Simulation of smoothness using discrete quantities

[Kharakov]

Some infinity required.

What are the primary purposes of infinite series in mathematics?

 

[steve_bank]

Some infinity required.

What are the primary purposes of infinite series in mathematics?

You will have to elaborate on the OP title, no clue what you mean.

There are therecal math uses for series.

In the practcal world series are used everywhere. Calculating trig functions and other functions are done using series expansions. The number of terms yield the number of digits. When you hit the sin key on your calculator it uses a series expansion. Maclaurin and Taylor series. In practice the Fourier Transform is calculated by a series.

You can go through the limks.

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

Trig functions

http://www.efunda.com/math/taylor_series/trig.cfm

 

[Kharakov]

I watched Jeopardy yesterday, if that explains the question after the answer.

Basically, I'm talking about how integers (discrete quantities) can be used to represent a continuum if you have an infinite series, but you technically need an infinite amount of them to represent smoothness... or other values in some cases.

\(\sum_{x=2}^\infty\,\,\,\sum_{n=2}^\infty\frac{1}{x^n}\,\,\,= ?\)

 

[steve_bank]

That gets into number systems. What is a digit in base 10 arithmetic? It is an integer. A real number is made of integer digits with positional value.
1.0, 10.0,100.0....

Multiplication, division, and subtraction are algorithms operating on digits.

In computer digital calculations and the lsb or least significant bit sets the numerical precision. The precision of hand calculations depends on howy digits you want to use for an application.

So in practice there is always a granularity or a roughness to representations. Infinite smoothness is not possible,

If you divide the number 1 by 8 digital bits the granularity is 1/2^8. If I go to 16 bits it is 1/2^16 and so on. All numbers are inherently quantized.

 

[steve_bank]

I watched Jeopardy yesterday, if that explains the question after the answer.

Basically, I'm talking about how integers (discrete quantities) can be used to represent a continuum if you have an infinite series, but you technically need an infinite amount of them to represent smoothness... or other values in some cases.

\(\sum_{x=2}^\infty\,\,\,\sum_{n=2}^\infty\frac{1}{x^n}\,\,\,= ?\)

You would probably have to express it as a Taylor or Maclaurin series, not something I can do off the top of my head. There are then techniques to determine if a series converges on on finite number or goes to infinity..

By inspection as x goes to ihinity for any n, 1/x^n goes asymptotically to xero. Approaches zero but never gets there.

 

[Kharakov]

Think  Riemann zeta function, and  geometric series constrained to integers >=2.

 

[steve_bank]

Think  Riemann zeta function, and  geometric series constrained to integers >=2.

I realized it after I logged off. x^0 = 1 x^1 = x.

I don't see how it could be zero or finite. It sums an infinite number of of increasinging small numbers.

 

[Kharakov]

Think  Riemann zeta function, and  geometric series constrained to integers >=2.

I realized it after I logged off. x^0 = 1 x^1 = x.

I don't see how it could be zero or finite. It sums an infinite number of of increasinging small numbers.

\(\sum_{n=1}^\infty \frac{1}{x^n} = \frac{1}{x} + \frac {1}{x^2} +\frac {1}{x^3} + \dots = \frac{1}{x-1}\)

so

\(\sum_{n=a+1}^\infty \,\,\,\,\, \frac{1}{x^n} = \frac{1}{x^{a+1}} + \frac {1}{x^{a+2}} +\frac {1}{x^{a+3}} + \dots = \frac{1}{ x^a (x-1)}\)


Riemann zeta (analytic continuation of the following series):

\(\zeta(s) =\sum_{x=1}^\infty\frac{1}{x^s}\,\,\,\,\) so \(\,\,\,\,\zeta(1) =\sum_{x=1}^\infty\frac{1}{x}\) \(\,\,\,\,\zeta(2) =\sum_{x=1}^\infty\frac{1}{x^2}\,\,\,\,\) etc.

The sum of all positive integer zetas = sum of all positive integer >=2 geometric series:

\( \zeta(1) \, = \,\sum_{x=2}^\infty \,\, \sum_{n=1}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=2}^\infty \, \frac{1}{x-1} \)

which is 1+ 1/2 +1/3 +1/4.....

Say you subtract the first term from every geometric series in the above equation... you get: 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4.... which is the same thing as shifting the starting point of n in the above equation from 1 to 2.

\(1 \, = \,\sum_{x=2}^\infty \,\, \sum_{n=2}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=2}^\infty \, \frac{1}{x (x-1)} \)

So that's the answer.





You can also see that:

\(\sum_{s=1}^\infty \, \zeta(s) \, = \,\sum_{x=1}^\infty \,\, \sum_{n=1}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=1}^\infty \, \frac{1}{x-1} \)

which gives us other ways of looking at the zeta function. remembering that 1/(1-1) = 1+1+1+1.... ("/" is reverse Cauchy series manipulation)

Show

1/(1-1)^2 = 1+2+3+4...; 1/(1-1)^3 = 1+3+6+10...; etc.

.

 

[steve_bank]

Think  Riemann zeta function, and  geometric series constrained to integers >=2.

I realized it after I logged off. x^0 = 1 x^1 = x.

I don't see how it could be zero or finite. It sums an infinite number of of increasinging small numbers.

\(\sum_{n=1}^\infty \frac{1}{x^n} = \frac{1}{x} + \frac {1}{x^2} +\frac {1}{x^3} + \dots = \frac{1}{x-1}\)

so

\(\sum_{n=a+1}^\infty \,\,\,\,\, \frac{1}{x^n} = \frac{1}{x^{a+1}} + \frac {1}{x^{a+2}} +\frac {1}{x^{a+3}} + \dots = \frac{1}{ x^a (x-1)}\)


Riemann zeta (analytic continuation of the following series):

\(\zeta(s) =\sum_{x=1}^\infty\frac{1}{x^s}\,\,\,\,\) so \(\,\,\,\,\zeta(1) =\sum_{x=1}^\infty\frac{1}{x}\) \(\,\,\,\,\zeta(2) =\sum_{x=1}^\infty\frac{1}{x^2}\,\,\,\,\) etc.

The sum of all positive integer zetas = sum of all positive integer >=2 geometric series:

\( \zeta(1) \, = \,\sum_{x=2}^\infty \,\, \sum_{n=1}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=2}^\infty \, \frac{1}{x-1} \)

which is 1+ 1/2 +1/3 +1/4.....

Say you subtract the first term from every geometric series in the above equation... you get: 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4.... which is the same thing as shifting the starting point of n in the above equation from 1 to 2.

\(1 \, = \,\sum_{x=2}^\infty \,\, \sum_{n=2}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=2}^\infty \, \frac{1}{x (x-1)} \)

So that's the answer.





You can also see that:

\(\sum_{s=1}^\infty \, \zeta(s) \, = \,\sum_{x=1}^\infty \,\, \sum_{n=1}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=1}^\infty \, \frac{1}{x-1} \)

which gives us other ways of looking at the zeta function. remembering that 1/(1-1) = 1+1+1+1.... ("/" is reverse Cauchy series manipulation)

Show

1/(1-1)^2 = 1+2+3+4...; 1/(1-1)^3 = 1+3+6+10...; etc.

.

Your math is more advanced then mine, so humor me.

Is the solution a finite 1 are is 1 a limit?

 

[fromderinside]

You might use http://tomlr.free.fr/Math%E9matique...ormulae%2031st%20Edition%20-%20Zwillinger.pdf

You are welcome

 

[Kharakov]

\(\sum_{n=1}^\infty \frac{1}{x^n} = \frac{1}{x} + \frac {1}{x^2} +\frac {1}{x^3} + \dots = \frac{1}{x-1}\)

so

\(\sum_{n=a+1}^\infty \,\,\,\,\, \frac{1}{x^n} = \frac{1}{x^{a+1}} + \frac {1}{x^{a+2}} +\frac {1}{x^{a+3}} + \dots = \frac{1}{ x^a (x-1)}\)


Riemann zeta (analytic continuation of the following series):

\(\zeta(s) =\sum_{x=1}^\infty\frac{1}{x^s}\,\,\,\,\) so \(\,\,\,\,\zeta(1) =\sum_{x=1}^\infty\frac{1}{x}\) \(\,\,\,\,\zeta(2) =\sum_{x=1}^\infty\frac{1}{x^2}\,\,\,\,\) etc.

The sum of all positive integer zetas = sum of all positive integer >=2 geometric series:

\( \zeta(1) \, = \,\sum_{x=2}^\infty \,\, \sum_{n=1}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=2}^\infty \, \frac{1}{x-1} \)

which is 1+ 1/2 +1/3 +1/4.....

Say you subtract the first term from every geometric series in the above equation... you get: 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4.... which is the same thing as shifting the starting point of n in the above equation from 1 to 2.

\(1 \, = \,\sum_{x=2}^\infty \,\, \sum_{n=2}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=2}^\infty \, \frac{1}{x (x-1)} \)

So that's the answer.





You can also see that:

\(\sum_{s=1}^\infty \, \zeta(s) \, = \,\sum_{x=1}^\infty \,\, \sum_{n=1}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=1}^\infty \, \frac{1}{x-1} \)

which gives us other ways of looking at the zeta function. remembering that 1/(1-1) = 1+1+1+1.... ("/" is reverse Cauchy series manipulation)

Show

1/(1-1)^2 = 1+2+3+4...; 1/(1-1)^3 = 1+3+6+10...; etc.

.

Your math is more advanced then mine, so humor me.

Is the solution a finite 1 or is 1 a limit?

1+1/2+1/3+1/4+1/5... - [1/2 +1/3 +1/4 + 1/5...] = 1

 

[steve_bank]

\(\sum_{n=1}^\infty \frac{1}{x^n} = \frac{1}{x} + \frac {1}{x^2} +\frac {1}{x^3} + \dots = \frac{1}{x-1}\)

so

\(\sum_{n=a+1}^\infty \,\,\,\,\, \frac{1}{x^n} = \frac{1}{x^{a+1}} + \frac {1}{x^{a+2}} +\frac {1}{x^{a+3}} + \dots = \frac{1}{ x^a (x-1)}\)


Riemann zeta (analytic continuation of the following series):

\(\zeta(s) =\sum_{x=1}^\infty\frac{1}{x^s}\,\,\,\,\) so \(\,\,\,\,\zeta(1) =\sum_{x=1}^\infty\frac{1}{x}\) \(\,\,\,\,\zeta(2) =\sum_{x=1}^\infty\frac{1}{x^2}\,\,\,\,\) etc.

The sum of all positive integer zetas = sum of all positive integer >=2 geometric series:

\( \zeta(1) \, = \,\sum_{x=2}^\infty \,\, \sum_{n=1}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=2}^\infty \, \frac{1}{x-1} \)

which is 1+ 1/2 +1/3 +1/4.....

Say you subtract the first term from every geometric series in the above equation... you get: 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4.... which is the same thing as shifting the starting point of n in the above equation from 1 to 2.

\(1 \, = \,\sum_{x=2}^\infty \,\, \sum_{n=2}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=2}^\infty \, \frac{1}{x (x-1)} \)

So that's the answer.





You can also see that:

\(\sum_{s=1}^\infty \, \zeta(s) \, = \,\sum_{x=1}^\infty \,\, \sum_{n=1}^\infty \frac{1}{x^n} \, = \, \, \sum_{x=1}^\infty \, \frac{1}{x-1} \)

which gives us other ways of looking at the zeta function. remembering that 1/(1-1) = 1+1+1+1.... ("/" is reverse Cauchy series manipulation)

Show

1/(1-1)^2 = 1+2+3+4...; 1/(1-1)^3 = 1+3+6+10...; etc.

.

Your math is more advanced then mine, so humor me.

Is the solution a finite 1 or is 1 a limit?

1+1/2+1/3+1/4+1/5... - [1/2 +1/3 +1/4 + 1/5...] = 1

Then I would have the same issue as om the infinity threads. The idea that an infinite summation of finite numbers can lead to a finite result. A staement, not opening debate, but I would ssay the series approches 1 as a limit.

 

[bilby]

1+1/2+1/3+1/4+1/5... - [1/2 +1/3 +1/4 + 1/5...] = 1

Then I would have the same issue as om the infinity threads. The idea that an infinite summation of finite numbers can lead to a finite result. A staement, not opening debate, but I would ssay the series approches 1 as a limit.

You can say that; But you would simply be wrong.

This is not a matter of opinion; It's mathematics, and so is subject to proof. If you cannot find an error in the proof, but still cannot grasp the proof, then mathematics clearly isn't for you.

Your opinion - and indeed, anyone else's - is completely irrelevant.

 

[Kharakov]

1+1/2+1/3+1/4+1/5... - [1/2 +1/3 +1/4 + 1/5...] = 1

Then I would have the same issue as om the infinity threads. The idea that an infinite summation of finite numbers can lead to a finite result. A staement, not opening debate, but I would ssay the series approches 1 as a limit.

The Riemann series theorem doesn't really apply here, as I'm assuming a 1 to 1 mapping of 1/2 -1/2 + 1/3-1/3 + 1/4-1/4+... which is essentially an infinite amount of 0s added up ~ 1 + 0 + 0+ 0+ 0+ 0+0...=1

If you want to rearrange the terms, you're playing fast and loose with the rules, to quote someone with a bit more education than me.

 

[ryan]

1+1/2+1/3+1/4+1/5... - [1/2 +1/3 +1/4 + 1/5...] = 1

Then I would have the same issue as om the infinity threads. The idea that an infinite summation of finite numbers can lead to a finite result. A staement, not opening debate, but I would ssay the series approches 1 as a limit.

Where I left off in my personal dealings with infinity is that there isn't really an infinite number of "halves" that the arrow has to travel to reach its end point. It goes through an arbitrarily large natural number of halves (whatever that means, and I believe this is the formal way of putting it).

So it seems as if it's something that isn't quite infinity but is larger than any fixed number.

That is how I sort of made peace with this - sort of.

 

[Keith&Co.]

in my personal dealings with infinity is that there isn't really an infinite number of "halves" that the arrow has to travel to reach its end point.

Then what stops it?

If you keep dividing the remaining distance to the target by halves, what forces you to stop doing that?

 

[untermensche]

Time

 

[Kharakov]

You have to think, it's such a small quantity, it has to be someone tiny, like Tim.

 

[ryan]

in my personal dealings with infinity is that there isn't really an infinite number of "halves" that the arrow has to travel to reach its end point.

Then what stops it?

If you keep dividing the remaining distance to the target by halves, what forces you to stop doing that?

It stops somewhere between an arbitrarily large number and infinity, where ever that is.

 

[Kharakov]

1+1/2+1/3+1/4+1/5... - [1/2 +1/3 +1/4 + 1/5...] = 1

Then I would have the same issue as om the infinity threads. The idea that an infinite summation of finite numbers can lead to a finite result. A staement, not opening debate, but I would ssay the series approches 1 as a limit.

Where I left off in my personal dealings with infinity is that there isn't really an infinite number of "halves" that the arrow has to travel to reach its end point. It goes through an arbitrarily large natural number of halves (whatever that means, and I believe this is the formal way of putting it).

So it seems as if it's something that isn't quite infinity but is larger than any fixed number.

Wut? It's pretty clear, in this case, that 1/2-1/2 + 1/3-1/3 + 1/4-1/4 ... = 0+0+0.... = 0

What are you talkin bout?