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

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?

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 Quote by 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?
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).

 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?

 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?

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

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