JorgeM said:
Summary: Hello there, I need to get the Taylor Series for f(r) and r is a function f(x,y,z))=r
Some people make a distinction between a Taylor Series and a McLaurin Series. Let's assume you mean you want an expression for ##f(x,y,z)## in powers of ##x,y,z## rather than in powers of ##(x-a),(y-b),(z-c)##
It is a strenuous exercise in LaTex to write it out higher order multi-variable expansions in ordinary notation. ( No wonder you only got hints - and I don't volunteer to write it out that way myself !)
You can find it written out in the "multi-index notation"
https://en.wikipedia.org/wiki/Multi-index_notation
I'm not skilled at reading the mult-index notation. I'll try some concrete examples.
Example: consider the 6th order expansion of ##f(x,y,z,w)## What is the coefficient of ##x^3 y^1 z^2 w^0## in that expansion?
It is:
##\frac{ 1}{3!\ 1!\ 2!\ 0!}## ## \frac{\partial^3 f}{\partial x^3} \frac{\partial^1 f}{\partial y^1} \frac{\partial^2 f}{\partial z^2} \frac{\partial^0 f}{\partial w^0}##
where the partial derivatives are evaluated at x = y = z = w = 0 and keeping in mind the definition 0! = 1! = 1 and using the convention that ##\frac{\partial^0 f}{\partial w^0} = 1##
The above expressions implicitly use the multi-index {3,1,2,0}.
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For a third order expansion in powers of ##f(x,y,z)##, we could employ some concise (and atrocious?) notation such as saying ##T(a,b,c)## will abbreviate ##\frac{1}{a! \ b! \ c!} \frac{\partial^a f}{\partial x^a} \frac{\partial^b f}{\partial y^b} \frac{\partial^c z}{\partial z^c} x^a y^b z^c## and define ##T(0,0,0) = f(0,0,0)##
## f(x,y,z) =##
## T(0,0,0) ##
## + T(1,0,0) + T(0,1,0) + T(0,0,1) ##
## + T(2,0,0) + T(1,1,0) + T(1,0,1) + T(0,2,0) + T(0,1,1) + T(0,0,2)##
## + T(3,0,0) + T(2,1,0) + T(2,0,1) + T(1,2,0) + T(1,1,1) + T(1,0,2) + T(0,3,0) + T(0,2,1) + T(0,1,2) + T(0,0,3)##
Someone should check my work.