Prove/Disprove: Existence of a Basis of P_3(F) w/ Degree 2 Polynomial

jaejoon89
Messages
187
Reaction score
0

Homework Statement



Prove or disprove: there exists a basis (p_0, p_1, p_2, p_3) of P_3 (F) such that one of the polynomials p_0, p_1, p_2, p_3 has degree 2.

Homework Equations



none really

The Attempt at a Solution



Is the following proof correct?

----

Let p_0, p_1, p_2, p_3 be elements of P_3(F) s.t.

p_o (x) = 1,
p_1 (x) = x,
p_2 (x) = x^2 + x^3,
p_3(x) = x^3.

None of the polynomials are degree 2 although (p_0,p_1,p_2,p_3) is clearly spanning P_3 (F) with dimP_3(F) = 4 and forms a basis. Hence proved.
 
Physics news on Phys.org
Looks good to me assuming the "one" in the statement of the problem should be "none".
 
I'm not sure that stating that it clearly spans it will suffice even if it is obvious. If you think that this suffices for you class, you're fine.

On the other hand, you could cook up a matrix that maps a degree three polynomial represented in the standard basis to it's representation in this basis pretty easily.
 
Typo: the original statement is supposed to be "none of the polynomials has degree 2." Thanks for pointing that out, LCKurtz.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top