$fracb+delta(b)a+delta(a)$ — Calculus for the Ambitious

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I am reading book titled "Calculus for the Ambitious" by TW Korner, he had made it available legally for free here: https://www.dpmms.cam.ac.uk/~twk/



In first chapter he is teaching about how small errors affect bigger values. On page number 7, he has this problem:



$fracb+delta(b)a+delta(a)$



He says, "if we stare at the problem long enough, the following idea may occur to us:



$u = frac1a$ and $u + delta(u) = frac1a+delta(a)$



and goes onto solving the equations for delta(u) and then finds this at the end:



$fracb+delta(b)a+delta(a)$ = $fracba - fracb * delta(a)a^2 + fracdelta(b)a$



enter image description here



enter image description here



I got 2 questions:



  1. I can stare long enough, 4 hours, but such ideas don't occur to me.

  2. why choose $u = frac1a$, I can also use: $u - delta(u) = a$, and it will simplify the answer to: $fracba - fracdelta(b)a$ . Why choose one over the other ?



calculus




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    You can type it better with math.meta.stackexchange.com/questions/5020/…
    – Calvin Khor
    2 days ago






  • 1




    Thanks for the link. It's so cool to learn MathJax :)
    – Arnuld
    2 days ago










  • You're welcome :)
    – Calvin Khor
    2 days ago














up vote
1
down vote

favorite












I am reading book titled "Calculus for the Ambitious" by TW Korner, he had made it available legally for free here: https://www.dpmms.cam.ac.uk/~twk/



In first chapter he is teaching about how small errors affect bigger values. On page number 7, he has this problem:



$fracb+delta(b)a+delta(a)$



He says, "if we stare at the problem long enough, the following idea may occur to us:



$u = frac1a$ and $u + delta(u) = frac1a+delta(a)$



and goes onto solving the equations for delta(u) and then finds this at the end:



$fracb+delta(b)a+delta(a)$ = $fracba - fracb * delta(a)a^2 + fracdelta(b)a$



enter image description here



enter image description here



I got 2 questions:



  1. I can stare long enough, 4 hours, but such ideas don't occur to me.

  2. why choose $u = frac1a$, I can also use: $u - delta(u) = a$, and it will simplify the answer to: $fracba - fracdelta(b)a$ . Why choose one over the other ?



calculus




share|cite|improve this question

















  • 1




    You can type it better with math.meta.stackexchange.com/questions/5020/…
    – Calvin Khor
    2 days ago






  • 1




    Thanks for the link. It's so cool to learn MathJax :)
    – Arnuld
    2 days ago










  • You're welcome :)
    – Calvin Khor
    2 days ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I am reading book titled "Calculus for the Ambitious" by TW Korner, he had made it available legally for free here: https://www.dpmms.cam.ac.uk/~twk/



In first chapter he is teaching about how small errors affect bigger values. On page number 7, he has this problem:



$fracb+delta(b)a+delta(a)$



He says, "if we stare at the problem long enough, the following idea may occur to us:



$u = frac1a$ and $u + delta(u) = frac1a+delta(a)$



and goes onto solving the equations for delta(u) and then finds this at the end:



$fracb+delta(b)a+delta(a)$ = $fracba - fracb * delta(a)a^2 + fracdelta(b)a$



enter image description here



enter image description here



I got 2 questions:



  1. I can stare long enough, 4 hours, but such ideas don't occur to me.

  2. why choose $u = frac1a$, I can also use: $u - delta(u) = a$, and it will simplify the answer to: $fracba - fracdelta(b)a$ . Why choose one over the other ?



calculus




share|cite|improve this question













I am reading book titled "Calculus for the Ambitious" by TW Korner, he had made it available legally for free here: https://www.dpmms.cam.ac.uk/~twk/



In first chapter he is teaching about how small errors affect bigger values. On page number 7, he has this problem:



$fracb+delta(b)a+delta(a)$



He says, "if we stare at the problem long enough, the following idea may occur to us:



$u = frac1a$ and $u + delta(u) = frac1a+delta(a)$



and goes onto solving the equations for delta(u) and then finds this at the end:



$fracb+delta(b)a+delta(a)$ = $fracba - fracb * delta(a)a^2 + fracdelta(b)a$



enter image description here



enter image description here



I got 2 questions:



  1. I can stare long enough, 4 hours, but such ideas don't occur to me.

  2. why choose $u = frac1a$, I can also use: $u - delta(u) = a$, and it will simplify the answer to: $fracba - fracdelta(b)a$ . Why choose one over the other ?




calculus





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share|cite|improve this question




share|cite|improve this question








edited 2 days ago
























asked 2 days ago









Arnuld

165




165







  • 1




    You can type it better with math.meta.stackexchange.com/questions/5020/…
    – Calvin Khor
    2 days ago






  • 1




    Thanks for the link. It's so cool to learn MathJax :)
    – Arnuld
    2 days ago










  • You're welcome :)
    – Calvin Khor
    2 days ago












  • 1




    You can type it better with math.meta.stackexchange.com/questions/5020/…
    – Calvin Khor
    2 days ago






  • 1




    Thanks for the link. It's so cool to learn MathJax :)
    – Arnuld
    2 days ago










  • You're welcome :)
    – Calvin Khor
    2 days ago







1




1




You can type it better with math.meta.stackexchange.com/questions/5020/…
– Calvin Khor
2 days ago




You can type it better with math.meta.stackexchange.com/questions/5020/…
– Calvin Khor
2 days ago




1




1




Thanks for the link. It's so cool to learn MathJax :)
– Arnuld
2 days ago




Thanks for the link. It's so cool to learn MathJax :)
– Arnuld
2 days ago












You're welcome :)
– Calvin Khor
2 days ago




You're welcome :)
– Calvin Khor
2 days ago










1 Answer
1






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$textsf(1)$ Long enough can sometimes mean too long to be sensible...but the time gets shorter when you see more problems.



$textsf(2)$ If you use $u=frac1a$ and try $u+delta u = a $ then $delta u = a - frac1a$ which can be very large, for instance if $a=100$ then $u = 0.01$, but $delta u = 99.99 $. So the approximation property
$$ u+delta u approx u$$
doesn't hold.






share|cite|improve this answer





















  • You used numbers (integers) to clarify the problem and I understood it very well. Now, author himself in the book, did not use any numbers. Should I assume, such kind of error-correction was already done using numbers while he was writing the book or while such error-formulas are being derived/found, they are always backed by practical proof using numbers.
    – Arnuld
    2 days ago










  • @Arnuld I think its correct to think of what he writes as being true for any (reasonable) collection of actual numbers, yes. In this case, the choice could have been motivated by the graph of $1/x$, and a graph is a large collection of such 'practical' numerical evidence.
    – Calvin Khor
    2 days ago











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1 Answer
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active

oldest

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active

oldest

votes






active

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votes








up vote
0
down vote













$textsf(1)$ Long enough can sometimes mean too long to be sensible...but the time gets shorter when you see more problems.



$textsf(2)$ If you use $u=frac1a$ and try $u+delta u = a $ then $delta u = a - frac1a$ which can be very large, for instance if $a=100$ then $u = 0.01$, but $delta u = 99.99 $. So the approximation property
$$ u+delta u approx u$$
doesn't hold.






share|cite|improve this answer





















  • You used numbers (integers) to clarify the problem and I understood it very well. Now, author himself in the book, did not use any numbers. Should I assume, such kind of error-correction was already done using numbers while he was writing the book or while such error-formulas are being derived/found, they are always backed by practical proof using numbers.
    – Arnuld
    2 days ago










  • @Arnuld I think its correct to think of what he writes as being true for any (reasonable) collection of actual numbers, yes. In this case, the choice could have been motivated by the graph of $1/x$, and a graph is a large collection of such 'practical' numerical evidence.
    – Calvin Khor
    2 days ago















up vote
0
down vote













$textsf(1)$ Long enough can sometimes mean too long to be sensible...but the time gets shorter when you see more problems.



$textsf(2)$ If you use $u=frac1a$ and try $u+delta u = a $ then $delta u = a - frac1a$ which can be very large, for instance if $a=100$ then $u = 0.01$, but $delta u = 99.99 $. So the approximation property
$$ u+delta u approx u$$
doesn't hold.






share|cite|improve this answer





















  • You used numbers (integers) to clarify the problem and I understood it very well. Now, author himself in the book, did not use any numbers. Should I assume, such kind of error-correction was already done using numbers while he was writing the book or while such error-formulas are being derived/found, they are always backed by practical proof using numbers.
    – Arnuld
    2 days ago










  • @Arnuld I think its correct to think of what he writes as being true for any (reasonable) collection of actual numbers, yes. In this case, the choice could have been motivated by the graph of $1/x$, and a graph is a large collection of such 'practical' numerical evidence.
    – Calvin Khor
    2 days ago













up vote
0
down vote










up vote
0
down vote









$textsf(1)$ Long enough can sometimes mean too long to be sensible...but the time gets shorter when you see more problems.



$textsf(2)$ If you use $u=frac1a$ and try $u+delta u = a $ then $delta u = a - frac1a$ which can be very large, for instance if $a=100$ then $u = 0.01$, but $delta u = 99.99 $. So the approximation property
$$ u+delta u approx u$$
doesn't hold.






share|cite|improve this answer













$textsf(1)$ Long enough can sometimes mean too long to be sensible...but the time gets shorter when you see more problems.



$textsf(2)$ If you use $u=frac1a$ and try $u+delta u = a $ then $delta u = a - frac1a$ which can be very large, for instance if $a=100$ then $u = 0.01$, but $delta u = 99.99 $. So the approximation property
$$ u+delta u approx u$$
doesn't hold.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered 2 days ago









Calvin Khor

7,93911032




7,93911032











  • You used numbers (integers) to clarify the problem and I understood it very well. Now, author himself in the book, did not use any numbers. Should I assume, such kind of error-correction was already done using numbers while he was writing the book or while such error-formulas are being derived/found, they are always backed by practical proof using numbers.
    – Arnuld
    2 days ago










  • @Arnuld I think its correct to think of what he writes as being true for any (reasonable) collection of actual numbers, yes. In this case, the choice could have been motivated by the graph of $1/x$, and a graph is a large collection of such 'practical' numerical evidence.
    – Calvin Khor
    2 days ago

















  • You used numbers (integers) to clarify the problem and I understood it very well. Now, author himself in the book, did not use any numbers. Should I assume, such kind of error-correction was already done using numbers while he was writing the book or while such error-formulas are being derived/found, they are always backed by practical proof using numbers.
    – Arnuld
    2 days ago










  • @Arnuld I think its correct to think of what he writes as being true for any (reasonable) collection of actual numbers, yes. In this case, the choice could have been motivated by the graph of $1/x$, and a graph is a large collection of such 'practical' numerical evidence.
    – Calvin Khor
    2 days ago
















You used numbers (integers) to clarify the problem and I understood it very well. Now, author himself in the book, did not use any numbers. Should I assume, such kind of error-correction was already done using numbers while he was writing the book or while such error-formulas are being derived/found, they are always backed by practical proof using numbers.
– Arnuld
2 days ago




You used numbers (integers) to clarify the problem and I understood it very well. Now, author himself in the book, did not use any numbers. Should I assume, such kind of error-correction was already done using numbers while he was writing the book or while such error-formulas are being derived/found, they are always backed by practical proof using numbers.
– Arnuld
2 days ago












@Arnuld I think its correct to think of what he writes as being true for any (reasonable) collection of actual numbers, yes. In this case, the choice could have been motivated by the graph of $1/x$, and a graph is a large collection of such 'practical' numerical evidence.
– Calvin Khor
2 days ago





@Arnuld I think its correct to think of what he writes as being true for any (reasonable) collection of actual numbers, yes. In this case, the choice could have been motivated by the graph of $1/x$, and a graph is a large collection of such 'practical' numerical evidence.
– Calvin Khor
2 days ago













 

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