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zmudz
 one year ago
If \(f(a + b) = f(a) + f(b)  2f(ab)\) for all nonnegative integers \(a\) and \(b\), and \(f(1) = 1\), compute \(f(1986)\).
zmudz
 one year ago
If \(f(a + b) = f(a) + f(b)  2f(ab)\) for all nonnegative integers \(a\) and \(b\), and \(f(1) = 1\), compute \(f(1986)\).

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4\[f(n) = f(1) + f(n1)  2f(1) f(n1)\] \[ \implies f(n) + f(n1) = 1\] f(2) = 0, f(3) = 1, f(4) = 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that looks really neat!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4is there proper mathese for this? ie a formal way to write it up?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2to me it looks good the way it is now lets ask @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I don't see a problem with what was said

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2what follows is really not necessary : \(f(n)+f(n1)=1\) is a recurrence relation so we can solve it formally clearly a particular solution is \(f(n)=\frac{1}{2}\) for homogeneous solution, the characteristic equation is \(r+1=0\\\implies r=1\) so the homogeneous solution is \(c*(1)^n\) therefore the complete solution is given by \(f(n)=\frac{1}{2}+c*(1)^n\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(c\) can be solved from the given initial condition, but i prefer just seeing the 1,0,1,0... pattern to the donkey work of solving the recurrence relation :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@IrishBoy123 I don't get how you link the problem to f(n) = f(1) +.....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4that what you mean?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Got it, it is a smart interpretation.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4i've tried googling "homogeneous solution, the characteristic equation " and all i'm getting are references to DE's.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4ie i am trying to work out @ganeshie8 's post

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4honestly, just a steer fpr reading later, if there is one

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2he was letting \(a=1\) and \(b=n1\) @Loser66 for sure i couldn't have figured out that substitution on my own

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1but the way he said n = n1+1 is a perfect way to determine the function f(n). wwwwwwwwoah!!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I love this site, I learn the new thing everyday.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 I don't think he set a =1, b = n1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1just f(n) in general and link it to f(1) then steer to a completely different from the original one. Solve the problem in general case, not just 1986 = 1+ 1985

freckles
 one year ago
Best ResponseYou've already chosen the best response.1http://openstudy.com/users/jagr2713#/updates/559d6afde4b0f93dd7c2b398 this problem reminds me of this one

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2at least thats how i interpreted his solution that induction idea seems interesting too

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yeah I interpret that to a=1 and b=n1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1OMG, so I am the only one person who has the weirdest interpretation? hehehe... It's ok, I am crazy originally.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[(f(a + b) = f(a) + f(b)  2f(ab)) \\ a=1 \text{ and } b=n1 \text{ gives } \\ f(1+(n1))=f(1)+f(n1)2f(n1)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2@IrishBoy123 i think that pattern \(\large x_{complete} = x_{particular} + x_{homogeneous}\) is seen everywhere in math including linear algebra, number theory, differential equations etc...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2as you know in linear algebra, to solve the system \(Ax=b\), the process is : 1) find a particular solution 2) find null solution then put the complete solution : \(\large x_{complete} = x_{particular} + x_{null}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2in number theory, to solve a diophantine equation like \(2a+3b=7\), the process is : 1) find a particular solution 2) find null solution then put the complete solution : \(\large x_{complete} = x_{particular} + x_{null}\) (diophantine equation is an equation that needs to be solved over "integers")

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2you mentioned about differential eqns and we have seen recurrence relations in this thread already.. looks that pattern is common in any area of math that deals with solving equations..
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