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

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 . 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)

2 Answers

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 

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).