little problem with CAS

01252015, 08:52 PM
Post: #1




little problem with CAS
\((6+1)*(6^2+1)*(6^4+1)*(6^8+1)*(6^{16}+1)0.2*6^{32} =?\)
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 HP50. Wolfram Alpha gives me "0.2". Am I wrong with CAS settings? Will be thankful for any prompt. 

01252015, 09:04 PM
(This post was last modified: 01252015 09:07 PM by Mark Hardman.)
Post: #2




RE: little problem with CAS
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: Ceci n'est pas une signature. 

01252015, 09:53 PM
Post: #3




RE: little problem with CAS
(01252015 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! 

01252015, 10:17 PM
Post: #4




RE: little problem with CAS
(01252015 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 \(a1\) which consecutively "eats up" the next factor: \[ \begin{align} (a1)(a+1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1) & \\ (a^21)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1) & \\ (a^41)(a^4+1)(a^8+1)(a^{16}+1) & \\ (a^81)(a^8+1)(a^{16}+1) & \\ (a^{16}1)(a^{16}+1) & \\ (a^{32}1) & \\ \end{align} \] Thus we end up with: \[\frac{a^{32}1}{a1}\frac{a^{32}}{a1}=\frac{1}{a1}\] This is \(\frac{1}{5}\) for \(a=6\). Cheers Thomas 

01252015, 10:51 PM
Post: #5




RE: little problem with CAS
I'm not sure which CAS you're talking about. If it's the one of the Prime I'll move the thread there.
katie 

01262015, 01:23 AM
Post: #6




RE: little problem with CAS
(01252015 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 Bob Prosperi 

01262015, 08:22 AM
Post: #7




RE: little problem with CAS  
01302015, 08:40 AM
Post: #8




RE: little problem with CAS
Though offtopic I have to ask how the 0.2*a^32 part has to be modified.
The result (a^321)/(a1) for the first part is clear and beautiful, but I can't figure out how 0.2*a^32 converts to a^32/(a1) ? Thank you 

01302015, 01:00 PM
(This post was last modified: 01302015 06:18 PM by Hlib.)
Post: #9




RE: little problem with CAS
(01302015 08:40 AM)Angus Wrote: ... but I can't figure out how 0.2*a^32 converts to a^32/(a1) ?\[\frac{a^{32}1}{a1}\frac{a^{32}*(a1)}{5*{(a1)}}\] \[\frac{a^{32}1}{a1}\frac{a^{32}*(61)}{5*{(a1)}}=\frac{1}{a1}\] \[\frac{a^{32}1}{a1}\frac{a^{32}}{a1}=\frac{1}{a1}\] Thomas Klemm forgot to write "5". This is a groundless mathematical trick, but the result is really true. 

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