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**Karrl****Guest**

How do I find taylor series of a function?Is there analysis method,formula or something else?

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,911

hi Karrl

Welcome to the forum.

Yes, there is a formula. Have a look at:

http://en.wikipedia.org/wiki/Taylor_series

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 94,482

Hi;

What is the function?

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**Karrl****Guest**

If you meant the first formula,then what are the variables x and a?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 94,482

Helps a lot to see the particular function you have in mind and what is the point of expansion.

x is the independent variable and a is the point of expansion.

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**Karrl****Guest**

So for f(p),p=x,and a is any point I want?And will there be taylor series for every function

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 94,482

Any function that has derivatives that exist at the point of expansion. You can pick any point but that does not mean it will converge.

Sometimes the usual method will not work and you must use another.

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**Karrl****Guest**

Could you show me an example of taylor series(not the e^x,I know that)and explain how the series is found.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 94,482

That is a Taylor series expanded around zero. When it is expanded around zero it is called a Mclaurin series.

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**Karrl****Guest**

What is the difference in expanding around different points?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 94,482

The formula or method used is slightly different. Here it is expanded around a and is called a Taylor series.

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

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**Karrl****Guest**

So,all values of a will estimate same?if not then what is the best value for a?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 94,482

The purpose of the series is to numerically evaluate a value of a function.

An example will make all of it clearer.

Supposing you want to evaluate sin(.1)? Taylor series usually only converge a small distance from the point of expansion. We choose zero because it is close to .1.

Now you plug into x the value that you are looking for. x = .1

The actual value of Sin(.1) is 0.09983341664682815

The approximation is a good one. This is an easy one. In practice they are usually trickier.

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**Karrl****Guest**

So the close a is to x,the better

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 94,482

Hi;

So the close a is to x,the better

Exactly. The Taylor polynomial has a limited range from the point of expansion.

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

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