How can you calculate the approximate value of #sqrt(17)# ?
4 Answers
Honest answer: get a calculator, type in
Explanation:
Find the largest integer which when squared is not greater than the number for which you are trying to find the square root. Since in this case the number for which we are trying to find the square root is
Set up as if doing long division:
"Bring down" 2 zeroes
Double the digits in the top row to form all but the last digit of the divisor
Estimate the digit required such that the divisor with this digit appended times this digit is a maximum but not greater than the remaining dividend.
Subtract to get the next remaining divisor
Repeat:
This process can be continued indefinitely to achieve a more accurate result but even with these few digits we have
Here we use a Bisection Method. Let us start by seeing which perfect squares
Initially we have:
# 16 lt 17 lt 25 => 4 lt sqrt(17) lt 5 #
.##
Iteration 1:
# alpha \ = (4+5)/2 = 4.5 => alpha^2 = 20.25 #
# 16 lt 17 lt 20.25 => 4 lt sqrt(17) lt 4.5 #
Iteration 2:
# alpha \ = (4+4.5)/2 = 4.25 => alpha^2 = 18.0625 #
# 16 lt 17 lt 18.0625 => 4 lt sqrt(17) lt 4.25 #
Iteration 3:
# alpha \ = (4 + 4.25)/2 = 4.125 => alpha^2 = 17.015625 #
# 16 lt 17 lt 17.015625 => 4 lt sqrt(17) lt 4.125 #
Iteration 4:
# alpha \ = (4 + 4.125)/2 = 4.0625 => alpha^2 = 16.50390625 #
# 16.50390625 lt 17 lt 17.015625 => 4.0625 lt sqrt(17) lt 4.125 #
So after
# 4.0625 lt sqrt(17) lt 4.125 #
And we can repeat this process, as required.
Calculator Value:
For comparison:
# sqrt(17) = 4.123105 ... #
Here's a method using generalised continued fractions...
Explanation:
There are at least
#sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))#
This works whether
This works particularly well for numbers like
#sqrt(17) = sqrt(color(blue)(4)^2+color(blue)(1)) = 4+1/(8+1/(8+1/(8+1/(8+...))))#
You can terminate this continued fraction early to get rational approximations to
#sqrt(17) ~~ 4+1/8 = 33/8 = 4.125#
#sqrt(17) ~~ 4+1/(8+1/8) = 4+8/65 = 268/65 = 4.123bar(076923)#
#sqrt(17) ~~ 4+1/(8+1/(8+1/8)) = 4+1/(8+8/65) = 4+65/528 = 2177/528 = 4.1231bar(06)#
The resulting approximation has about as many correct significant digits as the total number of digits in the quotient. So
Here's a method using integer sequences...
Explanation:
Another fun method uses integer sequences.
Theory
Suppose you have a positive integer
Consider the quadratic with zeros
#(x-a-sqrt(n))(x-a+sqrt(n)) = (x-a)^2-n#
#color(white)((x-a-sqrt(n))(x-a+sqrt(n))) = x^2-2ax-(n-a^2)#
From this quadratic, let us construct a sequence by recursive rules:
#{ (t_0 = 0), (t_1 = 1), (t_(k+2) = 2at_(k+1)+(n-a^2)t_k " for " k >= 0) :}#
It is possible (and not too difficult) to create a direct formula for
Note that as
Hence the ratio between successive terms will tend to
Application
Let
Define a sequence recursively by:
#{ (t_0 = 0), (t_1 = 1), (t_(k+2) = 2(color(blue)(4))t_(k+1)+(color(blue)(17)-color(blue)(4)^2)t_k = 8t_(k+1)+t_k) :}#
The first few terms are:
#0, 1, 8, 65, 528, 4289, 34840,...#
Since
#sqrt(17) ~~ 34840/4289-4 = 17684/4289 ~~ 4.123105619#
A calculator tells me that:
#sqrt(17) ~~ 4.12310562561766#
So our approximation was accurate to