-
Notifications
You must be signed in to change notification settings - Fork 0
/
chapter21.tex
162 lines (162 loc) · 9.51 KB
/
chapter21.tex
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
%%%%
%%
%%
%%%%
%%%% CHAPTER 21
%%%% CHAPTER 21
%%%%
%%
%%
%%%%
\section{Chapter 21 Solutions}
\begin{center}\hyperref[toc]{\^{}\^{}}\end{center}
\begin{center}\begin{tabular}{lllllllllllllllllllllllll}
\hyperref[problem1chapter21]{P1} & \hyperref[problem2chapter21]{P2} & \hyperref[problem3chapter21]{P3} & \hyperref[problem4chapter21]{P4} & \hyperref[problem5chapter21]{P5} & \hyperref[problem6chapter21]{P6}
\end{tabular}\end{center}
\setcounter{problem}{0}
\setcounter{solution}{0}
\begin{problem}\label{problem1chapter21}
Derive the preceding addition theorem $(7)$ by the method of Section 179. You may use results from the exercises at the end of Chapter 20.
\end{problem}
\begin{solution}
We seek an addition theorem for $\mathrm{cn}(u)$.
Now
$$\mathrm{cn}(u) = \dfrac{\theta_4}{\theta_2} \dfrac{\theta_2(u\theta_3^{-2})}{\theta_4(u\theta_3^{-2})}.$$
At the end of Exercise~5, Chapter~20 we obtained
$$\theta_4^2 \theta_4(x+4) \theta_4(x-y) = \theta_4^2(x) \theta_4^2(y) - \theta_1^2(x) \theta_1^2(y).$$
The next to last equation in Exercise~10, Chapter 20, is
$$\theta_2 \theta_4 \theta_2(x+y) \theta_4(x-y) = \theta_2(x) \theta_4(x) \theta_2(y) \theta_4(y) - \theta_1(x) \theta_3(x) \theta_1(y) \theta_3(y).$$
We divide the last equation above by the one preceding it to get
$$\begin{array}{ll}
\dfrac{\theta_2(x+y)}{\theta_4(x+y)} &= \dfrac{\theta_2(x) \theta_4(x) \theta_2(y) \theta_4(y) - \theta_1(x) \theta_3(x) \theta_1(y) \theta_3(y)}{\theta_4^2(x) \theta_4^2(y) - \theta_1^2(x) \theta_1^2(y)} \\
&=\dfrac{\frac{\theta_2(x)}{\theta_4(x)} \frac{\theta_2(y)}{\theta_4(y)} - \frac{\theta_1(x)}{\theta_4(x)} \frac{\theta_3(x)}{\theta_4(x)} \frac{\theta_1(y)}{\theta_4(y)} \frac{\theta_3(y)}{\theta_4(y)}}{1 - \frac{\theta_1^2(x)}{\theta_4^2(x)} \frac{\theta_2^2(y)}{\theta_4^2(y)}}.
\end{array}$$
Now use $x = u\theta_3^{-2}$ and $y = v\theta_3^{-2}$ to obtain
$$\dfrac{\theta_2}{\theta_4} \dfrac{\theta_2}{\theta_4} \mathrm{cn}(u+v) = \dfrac{\frac{\theta_2}{\theta_4} \mathrm{cn}(u) \frac{\theta_2}{\theta_4} \mathrm{cn}(v) - \frac{\theta_2}{\theta_3} \mathrm{\mathrm{sn}}(u) \frac{\theta_3}{\theta_4} \mathrm{dn}(u) \frac{\theta_2}{\theta_3} \mathrm{\mathrm{sn}}(v)\frac{\theta_3}{\theta_4} \mathrm{dn}(v)}{1 -\ \frac{\theta_2^2}{\theta_3^2} \mathrm{\mathrm{sn}}^2(u) \frac{\theta_2^2}{\theta_3^2}\mathrm{\mathrm{sn}}^2(v)}$$
or
$$\mathrm{cn}(u+v) = \dfrac{\mathrm{cn}(u)\mathrm{cn}(v) - \mathrm{\mathrm{sn}}(u)\mathrm{dn}(u) \mathrm{\mathrm{sn}}(v) \mathrm{dn}(v)}{1 - k^2 \mathrm{\mathrm{sn}}^2(u) \mathrm{\mathrm{sn}}^2(v)},$$
as desired.
\end{solution}
%%%%
%%
%%
%%%%
\begin{problem}\label{problem2chapter21}
Derive preceding $(8)$ by the method of Section 179, with the aid of results from the exercises at the end of Chapter 20.
\end{problem}
\begin{solution}
We know that
$$\mathrm{dn}(u) = \dfrac{\theta_4}{\theta_3} \dfrac{\theta_3(u\theta_3^{-2})}{\theta_4(u\theta_3^{-2})}, k = \dfrac{\theta_2}{\theta_3^2}.$$
From the last equation of Exercise~10, Chapter 20 we get
$$\theta_3 \theta_4 \theta_3(x+y) \theta_4(x-y) = \theta_3(x) \theta_4(x) \theta_3(y) \theta_4(y) - \theta_1(x) \theta_2(x) \theta_1(y) \theta_2(y)$$
and from the last equation of Exercise~8 Chapter 20 we get
$$\theta_4^2 \theta_4(x+y)\theta_4(x-y) = \theta_4^2(x) \theta_4^2(y) - \theta_1^2(x) \theta_1^2(y).$$
The two equations about at once yield
$$\begin{array}{ll}
\dfrac{\theta_3}{\theta_4} \dfrac{\theta_3}{\theta_4} \mathrm{dn}(u+v) &= \dfrac{\frac{\theta_3}{\theta_4} \mathrm{dn}(u) \frac{\theta_3}{\theta_4} \mathrm{dn}(v) - \frac{\theta_2}{\theta_3} \mathrm{\mathrm{sn}}(u)) \frac{\theta_2}{\theta_4} \mathrm{cn}(u) \frac{\theta_2}{\theta_3} \mathrm{\mathrm{sn}}(v) \frac{\theta_2}{\theta_4} \mathrm{\mathrm{sn}}(v)}{1 - \frac{\theta_2^2}{\theta_3^2} \mathrm{\mathrm{sn}}^2(u) \frac{\theta_2^2}{\theta_3^2} \mathrm{\mathrm{sn}}^2(v)}.
\end{array}$$
Hence we arrive at the desired result,
$$\mathrm{dn}(u+v) = \dfrac{\mathrm{dn}(u) \mathrm{dn}(v) - k^2 \mathrm{\mathrm{sn}}(u) \mathrm{cn}(u) \mathrm{\mathrm{sn}}(v) \mathrm{cn}(v)}{1 - k^2 \mathrm{\mathrm{sn}}^2(u) \mathrm{\mathrm{sn}}^2(v)}.$$
\end{solution}
%%%%
%%
%%
%%%%
\begin{problem}\label{problem3chapter21}
Show that $\displaystyle\int \mathrm{cn}^3(x) \mathrm{dn}(x) dx = \mathrm{\mathrm{sn}}(x) - \dfrac{1}{3} \mathrm{\mathrm{sn}}^3(x) + c.$
\end{problem}
\begin{solution}
We wish to evaluate
$$\displaystyle\int \mathrm{cn}^3(x) \mathrm{dn}(x) dx.$$
We know that
$$\dfrac{\mathrm{d}}{\mathrm{dx}} \mathrm{dn}(x) = \mathrm{cn}(x) \mathrm{dn}(x)$$
and that
$$\mathrm{cn}^2(x) =1-\mathrm{\mathrm{sn}}^2(x).$$
Hence
$$\begin{array}{ll}
\displaystyle\int \mathrm{cn}^3(x) \mathrm{dn}(x) dx &= \displaystyle\int [1-\mathrm{\mathrm{sn}}^2(x)] \mathrm{cn}(x) \mathrm{dn}(x) dx \\
&= \mathrm{\mathrm{sn}}(x) - \dfrac{1}{3} \mathrm{\mathrm{sn}}^2(x) +c.
\end{array}$$
\end{solution}
%%%%
%%
%%
%%%%
\begin{problem}\label{problem4chapter21}
Show that if $g(x)$ and $h(x)$ are any two different ones of the three functions $\mathrm{\mathrm{sn}}(x), \mathrm{cn}(x), \mathrm{dn}(x),$ and if $m$ is a non-negative integer, you can perform the integration
$$\displaystyle\int g^{2m+1}(x) h(x) dx.$$
\end{problem}
\begin{solution}
Consider
$$A = \displaystyle\int g^{2m+1}(x) h(x) dx.$$
We know that
$$\dfrac{\mathrm{d}}{\mathrm{dx}} \mathrm{\mathrm{sn}}(x) = \mathrm{cn}(x)\mathrm{dn}(x);$$
$$\dfrac{\mathrm{d}}{\mathrm{dx}} \mathrm{cn}(x) = -\mathrm{\mathrm{sn}}(x) \mathrm{dn}(x);$$
$$\dfrac{\mathrm{d}}{\mathrm{dx}} \mathrm{dn}(x) = -k^2 \mathrm{\mathrm{sn}}(x) \mathrm{cn}(x).$$
Now let $g(x)$ and $h(x)$ be any two different ones of $\mathrm{\mathrm{sn}}(x), \mathrm{cn}(x), \mathrm{dn}(x)$. Let $\psi(x)$ be the other of the three functions. Then
$$g(x) h(x) dx = c_1 d[\psi(x)],$$
in which $c_1=1, -1$ or $-K^{-2}$ according to whether $\psi(x)$ is $\mathrm{\mathrm{sn}}(x), \mathrm{cn}(x),$ or $\mathrm{dn}(x).$
Now we also know that
$$\mathrm{\mathrm{sn}}^2(x) + \mathrm{cn}^2(x) = 1,$$
$$k^2 \mathrm{\mathrm{sn}}^2(x) + \mathrm{dn}^2(x)=1,$$
$$\mathrm{dn}^2(x) - k^2 \mathrm{cn}^2(x) = x^2.$$
Therefore the square of any one of $\mathrm{\mathrm{sn}}(x), \mathrm{cn}(x), \mathrm{dn}(x)$ is a linear function of the square of any other one of them. Hence we may write
$$g^2(x) = a_1+b_1 \psi^2(x).$$
We know have
$$\begin{array}{ll}
\displaystyle\int g^{2m+1}(x) h(x) dx &= \displaystyle\int [g^2(x)]^m g(x) h(x) dx \\
&= c_1 \displaystyle\int [a_1 + b_1 \psi^2(x)]^m d \psi(x),
\end{array}$$
which is integrable as a sum of powers.
\end{solution}
%%%%
%%
%%
%%%%
\begin{problem}\label{problem5chapter21}
Obtain the result
$$\displaystyle\int \mathrm{\mathrm{sn}}(x) dx = \dfrac{1}{k} \log[\mathrm{dn}(x) - k \mathrm{cn}(x)] + c.$$
\end{problem}
\begin{solution}
Consider $\displaystyle\int \mathrm{\mathrm{sn}}(x) dx$. We may write
$$\begin{array}{ll}
\displaystyle\int \mathrm{\mathrm{sn}}(x) dx &= \displaystyle\int \dfrac{\mathrm{\mathrm{sn}}(x)[\mathrm{dn}(x)-k\mathrm{cn}(x)]}{\mathrm{dn}(x)-k\mathrm{cn}(x)} dx \\
&= \displaystyle\int \dfrac{\mathrm{\mathrm{sn}}(x) \mathrm{dn}(x) - k\mathrm{\mathrm{sn}}(x) \mathrm{cn}(x)}{\mathrm{dn}(x) - k\mathrm{cn}(x)} dx \\
&= \displaystyle\int \dfrac{-d[\mathrm{dn}(x)] + \frac{1}{k} d[\mathrm{dn}(x)]}{\mathrm{dn}(x) - k\mathrm{cn}(x)} \\
&= \dfrac{1}{k} \displaystyle\int \dfrac{d[\mathrm{dn}(x)]-k[\mathrm{cn}(x)]}{\mathrm{dn}(x) -k \mathrm{cn}(x)} \\
&= \dfrac{1}{k} \log[\mathrm{dn}(x) -k\mathrm{cn}(x)]+c.
\end{array}$$
\end{solution}
%%%%
%%
%%
%%%%
%%%%
%%
%%
%%%%
\begin{problem}\label{problem6chapter21}
Show that
$$\mathrm{dn}(2x) - k \mathrm{cn}(2x) = \dfrac{(1-k)[1+k \mathrm{\mathrm{sn}}^2(x)]}{1-k \mathrm{\mathrm{sn}}^2(x)}.$$
\end{problem}
\begin{solution}
We wish to show that
$$\mathrm{dn}(2x) - k \mathrm{cn}(2x) = \dfrac{(1-k) [1+k \mathrm{\mathrm{sn}}^2(x)]}{1-k \mathrm{\mathrm{sn}}^2(x)}.$$
Now from the addition formulas of Exercises~\ref{problem1chapter21} and \ref{problem2chapter21} we obtain
$$\mathrm{dn}(2x) = \dfrac{\mathrm{dn}^2(x) - k^2 \mathrm{\mathrm{sn}}^2(x) \mathrm{dn}^2(x)}{1 - k^2 \mathrm{\mathrm{sn}}^4(x)}$$
and
$$\mathrm{cn}(2x) = \dfrac{\mathrm{cn}^2(x) - \mathrm{\mathrm{sn}}^2(x) \mathrm{dn}^2(x)}{1 - k^2 \mathrm{\mathrm{sn}}^4(x)}.$$
Let us put each of the above in terms of $\mathrm{\mathrm{sn}}(x)$. We get
$$\mathrm{dn}(2x) = \dfrac{1-k^2 \mathrm{\mathrm{sn}}^2(x) - k^2 \mathrm{\mathrm{sn}}^2(x) [1-\mathrm{\mathrm{sn}}^2(x)]}{1 - k^2 \mathrm{\mathrm{sn}}^4(x)} = \dfrac{1 - 2k^2 \mathrm{\mathrm{sn}}^2(x) + k^2 \mathrm{\mathrm{sn}}^4(x)}{1 - k^2 \mathrm{\mathrm{sn}}^4(x)}.$$
$$\mathrm{cn}(2x) = \dfrac{1 - \mathrm{\mathrm{sn}}^2(x) - \mathrm{\mathrm{sn}}^2(x) [1-x^2 \mathrm{\mathrm{sn}}^2(x)]}{1 -k^2\mathrm{\mathrm{sn}}^4(x)} = \dfrac{1-2k^2\mathrm{\mathrm{sn}}^2(x) + k^2\mathrm{\mathrm{sn}}^4(x)}{1-k^2 \mathrm{\mathrm{sn}}^4(x)}$$
Then
$$\begin{array}{ll}
\mathrm{dn}(2x) - k \mathrm{cn}(2x) &= \dfrac{1 - k^2 \mathrm{\mathrm{sn}}^2(x) - k^2 \mathrm{\mathrm{sn}}^2(x) + k^2\mathrm{\mathrm{sn}}^4(x) - k + k \mathrm{\mathrm{sn}}^2(x) + k \mathrm{\mathrm{sn}}^2(k) - k^3 \mathrm{\mathrm{sn}}^4(x)}{1 - k^2 \mathrm{\mathrm{sn}}^4(x)} \\
&= \dfrac{1 + k\mathrm{\mathrm{sn}}^2(x) + k \mathrm{\mathrm{sn}}^2(x)[1+k \mathrm{\mathrm{sn}}^2(x)] - k[1+k \mathrm{\mathrm{sn}}^2(x)] - k^2 \mathrm{\mathrm{sn}}^2(x) [1+k\mathrm{\mathrm{sn}}^2(x)]}{[1-k\mathrm{\mathrm{sn}}^2(x)][1+k\mathrm{\mathrm{sn}}^2(x)]} \\
&= \dfrac{1 + k \mathrm{\mathrm{sn}}^2(x) - k - k^2 \mathrm{\mathrm{sn}}^2(x)}{1 - k \mathrm{\mathrm{sn}}^2(x)} \\
&= \dfrac{1 - k + k (1-k)\mathrm{\mathrm{sn}}^2(x)}{1-k\mathrm{\mathrm{sn}}^2(x)}.
\end{array}$$
Hence
$$\mathrm{dn}(2x)-k\mathrm{cn}(2x) = \dfrac{(1-k)[1+k\mathrm{\mathrm{sn}}^2(x)]}{1-k\mathrm{\mathrm{sn}}^2(x)}.$$
\end{solution}
\end{document}