0
$\begingroup$

Context:

I would like to proof that $C^2_0 p^0 (1-p)^2 + C^2_1 p^1 (1-p) + C^2_2 p^2 (1-p)^0 = 0$. When I do derivation for the equation, I stopped in the following point: $(1-p)^2 + 2p(1-p) + p^2$

I would like to find out the output of the first term, $(1-p)^0$, is it 1 or -1?

Ibrahim

$\endgroup$

closed as off-topic by Aqua, Martin Sleziak, B. Goddard, AugSB, GNUSupporter 8964民主女神 地下教會 Jan 9 '18 at 19:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Aqua, Martin Sleziak, B. Goddard, AugSB, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I have added the context. Would be possible to remove the on hold? $\endgroup$ –  Ibrahim EL-Sanosi Jan 11 '18 at 12:48
  • $\begingroup$ Hi Ibrahim, I voted to reopen, but you will get a better answer if you explain (in the question, not in comments) why you think it might be -1. $\endgroup$ – Matthew Towers Jan 11 '18 at 13:47
4
$\begingroup$

$$x^0=1\space(\forall)\space x\in\Bbb{R}^*$$

I did say $\Bbb{R}^*$, because $0^0$ is not unanimously agreed upon or might be context-dependent.

$\endgroup$
  • 1
    $\begingroup$ $0^0$ is less than context dependent, it can only be expressed as the limit of some function of x and thus defined as whatever that limit is. It’s exactly like $\frac00$. $\endgroup$ – Marcus Aurelius Jan 9 '18 at 17:33
  • 1
    $\begingroup$ @MarcusAurelius: There is a lot of discussion of this topic on this site. Often, $0^0$ is defined to be $1$ with the caveat that $x^y$ is not continuous at $(0,0)$. $\endgroup$ – robjohn Jan 9 '18 at 17:44
0
$\begingroup$

For every $x$, $x^0$ is $1$ except for $x=\infty$.

$\endgroup$
  • $\begingroup$ Yes, it makes sense, it is 1, because p is probability and p is always <=1. $\endgroup$ –  Ibrahim EL-Sanosi Jan 9 '18 at 17:14
-6
$\begingroup$

For all $1-p>0$ it's $1$, otherwise it's not defined.

$\endgroup$
  • 4
    $\begingroup$ Why is it undefined for negative numbers? $\endgroup$ – John Doe Jan 9 '18 at 17:08
  • 2
    $\begingroup$ $$1=k^n\cdot k^{-n}=k^0$$This doesn't have any specific requirement on the sign of $k$. $\endgroup$ – John Doe Jan 9 '18 at 17:17
  • 1
    $\begingroup$ After you have defined a branch of the $(\cdot)^\frac12$ function, you'd get $$i\cdot(-i)=1$$ $\endgroup$ – John Doe Jan 9 '18 at 17:21
  • 3
    $\begingroup$ $(-1)^0$ is also 1 $\endgroup$ – Plexus Jan 9 '18 at 17:25
  • 3
    $\begingroup$ Well, if you wish to square root negative numbers, of course we will end up in $\Bbb C$. In any case, I'm sure there are other ways to 'prove' that $(-k)^0=1$, other than the argument I used above - it was the first I thought of. The way I learned it, it was taken to be the definition that $k^0=1$ for any non-zero $k$, and then that it was convenient (for example in probability, and binomial expansions) to define $0^0=1$ also. $\endgroup$ – John Doe Jan 9 '18 at 17:28

Not the answer you're looking for? Browse other questions tagged or ask your own question.