Taylor's Formula in Higher Dimension/Higher order Total differentials

1. Feb 9, 2008

sinClair

1. The problem statement, all variables and given/known data
First write f(x,y) = x^2 + xy + y^2 in terms of powers of (x+1) and (y-1)
Then write the taylor's formula for f(x,y) a = (1,4) and p=3

2. Relevant equations
We write taylor's formula as:

f(x) = f(a) + sum[(1/k!)*D^(k)f(a;h)] + (1/p!)D^(p)f(c;h) where sum is from k=1 to p-1 and D^(k)f(a;h) is the kth total differential of f at a evaluated for change h=x-a.

3. The attempt at a solution
I'm not sure how to do the first part--I thought perhaps I could divide f(x,y) by x+1 or use the binomial theorem. Is there a calculational method to do this?

For the second part, By just plugging in to Taylor formula I have f(x,y) = 3 + D^(1)f(a;h) +1/2D^(2)f(a;h) + 1/6D^(3)f(c;h)

The problem is I'm not sure how to evaluate the n-order total derivative D^(n)f(a;h). For n=1, I thought Df(a;h) = df(a)/dx*h + df(a)/dy*k = h/2 + k/4.

Is this calculation right? Can someone help me with the evaluating of the total derivative?

2. Feb 9, 2008

HallsofIvy

Staff Emeritus
Yes, it is the Taylor's formula you give above! However I suspect you are supposed to use the "less sophisticated" method of taking u= x-1, v= y-1 so that x= u+1, y= v+1 and substituting. After you have it in terms of u and v, replace u with (x-1) and v with (y- 1).

3. Feb 9, 2008

sinClair

Thanks Ivy I got the first part, though I'm not exactly sure how that relates to taylor's theorem.

I couldn't see what you wrote for the second quote--mind repeating what you said?

4. Feb 10, 2008

sinClair

Can someone post a link to evaluating higher order total differentials or show how to do it?