little problem with CAS - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: little problem with CAS (/thread-2930.html) little problem with CAS - Hlib - 01-25-2015 08:52 PM $$(6+1)*(6^2+1)*(6^4+1)*(6^8+1)*(6^{16}+1)-0.2*6^{32} =?$$ I took this expression from the textbook on "Algebra" for secondary school to check calculator's CAS, but can't obtain the true answer neither on Prime (emulator) nor on HP-50. Wolfram Alpha gives me "-0.2". Am I wrong with CAS settings? Will be thankful for any prompt. RE: little problem with CAS - Mark Hardman - 01-25-2015 09:04 PM Introducing 0.2*6^32 causes the last term to be evaluated approximately. This results in a considerable rounding error when it is subtracted from the exact result of the first five terms. Try rewriting the expression without using an approximate term. For example: RE: little problem with CAS - Hlib - 01-25-2015 09:53 PM (01-25-2015 09:04 PM)Mark Hardman Wrote:  Introducing 0.2*6^32 causes the last term to be evaluated approximately. This results in a considerable rounding error when it is subtracted from the exact result of the first five terms. I have forgoten that decimal point cancels the exact mode. Thank you, Mark! RE: little problem with CAS - Thomas Klemm - 01-25-2015 10:17 PM (01-25-2015 08:52 PM)Hlib Wrote:  $$(6+1)*(6^2+1)*(6^4+1)*(6^8+1)*(6^{16}+1)-0.2*6^{32} =?$$ You can multiply the left product by $$a-1$$ which consecutively "eats up" the next factor: \begin{align} (a-1)(a+1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1) & \\ (a^2-1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1) & \\ (a^4-1)(a^4+1)(a^8+1)(a^{16}+1) & \\ (a^8-1)(a^8+1)(a^{16}+1) & \\ (a^{16}-1)(a^{16}+1) & \\ (a^{32}-1) & \\ \end{align} Thus we end up with: $\frac{a^{32}-1}{a-1}-\frac{a^{32}}{a-1}=\frac{-1}{a-1}$ This is $$\frac{-1}{5}$$ for $$a=6$$. Cheers Thomas RE: little problem with CAS - Katie Wasserman - 01-25-2015 10:51 PM I'm not sure which CAS you're talking about. If it's the one of the Prime I'll move the thread there. RE: little problem with CAS - rprosperi - 01-26-2015 01:23 AM (01-25-2015 10:51 PM)Katie Wasserman Wrote:  I'm not sure which CAS you're talking about. If it's the one of the Prime I'll move the thread there. Yes, this is a Prime topic RE: little problem with CAS - Hlib - 01-26-2015 08:22 AM (01-25-2015 10:17 PM)Thomas Klemm Wrote:   (01-25-2015 08:52 PM)Hlib Wrote:  $$(6+1)*(6^2+1)*(6^4+1)*(6^8+1)*(6^{16}+1)-0.2*6^{32} =?$$ You can multiply the left product by $$a-1$$ which consecutively "eats up" the next factor... Yes, thank you! It too very much was useful to me. RE: little problem with CAS - Angus - 01-30-2015 08:40 AM Though off-topic I have to ask how the -0.2*a^32 part has to be modified. The result (a^32-1)/(a-1) for the first part is clear and beautiful, but I can't figure out how 0.2*a^32 converts to a^32/(a-1) ? Thank you RE: little problem with CAS - Hlib - 01-30-2015 01:00 PM (01-30-2015 08:40 AM)Angus Wrote:  ... but I can't figure out how 0.2*a^32 converts to a^32/(a-1) ? Thank you $\frac{a^{32}-1}{a-1}-\frac{a^{32}*(a-1)}{5*{(a-1)}}$ $\frac{a^{32}-1}{a-1}-\frac{a^{32}*(6-1)}{5*{(a-1)}}=\frac{-1}{a-1}$ $\frac{a^{32}-1}{a-1}-\frac{a^{32}}{a-1}=\frac{-1}{a-1}$ Thomas Klemm forgot to write "5". This is a groundless mathematical trick, but the result is really true.