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use python 3. Wilkinson's Polynomial program, two versions. (12 points) Use this for 2nd version: ***20 -210****19 + 20615****18 - 1256850****17 +53327946****16 - 1672280820*x**15+ 40171771630*x**14
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3. Wilkinson's Polynomial program, two versions. (12 points) Use this for 2nd version: ***20 -210****19 + 20615****18 - 1256850****17 +53327946****16 - 1672280820*x**15+ 40171771630*x**14 - 756111184500*x**13+11310276995381****12 - 135585182899530*x**11 + 1307535010540395*x**10 - 10142299865511450****9 + 63030812099294896****8-311333643161390640****7 + 1206647803780373360*x**6 - 3599979517947607200*x**5 + 8037811822645051776****4 - 12870931245150988800*x**3 + 13803759753640704000*x**2- 8752948036761600000*x + 2432902008176640000 Use a-b> 1e-7 as your tolerance. Attach your code and a copy of the full output. 4. (2 points) of the twenty estimates your program made, which root estimate has the largest forward error? Compute the corresponding backward error (plug the estimate into the Python function that computes the factored version of the Wilkinson polynomial.) 3. Wilkinson's Polynomial program, two versions. (12 points) Use this for 2nd version: ***20 -210****19 + 20615****18 - 1256850****17 +53327946****16 - 1672280820*x**15+ 40171771630*x**14 - 756111184500*x**13+11310276995381****12 - 135585182899530*x**11 + 1307535010540395*x**10 - 10142299865511450****9 + 63030812099294896****8-311333643161390640****7 + 1206647803780373360*x**6 - 3599979517947607200*x**5 + 8037811822645051776****4 - 12870931245150988800*x**3 + 13803759753640704000*x**2- 8752948036761600000*x + 2432902008176640000 Use a-b> 1e-7 as your tolerance. Attach your code and a copy of the full output. 4. (2 points) of the twenty estimates your program made, which root estimate has the largest forward error? Compute the corresponding backward error (plug the estimate into the Python function that computes the factored version of the Wilkinson polynomial.) Step by Step Solution
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