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

• sinClair
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.
sinClair

## 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?

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?

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?

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

## 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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