Newton's Method

[ringokid]

Use Newton's method with initiatl approximation
a) x(o)=sqrt(3)
b) x(o)=1.7320508
c) x(o)=1.73205081
x(o) = x(to the 0 like it is when its a logbase)

Use those to solve the equation x^5-81x=0

Find the Exact Limits.

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ x^5-81x \ = \ x(x^4-81), \ \implies \ x \ = \ 0, \ x \ = \ \pm3\)

\(\displaystyle Hence, solutions \ are \ (0,0),(-3,0),(3,0)\)

\(\displaystyle Note: \ Descartes' \ Rule \ of \ Signs, \ to \ wit:\)

\(\displaystyle f(x) \ = \ x^5-81x, \ one \ change, \ f(-x) \ = \ -x^5+81x, \ one \ change, \ ergo \ only \ possibility \ is\)

\(\displaystyle -+ 0 \ \ I\)
\(\displaystyle 1 \ \ 1 \ 1 \ \ 2\)

\(\displaystyle As \ far \ as \ Newton's \ Method \ goes, \ have \ fun.\)

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