5=m^{1/m and proofs of
convergence}

By Brad Rodgers (P1930) on Thursday, May 11, 2000 - 03:34 am :

Are there any solutions to the equation 5=m^1/m ? My quess is that the only solution is infinity (I can show why), but before I turn you off from this, I would like to see your thoughts.
Moreover, are there any solutions for any rational (or not 1) n, where n=m^1/m ? Where n is greater than 1? Less than?

Brad

By Dan Goodman (Dfmg2) on Thursday, May 11, 2000 - 11:44 am :

It depends what sort of m you allow, if you want to have real solutions, then you need to consider how m^1/m behaves for m> 0. Well, it starts out at 0, increases to e^1/e at m=e, and then decreases to 0 as m tends to infinity. So there are two solutions if 0 < n < e^1/e , or one solution if n=e^1/e . Post again if you're interested in complex solutions as well, it's a bit more difficult to find them.

By Sean Hartnoll (Sah40) on Thursday, May 11, 2000 - 01:45 pm :

I think you will find it instructive to take logarithms of the equations... Also, the case for n=5 doesn't have any solutions, and at infinity the curves x=5 and x=m^1/m are diverging, so there isn't a solution at "infinity".

By Brad Rodgers (P1930) on Thursday, May 11, 2000 - 09:20 pm :

But, if 5=m^1/m , then you will find that m=5⁵ 5⁵ ...
As 5⁵ 5⁵ ... will equal infinity (I think), m=infinity.

I am a bit interested in complex solutions as well. By the way, is e irrational? I take from your statement above that it isn't, but all my books do is say "assume the value e=2.718282 correct to six decimal places". If so, is e^{1/ e} rational?

Thanks

Brad

By Sean Hartnoll (Sah40) on Thursday, May 11, 2000 - 10:54 pm :

This is one of those occasions in which you cannot "plug in" infinity, even though it looks like it's okay. I suggest you plot y=5 and y=x^1/x for x> 0, you will find that there is no intersection and furthermore that they do not get closer as x goes to infinity.

e is certainly irrational, this is not too difficult to prove, but I can't remember offhand.

By Brad Rodgers (P1930) on Thursday, May 11, 2000 - 11:38 pm :

I don't see how we plugged in infinity. We plugged in infinity no more than we do in the equation y=2+2+2+2+2......Certainly I have done something wrong- I examined the graph and I saw what you meant, but I still do not see what. Perhaps if you could explain to me how I plugged in infinity I would see my error. But, until then, I must simply accept this as something in the nature of m=n+n² +n³ .........

Brad

By Dan Goodman (Dfmg2) on Friday, May 12, 2000 - 12:08 am :

The problem is this:

You have said that IF $5 = m^{1 / m}$ THEN $m = 5^{5} 5^{5} \dots$ , which is true. Now, you have gone from $infinity = 5^{5} 5^{5} \dots$ to saying that ${infinity}^{1 / infinity} = 5$ . To see why this is wrong, think about solving $6 = m^{1 / m}$ , $infinity = 6^{6} 6^{6} \dots$ , so ${infinity}^{1 / infinity} = 6$ (by your logic), so $5 = 6$ . You have plugged in infinity when you put infinity into $m^{1 / m}$ .

Dealing with this sort of problem is a topic you'll do at university (if you decide to do maths at uni), called Analysis. The problem is that plugging in infinity is not allowed, otherwise you get contradictions like the above. The simplest one we can get from your example above. $y = 2 + 2 + 2 + \dots$ therefore $2 + y = 2 + 2 + 2 + 2 + \dots = y$ , therefore $2 = 0$ .

Here's a quick introduction to the ideas of Analysis. You're pobably familiar with the idea of a sequence, for instance $x_{n} = n$ is the sequence (1, 2, 3, 4, 5, ...). Or $x_{n} = 1 / n$ is the sequence (1,1/2,1/3,1/4,1/5,...). In Analysis we say that a sequence $x_{n}$ converges to $x$ if we can get as close to $x$ as we like, and stay at least that close. In other words, for the numbers $x_{n}$ to converge to $x$ , the following has to be possible. Imagine I choose a number $d$ which is very small, but bigger than 0. If $x_{n}$ converges to $x$ , there must be some integer $N$ , after which $x_{n}$ is within $d$ of $x$ . In other words, for $n \geq N$ , $x - d \leq x - n \leq x + d$ . If there is no number $x$ which has this property, we say that the sequence is divergent. Here is an example of a proof in Analysis.

Q. Prove that the sequence (1, 1/2, 1/4, 1/8, ...) converges to 0.

A. If $d > 0$ is a small number, we want to find an integer $N$ so that $1 / 2^{N} < d$ , or $2^{N} > 1 / d$ . Let $M$ be an integer bigger than $1 / d$ , we need to find an $N$ so that $2^{N} > M$ . But $2^{n} > n$ for all $n$ , so $2^{M} > M$ . Therefore $1 / 2^{M} < d$ . Also, the sequence $1 / 2^{n}$ is getting smaller (it halves each time), so if $n > M$ then $1 / 2^{n} < 1 / 2^{M} < d$ . So the sequence $1 / 2^{n}$ converges to 0.

If you followed that, then you're doing well, you might like to try these exercises (post the solutions if you do).

Exercises:

Does the sequence (1, 2, 3, 4, ...) converge? If so, what does it converge to? Prove your answer.
Does the sequence (1, 1/2, 1/3, 1/4, ...) converge? If so, what does it converge to? Prove your answer.
Does the sequence (1, -1, 1, -1, 1, -1, ...) converge? If so, what does it converge to? Prove your answer.
(Harder) Does the sequence $x_{n} = \sum_{i = 0}^{n} r^{n}$ where $0 \leq r < 1$ converge? If so, what does it converge to? Prove your answer.

What did all that stuff have to do with

m^{1 / m}

? Well, to find

{infinity}^{1 / infinity}

we need to find out if it exists, and what it is. One way to do this is to say

x_{n} = n^{1 / n}

, and test if the sequence

x_{n}

converges. If it does converge to

x

(say), we can (sort of) say that

{infinity}^{1 / infinity} = x

. If you try it out on your calculator, you'll see that

x_{n}

gets closer and closer to 1 as

n

gets large, so

{infinity}^{1 / infinity}

is sort of 1. I say ''sort of'' because you're not really allowed to say that, but it's not such a bad mistake.

I hope you found that interesting, and if you managed to follow it all, you're doing very well indeed. If you did manage to follow it all, just post and I'll write some more (if you're interested). If not, I'll try and explain a bit more carefully.

By Brad Rodgers (P1930) on Friday, May 12, 2000 - 02:26 am :

Ok, it's somewhat late and I need to go soon, but, could this be solved also by using a bit of set theory? As simple infinity is undefined, wouldn't there be a different "value" of infinity for the 5⁵ 5⁵ ... case and the 6⁶ 6⁶ ... case. I am still unsure entirely as to how to use set theory, but I think, used properly, it could allow one to "plug in" infinity and not have errors. If the above is true, then I still am not sure what is entirely wrong with what I did (which leads me to doubt that it isn't true as your last paragraph is very convincing-1/ infinity =0;0 quantities of infinity is still 1 (maybe not-things tend to change in the infinite)).

Is another way to talk of convergence, the difference between the current and preceding integer will eventually become 0? If so, I think I can answer your questions without much work.
Anyway, I'll get around to typing proofs to your questions later. Here are what I think the answers are if my definition of convergence (graphic convergence) is true.

no-it's linear
yes-0
no.

Haven't got around to 4 yet. I have an underlying feeling that this is not correct as you ask what does it converge to; if not, I'll try to read through what you said again-it's somewhat late; I'll probably understand it better tommorow.

Thanks,

Brad

By Sean Hartnoll (Sah40) on Friday, May 12, 2000 - 10:48 am :

I'll let Dan get back about the questions, I just wanted to say that the idea of convergence isn't quite the same as the difference between successive terms going to zero.

For example, consider the sequence a_n = Sum(i from 1 to n) 1/ i. Successive terms get closer and closer because a_n+1 - a_n = 1/(n+1) which goes to zero as n goes to infinity, but the sequence a_n does not converge!

A test that does work is that a sequence will converge a_n+1 /a_n is < 1 in the limit. However this test is not great because if the value in the limit is 1 then sometimes the sequence converges and sometimes it doesn't and the test isn't helpful in that case.

Sean

By Sean Hartnoll (Sah40) on Friday, May 12, 2000 - 11:21 am :

I've just thought of an easier example:

take a_n = ln(n). Then a_n+1 -a_n = ln((n+1)/n) which goes to 0 as n goes to infinity, but a_n again does not converge!

By Dan Goodman (Dfmg2) on Friday, May 12, 2000 - 12:10 pm :

Your answers to the first three questions are right, but I forgot to mention the point that Sean made above. Question 4 is a geometric series, which you might have already done, this should enable you to solve the problem.

By Brad Rodgers (P1930) on Friday, May 12, 2000 - 10:24 pm :

I am still not sure how you recieved the result in your proof. I follow it until the very end until you say, 1/2ⁿ converges to zero. By visualizing the series, I can see that it converges to zero however. So far the way I have been telling that a series converges is by seeing if it will eventually "settle down" at a number. I am not sure if this is correct however. I believe my problem may actually be in not understanding the notation used in series- I really have had no formal training in them, but I have read some about them. My other problem may be that I am still thinking of graphs when we speak of convergence. Is this what we are talking about?

Brad

By Brad Rodgers (P1930) on Friday, May 12, 2000 - 10:45 pm :

Also, how do we define what 2+2+2+2+2... equals if we cannot say that it equals infinity? I think that there must be different values of infinity.

By Dan Goodman (Dfmg2) on Friday, May 12, 2000 - 11:03 pm :

I'll deal with the second question first. The fundamental problem here, and it is a common one, is that infinity is not a number ! For the moment, don't let yourself say that anything equals infinity, you just can't say that, it always leads to problems. There are two things you can say concerning infinity (well, actually a lot more than two things, but anyway):

You can say that a set is infinite,
You can say that a sequence tends to infinity. You can say that a sequence tends to infinity if the following is true. For any large number M> 0, there is an integer N such that for any integer bigger than or equal to N (n> =N), x_n > M. In other words, however big a number you choose, the sequence always gets bigger than that number. Under this definition, we can say that the sum 2+2+2+2+... tends to infinity.

Now for the first question. I'll include the question and answer again, but this time I'll annotate the answer:

Q. Prove that the sequence (1,1/2,1/4,1/8,...) converges to 0.

If d> 0 is a small number, we want to find an integer N so that 1/2^N < d.: Why do we want to do this? Well, we know that 1/2ⁿ > =0 for all n, so we would expect that the number it tends to will also be > =0. Also, we know that it is smaller than 1/2, it is also smaller than 1/4, smaller than an 1/8, etc. If you were to tell me that you though the number was x, where x> 0, then I could prove you wrong by finding an integer N so that 1/2^N was less than x, so it couldn't be tending to x after all. However, if x=0, there is no N so that 1/2^N < x, so it must tend to x. The proof that it tends to 0 must then go as follows: Prove that for any number bigger than 0 it gets smaller than that number, and so can't converge to that number.
or 2^N > 1/d: (rearranging) .
Let M be an integer bigger than 1/d: (there must be an integer M that is bigger than 1/d) ,
we need to find an N so that 2^N > M: (because then 2^N > M> 1/d, so 1/2^N < d) .
But 2ⁿ > n for all n, so 2^M > M. Therefore 1/2^M < d. Also, the sequence 1/2ⁿ is getting smaller (it halves each time), so if n> M then 1/2ⁿ < 1/2^M < d.: Bearing in mind the previous comment in italics, you should now be able to see that the following statement does indeed follow:
So the sequence 1/2^n converges to 0.

Does that cover it?

P.S. Graphs are useful to visualise what is happening, keep thinking in terms of graphs, but when proving stuff, you can't use graphs.

By Brad Rodgers (P1930) on Saturday, May 13, 2000 - 01:15 am :

I think I am understanding convergence. Thanks. I am not quite sure I understand I understand the notation used in your 4th question though. Currently I haven't really used the sigma notation much, so I'll have to work to translate it. But, I think I can use it so you don't need to bother with writing another post.

Brad

By Brad Rodgers (P1930) on Saturday, May 13, 2000 - 01:23 am :

Actually, it may help if you were to just list the first few numbers in the sequence. I am not quite sure how sequence notation is used.

Brad

By Dan Goodman (Dfmg2) on Saturday, May 13, 2000 - 02:33 am :

Oops, yes, I shouldn't have used sigma notation. Basically, when I write

\sum_{i = 1}^{n} a_{i}

it means

a_{1} + a_{2} + \dots + a_{n}

. So, for Q. 4,

\sum_{i = 0}^{n} r^{i} = 1 + r + r^{2} + r^{3} + \dots + r^{n}

, so the sequence will look like

$1$

$1 + r$

$1 + r + r^{2}$

$1 + r + r^{2} + r^{3}$

...

Do you have answers to the second one (i.e. the one that converges)? If so, post it. If you manage to prove the 4th question converges and what it converges to, then you probably understand it pretty well.

By Brad Rodgers (P1930) on Saturday, May 13, 2000 - 03:44 am :

Here is my proof that the second one converges. as said before, it is somewhat simple to show that it converges merely by examining a graph, but this is not really a true proof. Basically by doing the exact same thing that I saw you do in your proof I came up with this. So don't be alarmed if it looks eerily similar.

given d> 0, we want to find 1/(n+1)1/d. Let M> d. Therefore n+1> M. For any n. Therefore, 1/(n+1) does become < d. And, as it becomes less each time(the divisor is greater),
1/(x+1) < 1/(n+1) < d. So it converges. I'm pretty sure that this is right.

I'll have to do a small number of calculations to do the next one, but I think that I can. I know right now that it does converge, but I haven't worked up a full proof yet; altough, I do have somewhat of a proof.

A semi-proof(it's enough for me to accept, but I do not know if it will pass for a real proof): From the original series, we may derive that 1+r+r² +r³ .....rⁿ =1-rⁿ⁺¹ /1-r. 1-r will stay a constant, so we are trying to prove that
rⁿ⁺¹ is convergent. We'll be multiplying by a fraction less than 1, and we've already shown that this is convergent in the example you proved( substitute 1/2 for 1/n -it still becomes correspondingly less each term). So, therefore by my logic the whole thing is convergent. It converges to 1/1-r.Sorry if this isn't very well explained, and I can try to explain it better, but it seems to make sense to me.Post again if this proof doesn't work very well.

Brad

By Dan Goodman (Dfmg2) on Saturday, May 13, 2000 - 01:12 pm :

Apart from a few typographical oddities in the above proof, I think that you have it right (e.g. I think you mean 1/(n+1) < d and M> 1/d, etc.) You've basically got the idea on the second part, actually a little bit more work needs to be done to show that if x_n converges and y_n converges and w is a constant, then wx_n converges and x_n +y_n converges. But this is quite easy to prove. You do have to be careful with this sort of thing though. I think that you've got the sense of it, but your proofs are not proper proofs yet. I wouldn't worry too much about that though.