ex03_solutions

Solutions

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0	0
0.5`	0.3032653298563167`
1.`	0.36787944117144233`
1.5`	0.33469524022264474`
2.`	0.2706705664732254`
2.5`	0.205212496559747`
3.`	0.14936120510359183`
3.5`	0.10569084197811475`
4.`	0.07326255555493671`
4.5`	0.04999048442209038`
5.`	0.03368973499542734`

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(*3*)poly = InterpolatingPolynomial[A, x] ; Expand[%]

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In the next excercise we'll do some tricks. The function Select takes as arguments a table and a criterion which it will apply to the table. In this case the criterion, #[[1]]==0.0& is in the form of a pure function, the argument of which, denoted by #, is a list. We test whether the first element of this list (which in this case is an element of A, equals to zero (or 1 or 2,5). The command & at the end just signifies that this really is a pure function. Select returns a list of elements of A which satisfy the criterion. We assume that there is only one element in A which satisfies our criterion and take the second element of the first element of the resulting list. This excercise can be done in a more simple way by just extracting A[[1,2]], A[[3,2]] and A[[6,2]].

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(*4*)Part[Select[A, #[[1]] 0.&], 1, 2] - poly/.x0.Part[S ... ] - poly/.x1.Part[Select[A, #[[1]] 2.5&], 1, 2] - poly/.x2.5

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(*5*)ListPlot[A, PlotStyle {PointSize[0.01], Hue[0.]}, DisplayFunctionᢃ ... DisplayFunctionIdentity] ; Show[%, %%, DisplayFunction$DisplayFunction]

[Graphics:../HTMLFiles/ex03_solutions_44.gif]

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$(*6*)Plot[poly - x Exp[-x], {x, 0, 5}, PlotRange {-6 * 10^(-6), 6 * 10^(-6)}]$

[Graphics:../HTMLFiles/ex03_solutions_47.gif]

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Created by Mathematica (April 10, 2007)