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

Jul 2010
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0
nyc
On p. 22 of "Applied Differential Geometry" by William Burke he says that Df(u) must be the unique linear operator satisfying

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

Why not just say:


\(\displaystyle
Df(u) \equiv \lim_{h\rightarrow0}||\frac{f(u+h) - f(u)}{h}||
\)

??

Then in the following example he says that Df(u) is a map.
 
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Opalg

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

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

Why not just say:


\(\displaystyle
Df(u) \equiv \lim_{h\rightarrow0}||\frac{f(u+h) - f(u)}{h}||
\)

??

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

You cannot define \(\displaystyle Df(u) =\displaystyle \lim_{h\rightarrow0}\left\|\frac{f(u+h) - f(u)}{h}\right\|\) 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 \(\displaystyle \mathbb{R}^n\) and h is a vector in \(\displaystyle \mathbb{R}^m\), and division of vectors is not defined.
 
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Jul 2010
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nyc
...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, \(\displaystyle f^{\mu}(u)\), but then that forms a vector since there are \(\displaystyle \mu\) 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.
 

Jose27

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

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


Why not just say:


\(\displaystyle
Df(u) \equiv \lim_{h\rightarrow0}||\frac{f(u+h) - f(u)}{h}||
\)

??

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

Your second limit makes no sense in a general setting, since \(\displaystyle h\) may very well be a vector as Opalg said (try some concrete examples for functions \(\displaystyle \mathbb{R} ^2 \rightarrow \mathbb{R}\) 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 \(\displaystyle f(u)=(f^1(u),...,f^n(u))\in \mathbb{R} ^n\)? \(\displaystyle f\) may have any space as domain and codomain (so long as the domain is an open subset of some euclidean space)
 
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Mar 2010
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Beijing, China
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.
 
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Jul 2010
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nyc
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!" (Rock)

\(\displaystyle \frac{df}{dp} = -1\)

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

No?

BTW: I am trying to learn Differential Geometry through the text, "Applied Differential Geometry", by a physicist well versed in mathematics, William Burke.
 
Mar 2010
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Beijing, China
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.
 
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May 2010
48
20
Texas
\(\displaystyle \frac{df}{dp} = -1\)
Since \(\displaystyle f:S^n \to S^n\) your differential operator is going to be an n x n matrix. Not a scalar. \(\displaystyle p \in S^n \)
 
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Feb 2010
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...TqN is the tangent plane to q, isomorphic to N...
(Naturally, he meant \(\displaystyle T_qN\) is isomorphic (as a linear space) to \(\displaystyle \mathbb{R}^n\), not N.)
 
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Jul 2010
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nyc
(Naturally, he meant \(\displaystyle T_qN\) is isomorphic (as a linear space) to \(\displaystyle \mathbb{R}^n\), not N.)
Isn't \(\displaystyle T_qN\) read as "the tangent plane to the manifold N at the point q (which is in N)."?

What is \(\displaystyle S^n\)? 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, \(\displaystyle g\)o\(\displaystyle f\).

(Should I post it here or start another thread showing the steps?)