The meaning of infinity

[Artemus]

You are introducing limits (from left field) to make use of the expression, not proving it has a final value.

Thus demonstrating again that you don't understand what 0.999... means.

 

[Jokodo]

You are introducing limits (from left field) to make use of the expression, not proving it has a final value.

Thus demonstrating again that you don't understand what 0.999... means.

He doesn't understand what "mean" means.

 

[untermensche]

You are introducing limits (from left field) to make use of the expression, not proving it has a final value.

Thus demonstrating again that you don't understand what 0.999... means.

You don't know what it means. It has no final value.

You only know what you can do to it.

It is an expression that has nothing to do with limits. It is limitless.

Why are you talking about limits and inventing statements containing limits?

 

[untermensche]

You don't know.

I am not arguing an equivalence where an infinite string by magic ends.

You are introducing limits (from left field) to make use of the expression, not proving it has a final value.

You are rounding off by definition when the difference is insignificant, not proving the 9's end.

You are not arguing anything. You're crying foul whenever something doesn't fit your preconceived conclusions, or whenever reality becomes too complex for you too understand (presumably mostly, because you aren't even trying; you probably could be a smart boy if you wanted).

You are not explaining what something is by transforming it or by applying operations to it.

You have a bag of tricks, no explanations of what things are.

You preform operations to entities with no final value and pretend you get a final product.

Total nonsense.

 

[DBT]

coincidence

Hilarious.

 

[steve_bank]

How so? A Reimann integration is a limut as dx goes to zero. It is an approximation.

No it isn't. It is an exact operation. Please cite any mathematical text or peer reviewed article that says it is an approximation.

Riemann integration is a limit as the interval of integration is divided into infinitesimal divisions, or dx. A limit as dx goes to zero. I went through this when I studied Numerical Methods in the 80s. The reason I got my first PC.

Calculate integral cos(x) over 0 pi/4. First use the exact solution int cos() = sin(). Then calculate using Rei,amm integration, or sums. As the dx or the number of intervals get small the numerical solution gets close to to the first integration. Get up to a few thousand intervalues and the result will equal the accuracy of the truncated series on a calculator.

https://en.wikipedia.org/wiki/Riemann_sum
https://en.wikipedia.org/wiki/Riemann_integral

"The basic idea of the Riemann integral is to use very simple approximations for the area of S. By taking better and better approximations, we can say that "in the limit" we get exactly the area of S under the curve."

We say it is exact because the error becomes infinitesimally small. Not because it is really exact.


The Fundamental Theorem results from a limit to infinity


https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

We say integral x^n = (x/n)^n+1 is an exact integral, but it is really an approximation. We say it is exact because by the fundamental theorem it reduces to an algebraic equation.,albeit in the limit. A limit that s never reached. An exact integral means not having to resort to numerical solutions.

 

[steve_bank]

That wont do. The answer must be a real number. If you cant come up with a real value then the error is EQUAL to 0. That is the definition of equal for reals.

You are not listening, an approximate real number.

Likewise limit x -> inf 1/[2 + 1/x] approaches 2 but never gets there. We take it as 2 in a calculation.

Infinite limits are a tool for analysis, but are nor real.

”Real number” means a point in a the continuum. A element if R, the set of Reals. Not a number for an actual measurement, those will by obvious reasons have finite accuracy.

But I leave you in the same sad abyss of ignirabce as your friend here, untermensche. You have too much arrogance to able to discuss the matter reasonably.. Bye.

Remember that the real number line has an infinite number of points. The word I am familiar with is instantiate, meaning reducing to a specific instance. 2.0 is a specific instantiation of real numbers. This prents a problem in statistics.

In a continuous(real number) distribution the probability of a single event is zero. Integrate the Gaussian Distribution between a and b owner a = b the result is zero. A probability can only be reduced to an interval, although the interval can be made arbitrarily small.

Practical problems with reals and infinities are more widespread than you may imagine. Treating math as an abstraction without seeing how the concepts are actually applied can be misleading.

 

[Juma]

A limit that is never reached.

Sorry for disturbing you again but must ask you about this. What does it mean?

It is true that the limit isnt reached for any finite number of terms of the series. But that is not a problem because the value we are after isnt for a finite number of terms. The value we want to calculate is the limit. Not one of the approximations by a using a finite number of terms. Thus the limit is an ACTUAL EXACT value. Not an approximation.

We are not actually summing all these terms, we are calculating the limit.

Do you realize the difference?

 

[steve_bank]

In an infinite limit that converges on a finite vale the error is infinitesimally small. Infinity can never be reached.

y = 1/(2 + 1/x) x can never go to zinfinity in reality, 1/x can never go to zero.. Computationally in practice 0.5 would be taken as exact.

1/x is 0 from certain points of view. If you divide a line into an infinite amount of pieces, each piece is a point. It shifts dimensions from 1d to 0d. You can do it with volumes of spacetime as well (shifting down a dimension by dividing infinitely).

The fact that it can be done mentally means that it is done by nature.... but that's a rabbithole of an argument if I've ever seen one. And I haven't. I've read and herd them.

From a practical view yes.

We say as x goes to infinity 1/x goes to 0 in the limit, but never gets there. Computationally I would use zero.

 

[steve_bank]

A limit that is never reached.

Sorry for disturbing you again but must ask you about this. What does it mean?
It is true that it isnt reached for any finite number of terms of the seires. But that is not a problem because the value we are after isnt for a finite number of terms. The value we want to calculate is the limit. Not one of the approximations by a using a finite number of terms. Thus the limit is the ACTUAL EXACT value. Not an approximation.
We are not actually summing all these terms, we are calculating the limit.

Did thus help you?

Juma

In the workplace I had a policy of never stealing someone's thunder unless it was necessary. Not my style to cut somebody down for no reason.

From all your posts I can only conclude you did not have the usual undergrad math or never learned any of it or applied any of it or it was a technology program not as rigorous as science and engineering.

I suggest getting your calc book and working some problems.

0.555.. . is a definition, an unchanging infinite number of 5s.

Converting to a geometric series and playing finite arithmetic is not the definition.

Infinite and finite are mutually exclusive. 0.999... can never equal a finite 1. ).333.. results in 1/3 with no incremting to a higher values, 1.3 = 0.333...

The method applied to 0.999.. yields the fraction 1/1 as an approximation. The fact that 0.999.. gets a rollover to 1 and 0.333... does not round up means that 0.999.. = 1/1 is not a proof that it is a finite 1. As syaed in the link geometric series applied to decimals yields a fractional approximation.

If you disagree you have to explain the discrepancy between the results for 0.333... and 0.999... Is there a difference between the two other than the fact that for 0.999.. the algorith can only yield 1/1, an artifact of the equation not a proof.

 

[Juma]

A limit that is never reached.

Sorry for disturbing you again but must ask you about this. What does it mean?
It is true that it isnt reached for any finite number of terms of the seires. But that is not a problem because the value we are after isnt for a finite number of terms. The value we want to calculate is the limit. Not one of the approximations by a using a finite number of terms. Thus the limit is the ACTUAL EXACT value. Not an approximation.
We are not actually summing all these terms, we are calculating the limit.

Did thus help you?

Juma

In the workplace I had a policy of never stealing someone's thunder unless it was necessary. Not my style to cut somebody down for no reason.

From all your posts I can only conclude you did not have the usual undergrad math or never learned any of it or applied any of it or it was a technology program not as rigorous as science and engineering.

I suggest getting your calc book and working some problems.

You dont realize how funny that post is...
you have obviously gained yourself a very skewed image of how limits works.
Top that with a ton of arrogance...
do you have any thought of why so many tells you that you are wrong?
And the fact that you dont bother to actually respond to the content of the post is telling...

Lets try again:

The exact sum S of an convergent infinite serie is DEFINED as the LIMIT of the partial sums when number of terms goes to infinity.
Each partial sum can be seen as an approximation of S.
But S isnt any of this partial sums, it is the limit of these sums.
The limit can be calculated in various ways without resorting to perform an actual infinite addition.
Thus it is not an approximation.

 

[steve_bank]

I stayed why I reject the conclusion.After this I will just be reating myself.

I have applied limits and series in real world problems. I'll go with my experience and analysis of the conclusion.

0.555.. . is a definition, an unchanging infinite number of 5s.

Converting to a geometric series and applying finite arithmetic is not the definition.

Infinite and finite are mutually exclusive. 0.999... can never equal a finite 1. )0333.. results in 1/3 with no incremting to a higher values, 1.3 = 0.333...

The method applied to 0.999.. yields the fraction 1/1 as an approximation. The fact that 0.999.. gets a rollover to 1 and 0.333... does not round up means that 0.999.. = 1/1 is not a proof that it is a finite 1. As stated in the link geometric series applied to decimals yields a fractional approximation.

0.999... = 1, 1 does not yield 0.999... and 0.333... = 1/3 = 0.333... An unresolved discrepancy. Conclusion the algorithm does not serve as a proof, it serves to provide fractional approximations.

If you disagree you have to explain the discrepancy between the results for 0.333... and 0.999... Is there a difference between the two other than the fact that for 0.999.. the algorithm can only yield 1/1, an artifact of the equation not a proof.

If you want to continue the debate, answer the above. I do not question the generation of a/1-r by a series, I question the conclusion of a proof. and interpretation of results. You are reading more into the method than it is intended to supply.

The series is by definition an approximation to 0.999... No different than a series approximation to trig functions,

 

[Juma]

I stayed why I reject the conclusion.After this I will just be reating myself.

I have applied limits and series in real world problems. I'll go with my experience and analysis of the conclusion.

0.555.. . is a definition, an unchanging infinite number of 5s.

Converting to a geometric series and applying finite arithmetic is not the definition.

Infinite and finite are mutually exclusive. 0.999... can never equal a finite 1. )0333.. results in 1/3 with no incremting to a higher values, 1.3 = 0.333...

The method applied to 0.999.. yields the fraction 1/1 as an approximation. The fact that 0.999.. gets a rollover to 1 and 0.333... does not round up means that 0.999.. = 1/1 is not a proof that it is a finite 1. As stated in the link geometric series applied to decimals yields a fractional approximation.

0.999... = 1, 1 does not yield 0.999... and 0.333... = 1/3 = 0.333... An unresolved discrepancy. Conclusion the algorithm does not serve as a proof, it serves to provide fractional approximations.

If you disagree you have to explain the discrepancy between the results for 0.333... and 0.999... Is there a difference between the two other than the fact that for 0.999.. the algorithm can only yield 1/1, an artifact of the equation not a proof.

If you want to continue the debate, answer the above. I do not question the generation of a/1-r by a series, I question the conclusion of a proof. and interpretation of results. You are reading more into the method than it is intended to supply.

The series is by definition an approximation to 0.999... No different than a series approximation to trig functions,

<definition>
A ”decimal number” is an ordered set of integers An (where n is all negative and positive integers and for each An < 9 and An >= 0.) and a corresponding value = Sum of An*10^n (for all n)
</definition>

Do you agree on this?

 

[Jokodo]

I stayed why I reject the conclusion.After this I will just be reating myself.

I have applied limits and series in real world problems. I'll go with my experience and analysis of the conclusion.

0.555.. . is a definition, an unchanging infinite number of 5s.

Converting to a geometric series and applying finite arithmetic is not the definition.

No conversion is going on. "an infinite number of 5s" is the definition of the string. Used as a label, the geometric series is what this string refers to, always did, without any conversion needed.

Infinite and finite are mutually exclusive. 0.999... can never equal a finite 1. )0333.. results in 1/3 with no incremting to a higher values, 1.3 = 0.333...

The method applied to 0.999.. yields the fraction 1/1 as an approximation. The fact that 0.999.. gets a rollover to 1 and 0.333... does not round up means that 0.999.. = 1/1 is not a proof that it is a finite 1. As stated in the link geometric series applied to decimals yields a fractional approximation.

0.999... = 1, 1 does not yield 0.999... and 0.333... = 1/3 = 0.333... An unresolved discrepancy. Conclusion the algorithm does not serve as a proof, it serves to provide fractional approximations.

What discrepancy? Both the number that's the sum of the series [3 * 10 ^ -i for i in range(inf)] and the number that's the sum of the series [9 * 10 ^ -i for i in range(inf)] are finite real and rational numbers. The fact that one of them can be shortened to 1.0 or 1 in decimal notation while the other cannot is an idiosyncracy of decimal notation. In duodecimal, 1/3 is 0.4, plain and simple and very much finite.

 

[bilby]

I stayed why I reject the conclusion.After this I will just be reating myself.

I have applied limits and series in real world problems. I'll go with my experience and analysis of the conclusion.

0.555.. . is a definition, an unchanging infinite number of 5s.

Converting to a geometric series and applying finite arithmetic is not the definition.

Infinite and finite are mutually exclusive. 0.999... can never equal a finite 1. )0333.. results in 1/3 with no incremting to a higher values, 1.3 = 0.333...

The method applied to 0.999.. yields the fraction 1/1 as an approximation. The fact that 0.999.. gets a rollover to 1 and 0.333... does not round up means that 0.999.. = 1/1 is not a proof that it is a finite 1. As stated in the link geometric series applied to decimals yields a fractional approximation.

0.999... = 1, 1 does not yield 0.999... and 0.333... = 1/3 = 0.333... An unresolved discrepancy. Conclusion the algorithm does not serve as a proof, it serves to provide fractional approximations.

If you disagree you have to explain the discrepancy between the results for 0.333... and 0.999... Is there a difference between the two other than the fact that for 0.999.. the algorithm can only yield 1/1, an artifact of the equation not a proof.

If you want to continue the debate, answer the above. I do not question the generation of a/1-r by a series, I question the conclusion of a proof. and interpretation of results. You are reading more into the method than it is intended to supply.

The series is by definition an approximation to 0.999... No different than a series approximation to trig functions,

You accept that 0.333... = 1/3

How can you possibly miss that multiplying by three gives:

0.999... = 3/3 ?

Or is the problem that you don't think 3/3 = 1 ?

Are you really that stupid, or are you just trolling?

 

[untermensche]

An undefined element, something that has no final value cannot rationally have an operation performed to it to produce a final value.

You cannot rationally multiply 0.333... X 3 and get a final answer. It is an infinite operation that never finishes.

You can pretend it can finish however and get a pretend final answer.

0.333... X 3 yields 0.99999.....3...

You can never reach the final 3 to perform an operation to the final 3.

In fact you cannot even approach the final 3. There is no final 3.

What infinity means is you cannot approach the end. No matter how far you travel you are no closer to the end.

No matter how many multiplications are performed you cannot perform the last. You cannot approach the last. You are always infinitely far from it.

0.3333... is never fully transformed to 0.9999... by multiplying it by 3. That would require being able to approach the last 3 to perform an operation on it.

Throwing a completed infinite operation into the mix is just pretending to have an answer.

These operations performed on infinite strings of numbers are all a bunch of pretending. A bunch of emperors new clothes.

And they violate the concept of infinity. You cannot approach the end of an infinite string to perform an operation on it. An operation performed to an infinite string never can finish. Unless you just pretend it finishes somehow.

 

[steve_bank]

I stayed why I reject the conclusion.After this I will just be reating myself.

I have applied limits and series in real world problems. I'll go with my experience and analysis of the conclusion.

0.555.. . is a definition, an unchanging infinite number of 5s.

Converting to a geometric series and applying finite arithmetic is not the definition.

Infinite and finite are mutually exclusive. 0.999... can never equal a finite 1. )0333.. results in 1/3 with no incremting to a higher values, 1.3 = 0.333...

The method applied to 0.999.. yields the fraction 1/1 as an approximation. The fact that 0.999.. gets a rollover to 1 and 0.333... does not round up means that 0.999.. = 1/1 is not a proof that it is a finite 1. As stated in the link geometric series applied to decimals yields a fractional approximation.

0.999... = 1, 1 does not yield 0.999... and 0.333... = 1/3 = 0.333... An unresolved discrepancy. Conclusion the algorithm does not serve as a proof, it serves to provide fractional approximations.

If you disagree you have to explain the discrepancy between the results for 0.333... and 0.999... Is there a difference between the two other than the fact that for 0.999.. the algorithm can only yield 1/1, an artifact of the equation not a proof.

If you want to continue the debate, answer the above. I do not question the generation of a/1-r by a series, I question the conclusion of a proof. and interpretation of results. You are reading more into the method than it is intended to supply.

The series is by definition an approximation to 0.999... No different than a series approximation to trig functions,

You accept that 0.333... = 1/3

How can you possibly miss that multiplying by three gives:

0.999... = 3/3 ?

Or is the problem that you don't think 3/3 = 1 ?

Are you really that stupid, or are you just trolling?

No. I accept that given 0.333... 1/3 is an integer approximation. Given 0.555... 5/9 is an integer approximation. Given 0.999... 9/9 is an integer approximation. The method does not serve as a proof, applied to repeating decimals it provides integer approximations.

0.999.. as an infinite defined infinite entity and a finite quantification are mutually exclusive.

That you are too shallow and inexperienced in applying math to follow my counter arguments is not my problem. Read, consider, and reply to my objections. What I have found is personal attack rises in proportion to lack of competence.

I new when you put me on ignore it would not last, a moth to a flame. If I an stupid put me on ignore and forget about me.

- - - Updated - - -

I stayed why I reject the conclusion.After this I will just be reating myself.

I have applied limits and series in real world problems. I'll go with my experience and analysis of the conclusion.

0.555.. . is a definition, an unchanging infinite number of 5s.

Converting to a geometric series and applying finite arithmetic is not the definition.

No conversion is going on. "an infinite number of 5s" is the definition of the string. Used as a label, the geometric series is what this string refers to, always did, without any conversion needed.

Infinite and finite are mutually exclusive. 0.999... can never equal a finite 1. )0333.. results in 1/3 with no incremting to a higher values, 1.3 = 0.333...

The method applied to 0.999.. yields the fraction 1/1 as an approximation. The fact that 0.999.. gets a rollover to 1 and 0.333... does not round up means that 0.999.. = 1/1 is not a proof that it is a finite 1. As stated in the link geometric series applied to decimals yields a fractional approximation.

0.999... = 1, 1 does not yield 0.999... and 0.333... = 1/3 = 0.333... An unresolved discrepancy. Conclusion the algorithm does not serve as a proof, it serves to provide fractional approximations.

What discrepancy? Both the number that's the sum of the series [3 * 10 ^ -i for i in range(inf)] and the number that's the sum of the series [9 * 10 ^ -i for i in range(inf)] are finite real and rational numbers. The fact that one of them can be shortened to 1.0 or 1 in decimal notation while the other cannot is an idiosyncracy of decimal notation. In duodecimal, 1/3 is 0.4, plain and simple and very much finite.

This is as far as I can take the debate on this problem.

 

[steve_bank]

I stayed why I reject the conclusion.After this I will just be reating myself.

I have applied limits and series in real world problems. I'll go with my experience and analysis of the conclusion.

0.555.. . is a definition, an unchanging infinite number of 5s.

Converting to a geometric series and applying finite arithmetic is not the definition.

Infinite and finite are mutually exclusive. 0.999... can never equal a finite 1. )0333.. results in 1/3 with no incremting to a higher values, 1.3 = 0.333...

The method applied to 0.999.. yields the fraction 1/1 as an approximation. The fact that 0.999.. gets a rollover to 1 and 0.333... does not round up means that 0.999.. = 1/1 is not a proof that it is a finite 1. As stated in the link geometric series applied to decimals yields a fractional approximation.

0.999... = 1, 1 does not yield 0.999... and 0.333... = 1/3 = 0.333... An unresolved discrepancy. Conclusion the algorithm does not serve as a proof, it serves to provide fractional approximations.

If you disagree you have to explain the discrepancy between the results for 0.333... and 0.999... Is there a difference between the two other than the fact that for 0.999.. the algorithm can only yield 1/1, an artifact of the equation not a proof.

If you want to continue the debate, answer the above. I do not question the generation of a/1-r by a series, I question the conclusion of a proof. and interpretation of results. You are reading more into the method than it is intended to supply.

The series is by definition an approximation to 0.999... No different than a series approximation to trig functions,

<definition>
A ”decimal number” is an ordered set of integers An (where n is all negative and positive integers and for each An < 9 and An >= 0.) and a corresponding value = Sum of An*10^n (for all n)
</definition>

Do you agree on this?

I do not agree that 0.999... and the series representaion are the same thing. By definition 0.999... is infinite and can not be reduced to a finite number. The only comclusion is that the method does not infer a finite 1 equality, it infers a fractional 9/9 approximation.

Beyond that we will be going in circles. For me the conflict between an infinite unchanging sequence 0.999.. and and a finite number can not be resolved. Therefore while the result of 09/9 is correct within the bounds of what the method provides, integer approximations to repeating decimals, conclusion is wrong.

 

[steve_bank]

limit as x -> inf 1/(1 + 1/x) equals exactly 1

limit as x -> inf 1/(1 + 1/x) approaches 1 but never gets there

Which statement is correct?

 

[Kharakov]

I do not agree that 0.999... and the series representaion are the same thing. By definition 0.999... is infinite and can not be reduced to a finite number. The only comclusion is that the method does not infer a finite 1 equality, it infers a fractional 9/9 approximation.

.9999.... is the equivalent of never subtracting anything from 1.

What is 1- .999...?
What is 1- .000...0001? nonsense


There is a reason the mathematicians don't touch this thread. Nobody pays attention to them.