Numerical Methods · Unit 3
Interpolation and Approximation
Exam-focused notes for Interpolation and Approximation (Numerical Methods, BIT203): what the TU syllabus asks and how it has actually been tested, with 8 solved past questions from this unit.
What this unit covers
- Lagrange interpolation formula and algorithm
- Newton's divided difference method
- Newton's forward difference formula
- Newton's backward difference table construction
- Interpolation versus regression
- Applications of interpolation
- Derivative estimation using interpolation formulas
Newton's divided difference method
Write an algorithm to compute the value of interpolation using Newton’s divided difference method.Write a program to compute the value of interpolation using Newton’s divided difference method.[5+5]
(a) Algorithm Newton's Divided Difference interpolation finds a polynomial passing through given data points (x₀,y₀), (x₁,y₁), ..., (xₙ,yₙ) and estimates the value at any point x. The interpolating polynomial is: Where divided differences are defined as: - ...
Full solved answer →Given the following set of data points. Obtain the table of divided difference and use that table to estimate the value of $f(1.5)$.
$$\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5 \ \hline f(x)=x^3-1 & 0 & 7 & 26 & 63 & 124 \ \hline \end{array}$$
[5]
$x$ 1 2 3 4 5 ------------------ $f(x)=x^3-1$ 0 7 26 63 124 Estimate: $f(1.5)$ --- First divided differences: $$f[1,2]=\frac{7-0}{2-1}=7,\quad f[2,3]=\frac{26-7}{3-2}=19$$ $$f[3,4]=\frac{63-26}{4-3}=37,\quad f[4,5]=\frac{124-63}{5-4}=61$$ Second divided dif...
Full solved answer →Derivative estimation using interpolation formulas
Evaluate $\frac{dy}{dx}$ at $x = 5$ using Newton's forward interpolation formula using the following table.
| X | 1 | 3 | 5 | 7 | 9 |
|---|---|---|---|---|---|
| y | -1.20 | 12.80 | 119.60 | 472.80 | 1302.80 |
[5]
X 1 3 5 7 9 ------------------ y -1.20 12.80 119.60 472.80 1302.80 - Uniform spacing: $h = 2$ - $x0 = 1$ - Evaluate $\dfrac{dy}{dx}$ at $x = 5$ $\Delta y$: - $12.80 - (-1.20) = 14.00$ - $119.60 - 12.80 = 106.80$ - $472.80 - 119.60 = 353.20$ - $1302.80 - 472...
Full solved answer →Divided Difference Table and Derivatives
Construct the divided difference table for the following data and find first and second order derivatives at $x=2$.
$$\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 4 & 8 & 10 \ \hline y & 0 & 1 & 5 & 21 & 27 \ \hline \end{array}$$
[5]
$x$ 1 2 4 8 10 ------------------ $y$ 0 1 5 21 27 $$f[x0,x1]=\frac{1-0}{2-1}=1$$ $$f[x1,x2]=\frac{5-1}{4-2}=2$$ $$f[x2,x3]=\frac{21-5}{8-4}=4$$ $$f[x3,x4]=\frac{27-21}{10-8}=3$$ $$f[x0,x1,x2]=\frac{2-1}{4-1}=\frac{1}{3}$$ $$f[x1,x2,x3]=\frac{4-2}{8-2}=\frac...
Full solved answer →Lagrange interpolation formula and algorithm
Question
What are the applications of interpolation? Differentiate between interpolation and regression. Consider the following data points estimate the $f(10)$ using Lagrange's interpolation.
$$\begin{array}{|c|c|c|c|c|}\hline x & 5 & 6 & 9 & 11 \ \hline y & 13 & 14 & 15 & 16 \ \hline \end{array}$$
[10]
- Estimating intermediate values: Finding function values between tabulated data points. - Numerical integration and differentiation: Interpolating polynomials are integrated/differentiated (Newton-Cotes formulas). - Computer graphics and CAD: Generating sm...
Full solved answer →Write an algorithm and program to compute the interpolation using Lagrange Interpolation.[10]
Lagrange Interpolation is a method to find a polynomial that passes through a given set of data points. Given n+1 data points $(x0, y0), (x1, y1), \ldots, (xn, yn)$, the interpolating polynomial is: $$P(x) = \sum{i=0}^{n} yi \cdot Li(x)$$ where $Li(x)$ is t...
Full solved answer →Interpolation versus regression
How interpolation differs from regression? Write down algorithm and program for Lagrange interpolation.[10]
--- Aspect Interpolation Regression --------- Definition Estimates the value of a function at a point within the given data range Finds the best-fit curve through a set of data points Data fit The curve passes exactly through all given data points The curve...
Full solved answer →Newton's backward difference table construction
Construct Newton's backward difference table for the given data points and approximate the value of $f(x)$ at $x=45$.
| $X$ | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| $f(x)$ | 0.173 | 0.342 | 0.5 | 0.643 | 0.766 |
[5]
x 10 20 30 40 50 ----------------------- f(x) 0.173 0.342 0.500 0.643 0.766 $h = 10$, and we want $f(45)$. x f(x) $\nabla f$ $\nabla^2 f$ $\nabla^3 f$ $\nabla^4 f$ -------------------------------- 10 0.173 20 0.342 0.169 30 0.500 0.158 -0.011 40 0.643 0.143...
Full solved answer →Make Unit 3 stick
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