Consider the function #f(x) = x^(1/4)# and develop it in Taylor series around #x = a# using only the first term, to have a linear approximation of the function:
As:
#f(a) = a^(1/4)#
#f'(x) = 1/4x^(-3/4)# so #f'(a) = 1/4a^(-3/4)#
we have:
#x^(1/4) = root(4)a+(x-a)/(4root(4)(a^3))+ R_2(x)#
Where #R_2(x)# is the sum of the terms we ignore, and so the approximation error.
Using now #a=81#, we have:
#root(4)81 = 3#
and:
#root(4)(81^3) = 27#, so:
#(80.5)^(1/4) ~= 3+(80.5-81)/(4*27)=3-0.5/108 ~= 2.9953#
To make sure that this approximation is correct to the fourth decimal place we can use Lagrange formula for the maximum error:
#abs(R_2 (x)) <= (x-a)^2/2 max_(xi in (x,a)) abs(f''(xi))#
So evaluate the second derivative:
#f''(x) = -3/16x^(-7/4)#
we can easily see that #abs (f''(x))# is strictly decreasing function, so in the interval #x in (80.5,81)# we have the maximum value for #x=80.5#.
Now, if we could compute #(80.5)^(-7/4)# we would not be doing this exercise in the first place. However we can note that:
#(80.5)^(-7/4) = 1/((80.5)^(1/4))^7#
and using the approximated value found above:
#(80.5)^(-7/4) ~=1/(2.9953)^7#
which means that to a good approximation the error is lower than:
#abs(R_2 (x)) <= (80.5-81)^2/2 (3/16) * 1/(2.9953)^7 = 0.00000541#
which gives us a good assurance that our approximation is good to the fourth decimal place.
