Two calculator exam problems - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: General Forum (/forum-4.html) +--- Thread: Two calculator exam problems (/thread-1421.html) Two calculator exam problems - Peter Murphy - 05-27-2014 01:47 PM I would greatly appreciate seeing solutions to the following problem: A. Let f(x) = 1 +x(2 - x(3 - x(4 - x(1 + x)))). Find the solution to f(f(x)) = 1/2 that is nearest to x = 0. And, especially if they would provide further insight, solutions to this: B. Let f(x) = sin (1/x) and define g(x) = f(f(f(f(x)))). Find the maximum value of g(x) for x in the range 102 to 103.2. Many thanks in advance. RE: Two calculator exam problems - Visu - 05-27-2014 02:48 PM Tailor series ? RE: Two calculator exam problems - CR Haeger - 05-27-2014 03:24 PM (05-27-2014 01:47 PM)Peter Murphy Wrote:  I would greatly appreciate seeing solutions to the following problem: A. Let f(x) = 1 +x(2 - x(3 - x(4 - x(1 + x)))). Find the solution to f(f(x)) = 1/2 that is nearest to x = 0. And, especially if they would provide further insight, solutions to this: B. Let f(x) = sin (1/x) and define g(x) = f(f(f(f(x)))). Find the maximum value of g(x) for x in the range 102 to 103.2. Many thanks in advance. Using HP Prime A. Defining f(x):=1 +x(2 - x(3 - x(4 - x(1 + x)))), then g(x):=f(f(x)) then fsolve(g(x)=1/2, x, -1..1) yields five roots between -1 to +1 with +0.21399.... closest to zero. Defining Function.F1:=g(x)-1/2 lets you view this interesting curve. B. Similarly. define f(x):=sin(1/x) then g(x):=f(f(f(f(x))). Defining Function.F2:=g(x) lets you graphically explore g(x) in the range 102-103.2 (assumed degrees?). Although 102 appears highest, 103.2 is max at 0.9459... - there seems to be a local max of ~1.00 at ~103.25 degrees.. Best, Carl RE: Two calculator exam problems - Peter Murphy - 05-28-2014 02:11 PM Carl, the Prime seems ideally suited for such problems. Can you (or anyone) show me how to set up and solve them on a 50g (in RPN mode) or a 48G? I've puzzled over this for a good while. The concept of the user-defined function (UDF) for the 48/50 calculators seems promising, but what I've done hasn't produced the structure that you exhibit in your Prime examples. I now find that David Hayden solved such a problem (problem 13) back in 2009: [url:http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv019.cgi?read=159915] but he's taking for granted what I cannot see at all, namely how one sets up such nested functions in the first place. RE: Two calculator exam problems - CR Haeger - 05-28-2014 05:02 PM Peter - Im not very expert at the HP50G so I hope others will provide you (and I) some tips on how to setup these problems. I will say that the HP Prime has been very nice to work with when in comes to symbolic/numerical solving and function graphing analysis tasks. RE: Two calculator exam problems - Thomas Klemm - 05-28-2014 06:30 PM (05-28-2014 02:11 PM)Peter Murphy Wrote:  but he's taking for granted what I cannot see at all, namely how one sets up such nested functions in the first place. Code: ```'f(x)=SIN(1/x)' ENTER DEFINE 'g(x)=f(f(f(f(x))))' ENTER DEFINE``` You can have a look at how these function are defined. Just push them on the stack. For instance f(x): Code: `\<< \-> x 'SIN(1/x)' \>>` Should you prefer RPL you could use this instead: Code: `\<< \-> x \<< INV SIN \>> \>>` HTH Thomas RE: Two calculator exam problems - dizzy - 05-28-2014 06:54 PM The first one is easy: Code: ```\<<   'F(X)'   '1 + X*(2 - X*(3 - X*(4 - X*(1 + X))))'   DUP UNROT = DEF   F   '1/2' -   SOLVEVX \>>``` Which yields: Quote:{X=0.213992196216 X=... } Similarly: Code: ```\<< 'F(X)=SIN(1/X)' DEF 'SIN(1/X)' F F F \>>``` You could graph this or use try to optimize it somehow. RE: Two calculator exam problems - Peter Murphy - 05-28-2014 09:29 PM Thanks Thomas and Dizzy for exactly what I was hoping to see. Now to sit down and work out your solutions; I'll report back tomorrow. Peter RE: Two calculator exam problems - Peter Murphy - 05-29-2014 02:32 PM Thanks again Thomas and Dizzy. Your suggestions helped me break through my mental barrier on this stuff. I still must explore Dizzy's solution to problem A, but I rendered f(x) as the corresponding coefficient array [ -1 -1 4 -3 2 1] 'F' STO. Then this programmed expression << F F * PEVAL PEVAL .5 - >> STEQ allowed me to plot g(x) and find x = 0.21399. Now for problem B. I am grateful for the existence of this Forum and for the generosity of its Users.