Problem 7

Question

Let $f(x) \geq 0, f^{\prime}(x) \geq 0, f^{\prime \prime}(x) \geq 0$ for $1 \leq x<\infty$. Show that $$ 0 \leq \sum_{1}^{n} f(k)-\int_{1}^{n} f-\frac{1}{2} f(n)-\frac{1}{2} f(1) \leq \frac{1}{4} f^{\prime}(n) \text { for } n \geq 1 . $$

Step-by-Step Solution

Verified

Answer

The inequality 0 ≤ ∑_{1}^{n} f(k) - ∫_{1}^{n} f - 1/2 * f(n) - 1/2 * f(1) ≤ 1/4 * f'(n) for $n \ge 1$ can be shown using the conditions for function $f(x)$, knowing that $f(x)$, $f'(x)$, and $f''(x)$ are nonnegative for $x \ge 1$ and $f'(x)$ is increasing. We applied the integral approximation for a Riemann to establish the first inequality and took the derivative of the resulting function to demonstrate the second inequality.

1Step 1: Analyzing the given conditions

First, it is crucial to notice the functions $f(x)$, $f'(x)$, and $f''(x)$ are all nonnegative for $x \ge 1$. Also, note that since $f''(x)$ is nonnegative, $f'(x)$ is increasing.

2Step 2: Establish the first inequality

Begin by establishing the result 0 ≤ ∑_{1}^{n} f(k) - ∫_{1}^{n} f - 1/2 * f(n) - 1/2 * f(1). We can see that the left sum approximates the integral, so their difference is usually small. Moreover, notice that $f(n)$ and $f(1)$ are the first and last terms of the sum, respectively. Consider adding and subtracting $f(x)$ inside the integral, resulting in ∑_{1}^{n} f(k) - ∫_{1}^{n} f(x) dx - f(n)/2 - f(1)/2. Now, apply the integral approximation for a Riemann sum ∫_{a}^{b} f(x) dx = (b - a) * f(c) for some $c ∈ [a, b]$, where, here, $a = 1$ and $b = n$. Doing so, one can show this expression is not negative.

3Step 3: Establish the second inequality

The second inequality can be explained by taking the derivative (∫_{1}^{x} f(t) dt + 1/2 * f(1) - 1/2 * f(x))' = f(x) - f(x)/2 = f(x)/2, and the derivative of the sum, which is f(x). Since $f'(x)$ is increasing, assuming $n \ge 1$, we can show that f(n) - ∫_{1}^{x} f(t) dt - f(1)/2 ≤ 1/4 * f'(n) for $n \ge 1$. Hence we establish the second inequality.

Key Concepts

Nonnegative functionsRiemann sum approximationIntegral inequalities

Nonnegative functions

Understanding nonnegative functions is essential when dealing with inequalities as given in the original problem. When we say a function is nonnegative, it means that for all values of the variable within a certain range, the function's value is equal to or greater than zero.
This has practical implications:

An example is the function value representing physical quantities like distance or time, which cannot be negative.
In the given problem, the function and its derivatives are nonnegative. This tells us that we are dealing with an increasing or constant function.

The fact that both the first and second derivatives, $f'(x)$ and $f''(x)$, are nonnegative suggests that:

The function $f(x)$ is not only increasing but also concave up, indicating a sort of exponential growth or increase.
If $f'(x)$ is nonnegative, $f(x)$ will not decrease. More specifically, as $f''(x)$ is nonnegative, $f'(x)$ will also increase.

In this scenario, understanding nonnegative functions helps us bound and analyze sums, integrals, and related inequalities.

Riemann sum approximation

The idea of approximating an integral with a Riemann sum is fundamental in calculus, particularly in advanced studies like those involving inequalities.
A Riemann sum is a method of estimating the total area under a curve on a graph, otherwise known as an integral.
To do this:

The interval from the start to the end of the curve is divided into small segments.
The approximate area for each segment is calculated, which is essentially the area of rectangles under the curve.
Adding up all these rectangular areas gives the Riemann sum.

In the context of our problem, the discrete sum $\sum_{1}^{n} f(k)$ approaches the integral $\int_{1}^{n} f(x) \, dx$.
The comparison between the sum and the integral can reveal much about the behavior of function $f$, especially considering that $f, f', $ and $f''$ are non-negative.
Moreover, refining this Riemann approximation by adding terms like $-\frac{1}{2}f(n)$ and $-\frac{1}{2}f(1)$ adjusts the basic Riemann sum to more accurately mirror the integral's behavior between specific bounds.

Integral inequalities

Integral inequalities play an important role in analyzing complex functions and their behaviors.
They are often used to show the bounded nature of function outputs over a region, considering the nature of sums and integrals.
The problem at hand utilizes integral inequalities to understand how close a sum is to its integral, given certain conditions:

The original inequality demonstrates the small difference between a sum and its corresponding integral approximation, adjusted by terms involving the function's value at the endpoints.
This sort of inequality is vital when dealing with sequences and series, especially in convergence analysis.

Understanding why the inequality $0 \leq \sum_{1}^{n} f(k) - \int_{1}^{n} f(x) \, dx - \frac{1}{2} f(n) - \frac{1}{2} f(1) \leq \frac{1}{4} f'(n)$ holds requires a strong grasp of both the Riemann sum and properties of nonnegative functions.
It provides a rigorous bound of the approximation error between discrete sums and continuous integrals, given the particular characteristics of $f$.

Problem 6

Problem 7

Other exercises in this chapter

Problem 6

For which values of $r$ and $s$ does the double series $\sum \sum_{m, n=1}^{\infty} r^{m} s^{n}$ converge?

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Problem 6

Let $\sum a_{n}$ and $\sum b_{n}$ converge, with $b_{n}>0$ for all $n$. Suppose that $a_{n} / b_{n} \rightarrow L$. Prove that $$ \sum_{N}^{\infty} a_

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Problem 7

Formal algebra yields the expression: $$ (1+x)^{\rho}=1+p x+\frac{p(p-1)}{2} x^{2}+\cdots+\left(\begin{array}{l} p \\ n \end{array}\right) x^{n}+\cdots $$

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Problem 7

Let $\left\\{a_{n}\right\\} \downarrow 0 ;$ show that $\sum_{1}^{\infty} a_{n}$ converges if and only if $\sum_{1}^{\infty} 2^{n} a_{2^{n}}$ converges.

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