Maclaurin series for multivariable

manenbu
Messages
101
Reaction score
0

Homework Statement



I got a few functions I need to expand to series using Maclaurin forumlas.

Homework Equations



http://mathworld.wolfram.com/MaclaurinSeries.html

The Attempt at a Solution



So here are the ones I managed to do:
f= \sqrt{1-x^2-y^2}
writing it in another form:
f= (1+(-x^2-y^2)^{\frac{1}{2}}
Then I use:
(1+x)^m = 1 + mx + ..
as a function of one variable where x = -x^2-y^2 and I get the correct answer.
same goes for
z=\frac{1}{1-x+2y}
and
p=\ln(1+x+y).
Basically - I found that whenever there is no multiplication involved, I can just treat the two variables as one big variable and it works (according to my given answers).
The problem comes when I got stuff like this:
g=\frac{\cos{x}}{\cos{y}}
or
v=e^{x}\cos{y}.
Expanding each part and then dividing or multiplying (as you would you do if it was a true single var function) doesn't work. Expanding with taylor series from the start works - but the point is to use the maclaurin series.
So where did I go wrong?
 
Physics news on Phys.org
?? The MacLaurin series is a Taylor's series, just evaluated at x= 0.
 
I know, but the MacLaurin series are given equations so you will not have to differentiate all over again. For example, for functions in the form of ex or siny. You just plug it in the formula.
My question is - can it be done for functions like f(x,y) = exsiny? If yes - how? This way the differentiation process can be avoided.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top