@TeresaLisbon Hello again. Actually I wanted to gain some personal feedback so asked this question here instead of cured. I posted a ad - math.meta.stackexchange.com/a/33723/876009 in math meta for community but it was downvoted. Should I let it be deleted or undelete it

@JitendraSingh I don't know how the ad part of the site works, so I'm sorry I can't be of any help here.

@TeresaLisbon no problem

@TeresaLisbon Hi

@JitendraSingh Which ad? Related to Math mind Grade 1-10?

@MathLover Hello!

Before I come to my question, I have wondered what senior agent, CBI means.

@MathLover Nothing really, Teresa Lisbon is a character in a serial : over there, she is a CBI inspector/agent.

@TeresaLisbon Remember how we derived the quadratic equation for graphs using CST 1 ma’am ?

@SrijanM.T I do remember.

@TeresaLisbon ah, I see :)

We had a condition for a + or -

why don’t we have for b ?

Oh. Because we had condition for a since a has to be > 0.

For b , it doesn’t matter .

Is it right ?

@SrijanM.T Yes, that's right.

@MathLover We can continue.

@TeresaLisbon 👍

@TeresaLisbon I have couple of posts I would like you to see and tell me if my thinking is correct. I felt the person did not really make any significant effort in solving it. First of all title is misleading. Then person asks for change of variable solution but the problem can be solved even without that. math.stackexchange.com/questions/4173949/…

@MathLover Ok , I'll let you know. If you have another post, you can drop it below. The weekend is coming ,and I usually keep the weekends for clearing doubts rapidly here, so I'll definitely look it up.

@TeresaLisbon yes the second one is this - math.stackexchange.com/questions/4176050/…

Sure, thanks.

I'll let you know my feedback soon enough.

@TeresaLisbon thanks

welcome!

@TeresaLisbon But ma’am , we made sure to shift the graphs by + b /2a and not - b / 2a.

Can we say it doesn’t matter. -b/2a or +b/2a

@SrijanM.T Let me take a look once again, something might have slipped because I'm doing two things at once.

@TeresaLisbon 👍

+b/2a means to left side.

@Wolgwang well ya is there any problem? Sorry if there is

@TeresaLisbon It is shifted on the - x axis.

Then , it is should be +b/2a but vertex is -b/2a

Getting confused

@SrijanM.T Ok, let me clarify once again.

Start with the graph of x^2, which we all know very well.

Yes

Now, we first have to accommodate b/2a.

Keeping a positive

@JitendraSingh I think there are only a few users on MSE who are in grade 1-10 so it will not be useful for the majority of MSE users. This might explain the reason for the downvote.

It can be downward or upward parabola.

If $b/2a$ is positive, the graph shifts to the left by $b/2a$.

So we get the graph for $(x+b/2a)^2$ from here.

Yes

Now, IF a is positive, then the graph retains its upward shape, and scales by $a$ .

IF a is NEGATIVE, then the graph flips upside-down and scales by $a$.

(By scales, I mean "vertical growth").

yes

So for example, if $a= \frac 12$, then the graph vertically goes up/down only 1/2 as fast as earlier.

With this, you will get to $a (x+b/2a)^2$.

Ok.

Now, calculate -D/4a : it is some constant.

@TeresaLisbon Yes.

The graph shifts up or down based on -D/4a. So if $-D/4a$ is positive, then the graph goes up. If -D/4a is negative, the graph goes down.

@TeresaLisbon Yes. Clear with this for D/4a.

Of course, it goes up by -D/4a / down by -D/4a respectively.

Which then gives us the graph for ax^2+bx+c

@TeresaLisbon Not clear this one.

Which is what was required.

@SrijanM.T Ok, I'll clarify this.

The point is to draw the curves of x^2 and 0.5 x^2. If you draw them, the graph of the second function is very similar to the first function, but it goes up slower and comes down slower. While drawing the function you have to capture this.

I remember this example you gave for -D/4a. When D has > or < 0.

Correct : in that case, $-D/4a$ is positive. So the graph shifts up by 3/16.

@TeresaLisbon If a * x^2 , then more is a. More is the graph curved and less is a , more it is expanded kind.

Still. I’m not getting what you mean to say for b

@SrijanM.T Correct : if $a$ is smaller, the graph is more expanded. If $a$ is big then the graph is very curved and looks very thin.

Shall we take a values example ?

Ok

Let's take $a=-2,b=-3,c=3$?

It’s like a = - 5 and + 5. b = -4 and +4. Ok. Then , for a <0. +b/2a is what we have as value but due to shifting towards left. It is gone on the -ve x axis. Then , +(-4)/2(-5) is wrong. Why ? You get 4/10 but you should be having -4/10 according to the negative x axis ?

@TeresaLisbon I took one maam. Just check this. Then , we can go onto yours.

Ok, I'll stick with your example.

You are taking $a=-5,b=-4$, is that right?

@TeresaLisbon yes

Since there is no condition for b

unlike a

See. It should be +b/2a as vertex.

But on -ve x axis.

@SrijanM.T You sem to be taking this -ve $x$ axis. Yes, that' correct : basically speaking ,whatever b/2a is (whether positive or negative), the vertex of the graph should be at -b/2a.

Similarly , it should be +b/2a on -ve x axis. Hence , proved.

Correct.

Voila.

If I took b= +4.

But .

Then , it is wrong.

If you take b=+4, then you have b/2a = 4/2(-5) = -4/10. Now, the vertex of the graph should be at 4/10, so the graph shifts to the right.

And that is wrong ?

Or right,

Since b has no condition. It’s right .

It's definitely correct. The general formula (without invoking any negative axes and so on) is this : find b/2a. The vertex of the graph must be at x = -b/2a. That's all.

So if b/2a = 4/10, then the vertex is at -4/10 i.e. the graph shifts left.

If b/2a = -4/10 then the vertex is at 4/10 i.e. the graph shifts right.

If b/2a = 0 , then the graph doesn't shift anywhere.

@TeresaLisbon ok. Therefore , Q.E= +ax^2 ± bx ± c

@TeresaLisbon ok.

Q.E. refers to what?

Quadratic equation

Oh ok : then the form mentioned is correct, except that a can be negative.

Ma’am. There is one thing.

For -D/4a on +Y axis.

When we get + b / 2a in which b=+ and a = +

Then ,how can on -ve X axis . We have +b2a

That’s wrong .

We can say b=+1 and a=+2

Ohk. So , here a is +ve.

Therefore , +Y axis,

Got to go, I'll come back and clarify.

Even though we get download curve . X axis is -ve

@TeresaLisbon k.

We will have to neglect +b. There is a condition. I’m just typing for later discussion

Therefore , there is a condition for b to be also +ve. Then , when b <0 and a>0. We do right shift. Since +b/2a is necessary. Therefore , - (-b)/2(+a)

Therefore , this is wrong. Q.E= +ax^2 ± bx ± c

@TeresaLisbon If you get anything ma’am , then just ping me.

@TeresaLisbon Got anything ma’am.

@SrijanM.T Yes, I definitely got, but I'm doing something else right now so I'll complete that and get back to you.

Sure.

Probably in the evening or tomorrow morning.

@TeresaLisbon K.

@TeresaLisbon Can I ask what the dot product means, but not by stating what the formula is?

@soupless Do you know when the dot product of two vectors is $0$ ? The most important application of the dot product is calculating angles and distances. I think you have already seen such exercises.

The dot product of two vectors is zero if both are perpendicular

Let me add, what I am asking is a definition of the dot product without stating the formula or resorting to vectors,

Is that possible?

I am not aware of a definition other than that it maps two vectors to a real number under some specific rules. It is usually defined by the angle between the two vectors and its lengths. I know no visualization method or how to grasp the concept intuitively without a formula.

Ok, thank you! I am trying to make some notes about matrices, that's why.

I enjoyed the one-line proofs of Thales' theorem , for example or even the cosine law. Very useful, this dot-product.

Wait, what should I define first: matrix multiplication or dot product? I am very much confused what to define first.

Never mind, I'll define dot product first.

@soupless: confer inner product

Inner product?

Yes

Correct me if I'm wrong, please, but from Wikipedia, "The inner product generalizes the dot product to abstract vector spaces over a field of scalars, being either the field of real numbers $\mathbb{R}$ or the field of complex numbers $\mathbb{C}$"

Hmm yes

I am stuck on this problem, answer given is option (A). Till now, I have narrowed the answer down to being either one of options of (A) and (C), but since they've asked the smallest such L, shouldn't the answer be (C), as it would be a subset of the language consisting of all possible words?

@ParasKhosla Good point, but there's something else I don't see. Isn't the language in (C) equal to the language in (A)? If I consider any word in (A), it's formed by finite concatenations of the letters alpha and beta. Putting the alphas and betas together mean that any word can be represented in such a form as described in (C), isn't it?

Of course the language in (C) is contained in the language in (A) : so I think both languages are the same.

@TeresaLisbon Yes, that makes sense, Thanks

@ParasKhosla You are welcome!

@TeresaLisbon Noticed my new project ? $$k=134901$$ gives at least $5$ primes of the form $$n^{n^2}+k$$

@Peter Noticed it, but can't quite pay attention to it as of yet. The weekend is coming, so I'm looking forward to working on it then.

And to answering some leftover questions here from previous discussions.

We had (and still have) a very hot day in Frankenthal. What was the temperature today where you live ?

83F in Bangalore today , with some rain

We had 96F (36C)

That's very hot!

We rarely , if ever, have such a hot day here.

Although in North India , such temperatures are common, I'm in South central India and this region is known for very stable temperatures.

@TeresaLisbon Once you're active, can I ask if $A^{\intercal}A = AA^{\intercal}$? What I got as of now is this.