The xth root of y on the hp 35s Quan Member Posts: 69 Threads: 5 Joined: May 2009 05-30-2009, 02:46 PM While quickly going through the first few chapters of the manual to get acquainted with the 35s, I came across an example on page 4-3 which shows the calculation of 3th root of -125 to be -5. This perked my interest since I was under the impression that taking any xth root of a negative number is not possible on the 35s. I performed the example, then tried calculating the xth of a handful of negative numbers. Conclusion? It only works if x is odd (and y can be any negative real numbers). If x is even, it will fail. The only condition mentioned on the page is that for y<0, x must be an integer. Can someone tell me why? My guess is that the condition should be for y<0, x must be an even integer. Thanks. Karl Schneider Posting Freak Posts: 1,792 Threads: 62 Joined: Jan 2005 05-30-2009, 04:22 PM Quan -- "x-th root of y" on the HP-35s (and its predecessors HP-33s and HP-32SII) has a real-valued range and real-valued domain of x and y. The negative real-valued odd root of a negative number can be obtained. However, complex-valued roots cannot be obtained with the function, and complex-valued inputs are invalid. The domain and range of "y^x" do encompass complex numbers, but there are some peculiarities. Please see the following archived posts: -- KS Edited: 30 May 2009, 11:12 p.m. Hal Bitton in Boise Senior Member Posts: 291 Threads: 43 Joined: Jun 2007 05-31-2009, 11:39 AM Hi Quan Since real valued roots are (apparently) all the 35S will return, and since a negative real number has 1 real odd root, that's what is returned. Since that same negative real has no real valued even roots (they're all complex), nothing is returned. All the even or odd roots of any number, real or complex, are relatively easy to come up with (DeMoivre's Theorum), but it would be nice if the calculator would just give them to you. If the 35s will solve for the zero's of a polynomial (I don't have a 35s, so I don't know), a workaround would be to transform to polynomial form and get the roots that way (cube root -8 would be (x^3 + 8), and the three cube roots (ie zero's) returned would be (in polar form) 2@60deg, 2@180deg (ie,-2), and 2@-60deg. Best regards, Hal Quan Member Posts: 69 Threads: 5 Joined: May 2009 06-01-2009, 12:59 PM Karl, Hal: Thanks for the informative responses. I sometimes forgot and expected more from HP calculators to give me the answers. Regards, Quan « Next Oldest | Next Newest »

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