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Newton

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Polynomials are the most common choice of interpolants because they are easy to: ... temperature corresponding to 754.8 ohms using the Newton Divided Difference ... – PowerPoint PPT presentation

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Title: Newton


1
Newtons Divided Difference Polynomial Method of
Interpolation
  • Electrical Engineering Majors
  • Authors Autar Kaw, Jai Paul
  • http//numericalmethods.eng.usf.edu
  • Transforming Numerical Methods Education for STEM
    Undergraduates

2
Newtons Divided Difference Method of
Interpolation http//numericalmethods.eng.us
f.edu
3
What is Interpolation ?

Given (x0,y0), (x1,y1), (xn,yn), find the
value of y at a value of x that is not given.
4
Interpolants
  • Polynomials are the most common choice of
    interpolants because they are easy to
  • Evaluate
  • Differentiate, and
  • Integrate.

5
Newtons Divided Difference Method
  • Linear interpolation Given
    pass a linear interpolant through the data
  • where



6
Example
  • Thermistors are based on materials change in
    resistance with temperature. A manufacturer of
    thermistors makes the following observations on a
    thermistor. Determine the temperature
    corresponding to 754.8 ohms using the Newton
    Divided Difference method for linear
    interpolation.

R (O) T(C)
1101.0 911.3 636.0 451.1 25.113 30.131 40.120 50.128
7
Linear Interpolation
8
Linear Interpolation (contd)
9
Quadratic Interpolation
10
Example
  • Thermistors are based on materials change in
    resistance with temperature. A manufacturer of
    thermistors makes the following observations on a
    thermistor. Determine the temperature
    corresponding to 754.8 ohms using the Newton
    Divided Difference method for quadratic
    interpolation.

R (O) T(C)
1101.0 911.3 636.0 451.1 25.113 30.131 40.120 50.128
11
Quadratic Interpolation (contd)


12
Quadratic Interpolation (contd)
13
Quadratic Interpolation (contd)
14
General Form
where


Rewriting
15
General Form
16
General form
17
Example
  • Thermistors are based on materials change in
    resistance with temperature. A manufacturer of
    thermistors makes the following observations on a
    thermistor. Determine the temperature
    corresponding to 754.8 ohms using the Newton
    Divided Difference method for cubic
    interpolation.

R (O) T(C)
1101.0 911.3 636.0 451.1 25.113 30.131 40.120 50.128
18
Example

19
Example
20
Example
21
Comparison Table
22
Actual Calibration
R (O) T (C) x(ln R) y(1/T)
1101.0 911.3 636.0 451.1 25.113 30.131 40.120 50.128 7.0040 6.8149 6.4552 6.1117 0.039820 0.033188 0.024925 0.019949
23
Actual Calibration
24
Actual Calibration
25
Actual Calibration
26
Actual Calibration
27
Additional Resources
  • For all resources on this topic such as digital
    audiovisual lectures, primers, textbook chapters,
    multiple-choice tests, worksheets in MATLAB,
    MATHEMATICA, MathCad and MAPLE, blogs, related
    physical problems, please visit
  • http//numericalmethods.eng.usf.edu/topics/newton_
    divided_difference_method.html

28
  • THE END
  • http//numericalmethods.eng.usf.edu
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