Approximate

Printable View

- September 4th 2010, 09:05 PMrazemsoft21Approximate Sqrt(2)
Approximate

- September 4th 2010, 09:12 PMEducated
- September 4th 2010, 09:23 PMrazemsoft21
Approximate to #.## and show how please.

- September 4th 2010, 09:33 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 that means .

Average these endpoints to get is a good starting guess.

.

So that means .

Average these out and you get a new guess.

Follow this process until you reach your desired level of accuracy. - September 4th 2010, 10:22 PMmr fantastic
- September 5th 2010, 12:29 AMrazemsoft21
- September 5th 2010, 01:43 AMEducated

razemsoft21, it helps if you tell us the full details in the**first**post. Saying Approximate 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. - September 5th 2010, 05:17 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

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

Since is between 1 and 2 . . .

We have the two decimal places.

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

1.41 gives an error of: .

1.42 gives an error of: .

Since 1.41 has the smaller error: .

- September 5th 2010, 05:42 AMgrgrsanjay
refer to this page Square root of 2 - Wikipedia, the free encyclopedia

- September 5th 2010, 05:46 AMSoroban
. . . An elementary approximation of

Let be your first approximation to

Substitute into: . .[1]

This produces , a better approximation.

For more accuracy, substitute into [1] . . . and so on.

__Example__: .

Let

We have: .

. . Check: .

Then: .

. . Check: .

Then: .

. . Check: .

Therefore: . .to 9 decimal places.

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

Why does [1] provide better and better approximations?

We want an approximation of

Our first guess is

We have factored into two factors: .

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: .

. . which simplfies to: . . . . . . ta-*DAA!*

- September 5th 2010, 06:05 AMggn
this is more accurate

1.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799 - September 5th 2010, 06:09 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 - September 5th 2010, 11:40 AMrazemsoft21
Thank you for help and especial THANKS for:

Prove It , Educated , Soroban - September 5th 2010, 05:36 PMHallsofIvy