Use Newton's method to approximate the given number correct to eight decimal places?

95^(1/95)

1 Answer
Dec 10, 2016

1.04910303 (8 d.p.)

Explanation:

Let x=95^(1/95)

x^95=95

x^95-95=0

Let f(x)=x^95-95

Then f'(x)=95x^94

Now we simply plug into the formula with a starting point.

x_0=1.2

x_1=x_0-(f(x_0))/(f'(x_0))

=1.2-(1.2^95-95)/(95(1.2)^94)

=1.18736845

Second iteration:

x_2=x_1-(f(x_1))/(f'(x_1))

=1.17486993

And so on, starting with 1.2 it took me 15 iterations to get to an approximation with 8 constant d.p. Answer should be 1.04910303 (8 d.p.)