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hmm, my method is to construct it such that each x value causes the unknown constants to zero out
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
if you can do a matrix augment, thats pretty snappy
yeah but i wanted the snappy method you used
....also interested in "snappy" method :D
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)
if there was another method that was snappier; you might have to refresh me memory how it looked to you
joes looks kinda legendre to me
@joemath314159, your method looks like some numerical methods stuff
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
It is legendre, learned it in Linear Algebra
its these lighntning fast clickers of mine, just makes it appear rico y suave lol
(0,-2) (1,1) (2,6) (3,19)
@satellite73, are you taking a course?
yes, at the school of amistre
\[a+bx+cx(x-1)+cx(x-1)(x-2)+dx(x-1)(x-2)\] actually i think this is slightly different then legrange, but maybe it is identical
(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
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
yeah legendre looks like what @joemath314159 wrote newtons i am not sure about
now i can sleep better. thanks!