Functional Equation: A, B, C Answers

[juantheron]

http://latex.codecogs.com/gif.latex?\hspace{-20}$%20A%20function%20$f:\mathbb{N}%20\rightarrow%20\mathbb{N}$%20and%20satisfies%20$f(ab)%20=%20f(a)+f(b)$.\\%20Where%20$a$%20and%20$b$%20are%20Coprime%20Natural%20no.\\%20and%20$f(c+d)%20=%20f(c)+f(d)\forall$%20prime%20no.%20$c$%20and%20$d$.%20Then\\%20(a)%20The%20value%20of%20$f(1)+f(2)+f(3)%20=%20$\\%20(b)%20$\frac{f(5)+f(7)}{f(4)}%20=$\\%20(c)%20$f(9)-f(6)+f(3)%20=%20$

 

[CaptainBlack]

http://latex.codecogs.com/gif.latex?\hspace{-20}$%20A%20function%20$f:\mathbb{N}%20\rightarrow%20\mathbb{N}$%20and%20satisfies%20$f(ab)%20=%20f(a)+f(b)$.\\%20Where%20$a$%20and%20$b$%20are%20Coprime%20Natural%20no.\\%20and%20$f(c+d)%20=%20f(c)+f(d)\forall$%20prime%20no.%20$c$%20and%20$d$.%20Then\\%20(a)%20The%20value%20of%20$f(1)+f(2)+f(3)%20=%20$\\%20(b)%20$\frac{f(5)+f(7)}{f(4)}%20=$\\%20(c)%20$f(9)-f(6)+f(3)%20=%20$

Since \(1\) is coprime to every natural \(f(1)=0\).

Also since for any prime \(c>2\) we have \(f(2c)=f(2)+f(c)\) and \(f(2c)=f(c+c)=f(c)+f(c)\) we conclude that \(f(c)=f(2)\), which is sufficient to allow us to answer (a), (b) and (c) in terms of \(f(2)\).

At present I don't see any means of evaluating \(f(2)\).CB

 

[Opalg]

Since \(1\) is coprime to every natural \(f(1)=0\).

Also since for any prime \(c>2\) we have \(f(2c)=f(2)+f(c)\) and \(f(2c)=f(c+c)=f(c)+f(c)\) we conclude that \(f(c)=f(2)\), which is sufficient to allow us to answer (a), (b) and (c) in terms of \(f(2)\).

CB

But that leads to something strange if you put $c=2$ and $d=3$, because it then follows that $f(2) = f(5) = f(2+3) = f(2) + f(3) = 2f(2).$ Thus $f(2)=0$ and hence $f(p)=0$ for every prime $p.$

 

[CaptainBlack]

But that leads to something strange if you put $c=2$ and $d=3$, because it then follows that $f(2) = f(5) = f(2+3) = f(2) + f(3) = 2f(2).$ Thus $f(2)=0$ and hence $f(p)=0$ for every prime $p.$

If that does not entail a contradiction then that gives us a full solution, including undefined for (b), it also answers the implied question I added to my post between you starting to reply and my seeing your reply :) . Alternativly there is a condition missing from the statement of the question.

CB

 

[CellCoree]

I would approach this functional equation by first examining its properties and implications. The equation states that for any two coprime natural numbers a and b, the function f will satisfy the equation f(ab) = f(a) + f(b). This means that the function is multiplicative, as the product of two numbers is equal to the sum of their individual values under the function.

Furthermore, the equation also states that for any two prime numbers c and d, the function will satisfy the equation f(c+d) = f(c) + f(d). This tells us that the function is also additive, as the sum of two prime numbers will have the same value under the function as the sum of their individual values.

Using these properties, we can now answer the three questions posed in the content.

(a) The value of f(1) + f(2) + f(3) can be determined by substituting in the values of a and b as 1 and 2, respectively. This gives us f(1*2) = f(1) + f(2), which simplifies to f(2) = f(1) + f(2). Since f(1) is equal to 0 (as 1 is coprime with all natural numbers), we can conclude that f(2) = f(2) + 0, and therefore f(1) + f(2) + f(3) = f(2) + f(3) = 2f(2). However, we cannot determine the exact value of f(2) without more information about the function f.

(b) The expression (f(5) + f(7))/f(4) can be simplified using the multiplicative property of the function. We know that f(5*7) = f(5) + f(7), and f(4) = f(2*2) = f(2) + f(2). Therefore, (f(5) + f(7))/f(4) = f(5*7)/f(2*2) = f(5) + f(7)/f(2) + f(2). Again, without more information about the function f, we cannot determine the exact value of this expression.

(c) Finally, we can use the additive property of the function to determine the value of f(9) - f