Interpolation Lec

8/13/2019 Interpolation Lec

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In the mathematical field of numerical

analysis, interpolationis a method of

constructing new data points within therange of a discrete set of known data

points.


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Many a times, a function y = f(x) is given only at discretepoints such as (xo, yo),(x1, y1),,(xn-1,yn-1), (xn,yn).How does one find the value of y at any other value of x ?A continuous function f(x) may be used to represent the

n+1 data values with f(x) passing through the n+1 points.Then one can find the value of y at any other value of x .This is called interpolation. Of course, if x falls outsidethe range of x for which the data is given, it is no longer

interpolation but instead is called extrapolation.


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Y

X(xo,yo)

(x1,y1) (x2,y2)

(x3,y3)fx


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Sir Edmund Whittaker, a professor of NumericalMathematics at the University of Edinburghfrom 1913 to 1923, observed the most common

form of interpolation occurs when we seek datafrom a table which does not have the exactvalues we want.

Throughout history, interpolation has beenused in one form or another for just about everypurpose under the sun.


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Speaking of the sun, some of the first surviving evidenceof the use of interpolation came from ancient Babylonand Greece.Around 300 BC, they were using not only linear, but also

more complex forms of interpolation to predict thepositions of the sun, moon, and the planets they knew of.Farmers, timing the planting of their crops, were theprimary users of these predictions.

Also in Greece sometime around 150 BC, Hipparchus ofRhodes used linear interpolation to construct a chordfunction, which is similar to a sinusoidal function, tocompute positions of celestial bodies.


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Farther east, Chinese evidence of interpolationdates back to around 600 AD.Liu Zhuo used the equivalent of second order

Gregory-Newton interpolation to construct anImperial Standard Calendar.In 625 AD, Indian astronomer and

mathematician Brahmagupta introduced amethod for second order interpolation of thesine function and, later on, a method for

interpolation of unequal-interval data.


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Direct Method of

Interpolation


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To maximize a catch of bass in a lake, it issuggested to throw the line to the depth of thethermocline. The characteristic feature of this

area is the sudden change in temperature. Weare given the temperature vs. depth data for alake in Table 1.


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Temperature, T (C) Depth, Z (m)

19.1 0

19.1 -1

19 -2

18.8 -3

18.7 -4

18.3 -5

18.2 -6

17.6 -7

11.7 -8

9.9 -9

9.1 -10


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Using the given data, we see the largest

change in temperature is between z = -8 m

and z = -7 m. Determine the value of thetemperature atz = -7.5 m using the direct

method of interpolation and a first order

polynomial.


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For first order polynomial interpolation (alsocalled linear interpolation), we choose thetemperature given by

T(z) = ao+ a1zzo= -8 , T(zo) = 11.7

z1= -7, T(z1) = 17.6

givesT(-8) = ao+ a1(-8) = 11.7T(-7) = ao+ a1(-7) = 17.6


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Writing the equations in matrix form, we have

1 - 8 ao 11.7

1 -7 a1 17.6

Solving the above two equations givesao = 58.9 and a1 = 5.9

=


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Hence,

T(z) = ao+ a1zT(z) = 58.9 + 5.9 zT(-7.5) = 58.9 + 5.9(-7.5)T(-7.5) = 14.65C


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For second order polynomial interpolation

(also called quadratic interpolation), we

choose the temperature given by

T(z) = ao+ a1z + a2z


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zo= -9 , T(zo) = 9.9z1= -8, T(z1) = 11.7

z2= -7, T(z2) = 17.6givesT(-9) = ao+ a1(-9) + a2(-9)= 9.9T(-8) = ao+ a1(-8) + a2(-8)= 11.7T(-7) = ao+ a1(-7) + a2(-7)= 17.6


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Writing the equations in matrix form, we have1 -9 81 ao 9.91 - 8 64 a1 11.71 -7 49 a2 17.6

Solving the above three equations gives

ao = 173.7, a1= 36.65, and a2= 2.05.

=


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Hence,

T(z) = ao+ a1z + a2z

T(z) = 173.7 + 36.65 z + 2.05 z

at z = -7.5

T(-7.5) = 173.7 + 36.65(-7.5) + 2.05(-7.5)= 14.138C


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The absolute relative approximate error | a |obtained between the results from the first

and second order polynomial is

| a | = 14.138 14.65 x 100%14.138

= 3.6251%


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For third order polynomial interpolation

(also called cubic interpolation), we

choose the temperature given by

T(z) = ao+ a1z + a2z + a3z


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zo= -9 , T(zo) = 9.9z1= -8, T(z1) = 11.7z2= -7, T(z2) = 17.6

z3= -6, T(z3) = 18.2givesT(-9) = ao+ a1(-9) + a2(-9) + a3(-9)= 9.9

T(-8) = ao+ a1(-8) + a2(-8) + a3(-8) = 11.7T(-7) = ao+ a1(-7) + a2(-7) + a3(-7)= 17.6T(-6) = ao+ a1(-6) + a2(-6) + a3(-6)= 18.2


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Writing the equations in matrix form, we have1 -9 81 ao 9.91 - 8 64 a1 11.7

1 -7 49 a2 17.61 -6 36 a3 18.2

Solving the above three equations givesao = -615.9, a1= -262.58, a2= -35.55 anda3= -1.5667

=


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Hence,

T(z) = ao+ a1z + a2z + a3z

T(z) = -615.9

262.58 z

35.55 z -1.5667 z

at z = -7.5

T(-7.5) = -615.9 -262.58(-7.5)

35.55(-7.5)-1.5667(-7.5)

= 14.725C


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The absolute relative approximate error | a |obtained between the results from the 2ndand 3rd order polynomial is

| a | = 14.725 14.138 x 100%14.725

= 3.9898%

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