Problem 15 - Calculus One Again

Question

Find all functions \(g: \mathbb{R} \rightarrow \mathrm{R}\) that satisfy \(g(x+y)+\) \(g(x-y)=2 x^{2}+2 y^{2}\).

Step-by-Step Solution

Answer

The function \( g(x) = x^2 \) satisfies the equation for all \( x \) and \( y \).

Step 1: Analyze the Functional Equation

We start by analyzing the given functional equation: \( g(x+y) + g(x-y) = 2x^2 + 2y^2 \). To find the potential form of \( g(x) \), we need to explore specific values for \( y \) that will simplify the equation.

Step 2: Substitute Specific Values

First, let \( y = 0 \), which gives \( g(x) + g(x) = 2x^2 \). This simplifies to \( 2g(x) = 2x^2 \) or \( g(x) = x^2 \). This suggests a potential solution, but we must verify it holds for all \( x \) and \( y \).

Step 3: Verify Solution

Now we verify if \( g(x) = x^2 \) satisfies the original equation: Substitute \( g(x) = x^2 \) into \( g(x+y) + g(x-y) \), which gives \( (x+y)^2 + (x-y)^2 = x^2 + 2xy + y^2 + x^2 - 2xy + y^2 = 2x^2 + 2y^2 \). This matches the right hand side of the given equation, so \( g(x) = x^2 \) is indeed a solution.

Key Concepts

Solution Verification

Solution verification is an essential step when dealing with functional equations. It ensures that the proposed solution satisfies the original problem. In our exercise, we suggested that the function \( g(x) = x^2 \) could be a potential solution. But how can we be sure it works for all values of \( x \) and \( y \)?

We proceed by substituting \( g(x) = x^2 \) back into the original equation \( g(x+y) + g(x-y) = 2x^2 + 2y^2 \). If our function holds, the left-hand side should simplify to match the right-hand side for all real numbers \( x \) and \( y \).

• Substitute \( g(x+y) = (x+y)^2 \) and \( g(x-y) = (x-y)^2 \).
• Calculate the expression: \((x+y)^2 + (x-y)^2 = x^2 + 2xy + y^2 + x^2 - 2xy + y^2\).
• Simplify it to \( 2x^2 + 2y^2 \).

This matches the right side exactly, verifying our solution. Thus, verifying solutions is crucial as it confirms correctness beyond our initial assumptions.

Substitution Method

The substitution method is a powerful tool for solving functional equations. It involves selecting specific values for the variables to simplify the equation and identify potential patterns or solutions. In our problem, this method plays a pivotal role.

Let's dissect how substitution was used effectively in our exercise:

• First, by setting \( y = 0 \) in the original equation, we simplify the expression to \( g(x) + g(x) = 2x^2 \).
• This aids us in determining that \( g(x) = x^2 \), as the equation simplifies to \( 2g(x) = 2x^2 \), and then \( g(x) = x^2 \).

By simplifying the problem with specific substitutions, we move closer to finding a valid solution. This method allows us to test hypotheses about the form of \( g(x) \) and see how these match the given conditions. Remember, choosing strategic values is key to going from a complex equation to a solvable one.

Quadratic Functions

Quadratic functions are a specific type of polynomial with the highest degree of two. In our exercise, we discovered that the function \( g(x) = x^2 \) behaves as a quadratic function. Understanding these properties helps decode why it fits the equation so perfectly.

Key aspects of quadratic functions include:

• Form: \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.
• Symmetry: Quadratic functions are symmetrical with a vertical axis of symmetry at \( x = -\frac{b}{2a} \).
• Graph: They graph as a parabola, opening upwards if \( a > 0 \) or downwards if \( a < 0 \).

In the case of \( g(x) = x^2 \), \( a = 1 \) and \( b = c = 0 \), simplifying to \( f(x) = x^2 \). This particular function is even and symmetric around the y-axis, offering us consistent results when substituted into the functional equation. Realizing this symmetry and simplicity often guides us in verifying and understanding why certain functions solve the equation at hand.