Calculate $ lim_{n to infty }a_n$
$begingroup$
Given $c>0$, $ a_1=frac{c}{2}$, $a_{n+1}=frac{c}{2}+frac{a_n^2}{2}$ $n=1,2,...$ Prove that$$ lim_{n to infty }a_n=begin{cases}
1-sqrt{1-c} & 0<c leq1 \
+infty & c>1
end{cases}
$$
What if $-3leq c<0$?
I'm wondering if there is clever ways to calculate the limit? As I think this can be done by showing the sequence is monotone and bounded/unbounded and then calculate the limit. For example,
Here $a_{n+1}=frac{c}{2}+frac{a_n^2}{2}>sqrt{ca_n^2}=sqrt{c}a_n>a_n $ if $c>1$. So this case, I think $a_n$ is increasing and diverge. I'm wondering other methods of calculating the limit.
sequences-and-series analysis
asked Dec 14 '18 at 3:15
nafhgoodnafhgood
1,805422
$endgroup$
add a comment |
$begingroup$
Given $c>0$, $ a_1=frac{c}{2}$, $a_{n+1}=frac{c}{2}+frac{a_n^2}{2}$ $n=1,2,...$ Prove that$$ lim_{n to infty }a_n=begin{cases}
1-sqrt{1-c} & 0<c leq1 \
+infty & c>1
end{cases}
$$
What if $-3leq c<0$?
I'm wondering if there is clever ways to calculate the limit? As I think this can be done by showing the sequence is monotone and bounded/unbounded and then calculate the limit. For example,
Here $a_{n+1}=frac{c}{2}+frac{a_n^2}{2}>sqrt{ca_n^2}=sqrt{c}a_n>a_n $ if $c>1$. So this case, I think $a_n$ is increasing and diverge. I'm wondering other methods of calculating the limit.
sequences-and-series analysis
asked Dec 14 '18 at 3:15
nafhgoodnafhgood
1,805422
$endgroup$
1
$begingroup$
If $a_nto x$, then $$x=dfrac{c}{2}+dfrac{x^2}{2},$$ this is equivalent to $$x^2-2x+c=0.$$ Applying quadratic formula and reducing we get $$x=1pmsqrt{1-c}.$$ If $cleq1$, and by assumption $a_nleq1,$ then $$a_{n+1}=dfrac{c}{2}+dfrac{a_n^2}{2}leqdfrac{1}{2}+dfrac{1}{2}=1,$$ so by order limit properties $xleq 1<1+sqrt{1-c},$ hence $x=1-sqrt{1-c}.$ This tells us for $cleq1$ it is enough to show that $a_n$ converges.
$endgroup$
– Melody
Dec 14 '18 at 3:42
$begingroup$
and when $3<c<0$, then the sequence has two converging sub-sequences converging to different limit.
$endgroup$
– nafhgood
Dec 14 '18 at 3:59
add a comment |
$begingroup$
Given $c>0$, $ a_1=frac{c}{2}$, $a_{n+1}=frac{c}{2}+frac{a_n^2}{2}$ $n=1,2,...$ Prove that$$ lim_{n to infty }a_n=begin{cases}
1-sqrt{1-c} & 0<c leq1 \
+infty & c>1
end{cases}
$$
What if $-3leq c<0$?
I'm wondering if there is clever ways to calculate the limit? As I think this can be done by showing the sequence is monotone and bounded/unbounded and then calculate the limit. For example,
Here $a_{n+1}=frac{c}{2}+frac{a_n^2}{2}>sqrt{ca_n^2}=sqrt{c}a_n>a_n $ if $c>1$. So this case, I think $a_n$ is increasing and diverge. I'm wondering other methods of calculating the limit.
sequences-and-series analysis
asked Dec 14 '18 at 3:15
nafhgoodnafhgood
1,805422
$endgroup$
Given $c>0$, $ a_1=frac{c}{2}$, $a_{n+1}=frac{c}{2}+frac{a_n^2}{2}$ $n=1,2,...$ Prove that$$ lim_{n to infty }a_n=begin{cases}
1-sqrt{1-c} & 0<c leq1 \
+infty & c>1
end{cases}
$$
What if $-3leq c<0$?
I'm wondering if there is clever ways to calculate the limit? As I think this can be done by showing the sequence is monotone and bounded/unbounded and then calculate the limit. For example,
Here $a_{n+1}=frac{c}{2}+frac{a_n^2}{2}>sqrt{ca_n^2}=sqrt{c}a_n>a_n $ if $c>1$. So this case, I think $a_n$ is increasing and diverge. I'm wondering other methods of calculating the limit.
sequences-and-series analysis
sequences-and-series analysis
asked Dec 14 '18 at 3:15
nafhgoodnafhgood
1,805422
asked Dec 14 '18 at 3:15
nafhgoodnafhgood
1,805422
asked Dec 14 '18 at 3:15
nafhgoodnafhgood
1,805422
asked Dec 14 '18 at 3:15
nafhgoodnafhgood
1,805422
asked Dec 14 '18 at 3:15
nafhgoodnafhgood
1,805422
1,805422
1
$begingroup$
If $a_nto x$, then $$x=dfrac{c}{2}+dfrac{x^2}{2},$$ this is equivalent to $$x^2-2x+c=0.$$ Applying quadratic formula and reducing we get $$x=1pmsqrt{1-c}.$$ If $cleq1$, and by assumption $a_nleq1,$ then $$a_{n+1}=dfrac{c}{2}+dfrac{a_n^2}{2}leqdfrac{1}{2}+dfrac{1}{2}=1,$$ so by order limit properties $xleq 1<1+sqrt{1-c},$ hence $x=1-sqrt{1-c}.$ This tells us for $cleq1$ it is enough to show that $a_n$ converges.
$endgroup$
– Melody
Dec 14 '18 at 3:42
$begingroup$
and when $3<c<0$, then the sequence has two converging sub-sequences converging to different limit.
$endgroup$
– nafhgood
Dec 14 '18 at 3:59
add a comment |
1
$begingroup$
If $a_nto x$, then $$x=dfrac{c}{2}+dfrac{x^2}{2},$$ this is equivalent to $$x^2-2x+c=0.$$ Applying quadratic formula and reducing we get $$x=1pmsqrt{1-c}.$$ If $cleq1$, and by assumption $a_nleq1,$ then $$a_{n+1}=dfrac{c}{2}+dfrac{a_n^2}{2}leqdfrac{1}{2}+dfrac{1}{2}=1,$$ so by order limit properties $xleq 1<1+sqrt{1-c},$ hence $x=1-sqrt{1-c}.$ This tells us for $cleq1$ it is enough to show that $a_n$ converges.
$endgroup$
– Melody
Dec 14 '18 at 3:42
$begingroup$
and when $3<c<0$, then the sequence has two converging sub-sequences converging to different limit.
$endgroup$
– nafhgood
Dec 14 '18 at 3:59
1
1
$begingroup$
If $a_nto x$, then $$x=dfrac{c}{2}+dfrac{x^2}{2},$$ this is equivalent to $$x^2-2x+c=0.$$ Applying quadratic formula and reducing we get $$x=1pmsqrt{1-c}.$$ If $cleq1$, and by assumption $a_nleq1,$ then $$a_{n+1}=dfrac{c}{2}+dfrac{a_n^2}{2}leqdfrac{1}{2}+dfrac{1}{2}=1,$$ so by order limit properties $xleq 1<1+sqrt{1-c},$ hence $x=1-sqrt{1-c}.$ This tells us for $cleq1$ it is enough to show that $a_n$ converges.
$endgroup$
– Melody
Dec 14 '18 at 3:42
$begingroup$
If $a_nto x$, then $$x=dfrac{c}{2}+dfrac{x^2}{2},$$ this is equivalent to $$x^2-2x+c=0.$$ Applying quadratic formula and reducing we get $$x=1pmsqrt{1-c}.$$ If $cleq1$, and by assumption $a_nleq1,$ then $$a_{n+1}=dfrac{c}{2}+dfrac{a_n^2}{2}leqdfrac{1}{2}+dfrac{1}{2}=1,$$ so by order limit properties $xleq 1<1+sqrt{1-c},$ hence $x=1-sqrt{1-c}.$ This tells us for $cleq1$ it is enough to show that $a_n$ converges.
$endgroup$
– Melody
Dec 14 '18 at 3:42
$begingroup$
and when $3<c<0$, then the sequence has two converging sub-sequences converging to different limit.
$endgroup$
– nafhgood
Dec 14 '18 at 3:59
$begingroup$
and when $3<c<0$, then the sequence has two converging sub-sequences converging to different limit.
$endgroup$
– nafhgood
Dec 14 '18 at 3:59
add a comment |
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1
$begingroup$
If $a_nto x$, then $$x=dfrac{c}{2}+dfrac{x^2}{2},$$ this is equivalent to $$x^2-2x+c=0.$$ Applying quadratic formula and reducing we get $$x=1pmsqrt{1-c}.$$ If $cleq1$, and by assumption $a_nleq1,$ then $$a_{n+1}=dfrac{c}{2}+dfrac{a_n^2}{2}leqdfrac{1}{2}+dfrac{1}{2}=1,$$ so by order limit properties $xleq 1<1+sqrt{1-c},$ hence $x=1-sqrt{1-c}.$ This tells us for $cleq1$ it is enough to show that $a_n$ converges.
$endgroup$
– Melody
Dec 14 '18 at 3:42
$begingroup$
and when $3<c<0$, then the sequence has two converging sub-sequences converging to different limit.
$endgroup$
– nafhgood
Dec 14 '18 at 3:59