# analytic quantity idea – Does this formulation coincide to a succession illustration of the Dirac delta duty \$delta(x)\$? Answer

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analytic quantity idea – Does this formulation coincide to a succession illustration of the Dirac delta duty \$delta(x)\$?

Consider the next formulation which defines a piece-wise duty which I consider corresponds to a succession illustration for the Dirac delta duty $$delta(x)$$. The parameter $$f$$ is the analysis frequency and assumed to breathe a constructive integer, and the analysis restrict $$N$$ should breathe chosen such that $$M(N)=0$$ the place $$M(x)=sumlimits_{nle x}mu(n)$$ is the Mertens duty.

(1) $$quaddelta(x)=underset{N,ftoinfty}{textual content{lim}} 2left.sumlimits_{n=1}^Nfrac{mu(n)}{n}sumlimits_{okay=1}^{f n} left(left{ commence{array}{cc} commence{array}{cc} cos left(frac{2 okay pi (x+1)}{n}privilege) & xgeq 0 cos left(frac{2 okay pi (x-1)}{n}privilege) & x<0 aim{array}$$

aim{array}
privilege.privilege.privilege),quad M(N)=0\$

The following design illustrates formulation (1) above evaluated at $$N=39$$ and $$f=4$$. The crimson discrete dots in design (1) under illustrate the analysis of formulation (1) at integer values of $$x$$. I consider formulation (1) at all times evaluates to precisely $$2 f$$ at $$x=0$$ and precisely to zero at different integer values of $$x$$.

Figure (1): Illustration of formulation (1) for $$delta(x)$$

Now deem formulation (2) under derived from the integral $$f(0)=int_{-infty}^{infty}delta(x) f(x), dx$$ the place $$f(x)=e^ xright$$ and formulation (1) above for $$delta(x)$$ was used to rate the integral. Formula (2) under can too breathe evaluated as illustrated in formulation (3) under.

(2) $$quad e^=1=underset{N,ftoinfty}{textual content{lim}} 4sumlimits_{n=1}^Nmu(n)sumlimits_{okay=1}^{f n}frac{n cosleft(frac{2 pi okay}{n}privilege)-2 pi okay sinleft(frac{2 pi okay}{n}privilege)}{4 pi^2 okay^2+n^2},,quad M(N)=0$$

(3) $$quad e^=1=underset{Ntoinfty}{textual content{lim}} mu(1)left(cothleft(frac{1}{2}privilege)-2right)+4sumlimits_{n=2}^Nfrac{mu(n)}{4 e left(e^n-1right) n}$$ $$left(-2 e^{n+1}+e^n n+e^2 n-e left(e^n-1right) left(e^{-frac{2 i pi }{n}}privilege)^{frac{i n}{2 pi }} B_{e^{-frac{2 i pi }{n}}}left(1-frac{i n}{2 pi },-1right)+e left(e^n-1right) left(e^{-frac{2 i pi }{n}}privilege)^{-frac{i n}{2 pi }} B_{e^{-frac{2 i pi }{n}}}left(frac{i n}{2 pi }+1,-1right)+left(e^n-1right) left(B_{e^{frac{2 i pi }{n}}}left(1-frac{i n}{2 pi },-1right)-e^2 B_{e^{frac{2 i pi }{n}}}left(frac{i n}{2 pi }+1,-1right)privilege)+2 eright),quad M(N)=0$$

The following desk illustrates formulation (3) above evaluated for a number of values of $$N$$ akin to zeros of the Mertens duty $$M(x)$$. Note formulation (3) above appears to converge to $$e^=1$$ because the magnitude of the analysis restrict $$N$$ will increase.

$$commence{array}{ccc} n & textual content{N=n^{th} zero of M(x)} & textual content{Evaluation of formulation (3) for e^} 10 & 150 & 0.973479, + i textual content{5.498812269991985solemn{ }*{}^{wedge}-17} 20 & 236 & 0.982236, – i textual content{5.786047752866836solemn{ }*{}^{wedge}-17} 30 & 358 & 0.988729, – i textual content{6.577233629689039solemn{ }*{}^{wedge}-17} 40 & 407 & 0.989363, + i textual content{2.6889189402888207solemn{ }*{}^{wedge}-17} 50 & 427 & 0.989387, + i textual content{4.472005325912989solemn{ }*{}^{wedge}-17} 60 & 785 & 0.995546, + i textual content{6.227857765313369solemn{ }*{}^{wedge}-18} 70 & 825 & 0.995466, – i textual content{1.6606923419056456solemn{ }*{}^{wedge}-17} 80 & 893 & 0.995653, – i textual content{1.1882293286557667solemn{ }*{}^{wedge}-17} 90 & 916 & 0.995653, – i textual content{3.521050901644269solemn{ }*{}^{wedge}-17} 100 & 1220 & 0.997431, – i textual content{1.2549006768893629solemn{ }*{}^{wedge}-16} aim{array}$$

Finally deem the next three formulation derived from the Fourier convolution $$f(y)=intlimits_{-infty}^inftydelta(x) f(y-x) dx$$ the place all three convolutions have been evaluated utilizing formulation (1) above for $$delta(x)$$.

(4) $$quad e^=underset{N,ftoinfty}{textual content{lim}} 4sumlimits_{n=1}^Nmu(n)sumlimits_{okay=1}^{f n}frac{1}{4 pi^2 okay^2+n^2} left(left{ commence{array}{cc} commence{array}{cc} n cosleft(frac{2 okay pi (y+1)}{n}privilege)-2 okay pi e^{-y} sinleft(frac{2 okay pi}{n}privilege) & ygeq 0 n cosleft(frac{2 okay pi (y-1)}{n}privilege)-2 okay pi e^y sinleft(frac{2 okay pi}{n}privilege) & y<0 aim{array}$$

aim{array}privilege.privilege), M(N)=0\$

(5) $$quad e^{-y^2}=underset{N,ftoinfty}{textual content{lim}} sqrt{pi}sumlimits_{n=1}^Nfrac{mu(n)}{n}$$ $$sumlimits_{okay=1}^{f n}e^{-frac{pi okay (pi okay+2 i n y)}{n^2}} left(left(1+e^{frac{4 i pi okay y}{n}}privilege) cosleft(frac{2 pi okay}{n}privilege)-sinleft(frac{2 pi okay}{n}privilege) left(textual content{erfi}left(frac{pi okay}{n}+i yright)+e^{frac{4 i pi okay y}{n}} textual content{erfi}left(frac{pi okay}{n}-i yright)privilege)privilege), M(N)=0$$

(6) $$quadsin(y) e^{-y^2}=underset{N,ftoinfty}{textual content{lim}} frac{1}{2} left(i sqrt{pi }privilege)sumlimits _{n=1}^{textual content{nMax}} frac{mu(n)}{n}sumlimits_{okay=1}^{f n} e^{-frac{(2 pi okay+n)^2+8 i pi okay n y}{4 n^2}} left(-left(e^{frac{2 pi okay}{n}}-1right) left(-1+e^{frac{4 i pi okay y}{n}}privilege) cosleft(frac{2 pi okay}{n}privilege)+privilege.$$ $$left.sinleft(frac{2 pi okay}{n}privilege) left(textual content{erfi}left(frac{pi okay}{n}+i y+frac{1}{2}privilege)-e^{frac{4 i pi okay y}{n}} left(e^{frac{2 pi okay}{n}} textual content{erfi}left(-frac{pi okay}{n}+i y+frac{1}{2}privilege)+textual content{erfi}left(frac{pi okay}{n}-i y+frac{1}{2}privilege)privilege)+e^{frac{2 pi okay}{n}} textual content{erfi}left(-frac{pi okay}{n}-i y+frac{1}{2}privilege)privilege)privilege),qquad M(N)=0$$

Formulas (4), (5), and (6) outlined above are illustrated within the following three figures the place the blue curves are the reference features, the orange curves delineate formulation (4), (5), and (6) above evaluated at $$f=4$$ and $$N=39$$, and the inexperienced curves delineate formulation (4), (5), and (6) above evaluated at $$f=4$$ and $$N=101$$. The three figures under illustrate formulation (4), (5), and (6) above appear to converge to the corresponding reference duty for $$xinmathbb{R}$$ because the analysis restrict $$N$$ is elevated. Note formulation (6) above for $$error(y) e^{-y^2}$$ illustrated in Figure (4) under appears to converge mighty sooner than formulation (4) and (5) above maybe as a result of formulation (6) represents an queer duty whereas formulation (4) and (5) each delineate plane features.

Figure (2): Illustration of formulation (4) for $$e^$$ evaluated at $$N=39$$ (orange round) and $$N=101$$ (inexperienced round) overlaid on the reference duty in blue

Figure (3): Illustration of formulation (5) for $$e^{-y^2}$$ evaluated at $$N=39$$ (orange round) and $$N=101$$ (inexperienced round) overlaid on the reference duty in blue

Figure (4): Illustration of formulation (6) for $$error(y) e^{-y^2}$$ evaluated at $$N=39$$ (orange round) and $$N=101$$ (inexperienced round) overlaid on the reference duty in blue

Question (1): Is it undoubted formulation (1) above is an instance of a succession illustration of the Dirac delta duty $$delta(x)$$?

Question (2): What is the category or house of features $$f(x)$$ for which the integral $$f(0)=intlimits_{-infty}^inftydelta(x) f(x) dx$$ and Fourier convolution $$f(y)=intlimits_{-infty}^inftydelta(x) f(y-x) dx$$ are each genuine when utilizing formulation (1) above for $$delta(x)$$ to rate the integral and Fourier convolution?

Question (3): Is formulation (1) above for $$delta(x)$$ an instance of what’s known as a tempered distribution, or is formulation (1) for $$delta(x)$$ extra common than a tempered distribution?

Formula (1) for $$delta(x)$$ above is predicated on the nested Fourier succession illustration of $$delta(x+1)+delta(x-1)$$ outlined in formulation (7) under. Whereas the Fourier convolution $$f(y)=intlimits_{-infty}^inftydelta(x) f(y-x) dx$$ evaluated utilizing formulation (1) above appears to converge for $$yinmathbb{R}$$, Mellin convolutions akin to $$f(y)=intlimits_0^inftydelta(x-1) fleft(frac{y}{x}privilege) frac{dx}{x}$$ and $$f(y)=intlimits_0^inftydelta(x-1) f(y x) dx$$ evaluated utilizing formulation (7) under sometimes appear to converge on the half-plane $$Re(y)>0$$. I’ll point to that in common formulation derived from Fourier convolutions evaluated utilizing formulation (1) above appear to breathe extra sophisticated than formulation derived from Mellin convolutions evaluated utilizing formulation (7) under which I think is not less than partially associated to the spare complexity of the piece-wise nature of formulation (1) above.

(7) $$quaddelta(x+1)+delta(x-1)=underset{N,ftoinfty}{textual content{lim}} 2sumlimits_{n=1}^Nfrac{mu(n)}{n}sumlimits_{okay=1}^{f n}cosleft(frac{2 okay pi x}{n}privilege),quad M(N)=0$$

The conditional convergence requirement $$M(N)=0$$ acknowledged for formulation (1) to (7) above is as a result of the nested Fourier succession illustration of $$delta(x+1)+delta(x-1)$$ outlined in formulation (7) above solely evaluates to zero at $$x=0$$ when $$M(N)=0$$. The situation $$M(N)=0$$ is required when evaluating formulation (7) above and formulation derived from the 2 Mellin convolutions outlined within the previous paragraph utilizing formulation (7) above, however I’m not positive it is actually needful when evaluating formulation (1) above or formulation derived from the Fourier convolution $$f(y)=intlimits_{-infty}^inftydelta(x) f(y-x) dx$$ utilizing formulation (1) above (e.g. formulation (4), (5), and (6) above). Formula (1) above is predicated on the analysis of formulation (7) above at $$|x|ge 1$$, so maybe formulation (1) above just isn’t as delicate to the analysis of formulation (7) above at $$x=0$$. Formula (1) above can breathe seen as taking formulation (7) above, slicing out the divest $$-1le x<1$$, after which gluing the 2 remaining halves collectively on the inception. Nevertheless I often rate formulation (1) above and formulation derived from the Fourier convolution $$f(y)=intlimits_{-infty}^inftydelta(x) f(y-x) dx$$ utilizing formulation (1) above at $$M(N)=0$$ because it would not damage something to limit the election of $$N$$ to this situation and I think this restriction could maybe result in sooner and/or extra constant convergence.

See this respond I posted to considered one of my avow questions on Math StackExchange for extra data on the nested Fourier succession illustration of $$delta(x+1)+delta(x-1)$$ and examples of formulation derived from Mellin convolutions utilizing this illustration. See my Math StackExchange query associated to nested Fourier succession illustration of $$h(s)=frac{i s}{s^2-1}$$ for data on the extra common theme of nested Fourier succession representations of different non-periodic features.

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