Method of Linear Interpolation

8/3/2019 Method of Linear Interpolation

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Method of Linear Interpolation:

This method is also known as themethod of false position, and bythe Latinized version regula falsi.It is also a very old method.

By: Ouday N. Ameen


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By: Ouday N. Ameen


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x3 = x2 f(x2) [(x2 x1)/(f(x2) f(x1)]

Where x3 represents the intersection of

the strait line with the x-axis. Wecalculate f(x3) and then test:

If f(x1) * f(x3) < 0 , then set x2 = x3

If f(x1) * f(x3) > 0 , then set x1 = x3 Repetition of this will give estimation of

the root.

By: Ouday N. Ameen


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Algorithm:

to determine the root of f(x) = 0, given values of x1 andx2 such that f(x1) and f(x2) are of opposite signs.

Repeat

Set x3 = x2 f(x2) [(x2 x1)/(f(x2) f(x1)] if f(x1) * f(x3) < 0, then Set x2 = x3 Else x1 = x3 End if Until abs(f(x3)) tolerance value

By: Ouday N. Ameen


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Example:

find the positive root of the cubicequation; f(x) = x3 + x2 -3x 3 = 0,starting with x1 = 1.0 and x2 = 2.0,and the tolerance equal to 0.001.

By: Ouday N. Ameen


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Note that although the false positionmethod aims to improve theconvergence speed over the bisection

method, it can not always achieve thegoal, especially when the curve of f(x)on [x1, x2] is not well approximated by

a strait line as depicted in figurebelow.

By: Ouday N. Ameen


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f(x) = tan( x) x = 0.

By: Ouday N. Ameen


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The Secant Method:This method is used to improve the method of linearinterpolation.

Instead of requiring that the function have opposite signs atthe two values used for interpolation, we can choose the two

values nearest the root and interpolate or extrapolate fromthese.

Usually the nearest values to the root will be the last twovalues calculated.

This makes the interval under consideration shorter andhence improves the assumption that the function can berepresented by the line through the two points.

By: Ouday N. Ameen


10/14By: Ouday N. Ameen


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x3 = x2 [f(x2) (x2 x1) / (f(x2) f(x1))]

Algorithm: given f(x) = 0, choose x1 andx2 close enough to the root.

Repeat Set x3 = x2 [f(x2) (x2 x1) / (f(x2) f(x1))] if abs(f(x2)) < abs(f(x1)) then

Set x1 = x2 End if Set x2 = x3 Until abs(f(x3)) tolerance value

By: Ouday N. Ameen


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Example:

find the positive root off(x) = x3 + x2 -3x 3, let

x1 = 1.0 and x2 = 2.0,tolerance = 0.001.

By: Ouday N. Ameen


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Solution:

#Ino. x1 x2 x3 f(x1) f(x2) f(x3)1 1.0 2.0 1.57142 -4.0 3.0 -1.364492 2.0 1.57142 1.70540 3.0 -1.36449 -0.247843 1.57142 1.70540 1.73513 -1.36449 -0.24784 -0.02924 1.70540 1.73513 1.73199 -0.24784 -0.0292 -0.000575 1.73513 1.73199 1.73205 -0.0292 -0.00057

*The positive root is x3= 1.73205 = 3.

By: Ouday N. Ameen


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Method of Linear Interpolation

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