Making my own table of logs by hand

[jwpaine]

Hi.

If I want to make my own table of logs in, lets say, base 2: for numbers 1 through 10....

for all numbers n: 1 through 10

\(\displaystyle log^{(n)}_2=x\)

2^x = 1
2^x = 2
2^x = 3
2^x = 4
2^x = 5
2^x = 6
2^x = 7
2^x = 8
2^x = 9
2^x = 10

How could I solve for each x to make a log base 2 table without using modern technology? how could I find an approximation for each x by hand? (some are easy...some are not)

How would they have done it for any base, back in the early years?

If I can make my own table of logs by hand, then my understanding of logs will be greatly benefited.

Thanks!

 

[tkhunny]

How would they have done it for any base, back in the early years?

Very, VERY carefully and in painstaking detail. For some, such things were a lifetime of work. Once published, they were a gem of accomplishment.

"without using modern technology" -- Why? What do you consider "modern"? Does your brain count? There is little doubt you know stuff of which your predeessors of 200 years ago had no idea. Clean paper and a smooth pencil? Those are rather modern. When you are through with this project, take a picture of the stick and the sand pile so all can amuse themselves.

Look up "Transcendental Numbers". Then ask again if you think there is a simple way to produce the tweeners.

2^0 = 1
2^1 = 2
2^2 = 4
2^{something between 1 and 2} = 3

That wasn't very many decimal places, was it?

 

[jwpaine]

"without using modern technology" -- Why? What do you consider "modern"?

Well... without doing the obvious Solve[2^x==3,x] etc....I consider that quite modern, and I believe that the CAS would use it's own table of known-logs for numbers in its computation.

I'll look into "Transcendental Numbers", thanks.

 

[tkhunny]

You might consider a geometrical solution. Those always are fun.

 

[pka]

Well... without doing the obvious Solve[2^x==3,x] etc....I consider that quite modern, and I believe that the CAS would use it's own table of known-logs for numbers in its computation.

This may be a futile effort. But I think that you need to have some of your misconceptions corrected.
Calculators and/or computers do not have built-in tables. Each time we ask such to give us the value of say ln(2) it is actually calculated. Calculated in much the same way it done in the seventeenth century. Now it we today we understand the process better today but it still much the same.

Here it is. We know that
\(\displaystyle \L \ln(1 - x) = - x - \frac{{x^2 }}{2} - \frac{{x^3 }}{3} - \cdots = - \sum\limits_{n = 1}^\infty {\frac{{x^n }}{n}} \quad ,\quad \left| x \right| < 1.\)
The machine just adds up a finite collection of the terms to give an approximation.

To find ln(2), let \(\displaystyle x = \frac{1}{2}\). Because we know \(\displaystyle \ln \left( {\frac{1}{2}} \right) = - \ln (2)\), we have:
\(\displaystyle \L \ln (2) = \frac{1}{2} + \frac{1}{8} + \frac{1}{{24}} + \cdots = \sum\limits_{n = 1}^\infty {\frac{1}{{n2^n }}} .\)

It turns out that most functions such as sine, cosine, exponential all have similar infinite sums. Therefore, calculators and/or computers just do what done by hand before the midpoint of the twentieth century.

So what TKH told you is correct. There a lot of work ahead of you.

 

[morson]

How could I solve for each x to make a log base 2 table without using modern technology?

As somebody said above, to solve 2^x = 3 without technology would take a very long time. People like Napier found ways to make it a lot easier, though.

 

[jwpaine]

Well:

\(\displaystyle \L 2^{x} = 3\)

\(\displaystyle \L ln(2^{x}) = ln(3)\)


\(\displaystyle \L xln(2) = ln(3)\)

\(\displaystyle \L x = \frac{ln(3)}{ln(2)}\)


and then just find both ln(3) and ln(2) using the summation series, as pka did? which would then give me an approximation for what \(\displaystyle \L log^{(3)}_2\) equaled....

I should just give up here lol.

 

[Count Iblis]

I don't understand what is so difficult about this assignment. When I was 13 years old I used a calculator with only addition, multiplication and division but no trignometric or logarithmic functions to calculate logarithms. I used the series expansion method. The series PKA gave must be improved to make it converge fast enough to be of practical use.

You take the slowly converging series:

\(\displaystyle \L \log(1+x) = \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n}}{n}\)

Note that we write Log(x) for the natural logarithm (ln(x) is an old fashioned notation that is nowadays rarely used in the mathematics and physics literature).

If you replace x by -x you get the series given by PKA. If you subract the two series from each other you get:


\(\displaystyle \L \log\left(\frac{1+x}{1-x}\right) = 2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\)

You can then calculate Log(y) by finding x such that:


\(\displaystyle \L\frac{1+x}{1-x} = y\)

So, you put:


\(\displaystyle \L x = \frac{y-1}{y+1}\)

in the above series. E.g. for y = 2, we have x = 1/3 and thus:


\(\displaystyle \L \log(2)= 2\sum_{n=0}^{\infty}\frac{1}{(2n+1)3^{2n+1}}\)

And you clearly see that this converges faster than the series PKA gave.

If you take large values for y, then x approaches 1 and the series will converge slowly. In such cases you should simply calculate the difference Log(y) - Log(y-1). Take, e.g. the case of y = 5. If you have already calculated Log(2), then you know what Log(4) is. So, you write Log(5) = Log(4) + Log(5/4). For y = 5/4 the value of x is 1/9, and the series will converge very rapidly.

To compute the logs for integers you only need the logs of prime numbers. If you start with 2, 3, 5, and proceed to larger primes you will always know log(p-1) when you need to compute log(p). So, in general, you'll need to calculate
Log(p/(p-1)). Taking y = p/(p-1), gives x = 1/(2p-1). So:


\(\displaystyle \L \log(p) = \log(p-1) + 2\sum_{n=0}^{\infty}\frac{1}{(2n+1)(2p-1)^{2n+1}}\)

 

[Count Iblis]

Of course, you can use some additional tricks. E.g. to compute Log(3) you can take p = 9 instead of p = 3:

\(\displaystyle \L\log(9) = \log(8) + \sum_{n=0}^{\infty}\frac{1}{(2n+1) 17^{2n+1}}\)

You use that log(8) = 3 log(2) and log(9) = 2 log(3) to extract log(3).

You only need the first five terms to get an accuracy of 10^(-12).

To compute log(5), use that 5^2 = 25 = 24+1 and that 24 = 2^3 * 3.

\(\displaystyle \L\log(25) = \log(24) + \sum_{n=0}^{\infty}\frac{1}{(2n+1) 49^{2n+1}}\)

To compute log(7) use that 7^4 = 2401 = 2400 + 1, and 2400 factorizes into numbers of which you already have computed the logarithms.

\(\displaystyle \L\log(2401) = \log(2400) + \sum_{n=0}^{\infty}\frac{1}{(2n+1) 4801^{2n+1}}\)