Let's discuss Mathematics

[Samson]

I was looking for a description, though
As in "the variable is restricted by x things in y ways" etc.

How about, specifically talking about change with time: "The variable changes with a rate that changes at a constant rate". This would be compared to linear: "The variable changes at a constant rate" or exponential: "The variable changes with a rate that is a product of current value of the variable".

 

[Kyriakos]

How about, specifically talking about change with time: "The variable changes with a rate that changes at a constant rate". This would be compared to linear: "The variable changes at a constant rate" or exponential: "The variable changes with a rate that is a product of current value of the variable".

Because the derivative of a function of degree 2 is itself a linear function. I actually hadn't thought of a definition on account of that
The simplest thing is to not form most of the understanding you are able to, from something you have some insight about.

But I was looking for a definition that directly implies/ties to the natural language in the problems which lead to second degree functions. Still, I am sure I can find it on my own, following the... angle... of focusing on the derivative.

 

[a pen-dragon]

Can you think of a simpler (by this I just mean having fewer lines, and employ a different approach; eg it could be through establishing the point where two functions connect - the critical step would be to express those two- or by using calculus) way to calculate what the radius of the smaller circle will be?

Yes.

I take N to be the center of the small circle, R=(radius of the big circle)=2, r=(radius of the small circle). I will use underlines to denote me meaning the line connecting both points.

In the following I will consider the shown diagonal.

By definition it has a length of \sqrt(2)*4. This length is equal to: 2*R + 2*r + 2*\sqrt(2)*r, since the line OB intersects the small circle exactly at the intersection of both circles and NB is \sqrt(2)*r by construction. (the small circle touches the square at two points and the big circle in one)

Thus \sqrt(2)*4 = 2*(R+r*(1+\sqrt(2))). Isolating r results in: 2*(\sqrt(2)-1)/(\sqrt(2)+1)=r.

 

[Kyriakos]

Thanks, but I am not following your notation... Also, did you redefine N for your solution? (since the way I set it, N is a point in the periphery of the smaller circle, and by homothety in analogy with Γ or Α).
If it is a different method to the one employed by myself, I am very interested, so pls explain

Here is the generalized version of what I did:

Trivia: in Greek, for historical reasons, we use comma for separation of the decimal and integer parts.

 

[a pen-dragon]

Ok, let me start over.

Construction of the problem:

Let there be a circle with radius R=2 with the center of that circle called O. Construct a square around that circle such that both touch (not intersect) in four points and the circle is completely contained within the square. Now construct a circle such that it touches the square twice, the circle once and is again completely contained in the square.


Preliminary considerations:

Notice that this problem has the same symmetry as a square, especially it is symmetric under mirroring on the diagonals. Thus the center of the second circle most lie on that diagonal. Call that center C (equivalent to what I called N before, but you mean something else, thus I rename it here).

Other points that I will reference is the point at which both circles touch (A as in your notation) as well as a corner of the square (B). There are four points each of A,B,C, due to symmetry. In the following I will choose one B, and the A and C that lie on the line OB this ensures that all points are on the same half-diagonal of the square.


Solution:

By construction the square has the side length OB=2R, thus the half diagonal has the length R*\sqrt(2).

OTOH from the construction it follows that OB = OA + AC + AB. OA connects the center of the big circle with a point of the circumference and thus has length R. AC similarliy has the length r.

Now we have to consider AB. to do this we consider the touching points of the small circle with the square. Since in both cases a line touches a circle the lines will be tangential and thus the radii pointing towards these points will be perpendicular on those lines. Thus it follows that the object made out of A,B and these two touching points is a square with side r and AB is the diagonal. This implies AB = \sqrt(2)*r.

Thus R\sqrt(2) = R + r*(1+\sqrt(2)), and r = R *(\sqrt(2)-1)/(\sqrt(2)+1).

 

[Kyriakos]

Ok, let me start over.

Construction of the problem:

Let there be a circle with radius R=2 with the center of that circle called O. Construct a square around that circle such that both touch (not intersect) in four points and the circle is completely contained within the square. Now construct a circle such that it touches the square twice, the circle once and is again completely contained in the square.


Preliminary considerations:

Notice that this problem has the same symmetry as a square, especially it is symmetric under mirroring on the diagonals. Thus the center of the second circle most lie on that diagonal. Call that center C (equivalent to what I called N before, but you mean something else, thus I rename it here).

Other points that I will reference is the point at which both circles touch (A as in your notation) as well as a corner of the square (B). There are four points each of A,B,C, due to symmetry. In the following I will choose one B, and the A and C that lie on the line OB this ensures that all points are on the same half-diagonal of the square.


Solution:

By construction the square has the side length OB=2R, thus the half diagonal has the length R*\sqrt(2).

OTOH from the construction it follows that OB = OA + AC + AB. OA connects the center of the big circle with a point of the circumference and thus has length R. AC similarliy has the length r.

Now we have to consider AB. to do this we consider the touching points of the small circle with the square. Since in both cases a line touches a circle the lines will be tangential and thus the radii pointing towards these points will be perpendicular on those lines. Thus it follows that the object made out of A,B and these two touching points is a square with side r and AB is the diagonal. This implies AB = \sqrt(2)*r.

Thus R\sqrt(2) = R + r*(1+\sqrt(2)), and r = R *(\sqrt(2)-1)/(\sqrt(2)+1).

Ok, I got what you were saying now - the dash ("\") threw me off the first time, which is why I spoke of notation...
But this is surely the same approach, no? You just argued that the circumscribed squares are analogous, while I did that for the circles. All such variations flow from effects of taking a parallel line to some part of the original construction, since then (by Thales theorem etc) the corresponding parts are in analogy.

 

[a pen-dragon]

In the end I replace your homothety argument, by another one, obtaining (obviously) the same result. It is not that different, but I replaced 2/3 of your handling of formulae with a geometric argument, which I believe to be shorter and more elegant. Doing the same with the homothety argument is possible.

 

[Kyriakos]

There was this question (as a type of informal test) posted on a math site in my FB feed today:

"Find (define their geometric location) the set of all points P for which the tangents to the circle form an angle of 45 degrees".
The basic part was forming KP and writing it as a function of the given circle's radius, which we can call r1 (which would obv be r1/sine of 22,5 degrees as K is on the perpendicular bisector of AB).
The more theoretical part was to explain why all P points would be in the periphery of a concentric to the original circle=> KP would be that new circle's radius. I just thought that all such points would be constructed if the original circle rotates for 360 degrees, so they'd be in a new circle (all other possible angles formed by tangents would be in different concentric circles) - but the person posting the test asked me to give a "better" explanation. Of course he is right that this wasn't formal at all
Any ideas for how to formalize that part?

 

[Samson]

This is the BBC bit about it, which is pretty much their silabus. The sort of way they solve it is illustrated in this picture:

They do not explicitly say it, but this is something like differentiation, and going the other way is something like integration, but the rules are not quite the same.

You can explain why the rules are the way they are quite visually, with wooden cubes making the shapes. It seems a small step to go from that to calculus, but I cannot find a description in those terms.

I have been thinking a bit more about this, and the question I think is what is the relationship between the derivative of f(x) (f'(x)) and f(x) - f(x-1). Here are some that I think illustrate what I mean. I shall use Δ(x) to mean f(x) - f(x-1), aka the difference between terms in the integer sequence:

f(x) = 1
f'(x) = 0
Δ(x) = 0

f(x) = x
f'(x) = 1
Δ(x) = 1

f(x) = x^2
f'(x) = 2x
Δ(x) = 2x - 1

f(x) = x^3
f'(x) = 3x^2
Δ(x) = 3x^2 - 3x + 1

f(x) = 2^x
f'(x) = ln(2)*2^x
Δ(x) = 2^x - 2^(x - 1)

You can derive the Δ(x) functions with simple algebra, not need for limits. It seems to make a useful segue into the rules of calculus, meaning they are less of a learn by rote thing. My impression, from a sample of 2 at opposite ends of the UK secondary education system, is that they teach them the rules for Δ(x) by rote, and do not make the link to calculus. I wonder if there is anything I could look at to make this link, in the context of tutoring an 11 year old.

 

[Kyriakos]

(@Samson )
You could present an introduction to calculus, using equations the kid will already be in the process of being taught (?) like the emblematic free fall acceleration. Although at least here that formula is first presented when you are 12 (second year of secondary education).
In the BBC's example, the rate of increase of the increase is stable (=2), which implies that this is a quadratic function (two derivatives reduce it to a number). The actual increase at each subsequent position varies, and due to the previous note this has to vary linearly (be some ax+b). From the increase of the increase being 2, you get that the first derivative was 2x+b, whose integral would be x^2+bx+c. Indeed, the function is x^2+1.

Are you willing to present the way you get the derivative (I don't mean what I wrote above; I mean the actual analysis that makes it clear why - say - the derivative of x^2 is 2x), to the kid? Obviously the Δx are there.

Maybe something like this, but with the "with calc" part elaborated

 

[Samson]

Are you willing to present the way you get the derivative (I don't mean what I wrote above; I mean the actual analysis that makes it clear why - say - the derivative of x^2 is 2x), to the kid? Obviously the Δx are there.

Maybe something like this, but with the "with calc" part elaborated

Yeah, I think something like this is the way to go.

 

[Kyriakos]

Just don't use Spivac's book (unless the kid is really hardcore) ^^
There was an old 3blue1brown intro to calculus series, which generally follows the approach of 'calculus for engineers' (quite the insult) books and consequently speaks of orders of magnitude twice removed and thus inconsequential.
Or, of course, you can try being as rigorous as possible - and maybe scaring your student away.

 

[Harv]

It looks like I was off by a factor of 2, because I assumed the radius of the circle was 1, and that would make the side of the square 2 instead of 4. So it would help me in the future to read the problem carefully. Here is what I got:

R is the radius of the big circle. (2 in the original problem, but shown as 1 below)
r is the radius of the small circle.
The circles intersect at [sqrt(0.5),sqrt(0.5)].
Inspecting the small circle, sqrt(0.5) + [1+sqrt(0.5)]r = 1
r = [1-sqrt(0.5)] / [1+sqrt(0.5)] = 0.171
Scaling by 2 gives 0.343.

Sorry about the text notation.

 

[Kyriakos]

It looks like I was off by a factor of 2, because I assumed the radius of the circle was 1, and that would make the side of the square 2 instead of 4. So it would help me in the future to read the problem carefully. Here is what I got:

R is the radius of the big circle. (2 in the original problem, but shown as 1 below)
r is the radius of the small circle.
The circles intersect at [sqrt(0.5),sqrt(0.5)].
Inspecting the small circle, sqrt(0.5) + [1+sqrt(0.5)]r = 1
r = [1-sqrt(0.5)] / [1+sqrt(0.5)] = 0.171
Scaling by 2 gives 0.343.

Sorry about the text notation.

Yes, I think you did this (for the highlighted step)

Which is correct, but imo counter-intuitive? (I like it though, as it doesn't bother with the bit past the diameter of the small circle) Although I recall you liked to use coordinates in other math stuff posted in the past. The counter-intuitive bit was expressing the radius' length both as a length and a diagonal to the confined set to measure (=the diagonal is also the radius, but has to be converted to the straight line analogue so as to count in the sum of the abscissa).
Of course this is also homothety; just neither of the circle nor the circumscribed square, but of the diagonal.

 

[Harv]

Although I recall you liked to use coordinates in other math stuff posted in the past.

I think the issue is I tend to think in terms of Cartesian Coordinates. x, y, z. This might gives some indication where my math skill peaked.

It makes me think of a chart that floats around the internet, plotting math skill (and knowledge) against time, expressed by stages of life. We learn arithmetic, algebra, basic calculus, differential equations, partial differential equations. I understand much of the universe is governed by partial differential equations. Based on this, I believe I peaked somewhere between algebra and basic calculus. Most likely, this is because I started to study calculus in high school. The punchline is the chart shows skill crashing as soon as we learn to use Microsoft Excel. (I wish I obtained a version of a program like Mathcad and developed proficiency using it.)

This also means I peaked in high school - and the first year of university. This coincides with Civilization coming out - but I never started playing that until the end of university.

 

[Kyriakos]

I think the issue is I tend to think in terms of Cartesian Coordinates. x, y, z. This might gives some indication where my math skill peaked.

It makes me think of a chart that floats around the internet, plotting math skill (and knowledge) against time, expressed by stages of life. We learn arithmetic, algebra, basic calculus, differential equations, partial differential equations. I understand much of the universe is governed by partial differential equations. Based on this, I believe I peaked somewhere between algebra and basic calculus. Most likely, this is because I started to study calculus in high school. The punchline is the chart shows skill crashing as soon as we learn to use Microsoft Excel. (I wish I obtained a version of a program like Mathcad and developed proficiency using it.)

This also means I peaked in high school - and the first year of university. This coincides with Civilization coming out - but I never started playing that until the end of university.

You are still alive => you can peak elsewhere

 

[Kyriakos]

I posted this yesterday, then took it down thinking it might be better if the image was more elegant (regarding the lines, and possibly I could just use millimeter paper for a specific case).
But ultimately I decided to repost pretty much the same now...

It is a very good example of how the same reality (a tautology) can acquire obvious meaning in one field (geometry, above) and obscure meaning in another (algebra, below). Both are about the square root of x.

Afaik this way to geometrically construct the square root was known already in the age of Archimedes. But Descartes shamelessly presents it in the start of his own book of geometry, without attribution, as if he invented it. Because Descartes was a jerk

 

[Harv]

Maybe I do not always think in terms of Cartesian Coordinates. This does not look like the way I would calculate a square root.

The picture makes me think of the last few times I attempted to calculate pi. I believe a linear (C = 2 x pi x r) approach is way too tedious and for ancient people only used this method as an exercise to build tolerance to tedious work. It looks like they would have to run 1.66 iterations to gain another digit. An area approach (A = pi x r²) will converge faster, 0.83 iterations per digit. This is the method I believe ancient people would have used, but dirt piles of archaeology say that I am wrong.

Another thing I do not know is how ancient people looked at triangles and trigonometry. Nowadays, we consider a unit circle and a right-angle triangle and define sine, cosine, and tangent. How did the Greeks do this, for example?

 

[Kyriakos]

Sine and cosine did not exist in ancient/hellenistic Greece. There were, of course, equivalent theorems - a very famous example is how proposition 12 of Book 2 of Euclid's Elements is equivalent to the law of cosines
And while area of triangle= 1/2 height(base) DID exist in ancient Greece (eg mentioned by Pappos of Alexandria, as a 'less refined, unit-based approach'), there was also the following method to calculate area of triangle:

 

[Kyriakos]

Sometimes the people who write the solutions to mathbooks' questions are clearly bored. Here is one example of writing a solution when you were asked to prove that a^2-ab+b^2>=0.

A bit more straightforward to just write a^2-ab+b^2>=0 <=>a^2-ab+b^2>=0=>2a^2-2ab+2b^2>=0=>(a-b)^2+a^2+b^2>=0 (obviously so, as they are all squares).