[ 3 / biz / cgl / ck / diy / fa / g / ic / jp / lit / sci / tg / vr / vt ] [ index / top / reports / report a bug ] [ 4plebs / archived.moe / rbt ]

/vt/ is now archived.Become a Patron!

/sci/ - Science & Math

[ Toggle deleted replies ]
File: 35 KB, 480x360, op_is_a_fag.jpg [View same] [iqdb] [saucenao] [google] [report]

$0^0 = 1$ and here is my quick proof:

[eqn] \displaystyle \lim_{x \rightarrow 0} x^x = \lim_{x \rightarrow 0} e^{x\ln{(x)}} = e^{\lim_{x \rightarrow 0} x\ln{(x)}} = \Bigg( \lim_{x \rightarrow 0} x \Bigg)\Bigg(\lim_{x \rightarrow 0} x\ln{(x)} \Bigg) [/eqn]

so by squeeze theorem:

[eqn] \displaystyle \Bigg( \lim_{x \rightarrow 0} x \Bigg)\Bigg(\lim_{x \rightarrow 0} x\ln{(x)} \Bigg) = 0 \implies \lim_{x \rightarrow 0} x^x = e^0 = 1 [/eqn]

Q.E.D :)

 >> Anonymous Sun Dec 30 14:58:05 2018 No.10254974 Your proof shows that [eqn]\lim_{x \rightarrow 0} x^x = 1[/eqn]. That part is not not disputed, I think. But that does not mean that 0^0 = 1.It is, mind you. But it doesn't follow from this limit.
 >> Anonymous Sun Dec 30 15:03:47 2018 No.10254986 File: 126 KB, 1131x622, 1534581478270.jpg [View same] [iqdb] [saucenao] [google] [report] Math is the shittiest meme to ever come out of humanity
 >> Anonymous Sun Dec 30 15:13:31 2018 No.10255016 >>10254963Okay, but where's your proof that exponentiation is continuous.
 >> Anonymous Sun Dec 30 15:17:13 2018 No.10255024 >>10254974how come it doesnt??>>10255016why do I need to proof that??
 >> Anonymous Sun Dec 30 15:29:54 2018 No.10255048 >>10255024>how come it doesnt??Why would it?
 >> Anonymous Sun Dec 30 15:31:15 2018 No.10255051 >>10255024x^x approaches 1 on the real axis, but not anywhere on the complex plane
 >> Anonymous Sun Dec 30 15:31:25 2018 No.10255052 >>10254974>He doesn't understand limits
 >> Anonymous Sun Dec 30 15:34:40 2018 No.10255055 >>10255052Do tell.
 >> Anonymous Sun Dec 30 15:37:46 2018 No.10255061 >>10255055im not this anon>>10255052$\displaystyle \lim_{x \to c} f(x) = a$ doesnt imply that $f(c) = a$thats why even though sin(x)/x approaches 1 as x approaches 0, sin(0)/0 isnt equal to one, its undefined.
 >> Anonymous Sun Dec 30 15:38:55 2018 No.10255063 >>10255061Indeed.
 >> Anonymous Sun Dec 30 15:51:43 2018 No.10255095 File: 12 KB, 244x241, happy.jpg [View same] [iqdb] [saucenao] [google] [report] >>10255061>partial sum definition of infinite series ignores thisand that's why its the worst way to define infinite series
 >> Anonymous Sun Dec 30 15:53:05 2018 No.10255098 >>10254963Isn't a^n just 1×a×...×a n times?So a^0 = 1 by defnAlso 0^0 is then 1 without ambiguity.Just like 0! = 1 by def
 >> Anonymous Sun Dec 30 15:54:53 2018 No.10255101   File: 1 KB, 157x85, wa.png [View same] [iqdb] [saucenao] [google] [report] >>10254963Wolframed
 >> Anonymous Sun Dec 30 15:55:12 2018 No.10255103 >>10255098Not quite.$a^0=1$ and $a^n=aa^{n-1}$ define exponentiation.
 >> Anonymous Sun Dec 30 18:26:34 2018 No.10255407 >>10255103$\frac{a^n}{a^n}=a^{n-n}=a^0=1$
 >> Anonymous Sun Dec 30 18:29:37 2018 No.10255417 >>10255407Yeah, but if you just define $a^ba^c=a^{b+c}$ you still have to define $a^1=a$, which is really short but still gives longer proofs.
 >> Anonymous Sun Dec 30 18:38:10 2018 No.10255437 >>10254963here is a definition of the exponent:a(n,0)=1a(n,k+1)=n*a(n,k)> every natural number is 0 or of the form k+1 for some k> this always halts for any input because the second argument decreases> the proof that a(0,0)=1 is obviousThis is why sometimes 0^0 = 1. Attempting to prove this with calculus is beyond retarded.
 >> Anonymous Sun Dec 30 19:47:36 2018 No.10255600 >>10255437>its 1 by definitionfaggot
 >> Anonymous Sun Dec 30 19:52:42 2018 No.10255613 >>10255024are you retarded? by definition the function doesn't have to be defined in the exact point.
 >> Anonymous Sun Dec 30 19:57:38 2018 No.10255632 >>10254974based and redpilled answer. Here is a real proof that 0^0=1. def: for cardinal numbers A, B, we define A^B as the number of functions from a set of size A to a set of size B. Since there is precisely one function from the empty set to itself, 0^0=1
 >> Anonymous Sun Dec 30 19:59:21 2018 No.10255634 >>10254963whoa dude! You're a genius!Doing a similar thing you can prove that:$\frac{1}{0}=infinity$
 >> Anonymous Sun Dec 30 20:00:09 2018 No.10255637 >>10255632ah yes, its obvious that 0^0 is 1 if you only work with the nonnegative integers
>>