Question

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

Answer 1

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

Answer 2

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

Answer 3

if you can do a matrix augment, thats pretty snappy

Answer 4

yeah but i wanted the snappy method you used

Answer 5

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

Answer 6

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)

Answer 7

Expanding this should work as well

Answer 8

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

Answer 9

joes looks kinda legendre to me

Answer 10

@joemath314159, your method looks like some numerical methods stuff

Answer 11

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

Answer 12

It is legendre, learned it in Linear Algebra

Answer 13

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

Answer 14

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

Answer 15

@satellite73, are you taking a course?

Answer 16

yes, at the school of amistre

Answer 17

lol

Answer 18

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

Answer 19

(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

Answer 20

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

Answer 21

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

Answer 22

now i can sleep better. thanks!

Answer 23

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