beero1000
Veteran Member
MSE is a good site, but is still mostly at a student level. If you want something to really blow your hair back, http://mathoverflow.net is the equivalent site for research-level mathematicians.
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MSE is a good site, but is still mostly at a student level. If you want something to really blow your hair back, http://mathoverflow.net is the equivalent site for research-level mathematicians.
George S
Veteran Member
By the way, for the programophobics, the "programs" in my prior post are identical in structure. Understand one, you got'm all.
Kharakov
Quantum Hot Dog
They're fun! But like manipulation of the alternating harmonic series can give any real..... there are probably manipulations of 1+2+3+4... that can give any fraction of a real.The divergent series are the invention of the devil, and it is a shame to base on them any demonstration whatsoever.
I just don't think the -1/12th for the series is justified. The analytic continuation of the Riemann zeta function is not the same thing as the series itself.
beero1000
Veteran Member
They're fun! But like manipulation of the alternating harmonic series can give any real..... there are probably manipulations of 1+2+3+4... that can give any fraction of a real.The divergent series are the invention of the devil, and it is a shame to base on them any demonstration whatsoever.
I just don't think the -1/12th for the series is justified. The analytic continuation of the Riemann zeta function is not the same thing as the series itself.
It is justified. It's just that the justification uses an extended definition of convergence. No one is arguing that the limit of the partial sums in the standard sense is -1/12, just like no one argued that there is a real number whose square is -1. Instead, there is a new kind of smoothed convergence and a new kind of complex number. These extended tools agree with all of the old calculations, and give new results for calculations that weren't possible before.
Kharakov
Quantum Hot Dog
They aren't the same. An analytic continuation of a function is not the same as the function itself. The gamma function is not just an analytic continuation of the factorial function- it is another function that has the factorial as a subset of itself.
That the Riemann zeta function has a higher order function that includes it as a subset is not a surprise. However, it is a subset of the higher order function, the higher order function is not just a continuation of it. The fact that people arrived at the analytic "continuation" of the function second... well, that means nothing. It's not a continuation- the "analytic continuation" is a higher order function that uses a specific value of another function to seed itself.
Fun with Grandi's series (it is all of the following):
(1+1+1+1...) - (1+1+1+1....)
1-2+3-4+5-6...
(1+2+3+4...) - (1+2+3+4...).
(.1 +.1 +.1 +.1...) - (.1+.1+.1+.1...)
To get any number you want, just change the order of terms around, since it's all
indeterminate form to me.beero1000
Veteran Member
They aren't the same. An analytic continuation of a function is not the same as the function itself. The gamma function is not just an analytic continuation of the factorial function- it is another function that has the factorial as a subset of itself.
That the Riemann zeta function has a higher order function that includes it as a subset is not a surprise. However, it is a subset of the higher order function, the higher order function is not just a continuation of it. The fact that people arrived at the analytic "continuation" of the function second... well, that means nothing. It's not a continuation- the "analytic continuation" is a higher order function that uses a specific value of another function to seed itself.
Fun with Grandi's series (it is all of the following):
(1+1+1+1...) - (1+1+1+1....)
1-2+3-4+5-6...
(1+2+3+4...) - (1+2+3+4...).
(.1 +.1 +.1 +.1...) - (.1+.1+.1+.1...)
To get any number you want, just change the order of terms around, since it's all
Huh? I never said they were the same; I said that one extended the other. 0 is not a counting number, 1/2 is not an integer, pi is not a rational, i is not a real - that doesn't stop us from completing the different number systems in order to solve more complicated problems. Anyone can choose to remain in a more rudimentary system, but they lose out on the benefits of extending to more advanced situations.
If math gets more mind blowing than MSE I don't know if I'd even notice the difference.
Math is one of those things that I'm curious to know more about, but it is so far removed from applicability to my life (maybe outside of statistics and data science) that it's about 60th on my priority list. Maybe I'll check it out in my fifties when I'm done reading about everything else.
Kharakov
Quantum Hot Dog
Yeah. The analytic solution is cool, but it is not the solution to the sum 1+2+3+4... That is different.
1-3+5-7+9.. +
1-1+1-1+1..
=
2-4+6-8+10...
=
1-2+3-4+5...+
1-2+3-4+5...
=
1-2+3-4+5...+
__1-2+3-4...
=
1-1+1-1+1...
Pretty neat to add a series to another and get the series back...
beero1000
Veteran Member
Except that it is... if you accept the continuation as extending the defined values of the function. No one is saying that the function and its continuation are the same, but they agree everywhere the original is defined, and the continuation is uniquely well-defined. For pretty much everyone who needs it, that is enough to justify calling the value of the continuation the "right" value for the original function (if one is not satisfied with merely saying divergent/undefined and must define a value).
Kharakov
Quantum Hot Dog
1+2+3+4+5...= 1+10+33+222+4+11+18+123+2... It doesn't = -1/12th. The "analytic continuation" of the zeta function does, and it's not the equivalent of the zeta function for real <=1.
It's not even an actual continuation of the function, it's just another function that maps to the same values for certain values of the zeta function.
I'm sure you can think of other functions that map to another function for certain intervals, and don't for others.
It's not even an actual continuation of the function, it's just another function that maps to the same values for certain values of the zeta function.
I'm sure you can think of other functions that map to another function for certain intervals, and don't for others.
Kharakov
Quantum Hot Dog
Take the following series:
\( \sum_{k=1}^\infty {\frac{({\frac{x-y}{x}})^k}{k}}\)
We can use the "analytic continuation" of this series, which happens to be log(x)-log
or log(x/y), which maps to the series for real x and y values for which the series converges.Say you take -(log(1) - log (2)). You get the alternating harmonic series.
So can someone claim the following divergent series is in fact log (3)?
\(\frac{2^1}{1} -\frac{2^2}{2}+\frac{2^3}{3}-\frac{2^4}{4}+\frac{2^5}{5} \cdots \)
In fact, can we then say that every natural number is equal to a specific divergent series:
\(n=\frac{(e^n-1)}{1} -\frac{(e^n-1)^2}{2}+\frac{(e^n-1)^3}{3}-\frac{(e^n-1)^4}{4}+\frac{(e^n-1)^5}{5} \cdots \)
Analytic continuation might be a bit silly, in the long run.
beero1000
Veteran Member
I don't know how to explain it any better. No one is claiming equality using the standard limit of partial sums definition of convergence, so I don't know who you're arguing against.
Terry Tao can definitely explain it better than me (without continuation too!): https://terrytao.wordpress.com/2010...tion-and-real-variable-analytic-continuation/
Terry Tao can definitely explain it better than me (without continuation too!): https://terrytao.wordpress.com/2010...tion-and-real-variable-analytic-continuation/
Kharakov
Quantum Hot Dog
It's not you. Someone claimed that the Dirichlet series was equal to the zeta function for real values of s<=1.
In other words they said:
\(\zeta(s) = \sum_{n=1}^{\infty} \frac {A_n}{n^s}\)
Which is not true for s<=1. Equivocating the values of the zeta function with actual values of the series seems unjustified.
beero1000
Veteran Member
Who claimed that?
Kharakov
Quantum Hot Dog
It was a meme about 1.5 years ago, and I just stumbled across it. It annoyed me, which is why I started posting about it here in the math thread.
Kharakov
Quantum Hot Dog
lpetrich
Contributor
Taking k terms, where the first term is for k = 0,
\( f(n,k) = \sum_{m=0}^{k} \frac{(1-n)(2-n) \cdots (k-n)}{k!} = \frac{(2-n)(3-n) \cdots (k+1-n)}{k!} \)
However,
\( f(n,\infty) = (1 - 1)^n = 0 \)
for n > 1.
I'm stumped.
\( f(n,k) = \sum_{m=0}^{k} \frac{(1-n)(2-n) \cdots (k-n)}{k!} = \frac{(2-n)(3-n) \cdots (k+1-n)}{k!} \)
However,
\( f(n,\infty) = (1 - 1)^n = 0 \)
for n > 1.
I'm stumped.
Kharakov
Quantum Hot Dog
The question may be misleading. I wanted to know if there is a solution for non-integer n>1 (in fact, what about tests for 0<n<1?). I don't know that there is an actual solution.
Integral test???
Integral test???
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beero1000
Veteran Member
The binomial series shows that \((1+x)^a = 1 + ax + \frac{a(a-1)}{2!}x^2 + \dots\).
Using your notation, \(nf(n) = n - \frac{n(n-1)}{2!} + \frac{n(n-1)(n-2)}{3!} - \dots = 1 - (1 - n + \frac{n(n-1)}{2!} - \dots) = 1-(1 - 1)^{n}=1\). IOW, \(f(n) - \frac{1}{n} = 0\) for all \(n> 0\).
Since the terms after the \(\lceil n\rceil\)-th all have the same sign and the sum converges to 0, the partial sums don't cross 0 after the \(\lceil n \rceil\)-th term.
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Kharakov
Quantum Hot Dog
So \(f(n)=\frac{1}{n}\) for positive n. In the following x>0, if y>0 n>=0, if y=0 then n>0.
\(g(n,x,y) = \frac{x-y}{x} \qquad + \qquad \frac{1-n}{2!} \, \times \, \Big ( \frac{x-y}{x}\Big)^2 \qquad + \qquad \frac{(1-n)(2-n)}{3!} \, \times \, \Big ( \frac{x-y}{x}\Big)^3 \qquad + \qquad \frac {(1-n)(2-n)(3-n)}{4!} \, \times \, \Big ( \frac{x-y}{x}\Big)^4 \qquad + \qquad \cdots\)
\(g(n,x,y) = \frac{1}{n} \, \times \, \Bigg (1 -\Big(\frac{x}{y}\Big)^{-n} \Bigg)\)
When n=0 && y>0, g(0,x,y) = log(x/y).
When n>0 && y>0, g(n,x,y) = 1/n *[1- (x/y)^(-n) ]
When n>0 && y=0, g(n,x,0) = 1/n.
So [1 - (x/y)^(-0)]/0 = log (x/y). This series analytically extends division by zero...
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