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Calculating Logarithms by hand in 1969

  1. Aug 29, 2011 #1
    This question is not about how to calculate the logarithm, but rather what method would be expected of someone in 1969.

    I am going through Apostol's Calculus, and in section 6.10 Apostol introduces polynomial approximations to the natural logarithm. Specifically, he introduces the following theorem:

    If [itex]0<x<1[/itex] and if [itex]m \ge 1[/itex] we have

    [itex]\ln{\frac{1+x}{1-x}}=2(x+\frac{x^3}{3}+...+\frac{x^{2m-1}}{2m-1})+R_m(x)[/itex]

    where the error term, [itex]R_m (x)[/itex], satisfies the inequalities

    [itex]\frac{x^{2m+1}}{2m+1}<R_m (x) \le \frac{2-x}{1-x} \frac{x^{2m+1}}{2m+1}[/itex]

    The first question of section 6.11 instructs the reader to use this theorem with [itex]x=\frac{1}{3}[/itex] and [itex]m=5[/itex] in order to calculate approvimations to [itex]\ln{2}[/itex]. It instructs to retain nine decimals in the calculations in order to obtain the inequalities [itex]0.6931460<ln{2}<0.6931476[/itex].

    Now, I know how to do this; what I mean is that I know the mechanical procedure required to arithmetically calculate this out by hand, but is that really what would be expected of someone in 1969 (when this book was published)? I have tried to do it out by hand three times and failed every time due to a simple mistake at some point. I know calculators were just being developed around that time, but I'm not sure to what extent I should use one, or if I should allow myself to use one at all. What were Apostol's expectations of his students here? I will say, however, that there are 4 questions like this, and given my current rate I may be stuck on these for a while.
     
  2. jcsd
  3. Aug 30, 2011 #2

    Borek

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    As far as I can tell nobody used such methods in practice - logarithms were either read from the tables

    numerki.jpg

    or from the slide rule

    bobg_birthday.jpg

    But then, I am about 10 years too young to be 100% sure.
     
  4. Aug 30, 2011 #3

    HallsofIvy

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    Yes, that is correct. No one was actually calculating logarithms by hand in the 1960s.

    Of course, someone had to write the first logarithm tables. Those were, if I remember correctly (I don't actually go back that far), by working "the other way"- finding powers of 10 for specific numbers.
     
  5. Aug 30, 2011 #4

    ehild

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    Yes, we used tables or the slide-rulers for the logarithm. I still have my slide ruler and even this time the tables are taught in the secondary schools in my country.
    At the College, and later at my first working place, in the sixties, I made the calculations which involved summations and multiplications with an electromechanical calculator. It was very noisy. And I used very heavy 10-digit tables to find sine, cosine, tangent and logarithm.

    ehild
     
  6. Aug 30, 2011 #5

    phyzguy

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    I think you're missing the point of the OP's question. It's a given that logarithm tables existed in the 1960's, and there was no need to calculate them by hand. The question is what did Apostol expect when he posed the homework question - did he expect the student to 'crank these out by hand'? I would say - yes he did expect it to be done by hand. It doesn't take that long - the first three terms are repeating decimals, and the last two terms are small enough that you only need the first few digits of the decimal expansion. Not that I like doing this sort of thing either, but I think it was expected.
     
  7. Aug 30, 2011 #6
    Thanks phyzguy, that is the sort of information I was looking for.

    My trouble was that I have been doing so much non-computational math, I forgot how to do something like this! Instead of doing divisions first, I was writing one big fraction which I would divide at the end. This certainly would work the same way, however in practice I would inevitably make mistakes along the way.

    Following your lead, I instead divided the fractions first and then multiplied by two - much easier, for me anyway. The other thing that the problem mentions is something I've never really understood - "correct" decimal places. The problem says to "Retain nine decimals in your calculations and obtain the inequalities 0.6931460 < ln2 < 0.6931476". In my calculations, the polynomial becomes
    2*(0.333333333 + 0.012345679 + 0.000823045 + 0.000065321 + 0.000005645) and the error term is between 0.000000513 and 0.000001282. This gives me the following inequality:
    0.693146359 < ln2 < 0.693147328
    Which, although it is inside Apostol's provided one and thus indicates that his is true as well, it leaves me wondering about the "correct" decimal places (in a later question, he asks you to tabulate the logs of values from 1 to 10 with as many "correct" decimal places as you can be certain of from the previous inequalities, and I am not sure how to do that).

    Here's my attempt at understanding "correct" decimal places:
    The numbers are written "correct" to 9 decimal places using long division. When adding, however, it's possible to be off by (almost) one decimal place on each one. For instance, if we expanded the last summand in the polynomial and it came out to 0.00000564699999999999989999... (obviously it couldn't be all nines repeating, because then it really would be equal to 0.000005647) then if we added only the nine digits we originally had to another number which was equally skewed (say 0.00006532199999999999989999...) we would get a sum which was actually incorrect in up to two digits. For instance, consider adding 5.9 and 4.9, first assuming you only had calculated 5 and 4 (giving 9) and then assuming you calculated it one further digit (yielding 10.8). Further additions and multiplications compound this problem.

    So, I recognize the issue with "correct" digits (I think), however I don't know what "rules" there are when adding and multiplying to get the right number of "correct" digits.
     
  8. Aug 30, 2011 #7

    Mark44

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    In English-speaking countries they are called slide rules.
     
  9. Aug 30, 2011 #8

    vela

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    One thing you can do is to round instead of simply truncating, so 5.9 -> 6 and 4.9 -> 5. Their sum, 11, has a much smaller error than the one you get by truncating.
    There are various factors you have to consider, like the order in which you add numbers, when trying to calculate an answer precisely as possible. You might want to look into the subject of numerical analysis to learn about minimizing errors and how errors propagate through calculations.
     
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