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How to explain powers of $(x+1)^2^n$ appearing in the Babylonian approximation of $sqrt x$?
Clash Royale CLAN TAG#URR8PPP
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I'm working with this iteration used for approximating square roots and trying to see what I can draw out from it, and in doing so I found something very strange that I can't logically explain. I'm looking for any insight into why this is the case or perhaps a proof of it. The iteration is as follows:
$$rho_n+1=frac(rho_n)^2+x2rho_n,rho_0=1$$
which approximates $sqrt x$
I'll list the first four general results here:
$$rho_1=fracx+12$$
$$rho_2=fracx^2+6x+14x+4$$
$$rho_3=fracx^4+28x^3+70x^2+28x+18x^3+56x^2+56x+8$$
$$rho_4=fracx^8+120x^7+1820x^6+8008x^5+12870x^4+8008x^3+1820x^2+120x+116x^7+560x^6+4368x^5+11440x^4+11440x^3+4368x^2+560x+16$$
The property I spotted was that in all cases, the co-efficients of $(x+1)^2^n$ appear in $rho_n$, with the co-efficients of even powers on the numerator and of odd powers on the denominator of $rho_n$, in such a way that they snake through the polynomial fractions. For example
$$(x+1)^8=x^8+8x^7+28x^6+56x^5+70x^4+56x^3+28x^2+8x+1$$ and a comparison with $rho_3$ shows my point nicely.
I can show the general results as sums of multiples of powers of $(x+1)$ and $x$, e.g.
$$rho_2=frac(x+1)^2+4x4(x+1)$$
$$rho_3=frac(x+1)^4+24x(x+1)^2+16x8(x+1)^3+32x(x+1)$$
however I don't know how much use this is in explaining the property I found.
Any ideas will be appreciated.
fixed-point-theorems
edited Jul 14 at 20:48
AccidentalFourierTransform
1,077625
asked Jul 14 at 16:00
Rhys Hughes
3,9281227
add a comment |Â
up vote
4
down vote
favorite
I'm working with this iteration used for approximating square roots and trying to see what I can draw out from it, and in doing so I found something very strange that I can't logically explain. I'm looking for any insight into why this is the case or perhaps a proof of it. The iteration is as follows:
$$rho_n+1=frac(rho_n)^2+x2rho_n,rho_0=1$$
which approximates $sqrt x$
I'll list the first four general results here:
$$rho_1=fracx+12$$
$$rho_2=fracx^2+6x+14x+4$$
$$rho_3=fracx^4+28x^3+70x^2+28x+18x^3+56x^2+56x+8$$
$$rho_4=fracx^8+120x^7+1820x^6+8008x^5+12870x^4+8008x^3+1820x^2+120x+116x^7+560x^6+4368x^5+11440x^4+11440x^3+4368x^2+560x+16$$
The property I spotted was that in all cases, the co-efficients of $(x+1)^2^n$ appear in $rho_n$, with the co-efficients of even powers on the numerator and of odd powers on the denominator of $rho_n$, in such a way that they snake through the polynomial fractions. For example
$$(x+1)^8=x^8+8x^7+28x^6+56x^5+70x^4+56x^3+28x^2+8x+1$$ and a comparison with $rho_3$ shows my point nicely.
I can show the general results as sums of multiples of powers of $(x+1)$ and $x$, e.g.
$$rho_2=frac(x+1)^2+4x4(x+1)$$
$$rho_3=frac(x+1)^4+24x(x+1)^2+16x8(x+1)^3+32x(x+1)$$
however I don't know how much use this is in explaining the property I found.
Any ideas will be appreciated.
fixed-point-theorems
edited Jul 14 at 20:48
AccidentalFourierTransform
1,077625
asked Jul 14 at 16:00
Rhys Hughes
3,9281227
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm working with this iteration used for approximating square roots and trying to see what I can draw out from it, and in doing so I found something very strange that I can't logically explain. I'm looking for any insight into why this is the case or perhaps a proof of it. The iteration is as follows:
$$rho_n+1=frac(rho_n)^2+x2rho_n,rho_0=1$$
which approximates $sqrt x$
I'll list the first four general results here:
$$rho_1=fracx+12$$
$$rho_2=fracx^2+6x+14x+4$$
$$rho_3=fracx^4+28x^3+70x^2+28x+18x^3+56x^2+56x+8$$
$$rho_4=fracx^8+120x^7+1820x^6+8008x^5+12870x^4+8008x^3+1820x^2+120x+116x^7+560x^6+4368x^5+11440x^4+11440x^3+4368x^2+560x+16$$
The property I spotted was that in all cases, the co-efficients of $(x+1)^2^n$ appear in $rho_n$, with the co-efficients of even powers on the numerator and of odd powers on the denominator of $rho_n$, in such a way that they snake through the polynomial fractions. For example
$$(x+1)^8=x^8+8x^7+28x^6+56x^5+70x^4+56x^3+28x^2+8x+1$$ and a comparison with $rho_3$ shows my point nicely.
I can show the general results as sums of multiples of powers of $(x+1)$ and $x$, e.g.
$$rho_2=frac(x+1)^2+4x4(x+1)$$
$$rho_3=frac(x+1)^4+24x(x+1)^2+16x8(x+1)^3+32x(x+1)$$
however I don't know how much use this is in explaining the property I found.
Any ideas will be appreciated.
fixed-point-theorems
edited Jul 14 at 20:48
AccidentalFourierTransform
1,077625
asked Jul 14 at 16:00
Rhys Hughes
3,9281227
I'm working with this iteration used for approximating square roots and trying to see what I can draw out from it, and in doing so I found something very strange that I can't logically explain. I'm looking for any insight into why this is the case or perhaps a proof of it. The iteration is as follows:
$$rho_n+1=frac(rho_n)^2+x2rho_n,rho_0=1$$
which approximates $sqrt x$
I'll list the first four general results here:
$$rho_1=fracx+12$$
$$rho_2=fracx^2+6x+14x+4$$
$$rho_3=fracx^4+28x^3+70x^2+28x+18x^3+56x^2+56x+8$$
$$rho_4=fracx^8+120x^7+1820x^6+8008x^5+12870x^4+8008x^3+1820x^2+120x+116x^7+560x^6+4368x^5+11440x^4+11440x^3+4368x^2+560x+16$$
The property I spotted was that in all cases, the co-efficients of $(x+1)^2^n$ appear in $rho_n$, with the co-efficients of even powers on the numerator and of odd powers on the denominator of $rho_n$, in such a way that they snake through the polynomial fractions. For example
$$(x+1)^8=x^8+8x^7+28x^6+56x^5+70x^4+56x^3+28x^2+8x+1$$ and a comparison with $rho_3$ shows my point nicely.
I can show the general results as sums of multiples of powers of $(x+1)$ and $x$, e.g.
$$rho_2=frac(x+1)^2+4x4(x+1)$$
$$rho_3=frac(x+1)^4+24x(x+1)^2+16x8(x+1)^3+32x(x+1)$$
however I don't know how much use this is in explaining the property I found.
Any ideas will be appreciated.
fixed-point-theorems
edited Jul 14 at 20:48
AccidentalFourierTransform
1,077625
asked Jul 14 at 16:00
Rhys Hughes
3,9281227
edited Jul 14 at 20:48
AccidentalFourierTransform
1,077625
edited Jul 14 at 20:48
AccidentalFourierTransform
1,077625
edited Jul 14 at 20:48
AccidentalFourierTransform
1,077625
1,077625
asked Jul 14 at 16:00
Rhys Hughes
3,9281227
asked Jul 14 at 16:00
Rhys Hughes
3,9281227
asked Jul 14 at 16:00
Rhys Hughes
3,9281227
3,9281227
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Possible fixed points of the iteration are $sqrt x$ and $-sqrt x$,
this suggests to look at
$$
fracrho_n+1 - sqrt xrho_n+1 + sqrt x =
fracrho_n^2 - 2 sqrt x rho_n + xrho_n^2 + 2 sqrt x rho_n + x =
left( fracrho_n - sqrt xrho_n + sqrt x right)^2 , .
$$
It follows that
$$
fracrho_n - sqrt xrho_n + sqrt x = left( frac1 - sqrt x1 + sqrt x right)^2^n , .
$$
This implies (quadratic) convergence of $rho_n$ to $sqrt x$ because
the fraction on the right-hand side is of absolute value less than one
if $x > 0$.
It also gives the explicit formula
$$
rho_n=frac(sqrt x+1)^2^n+(sqrt x-1)^2^n
(sqrt x+1)^2^n-(sqrt x-1)^2^nsqrt x
$$
which Lord Shark the Unknown found.
If we expand all expressions using the binomial formula then
all odd powers of $sqrt x$ cancel in the numerator, and all even
powers of $sqrt x$ cancel in the denominator. This gives the
expression (first found by achille hui in a now deleted answer):
$$
rho_n = fracsum_k=0^2^n-1 binom2^n2k x^ksum_k=0^2^n-1binom2^n2k+1 x^k
$$
where the coefficients of the binomial expansion of $(x+1)^2^n$
appear alternatingly in the numerator and denominator.
Note also that this is exactly Newton's method to find a zero
of $f(rho) = rho^2 - x$:
$$
rho_n+1 = rho_n - fracf(rho_n)f'(rho_n) =
rho_n - fracrho_n^2 - x2 rho_n = fracrho_n^2 + x2 rho_n , .
$$
answered Jul 14 at 16:29
Martin R
23.9k32743
Hi Martin, many thanks for the answer, but I'm not sure if it does, or exactly how it answers my question. I understand and appreciate that this iteration converges to $sqrt x$ (a quicker way to show this is to show that when $rho_k=sqrt x, rho_k+1=rho_k$ and we have a cut-off point). My question is asking why the co-efficients of the general result of $rho_n$ match those of $(x+1)^2^n$, and I'm not sure if/how your answer provides a solution to that.
– Rhys Hughes
Jul 14 at 17:49
@RhysHughes: See update. The new expression appeared first in an answer from achille hui and I did not want to copy it. Now that answer is deleted, so I have added it here.
– Martin R
Jul 14 at 18:07
add a comment |Â
up vote
4
down vote
From these formulae, it appears that
$$rho_n=frac(sqrt x+1)^2^n+(sqrt x-1)^2^n
(sqrt x+1)^2^n-(sqrt x-1)^2^nsqrt x$$
and once one has that formula, one may verify it by induction.
answered Jul 14 at 16:05
Lord Shark the Unknown
85.9k951112
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Possible fixed points of the iteration are $sqrt x$ and $-sqrt x$,
this suggests to look at
$$
fracrho_n+1 - sqrt xrho_n+1 + sqrt x =
fracrho_n^2 - 2 sqrt x rho_n + xrho_n^2 + 2 sqrt x rho_n + x =
left( fracrho_n - sqrt xrho_n + sqrt x right)^2 , .
$$
It follows that
$$
fracrho_n - sqrt xrho_n + sqrt x = left( frac1 - sqrt x1 + sqrt x right)^2^n , .
$$
This implies (quadratic) convergence of $rho_n$ to $sqrt x$ because
the fraction on the right-hand side is of absolute value less than one
if $x > 0$.
It also gives the explicit formula
$$
rho_n=frac(sqrt x+1)^2^n+(sqrt x-1)^2^n
(sqrt x+1)^2^n-(sqrt x-1)^2^nsqrt x
$$
which Lord Shark the Unknown found.
If we expand all expressions using the binomial formula then
all odd powers of $sqrt x$ cancel in the numerator, and all even
powers of $sqrt x$ cancel in the denominator. This gives the
expression (first found by achille hui in a now deleted answer):
$$
rho_n = fracsum_k=0^2^n-1 binom2^n2k x^ksum_k=0^2^n-1binom2^n2k+1 x^k
$$
where the coefficients of the binomial expansion of $(x+1)^2^n$
appear alternatingly in the numerator and denominator.
Note also that this is exactly Newton's method to find a zero
of $f(rho) = rho^2 - x$:
$$
rho_n+1 = rho_n - fracf(rho_n)f'(rho_n) =
rho_n - fracrho_n^2 - x2 rho_n = fracrho_n^2 + x2 rho_n , .
$$
answered Jul 14 at 16:29
Martin R
23.9k32743
Hi Martin, many thanks for the answer, but I'm not sure if it does, or exactly how it answers my question. I understand and appreciate that this iteration converges to $sqrt x$ (a quicker way to show this is to show that when $rho_k=sqrt x, rho_k+1=rho_k$ and we have a cut-off point). My question is asking why the co-efficients of the general result of $rho_n$ match those of $(x+1)^2^n$, and I'm not sure if/how your answer provides a solution to that.
– Rhys Hughes
Jul 14 at 17:49
@RhysHughes: See update. The new expression appeared first in an answer from achille hui and I did not want to copy it. Now that answer is deleted, so I have added it here.
– Martin R
Jul 14 at 18:07
add a comment |Â
up vote
3
down vote
accepted
Possible fixed points of the iteration are $sqrt x$ and $-sqrt x$,
this suggests to look at
$$
fracrho_n+1 - sqrt xrho_n+1 + sqrt x =
fracrho_n^2 - 2 sqrt x rho_n + xrho_n^2 + 2 sqrt x rho_n + x =
left( fracrho_n - sqrt xrho_n + sqrt x right)^2 , .
$$
It follows that
$$
fracrho_n - sqrt xrho_n + sqrt x = left( frac1 - sqrt x1 + sqrt x right)^2^n , .
$$
This implies (quadratic) convergence of $rho_n$ to $sqrt x$ because
the fraction on the right-hand side is of absolute value less than one
if $x > 0$.
It also gives the explicit formula
$$
rho_n=frac(sqrt x+1)^2^n+(sqrt x-1)^2^n
(sqrt x+1)^2^n-(sqrt x-1)^2^nsqrt x
$$
which Lord Shark the Unknown found.
If we expand all expressions using the binomial formula then
all odd powers of $sqrt x$ cancel in the numerator, and all even
powers of $sqrt x$ cancel in the denominator. This gives the
expression (first found by achille hui in a now deleted answer):
$$
rho_n = fracsum_k=0^2^n-1 binom2^n2k x^ksum_k=0^2^n-1binom2^n2k+1 x^k
$$
where the coefficients of the binomial expansion of $(x+1)^2^n$
appear alternatingly in the numerator and denominator.
Note also that this is exactly Newton's method to find a zero
of $f(rho) = rho^2 - x$:
$$
rho_n+1 = rho_n - fracf(rho_n)f'(rho_n) =
rho_n - fracrho_n^2 - x2 rho_n = fracrho_n^2 + x2 rho_n , .
$$
answered Jul 14 at 16:29
Martin R
23.9k32743
Hi Martin, many thanks for the answer, but I'm not sure if it does, or exactly how it answers my question. I understand and appreciate that this iteration converges to $sqrt x$ (a quicker way to show this is to show that when $rho_k=sqrt x, rho_k+1=rho_k$ and we have a cut-off point). My question is asking why the co-efficients of the general result of $rho_n$ match those of $(x+1)^2^n$, and I'm not sure if/how your answer provides a solution to that.
– Rhys Hughes
Jul 14 at 17:49
@RhysHughes: See update. The new expression appeared first in an answer from achille hui and I did not want to copy it. Now that answer is deleted, so I have added it here.
– Martin R
Jul 14 at 18:07
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Possible fixed points of the iteration are $sqrt x$ and $-sqrt x$,
this suggests to look at
$$
fracrho_n+1 - sqrt xrho_n+1 + sqrt x =
fracrho_n^2 - 2 sqrt x rho_n + xrho_n^2 + 2 sqrt x rho_n + x =
left( fracrho_n - sqrt xrho_n + sqrt x right)^2 , .
$$
It follows that
$$
fracrho_n - sqrt xrho_n + sqrt x = left( frac1 - sqrt x1 + sqrt x right)^2^n , .
$$
This implies (quadratic) convergence of $rho_n$ to $sqrt x$ because
the fraction on the right-hand side is of absolute value less than one
if $x > 0$.
It also gives the explicit formula
$$
rho_n=frac(sqrt x+1)^2^n+(sqrt x-1)^2^n
(sqrt x+1)^2^n-(sqrt x-1)^2^nsqrt x
$$
which Lord Shark the Unknown found.
If we expand all expressions using the binomial formula then
all odd powers of $sqrt x$ cancel in the numerator, and all even
powers of $sqrt x$ cancel in the denominator. This gives the
expression (first found by achille hui in a now deleted answer):
$$
rho_n = fracsum_k=0^2^n-1 binom2^n2k x^ksum_k=0^2^n-1binom2^n2k+1 x^k
$$
where the coefficients of the binomial expansion of $(x+1)^2^n$
appear alternatingly in the numerator and denominator.
Note also that this is exactly Newton's method to find a zero
of $f(rho) = rho^2 - x$:
$$
rho_n+1 = rho_n - fracf(rho_n)f'(rho_n) =
rho_n - fracrho_n^2 - x2 rho_n = fracrho_n^2 + x2 rho_n , .
$$
answered Jul 14 at 16:29
Martin R
23.9k32743
Possible fixed points of the iteration are $sqrt x$ and $-sqrt x$,
this suggests to look at
$$
fracrho_n+1 - sqrt xrho_n+1 + sqrt x =
fracrho_n^2 - 2 sqrt x rho_n + xrho_n^2 + 2 sqrt x rho_n + x =
left( fracrho_n - sqrt xrho_n + sqrt x right)^2 , .
$$
It follows that
$$
fracrho_n - sqrt xrho_n + sqrt x = left( frac1 - sqrt x1 + sqrt x right)^2^n , .
$$
This implies (quadratic) convergence of $rho_n$ to $sqrt x$ because
the fraction on the right-hand side is of absolute value less than one
if $x > 0$.
It also gives the explicit formula
$$
rho_n=frac(sqrt x+1)^2^n+(sqrt x-1)^2^n
(sqrt x+1)^2^n-(sqrt x-1)^2^nsqrt x
$$
which Lord Shark the Unknown found.
If we expand all expressions using the binomial formula then
all odd powers of $sqrt x$ cancel in the numerator, and all even
powers of $sqrt x$ cancel in the denominator. This gives the
expression (first found by achille hui in a now deleted answer):
$$
rho_n = fracsum_k=0^2^n-1 binom2^n2k x^ksum_k=0^2^n-1binom2^n2k+1 x^k
$$
where the coefficients of the binomial expansion of $(x+1)^2^n$
appear alternatingly in the numerator and denominator.
Note also that this is exactly Newton's method to find a zero
of $f(rho) = rho^2 - x$:
$$
rho_n+1 = rho_n - fracf(rho_n)f'(rho_n) =
rho_n - fracrho_n^2 - x2 rho_n = fracrho_n^2 + x2 rho_n , .
$$
answered Jul 14 at 16:29
Martin R
23.9k32743
edited Jul 14 at 18:05
answered Jul 14 at 16:29
Martin R
23.9k32743
answered Jul 14 at 16:29
Martin R
23.9k32743
answered Jul 14 at 16:29
Martin R
23.9k32743
23.9k32743
Hi Martin, many thanks for the answer, but I'm not sure if it does, or exactly how it answers my question. I understand and appreciate that this iteration converges to $sqrt x$ (a quicker way to show this is to show that when $rho_k=sqrt x, rho_k+1=rho_k$ and we have a cut-off point). My question is asking why the co-efficients of the general result of $rho_n$ match those of $(x+1)^2^n$, and I'm not sure if/how your answer provides a solution to that.
– Rhys Hughes
Jul 14 at 17:49
@RhysHughes: See update. The new expression appeared first in an answer from achille hui and I did not want to copy it. Now that answer is deleted, so I have added it here.
– Martin R
Jul 14 at 18:07
add a comment |Â
Hi Martin, many thanks for the answer, but I'm not sure if it does, or exactly how it answers my question. I understand and appreciate that this iteration converges to $sqrt x$ (a quicker way to show this is to show that when $rho_k=sqrt x, rho_k+1=rho_k$ and we have a cut-off point). My question is asking why the co-efficients of the general result of $rho_n$ match those of $(x+1)^2^n$, and I'm not sure if/how your answer provides a solution to that.
– Rhys Hughes
Jul 14 at 17:49
@RhysHughes: See update. The new expression appeared first in an answer from achille hui and I did not want to copy it. Now that answer is deleted, so I have added it here.
– Martin R
Jul 14 at 18:07
Hi Martin, many thanks for the answer, but I'm not sure if it does, or exactly how it answers my question. I understand and appreciate that this iteration converges to $sqrt x$ (a quicker way to show this is to show that when $rho_k=sqrt x, rho_k+1=rho_k$ and we have a cut-off point). My question is asking why the co-efficients of the general result of $rho_n$ match those of $(x+1)^2^n$, and I'm not sure if/how your answer provides a solution to that.
– Rhys Hughes
Jul 14 at 17:49
Hi Martin, many thanks for the answer, but I'm not sure if it does, or exactly how it answers my question. I understand and appreciate that this iteration converges to $sqrt x$ (a quicker way to show this is to show that when $rho_k=sqrt x, rho_k+1=rho_k$ and we have a cut-off point). My question is asking why the co-efficients of the general result of $rho_n$ match those of $(x+1)^2^n$, and I'm not sure if/how your answer provides a solution to that.
– Rhys Hughes
Jul 14 at 17:49
@RhysHughes: See update. The new expression appeared first in an answer from achille hui and I did not want to copy it. Now that answer is deleted, so I have added it here.
– Martin R
Jul 14 at 18:07
@RhysHughes: See update. The new expression appeared first in an answer from achille hui and I did not want to copy it. Now that answer is deleted, so I have added it here.
– Martin R
Jul 14 at 18:07
add a comment |Â
up vote
4
down vote
From these formulae, it appears that
$$rho_n=frac(sqrt x+1)^2^n+(sqrt x-1)^2^n
(sqrt x+1)^2^n-(sqrt x-1)^2^nsqrt x$$
and once one has that formula, one may verify it by induction.
answered Jul 14 at 16:05
Lord Shark the Unknown
85.9k951112
add a comment |Â
up vote
4
down vote
From these formulae, it appears that
$$rho_n=frac(sqrt x+1)^2^n+(sqrt x-1)^2^n
(sqrt x+1)^2^n-(sqrt x-1)^2^nsqrt x$$
and once one has that formula, one may verify it by induction.
answered Jul 14 at 16:05
Lord Shark the Unknown
85.9k951112
add a comment |Â
up vote
4
down vote
up vote
4
down vote
From these formulae, it appears that
$$rho_n=frac(sqrt x+1)^2^n+(sqrt x-1)^2^n
(sqrt x+1)^2^n-(sqrt x-1)^2^nsqrt x$$
and once one has that formula, one may verify it by induction.
answered Jul 14 at 16:05
Lord Shark the Unknown
85.9k951112
From these formulae, it appears that
$$rho_n=frac(sqrt x+1)^2^n+(sqrt x-1)^2^n
(sqrt x+1)^2^n-(sqrt x-1)^2^nsqrt x$$
and once one has that formula, one may verify it by induction.
answered Jul 14 at 16:05
Lord Shark the Unknown
85.9k951112
answered Jul 14 at 16:05
Lord Shark the Unknown
85.9k951112
answered Jul 14 at 16:05
Lord Shark the Unknown
85.9k951112
answered Jul 14 at 16:05
Lord Shark the Unknown
85.9k951112
85.9k951112
add a comment |Â
add a comment |Â
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