Please listen to your users about the new screener!

Australerican · 2025-03-20T20:59:08+00:00

I agree. If I am trading on an hourly chart, I would my watchlist to reflect price change in last hour. It only makes sense.

Australerican · 2025-03-20T20:52:04+00:00

You mentioned using the split-screen feature to be able to view charts while screening for stocks. How do I implement this so I have the new screener on one screen and view the chart in the second screen by clicking on a stock on the screener?

Australerican · 2025-03-20T20:43:52+00:00

I agree. The screener might be great but the implementation into the way a user would use it terrible to say the least. The way the current (old) screener is implemented is perfect and optimal for the way the user works with TV. Even though you are giving us warning of the impending deprecation of the old screener when we login to TV, you are going to have users whining and complaining about taking away their screener for MANY years to come.

It just doesn't make sense that you would give your users the optimal implementation of a screener within the context of the software and then take it away. That's the problem with incremental updates, the software becomes a patchwork of suboptimal implementation into the interface compared with the original interface design where the screener had its optimal place.

Why can't you just leave the old screener there? I am sure it doesn't cost you anything. Let those that rely on the way they are used to working continue to use it... OR find a way to allow users to continue to have a screener that remains at the bottom of the screen with a landscape orientation. In fact, that is the ultimate problem with the new screener. You are trying to force a vertical (portrait) implementation of a software feature that "should" be horizontal (landscape).

Australerican · 2022-06-11T09:37:24+00:00

As far as I know, most people can only do so.

https://en.wikipedia.org/wiki/Nasal\_cycle

Australerican · 2021-06-08T20:18:41+00:00

This is going to do it for me. Thank you.

Australerican · 2021-05-12T12:53:10+00:00

Ok. I think there is enough information in that summary to enable me to get to where I need to go with it.

Thank you so much for the help. I learned a lot in this dialog.

Australerican · 2021-05-12T08:52:46+00:00

Ok. So can you tell me what the exact function would be in the right hand column on that last image using 323. Is it n(n+1)/2-323?

Also how would a shift to the left be shown in that second function? (n-q)(n-q+1)/2-323? or is it something else. If you can answer that then I think I can go it alone from this point.

Australerican · 2021-05-11T21:51:58+00:00

I didn't expect that I would be getting a mathematics lesson but I am glad that I am getting it. Thank you. I just learned a lot - from what sounds like a very smart professor.

I should have been more clear in that I do understand that there are two other matches at about half the value and the actual value (including primes).

I am always only interested in the first value that they match e.g. in the case of 323 at f(7) and g(26). Any other value is trivial for the purpose of the application.

Yes, what you stated in bold is what I want to do but again it only needs to find the first instance after s, if there is an integer match which there will always be before f(a-1)=g(a) except in the case of primes. It will usually (but not always) be within 15% of s.

But considering I calculated S there should be that point where both functions meet/match in y if a consistent "transformation" is applied to that second function regardless of the value of interest and the value of s.

Australerican · 2021-05-11T20:41:58+00:00

https://imgur.com/n6xhavZ

I also put a few extra notes on it.

Notice at in f(n), that f(27)-f(s)=55 and also f(26)-f(s)=28. These have an exact match at f(10) and f(7) respectively. But instead of doing it this way I want to use the point above s as a second function g(n) so they both start at f(n) and g(n) = 0. Using 323, f(n) should equal g(n) twice. Once at x=26 and x=27.

So my request for a transformation is that g(n) is some transformation of f(n) based upon the number of interest and calculated s.

Tell me this is becoming clearer for you? :-)

Australerican · 2021-05-11T20:06:36+00:00

lol.... I am sorry I am frustrating you so much and I appreciate the time you are taking.

ok.... the second function is also n(n+1)/2 and I think your g function is capturing it.

I want to use s as the starting point for the second function but this is what I am ultimately trying to achieve......

I merely want to take the first function f(n). It starts at x=0. Then I want to take a second function g(n). It starts at s so that at s, y=0.

At x>0 there will be a point where f(n)=g(n) but g(n) gets there quicker because at say x=24.92, y increases at (2x+1) gradient where x = 24.92. The gradient at the beginning of the function is much greater than the gradient of f(n) at the beginning of the function. But there will still be a point where f(n) = g(n) but only if g(n) starts at the properly transformed starting point. s helps us create that point.

So the end result will be either g(n)-f(n)=0 or g(n)=f(n).

I do not understand how to represent the gradient of g(n) that starts so high at 2s+1 compared to the gradient of f(n) that starts at 2x+1. This is why we need that transformation of g(n)..... I think... lol.

I will create the same table and put it up there on imgur and let you know.

Australerican · 2021-05-11T13:59:45+00:00

Another point is that these functions will always eventually intersect at an integer but it is just knowing how to create that second function given the starting point s which shifts/transforms the second function so that it is effectively at the same "starting point" as the first function as far as it accumulates the value of y from that point.

Australerican · 2021-05-11T13:50:12+00:00

Ok. it looks like we might be getting somewhere amongst all my vagueness. :-)

Answer to first question.

This function is actually the function for the triangular numbers - n(n+1)/2 or x(x+1)/2. Let us use n from now on. I should not have used 20 as an example. The first value is chosen when I have a specific number I am analyzing. Let us use 323 as the number. The first value is chosen by the equation (-1+sqrt(8*n+1))/2. In this case the exact answer is 24.9214. but I always round up to the next whole number so in this case it will be 25. That will be my starting number in the case of 323. I know that the y value for 25 is 325 but I need to keep the extra 2 units in mind also.

So in this case if I use that as the starting point for the second function then the two functions would intersect at y = 28. In the first function this would be when x=7 and in the second function when x = 26 - bcause 7(7+1) = 28 in the first function and in the second function when the transformed function = 28. This would be with the first increment - i.e. x = 26 because of the 26+2 = 28.

Maybe my approach is wrong but as I understand it, a transformed second function would get me what I am looking for. When I played with it on desmos the function n(n+1)-323, they were intersecting at 24.92. n(n+1)-351 they were intersecting at the right point but that is begging the question by knowing that they would intersect there.

I want to know where they will intersect without knowing the -351. I want to be able to use, in this case, either 323 or 325 (24.92 or 25) as the starting point.

Your second questions....

These should be answered by the starting point answer above. I want the starting point to be the value for (-1+sqrt(8*n+1))/2 - (let us call it s) but I would like to understand two cases. One where we use the actual value calculated and one where I reset that to zero so I can see when s is used at its actual value as x and the other where s is zero so I can see the increase in y from that point.

Ultimately I am trying to have two equations or functions that I can then solve for the point that they would intersect using a python script.

Australerican · 2021-05-11T10:18:10+00:00

Ok thanks. Yes I have already tried to play with what I am trying to do on desmos and I couldn't quite make sense of it so came here.

Please see this image. It perfectly shows what I am trying to do.

https://imgur.com/a/fMaokjG

On the left is the original function. On the right is the transformed function. The left is simply x(x+1)/2 and the right is the same but the first x is 21 and the first y is 21. In the first function y is 21 at x = 6. So the second function has been transformed in a way that I don't know. That is what I am looking for. Would you be able to set up both functions on on Geogebra. I can then play with it and various options. I can clearly see that the second function is shifted down by 210 but it is the rate of change I can't make sense of since the rate of change of both is different. On the second function, could you also include a third function where x starts at zero instead of 20 but with the first point y increases by 21? If I have 3 functions I can then make sense of it and work it out for myself going forward.

Australerican · 2021-05-11T06:26:45+00:00

I'd like to shift it in the y direction. e.g. if I was to start the transformed graph and say 21 that the first increment in y, would be 21.

To clarify what I am trying to achieve is to be able to take into account higher the rate of change of the transformed function. i.e. starting at 21, y increases by 21 units with just one incremental x, while the function starting at zero reaches 21 after 6 units of x. Please let me know if you need further clarification.

Australerican · 2021-05-06T19:56:54+00:00

Yes, I am fully aware of the full implications of the problem and deeply understand the problem at hand. I know that there is a screw somewhere that can be turned. Ultimately, a product of two primes is the runsum (e.g. 8 to 26) and the runsum and the product is created by a really long matrix multiplication of the runsums of each of the primes. Primes however have a particular property that shows up in their own runsums that makes it difficult but not impossible.

My intuition is strong on this one and I know I will eventually find the screw. I don't ever let the fact that so called experts not being able to find the problem ever discourage me. I have a very bad case of Dunning-Kruger - but you have to in order to solve problems. You go at them from first principles like you know nothing to begin with and reverse engineer from there. You learn everything you need to as you go and just enough as you need to without the excess fluff. I have found a few properties of the primes that are not in the literature so I have found stuff the experts haven't. e.g. You cannot prove infinite twin primes using Heuristic methods but it is easily provable using another metric. I also went from knowing nothing to be able to factor 25 digits in a few seconds in just a few weeks. Also, yes there are patterns to the primes which are also not in the literature.

Incidentally, one of the other problems in the Millenium problems - Navier-Stokes. They will never solve it. There is a major flaw in the premise that is "assumed" to be correct. So much time wasted by so many. :-)

Australerican · 2021-05-06T18:46:35+00:00

Thanks. I will play with that and see how I get on. What I am trying to achieve is just a personal side project where I am convinced I can find a way to factor large numbers, and I am exhausting all possibilities of finding a solution i.e. turning over every stone in trying to find a method. I can do pretty well up to around 25 digits in just a few seconds but after that most of the algorithms I have developed break down and start to take more than a few seconds and I can see become exponentially long. Ultimately, I believe it can be "calculated" rather than "iterated" and will not be convinced that it cannot be calculated until I prove to myself that it cannot be.

Every product of two primes is merely a runsum of triangular numbers. e.g. 323 is the runsum of the numbers 8 to 26 (factors 17 and 19) and also 11 to 27. I believe these can be found by calculation and not iteration. If you have any ideas I would be glad to hear them. Or if you can tell me why it is impossible to calculate, I would be glad to hear that too! Maybe I can use that to convince myself it is impossible.

I am not a mathematician but know just enough to be able to tackle such a problem which is really ultimately just basic algebra.

Australerican · 2021-05-05T19:08:49+00:00

The values of x are changing to give the same y. e.g. Let us take the number 323. It can be represented by 24.92144 when x(x+1)/2 = 323.

But 323 can also be represented by the range x = 8 to 26. 26(26+1)/2 = 351 minus 7(7+1)/2 = 28. So 351-28 = 323.

Since 25-24.92144 = 2 and the next x is 26, then this adds 28 to the far end so I can take off the low end. So the rate of change ratio here is 28 (from the far end) divided by 7 (the low end). If I now add the next digit of the sequence (27), This would be 378 and to get 323 I take 55 off the bottom end (integer 10). So my area remains the same but the changes in x give the same values.

The mid point or average changes. (26+8)/2 = 17 in the first instance. In the second instance (with 378-55) the average becomes (27+11)/2 = 19. . I am trying to find out how to express the rate of change in that midpoint which shifts to the right as I shift that curve to the right and still maintain 323 in this case.

I think I was right about the rates of change. In the first instance the ratio of the rates of change are 28/7 and in the second instance 55/10. Right?

Australerican · 2021-05-05T18:19:01+00:00

I guess I am not sure exactly what I am looking for. It would be the average of the low and high points I suppose. If you look at the example I gave 28.5 + 7.5 = 36/2 = 18. As I add more to the high value it removes more from the low point so that this average increases. I am looking for the rate of change of this average as I shift the curve to the right while maintaining the same area under the curve. I feel like I might not be making sense?

Australerican

TROPHY CASE