Post Reply 
little problem with CAS
01-25-2015, 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 HP-50. Wolfram Alpha gives me "-0.2". Am I wrong with CAS settings? Will be thankful for any prompt.

[Image: 8b7c55a6c533t.jpg]
Find all posts by this user
Quote this message in a reply
01-25-2015, 09:04 PM (This post was last modified: 01-25-2015 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:

[Image: attachment.php?aid=1482]


Attached File(s) Thumbnail(s)
   

Ceci n'est pas une signature.
Find all posts by this user
Quote this message in a reply
01-25-2015, 09:53 PM
Post: #3
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!
Find all posts by this user
Quote this message in a reply
01-25-2015, 10:17 PM
Post: #4
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
Find all posts by this user
Quote this message in a reply
01-25-2015, 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

Visit this user's website Find all posts by this user
Quote this message in a reply
01-26-2015, 01:23 AM
Post: #6
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
Find all posts by this user
Quote this message in a reply
01-26-2015, 08:22 AM
Post: #7
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.
Find all posts by this user
Quote this message in a reply
01-30-2015, 08:40 AM
Post: #8
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
Find all posts by this user
Quote this message in a reply
01-30-2015, 01:00 PM (This post was last modified: 01-30-2015 06:18 PM by Hlib.)
Post: #9
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". Wink This is a groundless mathematical trick, but the result is really true.
Find all posts by this user
Quote this message in a reply
Post Reply 




User(s) browsing this thread: