MATH 472 002/992Final ExamDecember 16, 2020Read the following instructions carefully:•There areseven problemsand you have120 minutes.•Each submission must be asingle pdffile including your solutions in the order they wereassigned.•Do not use red ink or colorin your submission.•Write your name at the top right of each page.•Each problem should start on a fresh page in your submission pdf.•You may useone double-sided sheet(8.5×11inches) of notes in yourown handwriting.No other notes or books are allowed.•You may use a calculator, but all intermediate steps in your calculations must be compre-hensible from your notes. Simplify your results as far as possible.•All results have to bejustifiedby appropriateintermediate stepsand/orarguments.You can, however, use standard formulas and results known from the lecture without deriva-tion (unless explicitly stated otherwise).•You will have15 minutes to upload your submission to Canvas after the end ofthe exam.Good luck!Sheet 1 of 15
Problem 1(12 Points)Name:Assume that you are to use Chebyshev interpolation to find a degree 2 interpolating polynomialQ2(x)that approximates the functionf(x) =x1/3(=3√x)on the interval[3,4].(a)(5 Points) Write down the(x, y)points that will serve as interpolation nodes forQ2.(b)(4 Points) Find a worst-case estimate for the error|f(x)-Q2(x)|that is valid for allxinthe interval[3,4].(c)(3 Points) How many digits after the decimal point will be correct whenQ2is used toapproximatef?Solution:(a)For a2-degree interpolating polynomial, we needn= 3nodes.(1P)The Chebyshev inter-polation nodes on[-1,1]are given by¯x1:= cos(π6)=√32,¯x2:= cos(3π6)= cos(π2)= 0and¯x3:= cos(5π6)=-√32.(1P)Therefore, for the interval[a, b] = [3,4]the interpolating nodes arexi:=a+b2+b-a2¯xi=72+12¯xi,fori= 1,2,3.(2P)Their approximations arex1≈3.933,x2= 3.5andx3≈3.067.In total, the required nodes are(x1, f(x1))= (3.933,3√3.933) = (3.933,1.578528),(x2, f(x2))= (3.5,3√3.5) = (3.5,1.5183)and(1P)(x3, f(x3))= (3.067,3√3.067) = (3.067,1.4529).(b)The interpolation error formula givessupx∈[3,4]|f(x)-Q2(x)|= supx∈[3,4]n|(x-x1)(x-x2)(x-x3)|3!|f(3)(x)|o(1P)≤125×3!×0.019784≈0.000103.(1P)where we have used that|(x-x1)(x-x2)(x-x3)| ≤b-a2322=125(1P)andf(3)(x) =13x-2300=-29x-530=1027x-83with0≤f(3)(x)≤f(3)(3) = 0.019784forx∈[3,4](1P).