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Hi;
Exactly. The Taylor polynomial has a limited range from the point of expansion.
So the close a is to x,the better
The purpose of the series is to numerically evaluate a value of a function. 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.
So,all values of a will estimate same?if not then what is the best value for a?
The formula or method used is slightly different. Here it is expanded around a and is called a Taylor series.
What is the difference in expanding around different points?
That is a Taylor series expanded around zero. When it is expanded around zero it is called a Mclaurin series.
Could you show me an example of taylor series(not the e^x,I know that)and explain how the series is found.
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.
So for f(p),p=x,and a is any point I want?And will there be taylor series for every function
Helps a lot to see the particular function you have in mind and what is the point of expansion.
If you meant the first formula,then what are the variables x and a?
Hi;
hi Karrl
How do I find taylor series of a function?Is there analysis method,formula or something else? 