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Problem calculating an expression with roots
01-30-2015, 09:44 PM (This post was last modified: 01-30-2015 09:46 PM by retoa.)
Post: #1
Problem calculating an expression with roots
Hello.

I am trying to calculate a very simple expression with the Prime:

$ \frac{\sqrt{3}*\sqrt[3]{2*x}}{\sqrt[4]{3*x}} $

I don't know how to write that without latex but the attachment is an image of the Prime display.

As you can see the answer is not very useful, actually it should simply be $ \sqrt[4]{432*x} $
or $ 2^{1/3}*3^{1/4}*x^{1/12} $

If I try to simplify the prime goes in a loop...

What do I have to do to obtain the correct answer?

Thank you very much

reto


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01-30-2015, 09:57 PM
Post: #2
RE: Problem calculating an expression with roots
also here!

loop also with QPI program and using Simplify after Collect (that works)...
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01-30-2015, 10:02 PM
Post: #3
RE: Problem calculating an expression with roots
I also tried with collect and then simplify. Collect works, but the answer is still not very useful. Simplify loops.
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01-30-2015, 10:37 PM (This post was last modified: 01-30-2015 10:41 PM by Mark Hardman.)
Post: #4
RE: Problem calculating an expression with roots
Reformatted so that the Latex expressions work.

(01-30-2015 09:44 PM)retoa Wrote:  Hello.

I am trying to calculate a very simple expression with the Prime:

\( \frac{\sqrt{3}*\sqrt[3]{2*x}}{\sqrt[4]{3*x}} \)

I don't know how to write that without latex but the attachment is an image of the Prime display.

As you can see the answer is not very useful, actually it should simply be:

\( \sqrt[4]{432*x} \)

or

\( 2^{1/3}*3^{1/4}*x^{1/12} \)

If I try to simplify the prime goes in a loop...

What do I have to do to obtain the correct answer?

Thank you very much

reto

Are you sure?

If I define:

f:=(x)->√3*(2*x)^(1/3)/(3*x)^(1/4)

and

g:=(x)->(432*x)^(1/4)

And evaluate both of them at 1, I get 1.65814935214 and 4.55901411391 respectively.

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01-30-2015, 10:49 PM
Post: #5
RE: Problem calculating an expression with roots
You are right Mark !!

I did an error while writing the answer, the correct answer is

$ \sqrt[12]{432*x} $

How did you reformat that to let latex work?

Thank you very much !!

reto
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01-30-2015, 11:03 PM
Post: #6
RE: Problem calculating an expression with roots
(01-30-2015 10:49 PM)retoa Wrote:  You are right Mark !!

I did an error while writing the answer, the correct answer is

$ \sqrt[12]{432*x} $

How did you reformat that to let latex work?

Thank you very much !!

reto

Use the backslash, open parenthesis and backslash close parenthesis for delimiters rather than dollar sign ($).

For example:

Code:
\( \sqrt[12]{432*x} \)

Yields:

\( \sqrt[12]{432*x} \)

It still doesn't answer the question of how to simplify the expression.

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01-31-2015, 08:16 AM
Post: #7
RE: Problem calculating an expression with roots
It takes much time to simplify your expression because of the way the CAS handles fractional powers. Everything is converted to a unique algebraic extension, here of degree 24, that depends on x. On a desktop with Xcas, you get an impressive rootof over a big denominator after a few seconds.
a:=normal(sqrt(x)*(2x)^(1/3)*(3x)^(1/4))
then normal(normal(b^12)) returns 432*x^13, therefore a seems to be (432*x^13)^(1/12)
Of course, there are faster ways to handle expressions like that because there are only * and ^, but I did not implement anything for that, it's much too specific, while the general algorithm can handle +,-,inv,*,/ (but is too slow on a calc).

Just for fun, the denominator is
230844665274826752*x^21-1591866337624326144*x^20+427214488707388145664*x^19+12143550017802860494848*x^18​-383148224530214704644096*x^17+403326885144955994505216*x^16+24737181941124395515​772928*x^15+23100313665974376526774272*x^14-721159543494255466332020736*x^13-1879698125953101870690734592*x^12-1379343605380747369369610112*x^11-343081635027894091463265792*x^10-51960732759524362938918912*x^9-5636720287344516274888704*x^8-376051106333108239073280*x^7-15740221526669160087552*x^6-53246666561770487808*x^5
and the minimal polynomial for the algebraic extension:
[1,0,-12*x,-16*x,66*x^2-18*x,96*x^2,-220*x^3+148*x^2,-144*x^3-432*x^2,495*x^4+198*x^3+135*x^2,-384*x^4-10816*x^3,-792*x^5-2448*x^4+4104*x^3,2016*x^5+37632*x^4+4320*x^3,924*x^6+4476*x^5-22748*x^4-540*x^3,-4032*x^6-26688*x^5-43776*x^4,-792*x^7-648*x^6-183048*x^5+89640*x^4,4704*x^7-50912*x^6+48608*x^5-11232*x^4,495*x^8-6948*x^7+355338*x^6-258660*x^5+1215*x^4,-3456*x^8+104256*x^7-163584*x^6+350784*x^5,-220*x^9+9072*x^8-77640*x^7+710704*x^6+114372*x^5,1584*x^9-65664*x^8-116064*x^7+148608*x^6+3888*x^5,66*x^10-4842*x^9+35172*x^8-920100*x^7+804762*x^6-1458*x^5,-416*x^10+9024*x^9+164352*x^8-1139776*x^7+289440*x^6,-12*x^11+1044*x^10-40248*x^9+371496*x^8-876924*x^7+57348*x^6,48*x^11+3600*x^10+40608*x^9+610656*x^8-44496*x^7+11664*x^6,x^12-34*x^11-5529*x^10-115420*x^9-38561*x^8-29538*x^7+729*x^6]
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01-31-2015, 04:00 PM
Post: #8
RE: Problem calculating an expression with roots
(01-31-2015 08:16 AM)parisse Wrote:  ...Just for fun, the denominator is
230844665274826752*x^21-1591866337624326144*x^20+427214488707388145664*x^19+12143550017802860494848*x^18​-383148224530214704644096*x^17+403326885144955994505216*x^16+24737181941124395515​772928*x^15+23100313665974376526774272*x^14-721159543494255466332020736*x^13-1879698125953101870690734592*x^12-1379343605380747369369610112*x^11-343081635027894091463265792*x^10-51960732759524362938918912*x^9-5636720287344516274888704*x^8-376051106333108239073280*x^7-15740221526669160087552*x^6-53246666561770487808*x^5
and the minimal polynomial for the algebraic extension:
[1,0,-12*x,-16*x,66*x^2-18*x,96*x^2,-220*x^3+148*x^2,-144*x^3-432*x^2,495*x^4+198*x^3+135*x^2,-384*x^4-10816*x^3,-792*x^5-2448*x^4+4104*x^3,2016*x^5+37632*x^4+4320*x^3,924*x^6+4476*x^5-22748*x^4-540*x^3,-4032*x^6-26688*x^5-43776*x^4,-792*x^7-648*x^6-183048*x^5+89640*x^4,4704*x^7-50912*x^6+48608*x^5-11232*x^4,495*x^8-6948*x^7+355338*x^6-258660*x^5+1215*x^4,-3456*x^8+104256*x^7-163584*x^6+350784*x^5,-220*x^9+9072*x^8-77640*x^7+710704*x^6+114372*x^5,1584*x^9-65664*x^8-116064*x^7+148608*x^6+3888*x^5,66*x^10-4842*x^9+35172*x^8-920100*x^7+804762*x^6-1458*x^5,-416*x^10+9024*x^9+164352*x^8-1139776*x^7+289440*x^6,-12*x^11+1044*x^10-40248*x^9+371496*x^8-876924*x^7+57348*x^6,48*x^11+3600*x^10+40608*x^9+610656*x^8-44496*x^7+11664*x^6,x^12-34*x^11-5529*x^10-115420*x^9-38561*x^8-29538*x^7+729*x^6]

Although I must admit that seeing this result did make me smile, I cannot fully agree that this is fun. But thanks for taking the time to explain this.

--Bob Prosperi
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01-31-2015, 09:59 PM
Post: #9
RE: Problem calculating an expression with roots
Wow, how to complicate easy things!
Actually you only have to sum/subtract the exponents...

I'm comparing HP Prime and TI nspire CAS, I have to choose one to propose for the students in the school where I work, so I try some calculation we do at school with both of them.
The TI transforms

\( \frac{\sqrt{3}*\sqrt[3]{2*x}}{\sqrt[4]{3*x}} \)

to

\(2^{1/3}*3^{1/4}*x^{1/12} \)

in no time.

I will not say that I will choose TI because of that, the prime is better then TI in many other things, but I can not understand how a CAS calculator can not simplify something so easy. If the TI does it, then it should be possible also on the Prime hardware.
Does it mean that the Prime can not simplify any expressions with roots at the denominator or with negative fractional exponents? That would be really limiting for the use in a school...

Thank you for the answer.

reto
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01-31-2015, 11:08 PM
Post: #10
RE: Problem calculating an expression with roots
(01-31-2015 08:16 AM)parisse Wrote:  get an impressive rootof over a big denominator after a few seconds.

So then how does one go from the the "impressive" rootof to a human readable result?

TW

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02-01-2015, 09:17 AM (This post was last modified: 02-01-2015 09:19 AM by parisse.)
Post: #11
RE: Problem calculating an expression with roots
(01-31-2015 09:59 PM)retoa Wrote:  Wow, how to complicate easy things!
Actually you only have to sum/subtract the exponents...

I'm comparing HP Prime and TI nspire CAS, I have to choose one to propose for the students in the school where I work, so I try some calculation we do at school with both of them.
The TI transforms

\( \frac{\sqrt{3}*\sqrt[3]{2*x}}{\sqrt[4]{3*x}} \)

to

\(2^{1/3}*3^{1/4}*x^{1/12} \)

in no time.

I will not say that I will choose TI because of that, the prime is better then TI in many other things, but I can not understand how a CAS calculator can not simplify something so easy. If the TI does it, then it should be possible also on the Prime hardware.
Of course it's possible, but it's not a priority. It applies only to very specific expressions that you can do by hand. I did not make my CAS with the objective to solve american schoolbool exercices, because I don't know them. It does not mean I will not adapt, it depends on time constraints and priorities.

Quote:Does it mean that the Prime can not simplify any expressions with roots at the denominator or with negative fractional exponents? That would be really limiting for the use in a school...
It's exactly the reverse. The algorithm for simplifying fractional powers on the Prime can handle expressions with + (or -), it's therefore more powerful. Simplifying the example with this algorithm takes too much time on a calc, but it works on a desktop.
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02-01-2015, 02:40 PM
Post: #12
RE: Problem calculating an expression with roots
(01-31-2015 11:08 PM)Tim Wessman Wrote:  So then how does one go from the the "impressive" rootof to a human readable result?
There is no general way to do that, i.e. you must guess that a^12 has a simpler form, the CAS tries only equation of order 2 (and maybe bisquare equations). The goal is not to have something readable by a human, but have a normal form where the computer can determine if something is 0 or not.
Anyway, I have an improvement for regroup for these kind of specific expressions with only * / ^.
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02-01-2015, 11:56 PM
Post: #13
RE: Problem calculating an expression with roots
This, to me, is a problem...:

"Of course it's possible, but it's not a priority. It applies only to very specific expressions that you can do by hand. I did not make my CAS with the objective to solve american schoolbool exercices, because I don't know them. It does not mean I will not adapt, it depends on time constraints and priorities."

A large component of the market target for the Prime appears to be the American school classrooms! Given that the CAS offered by TI & CASIO produce easy to comprehend simplifications, the Prime must do the same in order to compete! Please keep this in mind, otherwise, the good works of Tim and Parisse on CAS will be passed by in the market place by the others. The market does not care about the greater raw ability of the XCAS in the Prime...it gravitates to the friendlier platform. If the greater raw ability of the XCAS, with it's potential to be a friendlier platform to it's target audience isn't taken advantage of, it will all be a waste and a huge shame!

Another area that the answers cry out for clarity is general solutions to differential equations...the plethora of constants that are produced in the answers are awful, and should be addressed!

Please...if attention is given to the readability of answers, intuitive entry interfaces, you will have a successful competitor that will be around long enough to then add all of the neat mathematical powers that will be exclusive to the Prime.
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02-02-2015, 11:49 AM (This post was last modified: 02-02-2015 11:52 AM by retoa.)
Post: #14
RE: Problem calculating an expression with roots
(02-01-2015 09:17 AM)parisse Wrote:  [quote='retoa' pid='26149' dateline='1422741551']

Quote:Does it mean that the Prime can not simplify any expressions with roots at the denominator or with negative fractional exponents? That would be really limiting for the use in a school...
It's exactly the reverse. The algorithm for simplifying fractional powers on the Prime can handle expressions with + (or -), it's therefore more powerful. Simplifying the example with this algorithm takes too much time on a calc, but it works on a desktop.

This remembers me the Pirelli Tyres slogan "Power is nothing without control". I don't find it very useful to have a powerful calculator that can not calculate easy things. It's like having a scientific calculator that only does roots and trigonometry, because you can do the additions and subtractions by yourself on a sheet of paper (actually it was so when we used slide rules).

"I did not make my CAS with the objective to solve american schoolbool exercices"
By the way I don't work in an american school, I live in Switzerland, but I think the problems with roots and fractional exponents are the same worldwide.

Thank you for the attention

reto
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02-02-2015, 12:53 PM
Post: #15
RE: Problem calculating an expression with roots
(02-02-2015 11:49 AM)retoa Wrote:  By the way I don't work in an american school, I live in Switzerland, but I think the problems with roots and fractional exponents are the same worldwide.
I don't think it's worldwide. I do not remember having to play with that kind of exercices in France. And it is perhaps because it does not bring anything more than playing with integral powers to understand power rules, while adding artifical complexity because the expression is not defined for all values of x (or ambiguous like the definition of x^(1/3) for x<0).
As I said, I'm not against improving the CAS with heuristics like that if it's not too hard to implement (and fortunately this one is not too complicated to implement), but I can't guess all kind of purely artificial/academic exercices and I don't think a CAS should have this as a priority, a CAS should concentrate on algorithms not heuristics.

lrdheat:
For the differential equations, I don't understand where there are too may constants, a first order equation has one constant of integration, a second order two constants. Or maybe you speak about linear differential equations with constant coefficients, where Xcas returns the general solution expressed with special constants: the value of the function and derivatives at x=0, in other words initial conditions at x=0. Then you may find there are too many time the same constant because you are used to express the general solution without connection to initial conditions.
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02-02-2015, 03:48 PM
Post: #16
RE: Problem calculating an expression with roots
On th 50G you can simplify this kind of equation with the sequence : EXPLN SIMPLIFY EXP2POW

but I don't found an equivalent of EXPLN on the Prime.

It would be very fine to have an equivalent of the 50G EQW on the Prime, so we could manipulate algebraic expressions (and sub expression)
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02-02-2015, 04:40 PM
Post: #17
RE: Problem calculating an expression with roots
expln is an Xcas command.
And yes, it would be very convenient to have something like the EQW on the Prime.
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02-02-2015, 06:08 PM
Post: #18
RE: Problem calculating an expression with roots
(02-02-2015 04:40 PM)parisse Wrote:  expln is an Xcas command.

Maybe I am misunderstanding something, but expln does not appear to be a command in xcas but rather an argument/identifier that can be used in conjunction with convert( )

TW

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02-02-2015, 06:10 PM
Post: #19
RE: Problem calculating an expression with roots
(02-02-2015 04:40 PM)parisse Wrote:  And yes, it would be very convenient to have something like the EQW on the Prime.

I agree Smile
almost, it would be very useful changing part of an equation tapping on it in the screen, without moving cursor...

Maybe...

Salvo

∫aL√0mic (IT9CLU) :: HP Prime 50g 41CX 71b 42s 39s 35s 12C 15C - DM42, DM41X - WP34s Prime Soft. Lib
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02-02-2015, 06:25 PM
Post: #20
RE: Problem calculating an expression with roots
(02-02-2015 06:08 PM)Tim Wessman Wrote:  
(02-02-2015 04:40 PM)parisse Wrote:  expln is an Xcas command.

Maybe I am misunderstanding something, but expln does not appear to be a command in xcas but rather an argument/identifier that can be used in conjunction with convert( )
Oops, yes, I have answered too fast. The commandname is trig2exp
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