The Iterative Degree

Algebra Level 5

If P P is a polynomial of degree 2 2 , and if f ( x ) f(x) is a function of degree 4 4 , the degree of f ( 19 ) ( P f ( 19 ) ( P f^{(19)}(P \cdot f^{(19)}(P )) can be expressed in the form a b a^{b} + + a c a^{c} , where a a , b b , and c c are coprime, positive integers. Find a + b + c a+b+c .

Note : The notation f ( n ) ( x ) f^{(n)}(x) means that f ( x ) f(x) is iterated , or repeatedly carried out, n n times. For example, f ( 3 ) ( x ) f^{(3)}(x) means f ( f ( f ( x ) ) ) f(f(f(x))) .


The answer is 118.

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1 solution

Rhoy Omega
May 2, 2014

Instead of letting P P = a 1 x 2 + b 1 x + c 1 a_{1}x^{2} + b_{1}x + c_{1} and f ( x ) f(x) = a 2 x 4 + b 2 x 3 + c 2 x 2 + d x + e a_{2}x^{4} + b_{2}x^{3} + c_{2}x^{2} + dx + e , we could let P P = x 2 x^{2} and f ( x ) f(x) = x 4 x^{4} , since we only want to find the degree after they are manipulated.

Now,

f ( x ) f(x) = x 4 x^{4}

f ( 2 ) ( x ) f^{(2)}(x) = ( x 4 ) 4 (x^4)^4 = x 4 2 x^{4^2}

f ( 3 ) ( x ) f^{(3)}(x) = ( x 4 2 ) 4 (x^{4^2})^4 = x 4 3 x^{4^3}

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f ( n ) ( x ) f^{(n)}(x) = x 4 n x^{4^n} , and this is what we will use to solve the problem.

Here is the one we have to solve:

f ( 19 ) ( P f ( 19 ) ( P f^{(19)}(P \cdot f^{(19)}(P ))

= f ( 19 ) ( x 2 f ( 19 ) ( x 2 f^{(19)}(x^2 \cdot f^{(19)}(x^2 ))

= f ( 19 ) ( x 2 ( x 2 ) ( 4 19 ) ) f^{(19)}(x^2 \cdot (x^2)^{(4^{19})})

= f ( 19 ) ( x 2 ( x 2 ) ( 2 38 ) ) f^{(19)}(x^2 \cdot (x^2)^{(2^{38})})

= f ( 19 ) ( x 2 ( x 2 39 ) ) f^{(19)}(x^2 \cdot (x^{2^{39}}))

= f ( 19 ) ( x 2 + 2 39 ) f^{(19)}(x^{2 + 2^{39}})

= ( x 2 + 2 39 ) 4 19 (x^{2 + 2^{39}})^{4^{19}}

= ( x 2 + 2 39 ) 2 38 (x^{2 + 2^{39}})^{2^{38}}

= x 2 39 + 2 77 x^{2^{39} + 2^{77}}

Thus, a = 2 a = 2 , b = 39 b = 39 , c = 77 c = 77 , and a + b + c a + b + c = = 2 + 39 + 77 2 + 39 + 77 = = 118 \boxed{118}

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