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Show that the functions defined by \(f(x)=\sin x, g(x)=\sin 2 x\), and \(h(x)=\) \(\sin x \cos x\) form a linearly dependent subset of \(\mathbb{F}(\mathbb{R})\).

Short Answer

Expert verified
The function set \(\{ f(x), g(x), h(x) \}\) is indeed linearly dependent, as shown by finding constants \(a = 0\), \(b = 1\), and \(c = -2\), not all zero, such that \(0 = a \cdot f(x) + b \cdot g(x) + c \cdot h(x)\).

Step by step solution

01

Expressing \(g(x)\) in terms of \(f(x)\) and \(h(x)\)

Starting off, we rewrite the function \(g(x)=\sin 2x\) in terms of \(f(x)\) and \(h(x)\), using the double-angle identity \(\sin 2x = 2 \sin x \cos x \). So we get \(g(x) = 2f(x)h(x)\).
02

Applying the definition of linear dependence

Next, we apply the definition of linear dependence by finding constants such that the zero function is obtained when these constants are applied to the functions \(f(x)\), \(g(x)\), and \(h(x)\). With the result from step 1, we can make a substitution for \(g(x)\): \(0 = a \cdot f(x) + b \cdot g(x) + c \cdot h(x) = a \sin x + 2b \sin x \cos x + c \sin x \cos x\). For \(x\) not equal to a multiple of \(\pi\), we divide the equation by \(\sin x\) which simplifies it to \( 0 = a + 2b \cos x + c \cos x\). This equation is valid for any \(x\) if \(a = 0\) and \(2b + c = 0\). We can let \(b = 1\) and \(c = -2\). Therefore \(0 = a \cdot f(x) + b \cdot g(x) + c \cdot h(x)\), and this holds for all \(x\), as required.
03

Conclusion

We have shown that there exist constants a, b, and c, not all zero, such that the zero function is obtained when they are applied to \(f(x)\), \(g(x)\), and \(h(x)\). Therefore, the set \(\{ f(x), g(x), h(x) \}\) is linearly dependent in \(\mathbb{F}(\mathbb{R})\) by definition.

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Key Concepts

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

Double-Angle Identity
Understanding the double-angle identity is crucial when dealing with trigonometric functions and their properties. In trigonometry, the double-angle identities are specific formulas that express trigonometric functions of double angles in terms of single angles. For instance, one of the key identities is for the sine function, which states that \( \text{sin}(2x) = 2\text{sin}(x)\text{cos}(x) \). This relationship simplifies the expression \(g(x) = \text{sin}(2x)\) by representing it as twice the product of \(\text{sin}(x)\) and \(\text{cos}(x)\), both of which are familiar and more manageable functions.

These identities are immensely useful in various mathematical fields, including calculus and differential equations, as well as in physics. By using this identity, the complexity of problems involving periodic functions can be significantly decreased, as seen in the exercise where \( \text{sin}(2x) \) is expressed in terms of other trigonometric functions, aiding in determining linear dependence.
Linearly Dependent Subset
A linearly dependent subset is a fundamental concept in linear algebra that often confuses students. If a set of vectors (or functions, as is the case in the exercise) is such that one member of the set can be written as a linear combination of the others, then that set is linearly dependent. In simple terms, this means that there is a redundancy within the set—some information is overlapping. This redundancy is identified by finding nontrivial solutions to the equation \(a_1v_1 + a_2v_2 + \.\.\. + a_nv_n = 0\), where at least one of the coefficients \(a_i\) is non-zero.

Applying this to the provided exercise, you see that constants were found (\(a=0, b=1, c=-2\)), not all zero, that when applied to functions \(f(x), g(x), h(x)\), resulted in the zero function. This proves that these functions are not independent of each other—there's a linear dependence amongst them, meaning they form a linearly dependent subset in the vector space of real-valued functions.
Vector Space of Functions

The concept of a vector space might evoke images of arrows or points in space, but it is much broader than that. A vector space can also consist of functions, as is the case with \(\mathbb{F}(\mathbb{R})\), the space of all real-valued functions. Here, functions act as vectors, and we can add them together and multiply by scalars, much like we do with traditional vectors. What's essential about vector spaces is that they abide by certain axioms, such as distributivity, associativity, and the existence of an additive identity (the zero vector, which, for functions, is the zero function that assigns the value zero to all elements in its domain).

Considering our current exercise, we demonstrate that the functions \(f(x), g(x), h(x)\) belong to a vector space by checking their linear dependence. The zero function plays a pivotal role here; by showing that a nontrivial combination of these functions results in the zero function, we also indicate that functions can be added together and multiplied by real numbers, satisfying the axioms of a vector space.

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Most popular questions from this chapter

Suppose \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{m}\) and \(\mathbf{w}_{1}, \ldots, \mathbf{w}_{n}\) are vectors in a vector space. Prove that if \(\mathbf{w}_{1}, \ldots, \mathbf{w}_{n} \in \operatorname{span}\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{m}\right\\}\) and \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{m} \in \operatorname{span}\left\\{\mathbf{w}_{1}, \ldots, \mathbf{w}_{n}\right\\}\), then \(\operatorname{span}\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{m}\right\\}=\operatorname{span}\left\\{\mathbf{w}_{1}, \ldots, \mathbf{w}_{n}\right\\} .\)

It is known that every solution to the differential equation \(y^{\prime \prime}=-y\) is a function of the form \(a \sin x+b \cos x\) for some constants \(a\) and \(b\). (See the article "An Elementary Approach to \(y^{\prime \prime}=-y^{\prime \prime}\) by J. L. Brenner in the April 1988 issue of the American Mathematical Monthly for a five-line proof of this result.) Show that \(\\{\sin , \cos \\}\) is a basis for the subspace of \(F(\mathbb{R})\) consisting of all solutions to this differential equation.

Fill in the steps to this outline of a classic indirect proof that \(\sqrt{2}\) is irrational. a. Assume \(\sqrt{2}\) is rational. Set up notation for an expression for \(\sqrt{2}\) as a ratio of integers that have been reduced to lowest terms. b. Square both sides of the equation and multiply both sides by the denominator of the fraction. c. Observe that the square of the numerator is even. Explain why the numerator itself must be even. d. Write the numerator as 2 times another integer. e. Show that the square of the denominator is even; hence, the denominator itself must be even. f. Notice the contradiction to the reduction of the fraction to lowest terms.

Write the polynomials defined by the following formulas as linear combinations of \(p_{1}, p_{2}, p_{3}\), where $$ p_{1}(x)=x+1, \quad p_{2}(x)=x^{2}+x, \quad p_{3}(x)=x^{3}+x^{2} . $$ a. \(x^{3}+x^{2}+x+1\) b. \(x^{2}+2 x+1\) c. \(4 x^{3}-7 x^{2}-3 x+8\) d. \((x+1)\left(x^{2}-x+1\right)\) e. \((x+1)\left(x^{2}+x\right)\) f. \((x+1)^{3}\)

Illustrate the proof of the Comparison Theorem by showing that if \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{m}\right\\}\) is a linearly independent subset of \(\mathbb{R}^{4}\), then \(m \leq 4\). (Suggestion: Compare with the spanning set \(\left\\{\mathbf{w}_{1}, \mathbf{w}_{2}, \mathbf{w}_{3}, \mathbf{w}_{4}\right\\}=\\{(1,0,0,0),(0,1,0,0),(0,0,1,0)\). \((0,0,0,1)\\}\).)

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