Hi all

 

I need to know if an algebraic equation is linear or fractional.
( fractional equation: an equation containing the unknown in the denominator of one or more terms )

 

3 * ( x - 1 ) + 5 = 0 // linear
3 / ( x - 1 ) + 5 = 0 // fractional

 

How do you think to solve this problem with Filemaker?

Assuming that the algebraic equation is being passed as text, you could write a custom function which would evaluate to TRUE if it were fractional, as follows: 

 

Fractional (myEquation)

 

defined as

 

Case(patterncount(myEquation;"/") > 0;True;False)

 

However, I suspect the issue is more difficult than simply looking for the / character as the algebra gets more complicated?

 

Brian

 

 

Thanks, it seems simple but isn't.

This one, for example,

 

1/2 + x = 0

 

contains the divisor symbol but isn't a fractional equation ( an equation containing the unknown in the denominator )

If there weren't any parentheses, then I would say that an equation is fractional if - and only if - it contains the pattern "/x".

 

However,  parsing an equation that may contain parentheses (esp. nested parentheses), using only Filemaker's text functions, is not something I would recommend.

 

 

Note also:

X^-1

Edited June 11, 201411 yr by comment

You could evaluate the equation for 3 values of x, calculate the line between the first 2 with two-point form, and check if the third point is on the same line. Then you'd just have to worry about avoiding values of x that might make the denominator zero — but if that does happen, you get a question mark for evaluating at that value, and you'd know that you have a fraction equation (or just a formula with a zero denominator).

You could evaluate the equation for 3 values of x, calculate the line between the first 2 with two-point form, and check if the third point is on the same line.

 

 

This solution is really interesting!

Recognizing if the equation is linear or fractional is in fact only a part of a bigger problem: finding the root (s) of an equation  in one unknown with a planned custom function:

Solve ( equation )

This solution is really interesting!

 

I agree. However, the graph of:

y = ( 3 * x  ) / ( 2 * x )

could not be more linear - though it still meets your condition of containing the unknown in the denominator of one or more terms.

y = ( 3 * x  ) / ( 2 * x )

 

this could be written:

 

x = 3*a/2*a = 3/2 ( with a  ≠ 0 )

 

the unknown ( x ) isn't in the denominator of any terms.

Yes, it could be written that way - but it wasn't. I think you need to define some constraints on the input, otherwise this is just too wide.

Yes, the constraint on the input must be:

 

whichever equation in one unknown ( x )

 

so that the CF will solve for x  

I am afraid we're not talking about the same thing.

y = ( 3 * x + a ) / ( 2 * x - a )

is non-linear for all values of a, except a=0. If it's legitimate to write:

y = ( 3 * x + 4 ) / ( 2 * x - 4 )

then:

y = ( 3 * x + 0 ) / ( 2 * x - 0 )

is also a legitimate input.

May be I didn't explain well myself or I do not remember well my algebraic studies.

 

y = ax + b

 

isn't an equation in one unknown but a function of x: f(x) = ax + b where there is a result for each value of the unknown.

 

The user must enter the equation in the format: ax + b = 0 or ax = - b or x = -b/a

An example:

 

Solve ( "6000/(1+x)^(61/360)+7000/(1+x)^(103/360)=12702.2" )

 

result: x ≈ 0.105

I was referring to Jeremy's suggestion to calculate 3 different points for 3 values of x. You cannot calculate 3 different points using an equation. You must use a function f(x) to calculate 3 y results for 3 values of x.

Ah, ok. Now I understand you.

BTW I'll post my "work in progress" custom function that use a "brute force" algorithm to search the root.

This solution is really interesting!

Recognizing if the equation is linear or fractional is in fact only a part of a bigger problem: finding the root (s) of an equation  in one unknown with a planned custom function:

Solve ( equation )

 

If root finding is your goal, there are plenty of techniques for that that don't require knowing if the function is linear or "fractional."

Great link.
My algorithm is nearly the same as: Bisection method.

But sometimes ( with fractional equation ) it fails to find two values with different signs.

Don't you want to try to create that CF ?

I did a few years ago just for kicks, but I never had an application for it, so it's relegated to the dustbin. I used a combination of bisection, secant, and false position methods.

Are you not going to publish it on Brian's site?

Are you not going to publish it on Brian's site?

 

If I come across an application where I need it, I'm sure I'll refactor it to suit my current tastes and post it somewhere. That isn't right now.