How does one simplify such a function?

Algebra Level 5

f ( 2 + x ) = f ( 2 x ) f ( 7 + x ) = f ( 7 x ) \large {f(2+x) = f(2-x) \\ f(7+x) = f(7-x)}

A function f f is defined for all real numbers and, for all real x x , satisfies the two equations above.

If f ( 0 ) = 0 f(0)=0 , what is the smallest number of solutions that f ( x ) = 0 f(x) = 0 could have in the interval 2000 x 2000 -2000 \leq x \leq 2000 ?


The answer is 801.

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3 solutions

Sharky Kesa
Jul 1, 2015

We have

f ( 2 + x ) = f ( 2 x ) = f ( 7 + ( 5 x ) ) = f ( 7 ( 5 x ) ) = f ( 12 + x ) . \begin{aligned} f(2+x) &= f(2-x)\\ &= f(7+(-5-x))\\ &= f(7-(-5-x))\\ &= f(12+x). \end{aligned}

Thus, f ( 2 + x ) = f ( 12 + x ) f(2+x)=f(12+x) for all real x x , which means that f ( t ) = f ( t + 10 ) f(t)=f(t+10) for all real t t .

Also

0 = f ( 0 ) = f ( 2 2 ) = f ( 2 + 2 ) = f ( 4 ) . 0 = f(0) = f(2-2) = f(2+2) = f(4).

which shows f ( 4 ) = 0 f(4)=0 .

Hence f ( 10 k ) = f ( 10 k + 4 ) = 0 f(10k) = f(10k+4) = 0 for all integers k k .

In the interval 2000 x 2000 -2000 \leq x \leq 2000 , there are 801 numbers that are equal to either 10 k 10k or 10 k + 4 10k+4 where k k is an integer.

Now consider the function f ( x ) f(x) such that f ( x ) = 0 f(x)=0 when x = 10 k x = 10k or x = 10 k + 4 x = 10k + 4 for some integer k k , and f ( x ) = 1 f(x)=1 for the other values of x x . It can be easily checked that this function satisfies f ( 2 + x ) = f ( 2 x ) f(2+x)=f(2-x) and f ( 7 + x ) = f ( 7 x ) f(7+x)=f(7-x) for all real x x .

Therefore, the answer is 801 \boxed{801} .

Moderator note:

How can we generalize this approach?

IE, what can we say if, for some constants a , b a, b , we have

f ( a + x ) = f ( a x ) f ( b + x ) = f ( b x ) f(a + x) = f(a - x) \\ f(b + x) = f (b - x ) \\

Nice solution.I did same.

Gautam Sharma - 5 years, 11 months ago

Since f(x) is symmetric about two lines x=a & x=b, we can say it is symmetric about infinitely many such lines.

Deepak Kumar - 5 years, 11 months ago
Gautam Sharma
Jul 2, 2015

My solution is the answer for challenge master note in Sharky's solution:

f ( a + x ) = f ( a x ) f ( b + x ) = f ( b x ) f(a + x) = f(a - x) \\ f(b + x) = f (b - x ) \\

Replacing ( a x ) (a-x) by t t in first we get f ( t ) = f ( 2 a t ) = f ( b ( t 2 a + b ) ) f(t)=f(2a-t)=f(b-(t-2a+b)) also f ( b ( t 2 a + b ) ) = f ( b + ( t 2 a + b ) ) = f ( t + 2 ( b a ) ) f(b-(t-2a+b))=f(b+(t-2a+b))=f(t+2(b-a)) So f ( t ) = f ( t + 2 ( b a ) ) f(t)=f(t+2(b-a))

Hence f ( x ) f(x) is periodic with period T = 2 ( b a ) T=2(b-a)

Remaining solution have been posted by Sharky.

Moderator note:

Not quite complete. What you have shown is that if f f is a function that satisfies the conditions, then 2 ( b a ) 2 ( b-a) is a period of the function.

What is more interesting, is to ask "What is the smallest value that is a period to all such functions that satisfy the conditions". So, we either need to show
1. There is a function that satisfies the function and has period 2 ( b a ) 2 (b-a) , or
2. There is a smaller value (maybe b a b-a ), which is a period of these functions.

As a further extension, when thinking about 2, also consider the case of a = 1 , b = 2 a = 1, b = \sqrt{2} .

Billy Sugiarto
Jul 4, 2015

It will be proven that the function f f have at least 801 real solution x x such that f ( x ) = 0 f(x) = 0 .

We have f ( 2 x ) = f ( 2 + x ) f(2-x) = f(2+x) and f ( 7 x ) = f ( 7 + x ) f(7-x) = f(7+x) for every x R x \in R . Therefore it is obvious that f ( x ) = f ( x + 10 ) = f ( x 10 ) x R f(x) = f(x+10) = f(x-10) \forall x \in R .

It is easy to show that f ( 0 ) = f ( 4 ) = 0 f(0) = f(4) = 0 . From the previous equality we have f ( 0 ) = f ( 10 k ) = f ( 10 k ) = 0 k N f(0) = f(10k) = f(-10k) = 0 \forall k \in N and f ( 4 ) = f ( 4 10 k ) = f ( 4 + 10 k ) = 0 k N f(4) = f(4-10k) = f(4+10k) = 0 \forall k \in N .

Therefore it is easy to check that there are exactly 801 integers k k satisfying the equality f ( k ) = 0 f(k) = 0 with k [ 2000 , 2000 ] k \in [-2000, 2000] . That is, if the only real number x x satisfying f ( x ) = 0 f(x) = 0 with x [ 3 , 2 ] x \in [-3, 2] is only x = 0 x = 0 .

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