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I have a question regarding the answer provided from this question, which is solving $\log$ without a calculator.

This was one of the solutions and I am having a hard time understanding it.

$$\log \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$ Let us apply it for computing the logarithm of $\frac 53 =1.666667$; this would correspond to $x=\frac 14$. So $$\log \left(\frac53\right)=2\left(\frac 14+\frac{1}{192}+ \frac{1}{5120}+\cdots\right)=\frac{3923}{7680}\approx 0.5108$$ Now, $$\frac {166}{100}=\frac 53\times\frac{249}{250}$$ So $$\log(1.66)=\log \left(\frac53\right)+\log \left(\frac{249}{250}\right)=\log \left(\frac53\right)+\log \left(1-\frac{1}{250}\right)\approx\log \left(\frac53\right)-\frac{1}{250}$$ $$\log(1.66)\approx 0.5108-0.0040=0.5068$$ while the exact value would be ... the same.
If you need less accuracy, just truncate the first part, that is to say $$\log \left(\frac53\right)=2\left(\frac 14+\frac{1}{192}+ \cdots\right)=\frac{49}{96}$$ $$\log(1.66)=\frac{49}{96}-\frac{1}{250}=\frac{6077}{12000}\approx 0.5064$$

And my first question is how does the equation $\frac{166}{100} = \frac{5}{3} \times \frac{249}{250} $ came out.

Second question is how does $\log(1- \frac{1}{250})$ becomes $-\frac{1}{250}$.

Thank you for your help.

soupless
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thom
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    It's an approximation. If $e$ is a really really small number then $\log(1\pm e)\approx \pm e$ (not exactly equal) just close. – fleablood Oct 17 '22 at 04:50
  • Here's an explanation https://planetmath.org/approximationofthelogfunction – fleablood Oct 17 '22 at 04:51
  • Ah! Here we go! https://math.stackexchange.com/questions/2320047/why-is-ln1-x-approx-x-when-x-is-small – fleablood Oct 17 '22 at 04:55
  • Thank you so much for the references. How about the first question? – thom Oct 17 '22 at 05:48
  • I mean, if you're asking why the answerer thought of expressing $166/100$ as $5/3 \cdot 249/250$, then I think it's because he wanted to use $\log{(1-x)} \approx -x$ for small $x$. – Accelerator Oct 17 '22 at 06:04
  • Well the question was what is $\log(1.66)$ and not what is $\log \frac 53$. But $\log \frac 53$ was "easy" to calculate. To get $\log(1.66)$ we can set $1.66 = \frac 53\times K$ and $\log 1.66 = \log\frac 53 + \log K$. The trick here is $K <<1$ so $K = 1-e$ and $\log K \approx -e$. So $1.66=\frac 53 \times K \implies K = \frac {1.66\times 3}5=\frac {166\times 3}{500}=\frac {249}{250} $ and that means $K=1-e\implies \frac {249}{250}=1-e\implies e = \frac 1{250}$ – fleablood Oct 17 '22 at 06:14
  • Ohh, now I get the purpose of using $\frac{1}{250}$. Thank you for your replies. – thom Oct 17 '22 at 06:21

1 Answers1

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There is formula the solver knew very well and was comfortable with.

$\ln(1+x) = x -\frac {x^2}2 + \frac {x^3}3- \frac {x^4}4 +......$

From this they conclude two things

$\ln \frac{1+x}{1-x}=\ln(1+x) - \ln (1-x) = (x -\frac {x^2}2 + \frac {x^3}3- \frac {x^4}4 +......) - (-x -\frac {x^2}2 - \frac {x^3}3- \frac {x^4}4 -......)=$
$2(x +\frac {x^3}3 + \frac {x^5}5 + ...... )$

And

If $|x|$ is very very small then all the $\frac {x^k}k$ terms are negligibly small and $\ln(1+x) = x -\frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4 + .... \approx x$.

They figure that as $ 1.66$ is close to $\frac 53$ and we can use $\frac 53 = \frac {1+\frac 14}{1-\frac 14}$ and use the second formula to approximate $\ln \frac 53$. (Although why they didn't they use $\frac 53 = 1 + \frac 23$ and use the first formula, isn't clear to me.)

But as they didn't want $\ln \frac 53$ but $\ln 1.66$. However they figured that as $ 1.66$ is smaller but very close to $\frac 53$ there must be some very small $e$ so that $1.66 = \frac 53(1-e)$.

If so, to solve $\ln 1.66=\ln\frac 53 (1-e)$ we get

$\ln \frac 53 + \ln (1-e)=$

$\ln \frac {1+\frac 14}{1-\frac 14} + \ln (1-e)$.

By the second formula we can figure out what $\ln \frac {1+\frac 14}{1-\frac 14}$ is.

And by the third approximation we can assume $\ln (1-e) \approx e$.

So what is $e$?

Well $1.66 = \frac 53(1-e)\implies e =\frac 1{250}$.

But they didn't actually do it that way. They did:

$1.66 =(1.66\cdot \frac 35)\times \frac 53 =$

$\frac {166\cdot 3}{500}\times \frac 53 =$

$\frac {83\cdot 3}{250}\times \frac 53=$

$\frac {249}{250}\times \frac 53=$

$\frac 35(1-\frac 1{250})$.

fleablood
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  • Although why [Claude] didn't they use 5/3=1+2/3 and use the [formula in positive integer powers $x^0, x^1, x^2, \ldots$ rather than the formula in odd powers, $x^1, x^3, x^5\ldots$] isn't clear to me. Claude mostly likely did this because, for a given number of terms $n$, the odd-power series has terms of higher orders, so they vanish faster wrt. $n$, making the series converges faster and the arithmetic simpler. Indeed, rough numerical estimates suggest that both series converge linearly wrt. $n$, but the odd-term series has rate $\mu=0.057$, faster than the all-term series' $\mu=0.65$ – Jam Oct 17 '22 at 07:54
  • @Jam Nice to know. (Although I suspect the original OP actually did want $\log \frac 53$ but thought it looked more grown up to use decimals). – fleablood Oct 17 '22 at 08:14