How do you find a one-decimal place approximation for #-root4 200#?

1 Answer
Oct 23, 2015

Answer:

Use some reasonings about #3^4#, #4^4# and #sqrt(200)# to choose approximation #3.8#, then use one step of Newton's method to refine the approximation, finding that #3.8# is good to one decimal place already.

Explanation:

#3^4 = 81# and #4^4 = 256#, so #root(4)(200)# lies somewhere between #3# and #4#.

More specifically, #200 ~~ 196 = 14^2# so #root(4)(200) ~~ sqrt(14)# and #sqrt(14) ~~ 4-1/4 = 3.75#, so #root(4)(200) ~~ 3.8#

To find the #4#th root of a number #n#, we could find its square root and then the square root of that, but instead let's choose a reasonable first approximation #a_0#, then use the following formula to iterate:

#a_(i+1) = a_i + (n - a_i^4)/(4a_i^3)#

Let #a_0 = 3.8#

Then:

#a_1 = a_0 + (200-3.8^4)/(4*3.8^3)#

#=3.8 + (200-208.5136)/219.488#

#=3.8 - 8.5136/219.488 ~~ 3.76#

So #3.8# is a good approximation of #root(4)(200)# to one decimal place and #-3.8# is a one decimal place approximation for #-root(4)(200)#