# Some Test Practice Questions

1. Nov 3, 2008

### altcmdesc

I've got a test coming up and I would like to know if I'm doing this stuff correctly. Thanks!

1. The problem statement, all variables and given/known data
1. Consider

$$f(x,y) = \left\{ \begin{array}{c l} \frac{xy}{x^{2}-y^{2}} & x \neq \pm y\\ 0 & x = \pm y \end{array} \right.$$

Is $$f$$ differentiable at $$(0,0)$$?

2. Consider the continuous function

$$g(x,y) = \left\{ \begin{array}{c l} \frac{xy}{\sqrt{x^{2}+y^{2}}} & (x,y) \neq (0,0)\\ 0 & (x,y) = (0,0) \end{array} \right.$$

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 $$f(b) \neq f(a)$$.

Last edited: Nov 3, 2008
2. Nov 4, 2008

### altcmdesc

Are any of these correct?

Especially the proofs?

3. Nov 4, 2008

### Office_Shredder

Staff Emeritus
I only looked at number 2... your answer was almost awesome, except that if you have the denominator bounded above, then plugging in that bound minimizes the fraction. For example, what if $$\sqrt{x^2+y^2$$ was really $$\delta^5$$ which is less than delta for small delta (of course, then xy would be smaller, but you didn't say that, which is why the result breaks down).

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook