Re: series expansion : a question

In article
<33119411.1186861057132.JavaMail.jakarta@xxxxxxxxxxxxx
forum.org> tommy1729 <tommy1729@xxxxxxxxx> writes:
> considering series expansion ..
>
> a power series is taylor
> an nth derivate series is taylor

You are doing things the wrong way. A Taylor series
exists for a function
that is analytical in a circle around the origin. In
that case the Taylor
series is convergent within that circle. When a
Taylor series does exist
we can get the n-th derivative of it by taking the
n-th derivative of the
terms, constructing a new Taylor series. And we can
integrate the function
by integrating the terms (where we have to insert
suitable constant terms).
A function for which the n-th derivative does not
exist at x = 0 does not
have a Taylor series expansion at all. (An example
is x^3.|x|, which has
first, second, third and fourth derivatives, but for
which the fifth
derivative at x = 0 does not exist, so it is not
analytical in x = 0 and
so does not have a Taylor series expansion.)

> what is the analogue for an integral ??
wich series can be expressed in the nth integral
l ??

See above. Although I do not understand the second
question.
--
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amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
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once agian i am misunderstood ...

i understand taylor series perfectly ...

radius of convergeance too...

thats NOT the issue here !!!

considering f(x+a) = f(a)+f'(a)x +...

(taylor ) it is clear that taylor is a power series ( infinite polynomial of x )

and a series of nth derivates ( f^n(a))

i am looking for the analogue of nth integrals ...

tommy1729
.

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