Can the following trick be expanded upon?

Question

Main Question

What is the expansion of $d^{1+\epsilon}?$

Background

I noticed the following trick (sometimes more laborious) to directly differentiate $ f(x) $ twice without differentiating it even once.

To show what I mean let $ f(x)=x^2 $

By Taylor expansion for any $f(x)$ :

Equation 1

$$ f(x+ \epsilon ) = f(x) + \epsilon f'(x) + \frac{\epsilon^2 f''(x)}{2!} + ... $$

Equation 2

$$ f(x+ \epsilon ) = (x+ \epsilon )^2 = ... + \epsilon^2 $$

Comparing the $ \epsilon^2 $ term in equation $1 $and $2 $ we get:

$$ \frac{f''(x)}{2!} = 1$$

Hence,

$$ f''(x) = 2 $$

Question

Can similar reasoning be applied for $ d^{1+\epsilon} g(x) $ where $g(x)$ is a dummy function to reveal the entire space of $ (d^n g(x)) $ on $g(x)$ ?

Answer 1

If I understand you correctly, you mean to expand

$$ d^{1+\epsilon} = \sum_{k=0}^\infty a_k d^k. $$

In general

$$ d^{1+\epsilon} = \big(1 + d - 1\big)^{1+\epsilon} = \sum_{\ell=0}^\infty \binom{1+\epsilon}{\ell} \big(d-1\big)^\ell $$

So

$$ d^{1+\epsilon} = \big(1 + d - 1\big)^{1+\epsilon} = \sum_{\ell=0}^\infty \binom{1+\epsilon}{\ell} \sum_{k=0}^\ell \binom{\ell}{k} (-1)^{\ell-k} d^k\\ = \sum_{\ell=0}^\infty \sum_{k=0}^\ell \binom{1+\epsilon}{\ell} \binom{\ell}{k} (-1)^{\ell-k} d^k\\ = \sum_{k=0}^\infty \left( \sum_{\ell=k}^\infty \binom{1+\epsilon}{\ell} \binom{\ell}{k} (-1)^{\ell-k} \right) d^k. $$

Answer 2

It is not clear to me what you are asking; I shall assume that you are asking for a method allowing one to find $f^{(n)} (x_0)$ without computing the derivatives of lower order.

Since your example uses Taylor series, let us restrict to the case when $f$ is real-analytic on $(a,b)$. Then $f$ admits an extension $F : D = \mathring D \subset \Bbb C \to \Bbb C$, with $(a,b) \subset D$ and $F$ complex-analytic. Now, choose a simple closed contour $\gamma \subset D$ surrounding $x_0 \in (a,b)$. Then, according to Cauchy's integral formula, $f^{(n)} (x_0) = \frac {n!} {2 \pi \Bbb i} \oint \limits _\gamma \frac {f(z)} {(z-x_0) ^{n+1}} \Bbb d z$ with $\gamma$ oriented counterclockwisely.

For $f$ being no longer analytic, but smooth and absolutely integrable on $\Bbb R$, another approach would be to use a basic idea from the theory of the Fourier transform. If $\widehat f (p) = \frac 1 {\sqrt {2 \pi}} \int \limits _{- \infty} ^\infty \Bbb e ^{- \Bbb i p x} f(x) \Bbb d x$ then one can prove that $\widehat {f^{(n)}} (p) = (\Bbb i p)^n \widehat f (p)$, so that $f^{(n)}$ can be obtained with the aid of the inverse Fourier transform: $f^{(n)} (x) = \frac 1 {\sqrt {2 \pi}} \int \limits _{- \infty} ^\infty \Bbb e ^{ \Bbb i p x} (\Bbb i p)^n \widehat f (p) \Bbb d p$.

A similar result could be obtained if one worked with the Laplace transform instead of the Fourier one.