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{\large ���wI���K�𓚗�@��12��@2007�N12��11���o�蕪} 

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�S�� ���n����CTA �O�p �~
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\noindent{\bf [1]}
(a)	
$x=r\cos\theta +1$, $y=r\sin\theta +1$ 
�ƕϐ��ϊ������
\begin{eqnarray*}
&{}&\int\int_{D} (x^2+y^2) dxdy
=\int\int_{\substack {0\le r\le 1 
\\ 0\le\theta <2\pi }} 
\{r^2 +2(\sin\theta +\cos\theta ) r+2
\} rdrd\theta \\
&=& \int_{0}^{1} dr \int_{0}^{2\pi} 
r\{r^2 +2+2r(\sin\theta +\cos\theta )
\} d\theta
= \int_{0}^{1} 2\pi r(r^2+2)dr=\frac{5}{2} \pi.
\end{eqnarray*}

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(b)	
$x=r\cos\theta $, $y=r\sin\theta $ 
�ƕϐ��ϊ������
\begin{eqnarray*}
&{}&\int\int_{\substack {x^2 +y^2 \le K \\
x, y>0 }} \frac{xy}{(1+x^2+y^2 )^3 } dxdy
= \int\int_{\substack {0<r\le \sqrt{K} \\
0<\theta <\pi /2}} 
\frac{r^2 \sin\theta\cos\theta }{(1+r^2 )^3} 
rdrd\theta \\
&=& \int_{0}^{\sqrt{K}} 
\frac{r^3 }{(1+r^2 )^3} dr \int_{0}^{\pi /2} 
\frac{1}{2} \sin 2\theta d\theta 
= \frac{1}{2} \int_{0}^{\sqrt{K}} 
\frac{r^3 }{(1+r^2 )^3} dr\\ 
&=& \frac{1}{4} \int_{1}^{1+K} 
\left( \frac{1}{u^2} -\frac{1}{u^3} 
\right) du 
= \frac{1}{8} \left( 1-\frac{2}{1+K} 
+\frac{1}{(1+K)^2} \right) \to \frac{1}{8} 
\quad (K\to\infty).  
\end{eqnarray*}
 
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\noindent{\bf [2]}  
(a)	
$y=x$, $y=1$, $x=0$��
�ʑ�$(x,y)\mapsto (u,v)$�ɂ����
���ꂼ��
������$u=0$, $v\ge 0$�A
������$u={v^2}/4 -1$�A
������$u\le 0, v=0$�Ɉڂ���A
�����ň͂܂��̈�$H$��$G$�̑��ƂȂ�B

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(b)	
$\left| 
\begin{array}{cc}
u_x & u_y \\
v_x & v_y \\
\end{array} 
\right| 
=4(x^2 +y^2 )$ ������
$dudv=4(x^2 +y^2 ) dxdy$�B������
\begin{eqnarray*}
&{}&\int\int_{G} \frac{x^4 -y^4 +2xy(x^2 +y^2 )}
{(1+x^2 -y^2 )^{1/2}} dxdy
= \frac{1}{4} \int\int_{H} 
\frac{u+v}{(1+u)^{1/2}} dudv \\
&=& \frac{1}{4} \int_{-1}^{0} du 
\int_{0}^{2\sqrt{1+u}} \frac{u+v}{(1+u)^{1/2}} dv
= \frac{1}{2} \int_{-1}^{0} (u+\sqrt{1+u})du 
=\frac{1}{12}. 
\end{eqnarray*}

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\noindent{\bf [3]}
$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} 
=\al >1$�Ɖ��肷��B���̂Ƃ�
$\al -\e >1$���݂����悤��
$\e >0$���Ƃ��B
����$\e$�ɑ΂���
�\���傫��$N$�����݂��āA
$n\ge N$�Ȃ��
$\left| \frac{a_{n+1}}{a_n} -\al 
\right| <\e $���݂����B
���̂Ƃ�����$\frac{a_{n+1}}{a_n} >\al -\e 
>1$�B
�]���āu$n\ge N \Rightarrow a_{n+1} > a_n $�v
($\ast$)�����藧�̂ŁA
\[
\sum_{n=1}^{\infty} a_n > \sum_{n=1}^{N-1} 
a_n +\sum_{n=N}^{\infty} a_N =\infty
\]
�ƂȂ�^����ꂽ���������͔��U����B

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(��) �Ɍ�����Ɍ��ɕς����ꍇ�͑�ӕs�����ł����B
�Ⴆ��$n$����̂Ƃ�$a_n ={1}/{3^{(n+1)/2}}$�A
$n$�������̂Ƃ�$a_n ={1}/{2^{n/2}}$�Ƃ����
����ɂȂ��Ă��܂��B����A�Ɍ������Ɍ��ɕς���
�ꍇ�ɂ��čl����ƁA($\ast$)�ɑΉ����镔�������l��
���藧���A��ӂ��������܂��B

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\noindent{\bf [4]}
$\displaystyle\sum_{n=1}^{\infty} \frac{1}{1+2^n } 
<\sum_{n=1}^{\infty} \frac{1}{2^n } =1$
���P�Ԗڂ̋����͎�������B

���ɁA$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{\al}}$
��$0<\al\le 1$�̂Ƃ����U�A$1<\al $�̂Ƃ�����
���鎖�ɒ��ӂ���΁A\\
$\displaystyle\sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} 
>\sum_{n=1}^{\infty} \frac{1}{2\sqrt{n}} 
=\infty $ ������Q�Ԗڂ̋����͔��U���A
$\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n}}{1+n^2 } 
<\sum_{n=1}^{\infty} \frac{1}{n^{3/2}} <\infty $
������R�Ԗڂ̋����͎�������B

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\noindent�E 
[3]��$\lim_{n\to\infty} a_{n+1} / a_n =1$
�̏ꍇ���l����ƁA$\sum_{n=1}^{\infty} a_n $�͔��U����ꍇ�Ǝ�������
�ꍇ������܂��B[4]�̂Q�ԖڂƂR�Ԗڂ̋����͂��̋�̗�ł��B

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\begin{flushright}(�𓚗�쐬�F�O�p)
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