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

by Emilijo
Tags: function
 P: 36 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
 P: 295 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.)
P: 2,251
 Quote by 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, ...
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.

HW Helper
Thanks
PF Gold
P: 7,663
Can I get a general function f(m,n)

 Quote by 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
$$f(m,n)=\left\{ \begin{array}{rl} 0&n=0\mod m\\ 1&\hbox{otherwise} \end{array}\right.$$
P: 2,251
 Quote by 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?
HW Helper
Thanks
PF Gold
P: 7,663
 Quote by LCKurtz $$f(m,n)=\left\{ \begin{array}{rl} 0&n=0\mod m\\ 1&\hbox{otherwise} \end{array}\right.$$
 Quote by rbj 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.
 P: 2,251 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.
 P: 2,251 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.
 P: 2,251 \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?
 P: 2,251 \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}