An Inconsistency in Numerical Approximation
up vote
6
down vote
favorite
Consider the expression
$$
10^5 - frac{10^{10}}{1+10^5}.
$$
Using the elementary properties of fractions we can evaluate the expression as
$$
10^5 - frac{10^{10}}{1+10^5} = frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = frac{10^5}{1+10^5}approx 1.
$$
Note that the approximation $10^5+1 approx 10^5$ is used in the last step. Now suppose we use the same approximation, but apply it before we perform the subtraction. We get
$$
10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} = 0.
$$
The same logic works for
$$
10^p - frac{10^{2p}}{1+10^p}
$$
for arbitrary large $p$, so it cannot be simply an issue with the accuracy of the approximation.
Is there an easy explanation of what's going on here?
arithmetic approximation fractions
edited Nov 29 at 11:14
Servaes
22.2k33793
asked Nov 28 at 19:37
SZN
2,713720
|
show 3 more comments
up vote
6
down vote
favorite
Consider the expression
$$
10^5 - frac{10^{10}}{1+10^5}.
$$
Using the elementary properties of fractions we can evaluate the expression as
$$
10^5 - frac{10^{10}}{1+10^5} = frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = frac{10^5}{1+10^5}approx 1.
$$
Note that the approximation $10^5+1 approx 10^5$ is used in the last step. Now suppose we use the same approximation, but apply it before we perform the subtraction. We get
$$
10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} = 0.
$$
The same logic works for
$$
10^p - frac{10^{2p}}{1+10^p}
$$
for arbitrary large $p$, so it cannot be simply an issue with the accuracy of the approximation.
Is there an easy explanation of what's going on here?
arithmetic approximation fractions
edited Nov 29 at 11:14
Servaes
22.2k33793
asked Nov 28 at 19:37
SZN
2,713720
4
en.wikipedia.org/wiki/Loss_of_significance might be a starting point
– Thomas
Nov 28 at 19:41
5
This is called "catastrophic cancellation"; see en.wikipedia.org/wiki/Loss_of_significance . The subject of numerical analysis is largely devoted to studying & combating this phenomenon, teaching in general how to calculate according to your first example.
– kimchi lover
Nov 28 at 19:42
relative to the numbers involved, the error is still pretty small, $(1-0)/10^5=10^{-5}$
– Vasya
Nov 28 at 19:48
@Vasya yes, but the difference between $1$ and $0$ leads to considerably different answers if this factor happens to be multiplying another and the approximation is applied improperly!
– SZN
Nov 28 at 19:50
1
Interestingly, the error compounds quickly; $$10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} approx 10^5 - frac{10^{10}}{10^5-1}approxldots approx 10^5 - frac{10^{10}}{2} approx 10^5 - frac{10^{10}}{1} approx-10^{10}.$$
– Servaes
Nov 29 at 1:17
|
show 3 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Consider the expression
$$
10^5 - frac{10^{10}}{1+10^5}.
$$
Using the elementary properties of fractions we can evaluate the expression as
$$
10^5 - frac{10^{10}}{1+10^5} = frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = frac{10^5}{1+10^5}approx 1.
$$
Note that the approximation $10^5+1 approx 10^5$ is used in the last step. Now suppose we use the same approximation, but apply it before we perform the subtraction. We get
$$
10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} = 0.
$$
The same logic works for
$$
10^p - frac{10^{2p}}{1+10^p}
$$
for arbitrary large $p$, so it cannot be simply an issue with the accuracy of the approximation.
Is there an easy explanation of what's going on here?
arithmetic approximation fractions
edited Nov 29 at 11:14
Servaes
22.2k33793
asked Nov 28 at 19:37
SZN
2,713720
Consider the expression
$$
10^5 - frac{10^{10}}{1+10^5}.
$$
Using the elementary properties of fractions we can evaluate the expression as
$$
10^5 - frac{10^{10}}{1+10^5} = frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = frac{10^5}{1+10^5}approx 1.
$$
Note that the approximation $10^5+1 approx 10^5$ is used in the last step. Now suppose we use the same approximation, but apply it before we perform the subtraction. We get
$$
10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} = 0.
$$
The same logic works for
$$
10^p - frac{10^{2p}}{1+10^p}
$$
for arbitrary large $p$, so it cannot be simply an issue with the accuracy of the approximation.
Is there an easy explanation of what's going on here?
arithmetic approximation fractions
arithmetic approximation fractions
edited Nov 29 at 11:14
Servaes
22.2k33793
asked Nov 28 at 19:37
SZN
2,713720
edited Nov 29 at 11:14
Servaes
22.2k33793
asked Nov 28 at 19:37
SZN
2,713720
edited Nov 29 at 11:14
Servaes
22.2k33793
edited Nov 29 at 11:14
Servaes
22.2k33793
edited Nov 29 at 11:14
Servaes
22.2k33793
22.2k33793
asked Nov 28 at 19:37
SZN
2,713720
asked Nov 28 at 19:37
SZN
2,713720
asked Nov 28 at 19:37
SZN
2,713720
2,713720
4
en.wikipedia.org/wiki/Loss_of_significance might be a starting point
– Thomas
Nov 28 at 19:41
5
This is called "catastrophic cancellation"; see en.wikipedia.org/wiki/Loss_of_significance . The subject of numerical analysis is largely devoted to studying & combating this phenomenon, teaching in general how to calculate according to your first example.
– kimchi lover
Nov 28 at 19:42
relative to the numbers involved, the error is still pretty small, $(1-0)/10^5=10^{-5}$
– Vasya
Nov 28 at 19:48
@Vasya yes, but the difference between $1$ and $0$ leads to considerably different answers if this factor happens to be multiplying another and the approximation is applied improperly!
– SZN
Nov 28 at 19:50
1
Interestingly, the error compounds quickly; $$10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} approx 10^5 - frac{10^{10}}{10^5-1}approxldots approx 10^5 - frac{10^{10}}{2} approx 10^5 - frac{10^{10}}{1} approx-10^{10}.$$
– Servaes
Nov 29 at 1:17
|
show 3 more comments
4
en.wikipedia.org/wiki/Loss_of_significance might be a starting point
– Thomas
Nov 28 at 19:41
5
This is called "catastrophic cancellation"; see en.wikipedia.org/wiki/Loss_of_significance . The subject of numerical analysis is largely devoted to studying & combating this phenomenon, teaching in general how to calculate according to your first example.
– kimchi lover
Nov 28 at 19:42
relative to the numbers involved, the error is still pretty small, $(1-0)/10^5=10^{-5}$
– Vasya
Nov 28 at 19:48
@Vasya yes, but the difference between $1$ and $0$ leads to considerably different answers if this factor happens to be multiplying another and the approximation is applied improperly!
– SZN
Nov 28 at 19:50
1
Interestingly, the error compounds quickly; $$10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} approx 10^5 - frac{10^{10}}{10^5-1}approxldots approx 10^5 - frac{10^{10}}{2} approx 10^5 - frac{10^{10}}{1} approx-10^{10}.$$
– Servaes
Nov 29 at 1:17
4
4
en.wikipedia.org/wiki/Loss_of_significance might be a starting point
– Thomas
Nov 28 at 19:41
en.wikipedia.org/wiki/Loss_of_significance might be a starting point
– Thomas
Nov 28 at 19:41
5
5
This is called "catastrophic cancellation"; see en.wikipedia.org/wiki/Loss_of_significance . The subject of numerical analysis is largely devoted to studying & combating this phenomenon, teaching in general how to calculate according to your first example.
– kimchi lover
Nov 28 at 19:42
This is called "catastrophic cancellation"; see en.wikipedia.org/wiki/Loss_of_significance . The subject of numerical analysis is largely devoted to studying & combating this phenomenon, teaching in general how to calculate according to your first example.
– kimchi lover
Nov 28 at 19:42
relative to the numbers involved, the error is still pretty small, $(1-0)/10^5=10^{-5}$
– Vasya
Nov 28 at 19:48
relative to the numbers involved, the error is still pretty small, $(1-0)/10^5=10^{-5}$
– Vasya
Nov 28 at 19:48
@Vasya yes, but the difference between $1$ and $0$ leads to considerably different answers if this factor happens to be multiplying another and the approximation is applied improperly!
– SZN
Nov 28 at 19:50
@Vasya yes, but the difference between $1$ and $0$ leads to considerably different answers if this factor happens to be multiplying another and the approximation is applied improperly!
– SZN
Nov 28 at 19:50
1
1
Interestingly, the error compounds quickly; $$10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} approx 10^5 - frac{10^{10}}{10^5-1}approxldots approx 10^5 - frac{10^{10}}{2} approx 10^5 - frac{10^{10}}{1} approx-10^{10}.$$
– Servaes
Nov 29 at 1:17
Interestingly, the error compounds quickly; $$10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} approx 10^5 - frac{10^{10}}{10^5-1}approxldots approx 10^5 - frac{10^{10}}{2} approx 10^5 - frac{10^{10}}{1} approx-10^{10}.$$
– Servaes
Nov 29 at 1:17
|
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
It is simply an issue of accuracy of approximation. Let me write $x = 10^p$. Then your expression is
$$ x - frac{x^2}{1+x}$$
Note that $$frac{x^2}{1+x} = frac{x}{1/x + 1} = x (1 - 1/x + O(1/x^2)) = x - 1 + O(1/x)$$
so that
$$ x - frac{x^2}{1+x} = x - (x - 1 + O(1/x)) = 1 + O(1/x)$$
In your second calculation you only evaluated $x^2/(1+x)$ to within $O(1)$, not $O(1/x)$, so naturally you have an error at the end that is $O(1)$.
answered Nov 28 at 19:50
Robert Israel
317k23206457
Thank you! This makes a lot of sense. I should have been clearer with my use of the word "approximation", which I took to be the magnitude of the error $$e = 1-frac{10^p}{1+10^{p}}$$ rather than the order to which this approximation effectively takes the Taylor series.
– SZN
Nov 28 at 20:03
add a comment |
up vote
1
down vote
The first approximation is fine. The second on is not, because, $10^5$ and $dfrac{10^{10}}{1+10^5}$ are large numbers with approximately the same size. You are saying that since $10,001$ is close to $10,000$, then $1$ is close to $0$.
answered Nov 28 at 19:43
José Carlos Santos
147k22117217
add a comment |
up vote
1
down vote
There is no paradox.
When you approximate $$frac{10^5}{1+10^5}=1-0.000099999000cdots$$ with $1$, the error is on the order of $10^{-5}$.
But in the second case, the same error is multiplied by $10^5$, so that it is no more negligible.
answered Nov 28 at 19:57
Yves Daoust
123k668219
I don't recall claiming there was a paradox, only a numerical issue.
– SZN
Nov 28 at 20:05
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
up vote
6
down vote
accepted
It is simply an issue of accuracy of approximation. Let me write $x = 10^p$. Then your expression is
$$ x - frac{x^2}{1+x}$$
Note that $$frac{x^2}{1+x} = frac{x}{1/x + 1} = x (1 - 1/x + O(1/x^2)) = x - 1 + O(1/x)$$
so that
$$ x - frac{x^2}{1+x} = x - (x - 1 + O(1/x)) = 1 + O(1/x)$$
In your second calculation you only evaluated $x^2/(1+x)$ to within $O(1)$, not $O(1/x)$, so naturally you have an error at the end that is $O(1)$.
answered Nov 28 at 19:50
Robert Israel
317k23206457
Thank you! This makes a lot of sense. I should have been clearer with my use of the word "approximation", which I took to be the magnitude of the error $$e = 1-frac{10^p}{1+10^{p}}$$ rather than the order to which this approximation effectively takes the Taylor series.
– SZN
Nov 28 at 20:03
add a comment |
up vote
6
down vote
accepted
It is simply an issue of accuracy of approximation. Let me write $x = 10^p$. Then your expression is
$$ x - frac{x^2}{1+x}$$
Note that $$frac{x^2}{1+x} = frac{x}{1/x + 1} = x (1 - 1/x + O(1/x^2)) = x - 1 + O(1/x)$$
so that
$$ x - frac{x^2}{1+x} = x - (x - 1 + O(1/x)) = 1 + O(1/x)$$
In your second calculation you only evaluated $x^2/(1+x)$ to within $O(1)$, not $O(1/x)$, so naturally you have an error at the end that is $O(1)$.
answered Nov 28 at 19:50
Robert Israel
317k23206457
Thank you! This makes a lot of sense. I should have been clearer with my use of the word "approximation", which I took to be the magnitude of the error $$e = 1-frac{10^p}{1+10^{p}}$$ rather than the order to which this approximation effectively takes the Taylor series.
– SZN
Nov 28 at 20:03
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
It is simply an issue of accuracy of approximation. Let me write $x = 10^p$. Then your expression is
$$ x - frac{x^2}{1+x}$$
Note that $$frac{x^2}{1+x} = frac{x}{1/x + 1} = x (1 - 1/x + O(1/x^2)) = x - 1 + O(1/x)$$
so that
$$ x - frac{x^2}{1+x} = x - (x - 1 + O(1/x)) = 1 + O(1/x)$$
In your second calculation you only evaluated $x^2/(1+x)$ to within $O(1)$, not $O(1/x)$, so naturally you have an error at the end that is $O(1)$.
answered Nov 28 at 19:50
Robert Israel
317k23206457
It is simply an issue of accuracy of approximation. Let me write $x = 10^p$. Then your expression is
$$ x - frac{x^2}{1+x}$$
Note that $$frac{x^2}{1+x} = frac{x}{1/x + 1} = x (1 - 1/x + O(1/x^2)) = x - 1 + O(1/x)$$
so that
$$ x - frac{x^2}{1+x} = x - (x - 1 + O(1/x)) = 1 + O(1/x)$$
In your second calculation you only evaluated $x^2/(1+x)$ to within $O(1)$, not $O(1/x)$, so naturally you have an error at the end that is $O(1)$.
answered Nov 28 at 19:50
Robert Israel
317k23206457
answered Nov 28 at 19:50
Robert Israel
317k23206457
answered Nov 28 at 19:50
Robert Israel
317k23206457
answered Nov 28 at 19:50
Robert Israel
317k23206457
317k23206457
Thank you! This makes a lot of sense. I should have been clearer with my use of the word "approximation", which I took to be the magnitude of the error $$e = 1-frac{10^p}{1+10^{p}}$$ rather than the order to which this approximation effectively takes the Taylor series.
– SZN
Nov 28 at 20:03
add a comment |
Thank you! This makes a lot of sense. I should have been clearer with my use of the word "approximation", which I took to be the magnitude of the error $$e = 1-frac{10^p}{1+10^{p}}$$ rather than the order to which this approximation effectively takes the Taylor series.
– SZN
Nov 28 at 20:03
Thank you! This makes a lot of sense. I should have been clearer with my use of the word "approximation", which I took to be the magnitude of the error $$e = 1-frac{10^p}{1+10^{p}}$$ rather than the order to which this approximation effectively takes the Taylor series.
– SZN
Nov 28 at 20:03
Thank you! This makes a lot of sense. I should have been clearer with my use of the word "approximation", which I took to be the magnitude of the error $$e = 1-frac{10^p}{1+10^{p}}$$ rather than the order to which this approximation effectively takes the Taylor series.
– SZN
Nov 28 at 20:03
add a comment |
up vote
1
down vote
The first approximation is fine. The second on is not, because, $10^5$ and $dfrac{10^{10}}{1+10^5}$ are large numbers with approximately the same size. You are saying that since $10,001$ is close to $10,000$, then $1$ is close to $0$.
answered Nov 28 at 19:43
José Carlos Santos
147k22117217
add a comment |
up vote
1
down vote
The first approximation is fine. The second on is not, because, $10^5$ and $dfrac{10^{10}}{1+10^5}$ are large numbers with approximately the same size. You are saying that since $10,001$ is close to $10,000$, then $1$ is close to $0$.
answered Nov 28 at 19:43
José Carlos Santos
147k22117217
add a comment |
up vote
1
down vote
up vote
1
down vote
The first approximation is fine. The second on is not, because, $10^5$ and $dfrac{10^{10}}{1+10^5}$ are large numbers with approximately the same size. You are saying that since $10,001$ is close to $10,000$, then $1$ is close to $0$.
answered Nov 28 at 19:43
José Carlos Santos
147k22117217
The first approximation is fine. The second on is not, because, $10^5$ and $dfrac{10^{10}}{1+10^5}$ are large numbers with approximately the same size. You are saying that since $10,001$ is close to $10,000$, then $1$ is close to $0$.
answered Nov 28 at 19:43
José Carlos Santos
147k22117217
answered Nov 28 at 19:43
José Carlos Santos
147k22117217
answered Nov 28 at 19:43
José Carlos Santos
147k22117217
answered Nov 28 at 19:43
José Carlos Santos
147k22117217
147k22117217
add a comment |
add a comment |
up vote
1
down vote
There is no paradox.
When you approximate $$frac{10^5}{1+10^5}=1-0.000099999000cdots$$ with $1$, the error is on the order of $10^{-5}$.
But in the second case, the same error is multiplied by $10^5$, so that it is no more negligible.
answered Nov 28 at 19:57
Yves Daoust
123k668219
I don't recall claiming there was a paradox, only a numerical issue.
– SZN
Nov 28 at 20:05
add a comment |
up vote
1
down vote
There is no paradox.
When you approximate $$frac{10^5}{1+10^5}=1-0.000099999000cdots$$ with $1$, the error is on the order of $10^{-5}$.
But in the second case, the same error is multiplied by $10^5$, so that it is no more negligible.
answered Nov 28 at 19:57
Yves Daoust
123k668219
I don't recall claiming there was a paradox, only a numerical issue.
– SZN
Nov 28 at 20:05
add a comment |
up vote
1
down vote
up vote
1
down vote
There is no paradox.
When you approximate $$frac{10^5}{1+10^5}=1-0.000099999000cdots$$ with $1$, the error is on the order of $10^{-5}$.
But in the second case, the same error is multiplied by $10^5$, so that it is no more negligible.
answered Nov 28 at 19:57
Yves Daoust
123k668219
There is no paradox.
When you approximate $$frac{10^5}{1+10^5}=1-0.000099999000cdots$$ with $1$, the error is on the order of $10^{-5}$.
But in the second case, the same error is multiplied by $10^5$, so that it is no more negligible.
answered Nov 28 at 19:57
Yves Daoust
123k668219
edited Nov 28 at 20:04
answered Nov 28 at 19:57
Yves Daoust
123k668219
answered Nov 28 at 19:57
Yves Daoust
123k668219
answered Nov 28 at 19:57
Yves Daoust
123k668219
123k668219
I don't recall claiming there was a paradox, only a numerical issue.
– SZN
Nov 28 at 20:05
add a comment |
I don't recall claiming there was a paradox, only a numerical issue.
– SZN
Nov 28 at 20:05
I don't recall claiming there was a paradox, only a numerical issue.
– SZN
Nov 28 at 20:05
I don't recall claiming there was a paradox, only a numerical issue.
– SZN
Nov 28 at 20:05
add a comment |
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4
en.wikipedia.org/wiki/Loss_of_significance might be a starting point
– Thomas
Nov 28 at 19:41
5
This is called "catastrophic cancellation"; see en.wikipedia.org/wiki/Loss_of_significance . The subject of numerical analysis is largely devoted to studying & combating this phenomenon, teaching in general how to calculate according to your first example.
– kimchi lover
Nov 28 at 19:42
relative to the numbers involved, the error is still pretty small, $(1-0)/10^5=10^{-5}$
– Vasya
Nov 28 at 19:48
@Vasya yes, but the difference between $1$ and $0$ leads to considerably different answers if this factor happens to be multiplying another and the approximation is applied improperly!
– SZN
Nov 28 at 19:50
1
Interestingly, the error compounds quickly; $$10^5 - frac{10^{10}}{1+10^5} approx 10^5 - frac{10^{10}}{10^5} approx 10^5 - frac{10^{10}}{10^5-1}approxldots approx 10^5 - frac{10^{10}}{2} approx 10^5 - frac{10^{10}}{1} approx-10^{10}.$$
– Servaes
Nov 29 at 1:17