# The Lumber Room

"Consign them to dust and damp by way of preserving them"

## The functional equation f(x+y) = f(x)f(y)

Suppose $f: \mathbb{R} \to \mathbb{R}$ satisfies $f(x+y) = f(x) f(y)$. What can we say about $f$?

Putting $y = 0$ gives

$\displaystyle f(x) = f(x+0) = f(x)f(0),$

which can happen if either $f(x) = 0$ or $f(0) = 1$. Note that the function $f$ which is identically zero satisfies the functional equation. If $f$ is not this function, i.e., if $f(x) \neq 0$ for at least one value of $x$, then plugging that value of $x$ (say $x^*$) into the equation gives $f(0) = 1$. Also, for any $x$, the equation $f(x^*) = f(x +x^* - x) = f(x)f(x^* - x)$ forces $f(x) \neq 0$ as well. Further, $f(x) = f(x/2 + x/2) = f(x/2)^2$ so $f(x) > 0$ for all $x$.

Next, putting $y = x$ gives $f(2x) = f(x)^2$, and by induction $f(nx) = f(x)^n$. Putting $\frac{x}{n}$ in place of $x$ in this gives $f(n\frac{x}{n}) = f(\frac{x}{n})^n$ which means $f(\frac{x}{n}) = f(x)^{\frac1n}$ (note we’re using $f(x) > 0$ here). And again, $f(\frac{m}{n}x) = f(x)^{m/n}$. So $f(\frac{m}{n}) = f(1)^{m/n}$, which completely defines the function at rational points.

[As $f(1) > 0$, it can be written as $f(1) = e^k$ for some constant $k$, which gives $f(x) = e^{kx}$ for rational $x$.]

To extend this function to irrational numbers, we need some further assumptions on $f$, such as continuity. It turns out that being continuous at any point is enough (and implies the function is $f(x) = f(1)^x$ everywhere): note that $f(x + m/n) = f(x)f(m/n) = f(x)f(1)^{m/n}$. Even being Lebesgue-integrable/measurable will do.

Else, there are discontinuous functions satisfying the functional equation. (Basically, we can define the value of the function separately on each “independent” part. That is, define the equivalence class where $x$ and $y$ are related if $y = r_1x + r_2$ for rationals $r_1$ and $r_2$, pick a representative for each class using the axiom of choice (this is something like picking a basis for $\mathbb{R}/\mathbb{Q}$, which corresponds to the equivalence class defined by the relation $y = r_1x$), define the value of the function independently for each representative, and this fixes the value of $f$ on $\mathbb{R}$. See this article for more details.)

To step back a bit: what the functional equation says is that $f$ is a homorphism from $(\mathbb{R}, +)$, the additive group of real numbers, to $(\mathbb{R}, \times)$, the multiplicative monoid of real numbers. If $f$ is not the trivial identically-zero function, then (as we saw above) $f$ is in fact a homomorphism from $(\mathbb{R}, +)$, the additive group of real numbers, to $(\mathbb{R_+^*}, \times)$, the multiplicative group of positive real numbers. What we proved is that the exponential functions $e^{kx}$ are precisely all such functions that are nice (nice here meaning either measurable or continuous at least one point). (Note that this set includes the trivial homomorphism corresponding to $k = 0$: the function $f(x) = 1$ identically everywhere. If $f$ is not this trivial map, then it is in fact an isomorphism.)

Written by S

Mon, 2013-04-08 at 11:24:08

Posted in mathematics

## Trajectory of a point moving with acceleration perpendicular to velocity

(Just some basic high-school physics stuff; to assure myself I can still do some elementary things. :P Essentially, showing that if a particle moves with acceleration perpendicular to velocity, or velocity perpendicular to position, then it traces out a circle. Stop reading here if this is obvious.)

Suppose a point moves in the plane such that its acceleration is always perpendicular to its velocity, and of the same magnitude. What is its path like?

To set up notation: let’s say the point’s position at time $t$ is $(p_x(t), p_y(t))$, its velocity is $(v_x(t), v_y(t)) = \left(\frac{d}{dt}p_x(t), \frac{d}{dt}p_y(t)\right)$, and its acceleration is $(a_x(t), a_y(t)) = \left(\frac{d}{dt}v_x(t), \frac{d}{dt}v_y(t)\right)$.

The result of rotating a point $(x,y)$ by 90° is $(-y, x)$. (E.g. see figure below)

So the fact that acceleration is at right angles to velocity means that $(a_x(t), a_y(t)) = (-v_y(t), v_x(t))$, or, to write everything in terms of the velocity,

\begin{aligned} \frac{d}{dt}v_x(t) &= -v_y(t) \\ \frac{d}{dt}v_y(t) &= v_x(t) \end{aligned}

where we can get rid of $v_x(t)$ by substituting the second equation (in the form $v_x(t) = \frac{d}{dt}v_y(t)$) into the first:

$v_y(t) = -\frac{d}{dt}v_x(t) = -\frac{d}{dt}\left(\frac{d}{dt}v_y(t)\right)$

or in other words

$v_y(t) = -\frac{d^2}{dt^2}v_y(t).$

By some theory about ordinary differential equations, which I don’t know (please help!) (but see the very related example you saw in high school, of simple harmonic motion), the solutions to this equation are $\sin(t)$ and $\cos(t)$ and any linear combination of those: the solution in general is

\begin{aligned} v_y(t) &= a \sin(t) + b \cos(t) \\ &= \sqrt{a^2 + b^2} \left(\frac{a}{\sqrt{a^2+b^2}}\sin(t) + \frac{b}{\sqrt{a^2+b^2}}\cos(t)\right) \\ &= R\sin (t + \alpha) \end{aligned}

where $R = \sqrt{a^2 + b^2}$ and $\alpha$ is the angle such that $\cos(\alpha) = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin(\alpha) = \frac{b}{\sqrt{a^2+b^2}}$. And the fact that $v_x(t) = \frac{d}{dt}v_y(t)$ gives $v_x(t) = R\cos(t + \alpha)$. So $(v_x(t), v_y(t)) = (R\cos(t + \alpha), R\sin(t + \alpha))$. Note that $(a_x(t), a_y(t)) = \left(\frac{d}{dt}v_x(t), \frac{d}{dt}v_y(t)\right) = (-R\sin(t+\alpha), R\cos(t+\alpha))$ is indeed perpendicular to $(v_x(t), v_y(t))$ as we wanted.

The actual trajectory $(p_x(t), p_y(t))$ can be got by integrating

$\left(\frac{d}{dt}p_x(t), \frac{d}{dt}p_y(t)\right) = (v_x(t), v_y(t)) = (R\cos(t + \alpha), R\sin(t + \alpha))$

to get $p_x(t) = R\sin(t + \alpha) + c_1$ and $p_y(t) = -R\cos(t + \alpha) + c_2$. This trajectory is a point moving on a circle centered at point $(c_1, c_2)$ and of radius $R$, with speed $R$ or unit angular speed. Note that velocity is also perpendicular to the point’s position wrt the centre of the circle, as velocity is tangential to the circle, as it should be.

With a suitable change of coordinates (translate the origin to $(c_1, c_2)$, then rotate the axes by $\frac{\pi}{2}+\alpha$, then scale everything so that $R = 1$), this is the familiar paremetrization $(\cos(t), \sin(t))$ of the circle.

Note: Just as we derived $(v_x(t), v_y(t)) = (R\cos(t + \alpha), R\sin(t + \alpha))$ from assuming that the acceleration is perpendicular to velocity, we can, by assuming that velocity is perpendicular to position, identically derive $(p_x(t), p_y(t)) = (R\cos(t + \alpha), R\sin(t + \alpha))$, i.e. that the point moves on a circle.

Written by S

Sun, 2013-04-07 at 23:38:01

Posted in mathematics

## Typing Kannada on Mac OS X

(Thanks to this and this.)

Turns out it’s very easy, and we can basically use the same input method (UIM) as in Linux.

1. Get MacUIM from its website
2. Install it.
3. Go to System Preferences -> Language & Text -> Input Sources System Preferences -> Keyboard -> Input Sources, and turn on MacUIM. (Click + and add “MacUIM (Roman)” under English.) Tick “Show Input menu in menu bar” too.
4. I now have three input methods: US, EasyIAST (see earlier post), and MacUIM (Roman).
5. Go to System Preferences -> MacUIM -> General, and in Input method, choose m17n-kn-itrans
6. Go to System Preferences -> MacUIM -> Helper, tick “Use Helper-Applet”, and in the list at the right, tick m17n-kn-itrans.
7. [Just for me] I have some changes to kn-itrans.mim, to make it closer to HK (and remove nonsense like “RRi” or whatnot just to type ಋ): download this file kn-itrans.mim, and remove the pdf extension. It goes into /Library/M17NLib/share/m17n/kn-itrans.mim

Written by S

Sun, 2013-04-07 at 01:14:35

Posted in compknow