Numerical Summation of a Series

Produce a table of the values of the series

 

Equation 1

for the 2001 values of x, x= 0.000, 0.001, 0.002, ..., 2.000. All entries of the table must have an absolute error less than 0.5e-12 (12 digits of precision). This problem is based on a problem from Hamming (1962), when mainframes were very slow by today's microcomputer standards.

Input

This problem has no input.

Output

The output is to be formatted as two columns with the values of x and y(x) printed as in the C printf or the Pascal writeln.

printf("%5.3f %16.12f/n", x, psix )		writeln(x:5:3, psix:16:12)

As an example, here are 4 acceptable lines out of 2001.

0.000   1.644934066848
...
0.500   1.227411277760
...
1.000   1.000000000000
...
2.000   0.750000000000

The values of x should start at 0.000 and increase by 0.001 until the line with x=2.000 is output.

Hint

The problem with summing the sequence in equation 1 is that too many terms may be required to complete the summation in the given time. Additionally, if enough terms were to be summed, roundoff would render any typical double precision computation useless for the desired precision.

To improve the convergence of the summation process note that

Equation 2

which implies y(1)=1.0. One can then produce a series for y(x) - y(1) which converges faster than the original series. This series not only converges much faster, it also reduces roundoff loss.

This process of finding a faster converging series may be repeated to produce sequences which converge more and more rapidly than the previous ones.

The following inequality is helpful in determining how may items are required in summing the series above.

Equation 3

 

 

解析：

要解決此問題要着重考慮兩點：精度和時間。如果單一用其中任何一個公式都面臨超時或精度不夠的問題。

所以，從公式2入手，將問題轉換成f(x)-f(1)=(1-x)/(k*(k+1)*(k+x))；而且其中f(1)=1；

這樣就增大了曲線到結果值的接近速度，k從1到10000計算後的精度可以達到小數點後九位。但是僅僅這樣是不夠的，

還需要用公式三，將剩餘項(1-x)/(k*(k+1)*(k+x))  k從10000到無窮大，即約數爲(1-x)/k³，通過公式三轉換

成(1-x)/(2*x1²)，其中x1 = 10000。

#include <iostream> #include <cstdio> using namespace std; int main() { double sum = 0; for (double x = 0.000; x <= 2.000; x += 0.001) { sum = 0; for (int k = 1; k < 10000; ++k) { sum += (1.0-x)/(k*(k+1)*(k+x)); } sum = sum + (1-x)/(2*10000*10000) + 1.0; printf("%5.3f %16.12f/n", x, sum); } return 0; }