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16-Pearl

## A question on the 'simplify' option

To all expert,
A simple question on the use of the ‘simplify’ option in mcad 15.

I was looking at trying to simplify some simple equation and after some simple manip I ended up with:

f*B / SQRT(f^3*A*C)

which can be then simplified to

(1/sqrt(f)) * (B/sqrt(A*C))                    I was interested in “extracting (B/sqrt(A*C))

Question: Why is mcad 15 not able to further simplify the term f/sqrt(f^3) ?

Thanks

Regards

1 ACCEPTED SOLUTION

Accepted Solutions
16-Pearl
(To:LucMeekes)

Thanks to all. Seems obvious to use 'assume, f>0' and in this context it is perfectly valid (f cannot be <0)

I will overlook the fact that my hand simplification was wrong to start with ! I forgot to multiply top and bottom by f so I should have ended up with: f^2 * B / (sqrt(f^4*A*C)

Mcad 15 did manage to simplify to B/sqrt(A*C)

4 REPLIES 4
24-Ruby V
(To:JBlackhole)

I

Question: Why is mcad 15 not able to further simplify the term f/sqrt(f^3) ?

Because sqrt(f^2) = f is only correct for f >= 0. Its not correct for  f<0 nor for non-real f.

So you have to tell Mathcad that f is positive:

Nevertheless there are many situations where Mathcad is not able to make even a very obvious simplification and cannot be talked into doing it by making assumptions. The reason for this is simply that Mathcads symbolic is far away from being perfect. After all it uses just a small part of a very old version of a program called muPad (which can't match with programs like Mathematica or Maple). Thats one of the reasons many users considers version 11 to be the best ever, as this version uses parts of Maple for its symbolic calculations. (MC12 and MC13 use Maple, too, but have other drawbacks).

17-Peridot
(To:JBlackhole)

M15 will also do it if you use indices rather than the square root symbol!

Alan

23-Emerald III
(To:JBlackhole)

And this is what Mathcad 11 (with Maple as symbolic engine) can make of it:

Success!
Luc

16-Pearl
(To:LucMeekes)

Thanks to all. Seems obvious to use 'assume, f>0' and in this context it is perfectly valid (f cannot be <0)

I will overlook the fact that my hand simplification was wrong to start with ! I forgot to multiply top and bottom by f so I should have ended up with: f^2 * B / (sqrt(f^4*A*C)

Mcad 15 did manage to simplify to B/sqrt(A*C)

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