1. ## newton's method?

Am i approximate 2 ^ (1/5) correctly with newton's method? (assuming x0 = 1, find x1,x2)

since 2 ^ (1/5) = x

i get the function f(x) = x^5-2
f'(x) = 5x^2

therefore, x0 = 1

x1 = 1- f(1)/f'(1) = = -2/5

but when i find x2 = -2/5 - f(2/5)/f'(2/5) = -18.67

i get a large negative number for x2 but x1 is like -2/5. Is this correct? The difference is so large :/

2. ## Re: newton's method?

Hi lc99

and also, substitutions a negative value in f(x) and f'(x) should include its negative sign....

My advice in newton's method is write the whole equation first:

${x_n} = {x_{n - 1}} - \frac{{f({x_{n - 1}})}}{{f'({x_{n - 1}})}}$
thus, in this case,
${x_n} = {x_{n - 1}} - \frac{{{x_{n - 1}}^5 - 2}}{{5{x_{n - 1}}^4}}$

then you can substitute your value in this equation....

3. ## Re: newton's method?

Opps!, i mistyped it, i had it written down as 5x^4 .

4. ## Re: newton's method?

when i found x1 = 6/5, i get really big numbers when i plug it back in to find x2. The numbers get very large. Am i suppose to be able to do this without a calculator cause this problem is part a midterm exam review. On the real exam, we don't get calculators

5. ## newton's method?

Hi lc99
if you not able to use calculator, then simplify the expression is the best way to solve it....
for example in this case,
${x_n} = {x_{n - 1}} - \frac{{{x_{n - 1}}^5 - 2}}{{5{x_{n - 1}}^4}}$
${x_n} = {x_{n - 1}} - \left( \frac{{{x_{n - 1}}^5}}{{5{x_{n - 1}}^4}} - \frac{2}{{5{x_{n - 1}}^4}} \right)$
${x_n} = {x_{n - 1}} - \left( \frac{{{x_{n - 1}}}}{5} - \frac{2}{{5{x_{n - 1}}^4}} \right)$
${x_n} = \frac{4}{5}{x_{n - 1}} + \frac{2}{{5{x_{n - 1}}^4}}$
${x_2} = \frac{4}{5}\left( {\frac{6}{5}} \right) + \frac{2}{{5{{\left( {\frac{6}{5}} \right)}^4}}}$

then u can evaluate it by simplify it even more simple to get the final answer in fraction form!

6. ## newton's method?

In addition, the number in the both numerator and denomenator are getting larger because the answer is getting more precise to the ${2}^{\frac{1}{5}}$ and this is not a rational number

7. ## Re: newton's method?

Hopefully, the ones on my exam will have cleaner numbers!!