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altcmdesc
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I've got a test coming up and I would like to know if I'm doing this stuff correctly. Thanks!
1. Consider
<br /> f(x,y) = \left\{<br /> \begin{array}{c l}<br /> \frac{xy}{x^{2}-y^{2}} & x \neq \pm y\\<br /> 0 & x = \pm y<br /> \end{array}<br /> \right.<br />
Is f differentiable at (0,0)?
2. Consider the continuous function
<br /> g(x,y) = \left\{<br /> \begin{array}{c l}<br /> \frac{xy}{\sqrt{x^{2}+y^{2}}} & (x,y) \neq (0,0)\\<br /> 0 & (x,y) = (0,0)<br /> \end{array}<br /> \right.<br />
Find \delta>0 so that |(x,y)|<\delta\Rightarrow|g(x,y)|<10^{-2}
3. Find the derivative of the mapping A\mapsto(I+A^{2})^{-1}.
4. Find N so that |z|>N\Rightarrow|p(z)|=|10z^{10}+9z^{9}+...+2z^{2}+z|>100
5. Suppose S\subset\Re and S is bounded above. Prove that \forall\epsilon>0 \exists y \inS with |supS-y|<\epsilon.
6. If f:\Re^{m}\rightarrow\Re^{n} has Df(a)\neq0, prove that \exists b\in\Re^{m} with f(a) \neq f(b)
2. The attempt at a solution
1. Along the x and y axes f(x,y)=0. Thus \frac{\partial f}{\partial x} = 0 and \frac{\partial f}{\partial y} = 0 at (0,0). From this, the directional derivative in any direction (v_{1}, v_{2}) should be 0. Using the limit definiton gives \lim_{h \rightarrow 0} \frac{\frac{h^{2}v_{1}v_{2}}{h^{2}(v_{1}-v_{2})}}{h} = \lim_{h \rightarrow 0} \frac{\frac{v_{1}v_{2}}{v_{1}-v{2}}}{h}. This limit does not exist for all v_{1}, v_{2}, so f is not differentiable at (0,0).
2. \sqrt{x^{2}+y^{2}}<\delta. Since x<\sqrt{x^{2}+y^{2}}<\delta and y<\sqrt{x^{2}+y^{2}}<\delta, xy<\delta^{2}. Now, |f(x,y)| = |\frac{xy}{\sqrt{x^{2}+y^{2}}}| < |\frac{\delta^{2}}{\sqrt{x^{2}+y^{2}}}| = |\frac{\delta^{2}}{\delta}| = |\delta| =\epsilon. Thus choosing \delta = 10^{-2} works.
3. Let F(A)=I+A^{2} and G(B)=B^{-1}. Then G\circF(A)=(I+A^{2})^{-1}. By the chain rule, D(G\circF(A))(H)=DG(F(A))DF(A)(H)=-(I+A^{2})^{-1}(DF(A)(H))(I+A^{2})^{-1}=-(I+A^{2})^{-1}(AH+HA)(I+A^{2})^{-1}.
4. If |10z^{10}|>|9z^{9}|+|8z^{8}|+...+|z|+100 then |p(z)|>100 follows. We have
|z|^{10}>9|z|^{9}, |z|^10>8|z|^{8}, |z|^10>7|z|^{7},...,|z|^{10}>100 since |z|^{10}>9|z|^{9} \Rightarrow |z|>9 is the most stringent, |z|>9 works.
5. If S is bounded above then \forally\inS y<supS. Thus \exists y_{n} in \overline{S} such that y_{n} \rightarrow supS as n\rightarrow\infty. For this sequence, we have \forall\epsilon>0 \existsN with n>N such that |y_{n}-supS|<\epsilon.
6. From the definition of the derivative at a: \lim_{h \rightarrow 0} \frac{|f(a+h)-f(a)-Df(a)h|}{|h|}=0. Let a+h=b, then \lim_{b \rightarrow a} \frac{|f(b)-f(a)-Df(a)(b-a)|}{|b-a|}=0 \rightarrow \lim_{b \rightarrow a} \frac{|f(b)-f(a)}{|b-a|}-\frac{|Df(a)(b-a)|}{|b-a|}=0. For this to be zero, \lim_{b \rightarrow a} \frac{|f(b)-f(a)}{|b-a|}\geq0 since \frac{|Df(a)(b-a)|}{|b-a|} is bounded as b \rightarrow a. Thus f(b) \neq f(a).
Homework Statement
1. Consider
<br /> f(x,y) = \left\{<br /> \begin{array}{c l}<br /> \frac{xy}{x^{2}-y^{2}} & x \neq \pm y\\<br /> 0 & x = \pm y<br /> \end{array}<br /> \right.<br />
Is f differentiable at (0,0)?
2. Consider the continuous function
<br /> g(x,y) = \left\{<br /> \begin{array}{c l}<br /> \frac{xy}{\sqrt{x^{2}+y^{2}}} & (x,y) \neq (0,0)\\<br /> 0 & (x,y) = (0,0)<br /> \end{array}<br /> \right.<br />
Find \delta>0 so that |(x,y)|<\delta\Rightarrow|g(x,y)|<10^{-2}
3. Find the derivative of the mapping A\mapsto(I+A^{2})^{-1}.
4. Find N so that |z|>N\Rightarrow|p(z)|=|10z^{10}+9z^{9}+...+2z^{2}+z|>100
5. Suppose S\subset\Re and S is bounded above. Prove that \forall\epsilon>0 \exists y \inS with |supS-y|<\epsilon.
6. If f:\Re^{m}\rightarrow\Re^{n} has Df(a)\neq0, prove that \exists b\in\Re^{m} with f(a) \neq f(b)
2. The attempt at a solution
1. Along the x and y axes f(x,y)=0. Thus \frac{\partial f}{\partial x} = 0 and \frac{\partial f}{\partial y} = 0 at (0,0). From this, the directional derivative in any direction (v_{1}, v_{2}) should be 0. Using the limit definiton gives \lim_{h \rightarrow 0} \frac{\frac{h^{2}v_{1}v_{2}}{h^{2}(v_{1}-v_{2})}}{h} = \lim_{h \rightarrow 0} \frac{\frac{v_{1}v_{2}}{v_{1}-v{2}}}{h}. This limit does not exist for all v_{1}, v_{2}, so f is not differentiable at (0,0).
2. \sqrt{x^{2}+y^{2}}<\delta. Since x<\sqrt{x^{2}+y^{2}}<\delta and y<\sqrt{x^{2}+y^{2}}<\delta, xy<\delta^{2}. Now, |f(x,y)| = |\frac{xy}{\sqrt{x^{2}+y^{2}}}| < |\frac{\delta^{2}}{\sqrt{x^{2}+y^{2}}}| = |\frac{\delta^{2}}{\delta}| = |\delta| =\epsilon. Thus choosing \delta = 10^{-2} works.
3. Let F(A)=I+A^{2} and G(B)=B^{-1}. Then G\circF(A)=(I+A^{2})^{-1}. By the chain rule, D(G\circF(A))(H)=DG(F(A))DF(A)(H)=-(I+A^{2})^{-1}(DF(A)(H))(I+A^{2})^{-1}=-(I+A^{2})^{-1}(AH+HA)(I+A^{2})^{-1}.
4. If |10z^{10}|>|9z^{9}|+|8z^{8}|+...+|z|+100 then |p(z)|>100 follows. We have
|z|^{10}>9|z|^{9}, |z|^10>8|z|^{8}, |z|^10>7|z|^{7},...,|z|^{10}>100 since |z|^{10}>9|z|^{9} \Rightarrow |z|>9 is the most stringent, |z|>9 works.
5. If S is bounded above then \forally\inS y<supS. Thus \exists y_{n} in \overline{S} such that y_{n} \rightarrow supS as n\rightarrow\infty. For this sequence, we have \forall\epsilon>0 \existsN with n>N such that |y_{n}-supS|<\epsilon.
6. From the definition of the derivative at a: \lim_{h \rightarrow 0} \frac{|f(a+h)-f(a)-Df(a)h|}{|h|}=0. Let a+h=b, then \lim_{b \rightarrow a} \frac{|f(b)-f(a)-Df(a)(b-a)|}{|b-a|}=0 \rightarrow \lim_{b \rightarrow a} \frac{|f(b)-f(a)}{|b-a|}-\frac{|Df(a)(b-a)|}{|b-a|}=0. For this to be zero, \lim_{b \rightarrow a} \frac{|f(b)-f(a)}{|b-a|}\geq0 since \frac{|Df(a)(b-a)|}{|b-a|} is bounded as b \rightarrow a. Thus f(b) \neq f(a).
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