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

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- IrishBoy123

\[f(n) = f(1) + f(n-1) - 2f(1) f(n-1)\]
\[ \implies f(n) + f(n-1) = 1\]
f(2) = 0, f(3) = 1, f(4) = 0

- IrishBoy123

and 1986 is even

- ganeshie8

that looks really neat!

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## More answers

- IrishBoy123

is there proper mathese for this?
ie a formal way to write it up?

- ganeshie8

to me it looks good the way it is now
lets ask @freckles

- IrishBoy123

great, thx!

- freckles

I don't see a problem with what was said

- ganeshie8

what follows is really not necessary :
\(f(n)+f(n-1)=1\) 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

\(c\) 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

@IrishBoy123 I don't get how you link the problem to f(n) = f(1) +.....

- IrishBoy123

a + b = 1 + n - 1

- IrishBoy123

that what you mean?

- Loser66

Got it, it is a smart interpretation.

- IrishBoy123

i've tried googling "homogeneous solution, the characteristic equation " and all i'm getting are references to DE's.

- IrishBoy123

ie i am trying to work out @ganeshie8 's post

- IrishBoy123

honestly, just a steer fpr reading later, if there is one

- ganeshie8

he was letting \(a=1\) and \(b=n-1\) @Loser66
for sure i couldn't have figured out that substitution on my own

- Loser66

but the way he said n = n-1+1 is a perfect way to determine the function f(n). wwwwwwwwoah!!

- Loser66

I love this site, I learn the new thing everyday.

- Loser66

@ganeshie8 I don't think he set a =1, b = n-1

- Loser66

just 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

http://openstudy.com/users/jagr2713#/updates/559d6afde4b0f93dd7c2b398
this problem reminds me of this one

- ganeshie8

at least thats how i interpreted his solution
that induction idea seems interesting too

- freckles

yeah I interpret that to a=1 and b=n-1

- Loser66

OMG, so I am the only one person who has the weirdest interpretation? hehehe... It's ok, I am crazy originally.

- freckles

\[(f(a + b) = f(a) + f(b) - 2f(ab)) \\ a=1 \text{ and } b=n-1 \text{ gives } \\ f(1+(n-1))=f(1)+f(n-1)-2f(n-1)\]

- ganeshie8

@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

as 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

in 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

you 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..

- IrishBoy123

ok got that! thx

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