@soupless No, this need not be true. Such matrices are called normal matrices (which commute with their transpose) and they are characterised by being "unitarily diagonalizable" i.e. under a unitary change of basis they are diagonal matrices.

@TeresaLisbon Helloo ma’am

Is it possible to discuss now ?

@TeresaLisbon In North India especially Punjab. I remember we would experience all kinds of weathers. Extreme cold , hot , rainy , windy.

@TeresaLisbon Do you have an idea how we can well approximate $$\sum_{j=k}^\infty \frac{1}{j^2\cdot \ln(j)}$$ with an easy function of $k$ ? I tried it with the corresponding integral , but unfortunately the expression has no antiderivate.

This is totally wrong now, right?

hi

Heh, didn't know all the CBI agents became mathematicians after the show ended :P

Hi, @lila. Thanks for checking out the many rooms that were sort of trolled by the user in question. I cast a couple of the flags.

@lila ! You moderate at Pets?

@amWhy Oh you are welcome, and yes, I moderate Pets.

That must be fun, but likely challenging, at times! Until I encountered you here, I hadn't even know there was a Pets site!!

I'll make a point of visiting the site!

@TeresaLisbon Hello ma’am

@SrijanM.T Hello! Let's discuss the issue. I will then move on the other issues which are pending.

@TeresaLisbon Thank god. I understand you’re really busy. Thanks

@SrijanM.T No, this is like my free period , Saturday night.

So thankfully I can compensate.

@TeresaLisbon Oh great.

Shall I write the whole Q in a clear way since I also understood I lot about my Q or have you got an answer ma’am ?

@SrijanM.T Let me see if I can get back the Q again : if you type it up at the same time it'll be nice as well.

@TeresaLisbon Ok. Perfect

Ill send a picture. Pls see if you can get it. Or I’ll write

The picture would be perfect, thanks.

If you’re wondering what is DBT1

I’ll send the pic.

Sure, it seems to be something important.

@Wolgwang Hi and no, it is correct, I really have only as little reputation, I gave it all away to my users for their top quality best answers to encourage participation.

@TeresaLisbon Did you get the Q ?

@SrijanM.T Yes, I got the picture.

When I wrote -Y but +1/2 . I meant that it should have been -ve negative I.e -1/2 maybe or sth.even though our variable is +b/2a .

@lila :-o That is amazing. Salute

I think what is meant to be said there, is that b/2a affects only how left or right the graph looks @SrijanM.T. The point is that a decides whether or not the graph looks upward or downward.

Yes.

a>0 means upward parabola.

And that's more important in the context of shape of a graph, than the situation of the vertex. That's what I think "b doesn't matter" means.

@TeresaLisbon I noticed one more thing.

@TeresaLisbon D/4a can decide whether it is going to +Y or - Y axis. But in end. It is always - ve X axis which is what is confusing

@TeresaLisbon If there is a way in maths to decide a value. Then , it has to be true for all condition or it should be written which are exception and also why. Since it is not inorganic chemistry.

@Wolgwang thanks!

@SrijanM.T The phrase "does not matter" is too subjective to be judged mathematically, so I won't wade in there. But regarding this whole negative X-axis thing : I'm just completely confused. The point is simple : the vertex of the parabola is always at -b/2a.

When you do left shift of b/2a. Then , you get + b/2a. Right. Doesn’t matter a is >0 or a<0 and same for b. The thing is. In end , +(b/2a ) value has to be always -ve since it is shifted on -ve X axis.

@TeresaLisbon Yes.

@SrijanM.T Not really : if the shift is a "left" shift along the negative x axis, then it technically falls on the positive x axis.

For example, -3 on the negative x axis is 3 on the positive x axis.

At least as far as I see it, this is the big trouble in the entire explanation, this "negative X-axis" part of things.

@TeresaLisbon Yes. But +3 on X axis is possible only when there is right shift

@TeresaLisbon Correct.

@TeresaLisbon Ma’am. I think it is also important that the values which we are writing satisfy as a quadratic condition.

@SrijanM.T I would say this : forget about the negative, positive x-axis etc. : the main point is that you calculate -b/2a , and put the vertex there on the x-axis, as a first step.

What if the condition falls into complex numbers , then we can’t solve it on this graph .

@TeresaLisbon Ok.

@SrijanM.T It wouldn't , because a,b,c are assumed real, so in fact we don't have to worry about that part of things.

In fact, all the CSTs are set up so that these heuristics work out.

@TeresaLisbon Ok.

Is it ok if I take a = 5 , b = 5 and c = 5 ?

It's ok, we can continue.

Sth wrong is happening. Let me show you the graph using conditions.

Ok

On + Y axis. I got -15/4

I used the condition correct.

This is perfect.

@TeresaLisbon This is not possible

@SrijanM.T I mean, the shape seems ok : the vertex is at (-1/2, 15/4), which I think it correct.

@TeresaLisbon I think that on + Y axis , there should be always a +ve value right.

@TeresaLisbon - 15/4

@SrijanM.T Oh yes, good point : it should be 15/4 : you are technically placing the vertex at y=-D/4a and x = -b/2a.

Not +15/4 I got ma’am

@TeresaLisbon oh yes. So , it will be +ve

But -X axis. My answer is wrong

No ,it's STILL correct. See, you found -b/2a : that equals -1/2. So your x coordinate of the vertex is -1/2.

Which you have drawn correctly.

@TeresaLisbon I’m confused 🤔 with CST now. 😅

@TeresaLisbon I found +b/2a

@SrijanM.T If you look at all the diagrams from your last photo, then you'll see that each time, the vertex is always at x = -b/2a.

So that part should be sorted : it doesn't matter what case it is, the centering of the parabola is always at x=-b/2a.

Q.E is never drawn or has a vertex at +ve X axis ?

@TeresaLisbon K. Yes

@TeresaLisbon Ok.

@TeresaLisbon Since I see that you online, I will remind that another room of yours has been inactive for some time. If a room has no messages for 14 days, it gets frozen.

@MartinSleziak Thanks, I'll respond there.

@TeresaLisbon I will look into it more. Understand CST better. Thanks a lot ma’am.

I have got the nitty-gritty of this concept.

@TeresaLisbon What are you doing on Father’s Day Tom ma’am ?

@SrijanM.T Sure. Actually, take this even simpler tip : JUST calculate -D/4a. The vertex is ALWAYS at (-b/2a , -D/4a).

@TeresaLisbon Ok.

@SrijanM.T Nothing much, apart from what I do everyday to be honest!

@TeresaLisbon Ok. Enjoy the day ma’am. Good night.

@SrijanM.T Good night!

In the recent 1 to 2 hours some people upvoted as much as 10 of my answers on Pets SE, I have to respect privacy and I will not reveal nicknames, but thanks to certain tools I could get a sense who those people were and the evidence points to this chat room - you made my day, big thanks!!!

@lila Oh I see! I wasn't one of them ,but either way it's good to know your posts received attention.

@lila More bounties?! Also, as a mod what you did was the best thing for a community :D

@TeresaLisbon Oh, by the way, sorry for starting a bit of off-topic in your room, I did not intend to de-rail the conversation, it just happened spontaneously; and also, I suck at math :D

@lila Completely fine by me!

@SafdarFaisal Once I get some more reputation, I will definitely continue and do some bounties in the future; I think you are right, I also came to realization that almost none of my privileges will go away no matter how much points I give away, and I have the impression that Pets SE taught me about 30 times as much knowledge as all of my answers collectively taught to other people, so it was almost like karmic destination of those points.