Test of Mathematics Solution Subjective 64 -Functional Equation

This is a Test of Mathematics Solution Subjective 64 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East-West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

If f(x) is a real-valued function of a real variable x, such that 2f(x) + 3f(-x) = 15 – 4x for all x, find the function f(x).

Solution:

2f(x) + 3f(-x) = 15 – 4x
Putting x =0, we gets
2f(0) + 3f(0) = 15
Or, f(0) = 3.
Let g(x) = f (x) – 3
2f(x) + 3f(-x) = 15 – 4x
Or, 2g(x) + 3g(-x) = - 4x …(i)
Now put x = - x
2g(-x) + 3g(x) = 4x …(ii)
(i) + (ii), g(x) + g(-x) = o
Or, g(x) = -g(-x) …(iii)

Now 2g(x) + 3g(-x) = - 4x
Or, –g(x) = - 4x [from (iii)]
Or, g(x) = 4x
Or, f(x) = 4x + 3

Useful Resources:

Functional Equation Problem from SMO, 2018 - Question 35

Try to solve this problem number 35 from Singapore Mathematics Olympiad, SMO, 2018 based on Functional Equation.

Problem - Functional Equation (SMO Entrance)

Consider integers ${1,2, \ldots, 10}$. A particle is initially -at 1 . It moves to an adjacent integer in the next step. What is the expected number of steps it will take to reach 10 for the first time?

Key Concepts

Functional Equation

Equation

Check the Answer

Answer : 81

Singapore Mathematical Olympiad

Challenges an Thrills - Pre - College Mathematics

Try with Hints

If you got stuck into this problem we can start taking an expected number of steps to be $g_{n}$. We need to remember at first the particle was in 1 then it will shift to the next step so for n no of position we can expressed it as n and n -1 where n = 2,3,4,........,100.

Now try the rest..............

Now let's continue after the last hint ............

Then $g_{n+1} = \frac {1}{2} (1+g_{n} + g_{n+1} )+ \frac {1}{2}$

which implies , $g_{n+1} = g_{n} + 2$

Now we know that,$g_{2} = 1$. Then $g_{3} = 3$, $g_{4}= 5$,..................,$g_{10}=17$

$g = g_{2}+g_{3}+g_{4}+....................+g_{10} = 1+3+.....................+17 = 81$[ Answer]

Functional Equations Problem | SMO, 2012 | Problem 33

Try this beautiful Problem from Singapore Mathematics Olympiad, 2012 based on Functional Equations.

Problem - Functional equations (SMO Test)

Let L denote the minimum value of the quotient of a 3- digit number formed by three distinct divided by the sum of its digits.Determine $\lfloor 10L \rfloor $.

Key Concepts

Functional Equation

Max and Min Value

Check the Answer

Answer: 105

Singapore Mathematical Olympiad, 2012

Challenges and Thrills - Pre - College Mathematics

Try with Hints

If you got stuck at first only here is the hint to begin with :

Anyway a three digit number we can be expressed as 100x + 10 y +z depending on the place values. and if we do minimize it :

F(x y z) = $\frac {100x + 10y + z}{x + y + z}$

Lets consider that for distinct x , y , z, F(x , y , Z) has the minimum value when x<y<z.

Again we can assume,

$ 0 < a < b < c \leq 9$

Note ,

F(x,y,z) = $\frac {100 x + 10 y + z }{x +y + z}$ = 1 + $\frac {99 x + 9 y }{x+y+z}$

Try the rest of the sum...............

From the last hint we can say

F(x y z ) is minimum when c = 9 (say)

F(x y 9) = 1+ $\frac {99x +9y }{x+y+9} = 1 + \frac {9(x + y + 9) + 90 x - 81}{x+y +9} = 10 + \frac {9(10x -9)}{x+y +9}$

Try to do for the next case for minimum value when b = 8..................

In the last hint we can do the next step which is b= 8:

F(x 8 9) = 10 + $\frac {9(10x -9)}{x+17}= 10 + \frac {90(a+17)- 1611}{x + 17} = 100 - \frac {1611}{x +17} $

which has the minimum value of x = 1 and so 10 L = 105.(answer)

Problem on Functional Equation | SMO, 2010 | Problem 31

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2010 based on functional equation.

Problem - Functional Equation (SMO Entrance)

Consider the identity $1+2+......+n = \frac {1}{2}n(n+1)$. If we set $P_{1}(x) = \frac{1}{2}x(x+1)$ , then it is the unique polynomials such that for all positive integer n,$p_{1}(n) = 1+2+..............+n$ . In general, for each positive integer k, there is a unique polynomial $P_{k} (x) $ such that :

$P_{k} (n) = 1^k + 2^ k+3^k +..................+n^k$ for each n =1,2,3...............

Find the value of $P_{2010} (-\frac {1}{2})$ .

Key Concepts

Polynomials

Functional Equation

Check the Answer

Answer : 0

Singapore Mathematics Olympiad

Challenges and Thrills - Pre College Mathematics

Try with Hints

If you got stuck in this question we definitely can start from here:

In the question given above say k is the positive even number :

so let $f(x) = P_{k} - P_(x-1)$

Then $f(n) = n^k $ for all integer $n\geq 2$ (when f is polynomials)

Like this then $f(x) = x^k$(again for all $ x \geq 2$ .

If you got stuck after first hint try this one

$P_{k} (-n + 1) - P_{k}(-n) = f(-n +1) = (n-1)^k$..................................(1)

Again, $P_{k} (-n + 2) - P_{k}(-n+1) = f(-n +2) = (n-2)^k$.......................................(2)

Now taking n = 1;The $eq^n$(1) becomes, $P_{k}(0) - P_{k}(-1) = f(0) = 0^{k}$,

And for $eq^(n)$ (2) ; $P_{k}(1) - P_{k}(0) = f(1) = 1^{k}$.

Now sum these equation and try to solve the rest...........

Summing this two equation we get , $P_{k}(1)-P_{k}(-n) = 1^{k} + 0^{k}+1^{k}+.......+(n-1)^k$.

so , $P_{k}(-n)+P_{k}(n-1)=0$

Again if $g(x) = (P_{k}(-x)+P_{k}(x-1)$

Then g(n) is equal to 0 for all integer $n\geq2$

As g is polynomial, g(x) =0;

So , $P_{k}(-\frac {1}{2}) + P_{k}(-\frac {1}{2}) = 0$

so $p_{k}(-\frac {1}{2}) = 0$ ............(Answer)

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.

Key Concepts

Functional Equation

Function

Check the Answer

Answer: 4

Singapore Mathematics Olympiad

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 $x = \pm 1$.

Functional Equation Problem | SMO, 2013 - Problem19 (Senior Section)

Try this beautiful problem from Singapore Mathematics Olympiad based on Functional Equation.

Problem - Functional Equation (SMO Exam)

Let f and g be functions such that for all real numbers x and y,

$ g (f (x+y)) = f( x ) + (x+y) g (y)$.

Find the value of $ g(0) + g (1) + ........................+ g (2013) $

Key Concepts

Functional Equation

Funcion

Arbitrary Numbers

Check the Answer

Answer: 0

Singapore Mathemaics Olympiad

Challenges and thrills

Try with Hints

We can start this problem by considering y = -x.

Then $ g (f (0) ) = f (x) $ for all x. This $f$ is is a constant function ; namely

$ f (x) = c $ for some c.

Try the rest of the sum ............................................................

For all value of x,y we have

$ (x+y) g(y) = g(f(x+y)) - f(x) = g(c) - c = 0 $

Since x + y is arbitrary , we must have $ g (y) = 0 $ for all y .Hence

$ g (0) + g ( 1 ) + ................................+ g(2013) = 0 $ (Answer).

Polynomial Functional Equation - Random Olympiad Problem

[et_pb_section fb_built="1" _builder_version="3.22.4" fb_built="1" _i="0" _address="0"][et_pb_row _builder_version="3.25" _i="0" _address="0.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.0.0"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" _i="0" _address="0.0.0.0"]

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: $latex P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x $.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.29.2" _i="0" _address="0.1.0.0.0"]

Unknown [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.29.2" _i="1" _address="0.1.0.0.1" open="off"]Functional Equation, Polynomials[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.29.2" _i="2" _address="0.1.0.0.2" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.29.2" _i="3" _address="0.1.0.0.3" open="off"]Excursion in Mathematics Challenges and Thrills in Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" _i="1" _address="0.1.0.1"]

Start with hints

[/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 $latex 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 $latex P(2P(x)) - 2P(P(x)) = P(x)^{2}$. Do the Degree trick now... You see it right? Yes, on the left it is $latex n^2 $ and on the RHS it is $latex 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: $latex 2n = n^2 $... i.e. P(x) is either a quadratic or a constant function. Case 2: $latex P(2P(x)) - 2P(P(x)) $ has coefficient zero till $latex x^2n$. We will study case 1 now. Case 1: $latex 2n = n^2$... i.e. P(x) is either a quadratic or a constant function. $latex P(2P(x)) - 2P(P(x)) = P(x)^{2}$ = $latex P(2y) - 2P(y) = y^{2}$ where $latex y = P(x) $. Now, expand using $latex P(x) = ax^2 + bx +c$, it gives $latex 2ay^2 -c = y^2 $... Now find out all such polynomials satisfying this property. For e.g. $latex \frac{x^2}{2}$ is a solution. If P(x) is constant, prove that $latex 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: $latex P(2y) - 2P(y) = y^{2}$. 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 $latex P(2P(x)) - 8P(P(x)) = P(x)^{2}$.
Find all polynomials $ P(cP(x)) - d P(P(x)) = P(x)^{2} $ depending on the values of c and d.

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Watch video

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Connected Program at Cheenta

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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" _i="8" _address="0.1.0.8"]

A functional inequation

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" _i="1" _address="0.0.0.1"]Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$f(x + y) + y \le f(f(f(x)))$
holds for all $x, y \in \mathbb{R}$ . [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" _i="0" _address="0.1.0.0.0" hover_enabled="0"]Benelux MO 2013 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" _i="1" _address="0.1.0.0.1" open="off" hover_enabled="0"]Functional Equations [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" _i="2" _address="0.1.0.0.2" open="off" hover_enabled="0"]Easy [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" _i="3" _address="0.1.0.0.3" open="off" hover_enabled="0"]Functional Equations by BJ Venkatachala [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" 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="3.27" _i="1" _address="0.1.0.2.1" hover_enabled="0"]Note that the RHS does not contain $latex y$. Thus it should be possible to play with different values of $latex y$ to get rid of the inconvenient $latex f\circ f\circ f$. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" _i="2" _address="0.1.0.2.2" hover_enabled="0"]Taking $latex y=f(f(f(x)))-f(x)$, we get $latex f(f(f(x)))\le f(x)$. Now show that $latex f(x)\le f(f(f(x)))$. Thus $f(f(f(x)))=f(x)$. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"]Taking $latex x=0$, we get $latex f(y)+y\le f(0)$. Now try to show that $latex f(y)+y\ge f(0)$. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]Taking $latex y=-x$ we get the desired inequality from hint 3. Thus $latex f(y)+y=f(0)\implies f(y)=f(0)-y$ for all $latex y$. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.26.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

Watch video

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Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

Functional equation dependent on a constant

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

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"]Find all real numbers $a$ for which there exists a non-constant function $f :\Bbb R \to \Bbb R$ satisfying the following two equations for all $x\in \Bbb R:$
i) $f(ax) = a^2f(x)$ and
ii) $f(f(x)) = a f(x).$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.26.6" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="3.26.6"]Baltic Way 2016[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.3.2" hover_enabled="0" open="on"]

Functional Equation

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.26.6" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.26.6" open="off"]Functional Equations by BJ Venkatachala[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.26.6" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.26.6"]Show that the choices $latex a=0,1$ work.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.26.6"]Show that $latex af(f(x))=a^2f(f(x))$. As we have already dealt with $latex a=0$, this gives $latex af(f(x))=f(f(x))$. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.26.6"]Hint 3 gives $latex (a-1)f(f(x))=0$. As $latex a=1$ has already been dealt with, we must consider the option $latex f(f(x))\equiv 0$.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.26.6"]Hint 3 gives $latex af(x)\equiv 0$. As $latex a\neq 0$, we have $latex f(x)\equiv 0$. This contradicts the fact that $latex f$ is non-constant. Hence, $latex a=0,1$ are the only options.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Watch video

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Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Find all arithmetic functions

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition:
$f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.23.3" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]

IMO longlist 1977 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Functional equations [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Medium [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" open="off" _builder_version="3.23.3" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Functional Equations by B J Venkatachala [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.23.3" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]Prove (by induction on $latex m$) that if $latex n\ge m$ then $latex f(n)\ge m$. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]Note that hint 1 implies that $latex f(n+1)>f(n)$. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Note that hint 2 implies that $latex n+1>f(n)$, i.e. $latex f(n)\le n$. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]Hint 3, combined with hint 1 gives $latex f(n)=n$ for all $latex n\in\mathbb{N}$. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Watch the video (Coming Soon)

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Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px"]

Problem

Solution:

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Problem - Functional Equation (SMO Entrance)

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Problem - Functional Equation (SMO Entrance)

Key Concepts

Check the Answer

Try with Hints

Other useful links

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Subscribe to Cheenta at Youtube

Problem - Functional Equation (SMO Test)

Key Concepts

Check the Answer

Try with Hints

Other useful links

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Problem - Functional Equation (SMO Exam)

Key Concepts

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Functional Equation

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