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Approximating $frac log(5/4)log(3/2)$ to a rational number
The Next CEO of Stack OverflowHermite Interpolation of $e^x$. Strange behaviour when increasing the number of derivatives at interpolating points.Approximating log of factorialNewton's Method, and approximating parameters for Bézier curves.Approximating a log-power functionApproximating Logs and Antilogs by handApproximating non-rational roots by a rational roots for a quadratic equationIs this formula for $frace^2-3e^2+1$ known? How to prove it?Does there exist infinitely many $mu$ which satisfy this:Approximating functions with rational functionsApproximating $log(X-Y)$
$begingroup$
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 an additional $50%$ increase at once. and type B gives $25%$.
What I want to know is how many times of upgrade $(x,y)$ provides the same additional increase. So what I've done is solve $(3/2)^x = (5/4)^y$ respect to $frac xy$.
Can you provide me a way to construct sequence $p_n$, $q_n$ which approximate the real number?
Thank you in advance.
approximation irrational-numbers
edited Mar 24 at 12:59
YuiTo Cheng
2,1862937
asked Mar 24 at 1:59
MrTanorusMrTanorus
31918
$endgroup$
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 an additional $50%$ increase at once. and type B gives $25%$.
What I want to know is how many times of upgrade $(x,y)$ provides the same additional increase. So what I've done is solve $(3/2)^x = (5/4)^y$ respect to $frac xy$.
Can you provide me a way to construct sequence $p_n$, $q_n$ which approximate the real number?
Thank you in advance.
approximation irrational-numbers
edited Mar 24 at 12:59
YuiTo Cheng
2,1862937
asked Mar 24 at 1:59
MrTanorusMrTanorus
31918
$endgroup$
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
$endgroup$
– amsmath
Mar 24 at 2:19
7
$begingroup$
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
Mar 24 at 2:21
2
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
Mar 24 at 2:21
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
Mar 24 at 2:23
2
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
Mar 24 at 2:32
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 an additional $50%$ increase at once. and type B gives $25%$.
What I want to know is how many times of upgrade $(x,y)$ provides the same additional increase. So what I've done is solve $(3/2)^x = (5/4)^y$ respect to $frac xy$.
Can you provide me a way to construct sequence $p_n$, $q_n$ which approximate the real number?
Thank you in advance.
approximation irrational-numbers
edited Mar 24 at 12:59
YuiTo Cheng
2,1862937
asked Mar 24 at 1:59
MrTanorusMrTanorus
31918
$endgroup$
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 an additional $50%$ increase at once. and type B gives $25%$.
What I want to know is how many times of upgrade $(x,y)$ provides the same additional increase. So what I've done is solve $(3/2)^x = (5/4)^y$ respect to $frac xy$.
Can you provide me a way to construct sequence $p_n$, $q_n$ which approximate the real number?
Thank you in advance.
approximation irrational-numbers
approximation irrational-numbers
edited Mar 24 at 12:59
YuiTo Cheng
2,1862937
asked Mar 24 at 1:59
MrTanorusMrTanorus
31918
edited Mar 24 at 12:59
YuiTo Cheng
2,1862937
asked Mar 24 at 1:59
MrTanorusMrTanorus
31918
edited Mar 24 at 12:59
YuiTo Cheng
2,1862937
edited Mar 24 at 12:59
YuiTo Cheng
2,1862937
edited Mar 24 at 12:59
YuiTo Cheng
2,1862937
2,1862937
asked Mar 24 at 1:59
MrTanorusMrTanorus
31918
asked Mar 24 at 1:59
MrTanorusMrTanorus
31918
asked Mar 24 at 1:59
MrTanorusMrTanorus
31918
31918
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
$endgroup$
– amsmath
Mar 24 at 2:19
7
$begingroup$
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
Mar 24 at 2:21
2
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
Mar 24 at 2:21
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
Mar 24 at 2:23
2
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
Mar 24 at 2:32
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
Mar 24 at 2:19
7
$begingroup$
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
Mar 24 at 2:21
2
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
Mar 24 at 2:21
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
Mar 24 at 2:23
2
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
Mar 24 at 2:32
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
$endgroup$
– amsmath
Mar 24 at 2:19
$begingroup$
I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
$endgroup$
– amsmath
Mar 24 at 2:19
7
7
$begingroup$
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
Mar 24 at 2:21
$begingroup$
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
Mar 24 at 2:21
2
2
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
Mar 24 at 2:21
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
Mar 24 at 2:21
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
Mar 24 at 2:23
$begingroup$
try 82/149 ........
$endgroup$
– Will Jagy
Mar 24 at 2:23
2
2
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
Mar 24 at 2:32
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
Mar 24 at 2:32
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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.
answered Mar 24 at 2:52
jmerryjmerry
16.9k11633
$endgroup$
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 $fraclogleft(frac54right)logleft(frac32right)$.
answered Mar 24 at 2:42
robjohn♦robjohn
270k27312640
$endgroup$
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
Mar 24 at 2:54
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.
answered Mar 24 at 2:27
Ross MillikanRoss Millikan
300k24200375
$endgroup$
2
$begingroup$
(+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 $frac1cq^2$ to the value approximated.
$endgroup$
– robjohn♦
Mar 24 at 3:07
add a comment |
$begingroup$
Use the following equation
$$log(1+frac1x)approxfrac3+6x6x^2+6x+1$$
so
$$log(frac54)=log(1+frac14)$$
$$log(frac32)=log(1+frac12)$$
the approximated value will be
$$frac log(5/4)log(3/2)approx frac333605=0.55041322$$
answered Mar 28 at 13:04
E.H.EE.H.E
16k11968
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Mar 24 at 2:52
jmerryjmerry
16.9k11633
$endgroup$
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.
answered Mar 24 at 2:52
jmerryjmerry
16.9k11633
$endgroup$
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.
answered Mar 24 at 2:52
jmerryjmerry
16.9k11633
$endgroup$
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 Mar 24 at 2:52
jmerryjmerry
16.9k11633
answered Mar 24 at 2:52
jmerryjmerry
16.9k11633
answered Mar 24 at 2:52
jmerryjmerry
16.9k11633
answered Mar 24 at 2:52
jmerryjmerry
16.9k11633
16.9k11633
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 $fraclogleft(frac54right)logleft(frac32right)$.
answered Mar 24 at 2:42
robjohn♦robjohn
270k27312640
$endgroup$
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
Mar 24 at 2:54
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 $fraclogleft(frac54right)logleft(frac32right)$.
answered Mar 24 at 2:42
robjohn♦robjohn
270k27312640
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$begingroup$
Thank you. A good addition to my answer.
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– Ross Millikan
Mar 24 at 2:54
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 $fraclogleft(frac54right)logleft(frac32right)$.
answered Mar 24 at 2:42
robjohn♦robjohn
270k27312640
$endgroup$
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 Mar 24 at 2:42
robjohn♦robjohn
270k27312640
answered Mar 24 at 2:42
robjohn♦robjohn
270k27312640
answered Mar 24 at 2:42
robjohn♦robjohn
270k27312640
answered Mar 24 at 2:42
robjohn♦robjohn
270k27312640
270k27312640
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
Mar 24 at 2:54
add a comment |
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
Mar 24 at 2:54
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
Mar 24 at 2:54
$begingroup$
Thank you. A good addition to my answer.
$endgroup$
– Ross Millikan
Mar 24 at 2:54
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.
answered Mar 24 at 2:27
Ross MillikanRoss Millikan
300k24200375
$endgroup$
2
$begingroup$
(+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 $frac1cq^2$ to the value approximated.
$endgroup$
– robjohn♦
Mar 24 at 3:07
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.
answered Mar 24 at 2:27
Ross MillikanRoss Millikan
300k24200375
$endgroup$
2
$begingroup$
(+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 $frac1cq^2$ to the value approximated.
$endgroup$
– robjohn♦
Mar 24 at 3:07
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.
answered Mar 24 at 2:27
Ross MillikanRoss Millikan
300k24200375
$endgroup$
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 Mar 24 at 2:27
Ross MillikanRoss Millikan
300k24200375
answered Mar 24 at 2:27
Ross MillikanRoss Millikan
300k24200375
answered Mar 24 at 2:27
Ross MillikanRoss Millikan
300k24200375
answered Mar 24 at 2:27
Ross MillikanRoss Millikan
300k24200375
300k24200375
2
$begingroup$
(+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 $frac1cq^2$ to the value approximated.
$endgroup$
– robjohn♦
Mar 24 at 3:07
add a comment |
2
$begingroup$
(+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 $frac1cq^2$ to the value approximated.
$endgroup$
– robjohn♦
Mar 24 at 3:07
2
2
$begingroup$
(+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 $frac1cq^2$ to the value approximated.
$endgroup$
– robjohn♦
Mar 24 at 3:07
$begingroup$
(+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 $frac1cq^2$ to the value approximated.
$endgroup$
– robjohn♦
Mar 24 at 3:07
add a comment |
$begingroup$
Use the following equation
$$log(1+frac1x)approxfrac3+6x6x^2+6x+1$$
so
$$log(frac54)=log(1+frac14)$$
$$log(frac32)=log(1+frac12)$$
the approximated value will be
$$frac log(5/4)log(3/2)approx frac333605=0.55041322$$
answered Mar 28 at 13:04
E.H.EE.H.E
16k11968
$endgroup$
add a comment |
$begingroup$
Use the following equation
$$log(1+frac1x)approxfrac3+6x6x^2+6x+1$$
so
$$log(frac54)=log(1+frac14)$$
$$log(frac32)=log(1+frac12)$$
the approximated value will be
$$frac log(5/4)log(3/2)approx frac333605=0.55041322$$
answered Mar 28 at 13:04
E.H.EE.H.E
16k11968
$endgroup$
add a comment |
$begingroup$
Use the following equation
$$log(1+frac1x)approxfrac3+6x6x^2+6x+1$$
so
$$log(frac54)=log(1+frac14)$$
$$log(frac32)=log(1+frac12)$$
the approximated value will be
$$frac log(5/4)log(3/2)approx frac333605=0.55041322$$
answered Mar 28 at 13:04
E.H.EE.H.E
16k11968
$endgroup$
Use the following equation
$$log(1+frac1x)approxfrac3+6x6x^2+6x+1$$
so
$$log(frac54)=log(1+frac14)$$
$$log(frac32)=log(1+frac12)$$
the approximated value will be
$$frac log(5/4)log(3/2)approx frac333605=0.55041322$$
answered Mar 28 at 13:04
E.H.EE.H.E
16k11968
answered Mar 28 at 13:04
E.H.EE.H.E
16k11968
answered Mar 28 at 13:04
E.H.EE.H.E
16k11968
answered Mar 28 at 13:04
E.H.EE.H.E
16k11968
16k11968
add a comment |
add a comment |
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I don't understand your game, but your number approximately $0.55034$ and thus $tfrac55034100000$ or $tfrac550310000$. What's wrong with that?
$endgroup$
– amsmath
Mar 24 at 2:19
7
$begingroup$
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
Mar 24 at 2:21
2
$begingroup$
You can take truncations of the continued fraction of that number. The first few of its values start like this.
$endgroup$
– user647486
Mar 24 at 2:21
$begingroup$
try 82/149 ........
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– Will Jagy
Mar 24 at 2:23
2
$begingroup$
Cool, a practical application of continued fractions. :)
$endgroup$
– Minus One-Twelfth
Mar 24 at 2:32