Ok. Where should I write this?

In general, these are not rotations in planes. There is a connection, but not the way you're thinking. It's because of the skew-symmetric matrices I referred to .

@dc3rd fit the definition into the limit exercise, of course.

Right.

So I arrive at $Df(a)h = \frac{a \cdot h}{\|a\|}$

So?

looking at it in terms of vectors $a \cdot h = a^{t} h$, so I concluded $\frac{a^{t}}{\|a\|}h$

and from here $Df(a) = \frac{a^{t}}{\|a\|}$

Yup!

Which will fit with the anteater example in the gradient section.

Isn't that sort of what I asked you at the start? Because this is the "linear map", I was trying to apply my linear algebra skills to establish this as linear map

i.e $T(a +b) = T(a) + T(b)$, but the denominator was messing me up

You're totally not understanding. As I said, you're confusing constants and variables.

@Govind75 I do not understand this question

@Thor, note $\dim \Lambda^2(\Bbb R^n)$.

oh Ted, just noticed your earlier comment -- yes, grades are irrelevant, but I mean colloquially an "easy A". meaning an easy class. I.e., I don't know if he gets to the good stuff during the semester

on the other hand - who wouldn't want to sit in on his lectures.. and yes, apparently, still around, still teaching. That's life for these people. They wouldn't know what else to do

what's the mistake I'm making in my understanding?

The linear map is $Df(a)$. $a$ is constant.

@XanderHenderson Thanks. I couldn't sleep last night till I had solved (or at least got to a point of nearly solving) that problem. And I think on reflection I should have called him out more on his reliance of the 'solve' feature on his calculator. It wasn't actually helping him get to the solution he needed (which was in terms of an algebraic expression)

Oh.....now it is starting to sink in.....So in general with our definition of a function being differentiable it's not the case that I could have a "general" linear map for the function? so in this specific example: $\frac{x}{\|x\|}$

how I'm thinking of it in the question I ask is through the lens of a derivative being a function itself

Hey Ted, can I ask you to check my sanity>

@dc3rd what do you mean a "general" linear map for a function?

given a cubic function $f(x)= x(x-3)^2$ depicted above, I had to find $a$ such that the areas $A$ and $B$ were equal

Please write a linear map explicitly with a formula for its value on a vector.

yea, my choice of words was not great

I've been complaining since the beginning, dc3rd.

@TedShifrin the "of time" is implied

Your formula is wrong, @Andrew

Yeah, should have cropped the image

that is from a section of notes I took while trying to guide my student :p

What is the area of $A$? Of $B$?

We are given that A+B = some number (I think 27/2 iirc)

oh cute

but we could easily determine that from the info provided bear with'

You said we want them equal?

so the idea is to find a line that cuts through the cubic that splits the area in two equal parts

(where the lower bound is the x axis, ie y=0)

Yes. So write down the integral for $A$ and the integral for $B$.

Hint: $a$ shows up in both.

No.

No?

Note that the area of the triangle is easy by inspection.

wait I may have made another mistake as well

Are they both wrong?

I've got it now. Sorry for frustrating you again

First one is right, but I don't want to integrate $g$. I just want to say what that area is.

This is an important hurdle to get over, @dc3rd.

oh right so putting B aside for the minute, you want to integrate the cubic then subtract the area of the triangle to $(a, f(a))$...

From $0$ to $a$, yes.

which would just be $\frac12 \Delta x \Delta y $

That same triangle gets added on To what integral for $B$?

I've truthfully been working on it. All up to yesterday I was reflecting on what areas I need to tighten up with my thinking....and I thought about you and my "stubborness" to reframing thins along with "lazy" thinking.

Still more work to do.....I'll get through

Don't introduce unnecessary notation, Andrew. Use what we already have.

@dc3rd It comes from years of sloppy math teaching, confusing a function and its value at $x$.

hold on I need a fresh piece of paper, though I'm not sure about the triangle being added to $B$. I'm not sure I'm seeing what you see.

I want to note that the triangle area is $\frac12 af(a)$, right?

Yes...I also have to take more time to read and process each word. Instead of glossing over it just because I've seen the familiar notation.

another note to add to my recent list.

@TedShifrin Yeah, of course! I see how changing notation has added to my confusion now

Today isn't even an analysis day, today is supposed to be stats. Well I've gotten that problem over the line and made some in roads on my thinking. ...back to the grind I go. Thanks for the assistance.

Especially with tutoring/teaching, extraneous notation screws students and teaches them bad habits.

So $A = \int_0^a{f(x)}dx - \frac12af(a)$

Take care, @dc3rd.

Right. And area of $B$?

and then $B = \int_0^3{f(x)}dx - A$

because$ A+B = \int_0^3{f(x)}dx$

OK. Or triangle area plus $\int_a^3$.

Ahhh nice

Oh now I see what you saw

and now it is just a matter of rearranging to get a

OK. They give us the same equation to solve.

man I hope i get a second shot at tutoring, though now I'm pretty sure I may have made a bigger mess than I thought

Send the kid a pdf explaining the right way.

yeah, I should type this up, and make it look proffesional

Tutoring is hard if you're not really on top of it.

should I give him all the steps?

I should probably leave out the numbers so he can fill in some blanks himself right

Explain what we went through. Let him try to finish from here. I haven't done the algebra to see if it's tricky.

actually just a quick comment I remember in Spivak that he talked about differentiating between $f$ and $f(x)$ and the problems that arise as you just talked about......

Integrating the cubic is sorta yuck algebra.

it is a bit, but he has a calculator that is basically handheld mathematica

Ah, but he needs skills, too.

I know right!

(he says after demonstrating ignorance yet again)

gtg, have some Tex to compose

Yes, Spivak has exercises in chapter 3 and 9 relating to this.

Bye, @Andrew.

Should I get a copy of Spivak?

this book? amazon.com/Calculus-4th-Michael-Spivak/dp/0914098918

and I did them too...........now what I need to bloody do is retrieve these sorts of things from my mind when working on exercises....I have had a bad bad habit in the past of doing exercises and then "forgetting" about them after they're done.

If you want to be a proofy mathematician, it’s a great book, but …

@dc3rd The issue to remember is that a matrix $A$ gives a linear map $T$ by $T(x) = Ax$.

I'll put it on my list. I need to pay of a car before I start buying books willy nilly :P

and that matrix $A$ is a bunch of constants.

Typically, although in our case it may depend on $a$.

well if we are not thinking of a vector space of functions. But it is clearer again now.....anyways...off I go. I'll probably pass in later

@JoeShmo I know nothing about astrophysics, so I don't have an answer to your question

@AndrewMicallef What value did you get for $a$? I calculated it by hand, but verified it & plotted it using SageMath, on my phone. script

@PM2Ring I hadn't found a value for a just yet (been faffing about writting up the steps instead :P)

overleaf.com/read/hwgxpfjnzvrk

Fair enough.

but that actually saves me the heartache of figuring out what tikz commands to make a nice image thanks :P

The first time I tried verifying it by hand I made a dumb mistake in the arithmetic for the area of B. Oop. Then I realised it's better to just do $a^2(a-3)^2$

@AndrewMicallef Sage can also do SVG plots, but it takes a little more overhead. But I guess the plain PNG version is adequate.

I will keep that in mind, was actually about to ask in the tex chat how to turn my sketch into a professional looking piece of math art :

Give me a few minutes & I'll do a simple SVG version. I just have to check my notes...

You don't have to do that

Presumably this is a calculator problem for the student to solve the quartic in $a$.

Perhaps I messed up, but I don't see where @PM2 got those exact values.

@AndrewMicallef Already done. :) Uncomment the final #print line to get the actual SVG code. SVG script

I think mine looks nicer :P

Though nasty in some ways

overleaf.com/read/hwgxpfjnzvrk

(I just guessed the values because the actual values are not the point of the sketch :P)

@AndrewMicallef Ok. I can do fills, and nice grids too. But the resulting code is probably too big to fit into a chat message. ;)

I do not agree with your polynomial at the end, @Andrew, although I may have messed up.

@TedShifrin wait what>

that was directly from the students book

I have $27=8a^3-2a^4$.

well now

I have more faith in Ted Shiffrin than I do Essendon Grammar School

@TedShifrin Well, there are 4 solutions. Two of them are zero, and the others are $(8\pm\sqrt{10})/3$, but $(8+sqrt{10})/3>3$, so it's invalid.

no wonder his calculator didnt agree!

I don't see where the quadratic roots are coming from, @PM2.

That quartic doesn't factor.

I'm getting a numerical solution of $a\approx 1.84\dots$.

I need to keep typing this up, but I should probably make a note that it doesn't get the answer his teacher is after... huh

I have $a^2(3a^2-16a+18)=0$. And my solutions for $a$ give equal values for the integral of the A area and twice the B area. I don't think I made a mistake...

Wait. Why twice the $B$ area? They're supposed to be equal.

Andrew and I worked out the equation earlier: $2\int_0^a f = \int_0^3 f + 2\Delta$, where $\Delta = af(a)/2$ is the area of the triangle.

Sorry, I meant the integral from 0 to a is twice the B area, so A equals B.

No, I don't agree with that. Where is the area of the triangle?

No, it's not right.

@Andrew: The teacher's diagram sure doesn't look like the areas of $A$ and $B$ are equal. :P

More evidence that his/her equation may be incorrect.

lol, the diagram is mine, but the stuff in the blue box came from his excercise book, verbatim

The function is $f(x)= x(x-3)^2$, right? So the integral is $\frac14 x^4-2x^3+\frac92x^2 +C$

also I updated to use your estimate of a

that also gets rid of the overshoot and the sliver of area not covered by the two shaded rejions :P so thanks

Yes, @PM2, I agree on the integral. And the integral from $0$ to $3$ is $27/4$.

Agreed. And the area of the B triangle is $\frac12 a^2(a-3)^2$, so $\int_0^a x(x-3)^2 dx$ needs to equal $a^2(a-3)^2$ for A & B to have equal areas.

@PM2Ring This equation is clearly wrong, by the way, as $0$ is a double root. I have no idea where you dug up imaginary roots.

$B$ is not a triangle.

$B$ is everything left over when you remove $A$.

@TedShifrin Huh? I never said anything about imaginary roots.

Oh, sorry. Well, anyhow, your equation is wrong. :)

So $\int_0^a f - \Delta = \Delta + \int_a^3 f$ is the correct equation. Do we agree on this? $\Delta = af(a)/2$.

@TedShifrin It is? I misunderstood Andrew's diagram then. I thought B was a right triangle, with the vertical side being that dotted line from (a, f(a)) to (a, 0)

No, $B$ is everything under the curve "to the right" of the hypotenuse of that right triangle.

@TedShifrin plus the area of the triangle

Ok. I'm getting too tired to redo it now.

That's "to the right" of the hypotenuse, @Andrew. We agree.

I'm confident that my equation is correct. The teacher messed up.

oh okay

Nice

well that makes me feel better about following up on this

The equation I wrote up 8-10 lines is the natural equation. Algebra turns it into the one you liked (which I put earlier).

I'm doing my best to add as much exposition to the method we went through just now

@AndrewMicallef The plot there makes it clear what B is supposed to be. I should've realised my error when I saw that plot. Oh well.

sorry which equations are you referring to as "the one I liked"?

This one.

yeah ok

Anyhow, I've had enough with this problem :P

hahah

It wasn't really that interesting.

I didn't mean it to become such a bg thing

Well, you asked me about it. So it's all your fault.

I was happy with where we left it 2 hours ago

LOL

Then after that you asked me about it.

I'll take that

I had fun. Even though I misunderstood the question. :)

Apparently, the teacher misunderstood it, too. :P Or else ...

@Clarinetist my point was that it's not an obscure overspecialization, if that's what you were getting at. It's an excuse to do, say, signal processing.

or Andrew misphotographed the question in question

anyway enough of this, I'm off, turns out math exposition is hard. Finding the right balance of clarity and brevity is a chore

ciao friends

@Andrew: That's why there's so much poor exposition! Ciao.