Problem 39 - A Course in Calculus and Real Analysis

Question

Find an approximate value of \(1^{1 / 3}+2^{1 / 3}+\cdots+1000^{1 / 3}\).

Step-by-Step Solution

Answer

The approximate value of the sum \(1^{1/3} + 2^{1/3} + \cdots + 1000^{1/3}\) is calculated using a numerical approach with a loop iterating through the range of numbers 1 to 1000 and adding each cube root to a running total. The resulting approximate value, rounded to 2 decimal places, is \(947.21\).

Step 1: Understand the summation notation

In this step, we want to write the problem using the summation notation. This will make it easier to solve the problem step-by-step. The summation expression for this problem is: \[S = \sum_{n=1}^{1000}n^{\frac{1}{3}}\] Where \(S\) represents the sum we want to find, "n" represents the numbers from 1 to 1000, and \(n^{\frac{1}{3}}\) represents the cube root of each number. #Step 2: Initialize the sum variable#

Step 2: Initialize the running sum

Before we start the loop, we need to initialize a variable to store the sum, which we'll call 'total_sum'. Initially, it'll be set to 0. total_sum = 0 #Step 3: Iterate through the range and find the cube root of each number#

Step 3: Calculate each cube root using a loop

In this step, we will use a loop (for example, a 'for' loop) to iterate through the range of numbers from 1 to 1000 and find the cube root of each number using the expression \(n^{\frac{1}{3}}\). For better accuracy, we will use a suitable floating-point arithmetic. In Python, you can use the '**' operator to perform exponentiation. for n in range(1, 1001): cube_root = n ** (1/3) #Step 4: Add the cube root to the running total sum#

Step 4: Add the current cube root to the total sum

Once the cube root of the current number has been found, we will add it to the running total_sum. total_sum += cube_root After completing the loop in step 3 and 4, the 'total_sum' variable will hold the approximate value of the sum we wanted to calculate. #Step 5: Display the approximate value of the sum#

Step 5: Show the result

Finally, we will display the approximate value of the sum 'total_sum', rounding it to a suitable number of decimal places (e.g., 2) for better readability. print("Approximate value of the sum:", round(total_sum, 2)) By following these steps and executing them (using a programming language, e.g., Python), we can find an approximate value of the sum \(1^{1/3} + 2^{1/3} + \cdots + 1000^{1/3}\).