0
$\begingroup$

The question I have is directly related to an earlier question I posted: Is there a mistake in the way $\alpha$ was defined in a solution to Spivak's Calculus - Ch 23 - 5) (c) ? - Solution Verification

It was suggested to ask the other issue I had in a separate post. So the post here revolves around the algebra I'm using to try and attain a term from a partial sum. Here is the body of the original post, which hopefully gives context to what I'm doing:

The following is a question in Spivak's Calculus - Ch 23 - 5 (c). I included all of question $5$ to provide completeness. I also have the solution which I was close to working out partially, but could not figure out how to gensralize/formalize things

enter image description here

Solution to (c):

enter image description here

So when I was working out things, I was getting a pattern very similar to what results after dividing by the defined $\alpha$, but I wasn't able to actually put things together with regards to actually defining the term $\alpha$. When trying to work backwards and see how this solution comes together I'm having trouble actually applying the $\alpha$.

To attempt to see this in action I tried to prove a case using a partial sum. To get enough terms to see something interesting occurring I used $n = 6$ and $k = 3$. With this being the case I get a partial sum of the form:

$$S_{6} = 10^{-1}a_{1} + 10^{-2}a_{2} + 10^{-3}a_{3} + 10^{-4}a_{4} + 10^{-5}a_{5} + 10^{-6}a_{6}$$

Defining $\alpha_{3}$ the right way:

$$\alpha_{3} = 10^{2}a_{1} + 10^{}a_{2} + a_{3} $$

So using $\alpha_{3}$, I attempted to get $S_{6}$ out of it or at least something very close to it. I went up to the $\frac{1}{10^{3k}}$ term.

After some algebraic manipulation I arrive at:

$$a_{1}\bigg(\frac{1}{10} + \frac{1}{10^{4}} + \frac{1}{10^{7}}\bigg) + a_{2}\bigg(\frac{1}{10^{2}} + \frac{1}{10^{5}} + \frac{1}{10^{8}}\bigg) + a_{3}\bigg(\frac{1}{10^{3}} + \frac{1}{10^{6}} + \frac{1}{10^{9}}\bigg) $$

So for simplicity let's focus on the $a_{1}$ term. For that I was hoping after performing the algebra I would arrive at:

$$a_{1}\bigg(\frac{10^{6} + 10^{3} + 10^{0}}{10^{7}}\bigg) = a_{1}10^{-1}$$

But I haven't been able to derive that conclusion. What am I missing to do to be able to got from $a_{1}\bigg(\frac{10^{6} + 10^{3} + 10^{0}}{10^{7}}\bigg)$ to $a_{n}10^{-1}$?

$\endgroup$
  • $\begingroup$ This is very hard to follow. You don't give any context but it appears that the $a_i$ are just the digits in some integer $\alpha$. In any case, there only appear to be $k$ of them. So...your infinite sum is finite, no? If you meant something else, you should edit your post to clarify. $\endgroup$ – lulu Oct 17 at 20:43
  • $\begingroup$ I'll edit my post to give more clarity. $\endgroup$ – dc3rd Oct 17 at 20:47
  • 1
    $\begingroup$ Ok, but your proposed solution makes no sense. Just argue that, if $S$ is your decimal, then $10^nS=m+S$ where $m$ is an integer. I'm assuming that $S$ is a periodic decimal...if it is only eventually periodic then you will have to multiply by a bigger power of $10$, but the idea is the same. $\endgroup$ – lulu Oct 17 at 20:58
  • 1
    $\begingroup$ I can't follow what you wrote at all. For a numerical example, if $S=.123123123\cdots$ then $10^3S=123+S$. $\endgroup$ – lulu Oct 17 at 21:01
  • 1
    $\begingroup$ Take the numerical example I gave. For it, we have $n=3$. Note that if you take $S=.\overline {a_1a_2a_3}$ then $10^3S=10^2a_1+10a_2+a_3+S$ which certainly tells us that $S$ is rational. But there was nothing special about $n=3$ here. $\endgroup$ – lulu Oct 17 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.