@satan29: look at the actual solution: $\frac\pi4e^{-2|t|}$. It's first derivative at $0$ is discontinuous. Is it surprising that the second derivative has problems?
Hello, I am trying to understand why if we are trying to prove a topology is contained in another, i.e., $\tau_1\subseteq\tau_2$ why is it sufficient to show that any basic open neighborhood of any point in $\tau_1$ contains an open neighborhood of this point in $\tau_2$?
So I can take a basic open set in the first topology, and show that it is open in the other topology - then I will be done, since any open sets are unions of basic open sets
So I take any basic open set, and any point in it. Then I need to show that that basic open set is open in the other topology. It is open in the other topology if it is a union of basic sets from that topology. If there is an open set containing that point in the other topology...then I am stuck
Do I first take an arbitrary point, or an arbitrary basic open set?
I think I got it. If you take any basic open neighborhood in the first topology, and for each point in it, there exists an open neighborhood of that point contained in that topology, then that basic open neighborhood must be open (as a union perhaps) of those contained neighborhoods
@shintuku you are scaring me a little :-). If $\delta\le 1$ then $|x^2+3-4| \le 2|x-1|$ and so $\delta = \min(1, { 1\over 2} \epsilon)$ will suffice. I do not understand where the square roots are coming from.
@copper.hat sorry, I didn't see this until just now. I have not read the Salmon of Doubt, but my wife has, and I intend to. Globetrotting sounds fun! Glad you were inspired.
I read the Hitchhiker trilogy when it was still only 3 books ;-)
If a professor told us how to do an answer of an physics assignment(which is a part of an active exam) and if I believe that there is a flaw in the logic of the professors answer and if everyone is writing professors answer for their assignment (except me) then: (i)Should I write down professor answer in my own assignment even though I believe that there is a flaw. (ii)Can I ask question about the flaw on this forum?
@shintuku this assignment is a part of an active public exam (countrywide) , and the school (the professor and a second examineer) will check it and submit it to the education board. I think that format of (1) and (2) might not be exceptable by school.
It is okay if I dont get marks for my answer if I am convinced that my answer is right.But it is kind of weird to me if I get full marks for an answer(provided by the professor during an active exam) which I "believe" is wrong.
you have no room to do all three above suggestions (at the same time, not only one or two of them)? also, are you not allowed to call/email the school to ask if you're allowed to correct the question?
Oh ok,I wont ask then :). By the way , I am not allowed to correct the question . I did ask about this dought of mine , but the professor still showed the same reasoning which I believe is false.
Now that I think of it , technically the professor himself is not allowed to show the students how to answer the assignment because it is a part of an active public exam. But the professor is still showing the answer (right or wrong).Isn't this also plagiarism?
I am stuck at this question: It is given that $a_1=a, b_1=b, a_{n+1}=\frac{a_n+b_n}2, b_{n+1}=\frac{a_{n+1}+b_n}2$. It is to be shown that $\lim a_n=\lim b_n$
I showed that: $a_{n+1}-a_n=\frac{b_n-a_n}2$ and $b_{n+1}-a_{n+1}=\frac{a_{n+1}-a_n}4=\frac{b_n-a_n}4$
whence it follows that $\lim (a_{n+1}-a_n)=0=\lim(b_{n+1}-a_{n+1})$
Considering that $a_{n+1}-a_n=\frac{b_n-a_n}2=\frac{b-a}{2.4^{n-1}}$ and assuming $b\le a$, it is clear to me that $(a_n)$ is decreasing but then I have problem proving it bounded from below to confirm its convergence
Once convergence of either $a_n$ or $b_n$ is confirmed then the convergence of the other one follows.
Any suggestions on this? Thanks.
Hi Leslie, how are you?
@shintuku @shin:if your $\epsilon\gt 0$ arbitrary and for each choice of $\epsilon$, you have a $\delta$ corresponding to it (to emphasize this point, let's say $\delta_{\epsilon}\gt 0$) then yes.
further to my earlier comment :why so? because due to arbitrary nature of epsilon, you can say let $\epsilon =\frac{\epsilon'}2$, where $\epsilon'\gt 0$. :)
Sorry to interrupt, but as I just realized, you can represent every larger power of $2^{-n}$ in terms of the smallest power of two that you desire. Then $\sum_{n=1}^\infty 2^{-n} = \sum_{n=0}^\infty 2^{n-M}\text{ for some constant } M\in\mathbb{Z}$.
$a_{n+1}$ is the average of $a_n$ and $b_n$. The average of two numbers is smaller than the larger of the two and larger than the smaller of the two. $b_{n+1}$ is a weighted average of $a_n$ and $b_n$ and the same observation applies. So, inductively, everything stays in the interval between $a$ and $b$.
(contd.) We can exploit this for $a$-bit dividend and $a$-bit divisor to compute quotients using the inequality $\sum_{n=0}^{\leq a} 2^{yn - M} \leq 1$ to compute the reciprocal $\frac{1}{y}$, then multiply by $x$ to approximate $\frac{x}{y}$.
This makes the divisors homogeneous, simplifying calculations and making it possible to parallelize the operation, and I haven't thought far enough into this yet so I'm uncertain about the requirements of the remainders for each individual term of the sum and how relevant they are.
yes, I have had LA during my first year at college.
I have figured out that question now professor Rob.
Instead of trying to analytically prove the whole thing, I should have plotted the given info. to see what it meant. I think that was the key to solve the exercise.
i think the idea is to set up recursion problems as application of a matrix to a state vector so that applying the matrix to the $n$th state vector gives you the $(n+1)$th state vector. Then the long term behavior of the sequence can be read off from the eigenvalues and eigenvectors of said matrix.
I tried to find nth term of $a_n$ and found it as $a_n=k\frac14 (\frac 1{4^{n-1}}-1)+a$, where $k$ is a constant. It follows that $a_n$ converges and therefore from relation deduced between $a_n$ and $b_n$ above, that $\lim b_n=\lim a_n$
@Quin i see, thanks. i have never seen that before :-(
The equations $$ \begin{align} a_{n+1}&=\frac{a_n+b_n}2\\ b_{n+1}&=\frac{a_{n+1}+b_n}2 \end{align} $$ are simply $$ \begin{align} a_{n+1}&=\frac{a_n+b_n}2\\ b_{n+1}&=\frac{a_n+3b_n}4 \end{align} $$ which gives the recursive relation $$ \begin{bmatrix}\frac12&\frac12\\\frac14&\frac34\end{bmatrix}\begin{bmatrix}a_n\\b_n\end{bmatrix}=\begin{bmatrix}a_{n+1}\\b_{n+1}\end{bmatrix} $$ The matrix has eigenvector $\begin{bmatrix}1\\1\end{bmatrix}$ with eigenvalue $1$ and eigenvector $\begin{bmatrix}-2\\1\end{bmatrix}$ with eigenvalue $\frac14$.
@robjohn I don't understand what is "eigenvector components from $[a,b]^T$" but I got the idea which I think is $Ax_n=x_{n+1}$ whence $x_{n+1}=A^n x_1$ and then distinct eigenvalues of A ensure that $A$ can be diagonalized hence there exists an invertible matrix $M$ such that $A=M \Lambda M^{-1}$, where $\Lambda$ is a diagonal matrix with eigenvalues and it is also to be noted that $A^n=M\Lambda^n M^{-1}$
And I also realize now that difference equations which you suggested in past also follows from this method.
I have now also determined $k$ here as $\frac 83(a-b)$ and therefore the expression obtained for $a_n$ is same as the $a_n$ expression obtained in method suggested by you above. And this is as expected as that difference method as I realize now is a a special case of this matrix approach.
@robjohn I don't understand what is "eigenvector components from $[a,b]^T$" but I got the idea which I think is $Ax_n=x_{n+1}$ whence $x_{n+1}=A^n x_1$ and then distinct eigenvalues of A ensure that $A$ can be diagonalized hence there exists an invertible matrix $M$ such that $A=M \Lambda M^{-1}$, where $\Lambda$ is a diagonal matrix with eigenvalues and it is also to be noted that $A^n=M\Lambda^n M^{-1}$
@robjohn wow wow wow wow. it's fantastic. I understand it now. So we had $x_{n+1}=Mx_n$ whence we got $ x_{n+1}=M^nx_1=M^n [a,b]^T=M^n(x [1,1]^T+y[-2,1]^T)$, where x and y are eigenvector components of $[a,b]^T$. And then we know that $M^nv=\lambda^n v$. And thus the last expression follows. Thanks a lot professor Rob. :)
I'm pretty sure I'm converting this recursive relation to an explicit relation incorrectly because I don't know how to get rid of $f\left(n-1\right)$. desmos.com/calculator/nwzbxc3kgt
I don't think the epsilonics are the interesting thing here. It's a question of understanding why the iterated limit is different from the joint limit. Can you do actual examples?
I would break the problem into pieces: Let $g(x)=\lim\limits_{y\to 0} f(x,y)$, and consider $\lim\limits_{x\to 0} g(x)$.
hm does the nested limit have it's own epsilon and delta in $\forall \epsilon \exists \delta: 0<|x|<\delta \implies |\lim \limits_{y\to0}f(x,y)-L| < \epsilon$?
the issue is that I'm trying to prove a specific theorem about precisely the epsilon, that $|f(x) - L| < g(\epsilon)$, where $\lim \limits_{y\to 0} g(y) = 0$ and $g(y) > 0$ implies $|f(x) - L| < \epsilon$
@shin: please note that in limit definition $\epsilon\gt 0$ is arbitrary so when you put $g(\epsilon)$, you are restricting the arbitrary nature of $\epsilon$ and if restricted nature of $\epsilon$ implies something, there is no reason to believe that the implication will hold for all arbitrary values of $\epsilon$.
@shintuku elaborating what Ted wrote (and before dealing with the $g(.)$ lemma), try showing that if $\lim_{x\to x_0} f(x) = L$ then for all $\epsilon>0$ there exists some $\delta>0$ such that if $|x-x_0| < \delta$ then $|f(x)-L| < {1 \over 5} \epsilon$.
is there a way with new notations to always be able to write (readable) a bigger number on the same area like a DIN A3 piece of paper or whatever size your favourite is.
the piece of paper must be finite (...)
but you of course always get a new paper if you want to write down a bigger number
you would need a more precise definition of notation, symbol, etc for this to have an answer. otherwise i can just introduce a new symbol each time i want to represent a bigger number.
Thanks for the suggestion, @robjohn. You may be right about looking for a generalization -- it feels like your answer and other people's answer before that have used contextual arguments to motivate $o\!\left((hu)^m\right)=u^mo\!\left(h^m\right)$. For example, you write that "...the article is looking at letting $h\to0$ in the estimate, while $u$ is integrated over $\mathbb{R}$...". This makes sense, however, with this argumentation it seems like one has determined $\textit{from context}$ what the definition of little-o is (i.e. which variable the limit should be taken with respect to). Thi…
that's one possible advantage of having kids around. most of my relatives who were frequently surrounded by kids lasted a lot longer (and in better mental health) than those who were more isolated.
