# Derive Multivariable Taylor Series

1. Aug 17, 2010

Hello all,

I am currently studying multivariable calculus, and I am interested in the Taylor series for two variable function.

I am not sure where to begin; I cannot understand any of the proofs (which are apparently sparse) on the internet; they all just state it using a sigma sum; not very helpful to try to learn how it came to be :)

I was wondering how the 2-variable Taylor series is derived? :)

Cheers,

2. Aug 17, 2010

### Petr Mugver

Basic idea:

Let f(x,y) be a sufficiently differentiable real function around the point (x0, y0). Consider the one-variable function

F(a) =f(x0 + a*nx, y0 + a*ny)

where a is a real number and (nx, ny) a real 2-vector. Try to calculate the one-dimensional Taylor expansion for the function F(a), and at the end put

nx = x - x0
ny = y -y0

You'll find the two-dimensional Taylor series for f(x,y).

3. Aug 17, 2010

### ross_tang

For a more direct approach, you may try this:

$$f(x+\Delta x, y+\Delta y) = \sum_{k=0}^{\infty} \frac{(\Delta x)^k}{k!} \frac{d^k}{dx^k} f(x,y+\Delta y)$$

$$= \sum_{k=0}^{\infty} \frac{(\Delta x)^k}{k!} \frac{d^k}{dx^k} \sum_{j=0}^{\infty} \frac{(\Delta y)^j}{j!} \frac{d^j}{dy^j} f(x,y)$$

$$= \sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \frac{(\Delta x)^k}{k!} \frac{(\Delta y)^j}{j!} \frac{d^k}{dx^k} \frac{d^j}{dy^j} f(x,y)$$

$$= \sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \frac{(\Delta x)^k}{k!} \frac{(\Delta y)^j}{j!} f^{(k,j)}(x,y)$$

Last edited: Aug 17, 2010
4. Aug 17, 2010

### arildno

After ross' eminent start, you should try to switch the order of summation, in an intelligent manner, in order to gain a nice formula.

5. Aug 18, 2010