Algebraic Testing of Numbers Yielding Boolean Responses
- Thread starter Fafnir665
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The standard way of seeing if a number is even or odd is with modulus or remainder division by 2. If the remainder is 0, the number is even.
I just thought of another way. Although trig functions are not algebraic, it might still be useful to you. Here it is:
$$\cos^2{(\frac{\pi}{2}n)}$$
You see if n is even, you will get 1, if n is odd, you get 0.
I just thought of another way. Although trig functions are not algebraic, it might still be useful to you. Here it is:
$$\cos^2{(\frac{\pi}{2}n)}$$
You see if n is even, you will get 1, if n is odd, you get 0.
That is awesome.
I spent an hour with my trig book in one of my classes today analyzing number patterns, and I got as far as realizing I needed a way to get some sort of boolean response in number format, but not in traditional boolean notation.
Heres the sequence of numbers I was trying to write a funtion for
1 3 7 9 13 15 19 21 25 etc
So, it would be, in mathlish, the sum of, k = 0 to n, 1 + ( ( 1 + whatyousaid) 2)
I could take a picture of the two pages I filled with stuff, but heres a small example
1 3 7 13 15 19 21 25
2 4 2 4 2 4 2
or
2 6 12 14 18 20 24
which are all even
then I noticed that even N add 4, while odd N add 2
which brings us to your boolean test.
This was all work that really had nothing to do with the class, the prof was just trying to demonstrate patterns or something, but it was a hell of a lot more interesting that the crap she was talking about...
Or was it a student doing some BS, I'm not entirely sure. Either way, youre awesome.
I spent an hour with my trig book in one of my classes today analyzing number patterns, and I got as far as realizing I needed a way to get some sort of boolean response in number format, but not in traditional boolean notation.
Heres the sequence of numbers I was trying to write a funtion for
1 3 7 9 13 15 19 21 25 etc
So, it would be, in mathlish, the sum of, k = 0 to n, 1 + ( ( 1 + whatyousaid) 2)
I could take a picture of the two pages I filled with stuff, but heres a small example
1 3 7 13 15 19 21 25
2 4 2 4 2 4 2
or
2 6 12 14 18 20 24
which are all even
then I noticed that even N add 4, while odd N add 2
which brings us to your boolean test.
This was all work that really had nothing to do with the class, the prof was just trying to demonstrate patterns or something, but it was a hell of a lot more interesting that the crap she was talking about...
Or was it a student doing some BS, I'm not entirely sure. Either way, youre awesome.
Or simply $$(-1)^n$$, which gives 1 if n is even and -1 if n is odd. If you want 0 and 1 you can use:
$$\Re \left( i^n \right)^2 = \frac{1}{2}\left[ 1 + (-1)^n \right]$$
Those are just due to the usual numerical errors: I'd expect both $$\pi x$$ and $$\cos(x)$$ are less accurately calculated/stored for large $$x$$.
If you need to test for even/oddness of integers in a computer program, just use the modulo function.
Or work in binary:
$$z = 2*\lfloor \frac{x}{2} \rfloor - x $$
In C you can use a bit mask to get the last binary digit (not really math, but oh well):
x & 1
(it looks like my compiler automatically substitutes expressions like "x%2" for "x&1" while optimizing)
Then I suppose there's stuff like this:
$$\int^{\infty}_{-\infty} f_n(x) \triangle_2(x) \mathrm{d}x$$
$$f_n(x) \: \equiv \: \left{ 1 \; \mathrm{if} \; x = n \\ 0 \; \mathrm{if} \; x \neq n \right.$$ (indicator function of n)
$$\triangle_2(x) \: \equiv \: \sum^{\infty}_{k = -\infty} \delta(x - 2k)$$ (Dirac comb of period 2)
$$\triangle_2(x) \: \equiv \: \sum^{\infty}_{k = -\infty} \delta(x - 2k)$$ (Dirac comb of period 2)
(n\2)AND(n/2)?
If the integer division isn't the same as the real division, then the division has yielded a uneven number.
That integal's value is zero since f_n(x) = 0 for all but 1 value.
Well it would give a boolean response, since it is the logical AND, 1 AND 1 = 1, 0 AND 1 = 0
It would work in programming languages, but perhaps it wasn't what you wanted.
Like this perhaps:
(((n/2)-(n\2)) + (1 - ((n/2)-(n\2)))) * (1*((n/2)-(n\2)))
Haha, I lost track somewhere along the line lol, perhaps it works, perhaps it doesn't...
$$\Re(z)$$ designates the real part of the complex number $$z$$. If you have a complex number $$z = a + b i$$ (where $$a$$ and $$b$$ are real), then $$\Re(z) = a$$.
$$i^n$$ cycles through $$i, \, -1, \, -i, \, 1$$, so $$\Re \left( i^n \right)$$ alternates between $$0$$ and $$\pm 1$$.
As it happens,
$$\Re \left( i^x \right) = \cos \left( \frac{\pi x}{2} \right)$$
Unless I'm being over-sloppy somewhere,
$$\int^{\infty}_{-\infty} f_n(x) \triangle_2(x) \mathrm{d}x$$
$$= \; \sum^{\infty}_{k = -\infty} \left[ \int^{\infty}_{-\infty} f_n(x) \delta(x - 2k) \mathrm{d}x \right]$$
$$= \; \sum^{\infty}_{k = -\infty} f_n(2k)$$
$$=$$ 1 if n is a multiple of 2, and 0 otherwise.
$$= \; \sum^{\infty}_{k = -\infty} f_n(2k)$$
$$=$$ 1 if n is a multiple of 2, and 0 otherwise.
(not that this is what I'd call algebraic...)
*****
EDIT: Found this condition:
Ah, well, gives me an excuse to make things more complicated.
I give you:
$$\lim_{\varepsilon \rightarrow 0} \left[ \int^{\infty}_{-\infty} f_n(x, \, \varepsilon) \triangle_2(x) \mathrm{d}x \right]$$, with $$f_n(x, \, \varepsilon) \: \equiv \: \left{ 1 + \frac{x-n}{\varepsilon} \; \mathrm{if} \; -\varepsilon < x - n \leq 0 \\ 1 - \frac{x-n}{\varepsilon} \; \mathrm{if} \; 0 < x - n \leq \varepsilon \\ 0 \; \text{otherwise} \right.$$
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Are you sure it's not the equality test you're thinking of? "n\2 == n/2.0" should return "true" if n is even and "false" otherwise.
Yes, in a programming language. In a circuit the logical AND would work (if the number is translated binary of course).
What do you mean by this?