Sturm-Louiville eigenvalues I am given this Sturm-Louiville problem: $$\frac{d^2y}{dx^2}+\lambda y=0$$

With boundary conditions: $y'(0) = 0, y(2) = 0$, where $n = 1, 2, 3, ...$

I am now trying to find the eigenvalues.

I tried using the general solution:

$y = A \cos(px) + B\sin(px)$, where $λ = -p^2 < 0$

I differentiated this and substituted the boundary conditions $y'(0) = 0$ to obtain that $B = 0$.

From this, I used the boundary condition $y(2) = 0$ to obtain:

$A\cos(2p) = 0.$

From this, I arrived at $p = \frac{n  \pi}{4}$.

I was hoping for some kind of confirmation if I am on the right track...I would also like some guidance of where to go from this to obtain the eigenvalues.

I tried to find the eigenvalues by simply finding $-p^2$, but I found that this was wrong.

Thank you for your time,
I appreciate any help.

$$\frac{d^2y}{dx^2}+\lambda y=0\tag1$$

For $\lambda =0$, the general solution is $y(x)=Ax+B$,  where  $A, B$ are arbitrary constants..

The given conditions are $y'(0) = 0, y(2) = 0$ gives $A=B=0$

Hence we get $y(x)=0$, which is the trivial solution.

For $\lambda\lt 0$, say $\lambda=-p^2$

then the general solution is $y(x)=Ae^{px}+Be^{-px}$, where  $A, B$ are arbitrary constants..

The given conditions are $y'(0) = 0, y(2) = 0$ gives

$pA-pB=0\implies A-B=0$

$Ae^{2p}+Be^{-2p}=0$ which gives $A=B=0$

Hence we again get $y(x)=0$, which is the trivial solution.

So we discard these two cases and take $\lambda\gt0$.

Let $\lambda=p^2 \gt 0$

The general solution is $y(x)=A \cos(px) + B \sin(px) $, where  $A, B$ are arbitrary constants.

Now the given conditions are $y'(0) = 0, y(2) = 0$

$y'(0) = 0\implies B=0$ and $y(2) = 0\implies A\cos(2p)=0$

If we take $A=0$, then we again get the trivial equation, so we take $A \ne 0$. This gives

$~\cos(2p)=0\implies 2p=(2n-1)\frac{\pi}{2}~~~~~\text{where $~n=1,~2,~3,~\cdots~$ }$
$~\implies ~p=(2n-1)\frac{\pi}{4}~~~~~\text{where $~n=1,~2,~3,~\cdots~$ }$

So the eigen value of the given differential equation $(1)$ is $$~\lambda=p^2=~(2n-1)^2\frac{\pi^2}{16}~~~~~\text{where $~n=1,~2,~3,~\cdots~$ }~$$

and the corresponding eigen functions are $$~y(x)~=~A~\cos\left\{(2n-1)\frac{\pi}{4}~x\right\}\qquad \text{where $~n=1,~2,~3,~\cdots~~~$and $~a~$ is a constant}~.$$

You can also take the following

$~\cos(2p)=0\implies 2p=(2n+1)\frac{\pi}{2}~~~~~\text{where $~n=0,~1,~2,~3,~\cdots~$ }$
$~\implies ~p=(2n+1)\frac{\pi}{4}~~~~~\text{where $~n=0,~1,~2,~3,~\cdots~$ }$

So the eigen value of the given differential equation $(1)$ is $$~\lambda=p^2=~(2n+1)^2\frac{\pi^2}{16}~~~~~\text{where $~n=0,~1,~2,~3,~\cdots~$ }~$$

and the corresponding eigen functions are $$~y(x)~=~A~\cos\left\{(2n+1)\frac{\pi}{4}~x\right\}\qquad \text{where $~n=0,~1,~2,~3,~\cdots~~~$and $~a~$ is a constant}~.$$

Note: In both the two cases the values of $~\lambda~$ are same, in the first one $~n~$ does not take the value $~0~$, and in the second case it takes.

Any eigenfunction must be a non-zero constant multiple of the solution of
$$
          y''+\lambda y =0,\;\; y(0)=1,\; y'(0)=0,
$$
which has unique solution
$$
        y(x) = \cos(\sqrt{\lambda}x).
$$
Then, in order to satisfy $y(2)=0$, $\lambda$ must satisfy
$$
           \cos(2\sqrt{\lambda})=0.
$$
 The general solution is
$$
                2\sqrt{\lambda} = (n+1/2)\pi ,\;\; n=0,1,2,3,\cdots, \\
                \lambda = (n+1/2)^2\pi^2/4,\;\;\; n=0,1,2,3,\cdots.
$$
The corresponding eigenfunctions are
$$
           y_n(x) = \cos((n+1/2)\pi x/2),\;\; n=0,1,2,3,\cdots.
$$


Proving that similarity transformation of state-space preserves the Euclidean norm I am aware that the state space realization of a dynamical system is not unique. So if we have a dynamical system:

$\dot{x} = Ax + Bu$

$y = Cx$

Then we can write it as

$\begin{bmatrix}
\dot{x}\\y
\end{bmatrix}
= G
\begin{bmatrix}x\\u\end{bmatrix}
$

where G is

$G = \begin{bmatrix}
A \ \ \ \ | & B\\
\hline
C \ \ \ \ | & 0
\end{bmatrix}
$

but one can also formulate it as

$G = \begin{bmatrix}
TAT^{-1} \ \ | & TB\\
\hline
CT^{-1} \ \  \ \ \ | & 0
\end{bmatrix}$

for any invertible matrix $T$.

I need to show that these two realizations of $G$ have the same Euclidean norm. I have found a video showing how similarity transformations preserve the trace and the determinant, but they did just an example problem, without doing the proof.

Edit:

It has been given that $G \in \mathcal{RH}_2$, where $\mathcal{RH}_2$ is a set of all real rational, strictly proper and stable transfer matrices. It is a Hilbert Space with the inner product:

$\langle F,G \rangle = \sup_{\sigma > 0} \left\{ \frac{1}{2\pi}\int_{-\infty}^{\infty} trace \left\{ F^* (\sigma+j\omega) G(\sigma+j\omega) \right\} d\omega \right\}$

and the corresponding norm given by

${{\left\Vert F \right\Vert}_{2}}^2 = \frac{1}{2\pi} \int_{-\infty}^{\infty} trace \left\{
F^*(j\omega)F(j\omega) \right\}d\omega$

The problem here is really the notation.

This is adding to the answer by obareey a bit more details. Given a linear system

$$
\begin{align}
\dot{x} &= A x + B u \\
y &= C x + D u \\
\end{align}
$$

In this question $D = 0$, but that doesn't really change anything. Then i.e. this is wrong:

$$
G = \left[  \begin{array}{c|c} A&B\\ \hline C&D \end{array} \right]
$$

As mentioned by obareey, $G$ is not a real matrix but instead a transfer function matrix. Often this is written instead like

$$
G(s) \sim \left[  \begin{array}{c|c} A&B\\ \hline C&D \end{array} \right]
$$

sometimes also written as

$$
G(s) \triangleq \left[  \begin{array}{c|c} A&B\\ \hline C&D \end{array} \right] \text{ or } G(s) \overset{s}= \left[  \begin{array}{c|c} A&B\\ \hline C&D \end{array} \right]
$$

Unfortunatly the notation varies and I have seen all of these. Some authors even use an equal sign, which makes things really confusing (or: wrong). However they really all mean the same, namely that $G(s) = C(s I - A)^{-1} B + D$ and $I$ is an identity matrix.

The $2$-norm of a general MIMO transfer function is

$$
\Vert G(s) \Vert_2 = \Big( \frac{1}{2 \pi} \int_{-\infty}^{\infty} \text{trace} \big[ G(j \omega)^H G(j \omega) \big] d \omega \Big)^{1/2}
$$

so of course $\Vert G(s) \Vert_2 = \Vert \widetilde{G}(s) \Vert_2$ when $G(s) = \widetilde{G}(s)$, assuming that $A$ is stable so that the integral exists, i.e. $\Vert G(s) \Vert_2 < \infty$. So all you need to show is that

$$
C(s I - A)^{-1} B + D = \widetilde{C}(s I - \widetilde{A})^{-1} \widetilde{B} + \widetilde{D}
$$

using an invertible $T$ and

$$
\begin{align}
\widetilde{A} &= T A T^{-1} \\
\widetilde{B} &= T B \\
\widetilde{C} &= C T^{-1} \\
\widetilde{D} &= D
\end{align}
$$

This is standard realization theory. You can check that both transfer functions are the same:

$$
\begin{align}
\widetilde{G}(s) &= \widetilde{C}(s I - \widetilde{A})^{-1} \widetilde{B} + \widetilde{D} \\
&= C T^{-1}(s I - T A T^{-1})^{-1} T B + D \\
&= C \big(T^{-1}(s I - T A T^{-1})T\big)^{-1}B + D \\
&= C \big(T^{-1} s I T - T^{-1} T A T^{-1} T\big)^{-1}B + D \\
&= C(s I - A)^{-1} B + D \\
&= G(s)
\end{align}
$$

because $T^{-1} s I T = s(T^{-1} I T)  = s(T^{-1} T) = s I$ since $T^{-1} T = I$ and the fact that $(K_1 K_2 K_3)^{-1} = K_3^{-1} K_2^{-1} K_1^{-1}$ for some invertible matrices $K_1, K_2, K_3$.

So because $G(s) = \widetilde{G}(s)$ their $2$-norms are the same.

It is not the case that the transformed matrix $G$ will have the same $2$-norm. For instance, consider
$$
A = \pmatrix{0&1\\0&0}, \quad B = \pmatrix{1\\0}, \quad C = \pmatrix{1&0}, \quad T = \pmatrix{2&0\\0&1}.
$$
We find that the matrices in question are
$$
G = \left[
\begin{array}{cc|c}
0&1&1\\
0&0&0\\
\hline
1&0&0
\end{array}
\right], \quad
G' = \left[
\begin{array}{cc|c}
0&2&2\\
0&0&0\\
\hline
\frac 12&0&0
\end{array}
\right].
$$
Verify that these two matrices do not have the same $2$-norm.

For an example of a stable system for which this occurs, it suffices to replace $A$ with the matrix
$$
\pmatrix{-1&1\\0&-1}
$$
in the above example.


Empty set as a singleton So, it is obvious that the empty set is a subset of every set, such that ∀A:∅⊆A.

Therefore, if the null set is an element of some set A, such that A = {∅}, is the empty set considered a singleton element of A?

Thank you.

Yes, the set $\{\varnothing\}$ is a set whose only element is the empty set $\varnothing$.

Divorce your mind from any thought that of an element being a set in its own right, from the role of the element as an element in another set.  If $x$ is the only element in $A$ then $x$ is a singleton element of $A$.  It doesn't matter what $x$ is and if $x$ just happens to be a set, its elements have nothing whatsoever to do with what kind of element $x$ is to $A$.

So $A=\{\emptyset\}$.  The $A$ has a single element.  That singleton element is $\emptyset$.

That's all that needs to be said.

.....

Um... I'm assuming you are taking "singleton" to mean the only element of  a set?


Independent events in a real life scenario The question:
Martin's journey home from work involves taking the bus to the station, catching the six o'clock train and then a short walk to his house. If the bus is late, he has to catch a later train.
On 10 days out of the past 50, the bus has not reached the station in time for him to catch the six o'clock train. Similarly on 10 days out of the past 50, the six o'clock train has arrived late at Martin's home station.
Is it more likely than not that Martin will get home on time tomorrow?
A. Yes, the buses and the trains each meet the requirements of
the timetable 40 out of 50 times respectively, so he should get home
on time tomorrow.
B. Yes, if the bus is late, the train may be too, so he would get home on time.
C. No, on 20 occasions over the 50 days either the train or bus has been late, so it is unlikely he will get home on time.
D. No, every day is different so predictions are no more than guesses and wholly unreliable.
In my opinion, B and C are obviously incorrect answer options. I chose D, because I thought whether or not the train / bus comes late tomorrow, is independent of past probabilities. But obviously, if you use the data given in the information, A is the correct answer. $ \frac{40}{50} \times \frac{40}{50} = 0.64 $
Could someone explain why the answer is A?
Like, let's say there are 4 white balls and 4 black balls in a bag. If you take one ball out and it was white, you put it back in the bag, and if you take one out again, the probability that the ball is white is still $\frac{1}{2}$. (Independent events).
Is this scenario different to the balls in a bag scenario?

We have $50$ samples for each event, so the quoted event counts can reasonably be taken as representatives of the overall punctuality of each service and a probability extracted. Furthermore, for the purposes of this question bus and train may be safely assumed as independent (you can't drive a bus and train at the same time, the two services run on different infrastructure and do not influence each other, etc.) so we have independent random variables and may multiply to arrive at $0.64$.

I chose D, because I thought whether or not the train / bus comes late tomorrow, is independent of past probabilities.

To choose D is to imagine that the bus or train drivers conspire to prevent Martin from reaching home on time, which violates Occam's razor.

I would do this: imagine 5000 such scenarios.  "10 out of every 50" is 1/5 so in 1000 such situations, the bus arrives at the station late, in 4000, the bus is on time.  In 1/5 of the times the bus is late, so 200, the train is also late so he can take it anyway.  There are 4000+ 200= 4200 times out of 5000 that he catches the train.  The probability he will get home on the first train is 4200/5000= 42/50= 84/100 or 84%.


Can i perform this step? a,b $\in$ R
$${5^{a-b}}\cdot{27^{-(a+b)}}={5^1}\cdot {27^1} \Rightarrow a^{2}-b^{2}=?$$

My Solution:
$a-b=1$
$a+b=-1$  then $a=0$ ; $b=-1$
Since $a$ and $b$ are real numbers, I believe, I cannot perform above solution. Am I right?

Note that for $(a,b)=(0,-1)$ we have
$$5^{a-b}\cdot27^{-(a+b)}=5^1\cdot27^1,$$
and $a^2-b^2=-1$, and that for $(a,b)=\left(\frac{\log(5^1\cdot27^1)}{\log(5/27)},0\right)$ we have
$$5^{a-b}\cdot27^{-(a+b)}=5^a\cdot27^{-a}=\left(\frac{5}{27}\right)^{\frac{\log(5^1\cdot27^1)}{\log(5/27)}}=5^1\cdot27^1,$$
and $a^2-b^2=\left(\frac{\log(5^1\cdot27^1)}{\log(5/27)}\right)^2\geq0$. So the value of $a^2-b^2$ is not uniquely determined by the given equation. In particular, this shows that the step you took is not valid.

In the above equation a and b are not whole numbers but real numbers
$a-b=1$

$a+b=-1$

We know that
$a^2-b^2=(a+b)(a-b)$

$a^2-b^2=(-1)(1)$
$a^2-b^2=-1$


Showing that $\frac{y}{x^2+y^2} \, dx - \frac{x}{x^2+y^2} \, dy = d\left(\tan^{-1}\left(\frac{x}{y}\right)\right)$ I'm trying to show that

$$
\frac{y}{x^2+y^2} \, dx - \frac{x}{x^2+y^2} \, dy = d\left(\tan^{-1}\left(\frac{x}{y}\right)\right)
$$

but am having trouble figuring out exactly how to approach the problem. I've tried a few tricks with switching to polar coordinates or trying to make use of right triangles, but have had no luck. Could anyone point me in the right direction here?

Hint:
$$ \frac{d}{dt}\tan^{-1}(t) = \frac{1}{1+t^2} $$
$$ d f(x,y) = \frac{\partial f(x,y)}{\partial x} dx + \frac{\partial f(x,y)}{\partial y} dy $$

Using the Hint from above:
\begin{align}
d\left(\tan^{-1}\left(\frac{x}{y}\right)\right) &= \frac{\partial \tan^{-1}\left(\frac{x}{y}\right)}{\partial\left(\frac{x}{y}\right)}\frac{\partial\left(\frac{x}{y}\right)}{\partial x} dx + \frac{\partial \tan^{-1}\left(\frac{x}{y}\right)}{\partial\left(\frac{x}{y}\right)}\frac{\partial\left(\frac{x}{y}\right)}{\partial y} dy \\
&= \frac{1}{1+\frac{x^2}{y^2}}\frac{1}{y}dx + \frac{1}{1+\frac{x^2}{y^2}}\left(-\frac{x}{y^2} \right)dy\\
&=\frac{y}{x^2+y^2}dx-\frac{x}{y^2+x^2}dy
\end{align}


What's the hard part of zero? A lot of textbooks said it was hard for human to accept zero when it was first introduced.

How could it be? It seems to me as natural as positive integer which represent there is no elements at all.

Of course it seems natural to you; you grew up in the modern world, where everyone accepts zero.  More importantly, people now accept the abstract concept of numbers and are capable of divorcing them from the things that they represent.  This is a sophisticated point of view.  From a more naive point of view, a number is a property of a collection of objects: when I say there are $2$ apples somewhere, that is a property of a collection of apples.  If there are no apples, then what is there to "hold" the corresponding property?  So instead of saying there are $0$ apples, people said there are no apples.

In computer science terms, apples.num() is not defined if there is no apples variable!

You also have to understand that "human" in your statement means (from my understanding) "Western mathematicians."  Indian mathematicians had no trouble with zero.

In many places, i.e., accounting 0 meaning nothing is represented by a line.

for example $3.—

for exactly 3 dollars. rather than $3.00

and sometimes by xx ... as written on a cheque:

---- three dollars ----- xx/100


Regarding eigenvalue equation in a matrix vector space I'm a physics grad student working, for my research, in a matrix vector space over the complex numbers. That is, both operators and the vectors they act upon are $n\times n$ complex matrices living in the same space. When it comes to the eigenvalue equation I've been using

$AB-BA=\lambda B$,

where $A$ and $B$ are $n\times n$ complex matrices. I can "justify" this choice based on the Heisenberg picture of Quantum Mechanics, but I would like to give it a more solid mathematical background. So in this regard I have several questions:

Is the choice above sensible? what other choices could I try? and last what is the relevant literature dealing with matrix vector spaces?
I mean, every semi-decent book mentions the space of matrices as a vector space, but that's just about it. I've found some valuable pieces of information scattered througout the problems in the Hoffman linear algebra text, but again, that's just about it. I have yet to find a text that treats matrix vector spaces in any depth. Any help would be really appreciated.

The linear operator $\text{ad}_A : B \mapsto AB - BA = [A, B]$ defines the adjoint representation of the Lie algebra $\mathfrak{gl}_n$, and what you are studying are its eigenvalues. You don't want a linear algebra book; you want a book on Lie theory. The classic reference (although it is not particularly directed towards physicists) is Fulton and Harris's Representation Theory: a First Course.

The important structure here is not the vector space but the action on it, which is why books on linear algebra won't say anything specific about "matrix vector spaces" because they are, after all, still vector spaces.

What other choices you have depends, of course, on what it is you're doing.

Here's something that might help.  Using Krnoecker products, we can write

$$AX-XA = 0$$

as

$$[(I \otimes A) - (A^T \otimes I)] \vec X = 0.$$

This is taken directly from Horn and Johnson's Topics in Matrix Analysis, pg. 257.


Testing Convergence of Integral How do I prove that the integral: $\int_0^1 x^p{} (1-x)^q dx$  converges or diverges?

Where do we have problems, and under what conditions they are vanished?
Hint :
Separate the integral into two integrals, one from $0$ to $1/2$ and the other from $1/2$ to $1$.
Then find for what values of $p$ and $q $ , both integrals converge.

HINT:

The integrand is the kernel of a  Beta$(p+1,q+1)$ density, and so the integral converges if $p>-1, q>-1$.

In that case, the value is
$$
  \int_0^1 x^p (1-x)^q dx
= B(p+1,q+1)
= \frac{\Gamma(p+1)\Gamma(q+1)}{\Gamma(p+q+2)},
$$
and in the case that $p$ and $q$ are integers, this is equal to $\frac{p!q!}{(p+q+1)!} = \frac{1}{\binom{p+q}{p}(p+q+1)}$


Open Thickening of Closed Balls - Homotopy Equivalent? Let us assume we have a finite union of closed ball with radius $\alpha$ around points $x$ in a Euclidean space
$$M:= \bigcup_{i \in \{1,...,q\}}B_\alpha(x_i)$$
Can I "thicken" them to open balls $B^\circ_{\alpha +\epsilon}(x_i)$ such that
$$M \text{   is homotopy equivalent to   } \bigcup_{i \in \{1,...,q\}}B^\circ_{\alpha+ \epsilon}(x_i)$$
It should work, but could you tell me the reason why? Thanks a lot!!

I think the concept underlying your question is that of a deformation retract. (See here: https://en.wikipedia.org/wiki/Retraction_(topology)).
The idea is that you can deformation retract on your union to get the same thing. Deformation retracts are special kinds of homotopy equivalence.


notation for precision or number of decimals If you have a formula like $x = y+2.1234$, is there a way to specify $y$ should have no more than 2 decimals? So if $y = 12.1234$ then you should use 12.12 when using this formula.

Same for the result, is there a way to specify you only should use the first 2 decimals of $y$ and discard the rest?

To clarify: it is for documentation purposes. I need to document a value should only have 2 decimal places and the rest should be discarded.
So when I have a complex formula $x = ....$, is there any notation possible which shows to use only 2 decimals of $x$?

You can write $12.1\bar{2}$ or $12.1\underline{2}$.

edit: I saw this notation on the Khan Academy videos for significant figures.
(He says that it's more common to write the answer in scientific notation and then the precision is implied by how many digits are after the decimal, though).


Homeomorphism on the Hilbert space We can consider two different topologies on the Hilbert space ; $l^{2}(\mathbb{N})$.
One is the topology deduced from the norm
\begin{equation*}
\|f\|=\sqrt{\sum_{n=1}^{\infty} f(n)^{2}},
\end{equation*}
and the other one is a subspace topology from $\mathbb{R}^{\omega}$ on $l^{2}(\mathbb{N})$.

I want to prove or disprove that the above two topological spaces are homeomorphic.

The only thing I found is the identity map is not open, so the two topologies are not equal.

Any help will be appreciated. Thank you.

It is enough to show that $(l^2(\omega), \tau)$ where $\tau$ is the product topology on $\mathbb{R}^{\omega}$ restricted to $l^2(\omega)$ is not completely metrizable. Suppose not. Then $l^2(\omega)$ must be a dense $G_{\delta}$ subset of $\mathbb{R}^{\omega}$. But then $l^2(\omega)$ is comeager in $\mathbb{R}^{\omega}$ which is easily refuted by observing that the sets $A_n = \{x \in l^2(\omega) : \sum_k (x(k))^2 < n\}$ are nowhere dense in $\mathbb{R}^{\omega}$.

Consider the sequence $e_k = \langle 0, 0, \dots, 1, 0, 0, \dots  \rangle$ where $1$ occurs at the $k$-th coordinate. This sequence converges to the constant sequence $\overline{0} = \langle 0, 0, 0, \dots \rangle$ in the (product) topology $\mathbb{R}^{\omega}$ simply because every basic open neighborhood of $\overline{0}$ contains $e_k$ for all sufficiently large $k$. But in the Hilbert space $l_2(\omega)$ this sequence has a constant distance $\sqrt{2}$ from $\overline{0}$. Hence the two topologies are different.


Path between vertices with an odd degree Is it true that in an undirected graph with exactly $2$ vertices with an odd degree there is a path that connects them and why?

It must be true but I cannot prove it.

Suppose there is no such path. Then the graph is not connected. Let $A$ and $B$ be connected components, $A$ with first vertex and $B$ with second. Then in each component we have, by Handshake lemma, at least two vertices with odd degree. So in total at least 4 vertices with odd degree. A contradiction.

graph
this is true as you can see in the above picture of graph. This is the only case of 2 vertices of odd degree so they must be connected.


Factorize : $P(x)=x^6+x^2+1$ Factorize in $\mathbb{R}[x]$ or $\mathbb{Q}[x]$ :
$$P(x)=x^6+x^2+1$$
I know all roots of $P$ are complex and I don't know if there a factorization
But My try as following :
$$P(x)=x^6+x^2+1=(x^3)^2+1+2x^3-2x^3+x^2$$
$$=(x^3+1)^2+x^2-2x^3$$
I don't know how I complete?

For factorisation in $\mathbb{Q}[X]$:

As you note $X^6+X^2+1$ has no real roots.

Now $X^6+X^2+1=(X^3+X+1)^2$ modulo $2$, and this cubic is irreducible modulo $2$.

Hence, if the polynomial factorises over $\mathbb{Q}$ - and so over $\mathbb{Z}$ - it factorises as the product of two cubics. But every cubic has a real root, so we can't have such a factorisation.


Number of ways to divide $160$ into three parts In how many ways can we divide $160$ into three parts, each with at least $40$ and at most $60$?

The amounts can only be integers. Please help me solve this, I don't really get exercises with at least and at most conditions.

SOLUTION:

This can be very easily solved by Partition method and then using generating function (other name: INTEGRAL EQUATION METHOD).

Let $\displaystyle x+y+z = \$160 \ \ $ as each gets atleast $\$40$  so just divide $\$40$ to each of them before hand.

Then the equation becomes $$\displaystyle a+b+c = \$40 \ \ $$ for some $a,b,c$.

Now we need to divide $\$20$ among $a,b,c$ as at max each can get $\$60$.

Thus By generating function or Integral equation method we can convert the following equation into this form :-

$$\displaystyle (1 + r^1 + r^2 + ... + r^{20})^3 = r^{40} $$

Where the coefficient of $ r^{40} $ determines the number of possible solutions to the equation. By GP-summation we convert this into:

$$\displaystyle \frac {(1-r^{21})^3}{(1-r)^3} = r^{40} \implies (1-r^{21})^3(1-r)^{-3} = r^{40} $$

Now if we ibserve it then we can see that the terms that matter to us in $({1-r^{21}})^3$ is just $({1-3r^{21}})$, as rest of the terms have exponents greater than $40$.

So, we get the coefficient of $r^{40}$ by $\displaystyle \binom {40 + 3-1}{3-1} - 3 \times \binom {19 + 3-1}{3-1}$, thus the answer is:

$$\displaystyle \binom {42}{2} - 3\binom {21}{2} = 231 .$$

All of the parts are greater than or equal to $40$,

So the sum will be at least $120$.

Now the remaining $40$ needs to be partitioned into $3$ groups $A,B,C$ so that none of the groups have more than $20$

The total ways to do this without the $20$ limit restriction is:

$A+B+C=40$, hence $\binom{42}{2}$ ways.

Now we have to subtract ways in which  $1$ group has more than $20$

Lets say that $A \gt 20$

So $B+C \lt 40-20$, or in other words, $B+C+D*=19$

This can be done in:  $\binom{21}{2}$ ways

Now any of $A,B,C$ could have been $\gt 20$

$3\cdot \binom{21}{2}$

And finally,

$\binom{42}{2}-3\cdot \binom{21}{2}$

*$D$ was added as another variable to account for the fact that if $B+C=18$, $D=1$, similarly if $B+C=12$, $D=7$ and so on


Prove $(n!)/n^n \leq 1/2^{k}$, where $k$ is the floor of $n/2$. I suppose the natural way to prove this is by induction. When I follow the rather natural steps $$\frac{(n+1)!}{(n+1)^{n+1}} = \frac{n!}{(n+1)^{n}} \leq \frac{n!}{(n)^{n}}$$ in order to apply the induction hypothesis, however, I obtain for $n+1$ exactly the same bound I assumed for $n$. This is not a problem if $n+1$ is odd, because then the floor of $n/2$ and $(n+1)/2$ coincide, but how can I "detect" the change in the case that $n + 1$ is even?

HINT:
$$\frac{n!}{n^n} = \left( \frac{1}{n} \cdot \frac{2}{n} \cdots \frac{n/2}{n} \right) \cdot  \cdots \frac{n-1}{n} \frac{n}{n}  \leq \left( \frac{1}{2} \cdots \frac{1}{2} \right) \cdot 1 \cdots 1 = \frac{1}{2^k}$$

This is obvious:
$$2^k\cdot n!=2(2\cdot 2)\cdots(2\cdot\lfloor \frac{n}{2}\rfloor) \cdot (\lfloor \frac{n}{2}\rfloor+1)\cdots n<n^n.$$


Zeroes of function with real exponents Let real $x \geq 0 $ and real $p > 2$.

Let $ f(x) = (x - 1)(x + 1)^{p - 1} - x^p + 1$.

Show that, for the given range of $x$ and $p$,  $f(x)=0$ only for $x=0$ and $x=1$.

Since $f(x)$ is not convex, I find it difficult to show that $f(x) < 0$ for $0<x<1$ and  $f(x) > 0$ for $x>1$.

EDIT (generalization): the same zeroes hold for all  $p>0$ ; $p \neq 1$ , $p \neq 2$.

We will  show that  $f(x) > 0$ for $x>1$ and $f(x) < 0$ for $0<x<1$.

Let's start with the first one,   $f(x) > 0$ for $x>1$.

For real $p > 2$, we can use Bernoulli's inequality (which holds for real exponents!):

$(1 + w)^r \ge 1 + r w $,  for real  $w > -1$ and real $r > 1$.

This gives for the first exponent in $f(x)$:

$(x+1)^{p-1} = x^{p-1} (1+1/x)^{p-1} \geq x^{p-1} (1 + \frac{p-1}{x})$

Inserting into $f(x)$ gives

$$f(x) = (x - 1)(x + 1)^{p - 1} - x^p + 1 \\\ge (x - 1)x^{p-1} (1 + \frac{p-1}{x}) - x^p + 1 \\= x^{p-1} (p-2) - x^{p-2} (p-1) + 1 = h(x)$$

with equality  for $x=1$. Now note that $h(x=1) = 0$. Further,

$h'(x) = (p-2)(p-1)x^{p-3} (x-1)$

so for $x>1$, $h'(x) > 0$ everywhere. Hence $h(x) \geq 0$ with equality  for $x=1$. This establishes that for $x\geq 1$, $f(x) \geq 0$ with equality  for $x=1$.

For the second case $0<x<1$, we will see now that this can be reduced to the first case $x>1$, so indeed there is nothing to do.

Let $y =\sqrt x$. Then the function  becomes:

$$
f(y = \sqrt x) = y^p \left[ (y - \frac1y)(y + \frac1y)^{p - 1} - y^p + \dfrac{1}{y^p} \right]
$$

We have shown above that for $y>1$, $f(y) >0$. I.e. we have that the following inequality holds:

$$
g(y) =  (y - \frac1y)(y + \frac1y)^{p - 1} - y^p + \dfrac{1}{y^p}  > 0 \qquad \bf\text{[ineq. 1]}
$$

Now, we want to show that for $0<y<1$, $f(y) <0$. Here, we need  to show

$$
g(y) =(-y + \frac1y)(y + \frac1y)^{p-1} -[- y^p + \dfrac{1}{y^p} ]> 0
$$
Now replacing $y$ by $1/y$ we get $y > 1$ and, in the replaced variable,
$$
\tilde g(y) =  (y - \frac1y)(y + \frac1y)^{p - 1} - y^p + \dfrac{1}{y^p} > 0
$$
So this is the same condition as before in ineq. 1, which has already been proven.

$\qquad \qquad \Box$

The term $x^p-1$ for $p \geq 1$ is zero for $x=1$ and therefore it has the factor $x-1$. It can easily be checked that this results in:
$$
x^p - 1 = (x-1) \sum_{k=0}^{p-1} x^k
$$
we can therefore rewrite the function $f(x)$ as
$$
f(x) = (x-1) \left[ (x+1)^{p-1} - \sum_{k=0}^{p-1} x^k\right]
$$
where we can now use the binomial expansion on the term $(x+1)^{p-1}$ to obtain
$$
f(x) = (x-1) \left[ \sum_{k=0}^{p-1} \binom{p-1}{k} x^k - \sum_{k=0}^{p-1} x^k\right]
$$
and if we combine the two sums we find
$$
f(x) = (x-1) \sum_{k=0}^{p-1} \left[ \binom{p-1}{k} - 1 \right] x^k
$$
The sum is a polynomial (of degree $p-2$ because the terms proportional to $x^0$ and $x^{p-1}$ vanish) with positive coefficients and hence has a positive value for all $x>0$. Therefore $f(x)$ has only a single root on $x>0$, which is $x=1$ and together with $f(0)=0$ concludes the proof.


Line integral segment of parabola Suppose
$$
\vec{F} = \nabla f(x,y) = 6y \sin (xy) \vec{i} + 6x \sin (xy) \vec{j},
$$
and C is the segment of the parabola $y = 5 x^2$ from the point $(2,20)$ to $(6,180)$. Then, what is
$$\int_C \vec{F} \cdot d\vec{r}\ ?$$

I can not fnd the right answer, tried many ways to do it! could not get so far on this

Parameterize the curve $C = C(t)$, say letting $t \in [2,6]$ and then $x(t) = t$ and $y(t) = 5t^2$, so
$$
d\vec{r} = d \pmatrix{t \\ 5t^2} = \pmatrix{1 \\ 10t} dt
$$
So now compute
$$
\int_{t=2}^{t=6} 6\sin(xy) \pmatrix{y \\ x} \cdot \pmatrix{1 \\ 10t} dt,
$$
plugging in the correct $x = x(t), y=y(t)$ as  we defined above.

Can you take it from here?


On a Sobolev embedding: every $u\in W^{1,p}(I)$ is Bounded Let $I\subset \mathbb R$ be a (not necessary bounded) interval and $p>1$. Show that there exists $C>0$ such that for  all $u\in W^{1,p}(I),$
$$
\|u\|_\infty\le C\|u\|_{W^{1,p}(I)}
$$

Recalling that $$\|u\|^p_{W^{1,p}(I)} = \|u\|^p_{L^{p}(I)}+\|u'\|^p_{L^{p}(I)}$$
I started with the following well known identity for  all $u\in W^{1,p}(I),$

$$u(y)-u(x)=\int_x^y u'(t) \, dt $$
This is true only on the real line. But I dont know how to proceed.

There might be a more elementary way to prove this, but the tools developed by Adams and Fournier in Sobolev Spaces make for a very easy proof (which can be used to verify that the proposition is correct, even in the case that the domain is unbounded). Note that for an interval $U\subset \mathbb{R}$, $U$ satisfies the strong local Lipschitz property (see page 83 of Sobolev Spaces). Therefore, for $p > 1$, we have the imbedding $W^{1,p}(U)\to C^{0,\gamma}(\overline{U})$ for $\gamma = 1-\frac{1}{p}$, i.e. for $u\in W^{1,p}(U)$, there is a version $u^*\in C^{0,\gamma}(\overline{U})$ of $u$ such that $\|u^*-u\|_{W^{1,p}(U)} = 0$ and $$\|u^*\|_{C^{0,\gamma}(\overline{U})}\lesssim_{U,p} \|u\|_{W^{1,p}(U)}$$ Of course, $$\|u^*\|_{C^{0,\gamma}(\overline{U})} = \|u^*\|_{C^0(\overline{U})}+\sup_{x,y\in \overline{U}} \frac{\lvert u^*(x)-u^*(y)\rvert}{\lvert x-y\rvert^{\gamma}}\geq \|u^*\|_{C^0(\overline{U})}$$ and $\|u^*\|_{C^0(\overline{U})} = \|u^*\|_{C^0(U)}$, so $$\|u^*\|_{C^0(U)}\lesssim_{U,p} \|u\|_{W^{1,p}(U)}$$ As $u^*$ is continuous, we have $\|u^*\|_{C^0(U)} = \|u^*\|_{L^{\infty}(U)}$, and as $u^* = u$ almost everywhere, we have $\|u^*\|_{L^{\infty}(U)} = \|u\|_{L^{\infty}(U)}$. Therefore, $$\|u\|_{L^{\infty}(U)}\lesssim_{U,p} \|u\|_{W^{1,p}(U)}$$

Let's first consider $I$ where the Lebesgue measure $\mu(I)<\infty$. We can use the following nice result (which can be derived from Holder's inequality):

If $I$ has finite (Lebesgue) measure, and we have $1<p\leq q\leq \infty$, $||f||_{L^p(I)}\leq\mu(I)^{1/p-1/q}||f||_{L^q(I)}$, so we have $L^q(I)\subset L^p(I)$. In particular, we have $L^{\infty}(I)\subset L^p(I)$ for $1\leq p<\infty$. Now note that $W^{1,p}(I)\subset L^p(I)$, so in particular, we have that $L^{\infty}(I)\subset W^{1,p}(I)$ for $1\leq p<\infty$.

The desired constant $C$ can just be set to $C=\mu(I)^{1/p}$ (since $q=\infty$ and $1/q=0$). This argument can be extended for Sobolev spaces with $k$ weak derivatives.

Now if $\mu(I)=\infty$, there is no containment among $L^p$ spaces.
For now, let's consider $I=(0,\infty)$, which certainly has infinite measure.
Consider $f(x)=1/\sqrt x$, $f\in L^1(I)$, but $f\notin L^2(I)$, likewise if we have $g(x)=x^{-1/3}$, $g\in L^1(I)$, but $g\notin L^3(I)$. In general, we can consider $x^{-1/p}$, which will be in $L^1(I)$ but not $L^p(I)$. Also, note that $x^{-1/p}$ is not bounded a.e. so $x^{-1/p}\notin L^{\infty}$, and so there cannot be such a constant $C$.


Is this the best route to go in this implicit derivation problem? I have this equation:

$$e^{\frac{x}{y}} = x - y$$

I seem to be going down the wrong path. Is this right so far?

$$\frac{dy}{dx} = e^{\frac{x}{y}} \cdot \frac{dy}{dx} (x \cdot y^{-1}) = 1 - y'$$

$$e^{\frac{x}{y}} ( x \cdot -y^{-2} \cdot y' + y^{-1}) = 1 - y'$$
$$ -e^{\frac{x}{y}} x \cdot y^{-2} \cdot y' + e^{\frac{x}{y}} y^{-1}) = 1 - y'$$

WolframAlpha has this as

$$y'(x) = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$

I don't know why your subtraction sign became an equals sign, but your goal, after implicitly differentiating, is to isolate the $y'$ terms so you can factor and divide:

$$e^{x/y} \left( x \cdot -y^{-2} \cdot y' + y^{-1}\right) = 1 - y'$$
$$e^{x/y} \cdot x \cdot -y^{-2} \cdot y' + e^{x/y} \cdot y^{-1} = 1 - y'$$
$$e^{x/y} \cdot x \cdot -y^{-2} \cdot y' + y' = 1 - e^{x/y} \cdot y^{-1}$$

Then factor:

$$y'\left(e^{x/y} \cdot x \cdot -y^{-2} + 1\right) = 1 - e^{x/y} \cdot y^{-1}$$

And divide. After some simplification, you should reach:

$$y' = {{y - e^{x/y} \over y} \cdot -{y^2 \over e^{x/y} x  - y^2}} = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$

Please check your line 2, it should be something like this:

$$e^{\frac{x}{y}}\times(y^{-1}-xy^{-2}y')=1-y'$$

Proceed using this to get the WolframAlpha  output


Zoom out fractals? (A question about selfsimilarity) It is well known that if we zoom in on the Mandelbrot set we get selfsimilarity.
So I wonder if $g$ is a fractal (in the complex plane) generated by a nonperiodic nonpolynomial entire function $f$

$g:=  f(f(...))$

Is it possible that the fractal is infinite in size and that when we zoom out , we get selfsimilarity too ? Lets call such a fractal a " zoom out fractal ".

As example Mandelbrot : $f(z)=z^2+1$, g$(z) [= f(f(...))]$ is the fractal.
The fractal has finite size (area or length ) since it diverges for $Re(z)>2$.
$f(z)$ is a nonperiodic nonpolynomial entire function. But when we zoom out we get no selfsimilarity. ( divergence is not considered valid as selfsimilar ) So Mandelbrot is NOT a zoom out fractal.

Does the existance of zoom out fractals require that the fractal is also a zoom in fractal ?

What is the formal way or term to express ' zoom out selfsimilarity ' or ' zoom out fractal ' , if any ?

A serious candidate for $f(z)$ is the following function : $\displaystyle f(z)=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{zt}}{t^t} \, dt $.

The reason is that the entire function $f(z)$ is bounded for $\mathrm{Im}(z)^2 > \pi$

This was given here : Entire function $f(z)$ bounded for $\mathrm{Re}(z)^2 > 1$?

See also : How fast does the function $\displaystyle f(x)=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{xt}}{t^t} \, dt $ grow?

It would be nice to see a plot.


Show that $\| T^{\ast}T \| = \| T \|^2$ Suppose that $T$ is any bounded operator. I want to show that $\| T^{\ast} T \| = \| T \|^2$. Can you please check my proof.

Consider that $\| T \|^2 = \sup_{\| x \| \leq 1} \| Tx \|^2 = \sup_{\| x \| \leq 1} \left| \langle x, T^{\ast}Tx \rangle \right| \leq \| T^{\ast}T \|$ and $\| T^{\ast} T \| = \sup_{\| x \| \leq 1} \| T^{\ast}Tx\| = \sup_{\| x \| \leq 1} \left| \langle x, T^2 x \rangle \right| \leq \| T \|^2$.

You make a similar gap in both inequalities you are trying to show.  Specifically, you claim that $\sup_{\|x\|\leq 1}|\left< x, Ax\right>| \leq \|A\|$. This claim requires justification.


Sequence which converges pointwise but not uniformly? it might be simple but I don't find a sequence $f_n: [0,1] \rightarrow \mathbb{R}, n \in \mathbb{N}$ that converges pointwise but not uniformly.

First I thought it could be $f_n(x) = \frac{x}{n}$ but it is not right, is it?

Thanks for help!

$f_n(x)=x^n, x\in[0,1)$ $f_n(1)=1$

It converges pointwisely to $f(x)=0, x\in[0,1)$ $f(1)=1$ but not uniformly since uniformly convergence preserves continuity.

Your example converges uniformly on $[0,1]$.

For a different solution, pick any derivative $f$ that is not continuous, for instance, the derivative of $$g(x)=\left\{\begin{array}{cl}x^2\sin(1/x)&x\ne0,\\0&x=0.\end{array}\right.$$ We can always find a sequence of continuous functions converging to $f$ pointwise, but the convergence cannot be uniform, or $f$, being the uniform limit of continuous functions, would be continuous as well.

To see that $f$ is limit of continuous functions, recall that $f=g'$, so for any $x$, $$f(x)=\lim_{n\to\infty}\frac{g(x+1/n)-g(x)}{1/n}.$$

(This requires a small adjustment if $g$ is only defined in $[0,1]$, as adding $1/n$ to points $x$ near $1$ may bump us out of the interval. But not even such an adjustment is needed for the example I suggest above.)


Mathematical difference between white and black notes in a piano The division of the chromatic scale in $7$ natural notes (white keys in a piano) and $5$ accidental ones (black) seems a bit arbitrary to me.

Apparently, adjacent notes in a piano (including white or black) are always separated by a semitone. Why the distinction, then? Why not just have scales with $12$ notes? (apparently there's a musical scale called Swara that does just that)

I've asked several musician friends, but they lack the math skills to give me a valid answer. "Notes are like that because they are like that."

I need some mathematician with musical knowledge (or a musician with mathematical knowledge) to help me out with this.

Mathematically, is there any difference between white and black notes, or do we make the distinction for historical reasons only?

The first thing you have to understand is that notes are not uniquely defined.  Everything depends on what tuning you use.  I'll assume we're talking about equal temperament here.  In equal temperament, a half-step is the same as a frequency ratio of $\sqrt[12]{2}$; that way, twelve half-steps makes up an octave.  Why twelve?

At the end of the day, what we want out of our musical frequencies are nice ratios of small lintegers.  For example, a perfect fifth is supposed to correspond to a frequency ratio of $3 : 2$, or $1.5 : 1$, but in equal temperament it doesn't; instead, it corresponds to a ratio of $2^{ \frac{7}{12} } : 1 \approx 1.498 : 1$.  As you can see, this is not a fifth; however, it is quite close.

Similarly, a perfect fourth is supposed to correspond to a frequency ratio of $4 : 3$, or $1.333... : 1$, but in equal temperament it corresponds to a ratio of $2^{ \frac{5}{12} } : 1 \approx 1.335 : 1$.  Again, this is not a perfect fourth, but is quite close.

And so on.  What's going on here is a massively convenient mathematical coincidence: several of the powers of $\sqrt[12]{2}$ happen to be good approximations to ratios of small integers, and there are enough of these to play Western music.

Here's how this coincidence works.  You get the white keys from $C$ using (part of) the circle of fifths.  Start with $C$ and go up a fifth to get $G$, then $D$, then $A$, then $E$, then $B$.  Then go down a fifth to get $F$.  These are the "neighbors" of $C$ in the circle of fifths.  You get the black keys from here using the rest of the circle of fifths.  After you've gone up a "perfect" perfect fifth twelve times, you get a frequency ratio of $3^{12} : 2^{12} \approx 129.7 : 1$.  This happens to be rather close to $2^7 : 1$, or seven octaves!  And if we replace $3 : 2$ by $2^{ \frac{7}{12} } : 1$, then we get exactly seven octaves.  In other words, the reason you can afford to identify these intervals is because $3^{12}$ happens to be rather close to $2^{19}$.  Said another way,

$$\log_2 3 \approx \frac{19}{12}$$

happens to be a good rational approximation, and this is the main basis of equal temperament.  (The other main coincidence here is that $\log_2 \frac{5}{4} \approx \frac{4}{12}$; this is what allows us to squeeze major thirds into equal temperament as well.)

It is a fundamental fact of mathematics that $\log_2 3$ is irrational, so it is impossible for any kind of equal temperament to have "perfect" perfect fifths regardless of how many notes you use.  However, you can write down good rational approximations by looking at the continued fraction of $\log_2 3$ and writing down convergents, and these will correspond to equal-tempered scales with more notes.

Of course, you can use other types of temperament, such as well temperament; if you stick to $12$ notes (which not everybody does!), you will be forced to make some intervals sound better and some intervals sound worse.  In particular, if you don't use equal temperament then different keys sound different.  This is a major reason many Western composers composed in different keys; during their time, this actually made a difference.  As a result when you're playing certain sufficiently old pieces you aren't actually playing them as they were intended to be heard - you're using the wrong tuning.

Edit:  I suppose it is also good to say something about why we care about frequency ratios which are ratios of small integers.  This has to do with the physics of sound, and I'm not particularly knowledgeable here, but this is my understanding of the situation.

You probably know that sound is a wave.  More precisely, sound is a longitudinal wave carried by air molecules.  You might think that there is a simple equation for the sound created by a single note, perhaps $\sin 2\pi f t$ if the corresponding tone has frequency $f$.  Actually this only occurs for tones which are produced electronically; any tone you produce in nature carries with it overtones and has a Fourier series

$$\sum \left( a_n \sin 2 \pi n f t + b_n \cos 2 \pi n f t \right)$$

where the coefficients $a_n, b_n$ determine the timbre of the sound; this is why different instruments sound different even when they play the same notes, and has to do with the physics of vibration, which I don't understand too well.  So any tone which you hear at frequency $f$ almost certainly also has components at frequency $2f, 3f, 4f, ...$.

If you play two notes of frequencies $f, f'$ together, then the resulting sound corresponds to what you get when you add their Fourier series.  Now it's not hard to see that if $\frac{f}{f'}$ is a ratio of small integers, then many (but not all) of the overtones will match in frequency with each other; the result sounds a more complex note with certain overtones.  Otherwise, you get dissonance as you hear both types of overtones simultaneously and their frequencies will be similar, but not similar enough.

Edit:  You should probably check out David Benson's "Music: A Mathematical Offering", the book Rahul Narain recommended in the comments for the full story.  There was a lot I didn't know, and I'm only in the introduction!

If you only had a repetitive series of keys on your piano it would be a bit difficult to visually get some reference points. I think this is the main reason wh


How fast would you need to move your hand to generate a sine wave? Lately I have been completely absorbed in the mathematics of sine waves, how they work, and how they are intertwined with many important maths principles that we follow today. I was thinking to myself "would it be possible to generate a near-perfect sine wave by hand using some rope". Now, when I say "generate a sine wave by hand", I mean to swing a piece of rope back and forth while the other end is tied to some fixed point in such a way that if you were to freeze the piece of rope in time and measure the amplitude of the wave, you would get 2 (-1 to +1), and get some wavelength ($\ \lambda$ ) that is similar to the value of $\ 2 \pi$. Now this may seem simple - yes. However, when you look into it that way of thinking rapidly falls apart.
Firstly, as we all know, $\ \pi $ is irrational. You cannot possibly write out all digits of $\ \pi $ on paper. This quickly terminates the thinking that the answer is "yes". But the answer isn't no either.
Well, why don't we ask a different question - How close can you get?
If we define our hand as an object with just one parameter - the speed through a defined medium (air) at which it can move (ie: The fastest speed at which a hand can move). I will define this speed as $\ \mu $. By changing $\ \mu $ we can modify how fast the hand can move.
If we take a look at the$\ \sin x$ function, we will see that it repeats its cycle every $\ 2 \pi $ units, and moves up and down between the numbers -1 and 1. So that gives us 2 variables to try and get close to. So if we want our hand to move back and forth at an amplitude of 2, we would need our hand to move up and down every 2 units (so if our units were centimetres (CM), the sine wave would need to reach its highest point in the first second, and its lowest point in the next). But then the sine wave needs to have $\ \lambda = 2 \pi $ (or something close to it).  So what would the value of $\ \mu $ be?

I worked really hard to solve this question, and I think this will work.
First of all, because the speed at which you move your hand is non-linear, I cannot represent it as a fixed value. It must be written as a function of $\ \mu (t) $. Here is what I have come up with:
$$\ \mu(t) = \frac{\int_{0}^t \sin x dx}{t}$$
Explanation:
We take the antiderivative of $\ \sin x$ from $\ 0 $ to $\ t $ since we want to get the distance travelled by the sine wave up until that point (the area underneath a sine wave will be the total distance travelled by the wave up until that point) and since the value of $\ t$ is also the time that we have been moving our hand, we then divide the distance by it (see: speed) to get the speed of your hand at that point. Voila! We get our speed that our hand must be moving at at that point in time.
If there are any mistakes, please comment on them and I will modify my answer!

You should not look at sin(x) but at sin (ax) for different a. so you get manny different waves. but a wave has a time and a space variable so a wave on a very long rope would have the form $$ y=A*sin({\frac{2\pi}{T}*t-}\frac{2*\pi}{c*T} *x)$$
T ist the period of one up and down, c the velocity of the propagation, cT the wavelength of the wave.
if the rope is is tied at one end you can only create a standing wave (look it up in wiki)
To move your Hand in a sin way  you take any turntable , put a marker on it and move your hand, so it follows the markers horizontal movement.


Plotting symbolic function using symsum I'm attempting to plot the temperature distribution with respect to time.

I found that the solution to the Heat Equation with initial temperature distribution of $f(x)=|x|$ on $[-\pi,\pi]$ is

$$u(x,t)=\frac{\pi}{2}+\sum\limits_{n=1}^\infty\frac{2}{\pi n^2}((-1)^n-1)e^{-n^2t}\cos(nx)$$

So far I was able to come up with the following Matlab code:

syms t;
r = symsum((2/(pi*n^2))*((-1^n)-1)*cos(n*x)*exp(-(t*n^2)),n,1,1000);
grid on
ezplot(r+(pi/2),[-1,1],[-1,1])

When I try to run this get the following error:

??? Undefined function or variable "X".

Error in ==> ezplot>ezimplicit at 252
[X,Y] = meshgrid(X,Y);

Error in ==> ezplot at 153
    hp = ezimplicit(cax,f{1},vars,labels,args{:});

Error in ==> sym.ezplot at 71
   h = ezplot(char(f),char(y),varargin{2:end});

Error in ==> fp at 26
ezplot(r+(pi/2),-1,1,-1,1)

I know that I have to adjust something small, since I was able to plot the infinite series $\sum\limits_{n=1}^\infty \frac{x^k}{k!}=e^x$ on $[0,1]$. The big difference between this example and my problem is that what I'm trying to plot has two variables (which I noticed Matlab didn't recognize per the error message), but I read online that I can specify the range for it in the argument of ezplot, yet I still came up with an error.

I'm wondering if there's an easier way of approaching this problem. I'm fairly new to Matlab and learn as I go.


Result of the limit: $\lim_{x \to \infty} \sqrt{x + \sin(x)} - \sqrt{x}$? From my calculations, the limit of
$\lim_{x \to \infty} \sqrt{x + \sin(x)} - \sqrt{x}$

Is undefined due to $sin(x)$ being a periodic function, but someone told me it should be zero.

I was just wondering if someone could please confirm what the limit of this function is? Thanks
Corey :)

I think the limit is $0$, since $sin(x)$ is bounded: At best, it is $1$ or $-1$, but if $x \rightarrow \infty$, a bounded number will look like a const; it won't influence the limit, so basically you have $\sqrt{x}- \sqrt{x}= 0$


Are "if" and "iff" interchangeable in definitions? In some books the word "if" is used in definitions and it is not clear if they actually mean "iff" (i.e "if and only if").
I'd like to know if in mathematical literature in general "if" in definitions means "iff".

For example I am reading "Essential topology" and the following definition is written:

  In a topological space $T$, a collection $B$ of open subsets of $T$ is said
  to form a basis for the topology on $T$ if every open subset of $T$ can be written as a union of sets in $B$.

Should I assume the converse in such a case?
Should I assume that given a basis $B$ for a topological space, every open set can be written as a union of sets in $B$?

This is just an example, I am not asking specifically about this sentence.

As it is a definition, the validity of the property (here "being a basis for the topology") must be defined by it in all cases. So implicitly all cases not mentioned do not have the property.

This convention is even stronger than the "if" meaning "iff" in definitions. Take as example the definition "an integer $p>1$ is called a prime number if it cannot be written as a product $p=ab$ of integers $a,b>1$". This says that $6$ is not a prime number, since $6=2\times 3$; this is an instance of the "iff" meaning in definitions. But it also says implicitly that $1$ is not a prime number, nor $-5$ nor $\pi$ nor $\exp(\pi\mathbf i/3)$ nor $\mathbf{GL}(3,\Bbb R)$, as none of these can be described as integers $p>1$; even a statement with "iff" would in itself not seem to state non-primality of those objects. But since it is a definition, anything that does not match its description is implicitly excluded from the property.

Without this implicit exclusion of cases not mentioned, it would be very hard indeed to give a complete definition of any property. Imagine (assuming the "everything is a set" philosophy) the ugliness of "a set $x$ is called a prime number if $x\in\Bbb Z$ and $x>1$ and ...".

"if" and "iff (if and only if)" have the difference. This difference is about their imperativeness. The "iff" is more imperative than just "if".

The analogous can be found in "needed and sufficient condition". "if" is analogue to "sufficient condition", whereas "iff" is analogue to "needed and sufficient condition".


What is one set of axioms which are sufficient for Calculus? I am curious to find out what are the "minimum axioms" needed in order to be able to have highschool level math. Another way to explain it: if we were visited by a super intelligent race of aliens (assume no language barriers), which axioms would we have to give them in order for them to be able to start completely from scratch and come up with calculus as rigorously as possible? Would the axioms of logic, ZF axioms, Peano axioms be enough?

In my experience, I think you would need at least Peano's Axioms, set theory and predicate logic. Using them, I was able to formally construct the positive real numbers as Dedekind cuts.

The set theory should include rules for: subsets, power sets, Cartesian products, set equality, n-tuple equality, functions, pairwise and arbitrary union and choice.

Your predicate logic should include rules for equality, and allow quantification over sets and functions.

If your set theory included some kind of axiom of infinity (as does ZFC), you might be able to derive the Peano Axioms.


Extended Euclidean Algorithm for Modular Inverse I'm currently learning how to find the inverse of a modulo with the Extended Euclid Algorithm and I stumbled upon a problem when finding an inverse when
the $m>p$ as for $m \equiv 1 \pmod{p}$

For example, in
$$
240x \equiv 1 \pmod{17}
$$
what is the inverse? What are the steps to find it?

You have to write
$$
1 = 240x+17y
$$
so
$$
240x\equiv 1\pmod{17}
$$

The Euclidean algorithm applied to $240$ and $17$ gives
\begin{align}
\color{red}{240} &= \color{red}{17}\cdot 14 + \color{red}{2} \\
\color{red}{17} &= \color{red}{2}\cdot 8 + \color{red}{1}
\end{align}
The successive remainders are colored red. Now start from the top:
$$
\color{red}{2}=\color{red}{240}-\color{red}{17}\cdot 14
$$
Go one line down:
$$
\color{red}{1}=\color{red}{17}-\color{red}{2}\cdot 8
$$
Substitute the value you have for $\color{red}{2}$:
$$
\color{red}{1}=\color{red}{17}-(\color{red}{240}-\color{red}{17}\cdot 14)\cdot 8
=\color{red}{17}\cdot(1+132)-\color{red}{240}\cdot 8
=\color{red}{240}\cdot(-8)+\color{red}{17}\cdot 133
$$
Thus you can take $x=-8$, or else $x=9$ since $-8\equiv 9\pmod{17}$. Indeed
$$
240\cdot9=2160=17\cdot127+1
$$
or
$$
240\cdot 9\equiv 1\pmod{17}.
$$

Just remember operating on the “red numbers” as if they were letters. Express each remainder in terms of the previous ones and substitute in the equations below the first. Only terms with $\color{red}{240}$ and $\color{red}{17}$ multiplied by integers will remain.

In this case, m > p. So find the remainder of m/p. m mod p = 240 mod 17 = 2.
so the inverse of 240 and the inverse of 2 (mod 17) are the same. The above answer stating the inverse is 9 is correct (2 * 9 = 18 and 18 mod 17 = 1)


Solving $\tan x = x + 1$ analytically. In my books a problem is given as follows:

  Prove that least positive value of x satisfying $\tan{x} = x+1$ lies in interval $(\frac{\pi}{4},\frac{\pi}{2} )$.

I tried to solve it as folllow

$$\tan{x} = x+1$$
$$\frac{\sin x}{\cos x} = x+1$$
$$\sin{x}= (x+1)\cos{x}$$

I differentiate the expression I

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

$$-(x+1)\sin x= 0$$

Here by $\sin{x} \neq 0$ because for this condition $x = \pi n$ which lies outside the interval and $x = -1$ is outright wrong

The book solves this kind of question using graph by studying the intersection of $y = \tan x$ and $y = x+1$.

SO WHY I DON'T WANT TO USE GRAPHS?

Because, the problem is we don't have graphs with us in exams and sometimes it's not feasible to draw graph by hands

So how I may go about these kind of trigonometrical equations?

Let $$f(x)=\tan(x)-x-1$$
Then
$$f(\frac{\pi}{4})=-\frac{\pi}{4} <0 \\
\lim_{x \to \frac{\pi}{2}} f(x)=+\infty
$$
and use the Intermediate Value Theorem...If you didn't cover that (yet), this is probably what the textbook is trying to argue.

To prove that this is the smallest positive root, you can use the fact that
$$f'(x)>0 \mbox{ on } (0, \frac{\pi}{2})$$

As for graphs, you should be able to draw the graphs of $\tan(x)$ and $x+1$ by hand in the exam. If the problem expects you to draw the graphs, the problem is chosen so that it is feasible to draw them by hand.

I am not an expert but this might be one of the way to know the root lie in the given interval or not


How to solve $\int _{\frac{\pi }{8}}^{\frac{\pi }{3}}\frac{dx}{1+\left( \tan x\right) ^{\sqrt{2}}}$ Calculate $$\int _{\pi/8}^{3\pi/8}\dfrac{dx}{1+\left( \tan x\right) ^{\sqrt{2}}}\,.$$

I thought.
If I put $t=\tan x$, it becomes $dx=\cos ^{2}xdt$ and $x$ remains.
So I can do integration by substitution.
And I thought that rationalize the denominator but I can't do.

This problem looks to be a trap.
If you look at the integrand in the range of integration, it is almost a straight line.
Expanding as a Taylor series around $x=\frac \pi 4$ (almost the middle of the interval), we have (I wrote a lot of terms knowing that most of them will not have any impact of the result)
$$\dfrac{1}{1+\left( \tan (x)\right) ^{\sqrt{2}}}=\frac{1}{2}-\frac{\sqrt{2}}{2}\left(x-\frac{\pi }{4}\right)+\frac{\sqrt{2}}{15}
   \left(x-\frac{\pi }{4}\right)^5+\frac{16\sqrt{2}}{315}  \left(x-\frac{\pi
   }{4}\right)^7+\frac{13\sqrt{2}}{405}  \left(x-\frac{\pi }{4}\right)^9+$$ $$\frac{736
   \sqrt{2} }{31185}\left(x-\frac{\pi }{4}\right)^{11}+\frac{135098 \sqrt{2}}{6081075}\left(x-\frac{\pi }{4}\right)^{13}+O\left(\left(x-\frac{\pi}{4}\right)^{15}\right)$$
On the other side
$$\int _{\frac\pi 8}^{\frac\pi3} \left(x-\frac{\pi }{4}\right)^{2n+1}\, dx=\frac{ \left(\left(\frac{4}{9}\right)^{n+1}-1\right) \pi ^{2 n+2}}{2^{6 n+7}(n+1)}$$ Cumulating these terms and converting them to decimals, we have the following results
$$\left(
\begin{array}{c}
 0.3272492347 \\
 0.3575394439 \\
 0.3574868754 \\
 0.3574819953 \\
 0.3574816063 \\
 0.3574815691 \\
 0.3574815645 \\
 0.3574815638
\end{array}
\right)$$ while numerical integration would give $0.3574815637$.
We can have a "nice looking" approximation builging the $[4,4]$ Padé approximant of the integrand
$$\dfrac{1}{1+\left( \tan (x)\right) ^{\sqrt{2}}}\sim\frac{\frac 12 + \sum_{m=1}^4 a_m \left(x-\frac{\pi }{4}\right)^m}{1+ \sum_{m=1}^4 b_m \left(x-\frac{\pi }{4}\right)^m }$$ where the $a_m$ are
$$\left\{-\frac{\sqrt{2}}{2},-\frac{8}{21},\frac{8 \sqrt{2}}{21},\frac{1}{15}\right\}$$ and the $b_m$ are
$$\left\{0,-\frac{16}{21},0,\frac{2}{15}\right\}$$ Using partial fraction decomposition we end with
$$\int _{\frac\pi 8}^{\frac\pi3}\dfrac{dx}{1+\left( \tan (x)\right) ^{\sqrt{2}}}=\frac{5 \pi }{48}-\frac{173}{28} \sqrt{\frac{5}{13}} \tanh ^{-1}\left(\frac{40
   \sqrt{130} \pi ^2}{483840-4160 \pi ^2+7 \pi ^4}\right)-$$ $$\frac{10\sqrt{2}}{7}
   \tanh ^{-1}\left(\frac{1600 \left(\pi ^2-273\right)}{651840-4160 \pi ^2+7 \pi
   ^4}\right)-\frac{40}{7} \sqrt{2} \coth ^{-1}(5)$$ which is $0.3574816071$.

As per the suggestion of @J.G the lower limit needs to be $\pi/6$, otherwise it is not doable by hand
Use the property that $$\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx.~~~~(1)$$
$$I=\int_{\pi/6}^{\pi/3} \frac{1}{1+\tan^{\sqrt{2}}x} dx~~~~(2)$$
Using (1), we get
$$I=\int_{\pi/6}^{\pi/3} \frac{1}{1+\tan^{\sqrt{2}}(\pi/2-x)} dx=\int_{\pi/6}^{\pi/3} \frac{1}{1+\cot^{\sqrt{2}}x} dx~~~~(2)$$
adding (2) and (3), we get
$$2I=\int_{\pi/6}^{\pi/3} \left(\frac{1}{1+\tan^{\sqrt{2}}x}+\frac{1}{1+\cot^{\sqrt{2}}x}\right)dx=\int_{\pi/6}^{\pi/3} 1 dx=\pi/6$$
$$\implies I=\pi/12$$


If $A=AA^{\top}$, show that $A^2=A$ I've been working trying to understand the following question:

Let n be a positive integer, let $F$ be a field, and let $A \in \mathrm{Mat}(n,F)$ satisfy the condition $A=AA^{\top}$. Show that $A^2=A$.

I haven't made much progress since my knowledge is pretty basic but I ran across this link and was wondering if this example was essentially the same?

Example

Since $A^T = (AA^T)^T = AA^T = A$, You have $A^2 = AA = AA^T = A$

$AA^T=A$

$A$ from the LHS to the RHS

$A^T=AA^{-1} \Rightarrow A^T=I\Rightarrow A=I$

$I^2=I \Rightarrow A^2=A$

Edit: It works only, if $det \ A\neq 0$


Increasing property of solution of heat equation Let $$u_t - \Delta u = f$$
with $u(t,0) = u(t,1) = 0$ and $u(0,x) = u_0(x)$ given, and $f(t,x)$ is also given. This is the heat equation on the interval $(0,1)$.

How can I choose $f$ and $u_0$ to ensure that $u$ is increasing with respect to time? I.e. if $t \geq s$ then $u(t,x) \geq u(s,x)$ for a.e. $x$?

If $f \equiv 0$, then $u$ is decreasing with respect to $t$. But I don't know how to ensure that it becomes increasing. Does anyone know?

For the Allen-Cahn equation, $u_t = \epsilon^2 u_{xx} + u - u^3$, it's possible to have $u(t, x) \geq u(s, x), (t \geq s)$. The AC equation can be viewed as the $L^2$ gradient of the Ginzburg-Landau free energy $E = \int_\Omega 1/2 \epsilon^2 |u_x|^2 + 1/4(1 - u^2)^2 \mathrm{d} x$. The second term is called a double-well potential. If $u_0 = c \in (0, 1)$ in $[a, b]$, then the solution $u$ will evolve to $1$ finally.


Techniques to solve such a PDE I have the eigenvalues problem on $[0,\pi] \times [0,2\pi]$
$$\left(\frac{1}{\sin\theta}\frac{\partial}{\partial \theta} \left[\sin\theta \frac{\partial}{\partial \theta}\right] + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\right)f (\theta,\phi) + g(\theta)f(\theta,\phi) = \lambda f(\theta,\phi).$$

The differential operator is the angular part of the laplacian in spherical coordinates and $g \in C^{\infty}$.
What I want to know is: Are there any other techniques of solving this PDE than using the particular ansatz: $f(\theta,\phi) = h(\theta) \frac{e^{in \phi}}{\sqrt{\sin(\theta)}}$?

Also another ansatz would be interesting. I am just interested in any technique that could be useful here.

If anything is unclear, please let me know.

I would ordinarily leave a comment in this case, because I don't have any real answers. But it's too long to leave in a comment.

For ODEs $-ay''+by'+cy=\lambda w y$, the common techniques are

Multiply on the left by $\rho$ to get $\rho(ay''+by')= ((\rho a) y')'$ which (essentially) uniquely determines $\rho$. This requires minimal smoothness of coefficients.
Change the dependent variable $y=\rho f$ to obtain a new equation where $a$, $w$ are unchanged, but you end up with selfadjoint form $-(ay')'+qy=\lambda wy$. This one is not unique because you can multiply the entire equation by any factor before applying the technique, which is very helpful, but leaves you without unique decisions. This method also requires more smoothness for the coefficients, which limits its theoretical applications, but does not affect you for this problem.
Starting with $-(ay')'+qy=\lambda wy$ with $a > 0$, change the independent variable by writing $y(x)=f(\int \frac{1}{a}\,dx)$ assuming $1/a$ is locally integrable on the interior of the interval. Then you get $-\frac{1}{a}f''+qy=\lambda wy$, but you must invert $\int\frac{1}{a}\,dx$ to write this as $-f''+Qy=\lambda W y$.

Using these forms you can get $-f''+Qy=\lambda y$ with enough differentiability for the coefficients of the original equation.

About the only thing left to try is $f(\theta,\phi)=F(\int_{\pi/2}^{\theta}\frac{1}{\sin\theta'}\,d\theta',\phi)$ to transform your equation to
$$
     \frac{1}{\sin^{2}\theta}\left[F_{11}+F_{22}\right]+g(\theta)F=\lambda F.
$$
You'll need to invert $u=\int_{\pi/2}^{\theta}\frac{1}{\sin\theta'}d\theta'$ to change $g(\theta)$ into $h(u)$, and $\sin^{2}(\theta)$ into $k(u)$. The new equation will be on $(-\infty,\infty)\times[0,2\pi]$. There may be something to be gained because of this becoming the Laplacian again, except that you'll have the new weighted space with weight $k(u)$ instead of $\sin\theta$, and some of the niceness of $g$ may be erased. I'm not hopeful because others have been down this road, and it sounds like you need something new.

Have you tried SLEIGN2 ? (Sturm-Liouville Eigenvalue/Eigenfunction package) This is a FORTRAN project by some impressive Mathematicians in the field. http://www.math.niu.edu/SL2/intro.pdf . Remove the intro.pdf from the link to get the *.f files.


What math do I need to learn to create a general equation solver? I'm interested in creating a general equation solving engine, akin to wolfram alpha (though it doesn't need to be quite as advanced). I'd like it to be able to solve any of the equations students encounter from grade school to early college; so including...

Simple real number polynomials
Equations of complex variables
Vector and Matrix equations
Equations with differentiation and integration
Differential equations
(Possibly) Boolean equations

Also, I'd like to be able to solve these equations axiomatically. For example if you wanted to solve $x + 3 = 5$, you could find the steps like...

Subtract $3$ on the right hand side

$$(x + 3) - 3 = 5 - 3$$

Associativity of addition

$$x + (3 - 3) = 5 - 3$$

Additive inverse

$$x + 0 = 5 - 3$$

Identity element

$$x = 5 - 3$$

Subtraction

$$x = 2$$

So I realize this is a lot, and before I get myself in too deep, I'm looking to know what math I should know before I would be able to create something like this.

So I'm assuming it'd be mostly algebra. I've studied group theory a bit, and touched on rings. Could someone with more knowledge, point me in the right direction? Is this even feasible for a single person?

Think for approaches yourself and then have a look at how others did it.

Back in my early undergraduate days I thought "Why not write a program that produces theorems in an endless fashion and then go pick some new to publish them?" Given that cute idea I wondered how math worked at core and how to formalize it. I noticed two things:

Math objects have a class hierarchy pattern. A group is a pair of a set and a binary operation on it. A pair is a two-element set etc. pp.
Functionals in a boarder sense, i.e. objects that assign objects to objects play a central role and just using functions/methods (in the programming sense) capsuled in classes would not be enough to imitate this.

So in the end I came to believe I would need an object-oriented functional language to bring math into the computer. Imagine my surprise when I found out about Coq and that it does exactly that.

Notice that, as opposite to Wolfram Alpha, Coq is open source, so you can learn a lot from it and also use it as backend.

In the end, I became more interested with formally proofing theorems than to find them (I also realized my idea is close to read interesting information out of $\pi$), so now I'm a Mizar enthusiast. Mizar is not open source, and it doesn't proof things for you but checks your proofs, very rigorously.


Limit comparison test on two same series? I am not quite sure how to phrase this question.
I have a series $$\sum_{n=1}^{\infty} |a_n|$$
I also have another series, for a natural number $N \geq 1$ $$\sum_{k=0}^{\infty}A_{k} = \sum_{n=N}^{\infty} |a_n|$$
Now, I am asked to write $\sum_{n=1}^{\infty} |a_n|$ in terms of $\sum_{k=0}^{\infty}A_{k}$
Since $\sum_{k=0}^{\infty}A_{k} = \sum_{n=N}^{\infty} |a_n|$, then (assuming I am correct) $$\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{N} |a_n| + \sum_{k=0}^{\infty}A_{k}$$
Now the question asks me to show that one series converges if and only if the other series converges.
First thing I thought was the limit comparison test
If I do limit comparison test on $\sum_{n=1}^{\infty} |a_n|$ and $\sum_{k=0}^{\infty}A_{k}$, does the starting point matter?($n=1 \quad n=N$)
If not, this will end up as $$\lim_{n \rightarrow \infty} \frac{|a_{n}|}{|a_{n}|} = 1$$
And since the limit exists, both series will have the same result for convergence test.
My concern is that I am not sure the steps I took were legit.

We have in fact that
$$\sum_{n=1}^{\infty} |a_n| = \sum_{r=1}^{N-1} |a_r| + \sum_{k=0}^{\infty}A_{k}$$
or using limits of the corresponding seuqneces of partial limits:
$$\lim_{n\to\infty}\sum_{j=1}^n|a_j|=\sum_{r=1}^{N-1}|a_r|+\lim_{K\to\infty}\sum_{k=0}^KA_k$$
Observe that the first term in the right side is a finite number as it is a finite sum.
And since all the series (finite and infinite) are of positive terms, comparison and arithmetic of limits come in handy: if $\;\sum\limits_{k=0}^\infty A_k=T\;$ converges, then the sequence $\;\left\{\,\sum\limits_{j=1}^n|a_j|\,\right\}_{n=}^\infty\;$ is monotonic ascending and bounded above, by the number $\;\sum\limits_{r=1}^{N-1} |a_r|+A\;$ and it thus converges. and something similar the other way around: if $\;\sum\limits_{j=}^\infty |a_j|=S\;$ converges, then by arithmetic of limits
$$\lim_{K\to\infty}\sum_{k=0}^KA_k=\lim_{n\to\infty}\sum_{j=2}^n|a_j|-\sum_{r=1}^{N-1}|a_r|=S-\sum_{r=1}^{N-1}|a_r|$$


The smallest positive period of the function $\sin{(k\cos{x})}$ let $$f(x)=\sin{(k\cos{x})},k>0$$
  Find the minimum of $T>0$,such
  $$f(x+T)=f(x),\forall x\in R$$

I have know $$f(x+2\pi)=\sin{(k\cos(x+2\pi))}=f(x)$$
maybe $T_{min}=2\pi$?,I can't proof why$2\pi$  is smallest? Thanks

Let $T$ be the minimum period of $f$. Since the period of $\sin(x)$ is $2\pi$, it must be true that

$$k\cos(x+T)=k\cos(x)+2\pi n\\
\implies \cos(x+T)=\cos(x)+\frac{2\pi n}{k}$$

for some $n \in \Bbb{Z}$.  The range of $k\cos(x+T)$ is $[-1,1]$. The range of $\cos(x)+\frac{2\pi n}{k}$ is $\big[-1+\frac{2\pi n}{k}, 1+\frac{2\pi n}{k}\big]$.  These can only be equal if $n=0$.  Hence,

$$\cos(x+T)=\cos(x)$$

But we know the minimum period of $\cos(x)$ is $2\pi$, so $T=2\pi$.

If
$f(x)=\sin(k\cos(x))
$
then
$f(x+2\pi)
=\sin(k\cos(x+2\pi))
=\sin(k\cos(x))
=f(x)
$
since
$\cos(x+2\pi)
=\cos(x)
$.


Prove that a sequence of reals together with its limit is compact by cover I need to prove that the set $K =\{X_n: n\in\Bbb N\}\cup \{l\} $ is compact by covers. Where  $ l=\lim_{n \to \infty} X_n$. Where $X_n$ is a sequence of values in $\mathbb R$.

I understand that there are 2 cases.
Case 1) K is countable and finite, where G = {$G_i$}$_{i\in I} $ is an arbitrary cove of K then $G^* = {G_i}_{n}$ is a finte subcover G.

I need help analyzing the other case.

hint

Assume $$K\subset \cup_{i\in I} O_i $$

then
$$\exists j\in I \;:\; l\in O_j .$$

but $$l=\lim_{n\to +\infty} X_n $$

thus $$\exists N\in\mathbb N \;:\;n> N \;\; \implies \;\;X_n\in O_j $$

and
$$K\subset (\cup_{p=0,1,2...N}O_p )\cup O_j $$

where $O_p $ is such that $$X_p\in O_p $$


Motivation for the Definition of Compact Space A compact topological space is defined as a space, $C$, such that for any set $\mathcal{A}$ of open sets such that $C \subseteq \bigcup_{U\in \mathcal{A}} U$, there is finite set $\mathcal{A'} \subseteq \mathcal{A}$ such that $C \subseteq \bigcup_{U'\in \mathcal{A'}} U'$.

Now, this definition leads to many interesting results, but if I were teaching someone about compact sets, how would I motivate this? Concepts like sequential compactness, open and closedness, and even connectedness are reasonably easy to motivate. I can not see how to motivate this definition. Compact spaces are often seen as generalizations of finite spaces. They are also seen as a generalization of boundedness and closedness. I can't see how to connect the definition with these concepts.

Alternatively, is there a definition of a compact set which is easier to motivate?

One of my favorite textbooks is Klaus Janich's Topology, and he has a nice motivation for compactness I feel, namely why we should care about. This is in addition to my comment about compact subsets of a Hausdorff space being essentially like finite point sets. But he writes:

  In compact spaces, the following generalization from "local" to "global" properties is possible: Let $X$ be a compact space and $P$ a property that open subsets of $X$ may or may not have, and such that if $U$ and $V$ have it, then so does $U\cup V$. Then if $X$ has this property locally, i.e. every point has a neighborhood with property $P$, then $X$ itself has property $P$.

This is nice, but it is slightly advanced, and he gives some examples that follow like a continuous/locally bounded map from a compact space to $\mathbb{R}$ is bounded, and some discussions of locally finite covers and manifolds (honestly, I like this book after the fact of learning topology, not to learn from).

Hope that helps somewhat.

Well I like to think of compactness like this because it gives me the idea of a compact set in terms of closed sets.

A collection $\mathcal{C}$ of subsets of $X$ is said to have the finite intersection property if for every finite subcollection

$$\{C_1,\cdots C_n\}$$
of $\mathcal{C}$ , $\bigcap_{i=1}^{n} C_i$ should be nonempty.

Now if in a topological space every collection $\mathcal{C}$ of closed sets in $X$ have the finite intersection property , then the space $X$ is compact.

Well instinctively you would always think of a compact set as a closed set(though we have examples where compact sets are not closed.. One lies in your backyard!!)  here is a definition involving closed
 sets.

Also you can use this definition to prove the tychonoff theorem


Differentiation of Taylor Series Let $g(x) = e^{-1/x^2}$ for $x$ not equal to zero, and $g(0) = 0$.

a) Please Show that $g^{(n)}(0) = 0$, for all $n = 0,1,2,3,4, \ldots$

Can someone please elaborate on the comments below for this one?

b) Please Show that the Taylor Series for $g$ about 0 agrees with $g$ only at $x = 0$.

I think this would be easy once I have part a, all I have to do is plug in n = 0?

Can someone please show how to do this?

A good way to do this is to prove something stronger.  Let $P$ be any polynomial.  Then show that the function $h$ defined by
$h(x) = P(1/x)e^{-1/x^2}$ for $x \ne 0$ and $h(x)=0$ has your property $h^{(n)}(0) = 0$ for all $n$.  Do it by showing: (1) $h$ is continuous; and (2) the derivative of such a function $h$ is another function of the same kind.

All you have to do is prove that $\dfrac{\rm{d}g(x)}{\rm{d}x} = (\text{something})\,g(x)$ so that the limit of any derivative when $x\rightarrow 0$ is zero, since g(0)=0.

I see that

$$\dfrac{\rm{d}g(x)}{\rm{d}x} = \dfrac{2}{x^3}\,g(x) $$


Tricky Logarithmic inequality I have tried proving this logarithmic inequality but I did not succeed. I tried to put every term on one side, I expanded and tried to use one of the properties of logarithms but the proof does not come out clearly mathematically.

$$\log_2(x) \leq (x-1)\log_2(e)$$

Thanks for your response

Hint

For $x \geq 1$ we have $ex \leq e^x$ (can you see why?). Then use monotonicity of loagrithm to take log of both sides with base $2$.

Actually this is easy.

Notice that we have $$ \log_2(x) = \frac{\log(x)}{\log(2)} $$

Plug it in both side and we have
$$
\frac{\log(x)}{\log(2)} \leq \frac{x-1}{\log(2)} \\
\log(x) \leq x-1
$$
which is trivial.


Zorich's proof on L’Hospital’s Rule The following proof on L’Hospital’s Rule is excerpted from Zorich's Mathematical Analysis I, I have questions concerning the selected part in the proof :

Why it is obviously possible to have $\dfrac {f\left( y\right) }{g\left( x\right) }\rightarrow 0$ and $\dfrac {g\left( y\right) }{g\left( x\right) }\rightarrow 0$ as $x\rightarrow a^{+}$ and $y\rightarrow a^{+}$, especially under the  hypothesis $2^0$ ? Anyone can explain it in detail?
The proof also requires $y\rightarrow a^{+}$, so I found the
conclusion the proof come to is NOT $\dfrac {f\left( x\right)
}{g\left( x\right) }\rightarrow A$  as $x\rightarrow a^{+}$, but
should also added with the condition $y\rightarrow a^{+}$, and since
$x<y$, then $x\rightarrow a^{+}$ could be ensured by $y\rightarrow
a^{+}$, so I think the proof only proved that $\dfrac {f\left(
x\right) }{g\left( x\right) }\rightarrow A$  as $y\rightarrow a^{+}$
( but not $x\rightarrow a^{+}$). In a nutshell, I think the proof
didn't achieve the goal of proving L’Hospital’s Rule, am I all right
?

As to your second question which you haven’t found a solution for, note that $\frac{f(x)}{g(x)}$ equals $\frac{f(y)}{g(x)}+\frac{f’(\zeta(y))}{g’(\zeta(y))}(1-\frac{g(y)}{g(x)})$ for every $y$, so you can make $y$ approach any quantity in the right side of this equality without affecting the left side.

Here is one way to prove the second part of L'Hospital's Rule which deals with the form "$\text{anything} /\infty $".

Let $\epsilon >0$ be arbitrary and then we choose a $\delta_1>0$ such that $$L-\epsilon<\frac{f'(x)} {g'(x)} <L+\epsilon\tag{1}$$ whenever $a<x<a+\delta_1$. Next choose a fixed number $x_0\in(a,a+\delta_1)$ and then we have the identity $$\frac{f(x)} {g(x)} =\frac{f'(c_x)} {g'(c_x)} \left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)}\tag{2}$$ for all $x$ with $a<x<x_0$ and some $c_x\in(x, x_0)$.

Since $g(x_0)/g(x)\to 0$ as $x\to a^+$ we can find another $\delta_2>0$ such that $$\left|\frac{g(x_0)}{g(x)}\right|<1\tag{3}$$ whenever $a<x<a+\delta_2$. Let $$\delta=\min(x_0-a,\delta_2)$$ and then we have the inequality $$(L-\epsilon) \left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)} <\frac{f(x)} {g(x)} < (L+\epsilon) \left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)}$$ whenever $a<x<a+\delta$ using $(1),(2),(3)$.

Now taking limits as $x\to a^+$ we get $$L-\epsilon\leq\liminf_{x\to a^+} \frac{f(x)} {g(x)} \leq\limsup_{x\to a^+} \frac{f(x)} {g(x)} \leq L+\epsilon$$ Since $\epsilon $ is arbitrary it follows that $f(x) /g(x) \to L$ as $x\to a^+$.

Note that $x_0$ is fixed in above proof and the variable $c_x$ lying in $(x, x_0)$ does not necessarily tend to $a$ as $x\to a^+$. Hence we can't the predict the limiting behavior of ratio $f'(c_x) /g'(c_x) $. But we know that this ratio lies between $L-\epsilon $ and $L+\epsilon$ and that is sufficient to predict the limiting behavior of $f/g$.

The same proof works with minor modifications if $L=\infty$ or $L=-\infty$.


How many commutative rings of order $p^2$ This question was part of my algebra quiz and I was unable to solve 1 option out of 4. So, I am asking it here.

There exists exactly 2 commutative rings of order $p^2$ are there?

I have a intuition that it's true as only two groups exists and as we are adding a multiplicative operation so that must be true but it's given false.
So, kindly tell what mistake I am making. I studied algebra from Joseph Gallian.

If the ring $R$ has characteristic $p^2$, then it must be $\mathbb Z/(p^2)$. Otherwise $R$ has characteristic $p$, so there’s a ring homomorphism $\mathbb F_p\to R$ (and so $R$ is an $\mathbb F_p$-algebra). Take any element $x$ not in the image of $\mathbb F_p$. Then there is an algebra homomorphism $\mathbb F_p[t]\to R$, $t\mapsto x$, necessarily surjective by comparing vector space dimensions. The kernel is generated by a monic polynomial $f$ of degree 2, and we see that there are three possibilities. Either $f$ is irreducible, or is a square, or it has two distinct factors. Up to isomorphism the quotient ring depends only on these three possibilities, not on the particular choice of $f$. This gives three more rings: a field, the ring $\mathbb F_p[t]/(t^2)$, or $\mathbb F_p\times\mathbb F_p$.

There are (at least) $3$ commutative rings of order $4$.
Aside from $\mathbb F_4$ and $\mathbb Z\big/2\mathbb Z\oplus \mathbb Z\big/2\mathbb Z$ we have the matrices of the form
$$
    \begin{bmatrix}
    x & y  \\
    0 & x  \\
  \end{bmatrix}
$$
Where $x,y \in \mathbb F_2$, the field of order $2$.
Shouldn't be difficult to demonstrate that these are all of them, though that doesn't matter for your problem.


Connectedness of set which is an intersection of some connected set Is a set which is an intersection of some connected set still connected? I think it is not true but could not think of an example.

I think that is true. Take an example in $\mathbb{R}$, 3 intervals [0,1],[$\frac{1}{2}$,$\frac{3}{2}$],[$\frac{5}{4}$,2], their intersection is [$\frac{1}{2}$,1] and [$\frac{5}{4}$,$\frac{3}{2}$] is separated


Can $\ln(\Gamma(e^s))\ln(\zeta(e^s))$ be expressed as an integral? Consider the known relation $$ \Gamma(s)\zeta(s)= \displaystyle \int_{0}^{\infty}\dfrac{x^{s-1}}{e^{x}-1}\,dx.$$

Can $$ \ln(\Gamma(e^s))\ln(\zeta(e^s)) $$ be expressed as an integral? Maybe there's some substitution that can be done? I haven't been able to express it as an integral yet, so any hints are welcome.

Things aren't magical. If $\Gamma(s)\zeta(s)$ has a simple expression it is not for nothing.

It doesn't apply to $\ln(\Gamma(e^s))\ln(\zeta(e^s))$.


How to show the action of $SL_2(\mathbb R)$ on the complex upper half plane is transitive? I've seen other answers here and here but I still am not understanding how to show that this action is transitive. I want to show that for all $z \in \mathbb H$ there exists a matrix $g \in SL_2(\mathbb R)$ such that $gi=z$.

If I start with $\frac{ai+b}{ci+d}=z=x+iy$, then I get
$$\frac{ai+b}{ci+d}=z \implies ai+b=z(ci+d) \implies i=\frac{dz-b}{a-cz}$$

I don't know what to do with this or what this tells me.

If I do $\frac{az+b}{cz+d}=i$ I don't get anywhere either.

How can I go about this?

Here's a hint. You can write

$$ z= a+bi = \frac{b\cdot i + a}{0\cdot i +1} ,$$

which is the transformation corresponding to the matrix

$$\begin{bmatrix} b & a \\ 0 & 1 \end{bmatrix} $$

evaluated on the point $i$ in $\mathbb{H}$. This matrix may not be in $SL_2 \mathbb{R} $, but you can easily fix that.

ANSWER:

If $A \in SL(2,\mathbb R)$, then

\begin{align*}
 A\cdot i&=\frac{ai+b}{ci+d}\\
 &=\frac{ai+b}{ci+d}\cdot \frac{(-ci+d)}{(-ci+d)}\\
 &=\frac{ac+bd+i(ad-bc)}{c^2+d^2}.
\end{align*}

We're in $SL(2, \mathbb R$), so $ad-bc=1$, and
$$\frac{ac+bd+i(ad-bc)}{c^2+d^2}=\frac{ac+bd}{c^2+d^2}+i\frac{1}{c^2+d^2}.$$

If $c=0$ then $a \ne 0$, $d\ne 0$ and $ad-bc=1$ becomes $ad=1$. So, $d=1/a$. Also,
$$\frac{ac+bd}{c^2+d^2}+i\frac{1}{c^2+d^2}=\frac{bd}{d^2}+i\frac{1}{d^2}=\frac{b}{d}+i\frac{1}{d^2}=ab+a^2i.$$

Since $z \in \mathbb H$, then $z=x+iy$ where $y>0$. Now $a^2=y \implies a=\sqrt y$. Therefore, $ab=x \implies b=\frac{x}{\sqrt y}$.

So, $\begin{pmatrix} \sqrt y & \frac{x}{\sqrt y} \\ 0 & \frac{1}{\sqrt y} \end{pmatrix} \in SL(2,\mathbb R)$ and
$$\begin{pmatrix} \sqrt y & \frac{x}{\sqrt y} \\ 0 & \frac{1}{\sqrt y} \end{pmatrix} \cdot i =z$$
implies transitive.


Why is the inverse of a sum of matrices not the sum of their inverses? Suppose $A + B$ is invertible, then is it true that $(A + B)^{-1} = A^{-1} + B^{-1}$?

I know the answer is no, but don't get why.

Let us look at the often-forgotten $1 \times 1$ matrices over $\mathbb{R}$, which is another name for the real numbers themselves. Then your statement translates to the statement that if $x,y$ are real numbers that aren't zero, then
$$ \frac{1}{x + y} = \frac{1}{x} + \frac{1}{y},$$
which is clearly wrong.

It's just that the operation of taking inverses isn't well-behaved when it comes to sums.

This isn't surprising, because taking inverses is defined in terms of multiplication and NOT in terms of addition. There is no reason really why inverses should act in some favorable way on sums.

For instance, if $A$ is some invertible matrix, then what is the inverse of $A+A$? Is it $A^{-1}+A^{-1}$? No. You have that $A+A=2A$, whose inverse is $1/2 A^{-1}$, and this is not the same as $A^{-1}+A^{-1}=2A^{-1}$.

What's more is that if $A$ and $B$ are invertible matrices, then their sum $A+B$ need not even be invertible, as others have pointed out: It is not even always possible to find $(A+B)^{-1}$, even though both $A^{-1}$ and $B^{-1}$ exist.

What does hold, which you probably already know, is that (for invertible $A$ and $B$), we have this nice multiplicative property:

$(AB)^{-1} = B^{-1}A^{-1}.$


Equation of a straight line in spherical coordinates I'm trying to prove the angle sum formula for a triangle on the surface of a sphere. In order to do this I wanted to create a general triangle on the sphere, with one vertex at $\theta = 0$  and one side along the angle $\phi = 0$ for simplicity. (I think this can be done without loss of generality?)

In order to do this I need to be able to express one of the other sides as a straight line in spherical coordinates. Intuition tells me that it is probably just $\theta = k\phi + \alpha$ and $\phi$ as a free parameter. I've tried proving this to myself by showing the the resulting curve is a great circle. I then tried to compute a path integral around this curve so I could check that it is equal to the circumference of my general sphere, however wolfram alpha tells me that I must upgrade to pro to get more computation time to do my integral. ($\int_{0}^{2\pi}\sqrt{k^2 + \sin{(kt + \alpha)}}$ dt)

I've also tried to show that in cartesian coordinates that any points resulting from this parameterization must be coplanar. I have had little success with this approach.

Am I completely wrong here? Is there some good way to see whether or not this is true?

I suppose you assumed the $r$-coordinate of your path is constant.
Since you're looking for a path on the surface of a sphere, that's appropriate.

When you're trying to describe an object in a coordinate system, it helps to have very clear notions about the properties of that object as expressed in
that coordinate system.
There are a few important facts to keep in mind about a great circle.
One is that every point on the great circle is exactly opposite another
point on the great circle, that is, if $(r,\phi,\theta)$ is on the
great circle then so is $(r,\pi-\phi,\theta + pi)$.

Another fact is that as you travel around the great circle, after you cross
a "line of latitude" going "upward", you must later cross that same
line of latitude at a different point while going "downward".

A third fact is that unless the great circle happens to be a
"line of longitude", it will not pass through the poles. That is,
each great circle has a minimum latitude angle $\phi_0$ and a maximum
latitude angle (which by the first fact above must be $\pi - \phi_0$),
where $0 < \phi_0 < \pi/2$.
The great circle only attains its minimum latitude at one point; likewise the maximum latitude.

So if you have an equation of a great circle in the form
$$f(\phi, \theta) = 0,$$
this equation must have exactly one solution with $\phi = \phi_0$
and one with $\phi = \pi - \phi_0$;
but for each $\phi$ such that $\phi_0 < \phi_0 < \pi - \phi_0$,
there must be exactly two solutions of the equation.

On further examination of a great circle drawn on a sphere,
it should be obvious that at the point on the great circle where $\phi = \phi_0$, the circle is tangent to a line of latitude,
and at that point $\frac{d}{d\theta}\phi = 0$.

Your proposed equation, $\theta = k\phi + \alpha$,
is equivalent to $\theta - k\phi - \alpha = 0$.
Now we can see several flaws.
First, there is nothing in this equation to require that $\phi$ have
a maximum or minimum value anywhere except possibly at $\phi=0$ and $\phi=\pi$.
Second, there is only one solution to the equation for any given $\phi$.
(There should be two solutions for most values of $\phi$.)
Third, there is no point at which $\frac{d}{d\theta}\phi = 0$.

Let's try a different approach. The great circle is the intersection of
the surface of the sphere and a plane through the center of the sphere.
The equation of that plane is
$$ax + by + cz = 0.$$

Now convert the Cartesian coordinates to spherical coordinates:
\begin{align}
x &= r \sin \phi \cos \theta, \\
y &= r \sin \phi \sin \theta, \\
z &= r \cos \phi.
\end{align}

Making these substitutions directly into the equation of the plane, we have
$$ar \sin \phi \cos \theta + br \sin \phi \sin \theta + cr \cos \phi = 0.$$

Take out the factor of $r$ and combine some terms and we have
$$(a \cos \theta + b \sin \theta)\sin \phi + c \cos \phi = 0.$$

To create a line in circular coordinates, you can just do it like this: $sin(\theta) = m cos(\theta) + c/r$, where $m$ is the gradient and $c$ is the y-intercept. This comes from the $y = mx + c$ definition.

The $y = mx$ part is straightforward, but it is not as obvious. To see how this works explicitly, consider that to your base equation $sin(\theta) = m cos(\theta)$, you are adding the equation for a horizontal line: $sin(\theta) = c/r$ which comes straight from geometry.

You can use similar principles to extend this to spherical coordinates.


Showing that complex exponentials of the Fourier Series are an orthonormal basis I am revisiting the Fourier transform and I found  great lecture notes by Professor Osgood from Standford (pdf ~30MB).

On page 30 and 31 he show that the complex exponentials form an orthonormal basis. I understand the result, but not his calculation. He shows that the inner product of two different exponentials $(e_n (t) = e^{2\pi int}, e_m(t)= e^{2\pi imt})$ with $m \neq n$ is $0$ (He uses round parenthesis to denote the inner product). So, he does the calculation:

$$ (e_n, e_m) = \int_0^1 e^{2\pi int} \overline{e^{2\pi imt}}dt = \dots = \frac{1}{2\pi i(n -m)} (e^{2\pi i(n - m)}  - e^0) = \frac{1}{2\pi i(n -m)} (1 - 1) = 0$$

So why is $e^{2\pi i(n - m)} = 1$? And why do I have to look at the case $ m = n$ separately and cannot also just plug it into the last step? (I am aware that I would not get a sensible result then, but it still seems strange). Does this have anything to do because I am using Lebesgue integration and complex numbers? I suppose I have to review some math basics ...

If $l=n-m\neq 0$ you have

$$e^{2\pi i(n-m)}=e^{2\pi i l}=\cos(2\pi l)+i\sin(2\pi l)=1+i\cdot 0=1$$

The case $m=n$ has to be considered separately because in the solution you showed you divide by $m-n$, which you can't do if $m-n=0$. For $m=n$ you simply get

$$\int_0^1e^{2\pi i nt}e^{-2\pi i nt}dt=\int_0^1dt=1$$

You can of course consider $m=n$ case separately, as Matt suggested.

Alternatively, you can use L'Hospitals rule, in the expression you found, and see that in the limit $m \to n,$ the integral indeed converges to $1.$


Non-isolated singularity points I am having trouble understanding non-isolated singularity points. An isolated singularity point I do kind of understand, it is when: a point $z_0$ is said to be isolated if $z_0$ is a singular point and has a neighborhood throughout which $f$ is analytic except at $z_0$. For example, why would $\text{tan}(1/z),\ \text{log}(z),\text{or even}\ \frac{1}{\sin(\frac{\pi}{z})}$ have a non-isolated singularity point?

$\tan(1/z)$ has a non-isolated singularity at $z=0$, which is the limit of the singularities at $\dfrac{2}{\pi}, \dfrac{2}{3\pi}, \dfrac{2}{5\pi}, \ldots$.

The singularity of $\log(z)$ at $z=0$ is a branch point: this is on a curve  where
any particular branch of $\log(z)$ is discontinuous (e.g. the negative real axis in the case of the principal branch).

A non-isolated singularity is where you can not have any further neighborhood without encountering another singularity


If both $P$ and $Q$ are true , how can I tell that $P$ implies $Q$? I am trying to understand the fundamentals of mathematical logic in order to be able to study discrete mathematics and computer science soon.

I have a big problem understanding Implication. I understand the idea intuitively very well; for example:

$$ x >0  \implies 2x>0 .$$

Or

$$ x \text{ is a prime number} \implies x \geq 2 .$$

I notice that there is always a (connection) between the hypothesis and the conclusion; they seem to be related and fall in the same context.

What I don't understand is: according to the truth table, any two propositions can be linked through an implication even if they are not related at all, or they belong to different contexts.

For example :

$$ \text{A day is 24 hours} \implies \text{A cat has four legs and a tail} .$$

which is logically or mathematically TRUE, because both statements are true and according to the truth table when when both inputs are true for any two statements then the implication is true.

How can that be true?

Another example:

$$ 2 \text{ is a prime number} \implies \text{An hour is 60 minutes}.$$

again, which is logically or mathematically TRUE, because both statements are true and according to the truth table when when both inputs are true for any two statements, then the implication is true.

How can that be true? That's my first question.

The second question could be the same:

  How can we use truth tables with implication anyway?

What I understand is truth tables are used to list the probabilities of the output based on the logical values of the inputs. So the value of the output of any line in the truth table depends (only) on the logical values of the inputs in that line and has nothing to do with the conditional connection between the two inputs.

How can we display a conditional statement in a truth table the same way we display a logic gate or so? In other words, how can I tell only from the values of $P$ and $Q$ that $P \implies Q$?

Here is the truth table for an implication:

$$ \begin{array}{ccccc}
   P & & Q & & P \to Q \\
   T & & T & & T \\
   T & & F & & F \\
   F & & T & & T \\
   F & & F & & T \end{array} $$

You can think of an implication as a conditional promise. If you keep the promise, it's true. If you break the promise, it's false.

If I tell my kids, "I'll give you a cookie if you clean up." Then they clean up. I better give them a cookie. If I don't, I've lied. However, if they don't clean up, I can either give them a cookie or not. I didn't promise either if they didn't keep up their end of the bargain.

So in other words, an implication is false only if the hypothesis is true and conclusion is false.

Logically $P \to Q$ is equivalent to $\neg P \vee Q$

I had a computer science professor who was fond of promising his kids things given a false premise. This way he wasn't compelled to follow through. Example: "If the moon is made of green cheese, I'll give you an x-box."

Our common thinking of implication ($P \implies Q$) is that P allows us to say wether Q is true. But if we already know that Q is true, this way of thinking does not make sense anymore.

Your confusion stems from the fact that you think P would have any implication on Q, which cannot be the case because Q is already given. The statement is true nevertheless, see here.

Take the example given in an other answer, where "clean up" implies "get cookie". I think it helps understanding a special case, but it is not a good one, because we can assume "clean up" would lead to "get cookie", if "get cookie" were false. But the statement holds true even for things that will never lead to "get cookie". In other words, if it is really decided, that "get cookie" is true, you can say anything in your if-sentence and the statement still holds:

If you crash the car, I will give you a cookie.

Our common understanding of this sentence is, that the cookie is the reward for crashing the car. But this does not make sense in this situation, so nobody would normally formulate it this way, but rather:

You can do anything, I will give you a cookie.


Evaluating $ \lim_{x \rightarrow 0}\left (\frac{\sin(x)}{x}\right)^\frac{1}{x^2}$ I have got this problem on limits
$$ \lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x}\right)^\frac{1}{x^2}$$
What I am doing here is that taking log and then applying L-Hospital's Rule
$$y= \lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x}\right)^\frac{1}{x^2}
$$
then $$\ln(y)=\lim_{x \rightarrow 0} \frac{1}{x^2}\ln\left(\frac{\sin(x)}{x}\right)$$ now it becomes $\frac{0}{0}$ type , but problem is how to take derivative here? when differentiating $\ln\left(\frac{\sin(x)}{x}\right)$ it will become $\frac{x}{\sin(x)}$ then we'll have to differentiate one more time like we do in case of $f(f(x))$ ? $\sin(x)$ and $x$ will be separately differentiated or together ?

Use the Taylor series to get
$$\lim_{x\to0}\frac1{x^2}\ln\left(\frac{\sin x}{x}\right)=\lim_{x\to0}\frac1{x^2}\ln\left(\frac{ x-x^3/6+o(x^3)}{x}\right)=\lim_{x\to0}\frac1{x^2}\ln(1-x^2/6+o(x^2))\\=\lim_{x\to0}\frac1{x^2}(-x^2/6+o(x^2))=-\frac16$$
hence the desired limit is $e^{-\frac16}$.

Using $$\displaystyle \sin (\pi z) = \pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)\;,$$ Then Put $\displaystyle \pi z= x\Rightarrow z=\frac{x}{\pi}$

So $$\displaystyle \lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}} = \lim_{x\rightarrow 0}\prod_{n=1}^{\infty}\left(1-\frac{x^2}{\pi^2n^2}\right)^{\frac{1}{x^2}}$$

Now we can Interchange Product and Limit, We get

$$\displaystyle \lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}} = \prod_{n=1}^{\infty}\lim_{x\rightarrow 0}\left(1-\frac{x^2}{\pi^2n^2}\right)^{\frac{1}{x^2}} = \prod_{n=1}^{\infty}e^{-\frac{1}{\pi^2n^2}} = e^{-\sum_{n=1}^{\infty}\frac{1}{\pi^2n^2}} = e^{-\frac{1}{6}}$$


Differentiate to the $1/2$ power Differentiate $ax^2 + b$ to the $1/2$ power. (Exact wording of question)

The answer is $2ax$.

I don't get the question. Is it ${d\over dx} \left(ax^2+b\right)^{1/2}$? If so I am getting
$$ax\over \sqrt{a x^2+b}$$
It's only when I don't do the half power thing that I get $2ax$.

If we really talk about fractional derivatives, then it is defined as following:$$D^ax^k = \frac{d^a}{dx^a} x^k = \frac{k!}{(k-a)!}x^{k-a} = \frac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a}$$So using $\Gamma$ function we can define fractional derivative. It has all the properties of linear operator and is an extension of regular derivative. For example for $n = a+b$: $$D^aD^bx^k\frac{d^a}{dx^a} \frac{d^b}{dx^b}x^k = \frac{d^n}{dx^n}x^k = D^nx^k$$. To make sure it is consistent with regular derivative we set that fractional derivative of constant is zero, i.e. $D^a(const) = 0$. Then $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x^k = \frac{\Gamma(k+1)}{\Gamma(k+\frac{1}{2})}x^{k-\frac{1}{2}}$$For $k=2$$$D^{\frac{1}{2}}x^2 = \frac{2}{\frac{3}{4}\sqrt{\pi}}x\sqrt{x}$$ So $$D^{\frac{1}{2}}(ax^2+b) = \frac{8}{3\sqrt{\pi}}ax\sqrt{x}$$