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4-Participant

## Symbolic evalution inapplicable to a certain problem?

Hello all,

I have question about the symbolic evaluation in Prime 6.0.

I thought it was simple to solve it by using the symbolic function, but it does not work well as I expected.

Am I wrong? Not sure how to solve this problem attached. Please you specialist how to solve this complicated equation about a variable. You can see the problem I am struggling.

Please let me know whether it is solvable or not, and how to solve this kind of problem.

I really thank you in advance.

Best regards,

JW

4 REPLIES 4
24-Ruby IV
(To:JW1)

You can use not symbolic but numeric math

4-Participant
(To:ValeryOchkov)
Do you mean the mathcad do not support such a problem finding a newly
arranged equation about a certain variable? Could you please explain the
reason more in detail?
--
--
Jinwon Park, Ph.D. / Systems Engineer and Analyst
Now, self-employed
C.P : +82-10-3880-4541
Email: jwpark1@gmail.com / QR code
24-Ruby V
(To:JW1)

@JW1 wrote:
Do you mean the mathcad do not support such a problem finding a newly
arranged equation about a certain variable?

Prime offers a couple of methods to solve an equation for a specific variable - a symbolic one and at least two numerical ones.
You tried the symbolic "solve" command which uses a third party engine integrated in Prime. As Luc already wrote there are even two engines integrated in Prime 6, because PTC is on the way to replace the Engine which was integrated in Mathcad 14 and 15 and also in Prime (this was MuPad, an older and stripped down version of the engine you also find in Matlab) for a new (cheaper) one. Unfortunately both are not capable enough to solve your equation and give you the solution using the Lambert-W function (which at the end also could only evaluated numerically).

As Luc showed, Mathcad 11, which used a subset of Maple for symbolic calculations, is able to give you a solution. I am not sure if this solution would make you happy as the Lambert-W function in Mathcad/Prime is a symbolic function only and can not evaluated by the numeric engine. So the result won't be good for a plot like the one Valery had shown (R->C(R)) unless you redefine the Lambert-W function yourself (examples for doing so can be found here in the forum).

The native numeric methods provided by Prime are the "root" command in its two flavours and the solve block with find. They will give you a numeric solution which also means that you will have to provide a value for R. Valery had shown how to set up a function which returns the C-value for every value of R.

The quality of the symbolic engines got worse every time the engine was changed.

A small change in your equation (omitting the square root and using just C alone) makes the equation solvable for the older engine Mupad which still is available in Prime 6, but the new engine can't solve this equation, too:

BTW, did you really mean "log"? You have to be aware that in Mathcad and Prime "log" means the base 10 logarithm. This is in contrast to other programs and publications where log is used for the natural logarithm.

It would be so easy and could avoid a lot of confusion if people would stick to the appropriate norm and only use "lg" or "ln".

23-Emerald III
(To:JW1)

Simple for Mathcad (11):

Note that:

- In your problem a = 0.242, I did not use that number to prevent a floating point result with a ridiculous number of digits.

- the function 'log' is the logarithm with base 10, if you intended the natural logarithm (base e), use 'ln' instead. The result then becomes:

- You get two possible solutions, possibly due to the fact that R and a could be complex numbers.

- the solution uses the function W(z), which is the Lambert W function which is the solution to a specific equation:

Also note that Prime6 gives you access to two different symbolic engines:

- The standard symbolic engine

- The alternative symbolic engine (MuPad), which was the standard (and only) symbolic engine used from Mathcad 14 to Prime 5.

Chances are that 'the alternative symbolic engine' of Prime 6 will find the solution with the Lambert W function.

Finally for a given (suppose) R=1 only one solution is found (a and R are now both real numbers):

(or, for the natural logarithm:)

Success!
Luc

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