General Method for Summing Divergent Series. Determination of Limits of Divergent Sequences and Functions in Singular Points v2

[http://vixra.org/abs/1511.0247-25-11-2015-general-method-for-summing-divergent-series-using-mathematica-and-a-comparison-to-other-summation-methods]

[http://vixra.org/abs/1612.0413-30-12-2016-area-of-torricelli’s-trumpet-or-gabriel’s-horn-sum-of-the-reciprocals-of-the-primes-factorials-of-negative-integers]

Abstract. In this work I am going to mention historical development of divergent series theory, and to give a number of different examples, as some of the methods for their summing. After that, I am going to introduce the general method, which I discovered, for summing divergent series, which we can also consider as a method for computing limits of divergent sequences and functions in divergent points, In this case, limits of sequences of their partials sums. Through the exercises, I am going to apply this method on given examples and prove its validity. Then I’m going to apply the method to compute the value of some divergent integrals.

http://vixra.org/abs/1502.0074 [v2]

— Algorithmization of the method —

Function {f,g}; EquationSolver solveEquation;

\displaystyle n\in N;

\displaystyle \lim^D_{z\rightarrow+\infty} z^n=\int_{-1}^0 z^n dz;

\displaystyle \lim^D_{z\rightarrow-\infty} z^n=\int_0^1 z^n dz;

\displaystyle \lim^D_{z\rightarrow 0^+} z^{-n}=\int_{-\infty}^{-1} z^{-n-2} dz;

\displaystyle \lim^D_{z\rightarrow 0^-} z^{-n}=\int_1^{+\infty} z^{-n-2} dz;

\displaystyle a, const\in C\bigcup\{\infty\},\ \alpha\in[0,2\pi);

if (({f} has a pole at {a})||({f} has a removable singularity at {a})) {

if (({g} has a pole at {a})||({g} has a removable singularity at {a})) {

\displaystyle \lim^D_{z\rightarrow a(\alpha)^\pm}(f(z)+g(z))=\lim^D_{z\rightarrow a(\alpha)^\pm}f(z)+\lim^D_{z\rightarrow a(\alpha)^\pm}g(z);

\displaystyle \lim^D_{z\rightarrow a(\alpha)^\pm}(c\cdot f(z))=c\cdot\lim^D_{z\rightarrow a(\alpha)^\pm}f(z);

}

if ({f} has a removable singularity at {a}) {

\displaystyle \lim^D_{z\rightarrow a} f(z)=\lim^D_{z\rightarrow a(\alpha)^\pm} f(z)=\lim_{z\rightarrow a} f(z);

}

if ({f} has a pole at {a}) {

if ({a==\infty}) {

\displaystyle f(z)=F_1(z)+F_2(z)=\sum_{k=-n}^{-1} c_k\cdot\frac{1}{z^k}+\sum_{k=0}^\infty c_k\cdot\frac{1}{z^k};

\displaystyle \lim^D_{z\rightarrow\infty(\alpha)^+}f(z)=\lim^D_{z\rightarrow+e^{i\alpha}\infty}f(z)=\lim^D_{r\rightarrow+\infty}f(r\cdot e^{i\alpha});

\displaystyle \lim^D_{z\rightarrow\infty(\alpha)^-}f(z)=\lim^D_{z\rightarrow-e^{i\alpha}\infty}f(z)=\lim^D_{r\rightarrow-\infty}f(r\cdot e^{i\alpha});

\displaystyle \lim^D_{z\rightarrow\infty(\alpha)}f(z)=\lim^D_{r\rightarrow\infty(0)}f(r\cdot e^{i\alpha});

\displaystyle \lim^D_{z\rightarrow\infty(\alpha)}f(z)=\frac{1}{2}(\lim^D_{z\rightarrow\infty(\alpha)^+}f(z)+\lim^D_{z\rightarrow\infty(\alpha)^-}f(z));

\displaystyle \lim^D_{z\rightarrow\infty}f(z)=\frac{1}{2\pi}\int_0^{2\pi}\lim^D_{z\rightarrow\infty(\alpha)}f(z)d\alpha;

\displaystyle \lim^D_{x\rightarrow\infty(0)^+}f(x)=\lim^D_{x\rightarrow+\infty}f(x);

\displaystyle \lim^D_{x\rightarrow\infty(0)^-}f(x)=\lim^D_{x\rightarrow-\infty}f(x);

\displaystyle \lim^D_{z\rightarrow \infty(\alpha)\pm}f(z)=\sum_{k=1}^{n} c_{k}\cdot e^{i\alpha k}\frac{(\pm 1)^k}{2(k+1)}+c_0;

\displaystyle \lim^D_{z\rightarrow \infty(\alpha)}f(z)=\sum_{k=1}^n c_k\cdot e^{i\alpha k}\frac{1+(-1)^k}{2(k+1)}+c_0;

}

if ({a\in C}) {

\displaystyle f(z)=F_1(z)+F_2(z)=\sum_{k=-n}^{-1} c_k\cdot(z-a)^k+\sum_{k=0}^\infty c_k\cdot(z-a)^k;

\displaystyle \lim^D_{z\rightarrow a(\alpha)^+}f(z)=\lim^D_{z\rightarrow a+e^{i\alpha}0}f(z)=\lim^D_{r\rightarrow 0^+}f(a+r\cdot e^{i\alpha});

\displaystyle \lim^D_{z\rightarrow a(\alpha)^-}f(z)=\lim^D_{z\rightarrow a-e^{i\alpha}0}f(z)=\lim^D_{r\rightarrow 0^-}f(a+r\cdot e^{i\alpha});

\displaystyle \lim^D_{z\rightarrow a(\alpha)}f(z)=\lim^D_{r\rightarrow 0(0)}f(a+r\cdot e^{i\alpha});

\displaystyle \lim^D_{z\rightarrow a(\alpha)}f(z)=\frac{1}{2}(\lim^D_{z\rightarrow a(\alpha)^+}f(z)+\lim^D_{z\rightarrow a(\alpha)^-}f(z));

\displaystyle \lim^D_{z\rightarrow a}f(z)=\frac{1}{2\pi}\int_0^{2\pi}\lim^D_{z\rightarrow a(\alpha)}f(z)d\alpha;

\displaystyle \lim^D_{x\rightarrow a(0)^+}f(x)=\lim^D_{x\rightarrow a^+}f(x);

\displaystyle \lim^D_{x\rightarrow a(0)^-}f(x)=\lim^D_{x\rightarrow a^-}f(x);

\displaystyle \lim^D_{z\rightarrow a(\alpha)^\pm}f(z)=\sum_{k=-n}^{-1} c_{k}\cdot e^{i\alpha k}\frac{(\pm 1)^k}{2(-k+1)}+c_0;

\displaystyle \lim^D_{z\rightarrow a(\alpha)}f(z)=\sum_{k=-n}^{-1} c_{k}\cdot e^{i\alpha k}\frac{1+(-1)^k}{2(-k+1)}+c_0;

}

\displaystyle \lim^D_{z\rightarrow a}f(z)=c_0;

}

} else if(({\lim^D_{z\rightarrow a(\alpha)\pm}f(z)} is not finite)||({\lim^D_{z\rightarrow a(\alpha)\pm}f(z)} does not exist)) { if(({f} can be expanded into a generalized Puiseux series about point {a})&&

(!({f(z)=\sum_{k=-n}^{+\infty}c_k(z)\cdot \frac{1}{z^k}})!=!({f(z)=\sum_{k=-n}^{+\infty}c_k(z)\cdot (z-a)^k)})) {

{const}=solveEquation.forConstantOfIntegration({\int d(c_0(z))=c_0(z)});

\displaystyle \lim^D_{z\rightarrow a(\alpha)\pm}f(z)=const;

} else {

\displaystyle c_0(z)=f(z);

{const}=solveEquation.forConstantOfIntegration({\int d(c_0(z))=c_0(z)});

\displaystyle \lim^D_{z\rightarrow a(\alpha)\pm}f(z)=const;

}

\displaystyle \lim^D_{z\rightarrow a}f(z)=const;

}

— The sum of some divergent series —

— Determination of values of some divergent integrals —

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2 thoughts on “General Method for Summing Divergent Series. Determination of Limits of Divergent Sequences and Functions in Singular Points v2

  1. Повратни пинг: General Method for Summing Divergent Series Using Mathematica and a Comparison to Other Summation Methods | м4т3м4т1к4

  2. Повратни пинг: Area of Torricelli’s Trumpet or Gabriel’s Horn, Sum of the Reciprocals of the Primes, Factorials of Negative Integers | м4т3м4т1к4

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