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Solution5

# Solution5 - ChE 132B HW5 Solutions 100pt Problem...

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ChE 132B HW5 Solutions 100pt Problem 1. (5pt) %HW5 #1 x0=-1;x1=1;x2=2; [email protected](x) (x-x1).*(x-x2)./((x0-x1).*(x0-x2)); [email protected](x) (x-x0).*(x-x2)./((x1-x0).*(x1-x2)); [email protected](x) (x-x0).*(x-x1)./((x2-x0).*(x2-x1)); x=-1:0.1:2; plot(x,L0(x),x,L1(x),x,L2(x)) legend( 'L0' , 'L1' , 'L2' ) xlabel( 'x' ) ylabel( 'y' ) title( 'HW5 #1' )

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Problem 2. (5pt) For N=2, the Chebyshev points are : 𝑥 0 = 1, 𝑥 1 = 0, 𝑥 2 = 1 (1) The Chebyshev polynomials are : 𝑇 0 = 1, 𝑇 1 = 𝑥 , 𝑇 2 = 2 𝑥 2 1 (2) Plug Eq.(2) into 𝑦 𝑁 = 𝑐 𝑖 𝑇 𝑖 ( 𝑥 ) 𝑁 𝑖=0 , 𝑦 = 𝑐 0 + 𝑐 1 𝑥 + 𝑐 2 (2 𝑥 2 1) (3) At x 0 , x 1 , and x 2 , the values of y(x) are : 𝑦 ( 𝑥 0 ) = 𝑦 0 = 𝑐 0 + 𝑐 1 + 𝑐 2 𝑦 ( 𝑥 1 ) = 𝑦 1 = 𝑐 0 − 𝑐 2 𝑦 ( 𝑥 2 ) = 𝑦 2 = 𝑐 0 − 𝑐 1 + 𝑐 2 (4) Solve the above system of equation with respect to the Chebyshev polynomial coefficients, 𝑐 0 = 𝑦 0 4 + 𝑦 1 2 + 𝑦 2 4 𝑐 1 = 𝑦 0 − 𝑦 2 2 𝑐 2 = 𝑦 0 − 𝑦 1 2 𝑦 0 − 𝑦 2 4 (5) Differentiate Eq.(3) and plug the Chebyshev polynomial coefficients into it, 𝑦 = 𝑐 1 + 4 𝑐 2 𝑥 (6) 𝑦 = 𝑦 0 2 𝑦 2 2 + �𝑦 0 2 𝑦 1 + 𝑦 2 4 � 𝑥 (7) At x 0 ,x 1 , and x 2 , the values of y’(x) are : 𝑦 ( 𝑥 0 ) = 3 2 𝑦 0 2 𝑦 1 + 𝑦 2 2 𝑦 ( 𝑥 1 ) = 𝑦 0 2 𝑦 2 2 𝑦 ( 𝑥 2 ) = 𝑦 0 2 + 2 𝑦 1 3 2 𝑦 2 (8)
In matrix form, 𝑦 0 𝑦 1 𝑦 2 = 3 2 2 1 2 1 2 0 1 2 1 2 2 3 2 𝑦 0 𝑦 1 𝑦 2 (9) Finally, the Chebyshev differentiation matrix for N=2 is : 𝐷 2 = 3 2 2 1 2 1 2 0 1 2 1 2 2 3 2 (10) Eq.(10) is confirmed using chebD.m >> chebD(2) ans = 1.5000 -2.0000 0.5000 0.5000 0 -0.5000 -0.5000 2.0000 -1.5000

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Problem 3. (20pt) % HW5 3.a (5pt) n=16; xfine = 0:.005:1; clf % a fine grid x = (0:n)/n; % x data values, equally spaced y = sin(2.*x).*cos(5.*x); % y values of data yfine = sin(2.*xfine).*cos(5.*xfine); % "exact" function on fine grid yfit = interp1(x,y,xfine, 'spline' ); plot(x,y, '.' , 'markersize' ,13) line(xfine,yfit) axis([0 1 -1 1]), title( 'HW5 #3.(a)' ) xlabel( 'x'
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