NA-PTA-NA Lab1 Hamming级数求和 [递推求级数和以减小误差]

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最新推荐文章于 2024-02-02 20:24:24 发布

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该博客介绍了如何通过递推方法解决Hamming级数求和问题，以减少计算误差并提高效率。针对3001个x值（0.0至300.00），要求每个结果的绝对误差小于10^-10。通过构造新级数，可以找到更快收敛的递推序列，从而减少所需计算的项数，降低舍入误差。博客中还提到了递推公式的设计及其误差稳定性，并提供了一个简单的实现示例。

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Produce a table of the values of the series
在这里插入图片描述

for the 3001 values of x, x = 0.0, 0.1, 0.2, …, 300.00. All entries of the table must have an absolute error less than 10^−10. This problem is based on a problem from Hamming (1962), when mainframes were very slow by today’s microcomputer standards.

Format of function:

void Series_Sum( double sum[] );

where double sum[] is an array of 3001 entries, each contains a value of ϕ(x) for x = 0.0, 0.1, 0.2, …, 300.00.

Sample program of judge:

#include <stdio.h>

void Series_Sum( double sum[] );

int main()
{
   
    int i;
    double x, sum[3001];

    Series_Sum( sum );

    x = 0.0;
    for (i=0; i<3001; i++)
        printf("%6.2f %16.12f\n", x + (double)i * 0.10, sum[i]);

    return 0;
}

/* Your function will be put here */

Sample Output:

0.00   1.644934066848
0.10   1.534607244904
...
1.00   1.000000000000
...
2.00   0.750000000000
...
300.00   0.020942212934

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
在这里插入图片描述
which implies ϕ(1)=1.0. One can then produce a series for ϕ(x)−ϕ(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 again on the second series to produce a third sequence, which converges even more rapidly than the second.

The following equation

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