On Jan 31, 2:30 pm, Nimo <azeez...@gmail.com> wrote: > On Jan 31, 6:57 pm, "Greg Neill" <gneil...@MOVEsympatico.ca> wrote: > > > > > Nimo wrote: > > > If I understand you correctly, you suggested me to use > > > union of 'cubic curves' for the solution. > > > > Exactly this is what I had done in my very first > > > post, > > > >> 2 + 4 + 5 + 6 + 8_______(1) > > >> total no.of terms(n) = 5 > > >> sum of all the 5 terms = 25 > > >> sum of first 4 terms = 17 > > >> sum of first 3 terms = 11 > > >> sum of first 2 terms = 6 > > >> sum of the first term = 2 > > >>> I assumed f(x) as X , > > >>> along with 5 un-knowns > > > >>> Int 0_5 (a+b+c+d+e) X = 25___(1) > > > >>> int 0_4 (a+b+c+d+e) X = 17____(2) > > >>> Int 0_3 (a+b+c+d+e) X = 11____(3) > > >>> int 0_2 (a+b+c+d+e) X = 6____(4) > > >>> Int 0_1 (a+b+c+d+e) X = 2____(5) > > >>> the equations are not giving me any solutions? > > > > instead of cubic, I assumed union of 'linears' > > > and I think that at the boundary conditions they > > > are very smooth which in turn give my values back on > > > Integrating. > > > Perhaps you can explain to us how two non-colinear line > > segments meeting at a point presents a smooth transition? > > The resulting curve is not differentiable at the > > intersection point (slope is different on "left" and > > "right" approach). > > > A spline approach fits a set of curves through the > > points so that they are "seamless" where they intersect > > at the fixed data points. Thus the resulting overall > > curve is continuously differentiable. > > > > and I just go on integrating the assumed function > > > to get the equations, > > > but sci.math users suggested me that I'm wrong. > > > This quoted text is your suggested answer for me. > > > >> If a large degree polynomial is not a solution for you, > > >> then try building a function that is the union (say) of > > >> natural cubic SPLINES (or cubic b-splines) such that the > > >> splines have the same derivatives at their endpoints (so your > > >> function is smooth). > > > > here you suggested me union of cubic curves, > > > > but,my question is why not linears which too are smooth > > > at boundary points ? > > > Linears are not smooth at the boundary (intersection) points. > > > > for you, I'm repeating the question again in an easy way > > > > Q) 1+2+5+9+10+12+13+18+...........+...n > > > a) I took them as like this > > > > sum of n terms = a > > > sum of n-1 terms = b > > > sum of n-2 terms = c > > > . > > > . > > > . > > > sum of 1 term = p > > > > Tool: Integrals > > > > Integral 0_n f(x) dx = a > > > > Integral 0_(n-1) f(x) dx = b > > > . > > > . > > > . > > > .Integral 0_1 f(x) dx = p > > > > In my first reply I took f(x) linear > > > i.e as X and just go on integrating > > > for equations, but sci.math users said > > > that it will not work in that way :( > > > You took one linear equation and tried to fit it to > > non-colinear points. How could you expect that to work? > > > At bare minumum you should at least have a set of independent > > linear functions to describe the curve in a piecewise continuous > > way, and ignoring differentiability issues at the intersection; > > just accept that the solution will be a peicewise one, too. > > > That is equivalent to "drawing" straight lines between > > successive data points, and writing the equation for each > > line. Differentiating is trivial -- the differential of > > each line segment is a step function of the same domain > > duration as the line segment, and suitable magnitude. > > > > Can you tell me what exactly your point is here ? > > > as you said that union of Cubic curves is enough > > > which are smooth at boundary points, in turn > > > which on integrating will give the values > > > a,b,c......p back. > > > Cubic splines will provide smoothness of transition > > (differentiability) through the data points. You then have > > a differentiable piecewise continuous curve that you can work > > with, rather than just a piecewise continuous one as with > > the linear fit I suggested above. > > You still will have as many **("terms")** in the solution as there > > > are data points. But at least they won't be high order > > polynomial terms. > > you mean by this time too, we are hopeless. > because of the disadvantage of these line > " You still will have as many **("terms")** in the solution as there > are data points "
See my earlier reply to you in this thread. It is IMPOSSIBLE to find a "simple" form for your function that will satisfy 20 or 30 arbitrary equalities. The "complexity" of the function will increase with the number of arbitrary equalities that it needs to satisfy. This is true whether you have a uniform function (such as a polynomial fit, with lots of terms) or a piecewise function (where each piece may be simpler but you need many of them, all separately defined).
The only way to avoid this increasing complexity is if the equalities to be satisfied are not arbitrary but have some special pattern.