Mixing
Prove $x_{n+1} = b + {aover x_n}$ converges and find its limit given $x_1>0, a > 0, b > 0, ninBbb N$
$begingroup$
Given a recurrence relation:
$$
x_{n+1} = b + {aover x_n} \
x_1 > 0 \
a, b > 0 \
nin Bbb N
$$
Prove ${x_n}$ converges and find its limit.
First of all notice that $forall ninBbb N: x_n > 0$. To show the sequence converges it suffices to show it is bounded and monotonic. In case the limit exists then it must equal to one of the fixed points of the recurrence:
$$
x = b + {aover x} \
x^2 = bx + a \
x^2 - bx -a = 0
$$
So the roots are given by:
$$
x = frac{b pm sqrt{b^2 + 4a}}{2}
$$
Since $x_n > 0$ the only possible fixed point is:
$$
x = frac{b + sqrt{b^2 + 4a}}{2}
$$
The whole sequence doesn't seem like a monotone one. But it contains monotone subsequences.
So apparently one has to consider two subsequences $x_{2k}$ and $x_{2k-1}$, then show both of them converge by monotone convergence theorem and then show the limits are equal, which would imply the convergence of $x_n$.
I've evaluated $x_{n+2}$ as:
$$
x_{n+2} = {aover x_{n+1}} + b = {aover {{aover x_{n}} + b}} + b \
= frac{ax_n + ab + b^2 x_n}{a+bx_n}
$$
Now I wan't to show:
$$
m le x_{2k+2} le x_{2k}
$$
Or:
$$
M ge x_{2k+2} ge x_{2k}
$$
which seems to be dependent on the initial conditions.
Where $m$ and $M$ are bounds to be found (are they just equal to the fixed point?). The same is related to odd terms of the sequence. Unfortunately I haven't been able to prove monotonicity of $x_{2k}$ and $x_{2k-1}$. How do I proceed from here?
Also please note this problem is given before the definition of a derivative. So the instrumentation is constrained accordingly.
Thank you!
real-analysis calculus sequences-and-series limits recurrence-relations
asked Jan 30 at 14:01
romanroman
2,43721226
$endgroup$
add a comment |
$begingroup$
Given a recurrence relation:
$$
x_{n+1} = b + {aover x_n} \
x_1 > 0 \
a, b > 0 \
nin Bbb N
$$
Prove ${x_n}$ converges and find its limit.
First of all notice that $forall ninBbb N: x_n > 0$. To show the sequence converges it suffices to show it is bounded and monotonic. In case the limit exists then it must equal to one of the fixed points of the recurrence:
$$
x = b + {aover x} \
x^2 = bx + a \
x^2 - bx -a = 0
$$
So the roots are given by:
$$
x = frac{b pm sqrt{b^2 + 4a}}{2}
$$
Since $x_n > 0$ the only possible fixed point is:
$$
x = frac{b + sqrt{b^2 + 4a}}{2}
$$
The whole sequence doesn't seem like a monotone one. But it contains monotone subsequences.
So apparently one has to consider two subsequences $x_{2k}$ and $x_{2k-1}$, then show both of them converge by monotone convergence theorem and then show the limits are equal, which would imply the convergence of $x_n$.
I've evaluated $x_{n+2}$ as:
$$
x_{n+2} = {aover x_{n+1}} + b = {aover {{aover x_{n}} + b}} + b \
= frac{ax_n + ab + b^2 x_n}{a+bx_n}
$$
Now I wan't to show:
$$
m le x_{2k+2} le x_{2k}
$$
Or:
$$
M ge x_{2k+2} ge x_{2k}
$$
which seems to be dependent on the initial conditions.
Where $m$ and $M$ are bounds to be found (are they just equal to the fixed point?). The same is related to odd terms of the sequence. Unfortunately I haven't been able to prove monotonicity of $x_{2k}$ and $x_{2k-1}$. How do I proceed from here?
Also please note this problem is given before the definition of a derivative. So the instrumentation is constrained accordingly.
Thank you!
real-analysis calculus sequences-and-series limits recurrence-relations
asked Jan 30 at 14:01
romanroman
2,43721226
$endgroup$
$begingroup$
this is not any related, btw, how do you get such questions? Which book are you doing now? :P
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:00
$begingroup$
@AbhasKumarSinha Volume $1$ of $3$ by Lev Kudryavtsev. ISBN 978-5-9221-0306-0. But i'm afraid it is only available in russian language.
$endgroup$
– roman
Jan 30 at 15:06
$begingroup$
I'm just started in High School, My teacher Gifted me "Calculus in One Variable" by IA Maron and "differential calculus for beginners" by Joseph Edwards, Both are awesome. Russians are really genetically genius in Calculus and Mechanics :P :) XD I love their writings and specially Irodov <3
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:09
$begingroup$
This problem can also be treated using generalized continued fractions $b+cfrac{a}{b +cfrac{a}{b +cfrac{a} {b + ddots}}} $
$endgroup$
– Jean Marie
Feb 2 at 14:50
add a comment |
$begingroup$
Given a recurrence relation:
$$
x_{n+1} = b + {aover x_n} \
x_1 > 0 \
a, b > 0 \
nin Bbb N
$$
Prove ${x_n}$ converges and find its limit.
First of all notice that $forall ninBbb N: x_n > 0$. To show the sequence converges it suffices to show it is bounded and monotonic. In case the limit exists then it must equal to one of the fixed points of the recurrence:
$$
x = b + {aover x} \
x^2 = bx + a \
x^2 - bx -a = 0
$$
So the roots are given by:
$$
x = frac{b pm sqrt{b^2 + 4a}}{2}
$$
Since $x_n > 0$ the only possible fixed point is:
$$
x = frac{b + sqrt{b^2 + 4a}}{2}
$$
The whole sequence doesn't seem like a monotone one. But it contains monotone subsequences.
So apparently one has to consider two subsequences $x_{2k}$ and $x_{2k-1}$, then show both of them converge by monotone convergence theorem and then show the limits are equal, which would imply the convergence of $x_n$.
I've evaluated $x_{n+2}$ as:
$$
x_{n+2} = {aover x_{n+1}} + b = {aover {{aover x_{n}} + b}} + b \
= frac{ax_n + ab + b^2 x_n}{a+bx_n}
$$
Now I wan't to show:
$$
m le x_{2k+2} le x_{2k}
$$
Or:
$$
M ge x_{2k+2} ge x_{2k}
$$
which seems to be dependent on the initial conditions.
Where $m$ and $M$ are bounds to be found (are they just equal to the fixed point?). The same is related to odd terms of the sequence. Unfortunately I haven't been able to prove monotonicity of $x_{2k}$ and $x_{2k-1}$. How do I proceed from here?
Also please note this problem is given before the definition of a derivative. So the instrumentation is constrained accordingly.
Thank you!
real-analysis calculus sequences-and-series limits recurrence-relations
asked Jan 30 at 14:01
romanroman
2,43721226
$endgroup$
Given a recurrence relation:
$$
x_{n+1} = b + {aover x_n} \
x_1 > 0 \
a, b > 0 \
nin Bbb N
$$
Prove ${x_n}$ converges and find its limit.
First of all notice that $forall ninBbb N: x_n > 0$. To show the sequence converges it suffices to show it is bounded and monotonic. In case the limit exists then it must equal to one of the fixed points of the recurrence:
$$
x = b + {aover x} \
x^2 = bx + a \
x^2 - bx -a = 0
$$
So the roots are given by:
$$
x = frac{b pm sqrt{b^2 + 4a}}{2}
$$
Since $x_n > 0$ the only possible fixed point is:
$$
x = frac{b + sqrt{b^2 + 4a}}{2}
$$
The whole sequence doesn't seem like a monotone one. But it contains monotone subsequences.
So apparently one has to consider two subsequences $x_{2k}$ and $x_{2k-1}$, then show both of them converge by monotone convergence theorem and then show the limits are equal, which would imply the convergence of $x_n$.
I've evaluated $x_{n+2}$ as:
$$
x_{n+2} = {aover x_{n+1}} + b = {aover {{aover x_{n}} + b}} + b \
= frac{ax_n + ab + b^2 x_n}{a+bx_n}
$$
Now I wan't to show:
$$
m le x_{2k+2} le x_{2k}
$$
Or:
$$
M ge x_{2k+2} ge x_{2k}
$$
which seems to be dependent on the initial conditions.
Where $m$ and $M$ are bounds to be found (are they just equal to the fixed point?). The same is related to odd terms of the sequence. Unfortunately I haven't been able to prove monotonicity of $x_{2k}$ and $x_{2k-1}$. How do I proceed from here?
Also please note this problem is given before the definition of a derivative. So the instrumentation is constrained accordingly.
Thank you!
real-analysis calculus sequences-and-series limits recurrence-relations
real-analysis calculus sequences-and-series limits recurrence-relations
asked Jan 30 at 14:01
romanroman
2,43721226
asked Jan 30 at 14:01
romanroman
2,43721226
edited Jan 30 at 14:14
roman
asked Jan 30 at 14:01
romanroman
2,43721226
asked Jan 30 at 14:01
romanroman
2,43721226
asked Jan 30 at 14:01
romanroman
2,43721226
2,43721226
$begingroup$
this is not any related, btw, how do you get such questions? Which book are you doing now? :P
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:00
$begingroup$
@AbhasKumarSinha Volume $1$ of $3$ by Lev Kudryavtsev. ISBN 978-5-9221-0306-0. But i'm afraid it is only available in russian language.
$endgroup$
– roman
Jan 30 at 15:06
$begingroup$
I'm just started in High School, My teacher Gifted me "Calculus in One Variable" by IA Maron and "differential calculus for beginners" by Joseph Edwards, Both are awesome. Russians are really genetically genius in Calculus and Mechanics :P :) XD I love their writings and specially Irodov <3
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:09
$begingroup$
This problem can also be treated using generalized continued fractions $b+cfrac{a}{b +cfrac{a}{b +cfrac{a} {b + ddots}}} $
$endgroup$
– Jean Marie
Feb 2 at 14:50
add a comment |
$begingroup$
this is not any related, btw, how do you get such questions? Which book are you doing now? :P
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:00
$begingroup$
@AbhasKumarSinha Volume $1$ of $3$ by Lev Kudryavtsev. ISBN 978-5-9221-0306-0. But i'm afraid it is only available in russian language.
$endgroup$
– roman
Jan 30 at 15:06
$begingroup$
I'm just started in High School, My teacher Gifted me "Calculus in One Variable" by IA Maron and "differential calculus for beginners" by Joseph Edwards, Both are awesome. Russians are really genetically genius in Calculus and Mechanics :P :) XD I love their writings and specially Irodov <3
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:09
$begingroup$
This problem can also be treated using generalized continued fractions $b+cfrac{a}{b +cfrac{a}{b +cfrac{a} {b + ddots}}} $
$endgroup$
– Jean Marie
Feb 2 at 14:50
$begingroup$
this is not any related, btw, how do you get such questions? Which book are you doing now? :P
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:00
$begingroup$
this is not any related, btw, how do you get such questions? Which book are you doing now? :P
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:00
$begingroup$
@AbhasKumarSinha Volume $1$ of $3$ by Lev Kudryavtsev. ISBN 978-5-9221-0306-0. But i'm afraid it is only available in russian language.
$endgroup$
– roman
Jan 30 at 15:06
$begingroup$
@AbhasKumarSinha Volume $1$ of $3$ by Lev Kudryavtsev. ISBN 978-5-9221-0306-0. But i'm afraid it is only available in russian language.
$endgroup$
– roman
Jan 30 at 15:06
$begingroup$
I'm just started in High School, My teacher Gifted me "Calculus in One Variable" by IA Maron and "differential calculus for beginners" by Joseph Edwards, Both are awesome. Russians are really genetically genius in Calculus and Mechanics :P :) XD I love their writings and specially Irodov <3
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:09
$begingroup$
I'm just started in High School, My teacher Gifted me "Calculus in One Variable" by IA Maron and "differential calculus for beginners" by Joseph Edwards, Both are awesome. Russians are really genetically genius in Calculus and Mechanics :P :) XD I love their writings and specially Irodov <3
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:09
$begingroup$
This problem can also be treated using generalized continued fractions $b+cfrac{a}{b +cfrac{a}{b +cfrac{a} {b + ddots}}} $
$endgroup$
– Jean Marie
Feb 2 at 14:50
$begingroup$
This problem can also be treated using generalized continued fractions $b+cfrac{a}{b +cfrac{a}{b +cfrac{a} {b + ddots}}} $
$endgroup$
– Jean Marie
Feb 2 at 14:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $M=frac{b+sqrt{b^{2}+4a}}{2}$ and suppose $x_{n}>M$.
Then $b+frac{a}{x_{n}}<b+frac{a}{M}=M$, so $x_{n+1}<M$ and in the same way $b+frac{a}{x_{n+1}}>b+frac{a}{M}=M$ so $x_{n+2}>M$. Thus the elements of the sequence $(x_{n})$ are alternating larger and smaller than $M$.
Now note that $x_{n}^{2}-bx_{n}-a>0$. Thus we find
$$x_{n+2}-x_{n}=frac{ax_{n}+ab+b^{2}x_{n}}{a+bx_{n}}-x_{n}=frac{ab+b^{2}x_{n}-bx_{n}^{2}}{a+bx_{n}}=bfrac{a+bx_{n}-x_{n}^{2}}{a+bx_{n}}.$$
As $b>0$, $a+bx_{n}>0$ and $a+bx_{n}-x_{n}^{2}<0$ we find that $x_{n+2}-x_{n}<0$ and hence $M<x_{n+2}<x_{n}$.
We can use a similar argument for $0<x_{n}<M$ using $x_{n}^{2}-bx_{n}-a<0$ to show that $x_{n}<x_{n+2}<M$. Taking wlog $0<x_{1}<M$ we find that $(x_{2n+1})$ and $(x_{2n})$ are both respectively monotonically increasing and monotonically decreasing bounded sequences converging respectively to some $0<L_{1}leq Mleq L_{2}$.
Finally suppose $L_{1}neq M$. Then
$$0=lim_{nrightarrowinfty}x_{2n+3}-x_{2n+1}=bfrac{a+bx_{2n+1}-x_{2n+1}^{2}}{a+bx_{2n+1}}=bfrac{a+bL_{1}-L_{1}^{2}}{a+bL_{1}}$$
which we know holds only if $L_{1}=M$ as $b>0$ and $a+bL_{1}>0$, which is a contradiction. One can rule out $L_{2}neq M$ similarly.
answered Jan 30 at 16:12
Floris ClaassensFloris Claassens
1,27628
$endgroup$
$begingroup$
Thank you! That is very helpful
$endgroup$
– roman
Jan 31 at 9:40
$begingroup$
There is a very efficient method for "homographic recurrences" using matrices that I sketch in my answer to this question : math.stackexchange.com/q/2445293
$endgroup$
– Jean Marie
Feb 2 at 14:34
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $M=frac{b+sqrt{b^{2}+4a}}{2}$ and suppose $x_{n}>M$.
Then $b+frac{a}{x_{n}}<b+frac{a}{M}=M$, so $x_{n+1}<M$ and in the same way $b+frac{a}{x_{n+1}}>b+frac{a}{M}=M$ so $x_{n+2}>M$. Thus the elements of the sequence $(x_{n})$ are alternating larger and smaller than $M$.
Now note that $x_{n}^{2}-bx_{n}-a>0$. Thus we find
$$x_{n+2}-x_{n}=frac{ax_{n}+ab+b^{2}x_{n}}{a+bx_{n}}-x_{n}=frac{ab+b^{2}x_{n}-bx_{n}^{2}}{a+bx_{n}}=bfrac{a+bx_{n}-x_{n}^{2}}{a+bx_{n}}.$$
As $b>0$, $a+bx_{n}>0$ and $a+bx_{n}-x_{n}^{2}<0$ we find that $x_{n+2}-x_{n}<0$ and hence $M<x_{n+2}<x_{n}$.
We can use a similar argument for $0<x_{n}<M$ using $x_{n}^{2}-bx_{n}-a<0$ to show that $x_{n}<x_{n+2}<M$. Taking wlog $0<x_{1}<M$ we find that $(x_{2n+1})$ and $(x_{2n})$ are both respectively monotonically increasing and monotonically decreasing bounded sequences converging respectively to some $0<L_{1}leq Mleq L_{2}$.
Finally suppose $L_{1}neq M$. Then
$$0=lim_{nrightarrowinfty}x_{2n+3}-x_{2n+1}=bfrac{a+bx_{2n+1}-x_{2n+1}^{2}}{a+bx_{2n+1}}=bfrac{a+bL_{1}-L_{1}^{2}}{a+bL_{1}}$$
which we know holds only if $L_{1}=M$ as $b>0$ and $a+bL_{1}>0$, which is a contradiction. One can rule out $L_{2}neq M$ similarly.
answered Jan 30 at 16:12
Floris ClaassensFloris Claassens
1,27628
$endgroup$
$begingroup$
Thank you! That is very helpful
$endgroup$
– roman
Jan 31 at 9:40
$begingroup$
There is a very efficient method for "homographic recurrences" using matrices that I sketch in my answer to this question : math.stackexchange.com/q/2445293
$endgroup$
– Jean Marie
Feb 2 at 14:34
add a comment |
$begingroup$
Let $M=frac{b+sqrt{b^{2}+4a}}{2}$ and suppose $x_{n}>M$.
Then $b+frac{a}{x_{n}}<b+frac{a}{M}=M$, so $x_{n+1}<M$ and in the same way $b+frac{a}{x_{n+1}}>b+frac{a}{M}=M$ so $x_{n+2}>M$. Thus the elements of the sequence $(x_{n})$ are alternating larger and smaller than $M$.
Now note that $x_{n}^{2}-bx_{n}-a>0$. Thus we find
$$x_{n+2}-x_{n}=frac{ax_{n}+ab+b^{2}x_{n}}{a+bx_{n}}-x_{n}=frac{ab+b^{2}x_{n}-bx_{n}^{2}}{a+bx_{n}}=bfrac{a+bx_{n}-x_{n}^{2}}{a+bx_{n}}.$$
As $b>0$, $a+bx_{n}>0$ and $a+bx_{n}-x_{n}^{2}<0$ we find that $x_{n+2}-x_{n}<0$ and hence $M<x_{n+2}<x_{n}$.
We can use a similar argument for $0<x_{n}<M$ using $x_{n}^{2}-bx_{n}-a<0$ to show that $x_{n}<x_{n+2}<M$. Taking wlog $0<x_{1}<M$ we find that $(x_{2n+1})$ and $(x_{2n})$ are both respectively monotonically increasing and monotonically decreasing bounded sequences converging respectively to some $0<L_{1}leq Mleq L_{2}$.
Finally suppose $L_{1}neq M$. Then
$$0=lim_{nrightarrowinfty}x_{2n+3}-x_{2n+1}=bfrac{a+bx_{2n+1}-x_{2n+1}^{2}}{a+bx_{2n+1}}=bfrac{a+bL_{1}-L_{1}^{2}}{a+bL_{1}}$$
which we know holds only if $L_{1}=M$ as $b>0$ and $a+bL_{1}>0$, which is a contradiction. One can rule out $L_{2}neq M$ similarly.
answered Jan 30 at 16:12
Floris ClaassensFloris Claassens
1,27628
$endgroup$
$begingroup$
Thank you! That is very helpful
$endgroup$
– roman
Jan 31 at 9:40
$begingroup$
There is a very efficient method for "homographic recurrences" using matrices that I sketch in my answer to this question : math.stackexchange.com/q/2445293
$endgroup$
– Jean Marie
Feb 2 at 14:34
add a comment |
$begingroup$
Let $M=frac{b+sqrt{b^{2}+4a}}{2}$ and suppose $x_{n}>M$.
Then $b+frac{a}{x_{n}}<b+frac{a}{M}=M$, so $x_{n+1}<M$ and in the same way $b+frac{a}{x_{n+1}}>b+frac{a}{M}=M$ so $x_{n+2}>M$. Thus the elements of the sequence $(x_{n})$ are alternating larger and smaller than $M$.
Now note that $x_{n}^{2}-bx_{n}-a>0$. Thus we find
$$x_{n+2}-x_{n}=frac{ax_{n}+ab+b^{2}x_{n}}{a+bx_{n}}-x_{n}=frac{ab+b^{2}x_{n}-bx_{n}^{2}}{a+bx_{n}}=bfrac{a+bx_{n}-x_{n}^{2}}{a+bx_{n}}.$$
As $b>0$, $a+bx_{n}>0$ and $a+bx_{n}-x_{n}^{2}<0$ we find that $x_{n+2}-x_{n}<0$ and hence $M<x_{n+2}<x_{n}$.
We can use a similar argument for $0<x_{n}<M$ using $x_{n}^{2}-bx_{n}-a<0$ to show that $x_{n}<x_{n+2}<M$. Taking wlog $0<x_{1}<M$ we find that $(x_{2n+1})$ and $(x_{2n})$ are both respectively monotonically increasing and monotonically decreasing bounded sequences converging respectively to some $0<L_{1}leq Mleq L_{2}$.
Finally suppose $L_{1}neq M$. Then
$$0=lim_{nrightarrowinfty}x_{2n+3}-x_{2n+1}=bfrac{a+bx_{2n+1}-x_{2n+1}^{2}}{a+bx_{2n+1}}=bfrac{a+bL_{1}-L_{1}^{2}}{a+bL_{1}}$$
which we know holds only if $L_{1}=M$ as $b>0$ and $a+bL_{1}>0$, which is a contradiction. One can rule out $L_{2}neq M$ similarly.
answered Jan 30 at 16:12
Floris ClaassensFloris Claassens
1,27628
$endgroup$
Let $M=frac{b+sqrt{b^{2}+4a}}{2}$ and suppose $x_{n}>M$.
Then $b+frac{a}{x_{n}}<b+frac{a}{M}=M$, so $x_{n+1}<M$ and in the same way $b+frac{a}{x_{n+1}}>b+frac{a}{M}=M$ so $x_{n+2}>M$. Thus the elements of the sequence $(x_{n})$ are alternating larger and smaller than $M$.
Now note that $x_{n}^{2}-bx_{n}-a>0$. Thus we find
$$x_{n+2}-x_{n}=frac{ax_{n}+ab+b^{2}x_{n}}{a+bx_{n}}-x_{n}=frac{ab+b^{2}x_{n}-bx_{n}^{2}}{a+bx_{n}}=bfrac{a+bx_{n}-x_{n}^{2}}{a+bx_{n}}.$$
As $b>0$, $a+bx_{n}>0$ and $a+bx_{n}-x_{n}^{2}<0$ we find that $x_{n+2}-x_{n}<0$ and hence $M<x_{n+2}<x_{n}$.
We can use a similar argument for $0<x_{n}<M$ using $x_{n}^{2}-bx_{n}-a<0$ to show that $x_{n}<x_{n+2}<M$. Taking wlog $0<x_{1}<M$ we find that $(x_{2n+1})$ and $(x_{2n})$ are both respectively monotonically increasing and monotonically decreasing bounded sequences converging respectively to some $0<L_{1}leq Mleq L_{2}$.
Finally suppose $L_{1}neq M$. Then
$$0=lim_{nrightarrowinfty}x_{2n+3}-x_{2n+1}=bfrac{a+bx_{2n+1}-x_{2n+1}^{2}}{a+bx_{2n+1}}=bfrac{a+bL_{1}-L_{1}^{2}}{a+bL_{1}}$$
which we know holds only if $L_{1}=M$ as $b>0$ and $a+bL_{1}>0$, which is a contradiction. One can rule out $L_{2}neq M$ similarly.
answered Jan 30 at 16:12
Floris ClaassensFloris Claassens
1,27628
answered Jan 30 at 16:12
Floris ClaassensFloris Claassens
1,27628
answered Jan 30 at 16:12
Floris ClaassensFloris Claassens
1,27628
answered Jan 30 at 16:12
Floris ClaassensFloris Claassens
1,27628
1,27628
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Thank you! That is very helpful
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– roman
Jan 31 at 9:40
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There is a very efficient method for "homographic recurrences" using matrices that I sketch in my answer to this question : math.stackexchange.com/q/2445293
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– Jean Marie
Feb 2 at 14:34
add a comment |
$begingroup$
Thank you! That is very helpful
$endgroup$
– roman
Jan 31 at 9:40
$begingroup$
There is a very efficient method for "homographic recurrences" using matrices that I sketch in my answer to this question : math.stackexchange.com/q/2445293
$endgroup$
– Jean Marie
Feb 2 at 14:34
$begingroup$
Thank you! That is very helpful
$endgroup$
– roman
Jan 31 at 9:40
$begingroup$
Thank you! That is very helpful
$endgroup$
– roman
Jan 31 at 9:40
$begingroup$
There is a very efficient method for "homographic recurrences" using matrices that I sketch in my answer to this question : math.stackexchange.com/q/2445293
$endgroup$
– Jean Marie
Feb 2 at 14:34
$begingroup$
There is a very efficient method for "homographic recurrences" using matrices that I sketch in my answer to this question : math.stackexchange.com/q/2445293
$endgroup$
– Jean Marie
Feb 2 at 14:34
add a comment |
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this is not any related, btw, how do you get such questions? Which book are you doing now? :P
$endgroup$
– Abhas Kumar Sinha
Jan 30 at 15:00
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@AbhasKumarSinha Volume $1$ of $3$ by Lev Kudryavtsev. ISBN 978-5-9221-0306-0. But i'm afraid it is only available in russian language.
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– roman
Jan 30 at 15:06
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I'm just started in High School, My teacher Gifted me "Calculus in One Variable" by IA Maron and "differential calculus for beginners" by Joseph Edwards, Both are awesome. Russians are really genetically genius in Calculus and Mechanics :P :) XD I love their writings and specially Irodov <3
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– Abhas Kumar Sinha
Jan 30 at 15:09
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This problem can also be treated using generalized continued fractions $b+cfrac{a}{b +cfrac{a}{b +cfrac{a} {b + ddots}}} $
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– Jean Marie
Feb 2 at 14:50