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For differentiable one-dimensional $f: \mathbb{R} \rightarrow \mathbb{R}$ or $f: \mathbb{C} \rightarrow \mathbb{C}$, after finding a root $x_0$ it's possible to split out a differentiable linear factor using $f(x) = f^*(x)(x - x_0)$, allowing to search for subsequent roots on $f^*$ since $f(x) = 0 \iff x = x_0 \vee f^*(x) = 0$, whilst $f^*(x_0) \neq 0$ unless $f'(x_0) = 0$.

Is a similar mechanism possible for the multidimensional case, e.g. for differentiable $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$, or at least for some class of multidimensional functions, so that a differentiable function $f^*$ can be derived from $f$ and a known root $x_0$ so that $f(x) = 0 \iff x = x_0 \vee f^*(x) = 0$ holds, but $f^*(x_0) \neq 0$ unless the Jacobian matrix of $f$ at $x_0$ is also zero.

ternary
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  • FYI, I removed the functional-analysis tag, since it is not appropriate. I'm guessing you didn't look at questions under that tag or google the term, and simply relied on the individual words "function[al]" and "analysis" to guide you to what it means. – Dave L. Renfro Jan 14 '23 at 14:44
  • Any requirements on $f^*$ like differentiabilty? – Kritiker der Elche Jan 14 '23 at 17:16
  • If you allow discontinuous $f^$, you can take $f^(x) = f(x)$ for $x \ne x_0$ and $f^*(x_0) = 1$. – Kritiker der Elche Jan 15 '23 at 10:33
  • @KritikerderElche You're right, for discontinuous $f^$ it's not very meaningful, so yes, it makes most sense if it's also differentiable, or at least continuous. I'm currently pondering whether it would be possible using https://math.stackexchange.com/questions/4619140/generalized-vector-multiplication/4619199 and continuously extending $f^(x) = g_n(f(x), k - x_0)$ to $x_0$. But maybe there is a better way. – ternary Jan 20 '23 at 23:28
  • I think you should require $f^*$ to be differentiable and edit your question. – Kritiker der Elche Jan 20 '23 at 23:33
  • Yes, already did that. – ternary Jan 22 '23 at 03:02

2 Answers2

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For a continuously differentiable function $f:\mathbb R^n\to\mathbb R^n$ and $x_0\in\mathbb R^n$, the fundamental theorem of calculus for $\varphi(t)= f(x_0+t(x-x_0))$ gives $$f(x)-f(x_0)=\varphi(1)-\varphi(0)=\int_0^1 \varphi'(t)dt=\langle g(x),x-x_0\rangle $$ for $g(x)=\int\limits_0^1 f'(x_0+t(x-x_0)dt$ by the chain rule.

Jochen
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Yes. If we know that $f(x_0) = 0$, then we can take $f^* = f$ and your requirement

$f(x) = 0 \iff x = x_0 \vee f^*(x) = 0$

is satisfied.

Kritiker der Elche
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  • You're right. I didn't properly specify what I'm at. The idea is to be able to somehow reduce $f$ in order to remove already found roots from it. I hope the updated question makes it more clear. – ternary Jan 14 '23 at 16:54