Taylor Series for any (x) = Function (x) for any (x) ?

When a Taylor Series is generated from a functions n derivatives at a single

point, then is that series for any value of x equal to the original function for

any value of x.

For example graph the original function (x) from x = 0 to x = 10.

Now plug into the Taylor expansion for x, values from 0 to 10 and graph each point.

Are the two plots approximate or equal ?

Re: Taylor Series for any (x) = Function (x) for any (x) ?

Quote:

Originally Posted by

**morrobay** When a Taylor Series is generated from a functions n derivatives at a single

point, then is that series for any value of x equal to the original function for

any value of x.

For example graph the original function (x) from x = 0 to x = 10.

Now plug into the Taylor expansion for x, values from 0 to 10 and graph each point.

Are the two plots approximate or equal ?

A function f(*) is said to be *entire* if it is analytic in the whole complex plane. In that case the Taylor expansion of f(*) around any x has radious of convergence infinity. Examples of entire functions are polynomials, circular functions sin and cos, exponential function, hyperbolic functions sinh and cosh, ...

Kind regards

Re: Taylor Series for any (x) = Function (x) for any (x) ?

Im asking a more basic question : Not asking for problem to be worked but :

Suppose F(x) = 6x^4 + 3x^3 - 4x^2 + 2x

Now generate a Taylor Expansion ( n derivatives at a point )

Graph the original function for x = 0 to 10

Now plug into this Taylor Expansions (x) , values from 0 to 10 and graph.

Are these two graphs geometrically equal ?

Re: Taylor Series for any (x) = Function (x) for any (x) ?

Re: Taylor Series for any (x) = Function (x) for any (x) ?

the answer is a somewhat qualified "yes, they are equal". not every Taylor series for a function x converges everywhere, and even if the Taylor series converges, it may not converge to f. but within the radius of convergence (the term radius is used because the natural setting for studying Taylor series is in the complex plane, where open intervals are replaced by open disks) the LIMIT of the series at a point x equals the value f(x). there are things that happen "off the real line" that account for some of the unexpected ill behavior of some Taylor expansions.

however, the point is that the n-th partial sum of a Taylor series is in fact a polynomial function (the n-th approximating polynomial of "best fit" of degree n). if we only take a finite partial sum (for some finite value of n, say 6 perhaps), there will be some discrepancy between f and the Taylor series sum for f (unless, of course, f is already a polynomial of degree ≤ n).

on this page: Taylor series - Wikipedia, the free encyclopedia you can see a picture of how well various Taylor series partial sums aproximate sin(x) (for n = 1,3,5,7,9,11 and 13). as you can see from the picture, even at a 13-th degree polynomial approximation (about 0, the point at which we are evaluating the derivatives) is only decent in the range (-5,5). if one desires the approximating partial sum to be within a certain range of accuracy (say, so many decimal places) one finds a bound for the n-th remainder term, so as to know how many terms one has to compute.

Re: Taylor Series for any (x) = Function (x) for any (x) ?

A function is said to be "analytic" at x= a if and only if its Taylor's series at x= a exists, **and** the value of the function is equal to the value of the Taylor's series in some neighborhood of a. An example of a smooth function that is NOT analytic is if x is not 0, 0 if x= 0. That function is infinitely differentiable for all x and every derivative is a rational function times if x is not 0, 0, if x= 0. That is, the Taylor's series at x= 0 is identically 0. But clearly for x not equal to 0. So while the Taylor's series exists and trivially converges for all x, the value of the function is not equal to the value of the Taylor's series for any x except x= 0.