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Jonask

  • one year ago

functional equations

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  1. Mertsj
    • one year ago
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    love em or leave em

  2. Jonask
    • one year ago
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    \[\color{brown}{f(x)+(x+1)^3=2f(x+1)}\]

  3. ParthKohli
    • one year ago
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    Find \(\color{#C00}{f(10)}\)

  4. Jonask
    • one year ago
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    lol thanks for posting the full question @ParthKohli

  5. ParthKohli
    • one year ago
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    \[f(-1) = 2f(0)\]

  6. Jonask
    • one year ago
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    f(0)+1=2f(1)

  7. Jonask
    • one year ago
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    wat to do next i 3 variables 2 equations??

  8. ParthKohli
    • one year ago
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    \[f(0) = 2f(1) - 1 \\ f(1) = 2f(2) - 8\]So\[f(0) = 2(2f(2) - 8) = 4f(2) - 16\]

  9. ParthKohli
    • one year ago
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    So hard...

  10. Jonask
    • one year ago
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    \[f(0)=4(f(2)-8)-1=4f(2)-16-1=4f(2)-17\]

  11. Jonask
    • one year ago
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    is f always of the from\[ax^2+bx+c\]

  12. ParthKohli
    • one year ago
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    Ah my bad, yes.

  13. ParthKohli
    • one year ago
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    No, it's a polynomial. It can be in that form, but we're not sure.

  14. Jonask
    • one year ago
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    so far we see \[f(10)+11^3=2f(11)\] \[f(9)+10^3=2f(10)\] ---------------------------------------------- \[f(10)=f(9)-2f(11)+11^3-10^3\]

  15. ParthKohli
    • one year ago
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    I think that it must be in the form \(ax^3 + bx^2 + cx + d\)

  16. Jonask
    • one year ago
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    \[f(x-1)+x^3=2f(x)\]

  17. Jonask
    • one year ago
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    we need a relation between f(9) and f(11)

  18. ParthKohli
    • one year ago
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    \[f(9) = 2f(10) - 1000\]\[f(10) = 2f(11) - 1331\]\[\iff f(9) = 2(f(11) - 1331 ) - 1000\]

  19. ParthKohli
    • one year ago
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    \[f(9) = 2f(11) - 2662 - 1000\]

  20. ParthKohli
    • one year ago
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    That's the relation.

  21. Jonask
    • one year ago
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    isnt this the solution then

  22. Jonask
    • one year ago
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    since \[f(9)-2f(11)=3662\] \[f(10)=3662+11^3-10^3\]

  23. ParthKohli
    • one year ago
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    OMG!

  24. ParthKohli
    • one year ago
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    But it's an integer between 0 and 999.

  25. Jonask
    • one year ago
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    okay so we found a relationship between numbers 2 units away from each other x-1 and x+1 \[\color{blue}{f(x-1)+x^3=2f(x).......... f(x)+(1+x)^3=2f(1+x)}\] \[f(x-1)+x^3=4f(x+1)+2(x+1)^3\]

  26. Jonask
    • one year ago
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    \[f(9)-4f(11)=2(11)^3-10^3\]

  27. Jonask
    • one year ago
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    earlier we used 2 not 4

  28. ParthKohli
    • one year ago
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    Ahhh.

  29. ParthKohli
    • one year ago
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    @Hero

  30. sauravshakya
    • one year ago
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    Let f(x)=a+bx+cx^2+dx^3 NOTE: f(x) must be a third degree polynomial

  31. sauravshakya
    • one year ago
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    Now, f(x)+(x+1)^3 =2f(x+1)

  32. Jonask
    • one year ago
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    \[\huge f(10)=f(9)-2f(11)+11^3-10^3=???\]

  33. sauravshakya
    • one year ago
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    use it to find a,b,c and d

  34. ParthKohli
    • one year ago
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    I see, yeah!

  35. sauravshakya
    • one year ago
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    u will get a=-13 b=9 c=-3 d=1

  36. ParthKohli
    • one year ago
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    Equating coefficients

  37. sauravshakya
    • one year ago
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    yep

  38. ParthKohli
    • one year ago
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    Yay

  39. sauravshakya
    • one year ago
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    now, once u know f(x) u can find f(10)

  40. Jonask
    • one year ago
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    help me i dont get where you get the coeeffients from

  41. ParthKohli
    • one year ago
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    @Jonask a technique.\[ax + by = 2x + 3y\]means\[a=2,b=3\]

  42. Jonask
    • one year ago
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    oh great

  43. sauravshakya
    • one year ago
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    f(x)+(x+1)^3 =2f(x+1) a+bx+cx^2+dx^3 +x^3+3x^2+3x+1 = 2{a+b(x+1) +c(x+1)^2 +d(x+1)^3

  44. sauravshakya
    • one year ago
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    can u continue from here?

  45. ParthKohli
    • one year ago
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    \[f(10) = 10^3 - 3(10)^2 +9(10) -13 = 1000 - 300 + 90 - 13 \\ = 777 \]

  46. sauravshakya
    • one year ago
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    thats correct

  47. Jonask
    • one year ago
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    \[\huge \color{green}{THANKS}\]

  48. sauravshakya
    • one year ago
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    welcome

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