How do you find a one-decimal place approximation for #sqrt 10#?

Question

Nathan Axel · Accepted Answer

Use one step of a Newton Raphson method to find:

#sqrt(10) ~~ 19/6 ~~ 3.2#

To find the square root of a positive number #n#, choose a first approximation #a_0# then apply the following iteration step as many times as you like, to get the accuracy you want: #a_(i+1) = (a_i^2 + n)/(2a_i)# In our case, #n = 10# and let #a_0 = 3#, since #3^2 = 9# is quite close to what we want. Next: #a_1 = (a_0^2 + n)/(2a_0) = (3^2 + 10)/(2*3) = (9+10)/6 = 19/6 = 3.1dot(6)# So to one decimal place #sqrt(10) ~~ 3.2#

Eli Assouline · Answer

Truncate the continued fraction expansion for #sqrt(10)# to find:

#sqrt(10) = [3;bar(6)] ~~ [3;6] = 3+1/6 = 3.1dot(6) ~~ 3.2#

The continued fraction expansion for #sqrt(n^2+1)# is #sqrt(n^2+1) = [n;bar(2n)] = n + 1/(2n+1/(2n+1/(2n+1/(2n+...))))# Right now #10 = 3^2 + 1# Thus #sqrt(10) = [3;bar(6)] = 3+1/(6+1/(6+1/(6+1/(6+...))))# So truncating at #[3;6]# we find: #sqrt(10) ~~ [3;6] = 3 + 1/6 = 3.1dot(6) ~~ 3.2# You can add more terms to get more accuracy, for example. #sqrt(10) ~~ [3;6,6,6] = 3+1/(6+1/(6+1/6)) = 3+37/228 ~~ 3.16228#

Avery Alexander · Answer

undefined To find a one-decimal place approximation for √10, we can use a method called the "guess and check" method. Start by guessing a number close to the square root of 10, such as 3. Now, square this number to see if it's close to 10. 3 squared is 9, which is close to 10 but slightly less. So, we need to adjust our guess slightly higher. Let's try 3.2. Squaring 3.2 gives us 10.24, which is too high. So, we need to adjust our guess slightly lower. Let's try 3.1. Squaring 3.1 gives us 9.61, which is too low. We can refine our guess by trying 3.15. Squaring 3.15 gives us 9.9225, which is closer to 10. Continuing this process, we find that 3.16 squared is 9.9856, which is very close to 10. Therefore, a one-decimal place approximation for √10 is approximately 3.2.