I just want a function f(x) = 1,0,0,0,0,0,... where x eof N

>>9047824
Why not?

>>9047824
0^x?

[math]f(0)=1, f(n)=f(n-1)*0 n>0[/math]

>>9047831
im searching for a solution thats not a recursion and does not use < or > or if

>>9047828
this is what im looking for but Wolfram alpha says 0^x is undefinded but google says its ok wat do

>>9047824
[eqn] f(n) = \frac{d^n}{dx^n} 1 [/eqn]

>>9047837
0^0 is undefined. You're in trouble if you want to describe your function without recursion or ifs. The last resort is using special functions such as floor, ie [math] f(n) = floor(1/(n+1)) [/math]

>>9047845
>0^0 is undefined.
What if I want to use it in a proof? Can I just define it or will people get pissed?

>>9047824
f(n)=1-min(1,n)

>>9047847
Try it out and see if they get pissed or not

>>9047824
>I just want a function f(x) = 1,0,0,0,0,0,... where x eof N
Well, a function is defined precisely once its domain is specified along with the value it assigns to each element of the domain.
Which is exactly what you've just done so there's no issue at all.

>>9047843
>>9047828
these are the only sane answers

floor is cheating, min is cheating

>>9047851
hardcoing values is cheating

>>9047855
>0^0
>sane

>>9047824
[math] f(x)={\begin{cases}1&{\mbox{if }}x\in \{ 0 \},\\0&{\mbox{if }}x\notin \{ 0 \}.\\\end{cases}} [/math]

>>9047867
Why use sets instead of equality (to zero) ?

>>9047867
no if allowed bro

sgn(x+1) - sgn(x)

>>9047824
[math]f(n) = \delta(n)[/math]

>>9047866
you must be atleast 18 years old to post on this board.

sin(2^(x-1)*pi)

>>9047942
Did I offend you?

>>9047950
Nice

>>9047845
>is undefined
No, [math] 0^0 = 1 [/math].

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542KB, 3000x2000px

>>9047950
my man

thanks

>>9047979
No, undefined. Consider for example:
[math]
0^2 : 0^2 = 0^0
[/math]

>>9047982
Go back to >>>/g/ you imbecile. You can't divide by 0.
[math] 0^0 [/math] is an empty product, so it's equal to 1.

>>9047982
That's like saying 0^3 is undefined because
[eqn] 0^5 : 0^2 = 0^3 [/eqn]

>>9047982
>>9047985
>dividing by 0

[math]\sum_{d|n+1}\mu(d)[/math]

>[math]0\in\mathbb{N}[/math]

i think this works
[eqn]f(x)=\lim_{k\to\infty}\exp\left({-\sum_{n=0}^{k}x^n }\right) [/eqn]

>>9047987
That's exactly what the second poster was pointing out.

[math]\displaystyle f(x)=\frac{\prod_{k}(x-k)}{-\prod_k k}[/math]

>>9047843
Bro your smart

>>9047979
>>9047982
>>9047866
>https://www.math.hmc.edu/funfacts/ffiles/10005.3-5.shtml

>>9048358
http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

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[math]f(x) = -\prod \limits_{i = 0}^x \left( i - 1 \right)[/math]

(where [math]0 \in \mathbb{N}[/math])

>>9050169
nice

[math]\displaystyle f(x)=\lim_{k\,\to\ x\pi}\frac{\sin k}{k}[/math]

this "functions can only be closed-form expressions" meme needs to die

>>9050200
you need to die

>>9050202
kys

Im sure theres a better way to do this, but
let [math]\mathcal{M_{xx}(1)}[/math] denote an [math]x\times x[/math] matrix populated by 1s. then let [math]f(x)=1-|\mathcal{M}_{xx}(1)|[/math]

>>9047824
This is the definition of delta_{1,x}, just use that.

Let f(x) be the probability that neither a head nor tails is obtained after x coin flips