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determine whether the given set of functions is linearly dependent or linearly independent. If they are linearly dependent, find a linear relation among them. $$ f_{1}(t)=2 t-3, \quad f_{2}(t)=t^{2}+1, \quad f_{3}(t)=2 t^{2}-t, \quad f_{4}(t)=t^{2}+t+1 $$

Short Answer

Expert verified
Provide a relation among the functions if they are linearly dependent. Answer: The given set of functions is linearly dependent. A linear relation among the functions is \(\frac{1}{2}(2t - 3) - 2(t^2 + 1) + 3(2t^2 - t) + (t^2 + t + 1) = 0\), which is valid for any non-zero constant \(c_4\).

Step by step solution

01

Set up a linear combination of the functions

Suppose we have a linear combination of these functions such that $$ c_{1}f_{1}(t) + c_{2}f_{2}(t) + c_{3}f_{3}(t) + c_{4}f_{4}(t) = 0 $$ where \(c_1, c_2, c_3, c_4\) are constants. We want to find if there exists a non-trivial solution (not all c's are zeros) to this equation. Substitute the given functions into the linear combination: $$ c_{1}(2t - 3) + c_{2}(t^2 + 1) + c_{3}(2t^2 - t) + c_{4}(t^2 + t + 1) = 0 $$
02

Simplify the linear combination of the functions

Combine like terms to obtain a simpler equation: $$ (c_{2} + c_{3} + c_{4})t^2 + (2c_{1} - c_{3} + c_{4})t + (-3c_{1} + c_{2} + c_{4}) = 0 $$ Now, this equation must hold for all values of \(t\). Therefore, we can set coefficients of like terms equal to zero and create a system of linear equations: $$ c_{2} + c_{3} + c_{4} = 0 \\ 2c_{1} - c_{3} + c_{4} = 0 \\ -3c_{1} + c_{2} + c_{4} = 0 $$
03

Solve the system of linear equations

Since we have 4 unknowns (c1, c2, c3, c4) and only 3 equations, we can set one of the constants (let's say, \(c_{4}\)) as the free variable. We'll then be able to solve the system of linear equations in terms of \(c_{4}\): $$ c_1 = \frac{1}{2}c_4 \\ c_2 = -2c_4 \\ c_3 = 3c_4 $$
04

Determine linear dependence or independence

As we have a non-trivial solution for the system of linear equations (all constants are nonzero if \(c_{4}\neq0\)), the given set of functions is linearly dependent.
05

Find a linear relation among the functions

A linear relation among the functions can be obtained by substituting the values of the constants \(c_1, c_2, c_3, c_4\) back into the original linear combination equation: $$ \frac{1}{2}c_{4}(2t - 3) + (-2c_{4})(t^2 + 1) + (3c_{4})(2t^2 - t) + c_{4}(t^2 + t + 1) = 0 $$ Divide both sides by \(c_4\) (assuming \(c_4 \neq 0\)): $$ \frac{1}{2}(2t - 3) - 2(t^2 + 1) + 3(2t^2 - t) + (t^2 + t + 1) = 0 $$ This is a linear relation among those functions, and it's valid if \(c_4 \neq 0\).

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Key Concepts

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

Linear Combination
A linear combination is a mathematical expression constructed by multiplying each term by a constant and adding the results. When we talk about a set of functions, like
\(f_{1}(t), f_{2}(t), f_{3}(t), f_{4}(t)\),
a linear combination of these would look like
\(c_{1}f_{1}(t) + c_{2}f_{2}(t) + c_{3}f_{3}(t) + c_{4}f_{4}(t)\).
The constants \(c_1, c_2, c_3, c_4\) can be real numbers, and they determine how much each function contributes to the combination. The concept of linear combinations is pivotal in understanding vector spaces and linear algebra, as it underlies the ideas of span, basis, and dependence.
System of Linear Equations
A system of linear equations consists of two or more linear equations that share a common set of variables and are solved simultaneously. In the context of our exercise, setting the linear combination to zero and organizing the terms equates to creating a system of equations:
\[\begin{align*} c_{2} + c_{3} + c_{4} &= 0 \ 2c_{1} - c_{3} + c_{4} &= 0 \ -3c_{1} + c_{2} + c_{4} &= 0\end{align*}\]
Here, we aim to find the values of \(c_1, c_2, c_3, c_4\) that satisfy all three equations. Solutions to such systems can be unique, infinite, or none, these outcomes help us to deduce properties like linear independence or dependence of the given functions.
Non-Trivial Solution
A non-trivial solution of a system of linear equations refers to a solution where not all of the variables are zero. In simpler terms, it means that there exists at least one variable which has a value other than zero. This concept is key in determining linear dependence. If a non-trivial solution exists for the equation
\(c_{1}f_{1}(t) + c_{2}f_{2}(t) + c_{3}f_{3}(t) + c_{4}f_{4}(t) = 0\)
with not all \(c_i\)'s being zero, it indicates that the functions are linearly dependent, as they can be combined in such a way that they cancel each other out.
Linear Relation
A linear relation among a set of functions or vectors indicates that the functions can be expressed as a linear combination equalling zero. Explicitly, for our functions
\(f_{1}(t), f_{2}(t), f_{3}(t), f_{4}(t)\),
a linear relation is given by a non-trivial solution to the equation
\(c_{1}f_{1}(t) + c_{2}f_{2}(t) + c_{3}f_{3}(t) + c_{4}f_{4}(t) = 0\).
Finding a linear relation, like in Step 5 of our exercise, illustrates the interconnections among the functions and how changes in one can be offset by changes in the others, hence proving their dependence.

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Most popular questions from this chapter

Find the general solution of the given differential equation. $$ y^{\mathrm{vi}}-3 y^{\mathrm{iv}}+3 y^{\prime \prime}-y=0 $$

Verify that the differential operator defined by $$ L[y]=y^{(n)}+p_{1}(t) y^{(n-1)}+\cdots+p_{n}(t) y $$ is a linear differential operator. That is, show that $$ L\left[c_{1} y_{1}+c_{2} y_{2}\right]=c_{1} L\left[y_{1}\right]+c_{2} L\left[y_{2}\right] $$ where \(y_{1}\) and \(y_{2}\) are \(n\) times differentiable functions and \(c_{1}\) and \(c_{2}\) are arbitrary constants. Hence, show that if \(y_{1}, y_{2}, \ldots, y_{n}\) are solutions of \(L[y]=0,\) then the linear combination \(c_{1} y_{1}+\cdots+c_{n} y_{n}\) is also a solution of \(L[y]=0 .\)

Determine the general solution of the given differential equation. \(y^{\mathrm{iv}}+2 y^{\prime \prime}+y=3+\cos 2 t\)

We consider another way of arriving at the proper form of \(Y(t)\) for use in the method of undetermined coefficients. The procedure is based on the observation that exponential, polynomial, or sinusoidal terms (or sums and products of such terms) can be viewed as solutions of certain linear homogeneous differential equations with constant coefficients. It is convenient to use the symbol \(D\) for \(d / d t\). Then, for example, \(e^{-t}\) is a solution of \((D+1) y=0 ;\) the differential operator \(D+1\) is said to annihilate, or to be an annihilator of, \(e^{-t}\). Similarly, \(D^{2}+4\) is an annihilator of \(\sin 2 t\) or \(\cos 2 t,\) \((D-3)^{2}=D^{2}-6 D+9\) is an annihilator of \(e^{3 t}\) or \(t e^{3 t},\) and so forth. Consider the problem of finding the form of the particular solution \(Y(t)\) of $$ (D-2)^{3}(D+1) Y=3 e^{2 t}-t e^{-t} $$ where the left side of the equation is written in a form corresponding to the factorization of the characteristic polynomial. (a) Show that \(D-2\) and \((D+1)^{2},\) respectively, are annihilators of the terms on the right side of Eq. and that the combined operator \((D-2)(D+1)^{2}\) annihilates both terms on the right side of Eq. (i) simultaneously. (b) Apply the operator \((D-2)(D+1)^{2}\) to Eq. (i) and use the result of Problem 20 to obtain $$ (D-2)^{4}(D+1)^{3} Y=0 $$ Thus \(Y\) is a solution of the homogeneous equation (ii). By solving Eq. (ii), show that $$ \begin{aligned} Y(t)=c_{1} e^{2 t} &+c_{2} t e^{2 t}+c_{3} t^{2} e^{2 t}+c_{4} t^{3} e^{2 t} \\ &+c_{5} e^{-t}+c_{6} t e^{-t}+c_{7} t^{2} e^{-t} \end{aligned} $$ where \(c_{1}, \ldots, c_{7}\) are constants, as yet undetermined. (c) Observe that \(e^{2 t}, t e^{2 t}, t^{2} e^{2 t},\) and \(e^{-t}\) are solutions of the homogeneous equation corresponding to Eq. (i); hence these terms are not useful in solving the nonhomogeneous equation. Therefore, choose \(c_{1}, c_{2}, c_{3},\) and \(c_{5}\) to be zero in Eq. (iii), so that $$ Y(t)=c_{4} t^{3} e^{2 t}+c_{6} t e^{-t}+c_{7} t^{2} e^{-t} $$ This is the form of the particular solution \(Y\) of Eq. (i). The values of the coefficients \(c_{4}, c_{6},\) and \(c_{7}\) can be found by substituting from Eq. (iv) in the differential equation (i).

Determine the general solution of the given differential equation. \(y^{\mathrm{vi}}+y^{\prime \prime \prime}=t\)

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