Hameeral: Write up the proof of symmetry of second partial derivatives.

2016-10-20T02:50:42Z

Write up the proof of symmetry of second partial derivatives.

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			== Symmetry of Second Partial Derivatives ==

			'''(Note: This is based off of the proof in the textbook, and may be slightly different from how it was presented in lecture.)'''

			=== Theorem ===

			Let <math>A</math> be an open subset of <math>\mathbb{R}^2</math>, and let <math>f: A \rightarrow \mathbb{R}</math> be of class <math>C^2</math>. Then, for all <math>c \in A</math>, we have that <math>D_1 D_2 f(c) = D_2 D_1 f(c)</math>.

			=== Proof ===

			==== Step 1 ====

			We begin by defining a function that will help us in our proof. Let <math>(a, b) \in A \subseteq \mathbb{R}^2</math> be an arbitrary point. We then define the function <math>\lambda</math> as follows:

			<math>\lambda(h, k) = f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)</math>.

			Why do we define such a function? In fact, we can define both of the partial derivatives in question in terms of <math>\lambda</math>:

			<math>
			\begin{align}
			D_1 D_2 f(a, b) & = \frac{\partial}{\partial x} \frac{\partial f}{\partial y} (a, b) \\
			& = \lim_{h \to 0} (\frac{1}{h}) [\frac{\partial f}{\partial y} (a + h, b)] - [\frac{\partial f}{\partial y} (a, b)] \\
			& = \lim_{h \to 0} (\frac{1}{h}) [\lim_{k \to 0} (\frac{1}{k}) f(a + h, b + k) - f(a + h, b)] - [\lim_{k \to 0} (\frac{1}{k}) f(a, b + k) - f(a, b)] \\
			& = \lim_{h \to 0} (\frac{1}{h}) [\lim_{k \to 0} (\frac{1}{k}) f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)] \\
			& = \lim_{h \to 0} \lim_{k \to 0} (\frac{1}{hk}) [f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)] \\
			& = \lim_{h \to 0} \lim_{k \to 0} (\frac{1}{hk}) \lambda(h, k)
			\end{align}
			</math>

			And similarly for <math>D_2 D_1 f(a, b)</math>, but with the two limits taken in the opposite order.

			==== Step 2 ====

			Now, let's make use of this function. Let us consider the square <math>Q = [a, a + h] \times [b, b + k]</math>, where <math>h</math> and <math>k</math> are so small that <math>Q \subseteq A</math>. We will show that there exist points <math>p, q \in Q</math> such that:

			<math>
			\begin{align}
			& \lambda(h, k) = D_2 D_1 f(p) \cdot hk \\
			& \lambda(h, k) = D_1 D_2 f(q) \cdot hk
			\end{align}
			</math>

			The proofs of the two equalities are symmetric, and thus we only explicitly prove the first one. We will prove this through a double application of the mean value theorem.

			Consider the function <math>\phi</math>, defined such that <math>\phi(s) = f(s, b + k) - f(s, b)</math>. Then <math>\phi(a + h) - \phi(a) = \lambda(h, k)</math>. By hypothesis, <math>D_1 f</math> exists at all points of <math>A</math>, so we can differentiate <math>\phi</math> with respect to the first variable of <math>f</math> on the interval <math>[a, a + h]</math>. By the mean value theorem, this means that there exists a point <math>s_0 \in [a, a + h]</math> such that <math>\lambda(h, k) = \phi(a + h) - \phi(a) = \phi^\prime(s_0) \cdot h = [D_1 f(s_0, b + k) - D_1 f(s_0, b)] \cdot h</math>.

			Now we want to apply the mean value theorem one more time. Consider the function <math>\gamma</math>, defined such that <math>\gamma(t) = D_1 f(s_0, t)</math>. By hypothesis, <math>D_2 D_1 f</math> exists at all points of <math>A</math>, so we can differentiate <math>\gamma</math> with respect to the second variable of <math>f</math> on the interval <math>[b, b + k]</math>. So, by the mean value theorem, there exists a point <math>t_0 \in [b, b + k]</math> such that <math>\gamma(b + k) - \gamma(b) = \gamma^\prime(t_0) \cdot k</math>. Thus:

			<math>
			\begin{align}
			\lambda(h, k) & = [D_1 f(s_0, b + k) - D_1 f(s_0, b)] \cdot h \\
			& = [\gamma(b + k) - \gamma(b)] \cdot h \\
			& = \gamma^\prime(t_0) \cdot hk \\
			& = D_2 D_1 f(s_0, t_0) \cdot hk
			\end{align}
			</math>

			==== Step 3 ====

			Now we can prove the theorem. Let <math>c = (a, b) \in A</math>, and let <math>t > 0</math> be so small that <math>Q_t = [a, a + t] \times [b, b + t] \subseteq A</math>. By what we have just shown, <math>\lambda(t, t) = D_2 D_1 f(p_t) \cdot t^2</math>, for some <math>p_t \in Q_t</math>.

			<math>t</math> is the length of the sides of the rectangle <math>Q_t</math>, so as <math>t \rightarrow 0</math>, <math>p_t \rightarrow c</math>. By hypothesis, <math>D_2 D_1 f</math> is continuous, and so as <math>p_t \rightarrow c</math>, <math>D_2 D_1 f(p_t) \rightarrow D_2 D_1 f(c)</math>. Thus:

			<math>\lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_2 D_1 f(c)</math>.

			We can use the same argument, and the second equality from step 2, to show that:

			<math>\lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_1 D_2 f(c)</math>.

			Therefore, by the uniqueness of limits, the two quantities must be equal. <math>\square</math>

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2016-10-15T16:22:05Z

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== Handwritten Lecture Notes in PDF ==
[[Media:1617-257(lecture14).PDF|MAT257 - Lecture14 (Oct 14)]]

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Hameeral: Write up the proof of symmetry of second partial derivatives.

Com16257: Created page with "{{1617-257/Navigation}} {{1617-257:Dror/Students Divider}} == Handwritten Lecture Notes in PDF == MAT257 - Lecture14 (Oct 14)"