11-Garnet

Solved

Symbolics without making explicit a function

Forum|Forum|6 years ago
December 18, 2019
6 replies
6094 views

Still using MC 11.2, which (for me) is fine.

I was wonder if a function which is not explicitly expressed, can be used in a symbolic operation. Maple and Mathematica I saw can do it quite well.

Below an example with derivatives.

It seems that only n.1. works while the others (I am in fact interested in a function f(y/x) do not. The n. 3. solution seems incomprehensible to me.

Does exist a trick to make possible to solve for instance function n.2 symbolically?

[[Sol: f(y/x)-[y*f'(y/x)]/x ]]

Thanks.

Calculus_Derivatives

Best answer by LucMeekes

Since you're still using Mathcad 11, try using the diff() function symbolically.

It is available due to a backdoor to Maple functionality.

Success!
Luc

L

LucMeekes

Answer

23-Emerald IV

Since you're still using Mathcad 11, try using the diff() function symbolically.

It is available due to a backdoor to Maple functionality.

Success!
Luc

A

anthonyQueen

Author

11-Garnet

@LucMeekes Thanks,

apart saying that "diff" does something, I cannot find for equation n.2 [i.e. z=f(x/y)*x] the solution. I don't know even how to interpret "&where" and the various t1, etc.

As I said the solution should be the one I gave in the first post.

F

Fred_Kohlhepp

23-Emerald I

Just a note, in your picture numbers 1 and 3 take derivatives WRT y, but number 2 asks for the derivative WRT x.

A

anthonyQueen

Author

11-Garnet

@Fred_Kohlhepp

>Just a note, in your picture numbers 1 and 3 take derivatives WRT y, but number 2 asks for the derivative WRT x.

Yes, I wanted just to try if there was a difference in processing the symbolic result to derive the variable at the nominator or at the denominator, guessing that the latter was much more difficult to handle.

L

LucMeekes

23-Emerald IV

You give the differential equation you are trying to solve as: f(y/x)-[y*f'(y/x)]/x

Assuming the * is meant as multiplication, not as convolution, that is the same as: f(y/x)-(y/x)*f'(y/x)]

And that, after substition t=y/x, is the same as f(t)-t*f'(t). The solution for that DE is f(t)=c*t, or after reverse substition, f(y/x)=c*y/x, where c is to be determined by a constraint.

Success!
Luc

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-MFra-

21-Topaz II

Hi,
you could do as in the photo:

A.Q. answer.jpg

A

anthonyQueen

Author

11-Garnet

@-MFra-

One point, may be helpful to stress that makes my question diverge somewhat from your example. Writing f(y/x) does not necessarily mean that f(x,y):=y/x, but it means that whatever function f is, it is dependent from the compound variable y/x.

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-MFra-

21-Topaz II

Hi,

I absolutely agree with you. Typically, in Mathcad, when defining a function, the arguments of the function are comma-separated variables and operators are not allowed. However, this is only possible when the function is already defined and a numerical value is desired.

You can see an application of the technique I suggested, in the attached example.

1_Example-of-MNA-based-on-the-incidence-matrix-09122012.zip

A

anthonyQueen

Author

11-Garnet

Just tried with MC 15 (and curious to know what MCPrime does).

With v.15, at least to me, the result seems much clearer than v. 11.2, and it does not show an error.

W

Werner_E

25-Diamond I

@anthonyQueen wrote:

(and curious to know what MCPrime does).

You will love it!

One of the few good things in Prime is that they have implemented the prime symbol (f ') as a fully working operator (in MC15 and below it was only available in solve block with odesolve).

So the symbolic results makes use of it and displays like you expected.

In Prime 6 they introduced a new sombolic (friCAS) which is supposed to replace muPad in one of the next releases. From what was posted here in the forum this new symbolic is less capable than muPad (which in turn was much less capable than Maple in mathcad 11).

Nevertheless given your symbolic task even the new symbolic returns the desired result: