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How do I simplify this expression with a floor function?

  1. Jul 28, 2015 #1
    I would want to change this expression to another one that would have P easily accessible.

    [itex]\frac{\left \lceil (n+1)^{P} \right \rceil}{\left \lceil n^{P} \right \rceil}[/itex]

    As an example of 'accessible'; given (n+1)^P / n^P, I can make P more accessible by using log(). Then it becomes P * (log(n+1)-log(n)), where P is just outside everything else and is easily accessible by just dividing by something.
  2. jcsd
  3. Jul 29, 2015 #2


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    I have seen two different approaches to this sort of problem.
    1) Make a conversion inside the ceiling function, such as ##e^{p \log (n+1) } ##
    2) Add in a dummy term 0<w<1 such that the ceiling (n+1)^p = (n+1)^p +w.
    The second option often lets you get sufficiently close to the answer you need.
  4. Jul 29, 2015 #3
    Regarding the 2nd option,

    I don't understand how to bring out P in the expression, ((n+1)^P + w1) / (n^P + w0)
  5. Jul 29, 2015 #4


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    Do you have any information about P or n, ## n \in \mathbb{N}?, P>0 \in \mathbb{R}?##
    What is the outcome you are looking for? Is it a bound of some sort, or a relation?
    You could say that ceiling(n+1)^p = (n+1+w)^p. w would still be bounded in [0,1), and would tend toward zero much more reliably than having it outside the exponent.
  6. Jul 29, 2015 #5
    P is a real number greater than 0
    n is a natural number (including 0)

    I'm trying to make a convergence acceleration formula for a (rounded up) sequence of numbers, then it eventually leads to this problem:

    seq(n) = ceil ((n+1)^P) / ceil (n^P) , for any real P>0

    which requires me to completely isolate P on its own to one side of an equation

    seq(n) isn't the sequence that I'm trying to accelerate the convergence on, it's just derived from something else.
    Last edited: Jul 29, 2015
  7. Jul 29, 2015 #6


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    If you use ##R = \frac{ (n+1+w )^p }{(n+v)^p} = \left( \frac{n+1+w }{n+v}\right)^p ##
    then, ##\ln R = p \ln \left( \frac{n+1+w }{n+v}\right). ##
    So, just in using the range [0,1) for w,v you can bound it by:
    ##\ln \left(\frac{ n+1 }{n+1}\right) < \frac{\ln R}{p}< \ln \left(\frac{n+2 }{n}\right) ##

    Is that a set of bounds you can work with? Or do you need finer resolution on how quickly w and v go to zero so you can estimate how close to ##p \ln \frac{n+1}{n}## the rate really is?
  8. Jul 29, 2015 #7
    I think the first step would actually be:

    R= ((n+1)^P + w) / (n^P + v)

    If it is, I have no idea how to go on from there..
  9. Jul 29, 2015 #8


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    Good point. Upon further investigation, the assumption that you could bring a small number inside the exponential fell apart pretty quickly.

    However, staying with the bounding idea,
    ##\frac{ (n+1)^p }{n^p +1 } <\frac{ \lceil (n+1)^p \rceil }{ \lceil (n)^p \rceil } < \frac{ (n+1)^p +1}{n^p }##
    Which letting ## Q =\frac{ (n+1)^p }{ (n)^p }## can be simplified to:
    ## \frac{Q}{1+ \frac{1}{n^p} } < R < Q + \frac{1}{n^p} ##
    The upper bound is pretty straightforward, the lower bound can be expressed as:
    ## \frac{Q}{1+ \frac{1}{n^p} } = Q - \frac{Q}{n^p} + \frac{Q}{n^{2p}}- \frac{Q}{n^{3p}} ...##

    Is this perhaps a form you could use to estimate your error bounds? Everything is expressed as a power of p, but not in such a form that you can just take the log of it straight away.
  10. Jul 29, 2015 #9
    That's very clever; I think I would be able to use it.

    Thanks for the help!

    What I decided to do next is to solve for the minimum and maximum values of P using the 2 bounds.

    I'm having trouble with solving for P in the expression:


    Is this even solvable?
    Last edited: Jul 29, 2015
  11. Jul 30, 2015 #10


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    I can't seem to get right to a solution.

    If p were an integer, there are ways to simplify it with polynomial methods.
    Otherwise, you can discuss the limits as n or p are large or small.
    For p close to zero and n > 0, this will tend toward 1/2.
    For p>1, as n grows this will tend toward 1.
    Note that for an integer p,
    ##(n+1)^p = \sum_{i=0}^p \begin{pmatrix} n\\i\end{pmatrix}n^i=n^p + 1 + \sum_{i=1}^{p-1} \begin{pmatrix} n\\i\end{pmatrix}n^i.##
    So, you could simplify
    ##\frac{(n+1)^p}{n^p + 1} = 1 + \frac{\sum_{i=1}^{p-1} \begin{pmatrix} n\\i\end{pmatrix}n^i}{n^p + 1}.##

    Generally, I would plot the log of convergence factor vs. n. The slope of this plot should be an approximation for your exponent for the convergence rate ##e^x##, with x being a negative value.
  12. Aug 4, 2015 #11
    Unfortunately, I have no idea about what you said (I don't have that much of a background in maths, sorry)

    But I have been searching around to look for ways to solve this and I found that perturbation theory could solve algebraic problems that are not exactly solvable; would that work?

    I would try it out myself but I don't know how to, all the examples I could find and (kinda) understand aren't applicable in my case.
  13. Aug 5, 2015 #12


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    The ##\begin{pmatrix} n\\i\end{pmatrix}## notation means the combination or "n choose i" operation which simplifies to:
    ##\begin{pmatrix} n\\i\end{pmatrix}=\frac{n!}{(n-i)!i!}##
    I made a mistake in the original formula:
    Should be
    for integer p,
    ##(n+1)^p = \sum_{i=0}^p \begin{pmatrix} p\\i\end{pmatrix}n^i=n^p + 1 + \sum_{i=1}^{p-1} \begin{pmatrix} p\\i\end{pmatrix}n^i.##

    If you are just looking to estimate, you could only keep the highest order term in the expansion:
    ##\frac{(n+1)^p}{n^p + 1} = 1 + \frac{\sum_{i=1}^{p-1} \begin{pmatrix} p\\i\end{pmatrix}n^i}{n^p + 1}\\
    =1 +\frac{pn^{p-1}}{n^p}+\mathcal{O}(n^{-2}) \\
    \approx 1+ \frac pn.##
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