Symbols!¶
Announcements:
- Lessons 7,8 due today
- HW 3 due Monday
- Midterm 1 starts tomorrow!
Today's Music:
- Door Into Summer, Jacob's Trouble; cover of the Monkey's song
In [1]:
import sympy
In [2]:
import matplotlib
In [3]:
x = 10
In [4]:
x + 2 = 10
Input In [4] x + 2 = 10 ^ SyntaxError: cannot assign to expression here. Maybe you meant '==' instead of '='?
In [5]:
x + 2 == 10
Out[5]:
False
In [6]:
x = sympy.symbols('x')
In [7]:
x
Out[7]:
$\displaystyle x$
In [8]:
[a,b,c] = sympy.symbols('a b c')
In [9]:
a
Out[9]:
$\displaystyle a$
In [10]:
a + b
Out[10]:
$\displaystyle a + b$
In [11]:
a + 30
Out[11]:
$\displaystyle a + 30$
In [12]:
a ** 4
Out[12]:
$\displaystyle a^{4}$
In [13]:
2 * x == x + x
Out[13]:
True
In [14]:
x + a == a + x
Out[14]:
True
In [15]:
2 x == 10
Input In [15] 2 x == 10 ^ SyntaxError: invalid syntax
In [16]:
2 * x == 10
Out[16]:
False
In [17]:
sympy.solve(2 * x == 10)
Out[17]:
[]
In [18]:
sympy.solve(sympy.Eq(2 * x,10))
Out[18]:
[5]
In [19]:
sympy.solve(sympy.Eq(x ** 2,5))
Out[19]:
[-sqrt(5), sqrt(5)]
In [20]:
sin(x)
--------------------------------------------------------------------------- NameError Traceback (most recent call last) Input In [20], in <cell line: 1>() ----> 1 sin(x) NameError: name 'sin' is not defined
In [21]:
sympy.sin(x)
Out[21]:
$\displaystyle \sin{\left(x \right)}$
In [23]:
sympy.solve(sympy.Eq(sympy.sin(x),1.2))
Out[23]:
[1.5707963267949 - 0.622362503714779*I, 1.5707963267949 + 0.622362503714779*I]
In [25]:
y = sympy.S('y')
In [29]:
sympy.solve(sympy.Eq(sympy.sin(y)*sympy.sin(x),sympy.cos(y)))
Out[29]:
[{x: pi - asin(1/tan(y))}, {x: asin(1/tan(y))}]
In [30]:
print(sympy.Eq.__doc__)
An equal relation between two objects.
Explanation
===========
Represents that two objects are equal. If they can be easily shown
to be definitively equal (or unequal), this will reduce to True (or
False). Otherwise, the relation is maintained as an unevaluated
Equality object. Use the ``simplify`` function on this object for
more nontrivial evaluation of the equality relation.
As usual, the keyword argument ``evaluate=False`` can be used to
prevent any evaluation.
Examples
========
>>> from sympy import Eq, simplify, exp, cos
>>> from sympy.abc import x, y
>>> Eq(y, x + x**2)
Eq(y, x**2 + x)
>>> Eq(2, 5)
False
>>> Eq(2, 5, evaluate=False)
Eq(2, 5)
>>> _.doit()
False
>>> Eq(exp(x), exp(x).rewrite(cos))
Eq(exp(x), sinh(x) + cosh(x))
>>> simplify(_)
True
See Also
========
sympy.logic.boolalg.Equivalent : for representing equality between two
boolean expressions
Notes
=====
Python treats 1 and True (and 0 and False) as being equal; SymPy
does not. And integer will always compare as unequal to a Boolean:
>>> Eq(True, 1), True == 1
(False, True)
This class is not the same as the == operator. The == operator tests
for exact structural equality between two expressions; this class
compares expressions mathematically.
If either object defines an ``_eval_Eq`` method, it can be used in place of
the default algorithm. If ``lhs._eval_Eq(rhs)`` or ``rhs._eval_Eq(lhs)``
returns anything other than None, that return value will be substituted for
the Equality. If None is returned by ``_eval_Eq``, an Equality object will
be created as usual.
Since this object is already an expression, it does not respond to
the method ``as_expr`` if one tries to create `x - y` from ``Eq(x, y)``.
This can be done with the ``rewrite(Add)`` method.
.. deprecated:: 1.5
``Eq(expr)`` with a single argument is a shorthand for ``Eq(expr, 0)``,
but this behavior is deprecated and will be removed in a future version
of SymPy.
In [31]:
type(x)
Out[31]:
sympy.core.symbol.Symbol
In [32]:
type(y)
Out[32]:
sympy.core.symbol.Symbol
In [33]:
e1 = sympy.S('x + y / 3')
In [34]:
type(e1)
Out[34]:
sympy.core.add.Add
In [35]:
e1
Out[35]:
$\displaystyle x + \frac{y}{3}$
In [36]:
sympy.solve(sympy.Eq(e1,10))
Out[36]:
[{x: 10 - y/3}]
In [37]:
f1 = sympy.S('x / 20')
In [38]:
f2 = sympy.S('1 / 10')
In [39]:
f1 + f2
Out[39]:
$\displaystyle \frac{x}{20} + \frac{1}{10}$
In [40]:
sympy.together(f1 + f2)
Out[40]:
$\displaystyle \frac{x + 2}{20}$
In [41]:
sympy.simplify(f1 + f2)
Out[41]:
$\displaystyle \frac{x}{20} + \frac{1}{10}$
In [42]:
x + x
Out[42]:
$\displaystyle 2 x$
In [43]:
sympy.factor(x ** 2 + a * x + a,a)
Out[43]:
$\displaystyle a \left(x + 1\right) + x^{2}$
In [44]:
sympy.factor(x ** 2 + a * x + a,x)
Out[44]:
$\displaystyle a x + a + x^{2}$
In [45]:
sympy.factor(x ** 2 + a * x + a)
Out[45]:
$\displaystyle a x + a + x^{2}$
In [46]:
sympy.plot(sympy.sin(x))
Out[46]:
<sympy.plotting.plot.Plot at 0x7f22336e0490>
In [51]:
sympy.plot(sympy.sin(x) + sympy.cos(x))
Out[51]:
<sympy.plotting.plot.Plot at 0x7f2222999e40>
In [52]:
import sympy as sp
In [53]:
sp.sin(x)
Out[53]:
$\displaystyle \sin{\left(x \right)}$
In [54]:
sympy.solve( [ sympy.Eq(2 * x + 3 * y, 10), sympy.Eq(4 * x - 8 * y , 10)])
Out[54]:
{x: 55/14, y: 5/7}
In [62]:
def howManyZeros(eqn):
return len(sp.solve(sp.S(eqn)))
In [59]:
sympy.solve(sympy.S('x * 3 + 4'))
Out[59]:
[-4/3]
In [61]:
sympy.solve(sympy.S('x ** 2 + 10 * x + 5'))
Out[61]:
[-5 - 2*sqrt(5), -5 + 2*sqrt(5)]
In [64]:
howManyZeros('x ** 2 + 10 * x + 5')
Out[64]:
2
In [ ]: