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Approximating irrational number to rational number
$lim_ntoinfty f(2^n)$ for some very slowly increasing function $f(n)$Hermite Interpolation of $e^x$. Strange behaviour when increasing the number of derivatives at interpolating points.Newton's Method, and approximating parameters for Bézier curves.Approximating Logs and Antilogs by handApproximating fractionsExistence of Irrational Number that has same $n$ digits of a given Rational Number.Finding Irrational Approximation for a given Rational Number.Atomic weights: rational or irrational?Does there exist infinitely many $mu$ which satisfy this:Approximating functions with rational functions
$begingroup$
I'm making a phone game, and I need to approximate $frac log(5/4)log(3/2)$ to a rational number $p/q$.
I wish $p$ and $q$ small enough. For example, I don't want $p$, $qapprox 10^7$; it's way too much for my code.
In the game, there's two way to upgrade ability. Type A gives additional $50%$ increase at once. and type B gives $25%$.
What I want to know is how many times of upgrade $(x,y)$ provides same additional increase. So what I've done is solve $(3/2)^x = (5/4)^y$ respect to $frac xy$.
Can you provide me way to construct sequence $p_n$, $q_n$ which approximate the real number?
Thank you in advance.
approximation
$endgroup$
add a comment |
$begingroup$
I'm making a phone game, and I need to approximate $frac log(5/4)log(3/2)$ to a rational number $p/q$.
I wish $p$ and $q$ small enough. For example, I don't want $p$, $qapprox 10^7$; it's way too much for my code.
In the game, there's two way to upgrade ability. Type A gives additional $50%$ increase at once. and type B gives $25%$.
What I want to know is how many times of upgrade $(x,y)$ provides same additional increase. So what I've done is solve $(3/2)^x = (5/4)^y$ respect to $frac xy$.
Can you provide me way to construct sequence $p_n$, $q_n$ which approximate the real number?
Thank you in advance.
approximation
$endgroup$
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
$endgroup$
– amsmath
47 mins ago
1
$begingroup$
The continued fraction expansion of your irrational number will produce the best rational approximation subject to a limit on size of the denominator.
$endgroup$
– hardmath
45 mins ago
1
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
45 mins ago
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
42 mins ago
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
33 mins ago
add a comment |
$begingroup$
I'm making a phone game, and I need to approximate $frac log(5/4)log(3/2)$ to a rational number $p/q$.
I wish $p$ and $q$ small enough. For example, I don't want $p$, $qapprox 10^7$; it's way too much for my code.
In the game, there's two way to upgrade ability. Type A gives additional $50%$ increase at once. and type B gives $25%$.
What I want to know is how many times of upgrade $(x,y)$ provides same additional increase. So what I've done is solve $(3/2)^x = (5/4)^y$ respect to $frac xy$.
Can you provide me way to construct sequence $p_n$, $q_n$ which approximate the real number?
Thank you in advance.
approximation
$endgroup$
I'm making a phone game, and I need to approximate $frac log(5/4)log(3/2)$ to a rational number $p/q$.
I wish $p$ and $q$ small enough. For example, I don't want $p$, $qapprox 10^7$; it's way too much for my code.
In the game, there's two way to upgrade ability. Type A gives additional $50%$ increase at once. and type B gives $25%$.
What I want to know is how many times of upgrade $(x,y)$ provides same additional increase. So what I've done is solve $(3/2)^x = (5/4)^y$ respect to $frac xy$.
Can you provide me way to construct sequence $p_n$, $q_n$ which approximate the real number?
Thank you in advance.
approximation
approximation
edited 44 mins ago
Rócherz
2,9863821
2,9863821
asked 1 hour ago
MrTanorusMrTanorus
1928
1928
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
$endgroup$
– amsmath
47 mins ago
1
$begingroup$
The continued fraction expansion of your irrational number will produce the best rational approximation subject to a limit on size of the denominator.
$endgroup$
– hardmath
45 mins ago
1
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
45 mins ago
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
42 mins ago
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
33 mins ago
add a comment |
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
$endgroup$
– amsmath
47 mins ago
1
$begingroup$
The continued fraction expansion of your irrational number will produce the best rational approximation subject to a limit on size of the denominator.
$endgroup$
– hardmath
45 mins ago
1
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
45 mins ago
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
42 mins ago
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
33 mins ago
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
$endgroup$
– amsmath
47 mins ago
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
$endgroup$
– amsmath
47 mins ago
1
1
$begingroup$
The continued fraction expansion of your irrational number will produce the best rational approximation subject to a limit on size of the denominator.
$endgroup$
– hardmath
45 mins ago
$begingroup$
The continued fraction expansion of your irrational number will produce the best rational approximation subject to a limit on size of the denominator.
$endgroup$
– hardmath
45 mins ago
1
1
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
45 mins ago
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
45 mins ago
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
42 mins ago
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
42 mins ago
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
33 mins ago
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
33 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The number you want to approximate is about $0.550339713213$. An excellent approximation is $frac 8911619approx 0.550339715873$. I got that by using the continued fraction. When you see a large value like $143$, truncating before it yields a very good approximation.
$endgroup$
add a comment |
$begingroup$
The continued fraction for $fraclogleft(frac54right)logleft(frac32right)$ is
$$
0;1,1,4,2,6,1,color#C0010,143,3,dots
$$
The convergents for this continued fraction are
$$
left0,1,frac12,frac59,frac1120,frac71129,frac82149,color#C00frac8911619,frac127495231666,frac383376696617,dotsright
$$
As Ross Millikan mentions, stopping just before a large continuant like $143$ gives a particularly good approximation for the size of the denominator; in this case, the approximation $frac8911619$ is closer than $frac1143cdot1619^2$ to $fraclogleft(frac54right)logleft(frac32right)$.
$endgroup$
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
12 mins ago
add a comment |
$begingroup$
Running the extended Euclidean algorithm to find the continued fraction:
$$beginarrayccx&q&a&b\
hline 0.55033971 & & 0 & 1\ 1 & 0 & 1 & 0\ 0.55033971 & 1 & 0 & 1\ 0.44966029 & 1 & 1 & -1 \ 0.10067943 & 4 & -1 & 2\ 0.04694258 & 2 & 5 & -9\ 0.00679426 & 6 & -11 & 20 \ 0.00617700 & 1 & 71 & -129 \ 0.00061727 & 10 & -82 & 149\ 4.31cdot 10^-6 & 143 & 891 & -1619 \
1.25cdot 10^-6 & 3 & -127495 & 231666endarray$$
The $q$ column are the quotients, that go into the continued fraction. The $a$ and $b$ columns track a linear combination of the original two that's equal to $x_n$; for example, $-11cdot 1 + 20cdot fraclog(5/4)log(3/2)approx 0.00679426$. The fraction $left|fraclog(5/4)log(3/2)right|$ is approximated by $fraca_n$, with increasing accuracy.
The formulas for building this table: $q_n = leftlfloor frac x_n-1x_nrightrfloor$, $x_n+1=x_n-1-q_nx_n$, $a_n+1=a_n-1-q_na_n$, $b_n+1=b_n-1-q_nb_n$. Initialize with $x_0=1$, $x_-1$ the quantity we're trying to estimate, $a_-1=b_0=0$, $a_0=b_-1=1$.
If you run the table much large than this, watch for floating-point accuracy issues; once the $x_n$ get down close to the accuracy limit for floating point numbers near zero, you can't trust the quotients anymore.
Now, how that accuracy increases is irregular. Large quotients go with particularly good approximations - see how that quotient of $143$ means that we have to go to six-digit numerator and denominator to do better than that $frac8911619$ approximation.
It is of course a tradeoff between accuracy and how deep you go. For your purposes in costing the two upgrades, I'd probably go with that $frac1120$ approximation.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The number you want to approximate is about $0.550339713213$. An excellent approximation is $frac 8911619approx 0.550339715873$. I got that by using the continued fraction. When you see a large value like $143$, truncating before it yields a very good approximation.
$endgroup$
add a comment |
$begingroup$
The number you want to approximate is about $0.550339713213$. An excellent approximation is $frac 8911619approx 0.550339715873$. I got that by using the continued fraction. When you see a large value like $143$, truncating before it yields a very good approximation.
$endgroup$
add a comment |
$begingroup$
The number you want to approximate is about $0.550339713213$. An excellent approximation is $frac 8911619approx 0.550339715873$. I got that by using the continued fraction. When you see a large value like $143$, truncating before it yields a very good approximation.
$endgroup$
The number you want to approximate is about $0.550339713213$. An excellent approximation is $frac 8911619approx 0.550339715873$. I got that by using the continued fraction. When you see a large value like $143$, truncating before it yields a very good approximation.
answered 39 mins ago
Ross MillikanRoss Millikan
300k24200374
300k24200374
add a comment |
add a comment |
$begingroup$
The continued fraction for $fraclogleft(frac54right)logleft(frac32right)$ is
$$
0;1,1,4,2,6,1,color#C0010,143,3,dots
$$
The convergents for this continued fraction are
$$
left0,1,frac12,frac59,frac1120,frac71129,frac82149,color#C00frac8911619,frac127495231666,frac383376696617,dotsright
$$
As Ross Millikan mentions, stopping just before a large continuant like $143$ gives a particularly good approximation for the size of the denominator; in this case, the approximation $frac8911619$ is closer than $frac1143cdot1619^2$ to $fraclogleft(frac54right)logleft(frac32right)$.
$endgroup$
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
12 mins ago
add a comment |
$begingroup$
The continued fraction for $fraclogleft(frac54right)logleft(frac32right)$ is
$$
0;1,1,4,2,6,1,color#C0010,143,3,dots
$$
The convergents for this continued fraction are
$$
left0,1,frac12,frac59,frac1120,frac71129,frac82149,color#C00frac8911619,frac127495231666,frac383376696617,dotsright
$$
As Ross Millikan mentions, stopping just before a large continuant like $143$ gives a particularly good approximation for the size of the denominator; in this case, the approximation $frac8911619$ is closer than $frac1143cdot1619^2$ to $fraclogleft(frac54right)logleft(frac32right)$.
$endgroup$
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
12 mins ago
add a comment |
$begingroup$
The continued fraction for $fraclogleft(frac54right)logleft(frac32right)$ is
$$
0;1,1,4,2,6,1,color#C0010,143,3,dots
$$
The convergents for this continued fraction are
$$
left0,1,frac12,frac59,frac1120,frac71129,frac82149,color#C00frac8911619,frac127495231666,frac383376696617,dotsright
$$
As Ross Millikan mentions, stopping just before a large continuant like $143$ gives a particularly good approximation for the size of the denominator; in this case, the approximation $frac8911619$ is closer than $frac1143cdot1619^2$ to $fraclogleft(frac54right)logleft(frac32right)$.
$endgroup$
The continued fraction for $fraclogleft(frac54right)logleft(frac32right)$ is
$$
0;1,1,4,2,6,1,color#C0010,143,3,dots
$$
The convergents for this continued fraction are
$$
left0,1,frac12,frac59,frac1120,frac71129,frac82149,color#C00frac8911619,frac127495231666,frac383376696617,dotsright
$$
As Ross Millikan mentions, stopping just before a large continuant like $143$ gives a particularly good approximation for the size of the denominator; in this case, the approximation $frac8911619$ is closer than $frac1143cdot1619^2$ to $fraclogleft(frac54right)logleft(frac32right)$.
answered 24 mins ago
robjohn♦robjohn
269k27311638
269k27311638
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
12 mins ago
add a comment |
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
12 mins ago
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
12 mins ago
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
12 mins ago
add a comment |
$begingroup$
Running the extended Euclidean algorithm to find the continued fraction:
$$beginarrayccx&q&a&b\
hline 0.55033971 & & 0 & 1\ 1 & 0 & 1 & 0\ 0.55033971 & 1 & 0 & 1\ 0.44966029 & 1 & 1 & -1 \ 0.10067943 & 4 & -1 & 2\ 0.04694258 & 2 & 5 & -9\ 0.00679426 & 6 & -11 & 20 \ 0.00617700 & 1 & 71 & -129 \ 0.00061727 & 10 & -82 & 149\ 4.31cdot 10^-6 & 143 & 891 & -1619 \
1.25cdot 10^-6 & 3 & -127495 & 231666endarray$$
The $q$ column are the quotients, that go into the continued fraction. The $a$ and $b$ columns track a linear combination of the original two that's equal to $x_n$; for example, $-11cdot 1 + 20cdot fraclog(5/4)log(3/2)approx 0.00679426$. The fraction $left|fraclog(5/4)log(3/2)right|$ is approximated by $fraca_n$, with increasing accuracy.
The formulas for building this table: $q_n = leftlfloor frac x_n-1x_nrightrfloor$, $x_n+1=x_n-1-q_nx_n$, $a_n+1=a_n-1-q_na_n$, $b_n+1=b_n-1-q_nb_n$. Initialize with $x_0=1$, $x_-1$ the quantity we're trying to estimate, $a_-1=b_0=0$, $a_0=b_-1=1$.
If you run the table much large than this, watch for floating-point accuracy issues; once the $x_n$ get down close to the accuracy limit for floating point numbers near zero, you can't trust the quotients anymore.
Now, how that accuracy increases is irregular. Large quotients go with particularly good approximations - see how that quotient of $143$ means that we have to go to six-digit numerator and denominator to do better than that $frac8911619$ approximation.
It is of course a tradeoff between accuracy and how deep you go. For your purposes in costing the two upgrades, I'd probably go with that $frac1120$ approximation.
$endgroup$
add a comment |
$begingroup$
Running the extended Euclidean algorithm to find the continued fraction:
$$beginarrayccx&q&a&b\
hline 0.55033971 & & 0 & 1\ 1 & 0 & 1 & 0\ 0.55033971 & 1 & 0 & 1\ 0.44966029 & 1 & 1 & -1 \ 0.10067943 & 4 & -1 & 2\ 0.04694258 & 2 & 5 & -9\ 0.00679426 & 6 & -11 & 20 \ 0.00617700 & 1 & 71 & -129 \ 0.00061727 & 10 & -82 & 149\ 4.31cdot 10^-6 & 143 & 891 & -1619 \
1.25cdot 10^-6 & 3 & -127495 & 231666endarray$$
The $q$ column are the quotients, that go into the continued fraction. The $a$ and $b$ columns track a linear combination of the original two that's equal to $x_n$; for example, $-11cdot 1 + 20cdot fraclog(5/4)log(3/2)approx 0.00679426$. The fraction $left|fraclog(5/4)log(3/2)right|$ is approximated by $fraca_n$, with increasing accuracy.
The formulas for building this table: $q_n = leftlfloor frac x_n-1x_nrightrfloor$, $x_n+1=x_n-1-q_nx_n$, $a_n+1=a_n-1-q_na_n$, $b_n+1=b_n-1-q_nb_n$. Initialize with $x_0=1$, $x_-1$ the quantity we're trying to estimate, $a_-1=b_0=0$, $a_0=b_-1=1$.
If you run the table much large than this, watch for floating-point accuracy issues; once the $x_n$ get down close to the accuracy limit for floating point numbers near zero, you can't trust the quotients anymore.
Now, how that accuracy increases is irregular. Large quotients go with particularly good approximations - see how that quotient of $143$ means that we have to go to six-digit numerator and denominator to do better than that $frac8911619$ approximation.
It is of course a tradeoff between accuracy and how deep you go. For your purposes in costing the two upgrades, I'd probably go with that $frac1120$ approximation.
$endgroup$
add a comment |
$begingroup$
Running the extended Euclidean algorithm to find the continued fraction:
$$beginarrayccx&q&a&b\
hline 0.55033971 & & 0 & 1\ 1 & 0 & 1 & 0\ 0.55033971 & 1 & 0 & 1\ 0.44966029 & 1 & 1 & -1 \ 0.10067943 & 4 & -1 & 2\ 0.04694258 & 2 & 5 & -9\ 0.00679426 & 6 & -11 & 20 \ 0.00617700 & 1 & 71 & -129 \ 0.00061727 & 10 & -82 & 149\ 4.31cdot 10^-6 & 143 & 891 & -1619 \
1.25cdot 10^-6 & 3 & -127495 & 231666endarray$$
The $q$ column are the quotients, that go into the continued fraction. The $a$ and $b$ columns track a linear combination of the original two that's equal to $x_n$; for example, $-11cdot 1 + 20cdot fraclog(5/4)log(3/2)approx 0.00679426$. The fraction $left|fraclog(5/4)log(3/2)right|$ is approximated by $fraca_n$, with increasing accuracy.
The formulas for building this table: $q_n = leftlfloor frac x_n-1x_nrightrfloor$, $x_n+1=x_n-1-q_nx_n$, $a_n+1=a_n-1-q_na_n$, $b_n+1=b_n-1-q_nb_n$. Initialize with $x_0=1$, $x_-1$ the quantity we're trying to estimate, $a_-1=b_0=0$, $a_0=b_-1=1$.
If you run the table much large than this, watch for floating-point accuracy issues; once the $x_n$ get down close to the accuracy limit for floating point numbers near zero, you can't trust the quotients anymore.
Now, how that accuracy increases is irregular. Large quotients go with particularly good approximations - see how that quotient of $143$ means that we have to go to six-digit numerator and denominator to do better than that $frac8911619$ approximation.
It is of course a tradeoff between accuracy and how deep you go. For your purposes in costing the two upgrades, I'd probably go with that $frac1120$ approximation.
$endgroup$
Running the extended Euclidean algorithm to find the continued fraction:
$$beginarrayccx&q&a&b\
hline 0.55033971 & & 0 & 1\ 1 & 0 & 1 & 0\ 0.55033971 & 1 & 0 & 1\ 0.44966029 & 1 & 1 & -1 \ 0.10067943 & 4 & -1 & 2\ 0.04694258 & 2 & 5 & -9\ 0.00679426 & 6 & -11 & 20 \ 0.00617700 & 1 & 71 & -129 \ 0.00061727 & 10 & -82 & 149\ 4.31cdot 10^-6 & 143 & 891 & -1619 \
1.25cdot 10^-6 & 3 & -127495 & 231666endarray$$
The $q$ column are the quotients, that go into the continued fraction. The $a$ and $b$ columns track a linear combination of the original two that's equal to $x_n$; for example, $-11cdot 1 + 20cdot fraclog(5/4)log(3/2)approx 0.00679426$. The fraction $left|fraclog(5/4)log(3/2)right|$ is approximated by $fraca_n$, with increasing accuracy.
The formulas for building this table: $q_n = leftlfloor frac x_n-1x_nrightrfloor$, $x_n+1=x_n-1-q_nx_n$, $a_n+1=a_n-1-q_na_n$, $b_n+1=b_n-1-q_nb_n$. Initialize with $x_0=1$, $x_-1$ the quantity we're trying to estimate, $a_-1=b_0=0$, $a_0=b_-1=1$.
If you run the table much large than this, watch for floating-point accuracy issues; once the $x_n$ get down close to the accuracy limit for floating point numbers near zero, you can't trust the quotients anymore.
Now, how that accuracy increases is irregular. Large quotients go with particularly good approximations - see how that quotient of $143$ means that we have to go to six-digit numerator and denominator to do better than that $frac8911619$ approximation.
It is of course a tradeoff between accuracy and how deep you go. For your purposes in costing the two upgrades, I'd probably go with that $frac1120$ approximation.
answered 13 mins ago
jmerryjmerry
15.8k1632
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$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
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– amsmath
47 mins ago
1
$begingroup$
The continued fraction expansion of your irrational number will produce the best rational approximation subject to a limit on size of the denominator.
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– hardmath
45 mins ago
1
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
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– user647486
45 mins ago
$begingroup$
try 82/149 ........
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– Will Jagy
42 mins ago
$begingroup$
Cool, a practical application of continued fractions. :)
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– Minus One-Twelfth
33 mins ago