n(x)=f^(n+1)(c) times product(x-x_i)/(n+1)!, where c is inside the smallest interval containing the nodes and x. The superscript denotes derivative order. At a node the error is zero. The nodal product replaces the Taylor remainder's power. If |f^(n+1)(t)| <= M throughout the interval, the absolute error is at most M/(n+1)! times |product(x-x_i)|. For a line through the values at endpoints a