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 2 years ago
@amistre64 please refresh my feeble memory
snap way to find polynomial through (2,4),(0,6),(4,70)
 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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amistre64
 2 years ago
Best ResponseYou've already chosen the best response.3hmm, my method is to construct it such that each x value causes the unknown constants to zero out

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i thought i had this, but i resorted to a two by two system then i tried writing \[45(x+2)+4x(x+2)\] which worked but it was agony finding those constants i though you had a snap way of finding them

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.3if you can do a matrix augment, thats pretty snappy

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1yeah but i wanted the snappy method you used

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0....also interested in "snappy" method :D

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.3i 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)

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1Expanding this should work as well

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.3if there was another method that was snappier; you might have to refresh me memory how it looked to you

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.3joes looks kinda legendre to me

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0@joemath314159, your method looks like some numerical methods stuff

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1no 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

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1It is legendre, learned it in Linear Algebra

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.3its these lighntning fast clickers of mine, just makes it appear rico y suave lol

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1(0,2) (1,1) (2,6) (3,19)

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0@satellite73, are you taking a course?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1yes, at the school of amistre

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1\[a+bx+cx(x1)+cx(x1)(x2)+dx(x1)(x2)\] actually i think this is slightly different then legrange, but maybe it is identical

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.3(0,2) (1,1) (2,6) (3,19) y = 2 + bx +cx(x1) + dx(x1)(x2) 1 = 2 + b; b= 3 y = 2 + 3x +cx(x1) + dx(x1)(x2) 6 = 2 +6 + 2c ; c=1 y = 2 + 3x +x(x1) + dx(x1)(x2) 19 = 2 + 9 + 6 + d(2); d=3 y = 2 + 3x +x(x1) + 3x(x1)(x2) if i mathed it right

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.3legendre 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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1yeah legendre looks like what @joemath314159 wrote newtons i am not sure about

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1now i can sleep better. thanks!

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.3http://nptel.iitm.ac.in/courses/Webcoursecontents/IITKANPUR/mathematics2/node109.html
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