Calculus
posted by Matt .
Great! I understand now that
C0=f(a)
c1=f'(a)
c2=f"(a)/2
Now, If I am to find the parabolization of the equation x^2x at x=2, then
c0=x^2x=2^22=2
c1=2x1=2(2)1=3
c2=2/2=1
So, the equation (taken from c0+c1(xa)+c2(xa)^2) is >>
2+3(x2)+1(x2)^2??
Is this correct?
Thanks
In calc, we are studying parabolization. It is the linerazation of a parabola. The linerazation of a standard line is L(x) = b0+b1(xa) when f(x) is at x=a
b0 = f(a)
b1=f'(a)
The parabolization of f(x) at x=a is given by the equation
P(x)=c0+c1(xa)+c2(xa)^2.
f(a) = P(a)
F'(a) = P'(a)
f''(a) = P''(a)
I need to find a formula for c0, c1, and c2 in terms of f(a), f'(a) and f"(a)
Im sure that I need to find the first and second derivitive of the equation c0+c1(xa)+c2(xa)^2. Im just not sure where to start...
Thanks!
Matt
Actually, linearization is approxiamtion by a straight line while Parabolization is approximation of a function by a parabola.
You start from:
f(a) = P(a)
f'(a) = P'(a)
f''(a) = P''(a)
Then you insert
P(x)=c0+c1(xa)+c2(xa)^2
in here.
So,
P(a) = C0
P'(x) = C1+2c2(xa) >
P'(a) = C1
P''(x) = 2C2 >
P''(a) = 2C2
This means that:
f(a) = P(a) = C0
f'(a) = P'(a) = C1
f''(a) = P''(a) = 2C2
And it follows that:
P(x) = C0+C1(xa)+c2(xa)^2 =
f(a) + f'(a)(xa) + f''(a)/2 (xa)^2
There is an easy way to check. If you parabolize a parabola, you should get the same thing back. So, let's see:
2+3(x2)+ (x2)^2 =
2 + 3x6 + (x^2  4x + 4) =
26+4 + 3x4x + x^2 =
x^2  x.
So, it's indeed the same!
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