Approximate $\displaystyle \sqrt{2}$

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- Sep 4th 2010, 08:05 PMrazemsoft21Approximate Sqrt(2)
Approximate $\displaystyle \sqrt{2}$

- Sep 4th 2010, 08:12 PMEducated
$\displaystyle \sqrt2\approx1.414$

- Sep 4th 2010, 08:23 PMrazemsoft21
Approximate $\displaystyle \sqrt{2}$ to #.## and show how please.

- Sep 4th 2010, 08:33 PMProve It
The Babylonian Method works well here. It makes use of the bisection method and the fact that if you divide a number by its square root, you get the square root.

You need to make an initial guess.

Since $\displaystyle \sqrt{1} < \sqrt{2} < \sqrt{4}$ that means $\displaystyle 1 < \sqrt{2} < 2$.

Average these endpoints to get $\displaystyle 1.5$ is a good starting guess.

$\displaystyle \frac{2}{1.5} = 1.33333333$.

So that means $\displaystyle 1.33333333 < \sqrt{2} < 1.5$.

Average these out and you get a new guess.

Follow this process until you reach your desired level of accuracy. - Sep 4th 2010, 09:22 PMmr fantastic
- Sep 4th 2010, 11:29 PMrazemsoft21
- Sep 5th 2010, 12:43 AMEducated

razemsoft21, it helps if you tell us the full details in the**first**post. Saying Approximate $\displaystyle \sqrt2$ isn't enough.

Also, there are*many*methods of solving squareroots. If you don't know*how*to solve it, then maybe asking for someone to tell you a method of solving squareroots.

Doing research would help too: Computing Square roots - Wikipedia. Here, it shows you many methods of how to solve a square root of a number. - Sep 5th 2010, 04:17 AMSoroban
Hello, razemsoft21!

Where did this problem come from?

It's unlikely that you would be assigned this problem

, ,being taught any approximation methods.*without*

Quote:

Approximate $\displaystyle \sqrt{2}$

The very worst you could do is guess-and-adjust.

Since $\displaystyle \sqrt{2}$ is between 1 and 2 . . .

$\displaystyle \begin{array}{ccccccc}

\text{Try 1.5:} & 1.5^2 &=& 2.25 & \text{too big} \\

\text{Try 1.4:} & 1.5^2 &=& 2.25 & \text{too small} \\

\text{Try 1.45:} & 1.45^2 &=& 2.1025 & \text{too big} \\

\text{Try 1.42:} & 1.42^2 &=& 2.0164 & \text{too big} \\

\text{Try 1.42:} & 1.41^2 &=& 1.9881 & \text{too small}

\end{array}$

We have the two decimal places.

Which is the "better" answer: 1.41 or 1.42 ?

1.41 gives an error of: .$\displaystyle 2 - 1.9881 \:=\:0.0119$

1.42 gives an error of: .$\displaystyle 2.0164 - 2 \:=\:0.0164$

Since 1.41 has the smaller error: .$\displaystyle \sqrt{2}\:\approx\:1.41$

- Sep 5th 2010, 04:42 AMgrgrsanjay
refer to this page Square root of 2 - Wikipedia, the free encyclopedia

- Sep 5th 2010, 04:46 AMSoroban
. . . An elementary approximation of $\displaystyle \sqrt{N}$

Let $\displaystyle a_1$ be your first approximation to $\displaystyle \sqrt{N}.$

Substitute into: .$\displaystyle \dfrac{N+a_1^2}{2a_1} \;=\;a_2$ .[1]

This produces $\displaystyle a_2$, a better approximation.

For more accuracy, substitute $\displaystyle a_2$ into [1] . . . and so on.

__Example__: .$\displaystyle \sqrt{2},\;\;N = 2$

Let $\displaystyle a_1 = 1.5$

We have: .$\displaystyle a_2 \:=\:\dfrac{2 + 1.5^2}{2(1.5)} \;=\;1.41666\hdots $

. . Check: .$\displaystyle (1.41666\hdots)^2 \:=\:2.006944443$

Then: .$\displaystyle a_3 \:=\;\dfrac{2 + (1.4166\hdots)^2}{2(1.41666\hdots)} \:=\:1.414215686$

. . Check: .$\displaystyle 1.414256862^2 \:=\:2.000007749$

Then: .$\displaystyle a_4 \:=\:\dfrac{2 + 1.41425686^2}{2(1.414256862)} \:=\:1.414213562 $

. . Check: .$\displaystyle 1.414213562^2 \:=\:1.999999999$

Therefore: .$\displaystyle \sqrt{2} \;\approx\;1.424213562$ .to 9 decimal places.

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

Why does [1] provide better and better approximations?

We want an approximation of $\displaystyle \sqrt{N}.$

Our first guess is $\displaystyle a_1.$

We have factored $\displaystyle N$ into two factors: .$\displaystyle a_1\,\text{ and }\,\dfrac{N}{a_1} $

If we are extremely lucky, the two factors will be*equal*.

. . (Think about it!)

Since it is unlikely that the two factors are equal,

. . a better approximation is the*average*of the two factors.

. . (The number exactly halfway between them.)

So we use: .$\displaystyle a_2 \;=\;\dfrac{a_1 + \frac{N}{a_1}}{2}

$

. . which simplfies to: .$\displaystyle a_2 \;=\;\dfrac{N+a_1^2}{2a_1}$ . . . . . ta-*DAA!*

- Sep 5th 2010, 05:05 AMggn
this is more accurate

1.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799 - Sep 5th 2010, 05:09 AMggn
we must appreciate the works of Robert Nemiroff he calculated to 1 million digits just check this out!

The Square Root of Two to 1 Million Digits - Sep 5th 2010, 10:40 AMrazemsoft21
Thank you for help and especial THANKS for:

Prove It , Educated , Soroban - Sep 5th 2010, 04:36 PMHallsofIvy