little problem with CAS
01-25-2015, 08:52 PM
Post: #1
 Hlib Member Posts: 242 Joined: Jan 2015
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 HP-50. Wolfram Alpha gives me "-0.2". Am I wrong with CAS settings? Will be thankful for any prompt.

01-25-2015, 09:04 PM (This post was last modified: 01-25-2015 09:07 PM by Mark Hardman.)
Post: #2
 Mark Hardman Senior Member Posts: 525 Joined: Dec 2013
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:

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Ceci n'est pas une signature.
01-25-2015, 09:53 PM
Post: #3
 Hlib Member Posts: 242 Joined: Jan 2015
RE: little problem with CAS
(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!
01-25-2015, 10:17 PM
Post: #4
 Thomas Klemm Senior Member Posts: 2,111 Joined: Dec 2013
RE: little problem with CAS
(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
01-25-2015, 10:51 PM
Post: #5
 Katie Wasserman Super Moderator Posts: 640 Joined: Dec 2013
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

01-26-2015, 01:23 AM
Post: #6
 rprosperi Super Moderator Posts: 6,398 Joined: Dec 2013
RE: little problem with CAS
(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

--Bob Prosperi
01-26-2015, 08:22 AM
Post: #7
 Hlib Member Posts: 242 Joined: Jan 2015
RE: little problem with CAS
(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.
01-30-2015, 08:40 AM
Post: #8
 Angus Member Posts: 212 Joined: Feb 2014
RE: little problem with CAS
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
01-30-2015, 01:00 PM (This post was last modified: 01-30-2015 06:18 PM by Hlib.)
Post: #9
 Hlib Member Posts: 242 Joined: Jan 2015
RE: little problem with CAS
(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.
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