result of simplify function
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Hi:
I am trying to use simplify() function.
If I define t1= sym('x+1/y+1/(x*y)=3'); t2=simplify(t1); then is it equivalent to t2=simplify('x+1/y+1/(x*y)-3')?
Thanks
Answers (1)
Walter Roberson
on 24 Mar 2011
0 votes
No. The first one creates a symbolic equation that is then simplified. The second one creates a symbolic expression that is then simplified.
The first one would come out like, (x^2*y+x+1)/(x*y) = 3 The second one would come out like, (x^2*y+x+1-3*x*y)/(x*y)
3 Comments
Huang Meng
on 24 Mar 2011
Walter Roberson
on 24 Mar 2011
Sorry, I'm working on theory here; theory and knowledge of a pretty similar computer program. I do not have the Symbolic Toolbox myself.
In the other program, the simply()'d version is:
-(1/54975581388800000000000000000000000000000000000000000000000000000000)*(-12884901888*C0^2*L0-2794227756815616035186997904500000000000000000000000*C0^5*L0^2-2147483648*L0^3+249156101152561171875*C0^3*L0-51870368705054935049759770874700000000000000000000000000000000*C0^5*L0^3+179398829795274998423971062520002440000000000000000000000000000000000000*C0^6*L0^3-6867007664798629656218750000000*C0^2*L0^3+1572261156935652588441050351000000000000000000000000*C0^4*L0^3+118140407280496373375428750000000000000000*C0^3*L0^3+83052033717520390625*C0*L0^3+2147483648*C0^3+1279185939839683623343750000000*C0^4*L0-332208134870081562500*C0^2*L0^2+157527740484329598232661250000000000000000*C0^4*L0^2-1967664285161703581260147853808461750080000000000000000000000000000000000000000000*C0^6*L0^4+610983299939981723372973776750000000000000000000000*C0^3*L0^4-836338272320550456725829424169330388000000000000000000000000000000000000*C0^5*L0^4+41496555030766781312265422860700000000000000000000000000000000*C0^4*L0^4+2036548401159578319890625000000*C0*L0^4-78761711009064468996703750000000000000000*C0^2*L0^4+2272087383959684089640625000000*C0^3*L0^2+35057939070727718549409982158352393929892480000000000000000000000000000000000000000000000000*C0^7*L0^4+10737418240*C0*L0^2)/(C0^7*L0^4) = 1055078633212981/17179869184
Walter Roberson
on 24 Mar 2011
[Trying that again!]
Sorry, I'm working on theory here; theory and knowledge of a pretty similar computer program. I do not have the Symbolic Toolbox myself.
In the other program, the simply()'d version is equivalent to the following:
-(1/549755813888/10^56)*(-12884901888*C0^2*L0 - 2794227756815616035186997904500000000000000000000000*C0^5*L0^2 - 2147483648*L0^3 + 249156101152561171875*C0^3*L0 - 51870368705054935049759770874700000000000000000000000000000000*C0^5*L0^3 + 17939882979527499842397106252000244*10^37*C0^6*L0^3 - 6867007664798629656218750000000*C0^2*L0^3 + 1572261156935652588441050351000000000000000000000000*C0^4*L0^3 + 118140407280496373375428750000000000000000*C0^3*L0^3 + 83052033717520390625*C0*L0^3 + 2147483648*C0^3+1279185939839683623343750000000*C0^4*L0 - 332208134870081562500*C0^2*L0^2 + 157527740484329598232661250000000000000000*C0^4*L0^2 - 1967664285161703581260147853808461750080000000000000000000000000000000000000000000*C0^6*L0^4 + 610983299939981723372973776750000000000000000000000*C0^3*L0^4 - 836338272320550456725829424169330388*10^36*C0^5*L0^4 + 41496555030766781312265422860700000000000000000000000000000000*C0^4*L0^4 + 2036548401159578319890625000000*C0*L0^4 - 78761711009064468996703750000000000000000*C0^2*L0^4 + 2272087383959684089640625000000*C0^3*L0^2 + 3505793907072771854940998215835239392989248*10^49*C0^7*L0^4 + 10737418240*C0*L0^2) / (C0^7*L0^4) = 1055078633212981/17179869184
with the difference that everywhere 10^ appears in the above, the other program had all of the zeros in-place. I had to convert to 10^N notation in order to fit some of the numbers in the width of the comment box, as Matlab Answers does not have a scroll-bar for comments.
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