Markov's Mouse and the Maze I recently started to learn about Markov Chains and had a problem regarding the expected time to absorption:

Problem:

Markov has an untrained mouse that he place in a maze. The mouse may move between adjoining rooms, or stay in the same room at any time-step. At the exit of the maze is an enormous piece of unscented cheese where the mouse may leave from and never return. Given the following transition matrix:

$$\begin{pmatrix}\frac 1{10}&\frac 3{10}&\frac 35&0&0&0\\\frac 12&\frac 12 &0&0&0&0\\\frac 3{10}&0&\frac 1{10}&\frac 35&0&0\\0&0&\frac 3{10}&\frac 1{10}&\frac 3{10}&\frac 3{10}\\0&0&0&\frac 12&\frac 12&0\\0&0&0&0&0&1\\ \end{pmatrix}$$

(a) What is the probability it leaves the maze, given it started in State $i \quad \forall \ i \in[0,5]$ ??

(b) How long before it leaves the maze, given it started in State $i \quad \forall \ i \in[0,5]$ ?

My Attempt:

Well first I made a diagram to get the understanding of the problem visually. I replace the number States to letters, i.e. State $1$ as State $A$ etc. and wrote down the transition probabilities from the matrix.

(a) I think because there exists an absorbing state at the exit, then the probability of exiting must be $1$ a.s..

(b) I used First-Step-Analysis to find the expected time for each $i$.

Let $v_i$ be the expected time to exit from starting position $i$, i.e. $v_i =\Bbb E_i[T_{\{5\}}]$. Then solving the system of equations:

$$
\left\{
\begin{array}{ll}
v_A &=1+\frac 1{10} v_A+\frac 3{10} v_B+\frac 35 v_C \\
v_B &=1+\frac 12 v_A +\frac 12 v_B\\
v_C &=1+\frac 3{10} v_A+\frac 1{10} v_C+\frac 35 v_D \\
v_D &=1+\frac 3{10} v_C+\frac 1{10} v_D+\frac 3{10} v_E+\frac 3{10} v_F \\
v_E &=1+\frac 12 v_D+\frac 12 v_E\\
v_F &=0
\end{array}
\right.
$$

So I get the following solution:
$$\Bbb E_A[T_{\{5\}}]=14; \quad \Bbb E_B[T_{\{5\}}]=16 \quad \Bbb E_C[T_{\{5\}}]=11 \frac 13 \quad \Bbb E_D[T_{\{5\}}]=8 \frac 13 \quad \Bbb E_E[T_{\{5\}}]= 10 \frac 13$$

I'm not $100\%$ sure about the part (b).

My Question(s):

I was wondering, is there another way to look at this problem without using Markov Chains or First-Step-Analysis? I feel like there are geometric random variables involved here.

Also what would happen if there were 2 exits, could I apply the exact same principles or is something changing?

Thank you in advance!

Here's another approach:

A random walk on a connected, undirected, unweighted graph spends the same time on each edge in each direction. Thus the proportion of time it spends at a vertex is the number of edges incident at the vertex divided by twice the total number of edges. This still holds true if you weight the edges and even if the walk is periodic. You can check all this by checking that the corresponding distributions are invariant under transitions (where the transition probabilities are proportional to the edge weights).

In your case, we need self-loops in the graph, which complicates things, but only slightly. A self-loop only has one “direction”, so it only counts once in the sum of the edge weights, whereas normal edges count twice.

The expected return time of a vertex (the time it takes for the random walk starting at the vertex to return to the vertex, possibly immediately through a self-loop) is the reciprocal of the proportion of time the walk spends at the vertex.

What does all this have to do with your problem? Add a vertex for the exit and assign edge weights to the doors so that they yield your transition matrix. If you assign weight $3$ to the self-loop at $B$ to make all the edge weights come out as integers, you can work your way from $B$ through the maze to arrive at these edge weights:

The numbers in the doorways are the weights of the proper edges, the numbers next to the room labels are the weights of the self-loops. The sum (with self-loops counted once and proper edges counted twice) is $112$.

The return time for a random walk on this graph starting at the exit is $1$ plus the exit time $v_D$ from $D$. Thus

$$
v_D=\frac{112}{12}-1=8\frac13\;,
$$

in agreement with your solution.

Now consider the subgraph comprising vertices $A$ through $D$, but without the self-loop at $D$. The sum of edge weights with multiplicities is $48$. The return time for a random walk on this subgraph starting from $D$ is $1$ plus the expected time to get from $C$ to $D$. Thus

$$
v_C=v_D+\frac{48}{12}-1=11\frac13\;,
$$

again in agreement with your solution. I think you can figure out the rest yourself.

Note that this requires us to solve only simple linear equations, no systems of equations (including the part of figuring out the edge weights). This works whenever the mouse is in a tree.

Your solution us fine, it would work the same for two exits and other approaches would lead to an equivalent calculation because you will get the solutions of this linear system in the end even if you do not write it down.


What's the Fourier transform of $f(x)/x$? What's the Fourier transform of $\frac{f(x)}{x}$? Is it even defined? Because I was thinking of the time dependent Schrödinger's equation for an electric potential:
$$ \frac{-\hbar^2}{2m} \nabla^2 \Psi + \frac{kq_1 q_2}{r} \Psi = i\hbar \frac{\partial \Psi}{\partial t} $$

So, I have to take the transform to solve it, but how do I take the transform of the potential part?

I think this might help.

Suppose $\Psi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\phi(k)e^{ikx}dk$.

So $\Psi(x)$ is the Fourier Transform of $\phi(k)$.

Take the derivative of either side with respect to $x$.

$\frac{\partial \Psi}{\partial x}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty ik\phi(k)e^{ikx}dk$

So we can get the transform of multiplying by a power of k if we just do the transform of $\phi$ then take the derivative of the transform.

This suggests integrating the transform with respect to $x$ to divide and get the transform of $\phi(k)/k$.

There are some restrictions on the method. The function which is to be integrated with respect to $k$ has to converge, so multiplying or dividing by powers of $k$ will only work for certain values of $\phi(k)$.


Can you determine small primes from larger primes? Suppose you are given the primes in the range $[n,n^2]$. Is there a known way to effectively reconstruct the primes less than $n$? Ideally, something that takes less calculation than figuring them out directly from sieving, as that would sort of defeat the purpose.

Heuristically, you can do pretty well by taking $n \pmod{p}$ where $n$ is composite and $p$ steps through your prime list, and then seeing what comes up most often, but I'm hoping for something better than that.

This is very slow compared to e.g. the sieve method. For example with $n=1000$ there are $\pi(10^6)-\pi(10^3) = 78330$ primes in $[n,n^2]$ but only $168$ primes less than $1000$.
So you would do $78330$ modulo operation with your method. For larger $n$ your method would even be worse.


Compare $2^{2016}$ and $10^{605}$ without a calculator So, I am supposed to compare $2^{2016}$ and $10^{605}$ without using a calculator, I have tried  division by $2$ on both sides  and then comparing $2^{1411}$ and $5^{605}$, and then substituting with $8,16,10$ and then raising to powers and trying to prove that but that did not go anywhere, I have also tried taking $\log$ of both sides, but did not help either. Also is there a more general approach to these kind of problems?

\begin{aligned}
2 ^ {2016} > 10 ^ {605} & \iff \\
e ^ {\ln (2 ^ {2016}) } > e ^ {\ln (10 ^ {605}) } & \iff \\
\ln (2 ^ {2016}) > \ln (10 ^ {605}) & \iff \\
2016 \cdot \ln (2) > 605 \cdot \ln (10) & \iff \\
\frac{2016}{605} > \frac{\ln 10}{\ln 2} & \iff \\
\frac{2016}{605} > \frac{\log 10}{\log 2} & \iff \\
\frac{2016}{605} > \frac{1}{\log 2} & \iff \\
3.33 \ldots > \frac{1}{0.301 \ldots} & \iff \\
3.33 \ldots > 3.32 \ldots
\end{aligned}

PS: the two boxed expressions (i.e $\boxed{ \frac{2016}{605} }$ and $\boxed{ \log 2}$ ) can be derived by hand: one can divide on paper and (as far as I can remember) during tests you can bring logarithmic tables


Formula for Conditional probability doesn't make sense to me conditional probability is defined us:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

But if (and there is the part where i might be missing something) $P(A \cap B) = P(A) \cdot P(B)$ then:

$$ \frac{P(A \cap B)}{P(B)} = \frac{P(A) \cdot P(B)}{P(B)} = P(A) $$

which doesn't make any sense. Why would the formula exist in the first place. I'm confused.

Thanks for helping me :).

The easiest way to picture conditional probability is by viewing a Venn diagram.  (Sorry, I am doing this quickly and have not learned to generate one in MathJAX yet.)  In a simple diagram with two sets A and B, think of conditional probability as the ratio of the area of the intersection of the two circles to the area of the "given" circle (for A given B, that circle is B).

Imagine redefining the universe to be the given set (B in the previous example); the conditional probability is then the probability of A in that universe.

Here's a more intuitive understanding that leads to the definition. (Sorry I'm not very good at latex.)
Take two events A and B, where they aren't necessarily independent. We seek to find the probability that both A and B occur. This turns out to be P(A and B).
Now, let's say A happens first. Therefore, P(A and B) = P(A)("something"). But since P(B|A) is defined as the probability of B occurring given A has already occurred, it is intuitively true that P(A and B) = P(A)P(B|A). Dividing P(A) on both sides yields the desired formula.
Note that when the A and B are independent events, P(B|A) = P(B), because the events don't affect each others' outcomes. This simplifies to the regular formula for probabilities of independent events.


Is the area of a circle ever an integer? Is the area of a circle ever an integer?

I was trying to answer someone else's question on yahoo answers today and I got thumbs down from people on my answer and have come here to get a thorough proof on it because now I just must know! :)

My assertion: Suppose we have two integers a and b. We can form a rational numbers out of each of these: $\frac{a}{1}$ and $\frac{b}{1}$. Now forming another rational by dividing them: $\frac{\frac{a}{1}}{\frac{b}{1}}$ = $\frac{a}{b}$. Now if we divide this by itself: $\frac{\frac{a}{b}}{\frac{a}{b}}$ = $\frac{ab}{ab}$ = 1. Therefore, we can conclude that given any rational number, we can always describe it using only integers thus always resulting in an integer number.

However, even if our radius is $\sqrt{\frac{1}{\pi}}$ then we will have $\pi\sqrt{\frac{1}{\pi}}^2$ = $\frac{\pi}{\pi}$, by definition of area = $\pi r^2$.

Now I think this the trickiest part. Does  $\frac{\pi}{\pi}$ = 1? Or can it only equal  $\frac{\pi}{\pi}$? How do we prove that  $\frac{\pi}{\pi} \neq$ 1? Or is my method sufficient at all? Thanks!

By definition, $\frac{1}{b}$ is the unique real number which, when multiplied by $b$, yields $1$.

By definition, $\frac{a}{b}$ is the product of $a$ and $\frac{1}{b}$.

Since $\frac{1}{\pi}$ is the unique real number that, when multiplied by $\pi$, yields $1$, then $\pi\left(\frac{1}{\pi}\right) = 1$. Hence, $\frac{\pi}{\pi}=1$.

If you allow any radius for a circle, then a circle has integer area if and only if its radius $r$ is the square root of an integer divided by $\pi$, that is, $r = \sqrt{a/\pi}$ for some nonnegative integer $a$.

Your first paragraph, though, is completely irrelevant.

This is not a proof; Arturo has said all there is to it. But physical world thought experiments and arguments hint strongly that yes, there should be a circle with area 1. I have the following reasons for this. The general argument is:

"All sizes have been created equal." Consider the counter-question: Why should no circle have an integer-sized area? What, in your opinion, distinguishes a real-world area of one size from an area of another size? Have you seen Vi Hart's math lesson rant about parabolas? At 2:50 she says:

  All parabolas are exactly the same, just bigger or smaller or moved
  around. Your teacher could just as well just hand you parabolas
  already drawn and have you draw coordinate grids on parabolas rather
  than parabolas on coordinate grids.

This still makes my eyes water because this perfectly ordinary math lesson is so sad and Vi Hart is so smart and the rant is so funny. All at the same time.—

Back to the issue: What's true for the conic section called parabola is even truer (ha!) for the conic section called circle.

The numerical value of a real-world circle's area depends on the unit you measure it with! Obviously you have no doubt that the area of a circle can be $\pi$; well, you see the point – if you change your area measuring unit so that one unit is the area of $\pi$ units in the old system, voilà: circle of area 1!  This is not a trick argument; our common length- and associated area units are completely arbitrary. There simply are no area sizes which are special. (This is the real-world isomorphism of Arturo's mathematical argument.)
You can just shrink an image with a circle of area $\pi$ by a linear factor of $\sqrt{\pi}$. Voilà: Area 1. (This is another real-world isomorphism of Arturo's mathematical argument.)
Say, you roll dough out; you can give it any shape you like, square, circle, whatever. Obviously you can have a square of area 1; there is no reason why it should be more difficult to shape it into a circle than any other square. (No, this is not the unsolvable "square a circle" (or vice versa) problem. I do not claim that the circle's radius would be rational.) Perhaps it's easier to think in terms of volumes: If you put a water balloon in a cubic liter container and fill it with water, you have a cube with the volume 1 (if you measure in liters, see first bullet point). Now take the balloon out; if it's a perfect balloon and weightless (because you throw it – what else would you do with a water balloon ??), it will be a sphere with the volume 1, if we consider water incompressible. This applies analogously to areas.


Name for Cayley graph of a semigroups? I did Google search and can't find a good answer. I thought I should ask experts here.

Cayley graph is defined for groups. My question is:

  Is there a special name for the Cayley graph of semigroups?

We call them Cayley graphs (it doesn't seem to be usual to say 'Cayley digraph'), and they are interesting.  I am doing research on semigroups, and quite often draw the Cayley graph of a semigroup to get an idea of what it's like.  I don't know how much background on semigroups you have, but one reason Cayley graphs for semigroups are interesting is that the $\mathcal{R}$-classes in a semigroup correspond to the strongly connected components in its (right) Cayley graph.

The difference between a semigroup and a group is the existence of an identity and inverses and it are just those properties that make Cayley graphs interesting ( read: possible ). There are no Cayley graphs for semigroups.


Topologies in a Riemannian Manifold I'm studying Differential Manifolds using Manfredo do Carmo's Book (Riemannian Geometry) and although I see no mention of this in Do Carmo's book, it's really easy to see a Riemannian Manifold as a metric space (if $M$ is the manifold, using $d_M(x,y) = \inf\{L(\gamma) \colon \gamma$ is a curve on $M$ and $a,b \in \gamma([0,1])\}$, where $L(\gamma)$ is the length of the curve $\gamma$.

Is the topology induced in $M$ (viewed as a set) by the metric $d_M$ the same of $M$ as topological space?

My problem to solve this self-proposed question was analyzing the open balls of $(M,d_M)$. References and Hints are appreciated.

First, you need to stipulate that $M$ is connected for the distance function to turn $M$ into a metric space.

When $M$ is connected, the answer is yes, the metric topology is the same as the given manifold topology. The proof boils down to showing that the Riemannian metric is uniformly comparable to the Euclidean metric in small coordinate balls. You can read a proof in my Introduction to Smooth Manifolds (2nd ed., Theorem 13.29).

If I remember correctly, what you are asking follows from Hopf-Rinow Theorm. Here is the link from wiki. But I recommend you to read another book by Do Carmo called Riemmanian Geometry.


How prove $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$ Let $a,b,c>0$,and such
  $$\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=\dfrac{1}{3}$$
  show that
  $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}$$

my try: since
$$\dfrac{1}{2+a^2}=\dfrac{1}{2}\left(1-\dfrac{a^2}{a^2+2}\right)$$
so
$$\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}=\dfrac{7}{3}$$
we only prove

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}$

and I want use Cauchy-Schwarz inequality to prove it,But I can't works,such
$$\left(\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}\right)(a^2+2+b^2+2+c^2+2)\ge (a+b+c)^2$$
$$(a^2+b^2+c^2+6)\ge \dfrac{3}{7}(a^2+b^2+c^2+2ab+2bc+2ac)$$
$$\Longrightarrow 4(a^2+b^2+c^2)+42\ge 6(ab+bc+ac)$$
and let $$p=a+b+c,q=ab+bc+ac,r=abc$$
so
$$2p^2+21\ge 7q$$
and we only prove
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ac}{abc}=\dfrac{q}{r}\le\dfrac{3}{\sqrt{7}}$$
maybe this is not true.

But this not usefull to solve this problem .

Thank you

This might not be the "proper" way, but it can be seen that both the expressions are symmetrical in all terms. And we need the maximum of the second eqn. This can happen only when all the terms contribute equally to the sum, if it were not so then, let 1/a<1/b<1/c, then we can make b and c to a, and get higher sum. So, putting it back in the first eqn. we get $ \frac{3}{a^2+2} = 1/3$, solving which we get $ a = \sqrt{7}$ or $-\sqrt{7} $. Choosing the positive root, as we need maximum values, we get $ \frac{1}{a} + \frac{1}{b}+\frac{1}{c} = \frac{3}{\sqrt{7}}$

Missed JJacquelin's answer, by a few minutes. It is more analytical, than mine.


Why the factor $e^{g(z)}$ in the Weierstrass factorization theorem? The Weierstrass factorization theorem is usually stated as such (quote from Wikipedia):

Let $f$ be an entire function, and let $\{a_n\}$ be the non-zero zeros of $f$ repeated according to multiplicity; suppose also that $f$ has a zero at $z = 0$ of order $m ≥ 0$ (...). Then there exists an entire function $g$ and a sequence of integers $\{p_{n}\}$ such that

$$f(z)=z^me^{g(z)}\prod_{n=1}^\infty E_{p_n}\left(\frac{z}{a_n}\right).$$

It seems to me that this combines two different theorems: that $f$ can be written as
$$f(z)=z^mh(z)\prod_{n=1}^\infty E_{p_n}\left(\frac{z}{a_n}\right)$$
for some entire function $h$ with no zeros, and that any entire function $h$ with no zeros can be written as $h(z)=e^{g(z)}$ for some entire function $g$. The latter is an interesting nontrivial statement in its own right, and, as far as I can tell, its proof is not connected with the rest of the proof of the Weierstrass factorization theorem.

So, what is the mathematical or historical reason for writing the theorem with $e^{g(z)}$ instead of $h(z)$?

I believe the reason is linked to the Argumente Principle. But the main property of writing e^h(z) is the fact that that function is never zero.


Evaluate $\lim_{n \to \infty} \int_{0}^1 [x^n + (1-x)^n ]^{1/n} \ \mathrm{d}x$ Evaluate
$$\lim_{n \to \infty} \int_{0}^1 [x^n + (1-x)^n ]^{1/n} \ \mathrm{d}x$$

I simplified the limit to
$$\dfrac{1}{2}\lim_{n \to \infty} \int_{0}^{1/2} \left[\left(\frac{1}{2}+x\right)^n + \left(\frac{1}{2}-x\right)^n \right]^{1/n} \ \mathrm{d}x$$
using properties of definite integrals, but I can't solve further.
Any help will be appreciated.
Thanks.

Hint: It looks like there may be a mistake in your proposed simplification step, but at any rate, you can break the limit up into $\lim_{n \to \infty} \int_0^{1/2} f_n dx + \int_{1/2}^1 f_n dx$ where $f_n = (x^n + (1 - x)^n)^{1/n}$. You should be able to show that in each piece of the broken up integral, either $x^n$ is hugely dominating or $(1-x)^n$ is hugely dominating (before you take the root). Since then you are taking the $n$th root before integrating, this will hopefully allow you to see what the limit should be.

$I=\lim_{n\to\infty}\int_0^1(x^n+(1-x)^n)^{1/n}\ dx=\lim_{n\to\infty}\int_0^1x(1+((1-x)/x)^n)^{1/n}dx$. Now $\lim_{n\to\infty}(1+(1-x)/x)ⁿ)^{1/n}=1$. This can be verified by taking log. Assuming uniform convergence we get the result as $\int_0^1xdx=1/2.$


Show that a special sequence converge How can I prove that $a_n = \sum_{i=1}^n \frac{1}{\sqrt{n + i}}$ converge. I know that if a sequence $(b_n)$ converge to $b$, then $(\sqrt{b_n})$ converge to $\sqrt{b}$ and $\frac{n}{\sqrt{2n}} \leq a_n \leq \frac{n}{\sqrt{n + 1}}$, but I am not sure how to finish this. Can you help?

Observe that $a_n = \sum_{i=1}^n \frac{1}{\sqrt{n^2+i}} \leq \sum_{i=1}^n \frac{1}{\sqrt{n^2+1}} = \frac{n}{\sqrt{n^2+1}}$. In the same way you will find that $a_n = \sum_{i=1}^n \frac{1}{\sqrt{n^2+i}} \geq \sum_{i=1}^n \frac{1}{\sqrt{n^2+n}} = \frac{n}{\sqrt{n^2+n}}$
Therefore your idea is correct and also Kaind's idea to use the sandwich theorem is good. The only thing is that both sequences converge to $1$, not to $0$. For the left one you could use the fact that $\sqrt{n^2+n} \leq n + \sqrt{n}$ and then you get
$$\frac{n}{\sqrt{n^2+n}} \geq \frac{n}{n + \sqrt{n}} = \frac{1}{1+\frac{1}{\sqrt{n}}}$$
and the last fraction obiously converges to $1$.
For the other one we find that
$$\frac{n}{\sqrt{n^2+1}} \leq \frac{n}{\sqrt{n^2}} = 1$$
So with the sandwich theorem we get $a_n \to 1$ as $n \to \infty$.

You're correct in your approach. Just observe that both $\frac{n}{\sqrt{n^2 + n}}$ and $\frac{n}{\sqrt{n^2 + 1}}$ converge to $1$ (substitute $n \rightarrow \frac{1}{x}$) and use sandwich theorem.


Trigonometric proof using trigonometric identies If $x=k\sec^2\theta +m\tan^2\theta$ and $y=l\sec^2\theta +n\tan^2\theta$ prove that $\frac{x-k}{k+m}=\frac{y-l}{l+n}$

Need help in solving this.
Can someone please post the entire working.

You are very very close.

$$x-k=(k+m)\tan^2\theta,$$
$$y-l=(l+n)\tan^2\theta.$$

$$x=k\sec^2\theta +m\tan^2\theta\\y=l\sec^2\theta +n\tan^2\theta\\$$ use $$\sec^2\theta=\frac{1}{cos^2\theta}=\tan^2\theta +1$$
$$x=k(\tan^2\theta +1) +m\tan^2\theta=\tan^2\theta (k+m)+k \to \frac{x-k}{k+m}=\tan^2\theta \\y=l(\tan^2\theta +1) +n\tan^2\theta =(l+n)\tan^2\theta +l \to \frac{y-l}{l+n}=\tan^2\theta \\$$ so
$$\frac{x-k}{k+m}=\tan^2\theta=\frac{y-l}{l+n} $$


show that $\mathbb{Z}$ is totally disconnected Show that $(\mathbb{Z},d)$ is totally disconnected (where $d$ is the metric induced by the Euclidean metric on $\mathbb{R}$).

I think that to prove this I should use contradiction but not really sure where to start

Hint: can you find neighbourhoods around each $x \in \mathbb Z$ such that the neighbourhoods are completely disjoint?

Note that a space is totally disconnected if its points are open, and in this case $\mathbb Z$'s points are open.  You can prove that by noting that $(n-1/2,n+1/2)\cap\mathbb Z=\{n\}$ for each $n\in\mathbb Z$.


is a one-by-one-matrix just a number (scalar)? I was wondering. Clearly, we cannot multiply a (1x1)-matrix with a (4x3)-matrix; However, we can multiply a scalar with a matrix. This suggests a difference.

On the other hand, I was, for example, in an econometrics lecture today, where we had for a (Tx1)-vector $\underline{û}=\left(
\begin{array}{c}
û_1\\
\vdots\\
û_T\end{array}\right)$:

  $S_{ûû}:= \sum_{i=1}^T û_i^2$ shall be minimized. We see see that
  $S_{ûû}=\underline{û}^T\underline{û}$.

Well, formally, shouldn't it be $(S_{ûû})=\underline{û}^T\underline{û}$ or $S_{ûû}=\det(\underline{û}^T\underline{û})$, to ensure that we stay in the space of matrices and not suddenly go to the space of scalars? So here, the professor (physicist) not only treats $\underline{û}^T\underline{û}$ like a scalar, but also calls it a scalar. Is this formally legit or a wrong simplification (though it does not seem to have any impact, and surely makes life easier)?

It's just a scalar in the sense that the ring of $1\times 1$ matrices over a field $K$ is isomorphic to $K$ (by the map $[x]\mapsto x$), but, as you observed, when you're considering the interaction of matrices of different sizes, then you have to treat them differently.

No, a one-by-one matrix is not a scalar.

Let's assume you want to multiply a one-by-one matrix $[c]$ with a 3x3 matrix $A$. To do this realize, that you can always extend a matrix to fit the dimensions by adding rows and columns of zeros. So you get
$$
\begin{bmatrix} c \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} &   a_{13} \\ a_{21} & a_{22} &   a_{23} \\ a_{31} & a_{32} &   a_{33}  \end{bmatrix} = \begin{bmatrix} c & 0 &   0 \\ 0 & 0 &  0 \\ 0 & 0 &   0  \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} &   a_{13} \\ a_{21} & a_{22} &   a_{23} \\ a_{31} & a_{32} &   a_{33}  \end{bmatrix} = \begin{bmatrix} ca_{11} & ca_{12} &   ca_{13} \\ 0 & 0 &   0 \\ 0 & 0 &   0  \end{bmatrix}
$$
But multiplying a scalar $c$ with $A$ is different. The scalar acts like a square diagonal matrix. So you get
$$
c  \begin{bmatrix} a_{11} & a_{12} &   a_{13} \\ a_{21} & a_{22} &   a_{23} \\ a_{31} & a_{32} &   a_{33}  \end{bmatrix} = \begin{bmatrix} c & 0 &   0 \\ 0 & c &  0 \\ 0 & 0 &   c  \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} &   a_{13} \\ a_{21} & a_{22} &   a_{23} \\ a_{31} & a_{32} &   a_{33}  \end{bmatrix} = \begin{bmatrix} ca_{11} & ca_{12} &   ca_{13} \\ ca_{21} & ca_{22} &   ca_{23} \\ ca_{31} & ca_{32} &   ca_{33}  \end{bmatrix}
$$

From this we see: A one-by-one matrix $[c]$  always acts like a matrix with a single eigenvalue $c$. A scalar $c$ on the other hand is a shortcut for a n-dimensional matrix with $n$ equal eigenvalues $c$.


Prove that $\int_0^\infty \frac{f(x) - f(x + 1)}{f(x)}\, dx$ diverges This is one of the problems I got in a test today, and one that i got no idea how to solve.

Let $f:[0,\infty) \to \mathbb{R}$ be strictly decreasing continuous function with $\displaystyle\lim_{x\to \infty} f(x) = 0$. Prove that $$ \int_0^\infty \frac{f(x) - f(x + 1)}{f(x)}\, dx$$ diverges.

I tries using the fact that $f$ has maximum and minimum on every closed intervals $[0,b]$  then limitting $b\to \infty$, $$\int_0^b \left( 1 - \frac{f(x + 1)}{f(x)}\right) dx  = b - \int_0^b \frac{f(x + 1)}{f(x)} dx \geq b - \int_0^b  f(0)/f_{\min}$$

But the integral on the right doesn't seem to converge as $b\to\infty$.

Any hint?

(done on my phone)

If $f'$ exists then
$f(x+1)-f(x) \approx f'(x)$
so the integrand is about
$-f'(x)/f(x) = (-\ln(f(x)))'$
so the integral out to $T$
is about $-\ln(f(T))
\to \infty$.

Not sure how to modify this
for general $f$.


How can I efficiently calculate the inverse of this symmetric near-tridiagonal matrix? I'm interested in calculating the inverse of the following matrix in order to solve a system of linear equations.

$$T=\begin{pmatrix}
b & a & 0 & 0 & \cdots & a\\
a & b & a & 0 & \cdots & 0\\
0 & a & b & a & \cdots & 0\\
0 & 0 & a & b & \cdots & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & a & b & a & 0\\
0 & 0 & 0 & a & b & a\\
a & 0 & 0 & 0 & a & b
\end{pmatrix}$$

This matrix is almost tridiagonal, except for entries in the top-right and bottom-left corners. I have seen how to calculate the inverse of tridiagonal matrix, but how would those corner-entries affect the inverse?

By the way, this matrix shows up in the Crank-Nicolson method applied to parabolic PDE's (diffusion equation particularly) with periodic boundary conditions.

I assume, that you want to solve $Ax = b$, and you are not iterested in the inverse $A^{-1}$ (which is dense).

You may perform the LQ factorization of $A$ (or QR of $A'$) using Givens rotations. In this case $n$ rotations are required. Then resulting $L$ matrix is lower triangular with two subdiagonals except the last row of $L$, which is dense and must be stored separately. The matrix $Q$ cannot be formed explicitly, since is dense. Required decomposition can be performed in $O(n^2)$ operations.

This can be solved efficiently by using the Thomas algorithm. The Wikipedia page does a pretty good job on explaining it. For your case of periodic boundary, you may need to spend a little bit more time to look for a better explanation.


$a^2\equiv1 \pmod n$ iff $a\equiv\pm\,1\pmod p$ for all $p\mid n$ (Not)if $a$ is an integer and $n$ a postive integer, then
  $a\equiv\pm 1\pmod p$ for all primes dividing n if and only if
  $$a^2\equiv 1\pmod n$$

$\Longrightarrow $ is wrong,Tonyk note count-example $a=4,p=3,n=9$

if $a^2\equiv 1\pmod n,\Longrightarrow (a,n)=1,$,since $p|n$,then we have
$$a^2\equiv 1\pmod p$$then we have
$$a\equiv \pm 1\pmod p?$$

My solution is right? if not,then How prove it?

How about using the definition of modular arithmetic:

Let $p$ be an odd prime.

So, we have $a^2 \equiv 1 \bmod p$. Let's subtract 1, we get: $$a^2-1 \equiv 0 \bmod p$$

We can use the basic rules of factoring here, and the definition of modular arithmetic:

$$p | a^2-1 \implies p | (a-1)(a+1)$$

Now, let us assume that we are dealing with an odd prime p, and it is easy to convince yourself that p cannot divide both $a+1$ and $a-1$, in which case it is one or the other. Either way, you are left with the result:

$$a \equiv \pm 1 \bmod p$$


the probability of guessing a number Choose any natural number.

For example I would choose:

3852011231231280130218920382342312420234801232321241231212131234 (and so for for another few bilions of digits)

What's the probability that somebody can guess it ?

Since there are infinite natural numbers I saw that the answer is 0.
(this is from a book from gregory chaitin)

is that correct ?

or is it more correct to say that it's very very very.... unlikely but still possible.

To be able to answer your question, you'd have to specify how someone would guess an arbitrary natural number. Since $\mathbb{N}$ is coutable infinite, there's no such thing as a uniform probability distribution over $\mathbb{N}$. Such a distribution would have to assign the same probability to each natural number (to be uniform), but that'd make the sum of all probabilities $+\infty$, not $1$ as you'd want for a probability distribution.

There are non-uniform distributions over $\mathbb{N}$ though, but to compute what the probability of 3852011231231280130218920382342312420234801232321241231212131234 under such a distribution would be, you first have to decide which distribution to use.


Given matrix $e^{tA}$, find $A$ I tried answering the following question in an old exam:

Given matrix:

$$ e^{tA} = \begin{bmatrix}\frac{1}{2}(e^t+e^{-t}) & 0 & \frac{1}{2}(e^t-e^{-t})\\0 & e^t & 0\\\frac{1}{2}(e^t-e^{-t}) & 0 & \frac{1}{2}(e^t+e^{-t})\end{bmatrix}$$

Calculate A.

Where $$e^{tA} = \sum_{k=0}^{\infty} \frac{t^k}{k!}A^k$$

Hope I can get some help with this, as I don't think I can find the right way to approach it myself.

The result given by José is fine if you know that the matrix  is for sure obtainable as $\exp(t A)$.

If you are doubtful (maybe it is a trick questions), you can use the alternative
$$A = e^{-t A}\left( \frac{d}{dt} e^{t A}\right) = \begin{pmatrix} 0& 0 & 1\\
0 & 1 &0\\
1 &0 &0 \end{pmatrix} $$
where you can check explicitly that $t$ drops out.


find the volume below the cone $z=2-\sqrt{x^2+y^2}$ and inside the cylinder $(x-1)^2+y^2=1$ I know I need to transform it into cylindrical system to integrate it but I don't know how to deal with r, since the cylinder is not center at (0,0)

Assuming the volume is bounded below by the $xy$ plane, you need to ask yourself what is the projection of the volume in the $xy$ plane. And indeed, cylindrical coordinates are the way to go.

So obviously, the projection is the disc $(x-1)^2+y^2\le 1$, which you need to convert to polar coordinates. Expand, and you get
$$
x^2-2x+1 +y^2 \le 1\quad \Rightarrow \quad r^2\le 2 r \cos \theta \quad \Rightarrow \quad 0 \le r\le 2  \cos \theta
$$
And careful with $\theta$, the cylinder is shifted along the $x$ axis, so
$$
-\pi/2 \le \theta \le \pi/2
$$

Can you take it from there?

$$\int_{0}^{2\pi}\int_0^1 (2-\sqrt[n]{x^2+y^2}) dx dy$$ and put

$$x=1+r\cos{\theta}$$ and $$y=r\sin{\theta}$$ and $$dxdy=rdrd{\theta}$$

answer:$2\pi$


What is the average minimum distance between two Sobol points? Having the first n points of a d-dimensional Sobol sequence, what is the average Euclidean distance from one arbitrarily point to its nearest neighbour?

Good question - the minimum distance is at least 0.5 * sqrt(d) / n, see "Quasi-random points keep their distance" on http://www.sciencedirect.com/science/article/pii/S0378475406002382


Why does the function f(|x|) look like this? If the image is f(x), why does f(|x|) look like two triangles above the x axis (basically the right side duplicated on the left)?

Ok, so let us call $g(x)=f(|x|)$. You agree that $g(x)=f(x)$ if $x\geq 0$, so the graph remains unchanged.

What happens when you try to calculate say $g(-1)$?. You get $f(|-1|)=f(1)=2$, it is the same for all $x$ values that are negative, so that is why you get a copy of the triangle.

Ok I think I should have put it in a better way like the other answers. Let $g(x)=f(|x|)$ and now $g(x)$ will look like the positive part of $f(x)$ flipped if $x<0$ since $$g(x)=\left\{ \begin{array}{ll} f(x) & x\geq0 \\ f(-x) & x<0\end{array}\right.$$
Now, $g(-2)=f(2)$ and so forth.


Real World 3D Geometry Question: Finding the Intersection Point of a Line and a Plane My team and I are developing a product which includes LASER systems and 3D geometry. We are stuck with a problem and I have tried my best to frame it into mathematical question. Here it goes,

$P_1=(x_1, y_1, z_1)$ and $P_2=(x_2, y_2, z_2)$ are two arbitrary points in space. Two lines with angles given with respective to $XY$, $YZ$ & $XZ$ planes $L_1(α_1, β_1, γ_1)$ and $L2(α_2, β_2, γ_2)$ pass through $P_1$ & $P_2$ respectively.

Lines $L_1$ and $L_2$ intersect a plane $R$ at points $R_1$ and $R_2$ respectively. The distance between $R_1$ and $R_2$ is $d$ and the plane $R$ is parallel to plane YZ.

Reference Image

Given, a third point $P_3=(x_3, y_3, z_3)$ & a line $L_3(α_3, β_3, γ_3)$ which intersects the plane $R$ at point $R_3$, find the $y$ and $z$ coordinates of $R_3=(x_r, y_r, z_r)$.

We are getting the precise values of $x_1$, $y_1$, $z_1$, $x_2$, $y_2$, $z_2$, $α_1$, $β_1$, $γ_1$, $α_2$, $β_2$, $γ_2$, $d$, $x_3$, $y_3$, $z_3$, $α_3$, $β_3$, $γ_3$ via sensors. Find $y_r$ & $z_r$.

I have written the solution in paper.
The key idea is that you first find out the equation of plane R. Then you find the point of intersection between L3 and R to obtain yr and zr.

The parametric equations for the lines and planes are included in the paper. Have a look!

First Page
Second Page


Helping my daughter with her homework: solving an algebra word problem. Three bags of apples and two bags of oranges weigh $32$ pounds.

Four bags of apples and three bags of oranges weigh $44$ pounds.

All bags of apples weigh the same. All bags of oranges weigh the same.

What is the weight of two bags of apples and one bag of oranges?

$x$: weight of a bag of apples (in pounds)

$y$: weight of a bag of oranges (in pounds)

First we "translate" the givens into algebraic equations:

$(1)$ "Three bags of apples and two bags of oranges weigh $32$ pounds." $\implies 3x + 2y = 32$.
$(2)$ "Four bags of apples and three bags of oranges weigh $44$ pounds." $\implies 4x + 3y = 44$

This gives us the system of two equations in two unknowns:
$$3x + 2y = 32\tag{1}$$
$$4x + 3y = 44\tag{2}$$

Ask your daughter to solve the system of two equations in two unknowns to determine the values of $x$ and $y$.

Hints for your daughter:

multiply equation $(1)$ by $3$, and multiply equation $(2)$ by $2$:

$$9x + 6y = 96\tag{1.1}$$
$$8x + 6y = 88\tag{2.1}$$

subtract equation $(2.1)$ from equation $(1.1)$, which will give the value of $x$.
Solve for $y$ using either equation $(1)$ or $(2)$ and your value for $x$.
Then determine what $2x + y$ equals. That will be your (her) solution.

Let one bag of apple weight $x$ pounds and one bag of orange weight $y$ pounds. We then have
\begin{align}
3x&+2y=32\qquad\\
4x&+3y=44\qquad
\end{align}

To find the value of $x$ and $y$:
\begin{align}
(3x+2y=32)  \cdot 3\\
(4x+3y=44)  \cdot 2
\end{align}
yields
\begin{align}
9x+6y=96 \qquad(1)\\
8x+6y=88 \qquad(2)
\end{align}
Subtracting $(2)$ from $(1)$, we get $x=8$ and $y = 4$. So the weight of $2$ bags of apples and $1$ bag of orange is
$$
2\cdot8  +  4 =16+4 =20\text{ pounds}.
$$


A few questions on nonstandard analysis I know that nonstandard analysis is analysis plus the existence of infinitesimal numbers. Does it mean that nonstandard analysis is the same theory as $ZF+\exists$infinitesimal numbers?

From what I read about it on Wikipedia there seem to be a few different approaches to it. There is "Robinson's method" which is "based on studying models". There is "Nelson's method" which is "Internal set theory" which is an extension of ZF. Are they the same? Or are there theorems that are true in one but not in the other?

If IST is an extension of ZF how much stronger than ZF is the "theory of nonstandard analysis"?

$\mathsf{IST}$ is a conservative extension of $\mathsf{ZFC}$: any internal formula that can be proved within internal set theory can be proved in $\mathsf{ZFC}$. This is proved in the Appendix to Edward Nelson’s original $1977$ paper in the Bulletin of the American Mathematical Society, ‘Internal Set Theory: A New Approach to Nonstandard Analysis’.

To understand the theory nonconstrutivly of Robinson, it's sufficient to use the model theory, in this link: https://hal.inria.fr/hal-01248379/ , we can find a new approach to nonstandard analysis without using t
he ultrafiltries.


number of elements on a finite vector space I know that there is no vector space over any field $\mathbb{F}$ having precisely $6$ elements. I would like to know if there is a theorem which characterizes those $n's \in \mathbb{N}$ such that there is a vector space having exactly $n$ elements over a field $\mathbb{F}$. For example: $n=p^{k}$ where $p$ is a prime number and $k\in \mathbb{N}$ ($k\neq 0$).

I think you can put a proof together from the following ingredients.

If $F$ is a finite field with $q$ elements, then $q$ is a power of a prime.

If $V$ is a finite vector space over a field of $q$ elements, then it has a finite basis, which we'll call $\lbrace\,b_1,b_2,\dots,b_r\,\rbrace$, and every element has a unique expression as a linear combination of basis elements, and the number of such linear combinations is $q^r$.

Are the "ingredients" things you already know?

You cannot have vector spaces with finite elements. Even if the field is finite having just finite elements
consider a vector space with basis $\{e_1, e_2, e_3,\ldots\}$
where
\begin{align}
e_1 &= (1,0,0,0,\ldots)\\
e_2 &= (0,1,0,0,\ldots)\\
e_3 &= (0,0,1,0,\ldots)\\
\vdots & = \vdots
\end{align}


How to solve a polynomial inequality? Ok guys, I need some more help with a question for my girlfriend. Basically she was given a problem on a test/quiz and the only way I know how to do it is with a method that she hasnt learned in class yet. So pretty much I want you guys to look it over and let me know if there is another method to the problem that she might know. The problem is...

$$2x^3 + 3x^2 \lt 11x + 6$$

She has to solve it and give the answer in interval notation. So I would first move everything from the right side to the left to have a third order polynomial. My next step would be to use the rational zero test to start finding a zero using synthetic division. After I found one zero, I would factor it out of the function, and i would be left with $(x-a)$*2nd orderpolynomial. I could factor out the polynomial and find my 3 zero's.

The only problem is, she didnt learn synthetic divison, rational zero test, or long division of polynomials. is there another way she could do this problem with knowledge she might have? i tried to briefly teach her my way and it went over her head

Factor like this
$$2x^3 +3x^2 - 11x - 6 = (x-2)(2x+1)(x+3).$$
Then draw a sign chart.

You could just move everything to one side and graph it to find approximate roots, then substitute them in.  Or you could feed it to Wolfram Alpha


No subgroup of order $8 $ in $S_5$ is abelian..True or False?? I may be heavily mistaken but these are my partial arguments in favour of the following statement

" No subgroup of order 8 in $S_5$ is abelian"

If not, let there be an abelian subgroup of order 8 in $S_5$ .Then it is isomorphic to either $\mathbb{Z}_8$, $\mathbb{Z}_4 \oplus \mathbb{Z}_2$ or $\mathbb{Z}_2\oplus \mathbb{Z}_2\oplus \mathbb{Z}_2$

case 1) The subgroup $H$ (say) is isomorphic to $\mathbb{Z}_8$.Then there should be a corresponding element of order 8 in $H$ which implies  $S_5$ has an element of order 8  which is impossible by considering the cycle type of different elements.

case 2) The subgroup $H$ is isomorphic to $\mathbb{Z}_4\oplus\mathbb{Z}_2$ .The latter group has $(1,0)$ and $(0,1)$ of order $4$ and $2$ respectively which commute and hence there are corresponding elements of order $4$ and $2$ in $H$. Now, The elements of order 4 in $S_5$ are of the cycle type $(a,b,c,d)$ and that of order 2 are of cycle type $(a,b)$ or $(a,b)(c,d)$. In any case , permutations of order 4 and 2 in $S_5$ will not commute since they do not have disjoint cycle types.Thus contradiction.

Now I am puzzled with the last case. Will there be some contradiction ?Are the above two cases handled properly? Is there rather a smart way to arrive at the result??

Sorry for a long list of questions but I need your help. Thanks for your ideas.

Let $S$ be an arbitrary subgroup of $S_5$, Since $|| = 23×3×5$ and $2^3|(G)$, $2^4\nmid o(G)$, therefore the number of $2$-sylow subgroups of $G$
of order $8$ is of the form $1+2k$ and $1+2k|o(G)$, that is, $1+2k|2^3×3× 5$. Therefore $1+2k|5×3$, the possible values for $k = 0, 1, 2$
and $7$. Therefore the maximum number of $2$-sylow subgroup of order $8$ is $15$ when $k = 7$.
Since there is no element of order $8$, so every element of $S$ must have order $1, 2 $and $4$. Can you finish from here?


Find polynomials $P(x)P(x-3) = P(x^2)$ Find all polynomials  $P \in \Bbb R[x]$ such that $P(x)P(x-3) = P(x^2) \quad \forall x \in \Bbb R$

I have a solution but I'm not sure about that. Please check it for me.

It is easy to see that $P(x) = 0, P(x) = 1$ satisfied.

Consider $P(x) \neq c$ :

We have $P(x+3)P(x) = P((x+3)^2)$

If $a$ is a root of $P(x) $ then $a^2$ and $(a+3)^2$ are roots of $P(x)$.

If $|a| > 1$, then $ |a^2| = |a||a| > |a| \Rightarrow P(x)$ has infinitely many roots.

If $0 < |a| < 1$, then $|a^2| = |a||a| < |a| \Rightarrow P(x)$ has infinitely many roots.

If $a = 0 \Rightarrow (a+3)^2 = 9$ is a root of $P(x)$, so  $P(x)$ has infinitely many roots.

Hence, every root $a$ of $P(x)$ has $|a| = 1$. So $1 = |(a+3)^2| = (|a+3|^2) \Rightarrow |a+3|=1$.

We have $a + 3 = \cos\alpha + i\sin\alpha + 3 \Rightarrow (\cos\alpha+3)^2 + \sin^2\alpha = 1 \Rightarrow \cos\alpha = -\frac{3}{2}\ (\text{impossible}).$

$P(x) = 0, P(x) = 1$ are all the results.

Considering

$$
P(x)P(x-b)=P(x^2),\ \ \ b \ne 0\ \ \ \ (1)
$$

As long as $P(x^2)$ is even then $P(x)P(x-b)$ is also even then

$$
P(b)P(0) = P(b^2)
$$

because making $x = b$ in $(1)$ we have $P(b)P(0) = P(b^2)$, As this is for a generic $b$ we have $P(x)P(0) = P(x^2)$

or

$$
P(x)P(0) = P(x^2) = P(-x)P(-(x+b))
$$

which implies that $P(x)$ is even as well as $P(x+b)$

This should be true for all $b \ne 0$ so the solution is $P(x) = C_0$

Finally, according to $C_0^2=C_0$ we conclude that $C_0 = 0$ or $C_0 = 1$

You point out that if $a$ is a root then $a^{2}$ is a root and $a+3$ is a root.

You don't need to divide into three cases to determine that means either infinitely or zero roots.

By induction  $a + 3k; k \in \mathbb N$ are roots.  If $j\ne k$ then $a+3j \ne a + 3k$ so if there is one root $a$ there are at least countably infinite distinct roots.  So there are either zero roots, of infinitely many roots.

Magnitude is irrelevant[1]

Only trivial constant polynomials have infinite or zero (complex) roots so that means $P(x) = c$ for some $c$ where $c^2 = P(x)P(x-3) =P(x^2) = c$.

[1]

Although if $a$ is a root then $a +3$ is a root.  And  if $|a+3| < 1$ then $|a+3| + |a|=|a+3| + |-a| \ge |(a+3)+(-a)| =|3|=3$ so $|a|\ge 3-|a+3| > 2$.  So at least one root is $\ge 1$.


Why do I get one extra wrong solution when solving $2-x=-\sqrt{x}$? I'm trying to solve this equation:
$$2-x=-\sqrt{x}$$
Multiply by $(-1)$:
$$\sqrt{x}=x-2$$
power of $2$:
$$x=\left(x-2\right)^2$$
then:
$$x^2-5x+4=0$$
and that means:
$$x=1, x=4$$

But $x=1$ is not a correct solution to the original equation.

Why have I got it? I've never got a wrong solution to an equation before. What is so special here?

This is because the equation $\;\sqrt x=x-2$  is not equivalent to $x=(x-2)^2$, but to
$$x=(x-2)^2\quad\textbf{and}\quad x\ge 2.$$
Remember $\sqrt x$, when it is defined, denotes the non-negative square root of $x$, hence in the present case, $x-2 \ge 0$, i.e. $x$ must be at least $2$.

Overshoot, with subsequent back-tracking to the destination, is sometimes the easiest path, a common example being the fact that, due to tricky lane changes and / or confusing roadway signage, it could well be easier to overshoot the destination and then wend your way back to it.  The phenomenon of such overshoot is just a special case of the fact that a larger footprint is easier than a smaller footprint. (cf: ‘collateral damage’, and cf: “I would have written a shorter letter, but I didn’t have the time.” -- Blaise Pascal) An example of mathematically necessary overshoot is the fact that the vertical coordinate of the brachistochrone overshoots the vertical coordinate of the destination, and, even more famously, that ‘the shortest distance between two truths in the real domain often passes through the complex domain’ – the passing through the complex domain being a stratospheric instance of overshoot, two examples of which are the explanation of the radius of convergence of 1/(1 + x^2), and the proof of the fact that the product of two numbers, each representable as the sum of two squares, is itself representable as the sum of two squares. A very common example is the squaring of both sides of an equation (in order to remove a radical), which may introduce an ‘extraneous’ root. You then ‘pull back’ to the actual solution set by inspecting each root of the second equation individually.


Simple Question On Relationship Between Cubes And Squares I'm new to this number theory business, not to mention terribly naive.  I wonder whether someone could explain the technique (assuming there is one) to show whether the expression

$12C - 3$

(where $C$ is any cube number greater than $1$) can ever be a square number.

In particular, I'm looking for a square number congruent to $3 \mod 6$, but I'll settle for just any square number for now.  A simple search of the first few hundred cube numbers turned up nothing (apart from the trivial $C = 1$)

Sorry if this question has you mathematicians rolling your eyes.

Thank you so much for your help.  By the way, if anyone can suggest a really gentle introduction to number theory, would you mind recommending a book to me?  Thanks.

Regards,

Mike.

First of all, $12C-3 = 6(2C-1)+3$, so $12C-3 \equiv 3 \pmod 6$ for any integer $C$. As Henning Makholm wrote, if a square is $3 \pmod 6$, then the root of the square is also $3 \pmod 6$, so if $12a^3-3 = k^2$, then $k = 6c+3$ for an integer $c$, and the relation can rewritten as
$$
12a^3-3 = (6c+3)^2,
$$
which is equivalent to
$$
a^3 = 3c^2+3c+1
$$
or
$$
a^3+c^3 = (c+1)^3.
$$
This is impossible for $c>0$ (it's a particular case of Fermat's Last Theorem for $n=3$; this case has been proven by Leonhard Euler), so $C=1$ is the only solution to your problem.

$1 ^ 3 + 2 ^ 3 = 3 ^ 2$. Likewise $1 ^ 3 + 2 ^ 3 + 3 ^ 3 = 6 ^ 2$. $1+2= 3$ and $1+2+3 = 6$. Like wise sum of cubes of first $4$ number is square of $10$ ($1+2+3+4$). How is it? You can check up to any number of your choice and prove whether I am right or wrong.


I need mathematical proof that the distance from zero to 1 is the equal to the distance from 1 to 2 I didn't know how to phrase the question properly so I am going to explain how this came about.

I know Math is a very rigorous subject and there are proofs for everything we know and use.  In fact, I am sure that if there was anything that we couldn't prove Mathematically, then we wouldn't use it any where in Math or Science.

Anyway, here is the story:
Not to long ago I was having a conversation with someone who insisted in telling me that any human knowledge based on Math is a sham, this of course includes Science.  He then proceed to tell me that he thought this was the case because according to him there is no proof that zero is a number and second he insisted even if we were to say that zero is a number,  that there is no proof that the distance from 0 to 1 is the same as the distance from 1 to 2.

According to this guy, there is no way to measure the distance from something that doesn't exist to something that does.  If I understood him correctly he said that is because he just can't see how we can measure from non-existence to something that does exist.  In this case, he said that there is now way to know for sure that the length or distance from zero or non-existent number to one would be the same as the distance or length from 1 to 2.

No matter what I said to him, he just disregarded offhandedly.  I am so upset that I want to punch him in the face, but I would rather get the proofs and show him that he is completely wrong.  To me this would be the equivalent of slapping him on the face with a gauntlet and I would have the satisfaction of knowing that Math and Science are in solid ground and that Mathematicians and/or Scientist haven't pulled a fast one on us.

So please post links to where I can read the Mathematical Proofs for these.  Or links that show that to be a fallacy.

Thanks.

The question was, is there a place where I can get proofs for both of those things.  First, for Zero being a number whether an integer, complex or any other kind.  And second, for a proof that would show that the distance from 0 to 1 is equal to the distance from 1 to 2.

But I guess what I should really have asked is whether or not these proof exist and where I can find them.

I am so upset that I want to punch him in the face

This would be an understandable (although inadvisable) reaction. Here are a few responses I recommend you give to him:

"Define what you mean by 'exists'."

"Define what you mean by '1'."

"Define what you mean by 'distance'."

"Prove that 1 exists."

"Prove that the distance from 1 to 2 is 1."

"What is the distance from 1 to 1?"

These questions are designed to get him a) actually thinking about what he's saying, and b) quiet.
In response to "Define what you mean by '1'" and "Prove that 1 exists", he will undoubtedly point to "one rock" or "one bird" or "one airplane", etc., and then you can respond,
"Ah, but those are all one of something. I am talking about the number 1." Here is where he will be a bit confused. The number 1 is a human abstraction from our experiences with one <object>, but it does not exist as a physical object. Here is my favorite quote regarding this (from  Linear Algebra by Fraleigh and Beauregard):

"Numbers exist only in our minds. There is no physical entity that is the number 1. If there were, 1 would be in a place of honor in some great museum of science, and past it would file a steady stream of mathematicians gazing at 1 in wonder and awe."

There are several fundamental issues that your adversary is missing:
Firstly, there is no agreed-upon definition of what it means for something to be a  "number". If someone said "I am only going to call even integers "numbers", the odd integers don't count", they are just as correct as someone who calls every mathematical construction humans have ever come up with a "number". Mathematicians have decided on definitions of "integer", "complex number", etc., and the statements that "0 is an integer" or "$\sqrt{2}$ is not an integer" are true because "integer" actually means something. The word "number" is a vague word that has no mathematical content.
Secondly, mathematics is a human construction:

"The mathematician is entirely free, within the limits of his imagination, to construct what worlds he pleases. What he is to imagine is a matter for his own caprice; he is not thereby discovering the fundamental principles of the universe nor becoming acquainted with the ideas of God."  - John William Navin Sullivan
"Mathematics is a game played according to certain rules with meaningless marks on paper." - David Hilbert
"[Mathematics] is an independent world/ Created out of pure intelligence." - William Wordsworth

So, what "exists" and what "doesn't exist" depends entirely on what we've set up for ourselves. Mathematicians pick a certain set of axioms, then follow them to logical conclusions. It makes no sense to say, separate from a previous decision about what axioms we're using, that "0 doesn't exist". Only after you've decided what axioms you are using, and what exactly it is that would characterize the object "0", can you decide whether your system includes such an object.

If you really want to get at him, here is a recommendation: Tell him, "I am going to prove directly to you that 0 exists." For several seconds, pretend to be deep in thought, trying to remember the details of the proof. Surely he will look on with anticipation, eager to attack whatever you are going to present. Then, appear to have recalled the key step, and very slowly and deliberately draw a massive "0" taking up the entire piece of paper. Look extremely pleased with yourself, as if you had made a difficult and convoluted argument.
The fact is, that is exactly what "0", or indeed anything in mathematics, is: a symbol we write down. What "0" means in mathematics can vary widely; perhaps it is the zero element of $\mathbb{Z}$ (the integers), or the function from $\mathbb{R}$ (the real numbers) to $\mathbb{R}$ that is the constant 0 function. Within a given context, e.g. specific axioms and structures, it makes sense to decide what properties an object you want to call the symbol "0" might have (for example, if there is a notion of "addition", you would probably ask for an object "0" to satisfy "$a+0=a$" for all objects $a$), and then, having decided what "0" should refer to if it did exist, actually find out whether it does, within your axioms / structures. But without deciding what "0" means beforehand (and remember, 0 isn't something like "no airplanes", since physical objects are irrelevant), all "0" means is the symbol 0, and it is quite easy to prove that that exists - just draw it!
Of course, this won't be satisfying to your adversary at all, but I guess I'm just trying to think of ways you can get your just revenge by making equally absurd arguments :)
Good luck confronting him, perhaps comment back to let us know how it goes!

Look into Abstract algebra, primary definitions of binary operation, neutral elements and groups. Continue with rings and fields. Then go on vector spaces. After some time you should reach metric spaces and norm. Then you will have whole definition of distance, in any dimension.

You can then say to him, in our world with addition, zero is neural element for addition, so it's created by definition. And by definition of metric space (and we live in metric space, where distance is measured by meter, feet, etc) distance between 0 and 1 is same as 1 and 2.


How old did the man live to be? On a man's tombstone, it is said that one sixth of his life was spent in childhood and one twelfth as a teenager. One seventh of his life passed between the time he became an adult and the time he married; five years later, his son was born. Alas, the son died four years before he did. He lived to be twice as old as his son did.
Question- How old did the man live to be?

(Assuming the end of teen life as adulthood)

Let the man lived for $x$ years.

Childhood: $\frac{x}{6}$;
Teenager: $\frac{x}{12}$;
From the end of teenage/adulthood to married: $\frac{x}{7}$;
From marriage till his son was born: $5$;
Till his son was dead: $y$;
Till he died: $4$.

Summation of all these is equal to $x$. The second equation comes from the fact that the father lived to be twice as old as his son.

$$\frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + y + 4 = x$$
$$x = 2y$$
Substituting $y$ from the second equation in first,
$$\frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + \frac{x}{2} + 4 = x$$
$$\frac{3x}{4} + \frac{x}{7} + 9 = x$$
$$\frac{x}{4} - \frac{x}{7} = 9$$
$$x = 84, y = 42$$

$84$ years

$\frac{x}{6}+\frac{x}{12}+\frac{x}{7} + 5 + y + 4 = 2y$. Simplifying this equation gives:
$\frac{11}{28} (x) + 9 = y$

Solve this for integer values of $x$ and $y$ using trial and error. Start with $x = 28$, $56$ and finally $84$ till you get an integer solution for $y$.

This gives $x = 84$ and $y = 42$


If $\cos^6(x)+\sin^6(x)=a+b\cos(4x)$. Find $ab$? As after solving above equation, I get $1-3\sin^2(x)\cos^2(x)$= $a+b\cos(4x)$.
Please help me to solve this problem.

Hint : What happens if you replace $x$ by $0$ or $\pi/4$?


Prove that the union of two ideals is an ideal only if one of the ideals is contained within the other. Here's my proof:

Let the ideal $I = \{i_1, i_2,...\}$ and $J=\{j_1, j_2,...\}$ Then $I \cup J$ is an ideal only if for all $\nu, \mu$, $i_{\nu}+{j_\mu} \in I \cup J$. Let us assume that there exists an element $i_{\nu}$ of $I$ that does not belong to $J$. Then, for any $j_{\mu}$, either $i_{\nu}+{j_\mu} \in J$, which is a contradiction since it would imply that $i_{\nu} \in J$, or $i_{\nu}+{j_\mu} \in I$, which implies that $j_{\mu} \in I$. Letting $\mu$ run through the index of $J$, we get that $J \subset I$. Thus, our proof is complete.

However, here are some things bothering me:

There seems something fishy about this proof, but I can't point it out. Is my proof correct?
Is this a constructive proof or a proof by contradiction?
Does this proof use the Axiom of Choice?

Thanks in advance for any help!

Your proof uses contradiction to show that $j_\mu\in J$.
It does not.

A less fishy proof:

Suppose that there are $i\in I\setminus J$, $j\in J\setminus I$. If $I\cup J$ is an ideal then $i+j\in I\cup J$, but $i+j\in I$ implies $j\in I$ and $i+j\in J$ implies $i\in J$. Both conclusions are contradictions, so there is no such pair $(i,j)$.

This theorem says that if A, B are ideals of ring R and A is subset of B, then the union of A and B is also ideal.
Notice that if A is subset of B, then the union of A and B is B itself. So, the union of A and B which is equal to B is also ideal, since B is ideal.


Follow-up question on conditionally convergent series. This is a follow-up from this question.
Hypothesis: If $\sum\limits_{n=0}^{\infty} a_n\ $ is a conditionally convergent series with $\ \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$, then is $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ convergent?
Maybe I'm wrong but I believe this isn't answered in the comments to the answer, because not only does it not allow for zeros to be part of the sequence, but also it's not clear to me how interweaving an absolutely convergent sequence answers this question.
$$$$
Actually I think I've found a sequence that disproves my hypothesis. For the following sequence, it doesn't matter what $a_0, a_1$ are: $a_0 =-1,\ a_1 =-1,\ a_2 = \frac{1}{2},\ a_3 = \frac{1}{3},\ a_4 = -\frac{1}{4},\ a_5 = -\frac{1}{5},\ a_6 = \frac{1}{6},\ a_7 = \frac{1}{7},\ a_8 = -\frac{1}{8},\ a_9 = -\frac{1}{9}, ...$
It's an alternating sequence of decreasing terms (which $\to 0$ ) when put into pairs $(a_0 + a_1, a_2+a_3, a_4+a_5,... ),$ so I think the corresponding series $\sum\limits_{n=0}^{\infty} a_n\ $ is conditionally convergent.
Also, the series $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ is $\geq $ the sum of the pairs: $\sum\limits_{ \text{n is even} }^{\infty} |a_n+a_{n+1}|\ \geq \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + ...,$ which famously diverges. So I believe this sequence disproves my hypothesis(but I'd like confirmation please).
$$$$
Assuming this is all correct, this then begs the modification to the original question:
If $\sum\limits_{n=0}^{\infty} a_n\ $ is a conditionally convergent series with $\ \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$, then is $\sum\limits_{n=0}^{\infty} a_n+a_{n+1}\ $ convergent?
But for this question, we can even get rid of the $\ \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$ condition and the answer is still yes, because
$\sum\limits_{n=0}^{\infty} a_n+a_{n+1}\ = lim_{n \to \infty} \left(a_0+2\left(a_1+a_2+a_3+...a_{n-1}\right) + a_n\ \right) = 2\sum\limits_{n=0}^{\infty} a_n\ - a_0, $ which converges because $\sum\limits_{n=0}^{\infty} a_n\ $ converges because it is conditionally convergent.

Conditional convergence means the sum converges, but not absolutely.
That is, $\sum_{n\in \Bbb N} a_n $ converges, but $\sum_{n\in\Bbb N} |a_n|$ does not.
Now: $\limsup_{n\to\infty} |\frac{a_{n+1}}{a_n}|=1$ tells us that $|a_{n+1}|$ and $|a_n|$ are very close in value as $n\to\infty$, which implies we can take: $$\sum_{n\in\Bbb N} |a_n+a_{n+1}|\approx\sum_{n\in\Bbb N} |2a_n|=2\sum_{n\in\Bbb N} |a_n|$$
which does not converge.
For the same reason: $$\sum_{n\in\Bbb N} a_n+a_{n+1}\approx \sum_{n\in\Bbb N} 2a_n=2\sum_{n\in\Bbb N} a_n$$ whenever $a_n$ and $a_{n+1}$ are sufficiently close, this implies convergence. While $\limsup|\frac{a_{n+1}}{a_n}|=1$ is a sufficient condition for this, it is  not necessary.


Proof that two norms $\lVert\cdot\rVert_1$ and $\lVert\cdot\rVert_2$ are equivalent Two norms $\def\norm#1{\lVert#1\rVert}\norm\cdot_1$ and $\norm\cdot_2$ are equivalent iff  $\;\exists\;c_1,c_2>0$ such that $c_1\norm x_1\le \norm x_2\le c_2\norm x_1$

Show that $\norm x_1=\sum_{i=1}^n \lvert x_i \rvert$ and $\norm x_2=\left ( \sum_{i=1}^n x_i^2 \right )^{1/2}$ are equivalent.

It looks like $c_2$ is $1$, and that this can be proven with induction. But what could $c_1$ be?

edit

I mean what I don't understand is the following: If I square both terms, and expand $(\sum_{i=1}^n \lvert x_i \rvert)^2 = \left ( \sum_{i=1}^n x_i^2 \right ) + \sum_{i=1}^n x_i(x_k+\dotsb+x_n)_{x_k\neq x_i}$. However, the second term grows with $n$, so how can $\frac{\norm x_1}{\norm x_2} \leq C$ at all?

You want to show that for a fixed $n$ the norms $\|\|_1,\|\|_2$ are equivalent in $\mathbb R^n$. The constant $c_1$ will depend on $n$.

Hint:
Cauchy Schwarz inequality.

Let $x=x_i$ for all $i$.  Then $||x||_1=x\dim x$, and $||x||_2=\sqrt{\sum_{i=1}^{\dim x}x^2}=\sqrt{x^2\dim x}=x\sqrt{\dim x}$.  Therefore $$ \lim_{\dim x\to\infty} \frac{||x||_1}{||x||_2}=\sqrt{\dim x}=\infty $$
So the two cannot be equivalent by your definition.


Can anyone explain intuitively why increasing the circumference by 1 meter always increases the radius by 15.9cm? I found the mathematical proof, and it is obviously correct. But how can the increase in radius be constant regardless of the starting circumference? With a very small circle, the increase should be huge but with a massive circle, the difference should end up miniscule shouldnt it? After all, the 1 m is getting distributed over a much larger circumference

Let the radius of the sphere be R and the new radius be R', hence

$2\pi R' = 2 \pi R + 1$

or,

$2\pi(R'-R) = 1$

or, the height $R' - R$ is $1/{2\pi} = 15.9 cms.$

With a small circle, the increase is huge with respect to the small circle. Basically, for a circle of radius 1cm, the 15.9cm increase is a lot ... with respect to the original circle.

To convince yourself, consider this: When you increase the radius of a 1m radius circle by 1m, the radius doubles, but if you do it to a 100m circle, it just increases by a tiny amount relative to the original radius. So in the case of the larger circle, the increase seems much less because the relative increase is less, but in both cases the actual increase is the same.

It's a similar situation with the 15.9 cm increase. The absolute increase is the same, but the relative increase (which is what our brains find easier to imagine) is much smaller

Because a circle with a circumference of 1m has a radius of 15.92cm.
So whenever you add 15.9cm to the radius of a circle you add 1m to the circumference.

Or whenever you add 1m to the circumference, you add 15,9cm to the radius.

Because $2\pi \times 0.159155 \approx 1$.


two inequalities, application of MVT? $$py^{p-1}(x-y) \leq x^p-y^p \leq px^{p-1}(x-y)$$
Where $0<y<x, \ 1\leq p <\infty$

I haven't been able to prove either of these inequalities. I tried subtracting the left from the middle and trying to show the whole thing is non-negative but haven't had any success.

Hint: If $y\le x$ and $m\le \varphi'(z)\le M$ for every $z\in[y,x]$, then $$(x-y)m\le \varphi(x)-\varphi(y)\le(x-y)M.$$

It is a typical application of the Mean Value Theorem to a convex function (i.e. a function whose derivative f' is an increasing function).

Mean Value Theorem: if f is continuous and its derivative exists in the interval (a,b), then

f(b)-f(a) = f'(c)(b-a)

for some c in (a,b).

If f' is an increasing function, then  a < c < b  implies  f'(a) < f'(c) < f'(b), but now (thanks to the Mean Value Theorem) you have an expression for f'(c). The last step is choosing the right convex function for your inequality. In fact, you can use the Mean Value Theorem to obtain many inequalities from many functions.


Getting into differential equations I'm just getting into differential equations now and I've got to show that the given $y(x)$ is a solution to the differential equation: $$u'+u = 0 \ , \ y(x) = Ce^{-x}$$

How do I tackle this? I know nothing about differential equations and my book has the strangest explanations.

The point is that you plug $y(x) = Ce^{-x}$ into the differential equation at the place of $u$ , and of course the derivative of $y(x)$ at the place of $u'$ (do you know how to differentiate $y(x)$?). Then you must show that the equality in the differential equation holds, that is $y'(x) + y(x) = 0$. So
\begin{align}
y'(x) + y(x) = -Ce^{-x} + Ce^{-x} = (-C+C)e^{-x} = 0\cdot e^{-x} = 0.
\end{align}
So it is, in fact, a solution since it equals zero as it should.

This is a first order differential equation so called because the highest derivative is the first derivative (that is u'). To solve in this instance you simply separate the variables.  Let dy/dx = u' and y = u ;  dy/dx  + y = 0 ====> dy/dx = -y. From here you see that multiplying both sides by dx/y yields dy/y = - dx. Apply the integral operator and you get ln|y| = -x + ln C. Taking the antilogarithm of both sides and remembering you're rules of logarithms gives you the answer. Y = Ce^(-x) the general solution


Why is Completeness not a Topological Property? I am trying to answer the question:
Show why completeness is not a topological property.

My answer:  $\mathbb{R}$ and the set $(0,1)$ are homeomorphic, but $\mathbb{R}$ is complete while $(0,1)$ is not.

My question to you all:  Does this answer the question?  I feel like I am not quite seeing what is going on with completeness and why it is not a topological property. Can someone give me another example?

Another way to see it is:  One can metrize the space $(0,1)$ in (at least) two different ways that generate the same topology.  One such metric is the ordinary Euclidean metric on $(0,1)$; a second metric is what you get when you pull the Euclidean metric on $\mathbb{R}$ back to $(0,1)$ via a homeomorphism.

In one of those metrics, $(0,1)$ is complete, but in the other it is not -- again, despite the fact that the two metrics generate the same topology. That shows that completeness depends not on the topology per se but rather on the metric one associates with the topological space.

The completenes property is dependent upon the metric. If you change the metric, the same Cauchy sequences which once converged may not under your new metric.  Conceptually, size or distance are not properties preserved under homeomorphism.  Topology is concerned with structure - limit point relations on the set - continuity, connectedness, compactness.


Infinitely many axioms of ZFC vs. finitely many axioms of NBG It is known that ZFC needs infinitely many axioms, but NBG (Neuman-Bernays-Gödel set theory) is finitely axiomatizable (as first-order theories of course). But both theories agree completely on the set part of their universe (as far as I have read).

How could this be? How can describing even more objects (proper classes in NBG) while keeping the complexity of some part can reduce the effort to describe this structure? Is there some plausible and evident explanation of this observation? Maybe some philosophical insight from someone who knows the proofs of these statements?

Maybe, is it because the proper classes allow NBG to quantify over predicates in some sense? Something for which ZFC usually needs axiom schemas? If so, why isn't NBG absolutely favorable to ZFC as foundation of math? I mean we also prefer set theory to Peano arithmetic because the latter one allows us to quantify over subsets (in some sense) despite it is a first-order theory (I know we prefer ZFC over PA for tons of other reasons too).

Note:

I know of this and this question, but I ask specifically why the finite axiom system is not a convincing reason for NBG.
However,the question on which to prefer, ZFC or NBG is secondary. Please concentrate on the finitely vs. infinitely many axioms part and how this can be.

I will discuss this in two parts:

Why is ZFC finitely axiomatizable in NBG.
Why is class comprehension finitely axiomatizable.

The first part is simply because, by class comprehensions, every first order predicate is some class. The axioms of ZFC is restricted to first order predicates, and therefore definable in NBG.

As for part 2. NBG allows as to quantify over first order predicates. It is a second order language, applied to discuss first order objects. Just as separation and replacement are restricted to first order object, is is class comprehension, and it is therefore definable with a second order formula.

As for why NBG is not favored. In a sense it is. Many statements in Set Theory discuss proper classes. However the basis of Set Theory is sets, and ZFC is more intuitive in this aspect. And as many classes can be described in terms of sets or formulas, many results in NBG are derivable from ZFC plus the intuitive assumption "There is such thing as a class." The two theories are indeed equiconsistent.


Is the integral of $\sin(2x)$ equal to both $\sin^2(x)+C$ and $\frac{1}{2}\cos(2x)+C$? This morning I encountered the integral $\int \sin(2x)$. My initial approach to solving it went something like this:

$\int \sin(2x)\ dx=\int 2\sin(x)\cos(x)\ dx$ (by the $\sin(2x)=2\sin(x)\cos(x)$ identity)

Let $u=\sin(x)$ and it's derivative $\frac{du}{dx}=\cos(x)$

Rewrite the integral with respect to $u$ and take out the constant:$\int 2u\ du = 2 \int u\ du$

Solve using the power rule: $2\int u\ du = 2 \cdot \frac{u^2}{2}+C = u^2 + C$

Resubstitute in terms of $x$: $\sin^2(x)+C$

I then tried to verify my results using various integral calculators (namely Wolfram Alpha), but they gave me the result $\frac 12\cos(2x)+C$, which does make sense if you instead of using the trigonometric identity just pick your subsitution to be $u=2x$. Graphing the results however shows this:
 It's quite clear that these results aren't equal to each other, no matter how you adjust the constants.

My question is, what is going on here? How come integrating the same functions using two different approaches yields results that are not the same function?

Lord Shark the Unknown spotted my mistake in the comments. The problem was that I had missed a minus sign when reading from Wolfram Alpha. The actual answer was $$\int \sin(2x)\ dx = -\frac{1}{2} \cos(2x)+C=\sin^2(x)+C$$ This makes the two solutions equal.

$$(\sin^2x)'=2\sin x\cos x=\sin 2x,$$

$$\left(\frac12\cos2x\right)'=-\sin 2x.$$


Limit of a Rational Trigonometric Function $\lim_{x \to 0} \frac{\sin5x}{\sin4x}$ When solving a trigonometric limit such as:

$$\lim_{x \to 0} \frac{\sin(5x)}{\sin(4x)}$$

we rework the equation to an equivalent for to fit the limit of sine "rule":

$$\lim_{x \to 0}\frac{\sin(x)}{x}=1$$

so, we move forward in such a manner as follows:

$$=\lim_{x \to 0} \frac{\frac{5\sin(5x)}{5x}}{\frac{4\sin(4x)}{4x}}$$

$$=\frac{5}{4}\lim_{x \to 0} \frac{\frac{\sin(5x)}{5x}}{\frac{\sin(4x)}{4x}}$$

$$=\frac{5}{4}\cdot\frac{1}{1}$$
$$L=\frac{5}{4}$$

From that mindset, I am trying to find this trigonometric limit:

$$\lim_{x\to 2} \frac{\cos(x-2)-1}{x^{2}+x-6}$$

I know the Limit "rule" for cosine is:

$$\lim_{x\to 0} \frac{\cos(x)-1}{x}=0$$

If you use direct substitution in the original function you end up with an equation that is same in value to the rule presented. So, from that I am assuming that the answer is $0$. I am just trying to prove that in a step-by-step manner as I did with the sine limit.

P.S. I searched through many many pages of questions and didn't find something that helped. So, if I am repeating a question, I apologize that I missed it.

Thanks for the help!

If you factor the denominator you would get:

$$\lim_{x\to 2} \frac{\cos(x-2)-1}{x^{2}+x-6}=\lim_{x\to 2} \frac{\cos(x-2)-1}{x-2}\frac{1}{x+3}=\ldots$$

For an easy evaluation you can use La'Hospital's rule.
$$
\lim_{x\to 2}\frac{\cos(x-2)-1}{x^2+x-6} = \lim_{x\to 2}\frac{-\sin(x-2)}{2x+1}=0
$$


Analysis Question from Berkeley Problems in Mathematics I'm wondering if the following is correct. The original question(1.1.21, Fa96) asks to prove that
\begin{align*}
f''(x) = \lim_{h\rightarrow0}\dfrac{f(x+h) - 2 f(x) +f(x-h) }{h^2}
\end{align*}

I thought it was straightforward and proposed the following solution:

\begin{align}
f'(x) &= \lim_{h\rightarrow0}\dfrac{f(x+h) - f(x)}{h}
\\
f'(x-h)&= \lim_{h\rightarrow0}\dfrac{f(x) - f(x-h)}{h}\\
f''(x) &=\lim_{h\rightarrow0} \dfrac{f'(x) - f'(x-h)}{h} \\
\end{align}

Substitute $f'(x)$ and $f'(x-h)$ and we'll end up with our desired identity. This solution is different from the one proposed in the book so I want to make sure it is correct.

Strictly speaking no your solution is not correct.

We have a function
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
$$
but note if we try to evaluate $f'(x-h)$ then the $h$ inputted and the $h$ in the limit are two different variables (one is being taken to $0$ while the other is being taken to be constant). So really what we have is
$$
f'(x-h') = \lim_{h \to 0} \frac{f(x-h'+h) - f(x-h')}{h}
$$
and now looking at $f''$ we get
$$
f''(x) = \lim_{h' \to 0} \frac{f'(x) - f(x-h')}{h'} = \lim_{h' \to 0} \frac{\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} - \lim_{h \to 0} \frac{f(x-h'+h) - f(x-h')}{h}}{h'}
$$
And now assuming sufficient differentiability of $f$ we get
$$
f''(x) = \lim_{h' \to 0} \lim_{h \to 0} \frac{f(x+h)-f(x)-f(x-h'+h)+f(x-h')}{h h'}
$$
but this requires justification in that you can interchange these limits (or take them both to $0$ at the same rate).

Hint: Another approach is to use Taylor series as

$$ f(x+h) \approx f(x)+f'(x)h+\frac{f''(x)}{2!}h^2 $$
$$ f(x-h) \approx f(x)-f'(x)h+\frac{f''(x)}{2!}h^2. $$

See a related problem.


Calculate $\int\frac{x\,dx}{\sqrt{x^2 + x + 1}}$ Calculate $\int\frac{x\,dx}{\sqrt{x^2 + x + 1}}$.

I tried Euler and hyperbolic substitutions, but both lead to complicated calculations and yet WolframAlpha is able to generate quite simple form of integral.

By completing the square in the denominator, this is equivalent to

$$\int\frac{x}{\sqrt{(x+\frac12)^2+\frac34}}\,dx.$$

Can you carry on from here?


Finding the minimum value of $|a+b\omega+c\omega^2|$ if $a,b,c$ are unequal integers where $\omega^3=1$ My try 1:
$$|a+b\omega+c\omega^2|\le\sqrt{|a+b+c||\underbrace{1+\omega+\omega^2}_0|}$$
Cauchy-Scwartz won't give us an upper bound since $a,b,c$ are nonequal integers.

My try 2:
$$|a+b\omega+c\omega^2|=\left|a+b\left(\frac{-1+i\sqrt3}{2}\right)+c\left(\frac{-1-i\sqrt3}{2}\right)\right|\\
=\sqrt{\left(a-\frac b2-\frac c2\right)^2+\frac34\left(b-c\right)^2}
\\=\sqrt{a^2+b^2+c^2-ab-bc-ca}
\\=\frac1{\sqrt2}\sqrt{(a-b)^2+(b-c)^2+(c-a)^2}$$

So it's just a kind of measure of dispersion of the numbers, a most suitable choice would be then the triplet $\{-1,0,1\}$ which gives an upper bound of $\sqrt3$ but it's not quite close (there's a more nearer one i.e.1)

Geometrically, the set
$$X =\{\; a + b\omega + c \omega^2 : (a,b,c) \in \mathbb{Z}^3\;\}$$
forms a triangular lattice in $\mathbb{C}$. Consider its subset
$$Y = \{\; a + b\omega + c \omega^2 : (a,b,c) \in \mathbb{Z}^3, a \ne b \ne c \ne a\;\}$$
For any complex number $z \ne 0$, let $\mathbb{R}z$ be the line $\{\; \lambda z : \lambda \in \mathbb{R} \;\}$ passing through $z$ and the origin.
It is easy to see

$$\begin{array}{ccr}
a \ne b &\implies& Y \cap \mathbb{R}\omega^2 = \emptyset\\
b \ne c &\implies& Y \cap \mathbb{R} = \emptyset\\
c \ne a &\implies& Y \cap \mathbb{R}\omega = \emptyset\\
\end{array}
$$
After we remove $\mathbb{R}, \mathbb{R}\omega, \mathbb{R}\omega^2$ from $X$, the closest candidate that can belong to $Y$ all lies at a distance $\sqrt{3}$ from $0$.
Since one of these candidates $1 - \omega^2$ can be rewritten as $3 + 2 \omega + \omega^2$, it belongs to $Y$ hence the minimum we seek is $\sqrt{3}$.

$\bf{My\; Solution::}$ Let $$\displaystyle f(a,b,c) = \left|a+b\omega+c\omega^2\right|\Rightarrow f^2(a,b,c) = \left|a+b\omega+c\omega^2\right|^2 $$

$$\displaystyle = \left(a+b\omega+c\omega^2\right)\cdot\left(a+b\omega^2+c\omega\right) $$

Above we have used the fact $$\left|z\right|^2=z\cdot \bar{z}$$ and We take $$\left[f(a,b,c)\right]^2 = f^2(a,b,c).$$

So We Get $$\displaystyle f^2(a,b,c)=a^2+ab\omega+ac\omega^2+ab\omega^2+b^2\omega^3+bc\omega^4+ac\omega+bc\omega^2+c^2\omega^3$$

Now Using the fact $$1+\omega+\omega^2 = 0$$ and $$\omega^3=1$$

We Get $$\displaystyle f^2(a,b,c) = a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$$

Now Given $a,b,c$ are Three Different Integers, and Diff. of Two Integer is Minimum

when we take as $a=n-1\;,b=n$ and $c=n+1$

We Get $$\displaystyle f^2(a,b,c) = \frac{1}{2}\left[1^2+1^2+2^2\right]=3$$

So We get $$\displaystyle f^2(a,b,c) = 3\Rightarrow f(a,b,c) = \sqrt{3}$$.

If $a,b$ and $c$ are $3$ unequal  Integers.


Solutions of $x^2 + 119 = 15 \cdot 2^n$ without trial and error I seen this equation at math.stack exchange

The equation $x^2 + 119 = 15 \cdot 2^n$ has only six solutions. Those are $(1,3) ,(11, 4), (19, 5), (29, 6), (61, 8)$ and other one is I don't know. This question I have seen in this site. There they given that, it has six solutions and they did not list these all solutions. I got $5$ of the solutions by my trail method. But, may be there is one more solution. Now my question is, how to find these solutions without using computer or calculator to determine these solutions and how one can conclude the number of solutions are six or some n?

Thank you for providing the last solution (-1, 3).
But, again by computation in trail and error I got the sixth solution. (701, 15).
Now, I understand that, the above equation has 6 solutions in positive integers. So, we can consider this equation as Diophantine equation. Now my question, how to generalize this equations for finding the solutions?

Gandhi, there is a proof and therefore I am not going to reproduce that and claim it is mine.  You may read it at here from the end of page 1 follow page 4. It shows there are only $6$ solutions.

look   because $x^2$   then it could be also -1,so  last solution is (-1,3)


How many combinations for a 5 digit code using 3 numbers. Can anyone please help here?
I have inherited a strange looking safe with only numbers 1 2 and 3.
The code to open it is 5 digits and the code uses all three numbers at least once.
Is there some formula I can apply to list all the combinations?
Thanks

Let $U$ be the set of all possible combinations with those three digits, $A$ the set of combinations that doesn't use $1$, combinations in $B$  doesn't use $2$, and in $C$ are the ones that don't use $3$.
You're trying to find the number of elements of $U/(A\cup B\cup C)$ where / is difference of sets. It's easy to discover the size of $U$, it is $3^5$. If we discover the size of $A\cup B\cup C$ then we're done.
We first observe that $\# (A\cup B\cup C) = \# A + \# B + \# C - \# (A \cap B) - \# (A \cap C) - \# (B \cap C) + \# (A \cap B \cap C)$, where $\#$ is the size of the set. Then we see that $\# A = \# B = \# C = 2^5$, and $\# (A \cap B) = \# (A \cap C) = \# (B \cap C) = 1$, the last thing is that $\# (A \cap B \cap C) = 0$ as there is no sequence that does not use any number. We conclude that there are $3^5 -3\times 2^5 + 3 = 150$ combinations.

Its easy see we can have repetition its just choosing 1 from 3 digits everytime now we want $5$ digits . Now for every digit we have single choices . So its $3.3.3.3.3=243$ Thats all.


If $ab+bc+ca=0$ then the value of $1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is If $ab+bc+ca=0$

then the value of

$1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is...

$ab+bc+ca=0$

$a(b+c)= -bc$

$b+c= -bc/a$

Adding both side $a$ then $b+c+a=a-bc/a$

$a(a+b+c)= a^2-bc$

$1/(a^2-bc)= 1/a(a+b+c)$

Similarly $1/(b^2-ca) = 1/b(a+b+c)$

and $1/(c^2-ab) = 1/c(a+b+c)$.

By adding all eqution we get

$1/a(a+b+c)+ 1/b(a+b+c)+ 1/c(a+b+c)$


Borderline case of interpolation of Banach spaces Let $B \subset A$ be Banach spaces with a continuous embedding.

Is the inequality
$$
\|b\|_B
\leq
C \sup_{t > 0} \inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \}
\quad
\forall b \in B
$$
valid for some $C \geq 1$?

What if the embedding is compact?

For example, take the sequence spaces $B := \ell_2$ and $A := \ell_\infty$ with norms $|\cdot|_\infty \leq |\cdot|_2$.
Let $b \in \ell_2$.
Let $K \geq 0$ be the $\sup \inf$. Then, for any $t > 0$,
the infimizer satisfies $|\tilde{b}|_\infty \leq K / t$,
and reduces each component of $b$ by at most $K/t$:
$$
\inf_{\tilde{b} \in \ell_2}
\{ |b - \tilde{b}|_2 + t |\tilde{b}|_\infty \}^2
\geq
\sum_{n} \max\{ 0, |b_n| - K/t\}^2
.
$$
This sum goes to $|b|_2^2$ as $t \to \infty$, implying the inequality with $C = 1$.

Edited.
Clearly the reverse inequality is always true with $C = 1$. Here is why:
$$\inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \}
\le \|b - 0\|_B + t \|0\|_A = \| b\|_B
$$
and hence
$$
\sup_{t > 0} \inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \}
\le \|b\|_B
$$
for all $b \in B$. Therefore the inequality in your question can only be correct if $C \ge 1$.

Edited again.
An incorreect attempt to prove the inequality has been withdrawn.


Uniqueness of Solution to a Boundary Value Problem Question

Let $f:\mathbb R_+ \to \mathbb R_+$ be a function twice continuously differentiable (with derivative $f'$ and second derivative $f''$), and $a$, and $b$ be parameters in $\mathbb R_+$. Consider the system

\begin{align}
\dfrac{dx(t)}{dt}=&\; f\left(x(t)\right)-y(t), \\[2ex]
\dfrac{dy(t)}{dt}=&\; ay(t)\left(f'(x(t))-b\right),
\end{align}

with $x(t)\ge 0$ and $y(t)\ge 0$ for all $t$, and  boundary conditions

\begin{equation}
x(0)= x_0, \qquad\text{and}\qquad\lim_{t\to\infty}e^{-bt}x(t)y(t)^{-a}=0.
\end{equation}

By choosing $f$ appropriately it is possible to show that this system, without the initial condition, can have multiple stationary points. I would like to show, that even if that is the case, the following conjecture it true:

  Conjecture $\;$ Given an $x_0$, there is a unique solution to the system that converges to one of the stationary points.

Why I think the conjecture is true

If $f'(0)>0$ and $f''(x)<0$ for all $x\in\mathbb R_+$, then there exists a unique stationary point and there is a simple proof of uniqueness of the solution given an arbitrary $x_0$ that involves drawing a phase diagram in the space $(x,y)$ and showing that there is a unique saddle path that the solution must be at all times (otherwise the second boundary condition would be violated) and that converges to the stationary point.

The reason why I think the conjecture is true is because, given an $f$ in $\mathcal C^2$, a similar phase diagram can be drawn. Here is a sketch of an example of a phase diagram with multiple stationary points:

Notice that the arrows flip between regions in a way that, given an $x_0$, there is only one saddle path (the lines in blue) that the solution could follow. My guess is that a proof for the conjecture would involve showing this is always the case.

Background

If $x$ denotes the capital stock, $y$ the consumption level, $f$ is the production function, $a$ the inverse of the intertemporal elasticity of substitution, and $b$ is the discount rate, then this system is describes the equilibrium allocation of a simple economic growth model.

Multiplying of the system equations leads to the equation in the form of
$$\left(f(x(t))-y(t)\right)\dfrac{dy}{dt} = ay(t)\left(f'(x(t))-b\right)\dfrac{dx(t)}{dt}.\tag1$$
Presentation of the function $y(t)$ as the superposition $y(x(t)),$
$$\dfrac{dy}{dx} = \dfrac{y'_t}{x'_t}$$
 allows to eliminate variable $t$ from $(1):$
$$(f(x)-y)\dfrac{dy}{dx} = ay\dfrac{d}{dx}(f(x)-bx).\tag2$$
Then
$$(y-f(x))\dfrac{d}{dx}(y-f(x))+(y-f(x))f'(x)+ay(f'(x)-b)=0,$$
$$(y-f(x))\dfrac{d}{dx}(y-f(x))+((a+1)f'(x)-ab)y -f(x)f'(x)=0,$$
$$(y-f(x))\dfrac{d}{dx}(y-f(x))+((a+1)f'(x)-ab)(y-f(x))+a(f'(x)-b)f(x)=0.\tag3$$
Equation $(3)$ allows the substitution
$$z(x)= y-f(x)\tag4,$$
$$\dfrac{dx}{dt}= -z(x(t)),$$
which changes the issue system to the form of
\begin{cases}
\dfrac{dt}{dx}= -\dfrac1z\\[4pt]
z\dfrac{dz}{dx}+((a+1)f'(x)-ab)z+a(f'(x)-b)f(x)=0,
\end{cases}
\begin{cases}
t= -\int\limits_{x_0}^x\dfrac{\,\mathrm d\xi}{z(\xi)}\\[4pt]
\left(\dfrac{dz}{dx}+((a+1)f'(x)-ab)\right)z+a(f'(x)-b)f(x)=0.\tag5
\end{cases}
Easy to see that the variable $t$ has the gaps in the points $x,$ where $z(x)= 0,$ or $y(x)=f(x).$

On the other hand, such points presents as the stationary points by the coordinate $t.$

Let us consider behavior of the equation $(5.2)$ at the simple case
$$f(x)= cx+d,\quad s=h(cx+d).\tag6$$
Then
$$zz'+(a(c-b)+c)z+a(c-b)(cx+d)=0,$$
$$\dfrac{dz}{dx}=\dfrac{dz}{ds}\dfrac{ds}{dx}= ch\dfrac{dz}{ds},$$
or
$$z\dfrac{dz}{ds}+gz+s=0,\tag7$$
where
$$\quad h = \sqrt{a\left(1-\dfrac bc\right)},\quad a(c-b)=ch^2,\quad  g=h+\dfrac1h.\tag8$$
Equation $(7)$ has the common solution
$$\begin{cases}
\ln|z^2+gsz+s^2|-\dfrac{2g}{\sqrt{4-g^2}}\arctan
\dfrac{2z+gs}{s\sqrt{4-g^2}}=c,\text{ if }|g|<2,\\
(s\pm z)\exp\dfrac{s}{s\pm z}=C,\text{ if }g=\pm2,\\
C_1|z-gs|^\alpha+C_2|z-gs|^\beta,\text{ if }|g|>2,\tag9
\end{cases}$$
wherein
$$|C_1|+|C_2|>0,\quad \alpha,\ \beta = -\dfrac12\pm\dfrac12\sqrt{g^2-4}.\tag{10}$$
Exact solution $(9)-(10)$ has two branches. The reason is that the equation $(5.2)$ can be written in the form of
$$\dfrac12\dfrac{dz^2}{dx}+((a+1)f'(x)-ab)\sqrt{z^2}+a(f'(x)-b)f(x)=0,$$
which allows existance of multiple solutions.


Using the Levi-Civita alternating tensor and suffix notation to concisely write the vector product rule. I am reading through a section on vector calculus in an electromagnetism book and it has started to use suffix notation and the Levi-Civita alternating tensor in order to prove some identities.  Some of the identities I am familiar with and others I am not.  The notation is new to me as are the concept of tensors and I am struggling to apply both to do things which I can already do.

As an example it is stated in the book that it is much simpler and more concise to write the product rule as

$$\left(\mathbf{A}\times\mathbf{B}\right)_i=\epsilon_{ijk}A_jB_k$$

and I know to work out the product rule I can use the determinant formula. What I am having trouble formulating in my head is how to actually read the above definition and get the correct expansion for the vector-product.  So to make sure I understood the notation I expanded it out on paper but ended up getting the wrong answer.  My interpretations and assumptions are as below:

$$\left(\mathbf{A}\times\mathbf{B}\right)_i = \epsilon_{ijk}A_jB_k$$
$$ = \epsilon_{ijk}A_yB_z\mathbf{i} + \epsilon_{jki}A_zB_x\mathbf{j} + \epsilon_{kij}A_xB_y\mathbf{k} + \epsilon_{jik}A_yB_z\mathbf{j} + \epsilon_{kji}A_zB_x\mathbf{k} + \epsilon_{ikj}A_xB_y\mathbf{i} $$

From here it can be seen that the first three terms are correct, but the last three are not.  I don't understand how the notation links back to the correct unit vector.  Here I have just take the first subscript letter in $\epsilon_{ijk}$ to also represent the relevant unit vector.  So I could really from getting the first three terms correct using the notation correctly write out the last three terms but that is just because I know what it should be.  What I do not understand is how the subscripts of the tensor link up to the subscripts of the two vectors.  I hope that is clear, if it isn't please leave a comment and I will try to remove anything confusing.

Actually, when you write $(A \times B)_i$, this is no longer a vector, but just one of its components. The vector itself is $$(A \times B) = ((A \times B)_1, (A \times B)_2, (A \times B)_3)$$
Having this in mind, the expression $$(A \times B)_i = \epsilon_{ijk}A_j B_k$$
means three equations, for $i$ ranges from $1$ to three. The equations are: $$(A \times B)_1 = \epsilon_{1jk}A_j B_k \\ (A \times B)_2 = \epsilon_{2jk}A_j B_k \\ (A \times B)_3 = \epsilon_{3jk}A_j B_k $$
For a better understanding of it, let's write everything for $(A \times B)_1$, say. I'll begin with the sum in $j$. So: $$\begin{align} (A \times B)_1 &= \epsilon_{\color {red}{11}k}A_1B_k + \epsilon_{12k}A_2B_k + \epsilon_{13k}A_3B_k  \\ &= \epsilon_{12k} A_2B_k + \epsilon_{13k} A_3 B_k
\end{align}$$
Notice that in the indices, repeats are zeros. Now, let's do the sum on $k$: $$\begin{align} (A \times B)_1 &= \epsilon_{\color{red}{1}2\color{red}{1}}A_2 B_1 + \epsilon_{1\color {red}{22}}A_2B_2 + \epsilon_{123}A_2B_3 + \epsilon_{\color{red}{1}3\color{red}{1}}A_3B_1 + \epsilon_{132}A_3B_2 + \epsilon_{1\color{red}{33}}A_3B_3 \\ &= \epsilon_{123}A_2B_3 + \epsilon_{132}A_3B_2 \\ &= A_2B_3 - A_3B_2 \end{align}$$

You can do the same for the other ones, to get used to it. Also, I reccomend getting a few basic properties of the cross product, and prove them using only this notation. I once asked a question related to it, you might find it helpful.

Lost 1 is correct what you wrote is totally absurd....it is utter nonsense.
Recall :

$$\epsilon_{ijk} = -\epsilon_{ikj} = -\epsilon_{jki} = -\epsilon_{kji}.$$

We realize that product vanishes, i.e. it becomes zero....which I must say it's wrong for we don't have enough info to justify that.

So we rewrite the product as follows.

$$C_x = (\vec{A} \times \vec{B}) = \epsilon_{ijk} A_j B_k = \epsilon_{xyz} A_y B_z + \epsilon_{xzy} A_z B_y$$
I have ignored the other expressions since they lead to zero, i.e.

$$\epsilon_{yzx} = \epsilon_{yxz} = \epsilon_{zyx} =\epsilon_{zxy} = 0,$$

since  in our expression we have $C_x$ and for the other components:

$$C_y = \epsilon_{yzx} - \epsilon_{yxz} \text{ and } C_z = \epsilon_{zxy} - \epsilon_{zyx}.$$


Understanding relationship between row space and null space I was given the following question:

If the null space of a $7\times 5$ matrix $A$ is 4-dimensional, what is the dimension of the row space of $A$?

I know the answer to this question is $1$ - which is the number of columns minus the dimensions of the null space. Does this mean that the dimensions of the null space plus the dimensions of the row space equal the number of columns of the matrix? What is the row space?

Does this mean that the dimensions of the null space plus the dimensions of the row space equal the number of columns of the matrix?

No. This sum is equal the dimension of the domain of the linear transformation given by the matrix $A$. Since the domain is $\mathbb{R}^5$, it is equal tot 5.

What is the row space?

It is the vector subspace in the domain of a linear transformation that is spanned by the row vectors of the matrix $A$.


Picard Theorem for Locally Lipschitz functions In order to apply the local Picard theorem in this question, one of the conditions I must satisfy is that $|f|\leq M$.

Which part of the question can this be inferred from?

According to the Hint, you need to show that there exists $h > 0$, such that the problem
$$
x' = f(x), \quad x(b) = p_0 \qquad \qquad (1)
$$
has a unique solution on the interval $(b-h, b+h)$.

Since $p_0 \in W$ and $W$ is open, there exists sufficiently small neighbourhood $U$ of the point $p_0 = x(b)$, such that $U \subset W$.

Since $f$ is continuously differentiable on $W$ and $U$ is a "strict" subset of $W$, there exists $M > 0$ such that $|f(x)| < M$ for all $x \in U$. Moreover, $f$ is, of course, Lipschitz in $U$.

Let us denote $D := \mathbb{R} \times U$. Then we can directly apply this theorem (all assumptions will be satisfied) and obtain some $h > 0$, such that (1) possesses a unique solution $x(t)$ on the interval $(b-h, b+h)$. This is the desired result.

That is why you don't need the boundedness of $f$ in the whole $W$ (and, in general, you surely don't have it).


Compute specific lim without l'hopital rule how can i compute this limit without L'hopital rule:

$\lim \limits _{z\to 0} \frac{1-cos z}{z^2}$

i believe the answer is 1/2 by the rule,

thanks

Remember that

$$1-\cos{z}=2\sin^2{z\over 2}$$

And so the limit we are looking for is

$$\lim_{z\to 0}{1\over 2}{\sin^2{z\over 2}\over \left({z\over 2}\right)^2}=\lim_{z\to 0}{1\over 2}\left({\sin{z\over 2}\over {z\over 2}}\right)^2$$

Now ${\sin{X}\over X}\to 1$ and so the limit is indeed $1/2$

Very easy, just use the fact that near $0$, the numerator $1 -\cos z$ is asymptotically equivalent to $\large{{1\over2}z^2}$ so we have:

$\Large{1 -\cos z \over z^2}$ $\sim $ $\Large{{1\over2}z^2 \over z^2}$ $=$ $\Large{1\over2}$ as $x \to 0$


Verify that T is a linear transformation Define
  $$ T\pmatrix  { \begin {bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} }=\begin {bmatrix} x_1+x_3 \\ x_2+x_3 \end{bmatrix}$$
  Verify that T is a linear transformation

I know that this is not the best format, and I understand if you are frustrated, but if you can resist, please don't post a comment just telling me that I should write my questions better.

I have just a few questions: Since there are three inputs, shouldn't there be three outputs as well? Is there a zero in there somewhere? Like, should I read it like this: T(x1, x2, x3) = (x1 + x2, 0, x2 + x3), where the zero is in the middle? How is it supposed to be read?

A function is a linear transformation if

$T(\mathbf x + \mathbf y) = T(\mathbf x) + T(\mathbf y)$ and $T(a\mathbf x) = aT(\mathbf x)$

Does this satisfy those two rules?

As for your other question, why can't a function have a different codomain from its domain?

Isn't $f(x,y) = x+y$ a valid function (and a linear map).

Going the other way, $f(t) = (t, 2t, 3t)$ is also a linear map.

$$T (1,2,3)=(4,5)$$

the image of a triplet is  a couple.


Problem based on proving equal triangles in area In the figure, $ABCD$ is a parallelogram. If $O$ be any point on $BD$ then prove that $$\triangle OAB=\triangle OAD+\triangle OAC$$

My Attempt
$$\triangle ADB=\triangle BDC$$
$$\triangle OAM=\triangle OMC$$

Now, please help me to move further..

Hint:

You have proved that $$\triangle OAM=\triangle OMC$$

Now note that $\overline{DO}+\overline{OM}=\overline {MB}$ so that :
$$
\triangle OAD+\triangle OAM=\triangle AMB
$$
because the triangles have the same height, and
$$
\triangle AOB=\triangle AOM+\triangle AMD
$$

$$
\triangle OAB=\triangle OAM+ \triangle MAB= \triangle OAM+\triangle OAM+ \triangle OAD=$$ $$= \triangle OAM+\triangle OMC+\triangle OAD=\triangle OAC+\triangle OAD
$$

Completing $AOC$ to a parallelogram centered at $M$, the diagram becomes a proof by dissection of the stated formula.

The line segments in the completed diagram divide $OAB$ into $3$ pieces that can be assembled into the other two triangles.

For the formula to hold for all positions of $O$ the areas should be taken with $\pm$ signs according to the orientation of the triangles.


Let $f:(0,1)→R$ be continuous. Pick out the statements which imply that $f$ is uniformly continuous. Let $f :(0, 1)→ R$ be continuous. Pick out the statements which imply
that $f$ is uniformly continuous.

a. $|f(x) − f(y)| ≤ \sqrt{|x − y|}, \text{ for all }x, y \in [0, 1].$

b.  $f\left(\frac{1}{n}\right)\rightarrow \frac{1}{2}$ and $f\left(\frac{1}{n^2}\right)\rightarrow \frac{1}{4}.$

Notice this

  $$ |f(x) − f(y)| ≤ \sqrt{|x − y|} < \epsilon \implies |x-y|< \epsilon^2 = \delta.  $$

For option $(b)$ does not  implies that uniform convergence.
Take $a_n=\frac{1}{n}$ and $b_n=\frac{1}{n^2}$.Here $a_n-b_n \rightarrow 0$ but $f(a_n)-f(b_n)\nrightarrow 0$ . Hence (b) does not implies uniform continuity.


Evaluating the complex integral $\int_{|z|=2}^{}\frac{1}{z^2+1}dz$ I tried finding the integral of $\int_{|z|=2}^{}\frac{1}{z^2+1}dz$ but not sure whether it is correct.
$\gamma(t)=2e^{it},t\in[0,2\pi]$

$$\int_{|z|=2}^{}\frac{1}{z^2+1}dz=\int_{0}^{2\pi}\frac{2ie^{it}}{4e^{2it}+1}dt$$

I cannot seem to take it forward from here. Is it possible to use Cauchy's integral formula to get an answer? Thanks

You can use the residue theorem to solve the problem:
$$I:=\int_{\left| z \right| = 2} {\frac{1}{{{z^2} + 1}}dz}  = 2\pi \left[ {\operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z = i} \right) + \operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z =  - i} \right)} \right].$$
We have
$$\operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z = i} \right) = \mathop {\lim }\limits_{z \to i} \left( {z - i} \right)\frac{1}{{{z^2} + 1}} = \mathop {\lim }\limits_{z \to i} \frac{1}{{z + i}} = \frac{1}{{2i}},$$
$$\operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z =  - i} \right) = \mathop {\lim }\limits_{z \to  - i} \left( {z + i} \right)\frac{1}{{{z^2} + 1}} = \mathop {\lim }\limits_{z \to  - i} \frac{1}{{z - i}} =  - \frac{1}{{2i}}.$$
So $I=0$.


Gauss Jordan elimination reduces to row-echelon form always? I am reading this text:

and I'm wondering if gauss-jordan elimination always leads to an identity matrix on the left? If so, that helps me understand this passage:

I'm trying to figure out why [A 0] can be rewritten as [I 0]. Why is this?

I'm not sure what kind of terminology is familiar to you, but one way to see this is by talking about the pivot positions (wikipedia link).  If $A$ is an invertible square matrix, then $A$ has a pivot position in every row.  This is because there is a unique solution to $Ax = b$ for every $b$.  But if $A$ is a square matrix with a pivot in every row, that means $A$ is row-equivalent to the identity matrix $I$.  That means $[A \;0]$ can be rewritten as $[I\; 0]$ using row operations.

Gauss-Jordan eliminition works if and only if an inverse exists. It doesn't work for the null element matrix (matrix of zeros).

They said that [A 0] can be rewritten as [I 0] using elementary row operations so there are operations so that works.