@sevdaicmis I frequently had students in my upper-division math courses who had passed basic calculus and linear algebra without having learned much of anything. I frequently complained that my colleagues' standards were woeful.
@sevdaicmis On the other hand, I had plenty of wonderful students, too, who were very much on the ball. And then there were the poor souls who chose to take many classes from me ...
Heading off to dinner. Y'all misbehave without me!
I am watching a lecture series on macroeconomics that is well above my prereq level (it is a graduate course and my experience with macro amounts to watching a shortish YT series), mostly in order to test learning from content far above my prereq level. In the context of an applied calculus of variations problem (which I also have only the most basic exposure, but the lecturer doesn't expect any), I found this slide: i.imgur.com/PYwl3eJ.png
I recognize this formula as a Laplace Transform or Fourier Transform, but the lecturer does not seem to point this out. How should I interpret this? It seems the discount rate $\theta$ is the frequency domain variable, but a discount rate doesn't seem analogous to a frequency description.
I guess just additional associations to hook into each concept. I recently discovered that moving averages are built on the same mathematics as some audio signal filters, for example.
@robjohn Am a bit late (it's well past morning now lol), but this here is a slightly different variation of the algorithm. You can see it's fairly linear in change from one term to the next in the exponent. Most of the cost is computing the exponent and then the necessary O(log n) time required to compute the final sum. In hardware, this would allow for arbitrary divides in O(n log n) which is optimal. The same circuit could be modified as well to compute both multiplies and divides.
What is the difference between method and algorithm in math, not in computer science? pls look at this question or share any link which clear the topic. I did not fine one.
i do not think that mathematicians have specific meanings attached to either beyond the usual English interpretation, but i also do not think the terms are interchangeable.
Because in mathematics we called these terms methods and algorithms. That's i want to know what is the difference. Is there any specific reason to be a method or algorithm.
suppose that you get an offer for a job that you really like but you are already employed at a company that pays you huge for the kind of job that you don't like. You wish to leave your current job soon as you have some plans. Due to the plans, you don't want to take up the new job offer also. How would you reply (declining the offer :( ) to the email for the new job offer?
that's the thing. I also want some things to be discussed in person. I can't meet them as to meet them I'll have to fly to the capital. And you know the current travel scenarios.
they also said -if I don't want to take up the job, they'll be pleased to know about my plans as they would be glad to help in any way possible for them.
And the ambiguity makes both interpretations funny, the first being that he's so bad at hide and seek that they die before he finds them, and the second as a reference to "you'll never catch me alive, copper!"
I'm stumped by a simple problem which is likely too trivial for a post. Sadly, Google turns up nothing... Given a vector, and a plane defined by three points, how can one determine the reflection of the vector in the plane (context: math.stackexchange.com/a/4348433/822508 ...which is what I would call a reflection "about" the plane)
so I have the sequence $x_n = \left ( 1+\dfrac{1}{n} \right)^n$ and $y_n = \left ( 1+\dfrac{1}{n} \right)^{n+1}$, since $y_n$ is decreasing a bounded below (by 0) and since $x_n = y_n \cdot (1+1/n)^{-1} = y_n \dfrac{n}{n+1} < y_n $ then $y_n$ is an upper bound for $x_n$ and hence $x_n$ has an upper bound and its increasing since convergent
God my brain went off for a minute.
I meant "hence $x_n$ has an upper bound and its increasing hence convergent"
That's correct for the convergence of $x_n$ right?
OK. So the sequence $(y_n)$ is decreasing and is bounded below by $0$. Hence $y_n$ is convergent that is $\lim y_n$ exists. Now $(1+\frac 1n)\to 1$. So by algebra: $x_n=\frac {y_n}{1+\frac 1n}$. Using limit rules (if $z_n=x_n/{y_n}$ and if $\lim x_n$ exists and $\lim y_n$ also exists but is non zero then $\lim z_n$ exists and equals $\frac{\lim x_n}{\lim y_n}$), it follows that $\lim x_n=\frac {\lim y_n}{\lim(1+\frac 1n)}$
Yeah, I think I wasnt clear with my question. What I tried to ask implicitely is, if $x_n<y_n$ for all $n$, and if $y_n$ is convergent, does $x_n$ has an upper bound?
@Odestheory12 I think that you shouldn't conclude directly from here. The problem is that $x_1<y_1, x_2<y_2$ etc. but what about $x_1<y_5$ for example. So you should show that for any m and n, $x_n\leq y_m$
Huh, y'know I feel kind of dumb now. Solving recurrence relations mod n is actually quite easy.
Just solve as though there weren't a recurrence relation, then apply modulus. This works assuming valid modular arithmetic identities since $x \equiv x \pmod{y}$.
Formally, if $a(n)$ is a recurrence relation which maps to a field, then it is true that $a(n) = a(n) \pmod{y}$.
err $a(n) \equiv a(n) \pmod{y}$
so, we can therefore say that $a(n + 1) \equiv a(n + 1) \pmod{y}$ given $a(0) = 2^{\lfloor\log_2(x)\rfloor + 1} - x$ and $a(n + 1) = 2^{f(x)} a(n)$ and $f(x) = \text{let it suffice to say that it's faster to write that it's a power of two of integer inputs and a given that the domain maps to a field LOL}$.
(An integer field specifically)
According to wolfram alpha, the explicit formula for this particular recurrence equation adjusted to be mod y is $a(n) = 2^{n f(x)} \left(2^{\lfloor\log_2(x)\rfloor + 1} - x\right) \pmod{x}$ for an integer $x$.
If this is correct, then the coefficient for any given term in my division algorithm is described by this formula.
In other words, we can now efficiently compute quotients to arbitrary precision. Only thing I will have to work out when I write an arbitrary precision math module is how many terms I need to achieve any given precision. Otherwise, we can determine this experimentally for 64-bit inputs and just compute that every time.
@robjohn Anything to say? (Is there any error per chance in my reasoning about recurrence relations mod y?)
Ok, so this $a(n)$ I've computed is actually the value of the numerator in the fractional part at the nth term in the division algorithm.
Or rather the coefficient in the numerator.
Computing the term coefficient we'll call it is then a matter of elementary fraction arithmetic.
As-is, this computes $\frac{2^x}{y}$ (which I mistakenly listed previously as $\lfloor\frac{2^x}{y}\rfloor$). We can easily convert it to a function of $\frac{x}{y}$ by substituting $x$ with $\log_2(x)$ and simplifying.
I will write down the expression as an infinite sum when I get the chance.
Now that I've solved my division problems, I'm going to look for that closed form definition for the Gamma function for complex arguments.
Me want fast binomial coefficients and real-time Mie scattering
Q : A sequence of integers $a_{1}+a_{2}+\cdots+a_{n}$ satisfies $a_{n+2}=a_{n+1}$ $-a_{n}$ for $n \geq 1$. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. Find the sum of the first 2002 terms.
My questions regarding this problem are:
What will be the 1st ...
Seems it isn't quite complete yet. Looks like the coefficient for $f_{lp2}(x)$ isn't purely linear. Alright, cool. Lemme just find that real quick. My bad.
Yep, seems to be the case. If you look at the fractions that we get incorrect results for (excluding powers of two), then you get that they are an integer multiple of the correct result. For example, I get $\frac{1}{36} = \frac{1}{4\cdot 9}$ for $f_{div}(53, 9) 2^{-53}$ (approximated using 16 terms of the series).
Now, continuation 1: cancelling fractional powers yields: $-\frac 52 (y''')^2+y''''y''=0$. From here: the order is $4$ and the degree is $1$. Continuation 2: removal of the fractional powers using square: $\frac {25}4(y''')^4(y'')^{-7}=(y'''')^2 (y'')^{-5}$ and taking $(y'')^{-7}$ to RHS suggests that the ode has order 4 and the degree equal to $2$.
@SayanChattopadhyay this only depends on properness and yes. in fact, you can show that any continuous map into a LCH space is proper if and only if it preserves nets going to infinity. Here, going to infinity means eventually being in the complement of any compact subspace, which is the same thing as having norm go to infinity on euclidean spaces.
Actually, in the case of monoidal's query, I think one doesn't need all of classification. Because of free abelianness, I think it's not hard to argue that the projection to the quotient splits.
I've created a simulation of a triangular billiard table with the ball bouncing off the wall and whatnot, but I'm not quite sure how to go about reproducing the Schwartz's result using my simulation.
If anyone has any suggestions, that would be much appreciated.
Here is an image of my simulation at work, if that inspires any ideas:
Actually, I'm not sure I'm making any sense -- are folks here familiar with the problem I'm describing, or would it help to clarify?
hello, i want to prove the existence and unicity of solution of the differential equation $y'=-x^2y^2+exp(-y^2)$ over $R=\{(x,y), 0\leq x\leq \frac12, |y|\leq 1\}$ using banach fixed point. My question is how to choose the complete normed space ?
i know that the operator is $T: E\to E$ defined by $Ty(x)=y_0+\int_{x_0}^x -s^2 y^2(s)+exp(-y^2(s)) ds $
can i choose $E=C([0,\frac12],[-1,1]),||.||_{\infty})$
@rb3652 I for one am not at all familiar with the problem, but given that you seem to have a clear goal and a functioning simulation, I think this would make for a good question to post on the main site (unless someone else here happens to be familiar).
In 2009, Richard Schwartz proved that any obtuse triangle whose largest angle is $≤100^{\circ}$ has a stable periodic billiard orbit. My question then, is:
How can I reproduce Schwartz's result using a simulation?
I've created a functioning simulation which produces paths in a triangular billia...
And now this very detailed and specific answer gets a downvote? With no comment from anyone? Seriously. (I was surprised the OP approved the other answer, since it used stuff above his pay grade, I thought. But that's fine.)
It's actually an argument that is usually done just by "path lifting in covering spaces," but it's easy enough to make an elementary argument for an undergraduate.
LOL, @PenAndPaper, you're practicing to be a politician.
What does it mean when we say function has minimum at x=a, does it mean the function has local minimum at x=a or the minimum value of function is obtained at x=a?
I want to find the order and degree of the ODE $\frac{\mathrm{d}^3}{\mathrm{dx^3}}(\frac{\mathrm{d}^2y}{\mathrm{dx^2}})^{-3/2}=0$?
I'm getting two different values of the degree of the ODE.
Simplifying gives: $-\frac 52 (y'')^{-\frac 72}(y''')^2+y'''' (y'')^{-5/2}=0$
We have two possible continua...
Order of an ode is: the highest order of the derivative that appears in the ode, whereas the degree of the ode is defined as the power of the highest order derivative.
By ode, I mean: $F(x,y,y’,y’’,…,y^{(n)})=0$.
I don’t know how to say this. But for writing degree we write the ode as powers (positive integral) of derivatives $y^{(i)}$.
That is, the degree, if exists, is a positive integer.
In 2009, Richard Schwartz proved that any obtuse triangle whose largest angle is $≤100^{\circ}$ has a stable periodic billiard orbit. My question then, is:
How can I reproduce Schwartz's result using a simulation?
I've created a functioning simulation which produces paths in a triangular billia...
