Proof related to minimum and maximum degree of vertices of an undirected graph I don't know how to proceed for this problem. I would appreciate any help. Thanks!

  Let $\delta$ and $\Delta$ be the minimum and maximum degree of the
  vertices of an undirected graph G.

  Show that $\delta \leq \frac{2\left | E(G) \right |}{\left | V(G)
\right |}\leq \Delta $.

HINT: Use the basic fact relating degrees of vertices to number of edges: $$\sum_{v\in V(G)}\deg v=2|E(G)|\;.$$ What are the minimum and maximum possible values of the sum?

Let $e$ be the no. of edges.
Let $v$ be the no. of vertices.
Let $mn$ be min degree of vertex in graph.
Let $mx$ be the max degree of vertex in graph.
 every vertex's degree must lie between this min and max limit
$\min\leq \deg(v)\leq \max$
$\min\leq 2e\leq \max$  // using the above hint
thus for n vertices in a graph,
$n(\min)\leq 2e\leq n(\max)$ and so
$\min \leq 2\frac{e}{n}\leq \max$


Can $ 2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $ for some $n>7$. Prove or disprove that there isn't any positive integer $n>7$ such that the linear congruence below is true.

$  2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $

Note that $2$ mod $7$ has order $3$ (since $2^2 \equiv 4 \bmod 7$ and $2^3 \equiv 1 \bmod 7$).

So $2^{3^m} \equiv 1 \bmod 7$ for any $m\geq 1$.


Size of outerplanar graph If $G$ is an outerplanar graph of order $n \geq 2$ and size $m$, show that $m \leq 2n -3$

[I can show the result for Hamiltonian outerplanar graphs, and I think its posible to extend the result, but my proof is very inelegant and tedious. Any ideas?]

We can assume that the graph $G$ is 2-connected, otherwise, for two adjacent blocks, let $v$ be their common vertex, in a planar embedding, we can see that we can add an edge $\{w_1,w_2\}$, where $w_1,w_2\in\Gamma(v)$, such that $G$ is still an outerplanar graph, but one block less.

Then, we assume that all inner faces are triangles. Hence, with these conditions, $2m=n+3(n-2)$, that is $m=2n-3$, as an $n$-gon is triangulated by $n-2$ triangles.

Note that what we are doing is just adding edges to make it Hamiltonian, and then triangulate it. Hence, back to the original graph, we have $m\leq 2n-3$.


If $\alpha= \frac{\pi}{7}$ and also if $\frac{3- \tan^2 (\alpha)}{1 - \tan^2 (\alpha)} = \lambda \cos(\alpha)$. Then find $\lambda$. Given that

$$\frac{3- \tan^2 \left(\dfrac{\pi}{7}\right)}{1 - \tan^2 \left(\dfrac{\pi}{7}\right)} = \lambda \cos\left(\dfrac{\pi}{7}\right)$$

Then find the value of $\lambda$.

Notice, setting the value of $\alpha=\pi/7$, one can simplify LHS as follows   $$LHS=\frac{3-\tan^2(\alpha)}{1-\tan^2(\alpha)}$$
$$=\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}$$

$$=\frac{1-\tan^2\frac{\pi}{7}+2}{1-\tan^2\frac{\pi}{7}}$$
$$=1+\frac{2}{1-\tan^2\frac{\pi}{7}}$$
$$=1+\frac{2\tan\frac{\pi}{7}}{\left(1-\tan^2\frac{\pi}{7}\right)\tan\frac{\pi}{7}}$$
$$=1+\left(\frac{2\tan\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}\right)\frac{1}{\tan\frac{\pi}{7}}$$
using $\color{blue}{\frac{2\tan A}{1-\tan^2A}=\tan2A}$,
$$=1+\frac{\tan\frac{2\pi}{7}}{\tan\frac{\pi}{7}}$$
$$=1+\frac{\sin\frac{2\pi}{7}\cos\frac{\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+\sin\frac{2\pi}{7}\cos\frac{\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
using $\color{blue}{\sin A\cos B+\sin B\cos A=\sin(A+B)}$,
$$=\frac{\sin\frac{3\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{\sin\left(\pi-\frac{3\pi}{7}\right)}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{\sin\frac{4\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{2\sin\frac{2\pi}{7}}{\sin\frac{\pi}{7}}$$
$$=\frac{2\cdot 2\sin\frac{\pi}{7}\cos\frac{\pi}{7}}{\sin\frac{\pi}{7}}$$
$$=4\cos \frac{\pi}{7}$$
now, $RHS=\lambda\cos(\alpha)=\lambda\cos\frac{\pi}{7}$,

Now, equating LHS & RHS, we have $$4\cos\frac{\pi}{7}=\lambda\cos\frac{\pi}{7}$$
 $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\lambda=4}}$$

Simply divide both sides of the equation by $\cos(\pi/7)$

$\lambda = \frac{3-\tan^2(\pi/7)}{\cos(\pi/7)(1-\tan^2(\pi/7)}$ = 3.999...


Are there numerical algorithms for Roman numerals? In positional number systems there are algorithms for performing certain operations, like long division, to name one of the simplest. This works for positional systems, whatever base. I realize in number theory there are very advanced algorithms, typically for working with Very Long Numbers. (disclaimer: except for a Fourier transform I don't know any of them, I'm not a mathematician.)

I was wondering how the Romans could do anything numerical with their odd Roman numerals. You can't divide MMDCCI by LXXIII using long division. So, question: are there numerical methods for Roman numerals, and if not, how did the Romans divide MMDCCI by LXXIII?

This web page has information about some possible approaches to doing arithmetic using Roman Numerals: http://turner.faculty.swau.edu/mathematics/materialslibrary/roman/

Are there numerical algorithms for Roman numerals?

Yes there are, all the algorithms for our current numbers can work on Roman Numerals ( as far as the numbers can be represented in Roman Numeral Form). All you have to do is to represent each number in it's Roman Form, why would dividing , adding or any other operation on numbers depend on how the number is represented?

You can't divide MMDCCI by LXXIII using long division

Yes you can, Here is one way to divide MMDCCI by LXXIII : Convert them to their decimal ( arabic numeral ) equivalants, carry out the long division as usual, at the end substitute roman numerals in place of every number appearing in the division. But this does not answer the question of how Romans did it, just that it is possible to divided MMDCCI by LXXIII using long division and arabic numerals as we are used to, the algorihm is a composite of Long Division and Converting numbers between two different types of their representation.

how did the Romans divide MMDCCI by LXXIII?

How did the Romans actually divide MMDCCI by LXXIII? One simple way would have been by adding LXXIII as many times to LXXIII until the results was greater than MMDCCI, a truely long divison algorithm, it is a way that although maybe not acceptable by todays standards but at least in principal they could do it.
This is on how it could be done, pondering about idiotic system can look this up youtube.com/watch?v=5PxXpDH8-2A


Does adding or/and dividing a random variable by a constant change its probability distribution? Suppose that we have a random variable X with probability distribution $PDF_X(\mu_x,\sigma_x)$

Consider random variable $Y=\frac {X-a}b$ , I
 know that mean and variance of Y would be:

$$\mu_y=\frac {\mu_x-a}b, \sigma_y=\sqrt{ \frac {\sigma_x^2} {b^2}  }$$

Would X and Y have the same type of probability distribution (Of course with different mean and variance)?

For example I know that if X is a Normal random variable, Y would be again a Normal random variable. Is this true for all the other probability distribution?

Thank you.

Informally, if we have a random variable $X$, and $Y=aX+b$, where $a$ and $b$ are constants and $b\ne 0$, then $X$ and $Y$ are close relatives.

But the distributions of $X$ and $Y$ need not have the same type of name. As you pointed out, if $X$ has normal distribution, then so does $Y$. Similarly, if $X$ has uniform distribution, so does $Y$.

However, if $X$ has binomial distribution, then $aX+b$ only has binomial distribution if $a=1$ and $b=0$.  Similar comments could be made about the hypergeometric, the Poisson, and many others.

The fact that if $X$ has binomial distribution, then (usually) $aX+b$ does not has no real mathematical significance. It just has to do with the kind of distributions we choose to call binomial.  The very close relationship between $X$ and $aX+b$ remains, even if we do not happen to give their distributions the same name.

$c\cdot X$ does not necessarily follows the same family of distributions as $X$; depends on the family of distribution $X$ belongs to (Not true e.g. for the Beta distribution, but true for Normal).


Calculate $ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{xy^{2}}{x^{4}+y^{2}} $ Im trying to calculate (or prove that the limit dosent exists)
$ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{xy^{2}}{x^{4}+y^{2}} $
Wolfram says the limit does not exists, but for every path I choose the limit ends up being 0.
If someone can tell if the limit exists or not, it would be helpful.
Thanks in advance


looking for some shortcut or faster method for solving a question involving square root If $x =\dfrac{ \sqrt 3 - \sqrt 2 }{ \sqrt 3 + \sqrt 2 } $ and $ y = \dfrac{\sqrt 3 + \sqrt 2 }{ \sqrt 3 - \sqrt 2}$, find the value of
$\dfrac{x^2 + xy + y^2}{x^2 - xy + y^2}.$

I obviously know the traditional method of solving this qn (by multiplying by conjugate).  I want to know if there is some faster method to solve this question.


Proving $\pi x(1-x)\le \sin \pi x \le 4x(1-x)$ for $x \in [0,1]$ The LHS of the following inequality
 $$\pi x(1-x)\le \sin \pi x \le 4x(1-x), ~~ x \in [0,1]~~~~(1)$$
can be proved by taking $$f(x)=\sin \pi x-\pi x(1-x) \implies f'(x)=\pi(\cos \pi x-1+2x), f''(x)=-\pi^2 \sin \pi x+2$$
Next, $$f'(x)=0 \implies \cos \pi x=1-2x$$ has exactly three real roots: $x=0, 1/2, 1$, it can be checked graphically that $y=\cos \pi x$  and $y=1-2x$ cut each other at these three points only.
As $f''(0)= f''(1)=2 >0$ and $f''(1/2)=2-\pi^2<0.$ Consequently, $$f(x) \ge f_{min}=f(0)\implies \sin \pi x\ge \pi x(1-x).$$

The question here is: How to prove the RHS of (1)?

If you replace $x$ by ${1 \over 2} - x$, the inequality you are trying to prove is
$$ \cos {\pi x} \leq 1 - 4x^2$$
You want to show this for $|x| \leq {1 \over 2}$. By evenness of the functions it suffices to show this on $[0,{1 \over 2}]$. The form of the right-hand side suggests one use  the double angle identity $\cos {\pi x} = 1 - 2\sin^2{\pi x \over 2}$. Making this substitution, we see that we must show
$$ 1 - 2\sin^2{\pi x \over 2} \leq 1 - 4x^2$$
This translates to simply
$$\sin{\pi x \over 2} \geq \sqrt{2} x$$
This is to be valid on $[0,{1 \over 2}]$. Note that the line $y = \sqrt{2} x$ intersects the graph of $\sin{\pi x \over 2}$ at $(0,0)$ and $({1 \over 2}, {\sqrt{2} \over 2})$, so you need to show that the graph of  $\sin{\pi x \over 2}$ lies above the line $y = \sqrt{2} x$ on $[0, {1 \over 2}]$. This follows from the concavity of $\sin{\pi x \over 2}$ on the interval.

Let $x=\frac12 - \frac{2t}\pi$. Then, the inequality
$$ \sin \pi x \le 4x(1-x), \>\>\>\>\> x\in [0,1]\tag 1$$

is equivalent to

$$f(t) = \frac{\sin t }t -  \frac{2\sqrt2}\pi \ge 0, \>\>\>\>\> t \in [-\frac\pi4,\frac\pi4]\tag 2$$

Due to the symmetry, consider only the positive domain $t\in[0,\frac\pi4]$. Note that,

$$f(\frac\pi4) = 0,\>\>\>\>\>
f'(t) = \frac{\cos t}{t^2} (t-\tan t)
=-\frac{\cos t}{t^2} \int_0^t \tan^2 udu \le 0$$

which means $f(t) \ge 0 $ for $t \in [0,\frac\pi4]$. Thus, the inequality (2) and its equivalent (1) hold.


possible values of $P(A|C)$ when we know that $P(A|B)=P(B|C) = 1/2$ What are possible values of $P(A|C)$ when we know that $P(A|B)=P(B|C) = \frac12$ ?

It is fairlt difficult for me, and I can't deal with it.  I know conditional probablity and bayesian probablity, however it doesn't help me.  Could you help me ?

Fix $t\in[0,1]$.  We can associate the sets $(A\cup B\cup C)^\complement$, $A\setminus (B\cup C)$, $B\setminus (C\cup A)$, $C\setminus (A\cup B)$, $(B\cap C)\setminus A$, $(C\cap A)\setminus B$, $(A\cap B)\setminus C$, and $A\cap B\cap C$ with the following probabilities: $$0\,,\,\, 0\,,\,\, \frac{t}{3}\,,\,\, \frac{1-t}{3}\,,\,\, \frac{1-t}{3}\,,\,\, \frac{t}{3}\,,\,\, \frac{1-t}{3}\,,\text{ and  }\frac{t}{3}\,,$$ respectively.  Then, $\text{Prob}(A|B)=\text{Prob}(B|C)=\frac{1}{2}$ and $\text{Prob}(A|C)=t$.

Hint:
use the Law of total probability,
https://en.wikipedia.org/wiki/Law_of_total_probability

and consider stochastic independence (if given).
Answer for more information^^


Prove $\rm AB = BA = 0$ if the set of nonzero eigenvalues of $\rm A + B$ is union of set of nonzero eigenvalues of $\rm A$ and $\rm B$. $A$ and $B$ are $n×n$ symmetric matrices, their non-zero eigenvalues are $(\lambda_{1},\ldots,\lambda_{r} )$，$( \mu_{1},\ldots,\mu_{s} )$.  If the nonzero eigenvalues of $A + B$ are  $(\lambda_{1},\ldots,\lambda_{r}，\mu_{1},\ldots,\mu_{s} )$,  show that $AB = BA = 0$

$\bf{Note}$ that the (non-zero) eigenvalues of a matrix form a multiset, so the above is a union of multisets.

Edit:
As stated this is not true. Take
 $$A=\begin{bmatrix}2 &0 &0\\0&1 &0 \\ 0 & 0&0\end{bmatrix} \text{ and } B= \begin{bmatrix}0 &0 &0\\0&1 &0 \\ 0 & 0&1\end{bmatrix}$$
then $AB \ne 0$, but $A,B, A+B$ fulfill the hypothesis.
So you can either assume that $A,B$ are positive semi-definite as orangeskid did or assume that the multiplicities add up as I will do:

What do we know about a diagonalizable matrix $M$? By looking at eigenspaces with non zero eigenvalue and the eigenspace with eigenvalue zero we get the decomposition
$$k^n= \operatorname{im} M \oplus \ker M$$

They are an isomorphism when restricted to their image.

By counting eigenvalues we know that
$$\operatorname{rank}{A}+ \operatorname{rank}{B} = \operatorname{rank}{(A+B)}.$$

This gives us $\dim\text{im}{A} + \dim \text{im}{B} =\dim \text{im}{(A+B)}$ by looking at the dimension of subspaces we know that the intersection must be $0$ and thus
$$\operatorname{im}{A} \oplus \operatorname{im}{B} =\operatorname{im}{(A+B)}.$$

In other word we get a decomposition
$$k^n = \operatorname{im}{A} \oplus \operatorname{im}{B} \oplus \ker (A+B).$$

How does
$\ker (A+B)$
looks like? We know that $\ker(A) \cap \ker(B) \subset \ker (A+B)$, but since $\operatorname{im}{A} $ and $\operatorname{im}{B} $ only intersect at zero we get equiality.

Thus
$$k^n = \operatorname{im}{A} \oplus \operatorname{im}{B} \oplus( \ker(A) \cap \ker(B)),$$

hence we can assume that $A+B$ is an isomorphism and diagonalize $A$ and $B$ on their images.

Let $O$ be orthonormal such that $A$ is diagonalized on $\operatorname{im} A$ and let $U$ be orthonormal such that $B$ is diagonalized on $\operatorname{im} B$

Choose appropriate $M,N,C$  such that
$$A+B=\begin{bmatrix}M &C\\ C^T& N\end{bmatrix} $$

then

$$T= \begin{bmatrix}O^T &0\\ 0& U^T\end{bmatrix}   \begin{bmatrix}M &C\\ C^T& N\end{bmatrix} \begin{bmatrix}O &0\\ 0& U\end{bmatrix} = \begin{bmatrix}O^T M O & O^T CU\\  U^T C^TO& U^TNU  \end{bmatrix} . $$

Now let us consider the trace of the square of $T$
\begin{align*} \operatorname{tr}(T^2) &=  \operatorname{tr} (O^T M^2O + O^TC C^TO)+ \operatorname{tr} (U^T N^2U + U^TC^T CU)  \\
&=  \sum \lambda_i + \operatorname{tr}(CC^T) + \sum \mu_i + \operatorname{tr}(C^TC)
\end{align*}

where
$$ \operatorname{tr}(T^2) = \operatorname{tr}(A+B)^2 =  \sum \lambda_i + \sum \mu_i $$
and $\operatorname{tr}(CC^T) >0,$
thus we can conclude that
$$ C=0.$$

Hence $T$ is diagonal, thus we have  $\operatorname{im}A \subset \ker B$ and $\operatorname{im}B \subset \ker A$.

As said by user60589 we need that

$$
r + s \leq n.
$$
$A,B$ and $A+B$ can be diagonalized, thus
$$
A = UD_{A}U^{T}
$$

$$
B =VD_{B}V^{T}
$$
where $U, V$ consist of eigenvectors to $A, B$ respectively. Algebraic and geometric manifold are the same because
we are studying symmetric matrices. Eigenvectors of different eigenvalues are always orthogonal to each other if we consider symmetric matrices because
$$
\lambda_{i} \langle x_{i}, x_{j} \rangle = \langle A x_{i}, x_{j} \rangle = \langle x_{i},A x_{j} \rangle = \lambda_{j} \langle x_{i}, x_{j} \rangle
$$
, from this and  $\lambda_{i} \neq \mu_{j}$ for all $i,j$ it follows
$$
V^{T}U = 0
$$
simply because $V,U$ consist of eigenvectors which must by assumption all be orthogonal to each other.
Thus
$$
BA\boldsymbol{v}=VD_{B}V^{T}UD_{A}U^{T}\boldsymbol{v} = 0
$$

What about the case $\lambda_{i} = \mu_{j}$ for some $i,j$ ?
So we have 2 linear independent eigenvectors $\boldsymbol{v} \neq \boldsymbol{w}$ to one eigenvalue $\lambda$ (with algebraic manifold 2). So we can choose 2 orthogonal eigenvectors $\boldsymbol{v},\boldsymbol{l}$ replacing our previous ones:
$$ \boldsymbol{l} = \boldsymbol{w}- \frac{ \langle \boldsymbol{v}, \boldsymbol{w} \rangle}{||\boldsymbol{v}||^{2}}\boldsymbol{v} $$
which is orthogonal to $\boldsymbol{v}$.

The eigenvalue equations read now:
$$
(A+B)\boldsymbol{v} = \lambda \boldsymbol{v}
$$
$$
(A+B)\boldsymbol{l}=(A+B)\boldsymbol{w}-(A+B)\frac{ \langle \boldsymbol{v}, \boldsymbol{w} \rangle}{||\boldsymbol{v}||^{2}}\boldsymbol{v} = \lambda \left(\boldsymbol{w} +\frac{ \langle \boldsymbol{v}, \boldsymbol{w} \rangle}{||\boldsymbol{v}||^{2}} \boldsymbol{v} \right) =\lambda \boldsymbol{l}.
$$
From here we can use the argumentation given above.


Prove that:$\int_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int_{f(a)}^{f(b)} f^{-1}(x) dx$ I just wanted to ask, if my proof is correct.
I haven't seen the equation before, but I think it's quite useful.

Let $f$ be an bijective differentiable function. Then the inverse function $f^{-1}$ exists and the following equation holds:

  $$\int\limits_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int\limits_{f(a)}^{f(b)} f^{-1}(x) dx$$

Proof.

$f$ is an bijective differentiable function and $F$ is an antiderivative of $f$.

First we need to find an antiderivative of $f^{-1}$.

$\int f^{-1}(x) dx$ with substitution $x = f(y)$ yields:

$$\int y \cdot f'(y) dy = y \cdot f(y) - \int f(y) dy = y \cdot f(y) - F(y)$$

resubstitution yields:

$$\int f^{-1}(x) dx = x \cdot f^{-1}(x) - F(f^{-1}(x))$$
hence $\int\limits_{f(a)}^{f(b)} f^{-1}(x) dx = \left[x \cdot f^{-1}(x) - F(f^{-1}(x)) \right ]_{f(a)}^{f(b)}$

$$=b \cdot f(b) - F(b) - (a \cdot f(a) - F(a)) = F(a) - F(b) + b \cdot f(b) - a \cdot f(a)$$
$$= \int\limits_{b}^{a} f(x)  dx + b \cdot f(b) - a \cdot f(a) = -\int\limits_{a}^{b} f(x)  dx + b \cdot f(b) - a \cdot f(a)$$

All in all:

$$\int\limits_{f(a)}^{f(b)} f^{-1}(x) dx = -\int\limits_{a}^{b} f(x)  dx + b \cdot f(b) - a \cdot f(a)$$

which is equal to

  $$\int\limits_{a}^{b} f(x) dx = b \cdot f(b) - a \cdot f(a) - \int\limits_{f(a)}^{f(b)} f^{-1}(x) dx$$

q.e.d.

Look at the area represented by the integrals.  For simplicity take the case $0 < a < b$ and $0 < f(a) < f(b)$. The first integral represents the area under the curve, above the $x$-axis, between $x=a$ and $x=b$.  The second integral represents the area to the left of the curve, to the right of the $y$-axis, between $y=f(a)$ and $y=f(b)$. When combined they give the region which could be described as the rectangle with width $b$ and height $f(b)$, less the rectangle with width $a$ and height $f(a)$, both in the first quadrant with one vertex at the origin.

It should be almost obvious that the identity in question is an immediate consequence of the formula for "integration by parts". Using integration by parts we have $$\int_{a} ^{b} f(x) \, dx=bf(b) - af(a) - \int_{a} ^{b} xf'(x) \, dx\tag{1}$$ and then we put $f(x) =t,x=f^{-1}(t)$ in the integral on right to get $$\int_{a}^{b}xf'(x)\,dx=\int_{f(a)}^{f(b)}f^{-1}(t)\,dt\tag{2}$$ Our formula follows from $(1)$ and $(2)$. The proof can also be carried out using Riemann-Stieltjes integral. This has the advantage that it avoids the derivative $f'(x) $ and establishes the identity under much weaker conditions on $f$ (namely that $f$ should be strictly monotone on $[a, b] $).

The identity in question has a nice geometric interpretation (see other answers here) which helps to convince us of the truth of the identity, but it does not constitute a proof (in the same way as geometric arguments can not be used to prove mean value theorem or intermediate value theorem).


Proving at least one number $\le 1/4$ How to prove that at least one of $\alpha(1-\beta),\beta(1-\gamma),\gamma(1-\alpha)$ is lesser than $1/4$ given that $\alpha,\beta,\gamma\in(0,1)$

The original problem was proving at least one triangle (except the middle one) formed by dividing an equilateral triangle into four triangles is at least 4 times smaller than the original which I simplified to above. Don't know how to proceed, any hints?

WLOG let $\alpha \ge \gamma$. Then we have:

$$\gamma(1-\alpha) \le \alpha(1-\alpha) = \alpha - \alpha^2 = \frac 14 - \left(\frac 12 - \alpha\right)^2 \le \frac 14$$

$\alpha(1-\beta)\times\beta(1-\gamma)\times\gamma(1-\alpha) = \alpha(1-\alpha)\times\beta(1-\beta)\times\gamma(1-\gamma)=X$

Since $\alpha,\beta,\gamma$ are independent we can find maximum value of $\alpha(1-\alpha)=1/4$ at $\alpha=1/2$. Similarly for other two.
Therefore $X \leq 1/16$.

If product of $\alpha(1-\beta),\beta(1-\gamma),\gamma(1-\alpha)$ is $\leq 1/16$, atleast one of $\alpha(1-\beta),\beta(1-\gamma),\gamma(1-\alpha)$ is $\leq1/4$ and atleast one of $\alpha(1-\beta),\beta(1-\gamma),\gamma(1-\alpha)$ is $\geq1/4$. $($Since at $\alpha=\beta=\gamma=1/2$ all three are $1/4)$


Maxima and minima involving exponential random variables I have a scenario where events can be modeled as exponentially distributed random variables, however, I need to consider logical combinations of these variables. For example, say I have three timers that go off according to an exponential distribution. I wait until one of two conditions are met: both timers 1 and 2 ring, or both timers 1 and 3 ring. I need to find the expected amount of time for this to occur. Eventually, I will earn some reward for the amount of time I wait, and want to figure out the expected reward. But right now I'm having issues with this probability. In general, I could have any number of timers, but my logical end conditions can always be written as the OR of ANDs.

My first thought would be to use conditional expectation to break it into pieces. But first, I think I need the probability at any given time none of my stopping criteria have been met. That is, for time $t$, $P(t<\min\{\max(1\land 2),\max(1\land3)\})$. Then integrate this from 0 to infinity, and call this probability $T$.

Let $T$ be the event that I have not yet stopped at time time $t$. That is, I'm trying to figure the probability that time $t$ has passed and my stopping criteria have not yet been met.

$E[T] = E[T|(1\land 2)]*P(1\land2) + E[T|(1\land 3)] *P(1\land 3)$

$= \frac{\lambda_1\lambda_2}{(\lambda_1+\lambda_2+\lambda_3)^2}E[P(t>\max(1,2))] + \frac{\lambda_1\lambda_3}{(\lambda_1+\lambda_2+\lambda_3)^2}E[P(t>\max(1,3))]$

Am I doing the conditional expectation correctly or is there a better approach that might scale with an arbitrary number of timers?

It may be easier to work with CDFs: If $X_1 \sim Exp(rate=\lambda_1)$
and $X_2 \sim Exp(rate = \lambda_2)$, then the CDF of $W_{12} = \max(X_1, X_2)$ has the CDF
$$F_{12}(t) = P(W_{12} \le t) = (1-e^{-\lambda_1 t})(1-e^{-\lambda_2 t}),$$
from which you can find $E(W_{12}).$

A somewhat similar formula expresses the CDF of the minimum of two
random variables.

For particular known rates $\lambda_i,$ it is easy to get a good
approximation of the overall problem using simulation. The
following simulation uses rates $\lambda_i = i,$ for $i = 1,2,3.$
From the simulation we get (correct to 2 or 3 significant digits)
$E(T) \approx 1.03,\,SD(T) \approx 0.98,\,$ and $P(T > 2) \approx 0.14.$

 m = 10^6;  lam1 = 1; lam2 = 2; lam3 = 3
 x1 = rexp(m, lam1); x2 = rexp(m, lam2);  x3 = rexp(m, lam3)
 w12 = pmax(x1,x2);  w13 = pmax(x1, x3)
 t = pmin(w12, w13)
 mean(t);  sd(t);  mean(t > 2)
 ## 1.035684    # aprx E(T)
 ## 0.9805355   # aprx SD(T)
 ## 0.136419    # aprx P(T > 2)

If this is an applied problem  with known $\lambda_i,$ simulation
may be all you need. In any case simulation can be a good way
to check a few particular instances of general analytic results.

Note: The maximum of exponentials is not exponential. The minimum
is exponential. If $W = max(X_1, X_2),$ where $X_1, X_2 \stackrel{iid}{\sim} Exp(\lambda),$ then
$E(W) = \frac{1}{2\lambda} + \frac{1}{\lambda};$ time $\frac{1}{2\lambda}$ for the first, and then (by the no-memory property)
time $\frac{1}{\lambda}$ for the second.


Lebesgue Continuity for nonmeasurable sets I proved this problem: Let $E =\bigcup\limits_{i=1}^{\infty} E_i$ where $E_i$ are Lebesgue measurable subsets of $[0,1]$, and $E_1 \subseteq E_2 \subseteq E_3 \subseteq ....$ Show that $m^*(E_i) \rightarrow m^*(E) $ as $i \rightarrow \infty$ where $m^*$ is Lebesgue outer measure.

But now, i need some hints how to spell out second part:
Is the conclusion above still true if the sets $E_i$ are not necessarily measurable?

I think the answer is No, and i try using the fact: that for any arbitrary set $E$, there exists a Lebesgue measurable set $A$ such that $E\subseteq A$ and $m^*(A) = m^*(E)$.

How do I continue from here? Thanks

If the sets $E_i$ are not Lebesque measurable then the question doesn't make any sense...the sequence $m^*(E_i)$ is not well-defined.


Expected value in dice game Consider the following game:

Successive rolls six six-sided dice are made.

In each roll some (o none) numbers are marked.

The goal is to mark the 6 numbers.

A number can be marked only if you have marked the previous number (in the current roll or earlier).

If $X$ is the number of rolls needed to win the game, what is the exact value of $\mathbf{E}[X]$?
In the two examples below they were obtained $X = 6$ and $X = 2$ (each row corresponds to a roll and green the marked numbers).

Win in six rolls:

Win in two rolls:

Start by computing the chance to get the next $n$ numbers you need.  The chance to get $0$ is $(\frac 56)^6$.  To get $1$, you need to get that one and not get the next one, so the chance is $6(\frac 16)(\frac 56)^5$.  To get $2$, the chance is ${6 \choose 2}(\frac 16)^2(\frac 56)^4$ and so on.  If you don't need that many, add in the chances of getting more.

Let $x_i$ be the expected number of rolls given that you already have $i$ numbers.  If you already have $5$, the expected number of rolls is $x_5=1+(\frac 56)^6x_5$ because if you don't get a $6$ you are back where you started.  $x_5\approx 1.5035$  If you already have $4$, we have $x_4=1+6(\frac 16)(\frac 56)^5x_5+(1-(\frac 56)^6-6(\frac 16)(\frac 56)^5)x_4$ because on the next roll you might get them both, you might get $1$ and need $x_5$ more tries, or you might get none and need $x_4$ more tries.  They get messier each step, but the approach is the same.  If you are happy with numerics, you can put it into a spreadsheet.

Of the $46,656$ possible dice rolls:

each individual number appears $31,031$ times.
A 2-run starting at any particular number appears $19,502$ times.
3-runs appear $11,340$ times
4-runs appear $5,880$ times
5-runs appear $2,520$ times
$720$ rolls contain 6-runs.

These numbers do not exclude longer runs: for instance, only $19,502 - 11,340 = 8,162$ rolls have a run that goes 2-3 but not 2-3-4.  So transitioning from state 1 to state 3 only has $8,162$ possible ways it can happen.

With this information we can determine the state transition matrix:

$$A = \frac{1}{46656}\left(\begin{matrix}
15,625 & 11,529 & 8,162 & 5,460 & 3,360 & 1,800 & 720 \\
0 & 15,625 & 11,529 & 8,162 & 5,460 & 3,360 & 2,520 \\
0 & 0 & 15,625 & 11,529 & 8,162 & 5,460 & 5,880 \\
0 & 0 & 0 & 15,625 & 11,529 & 8,162 & 11,340 \\
0 & 0 & 0 & 0 & 15,625 & 11,529 & 19,502 \\
0 & 0 & 0 & 0 & 0 & 15,625 & 31,031 \\
0 & 0 & 0 & 0 & 0 & 0 & 46,656
\end{matrix}\right)
$$

Finding the number of tries a typical game will take is then a matter of applying this matrix repeatedly to the start state, $(1, 0, 0, 0, 0, 0, 0)$; after $n$ applications, the value in the last slot will tell you the probability of the game having finished in $n$ or fewer rolls.

Unfortunately, the game is potentially infinite: there is no ironclad guarantee you'll ever see a particular number when you need it, so this appears (at first blush) to require an infinite number of additions.

But that's only at first blush.

The part that is potentially infinite here is simply this: that we will fail, potentially infinitely many times, to progress.  These are the $15,625$s along the diagonal: $(5/6)^6$ of the time we will not progress at all.  This is a geometric distribution!  On average it will take $\frac{46,656}{31,031}$ rolls to get a success.

With this we can make progress!

So the expected number of rolls to get from having the value $n$ to completing the task is $x_n$:

$x_6 = 0$

$x_5 = \frac{46,656 + 31,031x_6}{31,031}$

$x_4 = \frac{46,656 + 19,052x_6 + 11,529x_5}{31,031}$

$x_3 = \frac{46,656 + 11,340x_6 + 8,160x_5 + 11,529x_4}{31,031}$

etc.

This gets to some big fractions pretty quickly.  I've plugged it into Excel; it gave me $2.386$.  Honestly that feels low though!


Associativity of logical connectives According to the precedence of logical connectives, operator $\rightarrow$ gets higher precedence than $\leftrightarrow$ operator. But what about associativity of $\rightarrow$ operator?

The implies operator ($\rightarrow$) does not have the associative property. That means that $(p \rightarrow q) \rightarrow r$ is not equivalent to $p \rightarrow (q \rightarrow r)$. Because of that, the question comes op how $p \rightarrow q \rightarrow r$ should be interpreted.

The proposition $p \rightarrow q \rightarrow r$ can be defined in multiple ways that make sense:

$(p \rightarrow q) \rightarrow r$ (left associativity)
$p \rightarrow (q \rightarrow r)$ (right associativity)
$(p \rightarrow q) \land (q \rightarrow r)$

Which one of these definitions is used?

I could not locate any book/webpage that mentions about associativity of logical operators in discrete mathematics.

Please also cite the reference (book/reliable webpage) that you use to answer my question (as I'm planning to add this to wikipedia page about 'logical connectives').

Thanks.

PS: I got this question when I saw this problem:
Check if following compound  proposition is tautology or not:

$$ \mathrm{p} \leftrightarrow (\mathrm{q} \wedge \mathrm{r}) \rightarrow \neg\mathrm{r} \rightarrow \neg\mathrm{p}$$

Some logical operators are associative: both $\wedge$ and $\vee$ are associative, as a simple check of truth tables verifies. Likewise, the biconditional $\leftrightarrow$ is associative.
However, the implication $\rightarrow$ is not associative. Compare $(p\rightarrow q)\rightarrow r$ and $p\rightarrow(q\rightarrow r)$. If all of $p$, $q$, and $r$ are false, then $p\rightarrow (q\rightarrow r)$ is true, because the antecedent is false; but $(p\rightarrow q\rightarrow r$ is false, because $r$ is false, but $p\rightarrow q$ is true. They also disagree with $p$ and $r$ are false but $q$ is true: then $p\rightarrow(q\rightarrow r)$ is true because the antecedent is false, but $(p\rightarrow q)$ is true, $r$ false, so $(p\rightarrow q)\rightarrow r$ is false.
Since they take different values at some truth assignments, the two propositions are not equivalent, so $\to$ is not associative.

Note that since it's not a binary operator, but a binary relation, associativity does not always yield the same semantics, of reducing composition pairings to a queue. Suppose $p \implies q \implies r.$ But if $r$ is false, $p \implies q$ being true does not imply $r$ is true. Likewise, $p \implies q$ can be false while $p \implies (r \implies q)$ is true, so $(p \implies (q \implies r)) \implies (p \implies q), (q \implies r)$ is not necessarily true. Finally, if $(p \implies q) \implies r$, and $p \implies (q \implies r),$ and $p \equiv F,$ we have no reason to think $q \implies r$, and $p \implies q$ can still be true, which would mean $r$ is true regardless of whether $p \implies r$.


Formal Deduction (logic) Question: $\lnot C, (B \to \lnot C) \to A \vdash (A \to C) \to F$ I've been stuck on this question for around two hours now.

I'm trying to prove that:

$\lnot C, \ (B \to \lnot C)\to A \vdash (A \to C)\to F $

I'm trying to get my second last step to be:

$\lnot C, \ (B \to \lnot C)\to A, \ A \to C \vdash F $,

but I have no idea how to get started. I'm not sure what I can start off with, especially due to the $(B \to \lnot C)\to A)$ part.

I'm allowed to use 11 rules of formal deducibility/deduction.

If anyone can help me get started, or give me some hints, I would be extremely thankful!

The intuition to prove $\lnot C, \ (B \to \lnot C) \to A \vdash (A \to C) \to F$ is that from the assumption $\lnot C$ it follows that $B \to \lnot C$ (if $\lnot C$ holds, then in particular it holds under the additional hypothesis $B$).
From that, you can apply modus ponens with $(B \to \lnot C) \to A$ to get $A$, and then you can easily find a contradiction with the assumption $A \to C$ (to be discarded) and $\lnot C$. From a contradiction you can derive everything (principle of explosion or ex falso quodlibet), in particular $F$. Finally, you just have to discharge the hypothesis $A \to C$.

Formally, the derivation in natural deduction that you are looking for is the following:

\begin{align}
 \dfrac{\dfrac{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\dfrac{\lnot C \qquad \dfrac{[A \to C]^* \qquad \dfrac{(B \to \lnot C) \to A  \qquad \dfrac{\lnot C}{B \to \lnot C}\to_i}{A}\to_e\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!}{C} \lnot_e\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!}{\bot}\lnot_e\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!}{F}\bot_e\!\!\!\!\!\!\!\!\!}{(A \to C) \to F}\to_i^*
\end{align}

where the falsehood or absurdity (a formula that is always false) is denoted by $\bot$, and $[A]^*$ stands for an assumption $A$ that has been discharged by the rule $*$.

Here is one way to show that

  ¬C, (B→¬C)→A⊢(A→C)→F

hoping that it will provide some hints how one might proceed.

However, it uses the Law of Excluded Middle (LEM), Reiteration (R) and Modus Tolens (MT) besides introduction and elimination rules. Discussions of these can be found in forallx linked to below if one needs alternatives to them.

I assume "F" refers to False or "⊥". If not, since it is not referenced in the premises, one can derive "F" from "⊥" using Explosion.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/


Please prove that $\lim\limits_{n \to \infty} \int_0^3 \sqrt{\sin(n/x)+x+1}\,dx$ exists and evaluate it. Prove that $\displaystyle{\lim_{n \rightarrow \infty}} \int_0^3 \sqrt{\sin(n/x)+x+1}\,dx$ exists and evaluate it.

I'd love to take a calculus class in which such problems are assigned as homework. Not sure I would pass it, though.

First, let's consider the claim (made in another answer) that $\sin(n/x)\to 0$ as $n\to\infty$. Of course, this is false pointwise. But as a sequence in $L^\infty([0,3])$, $\sin(n/x)$ converges to $0$ in the weak* sense:  namely, for every $\varphi\in L^1([0,3])$ we have
$$\lim_{n\to\infty} \int_0^3 \sin(n/x)\, \varphi(x)\,dx =0 \tag1$$
One way to prove (1) is to approximate $\varphi$ by a smooth function with compact support  in $(0,3)$, change the variable $y=1/x$, and integrate by parts:
$$
\int_{1/3}^\infty \sin(ny)\, y^{-2}\varphi(1/y)\,dy = \frac{1}{n} \int_{1/3}^\infty \cos (ny )\, (y^{-2}\varphi(1/y))'\,dy \to 0
\tag2$$
since the integrand is bounded in $L^1$ norm.

If we did not have the square root, the story would end with
$$ \int_0^3 (\sin(n/x)+x+1)\,dx \to \int_0^3 (x+1)\,dx \tag3$$
by virtue of weak* convergence. But the square root changes everything: weak* limits do not commute with nonlinear operations such as squaring or taking root. We can scratch everything written so far.

Let's take a broader view: we are integrating the composition of continuous function $F(x,y)=\sqrt{y+x+1}$, defined on $[0,3] \times [-1,1]$, with the oscillatory function $y=\sin(n/x)$. Forgetting the particular form of $F$, what should the limit
$$\lim_{n\to\infty} \int_0^3 F(x,\sin(n/x))\,dx \tag4$$
be in terms of $F$? Being a positive linear functional of $F$, (4) must be of the form $\iint F(x,y)\,d\mu(x,y)$ for some positive measure $\mu$ on $[0,3] \times [-1,1]$. Let $\mu_n$ be the pushforward of the Lebesgue measure on $[0,3]$ under $x\mapsto (x,\sin(n/x))$. This is a measure of total mass $3$ supported on a wiggly curve. This measure $\mu$ will be the weak* limit (now in the sense of dual of $C(K)$) of $\mu_n$. A little computation involving $|\sin't|=\sqrt{1-\sin^2 t}$ reveals this limit:
$$\mu(x,y)=\frac{1}{\pi\sqrt{1-y^2}}\,dx\,dy \tag5$$
Since this is calculus homework, I shouldn't have to justify (5) in detail. I'll only say that this is the measure you get by projecting arclength measure on a circle to its diameter, and normalizing. It's also the weight for which Chebyshev polynomials are orthogonal, for this very  reason.

Now we are almost done: the limit is equal to
$$\int_{-1}^1 \int_0^3 \frac{\sqrt{y+x+1}}{\pi \sqrt{1-y^2}} \,dx\,dy = \frac{2}{3\pi}\int_{-1}^1 \frac{(y+4)^{3/2}-(y+1)^{3/2}}{\sqrt{1-y^2}}\,dy \tag6$$
Because I think that C- will be enough for me, I'll just stick (6) into Maple: the answer is
$$\boxed{\dfrac{4}{9\pi} \left(16\sqrt{5}\,E\left(\sqrt{2/5}\right)-3\sqrt{5}\,K\left(\sqrt{2/5}\right)-4\sqrt{2}\right)} \tag7$$
where $K$ and $E$ are complete elliptic integrals of the first and second kind, respectively. Having cheated with Maple, I wouldn't want to lose even more points for not putting my answer in a box.

Numerically, (7) is about $4.5957$, which agrees with direct numeric computations of the given integral.

For comparison, $\int_0^3 \sqrt{x+1}\,dx=14/3\approx 4.6667$. The square root is concave, which means that averaging it by means of oscillatory function stuck inside makes the integral smaller. This is why the answer is strictly less than $14/3$.

As sine is continous everywhere, we do not need to worry about using any special rules.

$\sin(n/x) \to 0$ as $n \to \infty$ and so

$\lim_{n \to \infty} \int^3_0 f(x) \mathrm{d} x \to \int^3_0 (x+1)^{1/2} \mathrm{d} x$ which just boils down to reversing the chain rule.


An algorithm to pair up words where a change has been made at a certain position? Say I have a bag of random 'words' of a given length such as:

$$ADHBCBAB$$

Now I make another bag and copy all the words except for making a random change in the 5th position of each word. (Each word is changed in the exact same position to a random letter).

Now I give these two bags of words to someone and they have to

1) Determine in what position the change was made

2) Pair up the words from each bag

or

3) Determine that such a pairing is impossible (i.e. the first person cheated)

I think this problem is harder than it sounds when you don't know in what position the change was made in. Do you know if this can be translated into a well known problem with a known solution. And weather that solution is Polynomial time?

I don't think it's possible to determine in general. Suppose you start with AB and BA, and the changed words are AA and BB. Which letter was changed?

The edit distance algorithm is a popular algorithm for pattern matching. You can read about the algorithm on Wikipedia: https://en.wikipedia.org/wiki/Edit_distance


Expected value of game involving 100-sided die The following question is from a Jane Street interview.

You are given a 100-sided die. After you roll once, you can choose to either get paid the dollar amount of that roll OR pay one dollar for one more roll. What is the expected value of the game? (There is no limit on the number of rolls.)

P.S. I think the question assumes that we are rational.

Let $v$ denote the expected value of the game. If you roll some $x\in\{1,\ldots,100\}$, you have two options:

Keep the $x$ dollars.
Pay the \$$1$ continuation fee and spin the dice once again. The expected value of the next roll is $v$. Thus, the net expected value of this option turns out to be $v-1$ dollars.

You choose one of these two options based on whichever provides you with a higher gain. Therefore, if you spun $x$, your payoff is $\max\{x,v-1\}$.

Now, the expected value of the game, $v$, is given as the expected value of these payoffs:
\begin{align*}
v=\frac{1}{100}\sum_{x=1}^{100}\max\{x,v-1\}\tag{$\star$},
\end{align*}
since each $x$ has a probability of $1/100$ and given a roll of $x$, your payoff is exactly $\max\{x,v-1\}$. This equation is not straightforward to solve. The right-hand side sums up those $x$ values for which $x>v-1$, and for all such values of $x$ that $x\leq v-1$, you add $v-1$ to the sum. This pair of summations gives you $v$. The problem is that you don't know where to separate the two summations, since the threshold value based on $v-1$ is exactly what you need to compute. This threshold value can be guessed using a numerical computation, based on which one can confirm the value of $v$ rigorously. This turns out to be $v=87\frac{5}{14}$.

Incidentally, this solution also reveals that you should keep rolling the dice for a $1 fee as long as you roll 86 or less, and accept any amount 87 or more.

ADDED$\phantom{-}$In response to a comment, let me add further details on the computation. Solving for the equation ($\star$) is complicated by the possibility that the solution may not be an integer (indeed, ultimately it is not). As explained above, however, ($\star$) can be rewritten in the following way:
\begin{align*}
v=\frac{1}{100}\left[\sum_{x=1}^{\lfloor v\rfloor-1}(v-1)+\sum_{x=\lfloor v\rfloor}^{100}x\right],\tag{$\star\star$}
\end{align*}
where $\lfloor\cdot\rfloor$ is the floor function (rounding down to the nearest integer; for example: $\lfloor1\mathord.356\rfloor=1$; $\lfloor23\mathord.999\rfloor=23$; $\lfloor24\rfloor=24$). Now let’s pretend for a moment that $v$ is an integer, so that we can obtain the following equation:
\begin{align*}
v=\frac{1}{100}\left[\sum_{x=1}^{v-1}(v-1)+\sum_{x=v}^{100}x\right].
\end{align*}
It is algebraically tedious yet conceptually not difficult to show that this is a quadratic equation with roots
\begin{align*}
v\in\left\{\frac{203\pm3\sqrt{89}}{2}\right\}.
\end{align*}
The larger root exceeds $100$, so we can disregard it, and the smaller root is approximately $87\mathord.349$. Of course, this is not a solution to ($\star\star$) (remember, we pretended that the solution was an integer, and the result of $87\mathord.349$ does not conform to that assumption), but this should give us a pretty good idea about the approximate value of $v$. In particular, this helps us formulate the conjecture that $\lfloor v\rfloor=87$. Upon substituting this conjectured value of $\lfloor v\rfloor$ back into ($\star\star$), we now have the exact solution $v=87\frac{5}{14}$, which also confirms that our heuristic conjecture that $\lfloor v\rfloor=87$ was correct.

All answers are correct, but maybe I can phrase it in a less complex manner. Say you in advance determine a value V, where if you roll V or some amount greater than V, you will stop the game. What is your expectation here? Given that you have rolled V or higher, you are uniformly likely to have rolled any value between V and 100. The expectation of this roll is 0.5*(V+100).
But how much did you pay to get to this point? On average, when an event occurs with probability p, where p is bounded between 0 and 1, then in expectation it takes 1/p independent trials to realize this event. The probability of rolling V or higher is (101-V)/100, so on average it will take 100/(101-v) rolls to achieve this outcome. Your expectation is gain - loss, which equals 0.5*(V+100) - 100/(101-V).
It is simple to see that by taking the derivative with respect to V and setting equal to zero, we can get the V that maximizes expectation. Assuming your opponent is perfectly rational, this is the only bid they will accept to play the game. As it turns out, we can solve the equation to get V equals approximately 87.


Gram Schmidt Process for finding orthonormal vector. I was just clearing my doubts on eigenvector and eigenvalues, but in one of the book, the eigenvector of degenerate eigenvalues had eigenvector to be linearly independent. And they made it to orthonormal eigenvector by Gram Schmidt process.
My question is

For any matrix with distinct eigenvalues are linearly independent eigenvectors. Then can we take those eigenvector and make it to be orthonormal by this process.

If that so then why we prefer or-tho normal basis to expand an vector just so that it can easily find us the constant associated with each eigenvector, because linearly independent eigenvector will do the work if we made them to normalize it use this process.

Edit- I know it will be absurd for large dimension, but am i thinking it in right way.

The process in general will change the property of being eigenvectors. However, we have this result: a real matrix has an orthonormal basis of eigenvectors if and only if it is a symmetric matrix.

Let me first introduce some terminology and notations. Let $A\in \mathbb{C}^{n\times n}$ be a square matrix. Consider for each $\lambda\in \mathbb{C}$ the space $E_{\lambda}=\{v\mid Av=\lambda v\}=\{v\mid (A-\lambda I)v=0\}$. If $\lambda$ is an eigenvalue, then $E_{\lambda}$ is called the corresponding eigenspace. The set of eigenvalues is denoted by $\sigma(A)$.
Assume now that $A$ is diagonalizable (there is a basis of eigenvectors). Equivalently, assume that $\mathbb{C}^n=\bigoplus_{\lambda\in \sigma(A)} E_{\lambda}$.
For each $\lambda\in \sigma(A)$, let $\{v_{\lambda,1}, \dots , v_{\lambda, d_{\lambda}}\}$ be a basis of $E_{\lambda}$ (so $d_\lambda=\dim(E_{\lambda}))$. Applying Gramm-Schmidt to each of these bases yields new orthonormal bases $\{v_{\lambda, 1}', \dots , v_{\lambda, d_{\lambda}}'\}$ for each $E_{\lambda}$, however, there is no reason why $E_{\lambda}$ should be orthogonal to $E_{\mu}$ for $\lambda\neq \mu$ and $\lambda,\mu\in \sigma(A)$.
If, for example, $A$ is a symmetric real matrix, then different eigenspaces are automatically orthogonal to each other. In that case, all you have to do to find an orthonormal basis of eigenvectors is simply applying the Gramm-Schmidt procedure to any basis of eigenvectors.
On the other hand, there are definitely matrices such that different eigenspaces are simply not orthogonal to each other. In that case, no orthonormal basis of eigenvectors can exist.
EDIT: Assume that $A\in \mathbb{R}^{n\times n}$ is diagonalizable. Let $\{v_1, \dots, v_n\}$ be a basis of eigenvectors such that $Av_i=\lambda_iv_i$ (I do not assume the $\lambda_i$'s to be distinct).
If your reasoning is correct, then applying the Gramm-Schmidt procedure would yield an orthonormal basis of eigenvectors $\beta=\{w_1, \dots ,w_n\}$ of $A$. Let $P$ be the matrix whose $i$'th column is precisely $w_i$. Then $PP^T=Id$ as $P$ is an orthogonal matrix. Furthermore, $P$ acts as the base change matrix $[Id]_{\beta}^{\text{st}}$ where $\text{st}$ is the standard basis. Hence $$A=PDP^{-1}=PDP^{T}$$ where $D$ is the diagonal matrix having the corresponding eigenvalues on it's diagonal. Now, note that
\begin{eqnarray}
 A^T &=& (PDP^T)^T\\
&=& PD^TP^T\\
&=& PDP^T\\
&=& A.
\end{eqnarray}
Thus, $A=A^T$. This shows that if your reasoning is correct, then $A$ must be symmetric!


Why does not differentiability at a point imply the existence of limit of derivative tending to the point What is wrong with the following argument?
Let $f$ be a real continuous function which is differentiable at every point $x\not=a$. By mean value theorem,
$$ \frac{f(a+h)-f(a)}{h} = f'(\xi) \qquad (h\not=0)$$
where $\xi$ is between $a$ and $a+h$. Letting $h\rightarrow0$, we obtain
$$ f'(a) = \lim\limits_{\xi\rightarrow a}f'(\xi) \mbox{.} $$
Therefore, f'(a) exists if and only if $\lim\limits_{\xi\rightarrow a}f'(\xi)$ exists.
That the only if statement is false can be seen from $f(x)=x^2\sin\frac{1}{x}$ for $x\not=0$, $f(0)=0$, which shows that $f'(0)$ exists but $\lim\limits_{x\rightarrow0}f'(x)$ does not. Why?

The problem is that $\xi\rightarrow a$ doesn't imply $h\rightarrow0$.
Observe that if $a<\xi<a+h$ or $a>\xi>a+h$, then $|h|>|\xi-a|$. Now examine the definition of limit, one can see at once that $|h|$ may very well be large given that $|\xi-a|$ is arbitrarily small.


Is this a modified Monty Hall problem (numbered doors)? On a job interview, I got this question:

Monty placed a car and two goats behind three identical doors (and the things do not move during the game). You receive the prize which is behind the door you picked in the second round.
You choose door number 1. Then, Monty opens door number 3 and you see a goat there. If you want to win a car, should you change your guess from door no 1 to door no 2?

I answered: YES and was told I was wrong. The interviewer explained me, that the difference between classical MH problem and this problem is that this problem clearly states, that Monty opens door number 3, which reduces the state space and after that my chance is 50/50.
I still believe chance of picking right door at first attempt is 1/3 and this probability is not changed by the information that Monty opens door number 3.
Am I missing something or was the interviewer wrong?

The interviewer is wrong in stating that Monty "clearly" picked door 3 and that therefore the chances are 50-50.  Monty picked door number 3  in this particular case but he didn't say why.

The interviewer is correct that if we do not know why Monty showed door number 3 (does he always show door number 3 no matter what is behind it or what you pick-- does he always pick a door at random that you did not pick no matter what is behind it-- does he always show a door with a goat you did not pick?) then the question can not really be answered. (But he is wrong in assuming that it is 50-50).  We have to make an assumption but... what assumptions are valid and which aren't?

Here are several possible rules Monty could be played by:

Classic:  Monty always shows you a goat you do not pick.  Strategy: switch. 2 out of 3 in your favor.

Random:  Monty picks a door you did not pick and this time it just randomly happened to be a goat.  Strategy: doesn't matter. 2 out of 4 whether you switch or stay.

Devious: If you pick the car Monty will show you a goat in the hopes that you will assume a classic game.  If you pick a goat he won't give you a choice.  Strategy: stay.  100% in your favor.

Tough luck: Monty will always show you the car if he can.  He'll only show you a goat if you pick the car.  Strategy: stay. 100% in your favor.

Warped: There is one goat with a spotted tail.  Monty will always show you a door you did not pick that does not have the spotted tail goat.  Strategy: stay.  If you switch it is 2 in 3 that you will get the goat with the spotted tail.  So it is 2 in 3 if you don't switch you get the car.

Hierarchical: If you pick the goat with spotted tail, Monty will show you the goat without the spotted tail.  If you pick the goat without the spotted tail Monty will show you the car.  If you pick the car Monty will show you the goat without the spotted tail. Strategy: 50-50.

etc.

Which is the more likely one he is playing?  We can't tell.  And obviously these are not the only strategies.

====

Actually what would be fair is if it were worded like this:

You are an a game show and where you have a chance to pick a car or two goats.  The hosts goal is to give you a goat and keep you from picking a car.  You pick door 1 and he shows you door 3 has a goat and offers you a chance to switch to door 2.  Should you?

Answer: it doesn't matter.  Whichever door you pick he will put the goat behind it  after you pick it.

This answer is based on statistics and variable change if the person changes his decision and choses door number two instead of door number one the probability changes from 66.6% to 33.33% ...


Prove that the inequality $(1+ \frac{1}{n})^n < n$ holds for all $n \geq 3$ First we need to prove the basis. If we let $n=3$, then $(1+ \frac{1}{3})^3 < 3$

$(\frac{3}{3}+ \frac{1}{3})^3 < 3$

$(\frac{4}{3})^3 < 3$

$(\frac{64}{27}) < 3$

The inequality statement is true

For $P(n), (1+ \frac{1}{n})^n < n$

We assume that $(1+ \frac{1}{n})^n < n$ is true for $P(n+1)$

$(1+ \frac{1}{n+1})^{n+1} < n+1$

$(1+ \frac{1}{n+1})^{n})(1+ \frac{1}{n+1})^{1}) < n+1$

And then I'm stuck afterwards. I know that there are a variety of problems that use induction and they have different methods, but I only know the ones that are similar to $1+2+3+...+n = n+2$  or $7^n-8^n$ is divisible by $8$. Is there any technique to tackle this type of problem?

As canaaerus points out in the comments, we can in fact prove
$$\left(1+\frac{1}{n}\right)^n<3$$
In fact there's a proof of this that doesn't even use induction! I think it's worth writing down here since it's so simple (and it won't be giving away the answer to the homework problem, since the teacher presumably expects induction). We will binomially expand and use the following facts:
$${{n}\choose{r}}=\frac{n\cdot(n-1)\cdot\ldots\cdot(n-r+1)}{r!}\leq\frac{n\cdot n\cdot\ldots\cdot n}{r!}=\frac{n^r}{r!}$$
and (for $r\geq1$)
$$\frac{1}{r!}=\frac{1}{1\cdot2\cdot3\cdot4\cdot\ldots\cdot r}\leq\frac{1}{1\cdot2\cdot2\cdot2\cdot\ldots\cdot 2}=\frac{1}{2^{r-1}}$$
Here's the proof:
$$\begin{align*}
\left(1+\frac{1}{n}\right)^n&=\sum^{n}_{r=0}{{n}\choose{r}}\frac{1}{n^r}\\
&\leq\sum^{n}_{r=0}\frac{n^r}{r!}\frac{1}{n^r}\\
&=\sum^{n}_{r=0}\frac{1}{r!}\\
&=1+\sum^{n}_{r=1}\frac{1}{r!}\\
&\leq1+\sum^{n}_{r=1}\frac{1}{2^{r-1}}\\
&<1+\sum^{\infty}_{r=1}\frac{1}{2^{r-1}}\\
&=1+2=3
\end{align*}$$

Your approach is totally right. Look for a way to relate what you have at the end to the induction hypothesis. For example, it's certainly true that
$$
\bigg( 1 + \frac1{n+1} \bigg)^n < \bigg( 1 + \frac1{n} \bigg)^n.
$$
So if you replace the former by the latter in your last desired inequality, does that help you?


Can't understand the logical structure of Euclid's infinitely many primes proof in Rosen's book. I'm reading the book Discrete Mathematics and It's Applications written by Rosen. At page 260, there is a proof using the technique similar to Euclid's. The following is my trying to understand the proof.

Assume that there are only finitely many primes, $p_1, p_2, ..., p_n$, I'll call it prime list later on. Let

$$Q = p_1p_2\cdots p_n+1.$$

By the fundamental theorem of arithmetic, $Q$ is prime or else it can be written as the product of two or more primes.

Let $p_j$ be an arbitrary prime in the prime list. Then $p_j$ can't divide $Q$, since:

If $p_j \mid Q$, then $p_j \mid (Q - p_1p_2\cdots p_n=1)$:

The is because $p_j \mid (p_1p_2\cdots p_n)$, so $p_j \mid (1\cdot Q+(-1)\cdot(p_1p_2\cdots p_n))$.
Since $p_j \mid (Q - p_1p_2\cdots p_n=1)$ is false, $p_j \not\mid Q$:

The only integer which divides all integer is $1$, but $p_j$ is a prime.

Since $p_j \not\mid Q$ also means that $Q$ is not $p_j$, $Q$ is not in the prime list. And since there is no prime in the prime list divides $Q$, $Q$ can't be written as the product of two or more primes. Finally, $Q$ has to be a prime, which is not in the prime list. We've arrived a contradiction.

I'm not sure whether it's correct, but that's how I understand the proof. And the following is the proof in my book:

I don't understand why it arrives the conclusion "Hence, there is a prime not in the list".

Please give me some hints...

If $p$ is a prime number such that $p\mid Q$ then, since $Q-p_1p_2\ldots p_n=1$, $p\notin\{p_1,p_2,\ldots,p_n\}$ because otherwise $p\mid1$.

I was going through checking it by arbitrary number.
 Let's  $Q= 9699691$ be the  prime number because as stated above   none of $p_j$ (i.e $2$ to $19$ prime numbers) divides $Q$ which can be represented as
$Q= 2\times3\times5\times7\times11\times13\times17\times19+1$ which must be prime number.
But $Q = 9699691=347\times27953$ which is a composite number.
The thing is that the statement "$Q$ is prime or else it can be written as a product of two or more primes" implies there exists a prime dividing $Q$ but it ain't at our list. Since none of the $p_j$ have this property, there has to be another prime not in the list which divides $Q$.. which is why
 $347$, $27953$ does... both are primes.

Hope, this clarifies everything.


Show that $\boldsymbol 1_{\{1/n\mid n\in\mathbb N^*\}\cup\{0\}}$ Riemann integrable over $[0,1]$. Let $$\mathcal D=\left\{\frac{1}{n}\mid n\in\mathbb N^*\right\}\cup\{0\}.$$
Show that $\boldsymbol 1_{\mathcal D}$ is Riemann integrable over $[0,1]$.

Attempts

Let $f_n(x)=\boldsymbol 1_{\mathcal A_n}$ where $A_n=[0,\frac{1}{2^n}]\cup\bigcup_{k=1}^n\left[\frac{1}{k}-\frac{1}{2^n},\frac{1}{k}+\frac{1}{2^n}\right]\cap[0,1]$. We have that $\bigcup_{n\in\mathbb N^*}A_n=\mathcal D$, and that $(f_n)$ is a sequence of step function that converge pointwise to $\boldsymbol 1_{\mathcal D}$ and $(f_n)$ is decreasing. In particular, $$0\leq \overline{S}(f) \leq \frac{(n+1)}{2^n}\to 0.$$
Since $\underline{S}(f)=0$, the claim follow. I recall that $\overline{S}(f)=\inf_{\sigma }\{\overline{S}_\sigma\mid \sigma \text{ is a subdivision of }[0,1] \} $ and $\underline{S}(f)=\sup_{\sigma }\{\underline{S}_\sigma\mid \sigma \text{ is a subdivision of }[0,1] \} $

and $\overline{S}_\sigma =\sum_{i}M_i(x_{i+1}-x_i)$ and $\underline{S}_\sigma =\sum_{i}m_i(x_{i+1}-x_i)$ where $m_i=\min_{[x_{i},x_{i+1}]}f$ and $M_i=\max_{[x_i,x_{i+1}]}f$.

Here an "easy" construction. Let $\varepsilon>0$ and $N$ s.t. $\frac{1}{N}<\varepsilon$. Let $$\sigma _k : 0<\frac{1}{kN}<\frac{2}{kN}<...<\frac{k}{kN},$$

$$\tilde \sigma _k : \frac{k}{kN}<\frac{k+1}{kN}<...<\frac{kN}{kN}=1,$$
and $$\tau_k: 0<\frac{1}{kN}<...<1.$$

If I denote $S^{\sigma _k}(a,b)$ is the upper Darboux sum on $[a,b]$, we have $$S^{\tau _k}(0,1)=S^{\sigma _k}\left(0,\frac{1}{N}\right)+S^{\tilde \sigma _k}\left(\frac{1}{N},1\right) \leq \frac{k}{kN}+S^{\tilde \sigma _k}\left(\frac{1}{N},1\right)< \varepsilon+S^{\tilde \sigma _k}\left(\frac{1}{N},1\right).$$

Now, you know that your function is piecewise continuous on $\left[\frac{1}{N},1\right]$ and thus integrable, and it's easy to prove that the integral is $0$. Therefore, $$\lim_{k\to \infty }S^{\tilde \sigma _k}\left(\frac{1}{N},1\right)=0,$$
and thus $$0\leq \bar S\leq \lim_{k\to \infty }S^{\tau_k}\leq \varepsilon+\lim_{k\to \infty }S^{\tilde \sigma _k}\left(\frac{1}{N},1\right)=\varepsilon.$$
We therefore proved that $\bar S=0,$ and the claim follow.

hint

Let $\epsilon>0$ given.

at $[\epsilon/2,1] $, $f $ is a step function, and it is integrable .
thus
there exist a subdivision $\sigma $ of $[\epsilon/2,1] $ such that

$U (f,\sigma)-L (f,\sigma)<\frac {\epsilon}{2} $

let $\sigma_1=\sigma \cup \{0\} $.
$$U (f,\sigma_1)-L (f,\sigma_1)=$$
$$\leq \frac {\epsilon}{2}+1.\frac {\epsilon}{2} $$

since $f $ is bounded at $[0,\frac {\epsilon}{2}] $.


How to prove that the curve $x^2+y^2-3=0$ has no rational points? There is a question 6.20 from the Book of Proof by Hammock that reads as follows:  We say that a point $P=(x,y) \in R^2$ is rational if both x and y are rational. More precisely, P is rational if $P=(x,y) \in Q^2$.  An equation $F(x,y)=0$ is said to have a rational pointif there exists $x_0, y_0 \in Q$ such that $F(x_0,y_0)= 0$.  For example, the cruve $x^2+y^2-1=0$ has rational point $(x_0,y_0)=(1,0)$.  Show that the curve $x^2+y^2-3=0$ has no rational points.
I can only take a guess at how to start a proof.  Doing algebra on both sides of $x^2+y^2-3=0$ yields $y = \sqrt{3-x^2}$.  Now, assume y is rational -- can x be rational?
Thus, let y = $a\over b$ for some integers a and b, $b \ne  0$,$gcd(a,b)=1$.  Then let y = $c\over d$ for some integers c and d, $d \ne  0$,$gcd(c,d)=1$.  Hence, ${a \over b} = \sqrt{3 - {{c^2} \over {d^2}}}$.
I think I may want to prove some sort of contradiction in terms of rational and irrational?  This as far as I got in my scratchwork for a proof.

Suppose $x=\frac{m}{n}$ and $y=\frac{a}{b}$, where both fractions are reduced. Then after a little algebra your equation says:
$$m^2b^2+n^2a^2=3b^2n^2$$
This is an integer equation. Two terms are divisible by $b^2$, so $b^2$ must divide the third term, $n^2a^2$. Since $a$ and $b$ are relatively prime, then actually $b^2$ divides $n^2$.
But similarly, $n^2$ divides $b^2$. So you may conclude $n=b$ and the equation reduces to
$$m^2+a^2=3n^2$$
The residues of squares mod $3$ are just $0$ and $1$, not $2$. So whne you look at that equation mod $3$:
$$m^2+a^2\equiv0$$
the only way it can work out is if $m^2\equiv a^2\equiv0$. So $3$ must divide both $m$ and $a$. But then look again at
$$m^2+a^2=3n^2$$
Now we know $9$ divides the left side. So $3$ must divide $n^2$. This is a contradiction, because now we are saying $3$ divides both $m$ and $n$, but $\frac{m}{n}$ was supposed to be a reduced fraction.
So the assumption that we could have a rational $x$ and $y$ was false.

$ y=\sqrt {3-x^2}$
For $y$ to be rational, we must have $ x^2\le3$ and $y=1\implies |x|\le\sqrt 3$ but there doesn't exist such a rational number $x$ except $|x|=1$ and for any $1<|x|<\sqrt 3, y\notin\Bbb{Q}$. Again for $x=\pm1, y=\sqrt 2\notin \Bbb{Q}$. Also $x$ cannot be $\pm\sqrt 3$.  Hence proved.


What is mathematical research like? I'm planning on applying for a math research program over the summer, but I'm slightly nervous about it just because the name math research sounds strange to me. What does math research entail exactly? For other research like in economics, or biology one collects data and analyzes it and draws conclusions. But what do you do in math? It seems like you would sit at a desk and then just think about things that have never been thought about before. I appologize if this isn't the correct website for this question, but I think the best answers will come from here.

Largely (very largely, so please take everything here with a grain of salt), there are two types of mathematical research, commonly referred to as 'theorem proving/problem solving' vs. 'theory building'. Typical characteristics of theorem proving/problem solving type research is to try and tackle a famous open problem, usually stated in the form of a conjecture as to the validity of a statement or the specification of a problem. Quite often this will entail spending a lot of time learning the relevant material, analyzing particular attempts at solutions, trying to figure out why they don't work, and hopefully come up with some improvement to an existing attempt, or a whole new attempt, that has a good change of working. Very famous open-standing questions include: The Riemann hypothesis and $P\ne NP$ (which are examples of theorem proving) and the solution of the Navier-Stokes equations (an example of problem solving), all three are in the Clay's Institute millennium problems list.

Theory building is a somewhat different activity that involves the creation of new structures, or the extension of existing structures. Usually, the motivation behind the study of these new structures is coming from a desire to generalize (in order to gain better insight or be able to apply particular techniques of one area to a broader class of problems) or there might be a need to these new structures to exist, due to some application in mind. Typical activities would include a lot of reading on relevant structures, understanding their global role, figuring out what generalizations or new structures would make sense, what the aim of the new theory will be, and then a long process of proving basic structure theorems for the new structures that will necessitate tweaking the axioms. A striking example of this kind of research is Grothendieck's reformalization of modern algebraic geometry. Cantor's initial work on set theory can also be said to fall into this kind of research, and there are many other examples.

Of course, quite often a combination of the two approaches is required.

Today, research can be assisted by a computer (experimentally, computationally, and exploratory). Any mathematics research will require extensive amount of learning (both of results and of techniques) and will certainly include long hours of thinking. I find the entire process extremely creative.

I hope this helps. As should be clear, this is a rather subjective answer and I don't intend any of what I said to be taken to be said with any kind of mathematical rigor.

We know that mathematics is the subject which is not invented it's just found out from nature and as we all know that research means doing a work which is already been searched or done just to get the more accurate results and minimise some how the approximate errors .mathematics research means some how researching the nature so that we are able to do any work more logically and get the most accurate results related to anything with the help of our previous knowledge in mathematics..


At what education level is this "simple" problem adequate? Given the following problem:

  Alice, Bob and Carl stand on a straight line.
  Alice is one rod away from Bob.
  Dana stands one rod away from both Bob and Carl.
  Carl is as far from Alice as Alice is from Dana.

  How far can Alice be from Carl?

  Answer using a well-formed sentence, with any distance expressed (in rod) with three significant digits (calculators allowed).

Question: At what education level could we expect that more than one in ten students give a mathematically correct answer, in exam conditions?

Note: I authored this problem, and know its solution.

Update: Spoiler with reference answer to the original problem follows; hover mouse to see it.

   Alice is either 1.62… or 0.618… rod away from Carl.

Update: Some have trouble with the problem's wording (which is an important aspect of the question), so here is a restatement in geometry terms; hover mouse to see it.

   $A$, $B$, $C$, $D$ are distinct points, with $A$, $B$, $C$ collinear,
 such that $|AB|=|BD|=|CD|=1$ and $|AC|=|AD|$.

 Give the set of possible $|AC|$ rounded to 3 significant digits.

Update: link to spoiler illustration; ctrl-click for new window or tab.

To be able to solve the problem I think a student would need to know

basic algebra
some trigonometry
some geometry, including the concept of circles, triangles, and equidistance

  At what education level...

Education level is a strange concept, but I think a bright student in an advanced secondary mathematics class should be able to solve this. It's somewhat tricky but it's not exceedingly difficult.

It's definitely well beyond the primary / elementary school student, and below the university / college level mathematics student. Probably above the level of the general high school mathematics student but within grasp of the advanced high school student.

Just my opinion, of course.

I might be being silly but it appears to me that the question is not written very well and it is mathematically inaccurate.

My main issue is that Dana might not be able to stand anywhere. How do we know that there is a place that is one rod away from both Bob and Carl? They might be at a distance of more than two rods away from each other, in which case there is no possible place for Dana to stand at all!


Show that if $f(0)=\displaystyle\lim_{x \rightarrow \infty}f(x)=0$ and $f$ is continuous then $f$ has a maximum in the interval $[0, \infty)$. Show that if $$f(0)=\displaystyle\lim_{x \to \infty}f(x)=0$$ and $f$ is continuous then $f$ has a maximum in the interval $[0, \infty)$.

It just seems so obvious! Yet I'm stuck... any hints would be much appreciated.

If $f$ is equal to $0$ everywhere, then the statement is trivial, so let's assume $f$ is not $0$ everywhere.

If $f(x)\leq 0$ for all $x$, then $f(0)$ is the maximum, so let's now see what happens if $f(x)>0$ for at least some $x$.

We then know there exists some $x_0>0$ for which $f(x_0)> 0$. Now, you can find such a value $M$ that for all $x>M$, you have $f(x) < \frac{f(x_0)}{2}$ (why?). You can also be sure that $f$ has a maximum on $[0,M]$ (why?), and that the maximum value on $[0,M]$ is actually the maximum on $[0,\infty)$ (why?)

Hint: Since $\lim_{x\to \infty} f(x) = M$ exists, for any $\epsilon > 0, \exists R > 0$ such that
$$
|f(x) - M| < \epsilon \quad\forall x \geq R
$$
Hence,
$$
|f(x)| \leq M+\epsilon \quad\forall x \in [R,\infty)
$$
Now what happens on $[0,R]$?

Now since $M = 0$, you can choose $\epsilon$ carefully so that the max is attained.


The Converse of the Pythagoras Theorem Let $ABC$ be a triangle in the plane. Suppose that $AB^2+AC^2=BC^2.$ Prove that:

$\angle BAC$ is right angle.

Remark: I believe this to be true. Now I have the following difficulty. I'm looking for a proof which a 10th grade student can understand, but any effort of mine is via proof by contradiction. I mean assuming that the given angle is not a right angle and then arrive on a contradiction. So I'm looking for a solution which could avoid the indirect method but if not then a solution by the indirect method.

Draw side by side the given triangle $\;\Delta ABC\;$ and a new triange $\,PQR\;$ as follows:

$$PQ=AB\;,\;\;PR=AC\;,\;\;\angle QPR=90^\circ$$

Now, we're given $\,AB^2+AC^2=BC^2\;$, and Pythagoras theorem also gives $\,PQ^2+PR^2=QR^2\;$ , so

$$BC^2=AB^2+AC^2=PQ^2+PR^2=QR^2\implies BC=QR$$

and we get $\,\Delta ABC\cong\Delta PQR\;$ by $\;S.S.S.\;$

Understand Converse of Pythagoras Theorem with a VIDEO explanation. Click on the link to WATCH the VIDEO: WATCH VIDEO

Converse of Pythagoras Theorem

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.
Given : A triangle ABC such that AB square plus BC square equals to AC square.
To prove : the triangle ABC is right - angled at B.
Construction: Construct a right - angled triangle PQR, right - angled at Q such that PQ equal to AB and QR equals to BC.
Prof: Appling the Pythagoras theorem to right - angled triangle PQR we have PQ square plus QR square equals to PR square or AB square plus BC square is equal to PR square since PQ is equal to AB and QR equals to BC by construction. Let this be equation 1 but it is given that AB square plus BC square equals to AC square.
From equations 1 and 2 we get PR square equals to AC square i.e., PR equals to AC.
Therefore, by SSS congruence criterion, we get triangle ABC is congruent to triangle PQR so we get angle B equal to angle Q but we know angle Q is equal to 90 degrees by construction.
Therefore, angle B equals to 90 degrees.
Hence, triangle ABC is a right triangle, right - angled at B.
Thus, the theorem is proved.


Probability of being Second Highest Suppose there are $n-1$ draws from a uniform distribution $[0, 1]$, then I get a draw from the distribution.  What is the probability that if I shout out an $x$, then $x$ will be exactly the second highest of all the draws?  I have a feeling its something like $n x^{n-1} (1-x)$, but I can't show it rigorously.

There are $n-1$ draws, then you draw $x$ on the $n$-th draw.

The probability that the $n$-th draw is the second-last order statistic of the $n$ draws, when given the value of that draw is $x$, is:

$$\frac{(n-1)!}{(n-2)!\;1!}x^{n-2}(1-x) \\[4ex] = (n-1)(1-x)x^{n-2}$$

Reasoning:  Of the $n-1$ draws preceding yours, the favoured space is formed by $n-2$ of them being no greater than yours (individual probability $x$ each), and $1$ of them being no less (probability $1-x$), and there are ${n-1\choose n-2}$ ways of arranging these favoured results.

The problem is equivalent to finding the probability that $x$ is the second last order statistic.
In that case, out of the remaining $(n-1)$ draws, there should be $n-2$ draws which are smaller than $x$ and 1 draw is bigger than $x$. The probability of a draw being smaller than $x$ is $F(x)=x$. There are in total $n$ independent draws, out of which $n-1$ are smaller than or equal to $x$ so each has probability $F(x)=x$, and the last draw is greater than $x$ having probability $1-F(x)=1-x$. Note that these draws form a Binomial distribution with parameters $n$ and $F(x)$.

So you have the required probability as $\dfrac{n!}{(n-1)!1!}$$x^{n-1}$$(1-x)$


What is $\bigcup\limits_{n=1}^\infty [0,1-\frac{1}{n}]$? This is probably a pretty dumb question, but I am confused by set theory again. The question is whether
$$\bigcup_{n=1}^\infty \left[0,1-\frac{1}{n}\right]$$
equals $[0,1]$ or $[0,1)$. However, I am looking for some explanation and not only the result, since I'd like to understand why it's the one or the other.

Remember that $x\in\bigcup\limits_{i\in I} A_i$ if and only if for some $i\in I$, $x\in A_i$.
So $1$ is in the union if and only if it appears in at least on of those intervals, so is it? No. It is not.

To the comment, don't think of $\bigcup_{i=1}^\infty$ as a limit in the calculus-sense of the word. Think of it as a logical operation which tells you that the index set is $\mathbb N$ (or some other set which is clear from context) and then use the above formula.
If you wish to think about it as $f(n)=\bigcup\limits_{k=1}^n [0,1-\frac1k]$, and think about the infinite union as $\lim\limits_{n\to\infty} f(n)$, then there are several caveats:

Limits will usually require some sort of topology, some underlying structure which tells us about convergence. How would you define the limit here? For every $\varepsilon>0$...? It makes no sense, since subsets of $\mathbb R$ do not have a natural metric function.

We can consider the following definition: $A$ is the limit of the sequence of $f(n)$ if and only if for every $x\in A$, there exists $n_0$ such that for all $n>n_0$, $x\in A_n$.
Observe, however, that this coincides with the definition above, that $x$ is in the union if and only if it appears in at least one of the functions. This definition, however, coincides with the above only because this sequence of sets is increasing.

Luckily, we can always think of an infinite union as an increasing sequence, but we would expect a definition for a limit to work for any sequence of sets, not just increasing unions.

We can, however, think of it as a limit of a sequence of characteristic functions, $$\lim_{n\to\infty}\chi_{\left[0,1-\frac1n\right]}=\chi_{[0,1)}$$
even as such limit, though, it is not "continuous" in the way you would like it to be, that is, $$\lim_{n\to\infty}\chi_{\left[0,1-\frac1n\right]}\neq\chi_{\left[0,1-\lim\limits_{n\to\infty}\frac1n\right]}$$

Note that, for this case, (with the "definition" of $\infty$ expanded)

$$\begin{aligned}
\bigcup_{n=1}^\infty \left[0,1-\frac{1}{n}\right] & \stackrel{\operatorname{def}}{=} \lim_{k\to\infty} \bigcup_{n=1}^k \left[0,1-\frac{1}{n}\right] \\
 & = \lim_{k\to\infty} \left[0,1-\frac{1}{k}\right] \\
 & = \left[0,1\right)
\end{aligned}$$


What is the answer to this infamous "Common Core" question? The following question (number 15 of this test) has become infamous as a poor "Common Core" question. What is the correct answer?

  Juanita wants to give bags of stickers to her friends. She wants to
  give the same number of stickers to each friend. She's not sure if she
  needs 4 bags or 6 bags of stickers. How many stickers could she buy so
  there are no stickers left over?

This "infamous" question is so poorly worded, but any mathematical answer that can be given would reinforce the perception that "the question is OK, see, it has a valid answer!"

The question is not OK.

Word problems should not be about mulling over "what did the author want to say?". Even the "trick" questions are normally about spotting some well-defined linguistic (or mathematical) trick, rather than based on complete lack of clarity. Plus, note it was not meant to be a trick question, but some standard question meant to assess how well students of certain age comprehend divisibility and common multipliers.

I can personally think about a dozen different ways to write a word problem which would boil down to the same mathematical problem and will be better worded.

As for this question, I hope if it was on some exam, that it did not affect anyone's passing or failure. I can relate to a poor methodical soul who got stuck for two hours on that question, trying to get their head around it and failing to comprehend it (and also wasting precious time for the following questions) not realising that it is not their fault.

The answer is 5.

Juanita cannot decide whether to buy "4" or "6" bags because she has 5 friends.   In common core lingo "5" is between "4" and "6."  And "4" and "6" are the clue numbers given in the problem--pay attention! LOL This is how we know Juanita has 5 friends.  Since Juanita doesn't seem to know how many stickers are in each bags, she simply buys 5 stickers.


Equivalence Relations, Partitions and Equivalence Classes Consider the partition P={{0},{−1,1},{−2,2},{−3,3},{−4,4},...} of Z. Describe the equivalence relation whose equivalence classes are the elements of P. I'm struggling to grasp what the question is asking and how to go about solving it.

Hint: What are the equivalence classes of the relation $\sim$ on $\mathbb Z$ defined by $x\sim y$ if and only if $x=\pm y$?

Consider set of integers $\Bbb Z$ With given any partition p then $x=\pm y$ if and only if {${x,y}$} $\in$ P


Geometrical meaning of dx, dy and dy/dx? I am also confused about whether these are symbols or have some meaning of their own.
PS- I know dy/dx geometrically represents the slope. But, I've come across dx/dy to make problems easier. What does dx/dy mean?

You should be very careful, those are merely notations.

$\frac{d\,y(x)}{d\,x}$ is the derivative of a variable $y$ with respect to $x$. It represents how much $y$ varies for small variations of $x$. If you draw the curve of $y(x)$, the derivative will be the slope, as you said.

The opposite works as well, if one can define $x$ as a function of $y$ (which is the same as inverting the function $y(x)$ - we must consider a subset of the image where it is injective to do so), you can differentiate it too: $\frac{d\,x(y)}{d\,y}$

But be careful! It is just a notation, it is not the division of $dx$ and $dy$, it is a limit as you can see here.

$\boxed{\frac{d\,y(x)}{d\,x} = \lim\limits_{h\rightarrow0} \frac{y(x+h)-y(x)}{h}}$

Here is some discussion about the meaning of $dx$ alone, pay attention to the commentary at the end: "Since I first posted this paper, two different people have emailed me to tell me that Real Mathematicians don't do this. Playing with dx in the ways described in this paper is apparently one of those smarmy tricks that physicists use to give headaches to mathematicians. "

In the terms dx and dy, the d is for delta or "change in". So they represent the change in y and the change in x as a function, usually in terms of each other but sometimes another parameter. So dy/dx as you said is the slope, or change in x divided by the change in y, dy/dx is simply the inverse slope.

The different between dy and ∆y or dx and ∆x is that dy is a function that can be solved at any point to give the change in y at that point in relation to another variable, where as ∆y is a numerical value representing the difference in y between two points.


Examples of mathematical discoveries which were kept as a secret There could be several personal, social, philosophical and even political reasons to keep a mathematical discovery as a secret.

For example it is completely expected that if some mathematician find a proof of $P=NP$, he is not allowed by the government to publish it as same as a usual theorem in a well-known public journal because of high importance and possible uses of this special proof in breaking security codes which gives an undeniable upper hand to the state intelligence services with respect to other countries. Also by some social reasons publishing such a proof publicly is not suitable because many hackers and companies may use it to access confidential information which could make a total chaos in the community and economy.

The example shows that in principle it is possible to have some very significant brilliant mathematical proofs by some genius mathematicians which we are not even aware of. But in some cases these "secrets" unfold by an accident or just because they lost their importance when some time passed and the situation changed.

  Question: What are examples of mathematical discoveries which were kept as a secret when they discovered and then became unfolded after a while by any reasons?

Niccolò “Tartaglia” Fontana invented the first general method to find the roots of an arbitrary cubic equation (based on earlier work by himself and others on how to solve cubics of particular forms), but kept his method secret so as to preserve his advantage in problem-solving competitions with other mathematicians.

He divulged the secret to his student Gerolamo Cardano on condition that Cardano keep the secret, and there was a bitter dispute between the two when Cardano broke his promise.

I think Gauss anticipated non-euclidian geometries, and kept it quite because he was afraid of how controversial it was. He actually had a dispute with Janos Bolyai, who thought he was stealing his idea.

There is information about this contained here


I think the mistake is from

$$\frac{y}{y-3}=e^{\frac{3}{2}t^2+3C}$$ to the next line. Multiplying by $$y-3$$ you will get

$$y=(y-3)A$$ so
$$y=\frac{3A}{A-1}$$ where $$A=e^{\frac{3}{2}t^2+3C}$$
For your Control: here is the Maple solution:
$$y \left( t \right) =3\, \left( 1+3\,{{\rm e}^{-3/2\,{t}^{2}}}{\it \_C1
} \right) ^{-1}
$$


Why $\mathbb P(\forall s\in [a,b], B_s\neq 0\mid B_a=x)=\mathbb P(\forall s\in [0,b-a], B_s\neq -x\mid B_0=0)$? I want to compute $$\mathbb P(\forall s\in (a,b), B_s\neq 0),$$
but I already don't see why $$\{\forall s\in (a,b), B_s\neq 0 \}$$
is $\mathcal F_b$ measurable. Also, I know that if $t>s$ then $$\mathbb P(B_t\in A\mid B_s=x)=\mathbb P(B_{t-s}\in A\mid B_0=x).$$
By why from this we can get $$\mathbb P(\forall t\in (a,b), B_t\neq 0\mid B_a=x)=\mathbb P(\forall t\in (0,b-a), B_t\neq -x\mid B_0=0) \ \ ?$$

First of all, mind that we are not dealing with classical conditional probabilities, that is, $$\mathbb{P}(A \mid B_s=x) \neq \frac{\mathbb{P}(A \cap \{B_s=x\})}{\mathbb{P}(B_s=x)}.\tag{1}$$

This is pretty obvious from the fact that $\mathbb{P}(B_s=x)=0$ for all $x \in \mathbb{R}$, $s>0$. The expression on the left-hand side of $(1)$ is defined as follows: It follows from properties of the conditional expectation that there exists a measurable mapping $g:\mathbb{R} \to \mathbb{R}$ such that $$\mathbb{P}(A \mid B_s) := \mathbb{P}(A \mid \sigma(B_s))=g(B_s) \quad \text{a.s.}$$  Now one defines $$\mathbb{P}(A \mid B_s = x) := g(x).$$ (This definition is not specific for Brownian motion; more generally, this is the way to define $\mathbb{P}(A \mid X)$ for an event $A$ and some random variable $X$).

Let's return to your original problem. Fix some Borel set $A$ and some $a<b$. Recall that the restarted process

$$W_t := B_{t+a}-B_a, \qquad t \geq 0,$$

is a Brownian motion which is independent of $(B_t)_{t \leq a}$. Clearly,

$$B_t(\omega) \neq 0 \, \, \text{for all $t \in (a,b)$} \iff W_t(\omega) \neq -B_a(\omega) \, \, \text{for all $t \in (0,b-a)$}.$$

If we set $\tau_x := \inf\{t>0; W_t=x\}$, we can write this equivalently as

$$B_t(\omega) \neq 0 \, \, \text{for all $t \in (a,b)$} \iff \tau_{-B_a(\omega)}(\omega)\geq b-a.$$

Hence,

$$\mathbb{P}\left(B_t \neq 0 \,  \, \text{for all $t \in (a,b)$} \mid B_a \right) = \mathbb{P}(\tau_{-B_a} \geq b-a \mid B_a).$$

Since $\omega \mapsto \tau_{-x}(\omega)$ is independent of $\mathcal{F}_a^B:=\sigma(B_s; s \leq a)$ for all $x \in \mathbb{R}$ and $B_a$ is $\mathcal{F}_a^B$-measurable, it follows that

$$\mathbb{P}(\tau_{B_a} \geq b-a \mid B_a) = g(B_a)$$

where

$$g(x) = \mathbb{P}(\tau_{-x} \geq b-a); \tag{2}$$

see e.g. the result which I stated in the first part of this answer. Combining the previous two identites, we get
$$\mathbb{P}\left(B_t \neq 0 \,  \, \text{for all $t \in (a,b)$} \mid B_a \right)=g(B_a)$$

for $g$ defined in $(2)$. Hence,

$$\mathbb{P}\left(B_t \neq 0 \,  \, \text{for all $t \in (a,b)$} \mid B_a=x \right)=g(x), \tag{3}$$

as I explained earlier. By definition,

\begin{align*} g(x) = \mathbb{P}(\tau_{-x} \geq b-a) &= \mathbb{P} \left(W_t \neq -x \, \, \text{for all $t \in (0,b-a)$}\right) \end{align*}

Using the fact taht both $(B_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ are Brownian motions, which means in particular that they are continuous with probability $1$ and have the same finite dimensional distributions, we conclude that

\begin{align*} g(x) = \mathbb{P} \left( \bigcap_{q \in \mathbb{Q} \cap (a,b)} \{W_q\neq-x\} \right) &= \mathbb{P} \left( \bigcap_{q \in \mathbb{Q} \cap (a,b)} \{B_q\neq-x\} \right) \\ &= \mathbb{P} \left(B_t \neq -x \, \, \text{for all $t \in (0,b-a)$}\right)\end{align*}

Plugging this into $(3)$ proves the assertion.

I can answer your second question:

$$\mathbb \{\forall t\in (a,b), B_t\neq 0\mid B_a=x\} =
\bigcap_{t \in (a, b)}\{B_t\neq 0\mid B_a=x\} =
\bigcap_{t \in (a, b)}\{B_t \in \{0\}^c\mid B_a=x\}.$$
Now, by translation invariance of $B_t$ we get that the law of $B_t$ started at $B_a=x$ is equal to the law of $B_{t-s}$ started at $B_0=x$, so
$$
\mathbf P \left( \cap_{t \in (a, b)}\{B_t \in \{0\}^c\}\mid B_a=x\right) =
\mathbf P \left(\cap_{t \in (a, b)}\{B_{t-a} \in \{0\}^c\}\mid B_0=x\right) $$
This last set is equal to
$$\bigcap_{t \in (0, b-a)}\{B_t \in \{0\}^c\mid B_0=x\} \\=
\{\forall t\in (0,b-a), B_t\neq 0\mid B_0=x\} \\=
\{\forall t\in (0,b-a), B_t - x\neq -x\mid B_0-x=0\} $$

Now, the process $B_t - x$ with $B_t$ started at $B_0 = x$ has the same distribution as another brownian motion $W_t$ started at $W_0 = 0$, so the set above has a probability equal to

$$\mathbf P(\forall t\in (0,b-a), W_t \neq -x\mid W_0 =0). $$


Prove that $(2n+1)+(2n+3)+\dots +(4n-1) = 3n^2$ by induction Note: This is for self study, the book is Elementary analysis by Kenneth. A. Ross

How to prove the following by mathematical induction, I am stuck


Prove $A \equiv B \vdash (\forall x)A \equiv (\forall x)B$ without relying on equivalence theorem. The given rules of inference for me to use are here: rules of inference

In addition, I was taught the deduction theorem, the ping-pong tautology, specialization and dual specialization (derived rules).

My solution:

Using Deduction Theorem, it suffices if we prove $$A \equiv B, (\forall x)A \vdash (\forall x)B.$$
Hilbert proof:

1) $A ≡ B\quad\langle\text{hypothesis}\rangle$

2) $(\forall x)A\quad\langle\text{hypothesis}\rangle$

3) $A\quad\langle\text{2 + Specialization}\rangle$

I couldn't move past step 3 because I did not know where to go. Any help on this question is appreciated.

Note: This is practice problem is for first year discrete mathematics for engineers course.

I did this, not sure if it's correct or not.


Soft question -- I need books and exercise books that will be working on my fundamental skills. I need help, urgently. I acquired a book called: Mathematics, Its Content, Method and Meaning. Now the problems is the book doesn't provide me with any exercises. I was searching for a book that would work on fundamental skills so basic skills and then gradually progressing to more advanced skills. I really need a book that will optimize all aspects of maths skills; I seem to be very rusty, and the last time I studied maths was at GCSE, but I seem to have lost many skills -- and these skills are basic! I need to gain these skills again. And again, I want to improve every aspect that there is. I want to see every single method, too. I'll give an example: Basic Mathematics by Serge Langey. I heard the book is marred with many errors, so I am reluctant to spend my money on a "flawed gem". If I seem to be slightly nebulous, please do forgive me, as I am in a rush. Thank you, and any response is greatly appreciated.

Having only just finished my gcse last year this book (http://www.amazon.co.uk/gp/aw/d/1446900185/ref=mp_s_a_1_3?qid=1452527349&sr=8-3&pi=SY200_QL40&keywords=gcse+maths+edexcel&dpPl=1&dpID=51TqDXPEC-L&ref=plSrch) really helped its a revison guide for all you need at gcse and it explains it all clearly and concicely. And at 3.99 at the moment its not going to break the bank, also there is another book that you can also buy with other questions in but the method is the same. Hope i helped!


Sequence converging to $\sqrt{3}$ Can someone help me with this:

pick $a_0 >0$ and define {$a_n$} by: $a_{n+1} = 2a_n/3 + 1/a_n.$

Explain why the sequence converges to $\sqrt{(3)}.$

I have calculated the first 10 elements and I can see that it's getting closer to $\sqrt{(3)}\approx1.7320$

It's obvious that $a_n>0$.

By AM-GM $$a_{n+1}=\frac{2}{3}a_n+\frac{1}{a_n}\geq2\sqrt{\frac{2}{3}a_n\cdot\frac{1}{a_n}}=\frac{2\sqrt2}{\sqrt3}>\frac{\sqrt{3}}{2}.$$
If $a_0>\sqrt3$ we obtain:
$$a_{n+1}-a_n=\frac{(\sqrt3-a_n)(\sqrt3+a_n)}{3a_n}<0$$
because $$a_{n+1}-\sqrt3=\frac{(a_n-\sqrt3)(2a_n-\sqrt3)}{3a_n}>0$$
by induction.

Thus, $a$ is a decreasing and $a_n>\sqrt3$.

Id est, there is a $\lim\limits_{n\rightarrow+\infty}a_n$.

Let $\lim\limits_{n\rightarrow+\infty}a_n=a$.

Thus, $a=\frac{2}{3}a+\frac{1}{a}$, which gives $a=\sqrt3$.

By the same way we can get that $a=\sqrt3$ for $0<a_0\leq\sqrt3$.

You first have to prove that it converges, if it converges it must converge to a fixed point:
$L=\frac{2}{3}L+\frac{1}{L}$

solving it you have, multiplying by $3L$

$3L^{2}=2L^{2}+3$

that leads to

$L^{2}=3\Rightarrow L=\sqrt{3}$ $ \quad$because your sequence is a sequence of positive numbers, because of $a_{0}>0$.

You can prove that the sequences converges by proving that it is a Cauchy sequence, for example.


Prove unit ball of $L^p(\mu)$ is strictly convex, when $1<p<\infty$ When $1<p<\infty$, $||f||_p=1$,$||g||_p=1$,$f\ne g$,then $\frac{1}{2} ||f+g||_p<1$.

I use parallelogram law

$||f+g||^2+||f-g||^2=||f||^2+||g||^2=4\\$

Since $f\ne g$,
$||f-g||^2>0$

then$||f+g||_p<1$

But my proof does not use $1<p<\infty$, and this obviously fail when p is 1 or $\infty$.
I just want to know where my proof is wrong.

I guess is this because the step Since $f\ne g$,$||f-g||^2>0$ ?I just use the definition 5.2 in rudin's book
(c)$||x||=0 $ implies $x=0$.

Note that norm satisfies the triangle inequality by the definition of norm, $|f|_p+|g|_p\geq |f+g|_p$

So if $|g|_p$, $|f|_p \leq 1$, then for $0< t< 1$, $
    1=(1-t) + t \geq |(1-t)f|_p+|tg|_p \geq |(1-t)f+tg|_p$


Closed form solution for $\rho$: $\frac{d}{d\tau}\sqrt{\left(\rho^{3}\frac{d^{2}\rho}{d\tau^{2}}+\rho\right)}=\nu\rho$ with initial conditions Consider the following differential equation for $\rho$:
$$\frac{d}{d\tau}\sqrt{\left(\rho^{3}\frac{d^{2}\rho}{d\tau^{2}}+\rho\right)}=\nu\rho$$
This equation can be rewritten as a system of 3 first-order differential equations:
$$\frac{d\rho}{d\tau}=A,$$
$$\frac{dA}{d\tau}=\frac{B^{2}-\rho}{\rho^{3}},$$
$$\frac{dB}{d\tau}=\nu\rho.$$
Therefore, it is possible to numerically solve this equation eg using a Runge-Kutta method.
However, I would like to know if it is possible to find a closed form solution to this equation with initial conditions: $\rho=1$, $A=0$ and $B=1$.
Any help is welcome.

Hint:

Let $u=\dfrac{d\rho}{d\tau}$ ,

Then $\dfrac{d^2\rho}{d\tau^2}=\dfrac{du}{d\tau}=\dfrac{du}{d\rho}\dfrac{d\rho}{d\tau}=u\dfrac{du}{d\rho}$

$\therefore\dfrac{d}{d\rho}\left(\sqrt{\rho^3u\dfrac{du}{d\rho}+\rho}\right)\dfrac{d\rho}{d\tau}=\nu\rho$

$\dfrac{u\left(\rho^3u\dfrac{d^2u}{d\rho^2}+\rho^3\left(\dfrac{du}{d\rho}\right)^2+3\rho^2u\dfrac{du}{d\rho}+1\right)}{2\sqrt{\rho^3u\dfrac{du}{d\rho}+\rho}}=\nu\rho$


Why do we still do symbolic math? I just read that most practical problems (algebraic equations, differential equations) do not have a symbolic solution, but only a numerical one.

Numerical computations, to my understanding, never deal with irrational numbers, but only rational numbers.

Why do mathematicians construct the real numbers and go through the pain of using their complicated properties to develop real analysis, etc. to be able to solve just a few cases symbolically?

Why do non-mathematicians deal with real numbers and symbolic calculation if the majority of cases must use rationals and numerical computation only?

Try to solve $x^{1/3} = 0$ with a "symbolic calculation". Now try to solve $x^{1/3} = 0$ with Newton's Method. The point is numerical methods can fail and anyone using numerical methods should have an understanding of mathematics so that they can detect when a numerical method is not working and perhaps even fix the problem. Also as this case shows sometimes the symbolic calculation is easier.

All these computational methods need to be designed and implemented. It takes an understanding of the underlying mathematics to do this.

There are too many sub-topics in math but it is split with two main topic theorical and applied mathematics. You can't prove 2n is an even number with numbers it goes to infinity and a simple question like ' a car try to reach a target. There is a road between target and car about 100 km. car is reaching half of way every time. 50,25,12.5,... when it is end ? Can car reach the target ? You cant answer questions like this without symbolic math.As a mathematician I can say numbers is just seen side of iceberg


Alternative notation for exponents, logs and roots? If we have

$$ x^y = z $$

then we know that

$$ \sqrt[y]{z} = x $$

and

$$ \log_x{z} = y .$$

As a visually-oriented person I have often been dismayed that the symbols for these three operators look nothing like one another, even though they all tell us something about the same relationship between three values.

Has anybody ever proposed a new notation that unifies the visual representation of exponents, roots, and logs to make the relationship between them more clear? If you don't know of such a proposal, feel free to answer with your own idea.

This question is out of pure curiosity and has no practical purpose, although I do think (just IMHO) that a "unified" notation would make these concepts easier to teach.

Always assuming $x>0$ and $z>0$, how about:
$$\begin{align}
x^y &={} \stackrel{y}{_x\triangle_{\phantom{z}}}&&\text{$x$ to the $y$}\\
\sqrt[y]{z} &={} \stackrel{y}{_\phantom{x}\triangle_{z}}&&\text{$y$th root of $z$}\\
\log_x(z)&={} \stackrel{}{_x\triangle_{z}}&&\text{log base $x$ of $z$}\\
\end{align}$$
The equation $x^y=z$ is sort of like the complete triangle $\stackrel{y}{_x\triangle_{z}}$. If one vertex of the triangle is left blank, the net value of the expression is the value needed to fill in that blank. This has the niceness of displaying the trinary relationship between the three values. Also, the left-to-right flow agrees with the English way of verbalizing these expressions. It does seem to make inverse identities awkward:
$\log_x(x^y)=y$ becomes $\stackrel{}{_x\triangle_{\stackrel{y}{_x\triangle_{\phantom{z}}}}}=y$. (Or you could just say $\stackrel{}{_x\stackrel{y}{\triangle}_{\stackrel{y}{_x\triangle_{\phantom{z}}}}}$.)
$x^{\log_x(z)}=z$ becomes $\stackrel{\stackrel{}{_x\triangle_{z}}}{_x\triangle_{\phantom{z}}}=z$. (Or you could just say $\stackrel{\stackrel{}{_x\triangle_{z}}}{_x\triangle_{z}}$.)
$\sqrt[y]{x^y}=x$ becomes $\stackrel{y}{\triangle}_{\stackrel{y}{_x\triangle_{\phantom{z}}}}=x$. (Or you could just say $\stackrel{}{_x\stackrel{y}{\triangle}_{\stackrel{y}{_x\triangle_{\phantom{z}}}}}$ again.)
$(\sqrt[y]{z})^y=z$ becomes $_{\stackrel{y}{_\phantom{x}\triangle_{z}}}\hspace{-.25pc}\stackrel{y}{\triangle}=z$. (Or you could just say $_{\stackrel{y}{_\phantom{x}\triangle_{z}}}\hspace{-.25pc}\stackrel{y}{\triangle}_z$.)
Having $3$ variables, I was sure that there must be $3!$ identities, but at first I could only come up with these four. Then I noticed the similarities in structure that these four have: in each case, the larger $\triangle$ uses one vertex (say vertex A) for a simple variable. A second vertex (say vertex B) has a smaller $\triangle$ with the same simple variable in its vertex A. The smaller $\triangle$ leaves vertex B empty and makes use of vertex C.
With this construct, two configurations remain that provide two more identities:
$_{\stackrel{y}{_\phantom{x}\triangle_{z}}}\hspace{-.25pc}\stackrel{}{\triangle_z}=y$ states that $\log_{\sqrt[y]{z}}(z)=y$.
$\stackrel{\stackrel{}{_x\triangle_{z}}}{_\phantom{x}\triangle_{z}}=x$ states that $\sqrt[\log_x(z)]{z}=x$.
I was questioning the usefulness of this notation until it actually helped me write those last two identities. Here are some other identities:
$$\begin{align}
\stackrel{a}{_x\triangle_{\phantom{z}}}\cdot\stackrel{b}{_x\triangle_{\phantom{z}}}&={}\stackrel{a+b}{_x\triangle_{\phantom{z}}}&
\frac{\stackrel{a}{_x\triangle_{\phantom{z}}}}{\stackrel{b}{_x\triangle_{\phantom{z}}}}&={}\stackrel{a-b}{_x\triangle_{\phantom{z}}}&
_{\stackrel{a}{_x\triangle_{\phantom{z}}}}\hspace{-.25pc}\stackrel{b}{\triangle}
&={}\stackrel{ab}{_x\triangle_{\phantom{z}}}\\
\stackrel{}{_x\triangle_{ab}}&={}\stackrel{}{_x\triangle_{a}}+\stackrel{}{_x\triangle_{b}}&
\stackrel{}{_x\triangle_{a/b}}&={}\stackrel{}{_x\triangle_{a}}-\stackrel{}{_x\triangle_{b}}&\stackrel{}{_x\triangle}_{\stackrel{b}{_a\triangle_{\phantom{z}}}}&=b\cdot\stackrel{}{_x\triangle}_{a}
\\
\stackrel{-a}{_x\triangle_{\phantom{z}}}&=\frac{1}{\stackrel{a}{_x\triangle_{\phantom{z}}}}&
\stackrel{1/y}{_x\triangle_{\phantom{z}}}&=\stackrel{y}{_\phantom{x}\triangle_{x}}&
\stackrel{}{_x\triangle_{1/a}}&=-\mathord{\stackrel{}{_x\triangle_{a}}}\\
\stackrel{}{_a\triangle_{b}}\cdot\stackrel{}{_b\triangle_{c}}&=\stackrel{}{_a\triangle_{c}}&
\stackrel{}{_a\triangle_{c}}&=\frac{\stackrel{}{_b\triangle_{c}}}{\stackrel{}{_b\triangle_{a}}}&
\stackrel{\stackrel{-n}{_y\triangle_{\phantom{z}}}}{_x\triangle_{\phantom{z}}}&=\stackrel{\stackrel{n}{_y\triangle_{\phantom{z}}}}{_\phantom{x}\triangle_{x}}&
\end{align}$$

How about this:

exp _b y = x

Basically, you would erase the word "log", which is completely non-descriptive of the process for newbies, and replace it with the word "exp" to represent that you are finding the exponent of base b that "gives you" y.  The dash represents subscript, so in this case it's a subscript variable b.

An example:

exp _2 8 = x

Thus, x = 3.

A newbie would read this from left to right as "exponent of base 2 that gives us 8", which is 3.

Edit:  This doesn't, unfortunately, unify the three ideas under a more indicative notation, but I think it does clarify the meaning of the logarithmic process.  One could, of course, alter the letters; perhaps "expn" would be a better mnemonic device, or even "exponent needed", the latter, of course, being used only when first learning about logarithms; it would be shortened to "expn" (or something) later.