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 Post subject: Lim and integration Delta function
PostPosted: Fri, 8 Jul 2011 17:09:08 UTC 
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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.


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 Post subject: Re: Lim and integration Delta function
PostPosted: Fri, 8 Jul 2011 19:05:13 UTC 
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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?

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 Post subject: Re: Lim and integration Delta function
PostPosted: Sat, 9 Jul 2011 00:57:50 UTC 
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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


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 Post subject: Re: Lim and integration Delta function
PostPosted: Sat, 9 Jul 2011 04:26:37 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
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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 X, so
\displaystyle
\int_X f\,\mathrm{d}\delta_x = f(x)\quad\forall f\in\mathbb{C}^X

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Wed, 13 Jul 2011 07:31:31 UTC 
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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


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 Post subject: Re: Lim and integration Delta function
PostPosted: Wed, 13 Jul 2011 08:57:13 UTC 
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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}


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 Post subject:
PostPosted: Wed, 13 Jul 2011 09:59:10 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
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).

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Lim and integration Delta function
PostPosted: Wed, 13 Jul 2011 10:00:48 UTC 
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User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
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}?

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Lim and integration Delta function
PostPosted: Wed, 13 Jul 2011 12:32:10 UTC 
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Member

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}}}


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 Post subject:
PostPosted: Wed, 13 Jul 2011 15:39:34 UTC 
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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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