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satellite73

  • 2 years ago

@amistre64 please refresh my feeble memory snap way to find polynomial through (-2,4),(0,-6),(4,70)

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  1. amistre64
    • 2 years ago
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    hmm, my method is to construct it such that each x value causes the unknown constants to zero out

  2. satellite73
    • 2 years ago
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    i thought i had this, but i resorted to a two by two system then i tried writing \[4-5(x+2)+4x(x+2)\] which worked but it was agony finding those constants i though you had a snap way of finding them

  3. amistre64
    • 2 years ago
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    if you can do a matrix augment, thats pretty snappy

  4. satellite73
    • 2 years ago
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    yeah but i wanted the snappy method you used

  5. Hero
    • 2 years ago
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    ....also interested in "snappy" method :D

  6. amistre64
    • 2 years ago
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    i believe the method you posted is what i used at first (-2,4),(0,-6),(4,70) y = a + bx + cx(x+2) ; (0,-6) -6 = a +0+0 y = -6 + bx + cx(x+2) ; (-2,4) 10 = -2b + 0 ; b=-5 y = -6 -5x + cx(x+2) ; (4,70) 70+6+20 = c4(4+2) 96/24 = c = 4 y = -6 -5x + 4x(x+2)

  7. joemath314159
    • 2 years ago
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    Expanding this should work as well

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  8. amistre64
    • 2 years ago
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    if there was another method that was snappier; you might have to refresh me memory how it looked to you

  9. amistre64
    • 2 years ago
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    joes looks kinda legendre to me

  10. Hero
    • 2 years ago
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    @joemath314159, your method looks like some numerical methods stuff

  11. satellite73
    • 2 years ago
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    no that was what i was looking for i guess that is what i did, somehow it looked easier when you did it. i guess the grass is always greener hello @Hero hello @joemath314159

  12. joemath314159
    • 2 years ago
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    It is legendre, learned it in Linear Algebra

  13. amistre64
    • 2 years ago
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    its these lighntning fast clickers of mine, just makes it appear rico y suave lol

  14. satellite73
    • 2 years ago
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    (0,-2) (1,1) (2,6) (3,19)

  15. Hero
    • 2 years ago
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    @satellite73, are you taking a course?

  16. satellite73
    • 2 years ago
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    yes, at the school of amistre

  17. Hero
    • 2 years ago
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    lol

  18. satellite73
    • 2 years ago
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    \[a+bx+cx(x-1)+cx(x-1)(x-2)+dx(x-1)(x-2)\] actually i think this is slightly different then legrange, but maybe it is identical

  19. amistre64
    • 2 years ago
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    (0,-2) (1,1) (2,6) (3,19) y = -2 + bx +cx(x-1) + dx(x-1)(x-2) 1 = -2 + b; b= 3 y = -2 + 3x +cx(x-1) + dx(x-1)(x-2) 6 = -2 +6 + 2c ; c=1 y = -2 + 3x +x(x-1) + dx(x-1)(x-2) 19 = -2 + 9 + 6 + d(2); d=3 y = -2 + 3x +x(x-1) + 3x(x-1)(x-2) if i mathed it right

  20. amistre64
    • 2 years ago
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    legendre i believe zeros out each constant term and omits one zero so that just a single constant is exposed at any time. Newtons method is similar to mine in that new information can be added as needed without have to reconstruct the whole equation

  21. satellite73
    • 2 years ago
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    yeah legendre looks like what @joemath314159 wrote newtons i am not sure about

  22. satellite73
    • 2 years ago
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    now i can sleep better. thanks!

  23. amistre64
    • 2 years ago
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    http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node109.html

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