Mixing
Finding $f(9999)$ if $f(1) = 2$ and $f(mn) = f(m) , f(n) - f(m+n) + 1001$ for $m$ or $n$ equal to $1$
$begingroup$
Given a function $f$ is defined for integers $m$ and $n$ as given:
$$f(mn) = f(m),f(n) - f(m+n) + 1001$$
where either $m$ or $n$ is equal to $1$, and $f(1) = 2$.
The problem itself is to prove that $$f(x) = f(x-1) + 1001$$ in order to find the value of $f(9999)$.
As such, what I've already tried is:
Replacing $n$ as $1$ and $m$ as $x$, and trying to solve from there.
$$f(x) = f(1) * f(x) - f(x+1) + 1001$$
$$f(x) = 2 * f(x) - f(x+1) + 1001,$$
Although I'm not sure how I should continue from here, it did occur that the negative sign could be manipulated in some way, but I'm fairly certain that $-f(x+1)$, cannot be rewritten as $f(-x-1)$, or anything of the sorts.
So if anyone has any thoughts or insight as how I should proceed, it would be greatly appreciated.
algebra-precalculus functions induction problem-solving functional-equations
edited Jan 3 at 1:41
Batominovski
1
asked Sep 26 '18 at 15:20
R.C.R.C.
182
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add a comment |
$begingroup$
Given a function $f$ is defined for integers $m$ and $n$ as given:
$$f(mn) = f(m),f(n) - f(m+n) + 1001$$
where either $m$ or $n$ is equal to $1$, and $f(1) = 2$.
The problem itself is to prove that $$f(x) = f(x-1) + 1001$$ in order to find the value of $f(9999)$.
As such, what I've already tried is:
Replacing $n$ as $1$ and $m$ as $x$, and trying to solve from there.
$$f(x) = f(1) * f(x) - f(x+1) + 1001$$
$$f(x) = 2 * f(x) - f(x+1) + 1001,$$
Although I'm not sure how I should continue from here, it did occur that the negative sign could be manipulated in some way, but I'm fairly certain that $-f(x+1)$, cannot be rewritten as $f(-x-1)$, or anything of the sorts.
So if anyone has any thoughts or insight as how I should proceed, it would be greatly appreciated.
algebra-precalculus functions induction problem-solving functional-equations
edited Jan 3 at 1:41
Batominovski
1
asked Sep 26 '18 at 15:20
R.C.R.C.
182
$endgroup$
$begingroup$
This is an oddly specific functional equation: "where either $m$ or $n$ is equal to $1$."
$endgroup$
– Batominovski
Sep 26 '18 at 18:32
add a comment |
$begingroup$
Given a function $f$ is defined for integers $m$ and $n$ as given:
$$f(mn) = f(m),f(n) - f(m+n) + 1001$$
where either $m$ or $n$ is equal to $1$, and $f(1) = 2$.
The problem itself is to prove that $$f(x) = f(x-1) + 1001$$ in order to find the value of $f(9999)$.
As such, what I've already tried is:
Replacing $n$ as $1$ and $m$ as $x$, and trying to solve from there.
$$f(x) = f(1) * f(x) - f(x+1) + 1001$$
$$f(x) = 2 * f(x) - f(x+1) + 1001,$$
Although I'm not sure how I should continue from here, it did occur that the negative sign could be manipulated in some way, but I'm fairly certain that $-f(x+1)$, cannot be rewritten as $f(-x-1)$, or anything of the sorts.
So if anyone has any thoughts or insight as how I should proceed, it would be greatly appreciated.
algebra-precalculus functions induction problem-solving functional-equations
edited Jan 3 at 1:41
Batominovski
1
asked Sep 26 '18 at 15:20
R.C.R.C.
182
$endgroup$
Given a function $f$ is defined for integers $m$ and $n$ as given:
$$f(mn) = f(m),f(n) - f(m+n) + 1001$$
where either $m$ or $n$ is equal to $1$, and $f(1) = 2$.
The problem itself is to prove that $$f(x) = f(x-1) + 1001$$ in order to find the value of $f(9999)$.
As such, what I've already tried is:
Replacing $n$ as $1$ and $m$ as $x$, and trying to solve from there.
$$f(x) = f(1) * f(x) - f(x+1) + 1001$$
$$f(x) = 2 * f(x) - f(x+1) + 1001,$$
Although I'm not sure how I should continue from here, it did occur that the negative sign could be manipulated in some way, but I'm fairly certain that $-f(x+1)$, cannot be rewritten as $f(-x-1)$, or anything of the sorts.
So if anyone has any thoughts or insight as how I should proceed, it would be greatly appreciated.
algebra-precalculus functions induction problem-solving functional-equations
algebra-precalculus functions induction problem-solving functional-equations
edited Jan 3 at 1:41
Batominovski
1
asked Sep 26 '18 at 15:20
R.C.R.C.
182
edited Jan 3 at 1:41
Batominovski
1
asked Sep 26 '18 at 15:20
R.C.R.C.
182
edited Jan 3 at 1:41
Batominovski
1
edited Jan 3 at 1:41
Batominovski
1
edited Jan 3 at 1:41
Batominovski
1
1
asked Sep 26 '18 at 15:20
R.C.R.C.
182
asked Sep 26 '18 at 15:20
R.C.R.C.
182
asked Sep 26 '18 at 15:20
R.C.R.C.
182
182
$begingroup$
This is an oddly specific functional equation: "where either $m$ or $n$ is equal to $1$."
$endgroup$
– Batominovski
Sep 26 '18 at 18:32
add a comment |
$begingroup$
This is an oddly specific functional equation: "where either $m$ or $n$ is equal to $1$."
$endgroup$
– Batominovski
Sep 26 '18 at 18:32
$begingroup$
This is an oddly specific functional equation: "where either $m$ or $n$ is equal to $1$."
$endgroup$
– Batominovski
Sep 26 '18 at 18:32
$begingroup$
This is an oddly specific functional equation: "where either $m$ or $n$ is equal to $1$."
$endgroup$
– Batominovski
Sep 26 '18 at 18:32
add a comment |
3 Answers
3
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oldest
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You got $$f(x)=2f(x)-f(x+1)+1001$$Now, let's find $f(x+1)$ using $f(x)$:$$f(x)=2f(x)-f(x+1)+1001iff f(x+1)=f(x)+1001$$Now, let $y=x+1, x=y-1$ you get $$f(y)=f(y-1)+1001$$
(Note, both $x$ and $y$ are just names of the variables, so if it is more convenient to you, you can rewrite the final line using $x$ instead of $y$)
answered Sep 26 '18 at 15:27
HoloHolo
5,60321030
$endgroup$
add a comment |
$begingroup$
Disclaimer. This is an attempt to see what happens if I drop the conditions that $f(1)=2$ and that at least one of $m$ and $n$ in the functional equation must be $1$. That is, this answer does not answer the OP's question, but I think it is a nice relevant question. Since the work is quite long, I have to put it here as an answer.
Fix $kinmathbb{C}$. Let $f:mathbb{Z}_{>0}tomathbb{C}$ be a function satisfying
$$f(mn)=f(m),f(n)-f(m+n)+k$$
for all $m,ninmathbb{Z}_{>0}$. Write $a:=f(1)$. Then, plugging in $m=1$, we have
$$f(n+1)=(a-1),f(n)+ktag{*}$$
for all $ninmathbb{Z}_{>0}$. We have three notable scenarios: $a=1$, $a=2$, and $anotin{1,2}$.
If $a=1$, then $f(n)=k$ for all $ninmathbb{Z}_{>1}$. That means
$$k=f(4)=f(2cdot2)=f(2),f(2)-f(2+2)+k=kcdot k-k+k,.$$
Thus, $k^2=k$, or $kin{0,1}$. Hence, there are two possible solutions with $f(1)=1$:
$k=0$, which gives $f(1)=1$ and $f(n)=0$ for all $ninmathbb{Z}_{>1}$;
$k=1$, which gives $f(n)=1$ for all $ninmathbb{Z}_{>0}$.
Now, we suppose that $a=2$. Thus, $f(n+1)=f(n)+k$ for all $ninmathbb{Z}_{>0}$. This leads to $f(n)=2+k(n-1)$ for all $ninmathbb{Z}_{>0}$. Consequently,
$$3k+2=f(4)=f(2cdot 2)=f(2),f(2)-f(2+2)+k=(k+2)^2-(3k+2)+k,,$$
yielding
$$k^2-k=0,.$$
Thus, $k=0$ or $k=1$. If $k=0$, then $f(n)=2$ for all $ninmathbb{Z}_{>0}$. If $k=1$, then $f(n)=n+1$ for all $ninmathbb{Z}_{>0}$.
Finally, we tackle the case $anotin {1,2}$. Then, the solution to the recurrence relation (*) is
$$f(n)=(a-1)^{n-1}left(a+frac{k}{a-2}right)-frac{k}{a-2},.$$
In particular, $f(2)=a(a-1)+k$ and $f(4)=a(a-1)^3+left((a-1)^2+(a-1)+1right)k$. Using the same equation $2,f(4)=big(f(2)big)^2+k$ from before, we get
$$2a(a-1)^3+2left((a-1)^2+(a-1)+1right)k=big(a(a-1)+kbig)^2+k,.$$
Thus,
$$a(a-2)(a-1)^2=k(k-1),.$$
This means $k=(a-1)^2$ or $k=-a(a-2)$. For the subcase $k=-a(a-2)$, we see that
$$f(n)=atext{ for all }ninmathbb{Z}_{>0},.$$
For the subcase $k=(a-1)^2$, we get
$$f(n)=a(a-1)^{n-1}+(a-1)^2left(frac{(a-1)^{n-1}-1}{a-2}right)text{ for each integer }n>0,.$$
For simplicity, write $b:=a-1$, so $k=b^2$ and $f(n)=(b+1)b^{n-1}+b^2left(dfrac{b^{n-1}-1}{b-1}right)$. We have
$$f(2)=(b+1)b+b^2,,,,f(3)=2(b+1)b^2,,$$ $$f(5)=(b+1)b^4+b^2(b^3+b^2+b+1),,$$ and $$f(6)=(b+1)b^5+b^2(b^4+b^3+b^2+b+1),.$$
Since $f(6)=f(2cdot3)=f(2),f(3)-f(2+3)+k=f(2),f(3)-f(5)+b^2$, we obtain
$$2b^6+2b^5+b^4+b^3+b^2=(2b^2+b)(2b^3+2b^2)-(2b^5+2b^4+b^3+b^2)+b^2,.$$
That is,
$$b^2(b^2-1)(2b^2-1)=2b^6-3b^4+b^2=0,.$$
Hence, $b=0$, $b=+1$, $b=-1$, $b=+dfrac1{sqrt{2}}$, or $b=-dfrac1{sqrt{2}}$.
Recall that $anotin{1,2}$. If $b=0$, then $a=1$, which is a contradiction. If $b=+1$, then $a=2$, another contradiction. If $b=-1$, then $a=0$, $k=1$, and
$$f(n)=frac{1+(-1)^{n}}{2}=left{begin{array}{ll}0,,&text{if }ntext{ is odd},,\
1,,&text{if }ntext{ is even},.
end{array}right.$$
For $b=+dfrac1{sqrt{2}}$, we get $a=1+dfrac1{sqrt{2}}$, $k=dfrac{1}{2}$, and
$$f(n)=1+frac1{sqrt{2}}text{ for all positive integers }n,.$$
For $b=-dfrac1{sqrt{2}}$, we get $a=1-dfrac1{sqrt{2}}$, $k=dfrac12$ and
$$f(n)=1-frac1{sqrt{2}}text{ for all positive integers }n,.$$
From the work above, we can now conclude our proposition. In particular, for $k=1001$, the proposition states that there are two solutions:
$$f(n)=1+10sqrt{10}text{i}text{ for each }ninmathbb{Z}_{>0}$$
and
$$f(n)=1-10sqrt{10}text{i}text{ for each }ninmathbb{Z}_{>0},.$$
That is, if you allow $m$ and $n$ in the functional equation to be any positive integer and drop the condition that $f(1)=2$, then the solutions look wildly different.
Proposition: Let $k$ be a complex number. Then, all solutions $f:mathbb{Z}_{>0}tomathbb{C}$ to the functional equation
$$f(mn)=f(m),f(n)-f(m+n)+ktext{ for all positive integers }mtext{ and }n$$
is given by the list below.
(a) If $k=-a(a-2)$ for some $ainmathbb{C}$, then $$f(n)=atext{ for all }ninmathbb{Z}_{>0},.$$ There are two values of $a$ for all $kinmathbb{C}setminus{1}$, namely, $a=1-sqrt{1-k}$ and $a=1+sqrt{1-k}$. For $k=1$, there is only one value of $a$, i.e., $a=1$.
(b) In the case $k=0$, there is one more solution apart from two solutions covered in (a), and this extra solution is given by
$$f(n)=left{begin{array}{ll}1,,&text{if }n=1,,\0,,&text{for }n=2,3,4,ldots,.end{array}right.$$
(c) In the case $k=1$, there are two more solutions apart from the one solution covered in (a). These solutions are the linear function
$$f(n)=n+1text{ for all }ninmathbb{Z}_{>0}$$ and the alternating function $$f(n)=1-(n,text{mod},2)=left{begin{array}{ll}0,,&text{if }ntext{ is odd},,\
1,,&text{if }ntext{ is even},.
end{array}right.$$
answered Sep 26 '18 at 20:45
BatominovskiBatominovski
1
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Given a function f is defined for integers m and n as given:
f(mn)=f(m)f(n)−f(m+n)+1001
where either m or n is equal to 1
As multiplication and addition is commutative we can assume, wolog $n = 1$ and that is just a convoluted way of simply saying $f(m) = 2f(m) -f(m+1) +1001$ or $f(m+1) = f(m) + 1001$. Which means, by induction, $f(m) = f(1) + 1001(m-1)$.
And as $f(1) = 2$ we have $f(m) = 2 + 1001(m-1)$ for all $min mathbb Z; m ne 1$.
So $f(9999) = 2 + 1001*9998=10008000$
answered Sep 26 '18 at 15:51
fleabloodfleablood
68.8k22685
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3 Answers
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$begingroup$
You got $$f(x)=2f(x)-f(x+1)+1001$$Now, let's find $f(x+1)$ using $f(x)$:$$f(x)=2f(x)-f(x+1)+1001iff f(x+1)=f(x)+1001$$Now, let $y=x+1, x=y-1$ you get $$f(y)=f(y-1)+1001$$
(Note, both $x$ and $y$ are just names of the variables, so if it is more convenient to you, you can rewrite the final line using $x$ instead of $y$)
answered Sep 26 '18 at 15:27
HoloHolo
5,60321030
$endgroup$
add a comment |
$begingroup$
You got $$f(x)=2f(x)-f(x+1)+1001$$Now, let's find $f(x+1)$ using $f(x)$:$$f(x)=2f(x)-f(x+1)+1001iff f(x+1)=f(x)+1001$$Now, let $y=x+1, x=y-1$ you get $$f(y)=f(y-1)+1001$$
(Note, both $x$ and $y$ are just names of the variables, so if it is more convenient to you, you can rewrite the final line using $x$ instead of $y$)
answered Sep 26 '18 at 15:27
HoloHolo
5,60321030
$endgroup$
add a comment |
$begingroup$
You got $$f(x)=2f(x)-f(x+1)+1001$$Now, let's find $f(x+1)$ using $f(x)$:$$f(x)=2f(x)-f(x+1)+1001iff f(x+1)=f(x)+1001$$Now, let $y=x+1, x=y-1$ you get $$f(y)=f(y-1)+1001$$
(Note, both $x$ and $y$ are just names of the variables, so if it is more convenient to you, you can rewrite the final line using $x$ instead of $y$)
answered Sep 26 '18 at 15:27
HoloHolo
5,60321030
$endgroup$
You got $$f(x)=2f(x)-f(x+1)+1001$$Now, let's find $f(x+1)$ using $f(x)$:$$f(x)=2f(x)-f(x+1)+1001iff f(x+1)=f(x)+1001$$Now, let $y=x+1, x=y-1$ you get $$f(y)=f(y-1)+1001$$
(Note, both $x$ and $y$ are just names of the variables, so if it is more convenient to you, you can rewrite the final line using $x$ instead of $y$)
answered Sep 26 '18 at 15:27
HoloHolo
5,60321030
answered Sep 26 '18 at 15:27
HoloHolo
5,60321030
answered Sep 26 '18 at 15:27
HoloHolo
5,60321030
answered Sep 26 '18 at 15:27
HoloHolo
5,60321030
5,60321030
add a comment |
add a comment |
$begingroup$
Disclaimer. This is an attempt to see what happens if I drop the conditions that $f(1)=2$ and that at least one of $m$ and $n$ in the functional equation must be $1$. That is, this answer does not answer the OP's question, but I think it is a nice relevant question. Since the work is quite long, I have to put it here as an answer.
Fix $kinmathbb{C}$. Let $f:mathbb{Z}_{>0}tomathbb{C}$ be a function satisfying
$$f(mn)=f(m),f(n)-f(m+n)+k$$
for all $m,ninmathbb{Z}_{>0}$. Write $a:=f(1)$. Then, plugging in $m=1$, we have
$$f(n+1)=(a-1),f(n)+ktag{*}$$
for all $ninmathbb{Z}_{>0}$. We have three notable scenarios: $a=1$, $a=2$, and $anotin{1,2}$.
If $a=1$, then $f(n)=k$ for all $ninmathbb{Z}_{>1}$. That means
$$k=f(4)=f(2cdot2)=f(2),f(2)-f(2+2)+k=kcdot k-k+k,.$$
Thus, $k^2=k$, or $kin{0,1}$. Hence, there are two possible solutions with $f(1)=1$:
$k=0$, which gives $f(1)=1$ and $f(n)=0$ for all $ninmathbb{Z}_{>1}$;
$k=1$, which gives $f(n)=1$ for all $ninmathbb{Z}_{>0}$.
Now, we suppose that $a=2$. Thus, $f(n+1)=f(n)+k$ for all $ninmathbb{Z}_{>0}$. This leads to $f(n)=2+k(n-1)$ for all $ninmathbb{Z}_{>0}$. Consequently,
$$3k+2=f(4)=f(2cdot 2)=f(2),f(2)-f(2+2)+k=(k+2)^2-(3k+2)+k,,$$
yielding
$$k^2-k=0,.$$
Thus, $k=0$ or $k=1$. If $k=0$, then $f(n)=2$ for all $ninmathbb{Z}_{>0}$. If $k=1$, then $f(n)=n+1$ for all $ninmathbb{Z}_{>0}$.
Finally, we tackle the case $anotin {1,2}$. Then, the solution to the recurrence relation (*) is
$$f(n)=(a-1)^{n-1}left(a+frac{k}{a-2}right)-frac{k}{a-2},.$$
In particular, $f(2)=a(a-1)+k$ and $f(4)=a(a-1)^3+left((a-1)^2+(a-1)+1right)k$. Using the same equation $2,f(4)=big(f(2)big)^2+k$ from before, we get
$$2a(a-1)^3+2left((a-1)^2+(a-1)+1right)k=big(a(a-1)+kbig)^2+k,.$$
Thus,
$$a(a-2)(a-1)^2=k(k-1),.$$
This means $k=(a-1)^2$ or $k=-a(a-2)$. For the subcase $k=-a(a-2)$, we see that
$$f(n)=atext{ for all }ninmathbb{Z}_{>0},.$$
For the subcase $k=(a-1)^2$, we get
$$f(n)=a(a-1)^{n-1}+(a-1)^2left(frac{(a-1)^{n-1}-1}{a-2}right)text{ for each integer }n>0,.$$
For simplicity, write $b:=a-1$, so $k=b^2$ and $f(n)=(b+1)b^{n-1}+b^2left(dfrac{b^{n-1}-1}{b-1}right)$. We have
$$f(2)=(b+1)b+b^2,,,,f(3)=2(b+1)b^2,,$$ $$f(5)=(b+1)b^4+b^2(b^3+b^2+b+1),,$$ and $$f(6)=(b+1)b^5+b^2(b^4+b^3+b^2+b+1),.$$
Since $f(6)=f(2cdot3)=f(2),f(3)-f(2+3)+k=f(2),f(3)-f(5)+b^2$, we obtain
$$2b^6+2b^5+b^4+b^3+b^2=(2b^2+b)(2b^3+2b^2)-(2b^5+2b^4+b^3+b^2)+b^2,.$$
That is,
$$b^2(b^2-1)(2b^2-1)=2b^6-3b^4+b^2=0,.$$
Hence, $b=0$, $b=+1$, $b=-1$, $b=+dfrac1{sqrt{2}}$, or $b=-dfrac1{sqrt{2}}$.
Recall that $anotin{1,2}$. If $b=0$, then $a=1$, which is a contradiction. If $b=+1$, then $a=2$, another contradiction. If $b=-1$, then $a=0$, $k=1$, and
$$f(n)=frac{1+(-1)^{n}}{2}=left{begin{array}{ll}0,,&text{if }ntext{ is odd},,\
1,,&text{if }ntext{ is even},.
end{array}right.$$
For $b=+dfrac1{sqrt{2}}$, we get $a=1+dfrac1{sqrt{2}}$, $k=dfrac{1}{2}$, and
$$f(n)=1+frac1{sqrt{2}}text{ for all positive integers }n,.$$
For $b=-dfrac1{sqrt{2}}$, we get $a=1-dfrac1{sqrt{2}}$, $k=dfrac12$ and
$$f(n)=1-frac1{sqrt{2}}text{ for all positive integers }n,.$$
From the work above, we can now conclude our proposition. In particular, for $k=1001$, the proposition states that there are two solutions:
$$f(n)=1+10sqrt{10}text{i}text{ for each }ninmathbb{Z}_{>0}$$
and
$$f(n)=1-10sqrt{10}text{i}text{ for each }ninmathbb{Z}_{>0},.$$
That is, if you allow $m$ and $n$ in the functional equation to be any positive integer and drop the condition that $f(1)=2$, then the solutions look wildly different.
Proposition: Let $k$ be a complex number. Then, all solutions $f:mathbb{Z}_{>0}tomathbb{C}$ to the functional equation
$$f(mn)=f(m),f(n)-f(m+n)+ktext{ for all positive integers }mtext{ and }n$$
is given by the list below.
(a) If $k=-a(a-2)$ for some $ainmathbb{C}$, then $$f(n)=atext{ for all }ninmathbb{Z}_{>0},.$$ There are two values of $a$ for all $kinmathbb{C}setminus{1}$, namely, $a=1-sqrt{1-k}$ and $a=1+sqrt{1-k}$. For $k=1$, there is only one value of $a$, i.e., $a=1$.
(b) In the case $k=0$, there is one more solution apart from two solutions covered in (a), and this extra solution is given by
$$f(n)=left{begin{array}{ll}1,,&text{if }n=1,,\0,,&text{for }n=2,3,4,ldots,.end{array}right.$$
(c) In the case $k=1$, there are two more solutions apart from the one solution covered in (a). These solutions are the linear function
$$f(n)=n+1text{ for all }ninmathbb{Z}_{>0}$$ and the alternating function $$f(n)=1-(n,text{mod},2)=left{begin{array}{ll}0,,&text{if }ntext{ is odd},,\
1,,&text{if }ntext{ is even},.
end{array}right.$$
answered Sep 26 '18 at 20:45
BatominovskiBatominovski
1
$endgroup$
add a comment |
$begingroup$
Disclaimer. This is an attempt to see what happens if I drop the conditions that $f(1)=2$ and that at least one of $m$ and $n$ in the functional equation must be $1$. That is, this answer does not answer the OP's question, but I think it is a nice relevant question. Since the work is quite long, I have to put it here as an answer.
Fix $kinmathbb{C}$. Let $f:mathbb{Z}_{>0}tomathbb{C}$ be a function satisfying
$$f(mn)=f(m),f(n)-f(m+n)+k$$
for all $m,ninmathbb{Z}_{>0}$. Write $a:=f(1)$. Then, plugging in $m=1$, we have
$$f(n+1)=(a-1),f(n)+ktag{*}$$
for all $ninmathbb{Z}_{>0}$. We have three notable scenarios: $a=1$, $a=2$, and $anotin{1,2}$.
If $a=1$, then $f(n)=k$ for all $ninmathbb{Z}_{>1}$. That means
$$k=f(4)=f(2cdot2)=f(2),f(2)-f(2+2)+k=kcdot k-k+k,.$$
Thus, $k^2=k$, or $kin{0,1}$. Hence, there are two possible solutions with $f(1)=1$:
$k=0$, which gives $f(1)=1$ and $f(n)=0$ for all $ninmathbb{Z}_{>1}$;
$k=1$, which gives $f(n)=1$ for all $ninmathbb{Z}_{>0}$.
Now, we suppose that $a=2$. Thus, $f(n+1)=f(n)+k$ for all $ninmathbb{Z}_{>0}$. This leads to $f(n)=2+k(n-1)$ for all $ninmathbb{Z}_{>0}$. Consequently,
$$3k+2=f(4)=f(2cdot 2)=f(2),f(2)-f(2+2)+k=(k+2)^2-(3k+2)+k,,$$
yielding
$$k^2-k=0,.$$
Thus, $k=0$ or $k=1$. If $k=0$, then $f(n)=2$ for all $ninmathbb{Z}_{>0}$. If $k=1$, then $f(n)=n+1$ for all $ninmathbb{Z}_{>0}$.
Finally, we tackle the case $anotin {1,2}$. Then, the solution to the recurrence relation (*) is
$$f(n)=(a-1)^{n-1}left(a+frac{k}{a-2}right)-frac{k}{a-2},.$$
In particular, $f(2)=a(a-1)+k$ and $f(4)=a(a-1)^3+left((a-1)^2+(a-1)+1right)k$. Using the same equation $2,f(4)=big(f(2)big)^2+k$ from before, we get
$$2a(a-1)^3+2left((a-1)^2+(a-1)+1right)k=big(a(a-1)+kbig)^2+k,.$$
Thus,
$$a(a-2)(a-1)^2=k(k-1),.$$
This means $k=(a-1)^2$ or $k=-a(a-2)$. For the subcase $k=-a(a-2)$, we see that
$$f(n)=atext{ for all }ninmathbb{Z}_{>0},.$$
For the subcase $k=(a-1)^2$, we get
$$f(n)=a(a-1)^{n-1}+(a-1)^2left(frac{(a-1)^{n-1}-1}{a-2}right)text{ for each integer }n>0,.$$
For simplicity, write $b:=a-1$, so $k=b^2$ and $f(n)=(b+1)b^{n-1}+b^2left(dfrac{b^{n-1}-1}{b-1}right)$. We have
$$f(2)=(b+1)b+b^2,,,,f(3)=2(b+1)b^2,,$$ $$f(5)=(b+1)b^4+b^2(b^3+b^2+b+1),,$$ and $$f(6)=(b+1)b^5+b^2(b^4+b^3+b^2+b+1),.$$
Since $f(6)=f(2cdot3)=f(2),f(3)-f(2+3)+k=f(2),f(3)-f(5)+b^2$, we obtain
$$2b^6+2b^5+b^4+b^3+b^2=(2b^2+b)(2b^3+2b^2)-(2b^5+2b^4+b^3+b^2)+b^2,.$$
That is,
$$b^2(b^2-1)(2b^2-1)=2b^6-3b^4+b^2=0,.$$
Hence, $b=0$, $b=+1$, $b=-1$, $b=+dfrac1{sqrt{2}}$, or $b=-dfrac1{sqrt{2}}$.
Recall that $anotin{1,2}$. If $b=0$, then $a=1$, which is a contradiction. If $b=+1$, then $a=2$, another contradiction. If $b=-1$, then $a=0$, $k=1$, and
$$f(n)=frac{1+(-1)^{n}}{2}=left{begin{array}{ll}0,,&text{if }ntext{ is odd},,\
1,,&text{if }ntext{ is even},.
end{array}right.$$
For $b=+dfrac1{sqrt{2}}$, we get $a=1+dfrac1{sqrt{2}}$, $k=dfrac{1}{2}$, and
$$f(n)=1+frac1{sqrt{2}}text{ for all positive integers }n,.$$
For $b=-dfrac1{sqrt{2}}$, we get $a=1-dfrac1{sqrt{2}}$, $k=dfrac12$ and
$$f(n)=1-frac1{sqrt{2}}text{ for all positive integers }n,.$$
From the work above, we can now conclude our proposition. In particular, for $k=1001$, the proposition states that there are two solutions:
$$f(n)=1+10sqrt{10}text{i}text{ for each }ninmathbb{Z}_{>0}$$
and
$$f(n)=1-10sqrt{10}text{i}text{ for each }ninmathbb{Z}_{>0},.$$
That is, if you allow $m$ and $n$ in the functional equation to be any positive integer and drop the condition that $f(1)=2$, then the solutions look wildly different.
Proposition: Let $k$ be a complex number. Then, all solutions $f:mathbb{Z}_{>0}tomathbb{C}$ to the functional equation
$$f(mn)=f(m),f(n)-f(m+n)+ktext{ for all positive integers }mtext{ and }n$$
is given by the list below.
(a) If $k=-a(a-2)$ for some $ainmathbb{C}$, then $$f(n)=atext{ for all }ninmathbb{Z}_{>0},.$$ There are two values of $a$ for all $kinmathbb{C}setminus{1}$, namely, $a=1-sqrt{1-k}$ and $a=1+sqrt{1-k}$. For $k=1$, there is only one value of $a$, i.e., $a=1$.
(b) In the case $k=0$, there is one more solution apart from two solutions covered in (a), and this extra solution is given by
$$f(n)=left{begin{array}{ll}1,,&text{if }n=1,,\0,,&text{for }n=2,3,4,ldots,.end{array}right.$$
(c) In the case $k=1$, there are two more solutions apart from the one solution covered in (a). These solutions are the linear function
$$f(n)=n+1text{ for all }ninmathbb{Z}_{>0}$$ and the alternating function $$f(n)=1-(n,text{mod},2)=left{begin{array}{ll}0,,&text{if }ntext{ is odd},,\
1,,&text{if }ntext{ is even},.
end{array}right.$$
answered Sep 26 '18 at 20:45
BatominovskiBatominovski
1
$endgroup$
add a comment |
$begingroup$
Disclaimer. This is an attempt to see what happens if I drop the conditions that $f(1)=2$ and that at least one of $m$ and $n$ in the functional equation must be $1$. That is, this answer does not answer the OP's question, but I think it is a nice relevant question. Since the work is quite long, I have to put it here as an answer.
Fix $kinmathbb{C}$. Let $f:mathbb{Z}_{>0}tomathbb{C}$ be a function satisfying
$$f(mn)=f(m),f(n)-f(m+n)+k$$
for all $m,ninmathbb{Z}_{>0}$. Write $a:=f(1)$. Then, plugging in $m=1$, we have
$$f(n+1)=(a-1),f(n)+ktag{*}$$
for all $ninmathbb{Z}_{>0}$. We have three notable scenarios: $a=1$, $a=2$, and $anotin{1,2}$.
If $a=1$, then $f(n)=k$ for all $ninmathbb{Z}_{>1}$. That means
$$k=f(4)=f(2cdot2)=f(2),f(2)-f(2+2)+k=kcdot k-k+k,.$$
Thus, $k^2=k$, or $kin{0,1}$. Hence, there are two possible solutions with $f(1)=1$:
$k=0$, which gives $f(1)=1$ and $f(n)=0$ for all $ninmathbb{Z}_{>1}$;
$k=1$, which gives $f(n)=1$ for all $ninmathbb{Z}_{>0}$.
Now, we suppose that $a=2$. Thus, $f(n+1)=f(n)+k$ for all $ninmathbb{Z}_{>0}$. This leads to $f(n)=2+k(n-1)$ for all $ninmathbb{Z}_{>0}$. Consequently,
$$3k+2=f(4)=f(2cdot 2)=f(2),f(2)-f(2+2)+k=(k+2)^2-(3k+2)+k,,$$
yielding
$$k^2-k=0,.$$
Thus, $k=0$ or $k=1$. If $k=0$, then $f(n)=2$ for all $ninmathbb{Z}_{>0}$. If $k=1$, then $f(n)=n+1$ for all $ninmathbb{Z}_{>0}$.
Finally, we tackle the case $anotin {1,2}$. Then, the solution to the recurrence relation (*) is
$$f(n)=(a-1)^{n-1}left(a+frac{k}{a-2}right)-frac{k}{a-2},.$$
In particular, $f(2)=a(a-1)+k$ and $f(4)=a(a-1)^3+left((a-1)^2+(a-1)+1right)k$. Using the same equation $2,f(4)=big(f(2)big)^2+k$ from before, we get
$$2a(a-1)^3+2left((a-1)^2+(a-1)+1right)k=big(a(a-1)+kbig)^2+k,.$$
Thus,
$$a(a-2)(a-1)^2=k(k-1),.$$
This means $k=(a-1)^2$ or $k=-a(a-2)$. For the subcase $k=-a(a-2)$, we see that
$$f(n)=atext{ for all }ninmathbb{Z}_{>0},.$$
For the subcase $k=(a-1)^2$, we get
$$f(n)=a(a-1)^{n-1}+(a-1)^2left(frac{(a-1)^{n-1}-1}{a-2}right)text{ for each integer }n>0,.$$
For simplicity, write $b:=a-1$, so $k=b^2$ and $f(n)=(b+1)b^{n-1}+b^2left(dfrac{b^{n-1}-1}{b-1}right)$. We have
$$f(2)=(b+1)b+b^2,,,,f(3)=2(b+1)b^2,,$$ $$f(5)=(b+1)b^4+b^2(b^3+b^2+b+1),,$$ and $$f(6)=(b+1)b^5+b^2(b^4+b^3+b^2+b+1),.$$
Since $f(6)=f(2cdot3)=f(2),f(3)-f(2+3)+k=f(2),f(3)-f(5)+b^2$, we obtain
$$2b^6+2b^5+b^4+b^3+b^2=(2b^2+b)(2b^3+2b^2)-(2b^5+2b^4+b^3+b^2)+b^2,.$$
That is,
$$b^2(b^2-1)(2b^2-1)=2b^6-3b^4+b^2=0,.$$
Hence, $b=0$, $b=+1$, $b=-1$, $b=+dfrac1{sqrt{2}}$, or $b=-dfrac1{sqrt{2}}$.
Recall that $anotin{1,2}$. If $b=0$, then $a=1$, which is a contradiction. If $b=+1$, then $a=2$, another contradiction. If $b=-1$, then $a=0$, $k=1$, and
$$f(n)=frac{1+(-1)^{n}}{2}=left{begin{array}{ll}0,,&text{if }ntext{ is odd},,\
1,,&text{if }ntext{ is even},.
end{array}right.$$
For $b=+dfrac1{sqrt{2}}$, we get $a=1+dfrac1{sqrt{2}}$, $k=dfrac{1}{2}$, and
$$f(n)=1+frac1{sqrt{2}}text{ for all positive integers }n,.$$
For $b=-dfrac1{sqrt{2}}$, we get $a=1-dfrac1{sqrt{2}}$, $k=dfrac12$ and
$$f(n)=1-frac1{sqrt{2}}text{ for all positive integers }n,.$$
From the work above, we can now conclude our proposition. In particular, for $k=1001$, the proposition states that there are two solutions:
$$f(n)=1+10sqrt{10}text{i}text{ for each }ninmathbb{Z}_{>0}$$
and
$$f(n)=1-10sqrt{10}text{i}text{ for each }ninmathbb{Z}_{>0},.$$
That is, if you allow $m$ and $n$ in the functional equation to be any positive integer and drop the condition that $f(1)=2$, then the solutions look wildly different.
Proposition: Let $k$ be a complex number. Then, all solutions $f:mathbb{Z}_{>0}tomathbb{C}$ to the functional equation
$$f(mn)=f(m),f(n)-f(m+n)+ktext{ for all positive integers }mtext{ and }n$$
is given by the list below.
(a) If $k=-a(a-2)$ for some $ainmathbb{C}$, then $$f(n)=atext{ for all }ninmathbb{Z}_{>0},.$$ There are two values of $a$ for all $kinmathbb{C}setminus{1}$, namely, $a=1-sqrt{1-k}$ and $a=1+sqrt{1-k}$. For $k=1$, there is only one value of $a$, i.e., $a=1$.
(b) In the case $k=0$, there is one more solution apart from two solutions covered in (a), and this extra solution is given by
$$f(n)=left{begin{array}{ll}1,,&text{if }n=1,,\0,,&text{for }n=2,3,4,ldots,.end{array}right.$$
(c) In the case $k=1$, there are two more solutions apart from the one solution covered in (a). These solutions are the linear function
$$f(n)=n+1text{ for all }ninmathbb{Z}_{>0}$$ and the alternating function $$f(n)=1-(n,text{mod},2)=left{begin{array}{ll}0,,&text{if }ntext{ is odd},,\
1,,&text{if }ntext{ is even},.
end{array}right.$$
answered Sep 26 '18 at 20:45
BatominovskiBatominovski
1
$endgroup$
Disclaimer. This is an attempt to see what happens if I drop the conditions that $f(1)=2$ and that at least one of $m$ and $n$ in the functional equation must be $1$. That is, this answer does not answer the OP's question, but I think it is a nice relevant question. Since the work is quite long, I have to put it here as an answer.
Fix $kinmathbb{C}$. Let $f:mathbb{Z}_{>0}tomathbb{C}$ be a function satisfying
$$f(mn)=f(m),f(n)-f(m+n)+k$$
for all $m,ninmathbb{Z}_{>0}$. Write $a:=f(1)$. Then, plugging in $m=1$, we have
$$f(n+1)=(a-1),f(n)+ktag{*}$$
for all $ninmathbb{Z}_{>0}$. We have three notable scenarios: $a=1$, $a=2$, and $anotin{1,2}$.
If $a=1$, then $f(n)=k$ for all $ninmathbb{Z}_{>1}$. That means
$$k=f(4)=f(2cdot2)=f(2),f(2)-f(2+2)+k=kcdot k-k+k,.$$
Thus, $k^2=k$, or $kin{0,1}$. Hence, there are two possible solutions with $f(1)=1$:
$k=0$, which gives $f(1)=1$ and $f(n)=0$ for all $ninmathbb{Z}_{>1}$;
$k=1$, which gives $f(n)=1$ for all $ninmathbb{Z}_{>0}$.
Now, we suppose that $a=2$. Thus, $f(n+1)=f(n)+k$ for all $ninmathbb{Z}_{>0}$. This leads to $f(n)=2+k(n-1)$ for all $ninmathbb{Z}_{>0}$. Consequently,
$$3k+2=f(4)=f(2cdot 2)=f(2),f(2)-f(2+2)+k=(k+2)^2-(3k+2)+k,,$$
yielding
$$k^2-k=0,.$$
Thus, $k=0$ or $k=1$. If $k=0$, then $f(n)=2$ for all $ninmathbb{Z}_{>0}$. If $k=1$, then $f(n)=n+1$ for all $ninmathbb{Z}_{>0}$.
Finally, we tackle the case $anotin {1,2}$. Then, the solution to the recurrence relation (*) is
$$f(n)=(a-1)^{n-1}left(a+frac{k}{a-2}right)-frac{k}{a-2},.$$
In particular, $f(2)=a(a-1)+k$ and $f(4)=a(a-1)^3+left((a-1)^2+(a-1)+1right)k$. Using the same equation $2,f(4)=big(f(2)big)^2+k$ from before, we get
$$2a(a-1)^3+2left((a-1)^2+(a-1)+1right)k=big(a(a-1)+kbig)^2+k,.$$
Thus,
$$a(a-2)(a-1)^2=k(k-1),.$$
This means $k=(a-1)^2$ or $k=-a(a-2)$. For the subcase $k=-a(a-2)$, we see that
$$f(n)=atext{ for all }ninmathbb{Z}_{>0},.$$
For the subcase $k=(a-1)^2$, we get
$$f(n)=a(a-1)^{n-1}+(a-1)^2left(frac{(a-1)^{n-1}-1}{a-2}right)text{ for each integer }n>0,.$$
For simplicity, write $b:=a-1$, so $k=b^2$ and $f(n)=(b+1)b^{n-1}+b^2left(dfrac{b^{n-1}-1}{b-1}right)$. We have
$$f(2)=(b+1)b+b^2,,,,f(3)=2(b+1)b^2,,$$ $$f(5)=(b+1)b^4+b^2(b^3+b^2+b+1),,$$ and $$f(6)=(b+1)b^5+b^2(b^4+b^3+b^2+b+1),.$$
Since $f(6)=f(2cdot3)=f(2),f(3)-f(2+3)+k=f(2),f(3)-f(5)+b^2$, we obtain
$$2b^6+2b^5+b^4+b^3+b^2=(2b^2+b)(2b^3+2b^2)-(2b^5+2b^4+b^3+b^2)+b^2,.$$
That is,
$$b^2(b^2-1)(2b^2-1)=2b^6-3b^4+b^2=0,.$$
Hence, $b=0$, $b=+1$, $b=-1$, $b=+dfrac1{sqrt{2}}$, or $b=-dfrac1{sqrt{2}}$.
Recall that $anotin{1,2}$. If $b=0$, then $a=1$, which is a contradiction. If $b=+1$, then $a=2$, another contradiction. If $b=-1$, then $a=0$, $k=1$, and
$$f(n)=frac{1+(-1)^{n}}{2}=left{begin{array}{ll}0,,&text{if }ntext{ is odd},,\
1,,&text{if }ntext{ is even},.
end{array}right.$$
For $b=+dfrac1{sqrt{2}}$, we get $a=1+dfrac1{sqrt{2}}$, $k=dfrac{1}{2}$, and
$$f(n)=1+frac1{sqrt{2}}text{ for all positive integers }n,.$$
For $b=-dfrac1{sqrt{2}}$, we get $a=1-dfrac1{sqrt{2}}$, $k=dfrac12$ and
$$f(n)=1-frac1{sqrt{2}}text{ for all positive integers }n,.$$
From the work above, we can now conclude our proposition. In particular, for $k=1001$, the proposition states that there are two solutions:
$$f(n)=1+10sqrt{10}text{i}text{ for each }ninmathbb{Z}_{>0}$$
and
$$f(n)=1-10sqrt{10}text{i}text{ for each }ninmathbb{Z}_{>0},.$$
That is, if you allow $m$ and $n$ in the functional equation to be any positive integer and drop the condition that $f(1)=2$, then the solutions look wildly different.
Proposition: Let $k$ be a complex number. Then, all solutions $f:mathbb{Z}_{>0}tomathbb{C}$ to the functional equation
$$f(mn)=f(m),f(n)-f(m+n)+ktext{ for all positive integers }mtext{ and }n$$
is given by the list below.
(a) If $k=-a(a-2)$ for some $ainmathbb{C}$, then $$f(n)=atext{ for all }ninmathbb{Z}_{>0},.$$ There are two values of $a$ for all $kinmathbb{C}setminus{1}$, namely, $a=1-sqrt{1-k}$ and $a=1+sqrt{1-k}$. For $k=1$, there is only one value of $a$, i.e., $a=1$.
(b) In the case $k=0$, there is one more solution apart from two solutions covered in (a), and this extra solution is given by
$$f(n)=left{begin{array}{ll}1,,&text{if }n=1,,\0,,&text{for }n=2,3,4,ldots,.end{array}right.$$
(c) In the case $k=1$, there are two more solutions apart from the one solution covered in (a). These solutions are the linear function
$$f(n)=n+1text{ for all }ninmathbb{Z}_{>0}$$ and the alternating function $$f(n)=1-(n,text{mod},2)=left{begin{array}{ll}0,,&text{if }ntext{ is odd},,\
1,,&text{if }ntext{ is even},.
end{array}right.$$
answered Sep 26 '18 at 20:45
BatominovskiBatominovski
1
edited Sep 26 '18 at 20:55
answered Sep 26 '18 at 20:45
BatominovskiBatominovski
1
answered Sep 26 '18 at 20:45
BatominovskiBatominovski
1
answered Sep 26 '18 at 20:45
BatominovskiBatominovski
1
1
add a comment |
add a comment |
$begingroup$
Given a function f is defined for integers m and n as given:
f(mn)=f(m)f(n)−f(m+n)+1001
where either m or n is equal to 1
As multiplication and addition is commutative we can assume, wolog $n = 1$ and that is just a convoluted way of simply saying $f(m) = 2f(m) -f(m+1) +1001$ or $f(m+1) = f(m) + 1001$. Which means, by induction, $f(m) = f(1) + 1001(m-1)$.
And as $f(1) = 2$ we have $f(m) = 2 + 1001(m-1)$ for all $min mathbb Z; m ne 1$.
So $f(9999) = 2 + 1001*9998=10008000$
answered Sep 26 '18 at 15:51
fleabloodfleablood
68.8k22685
$endgroup$
add a comment |
$begingroup$
Given a function f is defined for integers m and n as given:
f(mn)=f(m)f(n)−f(m+n)+1001
where either m or n is equal to 1
As multiplication and addition is commutative we can assume, wolog $n = 1$ and that is just a convoluted way of simply saying $f(m) = 2f(m) -f(m+1) +1001$ or $f(m+1) = f(m) + 1001$. Which means, by induction, $f(m) = f(1) + 1001(m-1)$.
And as $f(1) = 2$ we have $f(m) = 2 + 1001(m-1)$ for all $min mathbb Z; m ne 1$.
So $f(9999) = 2 + 1001*9998=10008000$
answered Sep 26 '18 at 15:51
fleabloodfleablood
68.8k22685
$endgroup$
add a comment |
$begingroup$
Given a function f is defined for integers m and n as given:
f(mn)=f(m)f(n)−f(m+n)+1001
where either m or n is equal to 1
As multiplication and addition is commutative we can assume, wolog $n = 1$ and that is just a convoluted way of simply saying $f(m) = 2f(m) -f(m+1) +1001$ or $f(m+1) = f(m) + 1001$. Which means, by induction, $f(m) = f(1) + 1001(m-1)$.
And as $f(1) = 2$ we have $f(m) = 2 + 1001(m-1)$ for all $min mathbb Z; m ne 1$.
So $f(9999) = 2 + 1001*9998=10008000$
answered Sep 26 '18 at 15:51
fleabloodfleablood
68.8k22685
$endgroup$
Given a function f is defined for integers m and n as given:
f(mn)=f(m)f(n)−f(m+n)+1001
where either m or n is equal to 1
As multiplication and addition is commutative we can assume, wolog $n = 1$ and that is just a convoluted way of simply saying $f(m) = 2f(m) -f(m+1) +1001$ or $f(m+1) = f(m) + 1001$. Which means, by induction, $f(m) = f(1) + 1001(m-1)$.
And as $f(1) = 2$ we have $f(m) = 2 + 1001(m-1)$ for all $min mathbb Z; m ne 1$.
So $f(9999) = 2 + 1001*9998=10008000$
answered Sep 26 '18 at 15:51
fleabloodfleablood
68.8k22685
answered Sep 26 '18 at 15:51
fleabloodfleablood
68.8k22685
answered Sep 26 '18 at 15:51
fleabloodfleablood
68.8k22685
answered Sep 26 '18 at 15:51
fleabloodfleablood
68.8k22685
68.8k22685
add a comment |
add a comment |
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$begingroup$
This is an oddly specific functional equation: "where either $m$ or $n$ is equal to $1$."
$endgroup$
– Batominovski
Sep 26 '18 at 18:32