Results 1 to 5 of 5

Math Help - What is the name of this method?

  1. #1
    Newbie
    Joined
    Jan 2013
    From
    Europe
    Posts
    20

    Unhappy What is the name of this method?

    Hello.

    If we take enough consecutive integer base numbers and raise them to the same power, we have (for example):

    1^4=1
    2^4=16
    3^4=81
    4^4=256
    5^4=625
    6^4=1296

    If we then subtract these numbers orderly, pair by pair, we get another vertical row of numbers, but now there is one number less than in original series (5 versus 6 numbers):

    16-1=15
    81-16=65
    256-81=175
    625-256=369
    1296-625=671

    If we repeat this process, we get:

    65-15=50
    175-65=110
    369-175=194
    671-369=302

    ...and...

    110-50=60
    194-110=84
    302-194=108

    ...and...

    84-60=24
    108-84=24

    So, in this example the end-point is 24.
    Had we chosen initial raising number to be 3, the end point would be 6. With raising number 2 it would be 2 and with raising number 5 we should have 120 as the end point.
    So, the rule appears to be: with raising number x we end to factorial of x.

    Can anyone tell me, what is the NAME OF THIS METHOD??? (not just this example-method which produces factorials, but this pair by pair subtraction method in general).

    I am very interested in this method, but have difficulties to find more information of it from internet and even from this math help forum.
    I have only found one name for it (from John H Conway & Richard K Guys book "The Book of Numbers" from 1996, pages 79-89). They gave it name "difference table".
    However, when I try this term in search-engines, I only find information of calculus, differentiation etc.

    Please, help me!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    652
    Thanks
    128

    Re: What is the name of this method?

    It comes under the general heading of 'Finite Difference Tables'.

    The example that you give is of the function f(x)=x^{4} for x=1 \text{  to  } 6 in steps of 1, (which is normally written as x=1(1)6 ).
    So long as the step length is constant, (and it needn't be an integer), any quartic will have a constant fourth difference column. So for example (taken at random), f(x)=2x^{4}-3x^{2}+4 tabulated \bold{exactly} for x=0.0(0.2)2.0 will have a constant fourth difference column. The method is linked to differential calculus, try differentiating your example and my example four times to arrive at a constant and see how it fits in with the constant difference in the fourth column.

    As you might guess, any linear function tabulated at equal steps will have a constant first difference column, a quadratic has a constant second difference column and so on.

    Finite difference tables can be used for Interpolation, Extrapolation (with caution), Numerical Differentiation (again with caution) and Numerical Integration.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2013
    From
    Europe
    Posts
    20

    Re: What is the name of this method?

    Thank You BobP for quick and informative answer!
    My humble high-school level mathematical education forces me to keep matters relatively simple, that is why I use integers preferably.
    However, I tried to remember calculus, which I have earlier used about 20 years ago, and took Your challenge.
    And indeed: X^4 gives following results while differentiating:
    4X^3, 12X^2, 24X^1 and finally 24X^0 = 24.
    I had no previous knowledge of this, so I am sure it will save a lot of my time, paper and pen in the future... so, thank You very much!

    With Your formula 2X^4-3X^2+4 I had more problems:

    Differentiating:

    8X^3-6X^1, 24X^2-6, 48X^1 and finally 48 is what I get.

    When I fed numbers 0 ; 0,2 ; 0,4 ; 0,6 ; 0,8 ; 1,0 ; 1,2 ; 1,4 ; 1,6 ; 1,8 and 2,0 in the original formula I get the following results:
    4 ; 3,8832 ; 3,5712 ; 3,1792 ; 2,8992 ; 3 ; 3,8272 ; 5,8032 ; 9,4272 ; 15,2752 and finally 24.

    But, when I performed that difference table with these numbers, the final result after 4 steps was uniform 0,0768...not 24 I was expecting!
    I did something wrong?
    In desperation I only found that 48/0,0768 = 625 = 5^4...but I guess that it is just a coincidence and has nothing to do with this challenge.

    Now with term "Finite Difference Tables" I can find more information of this subject, although most of it goes beyond my current capacities.

    The reason, why I am interested in difference tables is that we can use them to "open" large numbers written in form a^b where (in studies) a and b are both positive integers.
    I am especially interested in base number 2 based numbers. Why? During last 2 years I have intensively studied Pascalīs triangle. There horizontal row numbers while summarized obey rule 2^n, where n=row number.
    If n=prime number, then all numbers inside that horizontal row except lateral "1":s are divisible with that n. This is the basis of so called Chinese method to find prime numbers (albeit pseudoprimes make things more difficult). As far as I understand the method is not practical because formula (2^n-2)/n, which should give an integer if n=prime number, grows very rapidly even for computers to handle. But with difference tables, base number 2 (or base number 2 raised to another integer) and base number 10 we now have an easy (?) access to these large numbers. I think this method to have something akin to logarithms in this respect. So, in this method I use fixed integer base numbers raised to growing integer powers instead of method I presented in my question.
    What do You think about this approach?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2009
    Posts
    652
    Thanks
    128

    Re: What is the name of this method?

    You were very nearly there with your 5^4. Take the reciprocal of that and you get (0.2)^4, which is the step length raised to the power 4. The constant in the fourth difference column of a quartic is the (constant) 4th derivative multiplied by the step length raised to the power 4. So, for example, if you take your first example x^4 and step up 2 at a time, 2,4,6,... you should find that the number in the fourth column is 24*(2^4)= 384.
    For the example I gave, it would be 48*(0.2^4) = 0.0768.
    For a cubic, the constant in the third column would be the third derivative multiplied by the step length raised to the power three, and so on.

    I don't what it is that you are trying to do after that. The coefficients of Pascal's triangle (for bigger numbered rows) are usually calculated by the combination formula for n items taken r at a time nCr = n!/(r!(n-r)!).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2013
    From
    Europe
    Posts
    20

    Re: What is the name of this method?

    nCr method is familiar for me, but it requires opening of large factorials. As far as I know there is no easy way for this.

    In Pascalīs triangle we have outer diagonals comprised of number "1":s (=zero diagonals). The next diagonals (=diagonal 1) while we travel in inner parts of the triangle consists of natural numbers 1,2,3,4,...All prime numbers that there are in Pascalīs triangle reside in diagonal 1. Now, if we choose an arbitrary prime number, say 7, from Pascalīs triangle and look at horizontal row where it resides, they are: 1, 7, 21, 35, 35, 21, 7, 1. If we sum them up (1+7+21+35+35+21+7+1) the result is 128 = 2^7. And this is so with every horizontal row, not just those determined by a prime number. Prime number horizontal rows are, however, interesting in the respect that all "inner" numbers there (=all numbers except lateral "1":s) are divisible with that particular prime number. This is quite easy to prove through nCr definition: nCr=n!/(r!(n-r)!)...if n=prime p, then n! = 1*2*...*n = 1*2*...*p. As both r! and (n-r)! are smaller than n! (=p!), except in 7C0 and 7C7 cases in this example, then, by definition of a prime number you can never divide p out from the formula.
    So, now we know that in prime-horizontal row p all numbers summarized minus lateral number "1":s gives us a new number that is always divisible with p...alias: (2^p - 2)/p = integer.

    There are two problems in this method:
    1) pseudoprimes: Although all prime numbers give us an integer with formula (2^p - 2)/p, there are some other numbers which do the same thing too, which was only discovered in 19th century. The 1st one of these is 341 which, while giving an integer result, consists of multiplication of 11 and 31.
    2) practical problem: (2^p - 2) grows rapidly. My calculator fails to show integer-answers after 2^34 and while searching prime numbers, we must admit, that 34 or prime numbers around it (31 and 37) are not very large indeed!

    So, this practical problem was the reason I begin to find other methods to "open" large integer raisings of base number 2.

    The method I suggested is based on the following type of difference tables:

    Letīs choose a rigid integer base number, say 4, and raise it to consecutive integer exponents like that:

    4^0=1
    4^1=4
    4^2=16
    4^3=64
    etc.

    If we now perform the same kind of difference table as in my 1st question, we get not a rigid answer after certain number of steps, but another growing series:

    4-1=3
    16-4=12
    64-16=48

    and...

    12-3=9
    48-12=36

    and...36-9=27

    If we now look last numbers in all these groups, they are actually raisings of base number 3 (=4-1):
    1=3^0 ; 3=3^1 ; 9=3^2 and 27=3^3 etc.

    If, on the other hand, we perform similar kind of operation with base number 4 raisings, but instead of subtraction, use addition, the result is integer raisings of base number 5.

    So, we can change any integer base number, integer exponent, -series into a neighboring one with this method.

    Base number 10 is handy in this respect as in decimal number system we actually need not have to calculate their values: 10^5 is number "1" followed by 5 zeroes, 10^235=1000...(235 pieces of zero) etc.

    So, if instead of base number 4 in earlier example, we use base number 10 series, we have straight access to base number 9 or 11 series with difference table. From base number 9 series we have a straight access to base number 8 series through another difference table.

    ...but as 8=2^3 we then have already a substantial amount of knowledge of base number 2 series too, which we are interested in this Chinese method! If, say, we are interested, if 37 is a prime number, we should perform calculation (2^37 -2)/37. If the result is an integer, then we have quite a good change that 37 indeed is a prime number (pseudoprimes have to considered then anyway). But as my calculator fails beyond 2^34 we can not use this straight approach. Instead of that we can see, that a nearby number to 37, namely 36, is even, so it is divisible with 2, of course. So, now our strategy is to focus on number 2^36. But as 2^3=8^1, also 2^36=8^(36/3)=8^12. And when, via difference tables from base number 10, base number 9 and base number 8 we can "open" this 8^12, then we only have to multiply the result with 2 to find 2^37. Then we only subtract 2 out from it and divide the result with 37. If the result is now an integer, we have a fairly good change to find a new prime.

    In practice the real problem with more efficient computers is to find a number 2^x which resides close enought 10^y (where x and y are natural numbers). Of course they can never be the same numbers (according to fundamental theorem of arithmetic), nor can they be immediate natural number neighbours either (I guess it was Catalanīs conjecture which states that only 8 and 9 are such numbers as 2^3=8 and 3^2=9 are neighbours). But is there in reality some integer raising of base number 2 that is close enough to another base number 10 raising is not clear to me at the moment? 2^10=1024 demands use of 24 difference tables (or summation tables in this case) in order to find 10^3=1000 with this method.

    So, this is something I have been wondering...

    Sorry about lengthy explanation, but I do not know how to do it otherwise.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 6th 2010, 03:40 AM
  2. Replies: 5
    Last Post: January 22nd 2010, 05:50 AM
  3. Replies: 2
    Last Post: August 17th 2008, 12:02 PM
  4. Replies: 3
    Last Post: November 3rd 2007, 01:43 PM
  5. Replies: 0
    Last Post: January 4th 2007, 01:29 PM

Search Tags


/mathhelpforum @mathhelpforum