1617-257/Classnotes for Friday October 14: Difference between revisions
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(Write up the proof of symmetry of second partial derivatives.) |
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== Symmetry of Second Partial Derivatives == |
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'''(Note: This is based off of the proof in the textbook, and may be slightly different from how it was presented in lecture.)''' |
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=== Theorem === |
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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>. |
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=== Proof === |
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==== Step 1 ==== |
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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: |
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<math>\lambda(h, k) = f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)</math>. |
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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>: |
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<math> |
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\begin{align} |
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D_1 D_2 f(a, b) & = \frac{\partial}{\partial x} \frac{\partial f}{\partial y} (a, b) \\ |
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& = \lim_{h \to 0} (\frac{1}{h}) [\frac{\partial f}{\partial y} (a + h, b)] - [\frac{\partial f}{\partial y} (a, b)] \\ |
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& = \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)] \\ |
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& = \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)] \\ |
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& = \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)] \\ |
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& = \lim_{h \to 0} \lim_{k \to 0} (\frac{1}{hk}) \lambda(h, k) |
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\end{align} |
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</math> |
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And similarly for <math>D_2 D_1 f(a, b)</math>, but with the two limits taken in the opposite order. |
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==== Step 2 ==== |
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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: |
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<math> |
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\begin{align} |
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& \lambda(h, k) = D_2 D_1 f(p) \cdot hk \\ |
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& \lambda(h, k) = D_1 D_2 f(q) \cdot hk |
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\end{align} |
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</math> |
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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. |
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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>. |
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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: |
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<math> |
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\begin{align} |
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\lambda(h, k) & = [D_1 f(s_0, b + k) - D_1 f(s_0, b)] \cdot h \\ |
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& = [\gamma(b + k) - \gamma(b)] \cdot h \\ |
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& = \gamma^\prime(t_0) \cdot hk \\ |
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& = D_2 D_1 f(s_0, t_0) \cdot hk |
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\end{align} |
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</math> |
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==== Step 3 ==== |
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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>. |
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<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: |
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<math>\lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_2 D_1 f(c)</math>. |
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We can use the same argument, and the second equality from step 2, to show that: |
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<math>\lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_1 D_2 f(c)</math>. |
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Therefore, by the uniqueness of limits, the two quantities must be equal. <math>\square</math> |
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Latest revision as of 21:50, 19 October 2016
| Dror's notes above / Students' notes below |
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 be an open subset of , and let be of class . Then, for all , we have that .
Proof
Step 1
We begin by defining a function that will help us in our proof. Let be an arbitrary point. We then define the function as follows:
.
Why do we define such a function? In fact, we can define both of the partial derivatives in question in terms of :
![{\displaystyle {\begin{aligned}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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1e67111d9ecb026e1befb9ee801fe1e45baa851)
And similarly for , 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 , where and are so small that . We will show that there exist points such that:
![{\displaystyle Q=[a,a+h]\times [b,b+k]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6194030040349fcd9340c8d47a365e042e6d9532)
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 , defined such that . Then . By hypothesis, exists at all points of , so we can differentiate with respect to the first variable of on the interval . By the mean value theorem, this means that there exists a point such that .
![{\displaystyle [a,a+h]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f22e0d7755f1c29d2e577ab5a71ef94f668555e)
![{\displaystyle s_{0}\in [a,a+h]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d60a940f2a34b021a79cb4e763d7dfb7bffac28b)
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3f1db6c846e24a63fadd7b12313d1904ffb2de9)
Now we want to apply the mean value theorem one more time. Consider the function , defined such that . By hypothesis, exists at all points of , so we can differentiate with respect to the second variable of on the interval . So, by the mean value theorem, there exists a point such that . Thus:
![{\displaystyle [b,b+k]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77b1d5ef699fdd0f906e43bea744cf93952278b1)
![{\displaystyle t_{0}\in [b,b+k]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b003aa0358ce1e72719ea860bc9459252fb1b851)
![{\displaystyle {\begin{aligned}\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3148f35681565a319e8675720c6300de550e66ae)
Step 3
Now we can prove the theorem. Let , and let be so small that . By what we have just shown, , for some .
![{\displaystyle Q_{t}=[a,a+t]\times [b,b+t]\subseteq A}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f59bd646d20ae20690404ee0dcbfc63eff19713)
is the length of the sides of the rectangle , so as , . By hypothesis, is continuous, and so as , . Thus:
.
We can use the same argument, and the second equality from step 2, to show that:
.
Therefore, by the uniqueness of limits, the two quantities must be equal.
