# The Importance of Limits in the Development of Mathematics

1. Nov 6, 2007

### titaniumx3

"The Importance of Limits in the Development of Mathematics"

Hi,

I will be writing a project on "The Importance of Limits in the Development of Mathematics". This wasn't exactly my first choice of topic so I'm not totally sure where to start. The project is going to be about 4000 words and should be a blend of of mathematical and historical content. The idea is that it should be readable by anyone new to the topic at hand (or mathematics in general) but still "good enough" for a mathematician.

So, apart from the obvious (like calculus, etc) I would really appreciate some suggestions for avenues of research relating to limits or anything that has some relevance to it in mathematics.

Thank you! :)

2. Nov 6, 2007

3. Nov 7, 2007

### Gib Z

Or Solutions to Zeno's Paradox(s), thats a nice way for laymen to see what limits do.

4. Nov 7, 2007

### titaniumx3

I've done some brief reading on this and it seems to be exactly the type of thing I'm looking for. Would you know of any good books relating to topics like this? I would obviously need a solid understanding of the subject matter at hand before I write anything.

5. Nov 8, 2007

### Gib Z

I can't recommend books on these Paradoxes because I myself have only read about them online, although the internet does have some good information on them (and there isn't really that much to them anyway), so try Wikipedia, Planetmath has some stuff on it too, perhaps Mathworld will, and the random sites Google brings could be some help as well.

6. Nov 14, 2007

### zoki85

For nonegative integers n ,consider the limit:

(a) $$L=\sqrt{n+\sqrt{n+\sqrt{n+...}}}$$

(If $a_{1}=\sqrt{n},a_{k+1}=\sqrt{n+a_{k}}$ that's
another way of notation we are interested in
$L=\lim_{k\to\infty}a_{k}$)

Adding up n to the both sides of (a) we have

$$L+n=n+ \sqrt{n+\sqrt{n+...}}$$

By taking a square root from that obviously
$$\sqrt{L+n}=L$$

Then, by squaring this up
$$L+n=L^2$$

Finally ,rearranging gives:

$$n=L^2-L$$

Derived is the formula which shows how to calculate
number n if we want our limit to be number L.

For example ,if we want the limit that equals L= 2 we have
$$n=2^2-2=2$$.And it is true that

$$2=\sqrt{2+\sqrt{2 + \sqrt{2+...}}}$$

If we want n for L= 4 we have $$n=4^2-4=12$$
And it is true that

$$4=\sqrt{12+\sqrt{12+...}}$$

See what happens if we want our limit to be L=1.
For n we get $$n=1^2-1=0$$.
This would imply that

$$1=\sqrt{0+\sqrt{0+\sqrt{0+...}}}$$

And (obviously) this is WRONG.

Now the question.Where exactly is the error in the chain of conclusions?

I suppose paradoxes of this kind might be good examples
to present in your work (if you like challenges trickier than Zeno's)

7. Nov 14, 2007

### Math Jeans

Would you mind if I sent a copy of this problem to some of the members of my school as a challenge problem? I think that it is really cool.

8. Nov 15, 2007

### zoki85

Do with it whatever you want.
And I didn't find it online but,few years ago,I saw it in an article
discussing limits,operations with infinities and similar matters.
Maybe the article was from russian journal "Quant"(?),I'm not sure.
Anyway,I couldn't forget this example becouse I agree with you :it's "cool" .

1)Observe first the sequence $a_ {k}$ considered only for naturals n (including 0):

$$\sqrt{0},\sqrt{0+\sqrt{0}},\sqrt{0+\sqrt{0+\sqrt{0}}},...$$

Here,it's obviously true $$L=\lim_{k\to\infty}a_{k}=0$$

2)Than think of the limit constructed in reals $$\mathbb{R}$$:

$$L'=\lim_{x\to 0}\sqrt{x+\sqrt{x+\sqrt{x+...}}}$$

Compare two cases given becouse there is the difference!
What do you think,does the limit in (2) exist (what its' value L' equals to) ?
To resolve this paradox a full understanding of functions and their limits is necessary.
That's why it is harder to deal with than with Zeno's.

Last edited: Nov 15, 2007
9. Nov 15, 2007

### Math Jeans

>.< This is driving me crazy.

EDIT: I solved it earlier this morning. I feel so proud of myself .

Last edited: Nov 16, 2007
10. Dec 27, 2007

### titaniumx3

Sorry, for reviving such an old thread, but I just wanted to thank zoki85 for the excellent example! Now, I just need to solve it myself lol.

11. Dec 27, 2007

### yasiru89

zoki85's example is well and truly classic, dating back somewhere around James Gregory I think. I read about it first in a Michigan Reprint of Loney's Analytical Trigonometry as I remember.
You could also discuss the geometrical nature of limiting processes. I recommend borrowing Needham's Visual Complex Analysis. Practicalities will be interesting, like how renormalising processes are reduced to the theory of limits on occassion. Like the Cesaro definition and oscillating Abel sums.

12. Dec 27, 2007

### mathwonk

the first use of limits may be by the greeks, who used it to show that two pyramids with equal bases and equal heights have equal volume.

the analogous statement for triangles was demonstrated by finding a finite decomposition of one triangle into pieces that could be reassembled to form the other.

gauss remarked it was too bad such a proof had not been given for polyhedra and hilbert made the question of existence of congruent decompositions one of his famous problems.

max dehn soon solved it proving that finite decompositions cannot exist for all polyhedra of the same volume, e.g. a pyramid and a cube of same volume cannot have congruent decompositions.

this well described in hartshorne's book on geometry.

i have not studied it so may have rendered it here imperfectly, but it clear that limits as used in integral calculus for the computation of volumes, before the introduction of differentiation and antidifferentiation, is due to the ancient greeks, and they were masters of it.

see my thread who wants to be a mathematician for a recent post with more detail on archimedes masterful use of this technique.

Last edited: Dec 27, 2007
13. Dec 30, 2007

### titaniumx3

For (2), if I assume the limit exists then I get that $${L'}^{2}=L'$$ which means that L' is equal to 0 or 1, which is a contradiction, hence the limit does not exist.

I'm not sure how this fits in with the original problem though.

Last edited: Dec 30, 2007
14. Jan 7, 2008

### titaniumx3

Sorry, to bring this up again but I still can't get a conclusive answer to where the error is in the problem that zoki85 posted.

It clear to me that the two limits are different, where one is the limit of a sequence of 0's, where as the other is the limit of a function (which seems to be undefined). But I don't see how that helps with the original question (where is the error?).

15. Jan 7, 2008

### Math Jeans

hint: take the equation L=sqrt(L+n) and apply it over...and over.

16. Jan 7, 2008

### titaniumx3

You end up with the same limit. Since L^2 - L = 0 has two solutions than it is undefined?

17. Jan 7, 2008

### Math Jeans

No. What is under the last square root?

18. Jan 7, 2008

### titaniumx3

L+n ?

This is really doing my head in, I know there is something blatantly obvious about this whole thing, but my mind has just gone completely blank for some reason ...

Last edited: Jan 7, 2008
19. Jan 8, 2008

### Math Jeans

And the thought experiment is about what happens when L is equal to 1. So what happens if the value of the last square root is 1?

20. Jan 8, 2008

i got it
if iam right
in first one you found that value of the function by limiting the series of (0 )to infinity

but in second one you found the limit of function at 0

as a case limit need not to be equal to the value of the function at the point
hence 1 is limit ,where as 0 is value of the function

i thank the person for posting this paradox, it is certainly a good one i had in last 1 month

Last edited: Jan 8, 2008
21. Jan 8, 2008

### titaniumx3

Ok, so $1=\sqrt{0+\sqrt{0+\sqrt{0+...}}}$ is actually correct. So it's not really a limit at all, since you are really just solving L^2+L=0.

Also,

Isn't this actually undefined in the reals, since the left and right hand limits don't coincide? Though, if it were defined in the open interval (0, infinity) than it would equal to 1.

22. Jan 8, 2008

how it is undefined i dont know ,may be you can demonstrate with an example or proof

well what i had written was
L$$^{2}$$=n+L was a numerically equivalent expression to the one you mentioned

so L=$$\frac{1+\sqrt{1+4n}}{2}$$
so lim n=>0=1

but as by the series $$\sqrt{0}$$,$$\sqrt{0+\sqrt{0}}$$,.........so on
it tell that lim k=>$$\infty$$ is equal to the value of function at 0, not the limit of function

however on solving the L$$^{2}$$=n+L equation we get two general solutions
one we ignore as it gives negative value for natural number , the other we consider for sol. ,hence to calculate limit we use this (+) (one) solution .
as it cant have two limit
the other one when n=0 gives the value of function at n=0 which is different from limit

23. Jan 8, 2008

### Gib Z

He meant that you can't take the limit in the normal sense, instead we have the extra condition that x go through a sequence of decreasing positive numbers, ie approach zero from the right. We can not take it from the left because x< 0 is out of the domain of the expression, when dealing only in Real numbers.

24. Jan 8, 2008

### titaniumx3

This is what confused me. So say we were dealing with complex numbers too, and the expression was defined over all the reals would the limit not exist as x approaches 0?

But if you limited the domain to the open interval (0,infinity) wouldn't it be impossible to take the limit at 0 since 0 is not an element of (0, infinity)? [according to the formal definition of a limit]

25. Jan 8, 2008

### Gib Z

I havent worked it, but I think that when approaching from the left, you get some complex number as the limit, but from the right another limit. The limits arent equal.