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Chapter 7: Problem 54

Is it possible that \(\ln x=\left(\sqrt[3]{e x^{2}+(\pi-2 e) x+e-\pi}+\sqrt{\pi x^{2}+(\sqrt{2}-2 \pi) x+\pi-\sqrt{2}}\right)^{1 / 17}\) for all \(x>0\) ? Justify your answer.

Short Answer

Expert verified
The given equation is not true for all \(x > 0\), because when \(0 < x < 1\), \(\ln x\) is negative, while the right-side of the equation is always positive for \(x > 0\). This contradiction indicates that the equation cannot hold true for all \(x > 0\).

Step by step solution

01

Re-write the equation slightly for easier understanding

Let's rewrite the equation slightly: \(\ln x = \left(\sqrt[3]{f(x)} + \sqrt{g(x)}\right)^{\frac{1}{17}}\) where \(f(x) = ex^2 + (\pi - 2e)x + e - \pi\) \(g(x) = \pi x^2 + (\sqrt{2} - 2\pi)x + \pi - \sqrt{2}\)
02

Examine the behavior of f(x) and g(x) for x > 0

We need to determine the behavior of \(f(x)\) and \(g(x)\) for all \(x>0\). - For \(f(x)\), since every term is positive for \(x > 0\), it is always positive. - For \(g(x)\), since every term is positive for \(x > 0\), it is also always positive. Now, we know that both functions inside the cube root and square root are always positive for \(x>0\).
03

Check the range of \(\ln x\)

Recall the properties of natural logarithm: - If \(x > 1\), then \(\ln x > 0\) - If \(x = 1\), then \(\ln x = 0\) - If \(0 < x < 1\), then \(\ln x < 0\) However, the right side of the equation will always be positive for \(x > 0\) since both \(f(x)\) and \(g(x)\) are always positive, and we are taking the cube root and square root of them.
04

Identify contradiction in the equation

When \(0 < x < 1\), the natural logarithm, \(\ln x\), is negative. Since the right-side of the equation is always positive for \(x > 0\), this contradicts the given equation. Therefore, the given equation cannot be true for all \(x > 0\). The contradiction occurs when \(0 < x < 1\), where the left-side of the equation (\(\ln x\)) is negative, while the right-side of the equation is always positive. Hence, the given equation is not true for all \(x > 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Natural Logarithm
The natural logarithm, denoted as \( \ln x \), is a crucial concept in calculus. It represents the logarithm to the base \( e \), where \( e \) is approximately 2.71828. This mathematical function transforms multiplication into addition, making it easier to handle exponential growth or decay. It exhibits specific behavior based on the value of \( x \):
  • If \( x > 1 \), \( \ln x \) is positive because \( e^y = x \) where \( y > 0 \).
  • If \( x = 1 \), \( \ln x = 0 \) because \( e^0 = 1 \).
  • If \( 0 < x < 1 \), \( \ln x \) becomes negative since the exponential \( y \) must be less than one to result in a fraction less than 1.
Understanding these properties helps in various calculus problems, especially in determining an equation's feasibility when compared to increasing or decreasing functions over certain intervals.
Polynomial Functions Basics
Polynomial functions are expressions involving sums of powers in one or more variables multiplied by coefficients. The general form of a polynomial in one variable is \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \). The degree of the polynomial is the highest power of \( x \) that appears with a non-zero coefficient.
  • The polynomial \( f(x) = ex^2 + (\pi - 2e)x + e - \pi \) is a quadratic polynomial. Its behavior for \( x > 0 \) shows all terms contribute positively, ensuring \( f(x) > 0 \).
  • The polynomial \( g(x) = \pi x^2 + (\sqrt{2} - 2\pi)x + \pi - \sqrt{2} \) behaves similarly, being always positive for \( x > 0 \).
The positivity of these polynomials for \( x > 0 \) means they never reach zero or negative values within the domain, directly contributing to ensuring that their roots are valid inputs to other functions like square root or cube root.
Exploring Function Behavior
Understanding how a function behaves helps predict its output over a specific range of inputs. The behavior of the functions involved in this problem, \( f(x) \) and \( g(x) \), is significant:
  • Both \( f(x) \) and \( g(x) \) are strictly positive for \( x > 0 \) as expected from their polynomial terms that govern these functions' behaviors.
  • As a result, taking \( \sqrt{g(x)} \) or \( \sqrt[3]{f(x)} \) will always yield a positive result, ensuring that the right side of the given equation remains positive for all positive \( x \).
The expected behavior sets the ground for evaluating whether the original equation holds under all given conditions, invoking logical analysis and comparison to known behavioral patterns.
The Role of Mathematical Contradiction
A mathematical contradiction occurs when two statements or results directly oppose each other, proving that an assumption or equation cannot hold. In this case:
  • When \( 0 < x < 1 \), the function \( \ln x \) yields a negative value since it handles results less than \( e \). However, the expression \( (\sqrt[3]{f(x)} + \sqrt{g(x)})^{\frac{1}{17}} \) is always positive for \( x > 0 \).
  • This creates a situation of contradiction because the left-hand side of the equation can be negative, whereas the right-hand side remains strictly positive. Thus, the equation cannot be valid for all positive \( x \).
The identification of such contradictions is crucial, as it provides a clear cutoff point at which an equation or mathematical argument fails to hold universally, highlighting flawed assumptions or conditions in theoretical exercises.

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Most popular questions from this chapter

Prove the following for all \(x \in \mathbb{R}\) : (i) \(\sin 2 x=2 \sin x \cos x\), (ii) \(\cos 2 x=\cos ^{2} x-\sin ^{2} x=2 \cos ^{2} x-1=1-2 \sin ^{2} x\), (iii) \(\sin 3 x=3 \sin x-4 \sin ^{3} x\) (iv) \(\cos 3 x=4 \cos ^{3} x-3 \cos x\). Deduce that \(\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}, \quad \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}, \quad \cos \frac{\pi}{3}=\frac{1}{2}=\sin \frac{\pi}{6}\)

Show that (i) \(\int_{a}^{b} \ln x d x=b(\ln b-1)-a(\ln a-1)\) for all \(a, b \in(0, \infty)\), (ii) \(\int_{a}^{b} \exp x d x=\exp b-\exp a\) for all \(a, b \in \mathbb{R}\).

Let \(\alpha, \beta \in \mathbb{R}\). Suppose \(f, g: \mathbb{R} \rightarrow \mathbb{R}\) are differentiable functions such that $$ f^{\prime}=\alpha f+\beta g, \quad g^{\prime}=\alpha g-\beta f, \quad f(0)=0, \quad \text { and } \quad g(0)=1 $$ Show that \(f(x)=e^{\alpha x} \sin \beta x\) and \(g(x)=e^{\alpha x} \cos \beta x\) for all \(x \in \mathbb{R}\). (Hint: Consider \(h: \mathbb{R} \rightarrow \mathbb{R}\) given by \(h(x):=\left(f(x)-e^{\alpha x} \sin \beta x\right)^{2}+(g(x)-\) \(\left.e^{\alpha x} \cos \beta x\right)^{2}\). Find \(h^{\prime}\).) (Compare Exercise 6 of Chapter 4.)

(i) Show that for any \(n \in \mathbb{N}\), $$ \frac{1}{n+\frac{1}{2}} \leq \int_{n}^{n+1} \frac{d x}{x} \leq \frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right) $$ (ii) Let \(\left(a_{n}\right)\) be the sequence defined by \(a_{n}:=n ! e^{n} / n^{n} \sqrt{n}\) for \(n \in \mathbb{N}\). Show that $$ \ln \left(\frac{a_{n}}{a_{n+1}}\right)=\left(n+\frac{1}{2}\right) \ln \left(1+\frac{1}{n}\right)-1 $$ and hence $$ 1 \leq \frac{a_{n}}{a_{n+1}} \leq \exp \left(\frac{1}{4}\left[\frac{1}{n}-\frac{1}{n+1}\right]\right) \quad \text { for all } n \in \mathbb{N} $$ Deduce that \(\left(a_{n}\right)\) is a monotonically decreasing sequence of positive real numbers and it is convergent. Let \(\alpha:=\lim _{n \rightarrow \infty} a_{n}\). (iii) Use the inequalities in (ii) to show that $$ 1 \leq \frac{a_{n}}{a_{n+k}} \leq \exp \left(\frac{1}{4}\left[\frac{1}{n}-\frac{1}{n+k}\right]\right) \quad \text { for all } n, k \in \mathbb{N} $$ Taking the limit as \(k \rightarrow \infty\), deduce that \(\alpha>0\) and furthermore, \(1 \leq\left(a_{n} / \alpha\right) \leq \exp (1 / 4 n)\) for all \(n \in \mathbb{N}\) (iv) Show that the Wallis formula given in Exercise 58 can be written as \(\sqrt{2 \pi}=\lim _{n \rightarrow \infty} a_{n}^{2} / a_{2 n} .\) Deduce that \(\alpha=\sqrt{2 \pi}\) (v) Use (iii) and (iv) to show that for all \(n \in \mathbb{N}\), $$ (\sqrt{2 \pi}) n^{n+\frac{1}{2}} e^{-n} \leq n ! \leq(\sqrt{2 \pi}) n^{n+\frac{1}{2}} e^{-n+(1 / 4 n)} $$ and conclude that $$ \lim _{n \rightarrow \infty} \frac{n !}{(\sqrt{2 \pi n}) n^{n} e^{-n}}=1 $$ Thus, \(n ! \sim(\sqrt{2 \pi n}) n^{n} e^{-n}\) [Note: This result is known as Stirling's Formula.]

Let \(r_{0} \in \mathbb{R}\), and consider the function \(f_{0}: \mathbb{R} \rightarrow \mathbb{R}\) defined by $$ f_{0}(x)=\left\\{\begin{array}{ll} \sin (1 / x) & \text { if } x \neq 0 \\ r_{0} & \text { if } x=0 \end{array}\right. $$ (i) Show that \(f_{0}\) is not continuous at 0 . Conclude that the function \(x \mapsto\) \(\sin (1 / x)\) for \(x \in \mathbb{R} \backslash\\{0\\}\) cannot be extended to \(\mathbb{R}\) as a continuous function. (ii) (ii) If \(I\) is an interval and \(I \subset \mathbb{R} \backslash\\{0\\}\), then show that \(f_{0}\) has the IVP on \(I\). If \(I\) an interval such that \(0 \in I\), then show that \(f_{0}\) has the IVP on \(I\) if and only if \(\left|r_{0}\right| \leq 1\).
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