How to solve $(…((a - b) cdot y - b) cdot y ldots) cdot y geq 0$ if there are $n$ subtractions and...
$begingroup$
It's a programming task, but I'd like to know if there is a way to solve it mathematically.
I need to check $(...((a - b) cdot y - b) cdot y ldots) cdot y geq 0$ if we subtract $b$ and multiply on $y$, $n$ times.
I tried to represent it as
$$a cdot y^n geq b cdot frac{y^n - 1}{y - 1},$$
but I would like to get rid of big numbers such as $y^n$, because these are incomparable in terms of programming if $n leq 1 times 10^{18}$ and $y leq 1 times 10^{2}$. Do you have any idea how can it be represented differently?
inequality
edited Jan 22 at 20:07
EdOverflow
25119
asked Jan 22 at 19:14
Aut RerumAut Rerum
61
$endgroup$
add a comment |
$begingroup$
It's a programming task, but I'd like to know if there is a way to solve it mathematically.
I need to check $(...((a - b) cdot y - b) cdot y ldots) cdot y geq 0$ if we subtract $b$ and multiply on $y$, $n$ times.
I tried to represent it as
$$a cdot y^n geq b cdot frac{y^n - 1}{y - 1},$$
but I would like to get rid of big numbers such as $y^n$, because these are incomparable in terms of programming if $n leq 1 times 10^{18}$ and $y leq 1 times 10^{2}$. Do you have any idea how can it be represented differently?
inequality
edited Jan 22 at 20:07
EdOverflow
25119
asked Jan 22 at 19:14
Aut RerumAut Rerum
61
$endgroup$
$begingroup$
Well, if $y>0$, divide both sides by $y^n$ and approximate $1/y^n to 0$…
$endgroup$
– Macavity
Jan 22 at 20:04
$begingroup$
the equation corresponds to a continous (even differentiable) function being less or equal zero. you could try to calculate the boundary (=0) and either by between value or by differential calculus check which side of a point on the boundary is inner/outer of your solution set.
$endgroup$
– Max
Jan 22 at 20:16
add a comment |
$begingroup$
It's a programming task, but I'd like to know if there is a way to solve it mathematically.
I need to check $(...((a - b) cdot y - b) cdot y ldots) cdot y geq 0$ if we subtract $b$ and multiply on $y$, $n$ times.
I tried to represent it as
$$a cdot y^n geq b cdot frac{y^n - 1}{y - 1},$$
but I would like to get rid of big numbers such as $y^n$, because these are incomparable in terms of programming if $n leq 1 times 10^{18}$ and $y leq 1 times 10^{2}$. Do you have any idea how can it be represented differently?
inequality
edited Jan 22 at 20:07
EdOverflow
25119
asked Jan 22 at 19:14
Aut RerumAut Rerum
61
$endgroup$
It's a programming task, but I'd like to know if there is a way to solve it mathematically.
I need to check $(...((a - b) cdot y - b) cdot y ldots) cdot y geq 0$ if we subtract $b$ and multiply on $y$, $n$ times.
I tried to represent it as
$$a cdot y^n geq b cdot frac{y^n - 1}{y - 1},$$
but I would like to get rid of big numbers such as $y^n$, because these are incomparable in terms of programming if $n leq 1 times 10^{18}$ and $y leq 1 times 10^{2}$. Do you have any idea how can it be represented differently?
inequality
inequality
edited Jan 22 at 20:07
EdOverflow
25119
asked Jan 22 at 19:14
Aut RerumAut Rerum
61
edited Jan 22 at 20:07
EdOverflow
25119
asked Jan 22 at 19:14
Aut RerumAut Rerum
61
edited Jan 22 at 20:07
EdOverflow
25119
edited Jan 22 at 20:07
EdOverflow
25119
edited Jan 22 at 20:07
EdOverflow
25119
25119
asked Jan 22 at 19:14
Aut RerumAut Rerum
61
asked Jan 22 at 19:14
Aut RerumAut Rerum
61
asked Jan 22 at 19:14
Aut RerumAut Rerum
61
61
$begingroup$
Well, if $y>0$, divide both sides by $y^n$ and approximate $1/y^n to 0$…
$endgroup$
– Macavity
Jan 22 at 20:04
$begingroup$
the equation corresponds to a continous (even differentiable) function being less or equal zero. you could try to calculate the boundary (=0) and either by between value or by differential calculus check which side of a point on the boundary is inner/outer of your solution set.
$endgroup$
– Max
Jan 22 at 20:16
add a comment |
$begingroup$
Well, if $y>0$, divide both sides by $y^n$ and approximate $1/y^n to 0$…
$endgroup$
– Macavity
Jan 22 at 20:04
$begingroup$
the equation corresponds to a continous (even differentiable) function being less or equal zero. you could try to calculate the boundary (=0) and either by between value or by differential calculus check which side of a point on the boundary is inner/outer of your solution set.
$endgroup$
– Max
Jan 22 at 20:16
$begingroup$
Well, if $y>0$, divide both sides by $y^n$ and approximate $1/y^n to 0$…
$endgroup$
– Macavity
Jan 22 at 20:04
$begingroup$
Well, if $y>0$, divide both sides by $y^n$ and approximate $1/y^n to 0$…
$endgroup$
– Macavity
Jan 22 at 20:04
$begingroup$
the equation corresponds to a continous (even differentiable) function being less or equal zero. you could try to calculate the boundary (=0) and either by between value or by differential calculus check which side of a point on the boundary is inner/outer of your solution set.
$endgroup$
– Max
Jan 22 at 20:16
$begingroup$
the equation corresponds to a continous (even differentiable) function being less or equal zero. you could try to calculate the boundary (=0) and either by between value or by differential calculus check which side of a point on the boundary is inner/outer of your solution set.
$endgroup$
– Max
Jan 22 at 20:16
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Your correct expression is $ay^n ge byfrac {y^{n}-1}{y-1}$ because the first term of the geometric series is $by$ and the common ratio is $y$. Now if $n$ is large and $y gt 1$, $y^n$ will be huge, so we should divide it out and get
$$a ge byfrac {1-y^{-n}}{y-1}$$
If you ignore $y^{-n}$ because it is so small compared to $1$, you get
$$a ge byfrac {1}{y-1}$$
which has no especially large terms.
answered Jan 22 at 20:54
Ross MillikanRoss Millikan
298k23198371
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add a comment |
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1 Answer
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$begingroup$
Your correct expression is $ay^n ge byfrac {y^{n}-1}{y-1}$ because the first term of the geometric series is $by$ and the common ratio is $y$. Now if $n$ is large and $y gt 1$, $y^n$ will be huge, so we should divide it out and get
$$a ge byfrac {1-y^{-n}}{y-1}$$
If you ignore $y^{-n}$ because it is so small compared to $1$, you get
$$a ge byfrac {1}{y-1}$$
which has no especially large terms.
answered Jan 22 at 20:54
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
Your correct expression is $ay^n ge byfrac {y^{n}-1}{y-1}$ because the first term of the geometric series is $by$ and the common ratio is $y$. Now if $n$ is large and $y gt 1$, $y^n$ will be huge, so we should divide it out and get
$$a ge byfrac {1-y^{-n}}{y-1}$$
If you ignore $y^{-n}$ because it is so small compared to $1$, you get
$$a ge byfrac {1}{y-1}$$
which has no especially large terms.
answered Jan 22 at 20:54
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
Your correct expression is $ay^n ge byfrac {y^{n}-1}{y-1}$ because the first term of the geometric series is $by$ and the common ratio is $y$. Now if $n$ is large and $y gt 1$, $y^n$ will be huge, so we should divide it out and get
$$a ge byfrac {1-y^{-n}}{y-1}$$
If you ignore $y^{-n}$ because it is so small compared to $1$, you get
$$a ge byfrac {1}{y-1}$$
which has no especially large terms.
answered Jan 22 at 20:54
Ross MillikanRoss Millikan
298k23198371
$endgroup$
Your correct expression is $ay^n ge byfrac {y^{n}-1}{y-1}$ because the first term of the geometric series is $by$ and the common ratio is $y$. Now if $n$ is large and $y gt 1$, $y^n$ will be huge, so we should divide it out and get
$$a ge byfrac {1-y^{-n}}{y-1}$$
If you ignore $y^{-n}$ because it is so small compared to $1$, you get
$$a ge byfrac {1}{y-1}$$
which has no especially large terms.
answered Jan 22 at 20:54
Ross MillikanRoss Millikan
298k23198371
answered Jan 22 at 20:54
Ross MillikanRoss Millikan
298k23198371
answered Jan 22 at 20:54
Ross MillikanRoss Millikan
298k23198371
answered Jan 22 at 20:54
Ross MillikanRoss Millikan
298k23198371
298k23198371
add a comment |
add a comment |
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$begingroup$
Well, if $y>0$, divide both sides by $y^n$ and approximate $1/y^n to 0$…
$endgroup$
– Macavity
Jan 22 at 20:04
$begingroup$
the equation corresponds to a continous (even differentiable) function being less or equal zero. you could try to calculate the boundary (=0) and either by between value or by differential calculus check which side of a point on the boundary is inner/outer of your solution set.
$endgroup$
– Max
Jan 22 at 20:16