How many minutes would it take to make 100 toys? 7 machines take 7 min to make 7 identical toys. At the same rate how many minutes would it take for 100 machines to make 100 toys?
(A) 1 (B) 7
(C) 100 (D) 700

My trial: Given that
7 machines:7minutes:7 toys
1 machine:1minute:1 toy
100 machines:100 minutes:100 toys
therefore it would 100 minutes for 100 machines to make 100 toys.
But my book suggests that answer must be 7 minutes I don't know where I am wrong. I think there should be a correct method to solve such problems. Please give correct solution to this problem.
thanks

7 machines make 7 toys in - 7 minutes.

So, 7 machines make 100 toys in - $\frac {7} {7}.100 $ minutes = 100 minutes

Or, 100 machines make 100 toys in - $ 100. \frac {7}{100} = 7 $ minutes

The general way to proceed is to keep all quantities constant, except one, and see how it varies with the quantity you require.
Now the explanation is that same number of machines would take more time to make more toys, hence it is directly proportional. However, more machines would take less time to make same number of times,i.e, it is inversely proportional.

7 machines make 7 toys in - 7 minutes.

So, 7 machines make 100 toys in - 77.100 minutes = 100 minutes

Or, 100 machines make 100 toys in - 100.7100=7 minutes
Simple!!!!!!


Two Car Meeting At a Point Probability Question There is a car at point A. It can go either right or down. There is a another car at point B. This car can go either up or left. Obviously with this restrictions cars cannot go in reverse directions. If car at the point A goes to point B, and car at point B goes to point A with the same speed, what is the probability that these two cars meet at point Z.

(As an example one of the path that car at point A will follow is AKZMB or, one of the path that car at point B will follow is BMDNA.)

I have couple of answers but I can't be sure about which one is true. I need an explanation for right answer. Thank you for your time.

i think by counting, we have 6 ways A to B, 6 ways B to A
so Sample space =6*6=36
after 4 steps then A to B, and B to A. so two cars meet at point Z when they goes 2 steps.
A - Z - B have 4 way
B - Z - A have 4 way
so we have 16 cases  A goes to point B, and car at point B goes to point A, meet at point Z
 P = 16 /36 =4/9
I hope this solution is correct

edit

"assuming that at each intersection, the driver chooses the way to turn by flipping a fair coin." is not correct. I can Choose another distribution but the way A to B is unchanged, but with  your assuming , answer change.

Cool question, I like this problem!

You could consider the arrangements of points and lines as nodes and edges, and then consider the hole arrangement as a graph (as in graph theory).
Doing this, allows you to then construct a tree, featuring all possible ways for both cars to get from A to B.
When I did this, I found that there are 12 possible ways for each car to get to the other side (A to B, B to A), 10 from those ways cross the node Z, at some point during there way to the other side. Considering this, and the fact that you need both cars to take a route with Z being an equal distance fron there starting point, you will find that you have a 1/2 chance of both cars Meeting each other at Z.
This means a 50% change!

I hope this helped...
.


Rules for algebraically manipulating pi-notation? I'm a bit of a novice at maths and want to learn more about algebraically manipulating likelihoods in statistics.
There are a lot of equations that involve taking the product of a set of values given a model.

I know a few rules for manipulating sigma-notation (e.g., here and here).

What are the basic rules for manipulating sequences of products (i.e, $\prod_{i=1}^{I} ... $)? Is there a web page that you could direct me to?

e.g.,

$\prod_{i=1}^{I} x_i$
$\prod_{i=1}^{I} x_i y_i$
$\prod_{i=1}^{I} a + b x_i$
$\prod_{i=1}^{I} \exp x_i$

You just need to think about what the product notation means, and you can work out the rules yourself.

Take one of your examples:

$$\prod_{i=1}^I x_i y_i = \left(\prod_{i=1}^I x_i\right)\left(\prod_{i=1}^I y_i\right)$$

This rule works because the left side multiplies $x_1$ times $y_1$ times $x_2$ times $y_2$ etc., and the right side multiplies the x's first, then multiplies the result by the y's.  Clearly those are equal.

Another one of your examples:

$$\prod_{i=1}^I \exp x_i = \exp \left( \sum_{i=1}^I x_i \right)$$

Can you figure out why this rule works?

This might be inappropriate for an answer but I believe you tried yourself too hard at here. The $\prod $ sign just means multiplying some elements together, with a label in the bottom to denote the beginning element and a label on the top to denote the end element. The files you provided are very good and you should get a sense of what the general situation is. You will get better in using it if you read articles proving things using this symbol. But as any other short hand symbol writing the product of elements in this way does not really simplify anything; you should be able to write out any such products explicitly if you can write it in the $\prod$ sign.


How do I find the roots of $y=e^x-2$? I need to find the root of this: $y=e^x-2$.

I don't know how to find the roots of this function because I have two variables...
The teacher said the root is between $0$ and $1$.

When they talk about roots they want those $x$ values which makes $y=0$

So you want to find $x$ such that $$e^x -2 =0$$

That is solving $e^x=2$.

You can solve it by taking the natural logarithm of both sides.

You can solve it algebraically as above. There's also many  numerical approaches. Here's one.

Let $x_0$=1.

It can be shown that the sequence,  $x_{n+1}=x_n-\frac{e^{x_n}-2}{e^{x_n}}$, gets closer to the root for higher and higher iterations.

$x_{n+1}=x_n-1+\frac{2}{e^{x_n}}$


How to calculate sin(65) without a calculator. I know about the sum and difference formula but I can't think of two values which will be able to use for sin(65). Therefore, I come to the question: How to calculate sin(65) without a calculator.

$$65 = 45 + 20$$

So if you can figure out sine and cosine of 20, you're in good shape. Fortunately,  there are triple-angle formulas:

$$
\sin(3x) = 3\sin x-4\sin^3 x \\
\cos(3x) = \cos^3 x-3\cos x \sin^2 x = \cos^3 x-3\cos x (1 - \cos^2 x)
$$
Since you know $\sin(60) = \sqrt{3}/2$, you know that the sine of 20 -- call it $u$ -- satisfies the equation

$$
\sqrt{3}/2 = 3u-4u^3
$$
From this, you can solve for $u$ (using Cardano's formula for the solution of a cubic). To be honest, this is a complete pain in the neck, but at least it's a route to the solution.

Split it up, into sin 45 plus sin 30. You will get sin 75, then subtract sin 10(you can do by using sin(3x) formula). Then you will get sin 65


definite integral $\int_{-1}^1x^2\sqrt{4-x^2}dx$ I have solved the problem and obtained the value of the integral $2\pi/3-\sqrt{3}/2$.
In the actual solution of this problem I have seen that $x$ was substituted by $2\sin z$ and upper limit $1$ was substituted by $\pi/6$ and lower limit $-1$ by $-\pi/6$.
Now the question is why was the upper limit substituted by $\pi/6$ instead of $5\pi/6$ or anything equivalent to $\arcsin\tfrac12$?

Actually,$$\sin(\pi-y)=\sin y\implies\sin\tfrac{5\pi}{6}=\sin\tfrac{\pi}{6}\ne-\sin\tfrac{\pi}{6}.$$We need a $z$ satisfying $\sin z=-\sin\tfrac{\pi}{6}$. Since the sine function is odd, $z=-\tfrac{\pi}{6}$ will do.
Alternative argument: since the original integrand is even,$$\begin{align}\int_{-1}^1x^2\sqrt{4-x^2}dx&=2\int_0^1x^2\sqrt{4-x^2}dx\\&=2\int_0^{\pi/6}16\sin^2z\cos^2zdz\\&=\int_{-\pi/6}^{\pi/6}16\sin^2z\cos^2zdz,\end{align}$$as the new integrand is even too.

It must be $2 \sin{z} = \pm1 \Rightarrow z = \arcsin{\pm\frac{1}{2}} = \pm\frac{\pi}{6}$
That's all.


Proving that $(u+v)×w=u×w+v×w$ Let's $$(\overrightarrow{u}+\overrightarrow{v})\times\overrightarrow{w}=\overrightarrow{u}\times\overrightarrow{w}+\overrightarrow{v}\times\overrightarrow{w}$$ How to prove it?

Update: The problem is that I don't know how to do it because I don't know the number of dimensions. Is there a general formula?

This property should do it. Let $u,v,w\in V$ (where we take $V$ a presumably three dimensional inner product space). Let $\langle\cdot,\cdot\rangle$ be the inner product on $V$. Then $$\langle u,v\times w\rangle =\det[u,v,w]_{\mathcal B}$$

Where $\det[\,]_{\mathcal B}$ denotes the determinant of the matrix formed by $u,v,w$ in an orthonormal basis $\mathcal B = \{e_1,e_2,e_3\}$.

In this light let, $\gamma_1$ be the first coordinate of $\gamma = (a+b)\times w$. Then, $$\gamma_1 = \langle e_1,(a+b)\times w\rangle = \det[e_1,a+b,w] = \det[e_1,a,w] + \det[e_1,b,w] =  \\ =  \langle e_1,a\times w\rangle + \langle e_1, b\times w\rangle$$

Idem for $\gamma_2,\gamma_3$. Can you take it from here?

P.S. Take $\mathbb R^3$ with the usual scalar product if the notation here clouds things up.

The cross product is defined on two vectors of three dimensions as: $$\vec a \times \vec b =\begin{bmatrix} a_1 \\a_2\\a_3\end{bmatrix}\times \begin{bmatrix} b_1 \\b_2\\b_3 \end{bmatrix} = \begin{bmatrix} a_2b_3-b_3a_2 \\a_3b_1-a_1b_3\\a_1b_2-b_2a_1 \end{bmatrix}$$

The sum of two three-dimensional vectors is:
$$\vec c + \vec d =\begin{bmatrix} c_1 \\c_2\\c_3\end{bmatrix}+ \begin{bmatrix} d_1 \\d_2\\d_3 \end{bmatrix} = \begin{bmatrix} c_1+d_1 \\c_2+d_2\\c_3+d_3 \end{bmatrix}$$

Use this to show that: $$(\vec u+\vec v)\times \vec w = (\vec u\times \vec w)+(\vec v\times \vec w)$$


Find the triangular matrix and determinant. I have a 4x4 matrix and I want to find the triangular matrix (lower half entries are zero).

$$A=
\begin{bmatrix}
2 & -8 & 6 & 8\\
3 & -9 & 5 & 10\\
-3 & 0 & 1 & -2\\
1 & -4 & 0 & 6
\end{bmatrix}
$$

Here are the elementary row operations I performed to get it into triangular form.

row swap rows 1 and row 4

$r_2 - 3\cdot r_1$ replacing $r_2$

$r_3 + 3\cdot r_1$ replacing $r_3$

$r_4 - 2\cdot r_1$ replacing $r_4$

I get this matrix

$$A= -
\begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & -12 & 1 & 16\\
0 & 0 & 6 & -4
\end{bmatrix}
$$

I then did $4\cdot r_2 + r_3$ to replace $r_3$ and got

$$A= -
\begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 21 & -16\\
0 & 0 & 6 & -4
\end{bmatrix}
$$

I then did $-21\cdot r_4 + 6\cdot r_3$ to replace $r_4$ and got

$$A= -
\begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 21 & -16\\
0 & 0 & 0 & -12
\end{bmatrix}
$$

I am not sure if I did this correctly but the determinant of the matrix should be -36. When I multiply the diagonal entries it isn't -36. I can't figure out what I am doing wrong.

"I then did -21*row 4 + 6*row 3 to replace row 4 and got"

This is a determinant altering operation and not an elementary operation.

Don't write that $A$ equals something which isn't $A$.

Picking up where you errored and using the same idea you had one gets:

$$\begin{align} \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 21 & -16\\
0 & 0 & 6 & -4
\end{bmatrix}&\leadsto \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 6\cdot 21 & -6\cdot 16\\
0 & 0 & -21\cdot 6 & (-21)\cdot (-4)
\end{bmatrix}\\
&\leadsto \begin{bmatrix}
1 & -4 & 0 & 6\\
0 & 3 & 5 & -8\\
0 & 0 & 6\cdot 21 & -16\\
0 & 0 & 0 & -12
\end{bmatrix}_.\end{align}$$

Making the proper compensation yields
$$\det(A)=-\dfrac{1\cdot 3\cdot (6\cdot 21)\cdot (-12)}{-21\cdot 6}=-36.$$

"I then did −21⋅r4+6⋅r3 to replace r4 and got"..

Whenever we are doing Row operations on a particular row, whatever the co-efficients that we are multiplying on the same row should be taken as a divisor with sign outside of determinant.
For example,
You have a matrix A and it's determinant is |A|

If we are doing the below operation,
R3 -> 3 R2 - 5R3
The operation shout be processed by taking (-1/5) outside.
Concept is, we are indirectly multiplying the Row 3 by (-5) through this operation. We no need to worry about the multiplier 3 with R2, as it will not affect the determinant value(we are altering the Row 3, hence remaining Row's co-efficients will not affect |A|).

In your case, Till the final step, the Row operations didn't have co-efficients for the particular Row alterations.
In final step you did,
R4 -> -21 R4 + 6 R3
So, you should take (-1/21) outside.
After taking outside, the determinant calculation will be like:

|A| = - (-1/21) (1) (3) (21) (-12) = -36


Given $x+y$ and $x\cdot y$, what is $x^3+ y^3$ ? I have been looking at an assortment of high school number sense tests and I noticed a reoccurring problem that states what x+y is and what $x\cdot y$ is then asks for $x^3+ y^3$. I want to know how to work these problems. I have a couple of examples.

$x+y=5$ and $x\cdot y=1$, then $x^3+y^3=?$ [key says 110]

$x+y=-1$ and $x\cdot y=2$, then $x^3+y^3=?$ [key says 5]

$x-y=-1$ and $x\cdot y=2$, then $x^3 -y^3=?$ [key says -7]

$x+y=\frac{1}{3}$ and $x\cdot y=\frac{1}{9}$, then $x^3+y^3=?$ $\left[-\dfrac{2}{27}\right]$

If the problem cannot be reduced to the symmetric case then Groebner basis can be used to solve the problem. But the calculation by hand is very cumbersome so I use Maxima and got the following output. poly_reduced_grobnercalculates a Groebner base and poly_pseudo_dividedoes the division. The first element of the result list is the list of coefficients the second and the third are the numerator and the denomiantor of the remainder. This is the number we are looking for.

  (%i4)                            load(grobner)
  Loading maxima-grobner $Revision: 1.6 $ $Date: 2009-06-02 07:49:49 $
  (%i5) poly_reduced_grobner([-5+y+x,x*y-1],[x,y])
  (%o5) [y+x-5,y^2-5*y+1]
  (%i6) poly_pseudo_divide(y^3+x^3,%,[x,y])
  (%o6) [[y^2-x*y-10*y+x^2+5*x+25,15],110,1,7]
  (%i7) poly_reduced_grobner([1+y+x,x*y-2],[x,y])
  (%o7) [y+x+1,y^2+y+2]
  (%i8) poly_pseudo_divide(y^3+x^3,%,[x,y])
  (%o8) [[y^2-x*y+2*y+x^2-x+1,-3],5,1,7]
  (%i9) poly_reduced_grobner([1+y+x,2+x*y],[x,y])
  (%o9) [y+x+1,y^2+y-2]
  (%i10) poly_pseudo_divide(y^3+x^3,%,[x,y])
  (%o10) [[y^2-x*y+2*y+x^2-x+1,-3],-7,1,7]
  (%i11) poly_reduced_grobner([(-1)/3+y+x,x*y-1/9],[x,y])
  (%o11) [3*y+3*x-1,9*y^2-3*y+1]
  (%i12) poly_pseudo_divide(y^3+x^3,%,[x,y])
  (%o12) [[9*y^2-9*x*y-6*y+9*x^2+3*x+1,3],-2,27,7]


Is every commutative binary operation on a set with two elements associative? - Fraleigh pp27 2.28 Prove or disprove: Every commutative binary operation on a set having just $2$ elements is associative. This is from John Fraleigh's 7th edition of A First Course in Abstract Algebra.

First of all, how did they realize this is false?

Second, how did they construct this counterexample? How did they know to start with $(a * a) * b$ and $a * (a * b)$ and know the problem is with these two ? My work here looks too complicated.

The binary operation here is given as commutative, hence $\color{green}{a*b = b*a}$.
For $a * (a * b)$, WLOG, suppose  $\color{green}{a*b = b*a = a}$.  Then $a * \color{green}{(a * b)} = a*\color{green}{a}$.

For $(a * a) * b$, WLOG, suppose $\color{blue}{a*a = b}$. Then $\color{blue}{(a*a)} * b = \color{blue}{b} * b$.

Then $\color{blue}{(a*a)} * b \neq a *\color{green}{(a * b)} \implies \color{blue}{b} * b \neq a*\color{green}{a} $.
I already picked $\color{blue}{a*a = b}$ so need $\color{blue}{b} * b = a$ for a counterexample.

While the example can be constructed easily enough by reasoning of the sort that you used, there is another very straightforward possibility. Since the operation is commutative, we must have $a*b=b*a$. There are $2$ independent choices each for $a*a$, $a*b$, and $b*b$, and once these are made, the operation is defined. Thus, there are only $2^3=8$ commutative binary operations on $\{a,b\}$, and it’s possible simply to write them all out and check whether they are associative.

For $(a\cdot b)\cdot a$,let $a\cdot b=b\cdot a=a$,then $(a\cdot b)\cdot a=a\cdot a$
But $a\cdot(b\cdot a)=a\cdot a$.how say associative is false?


Show that: $f(x)$ is a constant function Let $f(x)$ be a continuous function on $\mathbb{R}$, such that for any real number $x$ we have:
$$\lim_{h\to 0}\dfrac{1}{h^3}\int_{-h}^{h}f(x+t)\cdot t\,dt=0.$$
Show that: $f(x)$ is a constant function.
Maybe use the following lemma? But I can't it, thanks.
Lemma. If $g$ is a continuous function, then
$$\lim_{h\to0}\frac{1}{2\,h}\int_{x-h}^{x+h}g(s)\,ds=g(x).$$
Proof. We may assume $h>0$.
$$
\left|g(x)-\frac{1}{2\,h}\int_{x-h}^{x+h}g(s)\,ds\right|=\frac{1}{2\,h}\left|\int_{x-h}^{x+h}(g(x)-g(s))\,ds\right|\le\frac{1}{2\,h}\int_{x-h}^{x+h}\left|g(x)-g(s)\right|\,ds.$$
Use that $g$ is continuous at $x$ to show that the last expression converges to $0$ as $h\to0$.

Define $F (x):= \int_{0}^x f(t)dt$
By integration by parts, we observe that:
$$\begin{align}\frac{1}{h}\int_{-h}^h f(x+t)tdt &= \frac{1}{h}\int_{0}^h \big[ f(x+t)-f(x-t)]tdt\\& = F(x+h)+F(x-h)  -  \frac{1}{h}\int_{0}^h \big[ F(x+t)+F(x-t) \big]dt
\end{align}$$
So the provided limit for f implies that for all $x \in \mathbb{R}$
$$T(F)(x)=0$$
Where  $T$ is an operator defined as
$$T(g)(x)=\lim_{h \rightarrow 0^+} \frac{1}{h^2} \left\{ g(x+h)+g(x-h)-\frac{1}{h}\int_{0}^h \big[ g(x+t)+g(x-t)\big]dt \right\} $$
for any continuous function $g$.
Note that $T(g)$ is not necessarily defined on all points in $\mathbb{R}$.
Let $\epsilon$ be any positive number , $(a,b)$ be any interval in $\mathbb{R}$. Define the functions
$$G(x)= F(x)-F(a)-\frac{F(b)-F(a)}{b-a}(x-a)+\epsilon(x-a)(x-b)$$
and
$$H(x)= F(x)-F(a)-\frac{F(b)-F(a)}{b-a}(x-a)\boldsymbol{-}\epsilon(x-a)(x-b)$$
We prove that $G(x) \le 0$ everywhere on $[a,b]$. If not then, because $G(a)=G(b)=0$, there is a point $c \in (a,b)$ at which $G$ attains its positive maximum. Observe that:
$$T(G)(c)= \lim_{h \rightarrow 0^+} \frac{1}{h^2} \left\{ G(c+h)+G(c-h)-\frac{1}{h}\int_{0}^h \big[ G(c+t)+G(c-t)\big]dt \right\} $$
Because $c$ is a maxima of $G$, the function $u(t):=G(c+t)+G(c-t)$ attains its local maximum  at $t=0$. So we can choose a strictly decreasing positive sequence $(h_n, n \in \mathbb{N})$ such that $h_n$ converge to $0$ and $h_n \in \text{argmin}_{ t \in [0,h_n]} u(t)$ for all $n$ . Thus,
$$ G(c+h_n)+G(c-h_n)-\frac{1}{h_n}\int_{0}^{h_n} \big[ G(c+t)+G(c-t)\big]dt \le 0$$
Thus $T(G)(c) \le 0$. Note that the limit $T(G)(c)$  exists because $T(F)$ exists and in fact, according to the definition, we see that:
$$T(G)(c)=0+0+\frac{4}{3}\epsilon>0 $$
Which is a contradiction to what we have just proven.
So $G(x) \le 0$ on $[a,b]$. Similarly, we can prove that $H(x) \ge 0$ everywhere on $[a,b]$. These give:
$$ \left|   F(x)-F(a)-\frac{F(b)-F(a)}{b-a}(x-a) \right| \le \epsilon (b-a)^2$$
for all $\epsilon>0$ and $x \in [a,b]$. Hence forth the conclusion.
Side note: The hypothesis is fairly tight as we can show that the Cantor function satisfies the condition on an almost everywhere set of $\mathbb{R}$ even though it is not constant.

If $f(x)$ is continuous in $\textbf{R}$, then we can write
$$
\int^{h}_{0}f(x+t)tdt-\int^{-h}_{0}f(x+t)tdt=\int^{h}_{0}(f(x+t)-f(x-t))tdt.
$$
Hence from remark below (and for all $x$ real) we have
$$
0=\lim_{h\rightarrow 0}\frac{1}{h^3}\int^{h}_{0}(f(x+t)-f(x-t))tdt=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x-h)}{3h}.\tag 1
$$
Clearly about (1) we have:
$$
\lim_{h\rightarrow 0}h^{-1}\int^{h}_{0}(f(x+t)-f(x-t))tdt=0
$$
Set
$$
F(y):=y^{-1}\int^{y}_{0}(f(x+t)-f(x-t))tdt,
$$
then $F(y)$ is continuous and differentiatable in $\textbf{R}$ with $F(0)=0$ and $F'(0)=0$. This last follows from the fact that $F(y)$ is differentiable in all $\textbf{R}-\{0\}$ and from problem condition:
$$
\lim_{y\rightarrow 0}\frac{F(y)}{y}=0.
$$
Also:
$$
\lim_{y\rightarrow 0}F'(y)=\lim_{y\rightarrow 0}\left(f(x+y)-f(x-y)-\frac{F(y)}{y}\right)=0.
$$
All that since
$$
\lim_{h\rightarrow 0}\frac{1}{h^3}\int^{h}_{0}\left(f(x+t)-f(x-t)\right)tdt=0.\tag 2
$$
Using the condition of the problem and Cauchy's mean value theorem (see below) we have
$$
0=\lim_{y\rightarrow 0^{+}}\frac{F(y)}{y^2}=\lim_{y\rightarrow 0^{+}}\frac{F'(y)}{2y}=
$$
$$
=\lim_{y\rightarrow 0^{+}}\left(\frac{f(x+y)-f(x-y)}{2y}-\frac{1}{2y^3}\int^{y}_{0}(f(x+t)-f(x-t))tdt\right)=
$$
$$
=\lim_{y\rightarrow 0^{+}}\frac{f(x+y)-f(x-y)}{2y}=(f_{s})'_{+}(x).
$$
Also in the same way
$$
\lim_{y\rightarrow 0^{-}}\frac{f(x+y)-f(x-y)}{2y}=(f_{s})'_{-}(x).
$$
and $(f_s)'_{-}(x)=(f_s)'_{+}(x)=f_s'(x)=\lim_{y\rightarrow 0}\frac{f(x+y)-f(x-y)}{2y}$ exists. However there exists the
Theorem.
I) If for a function $f(x)$ the left $f'_{-}(x)$ and right $f'_{+}(x)$ derivatives exists, then $f'_s(x)$ exist and $f'_s(x)=\frac{1}{2}(f'_{-}(x)+f'_{+}(x))$.
II) If $f_s'(x)$ exist then it is not necessary that $f'(x)$ exist.
Hence we arrive to the result that
$$
f'_s(x)=0\textrm{, }\forall x\in\textbf{R}.\tag 3
$$
However (3) is not so well established since (5) is not applied to functions that  change sign infinite times in a neighborhood of $0$. For example
$$
F(y)=y^3\sin\left(\frac{1}{y}\right).
$$
The true argument L'Hospital is that
$$
\textrm{liminf}_{h\rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}\leq 0\textrm{ and }\textrm{limsup}_{h\rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}\geq 0.\tag 4
$$
Notes.
From Cauchy mean value theorem in $[0,y]$, we have that exists always $\xi=\xi(y)$: $0<\xi(y)<y$ such that
$$
\frac{F(y)-F(0)}{y^2-0}=\frac{F'(\xi)}{2\xi}.
$$
Hence when $y\rightarrow 0^{+}$, from the continuity of
$$
\frac{F'(t)}{2t}=\frac{f(x+t)-f(x-t)}{2t}-\frac{1}{2t^3}\int^{t}_{0}(f(x+t_1)-f(x-t_1))t_1dt_1,
$$
in $(0,y)$ and knowing that $\lim_{y\rightarrow 0^{+}}\frac{F(y)}{y^2}=0$, $\lim_{y\rightarrow 0^{+}}\xi(y)=0$ we have
$$
0=\lim_{y\rightarrow 0^{+}}\frac{F(y)}{y^2}=\lim_{y\rightarrow 0^{+}}\frac{F'(\xi(y))}{2\xi(y)}=\lim_{t\rightarrow 0^{+}}\frac{F'(t)}{2t}.\tag 5
$$
The same way is used when $y\rightarrow0^{-}$


Can we express median in terms of standard operations? Given a list of integers $a_1,a_2,...,a_n$ where $n \geq 1$ is odd (may not be in ascending order), is it possible to express their median in terms of some well-known operations such as maximum or minimum?

For example, if we have three numbers, say $3,6,5$, then their median is always the one which is neither the maximum nor minimum. In the example, their median is $5$.

Yes! Write $n=2k+1$. Then the median is the maximum of the minima of all $k+1$ element subsets of the data.

For example, when $n=3$, then there are $\binom{3}2=3$ two element subsets of the data: $\{a_1,a_2\},\{a_1,a_3\},\{a_2,a_3\}$. The median is the max of all these minima:
$$
\text{median}(a_1,a_2,a_3)=\max\big(\min(a_1,a_2),\min(a_1,a_3),\min(a_2,a_3)\big)
$$
When $n=5$, there are $\binom53=10$ three element subsets of the data, so
$$
\text{median}=\max\Big(\min(a_1,a_2,a_3),\min(a_1,a_2,a_4),\min(a_1,a_2,a_5),\min(a_2,a_3,a_4),\dots,\min(a_3,a_4,a_5)\Big)
$$
As you can see, this formula quickly gets unwieldy. The number of terms is $\binom{n}{(n+1)/2}$, which grows exponentially.

$F(k)=\sum_{r=1}^n |k-a_r| $, (where k is a variabe , $a_r>a_{r+1}$)

If n is odd ,n=2m+1

$k>a_1\Rightarrow$ all mods open as $k-a_r$ and so derivative is >0(as you increase 'k' the value of function increase).

$a_1>k>a_2$ then one mod only open as $a_r-k$(r=1) ,  and others as $k-a_r$, hence still derivative >0

$a_m>k>a_{m+1} $ ,here m+1 mods open as $k-a_r$ and m mods open as $a_r-k$ ,so derivative is greater than 0

So obviusly derivative becomes negative if $k<a_{m+1}$ and derivative is greater than 0 if $k>a_{m+1}$ , hence minimum of the function is at $a_{m+1}$.

If n is even ,n=2m

You will find that derivative is >0 if , $k>a_{m}$

Derivative <0 if $k<a_{m+1}$

Derivative =0 if $ a_m>k>a_{m+1}$ , so the function is minimum for all such $k$(including $a_m$ and $a_{m+1}$).

So for even number of terms we define mode as $\frac{a_m+a_{m+1}}{2}.$


Proof that maximizing a function is equivalent to minimizing its negative The statement that maximizing a function over its argument is equivalent to minimizing that function over the same argument with a sign change seems to be accepted as trivial wherever I look (MSE, proofwiki, textbooks outside of optimization theory).

Intuitively, if you have some function of a single variable that has a global maximum, and you "flip it over" by changing the sign, the global maximum is now a global minimum.

However, it seems to me that math is all about a meticulous examination of surprising subtleties. Does anyone know of a good way to prove this statement?

$f$ is maximized at $x$ if $f(x)\geq f(y)$, for all $y$. Similarly, $−f$ is minimized at $x$ if $−f(x) \leq −f(y)$, for all $y$.

Let's now add to the left and right sides of the inequality $−f(x) \leq −f(y)$ respectively $f(y)-f(y) = 0$ and $f(x)-f(x) = 0$, that is

$$(f(y)-f(y))-f(x) \leq (f(x)-f(x))-f(y)$$

which can be written as

$$f(y) + (-f(y)-f(x)) \leq f(x) + (-f(x)-f(y))$$

we can remove $(-f(y)-f(x))$ from both sides to obtain

$$f(y) \leq f(x) $$

This proof followed from the ordering property of real numbers.

Mathematically, we can say $\min f(x) = - \max(-f(x))$


Calculate $\int \frac{dx}{x\sqrt{x^2-1}}$ I am trying to solve the following integral

$$\int \frac{dx}{x\sqrt{x^2-1}}$$

I did the following steps by letting $u = \sqrt{x^2-1}$ so $\text{d}u = \dfrac{x}{\sqrt{{x}^{2}-1}}$ then

\begin{align}
&\int \frac{\sqrt{x^2-1} \, \text{d}u}{x \sqrt{x^2-1}} \\
&\int \frac{1}{x} \text{d}u \\
&\int \frac{1}{\sqrt{u^2+1}} \text{d}u\\
\end{align}

Now, this is where I am having trouble. How can I evaluate that? Please provide only hints

Thanks!

EDIT:

The problem specifically states that one must use substitution with $u = \sqrt{x^2-1}$. This problem is from the coursera course for Single Variable Calculus.

$$
\int\frac{dx}{x\sqrt{x^2+1}} = \int\frac{x\,dx}{x^2\sqrt{x^2+1}} = \int\frac{1}{x^2\sqrt{x^2+1}}
\Big(x\,dx\Big)
$$

The big parentheses are of course a hint that what's inside them is to become $du$, or a constant times $du$.  But should $u$ be $x^2$ or $x^2+1$?  Either way, $\displaystyle\Big(x\,dx\Big)$ becomes $\displaystyle\Big( \frac12\,du\Big)$.  I think usually it's better to have the thing under the radical be simple, so I'll say $u=x^2+1$, and we have
$$
\frac12\int\frac{du}{(u-1)\sqrt{u}}.
$$
We can rationalize $\sqrt{u}$ by letting
\begin{align}
w & = \sqrt{u} \\
w^2 & = u \\
2w\,dw & = du
\end{align}
and we have
$$
\frac12\int\frac{2w\,dw}{(w^2-1)w} = \int\frac{dw}{w^2-1}.
$$
Then use partial fractions, getting
$$
\int\left(\frac{A}{w-1}+\frac{B}{w+1}\right)\,dw
$$
and you need to figure out what $A$ and $B$ are.

That works, but a trigonometric substitution also comes to mind.  The expression $\sqrt{x^2+1}$ should remind you of $\sqrt{\tan^2\theta+1}= \pm\sec\theta$, and if it doesn't remind you of that, that's something to work on.  Review some trigonometry and trigonometric substitutions.  If $x=\tan\theta$ then $dx=\sec^2\theta\,d\theta$, and we have
$$
\int\frac{\sec^2\theta\,d\theta}{\tan\theta\sec\theta} = \int\frac{\sec\theta\,d\theta}{\tan\theta} = \int\csc\theta\,d\theta.
$$
That's a hard one to do from scratch, but it's also one that you can look up in standard tables.

When $x =\sin u$, $\displaystyle\int\frac{dx}{x\sqrt{x^2 -1}} \, dx$ becomes $\displaystyle\int\frac{\cos u \, du}{\sin u\sqrt{-\cos^2u}}$.


Show that rank($A^{n+1}$) = rank($A^n$) Suppose $A$ is a $n \times n$ matrix i.e. $A \in \mathbb{C}^{n \times n}$, prove that rank($A^{n+1}$) = rank($A^n$). In other words, I need to prove that their range spaces or null spaces are equal. If it helps, $A$ is a singular matrix.
Note that, I don't want to use Jordan blocks to prove this. Is it possible to prove this without using Jordan form? I can use Schur's triangularization theorem. Also, it's not known if A is diagonalizable.

$\newcommand{\rg}{\operatorname{range}}$
Obviously for every $m$, $\rg A^{m+1}\subset\rg A^{m}$, so if $d_m=\dim\rg A^m$, $d_{m+1}\le d_m$. If $d_{m+1}=d_m$ for some $m$, then $\rg A^{m+1}=\rg A^{m}$ and therefore $\rg A^m=\rg A^{m+1}=\rg A^{m+2}=\dotsb{}$. That is, the sequence $d_0,d_1,\dots$ becomes constant once it stops descending.

Because $d_0= n$, the sequence must stop descending within $n$ terms.
Edit: For the problem you phrased in the comment, $\rg A^{m+1}=\{AA^{m}y:y\in \mathbb C^n\}=\{Ax:x=A^my\in\rg A^m\}=\{Ax:x\in\rg A^m\}$,
Therefore  $\rg A^{m}=\rg A^{m+1}\implies$
$\rg A^{m+1}=\{Ax:x\in\rg A^m\}=\{Ax:x\in\rg A^{m+1}\}=\rg A^{m+2}$.

Every matrix corresponds to a linear map. Suppose that $A=\mathcal{M}(T)$, where $T\in \mathcal{L}(V)$ and $dim (V)=n$. Using Theorem 8.4 in Axler,
$$Null(T^n)=Null(T^{n+1}).$$
According to the Fundamental Theorem of Linear Maps,
$$n=dim Null(T^n)+ dim Range(T^n)$$
$$n=dim Null(T^{n+1})+dim Range(T^{n+1})$$
$dim Range(T^n)=dim Range(T^{n+1})$ implies that $rank(A^n)=rank(A^{n+1})$.


How to tell if I'm good enough for graduate school? I'm concerned about my level of preparation for graduate school. I have decent grades (3.64 GPA, 4.0 in math), great recommendations lined up, and some OK research experience (no publications), but I really don't feel like I'm ready for the next step.

I'm interested in applied math, and was a physics/computer science double major for most of undergrad, switching into math just last semester. I'm taking Analysis 1 now as a first semester senior, and it is the only pure math class I've ever taken (though I've taken a bunch of applied courses and semi-pure courses, like advanced udergrad ODEs). I'm doing very well in the course, and I'm really enjoying it, but I feel that, because I'm so late in joining mathematics, I'll be really behind everyone when I get to graduate school, if I do get in somewhere decent. My scores on practice tests for the math GRE are in the mid 600's, and that makes me feel like a phony as well (although I haven't really studied that hard yet, maybe they'll go up? Also, its a good thing many applied programs don't care about the subject test...). Sure, I have some skills from my background that other students don't have, such as non-trivial software development and an ability to solve difficult physics problems, but when I look at the profiles of the 'good' people on this website, or talk to my friends that have been doing math olympiads since they were 3, I feel like there is just no way I can be good enough for a career in mathematics given my 'late' start.

Is it possible to catch up with others at this stage? Will I be doomed to a life of constantly being behind those who are smarter than me? Has anyone ever built a strong career from my kind of situation before?

I just wish I had some more time...

At my university we use baby Rudin, Dummit and Foote, and Munkres in the first year analysis, algebra, and topology courses (respectively), which is the average schedule for an incoming grad student.  You can also test out of these and move to the 2nd year if you happen to be good in one of those fields (or two, if you're really exceptional).  My opinion is that most math majors could handle this schedule given an appropriate amount of studying.  Worst case you can opt out at a master's if you decide you don't want to do research.

As far as being stupider than other people goes: of course you are.  Unless you're the smartest person on earth, there are smarter people, and if you spend your time sitting around thinking about who's smarter than who, they're all you're going to pay attention to. If you're getting a math degree, you're probably pretty smart, and that's good enough.

I embraced my own stupidity long ago, and in doing so, I gained the ability to ask questions without losing face.  Instead of worrying about other students knowing I'm not a genius, I can talk my problems out with them, gain their insight, and learn from my own mistakes.  For whatever reason, people in the mathematical field have this tendency to try to convince you that you're too dumb to do anything. Don't listen to them.  If I miss a proof, I'll get mad, reread the chapter, work book problems, harass the professor, and stop sleeping until I understand everything. Be stubborn, not smart. You've got to have grit.

You just take courses and if they are too hard you take them again or take easier courses. I will suggest if you ask for math you can prepare yourself with taking this course it probably the best out there it wont make you Gaus but it will soften the conceptual barrier which is what you want to do.

The course have started but if you have some time you can go through it and also Kieth is very nice, your goal should be to be conceptually prepared, but a university basically is an institute that suppose to arrange you the condition to overcome that state in its essence:
https://www.coursera.org/course/maththink


How can I find the integer solutions to $x^2+x-2y^2=0$? I enter this equation in Wolfram Alpha :   $x^2+x-2y^2=0$
and it gave me something like this :

and I am wondering how this solution is found and how to know if a given equation would guarantee to have integer solution.

Can anyone teach me what's behind the hood?

Thanks.

Hint:
Rewrite the equation: $x^2+x=2y^2$, that is
$$x(x+1)=2y^2$$
Notice that $(x,x+1)=1$, so we can write $y^2$ as $2a^2b^2$, here $(a,b)=1$ and $2\nmid b$.
Then we get
$$\left\{\begin{array}\\x=2a^2\\x+1=b^2\end{array}\right.\text{ or }\left\{\begin{array}\\x+1=2a^2\\x=b^2\end{array}\right.$$
Hence,
$$b^2-2a^2=\pm 1$$
So we get two Pell-like equations. And you know what's going on then?

Expanding on answerr by @lab bhattacharjee:

z^2 = 8.y^2 + 1 gives us equation of form x^2 = N.y^2 + 1

which can be done simply https://en.wikipedia.org/wiki/Chakravala_method.

started in 6th century CE India


Solve the following equation for x. $$\frac{x}{m} - \frac{1}{n} = \frac{3x}{mn}$$

I've tried $$\frac{x}{m} - \frac{1}{n} = \frac{3x}{mn} // \frac{x}{m} x \frac{n}{n} = \frac{3x}{mn} // \frac{xn - m}{mn} = \frac{3x}{mn}$$ This is where I get stuck. I need a tutor!

Your original equation is this and you are asked to isolate $x$;

$$\frac{x}{m} - \frac{1}{n} = \frac{3x}{mn}.$$

If you place the terms on the left hand side over a common denominator you will get this;

$$\frac{nx}{mn} - \frac{m}{mn} = \frac{3x}{mn}.$$

Since the terms on the left hand side have a common denominator, you can rewrite this as follows;

$$\frac{nx - m}{mn} = \frac{3x}{mn}.$$

You now have the left hand side and right hand side over a common denominator and can multiply through by $mn$ to get

$$nx - m = 3x.$$

Now, this is a lot simpler; collect the $x$ terms, factorise and isolate.

$$nx - m = 3x \implies nx - 3x = m$$

Factorising gives;

$$x(n - 3) = m.$$

Finally, divide through by $(n - 3)$;

$$x = \frac{m}{n -3}.$$

It is important to note that this is only valid for the case where $n \neq 3$ since this would give us an undefined answer.

Happy mathing!

You can easily see that $x= \frac{m}{n-3}$.


$\Bbb{Q}(\sqrt{2})$ is a field extension of $\Bbb{Q}$. Problem: $\Bbb{Q}(\sqrt{2})$ is a field extension of $\Bbb{Q}$.
Solution: Let $F=\Bbb{Q}$ be a field of rational numbers and take $p(x)=x^2-2$, irreducible over $\Bbb{Q}$ By Eisenstein's Criterion.
Define a map $\psi:F\to {F[x]}/{\left<x^2-2\right>}$ by $\psi(a)=a+\left<x^2-2\right>$, for $a\in F$.
This map makes sense. because $\left<x^2-2\right>$ is irreducible over $F$. Thus $\left<x^2-2\right>$ is a maximal ideal. Therefore ,${F[x]}/{\left<x^2-2\right>}$ is a field.
Clearly $\psi$ is injective ring homomorphism. Therefore, ${F[x]}/{\left<x^2-2\right>}$ is a field extension of $F$. Of course I know that $\Bbb{Q}(\sqrt{2})\cong \Bbb{Q}[x]/<x^2-2>$.
Is it enough to show that for extension?

I am writing this answer because it answers also two to three questions. It would be helpful for others who looks answers like$\Bbb{Q}(\sqrt{2})\cong \Bbb{Q}[x]/<x^2-2>$.
Let $F=\Bbb{Q}$ be a field of rational numbers and take $p(x)=x^2-2$, irreducible over $\Bbb{Q}$ By Eisenstein's Criterion.
Define a map $\psi:F\to {F[x]}/{\left<x^2-2\right>}$ by $\psi(a)=a+\left<x^2-2\right>$, for $a\in F$.
This map makes sense because $\left<x^2-2\right>$ is irreducible over $F$. Thus $\left<x^2-2\right>$ is a maximal ideal. Therefore ,${F[x]}/{\left<x^2-2\right>}$ is a field.
Clearly $\psi$ is injective ring homomorphism. If we'll show that ${F[x]}/{\left<x^2-2\right>}\cong F$. Then $\Bbb{Q}(\sqrt{2})$ is extension of $\Bbb{Q}$.
So first we will show  that $\Bbb{Q}(\sqrt{2})\cong \Bbb{Q}[x]/<x^2-2>$.
Define a map $\psi:\Bbb{Q}\sqrt{2}\to \Bbb{Q}[x]/<x^2-2>$ by $\psi(a+b\sqrt{2})=[a+bx]$ for any $a,b\in\Bbb{Q}$.
$\psi$ is homomorphism:
$\psi((a+b\sqrt{2})+(c+d\sqrt{2}))=\psi((a+c)+(b+d)\sqrt{2}))=[(a+c)+(b+d)x]=[a+bx]+[c+dx]=\psi(a+b\sqrt{2})+\psi(c+d\sqrt{2}$.
$\psi((a+b\sqrt{2})(c+d\sqrt{2})=\psi((ac+2bd)+(ad+bc)\sqrt{2}))=[(ac+2bd)+(ad+bc)x]$.
and,
$\psi((a+b\sqrt{2})\psi(c+d\sqrt{2})=[a+bx][c+dx]=[ac+adx+bcx+bdx^2]=[ac+(ad+bc)x+bdx^2]=[(ac+2bd)+(ad+bc)x]$.
Therefore, $\psi$ is homomorphism.
$\psi$ is injective:
Let, $\psi(a + b\sqrt{2}) = \psi(c + d\sqrt{2})$, then $[a + bx] = [c + dx]$. Since there is a
unique polynomial of degree $1$ or less for each congruence class $\operatorname{mod} p(x)$.
$\Rightarrow$  $a + bx = c + dx$ $\Rightarrow$ $ a = c$ and $b = d$. Hence $\psi$ is injective.
$\psi$ is surjective:
Let $y\in \Bbb{Q}[x]/<x^2-2>$ then $y=[l+mx]$ for some $l,m\in \Bbb{Q}$. Therefore $\psi(l+m\sqrt{2})=y=[l+mx]$.
$\Rightarrow$  $\psi$ is surjective.
And hence $\psi$ is isomorphism. Thus $\Bbb{Q}(\sqrt{2})\cong \Bbb{Q}[x]/<x^2-2>$. Thus $\Bbb{Q}(\sqrt{2})$ is extension of $\Bbb{Q}$.

$\Bbb{Q}(\sqrt{2})$ is a field extension of $\Bbb Q$ by definition.
The definition of it is that it is the field generated by adjoining $\sqrt{2}$ to $\Bbb Q$ and all sums products and inverses of it and the numbers it generates.


How to evaluate $\int_{0}^{\infty} \frac{x^{-\mathfrak{i}a}}{x^2+bx+1} \,\mathrm{d}x$ using complex analysis? We were told today by our teacher (I suppose to scare us) that in certain schools for physics in Soviet Russia there was as an entry examination the following integral given

$$\int\limits_{0}^{\infty} \frac{x^{-\mathfrak{i}a}}{x^2+bx+1} \,\mathrm{d}x\,,$$

where $a \in \mathbb{R}$, $b \in [0,2)$, and $\mathfrak{i}$ is the imaginary unit. And since we are doing complex analysis at the moment, it can, according to my teacher, be calculated using complex methods.

I was wondering how this could work? It seems hard to me to find a good curve to apply the residue theorem for this object, I suppose. Is there a trick to compute this integral?

$\phantom{}$ Dear MSE users, this is the new episode of Mister Feynman and Monsieur Laplace versus contour integration.

Tonight we have a scary integral, but we may immediately notice that
$$ \mathcal{L}(x^{-ia})(s) = s^{ia-1}\Gamma(1-ia),\qquad \mathcal{L}^{-1}\left(\frac{1}{x^2+bx+1}\right)(s) =\frac{e^{Bs}-e^{\overline{B}s}}{\sqrt{b^2-4}}$$
where $B$ is the root of $x^2+bx+1$ with a positive imaginary part. By the properties of the Laplace transform, the original integral is converted into
$$\frac{\Gamma(1-ia)}{\sqrt{b^2-4}}\int_{0}^{+\infty}s^{ia-1}\left(e^{Bs}-e^{\overline{B}s}\right)\,ds$$
which can be evaluated in terms of the $\Gamma$ function. Due to the reflection formula, the final outcome simplifies into
$$ \frac{\left(B^{-i a}-\overline{B}^{-i a}\right) \pi }{\sinh(\pi a)\sqrt{4-b^2}} $$
and  we may notice that $B=\exp\left[i\arccos\frac{b}{2}\right]$ allows a further simplification.

Outro: poor children.

In a complex analysis class, you’re more likely to come across the integral $$\int_{0}^{\infty} \frac{x^{\alpha}}{1+2x \cos \beta +x^{2}} \, dx = \frac{\pi \sin (\alpha \beta)}{\sin(\alpha \pi) \sin(\beta)},  \quad (-1<\alpha <1, \ 0 < \beta < \pi).$$

See this answer, for example, which uses a semicircle in the upper half-plane that is indented at the origin.

Alternatively, you could also use the keyhole contour that Shashi used.

Also see this question about the peculiar symmetry of this integral when it's written in a slightly different form.

The function on the right is not defined at $\alpha =0$. But you could show separately (by using the same contour or by completing the square) that the value of the integral at $\alpha =0$ is $$\lim_{a \to 0} \frac{\pi \sin (\alpha \beta)}{\sin(\alpha \pi) \sin(\beta)} = \frac{\beta}{\sin \beta}.$$

Now if we assign the function on the right the value $\frac{\beta}{\sin \beta}$ at $\alpha =0$, then right side of the equation is a holomorphic function for $-1 <\operatorname{Re}(\alpha) <1$ with $\beta$ fixed.

And since the integral on the left is absolutely convergent in the strip $-1 < \operatorname{Re}(\alpha) < 1$, we can use a property of the Mellin transform that states that the integral defines an holomorphic function in that strip.

(This is very similar to property of the Laplace transform mentioned here, and can be proved in essentially the same manner.)

So by the identity theorem, the formula holds for $-1 <\operatorname{Re}(\alpha) <1$.

Your integral is the case $\alpha=-ia$ and $\cos(\beta) = \frac{b}{2}$.

If $b\in [0, 2)$, then $\beta$ falls between means  $0$ and $\pi$, and we get

$$\begin{align} \int_{0}^{\infty} \frac{x^{-ia}}{x^{2}+bx+1} \, dx &= \frac{\pi \sin\left(-ia \arccos\left(\frac{b}{2} \right)\right)}{\sin(-ia \pi) \sin \left(\arccos \left(\frac{b}{2} \right) \right)} \\ &= \frac{\pi \sinh \left(a \arccos\left(\frac{b}{2} \right) \right)}{\sinh(a \pi)\frac{\sqrt{4-b^{2}}}{2}}. \end{align} $$


Problem in proof of: Show the order $d$ of $a$ modulo $m$ exists and $d\mid\phi(m)$ Theorem: Let $m\in\mathbb{N}$ and $a\in\mathbb{Z}$ satisfy $(a,m)=1$. Then the order $d$ of $a$ modulo $m$ exists, and $d\mid\phi(m)$.

Proof: By Euler's theorem, one has $a^{\phi(m)}\equiv 1\pmod{m}$, and so the order of $a$ modulo $m$ clearly exists.

Suppose then that $d$ is the order of $a$ modulo $m$, and further that $a^k\equiv 1\pmod{m}$.

Then it follows from the division algorithm that there exists integers $q$ and $r$ with $k=dq+r$ and $0\leq r<d$.

But then we obtain $a^k=(a^d)^qa^r\equiv a^r\equiv 1\pmod{m}$,

therefore $r=0$.

Thus we have $d\mid k$ and in particular $d\mid\phi(m)$

Point of contention: I understand that $a^k=(a^d)^qa^r$ and that $a^d\equiv 1\pmod{m}$ since it is the order of $a$ modulo $m$, therefore $(a^d)^q\equiv 1\pmod{m}$.

But I dont understand how $a^r\equiv 1\pmod{m}$

I understand the rest of the argument after this.

Since $(a^d)^q \equiv 1 \pmod{m}$ it follows that $a^r \equiv a^k \equiv 1 \pmod{m}$, where the second $\equiv$ is by hypothesis.

we have 1=a^d=a^k from def of k,d. And so 1=a^k=a^(dq+r)=a^(dq).a^r=1.a^r=a^r


Formula For Finding the Next Near Consecutive Perfect Square For any three consecutive members of a sequence, the first and third members are near consecutive.

1 squared is 1. 2 squared is 4. So 1 and 4 are consecutive perfect squares.

1 squared is 1. 3 squared is 9. So 1 and 9 are near consecutive perfect squares.

I want to verify this formula to go from any perfect square to the next near consecutive perfect square. Let a be any perfect square. Let c be the next near consecutive perfect square. Here is the formula:

a+4(1+$\sqrt a$)=c

Example:

49+4(1+$\sqrt 49$)=81

Did I get the formula right? Is my terminology ok?

Yes! Its right!
If your square is $a$ then the number that originate it is $\sqrt{a}$ thus the next near consecutive perfect square is $(\sqrt{a}+2)^2$, but $(\sqrt{a}+2)^2 = a + 4(\sqrt{a}+1)$

I have found there is a easier way to do this

(N+1)squared
And
(N-1)squared

Like for example
(1+1)squared is 4 then (4-1)squared is 9 then (9-1)squared is 18


Prove that $|e^{a}-e^{b}|<|a-b|$ Possible Duplicate:
  $|e^a-e^b| \leq |a-b|$

Could someone help me through this problem?
Let a, b be two complex numbers in the left half-plane. Prove that $|e^{a}-e^{b}|<|a-b|$

By mean value theorem,
$$ |e^a - e^b| \leqslant |a-b|\max_{x\in [a,b]} e^x $$
But $a$ and $b$ have a negative real part, and then all $x$ in $[a,b]$ also have a negative real part. Hence the $\max$ is less than one. And thus
$$ |e^a - e^b| &lt|a-b|. $$


What is the term for a factorial type operation, but with summation instead of products? (Pardon if this seems a bit beginner, this is my first post in math - trying to improve my knowledge while tackling Project Euler problems)

I'm aware of Sigma notation, but is there a function/name for e.g.

$$ 4 + 3 + 2 + 1 \longrightarrow 10 ,$$

similar to $$4! = 4 \cdot 3 \cdot 2 \cdot 1 ,$$ which uses multiplication?

Edit: I found what I was looking for, but is there a name for this type of summation?

The name for

$$
T_n= \sum_{k=1}^n k = 1+2+3+ \dotsb +(n-1)+n = \frac{n(n+1)}{2} = \frac{n^2+n}{2} = {n+1 \choose 2}
$$

is the $n$th triangular number. This picture demonstrates the reasoning for the name:

$$T_1=1\qquad T_2=3\qquad T_3=6\qquad T_4=10\qquad T_5=15\qquad T_6=21$$

$\hskip1.7in$

Not exactly a name, but note that
$$
\sum\limits_{k=1}^{n} k= \frac{n(n+1)}{2}={n+1 \choose 2}
$$


Uniform Random Variable I was given these two questions regarding Uniform Random Variable but I don't know how to think or approach these questions.

1) A stick of length 1 is uniformly distributed over [0,1]. There is a point 'p' on the stick, and we break the stick in two at a point 'u'. Determine the expected length of the piece that contains the point 'p'.

2) Your company must make a sealed bid for a construction project. If you succeed in winning the contract (by having the lowest bid), then you plan to pay another firm 100 thousand dollars to do the work. If you believe that the minimum bid (in thousands of dollars) of the other participating companies can be modeled as the value of a random variable that is uniformly distributed on (70, 140), how much should you bid to maximize your expected profit?

I was told these questions are similar, but I don't understand the approach. What type of questions are these regarded as?

Thank you!

A sensible interpretation of the first question is as follows. Let $0<p<1$ is a fixed point on the $[0,1]$ interval and let $U$ be uniformly distributed over $[0,1]$. Define $X$ like

$$X=\begin{cases}p&\text{ if } &U<p\\
1-p&\text{ if }& U\ge p.\end{cases}$$

The expectation of $X$ is

$$E[X]=p^2+(1-p)^2.$$

This value is minimal if $p$ is at the middle of the "stick."

Another interpretation is (Did):

$$X=\begin{cases}1-U&\text{ if } &U<p\\
U&\text{ if }& U\ge p.\end{cases}$$

In this case

$$E[X]=E[X\mid U<p]p+E[X\mid U\ge p](1-p)=$$
$$=E[1-U\mid U<p]p+E[U\mid U\ge p](1-p)=$$
$$=p-E[U\mid U<p]p+E[U\mid U\ge p]-E[U\mid U\ge p]p=$$
$$=p-p^2+\frac12$$

which takes its maximum at $\frac12$.

EDIT

As far as Question 2:

Let $x$ and $B$ denote our bid and the other bid, respectively. $B$ is uniformly distributed over $[70,140]$. Our profit is

$$PR=\begin{cases}
0&\text{ if }& B\leq x&\\
x-100&\text{ if }& B> x.&\end{cases}$$

The expected profit is then

$$E[PR]=(x-100)P(B>x)=(x-100)\frac{140-x}{70}$$
whose minimum can be easily calculated.


How can I speed up the search for a special number? A number $N$ is given. The object is to find the smallest nonnegative integer $k$, such that $N+k$ is the product of three distinct primes, each having the same number of decimal digits.

For example, for $N=33^{33}$, the smallest $k$ is $27484$ because $$33^{33}+27484$$ splits in three $17$-digit prime factors and no smaller number beyond $N$ has this property.

Of course, brute force finds such a number, and the first step is sieving out small factors, but is a significant speed-up possible ? In particular, I am interested in the solution for $N=44^{44}$


Laplace transform of noncentral chi-square distribution I'm interested in non central chi-square distribution. More specifically, i want to derive the laplace transform of noncentral chi-sqruae disribution or density function.

Let me know whether it exists or relevent book and paper.

Thanks in advance

If you mean the probability function for the chi-square distribution with $r$ degrees of freedom
$$P_r(x)=\frac{x^{-1+r/2}e^{-x/2}}{2^{r/2}\Gamma(r/2)}$$
The Laplace transform is :
$$\int_0^\infty P_r(x)e^{-sx}dx = \frac{(s+1/2)^{-r/2}\Gamma(r/2)}{2^{r/2}\Gamma(r/2)}= (2s+1)^{-r/2}$$

It is mentioned in the following paper:
http://link.springer.com/article/10.1007/BF02595410


Show that these metrics induces the same topology on X Let $X$ be the set of positive integers. Let $d_1$ be the usual metric space on $X$ and $d_2$ be the discrete metric on $X$. Define $d_3:X\times X \rightarrow R$ by $d_3(m,n)=|\frac{1}{m}-\frac{1}{n}|$ for $m,n\in X$. Prove that $d_3$ is also a metric on $X$ and $d_1,d_2,d_3$ all induce the same topology on $X$.

I've proved that $d_3$ is also metric.. But I could not proved that all of these induces the same topology. I know that, in order to prove that, these metrices induces that the same topology... I have to prove that all are equivalent metrices. I am stuck at this part.

I'll give one equivalence and edit this answer if OP struggles with the others:

Claim: $d_2 \tilde{} \space d_3$ with the notation as in the question.

Proof:

Observe that $d_3(m,n)= \vert \frac{1}{m}-\frac{1}{n} \vert \leq 1$ for all positive integers $m,n$. Therefore we have:

$$d_3(m,n) \leq d_2(m,n) \leq 2d_3(m,n)$$
since (w.l.o.g $n >m$):

$$ 2\vert \frac{1}{m}-\frac{1}{n} \vert=\frac{2}{m}-\frac{2}{n}=\frac{2(n-m)}{mn}= 1+\frac{2n-(mn+2m)}{mn}$$ and because

$$1 \geq \frac{m}{2}+ \frac{m}{n} \implies 2n \geq mn+2m$$

we can conclude the equivalence.

All metrics induce the discrete topology (i.e. all sets are open). Forget about any equivalence of metrics, this is a stronger requirement (which does not hold in this case) that is not needed for equivalence of topologies


Harmonics functions take minimum value on the boundary Let $u$ be harmonic on the bounded region $A$ and continuous on the closure of $A$. Then $u$ takes its minimum only on $\partial A$ unless $u$ is a constant.

How can we prove this result? Do we have to use the Maximum modulus principles? in my notes, the proof only says that considering $v = - u$, the conclusion follows. But, how? I dont see it.

$-u$ is a perfectly good harmonic function, so it obeys the mean value property. Also, it is clear that $-u$ attains a maximum when $u$ is at a minimum.

$\overline{A}$ is compact, so $v$ obtains a maximum on $\overline{A}$. Suppose $v:=-u$ attains this maximum at some point $z_0\in A^{\circ}$ (the interior of $A$). If $\varepsilon>0$ is small enough that $D(z_0,\varepsilon)\subset A$ (this is possible because $z_0$ is in the interior), then the mean value property dictates that

$$
v(z_0)=\frac{1}{2\pi}\int_{0}^{2\pi}v(z_0+\varepsilon e^{i\theta})\,d\theta.
$$

Now, the expression on the right-hand side is precisely the average value of $v$ over $\partial D(z_0,\varepsilon)$. $v(z_0)$ is assumed to be a local maximum, so it cannot possibly be the average value of other points in a neighborhood of it unless it is constant. Therefore $v$ is constant on $D(z_0,\varepsilon)$.

We just need to show that this extends to mean $v$ is globally constant. I believe this argument should work:

Let $z_1\in A^\circ$ be distinct from $z_0$. Then there exists some continuous curve $\gamma:(0,1)\to A^{\circ}$ such that $\gamma(0)=z_0$ and $\gamma(1)=z_1$ (here I'm assuming $A$ is connected--else the statement is not true). Let $m$ be the infimum of $\text{dist}(\gamma,\partial A)$. Since the image of $\gamma$ is compact, there exists a finite sequence of overlapping open disks $\{U_i\}$ with radius $\min\!\big(m/2,\varepsilon/2\big)$ such that $U_i\subset A^{\circ}$ and $\bigcup_iU_i$ covers the image of $\gamma$. If $U_0$ contains $z_0$, then $U_0\subset D(0,\varepsilon)$ and $v$ is constant on $U_0$. There is a further disk $U_1$ such that $U_0\cap U_1\neq\emptyset$ and, since $U_1$ has radius at most $\varepsilon/2$, the center of $U_1$ attains the same maximum. By the same argument $v$ is constant on $U_1$ as well. This process can be continued to show that $v$ is constant on every $U_i$, and hence is constant in a neighborhood of $z_1$. $z_1$ was chosen arbitrarily, so $v$ is constant on all of $A$.

The argument got a little technical, but the main idea is you take a finite sequence of disks--the first containing $z_0$ and the last containing $z_1$--each small enough so that they are all contained within $A$ and such that the center of each disk is contained within the previous disk (that way after showing $v$ is constant on one disk, it attains the maximum at the center of the subsequent disk, so the previous argument will immediately apply to show it is constant on that disk as well).

One could prove the minimum modulus principle via complex analysis and knowing it for holomorphic functions. I assume you realize the maximum modulus principle for holomorphic functions (or , if not, you can easily prove it knowing it for harmonic functions)
In order to prove this you should look at $g=\dfrac{1}{f}$ which is also holomorphic (since we assume $f\ne 0$) and use the maximum modulus principle.

Edit: I would take the downvotes as a hint for lack of explaination:
Harmonic functions can be viewed as a real or imaginary part of holomorphic functions. The maximum (minimum) modulus principles of holomorphic functions can be used to prove the same principles for harmonic functions.


If a coin is flipped 10 times, what percentage will start with heads If a coin is tossed 10x, out of all the possibilities, what percentage will start with heads?

Assuming you ignore the "land on its edge" cases, the answer is 50%.

If you mean heads plural then I would say $25$% cuz you have to get the first $2$ heads and that is $1/4$ chance.  The other $8$ don't really matter.


Limit with fractional part and greatest integer part Find $$\lim_{x\rightarrow 2-} \{x+(x-[x]^2)\}$$

For $x\to 2-$, $[x]=1$, i.e $[x]^2=1$,

so $\lim_{x\rightarrow 2-} \{x+(x-1)\}=\lim_{x\rightarrow 2-} \{2x-1\}=3.$

I don't know whether $\{\}$ symbolizes fractional part=$x-[x]$ or not for this particular question. I have assumed  $\{\}$ not as fractional part and solved. Please check my solution is correct or not.

  In case $\{\}$ symbolizes fractional part=$x-[x]$, then what will be the answer?

We claim that the limit as $x \to 2-$ is $1$; but:

If $0 < 2-x < 1/2$, then
$$
|\{ 2x - \lfloor x \rfloor^{2} \} - 1| = \lfloor 2x-1 \rfloor - 2x+2 = 4-2x = 2(2-x);
$$
given any $\varepsilon > 0$, we have $2(2-x) < \varepsilon$ if in addition $2-x < \varepsilon/2$. Hence, for every $\varepsilon > 0$, if $0 < 2-x < \min \{1/2, \varepsilon/2 \}$ then $|\{ 2x - \lfloor x \rfloor^{2} \} - 1| < \varepsilon$.

lim −{x+(x−[x]2)}=1, Because {2x-1} lies between 0 to 1, but in
x→2                  question
                     it was mentioned as 2 minus, hence it will
                     lead to 1.


Help With: $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ I have this problem, $2(2^x- 1)x^2 + (2^{x^2}-2)x = 2^{x+1} -2$, but am stuck on how to solve it. I need to find all off the real solutions, but I have really no idea how to start. I noticed that it is a quadratic (kind of) if you move all the terms to one side, with $2(2^x- 1)$ being a, $(2^{x^2}-2)$ being b, and $-2^{x+1}+2$ being c.
I still want to find the answer by myself, so can anyone help point me in the right direction?
Thank you.

$\DeclareMathOperator{\sign}{sign}$

  Lemma 1: $$\sign(2^z-1)=\sign(z)$$

Proof: $2^z-1$ is strictly increasing and $2^0-1=0$.

  Lemma 2: $$(2^a-1)b+(2^b-1)a=0\iff ab=0$$

Proof: By lemma 1, $\sign\left((2^a-1)b+(2^b-1)a\right)=\sign(ab)$.

  Claim: $$x=0,\pm 1$$

Proof: By algebraic manipulation, the original becomes:

\begin{align*}
2(2^x-1)x^2+(2^{x^2}-2)x&=2^{x+1}-2\\
(2^x-1)x^2+(2^{x^2-1}-1)x&=2^x-1\\
(2^x-1)(x^2-1)+(2^{x^2-1}-1)x&=0
\end{align*}

Therefore by lemma 2, $x(x^2-1)=0$.

Here's the graph of $2(2^x -1)x^2+(2^{x^2}-2)x - \left( 2^{x+1} - 2\right)$

$$x = \{ -1, 0, 1\}$$

Clearly the $x 2^{x^2}$ term dominates for large $|x|$, and guarantees that there are no more solutions, i.e., the set stated is maximal.


How is this a number sequence $58, 26, 16, 14, 10$ I recently had a IQ Test taken and we all got stuck on the same question. The question was:

What comes next in the following sequence?
$$58, 26, 16, 14,\_\_$$

The answer given in the answer sheet was $10$. My question is why? What pattern exists in those numbers?


Proof for the number of basic vectors are the same for any basis If $v_1, ... , v_m$ and $w_1, ... , w_n$ are both bases for the same vector space, then $m
=n$

Proof
Assume that there are more $w$'s than $v$'s, ie., $n>m$
$$
w_1=a_{11}v_1+....+a_{m1}v_m
$$
Similarly, each $w$ is a linear combination of the $v$'s
$$
\begin{bmatrix}w_1&w_2&\dots&w_n\end{bmatrix}=\begin{bmatrix}v_1&v_2&\dots&v_m\end{bmatrix}\begin{bmatrix}a_{11}&\dots&a_{1n}\\
\vdots&&\vdots\\
a_{m1}&\cdots&a_{mn}\end{bmatrix}\iff W=VA
$$
Then it says "$A$ is a short-wide matrix, since we assumed $n>m$. So $Ax= 0$ has a nonzero solution"
How come $n>m$ alone says anything about the system ?
I understand that if $rank(A)<n$(not Full rank), then $Ax=0$ must have nonzero solutions, but how can I know about the rank of the matrix $A$ from the given details ?
Reference: Page 170, Dimension of Vector Spaces, Introduction to Linear Algebra - Ed 5, Gilbert Strang

When row reducing $A$ to row echelon form, the number of pivots (which is the rank of $A$) can never exceed the number of columns or the number of rows of $A$. Hence, when $A$ is an $m\times n$ matrix with $m<n$, you can have at most $m$ pivot variables in $Ax=0$, leaving you at least $n-m>0$ free variables. Hence there is a non-zero solution to $Ax=0$ that can be found by choosing non-zero values for the free variables and solving for the pivot variables.

Hint: $Ax=0$ has a unique solution iff $A$ is square and has full rank. Can you proceed now?


Convergence of the spectrum under norm resolvent convergence Suppose $\{A_n\}$ is a sequence of self-adjoint operators in a Hilbert space $\mathcal H$, and $A$ is a self-adjoint operator, with $A_n \to A$ in norm resolvent sense.

Since $A_n \to A$ in strong resolvent sense also, if $\lambda \in \sigma(A)$ then there exists $\lambda_n \in \sigma(A_n)$ for all $n$, such that $\lambda_n \to \lambda$.

Since $A_n \to A$ in norm resolvent sense, if $\rho \not \in \sigma(A)$ then there exists $N$ such that $\rho \not \in \sigma(A_n)$ for all $n > N$.

Suppose now that we have $\lambda_n \in \sigma(A_n)$ for all $n$, and $\lambda_n \to \lambda$ for some $\lambda$. How can we see that $\lambda \in \sigma(A)$?

Assuming you mean that $\|A_n - A\| \to 0$, suppose that $\lambda \notin \sigma(A)$, then $(A-\lambda I)$ is invertible. The set of invertible elements is open, so $\exists \epsilon > 0$ such that if
$$
\|B - (A-\lambda I)\| < \epsilon
$$
then $B$ must also be invertible. Now choose $B$ to be of the form $A_n - \lambda_n I$ for large enough $n$. This would contradict the fact that $\lambda_n \in \sigma(A_n)$.


What is the mathematical nature of $i$? It is well known that $i$ is unit imaginary part of any complex number, but many uses of $i$ show that has others mathematical properties, for example in integration area, if I want to compute integral of $ix$ I will get $i \frac{x²}{2} $ then here $i$ is considered a constant. Also if I want to check divisors of $i$ I got only  $1$ then here $i$ has divisors however it is not integer.

Really I would like to know more about the nature of "unit imaginary part" $i$, or what the Mathematical nature of $i$ is?

Thank you for any help.

$i$ is an abbreviated form for $(0,1)$.

Its introduction is particularly useful because it lets you remember what the rules of complex product are.

E.g. $i^2=-1$ is more simple to remember than $(0,1) \dot (0,1) = (-1, 0)$
Also $(a+ib)(c+id)=ac-bd+i(ad+bc)$ is more simple to remember than $(a,b)\dot (c,d) = (ac-bd, ad+bc)$


Singular jacobian matrix? I have a series of questions, in various degrees of befuddled muddledness (and they are related to my previous questions: this and this)

First question: how do I do a change of variable if the determinant of the jacobian is singular?

The setting for this question is as follows: I have one $n$-dimensional standard gaussian random variable $u \sim N(0,I)$ and a fixed $v \in \mathbb{R}^n$. Then I define the random variable
$$z = u - \frac{u^Tv}{v^Tv}v$$
and I'd like to derive a density for $z$. So the Jacobian:
$$\frac{dz}{du} = I - \frac{vv^T}{v^Tv}$$
which turns out to be singular and so $|\frac{dz}{du}|=0$. Does this mean trying to do a change of variable is fundamentally wrong here? Or is there a way to do this?

Second question: Aside from the Jacobian, I'm not sure how to change a standard normal distribution on $u$ to a distribution on $z$. So if the density on $u$ is
$$\frac{1}{\sqrt{2\pi}}\exp(-u^Tu/2)$$
is there an inverse function $z^{-1}$ such that $z^{-1}(z) = u$? Then (I think)
$$\frac{1}{\sqrt{2\pi}}\exp(-u^Tu/2) du = \frac{1}{\sqrt{2\pi}}\exp(-z^{-T}z^{-1}/2) \left|\frac{dz}{du}\right| du$$
So, is there such a $z^{-1}$? And if there is, what is it?

Third question: ultimately, I'm trying to answer this question. Am I going about this the right way by asking the two questions above? (The person who replied to my question there says something about Haar measure which I'd never heard of before so it's not enlightening to me as a proof.)

I think for the second question... If $f$ is the density of v.a. $u$ with
$$f(u\;\vert\;\mu={\bf 0},\sigma={\bf I})$$
according to the transformation $z$, then
$$f(z\;\vert\;\tilde{\mu},\tilde{\sigma})=f\left(u(z)\;\vert\;\mu={\bf 0},\sigma={\bf I}\right)\left|\frac{\partial u}{\partial z}\right|$$
(respect to $dz$!!!)

P.D.: Excuse my English ;)


Questions of division formulas and their origin Hi guys I came across a few neat formulas to calculate the number of intergers in an interval divisable by another integer!
I was wondering why does it work? what's the mathematics behind them?
If for example I want to find the number of integers between 99 and 993 that are divisable by 2
I can do the following, choose the highest number divisable by x and the lowest number divisable by x and then
$\frac{\left(992-100\right)}{2}+1=447$
and also I can do
$⌊\frac{993-99}{2}⌋=447$
in the case of the floor fucntion if 2 is to divide both the upper and lower bound I would add 1 to the result!
Thank you so in advance guys!

You showed an example but not the original formula with variables but$\cdots$
If you mean how many numbers between $x$ and $y$ are divisible by $z$, the answer is that every $z^{th}$ number will be a multiple of $z$. for example, between $1$ and $21$ how many numbers will be divisible by $z$? Those numbers will be $3,6,9,12,15,18,21\quad$ seven.
In the "formula",
$$\bigg\lfloor\frac{21-1}{3}\bigg\rfloor=6\qquad\bigg\lceil\frac{21-1}{3}=7\bigg\rceil$$
however
$$\bigg\lfloor\frac{22-1}{3}\bigg\rfloor=7\qquad\bigg\lceil\frac{22-1}{3}=7\bigg\rceil$$
and
$$\bigg\lfloor\frac{23-1}{3}\bigg\rfloor=7\qquad\bigg\lceil\frac{23-1}{3}=8\bigg\rceil$$
I would say the formula is good for a estimate, $\pm1$.
If you choose  $0$ as the lowest number divisible by $z$ then you need no rounding but, for instance, between $10$ and $20$ where $12,15,18$ are divisible by $3$ we are "off" again
$$\frac{18-12}{3}=\frac{6}{3}=2$$

This formula is based on arithmetic progression.
Consider the interval $[a,b]$ and suppose $x,y$ are respectively the smallest and largest numbers divisible by $k$ in this interval.
Note that the multiples of $k$ in this interval would be
$$x,x+k,x+2k,\cdots x+(n-2)k,y$$
where $y=x+(n-1)k$ is the $nth$ multiple of $k$ in this interval. We are required to find $n$
A simple rearrangement of $y=x+(n-1)k$ would yield $n=\frac{y-x}{k}+1$ which is the formula you have used


How can I find $\lim_{x \to 0}\left(\frac{e^{2\sin x}-1}{x}\right)$ without l'Hopital's Rule? How do I evaluate
$$\lim_{x \to 0}\left(\frac{e^{2\sin x}-1}{x}\right)$$

I know it's the indeterminate form since the numerator and denominator both approach 0, but I can't use l'Hopital's rule so I'm not sure how to go about finding the limit.

Hint: Consider the function $f(x)=e^{2\sin x}$. What is the derivative of said function at $x=0$?

Well, there is an easy solution if you know the series expansion of $e^x$.

Near $x=0$,  $\sin(2x) \approx 2x$ therefor the given equation reduces to

$$\lim_{x\to0} \frac{e^{2x}-1}{x}$$

Now apply series expansion for $e^{2x}=1+x+...$ , neglecting higher powers as $x\to0$
There  for it reduces to

$$\lim_{x\to0}\frac{1+x-1}{x}=1$$


How to efficiently use a calculator in a linear algebra exam, if allowed We are allowed to use a calculator in our linear algebra exam. Luckily, my calculator can also do matrix calculations.

Let's say there is a task like this:

  Calculate the rank of this matrix:

  $$M =\begin{pmatrix} 5 & 6 & 7\\  12 &4  &9 \\  1 & 7 & 4
\end{pmatrix}$$

The problem with this matrix is we cannot use the trick with multiples, we cannot see multiples on first glance and thus cannot say whether the vectors rows / columns are linearly in/dependent.
Using Gauss is also very time consuming (especially in case we don't get a zero line and keep trying harder).

Enough said, I took my calculator because we are allowed to use it and it gives me following results:

$$M =\begin{pmatrix} 1 & 0{,}3333 & 0{,}75\\  0 &1  &0{,}75 \\  0 & 0 & 1
\end{pmatrix}$$

I quickly see that $\text{rank(M)} = 3$ since there is no row full of zeroes.

Now my question is, how can I convince the teacher that I calculated it? If the task says "calculate" and I just write down the result, I don't think I will get all the points. What would you do?

And please give me some advice, this is really time consuming in an exam.

There is a very nice trick for showing that such matrix has full rank, it can be performed in a few seconds without any calculator or worrying "moral bending". The entries of $M$ are integers, so the determinant of $M$ is an integer, and $\det M\mod{2} = \det(M\mod{2})$. Since $M\pmod{2}$ has the following structure
$$ \begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 0\end{pmatrix} $$
it is trivial that $\det M$ is an odd integer. In particular, $\det M\neq 0$ and $\text{rank}(M)=3$.

You're allowed to use your calculator.  So, if it were me on the test, I'd write something like this:

  $$
\pmatrix{5&6&7\\12&4&9\\1&7&4} \overset{REF}{\to}
\pmatrix{
1 & 0,3333 & 0,75\\  0 &1  &0,75 \\  0 & 0 & 1
}
$$
  because the reduced form of $M$ has no zero rows, $M$ has rank $3$.

REF here stands for row-echelon form.

Note: You should check with your professor whether or not this constitutes a sufficient answer.  It may be the case that your professor wants any matrix-calculations to be done by hand.  See Robert Israel's comment.

If that's the case, then you should do row-reduction by hand.  It only takes 3 row operations, though.


Convergence of the characteristic function Check that $$A_n \rightarrow A \iff 1_{A_n} \rightarrow 1_A$$ pointwise.
Is my proof correct?

Proof so far:
$A_n \rightarrow A \iff \limsup_{n \rightarrow \infty} A_n=\liminf_{n \rightarrow \infty}A_n=A$.
$\iff \limsup_{n \rightarrow \infty} 1_{A_n}=1_{\limsup_{n \rightarrow \infty} A_n}=1_A= 1_{\liminf_{n \rightarrow \infty} A_n}=\liminf
_{n \rightarrow \infty} 1_{A_n} \iff \lim_{n \rightarrow \infty}1_{A_n}=1_A \iff 1_{A_n} \rightarrow 1_A$ .

I'm not sure the problem makes sensea. It seems like you are asked to prove:

  Given a probability space $(\Omega, \mathscr F, \mathbb P)$, let $A_1, A_2, ...$ be events.

  If $\limsup A_n = \liminf A_n := A$ and $\limsup 1_{A_n} = \liminf 1_{A_n}$, then $\lim 1_{A_n} = 1_A$
  If $\lim 1_{A_n} = 1_A$, then $\limsup A_n = \liminf A_n := A$

If we change all your $\iff$ to $\rightarrow$, you've proven #1, but...

What does #2 even mean? By definition $\limsup A_n = \liminf A_n := A$.

I think #2 should be restated:

  If $1_{A_n} \to 1_{\limsup A_n}$ and $1_{A_n} \to 1_{\liminf A_n}$,
  then $\limsup A_n = \liminf A_n$

which I'm not even sure is true:

We have $\lim 1_{A_n} = 1_{\limsup A_n} = 1_{\liminf A_n}$

$\to 1_{\limsup A_n} = 1_{\liminf A_n}$

$\to \limsup A_n = \liminf A_n?$

I think we merely have $\limsup A_n = \liminf A_n$ a.s.


Relatively Prime Integers If $m$ and $n$ are relatively prime and $k\mid m$, show that $k$ and $n$ are also relatively prime.

I haven't really any idea where to start with this. I have that if k|m then m=km' but I'm not really sure where to go after that. Thanks!

There are already some good answers here. I'll try something basic.

Since $m$ and $n$ are relatively prime, $\gcd(m,n) = 1.$ Since $k \mid m$, $m = rk, \exists r \in \mathbb{Z}$. But then we can just write $\gcd(rk, n) = 1$.

This implies $\gcd(k, n) = 1.$ You can see this by contradiction.

Since the integers are a Principal Ideal Domain, we have a Bézout identity and can write $am+bn=1$ for some $a,b \in \mathbb{Z}$. Since $k\, |\, m$, we have $atk+bn=1$ for some $t \in \mathbb{Z}$. Now suppose for contradiction that $gcd(k,n) \neq 1$. Then, clearly $gcd(k,n) \, | \, atk+bn$ and moreover $gcd(k,n) \, | \, 1$. Since $gcd(k,n)$ is an integer, this implies that $gcd(k,n)=1$, a contradiction. Hence, $gcd(k,n)=1$.


Proof check that $\int\limits_0^1\left(\sum\limits_{k=0}^\infty a_kx^k\right)dx=\sum\limits_{k=0}^\infty\frac{a_k}{k+1}$ Prove the following:  If $\sum |a_{k}| < \infty$, prove that $\int\limits_{0}^{1} \left( \sum\limits_{k=0}^{\infty} a_{k}x^{k} \right)dx=\sum\limits_{k=0}^{\infty}\dfrac{a_{k}}{k+1}$.

$\textbf{My attempt}$: Fix $x \in [0,1]$. By the Weierstrass M-Test consider $\sum\limits M_{k}=|a_{k}| < \infty$; then it can be seen that $|a_{k}x^{k}| \leq M_{k}$,  $\forall x \in [0,1]$ and $k \in \mathbb{N}$. Therefore it follows that $\sum\limits_{k=0}^{\infty} a_{k}x^{k}$ converges uniformly on $[0,1]$.

Let $f(x)=\sum\limits_{k=0}^{\infty} a_{k}x^{k}=\sum\limits_{k=0}^{\infty} f_{k}(x)$. Since the sequence of functions converges uniformly, then $f_{k} \in \mathscr{R}[0,1], \forall k \in \mathbb{N}$.

Therefore, $f(x) \in \mathscr{R}[0,1]$ as well with the following:
\begin{align*}
\int_{0}^{1}f(x)dx &= \sum_{k=0}^{\infty}\int_{0}^{1}f_{k}(x)dx \\ \\
&= \sum_{k=0}^{\infty}\int_{0}^{1} \left[a_{k}x^{k} \right]dx \\ \\
&= \sum_{k=0}^{\infty} a_{k} \int_{0}^{1} \left[x^{k} \right]dx \\ \\
&= \sum_{k=0}^{\infty} \dfrac{a_{k}}{k+1}
\end{align*}
$\mathbf{QED}$

I was wondering if my proof is correct; I was wanting to get some feedback! Prepping for a final this Thursday, and our professor kind of skimmed through this section quickly.

You are complicating too much.  We know that $ \sum a_{k}x^{k} $ is a power series and so we can integrate a power series term by term inside its interval of convergence. Since $ \sum \mid a_{k}\mid $ converges, 1 is in the interval of convergence. Then, continue...

I don't thing they want you to prove the cited theorem.


Does there exist $u(x)\neq 0$ such that integral is equal to $0$? Let $\theta \in \mathbb R$. Does there a function $u(x)\neq 0$, not depending on $\theta$, such that
$$\int\limits_{0}^{\theta} u(x)(\theta-x) dx = 0 ?$$

I spend much time to try guessing $u(x)$, like $u(x)=x$, $u(x)=x^2$, $u(x)=x^3$, but the integral is not equal $0$.

So does there exist $u(x)\neq 0$ such that integral is equal to $0$?
Is there a way to find $u(x)$ without guessing?

If $\  \displaystyle\int_0^\theta u(x)(\theta-x)xdx =0\ $ for all values of $\ \theta\ $ and $\ u\ $ does not depend on theta, then
\begin{align}
0&=\frac{d}{d\theta} \int_0^\theta u(x)(\theta-x)dx\\
&=u(\theta)(\theta-\theta) + \int_0^\theta u(x)dx\\
&= \int_0^\theta u(x)dx\ .
\end{align}
Now differentiating the equation $\  \displaystyle\int_0^\theta u(x)dx=0\ $ with respect to $\ \theta\ $ gives
$
u(\theta)=0\
$.  If $\ u\ $ is continuous, this must be true for all values of $\ \theta\ $. If $\ u\ $ is not required to be continuous, it must still be true with the possible exception of for values of $\ \theta\ $ lying within some set of measure zero.

As $\theta-x$ is a straight line $\int_0^{\theta}\theta-xdx = \frac{\theta^2}{2}$.Now there exists a $c$ in $(0,\theta) $ such that $\int_0^{c}\theta-xdx = \int_c^{\theta}\theta-xdx $ so, now define $u(x) = 1$ for x in $[0,c]$ and $u(x)=-1$ in  $(c,\theta]$ and we have value of  that integral $0$.


Pigeon hole principle: Prove that any set of six positive integers whose sum is 13 must contain at least one subset whose sum is three. Prove that any set of six positive integers whose sum is 13 must contain at least one subset whose sum is three.

My work. I am trying by using the Pigeon hole principle. I have proved that at least two non-empty disjoint subsets have the same sum but can't go any further.

Hint. Let $1\leq a_1\leq a_2\leq a_3\leq a_4\leq a_5\leq a_6$.  We know that $\sum_{k=1}^n a_k=13$.

Now consider the following cases:

1) If $a_4\geq 4$ then $$10=13-1-1-1\geq 13-a_1-a_2-a_3\\=a_4+a_5+a_6\geq 4\cdot 3=12$$
which is a contradiction.

2) If $a_4=3$ then ...

3) If $a_4\leq 2$ then ...

I don't think that the Pigeonhole principle is strictly necessary here.

The set cannot contain the number 3.

Now consider how many of the numbers can be $\ge 4$. Clearly there can not be more than three of them, or the sum would be $\ge 13$. There can not be three either, because the sum of the other three numbers would be $\ge 3$ and the sum of all 6 would be $\ge 15$.

So there are at most 2 numbers $\ge 4$, i.e. there are at least 4 numbers $\le 2$.

If any of those four numbers is $1$, there is a subset $\{1,2\}$ or $\{1,1,1\}$ which sums to $3$.

Therefore, there are at least $4$ numbers equal to $2$, and the other numbers are all $\ge 4$.

But that is impossible, because if all six numbers are equal to $2$ the sum is $12$, and if four or five are equal to 2 the sum is $\ge 13$.


A function $f$ is increasing on the closed internal $[a,b]$ Let the function $f$ be increasing on the closed internal $[a,b]$. If $a \le f(a)$ and $f(b)\le b$, prove that:

  $\exists x_0\in [a,b]$, such that $f(x_0)=x_0$.

Thanks for your help.

Note that $f$ need not be continuous.

Assume otherwise. Especially, $f(a)>a, f(b)<b$. Let $u_0=a,v_0=b$.
Given an interval $[u_n,v_n]\subseteq [a,b]$ with $f(u_n)>u_n, f(v_n)<v_n$ we can find a subinterval $[u_{n+1},v_{n+1}]\subset [u_n,v_n]$ with
$$f(u_{n+1})>u_{n+1}, f(v_{n+1})<v_{n+1}\ \text{ and }\ |v_{n+1}-u_{n+1}|=\frac12|u_n-v_n|.$$
To do so let $w_n=\frac{u_n+v_n}{2}$. If $f(w_n)>w_n$, let $u_{n+1}=w$, $v_{n+1}=v_n$; otherwise let $u_{n+1}=u_n$, $v_{n+1}=w_n$.
Then these nested intervals converge towards a number $x_0\in[a,b]$.
Then $u_n<f(u_n)\le f(x_0)\le f(v_n)<v_n$ for all $n$ implies $f(x_0)=x_0$.

Let $g(x)=f(x)-x$.
(*)Prove that if $x\in[a,b]$ and $g(x)>0$ then there is an $\epsilon>0 $ such that $g(x)>0$ in $[x,x+\epsilon]$. For example for $x=a$ you could take
$\epsilon=f(a)-a$. Prove a similar property for $x\in[a,b]$ with $g(x)<0$.

Now if $r=\sup\{x\in[a,b]:g(x)>0\}$ (or $\inf\{x\in[a,b]:g(x)<0\}$) use (*) to show that $g(r)=0$.


How to evaluate the integral $ \int_{0}^{2\pi}|2e^{it}-1|^2 2ie^{it} dt $? How to evaluate the integral $$\int_{0}^{2\pi}|2e^{it}-1|^2 2ie^{it} dt $$

I have attempted it replacing $e^{it}$ with $\cos t +i\sin t$ but it doesn't seem to be working. How would I go about this?

Motto: Complex exponentials are often simpler to manipulate than their underlying cosine and sine parts.

Several answers are already posted, which rush to cosine-sine decompositions at the expense of overly long computations, so let us explain in details how to apply the general principle above to the present case.

1. We start with the identities $$|z|^2=z\cdot\bar z\qquad \overline{e^{it}}=e^{-it}$$ They yield at once $$|2e^{it}-1|^2e^{it}=(2e^{it}-1)(2e^{-it}-1)e^{it}=5e^{it}\color{red}{-2}-2e^{2it}\tag{$\ast$}$$
The good news is that, basically, the computation part is already over... To see why, let us continue.

2. Our second ingredient is that, for every nonzero integer $n$, $$\int_0^{2\pi}e^{int}dt=0$$ which can be proven in several different, elementary, ways, and that the integral for $n=0$ is simply $2\pi$. Thus, once integrated over $(0,2\pi)$, the only remaining term in the expansion $(\ast)$ is $\color{red}{-2}$.

3. Finally, using 1. and 2., $$\int_{0}^{2\pi}|2e^{it}-1|^2 2ie^{it} dt=2i\int_0^{2\pi}(\color{red}{-2})\,dt=2i\cdot(\color{red}{-2})\cdot2\pi=-8i\pi$$

Expanding the squared modulus,

$$((2\cos t-1)^2+\sin^2t)(\cos t+i\sin t)=(4\cos^2t-4\cos t+1+\sin^2t)(\cos t+i\sin t)=(3\cos^2t-4\cos t+2)(\cos t+i\sin t).$$

Now the integrals are over a full period and the functions can be replaced by their average (odd functions cancel out):

$$\cos^3t\to0,$$
$$\cos^2t=\frac{\cos 2t+1}2\to\frac12,$$
$$\cos t\to0,$$
$$\cos^2t\sin t\to\sin t-\sin^3t\to0,$$
$$\cos t\sin t\to\frac12\sin 2t\to 0,$$
$$\sin t\to0.$$

Hence
$$-4\cdot2i\cdot2\pi\cdot\frac12=-8i\pi.$$

A faster way:

$$2i|2e^{it}-1|^2e^{it}=2i(2e^{it}-1)(2e^{-it}-1)e^{it}=2i(2e^{it}-1)(2-e^{it}).$$

Then over a full period, only the constant term yields a nonzero contribution,

$$2i\cdot 2\pi\cdot(-2).$$


Looking for an intuitive explanation why the row rank is equal to the column rank for a matrix I am looking for an intuitive explanation as to why/how row rank of a matrix = column rank. I've read the proof at http://en.wikipedia.org/wiki/Rank_of_a_linear_transformation and I understand the proof, but I don't "get it". Can someone help me out with this ?

I find it hard to wrap my head around the idea of how the column space and the row space is related at a fundamental level.

You can apply elementary row operations and elementary column operations to bring a matrix $A$ to a matrix that is in both row reduced echelon form and column reduced echelon form. In other words, there exist invertible matrices $P$ and $Q$ (which are products of elementary matrices) such that
$$PAQ=E:=\begin{pmatrix}I_k\\&0_{(n-k)\times(n-k)}\end{pmatrix}.$$
As $P$ and $Q$ are invertible, the maximum number of linearly independent rows in $A$ is equal to the maximum number of linearly independent rows in $E$. That is, the row rank of $A$ is equal to the row rank of $E$. Similarly for the column ranks. Now it is evident that the row rank and column rank of $E$ are identical (to $k$). Hence the same holds for $A$.

Maybe this helps a bit with the intuition: When you transpose a matrix you don't change the dimension of the image.  But when you transpose a matrix, the column rank becomes the row rank and vice versa. As the dimension of the image is the column rank those are equal.


Find $z^5 +\frac{1}{z^5}$ given that $z > 0$ and $z^2 + \frac{1}{z^2}=14$ I have been struggling with this problem and I think I am nearly there. Here is my working so far:

Any hints?

As $z>0, z+\dfrac1z=4$

There is a typo in your $$\left(z+\dfrac1z\right)^3$$

The right hand side should be $$z^3+\dfrac1{z^3}+3\left(z+\dfrac1z\right)$$

which will give us the value of $z^3+\dfrac1{z^3}$

Finally,

$$\left(z^2+\dfrac1{z^2}\right)\left(z^3+\dfrac1{z^3}\right)=z^5+\dfrac1{z^5}+\left(z+\dfrac1{z}\right)$$

Replace the values of $z+\dfrac1{z},z^2+\dfrac1{z^2}, z^3+\dfrac1{z^3}$

See also : Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.


Congruence of a congruence $(N \bmod n) \bmod m$ Is there a general concept for the congruence of a congruence?

Suppose we have three arbitrary integers $N, m$ and $n$. My question is whether
$(N \bmod n) \bmod m$
corresponds to any known mathematical concept (beyond a double operation).

A simpler case, for $m=2$, would be:  Is $N \bmod n$ even or odd? Is there any result for this kind of operation?

There are two uses for "mod" and they often get confused.  The computer operation "mod" is a unary operation and it's just finding the remainder upon division.  This is the sense you're using "mod" and I wouldn't call that a "congruence."

The more math-y usage is as an equivalence relation and "mod" is really part of the $\equiv$ symbol (or a modifier of it.)  $a \equiv b (\bmod{n})$ is a "congruence" because you have two things and they're congruent to each other..

So your question is really "can I divide $N$ by $n$ and then divide the remainder by $m$?"  The answer is "sure."


Sum of closed convex set and unit ball in normed space Let $(X, \|.\|)$be a real normed space. Let $A$ be a closed convex suset of $X$ and $\mathbb{B}$ a unit ball in X, i.e.
$$
\mathbb{B}=\{x\in X: \|x\|\leq 1\}.
$$
I would like to ask whether $A+\mathbb{B}$ is still closed.

Edit: Here is a more complete solution.

  Proposition.  Suppose $X$ is a Banach space.  Then $A + \mathbb{B}$ is closed for every closed convex $A$ if and only if $X$ is reflexive.

Proof.  One direction I posted previously.  If $X$ is reflexive then $\mathbb{B}$ is weakly compact.  If $A$ is closed and convex, it is weakly closed.  Then by my answer here, applied to the weak topology, $A+\mathbb{B}$ is weakly closed, and in particular norm closed.

For the converse, James's Theorem asserts that if $X$ is not reflexive, there exists a continuous linear functional $f \in X^*$ which does not attain its norm, i.e. (after rescaling) $\|f\| = 1$ but there is no $x \in \mathbb{B}$ with $f(x) = 1$.  Set $A = \{ x : f(x) \ge 1\}$; $A$ is obviously closed and convex.  Also, there exist $x_n \in \mathbb{B}$ with $f(x_n) \to 1$; let $y_n = \frac{1}{f(x_n)} x_n$, so that $f(y_n) = 1$ and thus $y_n \in A$.  We have $\|y_n + (-x_n)\| = (f(x_n) - \frac{1}{f(x_n)}) \|x_n\| \to 0$, so that $0$ is a limit point of $A + \mathbb{B}$.  Now if $0 \in A + \mathbb{B}$, i.e. there are $x \in \mathbb{B}$ and $y \in A$ with $x+y=0$, then $y=-x$ and so $y \in A \cap \mathbb{B}$.  Since $y \in A$, we have $f(y) \ge 1$.  Since $y \in \mathbb{B}$, $f(y) \le 1$.  Hence $f(y)=1$, but by assumption no such $y$ exists.

For a simple example of a functional which does not attain its norm, consider $X = \ell^1$, $f(x) = \sum_n (1 - \frac{1}{n}) f(n)$.

Here is an example of an incomplete normed space where $A + \mathbb{B}$ need not be closed.

Let $X$ be a dense subspace of $C([0,1])$ which does not contain the constants.  (For example, take $X$ to be the set of all functions $f(x) = p(x) + e^{x}$, where $p$ is a polynomial with $p(0) = 0$.  $X$ is dense because by the Weierstrass theorem we can approximate $e^x-1$ by polynomials vanishing at 0, so the constants are in the closure of $X$, whence by Weierstrass again, the closure of $X$ is dense.)  $X$ is still equipped with the uniform norm.

Set $A = \{f \in X : f \ge 1\}$.  $A$ is easily seen to be closed and convex in $X$.  Since $X$ is dense in $C([0,1])$, for each $n$ we can choose $f_n, g_n \in X$ with $\|f_n - (1 + 1/n)\| < 1/n$, $\|g_n - (-1 + 1/n)\| < 1/n$.  Then $f_n \in A$ and $g_n \in \mathbb{B}$.  It is easy to see that $f_n + g_n \to 0$ uniformly, so 0 is a limit point of $A+\mathbb{B}$.  On the other hand, if $f \in A$ and $g \in \mathbb{B}$, then $f \ge 1$ and $g(x) \ge -1$, so $f+g \equiv 0$ only if $f \equiv 1$ and $g \equiv -1$, which is impossible.  Hence $0 \notin A+\mathbb{B}$ and so $A+\mathbb{B}$ is not closed.

One question remains: does this property fail in every incomplete normed space?  Another paper of James gives an example of an incomplete normed space where every continuous linear functional attains its norm, so the construction from the non-reflexive case will not carry over.

I think it's clear that $B$ is closed, because the norm function $X \rightarrow \mathbb{R}$ is continuous, and $B$ is the preimage of the closed unit interval.  Also, the addition function $+: X \times X \rightarrow X$ is continuous.  For any closed $A$ and $B$, $A + B = +[A \times B]$ is the continuous image of a closed set, and is therefore closed.  So the answer is yes.


A Lower bounds for exponential function How to find a "nonnegative" lower bounds for

$e^{uX-u^2Y} + e^{-uX-u^2Y} - 2 $?

where $u>0$, $Y \geq 0$ and $X \neq0$

I know I can bound above function by using the fact that $e^{uX} + e^{-uX} > 2$; i.e.,

$ (e^{uX} + e^{-uX})e^{-u^2Y} - 2 > 2(e^{-u^2Y} -1) $ but this yields negative value.

It looks to me like that expression can indeed become negative. Fix $u$ and $X$ and let $Y$ grow very large. The expression approaches $-2$. I would say your greatest lower bound is $-2$. You can see that this is the case because the expression cannot ever equal $-2$. If you set the expression equal to $-2$, you will get $e^{-u^2Y}\cosh uX = 0$ which is impossible.


Finding all of the solutions for a functional equation Let $P(x+2)+P(x-2) = 8x+6$ . Find all of the solutions for $P(x)$ . If we put $P(x) = ax + b $ the answer is obvious but how to determine all of the solutions ?

First get rid of the RHS by using $Q(x):=P(x)-4x-3$ so that $$Q(x+2)+Q(x-2)=P(x+2)-4x-8-3+P(x-2)-4x+8-3=0.$$

Then the solution of

$$Q(x+2)=-Q(x-2)$$ is of the form

$$Q(x)=(-1)^{\lfloor x/4\rfloor}Q(x\bmod 4),$$

where $Q(x)$ is absolutely arbitrary in $[0,4)$.

Hence,

$$\color{green}{P(x)=4x+3+(-1)^{\lfloor x/4\rfloor}Q(x\bmod 4)}.$$

Below, an example of a smooth solution:

and another, piecewise linear and discontinuous:

This difference equation is linear so it's solution can be reprasented as

$$
P(x) = P_h(x) + P_p(x)
$$

with

$$
P_h(x+2)+P_h(x-2) = 0\\
P_p(x+2)+P_p(x-2) = 8x+6
$$

for the homogeneous we propose $P_h(x) = \gamma^x$ and after substituting

$$
\gamma^x\left(\gamma^2+\frac{1}{\gamma^2}\right)=0
$$

and the solutions are

$$
\gamma_h = \left\{-\frac{1+i}{\sqrt{2}},\frac{1+i}{\sqrt{2}},\frac{1-i}{\sqrt{2}},-\frac{1-i}{\sqrt{2}}\right\}
$$

so the homogeneous solution is

$$
P_h(x) = C_1\sin\left(\frac{\pi x}{4}\right)+C_2\cos\left(\frac{\pi x}{4}\right)+C_3\sin\left(\frac{3\pi x}{4}\right)+C_4\cos\left(\frac{3\pi x}{4}\right)
$$

now for the particular making $P_p(x) = 4x+3$ we conclude

$$
P(x) = 4x+3+C_1\sin\left(\frac{\pi x}{4}\right)+C_2\cos\left(\frac{\pi x}{4}\right)+C_3\sin\left(\frac{3\pi x}{4}\right)+C_4\cos\left(\frac{3\pi x}{4}\right)
$$


How to develop intuition in topology? Is there any efficient trick (besides doing exercises) to develop intuition in topology?

The question is general but i would like to add my view of things.

I started to teach myself topology through several books a couple of months ago. I already passed the point of being overwhelmed by the amount definitions. Most of them I remember although sometime i check to remind myself (I have, after all, a terrible memory).

The point is most of the theorems and exercises I prove don't sink in and usually whenever I'm given a statement to prove I start with the definitions and work up from there. My feeling is that it's part of the nature of the subject. Browsing through Counterexamples in Topology really makes my head turn (all the one way implications... what ever happened to "if and only if"?)

This is in contrast to when I’m doing problems in analysis where i have a visual picture which tells me usually straight away if a given statement is true or false even before i start proving it.

I think that time here is a key element and intuition will inevitably develop at some point and so my question is:

Is there any efficient way to develop intuition in topology?

By intuition I mean a mental model that helps you see things more clearly for example:
If you’re given a space with certain properties (say first countable, countably compact hausdorff space) than your intuition tells you it has to have some other properties ($T_3$ in this case).

In my opinion, one of the best ways of developing intuition in topology is to study other branches of mathematics in which there are topological spaces. Many of these definitions, properties and theorems were imagined by people who were working in related branches of math, mostly analysis and geometry. These people stumbled upon spaces which had remarkable or singular properties, so they studied these properties. The examples came first.

Why does anyone care about compactness, say? The best way to answer this question is to ask the question: who were the first people to care about compactness, and why did they? What kind of spaces were they working with?

If you want to learn a new language, there is no point in reading the dictionary and the thesaurus. There is also not much point in learning all of the bizarre exceptions before you encounter them naturally. Instead, you should learn a few basic principles, and then go out and talk to people. Figure out how they speak, and refer to the thesaurus as you go along.

The way I developed intuition in topology is from advanced multivariable calculus. All the concepts of "open", "closed", "compact", "convergence", "connected", and so forth, have a very visual meaning behind them when you study these concepts from advanced calculus. So when you see the abstract definitions in topology they are similar to how it comes up in advanced calculus.


Probability of a particle penetrating through a shield Suppose that the probability that any particle is emitted by a
  radioactive material will penetrate a certain shield is 0.01. If 10
  particles are emitted, what is the probability that exactly one
  particle will penetrate the shield?

The answer is $10(0.01)(0.99)^9$, and I'm confused on why we need to multiply it by $10$ when the 9th power seems to constitute the $9$ failures and then $0.01$ for $1$ success.

You multiply by 10 because any of the 10 particles can be the penetrating particle.

In the case in which exactly one particle penetrates, there are (1 choose 10) ways to organize 1 penetrating particle and 9 non-penetrating particles.

For each particle in turn getting to be the one that penetrates the shield, while none of the others do:

$\begin{align}
\#1 &: 0.01 \times 0.99^9 \\
\#2 &: 0.99 \times 0.01 \times 0.99^8 \\
\#3 &: 0.99^2 \times 0.01 \times 0.99^7 \\
\#4 &: 0.99^3 \times 0.01 \times 0.99^6 \\
\#5 &: 0.99^4 \times 0.01 \times 0.99^5 \\
\#6 &: 0.99^5 \times 0.01 \times 0.99^4 \\
\#7 &: 0.99^6 \times 0.01 \times 0.99^3 \\
\#8 &: 0.99^7 \times 0.01 \times 0.99^2 \\
\#9 &: 0.99^8 \times 0.01 \times 0.99 \\
\#10 &: 0.99^9 \times 0.01  \\
\end{align}$

...all exclusive (selecting one option excludes all the other options). Obviously in each case the resulting chance is the same. Total probability then is the sum of these, $10\times  0.01 \times 0.99^9$

If the question were asking the chance that exactly two particles penetrated the shield, the same exercise for every possible pair would be rather dull and instead we should turn to the combinatorics option, using the binomial coeeficient directly: $\binom {10}{2} = \frac {10!}{2!\,8!} = \frac {10\cdot 9 }{2} = 45$ possible pairings, multiplied by the identical chance of each considered pair, $0.01^2\cdot 0.99^8$


sphere-filling curve Let $S^2$ denote the $2$-dim sphere in $\mathbb R^3$. I am interested in finding a space-filling curve, i.e. a map $\varphi: [0,1]\to S^2$ that is continuous and onto.

We know that there is such a space-filling curve onto $[0,1]^2$ from Peano's and Hilbert's results.

Know my idea was to consider a unit cube in $\mathbb R^3$. Then I want to take a path that traverses enough edges of the cube (or just take a Hamiltonian path). From an edge I want to fill its face with Peano's curve and then go back to the edge after filling. This construction yields a cube-filling curve. Then I blow the cube up to the sphere and I am done.

However, this seems to simple to me. Does that work or do I miss something?

Your construction seems doable. Here's an alternate rough sketch of a construction (I hope it's ok if I don't produce the details):

Choose space filling curves $f_{m, n} : [0, 1] \to [n, n+1] \times [m, m + 1]$ for each integer $m, n$ so that they glue to a continuous surjection $f : \Bbb R \to \Bbb R^2$ in the sense that for each unit interval $I_k = [k, k +1] \subset \Bbb R$ with $k$ an integer, $f|_{I_k}$ is $f_{m, n}$ for some choice of $m, n$ (depending on $k$). That is, glue a bunch of space filling curves on each $[n, n + 1] \times [m, m + 1]$ systematically so that the endpoints agree. Continuity of $f$ is guaranteed by gluing lemma since endpoints of $f_{m, n}$ agree.

Note that the above construction can be done because you can always choose a space filling curve $f : I \to I^2$ so that the endpoint $f(0)$ and $f(1)$ are whatever you want them to be (you can cook up a Peano or Hilbert like construction with endpoints given).

Staring a bit carefully tells you that you can actually choose the endpoints so that $f$ is proper (that is, glue $f_{m, n}$ in a way so that $\{f(k)\}$ heads off to infinity as $k \to +\infty$ or $k \to -\infty$). Once that is done, proper maps can be extended to one-point compactifications, so $f$ extends to a space filling curve $\tilde{f} : S^1\to S^2$. Now compose with the map $g : I \to S^1$, $g(x) = e^{2\pi ix}$ and you're done.

I would approach this by first trying to fill a 3D simplex with a space filling triangle like this: https://www.youtube.com/watch?v=pw_50szQfA0. Then subdivide the simplex to approximate a sphere.

We want to construct a function $f: [0,1) \mapsto S^2$ such that $f$ is bijective and preserves locality like Peano's and Hilbert's curves.

Let $v_k$ denote the vertices of a tetrahedron inscribed in a sphere of radius 1. That is $||v_k|| = 1, \forall k \in \{0,1,2,3\}$.

Define $N(v) = \cfrac{v}{||v||}$

Let's first define the inverse map:

Given a vector $v$

First, find out the closest 3 points of the tetrahedron in the sphere. Say they are $v_a, v_b, v_c$. The first digit of the answer will be the only one in $\{0,1,2,3\} - \{a,b,c\}$

Now we subdivide this section, adding the points $N(v_a +v_b), N(v_b+v_c), N(v_c + v_a)$ to form 4 new triangles in the sphere:

(0) $N(v_c + v_a), v_a , N(v_a + v_b)$;

(1) $N(v_a + v_b), N(v_b + v_c), N(v_c + v_a)$;

(2) $N(v_a + v_b), v_b, N(v_b + v_c)$;

(3) $N(v_b + v_c), v_c , N(v_c + v_a)$;

Then you find the 3 closest of these in the sphere and you know in what triangle you are, the next digit is given by the name of the triangle in brackets. Now rename the vertices of the inner triangle and repeat the process.

Define $f$ by:

Given $x \in [0,1)$

Let $x_k = \cfrac{\lfloor x\cdot 4^{k} \rfloor}{4}$ denote the k'th digit of x in base 4 after the radix.

Firstly, with $x_1$ we already know that we are in the only triangle that doesn't contain $v_{x_1}$, now we get the other vertices $a,b,c$ of the triangle and subdivide it like in the definition of the inverse, the next digit determines in which triangle we are, rename the vertices of the newly find triangle in the same order provided and continue with the next digit


Solving a System of Equations without a Square Matrix I am new to linear algebra.

I have:

a known Mx1 vector called y
a known MxN vector called X
an unknown Nx1 vector called a

such that:

$$
y = Xa
$$

I also know that

each row of X ($x1i ...x1N$, $x2i ...x2N$, etc.) is linearly independent
M < N (i.e. there are more columns than rows in X)

How can I go about solving this system of equations?  If the matrix were square, I think I could just invert X and solve accordingly for a.  However, I know you cannot invert a non-square matrix.  Intuitively I think there is a solution because the rows are both linearly independent and of a higher dimension than the number of elements in y.  Any tips would be greatly appreciated!

There is a solution because the rank of $A$ is $M$ (so the transformation is onto $\mathbb F^M$).

You could row reduce $A$ and then read off the solutions.

There are a number of ways of looking at this, you're going to have to look for a least squares solution.

Multiply throughout by $X^{T}$, this will give you a square system which you can solve.


How to plot a phase space I was trying to study the function $x(t)=\int_0^t e^{-s^2}ds$, which I did successfully using high school maths. After that, I decided that I wanted to try to study it using some multivariable calculus, but couldn't really do it, so I'm asking for the proper way to face this: the problem is equivalent to the ODE $x'(t)=e^{-t^2}$ given some initial value. Let $y(t)=t$ and therefore $y'(t)=1$. Now I want to study the phase space of the system $\begin{cases} x'=e^{-y^2}\\ y'=1 \end{cases}$. How can I do it? There are no equilibrium points and the Jacobian's only eigenvalue is $0$. Also, the linearized system is $\begin{cases}x'=(-2y_0e^{-y_0^2})\cdot y \\ y'=0\end{cases}\;$ which makes no sense since in $(0, 0)$ it gives $x'=0, y'=0$ whereas in the original one it is $x'=1, y'= 1$. What's the proper way to study this system?

If I'm interpreting your question correctly, you want to understand how to sketch a phase portrait from just knowing the dynamical system; in that situation, many techniques work, but I personally prefer to use nullclines and common sense. As it happens, it turns out this particular example is not well-suited for this technique, but I'll describe it anyways.

The first step is to try and figure out where/when the flow is only moving vertically or horizontally, which is equivalent to finding the roots of each component of the dynamical system.

In your case, $y' = 1$ which has no roots, so the flow will always have a vertical component of $1$ upwards (and so you always expect the flow to either be directly upwards or tilted upwards). For $x'$, you find $x' = 0 = e^{-y^2}$, which is only true for $y \rightarrow \pm \infty$. So how do you construct a sketch of the phase portrait from this info?

Well, because you know that there's no horizontal flow when $y = \pm \infty$, then the flow has to asymptotically approach arrows pointing straight up (which you know from $y' = 1$) as you move away from the x-axis. Then, calculating the flow directly on the x-axis tells me $x' = y' = 1$, so arrows there are pointing diagonally "northeast". Using these two results, the flow has to look like "northeast" pointing arrows slowly pointing more and more upwards as you move away from the x-axis, which is a succinct description of the whole phase plot!

If you're wondering why the flow can't suddenly change direction from northeast to northwest or something, well it's because that would imply there's a nullcline, which you know there isn't! From there on out it's up to understanding what the right "tilt" is, which really comes down to common sense.

Hope this helps!

Yes, there are no equilibrium points, so it has no meaning to compute the Jacobian at the fixed points. Also no meaning of linearization around a fixed point.

We have $\dot{x} = e^{-y^2}$ and $\dot{y} = 1$, we see that the $\dot{y}=1$ is independent of $x$ and $y$, so we have $y = t + c$, $c$ an integration constant. So substituting the expression of $y$ in $\dot{x} = e^{-y^2}$, we have $x = D \int_{0}^{t} e^{-s^2} ds$, $D$ is integration constant. This definite integral reminds me of the error function, whose plot if you have in mind then it would be handy.

In order to plot the phase space that is plot $x$ vs $y$ we need to have a plot of $D$ erf(t) vs $t$, same as the graph of erf(x).

So we have the vector field, phase plane as below for different initial conditions.