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ThudnBlunder
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Find a Function
« on: Nov 4th, 2003, 10:44am » |
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Find a function F: N+ -> N+ such that
(i) if x < y then F(x) < F(y)
(ii) F(yF(x)) = x2F(xy)
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| « Last Edit: Feb 29th, 2004, 4:44am by ThudnBlunder » |
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towr
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Re: Find a Function
« Reply #1 on: Nov 4th, 2003, 11:33am » |
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::quite a usefull function for making lenses and mirrors for telescopes..:: 
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ThudnBlunder
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Re: Find a Function
« Reply #2 on: Nov 4th, 2003, 9:48pm » |
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:Don't see how it concerns my refractor telescope. Or my contact lenses.
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towr
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Re: Find a Function
« Reply #3 on: Nov 5th, 2003, 12:58am » |
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I don't know about the first (because the term doesn't ring a bell), but it does concern the second.. (though that get's a lot more complecated by other considerations as well)
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Eigenray
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Re: Find a Function
« Reply #4 on: Nov 5th, 2003, 2:33am » |
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If F is a polynomial, condition (i) is extraneous (there is still a unique solution without it), so should we be looking for non-polynomial solutions?
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Dudidu
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Re: Find a Function
« Reply #5 on: Nov 5th, 2003, 3:17am » |
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Can anyone tell me what is wrong with this solution (if it is wrong)  :
F(n) = n2.
Quote: The suggested F fulfills this condition.
Quote:| if x < y then F(x) < F(y) |
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The suggested F fulfills this condition since 2 is strictly monotonically increasing when n[in]N+.
Quote:The suggested F fulfills this condition since F(yF(x)) = F (yx2) = y2x4 and x2F(xy) = x2x2y2 = y2x4.
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ThudnBlunder
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Re: Find a Function
« Reply #6 on: Nov 5th, 2003, 3:26am » |
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Quote:| Can anyone tell me what is wrong with this solution |
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I don't see a solution. Only an answer. 
on Nov 5th, 2003, 2:33am, Eigenray wrote:| If F is a polynomial, condition (i) is extraneous |
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Condition (i) is used in condition (ii) to arrive at a solution.
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| « Last Edit: Nov 5th, 2003, 3:30am by ThudnBlunder » |
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Eigenray
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Re: Find a Function
« Reply #7 on: Nov 5th, 2003, 8:15am » |
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on Nov 5th, 2003, 3:26am, THUDandBLUNDER wrote:| Condition (i) is used in condition (ii) to arrive at a solution. |
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I don't believe there are any polynomials F : N+ -> N+ that satisfy (ii) but not (i). Since you bothered to include (i) as a condition, it suggests to me that there are nonpolynomial solutions to (ii), but not (i), which should be excluded from the nonpolynomial solutions to (ii) that we should perhaps be trying to find, if that was your intention when asking this question. Or, more likely, including (i) makes the solution unique.
Is F(x)=x2 the only solution? So far I've figured out that you must have F(2k) = 4k, so it seems likely.
Edit: What I mean is that since the problem asked for a solution, (i) is irrelevent, unless we're looking for all solutions.
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| « Last Edit: Nov 5th, 2003, 6:59pm by Eigenray » |
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Icarus
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Re: Find a Function
« Reply #8 on: Nov 5th, 2003, 10:06am » |
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Functional equations such as (ii) often have a finite number (usually 1) of continuous solutions, but also have an infinite number of discontinuous ones. Whether this is the case for (ii) itself, I don't know. But in order for the solution to be unique, you generally have to give a second condition, such as "F is continuous" or "F is a polynomial" or "F is strictly increasing".
I do not see why either of the first two would be preferable to the third in the statement of the problem. In particular, "F is a polynomial" gives a considerable amount of additional information which I doubt is needed to obtain uniqueness - thereby weakening the final result.
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ThudnBlunder
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Re: Find a Function
« Reply #9 on: Nov 6th, 2003, 3:32am » |
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Quote:But in order for the solution to be unique, you generally have to give a second condition, such as "F is continuous" or "F is a polynomial" or "F is strictly increasing".
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Or, as in this case, "F is one-to-one".
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Icarus
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Re: Find a Function
« Reply #10 on: Nov 6th, 2003, 9:57am » |
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on Nov 4th, 2003, 10:44am, THUDandBLUNDER wrote:| i) if x < y then F(x) < F(y) |
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This would be "F is strictly increasing" - much stronger than "one-to-one".
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ThudnBlunder
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Re: Find a Function
« Reply #11 on: Feb 29th, 2004, 5:07am » |
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Solution:
We have the conditions
(i) If x < y then F(x) < F(y)
(ii) F(yF(x)) = x2F(xy)
(i) implies that F is one-to-one.
Putting x = 1 in (ii) we get
F(yF(1)) = F(y)
If F(1) > 1 then condition 1 is violated.
So F(1) = 1
Now putting y = 1 in (ii) we get
(iii) F(F(x)) = x2F(x)
By applying condition (ii) twice we get
F(F(x)F(y)) = y2F(yF(x)) = y2x2F(xy) = (xy)2F(xy)
Hence
(iv) F(F(x)F(y)) = (xy)2F(xy)
In (iv) replacing xy with x and 1 with y we get
F(F(xy)) is also equal to (xy)2F(xy)
But since F is one-to-one we have that
(v) F(x)F(y) = F(xy)
Hence
(a) F(F(x)) = x2F(x)
(b) F(x)F(y) = F(xy)
Now suppose F(x) > x2 for some x
Then
F(F(x)) > F(x2) = F(x)F(x)
That is
x2F(x) > F(x)F(x)
which means
x2 > F(x)
which contradicts F(x) > x2
A similar contradiction arises by assuming F(x) < x2
Therefore we must have that
F(x) = x2 for all x
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| « Last Edit: Feb 29th, 2004, 9:03am by ThudnBlunder » |
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MatrixFrog
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Re: Find a Function
« Reply #12 on: Mar 26th, 2004, 8:13pm » |
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on Nov 4th, 2003, 9:48pm, THUDandBLUNDER wrote:| :Don't see how it concerns my refractor telescope. Or my contact lenses. |
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y=x2 is an equation for a parabola. there is a particular point for every parabola, called the "focus"... if you put a light source at the focus of a parabolic mirror, every ray of light will end up reflecting off the parabola and moving parallel to the axis of symmetry... http://www.cut-the-knot.org/Curriculum/Geometry/ParabolaMirror.shtml
i meant to hide this, but my coding was a failure... sorry
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ThudnBlunder
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Re: Find a Function
« Reply #13 on: Mar 26th, 2004, 11:01pm » |
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Quote:| y=x2 is an equation for a parabola. there is a particular point for every parabola, called the "focus"... if you put a light source at the focus of a parabolic mirror, every ray of light will end up reflecting off the parabola and moving parallel to the axis of symmetry... |
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I mentioned refractor telescopes, not parabolic reflectors.
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Icarus
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Re: Find a Function
« Reply #14 on: Mar 27th, 2004, 6:47am » |
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I believe parabolic surfaces are preferable for lenses as well as for mirrors. The reason we usually use spherical surfaces is that they are much easier to produce. But with any spherical lens or mirror, you get "spherical aberation" - the failure of light to focus on a single point, causing the image to be slightly blurry.
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I wonder: Which is larger
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