I have this differential equation written in SymPy
diffeq = Eq(f(x).diff(x, x) - 2*f(x).diff(x) + f(x), sin(x))
where f(x) is a symbol Function symbol and x is a variable symbol.
When I solve it with this:
expr = dsolve(diffeq, f(x))
I get
f(x)=(C_1+C_2x)ex+12cos(x)
Which is the correct solution to this equation. But now I would like to evaluate this function in several points. I know I can substitute the x with the subs function, but is there a way to substitute the constant values C_1 and C_2 so I can evaluate the function?
There is an open PR for this on GitHub, that would add an ics flag to dsolve.
For now, you can substitute the values manually using subs, use solve to solve for C1 and C2, and use subs to substitute the values back into the solution.
For example, if f(0) = 1 and f'(0) = 0, you'd use something like
>>> p1 = expr.subs([(x, 0), (f(0), 1)])
>>> dexpr = Eq(expr.lhs.diff(x), expr.rhs.diff(x))
>>> p2 = dexpr.subs([(x, 0), (f(x).diff(x).subs(x, 0), 0)])
>>> p1
Eq(1, C1 + 1/2)
>>> p2
Eq(0, C1 + C2)
>>> C1, C2 = symbols('C1 C2')
>>> sol = solve([p1, p2], [C1, C2])
>>> sol
{C1: 1/2, C2: -1/2}
>>> expr.subs(sol)
Eq(f(x), (-x/2 + 1/2)*exp(x) + cos(x)/2)
You can get your constants from your expression. It's a bit messy, but it works:
v1 = expr.args[1].args[1].args[0].args[0]
v2 = expr.args[1].args[1].args[0].args[1].args[0]
expr.subs(v1,1).subs(v2,2)
Explanation:
Have a look at expr.args. It is a tuple the left and the right hand side of the equation. Here we want the second entry of the tuple, thus index 1.
We then get some sympy.core.add.Add. This we can again decompose using args and can continue until we reach our constants.
Related
I'm trying to solve the cosine formula using sympy. (A is angle and a, b, c is sides)
sol = solve(a**2-b**2-c**2+2*b*c*cos(pi/180*A), [A, a, b, c])
print(sol)
And I end up getting results like this:
[(90.0000000000000, 5.00000000000000, 4.00000000000000, 3.00000000000000), (270.000000000000, 5.00000000000000, 4.00000000000000, 3.00000000000000)]
But it's impossible for a triangle to have a 270º interior angle.
I tried to do A, B, C <= 180 and solve(a**2-b**2-c**2+2*b*c*cos(pi/180*A), [A<=180, a, b, c]).I also try to use exclude=[A<=180] in the solve(), but it don't work too.
Let's understand what's happening:
from sympy import *
a, b, c, A = symbols("a, b, c, A")
eq = a**2-b**2-c**2+2*b*c*cos(pi/180*A)
eq is a trigonometric equation involving the cosine function, which is periodic with a period of 2*pi. Suppose you want to solve cos(x) = y where y is contained in (-1, 1]: you will get two solutions! For example, suppose you want to solve cos(x) = 0.5: one solution is x=60deg and the other is x=300deg.
Now let's focus on your equation. You have one equation with 4 symbols. Let's assume that a,b,c are known quantities and asks Sympy to solve for A (the unknown, which I assume it is the angle in degrees).
We can filter out the solutions in two approaches.
First approach
Substitute the known values a, b, c into eq and then solve for A with a specified constraint:
# substitution dictionary
d = {a: 1, b: 1, c: 1}
# solve the equation asking to find the solution for which A <= 180
sol = solve([eq.subs(d), A <= 180], A)
print(sol)
# out: (-oo < A) & (A <= 180) & (Eq(A, 60) | Eq(A, 300))
Here, SymPy produced a boolean solution (meaning it is a combination of one or more solutions with the And and Or objects). Let's try to simplify it a bit:
sol2 = sol.as_set()
print(sol2)
# out: {60}
Now we can extract it with:
A_num = list(sol2)[0]
print(A_num)
# 60
Second approach
Obtain a fully symbolic solution, then substitute the known values and filter out according to your rules:
sol = solve(eq, A)
for i, s in enumerate(sol):
print("sol", i+1, ":", s)
# output:
# sol 1 : 180*(-acos((-a**2 + b**2 + c**2)/(2*b*c)) + 2*pi)/pi
# sol 2 : 180*acos((-a**2 + b**2 + c**2)/(2*b*c))/pi
# substitution dictionary
d = {a: 1, b: 1, c: 1}
sol = [s.subs(d) for s in sol]
# apply custom filter:
sol = [s for s in sol if s <= 180]
sol = sol[0]
print(sol)
# out: 60
I have to find the equilibrium points where the nullclines intersect. My code is as below.
>>> from sympy import symbols, Eq, solve
>>> A,M = symbols('A M')
>>> dMdt = Eq(1.05 - (1/(1 + pow(A,5))) - M)
>>> dAdt = Eq(M*1 - 0.5*A - M*A/(2 + A))
>>> solve((dMdt,dAdt), (M,A))
[]
Why is it not giving a solution?
You will see why as I work to get the solution.
I'm going to write the equations as e1 and e2 -- use of Eq without a second arg no longer works (or does so with a warning in the latest versions of SymPy):
>>> from sympy import solve, nsimplify, factor, real_roots
>>> from sympy.abc import A, M
>>> e1 = (1.05 - (1/(1 + pow(A,5))) - M)
>>> e2 = (M*1 - 0.5*A - M*A/(2 + A))
Solve for M using e1
>>> eM = solve(e1, M)[0]
Substitute into e2
>>> e22 = e2.subs(M, eM); e22
-0.5*A - 0.05*A*(21.0*A**5 + 1.0)/((A + 2)*(A**5 + 1.0)) + 0.05*(21.0*A**5 + 1.0)/(A**5 + 1.0)
Get the numerator and denominator
>>> n,d=e22.as_numer_denom()
Find the real roots for this expression (which depends only on A)
>>> rA = real_roots(n)
Find the corresponding values of M by substituting each into eM:
>>> [(a.n(2), eM.subs(A, a).n(2)) for a in rA]
[(-3.3, 1.1), (-1.0, zoo), (-0.74, -0.23), (0.095, 0.050)]
That root of A = -1 is spurious -- if you look at your denominator of e1 you will see that such a value causes division by zero. So that root can be ignored. The others can be verified graphically.
Why didn't solve give the solution? It couldn't give the solution for this high-order polynomial in closed form. Even if you factor the numerator described above (and make floats into Rationals with nsimplify) you have a factor of degree 7:
>>> factor(nsimplify(n))
-(A + 1)*(A**4 - A**3 + A**2 - A + 1)*(5*A**7 + 10*A**6 - 21*A**5 + 5*A**2 + 10*A - 1)/10
I am using solver to find the zeros in the following equation. Solver returns only the formula of each one. How can i make it return for me a list of all the values calculated.
from sympy import *
A, B, C, D, r_w, r_j, r, R = symbols('A B C D rw1 rj1 r R')
equation=-pi*r**2*(A + B/(r/r_j + 1)) + pi*r**2*(C + D/(r/r_w + 1))
substitued=equation.subs([(A,232),(B,9768),(C,590),(D,7410),(rj1,1),
(rw1,2),(r,1)])
x=diff(substitued.subs(r,R),R)
solve(x,R)
`
why do i get returned the equations and not the values as a list. Please HELP!
0,−2337716−2200747384492+23062973288369571462634880387069105√137648136+3800344321791238833224√3+2571462634880387069105√137648136+3800344321791238833224⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2−−2223856515229413562200747384492+23062973288369571462634880387069105√137648136+3800344321791238833224√3+2571462634880387069105√137648136+3800344321791238833224√3−2571462634880387069105√137648136+3800344321791238833224⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3−23062973288369571462634880387069105√137648136+3800344321791238833224√3+2200747192246⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2,−2337716−2200747384492+23062973288369571462634880387069105√137648136+3800344321791238833224√3+2571462634880387069105√137648136+3800344321791238833224⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2+−2223856515229413562200747384492+23062973288369571462634880387069105√137648136+3800344321791238833224√3+2571462634880387069105√137648136+3800344321791238833224√3−2571462634880387069105√137648136+3800344321791238833224⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3−23062973288369571462634880387069105√137648136+3800344321791238833224√3+2200747192246⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2,−2337716−−2571462634880387069105√137648136+3800344321791238833224⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3−23062973288369571462634880387069105√137648136+3800344321791238833224√3+2200747192246+2223856515229413562200747384492+23062973288369571462634880387069105√137648136+3800344321791238833224√3+2571462634880387069105√137648136+3800344321791238833224√3⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2+2200747384492+23062973288369571462634880387069105√137648136+3800344321791238833224√3+2571462634880387069105√137648136+3800344321791238833224⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2,−2337716+−2571462634880387069105√137648136+3800344321791238833224⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3−23062973288369571462634880387069105√137648136+3800344321791238833224√3+2200747192246+2223856515229413562200747384492+23062973288369571462634880387069105√137648136+3800344321791238833224√3+2571462634880387069105√137648136+3800344321791238833224√3⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2+2200747384492+23062973288369571462634880387069105√137648136+3800344321791238833224√3+2571462634880387069105√137648136+3800344321791238833224⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯2
It looks like things are a little scrambled here. But the short answer is that SymPy is a symbolic calculator and, except for combining numbers together automatically (like 1 + 2 -> 3), it doesn't compute sqrt(2) + 1 as 2.414... unless you tell it to. And one way to "tell it to do it" is nfloat:
>>> sol = solve(substitued.diff(r), r)
>>> nfloat(sol,3)
[0.0, -6.07 - 2.81*I, -6.07 + 2.81*I, -1.29, 0.386]
There is also nsolve if you want numerical approximations of your solutions, but for an equation like this where there are multiple roots, you will have to supply an initial guess to help the solver find the root in which you are interested. Here is a simpler equation to demonstrate:
>>> sol = solve(x**2+x-sqrt(3))
>>> sol
[-1/2 + sqrt(1 + 4*sqrt(3))/2, -sqrt(1 + 4*sqrt(3))/2 - 1/2]
>>> nfloat(sol, 3)
[0.908, -1.91]
>>> nsolve(x**2+x-sqrt(3), x, 1)
0.907853262086954
>>> nsolve(x**2+x-sqrt(3), x, -1)
-1.90785326208695
How can I evaluate the constants C1 and C2 from a solution of a differential equation SymPy gives me? There are the initial condition f(0)=0 and f(pi/2)=3.
>>> from sympy import *
>>> f = Function('f')
>>> x = Symbol('x')
>>> dsolve(f(x).diff(x,2)+f(x),f(x))
f(x) == C1*sin(x) + C2*cos(x)
I tried some ics stuff but it's not working. Example:
>>> dsolve(f(x).diff(x,2)+f(x),f(x), ics={f(0):0, f(pi/2):3})
f(x) == C1*sin(x) + C2*cos(x)
By the way: C2 = 0 and C1 = 3.
There's a pull request implementing initial/boundary conditions, which was merged and should be released in SymPy 1.2. Meanwhile, one can solve for constants like this:
sol = dsolve(f(x).diff(x,2)+f(x),f(x)).rhs
constants = solve([sol.subs(x,0), sol.subs(x, math.pi/2) - 3])
final_answer = sol.subs(constants)
The code returns final_answer as 3.0*sin(x).
Remarks
solve may return a list of solutions, in which case one would have to substitute constants[0], etc. To force it to return a list in any case (for consistency), use dict=True:
constants = solve([sol.subs(x,0), sol.subs(x, math.pi/2) - 3], dict=True)
final_answer = sol.subs(constants[0])
If the equation contains parameters, solve may or may not solve for the variables you want (C1 and C2). This can be ensured as follows:
constants = solve([sol.subs(x,0), sol.subs(x, math.pi/2) - 3], symbols('C1 C2'))
where again, dict=True would force the list format of the output.
I have a function named f = 0.5/(z-3). I would like to know what would the coefficients p and q be if f was written in the following form: q/(1-p*z) but unfortunately sympy match function returns None. Am I doing something wrong? or what is the right way of doing something like this?
Here is the code:
z = symbols('z')
p, q = Wild('p'), Wild('q')
print (0.5/(z-3)).match(q/(1-p*z))
EDIT:
My expected answer is: q=-1/6 and p = 1/3
One way of course is
p, q = symbols('p q')
f = 0.5/(z-3)
print solve(f - q/(1-p*z), p, q,rational=True)
But I don't know how to do that in pattern matching, or if it's capable of doing something like this.
Thanks in Advance =)
If you start by converting to linear form,
1 / (2*z - 6) == q / (1 - p*z)
# multiply both sides
# by (2*z - 6) * (1 - p*z)
1 - p*z == q * (2*z - 6)
then
from sympy import Eq, solve, symbols, Wild
z = symbols("z")
p,q = symbols("p q", cls=Wild)
solve(Eq(1 - p*z, q*(2*z - 6)), (p,q))
gives
{p_: 1/3, q_: -1/6}
as expected.
Edit: I found a slightly different approach:
solve(Eq(f, g)) is equivalent to solve(f - g) (implicitly ==0)
We can reduce f - g like simplify(f - g), but by default it doesn't do anything because the resulting equation is more than 1.7 times longer than the original (default value for ratio argument).
If we specify a higher ratio, like simplify(f - g, ratio=5), we get
>>> simplify(1/(2*z-6) - q/(1-p*z), ratio=5)
(z*p_ + 2*q_*(z - 3) - 1)/(2*(z - 3)*(z*p_ - 1))
This is now in a form the solver will deal with:
>>> solve(_, (p,q))
{p_: 1/3, q_: -1/6}
SymPy's pattern matcher only does minimal algebraic manipulation to match things. It doesn't match in this case because there is no 1 in the denominator. It would be better to match against a/(b + c*z) and manipulate a, b, and c into the p and q. solve can show you the exact formula:
In [7]: solve(Eq(a/(b + c*z), q/(1 - p*z)), (q, p))
Out[7]:
⎧ -c a⎫
⎨p: ───, q: ─⎬
⎩ b b⎭
Finally, it's always a good idea to use exclude when constructing Wild object, like Wild('a', exclude=[z]). Otherwise you can get unexpected behavior like
In [11]: a, b = Wild('a'), Wild('b')
In [12]: S(2).match(a + b*z)
Out[12]:
⎧ 2⎫
⎨a: 0, b: ─⎬
⎩ z⎭
which is technically correct, but probably not what you want.