Approximating irrational number to rational number
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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 $tfrac{55034}{100000}$ or $tfrac{5503}{10000}$. What's wrong with that?
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– amsmath
3 hours ago
1
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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
3 hours ago
1
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You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
3 hours ago
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try 82/149 ........
$endgroup$
– Will Jagy
3 hours ago
1
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Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
3 hours 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 3 hours ago
Rócherz
2,9863821
2,9863821
asked 3 hours ago
MrTanorusMrTanorus
1978
1978
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I don't understand your game, but your number approximately $0.55034$ and thus $tfrac{55034}{100000}$ or $tfrac{5503}{10000}$. What's wrong with that?
$endgroup$
– amsmath
3 hours 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
3 hours 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
3 hours ago
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
3 hours ago
1
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
3 hours ago
add a comment |
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac{55034}{100000}$ or $tfrac{5503}{10000}$. What's wrong with that?
$endgroup$
– amsmath
3 hours 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
3 hours 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
3 hours ago
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
3 hours ago
1
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
3 hours ago
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac{55034}{100000}$ or $tfrac{5503}{10000}$. What's wrong with that?
$endgroup$
– amsmath
3 hours ago
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac{55034}{100000}$ or $tfrac{5503}{10000}$. What's wrong with that?
$endgroup$
– amsmath
3 hours 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
3 hours 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
3 hours 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
3 hours ago
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
3 hours ago
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
3 hours ago
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
3 hours ago
1
1
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
3 hours ago
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
3 hours 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 $frac {891}{1619}approx 0.550339715873$. I got that by using the continued fraction. When you see a large value like $143$, truncating before it yields a very good approximation.
$endgroup$
$begingroup$
(+1) I wondered where you got $143$ until I actually computed the continued fraction. I'd hoped it was okay to expand upon your answer to show where that number came from. Also to mention that if $frac pq$ is a continued fraction approximation and $c$ is the next term in the conitnued fraction (which I think is called a continuant), then $frac pq$ is closer than $frac1{cq^2}$ to the value approximated.
$endgroup$
– robjohn♦
2 hours ago
add a comment |
$begingroup$
The continued fraction for $frac{logleft(frac54right)}{logleft(frac32right)}$ is
$$
{0;1,1,4,2,6,1,color{#C00}{10},143,3,dots}
$$
The convergents for this continued fraction are
$$
left{0,1,frac12,frac59,frac{11}{20},frac{71}{129},frac{82}{149},color{#C00}{frac{891}{1619}},frac{127495}{231666},frac{383376}{696617},dotsright}
$$
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 $frac{891}{1619}$ is closer than $frac1{143cdot1619^2}$ to $frac{logleft(frac54right)}{logleft(frac32right)}$.
$endgroup$
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
2 hours ago
add a comment |
$begingroup$
Running the extended Euclidean algorithm to find the continued fraction:
$$begin{array}{cc|cc}x&q&a&b\
hline 0.55033971 & & 0 & 1\ 1 & 0 & 1 & 0\ 0.55033971 & 1 & 0 & 1\ 0.44966029 & 1 & 1 & -1 \ 0.10067943 & 4 & -1 & 2\ 0.04694258 & 2 & 5 & -9\ 0.00679426 & 6 & -11 & 20 \ 0.00617700 & 1 & 71 & -129 \ 0.00061727 & 10 & -82 & 149\ 4.31cdot 10^{-6} & 143 & 891 & -1619 \
1.25cdot 10^{-6} & 3 & -127495 & 231666end{array}$$
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, $-11cdot 1 + 20cdot frac{log(5/4)}{log(3/2)}approx 0.00679426$. The fraction $left|frac{log(5/4)}{log(3/2)}right|$ is approximated by $frac{|a_n|}{|b_n|}$, with increasing accuracy.
The formulas for building this table: $q_n = leftlfloor frac {x_{n-1}}{x_n}rightrfloor$, $x_{n+1}=x_{n-1}-q_nx_n$, $a_{n+1}=a_{n-1}-q_na_n$, $b_{n+1}=b_{n-1}-q_nb_n$. 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 $frac{891}{1619}$ 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 $frac{11}{20}$ 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 $frac {891}{1619}approx 0.550339715873$. I got that by using the continued fraction. When you see a large value like $143$, truncating before it yields a very good approximation.
$endgroup$
$begingroup$
(+1) I wondered where you got $143$ until I actually computed the continued fraction. I'd hoped it was okay to expand upon your answer to show where that number came from. Also to mention that if $frac pq$ is a continued fraction approximation and $c$ is the next term in the conitnued fraction (which I think is called a continuant), then $frac pq$ is closer than $frac1{cq^2}$ to the value approximated.
$endgroup$
– robjohn♦
2 hours ago
add a comment |
$begingroup$
The number you want to approximate is about $0.550339713213$. An excellent approximation is $frac {891}{1619}approx 0.550339715873$. I got that by using the continued fraction. When you see a large value like $143$, truncating before it yields a very good approximation.
$endgroup$
$begingroup$
(+1) I wondered where you got $143$ until I actually computed the continued fraction. I'd hoped it was okay to expand upon your answer to show where that number came from. Also to mention that if $frac pq$ is a continued fraction approximation and $c$ is the next term in the conitnued fraction (which I think is called a continuant), then $frac pq$ is closer than $frac1{cq^2}$ to the value approximated.
$endgroup$
– robjohn♦
2 hours ago
add a comment |
$begingroup$
The number you want to approximate is about $0.550339713213$. An excellent approximation is $frac {891}{1619}approx 0.550339715873$. 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 $frac {891}{1619}approx 0.550339715873$. 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 3 hours ago
Ross MillikanRoss Millikan
300k24200374
300k24200374
$begingroup$
(+1) I wondered where you got $143$ until I actually computed the continued fraction. I'd hoped it was okay to expand upon your answer to show where that number came from. Also to mention that if $frac pq$ is a continued fraction approximation and $c$ is the next term in the conitnued fraction (which I think is called a continuant), then $frac pq$ is closer than $frac1{cq^2}$ to the value approximated.
$endgroup$
– robjohn♦
2 hours ago
add a comment |
$begingroup$
(+1) I wondered where you got $143$ until I actually computed the continued fraction. I'd hoped it was okay to expand upon your answer to show where that number came from. Also to mention that if $frac pq$ is a continued fraction approximation and $c$ is the next term in the conitnued fraction (which I think is called a continuant), then $frac pq$ is closer than $frac1{cq^2}$ to the value approximated.
$endgroup$
– robjohn♦
2 hours ago
$begingroup$
(+1) I wondered where you got $143$ until I actually computed the continued fraction. I'd hoped it was okay to expand upon your answer to show where that number came from. Also to mention that if $frac pq$ is a continued fraction approximation and $c$ is the next term in the conitnued fraction (which I think is called a continuant), then $frac pq$ is closer than $frac1{cq^2}$ to the value approximated.
$endgroup$
– robjohn♦
2 hours ago
$begingroup$
(+1) I wondered where you got $143$ until I actually computed the continued fraction. I'd hoped it was okay to expand upon your answer to show where that number came from. Also to mention that if $frac pq$ is a continued fraction approximation and $c$ is the next term in the conitnued fraction (which I think is called a continuant), then $frac pq$ is closer than $frac1{cq^2}$ to the value approximated.
$endgroup$
– robjohn♦
2 hours ago
add a comment |
$begingroup$
The continued fraction for $frac{logleft(frac54right)}{logleft(frac32right)}$ is
$$
{0;1,1,4,2,6,1,color{#C00}{10},143,3,dots}
$$
The convergents for this continued fraction are
$$
left{0,1,frac12,frac59,frac{11}{20},frac{71}{129},frac{82}{149},color{#C00}{frac{891}{1619}},frac{127495}{231666},frac{383376}{696617},dotsright}
$$
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 $frac{891}{1619}$ is closer than $frac1{143cdot1619^2}$ to $frac{logleft(frac54right)}{logleft(frac32right)}$.
$endgroup$
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
2 hours ago
add a comment |
$begingroup$
The continued fraction for $frac{logleft(frac54right)}{logleft(frac32right)}$ is
$$
{0;1,1,4,2,6,1,color{#C00}{10},143,3,dots}
$$
The convergents for this continued fraction are
$$
left{0,1,frac12,frac59,frac{11}{20},frac{71}{129},frac{82}{149},color{#C00}{frac{891}{1619}},frac{127495}{231666},frac{383376}{696617},dotsright}
$$
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 $frac{891}{1619}$ is closer than $frac1{143cdot1619^2}$ to $frac{logleft(frac54right)}{logleft(frac32right)}$.
$endgroup$
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
2 hours ago
add a comment |
$begingroup$
The continued fraction for $frac{logleft(frac54right)}{logleft(frac32right)}$ is
$$
{0;1,1,4,2,6,1,color{#C00}{10},143,3,dots}
$$
The convergents for this continued fraction are
$$
left{0,1,frac12,frac59,frac{11}{20},frac{71}{129},frac{82}{149},color{#C00}{frac{891}{1619}},frac{127495}{231666},frac{383376}{696617},dotsright}
$$
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 $frac{891}{1619}$ is closer than $frac1{143cdot1619^2}$ to $frac{logleft(frac54right)}{logleft(frac32right)}$.
$endgroup$
The continued fraction for $frac{logleft(frac54right)}{logleft(frac32right)}$ is
$$
{0;1,1,4,2,6,1,color{#C00}{10},143,3,dots}
$$
The convergents for this continued fraction are
$$
left{0,1,frac12,frac59,frac{11}{20},frac{71}{129},frac{82}{149},color{#C00}{frac{891}{1619}},frac{127495}{231666},frac{383376}{696617},dotsright}
$$
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 $frac{891}{1619}$ is closer than $frac1{143cdot1619^2}$ to $frac{logleft(frac54right)}{logleft(frac32right)}$.
answered 3 hours ago
robjohn♦robjohn
269k27311638
269k27311638
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
2 hours ago
add a comment |
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
2 hours ago
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
2 hours ago
add a comment |
$begingroup$
Running the extended Euclidean algorithm to find the continued fraction:
$$begin{array}{cc|cc}x&q&a&b\
hline 0.55033971 & & 0 & 1\ 1 & 0 & 1 & 0\ 0.55033971 & 1 & 0 & 1\ 0.44966029 & 1 & 1 & -1 \ 0.10067943 & 4 & -1 & 2\ 0.04694258 & 2 & 5 & -9\ 0.00679426 & 6 & -11 & 20 \ 0.00617700 & 1 & 71 & -129 \ 0.00061727 & 10 & -82 & 149\ 4.31cdot 10^{-6} & 143 & 891 & -1619 \
1.25cdot 10^{-6} & 3 & -127495 & 231666end{array}$$
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, $-11cdot 1 + 20cdot frac{log(5/4)}{log(3/2)}approx 0.00679426$. The fraction $left|frac{log(5/4)}{log(3/2)}right|$ is approximated by $frac{|a_n|}{|b_n|}$, with increasing accuracy.
The formulas for building this table: $q_n = leftlfloor frac {x_{n-1}}{x_n}rightrfloor$, $x_{n+1}=x_{n-1}-q_nx_n$, $a_{n+1}=a_{n-1}-q_na_n$, $b_{n+1}=b_{n-1}-q_nb_n$. 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 $frac{891}{1619}$ 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 $frac{11}{20}$ approximation.
$endgroup$
add a comment |
$begingroup$
Running the extended Euclidean algorithm to find the continued fraction:
$$begin{array}{cc|cc}x&q&a&b\
hline 0.55033971 & & 0 & 1\ 1 & 0 & 1 & 0\ 0.55033971 & 1 & 0 & 1\ 0.44966029 & 1 & 1 & -1 \ 0.10067943 & 4 & -1 & 2\ 0.04694258 & 2 & 5 & -9\ 0.00679426 & 6 & -11 & 20 \ 0.00617700 & 1 & 71 & -129 \ 0.00061727 & 10 & -82 & 149\ 4.31cdot 10^{-6} & 143 & 891 & -1619 \
1.25cdot 10^{-6} & 3 & -127495 & 231666end{array}$$
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, $-11cdot 1 + 20cdot frac{log(5/4)}{log(3/2)}approx 0.00679426$. The fraction $left|frac{log(5/4)}{log(3/2)}right|$ is approximated by $frac{|a_n|}{|b_n|}$, with increasing accuracy.
The formulas for building this table: $q_n = leftlfloor frac {x_{n-1}}{x_n}rightrfloor$, $x_{n+1}=x_{n-1}-q_nx_n$, $a_{n+1}=a_{n-1}-q_na_n$, $b_{n+1}=b_{n-1}-q_nb_n$. 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 $frac{891}{1619}$ 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 $frac{11}{20}$ approximation.
$endgroup$
add a comment |
$begingroup$
Running the extended Euclidean algorithm to find the continued fraction:
$$begin{array}{cc|cc}x&q&a&b\
hline 0.55033971 & & 0 & 1\ 1 & 0 & 1 & 0\ 0.55033971 & 1 & 0 & 1\ 0.44966029 & 1 & 1 & -1 \ 0.10067943 & 4 & -1 & 2\ 0.04694258 & 2 & 5 & -9\ 0.00679426 & 6 & -11 & 20 \ 0.00617700 & 1 & 71 & -129 \ 0.00061727 & 10 & -82 & 149\ 4.31cdot 10^{-6} & 143 & 891 & -1619 \
1.25cdot 10^{-6} & 3 & -127495 & 231666end{array}$$
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, $-11cdot 1 + 20cdot frac{log(5/4)}{log(3/2)}approx 0.00679426$. The fraction $left|frac{log(5/4)}{log(3/2)}right|$ is approximated by $frac{|a_n|}{|b_n|}$, with increasing accuracy.
The formulas for building this table: $q_n = leftlfloor frac {x_{n-1}}{x_n}rightrfloor$, $x_{n+1}=x_{n-1}-q_nx_n$, $a_{n+1}=a_{n-1}-q_na_n$, $b_{n+1}=b_{n-1}-q_nb_n$. 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 $frac{891}{1619}$ 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 $frac{11}{20}$ approximation.
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Running the extended Euclidean algorithm to find the continued fraction:
$$begin{array}{cc|cc}x&q&a&b\
hline 0.55033971 & & 0 & 1\ 1 & 0 & 1 & 0\ 0.55033971 & 1 & 0 & 1\ 0.44966029 & 1 & 1 & -1 \ 0.10067943 & 4 & -1 & 2\ 0.04694258 & 2 & 5 & -9\ 0.00679426 & 6 & -11 & 20 \ 0.00617700 & 1 & 71 & -129 \ 0.00061727 & 10 & -82 & 149\ 4.31cdot 10^{-6} & 143 & 891 & -1619 \
1.25cdot 10^{-6} & 3 & -127495 & 231666end{array}$$
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, $-11cdot 1 + 20cdot frac{log(5/4)}{log(3/2)}approx 0.00679426$. The fraction $left|frac{log(5/4)}{log(3/2)}right|$ is approximated by $frac{|a_n|}{|b_n|}$, with increasing accuracy.
The formulas for building this table: $q_n = leftlfloor frac {x_{n-1}}{x_n}rightrfloor$, $x_{n+1}=x_{n-1}-q_nx_n$, $a_{n+1}=a_{n-1}-q_na_n$, $b_{n+1}=b_{n-1}-q_nb_n$. 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 $frac{891}{1619}$ 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 $frac{11}{20}$ approximation.
answered 2 hours ago
jmerryjmerry
15.8k1632
15.8k1632
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$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac{55034}{100000}$ or $tfrac{5503}{10000}$. What's wrong with that?
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– amsmath
3 hours 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
3 hours 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
3 hours ago
$begingroup$
try 82/149 ........
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– Will Jagy
3 hours ago
1
$begingroup$
Cool, a practical application of continued fractions. :)
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– Minus One-Twelfth
3 hours ago