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satellite73 Group Title

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

  • one year ago
  • one year ago

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  1. amistre64 Group Title
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    hmm, my method is to construct it such that each x value causes the unknown constants to zero out

    • one year ago
  2. satellite73 Group Title
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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

    • one year ago
  3. amistre64 Group Title
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    if you can do a matrix augment, thats pretty snappy

    • one year ago
  4. satellite73 Group Title
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    yeah but i wanted the snappy method you used

    • one year ago
  5. Hero Group Title
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    ....also interested in "snappy" method :D

    • one year ago
  6. amistre64 Group Title
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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)

    • one year ago
  7. joemath314159 Group Title
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    Expanding this should work as well

    • one year ago
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  8. amistre64 Group Title
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    if there was another method that was snappier; you might have to refresh me memory how it looked to you

    • one year ago
  9. amistre64 Group Title
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    joes looks kinda legendre to me

    • one year ago
  10. Hero Group Title
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    @joemath314159, your method looks like some numerical methods stuff

    • one year ago
  11. satellite73 Group Title
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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

    • one year ago
  12. joemath314159 Group Title
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    It is legendre, learned it in Linear Algebra

    • one year ago
  13. amistre64 Group Title
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    its these lighntning fast clickers of mine, just makes it appear rico y suave lol

    • one year ago
  14. satellite73 Group Title
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    (0,-2) (1,1) (2,6) (3,19)

    • one year ago
  15. Hero Group Title
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    @satellite73, are you taking a course?

    • one year ago
  16. satellite73 Group Title
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    yes, at the school of amistre

    • one year ago
  17. Hero Group Title
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    lol

    • one year ago
  18. satellite73 Group Title
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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

    • one year ago
  19. amistre64 Group Title
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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

    • one year ago
  20. amistre64 Group Title
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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

    • one year ago
  21. satellite73 Group Title
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    yeah legendre looks like what @joemath314159 wrote newtons i am not sure about

    • one year ago
  22. satellite73 Group Title
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    now i can sleep better. thanks!

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

    • one year ago
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