(1)
%Wallis
int a;
int k;
b=1.0;
fprintf('wallis\n');
k=input('输入k');
for a=1:k;
b=(b*(2*a)/(2*a-1))*(2*a)/(2*a+1);
end;
vpa(2*b,100000)
(2)
1.
%taylor
s=-1;
int k;
n=1;
t=1;
ppi=0;
k=input('输入k的值');
for ii=0:k
n=2*ii+1;
s=-s;
t=s/n;
ppi=ppi+t;
end
vpa(4*ppi,100000)
2.%taylor改进
s1=0;s2=0;