Question

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

Answer 1

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)`

Answer 2

`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