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zmudz
 one year ago
The function \(f(n)\)takes the integers to the real numbers such that
\(f(m + n) + f(m  n) = 2f(m) + 2f(n)\)
for all integers \(m\) and \(n\) and \(f(4) = 16\). Find \(f(n)\).
zmudz
 one year ago
The function \(f(n)\)takes the integers to the real numbers such that \(f(m + n) + f(m  n) = 2f(m) + 2f(n)\) for all integers \(m\) and \(n\) and \(f(4) = 16\). Find \(f(n)\).

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freckles
 one year ago
Best ResponseYou've already chosen the best response.2so far I have come up with a value for f(0) and I came up with the function f is even: how I did this: ... m+n=4 mn=4  2m=8 m=4 when n=0 so \[f(4+0)+f(40)=2f(4)+2f(0) \\ f(4)+f(4)=2f(4)+2 f(0) \\ 0=2 f(0) \\ f(0)=0 \\ \text{ now I set } m=0 \\ \text{ and so we have } \\ f(0+n)+f(0n)=2f(0)+2f(n) \\ f(n)+f(n)=0+2f(n) \\ f(n)=f(n) \text{ which says } f \text{ is even }\] so maybe we can use these two facts somehow or maybe not

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1me too, i got what you got but didn't see the link of f(4) =16 to the problem :(

freckles
 one year ago
Best ResponseYou've already chosen the best response.2well we can come with f(8) using that fact with the already found stuff

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f(4+4)+f(44)=2f(4)+2f(4) \\ f(8)+f(0)=4 f(4) \\ f(8)+0=4(16) \\ f(8)=64\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1It seems the function is f(x) = x^2

freckles
 one year ago
Best ResponseYou've already chosen the best response.2lol let's test \[(m+n)^2+(mn)^2=2m^2+2n^2 \\ m^2+2nm+n^2+m^22nm+n^2=2m^2+2n^2 \\ \] the equation holds for f(x)=x^2

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1You are well organized. I just look at f(0) =0 , f(4) =16, f(8) =64 to conclude that f(x) =x^2

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1oh, we are done. hahaha. it asked us to find f(n) and we get f(x) = x^2 hence f(n) = n^2 right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1yeah!! I think so, with your argument!!! we prove that f(n) = n^2 . And by your testing, we have the expression hold for any m, n in Z
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