MATH256 Problem Sheet 6
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MATH256 Problem Sheet 6
6.1 (a) Use the Lagrange interpolation formula to find the quadratic polynomial P (x) which satisfies P (x) = ex for x = -1, x = 0 and x = 1.
(b) Use the Lagrange interpolation formula to find the quadratic polynomial Q(x) which satisfies Q(x) = ex for x = -^3/2, x = 0 and x =^3/2.
(c) Plot graphs of the three functions
|ex - P (x)|, |ex - Q(x)| and |ex - R(x)|
where R(x) is the three term Taylor series approximation to ex , i.e.
R(x) = 1 + x + x2
(d) Using your graphs, estimate the maximum absolute errors in the three approximations.
6.2 (a) Create a divided difference table for the following data:
x 
0.1 0.3 0.8 1.0
y 
-0.2 -0.4 1.0 0.2
Record each value in the table to six correct significant digits.
(b) Write down the Newton polynomial resulting from part (a) and calculate its value at x = 0.6.
(c) Add an extra row to the table from part (a), using the new data point x = 0.9, y = 0.4. Then evaluate the revised Newton polynomial at x = 0.6.
6.3 (a) Using Maple, generate an array containing 11 equally spaced nodes xj ,
with x0 = -1 and x10 = 1. Also generate an array of y values with yj = cos(πxj ), j = 0, 1, . . . , 10.
(b) Using the procedures in d taX)mw, generate the Newton polynomial P such that P (xj ) = yj for the eleven points in your array. Plot the absolute error in the approximation cos(πx) ≈ P(x) for -1 s x s 1.
(c) Add the extra point (x11 , y11 ) = /0.1, cos(0.1π)、 to the divided differ- ence table, and generate a new Newton polynomial Q(x) such that Q(xj ) = yj and Q(0.1) = cos(0.1π). Plot the absolute error in the approximation cos(πx) ≈ Q(x) for -1 s x s 1.
(d) Which of the approximations cos(πx) ≈ P(x) and cos(πx) ≈ Q(x) has the greater maximum error? Why do you think this happens?
6.4 (a) Use the fact that Tn (cos θ) = cos(nθ) to simplify the expression Tn /Tm (t)、.
(b) Use Maple to check your result for several different positive values of n and m.
Hint: the shmpihey command will convert a Chebyshev polynomial to explicit form; for example try
shmpihey( CfaXysfavT( 3 ( t ) )
6.5 Use the fact that Tn (cos θ) = cos(nθ) to simplify the expression Tn-1 (t) + Tn+1(t). Hence prove that the leading order coefficient of Tn (t) is 2n-1 , for n > 0.
6.6 (a) By differentiating the expression Tn (cos θ) = cos(nθ), find all points at
which T
(t) = 0, for n > 0 . Make sure you have the correct number of
results, and that none are repeated .
(b) Do the results for θ = 0 and θ = π contradict the results of §4 .8 .3?
Why (or why not)?
2023-05-06