Hi I've been playing with ln(Natural logarithm) since I first learn it in physic class and my goal is to calculate it without using the calculator. I done $$\ln(12,14,15,16,18,\ldots)$$
just fine but for
$$\ln(17,13,11)$$
I can't seen to divide them into anything other than $\ln(17)=(17\cdot1).$ I was wondering if there's a method to do this? Thank you :D
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$\begingroup$You might try $\ln(17)=\ln(16)+\int_{16}^{17} \frac1x dx$.
You can then estimate the integral as a rectangular area with width $1$ and height $\frac{1}{16.5}$.
This gives $\ln(17)\approx \ln(16)+\frac{2}{33}$
According to my calculator, $\ln(17)\approx 2.83321334$, and $\ln(16)+\frac{2}{33}\approx 2.833194783$.
For better accuracy, you could partition $[16,17]$ into two (or more) subintervals.
$\endgroup$ 1 $\begingroup$You can first calculate :
$$
\ln(50) = \ln2+2\ln5
$$
and then
$$
\ln(52) = \ln13+2\ln2
$$
After this you can get with precision to 3E-8 a value for ln(17) :
$$
\ln(17) \approx { {\ln50 + \ln52} \over 2} + ({ 1 \over 50}-{1 \over 52})/4 - \ln3
$$
If you calculate pretty exactly you can get a difference of only 0.000000037 with ln(17)
greatings, Daniel
From your comments, you were doing your calculations assuming that you already knew the value of $\ln(x)$ for certain fixed values of $x$. But how might you actually approximate $\ln(x)$ if you didn't actually have any values handed to you?
One way is with numerical integration. $\ln(x)=\int_1^x \frac{1}{x}dx$, and one can use various numerical integration techniques to evaluate this to any desired degree of accuracy.
Somewhat more usable, one can use the Taylor series
$$\ln(1+x)=\sum_{n=1}^{\infty} (-1)^{n+1}x^n/n \quad \text{ when } |x|<1$$
to calculate $\ln(a)$ for values of $a$ close to $1$ and write other numbers as a product of such numbers. For example, $2=(4/3)(3/2)$, so$$\ln(2)=\ln(1+1/3)+\ln(1+1/2)=\left(\frac{1}{3}-\frac{1}{2\cdot 3^2}+\frac{1}{3\cdot 3^3}+\cdots \right)+\left(\frac{1}{2}-\frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+\cdots \right).$$
You can take as many terms as needed to get your desired accuracy. However, there is a nice trick to get a faster converging series that doesn't require you to break up your number into a product of numbers close to $1$ (or to have precomputed $\ln(n)$ for a few big numbers, although both these tricks are still useful here).
If we compare the Taylor series of $\ln(1+x)$ and $\ln(1-x)$, they are very similar except that the first alternates $+$ and $-$ signs on the terms, while the other has $-$ signs on all the terms. If we subtract the two, we get lots of cancellation:
$$\ln\left( \frac{1+x}{1-x} \right)=\ln(1+x)-\ln(1-x)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right).$$
Let's use this to compute $\ln(17)$ two ways.
First, we solve $17=\frac{1+x}{1-x} \Leftrightarrow 17-17x=1+x \Leftrightarrow x=16/18=8/9$. Then we plug in:
$$ ln(17)=2\left((8/9)+\frac{(8/9)^3}{3}+\frac{(8/9)^5}{5}+\cdots\right).$$
Unfortunately, because $8/9$ is close to $1$, it takes a while for the powers of $8/9$ to become small, so you would need a lot of terms to get good accuracy. Here is a better way:
Since $17$ is close to $16=2^4$, we can write $17=2^4\cdot (17/16)$. Then $\ln(17)=4\ln(2)+\ln(17/16)$, and if we solve $\frac{1+x}{1-x}=17/16$, we get $x=1/33$, and so
$$\ln(17/16)=2\left((1/33)+\frac{(1/33)^3}{3}+\frac{(1/33)^5}{5}+\cdots\right)$$
Taking just the first 3 terms, we get something correct to 10 decimal places.
And if you didn't have a good value for $\ln(2)$? Solving $\frac{1+x}{1-x}=2$ gives $x=1/3$, and the series
$$\ln(2)=2\left((1/3)+\frac{(1/3)^3}{3}+\frac{(1/3)^5}{5}+\cdots\right)$$
gives about a digit of accuracy for every term, so taking the first 3 terms gives
$0.6930041152$ as opposed to a correct value of $0.6931471806$, an error of $\approx 1.4\times 10^{-4}$ or 0.02%. Taking 5 terms gives an error of $\approx 1.1\times 10^{-6}$. But if you were an engineer in an age before calculators, you could calculate $\ln(2)$ to 10 digits of accuracy by hand once, and then use it in other calculations.
$\endgroup$ $\begingroup$For calculating natural logarithm of any number, all you need is to remember $$\ln(10)=2.3026$$Now,
$$\begin{equation} \ln(17)=\ln(1.7 \times 10^1)=\ln(1.7)+\ln(10) \tag{1} \label{eq1} \end{equation} $$
$\ln(1.7)$ can be calculated from the Taylor expansion, When $ |x|<1$$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+...... $$Sometimes you might need the equation$$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-......$$
Now,$$\ln(1.7)=\ln(1+0.7)=0.7-\frac{0.7^2}{2}+\frac{0.7^3}{3}...$$Expand up to the precision you want.
From $\eqref{eq1}$, we can find$$\ln(17)=\ln(1.7)+\ln(10)=\ln(1.7)+2.3026 $$
As an another example;$$\ln(825)=\ln(0.825\times 10^3)=\ln(0.825)+(3\times\ln(10))$$And$$\ln(0.825)=\ln(1-0.175)$$Proceed by using the expansion for $\ln(1-x)$ upto an order corresponds to the precission you want.
$\endgroup$ $\begingroup$Here is an attempt to understand the rather mysterious answer of Daniel Pol (who gave an answer without an explanation).
Given a differentiable function $f(x)$, we can use differentials to approximate $f(x)$ close to known values. If $h$ is small, $f(a+h)\approx f(a)+hf'(a)$. In particular, $\ln(a+h)\approx \ln(a)+h/a$.
Since $17\cdot 3=51$, and since $52$ and $50$ both have small factors making them computable from your table of values, we can try to approximate $\ln(51)$ using differentials at $a=52$ or $a=50$.
$$\ln(51)=\ln(50+1)\approx \ln(50)+1/50, \qquad \ln(51)=\ln(52-1)\approx \ln(52)-1/52.$$
Taking the average of these two approximations yields almost Daniel's result, except he has a $1/4$ where we get a $1/2$.
We can use differential approximation in a slightly different way to recover this better constant.
$$\ln(51)-\frac{\ln(50)+\ln(52)}{2}=\frac{\ln((50)(52)+1)-\ln((50)(52)}{2}\approx \frac{1}{2}\frac{1}{(50)(52)}=\frac{1}{4}\left(\frac{1}{50}-\frac{1}{52}\right).$$
I do not know if this is at all related to Daniel's reasoning, but it gets the same result.
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