How to do a symbolic taylor expansion of an unknown function $f(x)$ using sympy

Question

In sage it is fairly easy to do a Taylor expansion of an unknown function f(x),

x = var('x')
h = var('h')
f = function('f',x)
g1 = taylor(f,x,h,2)

How can this be done in sympy?

---

Update

asmeurer points out that this is a feature which will be available soon in sympy from the pull request http://github.com/sympy/sympy/pull/1888. I installed the branch using pip,

pip install -e [email protected]:renatocoutinho/sympy.git@897b#egg=sympy --upgrade

However, when I try to calculate the series of f(x),

x, h = symbols("x,h")
f = Function("f")
series(f,x,x+h)

I get the following error,

TypeError: unbound method series() must be called with f instance as first argument (got Symbol instance instead)

Answer 1

As @asmeurer described, this is now possible with

from sympy import init_printing, symbols, Function
init_printing()

x, h = symbols("x,h")
f = Function("f")

pprint(f(x).series(x, x0=h, n=3))

or

from sympy import series
pprint(series(f(x), x, x0=h, n=3))

both returns

                                              ⎛  2        ⎞│                          
                                            2 ⎜ d         ⎟│                          
                                    (-h + x) ⋅⎜────(f(ξ₁))⎟│                          
                                              ⎜   2       ⎟│                          
                ⎛ d        ⎞│                 ⎝dξ₁        ⎠│ξ₁=h    ⎛        3       ⎞
f(h) + (-h + x)⋅⎜───(f(ξ₁))⎟│     + ──────────────────────────── + O⎝(-h + x) ; x → h⎠
                ⎝dξ₁       ⎠│ξ₁=h                2                                    

If you want a finite difference approximation, you can for example write

FW = f(x+h).series(x+h, x0=x0, n=3)
FW = FW.subs(x-x0,0)
pprint(FW)

to get the forward approximation, which returns

                                  ⎛  2        ⎞│
                                2 ⎜ d         ⎟│
                               h ⋅⎜────(f(ξ₁))⎟│
                                  ⎜   2       ⎟│
          ⎛ d        ⎞│           ⎝dξ₁        ⎠│ξ₁=x₀    ⎛ 3    2        2    3                 ⎞
f(x₀) + h⋅⎜───(f(ξ₁))⎟│      + ────────────────────── + O⎝h  + h ⋅x + h⋅x  + x ; (h, x) → (0, 0)⎠
          ⎝dξ₁       ⎠│ξ₁=x₀             2

Answer 2

There is no function for this in sympy, but it's rather easy to do it "by hand":

In [3]: from sympy import *
        x, h = symbols('x, h')
        f = Function('f')
        sum(h**i/factorial(i) * f(x).diff(x, i) for i in range(4))

Out[3]: h**3*Derivative(f(x), x, x, x)/6 + h**2*Derivative(f(x), x, x)/2 + h*Derivative(f(x), x) + f(x)

Note that sympy typically works with expressions (like f(x)) and not with bare functions (like f).