On nascent delta 'function'

[mmzaj]

greetings . i have two questions regarding the sinc function in the week limit , where it can be used as a nascent delta function.
the definition :
$$\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}} \phi(x)dx=\phi(x_{0})$$ is said to be valid for any smooth function $\phi(x)$ with compact support . does that mean that the following is not valid :
$$\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}}xdx=x_{0}$$

moreover . if we expand the sine function, we get :
$$\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\phi(x)\sum_{n=0}^{\infty}\frac{(-1)^n(x-x_{0})^{2n}}{(2n+1)!(\varepsilon)^{2n+1}}dx =\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}} \int_{-\infty}^{\infty}(x-x_{0})^{2n}\phi(x) =\phi(x_{0})$$

is it legit to perform the integration term by term ??

 

[Mute]

greetings . i have two questions regarding the sinc function in the week limit , where it can be used as a nascent delta function.
the definition :
$$\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}} \phi(x)dx=\phi(x_{0})$$ is said to be valid for any smooth function $\phi(x)$ with compact support . does that mean that the following is not valid :
$$\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}}xdx=x_{0}$$

You're missing a factor of epsilon. Your integral should be

$$\lim_{\epsilon \rightarrow 0} \frac{1}{\pi}\int_{-\infty}^{\infty} dx \frac{\sin\left(\frac{x-x_0}{\epsilon}\right)}{\frac{x-x_0}{\epsilon}} \phi(x)$$

If you make the change of variables $y = (x-x_0)/\epsilon$ you can show (under the assumption that you can exchange the limit and the integral) that the result is $\phi(x_0)$. Whether or not you can legitimately choose $\phi(x) = x$ depends on whether or not you can actually exchange the limit with the integral.

I don't think expanding the sine as a series will really get you anywhere. The key property of the nascent delta function is that you pick a function that integrates to 1 when you scale out the epsilon.

 

[mmzaj]

You're missing a factor of epsilon. Your integral should be

$$\lim_{\epsilon \rightarrow 0} \frac{1}{\pi}\int_{-\infty}^{\infty} dx \frac{\sin\left(\frac{x-x_0}{\epsilon}\right)}{\frac{x-x_0}{\epsilon}} \phi(x)$$

no , i'am not ... check (37) here "delta function"

as for the expansion of the sine , here is what i was trying to do :

$$\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}} (x-x_{0})^{2n} = \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}}\left(\sum_{k=0}^{2n}\binom{2n}{k}x^{2n-k} (-1)^{k} (x_{0})^{k} \right)$$

now define $\Phi(x_{0j})$ as a delta distribution such that :
$$\Phi(\textbf{X})=\sum_{j=1}^{m} \delta(x-x_{0j})$$

$$\Phi(\textbf{X})=\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}}\left(\sum_{k=0}^{2n}\binom{2n}{k}x^{2n-k} (-1)^{k} \sum_{j=1}^{m} (x_{0j})^{k} \right)$$
now, for a certain situation , we are able to compute $\sum_{j=1}^{m} (x_{0j})^{k}$ for any $k$, but not the $x_{0j}$ themselves. so i was hoping to extract the $x_{0j}$ by :
$$x_{0j}= \int x\Phi(\textbf{X}) dx$$ over some period that contains $x_{0j}$

 

[mmzaj]

mathematica is giving me some encouraging results . to spice things up, for a polynomial $f(x)$ , we can find $\sum_{j=1}^{m} (x_{0j})^k$ where $x_{0j}$ are the roots of the polynomial :
$$\sum_{j=1}^{m} (x_{0j})^k = \lim_{x \rightarrow 0}\frac{1}{q(0)k!}\left( \sum_{j=0}^{k}\binom{k+1}{j+1} \left( p(x) \hat{q}(x)^{j}\right)^{(k)}\right)$$
where :
$$p(x)=x^{m-1}f^{'}\left(\frac{1}{x} \right)$$
$$q(x)=x^{m}f\left(\frac{1}{x} \right)$$
$$\hat{q}(x)=-\frac{q(x)}{q(0)}$$
$$m= deg[f(x)]$$

so , for a root $z$ contained in a small enough contour $\kappa$ :

$$z(\kappa )=\lim_{\varepsilon \rightarrow 0}\lim_{s \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}}\left(\sum_{k=0}^{2n}\binom{2n}{k}\frac{(-1)^{k}}{{q(0)}k!}\left(\oint_{\kappa } {x}^{2n-k+1}dx\right)\sum_{j=0}^{k}\binom{k+1}{j+1}\left(p(s)\hat{q}(s)^{j}\right)^{(k)}\right)$$

i was able to calculate the roots of some polynomials this way . and i know i can take it to the next step , and come up with a general formula . furthermore, this technique could be easily extended to study the roots of some complex functions .

 

[Mute]

no , i'am not ... check (37) here "delta function"

You're right, my mistake.

as for the expansion of the sine , here is what i was trying to do :

$$\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}} (x-x_{0})^{2n} = \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}}\left(\sum_{k=0}^{2n}\binom{2n}{k}x^{2n-k} (-1)^{k} (x_{0})^{k} \right)$$

now define $\Phi(x_{0j})$ as a delta distribution such that :
$$\Phi(\textbf{X})=\sum_{j=1}^{m} \delta(x-x_{0j})$$

$$\Phi(\textbf{X})=\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}}\left(\sum_{k=0}^{2n}\binom{2n}{k}x^{2n-k} (-1)^{k} \sum_{j=1}^{m} (x_{0j})^{k} \right)$$
now, for a certain situation , we are able to compute $\sum_{j=1}^{m} (x_{0j})^{k}$ for any $k$, but not the $x_{0j}$ themselves. so i was hoping to extract the $x_{0j}$ by :
$$x_{0j}= \int x\Phi(\textbf{X}) dx$$ over some period that contains $x_{0j}$

It's still not clear to me what you're trying to accomplish here. Are you expanding the sine because you want to compute the integral

$$\int_{-\infty}^\infty dx (x-x_0)^{2n}\phi(x)$$

and then evaluate the infinite sum and see if it reduces to $\phi(x_0)$? If so, interchanging the integral with the infinite sum is only going to be valid in certain cases. If you chose $\phi(x) = x$ the integrals aren't well defined. Even if you replace the infinite limits with finite limits $\pm T$ which you take to infinity at the end, you'll have to be careful with the order of the epsilon and T limits. I think you probably get different results for functions $\phi(x)$ which are not smooth.

 

[mmzaj]

It's still not clear to me what you're trying to accomplish here. Are you expanding the sine because you want to compute the integral

$$\int_{-\infty}^\infty dx (x-x_0)^{2n}\phi(x)$$

and then evaluate the infinite sum and see if it reduces to $\phi(x_0)$? If so, interchanging the integral with the infinite sum is only going to be valid in certain cases. If you chose $\phi(x) = x$ the integrals aren't well defined. Even if you replace the infinite limits with finite limits $\pm T$ which you take to infinity at the end, you'll have to be careful with the order of the epsilon and T limits. I think you probably get different results for functions $\phi(x)$ which are not smooth.

i am expanding the sine , and then sum over $x_{0j}$ in order to isolate the term
$$\sum_{j=1}^{m} (x_{0j})^{k}$$

which i can calculate via :

$$\sum_{j=1}^{m} (x_{0j})^k = \lim_{x \rightarrow 0}\frac{1}{q(0)k!}\left( \sum_{j=0}^{k}\binom{k+1}{j+1} \left( p(x) \hat{q}(x)^{j}\right)^{(k)}\right)$$
where :
$$p(x)=x^{m-1}f^{'}\left(\frac{1}{x} \right)$$
$$q(x)=x^{m}f\left(\frac{1}{x} \right)$$
$$\hat{q}(x)=-\frac{q(x)}{q(0)}$$
$$m= deg[f(x)]$$

for some polynomial $f(x)$

if we plug this term into the expansion we get :

$$\Phi(\mathbf{X})=\sum_{j=1}^{m}\delta (x-x_{0j})=\lim_{\varepsilon \rightarrow 0}\lim_{s \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}}\left(\sum_{k=0}^{2n}\binom{2n}{k}\frac{(-1)^{k}}{{q(0)}k!} {x}^{2n-k}\sum_{j=0}^{k}\binom{k+1}{j+1}\left(p(s)\hat{q}(s)^{j}\right)^{(k)}\right)$$

if we integrate, we get to extract the roots of the polynomial :

$$\oint _{\kappa} x\Phi(\mathbf{X})dx = x_{i}$$

where $x_{i}$ is contained in the contour $\kappa$ .