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 March 6th, 2015, 06:11 AM #1 Newbie   Joined: Mar 2015 From: Lubbock, Texas Posts: 5 Thanks: 0 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.
 March 6th, 2015, 06:55 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra $\mathbb P _3$ 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 $r(x) = 1 + 2x + 3x^2$.
 March 6th, 2015, 08:07 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,941 Thanks: 2210 For (C), use e^x = cosh(x) + sinh(x) and e^(-x) = cosh(x) - sinh(x).
March 14th, 2015, 10:06 AM   #4
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Quote:
 Originally Posted by v8archie $\mathbb P _3$ 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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