the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example, The value of 0! is 1
@KartikWatwani 1, 2*1, 3*2*1, 4*3*2*1, 5*4*3*2*1, ... What is the relationship between 4! and 3! ? That is, what is the relationship between 4*(3*2*1) and 3*2*1?
I said what's the remainder, not what's the quotient.
do you know the relationship between remainders in division and modulo?
a=r mod n is true when r is the remainder of a divided by n. more generally a=b mod n is true if and only if a and b have the same remainder upon division by n. in fact, computer scientists and programmers treat modulo as a binary operation rather than a binary relation, and Mod[a,n] returns the remainder of dividing a by n.
oh I see it now (p-1)! = (p-1)(p-2)(p-3)! since p-1 = -1 mod p and p-2 = -2 mod p from Wilson's Theorem we have (p-1)! = (-1)(-2)(p-3)! = 2(p-3)! mod p
Thus 2(p-3)! +1 -> (p-1)! +1 then by Wilson's THeorem again p-1 = -1 mod p -1+1 0 mod p oml
If $A,B\subseteq\Bbb R^n$ are connected open sets such that $A\cup B=\Bbb R^n$, then $A\cap B$ is connected.
That's equivalent to the thing I was talking about yesterday, where there's a simple proof using homology but I'm not sure if there's a nice proof without it
If I know that people went to an english test and math test. And I know that 2/15 passed the math test but not the english test. 8/13 from the people that passed math test didn't pass english test. How can I find how many passed the math test (I mean in fractions as I did above)?
@Aldon Differentition doesn't really have an inverse, because of the $+C$ business. Like, is $(d/dx)^{-1}1$ equal to $x$, or $x+1$, or $x-\pi$, or… Those all differentiate to $1$.
really, though, $(d/dx)^{-1}$ is just notation i.e. a convenient way of writing out the mapping that @akiva gave. (it is quite useful notation in some cases, but it shouldn't be taken as more than it is)
I want to show the embedding $W^{1,p}(0,1) \subset C^0[0,1]$. So we have to pick an element $u \in W^{1,p}(0,1) $ and show that it also belongs to $C^0[0,1]$.
Since $u \in W^{1,p}(0,1)$, we have that in $(0,1)$ it holds $|u|_p< +\infty$ and $|u'|_p<+\infty$, right? How can we show that $u \in C^0[0,1]$ ?
@Semiclassical OK, now I'm awake. I should start writing soon (tomorrow?) but today my goal is to read (and hopefully understand?) about gluing maps in the context of Morse theory.
@robjohn It holds that $u(x+h)-u(x)=\int_x^{x+h} u'(t) dt$.
$f$ is continuous at the point $x_0$ if: $\forall \epsilon>0 \exists \delta>0$ such that $\forall y \in [x, x+h]$ with $|y-x_0|< \delta$ we have $|u(y)-u(x_0)|< \epsilon$.
Yes, so just a quick summary of the trick just because it's good not to step-jump: since that polynomial is divisible by $x = -3/2$, I can say $4(-3/2)^3 + 4(-3/2)^2 + a(-3/2) - 18 = 0$ (i.e. $-3/2$ is a root). Now I can solve for $a$.
You proved that for $f(x)$ to have no asymptote at $x = -3/2$, $a = -15$ must hold. But what's the guarantee that $f(x) = (4x^3 + 4x^2 - 15x - 18)/(2x + 3)$ has no other asymptotes?
It can have an asymptote somewhere other than a vertical asymptote at $x = -3/2$. You need to make sure there are none.
This is not a big issue, anyway. Note that graph of $f(x)$ is the same as graph of the polynomial you get by dividing $4x^3 + 4x^2 - 15x - 18$ by $2x + 3$ (except at $x = -3/2$ where there is a jump discontinuity). And polynomials have no asymptotes...
@BalarkaSen Do you mean other types of asymptotes like horizontal? Maybe I didn't mention but it asks "It is known that (...) hasn't vertical asymptotes. Find $a$."
There are thousands upon thousands of TeX packages and no source that contains absolutely all of them. There are some very standard things that everyone needs, and then there are much less common things that most people (all people?) will never, ever use.
$\uppi \textrtails$
Just two random things I got from Detexify that didn't look like they were from the standard packages. The first is a different way of drawing pi (it flares out the bottom of the letter a bit). The second was an S with a small symbol coming out of the bottom left of the S.
I think it's unlikely anyone here has tried to use those before me.
let $a^2 = 23+\sqrt{129}$ and $b^2 = 23- \sqrt{129}$. Your equation is $(a + b)$, which is the same as $\sqrt{(a + b)^2} = \sqrt{a^2 +2ab + b^2}$, which is my equation.
it's essential a difference in presentation rather than content. i prefer to do it as $x=a+b\to x^2=a^2+2ab+b^2$, simplify, then take the square root. but it's just as correct to write $x=\sqrt{x^2}$ directly.
I imagine the dilemma of deciding to get groceries is the of existential crisis inducing thing which Sarte says one "cannot but feel a certain anguish" about making a decision about. Therefore the only reasonable thing is to not get any ;P
