How prove this sequences have limts and find this value?
Clash Royale CLAN TAG#URR8PPP
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let $f:Rlongrightarrow R$,and $f(x)=x-dfracx^22$,and
$$L_1(x)=f(x),L_2(x)=f(f(x)),L_3(x)=f(L_2(x)),cdots,L_n(x)=f(L_n-1(x))$$
and let $$a_n=L_nleft(dfrac17nright)$$
prove that
$$displaystylelim_ntoinftyleft(na_nright)$$ is exsit,and find the value limit.
my idea: I want find this
$$L_n(x)=cdots?$$ But is very ugly,becasuse
$$L_2(x)=f(f(x))=left(x-dfracx^22right)-dfrac(x-dfracx^22)^22=-dfrac18x^4-dfrac12x^3-x^2+x$$
limits recurrence-relations functional-equations
edited Aug 13 '13 at 9:16
Willie Wong
54.4k8104207
asked Aug 13 '13 at 8:56
math110
32.3k454214
 |Â
show 3 more comments
up vote
3
down vote
favorite
let $f:Rlongrightarrow R$,and $f(x)=x-dfracx^22$,and
$$L_1(x)=f(x),L_2(x)=f(f(x)),L_3(x)=f(L_2(x)),cdots,L_n(x)=f(L_n-1(x))$$
and let $$a_n=L_nleft(dfrac17nright)$$
prove that
$$displaystylelim_ntoinftyleft(na_nright)$$ is exsit,and find the value limit.
my idea: I want find this
$$L_n(x)=cdots?$$ But is very ugly,becasuse
$$L_2(x)=f(f(x))=left(x-dfracx^22right)-dfrac(x-dfracx^22)^22=-dfrac18x^4-dfrac12x^3-x^2+x$$
limits recurrence-relations functional-equations
edited Aug 13 '13 at 9:16
Willie Wong
54.4k8104207
asked Aug 13 '13 at 8:56
math110
32.3k454214
2
Source?????????
– Gerry Myerson
Aug 13 '13 at 9:04
You could start looking for equilibrium points by setting $f(x_eq) = x_eq$ and check whether it's stable or not.
– user88595
Aug 13 '13 at 9:10
@GerryMyerson,My frend ask me, I think can find this $L_n(x)$
– math110
Aug 13 '13 at 9:27
1
Writing $L_n(x) = x_n$ we have the recursion $x_n+1=x_n - x_n^2/2$. With the transformation $x_n =: 2y_n$, the recursion becomes $y_n+1=y_n(1 - y_n)$. This is the special case $r=1$ of the en.wikipedia.org/wiki/Logistic_map with $y_n rightarrow 0$.
– gammatester
Aug 13 '13 at 9:47
1
@gammatester It's not quite though because the input value depends on $n$, there's not just one fixed starting value $x_0$ that you iterate off of. Right?
– snulty
Apr 26 '17 at 15:48
 |Â
show 3 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
let $f:Rlongrightarrow R$,and $f(x)=x-dfracx^22$,and
$$L_1(x)=f(x),L_2(x)=f(f(x)),L_3(x)=f(L_2(x)),cdots,L_n(x)=f(L_n-1(x))$$
and let $$a_n=L_nleft(dfrac17nright)$$
prove that
$$displaystylelim_ntoinftyleft(na_nright)$$ is exsit,and find the value limit.
my idea: I want find this
$$L_n(x)=cdots?$$ But is very ugly,becasuse
$$L_2(x)=f(f(x))=left(x-dfracx^22right)-dfrac(x-dfracx^22)^22=-dfrac18x^4-dfrac12x^3-x^2+x$$
limits recurrence-relations functional-equations
edited Aug 13 '13 at 9:16
Willie Wong
54.4k8104207
asked Aug 13 '13 at 8:56
math110
32.3k454214
let $f:Rlongrightarrow R$,and $f(x)=x-dfracx^22$,and
$$L_1(x)=f(x),L_2(x)=f(f(x)),L_3(x)=f(L_2(x)),cdots,L_n(x)=f(L_n-1(x))$$
and let $$a_n=L_nleft(dfrac17nright)$$
prove that
$$displaystylelim_ntoinftyleft(na_nright)$$ is exsit,and find the value limit.
my idea: I want find this
$$L_n(x)=cdots?$$ But is very ugly,becasuse
$$L_2(x)=f(f(x))=left(x-dfracx^22right)-dfrac(x-dfracx^22)^22=-dfrac18x^4-dfrac12x^3-x^2+x$$
limits recurrence-relations functional-equations
limits recurrence-relations functional-equations
edited Aug 13 '13 at 9:16
Willie Wong
54.4k8104207
asked Aug 13 '13 at 8:56
math110
32.3k454214
edited Aug 13 '13 at 9:16
Willie Wong
54.4k8104207
asked Aug 13 '13 at 8:56
math110
32.3k454214
edited Aug 13 '13 at 9:16
Willie Wong
54.4k8104207
edited Aug 13 '13 at 9:16
Willie Wong
54.4k8104207
edited Aug 13 '13 at 9:16
Willie Wong
54.4k8104207
54.4k8104207
asked Aug 13 '13 at 8:56
math110
32.3k454214
asked Aug 13 '13 at 8:56
math110
32.3k454214
asked Aug 13 '13 at 8:56
math110
32.3k454214
32.3k454214
2
Source?????????
– Gerry Myerson
Aug 13 '13 at 9:04
You could start looking for equilibrium points by setting $f(x_eq) = x_eq$ and check whether it's stable or not.
– user88595
Aug 13 '13 at 9:10
@GerryMyerson,My frend ask me, I think can find this $L_n(x)$
– math110
Aug 13 '13 at 9:27
1
Writing $L_n(x) = x_n$ we have the recursion $x_n+1=x_n - x_n^2/2$. With the transformation $x_n =: 2y_n$, the recursion becomes $y_n+1=y_n(1 - y_n)$. This is the special case $r=1$ of the en.wikipedia.org/wiki/Logistic_map with $y_n rightarrow 0$.
– gammatester
Aug 13 '13 at 9:47
1
@gammatester It's not quite though because the input value depends on $n$, there's not just one fixed starting value $x_0$ that you iterate off of. Right?
– snulty
Apr 26 '17 at 15:48
 |Â
show 3 more comments
2
Source?????????
– Gerry Myerson
Aug 13 '13 at 9:04
You could start looking for equilibrium points by setting $f(x_eq) = x_eq$ and check whether it's stable or not.
– user88595
Aug 13 '13 at 9:10
@GerryMyerson,My frend ask me, I think can find this $L_n(x)$
– math110
Aug 13 '13 at 9:27
1
Writing $L_n(x) = x_n$ we have the recursion $x_n+1=x_n - x_n^2/2$. With the transformation $x_n =: 2y_n$, the recursion becomes $y_n+1=y_n(1 - y_n)$. This is the special case $r=1$ of the en.wikipedia.org/wiki/Logistic_map with $y_n rightarrow 0$.
– gammatester
Aug 13 '13 at 9:47
1
@gammatester It's not quite though because the input value depends on $n$, there's not just one fixed starting value $x_0$ that you iterate off of. Right?
– snulty
Apr 26 '17 at 15:48
2
2
Source?????????
– Gerry Myerson
Aug 13 '13 at 9:04
Source?????????
– Gerry Myerson
Aug 13 '13 at 9:04
You could start looking for equilibrium points by setting $f(x_eq) = x_eq$ and check whether it's stable or not.
– user88595
Aug 13 '13 at 9:10
You could start looking for equilibrium points by setting $f(x_eq) = x_eq$ and check whether it's stable or not.
– user88595
Aug 13 '13 at 9:10
@GerryMyerson,My frend ask me, I think can find this $L_n(x)$
– math110
Aug 13 '13 at 9:27
@GerryMyerson,My frend ask me, I think can find this $L_n(x)$
– math110
Aug 13 '13 at 9:27
1
1
Writing $L_n(x) = x_n$ we have the recursion $x_n+1=x_n - x_n^2/2$. With the transformation $x_n =: 2y_n$, the recursion becomes $y_n+1=y_n(1 - y_n)$. This is the special case $r=1$ of the en.wikipedia.org/wiki/Logistic_map with $y_n rightarrow 0$.
– gammatester
Aug 13 '13 at 9:47
Writing $L_n(x) = x_n$ we have the recursion $x_n+1=x_n - x_n^2/2$. With the transformation $x_n =: 2y_n$, the recursion becomes $y_n+1=y_n(1 - y_n)$. This is the special case $r=1$ of the en.wikipedia.org/wiki/Logistic_map with $y_n rightarrow 0$.
– gammatester
Aug 13 '13 at 9:47
1
1
@gammatester It's not quite though because the input value depends on $n$, there's not just one fixed starting value $x_0$ that you iterate off of. Right?
– snulty
Apr 26 '17 at 15:48
@gammatester It's not quite though because the input value depends on $n$, there's not just one fixed starting value $x_0$ that you iterate off of. Right?
– snulty
Apr 26 '17 at 15:48
 |Â
show 3 more comments
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2
Source?????????
– Gerry Myerson
Aug 13 '13 at 9:04
You could start looking for equilibrium points by setting $f(x_eq) = x_eq$ and check whether it's stable or not.
– user88595
Aug 13 '13 at 9:10
@GerryMyerson,My frend ask me, I think can find this $L_n(x)$
– math110
Aug 13 '13 at 9:27
1
Writing $L_n(x) = x_n$ we have the recursion $x_n+1=x_n - x_n^2/2$. With the transformation $x_n =: 2y_n$, the recursion becomes $y_n+1=y_n(1 - y_n)$. This is the special case $r=1$ of the en.wikipedia.org/wiki/Logistic_map with $y_n rightarrow 0$.
– gammatester
Aug 13 '13 at 9:47
1
@gammatester It's not quite though because the input value depends on $n$, there's not just one fixed starting value $x_0$ that you iterate off of. Right?
– snulty
Apr 26 '17 at 15:48