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Calculus for sequences

  1. May 25, 2015 #1
    First, please take a look at http://www.purplemath.com/modules/nextnumb.htm (the second-order sequence problem)

    http://www.purplemath.com/modules/nextnumb.htm :
    "Since these values, the "second differences", are all the same value, then I can stop. It isn't important what the second difference is (in this case, "2"); what is important is that the second differences are the same, because this tells me that the polynomial for this sequence of values is a quadratic.(Once you've studied calculus, you'll be able to understand why this is so. For now, just trust me that this works.)"

    And now, I've studied introductory (basic differentiation and integration)
    But, I still have no idea what calculus for sequence is.

    Please explain me or give me a link to a web which clearly explains it.
     
  2. jcsd
  3. May 25, 2015 #2

    Merlin3189

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    Gold Member

    For a linear equation, y=mx +c, the first derivative ( y' = m ) is a constant.
    For quadratic, y=ax2 + bx + c, the first derivative ( y'= 2ax + b ) is linear and the second derivative (y''=2a) is constant.
    For cubic first deriv is quadratic, second deriv is linear, third deriv is constant,
    and so on.

    Derivatives are the rate of change of the function, so derivative of a constant is zero. There is no point in calculating more derivatives, because they are now all zero.
    For polynomials each differentiation reduces the powers by one and loses any constants (which is why you need to add a constant in integration.)


    In your table of differences, the first differences reflect the first derivative of the sequence function - the size of the difference is how much the numbers are changing term by term.
    The second difference reflects the second derivative, the third difference the third deriv, etc.
    Once your differences are constant you know how many differentiations are needed to eliminate all power terms, so you know the order of the polynomial.
     
  4. May 25, 2015 #3

    pasmith

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    If you consider the sequence [itex]x_n = n^k[/itex] for integer [itex]k[/itex], then you see that [itex]x_{n+1} - x_n = (n+1)^k - n^k = kn^{k-1} + \dots + 1[/itex]. Continue this and you'll find that the [itex]k[/itex]th difference is [itex]k!n^0 = k![/itex]. The (k+1)th difference is zero, since the difference of a constant sequence is zero.
     
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