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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 \(f(1986)\).

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  1. IrishBoy123
    • one year ago
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    \[f(n) = f(1) + f(n-1) - 2f(1) f(n-1)\] \[ \implies f(n) + f(n-1) = 1\] f(2) = 0, f(3) = 1, f(4) = 0

  2. IrishBoy123
    • one year ago
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    and 1986 is even

  3. ganeshie8
    • one year ago
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    that looks really neat!

  4. IrishBoy123
    • one year ago
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    is there proper mathese for this? ie a formal way to write it up?

  5. ganeshie8
    • one year ago
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    to me it looks good the way it is now lets ask @freckles

  6. IrishBoy123
    • one year ago
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    great, thx!

  7. freckles
    • one year ago
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    I don't see a problem with what was said

  8. ganeshie8
    • one year ago
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    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 \(f(n)=\frac{1}{2}+c*(-1)^n\)

  9. ganeshie8
    • one year ago
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    \(c\) 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 :)

  10. Loser66
    • one year ago
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    @IrishBoy123 I don't get how you link the problem to f(n) = f(1) +.....

  11. IrishBoy123
    • one year ago
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    a + b = 1 + n - 1

  12. IrishBoy123
    • one year ago
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    that what you mean?

  13. Loser66
    • one year ago
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    Got it, it is a smart interpretation.

  14. IrishBoy123
    • one year ago
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    i've tried googling "homogeneous solution, the characteristic equation " and all i'm getting are references to DE's.

  15. IrishBoy123
    • one year ago
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    ie i am trying to work out @ganeshie8 's post

  16. IrishBoy123
    • one year ago
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    honestly, just a steer fpr reading later, if there is one

  17. ganeshie8
    • one year ago
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    he was letting \(a=1\) and \(b=n-1\) @Loser66 for sure i couldn't have figured out that substitution on my own

  18. Loser66
    • one year ago
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    but the way he said n = n-1+1 is a perfect way to determine the function f(n). wwwwwwwwoah!!

  19. Loser66
    • one year ago
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    I love this site, I learn the new thing everyday.

  20. Loser66
    • one year ago
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    @ganeshie8 I don't think he set a =1, b = n-1

  21. Loser66
    • one year ago
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    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

  22. freckles
    • one year ago
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    http://openstudy.com/users/jagr2713#/updates/559d6afde4b0f93dd7c2b398 this problem reminds me of this one

  23. ganeshie8
    • one year ago
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    at least thats how i interpreted his solution that induction idea seems interesting too

  24. freckles
    • one year ago
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    yeah I interpret that to a=1 and b=n-1

  25. Loser66
    • one year ago
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    OMG, so I am the only one person who has the weirdest interpretation? hehehe... It's ok, I am crazy originally.

  26. freckles
    • one year ago
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    \[(f(a + b) = f(a) + f(b) - 2f(ab)) \\ a=1 \text{ and } b=n-1 \text{ gives } \\ f(1+(n-1))=f(1)+f(n-1)-2f(n-1)\]

  27. ganeshie8
    • one year ago
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    @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...

  28. ganeshie8
    • one year ago
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    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 : \(\large x_{complete} = x_{particular} + x_{null}\)

  29. ganeshie8
    • one year ago
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    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")

  30. ganeshie8
    • one year ago
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    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..

  31. IrishBoy123
    • one year ago
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    ok got that! thx

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