Re: Unstable Simultaneous Linear Equations!

On Apr 1, 9:34 am, monir <mon...@xxxxxxxxxxxx> wrote:
On Apr 1, 12:47 am, Dave Dodson <dave_and_da...@xxxxxxxx> wrote:





On Mar 31, 11:20 pm, monir <mon...@xxxxxxxxxxxx> wrote:

On Mar 31, 10:20 pm, Dave Dodson <dave_and_da...@xxxxxxxx> wrote:
On Mar 31, 9:01 pm, monir <mon...@xxxxxxxxxxxx> wrote:

2) One uses +ve or -ve sqrt for "y" depending on the value of the
independent variable "x" in relation to the centre of the curve and
the coordinates system.

So I give you an x. How do you know which sign, and therefore which y
to produce?

3) I haven't observed a "cusp" or a "sharp point" anywhere and
certainly not at or near P2 for any of the tested cases so far!  It
would be a serious setback if your observation could be substantiated.
I've enlarged the corresponding plot and zoomed on P2.  There's no
"cusp" or any abnormality to be identified anywhere!
Did you observe it visually or from the analysis of the provided
numerical table ??
Could you please elaborate?

In an early post in this thread, you stated that you had a tiny gap in
an analytic development between two points, and that you know the
points and the slopes at each. To me that implied that you have a
curve on each side of the gap, and you want to fill the gap with an
interpolating curve. If this is not the case, then my comment is
irrelevant.

Dave

Dave;

Thanks again for your prompt reply and for your interest in the
problem.

So I give you an x. How do you know which sign, and therefore which y to produce?

As I suggested earlier, there're 4 possible scenarios that one must
check after solving the 4 SLE for A, B, C, D in order to channel the
procedure to construct the correct elliptic interpolation curve:
given: point P1(x1,y1) & slope d1
       point P2(x2,y2) & slope d2

scen1: IF y1 >= k & y2 >= k  i.e.; both P1 and P2 on the upper surface
..........use "+" sqrt() for x1 <= x <= x2
scen2: IF y1 > k and y2 < k  i.e.; P1 on the upper and P2 on the lower
surface
..........use "+" sqrt() for x1 <= x <= h+a  on the upper surface
..........use "-" sqrt() for h+a > x >= x2  on the lower surface

Here is where I see the problem with your curve not being single
valued. If I give you an x > x2, where the curve has two values, how
do you know whether I want the upper one or the lower one so that you
know whether to use the + sign or the - sign?

scen3: IF y1 < k and y2 > k  i.e.; P1 on the lower and P2 on the upper
surface
..........use "-" sqrt() for x1 >= x <= h-a  on the lower surface
..........use "+" sqrt()for h-a < x <= x2  on the upper surface
scen4: IF y1 < k and y2 < k  i.e.; P1 and P2 on the lower surface
..........use "-" sqrt() for x1 >= x <= h-a  on the lower surface
..........use "+" sqrt() for h-a < x <= h+a  on the upper surface
..........use "-" sqrt() for h+a > x > x2  on the lower surface
(scen2 above applies to the worked example and the solution provided
earlier)

where:
elliptic interpolation curve: Ax^2 + By^2 + Cx + Dy = 1
A = (b^2/a^2)/[b^2 - k^2 - (b^2/a^2)h^2]
B = 1/[b^2 - k^2 - (b^2/a^2)h^2]
C = -2h(b^2/a^2)/[b^2 - k^2 - (b^2/a^2)h^2]
D = -2k/[b^2 - k^2 - (b^2/a^2)h^2]
with centre at (h,k)
ellipse axis along the X-axis = 2a
ellipse axis along the Y-axis = 2b

In an early post in this thread, you stated that you had a tiny gap in an
analytic development between two points, and that you know the points and
the slopes at each. To me that implied that you have a curve on each side
of the gap, and you want >to fill the gap with an interpolating curve..
If this is not the case, then my comment is irrelevant.

This is exactly the case!

Okay. If you trace along the ellipse from (x1,y1), you get to the
point (x2,y2), headed downward and to the left. The curve that you are
matching function value and slope is heading upward and to the right.
This creates a cusp. If, for example, you were marching along the
ellipse from (x1,y1), you would have to do an about-face at point
(x2,y2) in order to continue marching along the analytically developed
curve that begins at that point. I wouldn't consider this "smooth." Do
you?

Now, for an x > x2, it is possible that a vertical line cuts your
combined curve in 3 places: the upper curve of the ellipse, the lower
curve of the ellipse, and the analytically-developed curve starting at
(x2,y2). How do you know which of the three possible y values I want?

Am I beating a dead horse?

Dave- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Dave;

There's a good chance that I overlooked little details in describing
the procedure in my post which's causing such confusion!  This is
likely the case since I haven't had any problem (so far) in applying
the procedure for 10s of different scenarios.

Since the proof of the pudding in the eating, let me suggest the
following.
Elliptic Equation:
Ax^2 + By^2 + Cx + Dy = 1

1) Plot the numerical table provided earlier (Mar 31, 3:36 local) for:
...........x.........y........d=(dy/dx)
P1:: -0.0266255  0.9981449   -0.1593857
P2:: -0.0248588  0.9847863  132.5084940
A = -48.857138
B = -0.994123
C = -2.605710
D =  1.959330

2) Plot the following 2nd example:
..........x.........y......d=(dy/dx)
P1:: -0.057785  0.996125   0.496957
P2:: -0.048650  0.985900  43.111960
A = -2.284E+00
B = -1.021E+00
C = -2.540E-01
D =  2.014E+00

The full-range Elliptic interpolation results:
X..........Y results...(dY/dX)results
-0.057785  0.996125    0.496957
-0.057557  0.996232
-0.057328  0.996327
-0.057100  0.996409
-0.056871  0.996478
-0.056643  0.996536
-0.056414  0.996583
-0.056186  0.996618
-0.055957  0.996641
-0.055729  0.996654
-0.055500  0.996655
-0.055272  0.996644
-0.055043  0.996623
-0.054815  0.996590
-0.054586  0.996545
-0.054358  0.996490
-0.054129  0.996422
-0.053901  0.996342
-0.053672  0.996250
-0.053444  0.996145
-0.053215  0.996027
-0.052987  0.995896
-0.052758  0.995750
-0.052530  0.995590
-0.052301  0.995414
-0.052073  0.995222
-0.051844  0.995012
-0.051616  0.994783
-0.051387  0.994534
-0.051159  0.994262
-0.050930  0.993966
-0.050702  0.993642
-0.050474  0.993287
-0.050245  0.992895
-0.050017  0.992460
-0.049788  0.991971
-0.049560  0.991414
-0.049331  0.990763
-0.049103  0.989968
-0.048874  0.988904
-0.048646  0.986261
-0.048646  0.986261
-0.048646  0.986147
-0.048646  0.986100
-0.048647  0.986063
-0.048647  0.986033
-0.048648  0.986006
-0.048648  0.985982
-0.048649  0.985959
-0.048649  0.985938
-0.048649  0.985919
-0.048650  0.985900   43.111960

3) In the above two examples:
Do you actually observe any cusps, discontinuities, cross-overs, sharp
points, etc., anywhere along the "smooth" elliptic curve joining P1
and P2 ??
I don't !!

As I have been trying to explain, the cusp isn't within the ellipse.
It it where the ellipse joins the curve to the right at point P2. As
you traverse the ellipse from P1 to P2, you go down and to the right
until you get to the vertical tangent somewhere beyond x2. Then you
continue down, but go to the left until you get to P2. At P2, you
switch from the ellipse to the analytically derived curve that goes up
and to the right. The cusp is where you switch from going down and to
the left to going up and to the right, exactly at P2. Oh for a
picture!

4) If your answer is YES, then there's a problem with the procedure.
If your answer is NO, then it would be clear that I overlooked some
details in describing the method and I'll try to provide a complete
description.

Regards.
Monir
.

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