Integrals

[sanctus]

Don't really see your point (read all your links), with \Delta x tending to zero so does the error and integral being defined as the limit of \Delta x tending to zero...

I mean by definition a Riemann Summation is an approximation so no surprise that there is an error and also no surprise that there are better approximations..

[exchemist]

Well, yes the Reimann Summation that the Define Integral is equal to is wrong slightly. http://mathworld.wolfram.com/RiemannIntegral.html

 

Reimann Summation has Error just as the other rules it is just much smaller do you see how the error is .0001 that means at the milisegment of change it will be inaccurate being at the size of a milli the integral distance. so lets say from a to b then that would error out at .0001 of (a + b / 2) size of the object being integrated but taken to an infinity will greatly reduce the error but still it is there. http://www.math.montana.edu/courses/m171/documents/ExampleRiemann.pdf

 

There are several error rules for testing error the most accurate is Simpson's rule if it says there is error then there is error in your integral.

 

http://tutorial.math.lamar.edu/Classes/CalcII/ApproximatingDefIntegrals.aspx

 

Read all the links.

These links are saying there is an error in the Riemann summations, not in the integral.

 

The integral represents the limit, in which there is (in effect) a sum of an infinite number of infinitely small elements and, in that limit, the error reduces to zero.  That is what your links are saying, not that integrals themselves are always approximations. 

[sluggo]

The limit is the value if you performed the function (adding increments) without ever stopping. Since that is not possible in the real world, a truncated value/approximation is used. By definition there is always a small difference between the summation and the limit.

If the function ever equalled the limit, the limit statement would be false. The definition implies you can make the difference as small as needed by an extended application, but never equal to zero.

Those who can’t differentiate between an equality and a limit, believe the limit is the answer to the ‘infinite’ sum. The integral uses the same assumption.

If a limit was the same as an equality, why would there need to be another definition?

 

[exchemist]

 

The limit is the value if you performed the function (adding increments) without ever stopping. Since that is not possible in the real world, a truncated value/approximation is used. By definition there is always a small difference between the summation and the limit.

If the function ever equalled the limit, the limit statement would be false. The definition implies you can make the difference as small as needed by an extended application, but never equal to zero.

Those who can’t differentiate between an equality and a limit, believe the limit is the answer to the ‘infinite’ sum. The integral uses the same assumption.

If a limit was the same as an equality, why would there need to be another definition?

 

 

According to my understanding that is because the limit is a particular method of extrapolation, by which you reach a defined value for something that would, in the absence of such a limit extrapolation, be undefined, due to the presence of infinities and zeroes.

 

When we use calculus we do not in fact refer to limit notation, do we? We write d/dx (x² + c) = 2x. There is, in the d/dx operator, an implied process of extrapolation by infinitesimals tending to a limit, but we do not recite this, nor do we say the result is approximate. It is exact. Same with integration. 

 

Or am I misunderstanding you? 

[Super Polymath]

According to my understanding that is because the limit is a particular method of extrapolation, by which you reach a defined value for something that would, in the absence of such a limit extrapolation, be undefined, due to the presence of infinities and zeroes.

 

When we use calculus we do not in fact refer to limit notation, do we? We write d/dx (x² + c) = 2x. There is, in the d/dx operator, an implied process of extrapolation by infinitesimals tending to a limit, but we do not recite this, nor do we say the result is approximate. It is exact. Same with integration.

 

Or am I misunderstanding you?

 

That's basic calculus.

 

In real world applications of calculus you'd probably arrive at close enough, like some rounded fractional value that's for sure merely approximate.

 

That's because you're using iterated integral & differentiatial fractional calculus for non-smooth structures in nature, with computer aid of course.

Edited October 31, 2017 by Super Polymath

[sanctus]

Deriving velocity at a given point in time does not seem an approximation to me

[Super Polymath]

Deriving velocity at a given point in time does not seem an approximation to me

 

What if it's the average speed achieved when your velocity was dynamic during a trip? Wind pressure might have slowed you down infinitesimally, which lowers your average velocity for a given trip. That's a poor example but in the real world there's hardly anything but approximations. 

 

As for fractional calc, I'd just study up on what it was first used for in phone lines:

 

http://dangerousminds.net/comments/mandelbrot_fractals_hunting_the_hidden_dimension

 

P.S. Your velocity can be off by nanoseconds, or more. Measurables can always get smaller &, since space & time are interrelated, they can always get faster. Same works for integrals, you can always get larger, just because the CMB might have out-shined older structures doesn't mean there wasn't older structures. 

Edited November 1, 2017 by Super Polymath

[sluggo]

My argument is more specific to infinite sequences and not integration in general, so ignore my post.

[Vmedvil]

Deriving velocity at a given point in time does not seem an approximation to me

 

It's like Pi the Integral you can write Pi as 3.14 but that is not is actual value its actual value can go for trillions of digits, The Integrals value is the same way, How long of a distance is infinity, the closer to infinity the closer the integral just like a summation is to correct but there is no number that is infinity so it has to have a definite value at some point which is where you lose some accuracy when you round of digits to infinity just like if you were rounding off decimal values of pi. The Riemann Integral is the type of integral that has been found to be most accurate the Summation has Error thus if you don't take his Integrate from 0 to infinity on every Summation or Integral it will have error but the error manifests itself in the same way as not taking pi to enough digits.

Edited November 2, 2017 by Vmedvil

[exchemist]

It's like Pi the Integral you can write Pi as 3.14 but that is not is actual value its actual value can go for trillions of digits, The Integrals value is the same way, How long of a distance is infinity, the closer to infinity the closer the integral just like a summation is to correct but there is no number that is infinity so it has to have a definite value at some point which is where you lose some accuracy when you round of digits to infinity just like if you were rounding off decimal values of pi. The Riemann Integral is the type of integral that has been found to be most accurate the Summation has Error thus if you don't take his Integrate from 0 to infinity on every Summation or Integral it will have error but the error manifests itself in the same way as not taking pi to enough digits.

This analogy is wrong. The integral is the limit, not a series approaching it. 

 

Are you really saying that the integral of velocity with respect to time is only the approximate distance travelled? Don't be absurd. 

[Vmedvil]

This analogy is wrong. The integral is the limit, not a series approaching it. 

 

Are you really saying that the integral of velocity with respect to time is only the approximate distance travelled? Don't be absurd. 

 

Well, that is a bad example knowing that was shown to be wrong and replaced by SR's Velocity equation but Yes I tend to think that the Integral of velocity has some errors in classical physics. Being only a first integral it has fewer errors than the 5th integral being energy because the errors stack but still error. Doesn't that classical velocity equation say you can have a velocity greater than C being greater than the limit of the universe, I would say that is a error, but of a different type. There is error approximating that Summation is the error I have been speaking of though since velocity can only be taken to C and not Infinity I would say there is error in that equation to some degree since it is integrated with a limit below Infinity thus the bounds of the Summation being approximated is less than infinity too.

Edited November 3, 2017 by Vmedvil

[exchemist]

Well, that is a bad example knowing that was shown to be wrong and replaced by SR's Velocity equation but Yes I tend to think that the Integral of velocity has some errors in classical physics. Being only a first integral it has fewer errors than the 5th integral being energy because the errors stack but still error. Doesn't that classical velocity equation say you can have a velocity greater than C being greater than the limit of the universe, I would say that is a error, but of a different type.

What? Velocity is defined as the change of distance with respect to time, isn't it?

 

So far as I am aware, that definition holds irrespective of whether you talk about Newtonian kinematics, relativity or quantum theory. (In QM you may sometimes not be able to specify it, but that's another thing.) 

 

Can you direct me to an equation that tells me the integral of velocity with respect to time is NOT distance? 

Edited November 3, 2017 by exchemist

[Vmedvil]

What? Velocity is defined as the change of distance with respect to time, isn't it?

 

So far as I am aware, that definition holds irrespective of whether you talk about Newtonian kinematics, relativity or quantum theory. (In QM you may sometimes not be able to specify it, but that's another thing.) 

 

Can you direct me to an equation that tells me the integral of velocity with respect to time is NOT distance? 

 

No velocity is by nature with respect to time as a Displacement over Time you would have to go to a different property to change that in physics all things have their units which velocity will always be the change in Displacement over Time as defined but it does not say how that displacement should be measured just (X₁ - X₂ / t₁ - t₂) = Velocity. The issue is not that it is taking respect for time but rather is it taking the proper respect for time in any situation to which the equation could be applied. 

Edited November 3, 2017 by Vmedvil

[exchemist]

No velocity is by nature with respect to time as a Displacement over Time you would have to go to a different property to change that in physics all things have their units which velocity will always be the change in Displacement over Time as defined but it does not say how that displacement should be measured just (X₁ - X₂ / t₁ - t₂) = Velocity. The issue is not that it is taking respect for time but rather is it taking the proper respect for time in any situation to which the equation could be applied. 

So how was my example wrong, then? Show the correct formula I should have used. 

[Vmedvil]

So how was my example wrong, then? Show the correct formula I should have used. 

 

I don't have a better way to define it than Integrals which is why I use them but my point is integration is not perfect.

[exchemist]

I don't have a better way to define it than Integrals which is why I use them but my point is integration is not perfect.

I must say I think this is absurd. Do you also think that d/dx (x² + c) =2x is only an approximation? 

 

Because if you accept that the derivative is exact, then the fact that the integral of 2x gets you back to x² + c, i.e. exactly to the function you started with, must mean that the integration is just the inverse of the differentiation, in which case if one is exact, the other MUST be also.  

[Vmedvil]

I must say I think this is absurd. Do you also think that d/dx (x² + c) =2x is only an approximation? 

 

Because if you accept that the derivative is exact, then the fact that the integral of 2x gets you back to x² + c, i.e. exactly to the function you started with, must mean that the integration is just the inverse of the differentiation, in which case if one is exact, the other MUST be also.  

 

All of the functions generated by calculus via differentiation or Integration do have a small error, I do say that neither are Perfectly correct at every point, have you ever heard of holes in derivatives of functions.

 

Edited November 5, 2017 by Vmedvil