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Sequences-changing differences

  1. Oct 22, 2012 #1
    recently, I have been learning, arthimitic progression and series and also geometric progression and series. When searching the internet for extra study material I came across a formula for changing difference: nth term: [itex] a + (n - 1)d + ½(n - 1)(n - 2)c [/itex]. I would really like to know how this formula was derived cant find the proof on the internet, just the formula. I have even tried myself have no luck. I can see the first part [itex] a+(n-1)d [/itex] as AP but after that a bit of confusion I have a felling it has to do with a quadratic only because, the example the formula has been applied to is a quadratic sequence.

    I did read on purple maths, that to understand the real method on solving a quadratic sequence, you have to understand calculus to a certain extent, unfortunally I have only just started reading calculus so I lack allot of knowledge in that specific area, which makes me think the formula was derived from a calculus equation; correct or not ?

    I would really appreciate any insight into this formula, for example shown how it was derived, what the formula is called and if there is any text or web sites that I can be pointed to understand this more.

    Thanks in advanced for any input; much appreciated.
     
  2. jcsd
  3. Oct 22, 2012 #2

    Stephen Tashi

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    Give a link or explain what the formula represents. (It is anywhere in http://en.wikipedia.org/wiki/Arithmetic_progression ?) The phrase "changing difference" isn't familiar.

    if [itex] x[n] [/itex] is a sequence, the "first difference" operator is usually defined as
    [itex] \triangle x[n] = x[n+1] - x[n] [/itex]
    The "second difference" is
    [itex] \triangle^2 x[n] = \triangle (\triangle x[n]) = \triangle ( x[n+1] - x[n] ) = x[n+2] - x[n+1] - (x[n+1] - x[n]) = x[n+2] - 2 x[n+1] + x[n] [/itex]

    In a similar manner, one may define the "nth difference" operator.

    The mathematics of difference operators and finite summations is called "the calculus of finite differences" and you can understand most of it without knowing calculus. http://en.wikipedia.org/wiki/Finite_difference
     
  4. Oct 23, 2012 #3
    Thanks for coming back to me, I found it on an old gcse site I used, unfortunately they only give the formula, I will link the website below, to see if anyone can get the head around the formula.



    http://www.gcsemathstutor.com/sequences.php
     
  5. Oct 23, 2012 #4
    I might have just found something, it tells me the formula I have given is the general term for a quadratic, so I assume if I write a quadratic sequence out a, try solve algerbraiclly; I should be able to put into this form?

    Here is the link to the web site were I found the info:http://lgfl.skoool.co.uk/examcentre.aspx?id=636 [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Oct 23, 2012 #5

    MarneMath

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    You can represent any AP of the nth order as a nth degree polynomial, which is what Mr. Tashi is telling you more or less, you can see this via the method of finite difference.

    Let's take some sequence. Let's use one that website use.

    3, 8, 15, 24, 35

    Given that this can be written as a quadratic polynomial, we say that this sequence is equal to some 2nd degree polynomial.* Thus f_n = an^2 + bn + c.

    f(1) = a + b + c = 3
    f(2) = 4a+2b + c = 8
    f(3) = 9a + 3b + c = 15

    We can solve these system of equations. We find that a = 1 , b = 2 and c = 0.

    f_n = n^2 + 2n. So let's see if this works. f(4) = 24, which is indeed our 4th term in the sequence. You can use this method to solve for different arthimetic progressions of different order. Hope this provides some insight.

    *In case you're wondering how I know it's a 2nd degree polynomial. It's simple. It takes two difference for the change to be constant. However many times it takes for the difference to be constant, that is the degree of polynomial you'll need to represent a certain sequence.
     
  7. Oct 23, 2012 #6

    MarneMath

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    Also, i'm not ignoring your formula, if you were to perform this in general terms using 'foward difference' you'll actually end up with your formula. It helps to make a table if you decide to do it.
     
  8. Oct 23, 2012 #7
    Thanks for the response, it dose give me a better insight now. Once again thanks for the responses.
     
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