How do you find a one-decimal place approximation for #root5 26#?

1 Answer
Oct 21, 2015

Answer:

Use one step of Newton's method to find #root(5)(26) ~~ 1.9#

Explanation:

To find an approximation for the #5#th root of a number #n#, choose a reasonable first approximation #a_0# then iterate using the following formula:

#a_(i+1) = a_i + (n - a_i^5)/(5 a_i^4)#

In our case let #n = 26# and #a_0 = 2# (since #2^5 = 32# is close)

Then:

#a_1 = a_0 + (n - a_0^5)/(5 a_0^4) = 2 + (26-32)/(5*16) = 2 - 6/80 = 2 - 0.075 = 1.925#

So #root(5)(26) ~~ 1.9#

If we want more accuracy, just apply the formula again to get #a_2# from #a_1#, etc.

This method is a particular case of Newton's method for finding a zero of a continuous differentiable function #f(x)#.

In our case #f(x) = x^5 - n#, #f'(x) = 5x^4# and Newton's method says to iterate using the formula:

#a_(i+1) = a_i - f(a_i)/(f'(a_i)) = a_i - (a_i^5 - n)/(5a_i^4) = a_i+(n-a_i^5)/(5a_i^4)#