Newton's Method

[dear2009]

Hey Everybody,


Apply Newton's method to find a root of

f(x) = x3 – 10

in the interval (2, 3). If your initial guess is x0 = 2, then approximately
how far is x2 from the cube root of 10 (the actual root of the equation)?

This is how I set it up
x1 = x0 - f(x0)/ f ' (x0)
= 2 - 8 -10/ 3 - 4
= 2 + 1/6
= 13/6, which i then plug in again to get
x2 = x1 - f(x1) / f ' (x1)
f ' (x) = 3x^2

x2 = 13/6 - (2197 / 216) / (4.69444)

I dont know what to do afterwards, I think I am supposed to get 0.000068926

 

[Deleted member 4993]

Hey Everybody,


Apply Newton's method to find a root of

f(x) = x3 – 10

in the interval (2, 3). If your initial guess is x0 = 2, then approximately
how far is x2 from the cube root of 10 (the actual root of the equation)?

This is how I set it up
x1 = x0 - f(x0)/ f ' (x0)
= 2 - 8 -10/ 3 - 4
= 2 + 1/6
= 13/6, which i then plug in again to get
x2 = x1 - f(x1) / f ' (x1)
f ' (x) = 3x^2

x2 = 13/6 - (2197 / 216) / (4.69444)

I dont know what to do afterwards, I think I am supposed to get 0.000068926

On the top of the "writing window" - there are buttons to make subscripted or superscripted text. Use those to clarify your work.

As presented, it will take too much work on my part to decipher.

 

[Deleted member 4993]

Hey Everybody,


Apply Newton's method to find a root of

f(x) = x3 – 10

in the interval (2, 3). If your initial guess is x0 = 2, then approximately
how far is x2 from the cube root of 10 (the actual root of the equation)?

This is how I set it up
x1 = x0 - f(x0)/ f ' (x0)
= 2 - 8 -10/ 3 - 4
= 2 + 1/6
= 13/6, which i then plug in again to get
x2 = x1 - f(x1) / f ' (x1)
f ' (x) = 3x^2

x2 = 13/6 - (2197 / 216) / (4.69444) ? evaluate this expression

then use calculator to find cube-root of 10 (i.e. 10[sup:31m05blj]1/3[/sup:31m05blj]) - then find the difference from your answer

I dont know what to do afterwards, I think I am supposed to get 0.000068926

On the top of the "writing window" - there are buttons to make subscripted or superscripted text. Use those to clarify your work.

As presented, it will take too much work on my part to decipher.

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ f(x) \ = \ x^{3}-10, \ and \ you \ are \ to \ find \ a \ zero \ between \ (2,3).\)

\(\displaystyle Note, \ since \ f(x) \ is \ a \ polynomial, \ Descartes' \ rule \ of \ signs \ tell \ us \ there \ is \ one \ positive \ root\)

\(\displaystyle and \ two \ imaginary \ roots. \ Obviously, \ the \ positive \ root \ lies \ between \ (2,3).\)

\(\displaystyle Newton's \ Method: \ x_{n+1} \ = \ x_n-\frac{f(x_n)}{f'(x_n)}\)

\(\displaystyle f(x_n) \ = \ x_n^{3}-10, \ f'(x_n) \ = \ 3x_n^{2}, \ Let's \ start \ with \ x_n \ = \ x_1 \ = \ 2.5\)

\(\displaystyle Hence, \ x_2 \ = \ 2.5-\frac{2.5^{3}-10}{3(2.5)^{2}} \ = \ 2.2\)

\(\displaystyle x_3 \ = \ 2.2-\frac{2.2^{3}-10}{3(2.2)^{2}} \ = \ 2.15537190083\)

\(\displaystyle x_4 \ = \ 2.15537190083-\frac{2.15537190083^{3}-10}{3(2.15537190083)^{2}} \ = \ 2.1544350975\)

\(\displaystyle x_5 \ = \ 2.1544350975-\frac{2.1544350975^{3}-10}{3(2.1544350975)^{2}} \ = \ 2.15443469003\)

\(\displaystyle Ergo, \ with \ four \ iterations, \ we \ have \ the \ solution \ correct \ to \ 11 \ decimal \ places.\)

Note: In Newton's time he had people (known as computers) who would do the number crunching to arrive at the correct answer. Fortunately, today we have electronic computers to do our grunt work.