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Math problem where graphing calculator may slow you down... - CR Haeger - 08-14-2014 02:08 PM
I was messing around with a graphing calculator and thought of a math problem that may not be best (or quickly) served by a graphing calculator. It may be a bit contrived but oh well... Problems 1. Find the largest three values of x which are solutions to f(x)=g(x), where f(x) = sin(ln(x)) and g(x) = cos(ln(x)) in the range x = 0 to 50 and Radians are used. 2. Repeat problem 1, substituting x = 0 to 1E20 for the range and using Degrees. What do your machines and/or brains come up with? RE: Math problem where graphing calculator may slow you down... - Thomas Klemm - 08-14-2014 04:25 PM
1. \(e^\frac{\pi}{4} > e^\frac{-3\pi}{4} > e^\frac{-7\pi}{4}\) 2. \(e^{45} > e^{-135} > e^{-315}\) Cheers Thomas RE: Math problem where graphing calculator may slow you down... - CR Haeger - 08-14-2014 04:36 PM
(08-14-2014 04:25 PM)Thomas Klemm Wrote: 1. \(e^\frac{\pi}{4} > e^\frac{-3\pi}{4} > e^\frac{-7\pi}{4}\) Thanks Thomas and of course you are correct. I assume for you this was mainly brain and perhaps paper/pencil work? RE: Math problem where graphing calculator may slow you down... - Thomas Klemm - 08-14-2014 05:12 PM
(08-14-2014 04:36 PM)CR Haeger Wrote: I assume for you this was mainly brain and perhaps paper/pencil work? I used Free42 to find \(e^\frac{\pi}{4} < 50 < e^\frac{5\pi}{4}\) and similarly that \(e^{45} < 10^{20} < e^{225}\). And then I used WolframAlpha to plot the graph of the 1st problem. RE: Math problem where graphing calculator may slow you down... - Bunuel66 - 08-17-2014 10:25 PM
Well, it seems that a fully analytical solution is rather trivial: Using the variable transform u=ln(x) the problem reduces to tg(u)=1 whom solutions are: u=pi/4+k.pi, k in Z. Then searching u=ln(50) gives the upper bound which is for k=0. The following values are then for k=-1 and -2. Same solution for degrees but then u=ln(1e20). Nothing more than a pen, a 'small' paper and the first scientific calculator you can grasp around you (in that case a TI36X pro, shame on me ;-) RE: Math problem where graphing calculator may slow you down... - CR Haeger - 08-18-2014 03:10 PM
(08-17-2014 10:25 PM)Bunuel66 Wrote: Well, it seems that a fully analytical solution is rather trivial: Again, Bunuel66, you provided a pretty straightforward analytical solution before reaching for any calculator. BTW, I think the TI36X pro is also pretty straightforward. I should have expected the typical MoHPC users to have no issues with these problems. I was hoping to illustrate where relying on a graphing/solver calculator might cause confusion to students just learning Trig and/or Logarithms. RE: Math problem where graphing calculator may slow you down... - walter b - 08-18-2014 03:41 PM
(08-17-2014 10:25 PM)Bunuel66 Wrote: Nothing more than a pen, a 'small' paper and the first scientific calculator you can grasp around you (in that case a TI36X pro, shame on me ;-) Congratulations! Though I hope you'll grasp a WP 31S in the future instead. RE: Math problem where graphing calculator may slow you down... - Bunuel66 - 08-18-2014 06:59 PM
Thanks to all and, please accept my apologies for having used a TI. Just to say that I have a bunch of HP calculators, starting from a HP25 for which I have worked all a summer as a kid to be able to buy it (was a f...g amount of money at that time in France), in the crowd there is also a HP34C, probably one of my favorite, a couple of HP28S very powerful but a bit too complex, lot of others, the last one being a HP 39GII. I'm a bit disappointed with the last HP products (that's why I haven't got a Prime) due to buggy software and unconsistencies between modes. I have a HP35s who could be a great product but so much bugs. I want something I can trust. I like also the HP32 who is quite reliable and it is a product you can trust. Then, yes, I'm a HP lover but from the good old time when a calculator was a tool and not a toy. That said, in France, the question of using HP is a bit theoritical as they just vanished from the shops. TI and Casio rule. I have a slight bias toward Casio for education as I find them more straightforward to use than TI and more tolerant to the various ways kids input data and equations. Still on an educational side I like Geogebra who is really really great and can run on tablets with Android which is way cool for kids. Professionally I use Scilab and Python on whatever computer I get. But that is another story. BTW: I'm not a teacher..... RE: Math problem where graphing calculator may slow you down... - aurelio - 08-18-2014 09:34 PM
(08-18-2014 06:59 PM)Bunuel66 Wrote: Then, yes, I'm a HP lover but from the good old time when a calculator was a tool and not a toy.How could not be agree with you? HP 39GII and HP35s altough I don't know what they are ( ), I don't think they belong yet to the "toys" class, but I don't think your expression was referred to them The Prime.......... I must find actually the right time to dedicate to this calculator, but, looking to the big consideration and its success, let me say, in the dedicated section of this forum, it should be not so bad, considering it is not a RPN calculator RE: Math problem where graphing calculator may slow you down... - Thomas Klemm - 08-18-2014 09:47 PM
(08-17-2014 10:25 PM)Bunuel66 Wrote: Nothing more than a pen, a 'small' paper and the first scientific calculator you can grasp around you Here's how you can figure out with a 4-banger whether \(e^{\frac{5}{4}\pi} > 50\): AC 3.141592654 ÷ 1024 + 1 = × = × = × = × = × = × = × = × = M+ × = × = × MR = The result is: 50.44976377 As \((1+\frac{x}{n})^n \leq e^x\), we can conclude that indeed: \[ \begin{align} 50 & < 50.44976377 \doteq (1+\frac{\pi}{1024})^{1024+256} \\ & = (1+\frac{\pi}{1024})^{1024 \times 1.25} \\ & = ((1+\frac{\pi}{1024})^{1024})^{1.25} \\ & \leq (e^\pi)^{1.25} \\ & = e^{\pi \times 1.25} \\ & = e^{\frac{5}{4}\pi} \end{align} \] Cheers Thomas RE: Math problem where graphing calculator may slow you down... - Bunuel66 - 08-18-2014 10:26 PM
Very smart indeed, but even assuming that pi is known with the appropriate accuracy, the choice of 1024 and 256 are not random as they must insure the convergence of the approximation in a neighborhood small enough for being conclusive. May be some trials and errors? Very nice nevertheless. RE: Math problem where graphing calculator may slow you down... - Thomas Klemm - 08-18-2014 10:28 PM
(08-14-2014 05:12 PM)Thomas Klemm Wrote: similarly that \(e^{45} < 10^{20} < e^{225}\) Here's how you can show this with a 4-banger. For this we want to calculate \(\log(e)\) where \(\log\) is the common logarithm (base 10) while \(\ln\) shall denote the natural logarithm. Of course we know that \(\log(e)=\frac{\ln(e)}{\ln(10)}=\frac{1}{\ln(10)}\). To calculate \(ln(10)\) we can use the same trick as before but turn it around and calculate the roots: 10 \(\sqrt{x}\) \(\sqrt{x}\) \(\sqrt{x}\) \(\sqrt{x}\) \(\sqrt{x}\) \(\sqrt{x}\) \(\sqrt{x}\) \(\sqrt{x}\) \(\sqrt{x}\) \(\sqrt{x}\) − 1 × 1024 = This gives us: 2.305174528 Now instead of \(45\times\log(e)\) we calculate \(\frac{45}{\ln(10)}\) and get: 19.52129844 Thus indeed \(e^{45}<10^{20}\) and analogous for \(10^{20}<e^{225}\). Cheers Thomas RE: Math problem where graphing calculator may slow you down... - Thomas Klemm - 08-18-2014 10:52 PM
(08-18-2014 10:26 PM)Bunuel66 Wrote: Very smart indeed, but even assuming that pi is known with the appropriate accuracy, the choice of 1024 and 256 are not random as they must insure the convergence of the approximation in a neighborhood small enough for being conclusive. May be some trials and errors? Very nice nevertheless. You just have to use an n so that \(50 < (1+\frac{\pi}{n})^{\frac{5}{4}n}\). You could just as well use n = 512 but n = 256 would be too small. BTW: Who doesn't know \(\pi\) to 10 places? Ask Geir Isene. Cheers Thomas RE: Math problem where graphing calculator may slow you down... - Gilles - 11-15-2014 12:45 PM
With HP50G,approx mode (but exact mode is OK): Code: `'SIN(X)=COS(X)' SOLVEVX ` Code: `{ 'X=EXP(6.2832*n-2.3562)' ` Note that Lsub is a program I use for | (or SUBST) when equations are in a list (list processing don't work with | or subst) : Code: `« 1 ->LIST { | } + ->PRG MAP »` I also often use Lexr to get only the right part of equations in a list : Code: `« 1 « EXLR NIP » DOSUBS »` { 'X=EXP(6.2832*n-2.3562)' 'X=EXP(6.2832*n+0.7854)' } Lexr gives { 'EXP(6.2832*n-2.3562)' 'EXP(6.2832*n+0.7854)' } and for example to test n from -5 to 5, continue with STEQ {} -5 5 FOR a EQ 'n' a 2 ->LIST Lsub + NEXT XNUM SORT that gives { 2.1526E-15 4.9812E-14 1.1527E-12 2.6674E-11 6.1725E-10 1.4284E-8 3.3053E-7 7.6487E-6 0.0002 0.0041 0.0948 2.1933 50.7540 1174.4832 27178.3539 628925.9347 14553781.7413 336784589.923 7793428678.94 180345337617. 4.1733E12 9.6573E13 } In exact mode you get { 'EXP(2*ATAN(-1+√2)-2*PI)' '1/EXP(2*ATAN(1+√2))' 'EXP(2*ATAN(-1+√2))' } As you see the HP don't simplify the ATAN (but you can write easily your own program for trivial trig simplifications with a serie of MATCH commands. But it's not automatic and rather slow). I can post this if there is some intesrest with the Prime solve(sin(ln(x))=cos(ln(x))) returns directly the symbolic simplified general solution... ( 'principal' uncheck in CAS setting) It would fine if we could do : equation | n={-1,0,1,2} or equation | n={-5...5} to obtain a list of equations for each value. But this dont work .... So I do this: Edit the general solution to change n_xxx by n then f(n):=ANS concat(f(-1),f(0)) concat(ANS,f(1)) concat(ANS,f(2)) sort(ans) ~= and you get the result (approx or exact) |