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Discrete antiderivative of x^n

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data

    [tex]\sum_{x} x^k[/tex]
    for k ∈ Q

    2. Relevant equations

    [tex]\sum_{x} x=\frac{1}{2}x^2-\frac{1}{2}x[/tex][tex]\sum_{x} x^2=\frac{1}{3}x^3-\frac{1}{2}x^2+\frac{1}{6}x[/tex][tex]\sum_{x} x^{-1}=\Psi (x)[/tex]

    3. The attempt at a solution

    I don't know. There isn't way to compute the antiderivative of any function and I haven't theoretical knowledge about discrete calculus.
     
  2. jcsd
  3. Feb 10, 2014 #2

    Ray Vickson

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    Your notation is terrible, and is guaranteed to lead you into trouble later. You should never use the same symbol for a summation index and a summation limit. Also, you absolutely must specify the limits on the summation. So, you should write something like
    [tex] \sum_{k=1}^x k = \frac{1}{2}x^2-\frac{1}{2}x \\
    \sum_{k=1}^x k^2 = \frac{1}{3}x^3-\frac{1}{2}x^2+\frac{1}{6}x[/tex]
    and so forth. Here, ##x = 1,2,3, \ldots .##

    Your question is not very clear. Are you asking for a general formula for the summation
    [tex] \sum_{k=1}^x k^q, \; q \in \mathbb{Q}, \; x = 1,2,3, \ldots?[/tex]
    I very much doubt there is any such formula that uses only a finite number of 'elementary' operations and functions.
     
  4. Feb 10, 2014 #3
    My notation isn't terrible, terrible is omit the Δx in summation. The indefinte sum f[x] wrt to x is: ##\sum f[x]\Delta x## but in discrete calculus Δx is unitary (Δx=1) and one is the neutral element of multiplication, so it's omited ##\sum f[x]\Delta x=\sum f[x]\cdot 1= \sum f[x]##, if you have a function of 2 variables, the summation of f[x, y] is ##\sum f[x, y]## e thence rise the question: "summation wrt which variable?" So, an idiot decided to place "respective variable" below of sign of sumation, therefore ##\sum_{x}f[x]=\sum f[x]\Delta x##.

    Not necessarily. Like in integration, exist the definite sum and too the indefinite sum.

    So, backing to my question, I'd like to know the formula for the indefinite sum for x^k.

    In actually, I forgot of add the arbitrary constant C in indefinte summations above, the correct would be:

    2. Relevant equations

    [tex]\sum x \Delta x=\frac{1}{2}x^2-\frac{1}{2}x + C[/tex][tex]\sum x^2\Delta x=\frac{1}{3}x^3-\frac{1}{2}x^2+\frac{1}{6}x+C[/tex][tex]\sum x^{-1}\Delta x=\Psi (x)+C[/tex]
     
    Last edited: Feb 10, 2014
  5. Feb 10, 2014 #4

    Ray Vickson

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    Your screed about notation and "idiots" putting things below the summation sign are a sign of arrogance on your part. If you truly want help you need to scale back the insulting tone.

    Anyway, as regards a summation formula: I partially answered that already. However, I will expand my answer: if k is a positive integer, you can give a general formula---see http://mathworld.wolfram.com/PowerSum.html --- but for non-integer k (or for k < 0) there is probably no nice formula. Summation and integration are very different in that regard.
     
  6. Feb 10, 2014 #5
    [tex]\sum_{x} f[x][/tex] is the convention for "indefinite sum of f[x] wrt to x". The guy that created this conventions is an idiot yes, because he not perceived that is convention is essentialy damaged and mathematically wrong, by chance you already viewed ##\int f(x) dx## wrote like ##\int_{x} f(x)##? No because ##\int f(x) dx## is ##\lim_{\Delta x → 0} \sum f(x) \Delta x##, ##\sum_{x} f[x]## don't exist. And more, the definition ##\Delta f(x) = f(x+1)-f(x)## is another wrong and idiot defition too, IMO. Give to you f(x, y, z), Δf(x, y, z) is equal to f(x+1, y, z) - f(x, y, z) or f(x, y+1, z) - f(x, y, z) or f(x, y, z+1) - f(x, y, z)? Nothing this. The correct wold be ##\frac{\Delta f}{\Delta x}=f(x+1, y, z)-f(x, y, z)##, ##\frac{\Delta f}{\Delta y}=f(x, y+1, z)-f(x, y, z)## and ##\frac{\Delta f}{\Delta z}=f(x, y, z+1)-f(x, y, z)##. I'm saying this to you show something more, not to offend you.

    This answer is an stroke of luck of your part, the form how you formulate the limits
    doesn't serve for calculate the discrete antiderivative of other functions, like for example, e^k. But I appreciated your help!
     
  7. Feb 10, 2014 #6

    Mark44

    Staff: Mentor

    I agree with Ray that the notation is terrible. It's never a good idea to use the same letter for the index of a summation and as a limit.

    How would you write this sum in expanded form?
    $$\sum_{x} x^k, k~\in~Q$$
    Unlike the positive or nonnegative integers, there is no smallest rational number, so I'm at a loss as to how to expand the summation above.
     
  8. Feb 10, 2014 #7
  9. Feb 10, 2014 #8

    Mark44

    Staff: Mentor

    I really have no idea what you're saying. What does this integral have to do with the summation you posted? I.e., this one?
    $$\sum_{x} x^k $$
    for k ∈ Q

    The integral in the image is clear to me. The dummy variable in the integral is t, and the limits of integration are a and x. n is a constant. Ray's and my complaint was with the summation.
     
  10. Feb 10, 2014 #9
    i) What is an integration? Is infinitesimal summation.

    ii) And here be an good example where the same letter is used in the "index" and as a limit too.
    e48a88551eb7f5907007df368509cc53.png

    You viewed that the x variable is in the superior limit and is in the integrating too?
     
  11. Feb 10, 2014 #10

    D H

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    That is incorrect. The equivalent of the index variable in this case is the dummy variable t, not x.
     
  12. Feb 10, 2014 #11

    haruspex

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    That is not an example of the problem. The objection (which I share) is to using the same symbol to mean two different things. (Funnily enough, I've been having just this discussion with another HomeworkHelper.) E.g. in sin(x) = ∫cos(x).dx , the x on the right is a dummy variable, so cannot mean the same as the x on the left. What that equation intends is sin(x) = ∫xcos(x).dx, but that's still unclear because it contains a 'pun', i.e. using x with two different meanings. To unravel it clearly we should write e.g. sin(x) = ∫t=xcos(t).dt.
    In the equation you posted above, there is no pun. The dummy variable is t, and remains safely locked up inside the integral. Since the upper bound is x, the result of the integral is a function of x, as expressed by the left hand side.
     
  13. Feb 10, 2014 #12
    Thrut!
     
  14. Feb 10, 2014 #13

    haruspex

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    Gesundheit!
     
  15. Feb 11, 2014 #14

    D H

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    You are confusing the antiderivative (##\int \cos x\,dx = \sin x + C##) with the definite integral (##\int_0^x \cos t\,dt = \sin x##). They are rather different mathematical concepts that are connected to one another via the fundamental theorem of calculus.

    That is just as bad notation as we have been accusing Jhenrique of making. That doesn't make a bit of sense.

    ------------------------------------------------------------------------------------------

    Your notation is rather bizarre, Jhenrique.

    Nonetheless, I think I know what you are going after here, which is the concept of an inverse difference. First let's look at a difference table. I'll use forward differences, ##\Delta f_i = f_{i+1}-f_i##. Backward differences (##\nabla f_i = f_i - f_{i-1}##) and central differences (##\partial f_i = f_{i+1/2}-f_{i-1/2}##) yield analogous results.

    [tex]
    \begin{matrix}
    n & f_n & \Delta^1 f_n\\
    0 & f_0 & f_1 - f_0 \\
    1 & f_1 & f_2 - f_1 \\
    2 & f_2 & f_3 - f_2 \\
    3 & f_3 & f_4 - f_3 \\
    \cdots & \cdots & \cdots \\
    N & f_N & f_{N+1} - f_N \\
    \cdots & \cdots & \cdots
    \end{matrix}
    [/tex]

    This table can be extended to the right via ##\Delta^k f_n = \Delta^{k-1} f_{n+1} - \Delta^{k-1} f_n##. It can also be extended to the left, and this is what you want. Almost all that is needed is ##\Delta f^0 f_n##, and this is just the identity operator: ##\Delta^0 f_n##.

    That's not quite all that is needed. Just as ##\int f(x)\,dx## involves an arbitrary constant, so do their difference calculus equivalents. With these arbitrary constants, the difference table takes the form
    [tex]
    \begin{matrix}
    n & \cdots & \Delta^{-2} & \Delta^{-1} & \Delta^0 & \Delta^1 & \Delta^2 & \cdots \\
    0 & \cdots
    & c_2
    & c_1
    & f_0
    & f_1 - f_0
    & \Delta^1 f_1 - \Delta^1 f_0
    & \cdots \\
    1 & \cdots
    & c_2 + c_1
    & c_1 + f_0
    & f_1
    & f_2 - f_1
    & \Delta^1 f_2 - \Delta^1 f_1
    & \cdots \\
    2 & \cdots
    & c_2 + 2c_1 + f_0
    & c_1 + f_0 + f_1
    & f_2
    & f_3 - f_2
    & \Delta^1 f_3 - \Delta^1 f_2
    & \cdots \\
    3 & \cdots
    & c_2 + 3c_1 +2 f_0+f_1
    & c_1 + f_0 + f_1 + f_2
    & f_3
    & f_4 - f_3
    & \Delta^1 f_4 - \Delta^1 f_3
    & \cdots \\
    \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
    N & \cdots
    & \text{your job}
    & \text{your job}
    & f_N
    & f_{N+1} - f_N
    & \Delta^1 f_{N+1} - \Delta^1 f_N
    & \cdots \\
    \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots
    \end{matrix}
    [/tex]

    What you want is that ##\Delta^{-2}## column.
     
  16. Feb 11, 2014 #15

    haruspex

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    I stand by both of my statements, but I'll take it up with you in private mails.
     
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