Mathematics - 2016-11-15 (page 5 of 5)

Dartek12

21:00

that seems right

am I obliged to use '<' operator

or could it be '=<' one

?

Astyx

The =< one is perfectly fine here

Dartek12

ok, thank you :)

Astyx

My pleasure ! :)

Topologicalife

I'm asked to minimize the function f = $\displaystyle\sum_{i=1}^{i=n} x_i$ with the restriction $K=\displaystyle\prod_{k=1}^{k=n} x_i$. What should I do is just consider $F(x,\lambda) = \displaystyle\sum_{i=1}^{n} x_i - \lambda \left(\displaystyle\prod_{i=1}^{n} x_i - K \right)$, and make zero the partial derivatives, and then find the critical points, right?

Null

what is a neutral function? such that f(x)+e=f(x)?

what is e?

Topologicalife

21:06

$f(x) = 0 \forall x$

Null

could you look at this?

(A0) Closure under addition

$f(x)+g(x)$ can be expressed as $h(x)=f(x)+g(x)$, since $h(x)$ is then a function from $A$ to $R$, it is an element of $map(A,R)$. Which satisfies closure under addition.

(A1) Associativity of addition

$((f+g)+h)(x)=(f+(g+h))(x)$

$((f+g)+h)(x)=(f+g)(x)+h(x)=f(x)+g(x)+h(x)$

By symmetry associativity is given.

(A2) Commutativity of addition

This is given by $R$ being a ring.

(A3) Identity for addition

$f(x)=0\forall x$ satisfies this.

(A4) Inverse elements for addition

math.stackexchange.com/questions/2014698/…

Astyx

Your trying to demonstrate that the set of functions is a ring right ?

Mahmoud

Is there something special about $\frac{\sqrt{2}}{2}$, any infinite series converging to it ?

Astyx

There is an infinite number of infinite series converging to ${\sqrt 2\over2}$

Null

@Astyx exactly!

hi there :)

Mahmoud

21:18

Such as ?

Astyx

I can't give you any concrete example right now

But there are an infinite number of series converging to any real number

Null

search for a series which converges to squareroot 2, then devide by 2

there are prolly a bunch of them

Mahmoud

Yes, thanks.

Astyx

@Null I don't think your proof holds

You should work with images of functions

For insance take $f$ and $g$ two numeric functions

Mahmoud

Also does $a^m\times a^n\times a ^p=a^{m+n+p}$ ?

Null

21:20

@Astyx well, where do you think the proof is bad?

(i dont know anything about it)

i just followed the answerers method

Astyx

For instance for Associativity :
Let $f$, $g$, and $h$ be 3 functions.
Let $x\in E$ ($E$ is the domain of your functions)
Then $(f(x) + g(x)) + h(x) = f(x) + g(x) + h(x) = f(x) + (g(x) + h(x))$ (using associativity of reals - r whichever number set you're working on)
And since this holds for all $x\in E$ :
$(f+g) + h = f+(g+h)$
Which proves associativity

Null

but "+" is the same as in the ring R

Astyx

No it isn't

That's the point

Null

there "+" HAS tobe associative ?

Astyx

$f+g$ is not the same as $f(x) + g(x)$

Null

21:25

i asked my teacher, he said its the same "addition" like in R

i dont understnd it

Astyx

In a way it's "the same"

Null

you mean its not exactly the same because now, instead of numbers, hole functions get added?

Astyx

But $(\cdot+\cdot)$ over functions is a binary operator from $E^{\Bbb R} \times E^{\Bbb R}$ to $E^{\Bbb R}$
whereas over reals, it's a binary operator from ${\Bbb R}\times {\Bbb R}$ to ${\Bbb R}$

Or what you said

And that's what your expected to prove (or at least I think so)

That functions behave "the same way as reals" because of the proof I gave you

Because you can solve the property for any $x\in E$ (using associativity over reals) which allows you to conclude that, since this is true for all $x\in E$ we have equality between $(f+g)+h$ and $f+(g+h)$

Null

ah but only because a binary operator behaves a way in one set, it doesnt mean that it even makes sense in another set?

Astyx

Yes

An because something is true for $+$ over reals doesn't mean it's necessarily true over functions

And that's what you have to prove (even though it's trivial)

Null

21:31

but isnt f(x) a collection of elements of a codomain?

Astyx

No $f(x)$ is a real

Null

or is f(x) a collection of ordered tupels?

like (1,a),(2,b)

Astyx

Formally $f$ is a triplet $(\Gamma, E, F)$ where $E$ is the domain, $F$ the codomain, and $\Gamma \subset E\times F$ a graph, ie $\forall x\in E, \exists ! y \in F, (x,y)\in \Gamma$

Therefore adding you are definitely not saying the same thing when you say that $f+g = h$ and when you say $x+y = z$

(Where $f$, $g$ and $h$ are functions, whereas $x$, $y$ and $z$ a reals)

I hope I'm not confusing you in any way

Null

mmh, how does then f+g look like?

if we dont know which addition is meant

Astyx

We do know

That's conventionnaly defined as $f+g = (x\mapsto f(x)+g(x))$

Null

21:37

and h is therefore another valid function from one set to another

therefore its included in the set of all functions?

Astyx

Well $h$ is defined as function, but sure

Null

oh i mean

f+g=h, is another valid function

Astyx

Again, it is defined as a function

(not quite sure what you mean by "valid")

Null

i mean, thats basicly saying closure under addition is a given? why do we need a proof?

Astyx

Oh yes this is juste something you have to state

That the sum of two functions is a function therefore the sets of all functions is closed under addition

(given of course that they have the same domain/codomain and that an addition is defined over the codomain)

Null

21:41

ah, at least i think i understood that by now :)

and given that the codomain is a ring right?

Astyx

Not necessarily a ring

I edited

Null

but closed under addition at least

Astyx

Yup

@Null I really hope you're not confused because of what I said ^^

Steve

22:01

Hi guys, I know this isn't the place for it, but I can't ask in the right place since I don't have enough reputation. Any Java experts here who can help me out?

Astyx

I'm not an expert but tell us your problem

I've been wondering : how is SE sustainable ? Who funds it ?

Steve

Ok. I was tasked to design a graphical function plotter where I had to design an abstract class and a bunch of subclasses with following constructors:
SineFunction(double a, double b), PowerFunction(double a, double b), ExponentialFunction(double a, double b), PolynomialFunction(double[] a), SumFunction(Function[] f) and another subclass named FunctionComposition(Function[] f).

I keep getting this error on my class's homepage code reviewer: java.lang.IllegalAccessError: tried to access method

@DemCodeLines Hi, you wouldn't happen to be a Java expert?

Null

@Astyx is this ok? (A0) Closure under addition

The sum of two functions with the same domain/codomain is also a function. Since our codomain is a ring, this new function is guaranteed to also have the same domain/codomain.
Therefore the sets of all functions is closed under addition.

Astyx

@Steve perhaps Software-engineering-SE is more suited for your question ?

Steve

Probably. I'll give it a try there :) thanks

Astyx

22:14

@Null seems fine to me, although the fact that the codomain is a ring ensures that the addition makes sense, rather than ensuring the codomain is the same

@Steve, my pleasure, hope you'll find your answer

@Null As in : the sum would be well defined because addition is always well defined between two numbers in a ring

Null

@Astyx mmh yes

Steve

faded icons mean people are afk, right?

Mike Miller

yes

Null

@Astyx well if the codomain would only be $\{1\}$, then addition would fail, that was my thought

Hiro

@Astyx remember the probability question we discussed a few days ago, I am still trying to figure out the probability that when we stop the sum is odd~ My major problem is that we stop when the sum is even, so how could the sum be even and odd at the same time? Would that mean that if the sum is even, then we do not include it in the probability of getting an odd sum?

Astyx

22:19

@Null yes, but that would be a problem when defining $h:x \mapsto g(x)+f(x)$

Null

@Astyx but the same domain/codomain is necessary for it to be in $map(A,R)$ or not? otherwise it would be maybe some function, but not from A to R?

Astyx

Yes (although people tend to confuse function that coincide on the domain but have different domains, for instance $\begin{cases}\Bbb R \to \Bbb R \\ x\mapsto x^2\end{cases}$ would be considered the same as $\begin{cases}\Bbb R \to \Bbb R _+\\ x\mapsto x^2\end{cases}$)

Null

mmh, how is this called? all y in the codomain that are "hit"?

Astyx

Surjection

Null

eh no i mean the set of y that are hit

Alessandro

22:27

image (sometimes range) of the function

Null

ah

ah so my argument would be rather: the image of the function has to be a subset of R(or R itself). and the domain has to be the same as A?

meow-mix

hi everyone

Null

hi @meow-mix

meow-mix

hi @Null

i think i did ok on the AMC

Astyx

@Null Yes, according to the definition of a function I gave you earlier

@Hiro You need to draw another graph

Hiro

22:30

@Asyx what kind of graph?

Maks

Hi, if a have a power series of the form $ \sum \frac {(-1)^nx^n} {n+1} $
By using the ratio test I know that $ | x | < 1 $, so I have to test the interval extremes, so I try $ \sum \frac {(-1)^n(-1)^n} {n+1} $ why cant I apply the limit protery of absolute value ? And say $ \sum \frac {(-1)^n(-1)^n} {n+1} = \frac {1} {n+1} $ which tends to 0 when n tends to infinity and say it converges ??

Astyx

@Hiro The same I drew last time

Hiro

@Astyx or rather what values should be in the chain?

is it very similar to the previous one, what changes and what doesn't?

SteamyRoot

ummm

Astyx

You must add another state like F

Null

22:31

@Astyx it's all nice and dandy to have definitions but i always need some time to get what they are saying haha

Forever Mozart

when someone proves a nice theorem I get jealous and angry with myself

Astyx

Split F into 2

SteamyRoot

$\sum \frac{(-1)^n(-1)^n}{n+1} \neq \frac{1}{n+1}$

Hiro

so i'd have two circles f1 and f2?

SteamyRoot

$\sum \frac{(-1)^n(-1)^n}{n+1} = \sum \frac{1}{n+1}$ which tends to infinity because it's the harmonic series

meow-mix

22:32

i feel incompetent though,

Astyx

@Null take all the time you need :) But then again this definition only exists to prove one can formallise the idea of "function", it is important to understand that none of this really matters anyway

Maks

@SteamyRoot but $ \lim_{n\rightarrow \infty} \frac {1} {n+1} = 0$

Astyx

@Hori split F into E and O, where E is the final state where the sum is even, and O when it's odd

Now from 2 some numbers will point to E and some to O

SteamyRoot

$\lim_{n \to \infty} x_n = 0$ does NOT imply that that $\sum_n x_n < \infty$

Hiro

so now 1 will point to e and o, and 2 will point to e and o?

Maks

22:34

I got it, I was misunderstanding the theory, that limit tells me nothing about the converge, so I have to use a convergence test

Astyx

Every other number pointing to E (except enterring number 9) will still point to e

Maks

Only if the limit is not 0 it implies that the series diverges

Am I right ?

Astyx

@Hiro theorically yes

But try to build it and (if I'm not mistaken) you'll see there are no numbers pointing from 1 to O

Hiro

@Astyx 9 will point to O for just 1 roll right?

Null

@Astyx for commutativity i proceed like you in associativity or? (simply showing it by calculating)

Hiro

22:35

yeah i tried 1 to O, it all gives E

Astyx

@Astyx You need to proceed like I did to be concise

SteamyRoot

@Maks yes, indeed.

Astyx

@Hiro yup

@Hiro Why don't you try and draw it and let me check afterwards :)

Hiro

alright ill send a screenshot

Maks

@SteamyRoot In power series I can apply the same converge test than in normal series ?
Comparison, ratio, integral, root, etc ??

SteamyRoot

22:37

Yes, of course. However, some will be more useful than others

Maks

@SteamyRoot Last question, if I have a power series $ \sum c_n x^n $ And I know that the series converges on x = -4 and diverges on x = 6
Can I imply that $ \sum c_n $ , converges ? And how can I prove it

Because supposedly, that series converges on the interval [-4,6] and on $ \sum c_n $ the value of x = 0, which is inside the interaval, so it has to converge, but I dont know if that's enough to prove it

arctic tern

the series cannot converge on the interval [-4,6] and diverge at x=6

that's a contradiction

Maks

My bad, [-4,6)

arctic tern

the given information is not enough to say the power series converges on [4,6)

all it implies is that the power series converges on an interval (-r,r) with r greater than or equal to 4

such an interval includes x=1 which you can plug in to get \sum c_n converges

Hiro

@Astyx wanna go to a private room?

Astyx

22:44

Fine by me

Hiro

Alright, I am not sure how to invite lol

Maks

@arctictern How can you be sure of that ??

arctic tern

@Maks be sure of what? I said many things.

SteamyRoot

You have a power series with center $0$

(not sure if center is the right term)

Maks

@arctictern I marked the middle sentence, the $ (-r,r) $ with $ r \geq 4 $

SteamyRoot

22:46

but the idea is that, a general power series is of the form $\sum_n c_n(x-a)^n$

and you have $a=0$

So there is a radius of convergence around $a = 0$

arctic tern

it's a fact of analysis that any power series around a point converges for values inside a disk centered at the point and diverges for values outside of it. (one proves it with the ratio test)

SteamyRoot

Since your series converges at $x=4$, this radius is at least $4$.

Maks

Ohh ok

Null

@Astyx so $f:\mathbb{R}\to\mathbb{R}$ doesnt have to hit anything in the codomain, but it has to have a value for all $x\in\mathbb{R}$?

Astyx

Yes, and this value has to be in $\Bbb R$

Null

22:50

then $E$ is superfluous i think, A is the domain ;) (but ok, at least i got it!)

Astyx

What do you mean $E$ is superfluous ?

Null

well, the functions are in $map(A,R)$. So A is the domain of those functions

Astyx

And what do you call $E$ ?

Null

you called it that way chat.stackexchange.com/transcript/message/33531727#33531727

Astyx

$E$ was the domain, no ?

Null

22:54

mmh, no, those functions are just in $map(A,R)$. Therefore their domain has to be A or? (and the codomain some subset of R)

Astyx

Yeah, I didn't know you called their domain $A$, so I called it $E$, but it's the same thing really

Null

ah ok!

i just thought the first time that this would be necessary ;)

Astyx

Right :)

Maks

How can I find a power series that represents $ \int \frac {1} {1-x} $ ?

DHMO

not straightforward: realize that 1+x+x^2+.. = 1/(1-x)

(geometric series)

Null

23:01

@Astyx eh, i dont know what there is to show for commutativity. This seems pretty obvious, but what is there to show?

DHMO

straightforward: use the (extended) binomial theorem

Null

$ab=ba$

Astyx

exactly the same thing

Max

I am trying to calculate the extremas and monocity of function $arcsin(\frac{2x}{x^2+1})$ - I calculated the derivative which is equal to $\frac{-2}{x^2+1}$

Astyx

Show that for any element in the codomain, $g(x) +f(x)= f(x) + g(x)$

Max

23:01

which means that the derivative of this arcsin is always negative

Astyx

An dthus $f+g = g+f$

Null

$f(x)+g(x)=g(x)+f(x)$ what step inbetween this would there be?

Max

yet when i plot it in wolfram i get that it has local extremas:

m.wolframalpha.com/input/…

can any1 explain what i am doing wrong here?

DHMO

confirm your derivative with wolframalpha

Max

its the same

the derivative is always negative

for all x

Null

23:04

@Astyx or do i have to show: $(f+g)(x)=(g+f)(x)$ rather?

Max

yet the function is not always decreasing

Astyx

$=$ is symmetric

Oh yes I see what you mean

Yes you do

Null

and then i expand the LHS, and use symmetry as an argument?

Astyx

(but since $(f+g)(x) = f(x) + g(x)$ ... :p)

Yup

Not symmetry

Commutativity

@Null what was the link you sent to me for mathjax again ?

DHMO

@maks no it isnt

your derivative does not match

Maks

23:07

@DHMO I think you meant to write "max"

Null

@Astyx it was just your post with associativity

@Astyx ah

math.ucla.edu/~robjohn/math/mathjax.html

DHMO

are you playing on your name

Astyx

thanks a lot

DHMO

any way, it is a common misconception that sqrt(a^2) = a

Astyx

^

Null

23:08

@Astyx (A2) Commutativity of addition

Let $f,g$ be two functions in $map(A,R)$.
Let $x\in A$.
Then $(f+g)(x)=f(x)+g(x)$ which by symmetry is equal to $(g+f)(x)$.
And since this holds for all $x\in A$:
$f+g=g+f$
Which proves commutativity.

Astyx

Yes that's right

Null

something new everyday xd

DHMO

wait

Astyx

@Null However I would emphasize on the fact that addition is commutative on $R$, and that that's what's used here

DHMO

@Max see above

Max

23:10

ohh it is sqrt(a^2)=|a|

:p

DHMO

yes

Null

@Astyx "This uses the fact, that $R$ is a ring." would be sufficent or?

Forever Mozart

nj

Astyx

@Null Well then again i would emphasize that you are using commutativity in a ring !

But maybe that's just me

The idea is to be as concise as possible

Null

@Astyx ah, ok i understand

Astyx

23:12

Yes, $R$ is a ring so it's sufficient

But it's not necessary

Whereas commutativity for addition in $R$ is

Anyway I'm gonna go now

Good day/night to all of you

Null

@Astyx good sleep

Null

23:42

in a set $map(A,R)$, what is the additive inverse of a given function f(x)? (-f)(x) i guess, but how do we show that (-f)(x) is indeed again in $map(A,R)$?

R is a ring, A some nonempty set

arctic tern

@Null you already know the answer in your heart. what's the additive inverse of a real-valued function f(x)? it's -f(x), where we simply take the additive inverse pointwise. same applies to any function taking values in a set with an abelian group operation (like addition in a ring).

Null

@arctictern (A4) Inverse elements for addition: negative elements

Let $f$ be a function in $map(A,R)$.
Let $x\in A$.
Then $f(x)=y$, where $y\in R$.
Since $R$ is a ring, there exists an additive inverse of $y$.
Now we define $-f(x)$ in such a way that if $f(x)=y$ then $-f(x)=-y$ for all $y\in R$.
Therefore any function in $map(A,R)$ has at least one additive inverse. like that?

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$Mathematics$ Mathematics