Convergence or divergence of alternating series $\sum _{n=1}^{\infty }\:\frac{\left(-1\right)^nn^2+3}{n\sqrt{9n^2+2}}$ Trying to figure out this problem. Alternating series test doesn't seem to work because finding the limit seems to be an endless loop of L'Hopital's rule in the denominator. Ratio test is inconclusive because it turns out to be 1. Comparing seems difficult because of the same. Not to mention difficulty finding something that is less than or greater than the series for all values for comparison. I'm a bit stumped. I suppose that technically, the positive series convergence would be fine as well, as it would prove absolute convergence, but I can't even figure that out.

Hint: The general term does not tend to $0$ so the series is not convergent.

Your general term is
$$  \frac{(-1)^n n^2 + 3}{n \sqrt{9n^2 + 2}} = \frac{n^2}{n^2} \cdot \frac{(-1)^n+\frac{3}{n^2}}{\sqrt{9 + \frac{2}{n^2}}}  \text{.}  $$
It should be fairly clear that the terms of your series do not decrease to zero, so (first) you cannot apply the alternating series test and (second) the series does not converge because (for both) the limit of the general term is not zero.


Calculus made easy Exercise 9 Question 4 A piece of string 30 inches long has its two ends joined together
and is stretched by 3 pegs so as to form a triangle. What is the largest
triangular area that can be enclosed by the string?

I took P = a+b+c and Area = 1/2 (b * h) And found h in terms of a and b.Substituting this value to the h in the area equation I differentiated the Area wrt to b and equated to zero.Is this the right way to do it?
Can this be soved only using heron's forumla(not A = 1/2(base*height)?

If you are convinced without proof that there is a triangle of maximal area it is easy to see that the maximal triangle has to be equilateral: Keep two pegs $A$, $B$ fixed and use the third peg as a pencil to draw an arc of an ellipse $\gamma_{AB}$ with foci $A$, $B$. The triangle $ABC$ will have maximal area when $C$ is a minor vertex of $\gamma_{AB}$, and will have strictly smaller area for all other positions of $C$. It follows that the triangle of maximal area having a given perimeter is isosceles over all three sides, whence equilateral.

Look Thompson said that we can solve it by using the hints given on the last line of some page so I'll solve it using that only , using herons formula $A = \sqrt{S(S-a)(S-b)(S-c)}$ , $S$ here is half of perimeter $a,b$ and $c$ are the sides, now u need the maximum value of the part $(S-a)(S-b)(S-c)$. Adding these $(S-a) + (S-b) + (S-c) = 3S - (a+b+c) = 3(15) - (30) = 15$. Now u need to divide 15 in three parts in such a way that u get the maximum value when u multiply the three together, the hint on that page helps us in figuring out that value and its $(n+m+p)/3$ , so $15/3 = 5$. This means that $(S-a) = 5$ , $(S-b) = 5$ and $(S-c)=5$ too. $S=$ half of perimeter = $15$ . sub the values yourself to get $a=b=c=10$ , area of equilateral triangle=$\sqrt{3}/4  \text{(side)}^2$  which gives us the answer


Polyhedron's Representations and spanning the Euclidian space Let's say you have to different representations of the same polyhedron $P\neq \emptyset$:

$$P=\{x\in \mathbb{R}^n\;|\;h_i^Tx\leq c_i, i=1,...,k \} =\{x\in \mathbb{R}^n\;|\;g_j^Tx\leq d_i, j=1,...,l \}. $$

I'd like to show that given those representations, if I know that the vectors $h_1, . . . ,h_k$ are spanning $\mathbb{R}^n$, then so do also the vectors $g_1, ..., g_l$.

How would you do that ?

I was thinking about using the properties of the polyhedron, seen as geometric objects, but I ended up showing nothing.

Alright, I've found a way to do it :

Let's consider the matrix $A$ whose lines are the vectors $a_i^T$. It has to have full column-rank, since the $a_i$ are spanning $\mathbb{R}^n$ (we necessarily have $m\geq n$).

Since it has full column-rank, it means our polyhedron can't contain a line.
(I let you prove the lemma saying a polyhedron does not contain a line if and only if its matrix has full column-rank).

But now, since it doesn't contain a line, it has to have one extreme point (with $n$ active constraints), because if you take a point in P, choose a direction in which every active constraints remain active and move along this direction, you'll end up meeting another constraint, which will become active. (You have to met such a constraint, since $P$ can't contain a line.) Just repeat the procedure until you've got $n$ linearly independent active constraints. Actually, what you'll get is a basic feasible solution, in terms of Linear Programming.

Now you've got a point with $n$ linearly independent active constraints and you know you can represent it using the $h_i$ too. Let's do it. And that's it, you've got $n$ linearly independent vectors $h_j$, since $\mathbb{R}^n$ is of dimension $n$, you know that those $h_j$ are spanning the whole space.


Time Complexity Recursion I have this recursive function:
$$
F(n) = \left\{
\begin{array}{l l}
F(n-2) + 10 F\big(\lfloor\frac{n}{6}\rfloor\big)^2 + 6 F\big(\lfloor \frac{n}{7} \rfloor \big) + \frac{n^4}{5} & \text{if } n > 1\\
2 & \text{otherwise}\\
\end{array} \right.
$$

And I need to calculate its time complexity. Could you please guide me on how to do it? I assume time grows exponentially, maybe it is $ O(3^n)$  because each time this function calls itself 3 times more? (Given we implement this by simple recursion)

If we take the floor of $n/6$ and $n/7$, $O(3^n)$ looks too high to me.

One straight-forward algorithm is to calculate every point starting from $1$ to $n$ (and store all the results in an array), and then you get complexity of $O(n)$

EDIT

If you brutal force by implementing this using simple recursion approach, yes, it is $O(3^n)$, because you cannot re-use any results you calculated when arriving each single point of $F$.

In that case, for each $F(n)$, you need to recurse for $O(n)$ layers, and within each layer, you spawn three function calls that each will spawn another three function calls for the next layer of recursion calls. Thus it is $O(3^n)$


What is the physical meaning of the empty set? Take the example of rolling one $6$ sided dice. If an event $A$ represents a null set, what would be the outcome or the result of getting the event of an empty set?
For example if $A = \{1,2,7\}$ you can say that $A$ is the set of an outcome of $1, 2,$ or $7$ which the last is impossible. Or $A$ could be the sample space meaning the set of all outcomes, but if $A$ is the null set, does it mean that there are no outcomes, i.e. the dice was not rolled?
I was trying to find independent and not disjoint events for this die example. This led me to consider the null set, but I couldn't visualize what the meaning of the null set was for this scenario.

I think it helps to view events as payoff sets in games. For $A=\{1, 2, 7\}$, for example, we play the game where you roll a $6$-sided die, and I win if it comes up $1, 2$, or $7$.

The empty set, then, corresponds to the "fixed game": you roll a $6$-sided die, and I lose.

There are other "fixed game" events (that is, events with zero probability): e.g. $\{7\}$. But those are not obviously fixed (maybe one of the faces of the die is marked $7$! Or maybe I can't count very well!), while the empty set corresponds to a game which is fixed by definition.

This is very similar to the physical significance of any number.

I'll use $0$ as an example since it's more interesting from a physical perspective than the others. While certainly, negative numbers have no physical relevance.

If you just say $0$, it has no physical relevance. You can have $0$ chairs, $0$ cars, $0$ snow flakes, etc. But that depends entirely on us to give it meaning, and as long as we don't we can argue (equally validly) that either it has no physical meaning or that it has an infinite number of possible meanings which are not yet determined.

I guess a lot of pure math is like that.

So a null set is not a null set in relation to anything as long as there is no context.

To answer your question specifically, in the case of rolling dice, we could agree that this simply means the dice has not been rolled yet. Though, this interpretation is in no way fundamental or essential.


Why is this function continuous but not differentiable? I have a true and false on my practice exam about this function being continuous but not differentiable. The answer is true, but I don't understand how it is not differentiable.
$$
f(x) =
\begin{cases}
x\sin(1/x)  & : \text{if }x \neq 0 \\
0 & : \text{if }x=0
\end{cases}
$$

I know that when $x$ doesn't equal $0$, that it is continuous and differentiable since $x\sin(1/x)$ is composed of 2 continuous and differentiable functions.

I also found it to be continuous when $x=0$.
For it being differentiable at $x=0$, I took $\lim_{x\to 0}\frac{f(x)-0}{x-0}$ and got $\lim_{x\to 0}\frac{x\sin x}{x}$ and then the limit equals $0$. Does this not show it's differentiable? what am I doing wrong?

In the calculation of the limit you have substituted $r=1/x.$ The limit is not $\frac{r\sin r}r$ as $r$ goes to $0$ but (the same expression) as $r$ goes to $\pm\infty.$

Because $$sin (1/x)$$ is oscillating near 0 and the limit doesn't exist...
And just use the fact that
 $$x^msin(1/x^n)$$ is differentiable iff $$m  >1$$ .ofcourse the interval of consideration is $$[0,1]$$


find a solution of 9x = 24 (mod 21) I need help finding a solution of $9x\equiv {24}\pmod {21}$.

Here is what I tried, but it's wrong.

mod x is the positive value of x. mod $21 = 21.$

$9x\equiv {24}\pmod {21}$.

$9x = 24*21$

$x = 24*21/9 = 56$

I'm not sure what you mean by "mod x is the positive value of x", but you should not multiply $24$ and $21$.  What you want is a number $x$ such that $21$ will divide $9x - 24$.  We write this as $21 \mid 9x - 24$.  Note that $3$ divides both sides, so it must be that $21/3 \mid (9x - 24)/3$ or $7 \mid 3x - 8$.  From there you can try and calculate an inverse to $3$ modulo $7$ or you can just do a little guess and check to find that $x = 5$ works.

By trial and error method:
\begin{align*}
9x &\equiv24\mod21\\
\implies 3x&\equiv24 8\mod7  \text{ (dividing by} 3)\\
\implies 7 &| (3x-8)
\end{align*}
(Put the values: $x=0,1,2,3,4,5\cdots$ or $-1,-2,-3,-4,-5\cdots$ and check it.
So for the value $x=5$ or $-2$  it satisfies $7 | (3x-8).$)

Thus $x\equiv5\mod7$ or $x\equiv-2\mod7.$ This implies $x\equiv5\mod7.$

[Hint: For  entrance exams, put the given options directly and check it.]
©by Ramesh Bhat, Nakre


Infinite direct product of fields. Let $F$ be a field, and consider the infinite direct product$$F \times F \times F \times F \times \dots,$$i.e. $\prod_{i=0}^\infty F$, i.e. the direct product of a countable number of copies of $F$. The theory of bases for vector spaces shows that this direct product has a basis, and so is isomorphic to a direct sum of copies of $F$. How many copies of $F$ do I need $($i.e. what is the dimension of this direct product as an $F$-vector space$)$? Is it possible to exhibit this isomorphism between an infinite direct product and an infintie direct sum explicitly?

The dimension is the cardinality $|F|^{\aleph_0}$ according to the Theorem of Erdös-Kaplansky. See MO/49551 for a general formula of the dimension of an infinite family of vector spaces. All this belongs to "set-theoretic linear algebra", and no bases can be explicitly written down, since the axiom of choice plays a central role. Probably there are models of ZF where $\mathbb{Q} \times \mathbb{Q} \times \dotsc$ is no direct sum of copies of $\mathbb{Q}$; hopefully a set-theorist can add a reference for this.

To see the difference between infinite direct product and infinite direct sum (both involving countable number of copies) see their concrete realisations:
The vector space of all polynomials in $X$ over $F$, and the vector space of all formal power series (convergent or not) in $X$ over $F$. Now you can see what forms the basis in each case.


Are limits of integration defined with respect to the pronumeral or the differential? Consider the indefinite integral $$\int x^2 \,d(x^2).$$ It evaluates fairly easily to $\tfrac{x^4}{2} + C$. My question is about what happens when we start evaluating definite integrals with respect to these functions. In a specific example, how would $\displaystyle \int_0^2 x^2 \,d(x^2)$ be evaluated?
Would it be: $$\int_0^2 x^2 \,d(x^2)=\left[\frac{1}{2}\cdot\left(x^2\right)^2\right]_0^2 =\frac{4}{2}-0=2 $$ with the limits applying to $x^2$, or would it be: $$\int_0^2 x^2 \,d(x^2)=\left[\frac{1}{2}\cdot x^4\right]_0^2 =\frac{16}{2}-0=8 $$ with the limits applying to $x$.
Hope someone can clarify, thanks in advance!

Confusion
If this is a Riemann integral I think I have to disappoint you. I don't think there is something like
$$\int_a^b f(x)\,\mathrm{d}g(x) = \int_a^b f(x)\,\frac{g(x)}{x}\,\mathrm{d}x.$$
It just does not make sense to define something like that. What is the problem with the integral on the right hand side? Why do we need a new notation? If there is, I take everything back. But it is not widely known, you should have defined it.
A similar integral
In fact there is a well-defined integral that looks like yours. It's called the Stieltjes integral, a generic term for the Riemann-Stieltjes integral and the Lebesgue-Stieltjes integral.
$$\int_a^bf(x)\,\mathrm{d}g(x).$$
You may look up it's definition here. I will not go into detail but it has a nice property if $g(x)$ is absolutely continous
$$\int_a^b f(x)\,\mathrm{d}g(x) = \int_a^b f(x)g'(x)\,\mathrm{d}x.$$
Stieltjes integral of your problem
In your example take $f(x)=g(x)=x^2$. Since $g(x) = x^2$ is absolutely continous we can write
$$\int_0^2 x^2\,\mathrm{d}x^2 = 2\int_0^2 x^3\,\mathrm{d}x=\left[\frac{x^4}{2}\right]_{0}^{2} = 8.$$


Finding the Derivative of $\sqrt{x}$ How can I find the derivative of $\sqrt{x}$ using first principle. Specifically I'm having difficulty expanding $\sqrt{x + h}$ or rather $(x + h)^.5$.

Is there any generalized formula for the expansion of non integer exponents less than 1?

You can use the following
$$\begin{aligned}
\frac{\sqrt{x+h}-\sqrt{x}}{h}& =\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\& =\frac{1}{\sqrt{x+h}+\sqrt x}.
\end{aligned}$$

Notice, $$\frac{d\sqrt x}{dx}=\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$$
$$=\lim_{h\to 0}\frac{\sqrt x\left(1+\frac{h}{x}\right)^{1/2}-\sqrt{x}}{h}$$
Since, $h\to 0 \iff \frac{h}{x}\to 0$ hence using binomial expansion & neglecting higher power terms, we get
$$=\lim_{h\to 0}\frac{\sqrt x\left(1+\frac{1}{2}\frac{h}{x}\right)-\sqrt{x}}{h}$$
$$=\lim_{h\to 0}\frac{\sqrt x\left(1+\frac{h}{2x}-1\right)}{h}$$
$$=\lim_{h\to 0}\frac{\sqrt x\left(\frac{h}{2x}\right)}{h}$$
$$=\lim_{h\to 0}\frac{\left(\frac{h}{2\sqrt x}\right)}{h}$$
$$=\lim_{h\to 0}\frac{1}{2\sqrt x}=\frac{1}{2\sqrt x}$$


Analytic solution of a certain functional equation Here's my question: let $b_2$, ..., $b_d \in \mathbb{C}$ ($d$ is an integer greater than $2$), and consider the functional equation

$$V(z^d)=dz^{d-1} V(z)+(b_2 z^{d-2} + b_3 z^{d-3}+\ldots + b_d)$$

In this article: http://www.math.harvard.edu/~ctm/papers/home/text/papers/wp/wp.pdf , page 33, McMullen claims that it is clear this functional equation admits a unique analytic solution on $\{z$ $s.t. |z|>1\}$, and that the solution is

$$V(z)=\frac{-z}{d} \sum_{k=2}^d \sum_{n=0}^{\infty} \frac{b_k z^{-k d^n}}{d^n}$$

Of course I believe him =) , but I've been staring at the equation and the solution for some time, and it's not clear to me that the solution is unique (or how he found the solution). All I can do is check it is indeed a solution. So:

Question 1: why is it unique ?

Question 2: can you see a method on how the formula was obtained?

PS: I mentioned the article, but I believe there is no need to refer to it. This problem should be self contained.

Thanks in advance!

First, this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2308.pdf

Let $\begin{cases}z_1=\ln z\\V_1(z_1)=V(z)\end{cases}$ ,

Then $V_1(dz_1)=de^{(d-1)z_1}V_1(z_1)+\sum\limits_{k=2}^db_ke^{(d-k)z_1}$

Then, this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2303.pdf

Let $\begin{cases}z_2=\ln z_1\\V_2(z_2)=V_1(z_1)\end{cases}$ ,

Then $V_2(z_2+\ln d)=de^{(d-1)e^{z_2}}V_2(z_2)+\sum\limits_{k=2}^db_ke^{(d-k)e^{z_2}}$

$z_2\to z_2\ln d$ :

$V_2(z_2\ln d+\ln d)=de^{(d-1)e^{z_2\ln d}}V_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)e^{z_2\ln d}}$

$V_2((z_2+1)\ln d)=de^{(d-1)d^{z_2}}V_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)d^{z_2}}$

$\therefore V_{2,c}((z_2+1)\ln d)=de^{(d-1)d^{z_2}}V_{2,c}(z_2\ln d)$

$V_{2,c}(z_2\ln d)=\Theta_1(z_2)\prod\limits_{z_2}de^{(d-1)d^{z_2}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Indefinite_product#Rules,

$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}\left(\prod\limits_{z_2}e^{d^{z_2}}\right)^{d-1}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}\left(e^{\sum\limits_{z_2}d^{z_2}}\right)^{d-1}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Indefinite_sum#Antidifferences_of_exponential_functions,

$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}\left(e^{\frac{d^{z_2}}{d-1}}\right)^{d-1}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}e^{d^{z_2}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

Let $V_2(z_2\ln d)=d^{z_2}e^{d^{z_2}}W_2(z_2\ln d)$ ,

Then $d^{z_2+1}e^{d^{z_2+1}}W_2((z_2+1)\ln d)=de^{(d-1)d^{z_2}}d^{z_2}e^{d^{z_2}}W_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)d^{z_2}}$

$d^{z_2+1}e^{d^{z_2+1}}W_2((z_2+1)\ln d)=d^{z_2+1}e^{d^{z_2+1}}W_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)d^{z_2}}$

$W_2((z_2+1)\ln d)=W_2(z_2\ln d)+\sum\limits_{k=2}^d\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}$

$W_2(z_2\ln d)=\Theta_1(z_2)+\sum\limits_{z_2}\sum\limits_{k=2}^d\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$W_2(z_2\ln d)=\Theta_1(z_2)+\sum\limits_{k=2}^d\sum\limits_{z_2}\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Antidifference#Mueller.27s_formula,

$\because\lim\limits_{z_2\to+\infty}\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}=0$

$\therefore W_2(z_2\ln d)=\Theta_1(z_2)-\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^{z_2+n}}}{d^{z_2+n+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$V_2(z_2\ln d)=\Theta_1(z_2)d^{z_2}e^{d^{z_2}}-e^{d^{z_2}}\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^nd^{z_2}}}{d^{n+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$V_2(z_2\ln d)=\Theta_1(z_2)e^{z_2\ln d}e^{e^{z_2\ln d}}-e^{e^{z_2\ln d}}\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^ne^{z_2\ln d}}}{d^{n+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$V_2(z_2)=\Theta(z_2)e^{z_2}e^{e^{z_2}}-e^{e^{z_2}}\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^ne^{z_2}}}{d^{n+1}}$ , where $\Theta(z_2)$ is an arbitrary periodic function with period $\ln d$

$V(z)=\Theta(\ln\ln z)z\ln z-z\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^n\ln z}}{d^{n+1}}$ , where $\Theta(z)$ is an arbitrary periodic function with period $\ln d$

$V(z)=\Theta(\ln\ln z)z\ln z-z\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_kz^{-kd^n}}{d^{n+1}}$ , where $\Theta(z)$ is an arbitrary periodic function with period $\ln d$

The solution should be not unique.

So McMullen claims should be wrong. I think he should first make deep introspection about the form of the general solution of the functional equation $V(z^d)=f(z)V(z)+g(z)$ .


Why Does $f(x) = x\sqrt{x+3}$ Only Have One Critical Point? I am trying to find the critical points of the function $f(x) = x\sqrt{x+3}$, then by using the First Derivative Test, determine which ones are a local maximum, local minimum, or neither.

Using the product rule, we get $\frac{3(x+2)}{2\sqrt{x+3}}$. So $x=-2$ is a critical point. But isn't $x=-3$ also a critical point because if $x=-3$ then the derivative of $f(x)$ wouldn't exist.

To give an example of why I am thinking this, look at $g(x) = x^{2/3}(6-x)^{1/3}$.

The derivative of $g(x)$ is $\frac{4-x}{x^{1/3}(6-x)^{2/3}}$, where $x$ cannot equal $0$ or $6$.

The critical points for $g(x)$ are $0$, $6$, and $4$.

So why aren't $-2$ and $-3$ the critical points for $f(x)$?

In order for $a$ to be a critical point (at least, under the definition I'm familiar with), $a$ has to be in the interior of the domain of the function (that is, the function has to be defined on either side of $a$); for any $x<-3$, $f(x)$ isn't defined (exercise: why?), so $-3$ can't be a critical point.

you want $\sqrt{-3+3} = 0$ as another point right?

isn't a trying to square root a negative leading to something else entirely like complex numbers.


How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE? My solution:

In the word INVOLUTE, there are $4$ vowels, namely, I,O,E,U and $4$
consonants, namely, N, V, L and T.

The number of ways of selecting $3$ vowels out of $4 = C(4,3) = 4$.
The number of ways of selecting $2$ consonants out of $4 = C(4,2) = 6$.
Therefore, the number of combinations of $3$ vowels and $2$ consonants is
$4+6=10$.

Now, each of these $10$ combinations has $5$ letters which can be arranged among
themselves in $5!$ ways. Therefore, the required number of different words is
$10\times5! = 1200$.

But the answer is $2880$.

What am I doing wrong? Please explain.

The number of combinations of $3$ vowels and $2$ consonants should be $4\times 6=24$ instead of $4+6=10$. Since you are considering combinations, each set of $3$ vowels and each set of $2$ consonants form a combination, so you need to multiply and not add.

The combinations are $(\{I,O,E\},\{N,V\}),(\{I,O,E\},\{N,L\}),\dots$. You can try writing out all $24$ combinations as an exercise.

Then you will get $24\times 120=2880$.

IN THE WORD INVOLUTE, WE HAVE 4 VOWELS AND 4 CONSONANTS.

OUT OF 4 VOWELS WE CHOOSE 3 VOWELS THAT MAKES IT 4C3.
OUT OF 4 CONSONANTS WE CHOOSE 2 CONSONANTS THAT MAKES IT 4C2.

Therefore on calculation, we get

4C2*4C3*5!( we take 5! because we can arrange it in 5 factorial ways.)

Therefore,
6*4*5!= 24*120=2880


Minimum value of $(\cos\theta_1+\cos\theta_2+\cos\theta_3)$ If $\hat{x}$, $\hat{y}$ and $\hat{z}$ are three unit vectors in three-dimensional space, then the minimum value of\
$|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2$ is:

My Attempt : Let $\theta_1$ be angle between $\hat{x}$ and $\hat{y}$, $\theta_2$ be angle between $\hat{y}$ and $\hat{z}$ and $\theta_3$ be angle between $\hat{z}$ and $\hat{x}$.

$\therefore \theta_1,\theta_2$ and $\theta_3$ are the angles between any 2 edges of a tetrahedron from a single vertex.

$\therefore |\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2=(2+2\cos\theta_1)+(2+2\cos\theta_2)+(2+2\cos\theta_3)$.

$=6+2(\cos\theta_1+\cos\theta_2+\cos\theta_3)$

Can anyone please tell me what to do next?

You can rotate the picture and assume $\hat{x} = (\cos \theta, \sin \theta, 0)$ and $\hat{y} = (\cos \theta, -\sin \theta, 0)$ for some $0 \leq \theta \leq {\pi \over 2}$. Writing $z = (z_1,z_2,z_3)$ you have
$$\hat{x} \cdot \hat{y} + \hat{x} \cdot \hat{z} + \hat{y} \cdot \hat{z} = \cos^2 \theta - \sin^2 \theta + 2z_1\cos \theta $$
This is minimized for $z = (-1,0,0)$, in which case we have
$$\hat{x} \cdot \hat{y} + \hat{x} \cdot \hat{z} + \hat{y} \cdot \hat{z} = \cos^2 \theta - \sin^2 \theta - 2\cos \theta $$
$$= 2\cos^2 \theta - 2\cos \theta - 1$$
The function $2x^2 - 2x - 1$ has its minimum at $x = {1 \over 2}$, so $2\cos^2 \theta - 2\cos \theta - 1$ is minimized when $\theta = {\pi \over 3}$.

Thus we have $\hat{x} = ({1 \over 2}, {\sqrt{3} \over 2}, 0)$, $\hat{y} = ({1 \over 2}, -{\sqrt{3} \over 2}, 0)$, and $z = (-1,0,0)$, corresponding to
$\hat{x} \cdot \hat{y} + \hat{x} \cdot \hat{z} + \hat{y} \cdot \hat{z} = 2({1 \over 2})^2  - 2{1 \over 2}  + 1 = -{3 \over 2}$, so that
$$|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2 \geq 6 - 3$$
$$ = 3$$

$$\sum |\vec{x}+\vec{y}|^2=6+2\sum \vec{x}\cdot \vec{y}\geq 6-3\geq 3$$

Above  i am using

$$|\vec{x}+\vec{y}+\vec{z}|^2\geq 0\Longrightarrow 3+2\sum \vec{x}\cdot \vec{y}\geq 0$$

$$\Longrightarrow \sum \vec{x}\cdot \vec{y}\geq -\frac{3}{2}$$


Probability of different balls to different boxes There are $6$ different balls that need to be arranged in $4$ different boxes. If there is at least one ball in each box, what is the probability that there is a box with $3$ balls? (assuming that for every arrangement of the balls in the boxes there is the same same probability).

My attempt: is to choose $4$ balls and arrange them in the boxes and then to choose one box that will get the other 2 balls: $$\frac{\dbinom{6}{4}4!\cdot4}{4!\cdot\dbinom{4+6-1}{6}}$$

Is that correct?

Step one: Find the size of the sample space.  How many ways can you arrange six balls into four boxes such that every box gets at least one ball.

Approach via inclusion-exclusion.

Let the balls be labeled $1,2,3,4,5,6$ and the boxes be labeled $A,B,C,D$.

Number of ways placing balls in boxes can be thought of as counting how many functions there are from the set of balls to the set of boxes.  In this case, in particular we wish to count how many surjective functions there are.

Let $X$ be the universal set of all functions from six balls to four boxes.  $X_A$ denote the set of functions where box $A$ is empty (i.e. $A$ has no preimage).  $X_B$ denote the set of functions where box $B$ is empty, ... $X_{A,B}$ the set of functions where both $A$ and $B$ are empty, et cetera.

Then $|X_A^c\cap X_B^c\cap X_C^c\cap X_D^c|$ represents the number of ways of sending the balls to the boxes so that $A$ is not empty, $B$ is not empty, $C$ is not empty, and $D$ is not empty.

Now, expand this via inclusion exclusion: $|X_A^c\cap\dots X_D^c| = |X|-|X_A\cup X_B\cup X_C\cup X_D| \\= |X|-|X_A|-|X_B|-|X_C|-|X_D|+|X_{A,B}|+\dots-|X_{A,B,C}|-\dots+|X_{A,B,C,D}|$

How big is $|X|$?  $|X_A|$?  $|X_{A,B}|$?...

  $=4^6-4\cdot 3^6+6\cdot 2^6-4\cdot 1^6 + 0$

Now, the problem tells us to treat each of these possible ways as having equal probability, so to continue, this will be our denominator.

For the numerator, we notice that for there to be a box with three balls and all boxes to have at least one ball (and not be empty), it must be that there is exactly one box with three and exactly three boxes with one ball each.

Approach via multiplication principle:

Pick which box has three balls
Pick which three balls went into that box
Left to right, for each still empty box, pick a ball from those remaining to go into the box

   There are then $4\cdot \binom{6}{3}\cdot 3!$ good arrangements.

Combine this information using that probability in an unbiased sample space is given as $Pr(E)=\frac{|E|}{|\Omega|}$

Extremely new stuff: we are wrong!

Check this out: simulation disagrees with both of us!

Simulation says:

Numerator = 2880.
Denominator = 7200.
Probability = 0.4.

Go figure.

Simulation code:

import random

arrange_good_target = dict()
arrange_good = dict()

for i in range(0, 10000000):
    balls = [1,2,3,4,5,6]
    buckets = [[],[],[],[]]

    # assign all balls to buckets
    while len(balls) > 0:
        # choose ball and bucket
        ball_i = random.randrange(0,len(balls))
        bucket_i = random.randrange(0,4)

        # put chosen ball in chosen bucket
        buckets[bucket_i].append(balls.pop(ball_i))

    # test if configuration is good
    good = True
    for i in range(0, 4):
        if len(buckets[i]) == 0:
            good = False

    # test if configuration is good and meets target
    good_target = False
    for i in range(0, 4):
        if len(buckets[i]) == 3:
            good_target = True

    # update counters
    if good:
        arrange_good[str(buckets)] = 1

    if good and good_target:
        arrange_good_target[str(buckets)] = 1

# hell yeah dude
probability = float(len(arrange_good_target))/len(arrange_good)

print 'numerator is: ' + str(len(arrange_good_target))
print 'denominator is: ' + str(len(arrange_good))
print 'probability is: ' + str(probability)

#for key in arrange_good.keys():
#    print key

New stuff

My answer after thinking independently (thanks to JMoravitz for putting spoiler-guards):
\begin{equation}
  \frac{{4 \choose 1} \times [{6 \choose 3} \times 3 \times 2 \times 1]}{{4 \choose 1} \times [{6 \choose 3} \times 3 \times 2 \times 1] + {4 \choose 2} \times [{6 \choose 2} \times {4 \choose 2} \times 2 \times 1]}
\end{equation}

Turns out that my answer:

Has an identical numerator.
But my denominator is more straightforward to understand in my view

Let me rewrite mine to make it look tinier:
\begin{equation}
  \frac{{4 \choose 1} {6 \choose 3} 3!}{{4 \choose 1} {6 \choose 3} 3! + {4 \choose 2} {6 \choose 2} {4 \choose 2} 2!}
\end{equation}

Old stuff

let's see... there are only 4 possible configurations where a single box has 3 balls. So clearly the probability is this:
\begin{equation}
\Pr(\text{4 balls in a box}) = \frac{4}{\text{normalization stuff goes here}}
\end{equation}

Now, what are all the possible configurations of fitting six balls into four buckets? This is the value of our $\text{normalization stuff goes here}$.

Let's see. If three balls in a bucket, we get four possibilities. What if we allow for two balls in a bucket? We must have two balls in one bucket, and two other balls in another bucket, then have two buckets each with a single ball (so that we consume all the six balls). Here, we have ${6 \choose 2}$ possibilities.

Now, what if we allow for one ball in each box? Impossible, cause we won't use all six balls.

Now, what if we allow for four balls in a single box? Impossible too, cause we will have an empty bucket.

There you go buddy, answer is:
\begin{equation}
\Pr(\text{4 balls in a box}) = \frac{4}{4 + {6 \choose 4}}
\end{equation}


Find the value of integration following : $\int x^2\sqrt{x^2+1}dx$ Find the value of integration following :
$$\int x^2\sqrt{x^2+1} dx$$

Let $x=\sinh(t)$ and $dx=\cosh(t) \,dt$. So, $$I=\int x^2\sqrt{x^2+1}\ dx=\int \sinh^2(t)\cosh^2(t)\,dt=\frac 14 \int \sinh^2(2t)\,dt=\frac 18 \int (\cosh(4t)-1)\,dt$$ $$I=\frac 1{32}\sinh(4t)-\frac t8+C$$

$$
\frac{1}{4}(x^2+1)^{\frac{3}{2}}-\frac{x}{8}\sqrt{x^2+1}-\frac{1}{8}\mathrm{arcsinh}(x)
$$
Differentiating the above will give you $x^2\sqrt{x^2+1}$, proving it is indeed the antiderivative.


Understanding fractions in modular arithmetic I would like confirmation that my understanding of a congruence of two fractions is correct.

Suppose we have

$$\frac{2}{3} \equiv \frac{4}{6}\mod 5$$

What happened was multiplying both the numerator and denominator of the left fraction by $2$.

So $\frac{2}{3}$ is essentially the same as $3x\equiv 2 \mod 5$ and we can multiply both sides by $2$ since that is valid in modular arithmetic we get $6x \equiv 4\mod 5$ and rewriting that as a fraction we get $\frac{4}{6}$.

A fraction really represents that $x$ and since that $x$ remains unchanged via such manipulations then those fractions are essentially congruent.

Of course I am aware this is true only for valid fractions, i.e. where the denominator is invertible mod $5$.

Is my understanding of this correct?

Fractions work fine in modular arithmetic. They're not very commonly used, though.

If you want to use fractions, you have to stop thinking about $\frac23$ as $0.666\ldots$ and instead think of it as $4$ (at least in modulo $5$). Also, there is no simple way to just look at a modular fraction and know more or less what its value is, the way you can just know that $\frac{123456}{1000001}$ is very close to $0.123456$. In other words, small change in the denominator gives large changes in fraction value in modular arithmetic.

So there are a few mental roadblocks on the way to using fractions in modular arithmetic, and I suspect that that's why it isn't done more. That being said, if you do use fractions, the same rules that you learned in elementary school for regular (rational) fractions still all apply: How to expand / simplify fractions, how to add or multiply them, or how to check that they are equal. The technique known as "cross multiplication" still works: $\frac ab\equiv \frac cd\iff ad = bc$ (this does require that only invertible elements are allowed as denominators, though).

You can allow non-invertible elements as denominators, but then more care is needed. See, for instance, my answer here on allowing fractions with odd denominators (including $3$) in modulo $6$.

If you only allow invertible elements as denominators, then because they're already invertible, you're not really changing anything, as all fractions are elements that already exist as whole numbers. If you allow non-invertible elements, you basically get a smaller ring as a result (modulo $6$ with $3$ allowed as a denominator gives you what is essentially modulo $2$). Sadly, you can't get something "bigger" (the way $\Bbb Q$ is bigger than $\Bbb Z$), beause that requires you to have denominators which are neither invertible nor zero-divisors.

Edit: Since I apparently missed the real point of the question. Yes, those manipulations are fine. Whatever $\frac23$ is, give it the name $x$. By the definition of fractions, $3x = 2$. Multiplying both sides by $2$ can't make them not-equal, so $6x = 4$. Since $6$ is a valid denominator, by definition of fractions we get $x = \frac46$. What you have ultimately done is to expand the fraction $\frac23$ by $2$ and shown that this doesn't change the value of the fraction.

Yes, and just like you can do modular arithmetic to simplify ordinary arithmetic using integers, one can also do this for fractions. This looks impossible because of the multiple ways one can write a fraction modulo an integer representing a given number. But if it is known that the answer one is interested in, has a numerator between $0$ and $N$ and a denominator between $0$ and $D$ then a result modulo a number larger than $2 N D$ will allow one to reconstruct this rational number. As explained here, finding this fraction involves just the Euclidean algorithm.


Disproving complex differentiability I think I lack intution when dealing with disproving holomorphicity of functions like:

  a) $f(z) = \frac{z}{\lvert z \rvert}$
  b) $f(z) = \left \{ \begin{array} .\frac{z^5}{\lvert z^4 \rvert} &z\neq 0\\0 & z=0 \end{array} \right . $

For a), one way to prove it is that we have constant modulus, therefore $f$ can't be holomorphic since in that case it would be constant (using the Open Mapping Theorem or the Cauchy Riemann equations). Is there an alternative way of proving this conveniently?

For b), I am unsure, since we don't have constant modulus. We can write $z = |z|e^{i \varphi}$ using polar coordinates and get $$f(z) = \left \{ \begin{array} . \lvert z \rvert e^{5i \varphi} &z\neq 0\\0 & z=0 \end{array} \right .$$
This to me looks similar like the identity, besides the $5$ in the exponent. How do I prove this one isn't holomorphic (at $z=0$)?

In mathematics, it is often useful to simply go back to the definition. A function is complex differentiable if for all $z_0 \in \mathbb{C}$, $\frac{f(z)-f(z_0)}{z-z_0}$ has a limit when $z \to z_0$.

Here there is clearly a problem at $z_0 = 0$. So let's try to compute $\lim_{z \to 0} f(z)/z$.

For a), $\lim_{z \to 0} f(z)/z = (z/|z|)/z = 1/|z|$ which does not converge to anything as $t \to 0$. So the limit doesn't exist.

For b), it's a bit more subtle. A useful property in complex analysis is that if a limit exists, then it must be the same limit along all path leading to the point $z_0$. Let's try to compute the limit along a horizontal vertical line, i.e. with $z = t$, $t \in \mathbb{R}$ and $t \to 0$ (distinguishing the two cases $t>0$ and $t<0$).

For the first one, $f(t) = t^5/|t|^4$. If $t > 0$, this gives $f(t) = t$ and so $f(t)/t = 1$ and the limit is $1$. But if $t < 0$, then $f(t) = -t$ and $f(t)/t = -1$ and the limit is $-1$. But to be complex differentiable, the limit should have been the same, a contradiction. So the function is not complex differentiable.

Hint
Compute $$\lim_{t\to 0}\frac{f(it)}{it}\quad \text{and}\quad \lim_{t\to 0}\frac{f(t)}{t}$$


Derivative of $\|Xa\|_2 $ with respect to $X$ Can someone give me the answer to the following expression?

$\frac{\partial}{\partial X}\|Xa\|_2 =?$

$X$ is a square matrix and $a$ is a vektor of the apropriate size. $\|\cdot\|_2$ is the euclidean norm.
Thanks

Take any matrix $B$ and expand
$$
\|(X+tB)a\|_2^2 = \|Xa\|_2^2 + 2 t \langle Xa , Ba \rangle + t^2 \|Ba\|_2^2.
$$
Therefore the directional derivative of the square of your map in the direction $B$ is
$$
2 B^t Xa
$$
where $B^t$ is the transposed matrix of $B$.

If the following is correct, then how are the above answers related to it?

$$
\frac{\partial}{\partial X}\left[\|Xa\|_2\right] = \frac{\partial}{\partial X}\left[(a^TX^TXa)^{\frac{1}{2}}\right]
= \frac{1}{2}(a^TX^TXa)^{-\frac{1}{2}}\cdot\frac{\partial}{\partial X}\left[(a^TX^TXa)\right]
=\frac{1}{2}(a^TX^TXa)^{-\frac{1}{2}}X(aa^T + aa^T)
= (a^TX^TXa)^{-\frac{1}{2}}Xaa^T
= \frac{Xaa^T}{\|Xa\|_2}
$$


number of ways to obtain a pair Here is my attempt to find number of hands in a 5 cards poker game that gives one pair:

$ \binom{4}{2}\binom{48}{3} \cdot 13$ - 2pairs - full house
=1349088-123552-3744=1221792

But it is wrong. What did I do wrong? I know I have the right numbers for full house and two pairs. I dont need another solution for obtaining number of hands that holds one pair. I only need to get an explanation for why this approach to finding number of one pair is wrong.

My approach is to find any hand that has 1 pair and any other 3 cards and then subtract hands in that category that gives other poker combinations which should be full house and two pairs.

As far as I can see, your approach is:

Select a card value. Now, from the $4$ cards with that value in the deck, select $2$ of them to be in the hand. Next, any $3$ of the other $48$ cards in the deck can go in the hand.

Now since there are $13$ different values, multiply by $13$.

The hands which aren't pairs using this approach are 2 pair and full house, so subtract them off the final total.

Correct?

The problem seems to be that you're counting the hands with two pairs twice here.

Example: I select Ace. Now I have AA in hand. I select 553 as the other cards for the hand AA553.

But when I select 5, I can also select 55AA3 as a hand which is the same.

There aren't any further problems, so $1349088-2(123552)-3744=1098240$ is actually the correct answer.

You've counted all the hands that have a pair in, and then subtracted the hands with two pairs or a full house. You haven't subtracted the four-of-a-kind or triplet hands which also contain a pair.


Question about a proof involving polar decomposition of an operator Consider the following theorem in Murphy's book "$C^*$-algebras and operator theory":

Can someone explain why the marked equality is correct? I know $|u^*|^2 = uu^* = w |u|^2 w^*$ but can't see why this equals $(w|u|w^*)^2$.
Context:
$\Vert \cdot \Vert _1$ is the trace class norm.

In Murphy's book theorem 2.3.3 it is proven (see the implication $(4) \implies (2)$) that if $w$ is a partial isometry, then $w^*w$ is the projection on $\ker(w)^\perp =  \ker (u)^\perp  = \ker (|u|)^\perp = \overline{\operatorname{im}(|u|^*)}= \overline{\operatorname{im}(|u|)}$. Hence, $w^*w$ is the identity on the image of $|u|$ and consequently
$$(w|u|w^*)^2 = w|u|w^*w|u|w^* = w|u|^2w^*$$

$(w\vert u \vert w^*)^2 = (w \vert u \vert w^*)(w \vert u \vert w^*) = w \vert u \vert (w^*w) \vert u \vert w^* = \ldots$


Does $\sum{\ln n\over n^{4/3}}$ converge or diverge? Does $\displaystyle\sum{\ln n\over n^{4/3}}$ converge or diverge?
Which test should I use?
I tried the ratio test and root test but both of them are inconclusive.
I could try the comparison test but I don't know which function I have to compare with.

By the integral test you may compare your series with the following integral:
$$
\int_1^\infty \frac{\ln x}{x^{4/3}} dx=\left.\frac{x^{-4/3+1}}{-4/3+1}\ln x\right|_1^\infty-\frac1{-4/3+1}\int_1^\infty x^{-4/3+1}\frac1{x} dx=9<\infty
$$ giving the convergence of your series
$$
\sum_1^\infty\frac{\ln n}{n^{4/3}}.
$$

An other way is to compare $\ln(.)$ with $\sqrt{.}$ in fact $$\forall x \in \mathbb{R}, \ln(x) \le \sqrt{x}$$

Then $$\forall n \in \mathbb{N}, \ln(n) \le \sqrt{n}$$

So for $n> 1$ $$\left|\frac{\ln(n)}{n^{\frac{4}{3}}}\right|\le\left|\frac{\sqrt{n}}{n^{\frac{4}{3}}}\right|=\frac{1}{n^{\frac{6}{5}}}$$

And obviously $\frac{6}{5}>1$ so the series is convergent.


Where can I find the proof of this theorem about the smoothness of the initial data? What to follow is a reference from the text Differential forms by Victor Guillemin and Peter Haine

So clearly the theorems $2.2.4$ and $2.2.5$ follow directely by the Cauchy theorem for ordinary differential equation provided that the vector field if locally lipschitz or of class $C^r$ for $r\ge 1$ since any function of calss $C^1$ is locally lipschitz. However how prove the last theorem? In particular I know that if the vector field $\vec v$ is of class $C^r$ for $r\ge 1$ then the solution of the correspondent system of differential equations is of class $C^r$ too but unfortunately this did not help me to prove the theorem. So could someone idicates where I can find the proof of the last theorem, please? I point out I did NOT study lebesgue integration and measure theory so that I ask courteously to not use them, thanks.
MY PROOF ATTEMPT
So by the regularity theorem we know that the function $\gamma_p$ is of class $C^\infty$ so that if $A_p$ is an open neighborhood of any $p\in V$ contained in $V$ then the set $J:=\gamma^{-1}_p[A_p]$ is open an open interval containing $a\in I$ and so if $\pi$ is the projection of $V\times I$ onto $I$ the statement follows proving that
$$
h(q,t)=[\gamma_p\circ\pi](q,t)
$$
for any $(q,t)\in A_p\times J$ because the set $A_p\times J$ is open and the function $\gamma_p\circ\pi$ is of class $C^\infty$ being a composition of such functions. So how prove the last equality? Is effectively it hold?

There are quite a few alternative proofs of Theorem 2.2.6 and of many of its variants.
Incidentally, Theorem 2.2.6 is just a very particular case of the general situation (most notably, in general it is quite restrictive to assume that $I$ is the same for all $p$ as in Theorem 2.2.6 but it is fine in case we just care about a small neighborhood).
Unfortunately, there are no simple proofs (read short if you want since all proofs are classical), although in the present case most tend to be slightly simple since it is much harder to prove that $C^k$ implies $C^k$ for some finite $k$ then for $k=\infty$. The reason is that for a finite $k$ it is usually necessary a different argument to obtain say $C^k$ and not only $C^{k-1}$ (I am really simplifying this a lot since the theory is quite vast). Still all these proofs for $k=\infty$ are quite long.
Overall it is thus better to give a pointer to a canonical reference, such as Ordinary Differential Equations, by Jack K. Hale that unfortunately is not with us anymore. Have a look at Theorem 3.3 in page 21 (a bit too concise but one of the best places).


A mathematical expression that gives 0 if variable is different from constant and 1 if variable same as constant Example,
∆ = any mathematical expression;
c = constant;
x = variable;
c∆x=1 (if c = x)
c∆x=0 (if c ≠ x)
am just asking?

You can use an Iverson bracket and write $[\Delta=c]$, though you should explain it the first time that you use it.

What you are describing is the Dirac delta function $\delta(x)$, which is often used in theoretical physics.
In your case, the function $\delta(x-c)$ has the property that it has value $1$ if $x=c$, and value $0$ if $x \neq c$.
For more details, you may wish to refer to this Wikipedia article:
https://en.wikipedia.org/wiki/Dirac_delta_function


Is series convergent? Does this series converge?
$$\sum_{n=1}^{\infty }\frac{{(-1)}^{n}(n+2)}{{n}^{2}+4}$$
How can i step-by-step calculate it?

We have

$$\frac{{(-1)}^{n}(n+2)}{{n}^{2}+4}=\frac{(-1)^n}{n}+O\left(\frac{1}{n^2}\right)$$
and the series $\displaystyle\sum \frac{(-1)^n}{n}$ is convergent by the Leibniz theorem and the series $\displaystyle\sum O\left(\frac{1}{n^2}\right)$ is convergent by comparison with Riemann convergent series hence the given series is convergent.

$ | \frac{n+2}{n^2+4} |$ = $ \frac{n}{n^2+4} + \frac{2}{n^2+4} \leq \frac{1}{n^2} + \frac{2}{n^2} = \frac{3}{n^2}$ this implies $|a_n|$ is convergent therefore the sequence $a_n$ converges. I guess just go about it carefully


First Order Separable Differential Equation IVP Solution based on cases Consider the initial value problem
$$\ 4yy' = x, y(x_0)=1$$

a) Find the general solution to the ODE, solving for $y(x)$ explicitly.

b) Find the solutions to the initial value problem. What is the interval of existence? Note that you will need to separate your answer into cases, depending on $x_0$.

c) Sketch some solution curves in the $(x,y)$ plane, choosing at least one $x_0$ for each of the cases you identified in (b).

So far, I have solved part a), but I am unsure where to go from here. I plugged in $y=1$ and $x=x_0$ to solve for C, but I can't figure out what the interval of existence is/what the cases would be.
Thanks!

Note that

$$ y' = \frac{x}{4y} $$

is undefined at $y=0$. Thus, you cannot have a solution that contains $y=0$. Any solution must be either $y<0$, or $y>0$.

Since the given initial condition is $y(x_0)=1$, the solution must be $y > 0$, i.e.

$$ y(x) = \frac{\sqrt{x^2-x_0^2 + 1}}{2} $$

You can then find the domain of this function depending on $x_0$

$4 \times y \times \frac{dy}{dx} =x$

$4 \times y \times dy =x \times dx$

Integrating both sides, we get

$4 \times \frac{y^2}{2}=\frac{x^2}{2} + c$


Math used in industry positions Are there are any jobs in industry that involve more mathematics than just what a (good) undergrad program would cover: calculus, differential equations, real and complex analysis, some basic group theory, etc.? I'm stuck without an academic position but would love to do research, and I just don't see anything out there that isn't trivial. It's incredibly frustrating how little math there is available to me, and I have zero interest in doing mathematical modeling or becoming an actuary. So, what kinds of jobs--- salary, location, etc. are irrelevant--- involve math that's advanced enough to be interesting and challenging?

Depending on the areas of mathematics you find interesting, the answer may be an emphatic "Yes" or an emphatic "No".

Microsoft Research has a pretty strong group in Mathematics. Some members of that group, such as Michael Freedman, have background in very abstract mathematics and a world class research record, though their work at Microsoft is naturally geared more towards applications. The math involved in Microsoft Research's projects is certainly highly challenging, but whether it's interesting to you or not, only you can judge.

Google and Facebook also have strong research groups. I don't know what, exactly, you call "mathematical modeling", which is something you have zero interest in, but hopefully some of the focus areas of this group are of interest to you, and they certainly dedicate time for doing things that would qualify as research in academia.

The financial industry hires people with fairly deep mathematical background, usually deeper than what you'd get in a typical undergrad program. DE Shaw is known for having strong mathematical staff, and I have friends with math Ph.D.'s who were hired by Goldman Sachs and found both the interviews and the actual work very challenging. Renaissance Technologies is a pretty secretive firm but they are clearly hiring for mathematicians.

Another possible direction is bioinformatics. There's plenty of research being done in bioinformatics in industry, but again, I can't judge if this is an area you'd find appealing, or sufficiently "mathematical".

Are you interested in reservoir simulation?

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

Or finite element method?

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

Or aeronautical and aerospace engineering?


Closure and inclusions Let $A$ and $B$ be subsets of $\mathbb{R}^n$. If $A$ is open, and $B$ is arbitrary, does one always have the following inclusion $A \cap \operatorname{cl}(B) \subseteq \operatorname{cl}(A\cap B)$.

Is there a counter-example? I can't seem to find a way to prove it.

The statement is true. The proof just uses the relevant definitions.

Suppose $x \in A \cap \operatorname{cl} (B)$, i.e., $x \in A$ and $x \in \operatorname{cl}(B)$. The latter condition means that if $U$ is an open set containing $x$, then $U$ intersects $B$ (nontrivially).

Now, pick any open set $V$ containing $x$. Then $V \cap A$ is an open set containing $x$. Then by the preceding paragraph, $(V \cap A) \cap B \neq \emptyset$, which is equivalent to $V \cap (A \cap B) \neq \emptyset$. This means that $x \in \operatorname{cl}(A \cap B)$.

the closure is closed, and the intersection of a closed set and an open set is not always a closed set

a conter exemple may be: A=]0,2[ B=]1,3[


Once and for all - "Rational numbers" - because of ratio, or because they make sense? This is a question I'm sure was asked before but I can't find it. There are many sources claiming that the term "rational number" for the elements of $\mathbb{Q}$ comes from the word "ratio", since a rational number is the ratio of two integral numbers. However, there are also other sources that claim that it is because they "make sense" as opposed to irrational numbers.

None of the sources I've encountered were very much convincing as to why their claim is correct. Rational and irrational numbers were known already to the Greeks but I don't know if that's the terminology they used and what it meant then.

So my question is simple: What is the meaning of "rational" in rational number? What's more important is that any claim will be backed up with a reliable and convincing source.

This is not a complete answer, but too long for a comment.  I think that to definitively answer your question would require access to (or knowledge of) both Renaissance era math texts and books/papers from the 1700s (when math in Latin started to be translated to English).

Once upon a time the Greeks used the word "logos" to mean what we think of today as a ratio (a scaling factor; one quantity divided by another).  In the 1600's, Greek mathematical text was translated into Latin and the word "ratio" was used for "logos".  In Latin, "ratio" meant something that was reasoned out, calculated, or thought through.  You can perform all of these actions using logic.  But if you are reasoning out, calculating, or thinking through a numerical computation (like evaluating $\frac{a}{b}$), you might have what we today call a "ratio".

So I would say the answer is both.  Most recently, a "rational number" is what we today call a "ratio" - it's one number divided by another (specified to two whole numbers).  But if you look a little further back in the etymology, the reason that "one number divided by another" is today called a "ratio" is because that happens to be something that you would reason out.  And so with that underlying etymology, a rational number is a number that "makes sense" as the end result of some logical thought.

Just because, here are my two other favorite math etymology items.

"radical" comes from Latin for "root": "radix".  (Pronounced properly, this sounds a lot like "radish".)  So why is $\sqrt{}$ called a radical sign?  Probably because $\sqrt{2}$ is a root of $x^2-2$.  But why are zeros of polynomials called "roots"?  Does this have anything to do with other modern uses of "radical": applying to politics, ideas, chemistry, Chinese character sets?  Yes, it does. Think about squares and sides. In all these instances, something "radical" is "off to the side".  None of this this has anything to do with "radius", despite the apparent similarity.
"polygon" is often translated as a many-sided figure.  Certainly, "poly" means many.  But the "gon" actually means corner or angle.  In modern Greek, "goneis" means elbow.  So I like to think of "polygon" as a many-elbowed figure. "Ortho" means straight/direct (think orthodontia: straightening teeth, and orthodoxy:direct interpretation). So "orthogonal" means something like "having straight corners", which we would translate to "having right angles".

I was taught (decades ago in a mathematics course) that in Ptolemy's ancient Egypt, before Pythagoras discovered and proved the relationship between the sides of right triangles, the notion that there can exist numbers that cannot be expressed as a ratio of two integers was a crazy idea. Irrational, i.e., not able to be expressed as a ratio, eventually became a synonym for crazy.

It's easy to understand why people thought an irrational number was a crazy idea because any physical measurement of the distance between two points can be refined by magnification.  For example, a measurement that is part way between say 5/32 and 3/16 of an inch can be perhaps be more precisely expressed as 11/64 inch.  If that is not precise enough perhaps 21/128 or 23/128 will be the answer, and so forth.  But Pythagoras proved that not to be the case.

I have been seeking a reference that supports that version of the etymology.


Geometrical problem on semi-circles Given that $AE$ is the tangent to the small semi-circle at $D$ and that arc $CD$ : arc $DB$ = $3 : 10$, find arc $AE$ : arc $EB$.

How do I go about solving this? I do not know how to start.

connect BD, CD, BE, draw perpendicular line from D to AB as F, connect D with the centre of the smaller semi-circle O. from $\frac{CD}{BD}$, derive $\frac{CF}{BF}$. Together with CO=OB=OD, you can derive $\frac{OF}{OD}$ and $\frac{DF}{FO}. $ since OD is perpendicular to AE, BE is also perpendicular to AE, using similar triangle, $\frac{DF}{FO}$ =$\frac{AD}{DO}$=$\frac{AE}{BE}$, you get the answer.

Let $O$ be a center of the small circle.

Thus, $$\measuredangle COD:\measuredangle DOB=3:10$$ and since
$$\measuredangle COD+\measuredangle DOB=\pi,$$ we obtain:
$$\measuredangle COD=\frac{3\pi}{13},$$
$$\measuredangle EAB=\measuredangle DAO=\frac{\pi}{2}-\frac{3\pi}{13}=\frac{7\pi}{26}.$$
Can you end it now?


Application of Picard-Lindelöf to determine uniqueness of a solution to an IVP I am still struggling quite a lot with the Picard-Lindelöf-Theorem (also known as the Cauchy-Lipschitz-Theorem).

  Problem: Consider the following IVP with $\alpha \neq 1$  $$\begin{cases} y'&= t|y|^\alpha \\ y(0)&=1 \end{cases} $$
  And show that there exists a unique solution $f_\alpha$ on an Interval $I_\alpha$ with $0 \in I_\alpha$, also show that $f_\alpha (t) > 0$ for small $t$

So to use Picard-Lindelöf I want to show that $f(t,y)=t|y|^\alpha$ is Lipschitz-continuous with respect to its second variable. I could do this by computing the partial derivative with respect to $y$ and find an interval on which this expression is bound, then on said hypothetical interval the solution would exist and be unique.

My approach: $$\frac{\partial f}{\partial y}= t\alpha|y|^{\alpha-1} \cdot \frac{y}{\sqrt{y^2}} $$
At which point I already run into trouble. I have made use of the fact that $|x|'= x/\sqrt{x^2}$ because it gives correct results in regard of the direction (when approaching from the left and the right). But the above expression is not defined at $y=0$.

So I thought if I could come up with an argument that shows why $y \neq 0$, then at least this problem would be out of the way.

Assume that $y=0$ then $y'=0=f(t,y) \implies \frac{\partial f}{\partial y}=0$ which is bound on $\mathbb{R} \implies f$ is Lipschitz continuous with respect to $y$ but we have $y(0)=1$ and thus $$y'=0 \implies y=c $$
So $y$ is a constant function and $y(0)=1$ means that $c=1 \implies y=1 \neq 0$ which shows that $y$ can not be the 'zero' function (if that exists).

Questions:

Is the above argumentation correct?
How can I find the desired interval $I_\alpha$ with $0 \in I_\alpha$?

Update: I tried to continue a bit on my own and ignore the Picard-Lindelöf part and just focus on solving the IVP, apparently the step I have taken above was necessary so I can divide by $|y|^\alpha$, however I will land at $$\frac{dy}{|y|^\alpha}=tdx $$
So it seems like I have to come up with a condition that guarantees me that $y$ is positive in some Intervall.

The "argumentation" can hardly be called such; it's an inconclusive train of thought. So I'll leave it aside and address the problem: what does Picard-Lindelöf theorem actually say?

It says: consider the problem $y'=f(t,y)$ with $y(t_0)=y_0$, and suppose  $(t_0,y_0)$ has a neighborhood (a rectangle $R$ in $ty$-plane) in which $f$ is Lipschitz in $y$ and continuous in $t$. Then there is an interval $(t_0-\delta,t_0+\delta)$ in which a solution exists, is unique, and stays within the aforementioned rectangle.

In your case, $(t_0,y_0)=(0,1)$. Take $R$ to be $[-1,1]\times [1/2,3/2]$; all that matters is that we stay away from $y=0$. Check that $f$ satisfies the above properties on $R$; this isn't hard especially because $|y|=y$ (we have positive values of $y$ only). Apply the theorem. Done.

Nothing in the problem statement says that you have to produce an explicit interval of existence/uniqueness. But if you want  to do this, it's best to use the optimized version  which gives the interval of existence/uniqueness as $|t-t_0|\le \min(a,b/M)$. Here $a$ and $b$ describe the rectangle $R$ that is ours to choose; with my choice $a=1$ and $b=1/2$. The number $M$ is the maximum of $|f|$ on said rectangle, which turns out to be $(3/2)^\alpha$. So, in the interval $|t|\le 2^{-1}(2/3)^{\alpha}$ the solution exists and is unique.

You should work on the specific values of \alpha. For example, if \alpha >1, then we get the local existence of non-negative solution, which is blow-up at a certain time. It is of course that you also get the uniqueness in this case.
Other way, I propose you a different approach to your problem. You can approximate |y|^\alpha by (|y|+\delta)^\alpha. Then, you have no problem with the equation of y_\delta. After that you try to pass to the limit as \delta\rightarrow 0 in order to get the result.

Good luck


Solve $\frac{a+x-\sqrt{a^2-x^2}}{a+x+\sqrt{a^2-x^2}}=\frac{x}{b}$ for $x$ Can somebody please help me find $x$ in the following equation?

$$\dfrac{a+x-\sqrt{a^2-x^2}}{a+x+\sqrt{a^2-x^2}}=\dfrac{x}{b}$$

multiplication by $b$ and $a+x+\sqrt{a^2-x^2}$ and rearranging the variables we get
$$ab+bx-ax-x^2=\sqrt{a^2-x^2}(x+b)$$
now we obtain by squaring the equation $$2 x (a+x) \left(-2 a b+b^2+x^2\right)=0$$ which is easy to solve.


Is there any relation between index of matrix $AB$ and matrices $ A$, and $B$? The index of a matrix $A$ is the smallest nonnegative integer $k$ such that $\operatorname{rank}(A^{k+1})=\operatorname{rank}(A^k)$.

Now I want to know: what relation is there between $\operatorname{Index}(AB)$ and $\operatorname{Index}(A)$, $\operatorname{Index}(B)$?

No it is not true. Consider matrices $A=\begin{pmatrix}1& 0\\  0 & 0\end{pmatrix}$ and $A'=\begin{pmatrix}0& 1\\  0 & 0\end{pmatrix}$, the index of $A$ is one and the index of $A'$ is 2. You are right if the matrices are nonsingular.


Expected time of last bus left I came across the following problem in a probability exercise, and got confused. The problem is the following.

Problem: In a bus station, assume the expected time of next bus arriving is a exponential distribution $Exp(\lambda)$, where $\lambda$=0.1/min, and you arrive at the bus station in a uniform random time. Let $X$ be the amount of time that last bus has left before you arrive. What is $E(X)$?

I applied some probability properties, and got answer 0, which is obviously wrong. Could you please help me see where I am wrong? It would be even better if you could give me a correct solution. Thanks!

My solution: Assume starting at time 0, the buses arrive at time $X_1, X_2, \cdots$, and the interval between buses are $Z_1, Z_2, \cdots$, where $Z_i=X_{i}-X_{i-1}$. Then $Z_i \sim Exp(\lambda)$. Assume the person arrives at the station at time $Y$, and assume $X_k \leq Y < X_{k+1}$. Then
$$X=Y-X_k=Y-\sum_{i=1}^k Z_i.$$
Fix $Y$ and $k$, we have
$$E(X)=Y-k E(Z_1) = Y-k\frac{1}{\lambda}.$$
Fix $Y$, the number $k$ is the number of events between $[0, Y]$, and satisfies a Poisson  distribution with parameter $\lambda$. Thus
$$E(k)=\lambda Y.$$
Therefore we have
$$E(X)=Y-\lambda Y \frac{1}{\lambda}=0.$$

One cannot arrive at the bus station [at] a uniform random time because there is no uniform distribution on the positive real halfline hence the problem is ill posed.

The correct version of the problem is called the waiting time paradox, and asks about the limit of the expected time since the last bus left if one arrives at time $y$, when $y\to\infty$. The expected time since the last bus left if one arrives at time $y$ is $\frac1\lambda(1-\mathrm e^{-\lambda y})$ hence the limit when $y\to\infty$ is $\frac1\lambda$. See here for some more detailed explanations.

The most important thing about $Y$ is $X_k \le Y \lt X_{k+1}$.  Anything which happened before $X_k$ is irrelevant.

The obvious answer is that the expected time since the last bus is the expected time to the next bus, i.e. 10 minutes.

Let's prove this looking forward: the exponential distribution is memoryless, so knowing $X_k \le Y \lt X_{k+1}$ you find $\Pr(X_{k+1}-Y=x)=1 − e^{−\lambda x}$, leading to $E[X_{k+1}-Y] = 1/\lambda$.

Going backwards, you may decide the distribution is the similar by symmetry. Alternatively


A topology on $R$ defined by the filtration $\{I_n\}$ is Hausdorff if and only if the intersection of all $I_n$ is the zero ideal (0). A topology on $R$ defined by the  filtration $\{I_n\}$ is Hausdorff  if and only if the intersection of all $I_n$ is the zero ideal (0).

I tried but I could not get. Any help will be highly appreciated

The ideals in the filtration form a basis of neighborhoods of zero.

In general topological spaces, we have that $\overline{\{x\}}$ (the closure of the singleton $\{x\}$) is the intersection of the neighborhoods of $x$ or, as well, the intersection of the neighborhoods in a basis.

Since in a Hausdorff space singletons are closed, one of the implications is clear: if $R$ is Hausdorff, then the intersections of the filtration ideals is $\{0\}$.

For the converse implication, you need to know that $\{0\}$ being closed is equivalent to the topological ring $R$ being Hausdorff. This is even easier when the topology is given by a filtration. If $x\ne y$, then $x-y\ne0$, so there exists $n$ such that $x-y\notin I_n$. Then $(x+I_n)\cap(y+I_n)=\emptyset$: indeed, if $z\in(x+I_n)\cap(y+I_n)$ we have
$$
z=x+a=y+b
$$
with $a,b\in I_n$, so $x-y=b-a\in I_n$ contrary to the hypothesis.


Interesting rotating parabola proof So, I am reading linear algebra and cartesian geometry by Juan de Burgos for the first time, and came across this problem that I just can't figure out. Goes like this:
Consider a parabola, whose distance from the focus to the directrix is a fixed p, and that rotates while keeping its focus fixed.
We also consider a certain fixed direction in the parabola's plane, and consider the tangent to the parabola that has that given direction.
Find the equation of the geometric space that all the tangency points describe.
The original question is in Spanish but I have tried to translate it the best I can. If anyone speaks Spanish I can send you the original question.
Edit: Ill keep working on it and try to write my work in here, sorry that I didnt do from the beginning
Ok, so first off the book gives the soultion to a problem in the back, but without any explanation or none of that. it says:
**Origin at the focal point, << x axis>> in the given direction;
directrix $xcos(\theta)+ysin(\theta)=p$.
Parabola $x^2 sin^2(\theta)+y^2 cos^2(\theta)-2xysin(\theta)cos(\theta)+2p(xcos(\theta)+ysin(\theta))=p^2$
The polar of (0,1,0) is $xsen^2(\theta)-ysin(\theta)cos(\theta)+pcos(\theta)=0$.
Eliminating $\theta$ we get $p^2(x^2+y^2)=4y^4$.**
I get the part of creating a new coordinates taking the focal point to be F(0,0). as so:
see image here
The directrix in a standard parabola where the vertix is in the origin has directrix x=p/2, so because we are taking the focal point it would be now x=p. Because of the rotation described for the parabola, i thought of using rotation of coordinate axes, as so:
see image
And thats why the directrix would be  $xcos(\theta)+ysin(\theta)=p$ as described in the solution given by the book.
(Thanks to a commenter here who clarified to me that  $xcos(\theta)+ysin(\theta)=p$ or  $xcos(\theta)+ysin(\theta)=-p$ would make no difference, I was confused on that too).
Now, how in the world did they get the parabola described in the solution and the stuff that comes after?
I have read the chapter like three times but I dont know why i cant understand this one problem

HINT.
Instead of following the path suggested by the book, you could have the parabola fixed and rotate the direction of the tangent: it is not difficult to find a function $r=f(\phi)$, where $r=FP$ is the distance between focus $F$ and tangency point $P$, while $\phi$ is the angle between line $FP$ and the tangent at $P$.
It may also be useful to remember that the tangent line at $P$ is the bisector of $\angle FPH$, where $H$ is the projection of $P$ on the directrix.

I understood the problem in this way:
A parabola  (focus to directrix distance = 2p)  rotates about its focus  at origin. Find equation to the variably rotated parabola and its tangent as function of the rotation angle $\alpha$.
Answer:
I have tried to rotate the entire  parabola/tangent at vertex as one combination.
The given parabola equation:
$$ x^2 \sin ^2 t + y^2 \cos^2 t-2 x y \sin t \cos t+y^2 \cos ^2 t +2 p (x \cos t +y \sin t) = p^2 \tag 1 $$
For $t=0$ it reduces to
$$ 2 p x + y^2 = p^2 \tag 2 $$
Bring in parameter $t$ in a slightly different way
$$ x= \dfrac{p^2-t^2}{2p},   y = t  \tag 3$$
and then we try to  rotate the entire parabola through $\alpha$.
$$ x\to x \cos \alpha- y \sin \alpha , y\to x \sin \alpha +y \cos \alpha \tag 4 $$
Solving for $(x,y)$ from the rotated coordinates ( rather than directly)  with new parametrization $t$ included
$$
x \cos \alpha- y \sin \alpha =t, \; x \sin \alpha +y \cos \alpha = \dfrac{p^2-t^2}{2p} \tag 5 $$
Next put $t=0$ gives starting point $( x=0, y=p/2)$ that is the point of tangency of tangent $x= 2 p$ at parabola vertex is obtained by $\alpha$ rotation
$$ x \sin \alpha+y \cos \alpha = p/2 $$
are lines tangent to circle diameter $p$ not sketched for the  familiar tangent to circle tangent lines not to crowd up too much in the image.
Let the variable $ \cos \alpha = u $  for short.
$$ x u -y \sqrt{1-u^2}=t,\; x \sqrt{1-u^2} + y u =\dfrac{p^2-t^2}{2p} \tag 6$$
Solve for $(x,y)$  as functions of $(u,t)$ that respectively act upon to enable tangent point change as well as rotate the entire parabola.
$$ (x,y)=$$
$$ \left( \dfrac {t u + (p^2 - t^2) \sqrt{1 - u^2}} {2 p}, \dfrac {  u(p^2-t^2)  + 2 p t \sqrt {1 - u^2}}{2 p} \right) \tag 7 $$
The graph is drawn for $p=2$.
A non-vertex parameter $t$ point could be also taken for a parameter wlog.


Combinations - Solving recurrence relation in 2 variables without using generating functions I know that there are similar questions here solving 2 variable recurrence and combinations recurrence equation, but both of them use generating functions to solve this. Is there any other way to solve this problem:

$$\Psi(n,k) = \begin{cases}
                0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{ if } n = 0 \text{ and
                } k > 0\\
                1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{ if } n \geq 0 \text{ and
                } k=0\\
                \Psi(n-1,k) + \Psi(n-1,k-1)\;\;\text{ if } n > 0 \text{ and } k > 0\\
                0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{ otherwise}
                \end{cases}$$

I know the solution is $\binom{n}{k}$, but I would like to know a way to solve this without generating functions.

The most natural way to solve this is the following: I'd start by making a table up to $n=5$ or so and then say "Ah, these  are just the binomial coefficients". Then I'd set up an induction proof that this is indeed the case.

Following the idea of @Christian Blatter, I find $\Psi$ for $k=0\dots 2$:
$$\Psi(n,0) = 1$$
$$\Psi(n,1) = n = n[1]$$
$$\Psi(n,2) = \frac{n(n-1)}{2} = \frac{n-1}{2}[n][1]$$

My guess for $\Psi(n,3)$ will be:

$$\Psi(n,3) = \frac{(n-2)}{3}\left [\frac{n(n-1)}{2} \right]$$

The pattern seems to be:
$$\Psi(n, k) = \frac {n -(k-1)}{k} \Psi(n, k-1)\;\;\;(1)$$

From (1) recursion is easy to get the combinatorial formula:
$$\Psi(n, k) = \binom{n}{k}$$

Then make the proof by induction as @Christian Blatter said.

I wrote more details for this answer here.


Intuition behind using complementary CDF to compute expectation for nonnegative random variables I've read the proof for why $\int_0^\infty P(X >x)dx=E[X]$ for nonnegative random variables (located here) and understand its mechanics, but I'm having trouble understanding the intuition behind this formula or why it should be the case at all. Does anyone have any insight on this? I bet I'm missing something obvious.

For the discrete case, and if $X$ is nonnegative, $E[X] = \sum_{x=0}^\infty x P(X = x)$.  That means we're adding up $P(X = 0)$ zero times, $P(X = 1)$ once, $P(X = 2)$ twice, etc.  This can be represented in array form, where we're adding column-by-column:

$$\begin{matrix} P(X=1) & P(X = 2) & P(X = 3) & P(X = 4) & P(X = 5) & \cdots \\ & P(X = 2) & P(X = 3) & P(X = 4) & P(X = 5) & \cdots \\ & & P(X = 3) & P(X = 4) & P(X = 5) & \cdots \\ & & & P(X = 4) & P(X = 5) & \cdots \\ & &  &  & P(X = 5) & \cdots\end{matrix}.$$

We could also add up these numbers row-by-row, though, and get the same result.  The first row has everything but $P(X = 0)$ and so sums to $P(X > 0)$.  The second row has everything but $P(X =0)$ and $P(X = 1)$ and so sums to $P(X > 1)$.  In general, the sum of row $x+1$ is $P(X > x)$, and so adding the numbers row-by-row gives us $\sum_{x = 0}^{\infty} P(X > x)$, which thus must also be equal to $\sum_{x=0}^\infty x P(X = x) = E[X].$

The continuous case is analogous.

In general, switching the order of summation (as in the proof the OP links to) can always be interpreted as adding row-by-row vs. column-by-column.

Perhaps by considering this question with a concrete physical example,
it will provide some intuition.

Consider a beam of length $L = 10$ (you can pick your favorite units)
attached to a wall. Now, at positions $1, 2, \ldots, 9$
hang weights $w_1,w_2,\ldots,w_9$. For simplicity, let's assume
$\sum_{n=1}^9 w_n = 1$.

Then the center of mass of the beam is $c = \sum_{n=1}^9 n
w_n$. below is an example picture, with the weights in blue (heights
proportional to weight) and the center of mass in red.

In a probabilistic setting, our weights correspond to probabilities
and $c = \mathbb E X$ where $X$ takes on the values $1,2,\ldots,9$
with probabilities $w_1,w_2,\ldots,w_9$, respectively.

Now, to explain how $c = \mathbb E X = \sum_{n=0}^9 \mathbb P(X > n) = \sum_{n = 0}^9
\sum_{k=n+1}^9 w_k$ comes about, expanding out the latter sum we have
$$
c = (w_1 + \cdots + w_9) + (w_2 + \cdots + w_9) + \cdots + (w_9) \>,
$$
so, $w_1$ appears once, $w_2$ appears twice, $w_3$ appears three
times, etc. Hence $c = \sum_{n=1}^9 n w_n$.

In terms of the beam, we can think of the expression $\sum_{n=0}^9 \mathbb P(X
> n)$ in the following way. Standing at zero, look out to the right
and count up all the weights in front of you. Now, move one step to
the right and repeat this process, adding the result to your initial
sum. Continue this process until you get out to position 9, at which
point there are no more weights in front of you.

The resulting sum is the center of mass, or, in probabilistic terms,
the expectation $\mathbb E X$.

Extending this intuition to discrete random variables taking on non-integer values is straightforward. The extension to continuous variables is also not difficult.


Proving for polynomial: $P (1)=0\implies|P'(1)|\leq \frac{\deg(P)}{2}\max_{|z|=1}|P(z)|$. I wonder if it is true that any polynomial $P(z):\mathbb C\to\mathbb C$ with $P (1)=0$ satisfies
$$|P'(1)|\leq \frac{\deg(P)}{2}\max_{|z|=1}|P(z)|, $$ where the maximum is taken over the unit circle.
At least numerically it seems to be the case and the extremizer seems to be $P(z)=z^n-1$.
If $\deg(P)=1$, then $P(z)=z-1$. Hence, $P'(1)=1$ and $\max_{|z|=1}|P(z)|=2$.

The conjectured inequality is true, and a consequence of Lemma 2 in

Abdul Aziz and Q. G. Mohammad, Simple proof of a theorem of Erdős and Lax, Proc. Amer. Math. Soc. 80 (1980), 119-122.

Lemma 2. If $P(z)$ is a polynomial of degree $n$, then
$$|P'(z)| + |nP(z) - zP'(z)| \le n \max_{|z|=1}|P(z)|$$
for $|z| = 1.$

In our case is $P(1) = 0$, so that
$$
2|P'(1)| \le n \max_{|z|=1}|P(z)| \, .
$$

I don't think this can hold in general, take $P(z) = z^2-1$, this satisfies $P(\pm1)=0$, so that $\max_{|z|=1}|P(z)|=\max_{z\in\{-1,1\}}|P(z)|$. However, $P'(1) = 2$ which is a contradiction.


Are the equations $2x - 2y = 11, x = y - 2$ unsolvable? My 9th grade son had this math problem, which seemed unsolvable to me:

$$2x - 2y = 11$$
$$x = y - 2$$

So we can use substitution to come up with:
    $$2(y - 2) - 2y = 11$$

Now distribute:
$$2y - 4 - 2y = 11$$

Reduce
$2y$ and $-2y$ cancel each other out

Answer   $$-4 = 11$$

Did we do something wrong here, or is this math problem unsolvable?

I can't believe that nobody has said that the coordinates that satisfy these equations represent a pair of parallel lines:
$$y=x+2\text{ and }y=x-\frac{11}{2}.$$
Intersections are points that are on both lines and so satisfy both equations at the same time. However as the lines are parallel there is no intersection and therefore no solution of the simultaneous equations.

Equation 1:

2x - 2y = 11
==> 2 (x - y) = 11
==> x - y = 11 / 2
==> x = (11 / 2) + y

Now, submit the value of x in the same Equation 1 gives:

2 (11/2 + y) - 2y = 11 / 2
Gives y = 0

Equation 2:

Put this is y in eq 2

x = y - 2
==> x = 0 - 2
==> x = -2

Solution set:

X , Y = { -2 , 0 }


Show that $X\subset X\cup\left\{ X\right\}$ Show that If $X$ is an any set, then $X\subset X\cup\left\{ X\right\}$

Proof. Let $t\in X$. We must show $t\in X\cup\left\{ X\right\}$, that is we need to show either $t\in X$ or $t\in\left\{ X\right\}$, so we know that $t\in X$, hence we are done.

Can you check my proof?

Apparently, an answer on ground level of set theory is needed here. Since the inclusion $X \subseteq X \cup \{X\}$ is already discussed in full detail, I want to make a remark about the case $X \neq X \cup \{X \}$.

The Axiom of Regularity reads
$$ \forall x \left( x \neq \emptyset \Rightarrow \exists y \in x: y \cap x = \emptyset \right).$$

Let us prove, that $X \notin X$ for all sets $X$.

Soo, let $X$ be any set. Due to the axiom of pairing, $\{X\}$ is a set aswell and clearly not empty (well, $X$ is an element). But in consequence of the axiom of regularity, we must have $X \cap \{X\} = \emptyset$ since $X$ is the only element in $\{X\}$. Hence, $X \notin X$.

Exercise.
Show for all sets A and B, that A $\subseteq$ A $\cup$ B.

With that, your problem is just a special case.


If $f\in L^1(\mathbb R)$ and $f'\in L^1(\mathbb R)$, then $\lim_{x\to \infty }f(x)=0$. Let $f,f'\in L^1(\mathbb R)$. Prove that $$\lim_{x\to \infty }f(x)=0.$$

First of all, is $f'$ defined a.e. ? Because there are no assumption on the fact that $f$ is derivable. So, is $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
for almost every $x$ ?

My attempt for the statement : Suppose WLOG $f'>0$. I'm not sure if it's true, but I would say that : Let $\varepsilon >0$. Since $f'$ is $L^1$, there is a ball $[a,b]$ s.t. $$\int_{]-\infty ,a[\cup]b,+\infty [}f'(x)dx<\varepsilon .$$
In particular, if $x>y>b$, then $$f(x)-f(y)=\int_y^x f'(t)dt<\varepsilon,$$
but I just can conclude that $\lim_{x\to \infty }f(x)-f(y)<\varepsilon .$

I think the question is not properly posed. Normally the statement is interpreted as follows: if $f$ is integrable and differentiable almost everywhere with $f'$ also integrable then $f(x) \to 0$ as $ x \to \infty$. This statement is false. $f(x)=1$ when $x$ is an integer and $0$ otherwise gives a counterexample. Some additional assumptions on $f$ are necessary to prove that $f(x) \to 0$ as $ x \to \infty$.

The idea is to show that $f(x)-f(0)=\int_0^x f'$ hence $f$ has a limit in $\infty$, which must be $0$ because $f\in L^1$.


Determinant of block $n \times n$ matrix Let $$\det (A) = \det \begin{bmatrix} B & 0\\ 0 & I_m\end{bmatrix}$$ where $B$ is a square matrix and where $I_m$ means an $m \times m$ identity matrix.
I keep reading that it is obvious that we can view the RHS as a function of the $n$ rows of $A$ and that this function satisfies all three axioms for a determinant.
However, I fail to see that this is true.

$\det(\begin{bmatrix}B& 0\\ 0 &I_m\end{bmatrix}) = \det(\begin{bmatrix}B& 0\\ 0& I_m\end{bmatrix})$?
$F = tA$, then $\det F = \det(\begin{bmatrix}tB &0\\ 0& tI_m\end{bmatrix}) = t\det(A)$?
determinant of the identity matrix = 1. ?

The question marks mean that these facts are apparently so obvious to everyone else that no one proves them but just states them.
(Big picture. I'm hoping getting clear on this will help me on my bigger question: how to prove $\det(\begin{bmatrix}B& 0\\ C &D\end{bmatrix}) = \det(B)(\det(D))$ for square matrices $B$ and $D$ and a rectangular matrix $C$). I have found this question elsewhere (Determinant of a block lower triangular matrix) but it seems way too handwavy. Other answers seem way too convoluted with too little explanation.)

@larry:

1)
 $  A = \det(\begin{bmatrix}B& 0\\ 0& I_mI\end{bmatrix})$

seems wrong, did you mean

$ A = \det(\begin{bmatrix}B& 0\\ I_m& D\end{bmatrix})$ ?

2) if
$F = tA$ then $det(F) = t^m det(A)$
 this is easy to see if $A = I_m$, using the remark (2a) below

2a) the determinant of a diagonal matrix is just a product of diagonal elements, say
$\det(\begin{bmatrix}2 & 0 & 0 \\ 0 & 3 & 0 \\  0 & 0  & 4\end{bmatrix}) = 2*3*4 = 24$
and yes, $det(I_m) = 1$

3) I don't know a better way to prove
$\det A = \det(\begin{bmatrix}B& 0\\ C &D\end{bmatrix}) = \det(B)(\det(D))$
rather than using the definition of the determinant. This is tedious, and will take a while to set up here.

But you can get the general idea from the following example
$\det A = \det(\begin{bmatrix}b & 0\\ C &D\end{bmatrix}) = b \det(D)$, that is,
matrix B is now 1-by-1.

This should follow if you decompose the determinant using the first row. (Look up "cofactor decomposition", a 3-3 example is here: http://en.wikipedia.org/wiki/Determinant#3.C2.A0.C3.97.C2.A03_matrices).

3b) We can keep nibbling at the general formula by considering
$\det A = \det(\begin{bmatrix}I_m & 0\\ C &D\end{bmatrix}) =  \det(D)$, that is,
matrix B is now $I_m$. Now, when you try to decompose, you should get
 $\det A = 1\cdot 1 \cdot \dots 1 \cdot \det(D)$

4) Come to think of it, we can finish the proof, if you are familiar with Gaussian elimination. I will be a bit sketchy with details, though.

Step 1:
 Do row operations on matrix D to make it low triangular, say $L = TD$. The result of applying the same row operations on the whole matrix is
$\begin{bmatrix}B & 0\\ TC & L\end{bmatrix}$

BTW, row operations (take $Row_i + b*Row_j$, put into $Row_i$) do not change the determinant.

Step 2:
 Divide each of the rows of L (and also TC) by the element on the diagonal of L (this is like doing a second step of Gaussian elimination). Since det(L) is the product of its diagonal elements (you can see this by repeatedly using cofactor expansion), this is equivalent to dividing the whole thing by det(L) = det(D).

So far, we have
  $det(A) = det(D) * det(\begin{bmatrix}B & 0\\ FTC & FL\end{bmatrix})$
where FL has ones on the diagonal.

Step 3:
 Apply more row operations (which are equivalent by multiplying by some other matrix, G) to turn FL into $I_m$. Thus, you get

$det(A) = det(D)*det(\begin{bmatrix}B & 0\\ GFTC & I_m \end{bmatrix})$

Now you can go back to the note 3b from above.

QED!

By the way: $det(\begin{bmatrix}B & C\\ E & D \end{bmatrix})$ generally does not equal $det(B)det(D) - det(C) det(E)$. Go figure!

To understand that
$$\det A = \det(\begin{bmatrix}B& 0\\ C &D\end{bmatrix}) = \det(B)(\det(D))$$
try to use determinant definition
$$\det(A) = \sum_{for\ all\ (i_1,i_2,\dots,i_n)}(-1)^{inversionsCount((i_1,i_2,\dots,i_n))}a_{1i_1}a_{2i_2}\cdots a_{ni_n}$$
where $(i_1,i_2,\dots,i_n)$ transposition of $(1,2,\dots,n)$.


Prove that $\lim_{x \rightarrow \infty} x^ne^{-x}=0$ I'm trying to prove that $$\lim_{x \rightarrow \infty} x^ne^{-x}=0$$

So far I have said:

Let $\epsilon>0.$

Then $$|x^ne^{-x}-0|=|x^ne^{-x}| \leq |x^n...|$$

I'm not sure how to bound $e^{-x}$ though. How can I do this?

Don't try to prove it by definition if you don't have to.

Since $n$ is a fixed number in this limit we can apply L' Hospital rule n times.

$$
\lim_{x \rightarrow +\infty} x^ne^{-x}=\lim_{x \rightarrow +\infty} \frac{x^n}{e^{x}}=\lim_{x\rightarrow +\infty}\frac{n!}{e^x}=0
$$

Even though here it wasn't necessary note that since $x\rightarrow +\infty$ we can observe $x\ge n_0 , n_0\in \mathbb{N}$

Do expansion of e^x. You may know it. Then use either Mathematical Induction or Reasoning to prove that limit equals 0 ... but I prefer to apply L'Hospital's Rule, because that will prevent you from doing Hard Reasoning.


How many different ways can a number N be expressed as a sum of K different positive integers? How many different ways can a number $n \in N$ be expressed as a sum of $k$ different positive numbers?

We  show how  to  establish a  recurrene for  the  number $Q(n,k)$  of
partitions  of  $n$  into  $k$  different parts  where  order  is  not
important. (If  it were  and with  all parts  distinct  we  could just
multiply by  $k!.$) Using the  Polya Enumeration Theorem (PET)  we get
for the answer that it is given by

$$[z^n] Z(P_k)\left(\frac{z}{1-z}\right)$$

where  $Z(P_k)$  is the  cycle  index  of  the unlabled  set  operator
$\mathfrak{P}_{=k}$ with OGF

$$G(w) = \exp\left(a_1 w - a_2 \frac{w^2}{2} + a_3 \frac{w^3}{3} - \right)
\\ = \exp\left(\sum_{l\ge 1} a_l (-1)^{l-1} \frac{w^l}{l}\right).$$

Differentiating we obtain

$$G'(w) = G(w)
\left(\sum_{l\ge 1} a_l (-1)^{l-1} w^{l-1}\right).$$

Extracting coefficients on [w^k] we find

$$(k+1) Z(P_{k+1}) =
\sum_{q=0}^k [w^{k-q}] G(w)
[w^q] \sum_{l\ge 1} a_l (-1)^{l-1} w^{l-1}
\\ = \sum_{q=0}^k Z(P_{k-q}) a_{q+1} (-1)^q
= \sum_{q=1}^{k+1} Z(P_{k+1-q}) a_{q} (-1)^{q-1}.$$

We get the recurrence

$$Z(P_k) = \frac{1}{k} \sum_{q=1}^{k} Z(P_{k-q}) a_{q} (-1)^{q-1}$$

with $Z(P_0) = 1.$ This yields for the present case that the substituted
cycle index

$$L_k(z) = Z(P_k)\left(\frac{z}{1-z}\right)$$

has recurrence

$$L_k(z) = \frac{1}{k}
\sum_{q=1}^{k} L_{k-q}(z) \frac{z^q}{1-z^q} (-1)^{q-1}.$$

Extracting the coefficient on $[z^n]$ we obtain

$$Q_{n,k} = \frac{1}{k}
\sum_{q=1}^{k} [z^n] L_{k-q}(z) \frac{z^q}{1-z^q} (-1)^{q-1}
\\ = \frac{1}{k}
\sum_{q=1}^{k} [z^{n-q}] L_{k-q}(z) \frac{1}{1-z^q} (-1)^{q-1}
\\ = \frac{1}{k}
\sum_{q=1}^{k}  (-1)^{q-1}
\sum_{p=0}^{\lfloor n/q\rfloor - 1}
Q_{n-(p+1)q, k-q}.$$

The boundary conditions here are $Q_{0,0}  = 1$ and $Q_{n,0} = Q_{0,k}
= 0$  and $Q_{n,1} = 1.$  There is a non-essential  boundary condition
which is that $Q_{n,k} = 0$ when $n \lt \frac{1}{2} k(k+1).$

Observe that we  can obtain an alternate form of  the recurrence by
writing

$$\frac{1}{k}
\sum_{q=1}^{k}  (-1)^{q-1}
\sum_{p=1}^{\lfloor n/q\rfloor}
Q_{n-pq, k-q}$$

which is

$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{k}
\sum_{m=1}^n
\sum_{q|m\wedge q\le k} (-1)^{q-1} Q_{n-m, k-q}.}$$

This yields the following Maple routine:

with(numtheory);

Q :=
proc(n, k)
    option remember;

    if n=0 and k=0 then return 1 fi;
    if n=0 or k=0  then return 0 fi;

    if k=1 then return 1 fi;

    if n < 1/2*k*(k+1) then return 0 fi;

    1/k*add((-1)^(q-1)*
            add(Q(n-(p+1)*q, k-q), p=0..floor(n/q)-1),
            q=1..k);
end;

QX :=
proc(n, k)
    option remember;

    if k < 0 then return 0 fi;

    if n=0 and k=0 then return 1 fi;
    if n=0 or k=0  then return 0 fi;

    if k=1 then return 1 fi;

    if n < 1/2*k*(k+1) then return 0 fi;

    1/k*add(add((-1)^(q-1)*Q(n-m, k-q), q in divisors(m)),
            m=1..n);
end;

A := n -> add(Q(n,k), k=1..floor(sqrt(2*n)));
AX := n -> add(QX(n,k), k=1..floor(sqrt(2*n)));

This will produce the following  sequence for the number of partitions
into any number of distinct parts:

$$1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 22, 27, 32, 38, 46,
\\ 54, 64, 76, 89, 104, 122, 142, 165, 192, 222, 256, 296,\ldots $$

which  points  us  to OEIS  A000009  where
these data are confirmed. We get the following sequence for partitions
into seven  distinct parts starting at  $n=-1 + 1/2\times 7\times  8 =
27:$

$$0, 1, 1, 2, 3, 5, 7, 11, 15, 21, 28, 38, 49, 65, 82, 105, 131,
\\ 164, 201, 248, 300, 364, 436, 522, 618, 733, 860, 1009,\ldots$$

which   points  to   OEIS  A008636   which
again confirms the data and lists additional references.

Addendum. Starting from the generating function (first principles)
we  extract  coefficients  for  an  efficient  recurrence  for  strict
partitions into any number of parts:

$$P(x) = \prod_{q\ge 1} (1+x^q)$$

Differentiatinng we get

$$P'(x) = P(x) \sum_{q\ge 1} \frac{qx^{q-1}}{1+x^q}$$

and on extracting coefficients we find

$$(n+1) P_{n+1} = \sum_{m=0}^n P_{n-m} [x^m]
\sum_{q\ge 1} \frac{qx^{q-1}}{1+x^q}
\\ = \sum_{m=0}^n P_{n-m}
\sum_{q\ge 1} q [x^{m+1-q}] \frac{1}{1+x^q}
\\ = \sum_{m=0}^n P_{n-m}
\sum_{q=1}^{m+1} q [x^{m+1-q}] \frac{1}{1+x^q}
\\ = \sum_{m=0}^n P_{n-m}
\sum_{q=0}^{m} (q+1) [x^{m-q}] \frac{1}{1+x^{q+1}}
\\ = \sum_{q=0}^n (q+1)
\sum_{m=q}^{n}  P_{n-m}  [x^{m-q}] \frac{1}{1+x^{q+1}}.$$

This implies $m-q = p(q+1)$ with $p\ge 0$ and where
$m\le n$ or $p(q+1)+q \le n$ so that
$p\le \lfloor (n+1)/(q+1)\rfloor -1$ and we obtain

$$\sum_{q=0}^n (q+1)
\sum_{p=0}^{\lfloor (n+1)/(q+1)\rfloor -1}
P_{n-p(q+1)-q}
[x^{p(q+1)}] \frac{1}{1+x^{q+1}}
\\ = \sum_{q=0}^n (q+1)
\sum_{p=0}^{\lfloor (n+1)/(q+1)\rfloor -1}
P_{n-p(q+1)-q} (-1)^{p}
\\ = \sum_{q=0}^n (q+1)
\sum_{p=1}^{\lfloor (n+1)/(q+1)\rfloor}
P_{n-(p-1)(q+1)-q} (-1)^{p+1}
\\ = \sum_{q=0}^n (q+1)
\sum_{p=1}^{\lfloor (n+1)/(q+1)\rfloor}
P_{n+1-p(q+1)} (-1)^{p+1}
\\ = \sum_{q=1}^{n+1} q
\sum_{p=1}^{\lfloor (n+1)/q\rfloor}
P_{n+1-pq} (-1)^{p+1}.$$

This is

$$\sum_{k=1}^{n+1} P_{n+1-k}
\sum_{p|k} \frac{k}{p} (-1)^{p+1}
= \sum_{k=1}^{n+1} P_{n+1-k}
\sum_{p|k} p (-1)^{k/p+1}.$$

This finally yields a linear recurrence which is

$$\bbox[5px,border:2px solid #00A000]{
P_n = \frac{1}{n}
\sum_{k=1}^{n} P_{n-k}
\sum_{p|k} p (-1)^{k/p+1}}$$

where $P_0 = 1.$

Now for the inner sum its Dirichlet series is evidently
given by

$$-\zeta(s-1) \sum_{n\ge 1} \frac{(-1)^n}{n^s}
= -\zeta(s-1) \left(\frac{2}{2^s}-1\right)\zeta(s).$$

Multiplication by $\zeta(s)$ is summing  over the divisors in terms of
Dirichlet series  so we need to  examine the product of  the first two
terms. We get that for $n$  even the term $1/n^s$ appears twice, first
paired with $-2\times  n/2$ and second with $n$  for a contribution of
zero. For  $n$ odd the term appears  just once paired with  $n$, for a
contribution  of $n.$  Hence  the inner  sum  is the  sum  of the  odd
divisors of $k$ and we get

$$\bbox[5px,border:2px solid #00A000]{
P_n = \frac{1}{n}
\sum_{k=1}^{n} P_{n-k}
\sum_{p|k\wedge p\;\text{odd}} p.}$$

Number of integer partitions of a number n, denoted by p(n), is much harder to find and no easy formula available.

This wiki page gives various algorithms for finding p(n)

Approximate value for p(n) can be found using Hardy-Ramanujan Asymptotic Partition Formula
$p(n) \approx \dfrac{1}{4n\sqrt{3}}e^\left(\pi \sqrt{\dfrac{2n}{3}}\right)$

This tool generates number of integer partitions online and is very useful

This link is also is related to this.


Period of $\sin^2 x +\cos^4 x$? How do I approach this problem? I tried reducing it to $\sin ax + \cos bx$ form but can't. I converted it to 1+ $\sin^2 x \cos^2 x $, but I don't know if I can get the answer from it. Would someone please help me with this? The answer is $\pi$.

Since$$\sin\left(x+\frac\pi2\right)=\cos(x)\text{ and }\cos\left(x+\frac\pi2\right)=-\sin(x),$$it is clear that $\sin^2+\cos^4$ is periodic with period $\frac\pi2$. But is $\frac\pi2$ the smallest period? Yes, because\begin{align}(\sin^2+\cos^4)'(x)&=2\sin(x)\cos(x)-4\cos^3(x)\sin(x)\\&=2\sin(x)\cos(x)\times\bigl(1-2\cos^2(x)\bigr)\\&=2\sin(x)\cos(x)\times\bigl(\sin^2(x)-\cos^2(x)\bigr)\\&=-\sin(2x)\cos(2x)\\&=-\frac{\sin(4x)}2\end{align}and so obviously $\frac\pi2$ is the smallest period of $(\sin^2+\cos^4)'$.

Use Euler's formula :
$$cos^4(x)=\displaystyle\left(\frac{e^{ix}+e^{-ix}}2\right)^4$$
And you can develop using Newton's binomial formula


We start with two identities that we can call Lamé-type identites:
$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(x+y)(x+z)(y+z)(x^2+y^2+z^2+xy+xz+yz)=P$$
$$(x+y+z)^7 - (x^7+y^7+z^7) = 7(x+y)(x+z)(y+z)((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))=Q$$

Furthermore we have :

$$(a+b)(b+c)(c+a)=\frac{(a+b+c)^3-a^3-b^3-c^3}{3}$$

Or

With the identity of Gauss we have :

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

So we deduce this :

$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(\frac{(x+y+z)^3+(x+y+z)(xz+yz+yz-x^2-y^2-z^2)-3}{3})(x^2+y^2+z^2+xy+xz+yz)$$

And

$$(x+y+z)^7 - (x^7+y^7+z^7)=7(\frac{(x+y+z)^3+(x+y+z)(xz+yz+yz-x^2-y^2-z^2)-3}{3})((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))$$

your inequality is equivalent to with ($xyz=1$) :

$$(x^5+y^5+z^5)^2\ge 3(x^7+y^7+z^7)$$

Or :

$$(x^5+y^5+z^5-(x+y+z)^5+(x+y+z)^5)^2\ge 3(x^7+y^7+z^7-(x+y+z)^7+(x+y+z)^7)$$

Or :

$$(-P+(x+y+z)^5)^2\geq 3(-Q+(x+y+z)^7)$$

Or :

$$-P\geq (3(-Q+(x+y+z)^7))^{0.5}-(x+y+z)^5$$

But :

$$-P=-(x+y+z)^5 + (x^5+y^5+z^5) = 5(\frac{-(x+y+z)^3+(x+y+z)(-xz-yz-yz+x^2+y^2+z^2)+3}{3})((x+y+z)^2-(xy+xz+yz))$$

$$-Q=-(x+y+z)^7 + (x^7+y^7+z^7)=7(\frac{(x+y+z)^3+(x+y+z)(xz+yz+yz-x^2-y^2-z^2)-3}{3})(-(-(x+y+z)^2+xy+zx+yz)^2-xyz(x+y+z))$$

Now we remark that if we put $x+y+z=\alpha$$\quad$ so $xy+yz+zx$ is maximal for $x=y=z=\frac{\alpha}{3}$

Furthermore in $-Q$ all the $xy+yz+zx$ are positive and in $-P$ there are negative so we can minimizing and maximazing the LHS and the RHS respectively

So we can reduce the problem to a one variable problem wich is easily solvable if we remark that :

$$xyz=1 \implies x+y+z\geq 3$$


Why does $\lim_{n \to \infty} x^{1+1/(2n-1)}=|x|$? Consider the following limit for $x \in [-1,1]$ and $n \in \mathbb N$:

$$\lim_{n \to \infty} x^{1+\frac{1}{2n-1}}=|x|$$

I can't see where does the absolute value come from. Why isn't the answer simply $x$?

  This limit is available on page 156 of Abbott's Understanding Analysis.

EDIT: Sorry i found a dublicate here.

This can also be done this way :
$$\lim_{n\to\infty}x^{1+ \frac{1}{2n-1}}= \lim_{n\to\infty}x^{\frac{2n-1+1}{2n-1}}= \lim_{n\to\infty}x^{\frac{2n}{2n-1}}$$
$$= \lim\limits_{n\to\infty}\left(x^2\right)^{\frac{n}{2n-1}}= \left(x^2\right)^{ \lim\limits_{n\to\infty} \frac{n}{2n-1}}$$
$$= \left(x^2\right)^{ \lim\limits_{n\to\infty} \frac{1}{2-\frac1n}}= \left(x^2\right)^{\frac{1}{2-0}} = \left(x^2\right)^{\frac{1}{2}}=\sqrt{x^2}$$
Since, $x\in [-1,1]$
$$=|x|$$

Hence proved

|x| implies x < 0 0r x > 0, i.e. x has a possibility of being either. 1/(2n - 1) approaches 0 as n -> infinity and therefore the limit is x^1 since x^0 = 1, and then it follows that if x < 0, this limit is negative and positive is x > 0. All it means is that x can be positive or negative.


Prove this number fact Prove that $x \neq 0,y \neq 0 \Rightarrow xy \neq 0$.

Suppose $xy = 0$. Then $\frac{xy}{xy} = 1$. Can we say that $\frac{xy}{xy} = 0$ and hence $1 = 0$ which is a contradiction? I thought $\frac{0}{0}$ was undefined.

The proof is valid in any field $\rm\: K$ (though it might be circular depending on the context). Namely, $\rm\:0\ne x,y\in K\:$ $\:\Rightarrow\:$ $\rm\: 1/x,1/y\in K\:$ $\:\Rightarrow\:$ $\rm\:(1/x)(1/y) = 1/(xy)\in K\:$ $\:\Rightarrow\:$ $\rm\:xy\ne 0\:.$ The OP's proof is simply this proof recast into a proof by contradiction. To be precise the OP's proof is as follows:

As above, $\rm\: x,y\ne 0\ \Rightarrow\ z := 1/(xy)\in K\:,\:$ i.e. $\rm\:xyz = 1\:.\:$ So $\rm\ xy=0\ \Rightarrow\ 0 = 1\:,\:$ a contradiction.

That's precisely the OP's proof, except I've replaced $\rm\:xy/(xy)\:$  by $\rm\:xyz\:$ to avoid possible confusion.

This is a valid proof. The confusion stems from the fact that it is a proof by contradiction. Such proofs - by their very nature - may encounter all sorts of strange looking mathematical objects, such as the above expression of the form $\rm\: 0/0 = 1\:.\:$ This is just $\rm\:1/1 = 1\:$ in the trivial ring $\:\{0\}\:$ where $\rm\:0 = 1\:.\:$ However, the trivial ring is not a field, since $\rm\:0\ne 1\:$ by the definition of a field (or integral domain). So, as above, $\rm\:0 = 1\:$ is a common target for proofs by contradiction in a field.

Proofs by contradiction often prove immensely confusing to students when first encountered. Learning to wrap one's mind around the bizarre contradictory objects encountered in such proofs is skill that comes with practice. A striking example of such confusion is Euclid's classical proof that there are infinitely many primes. Although Euclid's proof was constructive, it is widely presented as a proof by contradiction (and falsely claimed that this was Euclid's proof). When presented in contradictory form this proof often leads to much confusion. There are hundreds of threads on sci.math permeated by such confusion. One can reach all sorts of contradictions to terminate Euclid's proof, e.g. $\rm\:0 = 1\:$ or $\rm\: 1\:$ is prime, or some integer is both prime and composite, etc. Indeed, one can deduce anything in a contradictory theory such as the integers with finitely many primes. Such contradictions often prove too much to grasp for many beginners. Apparently this is because  we have such strong intuition about integers that one contradiction easily implies many others, and this quickly grows too much to handle intuitively. This does not occur to the same degree when one works with more abstract structures, where real-world intuition has less chance to restrain logical thought processes. Such is the strange nature of proofs by contradiction.

Note $\ $ The OP has revealed the source as Proposition 1.16 in Rudin's Principles of Mathematial Analysis. I've appended it below. It is essentially as I surmised above.

consider that xy=0 from the  basic  axioms we know that  product is zero  when  at least one of the  multiplier is  zero  so is means that  x is zero or y is zero  or both x and y is equal zero  which contradicts  to given condition that neither x or y is zero so it means that their product is also  not equal to zero


Prove that zero multiplied by zero is equal to zero. This is my proof:

So,  $0\cdot0=0$

And we know that

$a-a=0$

By substitution,

We have $(a-a)(a-a)=0$

Then by simplifying, $a^2-a^2+a^2-a^2=0$

and the we have $0-0=0$,

Therefore, $0=0$.

I am not sure about my answer. Will you please show me another way of proving it or some way to improve my answer? Thank you!

$0 \cdot 0 = 0$ because $a \cdot 0 = 0$, which is an axiom of multiplication in Peano arithmetic.

i will try the proof with "Reductio ad absurdum": 0x0<>0

(a) If 0x0<>0 then (a.1) 0x0>0 or (a.2) 0x0<0.

   For example - a.1:
      0x0>0 => multiply by 1 both sides, we have the following equation
                (3) 1*0*0>1*0

    And for the same logic to a.2, we have the following:
                (4) 1*0*0<1*0

      For (3) and (4), we have that it's NOT possible that the same numbers:
      1*0*0  and 1*0 are less and greater simultaneously. Therefore, The only way is that the numbers are equal. For that, We Can conclude that 0*0=0.

 I don't pretty sure for the details of applying "Reductio ad absurdum" but I believe that could be a possible way to prove it.