Answer:
[tex](c_1,c_2,c_3)=(0,0,0)[/tex]
Linearly independent
Step-by-step explanation:
The given functions are:
[tex]f_1(x)=0,f_2(x)=x,f_3(x)=e^x[/tex]
[tex]g(x)=c_1f_1(x)+c_2f_2(x)+c_3f_3(x)[/tex]
We want to solve for [tex]c_1,c_2,[/tex] and [tex]c_3[/tex] so that [tex]g(x)=0[/tex] on the interval [tex](-\infty,\infty)[/tex].
Let us substitute to get:
[tex]c_1\cdot 0+c_2\cdot x+c_3\cdot e^x=0[/tex]
This implies that:
[tex]0+c_2x+c_3e^x=0\\\implies c_2x+c_3e^x=0[/tex]--->eqn1
Let us differentiate each ter wrt x to get:
[tex]c_2+c_3e^x=0[/tex]---->eqn2
Make [tex]c_2[/tex] the subject in equation (2) to get:
[tex]c_2=-c_3e^x[/tex]---->eqn3
Substitute [tex]c_2=-c_3e^x[/tex] into equation (1) to get:
[tex]-c_3xe^x+c_3e^x=0[/tex]
Factor nicely to get:
[tex]c_3(1-x)e^x=0[/tex]
Note that we are after constants that will this equation true for all x. The only way this could be possible is when [tex]c_3=0[/tex]
Put [tex]c_3=0[/tex] into eqn3 to get:
[tex]c_2=-\cdot 0\cdot e^x\\\implies c_2=0[/tex]
Let us substitute everything into g(x) to find [tex]c_1[/tex]
[tex]c_1\cdot 0+0\cdot x+0\cdot e^x=0[/tex]
This implies that:
[tex]c_1\cdot 0+0+0=0\\\implies c_1\cdot 0=0[/tex]
By the zero product principle , for r [tex]c_1\cdot 0=0[/tex], we must have either [tex]o=o[/tex] or [tex]c_1=0[/tex].
Since [tex]0=0[/tex] is true, [tex]c_1=0[/tex] is also true
Since [tex]c_1f_1(x)+c_2f_2(x)+c_3f_3(x)=0[/tex] for all x, if [tex]c_1=0,c_2=0,c_3=0[/tex], the functions [tex]f_1(x)=0,f_2(x)=x,f_3(x)=e^x[/tex] are linearly independent.
a) Only a trivial solution exists.
b) Since there is only a trivial solution, then we infer that the set of functions is linearly independent.
a) According to Linear Algebra, a set of elements is linearly independent if and only if a coefficient at least within a linear combination under the following form is not zero:
[tex]\Sigma\limits_{i=1}^{n} c_{i}\cdot v_{i} = 0[/tex] (1)
Where:
If we know that [tex]v_{1} = 0[/tex], [tex]v_{2} = x[/tex] and [tex]v_{3} = e^{x}[/tex], then we have the following expression:
[tex]c_{1}\cdot 0 + c_{2}\cdot x + c_{3}\cdot e^{x} = 0[/tex]
Given that [tex]v_{1} = 0[/tex], we conclude that [tex]c_{1}[/tex] can be any real number and we can reduce (1) by Compatibility with Addition:
[tex]c_{1}\cdot 0 + c_{2}\cdot x + c_{3}\cdot e^{x} = 0+0[/tex]
[tex]c_{2} \cdot x + c_{3}\cdot e^{x} = 0[/tex]
[tex]e^{x} = -\frac{c_{2}}{c_{3}} \cdot x[/tex]
[tex]e^{x}[/tex] is a trascendental function bounded between 0 and [tex]+\infty[/tex] and [tex]x[/tex] represents a polynomic function with all real numbers as domain. There is a trivial solution for [tex]x = 0[/tex], but nontrivial ones for all [tex]x \ne 0[/tex]. Hence, we have the following finding:
[tex]v_{1}[/tex], [tex]v_{2}[/tex] and [tex]v_{3}[/tex] have a set of the form [tex]\left\{\frac{c_{1}}{c_{3}}, \frac{c_{2}}{c_{3}}, 1 \right\}[/tex] for all [tex]c_{3}\ne 0[/tex].
Now if we consider [tex]c_{3} = 0[/tex] and [tex]c_{1}[/tex] represents any real number, then we reduce (1) into this:
[tex]c_{1}\cdot 0 + c_{2} \cdot x = 0[/tex]
And the only choice to meet this linear combination is considering that [tex]c_{2} = 0[/tex].
Hence, we have a trivial solution.
b) In consequence, the three functions are linearly independent since it is guaranteed that trivial solution is the only existing solution.
We kindly invite to see this question on linear independence: https://brainly.com/question/7193802
