The notation for matrices as lists of lists is very tedious to use for practical calculations. If you need to input a matrix with three columns and ten rows, you'll have to write a lot of square brackets!
There is another convention for specifying matrices which is more convenient.
If you have access to numpy on your computer, you can specify matrices
in the alternate notation
>>> from numpy import mat as M
>>> A = Matrix( M('1 2; 3 9') )
>>> A.inv() # call the inv method on A
[ 3, -2/3]
[-1, 1/3]
When computing determinants, we attribute particular attention to whether the the zero-vs-nonzero nature Of particular interest is the
adjugate
Consider the parallelepiped which has as its sides the vectors $\vec{u}=(1,2,3)$, $\vec{v}=(2,−2,4)$, and $\vec{w} =(2,2,5)$. If we place these vectors as the rows of matrix $M$, then $\det(M)$ is the volume of the parallelepiped:
>>> M = Matrix( [[1, 2, 3],
[2,-2, 4],
[2, 2, 5]] )
>>> M.det() # V = det( [u above v above w] )
2
>>> M.row(0).dot( M.row(1).cross( M.row(2) ) )
2 # verify using V = u\cdot( v \times w )
The volume of the parallelepiped is 2.
By default, sympy compute derivative or integral expression eagerly.
If you tell it to
Limit(sin(x)/x, x, 0) Derivative Integral
The student can then enter their answer in a textarea:
// x(t)
answer = function (t) { var tt,ttt; return x_i + integrate( v_i + integrate( a, (tt,0,ttt)), (ttt,0,t) ); }
You can use python as a basic calculator
>>> 2+2 4 >>> [1,2,3] [1,2,3]
Keep this in mind and always simplify things before you try to compare them.
>>> map(lambda e:e.subs({'a':1,'b':2,'c':-8}), solve(a*x**2+b*x+c,x) )
[2, -4]
>>>
exp2.rewrite(exp)
In [23]: print (E(I*x)).expand(complex=True) I*sin(x) + cos(x) »> ( (A*e1)[0] ).rewrite(exp) x,f=symbols('x,f',real=True) exp1=E(-I*2*pi*f*x)
exp2=exp1.expand(complex=True)
exp2 returns: -I*sin(2*pi*f*x) + cos(2*pi*f*x)
Is there a way to go from exp2 back to exp1?
I'm relatively new over here, but I think this should do.
>>> >>>exp(-2*I*pi*f*x)
>>> def axn_sin(x,n): # press SHIFT+ENTER for a newline
return (-1.0)**n * x**(2*n+1) / factorial(2*n+1)
>>> sum( [ axn_sin(forty,n) for n in range(0,10) ] ) >>> sin(forty).evalf()
>>> summation( sk, (k,1,n) )
>>> def an_ln2(n): return 1.0*(-1)**(n+1)/n >>> sum([ an_ln2(n) for n in range(1,1000) ])