Abstract Nonsense

Crushing one theorem at a time

Further Properties of the Total Derivative (Pt. II)


Point of Post: This is a continuation of this post.

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We showed all of these proofs because it’s helpful to go through them, but they are all generalized by the following theorem:

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Theorem: Let f_1,\cdots,f_k:U\to\mathbb{R}^m, where U\subseteq\mathbb{R}^n, be differentiable at a\in U and let K:\left(\mathbb{R}^m\right)^k\to\mathbb{R}^s be multilinear then M:U\to\mathbb{R}^s given by M(x)=K(f_1(x),\cdots,f_k(x)) is differentiable at a and

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\displaystyle D_M(a)(x)=\sum_{j=1}^{k}K\left(f_1(a),\cdots,D_{f_j}(a)(x),\cdots,f_k(a)\right)

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Proof: Unsurprisingly we mimic the special cases we’ve already done. In particular, we define F:\mathbb{R}^n\to\mathbb{R}^{mk} by

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F(x)=(f_{1,1}(x),\cdots,f_{1,m}(x),\cdots,f_{k,1}(x),\cdots,f_{k,m}(x))

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(where f_{i,j} is the j^{\text{th}} coordinate function of f_i). We know that F is differentiable at a since each of its coordinate functions is differentiable at a. Also, we know that K is differentiable since it’s multilinear. Thus, since M=K\circ F we know from the chain rule that M is differentiable at a and D_M(a)=D_{K}(F(a))\circ D_F(a). But, as always

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\displaystyle D_F(a)=\left(\begin{array}{ccc} & \text{Jac}_{f_{1,1}}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{f_{1,m}}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{f_{k,1}}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{f_{k,m}}(a) & \end{array}\right)

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and we know that

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\displaystyle D_{K}(F(a))(x)=\sum_{j=1}^{k}K\left(f_1(a),\cdots,x_j,\cdots,f_k(a)\right)

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where x_j\in\mathbb{R}^m. It then follows that for any x\in U one has

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\displaystyle \begin{aligned}D_M(a)(x) &= \left(D_K(F(a))\right)\left(D_F(a)(x)\right)\\ &= \left(D_K(F(a)\right)\left(\text{Jac}_{f_{1,1}}(a)(x),\cdots,\text{Jac}_{f_{1,m}}(x),\cdots,\text{Jac}_{f_{k,1}}(a)(x),\cdots,\text{Jac}_{f_{k,m}}(a)(x)\right)\\ &= \sum_{j=1}^{k}K\left(f_1(a),\cdots,\text{Jac}_{f_j}(a)(x),\cdots,f_k(a)\right)\\ &= \sum_{j=1}^{k}K\left(f_1(a),\cdots,D_{f_j}(a)(x),\cdots,f_k(a)\right)\end{aligned}

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since x was arbitrary the conclusion follows. \blacksquare

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As a last point in this post we generalize the ‘quotient rule’ to mappings \mathbb{R}^n\to\mathbb{R}. In other words:

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Theorem: Let g:U\to\mathbb{R} where U\subseteq\mathbb{R}^n be differentiable at a\in U with g(a)\ne 0, then \displaystyle \frac{1}{g} is differentiable at a and

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\displaystyle D_{\frac{1}{g}}(a)=\frac{-D_g(a)}{g(a)^2}

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Proof: We begin by noting that since g is continuous at a we have that g(a)\ne 0 implies there exists an open ball B_{\delta}(a) of a for which g is non-zero. Clearly then choosing \|h\|<\delta implies that g(a+h)\ne 0. Consequently it makes sense to consider limits as h\to\bold{0} with \displaystyle \frac{1}{g(a+h)} in them. Thus, consider then that for \|h\|<\delta one has that

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\displaystyle \frac{1}{g(a+h)}-\frac{1}{g(a)}+\frac{D_g(a)(h)}{g(a)^2}=\frac{\left(g(a+h)-g(a)\right)D_g(a)(h)}{g(a)^2g(a+h)}-\frac{g(a+h)-g(a)+D_g(a)(h)}{g(a)g(a+h)}

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so that for \|h\|<\delta

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\displaystyle \begin{aligned}\frac{ \displaystyle \left|\frac{1}{g(a+h)}-\frac{1}{g(a)}+\frac{D_g(a)(h)}{g(a)^2}\right|}{\|h\|} &\leqslant \frac{\left|(g(a+h)-g(a))D_g(a)(h)\right|}{\left|g(a)^2g(a+h)\right|\|h\|}+\frac{1}{\left|g(a)g(a+h)\right|}\frac{\left|g(a+h)-g(a)-D_g(a)(h)\right|}{\|h\|}\\ &\leqslant \frac{\left\|D_g(a)\right\|_\text{op}\left|g(a+h)-g(a)\right|}{\left|g(a)^2g(a+h)\right|}+\frac{1}{\left|g(a+h)g(a)\right|}\frac{\left|g(a+h)-g(a)-D_g(a)(h)\right|}{\|h\|}\end{aligned}

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(where \|\cdot\|_\text{op} is the operator norm )but since both terms go to zero, the first since the numerator goes to zero while the denominator goes to something non-zero and the second because it’s a term going to a non-zero constant times a term which goes to zero by the definition of differentiability, we may conclude that \displaystyle \frac{1}{g} is in fact differentiable at a and has the claimed derivative.\blacksquare

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Using the product rule we evidently get the following corollary:

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Corollary: Let f,g:U\to\mathbb{R}^m where U\subseteq\mathbb{R}^n be differentiable at a\in U. If g(a)\ne 0 then \displaystyle \frac{f}{g} is differentiable at a and $

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\displaystyle D_{\frac{f}{g}}(a)=\frac{D_f(a)g(a)-D_g(a)f(a)}{g(a)^2}

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References:

1. Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.

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May 26, 2011 - Posted by | Analysis | , , , , ,

1 Comment »

  1. […] first and last one we have proven before (here and here respectively) and the second can be proved thinking of multiplication as a bilinear map and […]

    Pingback by Complex Differentiable and Holmorphic Functions (Pt. II) « Abstract Nonsense | May 1, 2012 | Reply


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