How do you use Newton's method to find the two real solutions of the equation #x^4-x^3-3x^2-x+1=0#?

1 Answer
Dec 4, 2016

The roots of the equation #x^4-x^3-3x^2-x+1 =0# are #0.4221# and #2.3692# to 4dp.

Explanation:

First let us look at the graphs (not strictly needed but it gives s insight into the problem):

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We can see there are two solutions. There is one solution in the interval #0 < x < -1 # and a second solution in the interval # 2 < x < 3 #. We can find the solution to #f(x)-0# numerically, using Newton-Rhapson method.

Let

#\ \ f(x) = x^4-x^3-3x^2-x+1 => #
# f'(x) = 4x^3-3x^2-6x-1 #

and we will need to apply the Newton-Rhapson method twice to find both roots. We use the following iterative sequence

# { (x_0=alpha), ( x_(n+1)=x_n - f(x_n)/(f'(x_n)) ) :} #

To find the first root (arbitrarily) choose #alpha=1#

Then using excel (working to 6dp) we can tabulate the iterations as follows:

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To find the second root (arbitrarily) choose #alpha=3#
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We would need to vary #alpha# if the NR Method diverged or provided the incorrect root.

And so we conclude that the roots of the equation

#x^4-x^3-3x^2-x+1 =0#

are #0.4221# and #2.3692# to 4dp.