Solve q 4 please
#Polynomial
hollow rivet
whole brambleBOT
Solve q 4 please
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tranquil marsh
Solve q 4 please
Suppose the polynomial is f(x) = Ax^2 + Bx + C. Then you have three equations from which you can find the coefficients A, B, C.
hollow rivet
I did that and didn't get
tranquil marsh
What did you not get?
ivory ledge
Suppose the polynomial is f(x) = Ax^2 + Bx + C. Then you have three equations fr...
i think there's a certain step that not as lengthy as your method
tranquil marsh
Oh, without finding the coefficients? Hm...
Maybe, yeah. Though, at the moment, no idea on what it can be.
ivory ledge
yes some way we can substitute the "combination" of all the coefficients we get in f(a+b+c) instead of solving for them induvidually
ill check it out
tranquil marsh
Oh, actually.
I know how you can find the coefficients without solving the system.
hollow rivet
ivory ledge
I know how you can find the coefficients without solving the system.
cramer?
tranquil marsh
cramer?
That's still a method of solving a linear system.
One sec, I figured a method where you don't really need to make equations like that.
ivory ledge
for the sake of solving it i'd plug
a = -1
b= 0
c=1
and then solve for the coefficients lmao
hollow rivet
You will have 9 ways to do this method if it's just small digits like u said
hollow rivet
for the sake of solving it i'd plug
a = -1
b= 0
c=1
and then solve for the co...
Are you jee advance too by chance?
ivory ledge
You will have 9 ways to do this method if it's just small digits like u said
yeah just pick the easiest one
ivory ledge
Are you jee advance too by chance?
yes
hollow rivet
Nice I am repeat
ivory ledge
ive passed
tranquil marsh
Suppose f(x) = Ax^2 + Bx + C and y1 = f(x1), etc. Then you can easily see the following:
(Δy/Δx)(12) = (y2 - y1)/(x2 - x1) = A(x2 + x1) + B
(Δy/Δx)(23) = (y3 - y2)/(x3 - x2) = A(x3 + x2) + B
(Δ^2 y/Δx^2)(123) = ((Δy/Δx)(23) - (Δy/Δx)(12))/(x3 - x1) = A
So, you can find A using the second difference quotient, then find B from one of the first, and then find C from the function.
ivory ledge
anyways
hollow rivet
Oh ok
ivory ledge
Suppose f(x) = Ax^2 + Bx + C and y1 = f(x1), etc. Then you can easily see the fo...
Regarding y2-y1 / x2-x1
the C(1/x2 - 1/x1) term is missing??
tranquil marsh
Regarding y2-y1 / x2-x1
the C(1/x2 - 1/x1) term is missing??
Yes, because when you do y2 - y1, it disappears.
Same with B for (Δy/Δx)(23) - (Δy/Δx)(12).
tranquil marsh
aah im dividing by x2-x1 beofre subtracting y2-y1
What do you mean?
The first difference quotient is (y2 - y1)/(x2 - x1), so you find the differences first, then divide them.
ivory ledge
yess i made a mistake
polar pecan
?tag helping
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static pendant
Solve q 4 please
use Lagrange interpolation
you will get (THE SPOILER HAS THE ANSWER) ||f(x) = x^2 - (a+b+c)x + (ab+bc+ca)||
tranquil marsh
To be honest, can't see any shorter way than the two we discovered: solving a linear system (or Lagrange interpolation, which is pretty much the shortcut to that) or using finite differences.
ivory ledge
use Lagrange interpolation
what even is that 
static pendant
what even is that <:betterskull:1305605566886055997>
Google it
static pendant
it's really intuitive yeah
ivory ledge
get a fuinction which is zero at our required points
then start tailoring it to lift if towards our required point
damn
ill study it
ripe rapidsBOT
@hollow rivet
hollow rivet
I still haven't got it
Just post the dam soln
polar pecan
No, that's against the rules lol
hollow rivet
Atleast in dms smh
ivory ledge
Atleast in dms smh
brother
which one do you want to me to explain?
Lagrannge interpolation
or normal algebra?
hollow rivet
or normal algebra?
.
ivory ledge
who pinged