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Chapter 7: Problem 53

Determine whether the following functions are algebraic or transcendental: (i) \(f(x)=\pi x^{11}+\pi^{2} x^{5}+9\) for \(x \in \mathbb{R}\) (ii) \(f(x)=\frac{e x^{2}+\pi}{\pi x^{2}+e}\) for \(x \in \mathbb{R}\), (iii) \(f(x)=\ln _{10} x\) for \(x>0\), (iv) \(f(x)=x^{\pi}\) for \(x>0\).

Short Answer

Expert verified
The given functions can be classified as follows: (i) Algebraic (ii) Algebraic (iii) Transcendental (iv) Transcendental

Step by step solution

01

(i) Determine if \(f(x)=\pi x^{11}+\pi^{2} x^{5}+9\) is algebraic or transcendental

Since the function comprises the finite sum of \(x\) raised to some power (a polynomial) multiplied by constant coefficients, the given function is algebraic.
02

(ii) Determine if \(f(x)=\frac{e x^{2}+\pi}{\pi x^{2}+e}\) is algebraic or transcendental

In this function, we have a ratio of two polynomials, with constant coefficients being the transcendental numbers, e and 蟺. Since the function is a ratio of polynomials, it is considered algebraic as it can be expressed using algebraic operations involving polynomials.
03

(iii) Determine if \(f(x)=\ln _{10} x\) for \(x>0\) is algebraic or transcendental

The natural logarithm function, \(\ln _{10} x\), does not involve algebraic operations on a polynomial. Although it is possible to approximate the logarithm function using a polynomial, it cannot be accurately expressed as a finite algebraic combination of polynomials. Therefore, the given function is transcendental.
04

(iv) Determine if \(f(x)=x^{\pi}\) for \(x>0\) is algebraic or transcendental

The function \(x^{\pi}\) is similar to the power function but with an irrational exponent (蟺). The irrational exponent prevents the function from being expressed as a finite algebraic combination of polynomials, making the given function transcendental. In summary, we have the following classifications for each function: (i) Algebraic (ii) Algebraic (iii) Transcendental (iv) Transcendental

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Most popular questions from this chapter

(i) Show that for any \(n \in \mathbb{N}, \int_{0}^{\pi / 2} \sin ^{n} x d x=\frac{n-1}{n} \int_{0}^{\pi / 2} \sin ^{n-2} x d x\). (ii) Show that for any \(k \in \mathbb{N}\), $$ \int_{0}^{\pi / 2} \sin ^{2 k} x d x=\frac{(2 k-1)(2 k-3) \cdots 3 \cdot 1}{(2 k)(2 k-2) \cdots 4 \cdot 2} \cdot \frac{\pi}{2}=\frac{(2 k) !}{\left[2^{k} k !\right]^{2}} \cdot \frac{\pi}{2} $$ and $$ \int_{0}^{\pi / 2} \sin ^{2 k+1} x d x=\frac{2 k(2 k-2) \cdots 4 \cdot 2}{(2 k+1)(2 k-1) \cdots 3 \cdot 1}=\frac{\left[2^{k} k !\right]^{2}}{(2 k+1) !} . $$ (iii) For \(k \in \mathbb{N}\), let $$ \mu_{k}:=\frac{\int_{0}^{\pi / 2} \sin ^{2 k} x d x}{\int_{0}^{\pi / 2} \sin ^{2 k+1} x d x} $$ Show that \(1 \leq \mu_{k} \leq(2 k+1) / 2 k\) for each \(k \in \mathbb{N}\) and consequently that \(\mu_{k} \rightarrow 1\) as \(k \rightarrow \infty\). Deduce that $$ \sqrt{\pi}=\lim _{k \rightarrow \infty} \frac{(k !)^{2} 2^{2 k}}{(2 k) ! \sqrt{k}} $$ Thus, \(\pi \sim(k !)^{4} 2^{4 k} /[(2 k) !]^{2} k .\left(\right.\) Hint: \(\sin ^{2 k+1} x \leq \sin ^{2 k} x \leq \sin ^{2 k-1} x\) for all \(x \in[0, \pi / 2] .)\) [Note: This result is known as the Wallis formula.]

If \(x \in \mathbb{R}\) with \(x \neq k \pi\) for any \(k \in \mathbb{Z}\), then show that \(1+\cot ^{2} x=\csc ^{2} x, \quad(\cot )^{\prime} x=-\csc ^{2} x, \quad\) and \(\quad(\csc )^{\prime} x=-\csc x \cot x\)

Show that for \(x \in(0, \pi / 2)\), $$ \frac{2 x}{\pi}<\sin x<\min \\{1, x\\} \text { and } 1-\frac{2 x}{\pi}<\cos x<\min \left\\{1, \frac{\pi}{2}-x\right\\}, $$ whereas for \(x \in(-\pi / 2,0)\), $$ \max \\{-1, x\\}<\sin x<\frac{2 x}{\pi} \text { and } 1+\frac{2 x}{\pi}<\cos x<\min \left\\{1, \frac{\pi}{2}+x\right\\} . $$

Consider the function \(g: \mathbb{R} \backslash\\{0\\} \rightarrow \mathbb{R}\) defined by \(g(x):=\cos (1 / x) .\) Prove the following: (i) \(g\) is an even function. (ii) \(\lim _{x \rightarrow 0} g(x)\) does not exist, but \(\lim _{x \rightarrow 0^{+}}[g(x)-g(-x)]\) exists. Also, \(g\) cannot be extended to \(\mathbb{R}\) as a continuous function. (iii) For any \(\delta>0, g\) is not uniformly continuous on \((0, \delta)\) as well as on \((-\delta, 0)\), but it is uniformly continuous on \((\infty,-\delta] \cup[\delta, \infty)\). (iv) For any \(\delta>0, g\) is not monotonic, not convex, and not concave on \((0, \delta)\) as well as on \((-\delta, 0)\).

Prove that the secant function, the cosecant function, and the cotangent function are transcendental.
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