S（n）=（1/1 + 1/2 + 1/3 + 1/4 + ... + 1 / n） [英] S(n) = (1/1 + 1/2 + 1/3 + 1/4 + … + 1/n)

问题描述

第2部分的目标是开发一种算法并实现代码来计算倒数的串联和S（n）=（1/1 + 1/2 + 1/3 + 1/4 + ... + 1 / n ）使你的程序找到最小数字n，其系列倒数之和S（n）等于或超过用户输入的值。



例如，如果用户输入1.5，您的程序应显示以下内容：



最小的系列总和> = 1.5是S（2），其值为1.5。



我尝试过的事情：



我已经尝试过我所知道的所有事情并且可以'似乎让输出正确。

The goal of Part 2 is to develop an algorithm and implement code to calculate the series sum of reciprocals S(n) = (1/1 + 1/2 + 1/3 + 1/4 + … + 1/n) such that your program finds the smallest number n for which the series sum of reciprocals S(n) EQUALS OR EXCEEDS a value entered by the user.

For example, if the user enters 1.5 your program should display the following:

The smallest series sum >= 1.5 is S(2) which has the value 1.5.

What I have tried:

I have tried everything I know and can't seem to get the output to be correct.

推荐答案

对于大N，谐波系列有很多舍入误差。你的程序是否适用于N = 10？

对于N = 1,000,000 S = 14.393，相邻项之间的差异可能小于舍入。

使用补偿加法 Kahan求和算法 - 维基百科 [ ^ ]

并将每个术语存储在数组S []中。然后二进制搜索数组中的目标值。

The harmonic series has a lot of roundoff error for large N. Does your program work for N = 10?
For N = 1,000,000 S = 14.393, and the difference between adjacent terms may be less than roundoff.
Use compensated addition Kahan summation algorithm - Wikipedia[^]
and store each term in an array S[]. Then binary search the array for the target value.

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