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

Question

95^(1/95)

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Answer 1 · 2016-12-10T08:44:59

#1.04910303 (8 d.p.)#

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.)#