Solve the function equation (x+4)P(x) = xP(x+1)?
2 Answers
See below.
Explanation:
Given
then
or
or
so using this recurrence formula we can compute the due
P(x) = Kx(x+1)(x+2)(x+3) whereK is a constant
Explanation:
We have a functional equation:
(x+4)P(x) = xP(x+1) ..... [A]
We can immediately find two roots of
Put
x=0 => 4P(0))=0
Putx=-4 => 0 = -4P(-3)
So.
P(x) = x(x+3)P_1(x) , say, wheredeg(P_1) lt deg(P)
Substitute this form of
(x+4){x(x+3)P_1(x)} = x{(x+1)(x+4)P_1(x+1)}
:. (x+3)P_1(x) = (x+1)P_1(x+1) ..... [B]
As before, we can immediately find two roots of
Put
x=-1 => 2P_1(-1) = 0
Putx=-3 =>0 = -2P_1(-2)
So.
P_1(x) = (x+1)(x+2)P_2(x) , say, wheredeg(P_2) lt deg(P_1)
We now have:
P(x) = x(x+3)P_1(x)
\ \ \ \ \ \ \ \ = x(x+3)(x+1)(x+2)P_2(x)
\ \ \ \ \ \ \ \ = x(x+1)(x+2)(x+3)P_2(x) ..... [C]
Substitute this form of
(x+4){x(x+1)(x+2)(x+3)P_2(x)} = x{(x+1)(x+2)(x+3)(x+4)P_2(x+1) }
:. P_2(x) = P_2(x+1)
We now have a new functional equation:
P_2(x) = P_2(x+1)
Or equally
Put
x=0 => P_2(1) = P_2(0) = K, Say
Putx=1 => P_2(2) = P_2(1) = K
Putx=2 => P_2(3) = P_2(2) = K
\ \ \ \ \ vdots
Putx=n => P_2(n) = K \ AA n in NN
And if we apply the MVT we can see that either
Hence, we can deduce from, [C] that:
P(x) = Kx(x+1)(x+2)(x+3)
Quick Validation:
By direct substitution into the above derived result we find that:
Put
x=0 => P(0) = 0
Putx=1 => P(1) = K.1.2.3.4 = 24 K
Putx=2 => P(2) = K.2.3.4.5 = 120K
Putx=3 => P(3) = K.3.4.5.6 = 360K
And from the functional equation we find that:
Put
x=1 => 5P(1) = P(2) \ \ \ , which holds true
Putx=2 => 6P(2) = 2P(3) , which holds true