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maths

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Thread replies: 16
Thread images: 1

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Is there any function that can check if a nummber is even or not. By that i mean an algebraic function not a computer functions, that means for example modulus can't be used.
>>
i know my english is bad
>>
>>8083530
sin^2(pi*x)
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>>8083530
[math]\frac{1+(-1)^n}{2}[/math]
>>
>>8083533
Yes.

if x |-> f(x):=1, if a natural number n exists so that x=2n, 0 else.
>>
>>8083530
Sure, [math]n[/math] is an even integer if and only if [math]\zeta(-n) = 0[/math]. Here [math]\zeta[/math] refers to the

https://en.wikipedia.org/wiki/Riemann_zeta_function

Is this what you're looking for?
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>>8083549
obviously I meant "positive even integer"
>>
>>8083530
f(x) = x/2, where f(x) is an integer
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>>8083530
Isn't it just (1 + n)/2 or something?
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>>8083530
>modulus
>not an algebraic function
>>
In what kind of world, this function
>>8083549
is algebraic and taking the mod 2 isn't?
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>>8083530
(-1)^n is 1 if n is even and -1 if it is odd. Don't know if that's what you mean by algebeaic though.
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>>8083530
the function [math]f(n)=n[/math] is even when [math]n[/math] is even and odd when [math]n[/math] is odd
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>>8084011
>the function f(n)=n is even when n is even
A function being even or odd doesn't depend on any single input. Your function is an odd function
>>
>>8083530
f(x) = sin(pi*x/2)
f(x) = 0 iff x is even
>>
If n is an even integer, n/2 gives k such that k belongs to Z. So when the output of f(n) = n/2 is in Z, n is even.
Thread posts: 16
Thread images: 1


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