Side Topic: 'It's Not a Toom-ah' - Page 5

wblipp · 2015-11-13, 16:36

Quote:

Originally Posted by R.D. Silverman

FALSE. I was, for example, introduced to epsilon-delta proofs in high school

I agree that delta-epsilon proofs are not graduate level mathematics - I also encountered them in high school. I was referring to the mention of Hilbert spaces earlier in the thread. Although I was introduced to Hilbert Spaces as an undergraduate, I wouldn't expect business majors to be. I wouldn't even expect math majors taking a first calculus course to already be familiar with Hilbert Spaces.

Quote:

Originally Posted by R.D. Silverman

NO! NO! NO! This is NOT what it means. It means that the infinite sequence .3, .33, .333, .3333, ..... converges to 1/3 in the limit. All this blather about "closer and closer to 1/3" is informal HAND-WAVING.

Toom would not have expected the students to already know the formal definition of convergence. At least for me, it was part of the calculus curriculum that he was preparing to teach these kids. He would have been pleased to see hand-waving blather about closer and closer because they would then be prepared for how to formalize those ideas with delta-epsilon definitions. What he really got was students using their calculators to find the difference - a clear indication that their failure to grasp the ideas.

davar55 · 2015-11-13, 16:42

Quote:

Originally Posted by wblipp

I agree that delta-epsilon proofs are not graduate level mathematics - I also encountered them in high school. I was referring to the mention of Hilbert spaces earlier in the thread. Although I was introduced to Hilbert Spaces as an undergraduate, I wouldn't expect business majors to be. I wouldn't even expect math majors taking a first calculus course to already be familiar with Hilbert Spaces.
Toom would not have expected the students to already know the formal definition of convergence. At least for me, it was part of the calculus curriculum that he was preparing to teach these kids. He would have been pleased to see hand-waving blather about closer and closer because they would then be prepared for how to formalize those ideas with delta-epsilon definitions. What he really got was students using their calculators to find the difference - a clear indication that their failure to grasp the ideas.

And I would add, the definition of convergence to a limit (delta-epsilon) must be accepted as straight-forward
but not obvious. It takes a few days or weeks of doing epsilon-delta proofs (homework!) to accept the concept.

R.D. Silverman · 2015-11-13, 17:06

Quote:

Originally Posted by retina

I will try.

Very early in my schooling I learned to do long division. If I use that technique to divide 3 into 1 then the answer generates 0.3333... The remainder at each step after dividing 10 by 3 is always 1, and carrying down the next 0 gives back 10/3 to the next step.

And.....??

All you have done is show that .3, .33, .333, .3333 , ..... is a series of approximations to 1/3. One can argue
that no matter how far you carry the long division process you are STILL left with a remainder.

You need to PROVE that it equals 1/3 *IN THE LIMIT*. You have not addressed what "IN THE LIMIT" means,
all you have done is hand wave around it.

One needs to work from a precise definition of limit. Math is done and proofs are given as a series
of steps, where each step follows from its predecessor and each step is governed by logic.
But you need to start with a well founded set of AXIOMS and DEFINITIONS. Your "long division"
does none of this because no matter how far you carry out the definition you still have a non-zero
remainder.

R.D. Silverman · 2015-11-13, 17:18

Quote:

Originally Posted by alpertron

By definition:
$x = .3333... = \frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+...$ (1)

NO! This is more hand waving because you have not given a definition
to the ellipsis.........

You have made an assertion using notation that you have not defined,.
In fact, the ellipsis is just SHORTHAND notation for the *limit* of
the sequence .3, .33, .333, .......

A correct proof must use the formal definition of limit.

All people are doing is making informal assertions. I agree that
these assertions are convincing, but they are not PROOFS./

Quote:

Multiplying by 10:
$10 x = 3+\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+...$ (2)

Please give the axiom or prior theorem that allows one to multiply an
*infinite* sequence term by term and still get the correct answer.

This is just more hand waving.

Quote:

Subtracting (2) - (1) all terms after the 3 are cancelled:

9x = 3, so x=1/3

What do you mean by "cancelled"? You must PROVE that the subtraction on a digit by
digit basis for a number with an infinite decimal representation is valid.

This is just more informal hand waving.

Quote:

Another attempt without subtracting infinite terms is by considering the sequence .3, .33, .333, .3333, ...

The nth term is:
$x = .3333...3 = \frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+...+\frac{3}{10^{n-1}}+\frac{3}{10^n}$ (3)
$10x = 3+\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+...+\frac{3}{10^{n-2}}+\frac{3}{10^{n-1}}$ (4)

Subtracting (4) - (3), discarding all terms that are equal:
$9x = 3-\frac{3}{10^n}$

So for n->inf we get 9x=3, which implies x=1/3

This is much much better. It is very close to a complete and correct proof. But one still needs to prove that lim n-->oo of 3/10^n = 0.
One can't just assert it.

R.D. Silverman · 2015-11-13, 17:30

Quote:

Originally Posted by R.D. Silverman

This is much much better. It is very close to a complete and correct proof. But one still needs to prove that lim n-->oo of 3/10^n = 0.
One can't just assert it.

A digression. It is "mildly" interesting that lim n-->oo 1/10^n = lim n-->oo k/10^n for all real k.
For those of you who know:

Prove or disprove for real k.

If lim n-->oo f(n) exists, then lim n-->oo k*f(n) = k* lim n-->oo f(n)

alpertron · 2015-11-13, 18:04

Quote:

Originally Posted by R.D. Silverman

But one still needs to prove that lim n-->oo of 3/10^n = 0.
One can't just assert it.

First of all we can notice that 3/10^n is positive for all integer n by using induction. For n=0, 3/10^0 is positive, and for n: 3/10^n is positive, so we get that 3/10^(n+1) is positive also (the quotient of two positive numbers is also positive). (1)

if n > r we get 10^(n-r) > 1, and multiplying both sides by 3/10^n we get: 3/10^r > 3/10^n (2)

Then we select a small number eps. Selecting r=log(eps)/log(10)+1 we get that eps = 1/10^(r-1), so eps > 3/10^r, and from (2) we get eps > 3/10^n for all n>=r. So the limit is lower than any positive number but it is not negative due to (1). So the limit as n->inf must be zero.

R.D. Silverman · 2015-11-13, 18:22

Quote:

Originally Posted by alpertron

. So the limit is lower than any positive number but it is not negative due to (1). So the limit as n->inf must be zero.

Not quite. This is correct, but it requires the squeeze theorem. A full blown proof from the definition
of limit would be better.

alpertron · 2015-11-13, 18:38

Quote:

Originally Posted by R.D. Silverman

Not quite. This is correct, but it requires the squeeze theorem. A full blown proof from the definition
of limit would be better.

I have not used that theorem. The definition of limit is:

For each real number $\epsilon > 0$ , there exists a natural number $N$ such that, for every natural number $n > N$ , we have $|x_n - x| < \epsilon$ .

In my proof I used r for N and eps for $\epsilon$ .

R.D. Silverman · 2015-11-13, 18:54

Quote:

Originally Posted by alpertron

I have not used that theorem. The definition of limit is:

For each real number $\epsilon > 0$ , there exists a natural number $N$ such that, for every natural number $n > N$ , we have $|x_n - x| < \epsilon$ .

In my proof I used r for N and eps for $\epsilon$ .

No, especially since you have not defined x, and x_n..

Let f be a function over R. Then limit x-->c f(x) = L implies that for ALL epsilon > 0 there exists a delta
such that if 0 < | x-c| < delta then |f(x) - L| < epsilon.

R.D. Silverman · 2015-11-13, 19:00

Quote:

Originally Posted by R.D. Silverman

No, especially since you have not defined x, and x_n..

Let f be a function over R. Then limit x-->c f(x) = L implies that for ALL epsilon > 0 there exists a delta
such that if 0 < | x-c| < delta then |f(x) - L| < epsilon.

BTW, over C the absolute value condition is replaced by "lies in a circle of radius less than delta (resp) epsilon.

Let's get back to number theory. We can do real analysis some other time......

alpertron · 2015-11-13, 19:03

Quote:

Originally Posted by R.D. Silverman

No, especially since you have not defined x, and x_n..

Let f be a function over R. Then limit x-->c f(x) = L implies that for ALL epsilon > 0 there exists a delta
such that if 0 < | x-c| < delta then |f(x) - L| < epsilon.

The definition does not work if c is infinite, especially the part 0 < | x-c| < delta.

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