Taylor expansion with multi variables

Click For Summary
The discussion centers on the Taylor expansion for functions of multiple variables and the apparent absence of explicit derivatives in the provided formula. It is suggested that the derivatives are inherently included in the coefficients, evaluated at a specific point for approximation. The participants clarify that the general expression does not need to specify the point of approximation, as it serves to illustrate the concept. A participant expresses gratitude for the explanation, indicating it clarified their understanding of the multivariable Taylor expansion. Resources, including Wikipedia and a specific link, are shared for further exploration of the topic.
bubblewrap
Messages
134
Reaction score
2
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.
 

Attachments

  • 20160108_151010.jpg
    20160108_151010.jpg
    39.7 KB · Views: 517
Physics news on Phys.org
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.
 
blue_leaf77 said:
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.
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.
 
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.
 
blue_leaf77 said:
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.
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
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K