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In Problems, show directly that the given functions are linearly dependent on the real line. That is, find a nontrivial linear combination of the given functions that vanishes identically. $$ f(x)=5 p(x)=2-3 x^{2}, h(x)=10+15 x^{2} $$

Short Answer

Expert verified
The functions are linearly dependent, as demonstrated by the combination \(-4f(x) + 5p(x) + h(x) = 0\).

Step by step solution

01

Understanding Linear Dependence

To show that three functions are linearly dependent, we need to find constants \(c_1, c_2,\) and \(c_3\), not all zero, such that \(c_1 f(x) + c_2 p(x) + c_3 h(x) = 0\) for all \(x\).
02

Setting Up the Equation

Write the linear combination of the given functions: \(c_1 \cdot 5 + c_2 (2 - 3x^2) + c_3 (10 + 15x^2) = 0\). This simplifies to \(5c_1 + (2c_2 + 10c_3) + (-3c_2 + 15c_3)x^2 = 0\).
03

Solving for Coefficients

Equate the coefficients of like powers of \(x\) to zero: For constant term: \(5c_1 + 2c_2 + 10c_3 = 0\). For \(x^2\) term: \(-3c_2 + 15c_3 = 0\).
04

Finding a Nontrivial Solution

From \(-3c_2 + 15c_3 = 0\), solve for \(c_2 = 5c_3\). Substitute \(c_2 = 5c_3\) into \(5c_1 + 2c_2 + 10c_3 = 0\): \(5c_1 + 2(5c_3) + 10c_3 = 0\) simplifies to \(5c_1 + 20c_3 = 0\). So \(c_1 = -4c_3\).
05

Conclusion with Specific Coefficients

Choose \(c_3 = 1\) to find specific values: \(c_2 = 5\), \(c_1 = -4\). Therefore, the nontrivial linear combination \(-4f(x) + 5p(x) + h(x) = 0\) vanishes for all \(x\), proving the functions are linearly dependent.

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Key Concepts

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

Nontrivial Linear Combination
A nontrivial linear combination is a crucial concept when discussing linear dependence among functions or vectors. It refers to a combination where at least one of the coefficients is not zero. This is in contrast to a trivial linear combination, where all coefficients are zero. Finding a nontrivial linear combination is key because it shows that the involved functions or vectors are dependent and cannot stand alone.
To find a nontrivial linear combination, you need to determine constants (let's call them \(c_1, c_2, c_3\), etc.) such that the weighted sum of the functions results in zero. Here, it's important that not all constants are zero. In simpler terms, at least one of the constants will have a non-zero value, proving a dependency among the functions.
  • The overall goal is to express zero using a mix of given functions with specific, non-zero coefficients.
  • This is useful because it allows you to understand relationships and dependencies between functions or vectors.
  • The concept extends to solving systems of equations, which can show dependencies in complex mathematical structures.
Nontrivial linear combinations are essential in vector spaces, differential equations, and more as they reveal underlying structures and relationships.
Coefficient Comparison
Coefficient comparison is a technique used to solve equations involving variables. It involves lining up terms with similar powers or variables and setting their coefficients equal to one another. This method is particularly useful when dealing with polynomials or functions.
In our exercise, after setting up the equation, you compare the coefficients of terms such as constants, \(x^2\), and possibly other powers of \(x\).
  • By isolating terms of the same kind, you turn a complex equation into simpler ones.
  • This systematic approach allows you to manage multiple variables and find solutions efficiently.
For each type of term (constant, \(x^2\), etc.), you create separate equations that add clarity and simplify the overall problem. This step-by-step process makes it easier to handle algebraic expressions and is widely used in solving systems of equations and understanding polynomial identities.
Solving Linear Equations
Solving linear equations is a foundational skill in algebra and numerous mathematical applications. It involves finding the values of variables that satisfy given equations. The process typically includes isolating the variable and determining its specific value.
In our context, solving linear equations also involves substituting relationships between coefficients, like \(c_2 = 5c_3\), into another equation. This step-by-step method highlights logical relationships and simplifies complex equations.
  • You begin by aligning all terms on one side of the equation, aiming for a constant on the other.
  • Then, ensure all terms are expressed with known values, except the variable or unknowns you are solving for.
For the exercise, it implies deriving \(c_1 = -4c_3\) after substituting known relationships. This process of substitution and comparison transforms a system of equations into manageable steps, leading to a concrete solution.
Knowing how to solve linear equations is immensely useful for real-world problems, modeling scenarios, and further studies in mathematics, making this skill vital for students and professionals alike.

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