UNIX shell while [ "$yn" = "yes" -a "$yn" = "YES" ] until [

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UNIX shell while [ "$yn" = "yes" -a "$yn" = "YES" ] until [ "$yn" == "yes" -a "$yn" == "YES" ]
上面两句里面的 -a 和-o
和&& | 一样吗

-a 表示与 and
-o 表示或 or
-a 和 && 一样,-o 和||一样,但&&和||需要在方括号外使用：
while [ "$yn" != "yes" ] && [ "$yn" != "YES" ]
until [ "$yn" == "yes" ] || [ "$yn" == "YES" ]

UNIX shell while [ "$yn" = "yes" -a "$yn" = "YES" ] until [ 设Xn≤a≤Yn,lim(n→∞)《Yn-Xn》=0,则Xn与Yn X1=a>0,Y1=b>0,Xn+1=(Xn+Yn)/2,Yn+1=(Xn*Yn)^1/2,求证数列Xn,Yn收敛并求其 limxn=a lim(yn-xn)=0 则数列{yn} n趋于无穷 yn=1/2(yn-1+2/yn-1)的极限求差分方程通解Yn+2-Yn+1-12Yn=0 证明：若lim(n→∞)yn（数列yn）=A且A>0,则存在正整数N,当n>N时恒有yn>0. 设Xn≤a≤Yn,lim(n→∞)（Yn-Xn）=0,则Xn与Yn的收敛? 数列极限证明lim(n=无穷大）Yn=1,Yn=(n^2+a^2)1/2*n 第一题:求差分方程yn+1-yn=ln2n的通解;第二题:求差分方程yn+1-yn=arcsin(n^2)(这是我们明天大一高数问题:已知数列Yn有极限,且满足Yn+1(小1小n)=根号下2+Yn,则Yn的极限为? yn=cosx/(e^x+e^-x)求n趋于无穷大时,yn极限