Question

simple way to define a polynomial from given points.

Answer 1

say you have 3 point; \((x_o,y_o),(x_1,y_1),(x_2,y_2)\)

set up a polynomial of the form:\[y=y_o+c_1(x-x_o)+c_2(x-x_o)(x-x_1)\]

Answer 2

Does this include the normal straight line thingy too?

Answer 3

yes

Answer 4

i think this has something o do with derivatives too :/

Answer 5

to do*

Answer 6

i thought of it last night while playing about with series

Answer 7

But this demonstration is for 3 points.

Answer 8

for 2 points, use the first 2 terms and it becomes the line eqatuon\[y=y_o+c_1(x-x_o)\]
\[y-y_o=c_1(x-x_o)\]look familiar?

Answer 9

each term is the product of a constant and the zeroed out xs of the pther points

Answer 10

(2,3) (5,8)

y = 3 + c(x-2) ; when x=2, y=3

when x=5, we want y=8

8 = 3 + c(5-2)
8 - 3= c(3)
5/3= c

y = 3 + (5/3) (x-2)

Answer 11

my mind had wandered last night to those "find a quadratic equation that fits these 3 points" ... and its always a pain to go thru system of equations methods to me

Answer 12

i can also define any sequence of "given numbers" with a similar method

Answer 13

i Really like this,

Answer 14

Haha that's the point-slope form :-D

Answer 15

thanks, its a short tutorial but fun nonetheless lol

Answer 16

give me a list of say 3 or 4 random numbers and lets see if i can define an expression for them :)

Answer 17

ok so , i have three points,
\[\{(-1,1) , (1,1) , (1,3)\}\]

Answer 18

that will form a parabola that is "sideways" since you have 2 x values that are the same

Answer 19

just invert the point to (y,x) and itll work the same