When is a real-analytic function harmonic? I recently learnt that every harmonic function occurs as the real part of a complex analytic function. We also know that every harmonic function is real analytic. So, when is a real-analytic function harmonic ?

When $-\Delta u=-\sum_{k=1}^n\frac{\partial^2u}{\partial x_k^2}=0$

I believe an equivalent property is the mean integral property, I.e.:

$u(x)=\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}u(y)dS(y)=\frac{1}{\alpha(n)r^{n}}\int_{ B(x,r)}u(y)dy$, $\forall r\gt 0$.

Where $\alpha(n)$ is the volume on the unit sphere in $\Bbb R^n$.

A real valued function can never be analytic except a constant function.let it be a $~f(z)=x~$ so it have to satisfy CR equation. but here $~Ux=1~$ but $~Vy=0~$. So CR equation not satisfied  so not differentiable any where.so we can't talk about analytic but it is hormonic.


limit $\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}}$ Let $$\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}}$$

How do I show it's limit is $0$?

I thought about:

$$\frac{xy}{\|(x,y)\|} \le \frac{xy}{|x|} \le \frac{xy}{x} \le y \to 0$$

Is that correct?

It looks fine.

As I wrote in another post, usually it is useful to set $x = \rho \cos \theta$, $y = \rho \sin \theta$ and taking the limit as $\rho \to 0$. More specifically, if you find that
$|f(\rho \cos \theta, \rho \sin \theta)| \le g(\rho) \to 0$, then you showed that $f(x, y) \to 0$

In your case

$$|f(\rho \cos \theta, \rho \sin \theta)| = \frac{|\rho^2 \sin \theta \cos \theta|}{|\rho|} \le \rho = g(\rho) \to 0$$

so your limit is indeed $0$

Wrong answer

Actually,it has no limits.
From different directions to (0,0),it's limit is different.
 For example:(x,0) to (0,0),the limit is 0;(t,t) to (0,0),the limit is 1/sqrt(2)


Convergence of integral with parameter For which $t$ does the following integral converge?

$$\int_{0}^{1/2}\frac{(1-\cos x)^t(1-x)^3}{\ln^2(1-x)}\,dx$$ where $t\in \mathbb{R}$.

Hint. As $x$ tends to $0^+$, you have
$$\frac{(1-\cos x)^t(1-x)^3}{\ln^2(1-x)} \sim \frac{(x^2/2)^t}{x^2}=\frac1{2^t}x^{2t-2}$$ your integral converges iff $t>\frac12$.

Hint: Use series. It is necessary to see first two terms in $(1-x)^3/\ln^2(1-x)$.


Find all zeros of the equation $e^z = 1+2z$ in the unit disk. I tried solving the pair of equations
\begin{align}
e^x\cos y&=1+2x\\
e^x\sin y&= 2y
\end{align}
but I got stuck.

let $f(z) = e^{z}$ and $g(z) = -(1+2z)$ then on the boundary $e^{i\theta}$,

$$ |f(z)| = |e^{\cos \theta}| < \sqrt{5 + 4 \cos(\theta)} \tag{1}$$
Since $g(z)$ has  one root on $|z|<1$, by Rouché's Theorem $f(z)+g(z)$ has one root on $|z|< 1 $. In particular, $z=0$ satisfies the equation and it is only the root of it as well.

To show that $(1)$ holds, one can simply show that $h(x) = e^x - \sqrt{5 + 4x}$ is smooth function on $[1,1]$ has $h(x)$ has minimum value of $3-e$ on $x=1$.

Hint: one solution is quite easy.  It also happens to be the only one.


Rotating coordinates of a matrix Say I have a 2D image represented by a matrix $ A_{ij} $ with $i,j\in[0,..,N-1]$, so that to each entry $(i,j)$ corresponds its value $A(i,j)$. I have a new coordinates system with the indices being $u=i+j$ and $v=i-j$, and I want to get the matrix in this coordinate system $A_{u,v}$ by making:
\begin{eqnarray}
A_{uv}=\sum_{ij}R_{ui}^{\dagger}.A_{ij}.R_{jv}
\end{eqnarray}
I want to get the rotation matrices that follow the $u,v$ transformation of coordinates above. How is this achieved? I understand this corresponds to a change of basis from the canonical for $i,j$ to the one given by $u,v$, my problem is on how to get the entries of $A(u,v)$.

I am not sure I quite understand why you desire the form you mentioned, but we can compute a transformation matrix that given you the desired result. So, you are interested in finding a transformation as follows:
$$ \begin{bmatrix} u' \\ v'\end{bmatrix} = \mathbf{T} \begin{bmatrix} u \\ v\end{bmatrix} $$
where $u$ and $v$ are the original image coordinates and $u'$ and $v'$ are the transformed image coordinates.
You want a matrix so that $u' = u + v$ and $v' = u - v$. We do some algebra to find this matrix.
$$
\begin{align}
\begin{bmatrix} u+v \\ u-v \end{bmatrix} = \begin{bmatrix} \mathbf{T}_{00} & \mathbf{T}_{10} \\ \mathbf{T}_{01} & \mathbf{T}_{11} \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix}
\end{align}
$$
Then, multiplying through we have the following system of equations:
$$
\begin{align}
u + v &= \mathbf{T}_{00} u + \mathbf{T}_{10} v \\
u - v &= \mathbf{T}_{01} u + \mathbf{T}_{11} v
\end{align}
$$
Solving, you have the following:
$$
\begin{align}
\mathbf{T} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}
\end{align}
$$
Unfortunately, this is not a proper rotation matrix because $\det[\mathbf{T}] \neq 1$, but the transformation matrix exists.


Proof in Bartle's book I am studying real analysis by my own and stumble into the following theorem, that Bartle's proves in its appendix, however, I do not understand the case 2. Specifically I do not understand why does he define $h_1$, and the induction hypothesis is not supposed to be that there is not induction from $ N_m \rightarrow N_k$, so how does it implies that $h_1$ is not an injection. Can somebody please clarify this part, or explain it thoroughly, for me it is not easily seen as the book states. Thanks

Here is a few ideas.

$h_1$ is well defined because $h$ is well defined and no element in $\mathbb N_{m-1}$ is being sent to $k+1$.
$h_1$ is not an injection because $m > k + 1\implies \color{red} {m -1 > k}$ and by the induction step $h_1 : \mathbb N _{m-1} \to \mathbb N_k$ can't be an injection, notice in this case we are assuming $h(p) = k+1 $.

Finally, $h$ is not an injection because we have seen that $h_1$ isn't, and we know that $h(p) = k+1$. Notice that $\mathbb N_{m-1} \cup \{p\} = \mathbb N_m$ and $\mathbb N_k \cup \{k+1\} = \mathbb N_{k+1}$.

Well, the induction hypothesis is actually that there's no injection from $\mathbb{N}_m$ into $\mathbb{N}_k$.


How do I get the integer part of a number by using basic arithmetic? While it is trivial to simply remove the fractional part of an irrational or rational number, and in programming I could just use the floor() or ceil() functions, I was wondering how such calculation is done using basic arithmetic.

In other words, how can I get the integer part of a (base-10) number using merely addition, subtraction, multiplication and division?

Perhaps this is trivial, but after a bit of trying and a bit of googling, I couldn't quite get my head around it. Maybe it is not so trivial after all?

It's impossible.

If you consider the set of all functions $P$ containing $f(x) = x$ and when $f,g \in P$ we have the sum $f(x)+g(x) \in P$, etc. then

Theorem $\text{floor}$ is not contained in $P$.

proof: We prove this by finding a property that everything in $P$ has but $\text{floor}$ does not. All functions in $P$ are continuous, but floor is not.

It is possible to find the integer part of any number divided by two using basic arithmetic:

$$\text{Floor}\left(\frac{n}{2}\right) = \frac{n}{2} - \frac{1}{4} + \frac{(-1)^n}{4}.$$

I'm not aware of a solution for division by other numbers.


Finding out how out how much is 100% I know 14% is 41. So how much is 100%? I know it is simple math, but... you know how it is being out of school for a few years. If at all possible, I'd like to see the equation too.

Let $x$ be the total value, so we have:

$14\%x = 41 \implies 0.14x = 41$

From here, you should get the value of $x$ and that would be your 100$


Programming discrete fourier coefficients in matlab Alright so I am having the following issue:  I want to figure out how to find the fourier coefficients of the following function:
$$D(X)=\frac {a'(x)} {1+a'(x)^2}$$

Where $a(x)$ is an arbitrary function.  I already have a model for finding the fourier coefficients for $a(x)$ and $a'(x)$:

fc = fft(a) / Nfft;
fc = fftshift(fc);                % fft of a(x)
fc = conj(fc);                    % sign correction
aprimec = -i * [0:Dim2-1] .* fc;  % fc of derivative (definition)

The equation I am given to use is:
$$f_m=\frac 1 N \sum^Nf_ie^{+im2\pi x}$$

Which confuses me because of the $f_i$.  So does any one have any suggestions?

Additionally, I do not know how to define d

d = diff(a)/(1+diff(a)^2);

I do not think that this would work because doesn't diff(x) just take the difference between two consecutive components in the vector?

I would greatly appreciate any help.  Thanks!

The DFT is isomorphic, so if you have the DFT coefficients for $a'$, then all you need to do to get $a'$ is apply the inverse DFT:

D = ifft(aprimec) ./ (1 + ifft(aprimec).^2);

You can try to approximate $D$ using the ifft of $a'$, additionally your formula seems to be
$$ \hat{f}_m = \frac{1}{N} \sum_{j=0}^{N-1} f_j \omega_N^{mj} $$
where $\omega_N = \exp(\frac{2\pi i}{N})$.


Direction of line parallel to the x-axis in the complex plane How can  say the direction of a line parallel to the real axis as clockwise or anti-clockwise?

More precisely what's the direction of blue line as well as orange line ?

Which one is clockwise & which one is anti-clockwise ?

Update :

My actual question is : Suppose this is the whole contour where I want to integrate a complex function. The middle line is the negative real axis. For part $C_2$ the direction is anticlockwise (positively oriented). But for the curve $C_1$ and $C_3$ the direction is clockwise or anticlockwise so that I can use negative sign and positive sign in the evaluation of the integration on the whole contour ? How can I say that ?

A line doesn't have a direction.
It has an equation.

A vector has origin and direction.

The lines above would have equations
of the form
Im(z) = c
for some real c.

As vectors,
their specification would be
a+bt
where $t \ge 0$
and b is real.

If b > 0,
the vector goes to the right.

If b < 0,
the vector goes to the left.

I'd avoid using the terms "clockwise" or "anti-clockwise" since they don't apply in this context. These only refer to things moving in a circular way.

I would agree with marty cohen: we would say that these lines "go to the right/left", or "the direction of the line is to the right/left".

P.S. This question might be better off on english.stackexchange.com


Conjugate to σ in the alternating group There is $S_7$ and the subgroup $\langle σ\rangle = \{(1 \; 2\; 3\; 4 \; 5)^i | i \in \mathbb Z\}$ of $S_7$ generated by $σ = (1 \; 2\; 3\; 4 \; 5)$

Then, how many elements are conjugate to σ in $A_7$?

The order of σ is 5 and well.. I know that conjugation of symmetric group(preserve the cycle type) but how to find the number of elements?

Thanks.

The cycle decomposition of the given permutation $\sigma=(1,2,3,4,5)(6)(7)$ contains one $5$-cycle and two fixed points ($1$-cycles).  By the splitting criterion [1,2], because the cycle lengths $5$, $1$, and $1$ are not distinct odd numbers (there are two $1$-cycles in $\sigma$), the conjugacy class of $\sigma$ does NOT split when restricted to $A_{7}$ from $S_{7}$.

The process is: (hover to view)

   There are ${7\choose 5}\cdot4!=504$ [3] conjugates of $\sigma$ in $S_{7}$.
 By the splitting criterion, because the cycle lengths of $\sigma$ (namely $5$, $1$, and $1$) are not distinct odd numbers, the conjugacy class of $\sigma$ does not split.  So there are exactly $504$ conjugates of $\sigma$ in $A_{7}$.


Solving a system of linear equations! I am not sure how to solve a system of linear equations of this nature:

s(Y-T)= G-T-bi
M/P=h×Y-gi

Here's how I tried to solve it:

I'm doing something wrong but I do not know what.

I hope that someone can help me with this problem.

Thank you in advance!

What you got is the same as the answer because
$$\begin{align}\frac{1}{bh+gs}\left(gG-gT+gsT+\frac{bM}{p}\right)&=\frac{g}{bh+gs}\left(G-T+sT+\frac{bM}{gp}\right)\\&=\frac{g}{hb+gs}\left(\frac{bM}{gp}-T+sT+G\right)\\&=\frac{g}{hb+gs}\left(\frac{bM}{gp}-(1-s)\times T+G\right)\end{align}$$

Thank you very much for all your help and useful information!

Yes, this was macroeconomics stuff (it's really not my thing).


Use $\epsilon\ -\ \delta$ definition to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$ It's known that $\lim\limits_{x \rightarrow x_0}f(x) = A$, how to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = \sqrt[3]{A}$?

Here's what I've got now:

When $A = 0$, to prove $\lim\limits_{x \rightarrow x_0}\sqrt[3]{f(x)} = 0$: Since we have $\lim\limits_{x \rightarrow x_0}f(x) = A = 0$, so $|f(x)| < \epsilon$. => $|\sqrt[3]{f(x)}| < \epsilon_0^3 < \epsilon$

When $A \ne 0$, $|\sqrt[3]{f(x)} - \sqrt[3]{A}| = \frac{|f(x) - A|}{|f(x)^{\frac{2}{3}}+(f(x)A)^{\frac{1}{3}} + A^{\frac{2}{3}}|}$...

How can I deal with $(f(x)A)^{\frac{1}{3}}$? Thanks.

You might want to take a look at the limit of composite functions. This is a standard result in many calculus textbooks. Fishing around online immediate gives several proofs, for example here or here.

You need to bound the term

$$ \frac{1}{|f(x)^{\frac{2}{3}}+(f(x)A)^{\frac{1}{3}} + A^{\frac{2}{3}}|}\,, $$

and exploit the fact that $|x-x_0|<\delta \implies |f(x)-A|<\epsilon\,.$ See here for details.


Calculate $\lim_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\ \{n\sqrt{2}\} }$ $$\text{Calculate :}\lim_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} } . $$

Note:
Weyl's equidistributed criterion. The following are equivalent:
$$x_n\quad\text{is equivalent modulo 1}$$
$$\forall~ \text{continuous & 1-peridic} f: \quad\frac{1}{N}\sum_{n=1}^Nf(x_n)\rightarrow\int_0^1f $$
$$\forall~ k\in \mathbb Z^*:\quad \frac{1}{N}\sum_{n=1}^Ne^{2πikx_n}\rightarrow 0 $$

Background:
Im trying to approach this problem by weyl's criterion, so my thoughts so far are:
\begin{align}
&\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} }\\
&=\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\}\big)^{1/n}\\
&=e^{\frac{1}{n}\log\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\}\big) }\\
&=e^{\big(\frac{1}{n}\sum_{k=1}^n \log(\{k\sqrt{2}\})\big)}\\
\end{align}
So, since $\sqrt{2}$ is irrational  then its simple to prove that the sequence $x_n=\{ n\cdot \sqrt{2}\}$ is equidistributed $\text{mod}\ 1$.
Let us define the continuous & 1-periodic function $f(x):=\log(x-[x])$
by the weyl's criterion we get:
$$\begin{align*}
\frac{1}{n}\sum_{k=1}^n \log(\{k\sqrt{2}\})&\longrightarrow\int_0^1\log(\color{black}{\underbrace{\{x\}}_{=x-[x]}})dx\\[5pt]
&=\int_0^1\log(x)dx\\[5pt]
&=\bigg[x\log(x)\bigg]_0^1-\int_0^1dx\\[5pt]
&=-1\\[5pt]
\end{align*}$$
Hence
$$\lim\limits_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} }=e^{-1}  $$
Is there something wrong? Also, can we find this limit with some other way?
Let me know, thank you.

Since $\{ n\sqrt{2} \}$ is equidistributed modulo $2$, the limit could be rewritten as the limit of the expected value of the geometric average of $n$ uniform random variables. The integral for this would be $$\lim_{n \to \infty}\int_0^1 \int_0^1... \int_0^1 (x_1x_2...x_n)^{\frac{1}{n}} dx_n...dx_2dx_1$$
This can actually be rewritten as $$\lim_{n \to \infty}\left(\int_0^1 x^{\frac{1}{n}} dx \right)^n$$
since each $x_i$ is independent of the others. The inner integral is then equal to $\frac{n}{n+1} = 1-\frac{1}{n+1}$, so the limit is $$\lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)^{n}$$
which is clearly $e^{-1}$.

Let
$y_{n}=\{\sqrt{2}\}\{2\sqrt{2}\}\cdots \{n\sqrt{2}\}$
Assume the $\lim_{n\to+\infty} y_n $ exists ,
$\lim_{n\to+\infty}\sqrt[n]{y_n}=\lim_{n\to+\infty}\frac{y_{n+1}}{y_n}=\lim_{n\to+\infty}\{(n+1)\sqrt{2}\}=L$
$L$ does not exists.


How to solve linear congruence with two variable that is not system congruence? This problem from Number theory-James K. Strayer

The congruence is

$$2x+3y\equiv 4 \pmod{7}$$

since it is not system congruence the eliminate method does not work

Can anyone help me?

Note that $3\cdot 5\equiv 1\pmod{7}$, so your congruence is equivalent to
$$
5(2x+3y)\equiv 5\cdot4\pmod{7}
$$
that is $3x+y\equiv 6\pmod{7}$ and therefore to
$$
y\equiv 4x+6\pmod{7}
$$
Give $x$ any value between $0$ and $6$ and determine $y$ (up to congruence modulo $7$, of course).

$2x\equiv (4 -3y) mod 7$

$x\equiv (2-(3/2)y) mod 7$ (For $k=1$ since   $\frac{3+7k}{2}=5$ then $(3/2)\equiv 5 mod7 $

$x\equiv (2-5y) mod 7$.

Since $-5\equiv 2 mod 7$, we get $x\equiv (2+2y) mod 7$. Then $x\equiv (2(1+y)) mod 7$.

Then $y=\bar{-1}=\bar6$ for $x=\bar0$,  and $y=\bar{3}$ for $x=\bar1$ (Put $x=1$ in the $x=2(1+y)$ then we have $y=\frac{-1}{2}$, for $k=1$, $y=\frac{-1+7k}{2}=3 $), and  $y=\bar{0}$ for $x=\bar2$ (we can calculate by using same argument), and  $y=\bar{4}$ for $x=\bar3$, and $y=\bar{1}$ for $x=\bar4$, and $y=\bar{5}$ for $x=\bar5$, and $y=\bar{2}$ for $x=\bar6$.

So the answer is the set of binaries $\{(\bar0,\bar6), (\bar1,\bar3),(\bar2,\bar0),(\bar3,\bar4),(\bar5,\bar5),(\bar6,\bar2) \}$


Check condition normal subgroup in these three examples Is the subgroup H of G is a normal subgroup of G, for:
$$ i)\ G = S_5, \ H = \{id, (1,2)\} $$
$$ ii) \ G = (Sym(\mathbb{N}), \circ), \ H = \{f\in Sym(\mathbb{N}) : f(0) = 0 \}$$
$$ iii) \ G = S_4, \ H = \{id, (1,2,3), (1,3,2) \} $$

I know, that subgroup is normal, if:
$$ \forall g\in G \ \ gH = Hg  \ \ - \ \ \ H \lhd G $$

But I don't know how. Could you solve example number one and two?
Thanks for your help. I know that $$ S_5 = \{id, (1,2,3,4,5), ..., (5,4,3,2,1)\} \ \ \ \ 5! = 120  $$ but I don't understand it.

While the condition for a subgroup being normal is correct, it is often not the most efficient or at least intuitive to check this via using it.

Multiplying by $g^{-1}$ you get equivalently $gHg^{-1} = H$ for all $g \in G$.

So, to get you started you could just calculate $ghg^{-1}$ for a couple of $g\in G$ and $h \in H$ and see if all the elements are actually in $H$.

If you just find one where this is not true, you are done, and can conclude the group is not normal.

If you do not seem to find such a counterexample you might start to suspect the subgroup is actually normal and then try to show that indeed for all $g \in G$ and $h \in H$ you have $ghg^{-1} \in H$. If you can show this you are also done and the group is a normal subgroup. Note: what I just said means $gHg^{-1} \subset H$ for all $g \in G$ (not equality). But this is yet another condition for a subgroup to be a normal subgroup.


Need help on a calculation for a gravel pit I have a gravel pit $70$ inches in diameter on the inside. I want to make the pit $4$ inches deep in gravel. The bags come in $0.5$ cubic feet.

I can't even figure out how many cubic feet. How many bags would I need to do this? Any help would be greatly appreciated!

1. Get the volume of the pit: I guess the gravel pit is supposed to be in cylindric shape. The volume of a cylinder is (as in most geometric 3D-objects) given by

$$\text{volume}_{\text{cyl}}=(\text{area of base}) \cdot \text{height}.$$

The base is a circle which has area $r^2\pi$ where $r$ is the radius (which is half the diameter). Your height is 4 inches. Plug this in and you have the volume of the gravel pit in cubic inches you have to fill.

Computing that yields

   $\text{volume}_{\text{cyl}}=(35 ~\mathrm{in})^2\pi\cdot 4~\mathrm{in}\approx 15393.80~\mathrm{in}^3.$

2. Get the volume of 1 bag: This is already given in the question. But you need to have the same units, so you might want to switch from $0.5$ cubic feet to $$\text{volume}_{\text{bag}}=864 ~\mathrm{in}^3.$$

3. Get the amount of bags: This is easily done by

$$\text{amount of bags} = \frac{\text{volume}_{\text{cyl}}}{\text{volume}_{\text{bag}}}.$$

Computing that yields

   $\text{amount of bags} = \frac{15393.80~\mathrm{in}^3}{864 ~\mathrm{in}^3}\approx 17.82$

4. The aftermath: If the number you get from 3. is not an integer, you need to round that number up to the next integer since you can't buy fractions of bags. This means you'll need

   $\lceil 17.82 \rceil=18$

bags.

For ease u can do conversions from inches to feet and proceed by formula of volume is π rsquare h. And then calculate a 0.5 cubic feet bag takes


compact convex problem Would you please help me in how to show that:

If $T : X \to Y$ is a continuous linear map between two locally convex spaces and $K \subset X$ is compact convex, then:

The continuous image of compact convex space is a compact convex.
If $y$ ia an extreme point of $T(K)$,  then $\exists$  $x \in X$ an extreme in $K$ such that $Tx =y$

Thank you,
Saba

The first point is obvious:

Compactness is preserved by continuous maps so $T(K)$ is compact.
Convexity is preserved by linear maps, so $T(K)$ is convex.

As for the second point, we need to know a bit more, I think.

First of all, I assume that your definition of locally convex includes Hausdorff, otherwise there are silly counterexamples (see right at the end of the answer).

The idea is the following: If $y \in T(K)$ is extremal then $F = \{x \in K\,:\,T(x) = y\}$ is a face of $K$ and by the Kreĭn-Mil'mans theorem every face contains an extremal point. If you know what a face is and just want to see see why $F$ is a face, you can ignore the next section and jump right to the claim afterwards.

Also, I overlooked the homework tag, so maybe I shouldn't have given all away below, stop reading now if you want to think for yourself.

Let me elaborate on this:

Recall that a face $F$ of a convex set $K$ is a non-empty and closed subset of $K$ with the following property: If $x,y \in K$ are any points for which there is $0 \lt \alpha \lt 1$ such that $(1-\alpha)x + \alpha y \in F$ then we must have $x,y \in F$ already. (Think of the faces of a cube, for example).

Let me sketch the proof of Kreĭn-Mil'man's theorem which states:

  If $K$ is a compact convex set in a locally convex space, then it is the closed convex hulls of its extremal points.

Proof:

For $x \in K$ the set $\{x\}$ is a face if and only if $x$ is extremal.
If $\phi: X \to \mathbb{R}$ is a continuous linear functional and $F \subset K$ is a face, then $\{x \in F\,:\,\phi(x) \text{ is minimal}\}$ is a face again.
Order the faces of $K$ by reverse inclusion. The intersection of a chain of faces is a face, thus every face contains a minimal face by Zorn's lemma.
Combine 2. and Hahn-Banach to see that a minimal face cannot contain more than one point.

Taking 3. and 4. together we see that we have proved what we need for solving the exercise.

  Every face of a compact convex set contains an extremal point.

Since it would be a pity to stop here, let me finish the argument for Kreĭn-Mil'man:

Since $K$ is a face of itself, there is at least one extremal point. If the closed convex hull $C$ of the extremal points were not all of $K$ then there were a point $z$ and a continuous linar functional $\phi$ such that $\phi(z) \lt \phi(c)$ for all $c \in C$. But by applying 2. to $\phi$ we find a face disjoint from all extremal points, a contradiction.

  Claim: If $y \in T(K)$ is extremal, then $F = \{x \in K\,:\,T(x) = y\}$ is a face.

In particular, $F$ contains an extremal point by what I just said.

Proof of the claim: Obviously, $F$ is closed and non-empty. If $F$ were not a face, there would be $x_0,x_1 \in K \smallsetminus F$ and $0 \lt \alpha \lt 1$ such that $(1-\alpha)x_{0} + \alpha x_1 \in F$, so $(1-\alpha) T(x_0) + \alpha T(x_1) = y$, but $T(x_{0}) \neq y \neq T(x_1)$, so $y$ is not extremal, a contradiction.

To see that we need to assume at least that $X$ is Hausdorff, let $X = \mathbb{R}^2$ with the semi-norm $|(x,y)| = |x|$ and let $Y = \mathbb{R}$ with the usual topology. Let $T: X \to \mathbb{R}$ be projection on the first coordinate, $T(x,y) = x$. This map is continuous and obviously the $y$-axis $K$ in $X$ is compact and convex and contains no extremal point. However, $T(K) = \{0\}$ and the point $\{0\}$ is extremal. By construction there is no extremal point in $K$ contradicting the exercise.

I think you will find this easier if you recall that "convex" can be defined in terms of inequalities and that the linear map will preserve those features.


Square root of a negative number squared √x^2=|x|,
What about √-x^2 ?
If we use the number $5$ as en example,
would this evaluate to √-5^2 = √25 =5
OR do we need to get the imaginary number 'i' involved, resulting in √-5^2 = 5i
I have found many conflicting answers
Any clarification on the topic would be much appreciated

Kind Regards

In real numbers, the square root of a negative is not defined.

$$\color{red}{\sqrt{-25}}.$$

In complex numbers, the square root of a negative can indeed be defined as $i$ times the square root of the absolute value.

$$\sqrt{-25}=5i.$$

But you need to understand the concept of principal branch, as $-5i$ could also be an acceptable answer.

The answer is ±5.

$-5$ is converted to $5i^2$ then squared, resulting to $25i^4$, according to the PEMDAS rule.

The root of $25i^4$ is then evaluated by taking the roots of a complex number. Note that $z = r(\cos\phi + i\sin\phi)$, where $z$ is the $n$-th root.

$r$ is the real root part, but in this case is used to contain imaginary numbers as well. Therefore, $r = \sqrt{25i^4} = 5i^2$, without regard to sign.

$\phi = 360°/n$, where $n$ is the $n$-th root. Thus, $n = 2$, therefore $\phi = 180°$.

Using the argand diagram, there are two cases, $\phi = 0°$ and $\phi = 180°$. Evaluating $z$, we have $z = -5i^2$ ($\phi = 180°$) and $z = 5i^2$ ($\phi = 0°$), giving us $z = ±5$ upon simplification.

The only thing I am not so sure about is if it is allowed to use imaginary numbers in $r$, as this is counter-intuitive from its derivation, and I'm gonna confirm it with my math doctor.

Hope this helps.


Solving the equation $ f^{-1}(x)=f(x)$ I attempted to solve the equation given in the title for the function;

$$f: \mathbb R_{++} \to\mathbb R_{++}; \quad f(x)=x^2(x+2)$$

I understand that the problem is equivalent to solving $f(f(x))=x$
but since this seemed like too much work, I had a look at the solution and it stated that;

$$ f^{-1}(x)=f(x) \Longrightarrow  f(x)=x$$

I don't understand why this is the case. Can someone please explain this?

Thanks

Although $f(f(x))=x\implies f(x)=x$ is not always true, it is true if $f$ is an increasing function (even if only weakly increasing).

Suppose that $f(x)=y$ and $f(y)=x$. If $x\le y$, then $f(x)=y\le f(y)=x$, and vice-versa.
So either way, both $x\le y$ and $y\le x$ are true, and thus $x=y$ and $f(x)=x$.

So your book is not off the rails; this is a valid approach to the problem.

Answering this problem is simple.

f(x) = x^2(x+2)

Thus f(x) = a(x)b(x) where a(x) = x^2 and b(x) = x+2

The product rule tells us that:

f'(x) = 3x^2 + 4x

Thus solve for f(x) = f(x)':

x^2(x+2) = 3x^2 + 4x

x(x^2 -x - 4) = 0

Therefore x = 0; x = (1±sqrt[17])/2


A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. I have this question as an example in my maths school book. The solution given there is:-

E = the man reports six

P(S1)= Probability that six actually occurs = $\frac{1}{6}$

P(S2)= Probability that six doesn't occur= $\frac{5}{6}$

P(E|S1)= Probability that the man reports six when six has actually occurred = $\frac{3}{4}$

P(E|S2)= Probability that the man reports six when six has not occurred = $1-\frac 3 4=\frac 1 4$

Therefore, by Bayes' Theorem,
$P(S1|E)=\frac{(\frac{1}{6}\cdot\frac 3 4)}{[(\frac 1 6\cdot 3 4)+(\frac 5 6\cdot 1 4)]} =\frac 3 8  $

I have its solution but my teacher said that the solution given is incorrect and told that the actual solution would be something else:-

$P(S1|E)=\frac{\frac 1 6\cdot\frac3 4}{(\frac 1 6\cdot \frac 3 4)+(\frac 5 6\cdot \frac1 4\cdot\frac1 5)} = \frac 3 4$

So, I want to ask which one is correct.
Thank you.

The difference in solutions comes in the estimation of the probability that the man reports six when six has not occurred.

If the man randomly chooses a number to report when he lies (which seems like a reasonable statement), then the probability he chooses 6 is 1/5. If you multiply your calculation of P(E|S2) by this, you get your teacher's solution.

Let us define that,

$E_1=A$ speaks truth;      $E_2=A$ tells a lie;       $E= A$ reports a six

Given,

$P (E_1) =\frac{3}{4},~ P (E_2) =\frac{1}{4},~ P (E|E_1) =\frac{1}{6},~  P (E|E_2) =\frac{5}{6}.$

The required probability that actually there were six (by Bayes’) is

$P (E_1|E) = [P (E_1) \cdot P (E|E_1)] / [P (E_1) \cdot P (E|E_1) + P (E_2) \cdot P (E|E_2)] = \frac{3}{8}$.

(Gourav's answer is right)


linear space W which contains three subspaces $V_1$, $V_2$, $V_3$ $⊂W$ such that $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$ Give an example of a linear space W of finite dimension $d \geq 4$ which contains three subspaces $V_1$, $V_2$, $V_3$ $⊂W$ such that $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$.
Is that possible? I mean the only possibility of $W=V_1⊕V_2=V_1⊕V_3=V_2⊕V_3$ that I can think of is $V_1=V_2=V_3$ but then $V_1⊕V_2=V_1⊕V_3=V_2⊕V_3 = \emptyset$. It's probably something basic and I wasted way to much time thinking of it.

Does this work? Let $\{e_1,e_2,e_3,e_4\}$ be a basis for a 4-dimensional $V$, then take
$V_1=\langle e_1+e_2,e_1+e_3\rangle$
$V_2=\langle e_4+e_2,e_4-e_3\rangle$
$V_3=\langle e_1,e_4\rangle$.
It's not difficult to show that $V_i\oplus V_j=V$ for $i\neq j$. Also not difficult to show that $V_1\cap V_3=V_2\cap V_3=\emptyset$. The only difficult bit is to show that $V_1\cap V_2=\emptyset$, which I haven't done here.

Consider W =M2×3(R) with subspace V1,V2,V3 such that
V1={A€W:a11,a12&a13 are non zero and all other entries are 0s}
V2={A€W:a21,a22 & a23 are non-zero and all other entries are 0s}
V3={A€W: a23 ,a22& a13 are nonzero and all other entries are 0s}
We get dimV1=dimV2=dimV3=3
And their mutual intersection will give zero space.


Interesting problem in congruence of triangles While solving the exercises of my book I came across this interesting problem:

  $\triangle ABC$ is isosceles triangle with $AB=AC$. D is a point on base BC such that $AD$ perpendicular on $BC$. To prove that $\angle BAD=\angle CAD$ a student does as follows.
  Between $\triangle ABD$ and $\triangle ACD$,

$AB=AC$ (given)
$\angle B=\angle C$ (because $AB=AC$)
$\angle ADB=\angle ADC$ ($=90^\circ$).

Therefore $\triangle ABD\cong \triangle ACD$. So, $\angle BAD=\angle CAD$. Now what is the defect in these arguments?

Now what is the defect in this argument? Please try to solve it.

The proof is correct.

$\angle B = \angle C$ is very vague. He should have written $\angle ABC = \angle ACB$.

But we can see that $\angle ACD = \angle ACB = \angle ABC = \angle ABD$.

Here is a counterexample of the argument, showing that the argument is not generally valid:

$BC=BC$ (common side)
$\angle ACB = \angle BCD$ (common angle)
$\angle ABC=\angle BDC$ (= $90^\circ$)

By the same logic, $\triangle ABC \cong \triangle BDC$. But they aren't. They are only similar triangles.

The key to the argument's apparent validity is that there, indeed, are equal corresponding sides of the two triangles in the student's case, namely, $BD=CD$ and $AD=AD$. But in this case there aren't. And the argument itself shows seeming congruence whether or not there are equal corresponding sides.

If the student really wanted to prove the congruence between the two triangles, simply use Pythagoras theorem.

Edit:
Yes, the comments below are right.

In addition, if defect does not mean invalidity but imperfection, like indirectness, then there is. In the student's argument step 2 and step 3 are enough to prove $\angle BAD=\angle CAD$.


Inverse of finite squared matrices. I've usually used that given a square matrix $A$ with determinant $\det(A)\neq0$, then its inverse $A^{-1}$ is the matrix that meets:

$$A^{-1}A=\mathbb{I}$$

and

$$AA^{-1}=\mathbb{I}.$$

However, I've found in several sources, including Wikipedia that if $A$ is finite and square, then:

$$A^{-1}A=\mathbb{I} \Longrightarrow AA^{-1}=\mathbb{I}.$$

I'm struggling to find a proof. Does anybody know one?

If $f, \, g \; : \; \mathbb{R}^{n} \; \longrightarrow \; \mathbb{R}^{n}$ are two linear maps such that : $f \circ g = \mathrm{Id}$ then, it implies that :

$f$ is invertible
$g$ is invertible
$g \circ f = \mathrm{Id}$.

To prove that $f$ and $g$ are invertible, you only need to prove that $\mathrm{ker}(f) = \lbrace 0 \rbrace$ and $\mathrm{ker}(g)=\lbrace 0 \rbrace$. Let $x \in \mathrm{ker}(g)$. Then, $(f \circ g)(x) = x$ and $(f \circ g)(x) = f(0)=0$. So, $x=0$ and it follows that $\mathrm{ker}(g) = \lbrace 0 \rbrace$ and $g$ is invertible. You can prove as well that $\mathrm{ker}(f) = \lbrace 0 \rbrace$ (so $f$ is invertible).

Let $x \in \mathbb{R}^{n}$. Since $g$ is invertible, there exist $u \in \mathbb{R}^{n}$ such that $x = g(u)$. Therefore, $(g \circ f)(x) = g\big( (f \circ g)(u) \big) = g(u) = x$. As a consequence, $g \circ f = \mathrm{Id}$.

If $A^{-1}A$ is the identity, this means that $A$ must be injective. Since we are in finite dimension and $A$ is square, this means that $A$ is surjective too. So it is invertible: let's call $B$ its inverse. Then, if you have $CA=Id$, it follows that $CAB=B$ and then $C=B$, which is what you wanted: $C$ is actually the inverse af $A$.

Let me remind you that this applies only in finite dimension. In infinite dimension you can very easily have linear operators that are injective but not surjective (or surjective, but not injective).


Does there exist a spherical quadrilateral with all angles pi/2? Does there exist a spherical quadrilateral with all angles pi/2?

I do not think so but I am not sure. I am unable to really visualize this.

Please offer suggestions.
Thank you.

Wouldn't having all the angles pi/2 just make the figure a regular square because every angle would be 90 degrees?


Number of solutions to a differential equation The number of solutions of $$\frac{dy}{dx} = \frac{y+1}{x-1}$$ when $y(1) = 2$ is

Options are a) none   b) one   c) two   d) infinite

Now this can be solved via variable separable method which gives me
$$\frac{dy}{y+1} = \frac{dx}{x-1}$$
And on integrating both sides I got,
$\log (y + 1) = \log(x - 1) + \log c$
which gave me
$y + 1 = c(x - 1)$
On putting the given values as $x = 1$ and $y = 2$ gives
$3 = 0$ or $c = \frac{3}{0}$ which means no solution.

I just want to ask if my solution is correct because the book gave the answer as one solution. I checked few other places and they for some reason are solving by integrating as
$\log (y + 1) = \log(x - 1) - \log c$
which of course would give them the desired one solution as $x = 1$.

I am confused.

The problem is ill-posed since $x=1$ is a singular point. But even if you consider the given point to be a limit, you can see from the general solution that

$$ \lim_{x\to 1} y(x) = \lim_{x\to1} \big[c(x-1)-1 \big] = -1 $$

Clearly, there are no such functions where $\lim_{x\to1} y(x) = 2$

If you consider $x(y)$ not $y(x)$ you'll get $x = 1 + (1 + y)\cdot c$, and therefore substituting $y=2$ you'll get $c=0$ if you want $x=1$. The solutions is therefore $x(y)=1$.

Edit: also mind that $\int dx/x=\ln|x|$.


Derivation of the Compound Interest Formula Without Using Calculus I am a student but I am only in Algebra 2. I am doing a project on compound interest and I am trying to prove how we arrive at the formula $y=Pe^{rt}$ but I am stuck at the step where you get from $y=P(1+r/n)^{nt}$ to $y=P((1+1/n)^{n/r})^{rt}$. I would really appreciate some help with my algebra. Thanks!

Generally, $y=P(1+r/n)^{nt}$ does not imply $y=P((1+1/n)^{n/r})^{rt}$ (Try $P=1, n=2,  r=2, t=2$. It fails, since $P(1+r/n)^{nt}=1(1+\frac{2}{2})^{2 \cdot 2}=16 \neq P((1+1/n)^{n/r})^{rt}=1((1+\frac{1}{2})^{2/2})^{2\cdot 2}=\frac{81}{16})$.

It looks like something may have been mistyped here.

However, it is true that $$y=P(1+r/n)^{nt}=P((1+r/n)^{n/r})^{rt}.$$

There is a rule for exponentiation that states $(a^b)^c=a^{bc}$. Assuming that $r$ is nonzero, we can write $$y=P(1+r/n)^{nt}=P(1+r/n)^{\frac{n}{r}\cdot (rt)}=P((1+r/n)^{\frac{n}{r}})^{\cdot rt}.$$

For compound interest if the amount at time $t$ is given by $A_t$:
$$A_1 = A_0 + rA_0 = A_0(1 + r)$$
$$A_2 = A_0 + rA_0 + r(A_0 + rA_0) = A_0(1 + r)^2$$
So in general:
$$A_t = A_0(1 + r)^t$$
Where do you want to go from here?


Calculating area of quadrilateral when distance of vertices from an arbitrary point is known Given a convex quadrilateral $ABCD$ circumscribed about a circle of diameter $1$. Inside $ABCD$ there is a point $M$ such that $MA^2 + MB^2 +MC^2 + MD^2 =2$. Find the area of the quadrilateral.

My attempt at a solution:
I tried to solve it using coordinate geometry. I guessed $M$ to be on centre of circle and taking $ABCD$ as a square, the sides come out to be $1$. Hence, area comes out to be $1$ unit sq.

However I was looking for a proper solution for the question.

By AM-GM

$$\sum_{cyc}MA^2=\frac{1}{2}\sum_{cyc}\left(MA^2+MB^2\right)\geq\sum_{cyc}MA\cdot MB\geq$$
$$\geq2\sum_{cyc}S_{\Delta AMB}=2S_{ABCD}=r\sum_{cyc}AB,$$
where $r$ is a radius of our circle.

In another hand, by AM-GM again
$$\left(\sum_{cyc}AB\right)^2\geq4(AB+CD)(AD+BC)=4\sum_{cyc}AB\cdot AD=$$
$$=2\sum_{cyc}(AB\cdot AD+BC\cdot CD)\geq2\cdot4\cdot2S_{ABCD}=16S_{ABCD},$$
which gives
$$\sum_{cyc}AB\geq4\sqrt{S_{ABCD}}=4\sqrt{\frac{1}{2}r\sum_{cyc}AB}$$ or
$$\sum_{cyc}AB\geq8r.$$
Thus, $$\sum_{cyc}MA^2\geq r\sum_{cyc}AB\geq8r^2=2.$$
The equality occurs for $$\measuredangle DAB=\measuredangle ABC=\measuredangle BCD=\measuredangle CDA=90^{\circ}$$ and
$$\measuredangle AMB=\measuredangle BMC=\measuredangle CMD=\measuredangle DMA=90^{\circ},$$
which says that $ABCD$ is a square.

Id est, $$S_{ABCD}=1\cdot1=1.$$
Done!

You can't do that, as there is no single solution.

Take the degenerate quadrilateral $ABCD$ where $A=B$ and $C=D$, with $M=C$, you have fulfilled the equation, but the area is $0$.


What is the smallest possible value of $q$ such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$? If $p$ and $q$ are positive integers such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$ then the smallest possible value of $q$ is:
$(A)\quad 60;\quad (B)\quad 30;\quad (C)\quad 25;\quad (D)\quad 7$.

What is the correct way to solve this kind of problems? I have tried to simplify the given inequality:
$$0.70<\frac{p}{q}< \approx0.73 \quad\text{ or }\quad \frac{21}{30}<\frac{p}{q}<\frac{22}{30}$$
Thank you in advance!

The way to find the smallest denominator "from scratch" is with continued fractions.

Begin by rendering the proposed bounds thusly:

$\dfrac{7}{10}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{3}}}}$

$\dfrac{11}{15}=\dfrac{1
}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{1+\dfrac{1}{3}}}}}$

The upper layers of the continued fractions are identical but they eventually become different when we get down to the layers in blue.  We may now replace those entries with the smallest whole number lying between, thus

$1+\dfrac{1}{3}<2<3$

So the smallest denominator fraction meeting the betweenness criterion will be

$\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{2}}}}=\dfrac{5}{\color{blue}{7}}$

We can take the common denominator:
$$
\frac{21q}{30q}<\frac{30p}{30q}<\frac{22q}{30q},\qquad\text{or}\qquad 21q<30p<22q.
$$

Then you go from the smallest $q$:
$$
21\cdot7=147<30p<22\cdot7=154, \qquad \ldots 120,150,180\ldots\\
$$
We see that $30p=150$ lies between $147$ and $154$.


at least one of 100 consecutive integers is relatively prime to all natural numbers less or equal 100 For an arbitrary integer $n$ define $A_n=\{i|n \leq i \leq n+99  \text{ where }i\text{ is an integer}\}$ (i.e. $A_n$ is 100 consecutive integers)

Is it true that for any integer $n$ there is an element in $A_n$ which is relatively prime to all numbers in $A_1$?

Thanks

Here is an explicit construction of an $A_n$ which fails to have an element that is relatively prime to all the elements in $A_1$. The idea behind this construction is as follows. $$A_n = \{n, n+1, n+2, \cdots, n+99\}.$$ Now we want to "spread" the $25$ primes in $A_1$ well enough such that everything in $A_n$ is congruent to $0 \pmod p$ for atleast one prime in $A_1$. Furthermore, if we assign $n+k \equiv 0 \pmod p$ for some $0 \le k \le 99$, then for all $m \equiv k \pmod p$, $n+m \equiv 0 \pmod p$ since $$n+m \equiv n+k \equiv 0 \pmod p.$$ Once I have assigned $n+k \equiv 0 \pmod p$, I look at the next value of $k$ (say $k'$) that is not already covered and assign $n+k' \equiv 0 \pmod p'$ where $p'$ is the next largest prime in $A_1$ after $p$. My construction is as follows.

$$n \equiv 0 \pmod {2\cdot 3 \cdot 13 \cdot 17}, \quad n + 1 \equiv 0 \pmod 5$$
$$n + 7 \equiv 0 \pmod {11}, \quad n + 23 \equiv 0 \pmod {19}$$
$$n + 25 \equiv 0 \pmod {23}, \quad n + 35 \equiv 0 \pmod {29}$$
$$n + 37 \equiv 0 \pmod {31}, \quad n + 43 \equiv 0 \pmod {37}$$
$$n + 49 \equiv 0 \pmod {41}, \quad n + 53 \equiv 0 \pmod {43}$$
$$n + 55 \equiv 0 \pmod {47}, \quad n + 59 \equiv 0 \pmod {53}$$
$$n + 65 \equiv 0 \pmod {59}, \quad n + 67 \equiv 0 \pmod {61}$$
$$n + 77 \equiv 0 \pmod {67}, \quad n + 79 \equiv 0 \pmod {71}$$
$$n + 83 \equiv 0 \pmod {73}, \quad n + 97 \equiv 0 \pmod {79}.$$

Mathematica gives the smallest value of such an $n$ to be $2479447854547486799186954214354$ and the following Python program quickly checks that this $n$ does indeed work. We could possibly generalize this question to ask for what size of $A_1$, do the translates of $A_1$ have this "gcd" property? The user catfish showed that $4$ has this property while $100$ does not. Interestingly enough, if we start out with $n \equiv 0 \pmod 2$ in my construction, it is not possible to find an $n$ so there has to be an underlying story behind why the first term needs to have more than $1$ prime factor in $A_1$.

L = []
for k in range(2479447854547486799186954214354, 2479447854547486799186954214354+100):
    L.append(k)
P = [2, 3, 5, 7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97 ]

while len(P) > 0:
    for k in L:
        if  k% P[0] == 0:
            L.remove(k)
    P.remove(P[0])

print (len(L))

If we assume $0$ is relatively prime to all integers, perhaps we can use the Chinese remainder theorem to see more. Let $n\in\mathbb{Z}$. An element $m\in A_{n}$ will be relatively prime to all of $A_{1}$ iff it has nonzero remainder modulo every prime in $A_{1}$.

This gives a system of congruences
$$
m \equiv a_{1} (\bmod p_{1})\\
...\\
m \equiv a_{N} (\bmod p_{N})
$$

where $p_{1},...,p_{N}$ are the primes in $A_{1}$, and $a_{1},...a_{N}\neq 0$.
Let $M = p_{1}\cdots p_{N}$.
Then by the Chinese remainder theorem, $m = (a_{1}b_{1}\frac{M}{p_{1}} + ... + a_{N}b_{N}\frac{M}{p_{N}}) + tM$ for some $t\in\mathbb{Z}$. (here each $b_{i}$ is $a_{i}^{-1}$ taken in $\mathbb{Z}/p_{i}\mathbb{Z}$)

So I think a way to find the answer is to consider the set $B$ of smallest positive values of $a_{1}b_{1}\frac{M}{p_{1}} + ... + a_{N}b_{N}\frac{M}{p_{N}} + tM$ for all $a_{1},...a_{N}\neq 0$ and then make sure that $\forall b\in B$ $\exists c\in B$ with $|b-c|\leq 100$. If this is the case, then the answer is affirmative, if not, then the answer is negative (this is because $\forall b\in B$, $b+tM \not\equiv 0 (\bmod p_{i})$ $\forall i$, that is, $b+tM$ is relatively prime to all of $A_{1}$ $\forall t\in\mathbb{Z}$). For your definition of $A_{n}$, $|B| = (p_{1}-1)\cdots (p_{N}-1) = 277 399 690 427 737 839 953 078 806 118 400 000$ though...

We could do this for a much simpler scenario; suppose we define $A_{n} = \{i | n\leq i\leq n + 3\}$. The primes in $A_{1}$ are $2,3$, and the nonzero values of $a = (a_{1},a_{2})$ are $(1,1)$ and $(1,2)$, and $B = \{1,5\}$. Here $M = 2\cdot 3 = 6$. Since the distance between $1,5$ is $4$ and so not strictly greater than $4$, and since all translates of $1,5$ by multiples of $6$ will have nonzero remainder modulo all primes of $A_{1}$, all such translates will be relatively prime to every element of $A_{1}$. So for any $A_{n}$, there will be an element of $A_{n}$ relatively prime to all of $A_{1}$.


What is the difference between regression and classification? What is the difference between regression and classification, when we try to generate output for a training data set $x$?

Regression: the output variable takes continuous values.

Classification: the output variable takes class labels.

It is important to be clear when using terms like regression, classification and prediction to discriminate between the task you are performing and the method used to perform it. A classification task involves taking an input and labelling it as belonging to a given class, so the output is categorical. On the other hand, a prediction task involves predicting a continuous valued output.

Methods for achieving these tasks include regression, in which a continuous valued output is estimated (or, rather, the expected value of a distribution on a continuous variable is estimated, conditional on a given set of input values). This can be used to carry out a prediction task, as you would expect. It can also be used to carry out a classification task, for example using logistic regression to estimate the log odds of the input pattern belonging to a given class. In this case, the task is classification, the method is regression.

Classification methods simply generate a class label rather than estimating a distribution parameter. K nearest neighbour is a good example where the task and the method are both called classification.


Intersection of $x+y+z=0$ and $x^2+y^2+z^2=1$ I got a problem which is to show that the circle that is the intersection of the plane $x+y+z=0 $ and the sphere $x^2+y^2+z^2=1$ can be expressed as
$$x(t)=\frac{\cos(t)-\sqrt{3}\sin(t)}{\sqrt{6}},$$
$$y(t)=\frac{\cos(t)+\sqrt{3}\sin(t)}{\sqrt{6}}$$
and
$$z(t)= \frac{-2\cos(t)}{\sqrt{6}}.$$
The circle turns out like a $45$ degree circle in three dimensions. I know if the circle is in two dimensions like $x$ and $y$, the parametrization will be $(x(t),y(t))=(r\cos(t),r\sin(t))$. But in this question , what should I do?

Here is how I solved for the intersection of the plane $x+y+z=1$ and the sphere $x^2+y^2+z^2 = 1$, which is a little different with your question but the steps are typically the same.


why does e raised to the power of negative infinity equal 0? Why is it that e raised to the power of negative infinity would equal 0 instead of negative infinity? I am working on problems with regards to limits of integration, specifically improper integrals and a little confused as to what things approach infinity our negative infinity versus approaching zero

Note that: $e^{k} = \dfrac{1}{e^{-k}}$. Thus if $k \to -\infty \Rightarrow -k \to +\infty$, and $e^{-k} \to +\infty$ because $e^{-k} > -k$, and the as the latter approaches $+\infty$, so does the former, hence $e^k=\dfrac{1}{e^{-k}} \to 0$.

$$e^{-k} = \dfrac{1}{e^{k}}$$
Also $e^{\infty} = \infty$  and  $e^{-k}= \dfrac{1}{e^k} \Rightarrow e^{-\infty} = \dfrac{1}{\infty}= 0$

(anything divided by infinity is zero)


If $AB = I$ then $BA = I$ If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.

I do not understand anything more than the following.

Elementary row operations.
Linear dependence.
Row reduced forms and their relations with the original matrix.

If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?

P.S.: Please avoid using the transpose and/or inverse of a matrix.

Dilawar says in 2. that he knows linear dependence! So I will give a proof, similar to that of TheMachineCharmer, which uses linear independence.

Suppose each matrix is $n$ by $n$. We consider our matrices to all be acting on some $n$-dimensional vector space with a chosen basis (hence isomorphism between linear transformations and $n$ by $n$ matrices).

Then $AB$ has range equal to the full space, since $AB=I$. Thus the range of $B$ must also have dimension $n$. For if it did not, then a set of $n-1$ vectors would span the range of $B$, so the range of $AB$, which is the image under $A$ of the range of $B$, would also be spanned by a set of $n-1$ vectors, hence would have dimension less than $n$.

Now note that $B=BI=B(AB)=(BA)B$. By the distributive law, $(I-BA)B=0$. Thus, since $B$ has full range, the matrix $I-BA$ gives $0$ on all vectors. But this means that it must be the $0$ matrix, so $I=BA$.

The OP from 10 years ago said he understood
1.Elementary row operations.
2.Linear dependence.
3.Row reduced forms and their relations with the original matrix.

I'll use only a bit more than (1) and (3). Each elementary row operation corresponds to left multiplication by an elementary matrix and every elementary matrix is invertible. I won't use (2) but I will use
(4) If the rref of square matrix has a zero row, then it does not have a left inverse.

Proof. Let square $H$ have rref $R$ whose last row is zero. Setting a parameter equal to 1 and using $R,$ there is a nonzero column vector $X$ such that $HX={\bf 0}.$ Suppose $KH=I.$ Then
$$ X=IX=KHX=K(HX)=K{\bf{0}}={\bf 0},\ \text{a contradiction.}$$
Let $AB=I.$ By (4), $I$ is the rref of $B.$  Therefore, $CB=I,$ where $C$ is a product of elementary matrices. By taking, in reverse order, the product of the inverses of these elementary matrices, it follows that $C$ is invertible. $$C^{-1}=C^{-1}(CB)=IB=B$$
so that $BC=I.\tag 1$   $$A=AC^{-1}C=(AB)C=IC=C\tag 2 $$
Combining (1) and (2), we conclude that $BA=I.$
(I just noticed the P.S. of the OP. Sorry. Also, there's probably overlap with earlier proofs. I just wanted to give a short, simple proof that did not use any concepts from vector spaces.)


If $|\vec {A}.\vec {B}|=|\vec {A} \times \vec {B}|$, then find the resultant of $\vec {A}$ and $\vec {B}$ If $|\vec {A}.\vec {B}|=|\vec {A} \times \vec {B}|$, then find the resultant of $\vec {A}$ and $\vec {B}$

My Attempt:
$$|\vec {A}.\vec {B}|=|\vec {A} \times \vec {B}|$$
$$|\vec {A}|.|\vec {B}|.\cos \theta=|\vec {A}|.|\vec {B}|\sin \theta n^{\cap}$$
$$\cos \theta=\sin \theta n^{\cap}$$

How do I proceed further?

Assuming that $\vec A$ and $\vec B$ are non-zero:

$$\begin{array}{rcl}
|\vec A \cdot \vec B| &=& |\vec A \times \vec B| \\
|\vec A| |\vec B| \cos \theta &=& |\vec A| |\vec B| \sin \theta \\
\cos \theta &=& \sin \theta \\
\theta &=& 45^\circ
\end{array}$$

Therefore, $|\vec A + \vec B|^2 = |\vec A|^2 + |\vec B|^2 - 2|\vec A||\vec B|\cos45^\circ = |\vec A|^2 + |\vec B|^2 - \sqrt2|\vec A||\vec B|$.

Therefore, $|\vec A + \vec B| = \sqrt{|\vec A|^2 + |\vec B|^2 - \sqrt2|\vec A||\vec B|}$.


Solving for the value of $x$ I have the equation below which is correct:

$$\frac{(9650 - 9450)}{9650} \cdot 100 \cdot 5 = 10.36.$$

Supposed I want to solve for $9450$. So, substituting $x$ in:

$$\frac{(9650 - x)}{9650} \cdot 100 \cdot 5 = 10.36.$$

I tried solving the value of $x$ but I got $x = 9854.18$ which is wrong!

What will be the new formula to get the correct answer which should result in $x = 9450$?

Any help is greatly appreciated. Thanks in advance.

The correct answer is 9450.052. Here's how to solve it:

First, divide both sides by 100 and 5.
Second, multiply both sides by 9650, which is the denominator on the left side of the equation. This value (199.948) equals the difference between 9650 (in the numerator) and x. Symbolically, 9650-x=199.948. We add x to both sides to get 9650=199.948+x. Then, we subtract 199.948 from both sides to get x=9650-199.948. Putting this into the calculator, we have x=9450.052, which rounds to 9450.


$f(x)=|\cos x|+|\sin(2-x)|$ at which of the following point $f$ is not differentiable? $f(x)=|\cos x|+|\sin(2-x)|$ at which of the following point $f$ is not differentiable?

1.$\{(2n+1){\pi\over2}\}$

2.$\{n\pi\}$

3.$\{{n\pi\over 2}\}$

4.$\{n\pi+2\}$

in all cases $n\in\mathbb{Z}$

well, is there any easy trick to solve this type of probelem in an competitive examination? Do I need to pick $\pi \over 2$ from $1$ and cheked that $f$ is differentiable and same trick from each set? in some what I am lost.

A function isn't differentiable where it has sharp corners since the tangent line at that point is not well-defined. In this case, it fails to be differentiable when $\cos(x)$ and $\sin(2-x)$ change sign since the absolute value of a function has a sharp cusp when its argument changes sign.

Hint: The function $ g(y)=|y|$ isn't differentiable at $ y=0$, and therefore, $ g(\cos x)+g(\sin(2-x))$ isn't differentiable when it's arguments are $0$, meaning, when $ \cos x=0$ or when $ \sin (2-x)=0$


How many $3$-digit numbers can be formed so that the sum of two digits will be equal to the third digit? How many $3$-digit numbers can be formed so that the sum of two digits will be equal to the third digit?

I am confused in this question whether to take first 2 digits sum or any digit sum such that it is equal to 3rd digit.

For example, some combinations 112 can be written into 3 ways  112 112 121
such that in the last

Example 3rd number and first number sum = 2 (middle number).

Also, is there any better approach towards this problem rather than counting and arranging the 3 digit number?

To write the problem in somewhat mathematical terms, we have three digits $d_1 \in \{1, \dots, 9\}$, $d_2,d_3 \in \{0, 1, \dots, 9\}$ that form a $3$-digit number
$$x = d_1 \cdot 10^2 + d_2 \cdot 10 + d_3$$
Now, we want to find all possible values of $x$ such that at least one of the following holds:

$d_1 + d_2 = d_3$,
$d_1 + d_3 = d_2$,
$d_2 + d_3 = d_1$

Let's start with the first condition. We may pick $d_1$ freely, which gives us $9$ options. However, since $d_3 \in \{0, 1, \dots, 9\}$, we must always choose $d_2 \in \{0, \dots, 9-d_1\}$. Hence:
$$\# \{ x \mid d_1 + d_2 = d_3\} = \sum_{d_i = 1}^{9}(10-d_i) = 45$$
The same reasoning holds for the second condition:
$$\# \{ x \mid d_1 + d_3 = d_2\} = \sum_{d_i = 1}^{9}(10-d_i) = 45$$
However, the last condition is different! We pick $d_2$ freely which gives us $10$ options, and we must choose $d_3 \in \{0, \dots, 9-d_2\}$. However, we also have to exclude one case: $d_2 = d_3 = 0$, because then $d_1$ would also be $0$. Hence:
$$\# \{ x \mid d_2 + d_3 = d_1\} = \sum_{d_i = 0}^{9}(10-d_i)-1 = 54$$
We now have $45 + 45 + 54 = 144$ numbers $x$ such that either of the conditions hold. However, some numbers are counted twice. It is easy to see that a number is counted twice if and only if $d_i = d_j$ and $d_k = 0$ for $i,j,k$ being some permutation of $1,2,3$. Since $d_1 \neq 0$, the multiples we need to discard are those of the form $d_2 = 0$ or $d_3 = 0$. There are $9$ of each ($d0d$ and $dd0$ with $d = 1, \dots, 9$), so we remove $18$ multiples. There is no number for which all $3$ conditions hold, since that would require $d_1 = d_2 = d_3 = 0$.

The solution is then:
$$\#\{x\} = 144- 18 = 126.$$

I'm going to assume that the problem means that the sum of any two digits equals the third.

To start, let's find out how many 3 digit numbers have the last digit as the sum of the first two digits:

Here, the first digit cannot be $0$ or $9$.  It can't be $9$ because the sum of $9$ and any digit is a two digit number If the first digit is $1$ then second number can be any digit from $0$ to $8$.  So nine possibilities.  If the first digit is $2$ then the possibities for the second digit is $0$ through $7$.

So, the number of three digit numbers whose first two digits add up to the third is: $9+8+\cdots +1$.

Now, the total of all numbers in which the sum any two digits is 3 times the previous answer because the sum digit could be the first, second or third digit.


Symmetric matrix eigenvectors. I've noticed that if I have a square matrix of dimension n. Where the diagonal entries are all equivalent to one another and the off diagonal entries are also equivalent to one another (but not necessarily equivalent to the diagonals). Say: $$
\begin{bmatrix}
a & b & b \\
b & a & b \\
b & b & a
\end{bmatrix}
$$
I always get one eigenvector of the form
$$
\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}
$$

and the remainder with identical eigenvalues.

Is this a well known result? Can anyone direct me to a proof of such a result?

It's well known and not particularly remarkable (it's common in linear algebra tests, and I recall one question on SE that I answered that involved precisely this type of a matrix).

Basically, write your matrix as a sum of a scaled identity and a projection:

$$A=(a-b)I+b \vec{n}\otimes\vec{n}$$
where $\vec{n}$ is a vector of all ones. This means there is only one preferred (special) direction in this system: $\vec{n}$. So $\vec{n}$ is the only vector that can have a special eigenvalue, and all the vectors from its orthogonal complement are equivalent.

Extract the eigenvalue of $\vec{n}$:

$$A\vec{n}=(a-b)\vec{n}+b\vec{n}(\vec{n}\cdot\vec{n})=(a+2b)\vec{n}$$
where $|\vec{n}|^2=3$ is the only place where the dimension of the matrix comes in.

The other two eigenvalues must then be $(a-b)$:
$$A\vec{n}_\perp=(a-b)\vec{n}_\perp+0=(a-b)\vec{n}_\perp$$

Consider the identity matrix $I$ and the matrix $J$ with a $1$ in every cell. Your matrix $A$ is now $A = bJ + (a-b)I$.

Multiplying a matrix by $b \neq 0$ does not change the eigenspaces, it just multiplies their corresponding eigenvalues by $b$.

Adding $(a-b)I$ to a matrix does not change the eigenspaces, it just increments their corresponding eigenvalues by $(a-b)$.

So the eigenspaces of $A$ are exactly the eigenspaces of $J$.


Prove the following: Let $n$ be an integer. If $2|n^2$, then $4|n^2$. I came up with the following:

$2|n^2$ implies that $2|n*n$. We proved in class that if $q|b*p$, then $q|b$ or $q|p$. Therefore, if $2|n^2$, then $2|n$ or $2|n$.
So, $2|n$ implies $n=2k$, for some $k ∈ Z$.
So, $n^2|(2k)^2$,
    $n^2 = 4k^2$, where $k^2 ∈ Z$.

So, if $2|n^2$, then $4|n^2$ as desired.

What do you all think?

If $2|n^2$, $n^2$ is even. Note that  $n^2$ is even if and only if  $n$ is even (in $\Bbb Z$). So  $n=2m $ for some  $m \in \Bbb Z$. Hence  $n^2=4m^2$.

An import ant part of writing any proof is to determine from the outset what you can assume to be true.

If you can assume the fundamental theorem of arithmetic then you have the unique prime factorisation of $n=p_1^{i_2}\cdot p_2^{i_2}\cdot p_3^{i_3}\ldots$

Then the prime factorisation of $n^2=p_1^{2i_2}\cdot p_2^{2i_2}\cdot p_3^{2i_3}\ldots$

Then since every exponent is even, the power of every prime factor of $n^2$ is either $1$ or a power of $p^2$, which generalises the result you need to every prime.

(In $2^3=8$, 2 is the prime, 3 is the exponent and 8 is the power).


Justifying differentiation of function If a function say $g(x)$ is Continuous such that,

$$
g'(x) = 2x\sin \frac{1}{x} - \cos \frac{1}{x}  , \forall x \neq 0
$$

Will this $g$ be differentiable at $0$ ?

My attempt:
As the $\lim\limits_{x\rightarrow 0} g'(x)$ doesn't exists hence i think the $g$ won't be differentiable at $0$

Is there way to show it or prove otherwise?
Any hint will be appreciated...

For $x \neq 0$ it can be checked by differentiating that

$$
g(x) = \begin{cases}
x^2\sin(1/x) + C & x > 0, \\
x^2\sin(1/x) + D & x < 0,
\end{cases}
$$

where $C$ and $D$ are real constants. Now

$$
\lim_{x \to 0^+} g(x) = C \qquad \text{and} \qquad \lim_{x \to 0^-} g(x) = D,
$$

so for $g$ to be continuous we need $C = D$ and $g(0) = C$. Thus

$$
g(x) = \begin{cases}
x^2\sin(1/x) + C & x \neq 0, \\
C & x = 0.
\end{cases}
$$

We can then calculate

$$
\lim_{x \to 0} \frac{g(x) - g(0)}{x-0} = \lim_{x \to 0} \frac{x^2\sin(1/x)}{x} = \lim_{x \to 0} x\sin(1/x) = 0,
$$

which shows that $g$ is differentiable at $0$ with $g'(0) = 0$.

EDIT my answer regards the question if the function is continuously differentiable ( of class $\bf C^1$ )

The classification of functions into these classes is something you will often when studying more advanced analysis. So this answer attacks the question :

Is $g$ continously differentiable?

For it to be continously differentiable what we must do is to check that $g'(x)$ is continuous everywhere.

Both $\sin(1/x)$ and $\cos(1/x)$ will oscillate between -1 and 1 faster and faster no matter how small neighbourhood you pick. The first term will therefore be bounded by $|\pm2x|$ which you can prove will be squeezed to 0 but the second one only $|\pm 1|$.

Depending on the strictness of your course you may want to prove this with $\epsilon-\delta$ or some other method but it will probably be straight forward to do so. So neither left nor right limit for $g'(x)$ will exist for $x=0$, therefore it can not be continuous there. So the function is not in $\bf C^1$


How is defined the output of an exponential function when the input is not a rational number? Suppose $f$ is a function such that $f(x)= 2^x$.

Certainly, $f$ is defined when $x$ is an integer.

It is also defined when $x$ is a rational number such that $x=\frac mn$ ( with $m$ and $n$ integers), for in that case we have an exponent rule telling us that :

$2^\frac mn$ = $\sqrt[n]{2^m}$.

But since function $f$ is continuous on $R$ ( the set of real numbers), it must also be defined when the exponent $x$ is not a rational number, and hence cannot be written as a fraction.

So for example, how is $f(\pi)=2^\pi$ defined?

That's a good question and there are two standard answers.

1) Based on calculus.

Forget everything.

Note that that for $t > 1$ that $\frac 1t >0$ so that if $F(x)=\int_{1}^{x}\frac 1t dt$ when viewed as "the area under a curve for $\frac 1t$ from $t=1$ to $t=x$" is a continuous monotonically increasing function. Thus it is one-to-one/injective.  A little bit of work we can see that $\lim_{x\to \infty}F(x) =\int_{1}^{ \infty} \frac 1t dt =  \infty$ and $\lim_{x \to 0^+} F(x) = \lim_{x \to 0^+} \int_1^x \frac{1}{t} dt = - \infty$.

Thus $F: (0, \infty) \to \mathbb R$ is injective and surjective (so bijective and invertible) function.

We define $Ln(x) := F(x)$.  And we define $Exp(x) = Ln^{-1}(x)$, the inverse function of $Ln$.  It's easy to see $Ln(1) =0$ so $Exp(0)=1$ and that for $0< x<1$ then $Ln(x)<0$ so for $w<0$ that $0<Exp(w)< 1$ and for $x > 1$ and $w>0$ that $Ln(x) > 0$ and $Exp(w)> 1$.  And that there is a real number $e$ so that $Ln(e) =1$ and by the nature of $Ln(x) =\int_{1}^x \frac 1t dt$ that $e=\lim_{n\to \infty} (1+\frac 1n)^n$.

Using basic laws of derivatives and antiderivatives we can show that $Ln(xy)= Ln(x)+Ln(y)$ (assuming $x,y$ are positive) and that $Exp(x+ y)= Exp(x)Exp(y)$.

Then we define for any $b > 0$ that we be use as notation $b^x = Exp(Ln(b)*x)$.  From this we can show that if $x\in \mathbb N$ that $b^x=\underbrace{b\cdot....\cdot b}$ and by extension if $x\in \mathbb Q$ then $b^x$ is .... the exponential function we all know and love.

Also $e^x = Exp(Ln(e)*x)=Exp(1*x) = Exp(x)$ and if we define (for $b>0;b\ne 1$) the notation $log_b M$ as being the unique $k$ so that $b^k=M$ then $log_b M$ can be shown to be equal to $log_b M = \frac {Ln(M)}{Ln(b)}$

This may seem artificial and completely backwards (at least it does to me) but it is actually the most common definition and it has the advantage in that it works, and it very powerfully introduces the natural exponent and logarithm and immediately validates the obscure fact that the derivative of the natural log of $x$ is $\frac 1x$ and that the anti derivative of $\frac 1x=x^{-1}$ which ought to be $\lim_{h\to 0} \frac 1h x^h= \ln x$.

2) limits.

For every real number $x$, rational or irrational, there is a Cauchy sequence of rational numbers $\{q_n\}$ so that $\lim_{n\to \infty} q_n = x$.  (That's basically the definition of the real numbers.)

We can show that the sequence $\{b^{q_n}\}$ converges to a real number.  And we can show that if $\{q_n\}$ and $\{r_n\}$ are twoo sequences of rational numbers both converging to $x$ then the sequences $\{b^{q_n}\}$ and $\{b^{r_n}\}$ both converge to the same real number.

We define $b^x$ as $\lim_{n\to \infty} b^{q_n}$.

In my opinion this definiton makes the most intuitive sense.  However it is the less common definition.  Probably because it is a lot harder than it looks and also it takes quite a bit of work to figure out the natural exponents and logarithms or the notion that the derivative of the natural log is $\frac 1x$.

Post-script: As Martin Agerami points out in a comment. Another issue of the process of defining

$b^n; n\in \mathbb Z; n\ge 2$ as $\underbrace{b\times....\times b}_{n\text{ times}}$ which is clearly well defined (and which as a consequence means $b^{n+m}=b^nb^m$ and $(b^n)^m=b^{nm}$).

$b^0 = 1; b^1 =b$ for consistency and for negative values $b^{-n}=\frac 1{b^n}$ which is alwo well-defined.

And $b^{q}; q\in \mathbb Q; q=\frac mn; m,n\in \mathbb Z;n> 0$ as $\sqrt[n]{b}^m$ with the understanding that $b$ is either positive of if $b < 0$ that $m$ is odd; ... this has immediate issues (not insurmountable but issues nonetheless) that we have to prove roots actually exist (its not a given) and that if $q =\frac mn=\frac ac$ that $\sqrt[n]{b}^m=\sqrt[b]{b}^c$ (which turns out to be true if $b$ is positive but not if $b$ is negative)

It's a little weird in that the, to me, intuitive and straightforward definitions have issues, that to be solved requires knowledge of continuity.  But once one has the knowledge to address the issues, it turns out wiping the slate clean and creating a definition based on that knowledge but completely different in apperence is actually better and easier.

Let $\{a_n\}$ be a sequence of rational numbers such that $$\lim _{n\to \infty }  a_n=\sqrt 5$$

Define $$2^{\sqrt 5 } = lim _{n\to\infty} 2^{a_n}$$

It remains to show that the sequence is convergent and the limit does not depend on the initial rational sequence approaching $\sqrt 5$


If a function maps an input to its inverse, is it bijective? I read in my textbook that a function is a bijection if and only if it has an inverse. Is it the same thing to say a function $f: X → X$ is a bijection if $f(x) = x^{-1}$? If $a = x$ and $b = x^{-1}$, then I'd have $f(a) = b$, and in this situation $f(b) = a$. Would that make $f$ its own inverse and therefore mean it's bijective?

Your problem stems from the fact that the word "inverse" has (at least) two meanings. One describes the inverse of a function, another the inverse of a number (or element of some algebraic structure where that makes sense).

These meanings are essentially different unless you move up one level of abstraction, where you "multiply" functions by composing them. Then they can occur in the same problem - with their two meanings. That's exactly what you encountered in your example with permutations. As @DustanLevenstein points out, the function from permutations to permutations that assigns to each permutation its inverse (as a permutation thought of as a function) has an inverse (as a function).

(This is a good beginner's question.)

Your $x^{-1}$ is an example of a function with an inverse and therefore it can be a bijection but that isn't the meaning of having an inverse, or the notation.

Having an inverse means that there exists another function which reverses the function $f$, so the inverse of $x+1$ is $x-1$.  The inverse of $2x$ is $x/2$.  The correct notation to say there exists an inverse of $f$ is this: $\exists f^{-1}(x)$

By chance, your function $x^{-1}$ is unusual in that it is its own inverse, i.e. apply it twice and you get back where you started. This is written $f^{-1}(x)=f(x)=x^{-1}$.  That type of function is called an involution, but this is a distraction, it's not fundamental to the understanding of bijections.

A bijective function means you must define the domain and range, and both the function and its inverse map every element of one to exactly one element of the other, with no members of either space not mapped. So for example the function $x+1$ is bijective from the integers to the integers because every integer is mapped to another without exception.  Its inverse is $x-1$.


Show that matrix of $T$ has at least dim range $T$ nonzero entries. Suppose $V$ and $W$ are finite-dimensional and $T \in \mathcal{L}(V, W)$. Show that with respect to each choice of bases of $V$ and $W$, the matrix of $T$ has at least dim range $T$ nonzero entries.

I have no idea on how to solve this.

This is an exercise from Linear Algebra Done Right, so here's an answer in line with Axler's style: Choose bases $v_1,v_2,...,v_n$ of $V$ and $w_1,w_2,...,w_m$ of $W$ and let $\mathcal{M}(T)$ be the matrix of $T$ with respect to these bases. Column $j$ of $\mathcal{M}(T)$ will consist entirely of zeros iff $Tv_j=0$; equivalently column $j$ has at least one nonzero entry iff $Tv_j \not=0$. So out of the $n$ basis vectors of $V$ say $k \leq n$ of them are mapped to zero; assume without loss that $Tv_1,Tv_2,...,Tv_k=0$. Then $\mathrm{range} \: T = \mathrm{span}(Tv_{k+1},Tv_{k+2},...,Tv_n)$, which has dimension $\leq n-k$ (since these vectors need not be linearly independent.) Now for each $j=k+1,k+2,...,n$, $Tv_j \not=0$ so at least one of the scalars in its basis expansion with respect to the $w_i$'s is nonzero; hence the corresponding column of the matrix has at least one nonzero entry. So each of these $n-k$ vectors contributes at least one nonzero entry to the matrix, so we have at least $n-k$ nonzero entries (which is at least $\mathrm{dim} \: \mathrm{range} \: T$.)

Suppose we have a basis {$v_{1},...,v_{n}$} of $V$ and {$w_{1},...,w_{m}$} of $W$, and so $M(T)$ is an $m$ by $n$ matrix. Let $p$ denote the number of zero-columns in the matrix. Thus, we know the dimension of the null space is at least $p$. And so by the fundamental theorem of linear maps, $n = \dim \operatorname{ran}(T)+\dim \operatorname{null}(T) \geq \dim \operatorname{ran}(T)+p$. Therefore, $n-p \geq \dim \operatorname{ran}(T)$, and we are done.


Find the roots of the function find the roots of the function when $x$ goes from $0$ to $3$?

$$f(x) = \tan(2 x) - 1$$

I tried to set $f(x)=0$ but came up with $x=28$

root of trigonometric function is not unique

Just set $f(x) = 0$, that is,

$\tan(2x) - 1 = 0 \implies \tan(2x) = 1 \implies 2x = \arctan(1) \implies x = \frac{\arctan(1)}{2} \approx 0.3927$


Logic Statement Simplifiction For the following logical statements, find statements that are logically equivalent using fewer symbols.

P $ \land$ P

P $ \lor$ P

(P $\land$ ~P) $\lor$ Q

P $ \land$ P = P

P $ \lor$ P = P

(P $\land$ ~P) $\lor$ Q = Q

//Proved by truth tables

Your answer clearly works fine for what you wanted, so I've upvoted it and you can accept it in a few days if you wish.  However, the assumption of only two truth values seems quite strong for this problem.  The following formulas in Polish notation taken as axioms, with C-detachment and uniform substitution as the only rules of inference, allow for the deduction of the formulas EKxxx, EAxxx, and EAKxNxyy.  They all hold for an intuitionistic propositional logic which does not not have any adequate finite-valued model (semantics).

Axiom 1 CxCyx

Axiom 2 CCxCyzCCxyCxz

Axiom 3 CKxyx

Axiom 4 CxCyKxy

Axiom 5 CKxNxy

Axiom 6 CxAyx

Axiom 7 CAxyCCxzCCyzz

Axiom 8 CCxyCCyxExy

I just input corresponding parenthesized formulas with a predicate "P" for provable into Prover9 and after 93.38 seconds, it has produced each of the corresponding formulas.

The notation $\alpha$/$\beta$ indicates that $\alpha$ gets uniformly substitute with $\beta$ throughout a given formula.  The notation C$\alpha$-$\beta$ indicates that the formulas $\alpha$ and $\beta$ qualify as substitution instances of that form, with $\alpha$ already proved, and $\beta$ about to get detached.  You might want to note how the law of identity Cxx gets used in this proof.

Axiom                    1     CxCyx
Axiom                    2     CCxCyzCCxyCxz
Axiom                    3     CKxyx
Axiom                    4     CxCyKxy
Axiom                    5     CKxNxy
Axiom                    6     CxAyx
Axiom                    7     CAxyCCxzCCyzz
Axiom                    8     CCxyCCyxExy
2 z/x                    9     CCxCyxCCxyCxx
9 * C1-10               10     CCxyCxx
10 y/Cyx                11     CCxCyxCxx
11 * C1-12              12     Cxx
1 x/CKxNxy, y/z         13     CCKxNxyCzCKxNxy
13 * C5-14              14     CzCKxNxy
2 z/Kxy                 15     CCxCyKxyCCxyCxKxy
15 * C4-16              16     CCxyCxKxy
1 x/CxAyx, y/z          17     CCxAyxCzCxAyx
17 * C6-18              18     CzCxAyx
2 x/Axy, y/Cxz, z/CCyzz 19     CCAxyCCxzCCyzzCCAxyCxzCAxyCCyzz
19 * C7-20              20     CCAxyCxzCAxyCCyzz
2 x/Cxy, y/Cyx, z/Exy   21     CCCxyCCyxExyCCCxyCyxCCxyExy
21 * C8-22              22     CCCxyCyxCCxyExy
8 x/Kxy, y/x            23     CCKxyxCCxKxyEKxyx
23 * C3-24              24     CCxKxyEKxyx
20 x/KxNx               25     CCAKxNxyCKxNxzCAKxNxyCCyzz
14 z/AKxNxy, y/z        26     CAKxNxyCKxNxz
26 * C25-27             27     CAKxNxyCCyzz
22 x/Azy                28     CCCAzyyCyAzyCCAzyyEAzyy
18 x/CAzyy, x/y, y/z    29     CCAzyyCyAzy
28 * C29-30             30     CCAzyyEAzyy
16 y/x                  31     CCxxCxKxx
31 * C12-32             32     CxKxx

break:

1 x/Cxx                 33     CCxxCyCxx
33 * C12-34             34     CyCxx
2 x/AKxNxy, y/Cyz       35     CCAKxNxCCyzzCCAKxNxyCyzCAKxNxyz
35 * C27-36             36     CCAKxNxyCyzCAKxNxyz
24 y/x                  37     CCxKxxEKxxx
37 * C32-38             38     EKxxx
20 z/x                  39     CCAxyCxxCAxyCCyxx
34 y/Axy                40     CAxyCxx
39 * C40-41             41     CAxyCCyxx
2 x/Axy, y/Cyx, z/x     42     CCAxyCCyxxCCAxyCyxCAxyx
42 * C41-43             43     CCAxyCyxCAxyx
36 z/y                  44     CCAKxNxyCyyCAKxNxyy
34 y/AKxNxy             45     CAKxNxyCyy
45 * C44-46             46     CAKxNxyy
30 z/KxNx               47     CCAKxNxyyEAKxNxyy
47 * C46-48             48     EAKxNxyy
43 y/x                  49     CCAxxCxxCAxxx
34 y/Axx                50     CAxxCxx
49 * C50-51             51     CAxxx
30 z/x, y/x             52     CCAxxxEAxxx
52 * C50-53             53     EAxxx

Using OTTER, still an excellent assistant, the following got produced, which helped me to construct the above proof:

-----> EMPTY CLAUSE at   2.94 sec ----> 38809 [hyper,2,38788,14428,245] $F.

Length of proof is 20.  Level of proof is 7.

---------------- PROOF ----------------

1 [] -P(C(x,y))| -P(x)|P(y).

2 [] -P(E(A(p,p),p))| -P(E(A(K(p,N(p)),q),q))| -P(E(K(p,p),p)).

3 [] P(C(x,C(y,x))).

4 [] P(C(C(x,C(y,z)),C(C(x,y),C(x,z)))).

5 [] P(C(K(x,N(x)),y)).

6 [] P(C(K(x,y),x)).

7 [] P(C(x,C(y,K(x,y)))).

8 [] P(C(x,A(y,x))).

9 [] P(C(A(x,y),C(C(x,z),C(C(y,z),z)))).

10 [] P(C(C(x,y),C(C(y,x),E(x,y)))).

14 [hyper,1,4,3] P(C(C(x,y),C(x,x))).

16 [hyper,1,3,5] P(C(x,C(K(y,N(y)),z))).

20 [hyper,1,4,7] P(C(C(x,y),C(x,K(x,y)))).

28 [hyper,1,3,8] P(C(x,C(y,A(z,y)))).

36 [hyper,1,4,9] P(C(C(A(x,y),C(x,z)),C(A(x,y),C(C(y,z),z)))).

40 [hyper,1,4,10] P(C(C(C(x,y),C(y,x)),C(C(x,y),E(x,y)))).

44 [hyper,1,10,6] P(C(C(x,K(x,y)),E(K(x,y),x))).

60 [hyper,1,14,8] P(C(x,x)).

104 [hyper,1,36,16] P(C(A(K(x,N(x)),y),C(C(y,z),z))).

110 [hyper,1,40,28] P(C(C(A(x,y),y),E(A(x,y),y))).

156 [hyper,1,20,60] P(C(x,K(x,x))).

161 [hyper,1,3,60] P(C(x,C(y,y))).

165 [hyper,1,4,104] P(C(C(A(K(x,N(x)),y),C(y,z)),C(A(K(x,N(x)),y),z))).

245 [hyper,1,44,156] P(E(K(x,x),x)).

268 [hyper,1,36,161] P(C(A(x,y),C(C(y,x),x))).

324 [hyper,1,4,268] P(C(C(A(x,y),C(y,x)),C(A(x,y),x))).

14389 [hyper,1,165,161] P(C(A(K(x,N(x)),y),y)).

14428 [hyper,1,110,14389] P(E(A(K(x,N(x)),y),y)).

38716 [hyper,1,324,161] P(C(A(x,x),x)).

38788 [hyper,1,110,38716] P(E(A(x,x),x)).

38809 [hyper,2,38788,14428,245] $F.

------------ end of proof -------------

Perhaps of interest, the proof analysis indicates that the most common used formula 3 here corresponds to Axiom 2 CCxCyzCCxyCxz.  Also, formula 3 corresponds to Axiom 1 which gets used in two different ways, unlike say how formula 4 only gets used in one way.  Formula 10 corresponding to axiom 8 CCxyCCyxExy also gets used in two different ways.  Formula 8 corresponding to axiom 6 CxAyx only gets used in one way here, as one of the shorter significant expressions in a proof.  From the context of some multi-valued logic, or even infinite-valued logic formula 268 stands out, since disjunction Axy gets defined as CCxyy.  Also, CCxyy = CCyxx, so CAxyCCyxx is consistent with such.  I will conjecture that a shorter proof under condensed detachment can get constructed, since I find CxCyy suspicious for a shortest proof.


Finding the null space of symmetric matrix generated by outer product Let $p, q \in \mathbb{R}^n$ such that $\|p\| = \|q\| = 1$ and define $$A = pq^T + qp^T$$ Find the null space of $A$.

I am not having very much luck. I have managed to show that $p + q$ and $p - q$ are eigenvectors, but I can't seem to connect those two. I know that since $A$ is generated by an outer product, it is of rank one. That implies that the rank of the null space is $n - 1$, but I can't seem to get any further. Can anyone help?

First notice that any vector $w$ orthogonal to both $p$ and $q$ is the null space of $A$, since
$$Aw=p(q^Tw)+q(p^Tw)=0.$$
Thus the null space has dimension at least $n-2$. Since the eigenvectors $p\pm q$ correspond to eigenvalue $p^Tq\pm 1$, at least one of the two eigenvalues is non-zero, so the nullspace has dimension at most $n-1$.

Now we have two cases to cover :

$p,q$ are linearly independant : in this case the Cauchy-Schwarz inequality implies that both eigenvalues above are non-zero, and the null space has dimension $n-2$.
$p,q$ are not linearly independant : then since they have the same norm we must have $p=\pm q$, thus $\langle p,q\rangle$ has dimension $1$, and its orthogonal complement has dimension $n-1$.

In both case, the null space is the orthogonal complement of $\langle p,q\rangle$ because one is included in the other and their dimensions agree.

You can even see the result more easily, without even considering eigenvalues and eigenvectors. The first equation shows that $\langle p,q\rangle^{\perp}\subset Ker A$. For the reverse inclusion, just notice that if $p,q$ are linearly independant then
$$0=Aw=pq^Tw+qp^Tw\Rightarrow q^T w =0=p^Tw,$$and if they aren't then the dimensions agree (as explained above); or you can notice that $p=\pm q$ and thus $A=\pm 2 qq^T$, and thus
$$0=Aw=\pm 2qq^Tw\Rightarrow q^Tw=0.$$

I saw a very similar question from my textbook. In that question, it defines A as a nxn symmetric matrix. With this additional piece of information, this question becomes very easy. First of all, we can prove N(A) = N(A_TA), since A is symmetric, A_TA = I. N(A) = N(I) = {0}. Similarly, we have rank(A) = rank(A_T*A) = rank(I) = n


Is $\int_0^x\left|\sin\left(\frac{1}{t}\right)\right|\mathrm{d}t$ differentiable at $0$? Let $f(x)=\int_0^x\left|\sin\left(\frac{1}{t}\right)\right|\mathrm{d}t$ for $x\in\mathbb{R}$. Is $f$ differentiable at $x=0$?

This is not a routine problem. And the solution is tricky (I have seen it on MSE but my search capabilities are limited so I repeat the same here). Let me know if you need more details for any of the steps involved.

Consider the limit $$L=\lim_{x\to 0}\frac{1}{x}\int_{0}^x|\sin(1/t)|\,dt\tag{1}$$ By definition the given function is differentiable at $0$ if and only if the above limit exists. Putting $x=1/u$ and considering first $x\to 0^{+}$ we see that the above limit $L$ equals $$\lim_{u\to\infty} u\int_{0}^{1/u}|\sin(1/t)|\,dt=\lim_{u\to\infty} u\int_{u} ^{\infty} \frac{|\sin t|} {t^2}\,dt$$ Now split the interval of integration into an infinite number of intervals of type $[u+k\pi, u+(k+1)\pi]$ and thus we have $$L=\lim_{u\to\infty} u\sum_{k=0}^{\infty} \int_{u+k\pi} ^{u+(k+1)\pi}\frac{|\sin t|} {t^2}\,dt=\lim_{u\to\infty} u\sum_{k=0}^{\infty} I_k\tag{2}$$ Next use the inequality $$(u+k\pi) ^2\leq t^2\leq (u+(k+1)\pi)^2$$ to estimative the integral $I_k$ above as $$\frac{1}{(u+(k+1)\pi)^2}\int_{u+k\pi}^{u+(k+1)\pi}|\sin t|\, dt\leq\int_{u+k\pi} ^{u+(k+1)\pi}\frac{|\sin t|} {t^2}\,dt\leq\frac{1}{(u+k\pi)^2}\int_{u+k\pi}^{u+(k+1)\pi}|\sin t|\, dt$$ Thus we have $$\frac{2}{(u+(k+1)\pi)^2}\leq I_k\leq\frac{2}{(u+k\pi)^2}\tag{3}$$ We next need to estimate these lower and upper bounds using $$\frac{1}{(u+(k+1)\pi)^2}\geq \frac{1}{\pi}\int_{u+(k+1)\pi}^{u+(k+2)\pi}\frac{dt}{t^2}$$ Adding these from $k=0$ to $k=\infty$ we get $$\sum_{k=0}^{\infty} \frac{2}{(u+(k+1)\pi)^2}\geq \frac{2}{\pi}\int_{u+\pi}^{\infty} \frac{dt} {t^2}=\frac{2}{\pi(u+\pi)}\tag{4}$$ Similarly the inequality $$\frac{1}{(u+k\pi)^2}\leq \frac{1}{\pi}\int_{u+(k-1)\pi}^{u+k\pi}\frac{dt}{t^2}$$ gives us $$\sum_{k=0}^{\infty} \frac{2}{(u+k\pi)^2}\leq\frac{2}{\pi}\int_{u-\pi}^{\infty} \frac{dt} {t^2}=\frac{2}{\pi(u-\pi)}\tag{5}$$ From $(3),(4),(5)$ it follows that $$\frac{2}{\pi(u+\pi)}\leq\sum_{k=0}^{\infty} I_k\leq\frac {2}{\pi(u-\pi)}$$ Multiplying by $u$ and using Squeeze theorem we see that the limit $L$ in $(2)$ equals $2/\pi$.

The case for $x\to 0^{-}$ can be dealt by a simple substitution $x=-y$. Thus the derivative of given function at $0$ is $L=2/\pi$.

At the given formula $ f(x)=\int _{ 0 }^{ x }{ \left| \sin { \frac { 1 }{ t }  }  \right| dt }  $,
we can evaluate whether f '(0) exists.

Then we can think this problem with the definition of derivative.
$$\lim _{ h\rightarrow 0 }{ \frac { f(x+h)-f(h) }{ h }  } =\lim _{ h\rightarrow 0 }{ \frac { \int _{ 0 }^{ x+h }{ \left| \sin { \frac { 1 }{ t }  }  \right| dt\quad -\int _{ 0 }^{ x }{ \left| \sin { \frac { 1 }{ t }  }  \right| dt\quad  } \quad  }  }{ h }  } =\lim _{ h\rightarrow 0 }{ \frac { \int _{ x }^{ x+h }{ \left| \sin { \frac { 1 }{ t }  }  \right| dt\quad  }  }{ h }  } $$

Since we evaluate the function when x=0, so it is same as evaluating $$ \lim _{ h\rightarrow 0 }{ \frac { \int _{ 0 }^{ h }{ \left| \sin { \frac { 1 }{ t }  }  \right| dt\quad  }  }{ h }  } $$

The value of $ \left| \sin { \frac { 1 }{ x }  }  \right| $ does not exist when $x\rightarrow 0$.

Substitue 1/x=u, then $ lim_{ x\rightarrow 0 }{ \left| \sin { \frac { 1 }{ x }  }  \right| = }lim_{ u\rightarrow \infty  }{ \left| \sin { u }  \right|  } $, it oscillates between 0 and 1.

So the limit does not exist, therefore it is not differentiable.


How to find the values of $a$ that make $\det A= 0$? $$A=\begin{pmatrix}a & 6 & 7\\
a &-8 & 9\\
3 &-9 & a\end{pmatrix}$$

  Find the values of $a$ so that $\det A=0$

When I do it I get $a = 5.54$ or $a = -4.25$  but when I substitute those values back to find determinant I don't get $\det A=0$

Just solve the following equation.
$$a\cdot(-8)\cdot a+3\cdot6\cdot9+7\cdot a\cdot(-9)-3\cdot(-8)\cdot7-a\cdot(-9)\cdot9-a\cdot a\cdot6=0.$$
I got $$7a^2-9a-165=0,$$ which gives $$\left\{\frac{9+\sqrt{4701}}{14},\frac{9-\sqrt{4701}}{14}\right\}.$$
Now, if you'substitute these values, you'll get that the determinat is equal to $0$.

You got that $\det A\neq0$ because you took an approximated values of $a$.

For a matrix $$\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$$ the determinant is equal to $$a(ei-fh)-b(di-fg)+c(dh-eg)$$

For your example, $$A=\begin{pmatrix}a&6&7\\a&-8&9\\3&-9&a\end{pmatrix}$$

We put this into the formula and get

\begin{align}\det A &= a(-8a-(-9\times 9))-6(a\times a-(9\times 3))+7(-9a-(-8\times 3))\\
&=a(-8a+81)-6(a^2-27)+7(-9a+24)\\
&=-8a^2+81a-6a^2+162-63a+168\\
&=-14a^2+18a+330\end{align}

Now we need to solve this for $\det A=0$, so we say \begin{align}-14a^2+18a+330&=0\\
14a^2-18a-330&=0\end{align}

This is a quadratic equation, so we can solve this in the normal way to get \begin{align}a_1&=\frac1{14}(9-\sqrt{4701})\\
a_2&=\frac 1{14}(9+\sqrt{4701})\end{align}

If we round these to 2 decimal places, as you have done, then we get \begin{align}a_1&\approx-4.25\\
a_2&\approx5.54\end{align}

which are the values that you had found.

When checking the answers, you need to use the exact result, i.e. the ones with the square roots in them, otherwise you will encounter rounding errors which is presumably where your confusion is coming from.


Calculating size of an object based on distance So, say an object that is 10 feet tall is 100 feet away. If I hold up a ruler 3 feet away, then the object in the distance would correspond to about how many inches?

Tried using this guy: http://www.1728.org/angsize.htm
to calculate the angle, which ends up being 5.7248 degrees

Then, if I solve for size using 5.7248 degrees at a distance of 3 feet I get 0.3, or 4.8 inches.

The thing is is that that does not seem accurate to me. Perhaps my perception of distance is off, but 4.8 inches looks more like a 10 foot tall object at 50 feet to me...?

I mean, it is a simple ratio really..

x/3 feet = 10 feet/100 feet  right???

Thanks to the intercept theorem this is indeed a simple ratio:

$$\frac{x}{3\,\text{feet}}=\frac{10\,\text{feet}}{100\,\text{feet}}
\qquad\implies\qquad x=0.3\,\text{feet}$$

If you want to also involve the angles, you have

\begin{align*}
2\tan\frac\alpha2=\frac{10\,\text{feet}}{100\,\text{feet}}
\qquad&\implies\qquad \alpha=2\arctan0.05\approx5.7248°
\\
2\tan\frac\alpha2=\frac{x}{3\,\text{feet}}
\qquad&\implies\qquad x=3\,\text{feet}\times2\tan\frac\alpha2 = 0.3\,\text{feet}
\end{align*}

So the computations you did using that tool are correct. Anything that looks wrong is likely an optical illusion.

It seems to me that x/3 =10/100 shows that x= 0.3 right so far, but .3 of a foot is 3.6 inches not 4.8 right? So that may account for the difference in perceived size at 100' and 50'. Thanks


Solutions to $8^n = a^3+b^3+c^3-3abc$ Let $n$ be a positive integer. What is the number of solutions to the equation
$$8^n = a^3+b^3+c^3-3abc$$
with integers $a\geq b\geq c\geq 0$?

We have the factoring
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
meaning that both factors must be powers of $2$.

Note $$a^2+b^2+c^2-ab-ac-bc=(1/2)((a-b)^2+(a-c)^2+(b-c)^2)$$ Now suppose $a+b+c$ is even. Then all three variables are even, or else exactly one of them is even. If all three are even, then we can divide the original equation through by 8, getting a smaller solution. Repeat this enough times, and we get to a solution with exactly one even variable; let it be $a$. Then $a^2+b^2+c^2-ab-ac-bc$ is odd, but it's a power of 2, so it must be 1, so $(a-b)^2+(a-c)^2+(b-c)^2=2$. The only way to get three squares to sum to $2$ is for one of them to be zero and the others to be $1$, so $a-b=\pm1$, $a-c=\pm1$, $b=c$. Taking the plus sign, we get $3a-2=2^r$ for some $r$, so $a=(2^r+2)/3$, $b=c=(2^r-1)/3$. Then $$a^3+b^3+c^3-3abc=(1/27)((2^r+2)^3+2(2^r-1)^3-3(2^r+2)(2^r-1)^2)=2^r$$ so we need to take $r$ to be a multiple of $3$. For example, $r=6$ leads to $a=22$, $b=c=21$, $a^3+b^3+c^3-3abc=64=8^2$.

The reader may enjoy working through the case with the minus sign.

Strangely enough, the solution is finite.

for the equation:

$$X^3+Y^3+Z^3-3XYZ=q=ab$$

If it is possible to decompose the coefficient as follows:

$4b=k^2+3t^2$

Then the solutions are of the form:

$$X=\frac{1}{6}(2a-3t\pm{k})$$

$$Y=\frac{1}{6}(2a+3t\pm{k})$$

$$Z=\frac{1}{3}(a\mp{k})$$


Finding Homology of Koszul Complex Let $R = \mathbb{Q}[x,y,z]$ and $I = \langle x,y,z\rangle$. I have the free resolution of $R/I$ as

$K : 0 \rightarrow R \rightarrow R^3 \rightarrow R^3 \rightarrow R \rightarrow 0$

I need to find the $H_1(K)$ which is equal to $\ker (\phi_1)/ \operatorname{im} (\phi_2)$.

But I am stuck here. I know the maps explicitly but I am not being able to find what is it exactly. I am guessing as this is a regular sequence the homology would be zero, so that sequence would be exact. But how to proceed ?

Theorem [Commutative Ring Theory-H Matsumura] (see Page-128)

  Let $A$ be a ring, $M$ an $A$-module, and $x_1,\ldots,x_n$ an
  $M$-sequence; then $H_p(x,M)=0$ for $p > 0$ and $H_0(x,M)=M/xM$.


Simple Logical Simplification - Distributive Law I can't seem to understand the hopefully simple step of getting from :
$(a \land b \land c) \lor (a \land b \land \lnot c)$
to:
$(a \land b) \land (c \lor \lnot c)$
This step is in the answers and only has '(distribution)' written next to it, however even when breaking this down into several parts with Distributive Laws, I got nothing close to the expected solution (most likely my fault rather than teacher's). So any help explaining this step would very much be appreciated. Thank you!

Apply the distributive law to $d \wedge (c\vee \neg c)$ where $d=a\wedge b$. Then you get $$\begin{array}{rcl}d\wedge (c\vee \neg c)&=&(d\wedge c )\vee (d\wedge \neg c)\\&=&((a\wedge b)\wedge c)\vee ((a \wedge b)\wedge \neg c)\\&=&(a\wedge b\wedge c)\vee (a\wedge b \wedge \neg c). \end{array}$$Hope this helps!

It might help to put it into words:
I will use $TRUE/FALSE$ notation.
$A,B,C$ are $TRUE$ or $A,B$, NOT $C$ are $TRUE$. In either case $A$ and $B$ are $TRUE$.
Then either $C$ is $TRUE$ or NOT $C$ is $TRUE$ since this is an or statement and may be either one.


Transitivity on a set I have a question about equivalence relationships.
I know that a relationship is an equivalence relationship when the reflexive property, symmetric property and the transitive property exists on a set.
Ok...! ;-)
I have the following set
A = {0, 1, 2, 3, 4, 5, 6}

With the following relationship:
R = {(3, 3),(4, 0),(0, 0),(0, 4),(1, 1),(1, 6),(6, 1),(2, 2),(4, 4),(5, 5),(6, 6),(5, 3),(3, 5)}

With this data, we can draw the digraph:
Digraph
Then it's easy to check that the transitive property isn't exist in this case.
is it true?
and.., if is it true the question is... what kind of relationship is?

It is an equivalence relation, since it is reflexive, symmetric and transitive. If you think that it is not transitive, then please provide an example of elements $a,b,c\in\{0,1,2,3,4,5,6\}$ such that $(a,b),(b,c)\in R$, but $(a,c)\notin R$.

If you add for example (1, 0) and (0, 1) to R then it won't be transitive since (0, 1), (1, 6) ∈ R but (0, 6) ∉ R. That way it will not be an equivalence relation.


Solving for n in the equation $\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$ Solving for $n$ in the equation $$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$$

Can anyone show me a numerical method step-by-step to solve this?
Thanks

You seem to want the details of a numerical method to find an good approximation for $x$. Here is one way, using the Newton-Raphson method, which is well-known and uses fewer steps than other famous methods.

We want to find the zeros of

$$f(x)=0.25^x+0.5^x+0.75^x-1$$

(I rearranged the terms to be in ascending order, which satisfies my sense of style. I also changed the fractions to decimals to make it easier to type into a calculator.) We already know from comments and other answers that $x=1$ is too low and $x=2$ is too high. So let's choose the average, $1.5$, as our initial guess $x_0$. We then use the recurrence

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

This is what I typed into my graphing calculator (TI-Nspire CX):

$$x:=1.5$$
$$x:=x-\frac{0.25^x+0.5^x+0.75^x-1}{0.25^x\cdot\ln(0.25)+0.5^x\cdot\ln(0.5)+0.75^x\cdot\ln(0.75)}$$

That gave me the answer
$$1.71161782284$$

which is $x_1$. Repeated presses of the [enter] key repeated the recurrence, and I got

$$1.73037816962$$
$$1.73050735181$$
$$1.73050735786$$
$$1.73050735786$$

More [enter]'s just repeated the last value. Therefore, an excellent approximation to your desired value, probably to $12$ significant digits, is $1.73050735786$.

$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}= \frac{2^n}{4^n} +\frac{1^n}{4^n} +\frac{3^n}{4^n} = \frac{1^n+2^n+3^n}{4^n}=1,\: \therefore 1^n+2^n+3^n=4^n  $

I don't think this can be solved analytically,
you just have to use trial and error with n


Why can't I make the substitution $ u = \sin (ax + b) $ to evaluate $ \int \sin (ax + b) \cos (ax + b) dx$? Evaluate $ \int \sin (ax + b) \cos (ax + b) dx$?

To do this, I started of by substituting $ u = \sin (ax + b) $. That made $ du = cos (ax + b) \cdot a $ and wrote the integral as $ \frac 1a \int u \ du $ to get the final answer as: $$ \frac 1{2a} \sin^2 (ax + b) $$
This answer however, is wrong. My textbook uses a different method and arrives at a different answer. I understand how to arrive at the (right) answer but I want to know why I can't get the same answer by substitution here.

My textbook starts off by rewriting the expression as $ \frac {\sin 2 (ax + b)}{2} $ and then substitutes $ 2 (ax + b) = u $ to get this answer: $$ - \frac { \cos 2 (ax + b) } {4a } $$

Note that $$1-\cos2(ax+b)=2\sin^2(ax+b)$$

So the textbook gives almost the same as your outcome and the difference is constant.

That is allowed because to be found is an antiderivative (not an integral).

You can solve with the substitution
u=sin(ax+b) then,
du=acos(ax+b)dx so,
∫sin(ax+b)cos(ax+b)dx=∫(u/a)du=(u^2)/(2a)=sin^2(ax+b)/(2a)
which is same with your answer-1/4a


Can the difference of 2 undefined limits be defined? Is this limit defined or undefined?
$$\lim\limits_{x \to 0+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)$$
When I apply the rule of difference of limits, it's undefined. But, when I manipulate it, it gives me zero. And the graph of the function indicates it's defined on the right side.

By multiplying by $\frac{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$:
$$\lim\limits_{x \to 0+} \frac{\left( \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}} \, \right) \left(\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}} \, \right)}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$

$$=\lim\limits_{x \to 0+} \frac{\frac{1}{x}+2-\frac{1}{x}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$
$$=\lim\limits_{x \to 0+} \frac{2}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$
Then, we multiply by $\frac{\sqrt{x}}{\sqrt{x}}$:
$$=\lim\limits_{x \to 0} \frac{2\sqrt{x}}{\sqrt{1+2x}+1}$$
And, we substitute:
$$=\frac{2\sqrt{0}}{\sqrt{1+2\times0}+1} = 0$$
So, is this limit defined or not? and what's my error, if any?

Remember that the rule that you referred to, "the rule of difference of limits", is not just the equation
$$
\lim_{x\to a}(f(x)-g(x))=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)
$$
but rather the statement that, if both of the limits on the right side of this equation are real numbers, then the limit on the left side (is also a real number and) is given by this equation. So this rule does not apply to the limit in your question.

More generally, when learning rules (or theorems or principles or whatever they may be called), don't just learn formulas, but pay attention also to the words around them. The words are not just decoration but are essential for the correctness of the rule.

What you wrote,$$\lim\limits_{x \to 0+} \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}},$$is undefined, because it is the difference of$$\lim\limits_{x \to 0+} \sqrt{\frac{1}{x}+2},$$which is undefined, and a another quantity$$\sqrt{\frac{1}{x}},$$the latter being meaningful, although dependent on the variable $x$. What I think you meant to write is $$\lim\limits_{x \to 0+}\left( \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right).$$ This is not, as the title of your question says, "the difference of two undefined limits". Rather, it is a (single) limit of a quantity (that happens to be expressed as a difference). As long as you treat this as a single limit, which is what you have done quite correctly, no problem arises.

In general, it is a false move to split a limit of a difference into the difference of two limits. It works sometimes, but then it needs to be justified carefully. On the other hand, if two limits are separately well defined, then their difference can be expressed as the limit of a difference (assuming that the variable, domain, and endpoint of the limiting process is the same for both).


Discrete Math -- Sets I am struggling with thinking about this. Any help would be great!!

A medical research survey categorizes adults as follows:

by gender (male or female)
by age group (age groups are 18-25, 26-35, 36-50, 51+)
by income (less than 30k/year, 30k-60k/year, more than 60k/year)
for women only: by whether they have been pregnant (yes/no)
for men only: by frequency of undergoing prostate exams (frequently, rarely, never).

What minimum size of a set of adults will guarantee that there are two people in it with matching characteristics in all categories? You do not need to explain your answer.

Based on this

I believe you would have (choices)

(5)

Woman : Yes

Woman : No

Male  : Frequently

Male  : Rarely

Male  : Never
(4) by age group (age groups are 18-25, 26-35, 36-50, 51+)
(3) by income (less than 30k/year, 30k-60k/year, more than 60k/year)

$$5*4*3=60$$

By the product rule, there are 60 ways of answering this survey. Therefore if,

$$60+1=61$$

61 people were surveyed, two will have matching characteristics.

By the Pigeonhole Principle it would be one more than the number of categories, C(5,2)*4*3 = 10*4*3 = 120 => 120 +1 = 121.


Find non-increasing PDF after transformation of a continuous variable I'm looking for a continuous distribution for $x \in R$, such that after applying the uni-variate transformation for $y = \exp\{ax+b\} / (1 + \exp\{ax+b\})$, the resultant PDF $f_Y(y)$ becomes non-increasing in $y$.

I already tried a normal distribution for $x$, but $f_Y(y)$ turned out to be non-decreasing in $y$. Rather than a trial-and-error approach, is there any mechanism that I can narrow down my search for an appropriate distribution for $x$?

$$\frac{dy}{dx}=\frac{a\exp[ax+b]}{\left(1+\exp[ax+b]\right)^2}$$
$$f_Y(y)=f_X(x)\left|\frac{dx}{dy}\right|$$
$$\frac{df_Y(y)}{dy}=\frac{d}{dx}\left(f_X(x)\left|\frac{dx}{dy}\right|\right)\frac{dx}{dy}\le 0$$

$$\left[f'_X(x)\left|\frac{dx}{dy}\right|+f_X(x)\frac{d}{dx}\left|\frac{dx}{dy}\right|\right]\frac{dx}{dy} \le 0$$
$$\left[f'_X(x)\frac{\left(1+\exp[ax+b]\right)^2}{|a|\exp[ax+b]}+f_X(x)\frac{d}{dx}\frac{\left(1+\exp[ax+b]\right)^2}{|a|\exp[ax+b]}\right]\frac{\left(1+\exp[ax+b]\right)^2}{a\exp[ax+b]}\le 0$$
$$\left[f'_X(x)\frac{\left(1+\exp[ax+b]\right)^2}{\exp[ax+b]}+f_X(x)\frac{d}{dx}\frac{\left(1+\exp[ax+b]\right)^2}{\exp[ax+b]}\right]\frac{1}{a|a|}\le 0$$
$$\frac{1}{a|a|}\bigg\{f'_X(x)\frac{\left(1+\exp[ax+b]\right)^2}{\exp[ax+b]}$$
$$+f_X(x)\left[\frac{2(1+\exp[ax+b])a\exp[ax+b]}{\exp[ax+b]}-\frac{(1+\exp[ax+b])^2}{(\exp[ax+b])^2}\exp[ax+b]a\right]\bigg\}\le 0$$
$$\frac{1}{a}\bigg\{f'_X(x)\left(1+\exp[ax+b]\right)^2$$
$$+f_X(x)\left[2a\exp[ax+b]\left(1+\exp[ax+b]\right)-a\left(1+\exp[ax+b]\right)^2\right]\bigg\} \le 0$$
$$\frac{1}{a}\left\{f'_X(x)+f_X(x)\frac{a\exp[ax+b]-a}{1+\exp[ax+b]}\right\} \le 0$$
$$\frac{1}{a}\left[f'_X(x)+af_X(x)\tanh\frac{ax+b}{2}\right] \le 0$$
$$\frac{1}{a}\left[\frac{f'_X(x)}{f_X(x)}+a\tanh\frac{ax+b}{2}\right] \le 0$$
$$\frac{1}{a}\left[\frac{d}{dx}\ln f_X(x)+a\tanh\frac{ax+b}{2}\right] \le 0$$

Case 1: If $a>0$,
$$\frac{d}{dx}\ln f_X(x)+a\tanh\frac{ax+b}{2} \le 0$$
$$\frac{d}{dx}\ln f_X(x) \le -a\tanh\frac{ax+b}{2}$$
$$\ln f_X(x)-\ln f_X(x_0) \le \int_{x_0}^x -a\tanh\frac{ax+b}{2} dx$$
$$=-2\int_{x_0}^x \tanh\frac{ax+b}{2}d\frac{ax+b}{2}$$
$$=-2\ln\frac{\cosh\frac{ax+b}{2}}{\cosh\frac{ax_0+b}{2}}$$
$$f_X(x) \le f_X(x_0)\frac{\cosh^2\frac{ax_0+b}{2}}{\cosh^2\frac{ax+b}{2}}$$

Case 2: If $a<0$, similarly,
$$f_X(x) \ge f_X(x_0)\frac{\cosh^2\frac{ax_0+b}{2}}{\cosh^2\frac{ax+b}{2}}$$

If one chooses $x_0=-b/a$, then for $a>0$
$$f_X(x) \le f_X(-b/a)\cosh^{-2}\frac{ax+b}{2}$$
and for $a<0$,
$$f_X(x) \ge f_X(-b/a)\cosh^{-2}\frac{ax+b}{2}$$