How to transform quadtratic equation (in form y=ax^2+bx+c) to linear?DutchChemistryBox - 15-2-2014 at 04:42
What do you exactly want? Do you want to find the points of intersection?
Because if you really want quadratic equation > linear equation you are going to have a bad time. You can't make paraboles with a linear function.
[Edited on 15-2-2014 by DutchChemistryBox]Mildronate - 15-2-2014 at 10:13
I want to make parabola graph to straight line. If i have equation y=ax^2+c i can plot y=f(z) plot, where z=x^2, but how to do it with y=ax^2+bx+c is
it possible? Why tread was removed from computing sections?turd - 15-2-2014 at 12:54
I want to make parabola graph to straight line. If i have equation y=ax^2+c i can plot y=f(z) plot, where z=x^2, but how to do it with y=ax^2+bx+c is
it possible? Why tread was removed from computing sections?
If you set b'=b/2a and c'=c-b2/4a2 then y=ax2+bx+c becomes y=a(x+b')2+c'. Set z=(x+b')2 and you
get y=az+c'.
But seriously - this is so incredibly elementary that I don't think this is the right place.
Edit: and this has nothing to do with "approximation". If you want to approximate a parabola at a certain point, you calculate the tangent, which is
trivially done by derivation.
[Edited on 15-2-2014 by turd]unionised - 16-2-2014 at 08:37
The question isn't well enough defined for turd's algebra to answer.
For example, take the parabola y=x2
Over the range -.1 to .1 it's flat on average and the best approximation is something like y=0.005 (The linear term is zero)
Near x=10 the best linear approximation will be a steep upward line (The linear term is large + positive)
Near x= -10 it will slope downwards. (The linear term is large + negative)
So there's no simple way to say what the "best" approximation is.woelen - 17-2-2014 at 00:19
You only can get a reasonable linear approximation over a small range. Unionised's post explains why. For a given continuous and differentiable
function f(x), a decent approximation around a point x0 is A(x) = f(x0) + f'(x0)*(x-x0). Here A(x) is the approximation and f'(x0) is the first
derivative of the function f at the value x0. The approximation is exact in x0 and may be quite good near x0, but what is 'good' and 'near' depends on
your context. The quality of the approximation can also be estimated, this requires higher derivatives of the function f.
[Edited on 17-2-14 by woelen]turd - 17-2-2014 at 05:16
The question isn't well enough defined for turd's algebra to answer.
Actually I think post #319258 defined the question pretty unambiguously - the question was what variable substitution gets you from a second order
polynomial to a linear expression.
As I noted upthread the word "approximation" was a red herring. Thus the discussion on derivatives is off topic and not very enlightening, because I
would expect everybody with even the most basic scientific education to understand what a derivative is.woelen - 17-2-2014 at 07:20
@turd, now what you say might be true, but the question then is quite ill-posed. What is meant with "transform" when in the same sentence we have the
word "approximation".
The second order expression f(x) = axx + bx + c can be transformed to any linear form over part of its interval by simply solving the equation f(x) =
axx + bx + c = pX + q. Express x as function g of X and evaluate f(g(X)) and this will be linear in X. Expressing x as function of X is not very
difficult, just use the standard formula for solving quadratic equations.
The expressions are quite ugly, but they are exact. For general f(x) the thing can be much more difficult, because you need to find the inverse of
f(x) - pX + q with respect to x. Sometimes you can find good approximations by estimating f(x) by a linear or second order polynomial around a point
x0 and then solving for x. The approximation then only is valid for values of x near x0.turd - 17-2-2014 at 07:59
@turd, now what you say might be true, but the question then is quite ill-posed. What is meant with "transform" when in the same sentence we have the
word "approximation".
Of course the question was terrible. But the post I'm referring to (the third in this thread) does not leave much room for interpretation.
Quote:
The second order expression f(x) = axx + bx + c can be transformed to any linear form over part of its interval by simply solving the equation f(x) =
axx + bx + c = pX + q. Express x as function g of X and evaluate f(g(X)) and this will be linear in X. Expressing x as function of X is not very
difficult, just use the standard formula for solving quadratic equations.
Or simply note, as I did above, that any parabola y=ax2+bx+c can be written in the form y=a(x+b')2+c'. No need for the quadratic
formula. Of course it must read c'=c-b2/4a, not c'=c-b2/4a2 as I stated above.
That the question becomes more "interesting" for other analytical expressions or over finite rings is evident. woelen - 17-2-2014 at 10:15
@turd: You are right. The third post makes clear what the OP wanted. I should have read more carefully AJKOER - 18-3-2014 at 18:25
OK, just think short intervals.
The slope changes with each short linear interval being the 1st derivative of the underlying function or estimate there of at the midpoint of the
interval.The intercept is derived by requiring the next linear segment to pass through the last point of the prior segment:
Y -Yo = f'(Xm)*(X - Xo)
With enough short intervals, even a circle may still look like a circle .
.