Respected Sir ,
Good Morning : Sir is it possible to find natural logarithm of 200 digit number and its antilog also
If yes , then How to find it . Kindly tell.
OK this might take a while...
If you think about a calculator, if you asked it to evaluate $\displaystyle \ln{3}$ for example. Do you think that the calculator immediately knows how to calculate this? Of course not - it's had to have an algorithm programmed into it.
The ONLY operations that a calculator can do are addition, subtraction, multiplication, division and exponentiation. So you would need to tell the calculator what the correct combination of these would give you the value of $\displaystyle \ln{3}$ (say).
To do this, we have to use polynomials, since the general polynomial is the most general combination of the 5 operations.
Let's start with the exponential function (since I like exponentials more than logarithms). We should already know a few things about the exponential function, like:
$\displaystyle e^0 = 1$ and $\displaystyle \frac{d}{dx}(e^x) = e^x$.
Let's assume we can write it as a polynomial. Then that means
$\displaystyle e^x = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots$.
We almost have enough information to program the calculator. We just need to find out what the constants are...
Since we know $\displaystyle e^0 = 1$, if we let $\displaystyle x = 0$ in the polynomial that means
$\displaystyle e^0 = c_0 + 0c_1 + 0^2c_2 + 0^3c_3 + \dots$
$\displaystyle 1 = c_0$.
So we have managed to find one of the constants. Now we need to find a "process" that will reduce the powers and enable us to find the next constant. That process is differentiation. So we will take the derivative of both sides
$\displaystyle e^x = c_1 + 2c_2 x+ 3c_3x^2 + 4c_4x^3 + \dots$
and by letting $\displaystyle x = 0$ we find
$\displaystyle e^0 = c_1 + 2\cdot 0c_2 + 3\cdot 0^2c_3 + 4\cdot 0^3 c_4 + \dots$
$\displaystyle 1 = c_1$.
Take the derivative of both sides
$\displaystyle e^x = 2c_2 + 3\cdot 2c_3 x + 4\cdot 3c_4 x^2 + 5\cdot 4c_5 x^3 + \dots$.
Let $\displaystyle x = 0$ and we find
$\displaystyle 1 = 2c_2$
$\displaystyle \frac{1}{2} = c_2$.
Take the derivative of both sides
$\displaystyle e^x = 3\cdot 2c_3 + 4\cdot 3 \cdot 2 c_4x + 5\cdot 4 \cdot 3c_5x^2 + 6\cdot 5\cdot 4c_6 x^3 + \dots$
Let $\displaystyle x = 0$ to find
$\displaystyle 1 = 3\cdot 2c_3$
$\displaystyle c_3 = \frac{1}{3\cdot 2} = \frac{1}{3!}$.
Take the derivative of both sides
$\displaystyle e^x = 4\cdot 3\cdot 2c_4 + 5\cdot 4\cdot 3\cdot 2c_5x + 6\cdot 5\cdot 4\cdot 3c_6x^2 + \dots$
Let $\displaystyle x = 0$ to find
$\displaystyle 1 = 4\cdot 3\cdot 2c_4$
$\displaystyle c_4 = \frac{1}{4\cdot 3\cdot 2} = \frac{1}{4!}$.
Are you starting to see a pattern? If we were to keep going we'd find
$\displaystyle e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \frac{1}{5!}x^5 + \dots$
$\displaystyle e^x = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots$
$\displaystyle e^x = \sum_{n = 0}^{\infty}\frac{x^n}{n!}$.
See now you are able to simply substitute any value of $\displaystyle x$ you like, say 6, to find $\displaystyle e^6$, and then simplify the polynomial to whatever accuracy you like. You could do this for ANY value (even ones over 200 digits - it would just take a long time).
So, do you think you might be able to use the same process to find a polynomial (series) to find the logarithm of a number?
Hint: Since $\displaystyle \ln{x}$ is undefined and not differentiable at $\displaystyle x = 0$, you should use $\displaystyle \ln{(x + 1)}$ as your starting function. You should know that $\displaystyle \ln{1} = 0$ and $\displaystyle \frac{d}{dx}[\ln{(x + 1)}] = (x + 1)^{-1}$.
You can never find the exact natural logarithm of an integer greater than 1. But you can get it to any desired degree of accuracy as follows (assuming that you have an accurate enough method for finding the log to base 10 of a number between 1 and 10).
Suppose that x = abcde... is the decimal representation of an integer x consisting of n digits a, b, c, ... . Let L be the logarithm to base 10 of the number a.bcde... (with a decimal point after the a). Then the log to base 10 of x is n – 1 + L. To get the natural logarithm, you multiply the log to base 10 by ln(10).
Suppose you have a two hundred and one digit number that looks like this
$\displaystyle N = n_{201}n_{200}n_{199}n_{198}\ldots n_{2}n_{1}n_{0} = n_{201}.n_{200}\ldots n_{1} 10^{200} = n10^{200}$
$\displaystyle \ln(N) = \ln(n10^{200})=\ln(n)+\ln(10^{200}) = \ln(n)+200\ln(10) $
So now if you know how to calculate the natural log for numbers from 1 to 10, then you can easily calculate the natural log for any huge number.
If you want a very powerful, open source, lightweight calculator that can handle large numbers (actually it is a CAS which is more powerful than a calculator), I recommend PARI/GP. Use \p to set precision, as in "\p 100". If you're using windows, you can use notepad to save your number to a file, such as typing n = 1299271937... and saving as say "number.gp", then from the command prompt (cmd) you can type "gp number.gp" with the appropriate paths, then type log(n), or log(n)/log(b) if you want log in another base. If you're using linux then probably you don't require much tutorial. I have no idea how it works on macs.
Edit: Of course antilog is just exp(n) for given n, or b^n if you are using a base other than e. For a 200 digit number you will get overflow, because for example e^(10^5) is already approximately 2.8e43429.
Sir , thanks , u told abt peri/gp , its an excellent software......
but my project failed. Actually i a trying to find generic solution to factorize a large prime number without involving any group theory , but i failed.... i spend my continueous 45 hours on this .... still i didn't got solution.....
If you are trying to break the RSA encryption, good luck! Factoring a 200 digit number formed as a product of two 100 digit prime numbers should take you about a few million years on your computer. I don't get how taking the logarithm will help you find the factors!