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

2 Answers
Cesareo R.
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)#

Steve M
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

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