Mathematics, 11.03.2020 23:09 lays20001

Consider the following functions. f1(x) = 0, f2(x) = x, f3(x) = ex g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 so that g(x) = 0 on the interval (−[infinity], [infinity]). If a nontrivial solution exists, state it. (If only the trivial solution exists, enter the trivial solution {0, 0, 0}.) {c1, c2, c3} = Incorrect: Your answer is incorrect. Determine whether f1, f2, f3 are linearly independent on the interval (−[infinity], [infinity]). linearly dependent linearly independent Correct: Your answer is correct.

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Answer from: runaway173

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:

$\Sigma\limits_{i=1}^{n} c_{i}\cdot v_{i} = 0$ (1)

Where:

$c_{i}$ - i-th coefficient. $v_{i}$ - Set element.

If we know that $v_{1} = 0$ , $v_{2} = x$ and $v_{3} = e^{x}$ , then we have the following expression:

$c_{1}\cdot 0 + c_{2}\cdot x + c_{3}\cdot e^{x} = 0$

Given that $v_{1} = 0$ , we conclude that $c_{1}$ can be any real number and we can reduce (1) by Compatibility with Addition:

$c_{1}\cdot 0 + c_{2}\cdot x + c_{3}\cdot e^{x} = 0+0$

$c_{2} \cdot x + c_{3}\cdot e^{x} = 0$

$e^{x} = -\frac{c_{2}}{c_{3}} \cdot x$

$e^{x}$ is a trascendental function bounded between 0 and $+\infty$ and represents a polynomic function with all real numbers as domain. There is a trivial solution for x = 0 , but nontrivial ones for all $x \ne 0$ . Hence, we have the following finding:

$v_{1}$ , $v_{2}$ and $v_{3}$ have a set of the form $\left\{\frac{c_{1}}{c_{3}}, \frac{c_{2}}{c_{3}}, 1 \right\}$ for all $c_{3}\ne 0$ .

Now if we consider $c_{3} = 0$ and $c_{1}$ represents any real number, then we reduce (1) into this:

$c_{1}\cdot 0 + c_{2} \cdot x = 0$

And the only choice to meet this linear combination is considering that $c_{2} = 0$ .

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: link

Answer from: shifaxoxoxo

(c_1,c_2,c_3)=(0,0,0)

Linearly independent

Step-by-step explanation:

The given functions are:

f_1(x)=0,f_2(x)=x,f_3(x)=e^x

g(x)=c_1f_1(x)+c_2f_2(x)+c_3f_3(x)

We want to solve for c_1,c_2, and c_3 so that g(x)=0 on the interval $(-\infty,\infty)$ .

Let us substitute to get:

$c_1\cdot 0+c_2\cdot x+c_3\cdot e^x=0$

This implies that:

$0+c_2x+c_3e^x=0\\\implies c_2x+c_3e^x=0$ --->eqn1

Let us differentiate each ter wrt x to get:

c_2+c_3e^x=0 ---->eqn2

Make c_2 the subject in equation (2) to get:

c_2=-c_3e^x ---->eqn3

Substitute c_2=-c_3e^x into equation (1) to get:

-c_3xe^x+c_3e^x=0

Factor nicely to get:

c_3(1-x)e^x=0

Note that we are after constants that will this equation true for all x. The only way this could be possible is when c_3=0

Put c_3=0 into eqn3 to get:

$c_2=-\cdot 0\cdot e^x\\\implies c_2=0$

Let us substitute everything into g(x) to find c_1

$c_1\cdot 0+0\cdot x+0\cdot e^x=0$

This implies that:

$c_1\cdot 0+0+0=0\\\implies c_1\cdot 0=0$

By the zero product principle , for r $c_1\cdot 0=0$ , we must have either o=o or c_1=0 .

Since 0=0 is true, c_1=0 is also true

Since c_1f_1(x)+c_2f_2(x)+c_3f_3(x)=0 for all x, if c_1=0,c_2=0,c_3=0 , the functions f_1(x)=0,f_2(x)=x,f_3(x)=e^x are linearly independent.

Answer from: Quest

step-by-step explanation: divide the numerator by the denominator to get the decimal

it would still be 193

if divide by 100 then 1.93

i am not sure

Answer from: Quest

no because mercury has a density of 13.6 grams per centimeter squared while gold has 19.32 grams centimeter squared

step-by-step explanation:

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Consider the following functions. f1(x) = 0, f2(x) = x, f3(x) = ex g(x) = c1f1(x) + c2f2(x) + c3f3(x...

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