Where can I see proof for this? Also a few "simple" deductions (cos idk yet) of most simple Taylor series (cos x, sin x, e^x, etc).

A book with a straightforward explanation or something like that.

tks in advance

peter

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- Mar 13th 2012, 09:54 PMIsslerWhy can a function be expressed as a Taylor Series?
Where can I see proof for this? Also a few "simple" deductions (cos idk yet) of most simple Taylor series (cos x, sin x, e^x, etc).

A book with a straightforward explanation or something like that.

tks in advance

peter - Mar 16th 2012, 06:51 AMHallsofIvyRe: Why can a function be expressed as a Taylor Series?
First, if a function is "infinitely differentiable", we can certainly

**form**the Taylor's series for that function. But that is probably not the question you are really asking. You seem to be under to be under the impression that if a function**has**a Taylor's series then the Taylor series must be equal to the function- and that is not true. For one thing, it does not follow that the series**converges**. And even if the Taylor's series for a function converges for all x, it does**not**necessarily equal the function on any interval. For example, the function defined by $\displaystyle f(x)= e^{1/x^2}$ if x is not 0, f(0)= 0, is infinitely differentiable at x= 0. (Any derivative is 0 at x= 0 and the nth derivative for x not 0 is an exponential with exponent a polynomial in 1/x. That goes to 0 as x goes to 0.) That is, its MacLaurin series (Taylor's series at x= 0) is identically equal to 0 but the function itself is not 0 except at x= 0.

If there exist an interval about x= a on which the Taylor's series for a function, f, converges and is equal to the function, we say that f is "analytic" at x= a (sometimes "real analytic" to distinguish it from "analytic" for functions of a complex variable which is equivalent but typically defined differently.)

How we show that a given function is "analytic" at a particular x= a is highly dependent on the precise function. For example, the simplest way to show that the Taylor's series for sine, cosine, and exponential, for example, are analytic is to show that the Taylor's series satisfy the same differential equation and "initial values" as the functions themselves. Then, by the "existance and uniqueness theorem" for initial value problems, they must be equal. - Mar 16th 2012, 10:59 PMProve ItRe: Why can a function be expressed as a Taylor Series?
I interpreted the question the OP asked differently to how HallofIvy did. I am interpreting the question as "why would one WANT to express a function as a Taylor series?"

Let's use the function sin(x) as an example. From the unit circle definition, we know a few values of this function (e.g. sin(0) = 0), but we definitely can't use the unit circle to evaluate the sine function at every possible value of x (or even get a decimal approximation). You might say "well use a calculator", but how do you think the calculator is able to do it? All a calculator can do is add (and by extension, subtract, multiply, divide and exponentiate), so we need to find the right combination of addition, subtraction, multiplication, division and exponentiation to get the value we want. This is what the Taylor series is, it's the right combination of operations to be able to evaluate the function at every possible point.