@Daminark i think group theory won't end, every time i think i finished something i find that there is stuffs i should know. what your opinion about this?is that because group theory is huge?
With the second, one has $|x|=x$. But for the first it's $|x|=-x$.
In both cases, the left-hand side is positive. But the right-hand side is positive for the first one only when x>0, and positive for the second when x<0.
There are two aspects to that curve. On the one hand, if $x$ is large (positive or negative) then x^3 is the dominant term. Since this doesn't depend on a, we should have that the plot of $f(x)$ won't change much if we change $a$.
Suppose the fourier series coefficient of Y(t) is Cn then what will be the fourier series coefficint of Y(2t+1)?
My doubt regarding this is if we do first shifting and then scaling ans will be ejnwCnejnwCn but if we do first scaling and the shifting then it will be e0.5jnw)Cne0.5jnw)Cn why is my answer changing and what is the correct procedure?
I don't know anything about stacks or topoi, I just do it all memetically. I don't know what I'm going to end up going for, but I will say that I have less concern than most about things being practical to society, nice as that may be, and more about cool ideas coming together. And see you!
@SoumyoB So take an random variable $X$ such that $|X|\le1$ and $E(X) = 0$. First we see that $E(e^{tX}) \le e^{t^2/2}$ for all $t\in \Bbb R$ by convexity of the exponential. Now take $X_1, \dots, X_n$ centered independant random variables such that $|X_i|\le a_i$ for some $a_1, \dots , a_n$, and finally let $S_n = X_1 + \dots + X_n$. From what we just did we deduce $$E(e^{tS_n}) \le \exp\left({t^2\over2}\sum_{i=1}^n a_i^2\right).$$
Thus by Markov inequality, for $\epsilon,t\in \Bbb R_+^*$, we have $$P(|S_n|\ge \epsilon) \le 2\exp\left(-{\epsilon^2\over 2\sum_{i=1}^n a_i^2}\right).$$ We then apply this more specifically to a sequence $(X_n)_{n\ge 1}$ of random variables following the same Bernouilli law of parameter $p$ to find strictly positive reals $\epsilon_n$ such that $$P\left(\left|{S_n\over n}-p\right|\ge \epsilon_n\right) \le {2\over\sqrt n}$$ for $n\ge 1$.
This is just one example among many in an exercise sheet
The point of that topology is that it makes the currying a homeomorphism
You know there's a bijection between the continuous functions from $X\times Y$ to $Z$ and those from $X$ to (the space of continuous functions from $Y$ to $Z$) via this currying
So that's kinda the point of the compact open topology, to make that natural bijection into a homeomorphism
I mean that's my heuristic for believing that it's probably not the coarsest one, it just doesn't seem like that definition leads to it naturally, like it'd be a huge coincidence if it were
I have a bivariate function $f(a,b)$ that takes 2 positive integers as input and gives another as output. I do not know the "inner-workings" of the function — I can only see the value it returns when I give it any 2 variables. I would like to represent this function with an equation.
My naive at...
Have you tried making a table of $f(a,b)$ for $a=1,2,3,4,5...$ and $b=1,2,3,4,5...$ and then looking up the values in the inverse symbolic calculator: https://isc.carma.newcastle.edu.au/advanced
@Astyx An endlessly winding set of topics, from the question "do mathematical objects exist" to the nature of human consciousness, and how we as two individuals interface with the world around us.
And I'm glad to hear you're getting better. Low morale is something I struggle with frequently, I wouldn't wish it on anyone else.
That's my line of thought. I tend to think I overestimate how bad I actualy am, which is why I didn't bother at first. But these keep on happening regularily, so I might just as well try
What I am saying is, one of the things that's going on is that I don't want to feel better (presumably, better means being happy and functioning), if that's what lies on the complement of feeling bad.
