A croissant shop has plain, cherry, chocolate, almond, apple, and broccoli croissants. a) How many ways are there to choose a dozen croissants?

b) How many ways are there to choose three dozen croissants?

c) How many ways are there to choose two dozen croissants with at least two of each kind?

d) How many ways are there to choose two dozen croissants with no more than two broccoli croissants?

How I thought to do the problem:

a) $6^{12}$ because you can have repetition.

b) $6^{36}$ - same argument as a.

c) If you have to pick at least two of each kind, then you subtract a dozen croissants that you are allowed to choose because you have six different varieties. This gives $6^{12}$.

d) If you have two dozen croissants with no more than two broccoli croissants, then you can first take all the possibilities with zero broccoli croissants ($5^{26}$) + having one broccoli croissant ($5^{25}$) + having two broccoli croissants ($5^{24}$).

Am I correct in my thinking?

Given Question is of the form "selecting indistinguishable objects into distinguishable boxes" and the formula is given by  $$\binom{n+r-1}{r}$$ where $n=$number of distinguishable boxes and $r=$number of indistinguishable object.

a) $$\binom{6+12-1}{12} =6188$$
b  $$\binom{6+36-1}{36} =749398$$
c) $$\binom{6+(24-12)-1}{12} =6188$$
d) $$\underbrace{\binom{5+(24-0)-1}{24}}_\text{zero broccoli croissants} + \underbrace{\binom{5+(24-1)-1}{23}}_\text{one broccoli croissants} +  \underbrace{\binom{5+(24-2)-1}{22}}_\text{two broccoli croissants} =20475+17550+14950=52975$$


Mathematical subjects you wish you learned earlier I am learning geometric algebra, and it is incredible how much it helps me understand other branches of mathematics. I wish I had been exposed to it earlier.

Additionally I feel the same way about enumerative combinatorics.

What are some less popular mathematical subjects that you think should be more popular?

Category theory and algebraic geometry.

I spent a lot of time in undergrad studying things that were kinda nifty, but way too classical to be of any use/interest beyond "fun math". When I got to grad school, category theory was assumed and made some of my courses much harder than they should've been.

In the words of Ravi Vakil, "algebraic geometry should be learned slowly over a number of years". I currently NEED algebraic geometry, so I don't have this number of years. I wish I would've started that a long time ago. Additionally, both of these topics would've helped me learn the things I was thinking about anyways, in particular commutative algebra.

Linear Algebra for sure. I would also add Statistics, though I would rather learn both well and in good depth than necessarily early.

Would second the answer on Logic.


Definition of Ring Vs Rng When I took abstract algebra I learned that a ring was a set that is an abelian group under addition and monoid under multiplication (along with the distributive property).

In preperation to tutor someone in algebra I've noticed that some books present a ring as what I know as a "rng" or an abelian group under addition and a semi-group under multiplication.

Is there any reason to prefer one as the definition for a ring vs the other?

EDIT
And a very related question, is there any math authority or consensus that has dictated/specified that it is more correct to use the ring/ring with unity or the rng/ring definition?

There are no math authorities. There are just conventions, and as far as I can tell the convention that "ring" means "ring without identity" can only be traced back to people who learned algebra using Hungerford.

The main reason to prefer "ring" to mean "ring with identity" is that I am pretty sure it is the statistically dominant convention, although I don't have the statistics to actually back that up. (Unless this is not what you mean by "reason," in which case I'll guess another possible meaning: for most applications, your rings will have identities.)

The standard definition of a ring $R$ is that $R$ is an abelian group under addition and a semi-group under multiplication. The existence of multiplicative identity is generally not required. If rings are defined in the "monoid under multiplication" way we can not consider the set $\{0\}$ as the smallest possible ring. (I see Artin defines rings in the "monoid under multiplication" way where Dummit & Foote, Herstein do not. So there is inconsistency in the standard definition.)


How do I know if a second order nonlinear differential equation is analytically solvable? How can I test if a non linear second order ODE is analytically solvable?

for example:

$$y''+\frac{3}{x}y'=ay^3+by$$

where $a,b$ are constants, I've been trying to solve such an equation for a while but I'm starting to think that I should test if its solvable to begin with

Hint:

Assume $a\neq0$ for the key case:

Let $x=e^t$ ,

Then $t=\ln x$

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=\dfrac{1}{x}\dfrac{dy}{dt}=e^{-t}\dfrac{dy}{dt}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(e^{-t}\dfrac{dy}{dt}\right)=\dfrac{d}{dt}\left(e^{-t}\dfrac{dy}{dt}\right)\dfrac{dt}{dx}=\left(e^{-t}\dfrac{d^2y}{dt^2}-e^{-t}\dfrac{dy}{dt}\right)e^{-t}=e^{-2t}\dfrac{d^2y}{dt^2}-e^{-2t}\dfrac{dy}{dt}$

$\therefore e^{-2t}\dfrac{d^2y}{dt^2}-e^{-2t}\dfrac{dy}{dt}+3e^{-2t}\dfrac{dy}{dt}=ay^3+by$

$e^{-2t}\dfrac{d^2y}{dt^2}+2e^{-2t}\dfrac{dy}{dt}=ay^3+by$

$\dfrac{d^2y}{dt^2}+2\dfrac{dy}{dt}=ae^{2t}y^3+be^{2t}y$

Let $y=e^{nt}u$ ,

Then $\dfrac{dy}{dt}=e^{nt}\dfrac{du}{dt}+ne^{nt}u$

$\dfrac{d^2y}{dt^2}=e^{nt}\dfrac{d^2u}{dt^2}+ne^{nt}\dfrac{du}{dt}+ne^{nt}\dfrac{du}{dt}+n^2e^{nt}u=e^{nt}\dfrac{d^2u}{dt^2}+2ne^{nt}\dfrac{du}{dt}+n^2e^{nt}u$

$\therefore e^{nt}\dfrac{d^2u}{dt^2}+2ne^{nt}\dfrac{du}{dt}+n^2e^{nt}u+2e^{nt}\dfrac{du}{dt}+2ne^{nt}u=ae^{2t}e^{3nt}u^3+be^{2t}e^{nt}u$

$e^{nt}\dfrac{d^2u}{dt^2}+2(n+1)e^{nt}\dfrac{du}{dt}+n(n+2)e^{nt}u=ae^{(3n+2)t}u^3+be^{(n+2)t}u$

$\dfrac{d^2u}{dt^2}+2(n+1)\dfrac{du}{dt}+n(n+2)u=ae^{2(n+1)t}u^3+be^{2t}u$


Does $-x^2$ does mean $-(x^2)$ or $(-x)^2$? When I see something written as:  $-x^2$ does it mean $-(x^2)$ or $(-x)^2$ ?

..and why? I can't apparently apply a streight PEMDAS rule to it

If it helps, think of it as $0-x^2$.  The Exponent wins first, then the Subtraction.  Hence it means $-(x^2)$.

The products have always prioprity upon the sums. So $-x^2 = - x \times x = - (x^2)$.


Conclusion about limit definition of e^a for a sequence of real numbers {a_n} converging to a? I have seen this fact used in several demonstrations, but have never seen a proof of it.

I believe the statement is: If $\{a_n\}$ is a sequence of real numbers such that $a_n \rightarrow a$ finite, then $(1 + \frac{a_n}{n} )^n \rightarrow e^a $.

Any help or references appreciated!

A simple approach starts from the fact that $\mathrm e^{x-x^2}\leqslant1+x\leqslant\mathrm e^x$ for every $x\geqslant-\frac12$ and that $\frac{a_n}n\geqslant-\frac12$ for every $n$ large enough, hence, applying this to $x=\frac{a_n}n$ for every $n$ such that $\frac{a_n}n\geqslant-\frac12$ yields
$$
\mathrm e^{nx-nx^2}\leqslant\left(1+\frac{a_n}n\right)^n\leqslant\mathrm e^{nx},
$$
that is,
$$
\mathrm e^{a_n-a_n^2/n}\leqslant\left(1+\frac{a_n}n\right)^n\leqslant\mathrm e^{a_n}.
$$
Since $a_n\to a$ and $\frac{a_n^2}n\to0$, the LHS and the RHS both converge to $\mathrm e^a$.

This is  called sequential continuity, one of the equivalent forms of continuity that applies to Real numbers , that says : $x_n\rightarrow x$ , then $f(x_n)\rightarrow f(x) $ , i.e., if the sequence $x_n$ converges to $x$, then the sequence $f(x_n)$ converges to $f(x)$. Note that there are spaces where continuity and sequential continuity are not equivalent. Can you tell how it applies here?

EDIT: More formally: if $a_n \rightarrow a$, we have,  $Lim_{n\rightarrow \infty} (1+a_n/n)^n=e^{Lim_n\rightarrow \infty a_n}=e^a$

EDIT 2: The argument above is correct, but incomplete. I will include a justification
when I can make it rigorous, or delete if I cannot do so in a couple of days.

There is a straightforward argument for why $(1+a_n/n)^n=e^{Lim_n\rightarrow \infty a_n}=e^a$: basically, $a_n$ will converge to $a$, and then $a_n$ will become a constant, so that the situation reduces to that of $Lim_{n\rightarrow \infty}(1+x/n)^n=e^x$:

For given $\epsilon>0$ , we can find a positive integer $N$ with $|a_n-a|< \epsilon$ for all $n>N$ , and there is a positive integer $N'$ with $|(1+x/n)^n-e^x |<\epsilon$ for all $n>N'$. Now, choose $N_2>$ $Max${${N,N'}$}. Then $$(1+(a-\epsilon)/N_2)^{N_2} <(1+a/{N_2})^{N_2}< (1+(a+\epsilon)/N_2)^{N_2}$$ Now let $N_2$ become even larger, so that $\epsilon$ gets increasingly-closer to $0$, and the left- and right- sides of the inequality will squeeze the middle term into being $e^a$


Why is the square root of a sum not equal to the square root of each its addends? Example: Let's presume one was attempting to isolate m below:

A common mistake would be:
$k^2 = m^2 + n^2 \to k = m +n$

Even though: $k^2 = m^2 + n^2 \to k \neq m +n$

If you apply a square root to both sides of the equation, you will have an inequality.

Why is this true?

If you put the question in a broader context you may understand it better.

Suppose $f$ is some function you can apply to numbers. It might be squaring, or square-rooting, or inverting, or raising to some other power, or taking logarithms or $\sin$ or $\cos$ or just adding 15. In none of these cases is $f(x+y)$ the same as $f(x) + f(y)$ (you should check). In general, you would not expect that coincidence.

The special case in which it is true is the function "multiply by a fixed quantity". That's the distributive law:

$$
c \times (x + y ) = c \times x + c \times y .
$$

Many of the most common errors students make in algebra or precalculus come from thinking that those other functions behave this way too.

Edit: Just in case you missed @Rahul 's comment: What we need here is a cure for the “law of universal linearity”: Pedagogy: How to cure students of the "law of universal linearity"?

Otherwise
$\sqrt n = n$ for all natural number $n$.

More generally
$\sqrt q = q$ for all positive rational number $q$.

Homework:

Is it true that:
$\sqrt x  = x$ for all positive real number $x$?


If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$, is $a=b=c=0$? If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$ , then is it true that $a=b=c=0$ ? I was thinking of infinite descent but can't actually proceed , please help. Thanks in advance

The number $c$ would have to be even, $c = 2k$. Divide the equation by $2$ to get

$$4k^3 + 2 a^3 + b^3 = 6kab$$

This gives the infinite descent, since you can always deduce that the number with the coefficent $1$ is even.

$a=1$, $b=1$ and $c=1$ is a solution of $4a^3+2b^3+c^3=6abc$, so $a=b=c=0$ is not the only solution.


Showing algebraic dependence of meromorphic functions on a compact Riemann surface I have been given the following question to do: Let $f,g$ be meromorphic functions on a compact Riemann Surface $R$. Show that there is some polynomial such that $P(f,g) = 0$ (i.e. show that any two meromorphic functions
on $R$ are algebraically dependent). I have seen this result over the torus which follows from looking  at the Weierstrass $\wp$ function, however I have no idea how to generalise that to every compact Riemann Surface.

There is a hint which says I should let $d = m+n$ where $m,n$ are the valencies of $f,g$ respectively and consider $P(f,g) = \sum\limits_{j = 0}^d\sum\limits_{k = 0}^d a_{jk}f(z)^jg(z)^k$ and show that this has at most $d^2$ poles and that I can choose the $a_{jk}$ so that $P(f,g)$ has at least $d^2+2d$ roots and so is constant by the valency theorem.

Showing that there are at most $d^2$ poles is easy but I don't know how to select the $a_{jk}$ to get $d^2+2d$ roots. I don't see whether I should try and find them explicitly (seems hard) or use some indirect argument (but I can't see where to start). Any help is much appreciated.

You can prove that there exists two polynomial $p,q\in \mathbb{C}[x,y]$ such that

$ord_a(\frac{p(f,g)}{q(f,g)})\geq 0$ for every $a\in X$

so it is an olomorphic function on $X$ that is  a Compact R.S. so there exist a costant $c$ such that

$\frac{p(f,g)}{q(f,g)})=c$

Then

$p(f,g)-cq(f,g)=0$

But I don’t know how built these two polynomials.
We can indicate $\{a_1,\dots , a_n\}$ the set of the poles of $f$ and $\{b_1,\dots b_m\}$ the set of the poles of $g$.

If $l:=ord_b(g)<0$ and $ord_b(f)\geq 0$ then you have that

$ord_b((f-f(b))^l g)\geq 0$

So you can resolve the problem in this particular case.


Prove that $a^2(p-q)(p-r)+ b^2(q-r)(q-p)+ c^2(r-p)(r-q) =4(\delta)^2$ If $p$,$q$,$r$ are the perpendiculars drawn from the vertices of a triangle ABC upon any straight line meeting the sides externally in D,E,F.

where a,b,c are the sides opposite to angles A,B,C in Triangle ABC and delta represents the area of the triangle ABC.
Also prove that EF= $(2pq)/((p-q)(p-r))$
I have tried this problem in this way but couldn't solve the problem, can anyone help me from here??

Let G,H,I be the foot of perpendiculars from A,B and C. GH=x GI=y check the following pictures carefully.we have broke the figure into rectangles and applied Pythagoras theorem...breaking the figure and applying Pythagoras theorem is very important

]1


If there a formal definition for the sum of two irrationals is rational if it is only the additive inverse plus some rational? If $x$ and $y$ are two irrational numbers and $x + y$ is a rational number, I know one of them has to be the additive inverse with/without some rational term I believe. I know it is not always the case that the sum of two irrational numbers is irrational.
for example:
$(\sqrt2 -1) + (-\sqrt2 +4) = 3$
But I can only seem to come up with examples that are of this nature and that seems to be the general consensus. But is there a formal definition stating it has to be of this form or is there a counterexample I am ignorantly unaware of.
then if $x+y\in\mathbb{Q}$ would it be true that $x-y$ would always just produce a multiple of that irrational plus some rational implying that if $x$ and $y$ are irrational and $x+y\in\mathbb{Q}$ then $x-y\not\in\mathbb{Q}$

You would not look for a formal definition but for a theorem, and also let's not talk about consensus. (Namely, an argument in math is either right or wrong, regardless on how many people are "for" or "against" it.)
So if $\alpha, \beta$ are irrational and $\alpha+\beta=q$ - rational, then $\beta=-\alpha+q$ - i.e. indeed $\beta$ is the inverse of $\alpha$ plus a rational constant, as you have claimed. Then $\alpha-\beta=2\alpha-q$, which must then be irrational. (Because, otherwise you would add $q$ and conclude that $2\alpha$ would be rational, and then you would halve it and conclude that $\alpha$ would be rational - contradiction!)

Yes, but it is trivial.  Say $x,y$ are two rational numbers, but $x+y = r$ is rational.  Then the sum is of the form
$$
x + y = (x+0) + (-x+r) = r
$$


Discrete Math Quick Regular Expression question! For a regular expression (a | b)*, can I write this string: aaaaaaabbbbbbbbbbbabababbb. I was thinking this means that when I choose a, I can have multiple a's and then I can choose b too after choosing multiple a's. Or is it that once I chose a, I cannot go back and choose b after finishing with a?

Your first interpretation is correct. The second one would correspond to the regular expression $a^*b^*$. But in both cases, don't forget to consider the empty word as well...

$(a|b)*$ means $a$ or $b$ zero or more times. You can only pick $a$ or $b$ at any given iteration
so $bbbbb, aaaaa, ab,$ null, $abaab$ are all correct.


Fubini-Study measure on a product I read that if $\Omega_2$ is the Fubini-Study form on $\mathbb{P}_1\times\mathbb{P}_1$, and $\Omega$ the Fubini-Study form on $\mathbb{P}_1$, then for all $(x,y)\in\mathbb{P}_1\times\mathbb{P}_1$, one has $\Omega_2(x,y)=\Omega(x)+\Omega(y)$.

I wish to justify this equality. What is the exact definition of the Fubini-Study form on the product space?

In the language of Quantum Mechanics, if we can say that the ket $ |\psi> $ is separable then it can clearly be written as $ \psi> = |\psi_{S_{1}}>\otimes |\psi_{S_{2}}>$ for subspaces $S_{1}$ and $S_{2}$ then the full metric is the sum of the metrics on the subspaces $g = g_{S_{1}}+g_{S_{2}}$.


How to prove $\sqrt\frac{1 - \sin x}{1 + \sin x} = \frac{1}{\cos x} - \tan x$ Prove $\sqrt\frac{1 - \sin x}{1 + \sin x} = \frac{1}{\cos x} - \tan x$

I tried but I couldn't figure it out, give me a hint please.

Hint: Multiply the quantity under the root by $$\frac{1-\sin(x)}{1-\sin(x)}$$

$\sqrt\frac{1 - \sin x}{1 + \sin x} $

=$\sqrt\frac{(1 - \sin x)(1 - \sin x)}{(1 - \sin x)(1 + \sin x)}$

=$\sqrt\frac{(1 - \sin x)^2}{(1 - \sin^2 x)}$

=$\sqrt\frac{(1 - \sin x)^2}{(\cos^2x)}$           [because $sin^2 x + \cos^2 x =1$]

=$\frac{1-\sin x}{\cos x}$

=$\frac{1}{cosx}-\frac{\sin x}{\cos x}$

=$\frac{1}{cosx}-\tan x$

NOTE : This answer holds only if $sine$ and $cosine$ functions are in first quadrant.


Evaluate integral of $z/(z^2 + 2z + 5)\,dz$ around the unit circle Evaluate integral

$$\int_c\frac{z}{(z^2 + 2z + 5)}\,dz,$$
where $c$ is the unit circle

I take $u = z^2 +2z$, then $du = 2z+2$, but that not work

Pretty sure that the answer is zero. By the residue theorem (link), we need to find out how many poles the closed curve (unit circle) winds around. Looking for points where the function is undefined, solve $z^2 + 2z + 5 = 0$ by completing the square; this gives $z = -1 + 2i, -1 - 2i$ and both of these points are clearly outside the unit circle (height $±2i$ in the complex plane). So the curve winds around no poles and it follows that the integral must be zero.


Undirected graph 1 degree checking Given an undirected graph, which consists of n vertexes and m edges. Provided we can delete edges from the graph. Now we want to check is it possible to delete edges in the graph so that the degree of each vertex in the graph will be equal 1?

There can be multiple edges in the graph, but can not be any self loops.

EXAMPLE : Consider n=4 and m=6 and let edges be (1,2),(1,3),(1,4),(2,3),(2,4),(3,4).

Then answer would be "TRUE".As it is possible to delete edges to get every vertex with degree 1.

This basically boils down to figuring out whether or not the graph contains a perfect matching; a matching that uses every vertex.

If such a matching exists, then we simply remove every edge that isn't in the matching. Similarly, if we are able to remove edges so that every vertex has degree 1, we're necessarily left with a perfect matching.

In that Wikipedia article I linked to, there's a link to Edmond's blossom algorithm, which is used to find the maximum matching (i.e. matching with the largest number of edges) of a graph. To find whether a graph has a perfect matching, you could run that algorithm and see if the maximum matching is a perfect matching.

Just check whether the graph is 1-Regular or not. You can find more on Regular Graphs on Wikipedia Regular Graph


Limiting value of Beta function I would like to know the limiting value of the Beta function:
$$ B(a,b) = \int_0^1 x^{a -1}(1-x)^{b-1}\,d x~ \tag 1. $$

For instance, what  does (1) reduce to as

$ a \to \infty$
$ b\to \infty$
$ a\to \infty, b\to \infty$?

Knowing the above would allow me to ( I hope) among others that

$$ \lim_{b\to\infty} \frac{\left(\frac{a}{b}\right)^{a} e^{wa}\left( 1 + \frac{a}{b} e^w \right)^{-(a+b)}}{B(a , b)} =  \frac{a^a \exp(aw-ae^w)}{\Gamma(a)} \,,\,a,b \gt 0~,$$
where $$ \Gamma(a) = \int_0^\infty x^{a-1} e^{-x}~ dx~. $$

For $a$ or $b$ going to infinity, take the limit inside the integral and see that the integrand vanishes, so the integral term goes to zero. Same with when both go to infinity.

In more details:

Consider $$ B(a,b) = \int_0^1 x^{a -1}(1-x)^{b-1}\,d x~ \tag 1. $$

What happens when $a$ goes to infinity? This is equivalent to asking what happens in $\lim_{n\to\infty} \int_0^1 f_n(x)d x $ where, $f_n(x)=x^{a_n -1}(1-x)^{b-1}\,$ and $a_n\to\infty$.

Now, it would be really convenient if we could take the limit inside the integral, and it so happens that we can totally do that. We need to use a version of what is called the Monotone Convergence Theorem.

Once we take the limit inside, for any $x\in (0,1)$, for any $b$, $f_n(x)$ decreases monotonically to zero as $n\to\infty$. So, basically $f_n$ converges pointwise to the zero function. Now integration of the zero function of the interval is definitely zero.

Check this out if you want to know the technical details of the theorem we used to interchange limit and integration.

Monotone Convergence Theorem for non-negative decreasing sequence of measurable functions

Same happens for the other limits you want to know.

First note that when you write $a \to \infty$ what you are actually asking for is asymptotic behavior, i.e. $a >> b$  where $a$ is variable and $b$ is fixed

With that in mind, first note that
$$B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
We can thus get the cases $b \to \infty$ and $a \to \infty$ by using Stirling's Approximation. Since this is symmetric, we only really need to know one case. When you work it out, you should get $B(a,b) \stackrel{a \gg b}{\sim} \Gamma(b)a^{-b}$

When both go to infinity we have to start worrying about size. Are $a$ and $b$ increasing at the same rate? Is one going to infinity at twice the rate of the other? There is no definitive answer unless you put some conditions on growth rate. Nevertheless, using Stirling again we find that
$$B(a,b) \stackrel{a,b\gg 0}{\sim} \sqrt{2 \pi}\frac{a^{a-1/2}b^{b-1/2}}{(a+b)^{a+b-1/2}}$$
These formulas can be found here, on the Wikipedia page for the Beta Function


Find the probability of a 4 turning uppermost at least once in two tosses of a fair die? I am a bit confused about what this questions about.

Is it looking for the possibilities of getting 4 at least once in the toss,
or
The possibilities to get 4 as the highest number of the two toss at least once?

A = {{1,4},{2,4},{3,4},{4,4},{5,4},{6,4},{4,1},{4,2},{4,3},{4,5},{4,6}}

B = {{1,4},{2,4},{3,4},{4,4},{4,1},{4,2},{4,3},}

From the context of "at least once in two tosses", I would interpret "uppermost" to mean "shows on the upper face of".

Your alternative interpretation is not an "at least once" type of event, it is more of a "does or does not" kind of thing.

S = {1,2,3,4,5,6,1,2,3,4,5,6}

n(S) = 12

A is the event of getting 4

A = {4,4}
n(A) = 2

P(A) = n(A)/n(S)
         = 2/12
         = 1/6

Therefore, the probability of getting 4 atleast once in two tosses of a fair die is 1/6


Probability $\mathbb{P}[X_N = 1 | X_0 = 1]$ for a Markov Chain over $\mathbb{X} = \{1,2,3\}$ Exercise :

  Let $\{ X_n \}_{n \in \mathbb{N}_0}$ be a Markov Chain over the space $\mathbb{X} = \{ 1,2,3 \}$ with the following transition matrix :
  $$P = \begin{pmatrix} 0 & 1/2 & 1/2 \\ 1/3 & 1/3 & 1/3 \\ p & 2/3 -p & 1/3 \end{pmatrix}$$
  Calculate the probability $\mathbb{P}[X_n=1|X_0=1]$ for all $n \in \mathbb{N}$ for the cases of a) $p=0$, b) $p =1/6$, c) $p=2/3$. How would you calculate $\lim_{n\to \infty} P^n$ without many operations ?

Attempt :

First of all, I start by finding the eigenvalues and eigenvectors for $P$, which are :

$$\det(P-\lambda I) = 0 \Rightarrow \begin{vmatrix} 0 & 1/2 & 1/2 \\ 1/3 & 1/3 & 1/3 \\ p & 2/3 -p& 1/3 \end{vmatrix} = 0 \Leftrightarrow \dots \Leftrightarrow \lambda = \begin{cases} 1 \\ \frac{1}{6}\big(\pm\sqrt{6p-1}-1\big)\end{cases}$$

The problem I have though, is that for example, the case $p=0$ yields complex eigenvalues, which I don't know how to handle in terms of probabilities to continue on. Also, in the case $p=1/6$ we have a Jordan case, which still leaves us with a hard case for $P^n$. Any tips on how to calculate the given probability ?

I would guess its something like this:

$x_{n}$ = $A^{n}$ * $x_{0}$

$x_{n}$ = ($P^{-1} L^{n} P$) * $x_{0}$

Where P is the matrix made of all the eigenvectors of A, L is the matrix with the corresponding eigenvalues, and $x_{0}$ is the unit vector [1, 0, 0].

Since you only care about the probability Pr($X_{n}$ = 1 | $X_{0}$ = 1), you only care about the element $A^{n}_{0,0}$. Since you have complex eigenvectors, your final result should be real positive number.


Localization at a prime ideal is a reduced ring Here is the question that I came up with, which I am having trouble proving or disproving:

Let $A$ be a ring (commutative). Let $p \in Spec(A)$ such that $A_p$ is reduced. Then there exists an open neighborhood of $U \subset Spec(A)$ containing $p$ such that $\forall q \in U$, $A_q$ is reduced.

Here is some background to my question:

I am basically trying to prove that if the stalks at all closed points of a quasicompact scheme are reduced rings, then the scheme is reduced.

Since the closure of every point of a quasicompact scheme contains a closed point of that scheme, proving the above commutative algebra statement (if it is true) will yield a proof of this statement about reducedness of quasicompact schemes.

If the statement in bold is true, then I guess the neighborhood $Spec(A)-V(A-p)$ should suffice (this is just a guess), but I am running into some problems trying to use this neighborhood to show that the localization at every point of $Spec(A)-V(A-p)$ gives me a reduced ring. So there might be some other neighborhood of $p$ that I am missing, or the statement in bold is not true. Either way, some help would be appreciated (if the statement in bold is true, then I would appreciate hints and not complete answers).

Your boldface statement is false in general. Consider
$$A=F[X_1, \dots, X_n, ...., Y_1, \dots, Y_n, ....]/(X_n^2, X_nY_n)_n$$
over a field $F$. Denote by $x_n, y_n$ the images of $X_n, Y_n$ in $A$. Let $\mathfrak p$ be the ideal generated by the $x_n$'s. As $A/\mathfrak p$ is $F[y_1, \dots, y_n, ....]$ which is integral, $\mathfrak p$ is a prime ideal, generated by nilpotent elements, thus is the minimal prime ideal of $A$. As every $x_ny_n=0$, $\mathfrak pA_{\mathfrak p}=0$, hence $A_{\mathfrak p}$ is reduced.

If there were a reduced open neighborhood $U$ of $\mathfrak p$, then $U$ contains a non-empty reduced principal open subset $D(f)$. So $A_f$ is reduced and $\mathfrak p A_f=0$. Therefore for all $n\ge 1$, there exists $r_n\ge 1$ such that $f^{r_n}x_n=0$. Now it is easy to check that $f\in y_nA+\mathfrak p$ and $\cap_n (y_nA +\mathfrak p)=\mathfrak p$. Thus $f$ is nilpotent and $D(f)=\emptyset$. Contradiction.

If the localizations of $A$ at all of its maximal ideals are reduced, then the annihilator of any nilpotent element of $A$ is not contained in any maximal ideal, and so contains $1$.


Show a matrix is positive definite I was looking at UCLA's Department of Mathematics's Fall 2015 Basic Qualifying Exam, and question #12 is the following.

  Show that the following matrix is positive definite.

  $$M=\begin{pmatrix}
2 & 1 & 1 & \ldots & 1\\
1 & 3 & 1 & \ldots & 1\\
1 & 1 & 4 & \ldots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1& 1& \ldots &n+1
\end{pmatrix}$$

I wasn't super sure how one should proceed; showing $x^TMx>0$ for all $x \neq 0$ seems like a pain, and I was curious if there was a slick way to do this. Other ways would be to write $M = AA^T$ or go through the characteristic polynomial, but I'm not super sure about those either. If anyone can give some hints, that would be great!

Write
$$M=\begin{pmatrix}
1 & 0 & 0 & \ldots & 0\\
0 & 2 & 0 & \ldots & 0\\
0 & 0 & 3 & \ldots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0& 0& \ldots &n
\end{pmatrix}+
\begin{pmatrix}
1 & 1 & 1 & \ldots & 1\\
1 & 1 & 1 & \ldots & 1\\
1 & 1 & 1 & \ldots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1& 1& \ldots &1
\end{pmatrix}.$$

Now it's easy to see that the first of these matrices is positive-definite, while the second is positive-semidefinite (in particular, its eigenvalues are $n$ and $0$).

Maybe you can consult https://en.wikipedia.org/wiki/Sylvester%27s_criterion. Proceed by induction.


Prove that if $a$ is an integer, then $a(a^2+2)$ is divisible by $3$ I've been working on this problem for a while.

I've been trying a contrapositive contradiction proof, as the other examples of direct proofs I've seen aren't very clear.

For contrapositive, I assume $3$ does not divide $a(a^2 + 2)$, and $a$ is an integer. Then $a(a^2 + 2) = 3k + 1$ or $a(a^2 + 2) = 3k + 2.$

I'm thinking through simplifying/rearranging $a(a^2 + 2) = 3k + 1$ into $(a^2 + 2a - 2) / 3  = k$ might be sufficient as a contradiction, because since $a$ is an integer, the top portion is also an integer (since integers are closed under addition/subtraction. So an integer divided by $3$ is sometimes not an integer. But I'm not sure if I'm on the right track or not.

Thank you in advance for any help!

It's pretty easy to prove that if $3$ is not a factor of $a$, it is a factor of $a^2+2$ by noting that $a^2\equiv0,1\bmod3$ (see here)

The rest is simple.


Is $R$ a PID if every submodule of a free $R$-module is free? Let $R$ be a commutative ring. Before I proved that every submodule of a free $R$-module is free over a P.I.D.

  Now I'm trying to prove the reciprocal, if every submodule of a free $R$-module is free, is $R$ a PID?

I couldn't find my second question here, only the first one.

Another question regarding this if it's true is how much can we "stretch" the conditions, this is, could be concluded that $R$ is a PID knowing that some submodules of a free $R$-module verify some condition? "Some condition" may sounds vague, but I couldn't think of any that will do.

Yes, it is.

If $a\in R$, $a\ne 0$, then $aR$ is free (as a submodule of the free $R$-module $R$), so $a$ is a non-zero divisor. In order to prove this let $ax$ be an element of an $R$-basis of $aR$. If $ba=0$ for some $b\in R$ then $b(ax)=0$ and therefore $b=0$. (For a characterization of free ideals see here.) This shows that $R$ is necessarily an integral domain.
Now let $I$ be a non-zero ideal of $R$. Since $I$ is free (as a submodule of the free $R$-module $R$) it must have a basis with a single element (as you already know), so $I$ is principal.

(As you may noticed, I've only used the property for the free $R$-module $R$.)

We know that $R$ is a free $R$-module with basis $X=\{1\}$.
Now let $I$ be an ideal of $R$.(the submodules of a ring are exactly its ideals.)
By hypothesis $I$ has a basis $X$. We claim that $X$ has exactly one element. If $a$ and $b$ are two distinct elements of $X$, then $ab-ba=0$, since $R$ is commutative, a contradiction to the fact that $X$ is linearly independent. Therefore $X$ has only one nonzero element, i.e. $I$ is a principal ideal.


http://en.wikipedia.org/wiki/Continued_fraction#Finite_continued_fractions

The cross product of consecutive convergents in simple continued fractions are either $\pm 1.$ The first in the string is always  $\frac{0}{1}.$ The second in the string is always the nonsensical $\frac{1}{0}.$ The convergents for $\frac{4999}{1109}$ are
$$ \frac{0}{1}, \; \frac{1}{0}, \; \frac{4}{1}, \; \frac{5}{1}, \;  \frac{9}{2}, \;   \frac{293}{65}, \;  \frac{2353}{522}, \; \frac{4999}{1109}.$$

We get little 2 by 2 determinant $$ 2353 \cdot 1109 - 522 \cdot 4999 = -1.   $$

So $$ -2353 \cdot 1109 + 522 \cdot 4999 = 1.   $$

This is not a difficult method to learn and has other uses. Note that the letters $a_i,$ called quotients on wikipedia, are $$ \langle 4; 1,1,32,8,2 \rangle. $$
These are $ \langle a_0; a_1,a_2,a_3,a_4,a_5 \rangle. $ I did it with a calculator. Beginning with the number $x_0 = \frac{4999}{1109} \approx 4.5077664563,$ we begin with $$ a_0 = \lfloor x_0 \rfloor = 4.$$ The fractional part is $x_0 - \lfloor x_0 \rfloor \approx 0.5077664563,$ and we take the reciprocal to get
$$ x_1 = \frac{1}{x_0 - \lfloor x_0 \rfloor} \approx  \frac{1}{0.5077664563} \approx 1.969804617  $$

Then $$ a_1 = \lfloor x_1 \rfloor = 1.$$ The fractional part is $x_1 - \lfloor x_1 \rfloor \approx 0.969804617,$ and we take the reciprocal to get
$$ x_2 = \frac{1}{x_1 - \lfloor x_1 \rfloor} \approx  \frac{1}{0.969804617} \approx 1.031135532.  $$

Then $$ a_2 = \lfloor x_2 \rfloor = 1.$$ The fractional part is $x_2 - \lfloor x_2 \rfloor \approx 0.031135532,$ and we take the reciprocal to get
$$ x_3 = \frac{1}{x_2 - \lfloor x_2 \rfloor} \approx  \frac{1}{0.031135532} \approx 32.11764617.  $$

Then $$ a_3 = \lfloor x_3 \rfloor = 32.$$ The fractional part is $x_3 - \lfloor x_3 \rfloor \approx 0.11764617,$ and we take the reciprocal to get
$$ x_4 = \frac{1}{x_3 - \lfloor x_3 \rfloor} \approx  \frac{1}{0.11764617} \approx 8.500064218.  $$

Then $$ a_4 = \lfloor x_4 \rfloor = 8.$$ The fractional part is $x_4 - \lfloor x_4 \rfloor \approx 0.500064218,$ and we take the reciprocal to get
$$ x_5 = \frac{1}{x_4 - \lfloor x_4 \rfloor} \approx  \frac{1}{0.500064218} \approx 1.999743161.  $$ it is here that experience becomes critically important. We are seeing the effect of tiny calculator roundoffs. The actual value is $x_5 = 2,$ from which we get $a_5 = 2,$ and that concludes the s.c.f.

Let us put in the "quotients" where they belong in the calculation:
$$ \frac{0}{1}, \; \frac{1}{0}, \; \color{magenta}{\frac{4}{}}, \; \frac{4}{1}, \; \color{magenta}{\frac{1}{}}, \; \frac{5}{1}, \; \color{magenta}{\frac{1}{}}, \; \frac{9}{2}, \; \color{magenta}{\frac{32}{}}, \;  \frac{293}{65}, \; \color{magenta}{\frac{8}{}}, \; \frac{2353}{522}, \; \color{magenta}{\frac{2}{}}, \;\frac{4999}{1109}.$$

In every case, given two consecutive numerators (in black) followed by a quotient (in magenta), the first numerator plus the second numerator times the quotient is the following numerator. So, $$ 0 + 1 \cdot 4 = 4, \;  1 + 4 \cdot 1 = 5, \;  4 + 5 \cdot 1 = 9, \;  5 + 9 \cdot 32 = 293, \;  9 + 293 \cdot 8 = 2353, \;  293 + 2353 \cdot 2 = 4999.  $$

Same for denominators, $$ 1 + 0 \cdot 4 = 1, \;  0 + 1 \cdot 1 = 1, \;  1 + 1 \cdot 1 = 2, \;  1 + 2 \cdot 32 = 65, \;  2 + 65 \cdot 8 = 522, \;  65 + 522 \cdot 2 = 1109.  $$

In case you are worried by this: the exact version, with fractions instead of calculator decimal approximations, is exactly what you have been doing:
$$  x_0 = \frac{4999}{1109} = 4 + \frac{563}{1109},  $$
$$  x_1 = \frac{1109}{563} = 1 + \frac{546}{563},  $$
$$  x_2 = \frac{563}{546} = 1 + \frac{17}{546},  $$
$$  x_3 = \frac{546}{17} = 32 + \frac{2}{17},  $$
$$  x_4 = \frac{17}{2} = 8 + \frac{1}{2},  $$
$$  x_5 = \frac{2}{1} = 2.  $$

Write down the quotients backwards
8  32  1  1  4, leaving out the last quotient 8.
Then write
   8   32   1   1   4
1  8
Then multiply the first 32 in the top row by the 8 in the bottom row and add the 1 in the second row, getting 257, and write
   8   32    1   1   4
1  8   257
Then multiply the first 1 in the first row by the 257 in the second row and add the 8 in the second row, getting 265, and write
   8   32    1    1    4
1  8  257  265
Multiply the second 1 in the first row by the 265 and add 257, getting 522, and write
   8    32    1     1    4
1  8   257  265   522
Finally, multiply the 4 in the first row by 522 and add 265, getting 2353, and write
   8    32    1     1      4
1  8   257  265   522   2353
Now write a - sign under the 8 in the second row, a + under the 257, etc, getting
   8    32    1     1     4
1  8   257  265   522  2353
   -    +    -     +     -
Finally, write x under the 2353 and y under the 522, the result of which is
x = -2253, y - 522, and we have solved 1109x + 4999y = 1.

Art DuPre'


Find Area of Area Not Filled In By Circle If I have a square with four quarters of a circle, how can I find the area not filled in by the circles? What I'm asking is here:

My drawing is har from perfect, but I think you get what I mean. I'm trying to find the area of the space where the arrow is pointing. Here are my thoughts on the problem:
The square is 24 inches by 24 inches, meaning its area is 576. Four quarters of a circle make a circle, and the radius of the circle is 12. To find the area of the circle, I do pi*12^2, which is 144pi. So does that mean the area of the space in between the semicircles is 576 - 144pi, or can the answer be further simplified without doing 144 * pi?

Here u go,

This is made up of four 90 degrees sectors and a small part that u want to find the area.
So what u must do is...

||  AREA OF SQUARE - 4(AREA OF 90 DEGREES            SECTOR ||

So now lets find the value,

1.Finding the area of square :-

    24x24 = 576

2.Finding the area of 90 degrees sector :-

So as u mentioned the radius is 12 therefore,

\frac{90}{360}x\pixr^2
   = 113.0973355

3.Substitution

AREA OF SQUARE - 4(AREA OF 90 DEGREES                       SECTOR

576 - 4(113.0973355)

576 - 452.3893421= 123.6106579

Finally your answer is..
   123.6106579


What shape is a Calippo? The Calippo™ popsicle has a specific shape, that I would describe as a circle of radius $r$ and a line segment $l$, typically of length $2r$, that's at a distance $h$ from the circle, parallel to the plane the circle is on, with its midpoint on a perpendicular line that goes through the centre of the circle.
The shape itself consists of lines connecting the circle to line segment.

Do I need to specify how the points on the circle map to a point on the line segment?
Obviously, the points on the circle directly under the end points of the line segment should map "straight up" to those end points. The points on the circle half way between those, at $\tfrac{\pi}{2}$, should map to the mid point of the line segment.
But intuitively there should be a mapping that gives the "most outer" shape such that even if every point on the circle is connected to every point on the line segment, those lines never leave "the popsicle".
Thanks to a comment by Mark S. I now know this is called the convex hull of the circle and the line segment.
Given this description, how do I calculate the surface and the volume of this shape?
Does this shape have an official name?
It's not the round chisel as shown in this answer, since it lacks the edge in the shape of half an ellipse.

It's hard to tell from the picture, but from the description "The shape itself consists of lines connecting the circle to line segment", I might represent the shape thusly:

Suppose that the circle has radius $r$, and that the shape has height $h$. If the surface lines joining the segment to the circle lie in a plane perpendicular to that segment, then the figure looks like this:

Then, we can parameterize $P_\theta$ on the circle, and companion point $Q_\theta$ on the segment ...

$$P_\theta = (r \cos\theta, r \sin\theta, h) \qquad Q_\theta = ( r\cos\theta, 0, 0)$$

... so that the line between them has equation

$$P_\theta + t \; (Q_\theta - P_\theta) : \begin{cases} x = r \cos\theta \\ y = r (1-t) \sin\theta \\ z = h (1- t)\end{cases}$$

Eliminating the parameters $\theta$ and $t$ yields this formula for the surface:

  $$x^2 z^2 + y^2 h^2 = r^2 z^2 \tag{1}$$

Note that the equation (as well as the figure) indicates that the level curves of the surface are ellipses, with constant major radius ($r$) and linearly-varying minor radius ($z r/h$).

Now, the volume and surface area of the shape can be determined from this function:

  $$y = f(x,z) = \frac{z}{h} \sqrt{r^2-x^2}  \quad \tag{2}$$
  over the rectangular region determined by $-r \leq x \leq r$ and $0\leq z \leq h$.

Edit. OP has indicated, in comments with @YvesDaoust, that the desired shape is the conoid, which indeed fits the surface described by $(1)$. (The parameterization given in the Wikipedia article matches $(1)$, for $r = 1$ and $h = z_0$, and under the coordinate transformation $z \to z_0-z$.)

The Wikipedia entry states that the volume of the conoid is $\frac{\pi}{2}r^2 h$. This is easily confirmed from $(2)$ ...

  $$V = \frac{2}{h} \int_{0}^{h} z \int_{-r}^{r} \sqrt{r^2-x^2}\,dx dz = \frac{2}{h} \int_{0}^{h} \frac{\pi}{2} r^2 z dz = \frac{2}{h}\cdot \frac{\pi}{4} r^2 h^2 = \frac{\pi}{2}r^2 h \tag{3}$$

(where we have recognized the $x$ integral as giving the area of the half-circle of radius $r$).

For surface area, we note that
$$f_x = \frac{-xz}{h \sqrt{r^2-x^2}} \qquad\qquad f_z = \frac{1}{h}\sqrt{r^2-x^2}$$
so that
$$\begin{align}
S &= 2\;\int_{0}^{h} \int_{-r}^{r} \sqrt{(f_x)^2 + (f_z)^2 + 1\;}\; dx dz \\
&= \frac{2}{h}\;\int_{0}^{h} \int_{-r}^{r} \;\sqrt{\frac{
h^2 (r^2-x^2) + (r^2-x^2)^2 + x^2 z^2}{r^2 - x^2}\;} \; dx dz
\end{align}$$

This is a bit trickier to evaluate symbolically. I'll have to return to it.

Two cutting planes shown symmetric to cone axis meeting below its vertex before removing Region above the planes. What remains is like the Calippo shape of different geometric proportion.

EDIT1:

If $ 2r,2w, h $ ,concurrency of straight cone generators are indicated as being  given, then the truncated shape is uniquely given as part of cone of semi vertical angle= $ \tan^{-1}\dfrac{r-w}{h}.$

EDIT 2:

Oh, may be you are referring to a shape like a long toothpaste tube  right or left conoid? page 61, DJ Struik, ruled surface of parametrization

$$ (x,y,z)= ( r \cos\theta, r \sin \theta, f(\theta) \,) $$

$$ f(\theta) = a \theta + b $$

with metric

$$ ds^2 = dr^2 + (r^2+f^{\prime2})\, d\theta^2 $$

The special case $ \pi r = 2 w $ is easy, obtained by  isometrically  flattening a circle to a diameter during  manufacture.


Wrong Definition of a Limit What would this mean:

  $\exists \delta >0$ such that $\forall \epsilon > 0$ and $\forall x$ satisfying $0 < |x-a| < \delta$, then $|f(x)-L| < \epsilon$

I am pretty confused by the symbols too...

Here's I read it:

There exists a delta larger than zero such that for any epsilon larger than zero and for any $x$ satisfying $0 < |x − a| < \delta$, we will have $|f(x) − L| < \epsilon$.

Does this show that there simply exists an interval where $f(x)$ is a constant function?

Definitely it would mean $f$ is constant, and equal to $L$, on a neighborhood of radius $\delta$ about $a$.  If you write $\forall\varepsilon>0\ \exists\delta>0$ rather than the other way around, then it is a weaker assertion: that $\lim\limits_{x\to a}f(x)=L$.

Not really. f(x) is still the function itself - it does not become a constant function.

What it says is that if you select an appropriate δ (often small enough), you can get the distance to f(x) smaller than any ε. Very often δ is built on ε, been restricted to the f:=x relationship, the difference between f(x) and L is always smaller than ε. Now L is the limit, meaning you are close enough to f(x), but you are not at the point of (x, f(x)).


If a symmetric matrix $A$ has $m$ identical rows show that $0$ is an eigen value of $A$ whose geometric multiplicity is atleast $m-1$. If a symmetric matrix $A$ has $m$ identical rows show that $0$ is an eigen value of $A$ whose geometric multiplicity is atleast $m-1$.

If $A$ has $m$ identical rows then then by elementary row operation $A$ has $m-1$ zero rows.

But how to show that geometric multiplicity of $A$ is atleast $m-1$ from above?

I know that geometric multiplicity of an eigen value is ($\dim(ker(A-\lambda I))$ but how to compute it from above?

Please help.

Direct method.

By definition, the geo-multiplicity of $0$ is just the dimension of the space of solution for $\boldsymbol {Ax}= \boldsymbol 0$. By symmetry, $\boldsymbol A$ has $m$ identical columns. Suppose these columns are the $(k_j)_{j =1}^m$-th columns of $\boldsymbol A$. Let $\boldsymbol e_j$ be the $j$-th standard basis vector, then clearly at least
$$
\boldsymbol e_{k_1} - \boldsymbol e_{k_j} \quad [j =2, \ldots, m]
$$
are solutions of $\boldsymbol {Ax=0}$, thus $\dim(\mathrm{Ker} \boldsymbol A) \geqslant m-1$ as we desire.

It should be clear that if at least one entry of $A$ is non zero, then its rank is at greater or equal to $1$.

suppose that the rank of $A$ is strictly larger than $1$, then you can find $\pmb{x}$ and $\pmb{y}$ such that $A \pmb{x}$ and $A \pmb{y}$ are not collinear. This is impossible since all the rows of $A$ are the same (let say $\pmb{a}^T$), then $A \pmb{x} = \left( \sum x_i a_i \right) \pmb{1}$ and $A \pmb{x} = \left( \sum y_i a_i \right) \pmb{1}$ and so they are collinear.

Now we know that the $A$ have rank $1$. By your formula, the multiplicity is $\dim( \ker(A-0 I)) = \dim( \ker(A)) = m - rank(A) = m-1$.

EDIT : this is only true if $A$ is a $m\times n$ matrix for some $n$


Convolution Integral problem I'm having a hard time to do this exercise. I'm using the definition for convolution but I'm stuck at the integral. Thanks in advance.

For $t>0$ consider $f_a(x)=\frac{1}{\sqrt{4 \pi a}}e^{-\frac{x^2}{4a}}$.
Show that $f_a*f_b=f_{a+b}$

By definition: $(f*g)(x)$= $\int_{-\infty}^{+\infty} f(u)g(x-u) du$
One computes,
$(f_a*f_b)(x)=$ $\int_{-\infty}^{+\infty} \frac{1}{\sqrt{4 \pi a}}e^{-\frac{u^2}{4a}} \frac{1}{\sqrt{4 \pi b}}e^{-\frac{(x-u)^2}{4b}} du$
$=\frac{1}{\sqrt{4 \pi a}}\frac{1}{\sqrt{4 \pi b}} \int_{-\infty}^{+\infty} e^{-\frac{u^2}{4a}}e^{-\frac{(x-u)^2}{4b}}$
I can't solve this integral. I tried using Gaussian Integral but still cant solve it.

One of the most useful elementary tricks to know when dealing with Gaussian-type integrals is completing the square.

On that note, we calculate
\begin{align}
f_a*f_b(x)
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{y^2}{4a}-\frac{(x-y)^2}{4b}\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{1}{4a}\left(y^2+\frac{a}{b}(x-y)^2\right)\right)\,\mathrm{d}y\\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{1}{4a}\left(\left(1+\frac{a}{b}\right)y^2-\frac{2a}{b}xy+\frac{a}{b}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}\left(y^2-\frac{2a}{a+b}xy+\frac{a}{a+b}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}\left(\left(y-\frac{a}{a+b}x\right)^2+\frac{ab}{(a+b)^2}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}\left(y^2+\frac{ab}{(a+b)^2}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\exp\left(-\frac{x^2}{4(a+b)}\right)
\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}y^2\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\sqrt{\frac{4ab}{a+b}}\exp\left(-\frac{x^2}{4(a+b)}\right)
\int_{-\infty}^\infty e^{-z^2}\,\mathrm{d}z \\
&= \frac{1}{\sqrt{4\pi(a+b)}}\exp\left(-\frac{x^2}{4(a+b)}\right) \\
&=f_{a+b}(x).
\end{align}

Of course, using the Fourier transform, we can simplify this calculation somewhat – if you haven't read about the Fourier transform (and its action on Gaussians), I highly recommend you take a look. The basic techniques used in calculating the Fourier transform of a Gaussian are in fact very closely analogous to the calculations above.

https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables#Proof_using_convolutions
You can find the proof here using convolutions - just use $\sigma^2_X = 2a$ and $\sigma^2_Y = 2b$, $\mu_X = 0$ and $\mu_Y = 0$. It involves making a substitution involving $a$ and $b$, then completing the square.


Representation theory and neutral elements Let (G1,∗) and (G2,◦) be groups. A mapping f : G1 → G2 is said to be a group homomorphism if for all g,h∈G1  f(g∗h)=f(g)◦f(h)

It's true that a function between groups preserving the group operation automatically preserves neutral elements. That's the reason why the condition is sometimes omitted. In this case, I want a group homomorphism G → GL(hom(V, W)). And if I already knew that the action of g∈G on hom(V, W) is invertible (so that it is an element of GL(hom(V, W))), then I would fine be. But how do you show that?

In image is my answer to original question, which I did, but I don't have the answer how I show that action is invertible.

Original question

When you say that you have a map from $G$ into $GL(\hom(V,W))$, you are already claiming that the image of each $g\in G$ is invertible.


How to know if the solution for this angle ϕ is in units of radians or degrees? I am studying out of interest a modelling approximation for a plectrum plucking a string (eg. guitar, piano, harpsichord) as given by these articles:

http://recherche.ircam.fr/pub/dafx11/Papers/24_e.pdf (relevant part on page 3 of pdf)
https://stacks.stanford.edu/file/druid:wp454hs7976/Jack%20Perng%20Thesis-augmented.pdf (relevant part on page 75 of pdf)

The model works roughly like this:

The string is represented here as the yellow circle in cross-section. The red part on the left is the component the plectrum is attached to that moves (moving upwards along the y-axis here). The black bendy part is obviously the plectrum.
$F_p$ (the force from the plectrum) can be broken down in terms of $x$ and $y$ vector components as per:

Where E is the Young's Modulus, I is moment of inertia, and L' is how far along the length of the plectrum the string is contacting.
The only trouble I'm having in understanding this model is concerning the units of $\phi$. Is this angle solution going to be in radians or degrees? In C++, sin and cos take radians by default and you have to convert degrees to radians first if working from degrees. So I need to know which one I'm getting from this equation for $\phi$.
Is $\phi$ here in radians or degrees?
Thanks for any guidance.

Oddly enough, the angle should usually be given in radians. However, looking at the first referenced paper, I am 99% sure that they are using angles in degrees. The authors speak of ( after (6) )

For small-angle approximations ($\phi<<1$), this reduces [...].

If this would be in radians, this would mean roughly 57° which is not really a "small-angle approximation".
Also, they say that

when $\theta=90°$, there is no bending [...].

Which also indicates the use of angles in degrees. I would suggest contacting the authors. Usually, authors are very happy if someone is interested in their work and are willing to help. You could also try to replicate the given results since the authors seem to have provided all necessary data.


Solving $\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$ I need some tips with solving this algebraic equation
$$\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$$

I've tried subtracting: $$\frac{8x}{x+2}$$
and also setting $$\sqrt x = t$$
Where "t" is a substitution to make things simpler.
This is how it ends up:
$$x + 2 = 6\sqrt x - \frac{8x}{x+2}$$
or
$$t^2 + 2 = 6t - \frac{8t^2}{t^2+2}$$
Unfortunately, I couldn't find a way from here without getting polynomials of the fourth degree or equations with $$x\sqrt x$$

I'd just like to clarify that I'm not looking for the solution here. I'd just like it if I could have some pointers or tips on where to go from here, or even if I did something wrong.

Thanks in advance!!!

Middle term splitting will always remain my favorite method for factorising polynomials!
\begin{align*}
&\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x\\
\iff &\dfrac{(x+2)^2+8x}{x+2}=6\sqrt x\\
\iff &(x+2)^2-6\sqrt x(x+2)+8x=0\\
\iff &(x+2)^2-4\sqrt x(x+2)-2\sqrt x(x+2)+8x=0\\
\iff &(x+2-4\sqrt x)(x+2-2\sqrt x)=0
\end{align*}
Now apply the quadratic formula.

Start with
$$\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$$
Then,
$$x^2+12x+4=6\sqrt{x}(x+2)$$
Square both sides,
$$ x^4 + 24 x^3 + 152 x^2 + 96 x + 16 = 36x^3+144x^2+144x$$
Therefore we want to find the roots of the equation
$$x^4-12x^3+8x^2 - 48x +16 =0$$
Although you could use the quartic formula it's a real pain. Wolfram Alpha finds
$$x=6 \pm 4\sqrt{2} \ ; x= \pm 2i.$$


Calculating "pooled and unpooled" standard error Data given:

Treatment          Y^bar         n                 s

1                  18.1          9                 3.851

2                  20.9          23                9.854

a. Calculate the unpooled standard error for the difference of means.

b. Calculate pooled standard error for the difference of means.

how to do that? what do they mean by "pooled and unpooled?

Appreciate your help!

Do you know the difference between pooled and unpooled?

Hint: pooled tests assumes that two populations have equal variances. Use this to calculate standard error.


Assuming $AB=I$ prove $BA=I$ Possible Duplicate:
  If $AB = I$ then $BA = I$

Most introductory linear algebra texts define the inverse of a square matrix $A$ as such:

Inverse of $A$, if it exists, is a matrix $B$ such that $AB=BA=I$.

That definition, in my opinion, is problematic. A few books (in my sample less than 20%) give a different definition:

Inverse of $A$, if it exists, is a matrix $B$ such that $AB=I$. Then they go and prove that $BA=I$.

Do you know of a proof other than defining inverse through determinants or through using rref?

Is there a general setting in algebra under which $ab=e$  leads to $ba=e$ where $e$ is the identity?

Multiply both sides of $AB-I=0$ on the left by $B$ to get
$$
(BA-I)B=0\tag{1}
$$
Let $\{e_j\}$ be the standard basis for $\mathbb{R}^n$. Note that $\{Be_j\}$ are linearly independent: suppose that
$$
\sum_{j=1}^n a_jBe_j=0\tag{2}
$$
then, multiplying $(2)$ on the left by $A$ gives
$$
\sum_{j=1}^n a_je_j=0\tag{3}
$$
which implies that $a_j=0$ since $\{e_j\}$ is a basis. Thus, $\{Be_j\}$ is also a basis for $\mathbb{R}^n$.

Multiplying $(1)$ on the right by $e_j$ yields
$$
(BA-I)Be_j=0\tag{4}
$$
for each basis vector $Be_j$. Therefore, $BA=I$.

Failure in an Infinite Dimension

Let $A$ and $B$ be operators on infinite sequences. $B$ shifts the sequence right by one, filling in the first element with $0$. $A$ shifts the sequence left, dropping the first element.

$AB=I$, but $BA$ sets the first element to $0$.

Arguments that assume $A^{-1}$ or $B^{-1}$ exist and make no reference to the finite dimensionality of the vector space, usually fail to this counterexample.

For $A$ and $B$ square, $AB=I$ implies $(AB)^{-1} = B^{-1}A^{-1}=I$. Then multiply $B$ from the left side, $A$ from the right side. $BB^{-1}A^{-1}A=BIA$ implies $BA=I$.


Max of Brownian motion with drift is finite almost surely For $B_t$ Brownian Motion with drift $\mu<0$, I need to prove that the max value, $X = \max_{0<t<\infty}B_t$ is finite almost surely, ie $P(X<\infty)=1$.

Now, I know that because the mean is negative, it will go more and more negative, and it is also a supermartingale. But I don't know how to prove almost surely...

Appreciate any hints.

Hint: Try the strong law of large numbers.  What does it say about $\lim_{t \to \infty} B_t/t$?  What does this say about the sign of $B_t$ for large $t$?

Actually that is wrong in the sense that it is not finite in the limit.  Brownian Motion with a negative drift will wander off to -∞ almost surely. It will tend to go down in a linear fashion with the slope equal to the drift parameter.


How to calculate the mod value of a rational/irrational value? We have a course in network security this semester and we are being taught RSA algorithm.

I came across a typical math problem that I was unable to solve here.

$$D*E \equiv 1 \mod{\phi(n)}$$

This became $$D \equiv E^{-1}  \mod{\phi(n)}$$ How do you solve this ???

My lect gave this specific example

$$D \equiv 7^{-1} \mod{160}$$ and the solution for $D$ was $23$! How did she arrive at this.

Hint

$7D = 1 \mod 160$

Sol'n

  $7D = 1 \mod 160$ and $D = 23 \Longrightarrow 23*7 \mod 160$ satisfies this.

https://www.youtube.com/watch?v=mgvA3z-vOzc
Check this out. The problem can be solved using extended Euclidean Algorithm to be precise. For smaller numbers, the answers already mentioned are sufficient.


What is the symbol $\triangleq$? I came across this new symbol while reading a document about writing proofs, and I have never seen it before.

It’s is defined to be equal to; it’s my preferred symbol, but the most common one is $:=$, and I’ve also seen $\overset{\text{def}}=$.

  For each $x\in X$ there is an open nbhd $U_x$ of $x$ such that ... . Then $\mathscr{U}\triangleq\{U_x:x\in X\}$ is ...

The $\triangleq$ indicates that $\mathscr{U}$ is being defined to be $\{U_x:x\in X\}$: we are not saying that some previously defined $\mathscr{U}$ is equal to the collection of these sets $U_x$.

It actually is defined as "estimates", and also means that the content on the two sides correlates or conforms to each other.

For further reference, see here and here, the symbol is quite often used in (German?) academic context.


Prove $x_n$ converges IFF $x_n$ is bounded and has at most one limit point I'm not entirely sure how to go about proving this so hopefully someone can point me in the right direction. The definition I have for a limit point is "$a$ will be a limit point if for a sequence $x_n$ there exists a subsequence $(x_{n_k})$ such that $\lim_{k\to\infty} (x_{n_k})=a$".

In the forward direction if $x_n$ converges to a value $a$ then every convergent subsequence of $x_n$ must also converge to $a$. I'm not entirely sure where else to go from here. Intuitively I know that what I'm trying to prove must be true since if $x_n$ converges to $x$ then there exists an $N\in\mathbb{N}$ such that $\forall n\geq N$ we have infinitely many terms satisfying $\forall \varepsilon >0\Rightarrow x-\varepsilon<x_n$ which implies that $x$ is a limit point. I'm not entirely sure how to rule out a second limit point. Would it be sufficient to show that if $a$ and $b$ are limits of a convergent sequence $x_n$ then $a$ must equal $b$?

In the backwards direction if $x_n$ is bounded and has only one limit point then we know that there exists only one point $x$ such that we can find infinitely many terms satisfying $\forall \varepsilon >0$ $\left | x_n-x\right |<\varepsilon$ which implies that $x_n$ converges to $x$.

I'm quite concerned I am proving this incorrectly since I don't think I am using any facts about subsequences although they are mentioned directly in the definition for a limit point.

The arrow from right to left can be proven as follows. Suppose there is subsequence which does not converge to that unique limit point. There is an area around the unique limit point such that our subsequence has infinitely many points outside this area. This constitutes another subsequence which must have a limit point other than the unique one. Contradiction.

I'm assuming you're working in $\mathbb R$ here. Since the sequence is bounded it has a convergent subsequence and hence at least one limit point. Therefore by the condition it has exactly one limit point.

I claim that the whole sequence converges to this limit point. This is the same as saying every subsequence converges to this limit point. If this were not the case, meaning there were a subsequence that did not converge to this limit point, then this hypothetical subsequence itself has a limit point since it is bounded. This limit point is different from the unique limit point of the larger sequence by the choice of subsequence. However, being a limit point of the subsequence, this limit point is a limit point of the larger sequence, contradicting uniqueness.


Extending basis of a subspace (From Gilbert Strang's Linear Algebra(4e). Section 3.5 . Problem 22)

Question: Suppose S is a 5-dimensional subspace of $R^6$. True or false.

a) Every basis for S can be extended to a basis for $R^6$ by adding one more vector

b) Every basis for $R^6$ can be reduced to a basis for S by removing one vector

For part a), the basis for S contains 5-dimensional vectors. How can adding another 5-dimensional vector help create a 6-dimensional basis for $R^6$ ? Isn't there a dimension mismatch ? Similarly for part b) ?

You seem to think that if $v$ is a vector in a $5$ dimensional vector space, then $v$ has $5$ components, i.e., that it looks something like $v=(x_1,\cdots ,x_5)$. This is not the case. In fact, vectors in general need not look anything like a list of numbers. For instance, the space of all continuous functions $f:\mathbb R \to \mathbb R$ is a linear space. The dimension of a vector space is the number of vectors in a basis (after proving all bases of a vector space have the same size).

Further hints:

a) Any linearly independent set of vectors in a vector space can be extended to a basis.

b) think of $\mathbb R^5$ sitting inside $\mathbb R^6$. Can you find a basis of $\mathbb R^6$ such that none of the basis vectors is in $\mathbb R^5$?

Personally, I think you need a strict definition of that vector space with some conditions, e.g. relate to orthogonality to other vector subspaces. For example, if that vector space is linear function of other vector spaces, your questions would be false.

Please leave your comments.


Prove that if a|b and b|c then a|c using a column proof that has steps in the first column and the reason for the step in the second column. Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then
$$
$$
(i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$)
$$
$$
(ii) if $a$ | $b$ then $a$|$bc$ for all integers $c$;
$$
$$
(iii) if $a$ |$b$ and $b$|$c$, then $a$|$c$.

Prove that if $a$|$b$ and $b$|$c$ then $a$|$c$ using a column proof that has steps in the first column
and the reason for the step in the second column.

My book is really vague. I'm not really sure what to do..

First I was thinking something like this: $a\mid b\Rightarrow b=as$ and  $b\mid c\Rightarrow c=bt$ and $a\mid c\Rightarrow c = au$
$$
$$
$b + c + c = b + 2c$
$$
$$
$=as + 2(bt+au)$

Yea then I just got lost.

If you are allowed to use fractions then all inferences are immediate consequences of the fact that $\,\Bbb Z\,$ is closed under addition and multiplication - see below. If you are not allowed to use fractions then you can translate the below proofs to integers by clearing denominators, e.g. for $(i)$ we get $\ b=aj,\, c = ak\,\Rightarrow\, a(j + k) = b + c$

$( i)\quad \dfrac{b}a,\,\dfrac{c}a\in\Bbb Z\ \Rightarrow\ \dfrac{b}{a}+\dfrac{c}a = \dfrac{b+c}{a}\in\Bbb Z$

$(ii)\quad \dfrac{b}a,\ c\in\Bbb Z\ \Rightarrow\ c\, \dfrac{b}a = \dfrac{cb}a\in\Bbb Z$

$(iii)\ \ \ \dfrac{b}a,\,\dfrac{c}b\in\Bbb Z\ \Rightarrow\ \dfrac{b}a\dfrac{c}b = \dfrac{c}a\in\Bbb Z$

Note: $ $ if $\,b=0\,$ is allowed in $\,(iii)\,$ then you need to consider that case too.

According to the definition of division,
$b=as$  and $c=bt$

$$bc=asbt$$

$$c=\frac{asbt}b$$

$c=a(st) \implies s$ and $t$ are integers which makes $st$ also an integer
this means $a|c$


How many subgroups or order 8 an abelian Group of order 72 can have Let $G$ be an abelian group of order 72. How many subgroups of order 8 and 4 can it have?

I have listed all possible abelian groups there are 6. Then i said that if I'm looking for an abelian group of order 8 I have 3 possible abelian groups and for each 1 of the 6 first abelian groups must check if 1 of the 3 of the later ones are subgroups. But that is taking forever and I don't have any easy way to check if they are subgroups. I use "If A is a subgroup of G and B is a subgroup of H, then the direct product A × B is a subgroup of G × H". So I found that for every possible G there is 1 subgroup of order 8. But is that correct? If I have a subgroup of a direct product say A × B of a G SxS1xS2x...xSn then A must be subgroup of S and B of S1?

Hint: If $|G|=72$ and the subgroup $|H|=8$, then $[G:H]$ is an odd number. Thus $H$ is a Sylow $2$-subgroup of $G$. Those are known to all be conjugates of each other.

Of course, if you have not seen Sylow theory yet (a live possibility), then this was not very helpful to you. In that case, try and prove that
$$H=\{x\in G\mid 8x=0\}$$
is a subgroup. Because it contains all the elements of $G$ with order that is a power of two, it is unique. It is a bit harder to show that $H$ has eight elements, but that does follows from the structure theory of f.g. abelian groups.

If you have not seen the structure theorem of f.g. abelian groups (another live possibility), then do the following. Prove that $G=H\oplus K$, where $K$ consists of elements $y\in G$ such that $9y=0$. A hint for that: for all $x\in G$ we have $x=9x-8x$. Show that $9x\in H$, $8x\in K$.

By Sylow's first theorem :
      o(G)=72=2^3×3^2
Now take p=2,a prime  and n=3 then clearly
p^n divides o(G) but p^n+1 does not divide o(G)
Hence there exist subgroups of order p^n
I.e,of order 8 but in this case subgroup of order 8 are called Sylow 2-subgroup . That is all the subgroupsubgroups of order 8 in the group G of order 72 are must be Sylow 2-subgroups .
But by Sylow third theorem ,we know that the no of Sylow p- subgroups in G is of the form 1+kp ,where 1+kp divides o(G), k being a +ve integer .
Now here p= 2 . What we have to find is k .
Clearly k= 1( one can verify it ) and hence the no of subgroups of order 8 in the given group of order 72 is 1+1×2=3 .


Limit theorems, prove function has a limit at every point Suppose that $f:R\to R$ is a function such that $f(x+y)=f(x)+f(y)$ for all $x,y∈R$.

Assume that $f$ has a limit at $0$, $f(1)=1$.
Prove that $f(x)=x$ for all $x \in R$

Hint: Show first that $f$ is continuous at any point $c ∈ R$. Prove then that for any rational number $r$, $f(r) = r$, and deduce the statement using the continuity of $f$.

My attempt:

To show continuity at any point $c\in R$

$$\lim_{x \to c}f(x)=\lim_{ h\to0}f(c+h)=\lim_{h\to0}f(h)+f(c)=f(c)$$

Suppose $\lim_{x\to0}f(x)$ exists (as given). We need to show that for $c∈R$ arbitrary, also $\lim_{x\to c}f(x)$ exists.

We have
$$f(x)=f(x−c+c)=f(x−c)+f(c)$$

Letting $x→c$ and setting $y=x−c$ we see that $y→0$, so the limit in question exists by assumption and equals

$$\lim_{x→c}f(x)=f(c)+\lim_{y→0}f(y)$$

which implies $f(x)=x$ QED

My question is how to use $f(1)=1$

Then how to prove continuity for any rational number?

The posted question assumes $\lim_{x\to0}f(x)=0$, but that was not given.  What was given is that $\lim_{x\to0}f(x)$ exists.  That the limit is $0$ must be proved.

The functional equation $f(x+y)=f(x)+f(y)$ entails that $f(0)$ $=f(x+(-x))$ $=f(x)+f(-x)$; hence we must have $f(-x)=-f(x)$, i.e. this is an odd function.  If an odd function $f$ has a limit $L$ at $0$, then
$$
\lim_{x\downarrow0}f(x) =\lim_{x\to0} f(x) = \lim_{x\uparrow0} f(x),
$$
but the last one is
$$
\lim_{x\uparrow0} f(x) = \lim_{x\downarrow0} f(-x) = -\lim_{x\downarrow0} f(x) = -L.
$$
So $L=-L$, and therefore $L=0$.

As to using $f(1)=1$, notice that the function $g(x)=3x$ also satisfies the equation $g(x+y)=g(x)+g(y)$ (and similarly if we had used any other number besides $3$), so certainly that condition must be used.

What the two equalities $f(x+y)=f(x)+f(y)$ and $f(1)=1$ imply is that $f(x)=x$ whenever $x$ is rational.  Let's see that first for positive rational numbers.

\begin{align}
1 =f(1) & = f\left(\frac15+\frac15+\frac15+\frac15+\frac15\right) \\[10pt]
& = f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) = 5f\left(\frac 15 \right)
\end{align}
so $f\left(\dfrac15\right)= \dfrac15$.  And
$$
f\left(\frac35\right)=f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) = \frac15+\frac15+\frac15 = \frac 3 5
$$
and similarly for all other positive rational numbers.

If $x$ is irrational, look at a sequence of rational numbers approaching $x$ and then use continuity.

You only proved that $f$ is continuous at every point yet, that is
$$\lim_{x\to c}f(x)=f(c)$$
at every $c\in\Bbb R$. (So, your first line was enough here.)
First prove that $f(1/n)=1/n$ for all $n\in\Bbb N$. Here you will have to use $f(1)=1$.
Then it follows that $f(r)=r$ for all $r\in\Bbb Q$.
Finally use continuity to prove $f(x)=x$ for all $x\in\Bbb R$.


the mapping class group of the disk is trivial proof Proof :

Identify $D^2$ with the closed unit disk in $\mathbb{R}^2$. Let $\phi : D^2 \rightarrow D^2$ be a homeomorphism with $\phi_{\partial D^2}$ equal to the identity. We define,

$F(x,t) = \begin{cases} (1-t)\phi(\displaystyle\frac{x}{1-t}) & \text{$0 \leq |x| < 1 - t$} \\ x & \text{$1 - t \leq |x| \leq 1$}
\end{cases}$

for $0 \leq t < 1$ and we define $F(x,1)$ to be the identity map of $D^2$. The result is an isotopy $F$ from $\phi$ to the identity.

What exactly is going on here in this proof visually? What is $F(x,t)$ telling us? I am not sure what the proof is doing here.

Just imagine you extend $\phi$ to a homeomorphism $\Phi$ of the whole plane by setting $\Phi(z)=\phi(z)$ whenever $|z|\leq 1$ and $\Phi(z)=z$ whenever $|z|\geq 1$. Then $F(\cdot,t)$ is obtained by considering $\Phi$ on a large disk and scaling that disk to be of unit size. You basically compress the area where the original $\phi$ is active to a disk of radius $1-t$, outside of which $\Phi$ acts as the identity.

Every map defined on a contractible space is nullhomotopic. So$ MCG(D^n)$={$id^*$.}

*:Of course, you may work in different categories and you want your maps to be either
isotopic , diffeomorphic, etc. So,in the broadest sense,i.e,  up to homotopy alone, $$MCG(D^n)$$ is trivial.(If  we only mention a homotopy,  this is what we want ). Now, considering isotopies,  since $D^n$ is homeo. to $\mathbb R^n$ , and $MCG (\mathbb R^n)=$ {$\pm Id$} , it all comes down to deciding if $Id$ ~ -$Id$ . But, e.g., (for n=2 )$-Id \circ -Id$ (maybe unfortunate shorthand for the isotopy $z \rightarrow ze^{i\pi t} ; t \in [0,1]$) gives you an isotopy from $-Id$ to $Id$ , so you are then done.


Ratio of (Collatz "path" bits) / (starting number bits) $\approx 2$ The "Path Records" page by Eric Roosendaal related to the Collatz conjecture (see here), reports a "striking" feature of the path records.

If we define $\text{MaxSeq}(M)$ (in Roosendaal it is $\text{Mx}(M)$) the maximum value reached starting the Collatz sequence from a positive integer $M$, a positive integer $N$ is called a path record if for all $M$ < $N$ we have $\text{MaxSeq}(M) \lt \text{MaxSeq}(N)$.

The "striking" feature about path records is this:

$$\log_2{\text{MaxSeq}(N)}\approx 2\log_2{N} \tag{1}$$

meaning that the maximum number of bits to store the Collatz sequence values starting from a number with $k$ bits is about $2k$ and not much more than that.

So, now this is my comment, when we start from a path record $N$ we could think it has an "energy" of about $2$, because the number of its bits will be eventually doubled, but other bigger values in the sequence toward $\text{MaxSeq}(N)$ have a lower "energy", and lower and lower when they are more near to $\text{MaxSeq}(N)$. This $2$ factor is not so big after all.

My question is: someone can help giving some probabilistic/heuristic explanation for equation $(1)$?

I tried using the fact that each odd number in the sequence is on average $3/4$ of the previous one (see here), but with no success.

Try this and find out.
$$a,n,k,\ell \in \mathbb{N}$$ $$\forall\;n_0:=2^ka-1\;\exists\;\mathcal{C}^k(n_0):\;n_k=\frac{3^ka-1}{2^\ell}$$


Compact set of probability measures I think I can solve the following exercise if $X$ is assumed to be separable, otherwise I can't.

  Let $X$ be a (Hausdorff) locally compact space, $\pi\colon X \to Y$ a continuous map into a topological space $Y$ such that $Y$ is the union of a countable sequence of compact sets $(K_n)$, and such that $\pi^{-1} (K_n)$ is compact for each $n$. Let $\mu$ be a regular Borel probability measure on $Y$. Define the space $\mathcal{M}_\mu (X)_1$ consisting of all Borel probability measures $\nu$ on $X$ such that $\pi_* \nu = \mu$. Show that $\mathcal{M}_\mu(X)_1$ is compact with respect to the weak-* topology.

Now if $X$ is assumed to be separable, so is $C_0(X)$ and the same proof as in Helly-Bray's theorem works (mutatis mutandis). What if it's not?

Here is what I got for a compact metric space:

We must show that any sequence $\mu_n$ in $\mathscr{M}(X)$ has a $\omega^*$-convergent subsequence (this shows that $\mathscr{M}(X)$ is actually sequentially compact).

Let $\{f_i\}_{i=1}^{\infty}$ be a countable dense subset of functions in $C(X)$. For any sequence $\mu_n$ in $\mathscr{M}(X)$ we have that \begin{equation*} |\int_X f_1 d\mu_n| \leq \|f_1\|_\infty \end{equation*} for all $n$, hence the sequence $\int_X f_1 d\mu_n$ is bounded and therefore has a convergent subsequence which we will denote by $\int_X f_1 d\mu_{n}^{1}$.

Now consider the sequence  $\int_X f_2 d\mu_{n}^{1}$. It is again a bounded sequence of real numbers and so it has a convergent subsequence  $\int_X f_2 d\mu_{n}^{2}$.

We continue in this fashion and obtain, for each $i \geq 1$, \begin{equation*} ... \subset \mu_{n}^{i} \subset \mu_{n}^{i-1} \subset ... \subset \mu_{n}^{1} \end{equation*} such that $\int_X f_j d\mu_{n}^{i}$ converges for $1 \leq j \leq i$. Now consider the sequence $\mu_{n}^{n}$. Since, for $n \geq i$, $\mu_{n}^{n}$ is a subsequence of $\mu_{n}^{i}$, $\int_X f_i d\mu_{n}^{n}$ converges for every $i \geq 1$. \
We can now use the fact that $\{f_i\}_{i=1}^{\infty}$ is dense to show that $\int_X f d\mu_n$ converges for all $f \in C(X)$. For any $\epsilon > 0$, choose $f_i$ such that $\|f-f_i\|_\infty \leq \epsilon$. Since $\int_X f_i d\mu_{n}^{n}$ converges, there exists $N$ such that if $n,m \geq N$ then \begin{equation*} \lvert \int_X f_i d\mu_{n}^{n} - \int_X f_i d\mu_{m}^{m} \rvert < \epsilon. \end{equation*}

Thus if $n, m \geq N$ we have \begin{equation*}  | \int_X f d\mu_{n}^{n} - \int_X f d\mu_{m}^{m} | \leq \end{equation*} \begin{equation*}| \int_X f d\mu_{n}^{n} - \int_X f_i d\mu_{n}^{n} | + | \int_X f_i d\mu_{n}^{n} - \int_X f_i d\mu_{m}^{m} |  + | \int_X f_i d\mu_{m}^{m} - \int_X f d\mu_{m}^{m} | \leq 3 \epsilon. \end{equation*}

So, $\int_X fd\mu_{n}^{n}$ converges, as required.
To complete the proof, write \begin{equation*} \Lambda (f) = \lim_{n \rightarrow \infty} \int_X f d\mu_{n}^{n}. \end{equation*} It is easily verified that $\Lambda$ satisfies the hypothesis of the Riesz Representation Theorem. Therefore, there exists $\mu \in \mathscr{M}(X)$ such that \begin{equation*} \Lambda (f) = \int_X f d\mu.\end{equation*} We then have \begin{equation*} \int_X f d\mu_{n}^{n} \rightarrow \int_X f d\mu, \end{equation*} as $n \rightarrow \infty$, for all $f \in C(X)$, so we have proved that $\mu_{n}^{n}$ $\omega^*$-converges to $\mu$, as $n \rightarrow \infty$ and we are done. $\blacksquare$


Ring extensions $A \subseteq B$ such that $P^{ce}=P$, for every prime ideal $P$ of $B$ Let $A \subsetneq B$ be a ring extension, where $A$ and $B$ are commutative $k$-algebras, $k$ is a field of characteristic zero.
Let $P$ be a prime ideal of $B$.
Denote the contraction of $P$ to $A$ by $P^{c}=P \cap A$, and its extension to $B$  by $P^{ce}=(P^{c})^{e}=(P \cap A)^{e}=(P \cap A)B$.

  I am looking for special cases in which $P^{ce}=P$.

Remarks:

(1) If $f: U \to V$ is a surjective ring homomorphism, then for every ideal of $U$ (prime or not) we have $I^{ce}=I$, see here.

(2) If $A \subset B$ is faithfully flat, then for every prime ideal $p$ of $A$,
$p^{ec}=p$, see the answer to this question.

(3) Is it true that if $B$ is a finitely generated free $A$-module, then my question has a positive answer? (the condition "finitely generated" is necessary, since, for example, my question has a negative answer for $A=k$, $B=k[x,y]$).

I know it has been a while since the question was posted, anyway (if I'm right) the following should be an example of $\mathfrak p^{ce}=\mathfrak p.$

Let $f\colon A\longrightarrow B$ be an injective, flat ring homomorphism. Let $\mathfrak p\in\mathsf{Spec}(B)$ and $\mathfrak q:=f^{-1}(\mathfrak p)$. I claim that $$\mathfrak qB=\mathfrak p.$$

The ring map $B/\mathfrak qB\longrightarrow B/\mathfrak p$ is onto. We show that it is injective too. As $f$ is flat and $A/\mathfrak q\longrightarrow B/\mathfrak p$ is injective, tensorizing by -$\otimes_AB$ we get an injective ring homomorphism $$\phi\colon B/\mathfrak qB=B\otimes_A A/\mathfrak q\longrightarrow B\otimes_AB/\mathfrak p.$$ Now just notice that $\phi$ factorizes as $$B/\mathfrak qB\longrightarrow
B/\mathfrak p \longrightarrow B\otimes_AB/\mathfrak p.$$


Sum of convex and concave function can have non global minima So I just read that the sum of a convex with a concave function is neither convex nor concave in the general case. This would imply that I should be able to find two functions $a(x)$ and $b(x)$ of which one is convex and the other one concave such that a third non convex function $c(x) = a(x) + b(x)$ has at least two local minima of which only a single one is the global one. This would obviously increase the difficulty of our optimization problem as we can not be certain anymore that any local minimizer is the global one. However I can not come up with an example of $a(x)$ and $b(x)$, could somebody share some insights?

Thank you!

Hints:

$-f(x)$ is concave when $f(x)$ is convex

$x^{2n}$ is convex for $n \in \mathbb{N}^+$

$f(x + a)$ has the same convexity as $f(x)$ for $a \in \mathbb{R}$


no simple group of order $945$ I need to show that there are no simple groups of order $945$.

I've tried the regular method using the Sylow theorems.

$$|G|=945=3^3\cdot5\cdot7 $$

If $G$ is simple then there should be 7 Sylow-3 groups ; 21 Sylow-5 groups and 15  Sylow-7 groups. Even if they would all intersect trivially, there will still be no contradiction.

Any ideas?

If as you say there are 7 Sylow 3-subgroups, then the normalizer $N$ of one of these subgroups has index 7.  If $G$ is simple, then there is an injective permutation representation $G \hookrightarrow S_7$, and so $|G| = 945$ must divide $7!$, but this is not the case.

I think every simple group has even order.


Solving a cubic diophantine equation $ax^3+bx^2=cy^3$ I want to know how to solve the equation $ax^3+bx^2=cy^3$ in positive integers $x$ and $y$, assuming $a, b, c$ are positive integers and $gcd(a, b, c) = 1$. If there is not a general algorithm, I would like to solve more specific ones, such as when $a$ and/or $b$ and/or $c$ equal $1$. I have tried searching online, but could not find an algorithm to solve this.

Edit: How would the answer change if a and c are both 1, while b is a positive integer?

This is a special case of the generalized Fermat equation
$$
ax^p+by^q=cz^r
$$
for $(p,q,r)=(2,3,3)$. This case is spherical, because
$$
\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1.
$$
So there is either no non-trivial solution, or infinitely many solutions, which can be constructed by a finite set of polynomial parametrisations of the equation, see here:

F.Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.

Then you can specify the solutions satisfying $y=x$. For example, when $a=b=c=1$, then $x^2+x^3=y^3$ has only the trivial solution $x=y=0$. This fact also follows from this question:

Integer solutions of $x^3+y^3=z^2$

This equation gives you a (singular) genus 0 curve for any choice of $a,b,c$. Now, there is a famous theorem by Siegel which asserts that there are a finite number of integral solutions if the number of points at infinity (over $\mathbb{C}$) is 3 or larger. In you case the points at infinity are the points (in projective coordinates $[x:y:z]$) equal to
$[1: \xi: 0]$, with $\xi^3=(c/a)$, which are clearly 3.

So for any given $a$, $b$ and $c$ there is only a finite number of integer solutions.

I expect there is no general formula for this solutions.

Except may be some special values of $a$, $b$ and $c$, I guess that the equation will only have between 0 and 1 solution in coprime strictly positive integers ($x>0$ and $y>0$).

The usual "method" to find all the solutions is to do a search on "small solutions", and then try to show there are no others: either because you have a theorem telling you that all the solutions must be small, or by congruence considerations, or with arguments working over some cubic field.

Edit

For the special case $a=c=1$, observe that, if $b=r^3-s^3$ for some integers $r>s>1$, there is a "trivial solution" $x=s^3$ and $y=rs^2$. This can be extended to the case $b=t(r^3-s^3)$ for some integers $r>s>0$ and $t>0$, and we get the solution $(x,y)=(s^3t,rs^2t)$.

Guess: Any solution $(x,y)$ with strictly positive integers of $y^3=x^3+bx^2$, for $b>0$ integer, verifies that, if $t=gcd(x,y,b)$, $gcd(x,y)=ts^2$ for some $s>0$, and $x=s^3t$, $y=rs^2t$ for some $r>0$. In particular, if $gcd(x,y)=1$, then $x=1$.

I tested this guess for $b$ up to 800 and "small" solutions and it seems it holds.


Finding dominating function Let $f_n:(0,1)\rightarrow\mathbb{R}$ be $n$ for $x\in(0,1/n)$ and be $0$ otherwise. Then why this $f_n$ have no dominating function which is lebesgue integrable?
I want that dominating function condition cannot be cancelled from dominated convergence theorem

Suppose $g(x) \geq f_n(x)$ for all $x$ and all $n$. Then $g(x) \geq n$ for $\frac  1 {n+1} <x<\frac 1 n$. Hence $\int_0^{1} g(x) dx \geq \sum \int_{1/n+1}^{1/n} g(x)dx \geq \sum n(\frac  1 n-\frac  1 {n+1})=\sum \frac 1  {n+1}=\infty$.

So we have
$$\forall n, \int_{\mathbb{R}} f_n =1$$
and in the meaning of distribution we get Dirac distribution
$$ f_n \to \delta_0 $$
which is not Lesbegues integrable.

Note that if we get domination we would get $\delta_0$ integrable which is contradictory.


Spivaks Chain Rule Proof (Image of proof provided) Here is a proof of the chain rule in Spivaks Calculus. Note there is a second page, but I understand it, and this is the meat of the proof. I have a few questions.

$\textbf{1.}$ "It should be intuitively clear that $\phi$ is continuous at $0$." Do we care that it is continuous at zero so we will not have a division by zero since $g(a+h)-g(a)$ is in the denominator and could equal zero? I am not sure I understand why it is continuous at zero. I understand what he was saying but I was always under the impression continuity was when there was no breaks in the graph visually. Here, I am imagining $\phi(h)$ being continuous up to zero, then it jumping to another point when it is zero.

$\textbf{2.}$ At (2),I do not understand what we are trying to do here. We randomly switched to $h$ and are defining continuity I think. The switch back and forth from $k$ to $h$ is confusing me.

The even more intuitive reason why the result should be true is that:

$$\begin{split} (f \circ g)'(a) & = \lim_{h\to0} \frac{f(g(a+h)) - f(g(a)))}{h} \\
  & = \lim_{h\to0} \frac{f(g(a+h)) - f(g(a)))}{g(a+h) - g(a)} \frac{g(a+h) - g(a)}{h} \\
  & = (\lim_{g(a+h)\to g(a)} \frac{f(g(a+h)) - f(g(a)))}{g(a+h) - g(a)}) (\lim_{h\to0}\frac{g(a+h) - g(a)}{h}) \\
  & = f'(g(a)) g'(a)
\end{split}$$

Unfortunately this intuitive proof can fall apart when $g'(a) = 0$ because you may be multiplying and dividing by 0.  So he's creating an auxiliary function which is guaranteed well-defined, and has the limit he wants for the first part of that expression.  Then he's using that function to get around the potential division by 0 problem.

Editorial here.

I have always hated this proof.  I much prefer thinking in terms of the tangent line approximation in which $f(a+h) \approx f(a) + f'(a) h$.  If you define $\approx$ right, this leads to a definition which is fully equivalent to the usual limit definition.  And now the intuitive reason why this is true is that:

$$\begin{split}
f(g(a+h)) & \approx f(g(a)) + f'(g(a)) ( g(a+h) - g(a) ) \\
& \approx f(g(a)) + f'(g(a))(g(a) + g'(a)h - g(a)) \\
& = f(g(a)) + f'(g(a))g'(a)h\end{split}$$

Proving the first approximation does require an admittedly more technical proof.  However it leads to a correct proof of the chain rule that works without change for multi-variable calculus.  As well I think it important for people to really understand the tangent line approximation.

Spivak's is an introductory book, that in theory should be usable by people who have never seen any calculus before, so he is making a long story out of something simple (but profound).   His exposition is idiosyncratic in some places and is not necessarily worth trying to understand in every detail when clearer approaches to the same things are available. I'd suggest ignoring this forced attempt to make $f'(g(x))$ into a ratio, and reading the very short and easily understood proof by linear approximation (see btilly's answer) given in books on analysis such as Rudin.


Zeros of a holomorphic function Suppose $\Omega$ is a bounded domain in the plane whose boundary consist of $m+1$ disjoint analytic simple closed curves.

Let $f$ be holomorphic and nonconstant on a neighborhood of the closure of $\Omega$ such that

$$|f(z)|=1$$ for all $z$ in the boundary of $\Omega$.

If $m=0$, then the maximum principle applied to $f$ and $1/f$ implies that $f$ has at least one zero in $\Omega$.

What about the general case? I believe that $f$ must have at least $m+1$ zeros in $\Omega$, but I'm not able to prove it...

Thank you

This is true and it is a consequence of the Argument Principle; the Maximum Modulus Principle is not precise enough to prove it. Here is the proof for $m=0$ and it generalizes: We know that $$n(w)=\frac{1}{2\pi i}\int_{\partial \Omega}\frac{f'(z)}{f(z)-w}\, dz$$ gives the number of times $f$ equals $w$ in $\Omega$ and is continuous on $D=\{|w|<1\}$. Since $f$ is not constant, $n(w)\ge 1$ for some $w\in D$. By connectedness and continuity, we also have $n(0)\ge 1$. Therefore $f$ has at least one zero in $D$.

In the general case, let $\partial\Omega=\gamma_1\cup\cdots\cup \gamma_{m+1}$ with orientation $\gamma_1-\cdots- \gamma_{m+1}$ (so $\gamma_1$ is also the boundary of the unbounded compoent of ${\mathbb C}\backslash\Omega$). Using the Open Mapping Theorem, one can deduce that each $f(\gamma_k)$ traverses the unit circle at least once, with $f(\gamma_1)$ in the positive direction and $f(\gamma_k)$ in the negative direction for $k\ge 2$. So each point on the unit circle is assumed by $f$ at least $m+1$ times on $\partial\Omega$. Use this fact and a point $w$ close to the unit circle to see that there are at least $m+1$ points in $\Omega$ that has image $w$. So $n(w)\ge m+1$. Now use the same argument as the $m=0$ case to conclude that $n(0)\ge m+1$. (Or one can just use an analytic continuation of $f$ and do away with $w$ close to the unit circle.)

Since each of the $m+1$ simple closed curves are disjoint, it is clear that $\Omega$ has $m+1$ connected components. Let $U$ be any one of these.

Then $f(z)$ restricted to $U$ is a holomorphic function. Furthermore, it is non-constant and holomorphic on some neighborhood of the closure of $U$, because this closure is contained in $\Omega$. Furthermore, the boundary of $U$ is contained in the boundary of $\Omega$, so $|f(z)| = 1$ on the boundary of $U$.

Like you outlined in your explanation of the problem, if we assume that $f$ is non-zero in $U$ then both $f(z)$ and $1/f(z)$ are holomorphic in $U$, and thus must achieve their maximums on the boundary of $U$. But its given that $|f(z)| = 1$ on the boundary of $U$, and so we get

$|f(z)| \leq 1$ and $|1/f(z)| \leq 1$ for $z \in U$. Thus we must have that $|f(z)| \equiv 1$ on $U$, which is a contradiction as we assumed that $f(z)$ is non-constant. Thus $f$ has at least one zero in $U$. We assumed nothing about which connected component we used, so this is true in each component of $\Omega$, and therefore $f(z)$ has at least $m+1$ zeroes.


Closed subset $Z \subseteq \operatorname{Spec} A$ This is an argument in Gortz algebraic Geometry I text book pg85. I extracted what I don't understand.
He proves the following

Let $A$ be a unital commutative ring. $Z \subseteq  \operatorname{Spec} A$ a closed subset satisfying the following property: Given $s \in A$,
If $Z \subseteq V(s)$, then $s^n=0$ for some $n$.

He concludes

$V(s)=X$ and $Z=X$.

I don't understand this argument. What facts did we used here?
EDIT: The following is my proposed proof: $V(s)=X$. Since if $s$ is nilpotent, it lies in the nil radical, hence intersection of all primes.
Now $Z=V(I) = V(\sum_{i \in I} A i) = \bigcap_{i \in I} V(i) = X$.

I have the good idea. I would emphasize on the fact that $V(i)=X$ since $i$ is nilpotent by hypothesis.

Since $Z$ is closed, $Z=V(I)$, this implies that for every $s\in I$, $Z\subset V(I)\subset V(s)$, we deduce that $s$ is nilpotent and $I\subset nil(A)$, $X=V(nil(A))\subset V(I)$ implies that $V(I)=Z=X$.


Why isn't the union of the coordinate axes a manifold, using the formal definition of a manifold? Let me call $M$ a manifold iff locally it is given as the graph of a $C^1$ function. Then without appealing to "remove a point" type arguments, why isn't the union of the $x$ and $y$ coordinate axes (the solution set of $xy=0$) a manifold? That is, why isn't this locally a graph around the origin?

The formal definition of manifold I know is that it's a topological space that's locally homeomorphic to $\mathbb R^N$. Thus in order for the union of the axes to be a 1-manifold, there must be an open neighborhood (in the putative manifold) of $\langle 0,0\rangle$ that is homeomorphic to $\mathbb R$.

By definition, every such open neighborhood contains a ball $B_{2\delta}(\langle0,0\rangle)$ for some $\delta>0$, and therefore it contains the points $P\langle0,\delta\rangle$, $Q\langle-\delta,0\rangle$ and $R\langle\delta,0\rangle$.

These three points have the property that there's a path from $P$ to $Q$ not passing through $R$, a path from $Q$ to $R$ not passing through $P$, and a path from $R$ to $P$ not passing through $Q$. But there's no triple of points with this property in $\mathbb R$, so the neighborhood of $\langle 0,0\rangle$ cannot possibly be homeomorphic to $\mathbb R$.

The intersecting lines cannot be a manifold of higher dimension either, because every neighborhood of $\langle 0,0\rangle$ has an open subset that is homeomorphic to an interval, and $\mathbb R^d$ doesn't when $d\ge 2$.

A function is single-valued, but in any neighborhood of $0$ we have an infinite number of points of the form $(0,y)$ for different $y$.


This is what I wrote first: $a+b+c+d\ge 4$ because $abcd=1$ and arithmetic mean $\ge$ geometric mean. As pointed out by Feanor, this inquality doesn't help from where the OP left.

Alternative: If $a,b,c,d$ are positive, arithmetic mean > geometric mean > harmonic mean so $(a+b+c+d)/4>1>4/(1/a+1/b+1/c+1/d)$. So $1/a+1/b+1/c+1/d+9/(a+b+c+d)>4+9/4=25/4$.


maths for a strong visual spatial thinker with sequential processing weakness? Can someone with superior visual spatial thinking ability but weak sequential/word ability be successful in university maths and beyond?

For example, for real analysis, would definitions with lots of symbols and complex sentences be difficult to understand?

Is it true that someone who are both strong sequentially and visually have best potential to be a good mathematician?

also, what branches of maths would be best suited to very spatially oriented person?

edit: please note that I'm talking about someone who has poor sequential ability. Not just someone who are strong in both but more visiospatially oriented.

thanks

As for the first question: Most of the powerful tools we have nowadays are algebraic, and not visual. However some tools like euclidean geometry or calculus  do benefit from spacial thinking.

Of course someone who is good at spacial and sequencing would be better off, but that may be obvious.

However I believe it is a matter of practice, having great spacial ability is certainly a good tool in mathematics, and you would benefit even more from it if you tried to improve the other parts also.

Edit: Also maths are getting more and more intertwined every day. So mabye your great spacial skill can be keen to solve a combinatorics problem. While complex analysis can be useful for you geometry problems.


How prove $A^2=0$,if $AB-BA=A$ let $A_{2\times 2}$ matrix, and  The matrix $B$ is  order  square,such
  $$AB-BA=A$$ show that
  $$A^2=0$$

My idea: since $$Tr(AB)=Tr(BA)$$ so
$$Tr(A)=Tr(AB-BA)=Tr(AB)-Tr(BA)=0$$

Question:2

if $A_{n\times n}$ matrix,and the matrix $B$ is order square,such
$$AB-BA=A$$
then we also have
$$A^2=0?$$
and then I can't Continue .Thank you

You say that $tr(A) = tr(AB)-tr(BA)=0$. Therefore Cayley Hamilton equation tells us that $A^2 = \det(A) I_2$.

On the other hand we have $A^2= A^2B-ABA=ABA-BA^2$. Therefore $2A^2 = A^2B-BA^2=0$ since $A^2$ is a multiple of the identity.

$AB - BA = A \Rightarrow A^2B - ABA = A^2 \Rightarrow$ $$-ABA = A^2(I-B)$$
$AB - BA = A \Rightarrow ABA - BA^2 = A^2 \Rightarrow$ $$ABA = (I+B)A^2$$
Adding this identities: $0 = A^2(I-B) + (I+B)A^2$


How to calculate the gradient (vector) of a vector field? First question from me. I hope it's clear enough.

I'm trying to make a physics simulator of magnets, magnetic interactions and such.
I've followed the formulas in Wikipedia pages, and some first results are great (same newton force value between math model and reality test).

Now I have to translate everything in a software coding; I'm using c# inside grasshopper inside Rhinoceros.

I can calculate the magnetic field vector "B" of a magnetic dipole vector "m" with this formula: wiki 4 ,
Pic 1

Anyway,
here: wiki 1 , here: wiki 2 and here: wiki 3 it says that the force is this:
Pic 2

That is the gradient of the dot product of "m"(the magnetic dipole moment vector) and "B"(the local magnetic field vector).

I've understood that the gradient is somehow "like" the slope of a function (which you can "manually" calculate by evaluating two points really close and divide their Y difference by their X distance).

For every sample point I'd like to have one gradient vector (of the local resulting magnetic field) to multiply (dot product) with whatever dipole is passing by in that iteration.
(Does this make sense? I'm not native English...)

Can I calculate the gradient by evaluating 2 points near the target location in the same way?

How?

Other ideas?

Thanks in advance.

First of all, since the dipole $m$ on which the force acts is constant, the formula simplifies to
$$
F=\nabla(m\cdot B) = m^TJ_B = J_B^T m,
$$
where $J_B$ is the Jacobian matrix. See also here.

If you want to see the reason why, just work with coordinates and you find
$$
[\nabla(m\cdot B)]_i = \frac{\partial}{\partial x_i} \sum_{j=1}^n m_j B_j
= \sum_{j=1}^n m_j \frac{\partial B_j}{\partial x_i}
= m^T J_B.
$$

Regarding the question of how to compute $J_B$, there are several approaches:

if $B$ has a specific closed form expression, you can of course use it to compute explicitly its gradient;
you can use finite differences, as you mentioned;
you can use automatic differentiation to compute (a numeric approximation of) the gradient at the same time as you compute the field itself.

No,
$$\operatorname{grad}(\vec{m}\cdot\vec{B}) \neq m \cdot\operatorname{grad}(\vec{B})$$
But rather
$$\begin{align}
\operatorname{grad}(\vec{m}\cdot\vec{B})_i&=\partial_i (m_kB_k)\\
&=(\partial_im_k)B_k+(\partial_iB_k)m_k\\
&=J(\vec{m})_{ik}B_k+J(\vec{B})_{ik}m_k\\
&=(J(\vec{m})\cdot\vec{B})_i+(J(\vec{B})\cdot\vec{m})_i
\end{align}$$
Where $J(\vec{m})$ is the Jacobian of $\vec{m}$ and $J(\vec{B})$ is the jacobian of $\vec{B}$. And if $\vec{m}$ is a constant vector, then you have that
$$\operatorname{grad}(\vec{m}\cdot\vec{B})=J(\vec{B})\cdot\vec{m}$$


how to solve a First Order inhomogenous difference/recurrence equation with a constant on the right side Suppose we have the difference equation

$aq_n +bq_{n-1} = c$,

where $a,b, c$ are constants, and we want to find the function $q_n$.

My book says, first, we find the solution to the homogeneous equation

$aq_n +bq_{n-1} = 0$,

which is $q_n = q_0(\frac{-b}{a})^n$

Then we solve the inhomogenous  equation by guessing $q_n = d $ to be a constant, so we get

$ad +bd = c$,   $d=\frac{c}{a+b}$

Putting together these two solutions we get the general solution:

$q_n = C(\frac{-b}{a})^n +\frac{c}{a+b}$

My question is could this have been done another way (to be specified in a moment), and if yes, how are these two solutions equivalent?

Could I solve it by writing

$q_n = \frac{c-bq_{n-1}}{a} = \frac{c}{a} - \frac{b}{a}q_{n-1}$

let $\alpha = \frac{c}{a}$ and $\beta = \frac{b}{a}$

so,

$q_n = \alpha - \beta q_{n-1}$

but, recursively, we have

$q_n = \alpha - \beta *(\alpha - \beta q_{n-2}) = \alpha - \alpha \beta + \beta^2 q_{n-2}$

again, we get

$q_n = \alpha - \alpha \beta +\alpha \beta^2 - \beta ^3 q_{n-3}$

and if we keep going,

$q_n= \alpha (1 - \beta + \beta^2 -\beta^3 +\beta^4 - ... + (-1)^{n-1}\beta^{n-1}) + (-1)^{n} \beta^nq_0$

Is this a correct solution? and more importantly if yes, how is it equivalent to the first one, and if it's not correct, where did I go wrong?

I do not know if this is the kind of answer you are waiting for.

Let us make the problem a bit more general considering $$aq_n +bq_{n-1} = c+dn$$

To get rid of the problem of constants $c$ and $d$, start defining $$q_n=u_n+\alpha+\beta n$$ and replace in the original equation. This hould lead to $$a u_n+b u_{n-1}=(-a \alpha -\alpha  b+b \beta +c)+n (-a \beta -b \beta +d)$$ Cancelling the extra terms in the rhs leads to  $$-a \alpha -\alpha  b+b \beta +c =0$$ $$-a \beta -b \beta +d=0$$ leading to $$\alpha=\frac{a c+b c+b d}{(a+b)^2}\qquad , \qquad \beta=\frac{d}{a+b}$$ and you are just left with the recurrence equation $$a u_n+b u_{n-1}=0$$ Introducing, just as you did $\beta = \frac{b}{a}$ then you just need to focus on $$u_n+\beta u_{n-1}=0$$ the solution of which being $$u_n=c_1 (-\beta )^{n-1}$$ and then $$q_n=c_1 (-\beta )^{n-1}+\frac{a c+b c+b d}{(a+b)^2}+\frac{d}{a+b}n$$

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A 'particular solution' $\ds{\quad p_{s}\quad }$ is given by
$\ds{\quad a\,p_{s} + b\,p_{s} = c
\quad\imp\quad p_{s} = {c \over a + b}.\quad}$ Then, $\ds{q_{n} - p_{s}}$ satisfies the homogeneous
$\ds{\quad a\pars{q_{n} - p_{s}} + b\pars{q_{n - 1} - p_{s}} = 0}$
\begin{align}
\mbox{and}\quad q_{n} - p_{s} & = -\,{b \over a}\,\pars{q_{n - 1} - p_{s}} =
\pars{b \over a}^{2}\,\pars{q_{n - 2} - p_{s}} =
-\pars{b \over a}^{3}\,\pars{q_{n - 3} - p_{s}}
\\[5mm] & =
\cdots = \pars{-1}^{n}\pars{b \over a}^{n}\,\pars{q_{0} - p_{s}}
\end{align}

If $\ds{\,\,\,\verts{b \over a} < 1}$, $\ds{\lim_{n \to \infty}q_{n} = p_{s}}$. In any case, $\ds{q_{n}}$ is given by
$$
q_{n} =
{c \over a + b} +
\pars{-1}^{n}\,\pars{b \over a}^{n}\,\pars{q_{0} - {c \over a + b}}
$$