Polynomial Functions: Algebra, Wronskian Method, Vectors (Research Paper Sample)

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Linear algebra
Question One
For the set of vectors of Polynomials [ 1+2x-3x2, 4-2x-5x2, -4+ 12x- 2x2] in the space of the same polynomials for all x€│R, we shall have the following if we multiply the Polynomials by the constants, a, b, and c: a(1+2x-3x2)+b(4-2x-5x2) +c(-4+ 12x- 2x2)= 0
Then; a+ 2ax-3ax2 + 4b-2bx-5bx2 – 4c + 12cx – 2cx2 = 0
= - (3a+5b+2c) x2 + (2a-2b+12c) x + a+4b-4c = 0
From the equations above;
- 3a -5b-2c = 0……….i
2a-2b+12c =0……….ii
a + 4b-4c =0………iii
from eqn iii a= 4c-4b
sub in i. -3(4c-4b)-5b-2c= 0
also in ii. 2(4c-4b)-2b+12c =0
then 12c+ 12b-5b-2c= 0
8c-8b-2b+ 12c =0
7b-14c =0
-10b+20c=0
b-2c =0
b= 2, c=1, a=-4
thus a≠ b ≠c≠ 0 and hence the set is linearly dependent
Using Wronskian method
W= 1+ 2x-3x2 4-2x-5x2 -4+12x-2x2
2-6x -2-10x 12-4x
-6 -10 -4
-6 4-2x-5x2 -4+12x-2x2 + 10 1+ 2x-3x2 4-2x-5x2 + -4 1+ 2x-3x2 4-2x-5x2
-2-10x 12-4x 2-6x 12-4x 2-6x -2-10x

-6[(12-4x )(4-2x-5x2 )+ (2-10x)( -4+12x-2x2)] 10[(12-4x)( 1+ 2x-3x2 )-( 2-6x)( 4-2x-5x2)-4[(-2-10x)( 1+ 2x-3x2 )-( 2-6x)( 4-2x-5x2)= 0
It is possible to solve the above polynomial function
b). w= 1 sin 2 x/2 cosx
0 2-1/2(sin x/2) cosx/2 -sinx
 0 cosx/2 -cosx
= -cos x/2 1 -cosx - cosx 1 sin 2
0 -sin x 0 sin x/2 cos x/2

= (cosx sin x)/2 – cosx (sinx/2 cosx/2) =0
This implies that the set is linearly dependent
c). w= cosx cos 2x cos 3x
-sinx -2sinx -3sin3x
-cosx -4cos2x -9cos3x
 So that w = cos x -2sinx -3sin 3x -cos 2x -sin x -3sin3x
-4cos2x -9cos3x -cosx -9cos 3x
+ cos3x -sin x -2sin2x
cosx -4cos2x
Which breaks to cosx[18cos3xsinx- 12cos2xsin3x]- cos2x[9cos3xsinx-cosx3sin3x]+ cos3x[4cos2xsinx+cosx2sin2x] ≠0, hence linearly dependent.
Question Two
Let a,b, c represent elements of this set where P1= 1-2x-2x2, P2= -2+3x-x2, and P3= 1-x+6x2 so that;
a(1-2x-2x2) + b(-2+3x-x2) + c(1-x+6x2 ) =0
Factoring out the like terms becomes;
x2(-2a-b-6c) + x(-2a+3-c)+ a-2b+c =0
-2a-b-6c=0………..i
-2a+3-c= o……&...