Did you take my advise?

Hey guys, I didn't think this club actually existed! I know this is more of a math club, as stated by the name, but I'm currently studying the most effective way to teach mathematics to people. Does anyone have any advice?

At 3/31/11 10:10 AM, ChainsawNinjaZX wrote: I'm currently studying the most effective way to teach mathematics to people. Does anyone have any advice?

What kind of level of maths are we talking here? I guess the complexity would alter the best way of teaching it. I probably wont be much help, but I've sat through enough maths classes in my life to possibly have some suggestions.

All levels, but for the sake of specifics lets lay algebra-algebra 2.

At 3/31/11 10:10 AM, ChainsawNinjaZX wrote: Hey guys, I didn't think this club actually existed! I know this is more of a math club, as stated by the name, but I'm currently studying the most effective way to teach mathematics to people. Does anyone have any advice?

stick to the basics. only introduce important theorems (might want to go through the proofs as well) to them. typical questions on selected topics should be discussed while leaving challenging questions to them.

appropiate teaching sequence also helps students learn more effectively. for example, when teaching techniques to solve systems of linear equations, the properties of matrices should be covered beforehand; the concept of locus should be introduced before teaching students conic sections.

At 3/31/11 10:10 AM, ChainsawNinjaZX wrote: Hey guys, I didn't think this club actually existed! I know this is more of a math club, as stated by the name, but I'm currently studying the most effective way to teach mathematics to people. Does anyone have any advice?

Are you studying to become a math teacher? Do you have background in advanced math courses?

Say, anyone got a suggestion how to revive this club?
It would be awesome to have some basic math challenges to keep me sharp or discuss stuff.

Just today I revised Jordan form of a matrix, involving minimal polynomials and nilpotent matrices.

A basic exercise that took me surprisingly long as a student is to show that for a nilpotent matrix A the kernel of A^{k+1} does contain other vectors than those in the kernel of A^k, given A^k =/=0.

You can have a go at this one if you like.

At 4/7/11 04:22 PM, RubberTrucky wrote: Say, anyone got a suggestion how to revive this club?
It would be awesome to have some basic math challenges to keep me sharp or discuss stuff.

Just today I revised Jordan form of a matrix, involving minimal polynomials and nilpotent matrices.

A basic exercise that took me surprisingly long as a student is to show that for a nilpotent matrix A the kernel of A^{k+1} does contain other vectors than those in the kernel of A^k, given A^k =/=0.

You can have a go at this one if you like.

Any hints? For a final year mathematician, I'm surprisingly bad at the proof side of things. I kind of wish I'd honed that skill throughout my university years...

Just out of curiosity, how did you get into maths, RubberTrucky? Do you use a lot of maths in your job (and in everyday life)?

At 4/8/11 12:42 PM, HeartbreakHoldout wrote:

At 4/7/11 04:22 PM, RubberTrucky wrote:
A basic exercise that took me surprisingly long as a student is to show that for a nilpotent matrix A the kernel of A^{k+1} does contain other vectors than those in the kernel of A^k, given A^k =/=0.

You can have a go at this one if you like.

Any hints? For a final year mathematician, I'm surprisingly bad at the proof side of things. I kind of wish I'd honed that skill throughout my university years...

A way I started to deal with the problem is by isolating the kernel. So fix a basis of the vectorspace by taking a basis of it's kernel and then completing it with linearly independent vectors outside of the kernel. This way you split the space into two complementary subspaces.

Just out of curiosity, how did you get into maths, RubberTrucky? Do you use a lot of maths in your job (and in everyday life)?

Started out in Physics, and notice that maths are better than lab-work. Went through theoretical physics and then jumped to mathematics. took some courses in geometry and finally decided to write a phd in geometry, what I'm currently working at.

99.99% of the NG users (including me) knows nothing about abstract algebra.
(i'm just a first year undergraduate in mathematics+physics). how am i supposed to come up with a solution?

posting questions that do not involve college stuff would be nice, if you want to have any chance of reviving this thread.

problem 1:
(i came up with this proof last year, though many of you might have seen the inequality below already)
let x, y, z be three non-negative real numbers.
denote x^k+y^k+z^k by S_k,

(a) show that (a-b)(b^k-a^k)>=0. for any non-negative real numbers a,b and non-negative integer k. when does equality hold?
(b) by expanding (S_1)(S_k)-3(S_(k+1)), and using (a),
show that [(x+y+z)/3]^n<=(x^n+y^n+z^n)/3 for all n=0,1,2...
equality holds iff x=y=z
(in fact it can be generalized to m non-negative real numbers to obtain a more useful inequality)

more problems to come! high school stuff of course.

let x, y, z be three non-negative real numbers.
denote x^k+y^k+z^k by S_k,

(a) show that (a-b)(b^k-a^k)<=0. for any non-negative real numbers a,b and k=2,3,4.... when does the equality hold?
(b) by expanding (S_1)(S_k)-3(S_(k+1)), and using (a),
show that [(x+y+z)/3]^n<=(x^n+y^n+z^n)/3 for all n=2,3,4...
equality holds iff x=y=z

sorry about the typo. changes made.

At 4/12/11 03:24 AM, i-am-ghey wrote: let x, y, z be three non-negative real numbers.
denote x^k+y^k+z^k by S_k,

(a) show that (a-b)(b^k-a^k)<=0. for any non-negative real numbers a,b and k=2,3,4.... when does the equality hold?
(b) by expanding (S_1)(S_k)-3(S_(k+1)), and using (a),
show that [(x+y+z)/3]^n<=(x^n+y^n+z^n)/3 for all n=2,3,4...
equality holds iff x=y=z

sorry about the typo. changes made.

b) I presume only works in the same conditions as a) (x,y,z>=0), since (1-2)^3/27=-1/27, whereas (1-8)/3=-7/3 <-1/27.

Of course, induction seems strongly hinted but I'll let people look over it for a while.

At 4/12/11 03:24 AM, i-am-ghey wrote: let x, y, z be three non-negative real numbers.
denote x^k+y^k+z^k by S_k,

(a) show that (a-b)(b^k-a^k)<=0. for any non-negative real numbers a,b and k=2,3,4.... when does the equality hold?
(b) by expanding (S_1)(S_k)-3(S_(k+1)), and using (a),
show that [(x+y+z)/3]^n<=(x^n+y^n+z^n)/3 for all n=2,3,4...
equality holds iff x=y=z

An immediate extension
c) Proof that [(x_1+...+x_r)/r]^n <=(x_1^n+...+x_r^n)/r

Hmm, there just might be something hidden there about convergence of series.
Something along the way of E(X)^n<=E(X^n), where E(X) is the expectancy of a sequence of positive valued outcome of a discrete experiment.

288

Now proof the theorem as posted by i-am-ghey.

solution to problem 1:

(a) divide into three cases 1. a<b, 2. a=b, 3. a>b.
it is easy to see that LHS<RHS in case 1 and 3. LHS=RHS in case 2. so, strict inequality holds iff a does not equal to b. equality holds iff a=b.

(b) expand (x-y)(y^k-x^k)+(y-z)(z^k-y^k)+(x-z)(z^k-
x^z) to get xy^k+xz^k+yz^k+zx^k+zy^k+yx^k-2(x^(k+1)+
y^(k+1)+z^(k+1)).............. (*)
from (a) (*)<=0 for all x,y,z>=0. moreover, equality holds iff x=y=z.
note that (S_1)(S_k)-3(S_(k+1))= (*) <=0 from direct expansion, so S_1<=3(S_(k+1))/S_k if S_k=/=0....(**)

(if x,y,z not all 0)
let P(n) be the statement[ (x+y+z)/3]^n<=(x^n+y^n+z^n)/3.
P(1) holds trivially from (*).
assume P(k) is true, (i.e. (S_1)^k<=3^(k-1)(S_k)
then (S_1)^(k+1)<=3^(k-1)(S_k)(S_1) (induction assumption)
using (**), it is clear that (S_1)^(k+1)<=3^k(S_(k+1))
this completes the induction.
hence, result follows.

(if x,y,z are all 0. trivial)

actually, the same argument can be applied to the r numbers case.

________________________________________
_____________________
problem 2: (this one is relatively easy)

let f(x)=x^2+mx-m^2, where m>0. it is known that the equation f(x)=0 has exactly one positive root r_1 and one negative root r_2.
suppose further that m>r_1>0, and given a sequence ofpositivereal numbers {a_n} defined by the recurrence relation
a_1=(1/2)(r_1), a_(n+1)=(2ma_n+m^2)/(a_n+m) (for n>=1)

(a) using induction and monotonic sequence theorem, prove that {a_n} converges to r_1.
(b) let g(x)=x^2-mx-m^2, g(r_3)=g(r_4)=0, r_3<r_4.
suppose now a_1=(1/2)(r_3). without repeating the same steps in (a), prove than {a_n} convergers to r_3. (it may be necessary to define a new sequence and apply the result in (a))

you may ask for more hints. solution will be posted one week later. the next problem will be less computational and most likely be more challenging than problem 2.

I would like to join.

At 4/17/11 12:04 PM, germanturtle wrote: I would like to join.

thanks for your interest in the club. you are probably one of the very few members who are interested in maths.
there is no need for you sign up. just feel free to post whenever you want. any questions about math are welcome.

At 4/17/11 12:27 PM, i-am-ghey wrote:

At 4/17/11 12:04 PM, germanturtle wrote: I would like to join.

thanks for your interest in the club. you are probably one of the very few members who are interested in maths.
there is no need for you sign up. just feel free to post whenever you want. any questions about math are welcome.

OK

At 4/17/11 05:54 AM, i-am-ghey wrote:
problem 2: (this one is relatively easy)

let f(x)=x^2+mx-m^2, where m>0. it is known that the equation f(x)=0 has exactly one positive root r_1 and one negative root r_2.
suppose further that m>r_1>0, and given a sequence ofpositivereal numbers {a_n} defined by the recurrence relation
a_1=(1/2)(r_1), a_(n+1)=(2ma_n+m^2)/(a_n+m) (for n>=1)

(a) using induction and monotonic sequence theorem, prove that {a_n} converges to r_1.

A tricky one. I think, by the way that you mean that a_{n+1}=(ma_n+m^2)/(a_n+2m)=m (1-m/(a_n+2m))

In case of convergence, you need that the limit a satisfies a=(ma+m^2)/(a+2m) then you have that
a^2+2ma=ma+m^2, or a^2+ma-m^2=0. When a has to be a positive number, then it has to be r_1.
The tricky part is to show that a convergence is present. Note at that that if a_1 is positive, so is a_2 and so on, since m/(a_n+2m)<1 when m and a_n are positive. Furthermore, a_n lies between m and 0.
So a_n is a bounded positive series. Now you only have to show convergence.

To this avail, show that the map x->m(1-m/(x+2m)) is monotonous. It's derivative reads m^2/(x+2m)^2>0 for x a positive number, so the series is increasing on the positive axis. So an increasing positive sequence that has an upperbound converges to its supremum.
Note that the convergence is independent on a_1, as long as a_1 is positive.

(b) let g(x)=x^2-mx-m^2, g(r_3)=g(r_4)=0, r_3<r_4.
suppose now a_1=(1/2)(r_3). without repeating the same steps in (a), prove than {a_n} convergers to r_3. (it may be necessary to define a new sequence and apply the result in (a))

Given the previous problem, the sequence should become a_{n+1}=(ma_n+2m^2)/(a_n+m). A possible issue may arise because a_1<0.

But, it's possible I made a few mistakes here.

At 4/13/11 05:23 PM, RubberTrucky wrote: 288

Now proof the theorem as posted by i-am-ghey.

48 ÷ 2(9+3)
48 ÷ 2(12)
48 ÷24
= 2

wat

At 4/17/11 05:24 PM, Tokecat wrote:
I know this is an old question, but I'll ask it anyway. Is there any feasible way of dividing by zero?

Just curious.

The argument against, would say that if 0 has an inverse, call it a, then we would have

1*0=2*0, so (1*0)*a=1*(0*a)=1=(2*0)*a=2*(0*a)=2. which would be ridiculous.

Formally, however, projective geometry would allow 1/0 to be defined as infinity (outside conventional numbers)

At 4/17/11 06:30 PM, JaY11 wrote:
48 ÷ 2(9+3)
48 ÷ 2(12)
48 ÷24
= 2

wat

It's a stupid formality regarding the order of operations. mathematics solves this by writing fractions, so it's not really a mathematical problem as much as a syntax problem. Obviously
(a*b)/c=/=a/(b*c), hence the ambiguity of a/b*c. Syntax then says read from left to right.

Also, as the sequence goes: x_{n+1}=f(x_n) defines a convergent sequence, then it's limit point x satisfies x=f(x) at least when f is continues. But I've made a few mistakes showing the sequence is convergent.

At 4/17/11 06:46 PM, RubberTrucky wrote:
Also, as the sequence goes: x_{n+1}=f(x_n) defines a convergent sequence, then it's limit point x satisfies x=f(x) at least when f is continues. But I've made a few mistakes showing the sequence is convergent.

An interesting way to approach this problem can be stability analysis. So, suppose x_{n+1}=f(x_n)

First of all, consider x->f(x), what is its behaviour? A nice function can be increasing/decreasing monotonous. For example: f(x)=m(x+m)/(x+2m), i've shown to be strictly increasing (it's derivative is m^2(x+2m)^2), take away x=-2m.

Secondly, consider x-f(x). You can calculate where this equals 0, where x>f(x) or x<f(x). In this case,
You find (x^2+2mx-mx-x^2)/(x+2m)=(x^2+mx-x^2)/(x+
2m).
focus on x>-2m. Then x=f(x) in both r_2 and r_1. Between both values, f(x)>x, outside the two values, f(x)<x (with exception as x<-2m.

Finally we proof convergence to r_1, given that x_1>r_2. First of all, suppose that x_1 lies between r_2 and r_1. In this case, f(x_1)=x_2 is bigger than x_1 as f(x)>x in this part, but it's smaller than r_1 as f(x) is increasing and x_1 is smaller than r_1. (and f(r_1)=r_1) Now repeat the argument with x_2. So you obtain that r_2<x_1<x_2<x_3<...<x_n<r_1 is an increasing sequence bounded from above and has a limit. Because in this limit x_* we have f(x_*)=x_*, this limit is r_1.
Suppose now that x_1>r_1. in this case, f(x_1)<x_1, but bigger than r_1 for similar arguments. So now you obtain a decreasing series x_1>x_2>...>x_n>r_1. This will once again converge, since the sequence is decreasing and bounded from below. The limit will again be r_1.

I don't really get the second part. You have to proof convergence to r_3 of another sequence you have to build yourself?

in the second part, suppose the initial value of a_1 changed to a value between r_3 and 0.
also, the corresponding value of m is changed so that m<r_3<0 (i didn't type it in the question).
without using monotonic sequence theorem and any advaned mathematics, prove that the new sequence {a_n} converges to r_3.

wish they had a edit button.

shit. assume g(x)=x^2+mx+m^2.
can't get it right without being careless.

actually it is g(x)=x^2-mx-m^2. same as f(x) be with different m.

At 4/17/11 08:28 PM, i-am-ghey wrote: actually it is g(x)=x^2-mx-m^2. same as f(x) be with different m.

I let others think about this for a while.

Here's a math problem I took of the BBS a while ago:
Consider a plane which has a circle of radius R>0. Now take a second circle of radius r>0, which lies tangent to the first circle. Consider then the isosceles triangle which lies tangent to both the circles (does such a triangle exist?)
a) Given R, find as a function of r the height and area of the triangle, in case such a triangle exists. What happens when r<<R? And what when r>>R?
b) Given h_0>R, find a condition on r, such that the height h(r,R) obtained equals h_0.
c) Now, the same question, but instead of using circles in the plane, use spheres in the 3-space. And instead of triangles, use circular cones. (area becomes volume)

In this problem, all circles/spheres/triangles/cones/.. are of course the standard/classical concepts.

it seems that the first parts can be solved using similar triangles. i believe.
but i am not sure if i have the right picture.
does the smaller circle has to lie outside the larger circle?
and does the triangle has to be tangent to the larger circle at three points?

i will come back to the question once i have finished the presentation assignments this week.

At 4/18/11 08:08 AM, i-am-ghey wrote: it seems that the first parts can be solved using similar triangles. i believe.
but i am not sure if i have the right picture.
does the smaller circle has to lie outside the larger circle?
and does the triangle has to be tangent to the larger circle at three points?

i will come back to the question once i have finished the presentation assignments this week.

The circles touch on the outside and the triangle is obtained by taking both common tangents to the circles. The third side is then as economic as possible (tangent to the appropriate circle in such a way that the two previous tangents form the equal sides.
I think i could make it a bit more difficult by taking the 'best' triangle as the triangle that holds both circles, but has the least area.

At 4/18/11 10:50 AM, RubberTrucky wrote:
I think i could make it a bit more difficult by taking the 'best' triangle as the triangle that holds both circles, but has the least area.

Hmm, so to summarize: put 2 circles tangent to eachother (on the outside, take common tangent (there are 2, symmetrical around the axis of their centres) and then take the third side, orthogonal to the axis and tangent to one of the circles.

An interesting, but more difficult problem would be determining what triangle contains both circles and has the minimal area. Kudos to the person who can solve this by himself (I think it's feasible, even for someone with basic notions of analytic geometry). Something that might help is that each of the 3 sides of the triangle should be tangent to at least one of both circles. This is easy to see, because if one is not tangent, simply displace the side paralleled untill it is tangent. Because you move inward, the area will decrease (make a graph if you're not convinced). Furthermore, the smallest of both circles can have at most 2 tangent sides. Also easy to see graphically.
So essentially, one way to solve this problem is by picking three tangent points and start a minimalisation procedure. (though this might probably require some notion of partial derivatives).