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**Ronald****Guest**

How do i find a number's convergent,could you help me find sqrt(3)'s convergents

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,215

Hi Ronald;

Do you know how to get the continued fraction representation of that first?

The convergents are:

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**Ronald****Guest**

Yes,I know how to get the representation,I just want to know what is the method of finding convergents

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,215

You use two recurrences.

Call the representation above a1,a2,a3,a4,a5...

You start with the numerators of the a's and use this recurrence.

with the initial conditions of

The first few numerators are

The denominators are done in the same way.

with the initial conditions of

The first few denominators are;

For example the fifth convergent is 19 / 11.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**Ronald****Guest**

Are the h1,h2,k1,k2 defined for every representation,or are they depending on the representation?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,215

They are the numerators and the denominators of the first two of the cf representation.

Look at the cf:

See the 1,1 at the beginning? That can be expressed as 1 / 1 and 1 / 1.

So

h[1] = 1 and h[2] = 1. ( the numerators )

k[1] = 1 and k[2] = 1. ( the denominators )

Now isn't that the neatest thing you ever saw! Computational mathematics!

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**Ronald****Guest**

So for sqrt(.9),h1=1,h2=1,k1=1,k2=1?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,215

Is that .9?

Here is a shortcut method to get the initial conditions that prime those recurrences.

In some books you will see them start at a0 and h[0] rather than a1 and h[1]. Just shift all indices down by one in that case. I used the higher indices because programming languages start at 1 not 0 for an array.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.** **Thinking is cheating.**

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