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Jonask

functional equations

  • one year ago
  • one year ago

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  1. Mertsj
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    love em or leave em

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    • one year ago
  23. ParthKohli
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    OMG!

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

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

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

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

    • one year ago
  28. ParthKohli
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    Ahhh.

    • one year ago
  29. ParthKohli
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    @Hero

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

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

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

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

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

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

    • one year ago
  36. ParthKohli
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    Equating coefficients

    • one year ago
  37. sauravshakya
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    yep

    • one year ago
  38. ParthKohli
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    Yay

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

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

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

    • one year ago
  42. Jonask
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    oh great

    • one year ago
  43. sauravshakya
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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

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

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

    • one year ago
  46. sauravshakya
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    thats correct

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

    • one year ago
  48. sauravshakya
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    welcome

    • one year ago
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