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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 multivariable as the onevariable expansion to the function f not be
Can I apply Taylor's multivariable as the onevariable 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: 628 Thanks: 398 Math Focus: Dynamical systems, analytic function theory, numerics 
I have no idea what you are asking.

June 25th, 2018, 06:33 PM  #3 
Newbie Joined: Jun 2018 From: Viet Nam Posts: 3 Thanks: 0  Multivariable Taylor's theorem : $f\in C^k(D)$, open set $D \subset \mathbb{R}^n$, exits $c \in (x,x+u)$ such that: $$f(x+u)=f(x)+f'(x)(u)+...+\frac{1}{(k1)!}f^{(k1)}(x)(u)^{k1}+\frac{1}{k!}f^{(k)}(c)(u)^k\quad (1)$$ The proof used onevariable 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 $$g(1)=g(0)+g'(0)+...+\frac{1}{(k1)!}g^{(k1)}(0)+\frac{1}{k!}g^{(k)}(c) \quad (2)$$ 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 $f\in C^{k1}(D)$, $f$ 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: 6,770 Thanks: 700 
Your expression for multidimensional looks exactly like one dimensional. Could you clarify?

June 26th, 2018, 06:56 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 628 Thanks: 398 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 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  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 
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