Problem 78 - Calculus Single Variable

Question

\- A real-valued function \(f\) of a real variable \(x\) is said to be algebraic if there is a polynomial \(p(u, v)\) with integer coefficients such that \(p(x, f(x))=0\) for all \(x .\) For example, \(f(x)=\) \(2 x+\sqrt{x^{2}+1}\) is algebraic because $$ \begin{aligned} p(x, f(x))=& 3 x^{2}+\left(2 x+\sqrt{x^{2}+1}\right)^{2} \\ &-4 x\left(2 x+\sqrt{x^{2}+1}\right)-1 \\ \equiv & 0 \end{aligned} $$ for \(p(u, v)=3 u^{2}+v^{2}-4 u v-1 .\) A function that is not algebraic is said to be transcendental. Find a polynomial that shows that the given expression is algebraic. $$ 2+\sqrt{1+x / x} $$

Step-by-Step Solution

Answer

The polynomial is \((v-2)^2 - 2 = 0\).

Step 1: Identify Function Form

We are given the function \( f(x) = 2 + \sqrt{1 + \frac{x}{x}} \). Simplify it to \( f(x) = 2 + \sqrt{1 + 1} = 2 + \sqrt{2} \). This means the expression simplifies to a constant value \( 2 + \sqrt{2} \), which is independent of \( x \).

Step 2: Start with Polynomial Assumption

Assume the polynomial form \( p(x, f(x)) = 0 \) which can be rearranged to a presumed simpler form if needed. In this case, as \( f(x) \) is constant (\( f(x) = 2 + \sqrt{2} \)), the polynomial can be chosen to reflect that.

Step 3: Construct Polynomial with Constants

Given that \( f(x) = 2 + \sqrt{2} \), think of \( u = x \) and \( v = f(x) \). Consider the squared form \( (v - 2)^2 - 2 = 0 \) to eliminate the square root in \( \sqrt{2} \). This gives the polynomial form \( p(u,v) = (v-2)^2 - 2 \).

Step 4: Verification of Polynomial

Substitute \( v = 2 + \sqrt{2} \) back into the polynomial \( p(v) = (v-2)^2 - 2 \) to verify:\[(2 + \sqrt{2} - 2)^2 - 2 = (\sqrt{2})^2 - 2 = 2 - 2 = 0\]Thus, the polynomial \( p(u,v) = (v-2)^2 - 2 \) holds true for \( v = 2 + \sqrt{2} \).

Key Concepts

Polynomials

Polynomials are expressions consisting of variables arranged in terms of powers along with coefficients. They are one of the basic building blocks of algebra. A general polynomial in one variable can be written as:

• \[P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0\]
• Here, \(a_n, a_{n-1}, \ldots, a_0\) are the coefficients, and \(x\) is the variable.

When talking about polynomials with two variables, notation like \(p(u, v)\) is used, illustrating that the polynomial is a function of both \(u\) and \(v\).
In the original exercise, a polynomial \(p(u, v)\) was utilized to characterize the algebraic nature of the function \(f(x) = 2 + \sqrt{2}\). Here, the polynomial was simplified to \((v-2)^2 - 2\), which effectively demonstrates that this expression has algebraic properties, as it can satisfy the condition \(p(u, v) = 0\) when substituted with \(v = 2 + \sqrt{2}\). This example illustrates how polynomials can represent complex expressions in terms of simpler algebraic forms.

Square Roots

Square roots are a critical concept in algebra, allowing us to revert powers to their base numbers. The square root of a number \(a\), denoted \(\sqrt{a}\), is such that \((\sqrt{a})^2 = a\). For many numbers, particularly non-perfect squares, the result is an irrational number.
In our discussion, the square root arises naturally in the expression \(2 + \sqrt{2}\) when determining its algebraic nature. Here, the square root impacts whether or not our function is algebraic or transcendental. To determine algebraicity, it's beneficial to eliminate square roots, often through techniques such as squaring both sides of an equation to transform the expression into a polynomial form. This is a common tactic in algebraic operations aimed at simplifying expressions that initially contain square roots.

• This concept was applied effectively in the polynomial \((v-2)^2 - 2\), showing that squaring helps in providing an equation that resolves the square root complexity.

Real-Valued Functions

Real-valued functions are defined as functions that produce real numbers as outputs for each real number input. These functions map elements from the real number set \( \mathbb{R} \) to itself.
In simpler terms, if you input a real number into the function, you're guaranteed to receive a real number as output. This is the foundation for many functions considered in algebra and calculus.
The given function \( f(x) = 2 + \sqrt{2} \) is an example of a real-valued function. Though it appears to depend on the variable \(x\), it simplifies entirely to a constant real number, as seen in the solution \( f(x) = 2 + \sqrt{2} \). Because the function's value remains consistent regardless of \(x\), it emphasizes how real-valued functions can sometimes be independent of the input variable and still remain valid.

Transcendental Functions

Transcendental functions are distinct from algebraic functions in that they cannot be expressed as a finite combination of polynomials root operations (such as square roots). These functions often involve exponential, logarithmic, or trigonometric operations.
The crucial point that sets transcendental functions apart is that they do not satisfy polynomial equations with constant integer coefficients. Well-known examples of transcendental functions are \(\text{e}^x\), \( \sin(x) \), and \( \ln(x) \).
In the context of the original problem, an expression like \(2 + \sqrt{2}\) was shown to be algebraic, thus not transcendental because it could be reduced to a polynomial form: \((v-2)^2 - 2 = 0\). This highlights that not every mathematical expression involving square roots or constants qualifies as transcendental; verification through polynomial representation is essential. Understanding this distinction is crucial for determining the nature of functions in more complex analytical studies.