(12C) Newton's Method

03242018, 01:10 PM
(This post was last modified: 03272018 11:10 AM by Gamo.)
Post: #1




(12C) Newton's Method
This program is an adaptation of the HP 25 Solver program from our Member name Dwight Sturrock under the topic (12C) Numeric Solver
I make update to this program from Dieter advice as well. I have been looking for the Solver Program for HP12C for a long time, I didn't notice before that Dwight post this since 2013 This is a very good all around Solver for HP12C and this one work very well. Newton's Method Solver work extreamly fast on HP12C+ and I'm very satisfy with this program. The program uses registers 0, 1, 2, 3 and 4. >Put f(x) program to line 41 >End with GTO 19 Procedure: 1. Store initial guess to R1 (STO 1) 2. Store tolerance to R2 (STO 2) Program: Code:
Example: Solve for X^X=Y (X to power of X for the given Y) X^X=1000 Formula use to solve this equation: LN(X)*X – LN(Y) = 0 Press [g] [GTO] 40 and Switch to Program Mode: Program start at Line 41 LN LSTx x RCL 5 LN  GTO 19 (End) Setup equation: 1. Initial Guess for 4 > STO 1 2. Tolerance > EEX CHS 6 > STO 2 3 1000 > STO 5 (This number can be change to any number you like to solve) 4. R/S to solve. Result 4.555535705 RCL 1 is the store answer of the problem. RCL 4 to check for accuracy. [0.000000000] is the best result with ± estimation. Gamo 

03242018, 01:31 PM
(This post was last modified: 03242018 07:30 PM by Dieter.)
Post: #2




RE: (12C) Newton's Method
(03242018 01:10 PM)Gamo Wrote: This is a very good all around Solver for HP12C and this one work very well. ...if you remove line 33. ;) The sequence 2 ENTER x SQRT does not make sense, this always yields 2. Remove the line with the "2" and the program should work as intended (compare the absolute value of the last correction term with the tolerance in R2). This also means that all references to line 42 should be changed to line 41. There is an error in the instructions for the example: the tolerance has to be stored in R2. So it's EEX 6 CHS STO 2. I think you should also mention that x^x=y here is transformed to x · ln x – ln y = 0, which is how the function is coded. The program may have an error in that it stops the iteration not only if f(x)=0 but also if f(x+h)=0. OK. this can be fixed easily. And maybe the final step should better be GTO 00 instead of R/S so that the program can be restarted with a simple R/S. Here is my attempt at a corrected version: Code: 01 STO 1  store the tolerance in R2  enter a guess, start with R/S By the way, I have also tried a Regula Falsi / Illinois version for the 12C. This is slightly more effort as the 12C does not support subroutines and there are three different cases (as opposed to two for Newton) that have to be handled by the code that accepts the return value of the function call. But it is possible – in 70 steps. Dieter 

03252018, 03:05 AM
Post: #3




RE: (12C) Newton's Method
I took off the 2 and rearrange GTO lines and now the program is updated.
Retry the same X^X=Y equation. Now work flawlessly for this particular problem. I like to see the Regula Falsi Method if it possible to make it to 12C HP12C should include SOLVE like the higher model of the HP Financial Calculator. Gamo 

03252018, 06:26 PM
Post: #4




RE: (12C) Newton's Method
(03252018 03:05 AM)Gamo Wrote: I like to see the Regula Falsi Method if it possible to make it to 12C [x] Done. (12C) Solve f(x)=0 with modified Regula Falsi method. (03252018 03:05 AM)Gamo Wrote: HP12C should include SOLVE like the higher model of the HP Financial Calculator. I don't think it has to. Look, the 12C is not a universal tool, it is a specialized calculator for financial applications. That's why it has a very limited set of scientific functions: square root, power, natural log and e^x – and that's it. There is not even an x² key, let alone sin, cos, tan, common log and antilog and others. With the limited ressources of 1981 you could either have a financial calculator with a quite complete function set for such applications, or you could get a more universal scientific device like the 11C. Looking at 37 years of success this seems to have been a quite good decision. ;) Dieter 

03272018, 06:08 PM
(This post was last modified: 03272018 06:23 PM by Dieter.)
Post: #5




RE: (12C) Newton's Method
(03242018 01:31 PM)Dieter Wrote: Here is my attempt at a corrected version: Gamo, in the Regula Falsi thread you said that using the Newton solver is a bit cumbersome as – in the original version you posted – both the tolerance and an initial guess have to be prestored. The latter can be fixed easily (cf. the version I posted above), but also the exit condition can be simplified so that no user interaction is required. The above versions use a fixed tolerance that is prestored by the user. On the other hand my Regula Falsi program quits as soon at the last two approximation agree to display precision. Here is another option: the following Newton solver iterates until the relative difference between the last two approximations drops below 5 E–9. This should provide 8 correct digits, but since the Newton method converges quadratically, in most cases the result should be close to exact. Here is the code: Code: 01 STO 1 Try it with your example function 3^x – x^3 = 0. Code: g [GTO] 46 Now try this with an initial guess of 2: Code: 2 [R/S] => 2,478052683 You can always add one more iteration by pressing [R/S] again. In the above case the result remains the same, with 10 digit precision f(2,478052683) yields exactly 0. Now start with a guess of 2,5 which is even closer to the solution: Code: 2,5 [R/S] => 2,478052686 The result differs in the last digit. Now press [R/S] again and see what you get: Code: [R/S] => 2,478052680 See? The iteration oscillates between 2,478052680 and ...86. That's why it is not a good idea to continue until the last two approximations match exactly (which would have caused an infinite loop here). The method in the above program stops if they agree in at least 8 digits, which usually means that it will not get more accurate anyway. Nevertheless you can always press [R/S] again and see if the result improves. This way an infinite loops is avoided (well, at least in most cases). Try the function with a guess of 6 and you see that it would otherwise oscillate endlessly between 3,000000002 and 2,999999998. In detail: line 3435 divides the last correction term by h, and since h = x / 1E+4 this yields the current relative error times 1E+4. Dividing by h instead of x automatically handles the x=0 case for which h is set to 1E–4. Then 2E+5 is added and subtracted again, which shifts a sufficiently small error beyond the 10th digit so that the result is zero if the error is small enough. Selecting a constant with more or less digits (2E+6 or 2E+4 etc.) changes the error threshold, but I think 2E+5 is just right for a 10digit calculator while h = x / 1E+4. Dieter 

03282018, 04:23 AM
(This post was last modified: 03282018 10:22 AM by Gamo.)
Post: #6




RE: (12C) Newton's Method
The new updated user friendly version of the Newton's Method now work even better than the previous version just start your guess and go R/S !!!
I try the same X^X=Y Store Y in Register 5 (STO 5) X^X=1000 1000 [STO 5] 2 R/S > 4,555535706 3 R/S > 4,555535705 4 R/S > 4,555535705 150.25 [STO 5] 2 R/S > 3.773964520 3 R/S > 3.773964519 4 R/S > 3.773964520 5 R/S > 3.773964519 6 R/S > 3.773964519 123456789 [STO 5] 4 R/S > 8.640026440 5 R/S > 8.640026441 6 R/S > 8.640026437 7 R/S > 8.640026440 8 R/S > 8.640026440 9 R/S > 8.640026439 10 R/S > 8.640026440  Solve: LN(X) + 3X = 10.8074 Change to: LN(X) + 3X  10.8074 = 0 Program Code: LN LSTx 3 x + RCL 5  Solve Equation by first store 10.8074 to Register 5 [STO 5] 4 > R/S > 3.213360869 check tolerant gave exact 0.000000000 3 > R/S > 3.213360870 check tolerant gave 0.000000020 So far this version is very accurate and no need to manually change tolerant. Gamo 

03292018, 05:11 AM
Post: #7




RE: (12C) Newton's Method
I like it too. Gamo said it all. Key in your guess and press [R/S]. Well done


03292018, 05:04 PM
(This post was last modified: 03292018 05:25 PM by Dieter.)
Post: #8




RE: (12C) Newton's Method
(03292018 05:11 AM)Carsen Wrote: I like it too. Gamo said it all. Key in your guess and press [R/S]. Well done Thank you. But there is always a way to squeeze out a few bytes. Here is an optimized version with some changes. These include:
Here is a commented listing: Code: 01 STO 1 // store initial guess As usual, comments and suggestions are welcome. Dieter 

03292018, 05:23 PM
Post: #9




RE: (12C) Newton's Method
Great job as always.
I should turn you loose on my lifetime HP 41 Yahtzee game code. First wrote it back in 1985 and always looking to make it shorter, faster, better. ;) 

03302018, 01:15 PM
Post: #10




RE: (12C) Newton's Method
The latest update to only 40 line Newton's Method program work very good.
Find a root of a polynomial of f(x) = X^4  4X^3 + 8X^2 + 20X  65 Program for f(x) [ENTER] [ENTER] [ENTER] 4 [] [x] 8 [+] [x] 20 [+] [x] 65  5 > R/S > 2.236067978 4 > R/S > 2.236067978 3 > R/S > 2.236067977 2 > R/S > 2.236067977 1 > R/S > 2.236067978 Answer is 2.236067977 Gamo Can I check for accuracy on Register 2 

03302018, 02:00 PM
(This post was last modified: 03302018 02:54 PM by Dieter.)
Post: #11




RE: (12C) Newton's Method
(03302018 01:15 PM)Gamo Wrote: 5 > R/S > 2.236067978 The true result is right in the middle. Even with 13digit precision the answer is 2,236067977500. See below. (03302018 01:15 PM)Gamo Wrote: Can I check for accuracy on Register 2 In a way, yes. But let's be precise for a moment now. What is accuracy? How do you define it? The term may relate to two different things. On the one hand you may refer to the difference between the returned result and the actual true answer. For instance, if the true result is 2 and the program returns 1,999999999 this is accurate to 9 places, or it has an error of 1 ULP. On the other hand you may take a look at f(x) for the returned value. Does it return exactly zero, or 0,000000001 or something else? Both definitions may come to different conclusions. For instance, there are cases where a whole range of results yields f(x) = exactly 0. Remember the 3^x–x^3 case? Here anything between 2,478052679 and 2,478052683 will return exactly zero on a correctly working 10digit calculator. The true result is 2,4780526802883... On the other hand even a function result that is not zero does not mean that the result has a slight error. It may be exact to all 10 digits. Consider the function x^2–2=0. The exact 10digit result here is 1,414213562, but f(1,414213562) yields –0,000000001. The next higher answer 1,414213563 returns +0,000000002. So there simply is no 10digit result that returns exactly 0. You can also check by how much the final value of x has been corrected. This can be a measure for its accuracy. But again, even if the last two approximations agree and the correction term is zero, this does not neccessarily mean that the result is dead on. Back to the programs now. You said that in the original version of the program (cf. your initial post) RCL 4 could be used to check the accuracy of the result. In fact this is not possible. The original version calculates f(x) and stores this in R4. But then a "better x" is calculated and finally displayed. So R4 holds the function value of the secondtolast x, and not that of the finally returned x (!). In the last version of my program you may recall R2 to see the last correction term. If R2 is zero, f(x) returns zero as well. Nonzero values mean that f(x) of the previous (!) approximatino was not zero, but x should be at least as accurate as this. So if x is for example 3,5 and RCL 2 returns 3E–9 you know that the final correction was 3 units in the last digit. The program terminates if this update only affected the last one or two digits. In most cases this means that the new result is as good as it gets with 10digit precision. Due to the quadratic convergence of Newton's method, in a perfect world even 15 digits should be fine. There is an easy way to get another impression of the accuracy: press [R/S] again for an update of x. In some cases this may show how x oscillates between two values. Remember the example where the last two digits changed between ...80 and ...86? Here the calculator simply cannot do better, for instance because of limited accuracy in the function itself, and the last digit remains uncertain. He is an example for the polynomial in your last post. 5 [R/S] => 2,236067978 RCL 2 => 4,938E–10 So f(x) is not exactly zero, but the last correction was less than 5 units in the 11th digit, thus not affecting the 10digit result. In fact the true result is right in the middle between 2,236067977 and ...78. The program tried to correct it to 2,2360679775 but due to 10digit precision it could not do so. Dieter 

03312018, 01:15 PM
Post: #12




RE: (12C) Newton's Method
New equation: X = (e^X) solve for X
Change equation to (e^X)  X = 0 f(x) Program: STO 5 CHS e^x RCL 5  3 R/S > 0,567143290 2 R/S > 0,567143290 1 R/S > 0,567143290 X = 0,567143290 e^0,567143290 = 0,567143290 Gamo 

03312018, 05:27 PM
Post: #13




RE: (12C) Newton's Method  
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