# Taylor expansion with multi variables

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1. Jan 8, 2016

### bubblewrap

I was reading a book on differential equations when this(taylor expansion of multi variables) happened. Why does it not include derivatives of f in any form? The page of that book is in the file below.

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2. Jan 8, 2016

### blue_leaf77

I think the derivatives must have been absorbed into the definition of the coefficients, because in the end, the derivatives in Taylor expansion will be evaluated at the point around which the function is approximated.

3. Jan 8, 2016

### bubblewrap

Still, I have not seen anything like this in Taylor expansion. Perhaps the derivates were included in the coefficients but it doesn't say at which point it was evaluated and so on.

4. Jan 8, 2016

### blue_leaf77

That's the Taylor expansion for three variables, up to the second order it goes like
$$f(x_1,x_2,x_3) \approx f(a_1,a_2,a_3) + \sum_{i=1}^3 \frac{\partial f(a_1,a_2,a_3)}{\partial x_i} (x_i-a_i) + \frac{1}{2!}\sum_{i=1}^3\sum_{j=1}^3 \frac{\partial^2 f(a_1,a_2,a_3)}{\partial x_i \partial x_j} (x_i-a_i)(x_j-a_j)$$
The first term of the expansion of $\lambda$ in that book is equal to the first term in the above formula, the next three terms linear in $\pi$, $s$. and $p/n$ belong to the second term (the one containing single summation), and the rest belong to the double summation term. It needs not specify around which point the function is approximated when it only wants to give a general expression.

5. Jan 8, 2016

### bubblewrap

Ah thank you, I've been searching for multi variable taylor expansion bit wasn't able to find one like your explanation. It really all makes sense now, thanks

6. Jan 8, 2016