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EquisDeXD

Infinitely Accurate Integrals?

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No, there are different levels of infinite.

 

Here's an easy example. There are an infinite number of numbers between 0 and 1. There are also an infinite number of numbers between 0 and 100. Both an infinite number, but there are still more between 0 and 100.

 

Now, going with my example in the previous post, there are more irrational numbers than rational. This is because the rational numbers are countably infinite. That means, you can use logic and put them in an order, with a direct 1 to 1 correspondence with the natural numbers. http://en.wikipedia.org/wiki/Countable_set No such logic exists for irrational numbers, and hence they are considered uncountably infinite. http://en.wikipedia.org/wiki/Uncountable_set This also implies that there are more irrationals than rationals.

 

In terms of its rareness... if you really had a true random number generator that picked any and all numbers equally between 0 and 1, you would pick an irrational number far more often than a rational one.

 

In the same way, the polynomials (per the traditional definition of non-negative integer exponents) are also countable. Yes, there are an infinite number of them, but they are still countable. However, the set of all functions is uncountably infinite. Again, uncountable implies there are more than any countable set, even if both are infinite.

 

Again, if you had a true random function picker, something that generated a random function from all possible functions. The chances of it picking a polynomial function are really quite rate. I think you do not have a very good appreciation of the true definition of a function, nor an understanding that the very tiny number that we (mankind) have named are only a very, very, very small amount of all the function that are out there. Yes, there are an infinite number of Bessel Functions of the First Kind, and an infinite number of polynomials.... but that is such a small slice of all the available functions out there.

 

I will stick by my assertion that polynomials are very rare in the set of all functions. As well as stick with my assertion that the number of integrals that can be evaluated with infinite precision is also rare.

 

To wit, even if an integral is over a polynomial, the limits still have to be rational as well.

 

[math]\int^{\sqrt{3}}_{\sqrt{2}} 1 \, \mathrm{d} x[/math] isn't knowable to infinite accuracy, for example, despite being a very simple polynomial integrand. Because you can't know the two irrational endpoints to infinite accuracy.

 

Lastly, there is no reason to be rude and use derogatory phrases like "I would think that a math expert..." and then follow that up with a snide remark. I invited you to take up any personal issue you have with me via the mods or via the PM system. It also obviously wasn't a lead in to ask me to clarify -- which I obviously was willing to do per the paragraphs before this one.

 

I didn't just make that statement willy-nilly or create it out of thin air. There are formal definitions and concepts which back it up. I am happy to discuss these, but the attitude I am getting from your posts in this thread really make me think you aren't open to discussion. You have an attitude that you already know everything, and anyone who says something different isn't worth your time. If I am misreading this, I do apologize, but I would ask that you look over your posting style and review, because the impression you are giving off seems very aggressive and smug.

Edited by Bignose

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To wit, even if an integral is over a polynomial, the limits still have to be rational as well.

 

932bf780227971e7a6d5d496a8ce7624-1.png isn't knowable to infinite accuracy, for example, despite being a very simple polynomial integrand. Because you can't know the two irrational endpoints to infinite accuracy.

 

 

Which is what I've been telling the OP.

 

The only functions you can guarantee to integrate to 'infinite accuracy' are of the form

 

y = an integer constant betwen limits that are integers.

Edited by studiot

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Even if you try and fit infinite elements into a single set, the probability of any particular element being picked is 0.

 

EquisDeXD,

 

If Bignose corrects me please make my reputation -1.00000000000000000000000000000000000000000000000000003 approximately. When I see a probability of 0 it describes , no chance. If an element is an element with equal chance of being picked just like all elements there is a paradox. An element must be picked. Are you approximating?

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No, there are different levels of infinite.

 

Here's an easy example. There are an infinite number of numbers between 0 and 1. There are also an infinite number of numbers between 0 and 100. Both an infinite number, but there are still more between 0 and 100.

 

Now, going with my example in the previous post, there are more irrational numbers than rational. This is because the rational numbers are countably infinite. That means, you can use logic and put them in an order, with a direct 1 to 1 correspondence with the natural numbers. http://en.wikipedia....i/Countable_set No such logic exists for irrational numbers, and hence they are considered uncountably infinite. http://en.wikipedia....Uncountable_set This also implies that there are more irrationals than rationals.

 

In terms of its rareness... if you really had a true random number generator that picked any and all numbers equally between 0 and 1, you would pick an irrational number far more often than a rational one.

 

In the same way, the polynomials (per the traditional definition of non-negative integer exponents) are also countable. Yes, there are an infinite number of them, but they are still countable. However, the set of all functions is uncountably infinite. Again, uncountable implies there are more than any countable set, even if both are infinite.

 

Again, if you had a true random function picker, something that generated a random function from all possible functions. The chances of it picking a polynomial function are really quite rate. I think you do not have a very good appreciation of the true definition of a function, nor an understanding that the very tiny number that we (mankind) have named are only a very, very, very small amount of all the function that are out there. Yes, there are an infinite number of Bessel Functions of the First Kind, and an infinite number of polynomials.... but that is such a small slice of all the available functions out there.

 

I will stick by my assertion that polynomials are very rare in the set of all functions. As well as stick with my assertion that the number of integrals that can be evaluated with infinite precision is also rare.

 

To wit, even if an integral is over a polynomial, the limits still have to be rational as well.

 

[math]\int^{\sqrt{3}}_{\sqrt{2}} 1 \, \mathrm{d} x[/math] isn't knowable to infinite accuracy, for example, despite being a very simple polynomial integrand. Because you can't know the two irrational endpoints to infinite accuracy.

 

Lastly, there is no reason to be rude and use derogatory phrases like "I would think that a math expert..." and then follow that up with a snide remark. I invited you to take up any personal issue you have with me via the mods or via the PM system. It also obviously wasn't a lead in to ask me to clarify -- which I obviously was willing to do per the paragraphs before this one.

 

I didn't just make that statement willy-nilly or create it out of thin air. There are formal definitions and concepts which back it up. I am happy to discuss these, but the attitude I am getting from your posts in this thread really make me think you aren't open to discussion. You have an attitude that you already know everything, and anyone who says something different isn't worth your time. If I am misreading this, I do apologize, but I would ask that you look over your posting style and review, because the impression you are giving off seems very aggressive and smug.

 

But see, despite all this pointless bickering, you STILL are somehow incapable of answering the original topic question. I honestly don't give an s word if it comes up only rarely if that's somehow the case.

Your right that logically there are different types of infinity, such as different cardinal numbers, but infinite is still infinite, it's still not something you can count to, there are infinity functions to get infinitely accurate integrals and infinity functions to get non accurate integrals, infinity=infinity, so I don't see how you can say it's "more rare".

Also, sqrt(2) is an exact number, because your not approximating, so if I have an end result where the area under the curve with limits extending to irrational numbers can be noted as the square root of whatever number, that's fine by me, because the square root of two is itself exactly the square root of two even if we can't perfectly count it.

 

EquisDeXD,

 

If Bignose corrects me please make my reputation -1.00000000000000000000000000000000000000000000000000003 approximately. When I see a probability of 0 it describes , no chance. If an element is an element with equal chance of being picked just like all elements there is a paradox. An element must be picked. Are you approximating?

 

http://en.wikipedia....i/Almost_surely

Edited by EquisDeXD

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EquisDeXD,

 

I must be old school, my enlightener was even older school and 0 is zero. No chance is no chance. If an element must be picked and all elements have the same chance of being picked then the sum of the chances can't equal 0. The wiki can't rewrite my probability books.

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Also, sqrt(2) is an exact number, because your not approximating,

 

But, you cannot be infinite accurate with it. Because you don't know it to infinite decimal points.

 

This is your original question, that despite how many times you don't care for my answer, I have been trying to answer as best I can.

 

so I don't see how you can say it's "more rare".

 

Because the set of non-polynomial functions is uncountably infinite, whereas the set of polynomial functions are countably infinite.

 

Just like there are more numbers between 0 and 100 than there are between 0 and 1. Wouldn't you say that it is fair, that if you selected numbers at random across all of the reals, that it is more rare to select one between 0 and 1 than it is to select one between 0 and 100? Yet, there are infinite numbers between each of them.

 

This is how you can have something be infinite, and yet be rare.

 

The analogy is exactly the same for polynomials out of the set of all functions.

 

And exactly the same for integrals that are evaluatable to infinite precision out of the set of all polynomials. Yes, there are an infinite number of integrals that can be evaluated exactly. But, they are like the reals between 0 and 1 when picking from the entire set of all real numbers. They are rare.

 

I don't know how else to explain it; so please ask more questions. Please tell me where I lose you. I do want to help you understand, if you are willing to learn.

 

 

EquisDeXD,

 

If Bignose corrects me please make my reputation -1.00000000000000000000000000000000000000000000000000003 approximately. When I see a probability of 0 it describes , no chance. If an element is an element with equal chance of being picked just like all elements there is a paradox. An element must be picked. Are you approximating?

 

ACUV,

 

on a continuous distribution, the probability of picking any single individual number is 0. Because over the range of a distribution, there are infinite choices. This again is the uncountably infinite-ness of the real numbers. What you can talk about, however, it he probability of pulling a number between two numbers.

 

For example, consider the uniform distribution over 0 to 1. The probability of selecting exactly 0.72358723875365329853232875354876 is 0. But, the probability of selection a number between 0.7 and 0.8 is 10%.

 

It seems counterintuitive, but this is what happens over sets with an infinite number of elements. The way you think about probability with sets with finite counts of elements aren't always straightforward to extend to sets with infinite counts.

Edited by Bignose

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But, you cannot be infinite accurate with it. Because you don't know it to infinite decimal points.

 

This is your original question, that despite how many times you don't care for my answer, I have been trying to answer as best I can.

 

 

 

Because the set of non-polynomial functions is uncountably infinite, whereas the set of polynomial functions are countably infinite.

 

Just like there are more numbers between 0 and 100 than there are between 0 and 1. Wouldn't you say that it is fair, that if you selected numbers at random across all of the reals, that it is more rare to select one between 0 and 1 than it is to select one between 0 and 100? Yet, there are infinite numbers between each of them.

 

This is how you can have something be infinite, and yet be rare.

 

The analogy is exactly the same for polynomials out of the set of all functions.

 

And exactly the same for integrals that are evaluatable to infinite precision out of the set of all polynomials. Yes, there are an infinite number of integrals that can be evaluated exactly. But, they are like the reals between 0 and 1 when picking from the entire set of all real numbers. They are rare.

 

I don't know how else to explain it; so please ask more questions. Please tell me where I lose you. I do want to help you understand, if you are willing to learn.

 

 

Can you just stop posting? Because you STILL HAVEN"T ANSWERED IT!!.

Furthermore, no, I don't know what you don't get about "never ending" or "endless". You can have different cardinal infinities, but no matter what, endless=endless, there are endless numbers between 1 and 0 and endless numbers between 1-100 no matter how you want to put it, it doesn't matter if I can count "1,2,3..." because I will never actually count to the number infinity, because its not an actual value you can attain. I can go ".01, .001, .0001..." which still goes on never endingly between 1 and 0 OR 1 and 100.

EquisDeXD,

 

I must be old school, my enlightener was even older school and 0 is zero. No chance is no chance. If an element must be picked and all elements have the same chance of being picked then the sum of the chances can't equal 0. The wiki can't rewrite my probability books.

 

But the rules are different for an infinite set because infinity isn't a real number, you sort of have to bend the original rules to fit it in. It's no different than how grad students are taught the way to figure out the relative speed of two objects simply by adding the speeds or subtracting the speeds from an object at rest, but in reality there's plenty of circumstances where that same concept doesn't work.

Edited by EquisDeXD

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EquisDeXD

 

You don't get to choose who posts in the threads you have opened. And the fact that you have dismissed the answers of two highly qualified mathematicians in your threads in the maths forum makes me believe you should attempt to understand the answers you have been given rather than repudiate them

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EquisDeXD

 

You don't get to choose who posts in the threads you have opened. And the fact that you have dismissed the answers of two highly qualified mathematicians in your threads in the maths forum makes me believe you should attempt to understand the answers you have been given rather than repudiate them

 

How can I dismiss answers when there aren't any? How does saying "infinitely accurate integrals occur rarely" in any way shape or form answer the question "how do I find an infinite accurate integral?"? I doubt you even read this thread given you said that.

Edited by EquisDeXD

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ACUV,

 

on a continuous distribution, the probability of picking any single individual number is 0.

 

 

I'm supposing now that as the distribution is continuous, the choices are endless. If a number is to be picked, is the chance of picking a number equal to 0, because it is impossible to consider the choices?

 

This is a bit like saying the chances of stepping in dog dirt while I go for a walk are 0 if I don't go for a walk. It doesn't take into account the infinite number of places that dog dirt could be because I amn't actually going for a walk.

 

To me it seems that a value of "incalculable" or "undefined" chance of being chosen would be better suited.

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I'm supposing now that as the distribution is continuous, the choices are endless. If a number is to be picked, is the chance of picking a number equal to 0, because it is impossible to consider the choices?

 

This is a bit like saying the chances of stepping in dog dirt while I go for a walk are 0 if I don't go for a walk. It doesn't take into account the infinite number of places that dog dirt could be because I amn't actually going for a walk.

 

To me it seems that a value of "incalculable" or "undefined" chance of being chosen would be better suited.

 

No, it's definitely 0. The probability of any particular element being picked in an infinite set is 0, but if you pick something there has to be something that you pick up. Like if I have a probability of hitting a particular point on a scoreboard with a dart, a point is a one dimensional object and occupies no area, so the chances of hitting a specific point are 0, but there's an infinite number of 1 dimensional points that can fit into a 2D area, and when I throw a dart it has to hit one of them regardless, this is where the term "almost surely" comes from.

Edited by EquisDeXD

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My logical argument would be that if the chances of picking any particular item in an infinite amount of choices is 0, it is because one can't consider all choices, therefore one can't randomly select an item, the chance is 0 because one can't make a choice. It isn't because all choices have been considered.

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Because you STILL HAVEN"T ANSWERED IT!!.

 

I have answered it, since I have tried to explain to you many times, what you are asking cannot occur. It only occurs in special cases. So, the answer to your general question (paraphrasing here) "can you show me a process by which you can evaluate any integral to any precision?" is "no." Whether you accept it, or not.

 

This is why I tried to show you many times, that what you are asking for happens rarely. Because there is NO general answer to your question. Perhaps I should have been much more explicit. I really thought my post #15 was rather clear, however.

 

Now, if you want to modify your question, and talk about those you can evaluate more accurately, or limit yourself to talking about certain kinds of functions and integrals, we can. But, just because you don't like the answer, doesn't mean you can shout at me in all-caps.

 

My logical argument would be that if the chances of picking any particular item in an infinite amount of choices is 0, it is because one can't consider all choices, therefore one can't randomly select an item, the chance is 0 because one can't make a choice. It isn't because all choices have been considered.

 

I did put a little more detail into my answer in a different thread in this section. http://www.scienceforums.net/topic/69457-probability-in-an-infinite-set/page__view__findpost__p__707277

 

In short, when dealing with a set with infinite choices, it doesn't make sense to talk about choosing any specific element. You have to talk about selecting elements in a subset of the set (or a range).

Edited by Bignose

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what you are asking cannot occur. It occurs in special cases.

You just contradicted yourself, and as I said not only do I NOT CARE if it occurs rarely, but I've seen someone do it right infront of me, concluding at the integral of some arbitrary function I asked about was exactly 25/3

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The 'cannot' refers to the implied 'in every case' in your question. Your question was stated very generally, and the answer to the general case is 'no'. But there are special cases. And this isn't a thread about grammar, is it? I concede I may not be the best writer, but no one else who is posting seems to have much of a problem understanding what I am trying to get across.

 

Now, this next followup question about 25/3 is also unclear. Without more specifics, how can anyone actually answer this? How can we be expected to read your mind and know what your arbitrary function that was integrated to 25/3 is?

 

[math] \int^{\sqrt[3]{5}}_{0} x^2 \, \mathrm{d} x = \frac{25}{3} [/math] ?

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The 'cannot' refers to the implied 'in every case' in your question. Your question was stated very generally, and the answer to the general case is 'no'. But there are special cases. And this isn't a thread about grammar, is it? I concede I may not be the best writer, but no one else who is posting seems to have much of a problem understanding what I am trying to get across.

But that's not even a grammar problem, that's just you not being clear with what your saying.

 

Now, this next followup question about 25/3 is also unclear. Without more specifics, how can anyone actually answer this? How can we be expected to read your mind and know what your arbitrary function that was integrated to 25/3 is?

Where did I ask you to find a function that had an integral of 25/3? I hope you didn't waste a lot of time doing that because I certainly didn't ask that, all I did was point out that I've seen a perfectly accurate number as the value of an integral.

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that's just you not being clear with what your saying.

 

And this is not taken to be snide or rude at all, but I would similarly lobby that the question you have asked has also been not clear.

 

Where did I ask you to find a function that had an integral of 25/3? I hope you didn't waste a lot of time doing that because I certainly didn't ask that, all I did was point out that I've seen a perfectly accurate number as the value of an integral.

 

You didn't. I was just trying to factitiously show how unclear that last question is. And it does fulfill your requirement of a perfectly accurately evaluated integral. But, back to your question, with no specifics at all, I am not sure how anyone can help much more. You can either evaluate an integral using various integration techniques (again, integration by parts, trigonometric integration, etc.) or numerical techniques (most of which will approximate the value to a precision that is dependent upon the specific technique applied). Beyond that, I really am unsure what specific question is being asked.

 

And no, it didn't take much time at all to write up that integral. This forum's LaTeX capabilities are very easy to use.

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And this is not taken to be snide or rude at all, but I would similarly lobby that the question you have asked has also been not clear.

infinite accuracy = exact value.

 

 

 

You didn't. I was just trying to factitiously show how unclear that last question is. And it does fulfill your requirement of a perfectly accurately evaluated integral. But, back to your question, with no specifics at all, I am not sure how anyone can help much more. You can either evaluate an integral using various integration techniques (again, integration by parts, trigonometric integration, etc.) or numerical techniques (most of which will approximate the value to a precision that is dependent upon the specific technique applied). Beyond that, I really am unsure what specific question is being asked.

 

And no, it didn't take much time at all to write up that integral. This forum's LaTeX capabilities are very easy to use.

I'm fairly sure that if you can use a summation formula to get an infinitely accurate integral, even though not every function may have a direct summation formula, there's plenty of mathematical expressions that do. It matches my memory and it matches what ajb said, I don't get why you think an infinitely accurate integral is impossible all of a sudden, you already said it occurs but rarely, so if it does occur, why not just be done with that? What are you even arguing for? I already said multiple times that I don't care if it occurs rarely.

Edited by EquisDeXD

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I'm fairly sure that if you can use a summation formula to get an infinitely accurate integral,

 

This is exactly what all the numerical integration techniques do. But, except in some (rare) special cases, the sum is going to going to be from i = 1 to infinity. As I wrote above, this would take an infinite amount of time to compute, and hence you will never actually know its "exact value". The most straightforward numerical technique, Euler's method, is just the literal computation of a Reimann sum. Improvements on that method allow the computation to consist of fewer terms and/or greater accuracy to the approximation. And, yes I'm going to post it again, it is really only in very, very rare circumstances does the numerical approximation equal the actual exact evaluation of the integral. This is how they test numerical techniques, compare the numerical approximations with known solutions.

 

And, it is also rare that a summation would be amenable to analysis to be convergent to an exact or easily calculable value. Or to a summation that wouldn't require a large number of terms. Again, I say rare simply because it is almost wholly a function of the form the integrand takes. One can do the Reimann sum, and get this summation you are looking for, for nicely behaved integrand functions, like polynomials. But again, polynomials are exceptionally rare in the entire range of all functions. In other words, in the vast majority of cases, forming the Reimann sum will give you a summation formula that won't be easily analyzed. If the Reimann sum even exists, hence ajb's comment about Lesbegue measures.

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snapback.pngBignose, on 13 October 2012 - 04:48 AM, said:

 

And this is not taken to be snide or rude at all, but I would similarly lobby that the question you have asked has also been not clear.

Equisdxd

 

infinite accuracy = exact value.

 

 

There was more than this to your original question which I answered infinitely accurately (if concisely).

 

So I know the general approximation formula using the limit of the function with n numbers of DeltaX times the height that changes over n intervals of delta x, but what if I want to be infinitely accurate or get the exact amount of area and not just some approximation? Would I use an infinitely small deltaX? How?

 

As to your point about the size of the infinitesimals (delta x) for whole numebrs they it does not matter whether they are small or large for the proof. The sums in the integral are still exact.

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This is exactly what all the numerical integration techniques do. But, except in some (rare) special cases, the sum is going to going to be from i = 1 to infinity. As I wrote above, this would take an infinite amount of time to compute, and hence you will never actually know its "exact value". The most straightforward numerical technique, Euler's method, is just the literal computation of a Reimann sum. Improvements on that method allow the computation to consist of fewer terms and/or greater accuracy to the approximation. And, yes I'm going to post it again, it is really only in very, very rare circumstances does the numerical approximation equal the actual exact evaluation of the integral. This is how they test numerical techniques, compare the numerical approximations with known solutions.

 

And, it is also rare that a summation would be amenable to analysis to be convergent to an exact or easily calculable value. Or to a summation that wouldn't require a large number of terms. Again, I say rare simply because it is almost wholly a function of the form the integrand takes. One can do the Reimann sum, and get this summation you are looking for, for nicely behaved integrand functions, like polynomials. But again, polynomials are exceptionally rare in the entire range of all functions. In other words, in the vast majority of cases, forming the Reimann sum will give you a summation formula that won't be easily analyzed. If the Reimann sum even exists, hence ajb's comment about Lesbegue measures.

Ok, that's fine, there's many instances where you can't find an exact value as the integral and I don't have a problem with that, but there's still a theoretically infinite number of instances where you can. For some equations there's no summation formula to yield an exact value, but for some I know there are.

Edited by EquisDeXD

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Ok, that's fine, there's many instances where you can't find an exact value as the integral and I don't have a problem with that, but there's still a theoretically infinite number of instances where you can. For some equations there's no summation formula to yield an exact value, but for some I know there are.

 

I think we're finally on the same page, since this is what I've been trying to say form the very first post.

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I think we're finally on the same page, since this is what I've been trying to say form the very first post.

But I've been saying this whole time I DON"T CARE IF ITS RARE.

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But I've been saying this whole time I DON"T CARE IF ITS RARE.

 

So, you don't care that the question you posed doesn't have an answer in a very large majority of the cases? How very odd. I thought that that was rather relevant and important, really. Sometimes, the toughest thing with a problem is to show that it does or doesn't have an answer. For example, if you can prove that the Navier-Stokes equations have unique solutions, the Clay Mathematics Institute will give you $1 million. http://theconversation.edu.au/millennium-prize-the-navier-stokes-existence-and-uniqueness-problem-4244

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