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

In summary, the conversation discusses how to write a given equation in terms of powers of (x+1) and (y-1) and how to use Taylor's formula to find the value of a given function at a specific point. The conversation also includes a question about how to evaluate higher order total differentials.
  • #1
sinClair
22
0

Homework Statement


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

Homework 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.

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 into 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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  • #2
sinClair said:

Homework Statement


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

Homework 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.

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 into 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?
 
  • #3
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
Can someone post a link to evaluating higher order total differentials or show how to do it?
 

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

1. What is Taylor's formula in higher dimensions?

Taylor's formula in higher dimensions is a generalization of the one-dimensional Taylor's formula, which is used to approximate a function using its derivatives at a single point. In higher dimensions, the formula involves partial derivatives and is useful for approximating a multivariable function around a given point.

2. How does Taylor's formula in higher dimensions differ from the one-dimensional formula?

The main difference between Taylor's formula in higher dimensions and the one-dimensional formula is that it involves partial derivatives instead of just ordinary derivatives. This allows for a more accurate approximation of a multivariable function, taking into account its behavior in different directions.

3. What is the purpose of using Taylor's formula in higher dimensions?

The purpose of using Taylor's formula in higher dimensions is to approximate a multivariable function around a given point. This can be useful in many applications, such as optimization, numerical analysis, and physics. It allows us to better understand the behavior of a function in multiple dimensions and make more accurate predictions.

4. How is Taylor's formula in higher dimensions related to total differentials?

Taylor's formula in higher dimensions is closely related to total differentials, as it is used to approximate a multivariable function using its partial derivatives. The total differential of a function is the sum of all its partial differentials, and Taylor's formula allows us to approximate this sum using only a few terms.

5. Can Taylor's formula in higher dimensions be extended to higher order derivatives?

Yes, Taylor's formula in higher dimensions can be extended to higher order derivatives, just like the one-dimensional formula. This allows for an even more accurate approximation of a function, taking into account its behavior at higher orders of differentiation. However, the calculations become more complex as the number of dimensions and order of derivatives increase.

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