HP Forum Archive 14

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49g+ Cube roots Message #1 Posted by Kellie on 24 Feb 2004, 10:07 p.m. I just recently purchased a 49g+, and am trying to graph a cube root function. I am only able to generate the portion of the graph with y values greater than 0. Is there a way to get the other half of the graph as well?

Re: 49g+ Cube roots Message #2 Posted by andy on 24 Feb 2004, 10:20 p.m., in response to message #1 by Kellie 1. Go to the "plot window" screen (press and hold left-shift and press F2). 2. Check the value for "Indep Low". I think it defaults to zero. If you change it to a number less then zero then the 49g+ will graph from that number to whatever you have the "Indep High" value set to.

Re: 49g+ Cube roots Message #3 Posted by Dave Shaffer on 25 Feb 2004, 12:07 p.m., in response to message #1 by Kellie I suspect you've run into the "what is the cube root of a negative number" problem. For positive numbers, the cube root (one of them) is a positive number, too. These are the positive (i.e. y>0) values that you can plot now. For negative numbers, in most cases (except for numbers which are a perfect cube of some other negative number, such as -8 = -2^3) the cube root will be a complex number: cube root of -4 = 0.794 + 1.37i (as my '42S just gave me). I don't know for sure (I don't have one), but I'll bet that the 49g+ doesn't know what to do when it tries to plot that complex number - so it just gives up.

Re: 49g+ Cube roots Message #4 Posted by R Lion on 25 Feb 2004, 12:52 p.m., in response to message #3 by Dave Shaffer Not exactly: Any number has three complex roots (one of them is real) Try how -1,58740105197 is a cubic root of -4... The next 65 bytes program is xROOTy command for the 42s: LBL "xROOTy" EXITALL ROLLUP STO "A" ROLLUP STO "B" ROLLDOWN ROLLDOWN ENTER ENTER 2 MOD X=0? GTO B CLx + X<>Y SIGN LASTx ABS RCL ST Z GTO A LBL B CLx 1 STO ST T ROLLDOWN LBL A 1/X Y^X * RCL "B" STO ST Z CLx + RCL "A" STO ST T ROLLDOWN END Sure it is not a good program from the programmer point of view, but it does the work... Please: post if you don't get the expected results to correct the prg

Re: 49g+ Cube roots Message #5 Posted by bill platt on 25 Feb 2004, 12:59 p.m., in response to message #4 by R Lion Hi Raul! I want to try this in a not 42s. Could you tell me what the "MOD" command is so that I might find an equivalent on a 15 or 32 or some other? Best regards, Bill Platt plattdesign dot net

Re: 49g+ Cube roots Message #6 Posted by Valentin Albillo on 25 Feb 2004, 1:23 p.m., in response to message #5 by bill platt Hi Bill: Bill posted: "Could you tell me what the "MOD" command is so that I might find an equivalent on a 15 or 32 or some other?" You can use this expression to evaluate the MOD command, coded in your favorite programming language: X MOD Y = X - Y*INT(X/Y) so, for instance: 8 MOD 3 = 8 - 3*INT(8/3) = 8 - 3*INT(2.666+) = 8-3*2 = 2 Best regards from V.

Thanks V. , R.L. :^) (nt) Message #7 Posted by bill platt on 26 Feb 2004, 5:01 p.m., in response to message #6 by Valentin Albillo .

Valentín answered ;-) (nt) Message #8 Posted by R Lion on 25 Feb 2004, 1:30 p.m., in response to message #5 by bill platt

Re: 49g+ Cube roots - OOOOOPS Message #9 Posted by Dave Shaffer on 25 Feb 2004, 1:40 p.m., in response to message #4 by R Lion You are, of course, correct: "Any number has three complex roots (one of them is real)" (Although this should actually be "Any number has three complex [CUBE] roots (one of them is real)") I realized this about an hour after I posted. I fixated on the complex root which the 42S gave me - which I obtained by taking -4 to the 1/3 power. When I use the "x-th root of y" button to take the cube root on my 32SII, it gives me the -1.587 answer. Thus, Kellie's problem may or may not relate to how she is taking the cube root. Which of these modes does the 49g+ use?

Re: 49g+ Cube roots - OOOOOPS Message #10 Posted by R Lion on 25 Feb 2004, 1:51 p.m., in response to message #9 by Dave Shaffer Quote: (Although this should actually be "Any number has three complex [CUBE] roots (one of them is real)") You are, of course, correct ;-) About Kellie's problem... in my 48GX I get the correct graphic when I use cubic root, but only what Kellie says if I use x^(1/3) Raul L

Re: 49g+ Cube roots - OOOOOPS Message #11 Posted by Andrés C. Rodríguez (Argentina) on 25 Feb 2004, 5:38 p.m., in response to message #9 by Dave Shaffer Recently some of us had a thread on this matter (near the end of January), and we concluded that the 32 SII function "x root of y" answers with the real root (for instance: -2 as the cubic root of -8), while the "x to the 1/3 power" approach (as in the HP42S) answers with the so-called "primary" root, which is the one in the first quadrant of the complex plane.

Re: 49g+ Cube roots - OOOOOPS Message #12 Posted by Ernie Malaga on 25 Feb 2004, 10:01 p.m., in response to message #9 by Dave Shaffer [quote/ (Although this should actually be "Any number has three complex [CUBE] roots (one of them is real)")[/quote] Actually, that's not precise either, Dave. Any _real_ number will have one real cube root. But cube roots of complex numbers (a+bi, where a<>0 and b<>0) are unlikely to be real. Forgive me if you already know this, but there's a picture that helps remember this. Think of the Mercedes Benz symbol -- the points are at 120-degree intervals. Same with all 3 cube roots of a number. -Ernie

Re: 49g+ Cube roots - OOOOOPS Message #13 Posted by Dave Shaffer on 26 Feb 2004, 12:40 a.m., in response to message #12 by Ernie Malaga Ernie, I think you're wrong. There really are three cube roots for all numbers - even real ones. For sure, for negative, real numbers, (such as -2, -3, -4, etc.), there are the regular cube root (i.e. the same cube root as for the equivalent positive number, but with a minus sign in front) as well as those roots rotated by +/-120 degrees in the complex plane. (The Mercedes symbol has to be rotated by -90 degrees for positive numbers, and +90 degrees for negative numbers, where + implies counterclockwise rotation. The "simple" (non-imaginary) root lies along the +x axis for positive numbers and along the -x axis for negative numbers.) For example: the cube root of +4 returned by the 42S is 1.5874 (using the y^x button, with y=4 and x=1/3), i.e. along the +x (positive real numbers) axis. However, the complex roots -0.794-1.37i and -0.794+1.37i also give +4 when cubed (neglecting round-off, of course). Conversely, the cube root of -4 determined by the same set of button pushings gives +0.794+1.37i with the 42S. But, -1.5874^3 = -4, as does (0.794-1.37i)^3 . (Again, to within round-off) Similar results hold for any positive or negative (real; i.e. non-imaginary) number. Thus, Raul had it right (with my minor correction). The extension to any power is that there are n values for the n-th root, arranged as a Mercedes star with n points equally spaced in angle in the complex plane, with one point aligned with the +x (real) axis for positive numbers, and one point aligned with the -x (also real) axis for negative numbers. I leave complex values for our mathematician friends, with the expectation that one of the star points of the root system will be either aligned or anti-aligned with the original complex number. PS send me an e-mail direct

Cube roots -- clarification Message #14 Posted by Ernie Malaga on 26 Feb 2004, 3:05 a.m., in response to message #13 by Dave Shaffer Thanks for your reply, Dave. Here's the original wording: Quote: (Although this should actually be "Any number has three complex [CUBE] roots (one of them is real)") It has two parts. First "any number has 3 complex cube roots," with which I have no problem at all. The second part is "(one of them is real)" is the one I objected to. It should have been "one of them _may be_ real". Or change the first part to further qualify the second: "Any _real_ number has 3 complex cube roots (one of them is real)." Sorry I didn't make my meaning clear. -Ernie

Re: 49g+ Cube roots Message #15 Posted by Eddie Shore on 25 Feb 2004, 4:29 p.m., in response to message #1 by Kellie You might want to turn the Complex mode off in the CAS MODES menu.

Re: 49g+ Cube roots Message #16 Posted by Kellie on 26 Feb 2004, 8:04 p.m., in response to message #15 by Eddie Shore Thanks to everyone that responded to my question, I just recently switched to an HP calculator, and am still getting used to everything, all of the help was greatly appreciated!

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