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March 6th, 2015, 06:11 AM   #1
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3 parts

(A) Is

{x(x-1), (2x - 1)^2, 1}

a spanning set for P3?

(B) Are the matrices

1 1
1 1

0 -1
1 0

2 0
0 2

0 1
-1 -1

linearly independent?

(C) It can be shown that the solutions to the differential equation y'' = y are exactly all functions of the form y = ae^x + be^-x, for constants a and b. These solutions form a vector space. Show that

cosh(x) = (e^x + e^-x)/2 and sinh(x) = (e^x - e^-x)/2 form a basis for this vector space.
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March 6th, 2015, 06:55 AM   #2
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$\mathbb P _3$ is the set of all polynomials $$p(x) = \sum_{k = 0}^3 a_k x^k$$
Since your proposed basis has no terms in $x^3$ it is clearly not a basis of $\mathbb P_3$.

It might be a basis of $\mathbb P _2$, the quadratic polynomials. If it is, then any quadratic $p(x) = a_0 + a_1 x + a_2 x^2$ where the $a_k$ are not all zero can be written as a non-zero linear combination of the proposed basis vectors.

Writing $b_0(x) = 1$, $b_1(x) = (2x-1)^2 = 4x^2 -4x +1$ and $b_2(x) = x(x-1) = x^2 - x$ we see that any vector $$p b_0(x) + q b_1(x) + r b_2(x) = \big(p\big) + \big(4qx^2 - 4qx + q\big) + \big(rx^2 - rx\big) = (p + 4q) - (4q + r)x + (4q + r)x^2$$
In particular, any linear combination of the $b_k(x)$ is going to have the coefficient of $x$ being equal to minus the coefficient of $x^2$. That is $p(x) = a - bx + bx^2$. So, for example, we can't make $q(x) = 1 + x + x^2$ or $r(x) = 1 + 2x + 3x^2$.
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March 6th, 2015, 08:07 AM   #3
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For (C), use e^x = cosh(x) + sinh(x) and e^(-x) = cosh(x) - sinh(x).
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March 14th, 2015, 10:06 AM   #4
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Quote:
Originally Posted by v8archie View Post
$\mathbb P _3$ is the set of all polynomials $$p(x) = \sum_{k = 0}^3 a_k x^k$$
I agree but- some people use "P3" to mean the space of quadratic polynomials. The "3" because it is a vector space of dimension 3.

Quote:
Since your proposed basis has no terms in $x^3$ it is clearly not a basis of $\mathbb P_3$.

It might be a basis of $\mathbb P _2$, the quadratic polynomials. If it is, then any quadratic $p(x) = a_0 + a_1 x + a_2 x^2$ where the $a_k$ are not all zero can be written as a non-zero linear combination of the proposed basis vectors.

Writing $b_0(x) = 1$, $b_1(x) = (2x-1)^2 = 4x^2 -4x +1$ and $b_2(x) = x(x-1) = x^2 - x$ we see that any vector $$p b_0(x) + q b_1(x) + r b_2(x) = \big(p\big) + \big(4qx^2 - 4qx + q\big) + \big(rx^2 - rx\big) = (p + 4q) - (4q + r)x + (4q + r)x^2$$
In particular, any linear combination of the $b_k(x)$ is going to have the coefficient of $x$ being equal to minus the coefficient of $x^2$. That is $p(x) = a - bx + bx^2$. So, for example, we can't make $q(x) = 1 + x + x^2$ or $r(x) = 1 + 2x + 3x^2$.
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