f-x-is-a-polynomial-such-that-f-x-2-f-x-f-x-1-Find-f-x-such-that-f-x-0-

Question Number 830 by prakash jain last updated on 20/Mar/15

f (x) is a polynomial such that

f (x^{2}) = f (x) f (x - 1)

Find f (x) such that f (x) \neq 0 .

Commented by 123456 last updated on 23/Mar/15

f (x) = \underset{i = 0}{\sum^{n}} a_{i} x^{i}

f (x^{2}) = \underset{i = 0}{\sum^{n}} a_{i} x^{2 i} \equiv \underset{i = 0}{\sum^{2 n}} b_{i} x^{i}

f (x - 1) = \underset{i = 0}{\sum^{n}} a_{i} {(x - 1)}^{i} = \underset{i = 0}{\sum^{n}} c_{i} x^{i}

f (x) f (x - 1) = \underset{i = 0}{\sum^{n}} a_{i} x^{i} \underset{i = 0}{\sum^{n}} c_{i} x^{i} = \underset{i = 0}{\sum^{2 n}} d_{i} x^{i}

b_{i} = {\begin{cases} a_{i / 2} & i \equiv 0 (\mod 2) \\ 0 & i \equiv 1 (\mod 2) \end{cases}

{(x - 1)}^{i} = \underset{j = 0}{\sum^{i}} (\begin{matrix} i \\ j \end{matrix}) x^{j} {(- 1)}^{i - j}

\underset{i = 0}{\sum^{n}} a_{i} {(x - 1)}^{i} = \underset{i = 0}{\sum^{n}} a_{i} \underset{j = 0}{\sum^{i}} (\begin{matrix} i \\ j \end{matrix}) x^{j} {(- 1)}^{i - j} = \underset{i = 0}{\sum^{n}} c_{i} x^{i}

c_{i} \overset{?}{=} \underset{j = i}{\sum^{n}} (\begin{matrix} j \\ i \end{matrix}) {(- 1)}^{j - i} a_{j}

d_{i} = \underset{j = 0}{\sum^{i}} a_{j} c_{i - j}

d_{i} = b_{i}

Commented by 123456 last updated on 23/Mar/15

f (x) = a x + b

f (x - 1) = a (x - 1) + b = a x + (b - a)

f (x^{2}) = {a x}^{2} + b

{a x}^{2} + b = a^{2} x^{2} + a (2 b - a) x + b (b - a)

{\begin{cases} a = a^{2} & a = 0 \lor a = 1 \\ 0 = a (2 b - a) & a = 0 \lor 2 b = a \\ b = (b - a) b & b = 0 \lor b - a = 1 \end{cases}

a = 0 \Rightarrow {\begin{cases} 0 = 0 \\ 0 = 0 \\ b = b^{2} & b = 0 \lor b = 1 \end{cases}

a = 1 \Rightarrow {\begin{cases} 1 = 1 \\ 0 = 2 b - 1 & b = \frac{1}{2} \\ b = (b - 1) b & b = 0 \lor b = 2 \end{cases}

a = 2 b \Rightarrow {\begin{cases} 2 b = 4 b^{2} & b = 0 \lor b = \frac{1}{2} \\ 0 = 0 \\ b = - b^{2} & b = 0 \lor b = - 1 \end{cases}

a = b - 1 \Rightarrow {\begin{cases} b - 1 = {(b - 1)}^{2} & b = 1 \lor b = 2 \\ 0 = (b - 1) (b + 1) & b = - 1 \lor b = 1 \\ b = b \end{cases}

f (x) = 0 \lor f (x) = 1

Commented by 123456 last updated on 23/Mar/15

f (x) = {a x}^{2} + b x + c

f (x - 1) = a {(x - 1)}^{2} + b (x - 1) + c

= {a x}^{2} + (b - 2 a) x + (a - b + c)

f (x^{2}) = {a x}^{4} + {b x}^{2} + c

{\begin{cases} a = a^{2} \\ 0 = 2 a (b - a) \\ b = 2 a c + b^{2} - 3 a b + a^{2} \\ 0 = 2 (b - a) c - b^{2} + a b \\ c = c (a + c - b) \end{cases}

f (x) = 0 \lor f (x) = 1 \lor f (x) = x^{2} + x + 1

Commented by 123456 last updated on 23/Mar/15

suppose that f_{1} (x) is a solution and

f (x) = f_{1} (x) + f_{2} (x) is also a solution

f (x^{2}) = f (x) f (x - 1)

f_{1} (x^{2}) + f_{2} (x^{2}) = [f_{1} (x) + f_{2} (x)] [f_{1} (x - 1) + f_{2} (x - 1)]

= f_{1} (x) f_{1} (x - 1) + f_{1} (x) f_{2} (x - 1) + f_{2} (x) f_{1} (x - 1) + f_{2} (x) f_{2} (x - 1)

f_{2} (x^{2}) = f_{1} (x) f_{2} (x - 1) + f_{2} (x) f_{1} (x - 1) + f_{2} (x) f_{2} (x - 1)

f_{1} (x) = 0 \Rightarrow f_{2} (x^{2}) = f_{2} (x) f_{2} (x - 1)

f_{1} (x) = 1 \Rightarrow f_{2} (x^{2}) = f_{2} (x - 1) + f_{2} (x) + f_{2} (x - 1) f_{2} (x - 1)

Commented by prakash jain last updated on 23/Mar/15

Thanks . Then we can generalize the solution to

f (x) = {(x^{2} + x + 1)}^{n} where n \in N

Since raising to any power {won}^{'} t make

any difference in equality .

Commented by prakash jain last updated on 23/Mar/15

x^{4} + x^{2} + 1 = (x^{2} + x + 1) (x^{2} - 2 x + 1 + x - 1 + 1)

= (x^{2} + x + 1) (x^{2} - x + 1)