Show that the functions (c1+c2sin^2x+c3cos^2x) form a vector space. Find a basis of it. What is its dimension?

To prove that a set of functions forms a vector space, we need to show that it satisfies the ten properties of a vector space:

1. Closure under addition: Given any two functions in the set, their sum will also be in the set.
2. Closure under scalar multiplication: Given any function in the set, multiplying it by any scalar will also give us a function in the set.
3. Associativity of addition: The sum of functions is associative.
4. Commutativity of addition: The order of addition does not affect the result.
5. Identity element of addition: There exists an identity element such that adding it to any function does not change the function.
6. Existence of inverse elements for addition: For every function, there exists an additive inverse such that adding it cancels out the function.
7. Compatibility of scalar multiplication with field multiplication: Scalar multiplication distributes over field multiplication.
8. Identity element of scalar multiplication: There exists an identity element such that multiplying any function by it does not change the function.
9. Compatibility of scalar multiplication with vector addition: Scalar multiplication distributes over vector addition.
10. Associativity of scalar multiplication: Scalar multiplication is associative.

Let's go through each property to prove that the set of functions (c1 + c2*sin^2(x) + c3*cos^2(x)) forms a vector space:

1. Closure under addition:
Let f1(x) = c1 + c2*sin^2(x) + c3*cos^2(x)
Let f2(x) = c1' + c2'*sin^2(x) + c3'*cos^2(x)

Adding f1 and f2 gives us:
f1(x) + f2(x) = (c1 + c1') + (c2 + c2')*sin^2(x) + (c3 + c3')*cos^2(x)

The sum is also in the form of (c1 + c2*sin^2(x) + c3*cos^2(x)), which is in our set of functions. Therefore, closure under addition is satisfied.

2. Closure under scalar multiplication:
Let f(x) = c1 + c2*sin^2(x) + c3*cos^2(x)
Let a be a scalar

Multiplying f by a gives us:
a * f(x) = a * (c1 + c2*sin^2(x) + c3*cos^2(x)) = (a*c1) + (a*c2)*sin^2(x) + (a*c3)*cos^2(x)

This is also in the form of (c1 + c2*sin^2(x) + c3*cos^2(x)), which is in our set of functions. Therefore, closure under scalar multiplication is satisfied.

3. Associativity of addition:
The addition of functions is associative by the property of real numbers. Therefore, this property is satisfied.

4. Commutativity of addition:
The addition of functions is commutative by the property of real numbers. Therefore, this property is satisfied.

5. Identity element of addition:
The identity element for addition is the function f(x) = 0, which is in the form of (c1 + c2*sin^2(x) + c3*cos^2(x)). Therefore, this property is satisfied.

6. Existence of inverse elements for addition:
The inverse element for addition would be the function -f(x) = -c1 - c2*sin^2(x) - c3*cos^2(x), which is in the form of (c1 + c2*sin^2(x) + c3*cos^2(x)). Therefore, this property is satisfied.

7. Compatibility of scalar multiplication with field multiplication:
Scalar multiplication distributes over field multiplication, which is already satisfied by the trigonometric functions and the real numbers. Therefore, this property is satisfied.

Properties 8, 9, and 10 are also satisfied since scalar multiplication and vector addition follow the properties of real numbers, which already satisfy these properties.

Hence, the set of functions (c1 + c2*sin^2(x) + c3*cos^2(x)) forms a vector space.

To find a basis of this vector space, we need to find a set of functions that are linearly independent and span the entire space.

Let's start by considering the coefficients c1, c2, and c3 as variables. The basis functions will be the functions that form a linearly independent system for these variables. In this case, our basis functions can be chosen as:

1. f1(x) = 1
2. f2(x) = sin^2(x)
3. f3(x) = cos^2(x)

These functions are linearly independent, which means that no linear combination of them can be equal to zero unless all the coefficients are zero.

The dimension of this vector space is the number of basis functions, which is 3 in this case.

Therefore, a basis for the vector space of functions (c1 + c2*sin^2(x) + c3*cos^2(x)) is {1, sin^2(x), cos^2(x)}, and its dimension is 3.

To show that the given functions form a vector space, we need to verify that they satisfy the three defining properties of a vector space: closure under addition, closure under scalar multiplication, and the presence of an additive identity element.

1. Closure under Addition:
Let f(x) = c1 + c2sin^2(x) + c3cos^2(x) and g(x) = d1 + d2sin^2(x) + d3cos^2(x) be two arbitrary functions in the given set. Now, consider their sum:

f(x) + g(x) = (c1 + c2 sin^2(x) + c3 cos^2(x)) + (d1 + d2 sin^2(x) + d3 cos^2(x))
= (c1 + d1) + (c2 + d2) sin^2(x) + (c3 + d3) cos^2(x)

This new function is also in the given set since it can be expressed in the same form as c1 + c2sin^2(x) + c3cos^2(x) with different coefficients c1, c2, and c3. Thus, the set is closed under addition.

2. Closure under Scalar Multiplication:
Let f(x) = c1 + c2sin^2(x) + c3cos^2(x) be a function in the given set, and let r be a scalar. Consider the scalar multiplication:

r * f(x) = r * (c1 + c2sin^2(x) + c3cos^2(x))
= (r * c1) + (r * c2) sin^2(x) + (r * c3) cos^2(x)

This new function is also in the given set since it can be expressed in the same form as c1 + c2sin^2(x) + c3cos^2(x) with different coefficients c1, c2, and c3. Therefore, the set is closed under scalar multiplication.

3. Additive Identity:
The function with coefficients c1 = 0, c2 = 0, and c3 = 0 is the additive identity element. Let it be denoted by 0(x). For any function f(x) in the given set:

f(x) + 0(x) = (c1 + c2 sin^2(x) + c3 cos^2(x)) + (0 + 0) sin^2(x) + (0 + 0) cos^2(x)
= c1 + c2 sin^2(x) + c3 cos^2(x) = f(x)

So, the additive identity is present in the set.

Since all three properties are satisfied, the given set forms a vector space.

To find a basis of this vector space, we need to find a set of functions that are linearly independent and span the vector space.

In this case, one possible basis is {1, sin^2(x), cos^2(x)}.

To show that they are linearly independent, we need to prove that no nontrivial combination of these functions can equal the zero function. Suppose:

a * 1 + b * sin^2(x) + c * cos^2(x) = 0

This equation holds for all x if and only if a = b = c = 0. Hence, the functions {1, sin^2(x), cos^2(x)} are linearly independent.

Since the basis consists of three functions, the dimension of the vector space is 3.

Well, if you're looking to show that the functions (c1+c2sin^2x+c3cos^2x) form a vector space, you'll probably want to start by checking the three vector space axioms: closure under addition, closure under scalar multiplication, and the existence of an additive identity.

Closure under addition means that if we take any two functions from this set, their sum should also be in the set. In this case, if we take two functions f(x) = c1 + c2sin^2x + c3cos^2x and g(x) = k1 + k2sin^2x + k3cos^2x, their sum is (c1+k1) + (c2+k2)sin^2x + (c3+k3)cos^2x. Since the sum is still in the form (c1+c2sin^2x+c3cos^2x), this set is indeed closed under addition.

Closure under scalar multiplication means that if we take any function from the set and multiply it by a scalar, the result should still be in the set. Using the function f(x) = c1 + c2sin^2x + c3cos^2x and a scalar a, the scalar multiple is af(x) = ac1 + ac2sin^2x + ac3cos^2x, which is still in the form (c1+c2sin^2x+c3cos^2x), so the set is closed under scalar multiplication.

The zero vector or additive identity is the function where all the coefficients c1, c2, and c3 are zero. If we take this function, we get 0 + 0sin^2x + 0cos^2x, which is simply the constant zero function. It is indeed in the set, so we have closure under addition.

So far, we've shown that this set is closed under addition and scalar multiplication, and it has an additive identity. But to determine the dimension and find a basis, we need to check one more thing: linear independence.

To show that the functions are linearly independent, we need to find a set of scalars c1, c2, and c3 such that the only solution to the equation c1 + c2sin^2x + c3cos^2x = 0 is when c1 = c2 = c3 = 0. If we take x = 0, we get c1 + c3 = 0. If we take x = π/2, we get c1 + c2 = 0. And if we take x = π, we get c1 + c3 = 0 again. From these equations, we can see that the only solution is c1 = c2 = c3 = 0. Therefore, the functions (c1+c2sin^2x+c3cos^2x) are indeed linearly independent.

Since we have shown that this set is closed under addition, scalar multiplication, has an additive identity, and is linearly independent, we can conclude that these functions form a vector space. As for the basis and dimension, a basis is a set of vectors that is linearly independent and spans the entire vector space.

In this case, our basis can be simply taken as {1, sin^2(x), cos^2(x)}. These three functions form a linearly independent set, and any function in the given set can be written as a linear combination of these three basis functions. Therefore, the dimension of this vector space is 3.

I hope that answers your question! If not, I should have just stuck with clown jokes.