Question

How can you calculate the approximate value of `sqrt(17)` ?

Answer 1

Honest answer: get a calculator, type in `17` then hit the square root button; however for a manual mechanical process see below.

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 `17` the largest such integer is `4`.

Set up as if doing long division:
`color(white)("xxxxx|")ul(color(white)(1)4)`
`color(white)("xxxxx")|17.`
`color(white)("xxx")4|ul(16)`
`color(white)("xxxxx")|color(white)("x")1`

"Bring down" 2 zeroes
`color(white)("xxxxx|")ul(color(white)(1)4color(white)("xxx")`
`color(white)("xxxxx")|17.00`
`color(white)("xxx")4|ul(16)`
`color(white)("xxxxx")|color(white)("x")100`

Double the digits in the top row to form all but the last digit of the divisor
`color(white)("xxxxx|")ul(color(white)(1)4color(white)("xxx")`
`color(white)("xxxxx")|17.00`
`color(white)("xxx.")4|ul(16)`
`color(white)("xxxx.x")|color(white)("x")100`
`color(white)("xxx")8"_"|color(white)(".")ul(color(white)("XX")`

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.
`color(white)("xxxxx|")ul(color(white)(1)4 . color(white)("x")1`
`color(white)("xxxxx")|17.00`
`color(white)("xxx.")4|ul(16)`
`color(white)("xxxx.x")|color(white)("x")100`
`color(white)("xxx")81|color(white)(".")ul(color(white)("X")81)`

Subtract to get the next remaining divisor
`color(white)("xxxxx|")ul(color(white)(1)4 . color(white)("x")1`
`color(white)("xxxxx")|17.00`
`color(white)("xxx.")4|ul(16)`
`color(white)("xxxx.x")|color(white)("x")100`
`color(white)("xxx")81|color(white)(".")ul(color(white)("X")81)`
`color(white)("xxx"81)|color(white)("xx")19`

Repeat:
`color(white)("xxxxx|")ul(color(white)(1)4 . color(white)("x")1color(white)("XXX")`
`color(white)("xxxxx")|17.00color(white)("x") 00`
`color(white)("xxx.")4|ul(16)`
`color(white)("xxxx.x")|color(white)("x")100`
`color(white)("xxx")81|color(white)(".")ul(color(white)("X")81)`
`color(white)("xxx"81)|color(white)("xx")19color(white)("x")00`
`color(white)("xx")82"_"|color(white)("xx")ul(color(white)("XXXX"))`

`color(white)("xxxxx|")ul(color(white)(1)4 . color(white)("x")1color(white)("X")2`
`color(white)("xxxxx")|17.00color(white)("x") 00`
`color(white)("xxx.")4|ul(16)`
`color(white)("xxxx.x")|color(white)("x")100`
`color(white)("xxx")81|color(white)(".")ul(color(white)("X")81)`
`color(white)("xxx"81)|color(white)("xx")19color(white)("x")00`
`color(white)("xx")822|color(white)("xx")ul(16color(white)("x")44)`
`color(white)("xxxxx")|color(white)("xxxxx")2color(white)("x")56`

This process can be continued indefinitely to achieve a more accurate result but even with these few digits we have
`(4.12)^2=16.9744` which isn't too far off `17`

Answer 2

Here we use a Bisection Method. Let us start by seeing which perfect squares `17` lies between, then we will repeatedly bisect the interval and compare until we get a closer and closer approximation.

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` iterations we have established that::

`4.0625 lt sqrt(17) lt 4.125`

And we can repeat this process, as required.

Calculator Value:

For comparison:

`sqrt(17) = 4.123105 ...`

Answer 3

Here's a method using generalised continued fractions...

Explanation:

There are at least `25` methods for finding square roots, one of my favourites involving generalised continued fractions, using the fact that if `a > 0` then:

`sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))`

This works whether `b` is positive or negative, but in order to converge rapidly you want `b` to be small in comparison to `a`.

This works particularly well for numbers like `17 = 4^2+1` which are of the form `a^2+1` in that they give simple continued fractions which converge quickly...

`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)`, e.g.:

`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 `2177/528` is accurate to `7` significant digits.

Answer 4

Here's a method using integer sequences...

Explanation:

Another fun method uses integer sequences.

Theory

Suppose you have a positive integer `n` whose square root you want to find rational approximations for and an integer approximation `a` such that `abs(sqrt(n)-a) < 1`.

Consider the quadratic with zeros `a+sqrt(n)` and `a-sqrt(n)`, namely:

`(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 `t_k` in the form `A(a+sqrt(n))^k+B(a-sqrt(n))^k` for some constants `A, B`.

Note that as `k` increases `abs((a-sqrt(n))^k) -> 0`.

Hence the ratio between successive terms will tend to `a+sqrt(n)`

Application

Let `n=17` and `a=4`

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 `17=4^2+1` is close to `4^2`, the ratio will converge quite rapidly in this example and we find:

`sqrt(17) ~~ 34840/4289-4 = 17684/4289 ~~ 4.123105619`

A calculator tells me that:

`sqrt(17) ~~ 4.12310562561766`

So our approximation was accurate to `8` significant digits and nearly correct to `9` significant digits. If you want more correct significant digits, just calculate a few more terms of the integer sequence before dividing the last term by the penultimate one.