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Julli [10]
3 years ago
11

Are the functions f,g, and h given below linearly independent?

Mathematics
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

Linearly independent, x = 0

Step-by-step explanation:

- We are given three functions as follows:

                           f ( x ) = e^4^x\\\\g ( x ) = xe^4^x\\\\h( x ) = x^2e^4^x

- We are to determine the linear - independence of the given functions. We will use the theorem of linear independence which states that:

                           c_1*f(x) + c_2*g(x) + c_3*h(x) = 0

Where,

                       c1 , c2 , c3 are all zeroes then for all values of (x),

- The system of function is said to be linearly independent

- We will express are system of equations as such:

                           c_1*e^4^x + c_2*xe^4^x + c_3*x^2e^4^x = 0\\\\

- To express our system of linear equations we will choose three arbitrary  values of ( x ). We will choose, x = 0. then we have:

                          c_1*( 1 ) + c_2*(0) + c_3*(0 ) = 0\\\\c_1 = 0

- Next choose x = 1:

                          c_2*e^4 + c_3*e^4 = 0\\\\c_2 + c_3 = 0

- Next choose x = 2:

                          2*c_2*e^8 + 4*c_3*e^8 = 0\\\\c_2 + 2c_3 = 0

- Solve the last two equations simultaneously we have:

                          c_1 = c_2 = c_3 = 0     .... ( Only trivial solution exist )

Answer: The functions are linearly independent

- The only zero exist is x = 0.

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(a) Let R = {(a,b): a² + 3b &lt;= 12, a, b € z+} be a relation defined on z+)
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Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

The relation R is an equivalence if it is reflexive, symmetric and transitive.

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

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