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

• Jul 1st 2011, 11:47 PM
morrobay
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 ?
• Jul 2nd 2011, 12:00 AM
chisigma
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

$\chi$ $\sigma$
• Jul 2nd 2011, 12:29 AM
morrobay
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 ?
• Jul 2nd 2011, 01:10 AM
chisigma
Re: Taylor Series for any (x) = Function (x) for any (x) ?
Quote:

Originally Posted by morrobay
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 ?

Your $f(x)= 6\ x^{4}+ 3\ x^{3} -4\ x^{2} + 2\ x$ is a polynomial and that means that is 'well defined' for all the real or complex x, not only for $x \in [0,10]$. The fact that $f(*)$ is entire guarantees You that the Taylor expansion of $f(*)$ around any value of $x$ allows You to compute $f(x)$ for any other $x$. In case of a polynomial the Taylor expansion around $x=0$ is the polynomial itself and the fastest way to compute $f(x)$ is, of course, to use the 'original expression'...

Kind regards

$\chi$ $\sigma$
• Jul 2nd 2011, 01:43 AM
Deveno
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.
• Jul 2nd 2011, 09:32 AM
HallsofIvy
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 $f(x)= e^{-1/x}$ 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 $e^{-1/x}$ if x is not 0, 0, if x= 0. That is, the Taylor's series at x= 0 is identically 0. But clearly $e^{-1/x}\ne 0]$ 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.