Problem 1 - Advanced Calculus

Question

Prove that $$ 1+\frac{x}{2}-\frac{x^{2}}{8}<\sqrt{1+x}<1+\frac{x}{2} \quad \text { if } x>0 $$ In particular, show that \(1.375<\sqrt{2}<1.5\).

Step-by-Step Solution

Answer

To prove the inequalities, demonstrate that the expressions involving \(x\) are decreasing and increasing respectively for \(x > 0\), and then apply these results to the specific case where \(x = 1\). This verifies that \(1.375 < \sqrt{2} < 1.5\).

Step 1: Prove the upper bound

To prove the inequality \(\sqrt{1+x} < 1 + \frac{x}{2}\), consider the function \(f(x) = \sqrt{1+x} - (1 + \frac{x}{2})\). Find its derivative, \(f'(x) = \frac{1}{2\sqrt{1+x}} - \frac{1}{2}\), which is negative for all \(x > 0\), indicating that \(f(x)\) is decreasing. Since \(f(0) = 0\), \(f(x) < 0\) for all \(x > 0\), proving the upper bound.

Step 2: Prove the lower bound

To prove \(1 + \frac{x}{2} - \frac{x^2}{8} < \sqrt{1+x}\), define the function \(g(x) = \sqrt{1+x} - (1 + \frac{x}{2} - \frac{x^2}{8})\). Find the derivative, \(g'(x) = \frac{1}{2\sqrt{1+x}} - (\frac{1}{2} - \frac{x}{4})\). For \(x > 0\), \(g'(x) > 0\) since \(\frac{x}{4} > \frac{1}{2\sqrt{1+x}} - \frac{1}{2}\). Thus, \(g(x)\) is increasing for \(x > 0\). With \(g(0) = 0\), it implies \(g(x) > 0\), which proves the lower bound.

Step 3: Apply the inequalities to \(x = 1\)

Substitute \(x = 1\) into the inequalities to get \(1.375 = 1 + \frac{1}{2} - \frac{1}{8} < \sqrt{1+1} < 1 + \frac{1}{2} = 1.5\), which demonstrates that \(1.375 < \sqrt{2} < 1.5\).

Key Concepts

Calculus

Calculus is a branch of mathematics focusing on derivatives, integrals, and limits. It is instrumental in solving problems involving change and motion, and in our exercise, calculus principles are employed to investigate and prove an inequality involving a square root function. The problem effectively uses the concept of the derivative, a fundamental tool in calculus, to analyze the behavior of functions as they increase or decrease. By finding the derivatives of the functions in question and assessing their sign (positive or negative), we can determine whether the original functions are increasing or decreasing — which is central to proving the given inequalities.

Inequalities

Inequalities are mathematical expressions that demonstrate the relative size or order of two values. They are essential in a vast array of mathematical disciplines, including calculus. The symbols used in inequalities, such as <, >, \(\leq\), and \(\geq\), dictate the relationship between the expressions involved. The exercise presents a complex inequality involving a square root function, and the core objective is to prove that this function is bounded by two simpler functions, for all positive values of \(x\). Understanding how to manipulate and solve inequalities is key to approaching calculus problems like the one provided, where we must ascertain that one function is always less than or greater than another within a specific domain.

Derivative of a Function

The derivative of a function at a point is the rate at which the function value changes with respect to a change in its input value. It is a measure of the instantaneous rate of change or the slope of the function at that point. The exercise uses the derivative to analyze the changes in two functions, \(f(x) \text{ and } g(x)\), which are being compared to \(\sqrt{1+x}\). The signs of their derivatives indicate the direction of the changes in these functions over the interval (\(0, \infty\)). In our problem, proving the derivative of \(f(x)\) is negative shows that the function decreases, confirming that \(\sqrt{1+x}\) is below \(1 + \frac{x}{2}\). Similarly, proving that the derivative of \(g(x)\) is positive verifies that the function is increasing, thereby \(\sqrt{1+x}\) is above \(1 + \frac{x}{2} - \frac{x^2}{8}\).

Proof Techniques in Mathematics

Proof techniques are various methods used to verify the truth of mathematical statements. The exercise demonstrates a direct proof, where the properties of functions and their derivatives are utilized to establish the truth of the inequalities for all positive values of \(x\). The proof employs the fact that if a function's derivative is always negative, the function must be decreasing, and similarly, if the derivative is always positive, the function must be increasing. By setting up functions whose values are the difference between \(\sqrt{1+x}\) and the proposed bounds, their derivatives provide the necessary evidence for the direction of change, leading to a successful proof of the inequality.