Wikipedia:Reference desk/Archives/Mathematics/2006 September 2

Programming a calculator
I recently got a TI-84 Plus Silver Edition calculator because my old one was crushed. My friends and I were trying to figure out how to program formulas into the calculator and we were sucessful in doing so. One I can't get I don't like to do on paper is Point-slope formula. I use y=m(x-x1)+y1 to find this. I have to formula programmed in, I just don't know how to keep x just an x because when it asks me what I want x to equal I've tried a lot, and it hasn't worked. For example: I plug in y=5(x-5)+5 Anybody with a 7th grade education knows that it's 5x-20, but apparently my $100 calculator can't do it. As far as I can tell, the calculator plugs in 0 for x by default which gives me -20. I try x=x and that doesn't work. I already took a look at the relative aticles and around the TI internet site to no avail. If anybody can help me with this, that'd be great. Thanks. schyler 01:32, 2 September 2006 (UTC)
 * The calculator sees x as the number stored in it. If you want it to simplify expressions, I believe it's impossible, unless you download an application that does that. However, it can solve equasions. Just press [MATH], go all the way down to "Solver...", and enter the equasion. The press down, and then [Alpha][ENTER]. The answer will appear in the "x=" place. Any other questions?--Yanwen 03:01, 2 September 2006 (UTC)

That works great. Thanks. But how about if I had something like -10x+2y=10 and I wanted it to solve it for me. I know that's a really easy one, but it's just an example. Also, what about making a shortcut to ungroup the programs I have saved. Can I program "ALPHA" + "A" and that will automatically ungroup my group? Or take me straight to the equation solver? schyler 23:56, 2 September 2006 (UTC)
 * Do you want to make an equasion in standard form into point-slope form? You probably could make the shortcut but you would need to either modify the OS, or get an application to do it. I'm not too sure about this though because I only know how to program and use it, not download applications and modify it. --Yanwen 21:06, 5 September 2006 (UTC)

I don't like to brag but I'm a TI-83/84 master! ^_^ Firstly, the letters are "reals" - variables that hold numbers, not variables. If you want to solve for multiple variables, you'll have to set something up like

for(A, -10, 10) for(B, -10, 10) if (-10X+2Y = 10) PtOn(A,B)

or Disp A,B instead of point on. There are several good TI-BASIC guides floating around (several advanced algorithms and guides explaining them are available from my | profile at ticalc.org, a website you might want to get involved with) and you'll be able to pick up BASIC if you have any programming experience at all, you just have get used to the environment. When you feel you've mastered BASIC you can learn to give instructions directly to the processor instead of going through basic by using something called assembly language, which is natively supported on the 83 plus/SE and 84 plus/SE. Personally I never had much patience for ASM so I haven't published many things written in it, but you learn a lot about the real architecture of the thing and about embedded devices- for example, on a calculator, any given memory address is used for a single purpose (whether it's holding the contrast level of the screen or for BASIC program storage), something unheard of in higher-level applications like those on PCs. It can be quite valuable to explore and dedicate time to your new device. Good luck~ --Froth 20:26, 8 September 2006 (UTC)

Mathematical expression determinator
I've come across this, but forgotten what it's called or where it can be accessed. It takes a mathematical expression, say an infinite sum of terms or a definite integral, and gives an explicit formula by comparing calculated values of the expression against those of standard mathematical functions such as exp, log, sin. I think that's how it goes anyway - does anyone know what I mean?

86.132.237.108 18:16, 2 September 2006 (UTC)


 * I've never heard of such a device, but we were playing a game of reversing a taylor series here about a week ago (see above). - Rainwarrior 04:53, 3 September 2006 (UTC)


 * In any case, what you describe is something any decent computer algebra system can do (though they usually don't use the method you describe). Take a look at List of computer algebra systems - there are apparently available some free systems. My favorite proprietary system is Mathematica, but admittedly I haven't experimented with many others. Another interesting resource is the integrator, an online indefinite integrator based on Mathematica. -- Meni Rosenfeld (talk) 05:06, 3 September 2006 (UTC)
 * Oh, and there's Plouffe's Invertor - you can give it the decimal expansion of a number, and it returns possible analytic expressions for it. Not exactly what you asked for, but the method you mentioned of comparing with standard mathematical functions reminded me of it. -- Meni Rosenfeld (talk) 05:12, 3 September 2006 (UTC)

It was indeed the Invertor which I was thinking of - I'd forgotten that it worked only with numbers. Thanks for replies - 86.132.237.108


 * I think you were looking for the Inverse Symbolic Calculator. It works like Plouffe's inverter but also handles expressions. Fredrik Johansson 16:00, 3 September 2006 (UTC)

polynomials and symmetry
What polynomials (or other functions, I'm not that fussy) in x,y,z have T symmetry but not Th or Td? &mdash;Tamfang 18:40, 2 September 2006 (UTC)


 * It's conceptually easy but tedious to construct an example. The Roman surface has Td symmetry, which tends to occur naturally. Add decorations to break the reflection symmetry. Ab initio, take a suitable surface defined by a polynomial; make 12 copies of it rotated by each of the T group rotations; multiply those polynomials together.
 * For something found in nature, find a molecule with T symmetry. (These are rare, but do exist. For example, take a methane molecule, CH4, and replace each hydrogen by a radical with S symmetry. Or try 4C3+3C2.) Model each atom as a sphere (a quadratic polynomial), with different radii for different elements. Multiply all these spheres together. Or model each atom as an exponentially decaying potential function, and sum the potentials. --KSmrqT 21:04, 3 September 2006 (UTC)


 * For your symmetry group T, let points v1, v2, ..., v4 be an orbit of size 4 consisting of vertices of length 1. Then you can apply a linear isometry θ to R4 so that the image is x+y+z+w=0, and the image of v1 is parallel to (-3,1,1,1) and so on. Then the group T is A4 acting on the coordinates, the group Th is S4 acting on the coordinates, and group Td is A4 acting on the coordinates combined with -I.
 * Noether discovered that the invariant polynomials of T were generated by the 4 symmetric polynomials of x,y,z,w and the alternating polynomial (x-y)(x-z)(x-w)(y-z)(y-w)(z-w), the invariant polynomials of Th were the symmetric polynomials of x,y,z,w and the invariant polynomials of Td were generated by (x^2+y^2+z^2+w^2), (x+y+z+w)^2, (x+y+z+w)(x^3+y^3+z^3+w^3), (x^4+y^4+z^4+w^4) and the alternating polynomial.
 * Therefore, the lowest degree such polynomial is of degree 7 and is equal to (x+y+z+w)(x-y)(x-z)(x-w)(y-z)(y-w)(z-w).

--DoctorBozzballT 18:01, 7 September 2006 (EDT)

Random processes
Is there anything that can be learned about 2 (pseudo or apparently) random processes by noting thier cross ? correlation coefficient? e.g. say 'Bob' always happens to be in the pub when I turn up but the meetings are never planned--Light current 18:53, 2 September 2006 (UTC)


 * I would say that is evidence that they are dependent events. One may be dependent on the other (as if one person follows another) or they may be mutually dependent on a third event, like if you both go to the bar when you get off of work at 5:00 PM. StuRat 22:19, 2 September 2006 (UTC)

No theyre not dependent. Bob doest get off work at the same time as me. Say we're both random customers. Can I deduce anything about his frequency of visits by my seeing him there very often when I happen to turn up (also fairly randomly)?


 * If they were random, independent events, and you see Bob at the bar half the time you go, then that means he is at the bar about half the time. However, I doubt if they really are independent events.  If you both tend to go at the same times, this makes them dependent events. StuRat 23:25, 2 September 2006 (UTC)

Yeah but maybe I only go about half the time so what does that say about Bobs frequency. Is his probability of bengi there regardlesss of me still 0.5? Im very unclear on this.--Light current 23:29, 2 September 2006 (UTC)


 * Correct, your frequency of attendance has no effect on the calculation of his average rate of attendance. However, the more often you go, the more "sample points" are being taken, therefore the more accurate your calculation of his average attendance.  In an extreme case, if you are always there, you would know exactly what portion of the time he is there.  On the other hand, if you are never there, you would have no idea how often he is there. StuRat 23:36, 2 September 2006 (UTC)


 * I also suspect you may be recalling him there more often than he actually is, since seeing him there is an event you are likely to remember, but not seeing him is something forgettable. An example of this observation bias is that when we see people on TV talking about the lottery, it's almost always because they just won.  If we saw a million people on the TV telling us how they lost, there wouldn't be nearly as much interest in buying lottery tickets.  StuRat 23:36, 2 September 2006 (UTC)

Look I'm going there tomorrow and if I see Bob, Im going to ask him how often he's there OK?--Light current 23:44, 2 September 2006 (UTC)


 * You're heading to a bar to ask a man "do you come here often" ? This experiment could lead to unpredictable results. :-) StuRat 23:47, 2 September 2006 (UTC)

Its OK. I know him. Weve met many times (randomly I might add)--Light current 23:49, 2 September 2006 (UTC)

He wasnt there! So what do I do now?--Light current 04:33, 4 September 2006 (UTC)


 * If you want to approach it scientifically, record your exact times of arrival and departure, whether he is present, and his time of arrival and departure (if they occur when you are there). After you collect a few weeks of such data, you would be a better position to estimate how often he is at the bar.  Of course, you still haven't convinced me that your attendance and his are independent events.  To do that, you'd need to know his times of arrival and departure, even when you are not there. StuRat 07:53, 4 September 2006 (UTC)

Yeah well thats the problem! I could work out my statistics. If i knew those then mearured Bobs only whem Im there, would that tell me anything about Bobs true statistics ( not just how thet appear to me) TITQ 8-|--Light current 20:37, 4 September 2006 (UTC)


 * Yes, if they were truly independent events, and if you had enough sample points, then you could come up with a reliable estimate for what percentage of the time he's there. However, I'm almost certain you two must tend to go "at certain times" which correlate, making your attendance and his not independent events.  In this case, the statistics only show what percentage of the time he is there when you are there, and can't be extended to predict the percentage of the time when he is there, in general. StuRat 04:34, 5 September 2006 (UTC)

Agreed. But lets assume they are truly independent processes. Im sure I shold know how to do this, but Ive forgotten--Light current 04:39, 5 September 2006 (UTC)


 * Well, IF we assume the two events are independent, first total up the time you are both there together, then divide by the total time you are there (either with or without him). This will give you a good estimate of the portion of the time he is there.  Then multiply that by the number of hours the bar is open per week, and that should give you the approximate number of hours per week he is at the bar.  However, this also assumes an even distribution in both your hours of attendance (just as much at the start of the day as the middle and end).  I don't think we should try to go into standard deviations and confidence intervals, which are ways to measure how accurate your estimate is.  Suffice it to say, the more data you have, the more accurate your estimate will be.  StuRat 11:28, 5 September 2006 (UTC)

I need a drink after all that! 8-)--Light current 13:51, 5 September 2006 (UTC)