Can I get a general function f(m,n)

[Emilijo]

How can I get a general function f(m,n) that represents a series of 1 and 0, for example :
1,0,1,0,1,0,1,0...; but also
1,1,0,1,1,0,1,1,0...;
1,1,1,0,1,1,1,0,1...

where m is period and n nth number in certain period. In example two:
m=3 (...1,1,0...) f(3,4)=1

The function must be composed only by elementary function

 

[Millennial]

This might be a little desperate, but Fourier series should do the trick. I do not know if there is any other elementary function that would do the same, and I do not know if you consider Fourier series to be elementary (despite it involving only sine and cosine, which are both considered elementary.)

 

[rbj]

This might be a little desperate, but Fourier series should do the trick. I do not know if there is any other elementary function that would do the same, ...

oh c'mon.

consider

$$g(m) = \frac{1}{2} \left( 1 + (-1)^m \right)$$

for integer $m$

sure, find a Fourier series that, when sampled at integer values, gives you the values you want. but instead of sines and cosines, use the exponential version derived from Euler's formula:

$$e^{i \theta} \ = \ \cos(\theta) \ + \ i \sin(\theta)$$

and

$$-1 = e^{i \pi}$$

and

$$g(x) = \sum_{n=-\infty}^{+\infty} c_n \ e^{i \ n (2 \pi/P) x }$$

where $P$ is the period of the periodic function

$$g(x + P) = g(x)$$

for all $x$, and $c_n$ are the Fourier coefficients.

but you can start with the simple equation with $(-1)^m$ and imagine how you might put it together to get the gating functions you're looking for.

 

[LCKurtz]

How can I get a general function f(m,n) that represents a series of 1 and 0, for example :
1,0,1,0,1,0,1,0...; but also
1,1,0,1,1,0,1,1,0...;
1,1,1,0,1,1,1,0,1...

where m is period and n nth number in certain period. In example two:
m=3 (...1,1,0...) f(3,4)=1

The function must be composed only by elementary function

$$f(m,n)=\left\{ \begin{array}{rl}
0&n=0\mod m\\
1&\hbox{otherwise}
\end{array}\right.$$

 

[rbj]

$$f(m,n)=\left\{ \begin{array}{rl}
0&n=0\mod m\\
1&\hbox{otherwise}
\end{array}\right.$$

does that count as an "elementary function"? do the mod or floor operators count as elementary functions?

 

[LCKurtz]

$$f(m,n)=\left\{ \begin{array}{rl}
0&n=0\mod m\\
1&\hbox{otherwise}
\end{array}\right.$$

does that count as an "elementary function"? do the mod or floor operators count as elementary functions?

You will have to check with the Elementary Function Arbitration Committee to get a definitive answer.

 

[rbj]

good answer.

 

[rbj]

i think i figured out a possible answer, but it involves the dirac delta function (or more precisely, the dirac comb). the dirac comb is a periodic sequence of equally space delta functions and can be represented as an infinite series. maybe we can construction this thing with the sinc() function:

$$\ \operatorname{sinc}(x) \ = \ \frac{\sin(\pi x)}{\pi x}$$

and it has a removable singularity at zero so that

$$\ \operatorname{sinc}(0) \ = \ \lim_{x \rightarrow 0} \frac{\sin(\pi x)}{\pi x} \ = \ 1$$

it's also true that the sinc() function is 0 for all non-zero integers. does that count as an "elementary function"? does an infinite sum of these sinc() functions count as an "elementary function"?

if yes, i can assemble a general function. actually, this periodic sinc() function (the infinite sum) can be represented as a Fourier series with a finite number of terms, so i think that's where the answer is.

 

[rbj]

okay, so first we define

$$\ \operatorname{sinc}(x) \ = \ \frac{\sin(\pi x)}{\pi x}$$

and then we define this periodic function:

$$g_m(x) = \sum_{k=-\infty}^{+\infty} \operatorname{sinc}(x - mk)$$

where m is an integer (and so is k and so is mk). we know that for integer n that

$$g_m(n) = \begin{cases} 1 & \ \ \text{if }n = \text{ any multiple of }m \\ 0 & \ \ \text{if }n = \text{ is any other integer} \end{cases}$$

now, i am pretty sure that this is the case:

$$g_m(x) = \frac{1}{m} \sum_{k=1}^{m} \cos\left( 2 \pi \frac{k x}{m} \right)$$

even if it isn't, i think that this function has the property we need for when x is an integer n. it is zero for any integer n except when n is a multiple of m. since this is a harmonic series, you can get a closed form expression for it (someone else want to do it)? then subtract this from 1.

 

[rbj]

$$\begin{align} g_m(x) \ &= \ \frac{1}{m} \sum_{k=1}^{m} \cos\left( 2 \pi \frac{k x}{m} \right) \\ &= \ \frac{1}{2m} \sum_{k=1}^{m} e^{ i 2 \pi k x / m} \ + \ \frac{1}{2m} \sum_{k=1}^{m} e^{ -i 2 \pi k x / m} \\ &= \ \frac{e^{ i 2 \pi x / m}}{2m} \sum_{k=0}^{m-1} e^{ i 2 \pi k x / m} \ + \ \frac{e^{ -i 2 \pi x / m}}{2m} \sum_{k=0}^{m-1} e^{ -i 2 \pi k x / m} \\ &= \ \frac{e^{ i 2 \pi x / m}}{2m} \frac{ e^{ i 2 \pi m x / m} - 1}{e^{ i 2 \pi x / m} - 1} \ + \frac{e^{ -i 2 \pi x / m}}{2m} \frac{ e^{ -i 2 \pi m x / m} - 1}{e^{ -i 2 \pi x / m} - 1} \\ &= \ \frac{1}{2m} \frac{ e^{ i 2 \pi x} - 1}{1 - e^{ -i 2 \pi x / m}} \ + \frac{1}{2m} \frac{ e^{ -i 2 \pi x } - 1}{1 - e^{ i 2 \pi x / m}} \\ \end{align}$$

i'm getting tired, can someone else finish this?

 

[rbj]

$$\begin{align} g_m(x) \ &= \ \frac{1}{2m} \frac{ e^{ i 2 \pi x} - 1}{1 - e^{ -i 2 \pi x / m}} \ + \ \frac{1}{2m} \frac{ e^{ -i 2 \pi x } - 1}{1 - e^{ i 2 \pi x / m}} \\ &= \ \frac{e^{ i \pi x} \ e^{ i \pi x / m}}{2m} \frac{ e^{ i \pi x} - e^{ -i \pi x}}{e^{ i \pi x / m} - e^{ -i \pi x / m}} \ + \ \frac{e^{ -i \pi x} \ e^{ -i \pi x / m}}{2m} \frac{ e^{ -i \pi x} - e^{ i \pi x}}{e^{ -i \pi x / m} - e^{ i \pi x / m}} \\ &= \ \frac{e^{ i \pi (m+1) x / m}}{2m} \frac{ e^{ i \pi x} - e^{ -i \pi x}}{e^{ i \pi x / m} - e^{ -i \pi x / m}} \ + \ \frac{ e^{ -i \pi x (m+1) / m}}{2m} \frac{ e^{ i \pi x} - e^{ -i \pi x}}{e^{ i \pi x / m} - e^{ -i \pi x / m}} \\ &= \ \frac{ e^{ i \pi x} - e^{ -i \pi x}}{e^{ i \pi x / m} - e^{ -i \pi x / m}} \ \left( \frac{e^{ i \pi (m+1) x / m}}{2m} + \ \frac{ e^{ -i \pi x (m+1) / m}}{2m} \right) \\ &= \ \frac{ \sin(\pi x) }{m \sin(\pi x / m)} \ \cos\left(\pi (m+1) x / m \right) \\ &= \ \frac{ \sin(\pi x) }{m \sin(\pi x / m)} \ \cos(\pi x + \pi x / m) \\ &= \ \frac{ \sin(\pi x) }{m \sin(\pi x / m)} \ \left( \cos(\pi x) \cos(\pi x / m) - \sin(\pi x) \sin(\pi x / m) \right) \\ &= \ \frac{ \sin(\pi x) }{m} \ \left( \frac{\cos(\pi x)}{\tan(\pi x / m)} - \sin(\pi x) \right) \\ \end{align}$$

 

[rbj]

i see it's that Dirichlet kernel on the left. can someone check out to see that this is 1 for n being a multiple of m and 0 for all other integers?

maybe the bottom two or three equalities does not help us.