Show that this hyperplane is parallel to the null space of this linear functional. This is an exercise of my homework of "Functional Analysis".

  If $Y$ is a subspace of a vector space $X$ and $\dim(X/Y)=1$, then every element of $X/Y$ is called a hyperplane parallel to Y. Show that for any linear functional $f\neq 0$ on X, the set $H_1=\{x\in X: f(x)=1\}$ is a hyperplane parallel to the null space $N(f)$ of $f$.

According to the definition of "hyperplane parallel to $N(f)$", we just need to show that $H_1$ is an element of $X/N(f)$, i.e. it is of the form $H_1=x_1+N(f)$ for some vector $x_1\in X$.

First $H_1\neq\emptyset$. In fact, since $f\neq0$ there is $x\in X$ with $f(x)\neq0$. Therefore $f(x)^{-1}x$ satisfies $f(f(x)^{-1}x)=f(x)^{-1}f(x)=1$, i.e. $f(x)^{-1}x\in H_1$.

Let $x_1$ be any element of $H_1$, then we can show that $H_1=x_1+N(f)$. In fact, if $x\in H_1$ then $x=x_1+(x-x_1)$ where $f(x-x_1)=f(x)-f(x_1)=1-1=0$, i.e. $x-x_1\in N(f)$. Conversely, if $y=x_1+x\in N(f)$ then $f(y)=f(x_1+x)=f(x_1)+f(x)=1+0=1$, i.e. $y\in H_1$.

Therefore $H_1$ is one of the cosets in $X/N(f)$, i.e. it is a hyperplane parallel to $N(f)$.

I didn't understand your attempt, so I propose the following:

Since $\;0\neq f\in X^*\;$ , we know $\;H:=\ker f\;$ is a hyperplane or maximal subspace of $\;X\;$ , or in other words: $\;\dim X/H=1\;$ , which means this last vector space is spanned (algebraically, not only topologically) by any non-zero vector $\;\overline v\in X/H\;$.

But clearly $\;\overline v:=f(x)+H\;,\;\;f(x)\neq 0\;$  fulfills the condition...


Dimension of the tangent space in algebraic geometry In my algebraic geometry course, we are studying tangent space and there is something I don't really understand. We define the tangent space as follows.

Let $a \in X$ be an variety. By chosing an affine neighborhood of $a$ we assume that $X \subset \mathbb A^n$ and that $a = 0$ is the origin. Then
$$T_aX = V(f_1:f \in I(X))$$
is the tangent space of $X$ at $a$, where $f_1$ denotes the linear term of $f\in I(X)$.

Thereafter, my teacher said that in the above definition of the tangent space, we had embedded $X$ in $\mathbb A^n$ but there are many other ways to embbed $X$ in some $\mathbb A^k$ and it is not at all obvious that the tangent space has the same dimension in these different embeddings. And then we showed the following statement that gives a definition of the tangent space that is independant of the embedding we chosed.

The tangent space is isomorphic to be the dual space of $I_a/I_a^2$, where $I_a$ is the unique maximal ideal in $\mathcal O_{X, a}$ (the stalk of the regular function at $a$).

Here are my questions: Why does we need this statement to ensure that the dimension of the tangent space do not depend on the embedding ? The tangent space is an affine variety, so it exists on its own and does not depends on any embedding right ? Moreover, as it is a vector space, the dimensons (the "topological" one and the "vector space" one) coincide and the dimension as a vector space is strictly related to the space itself, not on any embedding.. Why isn't the invariance of the dimension obvious ?

Well in your definition the tangent space is constructed from an embedding of $X$, so there is no guarantee that the tangent space that is constructed from another embedding is isomorphic to this one.

Question: "Why does we need this statement to ensure that the dimension of the tangent space do not depend on the embedding?"
Answer: If $X:=V(I) \subseteq \mathbb{A}^n_k$ where $I \subseteq k[x_1,..,x_n]$ is your variety and if $X :=V(J) \subseteq \mathbb{A}^m_k$ with $J\subseteq k[y_1,..,y_m]$ it follows you have two embeddings
$$i: X \rightarrow \mathbb{A}^n_k\text{  and  }j: X \rightarrow \mathbb{A}^m_k$$
You get two definitions of the tangent space $T_x(X)$ at $x$: One using $I$ and one using $J$. Denote these two spaces by $T^I_x(X)$ and $T^J_x(X)$.
Example: Let $f(x,y):=x^2+y^2-1$ and let $I:=(f) \subseteq k[x,y]$. Let $p:=(1/\sqrt{2},1/\sqrt{2})\in S:=V(I)$ and make the change of coordinates
$$x:=u+1/\sqrt{2}, v:=y+1/\sqrt{2}$$
to get
$$f(u,v)=\sqrt{2}(u+v)+u^2+v^2$$
with $f^*:=\sqrt{2}(u+v)$. Let $I^*:=(f^*)=(u+v)\subseteq k[u,v]$. Define $C_p(S):=Spec(k[u,v]/I^*)$. With this construction you have made a coordinate
change and in the new coordinates $u,v$ it follows the original point $p$ corresponds to the origin $(0,0)$ and $C_p(S)$ is the "tangent line" of $S$ at $p$ viewed as an algebraic variety/scheme. There is an isomorphism
$$C_p(S) \cong Spec(k[t])\cong \mathbb{A}^1_k$$
since $k[u,v]/(u+v) \cong k[t]$ is an isomorphism of rings: map $t$ to the class of $u$ or $v$.
You may check that there is an isomorphism $\mathfrak{m}/\mathfrak{m}^2 \cong k\{t\}$ when $\mathfrak{m}:=(x-\frac{1}{\sqrt{2}}, y-\frac{1}{\sqrt{2}})$. Hence $Sym_k^*((\mathfrak{m}/\mathfrak{m}^2)^*) \cong k[x]$ with $x:=t^*$ the dual basis. Hence
$$Spec(Sym_k^*(\mathfrak{m}/\mathfrak{m}^2)^*)) \cong \mathbb{A}^1_k \cong C_p(S).$$
Note: You have defined the tangent space $T^I_x(X)$ as an algebraic "variety".
Given any vector space $V$ of dimension $n$ over $k$, we define the "affine space of $V$" (scheme theoretically) as
$$\mathbb{A}^n(V):=Spec(Sym_k^*(V^*)).$$
If $V:=k\{e_1,..,e_n\}$ is a basis for $V$ and $x_i$ the dual basis it follows
$Sym_k^*(V^*) \cong k[x_1,..,x_n]$ hence $Spec(Sym_k^*(V^*)) \cong \mathbb{A}^n_k$ is affine $n$-space over $k$.
When your lecturer says
"The tangent space is isomorphic to be the dual space of Ia/I2a, where Ia is the unique maximal ideal in OX,a (the stalk of the regular function at a)."
he means there is an isomorphism
$$T_a^I(X) \cong Spec(Sym_k^*(I_a/I_a^2)).$$
There cannot be an "isomorphism" $T_a^I(X) \cong (I_a/I_a^2)^*$ since $T_a(X)$ is an"algebraic variety" and $(I_a/I_a^2)^*$ is a $k$-vector space.
Since the local ring $\mathcal{O}_{X,x}$ is independent of embedding you get using this construction isomorphisms (let $I_a:=\mathfrak{m}_x$)
$$T^I_x(X) \cong Spec(Sym_k^*((\mathfrak{m}_x/\mathfrak{m}_x^2)^*)) \cong \mathbb{A}^d_k \cong T^J_x(X)$$
since $\mathfrak{m}_x/\mathfrak{m}_x^2$ has dimension $d$ over $k$.
It follows the two "tangent spaces" are isomorphic.
Question: "Why isn't the invariance of the dimension obvious?"
Answer: The definition of $T^I_x(X)$ involves the ideal $I$. It is not immediate that there is an isomorphism $T^I_x(X) \cong T^J_x(X)$.
Again: Mumford's book "The red book of varieties and schemes" devotes much space to explaining definitions in algebraic geometry. There is an intuitive explanation of the definition of the tangent space and tangent cone.


Why is the additive group of rational numbers indecomposable? A group $G$ is indecomposable if $G \neq \langle e \rangle$ and $G$ cannot be written as the direct product of two of its proper subgroups. Why is the additive group of rational numbers $(\mathbb{Q},+)$ indecomposable?

I suspect, what you want to ask is why $(\mathbb{Q},+)$ is indecomposable, ie. it cannot be written as the direct sum of two subgroups.

The answer is that two non-trivial subgroups must intersect non-trivially. If $\{0\}\neq H, K < \mathbb{Q}$, then choose non-zero $p/q \in H, a/b\in K$, then
$$
qa\frac{p}{q} = ap = pb\frac{a}{b} \in H\cap K\setminus \{0\}
$$

Because there are no additive primes. Every number is the sum of two other nontrivial numbers.


Does $f_n(x) = x^{1/n}$ converge? How would you prove that $f_n(x) = x^{1/n}$ converges pointwise to $1$ for $x \in (0, \infty)$? It's obvious if you take the limit as $n \to \infty$ but how would you prove it using epsilon?

$\lim_{n\rightarrow\infty} x^{1/n}=x^{\lim_{n\rightarrow\infty} 1/n}= x^0 = 1$ for any $x>0$ (by continuity of the exponential function).

The user @mathematician already gave a proof using limit's properties, let us now prove it by the epsilon-delta definition of limit.

I will start by writing it for the particular case when $x \to \infty $:

$$\lim_{x \to \infty} f(x) = L \iff \forall_{ \epsilon > 0}\ \exists_{\delta > 0}: x > \frac{1}{\delta } \Rightarrow |f(x) - L| < \epsilon $$

Let us fix some $\epsilon > 0$. We want to show that $f(x) \to 1$ as $x \to \infty $.

The condition for the existance of $\delta $ is equivalent to finding a lower bound $b $ such that $x > b \Rightarrow |f(x) - 1| < \epsilon $.

Pausing for a moment we realize that for $x > 1, \sqrt[x]{x} > 1$ thus if $f(x) $ is to fall under a distance shorter than $\epsilon $ of 1, we must have $f(x) < 1 + \epsilon $.

Now we just rewrite our inequality:

$$\sqrt[x]{x} < 1 + \epsilon \iff x < (1 + \epsilon)^x $$

This is trivially satisfied by $x = 1$, for any $\epsilon $, but surely $b \not = 1$ for any $\epsilon $. For example, taking  $\epsilon = 0.01$, surely we don't have $2 < (1 + 0.01)^2$.

What one sees is that the inequality $x < (1 + \epsilon)^x $ is satisfiable for very small $x $, and then for very big values of $x $. This means your lower bound $b $ is the value of $x $ for which any value above satisfies given inequality.

As a small example, we set $\epsilon = 0.001$ and ask mathematica to find our answer:

meaning your lower bound is $9123.13$ or that your delta is $\delta = \frac{1}{9123.13} $


Zeros of the decimal representation of $k!$ I'd like a hint for the question:

For how many positive integers $k$ does the ordinary decimal representation of the integer $k!$ end in exactly $99$ zeros?

Thanks.

Hint. A number ends in $0$ if and only if it is a multiple of $10$. A number ends in two zeroes if and only if it is a multiple of $100=2^2\times 5^2$. A number ends in three zeroes if and only if it is a multiple of $1000 = 2^3\times 5^3$. Etc. So, e.g.,  a number ends in exactly two zeroes if and only if it is a multiple of $100$ but not of $1000$.

In order that k! end in 99 zeroes, it must have 99 factors of 10 and so 99 2's and 99 5's. Since there are going to be fewer 5's than 2's, k! must have exactly 99 5's. We first get that for k= 99*5= 495 and then get 100 5's for k= 100*5= 500. Any number from 495 to 499 will have exactly 99 5's in its factorial. There are 5 such numbers.

Actually, the answer is correct but the process has a little problem. 25=5^2 provides double 5. 125=5^3 provides triple 5. However,100 and 99 minus 3(1+2 extra 5) at the same time. Hence, it has the same answer with the solution above.


Find complex number(s) $z$ for which $|z|$ has maximum and minimum value if $|z-2+2i|=1$ Find  complex number(s) $z$ for which $|z|$ has maximum and  minimum  value if $|z-2+2i|=1$

My try: I know that $|z-2+2i|=1$ is a circle centered at $(2,-2)$
and having unit radius. Also $|z|$ is the modulus of moving point on this circle and I have to maximize and then minimize $|z|$

By using $$|z+w|>=||z|-|w||$$ I managed to get maximum and minimum values of $|z|$ and they turned out to be $2√2+1$ and $2√2-1$ respectively. But I am not able to get $z$.

I hope somebody will help me and nobody would down vote me If i sound so ignorant because I am at verge of loosing the right of asking a question in this site.

Hint:

From a geometric point of view, the values of $z$ will be the affixes of the intersections of the circle with the line joining the origin to the centre of the circle.

It will simpler to calculate with the exponential form of complex numbers: $$2-2i=\sqrt2\mathrm e^{-\tfrac{i\pi}4},\enspace\text{hence }\quad z=\dotsm$$

put z = x+ iy .

The equation will become
| (x-2) + i (y+2) | = 1

You will get a circle with equation
(x-2)^2 + (y+2)^2 = 1

Parametric points of cirlce will be :
x= 2 +- cosp
y= -2 +- sinp

Use differentiation to get maximum and minimum values of |z| .


Find the complex roots of the equation $z^6+(1+i)z^3+i=0$ I am aware that my solution is not correct. Appreciate if someone could point out what I am doing wrong.

Let $w=z^3$, then

\begin{align}
0=z^6+(1+i)z^3+i&=w^2+(1+i)w+i\\
&=\Big(w+\frac{1+i}{2}\Big)^2+\frac{1}{4}-\frac{i}{2}+i\\
\Big(w+\frac{1+i}{2}\Big)^2&=-\frac{1}{4}+\frac{i}{2}-i\\
&=-\frac{1}{4}-\frac{i}{2}.
\end{align}
Set $$w+\frac{1+i}{2}=x+yi.$$
Then $$(x+yi)^2=x^2-y^2+2xyi=-\frac{1}{4}-\frac{i}{2}$$
$$\implies \quad(1) \quad x^2-y^2=-\frac{1}{4},\quad (2)\quad 2xy=-\frac{1}{2}$$

Futhermore $$|(x+yi)|^2=\sqrt{x^2+y^2}^2=x^2+y^2=\Big|-\frac{1}{4}-\frac{i}{2}\Big|=\frac{\sqrt{5}}{4}\quad(3)$$

This gives $(x^2+y^2)+(x^2-y^2)\iff 2x^2=\frac{\sqrt{5}-1}{4}\implies x=\pm\sqrt{\frac{\sqrt{5}-1}{8}}$.

Similarly $(x^2+y^2)-(x^2-y^2)\iff 2y^2=\frac{\sqrt{5}+1}{4}\implies y=\pm\sqrt{\frac{\sqrt{5}+1}{8}}.$ ???

Let $z^3=w$.

So we need to solve that
$$w^2+(1+i)w+i=0$$ or
$$(w+1)(w+i)=0$$


What is the counterexample for "The sequence {$d(x_n,y_n)$} converges as $n\rightarrow\infty$ only if $X$ is complete."? Let ($X,d$) be a metric space and let  $<x_n>$ and $<y_n>$ be  arbitrary Cauchy sequences in $X$.Which of the following statement is true?

a.The sequence {$d(x_n,y_n)$} converges as $n\rightarrow\infty$.

b.The sequence {$d(x_n,y_n)$} converges as $n\rightarrow\infty$ only if $X$ is complete.

Solution:
the proof for part (a) is same as amsmath's answer

What is the counterexample for (b) part?

Get $\mathbb{Q}$ with the Euclidean distance. Then if $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ are arbitrary Cauchy sequences in $\mathbb{Q}$, they are also Cauchy sequences in $\mathbb{R}$, so by the completeness of $\mathbb{R}$ they have limit in $\mathbb{R}$, say to $x\in\mathbb{R}$ and to $y\in\mathbb{R}$ respectively. So:
$$d(x_n,y_n)=|x_n-y_n|\rightarrow|x-y|, n\rightarrow\infty.$$
On the other hand, $\mathbb{Q}$ with the Euclidean distance is not a complete metric space, because for example the truncated decimal expansion of $\sqrt2$ is a Cauchy sequence in $\mathbb{Q}$ that doesn't admit limit in $\mathbb{Q}$.

Take

$$X=\Bbb Q$$
$$d=| \;\; |$$

$$x_n=1+\frac 11 +\frac{1}{2!}+...+\frac{1}{n!}$$

$$y_n=2x_n$$

$$\lim_{n\to+\infty}|y_n-x_n|=e\notin \Bbb Q$$

$(x_n)$ and $(y_n)$ are Cauchy in $\Bbb Q$ but the differene $(y_n-x_n)$ is not convergent in $\Bbb Q$.


Infinitesimal Unit of Measurement This is just a question that popped into my head which I lack the knowledge to answer (or even to know whether there is an answer, honestly). Does the idea of an infinitesimal unit of measurement even make sense theoretically? If I had a line segment, could it be of infinite length when measured in such units?

The answer is affirmative. Thus, the unit interval $[0,1]$ can be partitioned into $H$ subintervals of infinitesimal length. These are called hyperfinite partitions. Wallis in fact used the symbol $\infty$ in the sense of $H$ above but today it is not customary to use it this way so I used $H$ instead (Wallis was the one who introduced the infinity symbol in the first place).

For an infinite line segment, its length is infinite inches, infinite kilometers, infinite parsecs, or infinite angstroms - all are same. Angstroms and light-years does not make any difference as infinity + infinity = infinity * infinity = infinity

As infinity is not a number but a concept, it cannot be compared with any real number. A crude definition of infinite is 'a number larger than the largest real number' is also mathematically senseless.

Einstein said nothing is infinite except the universe and human stupidity.


Possible error at definition of essential singularity Let $\Omega \subseteq \mathbb{C}$ be open and $a\in\Omega$, with $f\in\mathcal{H}(\Omega\backslash\{  a\})$.

A singularity at $a$ is called essential, iff $\forall\, w\in\mathbb{C},\ \exists$ a sequence $(z_n)_{n}$ in $\mathbb{C}$ with $\lim_{n\to\infty}z_n=a$, so that $\lim_{n\to\infty}f(z_n)=w$.

I know that there are quite a few different defintions of an essential singularity, but they all amount to the fact, that with the possible exception of a point, every complex value is adopted in an arbitrary small neighborhood around the singularity. (Great Picard theorem) At least, if I am correct.

Now how does this first definition in my post make sense? This just seems completely wrong to me. I might live with something like

"A singularity is called essential, iff $\forall\, w\in\mathbb{C},\ \exists$ a sequence $(z_n)_{n}$ in $\mathbb{C}$ with $\lim_{n\to\infty}z_n=c,\, \ c\in\mathbb{C}$, so that $\lim_{n\to\infty}f(z_n)=w$."

But if $a$ is fixed and the singularity, how can this be true?

The usual definition says that an isolated singularity that is neither removable nor a pole is called an essential singularity. (And some authors also call non-isolated singularities, e.g. branch points, essential singularities.)

Then the Casorati-Weierstraß theorem says that the isolated singularity $a$ of $f\in \mathcal{H}(\Omega\setminus \{a\})$ is essential if and only if $f(U\setminus \{a\})$ is dense in $\mathbb{C}$ for every neighbourhood $U \subset \Omega$ of $a$. [It may be stated with only the direction "$a$ essential $\implies f(U\setminus \{a\})$ dense for all $U$", but since the only possibilities for an isolated singularity are to be removable, a pole, or essential, and neither removable singularities nor poles have this property, the other direction follows.]

The definition you were given is an alternative formulation of the characterisation of essential (isolated) singularities by the Casorati-Weierstraß theorem. For, if we choose a sequence $\{U_n : n \in \mathbb{N}\}$ of neighbourhoods of $a$ in $\Omega$ with $\lvert z - a\rvert < 1/n$ for all $z \in U_n$, given an arbitrary $w\in \mathbb{C}$, by the Casorati-Weierstraß theorem there is a $z_n \in U_n\setminus \{a\}$ with $\lvert f(z_n) - w\rvert < 1/n$ for all $n\in \mathbb{N}$. Since $z_n \in U_n$ implies $\lvert z_n - a\rvert < 1/n$, we have $z_n \to a$, and by construction we also have $f(z_n) \to w$. Conversely, if for every $w \in \mathbb{C}$ there is a sequence $(z_n)$ in $\Omega \setminus \{a\}$ with $z_n \to a$ and $f(z_n) \to w$, then $f(U\setminus \{a\})$ is dense in $\mathbb{C}$ for every neighbourhood $U$ of $a$. For, given such a neighbourhood, we have $z_n \in U$ for all large enough $n$ by the definition of $z_n \to a$, and consequently $w \in \overline{f(U\setminus \{a\})}$. Since $w$ was arbitrary, that means $f(U\setminus \{a\})$ is dense in $\mathbb{C}$.

If $f\in\mathcal{H}(\Omega\backslash\{  a\})$, then $a$ is called a isolated sinfgularity of $f$.

The singularity is $a$ is called essential, if $a$ is not a pole and $a$ is not a removable singularity of $f$.

THEOREM(Casorati- Weierstraß): The singularity $a$ is essential iff $\forall\, w\in\mathbb{C},\ \exists$ a sequence $(z_n)$ in $ \Omega \setminus \{a\}$ with $z_n \to a$ and $f(z_n) \to w$.


How can I show that a sequence of regular polygons with $n$ sides becomes more and more like a circle as $n \to \infty$? If we construct regular polygons with larger and larger numbers of sides, they will look more and more like circles.
That is intuitively true.
I hope you will help me to express and prove it mathematically.

Here's one serious approach: Let $f_n\colon [0,2\pi]\to \Bbb R_+$ be the function whose graph, in polar coordinates, is the regular $n$-gon centered at the origin with a vertex at $(1,0)$. Then $(f_n)$ converges uniformly to a constant function mapping any angle to $1$, whose graph is a circle.

We can also look at the limit of the area, a la Archimedes, and the perimeter.

For Polygons, as number of sides increases,the length of each side is reducing

so at ∞, the length will be equal to a point.

Each Side can be assumed to be shortened to the vertex(ie a point)

Now, the polygon becomes a locus of points which are actually the vertices of the polygon.

However, each vertex is equidistant from the centre of polygon, so the resulting polygon

becomes a locus of points, equidistant from a common centre point, which is a circle.

After Edit:
Now, what remains, is to prove that at n->∞, length of side tends to zero, ie a point.

It can be proved by contradiction.
1) Suppose at n->∞, length of each side is not equal to a point, but is L.for n-> ∞, total    length = n*L = ∞.

Now, total length is infinite,however, polygon is a closed figure, so its total length is
obviously finite.

Thus this contradicts that length is infinite, which in turn contradicts our original supposition.

Therefore, at n-> ∞, length of each side tends to 0, ie each side becomes a point.


Area of $\{(x,y)^T\in\mathbb{R}^2\mid x^2+y^2<r^2 \wedge -|x|<y<|x|\}$ What is the area of $\{(x,y)^T\in\mathbb{R}^2\mid x^2+y^2<r^2 \wedge -|x|<y<|x|\}$?
I have that $x\in[0,r)$ but what about $y$. There I have two restrictions.

First of all x is not in [0,r] but in [-r,r].You will have to see x in 4 cases.
i)$x\in [-r,-r/\sqrt 2]$ then $y^2<r^2-x^2$
ii)$x\in [-r/\sqrt 2),0]$ then x<y<-x
iii)$x\in [0,r/\sqrt 2),]$ then -x<y<x
iv)$x\in [r/\sqrt 2),r]$ then $y^2<r^2-x^2$


$f(x,y)=4y+3x^2$ between $y=2x,y=0$ and $x=2$. To find the surface area of the portion of the surface $f(x,y)=4y+3x^2$ between $y=2x,y=0$ and $x=2$.
The limits of $y$ is clear. What is the limit of $x$? Is it from $x=0$ to $x=2$

Yes. The lines $y=0$ and $y=2x$ meet at $(0,0)$. So, the region bounded by the previous two lines and by the line $x=2$ is the triangle whose vertices are $(0,0)$, $(2,0)$, and $(2,4)$. So, your integral is$$\int_0^2\int_0^{2x}4y+3x^2\,\mathrm dy\,\mathrm dx.$$

Presumably you want the area of the surface $z=f(x,y)$, also known as the graph of $f(x,y)$. Anyway, you got the correct limits. The region is a triangle defined by the inequalities $0\le y\le 2x, 0\le x\le2$.
The surface area $A$ of the graph of $f(x,y)$ with $(x,y)$ ranging over a region $S$ is calculated by the integral
$$
A=\int_S\sqrt{1+f_x^2+f_y^2}\,dy\,dx.
$$
Here $f_x(x,y)=6x$ and $f_y(x,y)=4$, so the integrand is $\sqrt{17+36x^2}$.  This doesn't depend on $y$ at all, which indicates that it may be simpler to integrate over $y$ first.
So
$$
A=\int_{x=0}^2\int_{y=0}^{2x}\sqrt{17+36x^2}\,dy\,dx.
$$
The inner integral is
$$
\int_{y=0}^{2x}\sqrt{17+36x^2}\,dy=2x\sqrt{17+36x^2}.
$$
Leaving the outer integral to you. You are expected to spot that the outer integrand is (up to a scalar multiple) of the form $g'(x)g(x)^\alpha$ for a suitable function $g(x)$ and exponent $\alpha$.


Graph Coloring and Complete Graph If a graph is k-colorable, then does it imply that it must have a k-complete graph as it's subgraph? For example if a graph has chromatic no = 5, then is this sufficient to imply that it must have K5 as its subgraph?

Basically I am trying to solve this question of MIT 6.042, problem set 4 (2010):

  Let ($s_1$, $s_2$, ..., $s_n$) be an arbitrarily distributed sequence of the number 1, 2, ..., n − 1, n. For instance, for n = 5, one arbitrary sequence could be (5, 3, 4, 2, 1).

  Define the graph G=(V,E) as follows:

  V = {v1, v2, ..., vn}
  e = (vi, vj ) ∈ E if either:

  a. j = i + 1, for 1 ≤ i ≤ n − 1

  b. i = $s_k$, and j = $s_{k+1}$ for 1 ≤ k ≤ n−1

  Prove that this graph is 4-colorable for any (s1, s2, ..., sn).

  Hint: First show that that a line graph is 2-colorable. Note that a line graph is defined as follows: The n-node graph containing n-1 edges in sequence is known as the line graph $L_n$

My approach: Trying to prove by contradiction, I will assume that graph G is not 4-colorable, then it requires atleast 5 colours, this implies that there must be K-5 as a subgraph in G (Now here I am using a strong statement about which I am not sure). Then I will show that K-5 is not possible under the definitions of graph G, hence a contradiction.
Is this approach correct or is there a better one?

I have been working on this same problem and found this older question on this forum.

Here is the link to the pdf for the problem set:

Problem Set 4

This is question 6 (a). Here is how I think you are supposed to answer it.

First you color the vertices 1 through n with 2 colors, say blue and red, just using the first set of edges like $(v_i,v_j)$ where j=i+1, 1<=i<=n-1

So now the graph from vertex 1 to vertex n looks like this:

R-B-R-B-....

Now you have to add the edges $(v_{S_k},v_{S_k+1})$, 1<=k<=n-1, one at a time adjusting the color of $v_{S_k+1}$ so it is different from the color of the vertices that it is connected to.

But there are only three possible vertices connected to the new vertex $v_{S_k+1}$. There are at most two from the original line and then also $v_{S_k}$. The two from the original line would be labeled $v_{(S_k+1)-1}$ and $v_{(S_k+1)+1}$.

So, as you add each edge $(v_{S_k},v_{S_k+1})$ you just have to pick at most a fourth color to color $v_{S_k+1}$, a different color from $v_{S_k}$, $v_{(S_k+1)-1}$ and $v_{(S_k+1)+1}$.

Lemma1: A line graph is 2 colorable.
The proof is fairly straight forward.
Lemma2: The subgraph containing the odd vertices(v1,v3,v5,...) forms one or more non intersecting line graphs(non intersecting in the sense there are no common nodes in two line graphs)
Proof: We need to show that there doesn't exist a cycle, and we can use this along with the fact that a node can have at most 2edges in the subgraph
Lemma3: Lemma2 for the subgraph with even vertices.
Thm: The graph is 4-colorable
Pf: First take all the odd vertices. by lemma 2 we can color the odd vertices using 2 colors(say Black and White). Now using lemma 3 we can color the even vertices using 2 colors(different from odd)(say Red and Yellow). This covers all the edges from an odd to an odd vertex and even to even vertex without violating the coloring. What is left is the edges from odd to even vertex. This is clearly not violating the vertex coloring since odd vertex can have either black/white and even vertex can have either red/yellow. Hence the graph is 4-colorable.


The sum of angles in a triangle is greater than 180? So basically the question goes:
Let $ABC$ be a triangle and that $\cos A=-\frac 3 5$ and $\cos B=\frac 5 {13}$, find $\cos C$.
What I did was I get the value of $A$ and $B$ by calculating the arccosine of $-3/5$ and $5/13$ and the sum of both of these angles was already greater than $180$, how can I calculate $\cos C$?

I suspect you're meant to calculate $\cos C$ by using the cosine of sums rule, not by computing inverse cosines.

With that in mind, and on the off chance that the minus sign on $-3/5$ is an error, and the cosine of angle $A$ is really meant to be $3/5$, I'll point out

$$
\arccos 3/5 \approx 53.13 \\
\arccos 5/13 \approx 67.38
$$

which leaves $C \approx 180-53.13-67.38 = 59.49$.  And indeed, we see that (assuming the minus sign is spurious)

\begin{align}
\cos(A+B) & = \cos A \cos B - \sin A \sin B \\
          & = \frac{3}{5} \times \frac{5}{13} - \frac{4}{5} \times \frac{12}{13} \\
          & = \frac{15}{65} - \frac{48}{65} = -\frac{33}{65}
\end{align}

That would mean that $\cos C = 33/65$ (because $A+B+C = 180$), and sure enough,

$$
\arccos 33/65 \approx 59.49
$$

You need to pay attention to the conditions of the problem, not just compute mechanically.

First, the fact that the angles belong to a triangle tells us that none of them is up to $180°$. Also, that $\cos A<0$ tells us that the angle at $A$ is obtuse. Similarly, the angle at $B$ is acute. This immediately tells us that we expect $\cos C$ to be positive, since otherwise $\hat C$ would be obtuse, which contradicts the condition $A+B+C=180°$.

We can now begin the computation with these facts in mind. You can use the relationships $$\tan^2 x=\frac{1} {\cos^2 x}-1$$ and $$\tan (A+B+C)=\frac{\tan A+ \tan B +\tan C -\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C - \tan C\tan A}=\tan 180°=0\implies \tan A+ \tan B +\tan C -\tan A\tan B\tan C=0$$ to get $\tan A=-4/3$ and $\tan B=12/5$, which when you substitute in the above should give you $\tan C$, which you can put back in the first relationship to give you $\cos C$(note, take the positive square root here).

My own computation (which may be mistaken) tells me that $$\cos C=63/65>0,$$ as expected.

PS. Looking back at my work, I see that $\tan C=-16/63<0$, contrary to expectation (since $C$ must be acute) so that the angles don't belong to a triangle as claimed. Nevertheless, the $\text{cosine}$ of the angle $C$ is a valid one since the trig relations used are valid for all angles.


Validity of my proof by contradiction of converse of Pythagorean Theorem. So in, $\triangle ABC$ it is given that $AB^2=AC^2+BC^2$. Let us assume that $\angle C\neq{90}^{\circ}$. And let us make a perpendicular $AD$ to $BC$.  Now, by the Pythagorean Theorem, in $\triangle ADB$ & $\triangle ADC$, $AB^2=AD^2+BD^2$ and $AD^2+DC^2=AC^2$. Adding both, we get $$AB^2+AC^2=2AD^2+BD^2+DC^2$$
By, $AD^2=AC^2-DC^2$,
$$AB^2+AC^2=2AC^2-2DC^2+BD^2+DC^2$$
$$AB^2-AC^2=BD^2-DC^2$$
By, $BC^2=AB^2-AC^2$ and $a^2-b^2=(a-b)(a+b)$
$$BC^2=(BD-DC)(BD+DC)$$
$$BD+DC=BD-DC$$
So, $DC=0$. So it means that $C$ and $D$ are coincident points. Which contradicts the fact that $\angle C\neq90^{\circ}$. Thus, proved by contradiction.

I know this is so easy proof, but what happened is that my teacher said that this proof is not in the book, so it is wrong. He didn't give me any marks in this question. Also, I didn't remember my book's proof that day, so I made my own and wrote. But it seems alright to me. So, is there really any mistake?

Your proof is nice but incomplete, since you are assuming $\angle C<90^\circ$ (that is point $D$ falling inside $BC$). To complete it, you should also examine the $\angle C>90^\circ$ case. However, if I were your teacher I would reward your effort with a fair mark instead of dismissing it.

Your proof is wrong as you said angle $C$ is not equal to $90^{\circ}$. But
when you derived $AB^2-AC^2$, you took it as $BC^2$, and so took $C$ as $90^{\circ}$, thereby contradicting your own assumption, hence you got $DC=0$.


arrangement of the word $"\bf{MATHEMATICS}"$ in which no two same letter occur together. Total number of arrangement of the word $"\bf{MATHEMATICS}"$ in which
no two same letter occur together.

$\bf{My\; Try::}$ Here word contain $\bf{2M,2A,2T,H,E,I,C,S}$
So first we will arrange $\bf{H,E,I,C,S}$ as $\bf{5!}$ ways
$$\bf{-H-E-I-C-S-}$$
No we have $6$ gap and we can arrange $\bf{2M,2T,2A}$ as $\displaystyle \frac{6!}{2!\cdot 2!\cdot 2!}$
So total number of ways is $\displaystyle \bf{5!\times \frac{6!}{2!\cdot 2!\cdot 2!}}$
Is my solution is right, If not then how can we calculate it.
Help me
Thanks

First, let's count the number of distinguishable arrangements of the word MATHEMATICS, which has eleven letters.  We can fill two of the eleven positions with an M in $\binom{11}{2}$ ways. We can fill two of the remaining nine positions with an A in $\binom{9}{2}$ ways.  We can fill two of the remaining seven positions with a T in $\binom{7}{2}$ ways.  The five remaining letters can be permuted in $5!$ ways.  Hence, the number of arrangements of the letters of the word MATHEMATICS is
$$\binom{11}{2}\binom{9}{2}\binom{7}{2} \cdot 5! = \frac{11!}{9!2!} \cdot \frac{9!}{7!2!} \cdot \frac{7!}{5!2!} \cdot 5! = \frac{11!}{2!2!2!}$$
From these arrangements, we must exclude those in which two adjacent letters are the same.

We use the Inclusion-Exclusion Principle.

Consider those arrangements in which two adjacent letters are the same.  Suppose, for example, that the two A's are consecutive.  Place them in a box.  We now have ten objects to arrange, the other nine letters in the word MATHEMATICS and the box containing the two A's.  We can select two of the ten positions for the M's in $\binom{10}{2}$ ways, two of the remaining eight positions for the T's in $\binom{8}{2}$ ways, and arrange the other six objects in $6!$ ways.  Thus, the number of arrangements in which the two A's are adjacent is
$$\binom{10}{2}\binom{8}{2} \cdot 6! = \frac{10!}{2!2!}$$
An analogous argument applies to the two M's and two T's.  Since there are three ways of choosing the pair of adjacent letters that are the same, the number of arrangements in which two consecutive letters are the same is
$$\binom{3}{1} \cdot \frac{10!}{2!2!}$$

Next, we count those arrangements in which two adjacent letters are the same and two other adjacent letters are the same.  Suppose, for example, that the two A's are adjacent and the two M's are adjacent.  Place the A's in an amber box and the M's in a maroon box.  We now have nine objects to arrange, the two boxes, two T's, and the other five letters.  We can select two of the nine positions for the T's in $\binom{9}{2}$ ways, then arrange the remaining seven objects in $7!$ ways.  Hence, the number of arrangements in which the two A's are adjacent and the two M's are adjacent is
$$\binom{9}{2} \cdot 7! = \frac{9!}{2!}$$
Since there are $\binom{3}{2}$ ways to select two adjacent letters that are the same and two other adjacent letters that are the same, the number of arrangements of the word MATHEMATICS in which two adjacent letters are the same and two other adjacent letters are the same is
$$\binom{3}{2} \cdot \frac{9!}{2!}$$
Finally, we count those arrangements in which the two A's are adjacent, the two M's are adjacent, and the two T's are adjacent.  Place the A's in an amber box, the M's in a maroon box, and the T's in a turquoise box.  We then have eight objects to arrange, the three different color boxes and the five other letters.  They can be arranged in $8!$ ways.

Thus, by the Inclusion-Exclusion Principle, the number of distinguishable arrangements of the word MATHEMATICS in which no two adjacent letters are the same is
$$\frac{11!}{2!2!2!} - \binom{3}{1} \cdot \frac{10!}{2!2!} + \binom{3}{2} \cdot \frac{9!}{2!} - 8!$$

My thanks to Barry Cipra for making me aware of the flaws in my first attempt to solve the problem.

Hint an alternative to check the correctness do parts it is safest .
\begin{align*}
\text{total arrangements} & - \text{ways where all letters M,A,T are together with the same letters}\\
& - \text{ways where M,T are together}\\
& - \text{ways where A,T are together}\\
& - \text{ways where A,M are together}\\
& - \text{ways where M are together}\\
& - \text{ways where A are together}\\
& - \text{ways where T are together}
\end{align*}
though its not healthy and competitive approach it ensures that all cases are taken.


Line integral $F(x,y,z) = \left( 2xyz + sin(x)\right)\mathbf i + x^2z \mathbf j +x^2 y \mathbf k $ Let $F(x,y,z) = \left( 2xyz + sin(x)\right)\mathbf i + x^2z \mathbf j +$ $x^2 y \mathbf k $. Evaluate
$$ \int_C F \cdot ds$$
where C is the parametrized curve $ c(t) = \left(cos^5(t),sin^3(t),t^4\right), 0\le t \le \pi $

I tried to calculate this line integral by calculating the derivative of the c and dot product of the F and c' as the integrand. But the calculation is so complicated, so I think there might be much easier way to do this.

Observe that

$$\text{curl}\,F:=\begin{vmatrix}i&j&\;\;k\\\frac\partial{\partial x}&\frac\partial{\partial y}&\;\;\frac\partial{\partial z}\\2xyz+\sin x&x^z&\;\;x^2y\end{vmatrix}=0$$


Is $\operatorname{Spec}(\mathbb{C}[x,y]/(y-x^2))$ the same as $\operatorname{Spec}(\mathbb{C}[x])$ ? Let $R=\mathbb{C}[x,y]/(y-x^2)$. We have that $$\operatorname{Spec}(R)=\{(0),(x-a,y-a^2),(y-x^2)\}. $$
But if we consider the quotient ring $\mathbb{C}[x,y]/(y-x^2) \simeq \mathbb{C}[x]$. But $\operatorname{Spec}(\mathbb{C}[x])=\{(0),(x-a)\}$. So we have isomorphic rings that have different $\operatorname{Spec}$... How can I solve this problem?

$(x-a,y-a^2)\equiv(x-a)$ since $y-a^2\equiv x^2-a^2\equiv(x-a)(x+a)$ in the quotient.

Deleting the $(0)$ (which looks rather like a typo) from the first list, your two lists match.

There is no problem: the element $a$ in your description of $\mathrm{Spec}(R)$ can be any complex number and different $a$'s lead to different prime ideals. Hence the spectrum is infinite, as well as the spectrum of the polynomial ring.


Arc length of $f(x)=x^2-\ln x$ over [1,e] This is how I learned to solve for arc length:
$$ \frac d{dx} f(x)=2x - \frac 1x$$
$$ \left(\frac d{dx}f(x)\right)^2 = \left(2x - \frac 1x\right)^2 = 4x^2-4+\frac1{x^2} $$
$$ 1+\left(\frac d{dx}f(x)\right)^2 = 4x^2-3+\frac1{x^2} $$
$$ \sqrt{1+\left(\frac d{dx}f(x)\right)^2} = \sqrt{4x^2-3+\frac1{x^2}}$$
So to solve for arc length, I have:
$$\int_1^e \sqrt{4x^2-3+\frac1{x^2}} ~dx $$
I already have the answer (from Wolfram alpha) but I cannot figure out how to solve this integral. It seems complicated since Wolfram cannot generate the steps for this integral.

Any suggestion is highly appreciated.

We have:

$$I=\int \sqrt{4x^2-3+x^{-2}}dx$$

doing substitution $t=x^2$ we have $dx=\frac{1}{2\sqrt{t}}dt$ and

$$\int \frac{\sqrt{4t-3+\frac{1}{t}}}{2\sqrt{t}}dt=\frac 1 2 \int \sqrt{4-\frac 3 t + \frac 1 {t^2}}dt=\frac 1 2 \int \frac 1 t \sqrt{4t^2-3t+1}dt=\frac 1 2 \int \frac{4t^2-3t+1}{t\sqrt{ 4t^2-3t+1}}dt$$

using the equality $4t^2-3t+1=\frac t 2 (8t-3) -\frac 3 2 t +1$:

$$\frac 1 4 \int \frac{8t-3}{\sqrt{4t^2-3t+1}}dt+\frac 1 4 \int \frac{2-3t}{t\sqrt{4t^2-3t+1}}dt=\frac 1 2 \sqrt{4t^2-3t+1} +\underbrace{\frac 1 2 \int \frac 1 {t\sqrt{4t^2-3t+1}}dt}_1 - \underbrace{\frac 3 4 \int \frac 1 {\sqrt{4t^2-3t+1}} dt}_2$$

Integral #1

$$\frac 1 2 \int \frac {dt}{t\sqrt{4t^2-3t+1}}$$

Doing substitution $u=\frac 1 t$, $du = -\frac{1}{t^2}dt$,

$$-\frac 1 2 \int \frac {du}{u\sqrt{\frac 4 {u^2} - \frac 3 u + 1}}=-\frac 1 2 \int \frac {du}{\sqrt{u^2-3u+4}}=-\frac 1 2 \int \frac{du}{\sqrt{(u-\frac 3 2)^2+\frac 7 4}}$$

Substitution $s=u-\frac 3 2$, $ds=du$:

$$-\frac 1 2 \int \frac{ds}{\sqrt{s^2+\frac 7 4}}=-\frac 1 2 \ln\left| s+\sqrt{s^2+\frac 7 4} \right|+C_1=-\frac 1 2 \ln\left| u-\frac 3 2+\sqrt{u^2-3u+4} \right|+C_1=-\frac 1 2 \ln\left| \frac 1 t-\frac 3 2+\sqrt{\frac{1}{t^2}-\frac 3 t+4} \right|+C_1=-\frac 1 2 \ln\left| \frac 1 {x^2}-\frac 3 2+\sqrt{\frac{1}{x^4}-\frac 3 {x^2}+4} \right|+C_1$$

Integral #2

$$\frac 3 4 \int \frac 1 {\sqrt{4t^2-3t+1}} dt=\frac 3 8 \int \frac 1 {\sqrt{t^2-\frac 3 4 t+\frac 1 4}} dt=\frac 3 8 \ln\left|t-\frac 3 8 + \sqrt{t^2-\frac 3 4 t+\frac 1 4}\right|+C_2=\frac 3 8 \ln\left|x^2-\frac 3 8 + \sqrt{x^4-\frac 3 4 x^2+\frac 1 4}\right|+C_2$$

Summary

$$I=\frac 1 2 \sqrt{4x^4-3x^2+1} -\frac 1 2 \ln\left| \frac 1 {x^2}-\frac 3 2+\sqrt{\frac{1}{x^4}-\frac 3 {x^2}+4} \right|  - \frac 3 8 \ln\left|x^2-\frac 3 8 + \sqrt{x^4-\frac 3 4 x^2+\frac 1 4}\right|+C$$

Calculating exact value for bounds $[1;e]$:

$$[I]_1^e=\frac 1 2 \sqrt{4e^4-3e^2+1} -\frac 1 2 \ln\left| \frac 1 {e^2}-\frac 3 2+\sqrt{\frac{1}{e^4}-\frac 3 {e^2}+4} \right|  - \frac 3 8 \ln\left|e^2-\frac 3 8 + \sqrt{e^4-\frac 3 4 e^2+\frac 1 4}\right|-
\frac 1 2 \sqrt{4-3+1} +\frac 1 2 \ln\left| 1-\frac 3 2+\sqrt{1-3+4} \right|  + \frac 3 8 \ln\left|1-\frac 3 8 + \sqrt{1-\frac 3 4+\frac 1 4}\right|
\\=\frac 1 2 \sqrt{4e^4-3e^2+1} -\frac 1 2 \ln\left| \frac{1+\frac{3}{2}e^2+\sqrt{1-3 e^2+4e^4}}{e^2}\right|  - \frac 3 8 \ln\left|\frac{8e^2-3 + 4\sqrt{4e^4-3 e^2+1}}{8}\right|-
\frac{\sqrt{2}}{2} +\frac 1 2 \ln\left| -\frac 1 2+\sqrt{2} \right|  + \frac 3 8 \ln\left|\frac{13}{8}\right|$$

Note

In my calculation I used Euler's substitution to get the following formula:

$$\int \frac{dx}{\sqrt{x^2+A}} = \ln\left|x+\sqrt{x^2+A}\right|+C$$

I think that the statement contains a misprint. Verify if it is not $x^{2}-%
\frac{1}{8}\ln x$ instead. If it is the case then
\begin{eqnarray*}
1+\left( f^{\prime }(x)\right) ^{2} &=&1+(2x-\frac{1}{8x})^{2} \\
&=&1+(2x)^{2}-2(2x)(\frac{1}{8x})+(\frac{1}{8x})^{2} \\
&=&1+(2x)^{2}-\frac{1}{2}+(\frac{1}{x})^{2} \\
&=&(2x)^{2}+(1-\frac{1}{2})+(\frac{1}{x})^{2} \\
&=&(2x)^{2}+\frac{1}{2}+(\frac{1}{x})^{2} \\
trick\ here &\rightarrow &=(2x)^{2}+2(2x)(\frac{1}{8x})+(\frac{1}{8x})^{2} \\
&=&(2x+\frac{1}{8x})^{2}.
\end{eqnarray*}

EDIT:
Assume that $f(x)$ is $ax^{2}-b\ln x,$ and ask whether $1+(f^{\prime
}(x))^{2}$ is a perfect square.
\begin{eqnarray*}
1+\left( f^{\prime }(x)\right) ^{2} &=&1+(2ax-\frac{b}{x})^{2} \\
&=&1+(2ax)^{2}-2(2ax)(\frac{b}{x})+(\frac{b}{x})^{2} \\
&=&(2ax)^{2}+(1-4ab)+(\frac{b}{x})^{2}
\end{eqnarray*}
this is a perfect square if
\begin{equation*}
1-4ab=4ab
\end{equation*}
and in this case
\begin{eqnarray*}
1-4ab &=&4ab \\
&=&2(2ax)(\frac{b}{x})
\end{eqnarray*}
then
\begin{eqnarray*}
1+\left( f^{\prime }(x)\right) ^{2} &=&(2ax)^{2}+2(2ax)(\frac{b}{x})+(\frac{b%
}{x})^{2} \\
&=&(2ax+\frac{b}{x})^{2}.
\end{eqnarray*}
So $ab=\frac{1}{8}.$ That is why, probably for your case $a=1$ then $b$
should be $\frac{1}{8}.$ But it is also possible that $ab=\frac{1}{8}$
for other combinations. $(a=\frac{1}{2},\ b=\frac{1}{4},\ ....)$

If you browse some textbook (for calculus courses), you will find that each
related exercise is constructed in this way. I mean each time $(1+\left(
f^{\prime }(x)\right) ^{2})$ is a perfect square!


Transpose of inverse vs inverse of transpose I can't seem to find the answer to this using Google.

Is the transpose of the inverse of a square matrix the same as the inverse of the transpose of that same matrix?

Is $(A^{-1})^T = (A^T)^{-1}$ you ask.

Well
$$\begin{align}
A^T(A^{-1})^T =  (A^{-1}A)^{T} = I^T = I \\
(A^{-1})^TA^T =  (AA^{-1})^{T} = I^T = I
\end{align}
$$

This proves that the inverse of $A^T$ is $(A^{-1})^T$. So the answer to your question is yes.

Here I have used that
$$
A^TB^T = (BA)^T.
$$
And we have used that the inverse of a matrix $A$ is exactly (by definition) the matrix $B$ such that $AB = BA = I$.

Let the inverse of $A$ be $B^T$ so that $B$ is the transpose of the inverse of$A$. Then by definition of the inverse

$$AB^{T}=I=B^{T}A.$$

Now taking the transpose

$$BA^T=I=A^TB,$$

which proves that $B$ is also the inverse of the transpose of $A$.


How to interpret the square root of inner product over an arbitrary field? In an inner product space, the norm $\|x\| = \sqrt{\langle x,x\rangle}$ is induced. I realise I have almost always dealt with either real or complex numbers, so took the square root for granted.
Reading the Inner Product entry on
Wolfram (https://mathworld.wolfram.com/InnerProduct.html), it says "A vector space together with an inner product on it is called an inner product space. This definition also applies to an abstract vector space over any field." This was said after introducing the axioms in the context of a space over $\mathbb{R}$.
I am confused because I do not know how to interpret the square root in an arbitrary field. I assume the obvious way would be to define it to be the element $a \in \mathbb{F}$ such that $a^2 = \langle x,x\rangle$. But the issue I am having is how we even know whether such an element exists in the field? Is this a standard result from ring theory?
My understanding has always been that inner product spaces (and normed spaces) are only defined over either the real or complex numbers. How do you construct them (or something equivalent) over some arbitrary field?

The claim doesn't make sense. On vector spaces $V$ over arbitrary field $k$ we have bilinear forms $b(x,y)$. When $k=\Bbb{C}$ we also look at sesquilinear forms, which means that the second argument is linear after applying an automorphism $\sigma$ of the field (the complex conjugaison). But then we can consider $V$ as a $k^\sigma$ vector space to make it linear, so assume that $b$ is truly linear.
$q(x) = b(x,x)$ is a quadratic form.
A first desirable property is that $b(x,y)=b(y,x)$ (when $char(k)\ne 2$ there is a one-to-one correspondence between quadratic forms and symmetric bilinear forms).
A second one is that $q(x)=0$ iff $x=0$. In that case $q$ is said anisotropic.
When $k$ is an ordered field there is a 3rd one: that $\forall x,q(x)\ge 0$. With the previous ones this is the definition of "$b$ is an inner product". When it is the case then $\|x\|=\sqrt{q(x)}$ is some kind of norm (when $k$ is not a subfield of $\Bbb{R}$ then $\|x\|$ is not real valued so this is a bit different). Do you think we always have $\|x+y\| \le \|x\|+\|y\|$ ?
$\sqrt{q(x)}$ is an element of the algebraic extension of $k$ obtained by adding all the square roots of elements $\ge 0$, it is ordered too, through $\sqrt{a}\ge \sqrt{b}$ iff $a\ge b$, then applying the law of orders.
Note that real valued norms exist other any fields, for example $\|x\| = 0$ if $x_1=x_2=0$ and $=1$ otherwise is a real-valued norm over $k^2$ for any field, a norm for the trivial absolute value  $|a|_{tr}= 0$ if $a=0$ and $=1$ otherwise, such that $\|ax\|=|a|_{tr} \|x\|$.

You do not have a norm defined for an arbitrary inner product space. For instance,  $\langle x, y\rangle = x_1 y_1 - x_2 y_2$ is an inner product on $\mathbb{R}^2$.
The norm will be defined on euclidian spaces or hermitian spaces. There are extra conditions ( positivity) required.


How to solve integration elegantly using contour integration Solve this integral analytically and prove the answer:

$$
\int_0^{\pi/2}\frac{d\psi}{1-\cos\theta \cdot \cos\psi} = \frac{\pi - \theta}{sin \theta}, \quad \theta \in (0,2\pi)
$$
To see the pic

I have already proved it numerically using Mathematica. But still can't prove it analytically.

This is a standard integral:

Let $\cos \theta = a$ (and $-1\leq a \leq 1$):

$$\int\limits_{\psi = 0}^{\pi/2} \frac{1}{1 - a \cos \psi}\ d\psi =\frac{2 \tan ^{-1}\left(\frac{a+1}{\sqrt{1-a^2}}\right)}{\sqrt{1-a^2}}$$


Square root of radicals Find the square root of $4^{1/3}+16^{1/3}+1$.
I tried to solve by supposing the square root to be $x$ and then cubing both sides but it didn't work.
I do not need exact value.
By hit and trial
I have seen that answer should involve $2^{1/3}+1....$
Thanks

I will assume you are interested only in real numbers and not complex numbers.  If you wish to simplify so that there are no nested radicals:

First express inner radicals as exponentials terms of the same base

$\sqrt{1+\sqrt[3]{4}+\sqrt[3]{16}}=\sqrt{1+2^{2/3}+2^{4/3}}=\sqrt{1+2\cdot 2^{1/3}+2^{2/3}}$

Above, we used that $2^{4/3}=2^{3/3}\cdot 2^{1/3}$.

Now, let $x=2^{1/3}$.  We recognize that the above can be written as

$\sqrt{1+2x+x^2}=\sqrt{(1+x)^2}=|1+x|$.  Remembering that $\sqrt[3]{2}$ is a positive real, we may replace this back in for $x$ and remove the absolute value sign to get that the original expression is equal to $1+\sqrt[3]{2}$


Finding solutions to $2^x+17=y^2$ Find all positive integer solutions $(x,y)$ of the following equation:
  $$2^x+17=y^2.$$

If $x = 2k$, then we can rewrite the equation as $(y - 2^k)(y + 2^k) = 17$, so the factors must be $1$ and $17$, and we must have $x = 6, y = 9$.

However, this approach doesn't work when $x$ is odd.

The only solutions are $x = 3, 5, 6, 9$. This is proved on pp. 148-152 of

  Tzanakis, N. (1983), On the diophantine equation $y^2-D = 2^k$, J. Number Theory 17, 144-64.

I think that there are not solutions for $x$ odd.

Indeed,

$y^2=2^x+17=2^{2k+1}+17=2 \cdot 4^k +17 \equiv 2+17 \equiv3 \mod4$, which cannot happen.

(If $x$ is of the form $x=2k$, then the equation can be easily written $(y-2^k)(y+2^k)=17$, and so we go on).

EDIT: Wrong solution!


Need help with integral and change the variable I have

$$\displaystyle \int_0^\pi \sqrt{x} ~\cos x ~dx$$

and I need to make change of the variable $u = \sin x$.

Write x as a function of u: you have $u=\sin(x)$ so $x=\arcsin(u)$. It gets a little sticky when you have to simplify $\cos(\arcsin(u))$; remember the identity $\sin^2\theta+\cos^2\theta=1$, letting $\theta=\arcsin u$ (arcsin gives you an angle, recall) you get $\cos(\arcsin(u))=\sqrt{1-u^2}$.

Actually, this integral doesn't have a closed form; are you sure the problem wasn't $\int \sqrt{x} \cos^2 x dx$


What's the integration of $\int \sin^4 (x)dx$? What's the integration of $\int \sin^4 x \,dx$? I don't see the approach to this question.

I have a issue with this question as well:
$$\int \sin x \cos x (\sin x+\cos x) \,dx.$$
I simplify this to $\sin^2 x \cos x + \sin x \cos^2 x$
and set $u= \sin x$, $du = \cos x \,dx$.
So I got
$$\int u^2 \,du + ???.$$
I don't get the second part now

Recall the identities
$$\sin^2(\theta) = \dfrac{1-\cos(2\theta)}2$$
and
$$\cos^2(\theta) = \dfrac{1+\cos(2\theta)}2$$
Hence,
$$\sin^4(x) = \left(\dfrac{1-\cos(2x)}2 \right)^2 = \dfrac{1 - 2 \cos(2x) + \cos^2(2x)}4 = \dfrac{1 - 2 \cos(2x) + \dfrac{1+\cos(4x)}2}4$$
Hence,
$$\sin^4(x) = \dfrac{3-4\cos(2x) + \cos(4x)}8$$
Now you should be able to integrate this term by term and obtain an answer.

For the second part, note that $$\int \sin^2(x) \cos(x) dx = \int \sin^2(x) d(\sin(x))$$ and $$\int \cos^2(x) \sin(x) dx = -\int \cos^2(x) d(\cos(x))$$

Here's how I solve problems like this. I find this far, far easier to remember than the methods presented in pretty much all textbooks. It's a bit more work, but there are no tricks, no cleverness, and it has the advantage that it always works.

Let $p$ and $q$ be integers.  Then:

$$\frac{d}{dx}(\sin^p x\, \cos^q x) = p\,\sin^{p-1} x\,\cos^{q+1} x - q\,\sin^{p+1} x\,\cos^{q-1} x$$

What I want you to notice is that the derivative of $\sin^p x\, \cos^q x$ is a linear combination of terms of the form $\sin^r x\, \cos^s x$, where $r+s=p+q$. Moreover, if $p$ is odd, then all of the $r$ are even, and vice versa, and the same relationship holds between $q$ and $s$.

It stands to reason that the opposite should happen if you integrate. So, for example, if $p$ and $q$ are both odd, then:

$$\int \sin^p x\, \cos^q x\,dx = \sum\limits_{i=0}^{(p+q)/2}\,\alpha_{i} \sin^{2i} x\,\cos^{p+q-2i} x + C$$

where $\alpha_i$ are constants.

Unfortunately, there is one gotcha. If $p$ and $q$ are both even, then you run into the problem that:

$$\sin^2 x + \cos^2 x = 1$$

This is Pythagoras' theorem, and it's one of exactly two trigonometric identities that you need memorise. (What's the other one and why do you only need two? Long story, and this post is getting long enough as it is.)

So if $p$ and $q$ are both even:

$$\int \sin^p x\, \cos^q x\,dx = \sum\limits_{i=0}^{(p+q)/2}\,\alpha_{i} \sin^{2i} x\,\cos^{p+q-2i} x + \beta x + C$$

I'll let you work out the case where one of $p,q$ is odd and the other is even. You don't have to come up with a general formula, just work out what will happen qualitatively.

To integrate $\int \sin^p x\, \cos^q x\,dx$, therefore, use the most general form including the unknown constants, take the derivative of both sides, simplify, match up coefficients, and then solve for the unknowns.

So for your first example, we would expect:

$$\int \sin^4 x\, \cos^0 x\,dx = \alpha_0 \sin x\, \cos^3 x + \alpha_1 \sin x\, \cos^3 x + \beta x + C$$

To handle $\beta$, we use the fact that:

$$1 = \sin^2 x + \cos^2 x$$

So taking the derivative of both sides:

$$\sin^4 x = \alpha_0 \frac{d}{dx}\sin x\, \cos^3 x + \alpha_1 \frac{d}{dx}\sin^3 x\, \cos x + \beta (\sin^2 x + \cos^2 x)^2$$

That is:

$$\sin^4 x = \alpha_0 (\cos^4 x - 3 \sin^2 x\,\cos^2 x) + \alpha_1 (3 \sin^2 x\,\cos^2 x - \sin^4 x) + \beta(\sin^4 x + 2 \sin^2 x\,\cos^2 x + \cos^4 x)$$

Rearranging gives:

$$\sin^4 x = (\alpha_0 + \beta) \cos^4 x + (- 3 \alpha_0 + 3 \alpha_1 + 2 \beta) \sin^2 x\,\cos^2 x + (- \alpha_1 + \beta) \sin^4 x$$

Matching up coefficients gives the system of linear equations:

$$\alpha_0 + \beta = 0$$
$$-3 \alpha_0 + 3 \alpha_1 + 2 \beta = 0$$
$$-\alpha_1 + \beta = 1$$

There are three equations and three unknowns. (Actually there are four unknowns, but $C$ is unconstrained.) Solve using your favourite method to find:

$$\alpha_0 = -3/2$$
$$\alpha_1 = 1/2$$
$$\beta = 3/2$$

That is:

$$\int \sin^4 x\,dx = -\frac{3}{2} \sin x\, \cos^3 x + \frac{1}{2} \sin^3 x\, \cos x + \frac{3}{2} x + C$$

As always, take the derivative to check your answer. (Disclaimer: I haven't done this.)

Note that this method is also slightly self-correcting, in that if you incorrectly guessed the form of the integral, you'll end up with a system of equations which you can't solve.

For your second question, by the way, you can superimpose the two integrals:

$$\int \sin^3 x\,\cos x + \sin x\,\cos^3 x\,dx = \alpha_0 \sin^4 x + \alpha_1 \sin^2 x\,\cos^2 x + \alpha_2 \cos^4 x + C$$

I haven't solved this one (left as an exercise), however in this case, I happen to know that you'll end up with a linear system without a unique solution. I know this because $C$ is a constant, and hence it's a multiple of $(\sin^2 x + \cos^2 x)^2$. You can either pick any solution, or simplify your algebra by setting one of the $\alpha_i$'s to zero before you start. In this case, setting $\alpha_1=0$ gives the answer more or less directly. There's a certain pretty symmetry here with the other case; previously, you have to add a term, and here you remove one.

As a general comment, integration is a great deal more systematic than modern textbooks would have you believe. If you're more comfortable deriving than memorising formulas (as I am), many problems which seem difficult can be solved in a straightforward way stepping back and looking at what happens if you take the derivative of functions of that form. Chances are pretty good that if you integrate, the opposite will happen.


Difference between two phase method and big M I am searching for differences between two phases and big M method for finding the solution for a linear problem with simplex method.
Also I realized that two phases method is algebraically more easier than big M method and as you see here, the two phase method breaks off big M function in two parts, first the real coefficients and second coefficients the the M's amount.
But I cant find any right answer for my question:" Why two phases method is algebraically more easier than big M  method?" I have just an intuitive answer and not any thing else...

Thanks for your help.

Big-M Method

The Big-M method of handling instances with artificial variables is the “commonsense approach”. Essentially, the notion is to make the artificial variables, through their coefficients in the objective function, so costly or unprofitable that any feasible solution to the real problem would be preferred, unless the original instance possessed no feasible solutions at all. But this means that we need to assign, in the objective function, coefficients to the artificial variables that are either very small (maximization problem) or very large (minimization problem); whatever this value,let us call it Big M.
In fact, this notion is an old trick in optimization in general; we simply associate a penalty value with variables that we do not want to be part of an ultimate solution(unless such an outcome is unavoidable). Indeed, the penalty is so costly that unless any of the  respective variables' inclusion is warranted algorithmically, such variables will never be part of any feasible solution. This method removes artificial variables from the basis. Here,  we assign a large undesirable (unacceptable penalty) coefficients to artificial variables from the objective function point of view. If the objective function (Z) is to be minimized, then a very large positive price (penalty, M) is assigned to each artificial variable and if Z is to be minimized, then a very large negative price is to be assigned. The penalty will be designated by +M for minimization problem and by –M for a maximization problem and also M>0.

Two-Phase Method
This method differs from Simplex method that first it is necessary to accomplish an auxiliary problem that has to minimize the sum of artificial variables. Once this first problem is resolved and reorganizing the final board, we start with the second phase, that consists in making a normal Simplex. Steps to solve a problem using two-phase simplex method:
n   Step 1
Modify the constraints so that the right-hand side of each constraint is nonnegative. This requires that each constraint with a negative right-hand side be multiplied through by -1.
n  Step 2
Identify each constraint that is now an = or ≥ constraint. In step 3, we will add an the artificial variable to each of these constraints.
n  Step 3
Convert each inequality constraint to standard form.
For ≤ constraint i, we add a slack variable si;
For ≥ constraint i, we add an excess variable ei;
n  Step 4
For now, ignore the original LP’s objective function. Instead solve an LP whose objective function is min w’= (sum of all the artificial variables). This is called the Phase I LP. The act of solving the phase I LP will force the artificial variables to be zero.
Since each ai≥0, solving the Phase I LP will result in one of the following three cases:
n  Case 1
The optimal value of w’ is greater than zero. In this case, the original LP has no feasible solution.
n  Case 2
The optimal value of w’ is equal to zero, and no artificial variables are in the optimal Phase I basis. In this case, we drop all columns in the optimal Phase I tableau that correspond to the artificial variables. We now combine the original objective function with the constraints from the optimal Phase I tableau. This yields the Phase II LP. The optimal solution to the Phase II LP is the optimal solution to the original LP.


If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$. If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.

I feel that the problem basically uses algebraic manipulation even though it's in a Number Theory textbook. I don't realize how to show $(a^2+b^2+c^2)^2$ as the sum of three squares. I have tried algebraic manipulation but this is the stage I have reached.

$$(b^2 + c^2)^2 + a^2(a^2 + b^2 + c^2 + b^2 + c^2)$$

Could you give me some hints on how to proceed with this question?

Thanks.

A theoretical and easy proof for $n$ odd.

According to a theorem of Gauss the positive integer $n$ is a sum of three squares if and only if $n$ is not of the form $4^a(8b-1)$; $a,b\in \mathbb Z$.

If $n^2=8B-1$ then $n^2\equiv -1 \pmod8$. This is not possible because $-1$ is not a square modulo $8$.

Thus the odd integer $n^2$ is always a sum of three squares, in particular when $n$ is itself a sum of three squares.

It should not be very hard the proof for n even, but I do not try because the constructive and nice proof of @individ above.


Can an integer be decomposed as $\sum_{k=1}^{m}k^3$? (and how to find $m$?) I have two questions:

I wonder how to check if an integer $n$ can be represented as:

$$n = \sum_{k=1}^{m}k^3$$

And if it can be decomposed this way, how to find $m$?

I tried to use an integral approximation like this: $k = \sqrt[3]{4x}$, but it is not correct, because I need an accurate (descrete) answer.

Since
$$\sum\limits_{k=0}^m k^3=\left(\frac{m(m+1)}{2}\right) ^2$$
(you can easily prove this by induction), $n$ necessarily has to be a perfect square $i^2$ because $\frac{m(m+1)}{2}\in\mathbb{N}$. Then we have
$$m(m+1)=2i\Leftrightarrow m^2+m-2i=0.$$
If this equation has a positive integer root $m_1$, then
$$\sum\limits_{k=0}^{m_1} k^3=n.$$
If such a root does not exist, the number $n$ cannot be represented in the desired way.

$n$ must be the square of a triangular number.

I.e. it must be a perfect square,

$$(\lfloor\sqrt{n}\rfloor)^2=n,$$

and $$(m(m+1))^2=4n,$$

where

$$m=\lfloor\sqrt[4]{4n}\rfloor.$$


Infinite Union/Intersection vs Infinite summation I have a question for you.

It's well known that I can't sum an infinite number of "number". Indeed, the series
$$
\sum_{k=1}^\infty a_k
$$
is defined as a Limit.

However, I can make the infinite union/intersection of sets. Someone can explain me why ?

Take this as the definition of infinite unions and intersections:

$x \in A_1 \cup A_2 \cup A_3 \dots $ if and only if $x \in A_i$ for at least one $i \in \mathbb N$.

$x \in A_1 \cap A_2 \cap A_3 \dots $ if and only if $x \in A_i$ for all $i \in \mathbb N$.

This makes perfect sense as it is. There is no need for a limiting procedure.

(In fact, there is even no need to assume that there are countably many $A_i$'s...)

The reason that sum of a series is different is that we define it inductively. Say we're trying to work out
$$ a_1 + a_2 + a_3 + a_4\dots $$
Well, let $S_i$ be the sum of the first $i$ terms. Then you define
$$ S_1 = a_1, \ \ \ \ S_2 = S_1+ a_2, \ \ \ \ S_3 = S_2 + a_3, \ \ \ \ S_4 = S_3 + a_4, \dots $$

But the problem with this is that, no matter how many steps you do, you will never reach $S_\infty$. This is why summing infinite series requires a special procedure.

The countable intersection and union are also limits. $$\bigcup_{n \in \mathbb{N}} A_n = \lim_{n \to \infty} \bigcup_{i=1}^n A_n$$

Set operations can be defined without reference to the size of the domain of "summation": $$\bigcup A := \{ x: x \in X \in A \}$$
$$\bigcap A := \{x : (\forall X \in A)(x \in x) \}$$
(no countability needed).

Note that infinite sums are only really defined when the domain of summation is countable (see this question). So infinite sums are really very restricted objects compared to sets.


Morphism of quasi affine varieties giving algebra homomorphism and back. $\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\vp}{\varphi}$ $\newcommand{\mr}{\mathscr}$ $\newcommand{\mc}{\mathcal}$
$\newcommand{\A}{\mathbb A}$
I am new to algebraic geometry and in the following I ask two related questions that have come to my mind given the material I have read so far.
I believe the notation and terminology I have used is standard.
However, I have supplied most of the relevant definitions and notations in the appendix.
Please feel free to correct and inaccuracies.
In what follows $k$ is an algebraically closed field.
Question 1
Let $X$ and $Y$ be affine varieties.
It is a fact that a morphism $\vp:X\to Y$ is an isomorphism if the corresponding map $\vp^*:\mr O(Y)\to \mr O(X)$ is an isomorphism of $k$-algebras.

Question.
Can we extend this fact to the case when $X$ and $Y$ are quasi affine varieties, that is, is a morphism $\vp:X\to Y$ an isomorphism if and only if the corresponding map $\vp^*:\mr O(Y)\to \mr O(X)$ is an isomorphism of $k$-algebras?

The problem I am facing in proving this is the following.
When $X$ and $Y$ are affine varieties in $\A^m$ and $\A^n$ respectively, then we have $\mr O(X) = k[x_1, \ldots, x_m]/\mr I(X)$ and $\mr O(Y) = k[y_1, \ldots, y_n]/\mr I(Y)$.
Now once we have a $k$-algebra homomorphism $\mr O(Y)\to \mr O(X)$, we get a morphism of varieties $X\to Y$ by using the induced map $\text{MaxSpec}(\mr O(X))\to \text{MaxSpec}(\mr O(Y))$ and then use the nullstellensatz to get a map $X\to Y$.
However, when $X$ and $Y$ are quasi affine varieties in $\A^m$ and $\A^n$ respectively, then one cannot think of $\mr O(X)$ or as a quotient of $k[x_1, \ldots, x_m]$ (and similarly for $\mr O(Y)$).
For instance, let $X=\A^1\setminus \set{0}$. Then $\mr O(X) = k[x]_{x}$ which is not a quotient of $k[x]$ but of $k[x, y]$.
So I am unable to identify $\text{MaxSpec}(\mr O(X))$ and $\text{MaxSpec}(\mr O(Y))$ as subsets of $\A^m$ and $\A^n$.
Question 2
Let $\mc V$ be the category of affine varieties and $\mc A$ be the category of finitely generated $k$-algebras that are integral domains.
We get an equivalence between two categories by associating to an affine variety its coordinate ring and to a morphism its corresponding map between the algebras, namely the pullback map.

Question.
Is there such a statement for the category of quasi affine varieties, that is, is there a nice algebraic category to which the category of quasi affine varieties equivalent?

Question.
Suppose we consider the category of subsets of $\A^n$, $n\geq 0$.
We can still define $\mr O(X)$ for any such subset and form a category with morphisms as maps which pullback regular functions to regular functions.
Is this category useful in any way? Is there a reason why it is not studied?

Appendix
Let $k$ be an algebraically closed field.
We write $\A^n$ to denote $k^n$ with the Zariski topology.
An affine variety is an irreducible closed subset of some $\A^n$.
Given an affine variety $X$ in $\A^n$, the coordinate ring $\mr A(X)$ of $X$ is defined as $k[x_1, \ldots, x_n]/\mr I(X)$, where $\mr I(X)$ is the set of all the polynomials in $k[x_1, \ldots, x_n]$ which vanish on each point of $X$.
It turns out that $\mr I(X)$ is a prime ideal.
A quasi affine variety is any open subset of an affine variety.
A regular function on a quasi affine variety $Y\subseteq \A^n$ is a map $\vp:Y\to k$ which can be locally written as a quotient of polynomials in $k[x_1, \ldots, x_n]$.
The set of all the regular functions on $Y$ is denoted by $\mr O(Y)$.
It turns out that $\mr O(Y)$ has a natural $k$-algebra structure.
It is a fact that if $X$ is an affine variety, then $\mr A(X)$ can be naturally identified with $\mr O(X)$.
A morphism of quasi affine varieties $X$ and $Y$ is a map $\vp:X\to Y$ which is continuous (in the Zariski topology) such that $\vp$ pulls back regular functions on $Y$ to regular functions on $X$.
Given a morphism $\vp:X\to Y$, we get the pullback map $\vp^*:\mr O(Y)\to \mr O(X)$, and it is easy to check that $\vp^*$ is a $k$-algebra homomorphism.

The problem with quasi-affine varieties is that, well, they are not affine. You should think of an affine variety as one that is ``determined by its global regular functions'' and this is not the case for arbitrary quasi-affine varieties.
Here is a classic example: let $X = \mathbb{A}^2 \setminus \{ (0,0) \}$. Then I claim $\mathcal{O}(X) \cong k[x,y]$ where the map $\mathcal{O}(\mathbb{A}^2) \to \mathcal{O}(X)$ induced by the inclusion $X \subset \mathbb{A}^2$ is an isomorphism. I highly recommend you try to prove this if you are not familiar with the example. It is often thought of as an algebraic version of the Harthogs's phenomenon.
This example shows that the answer to Question 1 is no. In Question 2, I don't know of any purely algebraic description and the category you suggest is useful it just does not give information about the quasi-affine varieties $X$ rather it tells you about the affine varieties $\mathrm{Spec}{(\mathcal{O}(X))}$.
Note: when you say subsets, I assume you mean locally closed subsets for the Zariski topology since an arbitrary subset of $\mathbb{A}^n$ need not have the structure of an algebraic variety.

"Question. Can we extend this fact to the case when $X$ and $Y$ are quasi affine varieties, that is, is a morphism $φ:X→Y$ an isomorphism if and only if the corresponding map $φ∗:O(Y)→O(X)$ is an isomorphism of k-algebras?"
Example 1. If $\phi: \mathbb{P}^1_k \rightarrow \mathbb{P}^d_k$ is the $d$-uple embedding with $d\geq 2$, it follows $\phi$ is not an isomorphism. But
$$\Gamma(\mathbb{P}^1_k, \mathcal{O}_{\mathbb{P}^1_k}) \cong \Gamma(\mathbb{P}^d_k, \mathcal{O}_{\mathbb{P}^d_k}) \cong k$$
Hence the global sections are isomorphic, but the varieties are not (a projective variety is quasi projective). Hence it seems the correspondence you mention can not be extended to quasi projective varieties. You find this construction in Hartshornes book, chapter I and II. The projective varieties $\mathbb{P}^1_k$ and $\mathbb{P}^d_k$ have different dimensions, but their structure sheaves $\mathcal{O}_{\mathbb{P}^1_k}$ and $\mathcal{O}_{\mathbb{P}^d_k}$ have the same number of global sections. In chapter I you find the definition of the $d$-uple embedding and in chapter II you find the definition of the structure sheaf.
Example 2. If $k$ is algebraically closed, $X:=Spec(k[x,y])$ and $U:=X-\{(0,0)\}$ it follows there is an isomorphism
$$i^{\#}: \Gamma(X, \mathcal{O}_X) \cong \Gamma(U, \mathcal{O}_U)\cong k[x,y]$$
(Hartshorne Exercise I.3.6) but $X$ and $U$ are not isomorphic. Hence the correspondence seems not to hold for quasi affine varieties either.
The map $i^{\#}$ is induced by the inclusion
$$ i: U \rightarrow X.$$
The scheme $U$ is not affine.
You may change the  notion "regular map" - this is discussed briefly here:
Regular rational functions over real projective lines and projective spaces.
Example 3. If $k:=\mathbb{R}$ is the field of real numbers and $f:=x^2+y^2$
it follows $Z:=V(f):=Spec(A)$ where $A:=k[x,y]/(f)$. $Z$ has $(0,0)$ as the only rational point: $x^2+y^2=0$ iff $(x,y)=(0,0)$. Why? Let
$$\phi: A \rightarrow k$$
be a map of $k$-algebras.It follows $\phi(x)=a, \phi(y)=b$ for $a,b\in k$.
Since the map $\phi$ is well defined it follows $\phi(x^2+y^2)=a^2+b^2=0$, and from this it follows $a=b=0$. Hence the only $k$-rational points of $Z$ is the point $(0,0)$. Hence we may claim that $Z=\{(0,0)\}$ is the "one point space"
with only rational point $(0,0)$. It is defined as closed subscheme of $X:=Spec(k[x,y])$ corresponding to the ideal $I:=(f)$, hence $Z$ is indeed a hypersurface in $X$.
Let $U:=X-Z:=X-\{(0,0)\}$. Since $U$ is the complement of the "hypersurface" $Z$, it follows $U\cong D(f)\cong Spec(B)$ where
$$B:=k[x,y,t]/(tf-1)$$
hence $U$ is affine. Hence over a non-algebraically closed field $k$ it follows $U:=\mathbb{A}^2_k-\{(0,0)\}$ may be an affine scheme.
Note: If $\mathfrak{m}:=(x,y)$ and $P:=Spec(k[x,y]/\mathfrak{m})$ we may realize $V(\mathfrak{m})$ as a closed subscheme of $Z$: There is a canonical surjective map
$$ A \rightarrow k[x,y]/\mathfrak{m}\cong  k.$$
Hence the hypersurface $Z$ is the rational point $(0,0)$ with a non-trivial scheme structure.
Example 4. This generalize as follows. Let $k$ be the field of real numbers and let $J:=(f_1,..,f_k)\subseteq A:=k[x_1,..,x_n]$ be an ideal generated by the polynomials $f_i$. Let
$$F(x_1,..,x_n):=f_1^2+\cdots +f_k^2 \in A.$$
It follows $F$ defines a hypersurface. Let $B_1:=A/J, B_2:=A/(F)$ and let $X_i:=Spec(B_i)$ for $i=1,2$.
Lemma: The two affine schemes $X_1,X_2$ have the same $k$-rational points.
Proof: Assume $\phi: B_1\rightarrow k$ is a map of $k$-algebras. It follows
$$ \phi(x_i)=a_i$$
for $i=1,..,n$ and since $\phi$ is well defined it follows $f_j(a_1,..,a_n)=0$
for all $j=1,..,k$. Hence the point $a:=(a_1,..,a_n)$ is a zero of the polynomials $f_j$. It follows
$$F(a_1,..,a_n):= \sum_j f_j(a_1,..,a_n)^2=0.$$
Conversely if $\phi: B_2\rightarrow k$ is a map of $k$-algebras with $\phi(x_i):=a_i\in k$ it follows
$$F(a_1,..,a_n)=0.$$
It follows
$$F(a_1,..,a_n):= \sum_j f_j(a_1,..,a_n)^2=0$$
and since $f_j(a_1,..,a_n)^2\geq 0$ it follows $f_j(a_1,..,a_n)=0$ for all $j=1,..,k$. This sets up a 1-1 correspondence between $k$-rational points of $X_1$ and $k$-rational points of $X_2$. QED.
Note: $X_1$ and $X_2$ have the same rational points but are not isomorphic as affine schemes in general. Over the field of real numbers, any "real affine algebraic variety" may set-theoretically be realized as the zero set of a single polynomial.
Since there is an inclusion of ideals $(F) \subseteq J$, we get a canonical map of affine schemes
$$\phi: X_2\rightarrow X_1$$
where $\phi$ induce a bijection $X_2(k)\cong X_1(k)$ at the level of $k$-rational points, but $\phi$ is not an isomorphism of schemes. Moreover the hypersurface $(F)$ depends on the choice of generators of the ideal $J$. If $X_i$ are "smooth algebraic varieties" we get an induced isomorphism of real smooth manifolds $(X_1)_k \cong (X_2)_k$.


Dual of each of these compound propositions. I'm doing my Discrete Structure assignment. The question asked to find the dual of each of these compound propositions. Note that T denotes a tautology and F denotes a contradiction. The questions are:

a) $p∧¬q∧¬r$

b) $(p∧q∧r)∨s$

c) $(p∨F)∧(q∨T)$

My answers:

a) $¬p∨q∨r$

b) $(¬p∨¬q∨¬r)∧¬s$

c) $(¬p∧T)∨(¬q∧F)$

Did I got it right?
Thank you!

What you have done is wrong. Why you have negated all the proposition and the then find the dual?

Like dual of (p ∧ ¬q) is (p ∨ ¬q) not (¬p ∨ q).

Here is the definition of dual of a compound proposition-
"The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T
by F, and each F by T. The dual of s is denoted by s∗."
(Reference:  Discrete Mathematics (7th Edition) Kenneth H. Rosen)

Example:
S  =(p ∧ q)∨ (¬p ∨ q)∨ F
          S* =(p ∨ q)∧ (¬p ∧ q)∧T  (dual of S is denoted by S*)

So the correct answer to your question is:

a) p∧¬q∧¬r

b) (p∧q∧r)∨s

c) (p∨F)∧(q∨T)

answers:

a) p∨¬q∨¬r

b) (p∨q∨r)∧s

c) (p∧T)∨(q∧F)

Yes.   All three check: okay.

You've negated all the propositions and dualled all the operators.

That is what you needed to do.


Continuity On $\mathbb{Q}$ I need a continuous function $f:\mathbb{Q}\rightarrow \mathbb{R}$ and discontinuous $g:\mathbb{R}\rightarrow \mathbb{R}$
s.t $f(x)=g(x)$ for all rational $x$ s. So if I say $f(x)=0$ and $g(x)=0$ for $x \in \mathbb{R}\setminus\{\sqrt2\}$ and $g(x)=1$ at $x=\sqrt 2$ .would I be right?

Yes, that will work.

More interesting examples would be

$$ g(x) =\begin{cases} 1 & x\ge \sqrt 2 \\ 0 & x < \sqrt 2\end{cases} $$

or

$$ g(x) = \begin{cases} 1 & x\notin\mathbb Q \\ 0 & x \in \mathbb Q \end{cases} $$

with $f$ in each case being the restriction of $g$ to $\mathbb Q$.

(Note, incidentally, the the first of these examples gives an example of a continuous function $\mathbb Q\to\mathbb R$ that cannot be extended to a continuous function $\mathbb R\to\mathbb R$).

This answer is unsalvageable wrong, I leave it so others could learn from my mistakes

Given $f : \mathbb Q \to \mathbb R$ we can extend it to a continuous function $\bar f : \mathbb R \to \mathbb R$ in the following way:

For every irrational $x$ having Cauchy sequence $(x_a)$ define $$\bar f(x) = (f(x_a)),$$ by continuity this is a Cauchy sequence hence a well defined real number. this is wrong you can not extend a continuous function $\mathbb Q \to \mathbb R$ to a continuous function $\mathbb R \to \mathbb R$, there is a counterexample above

In general metric spaces let $f : V \to X$ be a continuous function and $U$ a dense subset of $V$, then $f$ is defined by its restriction $f|_U : U \to X$. [For proof use the sequential definition of continuity]

That tells us that $\bar f$ is the unique continuous extension of $f$, so if $g : \mathbb R \setminus \mathbb Q \to \mathbb R$ is any function then the extension $$f_g(x) = \begin{cases} f(x), & x \in \mathbb Q \\ g(x), & x \not\in \mathbb Q \end{cases}$$ is discontinuous at $x \in \mathbb Q$ if there's a sequence $x_n \to x$ such that $g(x_n) \not \to f(x)$.

For example if we take $g(x) = \bar f(x) + 1$ then $f_g$ is discontinuous everywhere. If we take $g(x) = \bar f(x) + x$ then $f_g$ is discontinuous everywhere except $0$.


A question on linearly independent vectors in a Banach space Given a list of linearly independent vectors $\{x_1,...,x_n\}$ in a Banach space. If for each $1 \leq i\leq n$, there is a sequence of vectors $\{y_m^{(i)}\}_{m=1}^{\infty}$ converges to $x_i$. Then when $m$ is big enough, whether $\{y_m^{(i)}:1 \leq i\leq n\}$ is also linearly independent?

Thank you for all helps!

Yes. Suppose for the sake of convenience that the $y$'s are dependent for every $m$. Choose $c_m^i$ with $$\sum_{i=1}^nc_m^iy_m^i=0,$$and, say, $$\sum_{i=1}^n|c_m^i|=1.$$Now the vectors $c_m\in\Bbb R^n$ have a non-zero accumulation point...

This is not true, even in the simplest case of $\mathbb{R}^2$.  Take $x_1, x_2$ as the standard basis, $y^{(1)}_m=(1,1/m)$, $y^{(2)}_m=(1/m,1)$.


Dirac Delta definition in non-standard analysis? What is the definition of Dirac Delta in non-standard analysis?

I would define it either as a standard distribution with $\sigma=\epsilon$ or maximum equal to $\omega$. Which is the correct answer?

Just like in standard analysis where there are lots of ways to represent the Dirac delta distribution as a limit of a sequence, in nonstandard analysis there are lots of ways to represent the Dirac delta as a nonstandard function.

The simplest is probably to pick an infinite $H$ and set

$$ d(x) = \begin{cases} 0 & |x| > \frac{1}{2H}
\\ H & |x| \leq \frac{1}{2H} \end{cases} $$

Convolving $d$ with any standard continuous function gives

$$ \begin{align}\int_{-\infty}^{\infty} d(x) f(x) \, dx &= H \int_{-\frac{1}{2H}}^{\frac{1}{2H}} f(x) \, dx
\\&= H f(\epsilon) \int_{-\frac{1}{2H}}^{\frac{1}{2H}} 1 \, dx
\\&= H f(\epsilon) \frac{1}{H}
\\&= f(\epsilon)
\\&\approx f(0)
\end{align}$$

where I've used the mean value theorem to go from line one to line two. (one could instead use $|f(x) - f(0)| < e$ for $|x| < \frac{1}{2H}$ if they liked)

I found the answer.

$$\delta(x)=\frac{\omega e^{-(\omega x)^2}}{\sqrt{\pi }}$$

And its integral, Heaviside Theta is

$$\theta(x)=\frac12\operatorname{erf}(\omega x)+1/2$$

As such,

$$\delta(0)=\frac{\omega}{\sqrt{\pi}}$$


Primes as a difference of powers Find the smallest prime that cannot be written as
$$|3^a - 2^b|$$

EDIT: I forgot to mention that $a$ and $b$ are whole numbers.

  I tried to expand $3^a$ as $(2+1)^a$ using binomial theorem but I couldn't infer much. Please help. Thanks in advance!

Tried to find all the prime numbers by trial and error. Didn't get any combination for 41. So, it is a possible answer. Correct me if I am wrong!


Prove that the set of palindromes are not regular languages using the pumping lemma. Firstly I pick a language $xyz$ where $x = \epsilon$, $y = (abb)^{k}$, $z = (bba)^{k}$ where $|y| \ge$ the number of states in the automaton representing my language. Then $xyz = (abb)^k(bba)^k$ is a palindrome.

Now I split $y$ into $uvw$, where $v \ne \epsilon$ contains a state visited more  than once. If the language is regular then $xuv^iwz$ is within the language $\forall i \ge 0$.

I choose $i = 2$, then $uv^2w = (abb)^n$ where $n > k$.

Then $xuv^2wz = (abb)^n(bba)^k$ where $n > k$, which is not a palindrome.

Firstly, is my reasoning for this particular proof correct?  I'm unsure about how to properly apply the pumping lemma to this problem. And secondly the question asks me to state a proof for palindromes in general, whereas I only attempted to prove it for a single case. Is there an example I can choose which proves the conjecture for all palindromes? I assume not, seeing as I can't think of how that could be represented using regular expressions.

Your proof is incorrect, for instance you only handle a subset of palindromes (and regular languages can have non-regular subsets). I'm not quite sure what you're trying to do there, but for reference here's my version of the proof using the pumping lemma. I hope it opens up valuable insights.

In the proof, I will assume the alphabet to consist of at least two distinct symbols (as palindromes are a regular language if there is only one symbol in the alphabet).

My version of the proof

(using $s^R$ to denote the reverse of $s$)

Let $L = \{s : s = s^R\}$ or the set of all palindromes (strings that are equal to their reverse strings).

If $L$ is regular, there must be a pumping length $p \ge 1$ so that for each $s \in L$, if $|s| \ge p$ we can write $s$ as $xyz$ so these three conditions apply:

$y \neq \epsilon$ ($y$ has at least one character/symbol, in other words $|y| \ge 1$)
$|xy| \le p$
for all $i \in \mathbb{N} \cup \{0\}$, $xy^iz \in L$

Let $s = a^pba^p$. We can see $s$ is a palindrome, so $s \in L$. Also, $|s| = 2p + 1 \ge p$.

Because of the two first conditions, all possible $xyz$ divisions of $s$ are in the following form:

$x = a^k$
$y = a^m$
$z = a^{p-m-k}ba^p$

for some non-negative integers $k, m$ so that $k + m \le p$ and $m \ge 1$.

Now let's demonstrate that $xy^iz \in L$ doesn't hold for some values of $i$.

For example, let $i = 2$. Then $$xy^2z = xyyz = a^ka^ma^ma^{p-m-k}ba^p$$

which can be simplified further as $$a^ma^pba^p = a^{p+m}ba^p$$

The reverse string, $(xy^2z)^R$ is $a^pba^{p+m}$. Because we know that $m \ge 1$, it follows that $$a^pba^{p+m} \neq a^{p+m}ba^p$$ Therefore, $xy^2z \notin L$ under any possible divisions $xyz$ while following the three conditions ruled by the lemma. Hence, $L$ must be non-regular.

Two important notes

You can choose any string $s \in L$, as long as $|s| \ge p$. Make life easy for yourself and pick a string that makes the rest of the lemma easy to apply. I picked $a^pba^p$ simply to reduce the complexity of different $xyz$ divisions.
For a language to be non-regular, you must prove there are no possible divisions $xyz$ that are possible according to the three criteria. It's not enough to demonstrate that some $xyz$ division fails at the $xy^iz \in L$ phase.

Consider the language
$$
L = \left\{ w^p : w \in L \right\}, L \subseteq A^{*}.$$ Where $A$ is the
input alphabet of the automaton that recognizes $L$.

where $w = a_1a_2...a_n, \left| w \right| = n $ and $w_p = a_na_{n-1}...a_1, \left| w_p \right| = n $

From a general approach using Kleene's theorem we know that a language is regular iff it is rational. However, in order for language $L$ to be regular/recognisable from a finite automaton it must have an infinite set of states, which is impossible. Thus, Kleene's theorem guarantees that this language is NOT regular.

Things are getting a bit harder when we have to prove our claim using the pumping lemma.

By presuming the language is regular, there exists a number $p$ (length of pump) so that the following conditions are true:

$$ \forall i \geq 0, xy^iz \in L$$
$$ \left| y \right| >  0 $$
$$\left| xy \right| \leq p $$

You should try with the word $w_p$ and go through all 3 conditions of the pumping lemma. The key to the solution is condition Number 3 since the first condition cannot always lead to a contradiction.


What is the result of $2^{{2}^{{2}^{{2}^{0}}}}$? What is the result of $2^{{2}^{{2}^{{2}^{0}}}}$?

It seems simple but I don't understand it.

Edit: Whoever is down-voting, please leave a comment as to why...


Maximum number of triangles formed by points on a square? $N$ points are to be put on sides of a square. What's the maximum number of triangles which are formed by joining those points?

Hints would be appreciated.

Each triangle requires exactly three points, but not every set of three points defines a triangle. There are $n\choose 3$ sets of three points. You have to get rid of sets where all three points fall on the same side of the square. If $S_i$ points are on side $i$ of the square, $\sum_{i=1}^4 {{S_i}\choose 3}$ sets of three don't make triangles, and must be subtracted out. Since you're looking for the maximum number of triangles, minimize $\sum_{i=1}^4 {{S_i}\choose 3}$ subject to $S_1+S_2+S_3+S_4 = N$.

This is my own formula :  correct me if I'm wrong ,

n^2 C3 - 2n x nC3 - 2 * [ (n-1)C3+ (n-2)C3 +....3C3 ]


Cauchy Goursat, why triangles? Cauchy Goursat: Let
  $f$ be analytic in a simply connected domain $D$.If $C$ is a simple closed contour that lies in $D$ , then $$\int_C f(z) dz = 0.$$

I've been reading a lot of proofs on this theorem and all of them treats the contour $C$ as a triangle at first, but doesn't explain why it is sufficient to only show for triangles. Is it because every other simply connected curve is homotopic to a triangle?

Also, why are we allowed to assume, without any loss of generality, that one of the interior triangles is bigger than the other $3$?

First you prove cauchy-goursat for closed triangles and $f$ holomorph in an open set $U \subseteq \mathbb{C}$

Then you prove that if in adition $U$ is convex, then $f$ has a primitive in $U$.

Then you prove cauchy theorem for $U$ in convex set: you just use ths Leibniz-Newton formula in any closed path and it gives zero.

This way you get the result without having to triangulate the closed curve.


Is $p(x)=a\frac{x^2}{x}+b$ a linear polynomial? My question at first, may seem too simple,

  Is $p(x)=a\frac{x^2}x+b$ a linear polynomial?

But as we see the polynomial $p(x)$ is not continuous at $x=0$. So if it is not a polynomial, which function is this?
I got this definition on Wikipedia:

  "Polynomial functions are a class of functions having many important properties. They are all continuous, smooth, entire, computable, etc."

You are correct that it is NOT a polynomial. However it has a unique continuous extension to a polynomial ($ax+b$). However AS WRITTEN, it is not a polynomial.

Your question is philosophical haha. Really, the function is not defined in $x=0$. But... We can redefined f in $x=0$ because the limit exist! :).

In complex analysis this is a "singularity"


Finding a Smooth Function With Given Proprieties I need to find a function $f:\mathbb{R} \to \mathbb{R}$ such that

$\bullet$ $f$ is smooth ($C^\infty)$

$\bullet$ $|f'(x)| < 1$

$\bullet$ $f(x) \neq x \quad \forall x \in R$

What I've done so far:

I was able to find a function that seems to work (at least looking at the plot). I will show the function bellow:

$$f(x)=
\begin{cases}
e^\frac{-2}{x-2}  & \text{for $x>2$} \\
(x-2)e^{\frac{2}{x-2}} & \text{for $x<2$}\\
0 & \text{for $x=2$}
\end{cases}$$

Here is the Plot

But I have 2 problems:

1)I would like to know if there is a simpler function since I would prefer not to define things conditionally

2)I'm not sure I know how to show the function satisfies what I want in a rigorous way.

I'm not sure if it is possible, check https://en.wikipedia.org/wiki/Fixed_point_(mathematics)


What is the meaning of $dz$ in Complex Integrals? I'm sorry if I sound too ignorant, I don't have a high level of knowledge in math.

Interested in learning to integrate through other paths besides the real line in the complex plane, I searched how to do so in the internet. The most understandable "proof" of the "Line Integration Formula" was the following:

$C$ is a contour defined by $P(t)=R(t)+iI(t)$, $a<t<b$

$∫f(z)dz$ (through $C$) $=∫f(z(t))z'(t)dt$ (from $a$ to $b$)

since $z$ is in function of $t$, and $z'(t)=dz/dt$, so $dz$=$z'(t)dt$

The latter confuses me in the first step, when we state that the Integration of $C$ in $f$ $=∫f(z)dz$ (through $C$).

With the $∫f(z)$, I can imagine the sum of many (infinite) inputs of $C$ in $f$. But what is the meaning of $dz$?

In "real" integrals, $dx$ can be seen as $(b-a)/m$, where $m$ is the number of inputs we are adding, but I struggle when trying to see $dz$ in the same way since I can´t find the parallel of that $(b-a)$ in complex integrals. My first intuition was to think of $(b-a)$ as the length of the curve, but the "Line Integration Formula" seems to pay no attention on the length of the contour it is integrating.

Is there a way to approximate a complex integral in a similar way to the "real" integrals? I believe such approximation, if it exists, might give me an insight on the meaning of $dz$.

I think that my question is hard to explain, so if there is something that is confusing I would be happy to clarify.

I would really appreciate any explanation or insight!

Suppose we parametrize the contour $C$ as $\gamma(t),\; a \le t \le b$.
The contour integral $\int_C f(z)\; dz$ is the limit as $n \to \infty$ of "Riemann sums"
$$ \sum_{j=1}^n f(z_j)\; \Delta z_j$$
where $z_j = \gamma(t_j)$ and $\Delta z_j = z_j - z_{j-1}$, $a = t_0 < t_1 < \ldots < t_n = b$, if $\max(t_{j} - t_{j-1}) \to 0$.

Before explaining about dz, take L  as a Load/Force to be  a function of x, say L= 4x. If we now care to give this a meaning we can say that this function represents a man moving   in a straight line a distance x and as he does, he has to push against  a force y  which changes as he moves and described  as 4x.  Now, if we had to look at the work done by the man, we have to multiply the  Force he suffers by some distance so in this case we  first take a small distance  dx. So the work done
 by the man is given by Work  = L dx = 4x dx.   From this we can find  the  total work done.

The above problem was a man walking in a line so really it is  a one dimensional problem. If we have  to have the man walking  in an  area of two dimensions then we have to use the expression  z=x+jy.    and so dz= dx+jdy

dz in this case may be looked as a small distance in a  two dimensional plane when the man moves in the dx and the dy direction.  Now as before if  a man is made to face a force of   F(z)   as the walks on a two dimensional distance z=  x + jy,   then he too would be subjected to having to do work and so  one can look upon  F(z)dz  exactly the same as one looked at  P(x) dx.

Since z=x+jy  is a position vector then  any function F(z) will be a vector so
so here we have a situation where the poor man is subjected to a Load or a Force which is not only changing in magnitude but also in direction as he walks the plane x+jy.    All one needs to do is to imagine  the Load. Force F(z) acting at location z    and as z changes the change is dz.  If the man is drunk and walks through a contour then  the change of direction dz is tangential to the curve he walks   so     F(z)dz     is exactly the same as P(x)dx.  which, if taken between two points in the trajectory he walks,  could be interpreted as the work done by a  drunken man walking a curve on the flat plane  a distance z=x+jy from the origin and facing a force  F(z) which is always changing in magnitude and direction depending on the distance he is located from the origin of his walk. This integrated between  limits of the trajectory the man walks is the work done by the man.


Find the limit of $f(x)$ given the limit of $\frac{f(x)}{x}$ Given that:

$$\lim_{x\to 0}\frac{f(x)}{x}=1.$$

How do I evaluate $$\lim_{x\to 0}f(x)$$ and $$\lim_{x\to 0}\frac{f(x)}{g(x)} $$ given $\lim_{x\to 0}g(x)$ exists and is non zero.

Hint: $\lim_{x\to 0}{f(x)} = \lim_{x\to 0}{x\cdot \frac{f(x)}{x}}$.  What do you know about limits of products or quotients?

It's not a property, you are just multiplying and dividing a function by a number that is not zero(even though it tends to 0), so you are not changing the function.


Equation of the Tangent Line and Area of Parametric Equation I need to find the equation of the tangent line to the point (1,0) for the equation:

$x=e^{-0.1t}cos(t) \\
y=e^{-0.1t}sin(t)$

I also need to calculate the area in the first quadrant bounded on the outside by the curve for $t$ greater than or equal to $0$ and less than or equal to $\pi/2$.

I am not totally sure what to do for either one.

I know that $\frac{dy}{dx}$ is $\frac{dy}{dt}/\frac{dx}{dt}$.

Is the tangent line equation $y=\frac{dy}{dx}(x-1)$?  Once I calculate $\frac{dy}{dx}$ I'd just plug it in to the equation.

Hint:

For the first part, you can find the (unnormalized) tangent vector $T(t)$ at the point $p(t) = (1,0)$ on the curve by calculating the derivative with respect to the parameter, $t$: $$T(t) = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}$$

For the second part, find the polar form of the curve $r = f(\theta)$. Then the area is $$\iint f(\theta) \ r \ dr \ d\theta$$ You'll also need to figure out the appropriate limits of integration.


Alternative way to show $\sqrt{2}+ \sqrt{3}$ is algebraic over $\mathbb{Q}$ of degree 4. I want to show that $\sqrt{2}+ \sqrt{3}$ is algebraic over $\mathbb{Q}$ of degree 4. I am aware of the technique of setting $x = \sqrt{2} + \sqrt{3}$ and working backwards to obtain the fourth degree polynomial. I was wondering if there is a different way of doing this, maybe working with field extensions? Particularly, I was thinking that maybe showing that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] =4$ and that $\mathbb{Q}(\sqrt{2},\sqrt{3}) =  \mathbb{Q}(\sqrt{2} + \sqrt{3})$ would help me with this. Thanks for any suggestion!

There's a method I like quite a bit.  A spanning set (but not necessairly a basis, it may be too large) for $\mathbb Q(\sqrt{2},\sqrt{3})$ is undoubtedly $\lbrace 1,\sqrt{2},\sqrt{3},\sqrt{2}\sqrt{3}\rbrace$.  This is just "every multiplicative combination" of the bases for $\mathbb Q(\sqrt{2})$ and $\mathbb Q(\sqrt{3})$.

What we now want to do is find a matrix that has $\sqrt{2}+\sqrt{3}$ as an eigenvalue.
The way we do this is by first writing:
$$(\sqrt{2}+\sqrt{3})\begin{pmatrix} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6}\end{pmatrix}$$
Now, multiply out each row. As an example, the top row is $1\times (\sqrt 2+\sqrt 3) = \sqrt 2+\sqrt 3$.  The second row is $\sqrt 2(\sqrt 2+\sqrt 3) = 2+\sqrt 6$.  The third row is $\sqrt 3 (\sqrt 2+\sqrt 3) = \sqrt 6+ 3$, and the fourth row is $\sqrt 6(\sqrt 2+\sqrt 3) = 2\sqrt 3+3\sqrt 2$.
We can write this as:
$$(\sqrt 2+\sqrt 3) = \begin{pmatrix} 1 \\ \sqrt 2 \\ \sqrt 3 \\ \sqrt 6\end{pmatrix} = \begin{pmatrix} \sqrt 2+\sqrt 3 \\ 2+\sqrt 6 \\ 3+\sqrt 6 \\ 2\sqrt 3+3\sqrt 2\end{pmatrix}$$
We can "factor" this last vector as the following:
$$\begin{pmatrix} 0 & 1 & 1 & 0 \\ 2 & 0 & 0 & 1 \\3 & 0 & 0 & 1 \\0 & 3 & 2 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ \sqrt 2\\ \sqrt 3\\ \sqrt 6\end{pmatrix}$$
This matrix has $\sqrt 2+\sqrt 3$ as an eigenvalue, so its characteristic polynomial has this as a root.
This will be a degree $4$ polynomial.  It's not necessairly the minimal polynomial though, so you still have to check this.

Yes, that approach would work. In fact, we have:

A field extension $K\subseteq L$ is algebraic iff $[L:K]$ is finite.

where we say $K\subseteq L$ is algebraic if $\alpha$ algebraic over $K$ for any $\alpha \in L$.

Let's prove the backward direction, which is relevant to your question. Suppose $[L:K] = n$, and let $\alpha\in L$. Then we know that $1,\alpha,\ldots,\alpha^n$ must be linearly dependent. Therefore, we have $$ \alpha^n + a_{n-1}\alpha^{n-1} + \cdots + a_1\alpha + a_0 = 0$$ and so we have found a polynomial $f(x) = x^n + \cdots + a_0$ with $f(\alpha) = 0$, so $\alpha$ algebraic. Since $\alpha$ was arbitrary, the extension $K\subseteq L$ is algebraic.

Thus, since $[\mathbb{Q}(\sqrt2 + \sqrt3) : \mathbb{Q}]$ finite, we know that $\sqrt2 + \sqrt3$ algebraic.

In fact, we know more. We also know that $[K(\alpha):K] = deg\,p_{\alpha}$, where $p_{\alpha}$ is the minimal polynomial of $\alpha$. You can see this from the isomorphism of fields $$K(\alpha) \rightarrow \frac{K[x]}{(p_{\alpha})}$$So in your case, with $K=\mathbb{Q}$ and $\alpha = \sqrt2 + \sqrt3$, we find that the minimal polynomial of $\sqrt2 + \sqrt3$ has degree 4 since $[\mathbb{Q}(\sqrt2 + \sqrt3) : \mathbb{Q}] = 4$.


Finding the limit of recursive sequence Assuming that the solution of $e^{-x}=x$ is $c\in (0,1)$

And give the following sequence

$$a_{n+1}=e^{-a_n}$$

$$a_1=1$$

How can i prove that the sequence converge and that the limit is  $$\lim_{n \to \infty}a_n=c$$

I tried to define a new sequence $$b_n=a_n-c$$ and to prove it using the ratio test but without success

$f(x) = \mathrm{e}^{-x}$ is a contraction in the relevant range $(0, 1]$, thus $x_{n + 1} = f(x_n)$ converges if started inside that range.

Exponential function is continuous,then

$$
\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}e^{-a_n}=e^{-\lim_{n\to\infty}a_n}.
$$

So the limit of this sequence is the solution of $x=e^{-x}$ which is $c$.

To see the existence, first note by induction that $1\geq a_n\geq 0$. Now we claim that subsequence with odd terms decreasing and subsequence with even terms increasing. For $k=1$, $
a_{2k-1}\geq a_{2k+1}
$ is immediate and to see $a_{2k}\leq a_{2k+2}$ note that $1\geq a_n$ and $e$ is monotone. Thus, $
a_2=e^{-1}\leq a_4=e^{-a_3}.
$

Suppose now for some $k$, $a_{2k-1}\geq a_{2k+1}$ and $a_{2k}\leq a_{2k+2}$. Then

$$
e^{-a_{2k}}\geq e^{-a_{2k+2}}\implies a_{2k+1}\geq a_{2k+3},\\
e^{-a_{2k+1}}\leq e^{-a_{2k+3}}\implies a_{2k+2}\leq a_{2k+4}.
$$

This tells us that both $\lim_{k\to\infty}a_{2_k}=l_0$ and $\lim_{k\to\infty}a_{2k-1}=l_1$ exist by monotone convergence theorem. Now we claim that every odd term is bigger than every even term. Trivially for $k=1$, this claim holds. Suppose now that it is true for some $k$, $a_{2k-1}\geq a_{2k}$. Then

$$
a_{2k+1}=e^{-a_{2k}}\geq e^{-a_{2k-1}}=a_{2k}\implies e^{-a_2k}=a_{2k+1}\geq e^{a_{2k+1}}=a_{2k+2}
$$
which shows that the claim is true for $k+1$ and induction is completed. So we show that $a_{2k}\leq a_{2k+2}\leq a_{2k+1}\leq a_{2k-1}$. Note also that $l_0$ is supremum and $l_1$ is infimum for corresponding sequences. Then $l_1\geq l_0$.

By induction one can show $a_{2k+1}=a_{2k}^{1/e}\geq a_{2k+1}^{1/e}=a_{2k+2}$. Taking limit on both sides  gives us $l_0^{1/e}\geq l_1^{1/e}$ and $l_0\geq l_1$. Thus $l_0=l_1=l$ and $a_k\to l$.


Proof that two conditional random variables given a condition are independent How to show that the following condition implies random variables X and Y are independent?
$P(X > a | Y < b) = P(X > a)\ \forall\ a,\ b\ \in\ \mathbb{R}$

Let $F_X(a) = P(X \leq a)$, $F_Y(b) = P(Y \leq b)$ and $F_{X,Y}(a,b) = P(X \leq a \mbox{ and } Y \leq b)$ denote the cumulative distribution functions of $X$, $Y$ and $(X, Y)$, respectively. To show independence, we want to prove that $F_X(a)F_Y(b) = F_{X, Y}(a,b)$ for all $a, b \in \mathbb{R}$. Conditional probability tells us that: $$P(X > a | Y < b) = \frac{P(X>a \mbox{ and } Y<b)}{P(Y < b)}$$ and so by the assumptions of the question $$P(X > a) = \frac{P(X>a \mbox{ and } Y<b)}{P(Y < b)}$$ which tells us $$P(X > a)P(Y < b) = P(X>a \mbox{ and } Y<b)$$ for all $a, b \in \mathbb{R}$. Now since $P(X > a) = 1 - F_X(a)$ and $P(Y < b) = F_Y(b)$ we may rewrite the above equation as:
$$F_Y(b) - F_X(a)F_Y(b) = P(X > a \mbox{ and } Y < b)$$
and so $$F_X(a)F_Y(b) = F_Y(b) - P(X > a \mbox{ and } Y < b).$$
Finally, note that $$F_Y(b) = P(X > a \mbox{ and } Y < b) + P(X < a \mbox{ and } Y < b)$$ and we are done.

(>|<)=$\frac{P(X>a,Y<b)}{P(Y<b)}$ = P(X>a)
P(X>a,Y<b)=P(X>a).P(Y<b) so X & Y are independent


Finding the remainder after division by $3$ is simple. What about the quotient? A well-known simple trick for obtaining the REMAINDER after division by 3 is just the digital root of the dividend after division by 3, the digital root itself further simply obtained by first casting out 9's, then every digit pair that add up to 9, etc.

  In the same vein, is there a short-cut method to rapidly obtain the QUOTIENT after division by 3?

Not sure if we can call it shortcut,

in base $10,$
Using Why is $a^n - b^n$ divisible by $a-b$?
$$\dfrac{\sum_{r=0}^n10^ra_r}3=\sum_{r=0}^n\dfrac{10^r-1}3\cdot a_r+\dfrac{\sum_{r=0}^na_r}3$$

$$=\dfrac{\sum_{r=0}^na_r}3+\sum_{r=0}^n\underbrace{33\cdot33}_{r\text{ digits }}\cdot a_r$$


The difference between 87% and 71% of a number is 448. What is 59% of that number Today I came across an objective type question in that I have to choose from 4 optiona. I found myself struggling with this question since I have not been in touch with Math since I completed my graduation.

The question was:

The difference between 87% and 71% of a number is 448. What is 59% of that number?

1562
1652
1662
1552

I would really appreciate any help on this.

As mentioned in the comments, $0.87x - 0.71x = 448$. Then you can use the unitary method:

$$0.16x=448$$
$$0.01x=28$$
$$0.59x = \ ?$$

Use the rule of three.
$448$ is for $16$
like $x$ is for 59
then:
$$\frac{59 \times 448}{16} = ?$$