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# Configure Jupyter to display the assigned value after an assignment
%config InteractiveShell.ast_node_interactivity='last_expr_or_assign'
# import everything from SymPy.
from sympy import *
# Set up Jupyter notebook to display math.
init_printing()
The following displays SymPy expressions and provides the option of showing results in LaTeX format.
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from sympy.printing import latex
def show(expr, show_latex=False):
"""Display a SymPy expression.
expr: SymPy expression
show_latex: boolean
"""
if show_latex:
print(latex(expr))
return expr
Create a symbol for time.
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t = symbols('t')
If you combine symbols and numbers, you get symbolic expressions.
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expr = t + 1
The result is an Add object, which just represents the sum without trying to compute it.
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type(expr)
subs can be used to replace a symbol with a number, which allows the addition to proceed.
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expr.subs(t, 2)
f is a special class of symbol that represents a function.
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f = Function('f')
The type of f is UndefinedFunction
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type(f)
SymPy understands that f(t) means f evaluated at t, but it doesn't try to evaluate it yet.
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f(t)
diff returns a Derivative object that represents the time derivative of f
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dfdt = diff(f(t), t)
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type(dfdt)
We need a symbol for alpha
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alpha = symbols('alpha')
Now we can write the differential equation for proportional growth.
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eq1 = Eq(dfdt, alpha*f(t))
And use dsolve to solve it. The result is the general solution.
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solution_eq = dsolve(eq1)
We can tell it's a general solution because it contains an unspecified constant, C1.
In this example, finding the particular solution is easy: we just replace C1 with p_0
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C1, p_0 = symbols('C1 p_0')
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particular = solution_eq.subs(C1, p_0)
In the next example, we have to work a little harder to find the particular solution.
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r, K = symbols('r K')
Now we can write the differential equation.
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eq2 = Eq(diff(f(t), t), r * f(t) * (1 - f(t)/K))
And solve it.
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solution_eq = dsolve(eq2)
The result, solution_eq, contains rhs, which is the right-hand side of the solution.
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general = solution_eq.rhs
We can evaluate the right-hand side at $t=0$
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at_0 = general.subs(t, 0)
Now we want to find the value of C1 that makes f(0) = p_0.
So we'll create the equation at_0 = p_0 and solve for C1. Because this is just an algebraic identity, not a differential equation, we use solve, not dsolve.
The result from solve is a list of solutions. In this case, we have reason to expect only one solution, but we still get a list, so we have to use the bracket operator, [0], to select the first one.
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solutions = solve(Eq(at_0, p_0), C1)
type(solutions), len(solutions)
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value_of_C1 = solutions[0]
Now in the general solution, we want to replace C1 with the value of C1 we just figured out.
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particular = general.subs(C1, value_of_C1)
The result is complicated, but SymPy provides a method that tries to simplify it.
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particular = simplify(particular)
Often simplicity is in the eye of the beholder, but that's about as simple as this expression gets.
Just to double-check, we can evaluate it at t=0 and confirm that we get p_0
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particular.subs(t, 0)
This solution is called the logistic function.
In some places you'll see it written in a different form:
$f(t) = \frac{K}{1 + A e^{-rt}}$
where $A = (K - p_0) / p_0$.
We can use SymPy to confirm that these two forms are equivalent. First we represent the alternative version of the logistic function:
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A = (K - p_0) / p_0
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logistic = K / (1 + A * exp(-r*t))
To see whether two expressions are equivalent, we can check whether their difference simplifies to 0.
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simplify(particular - logistic)
This test only works one way: if SymPy says the difference reduces to 0, the expressions are definitely equivalent (and not just numerically close).
But if SymPy can't find a way to simplify the result to 0, that doesn't necessarily mean there isn't one. Testing whether two expressions are equivalent is a surprisingly hard problem; in fact, there is no algorithm that can solve it in general.
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Exercise: Use WolframAlpha to solve the quadratic growth model, using either or both forms of parameterization:
df(t) / dt = alpha f(t) + beta f(t)^2
or
df(t) / dt = r f(t) (1 - f(t)/K)
Find the general solution and also the particular solution where f(0) = p_0.
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