Is there a way of proving that a function has a particular number of fixed points.Show that $f$ has...

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Is there a way of proving that a function has a particular number of fixed points.


Show that $f$ has infinitely many fixed pointsNumber of fixed points of a meromorphic functionHow many fixed points are there for $f:[0,4]to [1,3]$Find 3 fixed points of function with 2 argumentshow do fixed points of a function help in finding it's root?How many fixed points are there?Fixed points of an iterated functionWhy does the recursion $x_{n+1} = f(x_n)$ with $f(x) = p^kcdotsqrt[k+1]{x}$ converge to $p^{k+1}$?How does one find all the fixed points of the operator defined on the factorial function and how it affects the definition of it?How to find fixed point without graph approach?













0












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From my understanding, a function is said to have a fixed point if $f(x) = x$.



Is there a way for finding how many fixed points a function has?










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therewillbecode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    $begingroup$
    In this level of generality? No. Given a specific function, maybe! Do you have one that you're interested in?
    $endgroup$
    – Billy
    Mar 9 at 17:35










  • $begingroup$
    Brouwer’s Fixed Point Theorem is a result of the existence of a fixed point. So is the Banach Fixed-Point Theorem. But there are some assumptions that go along with them.
    $endgroup$
    – Ekesh Kumar
    Mar 9 at 17:55










  • $begingroup$
    You are asking how one can find all zeros of function $phi$ defined by $phi(x):=f(x)-x$...
    $endgroup$
    – Jean Marie
    Mar 9 at 18:24










  • $begingroup$
    Basically you need to solve the equation f(x)-x=0. In some cases it is straight forward in some others in more complicated. For example: for x+2 you solve (x+2)-x=0 which tells you there is no such. On the other end x^2 has the solutions x^2-x+0 result x=0 and x=1 as fixed points (not more!).
    $endgroup$
    – Moti
    Mar 9 at 18:26
















0












$begingroup$


From my understanding, a function is said to have a fixed point if $f(x) = x$.



Is there a way for finding how many fixed points a function has?










share|cite|improve this question









New contributor




therewillbecode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    In this level of generality? No. Given a specific function, maybe! Do you have one that you're interested in?
    $endgroup$
    – Billy
    Mar 9 at 17:35










  • $begingroup$
    Brouwer’s Fixed Point Theorem is a result of the existence of a fixed point. So is the Banach Fixed-Point Theorem. But there are some assumptions that go along with them.
    $endgroup$
    – Ekesh Kumar
    Mar 9 at 17:55










  • $begingroup$
    You are asking how one can find all zeros of function $phi$ defined by $phi(x):=f(x)-x$...
    $endgroup$
    – Jean Marie
    Mar 9 at 18:24










  • $begingroup$
    Basically you need to solve the equation f(x)-x=0. In some cases it is straight forward in some others in more complicated. For example: for x+2 you solve (x+2)-x=0 which tells you there is no such. On the other end x^2 has the solutions x^2-x+0 result x=0 and x=1 as fixed points (not more!).
    $endgroup$
    – Moti
    Mar 9 at 18:26














0












0








0


1



$begingroup$


From my understanding, a function is said to have a fixed point if $f(x) = x$.



Is there a way for finding how many fixed points a function has?










share|cite|improve this question









New contributor




therewillbecode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




From my understanding, a function is said to have a fixed point if $f(x) = x$.



Is there a way for finding how many fixed points a function has?







functions fixed-point-theorems fixedpoints






share|cite|improve this question









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therewillbecode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




therewillbecode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Mar 9 at 17:37









Daniele Tampieri

2,3972922




2,3972922






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asked Mar 9 at 17:33









therewillbecodetherewillbecode

101




101




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New contributor





therewillbecode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






therewillbecode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    In this level of generality? No. Given a specific function, maybe! Do you have one that you're interested in?
    $endgroup$
    – Billy
    Mar 9 at 17:35










  • $begingroup$
    Brouwer’s Fixed Point Theorem is a result of the existence of a fixed point. So is the Banach Fixed-Point Theorem. But there are some assumptions that go along with them.
    $endgroup$
    – Ekesh Kumar
    Mar 9 at 17:55










  • $begingroup$
    You are asking how one can find all zeros of function $phi$ defined by $phi(x):=f(x)-x$...
    $endgroup$
    – Jean Marie
    Mar 9 at 18:24










  • $begingroup$
    Basically you need to solve the equation f(x)-x=0. In some cases it is straight forward in some others in more complicated. For example: for x+2 you solve (x+2)-x=0 which tells you there is no such. On the other end x^2 has the solutions x^2-x+0 result x=0 and x=1 as fixed points (not more!).
    $endgroup$
    – Moti
    Mar 9 at 18:26














  • 1




    $begingroup$
    In this level of generality? No. Given a specific function, maybe! Do you have one that you're interested in?
    $endgroup$
    – Billy
    Mar 9 at 17:35










  • $begingroup$
    Brouwer’s Fixed Point Theorem is a result of the existence of a fixed point. So is the Banach Fixed-Point Theorem. But there are some assumptions that go along with them.
    $endgroup$
    – Ekesh Kumar
    Mar 9 at 17:55










  • $begingroup$
    You are asking how one can find all zeros of function $phi$ defined by $phi(x):=f(x)-x$...
    $endgroup$
    – Jean Marie
    Mar 9 at 18:24










  • $begingroup$
    Basically you need to solve the equation f(x)-x=0. In some cases it is straight forward in some others in more complicated. For example: for x+2 you solve (x+2)-x=0 which tells you there is no such. On the other end x^2 has the solutions x^2-x+0 result x=0 and x=1 as fixed points (not more!).
    $endgroup$
    – Moti
    Mar 9 at 18:26








1




1




$begingroup$
In this level of generality? No. Given a specific function, maybe! Do you have one that you're interested in?
$endgroup$
– Billy
Mar 9 at 17:35




$begingroup$
In this level of generality? No. Given a specific function, maybe! Do you have one that you're interested in?
$endgroup$
– Billy
Mar 9 at 17:35












$begingroup$
Brouwer’s Fixed Point Theorem is a result of the existence of a fixed point. So is the Banach Fixed-Point Theorem. But there are some assumptions that go along with them.
$endgroup$
– Ekesh Kumar
Mar 9 at 17:55




$begingroup$
Brouwer’s Fixed Point Theorem is a result of the existence of a fixed point. So is the Banach Fixed-Point Theorem. But there are some assumptions that go along with them.
$endgroup$
– Ekesh Kumar
Mar 9 at 17:55












$begingroup$
You are asking how one can find all zeros of function $phi$ defined by $phi(x):=f(x)-x$...
$endgroup$
– Jean Marie
Mar 9 at 18:24




$begingroup$
You are asking how one can find all zeros of function $phi$ defined by $phi(x):=f(x)-x$...
$endgroup$
– Jean Marie
Mar 9 at 18:24












$begingroup$
Basically you need to solve the equation f(x)-x=0. In some cases it is straight forward in some others in more complicated. For example: for x+2 you solve (x+2)-x=0 which tells you there is no such. On the other end x^2 has the solutions x^2-x+0 result x=0 and x=1 as fixed points (not more!).
$endgroup$
– Moti
Mar 9 at 18:26




$begingroup$
Basically you need to solve the equation f(x)-x=0. In some cases it is straight forward in some others in more complicated. For example: for x+2 you solve (x+2)-x=0 which tells you there is no such. On the other end x^2 has the solutions x^2-x+0 result x=0 and x=1 as fixed points (not more!).
$endgroup$
– Moti
Mar 9 at 18:26










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