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 June 25th, 2018, 02:44 AM #1 Newbie   Joined: Jun 2018 From: Viet Nam Posts: 3 Thanks: 0 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 $C^1$ $f(x,y) = \begin{cases}(x^2+y^2) sin\left(\frac{1}{\sqrt{x^2+y^2}}\right)&(x,y) \ne (0,0) &\\0&(x,y)=(0,0)\end{cases}?$ June 25th, 2018, 04:45 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 645 Thanks: 408 Math Focus: Dynamical systems, analytic function theory, numerics I have no idea what you are asking. June 25th, 2018, 06:33 PM   #3
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 Originally Posted by SDK 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. June 26th, 2018, 01:12 PM #4 Global Moderator   Joined: May 2007 Posts: 6,823 Thanks: 723 Your expression for multi-dimensional looks exactly like one dimensional. Could you clarify? June 26th, 2018, 06:56 PM #5 Senior Member   Joined: Sep 2016 From: USA Posts: 645 Thanks: 408 Math Focus: Dynamical systems, analytic function theory, numerics 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. June 30th, 2018, 11:12 AM   #6
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 Originally Posted by Chloesannon 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 $\lim_{r\to 0}\frac{r^2 sin(1/r)}{r}= \lim_{r\to 0} r sin(1/r)= 0$. Tags apply, expansion, function, multivariable, onevariable, taylor Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post puppypower123 Calculus 3 March 6th, 2017 07:55 PM irunktm Algebra 6 September 7th, 2012 12:42 PM Chasej Calculus 7 September 16th, 2011 12:18 AM person1200 Calculus 1 September 12th, 2010 07:17 PM wetmelon Algebra 2 March 23rd, 2009 11:00 PM

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