@Hippalectryon Well my idea was $$\displaystyle p_{n+1}(x) = \frac{1}{2}p_{n}(x) + (1-\sqrt{x})p_n(x)$$ $$= p_n(x)\cdot\left(1-\sqrt{x} + \frac{1}{2}p_n(x)\right) \le p_n(x)\cdot\left(1-\sqrt{x} + \frac{1}{2}\sqrt{x}\right)$$ $$=p_n(x)\cdot\left(1- \frac{\sqrt{x}}{2}\right) \le p_0(x)\left(1- \frac{\sqrt{x}}{2}\right)^n$$ (by simple induction) .. now this would have been sufficient for the problem as $$\displaystyle \sqrt{x}\left(1- \frac{\sqrt{x}}{2}\right)^n \rightrightarrows 0\qquad (x\in[0,1])$$
@TedShifrin, the statistician is quick to point out the dangers of correlation $\Rightarrow$ causation. In this case, me joining the site might not be just coincidence. :P
I've downvoted people I don't know at all. One guy earlier today, who applied the divergence theorem to a non-closed surface. He finally admitted he'd misread and deleted.
"Computer cracks Erdős puzzle" wins grand prize for most-misleading headline of all time, unseating reigning champion "Is the Universe a Simulation?", and paving the way for a new class of painfully misguided coffee shop conversations.
@VibhavPant For this you need to prove anything. It's evident.
How about the simple fact that $$\int_0^{\pi/2} \sin^m(x) \ dx= \int_0^{\pi/2} \cos^m(x) \ dx$$ combined with the formula $\displaystyle \sin(2x)=2\sin(x)\cos(x)$?
@MrWho Im sure Eulers Formula will work, but I dont want to get a bad habit of pulling out complex exponential when anything remotely trig appears. (I already try to use L'Hospital rule everywhere)
@Chris'ssis That identity is geometrically intuitive, though I cant get to prove it
@Sawarnik define $x_1$ and $x_2$ such that $f(x_1)=f(x_2)$, so there must be a number $c$ between $x_1$ and $x_2$ for which $f'(c)$ = 0 (or $c$ is the local maxima)
@KajHansen That is the Cayley table. It is the internal grid part of a multiplication (or composition in this case) of the elements of $D_{100}$. It starts with rotations $S_{0}$ to $S_{99}$ and reflections (going anticlockwise through) from $R_{0}$ to $R_{99}$. If I had the reference rows and coloumns you would see that as the elements go along they are assigned a 100th (in this case) of colour spectrum. The resulting colours in the middle are the colours of the composition.
$\frac{\mathrm{d} }{\mathrm{d} x} \int_{u(x)}^{v(x)}f(t,x)dt=u'(x)f(u(x),x)-v'(x)f(v(x),x)+\int_{v(x)}^{u(x)}\frac{\partial }{\partial x}f(t,x)$ What is the proof?
@MrWho Consider $$F(u,v,x) = \int_u^v f(t,x)\,dt.$$ The partial derivatives of $F$ with respect to $u,v$, and $x$ are easy (or at least easier) to find. The rest is the chain rule. (And you have swapped $u$ and $v$ on one side of your formula.)
I'm looking for the differentiation of multivariable function integral $$\frac{\mathrm{d} }{\mathrm{d} x} \int_{u(x)}^{v(x)}f(t,x)dt=u'(x)f(u(x),x)-v'(x)f(v(x),x)+\int_{v(x)}^{u(x)}\frac{\partial }{\partial x}f(t,x)$$
Thank you!
@KarlKronenfeld The elements are ordered (usually lex) and then they each get split equally on the spectrum. The lowers would be redder and the highers would be purplerrer.
Have you tried alternative orders? Whenever you order a subgroup and then list it and each of its cosets in the corresponding order, you will get the mini cayley tables.
@BalarkaSen I am not doing the graphics part. There is one graphics part and that is done. All I have to do is program a group and it produces the cayley diagram by itself.
eh not quite like that. What I do is add elements to a set. then I define the operation. Then my well written structure checks it is really a group and then voila a group/
So really what I have to do is program something that maps element n and element m to some element k
@BalarkaSen No each element in any set is basically an integer. The group operation is a map from these two integers to another. However it acts like they were elements of that group
@BalarkaSen Right so to program a group $G$ I need it to be isomorphic to $\Bbb Z_{m}^{\circ}$ where $m$ is the order of $G$ and $\circ$ is some operator such that the isomorphic holds. This is how its programmed. The operator is the real defining property of the group. It takes two group elements and gives another. Now programmatic functions can behave differently to normal mathematically constructed ones. The function programmed behaves as if the elements were from $G$ if that makes sense?
I'm looking for the differentiation of multivariable function integral $$\frac{\mathrm{d} }{\mathrm{d} x} \int_{u(x)}^{v(x)}f(t,x)dt=u'(x)f(u(x),x)-v'(x)f(v(x),x)+\int_{v(x)}^{u(x)}\frac{\partial }{\partial x}f(t,x)$$
Thank you!
