Solve the function equation (x+4)P(x) = xP(x+1)?

2 Answers
Oct 3, 2017

See below.

Explanation:

Given p_n(x) = sum_(k=0)^n a_kx^k we have the general shift polynomial formula:

p_n(x+beta)=sum_(k=0)^n (p_n(beta)^((k)))/(k!)x^k where p_n(beta)^((k)) is the k derivative calculated at x=beta

then

(x+alpha)p_n(x) = x p_n(x+beta) rArr x sum_(k=0)^n a_kx^k+alpha sum_(k=0)^n a_kx^k = x sum_(k=0)^n (p_n(beta)^((k)))/(k!)x^k

or

sum_(k=0)^n(a_k-(p_n(beta)^((k)))/(k!))x^(k+1)+alpha sum_(k=0)^n a_kx^k =0

or

a_0=0
a_k-(p_n(beta)^((k)))/(k!)+alpha a_(k+1)=0

so using this recurrence formula we can compute the due a_k coefficients such that

(x+alpha)p_n(x) = x p_n(x+beta)

Oct 4, 2017

P(x) = Kx(x+1)(x+2)(x+3) where K is a constant

Explanation:

We have a functional equation:

(x+4)P(x) = xP(x+1) ..... [A]

We can immediately find two roots of P(x), by direct substitution into [A]:

Put x=0 => 4P(0))=0
Put x=-4 => 0 = -4P(-3)

So. P(x) has roots at x=0 and x=-3, hence by the factor theorem, P(x) has factors of (x) and (x+3), thus we can write P(x) in the form:

P(x) = x(x+3)P_1(x) , say, where deg(P_1) lt deg(P)

Substitute this form of P(x) into the original relationship [A] we get:

(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 P_1(x), by direct substitution into [B]:

Put x=-1 => 2P_1(-1) = 0
Put x=-3 =>0 = -2P_1(-2)

So. P_1(x) has roots at x=-1 and x=-2, hence by the factor theorem, P_1(x) has factors of (x+1) and (x+2), thus we can write P_1(x) in the form:

P_1(x) = (x+1)(x+2)P_2(x) , say, where deg(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 P(x) into the original relationship [A] we get:

(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 P_2(x+1) = P_2(x), from which we establish via direct substitution that:

Put x=0 => P_2(1) = P_2(0) = K, Say
Put x=1 => P_2(2) = P_2(1) = K
Put x=2 => P_2(3) = P_2(2) = K
\ \ \ \ \ vdots
Put x=n => P_2(n) = K \ AA n in NN

And if we apply the MVT we can see that either P_2(x) has an infinite number of turning points, or is a constant function. We also know that deg(P_2) lt deg(P_1) lt deg(P), and so it must be that P_2(x) = K, a constant

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
Put x=1 => P(1) = K.1.2.3.4 = 24 K
Put x=2 => P(2) = K.2.3.4.5 = 120K
Put x=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
Put x=2 => 6P(2) = 2P(3) , which holds true