The discussion focuses on finding an annihilator for the polynomial expression x² + 2x⁵. The participants confirm that the annihilator can be represented as D⁶, which effectively annihilates both x² and 2x⁵. The multiplication of individual annihilators, D³ for x² and D⁶ for 2x⁵, resulting in D¹⁸ is clarified, emphasizing that while multiple annihilators exist, the least common multiple is not necessary for this context. The key takeaway is that D⁶ is sufficient as an annihilator for the given polynomial.
PREREQUISITESStudents studying differential equations, mathematicians interested in operator theory, and educators teaching polynomial functions and their properties.
Doesn't the D6 operator also annihilate x2?magnesium12 said:Homework Statement
Find an annihilator of x2+2x5.
Homework Equations
xnecx -----> [D-c]n+1
[Q1(D)Q2(D)] = b1(x) + b2(x)
The Attempt at a Solution
[/B]
x2 -----> D3
2x5-----> D6
x2 + 2x5 ------> (D3)(D6) = D18
The answer should be D6.
Mark44 said:Doesn't the D6 operator also annihilate x2?
What does the last line mean? It's not clear to me.magnesium12 said:Homework Statement
Find an annihilator of x2+2x5.
Homework Equations
xnecx -----> [D-c]n+1
[Q1(D)Q2(D)] = b1(x) + b2(x)
Why are you multiplying the two annihilators?The Attempt at a Solution
[/B]
x2 -----> D3
2x5-----> D6
x2 + 2x5 ------> (D3)(D6) = D18
The answer should be D6.
Didn't notice this before -- is x3x6 = x18?magnesium12 said:(D3)(D6) = D18
Dn+1 annihilates xn and all lower powers of x. I don't see that this is related to the LCM in any way.magnesium12 said:I guess it does. So does that mean there can be multiple annihilators and they'll all work? Or should you always pick the least common multiple?
Heh...I didn't notice that either.Mark44 said:Didn't notice this before -- is x3x6 = x18?