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Nth term of an unknown equation

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Poll closed Nov 2, 2006.
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  1. Aug 4, 2006 #1
    Hello, About sometime ago, I came up with the following results, which might seem useless but it is very intresting from my point of view,
    I'd be glad to have your comments,:smile:

    I’ve found that there are possibilities for writing formulas for polynomial sequences which are independent of the law or the equation ruling the them & I was successful in writing the formulae, with its specific range which depends on the differentiated function’s equation or in other words the rate of changes between the terms. The method used is mainly following a numerical triangle which is due to the laws of calculus, in order to write the formula I had to propose the ‘M’ theorem too, which states that all types of series have the number “m” which is the key for finding the next term, (read *logic behind M theory* p.2)
    For better understanding about the value “m” let’s take an example,

    The sequence: 1, 8, 27, 64, 125… which follows an equation ( n^3 )

    The following chart should be plotted, (see dig1, attached)

    In this example the number “6” is reoccurring and if we keep on writing the next terms of the sequence we will again end up with the number “6”, which I called it “m”.
    The theory then states that the letter m + 24 + 61+125 = 216 = 6^3 = 6th term.

    By using the same method we can write an algebraic expression for the letter “m” and the 5th or 6th terms, in the following way;

    5th term = v 4th term = w 3rd term = z 2nd term = y 1st term = x

    Hence m = [ ( w – z ) – ( z – y ) ] – [ ( z – y ) – ( y –x ) ]

    m = w – 3z + 3y – x

    & the expression for the 5th term will be ;

    v = m + [ ( w – z ) – (z – y ) ] + ( w – z ) + w

    v = 4w – 6z + 4y – x

    Now by inserting any of the four consecutive terms of the sequence we can get the fifth but this formulae involves a range where the second derivative of the function should be zero or linear other wise the formula is not applicable. To improve the range we shall add the number of consecutive terms inserted in the formulae. But still there must be a way to improve the range without adding the variables where I have not yet been successful in finding the method…

    By using the same technique I have derived the formulae for the 6th term of a sequence by inserting any 5 consecutive terms; u = 6th term.

    u = 5v – 10w + 10z – 5y + x

    With a range, where the 3rd derivative of the function should be either linear or zero.

    After some observations we can conclude that the formulas for different terms are actually following the rule given below;

    nth term = (n – 1) C ( 1 ) (f(n-1)) – (n-1)C( 2 ) (f(n-2)) + ……………….. + (n-1) C (n-1) (f(1))

    To indicate which parts of the formulae should be negative or positive you have to use the following method,

    When “ r ” is even that specific part of the formula should be negative and if “ r “ is odd vice versa.

    As an example to write the formulae for “u” (the 6th term) using the law given above;

    u = (6-1) C 1(v) – 5 C 2(w) + 5 C 3(z) – 5 C 4(y) + 5 C 5(x)

    u = 5v – 10w + 10z -5y + x which means
    f(6) = 5f(5) – 10f(4) + 10f(3) – 5f(2) + f(1)

    By using the same formulae we can even get the nth term of the sequence as given below:

    First evaluate the 6th term and substitute the 6th term’s value in v (the fifth term) in other words use the terms; 2nd, 3rd, 4th, 5th & 6th to get the seventh and so on. We can do this because it doesn’t matters that the terms inserted are the first ones the important point is that we need consecutive terms.
    By following this technique we find that;

    f(6) = 5f(5) – 10f(4) + 10f(3) – 5f(2) + f(1)
    f(7) = 15f(5) – 40f(4) + 45f(3) – 24f(2) + 5f(5)
    f(8) = 35f(5) – 105f(4) + 126f(3) – 70f(2) + 15f(1)

    We can see that the factors of f(5),f(4),f(3)… are all connected to each and other by the equations which are

    Factors of f(5): 5n^2 – 55n + 155
    Factors of f(4): 17.5n^2 – 197.5n + 565
    Factors of f(3): 23n^2 – 264n + 766
    Factors of f(2): 13.5n^2 – 156.5n + 458
    Factors of f(1): 3n^2 – 35n+ 103

    Hence the final equation involving the variable ‘n’ is as follows;

    An = (5n^2–55n+155)f(5)-(17.5n^2–197.5n+565)f(4)+(23n^2–264n+766)f(3)-(13.5n^2–156.5n+458)f(2)+(3n^2–35n+103)f(1)

    Where ‘n’ is an Integer, In order to make n a real number we have to enter the terms with small delta x between them…

    **The Logic behind M theory;

    We can simply prove the theory using a little bit of calculus;
    Take An = n2 (see dig2, attached)

    By plotting the other points and connecting each consecutive point to the other by a straight line we can have their gradients so that we can then draw the graph for ‘d An / dn’ While we don’t mean the derivatives of the equation but the rate of change of terms.

    Now we can draw the graphs of ‘d An / dn’ & ‘d2An / dn2’ ;
    (see dig3)

    As shown above graph of ‘d An / dn’ is increasing with a constant rate of 2 units/n (shown in the graph
    d2n / dn2) so when n is 5 the graph ‘d An / dn’ will increase by 2 units/n (d an / dn axis will show 9), hence when n is 5 ‘An’ should be increased by 9 which means 9 + 16 = 25 = 52 = 5th term.
    When it comes to the numerical triangle, we are doing the same process done by calculus because when we are subtracting the two consecutive terms in the chart given on the first page, we are actually finding the gradient as given below;

    (see dig4)

    As a result; 0 + 2 + 7 + 16 = 25 = 5th term.

    Attached Files:

    Last edited: Aug 4, 2006
  2. jcsd
  3. Aug 4, 2006 #2
    Dig 4 is here,
    As the formula for 'An' is very long to be evaulated, My cousin wrote a computer program for it which is attached to the post,

    How to use the program?
    After extracting the .zip folder start the program by selecting the .exe file,
    you'll be provided with the instructions at the top of the window,
    you'll have to insert five consecutive terms of your sequence such that
    insert the 1st term and press enter
    second term and press enter......

    please inform me if the program is not correctly working...

    Attached Files:

    Last edited: Aug 4, 2006
  4. Aug 4, 2006 #3


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    You might want to look up "Newton's divided difference" formula. Any sequence of n terms can be produced by a polynomial of degree n-1.
  5. Aug 4, 2006 #4
    I'd suggest picking a name other than "M theory" - it's taken! :rofl:


    I didn't look over your work too carefully but it probably does work for polynomials - but if the general term of your sequence is defined by a polynomial, it'll always be easy to calculate it anyways :tongue2:

    As Halls mentioned you can also interpolate sequences of arbitrary (edit: finite, of course) length to polynomials by just solving systems of linear equations (but this'll only get you back the terms you already have!).
    Last edited: Aug 4, 2006
  6. Aug 4, 2006 #5
    The M theorem is taken for very strong fields...(wow!!) ;)
    Well data, What do you suggest?

    You are right, it's easy to find the equation itself with simple algabraic techniques, I just find it fast and intresting, What made me start my work on these polynomial sequences was the chapters related to patterns especialy when the next terms were asked,you really have to use the brain :) so I find it easier in this way... like you don't always have to enter 5 terms you can create the formula for 3 terms too (which will work for linear & quadratic expresions only) which is:

    An = (n-3)((f(3)-2f(2)+f(1))*(.5n-1)+(f(3)-f(2)))+f(3) where n>3

    Thankyou HallsOfIvy, I didn't know that...

    Last edited: Aug 4, 2006
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