A functional equation from a regional math olympiad (Dhaka regional, 2017)
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Question: For any rational numbers $x$ and $y$, $f(x)$ is a real number and:
$$f(x+y) = f(x)f(y) - f(xy) +1$$
Again, $f(2017) neq f(2018)$ and:
$$fleft(frac20172018right) = fracab$$
where $a$ and $b$ are co prime. Find $a-b$.
Attempts: I have tried for several days, but I don't have any clue on how to solve it. I tried substitution, and got some useless facts:
$$f(0) = 1$$
$$f(-x^2) = f(x)f(-x)$$
$$f(-1) = 0$$
$$f(2x) = [f(x)]^2 - f(x^2) + 1$$
Can anyone help with this problem please?
contest-math functional-equations
asked Aug 27 at 13:33
Hoque
503
 |Â
show 2 more comments
up vote
4
down vote
favorite
Question: For any rational numbers $x$ and $y$, $f(x)$ is a real number and:
$$f(x+y) = f(x)f(y) - f(xy) +1$$
Again, $f(2017) neq f(2018)$ and:
$$fleft(frac20172018right) = fracab$$
where $a$ and $b$ are co prime. Find $a-b$.
Attempts: I have tried for several days, but I don't have any clue on how to solve it. I tried substitution, and got some useless facts:
$$f(0) = 1$$
$$f(-x^2) = f(x)f(-x)$$
$$f(-1) = 0$$
$$f(2x) = [f(x)]^2 - f(x^2) + 1$$
Can anyone help with this problem please?
contest-math functional-equations
asked Aug 27 at 13:33
Hoque
503
How do you see that $f(-1)=0$? After all, the constant function $f(x)=1$ satisfies the functional equation (though not the condition that $f(2017)neq f(2018)$ of course).
– lulu
Aug 27 at 13:45
@Alphanerd Good catch, but that version adds the condition that $f(1)=2$ which makes the problem quite easy. Was that condition meant to be present here?
– lulu
Aug 27 at 13:52
@Alphanerd Thanks a lot. But what do I do with this question now? Am I supposed to delete this?
– Hoque
Aug 27 at 13:52
Once again: the older version is quite different from this one unless you intended to add the condition that $f(1)=2$.
– lulu
Aug 27 at 13:53
The condition f(1)= 2 was not given in the question
– Hoque
Aug 27 at 13:54
 |Â
show 2 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Question: For any rational numbers $x$ and $y$, $f(x)$ is a real number and:
$$f(x+y) = f(x)f(y) - f(xy) +1$$
Again, $f(2017) neq f(2018)$ and:
$$fleft(frac20172018right) = fracab$$
where $a$ and $b$ are co prime. Find $a-b$.
Attempts: I have tried for several days, but I don't have any clue on how to solve it. I tried substitution, and got some useless facts:
$$f(0) = 1$$
$$f(-x^2) = f(x)f(-x)$$
$$f(-1) = 0$$
$$f(2x) = [f(x)]^2 - f(x^2) + 1$$
Can anyone help with this problem please?
contest-math functional-equations
asked Aug 27 at 13:33
Hoque
503
Question: For any rational numbers $x$ and $y$, $f(x)$ is a real number and:
$$f(x+y) = f(x)f(y) - f(xy) +1$$
Again, $f(2017) neq f(2018)$ and:
$$fleft(frac20172018right) = fracab$$
where $a$ and $b$ are co prime. Find $a-b$.
Attempts: I have tried for several days, but I don't have any clue on how to solve it. I tried substitution, and got some useless facts:
$$f(0) = 1$$
$$f(-x^2) = f(x)f(-x)$$
$$f(-1) = 0$$
$$f(2x) = [f(x)]^2 - f(x^2) + 1$$
Can anyone help with this problem please?
contest-math functional-equations
asked Aug 27 at 13:33
Hoque
503
asked Aug 27 at 13:33
Hoque
503
asked Aug 27 at 13:33
Hoque
503
asked Aug 27 at 13:33
Hoque
503
503
How do you see that $f(-1)=0$? After all, the constant function $f(x)=1$ satisfies the functional equation (though not the condition that $f(2017)neq f(2018)$ of course).
– lulu
Aug 27 at 13:45
@Alphanerd Good catch, but that version adds the condition that $f(1)=2$ which makes the problem quite easy. Was that condition meant to be present here?
– lulu
Aug 27 at 13:52
@Alphanerd Thanks a lot. But what do I do with this question now? Am I supposed to delete this?
– Hoque
Aug 27 at 13:52
Once again: the older version is quite different from this one unless you intended to add the condition that $f(1)=2$.
– lulu
Aug 27 at 13:53
The condition f(1)= 2 was not given in the question
– Hoque
Aug 27 at 13:54
 |Â
show 2 more comments
How do you see that $f(-1)=0$? After all, the constant function $f(x)=1$ satisfies the functional equation (though not the condition that $f(2017)neq f(2018)$ of course).
– lulu
Aug 27 at 13:45
@Alphanerd Good catch, but that version adds the condition that $f(1)=2$ which makes the problem quite easy. Was that condition meant to be present here?
– lulu
Aug 27 at 13:52
@Alphanerd Thanks a lot. But what do I do with this question now? Am I supposed to delete this?
– Hoque
Aug 27 at 13:52
Once again: the older version is quite different from this one unless you intended to add the condition that $f(1)=2$.
– lulu
Aug 27 at 13:53
The condition f(1)= 2 was not given in the question
– Hoque
Aug 27 at 13:54
How do you see that $f(-1)=0$? After all, the constant function $f(x)=1$ satisfies the functional equation (though not the condition that $f(2017)neq f(2018)$ of course).
– lulu
Aug 27 at 13:45
How do you see that $f(-1)=0$? After all, the constant function $f(x)=1$ satisfies the functional equation (though not the condition that $f(2017)neq f(2018)$ of course).
– lulu
Aug 27 at 13:45
@Alphanerd Good catch, but that version adds the condition that $f(1)=2$ which makes the problem quite easy. Was that condition meant to be present here?
– lulu
Aug 27 at 13:52
@Alphanerd Good catch, but that version adds the condition that $f(1)=2$ which makes the problem quite easy. Was that condition meant to be present here?
– lulu
Aug 27 at 13:52
@Alphanerd Thanks a lot. But what do I do with this question now? Am I supposed to delete this?
– Hoque
Aug 27 at 13:52
@Alphanerd Thanks a lot. But what do I do with this question now? Am I supposed to delete this?
– Hoque
Aug 27 at 13:52
Once again: the older version is quite different from this one unless you intended to add the condition that $f(1)=2$.
– lulu
Aug 27 at 13:53
Once again: the older version is quite different from this one unless you intended to add the condition that $f(1)=2$.
– lulu
Aug 27 at 13:53
The condition f(1)= 2 was not given in the question
– Hoque
Aug 27 at 13:54
The condition f(1)= 2 was not given in the question
– Hoque
Aug 27 at 13:54
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
Suppose $f(1)=r$. Also, plugging $y=1$ gives $f(x+1)=f(x)cdot(r-1)+1$, so
$$f(2) = 1-r+r^2\
f(3) = 2r-2r^2+r^3\
f(4) = 1-2r + 4r^2 - 3r^3 + r^4$$
Now plug $x=y=2$ and we'll end up with an equation on $r$.
$$2f(4) = f^2(2)+1\
2(1-2r + 4r^2 - 3r^3 + r^4) = (1-r+r^2)^2+1\
-2r + 5r^2 - 4r^3 + r^4 = 0\
r(r-1)^2(r-2)=0
$$
Now,
- $r=0$ gives a pseudo solution $f(2n)=1,;f(2n+1)=0$ which leads to a contradiction later on;
- $r=1$ gives a possible solution of $f(n)=1$ which is ruled out by the condition $f(2017)ne f(2018)$;
- $r=2$ seems to be the only possible option, and also the one you know how to deal with.
answered Aug 27 at 14:21
Ivan Neretin
8,35821433
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Suppose $f(1)=r$. Also, plugging $y=1$ gives $f(x+1)=f(x)cdot(r-1)+1$, so
$$f(2) = 1-r+r^2\
f(3) = 2r-2r^2+r^3\
f(4) = 1-2r + 4r^2 - 3r^3 + r^4$$
Now plug $x=y=2$ and we'll end up with an equation on $r$.
$$2f(4) = f^2(2)+1\
2(1-2r + 4r^2 - 3r^3 + r^4) = (1-r+r^2)^2+1\
-2r + 5r^2 - 4r^3 + r^4 = 0\
r(r-1)^2(r-2)=0
$$
Now,
- $r=0$ gives a pseudo solution $f(2n)=1,;f(2n+1)=0$ which leads to a contradiction later on;
- $r=1$ gives a possible solution of $f(n)=1$ which is ruled out by the condition $f(2017)ne f(2018)$;
- $r=2$ seems to be the only possible option, and also the one you know how to deal with.
answered Aug 27 at 14:21
Ivan Neretin
8,35821433
add a comment |Â
up vote
6
down vote
accepted
Suppose $f(1)=r$. Also, plugging $y=1$ gives $f(x+1)=f(x)cdot(r-1)+1$, so
$$f(2) = 1-r+r^2\
f(3) = 2r-2r^2+r^3\
f(4) = 1-2r + 4r^2 - 3r^3 + r^4$$
Now plug $x=y=2$ and we'll end up with an equation on $r$.
$$2f(4) = f^2(2)+1\
2(1-2r + 4r^2 - 3r^3 + r^4) = (1-r+r^2)^2+1\
-2r + 5r^2 - 4r^3 + r^4 = 0\
r(r-1)^2(r-2)=0
$$
Now,
- $r=0$ gives a pseudo solution $f(2n)=1,;f(2n+1)=0$ which leads to a contradiction later on;
- $r=1$ gives a possible solution of $f(n)=1$ which is ruled out by the condition $f(2017)ne f(2018)$;
- $r=2$ seems to be the only possible option, and also the one you know how to deal with.
answered Aug 27 at 14:21
Ivan Neretin
8,35821433
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Suppose $f(1)=r$. Also, plugging $y=1$ gives $f(x+1)=f(x)cdot(r-1)+1$, so
$$f(2) = 1-r+r^2\
f(3) = 2r-2r^2+r^3\
f(4) = 1-2r + 4r^2 - 3r^3 + r^4$$
Now plug $x=y=2$ and we'll end up with an equation on $r$.
$$2f(4) = f^2(2)+1\
2(1-2r + 4r^2 - 3r^3 + r^4) = (1-r+r^2)^2+1\
-2r + 5r^2 - 4r^3 + r^4 = 0\
r(r-1)^2(r-2)=0
$$
Now,
- $r=0$ gives a pseudo solution $f(2n)=1,;f(2n+1)=0$ which leads to a contradiction later on;
- $r=1$ gives a possible solution of $f(n)=1$ which is ruled out by the condition $f(2017)ne f(2018)$;
- $r=2$ seems to be the only possible option, and also the one you know how to deal with.
answered Aug 27 at 14:21
Ivan Neretin
8,35821433
Suppose $f(1)=r$. Also, plugging $y=1$ gives $f(x+1)=f(x)cdot(r-1)+1$, so
$$f(2) = 1-r+r^2\
f(3) = 2r-2r^2+r^3\
f(4) = 1-2r + 4r^2 - 3r^3 + r^4$$
Now plug $x=y=2$ and we'll end up with an equation on $r$.
$$2f(4) = f^2(2)+1\
2(1-2r + 4r^2 - 3r^3 + r^4) = (1-r+r^2)^2+1\
-2r + 5r^2 - 4r^3 + r^4 = 0\
r(r-1)^2(r-2)=0
$$
Now,
- $r=0$ gives a pseudo solution $f(2n)=1,;f(2n+1)=0$ which leads to a contradiction later on;
- $r=1$ gives a possible solution of $f(n)=1$ which is ruled out by the condition $f(2017)ne f(2018)$;
- $r=2$ seems to be the only possible option, and also the one you know how to deal with.
answered Aug 27 at 14:21
Ivan Neretin
8,35821433
edited Aug 27 at 14:30
answered Aug 27 at 14:21
Ivan Neretin
8,35821433
answered Aug 27 at 14:21
Ivan Neretin
8,35821433
answered Aug 27 at 14:21
Ivan Neretin
8,35821433
8,35821433
add a comment |Â
add a comment |Â
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How do you see that $f(-1)=0$? After all, the constant function $f(x)=1$ satisfies the functional equation (though not the condition that $f(2017)neq f(2018)$ of course).
– lulu
Aug 27 at 13:45
@Alphanerd Good catch, but that version adds the condition that $f(1)=2$ which makes the problem quite easy. Was that condition meant to be present here?
– lulu
Aug 27 at 13:52
@Alphanerd Thanks a lot. But what do I do with this question now? Am I supposed to delete this?
– Hoque
Aug 27 at 13:52
Once again: the older version is quite different from this one unless you intended to add the condition that $f(1)=2$.
– lulu
Aug 27 at 13:53
The condition f(1)= 2 was not given in the question
– Hoque
Aug 27 at 13:54