Delta Function¶

Delta function $\delta(x)$ is defined such that this relation holds:

(1) $\int f(x)\delta(x-t)\d x=f(t)$

No such function exists, but one can find many sequences “converging” to a delta function:

(2) $\lim_{\alpha\to\infty}\delta_\alpha(x) = \delta(x)$

more precisely:

(3) $\lim_{\alpha\to\infty}\int f(x)\delta_\alpha(x)\d x = \int f(x)\lim_{\alpha\to\infty}\delta_\alpha(x)\d x = f(0)$

one example of such a sequence is:

$\delta_\alpha(x) = {1\over\pi x}\sin(\alpha x)$

It’s clear that (3) holds for any well behaved function $f(x)$ . Some mathematicians like to say that it’s incorrect to use such a notation when in fact the integral (1) doesn’t “exist”, but we will not follow their approach, because it is not important if something “exists” or not, but rather if it is clear what we mean by our notation: (1) is a shorthand for (3) and (2) gets a mathematically rigorous meaning when you integrate both sides and use (1) to arrive at (3). Thus one uses the relations (1), (2), (3) to derive all properties of the delta function.

Let’s give an example. Let ${\bf\hat r}$ be the unit vector in 3D and we can label it using spherical coordinates ${\bf\hat r}={\bf\hat r}(\theta,\phi)$ . We can also express it in cartesian coordinates as ${\bf\hat r}(\theta,\phi)=(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)$ .

(4) $f({\bf\hat r'})=\int\delta({\bf\hat r}-{\bf\hat r'})f({\bf\hat r})\,\d {\bf\hat r}$

Expressing $f({\bf\hat r})=f(\theta,\phi)$ as a function of $\theta$ and $\phi$ we have

(5) $f(\theta',\phi')=\int\delta(\theta-\theta')\delta(\phi-\phi') f(\theta,\phi)\,\d\theta\d\phi$

Expressing (4) in spherical coordinates we get

$f(\theta',\phi')=\int\delta({\bf\hat r}-{\bf\hat r'}) f(\theta,\phi)\sin\theta\,\d\theta\d\phi$

and comparing to (5) we finally get

$\delta({\bf\hat r}-{\bf\hat r'})={1\over\sin\theta} \delta(\theta-\theta')\delta(\phi-\phi')$

In exactly the same manner we get

$\delta({\bf r}-{\bf r'})=\delta({\bf\hat r}-{\bf\hat r'}) {\delta(\rho-\rho')\over\rho^2}$

See also (6) for an example of how to deal with more complex expressions involving the delta function like $\delta^2(x)$ .

When integrating over finite interval, this formula is very useful:

$\int_a^b f(x)\delta(x-t)\d x = f(t) \theta(b-t) \theta(t-a)$

in other words, the integral vanishes unless $a < t < b$ . In the limit $a\to -\infty$ and $b\to\infty$ we get:

$\theta(b-t) \to \theta(\infty - t) = 1 \theta(t-a) \to \theta(t - (-\infty)) = \theta(t+\infty) = 1$

Distributions¶

Some mathematicians like to use distributions and a mathematical notation for that, which I think is making things less clear, but nevertheless it’s important to understand it too, so the notation is explained in this section, but I discourage to use it – I suggest to only use the physical notation as explained below. The math notation below is put into quotation marks, so that it’s not confused with the physical notation.

The distribution is a functional and each function $f(x)$ can be identified with a distribution $\mathnot{T_f}$ that it generates using this definition ( $\varphi(x)$ is a test function):

$\mathnot{T_f(\phi(x))} \equiv \int f(x)\varphi(x) \d x \equiv \mathnot{f(\varphi(x))} \equiv \mathnot{(f(x), \varphi(x))}$

besides that, one can also define distributions that can’t be identified with regular functions, one example is a delta distribution (Dirac delta function):

$\mathnot{\delta(\phi(x))} \equiv \phi(0) \equiv \int \delta(x) \phi(x) \d x$

The last integral is not used in mathematics, in physics on the other hand, the first expressions ( $\mathnot{\delta(\phi(x))}$ ) is not used, so $\delta(x)$ always means that you have to integrate it, as explained in the previous section, so it behaves like a regular function (except that such a function doesn’t exist and the precise mathematical meaning is only after you integrate it, or through the identification above with distributions).

One then defines common operations via acting on the generating function, then observes the pattern and defines it for all distributions. For example differentiation:

$\mathnot{{\d\over\d x}T_f(\varphi)} = \mathnot{T_{f'}(\varphi)} = \int f'\varphi \d x = -\int f\varphi' \d x = \mathnot{-T_f(\varphi')}$

so:

$\mathnot{{\d\over\d x}T(\varphi)} = \mathnot{-T(\varphi')}$

Multiplication:

$\mathnot{gT_f(\varphi)} = \mathnot{T_{gf}(\varphi)} = \int gf\varphi \d x = \mathnot{T_f(g\varphi)}$

so:

$\mathnot{gT(\varphi)} = \mathnot{T(g\varphi)}$

Fourier transform:

$\mathnot{FT_f(\varphi)} = \mathnot{T_{Ff}(\varphi)} = \int F(f)\varphi \d x = =\int\left[\int e^{-ikx} f(k) \d k\right] \varphi(x) \d x =\int f(k)\left[\int e^{-ikx} \varphi(x) \d x\right] \d k =\int f(x)\left[\int e^{-ikx} \varphi(k) \d k\right] \d x = = \int f F(\varphi) \d x = \mathnot{T_f(F\varphi)}$

so:

$\mathnot{FT(\varphi)} = \mathnot{T(F\varphi)}$

But as you can see, the notation is just making things more complex, since it’s enough to just work with the integrals and forget about the rest. One can then even omit the integrals, with the understanding that they are implicit.

Some more examples:

$\int \delta(x-x_0)\varphi(x) \d x = \int \delta(x)\varphi(x+x_0) \d x = \varphi(x_0) \equiv \mathnot{\delta(\varphi(x+x_0))}$

Proof of $\delta(-x) = \delta(x)$ :

$\int_{-\infty}^\infty \delta(-x)\varphi(x)\d x = -\int_{\infty}^{-\infty} \delta(y)\varphi(-y)\d y = \int_{-\infty}^\infty \delta(x)\varphi(-x)\d x \equiv \mathnot{\delta(\varphi(-x))} = \varphi(0) =\mathnot{\delta(\phi(x))} \equiv \int_{-\infty}^\infty \delta(x)\varphi(x)\d x$

Proof of $x\delta(x) = 0$ :

$\int x\delta(x)\varphi(x)\d x = \mathnot{\delta(x\varphi(x))} = 0 \cdot \varphi(0) = 0$

Proof of $\delta(cx) = {\delta(x)\over |c|}$ :

$\int \delta(cx)\varphi(x)\d x = {1\over|c|}\int\delta(x)\varphi\left({x\over c}\right)\d x =\mathnot{\delta\left(\varphi\left({x\over c}\right)\over|c|\right)} ={\delta(0)\over|c|} =\mathnot{{\delta(\varphi(x))\over|c|}} = \int {\delta(x)\over |c|}\varphi(x)\d x$

Variations and Functional Derivatives¶

Variations and functional derivatives are generalization of differentials and partial derivatives to functionals. It is important to master this subject just like regular differentials/derivatives in calculus.

Functions of One Variable¶

Let’s first review differentials and derivatives of functions of one variable. We will use an approach that directly generalizes to multivariable functions and functionals. The differential $\d f$ is defined as:

$\d f \equiv \lim_{\varepsilon\to0} {f(x+\varepsilon h) -f(x)\over\varepsilon} = a h$

Last equality follows from the fact, that the limit is a linear function of $h$ :

$\lim_{\varepsilon\to0} {f(x+\varepsilon h) -f(x)\over\varepsilon} = \lim_{\eta\to0} {f(x+\eta) -f(x)\over\left(\eta\over h\right)} = \left(\lim_{\eta\to0} {f(x+\eta) -f(x)\over\eta}\right) h$

Where we used the substitution $\eta = \varepsilon h$ . We define the derivative $\d f\over \d x$ as:

${\d f\over \d x} = a$

To get a formula for ${\d f\over \d x}$ , we set $h=1$ and get:

${\d f\over \d x} = a = a \cdot 1 = \lim_{\varepsilon\to0} {f(x+\varepsilon \cdot 1)-f(x)\over\varepsilon} = \lim_{\varepsilon\to0} {f(x+\varepsilon)-f(x)\over\varepsilon}$

Using the formulas above we get an equivalent expression for the differential:

$\left.{\d\over\d\varepsilon}f(x+\varepsilon h) \right|_{\varepsilon=0} = \left.\lim_{\eta\to0} {f(x+(\varepsilon+\eta) h) -f(x+\varepsilon h)\over\eta}\right|_{\varepsilon=0} = = \lim_{\eta\to0} {f(x+\eta h) -f(x)\over\eta} = = \lim_{\varepsilon\to0} {f(x+\varepsilon h) -f(x)\over\varepsilon}$

So we get a general formula (the analogy of which we will use later):

$\d f \equiv\left.{\d\over\d\varepsilon}f(x+\varepsilon h) \right|_{\varepsilon=0} = \lim_{\varepsilon\to0} {f(x+\varepsilon h) -f(x)\over\varepsilon} = a h$

The variable $x$ can be treated as a function (a very simple one):

$x = g(x)$

So we define $\d x$ as:

$\d x \equiv \d g = {\d g\over\d x} h = h$

As such, $\d x$ can have two meanings: either $\d x = h = x-x_0$ (a finite change in the variable $x$ ) or a differential (if $x$ depends on another variable, thanks to the chain rule everything will work). With this understanding, for all calculations, we only need the following two formulas — the definition of the differential (using a limit):

$\d f = \lim_{\varepsilon\to0} {f(x+\varepsilon \d x) -f(x)\over\varepsilon}$

and the definition of the derivative (using the differential):

$\d f = {\d f\over \d x} \d x$

where $\d x$ is either a differential or a finite change in the variable $x$ .

If for example $x=p(y)$ is a function of $y$ then in the above $\d x$ is a differential and we get:

$\d f = {\d f\over \d x} \d x = {\d f\over \d x}{\d x\over \d y} \d y$

Thanks to the chain rule, this can also be written as:

$\d f = {\d f\over \d y} \d y$

and so the notation is consistent.

Functions of several variables¶

Let’s have $\x=(x_1,x_2,\dots,x_N)$ . The function $f(\x)$ assigns a number to each $\x$ . We define a differential of $f$ in the direction of ${\bf h}$ as:

$\d f \equiv\left.{\d\over\d\varepsilon}f({\bf x}+\varepsilon{\bf h}) \right|_{\varepsilon=0} = \lim_{\varepsilon\to0} {f({\bf x}+\varepsilon{\bf h}) -f(\x)\over\varepsilon} = {\bf a} \cdot {\bf h}$

The last equality follows from the fact, that $\left.{\d\over\d\varepsilon}f({\bf x}+\varepsilon{\bf h}) \right|_{\varepsilon=0}$ is a linear function of ${\bf h}$ . We define the partial derivative ${\partial f\over\partial x_i}$ of $f$ with respect to $x_i$ as the $i$ -th component of the vector ${\bf a}$ :

${\bf a} \equiv \left({\partial f\over\partial x_1}, {\partial f\over\partial x_2}, \dots, {\partial f\over\partial x_N}\right) \equiv \nabla f$

This also gives a formula for computing ${\partial f\over\partial x_i}$ : we set $h_j=\delta_{ij}h_i$ and

${\partial f\over\partial x_i} = a_i = {\bf a}\cdot{\bf h} = \left.{\d\over\d\varepsilon} f(\x+\varepsilon(0,0,\dots,1,\dots,0)) \right|_{\varepsilon=0} = = \lim_{\varepsilon\to0} {f(x_1,x_2,\dots,x_i+\varepsilon,\dots,x_N) -f(x_1,x_2,\dots,x_i,\dots,x_N) \over\varepsilon}$

The usual way to define partial derivatives is to use the last formula as the definition, but here this formula is a consequence of our definition in terms of the components of ${\bf a}$ . Every variable can be treated as a function (very simple one):

$x_i=g(x_1,\dots,x_N)=\delta_{ij}x_j$

and so we define

$\d x_i\equiv\d g=\d(\delta_{ij}x_j)=h_i$

and thus we write $h_i=\d x_i$ and ${\bf h}=\d\x$ and

$\d f={\d f\over\d x_i}\d x_i = (\nabla f) \cdot \d {\bf x}$

So $\d{\bf x}$ has two meanings — it’s either ${\bf h}={\bf x}-{\bf x}_0$ (a finite change in the independent variable ${\bf x}$ ) or a differential, depending on the context. The above is a detailed explanation why things are defined the way they are and what the exact meaning is. With this understanding, the only things that are actually needed for any calculations are the following – the definition of a differential:

$\d f = \left.{\d\over\d\varepsilon}f({\bf x}+\varepsilon\d {\bf x}) \right|_{\varepsilon=0}$

Only a regular derivative (defined in the previous section) is needed for this definition. The definition of a partial derivative (and a gradient):

$\d f={\d f\over\d x_i}\d x_i = (\nabla f) \cdot \d {\bf x}$

And finally the understanding that $\d{\bf x}$ means either ${\bf h}={\bf x}-{\bf x}_0$ or a differential depending on the context. That’s all there is to it.

Functionals¶

Let’s now define functional derivatives and variations. Functional $F[f]$ assigns a number to each function $f(x)$ . The variation is defined as

$\delta F[f]\equiv\left.{\d\over\d\varepsilon}F[f+\varepsilon h] \right|_{\varepsilon=0}=\lim_{\epsilon\to0}{F[f+\epsilon h]-F[f]\over\epsilon}= \int a(x)h(x)\d x$

We define ${\delta F\over\delta f(x)}$ as

$a(x)\equiv{\delta F\over\delta f(x)}$

This also gives a formula for computing ${\delta F\over\delta f(x)}$ : we set $h(y)=\delta(x-y)$ and

${\delta F\over\delta f(x)}=a(x)=\int a(y)\delta(x-y)\d y= \left.{\d\over\d\varepsilon}F[f(y)+\varepsilon\delta(x-y)] \right|_{\varepsilon=0}= =\lim_{\varepsilon\to0} {F[f(y)+\varepsilon\delta(x-y)]-F[f(y)]\over\varepsilon}$

Sometimes the functional derivative is defined using the last formula, here this formula just follows from our definition. Every function can be treated as a functional (although a very simple one):

$f(x)=G[f]=\int f(y)\delta(x-y)\d y$

and so we define

$\delta f\equiv\delta G[f]= \left.{\d\over\d\varepsilon}G[f(x)+\varepsilon h(x)] \right|_{\varepsilon=0}= \left.{\d\over\d\varepsilon}(f(x)+\varepsilon h(x)) \right|_{\varepsilon=0}= h(x)$

thus we write $h=\delta f$ and

$\delta F[f]=\int {\delta F\over\delta f(x)}\delta f(x)\d x$

so $\delta f$ have two meanings — it’s either $h(x)=f(x) - f_0(x)$ (a finite change in the function $f$ ) or a variation of a functional, depending on the context. It is completely analogous to $\d\x$ . Let’s summarize the only formulas needed in actual calculations – the definition of a variation (using a regular derivative):

$\delta F[f] = \left.{\d\over\d\varepsilon}F[f+\varepsilon \delta f] \right|_{\varepsilon=0}$

the definition of the functional derivative:

$\delta F[f]=\int {\delta F\over\delta f(x)} \delta f(x) \d x$

and the understanding that $\delta f$ means either $h(x)=f(x) - f_0(x)$ or a variation.

The correspondence between the finite and infinite dimensional case can be summarized as:

$\begin{eqnarray*} f(x_i) \quad&\Longleftrightarrow&\quad F[f] \\ \d f=0 \quad&\Longleftrightarrow&\quad \delta F=0 \\ {\partial f\over\partial x_i}=0 \quad&\Longleftrightarrow&\quad {\delta F\over\delta f(x)}=0 \\ f \quad&\Longleftrightarrow&\quad F \\ x_i \quad&\Longleftrightarrow&\quad f(x) \\ x \quad&\Longleftrightarrow&\quad f \\ i \quad&\Longleftrightarrow&\quad x \\ \end{eqnarray*}$

More generally, $\delta$ -variation can by applied to any function $g$ which contains the function $f(x)$ being varied, you just need to replace $f$ by $f+\epsilon h$ and apply ${\d\over\d\epsilon}$ to the whole $g$ , for example (here $g=\partial_\mu\phi$ and $f=\phi$ ):

$\delta\partial_\mu\phi = \left.{\d\over\d\varepsilon}\partial_\mu(\phi+\varepsilon h) \right|_{\varepsilon=0} = \partial_\mu\left.{\d\over\d\varepsilon}(\phi+\varepsilon h) \right|_{\varepsilon=0} =\partial_\mu h =\partial_\mu \delta\phi$

This notation allows us a very convinient computation, as shown in the following examples. First, when computing a variation of some integral, when can interchange $\delta$ and $\int$ :

$F[f]=\int K(x) f(x) \d x$

$\delta F=\delta \int K(x) f(x) \d x = \left.{\d\over\d\varepsilon}\int K(x) (f+\varepsilon h)\d x\right|_{\varepsilon=0}= \left.\int{\d\over\d\varepsilon} (K(x) (f+\varepsilon h))\d x\right|_{\varepsilon=0}=$

$=\int\delta(K(x) f(x))\d x$

In the expression $\delta(K(x) f(x))$ we must understand from the context if we are treating it as a functional of $f$ or $K$ . In our case it’s a functional of $f$ , so we have $\delta(K f)=K\delta f$ .

A few more examples (notice that one can do each calculation either in terms of the functional derivative or the variation, and the variation version is usually simpler):

${\delta\over\delta f(t)}\int\d t'f(t')g(t')= \left.{\d\over\d\varepsilon}\int\d t'(f(t')+\varepsilon\delta(t-t'))g(t') \right|_{\varepsilon=0}=g(t) \delta \int \d t f(t)g(t) = \left.{\d\over\d\varepsilon}\int\d t(f(t)+\varepsilon \delta f(t))g(t) \right|_{\varepsilon=0} = \int g(t) (\delta f(t)) \d t$

${\delta f(t')\over\delta f(t)}= \left.{\d\over\d\varepsilon}(f(t')+\varepsilon\delta(t-t')) \right|_{\varepsilon=0}=\delta(t-t')$

${\delta f(t_1)f(t_2)\over\delta f(t)}= \left.{\d\over\d\varepsilon}(f(t_1)+\varepsilon\delta(t-t_1)) (f(t_2)+\varepsilon\delta(t-t_2)) \right|_{\varepsilon=0}=\delta(t-t_1)f(t_2)+f(t_1)\delta(t-t_2)$

$\delta (f(t_1)f(t_2)) = \left.{\d\over\d\varepsilon}(f(t_1)+\varepsilon\delta f(t_1)) (f(t_2)+\varepsilon\delta f(t_2)) \right|_{\varepsilon=0} =(\delta f(t_1))f(t_2)+f(t_1)(\delta f(t_2))$

${\delta\over\delta f(t)}\half\int\d t_1\d t_2K(t_1,t_2)f(t_1)f(t_2)= \half\int\d t_1\d t_2K(t_1,t_2){\delta f(t_1)f(t_2)\over\delta f(t)}= =\half\left(\int\d t_1 K(t_1,t)f(t_1)+\int\d t_2 K(t,t_2)f(t_2)\right) =\int\d t_2 K(t,t_2)f(t_2)$

$\delta \half\int\d t_1\d t_2K(t_1,t_2)f(t_1)f(t_2)= \half\int\d t_1\d t_2K(t_1,t_2)(\delta f(t_1)f(t_2))= = \half\int\d t_1\d t_2K(t_1,t_2) ((\delta f(t_1))f(t_2) +f(t_1)(\delta f(t_2))= = \int\d t_1\d t_2 K(t_1,t_2) f(t_2) (\delta f(t_1))$

The last equality follows from $K(t_1,t_2)=K(t_2,t_1)$ (any antisymmetrical part of a $K$ would not contribute to the symmetrical integration).

Another example is the derivation of Euler-Lagrange equations for the Lagrangian density $\L=\L(\eta_\rho, \partial_\nu \eta_\rho, x^\nu)$ :

$0 = \delta I = \delta \int \L \,\d^4x^\mu = \int \delta \L \,\d^4x^\mu = \int { \partial \L\over\partial \eta_\rho}\delta\eta_\rho + { \partial \L\over\partial (\partial_\nu \eta_\rho)} \delta(\partial_\nu\eta_\rho) \,\d^4x^\mu = = \int { \partial \L\over\partial \eta_\rho}\delta\eta_\rho + { \partial \L\over\partial (\partial_\nu \eta_\rho)} \partial_\nu(\delta\eta_\rho) \,\d^4x^\mu = = \int { \partial \L\over\partial \eta_\rho}\delta\eta_\rho - \partial_\nu\left( { \partial \L\over\partial (\partial_\nu \eta_\rho)} \right) \delta\eta_\rho \,\d^4x^\mu +\int \partial_\nu \left( { \partial \L\over\partial (\partial_\nu \eta_\rho)} \delta\eta_\rho \right) \,\d^4x^\mu = = \int \left[{ \partial \L\over\partial \eta_\rho} - \partial_\nu\left( { \partial \L\over\partial (\partial_\nu \eta_\rho)} \right) \right] \delta\eta_\rho \,\d^4x^\mu$

Another example:

${\delta\over\delta f(t)}\int f^3(x)\d x= \left.{\d\over\d\varepsilon}\int(f(x)+\varepsilon\delta(x-t))^3\d x \right|_{\varepsilon=0}=$

$=\left.\int3(f(x)+\varepsilon\delta(x-t))^2\delta(x-t)\d x \right|_{\varepsilon=0}=\int3f^2(x)\delta(x-t)\d x=3f^2(t)$

One might thing that the above calculation is incorrect, because $\delta^2(x-t)$ is undefined. In case of such problems the above notation automatically implies working with some sequence $\delta_\alpha(x) \to \delta(x)$ (for example $\delta_\alpha(x) = {1\over\pi x}\sin(\alpha x)$ ) and taking the limit $\alpha\to\infty$ :

${\delta\over\delta f(t)}\int f^3(x)\d x= \left.\lim_{\alpha\to\infty}{\d\over\d\varepsilon}\int(f(x)+\varepsilon\delta_\alpha(x-t))^3\d x \right|_{\varepsilon=0}=$

$=\left.\lim_{\alpha\to\infty}\int3(f(x)+\varepsilon\delta_\alpha(x-t))^2\delta_\alpha(x-t)\d x \right|_{\varepsilon=0}=\lim_{\alpha\to\infty}\int3f^2(x)\delta_\alpha(x-t)\d x=$

(6) $=\int3f^2(x)\lim_{\alpha\to\infty}\delta_\alpha(x-t)\d x= \int3f^2(x)\delta(x-t)\d x=3f^2(t)$

As you can see, we got the same result, with the same rigor, but using an obfuscating notation. That’s why such obvious manipulations with $\delta_\alpha$ are tacitly implied.

Another example with a metric as a function of coordinates $g_{\mu\nu} = g_{\mu\nu}(x^\mu)$ :

$\delta g_{\mu\nu} = \delta g_{\mu\nu}(x^\mu) = \left.{\d\over\d\varepsilon} g_{\mu\nu}(x^\mu+\varepsilon (\delta x^\mu)) \right|_{\varepsilon=0} = \left.{\d\over\d\varepsilon} (x^\sigma+\varepsilon(\delta x^\sigma)) \right|_{\varepsilon=0} \partial_\sigma g_{\mu\nu} = (\delta x^\sigma) \partial_\sigma g_{\mu\nu}$

And an example of varying with respect to a metric:

$\delta \sqrt{ |\det g_{\mu\nu}| } =\sqrt{ |\det g_{\mu\nu}| }\, \delta \log \sqrt{ |\det g_{\mu\nu}| } =\half \sqrt{ |\det g_{\mu\nu}| }\, \delta \log |\det g_{\mu\nu}| = =\half \sqrt{ |\det g_{\mu\nu}| }\, \delta \Tr \log g_{\mu\nu} =\half \sqrt{ |\det g_{\mu\nu}| }\, \Tr \delta \log g_{\mu\nu} = =\half \sqrt{ |\det g_{\mu\nu}| }\, \Tr g^{\mu\nu} \delta g_{\mu\nu} =\half \sqrt{ |\det g_{\mu\nu}| }\, g^{\mu\nu} \delta g_{\mu\nu} = =-\half \sqrt{ |\det g_{\mu\nu}| }\, g_{\mu\nu} \delta g^{\mu\nu}$

Another example (varying energy functional):

$E[\rho] = 4\pi\int {a \rho(r)\over b + r_s(r)} r^2 \d r r_s(r) = \left(3\over 4\pi (-\rho)\right)^{1\over 3} {\d r_s\over\d \rho} = {1\over 3}\left(3\over 4\pi (-\rho)\right)^{-{2\over 3}} {3\over 4\pi \rho^2} = -{1\over 3\rho}\left(3\over 4\pi (-\rho)\right)^{1\over 3} = -{r_s\over 3\rho} \delta E[\rho] = 4\pi \delta \int {a \rho\over b + r_s} r^2 \d r = = 4\pi \int\left({a \delta \rho\over b + r_s} - {a \rho\over (b + r_s)^2 }\delta r_s\right) r^2 \d r = = 4\pi \int\left({a \delta \rho\over b + r_s} - {a \rho\over (b + r_s)^2 }\left(-{r_s\over 3\rho}\right) \delta\rho\right) r^2 \d r = = 4\pi \int\left({a \over b + r_s} +{1\over3} {a r_s\over (b + r_s)^2 }\right) (\delta\rho) r^2 \d r {\delta E[\rho]\over\delta\rho} = 4\pi r^2 \left({a \over b + r_s} +{1\over3} {a r_s\over (b + r_s)^2 }\right)$

Another example (Hartree energy):

$E[n] = \half \int {n({\bf r}') n({\bf r}'')\over | {\bf r}' - {\bf r}''| } \d^3 r' \d^3 r'' \delta E[n] = \half \delta \int {n({\bf r}') n({\bf r}'')\over | {\bf r}' - {\bf r}''| } \d^3 r' \d^3 r'' = = \half \int { (\delta n({\bf r}')) n({\bf r}'') + n({\bf r}') (\delta n({\bf r}''))\over | {\bf r}' - {\bf r}''| } \d^3 r' \d^3 r'' = = \int { n({\bf r}') \over | {\bf r}' - {\bf r}''| } (\delta n({\bf r}'')) \d^3 r' \d^3 r'' = = \int { n({\bf r}') \over | {\bf r} - {\bf r}'| } (\delta n({\bf r})) \d^3 r' \d^3 r {\delta E[n]\over \delta n({\bf r})} = \int { n({\bf r}') \over | {\bf r} - {\bf r}'| } \d^3 r'$

Dirac Notation¶

The Dirac notation allows a very compact and powerful way of writing equations that describe a function expansion into a basis, both discrete (e.g. a fourier series expansion) and continuous (e.g. a fourier transform) and related things. The notation is designed so that it is very easy to remember and it just guides you to write the correct equation.

Let’s have a function $f(x)$ . We define

$\begin{eqnarray*} \braket{x|f}&\equiv& f(x) \\ \braket{x'|f}&\equiv& f(x') \\ \braket{x'|x}&\equiv&\delta(x'-x) \\ \int\ket{x}\bra{x}\d x&\equiv&\one \\ \end{eqnarray*}$

The following equation

$f(x')=\int\delta(x'-x)f(x)\d x$

then becomes

$\braket{x'|f}=\int\braket{x'|x}\braket{x|f}\d x$

and thus we can interpret $\ket{f}$ as a vector, $\ket{x}$ as a basis and $\braket{x|f}$ as the coefficients in the basis expansion:

$\ket{f}=\one\ket{f}=\int\ket{x}\bra{x}\d x\ket{f}= \int\ket{x}\braket{x|f}\d x$

That’s all there is to it. Take the above rules as the operational definition of the Dirac notation. It’s like with the delta function - written alone it doesn’t have any meaning, but there are clear and non-ambiguous rules to convert any expression with $\delta$ to an expression which even mathematicians understand (i.e. integrating, applying test functions and using other relations to get rid of all $\delta$ symbols in the expression – but the result is usually much more complicated than the original formula). It’s the same with the ket $\ket{f}$ : written alone it doesn’t have any meaning, but you can always use the above rules to get an expression that make sense to everyone (i.e. attaching any bra to the left and rewriting all brackets $\braket{a|b}$ with their equivalent expressions) – but it will be more complex and harder to remember and – that is important – less general.

Now, let’s look at the spherical harmonics:

$Y_{lm}({\bf\hat r})\equiv\braket{{\bf\hat r}|lm}$

on the unit sphere, we have

$\int\ket{\bf\hat r}\bra{\bf\hat r}\d{\bf\hat r}= \int\ket{\bf\hat r}\bra{\bf\hat r}\d\Omega=\one$

$\delta({\bf\hat r}-{\bf\hat r'})=\braket{{\bf\hat r}|{\bf\hat r'}}$

thus

$\int_0^{2\pi}\int_0^{\pi} Y_{lm}(\theta,\phi)\,Y^*_{l'm'}(\theta,\phi)\sin\theta\,\d\theta\,\d\phi = \int\braket{l'm'|{\bf\hat r}}\braket{{\bf\hat r}|lm}\d\Omega= \braket{l'm'|lm}$

and from (?) we get

$\braket{l'm'|lm}=\delta_{mm'}\delta_{ll'}$

now

$\sum_{lm}Y_{lm}(\theta,\phi)Y_{lm}^*(\theta',\phi')= \sum_{lm}\braket{{\bf\hat r}|lm}\braket{lm|{\bf\hat r'}}$

from (?) we get

$\sum_{lm}\braket{{\bf\hat r}|lm}\braket{lm|{\bf\hat r'}}= \braket{{\bf\hat r}|{\bf\hat r'}}$

so we have

$\sum_{lm}\ket{lm}\bra{lm}=\one$

so $\ket{lm}$ forms an orthonormal basis. Any function defined on the sphere $f({\bf\hat r})$ can be written using this basis:

$f({\bf\hat r}) =\braket{{\bf\hat r}|f} =\sum_{lm}\braket{{\bf\hat r}|lm}\braket{lm|f} =\sum_{lm}Y_{lm}({\bf\hat r})f_{lm}$

where

$f_{lm}=\braket{lm|f}=\int\braket{lm|{\bf\hat r}}\braket{{\bf\hat r}|f}\d\Omega =\int Y_{lm}^*({\bf\hat r}) f({\bf\hat r})\d\Omega$

If we have a function $f({\bf r})$ in 3D, we can write it as a function of $\rho$ and ${\bf\hat r}$ and expand only with respect to the variable ${\bf\hat r}$ :

$f({\bf r})=f(\rho{\bf\hat r})\equiv g(\rho,{\bf\hat r})= \sum_{lm}Y_{lm}({\bf\hat r})g_{lm}(\rho)$

In Dirac notation we are doing the following: we decompose the space into the angular and radial part

$\ket{{\bf r}}=\ket{{\bf\hat r}}\otimes\ket{\rho} \equiv\ket{{\bf\hat r}}\ket{\rho}$

and write

$f({\bf r})=\braket{{\bf r}|f}=\bra{{\bf\hat r}}\braket{\rho|f}= \sum_{lm}Y_{lm}({\bf\hat r})\bra{lm}\braket{\rho|f}$

where

$\bra{lm}\braket{\rho|f}= \int\braket{lm|{\bf\hat r}}\bra{{\bf\hat r}}\braket{\rho|f}\d\Omega =\int Y_{lm}^*({\bf\hat r}) f({\bf r})\d\Omega$

Let’s calculate $\braket{\rho|\rho'}$

$\braket{{\bf r}|{\bf r'}}=\bra{\bf\hat r}\braket{\rho|\rho'}\ket{{\bf\hat r'}} =\braket{{\bf\hat r}|{\bf\hat r'}}\braket{\rho|\rho'}$

so

$\braket{\rho|\rho'} ={\braket{{\bf r}|{\bf r'}}\over\braket{{\bf\hat r}|{\bf\hat r'}}} ={\delta(\rho-\rho')\over\rho^2}$

We must stress that $\ket{lm}$ only acts in the $\ket{{\bf\hat r}}$ space (not the $\ket\rho$ space) which means that

$\braket{{\bf r}|lm} =\bra{\bf\hat r}\braket{\rho|lm} =\braket{{\bf\hat r}|lm}\bra{\rho} =Y_{lm}({\bf\hat r})\bra{\rho}$

and $V\ket{lm}$ leaves $V\ket\rho$ intact. Similarly,

$\sum_{lm} \ket{lm}\bra{lm}=\one$

is a unity in the $\ket{\bf\hat r}$ space only (i.e. on the unit sphere).

Let’s rewrite the equation (?):

$\sum_m\braket{{\bf\hat r}|lm}\braket{lm|{\bf\hat r'}}= {4\pi\over 2l+1} \braket{{\bf\hat r}\cdot{\bf\hat r'}|P_l}$

Using the completeness relation (?):

$\sum_l {2l+1\over2}\braket{x'|P_l}\braket{P_l|x}=\braket{x'|x}$

$\sum_l \ket{P_l}{2l+1\over2}\bra{P_l}=\one$

we can now derive a very important formula true for every function $f({\bf\hat r}\cdot{\bf\hat r'})$ :

$f({\bf\hat r}\cdot{\bf\hat r'})=\braket{{\bf\hat r}\cdot{\bf\hat r'}|f}= \sum_l \braket{{\bf\hat r}\cdot{\bf\hat r'}|P_l}{2l+1\over2}\braket{P_l|f}= \sum_{lm}\braket{{\bf\hat r}|lm}\braket{lm|{\bf\hat r'}}{(2l+1)^2\over8\pi} \braket{P_l|f}=$

$=\sum_{lm}\braket{{\bf\hat r}|lm}f_l \braket{lm|{\bf\hat r'}}$

where

$f_l={(2l+1)^2\over8\pi}\braket{P_l|f} ={(2l+1)^2\over8\pi}\int_{-1}^1 \braket{P_l|x}\braket{x|f}\d x ={(2l+1)^2\over8\pi}\int_{-1}^1 P_l(x)f(x)\d x$

or written explicitly

(7) $f({\bf\hat r}\cdot{\bf\hat r'})= \sum_{l=0}^\infty\sum_{m=-l}^l Y_{lm}({\bf\hat r}) f_l Y_{lm}^*({\bf\hat r'})$

Homogeneous functions¶

A function of several variables $f(x_1, x_2, \dots) \equiv f(x_i)$ is homogeneous of degree $k$ if

$f(\lambda x_i) = \lambda^k f(x_i)$

By differentiating with respect to $\lambda$ :

$x_i {\partial f(\lambda x_i)\over\partial x_i} = k\lambda^{k-1} f(x_i)$

and setting $\lambda=1$ we get the so called Euler equation:

$x_i {\partial f(x_i)\over\partial x_i} = k f(x_i)$

in 3D this can also be written as:

${\bf x} \cdot \nabla f({\bf x}) = k f({\bf x})$

Example 1¶

The function $f(x, y, z) = {xy\over z}$ is homogeneous of degree 1, because:

$f(\lambda x, \lambda y, \lambda z) = {\lambda x\lambda y\over \lambda z} = \lambda {xy\over z} = \lambda f(x, y, z)$

and the Euler equation is:

$x{\partial f\over\partial x} + y{\partial f\over\partial y} + z{\partial f\over\partial z} = f$

or

$x{y\over z} + y{x\over z} + z\left(-{xy\over z^2}\right) ={xy\over z}$

Which is true.

Example 2¶

The function $V(r) = -{Z \over r}$ is homogeneous of degree -1, because:

$V(\lambda r) = -{Z\over \lambda r} = \lambda^{-1} V(r)$

and the Euler equation is:

$r{\d V\over\d r} = -V$

or

$r{Z\over r^2} = -\left(-{Z\over r}\right)$

Which is true.

Green Functions¶

Green functions are an excellent tool for working with a solution to any ODE or PDE. In this text we explain how it works and then show how one can calculate them using FEM.

Introduction¶

Let’s put any ODE or PDE in the form:

(8) $Lu(x)=f(x)$

Here $L$ is a differential operator and $x$ can have any dimension, e.g. 1D (ODE), 2D, 3D or more (PDE). Then we can express the solution as

(9) $u(x) = L^{-1}f(x) = \int G(x, x')f(x')\d x'$

where $G(x, x')$ is a Green function, that needs to satisfy the equation:

(10) $LG(x, x')=\delta(x-x')$

Remember, that $L$ acts on $x$ only, so we can check, that (9) indeed solves the PDE (8):

$Lu(x) = L\int G(x, x')f(x')\d x' = \int LG(x, x')f(x')\d x' = \int \delta(x-x')f(x')\d x' = f(x)$

Boundary Conditions¶

The equation (10) doesn’t determine the Green function uniquely, because one can add to it any solution of the homogeneous equation $Lu(x)=0$ . We can use this freedom to solve (10) for any boundary condition. So we prescribe a boundary condition and find the Green function (by solving (10)) that satisfies the boundary condition. It can be shown, that $u(x)$ determined from (9) then also needs to satisfy the same boundary condition.

Symmetry¶

We write the equation for Green functions at two different points $x_1$ and $x_2$ :

$L G(x, x_1) = \delta(x-x_1) L G(x, x_2) = \delta(x-x_2)$

and multiply the first equation by $G(x, x_2)$ , second by $G(x, x_1)$ :

$G(x, x_2) L G(x, x_1) = \delta(x-x_1) G(x, x_2) G(x, x_1) L G(x, x_2) = \delta(x-x_2) G(x, x_1)$

substract them and integrate over $x$ :

$G(x, x_2) L G(x, x_1) - G(x, x_1) L G(x, x_2) = \delta(x-x_1) G(x, x_2) - \delta(x-x_2) G(x, x_1) \int \left(G(x, x_2) L G(x, x_1) - G(x, x_1) L G(x, x_2) \right) \d x = \int \left(\delta(x-x_1) G(x, x_2) - \delta(x-x_2) G(x, x_1) \right)\d x \int \left(G(x, x_2) L G(x, x_1) - G(x, x_1) L G(x, x_2) \right) \d x = G(x_1, x_2) - G(x_2, x_1)$

Assuming that the operator $L$ is Hermitean, we get:

$\int \left((L G(x, x_2)) G(x, x_1) - G(x, x_1) L G(x, x_2) \right) \d x = G(x_1, x_2) - G(x_2, x_1) 0 = G(x_1, x_2) - G(x_2, x_1)$

So the Green function is symmetric for Hermitean operators $L$ .

Examples¶

Poisson Equation in 1D¶

Poisson equation:

$-{\d^2\over\d x^2} u(x) = f(x)$

We calculate the Green function using the Fourier transform:

$-{\partial^2\over\partial x^2} G(x, x') = \delta(x-x') -(ik)^2 \tilde G(k, x') = {e^{ikx'}\over\sqrt{2\pi}} \tilde G(k, x') = {e^{ikx'}\over\sqrt{2\pi}\,k^2} G(x, x') = -\half (x-x') \sign(x-x') = -\half (x-x') (2H(x-x')-1) = H(x-x') (x'-x) +\half(x-x')$

Check:

${\partial\over\partial x} G(x, x') = \delta(x-x') (x'-x) + H(x-x') (-1) + \half = - H(x-x') + \half {\partial^2\over\partial x^2} G(x, x') = - \delta(x-x')$

Then:

$u(x) = \int G(x, x') f(x') \d x' = \int \left(H(x-x') (x'-x) +\half(x-x')\right) f(x') \d x'$

The green function can also be written using $x_<=\min(x, x')$ and $x_> = \max(x, x')$ :

$G(x, x') = H(x-x') (x'-x) +\half(x-x') = \half(x_< - x_>)$

Radial Poisson Equation¶

Let’s write $r_>$ and $r_<$ using the Heaviside step function:

$r_> = \max(r, r') = \begin{cases} r&\quad\mbox{for $r>r'$}\cr r'&\quad\mbox{for $r<r'$}\cr \end{cases} = H(r-r') r + H(r'-r)r' = = H(r-r') r + (1-H(r-r'))r' = H(r-r') (r-r') + r'$

and:

$r_< = \min(r, r') = \begin{cases} r'&\quad\mbox{for $r>r'$}\cr r&\quad\mbox{for $r<r'$}\cr \end{cases} = H(r-r') r' + H(r'-r)r = = H(r-r') r' + (1-H(r-r'))r = H(r-r') (r'-r) + r$

Then we can differentiate:

${\partial\over\partial r} r_> = \delta(r-r')(r-r') + H(r-r') = H(r-r') {\partial^2\over\partial r^2} r_> = \delta(r-r') {\partial\over\partial r} r_< = \delta(r-r')(r'-r) - H(r-r') + 1 = 1-H(r-r') = H(r'-r) {\partial^2\over\partial r^2} r_< = -\delta(r-r')$

Given:

(11) $u(r) = \int_0^\infty {f(r') r'^2 \over r_>} \d r'$

The Green function is

$G(r, r') = {r'^2\over r_>}$

Let’s differentiate:

${\partial\over\partial r} G(r, r') = -{r'^2\over r_>^2} {\partial\over\partial r} r_> = -{r'^2\over r_>^2} H(r-r') = -{r'^2\over r^2} H(r-r')$

and

${\partial^2\over\partial r^2} G(r, r') = +{2\over r}{r'^2\over r^2} H(r-r') -{r'^2\over r^2} \delta(r-r')$

So we get:

$-{\partial^2\over\partial r^2} G(r, r') -{2\over r}{\partial\over\partial r} G(r, r') =-{2\over r}{r'^2\over r^2} H(r-r') +{r'^2\over r^2} \delta(r-r') +{2\over r}{r'^2\over r^2} H(r-r') ={r'^2\over r^2} \delta(r-r') = \delta(r-r')$

So $u(r)$ from (11) is a solution to the radial Poisson equation:

$-{\d^2\over\d r^2} u(r) - {2\over r}{\d\over\d r}u(r) = f(r)$

Helmholtz Equation in 1D¶

$\left({\d^2\over\d x^2}+1\right)u(x)=f(x) \left({\d^2\over\d x^2}+1\right)G(x, x')=\delta(x-x')$

with boundary conditions $u(0)=u({\pi\over2})=0$ . We use the Fourier transform:

$\left((ik)^2+1\right)\tilde G(k, x')={e^{ikx'}\over\sqrt{2\pi}} \tilde G(k, x')={e^{ikx'}\over \sqrt{2\pi} (1-k^2)} G(x, x') = \half\sign(x-x')\sin(x-x')$

Check:

${\partial\over\partial x} G(x, x') = \delta(x-x')\sin(x-x') +\half\sign(x-x')\cos(x-x') = = \half\sign(x-x')\cos(x-x') {\partial^2\over\partial x^2} G(x, x') = \delta(x-x')\cos(x-x') -\half\sign(x-x')\sin(x-x') = = -\half\sign(x-x')\sin(x-x') + \delta(x-x') {\partial^2\over\partial x^2} G(x, x') +{\partial\over\partial x} G(x, x') = \delta(x-x')$

The general solution of the homogeneous equation is:

$u(x) = C_1 \sin x + C_2 \cos x$

so the general Green function is:

$G(x, x') = \half\sign(x-x')\sin(x-x') + C_1 \sin (x+x') + C_2 \cos (x+x')$

Satisfying the boundary conditions (for all $x' \ne 0$ ):

$G(0, x') = G({\pi\over 2}, x') = 0$

we get:

$C_1 & = -\half \\ C_2 & = 0$

and:

$G(x, x') = \half\sign(x-x')\sin(x-x') -\half \sin (x+x') = =-H(x'-x)\sin x\cos x' - H(x - x') \cos x \sin x' = = \begin{cases} -\sin x\cos x'&\quad x<x'\cr -\cos x\sin x'&\quad x>x'\cr \end{cases} =-\sin x_< \cos x_>$

and

$u(x) = \int G(x, x')f(x')\d x' =-\cos x\int_0^xf(x')\sin x'\d x' -\sin x\int_x^{\pi\over2}f(x')\cos x'\d x'$

To show that this really works, let’s take for example $f(x)=3\sin2x$ . Then

$u(x) =-\cos x\int_0^x3\sin 2x'\sin x'\d x' -\sin x\int_x^{\pi\over2}3\sin 2x'\cos x'\d x'$

We can use SymPy to evaluate the integrals:

In [1]: u = -cos(x)*integrate(3*sin(2*y)*sin(y), (y, 0, x)) - \
    sin(x)*integrate(3*sin(2*y)*cos(y), (y, x, pi/2))

In [2]: u
Out[2]:
-(cos(x)*sin(2*x) - 2*cos(2*x)*sin(x))*cos(x) - (sin(x)*sin(2*x)
    + 2*cos(x)*cos(2*x))*sin(x)

In [3]: simplify(u)
Out[3]:
     2                  2
- cos (x)*sin(2*x) - sin (x)*sin(2*x)

In [4]: trigsimp(_)
Out[4]: -sin(2*x)

And we get

$u(x)=-\sin 2x$

We can easily check, that $u''+u=3\sin2x$ :

>>> u = -sin(2*x)
>>> u.diff(x, 2) + u
3*sin(2*x)

and since $f(x) = 3\sin2x$ , we have verified, that $u(x)=-\sin 2x$ is the correct solution.

Poisson Equation in 2D¶

Let ${\bf x}=(x, y)$ and we want to solve:

$\nabla^2u({\bf x})=f({\bf x})$

So we have:

$\nabla^2G({\bf x}, {\bf x'})=\delta({\bf x}-{\bf x'})$

The solution is:

$G({\bf x}, {\bf x'}) ={1\over2\pi}\log|{\bf x}-{\bf x'}| ={1\over4\pi}\log|{\bf x}-{\bf x'}|^2 ={1\over4\pi}\log((x-x')^2+(y-y')^2)$

Poisson Equation in 3D¶

$\nabla^2u(x)=f(x) \nabla^2G(x, x')=\delta(x-x')$

with boundary condition $G(x) = 0$ at infinity. Then:

$G(x, x')=-{1\over4\pi}{1\over|x-x'| }$

and

$u(x) = -{1\over4\pi}\int {f(x')\over|x-x'| }\d x'$

Helmholtz Equation in 3D¶

$(\nabla^2+k^2)u(x)=f(x) (\nabla^2+k^2)G(x, x')=\delta(x-x')$

with boundary condition $G(x) = 0$ at infinity. Then:

$G(x, x')=-{1\over4\pi}{e^{ik|x-x'| }\over|x-x'| } u(x) = -{1\over4\pi}\int {f(x')e^{ik|x-x'| }\over|x-x'| }\d x'$

Finite Element Method¶

Let’s show it on the Laplace equation. We want to solve:

$\nabla^2G(x, x')=\delta(x-x')$

We will treat $x'$ as a parameter, so we define $g_{x'}(x)\equiv G(x, x')$ :

$\nabla^2g_{x'}(x)=\delta(x-x')$

We set $g_{x'}(x)=0$ on the boundary and we get:

$-\int\nabla g_{x'}(x) \cdot \nabla v(x) \d x = \int v(x)\delta(x-x')\d x -\int\nabla g_{x'}(x) \cdot \nabla v(x) \d x = v(x')$

So we choose $x'$ and then solve for $g_{x'}(x)$ using FEM and we get the Green function $G(x, x')$ for all $x$ and one particular $x'$ . We can then evaluate the integral (9) numerically – one would have to use FEM for all $x'$ that are needed in the integral, so that is not efficient, but it should work. One will then be able to play with Green functions and be able to calculate them numerically for any boundary condition (which is not possible analytically).

Binomial Coefficients¶

For $n$ and $k$ integers, the binomial coefficients are defined by:

$\binom{n}{k} = {n!\over k! (n-k)!} = {n (n-1) \cdots (n-k+1)\over k!}$

For $r$ real, one just uses the second formula as a definition:

$\binom{r}{k} = {r (r-1) \cdots (r-k+1)\over k!}$

Example I:

$\binom{-n}{k} = {(-n) (-n-1) \cdots (-n-k+1)\over k!} = (-1)^{k} {n (n+1) \cdots (n+k-1)\over k!} = (-1)^{k}\binom{n+k-1}{k}$

Example II:

$\binom{k-\half}{k} = {(k-\half) (k-\half-1) \cdots (k-\half-k+1)\over k!} = {(k-\half) (k-\half-1) \cdots \half\over k!} = = {(2k-1) (2k-3) \cdots 1\over 2^k k!} = {(2k-1)!!\over 2^k k!} = {(2k)!\over (2^k k!)^2} = {1\over 4^k}\binom{2k}{k}$

The binomial formula is for $n$ integer:

$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$

and for $r$ real and $|x| < |y|$ this can be generalized to:

$(x+y)^r = \sum_{k=0}^\infty \binom{r}{k} x^k y^{r-k}$

Example: (for $|x| < 1$ )

$(1 + x)^{-n} = \sum_{k=0}^\infty \binom{-n}{k} x^k = \sum_{k=0}^\infty (-1)^k\binom{n+k-1}{k} x^k = \sum_{k=0}^\infty \binom{n+k-1}{k} (-x)^k$

so:

$(1 - x)^{-n} = \sum_{k=0}^\infty \binom{n+k-1}{k} x^k (1 - x)^{-1} = \sum_{k=0}^\infty \binom{1+k-1}{k} x^k = \sum_{k=0}^\infty \binom{k}{k} x^k = \sum_{k=0}^\infty x^k (1 - x)^{-\half} = \sum_{k=0}^\infty \binom{\half+k-1}{k} x^k = \sum_{k=0}^\infty \binom{k-\half}{k} x^k = \sum_{k=0}^\infty {1\over 4^k}\binom{2k}{k} x^k$

Another example:

${1\over\sqrt{1-2xt+t^2}} = \left(1-(2xt-t^2)\right)^{-\half} = \sum_{n=0}^\infty {1\over 4^n}\binom{2n}{n} (2xt-t^2)^n = = \sum_{n=0}^\infty {1\over 4^n}\binom{2n}{n} \sum_{k=0}^n \binom{n}{k} (-t^2)^k (2xt)^{n-k} = = \sum_{n=0}^\infty {1\over 4^n}\binom{2n}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k t^{2k} (2x)^{n-k} t^{n-k} = = \sum_{n=0}^\infty {1\over 4^n}\binom{2n}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k t^{n+k} (2x)^{n-k} = = \sum_{n=0}^\infty \sum_{k=0}^{\left\lfloor n\over 2 \right\rfloor} {1\over 4^{n-k}}\binom{2n-2k}{n-k} \binom{n-k}{k} (-1)^k t^{n} (2x)^{n-2k} = = \sum_{n=0}^\infty \left( {1\over 2^n} \sum_{k=0}^{\left\lfloor n\over 2 \right\rfloor} (-1)^k \binom{n}{k} \binom{2n-2k}{n} x^{n-2k} \right) t^{n} = = \sum_{n=0}^\infty \left( {1\over 2^n} \sum_{k=0}^{\left\lfloor n\over 2 \right\rfloor} (-1)^k \binom{n}{k} {1\over n!}{\d^n \over \d x^n} x^{2n-2k} \right) t^{n} = = \sum_{n=0}^\infty \left( {1\over 2^n} \sum_{k=0}^n (-1)^k \binom{n}{k} {1\over n!}{\d^n \over \d x^n} x^{2n-2k} \right) t^{n} = = \sum_{n=0}^\infty \left( {1\over 2^n n!} {\d^n \over \d x^n} (x^2 - 1)^n \right) t^{n} = = \sum_{n=0}^\infty P_n(x) t^{n}$

where we used:

$\sum_{n=0}^\infty \sum_{k=0}^n a_{n k} = \sum_{n=0}^\infty \sum_{k=0}^{\left\lfloor n\over 2 \right\rfloor} a_{n-k, k}$

and

$\binom{2n-2k}{n-k} \binom{n-k}{k} = \binom{2n-2k}{n} \binom{n}{k}$

The $P_n(x)$ are Legendre Polynomials.

Triangle Inequality¶

Triangle inequality (condition) means that none of the three quantities $a$ , $b$ , $c$ is greater than the sum of the other two:

(12) $a + b \ge c b + c \ge a c + a \ge b$

This is equivalent to just one equation:

(13) $|a-b| \le c \le a+b$

we can do any permutation of the symbols, i.e. the above equation is equivalent to any of these:

$|b-c| \le a \le b+c |c-a| \le b \le c+a$

So instead of stating the three inequalities (12) it is more convenient to just write (13), using any permutation that we like.

To show, that (12) implies (13) we rewrite (12):

$a + b \ge c c \ge a-b c \ge b-a$

so

$a + b \ge c c \ge |a-b|$

and we get (13). To show, that (13) implies (12) we rewrite (13) for $a\ge b$ first:

$a \ge b |a-b| \le c \le a+b$

so:

$a \ge b a-b \le c \le a+b$

rearranging:

$a + b \ge c b + c \ge a a \ge b$

since $c$ is positive, if $a\ge b$ then also $c+a\ge b$ and we get (12). Finally, for $a < b$ :

$a < b |a-b| \le c \le a+b$

so:

$a < b -(a-b) \le c \le a+b$

rearranging:

$a + b \ge c b > a c + a \ge b$

since $c$ is positive, if $b > a$ then also $b+c\ge a$ and we get (12).

Gamma Function¶

The Gamma function $\Gamma(x)$ is defined by the following properties for $x > 0$ :

(14) $\Gamma(1) = 1$

(15) $\Gamma(x+1) = x \Gamma(x)$

(16) $\log\Gamma(x) \quad\mbox{is convex}$

It can be shown that this determines the function uniquely for $x > 0$ (this is called the Bohr-Mollerup theorem) and then it can be extended analytically to the whole complex plane.

The most common formula for $\Gamma(z)$ that satisfies (14), (15) and (16) is:

(17) $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \d t$

It satisfies (14) because:

$\Gamma(1) = \int_0^\infty t^{1-1} e^{-t} \d t = \int_0^\infty e^{-t} \d t = [-e^{-t}]_0^\infty = 1$

It satisfies (15) by integrating by parts:

$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \d t = (z-1)\int_0^\infty t^{z-2} e^{-t} \d t-[t^{z-1}e^{-t}]_0^\infty = (z-1)\Gamma(z-1)$

Finally it satisfies (16) by verifying the convex condition directly ( $x, y > 0$ and $0\le \lambda \le 1$ ):

$\log\Gamma(\lambda x + (1-\lambda) y) = \log\int_0^\infty t^{\lambda x + (1-\lambda) y - 1} e^{-t} \d t = = \log \int_0^\infty (t^{x-1} e^{-t})^\lambda (t^{y-1} e^{-t})^{1-\lambda} \d t \le \le \log \left( \left(\int_0^\infty t^{x-1} e^{-t} \d t\right)^\lambda \left(\int_0^\infty t^{y-1} e^{-t} \d t\right)^{1-\lambda}\right) = = \lambda \log\Gamma(x) + (1-\lambda) \log \Gamma(y)$

And thus (17) uniquely determines the Gamma function. We can use (17) to calculate $\Gamma(\half)$ :

$\Gamma(\half) = \int_0^\infty t^{{1\over2}-1} e^{-t} \d t = \int_0^\infty {e^{-t}\over\sqrt t} \d t = \int_0^\infty {e^{-x^2}\over x} 2x\, \d x = 2\int_0^\infty e^{-x^2} \d x = = \int_{-\infty}^\infty e^{-x^2} \d x = \sqrt{ \int_{-\infty}^\infty e^{-x^2} \d x \int_{-\infty}^\infty e^{-y^2} \d y } = \sqrt{2\pi \int_0^\infty e^{-r^2} r\d r} = = \sqrt{2\pi \int_0^\infty e^{-u} \half\d u} = \sqrt\pi$

From this and the definition of the Gamma function we get for integer $n$ :

(18) $\Gamma(n+1) = n\Gamma(n) = n(n-1)\Gamma(n-1) = n(n-1)(n-2)\cdots 2\cdot1\cdot\Gamma(1) = = n(n-1)(n-2)\cdots 1 = n!$

and

(19) $\Gamma(n+\half) = (n-\half)\Gamma(n-\half) = (n-\half)(n-1-\half)\Gamma(n-1-\half) = (n-\half)(n-1-\half)\cdots\half\Gamma(\half) = = {2n-1\over2}{2n-3\over2}{2n-5\over2}\cdots{1\over2}\Gamma(\half) = {(2n-1)!!\over 2^n}\Gamma(\half) = {(2n-1)!!\over 2^n}\sqrt\pi$

Factorial¶

The factorial $n!$ is defined as

$n! = n(n-1)(n-2)\cdots 3\cdot2\cdot 1$

By (18) it can be written using the Gamma function as:

$n! = \Gamma(n+1)$

Double Factorial¶

The double factorial $n!!$ is defined as:

$n!! = \begin{cases} n(n-2)(n-4)(n-6)\cdots 5\cdot3\cdot1\quad\quad\mbox{for odd $n=2k+1$} \\ n(n-2)(n-4)(n-6)\cdots 6\cdot4\cdot2\quad\quad\mbox{for even $n=2k$} \end{cases}$

One can rewrite double factorial using a factorial as:

$(2k)!! = 2\cdot4\cdot6\cdots (2k) = 2^k (1\cdot2\cdot3\cdots k) = 2^k k! (2k-1)!! = 1\cdot3\cdot5\cdots(2k-1) = {1\cdot2\cdot3\cdot4\cdot5\cdots(2k) \over 2\cdot4\cdot6\cdots (2k)} = {(2k)!\over (2k)!!} = {(2k)!\over 2^k k!}$

For odd $n$ it can be written using the Gamma function, see (19):

$(2k-1)!! = {1\over\sqrt\pi} 2^k \Gamma\left(k+\half\right)$

Example¶

$A(n) = {1\cdot3\cdot5 \cdot \dots \cdot (2n-1) \over 1\cdot 2\cdot 3\cdot \dots \cdot n} = {(2n-1)!!\over n!} = {(2n)!\over 2^n (n!)^2} = {1\over 2^n}\binom{2n}{n} B(n) = {1\cdot3\cdot5 \cdot \dots \cdot (2n-1) \over 2\cdot 4\cdot\cdot6 \dots \cdot 2n} = {(2n-1)!!\over (2n)!!} = {(2n)!\over (2^n n!)^2} = {1\over 4^n}\binom{2n}{n}$

Delta Function¶

Distributions¶

Variations and Functional Derivatives¶

Functions of One Variable¶

Functions of several variables¶

Functionals¶

Dirac Notation¶

Homogeneous functions¶

Example 1¶

Example 2¶

Green Functions¶

Introduction¶

Boundary Conditions¶

Symmetry¶

Examples¶

Poisson Equation in 1D¶

Radial Poisson Equation¶

Helmholtz Equation in 1D¶

Poisson Equation in 2D¶

Poisson Equation in 3D¶

Helmholtz Equation in 3D¶

Finite Element Method¶

Binomial Coefficients¶

Triangle Inequality¶

Gamma Function¶

Factorial¶

Double Factorial¶

Example¶

Table Of Contents

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Delta Function¶

Distributions¶

Variations and Functional Derivatives¶

Functions of One Variable¶

Functions of several variables¶

Functionals¶

Dirac Notation¶

Homogeneous functions¶

Example 1¶

Example 2¶

Green Functions¶

Introduction¶

Boundary Conditions¶

Symmetry¶

Examples¶

Poisson Equation in 1D¶

Radial Poisson Equation¶

Helmholtz Equation in 1D¶

Poisson Equation in 2D¶

Poisson Equation in 3D¶

Helmholtz Equation in 3D¶

Finite Element Method¶

Binomial Coefficients¶

Triangle Inequality¶

Gamma Function¶

Factorial¶

Double Factorial¶

Example¶

Table Of Contents

Previous topic

Next topic

This Page

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