调整序列求和的计算，以便在求和的每一步除以n

Question

我以前问过this问题，该问题引用了对一系列求和以通过公式计算pi的问题

pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 ...

在计算中，

A = Sum of a_i from i=1 to N

和

a_i = (-1)^(i+1)/(2i-1)

我正在编写此计算，以分析每N个求和计算pi所需时间的性能。我想了解随着N的增加，T_N和N之间的关系。

T_N = the total elapsed time it takes to do N summations.

N = the number of summations.

我的Matlab程序将T_N绘制为N的函数。在该程序中，我定义了变量以查找以下线性方程的a和b的值（其中a是y轴截距，b是斜率）

T_N=a + b*N

为此，我在下面编写了以下Matlab程序（初始测试为N = 1000）

clear all;
n=1000;
f=[];
telapsed = zeros(1,n);
tic
for jj=1:n
    ii=1:jj;
    f=[f 4*sum( ((-1).^(ii+1))./(2.*ii-1)  )];
    telapsed(jj) = toc;
end

hold on
plot(1:n,telapsed)
title('Time it takes to sum \pi using N summations')
xlabel('Number of summations (N)') 
ylabel('Total Time (T_N)')
p = polyfit(1:n,telapsed,1);
slope=p(1)      % slope of t_N = a + b*N or b
intercept=p(2)  % y-intercept or a

运行该程序将产生以下图形

因此，此代码返回N = 1000的a和b的值。该代码返回

a = slope = 9.3447e-06
b = intercept = 0.0011

所以，在N = 1000时，我发现了

T_n = a + b*N = 9.3447e-06 + 0.0011*N

我现在正在尝试从绘图中调整代码

T_N as a function of N

绘图

(T_N − a)/N as a function of N

为此，我进行了以下调整

clear all;
n=1000;
f=[];
g=[];
telapsed = zeros(1,n);
telapsed2 = zeros(1,n);
tic

% The original for loop computes T_N as a function of N
for jj=1:n
    ii=1:jj;
    f=[f 4*sum( ((-1).^(ii+1))./(2.*ii-1)  )];
    telapsed(jj) = toc;
end

% I think that I need to make a new for loop to compute (T_N − a)/N as a function of N
for jj=1:n
    ii=1:jj;
    g=[g 4*sum( ((-1).^(ii+1))./(2.*ii-1)  )];
    telapsed2(jj) = toc/jj; % This must be wrong, I think that I need to adjust to compute (T_N − a)/N
end

hold on
plot(1:n,telapsed) % the plot for T_N as a function of N
plot(1:n,telapsed2) % the plot for (T_N − a)/N as a function of N
title('Time it takes to sum \pi using N summations')
xlabel('Number of summations (N)') 
ylabel('Total Time (T_N)')
legend('T_N as a function of N','(T_N − a)/N as a function of N')

p = polyfit(1:n,telapsed,1);
slope=p(1)      % slope of t_N = a + b*N or b
intercept=p(2)  % y-intercept or a

p2 = polyfit(1:n,telapsed2,1);
slope2=p2(1)      % Adjusted slope
intercept2=p2(2)  % Adjusted y-intercept

产生以下图形

这不是将（T_N − a）/ N绘制为N的函数的正确方法。我在定义

时必须有一个错误

for jj=1:n
    ii=1:jj;
    g=[g 4*sum( ((-1).^(ii+1))./(2.*ii-1)  )];
    telapsed2(jj) = toc/jj; % This must be wrong, I think that I need to adjust to compute (T_N − a)/N
end

为了计算

(T_N − a)/N as a function of N

我可以对上面的Matlab代码进行调整，以便在该行计算（T_N-a）/ N

telapsed2(jj) = toc/jj;

我不确定如何更正我的代码，以便此行计算

slope = slope of computation per N summations
telapsed(jj)= (toc-slope)/n

我也许可以在f的第一个for循环中添加它，但是我不确定如何添加。

问题：如何调整Matlab代码，以便可以成功地从

更改计算

T_N as a function of N

到

(T_N − a)/N as a function of N

Answer 1

我认为我找到了部分解决方案。我最初编写了以下代码，将T_N绘制为N的函数。

clear all;
n=50;
f=[];
telapsed = zeros(1,n);
tic
for jj=1:n
    ii=1:jj;
    f=[f 4*sum( ((-1).^(ii+1))./(2.*ii-1)  )];
    telapsed(jj) = toc;
end

hold on
plot(1:n,telapsed)
title('Time it takes to sum \pi using N summations')
xlabel('Number of summations (N)') 
ylabel('Total Time (T_N)')
p = polyfit(1:n,telapsed,1);
slope=p(1)      % slope of t_N = a + b*N or b
intercept=p(2)  % y-intercept or a

运行该程序将产生以下图形

其中

slope = 1.3802e-05
intercept = 0.0021

所以

T_N = a + b * N = 0.0021 + 1.3802e-05 * N

为了将（T_N − a）/ N绘制为N的函数，我进行了以下更改

clear all;
n=50;
f=[];
telapsed = zeros(1,n);
tsum = zeros(1,n);
tic

for jj=1:n
    ii=1:jj;
    f=[f 4*sum( ((-1).^(ii+1))./(2.*ii-1)  )];
    telapsed(jj) = toc;
    psum = polyfit(jj,toc,n);
    pintercept = psum(2);
    tsum(jj) = (telapsed(jj)-pintercept)/jj;
end

hold on
plot(1:n,telapsed) % the plot for T_N as a function of N
plot(1:n,tsum) % the plot for (T_N - a)/N as a function of N
title('Time it takes to sum \pi using N summations')
xlabel('Number of summations (N)') 
ylabel('Total Time (T_N)')
legend('T_N as a function of N','(T_N - a)/N as a function of N')

p = polyfit(1:n,telapsed,1);
slope=p(1)      % slope of t_N = a + b*N or b
intercept=p(2)  % y-intercept or a

产生以下情节

我期望（T_N − a）/ N的值接近一个常数。从上图可以清楚地看到，该值接近y截距。数学会告诉我

T_N = a + b*N => (T_N-a)/N = b

因此，似乎已对

进行了调整

for jj=1:n
    ii=1:jj;
    f=[f 4*sum( ((-1).^(ii+1))./(2.*ii-1)  )];
    telapsed(jj) = toc;
    psum = polyfit(jj,toc,n);
    pintercept = psum(2);
    tsum(jj) = (telapsed(jj)-pintercept)/jj;
end

是正确的。