-1
$\begingroup$

This is just continuation of my previous post.

$$ A = a_0 + \left(\frac{1}{a_1}\right)^k + \left(\frac{1}{a_2}\right)^k + \left(\frac{1}{a_3}\right)^k +\ldots$$ Where $i\ge1$ and the recurrence relation $a_{i+1}\ge a_i\ge2$.

I just raised for each term by $k$. here $k\ge1$ and $A$ is any given real number. Then the above series will exist. Can we generalize this series of some real number with those initial conditions? If yes, kindly discuss...

edit The first term is $a_0$ but not $a_1$. Of course I edited now.

Edited and extended $$ A = a_0 + \left(\frac{1}{a_1}\right)^k + \left(\frac{1}{a_1}\right)^k \left(\frac{1}{a_2}\right)^k + \left(\frac{1}{a_1}\right)^k \left(\frac{1}{a_2}\right)^k\left(\frac{1}{a_3}\right)^k +\ldots$$

Where $a_1$ = 2, $a_i$$\ge1$ and 2 $\ge$ $a_i$ for $i\ge2$ with $a_i$ = 2. Again, here k $\ge1$ and k is some fised real number. Can you generalize with an example of this modified series?

  • 0
    What does $k$ add? From the looks of it I can always choose $k=1$ and just pick my constants differently to "simulate" any power of some other series that I like.2012-08-30
  • 0
    @Robert Mastragostino! not exactly k =1. here the statement is valid for all k > or = 1. Of course, k is fixed real number.2012-08-30
  • 0
    Maybe you could link to your previous post?2012-08-30
  • 0
    @celtschk! you can see the link: http://math.stackexchange.com/questions/187774/real-numbers-expressing-in-terms-of-series2012-08-30
  • 0
    @vidyaojal: Thank you. I've edited it into the question, so that others don't need the comments to find out.2012-08-30

2 Answers 2

2

Given that from your condition it is allowed that $a_{i+1}=a_i$, if $k$ is integer, you can reduce the problem for $k>1$ to the case $k=1$ by repeating $1/a_i^k$ for $a_i^{k-1}$ times before using a new value. By doing that repetition, you get a sum of $a_i^{k-1}/a_i^k = 1/a_i$.

Example:

Say your $k=1$ expansion is $$A = 1 + 1/2 + 1/8 + \dots$$ and $k=2$. Then your new expansion reads $$\begin{align} A &= 1 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2\\ &\phantom{= 1} + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2\\ &\phantom{= 1} + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{8}\right)^2 +\dots \end{align}$$

Since every real number can be written as the product of an integer and a real number in the range $[1,2)$, the same strategy can be used to reduce the problem for arbitrary non-integer $k>1$ to the problem $1.

  • 0
    Although in the comments above, the poster has stated that $k$ is a fixed real number, not necessarily an integer.2012-08-30
  • 0
    @Old John: I didn't notice that. For me, the name choice of $k$ together with it being an exponent suggested that it's an integer (well, strictly speaking, he also didn't specify that the $a_i$ are integers, therefore one could just choose $a_0$ as appropriate real number, no matter what the rest of the series converges to ...2012-08-30
  • 0
    I agree that the question needs to be defined much more clearly!2012-08-30
  • 0
    OK, I now edited the answer so that I make explicit that this method is only for integer $k$ (and added that it also can be used to reduce the general problem of non-integer $k$ to the problem of $1).2012-08-30
  • 0
    @celtschk! Thank you for your post and editing. I am looking for generalization of the statement for all k > or = 1. I am not looking for an example. Kindly do the generalization of A for K > = 1.2012-08-30
  • 1
    @vidyaojal: I've not just given an example, I've added an example for the *general rule* for integer $k>1$ I've given (that is, I've generalized it for all integer $k$ with $k>1$), and I also said that for non-integer $k$ the same strategy can be used to reduce the prtoblem to the (probably simpler) case $1 (this case I didn't solve, however). That is, I've solved the problem completely for integer $k$ and gave a possible step in solving it for non-integer $k$. That's definitely more than just giving an example. I admit it is not a *complete* solution for your problem, though.2012-08-30
  • 1
    @vidyaojal: BTW, I just now notice that with the conditions as given, no $A<2$ can be written that way (because your condition $a_i\ge 2$ doesn't exclude $a_0$)2012-08-30
  • 0
    @celtschk! you are right. I just modified the same with little more terms. Can you generalize with your example? Please... also, thank you so much for your QUICK attempt.2012-08-30
  • 0
    @celtschk!please look at the re-edited one and answer...2012-08-30
  • 0
    @sorry, I do not have enough points to chat. So, discuss for my new extended post.2012-08-30
  • 0
    @celtschk! Are you there Sir? Please see my new edited post and respond strictly on it.2012-08-30
2

If I am interpreting your question and intent correctly -- which is probably not the case -- then yes. In fact, all real numbers can be defined in terms of (Cauchy-) converging sequences of rational numbers.

I am sure that you already know that a series is a just sequence of partial sums so that, in a given base, the "power-series" representation of a number is just a converging sequence of rational numbers. For example, the decimal expansion of pi is 3.1459... . We could view this as the limit of the rational sequence {3, 3.1, 3.14, 3.145, ...}.

A classic way to define a real number is by a sequence of rational numbers whose successive differences converge to zero ("Cauchy convergence").

  • 0
    ! you are right and I studied the same after your post. Can you look my series and generalize?2012-08-30
  • 0
    ! can you apply the same for my post?2012-08-30
  • 0
    Koarik! r u there? Please see my new edited post and respond strictly on it.2012-08-30