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- Feb 5, 2012

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**Marie on Facebook writes:**

I was wondering how to find square root for numbers like 128. Can you do it step by step please?

- Thread starter Sudharaka
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- Thread starter
- #1

- Feb 5, 2012

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I was wondering how to find square root for numbers like 128. Can you do it step by step please?

- Mar 1, 2012

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I would use Prime Factorisation to find the prime roots and then take advantage of the rule $\sqrt{ab} = \sqrt{a}\sqrt{b}$. The numbers are positive so this is allowed. To that end if decimal approximations are needed, it's best to learn some of the simple prime roots like $\sqrt{2},\ \sqrt{3}\ \sqrt{5}$Marie on Facebook writes:

I was wondering how to find square root for numbers like 128. Can you do it step by step please?

Using 128 as an example (this one is relatively simple as an integer power of 2)

$128 \div 2 = 64$ -- so 2 is one prime factor

$64 \div 2 = 32$

and so on until

$4 \div 2 = 2$

Thus we can say that $128 = 2^7$ and so $\sqrt{128} = \sqrt{2^7}$

Using the rule above I then say that $\sqrt{2^7} = \sqrt{2^6}\sqrt{2} = 2^3\sqrt{2} = 8\sqrt{2}$

I would leave it as $8\sqrt{2}$ for an exact answer

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- #3

i) $\displaystyle x_{n+1}=\frac{x_n+2}{x_n+1}$

ii) $\displaystyle x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}$

You simply need to "seed" both recursions with some initial guess.

We know $\displaystyle 1<\sqrt{2}<2$ so $\displaystyle \frac{3}{2}$ will work just fine.

Here's another recursive approximation for a square root.

Suppose we want [tex]\sqrt{N}.[/tex]

Let [tex]a_1[/tex] be the first approximation to [tex]\sqrt{N}.[/tex]

Then: [tex]a_2 \:=\:\frac{N+a_1^2}{2a_1}[/tex] is an even better approximation.

Repeat the process with [tex]a_2[/tex] . . . and so on.

This procedure converges

. . depending on [tex]a_1.[/tex]

Find [tex]\sqrt{3}[/tex], using [tex]a_1 = 1.7[/tex]

[tex]a_2 \:=\:\frac{3 + 1.7^2}{2(1.7)} \:=\:1.732352941 \:\approx\:1.732353[/tex]

[tex]a_3 \:=\:\frac{3 + 1.732353^3}{2(1.732353)} \:=\:1.732050834[/tex]

ChecK: .[tex]1.732050834^2 \:=\:3.000000091\;\;\checkmark[/tex]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here is the reasoning behind this procedure.

We want [tex]\sqrt{N}.[/tex]

We approximate the square root, [tex]a_1.[/tex]

Suppose our approximation is correct.

Then [tex]a_1[/tex] and [tex]\tfrac{N}{a_1}[/tex] would be equal, right?

Chances are, they are

One is larger than the square root, one is smaller.

What is a better approximation of the square root?

. . . the

Hence: .[tex]a_2 \:=\:\frac{a_1 + \frac{N}{a_1}}{2} \:=\:\frac{\frac{a_1^2 + N}{a_1}}{2}[/tex]

Therefore: [tex]a_2 \:=\:\frac{N +a_1^2}{2a_1}[/tex]

- Jan 26, 2012

- 644

Its general form is:

$$a_{n + 1} = a_n - \frac{f(a_n)}{f'(a_n)}$$

To converge to a root of $f(x)$. Which root it converges to is dependent on the initial guess $a_1$.

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- #6

- Mar 5, 2012

- 9,416

Generally, Newton-Raphson finds a root of $f(x)=0$.

The algorithm is:

$x_{k+1} = x_k - \dfrac{f(x_k)}{f'(x_k)}$

With $f(x)=x^2-N$ this becomes:

$x_{k+1} = x_k - \dfrac{x_k^2-N}{2x_k}$

$x_{k+1} = \dfrac{x_k}{2} + \dfrac{N}{2x_k} \quad$ or $\quad x_{k+1} = \dfrac{N + x_k^2}{2x_k}$

$x_{k+1} = \dfrac{x_k}{2} + \dfrac{N}{2x_k} \quad$ or $\quad x_{k+1} = \dfrac{N + x_k^2}{2x_k}$

The initial guess should be above the root for fastest results.

If the initial guess is below the root, the first iteration will jump to the other side of the root and it will worsen the approximation.

The second form is the one soroban presented.

If we pick N=2 as Mark did, the first form becomes:

$x_{k+1} = \dfrac{x_k}{2} + \dfrac{1}{x_k}$

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- #7