The Babylonian method for approximating square roots is the recursive method given by:
Given:
x0≈k−−√
x
0
≈
k
Then:
xn+1=x2n+k2xn
x
n
+
1
=
x
n
2
+
k
2
x
n
where k
k
is the number for which we are approximating the square root.
So, armed with this formula, let's now answer the given questions:
1.) Fill in the missing info in the problem to find square root 130 by using the Babylonian method. Let 11.5 be your initial guess.
We will use 15 digits of precision here since it is not stated how many iterations to carry out:
x0=11.5
x
0
=
11.5
x1≈11.4021739130435
x
1
≈
11.4021739130435
x2≈11.4017542587143
x
2
≈
11.4017542587143
x3≈11.4017542509914
x
3
≈
11.4017542509914
x4≈11.4017542509914
x
4
≈
11.4017542509914
Thus, with 15 digits of accuracy, we may state:
130−−−√≈11.4017542509914
130
≈
11.4017542509914
2.) Use the Babylonian method for your next estimate of square root 115 by using 10.7262 as your guess what's your next result?
The next result here is approximately:
x1=10.72622+1152⋅10.7262≈10.7238055620816
x
1
=
10.7262
2
+
115
2
⋅
10.7262
≈
10.7238055620816
3.) Nails wants to use the Babylonia method to estimate square root 58 to the nearest hundredth her initial estimate is 7.8. What is her estimate after she correctly completes the Babylonian method once?
The next result here is:
x1=7.82+582⋅7.8=7.6179487¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
x
1
=
7.8
2
+
58
2
⋅
7.8
=
7.6
179487
¯
4.) If you use the Babylonian method to estimate square root 75 to the nearest hundredth, starting with the estimate 8. What is the next estimate?
The next result here is:
x1=82+752⋅8=8.6875
x
1
=
8
2
+
75
2
⋅
8
=
8.6875