#analyticcontinuation — Public Fediverse posts

@zvavybir
It does diverge. It has no sum.
However, the uniquely valued #Riemann #ZetaFunction can be analytically continued into the left half-plane where we find zeta(-1)=-1/12 (which 'looks like' 1+2+...). #Cesàro #summation will get you part of the way there also, and, as you say, yields the same result; presumably due to some ultimate cosmic logical rightness :-)
I very strongly recommend BP's superb exposition of this issue
https://www.youtube.com/watch?v=YuIIjLr6vUA
#maths #AnalyticContinuation #Ramanujan

@zvavybir
It does diverge. It has no sum.
However, the uniquely valued #Riemann #ZetaFunction can be analytically continued into the left half-plane where we find zeta(-1)=-1/12 (which 'looks like' 1+2+...). #Cesàro #summation will get you part of the way there also, and, as you say, yields the same result; presumably due to some ultimate cosmic logical rightness :-)
I very strongly recommend BP's superb exposition of this issue
https://www.youtube.com/watch?v=YuIIjLr6vUA
#maths #AnalyticContinuation #Ramanujan

@zvavybir
It does diverge. It has no sum.
However, the uniquely valued #Riemann #ZetaFunction can be analytically continued into the left half-plane where we find zeta(-1)=-1/12 (which 'looks like' 1+2+...). #Cesàro #summation will get you part of the way there also, and, as you say, yields the same result; presumably due to some ultimate cosmic logical rightness :-)
I very strongly recommend BP's superb exposition of this issue
https://www.youtube.com/watch?v=YuIIjLr6vUA
#maths #AnalyticContinuation #Ramanujan

@zvavybir
It does diverge. It has no sum.
However, the uniquely valued #Riemann #ZetaFunction can be analytically continued into the left half-plane where we find zeta(-1)=-1/12 (which 'looks like' 1+2+...). #Cesàro #summation will get you part of the way there also, and, as you say, yields the same result; presumably due to some ultimate cosmic logical rightness :-)
I very strongly recommend BP's superb exposition of this issue
https://www.youtube.com/watch?v=YuIIjLr6vUA
#maths #AnalyticContinuation #Ramanujan

For all \( a > 0 \) we have \[ \int_{-\infty}^\infty e^{-a x^2} = \sqrt{\frac{\pi}{a}} \]. Let's ignore the condition on \( a \) and say \( a \) is \( -1 \), then we have \[ \int_{-\infty}^\infty e^{x^2} = \sqrt{\pi} i \]. The left side obviously diverges and is never even slightly imaginary, but the right side is finite and purely imaginary. What is the connection between this integral and it's "value" (I'm pretty sure there is one, like how \( 1 + 2 + 4 + \dots = -1 \) makes sense over the 2-adic numbers)? Under what weird interpretation of integration is this correct? How can we find more of these weird integrals and their value?

#math #maths #mathematics #extendingMath #integration #analyticContinuation?

For all \( a > 0 \) we have \[ \int_{-\infty}^\infty e^{-a x^2} = \sqrt{\frac{\pi}{a}} \]. Let's ignore the condition on \( a \) and say \( a \) is \( -1 \), then we have \[ \int_{-\infty}^\infty e^{x^2} = \sqrt{\pi} i \]. The left side obviously diverges and is never even slightly imaginary, but the right side is finite and purely imaginary. What is the connection between this integral and it's "value" (I'm pretty sure there is one, like how \( 1 + 2 + 4 + \dots = -1 \) makes sense over the 2-adic numbers)? Under what weird interpretation of integration is this correct? How can we find more of these weird integrals and their value?

#math #maths #mathematics #extendingMath #integration #analyticContinuation?

For all \( a > 0 \) we have \[ \int_{-\infty}^\infty e^{-a x^2} = \sqrt{\frac{\pi}{a}} \]. Let's ignore the condition on \( a \) and say \( a \) is \( -1 \), then we have \[ \int_{-\infty}^\infty e^{x^2} = \sqrt{\pi} i \]. The left side obviously diverges and is never even slightly imaginary, but the right side is finite and purely imaginary. What is the connection between this integral and it's "value" (I'm pretty sure there is one, like how \( 1 + 2 + 4 + \dots = -1 \) makes sense over the 2-adic numbers)? Under what weird interpretation of integration is this correct? How can we find more of these weird integrals and their value?

#math #maths #mathematics #extendingMath #integration #analyticContinuation?

For all \( a > 0 \) we have \[ \int_{-\infty}^\infty e^{-a x^2} = \sqrt{\frac{\pi}{a}} \]. Let's ignore the condition on \( a \) and say \( a \) is \( -1 \), then we have \[ \int_{-\infty}^\infty e^{x^2} = \sqrt{\pi} i \]. The left side obviously diverges and is never even slightly imaginary, but the right side is finite and purely imaginary. What is the connection between this integral and it's "value" (I'm pretty sure there is one, like how \( 1 + 2 + 4 + \dots = -1 \) makes sense over the 2-adic numbers)? Under what weird interpretation of integration is this correct? How can we find more of these weird integrals and their value?

#math #maths #mathematics #extendingMath #integration #analyticContinuation?

A representation of \(18\) using an analytic continuation of the Dirichlet series and the numbers \(0, 1,2,3,4,5,6\).
\[18=\left|\dfrac{1}{\zeta^2(0)}\left(\dfrac{\mathcal{H}(-6)}{\zeta(-5)}-\dfrac{\mathcal{H}(-2)}{\zeta(-1)}\right)\dfrac{\mathcal{H}(-4)}{\zeta(-3)}\right|\]
where \(\displaystyle\zeta(z)=\sum_{n\geq1}\dfrac{1}{n^z}\) denotes the Riemann zeta function, and \(\displaystyle\mathcal{H}(z)=\sum_{n\geq1}\dfrac{H_n}{n^z}\) denotes the harmonic zeta function.

#ZetaFunction #RiemannZetaFunction #HarmonicZetaFunction #AnalyticContinuation #DirichletSeries #Series #Numbers #Zeta #Harmonic #HarmonicNumbers #Representation #Function #Expression

A representation of \(18\) using an analytic continuation of the Dirichlet series and the numbers \(0, 1,2,3,4,5,6\).
\[18=\left|\dfrac{1}{\zeta^2(0)}\left(\dfrac{\mathcal{H}(-6)}{\zeta(-5)}-\dfrac{\mathcal{H}(-2)}{\zeta(-1)}\right)\dfrac{\mathcal{H}(-4)}{\zeta(-3)}\right|\]
where \(\displaystyle\zeta(z)=\sum_{n\geq1}\dfrac{1}{n^z}\) denotes the Riemann zeta function, and \(\displaystyle\mathcal{H}(z)=\sum_{n\geq1}\dfrac{H_n}{n^z}\) denotes the harmonic zeta function.

#ZetaFunction #RiemannZetaFunction #HarmonicZetaFunction #AnalyticContinuation #DirichletSeries #Series #Numbers #Zeta #Harmonic #HarmonicNumbers #Representation #Function #Expression

@Archimage

ζ(-1)=-1/12

#maths #AnalyticContinuation

@Archimage

ζ(-1)=-1/12

#maths #AnalyticContinuation

@Archimage

ζ(-1)=-1/12

#maths #AnalyticContinuation

@Archimage

ζ(-1)=-1/12

#maths #AnalyticContinuation

Riemann zeta function \(\zeta(s)\) and \(\displaystyle\sum_{n=1}^\infty n=1+2+3+\cdots=-\dfrac{1}{12}\)

Have you ever heard that the sum of all natural numbers is \(-1/12\)?🤔 Of course not; this doesn't make sense in the usual sum, but using a summation method based on analytic continuation of the Riemann zeta function leads to the following result.

The Riemann zeta function is defined as:
\[\zeta(s)=\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^s}=\dfrac{1}{1^s}+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\cdots\]
for \(s\in\mathbb{C}\) such that \(\Re(s)>1\).
It can be extended to a meromorphic function with only a simple pole at \(s=1\), using analytic continuation and the following functional equation:
\[\zeta(1-s)=2^{1-s}\pi^{-s}\cos\left(\dfrac{\pi s}{2}\right)\Gamma(s)\zeta(s)\]
For \(s=2\), this gives \(\zeta(-1)=\displaystyle\sum_{n=1}^\infty n=-\dfrac{1}{2\pi^2}\zeta(2)=-\dfrac{1}{2\pi^2}\cdot\dfrac{\pi^2}{6}=-\dfrac{1}{12}\), which is a reason for assigning a finite value to the divergent sum/series (zeta function regularization). That is, \(\displaystyle\sum_{n=1}^\infty n=1+2+3+\cdots=-\dfrac{1}{12}\).
#RiemannZetaFunction #ZetaFunction #Riemann #DivergentSum #DivergentSeries #FiniteValue #ZetaRegularization #ZetaFunctionRegularization #NegativeFraction #MeromorphicFunction #AnalyticContinuation

“A mathematician, like a painter or poet, is a maker of patterns. If his patterns are more permanent than theirs; it is because they are made with ideas.”
― G.H. Hardy, A Mathematician's Apology

Let's define something...
Difference transform
\[\boxed{\boxed{\mathcal{D}\{f\}(a,b):=\int_a^b\left(f(\lfloor x\rfloor)-f(x)\right)\ \mathrm{d}x}}\]

Here an operator \(\mathcal{D}\) is acting on a function \(f\), describing the difference between an indiscrete sum and a discrete sum on a certain interval \(a\) to \(b\). This operator may play a pivotal role in asymptotic expansions, analytic continuation, etc.

#DifferenceTransform #DifferenceTransformOperator #TranformOperator #DifferenceTransformation #AsymptoticExpansion #AnalyticContinuation #Mathematics #GHHardy #Mathematicians #Definition #Prologue #Maths #Poet #Painter #Patterns #Beauty #BeautyOfPatterns