Use Newton's method to approximate the given number correct to eight decimal places. #49^(1/6)# ?

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Answer 1 · 2018-04-07T23:50:33

#49^(1/6) ~~ 1.91293118#

Let:

#f(x) = x^6-49#

Then:

#f'(x) = 6x^5#

If #a_i# is an approximation to a zero of #f(x)#, then a better one is:

#a_(i+1) = a_i - (f(a_i))/(f'(a_i)) = a_i - (a_i^6-49)/(6a_i^5) = (5a_i^6+49)/(6a_i^5)#

For our first approximation we could choose #a_0 = 2#

Putting the formula into a spreadsheet, I found:

#a_0 = 2#

#a_1 = 1.921875#

#a_2 ~~ 1.91303459383222#

#a_3 ~~ 1.91293119674636#

#a_4 ~~ 1.91293118277239#

#a_5 ~~ 1.91293118277239#

So correct to #8# decimal places we find:

#49^(1/6) ~~ 1.91293118#

Notes

Alternatively, we could notice that #49=7^2#, so:

#49^(1/6) = (7^2)^(1/6) = 7^(1/3)#

So we could define:

#g(x) = x^3-7#

Then:

#g'(x) = 3x^2#

and use the iterative formula:

#a_(i+1) = a_i - (g(a_i))/(g'(a_i)) = a_i - (a_i^3 - 7)/(3a_i^2) = (2a_i^3+7)/(3a_i^2)#

This converges slightly more rapidly, giving:

#a_0 = 2#

#a_1 ~~ 1.91666666666667#

#a_2 ~~ 1.91293845830708#

#a_3 ~~ 1.91293118280006#

#a_4 ~~ 1.91293118277239#

#a_5 ~~ 1.91293118277239#