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- satellite73

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

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- satellite73

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

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- amistre64

hmm, my method is to construct it such that each x value causes the unknown constants to zero out

- anonymous

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

- amistre64

if you can do a matrix augment, thats pretty snappy

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- anonymous

yeah but i wanted the snappy method you used

- Hero

....also interested in "snappy" method :D

- amistre64

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)

- anonymous

Expanding this should work as well

- amistre64

if there was another method that was snappier; you might have to refresh me memory how it looked to you

- amistre64

joes looks kinda legendre to me

- Hero

@joemath314159, your method looks like some numerical methods stuff

- anonymous

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

- anonymous

It is legendre, learned it in Linear Algebra

- amistre64

its these lighntning fast clickers of mine, just makes it appear rico y suave lol

- anonymous

(0,-2) (1,1) (2,6) (3,19)

- Hero

@satellite73, are you taking a course?

- anonymous

yes, at the school of amistre

- Hero

lol

- anonymous

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

- amistre64

(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

- amistre64

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

- anonymous

yeah legendre looks like what @joemath314159 wrote
newtons i am not sure about

- anonymous

now i can sleep better. thanks!

- amistre64

http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/mathematics-2/node109.html

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