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Math Help - How to find natural logarithm of 200 digit and above numbers

  1. #1
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    How to find natural logarithm of 200 digit and above numbers

    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.
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    Quote Originally Posted by krishna View Post
    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.
    How many digits of the log are you interested in?

    CB
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    Sir , i am doing some project and i need to find the exact log of 200 and above digits number , is it possible to find log of such a big number
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    You can always use the series expansions of the logarithmic and exponential functions...
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    To find log of 200 digit number and above

    Quote Originally Posted by Prove It View Post
    You can always use the series expansions of the logarithmic and exponential functions...
    Sir , i am not a mathematician , so if possibel kindly explain in detail how to find it and whether it is feasible to find log of such a big number of 200 digits and above

    Thanks
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    Quote Originally Posted by krishna View Post
    Sir , i am not a mathematician , so if possibel kindly explain in detail how to find it and whether it is feasible to find log of such a big number of 200 digits and above

    Thanks
    OK this might take a while...

    If you think about a calculator, if you asked it to evaluate \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 \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:

    e^0 = 1 and \frac{d}{dx}(e^x) = e^x.


    Let's assume we can write it as a polynomial. Then that means

    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 e^0 = 1, if we let x = 0 in the polynomial that means

    e^0 = c_0 + 0c_1 + 0^2c_2 + 0^3c_3 + \dots

    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

    e^x = c_1 + 2c_2 x+ 3c_3x^2 + 4c_4x^3 + \dots

    and by letting x = 0 we find

    e^0 = c_1 + 2\cdot 0c_2 + 3\cdot 0^2c_3 + 4\cdot 0^3 c_4 + \dots

    1 = c_1.


    Take the derivative of both sides

    e^x = 2c_2 + 3\cdot 2c_3 x + 4\cdot 3c_4 x^2 + 5\cdot 4c_5 x^3 + \dots.

    Let x = 0 and we find

    1 = 2c_2

    \frac{1}{2} = c_2.


    Take the derivative of both sides

    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 x = 0 to find

    1 = 3\cdot 2c_3

    c_3 = \frac{1}{3\cdot 2} = \frac{1}{3!}.


    Take the derivative of both sides

    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 x = 0 to find

    1 = 4\cdot 3\cdot 2c_4

    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

    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

    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

    e^x = \sum_{n = 0}^{\infty}\frac{x^n}{n!}.


    See now you are able to simply substitute any value of x you like, say 6, to find 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 \ln{x} is undefined and not differentiable at x = 0, you should use \ln{(x + 1)} as your starting function. You should know that \ln{1} = 0 and \frac{d}{dx}[\ln{(x + 1)}] = (x + 1)^{-1}.
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    Quote Originally Posted by krishna View Post
    Sir , i am doing some project and i need to find the exact log of 200 and above digits number , is it possible to find log of such a big number
    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).
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    Suppose you have a two hundred and one digit number that looks like this

    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}

    \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.
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    Thank you all in contributing your support by replying my query.
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    Quote Originally Posted by krishna View Post
    Sir , i am doing some project and i need to find the exact log of 200 and above digits number , is it possible to find log of such a big number
    You had better explain why you want to do this since this is either trivial, trivial but requiring a bit more knowledge or impossible and we can't give a sensible answer without knowing the background.

    CB
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    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.
    Last edited by undefined; August 21st 2010 at 06:29 AM.
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    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.....
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    How can you factorise a prime number? By definition, a prime number only has itself and 1 as factors...
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    I am sorry , i mean to say , that in my project i am trying to find a method by which we can find the factor of a number which is product of 2 prime number and we dont know those two number but we we know their product....
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    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!
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