Results 1 to 11 of 11

Math Help - App. Diff. Geom.: Df(u) <=> f'(x_0)??

  1. #1
    Newbie
    Joined
    Jul 2010
    From
    nyc
    Posts
    12

    App. Diff. Geom.: Df(u) <=> f'(x_0)??

    On p. 22 of "Applied Differential Geometry" by William Burke he says that Df(u) must be the unique linear operator satisfying

    \lim_{h\rightarrow0}\frac{||f(u+h) - f(u) - Df(u)||}{||h||} = 0

    Why not just say:


    <br />
Df(u) \equiv \lim_{h\rightarrow0}||\frac{f(u+h) - f(u)}{h}||<br />

    ??

    Then in the following example he says that Df(u) is a map.
    Last edited by livingdog; July 9th 2010 at 04:29 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by livingdog View Post
    On p. 22 of "Applied Differential Geometry" by William Burke he says that Df(u) must be the unique linear operator satisfying

    \lim_{h\rightarrow0}\frac{||f(u+h) - f(u) - Df(u)||}{||h||} = 0

    Why not just say:


    <br />
Df(u) \equiv \lim_{h\rightarrow0}||\frac{f(u+h) - f(u)}{h}||<br />

    ??

    Then in the following example he says that Df(u) is a map.
    If f is a function from \mathbb{R}^m to \mathbb{R}^n, and u\in\mathbb{R}^m then Df(u) is a linear map (or operator if you prefer) from \mathbb{R}^m to \mathbb{R}^n. If h\in\mathbb{R}^m then Df(u)h\in\mathbb{R}^n.

    You cannot define Df(u) =\displaystyle \lim_{h\rightarrow0}\left\|\frac{f(u+h) - f(u)}{h}\right\| because if the right-hand side made sense at all it would be a scalar, not a linear map. But it does not make sense at all, because f(u+h) – f(u) is a vector in \mathbb{R}^n and h is a vector in \mathbb{R}^m, and division of vectors is not defined.
    Last edited by Opalg; July 10th 2010 at 12:10 AM. Reason: previous version was full of mistakes
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2010
    From
    nyc
    Posts
    12

    Question

    Quote Originally Posted by Opalg View Post
    ...But it does not make sense at all, because f(u+h) – f(u) is a vector ...
    That's where you lost me - how is f(u), a function, a vector?? A vector has magnitude and direction. 'f(u)' is a scalar. In his later explanation he mentions that there may be many functions, f^{\mu}(u), but then that forms a vector since there are \mu components. ... wait, or are you saying that f(u) is a 1D vector??

    Please note that I am illiterate when it comes to modern math. I just started reading "Applied Differential Geometry" by Burke and am wadding through it like a hunter with a dull machete.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by livingdog View Post
    On p. 22 of "Applied Differential Geometry" by William Burke he says that Df(u) must be the unique linear operator satisfying

    \lim_{h\rightarrow0}\frac{||f(u+h) - f(u) - Df(u)||}{||h||} = 0


    Why not just say:


    <br />
Df(u) \equiv \lim_{h\rightarrow0}||\frac{f(u+h) - f(u)}{h}||<br />

    ??

    Then in the following example he says that Df(u) is a map.
    Notice that the first limit you have should be \lim_{h\rightarrow 0} \frac{\| f(u+h)-f(u)-Df(u)h\| }{\| h\| } =0. The thing is that Df(u) is a linear transformation for every u.

    Your second limit makes no sense in a general setting, since h may very well be a vector as Opalg said (try some concrete examples for functions \mathbb{R} ^2 \rightarrow \mathbb{R} and see what the derivative represents and why your definition fails).

    That's where you lost me - how is f(u), a function, a vector?? A vector has magnitude and direction. 'f(u)' is a scalar. In his later explanation he mentions that there may be many functions, , but then that forms a vector since there are components. ... wait, or are you saying that f(u) is a 1D vector??
    What if we were to DEFINE f(u)=(f^1(u),...,f^n(u))\in \mathbb{R} ^n? f may have any space as domain and codomain (so long as the domain is an open subset of some euclidean space)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    You need to understand that differential is a linear map in the tangent plane. Note that a smooth function on R^n is a smooth map between two manifolds R^n and R. So if you can understand differential of a general smooth map, it will helps you to understand differential of a smooth function, which is a real-valued map.
    Take an example, suppose S^n is the n-dimensional sphere, and f: S^n -> S^n is the anti-podal map f(p) = -p. How do you define its differential?

    See Differential of a function - Wikipedia, the free encyclopedia for more to understand that differential is a map.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2010
    From
    nyc
    Posts
    12
    Quote Originally Posted by xxp9 View Post
    You need to understand that differential is a linear map in the tangent plane. Note that a smooth function on R^n is a smooth map between two manifolds R^n and R. So if you can understand differential of a general smooth map, it will helps you to understand differential of a smooth function, which is a real-valued map.
    Take an example, suppose S^n is the n-dimensional sphere, and f: S^n -> S^n is the anti-podal map f(p) = -p. How do you define its differential?

    See Differential of a function - Wikipedia, the free encyclopedia for more to understand that differential is a map.
    Part of the problem is that I am coming to math from a physics background. The two disciplines are profoundly different - math being utterly (rightly) abstract, and physics being utterly (rightly) real (aka 3D). So when a mathematician says that "f(u)" is a vector they are using it in a most abstract sense. Whereas a physicist uses subscripts and means only a specific thing by "vectors." Granted Hilbert space uses functions as vectors, but still the labeling is far more identifiable in physics than in math. As my friend told me, who knows GR and String Theory, "mathematicians are 'cheap' with their notation." Indeed, so f(u) is meant to be a vector in a very broad sense and I have to either get used to this general way of thinking or give up studying physics ... to which I say: "f(u) is a vector!"

    \frac{df}{dp} = -1

    And, if f:R^n \rightarrow R^n,
    then df:R^n \rightarrow TR^{n-1}.

    No?

    BTW: I am trying to learn Differential Geometry through the text, "Applied Differential Geometry", by a physicist well versed in mathematics, William Burke.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    differential is the induced linear map in the tangent space. Suppose f is a smooth map from M (m-dimensional) to N( n-dimensional). q = f(p), TpM is the tangent plane to p, which is isomorphic to R^m, TqN is the tangent plane to q, isomorphic to R^n. u(t) is a smooth curve in M, u(0) = p. X is the velocity vector of u at p, X is in TpM. Then w(t) = f(u(t)) is a smooth curve in N with w(0) = f(p) = q, and Y is the velocity vector of w at q, Y in TqN. Then the differential Df is defined to be Df(X) = Y.
    Last edited by xxp9; July 10th 2010 at 10:01 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    May 2010
    From
    Texas
    Posts
    48
    Quote Originally Posted by livingdog View Post

    \frac{df}{dp} = -1
    Since f:S^n \to S^n your differential operator is going to be an n x n matrix. Not a scalar.  p \in S^n
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Feb 2010
    Posts
    422
    Quote Originally Posted by xxp9 View Post
    ...TqN is the tangent plane to q, isomorphic to N...
    (Naturally, he meant T_qN is isomorphic (as a linear space) to \mathbb{R}^n, not N.)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jul 2010
    From
    nyc
    Posts
    12
    Quote Originally Posted by maddas View Post
    (Naturally, he meant T_qN is isomorphic (as a linear space) to \mathbb{R}^n, not N.)
    Isn't T_qN read as "the tangent plane to the manifold N at the point q (which is in N)."?

    What is S^n? Is it standard notation for the n-dimensional hyper-surface of an "(n+1)-sphere"?

    I know I don't seem to get what is going on in this thread, but I think I correctly deduced the correct differential for a composite map, go f.

    (Should I post it here or start another thread showing the steps?)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Mar 2010
    From
    Beijing, China
    Posts
    293
    Thanks
    23
    thank maddas for the correction.
    S^n is the n-sphere, whose standard embedding is S^n=\{\mathbf{p} \in R^{n+1}| \|\mathbf{p}\|=1\}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. analytic geom
    Posted in the Geometry Forum
    Replies: 2
    Last Post: January 10th 2010, 08:16 PM
  2. PURSUIT CURVE (differential geom)
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 27th 2008, 01:28 PM
  3. Alg. Geom. Proofs
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 27th 2008, 02:14 AM
  4. diff. eq.
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: July 13th 2008, 09:00 PM
  5. analytic geom and calc 1
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 17th 2005, 05:58 PM

Search Tags


/mathhelpforum @mathhelpforum