zmudz
  • zmudz
If \(f(a + b) = f(a) + f(b) - 2f(ab)\) for all nonnegative integers \(a\) and \(b\), and \(f(1) = 1\), compute \(f(1986)\).
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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IrishBoy123
  • 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
  • IrishBoy123
and 1986 is even
ganeshie8
  • ganeshie8
that looks really neat!

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IrishBoy123
  • IrishBoy123
is there proper mathese for this? ie a formal way to write it up?
ganeshie8
  • ganeshie8
to me it looks good the way it is now lets ask @freckles
IrishBoy123
  • IrishBoy123
great, thx!
freckles
  • freckles
I don't see a problem with what was said
ganeshie8
  • 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
  • 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
  • Loser66
@IrishBoy123 I don't get how you link the problem to f(n) = f(1) +.....
IrishBoy123
  • IrishBoy123
a + b = 1 + n - 1
IrishBoy123
  • IrishBoy123
that what you mean?
Loser66
  • Loser66
Got it, it is a smart interpretation.
IrishBoy123
  • IrishBoy123
i've tried googling "homogeneous solution, the characteristic equation " and all i'm getting are references to DE's.
IrishBoy123
  • IrishBoy123
ie i am trying to work out @ganeshie8 's post
IrishBoy123
  • IrishBoy123
honestly, just a steer fpr reading later, if there is one
ganeshie8
  • 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
  • Loser66
but the way he said n = n-1+1 is a perfect way to determine the function f(n). wwwwwwwwoah!!
Loser66
  • Loser66
I love this site, I learn the new thing everyday.
Loser66
  • Loser66
@ganeshie8 I don't think he set a =1, b = n-1
Loser66
  • 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
  • freckles
ganeshie8
  • ganeshie8
at least thats how i interpreted his solution that induction idea seems interesting too
freckles
  • freckles
yeah I interpret that to a=1 and b=n-1
Loser66
  • Loser66
OMG, so I am the only one person who has the weirdest interpretation? hehehe... It's ok, I am crazy originally.
freckles
  • 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
  • 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
  • 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
  • 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
  • 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
  • IrishBoy123
ok got that! thx

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