# Will this method find the taylor expansion of ANY function $f(x)$?

Polynomials are themselves Taylor expansions, correct?

ex. $4x+5x^2+3 = 3+4x+5x^2 +0x^3 +0x^4 + \dots$ I'm assuming has no closed form besides $\sum_{n=0}^{2}(3+n)x^n + \sum_{n=3}^{\infty}0x^n$ but we were lucky with the consecutive coefficients and powers.

But for example, we can get the taylor expansion of $e^x$ by differentiating and equating sides at a convenient value:

$e^x = a_0 +a_1x +a_2x^2+a_3x^3 + \dots\\ \implies e^0 =1= a_0.$

$\frac{d}{dx}e^x=e^x=a_1+2a_2x+3a_3x^2+\dots\\ \implies e^0=1=a_1$

$\frac{d^2}{dx^2}e^x=e^x=2a_2+(2*3)a_3x+\dots\\ \implies e^0=1=2a_2\implies a_2=\frac{1}{2}$

$\frac{d^3}{dx^3}e^x=e^x=(2*3)a_3+\dots\\ \implies e^0=1=(2*3)a_3\implies a_3=\frac{1}{2*3}=\frac{1}{3!}$

Gives $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +\dots = \sum_{n=0}^{\infty}\frac{x^n}{n!}$.

Can the method find any taylor expansion? Do they all have a closed form? If not, what is the alternative?

• What is the question? – lcv Oct 21 '14 at 7:00
• The idea of a Taylor expansion is to express an arbitrary function as a sum of powers of $x$. Obviously, a polynomial already is a sum of powers of $x$, so you can't really speak of an 'expansion'. I'm not really sure what you are trying to say with this part. For the rest you're idea is correct, as the exponents are defined as $a_n=f^{(n)}(0)/n!$ when you are evaluating around $x=0$. – SPK.z Oct 21 '14 at 7:06
• You can also get Taylor series from other Taylor series by addition, term by term differentiation, term by term integration. – André Nicolas Oct 21 '14 at 7:07
• You can define a "Taylor expansion" for any smooth function $f$ but it is not true that $f$ and its expansion will coincide. That happens only if $f$ is analytical. There are examples of non analytical functions, i.e. here: en.wikipedia.org/wiki/Non-analytic_smooth_function – marco trevi Oct 21 '14 at 7:11