Mathematics - 2017-03-18 (page 2 of 4)

Secret

12:03

I think, naively speaking, we can find 2 charts to cover it, one for the top and one for the bottom, in a similar manner to the icing on a donut partitioning the donut into 2 regions

Alessandro Codenotti

@BalarkaSen hmm, I can't really visualize it but I'm not entirely convinced

Balarka Sen

@Alessandro It's like a tubular neighborhood of the 1-skeleton of the torus (that like two transverse strips) with a line from the boundary circle to the 1-skeleton removed. I think it doesn't work; it's not simply connected.

It seems you need to remove two segments like that, from each strip.

So back to 3 again :(

Secret

Not sure if I understood charts correctly, for example, do the chocolate icing form the domain of one chart for the torus, and the cake part forms the domain of the 2nd chart?

Alessandro Codenotti

No, they have a hole

@BalarkaSen aha, ok, I see it now

Balarka Sen

@Alessandro What if I join a segment from the boundary circle to the wedge point of the 1-skeleton though?

Nah, still doesn't work.

Secret

12:16

So you mean this shape cannot be a chart because it has a hole and thus the underlying set is not open?

Balarka Sen

It is open. It is not R^2.

Sha Vuklia

Hi guys. My book says that each subgroup can be the image of a homomorphism (inclusion mapping), however, not each subgroup can be the kernel of a homomorphism. I don't see why this should be a problem. Consider $H\subset G$ a subgroup. We know that $H$ is not empty. Now consider the mapping $f\colon H\to G:x\mapsto e$. This way, $H$ is the kernel of $f$, and $f$ is a homomorphism, because $f(a+b)=e=e+e=f(a)+f(b)$. Could someone explain their statement to me?

Secret

Ah right

Astyx

Hi chat

Sha Vuklia

Hi Astyx

Astyx

12:21

Doesn't a homomorphism need to be bijective ? @ShaV

Sha Vuklia

That would be an isomorphism

Alessandro Codenotti

I think you want G as domain for H to be the kernel

Astyx

Oh yeah right

Sha Vuklia

If $G$ becomes the domain, then $G$ is the kernel, and not $H$, right

we specifically want $H$ to be a kernel, I would think?

Alessandro Codenotti

Yes but I think they want H to be the kernel of a map G->something

Balarka Sen

12:23

@Alessandro Think of a tubular nbhd of the 1-skeleton in the torus as a cross ([-1, 1] x [-2, 2] cup [-2, 2] x [-1, 1]) with the ends identified appropriately. Remove the diagonals from [-1, 1] x [-1, 1]. That's a disk, correct? Now look at a tubular neighborhood of that pair of diagonals inside the cross.

Sha Vuklia

ah like that

Balarka Sen

That's also a disk. Doesn't those cover my thing?

Sha Vuklia

however @Alessandro I would have one objection for that

Alessandro Codenotti

can you do something like this @Balarka?

Sha Vuklia

They do consider the inclusion mapping $f\colon H\to G:x\mapsto x$ to argue that each $H$ can be the image of a homomorphism

But I'll ask my teacher I guess.. it's his syllabus after all:p

Balarka Sen

12:26

@Alessandro It sounds close to what I said above.

Ali Caglayan

@ShaVuklia The kernel is a normal subgroup. Not all subgroups are normal.

Balarka Sen

Yeah, that's basically it. My picture is yours if you bring those "straight" segments togather to form a diagonal.

So you just have two diagonals

Sha Vuklia

@AliCaglayan I don't know what a normal subgroup is, and I can't follow the wikipedia intro on it

I think we will get that in the next chapter

we haven't had conjugations and the like

Alessandro Codenotti

@BalarkaSen I agree. So we're back to 2

Balarka Sen

Coolio.

Ali Caglayan

12:32

@ShaVuklia The point is, for non-abelian (non-commutative) subgroups this will fail

So to construct a counter example try and find a group with a non-abelian subgroup

Then you will have a hard time realising it as the kernel of some homomorphism

Alessandro Codenotti

Must a nonabelian group with proper nontrivial subgroups have a nonabelian one?

Ali Caglayan

@AlessandroCodenotti not necesserily

just needs to be big enough

Take A5 for example

all its proper subgroups are abelian

But S5 has A5 as a subgroup

Balarka Sen

Or Q8

Alessandro Codenotti

@ShaVuklia when you want to construct the subgroups of G as the image of something you should have G as the codomain and are free to choose the domain of your morphism

Balarka Sen

I think that's the least order group for which it holds.

Alessandro Codenotti

12:36

@AliCaglayan ah, I see

Neat

(Let me google what Q8 is)

Balarka Sen

Quaternions

Ali Caglayan

If $H$ is a non-abelian subgroup of $G$. And $\varphi$ is some homomorphism, then for $g, h \in H$ we have $\varphi(gh)=\varphi(g)+\varphi(h)=0+0=\varphi(h)+\varphi(g)=\varphi(hg)$ contradicting non-abelianness of H. Or something like that @ShaVuklia

Obviously $\varphi$ has kernel $H$ and maps from $G$ to some group

@AlessandroCodenotti Unit quaternions to be precise

think of 1, i, j, k and negatives

the 8 comes from the fact that there are 8 elements

and usual rules of smashing i's and j's together apply

i^2=j^2=k^2=ijk=-1

Sha Vuklia

So... if we want to check that a subset $H\subset G$ is mapped as the image of some function, we need $G$ to be the codomain, and when we want to check if $H$ is the kernel of some function, we need $G$ as the domain

Alessandro Codenotti

hm, I see

we'll have a non mandatory lecture on quaternions next week, followed by one about slice functions on quaternions the week after that (I have no idea what those are, half of the results from google are papers by the professor who's holding the lectures)

Ali Caglayan

@AlessandroCodenotti What is your professors name?

Alessandro Codenotti

12:45

Riccardo Ghiloni (he's not my professor right now, he was last semester)

Ali Caglayan

I think slice functions are an attempt to get useful things from quarternions

I guess its trying to build up analysis for quarternions

Algebraically I found quarternions to be disappointing most of the time

Alessandro Codenotti

I know nothing about them, I'll find out I guess

Ali Caglayan

Once it fails to be a field all the interesting results about them are kind of partial

But I am just mouthing off about them because they didn't do what I thought they would this one time and I'm still sour about it probably.

Secret

One-dimensional symmetry group

A one-dimensional symmetry group is a mathematical group that describes symmetries in one dimension (1D). A pattern in 1D can be represented as a function f(x) for, say, the color at position x. The only nontrivial point group in 1D is a simple reflection. It can be represented by the simplest Coxeter group, A1, [ ], or Coxeter-Dynkin diagram . Affine symmetry groups represent translation. Isometries which leave the function unchanged are translations x + a with a such that f(x + a) = f(x) and reflections a − x with a such that f(a − x) = f(x). The reflections can be represented by the affine Coxeter...

What one needs when thinking about time crystals

Ali Caglayan

@ShaVuklia do you understand the argument why not every subgroup can be a kernel?

Sha Vuklia

12:50

Yes partly

I see that the non-abeliannes is violated

however, I still wonder about one example

consider $f\colon x\mapsto e$

I'm wondering if the non-abelianness is still violated here?

oh wait

I'm not allowed to have $H\subset G$ as the domain right?

Ali Caglayan

what does $f$ map from ?

Sha Vuklia

well, initially I wanted to write $H$, but I am being told that's not the convention

we need $G$ as the domain, to check if $H$ is the kernel

Ali Caglayan

Well $f$ should be a homomorphism $f:G \to$ whatever

Sha Vuklia

but then $f$ is not a homomorphism with the property that $H$ is the kernel, so my counterexample fails

yea, that's something I really didn't know

Ali Caglayan

then the kernel of $f$ which is $H$ would sit inside $G$

Sha Vuklia

12:53

but it's all cleared up now!

and thanks for the little proof about the non-abelianess

that was useful for me:)

Ali Caglayan

no prob

Secret

we have showed $H$ to be a abelian subgroup, but we need to do something more to show they are normal subgroups?

Ali Caglayan

@Secret we are not trying to show its normal

we are showing that not every subgroup is a kernel of some homomorphism

Sha Vuklia

May I ask what year you are in @AliCaglayan ? Mere curiosity

Ali Caglayan

because if it were then it would be abelian

@ShaVuklia first year I suppose

Secret

12:59

I see, so we can have not normal subgroups that are abelian being a kernel of a homorphism?

Ali Caglayan

well abelian subroups are normal :P

because if its abelian then left and right cosets are trivially the same

hence the thing is normal

Sha Vuklia

bachelor/master/phd? @AliCaglayan

Ali Caglayan

I guess you would call it the first

@Secret The point I am trying to make is that a kernel needs to be abelian. So if I choose a subgroup of a group to make into a kernel it clearly has a problem if the subgroup isn't abelian.

Secret

Yup I can see that

Sha Vuklia

ah cool @AliCaglayan

Ali Caglayan

13:02

@Secret all abelian subgroups are normal but not always the other way round

S5 is a good example

A5 is a normal subgroup

but it is not abelian

BAYMAX

hey guys a puzzle?

Secret

ok

Ali Caglayan

go for it @BAYMAX

BAYMAX

In a farewell meet there are n students and they have groups and it can be like 2 persons in a group , 3 persons in a group and so on and there is a car which goes from Point A to Point B and in that car maximum 4 persons can sit and hence go from A to B.

Then the minimum number of trips from A to B so that all of the n students go from A to B , another thing there can be mixing that is mixing of groups can take place , like a group of 2 and another group of 2 / a single 1 and a group of 3 ... any1?

yes@AliCaglayan

ChemiCalChems

i'm going crazy with the collatz conjecture

it's so simple yet so insanely difficult

Don't disturb

13:15

@MeowMix I'm self-educated in math.

ChemiCalChems

i should be studying, but i'm trying to convert v2(x) to an elementary function

Don't disturb

I think I stayed enough here (which is a few minutes)

Nick Pavini

14:18

1

Q: Three Distinct Points and Their Normal Lines

Suppose That three points on the graph of $y=x^2$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$-coordinates is $0$. I have a lot going but can not finish it. Proof: Let $(a,a^2)$, $(b,b^2)$, and $(c,c^2)$ be three distinct points on $...

Shahar Shalev

Hey i need help ,I have 2 vectors $\overrightarrow{a}$ and $\overrightarrow{b}$
It stated $\left|\left|\overrightarrow{a}\right|\right|=4,\left|\left|b\right|\right|=1$
and the degree between a and b is 120

$ \left(\left(2\overrightarrow{a}+\overrightarrow{b}\right)\,x\,\left(\overrightarrow{a}+2\overrightarrow{b}\right)\right)^{2}=?
$

Secret

14:43

what is $x$?

Shahar Shalev

vector product

Secret

have you tried expanding the vector product and use the trigonometric version of the vector product formula?

Little Rookie

15:55

Hello all, i have a question on phase portraits, it was written : Also the direction of the vectors give the direction of the trajectory as t increases so we can show the time dependence of the solution by adding in arrows to the trajectories.

Im aware that the direction of the vectors in the direction field corresponds to the derivative of the dependent variables. How does this relate to the time dependence of solution?

Tim The Enchanter

The derivatives are all time dependant. The derivatives are positive when the system moves towards an increase in the particular variables with time. So the arrows on a phase portrait portray the direction in which the system moves with time.

Little Rookie

The phase portrait does not have a $t$ axis, how can u indicate the change of derivative with time on the phase portrait

Tim The Enchanter

@LittleRookie It however does have 'axes' for the time dependant parameters. We do not indicate the change of derivatives. We use the derivatives to indicate the change of the target parameters (which do have 'axes').

Little Rookie

Yea, so on the direction field, you cant know for sure at a certain directed line segment, it belongs to the derivatives at a known $t$

Maple SE - Area 51 Proposal

Hello Everyone, I am confused about the transformation used in method of characteristics. To have change of coordinates, we use the slope dy/dx (=y/1 say) and get eta=xy. But for the second one, in most cases, it is assume that zeta=x out of the blue. My question is, how we arrived at zeta=x? any thoughts?

Tim The Enchanter

16:12

@LittleRookie Just to clarify, you are talking about an autonomous system right? Because then, you cannot solve a point in the phase portrait for an explicit time (in the general case) but the arrow says nothing about that. All it says is given a system in this state at some time, it will move in the direction of the arrow, which is independent of the time.

Meow Mix

Hi

Tim The Enchanter

@MeowMix Hey

Meow Mix

Anyone want a challenging problem?

Alessandro Codenotti

Hi Zach

Meow Mix

Hi @AlessandroCodenotti

Little Rookie

16:17

Yes im refering to autonomous system

@TimTheEnchanter

Alessandro Codenotti

Are you still here @Balarka? I might need an hint with an exercise from Hatcher

@MeowMix maybe, what is it about?

Meow Mix

you probably heard it already

For every $k$, find a circle lying on exactly $k$ lattice points

Alessandro Codenotti

Ah, yeah, Balarka told me about that problem already

I didn't think about it though

Meow Mix

for $8 |(k - 4 )$ the problem is pretty trivial

Alessandro Codenotti

I'm thinking about another one with circles without rationals point at the moment though

Meow Mix

16:20

which one would that be?

Evinda

Hello!!!

Suppose that we have the multiplicative group $\mathbb{F}_p^{\ast}$.
Does it hold that $a$ is a generator of the group iff ord(a)=p-1 ?

Hey @DanielFischer
Long time no see!
Do you maybe have an idea?

Alessandro Codenotti

That's true @Evinda

Evinda

Why? How can we show this?

Alessandro Codenotti

@MeowMix showing that the fundamental group of $\Bbb R^2\setminus \Bbb Q^2$ is uncountable. I'm not sure my approach with circles is a good one

Tim The Enchanter

@Evinda Follows from fermat's little theorem I believe.

Meow Mix

16:25

the fundamental group of points with at least one irrational coordinate?

Alessandro Codenotti

Yes

Meow Mix

fundamental group is like the group of words made by "letters" which are elements of $\Bbb R^2 \setminus \Bbb Q^2$, right?

Alessandro Codenotti

@Evinda remember that ord(a) is the cardinality of the subgroup it spans

Meow Mix

whose group operation is concatenation / juxtaposition of "words"

Evinda

@TimTheEnchanter How does it follow from fermat's little theorem?

Alessandro Codenotti

16:30

@MeowMix the fundamental group is the group of equivalence classes of loops based at a point (with the relation being homotopy equivalence), it is usually written as words with letters corresponding to the generators

Tim The Enchanter

@Evinda Correct me if I'm wrong, but is $\mathbb F_p$ the set {1,2,3,...p-1} where p is prime, because then $a^{p-1}=1$ iff a and p are coprime, which would make $a$ a generator of the group.

Meow Mix

@AlessandroCodenotti I learned it as "words" and "letters"

anyways, I'm not smart so I don't know the answer

I'm just going to work on some LA

Alessandro Codenotti

I have a very nice LA problem for you but I don't remember it exactly, I'll look it up later

Do you know something about finite fields and vector spaces over them?

Meow Mix

maybe a bit

Balarka Sen

16:58

@AlessandroCodenotti Hmm?

Meow Mix

Hi @Balarka

Balarka Sen

yo

Meow Mix

whats up?

Balarka Sen

stuff

CausingUnderflowsEverywhere

MEOW MIX?! how dare you impersonate my friend

Alessandro Codenotti

17:02

@BalarkaSen I want to prove that $\pi_1(\Bbb R^2\setminus\Bbb Q^2)$ is uncountable

Meow Mix

@CausingUnderflowsEverywhere Whaaa?

CausingUnderflowsEverywhere

Dat my friend Zach's picture

Balarka Sen

@Alessandro In particular proving that there are uncountably many non-null loops is sufficient, correct?

Hint: Try to construct lots of piece-wise linear loops. There are lines which do not hit any rational lattice point.

Alessandro Codenotti

So I pick a point $x$ in that set ($X$ from now on) to calculate the fundamental group at this point and another point $y$ with irrational coordinates linearly independent over $\Bbb Q$, the circle with center $y$ through $x$ contains no rational points

Balarka Sen

You could do that.

Meow Mix

17:08

@CausingUnderflowsEverywhere Yeah, I am Zach :P

Alessandro Codenotti

I need to show that there are uncountably many non homotopic circles like that though

I suspect that 2 continuous maps $S^1\to X$ whose image is a circle either have the same image or are not homotopic

CausingUnderflowsEverywhere

thats what any impersonator would say... I'll be keeping my eye on you... (just one. Im not going to dedicate all my vision to you)

Balarka Sen

@Alessandro Intuitively, two such loops would contain a lot of points with rational coordinates, which are missing from the space

Alessandro Codenotti

I meant loops whose image is an actual circle in that space, one without rational points

Balarka Sen

I have no idea what you mean. I am saying why two such circles should not be homotopic.

Alessandro Codenotti

17:18

If I fix a point in R^2\Q^2 there are uncountably many circles passing through it that don't pass through any rational point

Balarka Sen

I never disagreed with this. You want to prove that they are not homotopic, right? That's your question.

Alessandro Codenotti

Yes

Balarka Sen

That's what I was replying to. Intuitively, any two such circles contain a rational point - missing! - in the annulus they bound in the plane.

Alessandro Codenotti

So I realize those circles as the image of a map $S^1\to X$

@BalarkaSen aha, in the annulus, not in the circles themselves

Balarka Sen

Right. So you got to make that rigorous, somehow

Alessandro Codenotti

17:21

Yeah, I was thinking about this, if there is an homotopy between them we have $S^1\times I$ mapped to something which is a mess

Secret

@CausingUnderflowsEverywhere Hint: search zach in the search bar and then click on Zach's avatar and see which profile page is directed to

Alessandro Codenotti

I have to go now, I'll think about it later, thanks

Balarka Sen

@Alessandro If they are homotopic in R^2 - Q^2 they in particular are homotopic in R^2 - p where p is a rational point they bound.

That's a contradiction

I got to go get dinner too.

Alessandro Codenotti

Ohh, right

Neat

Buon appetito then

Daminark

Hey guys

SoumyoB

17:30

hey

Learning user

Hi,I have a percentage based Question problem doubt, who will help me

Evinda

@TimTheEnchanter Ah I see

SoumyoB

I think I've found the ultimate life changing hack for me

Evinda

Thank you @TimTheEnchanter :)

Learning user

Shall i ask question,anyone interested

SoumyoB

17:32

@Learninguser you just asked one

Learning user

No now new one @SoumyoB

SoumyoB

@Learninguser sorry didn't quite get you

Learning user

shall i ask now

SoumyoB

you just asked another question

@Learninguser I'm just messing with you, go ahead

Learning user

can i ask here

ok

A shopkeeper every once in a while raises his price by a% and then while reduces all the new prices by a%.After one such up down cycle ,the price of an article goes down by rs 441,After a second down up cycle , the article was sold for rs 1944.81.what was the original price of the article?

@SoumyoB please answer for this question

SoumyoB

17:42

just add 441 to 1944.81

I'm too lazy to do that

CausingUnderflowsEverywhere

thats easy

Learning user

anyone please answer for this

CausingUnderflowsEverywhere

2385.81

I just added in my head

not saying its correct I just executed soumyoB's calculation thats all

Learning user

ok but we need to work

Daminark

@SoumyoB Just wondering, are you an undergrad?

Can't remember if you were an undergrad or grad student

SoumyoB

17:46

I'm almost done with my masters @Daminark

Daminark

Ah, nice

Learning user

hello please guide me the steps for solution

Daminark

I don't understand the question as worded

Learning user

which one

?

Daminark

"raises his price by a% and then reduces all the new prices by a%"

Learning user

17:49

yes in question like that only is there

Daminark

I'm saying like, you just brought the price of something up and back down again? I mean that'll yield a slight change but something's iffy

Learning user

I dont know thats why i asked you

Semiclassical

"rs"?

Daminark

I'm not sure, like I haven't seen this question so I'm a bit in the dark. Like there's still some change for sure

If you start at $100, and then increase by 50%, then decrease by 50%, you get $75

So if we're rolling with that then maybe

I'm still suspicious that I'm interpreting this correctly

Learning user

I also thought we can start with 100, but prices is reduced by 441

Semiclassical

17:55

Well, a change of 441 is pretty big compared with 1944.81

Daminark

Well this is a double change

SoumyoB

@Semiclassical he means rupees

Daminark

So if $x$ is the original price

Semiclassical

Ahh.

SoumyoB

Indian currency

Daminark

17:57

Then we know that $x - (1+\frac{a}{100})(1-\frac{a}{100})x = 441$

And that $(x-441)(1+\frac{a}{100})(1-\frac{a}{100}) = 1944.81$

Semiclassical

The first one does simplify, which is handy

Learning user

I dont understand about the fract{a} like that, please edit the steps in simple way

'@Daminark

Daminark

The question was given as a percentage

2017

@Learninguser Enable mathjax on your browser!

Semiclassical

Use the Latex in chat link in the room desc to make that render properly

Learning user

18:00

How to enable ?

Daminark

So numerically this would translate to a/100

2017

19

A: Any chance of MathJax in chat?

As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

Daminark

If you look at the right side of the screen, you'll see it talk about $L^AT_EX$ in chat

Semiclassical

...like I said, use the link in the room desc

Daminark

But yeah, if I didn't get this horribly wrong, which is not impossible, then we have that $xa^2 = 4410000$

Which you could use in the second expression to make your life better

Learning user

18:03

I can see this LaTeX

then

how to enable

Lucas

Hello all. Question : If an endomorphism f is diagonalizable with 0 as eigenvalue of multiplicity 1 ; is always Ker f (+) Im f = Espace ?

Where (+) = direct sum

Learning user

ok

anybody got solution for my question

Lucas

space*

Daminark

And yet at least when I crunched out the numbers I still have a floating $x$ and $a$

Like, I don't think you can extract a numerical value

Unless I'm screwing up or we're not interpreting the question correctly

Semiclassical

The simple thing to do is to rewrite the second condition as $x(1-a/100)^2(1+a/100)^2=1944.81$

thats what I'm assuming.

Learning user

18:13

Question is correct only

Semiclassical

Right.

Daminark

Oh wait you squared things

I didn't notice that

Semiclassical

(Am typing from phone, incidentally. So I can't say much.)

Daminark

But yeah this would give us $x(1 - \frac{a^2}{10000})^2 = 1944.81$

@Learning maybe, but still somehow this feels like a bit of a contrived/bizarre example

Semiclassical

Right. From that and the first eq you can eliminate $x$ and solve for $a$

then use that value of $a$ to find $x$.

I think you end up with around (not exactly!) 30%

Learning user

18:19

what you are talking, i am not able to understand,after finding answer tell me

Daminark

Wait I'm not sure if what I said there was diagonalizable

Yeah it totally wasn't

Semiclassical

@Learninguser 1) translate the conditions in the text into equations. @daminark did that above.

Daminark

@Lucas I think your result should hold without knowing that you have an eigenvalue of $0$ with multiplicity $1$

Learning user

ok

Semiclassical

2) solve the equations for the unknowns. The best thing in this case is to first solve for the unknown percentage, and from this deduce what the initial price is.

Daminark

18:27

An operator is diagonalizable if and only if there exists some eigenbasis

Semiclassical

You should find a percent change which is about 30%. (That's not the exact figure, which you should find yourself.)

Learning user

ok , if prices is increases and reduces means we have to multiply

Semiclassical

An increase of 30% means you multiply the initial price by 1.30

Learning user

You are taking initial price as 1

Semiclassical

No. I'm saying you'd multiply the initial price by 1.3, whatever that number is

Learning user

18:32

why we have to multiply

Meow Mix

hi

Semiclassical

because that's how percent changes work.

Daminark

Hey @Meow

Semiclassical

Review that if it's not clear.

Meow Mix

i am in the process of writing a word game

and i used math in it

Daminark

18:34

Neat

Learning user

suppose if we consider as initial price as 1 and then prices is increases by a means 1+ a/100

?

@Semiclassical

0

A: Percentage based problems::::

When he raises the price and then reduces it, the price is multiplied by $(1+\frac a{100})(1-\frac a{100})=1-\frac {a^2}{10000}$ You have two sentences that each give you one equation. If the original price is $p$, what is the reduction in the first cycle? What is the final price? Two equatio...

please see this

Semiclassical

That's correct.

Learning user

how he is using 1+a/100

Semiclassical

a% of your initial price is a/100 * initial price, yes?

Learning user

yes

that means initial price as 1

he took

Semiclassical

18:38

So if you want to increase by a%, you add this to the initial price.

Learning user

yes

Semiclassical

So that's (initial price)+a/100*(initial price).

Learning user

ok

then again price reducing and price increasing , how he multiples

Semiclassical

Which is the same algebraically as (1+a/100)*(initial price).

x+a/100*x = (1+a/100)*x.

Learning user

now understood

Semiclassical

18:42

ok. If you decrease by a% instead, you subtract rather than add.

Learning user

ok

Semiclassical

So all that changes is you multiply by 1-a/100 instead

Learning user

why we need to multiply'

Semiclassical

What?

Learning user

(1+a/100)*x+ (1-a/100)*x

this correct

Semiclassical

18:46

No. the decrease by a% applies to the price after the a% increase.

Learning user

ok

Semiclassical

you only need x once.

Learning user

ok

Now understood

then we get

x*(1-a/10000)

which will be equal to

Semiclassical

Almost. (1-a/100)(1+a/100)x=(1-a^2/10000)x

Learning user

correct

after that

Semiclassical

18:50

So the final price after the cycle is x-a^2/10000^x

So the price has gone down by how much?

Learning user

441

Semiclassical

That's in terms of your givens. What about in terms of x,a?

Romantic Electron

Hello

Semiclassical

I.e. In terms of what I just wrote above

Learning user

x-441

Semiclassical

18:53

No.

that's the final price.

Learning user

one such up down a cycle less than 441

from initial price

Semiclassical

You're missing my point. If the initial price is x, and the final price is x*(1-a^2/10000), what is the change in price in terms of x and a?

Learning user

yes that is the final price

correct

Mike Miller

@BalarkaSen Yeah you're right you still max out at n+1. If you can always get rid of n-cells in a CW decomposition, then you only need n.

Semiclassical

Until you can give an answer in terms of x and a, you are not done with this part of the problem.

Learning user

18:58

after first down a cycle what will happen frame a equation

Semiclassical

There is really nothing more I can say. Reread what I've said; until you can answer what I've asked, I've nothing more to say.

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$Mathematics$ Mathematics