# Taylor series

1. Jun 10, 2008

### fresnelspot

Hi

If derivative of a function (f(x)) is another function(g(x)) then, this holds for the series terms of the functions.
My question is If one knows this derivation relation Just two equal labelled series terms of two other functions. And one term is exactly the differential of the other Then can we generelize this relation for the "big " functions?
Thanx

2. Jun 10, 2008

### mathman

Your message is confusing. Among other things what are "big" functions? Also what do mean by "equally labelled series"?

3. Jun 10, 2008

### fresnelspot

:)

by "big " functions i mean the sum of all terms .
by "equal labelled" i mean they both will be the nth terms of the series expansions they belong.

4. Jun 10, 2008

### Defennder

Let me see if I understand this. You're saying that a function f(x) is a derivative of another function g(x) if the nth term of f(x) is a derivative of the (n+1)th term of the power series of g(x)? Or are you asking if the converse is true? Or something else entirely?

5. Jun 11, 2008

### rbj

just to be clear with Defennder's question, if

$$f(x) = \sum_{n=0}^{+\infty} a_n x^n$$

and

$$g(x) = \sum_{n=0}^{+\infty} b_n x^n$$

and $f(x) = (d/dx)g(x)$

then we know that

$$a_n = (n+1)b_{n+1} \ \ \ \forall n \ge 0$$

knowing all of the an is sufficient to tell us what the bn are except for b0, which is information that is lost forever when g(x) is differentiated.

dunno if that is the question that was asked.

6. Jun 11, 2008

### fresnelspot

I am very sorry for my verbal skills .The problem is confusing for me so it becomes more difficult to explain it in words.
My problem is this :
The concept is clear when we see the relation between f and g and reflect this relation to the series terms.

What if we know there is a derivation relation between two terms ( we only know these terms . For example we have experient data of these two functions for a limited region . they are both kth series terms of their function ) . Can we generalize this derivation relation to the functions

7. Jun 11, 2008

### rbj

no, only knowing that

$$a_n = (n+1)b_{n+1}$$

for some particular n is not sufficient to say that $f(x) = (d/dx)g(x)$. you must know that

$$a_n = (n+1)b_{n+1} \ \ \ \forall \ n \ge 0$$

which means for all $\ n \ge 0 \$ to be able to conclude that $f(x) = (d/dx)g(x)$.

8. Jun 11, 2008

thanx