Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #1 20121027 02:13:51
Taylor seriesHow do I find taylor series of a function?Is there analysis method,formula or something else? #2 20121027 02:20:00
Re: Taylor serieshi Karrl You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #4 20121027 02:30:28
Re: Taylor seriesIf you meant the first formula,then what are the variables x and a? #5 20121027 02:31:19
Re: Taylor seriesHelps a lot to see the particular function you have in mind and what is the point of expansion. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #6 20121027 03:00:05
Re: Taylor seriesSo for f(p),p=x,and a is any point I want?And will there be taylor series for every function #7 20121027 03:04:51
Re: Taylor seriesAny function that has derivatives that exist at the point of expansion. You can pick any point but that does not mean it will converge. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #8 20121027 03:29:02
Re: Taylor seriesCould you show me an example of taylor series(not the e^x,I know that)and explain how the series is found. #9 20121027 03:38:18
Re: Taylor seriesThat is a Taylor series expanded around zero. When it is expanded around zero it is called a Mclaurin series. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #11 20121027 03:48:51
Re: Taylor seriesThe formula or method used is slightly different. Here it is expanded around a and is called a Taylor series. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #12 20121027 03:57:41
Re: Taylor seriesSo,all values of a will estimate same?if not then what is the best value for a? #13 20121027 04:07:23
Re: Taylor seriesThe 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. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #15 20121027 14:25:25
Re: Taylor seriesHi;
Exactly. The Taylor polynomial has a limited range from the point of expansion. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 