Problem 77 - 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. $$ \sqrt{x}+\sqrt{3} $$

Step-by-Step Solution

Answer

The polynomial is \(p(u, v) = v^4 - (2u + 6)v^2 + u^2 - 6u + 9\).

Step 1: Understand the Task

We need to find a polynomial with integer coefficients in terms of variables \(u\) and \(v\) such that when we substitute \(u = x\) and \(v = \sqrt{x} + \sqrt{3}\), the polynomial evaluates to zero for all real numbers \(x\).

Step 2: Express \(\sqrt{x} + \sqrt{3}\) as the Function \(v\)

Set \(v = \sqrt{x} + \sqrt{3}\). Our goal is to eliminate the radicals and find a relationship using a polynomial equation between \(x\) and \(v\).

Step 3: Isolate and Square the Expression

Start by isolating one of the radicals, for example, \(\sqrt{x} = v - \sqrt{3}\). Square both sides to eliminate the square root: \((\sqrt{x})^2 = (v - \sqrt{3})^2\), which simplifies to \(x = v^2 - 2v\sqrt{3} + 3\).

Step 4: Eliminate the Remaining Square Root

To further simplify, isolate \(\sqrt{3}\) in terms of \(x\) and \(v\): \(2v\sqrt{3} = v^2 + 3 - x\). Rearrange to get \(\sqrt{3} = \frac{v^2 + 3 - x}{2v}\). Now, square both sides to get rid of the square root: \(3 = \frac{(v^2 + 3 - x)^2}{4v^2}\).

Step 5: Multiply Through to Clear the Denominator

Multiply the equation by \(4v^2\) to eliminate the fraction: \(3 \cdot 4v^2 = (v^2 + 3 - x)^2\). Simplifying the left side gives \(12v^2 = (v^2 + 3 - x)^2\).

Step 6: Expand the Right Side and Combine Terms

Expand the right side: \((v^2 + 3 - x)^2 = (v^2 + 3 - x)(v^2 + 3 - x) = v^4 + 6v^2 + 9 - 2xv^2 - 6x + x^2\). Set the equation: \(12v^2 = v^4 + 6v^2 + 9 - 2xv^2 - 6x + x^2\).

Step 7: Rearrange to Form a Polynomial

Rearrange the terms to form a polynomial: \(v^4 - 2xv^2 + x^2 - 6x + 9 - 6v^2 = 0\). Combine like terms: \(v^4 - (2x + 6)v^2 + x^2 - 6x + 9 = 0\).

Step 8: Identify the Polynomial \(p(u,v)\)

This gives us the polynomial: \(p(u,v) = v^4 - (2u + 6)v^2 + u^2 - 6u + 9\), where \(u = x\) and \(v = \sqrt{x} + \sqrt{3}\).

Key Concepts

Polynomial Equations

Polynomial equations are mathematical expressions involving a sum of powers of a variable with corresponding coefficients. When we discuss these types of equations, an essential component is understanding terms like degree and coefficients. The degree of the polynomial is the highest power of the variable present in the expression.
For example, in the equation \(p(u,v) = v^4 - (2u + 6)v^2 + u^2 - 6u + 9\), the highest degree is 4, making it a fourth-degree polynomial. Each term is constructed by multiplying powers of the variables \(u\) and \(v\) by their coefficients. The equation is structured to provide a relationship between variables, often represented by \(u\) and \(v\) in algebraic contexts.
Polynomial equations are pivotal in mathematical modeling because they can represent a wide range of phenomena with real-world applications. Learning how to create and manipulate such equations is crucial for problem-solving in various fields.

Radical Functions

Radical functions involve roots, such as square roots, cube roots, etc. They can often appear complex due to the presence of radical symbols, like \(\sqrt{x}\) or \(\sqrt{3}\). These radicals can be tackled by transforming them into polynomial equations through various algebraic operations.
In the exercise, we began with \(v = \sqrt{x} + \sqrt{3}\). To handle such expressions, a common technique is to isolate the radicals and apply squaring to eliminate the roots. Take \(\sqrt{x}\), which was isolated as \(\sqrt{x} = v - \sqrt{3}\), and then squared both sides to remove the radical:

• Square both sides: \((\sqrt{x})^2 = (v - \sqrt{3})^2\), simplifying to \(x = v^2 - 2v\sqrt{3} + 3\).

By repeating similar steps for other radicals, you progressively simplify the expression. It demonstrates how algebraic manipulation can transform complex radical expressions into a more tractable polynomial form.

Transcendental Functions

Transcendental functions represent opposite concepts to algebraic functions. They are not solutions of polynomial equations with integer coefficients and often involve exponential functions, logarithms, or trigonometric functions. These functions do not satisfy any algebraic equation like \(p(x, f(x)) = 0\).
Understanding transcendental functions is essential because they frequently appear in higher mathematics and practical applications, ranging from calculus to engineering solutions. An example of a transcendental function could be \(f(x) = e^x\) or \(f(x) = \ln(x)\). These functions cannot be expressed as a solution of any polynomial equation.
While the exercise's aim was to show the given function \(\sqrt{x} + \sqrt{3}\) as algebraic by finding a polynomial, comprehending the distinction between algebraic and transcendental is fundamental. Such understanding allows us to identify which mathematical approaches can be used for problem-solving, highlighting the boundaries between what can be solved algebraically and what requires transcendental analysis.