Lim and integration Delta function

Ipie · **Joined:** Fri, 8 Jul 2011 16:38:49 UTC **Posts:** 10

Please help me to find out:
1. $f(\beta)=\lim_{\beta \to\infty}{\frac{e^{-\beta(b-a)^2}}{\sqrt{\frac{\pi}{\beta}}e^{\frac{-k^2}{4 \beta}}}$
2. Find Delta function $\delta(x-a)$ :
$\int_{-\infty}^{+\infty} {e^{-\beta(x-a)^2}\delta(x-a)} dx =1$
Thank alot.

Shadow · **Posted:** Fri, 8 Jul 2011 19:05:13 UTC

Ipie wrote:

Please help me to find out:
1. $f(\beta)=\lim_{\beta \to\infty}{\frac{e^{-\beta(b-a)^2}}{\sqrt{\frac{\pi}{\beta}}e^{\frac{-k^2}{4 \beta}}}$
2. Find Delta function $\delta(x-a)$ :
$\int_{-\infty}^{+\infty} {e^{-\beta(x-a)^2}\delta(x-a)} dx =1$
Thank alot.

Um, that first one makes no sense, there is no dependence on beta with that limit there.

The second one, what do you mean "find Delta function"? Are you supposed to define a function $\delta$ such that it satisfies that integral equation?

Ipie · **Joined:** Fri, 8 Jul 2011 16:38:49 UTC **Posts:** 10

Shadow wrote:

Ipie wrote:

Please help me to find out:
1. $f(\beta)=\lim_{\beta \to\infty}{\frac{e^{-\beta(b-a)^2}}{\sqrt{\frac{\pi}{\beta}}e^{\frac{-k^2}{4 \beta}}}$
2. Find Delta function $\delta(x-a)$ :
$\int_{-\infty}^{+\infty} {e^{-\beta(x-a)^2}\delta(x-a)} dx =1$
Thank alot.

Um, that first one makes no sense, there is no dependence on beta with that limit there.

The second one, what do you mean "find Delta function"? Are you supposed to define a function $\delta$ such that it satisfies that integral equation?

First one is a kind of Gaussian function. When beta becomes very high that function approaches Delta function.
'Delta function is a generalized function that can be defined as the limit of class delta sequences'-Mathworld. and the functional nature of the delta function is often suppressed.
Delta function has fundamental property:
$\int_{-\infty}^{+\infty} {f(x)}\delta(x-a)} dx =f(a)$
So I need to find delta function

outermeasure · **Posted:** Sat, 9 Jul 2011 04:26:37 UTC

Ipie wrote:

Shadow wrote:

Ipie wrote:

Please help me to find out:
1. $f(\beta)=\lim_{\beta \to\infty}{\frac{e^{-\beta(b-a)^2}}{\sqrt{\frac{\pi}{\beta}}e^{\frac{-k^2}{4 \beta}}}$
2. Find Delta function $\delta(x-a)$ :
$\int_{-\infty}^{+\infty} {e^{-\beta(x-a)^2}\delta(x-a)} dx =1$
Thank alot.

Um, that first one makes no sense, there is no dependence on beta with that limit there.

The second one, what do you mean "find Delta function"? Are you supposed to define a function $\delta$ such that it satisfies that integral equation?

First one is a kind of Gaussian function. When beta becomes very high that function approaches Delta function.
'Delta function is a generalized function that can be defined as the limit of class delta sequences'-Mathworld. and the functional nature of the delta function is often suppressed.
Delta function has fundamental property:
$\int_{-\infty}^{+\infty} {f(x)}\delta(x-a)} dx =f(a)$
So I need to find delta function

Delta function isn't a function, but a measure. What exactly are you trying to find here? If you are trying to find the measure, then it is simply
$\displaystyle \delta_x\colon\mathcal{P}(X)\to\mathbb{R};\quad \delta_x(A)=\begin{cases} 1 & x\in A\\ 0 & x\notin A\end{cases}$
for any set

, so
$\displaystyle \int_X f\,\mathrm{d}\delta_x = f(x)\quad\forall f\in\mathbb{C}^X$

Ipie · **Joined:** Fri, 8 Jul 2011 16:38:49 UTC **Posts:** 10

Thanks to @outermeasure. Do you know about Delta function applications?
Please solve the initial-value problem:
$x' + 2 x = \delta(t-t_0)$
$(0 < t < \infty)$
with initial value x(0)=3

Ipie · **Joined:** Fri, 8 Jul 2011 16:38:49 UTC **Posts:** 10

Shadow wrote:

Ipie wrote:

Please help me to find out:
1. $f(\beta)=\lim_{\beta \to\infty}{\frac{e^{-\beta(b-a)^2}}{\sqrt{\frac{\pi}{\beta}}e^{\frac{-k^2}{4 \beta}}}$
2. Find Delta function $\delta(x-a)$ :
$\int_{-\infty}^{+\infty} {e^{-\beta(x-a)^2}\delta(x-a)} dx =1$
Thank alot.

Um, that first one makes no sense, there is no dependence on beta with that limit there.

The first one: $f(\beta)=\lim_{\beta \to\infty}{\frac{\sqrt{\frac{\beta}{\pi}}}{e^{\beta(b-a)^2} e^{\frac{-k^2}{4 \beta}}}$

Apply Taylor series expansion for $g(\beta)=e^{\beta(b-a)^2}$

$g(\beta)=1+\beta(b-a)^2 + ...$

So $f(\beta)=\lim_{\beta \to\infty}{\frac{\sqrt{\frac{\beta}{\pi}}}{[1+\beta(b-a)^2] e^{\frac{-k^2}{4 \beta}}} \to{0}$

outermeasure · **Posted:** Wed, 13 Jul 2011 09:59:10 UTC

Ipie wrote:

Thanks to @outermeasure. Do you know about Delta function applications?
Please solve the initial-value problem:
$x' + 2 x = \delta(t-t_0)$
$(0 < t < \infty)$
with initial value x(0)=3

You have $\dfrac{\mathrm{d}}{\mathrm{d}t}(e^{2t}x(t))=e^{2t}\delta(t-t_0)$ (where the derivative is the weak derivative), so
$\begin{aligned} e^{2t}x(t)-3 &=\int_0^t \frac{\mathrm{d}}{\mathrm{d}\tau}(e^{2\tau}x(\tau))\,\mathrm{d}\tau\\ & = \int_0^t e^{2\tau}\,\mathrm{d}\delta_{t_0}(\tau)\\ & =\begin{cases}e^{2t_0} & t_0\in(0,t)\\ 0 & \text{otherwise}\end{cases} \end{aligned}$
(remember you can change the value on a Lebesgue null set without affecting the weak derivative).

outermeasure · **Posted:** Wed, 13 Jul 2011 10:00:48 UTC

Ipie wrote:

Shadow wrote:

Ipie wrote:

Please help me to find out:
1. $f(\beta)=\lim_{\beta \to\infty}{\frac{e^{-\beta(b-a)^2}}{\sqrt{\frac{\pi}{\beta}}e^{\frac{-k^2}{4 \beta}}}$
2. Find Delta function $\delta(x-a)$ :
$\int_{-\infty}^{+\infty} {e^{-\beta(x-a)^2}\delta(x-a)} dx =1$
Thank alot.

Um, that first one makes no sense, there is no dependence on beta with that limit there.

The first one: $f(\beta)=\lim_{\beta \to\infty}{\frac{\sqrt{\frac{\beta}{\pi}}}{e^{\beta(b-a)^2} e^{\frac{-k^2}{4 \beta}}}$

Apply Taylor series expansion for $g(\beta)=e^{\beta(b-a)^2}$

$g(\beta)=1+\beta(b-a)^2 + ...$

So $f(\beta)=\lim_{\beta \to\infty}{\frac{\sqrt{\frac{\beta}{\pi}}}{[1+\beta(b-a)^2] e^{\frac{-k^2}{4 \beta}}} \to{0}$

It still doesn't make any sense --- you have committed the cardinal sin of using the same variable for two different things. Which $\beta$ is from the $f(\beta)$ and which $\beta$ is your dummy $\displaystyle\lim_{\beta\to\infty}$ ?

Ipie · **Joined:** Fri, 8 Jul 2011 16:38:49 UTC **Posts:** 10

outermeasure wrote:

Ipie wrote:

Shadow wrote:

Ipie wrote:

Please help me to find out:
1. $f(\beta)=\lim_{\beta \to\infty}{\frac{e^{-\beta(b-a)^2}}{\sqrt{\frac{\pi}{\beta}}e^{\frac{-k^2}{4 \beta}}}$
2. Find Delta function $\delta(x-a)$ :
$\int_{-\infty}^{+\infty} {e^{-\beta(x-a)^2}\delta(x-a)} dx =1$
Thank alot.

Um, that first one makes no sense, there is no dependence on beta with that limit there.

The first one: $f(\beta)=\lim_{\beta \to\infty}{\frac{\sqrt{\frac{\beta}{\pi}}}{e^{\beta(b-a)^2} e^{\frac{-k^2}{4 \beta}}}$

Apply Taylor series expansion for $g(\beta)=e^{\beta(b-a)^2}$

$g(\beta)=1+\beta(b-a)^2 + ...$

So $f(\beta)=\lim_{\beta \to\infty}{\frac{\sqrt{\frac{\beta}{\pi}}}{[1+\beta(b-a)^2] e^{\frac{-k^2}{4 \beta}}} \to{0}$

It still doesn't make any sense --- you have committed the cardinal sin of using the same variable for two different things. Which $\beta$ is from the $f(\beta)$ and which $\beta$ is your dummy $\displaystyle\lim_{\beta\to\infty}$ ?

I am very sorry. It's only f, not $f(\beta)$ . So it should be:
$f=\lim_{\beta \to\infty}{\frac{e^{-\beta(b-a)^2}}{\sqrt{\frac{\pi}{\beta}}e^{\frac{-k^2}{4 \beta}}}$

Ipie · **Joined:** Fri, 8 Jul 2011 16:38:49 UTC **Posts:** 10

outermeasure wrote:

Ipie wrote:

Thanks to @outermeasure. Do you know about Delta function applications?
Please solve the initial-value problem:
$x' + 2 x = \delta(t-t_0)$ (1)
$(0 < t < \infty)$
with initial value x(0)=3

You have $\dfrac{\mathrm{d}}{\mathrm{d}t}(e^{2t}x(t))=e^{2t}\delta(t-t_0)$ (where the derivative is the weak derivative), so
$\begin{aligned} e^{2t}x(t)-3 &=\int_0^t \frac{\mathrm{d}}{\mathrm{d}\tau}(e^{2\tau}x(\tau))\,\mathrm{d}\tau\\ & = \int_0^t e^{2\tau}\,\mathrm{d}\delta_{t_0}(\tau)\\ & =\begin{cases}e^{2t_0} & t_0\in(0,t)\\ 0 & \text{otherwise}\end{cases} \end{aligned}$
(remember you can change the value on a Lebesgue null set without affecting the weak derivative).

The general solution of (1) is:

$x(t) = c R(t) + R(t) \int_0^t{Q(\tau) R^{-1} (\tau)} d\tau$

where, $R(t) = e^{-\int_0^t2d\tau}$

or $R(t) = e^{-2t}$

and $Q(\tau) = \delta(\tau-t_0)$

so $x(t) = c e^{-2t} + e^{-2t} \int_0^t{e^{2\tau} \delta(\tau-t_0) d\tau$ (2)

With the initial condition x(0)=3, we get:

$x(0)=3 = c .1 + 1. \int_0^0{e^{2\tau} \delta(\tau-t_0) d\tau$
So c = 3

Also from the delta function definition:

$\int_ {t_1}^{t2}{f(t) \delta(t-t_0)d(t) =\begin{cases}f(t_0) & t_0\in(t_1,t_2)\\ 0 & \text{otherwise}\end{cases}$

Substitute into (2)m yields,

$x(t) = \begin{cases}3e^{-2t}+e^{-2(t-t_0)} & t_0\in(0,t)\\ 3e^{-2t} & \text{otherwise}\end{cases}$
Is this procedure OK?

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Lim and integration Delta function

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