Question

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

Answer 1

\[f(n) = f(1) + f(n-1) - 2f(1) f(n-1)\]
\[ \implies f(n) + f(n-1) = 1\]

f(2) = 0, f(3) = 1, f(4) = 0

Answer 2

and 1986 is even

Answer 3

that looks really neat!

Answer 4

is there proper mathese for this?

ie a formal way to write it up?

Answer 5

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

Answer 6

great, thx!

Answer 7

I don't see a problem with what was said

Answer 8

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 \(f(n)=\frac{1}{2}+c*(-1)^n\)

Answer 9

\(c\) 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 :)

Answer 10

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

Answer 11

a + b = 1 + n - 1

Answer 12

that what you mean?

Answer 13

Got it, it is a smart interpretation.

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

@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...

Answer 28

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 : \(\large x_{complete} = x_{particular} + x_{null}\)

Answer 29

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")

Answer 30

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..

Answer 31

ok got that! thx