%latex
7.13 (a)
The \emph{k}th Chebyshev polynomial of the first kind is defined on the interval [-1,1] by \\
\begin{align}
T_k(t) & = \cos ({k \arccos {(t)}}).
\end{align}
Given that \\
\begin{align}
T_0(t) & = 1 \\
T_1(t) & = t
\end{align}
The three-term recurrence for the \emph{k}th Chebyshev polynomial of the first kind is
\begin{align}
T_{k+1}(t) & = 2 \, t \, T_{k}(t) - T_{k-1}(t).
\end{align}
Proof: \\
Basis: (k = 1) \\
\begin{align}
T_{1+1}(t) & = 2 \, t \, T_{1}(t) - T_{1-1}(t) \\
T_{2}(t) & = 2 \, t \, T_{1}(t) - T_{0}(t) \\
& = 2 \, t \, \bullet \, t - 1 \\
& = 2 \, t ^ 2 - 1 \\
& = \cos ({2 \arccos {(t)}})
\end{align}
Induction step: (k = n + 1) \\
If
\begin{align}
T_{n}(t) & = 2t T_{n-1}(t) - T_{n-2}(t) \\
& = 2t \cos ({(n-1) \arccos {(t)}}) - \cos ({(n-2) \arccos {(t)}}) \\
& = \cos ({n \arccos {(t)}}),
\end{align}
then
\begin{align}
T_{n+1}(t) & = 2t T_{n}(t) - T_{n-1}(t) \\
& = 2t \cos ({n \arccos {(t)}}) - \cos ({(n-1) \arccos {(t)}}) \\
& = 2t \cos ({n \arccos {(t)}}) - \cos ({n \arccos {(t)} - \arccos {(t)}}) \\
& = 2t \cos ({n \arccos {(t)}}) - (\cos ({n \arccos {(t)}}) \cos ({\arccos {(t)}}) + \sin ({n \arccos {(t)}}) \sin ({\arccos {(t)}})) \\
& = 2t \cos ({n \arccos {(t)}}) - \cos ({n \arccos {(t)}}) \cos ({\arccos {(t)}}) - \sin ({n \arccos {(t)}}) \sin ({\arccos {(t)}}) \\
& = 2t \cos ({n \arccos {(t)}}) - t \cos ({n \arccos {(t)}}) - \sin ({n \arccos {(t)}}) \sin ({\arccos {(t)}}) \\
& = t \cos ({n \arccos {(t)}}) - \sin ({n \arccos {(t)}}) \sin ({\arccos {(t)}}) \\
& = \cos ({n \arccos {(t)}}) \cos ({\arccos {(t)}}) - \sin ({n \arccos {(t)}}) \sin ({\arccos {(t)}}) \\
& = \cos ({n \arccos {(t)} + \arccos {(t)}}) \\
& = \cos ({(n+1) \arccos {(t)}})
\end{align}
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%latex
7.13 (b)
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%sage
for k in range(6):
print "cos({0}*arccos(t)) =".format(k),cos(k*arccos(var('t'))).full_simplify()
print "Plot of cos({0}*arccos(t)) =".format(k)
plot(cos(k*arccos(var('t'))), (-1,1))
cos(0*arccos(t)) = 1 Plot of cos(0*arccos(t)) = cos(0*arccos(t)) = 1 Plot of cos(0*arccos(t)) = |
cos(1*arccos(t)) = t
Plot of cos(1*arccos(t)) =
cos(2*arccos(t)) = 2*t^2 - 1
Plot of cos(2*arccos(t)) =
cos(3*arccos(t)) = 4*t^3 - 3*t
Plot of cos(3*arccos(t)) =
cos(4*arccos(t)) = 8*t^4 - 8*t^2 + 1
Plot of cos(4*arccos(t)) =
cos(5*arccos(t)) = 16*t^5 - 20*t^3 + 5*t
Plot of cos(5*arccos(t)) =
%latex
7.13 (c)
The \emph{k} roots of the \emph{k}th Chebyshev polynomial of the first kind are the values of
\begin{align}
t_i \mbox{ , } i = 0,1,\ldots,k
\end{align}
such that \\
\begin{align}
\cos ({k \arccos {(t_i)}}) & = 0 \\
k \arccos {(t_i)} & = \frac {(2i-1) \pi} {2} \mbox{ , } i = 1,\ldots,k \\
\arccos {(t_i)} & = \frac {(2i-1) \pi} {2k} \mbox{ , } i = 1,\ldots,k \\
t_i & = \cos \left ( {\frac {(2i-1) \pi} {2k}} \right ) \mbox{ , } i = 1,\ldots,k \\
\end{align}
The \emph{k} + 1 extrema (including the endpoints) of the \emph{k}th Chebyshev polynomial of the first kind are the values of
\begin{align}
t_i \mbox{ , } i = 0,1,\ldots,k
\end{align}
\begin{align}
| \cos ({k \arccos {(t_i)}}) | & = 1 \\
k \arccos {(t_i)} & = i \pi \mbox{ , } i = 0,1,\ldots,k \\
\arccos {(t_i)} & = \frac {i \pi} {k} \mbox{ , } i = 0,1,\ldots,k \\
t_i & = \cos \left ( {\frac {i \pi} {k}} \right ) \mbox{ , } i = 0,1,\ldots,k
\end{align}
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%latex
8.10 (a)
The \emph{n} Chebyshev nodes in the interval [-1,1] are
\begin{align}
x_i & = \cos \left ( \frac { 2 i - 1 } { 2 n } \pi \right ) \mbox{ , } i = 1,\ldots,n.
\end{align}
For a three-point Chebyshev quadrature rule on the interval [-1, 1], we need to determine the weight \emph{w} and the nodes
\begin{align}
x_1 \mbox{ , } x_2 \mbox{ , } x_3.
\end{align}
%latex
Nodes: \\
\begin{align}
x_1 & = \cos \left ( \frac { 2 \bullet 1 - 1 } { 2 \bullet 3 } \pi \right ) \\
& = \cos \left ( \frac { \pi } { 6 } \right ) \\
x_1 & = \frac { \sqrt 3 } {2}
\end{align}
\begin{align}
x_2 & = \cos \left ( \frac { 2 \bullet 2 - 1 } { 2 \bullet 3 } \pi \right ) \\
& = \cos \left ( \frac { 3 \pi } { 6 } \right ) \\
x_2 & = 0
\end{align}
\begin{align}
x_3 & = \cos \left ( \frac { 2 \bullet 2 - 1 } { 2 \bullet 3 } \pi \right ) \\
& = \cos \left ( \frac { 5 \pi } { 6 } \right ) \\
x_3 & = - \frac { \sqrt 3 } {2}
\end{align}
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%latex
Moment Equations: \\
\begin{align}
w_1 \bullet 1 + w \bullet 1 + w \bullet 1 & = \int_{ -1 }^{ 1 } \! 1 \, dx \\
& = x \mid_{ -1 }^{ 1 } \! \\
& = 1 - ( -1 ) \\
& = 1 + 1 \\
& = 2
\end{align}
\begin{align}
w_1 \bullet x_1 + w \bullet x_2 + w \bullet x_3 & = \int_{ -1 }^{ 1 } \! x \, dx \\
& = \frac { x ^ 2 } { 2 } \mid_{ -1 }^{ 1 } \! \\
& = \frac { 1 ^ 2 - ( -1 ) ^ 2 } { 2 } \\
& = \frac { 1 - 1 } { 2 } \\
& = 0
\end{align}
\begin{align}
w_1 \bullet x_1^2 + w \bullet x_2^2 + w \bullet x_3^2 & = \int_{ -1 }^{ 1 } \! x^2 \, dx \\
& = \frac { x ^ 3 } { 3 } \mid_{ -1 }^{ 1 } \! \\
& = \frac { 1 ^ 3 - ( -1 ) ^ 3 } { 2 } \\
& = \frac { 1 - ( -1 ) } { 2 } \\
& = \frac { 1 + 1 } { 2 } \\
& = \frac { 2 } { 2 } \\
& = 1
\end{align}
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print 'In matrix form:'
print Matrix(RR,
[
[1 , 1 , 1 , 2],
[sqrt(3) / 2 , 0 , - sqrt(3) / 2 , 0],
[(sqrt(3) / 2) ^ 2 , 0 , (- sqrt(3) / 2) ^ 2 , 1]
]
)
print 'Solving by Gauss-Jordan:'
print Matrix(RR,
[
[1 , 1 , 1 , 2],
[sqrt(3) / 2 , 0 , - sqrt(3) / 2 , 0],
[(sqrt(3) / 2) ^ 2 , 0 , (- sqrt(3) / 2) ^ 2 , 1]
]
).echelon_form()
In matrix form: [ 1.00000000000000 1.00000000000000 1.00000000000000 2.00000000000000] [ 0.866025403784439 0.000000000000000 -0.866025403784439 0.000000000000000] [ 0.750000000000000 0.000000000000000 0.750000000000000 1.00000000000000] Solving by Gauss-Jordan: [ 1.00000000000000 0.000000000000000 0.000000000000000 0.666666666666667] [0.000000000000000 1.00000000000000 0.000000000000000 0.666666666666667] [0.000000000000000 0.000000000000000 1.00000000000000 0.666666666666667] In matrix form: [ 1.00000000000000 1.00000000000000 1.00000000000000 2.00000000000000] [ 0.866025403784439 0.000000000000000 -0.866025403784439 0.000000000000000] [ 0.750000000000000 0.000000000000000 0.750000000000000 1.00000000000000] Solving by Gauss-Jordan: [ 1.00000000000000 0.000000000000000 0.000000000000000 0.666666666666667] [0.000000000000000 1.00000000000000 0.000000000000000 0.666666666666667] [0.000000000000000 0.000000000000000 1.00000000000000 0.666666666666667] |
%latex
Weight: \\
\begin{align}
w & = \frac {2} {3}
\end{align}
Final Answers: \\
\begin{align}
x_1 & = \frac {\sqrt 3 } {2} \\
x_2 & = 0 \\
x_3 & = - \frac {\sqrt 3 } {2} \\
w & = \frac {2} {3}
\end{align}
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%latex
8.10 (b)
Since the resulting rule uses 3 polynomial basis functions, said quadrature rule will then integrate any polynomial of degree 2 exactly. \\
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