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## Functional Equation Problem from SMO, 2013 – Senior Section

Try this beautiful Functional Equation Problem from SMO, Singapore Mathematics Olympiad, 2013.

## Problem – Functional Equation (SMO Test)

Let M be a positive integer .It is known that whenever $|ax^2 + bx +c|\leq 1$ for all

$|x|\leq 1$ then $|2ax + b |\leq M$ for all $|x|\leq 1$. Find the smallest possible value of M.

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### Key Concepts

Functional Equation

Function

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

We cant this sum by assuming a,b,c as fixed quantity.

Let $f(x) = ax^2 + bx + c$.

Then $f(-1) = a – b + c$ ; $f(0) = c$ ; $f(1) = a + b + c$ ;

Try to do the rest of the sum …………………………..

Suppose $|f(x)|\leq 1$ for all $|x|\leq 1$ . Then

$|2ax + b| = | (x – \frac {1}{2} ) f(-1) – 2 f(0) x + (x+\frac {1}{2} f(1) |$

$\leq |x – \frac {1}{2}| + 2 |x| + |x + \frac {1}{2}|$

$\leq |x – \frac {1}{2} | + |x+\frac {1}{2}| + 2$

$\leq 4$

Now I guess you have already got the answer but if not ………….

From the last step we can conclude ,

$|2 x^2 – 1|\leq 1$ whenever $|x|\leq 4$ and $|2x| = 4$

is achieved at $x = \pm 1$.

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# Understand the problem

[/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway||||||||” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″ _i=”1″ _address=”0.0.0.1″]Find all the real Polynomials P(x) such that it satisfies the functional equation: $P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x$.

[/et_pb_text][et_pb_tabs active_tab_background_color=”#0c71c3″ inactive_tab_background_color=”#000000″ _builder_version=”3.29.2″ tab_text_color=”#ffffff” tab_font=”||||||||” background_color=”#ffffff” hover_enabled=”0″ _i=”2″ _address=”0.1.0.2″][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″ _i=”0″ _address=”0.1.0.2.0″]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”3.29.2″ _i=”1″ _address=”0.1.0.2.1″]Well, it is really good that the information polynomial is given! You should use that. What is the first thing that you check in a Polynomial Identity? Degree! Yes, check whether the degree of the Polynomial on both the LHS and RHS are the same or not. Yes, they are both the same $n^2$.  But did you observe something fishy?  [/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”3.29.2″ _i=”2″ _address=”0.1.0.2.2″]Now rewrite the equation as $P(2P(x)) - 2P(P(x)) = P(x)^{2}$. Do the Degree trick now… You see it right? Yes, on the left it is $n^2$ and on the RHS it is $2n$. So, there are two cases now… Figure them out!

[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”3.29.2″ _i=”3″ _address=”0.1.0.2.3″]

Case 1: $2n = n^2$… i.e. P(x) is either a quadratic or a constant function. Case 2: $P(2P(x)) - 2P(P(x))$ has coefficient zero till $x^2n$. We will study case 1 now. Case 1: $2n = n^2$… i.e. P(x) is either a quadratic or a constant function. $P(2P(x)) - 2P(P(x)) = P(x)^{2}$ = $P(2y) - 2P(y) = y^{2}$ where $y = P(x)$. Now, expand using $P(x) = ax^2 + bx +c$, it gives $2ay^2 -c = y^2$… Now find out all such polynomials satisfying this property. For e.g. $\frac{x^2}{2}$ is a solution. If P(x) is constant, prove that $P(x) = 0 / \frac{-1}{2}$.  [/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”3.29.2″ _i=”4″ _address=”0.1.0.2.4″]Case 2: $P(2y) - 2P(y) = y^{2}$. Assume a general form of P(x) = $latex$and show that P(x) must be quadratic or lesser degree by comparing coefficients as you have a quadratic on RHS and n degree polynomial of the LHS.  Now, we have already solved it for quadratic or less degree.  [/et_pb_tab][et_pb_tab title=”Techniques Revisited” _builder_version=”3.29.2″ _i=”5″ _address=”0.1.0.2.5″]

• Always Compare the Degree of Polynomials in identities like this. It provides a lot of information.
• Compare the coefficients of Polynomials on both sides to equalize the coefficient on both sides.
[/et_pb_tab][et_pb_tab title=”Food for Thought” _builder_version=”3.29.2″ hover_enabled=”0″ _i=”6″ _address=”0.1.0.2.6″]
• Find all polynomials $P(2P(x)) - 8P(P(x)) = P(x)^{2}$.
• Find all polynomials $P(cP(x)) – d P(P(x)) = P(x)^{2}$ depending on the values of c and d.