Linked Questions
21 questions linked to/from On the functional square root of $x^2+1$
6
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257
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which functions can be obtained as a composition of a continuous function with itself? [duplicate]
let $f(x)=x^2$, then $f(f(x))=x^4$, so $x^4$ is a continuous function from $\Bbb R$ to $\Bbb R$ which can be obtained as $f\circ f$ for a continuous $f\colon \Bbb R\to \Bbb R$.
general example: for $f(...
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1
vote
0
answers
472
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Solving $f(f(x))=g(x)$ equations [duplicate]
Possible Duplicate:
Square root of a function (in the sense of composition)
I'm interested in solving equations of the form $f(f(x))=g(x)$ for $x\in\mathbb{R}$ where $g(x)$ is a known function.
...
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5
votes
1
answer
161
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If $q(x)=x^2+1$, does $q^{\circ 1/2}$ exist? [duplicate]
I've been doing a lot of research about functional half-iteration, and I posed the following question to myself:
Consider the function $q:\mathbb R\mapsto\mathbb R$ defined as
$$q(x)=x^2+1$$
...
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7
votes
0
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195
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Any chance for a solution of $f[f(x)]=x^2+1$? [duplicate]
Not sure if this is a closed or open question. But the question is suppose $f[f(x)]=x^2+1$ then what is $f(x)$? Though the question does not refer to the domain let us suppose it is defined on $\Bbb R$...
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11
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3
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How to calculate $f(x)$ in $f(f(x)) = e^x$?
How would I calculate the power series of $f(x)$ if $f(f(x)) = e^x$? Is there a faster-converging method than power series for fractional iteration/functional square roots?
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33
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1
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Characterising functions $f$ that can be written as $f = g \circ g$?
I'd like to characterise the functions that ‘have square roots’ in the function composition sense. That is, can a given function $f$ be written as $f = g \circ g$ (where $\circ$ is function ...
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8
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2
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if $\ f(f(x))= x^2 + 1$ , then $\ f(6)= $?
I want to know how to solve this type of questions. How can I find $\ f(x)$ from $\ f(f(x))$
Suppose, $\ f(f(x)) = x$ , then $\ f(x)=x$ or $\ f(x)=\dfrac{(x+1)}{(x-1)}$
how to find these ...
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14
votes
2
answers
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The 'Square root' Function
$G := \{f : f:[0,1] \rightarrow [0,1]$ such that it is bijective function and strictly increasing }
Now the question is
For any $ h \in G,$does there exist $g \in G$ such that $h=g \circ g $?
Is ...
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0
votes
2
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Functional square root of $f(x)=x^2-1$???
What I'm looking for is a function $\phi(x)$ such that $\phi(\phi(x))=f(x)$, where $f(x)=x^2-1$.
I am aware of this stack exchange post:
Square root of a function (in the sense of composition)
and ...
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15
votes
0
answers
876
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Compositional "square roots"
Let $A$ be a set and $f\colon A\to A$ a function.
The primary and general question is: What conditions are necessary so that there exists $g$ such that for each $x$ in $A$, $g(g(x))=f(x)$?
This is a ...
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1
vote
2
answers
444
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What is known about the functional square root of the Riemann Zeta function?
Let us consider the Riemann zeta function $\zeta(s)$ for $Re(s) > 1$:
$$\zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^{s}} .$$
I wonder what is known about the functional square root(s) of the ...
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15
votes
1
answer
526
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$f(f(x)) = 1 + x^2$, then what is f(1)?
I get $f(f(a)) = a^2 + 1 = f(f(-a))$, and so $f(a)^2 + 1 = f(a^2 + 1) = f(-a)^2 + 1$, so $f(a) = f(-a)$ or $f(a) = -f(-a)$, but then I donot know what to do next. Thanks for any help.
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4
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1
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333
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There exist a function such that $f\circ f(x)=e^x$? [duplicate]
Based on this question: How to calculate $f(x)$ in $f(f(x)) = e^x$?
I would like to know if I can get a function such that $f:\mathbb R \to \mathbb R^+$, defined by $f\circ f(x)=e^x$. My guess is no, ...
- 24.4k
5
votes
2
answers
399
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Finding the “root” of a monotone function (in the sense of composition)
Let $f:[0,\infty)\rightarrow [0,\infty) $ be a smooth and monotone function s.t $f(0)=0$. Let $N\in\mathbb{N}$. Can we find a function $g: [0,\infty) \rightarrow [0,\infty) $ s.t $g\circ\cdots\circ g$ ...
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2
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3
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294
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"Compositional roots" of functions, how to define them and how many are there?
Assume I am interested in solving $$(\underset{k \text{ times}}{\underbrace{g\circ \cdots \circ g)}}(x) = g^{\circ k}(x) = f(x)$$
That is, $g$ is in some sense a function which is a $k$:th root to ...
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