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Y
yfleury1-Visitor
1-Visitor
June 14, 2017
Question

Using a predefine function into a function

  • June 14, 2017
  • 2 replies
  • 1905 views

Hi everyone,

I need to resolve a 1-variable equation into a function. For that, I want to use the function 'Trouver', which might be 'Find' in English, to resolve my equation.

The problem is that the function 'Trouver' is 'unknown' when I use it into my own function.

'Soit' means 'Suppose that'.

Does anyone as an idea of what's wrong in my function ?

TIA

2 replies

L
LucMeekes23-Emerald IV
23-Emerald IV
June 14, 2017

Mathcad does not know the functions Trouver() or repartition(), nor the term or keyword Soit.

What is repartition() supposed to do?

It works like this:

That is to say: will find if the Mathcad built-in root can find it...

Success!

Luc

Y
yfleury1-VisitorAuthor
1-Visitor
June 15, 2017

Thank you for your answer (again) Luc !

Repartition is a function that I coded, it calculate the cumulative density function of a pdf.

Also, Soit and Trouver are 2 french keyword that we can use like that :

It almost works now, just a few things to handle before everything is ok.

Thanks again Luc!

Yvan

L
LucMeekes23-Emerald IV
23-Emerald IV
June 15, 2017

Ah, I see. 'Soit' is in English the 'Given' statement that is used to start a so-called 'Solve-block'.

You cannot use 'Soit' (='Given') in a (user defined) function, but you can define the output of a solve-block to be a function.

Bonne chance!
Luc

W
Werner_E25-Diamond I
25-Diamond I
June 15, 2017

I guess these few lines should do the job:

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