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luc037.c

Problem Statement

The natural logarithm can be approximated by the following series. (x-1)/x + 1/2 ((x-1)/x)^2 + 1/2 ((x-1)/x)^3 + 1/2 ((x-1)/x)^4 + ... If x is input through the keyboard, write a program to calculate the sum of the first seven terms of this series.

Metadata

Property Detail
Author Amit Dutta (amitdutta4255@gmail.com)
License MIT
Difficulty Beginner (index: 2 / 10)

Concepts

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  • Pointers
  • Recursion
  • Sorting (possible)
  • Iteration

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Source Code

#include <stdio.h>
#include <math.h>

double series(double x) // made this fn only for fun, making a fn was not necessary
{
    double result = (x - 1) / x;
    int i;
    for (i = 2; i <= 7; i++)
    {
        result += 0.5 * pow(((x - 1) / x), i);
    }
    return result;
}

int main()
{
    double x;
    printf("Enter the value for x : ");
    scanf("%lf", &x);
    printf("\nResult : %g", series(x));
    return 0;
}

Explanation

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    You are explaining a C programming code to a beginner.

    STRICT RULES:

    - Only use the given code. Do NOT assume anything not present.

    - Do NOT add extra examples.

    - Keep explanation clear and short.

    - If something is unclear, say "Not clear from code".

    - Follow the exact format below. Do NOT change headings.

    FORMAT:

    [START]

    ## What it does

    (Explain the overall purpose in 1-2 sentences)

    ## Step-by-step

    (Explain how the code works in steps, simple language)

    ## Key Concepts

    (List concepts like loop, condition, function, etc.)

    ## Notes

    (Mention any limitations, errors, or assumptions)

    [END]

    CODE (luc037.c):

    #include <stdio.h>
    #include <math.h>

    double series(double x) // made this fn only for fun, making a fn was not necessary
    {
        double result = (x - 1) / x;
        int i;
        for (i = 2; i <= 7; i++)
        {
            result += 0.5 * pow(((x - 1) / x), i);
        }
        return result;
    }

    int main()
    {
        double x;
        printf("Enter the value for x : ");
        scanf("%lf", &x);
        printf("\nResult : %g", series(x));
        return 0;
    }