Once again thank you all... I always afraid to ask question here (You guys are strict teachers, exceptions are there). But i cleared my many confusions here.
@TedShifrin You are right , i made many mistakes. I asked many silly questions i know that. Because there are lot of lot of terms which mixed and has many meanings in same and different context. That confused me. Then i asked questions here.
I would like to add to Ted's previous comment that: definition of transcendental/algebraic does not seem to be a universal one. For example: if we define algebraic functions as roots of "polynomial equations", then Dirichlet function is algebraic but if we define algebraic function as "the functions which can be expressed in terms of x using finitely many operations" then Dirichlet's function is transcendental.
The latter can be seen in Terry Tao's analysis book.
So what's an easy proof that $\sin$ is transcendental, @Koro? I guess I can use the fact that a line in $\Bbb K^2$ cannot meet the zero-set $P(x,y)=0$ of a polynomial in infinitely many points unless that zero-set contains the line.
No, I don't think so, @Koro. I think there's a lot of language confusion.
The easiest proof for sin being transcendental (using the polynomial equation def.) is the fact that no periodic function (with infinite range) is algebraic. I read that some time back. I even posted on mse as I had one confusion about algebraic nature of periodic functions. I'll share the link shortly.
yes, there seems to be a lot of language confusion above.
Right, an algebraic function cannot take on a value more than finitely many times. I know this, as @leslie pointed out, with a bit of algebra (unfortunately), but ...
Definition: A function $f(x)$ is said to be algebraic if it satisfies the following equation in $y$:
$y^m+R_1(x)y^{m-1}+R_2(x)y^{m-2}+\cdots+R_{m-1}y+R_m(x)=0$, where $R_i(x)'s$ are rational functions i.e. of the form $\frac {p(x)}{q(x)}$, where $p(x),q(x)\ne 0 \forall x\in \mathbb R$ are polyn...
The Wallis productsuggests that pi is irrational. If pi were rational, then we could cancel terms in the Wallis product and make both the numerator & denominator finite. But there's an infinite number of factors of 2 in the numerator and every number in the denominator is odd, so we can't can cancel any of those twos. (That's not supposed to be a proof, just a suggestive heuristic).
Suppose that $\displaystyle A=(a_{i,j})_{n\times n}=\begin{cases}k,\quad i=j,\\1,\quad i\not=j\end{cases}, k\in\mathbb{R}$. What's about of the $\det(A)$? I'm trying to find a closed form using row operations or Laplace expansions but it seems doesn't work well. Any hint?
@TedShifrin nice, yes we can find that $\det A=(k-1)^{n-1}(k-(n-1)1)$. For example $\det A_2 =(k-1)^{n-1}(k-(2-1)1)$, $\det A_3= (k-1)^{3-1}(k-(3-1)1)$, $\det A_4=(k-1)^{4-1}(k-(4-1)1)$, etc. Now I suppose I need to prove it using induction.
Lol, yeah. Still tryna wrap my head around how we're getting from point A to point B. The intuition for why this could be true hasn't immediately jumped out to me.
1. Even functions: f(-x) = f(x) 2. Odd functions: f(-x) = -f(x) 3. f' >= 0 implies increasing 4. f'' >= 0 implies concave up 5. Derivative of even function is odd, derivative of odd function is even
I mean, I guess I have to think more about why this might be helpful. But I'm pretty sure if the degree of f is even then f approaches infinity as x -> +- infinity
@Derivative Absolutely. Most of us really do like to hear from our students (even if they are wrong). It means that they are engaged and learning. And if you are taking a class from one of the rare professors who does not like to hear from students, better to know now.
@TedShifrin Wait, is that earlier statement true also for odd degree functions?
I don't know why I was thinking it was the opposite
The derivative of an even function is odd and the derivative of an odd function is even. As f >= 0, the coefficient of each derivative will be positive. Limit laws taking the limit of each polynomial is the same as taking the limit of the entire sum. So the end behavior of the entire limit is x->infty.
Suppose that $u'=\phi'\bar{\phi}+\phi \bar{\phi'}+\phi'' \bar{\phi'}+\phi' \bar{\phi''}$, where u, $\phi$ are functions. Then how can it be true that: $|u'(x)|\leq 2|\phi(x)||\phi'(x)|+2|\phi'(x)||\phi''(x)|$?
I mean by triangle inequality, we should get: $|u'(x)|=|\phi'(x)\bar{\phi}(x)+\phi (x)\bar{\phi'}(x)+\phi'' (x)\bar{\phi'}(x)+\phi' (x)\bar{\phi''}(x)|\leq |\phi'(x)\bar{\phi}(x)|+|\phi (x)\bar{\phi'}(x)|+|\phi'' (x)\bar{\phi'}(x)|+|\phi' (x)\bar{\phi''}(x)|$. From here I can't use AM>=GM.
I was going to say that since f goes up at both ends, there's an absolute min. f is always greater than the min. f' = 0 at the min. So f + f' >= 0. But I don't know if that fact holds for f'' + f''' + ...
This is badly written. So I know some this might be jumping the gun.
xander the problem with my investment project is that tonelli's went out of business and was replaced with another business. so, a substantially larger up-front cost on that project.
So you can write equations for the configuration space and figure out where it fails to be a manifold, etc. I don't think this has anything to do with epicycloids.
@TedShifrin BTW, I'll be presenting my thesis over Zoom in the spring. Any interest in attending virtually? I was hoping you might come since I ended up figuring out the circle discussions based on ur feedback a year or two ago
awesome! also, would you like the family xmas card? we are sending it out late, but i'm gonna be mailing them today or tomorrow. so if you would, just shoot me an email with your address and i can send you one :)
my grand^nth father was the gardener at muckross abbey. i've dropped this in every interview and have yet to see it move the needle either way, but i have hope.
Oh duh, g(x) = f + g'(x) is even degree. So absolute min. at some point p. g(x) is everywhere greater than g(p) and g'(p) = 0 at this point. So we know g(x) > g(p). Observe also that g(p) = f(p) + g'(p) = f(p). And as f(x) >= 0, we have g(x) > f(p) >= 0.
Done, @Ted.
Well, could be written out better, but I feel pretty good about the reasoning
Hello, I have this recurrent sequence $u_{n+1}=u_n+1/u_n$ with $ u_0=1$ the question is to find the nature of the sequence. I say as $u_{n+1}= f(u_n) $ with $f(x)= x+1/x$ f is increasing and not bounded then the sequence is divergent
But the hint says prove that $lim u_n=+\infty$ can we see this without using f ?
ok, here is another way since you are not doing the basics
$f(x) - x = {1 \over x}$ and $f$ maps $(0,\infty)$ into itself, so $f(x) >x$ and hence the sequence is increasing. if it is bounded then $f(x) = x$ for some $x$ which means what?
@UnderMathUate Yes, you got it. I would use $\ge$ instead of $>$, because you don't actually know $g(x)>g(p)$ for $x\ne p$ (and, besides, you don't have to keep saying $x\ne p$).
A problem in your book involving a 3 digit number which when divided by the product of digits yields the hundred's digit. I have proved it but by truly laborious means. Anyway, I will try more but may come running.
I'm pretty sure this is the case when x is the length of an arc of a circle from $|\sin(x)| \leq x$ inequality but this should also be the case here, I've tried on extreme examples
Fun fact that has the potential to be a theorem if I can prove it: for any natural $n$, and natural $b > 1$, $\lfloor\frac{b^{\lfloor\log_b(n)\rfloor + 1}}{n}\rfloor = b - 1$ iff $n$ satisfies $b^{\lfloor\log_b(n)\rfloor} < n < b^{\lfloor\log_b(n)\rfloor + 1}$.
Actually, now that I think about it, this should be easy to prove.
Huh, and if we think even more about it, this actually could be useful in certain applications (a relationship to factorial comes to mind).
Seems it only holds for $b=2$, so it makes more sense to try and prove $1\leq \lfloor\frac{b^{\lfloor \log_b(n)\rfloor + 1}}{n}\rfloor \leq b - 1$ instead.
Well, I was commenting on the reference to the "state of thinking". It seemed to me that you were saying "scratch your head" as something like "I'm thinking." I have also seen some mathematicians scratch their beards while thinking and since mathematicians think a lot, many mathematicians like beards. It was in that sense my comment
Definition: A function $f(x)$ is said to be algebraic if it satisfies the following equation in $y$:
$y^m+R_1(x)y^{m-1}+R_2(x)y^{m-2}+\cdots+R_{m-1}y+R_m(x)=0$, where $R_i(x)'s$ are rational functions i.e. of the form $\frac {p(x)}{q(x)}$, where $p(x),q(x)\ne 0 \forall x\in \mathbb R$ are polyn...
