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How to assume the sum of some variables is equal to a constant?

asked 2017-08-05 08:08:05 +0200

ensaba gravatar image

Suppose I have this expression:

$$\frac{a {\left(t_{1} + t_{2} + t_{3} - 1\right)}}{b} $$

This can be simplified to 0 if we assume $t_1 + t_2 + t_3 = 1$.

How can I accomplish this in Sage? I've tried:

var('t1', 't2', 't3', 'a', 'b')
expr1 = ((t1 + t2 + t3 - 1)*a)/b
expr1.full_simplify()

assume(t1 + t2 + t3 == 1)
expr1.full_simplify()

I expected the second call tofull_simplify() to return 0 but it returned the same result as the first call, which is $$\frac{a t_{1} + a t_{2} + a t_{3} - a}{b}$$

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Comments

Note that you can work ( in the sense of algebraic geometry) as follows:

sage: R.<t1,t2,t3,a,b> = PolynomialRing(QQ)
sage: Q = R.quotient( t1+t2+t3-1 ).fraction_field()
sage: expr1 = ((t1 + t2 + t3 - 1)*a)/b
sage: expr1
(t1*a + t2*a + t3*a - a)/b
sage: expr1.parent()
Fraction Field of Multivariate Polynomial Ring in t1, t2, t3, a, b over Rational Field
sage: Q( expr1 )
0

Working in the quotient ring modulo the "assumed relations" will always give the reduced expressions without contorsions.

dan_fulea gravatar imagedan_fulea ( 2017-08-05 12:31:56 +0200 )edit

Thanks for the suggestion. Do you have any idea why assume() didn't work?

ensaba gravatar imageensaba ( 2017-08-06 02:14:37 +0200 )edit
1

Asking for ?assume we see for instance in

   EXAMPLES:

   Assumptions are typically used to ensure certain relations are
   evaluated as true that are not true in general.

that the command is not designed to work with simplifications hand in hand, but rather with special, unstructured conditions that make an integral, a square root, etc. be defined, or to have evaluations of special functions like sin and cos. To give an unrelated example where we would also "expect more", let us consider...

sage: var('k'); assume( k, 'integer' );
sage: cos( k*pi ) == (-1)^k
cos(pi*k) == (-1)^k
sage: bool( cos( k*pi ) == (-1)^k )
True

We have to insist to get the whole information.

In your case just insist to get the information: bool( expr1 == 0 ) .

dan_fulea gravatar imagedan_fulea ( 2017-08-06 15:19:04 +0200 )edit

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answered 2017-08-05 09:38:02 +0200

ndomes gravatar image
var('t1', 't2', 't3', 'a', 'b')
expr1 = ((t1 + t2 + t3 - 1)*a)/b
equ = t1 + t2 + t3 == 1
expr1.subs(solve(equ,t1))
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Comments

Any idea why assume() didn't work?

ensaba gravatar imageensaba ( 2017-08-06 02:13:58 +0200 )edit

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Asked: 2017-08-05 08:08:05 +0200

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Last updated: Aug 05 '17