lecture 6 - CE 30125 Lecture 6 LECTURE 6 LAGRANGE...

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CE 30125 - Lecture 6 p. 6.1 LECTURE 6 LAGRANGE INTERPOLATION • Fit points with an degree polynomial = exact function of which only discrete values are known and used to estab- lish an interpolating or approximating function = approximating or interpolating function. This function will pass through all specified interpolation points (also referred to as data points or nodes ). N 1 + N th f 1 x 0 g(x) f(x) f 0 f 2 f 3 f 4 f N x 1 x 2 x 3 x 4 x N ... fx  N 1 + gx N 1 +
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CE 30125 - Lecture 6 p. 6.2 •Th e interpolation points or nodes are given as: : • There exists only one degree polynomial that passes through a given set of points. It’s form is (expressed as a power series): where = unknown coefficients, ( coefficients). • No matter how we derive the degree polynomial, • Fitting power series • Lagrange interpolating functions • Newton forward or backward interpolation The resulting polynomial will always be the same! x o fx o  f o x 1 1 f 1 x 2 2 f 2 x N N f N N th N 1 + gx a o a 1 xa 2 x 2 a 3 x 3 a N x N + +++ + = a i i 0 N = N 1 + N
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CE 30125 - Lecture 6 p. 6.3 Power Series Fitting to Define Lagrange Interpolation must match at the selected data points : : • Solve set of simultaneous equations • It is relatively computationally costly to solve the coefficients of the interpolating func- tion (i.e. you need to program a solution to these equations). gx  fx o f o = a o a 1 x o a 2 x o 2 a N x o N +++ + f o = 1 f 1 = a o a 1 x 1 a 2 x 1 2 ++ a N x 1 N f 1 = N f N = a o a 1 x N + a 2 x N 2 a N x N N + f N = 1 x o x o 2 x o N 1 x 1 x 1 2 x 1 N  1 x N x N 2 x N N a o a 1 : a N f o f 1 : f N =
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CE 30125 - Lecture 6 p. 6.4 Lagrange Interpolation Using Basis Functions • We note that in general •L e t where = polynomial of degree associated with each node such that • For example if we have 5 interpolation points (or nodes) Using the definition for : ; ; ; ; ,we have: gx i  f i = f i V i x i 0 = N = V i x N i V i x j 0 i j 1 i = j 3 f o V o x 3 f 1 V 1 x 3 f 2 V 2 x 3 f 3 V 3 x 3 f 4 V 4 x 3 ++++ = V i x j V 0 x 3 0 = V 1 x 3 0 = V 2 x 3 0 = V 3 x 3 1 = V 4 x 3 0 = 3 f 3 =
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CE 30125 - Lecture 6 p. 6.5 • How do we construct ? • Degree • Roots at (at all nodes except ) •L e t • The function is such that we do have the required roots, i.e. it equals zero at nodes except at node • Degree of is • However in the form presented will not equal to unity at • We normalize and define the Lagrange basis functions V i x  N x o x 1 x 2 x i 1 x i 1 + x N  x i V i x i 1 = W i x xx o 1 2 i 1 i 1 + N = W i x o x 1 x 2 ... , x N x i W i x N W i x x i W i x V i x V i x o 1 2 i 1 i 1 + N x i x o x i x 1 x i x 2 x i x i 1 x i x i 1 + x i x N ------------------------------------------------------------------------------------------------------------------------------------------------------- =
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CE 30125 - Lecture 6 p. 6.6 • Now we have such that equals: • We also satisfy e.g.
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lecture 6 - CE 30125 Lecture 6 LECTURE 6 LAGRANGE...

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