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00:16
Hi guys, how can I approximate the value of ln(1.4) using taylor series and lagrange formula ??
Because as f(x) = $ln(1.4)$, $f^{(n)}(x) = 0 $
And lagrange formula is always 0
My a is 0 , and my error margin is $ 5*10^-5 $
00:39
ughhh
im at a standstill right now
01:03
@meow-mix Why's that?
too young to buy book on my own
01:26
That's unfortunate.
Artin's "Algebra"
so i have to wait until i get that
01:41
I'm trying to write a CV for grad school apps... What needs to go on one?
Hey everyone, thanks in advance if you help me out. I'm a bit confused, how do I calculate a sin of an angle when a required side of the triangle is unknown? I mean I can't be expected to memorize every ratio for every angle, there's some formula for determining the ratio, then I can use the ratio to find my missing side length?
For example this website https://www.mathsisfun.com/sine-cosine-tangent.html

Has a boat problem near the bottom of the page. Sine is useful to find how deep the anchor sunk when we know the angle and length of the chain, just use a calculator to get sin(angle) and use substitution! HA what a effin joke. The whole point of doing it manually and not inputting the problem into a calculator/program is it do it manually, not to use a calculator for HALF of the problem...
@CausingUnderflowsEverywhere
remember SOHCAHTOA
Sine = opposite over hypotenuse
we know the hypotenuse is 30
so, if we let o denote the opposite
we know that sin of the angle = o/30
because, opposite over hypotenuse
and we also know the angle is 39
@CausingUnderflowsEverywhere There are only so many angles that we can memorize or use rules for. It's not a joke that we have to use calculators for the other angles.
01:53
so sin 39 = o/30
now, multiply each side by 30 to get
o = 30 sin 39
One does not simply use lowercase o as a variable in math...
3
then you calculate that using your calculator
Shhh ardi meow is explaining something cool
meow-mix is just saying everything that's already said in the linked page
ardi?
I dont have a calculator. Every calculator I touch encounters a stack underflow. This is why I need to know how to do it manually without one.
01:55
umm
taylor series approximation?
thats the best you could do
if you have the time to do the mental math
Asking to figure out sines of random angles like 39 manually, well, that's either a joke or a completely unnecessary but perhaps interesting personal challenge (like being able to multiply ten digit numbers mentally).
Well whats the source code of a calculator? Does it use recursive calculations or something?
some calculators use taylor series approximation
I just (insert word here) to pass a course on trig without truly understanding how a calculator does it.. thanks Ill research that!
02:00
You are not a calculator, you are a human. It takes a lot of work and background to compute sines of angles to good accuracy. Large amounts of tedious calculations is exactly the sort of thing we invented calculators and computers to do in the first place.
3
Wait a second, can't I use a compound angle formula then plot it on a unit circle? That would require a protractor for accuracy
Or, my bad I dont need the compound do I if I just plot it
Let's dissasemble calc.exe and extract Microsoft's way of of calculating angles! To the disssassembly chat exchange!
Please star my underflow joke message too ;)
To do: write a program in C++ to approximate angles!
I've gained some additional knowledge here today. Thank you @arctictern @meow-mix
02:26
that was a lot of work
now I can do my research again
02:46
Any hint to parametrize $x^2 = (y^2 + z^2)^2$
use cylindrical coordinates
Then should I do two parametrizations?
I am not allowed to take the positive $x$ and negative $x$.
03:03
Why does $ \sum n^n * 0 $ diverges ?
03:32
@Maks Is that $(\sum n^n)*0$ (which is $\infty\cdot0$ and thus undefined), or is that $\sum(n^n*0)$ (which converges to $0$)?
its $ \sum (n^n * 0) $
Wolfram says it diverges
What are your indices?
From where does $n$ start?
$\sum_{n=0}^\infty$ or $\sum_{n=1}^\infty$? @Maks
From 1 to infinity
Then Wolfram says it converges…?
I typed in sum_(n=1)^infinity n^n * 0, and it correctly said it converges
Yes, seems like it was my bad
Thank you
03:51
Is this paper arxiv.org/abs/1611.05119 not loading for anyone else?
I'm doesn't work for me.
04:07
@MikeMiller I emailed the author. Figured he'd like to be informed about something like that.
Hey quick question: I'm looking to numerically approximate f(x) = log(e^x + 1) - for large x the exponential can overflow. Any ideas?
HOLY CRAP
I just realized
if you take any product of 9, so long as it isn't a multiple of 99, and add all the digits
You get 9
@arctictern Why give a calculator to a person who can figure it out?
 
2 hours later…
06:13
@Cam.Davidson.Pilon If your exponent is large enough to cause an overflow and you just want an approximation, sounds like you can drop the "1" and compute f(x)=x log(e) for whatever base you're using.
 
1 hour later…
07:22
NB the first time I came across digital root is from a japanese visual novel
@Cam.Davidson.Pilon $$\begin{align} \log\left(e^x+1\right) &=x+\log\left(1+e^{-x}\right)\\ &=x+e^{-x}-\frac12e^{-2x}+\frac13e^{-3x}-\frac14e^{-4x}+\dots \end{align}$$ If $x$ is large, $e^{-x}$ is very small.
07:50
$a_0=x-1$ and $a_{n+1}=\frac{2a_n}{1+\sqrt{1+a_n/2^n}}$. What is $\lim\limits_{n\to\infty}a_n$?
08:03
I rewrote my last night problem: Now it is formal: for Ted and others who where interested in it.
Simply is: $\min \left\{\sum_{i=1}^nx_i|x_i\in \mathbb{R}^+,\,K=\displaystyle\prod_{i=1}^{i=n}{x_i},\, n\in \mathbb{N}\right\}=\displaystyle\min_{n\in \mathbb{N}}\left\{\min\left\{\sum_{i=1}^nx_i|x_i\in \mathbb{R}^+,\,K=\displaystyle\prod_{i=1}^{i=n}{x_i}\right\}\right\}$
Find that min.
Solved.
08:19
Does that min exist?
Oh, wait, I see
I initially misread $\sum_{i=1}^nx_i|x_i$ as sum of all $x_i$ that is divisible by itself
08:36
@Topologicalife It should be $nK^{1/n}$ is it not?
08:49
@Secret yay, number theory
generalized theorem: in a base-n notation, a number is divisible by (n-1) iff its n-digital sum is divisible by (n-1)
Let $f: [a,b] \rightarrow \mathbb{R}$ and $g: [c,d] \rightarrow \mathbb{R}$.
So that the composition $g\circ f$ is a function the only condition is that the image of $f$ is a subset of the domain of $g$, so $f \left( D_f \right) \subseteq D_g \Rightarrow f([a,b])\subseteq [c,d]$, or not?
Or is there also an other condition?
@MaryStar we usually say the image of $f$ is equal to the domain of $g$
but i don't think you are wrong
@DHMO So, only when they are equal the composition is defined?
i am not sure about this
@DHMO No, that is stronger than necessary for this
08:53
Nah, just image being contained in the domain is enough, @MaryStar
@BalarkaSen said that they must be equal right
@BalarkaSen Ah ok... And is this the only condition, right?
@DHMO No. I said it's not really $g \circ f$ that you're doing, but $g|_{f([a, b])} \circ f$. But it still makes sense.
@MaryStar Yep.
@BalarkaSen wow...
Ok. Thank you!! :-)
08:56
is there any interesting example of the composition of two uncontinuous functions resulting in a continuous function?
@DHMO depends on what you mean by interesting
give me anything
$f(x) = 0$ for all $x \neq 0$, $f(0) = 1$. $g(x) = 0$ for all $x \neq 0, 1$, $g(0) = g(1) = 1$. $g \circ f$ is continuous.
is there any function $f: \Bbb R \mapsto \Bbb R$ such that $f$ is nowhere continuous but $f(2f^{-1}(x))$ is everywhere continuous?
@BalarkaSen nice
Define $f(x) = 1/x$ if $x \neq 0$ and $f(0) = 0$.
Then $f \circ f$ is the identity, but $f$ is discontinuous
09:00
@SteamyRoot nice!
a nowhere continuous function $f$ .... hmmm
just discontinuous if it is too difficult
Meh - coffee break time, I'll think about it later
I don't know if it is possible
@BalarkaSen is it possible?
What is possible?
09:05
5 mins ago, by DHMO
is there any function $f: \Bbb R \mapsto \Bbb R$ such that $f$ is nowhere continuous but $f(2f^{-1}(x))$ is everywhere continuous?
Is there a name for the element $a$ obeying the following property?
$$\exists a : \exists x, a+x=1$$?
@Secret In most interesting contexts where that makes sense, all elements satisfy it
But I suppose you care about the semiring with unity case
Yeah, I am kinda wondering what's the + analogue of a zero divisor. I have seen the term zerosum, which is a+b=0 but that basically means a an b are additive inverses of each other
so it's nothing new
if you need to, unitysum lol
@Secret This would not be analogous to a zero-divisor since the unity is not absorbing for addition
09:08
I see
the unity has no distinguished properties with respect to the addition in general
@DHMO Shrug. Not thinking about it much though.
ok
i'm thinking that it is impossible since R is dense (wtf is this argument)
So you have reasons to believe it exists? Everywhere discountinous and bijective functions are pretty ugly
i don have
so i'm asking you
09:11
(Every topological space is dense in itself)
i know
Or did you mean dense as in the linear order on R is dense?
never mind that line of argument
In either case I don't think that's relevant
by the way, $\exists f$ such that $f$ is discotinuous everywhere but $f^{-1}$ continuous everywhere?
09:13
No
why not?
A continuous bijection R --> R automatically has a continuous inverse, even
i see
@BalarkaSen Which of the properties of the reals are we using for that?
Graphically for the case $\mathbb{R}^2$ you cannot flip a function about y=x and causing that to go from continuous to discontinuous or vise versa
09:16
(I can never recall which ones are sufficient for such statements)
@Tobias Such a map has to be monotonically increasing/decreasing.
any non-trivial examples for nowhere continuous functions f and g such that their composition is continuous
So we're using the order on R I guess
@Secret i do not trust graphs
In any case it's even true for R^n by invariance of domain
09:17
@BalarkaSen Ahh
Let f(x)=0 if x is in R\Q and 1 otherwise, f(f(x)) is constant and equal to 1 @DHMO
:o
not interesting
but quite neat
Could you take a look at math.stackexchange.com/questions/2017907/… ? I found in wikipedia en.wikipedia.org/wiki/Perpetuity#Detailed_description why do we have this formula for the present value of a perpetual annuity?
@Alessandro Is there a continuous bijection with everywhere discontinuous inverse for an arbitrary map between two topological spaces though? I can't come up with one.
I don't think we know many nowhere-continuous functions
09:24
@BalarkaSen Hmm, I hardly ever thing of "local" continuity for arbitrary spaces
@Balarka what do you mean "for an arbitrary map"?
@BalarkaSen But I suppose we can take a map from discrete to indiscrete
In mathematics, a nowhere continuous function, also called an everywhere discontinuous function, is a function that is not continuous at any point of its domain. If f is a function from real numbers to real numbers, then f(x) is nowhere continuous if for each point x there is an ε > 0 such that for each δ > 0 we can find a point y such that 0 < | x − y | < δ and | f(x) − f(y) | ≥ ε. Therefore, no matter how close we get to any fixed point, there are even closer points at which the function takes not-nearby values. More general definitions of this kind of function can be obtained, by replacing the...
Maybe something with discrete domain and trivial codomain? I'll think about it after the lecture I'm supposed to be following right now @Balarka
@tobias $f$ locally cont. at $x$: the preimage of all nieghbourhoods of $f(x)$ is a nieghbourhood of $x$
09:25
@s.harp Yeah, I know the definition, I just never think about the concept
@Alessandro @Tobias Yeah, maybe something like that.
@s.harp I am not sure what's unclear.
where does "an arbitrary map" come into play? You are looking for a continuous bijection with everywhere discontinuous inverse, how can this depend on an arbitrary map?
Actually, it should be the other way around of course (from indiscrete to discrete)
The same question was asked previously for a map R --> R. I wrote that to emphasize I am asking for a map between two general topological spaces.
ohh, no it should not, I just misremembered which was to be continuous
09:27
@Secret I think that is the only one we kno
Right, @Tobias.
@BalarkaSen So in fact the identity map works.
@BalarkaSen, ah ok so you meant that one should see if one can do it for every space?
@s.harp Huh? I just asked for X, Y and a map f : X --> Y such that the hypothesis in the qn is satisfied. It can't be done for X = R, Y = R.
Or X = R^n, Y = R^n.
09:29
alright^
Well, you can also construct a rather crazy one such as $f(x)=\{x\in \mathbb{R}: \textrm{Rand}(x)\neq \lim_{\epsilon\rightarrow 0}\textrm{Rand}(x+\epsilon)\}$
The obvious question now is if one can find a metric example. Shrug.
The domain can't be compact if both spaces have to be metrizable
Agreed.
@Secret What is that supposed to define?
09:34
A function over $\mathbb{R}$ that sample values randomly, with the only constraint that no two neighbouring x evaluates to the same value (thus discontinous everywhere). Kinda the most extreme generalisation of Thomas function, I guess...?
@Secret That does not define a function at all
nor does it even actually make sense
@Balarka for a metric example let $X$ be a metric space without isolated points and $\tilde X$ the discrete metric space on $X$. Then the identity $\tilde X\to X$ is continuous but the inverse nowhere continuous.
That works. Cute.
@TobiasKildetoft I have no idea how to define or whether it is possible to define, but my conception is basically This diagram but with the points not forming any nice pattern, thus discontinious everywhere
so in a sense is a random distribution of points on the xy plane
Nice @s.harp
We know that there are nowhere continuous functions satisfying f(x+y)=f(x)+f(y) if you want nontrivial examples of everywhere discontinuous functions @DHMO
09:40
@Alessandro for example?
So, found an everywhere discontinuous $f$ such that $f \circ 2f^{-1}$ is everywhere continuous yet?
@SteamyRoot of course not, lol
it seems that it is impossible
@tobias This might be a better example on what I have in mind, except only x is the independent variable
Everywhere discontinuous is rather painful I guess
09:40
If you have a bijection you can always view it as the identity on some set, then the question is whether the topology of the starting space is strictly finer on a local level than topolgy on the target space
I'm pretty sure I can do it for functions $\mathbb{R}_0 \to \mathbb{R}_0$
@DHMO you can't write one explicitely (you actually need some AC to prove they exist)
@Alessandro what....
You need a basis of R as a vector space on Q to construct that kind of functions
I should get back to my schoolwork now that I'm done with idle point-set topological ponderings.
09:44
that basis cannot be constructed explicitly because you need the AC to prove it exists (Zorn's lemma)
What's your schoolwork about?
$\mathbb R$ as a $\mathbb Q$ vector space is intersting:
It is a (metrisiable, so Hausdorff) topological vector space of uncountable dimension
It has a dense one dimensional subspace ($\mathbb Q$)
Every subspace is dense
@Alessandro it isn't a schoolwork
@Alessandro Literally everything I have taken this semester. Gonna study chemistry though.
@DHMO $f: \mathbb{R}_0 \to \mathbb{R}_0: x \mapsto \left\{\begin{array}{ll}x&x \notin \mathbb{Q}\\ -x & x \in \mathbb{Q} \end{array} \right.$
I'm pretty sure that satisfies $f \circ 2f^{-1}$ is everywhere continuous but $f$ is nowhere continuous
of course, it's not defined on all of $\mathbb{R}$, sadly
09:47
wow...
that is very nice
Meh... :P
It just kind of uses that $2f(x) = f(2x)$, so it's not that nice to be honest
And I don't think you can extend or "adapt" it to all of $\mathbb{R}$
@STeamyRoot its not discontinuous at $0$
and since you are sending to $-x$ I think you want your spaces to be $\mathbb R$ instead of $\mathbb R_0$
($\mathbb R_0 = \mathbb R_{>0}$?)
@s.harp it means $\Bbb R \setminus \{0\}$
Yeah - I had to exclude $0$ to get discontinuous everywhere
Hello @DanielFischer !! Do you maybe have an idea for math.stackexchange.com/questions/2017907/… ? In wikipedia I found the formula en.wikipedia.org/wiki/Perpetuity#Detailed_description Do we use this one? Why does this formula hold?
@MaryStar The question you have does not seem to include an actual interest rate but a fixed payment
10:16
The statement is in german:
Bestimmen Sie den Barwert einer ewigen Rente, die dem Inhaber zu jedem Jahresende eine Zinszahlung in Höhe von 1 € einbringt. Nehmen Sie hierbei einen kalkulatorischen Zinsfuß in Höhe von r an. Der erste Zeitpunkt sei t=0 und die erste Auszahlung erfolge zum Zeitpunkt t=1.

Do you speak german? @TobiasKildetoft
@MaryStar A bit
What of them do we have, an actual interest rate or a fixed payment? @TobiasKildetoft
No idea. My German is not strong enough to be sure
@TobiasKildetoft Ah ok...
@DanielFischer Do you maybe know if we have an actual interest rate or a fixed payment?
Sorry, economics is a language I don't speak. Can't tell it from Burmese.
10:22
@DanielFischer Ah ok...
10:53
@TobiasKildetoft Do you maybe know how we get the formula for PV at en.wikipedia.org/wiki/Perpetuity ? Why is this equal to this fraction?
11:06
@DHMO $$f: \mathbb{R} \to \mathbb{R}: x \mapsto \left\{\begin{array}{cc}
\frac{1}{x} & x \in \mathbb{Q}_0\\
-\frac{1}{x} & x \notin \mathbb{Q}\\
0 & x = 0
\end{array}\right.$$
@SteamyRoot brilliant
Can functions on $\mathbb{Q}$ be plotted?
to those who don't know what is going on: I was trying to find a function $f$ such that $f\circ 2f^{-1}$ is continuous everywhere but $f$ is discontinuous everywhere
@Secret technically every function we plot is a function on $\Bbb Q$
because we can never access $\Bbb R \setminus \Bbb Q$
but which function do you have in mind?
well I am just wondering how steamyrobot's can be plotted cause I am not good at $\mathbb{Q}$ functions
@DHMO Of course we can access non-rationals
11:09
@Secret since $\Bbb Q$ is dense and $\Bbb R\setminus \Bbb Q$ is dense, you can treat it as the "union" of the graph of $\dfrac1x$ and $-\dfrac1x$
@TobiasKildetoft not on computers
each pixel represents a rational coordinate
@DHMO Only if that is the way we normalize it
@TobiasKildetoft how do you normalize it?
note that there is nothing between pixels
the continuity is an illusion
Whichever way is more convenient
you are right, each pixel can represent $\sqrt{2}$
Looks like a star with many invisible holes
11:13
@Secret there are no holes
because both sets are dense
i really need to beef up my tolopogy else function discussions are going to go wway over my head
11:26
They're dense but not complete so they are arguably full of holes
 
3 hours later…
14:52
@anon @arctictern creepy...
Hi chat
uncreeps
That's much better. What're you upto?
waking up
super linear algebra and 2-categories
strange dude; i usually wake up with coffee, not super linear algebra and 2-categories
14:59
don't like coffee. stopped drinking sodas and fruit juices.
I don't drink coffee on a regular basis either. Find tea much more refreshing.
Does anyone know the Knuth algorithm for solving a mastermind game ?
Should I, as a beginner, be proud of having asking a question on Mathoverflow with 10 upvotes, 2 favorites and no answers in sight?
@abenthy yes
much pride
15:11
@abenthy It definitely does mean your question was good.
Whether or not you want to be proud of it, however, depends entirely on you :) You can be, for sure.
I'm more annoyed than proud of my unanswered question, because I know it's a doubt I won't solve any time soon...
@SirCumference I can multiply ten-digit numbers on paper, that doesn't mean I will abstain from calculators. I can read books and talk to people in real life, that doesn't mean I will abstain from the internet.
multiply this explanation-by-example by a hundred as desired
@abenthy The answer is almost certainly no, but there is a dearth of examples.
one time I was going to use the word dearth but didn't because I thought I was just imagining it was a word
15:41
@abenthy Borel proved that an aspherical manifold whose fg has trivial center cannot carry an action of a positive dimensional compact Lie group, and that any finite groups acting effectively on M must act effectively as outer automorphism of pi_1. That is, the map G -> Out(pi_1) is injective.
I suggest finding an aspherical centerlesd manifold with Out(pi_1) = Z/2, such that you can find a smooth involution. Then I suggest connect summing an exotic sphere. It is unlikely the involution can still be made smooth.
Hello, could someone help me with: Prove that for any complex matrix A there exists a polynomial f: R->R such, that f(A)=0.
@user379685 how can you apply a function R->R to a matrix A?
You can apply a polynomial to it
Basically, force it so that you can only have an involution, and then smoothly break that involution.
@SteamyRoot but the domain is R not matrices
15:45
@MikeMiller Sounds like a good plan
Well, the domain is a bit wrong in what he wrote, you're right about that
@user379685 the polynomial you want is the one that gives you eigenvalues
I should really know what it's called >.<
oooh, ooh.. characteristic polynomial
Characteristic polynomial @steamy
Or the minimal one, not sure which one you wanted, but either should work there
In linear algebra, the Cayley–Hamilton theorem (named after the mathematicians Arthur Cayley and William Rowan Hamilton) states that every square matrix over a commutative ring (such as the real or complex field) satisfies its own characteristic equation. If A is a given n×n matrix and In is the n×n identity matrix, then the characteristic polynomial of A is defined as p ( λ ) = det ( λ I n − A ) , ...
thanks for the tips guys c:
M_n(R) is finite-dimensional, R[T] is infinite-dimensional, the map R[T]->M_n(R) given by evaluation-at-A is a linear map which must have nontrivial kernel. (this gives the minimal polynomial in fact.)
15:48
yeah otherwise the question seems a bit weird
Take $X\mapsto X-A$
@SteamyRoot thanks for the theorem
Boom
do you guys have any example of a differential equation involving composition (that has a solution)?
A differential equation involving composition can always be turned into one not involving composition.
You can also think about matrices as a field extension of R and take the minimal polynomial if you prefer a more algebraic approach
15:50
@MikeMiller I'm interested, how?
the chain rule?
oh
the right hand side of the chain rule still has composition?
I'm bad
@Alessandro matrices do not form a field
@DHMO start with a function, compute some derivatives, find a nonlinear relation between them
Field extensions can be rings
15:57
@arctictern sure, but any interesting ones?
shrugs
The diagonal matrices are a subring of the ring of matrices isomorphic to the field you take the coefficents from
no
you mean scalar matrices
Yes, I forgot with all entries equal
field extensions must be fields
15:58
Didn't know they have a name, nice
and in particular k[A] may not be a field extension of k, where A is a matrix

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