\documentclass{amsart}
\usepackage{amsmath,amssymb}

\DeclareMathOperator{\im}{Im}

\newcommand{\R}{{\mathbb R}}
\newcommand{\D}{{\boldsymbol D}}
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\title{Math 23A SOLUTION SET 7}
\author{George Kirkup}

\begin{document}
\maketitle

1.7.9)  This problem is essentially trivial, so since everyone did quite
well on it I will not say any more.

1.7.12a)  The claim is that $L(M) = (\D f(A))M = A^2M + MAM + MA^2$.  I will
skip steps in this calculation.
\begin{equation}
\begin{split}
\lim_{H \ra 0} \frac{|f(A+H) - f(A) - L(H)|}{|H|} &=
\lim_{H \ra 0} \frac{|H^2A + HAH + AH^2 + H^3|}{|H|}\\
& \leq \lim_{H \ra 0} \frac{|H^2A|+|HAH|+|AH^2|+|H^3|}{|H|}\\
& \leq \lim_{H \ra 0} \frac{|H|^2|A|+|H||A||H|+|A||H|^2+|H|^3}{|H|}\\
& = \lim_{H \ra 0} (|H||A|+|H||A|+|A||H|+|H|^2)\\
& = 0
\end{split}
\end{equation}
Of course, the above limit is at least $0$ so we are done.

1.7.16b)  Evaluating the limit of $f$ along the line $y=x$, we have
$$\lim_{x\ra 0}\frac{\sin(x^2)}{2x^2}$$ which we can easily see is $1/2$
using L'Hospital's Rule if nothing else.  This disagrees with $f(0) = 0$ so
the function is not continuous much less differentiable.

1.7.17d)  This was also an exercise in partial derivatives.

1.7.21)  Let $f(x_1,x_2,x_3,x_4)=x_1x_4 - x_2x_3$ be a map from $\R^4$ to
$\R$.  Calculating the partial derivatives at $(1,0,0,1)$ we see that
$$\D f(1,0,0,1) = \begin{pmatrix} 1& 0 & 0 & 1\end{pmatrix}$$ so
$$(\D f(1,0,0,1))\begin{pmatrix} a\\b\\c\\d \end{pmatrix} = a + d.$$
Note that this is equivalent to the desired result.

1.8.3)  Interestingly enough the easiest way to do this problem was by
composing the functions to get
$$(f \circ \gamma)(t) = \sum_1^{n-1}t^{2i+1}$$
so it is clear that the derivative is
$$\sum_1^{n-1} (2i+1)t^{2i}.$$

1.8.4)  Suppose there is a smooth map $g$ with 
$$g(\begin{pmatrix} 1\\1\end{pmatrix}) = \begin{pmatrix} 0\\0\end{pmatrix}$$
and 
$$(f \circ g)(\begin{pmatrix} x\\y\end{pmatrix}) = \begin{pmatrix}  
y\\x\end{pmatrix}.$$
By the chain rule we have
$$(\D f(\begin{pmatrix} 0\\0\end{pmatrix}))(\D g(\begin{pmatrix}  
1\\1\end{pmatrix})) = \begin{pmatrix} 0&1\\1&0\end{pmatrix}$$
but 
$$(\D f(\begin{pmatrix} 0\\0\end{pmatrix})) = \begin{pmatrix}
1&1\\1&1\end{pmatrix}$$
so the above product is impossible.  The easiest way to see this is that
$$\begin{pmatrix} 0&1\\1&0\end{pmatrix}$$ is invertible while 
$$\begin{pmatrix} 1&1\\1&1\end{pmatrix}$$ is not.

\end{document}
