Mixing
Numerical Accuracy: How many digits are correct?
$begingroup$
I'm really not too sure how to go on about this:
Problem. Julia (or take MatLab etc. (I guess.)) computes
In [1]: exp(pi*sqrt(67)/6) - sqrt(5280)
Out[1]: 6.121204876308184e-8
Only by looking closely: How many digits are correct? (What could you do to compute $16$ decimals correctly?)
So I know that subtraction is somewhat dangerous numerically - though how do I solve this problem in a precise manner?
numerical-methods floating-point
edited Jan 3 at 20:08
LutzL
57.2k42054
asked Jan 3 at 19:12
KezerKezer
1,313421
$endgroup$
add a comment |
$begingroup$
I'm really not too sure how to go on about this:
Problem. Julia (or take MatLab etc. (I guess.)) computes
In [1]: exp(pi*sqrt(67)/6) - sqrt(5280)
Out[1]: 6.121204876308184e-8
Only by looking closely: How many digits are correct? (What could you do to compute $16$ decimals correctly?)
So I know that subtraction is somewhat dangerous numerically - though how do I solve this problem in a precise manner?
numerical-methods floating-point
edited Jan 3 at 20:08
LutzL
57.2k42054
asked Jan 3 at 19:12
KezerKezer
1,313421
$endgroup$
$begingroup$
Try writing it as $$frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}$$
$endgroup$
– Andrei
Jan 3 at 19:19
add a comment |
$begingroup$
I'm really not too sure how to go on about this:
Problem. Julia (or take MatLab etc. (I guess.)) computes
In [1]: exp(pi*sqrt(67)/6) - sqrt(5280)
Out[1]: 6.121204876308184e-8
Only by looking closely: How many digits are correct? (What could you do to compute $16$ decimals correctly?)
So I know that subtraction is somewhat dangerous numerically - though how do I solve this problem in a precise manner?
numerical-methods floating-point
edited Jan 3 at 20:08
LutzL
57.2k42054
asked Jan 3 at 19:12
KezerKezer
1,313421
$endgroup$
I'm really not too sure how to go on about this:
Problem. Julia (or take MatLab etc. (I guess.)) computes
In [1]: exp(pi*sqrt(67)/6) - sqrt(5280)
Out[1]: 6.121204876308184e-8
Only by looking closely: How many digits are correct? (What could you do to compute $16$ decimals correctly?)
So I know that subtraction is somewhat dangerous numerically - though how do I solve this problem in a precise manner?
numerical-methods floating-point
numerical-methods floating-point
edited Jan 3 at 20:08
LutzL
57.2k42054
asked Jan 3 at 19:12
KezerKezer
1,313421
edited Jan 3 at 20:08
LutzL
57.2k42054
asked Jan 3 at 19:12
KezerKezer
1,313421
edited Jan 3 at 20:08
LutzL
57.2k42054
edited Jan 3 at 20:08
LutzL
57.2k42054
edited Jan 3 at 20:08
LutzL
57.2k42054
57.2k42054
asked Jan 3 at 19:12
KezerKezer
1,313421
asked Jan 3 at 19:12
KezerKezer
1,313421
asked Jan 3 at 19:12
KezerKezer
1,313421
1,313421
$begingroup$
Try writing it as $$frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}$$
$endgroup$
– Andrei
Jan 3 at 19:19
add a comment |
$begingroup$
Try writing it as $$frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}$$
$endgroup$
– Andrei
Jan 3 at 19:19
$begingroup$
Try writing it as $$frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}$$
$endgroup$
– Andrei
Jan 3 at 19:19
$begingroup$
Try writing it as $$frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}$$
$endgroup$
– Andrei
Jan 3 at 19:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using a multi-precision package like the CAS Magma you can compute a value where perhaps the last 10 of 200 digits are polluted by truncation and rounding errors. The loop using different target digit numbers
for N in [0..5] do
RR := RealField(40+32*N);
Exp(Pi(RR)*(RR!67)^(1/2)/6) - (RR!5280)^(1/2);
end for;
gives the result
0.000000061212043348401464497838533606
0.00000006121204334840146449783853360999610568722086907308823071985234
0.0000000612120433484014644978385336099961056872208690730882307198520281414760909167862633336012797951
0.00000006121204334840146449783853360999610568722086907308823071985202814147609091678626333360127979547576192009412710141319214385241
0.00000006121204334840146449783853360999610568722086907308823071985202814147609091678626333360127979547576192009412710141319214385238281932550279412177187787689095831
0.0000000612120433484014644978385336099961056872208690730882307198520281414760909167862633336012797954757619200941271014131921438523828193255027941217718778768909589395497833016775104278900532217412
This shows empirically that all but the last 5 digits are correct. Compared to your floating point result
0.000000061212048763081839
we see that 7 digits are correct. Using the modified formula by Andrei
$$
frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}
$$
gives the floating point result
0.000000061212049284920693
which is further away from the true result, it does not cure the catastrophic cancellation.
If you have two numbers $a,b$ that coincide in $d$ leading digits, then for any smooth function $f$ you will also get coincidence in about $d$ leading digits of $f(a)$ and $f(b)$. This means that the difference $f(a)-f(b)$ has $15-d$ correct digits. There is no way to force more precision, as that is not present in the original floating point values.
Note that this is different from the situation where you want to compute $f(a+h)-f(a)$ where $h$ is explicitly known. Then you can apply Taylor formulas or just a mean value argument like $f(a+h)-f(a)approx f(a+h/2)cdot h$. This will have almost as many correct digits as $h$ has for $|h|<10^{-5}cdot |a|$.
answered Jan 3 at 19:51
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
$begingroup$
To assess the accuracy of your computation analytically you need to understand the errors in each piece. The subtraction is a problem because the numbers are represented by a fixed number of bits of mantissa. This represents a fractional error in the number of about $2^{-n}$ where $n$ is the number of bits. If we know $n$ for the computer in use we know what is left after the subtraction. The math functions should be accurate to (just about) the same number of bits.
We can compute $sqrt{5280} approx 72.6636$ so the size of the difference is about $frac {6.212cdot 10^{-8}}{72.6636}approx 8.5cdot 10^{-10}$ of the numbers themselves. The base $2$ log of this is about $-30$ so you are losing $30$ bits to the subtraction. If you use $64$ bit floats according to the IEEE specification, that leaves $23$ bits of accuracy. As $2^{-23} approx 10^{-7}$ you have about seven decimal digits of accuracy.
answered Jan 3 at 20:19
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060893%2fnumerical-accuracy-how-many-digits-are-correct%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using a multi-precision package like the CAS Magma you can compute a value where perhaps the last 10 of 200 digits are polluted by truncation and rounding errors. The loop using different target digit numbers
for N in [0..5] do
RR := RealField(40+32*N);
Exp(Pi(RR)*(RR!67)^(1/2)/6) - (RR!5280)^(1/2);
end for;
gives the result
0.000000061212043348401464497838533606
0.00000006121204334840146449783853360999610568722086907308823071985234
0.0000000612120433484014644978385336099961056872208690730882307198520281414760909167862633336012797951
0.00000006121204334840146449783853360999610568722086907308823071985202814147609091678626333360127979547576192009412710141319214385241
0.00000006121204334840146449783853360999610568722086907308823071985202814147609091678626333360127979547576192009412710141319214385238281932550279412177187787689095831
0.0000000612120433484014644978385336099961056872208690730882307198520281414760909167862633336012797954757619200941271014131921438523828193255027941217718778768909589395497833016775104278900532217412
This shows empirically that all but the last 5 digits are correct. Compared to your floating point result
0.000000061212048763081839
we see that 7 digits are correct. Using the modified formula by Andrei
$$
frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}
$$
gives the floating point result
0.000000061212049284920693
which is further away from the true result, it does not cure the catastrophic cancellation.
If you have two numbers $a,b$ that coincide in $d$ leading digits, then for any smooth function $f$ you will also get coincidence in about $d$ leading digits of $f(a)$ and $f(b)$. This means that the difference $f(a)-f(b)$ has $15-d$ correct digits. There is no way to force more precision, as that is not present in the original floating point values.
Note that this is different from the situation where you want to compute $f(a+h)-f(a)$ where $h$ is explicitly known. Then you can apply Taylor formulas or just a mean value argument like $f(a+h)-f(a)approx f(a+h/2)cdot h$. This will have almost as many correct digits as $h$ has for $|h|<10^{-5}cdot |a|$.
answered Jan 3 at 19:51
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
$begingroup$
Using a multi-precision package like the CAS Magma you can compute a value where perhaps the last 10 of 200 digits are polluted by truncation and rounding errors. The loop using different target digit numbers
for N in [0..5] do
RR := RealField(40+32*N);
Exp(Pi(RR)*(RR!67)^(1/2)/6) - (RR!5280)^(1/2);
end for;
gives the result
0.000000061212043348401464497838533606
0.00000006121204334840146449783853360999610568722086907308823071985234
0.0000000612120433484014644978385336099961056872208690730882307198520281414760909167862633336012797951
0.00000006121204334840146449783853360999610568722086907308823071985202814147609091678626333360127979547576192009412710141319214385241
0.00000006121204334840146449783853360999610568722086907308823071985202814147609091678626333360127979547576192009412710141319214385238281932550279412177187787689095831
0.0000000612120433484014644978385336099961056872208690730882307198520281414760909167862633336012797954757619200941271014131921438523828193255027941217718778768909589395497833016775104278900532217412
This shows empirically that all but the last 5 digits are correct. Compared to your floating point result
0.000000061212048763081839
we see that 7 digits are correct. Using the modified formula by Andrei
$$
frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}
$$
gives the floating point result
0.000000061212049284920693
which is further away from the true result, it does not cure the catastrophic cancellation.
If you have two numbers $a,b$ that coincide in $d$ leading digits, then for any smooth function $f$ you will also get coincidence in about $d$ leading digits of $f(a)$ and $f(b)$. This means that the difference $f(a)-f(b)$ has $15-d$ correct digits. There is no way to force more precision, as that is not present in the original floating point values.
Note that this is different from the situation where you want to compute $f(a+h)-f(a)$ where $h$ is explicitly known. Then you can apply Taylor formulas or just a mean value argument like $f(a+h)-f(a)approx f(a+h/2)cdot h$. This will have almost as many correct digits as $h$ has for $|h|<10^{-5}cdot |a|$.
answered Jan 3 at 19:51
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
$begingroup$
Using a multi-precision package like the CAS Magma you can compute a value where perhaps the last 10 of 200 digits are polluted by truncation and rounding errors. The loop using different target digit numbers
for N in [0..5] do
RR := RealField(40+32*N);
Exp(Pi(RR)*(RR!67)^(1/2)/6) - (RR!5280)^(1/2);
end for;
gives the result
0.000000061212043348401464497838533606
0.00000006121204334840146449783853360999610568722086907308823071985234
0.0000000612120433484014644978385336099961056872208690730882307198520281414760909167862633336012797951
0.00000006121204334840146449783853360999610568722086907308823071985202814147609091678626333360127979547576192009412710141319214385241
0.00000006121204334840146449783853360999610568722086907308823071985202814147609091678626333360127979547576192009412710141319214385238281932550279412177187787689095831
0.0000000612120433484014644978385336099961056872208690730882307198520281414760909167862633336012797954757619200941271014131921438523828193255027941217718778768909589395497833016775104278900532217412
This shows empirically that all but the last 5 digits are correct. Compared to your floating point result
0.000000061212048763081839
we see that 7 digits are correct. Using the modified formula by Andrei
$$
frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}
$$
gives the floating point result
0.000000061212049284920693
which is further away from the true result, it does not cure the catastrophic cancellation.
If you have two numbers $a,b$ that coincide in $d$ leading digits, then for any smooth function $f$ you will also get coincidence in about $d$ leading digits of $f(a)$ and $f(b)$. This means that the difference $f(a)-f(b)$ has $15-d$ correct digits. There is no way to force more precision, as that is not present in the original floating point values.
Note that this is different from the situation where you want to compute $f(a+h)-f(a)$ where $h$ is explicitly known. Then you can apply Taylor formulas or just a mean value argument like $f(a+h)-f(a)approx f(a+h/2)cdot h$. This will have almost as many correct digits as $h$ has for $|h|<10^{-5}cdot |a|$.
answered Jan 3 at 19:51
LutzLLutzL
57.2k42054
$endgroup$
Using a multi-precision package like the CAS Magma you can compute a value where perhaps the last 10 of 200 digits are polluted by truncation and rounding errors. The loop using different target digit numbers
for N in [0..5] do
RR := RealField(40+32*N);
Exp(Pi(RR)*(RR!67)^(1/2)/6) - (RR!5280)^(1/2);
end for;
gives the result
0.000000061212043348401464497838533606
0.00000006121204334840146449783853360999610568722086907308823071985234
0.0000000612120433484014644978385336099961056872208690730882307198520281414760909167862633336012797951
0.00000006121204334840146449783853360999610568722086907308823071985202814147609091678626333360127979547576192009412710141319214385241
0.00000006121204334840146449783853360999610568722086907308823071985202814147609091678626333360127979547576192009412710141319214385238281932550279412177187787689095831
0.0000000612120433484014644978385336099961056872208690730882307198520281414760909167862633336012797954757619200941271014131921438523828193255027941217718778768909589395497833016775104278900532217412
This shows empirically that all but the last 5 digits are correct. Compared to your floating point result
0.000000061212048763081839
we see that 7 digits are correct. Using the modified formula by Andrei
$$
frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}
$$
gives the floating point result
0.000000061212049284920693
which is further away from the true result, it does not cure the catastrophic cancellation.
If you have two numbers $a,b$ that coincide in $d$ leading digits, then for any smooth function $f$ you will also get coincidence in about $d$ leading digits of $f(a)$ and $f(b)$. This means that the difference $f(a)-f(b)$ has $15-d$ correct digits. There is no way to force more precision, as that is not present in the original floating point values.
Note that this is different from the situation where you want to compute $f(a+h)-f(a)$ where $h$ is explicitly known. Then you can apply Taylor formulas or just a mean value argument like $f(a+h)-f(a)approx f(a+h/2)cdot h$. This will have almost as many correct digits as $h$ has for $|h|<10^{-5}cdot |a|$.
answered Jan 3 at 19:51
LutzLLutzL
57.2k42054
edited Jan 3 at 20:05
answered Jan 3 at 19:51
LutzLLutzL
57.2k42054
answered Jan 3 at 19:51
LutzLLutzL
57.2k42054
answered Jan 3 at 19:51
LutzLLutzL
57.2k42054
57.2k42054
add a comment |
add a comment |
$begingroup$
To assess the accuracy of your computation analytically you need to understand the errors in each piece. The subtraction is a problem because the numbers are represented by a fixed number of bits of mantissa. This represents a fractional error in the number of about $2^{-n}$ where $n$ is the number of bits. If we know $n$ for the computer in use we know what is left after the subtraction. The math functions should be accurate to (just about) the same number of bits.
We can compute $sqrt{5280} approx 72.6636$ so the size of the difference is about $frac {6.212cdot 10^{-8}}{72.6636}approx 8.5cdot 10^{-10}$ of the numbers themselves. The base $2$ log of this is about $-30$ so you are losing $30$ bits to the subtraction. If you use $64$ bit floats according to the IEEE specification, that leaves $23$ bits of accuracy. As $2^{-23} approx 10^{-7}$ you have about seven decimal digits of accuracy.
answered Jan 3 at 20:19
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
To assess the accuracy of your computation analytically you need to understand the errors in each piece. The subtraction is a problem because the numbers are represented by a fixed number of bits of mantissa. This represents a fractional error in the number of about $2^{-n}$ where $n$ is the number of bits. If we know $n$ for the computer in use we know what is left after the subtraction. The math functions should be accurate to (just about) the same number of bits.
We can compute $sqrt{5280} approx 72.6636$ so the size of the difference is about $frac {6.212cdot 10^{-8}}{72.6636}approx 8.5cdot 10^{-10}$ of the numbers themselves. The base $2$ log of this is about $-30$ so you are losing $30$ bits to the subtraction. If you use $64$ bit floats according to the IEEE specification, that leaves $23$ bits of accuracy. As $2^{-23} approx 10^{-7}$ you have about seven decimal digits of accuracy.
answered Jan 3 at 20:19
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
To assess the accuracy of your computation analytically you need to understand the errors in each piece. The subtraction is a problem because the numbers are represented by a fixed number of bits of mantissa. This represents a fractional error in the number of about $2^{-n}$ where $n$ is the number of bits. If we know $n$ for the computer in use we know what is left after the subtraction. The math functions should be accurate to (just about) the same number of bits.
We can compute $sqrt{5280} approx 72.6636$ so the size of the difference is about $frac {6.212cdot 10^{-8}}{72.6636}approx 8.5cdot 10^{-10}$ of the numbers themselves. The base $2$ log of this is about $-30$ so you are losing $30$ bits to the subtraction. If you use $64$ bit floats according to the IEEE specification, that leaves $23$ bits of accuracy. As $2^{-23} approx 10^{-7}$ you have about seven decimal digits of accuracy.
answered Jan 3 at 20:19
Ross MillikanRoss Millikan
293k23197371
$endgroup$
To assess the accuracy of your computation analytically you need to understand the errors in each piece. The subtraction is a problem because the numbers are represented by a fixed number of bits of mantissa. This represents a fractional error in the number of about $2^{-n}$ where $n$ is the number of bits. If we know $n$ for the computer in use we know what is left after the subtraction. The math functions should be accurate to (just about) the same number of bits.
We can compute $sqrt{5280} approx 72.6636$ so the size of the difference is about $frac {6.212cdot 10^{-8}}{72.6636}approx 8.5cdot 10^{-10}$ of the numbers themselves. The base $2$ log of this is about $-30$ so you are losing $30$ bits to the subtraction. If you use $64$ bit floats according to the IEEE specification, that leaves $23$ bits of accuracy. As $2^{-23} approx 10^{-7}$ you have about seven decimal digits of accuracy.
answered Jan 3 at 20:19
Ross MillikanRoss Millikan
293k23197371
answered Jan 3 at 20:19
Ross MillikanRoss Millikan
293k23197371
answered Jan 3 at 20:19
Ross MillikanRoss Millikan
293k23197371
answered Jan 3 at 20:19
Ross MillikanRoss Millikan
293k23197371
293k23197371
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060893%2fnumerical-accuracy-how-many-digits-are-correct%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Try writing it as $$frac{e^{2pisqrt{67}/6}-5280}{e^{pisqrt{67}/6}+sqrt{5280}}$$
$endgroup$
– Andrei
Jan 3 at 19:19