little problem with CAS - Printable Version

little problem with CAS - Printable Version

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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:

[Image: attachment.php?aid=1482]

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". Wink

This is a groundless mathematical trick, but the result is really true.