Suppose we have functions $$f(x) = \sqrt{x}, \space g(f) = \frac{df}{dx}+\frac{d^2f}{dx^2}+\frac{d^3f}{dx^3}\space ...$$

Applying function f(x) to g(f) we get: $$g(f(x))=\frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{4}x^{-\frac{3}{2}} + \frac{3}{8}x^{-\frac{5}{2}} - \frac{15}{16}x^{-\frac{7}{2}}$$

Is it possible to tell if it's converging? If it is, what is it converging to?

  • $\begingroup$ For any value $x\notin (-1,1)$ the series diverges, that's pretty obvious. $\endgroup$ – Scippy Jul 22 '14 at 21:50

$$ g(f(x)) = \sum_{n=1}^{\infty} \left(\Pi_{k=1}^{n-1}(2n - 2k-1)\right)\frac{(-1)^{n+1}}{2^n} x^{-\frac{2n-1}{2}} $$

Applying the so-called ratio test: $$ \frac{1}{2}(2n-1)|x|^{-1} \to \infty \, \, \, \forall x \neq 0 $$ Hence the radius of convergence is $0$ and $g(f(x))$s domain (as user2804303 kindly corrected) is $(-\infty,0) \cup (0, \infty)$, so it actually doesn't ever converge.

  • $\begingroup$ Actually the function is not defined for $x=0$. Therefore the function diverges $\forall x\in \mathbb{R}\setminus \{0\}$. $\endgroup$ – Scippy Jul 23 '14 at 17:47
  • $\begingroup$ Oh, duh. facepalm $\endgroup$ – Austin Stromme Jul 23 '14 at 17:52

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