desmos.com/calculator/m2qy0ms7p0

i just saw air force one, i think. we live in a LGB flight path.

I think he’s been here (LB) a while.

they certainly shut down a lot of streets by noon.

a local article says he was scheduled to arrive 'a little before 7 pm.'

anyway, it wasn't the usual southwest flight that comes in at 6.

maybe they're flying in biden's body double or robot clone.

Meanwhile, complex analysis is all “differential geometry.”

Is there any finite combination of elementary functions that grows faster that x^x?

than*

Combination ?

e^x times x^2 times ...

why not something like e^(whatever you have in mind)

and then e to that

I don't have anything in mind. I suspect that no such combination exists

the question may be what you mean by 'combination' (and for example whether something like x^(x^(x^(x^x))) and further iterations of that theme would count)

yeah, I do have to draw a line between what I'm allowing into the combinations

like, no triple arrow notation

no ackermann functions

and nothing of the form x^x of course, because that's what to compare

what sort of math goes into modeling covid spread? or is it more of a programming thing?

difference/differential equations look up SEIR model.

i do not know how the models are validated, its not exactly physics.

thanks!

frankly, most of the covid papers that deal with diseases spread seem to be more focused on optics rather than critical analysis.

also, there are good studies out there but the press is very selective in what it reports.

@psitae What about $e^{x^2}$?

You’re way too vague for my taste.

I am wondering can someone help with this question: cs.stackexchange.com/questions/143909/…

i have no idea what I-ORD means. grammars have changed since my day :-).

Evening gentlemen...hope y'all are doing well in these wild times...looking like Newsome is going to take it?...the election race is even popping up in my BBC feed

most polls suggest that, but it's too close for my taste.

Was my first time reading up on Larry Elder.........trash man imho

california should reflect on a recall process that could allow a significant minority of voters to re-decide an election for no reason other than they didn't like the way it came out.

there's going to be one of these after every election now.

what was the "official" reason used to call the new election?

as a practical matter you don't need one. you just need a fairly small number of signatures to kick the process off. formally the petition does need to state 'reasons' but they can be anything. this one was started before covid and just mentions a laundry list of complaints about taxation, homelessness, and i believe immigration policy.

i.e. it could have been written any time in the last 20 years about any governor of california or indeed many states. pure political disagreement is enough. whatever the standard is for impeachment of a president, it's far below that.

elder would not be able to do a lot with state law because of supermajorities in both californian legislative houses. but he could muck up a lot on the law enforcement side and generally deciding to prioritize some issues over others. and, notably, appoint a US senator if our ridiculously old senior senator leaves office before the end of her term.

my wife met newsom once. she was working for SF's department of public health when he was the mayor and they did an event to promote flu vaccination in chinatown.

somewhere i have a vhs tape of my wife being given a flu vaccine on our local mandarin language news broadcast.

strikingly, newson won with 7.4 million votes vs. 4.7 million votes and all that is needed to initiate a recall vote is 1.5 million signatures

cool, isn't it

And Elder’s already got people claiming fraud on his website. The Tromp legacy — the death of democracy.

madness....

rump was taking digs at 43.

sorry, i meant trectum. damn autocorrect, i mean trump.

i confirmed via the timing of arrival pics appearing on twitter that i did indeed see biden's plane fly over my house a little after 6pm today.

i forgot to wave. i was trying to figure out what airline had livery like that

undergoing a bit of conceptual confusion: $\vec v = x\hat\imath + y\hat\jmath$ does not denote the ordered pair $\langle x, y\rangle$, right? what's the precise relationship between the two?

but nevertheless we say both $\vec v \in \mathbb{R}^2$ and $\langle x,y \rangle \in \mathbb{R}^2$

those usually do represent the same thing, although people who use one notation tend not to use the other at the same time.

there's some sort of commutative diagram holding there, right?

just got confused in a proof where i accidentally changed from one notation to the other

gosh, i hope not.

i is <1,0>. j is <0,1>. <x.y> = x<1,0> + y<0,1> follows from the definitions of addition and scalar multiplication.

ah there you go

thanks

i think the main reason for using i, j, and k is that it makes clear how you can compute cross products knowing only how to multiply i j and k with one another, and relates it to the kind of algebraic manipulation people have learned since the very beginning of algebra (with the twist of noncommutativity). on the pure math side i don't think it's very helpful for any other purpose.

I hate the i,j,k notation....

for example it makes it hard to distinguish between the use of the notation in R^2, and the use of the notation in R^3 when the k component is zero. that's not the kind of identification you want to make implicitly or automatically in pure math. you want to think about that stuff.

i was verifying the identity of the solution sets of $\langle k, -k/2 \rangle$ and $\langle -2k, k \rangle$ and spiraled into catatonic stupor when i accidentally switched to ijk notation

@dc3rd i am not a fan either, as maybe is evident from the above. there are enough letters flying around without adding more. boldfacing, underlining, and hats i can sometimes get behind to notationally distinguish vector from scalar although i would never use it outside of teaching people who might not yet be able to infer type from context

shintuku, at least you had an identifiable reason for spiraling into a catatonic stupor. i just do that from time to time.

no reason behind it

unpredictability, perfect way to throw your enemies off

@psitae $e^{e^x}$ is faster

In fact, I think $e^{x^2}$ is faster

Yes, because $x<e^x$ so $(x)^x<(e^x)^x$

i much prefer $e_1,e_2,e_3$.

On the other hand, I do believe it grows faster than all polynomials in $\{x,e^x\}$

some of my daughter's friends dropped off some cookie dough. one of them 'discovered' me on mse. was fascinated that i was banned for a day.

In other words, take a two-variable polynomial like $2+x+xy^2+y^3$

and let $y=e^x$

to get $2+x+xe^{2x}+e^{3x}$

I believe $x^x$ should be faster than all of those

because $x^x=(e^x)^{\ln x}$

so for $x>e^n$, $~x^x>(e^x)^n$

So $e^{x^2}$ is faster than $x^x$, but polynomials in $\{x,e^x\}$ are not

polynomial in {$x,e^x$}?

Like a polynomial in $\{x,y\}$ (basically a standard two-variable polynomial) but you then substitute in $y=e^x$. I defined it above

Maybe "over" is the right word, rather than "in"? I don't think it's standard terminology

what is motivating the need for speed?

…answering @psitae's question?

Sidenote: prove that $(\ln x)^{\ln x}$ grows faster than all polynomials

i was wondering about @psitae's motivation

Let $p(x)$ be a polynomial. What are the possible values of $\lim_{x\to\infty}\sqrt[{\Large x}]{p(x)}$?

are you asking a question or posing a problem for someone?

I think p(x) should be carefully picked here to avoid complex values.

Let's make it $\lim_{x\to\infty}\sqrt[{\Large x}]{|p(x)|}$

@copper.hat I just thought of the question just now, but I know the answer

I think it should be 1 or 0.

how is it defined for $p=0$?

The limit is 0; when p= zero polynomial. @AkivaWeinberger

of course not fully understanding and trying to skip over dual spaces would come back to haunt me with me learing about the adjoint of an operator.....back to "relearning" that section I go....😤

can anyone help me in understanding why the roots of linear differential equation or the solution of the nth order differential equation come in the form of $y=c_ne^nx+c_n-1e^{n-1}x...=0$?

why my book is assuming $y=ce\fracx$ it be trivial while deriving the solution of the equation

that's just the solution to such differential equations

you can verify it through substitution

you can also verify that it is unique

how?

to verify it is the solution, substitute it in whatever differential equation it is meant as a solution to

it'll take a while to solve but you'll be able to solve it using that solution

i'm saying: substitute the general form of the solution into the general form of the differential equation

@psitae $e$ is intimately connected with $x^x$. In particular, $x^x$ has a minimum at $1/e$. Also, for $n>0$, let $x=(1+1/n)^n$ and $y=(1+1/n)^{n+1}$. So $x<e<y$, with equality at $\lim_{n\to 0}$. Then $x^y=y^x$, hence $(1/x)^{1/x}=(1/y)^{1/y}$

ok let differential equation to be

$2y^2+y-1$

where y is dydx

and y^2 be d^2ydx

what is it equal to? what's the equation?

zero

considering it to be auxillary equation

you want the solution to $2y'' + y' - 1 = 0$?

yes

Nice detective work, shintuku. ;)

@PM2Ring morning sir

@JackRod differentiate the solution once to get $y'$, and differentiate it again to get $y''$, then substitute it in

@JackRod Greetings. (It's 4:23 PM in my time zone). It can be tricky doing MathJax in chat. You can use an answer on the main site to edit your MathJax and see a preview. Just be careful to not accidentally post that answer! :)

there's also stackedit.io i always use to preview my stuff before posting in chat

I get two real root of this equation

then what?

I always have the Maths, Physics, & Astronomy sites open in my browser, so I find it easy enough to use one of them for my MathJax editing. (For fancier stuff, I use Sage, which lets me print & preview MathJax from Python-like code).

@JackRod here's a tutorial tutorial.math.lamar.edu/classes/de/RealRoots.aspx

look at the examples

@robjohn can u help me going through it

once you have the two roots, you just substitute them into the solution equation

it's a recipe

I have a confusion is that what is the y u get?@shintuku

what is the solutionu get

And when you have the solution, check it! That is, find the derivatives and plug them into the original differential equation to make sure they work.

@JackRod $y(x) = c_1 e^{-x/2} + c_2 + x$

is it satisfying the differential equation abve?

try it

get $y'(x)$ and $y''(x)$ and substitute it in

By trying it, you find out if it works. But more importantly, it helps you "digest" why this recipe works.

Don't worry, it's normal to find this process a bit strange at first. You just need to practice it. Remember, it took some of the greatest mathematicians ever to invent this stuff. And they had to practice it too, until it became natural to them. :)

I'm trying to show Borsuk Ulam theorem for Torus : For every continuous map $f:S^1\times S^1\to\Bbb R^1$, there is $(x,y)\in S^1\times S^1$ such that $f(x,y) = f(-x,-y)$. I tried to relate this to the original Borsuk Ulam for $S^1$ but failed. Can somebody help?

Ok : Define a map $\tilde{f}:S^1\times S^1\to\Bbb R^1$ by $(x,y)\mapsto f(-x,-y)$ which is continuous. Let $g = f-\tilde{f}$. Then $g(x,y) = -g(-x,-y)$. If $g(x,y)>0$ then $g(-x,-y)<0$ so by the intermediate value theorem, there is $(x,y)\in S^1\times S^1$ such that $g(x,y) =0$ i.e. $f(x,y) = f(-x,-y)$.

Is my proof ok?

@Rithaniel what is the definition of $\|\cdot\|$? (that is, the rightmost norm)

Operator norm. So one definition is $$\sup_{\substack{\|x\|=1\\x\in X}}\vert\varphi(x)-\varphi_0(x)\vert$$

@robjohn I've still got nothing, by the way. I have an inequality in one direction for which my proof might be incorrect

Well, nah, it's correct

Right now I'm trying to get $$\inf_{\substack{\varphi_0\in Y^\perp}}\|\varphi-\varphi_0\|\leqslant\|\varphi\|_Y$$ However, this whole problem is really slippery

Why is it that, if the pivot columns of the echelon form of matrix $A$ are linearly independent, the columns corresponding to those in non-echelon form are also linearly independent?

@shintuku math.stackexchange.com/questions/354362/…

if a column vector $\vec v$ is linearly dependent on another $\vec w$, then their ith component is related by scalar multiplication: $v_i = kw_i$. but elementary row operations preserve that equality

I have a vector $l$ that is orthogonal to any linear combination of the column vectors of $A$, and it seems to me this could somehow mean that it forms part of the basis for the null space of $A$. am i totally wrong?

i say this since it's not actually a linear map, but it sort of maps any vector in the image (column space) of $A$ to $0$ (through the dot product)

mainly i'm trying to figure out an interpretation for the fact that if $E$ is the product of elementary row operations that reduce a matrix $A$ to echelon form $A^*$ (s.t. $EA = A^*$), if the row $r$ of $E$ corresponds to a $0$ row of $A^*$, then it is a basis for the null space of $A^T$

@shintuku figured it out

@copper.hat $0$ to any (positive) number is zero

@Koro Yeah

@love_sodam yes

the same type of proof immediately generalizes to show that if $X$ is a top space with an involution $f$ and $g\colon X\rightarrow\mathbb{R}$ is continuous, then there is an $x\in X$ s.t. $g(x)=g(f(x))$

however, Borsuk-Ulam fails for the torus if you make the codomain $\mathbb{R}^2$, which one might argue is the more natural generalization

@Thorgott if $X$ is connected?

right, thanks

Hello, a multigraph $G = <V,E>$ is one that has multiple undirected edges between its vertices and self loops. Say $<u,v>$ are 2 vertices then, we could have multiple edges between them. Now one way to remove multiple edges and keep only one is to keep an adjacency matrix that will record $A[i][j] = true $ in case we have an edge so that the same vertex $j$ wont be added twice as in case we have another edge. What do you think about this approach please? Would you recommend a better one?

This needs $O(V^2 + E)$

@Thorgott thanks!

Is the goal of stack exchange to train an AI to answer questions?

if $EA = R_A$, where $R_A$ is the row echelon form of $A$ (so $E$ is the product of elementary row operations on $A$ that reduce it to echelon form), what do we check to determine whether $E$ shares row or column space with $A$?

Or maybe it is to train an AI to pose questions that could in principle be answered by someone learning a particular subject. It seems to me that a "good" question according to the edicts of stack exchange is a question that a tester would pose on an exam.

And the goal is also to train an AI to grade the answer?

Is there an easy way of proving that $\frac{xy^3}{x^2+y^6}$ is not differentiable at the origin?

other than proving that the limit doesn't exist

@leslietownes Thanks Leslie

@Derivative the limit definition is usually how people do those. with rational functions, sometimes it's helpful to switch to polar coordinates; sometimes it makes clear that the limit as you approach zero depends on the angle of approach.

it's always the origin. if i write a multivar calc book, i am going to break with precedent and have someone determine differentiability of something somewhere else.

@leslie How's the troublemaker doing?

@shintuku What do you mean "what do we check"? Column space is obviously unrelated. And row space? We talked about this, didn't we?

she doesn't seem to be in pain anymore. completely immobilizing the leg seems to have fixed that. unfortunately this means she has more energy to misbehave.

@Derivative What limit? Do you mean checking the function is not continuous? Or do you mean finding the only possible derivative and showing it doesn't work?

today's garbage day and she keeps saying that she's going to put olivia in the 'trash truck' so they can 'take her away.' when my wife told her that we wouldn't be doing that, she told my wife that she was going to wind up in the trash truck too.

@leslie Well, face it. The acorn didn't fall far from the tree.

Oy. I would have been spanked for that.

fair point.

my daughter likes really weird threats. last night she kept telling us that we were going to eat ketchup.

Is that a bad thing? I prefer chili sauce on my hamburgers and hot dogs, but ...

i don't know. her tone suggested that it was a threat. we didn't know what to make of it.

Maybe she's practicing to be one of the Proud Boys.

my internet speed is 1 kbps

how retro

I have 512 MB RAM

are you running an early raspberry pi?

no

Don't give the troublemaker any ideas. She'll start blowing raspberries.

I just noticed that linus wrote some nice comments in linux's first version

github.com/zavg/linux-0.01/blob/master/include/errno.h

 * Hopefully these are posix or something. I wouldn't know (and posix
 * isn't telling me - they want $$$ for their f***ing standard).

i kind of liked my first modem (1.2 kbps). you could see it type out the screen. added a little bit of suspense. what's the next paragraph going to be?

TedShifrin trying to figure out whether row space is related. I know the matrix $E$ designates row operations, but I can't figure out what sort of strict relation would hold with the row space of whatever matrix it is applying row operations to

(am reviewing the lectures)

Remember we discussed why row operations have to be invertible @shin.

@TedShifrin Second one

I ended up turning it in like this i.imgur.com/LHs6tbg.png

hope you can read Portuguese

hm, but those properties only guarantee that the row spaces of $A$ and its row echelon form are identical, no? or am i missing something obvious about how this also applies to $E$?

apparently my discrete geometry class is going to be TDA somehow :/

now I gotta reread hatcher fml

I thought we were just gonna implement normie algorithms like voronoi diagrams and hull tricks etc :'(

TDA?

topological data analysis

ok 👍

It's supposed to print money or something

Oh, wait, I didn't read carefully. $E$ doesn't even have the right shape for you to be asking these questions. The entries of $E$ are meaningless except that you can read off the left nullspace $N(A^\top)$ from certain rows of $E$. Why are you thinking about this?

although Idk if it's out of style now O.o

also, who gave u permission to think about this

@Derivative: The first step is to identify $\alpha$, $\beta$, if the derivative exists. Did you actually do that?

Also, I think you have small errors in there. It seems like you put $\|(h,k)\| = h^2+k^2$, for example.

I'm trying to understand why the entries of $E$ matching with the zeroes in the echelon form of $A$ make a basis for $N(A^T)$. I thought I had figured out but I ended up only finding out why those entries are inside $N(A^T)$, not why they form a basis for it

Oh, well, that's a good question.

I get the linear independence from the invertibility of $E$, but have a hard time with the spanning

Are you granting we know the dimension?

If not, you can still argue. Could there be any further relations among the rows?

hm, well if we can do it without knowing the dimensions that would be interesting

for the relationship among the rows: orthogonality with the column space of $A$

only those row vectors of $E$ corresponding with zeroes in the echelon form $R_A$ of $A$ will be orthogonal to the column space of $A$

since their dot product with any linear combination of all column vectors of $A$ is $0$

The idea is that any additional relation among the rows would have to come from the nonzero rows of the echelon form, but that can't happen. Formally, you have $EA=U$, so $A=E^{-1}U$, so now look at $A^\top x = 0$.

what are we aiming at with additional relation among the rows? (of $E$?)

from $A^T\vec x = 0$ we get $U^T(E^T)^{-1}\vec x = 0$

Relationship between elements of matrix $B$ and $B^T$ please?

Right, so $x=E^\top y$ for some $y\in N(U^\top)$. But we know a basis for $N(U^\top)$.

$BB^T(i,j) = \sum_{k\in E}{}(b_{ik}b_{jk})$?

hm, how are you getting $x = E^Ty$ for some $y \in N(U^T)$? if I suppose $x \in N(A^T)$ I can get to $x \in N(U^T(E^T)^{-1})$

Do the next step.

alright meditating on it

Make it a little simpler. What is the nullspace of $AB^{-1}$?

hm, i think my brain has melted. i've been doing linear algebra all day

i'll take a break and get back at you, my bad

No prob.

@shintuku. This is linear algebra 1 or 2 please?

@Avra it's from here: youtube.com/…

Oh from Prof Ted!

I see

This is linear algebra I

This is the most fun topic in math for me

Better than abstract algebra and functional analysis nightmares

boo

tells munchkin to throw a raspberry at leslie

sher's way ahead of you.

I figured.

does anyone like pop-electronic-disco music

trying to identify what genre this song is

I think it's classified under electronic-pop

throw the link @geocalc33

Any idea why the shortest path $\sigma(s,v) \le \sigma(s,u) + 1$ please for edge $(u,v) \in E$ of $G=<V,E>$?

@RonaldVilliers okay I will

avra, the shortest path from s to u, followed by the edge from u to v, is a path from s to v. is there more to it?

@leslietownes. Thanks. Since $u$ is after $v$ you mean, so obviously the path to $s$ is shorter than the path to $u$ please?

If that is the case, why 1 is there please σ(s,u)+1?

i assume \sigma(a,b) is a notation for the length of the shortest path in the graph from a to b? is that right?

What's $G$?

is that a vector with sets?

i'm assuming G is a graph, with V the set of vertices and E the set of edges (presumably non oriented) but i could be wrong

Yes exactly

This is the setup

so the shortest path from s to u, concatenated with the edge from u to v, is a path from s to v

its length is sigma(s,u) + 1. the length of the path from s to u, plus 1 for that final edge

sigma(s,v) is, by definition, \leq the length of any path from s to v, so in particular, the length of that one

soundcloud.com/john-zimmerman-960525209/enough-time-1

whats the sigma notation for

just means path?

@RonaldVilliers

@RonaldVilliers. $\sigma$ means shortest path

or rather, the length of it?

@leslietownes. No the shortest distance

yeah, the length of the shortest path

point being only that sigma( ) is not a path but a length

a number

where is s in relation to u,v? is s in E as well?

otherwise i would agree that i'm not sure what it means to add the number 1 to a path

@leslietownes. So, again please, the shortest distance $s$ to $v$ is the shortest and say it's 5, then the shortest path from $s$ to $u$ is the shortest and say it's $5+1$ since $u,v$ are one edge, where 5 is the shortest from $s$ to $v$, so my question why $1$ is added please?

because there is a path of length 5 + 1 from s to v. namely, the shortest path to from s to u [length 5], followed by the one step along one edge from u to v [length 1]

if i can get from s to v in 6 steps, then the shortest path from s to v can't be any longer than that

Kinda sounds like old school techno/trance @geocalc33

@leslietownes. Oh! Your reading for the equation is nice. Thanks.

@RonaldVilliers oh no! 🤦‍♂️

avra i'd note that you can certainly have < for some graphs (i.e., going from s to v via u might not be the shortest way of getting there). you see this even with like an equilateral triangle graph with verticles s, v, u

it sounds good though, definitely more pop-sounding

than club/dance music

@leslietownes. I really love the way you read it! It says it all. Wording is enough. Thanks. So the shortest path to $v$ from $s$ can not be larger than shortest path from $s$ to $u$ +1

@leslietownes. Appreciate it

isnt shortest distance s,v s,u dependent on where s is in relation to u,v leslie

oh nevermind :p

it's < or equal to

@RonaldVilliers thanks :)

consider a real analytic dim. 1 foliation of a real analytic Riemannian manifold of dim. 2.

can you use schramm loewner evolution to smoothly randomize the real analytic foliation by interpreting it as an SLE curve?

@TedShifrin Wow, author and youtuber! Are you a youtube influencer? :-)

i know ted mainly for his series of dances on tiktok

@amWhy Hell, no!

Given that $G=\{z\in \mathbb C: z^n =1 \text{ for some positive integer n}\}$

I think that G is an infinite group (a group of all nth roots of unity) such that every element of G has a finite order.

But in a test that I took today, it was an MCQ question and I marked as per what I thought was right. But my points deducted because of that choice because apparently the correct answer (according to the test) is G is a finite group.

I disagree with the answer as if so the answer were correct G should have been defined as: For some integer $n, G=\{z\in \mathbb C: z^n=1\}$

you're right but as you seem to note the order of quantifiers is very important. one would need to see the exam question verbatim to be sure of the answer.

i don't even like "z^n = 1 for some positive integer n," it's unambiguous in context but i dislike using a symbol before explaining what it is

Is the ratio OB : OA = CB : CA, OB' : OA' = C'B' : C'A' some known thing? I've never seen before

OD, OD' are tangent lines

@leslietownes Let $G=\{z\in \mathbb C: z^n=1 \text{ for some positive integer $n$ } \}$. Then under multiplication of complex numbers,

a) G is a group of finite order. b) G is a group of infinite order, but every element of G has finite order. c) G is a non-cyclic abelian group. d) None of the above

The correct answer (as per the test) is a)

which I'm in disagreement with.

they're confused with quantifiers, as you pointed out above

this happened in earlier test also :(

but note also that if G is the group of all roots of unity, both b) and c) are true. which is a sign that something is up

You're right!! I didn't think of c) at all as this question was under the section "only one correct answer"

what a mess. good luck with that

and this is nothing. One day there was an integer answer type question. So some integer value was to be input in _______

I put the integer value and it was marked wrong as the correct answer was " the value is ....."

So points got deducted because the phrase "the value is" was missing

😅

stuff like that made me want to punch through a wall when i was a student

i got an awful lot of that in high school from math teachers who had the excuse of not knowing anything about math

weird but not unheard of to see it happening with more advanced material

when someone complains about it, they are told that don't worry this is for practice only. Just keep your concepts clear etc..

haha

I read a news today that some law firm is gonna sue Apple as their latest M1 pro laptops have screens which crack during even normal usage. I'm worried as I recently purchase one M1 chip pro laptop.

appleinsider.com/articles/21/09/09/…

interesting.

So far, I've not faced any issue with the laptop though :)

@leslietownes I mean, I f*ck up when writing exams from time to time. It happens to the best of us.

On the other hand, I don't ever write multiple choice questions, so my f*ck ups are a little harder to spot.

it's not so much the f'up as the response. "don't worry about it"

or not acknowledging it at all

everyone makes mistakes, but it takes a bad teacher to do it in a way that has students questioning their sanity/understanding

and taking off points for X instead of "the value is X," i'm sorry, just no

koro, if you're ever teaching, and have the time (and competent colleagues) it is always a good idea to have a colleague take a test that you write before you give it to students. there often isn't time for this, but when you can do it, they catch things that you don't see.

@leslietownes noted. Thanks sir. :)

Oh, I misunderstood the question. The correct question is : Let $O$ be a point outside of an ellipse and $A,A'$ be distinct points on an ellipse. Let $B,B'$ be two points on an ellipse obtained by intersection of lines passing through $OA, OA'$ respectively. If two points $C,C'$ satisfy $OB : OA = CB : CA, OB' : OA' = C'B' = C'A'$ then $D,D'$ are tangent points.