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Numerical Analysis Lecture 28
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Chapter 7 Numerical Differentiation and Integration
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INTRODUCTION DIFFERENTIATION USING DIFFERENCE OPREATORS DIFFERENTIATION USING INTERPOLATIONRICHARDSON’S EXTRAPOLATION METHOD NUMERICAL INTEGRATION
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NEWTON-COTES INTEGRATION FORMULAE THE TRAPEZOIDAL RULE ( COMPOSITE FORM ) SIMPSON’S RULES ( COMPOSITE FORM ) ROMBERG’S INTEGRATION DOUBLE INTEGRATION
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DIFFERENTIATION USING DIFFERENCE OPREATORS: Applications: Remember Using forward difference operator ∆, the shift operator, the backward difference operator and the average difference operator, we obtained the following formulae: Remember Using forward difference operator ∆, the shift operator, the backward difference operator and the average difference operator, we obtained the following formulae:
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Recall from what we mentioned in the last lecture that for calculating the second derivative at an interior tabular point, we use the equation while for computing the first derivative at an interior tabular point, we in general use another convenient form for D, which is derived as follows. Multiply the right hand side of another convenient form for D, which is derived as follows. Multiply the right hand side of
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by
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which is unity and noting the Binomial expansion
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we get simplification we get
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Therefore the equation can also be written in another useful form as
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The last two equations y” and y’ respectively are known as Stirling’s formulae for computing the derivatives of a tabular function. Similar formulae can be derived for computing higher order derivatives of a tabular function.
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The equation for y’ can also be written as
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In order to illustrate the use of formulae derived so far, for computing the derivatives of a tabulated function, we consider the Following example :
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Example: Compute and from the following tabular data.
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1.00 1.00 1.16 1.16 3.56 3.56 13.96 13.96 41.96 41.96101.00 0.00.20.40.60.81.0
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Solution Since x = 0 and 0.2 appear at and near beginning of the table, it is appropriate to use formulae based on forward differences to find the derivatives. The difference table for the given data is depicted below:
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0.0 1.00 0.2 1.16 0.4 3.56 0.6 13.96 0.8 41.96 1.0 101.00 0.16 2.40 10.40 28.00 59.04 2.24 8.00 17.60 31.04 5.76 9.60 13.44 3.84 3.84 0.0
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Using forward difference formula for i.e
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We obtain We obtain
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Also, using the formula for we have we haveHence,
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Example Find and from the table x y ( x ) 1.4 4.0552 1.6 4.9530 1.8 6.0496 2.0 7.3891 2.2 9.0250
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Solution: Since x=2.2 occurs at the end of the table, it is appropriate to use backward difference formulae for derivatives. The backward difference table for the given data is shown below:
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1.41.61.82.02.2 4.05524.96306.04967.38919.0250 0.8978 1.0966 1.3395 1.6359 0.19880.24290.2964 0.04410.0535 0.0094
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Using backward difference formulae for and we have
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Therefore,Also
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Therefore,
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Case IV: Derivation of Two and three point formulae: Retaining only the first term in equation :
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we can get another useful form for the first derivative as
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Similarly, by retaining only the first term in Eqn.
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Adding the last two equations, we have
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These Equations constitute two-point formulae for the first derivative. By retaining only the first term in Equation,
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We get
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Similarly, we have
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While retaining only the first term in the expression for y” in terms of we obtain
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The last three equations constitute three-point formulae for computing the second derivative. We shall see later that these two- and three-point formulae become handy for developing extrapolation methods to numerical differentiation and integration.
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Numerical Analysis Lecture 28
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