So the close a is to x,the better

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

]]>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.

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

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

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

]]>What is the function?

]]>Welcome to the forum.

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

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

Bob

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