Inverse of $f(x)= x+\sin(x)$? How to find the inverse of $f(x) = x+\sin(x)$, analytically?

Well how should I proceed to find the inverse of $f(x)$? Basically I have applied graphical approach to solve the equation, but I want to know the inverse equation by analytical method.

Let $y = x + \sin x = f(x)$;
So, $x = f^{-1}(y)$.
Differentiating the first expression according to $y$ leads to $x' + x'\cos(x) = 1$, or $x'(1+\cos x) = 1$.
Let us put $x=\arccos t$, so $x' = -t'/\sqrt{1-t^2}$.
This leads to $-t'(1+t)/\sqrt{1-t^2} = 1$, or $t'=-\sqrt{\frac{1-t}{1+t}}$.
I believe that this is a standard differential equation. I don't know if it is solvable. If it is not, then the problem has no analytical closed expression.

Another way:
Using the formula $\cos x =  \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$, we are lead to the differential equation $x' = \frac{1+\tan^2(x/2)}{2}$. We can put $u = \tan(x/2)$, so $u'=\frac{1}{2}x'(1+u^2)$, and $x' = \frac{2u'}{1+u^2}$. So, the equation becomes $u' = \frac{(1+u^2)^2}{4}$. If I'm not wrong, there is a method to handle somehow this kind of equations. I will try tomorrow.


Find the area under the graph of the function $f(x)=4x-7$ between $x=-2$ and $x=0$ Section 5.4

Can somebody verify this solution for me? Thanks!

Find the area under the graph of the function $f(x)=4x+7$ between $x=-2$ and $x=0$

Well, the area under the graph of $f(x)=4x+7$ between $x=-2$ and $x=0$ is exactly equal to:

$\int_{-2}^0 4x+7dx$

$= [\frac{4}{2}x^2+7x] |_{-2}^0$

$= [2x^2+7x] |_{-2}^0$

$=(2(0)^2+7(0))-(2(-2)^2+7(-2))$

$=0-(2(4)-14)$

$=6$

You can easly check it your self in this case. Just calulate the area of triangles:

You don't need integral since it's a linear equation. You can just find the area under the triangle as @Aqua mentioned. Base is -2, and height is 7 since $f(x)$ is 0. Then multiply and find the area.
$$
\frac{-2\times7}{2}=-7
$$


A few basic questions in logic from Zorich's mathematical analysis Let $A$ and $B$ be two mathematical statements, each of which can either be true, or false. The first problem is to show that:

  If $A$ is false, then $A \implies B$ is always true $\cdots\cdots(\star)$

Now I can consider $A:\{x>0\}$ and $B:\{x>5\}$. Now suppose we observe that $x=-2$. Then $A$ is false. But we never have $A \implies B$. Is the problem wrongly stated or am I missing something?

Two problems in the same line are to show that:

$(i)~~(A \implies B)\implies (not~~A)~~or~~B$

$(ii)~~not~~(A \implies B)\implies A~~and~~(not~~B)$

I can solve these two if I can show $(\star)$. I can sense that there is some vacuously true argument coming up but the counter example (?) is bugging me. Thanks in advance for any help.

Part of the problem is that you are parameterizing your propositions, which is a bit premature [this is only propositional calculus - later comes predicate calculus].  What I mean is: When you write 'A : {x > 0}' you are (I assume) asserting that A is the property 'x > 0'.  In other words, you are parameterizing A - what you really want to write is A(x).  But this is neither true nor false until you specify exactly what x is [similarly for B(x)].  So let us take a few examples:

x = -2 : Then A(-2) is the property '-2 > 0' and B(-2) is the property '-2 > 5'.  Neither of these is true.  So 'A(-2) ==> B(-2)' is vacuously true.

x = 2 : Then A(2) is '2 > 0' [true] and B(2) is '2 > 5' [false]  Hence 'A(2) ==> B(2)' is false.

x = 6 : Then A(6) is '6 > 0' [true] and B(6) is '6 > 5' [true]  Hence 'A(6) ==> B(6)' is true.

[I assume the cases like 'x=2' leads to your assertion 'we never have A==>B' but the cases like 'x=6' argue against that assertion]

To reiterate: When you try to define a property as a relation involving an unknown, you are not defining a proposition.  For the definition to be a proposition [i.e to have a definite truth-value], it must be completely defined.


Integration with $e$ What are the steps in this integration :

$$
\int4e^{t(u-4)}dt = \frac{-4}{u-4}e^{t(u-4)}
$$

I've calculated simpler integration formulas such as
$$
\int x^2dt = \frac{x^3}{3}
$$

but I'm unsure how integral above is calculated as there is an $e$ involved.

$$I=\int4e^{(u-4)t}dt=4\int e^{(u-4)t}dt$$
Substitute $$w=(u-4)t\\\Rightarrow dw=(u-4)dt\\\Rightarrow \frac{dw}{u-4}=dt$$
Hence $$I=4\int e^{w}\frac{dw}{u-4}=\frac{4}{u-4}\int e^wdw$$
Since $\frac{d}{dx}e^x=e^x$,
$$I=\frac{4}{u-4}e^w+C$$
Re-substitute:
$$I=\frac{4}{u-4}e^{(u-4)t}+C$$

Substituting $$m=t(u-4)$$ then you will get $$dm=(u-4)dt$$


difference in the definiton of modular arithmetic When i look at my textbook , i realized that there is a little difference in the definiton of modular artihmetic among books.
For example, wikipedia says that $a\equiv b\pmod n$ where $a,b \in Z,n \in Z^+$ and $n \gt1$ .
On the other hand, some books and websites says that $a\equiv b\pmod n$ where $a,b \in Z,n \in Z^+$ but there is not any restriction for $n$ ,i.e, $n$ can be $1$.
I know that it is not a big difference and does not affect results generally.However, i want to know the exact definiton of it.
Briefly, which definiton is correct ? If they are same , then why do books ignore $n=1$

The modulus $n=1$ is not interesting because $a \equiv b \bmod 1$ is always true.
I prefer the general definition that allows all $n \in \mathbb Z$ as modulus.
The only other singular modulus is $n=0$ because $a \equiv b \bmod 0$ iff $a=b$.

If $n>1$ then the ring $\mathbb{Z}_n$ is a "ring with unity" which means that it has a multiplicative identity distinct from the additive identity.   That is not the case if $n=1$.  I can't think of an example off the top of my head, but there must be some theorems about rings with unity that need the fact that $0 \neq 1$.  So you allow $n=1$, then every time to need to invoke some properties of a ring with unity, then you have to say "$n>1$".
So it might make sense to just specify $n>1$ at the outset and thereby exclude the "bo-ring."


Does $\sum_{n=1}^\infty \frac{\sqrt[n]e-1}{n}$ converge? Does $$\sum_{n=1}^\infty \frac{\sqrt[n]e-1}{n}$$ converge or diverge?
Could you please help me out with this one? I tried the ratio test, but it's inconclusive...
I also tried to use the inequality $e^x \ge x+1$, but it doesn't get me anywhere.

The series is convergent. $e^{x}-1-2x$ is decreasing in $(0, \ln 2)$ since its derivative is negative here. Since this function vanishes at $0$ it follows that $e^{x}-1-2x <0$ in this interval. Can you take over from here?

We have that
$$\sqrt[n]e=e^{\frac1n}= 1+\frac1n+O\left(\frac1{n^2}\right)$$
then
$$\frac{\sqrt[n]e-1}{n} = \frac1{n^2}+O\left(\frac1{n^3}\right)$$
and we can conclude by limit comparison test.


All translations of classical logic into intuitionistic logic What are all possible ways of translating classical logic into intuitionistic logic? That is, if $S$ is the collection of sentences of first order logic, what are all the functions $f : S \to S$ such that $\Gamma \vdash_c \phi$ if and only if $f(\Gamma) \vdash_i f(\phi)$? I know that there is a couple of variants of double negation translation. So, two questions: is there a classification of all possible translations? Are they all logically equivalent to double negation translation?

Remember that the set of theorems of intuitionistic logic is a proper subset of the set of theorems of classical logic. So, if you want to translate every classical theorem to an intuitionistic version you will fail in general.


Show $f$ is continuous at $(0,0)$ if $f(x,y) = \frac{x^3\sin(x+y) - y^4\ln(x^2+y^2)}{x^2+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0)=0$ A function $f: \mathbb R^2  \to \mathbb R$ is defined as $$f(x,y) = \frac{x^3\sin(x+y) - y^4\ln(x^2+y^2)}{x^2+y^2}$$

if $(x,y) \neq  (0,0)$ and $f(0,0)=0$.

How can I show it is continuous at $(0,0)$? Perhaps by using the Squeeze theorem?

Note that $\displaystyle f(x,y) = \frac{x^3 \sin(x+y)}{x^2 + y^2} - \frac{y^4 \ln(x^2 + y^2)}{x^2 + y^2}$.

So it's enough to show that each of these tends to $0$.

Applying the squeeze theorem on the first

$ \displaystyle | \frac{x^3 \sin(x+y)}{x^2 + y^2}| \le \frac{|x|}{1 + \frac{y^2}{x^2}} \le |x|$.

Next $\displaystyle |\frac{y^4 \ln(x^2 + y^2)}{x^2 + y^2}| = |\frac{y^2 \ln(x^2 + y^2)}{\frac{x^2}{y^2} + 1} | \le |y^2 \ln(x^2 + y^2)|$.

Now set $x = r\cos(\theta)$ and $y = r \sin(\theta)$.  Then the limit can be estimated by $|r^2 \ln(r^2)| \to 0$ as $r \to 0$ (say by L'Hopital) independent of $\theta$.

you just need to prove $f(x,y)$ differentiable at $(0,0)$ then you can use theorem "if $f$ differentiable at $c$ then $f$ continuos at $c.$"

you can see the theorem at any analysis books


Solving an equation with hyperbolic functions I'm trying to prove that for a given $s,t\in\mathbb{R}$ there exists $w\in\mathbb{R}$ such that

$\cosh(t)e^{i(s+w)}+\sinh(t)e^{i(s-w)}\in\mathbb{R}$.

How to solve this?

The imaginary part must be zero.
$$\cosh(t)\sin(s+w)+\sinh(t)\sin(s-w)=0$$
$$\cosh(t)\sin(s)\cos(w)+\cosh(t)\cos(s)\sin(w)+\sinh(t)\sin(s)\cos(w)-\sinh(t)\cos(s)\sin(w)=0$$
$$\tan(w)=\frac{\sinh(t)\cos(s)-\cosh(t)\cos(s)}{\cosh(t)\sin(s)+\sinh(t)\sin(s)}=-e^{-2t}\cot(s).$$

Try this:

1) divide by $\sinh(t)$ by eliminating from the domain the $t$ values for which $\sinh(t) = 0$.

2) divide by $e^{i(s-w)}$

3) Arrange the exponential to obtain in the end

$$e^{2iw} + \tanh(t)$$

Now proceed.


Help solving $\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$ Reading an article online I came across this integral I've been trying to solve since yesterday morning and I literally tried every integration method I know and I still can't solve it. I am a beginner to integrals so help will be appreciated!
I tried substitution, integration by parts and I can't seem to find the correct way!
$$\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$$
Does anyone know how to proceed from here? What to do? Because I am clueless. Thanks!


How does this inequality follow? This from the book Elliptic Partial Differential equations by Han and Lin.

Lemma 1.32 Suppose $u$ is a non-negative harmonic function in $B_1$. Then there holds
$$\sup\limits_{B_{1/2}}|D\log{u}|\leq C$$
where $C=C(n)$ is a positive constant.

Proof: We may assume $u>0$ in $B_1$. Set $v=\log{u}$. Then direct calculation shows
$$\Delta v=-|Dv|^2.$$

We need the interior gradient estimate on $v$. Set $w=|Dv|^2$. Then we get
$$\Delta w+2\sum\limits_{i=1}^{n}D_ivD_iw=2\sum\limits_{i,j=1}^n(D_{ij}v)^2.$$

We need a cutoff function. First note

$$\sum\limits_{i,j}^n (D_{ij}v)^2\geq\sum\limits_{i=1}^n(D_{ii}v)^2\geq \frac{1}{n}(\Delta v)^2=\frac{|Dv|^4}{n}=\frac{w^2}{n}$$

Taking a nonnegative function $\phi\in C_0^1(B_1)$. We obtain by the Holder inequality
\begin{align}
&\Delta(\phi w)+\sum\limits_{i=1}^nD_ivD_i(\phi w)\\
&=2\phi\sum\limits_{i,j=1}^n(D_{ij}v)^2+4\sum\limits_{i,j=1}^nD_i\phi D_jvD_{ij}v+2w\sum\limits_{i=1}^nD_i\phi D_iv+(\Delta\phi)w\\
& \geq \phi\sum\limits_{i,j=1}^n(D_{ij}v)^2-2|D\phi||Dv|^3-\left(|\Delta \phi|+C\frac{|D\phi|^2}{\phi}\right)|Dv|^2
\end{align}
if $\phi$ is chosen such that $|D\phi|^2/\phi$ is bounded in $B_1$. Choose $\phi=\eta^4$ for some $\eta \in C_0^1(B_1)$. Hence for such fixed $\eta$ we obtain

\begin{align}
&\Delta(\eta^4w)+ 2\sum\limits_{i=1}^n D_ivD_i(\eta^4w)\\
&\geq \frac{1}{n} \eta^4|Dv|^4-C\eta^3|D\eta||Dv|^3-4\eta^2(\eta\Delta\eta+C|D\eta|^2)|Dv|^2\\
&\geq \frac{1}{n}\eta^4|Dv|^4-C\eta^3|Dv|^3-C\eta^2|Dv|^2
\end{align}
where $C$ is a positive constant depending only on $n$ and $\eta$. Hence we get by the Holder inequality
$$\Delta(\eta^4w)+2\sum\limits_{i=1}^nD_ivD_i(\eta^4w)\geq\frac{1}{n}\eta^4w^2-C$$
where $C$ is a positive constant depending only on $n$ and $\eta$.
Suppose $\eta^4w$ attains its maximum at $x_0\in B_1$. Then $D(\eta^4w)=0$ and $\Delta\eta^4w\leq0$ at $x_0$. Hence there holds
$$\eta^4w^2(x_0)\leq C(n,\eta).$$
If $w(x_0)\geq 1$, then $\eta^4w(x_0)\leq C(n)$. Otherwise $\eta^4w(x_0)\leq w(x_0)\leq \eta^4(x_0)$. In both cases we conclude
$$\eta^4w\leq C(n,\eta)$$ in $B_1$.

My questions are

How is Holder inequality used here (the two instances mentioned in the proof)?
How do we infer the result from the last inequality?

The Holder inequality is

$$ 2ab \le a^2 + b^2.$$

A more useful one is

$$ 2ab = 2(\sqrt\epsilon a)(b/\sqrt\epsilon) \le \epsilon a^2 + b^2 /\epsilon.
$$

for all $\epsilon >0$. Then one has

\begin{align*}
4 D_i\phi D_jvD_{ij}v &= 2\left( 2 D_i\phi D_jv/\sqrt{\phi}\right) (\sqrt\phi D_{ij}v) \\
&\le  \epsilon \phi |D_{ij } v|^2 + \frac{4}{\epsilon}\frac{|D_i\phi|^2}{\phi} |D_jv|^2 \\
\Rightarrow 4 \sum_{i,j=1}^n D_i\phi D_jvD_{ij}v &\ge - \epsilon \phi \sum_{i,j=1}^n |D_{ij } v|^2 - \frac{4}{\epsilon}\frac{|D\phi|^2}{\phi} |Dv|^2
\end{align*}

Now choose $\epsilon = 1$. Also use

$$2|D v|^2 \sum_{i=1}^n D_i\phi D_i v= 2|D v|^2 D\phi \dot Dv \ge -2 |Dv|^2 |D\phi| |Dv|.$$

The second one could be done this way: use

\begin{align}
\eta^2 |Dv|^2 &\le \frac 12(\epsilon_1 \eta^4|Dv|^4 +1/\epsilon_1),\\
\eta^3 |Dv|^3 &= (\eta^2 |Dv|^2)(\eta |Dv|) \\
&\le \frac 12 (\epsilon_2 \eta^4 |Dv|^4 + \eta^2|Dv|^2/\epsilon_2) \\
&\le \frac 12 (\epsilon_2 \eta^4 |Dv|^4 + \frac 1{2\epsilon_2} (\epsilon_3\eta^4|Dv|^4+ 1/\epsilon_3))
\end{align}

Then we choose $\epsilon_1, \epsilon_2, \epsilon_3$ small so that
$$-C\eta^3|Dv|^3-C\eta^2|Dv|^2 \ge -\frac {1}{2n} \eta^4 |Dv|^4-C$$
and this gives
$$\frac {1}{n} \eta^4 |Dv|^4-C\eta^3|Dv|^3-C\eta^2|Dv|^2 \ge \frac {1}{2n} \eta^4 \omega^2-C.$$

This is slightly different from your inequality, but the argument is the same.

The last inequality follows from the fact that $x_0$ is the maximum of $\eta^4 \omega^2$. So

$$ \eta^4 \omega^2 (x) \le \eta^4 \omega(x_0)$$

for all $x$.


Questions about the fundamental theorem of arithmetic The fundamental theorem of arithmetic states that every positive integer greater than 1 is either a prime or a product of primes.

First question: why "either a prime or a product of primes", if every prime is in fact a product of primes with one factor? Wouldn't it be simpler to just say that every positive integer greater than 1 can be written as a product of primes?

The fundamental theorem of arithmetic is proved using strong induction. The formal definition of strong induction (in its transfinite version) is: $\forall n [\forall k [k<n \rightarrow P(k)] \rightarrow P(n)] \rightarrow \forall nP(n)$. In this case, $P(x)$ is substituted by "if x>1, then x is prime or x is a product of primes".

Second question: which are the formal definitions of x is prime and of x is a product of primes? Velleman says that x is not prime is the same of $\exists a\exists b[n=ab \land a<n \land b<n]$. Therefore, I would assume that x is prime would be $\forall a\forall b[n=ab \rightarrow a \geq n \lor b \geq n]$. Is it correct? How about x is a product of primes?

Now, expanding the inductive hypothesis  inside the scope of the natural numbers and excluding the vacuosly true statements, we end up with the following result:

$P(2) \land [P(2) \rightarrow P(3)] \land[P(2)\land P(3) \rightarrow P(4)] \land ... \rightarrow \forall n P(n)$

Third question: how is it possible that if 2 is prime or a product of primes implies that 3 is prime or a product of primes?

There is nothing wrong with the phrasing, even allowing "product of primes" to be understood to include the case of a product with one factor (which most mathematicians allow). Phrasing need not be the shortest possible, nor does the "or" need to be exclusive: the most important thing is that it not be misunderstood. You could just as well phrase the existence part of the Fundamental Theorem of Arithmetic as:

  Every positive integer is a (possibly empty) product of primes.

Why do we not phrase it that way? In part, for historical reasons. Euclid excluded $1$ (it wasn't a "number"); Gauss phrases the theorem as follows (Disquisitiones Arithmeticae, Section II.16, from the English edition translated by Arthur A. Clarke):

  Theorem. A composite number can be resolved into prime factors in only one way.

thus putting primes themselves on a separate category. Traditionally, people said numbers were either "primes or products of primes", before allowing $1$ to be considered a "number." The idea of having products with one or fewer factors is relative new.

I will note that your phrasing is incomplete: an oft-ignored part of the FTA is the uniqueness of the decomposition (up to order and, in modern parlance, associates). In fact, that's the part that Gauss is concerned with, saying

  "It is clear from elementary considerations that any composite number can be resolved into prime factors, but it is often wrongly taken for granted that this cannot be done in several different ways."

Gauss was the first to prove it explicitly (Euclid only proves it for square-free composite integers).

There are other ways to phrase it; you can specify that the uniqueness is only "up to order of the factors" if you want to make sure people don't think that $21$ can be factored two ways (as $3\times 7$ and as $7\times 3$). You can specify "up to associates" if you suddenly discover that negative numbers exist and that if $p$ is a (positive) prime number, then so is $-p$ (under the divisibility definition). Etc.

The formal definition of "$x$ is a prime" is

Definition. A (positive) integer $p$ is prime if and only if it is not $0$, not $1$ (or $-1$), and whenever $p$ divides a product $ab$, it divides one of the factors.

This is not the classical definition. Euclid defines a prime as a number (hence greater than $1$) which is divisible only by $1$ and itself; in modern parlance, this would be an irreducible, not a prime (though in the integers, the two concepts coincide).

"A product of primes" would be harder; you would want to have an operator that takes an arbitrary but finite number of integers and assigns an integer to it (their product). Or you could say that $m$ is a product of primes if

  there exists a positive integer $n$ [greater than $1$ if you want to exclude the one-factor product] and positive integers $p_1,\ldots,p_n$ such that $p_i$ is prime for each $i$ and $m=p_1p_2\cdots p_n$.

In fact, the inductive step is usually proven as follows:

Assuming that for any $k$, $1\lt k\lt n$, $k$ is either prime or a product of primes, consider $n$. If $n$ is prime, we are done. If $n$ is not prime, then there exist $a,b$, $1\lt a,b\lt n$ such that $n=ab$ (*). Applying the inductive argument to $a$ and $b$....  etc.

Now, (*) is not the definition of prime, but it is an equivalent property to "prime" in the integers.

Theorem. let $m$ be a positive integer, $m>1$. The following are equivalent:

If $m=ab$ with $a,b$ positive integers, then $a=1$ or $b=1$.
If $m=ab$ with $a,b$ positive integers, then $a=m$ or $b=m$.
If $m|ab$ with $a,b$ integers, then $m|a$ or $m|b$.
If $1\leq k\lt m$, $k$ an integer, then $\gcd(k,m)=1$.
For any integer $a$, $\gcd(a,m)=1$ or $\gcd(a,m)=m$.

And so the argument uses 1 or 2 to guarantee the existence of $a$ and $b$.

The implication "$2$ is a prime or a product of primes $\Rightarrow$ $3$ is a prime or a product of primes" holds because the consequent is true. For that matter, the implication "$6$ is a prime $\Rightarrow$ $7$ is a prime" is also true... because the consequent is true.

The wording is proper.  A prime (say $7$) is not a product of primes.  Since $1$ is not a prime, $7$ is not a product of primes--that is, one that requires the operation of multiplication.  What product could that possibly be??  (The "empty product" is explicitly not a product.) Taken alone, $7$ is not a product, any more than $28$ is a "product."

Such a weird attempt then makes every number prime: $28 = \prod\limits_{i\in I}$ where $I = \{ 28 \}$.  What an absurdity and mess!!!

Attempts to define a prime by some novel interpretation of product that has no argument, such as $7 = \prod\limits_{i\in I}$ where $I = \{ 7 \}$, have no justification.  The product function must take an argument, and this ad-hoc approach (which I've never seen in 40 years of working in mathematics) requires a redefinition of what a function is...  i.e., something that takes no argument.

From the Dictionary of mathematics:

product: The result of multiplication or any operations on mathematical objects deemed close enough in resemblance to multiplication. (Such as the scalar product and the vector product in vectors.)

The dictionary highlights the key property of "product" which is of course multiplication.  The ad-hoc attempts such as $7 = \prod\limits_{i\in I}$ where $I = \{ 7 \}$, involve no multiplication whatsoever, despite the symbol.

If you want to re-define prime in this way, you're forced to redefine product (and likely other foundational terms) in weird ways and admit that now every integer can be defined as a prime.

Good luck with that!


Derive the Trigonometric Functions How can the Sine Function be derived?

Given $\angle{A}$ as input, derive a function that would give $\frac{a}{c}$ as output.

$$$$
How can the Cosine Function be derived?

Given $\angle{A}$ as input, derive a function that would give $\frac{b}{c}$ as output.

$$$$

How can the Tangent Function be derived?

Given $\angle{A}$ as input, derive a function that would give $\frac{a}{b}$ as output.

$$$$

I am looking for either of the following:

The historical way to calculate the trigonometric functions as well as a proof that it works for a right-angled triangle
Any other way to calculate the trigonometric functions as well as a proof that it works for a right-angled triangle

In other words, an algorithm on its own would not be enough, you have to prove that it works for a right-angled triangle.
$$$$
Side note:

I am aware of the Taylor-series expansion of the trigonometric functions. $$$$

I am also aware of the exponential definition of the trigonometric functions.$$$$
 $$$$

If you could geometrically prove how any of these trigonometric identities work for a right-angled triangle, that would answer my question as well.

Another side note

I do not believe this question belongs in The History of Science and Mathematics-Stack Exchange. That forum focuses on where and when certain Mathematical concepts were created, which is not my question.

I'll just make a side note that I find it more natural to think about trigonometric functions as functions on unit circle. Once you know that $\theta \mapsto e^{i \theta}$ is surjective mapping of $\mathbb R$ in unit circle you can geometrically define $\cos \theta$ as projection of the point $e^{i \theta}$ on $x$ axis and $\sin \theta$ as projection on $y$ axis. This is way of thinking that I think is most natural and useful in practice (for example in calculations involving trig functions.) It's also geometric, which as I understand is whole point here. Once you have this definitions you can easily see the relation of these trig functions with right-angled triangles. This has great advantage that once you have that you can easily evaluate your functions in terms of exponentials (which means that you also automatically have Taylor expansion.) In this way you not only obtain a way to calculate these rations in right-angled triangle (i.e. answer original question) but also establish beautiful and profound correspondence between simple geometry and mathemathical analysis.

This is an expansion of a previous answer by Raj (an answer which was deleted), which I thought was note-worthy enough to keep.
But it is by no means a complete.

1) Assume there exists a function that satisfies the definition of $\sin(A)$. (Call this function $\sin(A)$). In other words, there exists a function that takes $\angle{A}$ as input and gives $\frac{a}{c}$ as output.

2) Assume the same for $\cos(A)$.

3) Prove that  $\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$

4) Prove that $\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$

5) Prove the half angle formula and the double angle formula for $cosA$ and $sinA$

6) Physically measure that $\sin30^\circ$ = $\frac{1}{2}$. Take this as your reference point.

7) Using the half-angle, double-angle, and sum formula, it would now be possible to calculate $\sin(A)$ using the reference point.

8) $\cos60^\circ$ = $\frac{1}{2}$ could be the reference point for the cosine function.


Probability of a chess game I have been asked to solve this problem and have literally NO idea where to begin. Any insight would be great!

Vlad is to play a 2-game chess match with Gary and wishes to maximize his chances of winning. To do this, he may select a strategy right before he plays each game: timid or bold. (Note: this means that Vlad may choose his strategy in the second game after he knows the result of the first game.)

(A few remarks on chess matches: a 2-game match means that the players play exactly two games, with a win counting 1, a draw counting 0.5, and a loss counting 0. The player with more points wins the match. Note that the match may end in a tie, if, for example, each player wins one game.)

Unfortunately, Gary is the superior player. If Vlad plays timidly, Gary will still win 10 percent of those games, and the rest will be draws. If Vlad plays boldly, Gary will win 5/9 of those games, and lose the rest.

Assume Vlad selects his strategies optimally, and find (a) the probability that he wins the match, (b) the probability that the match ends in a tie, and (c) the probability that Gary wins the match. (N.B. The problem requires that you determine Vlad's optimal approach to the match.)

Vlad has three decisions, which leads to eight possible global strategies:

What strategy to use in the first game
What strategy to use if he loses first
What strategy to use if he doesn't lose first.

One of the eight global strategies is to play timidly in all three situations.
Run through the eight global strategies.
In at least one of them, Vlad has a better chance of winning than Gary does.

The expected score for the bold strategy is higher, so the optimal strategy is for Vlad to play bold.

Under the timid strategy, each match ends D (for draw) or L (for Vlad loses) with probability $.9$ and $.1$, respectively. So the 2-game match ends DD (tie) with probability $.9^2 = .81$ and LL, DL, or LD (loss) with total probability $1-.81=.19$. The expected score is $.81\cdot .5 + .19\cdot 0= 0.405$.

Under the bold strategy, each match ends W (for Vlad wins) or L (for Vlad loses) with probability $4/9$ and $5/9$, respectively. So the 2-game match ends WW (win) with probability $(4/9)^2 =.1975$, ends WL or LW (tie) with total probability $40/81 \approx .4938$, and LL (loss) with probability $(5/9)^2 \approx .3086$. The expected score is about $.1975 \cdot 1 + .4938\cdot .5 + .3086\cdot 0= 0.4444$.


Why some people don't like proofs by contradiction Possible Duplicate:
  Are the “proofs by contradiction” weaker than other proofs?

I have been active on this site for two months and on a few occasions I noticed that some people judge contradiction proofs as being less direct(and less elegant) than proofs which do not use contradiction.

In my first year of college I gave headaches to my seminar teacher and my colleagues because I used most of the time proof by contradiction. For me it seemed so natural to argue by contradiction whenever I didn't have any idea to how to proceed in solving the problem directly. At least when you prove something by contradiction, you have a start point, a supplementary hypothesis on which you can develop the following arguments searching for a contradiction with the hypothesis or previous work (theorems, problems, etc.).

Therefore, my question is:

  Why do some consider that contradiction proofs are not that good as direct proofs?

I find it harder to read and proofread proofs by contradiction for the following reason: In an ordinary proof, one is trying to show $P \implies Q$. When I read it, I will have in my head a few examples of different things that obey $P$, and I'll check that each step in the proof is consistent with my examples. Hopefully, the last line of the proof will be "$Q$", and everything will work.

In a proof by contradiction, we start out assuming $P \wedge NOT(Q)$. If the theorem is true, there are no examples of things of things which obey $P$ and not $Q$, so I can't think of any examples.

Note that this problem does not arise in proofs which start out by assuming $NOT(Q)$ and deduce $NOT(P)$ at the end, since I can think of examples of things which obey $NOT(Q)$. I have taken to starting these proofs by writing "We establish the contrapositive, that $NOT(Q)$ implies $NOT(P)$" rather than my standard "This proof proceeds by contradiction. Assume...". My hope is that this will help my readers understand that there are still examples available for them to think about.

To prove something is impossible, you probably can't have a constructive proof and it's ease to understand the counterproof: You just assume that it's possible and derive a contradiction and you're done. Any other way to prove an impossibility?


Why do bell curves appear everywhere? Why do most probability graphs show a bell curve? I've been wondering why... Is it just something natural, like the fibonacci sequence?

Why do most probability graphs show a bell curve?

As you suspect, there is a natural tendency for distributions to be bell-shaped.

There are some distributions that are not bell-shaped at all. For example, the outcome of a roll of one fair die is a discrete uniform distribution:

By IkamusumeFan - Own work

This drawing was created with LibreOffice Draw, CC BY-SA 3.0, Link

The roll of one die is a pretty simple process. What about the sum of two dice? The Wizard of Odds illustrates:

Starting to look a little like a bell, right? What about the totals of three, or four dice? Wolfram MathWorld provides a nice illustration:

You can see where this is headed. Nature is full of complex processes. How tall are you? Well, it depends on genetics, nutrition, exercise, injuries, bone loss, and so many more things. The central limit theorem shows (see symplectomorphic's comment below) that when adding the sum of a large number of things together, the resulting distribution is not just any bell-looking curve, but specifically the normal distribution. Or for things with multiplicative combination, the log-normal distribution.

Why does this happen? mathreadler's answer hints it has to do with convolving distributions. The probability density function of a single die is a rectangular function (technically discrete, but let's pretend it's continuous). The sum of two rolls together is then the convolution of two rectangular functions.

By Convolution_of_box_signal_with_itself.gif: Brian Amberg
derivative work: Tinos (talk) - Convolution_of_box_signal_with_itself.gif, CC BY-SA 3.0, Link

Notice how the result (the black triangle) looks like the case of two dice above. If then convolve this triangle with another rectangle, you get three dice. The more times you do this, the closer the result gets to a normal distribution.

The probability density function of the normal distribution is a Gaussian function, which have some elegant properties:

A Gaussian convolved with a Gaussian is another Gaussian.
The product of two Gaussians is a Gaussian.
The Fourier transform of a Gaussian is a Gaussian.

From this you might intuitively see as things converge towards normal distributions, they "want" to stay as normal distributions since their "Gaussianess" is preserved under many operations.

Of course not everything is so simple as a single die roll, nor as complex as the determination of the height of a human. So there are a large number of distributions that look like a bell, but on careful examination aren't the normal distribution. Some of them exist in nature, and some find application as mathematical tools for some purpose. Looking through  Wikipedia's list of probability distributions you can see bell-like shapes are quite common, even if they aren't exactly the normal distribution.

But if you combine these two things:

The central limit theorem means the normal distribution is common, and
many distributions look like bells but aren't the normal distribution,

you might conclude most probability graphs show a bell curve.

A bell curve is a Gaussian distribution curve. It occurs very frequently. Like a pile of sand you slowly let fall from your hand on the beach, the shape of the pile, in one plane, is such a curve. It's everywhere from the beach to the atom.

But with all this theorising above, don't lose sight of the fact that mathematics is merely a human construct that reflects the way that matter and spacetime interact with each other and themselves. Mathematics does not dictate the way the Universe behaves; it merely reflects it.


How to show that an odd function always goes through zero? I have the standard definition of an odd-function from wikipedia:

  Again, let f(x) be a real-valued function of a real variable. Then f
  is odd if the following equation holds for all x and -x in the domain
  of
  $-f(x) = f(-x)$

Can anyone help me how to do this?
Do I have to show, that it converges to zero?

If $-f(x)=f(-x)$ for all $x$ in the domain for which $-x$ is in the domain, then if $0$ is in the domain
$$-f(0)=f(-0)=f(0),$$
which means $2f(0)=0$, and so $f(0)=0$.

The way I think of this property is through the intermediate value theorem. Assume that the odd function $f$, is continuous. Then choose some point $ a > 0 $. Suppose without loss of generality that $ f(a) > 0 $. Then since $ f(-a) = -f(a) <0 $, by the IVT, there must be a point $ b $ in $ (-a,a) $ such that $ f(b) = 0 $. Note that this only works if $ f $ is continuous.


Counting the number of subsets of a set of 2n elements satisfying some conditions. Let $X =\{v_1, v_2,\cdots, v_n, v_{n+1},\cdots, v_{2n}\}$ be a set of $2n$ elements. I want to find the number of subsets of $X$ with $n$ elements such that both $v_i$ and $v_{n+i} $ are not together in this subset.

I tried some small cases with $n=2,3,4$ and think it should be $2^n$ but I would like to prove it rigorously. I'm not sure how to use induction here. Is there a good combinatorial argument that can be used? Is my guess even correct?

Thank you.

You have set the requirement for subsets that do not contain both $v_i$ and $v_{n+i} $.

However, in order for such a subset to have $n$ elements, it is clear that it must contain one of these.

As such, every subset can be generated by combining all those choices between $v_i$ and $v_{n+i} $. Each of those choices is binary, and there are $n$ choices, giving a total of $2^n$ alternative subsets that meet the conditions.

WLOG, $\{v_1,..v_2n\}=\{0,...,2n-1\}=2n$ (by the Von-Neumann ordinal convention of $k=\{0,..,k-1\}$).

Our chosen set is determined by its intersection with $n$ which is an arbitrarily chosen element of the powerset $P(n)$ with cardinality $2^n$.


Why the sum of residuals equals 0 when we do a sample regression by OLS? That's my question, I have looking round online and people post a formula by they don't explain the formula. Could anyone please give me a hand with that ? cheers

If the OLS regression contains a constant term, i.e. if in the regressor matrix there is a regressor of a series of ones, then the sum of residuals is exactly equal to zero, as a matter of algebra.

For the simple regression,
specify the regression model
$$y_i = a +bx_i + u_i\,,\; i=1,...,n$$

Then the OLS estimator $(\hat a, \hat b)$ minimizes the sum of squared residuals, i.e.

$$(\hat a, \hat b) : \sum_{i=1}^n(y_i - \hat a - \hat bx_i)^2 = \min$$

For the OLS estimator to be the argmin of the objective function, it must be the case as a necessary condition, that the first partial derivatives with respect to $a$ and $b$, evaluated at $(\hat a, \hat b)$ equal zero.  For our result, we need only consider the partial w.r.t. $a$:

$$\frac {\partial}{\partial a} \sum_{i=1}^n(y_i -  a - bx_i)^2 \Big |_{(\hat a, \hat b)} = 0 \Rightarrow -2\sum_{i=1}^n(y_i -  \hat a - \hat bx_i) = 0 $$

But $y_i -  \hat a - \hat bx_i = \hat u_i$, i.e. is equal to the residual, so we have that

$$\sum_{i=1}^n(y_i -  \hat a - \hat bx_i) = \sum_{i=1}^n\hat u_i = 0 $$

The above also implies that if the regression specification does not include a constant term, then the sum of residuals will not, in general, be zero.

For the multiple regression,
let $\mathbf X$ be the $n \times k$ matrix containing the regressors, $\hat {\mathbf u}$ the residual vector and $\mathbf y$ the dependent variable vector. Let $\mathbf M  = I_n-\mathbf X(\mathbf X'\mathbf X)^{-1}\mathbf X'$ be the "residual-maker" matrix, called thus because we have

$$\hat {\mathbf u} = \mathbf M\mathbf y$$

It is easily verified that $\mathbf M \mathbf X = \mathbf 0$. Also $\mathbf M$ is idempotent and symmetric.

Now, let $\mathbf i$ be a column vector of ones. Then the sum of residuals is

$$\sum_{i=1}^n \hat u_i = \mathbf i'\hat {\mathbf u} =\mathbf i'\mathbf M\mathbf y = \mathbf i'\mathbf M'\mathbf y = (\mathbf M\mathbf i)'\mathbf y = \mathbf 0' \mathbf y = \mathbf 0$$

So we need the regressor matrix to contain a series of ones, so that we get $\mathbf M\mathbf i = \mathbf 0$.

take the estimated values from the line of best fit and use these y values to subtract from the original y values then add them up. if it is a good line of best fit then it should approach zero, but bad lines of best fit will be much less or more than zero


Prove that $9\mid (4^n+15n-1)$ for all $n\in\mathbb N$ First of all I would like to thank you for all the help you've given me so far.
Once again, I'm having some issues with a typical exam problem about divisibility. The problem says that:

Prove that $\forall n \in \mathbb{N}, \ 9\mid4^n + 15n -1$

I've tried using induction, but that didn’t work. I've tried saying that:

$4^n + 15n-1 \equiv 0 \pmod{9}$. Therefore, I want to prove that $4^{n+1} + 15(n+1) -1 \equiv 0 \pmod{9}$.
I've prooved for $n=1$, it's $18\equiv 0 \pmod{9}$, which is OK.

But for the inductive step, I get:

$4\cdot4^n + 15n+15-1 \equiv 0 \pmod{9}$

And from there, I don't know where to replace my inductive hypothesis, and therefore, that's why I think induction is not the correct tool to use here. I guess I might use some tools of congruence or divisibility, but I'm not sure which ones.
I do realize that all $n\in \mathbb{N}/ \ 3 \ |\ n \Rightarrow 4^n \equiv 1 \pmod{9} \text{ and } 15n \equiv 0 \pmod{9}$. In that case, where 3 divides n, then I have prove that $4^n + 15n-1 \equiv 0 \pmod{9}$. But I don't know what to do with other natural numbers that are not divisible by 4, that is, all $n \in \mathbb{N} / n \equiv 1 \pmod{3} \text{ or } n \equiv 2 \pmod{3}$.
Any ideas? Thanks in advance!

By the Inductive Hypothesis, $4^n + 15n -1 \equiv 0$ so $4^n \equiv 1-15n$ and thus
$$
4^{n+1}+15(n+1)-1 = 4 \cdot 4^n + 15n + 14 \equiv 4 \cdot (1-15n) + 15n + 14 = 18 -45n \equiv 0
$$
since both $18$ and $45$ are divisible by 9.


prove that $f^{-1} = f*$ as follow edit:
Let $f: A \to B$. Let $f*$ be the inverse relation, i.e.
\begin{equation*}
f* = \{(y,x) \in B \times A \mid f(x)=y \}.
\end{equation*}
Show that if $f* : B \to A$ is a function,
then $f^{-1} = f*$.
Attempt:
Let $f*: B \to A$ be a function.
Then, $f$ is bijective and hence $f$ is invertible, i.e. $f$ have an inverse, say $f^{-1}$.
It is clear that $f^{-1} \circ f = i_A$
and $f \circ f* = i_B$ where $i$ is the identity function. Hence,
\begin{equation*}
f^{-1} = f^{-1} \circ i_B = f^{-1} \circ (f \circ f*) = (f^{-1} \circ f) \circ f* = i_A \circ f* = f*.
\end{equation*}
Thus, $f^{-1} = f$.
Is the above correct?

First, suppose that $f*$ is a function. That means that each for each $y \in B, \exists!x \in A$ such that $f^*(y)=x.$ Then $(f^* \circ f)(x) = f^*(f(x)) = f^*(y)=x,$ and similarly, $(f \circ f^*)(y) =f(f^*(y))= f(x)=y$, so $f*$ is the inverse of $f$. Or perhaps I should say $f*$ is an inverse of $f$, since you've also been tasked with proving the uniqueness of the inverse.
To prove that the inverse of f is unique, suppose that f has two inverses, g and h. We need to show that g=h. Since g is the inverse of f, that means $g∘f=id_A$  and $f∘h=id_B$. Then $g=g∘id_B=g∘(f∘h)=(g∘f)∘h=id_A∘h=h$.


What are the leading alternative foundations for mathematics? I know that all of mathematics can be recast in terms of set theory. There are multiple choices for this set theory (some form of ZFC, NBG, NF, etc.), and so multiple possible set theoretic foundations for mathematics.

I am told (though I know so little about the subject that I cannot verify) that category theory also provides us with a foundation for mathematics.

What are the other well-known candidates for the foundations of mathematics, if any?

There is much to be said about any of the foundational systems below, so the following is necessarily only a very rough sketch on their main ideas and differences.

The category theoretic foundation you where told about is probably Lawvere's Elementary Theory of the Category of Sets (ETCS). Like classical set theory, it is a first order theory, but the basic concepts/ideas it builds upon are those of an object/set and a morphism between such, instead of sets and elements of sets as in classical set theory.

For example, see http://ncatlab.org/nlab/show/ETCS and http://ncatlab.org/nlab/show/Trimble+on+ETCS+I A nice introduction to elementary toposes, of which ETCS is the first-order axiomatization, is given in Maclane-Moerdijk, Sheaves in Geometry and Logic

Note, however, that like in classical set theory, we have a clear separation between the underlying logic and the theory build upon of it.

In type theory, this distinction is blurred: you are dealing with syntactic entities of types and terms, which may either be thought of as the domains of discourse and their elements, or as propositions about these and their proofs. Together with decidability of judgements "This term belongs to that type" which can be (and has been) implemented on a computer, this gives rise to a proof-sensitive and machine-checkable (and even machine-assisted) foundational formal system suitable for the formalization of both classical and intuitionistic reasoning in both first-order and higher-order logic.

For example, see http://ncatlab.org/nlab/show/type+theory and the references therein. In particular, you might want to have a look at homotopy type theory http://ncatlab.org/nlab/show/homotopy+type+theory, a type theoretic foundation  of mathematics under development building on the concepts of space and homotopy as the basic concepts.

Besides ZFC, I find the work of Brian Rotman interesting: A more behaviouristic, linguistic, physical approach:

"Mathematics as Sign"

Writing, Imagining, Counting
http://www.sup.org/books/title/?id=415


Prove by test integrabilty that: $\displaystyle\int_0^{+\infty} \left|\frac{\ln x}{x^2-1}\right|\, dx < +\infty$ how we prove that the function $\dfrac{\ln x}{x^2-1}$ is in $L^1(]0,+\infty[)$ using the tests of convegence integral? I calculate is and i found that $$\displaystyle\int_0^{+\infty} \left|\frac{\ln x}{x^2-1}\right|\, dx < +\infty$$, but i want to prove that this function is in $L^1(]0,+\infty[)$ using the tests of integrability of integral.

Thank you in advance

At infinity use

$$\lim_{x\to\infty} \frac{x^{3/2}\ln x}{x^2-1} = \lim_{x\to\infty} \frac{x^{2}}{x^2-1}\cdot\frac{\ln x}{x^{1/2}} =\lim_{x\to\infty} \frac{1}{2}\cdot\frac{\ln x^{1/2}}{x^{1/2}} = 0$$
That is there exists $A>0$ such that $x>A$ implies

$$ |\frac{\ln x}{x^2-1}|\le \frac{1}{x^{3/2}}$$

  At $x=0$ use

$$\lim_{x\to0} \frac{x^{1/2}\ln x}{x^2-1} = \frac12 \lim_{x\to0}x^{1/2}\ln x^{1/2} = 0$$
 That is there exists $a>0$ such that $0<x<a$ implies

$$ |\frac{\ln x}{x^2-1}|\le \frac{1}{x^{1/2}}$$

  At $x=1$ there singularity is fake since,

$$\lim_{x\to 1} \frac{\ln x}{x-1} = (\ln x)'|_{x=1} =1$$

Notice that $|\frac{\ln x}{x^2-1}| \leq \min\{x^{1/2}, x^{3/2}\}$ and that the RHS is integrable because $p < 1$, $p > 1$ respectively. So by the comparison test, the integral in question is finite, hence $\frac{\ln x}{x^2-1} \in L^1$.


Find the complex number $z$ for which the following is true: $|z-1|=|z|$ and $\arg(z^3)=\frac{\pi}{3}+ \arg(z)$ Find the complex number $z$ for which the following is true: $|z-1|=|z|$ and $\arg(z^3)=\frac{\pi}{3}+ \arg(z).$ The answer is given in, but in the very beginning, I am having trouble seeing how this deduction is made:

Let's write $z=r\cos\varphi+ir\sin\varphi$.

The first condition gives:
$$(r\cos\varphi-1)^2+(r\sin\varphi)^2=r^2$$ How does the first condition equivalent to this, is my question!

Writing $z = r e^{i \theta}$ we have from the second condition that $3 \theta = \frac{\pi}{3} + \theta$ so $\theta = \frac{\pi}{6}$.

Then use your condition to deduce $r$.


Solve a Diophantine equation with three variables How can I find all integral solutions of
$$4xy-x-y+20=z^2$$
I know two solutions are
$(x,y,z)=(-1,-3,\pm6)$. Is there a way to find new, or all integral solutions from this one?

This is not exhaustive. It is only a way given one solution, generate a family of solutions.

Let $\Lambda(x,y,z)$ be the $2\times 2$ symmetric matrices $\begin{bmatrix}4x - 1 & 2z \\ 2z & 4y - 1\end{bmatrix}$.

The equation at hand can be rewritten as
$$\det \Lambda(x,y,z) = (4x-1)(4y-1)-(2z)^2 = -79\tag{*1}$$

For any $2\times 2$ matrix $P$ with integer coefficients and $\det P = \pm 1$, $P^T \Lambda(x,y,z) P$ will be a $2 \times 2$ symmetric matrix with integer coefficients. It has same determinant as $\Lambda(x,y,z)$.
If $(x,y,z)$ is a solution of  $(*1)$ and we can find integers $x',y',z'$ such that $P^T \Lambda(x,y,z) P = \Lambda(x',y',z')$, then $(x',y',z')$ will be another solution for $(*1)$.

This prompts us to look for suitable $P$ from $SL(2,\mathbb{Z})$.
$SL(2,\mathbb{Z})$ are generated by following matrices:
$$
L = \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix},\quad
U = L^T = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}
\quad\text{ and }\quad
J = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$$
It is easy to check $J^T \Lambda(x,y,z) J = \Lambda( y, x, -z )$, so $J$ doesn't give us anything interesting.
However, $L$ and $U$ don't disappoint us. We find

$$\begin{align}
U^{2k} \Lambda(x,y,z) L^{2k} &= \Lambda(x + 2kz + k^2(4y-1), y, z + k(4y-1))\\
L^{2k} \Lambda(x,y,z) U^{2k} &= \Lambda(x, y+2kz + k^2(4x-1), z + k(4x-1))
\end{align}
$$

This means if $(x,y,z)$ is an integral solution for $(*1)$, then for any integer $k$, Both $$\begin{align}
& (x + 2kz + k^2(4y-1), y, z + k(4y-1))\\
\text{ and }\quad&(x, y+2kz + k^2(4x-1), z + k(4x-1))
\end{align}
$$ are solutions of $(*1)$.

For example, if one start from the solution $(-1,-3,6)$, we immediately obtain
following two parametric families of solution:

$$(-1 + 12k -13k^2, -3, 6-13k)\quad\text{ and }\quad (-1, -3+12k-5k^2, 6-5k)$$

More solutions can be constructed in this manner. However, I doubt this type of
construction cover all possible solution. I hope this is at least a start.

Solving for $y$, we get
$$ y = \frac{z^2+x-20}{4x-1}$$
That is, we want $z^2 + x - 20$ to be divisible by $4x-1$.  If $m = 4x-1$,
$z^2 + x - 20 = ((2 z)^2 + m - 79)/4$, so what you need is that $79$ is a square mod $m$, where $m \equiv -1 \mod 4$.

The first few $m$ are
$ 3, 7, 15, 27, 35, 39, 43, 47, 59, 63$, corresponding to $x = 1, 2, 4, 7, 9, 10, 11, 12, 15, 16$.  Then for example, for $x = 16$, $m = 63$, $2z \equiv \pm 4$ or $\pm 31 \mod 63$, making $z \equiv \pm 2$ or $\pm 16 \mod 63$.  If you take $z = 16$, $y = (16^2 + 16 - 20)/63 = 4$.

In particular, if $m$ is a prime $\equiv -1 \mod 4$, by quadratic reciprocity $79$ is a square mod $m$ if and only if $m$ is $79$ or is not a square mod $79$.  By Dirichlet's theorem we get infinitely many primes of this form.


integrate $\int \frac{1}{\sqrt{x^2 - 1}} \, dx$ I am solving an ODE

$$y^2 \, dx + \left(x\sqrt{y^2 - x^2} - xy\right)dy=0$$

I let $y=xu$ and did some algebra, and I ended up with this

$$\ln(x)+C=\int \frac{1}{\sqrt{u^2 - 1}} \, du - \int \frac{1}{u} \, du$$

I don't know how to solve the first right hand side integral. I tried to let $u^2 - 1 = t$, so it becomes

$$\int\frac{1}{\sqrt{u^2 - 1}}=\int\frac{1}{\sqrt{t}}\frac{1}{2u} \, dt = \int \frac{1}{\sqrt{t}} \frac{1}{2} \frac{1}{\sqrt{t-1}} \, dt,$$

which doesn't seem to work. The correct answer seems to be

$$\ln \left(\sqrt{x^2 - 1} - x\right)$$

But how do we get this?

Use a substitution of $u=\sin(t)$, differentiate to get $du$ then sub into that 2nd integral.

Then evaluate that integral & should progress towards the answer.


Proof that finite group contains an element of prime order I tried to prove the following claim but it seemed a bit too easy:

  Let $G$ be a finite group. Then $G$ contains an element of prime order.

Please could someone tell me if my proof is correct or if I'm missing something?

Let $g$ be any non-identity element of $G$. Let $p$ be any prime factor of $|g|$. Then $g^{|g|\over p}$ has prime order $p$. $\Box$

Yes. This is right. To ensure it:

If $|g|$ is a prime, then we are done. Let $p$ be a prime factor of $|g|$, so that $|g|>p$.

Let $h=g^{\frac{|g|}{p}}$. Then $h^p=g^{|g|}=1$, which implies that order of $h$ divides $p$, so it is $1$ or $p$.

If order of $h$ is $1$ then this means, $h=1$ i.e. $g^{|g|/p}=1$, i.e. $|g|\leq |g|/p$, this is contradiction.

No the proof is not correct. It is possible that $g^{\frac{|g|}{p}}=e$. The theorem you are stating is commonly called Cauchy's theorem.  There is an elementary proof here using the class equation.


Is this inequality true? $ (x + y)^{\alpha} < x^{\alpha} + y^{\alpha} $, for positive $x$ & $y$, and for $0 < \alpha < 1$ If $0 < \alpha < 1$, then

  $$ (x + y)^{\alpha} < x^{\alpha} + y^{\alpha} $$

  for $x$, $y$ positive.

Is this inequality true in general?

I tried using Young's Inequality: For $z,t > 0$, and for $n$, $m$ such that $n+m=1$, then

$$ z^n + t^m \leq nz + mt $$

So, using this we have

$$ (x+y)^{\alpha} \cdot 1^{1 - \alpha} \leq \alpha(x+y) + (1-\alpha) = \alpha x + \alpha y + (1-\alpha)$$

which is not as tight as I want.

Hint:  try rewriting the inequality as
$$x^\alpha \left( 1 + t \right)^\alpha \le x^\alpha \left( 1 + t^\alpha \right)$$
where $t = y/x$.

Of course the case $\alpha=0$ or $\alpha=1$ is trivial, so suppose $0<\alpha<1$. Dividing both sides by $(x+y)^\alpha$ reduces the proof to the case $x+y=1$ (to see this, just do the substitution $x'=\frac{x}{x+y}, y'=\frac{y}{x+y}$). So suppose $x+y=1$. Then $0<x<1$, $0<y<1$, implying $x<x^\alpha$ and $y<y^\alpha$. Hence $$(x+y)^\alpha=1=x+y<x^\alpha+y^\alpha.$$


Permutation of Dresses Possible Duplicate:
  In how many ways we can put $r$ distinct objects into $n$  baskets?

How many number of ways I can wear four dresses for n days without wearing the same dress for two consecutive days  and the dress on first and nth day should also not be same. Repetition of dresses is allowed.

For $n=3$   It'll be  $4\times3\times2 = 24$.

Can someone please tell me what will be the answer of $n\ge5$ and also how to solve such questions where objects to arrange are lesser than spaces.

A sample scenario for

$n=5$  A1,A2,A1,A2,A3

$n=6$  A1,A2,A1,A2,A1,A2

Edited:
for the dresses question.

$4*3^{n-1} -9$


Show that V is a subspace of M2x2 Matrices and Determine a basis A bit of information to start us off: Let V denote the set of all 2x2 matrices with equal column sums.

Show $V$ is a subspace of $M_{2\times 2}$ matrices:

and....

Determine a basis for $V$:

So for the first bit... $M$ is the set of all $2\times 2$ matrices, I used the definition of subspaces... A is a subset of B if for every element "a" in A, "a" is also in B. Am I able to then say, since $V$ is the set of all $2\times 2$ matrices with equal column sums, "$a$" being an element of $V$. Then "$a$" is a $2\times 2$ matrix. Since "$a$" is $2\times 2$ matrix, "$a$" is in $M_{2\times 2}$ as well. Therefore the set of all $2\times 2$ matrices with equal column sums is a subset of $M_{2\times2}$?

What I am really unsure of is how to determine a basis. I'm used to being given a spanning set and working from there. Thanks!

Note that a $2\times2$ matrix A has equal column sums iff $e^T A (e_1-e_2) = 0$, where $e=(1,1)^T$. The operator $L(A) = e^T A (e_1-e_2)$ is a linear function of $A$, hence $V = \ker L$, and so is a subspace (The kernel of a linear operator is automatically a subspace).

Since $\dim \mathbb{R}^{2 \times 2} = 4 = \dim \ker A + \dim {\cal R} A$, we see that $\dim V = \dim \ker A = 4-1 =3$.

Intuitively, note that if you pick 3 elements of a $2 \times 2$ matrix, you can always choose the fourth to satisfy the equation.

So, let the $A_{22}$ element be the 'special' element. A standard basis for $\mathbb{R}^{2 \times 2}$ is $e_i e_j^T$ for $i,j = 1,2$. So, pick one of these and figure out what the $22$ element must be to satisfy $L(A) = 0$.

Working through the computation gives $L(e_i e_j^T-\alpha e_2 e_2^T) = \delta_{1j}-\delta_{2j} + \alpha$, hence $\alpha = \begin{cases} -1 & j=1 \\
+1 & j = 2\end{cases}$. This construction produces three non-zero matrices that lie in $V$ by construction. It is easy to show that these are linearly independent, hence the form a basis.

More simplistically...

You must answer these three questions:

is $0$ (in this case, this means the zero matrix) in $V$?
if $A$ and $B$ are two matrices in V, is $A + B$ also in $V$?
if $r \in \mathbb{R}$ (or whatever field it is referring to, probably $\mathbb{R}$) and $A \in V$, then is $rA$ also in $V$?

The answer to the first question is yes (what are the column sums of the zero matrix?)

If the column sum of $A$ is $\alpha$ and the column sum of $B$ called $\beta$, what will the column sum of $A + B$ be?

Similarly, what will the column sum of $rA$ be?

For these questions, the "show it is a subspace" part is the easier part. Once you've got that, maybe try looking at some examples in your note for the basis part and try to piece it together from the other answer.


Selection of at least two colours Let there be three sketch pens of different colours. A regular pentagon is to be drawn using these colours where exactly one of the colours is used to draw a side and at least two colours are used to make the complete pentagon. In how many ways can this be done?
My attempt:
Each side can be drawn by using any of the three colours , so $3$ choices are available for each side. There are $5$ sides in total so they can be drawn in $3^{5}$ ways and the number of ways in which we can use only one colour is $3$ so the required number of ways $= 3^5 - 3$.
Did i miss anything?

Yes! A pentagon has 5 sides.  You say "they can be drawn in $3^5$ ways" and then include another way to color some, apparently, sixth way side!  What I would do is argue that there are $3^5$ ways to color the five sides but three of those ways include all five sides the same so there are $3^56- 3= 3(3^4-1)= 3(80)= 240 ways to do this.

since it is regular pentagon so anticlockwise and clockwise direction would be same so divide your answer by $5$ that is number of sides.


Problem based on area of triangle In the figure, E,C and F are the mid points of AB, BD and ED respectively. Prove that: $8\triangle CEF=\triangle ABD$

From the given,

$ED$ is the median of $\triangle ABD$
So, $\triangle AED=\triangle BED$

Also, by mid point Theorem
$EC||AD$ and $CF||AB$.

Now, what should I do next?

Let (area of) $\triangle ECF = x$.

Then $\triangle FCD = \triangle ECF = x$ (same base, same height - I'll just call this SBSH for brevity).

So $\triangle ECD = 2x$ and $\triangle EBC = \triangle ECD = 2x$ (SBSH)

Hence $\triangle EBD = 4x$ and $\triangle AED = \triangle EBD = 4x$ (SBSH)

Therefore $\triangle ABD = 8x = 8\triangle CEF \ \ (QED)$

ED is the median of △(ABD) and therefore △(AED)=△(BED)=1/2 x △(ABD).
EC is the median of △(BED) and therefore △(EBC)=△(ECD)=1/2 x △(BED).
CF is the median of △(ECD) and therefore △(CEF)=△(CDF)=1/2 x △(ECD).

From the three equations above △(CEF)=1/8 x △(ABD)
or 8 x △(CEF)=△(ABD)


$l^2$ with the $1$-norm is Banach? I have to show that $l^2$ is Banach when endowed with the $1$-norm. To this regard, I take a Cauchy sequence $(x_n)_{n \in \mathbb{N}}$. For each component, I have
$$
|x_{n,k}-x_{m,k}|=(|x_{n,k}-x_{m,k}|)^{\frac{1}{2}} \le (\sum\limits_{j=0}^\infty |x_{n,j}-x_{m,j}|)^{\frac{1}{2}}=||x_n-x_m||_2<\infty
$$
So, $x_n \rightarrow x$ for some sequence $x$. In order to prove that $x \in l^2$, I just observe that $x_n$ converges to a unique limit in $l^2$, and this forces $x$ to be such a limit.
I have only a doubt. If $(x_n)_{n \in \mathbb{N}} \in l^2$ is Cauchy with respect to the distance induced by $1$-norm, is it also Cauchy with respect to distance induced by $2$-norm?
Is my argument correct?

The $l^1$ norm is not a norm on $l^2$, because there are elements in $l^2$ that are not in $l^1$, e.g.,
$$
(1,\frac12, \frac13,\dots).
$$
The Cauchy sequence property is not a problem since $\|x\|_{l^2} \le \|x\|_{l^1}$ for all $x\in l^1$:
$$
\sum_{i=1}^N x_i^2 \le \left(\sum_{i=1}^N |x_i| \right)^2 \le \|x\|_{l^1}^2,
$$
which is true for all $N$.

The norms are equivalent down to a constant, see more details here, and from there you have what you need, the "Cauchy"-ness of the sequence remains.


propagation of measurement uncertainty in numerical integration (trapezoidal rule) Question: How does one calculate the propagated measurement uncertainty to a value that was determined through integration by use of the trapezoidal rule?

My problem:  I measure a value $p$ which has some associated absolute uncertainty $\delta p$, i.e., I state the result of my measurement as the best estimate of the quantity $p$ and the range within which I am confident the quantity $p$ lies: $p\pm \delta p$.  For this value of $p$ I want to calculate the value $m(p)$, which is calculated as
$$\tag{1} m(p)=\int_{p_o}^p \frac{\rho(p)}{\mu(p)} \ dp$$

For physical context, $p$ is fluid pressure, $\rho$ is fluid density, and $\mu$ is fluid viscosity.  The values of density and viscosity are pressure dependent as indicated in Eqn(1). For the definite integral in Eqn(1), the integration is from some "base pressure" $p_o$ to the pressure of interest $p$ (where $p>p_o$).

What I have tried: The trapezoidal rule integration formula is
$$\tag{2} \int_a^b f(x) \ dx \approx \frac{1}{2} \sum_{i=1}^n (x_i-x_{i-1})(y_i+y_{i-1})$$

Combining (1) and (2) we obtain
$$\tag{3} m(p)=\frac{1}{2} \sum_{i=1}^n (p_i-p_{i-1})[(\rho /\mu)_i+(\rho /\mu)_{i-1}]$$

The general equation for the propagation of uncertainty is
$$\tag{4} \delta q = \sqrt{\left(\frac{\partial q}{\partial x}\delta x\right)^2+...+\left(\frac{\partial q}{\partial z}\delta z\right)^2}$$
where $x,...,z$ are quantities measured with uncertainties $\delta x,...,\delta z$ and the measured values are used to compute the function $q(x,...,z)$. The uncertainties in $x,...,z$ are independent and random.

As we calculate the partial sums of Eqn(3) we note that the values used for $p_i$ and $p_{i-1}$ have no uncertainty. So our focus is on the uncertainties of calculated values for $(\rho /\mu)$. Denoting the quotient $(\rho /\mu)$ as $R$, the uncertainty in $R$, per Eqn(4), is:
$$\tag{5} \delta R = \sqrt{\left(\frac{\partial R}{\partial \rho}\delta \rho \right)^2+\left(\frac{\partial R}{\partial \mu}\delta \mu \right)^2}$$

Thus, we have the uncertainty in the partial sum $S$ computed as
$$\tag{6} \delta S =\sqrt{\delta R_i^2 + \delta R_{i-1}^2}$$

Per Eqn(3) we will need to perform this computation for each value of $i$, from $i=1$ to $i=n$, and then compute the summation of all these values.  And per Eqn(4) we would calculate the uncertainty in $m(p)$ as
$$\tag{7} \delta m(p) = \frac{1}{2} \sqrt{\sum_{i=1}^n (\delta S^2)_i}$$

Not sure if this all makes sense and therefore why I have asked my question: How does one calculate the propagated measurement uncertainty to a value that was determined through integration by use of the trapezoidal rule?

You're logic seems solid except for in between (4) and (5). You state that $p_i$ doesn't have any error but in the initial problem statement it seems that each $p_i$ has an uncertainty $\delta p_i$.
This link might help:  https://stats.stackexchange.com/questions/214850/propagate-errors-in-measured-points-to-simpsons-numerical-integral


probablity of random pick up three points inside a regular triangle which form a triangle and contain the center what is  the probablity of random pick up three points inside a regular triangle
which form a triangle and contain the center of the regualr triangle

the three points are randomly picked within the regular triangle and then form a new triangle
and the new triangle have to contain the center of the original regular triangle
what is the probability

Let $R$ be a convex region of the plane with unit area, and choose $n\ge 3$ points at random from the region.  We will derive a general expression for the probability $P_{n}$ that the convex hull of these points contains a particular point $\tau \in R$ (which we will take as the origin of our coordinate system). We can then evaluate that expression for the particular question posed (in which $n=3$, $R$ is an equilateral triangle, and $\tau$ is its center).

Suppose the hull does not contain $\tau$, and assume that $\tau$ is not collinear with any two of the points (this is true with probability 1).  Then we can draw a ray from the center point through one of the $n$ points (the "leftmost" point) such that the remaining $n-1$ points are to the right of the ray.  (Note that this condition is both necessary and sufficient.  Note also that if the convex hull does contain $\tau$, then there is no "leftmost" point.)  We can calculate the desired probability by integrating over all such configurations.  In particular, take $f(\varphi)$ to be the area of the subregion that lies to the right of the radial ray at angle $\varphi$, and take $r(\varphi)$ to be the distance from $\tau$ to the boundary of $R$ along that ray.  Then the probability that a particular point will lie in the angular interval $[\varphi, \varphi+d\varphi]$ is $da = \frac{1}{2}r(\varphi)^2 d\varphi$, and the probability that each of the other $n-1$ points will lie to its right is $f(\varphi)$.  Since there are $n$ points to choose as the leftmost, the overall probability of not capturing $\tau$ in the convex hull is
$$
1-P_{n} = n \int f(\varphi)^{n-1} da = n \int_{-\pi}^{\pi} \frac{1}{2} f(\varphi)^{n-1} r(\varphi)^2 d\varphi,
$$
We can make a few observations at this point.  First, if the region $R$ is symmetric under reflection through $\tau$, then $f(\varphi)$ is identically $1/2$: each radial line splits the region in half.  The result then is
$$
P_{n} = 1 - \frac{n}{2^{n-1}},
$$
which gives the known result $P_{3} = 1/4$ for polygons with an even number of sides.  Second, if the region $R$ instead has bilateral symmetry across the $x$-axis, then $r(\varphi)$ is an even function, and $f(\varphi) = 1/2 + f^{-}(\varphi)$, where $f^{-}$ is an odd function.  Then terms in the integral with odd powers of $f^{-}$ must vanish: in particular, we have
$$
P_{3} = \frac{1}{4} - 3 \int{{f^{-}(\varphi)}^2 da} = \frac{1}{4} - 3\int_{-\pi}^{\pi}\frac{1}{2}{f^{-}(\varphi)}^2 r(\varphi)^2 d\phi \le \frac{1}{4},
$$
and interestingly the relation $P_{4} = 2P_{3}$ continues to hold.

It remains to evaluate $f^{-}(\varphi)$, $r(\varphi)$, and the resulting integral in the case of an equilateral triangle.  The figure above shows an equilateral triangle with vertices at $(-x,0)$ and $(x/2, \pm x\sqrt{3}/2)$.  For $\varphi \in [0,\pi/3]$, the function $f^{-}(\varphi)$ is the area of the blue triangle minus the area of the red triangle,
$$
f^{-}(\varphi) = \frac{x^2}{8}\tan\varphi - \frac{x^2}{2}\frac{1}{\cot\varphi + \cot\frac{\pi}{6}} = \frac{x^2}{8}\tan\varphi \left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right);
$$
and the radius $r(\varphi) = \frac{1}{2}x\sec\varphi$ over the same domain. By symmetry, the full integral is just six times its value over $[0,\pi/3]$:
$$
\begin{eqnarray}
P_3 &=& \frac{1}{4} - 18\int_{0}^{\pi/3}\frac{x^6}{512}\tan^2\varphi\left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right)^2 \sec^2\varphi d\varphi \\
&=& \frac{1}{4} - \frac{1}{36\sqrt{3}}\int_{0}^{\pi/3}\tan^2\varphi\left(1 - \frac{4}{1 + \sqrt{3}\tan\varphi}\right)^2 \sec^2\varphi d\varphi \\
&=& \frac{1}{4} - \frac{1}{12}\int_{0}^{1} u^2\left(1 - \frac{4}{1+3u}\right)^2 du,
\end{eqnarray}
$$
where we have used $x = 2/\sqrt{3\sqrt{3}}$ (in order for the triangle to have unit area) and introduced the transformed variable $u = \tan\varphi / \sqrt{3}$ (so $du = (\sec^2\varphi / \sqrt{3}) d\varphi$).  The final integral is a straightforward exercise for the reader; the result is
$$
P_3 = \frac{1}{4} - \frac{1}{324}\left(57 - 80\ln{2}\right) = \frac{2}{81}\left(3 + 10\ln{2}\right) = 0.2452215...,
$$
in agreement with the brute-force and numerical results already given.  This result can be generalized easily to the regular $m$-gon for any odd $m$ (changing only the values of some constants), and to the case where $n>3$ (complicating the final integral).

If you ask for an beautiful solution. Then you should totally check this out.
Any point within the triangle can represent how the stick is broken into three pieces. For instance, if point P is selected, that represents that the stick is being broken into pieces with length a, b, and c. By the altitude theorem, it is known that a + b + c = x. Here x is the altitude
In this example it would not be possible to make a triangle, because a + b < c.
To find the probability that the two smaller sides will sum to be larger than the third side it is helpful to divide the equilateral triangle into four smaller equilateral triangles, by connecting the midpoints of the sides.
As long as point P is selected in the middle triangle, it will be possible to create a triangle from the three pieces.
The probability that randomly selected P is in the middle triangle is 1/4 so the probability that a triangle can be formed.
This is a beautiful Geometric approach


How I can prove that the derivative of $\sin x$ is $\cos x$? I want to prove that the derivative of $\sin x$ is $\cos x$, but how? Is this proposition true?

$$(\sin x) ^{\prime}=\cos x \Longleftrightarrow \sin x ^{2}+\cos x ^{2}=1 , \forall x$$
Thank you.

$$f(x)=\sin x$$
Using first principle
$$f'(x)=\lim_{h\to 0}\frac{\sin(x+h) -\sin x}{h}$$
$$=\lim_{h\to 0}\frac{\sin\left(\frac{h}{2}\right)\cos\left(\frac{2x-h}{2}\right)}{\frac{h}{2}}$$

We know that $$ f'(x) = \frac{d}{dx} f(x) = \lim_{h\to 0}\frac{f(x+h) -f(x)}{h}$$
Take $$f(x)=\sin x \Leftrightarrow f(x+h) = \sin(x+h) $$
Substitute them in the fundamental relation of derivative of function with limits.
$$\frac{d}{dx} \sin x =\lim_{h\to 0}\frac{\sin(x+h) -\sin x}{h}$$
$$=\lim_{h\to 0}\frac{2\sin \Big(\frac{h}{2}\Big)\cos\Big(\frac{2x-h}{2}\Big)}{h}$$
$$=\lim_{h\to 0}\frac{\sin \Big(\frac{h}{2}\Big)\cos\Big(\frac{2x-h}{2}\Big)}{\frac{h}{2}}$$
$$=\lim_{\frac{h}{2}\to 0}\frac{\sin \Big(\frac{h}{2}\Big)}{\frac{h}{2}} \times \lim_{h \to 0} \cos\Big(\frac{2x-h}{2}\Big)$$
$$= 1 \times \cos x = \cos x $$

You can learn the proof much clearly here

http://www.mathdoubts.com/calculus/differentiation/identity/derivative-of-sinx/


What is the probability to get a sum of 9 exactly once when tossing 2 die in 3 throws? So I am throwing 2 die at the same time, the probability for me to get a sum of 9 is $\frac{1}{9}$. So when I throw them 3 times, the chances of 9 occuring once should be $\frac{1}{3}$. Then $\frac{1}{9}$x$\frac{1}{3}$=$\frac{1}{27}$

Is this right?
Also how could I calculate it if the sum of 9 occurs at least once in three throws?

I don't know why but I feel that the answer is wrong but I am not sure with what I'm doing wrong.

To begin, we calculate the probability that within a throw of a pair of dice we get the sum of nine.  It helps to consider the dice as being different colors.  We have the thirty-six equally likely results for pairs of dice being $\{(\color{red}{1},\color{blue}{1}),(\color{red}{1},\color{blue}{2}),(\color{red}{1},\color{blue}{3}),(\color{red}{1},\color{blue}{4}),(\color{red}{1},\color{blue}{5}),(\color{red}{1},\color{blue}{6}),(\color{red}{2},\color{blue}{1}),\dots,(\color{red}{6},\color{blue}{6})\}$

(Note: the outcome of $(\color{red}{1},\color{blue}{2})$ is considered to be different than the outcome of $(\color{red}{2},\color{blue}{1})$)

Of these thirty-six equally likely results, exactly four of them yield a sum of $9$, namely $(\color{red}{3},\color{blue}{6}),(\color{red}{4},\color{blue}{5}),(\color{red}{5},\color{blue}{4})$ and $(\color{red}{6},\color{blue}{3})$.  Taking the ratio of the number of favorable outcomes to the total number of possible outcomes (in an equiprobable sample space) gives the probability $\frac{4}{36}$ which simplifies to $\frac{1}{9}$.  You calculated this correctly.

The question you asked was for calculating the probability that at least once in three independent attempts you roll a sum of nine.  (This is distinctly different than the sum being nine on the first and only the first attempt like so many other answers sadly calculated)

To calculate this, you may approach several different ways.

Method 1: Split into cases

We could have the sum be nine at least once in the following seven ways:

The first sum is nine, the second sum is not nine, the third sum is not nine
The first sum is not nine, the second sum is nine, the third sum is not nine
The first sum is not nine, the second sum is not nine, the third sum is nine
The first sum is nine, the second sum is nine, the third sum is not nine
$\vdots$
The first sum is nine, the second sum is nine, and the third sum is nine

Each of these cases are disjoint, and so their probabilities may be added.  Each can be calculated by multiplication principle as they are independent events.  The probability for the first for example being $\frac{1}{9}\times\frac{8}{9}\times\frac{8}{9}$ while the probability for the second is $\frac{8}{9}\times\frac{1}{9}\times\frac{8}{9}$ and the probability for the fourth is $\frac{1}{9}\times\frac{1}{9}\times\frac{8}{9}$, etc...

(The error in several other answers was in only considering the first of these cases and none of the others)

This is horribly tedious but will arrive at a final correct answer if you are careful enough.

Method 2: Split into cases in a smarter way

We could have the sum be nine at least once in the following three ways:

Exactly one of the throws summed to nine
Exactly two of the throws summed to nine
Exactly three of the throws summed to nine

To calculate these, we can apply the binomial distribution.  Exactly $k$ successes in $n$ attempts where each attempt is independent and has success with probability $p$ will occur with probability $\binom{n}{k}p^k(1-p)^{n-k}$

For this, your first would occur then with probability $\binom{3}{1}(\frac{1}{9})^1(1-\frac{1}{9})^{3-1}$, etc...  We can then sum these as well to arrive at a final answer.  Still a bit tedious for my liking.

Method 3: Break into cases even smarter and use complements

The following two outcomes are possible
 - At least one of the throws summed to nine
 - None of the throws summed to nine

As these are disjoint and cover all cases, their probability must sum to $1$, so to calculate that at least one of the throws summed to nine, it suffices to calculate the probability that none of the throws summed to nine and subtract it from $1$.

None of them summed to nine with probability $(\frac{8}{9})^3$, so the answer to your original question is $1-(\frac{8}{9})^3$

Method 4: Inclusion-exclusion

Letting $A,B,C$ represent the events that the first, second, and third throws ended with a sum of nine respectively, you are asking to calculate $Pr(A\cup B\cup C)$

Expanding via inclusion-exclusion we get

$Pr(A\cup B\cup C)=Pr(A)+Pr(B)+Pr(C)-Pr(A\cap B)-Pr(A\cap C)-Pr(B\cap C)+Pr(A\cap B\cap C)$

Then using that the events are independent, we expand a bit further to get a final answer of

$\frac{1}{9}+\frac{1}{9}+\frac{1}{9}-\frac{1}{9}\times\frac{1}{9}-\frac{1}{9}\times\frac{1}{9}-\frac{1}{9}\times\frac{1}{9}+\frac{1}{9}\times\frac{1}{9}\times\frac{1}{9}$

(The error in another answer was in only using the first few terms of this expansion and ignoring all of the other still relevant terms)

So the possible combinations of die rolls that will sum to 9 are 6,3 3,6 4,5 and 5,4. The probability of getting these exact combinations is clearly $\frac{1}{36} $ each. So the probability of getting 9 on a given throw of 2 dies is 1/9.
Obviously, the probability of not getting 9 is $ \frac {8}{9} $. So in three throws, you want 9, not 9 and not 9 in any given order. The probability of this is the product of all 3 probabilities times 3 ( so 9 miss miss , miss 9 miss and miss miss 9)

$ 3x \frac{1}{9} x \frac{8}{9} x \frac{8}{9} = \frac{64}{243} = 0.2634 $

This method can easily be applied to any other number by calculating its possible combinations.


Polynomial factorisation - absolute value of coefficients This question takes the factorisation of a polynomial $p(x)=q(x)r(x)$, where $p$ (and for my purpose here $q$ and $r$) have integer coefficients and asks if the maximum absolute value of the coefficients of $q,r$ can ever be greater than the maximum absolute value of the coefficients of $p$.

That is answered by a factor of $x^{105}-1$ which has a coefficient of absolute value $2$ and there are other examples of higher degree amongst the cyclotomic polynomials.

Also note that $x^4+1=(x^2+1)^2-2x^2$ has the factorisation $(x^2+\sqrt 2 x + 1)(x^2-\sqrt 2 x +1)$ which isn't integral, but does suggest that lower degree examples may exist.

But what is the lowest degree of an integer polynomial where $q,r$ have integer coefficients which are not bounded by the maximum absolute value of the original polynomial?

(apologies for the clumsy explanation - I couldn't find a neater way to ask what I wanted).

I supposed that @Ewan Delanoy's example were minimal, and on trying to prove that, I found the following very similar degree-3 factorization:
$$X^3-X^2-X+1 = (X^2-2X+1)(X+1)$$
which must be minimal because at least one factor must have degree at least $2$ because the leading and trailing coeffs of the factors are bounded by those of the product.

Kudos go to Ewan, as without his example, I would not have constructed the other one.

P. S.: If you want a factorization into irreducibles, consider
$$X^3+2X^2-2X-1 = (X^2+3X+1)(X-1)$$

No, there is no lowest degree of factorization if coeffecients can be irrational. Every binomial can be written in the form $a+b$ or $a-b$ ,both of which can be factorized.

$a+b=(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)$

$a-b=(\sqrt a +\sqrt b-\sqrt (2\sqrt (ab))(\sqrt a+\sqrt b+\sqrt (2\sqrt (ab))$

Each factor found can further be factorized using the same identities. $(\sqrt a+\sqrt b+\sqrt (2\sqrt (ab))$ will have to be written as $(\sqrt a+\sqrt b)+(\sqrt (2\sqrt (ab))$ for the $a+b$ identity to be used. (and similarly for the other factor.

If factors have to have to be integers, the lowest possible degree of a factor is $1/2$, that is when the factor is of the form $a+b$, it can be further factorized using the identity exactly once, keeping the coeffecients as integers. However, there has to be atleast 1 factor with degree 1.


If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles. If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles.

Here is my solution...

To Prove - Triangle ABC is isosceles or AB = AC.

$BD = CD$ (Given)
$\angle BAD = \angle CAD$ (Given)
$\angle ABD = \angle ACD$ ($AD$ is a common side, angles opposite equal sides are equal)
$\triangle ABD$ and $\triangle ACD$ are congruent as per AAS postulate.
And therefore, $AB = AC$.

Is this a right answer or am I wrong somewhere ? I've seen solutions for this question but all of them have solved through constructions. I feel this is a shorter and logical way. Am I right or wrong in this approach ?

Your proof is wrong because you did not prove that $\Delta ABD\cong\Delta ACD.$

The hint:

Let $E$ be placed on the line $AD$ such that $D$ is a mid-point of $AE$.

Thus, $$\Delta ADC\cong\Delta EDB,$$ $$\measuredangle BAD=\measuredangle BED.$$
Can you take it from this?

Your answer is wrong as what you were asked to prove, the same you are taking as a statement which is actually not allowed.
To prove it you have to draw a line from C which is parallel to AD and meeting BA at E. Now prove it with the help of converse of MPT.


Complex Integration poles real axis In class my professor said that
$$
\int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 - b^2}dx = -\frac{2\pi}{b}\sin(ab)
$$
where $a,b > 0$.
However, since the poles are on the real axis, isn't the integral equal to
$$
\pi i\sum_{\text{real axis}}\text{Res}(f(z); z_j)\mbox{?}
$$
If that is the case, the integral is
$$
-\frac{\pi}{b}\sin(ab).
$$

Define

$$f(z):=\frac{e^{iaz}}{z^2-b^2}\;,\;a,b\in\Bbb R^+\;,$$

$$C_R:=[-R,-b-\epsilon]\cup\gamma_{-b,\epsilon}\cup[-b+\epsilon,b-\epsilon]\cup\gamma_{b\epsilon}\cup[b+\epsilon,R]\cup\Gamma_R$$

with

$$\gamma_{r,s}:=\{r+se^{it}\;;\;0\le t\le \pi\}\;,\;r,s\in\Bbb R^+\;,\;\Gamma_R:=\{Re^{it}\;;\;0\le t\le \pi\}\;,\;\;R\in\Bbb R^+$$

Since $\;f(z)\;$ analytic on and within $\;C_R\;$ , we get

$$\oint\limits_{C_R}f(z)\,dz=0$$

By the corollary to the lemma in the second answer here we get

$$\begin{align*}\lim_{\epsilon\to 0}\int\limits_{\gamma_{-b,\epsilon}}f(z)dz&=i\pi\,\text{Res}\,(f)_{z=-b}=i\pi\frac{e^{-iab}}{-2b}=-\frac{\pi i}{2b}e^{-iab}\\
\lim_{\epsilon\to 0}\int\limits_{\gamma_{b,\epsilon}}f(z)dz&=i\pi\,\text{Res}\,(f)_{z=b}=i\pi\frac{e^{iab}}{2b}=\frac{\pi i}{2b}e^{iab}\end{align*}$$

And applying Jordan's Lemma the integral on $\;\Gamma_R\;$ goes to zero when $\;R\to \infty\;$ , so in the end

$$0=\lim_{R\to\infty\,,\,\epsilon\to 0}\oint\limits_{C_R}f(z)dz=\int\limits_{-\infty}^\infty f(x)dx-\frac{\pi i}{2b}\left(e^{iab}-e^{-iab}\right)\implies$$

$$\implies \int\limits_{-\infty}^\infty f(x)dx=-\frac\pi b\frac{e^{iab}-e^{-iab}}{2i}=-\frac\pi b\sin(ab)$$

and I get the same as you did...

This integral is unsolvable in real domain, because the singularities are not removable so integral should diverge.but you can take contours in complex plane which connect negative and positive infinitis and then you get the above result.


By differentiation or otherwise, prove that $(c(z))^2 +(s(z))^2 = 1$ Assume that the two power series $s(z) = \sum a_nz^n$ and $c(z) =\sum b_nz^n$ are convergent for all $z\in \mathbb{C},$ and that they satisfy the relations $s'(z) = c(z), c'(z) =-s(z).$ Deduce the identities
$$ a_n = -a_{n-2}/(n(n-1)) , b_n=-a_{n-2}/(n(n-1))
$$(To me, it looks like there is a typo for $a_n$)
If further $s(0)=0, c(0) =1,$ determine $s(z)$ and $c(z)$ completely. By differentiation or otherwise, prove that $(c(z))^2 +(s(z))^2 = 1$
To me it looks like $s(z) = sin(z)$ and $c(z) = cos(z)$ but I don't know what to do with this information.
Also while taking the derivate of the power series in an attempt to deduce the identities, I have
$(n+1)a_{n+1}=b_n$ and it does not seem like I have the correct identity

hint
if $f'(z) = 0$ for all $z$ then $f(z) = $ a constant .
this means that for all $z, f(z) = f(0).$
let $f(z) = [s(z)]^2 + [c(z)]^2.$

Addendum
Per request of OP

determine $s(z)$ and $c(z)$ completely.

Actually, the necessary analysis has already been provided by the comments of leslie townes.  All I need to do is fill in all of the ground work.
First of all, I completely agree with the response of Hagen von Eitzen, so following
through with the first part of my answer (before the Addendum) is the place to start.
$f'(z) = s(z)s'(z) + c(z)c'(z) = s(z)[c(z)] + c(z)[-s(z)] = 0.$
This establishes that for all $z, f(z) = f(0) = 1,$ which completely resolves
the question in the title of the query.
Next, since $s'(z) = c(z)$ and $c'(z) = -s(z)$, you have that
$s''(z) = c'(z) = -s(z)$ and that $c''(z) = -s'(z) = -c(z).$
Therefore,
$$\sum_{n=2}^\infty (n)(n-1)[a_n]z^{(n-2)} = s''(z) = -s(z) =
\sum_{n=0}^\infty -a_nz^n = \sum_{n=2}^\infty -a_{(n-2)}z^{n-2}.$$
As a result, you have that for each $n \in \{2,3,4,\cdots\}$
$(n)(n-1)a_n = -a_{(n-2)}.$
By identical analysis, for each $n \in \{2,3,4,\cdots\}$
$(n)(n-1)b_n = -b_{(n-2)}.$
All that remains is to compute $a_0, a_1, b_0, b_1.$
Since $a_0 = s(0) = 0$ and $b_0 = c(0) = 1$,
it is immediate that $a_0 = 0$ and $b_0 = 1.$
Also, remember that for all $z$, you have that $s'(z) = c(z)$ and $c'(z) = -s(z).$
Therefore, since $a_1 = s'(0) = c(0) = 1$, and $b_1 = c'(0) = -s(0) = 0$,
it is immediate that $a_1 = 1$ and $b_1 = 0.$

Hint : Let
$$e : z \mapsto c(z) + i s(z)$$
You get $e'(z)=ie(z)$. Can you proceed further ?


The presentation of a group I am confused with the following definitions from Dummit & Foote

"In general, if some group $G$ is generated by a subset $S$ and there is some collection of relations, say $R_1,\ldots, R_m$(here each $R_i$ is an equation in the elements from $S \cup \{1\}$ such that any relation among the elements of $S$ can be deduced from these, we shall call these generators a presentation of $G$"

I have two questions:

1.Shouldn't the relation come from $S \cup \{1\} \cup$ all the inverses of $S$ rather than only from $S \cup \{1\}$.

2.What does it mean by saying that one relation can be deduced from a set of relations.

Yes
Given an arbitrary relation $R$ in $G$, $R_1,\dots, R_m$ imply $R$


Determine the order of the permutation in Discrete math I have this question as a homework so please no full answers. I just need help on part b. we learned very briefly about permutation, mainly it's definition and how cycles works but never heard of the identity permutation and the order of the permutation so i actually don't understand what they're asking for. Can anyone help or give an example?

The identity permutation is the permutation that does nothing at all - it sends every number to itself. You could write this in two-row notation (for $n=9$) as
$$\begin{pmatrix} 1& 2& 3 &4& 5 &6 &7&8&9\\ 1& 2 &3 &4 &5& 6&7&8&9\end{pmatrix}$$for instance.

The order is defined in the question - see if you can figure out what it means using the identity here.

a)  The permutation maps 1 into 2 then 3 into 3 then 3 into 1.  (123) is a cycle of order 3. Doing it 3 times maps 123 to 123. The permutation maps 4 into 5 and 5 into 4. (45) is a cycle of order 2. Doing it twice maps 45 to 45.  The permutation maps 6 into 7, 7 into 8, and 8 into 9.  (6789) is a cycle of order 4.  Doing it 4 times maps 6789 to 6789.

b)  The least common multiple of 2, 3, and 4 is 12.  Repeating this permutation twelve times will map 123 to 123 three times but ends back at 123.  Repeating this permutation 12 times will map 45 to 45 6 times but ends back at 45.  Repeating this permutation will map 6789 to 6789 three times but ends back at 6789. 123456789 permuted 12 times winds up back at 123456789.


negate the quantified statement ∃y(Q(y) ∧ ∀x¬R(x, y)) I cant remember if i am doing it correctly. I am having trouble finding the negation.

Hint:

$$\neg\forall x\; P(x) =\exists x\;\neg P(x)$$
$$\neg\exists x\; P(x) =\forall x\;\neg P(x)$$

And also:

$$\neg (P \wedge Q)=\neg P\vee \neg Q$$

The negation is simply

$$\forall y \;\;(\lnot Q (y) \lor \exists x : R (x,y)) $$

Once again I know who downvoted, so good night .


Doubt on Lebesgue integration How to prove that $\int_{\Omega}(a+b|u|^{(p-1)})^{q}\le C\int_{\Omega}(1+|u|^{(p-1)q})$ where $ u\in L^{p}(\Omega)$, and a,b are constants. q is a conjugate exponent  of p

Note:
This answer may show how little I know
about Lebesgue integration.
I am treating this just as
an inequality between the integrands
and assuming some bounds on the parameters.

This is true if
$(1+x^u)^v
< C(1+x^{uv})
$,
where
$x > 0$
and
$C$ depends on $u$ and $v$
and $u = [-1$ and $v = q$.

If we let $x^u = y$,
this becomes
$(1+y)^v < C(1+y^v)$,
or
$1+y < C(1+y^v)^{1/v}$,
where the $C$'s are different
and depend on $u$ and $v$.

By the arithmetic means inequality,
if $v > 1$,
$\dfrac{1+y}{2}
< \left(\dfrac{1+y^v}{2}\right)^{1/v}
$
or
$1+y
< 2^{1-1/v}(1+y^v)^{1/v}
$

So, if $p>1$ and $q > 1$,
this seems to be true.


Translating a sentence into predicate logic "A dragon is happy only if it has a green child".

Have I translated this statement correctly into logic (below)?

$\forall X \cdot dragon(X) \wedge happy(X) \Rightarrow \exists Y \cdot childOf(Y,X) \wedge green(Y)$

$childOf(Y,X)$ is true when $Y$ is a child of $X$.

This is correct, if you make sure that $\forall X$ quantifies the whole statement, i.e. if you write:
$$\forall X\big(dragon(X)\land happy(X)\to \exists Y\left( childOf(Y,X)\land green(Y)\right)\big)$$

I think it is necessary to be more careful. Here we are dealing with 'only if' and not 'if'.
The correct translation is the following.

Dx: x is a dragon

Cxy: x is a child of y

Gx: x is green

Hx: x is happy

$$  \forall x ((Dx \wedge \exists y (Cyx \wedge Gy))  \to Hx) $$


Equivalences of regular expressions - which one is wrong? Let $\alpha,\beta \in \text{RA}(\left\{0\right\})$ where $RA(\Sigma)$
  is the set of all regular expressions of the alphabet $\Sigma$.

  The following two equivalences are given:

  $\alpha\beta \sim \beta\alpha$
  $(\alpha + \beta)^* \sim \alpha^* \beta^*$

One of them is wrong, but even after brooding over it I don't find a solution.

The first one should be true because no matter what to show first - the numver of zeros is always the same (x + y zeros = y + x zeros).

I tried to think about the second one less worse in form
$$\left[\left[(\alpha + \beta)^*\right]\right] = \left[\left[(\alpha + \beta)\right]\right]^*  = \left( \left[\left[\alpha\right]\right] \cup \left[\left[\beta\right]\right] \right)^* = \; ?$$
but as you can see I didn't find any mathematical proof for it (or am just too stupid to see it). Might this be the wrong one? I would have found a counterexample if the language wouldn't contain only $0$'s.

Could you please help me find the solution and identifying my mistakes?

Thanks in advance!

[Edit] We were given the following semantic rules:

Question was answered by Henning Makholm above


Real world examples of continuous uniform distribution on [0,1] Can someone give me real world examples of uniform distribution on [0,1] of a continuous random variable, because I could not make out one.

Consider a "spinner": an object like an unmagnetized compass needle that can pivots freely around an axis, and is stable pointing in any direction.  You give it a spin and see where it comes to rest, measuring the resulting angle (divided by $2\pi$) as a number from $0$ to $1$.

If there are N raffle tickets sold to N different people, then each ticket holder has the same probability of winning the prize. The distribution is uniform.


Applying Geometric Brownian Motion itself into Ito's Lemma I am new to this forum as well as partial differentiation. I would like to ask the following question.

Given a geometric brownian motion:

$$dS_t=(\mu S_t)dt+(\sigma S_t)dz_t$$

I would like to apply Ito's Lemma to $dS_t$ itself such that I would expect it will still give the same $dS_t$.

To do this, We need to know $\partial S_t \over \partial S_t$, $\partial^2 S_t \over \partial S_t^2$, and $\partial S_t \over \partial t$.

I have no problem of finding the first 2 derivatives, but how to calculate $\partial S_t \over \partial t$?

Thanks.

Thanks @chris-janjigian. Let me try to find out $\partial S_t \over \partial t$ by using the integral form of $dS_t$. I got the integral form after searching from Google: $$S_t = S_0 + \int_0^t (\mu S_u)du+\int_0^t(\sigma S_u)dz_u$$

As $t$ is just representing "t-times" in this form, ${\partial S_t \over \partial t}=0+0+0=0$. We also know that ${\partial S_t \over \partial S_t}=1$ and ${\partial^2 S_t \over \partial S_t^2}=0$. Therefore, using Ito's Lemma: $$dS_t=[{\partial S_t \over \partial S_t}(\mu S_t)+{\partial S_t \over \partial t}+{1 \over 2}{\partial^2 S_t \over \partial S_t^2}(\sigma S_t)^2]dt+[{\partial S_t \over \partial S_t}(\sigma S_t)]dz_t$$ $$dS_t=[1(\mu S_t)+0+0]dt+[1(\sigma S_t)]dz_t$$ $$dS_t=(\mu S_t)dt+(\sigma S_t)dz_t \Box$$


Limit Question (Calculus 2) Simply put, is there a way to show that $$\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} = 1$$

where $$a_n = \frac {(n^2-n)^{1+1/n}}{(1+\frac {1}{n})^\sqrt{(n^2-n)}}$$

This is a problem I may incorporate into a project for AP Calculus BC/Calculus 2, so using methods with that skill level in mind is preferable (if not possible, I will still appreciate a proof nonetheless).

The most progress I've made was through taking the natural logarithm and using those properties to arrive at $$\lim \limits_{n \to \infty} ln\left(\frac{a_{n+1}}{a_n}\right) = \lim \limits_{n \to \infty}\left[\left(1 + \frac{1}{n+1}\right)\ln\left(n^2+n\right)+\ln\left(1+\frac{1}{n}\right)\sqrt{n^2-n}-\left(1+\frac{1}{n}\right)\ln(n^2-n)-\ln\left(1+\frac{1}{n+1}\right)\sqrt{n^2+n}\right]$$

and although I can deal with the first and third terms using L'Hôpital's Rule, I was unable to finish off the second and fourth terms.

(Also, apologies if anything seemed confusing in the equations. I'm new to MathJax and the website in general).

Thank you kindly!

If you already know and use Taylor series, there are some simple possible approches.

Considering $$a_n = \frac {(n^2-n)^{1+1/n}}{(1+\frac {1}{n})^\sqrt{n^2-n}}$$ as you did, $$\log(a_n)=\left(1+\frac 1n\right)\log(n^2-n)-\sqrt{n^2-n}\log\left(1+\frac 1n\right)$$ $$\log(a_n)=\left(1+\frac 1n\right)\left(2\log(n)+\log(1-\frac 1n)\right)-n\sqrt{1-\frac 1n}\log\left(1+\frac 1n\right)$$ Now, using the classical Taylor series, you could get $$\log(a_n)=\left(2 \log \left(n\right)-1\right)+\frac{2 \log
   \left(n\right)}{n}+O\left(\frac{1}{n^2}\right)$$ So, limiting to the first order $$\log(a_{n+1})-\log(a_n)=2\log\left(\frac{n+1}n\right)+O\left(\frac{1}{n}\right)=2\log\left(1+\frac{1}n\right)+O\left(\frac{1}{n}\right)= \frac 2 n+O\left(\frac{1}{n^2}\right)$$ Now, using $x=e^{\log(x)}$ and Taylor again $$\frac{a_{n+1}}{a_n}=1+\frac 2 n+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.

Playing with my old pocket calculator $a_{10}\approx 57.1454$, $a_{11}\approx 67.7086$ which makes a ratio $\approx 1.18485$ while the Taylor expansion gives $1.2$. Similarly, $a_{100}\approx 4032.93$, $a_{101}\approx 4111.06$ which makes a ratio $\approx 1.01937$ while the Taylor expansion gives $1.02$

You can use this identity - If $\lim_{n \to \infty }\frac{a_{n+1}}{a_n}=l$ then $\lim_{n\to\infty} (a_n)^{1/n}=l.$ So you just havr to find the limit
$$L=\lim_{n\to\infty} (a_n)^{1/n}=\lim_{n\to\infty}[{ \frac {(n^2-n)^{1+1/n}}{(1+\frac {1}{n})^\sqrt{(n^2-n)}}}]^{\frac {1}{n}}$$

So now just take ln both the sides and put $n=\frac {1}{t}$, so limit will change to $$lnL=\lim_{t\to\ 0}t[(1+t)ln(\frac{1-t}{t^2})-\frac{\sqrt{1-t}}{t}ln(1+t)]$$

Now, you can solve this limit very easily which is equal to zero. So

$$lnL=0$$ So $$L=1$$
Hence Proved


Finding a couple of numbers which satisfy $(a + b)|(a + c)$ and $(a + b)|(2c)$ using a "systematic" method I want to find a couple of integers ($a$, $b$, $c$) which satisfy the following:

$(a+b) |(a + c)$ - "$(a+c)$ divisible by $(a + b)$"

$(a+b)|(2c)$ - "$(2c)$ divisibile by $(a+b)$"

I can try and test random numbers and eventually get something out of it, but I want to take a systematic (not random) approach towards it.

I tried to use the definition of divisibility ($a|b \Rightarrow b=k.a$) to no avail. I'm aware that there probably is an endless number of answers to this question, but I want to get a couple of distinct answers NOT in a hit or miss way.

How do I do that?

P.S: There actually might not be any answers to such expressions so I'm also interested in knowing how to find whether there is an answer or not.

Hint $ $ The statement $\,P\,$ has scaling symmetry $\,P(an,bn,cn)\iff P(a,b,c),\,$ thus cancelling $\,(a,b,c)$ reduces to case $\,(a,b,c) = 1$. Let $\, d = a\!+\!b,\,$  $\,\color{#c00}{(d,a,c)} = (a\!+\!b,a,c) = (b,a,c) = 1,\,$ so

$$d\mid a\!+\!c,2c\,\Rightarrow\, d\mid2a = 2(a\!+\!c)\!-\!2c\,\Rightarrow\, d\mid(2a,2c) = 2(a,c) \,\Rightarrow\, d\mid 2,\ \ {\rm by} \  \color{#c00}{(d,a,c)}  = 1$$

Thus we have reduced to the case of $\,(a,b,c) = 1\,$ and $\,a\!+\!b\mid 2,\,$ which is easy to  finish.

One thing to note is that if you have one solution, you can multiply $a,b,c$ by any number and get another.  You can subtract the two to get $(a+b)|(c-a)$ and again to get $(a+b)|2a$.  This shows we need $a=b$

The easiest is to take $a=b$ and $c=ka$ for $k$ odd.  You didn't say they have to be distinct, but may have been thinking it.


Does $p$-value change with sample size? I was wondering how $p$-values change with sample size or is their any relation between the two. To my knowledge, a $p$-value denotes the probability of finding observed or more extreme results than the null hypothesis claims (typically no difference). Based on the following example, let your null hypothesis be that there exists no difference in the amount of heads and tails you flip on a fair coin, that is you flip the same exact amount and your alternative be that a difference does exist. You flip a fair coin $n = 10$ times and get $7$ heads and $3$ tails, which suggests a relatively low $p$-value. But as you flip this coin more times (say $n = 100$) and now get $45$ heads and $55$ tails, your $p$-value increases - which results in you being more likely to fail to reject the null over the alternative hypothesis.

Thus, does increasing sample size, increase $p$-values in general?

In general, increasing the sample size changes the p-value (not necessarily increasing it). There is a lot to discuss as to why it is so and also why ''p-hacking'' is a thing, but there is a youtube video where it is explained in simple terms:

https://www.youtube.com/watch?v=42QuXLucH3Q

"the probability of finding observed or more extreme results than the null hypothesis claims" is not right. The p-value is the probability that a result at least as extreme as that which was actually observed would occur given that the null hypothesis is true.

If the coin you're tossing is fair, that being what the null hypothesis says, then with very high probability the p-value will be large when the number of trials is large. If the null hypothesis is false, then the probability is $1$ that the p-value will approach $0$ as the sample size grows.


Is this sequence decreasing? If a sequence $b_n>0$ and $b_n$ converges to $0$, can we say it is eventually decreasing? This problem bumps up when I am trying to something bigger. However, I am very unsure of this. If this is not true, can anyone give some counter examples please?

Here's one way to construct lots of counterexamples.  Take any positive term sequence, $(a_n)$ that is strictly decreasing to $0$.  Then form the new sequence $a_2,a_1,a_4,a_3,a_6,a_5,\dots$. (I.e., you swap each even-indexed term with the preceding odd-indexed term.)

If a series is convergent then it will satisfy the following:

$$
\lim_{n \to \infty} a_n = 0 \\
\lim_{n \to \infty} \frac {a_{n+1}}{a_n} \le 1
$$

So for some very large $n$ you could say that the series is decreasing, because $a_{n+1} \le a_n$, but since consecutive terms might be equal it depends on what "decreasing" is defined to mean.


Under what conditions $z=x+iy$ in $3^{rd}$ quadrant implies $\frac{\bar{z}}{z}$ in $3^{rd}$ quadrant? If $z=x+iy$ is in the $3^{rd}$ quadrant, then $\frac{\bar{z}}{z}$ also lies in the $3^{rd}$ quadrant if ?

How do I obtain the solution which is given as $x<y<0$.

My Attempt:

$x=|z|\cos\theta<0$ & $y=|z|\sin\theta<0$

Since $\frac{\bar{z}}{z}=\frac{|\bar{z}|}{|z|}[\cos(-2\theta)+i\sin(-2\theta)]$ is in the $3^{rd}$ quadrant,
$$
\cos(-2\theta)<0 \text{ & } \sin(-2\theta)<0\implies \cos^2\theta-\sin^2\theta<0\text{ & }\sin\theta\cos\theta<0
$$
From $\sin\theta\cos\theta<0$ we only take $x<0$ & $y<0$ since it is already stated. Now,
$$
\cos^2\theta-\sin^2\theta<0\implies\cos^2\theta<\sin^2\theta\implies|\cos\theta|<|\sin\theta|\implies|x|<|y|
$$
Since both $x,y<0$, I think $|x|=-x$ & $|y|=-y$
$$
|x|<|y|\implies-x<-y\implies x>y
$$
So the solution I obtain is $y<x<0$.

Is this the right way to approach the inequalities of this kind ?. What am I doing wrong ?

Note that a complex number $a+ib$ is in the third quadrant (not including axes) iff $a,b<0$. Therefore we can assume that $x,y<0$ in order to have $z$ in the third quadrant, and we want to find conditions so that
$\frac{\bar{z}}{z}$ is in the third quadrant. Now
$$
\frac{\bar{z}}{z} = \frac{x-iy}{x+iy} = \frac{x^2-y^2-ixy}{x^2+y^2}
= \frac{x^2-y^2}{x^2+y^2} +i\frac{-xy}{x^2+y^2},
$$
so we need to have $x^2-y^2<0$ and $-xy<0$. The latter is true because $x,y<0$, and the former is equivalent to $y<x$ since we know $x$ and $y$ are negative.

We conclude that $y<x<0$ are the desired conditions.

The conditions $x<y<0$, which appear in the question, don't work. Indeed, consider $z=-2-i$. Then $z$ is in the third quadrant, but
$$
\frac{\bar{z}}{z} = \frac{-2+i}{-2-i} = \frac{3-4i}{4+1} = \frac{3}{5}-\frac{4}{5}i
$$
is in the fourth quadrant.

You cannot, since $x<y<0$ is wrong.
Geometrically, what it tells you that $z = x + iy$ when written in form $z = re^{i\varphi}$ satisfies $\varphi\in(\pi,\frac{5\pi}4)$. Here's a graph to support the claim:

Note that $\frac{\bar z}z = \frac{re^{-i\varphi}}{re^{i\varphi}} = e^{-2i\varphi}$. Now, $\pi<\varphi<\frac{5\pi}4$ implies $-\frac{5\pi}2<-2\varphi<-2\pi$, and thus $\frac{\bar z}z$ is in fourth quadrant, not third.
Thus, your solution is correct, i.e. $y < x < 0$, and you can prove it similarly to what I've done here.


How to rotate an orientation (Euler angles) If I have an orientation defined by Euler angles and I want to simulate a rotation of the coordinate system about the origin (doesn't matter to me how the rotation is specified), how would I get the new Euler angles? I understand how to transform a point via a rotation matrix, but I'm not sure how to approach this in terms of Euler angles.

For example, I'm thinking of an airplane's orientation in terms of pitch, yaw, and roll, and say I want to simulate a random rotation of the airplane about its center.

Let us assume that the rotation is known by a $3\times3$ matrix, and let us focus on the one specified as $X_1Z_2X_3$ in the Wikipedia article.

You get $\alpha$ from $\tan\alpha=\dfrac{s_1s_2}{c_1s_2}$.

You get $\beta$ from $\tan\beta=\dfrac{\sqrt{(s_2c_3)^2+(s_2s_3)^2}}{c_2}$.

You get $\gamma$ from $\tan\gamma=\dfrac{s_3s_2}{c_3s_2}$.

As expected, $\beta=0$ creates an indeterminacy, but this case is easy to deal with (it reduces to a single rotation $\alpha+\gamma$).

You can probably improve the accuracy by a least-squares fitting of the matrix derived from the angles to the given matrix, but this will require the Levenberg-Marquardt algorithm as this is a non-linear problem.

I guess that you can also address the computation of the angles by using three well-chosen Givens rotations that will transform the input matrix to a unit one (backwards, $X_1Z_2X_3\to X_1Z_2\to X_1\to I$, corresponding to angles $(\alpha,\beta,\gamma)\to(\alpha,\beta,0)\to(\alpha,0,0)\to(0,0,0)$).