Let V={f(x)=c0 + c1x + c2x2 : �ç01 f(x)dx=1}. In other words, V is the set of all polynomials of degree 2 or less such that their integral from 0-1 is equal to 1.
a)Show that the sum of two polynomials in V is not in V
b)Show that an affine combination of two polynomials in V is in V
c)Give a parametric description of the polynomials in V
a) To show that the sum of two polynomials in V is not in V, let's take two arbitrary polynomials f(x) and g(x) in V. We need to show that the integral of f(x) + g(x) from 0 to 1 is not equal to 1.
Let's express f(x) and g(x) as follows:
f(x) = c0 + c1x + c2x^2
g(x) = d0 + d1x + d2x^2
Now let's find the integral of f(x) + g(x) from 0 to 1:
∫[0,1] (f(x) + g(x)) dx
= ∫[0,1] (c0 + c1x + c2x^2 + d0 + d1x + d2x^2) dx
= ∫[0,1] (c0 + d0) dx + ∫[0,1] (c1 + d1)x dx + ∫[0,1] (c2 + d2)x^2 dx
Integrating each term, we get:
= (c0 + d0)(x) ∣[0,1] + (c1 + d1)(x^2/2) ∣[0,1] + (c2 + d2)(x^3/3) ∣[0,1]
= c0 + d0 + (c1 + d1)/2 + (c2 + d2)/3
Since c0, d0, (c1 + d1)/2, and (c2 + d2)/3 are constants, their sum will also be a constant. Therefore, the integral of f(x) + g(x) will not equal 1 in general.
b) To show that an affine combination of two polynomials in V is in V, let's take two arbitrary polynomials f(x) and g(x) in V, and consider their affine combination h(x) = af(x) + bg(x), where a and b are constants.
We need to show that the integral of h(x) from 0 to 1 is equal to 1.
Let's integrate h(x) from 0 to 1:
∫[0,1] h(x) dx
= ∫[0,1] (af(x) + bg(x)) dx
= a∫[0,1] f(x) dx + b∫[0,1] g(x) dx
Since f(x) and g(x) are in V and their integrals from 0 to 1 are both equal to 1, we have:
∫[0,1] h(x) dx
= a(1) + b(1)
= a + b
Since a and b are constants, their sum will also be a constant. Therefore, the integral of h(x) will be equal to 1, and h(x) is in V.
c) To give a parametric description of the polynomials in V, we can express each polynomial in V as a linear combination of a set of basis polynomials.
Let's consider the basis polynomials for V:
{1, x, x^2}
Any polynomial f(x) in V can be written as:
f(x) = c0(1) + c1(x) + c2(x^2)
Where c0, c1, and c2 are parameters that determine the specific polynomial in V.
Therefore, the polynomials in V can be parameterized by the values of c0, c1, and c2.
To answer these questions, we need to examine the properties of the set V and understand what it means for a polynomial to be in V.
a) Showing that the sum of two polynomials in V is not in V:
We know that the integral of a polynomial is linear with respect to addition. Therefore, if we have two polynomials f(x) and g(x) in V, their sum (f(x) + g(x)) will be in V if and only if the integral of (f(x) + g(x)) from 0 to 1 is equal to 1.
However, in general, the sum of two polynomials of degree 2 or less will result in a polynomial of degree 2 or less. The integral of this sum may not be equal to 1. Therefore, it is possible to find two polynomials in V whose sum is not in V.
b) Showing that an affine combination of two polynomials in V is in V:
An affine combination of two polynomials f(x) and g(x) is defined as h(x) = αf(x) + βg(x), where α and β are constants such that α + β = 1.
To show that h(x) is in V, we must prove that the integral of h(x) from 0 to 1 is equal to 1.
∫[0,1] (αf(x) + βg(x)) dx = α∫[0,1] f(x) dx + β∫[0,1] g(x) dx
Since f(x) and g(x) are in V, we know that their integrals from 0 to 1 are equal to 1. Therefore, α∫[0,1] f(x) dx + β∫[0,1] g(x) dx = α(1) + β(1) = α + β = 1.
Hence, h(x) is in V.
c) Giving a parametric description of the polynomials in V:
To describe the polynomials in V, we need to determine the values of the coefficients c0, c1, and c2.
Let's consider a general form of a polynomial in V: f(x) = c0 + c1x + c2x^2.
We know that the integral of f(x) from 0 to 1 is equal to 1:
∫[0,1] (c0 + c1x + c2x^2) dx = 1
Integrating this expression gives us:
c0x + c1/2x^2 + c2/3x^3 | [0,1] = 1
Plugging in the limits of integration, we have:
(c0 + c1/2 + c2/3) - (0 + 0 + 0) = 1
Simplifying, we get:
c0 + c1/2 + c2/3 = 1
This equation represents a constraint on the coefficients c0, c1, and c2 of the polynomials in V. By solving this equation, we can find the parametric description of the polynomials.
For example, let's express c2 in terms of c0 and c1:
c2 = 3(1 - c0 - c1/2)
By taking any values for c0 and c1 that satisfy the constraint, we can generate different polynomials in V.
Therefore, the parametric description of the polynomials in V is:
f(x) = c0 + c1x + 3(1 - c0 - c1/2)x^2, where c0 and c1 are arbitrary constants that satisfy the constraint c0 + c1/2 + c2/3 = 1.