/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Find an approximate value of \(1... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 39

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

Short Answer

Expert verified
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 by step solution

01

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#
02

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#
03

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#
04

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#
05

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 \(1^{1/3} + 2^{1/3} + \cdots + 1000^{1/3}\).

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Most popular questions from this chapter

(Leibniz's Rule for Integrals) Let \(f\) be a continuous function on \([a, b]\) and \(u, v\) be differentiable functions on \([c, d] .\) If the ranges of \(u\) and \(v\) are contained in \([a, b]\), prove that $$ \frac{d}{d x} \int_{u(x)}^{v(x)} f(t) d t=\left[f(v(x)) \frac{d v}{d x}-f(u(x)) \frac{d u}{d x}\right] $$

Evaluate the following limits. (i) \(\lim _{h \rightarrow 0} \frac{1}{h} \int_{x}^{x+h} \frac{d u}{u+\sqrt{u^{2}+1}}\), (ii) \(\lim _{x \rightarrow 0} \frac{1}{x^{3}} \int_{0}^{x} \frac{t^{2} d t}{t^{4}+1}\), (iii) \(\lim _{x \rightarrow 0} \frac{1}{x^{6}} \int_{0}^{x^{2}} \frac{t^{2} d t}{t^{6}+1}\), (iv) \(\lim _{x \rightarrow x_{0}} \frac{x}{x-x_{0}} \int_{x_{0}}^{x} f(t) d t\) (v) \(\lim _{x \rightarrow x_{0}} \frac{x}{x^{2}-x_{0}^{2}} \int_{x_{0}}^{x} f(t) d t\), provided \(f\) is continuous at \(x_{0}\).

Let \(f:[a, b] \rightarrow \mathbb{R}\) be a bounded function. Without using Lemma \(6.30\), show that \(f\) is Riemann integrable if and only if there is \(r \in \mathbb{R}\) satisfying the following condition: Given \(\epsilon>0\), there is a partition \(P_{\epsilon}\) of \([a, b]\) such that \(|S(P, f)-r|<\epsilon\), where \(P\) is any refinement of \(P_{\epsilon}\) and \(S(P, f)\) is any Riemann sum for \(f\) corresponding to \(P\).

Let \(a, b, c \in \mathbb{R}\) with \(a

Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable, and \(g:[a, b] \rightarrow \mathbb{R}\) be a bounded function such that the set \(\\{x \in[a, b]: g(x) \neq f(x)\\}\) is of content zero. Show that \(g\) is integrable and $$ \int_{a}^{b} g(x) d x=\int_{a}^{b} f(x) d x $$ (Compare Proposition 6.12.)
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