How can I plot 0 with this function?

Question

Hello I got this piece of code and I achieved to plot in the for-function n=1,5,10. Now I should plot the zero as well. If I put for n_val in (0,1,5,10): ... , unfortunately I get the error x and y must have same first dimension, but have shapes (1000,) and (1,). Thank for helping me!

x = sp.symbols("x")
k = sp.symbols("k")
n = sp.symbols("n")
b = sp.Sum(((-1) ** k) * (x ** (2 * k + 1)) / sp.factorial(((2 * k + 1))), (k, 0, n))
c = sp.diff(b,x, 1)
a = sp.simplify(c)

for n_val in (1,5,10):
    a_np = sp.lambdify(x, a.subs(n, n_val).doit())
    x_vals = np.linspace(0, 10, 1000)        
    plt.plot(x_vals, a_np(x_vals), label=n_val)

plt.ylim(-2, 2)
plt.margins(x=0)
plt.legend(title='n', bbox_to_anchor=[1.02, 1.02], loc='upper left')
plt.tight_layout()
plt.show()

Answer 1

You could use the approach of Sympy: lambdify such that operations on arrays always result in arrays, also for constants?, which tests whether the function is constant and creates a constant numpy function in that case:

import sympy as sp
import numpy as np
import matplotlib.pyplot as plt

def np_lambdify(varname, func):
    lamb = sp.lambdify(varname, func, modules=['numpy'])
    if func.is_constant():
        return lambda t: np.full_like(t, lamb(t))
    else:
        return lambda t: lamb(np.array(t))

x = sp.symbols("x")
k = sp.symbols("k")
n = sp.symbols("n")
b = sp.Sum(((-1) ** k) * (x ** (2 * k + 1)) / sp.factorial(((2 * k + 1))), (k, 0, n))
c = sp.diff(b, x, 1)
a = sp.simplify(c)

for n_val in (0, 1, 5, 10):
    a_np = np_lambdify(x, a.subs(n, n_val).doit())
    x_vals = np.linspace(0, 10, 1000)
    plt.plot(x_vals, a_np(x_vals), label=n_val)

plt.ylim(-2, 2)
plt.margins(x=0)
plt.legend(title='n', bbox_to_anchor=[1.02, 1.02], loc='upper left')
plt.tight_layout()
plt.show()