Splitting result into different varibles for Maclaurin expansion

This is a discussion on Splitting result into different varibles for Maclaurin expansion within the C++ Programming forums, part of the General Programming Boards category; I have another problem which I cant seem to figure out, I was wondering how it would be possible to ...

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    Splitting result into different varibles for Maclaurin expansion

    I have another problem which I cant seem to figure out,

    I was wondering how it would be possible to calculate a Euler's number using maclaurin's series up to 30~35 digits, until the 30th power.

    f(x)=f(0)+f'(0)+f''(0)*x^2/2!+f'''(0)*x^3/3!+f''''(0)*x^4/4!+...+f^n(0)*x^n/n!

    Is the basic formula, but as the integers in c++ have a certain digit limit It cannot show until up to 30~35 digits.
    That is kinda hard to understand but as i cannot post urls now youd have to google a better one



    I was thinking to split up the value storing the digits by 5 each using series [x] but I cannot figure out how the hell to split up something inside there. As for the calculation of the formula, its not hard as a outer function lets me calculate the factorials.

    as an example,
    for the power being 1,
    i would need a variable x to store the number of that certain power in the expansion. Since the power is 1,
    x=f'(0)=1.0000000000~
    and a variable e which stores the total number calculated thus far.

    As a general form it should go like this.
    pow=1, x=1.000000000~ e=2.0000000000~
    pow=2, x=0.500000000~ e=2.5000000000~
    pow=3, x=1.666666666~ e=2.6666666666~

    So yeah, I know how to use the split storage in order to make the operation between 2 16 digit numbers split into 8 each but in that case, you input the values by yourself, while in this the program should run by itself storing the result of the division.

    Any help would be appreciated.

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    and the Hat of Guessing tabstop's Avatar
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    I'm not sure I know what the difference is you're thinking of. If you know how to split up a 32-bit integer into two 16-bit integer, it doesn't matter where the number comes from? You will have to write routines to operate on these pairs-of-numbers (i.e. you'll have to do carry and things yourself).

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    It sounds like you're looking for a big number library. There are several available, no need to brew your own.

    But I'm not sure that's the best solution. Why are you using integers and not doubles? There are some gotcha's with using floating point numbers, but I suspect for your application they will be sufficient. What are typical high values for x and n that you expect your program to work with?
    It is too clear and so it is hard to see.
    A dunce once searched for fire with a lighted lantern.
    Had he known what fire was,
    He could have cooked his rice much sooner.

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    Thanks for the answer,

    Basically I need to split the already calculated result into different variables.

    The fact that I used INT is because of the following.

    Lets say that I have declared int x[6]
    And that the result for euler's number expansion of the third exponent is 0.166666666666666666666666666666

    And I want to split it so

    x[0]=0
    x[1]=16666
    x[2]=66666
    and so on
    So I dont really need the use of double because I wont be having anything requiring doubles
    Then I want to add this number to the total of the expanded Euler's number until now.

    The results for the first and second exponential were :
    1.000000000000000000000
    0.500000000000000000000
    those will be added to the total of Euler's number which is I have declared as int e[6].

    In the case of expanding until the third power we get
    e=2.6666666666666666666666
    Which is basically by adding all what we get from the formula.
    This too will be split into numbered variables like x was
    What I am thinking is that there could be added some variation to the division in the formula to gain the exact number I need for each part, but i still don't get it

    Adding up the sum is something i have done and Is possible, but what I really want is how to get the specific value stored in the specific variable slot

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    and the Hat of Guessing tabstop's Avatar
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    So you don't want bignum, or even splitting(!), but rather just fixed point. I forget right now whether GMP has such (or equivalent high-precision float). Note also that division isn't going to get you from 0.16666 to 16666....

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    This was supposed to be achievable by only having knowledge about series, but I dont understand how the hell to split these

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    So I guess the question is: why are you using int? Why aren't you using long double? Are you actually supposed to have 30 decimal digits of accuracy, or will you settle for 18 (which is what long double guarantees on my machine)?

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    Edit: I am trying to understand this problem likely will be doing Taylor Series (super set of Maclaurin expansion) this fall.

    Links that help me understand what is being asked; still have no understanding of how to do it.

    Euler's number
    e (mathematical constant) - Wikipedia, the free encyclopedia

    Note: To increase accuracy you can add the smaller terms first when calculating "Euler's number (e)"; but, I have no idea if this applies directly with Maclaurin expansion method.

    http://en.wikipedia.org/wiki/Taylor_series
    http://en.wikipedia.org/wiki/Exponential_function


    Big Integer Library; still have only a slight understanding why OP wants to do it using Integers; never heard/used of this library before.
    https://mattmccutchen.net/bigint/


    Tim S.
    Last edited by stahta01; 07-13-2011 at 10:52 AM. Reason: Re-wrote it from scratch

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    For ease of understanding, you could do an array of characters. Each character holds a single digit 0 though 9. Do math on the arrays like you would by hand - add/subtract just requires you to carry or borrow digits between each element :

    Code:
    int carry = 0;
    for (int i = 0; i < ARRAY_LEN; i++)
    {
         int sum = op1[i] + op2[i] + carry;
         result[i] = sum % 10;
         carry = sum / 10;
    }
    if (carry)
       printf ("Overflow!");
    Subtraction is similar. Assuming you're not limited on time, multiplication is just repeated addition. Division is repeated subtraction with some hacks to get the remainder into a fractional form. Search for "arbitrary precision multiplication" and "arbitrary precision division" for quicker ways to do either.

    Using your idea of putting 5 decimal digits per int is similar - instead of 10 in my code you'll want to mod & divide by 10000. And doing this in fixed point will make things even more complex - you'll have to make a decision up front how many digits you need on either side of the decimal place. Sounds easy since you know the result is 2.x, but some of the intermediate math may produce results with a much larger integer portion.

    Are you sure you're understanding the question correctly? Normally these types of questions are to make you understand that even though x^N and N! individually will overflow, you can reuse part of the previous step to calculate it so that it doesn't. You just start with X, then each time through the loop multiply by x and divide by N instead of calculating x^N and N! each iteration. This won't give you 30 digits of precision but should give you close to the limits of precision of whatever floating point value you're using to hold the final result.

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    This is supposed to make you use series like x[n] in order to divide a result into multiple parts and use functions to calculate the results and such.

    According to it,
    I need to find the specific part of the Maclaurin's Expansion and calculate it.

    So the X in e=1+x+(1/2!)*x and so on is always 1
    Giving us e=1+1+1/2!+1/3!+1/n! to calculate

    The program should calculate it in order of the N

    so If N is 1 it will calculate only the corresponding factorial division part;
    meaning that one part of the variable will hold the result of the calculation which will be x=1.00000000~ and the other will hold the actual sum up until now which is e=2.000000~

    For N=2
    x=1/2!, e=previous e+x

    for N=3
    x=1/3!, e=previous e+x

    each time the result is calculated, it needs to hold all the numbers after the dot into separate variables like x[1] x[2] x[3] until all the 30~35 digits are filled with them.
    so when printing out, in the case of N=2

    x[0].x[1]x[2]x[3]~
    should come out as
    0.50000000000000000000
    where x[0] should hold the value above the dot and x[1~3] would be holding the rest in 5 digits each.

    Well yeah Sorry if my explanation sucks but This is what its asking.
    And I cant use bigint as it defeats the purpouse

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    So everything is going great until you get to here:
    each time the result is calculated, it needs to hold all the numbers after the dot into separate variables like x[1] x[2] x[3] until all the 30~35 digits are filled with them.
    which (a) makes no sense and (b) directly contradicts everything before it. Therefore I suspect that sentence to be false.

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    Thats what it says,
    Im supposed to make something that does different calculations and makes them up into one whole 30~35 digit number

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    Quote Originally Posted by Brother_Sharp View Post
    Thats what it says,
    Im supposed to make something that does different calculations and makes them up into one whole 30~35 digit number
    Then do so. That has nothing to do with trying to make x[0] the part before the decimal point, x[1] the first five digits after the decimal point, etc -- in fact it directly contradicts it (unless you're just using "series" as a substitute for any and every noun you come across).

    Is the 30-35 bit supposed to be the number of digits of accuracy, or the number of iterations in your series?

    EDIT: Forgot to write the rest of it: if you do need 30-35 decimal digits of accuracy, then forget about series and MacLaurin expansions and Euler's number, and just write code to make these numbers, add with them, multiply and divide with them, because it's not as though they exist already in C. That will keep you busy for a while, and you'll need it to actually try and do something useful with them anyway.
    Last edited by tabstop; 07-13-2011 at 09:30 PM.

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    Well, I need to expand euler's number for this so not using them is not really an option.
    And by 30~35 digits I mean accuracy.
    This is just one of the last-year problems that I randomly found at my university, and I am really interested on this could be solved with only series.

  15. #15
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    My point is that just getting + and * working for fixed point is probably going to take you several hours, and if you don't have + yet there's no reason to do anything more complicated. So start by getting + to work.

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