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George Weigt 2005-11-25 13:53:29 -07:00
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Example 1. Prove that $\lim(1/n^2)=0$.
1. Let $\epsilon>0$.
2. Let $N=1/\sqrt\epsilon$. (As follows: Start with $1/N^2=\epsilon$. Flip both sides to get
$N^2=1/\epsilon$. Square root of both sides to get $N=1/\sqrt\epsilon$.)
3. Now take the assertion $n>N$ and manipulate as follows:
4. Substitute $1/\sqrt\epsilon$ for $N$: $n>1/\sqrt\epsilon$
5. Square both sides: $n^2>1/\epsilon$
6. Multiply both sides by $\epsilon/n^2$: $\epsilon>1/n^2$
7. Flip end-to-end: $1/n^2<\epsilon$
8. Dress it up a little: $|1/n^2-0|<\epsilon$
9. Thus $n>N$ implies $|1/n^2-0|<\epsilon$
10. This proves that $\lim(1/n^2)=0$.
\end

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\magnification=1200
\parindent=0pt
Problem 8.1 (c) on page 32.
$$c_n={4n+3\over7n-5}$$
1. Limit looks to be $c={4\over7}$, set up the relation:
$$\left|{4n+3\over7n-5}-{4\over7}\right|<\epsilon$$
2. Combine denominators to get:
$$\left|{7(4n+3)-4(7n-5)\over7(7n-5)}\right|<\epsilon$$
3. Multiply out and simplify to get:
$$\left|{28n+21-28n+35\over49n-35}\right|
=\left|{41\over49n-35}\right|<\epsilon$$
4. Term is positive so drop absolute value:
$${41\over49n-35}<\epsilon$$
5. Multiply both sides by $(49n-35)/\epsilon$ to get:
$${41\over\epsilon}<49n-35$$
6. Add 35 to both sides and divide by 49 to get:
$${41\over49\epsilon}+{5\over7}<n$$
7. Set $N$ to $41/49\epsilon+5/7$. Then $N<n$ implies that
$${41\over49\epsilon}+{5\over7}<n$$
8. Subtract ${5\over7}$ from both sides to get:
$${41\over49\epsilon}<n-{5\over7}$$
9. Multiply both sides by $49\over41$ to get:
$${1\over\epsilon}<{49n-35\over41}$$
10. Mutiply both sides by inverses to get:
$${41\over49n-35}<\epsilon$$
11. This is the same as the result in step 4. We can work backwards from there
to get the equation in step 1 so the limit is proved.
\end

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\magnification=1200
\parindent=0pt
Problem 8.2 (c) on page 32.
$$c_n={4n+3\over7n-5}$$
1. Limit looks to be $c={4\over7}$, set up the relation:
$$\left|{4n+3\over7n-5}-{4\over7}\right|<\epsilon$$
2. Combine denominators to get:
$$\left|{7(4n+3)-4(7n-5)\over7(7n-5)}\right|<\epsilon$$
3. Multiply out and simplify to get:
$$\left|{28n+21-28n+35\over49n-35}\right|
=\left|{41\over49n-35}\right|<\epsilon$$
4. Term is positive so drop absolute value:
$${41\over49n-35}<\epsilon$$
5. Multiply both sides by $(49n-35)/\epsilon$ to get:
$${41\over\epsilon}<49n-35$$
6. Add 35 to both sides and divide by 49 to get:
$${41\over49\epsilon}+{5\over7}<n$$
7. Set $N$ to $41/49\epsilon+5/7$. Then $N<n$ implies that
$${41\over49\epsilon}+{5\over7}<n$$
8. Subtract ${5\over7}$ from both sides to get:
$${41\over49\epsilon}<n-{5\over7}$$
9. Multiply both sides by $49\over41$ to get:
$${1\over\epsilon}<{49n-35\over41}$$
10. Mutiply both sides by $41\epsilon/(49n-35)$ to get:
$${41\over49n-35}<\epsilon$$
11. This is the same as the result in step 4. We can work backwards from there
to get the equation in step 1 so the limit is proved.
\end

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\parindent=0pt
First, define two vector functions F and G.
\bigskip
{\tt\obeylines
> F=(FX(),FY(),FZ())
> G=(GX(),GY(),GZ())
}
\bigskip
Now verify the following vector identities.
$$\mathop{\rm div}(\mathop{\rm curl}F)=0\eqno1$$
{\tt\obeylines
> div(curl(F))
0
}
$$\mathop{\rm curl}(\mathop{\rm grad}f)=0\eqno2$$
{\tt\obeylines
> curl(grad(f()))
(0,0,0)
}
$$\mathop{\rm div}(\mathop{\rm grad}f)=\nabla^2f\eqno3$$
{\tt\obeylines
> div(grad(f()))-laplacian(f())
0
}
$$\mathop{\rm curl}(\mathop{\rm curl}F)=\mathop{\rm grad}
(\mathop{\rm div}F)-\nabla^2F\eqno4$$
{\tt\obeylines
> curl(curl(F))-grad(div(F))+laplacian(F)
(0,0,0)
}
$$\mathop{\rm grad}(fg)=f\mathop{\rm grad}g+g\mathop{\rm grad}f\eqno5$$
{\tt\obeylines
> grad(f()*g())-f()*grad(g())-g()*grad(f())
(0,0,0)
}
$$\mathop{\rm grad}(F\cdot G)=(G\cdot{\rm grad})F+(F\cdot{\rm grad})G
+G\times\mathop{\rm curl}F+F\times\mathop{\rm curl}G\eqno6$$
{\tt\obeylines
> grad(dot(F,G))-dot(grad(F),G)-dot(grad(G),F)-cross(G,curl(F))-cross(F,curl(G))
(0,0,0)
}
\bigskip
It turns out that the notation $(G\cdot{\rm grad})F$ actually means
$({\rm grad}\,F)\cdot G$.
Note: {\tt dot(grad(F),G)} is different from {\tt dot(G,grad(F))}! Why?
Because {\tt grad(F)} is a square matrix, not a vector.
Only the dot product of vectors is guaranteed to be commutative.
$$\mathop{\rm div}(fF)=f\mathop{\rm div}F+\mathop{\rm grad}f\cdot F\eqno7$$
{\tt\obeylines
> div(f()*F)-f()*div(F)-dot(grad(f()),F)
0
}
$$\mathop{\rm div}(F\times G)=G\cdot\mathop{\rm curl}F-F\cdot\mathop{\rm curl}G\eqno8$$
{\tt\obeylines
> div(cross(F,G))-dot(G,curl(F))+dot(F,curl(G))
0
}
$$\mathop{\rm curl}(fF)=f\mathop{\rm curl}F+\mathop{\rm grad}f\times F\eqno9$$
{\tt\obeylines
> curl(f()*F)-f()*curl(F)-cross(grad(f()),F)
(0,0,0)
}
$$\mathop{\rm curl}(F\times G)
=F\mathop{\rm div}G
-G\mathop{\rm div}F
+(G\cdot\mathop{\rm grad})F
-(F\cdot\mathop{\rm grad})G
\eqno10$$
{\tt\obeylines
> curl(cross(F,G))-F*div(G)+G*div(F)-dot(grad(F),G)+dot(grad(G),F)
(0,0,0)
}
\end

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\parindent=0pt
\beginsection Feynman
This is a problem from Feynman's ``Quantum Electrodynamics.''
Show that $\nabla_\mu A_\mu=0$ implies that $k\cdot e=0$.
The easiest way to do this is to write down all the components.
$$\nabla_\mu=(\partial/\partial t, -\partial/\partial x, -\partial/\partial y, -\partial/\partial z)$$
$$\nabla_\mu A_\mu=\nabla\cdot A={\partial A_t\over\partial t}+{\partial A_x\over\partial x}+{\partial A_y\over\partial y}+{\partial A_z\over\partial z}$$
$$A_t=e_t\exp[-i(k_tt-k_xx-k_yy-k_zz)]$$
$$A_x=e_x\exp[-i(k_tt-k_xx-k_yy-k_zz)]$$
$$A_y=e_y\exp[-i(k_tt-k_xx-k_yy-k_zz)]$$
$$A_z=e_z\exp[-i(k_tt-k_xx-k_yy-k_zz)]$$
$${\partial A_t\over\partial t}=(-ik_t)e_t\exp[-i(k_tt-k_xx-k_yy-k_zz)]$$
$${\partial A_x\over\partial x}=(+ik_x)e_x\exp[-i(k_tt-k_xx-k_yy-k_zz)]$$
$${\partial A_y\over\partial y}=(+ik_y)e_y\exp[-i(k_tt-k_xx-k_yy-k_zz)]$$
$${\partial A_z\over\partial z}=(+ik_z)e_z\exp[-i(k_tt-k_xx-k_yy-k_zz)]$$
$$\nabla_\mu A_\mu=-i(k\cdot e)\exp[-i(k_tt-k_xx-k_yy-k_zz)]$$
\beginsection Stefanovich
I stumbled across Stefanovich's web site and he has an interesting program
for doing QED calculations.
Here is Stefanovich's ``V'' file (shown with line numbers):
{
\tt
{\it\ 01}\ \char49
\
\
\char49
\
\
\char48
\
\
\char49
\
\
\char45
\char51
\
\
\char50
\
\
\char48
\
\char48
\ \par
{\it\ 02}\ \char120
\ \par
{\it\ 03}\ \char43
\char91
\char43
\char79
\char122
\char40
\char43
\char107
\char42
\char41
\char93
\ \par
{\it\ 04}\ \char120
\ \par
{\it\ 05}\ \char43
\char69
\char105
\char43
\char91
\char45
\char79
\char109
\char40
\char43
\char112
\char42
\char43
\char107
\char42
\char41
\char43
\char79
\char109
\char40
\char43
\char112
\char42
\char41
\char43
\char79
\char115
\char40
\char43
\char107
\char42
\char41
\char93
\char116
\ \par
{\it\ 06}\ \char120
\ \par
{\it\ 07}\ \char65
\char94
\char40
\char43
\char112
\char42
\char41
\ \par
{\it\ 08}\ \char49
\
\char48
\ \par
{\it\ 09}\ \char65
\char95
\char40
\char43
\char112
\char42
\char43
\char107
\char42
\char41
\ \par
{\it\ 10}\ \char50
\
\char48
\ \par
{\it\ 11}\ \char67
\char94
\char40
\char43
\char107
\char42
\char41
\ \par
{\it\ 12}\ \char49
\
\char50
\ \par
{\it\ 13}\ \char120
\ \par
{\it\ 14}\ \char49
\
\
\char49
\
\
\char48
\
\
\char49
\
\
\char45
\char51
\
\
\char50
\
\
\char48
\
\char48
\ \par
{\it\ 15}\ \char120
\ \par
{\it\ 16}\ \char43
\char91
\char43
\char79
\char122
\char40
\char43
\char107
\char42
\char41
\char93
\ \par
{\it\ 17}\ \char120
\ \par
{\it\ 18}\ \char43
\char69
\char105
\char43
\char91
\char45
\char79
\char109
\char40
\char43
\char112
\char42
\char45
\char107
\char42
\char41
\char43
\char79
\char109
\char40
\char43
\char112
\char42
\char41
\char45
\char79
\char115
\char40
\char43
\char107
\char42
\char41
\char93
\char116
\ \par
{\it\ 19}\ \char120
\ \par
{\it\ 20}\ \char65
\char94
\char40
\char43
\char112
\char42
\char41
\ \par
{\it\ 21}\ \char49
\
\char48
\ \par
{\it\ 22}\ \char65
\char95
\char40
\char43
\char112
\char42
\char45
\char107
\char42
\char41
\ \par
{\it\ 23}\ \char50
\
\char48
\ \par
{\it\ 24}\ \char67
\char95
\char40
\char43
\char107
\char42
\char41
\ \par
{\it\ 25}\ \char49
\
\char50
\ \par
{\it\ 26}\ \char120
\ \par
\end
}
\end

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From Math World, {\tt http://mathworld.wolfram.com/HilbertSpace.html}
\bigskip
A Hilbert space is a vector space $H$ with an inner product
$\langle f,g\rangle$ such that the norm defined by
$$|f|=\sqrt{\langle f,g\rangle}$$
turns $H$ into a complete metric space. If the inner product does not so
define a norm, it is instead known as an inner product space.
\bigskip
Examples of finite-dimensional Hilbert spaces include
1. The real numbers $R^n$ with $\langle v,u\rangle$ the vector dot product of
$v$ and $u$.
2. The complex numbers $C^n$ with $\langle v,u\rangle$ the vector dot product
of $v$ and the complex conjugate of $u$.
\end

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\magnification=1200
\parindent=0pt
\beginsection{December 24, 2002}
14.1(a) $\sum n^4/2^n$ Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={(n+1)^4/2^{n+1}\over n^4/2^n}
={(n+1)^4\over2^{n+1}}\times{2^n\over n^4}
={(n+1)^4\over2n^4}\rightarrow{1\over2}<1}$
\medskip
So by the ratio test $\sum n^4/2^n$ converges.
\bigskip
14.1(b) $\sum2^n/n!$ Try the ratio test.
\medskip
$\displaystyle{{a_{n+1}\over a_n}={2^{n+1}/(n+1)!\over 2^n/n!}
={2^{n+1}\over(n+1)!}\times{n!\over2^n}={2\over n+1}<1}$
\medskip
So by the ratio test $\sum2^n/n!$ converges.
\bigskip
14.1(c) $\sum n^2/3^n$ Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={(n+1)^2/3^{n+1}\over n^2/3^n}
={(n+1)^2\over 3^{n+1}}\times{3^n\over n^2}
={(n+1)^2\over3n^2}\rightarrow{1\over3}<1
}$
\medskip
So by the ratio test $\sum n^2/3^n$ converges.
\bigskip
14.1(d) $\sum n!/(n^4+3)$ Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={(n+1)!/((n+1)^4+3)\over n!/(n^4+3)}
={(n+1)!\over((n+1)^4+3)}\times{n^4+3\over n!}
\rightarrow n+1>1
}$
\medskip
So by the ratio test $\sum n!/(n^4+3)$ diverges.
\bigskip
14.1(e) $\sum\cos^2 n/n^2$ Use the comparison test.
Since $\sum1/n^2$ converges and
$|\cos^2 n/n^2|\le1/n^2$ for all $n$, $\sum\cos^2 n/n^2$ must also converge.
\bigskip
14.1(f) $\sum_{n=2}^\infty1/(\log n)$ Use the comparison test.
Since $\sum(1/n)$ diverges and $1/(\log n)>1/n$ for all $n>1$,
$\sum_{n=2}^\infty1/(\log n)$ must also diverge.
\beginsection{December 26, 2002}
Zeno's Paradox: The solution is that the infinite sum converges.
\bigskip
Proof of the Root Test. Let $\alpha=\lim\sup|a_n|^{1/n}$.
We can select an $\epsilon$ so that $\lim\sup|a_n|^{1/n}<\alpha+\epsilon$.
Next we can move the exponent around and put $|a_n|=(\alpha+\epsilon)^n$.
Now if $\alpha<1$ then we can choose a suitably small $\epsilon$ so that
$\alpha+\epsilon<1$. Since the geometric series $\sum(\alpha+\epsilon)^n$
converges, $\sum a_n$ must converge by the Comparison Test.
So if $\alpha<1$ then $\sum a_n$ converges.
\bigskip
Even simpler: If $\alpha<1$ then $\sum \alpha^n$ converges
and so $\sum a_n$ converges by the Comparison Test.
\bigskip
14.2(a) $\sum(n-1)/n^2$. Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={n\over(n+1)^2}\times{n^2\over n-1}
={n^3\over n^3+n^2-n-1}<1?
}$
\medskip
Not so sure. It gets closer and closer to 1 so is $\lim\sup=1$?
If we consider $\sum(n-1)/n^2=\sum(1/n-1/n^2)$ it looks divergent.
Is it fair to say that since
\medskip
$\displaystyle{\lim\sup{n^3+n^2-n-1\over n^3}=1}$ that
$\displaystyle{\lim\sup{n^3\over n^3+n^2-n-1}=1}$ also?
\bigskip
14.2(b) $\sum(-1)^n$ The sum is 0 for even $n$ and $-1$ for odd $n$.
So it does not converge but it is bounded.
\bigskip
14.2(c) $\sum3n/n^3$ This is the same as $\sum3/n^2$ which converges.
(Can't find the theorem that says if $\sum a_n$ converges then
$r\sum a_n$ also converges.)
\bigskip
14.2(d) $\sum n^3/3^n$ Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={(n+1)^3\over3^{n+1}}\times{3^n\over n^3}
\rightarrow{1\over3^n}<1
}$ so it converges.
\bigskip
14.2(e) $\sum n^2/n!$ Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={(n+1)^2\over(n+1)!}\times{n!\over n^2}
\rightarrow{1\over n+1}<1}$ so it converges.
\bigskip
14.2(f) $\sum1/n^n$ Converges by the Comparison test since $1/n^n<1/n^2$.
\bigskip
14.2(g) $\sum n/2^n$ Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={n+1\over2^{n+1}}\times{2^n\over n}
\rightarrow{1\over2}<1}$ so it converges.
\bigskip
14.3(a) $\sum1/\sqrt{n!}$ Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={1\over\sqrt{(n+1)!}}\times{\sqrt{n!}\over1}
={1\over\sqrt{n+1}\cdot\sqrt{n!}}\times{\sqrt{n!}\over1}
={1\over\sqrt{n+1}}<1}$ so it converges.
\beginsection{December 27, 2002}
Question: $\sum1/n$ seems to satisfy the Cauchy criterion.
For every $\epsilon$ there is an $n$ such that $1/n<\epsilon$.
But $\sum1/n$ does not converge.
What am I missing?
Ok, I think I see the problem.
The Cauchy criterion is
$|\sum_{k=m}^n a_k|<\epsilon$ for {\it all} values of $n\ge m>N$.
Consider the case where $m=N+1$ and $n=+\infty$.
In this case we have to add up
all the terms all the way out to infinity and have it come up less than
$\epsilon$.
\bigskip
Question about
$\displaystyle{\lim{n^3\over n^3+n^2-n-1}=1?}$
Solution: Multiply through by $1/n^3$:
\medskip
$\displaystyle{
\lim{n^3\over n^3+n^2-n-1}
=\lim{1\over1+{1\over n}-{1\over n^2}-{1\over n^3}}
=\lim{1\over1+0+0+0}=1
}$
\beginsection{December 30, 2002}
Page 89, example 1. Show that $f(x)=2x^2+1$ is continuous using the
$\epsilon$-$\delta$ property.
We want to show that $|f(x)-f(x_0)|<\epsilon$ when $|x-x_0|<\delta$.
Start by evaluating $|f(x)-f(x_0)|$ as follows:
\medskip
$\displaystyle{
|f(x)-f(x_0)|=|2x^2+1-(2x_0^2+1)|=|2x^2-2x_0^2|=2|x-x0|\cdot|x+x_0|
}$
\medskip
Ok, I get that, except for maybe the last step that distributes the absolute
value function. The book says, ``We need to get a bound for $|x+x_0|$ that
does not depend on $x$.'' Why?
Well, I think basically it's to get rid of $x^2$, to have a relation that's
linear in $x$.
Ok, onwards. Consider the case of $\delta<1$ which seems pretty typical
since $\delta$ is allowed to get infinitely small.
So $\delta<1$ means that $|x-x_0|<1$.
Now watch this. The book says this means that $|x|<|x_0|+1$.
It's true and here is why.
Suppose $|x|$ is already less than $|x_0|$.
Then it is certainly less than $|x_0|+1$.
Now suppose $|x|$ is greater than $|x_0|$.
Then by $\delta<1$ the difference has to be less than 1.
(I'm impressed by the creativity here. Or is this a common trick to use?)
So now that we have $|x|<|x_0|+1$, then we can put
$|x+x_0|\le|x|+|x_0|<2|x_0|+1$. Whew!
(How? By the triangle inequality and then by adding $|x_0|$ to both sides
of $|x|<|x_0|+1$.)
Now that we have $|x+x_0|<2|x_0|+1$ we can put
\medskip
$\displaystyle{
|f(x)-f(x_0)|<2|x-x_0|(2|x_0|+1)
}$
\medskip
with the caveat $\delta<1$.
Now we want $2|x-x_0|(2|x_0|+1)<\epsilon$ which leads to
\medskip
$\displaystyle{
|x-x_0|<{\epsilon\over2(2|x_0|+1)}
}$
\medskip
Since $|x-x_0|<\delta$ we can choose the following $\delta$:
\medskip
$\displaystyle{
\delta={\epsilon\over2(2|x_0|+1)}
}$
\medskip
and since $\delta<1$ we have
\medskip
$\displaystyle{
\delta=\min\left\{1,{\epsilon\over2(2|x_0|+1)}\right\}
}$
\medskip
This final relation shows that $|x-x_0|<\delta$ implies
$|f(x)-f(x_0)|<\epsilon$, which is what we wanted in the first place.
(Really? How?)
\beginsection{December 31, 2002}
From yesterday: $|x^2-x_0^2|=|x-x_0|\times|x+x_0|$. How?
First, factor it: $x^2-x_0^2=(x-x_0)(x+x_0)$.
Then apply the identity $|AB|=|A|\times|B|$.
\end

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\magnification=1200
\parindent=0pt
14.1(a) $\sum n^4/2^n$ Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={(n+1)^4/2^{n+1}\over n^4/2^n}
={(n+1)^4\over2^{n+1}}\times{2^n\over n^4}
={(n+1)^4\over2n^4}\rightarrow{1\over2}<1}$
\medskip
So by the ratio test $\sum n^4/2^n$ converges.
\bigskip
14.1(b) $\sum2^n/n!$ Try the ratio test.
\medskip
$\displaystyle{{a_{n+1}\over a_n}={2^{n+1}/(n+1)!\over 2^n/n!}
={2^{n+1}\over(n+1)!}\times{n!\over2^n}={2\over n+1}<1}$
\medskip
So by the ratio test $\sum2^n/n!$ converges.
\bigskip
14.1(c) $\sum n^2/3^n$ Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={(n+1)^2/3^{n+1}\over n^2/3^n}
={(n+1)^2\over 3^{n+1}}\times{3^n\over n^2}
={(n+1)^2\over3n^2}\rightarrow{1\over3}<1
}$
\medskip
So by the ratio test $\sum n^2/3^n$ converges.
\bigskip
14.1(d) $\sum n!/(n^4+3)$ Try the ratio test.
\medskip
$\displaystyle{
{a_{n+1}\over a_n}
={(n+1)!/((n+1)^4+3)\over n!/(n^4+3)}
={(n+1)!\over((n+1)^4+3)}\times{n^4+3\over n!}
\rightarrow n+1>1
}$
\medskip
So by the ratio test $\sum n!/(n^4+3)$ diverges.
\bigskip
14.1(e) $\sum\cos^2 n/n^2$ Use the comparison test.
Since $\sum1/n^2$ converges and
$|\cos^2 n/n^2|\le1/n^2$ for all $n$, $\sum\cos^2 n/n^2$ must also converge.
\bigskip
14.1(f) $\sum_{n=2}^\infty1/(\log n)$ Use the comparison test.
Since $\sum(1/n)$ diverges and $1/(\log n)>1/n$ for all $n>1$,
$\sum_{n=2}^\infty1/(\log n)$ must also diverge.
\end

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\magnification=1200
\parindent=0pt
Veltman, p. 23.
\bigskip
$$a=\left(\matrix{0&1&0\cr0&0& \sqrt2\cr0&0&0}\right)\quad
\tilde a=\left(\matrix{0&0&0\cr1&0&0\cr0&\sqrt2&0}\right)$$
\bigskip
Work out the products $a^2$, $\tilde a^2$, $\tilde aa$, $a\tilde a$, $\tilde aa-a\tilde a$, etc.
\bigskip
\tt\obeylines
> a=((0,1,0),(0,sqrt(2),0),(0,0,0))
> at=transpose(a)
> dot(a,a)
((0,0,2**(1/2)),(0,0,0),(0,0,0))
> dot(at,at)
((0,0,0),(0,0,0),(2**(1/2),0,0))
> dot(at,a)
((0,0,0),(0,1,0),(0,0,2))
> dot(a,at)
((1,0,0),(0,2,0),(0,0,0))
> dot(at,a)-dot(a,at)
((-1,0,0),(0,-1,0),(0,0,2))
\end

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\noindent
25. Show that the non-zero residue classes modulo $p$ form a group with
respect to multiplication if and only if $p$ is prime.
\medskip
As a counter-example, for $p=4$ we have
$S=\{[1],[2],[3]\}$.
The problem is that some products have residue 0.
For example, $2\times2=4$ has residue zero which is not in $S$.
What we need is $p$ with no factors so that we
cannot generate a multiple of $p$ which has residue 0.
So $p$ must be prime.
\end

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\noindent
1. A square acre measures approximately $208.71$ feet on each side.
What is the dimension exactly?
\medskip
640 acres is a square mile. One side of a square mile is 5280 feet.
Now we consider how many acre plots can we line up on one side of
a square mile.
Since there are 640 acres total, there must be $\sqrt{640}$ on each side.
Therefore one side of an acre must be $5280/\sqrt{640}$ feet
which is equivalent to $5280/(8\sqrt{10})$ feet
which is equivalent to $660/\sqrt{10}$ feet.
\end

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\noindent
2. From Quantum magazine, p. 18.
Prove the following theorem (the ``horse'' theorem).
If a function is defined on a circle
and is continuous, there exist
two diametrically opposite points
on the circle where this function
takes on equal values.
\medskip
Let $v$ be a continuous function defined on a circle.
$$v(\theta)$$
Here are its endpoints.
$$v(0)=v(2\pi)=A$$
Now consider the following function $u$.
$$u(\phi)=v(\phi+\pi)-v(\phi)$$
Evaluate $u$ at zero and $\pi$.
$$\eqalign{
u(0)&=v(\pi)-v(0)=v(\pi)-A\cr
u(\pi)&=v(2\pi)-v(\pi)=A-v(\pi)
}$$
We see that $u(0) = -u(\pi)$. The significance of this is that the function
$u(\phi)$
goes from positive to negative, so $u(\phi)=0$ must exist
somewhere.
Since $u(\phi)=0$ exists then by our definition of $u(\phi)$ we have
$$v(\phi+\pi)=v(\phi)$$
for some $\phi$.
\end