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June 25th, 2018, 03:44 AM   #1
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Apply Taylor's multi-variable as the one-variable expansion to the function f not be

Can I apply Taylor's multi-variable as the one-variable expansion to the function f not be in $C^1$
$f(x,y) = \begin{cases}(x^2+y^2) sin\left(\frac{1}{\sqrt{x^2+y^2}}\right)&(x,y) \ne (0,0) &\\0&(x,y)=(0,0)\end{cases}?$
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June 25th, 2018, 05:45 PM   #2
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I have no idea what you are asking.
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June 25th, 2018, 07:33 PM   #3
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Originally Posted by SDK View Post
I have no idea what you are asking.
Multivariable Taylor's theorem :
$f\in C^k(D)$, open set $D \subset \mathbb{R}^n$, exits $c \in (x,x+u)$ such that:
$$f(x+u)=f(x)+f'(x)(u)+...+\frac{1}{(k-1)!}f^{(k-1)}(x)(u)^{k-1}+\frac{1}{k!}f^{(k)}(c)(u)^k\quad (1)$$
The proof used one-variable Taylor's theorem, $g(t):=f(x+tu),t\in[0,1]$ so $g \in C^k$ then exits $c \in(0,1)$ such that
$$g(1)=g(0)+g'(0)+...+\frac{1}{(k-1)!}g^{(k-1)}(0)+\frac{1}{k!}g^{(k)}(c) \quad (2)$$
and we repalce $g^{(r)}(t)=f^{(r)}(x+tu)(u)^r,r=1,...,k$.(3)

I'm wondering whether multivariable Taylor expansion in the case $f\in C^{k-1}(D)$, $f$ is $k^{th}$-differentiable (not not necessary continuos) in D can be the same.

So I want to check with the function above.
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June 26th, 2018, 02:12 PM   #4
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Your expression for multi-dimensional looks exactly like one dimensional. Could you clarify?
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June 26th, 2018, 07:56 PM   #5
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This is not the multivariable Taylor theorem. At least not without explaining what you mean by the expressions of the form $f^{(k)}(x)$ and $(u)^{k}$. In the first case, I assume you mean "take $k$-many derivatives of $f$. However, in this case this derivative you speak of is tensor with $k$-many covariant indices. On the other hand, $x$ is a vector so you need to properly make sense of what $f^{(k)}(x)$ means. In the notation you have used it is typically meaningless.

You have a similar problem in the second case since $u$ is a vector so $u^k$ is not meaningful in the usual notation.

Most importantly though, you still haven't asked a question in any manner which I can understand what you are after.
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June 30th, 2018, 12:12 PM   #6
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Quote:
Originally Posted by Chloesannon View Post
Can I apply Taylor's multi-variable as the one-variable expansion to the function f not be in $C^1$
$f(x,y) = \begin{cases}(x^2+y^2) sin\left(\frac{1}{\sqrt{x^2+y^2}}\right)&(x,y) \ne (0,0) &\\0&(x,y)=(0,0)\end{cases}?$
Your basic error is in thinking that this is NOT in $C^1$. In polar coordinates this is $f(r, \theta)= r^2 sin(1/r)$ for r not o, $f(0,\theta)= 0$. Its derivative at r= 0 is $\lim_{r\to 0}\frac{r^2 sin(1/r)}{r}= \lim_{r\to 0} r sin(1/r)= 0$.
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