Calculus - Newton's Method

Question

How would I use Newton's method to find all roots of this equation to 6 decimal places?

e^x = 3-2x

Answer 1

If e^x = 3 - 2x, then f(x) = e^x + 2x - 3 = 0.

Note that f(x) is monotone increasing, f(0) = -2 and, as e is approximately 2.71, f(1) = 1.71.

ln(2) = .69, so .7 will be too big (2 + 1.4 - 3 = .4);√2.71 is approximately 1.65, so .5 will be too small (1.65 + 2 * .5 - 3 = -.35); let's start at .6.

f'(x) = e^x + 2. This will be greater than 3 near our solution so when our function value is accurate to 6 digits, our root will be as well. Incidentally, f'(x) is monotone increasing tells us that, if we are to the right of the root, we will stay to the right of the root, but if we are to the left, we will go to the right in our first step.

Our Newton step is x_n+1 = x_n - f(x_n)/f'(x_n), or x_n+1 = x_n - (e^xn + 2x_n - 3)/(e^xn + 2)

Then, writing each row as x, f(x), and f'(x), and -f(x)/f'(x), the amount that we change x, and starting at x = .6, we get

0.600000, 0.022119, 3.822119, -0.005787

0.594213, 0.000030, 3.811605, -0.000008

0.594205, 0.000000, 3.811590, 0.000000

As our values are rounded off for display but calculated exactly, if we go back and let the third value be exactly .594205, we get 0.594205, 0.000000, 3.811590, 0.000000, which looks the same. Note how we stayed to the right as we stated we would. Also, displaying more digits or starting a little further from the solution would allow us to see the quadratic convergence.

Answer 2

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