Is there a direct way to prove $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n=\sum\limits_{n=0}\frac{1}{n!}$? How to prove the equality of the two definitions of $e$ directly?
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\sum_{n=0}\frac{1}{n!}$$
In other words, I want to know a proof that is not "prove $e^x=\sum\limits_{n=0}^\infty\frac{1}{n!}x^n$ and then substitute $x=1$". I guess we do not need $e^x$ to prove this equality.

There are answers in other questions like this:
$$\begin{aligned}
\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n &= \lim_{n\to\infty}\sum_{k=0}^n \frac{1}{k!} \prod_{m=0}^{k-1}\left(1-\frac{m}{n}\right)\\
&=^?\sum_{k=0}^\infty\left[\frac{1}{k!}\lim_{n\to\infty}\prod_{m=0}^{k-1}\left(1-\frac{m}{n}\right)\right]\\
&=\sum_{k=0}^\infty\frac{1}{k!}
\end{aligned}$$
However, I think this operation of limits is wrong. For example, $\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\frac{1}{n}=1$ and $\sum\limits_{k=1}^\infty\lim\limits_{n\to\infty}\frac{1}{n}=0$.

None of the answers to Prove the definitions of e to be equivalent answers this question. There are 3 answers to it:

https://math.stackexchange.com/a/69825/769808
https://math.stackexchange.com/a/69809/769808
https://math.stackexchange.com/a/69957/769808

1 is begging the question. It assumes $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$.
2 is wrong. The operation of limits is incorrect.
3 is not direct proof.

A nice direct proof comes from Rudin:
Let
$$
s_n=\sum_{k=0}^n\frac{1}{k!}\;,\;\;t_n=\left(1+\frac{1}{n}\right)^{n}
$$
First, $s_n$ converges, as it is an increasing sequence that is bounded above by $1 + \sum_{k=0}^{n-1}\frac{1}{2^k} \le 3$.
Let $e:=\lim_{n\rightarrow\infty}s_n$. By the binomial theorem,
$$
\small{t_n = 1 + 1+\frac{1}{2!}\left(1-\frac{1}{n}\right) + \frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) +\; \cdots \;+  \frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{n-1}{n}\right)}
$$
Hence, $t_n\le s_n$, so that $$\limsup_{n\rightarrow\infty}t_n\le e$$
Next, if $n\ge m$,
$$
t_n\ge 1 + 1 + \frac{1}{2!}\left(1-\frac{1}{n}\right)+ \cdots +  \frac{1}{m!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)
$$
Let $n\rightarrow\infty$, keeping $m$ fixed. We get
$$
\liminf_{n\rightarrow\infty}t_n\ge 1+1+\frac{1}{2!} + \cdots + \frac{1}{m!}
$$
so that $s_m\le\liminf_{n\rightarrow\infty} t_n$. Letting $m\rightarrow\infty$, we finally get:
$$
e\le\liminf_{n\rightarrow\infty}t_n
$$

In the original series you can enter with $\lim$ under summation because it is dominated by $\sum_{k=1}^\infty\frac{1}{k!}$ which is convergent. Read more about domination here: https://en.wikipedia.org/wiki/Dominated_convergence_theorem (and the simplification for series: Discrete version of dominated convergence thm ).
But yes, this is not as trivial as other answers present. I'm not aware of an easier solution.


Can someone verify this calculus problem solution for me? From the origin point O, a particle moves with velocity 4m/min, in the direction $30$ degrees north-east. After one minute, another particle B moves from O on the line $x + \sqrt3y = 0$, with velocity 6m/min, in the direction (BOA), which is an acute angle. Evaluate the rate of change of the distance between A and B after 2 mins, from moving the particle B.

I think X= 12 because
X=x○ + dx/dt ×t
= 0+4×(2+1) = 4×3 =12
because it says after moving B and B moves after A with 1 min .
then Y = y○ + dy/dt ×t
= 0+6×2 = 12
So dS/ dt = 5(root)3
am I right ???


Do nested convex bodies have increasing "Volume/Surface Area" ratios? Suppose we have two convex bodies $A$ and $B$, where $A \subseteq B$.  Is it always true that $\frac{\mathrm{Vol}(A)}{\mathrm{SurfaceArea}(A)}\leq \frac{\mathrm{Vol}(B)}{\mathrm{SurfaceArea}(B)}$?

It's true in all the examples I've tried, but I'm not sure how to prove the general case, or whether the general case is even true.

Counterexample:

Let $B$ be a unit square. It has area $1$ and perimeter $4$.

Truncate one corner of $B$ by chopping off an isosceles right triangle of base and height $\delta$; call this new pentagon $A$. It has area $1-\frac12\delta^2$ and perimeter $4 - 2\delta + \sqrt2\delta$.

We have $A\subseteq B$ but $$\frac{\text{Area}(A)}{\text{Perimeter}(A)} = \frac{1-\frac12\delta^2}{4 - 2\delta + \sqrt2\delta} > \frac14 = \frac{\text{Area}(B)}{\text{Perimeter}(B)}$$
for any $\delta<1-\frac1{\sqrt2}\approx0.293$.

I imagine a similar strategy works in higher dimensions as well.

One might note that $V=\int \vec r \cdot \mathrm{d}S$,  for each element of S.  If one figure contains a second figure, then this ratio increases, because in the containing figure $R \ge r$, and so the integral of R is greater than r.

This is a dot product, and it is worth noting that for polyhedra (bounded by flat planes), and cylinders, the dot product is basically based on the tangential insphere.

So if we mean to find a counter-example, we might suppose that the outer figure has number of small inspheres, like a disphenoid tetrahedron, formed, eg by a square, with the diagonals of 1.4 (which would pull it in the fold).

The second figure would fit into this, but not venture out to the edges of the square.  The figure would need to exceed the insphere of the greater one, but there's plenty of room to fill out here.

So it could be possible.


Any examples of unbounded linear operators between $\ell^\infty$ and $\ell^\infty$? $\ell^p$ and $\ell^p$? $\ell^p$ and $\ell^\infty$? Let $X$ and $Y$ be normed spaces, either both real or both complex. Let $f \colon X \to Y$ be a linear operator. Then $f$ is said to be bounded if there exists a real number $r > 0$ such that
$$ \lVert  f(x) \rVert_Y \leq r \lVert x \rVert_X \mbox{ for every } x \in X. $$
If there is no such $r$, then $f$ is said to be unbounded.

Now my question is, can we find an example of an unbounded linear operator (i) $f \colon \ell^\infty \to \ell^\infty$? (ii) $f \colon \ell^p \to \ell^p$, where $p$ is such that $1 \leq p < +\infty$? (iii) $f \colon \ell^\infty \to \ell^p$? (iv) $f \colon \ell^p \to \ell^\infty$? (v) $f \colon \mathrm{C}[a, b] \to \mathrm{C}[a, b]$?

Or, examples of unbounded linear operators between any other pairs of these normed spaces?

By definition, $\ell^\infty$ is the normed space of all the bounded sequences of (real or complex) numbers, with the normed defined by
$$ \left\lVert \left( \xi_n \right)_{n \in \mathbb{N} } \right\rVert \colon= \sup \left\{ \ \left\lvert \xi_n \right\rvert \ \colon \ n \in \mathbb{N} \ \right\}. $$

For any real number $p$ such that $1 \leq p < +\infty$, the normed space $\ell^p$, by definition, is the vector space of all the sequences $\left( \xi_n \right)_{n \in \mathbb{N} }$ of (real or complex) numbers, for which the series $\sum \left\lvert \xi_n \right\rvert^p$ converges, that is,
$$ \sum_{n=1}^\infty \left\lvert \xi_n \right\rvert^p < +\infty,  $$
with the norm given by the formula
$$ \left\lVert \left( \xi_n \right)_{n \in \mathbb{N} } \right\rVert \colon= \sqrt[p]{ \sum_{n=1}^\infty \left\lvert \xi_n \right\rvert^p }. $$

For any real numbers $a$ and $b$ such that $a < b$, the space $\mathrm{C}[a, b]$ is the normed space of all the real or complex-valued functions defined and continuous on the closed interval $[a, b]$ of the real line, with the norm defined by $$ \lVert x \rVert \colon= \max \{ \ \lvert x(t) \rvert \ \colon \ a \leq t \leq b \ \}.  $$

Right now, the only example of an unbounded linear operator that I can recall is that of the differentiation operator of the normed space of all the continuously differentiable functions on a closed interval $[a, b]$ with the maximum norm into this normed space itself.

So any other examples, please?

I would appreciate references to some elementary-level text on analysis where examples of such unbounded linear operators can be found.

We have no examples of unbounded linear operators between these spaces. The Axiom of Choice tells us there are many such operators, but it does not tell us what they are, so we don't have any way to write down such an operator.

Wikipedia goes into this subject too:

  As noted above, the axiom of choice (AC) is used in the general existence theorem of discontinuous linear maps. In fact, there are no constructive examples of discontinuous linear maps with complete domain (for example, Banach spaces).
  [...] it is consistent with set theory without AC that there are no discontinuous linear maps on complete spaces. In particular, no concrete construction such as the derivative can succeed in defining a discontinuous linear map everywhere on a complete space.

With AC, the proof of existence of an unbounded linear map $T:X\to Y$, for any $X, Y$, can proceed like this: let $\{v_\alpha\}$ be a Hamel basis of $X$ and pick any nonzero vector $w\in Y$. Take any unbounded collection of numbers $\{c_\alpha\}$ indexed by the same set as $\{v_\alpha\}$. Define $T(v_\alpha) = c_\alpha \|v_\alpha\| w$ and extend by linearity (which is possible since $\{v_\alpha\}$ is a basis).

The above does not produce an example of an unbounded linear operator on an infinite-dimensional Banach space because there are no examples of a Hamel basis for such a space.

$1.$ Let $T : \ell^{\infty} \rightarrow \ell^{\infty}$ be given by $T((x_1, x_2, , x_3\ldots) = (1 \cdot x_1, 2 \cdot x_2, 3 \cdot x_3, \ldots)$.

Then, for $e_{n} = (0, \ldots,0, 1, 0,\ldots, 0, \ldots)$ $n \in \mathbb{N}$, (where $1$ appears at the $n$th position),  we get $T(e_{n}) = n e_{n}$. Now
 $\|T(e_{n})\|_{\infty}= n \|e_{n}\|_{\infty} = n \cdot 1 = n.$ Now as $n \rightarrow \infty$, the sequence of real numbers $\|T(e_n)\|$ goes to $\infty$. So, $\|T\| = \sup \{ \|T(x)\|_{\infty}: \|x\|_{\infty} =1 \} = \infty.$

$2.$ Let  $T : \ell^{p} \rightarrow \ell^{p}$ be given by $T((x_1, x_2, , x_3\ldots) = (1 \cdot x_1, 2 \cdot x_2, 3 \cdot x_3, \ldots)$.

Then, for $e_{n} = (0, \ldots,0, 1, 0,\ldots, 0, \ldots)$ $n \in \mathbb{N}$, (where $1$ appears at the $n$th position),  we get $T(e_{n}) = n e_{n}$. Now
 $\|T(e_{n})\|_{p}= \| (0, \ldots, 0, n, 0, \ldots)\|_{p} = (n^p)^{1/p}=n$. Now as $n \rightarrow \infty$, the sequence $\|T(e_n)\|_{p}$ goes to $\infty$. So, $\|T\| = \sup \{ \|T(x)\|_{p}: \|x\|_{p} =1 \} = \infty.$

$3.$  Same example works for an unbounded operator between $\ell^p$ to $\ell^{\infty}$ and vice-versa.


Why are countable sets considered small? Why are countable sets considered as small?

It is something that I never really understood. In fact, I would rather say they are big since they contain infinitely many elements. Any explanation would be welcome.

The term "small" is subjective. If you are comparing countably infinite sets to finite sets, then yes, they are big. But if you are comparing countably infinite sets to other infinite sets, then they are small. For example, the set of real numbers is uncountable. This is shown by Cantor's Diagonal Argument.

I've never really heard anyone call a countably infinite set "small".  I've only heard people call uncountable sets "bigger".  Which makes sense in the intuitive sense that now matter how you try to shove an uncountable set into a countable number of slots you'll always have an uncountable number that don't fit so it makes sense to assume that an uncountable set has "more" elements.

Of course all of this language is casual and based an intuition developed in a context that simply no longer applies.  It'd be best to avoid using terms like "bigger" or "more" or "small" or "large" unless you specifically make it clear that a set $A$ being "larger" than $B$ means that no injective function from $B$ to $A$ nor surjective function from $A$ to $B$ can possibly exist.

In that sense calling countable sets "small" does make perfect sense as among all the possible cardinalities of infinity, countable is the smallest.  The only sets that are smaller than a countable set would be finite sets.

Which I suppose sounds a bit like saying the only things slower than the speed of sound are, pffft, subsonic activity, and scoff they barely count.

Everything is relative.

....

There are times that I convince myself that "more" is not the issue but that "thicker" is more appropriate.  If I take a bag of integers and dump them onto a geometric plane or line or number line... whoo-ee... there sure are a lot of them all over the place like a bunch of specks of dust.  But they sure are thin.  I can poke my fingers between them and push them apart. But the points in the plane or line....  They are THICK.  So thick I can stand on a plane and not fall through.

(Ain't language fun?)


Showing $5a + 2b \equiv 0 \pmod 7$ is symmetric, pictorially I have a problem regarding the understanding of modulus.
So someone proved $5a + 2b \equiv 0\pmod 7$ is symmetric using the modulo circle,
and my brain cant comprehend how he meant that. I'm talking about circles like this:

I know there is a other way to prove it with $7 \mid 5a + 2b$

The relation is defined on $\Bbb Z$ like this: $a \sim b$ if and only if $5a + 2b \equiv 0 \pmod 7$. On the assumption this holds, we wish to show $b \sim a$, i.e. $5b+2a \equiv 0 \pmod 7$.
Using the definition of modular congruence, then, $\exists k \in \Bbb Z$ such that
$$\frac{5a+2b-0}{7} = k \iff 5a + 2b = 7k$$
Subtract $7a$ and add $7b$ to both sides. Then you have
$$-2a - 5b = 7(k - a + b)$$
Now multiply both sides by $-1$:
$$2a + 5b = 7(-k+a-b)$$
Thus, there an $\ell \in \Bbb Z$ (namely $\ell = -k+a-b$) such that
$$\frac{5b+2a}{7} = \ell$$
ensuring that $5b + 2a \equiv 0 \pmod 7$, i.e. $b \sim a$. Thus, symmetry.

Now, with this in mind, consider the "circle" for equivalence modulo $7$:

If $5a + 2b \equiv 0 \pmod 7$, then the sum $5a+2b$ is in that topmost notch. Moreover, adding multiples of seven to it will make $5a+2b$ cycle back to that same place. So what we do, then, is add $-7a-7b= 7(-a-b) \in 7 \Bbb Z$ to it. This gives us $-2a-5b \equiv 0 \pmod 7$. However, in this notch of the diagram, if $x$ is in it, so is $-x$. (That is, $x \equiv -x \pmod 7$. I leave it to you to justify why.) Thus, $-(-2a-5b)= 2a +5b \equiv 0 \pmod 7$ as desired.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.

modular arithmetic is basically arithmetic done using linear polynomials:$$y\equiv c\bmod m\iff y=mx+c$$ for some integer values of x,m,c, and y. So it's basically saying, in this case that: $$5a+2b=7x$$ if we allow $a=b=x=n$ where n is integer, then we have a solution.  Edit more to the point $5=7+(-2)$ so if the original has an answer so will $-2a+2b=7x$


Topology of $\mathbb R^n$ question I have the following problem:
$\newcommand{Rn}{\mathbb R^n}$
$\newcommand{U}{\bigcup}$
$\newcommand{a}{{a \in A}}$
$\newcommand{b}{{b \in B}}$
$\newcommand{es}{\emptyset}$
$\newcommand{ss}{\subset}$
$\newcommand{sm}{\setminus}$
$\newcommand{d}{\partial}$
$\newcommand{e}{\varepsilon}$

  Let $X \ss \Rn$, and let $A$ and $B$ be disjoint open subsets of $X$ with the subspace topology. Show that there exist disjoint open subsets $U$ and $V$ of $\Rn$ such that $U \cap X = A$ and $V \cap X = B$.

My partial solution is as follows:

Since $A$ is open in $X$, for every $\a$ there exists $\e_a > 0$ such that $A = \U_\a \beta_X(a, \e_a)$, where $\beta_X(x, \e)$ is the open ball relative to $X$, centered at $x$, of radius $\e$. Now let $P = \U_\a \beta(a, \e_a)$. Clearly, $P$ is open in $\Rn$ and $P \cap X = A$. Construct $Q$ from $B$ and $X$ in a similar way.

Clearly, $P \cap Q \cap X = A \cap Q = B \cap P = A \cap B = \es$. However, $P \cap Q$ isn't necessarily empty. It is merely guaranteed that their intersection lies outside of $X$, and hence outside of $A$ and $B$. That's why I haven't called them $U$ and $V$ in the first place.

Now, I want to show that $U = P \sm \bar Q$ and $V = Q \sm \bar P$ satisfy the given requirement. Clearly, $U$ and $V$ are both open in $\Rn$, and $U \cap X = (P \sm \bar Q) \cap X = (P \cap X) \sm (Q \cup \d Q) = A \sm \d Q$ (and similarly for $B$ and $V$). I just need to show that $A \cap \d Q = \es$. How do I do this? Or, is my idea completely wrong?

The proposition is false if $\mathbb{R}^{n}$ is replaced by a general
topological space (or the metric topology is replaced by an arbitrary
topology).

For example. Let $\mathbb{R}$ be equipped with the topology $\{\emptyset,\mathbb{R},\{1\}^{c},\{2\}^{c},\{1,2\}^{c}\}$
and let $X=\{1,2\}\subseteq\mathbb{R}$. Let $A=\{1\}$ and $B=\{2\}$,
Note that $A$ and $B$ are open subsets of $X$ because $A=\{2\}^{c}\cap X$
and $B=\{1\}^{c}\cap X$. Clearly $A\cap B=\emptyset$. However, we
cannot find disjoint open sets $U,V$ of $\mathbb{R}$ such that $A=U\cap X$
and $B=V\cap X$.

Therefore, to prove the proposition (if it is true), one must use
properties of metric topology on $\mathbb{R}^{n}$. Any proof that just uses set-theoretic construction on general setting (without anything specific to the metric topology) must be invalid.

Your quoted statement seems to be true by definition? If $A \in X$ is open, then there is an open $U \in \mathbb{R}$ so that $A =U\cap X$. See subspace topology. Or otherwise, you shall define openness in subspace another way.

It remains to show $U \cap V =\varnothing$. If not, then remove the common part of $U$ and $V$. Indeed, if we define $U' =U -\overline{U \cap V}$, then $U'$ is open. Similarly $V' =V -\overline{U \cap V}$, and $V'$ is open. Since $U \cap V \cap X =\varnothing$, we see still $A =U' \cap X$ and $B =V' \cap X$.


Increasing Convex Function Increasing Second Derivative? Let $f(x)$ be increasing and convex on a certain interval. Then, $f’’(x)\geq 0$ on this interval. I was wondering if it could be further said that $f’’(x)$ is increasing (or at least non-decreasing) on this interval.

No it can not be said. The second derivative gives you strength of the curvature but since a convex function can have decrasing curvature too the answer is no.
Imagine x^2 beeing cutoff at 1 and then continuoued with constant increase. Then f''(x) = 0 for x >1


Calculating the volume of a restaurant take-away box that is circular on the bottom and square on the top Having a bit of a problem calculating the volume of a take-away box:

I originally wanted to use integration to measure it by rotating around the x-axiz, but realised that when folded the top becomes a square, and the whole thing becomes rather irregular. Since it differs in circumference I won't be able to measure it like I planned.

Is there any method or formula that can be used to measure a shape like this, or do I just have to approximate a cylinder and approximate a box and add those two together?

A different approximation could be to take the cross-section at each height not as a linear interpolation between top and bottom surface, but as squares with rounded corners. This fits the photograph more closely, as the linear interpolation approach has a discontinuity in the radius of curvature at the bottom: The point that at the top is an edge of the square is modelled a crease along the entire side, while in the photograph, no such crease can be seen.

If we take the cross-section to have the above shape of a rounded square, we can calculate its area by using the formulas for circles and rectangles:

$$A = \pi \cdot r^2 + 2 \cdot a \cdot b - b^2$$

Using $b=a-2\cdot r$, we can eliminate that variable:

$$A = a^2 + (\pi -4) \cdot r^2$$

Now we assume that those variables vary linearly between top and bottom: $r$ goes from $r_0$ at the bottom to $0$ at the top, and $a$ goes from $2r_0$ at the bottom to $a_0$ at the top, giving us:

$$r(h) = \left(1-\frac{h}{h_0}\right)\cdot r_0$$

$$a(h) = \left(1-\frac{h}{h_0}\right)\cdot 2r_0 + \frac{h}{h_0} \cdot a_0$$

and thus:

$$A(h) = \left(1-\frac{h}{h_0}\right)^2 \cdot \pi \cdot r_0^2 + \left(1-\frac{h}{h_0}\right) \cdot \frac{h}{h_0} \cdot 4 \cdot r_0 \cdot a_0 + \left(\frac{h}{h_0}\right)^2 \cdot a_0^2$$

or, better ordered for integration:

$$A(h) = \pi \cdot r_0^2 + \frac{h}{h_0} \cdot \left( 4\cdot r_o \cdot a_0 - 2 \cdot \pi \cdot r_0^2 \right) + \left(\frac{h}{h_0}\right)^2 \cdot \left( a_0^2 - 4\cdot r_0 \cdot a_0 + \pi \cdot r_0^2\right) $$

Which gives us then:

$$ V = \int_0^{h_0} A(h) \mathrm{d}h= h_0 \cdot \pi \cdot r_0^2 + \frac{1}{2} \cdot \frac{h_0^2}{h_0} \cdot \left( 4\cdot r_o \cdot a_0 - 2 \cdot \pi \cdot r_0^2 \right) + \frac{1}{3}\cdot\frac{h_0^3}{h_0^2} \cdot \left( a_0^2 - 4\cdot r_0 \cdot a_0 + \pi \cdot r_0^2\right)$$

$$=h_0\cdot\left(\frac{1}{3}\pi r_0^2 + \frac{2}{3} r_0 a_0 + \frac{1}{3} a_0^2\right)$$

And here is a quick render of what this looks like in 3d:

This result can be obtained classically without any integrals as well, using only the formulas for the volume of conic solid and for cuboids. To do this, we split the solid into nine parts:

The central, yellow part is simply a square pyramid, and has a volume of $\frac{1}{3}\cdot h_0\cdot a_0^2$. The four magenta pieces are oblique cones with a quarter-circle as base. Again, using the formula for conic solids, their volume is each $\frac{1}{12}\cdot h_0\cdot\pi\cdot r_0^2$

And the four cyan pieces are irregularly shaped tetrahedrons, but we can determine their volume by adding a few pieces to see how they fit into a cuboid of the measurements $a_0\times r_0 \times h_0$:

In the more general case of $\frac{a_0}{2} \neq r_0$, there will be a fourth green piece needed (which would go in front and block our view of everything). However, the green pieces together form a prisma of height $a_0$ and a top surface area of $\frac{1}{2} \cdot r_0 \cdot h_0$, and the two blue-grey pieces are two pyramids with height $h_0$ and rectangular bases of size $\frac{1}{2} a_0 \times r_0$. Therefore, the volume of the tetrahedron is:

$$a_0 \cdot h_0 \cdot r_0 - a_0 \cdot \frac{1}{2} \cdot r_0 \cdot h_0 - 2 \cdot \frac{1}{3} \cdot h_0 \cdot \frac{1}{2} a_0 \cdot r_0 = \frac{1}{6} h_0 \cdot r_0 \cdot a_0$$

Putting it all together, we get the volume as:

$$V=\frac{1}{3}\cdot h_0\cdot a_0^2 + 4\cdot \frac{1}{12}\cdot h_0\cdot\pi\cdot r_0^2  + 4 \cdot \frac{1}{6} h_0 \cdot r_0 \cdot a_0$$
$$=h_0\cdot\left(\frac{1}{3}\pi r_0^2 + \frac{2}{3} r_0 a_0 + \frac{1}{3} a_0^2\right)$$

A more smooth model along the same lines would be to interpolate superellipses between the bottom and top, which similarly would give a creaseless change in the radius of curvature beneath the corners. However, superellipse areas have a formula involving the gamma function, and are therefore not easy to integrate again to get a volume.

A more general approach
Discussions in the comments have identified two conditions which should be fulfilled for a valid model of the box given that it is constructed from cardboard:

The lateral surface has a Gaussian curvature of 0, or more informally, can be unrolled into a flat surface without additional cuts
If the radius of the bottom circle or the side length of the top square are set to 0, the formula should turn into the known formulas for the volume of a pyramid or cone, respectively. (This actually follows from the first condition, because those shapes are the only ones fulfilling it in this degenerate case)

A class of objects which automatically fulfil these conditions are prismatoids, which are polyhedral solids whose vertices lie on two parallel planes:

Their lateral faces consist of triangles or trapezoids, which are automatically flat, and since there are no vertices in between the top and bottom surface, and therefore any lateral face can only connect to two neighbouring lateral faces, we can also unfold its lateral surface into a flat shape by "hinging" along the edges.
The prismatoid model can also be used as approximation for not polygonal top and bottom surfaces, by approximating them by increasingly fine polygons. For example, the model used by Aretino and J. M. can be approximated by a prismatoid by creating vertices at top and bottom positions corresponding to evenly spaced values of $\theta$. The model examined in my other answer can be approximated by taking evenly spaces vertices along the circle and connecting each to the closest corner of the square.
The good thing about prismatoids is that they have a very simple formula to calculate their volume, which requires one to know only the values of the area of the cross-sections at the top ($A_1$), bottom ($A_3$) and at half height ($A_2$). Then, the total volume of the solid is:
$$V= \frac{h}{6}(A_1 + 4 A_2 + A_3)$$
Since $A_1$ and $A_3$ are fixed by the nature of the problem as $\pi\cdot r^2$ and $a^2$, respectively, the choice of model for the shape of the solid is merely a choice of what value to assume for $A_2$.
In fact, by following the second condition to its conclusion, it can be shown that any model using the prismatoid approach must have in the end a volume formula of the shape:
$$V = h\left(\frac{1}{3}\pi r^2 + \gamma a r + \frac{1}{3}a^2\right)$$
In which there is only the choice of $\gamma$ remaining as free variable. While any particular model can then make additional claims about why it, and therefore its choice of $\gamma$ is the most valid one, as a more general overview the question can be posed: What is the possible valid range of $\gamma$? Or equivalently, what is the possible range for $A_2$?
As it turns out, the maximum $A_2$ is actually the one given by my other answer, as the described shape is actually the convex hull of the points on the circle and the square (something I'll state without proof because it should be easy enough for yourself to convince yourself of that fact), which has a $\gamma$ of $\frac{2}{3}$.
And for the prismatoid with the minimal $A_2$ (excluding self-intersecting or twisted models, because they don't agree with the problem as posed), I nominate this one:

And its cross-section looks like this:

I can't prove that connecting each circle point to the closest midpoint of the square gives the least area, but it seems likely, as moving any one connecting appears to increase the cross-section.
The area, as can be shown by simple calculation of the square and circle segment areas, is:
$$A_2 = \frac{1}{4}a^2 + \frac{1}{\sqrt{2}} a r + \frac{\pi}{4} r^2$$
Plugging this into the volume formula gives:
$$V_{min} = h\left(\frac{1}{3}\pi r^2 + \frac{\sqrt{2}}{3} a r + \frac{1}{3}a^2\right)$$
With therefore a minimal $\gamma$ of $\frac{\sqrt{2}}{3}$ and a maximal $\gamma$ of $\frac{2}{3}$, we can ask the question:
How much does choosing the model then actually matter?
By taking a "cube" box as example with $h = 1$, $a=1$ and $r=\frac{1}{2}$, we can calculate the volumes and their differences:
$$\Delta V = V_{max} - V_{min} = \left(\frac{1}{12}\pi + \frac{2}{3}\right) - \left(\frac{1}{12}\pi + \frac{1}{3\sqrt{2}} + \frac{1}{3}\right) = \frac{2-\sqrt{2}}{6}$$
Or, in floating point units:
$$V_{max} \approx 0.928466$$
$$V_{min} \approx 0.830835$$
$$\Delta V \approx 0.097631$$
Or in other words, the choice of exact model changes the result in volume by at most 10%.
There is technically a "best" model among all possible choices of prismatoids, which is the one whose lateral surface is unrolled into the closest approximation of a circle ring sector, which is the lateral net of a truncated cone, the shape of the takeaway box before the top is folded, and a deformation of the paper which does not tear it has to preserve the net. However, I will not attempt to find which $\gamma$ this then results in, and only, based on the shapes of the models examined already, assume it is rather close to $\frac{2}{3}$.


Prove that $A$ and $B$ are invertible and $B=A^{-1}$ Suppose $A$ and $B$ are $n \times n$ matrices.  Assume $AB=I$. Prove that $A$ and $B$ are invertible and that $B=A^{-1}$.

Please let me know whether my proof is correct and if there are any improvements to be made.

Assume $AB=I$.  Then $(AB)A=IA=A$. So, $A(BA)=AI=A$.  Then $BA=I$.  Therefore $AB=BA=I$.  Thus  $A$ and $B$ are invertible.  And by definition $B=A^{-1}$, so $AB=AA^{-1}=I$.

It doesn't seem quite correct.  Thanks in advance for any advice.

Proof #1: (along the lines mentioned in the comments)

As $AB=I$, you know that $A$ is onto as a linear transformation, because $x=Ix=ABx=A(Bx)$ for any $x\in\mathbb{R}^n$. This implies that $A$ is bijective, being a surjective linear transformation in a finite-dimensional space. So there exists $A^{-1}$. Now
$$
A^{-1}=A^{-1}I=A^{-1}AB=B.
$$

Proof #2 (using determinants)

Since $AB=I$, we have
$$
1=\det I=\det AB=\det A\,\det B.
$$
So $\det A\ne0$ and $A$ is invertible, and again we can do
$$
B=IB=A^{-1}AB=A^{-1}I=A^{-1}.
$$

Suppose $AB = I$.  ($A$ and $B$ are $n \times n$ matrices.)

First note that $R(A) = \mathbb{R}^n$.  (If $y \in \mathbb{R}^n$, then
$y = Ax$, where $x = By$.)

It follows that $N(A) = \{0\}$.

We wish to show that $BAx = x$ for all $x \in \mathbb{R}^n$.
So let $x \in \mathbb{R}^n$, and let $z = BAx$.
Then $Az = A(BAx) = (AB)Ax = Ax$, which implies that
$z = x$ because $N(A) = \{ 0 \}$.  So $BAx = x$.


Probability that the second and third card extracted are both aces? I would like to know if the following reasoning is correct: the situation is the extraction without replacement of 3 cards out of 40, where there are 4 aces. I would say the cardinality of the sample space is $\frac {40!}{(40-3)!}$ because order matters here, and P(C2 and C3 are aces) = P(C2 is an ace) × P(C3 is an ace | C2 is an ace) = $\frac{4\cdot \binom {39}{1}}{40\cdot39\cdot38} \frac{3\cdot \binom {38}{1}}{40\cdot39\cdot38} $

If you need exactly two aces (second and third card only) then first pick the non-ace ($36$) then pick the two aces in order $(4\cdot 3)$.

If you need at least two aces (second and third cards definitely, but the first can be an ace also) then choose the second and third cards as aces $(4 \cdot 3)$ then choose the first card as anything else $(38)$.

You have got is just slightly wrong.  The probability that the second card is an ace will be $\frac3{40}$ because it is $\frac3{39}$ times the probability that the first card was not an ace, plus $\frac2{39}$ times the probability that the first card was an ace:
$$
\frac3{39}\frac{37}{40}+\frac2{39}\frac3{40} = \frac3{39}\frac{37}{40}+\frac2{39}\frac3{40} = \frac3{39}\frac{37}{40}+\frac3{39}\frac2{40} =
\frac3{39}\left(\frac{37}{40}+\frac2{40} \right)=\frac3{40}
$$
Then the probability that the third card is also an ace will be $\frac2{39}$ to the answer probability is $\frac1{260}$.

If you want to do it by counting cases, the number of success cases is
$3$ (which A is in slot 2) times $2$ (which other A is in slot 3) times $388$ (which card is in slot $1$).
$$
\frac{3\cdot 2\cdot 38}{40\cdot 39 \cdot 38} = \frac1{260}$$


Is there a reliable reference that shows that Bayes theorem holds for measures, densities, masses, or combinations of these? I am looking for a book or a reliable reference that I can site in a research paper or thesis that shows that Bayes' theorem/rule holds for both probability measures (i.e. events), densities, mass function, and combinations of densities and mass functions (depending on whether the involved random variables are continuous or discrete)? Something like these notes (even more detailed or formal, i.e. it's ok that the book is based on measure theory, provided that it shows these things), but that is a reliable book or research paper that I can cite in a research paper.
Please, do not tell me to derive these things myself, because my knowledge of measure theory is not very good (and I don't have time now to study it), or that I can cite the notes, because I am looking for a reliable book or research paper. Most books I have looked at only present Bayes theorem by considering measures and events.

As announced in the comment, I'd cite Theorem 1.31 in this book. It says:

Suppose that $X$ has a parametric
family $\mathcal{P}_0$ of distributions with parameter space $\Omega$. Suppose that $P_\theta\ll \nu$
for all $\theta\in\Omega$, and let $f_{X|\Theta}(x|\theta)$ be the conditional density (with respect to
v) of X given $\Theta = \theta$. Let $\mu_\Theta$ be the prior distribution of $\Theta$. Let $\mu_{\Theta|X}(\cdot|x)$ denote the conditional distribution of $\Theta$ given $X = x$. Then $\mu_{\Theta|X}\ll\mu_\Theta$,
a.s. with respect to the marginal of $X$, and the Radon-Nikodym derivative is
$$
\frac{d\mu_{\Theta|X}}{d\mu_\Theta}(\theta|x)=\frac{f_{X|\Theta}(x|\theta)}{\int_\Omega f_{X|\Theta}(t|\theta)\,d\mu_\Theta(t)}
$$
for those $x$ such that the denominator is neither $0$ nor infinite. The prior
predictive probability of the set of $x$ values such that the denominator is
$0$ or infinite is $0$, hence the posterior can be defined arbitrarily for such $x$ values.

In your cases of interest $\nu$ is the Lebesgue measure.

Bayes' Rule is of course just an application of the definition of conditional probability/ density.  But if you'd like something to refer to, it is stated for general distributions (i.e., encompassing both "densities" and "mass functions") in the introduction (see p. 7 of 2nd edition) of Bayesian Data Analysis by Gelman, Carlin, Stern, and Rubin.


If $\lim_{n \rightarrow \infty} (a_{n+1}-\frac{a_n}{2})=0$ then show $a_n$ converges to $0$. I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:

Attempt 1:

Write out the terms of $b_n$:

$$b_1=a_{2}-\frac{a_{1}}{2}$$
$$b_2=a_{3}-\frac{a_{2}}{2}$$
$$b_3=a_{4}-\frac{a_{3}}{2}$$
$$\cdots$$
$$b_n=a_{n+1}-\frac{a_{n}}{2}$$

Adding up the terms you get:

$$\sum_{i = 1}^n b_i=a_{n+1}+\frac{a_n}{2}+\frac{a_{n-1}}{2}+\cdots+\frac{a_2}{2}-\frac{a_1}{2}.$$

But a dead end here.

Attempt 2:

For $ε=\dfrac{1}{2}$, $\exists K$ such that $\forall n>K$, $$\left|a_{n+1}-\frac{a_n}{2}\right|<\frac{1}{2}.$$

Now I attempt to prove $\{a_n\}$ is Cauchy and hence converges.

For $m>n>K$,
\begin{align*}
|a_m-a_n|&=\left|a_m-\frac{a_{m-1}}{2}+\frac{a_{m-1}}{2}-\frac{a_{m-2}}{2^2}+\cdots -+\frac{a_{n+1}}{2^{m-n-1}}-a_n\right|\\
&\leq \left|a_m-\frac{a_{m-1}}{2}\right|+\frac{1}{2}\left|a_{m-1}-\frac{a_{m-2}}{2}\right|+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&\leq \frac{1}{2}+\frac{1}{2} × \frac{1}{2}+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&<1+\left|\frac{a_n}{2^{m-n}}-a_n\right|,
\end{align*}
and a dead end.

Let $b_n = a_{n+1} - \frac{1}{2}a_n$. Then

$$ a_n = b_{n-1} + \frac{b_{n-2}}{2} + \cdots + \frac{b_1}{2^{n-2}} + \frac{a_1}{2^{n-1}}. $$

Since $(b_n)$ converges, there exists $M$ such that $|a_1| \leq M$ and $|b_n| \leq M$ for all $n$. Thus for any fixed $m$ and for any $n > m$, we have

$$ |a_n| \leq \Bigg| \underbrace{b_{n-1} + \cdots + \frac{b_{n-m}}{2^{m-1}}}_{\text{(1)}} \Bigg| + \underbrace{\frac{|b_{n-m-1}|}{2^m} + \cdots \frac{|a_1|}{2^{n-1}}}_{(2)}.$$

Note here that

$\text{(1)}$ consists of fixed number of terms, each tending to zero as $n\to\infty$.
$\text{(2)}$ is uniformly bounded by $\frac{M}{2^m} + \frac{M}{2^{m+1}} + \cdots = \frac{M}{2^{m-1}}$.

So, taking limsup as $n\to\infty$ yields

$$ \limsup_{n\to\infty} |a_n| \leq \frac{M}{2^{m-1}}. $$

Since the LHS is a fixed number and $m$ is arbitrary, letting $m\to\infty$ proves the claim.

Hint :
If $L_t n > \infty$ $\frac{A_{n+1}}{A_{n}}$ is less than $1$. Then the sequence $A_n$ must converge.


Proving a limit exists using Squeeze Theorem Let $g(x) = \ln(1+x)$ for $x>-1$.
Use the inequality $\lim_{x \to 0}{x-\frac{x^2}{2}} \leq \lim_{x \to 0}{\ln(1+x)} \leq \lim_{x \to 0}x$ to prove that $g$ has a limit at $x = 0$.

For the left side of the inequality:
Let $\epsilon>0$. Take $\delta = \min(\epsilon, \sqrt{2\epsilon})$.
Then, we have that $|x| < \delta$. We note that, if $x$ is in the interval $(-1, 1)$, $|x-\frac{x^2}{2}| \leq |x| < \delta = \epsilon$. Now if $x$ is in the complement of that interval, we note that $|x-\frac{x^2}{2}| \leq |\frac{x^2}{2}|.$ Moreover, $|x|<\sqrt{2\epsilon} \implies |\frac{x^2}{2}| < \epsilon $. Thus, this function has a limit at x = 0

For the right side of the inequality: Let $\epsilon > 0$. Take $\delta = \epsilon$. We have that $|x| < \delta = \epsilon$.  Thus the limit of this function exists at $x = 0$.
By the squeeze thereom, $g$ has a limit at $x = 0$

Is my proof correct?

There is an easier way to do this using the Squeeze Theorem.

It seems you are trying to prove that $\lim_{x \to 0}  \ln(1+x) = 0$ via the definition of a limit. But using the inequality that was given $$\lim_{x \to 0} x- \frac{x^2}{2} \leq \lim_{x \to 0} \ln(1+x) \leq \lim_{x \to 0} x$$

and the knowledge that $x- \frac{x^2}{2}$ and $x$ are continuous functions (since they are polynomial functions as Gabrielek points out above) we know that $\lim_{x \to 0} x- \frac{x^2}{2} = 0 - \frac{0^2}{2} = 0$ and $\lim_{x \to 0} x = 0$. Can you now see how to use the Squeeze Theorem and the above inequality to conclude that $\lim_{x \to 0}  \ln(1+x) = 0$?

We assume as verified the continuity of $x-\frac{x^2}{2}$, $\ x$, $\ \text{log}(1+x)$ in a neighborhood of $0$.

In particular $x-\frac{x^2}{2}$, $x$ are both polynomials and their limits as $x$ approaches $0$ is of course $0$.

To conclude using the squeeze theorem we have to demonstrate that in a neighborhood of $0$ hold the following inequalities: $$ x-\frac{x^2}{2} \le \text{log}(1+x) \le x$$

Because $x \rightarrow 0$ , we can expand $\text{log}(1+x)$ in series:
$$ \text{log}(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} = x -\frac{x^2}{2} + \frac{x^3}{3} + o(x^3)$$

And this concludes the left inequality. The right one follows as the other.


All right angles are equal to each other Why is it that

  All right angles are equal to each other

-a postulate in Euclid's Elements (Wikipedia). Shouldn't it be a congruence rather than an equivalence? Isn't this just a special case of the definition of congruent angles?

After looking at an article recently I found the answer. Throughout The Elements, Euclid only uses the measures of multiples of right angles. Although he says angles can be bigger or smaller he does not actually measure them.

Source: http://blogs.scientificamerican.com/roots-of-unity/whate28099s-the-deal-with-euclide28099s-fourth-postulate/

They are equal angles because they all are 90 degrees. If the right triangle was included in a triangle, then the triangles with right angles would be similar.


Proving $A$={$(x,y,z)\in \mathbb{R}^3 \mid |x^3+y+z|>3$} is open Having little trouble showing this, even though intuitively it is pretty clear. Definition of open set:

  $U$ is open if every point in $U$ has a neighborhood contained in $U$.

We have done few of these proofs, but they were bounded and this is clearly unbounded set, so I have no idea how to choose distance or anything.

Let's prove the function is continuous at each point in $A$.

Let $f(x,y,z) = x^{3} + y + z$.  Let $(x_0,y_0,z_0) \in \Bbb R^{3}$.  We want to show $f$ is continuous at $(x_0,y_0,z_0)$.

Okay, so for each $\epsilon > 0$, we must find $\delta > 0$ such that $|| (x,y,z) - (x_0,y_0,z_0)|| \leq \delta$ implies $|f(x,y,z) - f(x_0,y_0,z_0)| < \epsilon$.

Note that the norm $||\cdot||$ we are using on the domain here can be any norm on $\Bbb R^{3}$ since for any fixed natural number $n$, all norms on $\Bbb R^{n}$ are equivalent.  The norm on the range $| \cdot |$ is just the normal Euclidean norm on $\Bbb R$ (basically, it's absolute value).

Okay, so let's choose the norm $||(a,b,c) - (d,e,f)|| = |d - a| + |e- b| + |f - c|$, which is the sum of the (positive) differences in components.

So, let $\epsilon > 0$.  Does there exist a $\delta > 0$ such that $|x-x_0| + |y-y_0| + |z-z_0| < \delta$ implies $|x^{3} + y + z - (x_0^{3} + y_0 + z_0)| < \epsilon$?

Well, we know by the triangle inequality that $|x^{3} + y + z - (x_0^{3} + y_0 + z_0)| \leq |x^{3} - x_0^{3}| + |y - y_0| + |z - z_0|$.

Also, $h(x) = x^{3}$ is continuous.

So, let $y$ satisfy $|y - y_0| < \epsilon/3$ and $z$ satisfy $|z - z_0| < \epsilon/3$. We know by the continuity of $x^{3}$ that there is some $\delta_0>0$ such that $|x - x_0| < \delta$ implies $|x^3 - x_0^3| < \epsilon/3$.

So, finally, after all that work, let's choose $\delta = \min\{\delta_0, \epsilon/3 \}$.

Then $||(x,y,z) - (x_0, y_0, z_0)|| \leq \delta$ implies $|f(x,y,z) - f(x_0,y_0,z_0)| < \epsilon$.

So, $f(x,y,z)$ is continuous on $\Bbb R^{3}$.  Also, $r(x,y,z) := |f(x,y,z)|$ is continuous since the absolute value function is continuous, and so $r$ is the composition of continuous functions. Now that we know $r$ is continuous, we can rewrite the set $A$ in the question as $r^{-1}(3,\infty)$, i.e., the preimage of $(3,\infty)$.  Since $(3,\infty)$ is open and $r$ is continuous, this means $A$ is open.

That the set is unbounded doesn't really matter.  Where things are interesting is what is going on around the inner boundary.

For any $(x,y,z) \in A, |x^3 + y+ z|>3$

let $\delta = \min (\frac 18 ||x^3 + y+ z|-3|, 1)$

Let $U$ be the open ball around $(x,y,z)$ of radius $\delta$

Every element of $U$ is in $A$


modulo of series summation I have trouble with computing modulo.
First, I have a summation of series like this:
$$1+3^2+3^4+\cdots+3^n$$
And this is the formula which can be used to compute the series:
$$S=\frac{1-3^{n+2}}{1-3^2}=\frac{3^{n+2}-1}{8}$$
Then I want to compute $S \mod 1000000007$. But if $n$ is a large number, it's really hard for me to compute it. The only thing I know is $3^{n+2}-1$ divisible by 8 (n is an even number). Could anyone give me some good hint to solve this problem?

Update
My intention is computing $$M=\frac{3^{n+2}-1}{8} \mod 1000000007$$
For example: If $n=4000$ I must split $3^{4000+2}$ into $3^{40}3^{40}...3^{40}3^2$ and compute modulo for each part to improve speed like this:
$$(3^{40}\mod 1000000007)(3^{40}\mod 1000000007)...(3^{40}\mod 1000000007)3^2$$
But how can I compute M with the above idea.

Update more
It seems related to inverse modulo. I think the problem was solved like this
$$I=\frac{(1000000007+1)}{8}=125000001$$
$$\frac{3^{n+2}-1}{8} \mod 1000000007=(3^{n+2}I-I)\mod 1000000007$$
Is it right?

The modulo value of a geometric sum

1 + a + a^2 + ... +a^n mod m

can be calculated based on the identity (if the highest exponent is odd)

1 + a + a^2 + ... +a^(2x+1) = (1 + a)(1 + a^2 + (a^2)^2 + ... + (a^2)^x)

A complete algorithm which can deal with odd and even highest exponents looks like this in Mathematica:

geometricSeriesMod[a_, n_, m_] :=
   Module[ {q = a, exp = n, factor = 1, sum = 0, temp},

   While[And[exp > 0, q != 0],
     If[EvenQ[exp],
       temp = Mod[factor*PowerMod[q, exp, m], m];
       sum = Mod[sum + temp, m];
       exp--];
     factor = Mod[Mod[1 + q, m]*factor, m];
     q = Mod[q*q, m];
     exp = Floor[ exp /2];
   ];

   Return [Mod[sum + factor, m]]
]

Parameters:

a is the "ratio" of the series. It can be any integer (including zero and negative values).
n is the highest exponent of the series. Allowed are integers >= 0.
mis the integer modulus != 0

Note: The algorithm performs a Mod operation after every arithmetic operation. This is essential, if you transcribe this algorithm to a language with a limited word length for integers.

You can use the property of modulus,i.e., $$(a+b)\%m = (a\%m+b\%m)\%m$$

So to calculate
$$S=1+3+3^2+3^3+...$$

you can use

$$S\%M = (1\%M + 3\%M +3^2\%M +......)\%M$$

This will speed the process and even save you from integer overflow.


Prove existence of $n$ for an infinite product I've been studying a paper in which the author says:
Fix $n$ such that $m^n \prod_{j=1}^n \frac{j}{j+\delta} > 1$, where $1<m<\infty$, and $\delta >0$.
I seem not to be able to show why such $n$ must exist. I tried rewriting it this way:
$m^n \prod_{j=1}^n \frac{j}{j+\delta} =m^n \prod_{j=1}^n (1-\frac{\delta}{j+\delta})= m^n \exp\bigg[\sum_{j=1}^n \log(1-\frac{\delta}{j+\delta})\bigg] \geq m^n\bigg[1+\sum_{j=1}^n \log(1-\frac{\delta}{j+\delta})\bigg]$
but after that I'm stuck. In one part the author writes
$m^n \prod_{j=1}^n (1-\frac{\delta}{j+\delta}) \geq m^n e^{c_0} \exp\bigg[-\delta\sum_{j=1}^n \frac{1}{j+\delta}\bigg]$, where $c_0$ is some constant,
but then I still don't know how this guarantees that there exists $n$ such that $m^n \prod_{j=1}^n \frac{j}{j+\delta} > 1$. Does anyone have any idea on how to prove it?

Find an $N$ such that $\frac{mN}{N+\delta}>1+\eta$ for some $\eta>0$. Then take $M$ such that $(1+\eta)^M>\left(\prod_{j=1}^N\frac{j}{j+\delta}\right)^{-1}$. Then take $n=N+M$.


How do I solve this analytically $3^x=9x$ One of my friends ask me how to solve this equation analytically $3^x=9x$. Looking at it I guess 3 is the answer and I also plot a graph of line $9x$ and the curve $3^x$, they intersect at 3.

But, what I want is to give an analytical solution of the equation. I started

$3^x=9x$

$3^{x-2}=x$

$(x-2)\ln3=\ln x$

How can I continue?

$$3^x=9x \Rightarrow 1 =\frac {9x}{3^x} \Rightarrow 1= 9x \cdot e ^{-x\ln 3}\Rightarrow \frac{1}{9}=x \cdot e ^{-x\ln 3} \Rightarrow$$

$$\Rightarrow \frac{-\ln 3}{9}=(-x \ln 3)\cdot e^{-x\ln 3}\Rightarrow W\left(\frac{-\ln 3}{9}\right)=-x \ln 3\Rightarrow$$

$$x= \frac{-W\left(\frac{-\ln 3}{9}\right)}{\ln 3}$$

where W is Lambert W function .

I am going to give three methods of solving this problem and they are:
$1$. Differentiation -                                                                               It can be broken down into a simpler form by finding  the derivative of both sides.
$\frac{d(3^x)}{dx} = \frac{d(9x)}{dx}$
$x(3^x - 1) = 9$
$x = 9$
$3^{x - 1}=9$
$3^{x - 1} = 3^2$
Equating the exponents,
$x - 1 = 2$
$x = 3$
Thus $x = 3 \ or \ 9$.
But since $9$ does not satisfy the equation.
$x = 3$.
$2$. Graphical approach -
This can be done by plotting a graph of $y=9x$ and $y=3^x$ and the $x$ component of the intersection is going to be $3$.
Thus $x=3$.
3. Value analyzing - This done by replacing the variable with figures starting from zero to see which one satisfies the equation and you will find out the equation is true when $x$ is $3$.
Thus $x=3$.


Hoffman and Kunze Chapter 6 Theorem 13 I am completely stumped at the proof of Theorem 13 in Chapter 6, Hoffman and Kunze. The theorem goes:

Let $T$ be a linear operator on the finite-dimensional vector space $V$ over field $F$. Suppose that the minimal polynomial for $T$ decomposes over $F$ into a product of linear polynomials. Then there is a diagonalizable operator $D$ on $V$ and a nilpotent operator $N$ on $V$ such that

(i) $T = D + N$

(ii) $DN = ND$

The diagonalizable operator $D$ and the nilpotent operator are uniquely determined by (i) and (ii) and each of them is a polynomial in $T$.

The crux was to prove the uniqueness. In the proof given, there was a sentence: "$D'$ and $N'$ commute with any polynomial in $T$; hence they commute with $D$ and with $N$"

Does this mean $DD' = D'D$ and $NN' = N'N$? How did this come about? I am definitely overlooking something obvious...

First a comment: Uniqueness is fairly routine; the crux is in the existence which requires Chinese Remainder Theorem.

Now to your question: yes, $DD=D'D$ and $NN'=N'N$. The theorem constructs $D$ and $N$ as polynomials  $T$. So, if we have possible second candidate, then two polynomials in $T$ which give $D$ and $D'$ will commute.

$N^2=0$  where $N$ is a square matrix order $2$, $N$ will be similar  with which two matrices and how?


Cohomology ring of $\mathbb R P^\infty$ with $\mathbb Z_{2k}$ coefficients Let $k$ be a positive integer. I am trying to show that as rings, $H^*(\mathbb RP^\infty ; \mathbb Z_{2k}) \cong \mathbb Z_{2k}[a,b]/(2a  , 2b  , a^2 - kb)$. This is exercise 3.2.5 in Hatcher. The hint is to "Use the coefficient map $\mathbb Z_{2k} \rightarrow\mathbb Z_2$ and the proof of theorem 3.12". I tried to adapt the proof of theorem 3.12 (the computation of $H^*(\mathbb RP^\infty, \mathbb Z_{2k})$) but was unable to do so: the proof basically shows that cup products of a generator of $H^1$ generate all of the higher cohomology groups. In the $\mathbb Z_{2k}$ case the cohomology ring doesn't have such a simple description, so I couldn't find a way to make an inductive proof work. Does anyone have a proof?

Let's use the proof of the theorem as Hatcher suggested. It is easy to see that in the diagram below the map $H^1(P^n) \to H^1(P^1)$  is a multiplication by $k$ ( injectivity and $H^1(P^1) = Z_{2k}$). The same applies to the two downwards maps on the right
Proceeding as in the Hatcher's proof, we conclude that $H^1(P^n,P^n-P^1) \to H^1(R^n,R^n - R^1)$ has to be a multiplication by $k$ in order to make the diagram commute:

Therefore in the first cohomology the left downwards map is a multiplication by $k$.

So $a\cup a = kb$

I realize this question is old but, since nobody ever answered, I decided to post a solution.

Consider the short exact coefficient sequence
$$0\to \mathbb{Z}_2\overset{\times k}{\to} \mathbb{Z}_{2k}\to \mathbb{Z}_{2}\to 0\;.$$
This induces a long exact Bockstien sequence
$$... \to H^*(\mathbb{R} P^{\infty};\mathbb{Z}_2)\to H^*(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to H^*(\mathbb{R} P^{\infty};\mathbb{Z}_2)\overset{\beta}{\to} H^{*+1}(\mathbb{R} P^{\infty};\mathbb{Z}_2)\to...$$
Recall that
$$H^*(\mathbb{R}P^{\infty};\mathbb{Z}_{2})\simeq \mathbb{Z}_2[t]$$
This gives the exact segment
$$\mathbb{Z}_2(t^n)\to H^n(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to \mathbb{Z}_2(t^n) \to \mathbb{Z}_2(t^{n+1})\;.$$
Since $\beta$ satisfies the Leibniz rule and we are working over $\mathbb{Z}_2$, we have $\beta(t^{n})=0$ is $n$ is even and $t^{n+1}$ if $n$ is odd. This gives two possibilities. In the odd case, we have an iso
$$0\to \mathbb{Z}_2(t^{2n+1})\to H^{2n+1}(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to 0\;$$
and in the even case, we have the iso
$$0\to H^{2n}(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to \mathbb{Z}_2(t^{2n}) \to 0\;.$$
By the first isomorphism we conclude that $H^1$ is generated by a single element $a=kt$ satisfying $2a=0$. From the second isomorphism, we conclude that $H^2$ is generated by a single element $b$, satisfying $2b=0$. Since $b$ mod $k$ must be $t^2$, we have $b=kt^2$ and $kb=k^2t^2=a^2$. The full ring structure can be deduced from these two cases.


Solve $ \log_3x\log_4x(\log_5x-1)$=$\log_5x(\log_4x+\log_3x)$ Solve $ \log_3x\log_4x(\log_5x-1)$=$\log_5x(\log_4x+\log_3x)$ for $x>0$.

The constants $3$, $4$ and $5$ are meant to be the bases of the logs.

The trick, as N.S.JOHN already commented,is to use a common base.

Let us make the problem even more complex, looking for the solutions of $$\log_a(x)\log_b(x)(\log_c(x)-d)-\log_e(x)(\log_f(x)+\log_g(x)=0$$ and apply for each term $$\log_a(x)=\frac{\log (x)}{\log (a)}$$ After simplification, this write $$\log ^2(x) \left(\frac{\log (x)-d \log (c)}{\log (a) \log (b) \log
   (c)}-\frac{\frac{1}{\log (f)}+\frac{1}{\log (g)}}{\log (e)}\right)=0$$ So, the first solution is $$\log(x)=0\implies x=1$$ and the second one (from the term in parentheses) is given by $$\log(x)=\log (c) \left(\frac{\log (a) \log (b) (\log (f)+\log (g))}{\log (e) \log (f) \log
   (g)}+d\right)\implies x= $$

$$\sum\dfrac{(\log x)^2}{\log3\log4}=\dfrac{(\log x)^3}{\log3\log4\log5}$$

If $\log x=0\iff x=1$

Else $$\sum_{\text{cyc}(3,4)}\dfrac1{\log3\log4}=\dfrac{\log x}{\log3\log4\log5}$$

$$\iff\log x=\sum\log3$$

See Change of Base of Logarithm


Number of ways to arrange people around a table if men and women sit in alternate seats I have a problem counting the number of ways one can seat people around a circular table.

When ordering does not matter, $n$ people can sit around the table $n!$ ways, and then, because rotations are over counted and there are $n$ rotations of the table, we have that the number of ways is $(n-1)!$.

Now, when saying that if $n$ was an even number with half women and half men, how many ways to arrange them if no one can sit beside someone of the same sex?

I have read answers online, and they all say arrange the $(n/2)!$ women in a circle, divide by the number of rotations so you get $(n/2-1)!$ and then arrange the men so that you get $(n/2-1)!(n/2)!$.

My question here is: Why do we not divide by $n$ as before, if we once again have all those extra rotations?

In how many ways can $k$ men and $k$ women be seated at a round table if the men and women alternate seats?

Let Amanda be one of the women.  We seat Amanda first.  This determines which $k - 1$ seats are available to the other women.  The remaining $k - 1$ women can be seated in these seats in $(k - 1)!$ ways as we proceed clockwise around the circle relative to Amanda.  The men can be seated in $k!$ ways as we proceed clockwise around the circle relative to Amanda.  Hence, the number of permissible seating arrangements is
$$(k - 1)!k!$$

  You asked why the answer is not
  $$\frac{k!k!}{2k} = \frac{(k - 1)!k!}{2}$$

Observe that if there is only one man and one woman at the table, this answer yields $1/2$ seating arrangements.

What went wrong?

Suppose Amanda is one of the women, and she is seated first.  Since the men and women must alternate, seating Amanda determines in which $k$ seats the women will sit.  Hence, there are only $k$ rotations that preserve a given arrangement of the women.  Moreover, once the women are seated, the men cannot be rotated without changing the relative order of the people in the seating arrangement. Hence, the number of distinguishable seating arrangements is
$$\frac{k!k!}{k} = (k - 1)!k!$$

After getting this wrong, I now realize that a choice is made by virtue of seating a man first, or a woman, and this choice has consequences.  If a woman is seated first, the other (n/2 -1) women can be seated in the remaining (n/2 - 1) female chair slots, so there are (n/2  -1)! possibilities.  The (n/2) men can then occupy the (n/2) remaining chairs freely, multiplying the seating arrangements by (n/2)!.

The rotations are not overcounted, as no choice was counted for the first-seated  person's choice of chair, even though there were n empty chairs.  Answer:

(n/2 - 1)! * (n/2)! seating positions


Meaning of $dx$ in integration and differentiation In differentiation, we denote $dx$ as in infinitesimal change in $x$ and $dy$ the corresponding change.But in integration, we divide $x$ into infinitely many equal intervals and name them $dx$. Then $f(x)dx$ represents the area of an infinitely thin rectangle. From differentiation we know, $dy = f(x)dx$.Why do we substitute then $dy$ in place of $f(x)dx$? I mean $dy$ is change in differentiation. But here in integration we take $dy$ to be the area of each infinitesimal rectangle. Why do we mix them up? Please someone explain with clarity.

$dx$ and $dy$ are not infinitesimal changes, this is an old, obsolete legend. Here is the truth: https://en.wikipedia.org/wiki/Differential_of_a_function.

The answer is "the fundamental theorem of calculus".  The fundamental theorem of calculus shows the deep connection between slopes and areas.  Because of the fundamental theorem of calculus, $dy$ is both the $y$-component of the slope of $f(x)$ and the area of the individual boxes $f'(x)\,dx$.  That's the fundamental theorem of calculus.
However, I think viewing calculus as fundamentally about slopes and areas is a wrong way of going about it.  I think a simpler way to think about it is this:

Differentiation is about infinitesimal changes
Integration is about adding together infinitesimal changes

Here, you can easily see why integration works: you are adding up changes to get the total change (and the total change will be off from the `true value' by a constant, $+C$).
That is, if I differentiate $x^2$, I will get $2x\,dx$.  This means that the value of the change in $x^2$ depends on both (a) the current value of $x$, and (b) the amount of (infinitesimal) change in $x$.  If I add up all of those changes (I integrate it), I will get the total change.  In other words,
$$\text{total change in }x^2 = \int 2x\,dx$$
If I make a graph of this, it will be the same as the original graph, but perhaps off by a constant - which is precisely what integration does!  So integration gives us a mechanism to add together an infinite number of infinitely small values.
So, how does this relate to areas?
The area under a curve is simply adding together all of the infinitely small rectangles under it.  Each rectangle is $y$ high and $dx$ wide, or $y\,dx$.  Therefore, the sum of all the individual rectangles is $\int y\,dx$.
Now, we know enough to find the relationship between slopes and areas.  To find the slope of an equation, we take the differential and divide by $dx$.  To find the area under the curve of an equation for $y$, we multiply by $y$ and integrate.  Since integration and differentiation are opposites of each other, the process of finding areas is the reverse of the process of finding slopes.


What does locus of $\operatorname{arg}(z-i)=\pi/4$ on an Argand diagram? By thinking of the question in terms of vectors, any point lying on the locus, would satisfy that the vector from $i$ to $z$ would make an angle of $\pi/4$ with the horizontal line, like this:

But converting the equation into Cartesian co-ordinates, I get $y=x+1$, which as you know looks like so:

But if the part of the curve with negative x-values is included, then the $\theta$ would no longer be $\pi/4$, as shown here:

So by choosing a $z$ such that Re($z$)$<0$, arg($z$)$\neq \pi/4$
So my question is, which graph is correct, pink or green? And where have I gone wrong

In general, $$\arg(\alpha+i\beta)=\frac{\pi}{4}$$ implies $$\frac{\beta}{\alpha}=\arctan(\frac{\pi}{4})=1\Rightarrow \alpha=\beta,\color{red}{\text{ and }\alpha,\beta\ge0}$$
Assume $z=x+iy$. Therefore, $$z-i=x+iy-i=x+i(y-1)$$ Then $\arg(z-i)=\pi/4$ means the real part is equal to the imaginary part, i.e. $x=y-1$ Hence, $$y=x+1,\color{red}{\text{ and }x\ge0,y-1\ge0\Rightarrow y\ge1}$$
is the locus.

pink, as it is the way arg is usually presented. the negative values represent a different arg -(3pi/4)


Can the identity matrix be negative? I got the following question:
Find, if possible, the inverse of the matrix:
$\begin{bmatrix}3, -1\\2, -2\end{bmatrix} $

and I did the following:
$\begin{bmatrix}3, -1\\2, -2\end{bmatrix}^{-1} = \frac{1}{ad-bc}\begin{bmatrix}-2, 1\\-2, 3\end{bmatrix} = \frac{1}{4} \begin{bmatrix}-2, 1\\-2, 3\end{bmatrix}$

Test: $( A * A^{-1} = I)$
$\begin{bmatrix}3, -1\\2, -2\end{bmatrix} *\begin{bmatrix}-0.5,0.25\\-0.5, 0.75\end{bmatrix} =  \begin{bmatrix}-1, 0\\0, -1\end{bmatrix}$

Is this correct?

No, the identity matrix cannot be negative.  If your check yields $AA^{-1} = -I$ then something must have gone wrong.

La faute est dans le calcul du déterminant, on a : ad-bc=-4 et non 4

The error is in the calculation of the determinant. We have $ad-bc=-4$ and not $4$.


There is relation that is symmetric and transitive but not reflexive? Let $L=\left\{R\right\}$ be a language with only one relation symbol.
Consider these formulas:
$\Psi _1\:=\:\forall x\left(R\left(x,x\right)\right)$
$\Psi _2\:=\:\forall x\forall y\left(R\left(x,y\right)\rightarrow R\left(y,x\right)\right)\:$
$\Psi _3\:=\:\forall x\forall y\forall z\left(R\left(x,y\right)\wedge R\left(y,z\right)\rightarrow R\left(x,z\right)\right)\:$
I need to show $Mod\left\{\Psi _1,\Psi _2,\Psi _3\right\}\ne Mod\left\{\Psi \:_2,\Psi \:_3\right\}$.

So far, i understand the meaning of each formula, but i wonder which relation on any domain, can show that  $R$ is symmetric and transitive but not reflexive. Any ideas?

EDITED: Well, i looked at If a relation is symmetric and transitive, will it be reflexive? and it really helpfull! but i want to find other examples just from curiosity ..

The empty relation on a nonempty domain is vacuously symmetric and transitive, but not reflexive.

Quite generally, if the relation $R(x,y)$ holds, then from symmetry you have $R(y,x)$ and then from transitivity you have $R(x,x)$. So on the face of it symmetry+transitivity indeed imply reflexivity. However, note that we implicitly assumed that there exists such $y$.

So to conclude your question:

  Theorem: For a relation that's symmetric and transitive, and that for every $x$
  there's a $y$ such that $R(x,y)$, then indeed $R$ is also
  reflexive.


True or False: If $f>0$ is integrable on $[a,b]$ then $\sqrt{f}$ is integrable I'm having trouble proving or finding a counterexample for the following statement:

  If $f>0$ is integrable on $[a,b]$ then $\sqrt{f}$ is integrable

We're using the Riemann integral definition:

If $f$ is integrable on $[a,b]$ then given $\epsilon>0$ there exists a $\delta>0 \ $ s.t. if $P$ is a partition of $[a,b]$ and $\lambda(P)<\delta$ then $|S-I|<\epsilon \ $ (where $S$ is the Riemann sum and $I$ is the integral's value).

I tried using the fact that $\sqrt{x}$ is uniformly continuous on $(0,\infty)$ which means that if $f$'s oscillation gets very small, so does $\sqrt{f}$'s, but I wasn't able to rigorously prove it.

Is this statement actually true? is the uniformly continuous angle of any help?

Much appreciated.

By the Riemann-Lesbesgue criterion, we have that a function $g$ on $[a,b]$ is Riemann-integrable iff it is bounded and it's set of discontinuities has measure zero. So that should get it figured out for ya.

I will write it here, and if its wrong, I will delete it.

Let $g=\sqrt{f}$, we have that $f>0$, hence $g$ is monotonic, using theorem, that if you have a monotonic function on close an bounded interval, then monotonic function is integrable.


Prove that A is invertible if $A^2 - 4A -7I = 0$. The $2 \times 2$ matrix $A$ satisfies

$$A^2 - 4A -7I = 0,$$

where $I$ is the identity matrix. Prove that $A$ is invertible.

I'm not sure how to do this. Help would be appreciated.

Observe that
$$
\left(\frac17(A-4I)\right)A=I
$$

you can rewrite $A^2 - 4A - 7I = 0$ as $$A\left(\frac17A-\frac47I\right) = I.$$  therefore $A$ is invertible and $$A^{-1} = \frac17A-\frac47I.  $$


$f>0$ on real line ; $f(x+y)\le f(x)f(y) , \forall x,y \in \mathbb R$ ; $f([0,1])$ is bounded set ; does $\lim_{x \to \infty}(f(x))^{1/x}$ exist? Let $f: \mathbb R \to (0,\infty)$ be a function such that $f(x+y)\le f(x)f(y) , \forall x,y \in \mathbb R$ and $f$ is bounded on $[0,1]$ ; then does the limit $\lim_{x \to \infty}(f(x))^{1/x}$ exists ?

What I have found is $f(x)\le f(x/n)^n , \forall x \in \mathbb R , \forall n \in \mathbb N$ ; so say if $f(x) < M , \forall x \in [0,1]$ then we

get $f(x)\le f\Big(\dfrac x{[x]+1}\Big)^{[x]+1}\le M^{[x]+1} \le M^{2x} , \forall x>1$ ; so that $(f(x))^{1/x}$ remains bounded for large $x$ .

Please help . Thanks in advance .

The answer is yes.

Let $g(x)=\log f(x)$. We want to prove that $\lim\limits_{x\to\infty}\dfrac{g(x)}{x}$ exists.

Let $M:=\max\left(0,\log\sup_{[0,1]}f\right)$.
We have $g(x+y)\le g(x)+g(y)$ for $x,y\ge0$ and $g(x)\le M$ for $0\le x\le 1$.
By repeating the sub-additive property we can see that
$$ g(kx) \le k \cdot g(x) \quad \text{for $k=1,2,\ldots$}; $$
for $k=\lfloor x\rfloor+1$ this provides
$$
g(x) \le \Big(\lfloor x\rfloor+1\Big) \cdot g\left(\frac{x}{\lfloor x\rfloor+1}\right) \le (x+1)M
\quad\text{for all $x\ge 0$.}
$$

Take two sequences, $a_1,a_2,\ldots$ and $b_1,b_2,\ldots$ of positive reals such that $a_n\to\infty$, $b_n\to\infty$,
$\dfrac{g(a_n)}{a_n}\to\liminf\limits_{x\to\infty}\dfrac{g(x)}{x}$ and
$\dfrac{g(b_n)}{b_n}\to\limsup\limits_{x\to\infty}\dfrac{g(x)}{x}$.

Consider an arbitrary pair $n,m$ of indices. Let $K=K_{n,m}=\left\lfloor\frac{b_n}{a_m}\right\rfloor$, so $0\le b_n-K a_m<a_m$. Then
$$
g(b_n)
\le g(K a_m) + g(b_n-K a_m)
\le K g(a_m) + (b_n-Ka_m+1)M
\le K g(a_m) + (a_m+1) \cdot M.
$$
(If $K=0$ then
$g(b_n) \le (b_n+1)M \le (a_m+1) \cdot M$.)
Dividing by $b_n$,
$$
\frac{g(b_n)}{b_n}
\le \frac{K g(a_m) + (a_m+1) \cdot M}{b_n}
= \frac{K_{n,m} a_m}{b_n} \cdot \frac{g(a_m)}{a_m} + \frac{(a_m+1) \cdot M}{b_n}.
$$
Now fix $m$ and take limits with $n\to\infty$. (Update: some explanation is added:) On the LHS, by the definition of $b_n$,
$\frac{g(b_n)}{b_n}\to\limsup_{x\to\infty}\frac{g(x)}{x}$.
Since $a_m$ is fixed, we have $\frac{b_n}{a_m}\to\infty$, so
$\frac{K_{n,m} a_m}{b_n}=\frac{\lfloor b_n/a_m\rfloor}{b_n/a_m}\to1$. In the last fraction the numerator $(a_m+1) \cdot M$ is fixed, the denominator $b_n$ tends to $\infty$. Therefore,
$$
\limsup_{x\to\infty}\frac{g(x)}{x} \le \frac{g(a_m)}{a_m}.
$$
Now take $m\to\infty$ to get
$$
\limsup_{x\to\infty}\frac{g(x)}{x}
\le
\liminf_{x\to\infty}\frac{g(x)}{x}.
$$
Done.

I had a lengthy answer and it just crashed, so I will sketch the proof.

$g(x) = \ln f(x), g(x+y) \leq g(x)+g(y)$
show that $\lim_\limits{x\to\infty} g(x)/x$ exists.
$g(x)/x$ is easier to work with than $f(x)^{1/x}$

$g(x)/x$ is bounded.  There exists a sequence of $x_0<x_1<x_3\dots$ such that the sequence of $g(x_n)/x_n$ converges. (Bolanzo-Weierstrass)

Now we have shown that $|g(x_n)/x_n - L| < \epsilon$ for infinitely many x_n. We must show that $|g(x)/x - L|<\epsilon$ when $x\ne x_n$

Knowing that there is an upper bound for $g(x)/x$, you should be able to show that there is a maximal distance $d$ that $g(x)$ can possibly get way from the line $y = Lx$ and use that to show that when $x$ gets large $d/x$ goes to $0.$

continued....

How do we bound $d$?  If $M$ is the upper bound of $g(x)$ in $[0,1]$ then $g(x)-g(x_n)\leq M(x-x_n)+M$
or
$g(x_n+a) \leq M(a+1)$

And $a<x_{n+1}-x_n$

$\frac{g(x_n+a)}{x_n+a} - L \leq \frac{(M-L)a+M}{x_n+a}$

$\frac{g(x)}{x} - L \leq \frac{(M-L)a+M}{x}$

$(M-L)a+M$ is finite, and when $x$ get to be large enough,
$|\frac{g(x)}{x} - L|<\epsilon$ for any epsilon.


Complex functions and residues I know that, since the complex function
$$f(z)=\frac{1}{z(e^z-1)}$$
have a pole of 2nd order in z=0, I should be able to represent it as:
$$f(z)=\frac{g(z)}{(z-0)^2}$$
Being g(z) a function that is both analytic and different from zero when z=0.

But the only expression for g(z) that I was able to get to was
$$g(z)=\frac{z}{e^z-1}$$
wich doesn't meets the criteria explained before.

Can anyone say what am I getting wrong?
Does f(z) really have a pole of 2nd order in zero?

I'm just taking my first complex analysis but from my understanding you are correct in the fact that pole z= 0 is of second order.

Using the formula for higher order poles.

$$Res_{z \rightarrow x} = \frac{1}{(n-1)!} lim_{z\rightarrow x} \frac{d^{n-1}}{dz^{n-1}}(z-x)^n f(z)$$

Where x is the pole 0 and n is the order.

Using this formula:

$$Res_{z \rightarrow 0} = \frac{1}{(2-1)!} lim_{z\rightarrow 0} \frac{d^{2-1}}{dz^{2-1}}(z-0)^2 f(z)$$

$$Res_{z \rightarrow 0} = \frac{1}{(1)!} lim_{z\rightarrow x} \frac{d}{dz}(z-0)^2 \frac{1}{z(e^z - 1)}$$

$$  lim_{z\rightarrow 0} \frac{d}{dz}(z)^2 \frac{1}{z(e^z - 1)}$$
 $$  lim_{z\rightarrow 0} \frac{d}{dz}\frac{z }{(e^z - 1)}$$
 $$  lim_{z\rightarrow 0} \frac{(e^z - 1) - (e^z)z}{(e^z - 1)^2}$$

Now just do a couple L'Hopitals and you should get:
Res at z=0:
$$= - \frac{1}{2}$$


How to factorize polynomials to the 5th degree? I have the polynomial:
$$2x^5-x^4+10x^3-5x^2+8x-4$$
and I know that the final result is:
$$(2x-1)(x^4+5x^2+4) = (2x-1)(x^2+1)(x^2+4)$$
But how would you do it step by step? I've seen a couple of videos and blogs about it, but they mostly use examples, where their is a common factor between the expressions but in this case their are none.

Every now and then, you find a polynomial of higher degree that can be factored by inspection. In this case, there's a way to just "see" one step of the factorization:

$$2x^5-x^4+10x^3-5x^2+8x-4$$

Notice that the coefficients, when grouped in pairs, are all proportional: $2, -1$ are in the same ratio as $10,-5$ and also $8,-4$. That's uncommon, but it means you can "factor by grouping":

$$(2x^5-x^4)+(10x^3-5x^2)+(8x-4)$$
$$=x^4(2x-1)+5x^2(2x-1)+4(2x-1)$$
$$=(x^4+5x^2+4)(2x-1)$$

The first step is to take the gcf out of each group, and then, since the remaining parts match ($(2x-1)$), that can be factored out. Now we're left with something that is quadratic in form, and can be factored further:

$$x^4+5x^2+4=(x^2+4)(x^2+1)$$

Now, in most cases, nice patterns like this do not occur. Then, you're left with the rational roots theorem. From the fact that our lead coefficient is $2$ and our constant term is $4$, we conclude that any linear factor will correspond to a zero of the form $\pm\frac{p}{q}$, where $p$ is a factor of $4$ and $q$ is a factor of $2$. That gives us a few options: $\pm 4, \pm 2, \pm 1, \pm \frac12$. Trying a few of these, we see that plugging in $x=\frac12$ actually produces a value of $0$. That tells us that $(x-\frac12)$ is a factor, or clearing denominators, $(2x-1)$. You can now factor this linear piece out by long division, and end up just where we were after factoring by grouping.

The question slightly wrong in my opinion. Please try multiplying the factors or the eqn. and lets see if you will get back the original question.

The correct equation is $P(x): 2x^5-x^4-10x^3+5x^2+8x-4$

Group and Factorize: $(2x^5-x^4) (-10x^3+5x^2) (8x-4) =0$

                 $x^4(2x-1)   -5x^2(2x-1)  4(2x-1)$

        Factor out $(2x-1)$ then we have $x^4-5x^2+4$

Our first factor of the equation is $(2x-1)$

This looks like a quadratic form but our first coefficient has 4th degree

look let's expand $x^4-5x^2+4$ and we have $(x^2-4)$ and $(x^2-1)$. $-4$ and $-1$ when added will be $-4-1 = -5$ and when multiply $-4x-1 = 4. x^2 x x^2 = x^4$
so its just splitting them and distributing them on $-4$ and $-1$ which are the product of the constant and sum of the middle term.

Now with $(x^2-4)$ and $(x^2-1)$ we can have $(x-2) (x+2)$ and $(x-1) (x+1)$.

so our final factors of the polynomial are $(2x-1) (x-2) (x+2) (x-1) (x+1)$

Therefore; the zeros of the polynomial are $1/2, 2, -2, 1, -1$.


Locus of a point on a fixed-length segment whose endpoints slide along orthogonal lines Suppose we have some segment $AB$ of constant length that slides in such a way that its endpoints are moving along orthogonal lines. Let $P$ be a point in the segment so that $|AP| = a$ and $|PB| = b$. How can we find the curve along which $P$ moves?

I was trying to write lines $ax + by = c$ since this is orthogonal to the other line, we know normal to the orthogonal one should be $N = (-b,-a)$ and so the equation of the other line is of the form $-b x - a y + d = 0$. Now, point $A$ lie in one of them and $B$ lies on the other one line. should I solve the equation $d(P,A) + d(P,B) = a+ b $? Am I on the right track?

Here's an illustration of my comment, although I've swapped the positions of $A$ and $B$ (for notational reasons that should be clear shortly).

We can use the angle ($\theta$) that the segment makes with the $x$-axis to parameterize the coordinates of the endpoints. Of course, what's important are the coordinates of point $P$, namely ...

  $$P = (a \cos\theta, b \sin\theta) \tag{1}$$

One might (should?) recognize this as the parameterization of the origin-centered ellipse with radii $a$ and $b$.

The non-parametric form of this equation is, of course,

  $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \tag{2}$$

Generalization to the case of an arbitrary pair of perpendicular lines is straightforward. Let the lines meet at $Q = (x_0, y_0)$, and let the line containing sliding point $B$ make angle $\phi$ with the $x$-axis. We can transform equation $(2)$ into the equation of this ellipse with substitutions corresponding to rotation by $\phi$, followed by translation by $x_0$ and $y_0$.

$$\text{rotation:}\;\begin{cases} x \to \phantom{-}x \cos\phi + y \sin \phi \\ y \to -x \sin\phi + y \cos\phi \end{cases} \qquad\qquad
\text{translation:}\;\begin{cases} x \to x - x_0 \\ y \to y - y_0 \end{cases}$$

That is,

  $$\frac{\left(\;(x-x_0)\cos\phi + (y-y_0)\sin\phi\;\right)^2}{a^2}
+ \frac{\left(\;-(x-x_0)\sin\phi + (y-y_0)\cos\phi\;\right)^2}{b^2} = 1 \tag{$2^\prime$}$$

where expansion and simplification is left as an exercise to the reader.

Alternatively, noting that the lines have unit direction vectors $(\cos\phi, \sin\phi)$ and $(-\sin\phi, \cos\phi)$, we easily adapt $(1)$ to get this parametric form of the locus:

  $$P\left(\; x_0 + a\cos\theta \cos\phi - b \sin\theta\sin\phi\;,\;y_0 + a \cos\theta\sin\phi + b\sin\theta\cos\phi\;\right) \tag{$1^\prime$}$$


Determining Type of Probability Distribution Suppose that pollen spores are randomly scattered in a home, at a density of $8$ sports per cubic cm. What is the probability of finding at least two spores in a space of $0.2$ cubic cm?

Here is how I approached it:

Let's call the number of spores per cm$^3$ $n$. Then we are given $\mu=8$. Finding $2$ spores in an area of 0.2 cm$^3$ is equivalent to finding 10 spores in an area of $1$ $cm^3.$ So we want to find $\Pr(n\geq10),$ or $1 - \Pr(n \leq 9)$.

I'm confused about what sort of distribution this would be. Poisson?

And am I expected to then compute that distribution's values for values $n=0, 1, ..., 9?$

The usual modelization of this kind of real-life (?) situation is that the spores are spread according to a Poisson process. The intensity $\varrho$ of the process corresponds to the mean number of spores per volume unit. Then the number of spores in any volume $v$ is Poisson with parameter $\lambda=v\varrho$ hence the probability to find at least $2$ spores in the volume $v$ is $p=P(X_\lambda\geqslant2)$ where $X_\lambda$ is Poisson of parameter $\lambda$, that is, $p=1-(1+\lambda)\mathrm e^{-\lambda}$.

Here, $\varrho=8\,\mathrm{cm}^{-3}$ and $v=0.2\,\mathrm{cm}^{3}$ hence $v\varrho=1.6$ and $p=1-2.6\,\mathrm e^{-1.6}=47.5\%$ (and the parameter $\lambda=v\varrho$ is unitless, as it should).

Note that asking to find at least $2$ spores in a volume $v$ or at least $20$ spores in a volume $10v$ are not equivalent. The second happens with probability $q=1-\mathrm e^{-16}\sum\limits_{n=0}^{19}(16)^n/n!$, hence $q\ne p$ (and additional considerations allow to show without computations that $q\lt p$).

I think it would be poisson with $\mu = 8$.

$1 \ \text{cm}^{3}$ has a volume of $1$. 0.2 cm ^3 has volume 0.008. So $2/.008 = 250$.


How to find the probability of truth? A and B are independent witness in a case. The probablity that A
  speaks the truth is 'x' and that of B is 'y'.If A and B agree on a
  certain statement, how to find the probability that the statement is
  true ?

$ P(A)=x $ and $ P(B)=y $,

$A$ and $B$ are independent, so $P(A \cap B)=P(A)P(B)$

Therefore, the probability that both speak the truth will be $P(A \cap B)=P(A)P(B)=xy$.

And then, the probability that they agree on a certain statement is, they both speak the truth or they both tell lie, which would be:

$$xy + (1-x)(1-y) $$

As a result, the probability that the statement is true is:

$$ \frac{xy}{xy+(1-x)(1-y)} $$.

Say the fact is True or False (T/F), and independent statements A, B are binary.

We are given P(A=1|T)=x and P(B=1|T)=y.
By convexity P(A=0|T)=1-x and P(B=0|T)=1-y.

We can surmise P(A=0|F)=x and P(B=0|F)=y.
Also by convexity P(A=1|F)=1-x and P(B=1|F)=1-y.

We want P(T|A=1,B=1).

By Bayes rule P(T|A=1,B=1) = P(A=1,B=1|T)P(T)/P(A=1,B=1).

By independence P(T|A=1,B=1) = P(A=1|T)P(B=1|T)P(T)/(P(A=1)P(B=1)).

Plugging in, the numerator is xyP(T).

The denominator requires P(A=1) = P(A=1|T)P(T) + P(A=1|F)P(F) = xP(T) + (1-x)P(F), same for P(B=1).

So I don’t think we are identified without knowing the marginal probability of the truth P(T).

Of course, for a given x and y you can provide P(T|A=1,B=1) as a function of P(T) between 0 and 1.


"Cancelling Out" Differential Forms in Chain Rule Let's think about$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}.$$ Here function $u(x)$ is differentiable at $x=x_0$ and $y(u)$ is differentiable at $u=u_0=u(x_0)$.
I know this is a common question, but trying to learn about differential geometry and about differential forms in particular, I thought about the fundamental reason why we can't cancel out the $du$'s. I want you to check if this is correct.
Say that the tangent vector at $x=x_0$ is $\vec{a} = (a_1, a_2)$, and the tangent vector at $u=u_0$ is $\vec{b} = (b_1, b_2)$. Then the $du$ in the former is $du(\vec{a}) = a_2$, a function of $\vec{a}$, and the $du$ in the latter is $du(\vec{b}) = b_1$, a function of $\vec{b}$.
Unless it is a very special case that $a_2 = b_1$, the $du$'s have different values so we can't "cancel out" them (and I personally consider setting $a_2 = b_1$ a bit meaningless and just harms generality).
Is this logic correct?

I find the entire premise of the question is a little flawed. You want to regard derivatives as ratios of the corresponding differentials, but this is simply not true (or at best a bit circular). It's important to keep in mind that $df/dx$ is just notation, and shouldn't really be regarded as a true ratio of well-defined quantities. Differential forms are concepts based on derivatives, not the other way around.
I would go as far as to say that trying to regard the chain rule as "cancellation" is not just wrong, but entirely senseless. However, it is very suggestive nonsense, so it's perhaps worth exploring why it's doomed to failure.
A slight aside first. The way you regard differential forms is not quite correct. Your functions are one-dimensional (judging from context, since you never specified domain), but you take your differential forms to be functions of two-dimensional vectors. This is not how you define differential forms. In your notation, the correct thing would be $du(a_1) = a_2$.
Just to avoid confusion, let us first formalize the concept of a differential form. In one dimension, the concept of a differential form is little more than notation. We define $df_x:\mathbb{R}\rightarrow \mathbb{R}$ as the linear map
$$df_x(r) = f'(x)r,$$
for all $r \in \mathbb{R}$. Moreover, this is an $x$-dependent linear map. Due to the fact that a linear map in one-dimension is given by a single real number (a $1\times 1$ matrix, if you will), the concept of a differential form here is completely equivalent to the derivative. You gain nothing in one dimension by working with differential forms, which is why you almost never see it used in single variable calculus. The concept of a differential form is non-trivial in higher dimensions however, because the resulting linear maps are no longer trivial.
Now, we could define ratios of differential forms to see what we get. The first thing that goes immediately wrong is that this ratio will be undefined when the form in the denominator is at a critical point. This is already damning in my opinion, but let's continue regardless. Maybe it will work everywhere the ratio is well-defined. Let's use your set up where we have functions $f$ and $u$. Let me call $g = f\circ u$. This brings us to our second issue, which is the fact that the "$df$" on the left-hand side should really be "$dg$". The chain rule written with unambiguous notation should really be
$$\frac{dg}{dx} = \frac{df}{du}\frac{du}{dx},$$
but of course this is not nearly as memorable. The mnemonic for the chain rule relies on the crucial overloading of notation, where we regard $df/dx$ not as the derivative of $f(x)$ but of $(f\circ u)(x)$. When you try to write everything down carefully and unambiguously, you find immediate issues with the ratio interpretation of derivatives. The notation $df/dx$ is not only somewhat misleading, but ultimately ambiguous. This issue is only compounded in higher dimensions with the introduction of various partial derivatives. There, you have more explicit breakdowns, like the triple product rule.

If
$$lim_{dx \to 0,dy\to 0}\frac{dy}{dx}= \dfrac{lim_{dy \to 0,dt\to 0}\frac{dy}{dt}} {lim_{dx \to 0,dt\to 0}\frac{dx}{dt}}$$
can be abbreviated to symbolically denote a ratio of two limits
$$\frac{dy}{dx}= \frac{dy}{dt} / \frac{dx}{dt} \tag1$$
then $$ {lim_{dy \to 0,dt\to 0}\frac{dy}{dt}} \cdot {lim_{dx \to 0,dt\to 0}\frac{dt}{dx}}$$
can as well  be abbreviated to symbolically denote a product of two limits
$$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \tag2 $$
(1) and (2) are obtained by cancellation of differentials . They must be together correct or together wrong.


Joint Probability Distribution Function For some Joint Probability Function on two continuous random variables, $f(x,y)$, is it true that the integral of $f(t,t)dt$ from negative infinity to infinity is necessarily upper bounded by one? I thought that $4xy$ is a counter example, but I can't seem to visualize what I'm doing.

Assuming that $f(x,y)$ is the joint density of jointly continuous random variables $X$ and $Y$, it is straightforward to show that the density
of $Z = Y-X$ is given by
$$g(z) = \int_{-\infty}^\infty f(x, z+x)\,\mathrm dx.$$
Thus your integral
$\displaystyle \int_{-\infty}^\infty f(t,t)\,\mathrm dt$ can be recognized
as $g(0)$. I will leave it up to you to decide whether it is
true that $g(0) \leq 1$ for all possible pairs of jointly continuous
random variables.

Note that if your variables are $X,Y$ then
$$
\int_\mathbb{R} f(t,t)dt = \mathbb{P}[X=Y=t] \leq 1
$$


Equation solving involving terms inside roots How do I solve

$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x$$

My Try I tried to take the first term as $t$ but then I had to square both sides twice and that led to a complex bi quadratic. I'm not sure even that'll solve the problem.

guys please let me know what are different possible ways to tackle problems like these, and how would you go about on solving this particular one.

$$\left(x-\frac{1}{x}\right)^{1/2}-\left(1-\frac{1}{x}\right)^{1/2} = \frac{\left(x-\frac{1}{x}\right)-\left(1-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2}}=\frac{x-1}{x}=1-\frac1x,$$
adding both equations gives $$2\left(x-\frac{1}{x}\right)^{1/2}=x-\frac{1}{x}+1.$$ That means $$x-\frac{1}{x}=1,$$ so we have $x=\phi$ (the Golden Ratio). The other solution $x=-\phi^{-1}$ does not satisfy the original equation, we must have $x>0$.


$Ax^2+Bxy+Cy^2=1$ with $A=C=1$ and $B=2$ should be a parabola (because $B^2 = 4 AC$). Instead, it represents parallel lines. What went wrong? My calculus book says

  The equation $Ax^2+Bxy+Cy^2=1$ produces a hyperbola if $B^2>4AC$ and
  an ellipse if $B^2<4AC$. A parabola has $B^2=4AC$.

When I set $B=2,A=C=1$, the equation becomes $x^2+2xy+y^2=1$, which is equivalent to $(x+y)^2=1$ or $y=-x\pm1$. This is not a parabola but two parallel lines. What went wrong?

You are right: a nondegenerate parabola doesn't admit an equation of the form $Ax^2+2Bxy+Cy^2=1$.

Indeed, if $B^2=AC\ne0$, we can multiply all coefficients by $A$, finding
$$
A^2x^2+2ABxy+B^2y^2=B
$$
that becomes $(Ax+By)^2=B$. If $B>0$ this factors as
$$
(Ax+By-\sqrt{B})(Ax+By+\sqrt{B})=0
$$
which is a pair of parallel lines.

If $B<0$, the factorization is over the complex numbers and we get no real point.

If $B=0$, then either $A=0$ or $C=0$, so the equation becomes either $Cy^2=1$ or $Ax^2=1$, again a situation as before.

Only nondegenerate conics with a center admit an equation of the given form, which is obtained by translating the center of the conic in the origin.

$Ax^2+Bxy+Cx^2=1$ is symmetric under the $\pi$ rotation.  But parabola has no centre of symmetry.

You need at least one linear term.  Simplest example is $x^2-y=0.  A=1,B=C=0.$


How to find median from a histogram? I am doing a course on machine learning and as part of it i am also learning statistics.

I came across one question in which i have to find the median basing on a histogram.

Median is the th  element.

But in the histogram the hint is confusing me. What does that mean
43 is the median of the frequencies, but it's not the median of the values.

For the median of the values, if you sum up all the frequencies below the median, and all the frequencies above the median, you should get the same number.

Please help.

Use the formula:
$$\text{Median}=l+\frac{\frac{n}{2}-F}{f}\cdot w=25+\frac{\frac{403}{2}-159}{82}\cdot 5=27.59$$
where $l$ is the lower border of the median group, $F$ is the cumulative frequency up to the median group, $f$ is the frequency of the median group, $w$ is the width of the median group. Also, the median group is $25-30$, because the median position $\frac{403}{2}=201.5$ is greater than $36+54+69=159$ and less than $36+54+69+82=241$.

Actually to find median from histogram you have to draw cumulative frequency more than type and cumulative frequency less than type in form of frequency curves. Then from the point of intersection you drop a perpendicular to X axis . Point of intersection with X axis is median.


Puzzle about technique of fair using of unfair coin There is an unfair coin. It tends to land on one side more than on the other. It is unknown which side is it.

There is Mr. A and Mr. B. They argue about something and they want to use that coin to decide who is right.

Is there any technique of using that unfair coin to get fair result?

My solution:
I can think of only that there is going to be 2 rounds. First time Mr. A chooses HEAD, than Mr. B chooses HEAD if any of them wins twice they win.

I think you need an unbounded number of coin tosses, however the expected number of tosses to settle the issue will be finite: Toss twice. If it comes up heads, then tails, A wins. If it comes up tails, then heads, B wins. If the two tosses come up the same, start over. If the coin is extremely unfair, this will probably take a long time, but if it is only a little unfair, you should be settled soon enough.

There's a difference here between randomness and predictability.  If the coin is only tossed one time, the answer is unpredictable, even though the result is biased.  Ask one of them to pick heads or tails, and toss the coin.

The more often you toss, the more the bias will become evident, and predictability becomes an issue.  Fairness is based on unpredictability, not randomness.


System of non-linear ODE's do you have any suggestions to solve analytically the Non-linear ODE system

$\dot x=18 x^2 y-3p x^2+6p xy$

$\dot y=18 x^2 y-6p xy $

where $p$ is a real constant.

Thank you very much

cheers

There are no explicit solutions for most of the nonlinear systems. Therefore, they are linearized near the fixed points. The eigenvalues near the fixed points mostly give idea about the phase portrait solutions of such systems. For this system fixed points are $(0,0)$ and $(p/3,p/12)$. The eigenvalues near the $(p/3,p/12)$ are:

$\lambda_{1,2}=\frac{-p^{2}}{4}\mp\frac{p^{2}}{2}\sqrt{\frac{31}{4}}i$

which means this is an asymptothycally stable spiral point.

However, the eigenvalues near the origin are found to be zero. This means linearization does not describe the system. Therefore, I don't think you can find an easy solution for this system even when the linearization does not work.

Solutions to these types of  equations .....(dx/dt) =F1(x,y.z) ;(dy/dt) =F2(x,y,z);(dz/dt) =F3(x,y,z) , ....which are called State Space Equations , are found in Non-linear Control System Engg. and Advanced Control System Engg. books.


Computing the Lagrange multiplier in constrained optimization problem As I know the value of the Lagrange multiplier does not have a valuable meaning in most cases. Our goal is to compute the values of the parameters.
Here is an example for constrained optimization: https://www.svm-tutorial.com/2016/09/duality-lagrange-multipliers/#comment-11576
$$\text{minimize }\; f(x,y)=x^2+y^2 ~\text{ s.t. }~  g(x,y)= x+y-1=0$$
He constructs the Lagrange multipliers and computes the value of $x$ and $y$. But at the end and after having the values of $x$ and $y$, he also computes the value of $\lambda$, which I don't understand why?
What it is the purpose to compute the value of $\lambda$? Why do we need its value at the very end?

You can simplify your problem writing

$$f(x,1-x)=x^2+(1-x)^2$$
If we denote

$$g(x)=2x^2-2x+1$$
and with
$$g'(x)=4x-2$$ you will find your Minimum.
You can also consider

$$f(x,y)+\lambda(x-y-1)=F(x,y,\lambda)$$
$$\frac{\partial F(x,y,\lambda)}{\partial x}=2x+\lambda$$
$$\frac{\partial F(x,y,\lambda)}{\partial y}=2y+\lambda$$
$$\frac{\partial F(x,y,\lambda)}{\partial \lambda}=x+y-1$$


Sum of angles in cyclic pentagon My son got stuck on the March 9th puzzle from Corbett's conundrums (a website of maths questions designed for school children):

Unfortunately, I don't know how to help him solve this, can anyone here help?

We've found that opposite angles in a cyclic quadrilateral add up to 180 degrees, and this lets you write down various equations, but we just end up going round and round without progress.

My feeling is that we have not properly understood the question as it feels like we can move the north-east vertex around the circle and change the value of x without changing the value of y, so I don't see how there could be a specific answer for x+y.

Can anyone propose either a solution, or perhaps suggest the most likely missing piece of information that would make the problem solvable?

Yes, you can't get a numeric value for $x+y$ on the information given.

If you have the value of angle $a$ in the diagram below, you can infer the value of $x+y$ from that via $z=180-y$ and $w=180-x$.

To get answer use property of circle ,
The measure of an inscribed angle is equal to half the measure of the intercepted arc
As per above circle ,
Answer is
X+Y=180°+a


Apply $D(x,y)=(cx,cy)$ to the graph of $y=x^2$ and find the equation of the transformed graph. Hint: Let $u=cx$ and write $v=cx^2$ in terms of $u$.

What I think the answer is:

$y=x^2$ is the same as $(x, x^2)$ therefore, applying $D$ to the equation gives us $(cx, (cx)x=cx^2)$, that is, $y=cx^2$.

According to the given hint, $x = \dfrac {u}{c}$.

Expressing v in terms of u, we should have $v = cx^2 = c (\dfrac {u}{c})^2 = \dfrac {u^2}{c}$. Then, $y = \dfrac {x^2}{c}$.


Check the limits using Taylor expansion. I want to check the below limit using Taylor expansion.
$\displaystyle{\lim_{n\rightarrow +\infty}\frac{1}{n}\sqrt[n]{n^n-1}}$
For that how many terms do we have to consider at the expansion?
Also do we expand each term separately? I mean do we expand the whole function $\frac{1}{n}\sqrt[n]{n^n-1}$ ? Or do we expand $\frac{1}{n}$ and $\sqrt[n]{n^n-1}$ separately?

We can use that
$$\frac{1}{n}\sqrt[n]{n^n-1}=\frac{1}{n}\sqrt[n]{n^n}\sqrt[n]{1-\frac1{n^n}}=\sqrt[n]{1-\frac1{n^n}}=e^{\frac{\log\left(1-\frac1{n^n}\right)}{n}} \to 1$$
indeed by standard limits
$$\frac{\log\left(1-\frac1{n^n}\right)}{n}=\frac{\log\left(1-\frac1{n^n}\right)}{\frac1{n^n}}\frac1{n^{n+1}} \to -1\cdot 0=0$$


Solve for $x$ in the $80^\circ$-$80^\circ$-$20^\circ$ triangle I've been solving this for days but I still I couldn't solve this. Can I know how to solve $x$ ?

I fail to see the importance and difficulty of Langley’s Problem mentioned by Justin Benfield in his answer here apart perhaps some historical mention or maybe some interesting applications. In effect, here I give an EASY generalization of the problem proposed by the OP. I would appreciate any substantiated comment against (am I wrong?).

Let $\triangle{ABC}$ not necessarily isosceles. Given side $a$ and angles $\alpha,\beta,\gamma,\delta$ determine the angles $\phi_1,\phi_2,\phi_3,x,y$ as well as the segments $BE,BD,CD$ and the  said difficulty would be to determine $x$ or $y$ since $\phi_1,\phi_2,\phi_3$  are  immediate to calculate

We have
$$BE=\frac{\sin \beta}{\sin \phi_1}\cdot a\\BD=\frac{\sin \phi_3}{\sin \phi_2}\cdot a\\CD=\frac{\sin \alpha}{\sin \phi_2}\cdot a\\ED=BE\cdot\frac{\sin \delta}{\sin x}=\frac{\sin \beta\sin\delta}{\sin \phi_1\sin x}\cdot a $$ It follows the easy system
$$\begin{cases}x+y=\phi_3\\\space\\\dfrac{\sin (\phi_1+\phi_3-x)}{\sin x}=\dfrac{\sin\phi_1\sin\phi_3}{\sin\beta\sin\phi_2}\end{cases}$$ This gives an elementary equation of the form $$A\sin x+B\cos x=C$$ one of the quite popular way to solve is the following:
$$A\sin x+B\cos x=C\iff\frac{A\sin x}{\sqrt{A^2+B^2}}+\frac{B\cos x}{\sqrt{A^2+B^2}}=\frac{C}{\sqrt{A^2+B^2}}$$ so $$x+\theta= \arcsin{\frac{C}{\sqrt{A^2+B^2}}}\space\text{where } \space \theta=\arccos{\frac{A}{\sqrt{A^2+B^2}}}$$

Hey I got $\angle x=20°$

I did it by constructing triangle as per your requirement

(Not the correct way)


Why are integral and differential operators commutative? For instance, let's assume a constant 3D surface over time $S$.
$$
\frac{d}{dt}\iint_S \mathbf B \cdot \mathbf{ds}
\quad=\quad
\iint_S\frac{\partial \mathbf B}{\partial t}\cdot \mathbf{ds}
$$

Why is that? Why can I commute $\frac{d}{dt}$ and $\iint_S$?

I would appreciate formal answers. If you can provide references too, I would like.

You can do it when the function $\boldsymbol B$, and the function $\dfrac{\partial \boldsymbol B}{\partial t}$ are continuous.

Why? Here is the answer.

Here is a short answer. Integral are sums (Riemann sums) and derivatives are differences (to the limit). Sums and differences are commutative.


How to proceed with this Second Order PDE? I would really like to solve the following with the method you will see below. I have done some work but I have no idea how to continue. Any help would be greatly appreciated.

$$
\frac{\partial u(x, t)}{\partial t} = -k(x)u(x, t)
+A\frac{\partial u(x, t)}{\partial x}+B\frac{\partial^{2}u(x, t)}{\partial x^{2}}
$$

$$ u(x, 0) = 1 $$

Where ... lets say:

$A = 0.05, \ B = 0.5, \ k(x) = 0.5x^{2}$

and Boundary Conditions:

$ u(0, t) = 0, \ u(5, t) = 1 $

$ \\ $

Assume: $ u(x, t) = \phi (x)T(t) \\ $

$
\frac{\partial u(x, t)}{\partial t} = \frac{\partial(\phi (x)T(t))}{\partial t}
 = \phi (x)T^{\prime}(t)
$

$
A\frac{\partial u(x, t)}{\partial x} = A\frac{\partial(\phi (x)T(t))}{\partial x}
 = A\phi ^{\prime}(x)T(t)
$

$
B\frac{\partial^{2}u(x, t)}{\partial x^{2}} = B\frac{\partial^{2}(\phi (x)T(t))}{\partial x^{2}}
 = B\frac{\partial(\phi ^{\prime}(x))}{\partial x}T(t) = B\phi ^{\prime\prime}(x)T(t) \\
$

Once, all terms are defined - write out the whole equation:

$
\phi (x)T^{\prime}(t) = -k(x)\phi (x)T(t)
+A\phi ^{\prime}(x)T(t)+B\phi ^{\prime\prime}(x)T(t) \\
$

Manipulate it a little bit, set it equal to some constant:

$
\frac{T^{\prime}(t)}{T(t)} = \frac{-k(x)\phi (x)}{\phi (x)}+A\frac{\phi ^{\prime}(x)}{\phi (x)} +B\frac{\phi ^{\prime\prime}(x)}{\phi (x)}
$

$
\frac{T^{\prime}(t)}{T(t)} =
\frac{-k(x)\phi (x)+A\phi ^{\prime}(x)+B\phi ^{\prime\prime}(x)}{\phi (x)}
 = -\lambda
$

We end up with two separate equations:

$
T^{\prime}(t)+\lambda T(t) = 0
$

and

$
-k(x){\phi (x)}+A\phi ^{\prime}(x)+B\phi ^{\prime\prime}(x)
+\lambda \phi (x) = 0
$

I do not know how to continue beyond this point...

First observation: your boundary conditions are not consistent. If your left boundary condition is $u(0,t) = 0$, while the initial condition is $u(x,0) = 1$, then these conditions do not match at $(x,t) = (0,0)$. So, let's take a look at two situations: a) ignoring the initial condition, and b) ignoring the left boundary condition. We're going to try tackle both situations with separation of variables. Following the analysis in your question (which is entirely correct), we end up with two equations:
\begin{equation}
\frac{1}{2}\phi'' + \frac{1}{20} \phi' - \left(\frac{x^2}{2}-\lambda\right) \phi = 0 \tag{1}
\end{equation}
and
\begin{equation}
 T' + \lambda T = 0. \tag{2}
\end{equation}
The solution to $(2)$ is very easy: we obtain $T(t) = T_0 e^{-\lambda t}$. The solution to the second ODE can be expressed in terms of the Hermite polynomial and the hypergeometric function $_1F_1$, as
\begin{equation}
 e^{-\frac{x^2}{2}-\frac{x}{20}} \left[c_1 \,H_{-\lambda-\frac{401}{800}}(x) + c_2 \phantom{1}_1 F_1(\frac{401}{1600}+\frac{\lambda}{2},\frac{1}{2},x^2)\right]. \tag{3}
\end{equation}

a) Ignoring the initial condition, we have to find a way to incorporate the boundary conditions $u(0,t) = 0$ and $u(5,t) = 1$. From $u(x,t) = T(t) \phi(x)$, the left boundary condition gives $\phi(0) = 0$, since $T(t) = T_0 e^{-\lambda t}$ is only zero if $T_0 = 0$, which would yield the trivial solution $u = 0$; this trivial solution obviously does not satisfy the right boundary condition $u(5,t) = 1$.

This right boundary condition yields for the general solution $T(t) \phi(5) = 1$. This implies that $T(t)$ is constant in $t$, which in turn implies that $T_0=0$, violating the right boundary condition.

We conclude that a solution obtained by separation of variables cannot obey the right boundary condition $u(5,t) = 1$. Note that changing this boundary condition to $u(5,t) = 0$ could be a way forward, since the equations $\phi(0) = 0$ and $\phi(5) = 0$ can be used to determine the constants $c_1$ and $c_2$ in $(3)$.

b) Ignoring both boundary conditions for the moment, we try to incorporate the initial condition $u(x,0) = 1$. In terms of the solution obtained by separation of variables, this means that $T(0) \phi(x) = T_0 \phi(x) = 1$. This means that $\phi(x)$ is constant in $x$. One way to make sure this happens is to choose $c_1 = c_2 = 0$, but this would yield $\phi(x) = 0$, violating the initial condition. Therefore, the alternative would be to look for $\lambda$-values for which both the Hermite polynomial and the hypergeometric function are equal to $e^{\frac{x^2}{2}+\frac{x}{20}}$. However, from the ODE $(1)$, we already see that the only constant solution is the zero solution: if $\phi(x) = \frac{1}{T_0}$, then $\phi' = 0$ and $\phi'' = 0$, yielding (see $(1)$)
\begin{equation}
\left(\frac{x^2}{2}-\lambda\right)\phi = 0,
\end{equation}
which can only be true for all $x \in (0,5)$ if $\phi = 0$. So, a solution obtained by separation of variables cannot obey the initial condition $u(x,0) = 1$.

To conclude: If you want have a chance to obtain a solution to this equation by separation of variables, choose different boundary conditions.

The PDE isn't nonlinear; it is linear. It is only non-linear if $u$ shows up in some non-linear function. Like if there was a $u^2$ or an $e^u$ or $u \frac{\partial u}{\partial t}$ then it would be nonlinear. But this equation has nothing like that.

From where you are, you have $$B\phi'' + A\phi' + \lambda \phi = \tfrac 1 2 x^2, \,\,\,\,\,\, \phi(0) = 0, \phi(5) = 1.$$ To solve this, you need a particular solution and a homogeneous solution. For the particular solution, since you have a polynomial on the right hand side, you can just guess $$\phi_p(x) = \alpha x^2 + \beta x + \gamma,$$ plug it in and just solve for $\alpha, \beta, \gamma$ in terms of $A,B,\lambda$. For the homogeneous solution, you can guess $$\phi_h(x) = Ce^{mx},$$ where $C$ is an arbitrary constant. When you plug this in, you get a quadratic equation for $m$; solving this gives two roots, $m_+$ and $m_-$ (where the $\pm$ corresponds to the $\pm$ in the quadratic formula). Then your full solution for $\phi$ is $$\phi(x) = \phi_p(x) + \phi_h(x) = \alpha x^2 + \beta x + \gamma + C_1e^{m_+x} + C_2 e^{m_-x}$$ and you need to pick $C_1, C_2$ so that the boundary conditions are satisfied.

For $T$ the procedure is similar, but the equation is much easier: $$T' + \lambda T = 0, \,\,\,\, T(0) = 1$$ gives $T(t) = e^{-\lambda t}.$