Sample Problems, 4.8

Sample Problems, Lesson 4.8

Sample Problem: #8, Lesson 4.8

Use Newton's method to approximate the given number correct to eight decimal places:

Given problem: #8, Lesson 4.8

With some simple algebra, we can get this into an equation set equal to zero.

We also need the derivative so that Newton's method can be used.

x	f(x)	f'(x)
3	1187	5103
2.76739173	243.0668	3144.285
2.690087405	19.4472	2652.75
2.682756447	0.158271	2609.67
2.682695799	1.07E-05	2609.316
2.682695795	0	2609.316
2.682695795	0	2609.316

We use Newton's method as contained in the formula above. I used a spreadsheet to calculate. We have arrived at 8-digit accuracy when two consecutive values for x agree to all 8 digits.

The approximate value of is: 2.682695795

Sample Problem, #12, Lesson 4.8

Use Newton's method to find all the roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations.

Given Problem, #12, Lesson 4.8

We simplify the equation by multiplying out the left side, and then multiplying both sides by the denominator on the right side.

We plot the graph of the left side of the above equation. There appear to be 4places where this graph has a value of zero. The appear to be at approximately -2, -0.9, +0.9 and 2.

x	f(x)	f'(x)
-2	-4	80
-1.95	-0.39335	64.59187
-1.94391019	-0.00533	62.84649
-1.9438254	-1E-06	62.82239
-1.94382538	-4.1E-14	62.82238
-1.94382538	0	62.82238

We use Newton's method as contained in the formula above. I used a spreadsheet to calculate. We have arrived at 8-digit accuracy when two consecutive values for x agree to all 8 digits.

The approximation for the 1st root is -1.943825383

x	f(x)	f'(x)
-0.9	0.676859	-12.4051
-0.84543686	0.026551	-11.4234
-0.84311258	4.97E-05	-11.3806
-0.84310821	1.76E-10	-11.3805
-0.84310821	0	-11.3805

The approximation for the 2nd root is -0.84310821

x	f(x)	f'(x)
0.9	0.676859	12.40506
0.845436862	0.026551	11.42341
0.84311258	4.97E-05	11.3806
0.84310821	1.76E-10	11.38052
0.84310821	0	11.38052

The approximation for the 3rd root is 0.84310821

x	f(x)	f'(x)
2	-4	-80
1.95	-0.39335	-64.5919
1.943910186	-0.00533	-62.8465
1.943825399	-1E-06	-62.8224
1.943825383	-4.1E-14	-62.8224
1.943825383	0	-62.8224

The approximation for the 4th root is 1.943825383

Sample problem, #18, Lesson 4.8

Use Newton's method to find all the roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations.

Sample problem, #18, Lesson 4.8

We rearrange the terms so that the solutions to the equation will be x-intercepts on the graph.

We also calculate the derivative for use with Newton's Method.

We plot the graph of the left side of the above equation. There appear to be 2 places where this graph has a value of zero. The appear to be at approximately -2, and 1.3.

We note that the domain is limited to [-3,3].

x	f(x)	f'(x)
-2	-0.05103	6.668826
-1.99234826	-0.01291	6.86109
-1.99046632	-0.00329	6.910061
-1.98998953	-0.00084	6.922577
-1.98986783	-0.00022	6.925778
-1.98983671	-5.5E-05	6.926598
-1.98982875	-1.4E-05	6.926807
-1.98982671	-3.6E-06	6.926861
-1.98982619	-9.2E-07	6.926875
-1.98982606	-2.4E-07	6.926878
-1.98982602	-6.1E-08	6.926879
-1.98982601	-1.5E-08	6.926879
-1.98982601	-4E-09	6.926879

We use Newton's method as contained in the formula above. I used a spreadsheet to calculate. We have arrived at 8-digit accuracy when two consecutive values for x agree to all 8 digits.

The approximation for the 1st root is: -1.98982601

x	f(x)	f'(x)
1.3	0.898401	13.49432
1.233423744	0.116047	8.675818
1.220047849	-0.00755	8.025003
1.220988546	0.000858	8.068322
1.220882173	-9.5E-05	8.063406
1.220893896	1.04E-05	8.063948
1.2208926	-1.2E-06	8.063888
1.220892743	1.28E-07	8.063894
1.220892727	-1.4E-08	8.063894
1.220892729	1.56E-09	8.063894
1.220892729	-1.7E-10	8.063894

The approximation for the 2nd root is: 1.220892729

	*Assignment*
1, 3, 5, 11, 17
Lesson 4.8, Pages 327-329