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 ▼ Edward McNally Junior Member Posts: 4 Threads: 1 Joined: Jan 1970 11-20-2007, 03:50 PM In preparing for a calculus exam, the text showed a graph of the function x^4/3 - 4x^1/3. The calculator showed no data for x<0 although the text had a graph extending from -4> use the nth root button instead of x^(1/n) >> rewrite x^(4/3) as (x^4)^(1/3) >> on the HP35s write a negative number as -5i0, and then ^(1/n) ``` Good luck. But remember, you always need to be smarter than your calculator. It similar to the question of finding the all the solutions to the equation x^6-e^(.0001x) = 0 (or graphing the function on the left). Don't trust your calculator. Cheers. Edited: 20 Nov 2007, 8:58 p.m. ▼ Edward McNally Junior Member Posts: 4 Threads: 1 Joined: Jan 1970 11-21-2007, 12:38 AM Thank you. I do realize the calculator must sometimes make a choice between various solutions. I will use your thoughts for further study, and to further understand my calculator - which is an almost unbelievable machine. Thanks again. Hal Bitton in Boise Senior Member Posts: 291 Threads: 43 Joined: Jun 2007 11-21-2007, 02:39 AM Hello Gentlemen, FYI: My 50G plots the odd root of a negative number with no problem. (I plotted X^(1/3) from x=-14 to 14 and the result was plotted on both sides of the y axis (quadrants I and III), reflected about the origin. My 48G however, does not. It only plotted the values of x>0. With the 50G in exact mode, manually solving for -8^(1/3) on the stack gives me 2*e^(i*pi/3). Forcing a numerical evaluation gives me the vector [.9999, 1.732]. Interesting that the 50G will plot -2 as the cube root of -8, but will not display it when solved on the stack. Best regards, Hal ▼ Hal Bitton in Boise Senior Member Posts: 291 Threads: 43 Joined: Jun 2007 11-21-2007, 02:46 AM Quote: With the 50G in exact mode, manually solving for -8^(1/3) on the stack gives me 2*e^(i*pi/3). Forcing a numerical evaluation gives me the vector [.9999, 1.732]. My appologies, I meant to say the complex number [.99999, 1.732] :') Hal Meenzer Member Posts: 150 Threads: 6 Joined: Sep 2007 11-21-2007, 03:41 AM Quote: Hello Gentlemen, FYI: My 50G plots the odd root of a negative number with no problem. (I plotted X^(1/3) from x=-14 to 14 [...] My 48G however, does not. Neither my 50G nor my 48G plot x^(1/3) in quadrant III. I suppose we have different flags set or otherwise different setups. Could you enlightend me on how to do it? Thanks in advance. I can however plot y= 3rd root of x in quadrants I and III on both the 48G and the 50G. Edited: 21 Nov 2007, 4:31 a.m. Giancarlo (Italy) Member Posts: 223 Threads: 19 Joined: Jun 2005 11-21-2007, 04:24 AM Hi Hal. Quote: With the 50G in exact mode, manually solving for -8^(1/3) on the stack gives me 2*e^(i*pi/3) On my 50G, solving for (-8)^(1/3) in exact mode gives me 2*e^(i*pi/3)... Best regards. Giancarlo Edward McNally Junior Member Posts: 4 Threads: 1 Joined: Jan 1970 11-22-2007, 04:29 AM Thanks for your help. First, using the root function helped. When I created Y(X)=1/3Root(x^4)-4*1/3Root(x), the calculator returns the real number answer for all x. However, it would still not graph for x<0. But then I did get "smarter" than the machine, to use your phrase. I defined two functions, Y1x)=1/3Root(x^4) and Y2(x)=-4*1/3Root(x). These both plotted correctly for x<0,x>0. Then, I created Y3(x)=Y1(x)+Y2(x)and, lo and behold, I got the complete graph. Thanks again. Stefan K. Junior Member Posts: 13 Threads: 2 Joined: Nov 2005 11-27-2007, 08:45 AM I had to check this problem with Derive for Dos (running on an Poqet PC Plus, btw a good replacement for a TI 92 if you allow non RPN every now and then): ```(-8)^(3/4) simplifies to 1 + sqrt(3) i ``` but ```solve(x^3=-8,x) will give you [x=-2,x=1+sqrt(3)i,x=1-sqrt(3)i] ``` ie. you can't plot x^(1/3) for negative values straight forward. Interestingly, (-8)^(4/3) is simplified to -8-8 sqrt(3)i, but no solution for x^(3/4)=-8 is found. I wish there would be calculator with complete implementation of complex numbers, and a way to specify the domains for the calculation. Till then you still need to use your brain... Meenzer Member Posts: 150 Threads: 6 Joined: Sep 2007 11-21-2007, 01:27 AM Quote: But I have no clue as to why I am getting these seemingly arbitrary "vector components" They are not "vector components", but complex numbers with real and imaginary part that should read 1+1.732*i and 1.5+2.598*i - I'm sure you knew that. Edited: 21 Nov 2007, 1:59 a.m. Karl Schneider Posting Freak Posts: 1,792 Threads: 62 Joined: Jan 2005 11-21-2007, 02:31 AM Edward -- The so-described "smart" calculators are returning the primary roots, whose polar-coordinate angles -- as measured counter-clockwise from the positive real axis of the complex plane -- are the smallest. The primary root of a negative number is complex-valued. Certainly, the negative real-valued root is also of interest if the range of the function is strictly real-valued. Calculators such as the HP-35s, HP-33s, and HP-33SII can also return the real cube roots of negative numbers, using the "x-th root of y" or cube-root functions. -- KS Edited: 21 Nov 2007, 3:21 a.m. ▼ Rodger Rosenbaum Senior Member Posts: 305 Threads: 17 Joined: Jun 2007 11-21-2007, 04:26 AM Quote: The so-described "smart" calculators are returning the primary roots, whose polar-coordinate angles -- as measured counter-clockwise from the positive real axis of the complex plane -- are the smallest. The primary root of a negative number is complex-valued. I've always seen it called the "principal" root. ▼ Karl Schneider Posting Freak Posts: 1,792 Threads: 62 Joined: Jan 2005 11-25-2007, 04:24 AM Hi, Rodger -- Oops! A "misremembering" of accepted terminology on my part. "Principal" does make more sense than "primary", when one considers it. "Principal" = "chief of, or among, many", as in principal of a school, or a Principal Engineer. "Primary" = "first stage", as in primary education, primary election, or primary sewage treatment. Of course, the principal root is also the first, or primary, root in the order of identification (lowest positive angle in the complex plane). Less than a year ago, I got the term right: -- KS Rodger Rosenbaum Senior Member Posts: 305 Threads: 17 Joined: Jun 2007 11-21-2007, 04:37 AM There's an easy way to see all the nth roots of a negative number on the HP48G and descendants such as the HP50G. Let's say you want the nth roots of -i. Just solve the equation x^n + i = 0. To do this you use the PROOT (polynomial solver) function. Remember to insert zeroes for the missing powers of x. Say you want the 7th roots of -3, of which there are 7. You need to solve; x^7 + 0^6 + 0^5 + 0^4 + 0^3 + 0^2 + 0^1 + 3 = 0 The "0" character is a zero. PROOT expects the coefficients of the polynomial in a vector, so type in [ 1 0 0 0 0 0 0 3 ] and then execute PROOT. To find the 3 cube roots of -8, type [1 0 0 8] PROOT. Meenzer Member Posts: 150 Threads: 6 Joined: Sep 2007 11-21-2007, 05:42 AM Quote: In preparing for a calculus exam, the text showed a graph of the function x^4/3 - 4x^1/3. The calculator showed no data for x<0 although the text had a graph extending from -4

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