An Inconsistency in Numerical Approximation











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6
down vote

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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?










share|cite|improve this question




















  • 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

















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?










share|cite|improve this question




















  • 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















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?










share|cite|improve this question















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






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share|cite|improve this question













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edited Nov 29 at 11:14









Servaes

21.8k33792




21.8k33792










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
















  • 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












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 $O(1)$.






share|cite|improve this answer





















  • 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




















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 $0$.






share|cite|improve this answer




























    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.






    share|cite|improve this answer























    • I don't recall claiming there was a paradox, only a numerical issue.
      – SZN
      Nov 28 at 20:05











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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 $O(1)$.






    share|cite|improve this answer





















    • 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

















    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 $O(1)$.






    share|cite|improve this answer





















    • 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















    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 $O(1)$.






    share|cite|improve this answer












    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 $O(1)$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 at 19:50









    Robert Israel

    315k23206455




    315k23206455












    • 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






    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












    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 $0$.






    share|cite|improve this answer

























      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 $0$.






      share|cite|improve this answer























        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 $0$.






        share|cite|improve this answer












        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 $0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 19:43









        José Carlos Santos

        144k20113214




        144k20113214






















            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.






            share|cite|improve this answer























            • I don't recall claiming there was a paradox, only a numerical issue.
              – SZN
              Nov 28 at 20:05















            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.






            share|cite|improve this answer























            • I don't recall claiming there was a paradox, only a numerical issue.
              – SZN
              Nov 28 at 20:05













            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.






            share|cite|improve this answer














            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 28 at 20:04

























            answered Nov 28 at 19:57









            Yves Daoust

            123k668218




            123k668218












            • 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




            I don't recall claiming there was a paradox, only a numerical issue.
            – SZN
            Nov 28 at 20:05


















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