Value of an expression as K tends to infinity

[dee...@pmail.ntu.edu.sg]

Let A and B be two fixed reals, where B is not equal to 0.

THEOREM: As K tends to positive infinity, the value of (A+(B/K)) is A, if and only if, A is not equal to 0.
PROOF: Consider the expression (A+(B/K)), which is (AK+B)/K. As K is extremely large, we can neglect B and this expression is approximately equal to AK/K = A, if and only if, A is not equal to 0.

Look forward to your comments on my THEOREM.

[Ben Bacarisse]

You have some unnecessary restrictions. The statement is also true when
either A or B (or both) are zero. The proof is a little odd. If you
are going use arguments like "we can neglect B", it's simpler just to
say "we can neglect B/K" but the real problem is that both are a little
too vague. For a simple theorem like this, I'd make the proof relate
directly to the definition of a limit.

Was this something you were asked to prove for class? If so, I am sure
you should use, directly, the definition of a limit.

--
Ben.

[hidy ho]

<dee...@pmail.ntu.edu.sg> wrote in message
news:e7b08c8f-acdf-435d...@googlegroups.com...

> Let A and B be two fixed reals, where B is not equal to 0.

take out "fixed"

>
> THEOREM: As K tends to positive infinity, the value of (A+(B/K)) is A, if
> and only if, A is not equal to 0.

wrong. K could go to negative infinity, also A being 0 has no special
requirement (if and only if) on (A+(B/K))

> PROOF: Consider the expression (A+(B/K)), which is (AK+B)/K. As K is
> extremely large, we can neglect B and this expression is approximately
> equal to AK/K = A, if and only if, A is not equal to 0.

this is not a proof, rather a spoof.

>
> Look forward to your comments on my THEOREM.

see above

[Patricia Shanahan]

This is more of a hand wave than a proof. I would start from a
definition. See, for example,
https://en.wikipedia.org/wiki/Limit_of_a_function#Limits_involving_infinity

I would aim to isolate the claimed limit, in this case A, and then show
that, for any positive epsilon, I can find a value S such that the
absolute value of the remaining term, the difference between the claimed
limit and the actual value of the function, is less than epsilon for all
K greater than S.

You claim "if and only if" for the A limit. To prove that you would also
have to prove that the limit does not tend to A as K tends to positive
infinity if A is zero. What do you think it does if A is zero?

If you did not intend to prove anything about the A=0 case you should
have put that limitation somewhere in the conditions at the start of the
theorem.

Patricia

[dee...@pmail.ntu.edu.sg]

My Theorem intended to have the "as K tends to positive infinity" phrase completely outside the if and only if part (I was already aware that this works even if K tends to negative infinity). My Theorem also intended to use the phrase "value tends to" instead of "value is".

Fine, I agree it makes sense to remove the word "fixed".

Let me rephrase my Theorem as follows (and simultaneously I re-word the proof):

Definition: Let A and B be two reals, where B is not equal to 0. Let K tend to positive infinity.

THEOREM: (A+(B/K)) tends to A, if and only if, A is non-zero.
PROOF: Given two Boolean statements P and Q, then one way of proving "P if and only if Q" is to first prove Q implies P, and to then prove not-Q implies not-P. So lets first prove that A is non-zero implies that (A+(B/K)) tends to A. (A+(B/K)) = (AK+B)/K. Here B can be neglected as AK is very much larger in magnitude. So (AK+B)/K tends to AK/K = A, if A is non-zero. So now lets focus on proving that A is zero implies that (A+(B/K)) cannot tend to 0. Let us assume that B/K tends to 0. Let us introduce a large positive L where L is much greater than K. Multiplying both sides of B/K = 0 by L, means that BL/K can be approximated as 0, which is a contradiction. This means that A is zero implies that (A+(B/K)) cannot tend to 0.
HENCE PROVED

Note what happens if my argument of multiplying both sides by L is carried out for the case of A not equal to 0. Assuming that (A+(B/K)) tends to A. So with this assumption, multiplying both sides by L means that L(A+(B/K)) tends to LA, which means that (L(KA+B))/K tends to LA. We are allowed to neglect B, so (L(KA+B))/K tends to (L(KA))/K, which is equal to LA. So we do not get any contradiction in this case.

[dee...@pmail.ntu.edu.sg]

On Monday, September 10, 2012 8:28:38 PM UTC+5:30, Patricia Shanahan wrote:
> You claim "if and only if" for the A limit. To prove that you would also
> have to prove that the limit does not tend to A as K tends to positive
> infinity if A is zero. What do you think it does if A is zero?
>
> If you did not intend to prove anything about the A=0 case you should
> have put that limitation somewhere in the conditions at the start of the
> theorem.
>
> Patricia

Please see my post made a few minutes ago. I am proving "if-and-only-if". That means as K tends to infinity, we cannot say that 1/K tends to 0 (though I agree that 1/K will come closer and closer to 0 as K tends to infinity).

Again, I am still not sure whether it is right for me to use the word "tends" here. But I really do not know what other word to use.

Please read the PROOF I wrote in my post a few mins ago, and see if you can get my meaning.

[Ben Bacarisse]

dee...@pmail.ntu.edu.sg writes:

> On Monday, September 10, 2012 7:08:51 PM UTC+5:30, hidy ho wrote:
>> <dee...@pmail.ntu.edu.sg> wrote in message
>> news:e7b08c8f-acdf-435d...@googlegroups.com...
>> > Let A and B be two fixed reals, where B is not equal to 0.
>> take out "fixed"
>> >
>> > THEOREM: As K tends to positive infinity, the value of (A+(B/K)) is A, if
>> > and only if, A is not equal to 0.
>>
>> wrong. K could go to negative infinity, also A being 0 has no special
>> requirement (if and only if) on (A+(B/K))
>>
>> > PROOF: Consider the expression (A+(B/K)), which is (AK+B)/K. As K is
>> > extremely large, we can neglect B and this expression is approximately
>> > equal to AK/K = A, if and only if, A is not equal to 0.
>>
>> this is not a proof, rather a spoof.
>>
>> > Look forward to your comments on my THEOREM.
>>
>> see above
>
> My Theorem intended to have the "as K tends to positive infinity"
> phrase completely outside the if and only if part (I was already aware
> that this works even if K tends to negative infinity). My Theorem also
> intended to use the phrase "value tends to" instead of "value is".

There was no problem in how you had it before. Moving the phrase "Let K
tend to positive infinity" from where you had it, just makes thing less
clear. Also, what problem is that you see when K tends to negative
infinity? Why insist on +oo?

> Fine, I agree it makes sense to remove the word "fixed".
>
> Let me rephrase my Theorem as follows (and simultaneously I re-word
> the proof):
>
> Definition: Let A and B be two reals, where B is not equal to 0. Let K
> tend to positive infinity.
>
> THEOREM: (A+(B/K)) tends to A, if and only if, A is non-zero.

I'd write:

For all real a, b, lim k->oo (a+b/k) = a.

But then I probably would not bother at all. The more general point is
that the limit of f(k)+g(k) is the sum on the limits of f(k) and of
g(k).

> PROOF: Given two Boolean statements P and Q, then one way of proving
> "P if and only if Q" is to first prove Q implies P, and to then prove
> not-Q implies not-P. So lets first prove that A is non-zero implies
> that (A+(B/K)) tends to A. (A+(B/K)) = (AK+B)/K. Here B can be
> neglected as AK is very much larger in magnitude. So (AK+B)/K tends to
> AK/K = A, if A is non-zero. So now lets focus on proving that A is
> zero implies that (A+(B/K)) cannot tend to 0. Let us assume that B/K
> tends to 0. Let us introduce a large positive L where L is much
> greater than K. Multiplying both sides of B/K = 0 by L, means that
> BL/K can be approximated as 0, which is a contradiction. This means
> that A is zero implies that (A+(B/K)) cannot tend to 0. HENCE PROVED

The second part is wrong. BL/K tends to 0 just as B/K does. In fact, BL
is just a different choice for B.

The first part is not really a proof. "What can be neglected" can only
be used in sketches of proofs and can't be used here at all where the
theorem is so simple. If you still think it's worth proving, do it from
first principles.

> Note what happens if my argument of multiplying both sides by L is
> carried out for the case of A not equal to 0. Assuming that (A+(B/K))
> tends to A. So with this assumption, multiplying both sides by L means
> that L(A+(B/K)) tends to LA, which means that (L(KA+B))/K tends to
> LA. We are allowed to neglect B, so (L(KA+B))/K tends to (L(KA))/K,
> which is equal to LA. So we do not get any contradiction in this case.

--
Ben.

[dee...@pmail.ntu.edu.sg]

On Saturday, September 15, 2012 6:35:18 PM UTC+5:30, Ben Bacarisse wrote:
> The second part is wrong. BL/K tends to 0 just as B/K does. In fact, BL
> is just a different choice for B.
>

> Ben.

Here L is not a fixed real, but it is a function of K that is greater than K. For example, consider L = K^2. Hence, BL is not a different choice of B that is a fixed real.

I am not in any way, trying to contradict the well-known fact that -----
Limit (K tends to infinity) (1/K) = 0.

In fact, as mentioned in my previous posting on the public thread, I did mention that I agree that as K tends to infinity, then 1/K approaches 0.

It is just that I am currently unable to find the right words to express myself.

I will repost on the Thread, once I get my definitions straight.

[Ben Bacarisse]

dee...@pmail.ntu.edu.sg writes:

> On Saturday, September 15, 2012 6:35:18 PM UTC+5:30, Ben Bacarisse wrote:
>> The second part is wrong. BL/K tends to 0 just as B/K does. In fact, BL
>> is just a different choice for B.
>>
>> Ben.
>
> Here L is not a fixed real, but it is a function of K that is greater
> than K. For example, consider L = K^2. Hence, BL is not a different
> choice of B that is a fixed real.

OK, but then your argument about "multiplying by L" becomes spurious.

> I am not in any way, trying to contradict the well-known fact that
> ----- Limit (K tends to infinity) (1/K) = 0.

...or that limit (L => oo) (A+B/K) = 0 when A = 0? If not, just remove
the condition that A =/= 0 from your original statement.

> In fact, as mentioned in my previous posting on the public thread, I
> did mention that I agree that as K tends to infinity, then 1/K
> approaches 0.
>
> It is just that I am currently unable to find the right words to
> express myself.
>
> I will repost on the Thread, once I get my definitions straight.

If your words end up suggesting that the limit is A only when A is not
zero then there is something wrong with them.

--
Ben.

[dee...@pmail.ntu.edu.sg]

> I will repost on the Thread, once I get my definitions straight.

I got a better way of expressing my idea. Please see below.

DEFINITIONS:
Let A and B be two reals, where B is not equal to 0.
Let W = Limit(K tends to infinity) ( (B/K) / (A+(B/K)) )

THEOREM: (W = 0) if and only if (A is non-zero)

PROOF: If A is non-zero, W = Limit(K tends to infinity) ( B/(AK+B) ) = 0. If A is zero, W = Limit(K tends to infinity) ( (B/K)/(B/K) ) = 1. HENCE PROVED

My DEFINITIONS & THEOREM in their present exposition, do not contradict any of the well-known facts on Limits.

I look forward to your comments

[Patricia Shanahan]

On 9/16/2012 12:46 AM, dee...@pmail.ntu.edu.sg wrote:
>> I will repost on the Thread, once I get my definitions straight.

You still say nothing about the key definition, what you mean by
"limit". Does it match the first definition at
http://en.wikipedia.org/wiki/Limit_of_a_function#Limits_involving_infinity,
or do you mean something else?

>
>
> I got a better way of expressing my idea. Please see below.
>
> DEFINITIONS:
> Let A and B be two reals, where B is not equal to 0.
> Let W = Limit(K tends to infinity) ( (B/K) / (A+(B/K)) )
>
> THEOREM: (W = 0) if and only if (A is non-zero)
>
> PROOF: If A is non-zero, W = Limit(K tends to infinity) ( B/(AK+B) ) = 0. If A is zero, W = Limit(K tends to infinity) ( (B/K)/(B/K) ) = 1. HENCE PROVED
>
> My DEFINITIONS & THEOREM in their present exposition, do not contradict any of the well-known facts on Limits.
>
> I look forward to your comments
>

Still hand-waving, not proof. Hand-waving lets you proof false
statements just as easily as true ones. See your earlier effort. I think
this is a consequence of the first problem. Without a clear definition,
of "limit", it is not possible to prove anything about limits.

Patricia

[Ben Bacarisse]

dee...@pmail.ntu.edu.sg writes:

>> I will repost on the Thread, once I get my definitions straight.
>
>

> I got a better way of expressing my idea. Please see below.

What you have below is a completely different limit. If write about x+y
and then say that x/y is a better way of "expressing your idea" the
people who commented on x+y might feel that you wasted their time.

> DEFINITIONS:

>
> Let A and B be two reals, where B is not equal to 0.
>

> Let W = Limit(K tends to infinity) ( (B/K) / (A+(B/K)) )
>
> THEOREM: (W = 0) if and only if (A is non-zero)
>
> PROOF: If A is non-zero, W = Limit(K tends to infinity) ( B/(AK+B) ) = 0. If A is zero, W = Limit(K tends to infinity) ( (B/K)/(B/K) ) = 1. HENCE PROVED
>

> My THEOREM in its present exposition, does not contradict any of the well-known facts.

>
> I look forward to your comments

There's lots of things to say, but you might decide tomorrow that a
completely different limit is a better way of expressing your idea.

--
Ben.

[Patricia Shanahan]

On 9/16/2012 4:32 AM, Ben Bacarisse wrote:
> dee...@pmail.ntu.edu.sg writes:
>
>>> I will repost on the Thread, once I get my definitions straight.
>>
>>
>> I got a better way of expressing my idea. Please see below.
>
> What you have below is a completely different limit. If write about x+y
> and then say that x/y is a better way of "expressing your idea" the
> people who commented on x+y might feel that you wasted their time.

...

I think your comments remain relevant. The attempted proof is still,
unfortunately, a matter of vague assertions, not grounded in any
definition of what exactly the OP means by limit.

That proof method, as illustrated by your comments on earlier articles,
can be used to "prove" just about anything, regardless of whether or not
it could be otherwise proved or disproved.

Patricia

[Dike Melanis]

<dee...@pmail.ntu.edu.sg> wrote in message
news:68f49d3e-381a-4351...@googlegroups.com...

>> I will repost on the Thread, once I get my definitions straight.
>
>

> I got a better way of expressing my idea. Please see below.
>

> DEFINITIONS:

> Let A and B be two reals, where B is not equal to 0.

> Let W = Limit(K tends to infinity) ( (B/K) / (A+(B/K)) )
>
> THEOREM: (W = 0) if and only if (A is non-zero)

Brackets are not acceptable.
"tends" is the wrong word, "approches" is correct

stated clearly your theorm is
"the limit of (B/K) / (A+(B/K)) aproches 0 as K approches infinity"

you should simplify (B/K) / (A+(B/K)) = B/(B+KA)

again, the values of A, B are not needed.

try again



>
> PROOF: If A is non-zero, it is easy to see that W = Limit(K tends to
> infinity) ( (B/K) / A ) = 0. If A is zero, it is easy to see that W =
> Limit(K tends to infinity) ( (B/K) / (B/K) ) = 1.

> HENCE PROVED
>
> My THEOREM in its present exposition, does not contradict any of the
> well-known facts.
>

> I look forward to your comments.

[Patricia Shanahan]

On 9/20/2012 4:13 PM, Dike Melanis wrote:
>
> <dee...@pmail.ntu.edu.sg> wrote in message
> news:68f49d3e-381a-4351...@googlegroups.com...
>>> I will repost on the Thread, once I get my definitions straight.
>>
>>
>> I got a better way of expressing my idea. Please see below.
>>
>> DEFINITIONS:
>> Let A and B be two reals, where B is not equal to 0.
>> Let W = Limit(K tends to infinity) ( (B/K) / (A+(B/K)) )
>>
>> THEOREM: (W = 0) if and only if (A is non-zero)
>
> Brackets are not acceptable.
> "tends" is the wrong word, "approches" is correct
>
> stated clearly your theorm is
> "the limit of (B/K) / (A+(B/K)) aproches 0 as K approches infinity"
>
> you should simplify (B/K) / (A+(B/K)) = B/(B+KA)
>
> again, the values of A, B are not needed.

If A is zero then B/(B+KA) is B/B. For non-zero B, it is one. For zero B
it is 0/0.

If A is non-zero then the limit is zero regardless of the value of B.

Patricia

[Dike Melanis]

"Patricia Shanahan" <pa...@acm.org> wrote in message
news:yIudnXxwp83iCcHN...@earthlink.com...

which in this case can be proven to be 1.
so the value of B is not needed

>
> If A is non-zero then the limit is zero regardless of the value of B.

true. So A must be known to be non zero.

>
> Patricia
>

[Barb Knox]

In article <k3np4n$7ge$1...@news.albasani.net>,

"Dike Melanis" <inv...@invalid.com> wrote:

> "Patricia Shanahan" <pa...@acm.org> wrote in message
> news:yIudnXxwp83iCcHN...@earthlink.com...
> > On 9/20/2012 4:13 PM, Dike Melanis wrote:
> >>
> >> <dee...@pmail.ntu.edu.sg> wrote in message
> >> news:68f49d3e-381a-4351...@googlegroups.com...
> >>>> I will repost on the Thread, once I get my definitions straight.
> >>>
> >>>
> >>> I got a better way of expressing my idea. Please see below.
> >>>
> >>> DEFINITIONS:
> >>> Let A and B be two reals, where B is not equal to 0.
> >>> Let W = Limit(K tends to infinity) ( (B/K) / (A+(B/K)) )
> >>>
> >>> THEOREM: (W = 0) if and only if (A is non-zero)
> >>
> >> Brackets are not acceptable.
> >> "tends" is the wrong word, "approches" is correct
> >>
> >> stated clearly your theorm is
> >> "the limit of (B/K) / (A+(B/K)) aproches 0 as K approches
> >> infinity"
> >>
> >> you should simplify (B/K) / (A+(B/K)) = B/(B+KA)
> >>
> >> again, the values of A, B are not needed.

> > If A is zero then B/(B+KA) is B/B. For non-zero B, it is one. For zero B
> > it is 0/0.
>
> which in this case can be proven to be 1.

No, it cannot. If you believe it can, please provide at least the
outline of a proof.

> so the value of B is not needed


> > If A is non-zero then the limit is zero regardless of the value of B.
>
> true. So A must be known to be non zero.

"Must"? Says who? There are no constraints in the given problem for A
or B.


> > Patricia
--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum videtur.
| BBB aa a r bbb |
-----------------------------

[Barb Knox]

In article <see-BA9252.1...@news.eternal-september.org>,

Oh bother. The problem does state "B is not equal to 0".