Is there an alternative encoding scheme to binary where similarity of pattern correlates with size of number? If I compare binary for 7

111

and binary for 8

1000

there is no correlation between these two patterns that suggests that the numbers are sequentially close to each other.

Is there a recognized, efficient encoding scheme using 1s and 0s for base ten numbers in which the more similar two patterns are, the closer the base ten numbers are?

Of course you could just have

1 -> 1
2 -> 11
3 -> 111
4 -> 1111

etc, but there must be more compact ways to do it!

Yeah, this is what I'm asking. So essentially Gray code is what I'm envisioning, with 0's substituting decimals which are only applied after a given number to separate from the next. I don't see why it is necessary to compact if your goal is not for speed, but rather for truth. I mean couldn't we use a dual system where binary code is used for communication purposes and "gray" code is used for information(facts). I don't agree that we live in an Information Age, or atleast a logical Information Age...perhaps we live in an emotional Information Age as evidenced by blogs, news, and other mediums from which all we are doing is translating our thoughts from human language to binary to back to human language. I'm a hard believer in facts and principals (principles?) and would be nice to segregate both. I believe philosophically that the universe is infinite and that as we move forward to quantifying our problems we should realize that everything is a division of that whole infinity and so instead of taking a scientific bottom-up approach, we should take a top-bottom approach. Creating an established system that records only fact will allow us to then quantify quicker or atleast verify or recognize falsehoods. One interesting thing that you should note is that as my philosophy repeatedly calls on the truth, it acknowledges that falsehoods are not the opposite of truths, but rather the result of truths overlapping. Picture a binagram, Venn diagram with two truths overlapping making the middle a falsehood.

I think that by putting this system in place, we will be solving entscheidungsproblem in terms of effectively calculating questions with a universal language. Call it the mega algorithm.

I've read Freeman Dysons: Scientist as a Rebel
and he was ragging on reductionism and those who followed it like Einstein later in life and Hilbert also later in life, but i think that there is some real meat to what they were doing trying to find a way to theoretically find all the answers to the universe, like Einstein's grand unified theory.


Yet another difficult integration question For $r \in (0,1)$ and $k \in \mathbb Z^+$, prove
$$ \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(u)\right) \ln \left( 1 + r \cos(3^k u)\right) du = \left(\ln\left(\frac{2(1-\sqrt{1-r^2})}{r^2}\right)\right)^2.$$
Given that for any $n \in \mathbb Z^+$
$$ - \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(nu)\right) du = \ln\left(\frac{2(1-\sqrt{1-r^2})}{r^2}\right).$$
In my mind, the strategy should be to use the fact that $\cos(u)$ and $\cos(3^ku)$ are orthogonal. Write $[0,\pi]$ as a product space where $\cos(u)$ and $\cos(3^ku)$ are constant along the (orthogonal) fibres. Then use Fubini's theorem. But I am having difficulty in putting the pieces together. Thanks for your help.

For a positive integer $s$, and $r\in(0,1)$, let $I_s(r)$ be defined by
$$
I_s(r)=\frac{1}{\pi}\int_0^\pi\ln(1+r\cos(t))\ln(1+r\cos(s t))dt.
$$
We will prove the following result:

  Proposition. For every positive integer $s$, and every $r\in(0,1)$ we have
  $$
I_s(r)=\left(\ln\left(\frac{2(1-\sqrt{1-r^2})}{r^2}\right)\right)^2
+\frac{2}{s}{\rm Li}_2\left(\left(\frac{\sqrt{1-r^2}-1}{r}\right)^{s+1}\right)
$$
  where ${\rm Li}_2$ is the well-known Dilogarithm.

This result shows that the proposed formula in the question is wrong.

${\it Proof.}$ In what follows $z=\frac{1-\sqrt{1-r^2}}{r}$. $z$ is the root that belongs to $(0,1)$ of the equation $z^2-\frac{2}{r}z+1=0$.

Now
$$\eqalign{
\ln(1+z^2+2z \cos t)&=\ln(|1+z e^{it}|^2)=2{\rm Re}\,{\rm Log}(1+ze^{it})\cr
&=\sum_{n=1}^\infty\frac{2(-1)^{n-1}z^n}{n}\cos(nt)
}
$$
using the fact that $\frac{2z}{1+z^2}=r$ we conclude that
$$
\ln(1+r \cos t)=-\ln\left(\frac{2z}{r}\right)+\sum_{n=1}^\infty\frac{2(-1)^{n-1}z^n}{n}\cos(nt)\tag{1}
$$
Replacing $t$ by $st$ we get also
$$
\ln(1+r \cos (st))=-\ln\left(\frac{2z}{r}\right)+\sum_{n=1}^\infty\frac{2(-1)^{n-1}z^n}{n}\cos(sn t)\tag{1}
$$
Using Parseval's formula we conclude that
$$
\frac{1}{2\pi}\int_{-\pi}^{\pi}
\ln(1+r \cos (t))\ln(1+r \cos (st))dt =\left(\ln\left(\frac{2z}{r}\right)\right)^2
+\frac{2}{s}\sum_{n=1}^\infty
\frac{(-z)^{(s+1) n}}{n^2}
$$
Or, equivalently
$$
I_s(r)=\left(\ln\left(\frac{2z}{r}\right)\right)^2
+\frac{2}{s}\,{\rm Li}_2((-z)^{s+1}).
$$
which is the announced result.$\qquad\square$

Finally, note that the proposed integral corresponds to $s=3^k$.

Lemma 1: Let $n \in \mathbb{Z}\backslash\{0\}$ and let $f$ be a function such
  that $f(x+T)=f(x)$ for all $x$. Then $$
     \int_0^T f(nx) \mathrm{d}x =  \int_0^T f(x) \mathrm{d}x  $$

${}$

  Lemma 2: Let $f$ be any function such that $\int_0^{\pi} f(x)\,\mathrm{d}x  $ converges, then $$\int_0^{\pi} f(\cos x)\,\mathrm{d}x=\frac{1}{2}\int_0^{2\pi} f(\cos x)\,\mathrm{d}x$$

Using these two results it follows immediately that

\begin{align*}
I_n(r)
& = - \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(nu)\right) du \\
& = - \frac{1}{2\pi} \int_{0}^{2\pi} \ln\left(1 + r \cos(nu)\right) du \\
& = - \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos u\right) du \\
& = I_1(r) = I(r)
\end{align*}

The next step is to differentiate under the integral sign. I will come back to the conditions where this holds, so

\begin{align*}
I'(r)
=
- \frac{1}{2\pi} \int_{0}^{2\pi} \frac{\partial }{\partial r} \ln\left(1 + r \cos u\right)\,\mathrm{d}u
= \frac{-1}{2\pi} \int_{0}^{2\pi} \frac{\cos u}{1 + r \cos u}\,\mathrm{d}u
=  \frac{1}{r} - \frac{1}{r\sqrt{1-r^2}}\end{align*}

Also note that $I(0) = 0$, so integrating the above expression from $0$ to $r$ gives

\begin{align*}
\int_0^r I(\tau) \mathrm{d}\tau & = \int_0^r \frac{1}{\tau} - \frac{1}{\tau\sqrt{1-\tau^2}} \mathrm{d}\tau \\
I(r) - I(0) & = \log \left( \frac{2}{r} \right) - \operatorname{arctanh}\left( \frac{1}{\sqrt{1-r^2}} \right)
\end{align*}

This is valid since we assume that $|r|< 1$. Hence

$$
I_n(r)
 = - \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(nu)\right) du
 = \log \left( \frac{2}{r} \right) - \operatorname{arctanh}\left( \frac{1}{\sqrt{1-r^2}} \right)
$$

The proof of the second lemma is trivial, and the evaluation of the trigonometric integral can be done something along the lines of
$$
\frac{1}{r}\int_0^{2\pi} \frac{r\cos x}{1 + r \cos x} \mathrm{d}x
= \frac{1}{r}\int_0^{2\pi} 1 - \frac{\mathrm{d}x}{1 + r \cos x}
= \frac{2\pi}{r} - \int_{-\infty}^\infty \frac{2\mathrm{d}u}{(1-r)u^2 + 1+r}
$$
Then use something like $u \mapsto u\frac{1+r}{\sqrt{1-r}}$ to complete the work.
The proof of the first integral is also elementary note
\begin{align}
          \int_0^{T} f(kx) \mathrm{d}x
 = \frac{1}{k}\int_{0}^{kT} f(x)\mathrm{d}x
 = \frac{1}{k}\sum_{n=0}^{k-1} \int_{nT}^{(n+1)T} f(x)\mathrm{d}x
 = \frac{1}{k}\sum_{n=0}^{k-1} \int_{0}^{T} f(u + nT)\mathrm{d}u
\end{align}
This completes the proof since, $f$ is has a period of $T$ then $f(u+nT)=f(u)$.
Eg intuitively we have $k$ integrals $\int_0^T + \int_T^{2T} + \cdots + \int_{(k-1)T}^{kT}$ equally big.


Fourier Transform of the Weierstrass Function I'm looking to compute the Fourier Transform of the Weierstrass Function over the interval $[0,2\pi]$:
$$W(t) = \sum_{n=1}^\infty a^n \cos(b^n t) $$
where $0<a<1$, $b$ is a positive integer such that and $ab > 1+\frac{3\pi}{2}$.
In computing the integral
$$\int_0^{2\pi} \sum_{n=1}^{\infty} a^n\cos(b^nt)e^{2\pi i st}dt$$
I've tried writing $\cos(b^nt) = \frac{e^{ib^nt }+e^{-ib^nt}}{2}$. I use geometric series to get rid of the sum, but then I get the integral
$$\int_0^{2\pi} \frac{e^{2\pi i s t}}{2}e^{ibt} \left(\frac{1}{1/a-e^{ibt}} + \frac{1}{1/a-e^{-ibt}} \right)dt $$
which I don't know how to solve, even when I put it under a common demoninators and right the bracketed expression in terms of $\cos(bt)$.
I've also tried interchanging the sum and the integral. However, I get an unwiedly sum.
Is there a clever way of going about this or am I missing something in one of these methods.

For $b$ an integer $\ne -1,0,1$ and $|a|<1$ then $$W(t)=\sum_{n=1}^\infty a^n \cos(b^n t)$$ is already a $2\pi$-periodic Fourier series and $$c_k=\frac1{2\pi}\int_0^{2\pi} W(t)e^{-ikt}dt = \cases{\frac12 a^{n} \ if \ |k| = |b|^n\\ 0 \ otherwise}$$
$$W(t)=\sum_{k=-\infty}^\infty c_k e^{ikt}$$


Getting wrong answer trying to evaluate $\int \frac {\sin(2x)dx}{(1+\cos(2x))^2}$ I'm trying to evaluate $$\int \frac {\sin(2x)dx}{(1+\cos(2x))^2} = I$$

Here's what I've got:

$$t=1+\cos(2x)$$
$$dt=-2\sin(2x)dx \implies dx = \frac {dt}{-2}$$
$$I = \int \frac 1{t^2} \cdot \frac {-1}2 dt = -\frac 12 \int t^{-2}dt = -\frac 12 \frac {t^{-2+1}}{-2+1}+C$$
$$=-\frac 12 \frac {t^{-1}}{-1}+C = \frac 1{2t} +C = \frac 1{2(1+\cos(2x)}+C$$

I need to find what went wrong with my solution. I know the answer is correct but the line $dt=-2\sin(2x)dx$ concludes that $dx=\frac {dt}{-2}$ is false.  But why?  What should it be instead?


What are some mathematical problems which have been forgotten? As mathematicians continue to study mathematics, often times they run into a problem which takes a considerable amount of effort to solve. For instance, trying to factor polynomials has lead to a large portion of group theory and Galois theory especially. Another example, which I have heard reference from one of my professors, is that the study of elliptic functions has lead to a large amount of Algebraic Geometry. The original goal being how to integrate elliptic curves turned into quite an interesting field over the years.

My question, although "broad", is this:

What are some mathematical problems that have been forgotten?

In the process of trying to solve one question, often times many more questions are discovered. I find it hard to believe that no question has been forgotten. As in, sometimes the cutting-edge work of mathematics has trumped the questions that seemed of lesser importance at the time of research, and after a while most everyone has forgotten about them. Surely some of us must remember! Or, are there any stories of such problems like this that have sparked new mathematical research over the years?

The n-body problem dates back to the ancient Greeks and was once considered the key to understanding the movement of the planets, and, by extension, the very nature of the Universe.

A "geometrical" (i.e. exact) solution was always hoped for and expected, but the desire to compute orbital motions in practice led to the development of numerical methods; in the meanwhile, it was discovered that the case $n>2$ is chaotic (an early motivation for chaos theory).

Both numerical analysis and chaos theory have since dwarfed the original problem in terms of the amount of research effort dedicated to them, and while it's difficult to say that the n-body problem is "forgotten" (if it were, we couldn't be discussing it) I'm not aware of any substantial theoretical work (let alone progress) done on it for several decades now beyond what is already covered by the more general field of numerical methods.

Add to this the fact that the relativistic version of the n-body problem is fundamentally different from the classical formulation, leading to a solution for the classical problem being of much lesser interest in today's context of large-scale celestial mechanics than it would have been two centuries ago, and you have this once monumental problem being now reduced almost to its historical importance alone.

I suggest that you look up "unsolved" problems in the atlas of unsolved problems posted online. Another way to find "them" is to find your area first, then look up top mathematicians pages and in there you can find some post about problems they are still having trouble resolve.


Fredholm Index: Finite Corank $\Rightarrow$Closed Range Obviously closed subspaces turn quotient spaces into normed spaces rather than just merely vector spaces. However the dimension involved in Freholm's index are purely algebraic.

Why do we thus require the range to be closed?
Or is a subspace with finite codimension necessarily closed?

Moreover, what could happen if the range wouldn't be closed but would be of finite codimension though?

I'm thinking of some example like: $l^p_0\subsetneq l^p,1\leq p\leq\infty$

A bounded operator with finite cokernel has closed range (see Abramovich and Aliprantis, An Invitation to Operator Theory, Corollary 2.17).

You need the range to be closed in order to make sense of the cokernel. If $T:X \to Y$ then
$$ \text{coker}(T) = Y / \text{range}(T).$$
In order for  $Y / \text{range}(T)$ to be a Banach space you require $\text{range}(T)$ to be closed.


Why is $1^{\infty}$ considered to be an indeterminate form From Wikipedia: In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form.

The indeterminate forms include $0^{0},\frac{0}{0},(\infty - \infty),1^{\infty}, \ \text{etc}\cdots$

My question is can anyone give me a nice explanation of why $1^{\infty}$ is considered to be an indeterminate form? Because, i don't see any justification of this fact. I am still perplexed.

Forms are indeterminate because, depending on the specific expressions involved, they can evaluate to different quantities.  For example, all of the following limits are of the form $1^{\infty}$, yet they all evaluate to different numbers.

$$\lim_{n \to \infty} \left(1 + \frac{1}{n^2}\right)^n = 1$$

$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$$

$$\lim_{n \to \infty} \left(1 + \frac{1}{\ln n}\right)^n = \infty$$

To expand on this some (and this thought process can be applied to other indeterminate forms, too), one way to think about it is that there's a race going on between the expression that's trying to go to 1 and the expression that's trying to go to $\infty$.  If the expression that's going to 1 is in some sense faster, then the limit will evaluate to 1.  If the expression that's going to $\infty$ is in some sense faster, then the limit will evaluate to $\infty$.  If the two expressions are headed toward their respective values at essentially the same rate, then the two effects sort of cancel each other out and you get something strictly between 1 and $\infty$.

There are some other cases, too, like
$$\lim_{n \to \infty} \left(1 - \frac{1}{\ln n}\right)^n = 0,$$
but this still has the expression going to $\infty$ "winning."  Since $1 - \frac{1}{\ln n}$ is less than 1 (once $n > 1$), the exponentiation forces the limit to 0 rather than $\infty$.

$$
\lim_{n\to\infty} \left( 1 + \frac a n\right)^n = e^a.
$$

This limit depends on $a$. In other words, if the base approaches $1$ and the exponent approaches $\infty$, that's not enough to tell you what the limit is.


The product of two consecutive even numbers is $3248$, what is the larger number? The product of two consecutive even numbers is $3248$. Actually, I am interested in finding the larger number out of the two.

Say one of the even numbers is $a$ and the consecutive even number is $a+2$, then
$a(a+2) = 3248$, so we can get the larger number by finding the roots of the above quadratic equation.

I was also thinking of another method of applying the product and the sum of the roots!. from which I conclude that one of the roots is negative and the other root is positive.

Any other method of obtaining the larger number without calculating the roots of the quadratic equation?

You can also take two consecutive numbers as $a-1$ and $a+1$ the we have:

$a^2-1=3248$

$a^2=3249$

etc...

let $x$ & $x+2$ be $2$ consecutive even numbers
$$
\begin{align}
(x)(x+2)&=3248 \\
x^2 + 2x -3248 &= 0 \\
x^2 + 58x - 56x - 3248 &= 0 \\
x(x + 58) - 56(x + 58) &= 0 \\
(x - 56)(x + 58) &= 0 \\
\end{align}
$$
$x = 56$ since the value could not be in negative.
$x + 2 = 58$ which is the largest number.


What are the prime ideals of $K[x]/\langle x^n\rangle$? What are the prime ideals of $K[x]/\langle x^n\rangle$, where $K$ is a field?

I have tried it like this: suppose its prime ideal be $P/\langle x^n\rangle $. Then $x^n$ belongs to $P$, which implies that $x$ is in $P$ since $P$ is a prime ideal of $K[x]$. Now i am confused further how to solve. Is $P= K$?

Consider the general case: $K[x]/\langle f(x)\rangle$.

There is a bijection between the ideals in $K[x]/\langle f(x)\rangle$ and the ideals in $K[x]$ that contain $\langle f(x)\rangle$. This correspondence respects prime ideals.

Since $K[x]$ is a PID, the ideals that contain  $\langle f(x)\rangle$ are exactly those of the form $\langle p(x)\rangle$, where $p$ is an irreducible divisor of $f$.

Therefore, the prime ideals $K[x]/\langle x^n\rangle$ correspond to the prime divisors of $x^n$, which is just $x$.

Recall that $Spec A/\alpha =\{ p\in Spec A \mid \alpha \subset p\}$.So we can get $Spec k[x]/<x^n> =Spec k[x]/<x>$.we know $<x>$ is a maximal ideal of $k[x]$,so $Spec k[x]/<x^n>$ contains only one element.


Why is $\frac{\ln\infty}{\infty}$ equal to $\frac\infty\infty$? For me there is no intuitive explanation for this. Yes, i get that if you want to find the natural logarithm of a very high number (infinity) that the ln would be high too. But it does not grow nearly as fast as infinity itself, for example the natural logarithm of $\ln{100} \approx 4.60$ that number is 21 times as small as the number we are taking ln off. But anyhow, here comes the equation making my head jittery.

$$ \lim_{x\to\infty} \frac{\ln{\left(x\right)}}{x}=\frac{\infty}{\infty}$$

How, is this possible. Doesn't $\ln$ infinity grow at a much slower speed than infinity itself?

I know how to do derivatives. I'm asking for an intuitive explanation. Of why we use that specific convention.

In other words, why do we use an indeterminate form, and is there an intuitive explanation for why we might use infinity over infinity, or zero over zero?

Khans example. Showing that $\frac{ln\left(x\right)}{x} = \frac{\infty}{\infty}$

$\dfrac{\infty}{\infty}$ is a symbol that in this context just means the numerator and denominator both "approach infinity" (grow without bound). It's true that the numerator approaches infinity far slower than the denominator, which is why ${\displaystyle \lim_{x\to\infty}}\dfrac{\ln x}{x}=0$.

$\dfrac{\infty}{\infty}$ is called an "indeterminate form" because it doesn't tell you enough information to determine what the limit is.

If you use integration to find the average value of $y$ of the function $y = x^{-1}$ where $x$ is between $1$ and infinity, the result evaluates to $\frac{\ln(\infty)}{\infty - 1}$ and since the highest value of $y$ in this range is $\frac 1 1 = 1$, and the lowest value is $\frac 1 \infty = 0$, the value of $\frac {\ln(\infty)}{\infty - 1}$ lies between $0$ and $1$ (but not including $1$) so $\frac {\ln(\infty)}{\infty}$ is less than $1$.


Why eliminate radicals in the denominator? [rationalizing the denominator] Why do all school algebra texts define simplest form for expressions with radicals to not allow a radical in the denominator. For the classic example, $1/\sqrt{3}$ needs to be "simplified" to $\sqrt{3}/3$.

Is there a mathematical or other reason?

And does the same apply to exponential notation -- are students expected to "simplify" $3^{-1/2}$ to $3^{1/2}/3$ ?

The usual reason I've heard is that dividing by integers is computationally easier -- it's easier to find, say, $(5\sqrt{3})/3$ by computing $5 \times \sqrt{3} \approx 8.66 $ and then dividing by $3$ to get $2.89$ then to find $5/\sqrt{3}$ by dividing $5/1.73$ directly.

The slightly cynical teacher in me wants to say that the reason for demanding no radicals in the denominator is so there is only one right answer to each question, which simplifies grading.

Don't blame your high school teacher

They are right

$\frac{1}{\sqrt{3}}$ is correctly   "simplified" to $\frac {\sqrt{3}}{3}$.

Its difficult to find value $\frac{1}{\sqrt{3}}$ as $\sqrt{3}$ is irrational number

Its easy to find value of $\frac {\sqrt{3}}{3}$ as here we are dividing it with 3


Given the cumulative distribution function find a random variable that has this distribution. We are given a Cumulative distribution function $(CDF)$ how do we find a random variable that has the given distribution? Since there could be a lot of such variables we are to find anyone given that the function is differentiable and increasing.

I read somewhere that we could simply take inverse of the function(if it exists) and that would be the Random variable you are looking for, but I don't understand it, even if it's true(?).

If that's true, why? Else, how do we find a random variable for the distribution?

If you are studying elementary probability theory, allow me to reformulate your question as "how can I represent a random variable $X$ with a given CDF $F_X$ in terms of a uniform random variable $U$ on $(0,1)$?" The answer to that is the quantile function: you define

$$G_X(p)=\inf \{ x : F_X(x) \geq p \}$$
and then define $X$ to be $G_X(U)$.

Note that if $F_X$ is invertible then $G_X=F_X^{-1}$, otherwise this is "the right generalization". One can see this by looking at the discrete case: if $P(X=x)=p$ then $P(G_X(U)=x)=p$. This is because a jump of height $p$ in $F_X$ corresponds to a flat region of length $p$ in $G_X$, and the uniform distribution on $(0,1)$ assigns each interval a probability equal to its length.

The natural question is now "what's a uniform random variable on $(0,1)$?" Well, it has $F_U(x)=\begin{cases} 0 & x<0 \\ x & x \in [0,1] \\ 1 & x>1 \end{cases}$. But otherwise such a thing is a black box from the elementary point of view.

If you are studying measure-theoretic probability theory then the answer is a bit more explicit. A random variable with CDF $F_X$ is given by $G_X : \Omega \to \mathbb{R}$ where $G_X$ is the quantile function as defined before, $\Omega=(0,1)$, $\mathcal{F}$ is the Borel $\sigma$-algebra on $(0,1)$, and $\mathbb{P}$ is the Lebesgue measure. Note that on this space the identity function is a uniform random variable on $(0,1)$, so this is really the same construction as the one described above.

In any case these constructions can be generalized to finitely many random variables by looking at the uniform distribution on $(0,1)^n$ instead of $(0,1)$.

The $\text{cdf}$ of a random variable tells the probability that its value falls below a given bound,

$$\text{cdf}_X(x)=\mathbb P(X<x).$$

For a uniform variable $U$, we have

$$\text{cdf}_U(x)=\mathbb P(U<x)=x$$ for $x\in[0,1]$.

Now if you want to create a random variable with an imposed $\text{cdf}$, let $h(x)$, you want to achieve

$$\mathbb P(X<x)=h(x).$$

If $X$ derives from a uniform variable via some transformation, let $X=g(U)$, you have

$$\mathbb P(X<x)=\mathbb P(g(U)<x)=h(x).$$

But also, assuming $g$ invertible, and using the fact that $U$ is uniform,

$$\mathbb P(g(U)<x)=\mathbb P(U<g^{-1}(x))=g^{-1}(x).$$

Putting these facts together,

$$g^{-1}(x)=h(x)$$ or $$g(x)=h^{-1}(x).$$

For a simple example, take a negative exponential distribution,

$$\text{cdf}_X=h(x)=1-e^{-x}.$$

Then by inversion,

$$g(x)=-\ln(1-x)$$ which maps $0$ to $0$ and $1$ to $\infty$. By looking at the plot of $g$, you will understand that the "density" of the transformed points goes decreasing, because the slope increases, resulting in a bias in favor of the small values.

We can even play the exercise when $U$ is not uniform but has a known $\text{cdf}$, let $f(x)$. Then it suffices to adapt the equations, and

$$\mathbb P(X<x)=\mathbb P(g(U)<x)=h(x)=\mathbb P(g(U)<x)=\mathbb P(U<g^{-1}(x))=f(g^{-1})(x)$$ and we need

$$g(x)=h^{-1}(f(x)).$$


If $e^{f(z)}$ is analytic then $f(z)$ is analytic I want to know how to prove that if $e^{f(z)}$ is analytic in some open disk then $f(z)$ is analytic in that open disk. Can someone guide me through this? thanks!

Let $g(z)=e^{f(z)}$ where $g$ is analytic. Therefore, $$g'(z)=f'(z)e^{f(z)}\implies f'(z)=\frac{g'(z)}{e^{f(z)}}$$ is analytic (because quotient of two analytic function and $e^{f(z)}\neq 0$ for all $z$). Therefore $f'$ is analytic and thus $f$ is analytic.


May seem like a noob question: really, why can't we divide by 0? Yes, I know, can't be answered, blah, blah, blah.... but here are a few of my theories. I know, plenty of other questions like this, but before marking this as a duplicate, consider this, my mathematical friends:

We know that $ x/x = 1 $.
We also know that $ 0/x = 0 $.

Then, considering these to facts, $$ 0/0 $$ could be

$ 1 $, because $ x/x = 1 $
or $ 0 $, because $ 0/x = 0 $

Then, of course we can consider $ x/0 $, where $ x $ can be any real number. Then $ 0/0 $ can be $ 0 $ or $ 1 $, and then the rest according to theory, is $ \infty $ or $ \text {undefined} $. Then how come we say $ \dfrac {x}{0} = \infty $? Or is this only true being $ x \neq 0 $?

Or is it that just for practical sense, we just say that $ x/0 = \infty $? Is it because it is just because it is not computationally possible? Or is it just because $$ \dfrac {5}{\dfrac {1}{2}} = 5(\dfrac {2}{1}) = 10 $$  and then $$ \dfrac {5}{\dfrac {1}{100}} = 500 $$ and then $$ \dfrac {5}{\dfrac {1}{100000}} = 500000 $$ and so the numbers keep on going to $ \infty $ as we get closer to $ 0 $?

I know there are lots of possible answers but then the theory that $ \dfrac {x}{0} = \infty $ even though $ \dfrac {0}{0} $ could be 1 or 0, that just does not make sense to me. I will appreciate any answers / at lease possible answers, because I understand that this is just a very controversial topic of mathematics.

This is the way I convinced myself :

$$any number \times 0 = 0\; \rightarrow  \! \frac{0}{0}=any number$$

I thinks this is why it is unidentified.

The binary operator, division, can be thought of as a statement which asks how many times can "b" go into "a", a/b. If b is zero, I.e. nothing, the operator then asks how many times can nothing go into something?; Well, an infinite number of times. If a and b are zero, then how many times can nothing go into nothing, an infinite number as well, however, this case is less defined then simple division by zero. And, infinity is not a number, it's a concept.


Exponential decay and time constants The time constant τ is the amount of time that
an exponentially decaying quantity takes to decay by a factor of 1/e. Because 1/e
is approximately 0.368, τ is the amount of time that the quantity takes to decay to
approximately 36.8% of its original amount.
(a) How are the time constant τ and the decay rate λ related?

I used the decay model:

$r = r_0e^{\lambda t} $

Where t = time

r = amount of the particular exponentially decaying matter at time t

$-\lambda$ = decay rate

$e^{-\lambda t} = \frac{r}{r_0}$

Then I let $\frac{r}{r_0} = \frac{1}{e}$ and $t = τ$

$e^{-\lambda τ} = \frac{1}{e}$

Simplifying I got, $τ = \frac{1}{\lambda}$

Is this the correct way to show that the time constant is inversely proportional to the decay rate?

(b) Express the time constant in terms of the half-life.

I'm not sure how to do this, do I just set $\frac{r}{r_0}  = 0.5$?

Half life=$ln 2$*$\tau$ and $\tau$=1/$\lambda$

In both cases, you are absolutely correct. You can prove this here


Quantifiers. Translate english statements into Symbolic Form. 1. All natural numbers are Integer.
So I know natural number is N and integer is Z but how do I translate this statement into Symbolic Form ?

$$\forall x\in\mathbb N\space\space\space x\in \mathbb Z$$


I draw a card from a deck, and face it down. I draw another two cards of spades. What is the probability my first card is also a spade? Do I consider the probability before drawing both cards or after?
Question more clearly:
A single card is removed at random from a deck of $52$ cards. From the remainder we draw $2$ cards at random and find that they are both spades. What is the probability that the first card removed was also a spade?

The trick here is to realize that the situation described is equivalent to removing two spades from the deck and then drawing a card at random from what remains. If you can calculate the probability of a spade occurring as the third card in that simpler setting, you're good to go.
Many probability problems can be similarly simplified by realizing that the specific order in which certain things are said to happen is irrelevant.  That's not to say the order in which things happen is never relevant, just that the meanings of the letter strings (aka words) "first," "second," "third," etc., are sometimes interchangeable.

You are calculating the probability of the first card being a spade, therefor the next two additional draws mean nothing. The chance of drawing a spade from a deck of 52 would be 25%. It's as simple as that.
I hope this helps.


My son's Sum of Some is beautiful! But what is the proof or explanation? My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

  Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.
  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.
  If I then skip three ones and read the fourth, that one equals my sum!


Vertical Lines & Horizontal Lines and Their Slopes Hello, in my discrete mathematics for Information Technology, my instructor asked this question of us:
Discuss in your own words if it makes sense that a vertical line has an infinite slope and a horizontal line has a slope of zero.

My answer is:

The slope of a vertical line is undefined.

The slope of a horizontal line is zero.

Think about what slope is and it will make sense. Slope is rise over run and and be found by using 2 points on the line, (x1,y1) and (x2,y2)

slope = (y2 - y1) / (x2 - x1)

On a horizontal line all the y values are the same so y2-y1 will always be zero so the slope will be zero regardless of what x2-x1 equals.

On a vertical line all the x values are the same so x2-x1 will always be zero. Dividing by zero is undefined so the slope is undefined.

My brother has told me this is incorrect, but refuses to explain it. Can someone please confirm this?

It's a good question from your instructor because it prompts you to think about infinity, and by thinking about infinity sufficiently precisely you can argue for both answers.

If this is to make any sense, it needs to be considered as a limiting case. But a vertical line is a limiting case of both a line with slope $+\infty$ and a line with slope $-\infty$. For instance, we can make this precise by considering a line with slope $m$ given by

$$\vec r(s)=\vec r_0+\frac s{\sqrt{1+m^2}}\pmatrix{1\\m}\;,$$

where $s$ is the distance along the line. If we take $m$ to $\pm\infty$, $\vec r(s)$ converges pointwise to $\vec r_0\pm s\vec e_z$, so the locus of the limits is a vertical line. Thus a vertical line can equally well be described by a slope of $+\infty$ or a slope of $-\infty$, so it makes no sense to say that it has slope $+\infty$.

But who said that "infinity" means $+\infty$? Consider the slope represented not on the real line, by the number $m$, but on the unit circle, by the point at which the line through $(0,m)$ and $(1,0)$ meets the unit circle. Now taking $m$ to $+\infty$ or $-\infty$ leads to the same limit point $(1,0)$. Nothing forces us to add two different infinities to the real line, one at each "end". Though this is useful for many purposes, it's not useful when we consider the real line as representing the set of possible slopes. The construction indicates how we can consistently add a single infinity to that set and consider both $m\to+\infty$ and $m\to-\infty$ as the same limit where the slope, represented on the unit circle, goes to the point $(1,0)$. In this sense, it makes sense to say that a vertical line has infinite slope.

If a line had an undefined slope I would not be able to draw it.  I can draw a vertical line.  It must therefore have a defined slope.


Suppose $y(x)$ is continuous and $y'(x)=0$ has uncountable many solutions but $y(x)$ is not constant on any interval. Is this possible? Suppose $y(x)$ is continuous and $y'(x)=0$ has uncountable many solutions but $y(x)$ is not constant on any interval. Is this possible?

Take a Cantor set $C\subseteq [0,1]$ and let $d(x)$ be the distance function from $x$ to $C$.  Then $(d(x))^2$ is differentiable with derivative vanishing at all points of $C$ hence uncountably many.

Yes, it is true. A classic example of such an odd function is the Cantor function. It is defined like this: for $x$ in $[0,1]$

1.Express x in base 3.

2.If x contains a 1, replace every digit after the first 1 by 0.

3.Replace all 2s with 1s.

4.Interpret the result as a binary number. The result is c(x).

This function has derivative $0$ almost everywhere (i.e. the places where it doesn't have $0$ derivative has measure $0$) and yet somehow manages to increase in value from $0$ at $x=0$ to $1$ at $x=1$.

There is plenty written about the Cantor function. Perhaps wikipedia would be the best starting place: https://en.wikipedia.org/wiki/Cantor_function


Symmetrical combinatoral block design Suppose that $(V,\mathcal{B}$) is a symmetric block design of type $2-(v,k,\lambda)$. Pick an arbitrary block $B \in \mathcal{B}$. Let $B_1,...,B_{v-1}$ be a list of the remaining blocks and let $n_1,...,n_{v-1}$ be respectively the number of elements that $B$ has in common with these blocks. Show that $$n_1+...+n_{v-1} = (v-1)\lambda$$

I'm quite confused as to where to start this problem. I know that for a symmetric block design, $b=v$, so we know that the sum is equal to $(b-1)\lambda$, which essentially means that the each intersection has $\lambda$ elements. I'm unsure as to why that's true, or how to go about proving it. Any hints?

So we have $v$ blocks (in the set $V$ with $v$ elements), each having $k$ elements and each element in $V$ apears in $k$ blocks and every pair of elements $\{x,y\}$ appears in exactly $\lambda $ blocks.

So if we make a double counting between the blocks $B_1,...B_{v-1}$ and the elements in $B$ we have $$n_1+...n_{v-1} = (k-1)k$$
since each of $k$ elements in $B$ appears exactly $k$ times (once in $B$ and the rest in $k-1$ sets).

But on the other side each pair $\{x,y\}$ appears in exatly $\lambda $ blocks, so we have again by double counting between the set of all pairs of elements in $V$ and the set $\mathcal{B}$: $${v\choose 2}\cdot \lambda = v\cdot {k\choose 2}\implies (v-1)\cdot \lambda = k(k-1)$$

and thus the conclusion.

I would interpret the left hand side of the equation as counting all pairs $(B_i, p)$ such that $p\in B_i$ and $p\in B$. The left hand side of the equation does this by considering all blocks $B_i$, besides $B$, and counting their intersections with $B$. Another way to count these pairs is to instead consider all points $p\in B$, and then count how many blocks besides $B$ contain $p$. If you do that, and recall the few equalities associated with block designs and symmetric block designs in particular, you should get your answer.


When does :$\sigma(\sigma(2n))=\sigma(\sigma(n))$ and $\sigma(n)$ is sum divisors of the positive integer $n$? Is there someone who can show me When does: $$\sigma(\sigma(2n))=\sigma(\sigma(n))$$ where : $\sigma(n)$ denotes the sum of divisors of the positive integer $n$ ?

Note (1) : I came across this problem when I read some papers  about

"Iterating of the sum divisors of sigma function ".

Note(2) :${\sigma}^{0}(n)=n$ and ${\sigma}^{m}(n)=\sigma({\sigma}^{m-1}(n))$ and $m \geq 1$

Thank you for any help!

This is not a complete solution, but only a partial answer to the question.

Without loss of generality, let $n = {2^k}m$ where $k \geq 0$ and $\gcd(2,m)=1$.

Then we have:
$$\sigma(2n) = \sigma(2^{k+1})\sigma(m) = (2^{k+2} - 1)\sigma(m)$$
$$\sigma(\sigma(2n)) = \sigma((2^{k+2} - 1)\sigma(m))$$

$$\sigma(n) = \sigma(2^k)\sigma(m) = (2^{k+1} - 1)\sigma(m)$$
$$\sigma(\sigma(n)) = \sigma((2^{k+1} - 1)\sigma(m)).$$

Therefore, the equality
$$\sigma(\sigma(2n)) = \sigma(\sigma(n))$$
holds when there exist $k \geq 0$ and an odd $m$ such that
$$\sigma((2^{k+2} - 1)\sigma(m)) = \sigma((2^{k+1} - 1)\sigma(m)).$$

Lastly, using the inequalities
$$a\sigma(b) \leq \sigma(ab) \leq \sigma(a)\sigma(b)$$
we get
$$(2^{k+2} - 1)\sigma(\sigma(m)) \leq \sigma((2^{k+2} - 1)\sigma(m)) = \sigma((2^{k+1} - 1)\sigma(m)) \leq \sigma(\sigma(2^k))\sigma(\sigma(m)).$$

Hence,
$$2^{k+2} - 1 \leq \sigma(\sigma(2^k)).$$
We are therefore sure that
$$\sigma(2^k) = 2^{k+1} - 1$$
is not a (Mersenne) prime.

Additionally, we have the lower bound
$$3 < 4 - \frac{1}{2^k} \leq \frac{\sigma(\sigma(2^k))}{2^k},$$
so that $2^k$ is not an (even) superperfect number.  This agrees with our earlier finding that $2^{k+1} - 1$ is not prime.

$\sigma(2n)=3\times\sigma(n)$

if $m=\sigma(n)$
we should find $\sigma(3 m)=\sigma(m)$

$\sigma(3m)=4\times\sigma(m)$

then  $4\sigma(m)=\sigma(m)$

contradiction so there is no solution.


Area of Semi Circle With an Inscribed Triangle Suppose we have a triangle with a right angle at its height, with side a, 10 inches, side b, unknown, and side c, 24 inches; inscribed in a semi-circle.

Now, I'm asked to find the area of the semi circle only, in terms of $\pi$. I was already given the answer, which is $84.5\pi -120 \ in^2$, but I don't understand how to get this conclusion.

I tried finding the area this way, but I get a pretty different answer:
$A=\frac{\pi r^2}{2}, r=12 \ inches$

$$A=\frac{144\pi}{2}$$

$$A=72\pi \ in^2$$

Then I know I have to subtract the triangle because I only want the area of the shaded region (the semi-circle), so

$$10^2+B^2=24^2$$

$$2\sqrt{119}$$

So I end up getting an answer of $A=72\pi-2\sqrt{119} \ in^2$ which is clearly false. Could anyone explain where I went wrong?

The question is poorly worded - it is not clear which side of the triangle is the longest (the hypotenuse).

You are assuming that the 24 inch side of the triangle is the longest side and is therefore the diameter of the semi circle.

The book answer is assuming that the unknown side of the triangle is the longest side. In this case the length of this side is $\sqrt{10^2+24^2}=26$ inches, the radius of the semicircle is $13$ inches, the area of the triangle is $\frac{10 \times 24}{2} = 120$ square inches and the area of the semicircle outside of the triangle is

$\frac{13^2\pi}{2}-120$ square inches.

ans = 2533.3
first of all finding the area of the triangle which is equal 120.
then finding the hypotenuse which is also the diameter, so having gotten the diameter, i used to get the area of the semicircle which gave me 2653.3, then subtracting the area of the triangle from the area of the circle gives me (2533.3).


How to prove that these two angles are equal? In the above figure, how to prove that $\angle$ $Oab$ and $\angle$ $OAB$ are equal in measure?

Angles $Oab$ and $Oba$ are equal as $\triangle Oab$ is isosceles. That, plus the fact that the sum of the angles in a triangle is $180^\circ$, gives you:
$$\angle Oab=\frac12(180^\circ-\angle aOb)$$
Similarly, $\angle OAB=\frac12(180^\circ-\angle AOB)$. Now just note that $\angle aOb=\angle AOB$.

Oab and OAB are similar triangles . So, the corresponding sides have same ratios, this automatically proves that these two angles are equal.


Explain the theorem regarding convergent sequences (read details) . Theorem is:

Let {$p_n$} be a sequence in a metric space X.
{$p_n$} converges to p $\in$ X if and only if every neighbourhood of p contains all but finitely many of the terms of {$p_n$}.

I understood that every neighbourhood of p contains finite terms of {$p_n$} and also other way.
Please correct me if I am wrong.

But if we take {1/n : n$\in$ N}, where X is R, then there could be some neighbourhood of x=0(which is point of convergence of sequence) and it has infinitely many points.

I am not getting where I am making mistakes, Please find and correct me.

Explain proof of this theorem or give a hint or any intuitive way to think about this theorem.

NO : all but means : only a finite number of $p_i$ are "outside" the neighbourhood of $p$.

This means that we can find an $N$ such that :

  for all $n > N$, all $p_n$ are "inside" the neighbourhood.

N  is all of definite  equations  expandation  N>n이 성립하므로 {p}는 n의 부분적인  원소집합으로  치환가능하기때문에 N은 {p}

Bot translation of Korean text:

  This is because the establishment of the {p} is a set of n elements in a partial substitution is possible because N is {p}


Determine the convergence of a recursive sequence Determine the convergence of the sequence defined by $\sum_{n \in \mathbb{N}}a_n$ such as
$$a_1=1,\quad a_{n+1}=\frac{2+\cos(n)}{\sqrt{n}}a_n$$

By the test of the general term, we have
$$\frac{1}{\sqrt{n}} \le \frac{2 + \cos{(n)}}{\sqrt{n}} \le \frac{3}{\sqrt{n}}  $$
So, by the theorem of the sandwich,
$$ \lim_{n \rightarrow \infty}\frac{1}{\sqrt{n}}
=\lim_{n \rightarrow \infty}\frac{2 + \cos{(n)}}{\sqrt{n}}
=\lim_{n \rightarrow \infty}\frac{3}{\sqrt{n}} $$
We get, $\lim_{n \rightarrow \infty}\frac{2+\cos(n)}{\sqrt{n}}a_n=0$.

How to a continue my proof?

If we use the ratio test,
$$
\left| \frac{a_{n+1}}{a_n} \right|
= \left|\frac{2+\cos{(n)}}{\sqrt n} \right|
\leq \frac{3}{\sqrt{n}} \to 0.
$$
So the series converges absolutely.


Why are components of vectors scalars? Lets say we have a vector 1i+2j, I saw that the components of the vectors are said to be scalars, how can that be possible, doesnt 1i have a direction, also if they were a scalar why are the following vector laws of addition,

The components in this context aren't $1\mathbf i$ and $2\mathbf j$, but $1$ and $2$.
And indeed, $\mathbf i$ and $\mathbf j$ are vectors, which can be (and are) scaled to make linear combinations.

The components of $1 \hat{i}+2\hat{j}$ are $1,2$ and not $1  \hat{i}, 2\hat{j}$

Components of vectors are not scalars, and not vectors- they are simply components of vectors.
Scalars are objects who do not change under change of coordinates (e.g. if you rotate the axes by some angle). The components of a vector do change.
Example:
if we take some cartesian coordinates $x,y$, and the vector $\vec{V}= 2 \hat{x}+5\hat{y}$, if we take $x',y'$ to be coordinates which are rotated by $\pi/2$ we get $\vec{V}= 5 \hat{x'}-2\hat{y'}$. The components changed.  The vector length, hoever, is a scalar, as you can check.


Suppose that $x, y\in\mathbb{R}$ with $x^2+ y^2= 1$ and $y\geq 0$. Show that there exists $0\leq t\leq\pi$ such that $x= \cos t$ and $y= \sin t$ Suppose that $x, y\in\mathbb{R}$ with $x^2+ y^2= 1$ and $y\geq 0$.Show that there exists $0\leq t\leq\pi$ such that $x= \cos t$ and $y= \sin t$
This is an analysis question and I can only use the fundamental definition of $\sin$ and $\cos$ that is the infinite sum. I can also use the Taylor theorem, but I am stuck and clearly do not understand what I need to assume and what I need to prove. I can assume that $\sin$ and $\cos$ are infinitely differentiable. $\pi$ is defined as$:\quad\pi= 2\inf\{x\geq 0:\quad cos x= 0\}$.
Intuitively I should be using the Intermediate Value Theorem but I am not sure on which function.

So all you have is the power series for $\sin x$ and $\cos x$, and you know that $\frac \pi2$ is the smallest positive zero of $\cos x$.
Lemma. Let $f\colon\Bbb R\to\Bbb R$ be a twice differentiable function with
$$ f''(t)+f(t)=0\text{ for all }t,\quad f(0)=0,\quad f'(0)=0. $$
Then $f$ is identically $0$.
Proof. Indeed,
$$ \frac{\mathrm d}{\mathrm dx}(f(x)^2+f'(x)^2)
=2f'(x)f(x)+2f''(x)f'(x)=2f'(x)(f(x)+f''(x))=0$$
so that $f(x)^2+f'(x)^2$ is constant, and by evaluating at $0$, is constant zero.  But as squares of reals are non-negative, this implies that also $f$ is identically zero. $\square$
From the power series, you readily find that $$\sin't=\cos t,\qquad\cos't=-\sin t,$$ hence $$\sin''t=-\sin t,\qquad \cos''t=-\cos t.$$
As $\cos0=1$ and $\cos\frac\pi2=0$, the Intermediate Value Theorem gives us some $a\in[0,\frac\pi2]$ where $\cos a=|x|$ and then necessarily $|\sin a|=y$.
Note that $\sin t>0$ for  $t\in(0,\frac\pi2]$ (and in particular, $\sin\frac\pi2=+1$) as otherwise Rolle's theorem would give us a critical point of $\sin t$ and hence a zero of $\cos t$  in $(0,\frac\pi2)$, contradicting the definition of $\pi$. We conclude that in fact $\sin a=y$. Thus
Result 1. If $x^2+y^2=1$, $x\ge 0$, $y\ge 0$, then there exists $a\in[0,\frac\pi2]$ with $\sin a=y$, $\cos a=x$.
By switching some roles, this also gives us
Result 2. If $x^2+y^2=1$, $x\le 0$, $y\ge 0$, then there exists $b\in[0,\frac\pi2]$ with $\sin b=-x$, $\cos b=y$.
If the $x$ in our original problem is $\ge0$, result 1 gives us the desired answer. So assume $x<0$ and let $b$ be as in result 2.
Define $f(t)=\sin t+\cos(t+\frac\pi2)$. Then

$f(0)=\sin 0+\cos\frac\pi2=0+0=0$
$f'(0)=\cos 0-\sin\frac\pi2=1-1=0$
$f''(t)=-\sin t-\cos(t+\frac\pi2)=-f(t)$

Hence by the lemma, $f(t)=0$ for all $t$, i.e., $$\cos(t+\tfrac\pi2)=-\sin t$$
and similarly
$$\sin(t+\tfrac\pi2)=\cos t$$ for all $t\in\Bbb R$. In particular, with $c:=b+\frac\pi 2\in [0,\pi]$, we have
$$ \sin c = \cos b = y,\qquad \cos c = -\sin b = x$$
as desired.

$y=+\sqrt{1-x^2}>0.$
Let $x=\cos t ,t\in[0,\pi] \implies y=\sin t >0$


If $x+\frac{1}{y+\frac{1}{z+\frac13}}=\frac{118}{51}$ find $xyz$. I'm having a difficult time figuring out how to solve this algebra problem.

If $$x+\cfrac 1 {y+\cfrac 1 {z+\cfrac 1 3}}=\frac{118}{51}$$
find $xyz$.

I will presume that $x,y,z$ are intended to be positive integers.
\begin{align}
\frac{118}{51} & = 2 + \frac{16}{51} = 2 + \frac 1 {\left( \frac{51}{16} \right)} = 2 + \cfrac 1 {3 + \cfrac 3 {16}} \\[15pt]
& = 2 + \cfrac 1 {3 + \cfrac 1 {\left( \frac{16} 3 \right)}} = 2 + \cfrac 1 {3 + \cfrac 1 {5 + \cfrac 1 3}}.
\end{align}

Write:
$\begin{align*}
   x + \frac{1}{y + \frac{1}{z + \frac{1}{3}}} - \frac{118}{51}
     &= 0 \\
   \frac{((153 x - 354) y + 153) z + (51 x - 118) y + 153 x - 303}
        {153 y z + 51 y + 153}
     &= 0
\end{align*}$
If the denominator isn't zero, the numerator must be:
$\begin{align*}
  ((153 x - 354) y + 153) z + (51 x - 118) y + 153 x - 303
    &= 0 \\
  153 x y z - 354 y z + 153 z + 51 x y - 118 y + 153 x - 303
    &= 0
\end{align*}$
It is seen that this can't be written as a function only of $x y z$. You can e.g. set any pair of variables to zero and get a matching value for the third, and that doesn't always make the denominator zero.
This has no solution.


Applications of the Mean Value Theorem What are some interesting applications of the Mean Value Theorem for derivatives? Both the 'extended' or 'non-extended' versions as seen here are of interest.

So far I've seen some trivial applications like finding the number of roots of a polynomial equation. What are some more interesting applications of it?

I'm asking this as I'm not exactly sure why MVT is so important - so examples which focus on explaining that would be appreciated.

There are several applications of the Mean Value Theorem. It is one of the most important theorems in analysis and is used all the time. I've listed $5$ important results below. I'll provide some motivation to their importance if you request.

$1)$ If $f: (a,b) \rightarrow \mathbb{R}$ is differentiable and $f'(x) = 0$ for all $x \in (a,b)$, then $f$ is constant.

$2)$ Leibniz's rule: Suppose $ f : [a,b] \times [c,d] \rightarrow \mathbb{R}$ is a continuous function with $\partial f/ \partial x$ continuous. Then the function $F(x) = \int_{c}^d f(x,y)dy$ is derivable with derivative $$ F'(x) = \int_{c}^d \frac{\partial f}{\partial x} (x,y)dy.$$

$3)$ L'Hospital's rule

$4)$ If $A$ is an open set in $\mathbb{R}^n$ and $f:A \rightarrow \mathbb{R}^m$ is a function with continuous partial derivatives, then $f$ is differentiable.

$5)$ Symmetry of second derivatives: If $A$ is an open set in $\mathbb{R}^n$ and $f:A \rightarrow \mathbb{R}$ is a function of class $C^2$, then for each $a \in A$, $$\frac{\partial^2 f}{\partial x_i \partial x_j} (a) = \frac{\partial^2 f}{\partial x_j \partial x_i} (a)$$

An application which is used a lot in any calculus course:

The derivative of a differentiable function vanishes at an extremum(ie maximum or minimum) point.


Please help me solve this complex integral. $$\oint_C \frac{\cos(z-a)}{(z-a)}\mathrm{d}z$$

Such that $a\in \Bbb R^2$ and $C$ is a single closed curved defined by $|z-a|=\frac{|a|}{2}$

Here $z=x+iy$ is a complex number. Please solve the above integral. Thanks

Let $$f(z) = \cos(z - a)$$

Then you use the Cauchy Integral Formula:

$$f(z_0) = \frac1{2 \pi i}\oint_C \frac{f(z)}{(z-z_0)}dz$$

Let $z_0 = a$. Therefore this becomes:

$$f(a)=\frac1{2 \pi i}\oint_C \frac{\cos(z-a)}{(z-a)}dz = \cos(a-a) = \cos(0) = 1$$

Rearranging for your integral:

$$\oint_C \frac{\cos(z-a)}{(z-a)}dz = 2 \pi i \cos(0) = 2 \pi i$$


Wedge product of maps If $V$ and $W$ are $\mathbb{F}$ vector spaces, A $k$-multilinear alternating map $V^k \to W$ induces a unique linear map $f: \bigwedge^k V \to W$.

In the special case $W = \mathbb{F}$ and $\dim V < \infty$ (correct me if I'm wrong, I believe this last definition is necessary), we can write down $f$ as a sum of elements like $f_1 \wedge \dotsb \wedge f_k$, where $f \in V^*$. When $W \neq \mathbb{F}$, is there any notion of the wedge product of maps $\phi: V \to W$? How would it be evaluated?

Or is the best we can do to write
$$
f = \sum_{i=1}^n ((f_1^i \wedge \dotsb \wedge f_k^i)\otimes w_i), \quad \text{for } f_j^l \in V^*?
$$

Let $\{w_i\}_{i\in I}$ be a basis for $W$.

Let $\{w_i^*\in W^*\}_{i\in I}$ be the dual set, i.e., $w_i^*:W\rightarrow\mathbb{F}$ is  a linear functional s.t. $w_i^*(w_j)=\delta_{ij}$.

Since every $w\in W$ can be written as a finite linear combination of the elements of this basis then $w=\sum_{i\in I}w_i^*(w)w_i$ (Notice that this is a finite sum).

Thus, you can write the linearization $F:\bigwedge^kV\rightarrow W$ as $F(r)=\sum_{i\in I}w_i^*(F(r))w_i$.

Now, the map $w_i^*\circ F: \bigwedge^kV\rightarrow \mathbb{F}$  can be described as you know.


Proof there is an irrational number $r$ in every intervall $a < r < b$ Proof that for $a,b \in \mathbb{R}$ there is an irrational number $r$ so that
  $a < r < b$. Basically, proof, that between any two irrationals, there is another irrational r.

I'm sure there are already many ways out there how to do it, however I have troubles proving it in the following way:

(1) For every $x,y \in \mathbb{R}$ there is a bijective function between $[0,1]$ and $[x,y]$ (already proven)

(2) $\frac{\sqrt{2}}{2} \in ]0,1[$

(3) Now  when mapping $[0,1]$ onto $[x,y]$ $\frac{\sqrt{2}}{2}$ will also be mapped into the new intervall, therefore there has to be an irrational number in $[x,y]$

Now the problem I see is, that for example $\frac{\sqrt{2}}{2}$ could be mapped onto a rational number and therefore I'd have to proof, that there is a different irrational in $[x,y]$. It'd be nice if you could help me complete the proof.

If there are only rational numbers in $[x, y]$, then $[x, y]$ is countable and is in bijection with $[0, 1]$ which is uncountable, which is impossible.

If you want to continue your proof, then define $m = \frac{a + b}{2}$, $n = \frac{(a + b)\sqrt{2}}{2}$. You can show that $a < m < n < b$. Then either:

$m$ is irrational, and you're done.
$m$ is rational, then $n = \sqrt{2}m$ is the product of a rational and an irrational, and so is irrational, and you're done.


A formula with only one $0$ that evaluates to a given integer Just found this math puzzle online:

  Original puzzle: Using one zero (and no other numbers) and
  mathematical (including trigonometric and advanced) operations, how
  can you achieve a final result of 6?

  Advanced one: Using only trigonometric functions and a single instance
  of the number zero, derive a forumula to calculate any positive
  integer n.

I am sure there is an answer somewhere online but I can't find one - very interesting.

And the answer I can think of for the original one is

  $$\lim_{x\to0}\;(a^x+ a^x+ a^x+ a^x+ a^x+ a^x)$$

This may not be the most efficient solution for 6, but it works in general.  First, note that $cos(0) = 1$, so we can start with $1$.  The important formulas are then

(a) $\tan(\cot^{-1}(x)) = \frac{1}{x}$
(b) $\sin(\cot^{-1}(x)) = \frac{1}{\sqrt{1 + x^2}}$

These can be checked by drawing a triangle with the appropriate sides lengths.  Then, starting with $\sqrt{n}$, applying (b) and then (a) gives $\sqrt{n+1}$.  Thus, starting with $\sqrt{n} = n = 1$, repeated applications of (a) and (b) can give the square root of any natural number.  Since every natural number is the square root of another natural number, we can get any natural number this way.
Edit: After looking at the link Eugene gave below, this same process can be done more simply by iterating

$\sec( \tan^{-1}(x)) = \sqrt{x^2+1}$


How to factorize $n(n+1)(n+2)(n+3)+1$? How to factorize $n(n+1)(n+2)(n+3)+1$ ?

It's turned into the lowest power of $n$ ever possible but the question wants me to factorize it.

How can I do that ?

Just by running it through Wolfram, I think the simplest way to write it is $(n^2+3n+1)^2$. Many of the other ways still had a "$+1$" at the end of them.


What is the chance that a 1 in 1000 event will happen within the first 500 tries? Say you have a thousand-sided die.  You win if the die lands on 27, so a one in one thousand chance.  I was curious as to what the chances are of winning if you roll the die 500 times.  My gut told me that the odds were in your favor.  I computed the probability as:

1 - (999/1000) ^ 500

which gave me 0.3936210551388153.

Did I solve this correctly?  And if so, can anyone give me an intuitive explanation as to why you have less than a 40% chance of winning?  The only explanation I can think of is that if you rolled 1000 times, you certainly wouldn't have a 100% chance of winning, but I'm wondering if there's any other angle to think of.

Edit:  I understand basic probability concepts and am good at math, but so far most of the answers and comments are beyond my level of understanding.  I'm really looking for a more intuitive understanding than complex formulas explaining this... while I appreciate the effort I don't really understand how the more complicated formulas better explain the probability than my simplistic formula above (assuming it is correct in the first place, which it seems to be based on responses)

Okay... for the time, ignore the phrase expected value (though it is in effect what makes the following argument work).  This is just an attempt to try to build intuition.

Imagine for the time being that we have 100 boxes and exactly 50 balls.  We will choose to distribute these balls randomly among the boxes.

Notice that in this scenario, since we have 50 balls, at most 50 of the boxes will have at least one ball in it.  This implies that at least 50 of the boxes will have no balls in.

Now... if all 50 of the balls happened to be in different boxes, then sure, there will have been exactly 50 boxes with balls and exactly 50 boxes without balls.

Notice though that as we distribute the balls randomly, it is possible for some box or boxes to receive more than one ball.  In the case that there is some box or boxes with strictly more than one ball then we will have strictly fewer than 50 boxes with at least one ball and strictly more than 50 boxes with no balls.

Now... suppose we ask the question, "what is the probability that we picked a box with at least one ball in it", the answer will be strictly less than $50\%$.

Recognize that this is a metaphor (albeit an imperfect one) for your scenario in your question.  Rather than boxes we have "experiment trials" and rather than balls we have "27"'s.

This is $1-(1-1/n)^{n/2}$ for $n=500$. For large $n$, $(1-1/n)^{kn}$
is approximately $e^{-k}$. So here, you expect the answer to be approximately $1-e^{-1/2}\approx 0.393$


Solve $2^x+2^{-x} = 2$ Need to solve:

$$2^x+2^{-x} = 2$$

I can't use substitution in this case. Which is the best approach?

Event in this form I do not have any clue:

$$2^x+\frac{1}{2^x} = 2$$

Elucidate the problem by using the substitution $u = 2^x$, then you have $$u + \frac{1}{u} = 2$$

Multiply throughout by $u \neq 0$ to get $$u^2 +1 = 2u \iff u^2 - 2u + 1 = 0$$

This is an easy quadratic to solve, you should get $u = 1$ and hence you need only solve $2^x = 1 \iff x = 0$.


The Laplace transform of any null function is 0? First let N(t) be a function such that $\int_0^a N(t)dt=0\ \forall a>0$, this is what I call a null function.

I must prove that, for every null function N(t), $\mathcal{L}\{N(t)\}=0$ (supposing that every null function can have a Laplace transform).
I tried to prove it using the definition of the Laplace transform and partwise integration, and just want to check if something is wrong.

My attempt:

$\mathcal{L}\{N(t)\}=\lim_{b\to\infty}\int_0^b e^{-st}N(t)\,dt$

If we make $u=e^{-st}, dv=N(t)\,dt$, then $du=-se^{-st}dt, v=\int_0^b N(t)\,dt=0$. So we have:

$\mathcal{L}\{N(t)\}=\lim_{b\to\infty}[(0\cdot u)\mid_0^b - \int_0^b 0\cdot du]=\lim_{b\to\infty}[0]=0$

The Laplace transform, either unilateral or bilateral, of $f(t)=0$ is $F(s)=0$, simply because of linearity, by multiplying any known Laplace pair by the scalar $0$.


Can I get a percent of number without using decimals? I am programming in a language that does not support decimal numbers. The language only supports positive or negative whole numbers. Is there a way to calculate the percent of a number only using whole numbers?
The language automatically rounds the number down in the case of a remainder. I.e. 97 / 33 = 2
I have two values:
The percent and the number I am trying to get the percent of.
For example, lets say I have the number 126,345 and I want to get 13 % of that number.
The result of that normally would be 16,424.85. If this is possible the language would cut off the .85 remainder which is fine.
Thanks for your time and help!

As said by @magdiragdag in the comments, multiply by $13$ and then divide by $100$.
If your division rounds down (so that $160/100=1$) and you want to round to the nearest whole number instead, multiply by $13$, then add $50$, and then divide by $100$.


Intuition for probability density function as a Radon-Nikodym derivative If someone asked me what it meant for $X$ to be standard normally distributed, I would tell them it means $X$ has probability density function $f(x) = \frac{1}{\sqrt{2\pi}}\mathrm e^{-x^2/2}$ for all $x \in \mathbb{R}$.

More rigorously, I could alternatively say that $f$ is the Radon-Nikodym derivative of the distribution measure of $X$ w.r.t. the Lebesgue measure on $\mathbb{R}$, or $f = \frac{\mathrm d \mu_X}{\mathrm d\lambda}$.  As I understand it, $f$ re-weights the values $x \in \mathbb{R}$ in such a way that
$$
\int_B \mathrm d\mu_X = \int_B f\, \mathrm d\lambda
$$
for all Borel sets $B$.  In particular, the graph of $f$ lies below one everywhere:

so it seems like $f$ is re-weighting each $x \in \mathbb{R}$ to a smaller value, but I don't really have any intuition for this.  I'm seeking more insight into viewing $f$ as a change of measure, rather than a sort of distribution describing how likely $X$ is.

In addition, does it make sense to ask "which came first?"  The definition for the standard normal pdf as just a function used to compute probabilities, or the pdf as a change of measure?

Your understanding of the basic math itself seems pretty solid, so I'll just try to provide some extra intuition.

When we integrate a function $g$ with respect to the Lebesgue measure $\lambda$, we find its "area under the curve" or "volume under the surface", etc... This is obvious since the Lebesgue measure assigns the ordinary notion of length (area, etc) to all possible integration regions over the domain of $g$. Therefore, I say that integrating with respect to the Lebesgue measure (which is equivalent in value to Riemannian integration) is a calculation to find the "volume" of some function.

Let's pretend for a moment that when performing integration, we are always forced to do it over the entire domain of the integrand. Meaning we are only allowed to compute
$$\int_B g \,d\lambda\ \ \ \ \text{if}\ \ \ \ B=\mathbb{R}^n$$
where $\mathbb{R}^n$ is assumed to be the entire domain of $g$.

With that restriction, what could we do if we only cared about the volume of $g$ over the region $B$? Well, we could define an indicator function for the set $B$ and integrate its product with $g$,
$$\int_{\mathbb{R}^n} \mathbf{1}_B g \,d\lambda$$

When we do something like this, we are taking the mindset that our goal is to nullify $g$ wherever we don't care about it... but that isn't the only way to think about it. We can instead try to nullify $\mathbb{R}^n$ itself wherever we don't care about it. We would compute the integral then as,
$$\int_{\mathbb{R}^n} g \,d\mu$$
where $\mu$ is a measure that behaves just like $\lambda$ for Borel sets that are subsets of $B$, but returns zero for Borel sets that have no intersection with $B$. Using this measure, it doesn't matter that $g$ has value outside of $B$, because $\mu$ will give that support no consideration.

Obviously, these integrals are just different ways to think about the same thing,
$$\int_{\mathbb{R}^n} g \,d\mu = \int_{\mathbb{R}^n} \mathbf{1}_B g \,d\lambda$$
The function $\mathbf{1}_B$ is clearly the density of $\mu$, its Radon–Nikodym derivative with respect to the Lebesgue measure, or by directly matching up symbols in the equation,
$$d\mu = f\,d\lambda$$
where here $f = \mathbf{1}_B$. The reason for showing you all this was to show how we can think of changing measure as a way to tell an integral how to only compute the volume we care about. Changing measure allows us to discount parts of the support of $g$ instead of discounting parts of $g$ itself, and the Radon–Nikodym chainrule formalizes their equivalence.

The cool thing about this, is that our measures don't have to be as bipolar as the $\mu$ I constructed above. They don't have to completely not care about support outside $B$, but instead can just care about support outside $B$ less than inside $B$.

Think about how we might find the total mass of some physical object. We integrate over all of space (the entire domain where particles can exist) but use a measure $m$ that returns larger values for regions in space where there is "more mass" and smaller values (down to zero) for regions in space where there is "less mass". It doesn't have to be just mass vs no-mass, it can be everything in between too, and the Radon–Nikodym derivative of this measure is indeed the literal "density" of the object.

So what about probability? Just like with the mass example, we are encroaching on the world of physical modeling and leaving abstract mathematics. Formally, a measure is a probability measure if it returns 1 for the Borel set that is the union of all the other Borel sets. When we consider these Borel sets to model physical "events", this notion makes intuitive modeling sense... we are just defining the probability (measure) of anything happening to be 1.

But why 1? Arbitrary convenience. In fact, some people don't use 1! Some people use 100. Those people are said to use the "percent" convention. What is the probability that if I flip this coin, it lands on heads or tails? 100... percent. We could have used literally any positive real number, but 1 is just a nice choice. Note that the Lebesgue measure is not a probability measure because $\lambda(\mathbb{R}^n) = \infty$.

Anyway, what people are doing with probability is designing a measure that models how much significance they give to various events - which are Borel sets, which are regions in the domain; they are just defining how much they value parts of the domain itself. As we saw before with the measure $\mu$ I constructed, the easiest way to write down your measure is by writing its density.

Fun to note: "expected value" of $g$ is just its volume with respect to the given probability measure $P$, and "covariance" of $g$ with $h$ is just their inner product with respect to $P$. Letting $\Omega$ be the entire domain of both $g$ and $h$ (also known as the sample space), if $g$ and $h$ have zero mean,
$$\operatorname{cov}(g, h) = \int_{x \in \Omega}g(x)h(x)f(x)\ dx = \int_{\Omega}gh\ dP = \langle g, h \rangle_P$$

I'll let you show that the correlation coefficient for $g$ and $h$ is just the "cosine of the angle between them".

Hope this helps! Measure theory is definitely the modern way of viewing things, and people began to understand "weighted Riemannian integrals" well before they realized the other viewpoint: "weighting" the domain instead of the integrand. Many people attribute this viewpoint's birth to Lebesgue integration, where the operation of integration was first (notably) restated in terms of an arbitrary measure, as opposed to Riemnnian integration which tacitly always assumed the Lebesgue measure.

I noticed you brought up the normal distribution specifically. The normal distribution is special for a lot of reasons, but it is by no means some de-facto probability density. There are an infinite number of equally valid probability measures (with their associated densities). The normal distribution is really only so important because of the central limit theorem.

Your intuition for the measure $d\mu = f d \lambda$ is very reasonable and it is interesting to note how the function is distributed. How much area it has acumulated between $-1$ and $1$ at which point $f(x) = e^{-x^2/2}/{\sqrt{2\pi}}$  goes from concave to convex (f''(x)=0). Since
$$f''(x) = \frac{1}{\sqrt{2\pi}} \bigg(-e^{-x^2/2} + x^2e^{-x^2/2}\bigg) = 0 $$

this means that at $x=1$ the curvature changes sign. Can you notice it in the graph?

It is also interesting to note that between $-2$ and $2$ you have accumulated more than $95\%$ of the total mass.

The fact that it is bounded by $1$ ($\frac{1}{\sqrt{2\pi}}$ to be precise) is also remarkable, but as you might have heard there is a family of normal variables and they are not bounded by $1$.

Lastly, the normal distribution was discovered by de moivre in 1733 and was published on his Doctrines of chances https://en.wikipedia.org/wiki/The_Doctrine_of_Chances. So the Normal function was first considered a concrete probability object, and not a measure.


Elementary Level Algebra Question I'm trying to solve the following homework problem:

If $a\neq b$, $a^3-b^3 = 19x^3$ and $a-b=x$, which of the following conclusions is correct?

\begin{align}
\text{(1) }& a = 3x \\
\text{(2) }& a = 3x \text{ or } a = -2x \\
\text{(3) }& a = -3x \text{ or } a = 2x \\
\text{(4) }& a = 3x \text{ or } a = 2x
\end{align}

I am getting option (4): $a = 3x$ or $a=2x$ as the answer, which is incorrect. The correct answer is $a=3x$ or  $a=-2x.$

My work:-

$$ (a-b)^3 + 3ab(a-b)=19x^3$$
$$x^3 + 3ab(x) = 19x^3$$
$$ab=6x^2$$

Hence by comparing we have,

$$a * b = 3x * 2x \text{  or  } a * b = 2x * 3x$$

So a can be either $2x$ or $3x$.

Why is my answer wrong?

Use difference of cubes.
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$

$$\implies (a-b)(a^2+ab+b^2) = 19x^3$$

Set $\color{purple}{a-b = x}$.

$$\implies \color{purple}{x}(a^2+ab+b^2) = 19x^3 \implies a^2+ab+b^2 = 19x^2$$

Set $\color{blue}{b = a-x}$.

$$\implies a^2+a\color{blue}{(a-x)}+\color{blue}{(a-x)}^2 = 19x^2$$

Move $19x^2$ to the LHS, expand, and simplify.

$$\implies a^2+a^2-ax+a^2-2ax+x^2-19x^2 = 0$$

$$\implies 3a^2-3ax-18x^2 = 0$$

$$\implies a^2-ax-6x^2 = 0$$

Factor the trinomial.

$$\implies (a-3x)(a+2x) = 0$$

Set either factor equal to $0$.

$$a = 3x \text{ or } a = -2x$$

Edit: You’ve asked where your error is. Your comparison part wasn’t correct.

$$ab = 6x^2$$

You can’t just jump to conclusions on what $a$ and $b$ can be. Here, make the substitution $\color{blue}{b = a-x}$.

$$a\color{blue}{(a-x)} = 6x^2$$

$$a^2-ax = 6x^2 \implies a^2-ax-6x^2 = 0$$

This leads to the same answer.

The work you did till you reached ab=6x^2 is all good
But thats when you went wrong you chose a=3x and a=2x
For a=3x b=2x its all good but for a=2x b=a-x=x so ab=2x^2 in this case you are wrong so what solves this is a=-2x
In general for you to be able to solve such equation without guessing you just substitute b by its value:
a(a-x)=6x^2
a^2-ax=6x^2
a^2-ax-6x^2=0
You want the value of a so you solve this equation for a:
Delta=b^2-4ac=x^2+24x^2=25x^2
So a=3x or a=-2x


How do you divide complex numbers in polar form? A question in my textbook asks:

Find $\frac{z_1}{z_2}$ if $z_1=2\left(\cos\left(\frac{\pi}3\right)+i\sin\left(\frac{\pi}3\right)\right)$ and $z_2=\cos\left(\frac{\pi}6\right)-i\sin\left(\frac{\pi}6\right)$. I converted $z_2$ to $\cos\left(-\frac{\pi}6\right)+i\sin\left(-\frac{\pi}6\right)$ as I initially thought it would be easier to use Euler's identity (which it is) but the textbook hadn't introduced this yet so it must be possible without having to use it. As a result, I am stuck at square one, any help would be great. Thanks.

Just an expansion of my comment above: presumably you know how to do
$$
\alpha(a+bi)(c+di)\quad\text{here}\quad i=\sqrt{-1}; a,b,c,d,\alpha\in\mathbb{R}.
$$
Then for $c+di\neq 0$, we have
$$
\frac{a+bi}{c+di}=\alpha(a+bi)(c-di)\quad\text{with}\quad\alpha=\frac{1}{c^2+d^2}.
$$
In fact, this is usually how we define division by a nonzero complex number.

In your case, $a,b,c$ and $d$ are all given so just plug in the numbers. And with $a,b,c$ and $d$ being trig functions, I'm sure some simplication is going to happen.


Compute this line integral $\int_{C}(x+y){\mathbf{i}}+(x-y){\mathbf{j}}\ d\alpha $ Calculate the line integral of the function $$f(x,y)=(x+y){\mathbf{i}}+(x-y){\mathbf{j}}$$  around the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ counter to clockwise.
My approach is the following: The equation can be written in the form $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ then we can parametrize by taking $x=a\cos t$ and $y=b\sin t$ for $0\le t\le 2\pi$. So i have $\alpha(t)=(a\cos t)\mathbf{i}+(b\sin t)\mathbf{j}$. And to integrate i use the formula $$\int_{C}f\cdot d\alpha=\int_{a}^{b}f(\alpha(t))\cdot a'(t)dt$$ which is in this case $$\int_{0}^{2\pi}((a\cos t+b\sin t)\mathbf{i}+(a\cos t-b\sin t)\mathbf{j})\cdot ((-a\sin t)\mathbf{i}+(b\cos t)\mathbf{j})dt $$ which is equal to $$\int_{0}^{2\pi}-(a^2+b^2)\sin t\cos t-ab(\sin^2 t-\cos^2 t)dt$$ My problem is that this integral is not zero (which is the correct answer according to the book). Where is my error?

Or use Stokes' Theorem:  if $G$ is a vector field, then

$\int_\Omega \nabla \times G dA = \int_\Gamma G \cdot d\alpha, \tag{1}$

which holds for any surface $\Omega$ bounded by the curve $\Gamma$, where $dA$ is the area element in $\Omega$ and $d\alpha$ is the line element along $\Gamma$.  If we take

$G(x, y) = f(x, y) = (x + y)\mathbf i + (x - y) \mathbf j, \tag{2}$

we can compute that

$\nabla \times f(x, y) = \nabla \times ((x + y)\mathbf i + (x - y) \mathbf j)$
$= -\partial / \partial z (x - y) \mathbf i + \partial / \partial z (x + y) \mathbf j + (\partial / \partial x (x - y) - \partial / \partial y (x + y)) \mathbf k$
$= 0 \mathbf i + 0 \mathbf j + (1 - 1)\mathbf k = 0; \tag{3}$

this shows the left hand side of (1) must be zero when $G(x, y) = f(x, y)$.  Thus

$\int_\Gamma f \cdot d\alpha = 0 \tag{4}$

as well.

Hope this helps.  Cheerio,

and as always,

Fiat Lux!!!

f(t) = (acost,bsint); f'(t) = (-asint,bcost) so that

$f \cdot f' = -a^2costsint  + b^2sintcost$

Both these terms are periodic in 2$\pi$ so when you integrate from 0 to $2\pi$ you get zero for each term.


large x and small x expansion for gamma-like function Find two approximations for the integral ($x>0$)
  \begin{equation}
    I(x) = \frac{1}{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
      e^{x \cos^2(\theta)}d\theta
  \end{equation}
  one for small $x$ (keeping up to linear order in $x$) and one for large values of $x$ (keeping only the leading order term).

For the small x expansion, I tried using a Maclaurin series for the function in the exponential, and it seemed to work...I am unsure what to do for large x or how to check my answer.  I know this is likely related to the Gamma/Gaussian function, but am fairly novice at problems of this form.  Any help deeply appreciated!

For large $x$, Laplace's method seems like the best option. Or its counterpart, the Watson's lemma.

Let's transform the integral first:

$$ I(x) = \frac{1}{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
      e^{x \cos^2(\theta)}d\theta =\frac{1}{\pi}\int_0^{\frac{\pi}{2}}
      e^{x \cos^2(\theta)}d\theta=\frac{e^{x}}{\pi} \int_0^{\frac{\pi}{2}}
      e^{-x \sin^2(\theta)}d\theta= \\ =\frac{e^{x}}{\pi} \int_0^1
      \frac{e^{-x s^2}}{\sqrt{1-s^2}} d s=\frac{e^{x}}{2\pi} \int_0^1
      \frac{e^{-x t}}{\sqrt{t}\sqrt{1-t}} d t$$

Now for the latter integral, the main contribution for $x \to \infty$ will be given by $t \to0$. And the form of it allows us to use the Watson's lemma directly (see the Wikipedia link).

Using the notation from the article, we have:

$$\phi(t)=\frac{1}{\sqrt{t}\sqrt{1-t}}=t^{-1/2} g(t)$$

Where $g(t)$ can be expanded into Taylor series around $0$. Additionally, $\int_0^1 |\phi(t)| dt=\pi<\infty$. So the lemma conditions are satisfied.

Then we can represent:

$$\int_0^1
      \frac{e^{-x t}}{\sqrt{t}\sqrt{1-t}} d t \asymp \sum_{n=0}^\infty \frac{\Gamma(n+1/2) g^{(n)}(0)}{n! ~x^{n+1/2}} $$

How about this:
$$I(x)=\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}e^{x\cos^2\theta}d\theta=\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}e^{x(1-\sin^2\theta)}d\theta$$
$$\theta=\pi/2-\phi$$
$$d\theta=-d\phi$$
$$I(x)=\frac{1}{2\pi}\int_0^\pi e^{x\sin^2\phi}d\phi$$

EDIT:

taking the other approach, I have:
$$I'(x)=\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}\cos^2\theta e^{x\cos^2\theta}d\theta$$
then using $u=cos\theta$ you can obtain:
$$I'(x)=\frac1\pi\int_0^1\frac{u^2}{\sqrt{1-u^2}}e^{xu^2}du$$
now using $v=\sqrt{1-u^2}$ we can get:
$$I'(x)=\frac1\pi\int_0^1\sqrt{1-v^2}e^{x(1-v^2)}dv$$
which can be re-written as:
$$I'(x)=\frac1\pi\int_0^1\sqrt{1-v^2}e^{-(\sqrt{x}v)^2}dv$$
if you continue and differentiate once again you get:
$$I''(x)=\frac{1}{2\pi x^2}\int_0^x\beta e^\beta d\beta$$
so:
$$I(x)=\frac1{2\pi}\iint\frac{(x-1)e^x+1}{x^2}dxdx$$


Enneper surface is not injective I'm having trouble proving the following statement:

  $x(u, v) = (u − u^ 3/ 3
 + uv^2 , v − v^ 3/ 3
 + u^ 2 v, u^2 − v^ 2 )$ is a minimal surface and x is not injective

Proving that $x(u,v)$, which is also known as the Enneper surface, is minimal is not a problem. However, I can't prove that $x$ is not injective.

Is there any smart way to do this rather than trying some couples $ (a,b)$ and (c,d) hoping that $x(a,b)$ will be equal to $x(c,d)$ with $(a,b)$ and $(c,d) $ different couples?

One way is to look for symmetries. The second and third components are even functions of $u$, while the first is an odd function of $x$. So, if $(u_0, v_0)$ is a point such that the first component of $x$ is zero, then $x(u_0,v_0)=x(-u_0,v_0)$.

It's not hard to find a solution of $u−u^3/3+uv^2=0$ with nonzero $u$: for example, choose $v=0$ and solve for $u$.

Just use the counter example...  i.e.
$$x( \sqrt3 , 0) = (0,0,3)= x(- \sqrt3 , 0)$$
With
$$( \sqrt3 , 0) \ne (- \sqrt3 , 0)$$


Examine my argument that $\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$ Setting
Let $\mathcal{L} = \{+,0\}$. I want to show that $$\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$$. Note $\equiv$ means elementary equivalence in this question.

Updated Problem

My issue is that given the four ring axioms for addition and identity:

$$ \forall x \forall y \forall z ~  (x + y) + z = x + (y + z)\\
 \forall x        x + 0 = 0 + x = x\\
 \forall x \exists y     x + y = y + x = 0\\
 \forall x \forall y     x + y = y + x$$

it seems like $\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$. As someone commented below if I can use multiplication then I may prove $\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$, but multiplication is not in $\mathcal{L}$.

Here is Brian Scott's sentence, slightly simplified.

$$\forall x\forall y\exists z ((z+z=x)\lor (z+z=y)\lor (z+z=x+y))$$

This is true in $\mathbb Z$ - it essentially says that given $x,y$, at least one of $x,y,x+y$ is even.

It is not true for $\mathbb Z\oplus\mathbb Z$, for example if $x=(1,0),y=(0,1)$.

I am not quite what the question is here. But I will answer the question

Show that there does not exist a bijective group homomorphism (isomorphism) between $Z$ and $Z \times Z.$

Note that every element of $Z$ can be written as $n.1$ or $-(n.1)$ where $n.$ means add $n$ copies of together and $-$ means take the additive inverse. i.e. $1$ generates the group.

$Z\times Z$ does not have this property for any element of it. If we take $(a,0)$ with $a \neq 0,$ as our
generator we never get non-zero in the second coordinate. If we take $(0,b)$ we never get it the first, If we take $(a,b)$ with both non-zero, we'll never zero in either apart from $(0,0).$

So no element generates.

Since the property of having a group element generate is defined purely in terms of the group actions, it would be preserved by an isomorophism. Hence no isomorphism exists.

(I discuss more of these sorts of arguments in my forthcoming book Proof Patterns. )


Solving Discrete Markov Chain with diagonal band matrix. I am trying to model a certain process as a Discrete Markov Chain.
My system has $N+1$ states: $X=0, \ldots N$, and I can assume that the $(N+1)\times (N+1)$ transition matrix $T$ has the general form

$$T=\left(
        \begin{matrix}
        1 - p_0 & p_0 & 0 & \cdots & \cdots&\cdots & 0  \\
        0 & 1-p_1 & p_1 & 0 & \cdots & \cdots& 0 \\
       0 & 0 & \ddots & \ddots & 0 & \cdots&0  \\
       0 & 0 &  0 &1-p_n & p_n & 0 & 0 \\
       0 & \cdots & \cdots & 0 & \ddots & \ddots&0  \\
       0 & \cdots & \cdots & \cdots & 0 & 1-p_{N-1}&p_{N-1} \\
       0 & \cdots & \cdots & \cdots & \cdots & 0 &1  \\
        \end{matrix}
\right)
$$

where the $p_n$ are estimated from experimental data, and in principle (though it is unlikely) for some $j$ it might be $p_j=0$ or $p_j = 1$.

What I am interested for my evaluation in is the expected value of the distribution at the $m-$th step, $m \le N$, assuming that my process start from state $X=0$.

I am new to the study of Markov chains, so I'd like some advices on how to proceed correctly. And possibly some smart trick to solve the chain.

As far as I have understood I have to evaluate $T^m$. The first row of $T^m$ is then the distribution of probability at m-th time step assuming that my process starts from the state $X=0$.

The particular structure of the matrix would help me in solving the calculations? How to treat the cases $p_j=0$ and $p_j=1$?

Thanks a lot for the advices and help!

Diagonalize $T$ with the diagonal matrix $D$ and change of basis matrix $C$. To find the diagonal matrix $D$, find the eigenvalues of $T$, then $D$ is the diagonal matrix with its entries as its eigenvalues. Then matrix $C$ has its columns as the eigenvectors of $T$.

Then you'll have $T = CDC^{-1}$. Then $T^m=(CDC^{-1})(CDC^{-1})\cdots(CDC^{-1})$ (you're multiplying $CDC^{-1}$ by itself m times. Observing the left side we see be associativity $CD(C^{-1}C)D(C^{-1}C)\cdots(C^{-1}C)DC^{-1}$ Therefore we have the Identity matrix in every parenthesis and we have $T^m = CD^mC^{-1}$ which is a standard result for any diagonalizable matrix $T$.


How to solve an equation involving the floor of a number? I'm looking for a solving procedure for this type of exercises.

  If $[x]$ represents the floor of $x$, solve the equation:

  $$\left[\frac{6x+5}8\right]=\frac{15x-7}5$$

  Choose the correct answer:

  a) {$\frac{4}5$}

  b) {$\frac{3}4$}

  c) {$\frac{7}{15}$,$\frac{4}5$}

  d) {$\frac{6}{15}$}

  e) {$\frac{1}{2}$,$\frac{3}4$}

  f) {$\frac{1}{2}$,$\frac{4}5$}

Can someone please explain how I can solve this equation? Thank you very much!

ps. I asked a similar question, unfortunately, the same method doesn't seem to work.

If you really want to solve it without using the fact that it's multiple-choice, notice that the fractional part of $\frac{6x+5}{8}$ is $Q=\frac{6x+5}{8}-\frac{15x-7}{5}$. Since $Q$ is a fractional part, $0 \leq Q < 1$; you should be able to work backwards to find bounds on $x$. Once you have those bounds, Ilya's observation that $3x-7/5$ is an integer means there are only finitely many values of $x$ you have to worry about checking, so just try them all (or write a program to do it for you).

When you have to chose between different answers, you don't need to solve the equation, you just need to test the proposed answers.


no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$ The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $

$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}$

So $$\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$$ Then Differentiate

both side w . r to $x\;,$ We get $$\displaystyle f'(x)=1+x+x^2+x^3+..........+x^6$$

Now for max. and Minimum Put $$f'(x) = 0\Rightarrow 1+x+x^2+x^3+x^4+x^5+x^6 = 0$$

We can write $f'(x)$ as  $$\displaystyle \left(x^3+\frac{x^2}{2}\right)^2+\frac{3}{4}x^4+x^3+x^2+x+1$$

So $$\displaystyle f'(x) = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[3x^4+4x^3+4x^2+4x+4\right]$$

So $$\displaystyle f'(x) =  = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[\left(\sqrt{2}x^2+\sqrt{2}x\right)^2+(x^2)^2+2(x+1)^2+2\right]>0\;\forall x\in \mathbb{R}$$

So $f'(x) = 0$ does not have any real roots. So Using $\bf{LMVT}$ $f(x) = 0$ has at most one root.

In fact $f(x) = 0$ has exactly one root bcz $f(x)$ is of odd degree polynomial and it

will Cross $\bf{X-}$ axis at least one time.

My question is can we solve it any other way, i. e without using Derivative test.

Help me , Thanks

$f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$ So, as you note,

$f'(x)=1+x+x^2+x^3+..........+x^6$. Now consider $x$ in three ranges

For $x \ge 0$ then clearly $f'(x) >0$
For $ -1 \le x < 0$, $f'(x)=(1+x) + (x^2+x^3) + (x^4+x^5) + x^6$. Each bracketed term is non-negative, and $x^6$ is positive, so $f'(x) >0$
For $x < -1$,  $f'(x)=1 +(x + x^2) +(x^3 + x^4) +( x^5 + x^6)$. Again, each bracketed term is positive so $f'(x) >0$

From that conclude that $f(x) $ is a monotonic increasing function and can therefore have at most 1 zero.

Finally, as $x \to -\infty$, $f(x) \to -\infty$ and as $x \to +\infty$, $f(x) \to +\infty$ so that by continuity $f(x) $ must have at least one zero, and from the previous it therefore has exactly one zero.

An odd polynomial has at least one real root is due to the fact that complex roots come in conjugate that is if $\alpha$ is a complex root of a polynomial then $\overline{\alpha}$ is a root too.


Contour Integral with Many Singularities How would one compute?
$$ \oint_{|z|=1} \frac{dz}{\sin \frac{1}{z}} $$

Can one "generalize" the contour theorem and take the infinite series of the residues at each singularity?

Another method for this case: Think of the outside of the unit disk as your domain.  Go around the contour in the opposite direction.  Outside the unit disk, the function
$$
\frac{1}{\sin\frac{1}{z}}
$$
is meromorphic.  It has only one singularity in that region: a pole with residue $-1/6$ at $\infty$.  So the value of the integral is
$$
(-2\pi i)\left(\frac{-1}{6}\right) = \frac{i\pi}{3} .
$$

If you want to generalize the residue theorem for infinitely many singularities, understand how the residue theorem is proven first. One replaces the loop by a circuit that encloses the singularities plus some arcs that are integrated in both directions.

In the case of infinitely many singularities you can do that for all singularities at once, but you can enclose more and more.

The extra point that you need is to show is that integral over the loop enclosing the non-isolated singularity must tend to zero as you make the loop smaller. Or more general, show that the limit exist. Then the integral is equal to the series of the residues at the isolated singularities, plus the limit at the non-isolated one.

Now, for that specific integral one doesn't have to go that way.

$$\oint_{|z|=1}\frac{dz}{\sin\left(\frac{1}{z}\right)}=\oint_{|z|=1}\frac{dz}{\sin{\overline{z}}}$$.

You can expand in series and compute $\oint_{|z|=1}\overline{z}^kdz$.


Let $H$ be a subgroup of $G$. Let $K = \{x \in G: xax^{-1} \in H \iff a \in H\}$. Prove that $K$ is a subgroup of $G$. Closure under multiplication:

$\to$

Let $a, x, y \in H.$  Since $H$ is closed under inverses and multiplication, $xax^{-1} \in H$ and $yay^{-1} \in H.$ Since $H$ is closed under multiplication, $xyax^{-1}y^{-1} \in H.$

$\leftarrow$

Suppose $xyax^{-1}y^{-1} \in H$, then $a \in H$ since $H$ is closed under multiplication. So, $xy \in K.$

Closure under inverses:

Since $H$ is closed under multiplication and inverses, if $a \in H,$ then $x^{-1}ax \in H.$ If $x^{-1}ax \in H,$ then $a \in H$ because $H$ is a group. So, $x^{-1} \in K.$

Identity:

Suppose $a \in H.$ Then $a = eae^{-1} \in H.$ So, $e \in H.$

I'd like to see how I can fix/improve this proof.

You need to show just two things: $e \in K$ and $x \in K \implies x^{-1} \in K$.

Since $H$ is a group, the identity element $e$ is in $H$. Also, for any $x$ in $G$, $xex^{-1}=e\in H$ so for any $x\in G, x\in K$. In other words $G\subseteq K$, but also by definition, $K\subseteq G$. So $K=G$.


Is there a proof for the fact that, if you perform the same operation on both sides of an equality, then the equality holds? Is there a proof for this, or is it just taken for granted?

Does one need to prove it for every separate case (multiplication, addition, etc.), or only when you are operating with different elements (numbers, matrixes, etc.)?

When we say "$a = b$", what we are claiming is that $a$ and $b$ represent the same object. If you apply the same operation to the same object, then you get the same result, provided that operation is well-defined.

Just to be clear, there do exist operations that are not well-defined, so sometimes you do have to prove that $a=b \implies f(a) = f(b)$. For example, take $q \in \mathbb{Q}$, represented by $q= a/b$. Then $f(a/b) := a$ is not well-defined, as $f(2a/2b) = 2a$, but $\frac{a}{b} = \frac{2a}{2b}$.

If only the operation is a single-result operation.


Doubt in finding periods of sum of two functions It has been said that the period of sum of two functions f and g is the LCM of their periods if they exist. Further it has been given that even if LCM exist, it need not be the fundamental period.

Then, is there any algorithm to find the fundamental period, other than hit and trial method?

Please guide me in this regard.

Let, $h(x)=f(x)+g(x)$

Let the fundamental periods of $f$ and $g$ be $T_1$ and $T_2$ respectively.

To find the fundamental period of $h(x)=f(x)+g(x)$, find the L.C.M. of $T_1$ and $T_2$.

Let L.C.M. ($T_1$,$T_2$) $= T$

If L.C.M exists then $h$ is periodic with $T$ as a period but $T$ need not be fundamental. So check whether the submultiples of $T$ are satisfying the condition for periodicity.

For example, if $h(x+\frac{T}{2})=h(x)$ then check whether $h(x+\frac{T}{4})=h(x)$. If yes then proceed finding further submultiples of $T$. If not then $\frac{T}{2}$ is the fundamental period.


which axiom(s) are behind the Pythagorean Theorem There are many elementary proofs for the Pythagorean Theorem, but no matter they use areas, similarities, even algebraic proofs, it is not straightforward to tell why it is true tracing back to the (Euclidean geometry) axioms. Are all these proofs equivalent? Do they all track back to the same axioms?

Sure, the Pythagorean theorem is an item in the theory of Euclidean geometry, and it can be derived from the modern axioms of Euclidean geometry.

A full set of Euclidean geometry axioms contains the information about similarity and area that are sufficient to prove the Pythagorean theorem "synthetically," that is, directly from the axioms. The algebraic proofs are a little different, though!

It turns out that after defining the real numbers and basic algebra, you can create a model of Euclidean geometry in $\Bbb R\times \Bbb R$ which obeys all the Euclidean axioms. The algebraic operations in an algebraic proof reflect the synthetic axioms being used, but the direct connections are not obvious. You are still indirectly using the synthetic axioms, but they are all hidden assumptions about $\Bbb R\times\Bbb R$ and coordinate geometry.

All proofs I have seen so far trackback to the parallel axiom, or the axioms of similarity(which I think are equivalent). I'd be interested to see these "algebraic" proof though. Anything proved by pure algebra alone must be universally true, which is not so in the case of the Pythagorean theorem. It must use (euclidean) geometry in some form or the other.


Metrics w.r.t which $\mathbb R$ is not complete I have seen some examples of metrics on $\mathbb R$ which make it incomplete, like $d(x,y)=|e^{-x}-e^{-y}|$ and $d(x,y)=\arctan x - \arctan y$ where $1,2,3,\ldots$ is a Cauchy sequence which doesn't converge. I am curious about what other examples there are. Is it always infinity which turns out to be 'missing' or are there metrics with Cauchy sequences that do not converge but remain bounded?

That depends on how much leeway one has in specifying the metric.  Consider for example the fact that there exists a bijective map $\phi\colon\mathbb R \to \mathbb R \setminus \mathbb Q$.  If you define the metric on $\mathbb R$ by setting $d(x,y)=|\phi(x)-\phi(y)|$, it will be a metric alright, and clearly not complete.

Cauchy sequences always remain bounded:

If $\epsilon = 1$, there is $N$ such that for all $n,m>N$
$$d(x_n,x_m)<1$$

Therefore, $d(x_n,0)\leq d(0,x_{N+1})+d(x_{N+1},x_n)<d(0,x_{N+1})+1$, for $n>N$ and a fixed. This means that the whole sequence is inside the ball with center $0$ and radius $\max(d(0,x_{N+1})+1, d(x_1,0),d(x_1,0),...,d(x_N,0))$.

In your examples, $1,2,3,...$ is bounded.


Within Geometric Algebra, is there a way to express a linear operator within the language of geometric algebra? Thanks for your time and effort.  I appreciate your help.

I'm new to geometric algebra and I get that it supersedes linear algebra.

I was wondering though how I could learn to take an operator in linear algebra and express it in terms of geometric algebra?

For example,

Suppose I had the matrix operator:

\begin{vmatrix}
\mathbf{1} & \mathbf{ 1} & \mathbf{-1} \\
\mathbf{0} & \mathbf{ 1} & \mathbf{-1} \\
\mathbf{0} & \mathbf{-1} & \mathbf{ 1} \
\end{vmatrix}

How would I express this same function in the language of geometric algebra?

Thanks for your help.

You could express it as a function.  Let your operator be $\underline T$.  It could be described by

$$\begin{align*}\underline T(e_1) &= e_1 \\ \underline T(e_2) &= e_1 + e_2 - e_3 \\ \underline T(e_3) &= -e_1 -e_2 + e_3\end{align*}$$

You could instead use dot products to combine this into a single expression.  Let $a$ be an arbitrary vector, and you have

$$\underline T(a) = (a \cdot e_1) e_1 + (a \cdot e_2) (e_1 + e_2 - e_3) + (a \cdot e_3) (-e_1 - e_2 + e_3)$$

In particular, notice that the last column is just the negative of the second column, so the expression simplifies to

$$\underline T(a) = (a \cdot e_1) e_1 + (a \cdot e_2 - a \cdot e_3)(e_1 + e_2 - e_3)$$

There is (so far) nothing inherently GA-like to expressing a linear operator this way, but it is a bit more amenable to some of the operations you might be asked to perform that come from GA.

Edit: about tensor products.  The GA is sometimes referred to as a quotient of the tensor algebra.  That is, there is some loss of information in using geometric products instead of tensor products.  Let me illustrate:  let $a$ and $b$ be two vectors.  The tensor product is

$$a \otimes b =\sum_{i} (a^i b^i)(e_i \otimes e_i) + \sum_{i \neq j} (a^i b^j) (e_i \otimes e_j)$$

In GA, the corresponding use of the geometric product gives

$$a b = a \cdot b + a \wedge b = \sum_i a^i b^i + \sum_{i \neq j} a^i b^j (e_i \wedge e_j)$$

The tensor product keeps $e_1 \otimes e_1$ separate from $e_2 \otimes e_2$ and so on.  Thus, the GA can't keep information about those terms distinct.  It doesn't need to because these are all scalar terms when interpreted in the context of geometry.

When using tensor products to build up linear maps, however, GA can replicate this rather well.  You should be able to see that $a \otimes b$ interpreted as a map becomes $\underline M(c) = a (b \cdot c)$.

A tensor product of 2 matrices can probably be interpreted in some similar way.

The key idea of Geometric Algebra is that it converts abstract algebraic concepts into concrete spatial shapes, and the shapes are meaningful to the concept represented.

In the case of your matrix, that can be viewed as three spatial vectors of (x,y,z) each, sticking out of the origin in three directions. You can now multiply this structure with a scalar, a vector, bivector, or another trivector and it will perform a logically reasonable spatial transformation on its operand. The shape of the structure is meaningful. For example the wedge product between the three vectors represents the volume spanned by them, and that turns out to be equal to the determinant of the matrix, and its volume goes to zero as the three vectors become parallel, as explained on the Wiki page on Exterior Algebra

In general, I find the association of familiar algebraic concepts with novel spatial concepts to be the most interesting. If you multiply a vector by -1 it points in the opposite direction: Negation is Reflection!  Rotation by 90 degrees == x sqrt(-1), because if you do it twice, you get to 180 degrees.  Clifford multiplication between vectors produces a scaling and a rotation. The rotation also works like reflection. Vectors a*b = bivector ab (plus a scalar), which represents a twist from the direction of a to the direction of b. If you multiply a third vector c*a*b it will rotate c through the same angle as the one from a to b.

It is imagining the spatial interpretation of the algebraic expression that is the most interesting aspect of Clifford Algebra. Remember when you learned parallel lines crossed by an angled line, =/= your teacher showed you which angles are equal, and which are complementary, etc. but he never had to prove it, he merely had to point it out to you, because its truth is self-evident by inspection.  Clifford Algebra does that for algebra. Instead of manipulating abstract symbols back and forth across the =equals= sign, in Clifford Algebra you imagine reflections and rotations of vectors about the origin.


App for stable marriage problem? Today I spoke with a theater manager who needs to match 7 directors with 7 playwrights. Clearly, this is an instance of the "stable marriage problem", which can be addressed using the Gale–Shapley algorithm. My question is, what would be the easiest practical way for the theater manager to accomplish this? Is there an app into which each director and playwright can enter their preferences, which will then come back with 7 stable pairings?

https://en.wikipedia.org/wiki/Stable_marriage_problem

I guess it will be helpful for you. You need only to insert your names.
https://github.com/TelychkoVitalii/JS_algorithms/blob/master/game_theory/stable_marriage_problem.js