Paradox or Pair of Ducks?

D

Dale10101

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This is probably not a paradox but just me looking at the situation cross eyed but …

Suppose a set of circumstances seemingly require me to measure and supply three physical constants to satisfy f(x) = Ax^2 + Bx +C which is of course the equation for a parabola.

But the same parabola can be described by f(x) = (x-m)(x-n) which requires only two pieces of information.

Does this mean that the need to supply three pieces of information was in fact only apparent? That in fact had I used the second form of f(x) I could have made only two measurements of some other physical attributes of the circumstances that I was describing?

Putting it another way, when I go about measuring A, B, and C, should I be aware that they are not independent even if their dependency is not obvious?

This had been bugging me for a couple of days now. :sad:
 
Dale

I suspect you have got yourself tied up again in confusion about functions and equations.

Consider two points in the plane. They can be placed on the same line, which is not a parabola at all. Now it is true that there are parabolas that include those two points, in fact an infinite number of them. You do need three points or three parameters to identify a parabola uniquely.

Your assumption that any parabola can be decomposed into (x - a)(x - b) is simply wrong. Try doing so

with \(\displaystyle f(x) = 2x^2 - 8x - 24 = 2(x + 2)(x - 6)\). Three parameters.

When, however, you are thinking about the equation

\(\displaystyle 2x^2 - 8x - 24 = 0 \implies x^2 - 4x - 12 = 0 \implies (x + 2)(x - 6) = 0.\) Two roots.

It's very late, and I am sleepy. If this is not clear, let me know and I shall follow up tomorrow.
 
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This is probably not a paradox but just me looking at the situation cross eyed but …

Suppose a set of circumstances seemingly require me to measure and supply three physical constants to satisfy f(x) = Ax^2 + Bx +C which is of course the equation for a parabola.

But the same parabola can be described by f(x) = (x-m)(x-n) which requires only two pieces of information.

No - you should write

f(x) = A*(x-m)*(x-n)

Otherwise, you are imposing the condition → A = 1

Does this mean that the need to supply three pieces of information was in fact only apparent? That in fact had I used the second form of f(x) I could have made only two measurements of some other physical attributes of the circumstances that I was describing?

Putting it another way, when I go about measuring A, B, and C, should I be aware that they are not independent even if their dependency is not obvious?

This had been bugging me for a couple of days now. :sad:
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Dale

I want to justify some things I said in my first point

Consider 2 distinct points: \(\displaystyle (p,\ q)\ and\ (r,\ s),\ p \ne r.\)

\(\displaystyle Let\ A = \dfrac{B(p - r) + s - q}{r^2 - p^2}\ and\ C = q - \dfrac{p^2\{B(p - r) + s - q\}}{r^2 - p^2} - Bp.\)

\(\displaystyle f(x) = Ax^2 + Bx + C = \left(\dfrac{B(p - r) + s - q}{r^2 - p^2}\right)x^2 + Bx + \left(q - \dfrac{p^2\{B(p - r) + s - q\}}{r^2 - p^2} - Bp\right) =\)

\(\displaystyle q + B(x - p) + \dfrac{(x^2 - p^2)\{B(p - r) + s - q\}}{r^2 - p^2} = q + (x - p)\left(B + \dfrac{(x + p)\{B(p - r) + s - q\}}{r^2 - p^2}\right).\)

\(\displaystyle f(p) = q + (p - p)\left(B + \dfrac{(p + p)\{B(p - r) + s - q\}}{r^2 - p^2}\right) = q + 0 = q.\)

\(\displaystyle f(r) = q + (r - p)\left(B + \dfrac{(r + p)\{B(p - r) + s - q\}}{r^2 - p^2}\right) = q + B(r - p) + \dfrac{(r^2 - p^2)\{B(p - r) + s - q\}}{r^2 - p^2} \implies\)

\(\displaystyle f(r) = q - q + B(r - p + p - r) + s = s.\)

But B is a free variable. Consequently, there are an infinite number of parabolas going through the specific distinct points (p, q) and (r, s). This also semi-justifies my comment that a straight line is one of those infinite parabolas, albeit of an unusual kind.

Switching to equations. Many equations can be arranged in the form of \(\displaystyle f(x) = 0.\)

That is why the old mathematicians were so interested in roots: they were solutions to equations. But the equation imposes restrictions on the function so it requires care to go back and forth from the function embedded in the equation and the equation itself.
 
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Ding dang

Denis said:
Test

Post#1

I get same thing Mark; did not "quote"...but only the 1st post...others ok:
perhaps Dale is playing a trick on you :rolleyes:
Btw Dale, should be Pair of Docs !
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Ducks ... docs ... lol much wheezy laughter, now I am getting my jokes corrected (sorry Denis you've been punked), such an educational site! :D

I do see what you all are saying, but I am still connecting certain dots .... this is all a result of the "the Crispy ones" Tricky Quadratic problem and M. Subhotash Khan's last reply which I follow and can derive but am failing to integrate into my world view. :sad: What!

-and I don't do tricks, only disasters. :cool:
 
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JeffM

I had just finished my last post when I discovered your latest. Thank you, it will take a while to digest, I feel a twinge of guilt ... such a large of amount of work and effort on your part. Will respond later no doubt. Good day to you sir.:)
 
Dead ducks.

Thank you M.s JeffM and Subhotash Kahn.

JeffM, I was not quite unaware of the difference between a function and an equation and in fact I had recently begun thinking (provisionally) of quadratic “equations” as the intersection of a parabolic function and a horizontal line y = k. This led me to the realization that the roots of the resulting equation (after being put in the standard f(x) = 0 form) give the x-coordinates of that intersection, (x1,k) and (x2,k). And yes, the equation f(x) = 0, once you have it, cannot be used to reconstitute the original parabolic function since it (the equation) could have been the result of the intersection of any number of different parabolic functions and/or values of k.

I had originally thought that the roots delivered by the quadratic formula in solving Ax^2+Bx+C = 0 could be inserted into f(x) = (x-m)(x-n) to reconstitute Ax^2+Bx+C = 0. Doing so however delivers x^2 + (B/A)x + (C/A) = 0, which is in fact is the equation that forms the basis of deriving the quadratic formula once you have take the initial step of dividing Ax^2+Bx+C = 0 by A.

I can see now, in two ways, that f(x) = (x-m)(x-n) cannot possibly deliver Ax^2+Bx+C = 0 .
1) Multiplying out (x-m)(x-n) will always result in and A =1, i.e. x^2.
2) Multiplying (x-m)(x-n)=0 by any constant will not change the solution set but multiplying the right side of f(x) = (x-m)(x-n) will change the shape of the parabola, so the two cannot be interchangeable.

Subhotash Khan pointed out the error most succinctly when he changed f(x) = (x-m)(x-n) to
f(x) = A(x-m)(x-n).

I think the most important thing that I now realize is that each step in a chain of transformation is not uniquely reversible, that is, any given statement could have descended from any number of different previous statements. The one constant is the solution set.

Also I see now that an equation containing a solution set is infinitely variable and that the standard forms are only selected because their form provides groupings of equation constants that are useful extractable information, like roots, or vertices, or origin offsets etc.

Other problems are press forward but this post is long enough. Thanks all

(I knew my apparent paradox was a quack but had to ask.)

P.S JeffM, worked through you derivation but am reserving further comment because the point of your exercise intersects with confusions I am having about types of conditions that can be put on a function (conditions on the function vs conditions on its coefficients vs conditions on the roots ... still formulating my confusion, but this post did clear up in area. Good, thanks.
 
Dale10101 said:
Thank you M.s JeffM and Subhotash Kahn.

JeffM, I was not quite unaware of the difference between a function and an equation and in fact I had recently begun thinking (provisionally) of quadratic “equations” as the intersection of a parabolic function and a horizontal line y = k. This led me to the realization that the roots of the resulting equation (after being put in the standard f(x) = 0 form) give the x-coordinates of that intersection, (x1,k) and (x2,k). And yes, the equation f(x) = 0, once you have it, cannot be used to reconstitute the original parabolic function since it (the equation) could have been the result of the intersection of any number of different parabolic functions and/or values of k.
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I did not mean to imply that you were unaware of the distinction. I was merely pointing out that keeping the distinction clear can be quite helpful. This is what I was driving at even in your prior thread involving your "super-exponential."

As you say, if you have an equation in the form f(x) = c, a constant, the real solution set will be the intersection of the horizontal line
y = c, a real constant, and the graph of the real function f(x). This means of course that you can define the function g(x) = f(x) - c, and the solution set for the equation g(x) = 0 is the same as the solution set for f(x) = c. Hence, the interest in roots of functions. So let's stop thinking about any constant except 0.

The solutions to g(x) = 0 are generally constrained to far fewer than the domain of g(x) and to far fewer than the domain of z(x) = 0. When finding the solution set, you can use properties true of g(x) provided you simultaneously demand properties true of z(x) and vice versa. I at least find this to be helpful in keeping straight the role of functions versus the role of equations. May not tickle your fancy of course. It's just one way to think about things; may not be the most useful.

Dale10101 said:
Subhotash Khan pointed out the error most succinctly when he changed f(x) = (x-m)(x-n) to
f(x) = A(x-m)(x-n).Yes he did.

I think the most important thing that I now realize is that each step in a chain of transformation is not uniquely reversible, that is, any given statement could have descended from any number of different previous statements. I'd prefer to say not necessarily uniquely reversible. Sometimes they are; sometimes they are not. If you ever do a proof by Proclus's method, unique reversibility is the key.

The one constant is the solution set. Also I see now that an equation containing a solution set is infinitely variable and that the standard forms are only selected because their form provides groupings of equation constants that are useful extractable information, like roots, or vertices, or origin offsets etc. I am not sure I know what this means, but it seems to me to be muddying up once again the concepts of an equation, a function, and a family of functions. f(x) = 2x^2 - 8x - 24 is a specific function,
f(x) = Ax^2 + Bx + C may represent a specific but as yet unspecified function or an infinite family of similar functions,
2x^2 - 8x - 24 = 0 is an equation as is Ax^2 + Bx + C = 0.
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OK

Yes, thank you, I take the meaning of most of what you are saying. Things get deep fast and my thoughts are at best half formed which indeed makes things muddy. All that I can say is that my struggle is to develop both an understanding of how to do things and a context to put them in, the latter being often the most difficult. I think I will try and stick more to specific problems as it is easier to keep what I thinking, saying, and meaning to convey more consistent. I can say, that I do see things more clearly then a couple days ago. Thanks for your comments and insights.
 
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