Harry’s magic wand breaks at a random point (location of the point is uniform along the length of the stick, which is $40\ cm$ long). Suppose the piece of the stick that Harry is left with is $X\ cm$ long. Unfortunately, the next day part of the stick is accidentally burnt while casting a spell. After this accident, the length of the stick is reduced to $D\ cm$, where $D$ is uniformly distributed between $[0, X]$. Find $f_D (d)$.
Here's the solution: i.imgur.com/YuPidi8.png What I don't understand is why the bounds of integration in the final step go from $d$ to $40$. Why wouldn't it go, for example, from $0$ to $d$, or from $0$ to $40$?
If $f(x)$ and $g(x)$ are periodic functions with period $7$ and $11$ respectively. Then the period of $F(x)= f(x) g(x/5) - g(x) f(x/5)$ is
Let’s call $h(x) = f(x) g(x/5)$, and we know the period of $f(x)$ is 7 and $g(x/5)$ is 11/5. Then, the period of $h(x)$ is $$ T_1 = k’ \times 7 = k’’ \times 11 \implies \frac{k’}{k’’}= \frac{11}{7} \\ \therefore k’= 11 ~~~, ~~~k’’ = 7 \\ T_1= 77$$
Let’s call $p(x) = g(x) f(x/3)$ and doing the same thing as before $$ T_2 = k’ ~ 11 = k’’ ~7/3 \\ \frac{k’}{k’’} = \frac{7}{33} \\ k’ = 7 ~and~ k’’ = 33 \\ T_2 = 77$$
Therefore, the period of $F(x)$ is $77~n$ (where n is any integer). But the answer given is $1155$ (which is not divisible by 77). Where’s my mistake?
I posted a question a few moments back, to which I've accepted an answer(the answer is enlightening, and already has 4 upvotes). Here's the link: math.stackexchange.com/questions/3817624/…
Just a moment ago I discovered the same question with a very mild change. Should I consider deleting my question? (Here's the link to that similar question: math.stackexchange.com/questions/508733/…)
Hmmm... to give you an example, on a different sample I got this question “Find $f(x)$ given that $f’(x) = f(x) + \int_{0}^{2} f(x)dx$ and $f(0)= e^2$”
I couldn’t do it when clock was ticking, so stopped everything and went out to my balcony.
After washing my face and looking out, I contemplated on it for few minutes and realised it was just a simple linear differential equation. I just ...
So I highly recommend trying to do a test exam again with the questions.
And this time being a bit more lenient with skipping.
Remember you have 100 questions, and you only need 35 right ones (+ more if you make mistakes).
@TedShifrin I had a question for you: if I have a surjective map from the closed disk $D\to\mathbb C$, does it necessarily have a fixed point? The case where the codomain is $D$ again is negative, but every map I can think of for this case has a fixed point, but I don't have a good reason for why this might be.
I need your advice on this. How about if I do just 25-30 of them and do some guess work wi the remaining questions? That is, ticking all the remaining questions with option 3.
Four answers, and 1/4 is right on average, 3/4 is wrong. In the case you are right, you get 4 points for it, otherwise you get -1. So it's 70*(1/4)*4 - 70*(3/4).
why did I get the talkative badge when no one responds to me anymore presumably because some psychology major told them that doing so is just enabling me
@TedShifrin I was thinking the Riemann mapping theorem to get a biholomorphic map $D$ to the open upper-half plane, then shifting it around and stuff and applying rotations. But the application of a rotation undoes the shifting for some points (courtesy of the fundamental theorem of algebra) and you get a fixed point when you go back.
By rotation, I mean something like $z^n$ for $n\geqq 2$.
@geocalc33 I don't know I think it's all with good intentions like maybe giving younger users that kind of emotional boost or praise for all kinds of stuff actually has a positive outcome there and I just see the badges as dumb because im a millennial ie im already dead inside lol
but serious question, suppose you want to prove that an equivalence relation exists on an infinitely large domain, like for example the natural numbers, you would need to prove that all of the equivalence classes for the relation are also infinite in size right?
ok well there is no such guarantee that for all $n$, the sequence $a_n$ increments in either four, five or six consecutive equal values, just because that's what it's enumeration implies in terms of inductive reasoning, there could be values of $n$ that don't fall within those three definitions
