3.10. Periodic Functions

A function f(x) is periodic with period T:

f(x+T) = f(x)

Then you can shift the integration limits by the period T:

\int_a^b f(x) \d x = \int_a^b f(x+T) \d x = \int_{a+T}^{b+T} f(x) \d x

If you integrate f(x) from 0 to T, you can shift x in f(x) by any constant \alpha:

\int_0^T f(x+\alpha) \d x = = \int_\alpha^{T+\alpha} f(x) \d x = = \int_\alpha^0 f(x) \d x + \int_0^T f(x) \d x + \int_T^{T+\alpha} f(x) \d x = = -\int_0^\alpha f(x) \d x + \int_0^T f(x) \d x + \int_0^\alpha f(x) \d x = = \int_0^T f(x) \d x

3.11. Polar Coordinates

Polar coordinates (radial, azimuth) (r,\phi) are defined by

\begin{eqnarray*} x&=&r\cos\phi \\ y&=&r\sin\phi \\ \end{eqnarray*}

3.11.1. Example

When evaluating integrals of the type:

l(x, y) = \int_0^{2\pi} \sqrt{(x-r\cos\phi)^2 + (y-r\sin\phi)^2} \,\d \phi

we write x and y using polar coordinates:

x = r' \cos \phi' y = r' \sin \phi'

and then use the 2\pi periodicity of \cos x:

l(x, y) = \int_0^{2\pi} \sqrt{(x-r\cos\phi)^2 + (y-r\sin\phi)^2} \,\d \phi = = \int_0^{2\pi} \sqrt{x^2 + y^2 + r^2 - 2r(x\cos\phi + y\sin\phi)} \,\d \phi = = \int_0^{2\pi} \sqrt{r'^2 + r^2 - 2rr'(\cos\phi'\cos\phi + \sin\phi'\sin\phi)} \,\d \phi = = \int_0^{2\pi} \sqrt{r'^2 + r^2 - 2rr'\cos(\phi-\phi')} \,\d \phi = = \int_0^{2\pi} \sqrt{r'^2 + r^2 - 2rr'\cos\phi} \,\d \phi =

comparing to:

l(0, y) = \int_0^{2\pi} \sqrt{y^2 + r^2 - 2ry\sin\phi} \,\d \phi

we can see that because the integral is symmetric, we can just set x=0 and then replace y \to r'. The above method does everything algebraically, but you can use this symmetry argument to remember what to do, or even skip the calculation if you are sure that you didn’t make a mistake in the “symmetry argument”.

3.12. Spherical Coordinates

Spherical coordinates radial (\rho), zenith (\theta), azimuth (\phi):

(3.12.1)x &= \rho\sin\theta\cos\phi \\ y &= \rho\sin\theta\sin\phi \\ z &= \rho\cos\theta \\

Note: this meaning of (\theta,\phi) is mostly used in the USA and in many books. In Europe people usually use different symbols, like (\phi,\theta), (\vartheta,\varphi) and others.

The motivation is to first write x and y using polar coordinates:

x = \rho_{xy}\cos\phi y = \rho_{xy}\sin\phi

and then write z and the projection \rho_{xy} of \rho onto the plane x-y using polar coordinates:

z = \rho\cos\theta \rho_{xy} = \rho\sin\theta

so by combining these two we get:

x = \rho_{xy}\cos\phi = \rho\sin\theta\cos\phi y = \rho_{xy}\sin\phi = \rho\sin\theta\sin\phi z = \rho\cos\theta

3.12.1. Example I

To transform differential operators such as \d\over\d x into spherical coordinates, we make use of the chain rule:

\frac{\partial}{\partial x}= \frac{\partial \rho}{\partial x} \frac{\partial}{\partial \rho} +\frac{\partial\theta}{\partial x} \frac{\partial}{\partial \theta} +\frac{\partial\phi}{\partial x} \frac{\partial}{\partial \phi}

where r, \theta and \phi are functions of x, y, z to be expressed by inverting (3.12.1):

\rho(x, y, z) &= \sqrt{x^2+y^2+z^2} \\ \theta(x, y, z) &= \arccos{z\over\sqrt{x^2+y^2+z^2}} \\ \phi(x, y, z) &= \arctan{y\over x} \\

At the end, the derivatives are expressed using \rho, \theta, \phi again. For example

{\partial\rho\over\partial x} = {\partial\sqrt{x^2+y^2+z^2}\over\partial x} = = {x \over \sqrt{x^2+y^2+z^2}} = = {\rho\sin\theta\cos\phi \over \rho} = \sin\theta\cos\phi

Finally we obtain

(3.12.1.1)\frac{\partial}{\partial x}= \sin\theta\cos\phi \frac{\partial}{\partial\rho} +\frac{\cos\theta\cos\phi}{\rho} \frac{\partial}{\partial \theta} - \frac{\sin\phi}{\rho\sin\theta} \frac{\partial}{\partial \phi} \frac{\partial}{\partial y}= \sin\theta\sin\phi \frac{\partial}{\partial\rho} +\frac{\cos\theta\sin\phi}{\rho} \frac{\partial}{\partial \theta} + \frac{\cos\phi}{\rho\sin\theta} \frac{\partial}{\partial \phi} \frac{\partial}{\partial z}= \cos\theta \frac{\partial}{\partial\rho} -\frac{\sin\theta}{\rho} \frac{\partial}{\partial \theta}

These expressions can be combined to obtain more complicated objects such as Laplacian (in spherical coordinates). However straightforward this approach is, it is also rather cumbersome; an alternative is discussed in the Spherical Coordinates section of differential geometry (where it is shown, that the coefficients in (3.12.1.1) are simply the matrix elements of the inverse Jacobian).

3.12.2. Example II

When evaluating integrals of the type:

l(x, y, z) = \int_0^\pi\d\theta \int_0^{2\pi} \d\phi \sqrt{(x-r\sin\theta\cos\phi)^2 + (y-r\sin\theta\sin\phi)^2 + (z-r\cos\theta)^2 } \,\sin\theta

we write x and y using polar coordinates:

x = \rho_{xy} \cos \phi' y = \rho_{xy} \sin \phi'

and simplify:

l(x, y, z) = \int_0^\pi\d\theta \int_0^{2\pi} \d\phi \sqrt{(x-r\sin\theta\cos\phi)^2 + (y-r\sin\theta\sin\phi)^2 + (z-r\cos\theta)^2 } \,\sin\theta = = \int_0^\pi\d\theta \int_0^{2\pi} \d\phi \sqrt{x^2 + y^2 + z^2 + r^2 -2r( x\sin\theta\cos\phi +y\sin\theta\sin\phi +z\cos\theta ) } \,\sin\theta = = \int_0^\pi\d\theta \int_0^{2\pi} \d\phi \sqrt{\rho_{xy} + z^2 + r^2 -2r( \rho_{xy}\cos\phi'\sin\theta\cos\phi +\rho_{xy}\sin\phi'\sin\theta\sin\phi +z\cos\theta ) } \,\sin\theta = = \int_0^\pi\d\theta \int_0^{2\pi} \d\phi \sqrt{\rho_{xy} + z^2 + r^2 -2r( \rho_{xy}\cos(\phi-\phi')\sin\theta +z\cos\theta ) } \,\sin\theta = = \int_0^\pi\d\theta \int_0^{2\pi} \d\phi \sqrt{\rho_{xy} + z^2 + r^2 -2r( \rho_{xy}\cos\phi\sin\theta +z\cos\theta ) } \,\sin\theta

comparing to:

l(0, 0, z) = \int_0^\pi\d\theta \int_0^{2\pi} \d\phi \sqrt{z^2 + r^2 -2rz\cos\theta } \,\sin\theta =

we can see that because the integral is symmetric, we can just set x=0, y=0 and then replace z \to \rho.

3.13. Argument function, atan2

Argument function \arg(z) is any \varphi such that

z = r e^{i\varphi}

Obviously \arg(z) is unique up to any integer multiple of 2\pi. By taking the principal value of the \arg(z) function, e.g. fixing \arg(z) to the interval (-\pi, \pi] (so that the branch cut is on the negative x-axis, as usual), we get the \Arg(z) function:

-\pi < \Arg z \le \pi

then \arg z = \Arg z + 2\pi n, where n=0, \pm 1, \pm 2, \dots. We can then use the following formula to easily calculate \Arg z for any z=x+iy (except x=y=0, i.e. z=0, where it is not defined):

\Arg(x+iy) =\begin{cases}\pi&y=0;x<0;\cr 2\,\atan{y\over\sqrt{x^2+y^2}+x}&\rm otherwise\cr\end{cases}

Finally we define \atan2(y, x) as:

\atan2(y, x) = \Arg(x+iy) = \begin{cases}\pi&y=0;x<0;\cr 2\,\atan{y\over\sqrt{x^2+y^2}+x}&\rm otherwise\cr\end{cases}

The angle \phi=\atan2(y, x) is the angle of the point (x, y) on the unit circle (assuming the usual conventions), and it works for all quadrants (\phi=\atan({y\over x}) only works for the first and fourth quadrant, where \atan({y\over x})=\atan2(y, x), but in the second and third qudrant, \atan({y\over x}) gives the wrong angles, while \atan2(y, x) gives the correct angles). So in particular:

\atan2(0, 1) = 2\,\atan{0\over\sqrt{1^2+0^2}+1} = 0 \atan2(0, -1) = \pi \atan2(1, 0) = 2\,\atan{1\over\sqrt{0^2+1^2}+0} = 2\,\atan 1 = {\pi\over 2} \atan2(-1, 0) = 2\,\atan{-1\over\sqrt{0^2+1^2}+0} = -2\,\atan 1 = -{\pi\over 2}

This convention (\atan2(y, x)) is used for example in Python, C or Fortran. Some people might interchange x with y in the definition (i.e. \atan2(x, y)= \Arg(y+ix)), but it is not very common.

The following useful relations hold:

\sin\atan2(y, x) = {y\over \sqrt{x^2+y^2}} \quad\quad\quad\mbox{except $x=y=0$} \cos\atan2(y, x) = {x\over \sqrt{x^2+y^2}} \quad\quad\quad\mbox{except $x=y=0$} \tan\atan2(y, x) = {y\over x} \quad\quad\quad\mbox{for $x\neq 0$} \atan2(ky, kx) = \atan2(y, x) \quad\quad\quad\mbox{for $k>0$} \atan2(\sin x, \cos x) = x + 2\pi \left\lfloor \pi-x \over 2\pi \right\rfloor \atan2(-y, x) = -\atan2(y, x) + 2\pi \left\lfloor \atan2(y, x)+\pi \over 2\pi \right\rfloor {\partial \over \partial y} \atan2(y, x) = {x\over x^2 + y^2} {\partial \over \partial x} \atan2(y, x) = -{y\over x^2 + y^2}

We now prove them. The following works for all x, y except for x=y=0:

\sin\atan2(y, x) =\begin{cases}\sin\pi&y=0;x<0;\cr \sin\left(2\,\atan{y\over\sqrt{x^2+y^2}+x}\right) &\rm otherwise\cr\end{cases} = =\begin{cases}0&y=0;x<0;\cr {y\over \sqrt{x^2+y^2}}&\rm otherwise\cr\end{cases} = =\begin{cases}{y\over \sqrt{x^2+y^2}}&y=0;x<0;\cr {y\over \sqrt{x^2+y^2}}&\rm otherwise\cr\end{cases} ={y\over \sqrt{x^2+y^2}} \cos\atan2(y, x) =\begin{cases}\cos\pi&y=0;x<0;\cr \cos\left(2\,\atan{y\over\sqrt{x^2+y^2}+x}\right) &\rm otherwise\cr\end{cases} = =\begin{cases}-1&y=0;x<0;\cr {x\over \sqrt{x^2+y^2}}&\rm otherwise\cr\end{cases} = =\begin{cases}{x\over \sqrt{x^2+y^2}}&y=0;x<0;\cr {x\over \sqrt{x^2+y^2}}&\rm otherwise\cr\end{cases} ={x\over \sqrt{x^2+y^2}}

Tangent is infinite for \pm{\pi\over 2}, which corresponds to x=0, so the following works for all x\neq 0:

\tan\atan2(y, x) =\begin{cases}\tan\pi&y=0;x<0;\cr \tan\left(2\,\atan{y\over\sqrt{x^2+y^2}+x}\right) &\rm otherwise\cr\end{cases} = =\begin{cases}0&y=0;x<0;\cr {y\over x}&\rm otherwise\cr\end{cases} = =\begin{cases}{y\over x}&y=0;x<0;\cr {y\over x}&\rm otherwise\cr\end{cases} ={y\over x}

Finally:

\atan2(\sin x, \cos x) =\begin{cases}\pi &x=\pi;\cr 2\,\atan{\sin x\over\sqrt{\cos^2x+\sin^2x}+\cos x}& \rm otherwise\cr\end{cases} = =\begin{cases}\pi &x=\pi;\cr 2\,\atan{\sin x\over1+\cos x}& \rm otherwise\cr\end{cases} = =\begin{cases}x &x=\pi;\cr 2\,\atan\left(\tan {x\over 2}\right)& \rm otherwise\cr\end{cases} = x + 2\pi \left\lfloor \pi-x \over 2\pi \right\rfloor

In the above, we used the following double angle formulas:

\sin 2x = {2\tan x\over 1+\tan^2 x} \cos 2x = {1-\tan^2x\over 1+\tan^2 x} \tan 2x = {2\tan x\over 1-\tan^2 x}

to simplify the following expressions:

\sin\left(2\,\atan{y\over\sqrt{x^2+y^2}+x}\right) = {2\tan\atan{y\over\sqrt{x^2+y^2}+x}\over1+\tan^2\atan{y\over\sqrt{x^2+y^2}+x}} = = {2{y\over\sqrt{x^2+y^2}+x}\over1 +\left({y\over\sqrt{x^2+y^2}+x}\right)^2} = {2y\left(\sqrt{x^2+y^2}+x\right)\over \left(\sqrt{x^2+y^2}+x\right)^2+y^2} = = {y\left(\sqrt{x^2+y^2}+x\right)\over x^2+y^2+x\sqrt{x^2+y^2}} = {y\left(\sqrt{x^2+y^2}+x\right)\over \sqrt{x^2+y^2}\left(\sqrt{x^2+y^2}+x\right)} = = {y\over\sqrt{x^2+y^2}} \cos\left(2\,\atan{y\over\sqrt{x^2+y^2}+x}\right) = {1-\tan^2\atan{y\over\sqrt{x^2+y^2}+x}\over1+\tan^2\atan{y\over\sqrt{x^2+y^2}+x}} = = {1 -\left({y\over\sqrt{x^2+y^2}+x}\right)^2\over 1 +\left({y\over\sqrt{x^2+y^2}+x}\right)^2} = {\left(\sqrt{x^2+y^2}+x\right)^2-y^2\over \left(\sqrt{x^2+y^2}+x\right)^2+y^2} = = {x\left(\sqrt{x^2+y^2}+x\right)\over x^2+y^2+x\sqrt{x^2+y^2}} = {x\left(\sqrt{x^2+y^2}+x\right)\over \sqrt{x^2+y^2}\left(\sqrt{x^2+y^2}+x\right)} = = {x\over\sqrt{x^2+y^2}} \tan\left(2\,\atan{y\over\sqrt{x^2+y^2}+x}\right) = {2\tan\atan{y\over\sqrt{x^2+y^2}+x}\over1-\tan^2\atan{y\over\sqrt{x^2+y^2}+x}} = = {2{y\over\sqrt{x^2+y^2}+x}\over1 -\left({y\over\sqrt{x^2+y^2}+x}\right)^2} = {2y\left(\sqrt{x^2+y^2}+x\right)\over \left(\sqrt{x^2+y^2}+x\right)^2-y^2} = = {y\left(\sqrt{x^2+y^2}+x\right)\over x\left(\sqrt{x^2+y^2}+x\right)} = {y\over x}

Finally, for all k>0 we get:

\atan2(ky, kx) = \Arg(kx + iky) =\begin{cases}\pi&y=0;x<0;\cr 2\,\atan{ky\over\sqrt{(kx)^2+(ky)^2}+kx}&\rm otherwise\cr\end{cases} = =\begin{cases}\pi&y=0;x<0;\cr 2\,\atan{y\over\sqrt{x^2+y^2}+x}&\rm otherwise\cr\end{cases} = \Arg(x+iy) = \atan2(y, x)

The symmetry property can be proven by:

\atan2(-y, x) =\begin{cases}\pi&y=0;x<0;\cr 2\,\atan{-y\over\sqrt{x^2+(-y)^2}+x} &\rm otherwise\cr\end{cases} = =\begin{cases}\pi&y=0;x<0;\cr -\left(2\,\atan{y\over\sqrt{x^2+y^2}+x}\right) &\rm otherwise\cr\end{cases} = =-\atan2(y, x) + 2\pi \left\lfloor \atan2(y, x)+\pi \over 2\pi \right\rfloor

To prove the derivatives, we do:

{\partial \over \partial y} \atan2(y, x) = 2{\partial \over \partial y} \atan{y\over\sqrt{x^2+y^2}+x} = {x\over x^2 + y^2} {\partial \over \partial x} \atan2(y, x) = 2{\partial \over \partial x} \atan{y\over\sqrt{x^2+y^2}+x} = -{y\over x^2 + y^2}

Code:

>>> from sympy import atan, sqrt, var
>>> var("x y")
(x, y)
>>> (2*atan(y/(sqrt(x**2+y**2)+x)).diff(y)).simplify()
x/(x**2 + y**2)
>>> (2*atan(y/(sqrt(x**2+y**2)+x)).diff(x)).simplify()
-y/(x**2 + y**2)

An example of an application:

A\sin x + B\cos x = \sqrt{A^2+B^2}\left( {A\over\sqrt{A^2+B^2}}\sin x + {B\over\sqrt{A^2+B^2}}\cos x\right) = = \sqrt{A^2+B^2}\left( \cos\delta\sin x + \sin\delta\cos x\right) = \sqrt{A^2+B^2}\sin(x+\delta) = = \sqrt{A^2+B^2}\sin(x+\atan2(B, A))

where

\delta = \atan2\left({B\over\sqrt{A^2+B^2}}, {A\over\sqrt{A^2+B^2}}\right) =\atan2(B, A)

Another application

\atan2(\cos x, -\sin x) = \atan2\left(\sin\left(x+{\pi\over2}\right), \cos\left(x+{\pi\over2}\right)\right) = x + {\pi\over2}

3.14. Multiple Argument Formulas

3.14.1. sin(a x)

Systematic way to derive all multiple argument formulas is to use the following relation:

\sin(ax) = U_{a-1}(\cos x) \sin x

where U_n(x) are the Chebyshev polynomials of the second kind, first few are:

U_{-3}(x) = -2x U_{-2}(x) = -1 U_{-1}(x) = 0 U_{- {1\over2}}(x) = {1\over \sqrt 2 \sqrt{x+1}} U_0(x) = 1 U_{1\over2}(x) = {2x+1\over \sqrt 2 \sqrt{x+1}} U_1(x) = 2x U_2(x) = 4x^2 - 1 U_3(x) = 8x^3 - 4x U_4(x) = 16x^4 - 12x^2 + 1 U_5(x) = 32x^5 - 32x^3 + 6x U_6(x) = 64x^6 - 80x^4 + 24x^2 - 1

Code:

>>> from sympy import chebyshevu, var
>>> var("x")
>>> for i in range(7): print "U_%d(x) = %s" % (i, chebyshevu(i, x))
U_0(x) = 1
U_1(x) = 2*x
U_2(x) = -1 + 4*x**2
U_3(x) = -4*x + 8*x**3
U_4(x) = 1 - 12*x**2 + 16*x**4
U_5(x) = 6*x - 32*x**3 + 32*x**5
U_6(x) = -1 + 24*x**2 - 80*x**4 + 64*x**6

One can then use this to calculate:

\sin (-2x) = U_{-3}(\cos x) \sin x = -2\cos x\sin x \sin (-x) = U_{-2}(\cos x) \sin x = -\sin x \sin 0 = U_{-1}(\cos x) \sin x = 0 \sin {x\over 2} = U_{-{1\over2}}(\cos x) \sin x = {\sin x\over\sqrt 2\sqrt{\cos x + 1}} = {\sqrt{1-\cos^2x}\over\sqrt 2\sqrt{\cos x + 1}} = {\sqrt{1-\cos x}\over\sqrt 2} \sin x = U_0(\cos x) \sin x = \sin x \sin {3x\over 2} = U_{1\over2}(\cos x) \sin x = {(2\cos x+1)\sin x\over\sqrt 2\sqrt{\cos x + 1}} = {(2\cos x+1)\sqrt{1-\cos^2x}\over\sqrt 2\sqrt{\cos x + 1}} = {(2\cos x+1)\sqrt{1-\cos x}\over\sqrt 2} \sin 2x = U_1(\cos x) \sin x = 2\cos x\sin x \sin 3x = U_2(\cos x) \sin x = (4\cos^2 x-1)\sin x

Code:

>>> from sympy import chebyshevu, var, sin, cos
>>> var("x")
>>> for n in range(1, 7): print "sin(%d*x) = %s" % (n, chebyshevu(n-1, cos(x))*sin(x))
sin(1*x) = sin(x)
sin(2*x) = 2*cos(x)*sin(x)
sin(3*x) = -(1 - 4*cos(x)**2)*sin(x)
sin(4*x) = (-4*cos(x) + 8*cos(x)**3)*sin(x)
sin(5*x) = (1 - 12*cos(x)**2 + 16*cos(x)**4)*sin(x)
sin(6*x) = (6*cos(x) - 32*cos(x)**3 + 32*cos(x)**5)*sin(x)

3.14.2. cos(a x)

Similarly as above, we use:

\cos(ax) = T_a(\cos x)

where T_n(x) are the Chebyshev polynomials of the first kind, first few are:

T_0(x) = 1 T_{1\over2}(x) = {\sqrt{x+1}\over \sqrt 2} T_1(x) = x T_{3\over2}(x) = {(2x-1)\sqrt{x+1}\over \sqrt 2} T_2(x) = 2x^2 - 1 T_3(x) = 4x^3 - 3x T_4(x) = 8x^4 - 8x^2 + 1 T_5(x) = 16x^5 - 20x^3 + 5x T_6(x) = 32x^6 - 48x^4 + 18x^2 - 1

Code:

>>> from sympy import chebyshevt, var
>>> var("x")
>>> for i in range(7): print "T_%d(x) = %s" % (i, chebyshevt(i, x))
T_0(x) = 1
T_1(x) = x
T_2(x) = -1 + 2*x**2
T_3(x) = -3*x + 4*x**3
T_4(x) = 1 - 8*x**2 + 8*x**4
T_5(x) = 5*x - 20*x**3 + 16*x**5
T_6(x) = -1 + 18*x**2 - 48*x**4 + 32*x**6

One can then use this to calculate:

\cos 0 = T_0(\cos x) = 1 \cos {x\over 2} = T_{1\over 2}(\cos x) = {\sqrt{1+\cos x}\over\sqrt 2} \cos x = T_1(\cos x) = \cos x \cos {3x\over 2} = T_{3\over2}(\cos x) = {(2\cos x-1)\sqrt{1+\cos x}\over\sqrt 2} \cos 2x = T_2(\cos x) = 2\cos^2 x - 1 \cos 3x = T_3(\cos x) = 4\cos^3 x - 3\cos x

Code:

>>> from sympy import chebyshevt, var, cos
>>> var("x")
>>> for n in range(7): print "cos(%d*x) = %s" % (n, chebyshevt(n, cos(x)))
cos(0*x) = 1
cos(1*x) = cos(x)
cos(2*x) = -1 + 2*cos(x)**2
cos(3*x) = -3*cos(x) + 4*cos(x)**3
cos(4*x) = 1 - 8*cos(x)**2 + 8*cos(x)**4
cos(5*x) = 5*cos(x) - 20*cos(x)**3 + 16*cos(x)**5
cos(6*x) = -1 + 18*cos(x)**2 - 48*cos(x)**4 + 32*cos(x)**6