Just a couple of polynomials
P and Q are two distinct, non-constant polynomial functions such that ( P ∘ Q ) ( x ) = P ( x ) Q ( x ) and P ( 1 ) = P ( − 1 ) = 2 0 1 8 .
Find ( Q ∘ P ) ( − 1 ) .
Note: Function composition ( P ∘ Q ) ( x ) is sometimes denoted as P ( Q ( x ) ) .
The answer is 4072324.
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5 solutions
Thanks for your solution Mark and for reporting a lack of precision in the first draft (namely that P ( 1 ) = 2 0 1 8 could have led to an infinite number of solutions.) Apologies for those who have been misled.
This solution can be simplified: because Q is nonconstant, it has at least one complex zero. Substituting this zero into the given identity yields P(0) = 0, and we can then easily fit a quadratic P to the three points (-1, 2018), (0, 0), and (1, 2018).
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You are assuming that if the property holds true on all real numbers x , it must also hold true for complex x . This is true for polynomial functions, but not trivial.
When I first solved the problem, the boundary conditions were incorrect, and needed to be cleared up. I therefore solved the equation P ∘ Q = P Q in complete generality, and then had some discussion with Romain to determine which solution was wanted.
Your observation about a zero of Q is an alternative way to show that c = 0 , which was the first important deduction of my proof.
Could someone explain the (m-1)(n-1)=1 step. Thanks in advance!
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m n = m + n m n − m − n = 0 m n − m − n + 1 = 1 ( m − 1 ) ( n − 1 ) = 1 The factors must be whole numbers and non-negative. Leaving ( m − 1 ) ( n − 1 ) = 1 ⋅ 1 .
Can we say that P(x) is a linear function as P(-1)=p(1)??
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No. P ( x ) = x + 2 (for example) but does not satisfy P ( 1 ) = P ( − 1 ) . Since P is quadratic, we can say that P has no term in x (so that b = 0 ).
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And sir I had another doubt that why is it so that mn=m+n??
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@Erica Phillips – m n is the degree of P ∘ Q , while m + n is the degree of P Q .
Couldn’t understand the step “which implies that c = 0”. How did you arrive at that conclusion from the line above it? P(x)Q(x) can have a constant component.
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Because on the right side of the equation there is no constant term which means that c=0.
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Is it necessarily true? Since we know the polynomials are quadratic, let P ( x ) = a 1 x 2 + b 1 x + c 1 and Q ( x ) = a 2 x 2 + b 2 x + c 2 . Multiplying them would result in the constant c 1 c 2 , which may or may not be zero. I suppose I am missing something here. Could you point me to it?
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@Venkatesh G – Over here when I meant constant on the left side of the equation I meant constants of a polynomial where x=Q(x) hence x^2 term in a polynomial is replaced by (Q(x))^2 and on the other hand there are no coefficients of (Q(x))^2 and the only coefficient of Q(x) is P(x),with no constants finally added to the right side of the equation where x=Q(x) in the standard equation ax^2+bx+c. Hope this helps!!!!!If u have further doubts feel free to ask!!!
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@Erica Phillips – It's not necessarily true. According to your logic, since Q ( x ) 2 doesn't have a term on the right-hand side, a (the coefficient of Q ( x ) 2 on the left-hand side) should be zero too. I suppose your method was right but the thought process was a bit off.
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@Venkatesh G – Note that a Q ( x ) 2 + b Q ( x ) and P ( x ) Q ( x ) are both multiples of Q ( x ) . This means that c must also be a multiple of Q ( x ) . The only way this can happen is when c = 0 .
The equation implies that de g P ⋅ de g Q = de g P + de g Q , with only possible solutions de g P = de g Q = 2 . Thus both functions are quadratic.
Since P ( 1 ) = P ( − 1 ) , the function P must be symmetric about x = 0 , hence of the form P ( x ) = a x 2 + b with a + b = 2 0 1 8 . The equation reduces to a Q ( x ) 2 + b = ( a x 2 + b ) Q ( x ) , a Q ( x ) + Q ( x ) b = a x 2 + b . Since the right side is a polynomial, so is the left side; therefore b = 0 . This leaves (after division by a ) Q ( x ) = x 2 so that Q ( P ( − 1 ) ) = Q ( 2 0 1 8 ) = 2 0 1 8 2 = 4 0 7 2 3 2 4 .
why does b=0? couldn't this be allowed if Q(x) had constant term also?
Could someone explain why it can draw a conclusion that degP·degQ=degP+degQ?
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If f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 0 , g ( x ) = b m x m + b m − 1 x m − 1 + ⋯ + b 0 then f ( x ) ⋅ g ( x ) = a n b m x n + m + ( a n b m − 1 + a n − 1 b m ) x n + m − 1 + ⋯ + a 0 b 0 f ( g ( x ) ) = a n ( b m x m + ⋯ ) n + ⋯ + a 0 = a n b m x m ⋅ n + ⋯ Thus the highest powers of x are x n + m for f ⋅ g , and x n ⋅ m for f ∘ g .
Using the "deg" notation, we find de g f = n , de g g = m de g ( f ⋅ g ) = n + m = de g f + de g g , de g ( f ∘ g ) = n ⋅ m = de g f ⋅ de g g . Finally, if ( f ⋅ g ) ( x ) = ( f ∘ g ) ( x ) then de g f + de g g = de g f ⋅ de g g .
if the right side is a polynomial for x, why does the left side have to be a polynomial for q(x) ?
The other solutions given are more holistic and actually find f(x), but for this particular question that is not reqd and takes 5 secs as the value of f(x) for x=-1 is already given in the question. So Q(P(-1)) = Q(-1)P(-1) = 2018*2018 = 4072324
QED
How did you figure out the value Q ( − 1 ) though ?
This solution requires to demonstrate that P(Q(x))=Q(P(x)).
The value of Q ( − 1 ) is not given explicitly in the question.
As you can see from the 'holistic' solutions, Q(x) = x^2 so Q(-1) = 1, not 2018. You have assumed Q(P(x)) = P(Q(x)) too. In this case, however, two wrongs do make a right.
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Why Not Q(x)=-x And P(x)=2018x2 so the answer is2018
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L=P(Q(x))=2018x^2 R=P(x)Q(x)=-2018x^3 L is not equal to R which violates the given condition
I’m glad that you are good at guessing, but you did not solve the problem. You misread it.
Wow, 5 upvotes. This is completely wrong.
What subject (in math) would I have to study in order to learn how to do problems such as these?
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I guess linear algebra and more specifically polynomials.
In time-limited competition contexts, sometimes it is worth assuming that the question has an answer. It isn't quite as satisfying, since you are building an argument conditional on an unproven premise. It can unlock questions quickly, though, as follows:
If Q ( P ( − 1 ) has a value, then any P ( x ) and Q ( x ) that meet the criteria should do the trick.
P ( x ) = 2 0 1 8 x 2 and Q ( x ) = x 2 meet the criteria, since P ( − 1 ) = P ( 1 ) = 2 0 1 8 and P ( Q ( x ) ) = 2 0 1 8 x 4 = 2 0 1 8 x 2 × x 2 .
Using these, Q ( P ( − 1 ) ) = Q ( 2 0 1 8 ) = 2 0 1 8 2 = 4 0 7 2 3 2 4
**PοQ(x)=P(x)×Q(x)----(1)
let Q(x)=1
P(1)=P(x)
P(x)=2018
∵P(±1)=2018
By comparison,
∴x=±1
Substitute x=±1 ,Q(x)=1 into (1)
P(1)=P(1)×Q(±1)
Q(±1)=1
∴Q(x)=x^n (n must be even number)
degPοQ=degP+degQ
degP×degQ=degP+degQ
∴The answer is degP=degQ=2
∴function P and Q is quadratic function
∴Q(x)=x^{2}
QοP(-1)=Q(2018)
=2018^{2}
=4072324**
Your arguments are not rigourous - or always accurate - here. With a little re-arrangement, you can prove that if there is an x such that Q(x) = 1 then x=±1 but you then assume that Q(-1) = 1 and Q(1) = 1, when all you have proved is that one or other is the case. Even if Q(±1)=1 is true (which turns out to be the case), your next assumption is wrong since, for example, Q(x) = ax^2 + 1 - a also satisfies the condition.
We prove that the vale of q(x)=ax2+bx+c and p(x)=a1x2+b1x+C1 So The Vale Of aa1=0 c1=0 b=0 c=0 and the value of b1=-1
If P ∘ Q = P Q , and if P , Q have degree m , n respectively, then m n = m + n , so that ( m − 1 ) ( n − 1 ) = 1 , and hence (since P , Q are not constant) m = n = 2 . If we write P ( x ) = a x 2 + b x + c , then a Q ( x ) 2 + b Q ( x ) + c = ( P ∘ Q ) ( x ) = P ( x ) Q ( x ) which implies that c = 0 , and that a Q ( x ) + b = P ( x ) . Thus we have P ( x ) = a x 2 + b x Q ( x ) = a a x 2 + b x − b for a = 0 , and these polynomials satisfy the conditions, and are distinct, unless a = 1 and b = 0 .
Since P ( 1 ) = P ( − 1 ) = 2 0 1 8 , we deduce that a = 2 0 1 8 and b = 0 , so that P ( x ) = 2 0 1 8 x 2 and Q ( x ) = x 2 . Thus ( Q ∘ P ) ( − 1 ) = Q ( 2 0 1 8 ) = 2 0 1 8 2 = 4 0 7 2 3 2 4 .