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I want to know if there is a way to find for example $\ln(2)$, without using the calculator ?

Thanks

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5  
Yes:) Look it up in a logarithm table. –  gammatester Oct 18 '13 at 12:28
    
$$\frac{1-2^{1-1.1}}{1.1-1}=0.66967$$ $$\frac{1-2^{1-1.01}}{1.01-1}=0.69075$$ $$\frac{1-2^{1-1.001}}{1.001-1}=0.692907$$ $$\frac{1-2^{1-1.0001}}{1.0001-1}=0.693123$$ $$\frac{1-2^{1-1.00001}}{1.00001-1}=0.693145$$ $$\frac{1-2^{1-1.000001}}{1.000001-1}=0.693147$$ $$\frac{1-2^{1-1.0000001}}{1.0000001-1}=0.693147$$ $$\log (2)=0.693147$$ –  Mats Granvik Oct 18 '13 at 12:34
    
Thank you so much !! –  user2849967 Oct 18 '13 at 12:40
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No, I can't explain anything. Edit: There is a way of matrix multiplication (or row sums of a lower triangular array) which explains it in terms of the Riemann zeta function at the pole, which connects it to known results. math.stackexchange.com/questions/46378/… –  Mats Granvik Oct 18 '13 at 12:50
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@bluesh34: If $f(x)=2^x$, Mats Granvik's quotients are successive approximations to $f'(0)=\ln(2)$, taking $h=-0.1$, $h=-0.01$, etc. –  user86418 Oct 18 '13 at 13:45

6 Answers 6

And let's not forget this method (read off of the Ln scale).

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Image source

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$$\log 2 = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$$ In the general case $$\log \frac{1+x}{1-x} = 2(x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\ldots)$$

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The operations that are relatively easy to compute by hand are addition, multiplication, and their inverses, subtraction, and division. With these operations we can compute all rational functions, e.g. $\frac{2x^2-1}{x^3+x-1}$.

We know that $$\ln(x)=\sum_{k=1}^{\infty}(-1)^k\frac{(x-1)^k}{k}$$

for values of $x$ close to $1$. So, if we take partial sums of this series we get approximations to logarithm that only require multiplications and sum and subtractions.

Notice that we only need to be able to compute values of logarithm for numbers close to $1$, since using $\ln(e^kx)=k+\ln(x)$ can allow us to reduce to this case.

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It's also worthwhile to note that for the stated series (modulo the missing factor of $(-1)^{k-1}$), the convergence is much faster by taking $x = 1/2$ (and using $\ln(1/2) = -\ln(2)$) than by setting $x = 2$. :) –  user86418 Oct 18 '13 at 13:29

How precise do you need the calculation to be?

As a quick and dirty approximation, we know that $2^3 = 8$ and $e^2 \approx 2.7^2 = 7.29$, and so $\ln(2)$ should be just over $\frac{2}{3} \approx 0.67$. Contiuing to match powers, we find $2^{10} = 1024$, and $e^7 \approx (2.7)^7 = (3 - 0.3)^7 = 3^7 -7(3)^6(.3) + 21(3)^5(.3)^2 - 35(3)^4(.3)^3 \dots$ $= 3^7 (1 - .7 + .21 - .035 \dots)$ $\approx 2187(.475) = 1038.825$. Therefore, $e^7 \approx 2^{10}$ and so $\ln(2)$ should be just under $0.7$.

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$$\log2=\frac{2}{3}\left(1+\frac{1}{27}+\frac{1}{405}+\frac{1}{5103}+\frac{1}{59049}+\frac{1}{649539}+...\right)$$

The denominator is $(2k+1)9^k$.

http://oeis.org/A155988

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One can use the fact that $$ \log x=\lim_{n\to\infty}n\left(1-\frac{1}{\sqrt[n]{x}}\right) $$ For $\log2$ a good approximation is $$ 1048576\left(1-\frac{1}{\sqrt[1048576]{2}}\right) $$ where $$ \sqrt[1048576]{x} $$ can be computed by pressing twenty times the SQRT key on a pocket calculator, since $1048576=2^{20}$ (or computing it by hand, with much patience and time to spend).

What I get doing those computations is $0.6931469565952$, while a real computer gives $0.69314718055994530941$, so we have five exact decimal digits. Of course bigger numbers won't do, since the $2^{20}$-th root of it will be too near $1$ and the necessary digits would have already been lost.

(Note: $\log$ is the natural logarithm; I refuse to denote it in any other way. ;-))

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The task was to avoid a pocket calculator –  Hagen von Eitzen Oct 18 '13 at 15:30
    
@HagenvonEitzen One can compute square roots by hand. ;-) For this you need a pocket calculator with the four operations and the square root, not a full fledged calculator. With only four square roots ($n=16$) I get $0.67834750882272$; with eight ($n=256$) it's $0.69220964211968$. This is easier than using the Taylor expansion. Completely by hand: Taylor in the form $\log\frac{1+x}{1-x}$ –  egreg Oct 18 '13 at 15:43

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