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 April 17th, 2009, 11:15 AM #1 Newbie   Joined: Apr 2009 Posts: 2 Thanks: 0 Reverse solve f(x)? I don't even know if this is possible, but if I'm given f(1)=4; f(=37; f(3=265; f(66)=529; f(100)=922; Is there a way I can determine was the equation is? Or do I just not have enough information?
 April 17th, 2009, 01:29 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Reverse solve f(x)? If these are observations from an experiment, you need a theory as to what kind of function you are dealing with. You can always find a polynomial which passes exactly through all points. If f(p) = P, f(q) = Q, f(r) = R, et cetera, up to n points, then solving the following for a, b, c, ... $\begin{bmatrix}1=&p=&p^2=&p^3=&\dots=&p^{n-1}\\1=&q=&q^2=&q^3=&\dots=&q^{n-1}\\1=&r=&r^2=&r^3=&\dots=&r^{n-1}\\\\\vdots=&\vdots=&\vdots=&\vdots=&\vdots=$ yields the polynomial $f(x)= a+bx+cx^2+...$ which passes through the points. For your problem, there is the exact solution a = 2690755825098139/9676163421751760 b = 28995251727231868/8085932936232345 c = 4132022610192845/30031568564477792 d = -326183591/208301421328 e = 7844227/1041507106640 or roughly $f(x)= 0.27808+3.585888x+0.137589x^2-0.00156592x^3+0.00000753161x^4$ This method (one of many) is unsuitable if the number of points is large.
 April 17th, 2009, 02:07 PM #3 Newbie   Joined: Apr 2009 Posts: 2 Thanks: 0 Re: Reverse solve f(x)? Okay, I see where you are going with it thanks!
 April 17th, 2009, 02:43 PM #4 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Reverse solve f(x)? Polynomial regression, using "Graph" freeware, available from http://www.padowan.dk : f(x) = 0.035001795*x^2+5.7852048*x-6.1104195
 April 17th, 2009, 05:23 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,833 Thanks: 2161 Hi NumberA, where did your values come from? Notice that the above methods are not very good at giving you an "appropriate" value for f(0); there's little scope for improvement with so little information.
April 19th, 2009, 11:03 AM   #6
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Re: Reverse solve f(x)?

Quote:
 Originally Posted by NumberA I don't even know if this is possible, but if I'm given f(1)=4; f(=37; f(3=265; f(66)=529; f(100)=922; Is there a way I can determine was the equation is? Or do I just not have enough information?
Yeah, you could possiblily determine what the equation is, and employing idea of the 2nd post could proper to a technique for such an equation. For example, sine and cosine can be use as orthogonal basis for continuous function with finite discontinuities which satify it left and right approaching limit condition, thus one can insert them as the x in the respective polynomial to mimic all such possibility between the concerned points. By allowing more powerful techniques (for example the weilding of these points to examified an exponental-like function or the usage of double or triple limiting processes), the conditions on f soften* and the key here is the word "equation," and as intended, a formulation of known functions. However, the equation involved had many variables that are yet to be chosen and the polynomial stated is a special case when the rest are set to null.

*one should note that the smoothness of f is sometime harder than just it plain continuity or even worse just the existence of these points alone. For example, what is the algebraic equivalence of the infinite produce as a varies over R\{38} of (x - a) in the denominator of a rational function? Surely one approach is the use of multiple limit, double limits over the natural as our concern is rather of numerical by nature and hence the usage of exp to relay it in addition as negative power of e. Yet, one right the way "notice a jump" since e^(x) is continous in the real and that what concerned is not even in the domain to begin with. Hence one deduce that by the real definition of function itself the question is not well-defined in a sense but one can continue base on the fact of function composition and its related continuity. Since e is continuous the outer limit, both of them, pass thru sequentially and one naturally have a integral-like summation. One should also take a note of the series digital representation of the a's, thereby exchanging index for another sum. Thus one is essentially taking the triangle-spike integration of ln(x), that is, again taking advantage of the topology in the domain to have a fat triagular base as in "1/2 of a dx." Note, at this point one already have an equation that goes nowhere with the exception of the indicated points. Put is not wary of finding a "honest" answer as one notice of a familar trick of e to an integral in O.D.E.

Edit: as one already know the nature of singularity of natural log, however, exp(ln(1/x)) as x going to zero take advantage of the nature of infinity and use it, this is not really consider a technique per say but can be study in depth as its generality. By inserting it formulation against the background of an O.D.E. one hope it lead to something that is more concrete from what one already have rather something that is "out of the blue." For example, on can have a surface solution that have a missing string with some left-over attachment.
Edit 2: the nature of how one label each symbol is important, for example, formally one set f := the produce of the mentioned denominator but actually work with f(k, n) and its pointwise convergent in the one-dimensional setting. I have not work with it personally but it has two "seeable results" that are concrete one is an one variable expression which has most of it points going toward infinity and the other is the mentioned two variables overlay, which can be similfy by setting one to a particular constant. Thus to this end, there are more than on way to approach the former as one can use orthogonal function to approximate an expanding outward to infinity of constant functions and it horizontal expand as it mimic; as if trading the lack of expression for two limits within the real. Still yet, this poster will undertake another method.
Edit 3: for example the constant function 1 with a hole at zero has the algebraic close form of x/x, however it is analytically not as useful as the function define by 1/abs(x)^(Monster(8^(Monster(abs(x))abs(x)ln(abs(x + 1))) even though it is clearly not equal as one get near the hole and whereas the latter is managable (as it is differentiable without disappearing, add and substract and what else without major loss of structural upheld,...) and it former is still an abstraction of the mind.
Edit 4: I guess part of my personality is not being slick and mathematics is just a hobby. But notice the inner function in the denominator. Even though the example in Edit 3 seem like it has not much to do with the question at hand but it does. Indeed, one can use a central expression as a focal point and arrive at another expression to which one can use a system of 2 by 2 (note one have to get pass the equated equality, it is in actuality a 4 expression finder strictly speaking of course) to simlify what is asked in Edit 2 beside the whatever else methods one now have at will. And this is the method I had in mind in Edit 2, to literately push the parallel-horizons to negative or positive infinite where theirs vertical slope smash at a single point; that is, one part of the rational function is going to negative infinity and next-to-it-right is going to positive infinity such that the concerned curvature meet. Thus, in a sense the door is open to find a concrete managable function I think this poster asked.
Edit 5: I think I know why he/she/they asked it. Perhaps to write an analytical-program base language perhaps.

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