CODEWITHSUNDEEP
maxsympycomputer-algebra-systems
Kattern
Kattern

Reputation: 3109

How to find the maximum value of a function in Sympy?

These days I am trying to redo shock spectrum of single degree of freedom system using Sympy. The problem can reduce to find maximum value of a function. Following are two cases I cannot figure out how to do.

The first one is 

tau,t,t_r,omega,p0=symbols('tau,t,t_r,omega,p0',positive=True)
h=expand(sin(omega*(t-tau)))
f=simplify(integrate(p0*tau/t_r*h,(tau,0,t_r))+integrate(p0*h,(tau,t_r,t)))

The final goal is to obtain maximum absolute value of f (The variable is t). The direct way is 

df=diff(f,t)
sln=solve(simplify(df),t)
simplify(f.subs(t,sln[1]))

Here is the result, I tried many ways, but I can not simplify any further.

Result

Therefore, I tried another way. Because I need the maximum absolute value and the location where abs(f) is maximum happens at the same location of square of f, we can calculate square of f first.

df=expand_trig(diff(expand(f)**2,t))
sln=solve(df,t)
simplify(f.subs(t,sln[2]))

It seems the answer is almost the same, just in another form.

Result

The expected answer is a sinc function plus a constant as following:

Expected

Therefore, the question is how to get the final presentation.

The second one may be a little harder. The question can be reduced to find the maximum value of f=sin(pi*t/t_r)-T/2/t_r*sin(2*pi/T*t), in which t_r and T are two parameters. The maximum located at different peak when the ratio of t_r and T changes. And I do not find a way to solve it in Sympy. Any suggestion? The answer can be represented in following figure.

Max

Upvotes: 6

Views: 12001

Answers (1)

asmeurer
asmeurer

Reputation: 91580

The problem is the log(exp(I*omega*t_r/2)) term. SymPy is not reducing this to I*omega*t_r/2. SymPy doesn't simplify this because in general, log(exp(x)) != x, but rather log(exp(x)) = x + 2*pi*I*n for some integer n. But in this case, if you replace log(exp(I*omega*t_r/2)) with omega*t_r/2 or omega*t_r/2 + 2*pi*I*n, it will be the same, because it will just add a 2*pi*I*n inside the sin.

I couldn't figure out any functions that force this simplification, but the easiest way is to just do a substitution:

In [18]: print(simplify(f.subs(t,sln[1]).subs(log(exp(I*omega*t_r/2)), I*omega*t_r/2)))
p0*(omega*t_r - 2*sin(omega*t_r/2))/(omega**2*t_r)

That looks like the answer you are looking for, except for the absolute value (I'm not sure where they should come from).

Upvotes: 5

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