In [1]:

    

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sympy import symbols, sqrt, Rational, lambdify, init_printing, solve, symbols, Subs

%matplotlib inline
init_printing(use_latex=True)


    

In [2]:

    

a,b,c,d,x  = symbols('a b c d x')
solve(a*x**2+b*x+c,x)


    

    
Out[2]:





$$\left [ \frac{1}{2 a} \left(- b + \sqrt{- 4 a c + b^{2}}\right), \quad - \frac{1}{2 a} \left(b + \sqrt{- 4 a c + b^{2}}\right)\right ]$$

In [7]:

    

from sympy import(    diff,
    integrate,
    Integral,
    Sum,
    Product,
    Eq )

from IPython.display import *
from sympy.plotting import (
    plot, plot_implicit)

x, y, a, b, c = symbols("x y a b c")

expr = x**2*a + y**2*b
expr = expr.subs(a, 2)
expr = expr.subs(b, 2)

display(expr)

eq1 = Eq(expr, 5)
p1 = plot_implicit(eq1,  (x, -3, 3), (y, -3, 3))


    

    






$$2 x^{2} + 2 y^{2}$$






    

In [8]:

    

display(Sum(expr, (x,1,10)))
display(Product(expr, (x,0,10)))

display (solve(expr))


    

    






$$\sum_{x=1}^{10} \left(2 x^{2} + 2 y^{2}\right)$$






    






$$\prod_{x=0}^{10} \left(2 x^{2} + 2 y^{2}\right)$$






    






$$\left [ \left \{ x : - i y\right \}, \quad \left \{ x : i y\right \}\right ]$$

In [9]:

    

from sympy import Matrix

b11, b12, b13, b21, b22, b23, b31, b32, b33 = symbols ('b11 b12 b13 b21 b22 b23 b31 b32 b33')
c11, c21, c31 = symbols('c11 c21 c31')

mat_symbolic_1 = Matrix([[b11, b12, b13],[b21, b22, b23],[b31, b32, b33]])
mat_symbolic_2 = Matrix([[c11],[c21],[c31]])

mat_symbolic_1 * mat_symbolic_2


    

    
Out[9]:





$$\left[\begin{matrix}b_{11} c_{11} + b_{12} c_{21} + b_{13} c_{31}\\b_{21} c_{11} + b_{22} c_{21} + b_{23} c_{31}\\b_{31} c_{11} + b_{32} c_{21} + b_{33} c_{31}\end{matrix}\right]$$

In [88]:

    

mat = Matrix([[1.0, 2.0],[-(1.0 / 2.0), 1.0]])

print ('get eigenValues')
pprint ( Matrix(mat).eigenvals())

print ('\nget eigenVector')
pprint (Matrix(mat).eigenvects())

mat


    

    




get eigenValues
{1 - ⅈ: 1, 1 + ⅈ: 1}

get eigenVector
⎡⎛1.0 - 1.0⋅ⅈ, 1, ⎡⎡2.0⋅ⅈ⎤⎤⎞, ⎛1.0 + 1.0⋅ⅈ, 1, ⎡⎡-2.0⋅ⅈ⎤⎤⎞⎤
⎢⎜                ⎢⎢     ⎥⎥⎟  ⎜                ⎢⎢      ⎥⎥⎟⎥
⎣⎝                ⎣⎣ 1.0 ⎦⎦⎠  ⎝                ⎣⎣ 1.0  ⎦⎦⎠⎦






    
Out[88]:





$$\left[\begin{matrix}1.0 & 2.0\\-0.5 & 1.0\end{matrix}\right]$$

In [27]:

    

from sympy import Function, Eq, dsolve,pprint, cos, sin, Lambda, Derivative, var, latex


x,y,z,t = symbols('x y z t')
F = Function('F')

f = Eq(x**2 - x, 0)
f2 = Eq(x**2 + 2, x)
pprint (f)
pprint (f2)


    

    




 2        
x  - x = 0
 2        
x  + 2 = x

In [21]:

    

a,b,k = symbols('a b k')

display(cos(x*y) - sin(z*x))
display(Derivative(f, x))
display(Derivative(f, x).doit())
display(Derivative(f, x, 2))


    

    






$$- \sin{\left (x z \right )} + \cos{\left (x y \right )}$$






    






$$\frac{d}{d x} x^{2} - x = 0$$






    






$$\frac{d}{d x} x^{2} - x = 0$$






    






$$\frac{d^{2}}{d x^{2}}  x^{2} - x = 0$$

In [22]:

    

display(Integral(f,x))
display(Integral(f, (x, a,b)))
display(Integral(f, (x, a,b)).doit())


    

    






$$\int x^{2} - x\, dx = \int 0\, dx$$






    






$$\int_{a}^{b} x^{2} - x\, dx = \int_{a}^{b} 0\, dx$$






    






$$- \frac{a^{3}}{3} + \frac{a^{2}}{2} + \frac{b^{3}}{3} - \frac{b^{2}}{2} = 0$$

In [12]:

    

eq_1 = Eq(F(x).diff(x), x / F(x))
display(eq_1)
eq_solved = dsolve(eq_1)
display(eq_solved)


    

    






$$\frac{d}{d x} F{\left (x \right )} = \frac{x}{F{\left (x \right )}}$$






    






$$\left [ F{\left (x \right )} = - \sqrt{C_{1} + x^{2}}, \quad F{\left (x \right )} = \sqrt{C_{1} + x^{2}}\right ]$$

In [25]:

    

var('a b t k C1')
u = Function("u")(t)
de = Eq(u+u.diff(t) * k)
display(de)

des = dsolve(de,u)
display(des)
des = des.subs(C1,0)
display(des)


    

    






$$k \frac{d}{d t} u{\left (t \right )} + u{\left (t \right )} = 0$$






    






$$u{\left (t \right )} = e^{\frac{1}{k} \left(C_{1} - t\right)}$$






    






$$u{\left (t \right )} = e^{- \frac{t}{k}}$$

In [28]:

    

f = Lambda((t,k,a,b),(a-b) * des.rhs + b)
display(Latex('$f(t,k,a,b) = ' + str(latex(f(t,k,a,b))) + '$'))
x = np.linspace(0,5,100)


    

    






$f(t,k,a,b) = b + \left(a - b\right) e^{- \frac{t}{k}}$

In [29]:

    

plt.grid(True)
plt.xlabel('Time t seconds',fontsize=12)
plt.ylabel('$f(t,1,4,8)$',fontsize=16)
plt.plot(x,[f(t,1,4,8) for t in x],color='#008000')
plt.show()


    

In [30]:

    

var('a b t k C1')
u = Function("u")(t)
de = Eq(u+u.diff(t) * k)
des = dsolve(de,u).subs(C1,0)
f = Lambda((t,k,a,b),(a-b) * des.rhs + b)
f


    

    
Out[30]:





$$\left( \left ( t, \quad k, \quad a, \quad b\right ) \mapsto b + \left(a - b\right) e^{- \frac{t}{k}} \right)$$

In [31]:

    

fig = plt.gcf()
fig.set_size_inches(8,5)
x = np.linspace(0,5,100)
plt.grid(True)
plt.xlabel('Time t seconds',fontsize=12)
plt.ylabel('Voltage',fontsize=12)
plt.title('R-C circuit decay profile',fontsize=16)
r = 10000 # resistor value Ω
c = 100e-6 # capacitor value farads
plt.text(2.1,7,'R = %.1f $\Omega$' % r,fontsize=14)
plt.text(2.1,6.3,'C = %.1f $\mu$f' % (c * 1e6),fontsize=14)
plt.plot(x,[f(t,r * c,0,10) for t in x],color='#008000')
plt.show()


    

In [ ]:

    




    

In [32]:

    

from sympy import Limit, sin, oo
x = symbols('x')

print('Range approaching: ', Limit(sin(x) / x, x, 0).doit())
Limit(sin(x) / x, x, 0)


    

    




Range approaching:  1






    
Out[32]:





$$\lim_{x \to 0^+}\left(\frac{1}{x} \sin{\left (x \right )}\right)$$

In [33]:

    

print('Range approaching: ', Limit(1/x, x, oo).doit())
Limit(1/x, x, oo)


    

    




Range approaching:  0






    
Out[33]:





$$\lim_{x \to \infty} \frac{1}{x}$$

In [65]:

    

print('Range approaching: ', Limit(1/x, x , 0, dir = '+').doit())
Limit(1/x, x , 0, dir = '+')


    

    




Range approaching:  oo






    
Out[65]:





$$\lim_{x \to 0^+} \frac{1}{x}$$

In [66]:

    

print('Range approaching: ', Limit(1/x, x , 0, dir = '-').doit())
Limit(1/x, x , 0, dir = '-')


    

    




Range approaching:  -oo






    
Out[66]:





$$\lim_{x \to 0^-} \frac{1}{x}$$

In [69]:

    

print('Range approaching: ', Limit(1/x**2, x , - oo).doit())
Limit(1/x**2, x , - oo)


    

    




Range approaching:  0






    
Out[69]:





$$\lim_{x \to -\infty} \frac{1}{x^{2}}$$

In [35]:

    

from sympy import symbols, sqrt, Rational, lambdify

x, y = symbols("x y")
list1 = np.arange(1,1000)
list2 = pd.Series(list1)

def sympy_expr(x_val):
    expr = x**2 + sqrt(3)*x - Rational(1,3)
    return expr.subs(x, x_val)

expr_1 = [sympy_expr(item) for item in list2]
#print (expr_1)

%timeit [expr_1]

lambdifyFunction = f = lambdify(x, expr_1)

%timeit lambdifyFunction(list1)
%timeit lambdifyFunction(list2)


    

    




10000000 loops, best of 3: 101 ns per loop
10000 loops, best of 3: 206 µs per loop
1000 loops, best of 3: 205 µs per loop

In [29]:

    




    

In [43]:

    

from sympy import var,E, pi, exp

var('x')
f = Lambda(x,E**-x**2)
x = np.linspace(-3,3,100)
y = np.array([f(v) for v in x],dtype='float')
a = 40
b = 60


    

In [40]:

    

import numpy as np
import matplotlib.pyplot as plt

from sympy.plotting import (
    plot, 
    plot_parametric, 
    plot3d,
    plot3d_parametric_line, 
    plot3d_parametric_surface
)

%matplotlib inline

var('i')
plot(sin(i),(i,0, 2*pi))


    

    













    
Out[40]:





<sympy.plotting.plot.Plot at 0x1a05b2dcb38>

In [44]:

    

var('i, j')
plot3d((exp(-(i**2+j**2))),(i,-3,3),(j,-3,3))


    

    













    
Out[44]:





<sympy.plotting.plot.Plot at 0x1a05af88dd8>

In [45]:

    

fig = plt.gcf()
fig.set_size_inches(8,5)
plt.grid(True)
plt.fill_between(x[:a+1],y[:a+1],0,color='#ffff40',alpha=.3)
plt.fill_between(x[a:b],y[a:b],0,color='#40ff40',alpha=.3)
plt.fill_between(x[b-1:],y[b-1:],0,color='#ffff40',alpha=.3)
plt.plot(x,y,color='black')
plt.show()


    

In [46]:

    

x, b1, b2 = symbols("i b1 b2")

f = x/(x+exp(b1-b2*i))
res = {b1:29.3930964972769,b2:0.327159886574049}

plot(f.subs(res), (i, 0, 100))


    

    













    
Out[46]:





<sympy.plotting.plot.Plot at 0x1a05b000748>

In [47]:

    

t = symbols('t')
x = 0.05*t + 0.2/((t - 5)**2 + 2)
lam_x = lambdify(t, x, modules=['numpy'])

x_vals = np.linspace(0, 10, 100)
y_vals = lam_x(x_vals)

plt.plot(x_vals, y_vals)
plt.ylabel("Speed")
plt.show()


    

In [74]:

    

from IPython.display import HTML

init_printing(use_latex=True)


    

In [71]:

    

def wrap_tag(t,d):
   return '<%s style="border-color:white">%s</%s>' % (t,d,t) 
desc = (
  'Population at time t','Population at time zero',
  'Time','Population growth rate'
)

# source: http://mathworld.wolfram.com/PopulationGrowth.html
# problem: for a given multi-variable equation, generate all its forms
# create symbols
vl = (N,N_0,t,r) = symbols('N N_0 t r')
# base equation
b = log(N/N_0) - r*t


    

In [72]:

    

s = 'Variables:'
ss = ''
for n,x in enumerate(vl):
  ss += wrap_tag('td',' ')
  ss += wrap_tag('td','$%s$' % str(x))
  ss += wrap_tag('td',': %s' % desc[n])
  ss = wrap_tag('tr',ss)
s += wrap_tag('table',ss)
s += 'Equations:'
ss = ''
for v in vl:
  soln = solve(b,v)[0]
  ss += wrap_tag('td',' ')
  ss += wrap_tag('td','$%s$' % str(v))
  ss += wrap_tag('td','$\displaystyle = %s$' % latex(soln))
  ss = wrap_tag('tr',ss)
s += wrap_tag('table',ss)


    

In [77]:

    

def makeLatexToHTML(latex):
    return HTML(latex)
makeLatexToHTML(s)


    

    
Out[77]:




Variables:
 
$N$
: Population at time t
 
$N_0$
: Population at time zero
 
$t$
: Time
 
$r$
: Population growth rate
Equations:
 
$N$
$\displaystyle = N_{0} e^{r t}$
 
$N_0$
$\displaystyle = N e^{- r t}$
 
$t$
$\displaystyle = \frac{1}{r} \log{\left (\frac{N}{N_{0}} \right )}$
 
$r$
$\displaystyle = \frac{1}{t} \log{\left (\frac{N}{N_{0}} \right )}$

In [76]:

    

[solve(b,v)[0] for v in vl]


    

    
Out[76]:





$$\left [ N_{0} e^{r t}, \quad N e^{- r t}, \quad \frac{1}{r} \log{\left (\frac{N}{N_{0}} \right )}, \quad \frac{1}{t} \log{\left (\frac{N}{N_{0}} \right )}\right ]$$

In [ ]: