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Chapter 3: Problem 16

Since \((1) 0+(0) x+(0) e^{x}=0\) the set of functions is linearly dependent. A similar argument shows that any set of functions containing \(f(x)=0\) will be linearly dependent.

Short Answer

Expert verified
Any set of functions including the zero function is linearly dependent.

Step by step solution

01

Understanding Linear Dependence

A set of functions \( f_1(x), f_2(x), ext{...}, f_n(x) \) is considered linearly dependent if there exist constants \( c_1, c_2, ext{...}, c_n \) not all zero such that: \[ c_1 f_1(x) + c_2 f_2(x) + ext{...} + c_n f_n(x) = 0 \] for all \( x \). This implies that at least one of the functions can be written as a linear combination of the others.
02

Analyzing the Given Functions

In the problem, we have an expression: \( (1) imes 0 +(0) imes x +(0) imes e^{x} = 0 \). Notice that the set of functions only includes the function \( f(x) = 0 \), i.e., the zero function.
03

Establishing Linear Dependence

For any function set containing the zero function, say \( g(x), \) the equation: \( c_1 imes 0 + c_2 imes g(x) = 0 \) always holds as \( c_1 \) can be any non-zero constant ensuring the equation holds true. Therefore, a non-trivial linear combination exists, confirming linear dependence.

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Key Concepts

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

Zero Function
The zero function is one of the simplest functions in mathematics. It is defined as a function that returns zero for every input. In other words, for all values of \( x \), the zero function \( f(x) \) can be written as:\[f(x) = 0\]This may seem straightforward, but its role in understanding concepts like linear dependence is crucial. A function set that includes the zero function is always linearly dependent. Why? Because you can always find a combination of not-all-zero constants to express the zero function as a part of the set.Every function set involving the zero function is linearly dependent as it can be expressed using constants in a non-trivial way.
Linear Combination
The concept of a linear combination revolves around forming a function or expression by scaling and adding other functions or numbers. Mathematically, we say that a function \( f(x) \) is a linear combination of functions \( f_1(x), f_2(x), ... , f_n(x) \) if:\[c_1 f_1(x) + c_2 f_2(x) + ... + c_n f_n(x) = f(x)\]where \(c_1, c_2, ..., c_n\) are constants. These constants help scale the functions before adding them together.
  • When talking about functional linear dependence, a linear combination involving coefficients that are not all zero can express one function in a set using others in the set.
  • This forms a crucial part of studying systems and solving equations because it shows relationships between functions.
Functions Set
In mathematics, a set of functions refers to a collection of multiple functions that can be analyzed together. These functions can be constants, polynomials, exponentials, or other types. A set of functions can be represented as \( \{ f_1(x), f_2(x), ..., f_n(x) \} \).Analyzing a functions set often involves exploring properties like linear dependence or independence.To check if a set of functions is dependent, one seeks to express one function as a linear combination of others in the set. If at least one such non-trivial combination exists, the functions are not independent. This kind of analysis is vital in many mathematical and applied contexts.
Constants Coefficients
Constants coefficients play a vital role in forming linear combinations. They are the multipliers used to scale each function in a combination.For example, consider the expression:\[c_1 f_1(x) + c_2 f_2(x) + ... + c_n f_n(x) = 0\]Here, \(c_1, c_2, ..., c_n\) are constants. The solution's linear dependence rests on these constants being not all zero.
  • Understanding constants is crucial for determining linear dependencies.
  • If at least two of these constants are non-zero, it implies that the functions are linearly dependent.
  • When discussing linear dependence, these coefficients allow analyzing how functions relate to one another.

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Most popular questions from this chapter

(a) \(\quad y_{\max }=y(L)=w_{0} L^{4} / 8 E I\) (b) Replacing both \(L\) and \(x\) by \(L / 2\) in \(y(x)\) we obtain \(w_{0} L^{4} / 128 E I,\) which is \(1 / 16\) of the maximum deflection when the length of the beam is \(L\). (c) \(y_{\max }=y(L / 2)=5 w_{0} L^{4} / 384 E I\) (d) The maximum deflection in Example 1 is \(y(L / 2)=\left(w_{0} / 24 E I\right) L^{4} / 16=w_{0} L^{4} / 384 E I,\) which is \(1 / 5\) of the maximum displacement of the beam in part c.

Let \((x, y)\) be the coordinates of \(S_{2}\) on the curve \(C .\) The slope at \((x, y)\) is then \\[ d y / d x=\left(v_{1} t-y\right) /(0-x)=\left(y-v_{1} t\right) / x \\] or \(x y^{\prime}-y=-v_{1} t\) Differentiating with respect to \(x\) and using \(r=v_{1} / v_{2}\) gives \\[ \begin{aligned} x y^{\prime \prime}+y^{\prime}-y^{\prime} &=-v_{1} \frac{d t}{d x} \\ x y^{\prime \prime} &=-v_{1} \frac{d t}{d s} \frac{d s}{d x} \\ x y^{\prime \prime} &=-v_{1} \frac{1}{v_{2}}(-\sqrt{1+\left(y^{\prime}\right)^{2}}) \\ x y^{\prime \prime} &=r \sqrt{1+\left(y^{\prime}\right)^{2}} \end{aligned} \\] Letting \(u=y^{\prime}\) and separating variables, we obtain \\[ \begin{aligned} x \frac{d u}{d x} &=r \sqrt{1+u^{2}} \\ \frac{d u}{\sqrt{1+u^{2}}} &=\frac{r}{x} d x \\ \sinh ^{-1} u &=r \ln x+\ln c=\ln \left(c x^{r}\right) \\ u &=\sinh \left(\ln c x^{r}\right) \\ \frac{d y}{d x} &=\frac{1}{2}\left(c x^{r}-\frac{1}{c x^{r}}\right) \end{aligned} \\] At \(t=0, d y / d x=0\) and \(x=a,\) so \(0=c a^{r}-1 / c a^{r} .\) Thus \(c=1 / a^{r}\) and \\[ \frac{d y}{d x}=\frac{1}{2}\left[\left(\frac{x}{a}\right)^{r}-\left(\frac{a}{x}\right)^{r}\right]=\frac{1}{2}\left[\left(\frac{x}{a}\right)^{r}-\left(\frac{x}{a}\right)^{-r}\right] \\] If \(r>1\) or \(r<1,\) integrating gives \\[ y=\frac{a}{2}\left[\frac{1}{1+r}\left(\frac{x}{a}\right)^{1+r}-\frac{1}{1-r}\left(\frac{x}{a}\right)^{1-r}\right]+c_{1} \\] When \(t=0, y=0\) and \(x=a,\) so \(0=(a / 2)[1 /(1+r)-1 /(1-r)]+c_{1} .\) Thus \(c_{1}=a r /\left(1-r^{2}\right)\) and \\[ y=\frac{a}{2}\left[\frac{1}{1+r}\left(\frac{x}{a}\right)^{1+r}-\frac{1}{1-r}\left(\frac{x}{a}\right)^{1-r}\right]+\frac{a r}{1-r^{2}} \\] To see if the paths ever intersect we first note that if \(r>1\), then \(v_{1}>v_{2}\) and \(y \rightarrow \infty\) as \(x \rightarrow 0^{+} .\) In other words, \(S_{2}\) always lags behind \(S_{1}\). Next, if \(r<1\), then \(v_{1}

(a) A solution curve has the same \(y\) -coordinate at both ends of the interval \([-\pi, \pi]\) and the tangent lines at the endpoints of the interval are parallel. (b) For \(\lambda=0\) the solution of \(y^{\prime \prime}=0\) is \(y=c_{1} x+c_{2}\). From the first boundary condition we have $$y(-\pi)=-c_{1} \pi+c_{2}=y(\pi)=c_{1} \pi+c_{2}$$ or \(2 c_{1} \pi=0 .\) Thus, \(c_{1}=0\) and \(y=c_{2} .\) This constant solution is seen to satisfy the boundary-value problem. For \(\lambda=-\alpha^{2}<0\) we have \(y=c_{1} \cosh \alpha x+c_{2} \sinh \alpha x .\) In this case the first boundary condition gives $$\begin{aligned} y(-\pi) &=c_{1} \cosh (-\alpha \pi)+c_{2} \sinh (-\alpha \pi) \\ &=c_{1} \cosh \alpha \pi-c_{2} \sinh \alpha \pi \\ &=y(\pi)=c_{1} \cosh \alpha \pi+c_{2} \sinh \alpha \pi \end{aligned}$$ or \(2 c_{2} \sinh \alpha \pi=0 .\) Thus \(c_{2}=0\) and \(y=c_{1} \cosh \alpha x .\) The second boundary condition implies in a similar fashion that \(c_{1}=0 .\) Thus, for \(\lambda<0\), the only solution of the boundary-value problem is \(y=0\). For \(\lambda=\alpha^{2}>0\) we have \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x .\) The first boundary condition implies $$\begin{aligned} y(-\pi) &=c_{1} \cos (-\alpha \pi)+c_{2} \sin (-\alpha \pi) \\ &=c_{1} \cos \alpha \pi-c_{2} \sin \alpha \pi \\ &=y(\pi)=c_{1} \cos \alpha \pi+c_{2} \sin \alpha \pi \end{aligned}$$ or \(2 c_{2} \sin \alpha \pi=0 .\) Similarly, the second boundary condition implies \(2 c_{1} \alpha \sin \alpha \pi=0 .\) If \(c_{1}=c_{2}=0\) the solution is \(y=0 .\) However, if \(c_{1} \neq 0\) or \(c_{2} \neq 0,\) then \(\sin \alpha \pi=0,\) which implies that \(\alpha\) must be an integer, \(n\) .Therefore, for \(c_{1}\) and \(c_{2}\) not both \(0, y=c_{1} \cos n x+c_{2} \sin n x\) is a nontrivial solution of the boundary-value problem. since \(\cos (-n x)=\cos n x\) and \(\sin (-n x)=-\sin n x,\) we may assume without loss of generality that the eigenvalues are \(\lambda_{n}=\alpha^{2}=n^{2},\) for \(n\) a positive integer. The corresponding eigenfunctions are \(y_{n}=\cos n x\) and \(y_{n}=\sin n x\) c.

Define \(y=u(x) \cdot 1\) so $$y^{\prime}=u^{\prime}, \quad y^{\prime \prime}=u^{\prime \prime} \quad \text { and } \quad y^{\prime \prime}+y^{\prime}=u^{\prime \prime}+u^{\prime}=1$$ If \(w=u^{\prime}\) we obtain the linear first-order equation \(w^{\prime}+w=1\) which has the integrating factor \(e^{\int d x}=e^{x} .\) Now $$\frac{d}{d x}\left[e^{x} w\right]=e^{x} \quad \text { gives } \quad e^{x} w=e^{x}+c$$ Therefore \(w=u^{\prime}=1+c e^{-x}\) and \(u=x+c_{1} e^{-x}+c_{2} .\) The general solution is $$y=u=x+c_{1} e^{-x}+c_{2}$$

(a) From \(100 x^{\prime \prime}+1600 x=1600 \sin 8 t, x(0)=0,\) and \(x^{\prime}(0)=0\) we obtain \(x_{c}=c_{1} \cos 4 t+c_{2} \sin 4 t\) and \(x_{p}=-\frac{1}{3} \sin 8 t\) so that by a trig identity \\[x=\frac{2}{3} \sin 4 t-\frac{1}{3} \sin 8 t=\frac{2}{3} \sin 4 t-\frac{2}{3} \sin 4 t \cos 4 t\\] (b) If \(x=\frac{1}{3} \sin 4 t(2-2 \cos 4 t)=0\) then \(t=n \pi / 4\) for \(n=0,1,2, \ldots\) (c) If \(x^{\prime}=\frac{8}{3} \cos 4 t-\frac{8}{3} \cos 8 t=\frac{8}{3}(1-\cos 4 t)(1+2 \cos 4 t)=0\) then \(t=\pi / 3+n \pi / 2\) and \(t=\pi / 6+n \pi / 2\) for \(n=0,1,2, \ldots\) at the extreme values. Note: There are many other values of \(t\) for which \(x^{\prime}=0\) (d) \(x(\pi / 6+n \pi / 2)=\sqrt{3} / 2 \mathrm{cm}\) and \(x(\pi / 3+n \pi / 2)=-\sqrt{3} / 2 \mathrm{cm}\).
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