Proving ƒ(x) is the Identity Function

In summary, the conversation discusses a former Putnam exam question about determining and proving the nature of a polynomial function ƒ with real coefficients that has the property ƒ(g(x)) = g(ƒ(x)) for all polynomial functions g with real coefficients. The attempt at a solution involves using a proof by contradiction to show that ƒ(x) must be the identity function, with a consideration of the polynomial function k(x) to demonstrate a contradiction. The first part of the attempt could be simplified.
  • #1
Derek Hart
14
1

Homework Statement


I have been going through a textbook trying to solve some of of these with somewhat formal proofs. This is a former Putnam exam question. (Seemingly the easiest one I have attempted, which worries me).

Consider a polynomial function ƒ with real coefficients having the property ƒ(g(x)) = g(ƒ(x)) for all polynomial functions g with real coefficients. Determine and prove the nature of ƒ.

Homework Equations


No equations, just the tricky fact that constant functions are polynomials.

The Attempt at a Solution


A proof by contradiction that ƒ(x) is the identity function:

We can consider the function h(x) = 1 and its derivative, which I will denote as g, namely to state the fact that ƒ((g+h)(x)) = ƒ(1) = g(ƒ(x)) + h(ƒ(x)), which then = ƒ(0) + ƒ(1); therefore ƒ(0) = 0.
Now, suppose that ƒ is not the identity function, or, specifically, that ƒ(c) ≠ c. We can consider the polynomial function k(x) which has a zero at c but not at ƒ(c). Then ƒ(k(c)) = ƒ(0) = 0 ≠ k(ƒ(c)); a contradiction occurs. This proves that ƒ must be the identity function.
 
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  • #2
Derek Hart said:

The Attempt at a Solution


A proof by contradiction that ƒ(x) is the identity function:

We can consider the function h(x) = 1 and its derivative, which I will denote as g, namely to state the fact that ƒ((g+h)(x)) = ƒ(1) = g(ƒ(x)) + h(ƒ(x)), which then = ƒ(0) + ƒ(1); therefore ƒ(0) = 0.
Now, suppose that ƒ is not the identity function, or, specifically, that ƒ(c) ≠ c. We can consider the polynomial function k(x) which has a zero at c but not at ƒ(c). Then ƒ(k(c)) = ƒ(0) = 0 ≠ k(ƒ(c)); a contradiction occurs. This proves that ƒ must be the identity function.

I can't see what you are doing in the first part. There is a very simple way to show that ##f(0) = 0##.

The second part is sound. Although, this could be simplified by doing a straight proof, as opposed to using contradiction.
 

FAQ: Proving ƒ(x) is the Identity Function

1.

What does it mean for a function to be the identity function?

A function is considered the identity function if its output is equal to its input for all values of x. In other words, the function does not change the value of the variable and simply maps each input to itself.

2.

How do you prove that ƒ(x) is the identity function?

To prove that ƒ(x) is the identity function, you need to show that ƒ(x) = x for all values of x. This can be done by substituting x for ƒ(x) in the function and simplifying the equation to show that it equals x.

3.

What is the importance of proving that a function is the identity function?

Proving that a function is the identity function is important because it establishes the relationship between the input and output values. This can be useful in solving equations and understanding the behavior of the function.

4.

Can a function be the identity function for only some values of x?

No, for a function to be the identity function, it must hold true for all values of x. If there are any values of x for which the function does not equal x, then it is not the identity function.

5.

What are some common methods used to prove that ƒ(x) is the identity function?

There are several methods that can be used to prove that ƒ(x) is the identity function, including substitution, algebraic manipulation, and graphing. It is important to choose a method that is appropriate for the specific function and its properties.

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