1. Functional equations in one variable
1. Functional equations in one variable
1. Functional equations in one variable
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FUNCTIONAL EQUATIONS<br />
<strong>Functional</strong> <strong>equations</strong> are <strong>equations</strong> for unknown functions <strong>in</strong>stead of unknown numbers. In this<br />
chapter, we will try to explore how we can f<strong>in</strong>d the unknown function when we know that the<br />
conditions it satisfies.<br />
<strong>1.</strong> <strong>Functional</strong> <strong>equations</strong> <strong>in</strong> <strong>one</strong> <strong>variable</strong><br />
<strong>Functional</strong> <strong>equations</strong> <strong>in</strong> <strong>one</strong> <strong>variable</strong> are usually easier to solve. Although there is no def<strong>in</strong>ite<br />
method to solve functional <strong>equations</strong>, there are some tips.<br />
Transformation of <strong>variable</strong>s<br />
It is <strong>one</strong> of the most common approaches to solve a functional equation. When we apply this<br />
technique, we replace <strong>one</strong> <strong>variable</strong> by another (remember that the doma<strong>in</strong> of the orig<strong>in</strong>al <strong>variable</strong><br />
should NOT be affected) so that a new functional equation is obta<strong>in</strong>ed. Sometimes it is easier for us<br />
to f<strong>in</strong>d the unknown function.<br />
Example <strong>1.</strong><strong>1.</strong><br />
2<br />
If f( x+ 7) = x − 5x+ 2, f<strong>in</strong>d f ( x ).<br />
Solution.<br />
Let t = x+ 7 , then x = t − 7 . Directly substitution yields,<br />
2 2<br />
2<br />
f( t) = ( t− 7) + 5( t− 7) + 2= t − 9t+ 16. Thus f( x) = x − 9x+ 16.<br />
Example <strong>1.</strong>2.<br />
If<br />
2<br />
⎛ x+ 1⎞ x + 1 1<br />
f ⎜ ⎟ = +<br />
⎝ x ⎠ x x<br />
, f<strong>in</strong>d f ( x ).<br />
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Solution.<br />
x + 1<br />
Let t = , then<br />
x<br />
1<br />
x = t − 1<br />
. Directly substitution yields,<br />
()<br />
f t<br />
2<br />
⎛ 1 ⎞<br />
⎜ ⎟ + 1<br />
⎝t<br />
−1⎠<br />
1<br />
⎛ 1 ⎞ 1<br />
⎜ ⎟<br />
⎝t<br />
−1⎠<br />
t −1<br />
2<br />
= + = −<br />
2<br />
t<br />
t+ <strong>1.</strong><br />
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Thus<br />
2<br />
f ( x) = x − x+ <strong>1.</strong><br />
Example <strong>1.</strong>3.<br />
2<br />
If f (ln x) = x + x+ 1, where x > 0 , f<strong>in</strong>d f ( x ).<br />
Solution.<br />
Let<br />
t = ln<br />
x, then<br />
x<br />
t<br />
= e . Directly substitution yields,<br />
t<br />
( ) 2<br />
t<br />
2x<br />
x<br />
f() t = e + e + <strong>1.</strong> Thus f( x) = e + e + <strong>1.</strong><br />
In general, if we have f ( g( x)) = h( x)<br />
and g( x)<br />
has an <strong>in</strong>verse function, then we may replace<br />
g<br />
1 ( x)<br />
and get<br />
f ( x) = h( g ( x))<br />
.<br />
− −1<br />
x by<br />
Solv<strong>in</strong>g equation<br />
Sometimes after perform<strong>in</strong>g transformation of <strong>variable</strong>s, we can arrive at simultaneous<br />
<strong>equations</strong>. We can f<strong>in</strong>d the unknown function after solv<strong>in</strong>g the simultaneous <strong>equations</strong>. We may also<br />
treat the unknown function as a <strong>variable</strong> <strong>in</strong> ord<strong>in</strong>ary <strong>equations</strong> and solve it.<br />
Example <strong>1.</strong>4.<br />
If<br />
f ( x) 4+<br />
x<br />
=<br />
2<br />
3 + f ( x)<br />
x<br />
2<br />
, f<strong>in</strong>d f ( x ).<br />
Solution.<br />
It is equivalent to<br />
2 2<br />
x f x x f x<br />
( ) = (4 + )(3 + ( )) . Simplify<strong>in</strong>g it, we have<br />
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2 2 2<br />
x f x = + x + + x f x<br />
( ) 3(4 ) (4 ) ( )<br />
− = +<br />
2<br />
4 f( x) 3(4 x )<br />
f( x)<br />
=−<br />
2<br />
3(4 + x )<br />
4<br />
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Example <strong>1.</strong>5.<br />
If<br />
⎛ x −1⎞<br />
f ( x) + f ⎜ ⎟=<br />
1+ x, f<strong>in</strong>d f ( x ).<br />
⎝ x ⎠<br />
Solution.<br />
Let<br />
x −1<br />
t = , then<br />
x<br />
1<br />
x = 1 − t<br />
. Direct substitution yields<br />
f<br />
⎛ 1 ⎞ 1<br />
⎜ ⎟ + f() t = 1 +<br />
⎝1−<br />
t⎠<br />
1−<br />
t<br />
(<strong>1.</strong>1)<br />
⎛ 1 ⎞ 1<br />
f ⎜ ⎟ + f( x) = 1 +<br />
⎝1−<br />
x ⎠<br />
1−<br />
x<br />
Let<br />
1<br />
t = 1 − x<br />
, then<br />
t −1<br />
x = . Direct substitution yields<br />
t<br />
(<strong>1.</strong>2)<br />
⎛t−1⎞ ⎛ 1 ⎞ t−1<br />
f ⎜ ⎟+ f ⎜ ⎟= 1+<br />
⎝ t ⎠ ⎝1−<br />
t⎠<br />
t<br />
⎛ x −1⎞ ⎛ 1 ⎞ x −1<br />
f ⎜ ⎟+ f ⎜ ⎟= 1+<br />
⎝ x ⎠ ⎝1−<br />
x⎠<br />
x<br />
On the other hand, by substitut<strong>in</strong>g it <strong>in</strong>to (1), we have<br />
⎛ x −1⎞<br />
(<strong>1.</strong>3)<br />
f ( x) + f ⎜ ⎟= 1+x<br />
⎝ x ⎠<br />
((<strong>1.</strong>1) + (<strong>1.</strong>2) + (<strong>1.</strong>3))/2 :<br />
⎛ x−1⎞ ⎛ 1 ⎞ 1⎛<br />
1 x−1<br />
⎞<br />
(<strong>1.</strong>4)<br />
f ( x) + f ⎜ ⎟+ f ⎜ ⎟= ⎜3+ + + x ⎟⎠<br />
⎝ x ⎠ ⎝1−x⎠ 2⎝<br />
1−x x<br />
By subtract<strong>in</strong>g (<strong>1.</strong>2) from (<strong>1.</strong>4), we have<br />
3 2<br />
1⎛ 1 x−1 ⎞ ⎛ x−1⎞ 1⎛ 1 x−1⎞<br />
− x + x + 1<br />
f( x) = ⎜3+ + + x⎟− ⎜1+ ⎟= ⎜1+ + x− ⎟=<br />
2⎝ 1−x x ⎠ ⎝ x ⎠ 2⎝ 1−x x ⎠ 2 x(1 −x)<br />
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Method of undeterm<strong>in</strong>ed coefficients<br />
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When we know that the unknown function satisfies certa<strong>in</strong> conditions, say it is a quadratic or<br />
2<br />
cubic function, we can immediately set up <strong>variable</strong>s (e.g. let f ( x)<br />
= ax + bx+ c if f ( x ) is a<br />
quadratic polynomial) and solve for them.<br />
Example <strong>1.</strong>6.<br />
If f ( x)<br />
is a quadratic function such that f ( x+ 1) − f( x) = 8x+ 3 and f (0) = 5 , f<strong>in</strong>d f ( x ).<br />
Solution.<br />
Let<br />
2<br />
f ( x)<br />
ax bx c<br />
Simplify<strong>in</strong>g gives<br />
= + + , then ax ( + 1) 2 + bx ( + 1) + c−ax 2 −bx− c= 8x+3.<br />
After solv<strong>in</strong>g, we have a = 4 and b =−<strong>1.</strong><br />
Putt<strong>in</strong>g<br />
x = 0, we have c = 5 .<br />
2ax + a + b = 8x<br />
+ 3.<br />
Therefore<br />
2<br />
f ( x) = 4x − x+ 5.<br />
2. <strong>Functional</strong> <strong>equations</strong> <strong>in</strong> more than <strong>one</strong> <strong>variable</strong><br />
For functional <strong>equations</strong> with more than <strong>one</strong> <strong>variable</strong>, we can also apply the methods<br />
menti<strong>one</strong>d above. Besides we can also try to substitute some special values, say x = y = 0 <strong>in</strong>to the<br />
given condition given to obta<strong>in</strong> some results. As it is very difficult to describe these techniques <strong>in</strong><br />
words, we will try to see their application <strong>in</strong> various examples below.<br />
Example 2.<strong>1.</strong><br />
If f : →<br />
satisfies<br />
<strong>1.</strong> f (1) = 2 ,<br />
2. For all xy∈ , , f( xy) = f( x) f( y) − f( x+ y) + 1, f<strong>in</strong>d f ( x ).<br />
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Solution.<br />
Putt<strong>in</strong>g<br />
y = 1, then<br />
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f( x) = f( x) f(1) − f( x+ 1) + 1<br />
= 2 f( x) − f( x+ 1) + 1<br />
f( x+ 1) = f( x) + 1<br />
Therefore apply<strong>in</strong>g condition 1 and by mathematical <strong>in</strong>duction, for all <strong>in</strong>teger x, we have<br />
f ( x) = x+ <strong>1.</strong><br />
m<br />
m<br />
For any rational number, let x = where m, n are <strong>in</strong>tegers and n is not zero. Putt<strong>in</strong>g x = , y = n,<br />
n<br />
n<br />
⎛<br />
then ( ) m ⎞ ⎛m<br />
⎞<br />
f m = f ⎜ ⎟( n+ 1) − f ⎜ + n⎟<br />
1<br />
⎝ n ⎠ ⎝ n ⎠ + .<br />
S<strong>in</strong>ce f( x+ 1) = f( x)<br />
+ 1 for ∀x<br />
∈ , we have<br />
⎛m<br />
⎞ ⎛m⎞<br />
f ⎜ + n⎟= f ⎜ ⎟+n.<br />
⎝ n ⎠ ⎝ n ⎠<br />
⎛m⎞ ⎛m⎞<br />
Substitut<strong>in</strong>g this <strong>in</strong>to the orig<strong>in</strong>al equation, we have m+ 1 = f ⎜ ⎟( n+ 1) − f ⎜ ⎟−n+ <strong>1.</strong> Thus<br />
⎝ n ⎠ ⎝ n ⎠<br />
⎛m⎞ m<br />
f ⎜ ⎟ = + <strong>1.</strong><br />
⎝ n ⎠ n<br />
So we have f ( x) = x+ 1 ∀x<br />
∈ .<br />
Remark.<br />
In this question we can see that we first solve the functional equation for a special case (we found<br />
f ( x ) when x ∈ ), then we solve for the more general case. (we found f ( x ) when x ∈ ). Indeed,<br />
this method is used <strong>in</strong> many occasions. For<br />
f<strong>in</strong>d<br />
when<br />
f ( x ) when<br />
x∈<br />
about this later.<br />
x ∈ . Then, f<strong>in</strong>d f ( x ) when x ∈ by substitut<strong>in</strong>g<br />
f : →<br />
, we can apply this method similarly. First,<br />
m<br />
x = . F<strong>in</strong>ally, f<strong>in</strong>d f ( x )<br />
n<br />
by the density of rational numbers (for cont<strong>in</strong>uous functions only). We shall see more<br />
Example 2.2.<br />
If<br />
2 2<br />
( x y) f( x y) ( x y) f( x y) 4 xy( x y )<br />
− + − + − = − for all x,<br />
y , f<strong>in</strong>d f ( x ).<br />
Solution.<br />
The given condition is equivalent to<br />
Page 5 of 10
f( x+ y) f( x−<br />
y)<br />
− = 4 xy = ( x+ y) −(<br />
x−<br />
y)<br />
x+ y x−<br />
y<br />
2 2<br />
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f( x+ y) 2 f( x−<br />
y)<br />
2<br />
− ( x + y) = −(<br />
x−<br />
y ) for all x,<br />
y .<br />
x+ y x−<br />
y<br />
Thus<br />
f( x)<br />
x<br />
2<br />
− x is a constant. Let<br />
f( x)<br />
x<br />
2<br />
− x = k , then<br />
3<br />
f ( x)<br />
x kx<br />
= + .<br />
Remark.<br />
In this question, we have a symmetric condition. By us<strong>in</strong>g the symmetry, we reduce the equation to<br />
a <strong>one</strong>-<strong>variable</strong> functional equation. This is a useful technique for symmetric functional <strong>equations</strong>.<br />
Example 2.3.<br />
If f<br />
: → satisfies<br />
2<br />
f ( x + f( y)) = y+ xf( x)<br />
for all xy∈ , , f<strong>in</strong>d f ( x ).<br />
Solution.<br />
Putt<strong>in</strong>g<br />
x = 0 , then f ( f( y))<br />
= y. Thus we have,<br />
(2.1)<br />
f y xf x f f x f y x f y<br />
2 2<br />
( + ( )) = ( ( + ( ))) = + ( )<br />
Now replace x by f ( x ),<br />
2<br />
f ( y+ ( f( x) f( f( x))) = ( f( x)) + f( y)<br />
. Remember that f ( f( y))<br />
= y, so<br />
(2.2)<br />
f y xf x f x f y<br />
2<br />
( + ( )) = ( ( )) + ( )<br />
Compar<strong>in</strong>g (2.1) and (2.2), we have<br />
(2.3) ( ) 2 2<br />
f ( x)<br />
= x<br />
Now replace<br />
y by f ( y ) <strong>in</strong> the orig<strong>in</strong>al equation, we have<br />
2<br />
f ( x + y) = f( y) + xf( x)<br />
. Squar<strong>in</strong>g both sides, we have<br />
( ) 2<br />
2 2 2 2 2 2 4 2<br />
( x + y) = f( x + y) = f( y) + x f( x) + 2 xf( x) f( y) = x + y + 2 xf( x) f( y)<br />
From this, we have<br />
(2.4) xy = f( x) f( y)<br />
From (2.3), we have f ( x ) = x or f ( x)<br />
=− x.<br />
If f ( x ) = x, from (2.4) we have f ( x ) = x for x∈R .<br />
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If f ( x ) =−x, from (2.4) we have f ( x)<br />
= − x for x∈R .<br />
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Therefore f ( x ) = x or f ( x)<br />
=− x.<br />
Remark.<br />
In this question, we replace (or substitute) <strong>variable</strong>s by other th<strong>in</strong>gs. (we replaced x by f ( x ) <strong>in</strong> the<br />
question). This is a very useful technique. Some common replacements and substitutions <strong>in</strong>clude:<br />
replac<strong>in</strong>g x by f ( x ); replac<strong>in</strong>g x by f ( f( x )); substitut<strong>in</strong>g x = 0; substitut<strong>in</strong>g x = y = 0; substitut<strong>in</strong>g<br />
x = 1, etc.<br />
Example 2.4.<br />
If f : →<br />
<br />
+<br />
satisfies<br />
<strong>1.</strong> f ( xf ( y)) = yf ( x)<br />
for all xy , ∈<br />
2. f( x) → 0 as x →∞, f<strong>in</strong>d f ( x ).<br />
+<br />
Solution.<br />
First, let us show that f ( x ) is surjective.<br />
Let<br />
x ⎛ ⎛ x ⎞ ⎞ x<br />
y = , then f ⎜xf ⎜ ⎟⎟<br />
= f ( x)<br />
= x . Therefore f ( x ) is surjective, i.e. for all y<br />
f ( x)<br />
⎝ ⎝ f( x) ⎠⎠<br />
f( x)<br />
there exists an x such that f ( x)<br />
Assume<br />
= y.<br />
f( y ) = <strong>1.</strong> Putt<strong>in</strong>g x = 1, then<br />
f (1) = f(1 f( y)) = yf(1)<br />
. So y = 1 and f (1) = <strong>1.</strong><br />
Now putt<strong>in</strong>g x = y , then f ( xf ( x)) = xf ( x)<br />
. Thus xf( x ) is a fixed po<strong>in</strong>t for the function. Let us<br />
show two th<strong>in</strong>gs about fixed po<strong>in</strong>t now.<br />
(i) If a, b are fixed po<strong>in</strong>ts of f, then f ( ab) = f ( af ( b)) = bf ( a)<br />
= ab . Thus ab is also a fixed<br />
(ii)<br />
po<strong>in</strong>t.<br />
⎛ 1⎞ ⎛1<br />
⎞ ⎛1⎞<br />
If a is a fixed po<strong>in</strong>t of f, then 1 = f(1) = f ⎜ai<br />
⎟= f ⎜ f( a)<br />
⎟=<br />
af ⎜ ⎟<br />
⎝ a⎠ ⎝a<br />
⎠ ⎝a⎠ . Thus ⎛1⎞ 1<br />
f ⎜ ⎟ =<br />
⎝a⎠<br />
a<br />
,<br />
which means 1 a<br />
is also a fixed po<strong>in</strong>t.<br />
Now, if xf ( x ) > 1 , then (( ( )) n<br />
n<br />
f xf x ) = ( xf ( x))<br />
. As n ,<br />
n<br />
n<br />
f (( xf ( x)) ) = ( xf ( x))<br />
. This contradicts with condition 2.<br />
→∞ [ ]<br />
xf ( x) n<br />
→∞ while<br />
Page 7 of 10
If<br />
xf ( x ) < 1, then by (ii) we have<br />
condition 2.<br />
Thus, we must have<br />
Remark.<br />
1<br />
xf( x)<br />
xf ( x ) = <strong>1.</strong> This means<br />
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is a fixed po<strong>in</strong>t larger than <strong>1.</strong> This aga<strong>in</strong> contradicts with<br />
f( x)<br />
1<br />
= .<br />
x<br />
The concept of fixed po<strong>in</strong>t is <strong>in</strong>troduced <strong>in</strong> this question. If f ( x)<br />
of f. It is not so commonly used but it is also a useful technique.<br />
= x, then x is called a fixed po<strong>in</strong>t<br />
3. Some famous functional <strong>equations</strong><br />
In this part, we will <strong>in</strong>troduce some famous functional <strong>equations</strong>. We may quote them directly<br />
<strong>in</strong> competitions.<br />
Theorem 3.<strong>1.</strong> (Cauchy equation)<br />
If f is a cont<strong>in</strong>uous function such that f ( x+ y) = f( x) + f( y)<br />
for all xy∈ , , then f ( x)<br />
= cx<br />
where c is a constant.<br />
Proof.<br />
We may use the technique menti<strong>one</strong>d <strong>in</strong> example 2.<strong>1.</strong><br />
First, put<br />
f ( x)<br />
y = 1 and let x be a positive <strong>in</strong>teger. Then f ( x+ 1) = f ( x)<br />
+ c, where c = f(1)<br />
. Thus<br />
= cx for positive <strong>in</strong>tegers x. It is easy to verify that (0) 0<br />
orig<strong>in</strong>al equation. For negative <strong>in</strong>tegers, we replace x by –x and get<br />
f = by putt<strong>in</strong>g x = y = 0 <strong>in</strong>to the<br />
f ( − x+ 1) = f( − x)<br />
+ c. Thus<br />
f ( − x)<br />
=−cx for positive <strong>in</strong>tegers x. Therefore we conclude that f ( x)<br />
= cx for <strong>in</strong>tegers x.<br />
Let<br />
m<br />
x = , where m, n are <strong>in</strong>tegers. Then we have<br />
n<br />
⎛m+<br />
1⎞ ⎛m⎞ ⎛1⎞<br />
f ⎜ ⎟= f ⎜ ⎟+<br />
f ⎜ ⎟<br />
⎝ n ⎠ ⎝ n ⎠ ⎝n⎠ , thus ⎛m<br />
⎞ ⎛1<br />
⎞<br />
f ⎜ ⎟=<br />
mf ⎜ ⎟<br />
⎝ n ⎠ ⎝n⎠ . However, 1 ⎛1⎞<br />
c = f ( ⋅ n ) = nf ⎜ ⎟<br />
n ⎝n⎠<br />
⎛1 ⎞ c<br />
f ⎜ ⎟ =<br />
⎝n⎠<br />
n<br />
and we have<br />
⎛1 ⎞ c<br />
f ⎜ ⎟ =<br />
⎝n⎠<br />
n<br />
. This implies f ( x)<br />
= cx for rational numbers x.<br />
. So<br />
Page 8 of 10
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Now, as f is cont<strong>in</strong>uous, we can always bound an irrational number by two rational numbers. (For<br />
example, we may bound π by 3, 3.1, 3.14, 3.141 and 4, 3.2, 3.15, 3.142 respectively). Let<br />
{ xn},<br />
xn∈ be such a sequence with lim x n<br />
= x , then by cont<strong>in</strong>uity of f, we have<br />
n→∞<br />
f ( x) = f(lim x ) = lim f( x ) = lim cx = cx. Therefore f ( x)<br />
n n n<br />
n→∞ n→∞ n→∞<br />
= cx for all x∈R .<br />
Q.E.D.<br />
Corollary 3.2.<br />
If f is a cont<strong>in</strong>uous function and for all xy∈ , ,<br />
(i) f ( x+ y) = f( x) f( y)<br />
then f ( x)<br />
= c<br />
(ii) f ( xy) = f ( x) + f ( y)<br />
then f ( x) = clnx<br />
(iii) f ( xy) = f ( x) f ( y)<br />
then f ( x)<br />
= x<br />
where c is a constant.<br />
x<br />
c<br />
Example 3.3.<br />
If<br />
f :(1, +∞)<br />
→ is a cont<strong>in</strong>uous function such that f ( xy) = xf ( y) + yf ( x)<br />
for all1 < xy , ∈ , f<strong>in</strong>d<br />
f ( x ).<br />
Solution.<br />
It is equivalent to f ( xy) = f ( x) + f ( y)<br />
. If we let<br />
xy x y<br />
gxy ( ) = gx ( ) + gy ( )<br />
f ( x)<br />
gx ( ) = , then the equation becomes<br />
x<br />
which is simply (ii) of corollary <strong>1.</strong>32 (although there is a m<strong>in</strong>or difference<br />
between them). So we have g( x) = clnxand<br />
f ( x) = xg( x) = cxlnx.<br />
Page 9 of 10
4. Exercises<br />
<strong>1.</strong> Solve the functional equation<br />
2<br />
f ( x+ 2) = x + 4x+ 6.<br />
Mathematical Database<br />
2. If<br />
⎛1<br />
⎞<br />
3 f ( x) + 2f<br />
⎜ ⎟=<br />
4x, f<strong>in</strong>d f ( x ).<br />
⎝ x ⎠<br />
3. If af ( x − 1) + bf (1 − x)<br />
= cx where a, b, c are constants, f<strong>in</strong>d f ( x ).<br />
4. Solve the functional equation xf ( x) + 2 xf ( − x) =− <strong>1.</strong><br />
5. F<strong>in</strong>d all cont<strong>in</strong>uous functions f for x > 0 such that<br />
f( x+ y)<br />
=<br />
f ( x) f( y)<br />
.<br />
f ( x) + f( y)<br />
6. F<strong>in</strong>d all cont<strong>in</strong>uous functions f such that f ( x+ y) = f( x) + f( y) + f( x) f( y)<br />
.<br />
7. If f ( x+ y) + f( x− y) = 2 f( x)cosy, f<strong>in</strong>d f ( x ).<br />
Harder problems<br />
8. (IMO 1982) The function f ( n ) is def<strong>in</strong>ed for all positive <strong>in</strong>tegers n and takes on non-negative<br />
<strong>in</strong>tegral values.<br />
Also, for all m, n<br />
Determ<strong>in</strong>e f (1982) .<br />
f ( m+ n) − f( m) − f( n) = 0 or 1<br />
f (2) = 0, f (3) > 0, and f (9999) = 3333.<br />
9. F<strong>in</strong>d all pairs of functions f, g:<br />
→<br />
such that<br />
(a) if x < y , then f ( x) < f ( y)<br />
(b) for all xy∈ , , g( y) f( x) + f( y) = f( xy)<br />
.<br />
10. F<strong>in</strong>d all f : Z→<br />
Z such that<br />
(a) f (1) = 1<br />
(b) for all mn∈Z , , f ( m+ n)( f( m) − f( n)) = f( m− n)( f( m) + f( n))<br />
.<br />
References:<br />
<strong>1.</strong> Notes on functional <strong>equations</strong>, by Mr. Kwok Ka Keung<br />
2. 華 羅 庚 學 校 數 學 課 本 ( 高 一 年 級 ), 出 版 社 : 中 國 大 百 科 全 書 出 版 社<br />
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