Calculating the asymptotic of integrals For an integral like
$$D_{n}(x) \equiv \int_{0}^{x} \frac{t^{n}}{e^{t}-1} d t$$
The asymptotic values are given as
$$D_{n}(x) \simeq\left\{\begin{array}{ll}
n ! \zeta(n+1)-x^{n} e^{-x}+O\left(x^{n} e^{-2 x}\right), & x \rightarrow \infty \\
x^{n} / n-x^{n+1} / 2(n+1)+O\left(x^{n+2}\right), & x \rightarrow 0
\end{array}\right.$$
For $x\to 0 $, I can just Taylor expand $\frac{1}{e^t-1}$ as $\frac{1}{t}-\frac{1}{2}+\frac{t}{12}-\frac{t^3}{720}+O\left(t^5\right)$ and do term by term integration to to get the given expression.
For $x\to \infty $ the first term is simply gotten by replacing the upper bound of the integral with $\infty$ but how are the correction terms calculated?

The asymptotics you gave is not completely correct. I am going to derive the correct asymptotic form. First, we write
\begin{align*}
D_n (x) & = n!\zeta (n + 1) - \int_x^{ + \infty }  \frac{{t^n }}{{e^t  - 1}}dt
 \\ & \mathop  = \limits^{t = x(1 + s)} n!\zeta (n + 1) - x^{n + 1} e^{ - x}  \int_0^{ + \infty }  e^{ - xs} \frac{(1 + s)^n }{1 - e^{ - x(1 + s)}}ds .
\end{align*}
Now, by the geometric series,
$$
\int_0^{ + \infty } {e^{ - xs} \frac{{(1 + s)^n }}{{1 - e^{ - x(1 + s)} }}ds}  = \int_0^{ + \infty } {e^{ - xs} (1 + s)^n ds}  + \int_0^{ + \infty } {e^{ - xs} (1 + s)^n \mathcal{O}(e^{ - x(1 + s)} )ds} .
$$
By Watson's lemma,
$$
\int_0^{ + \infty } {e^{ - xs} (1 + s)^n ds}  = \int_0^{ + \infty } {e^{ - xs} (1 + s)^n ds}  = \frac{1}{x} + \mathcal{O}\left( {\frac{1}{{x^2 }}} \right)
$$
and
$$
\int_0^{ + \infty } {e^{ - xs} (1 + s)^n \mathcal{O}(e^{ - x(1 + s)} )ds}  = \mathcal{O}(1)e^{ - x} \int_0^{ + \infty } {e^{ - 2xs} (1 + s)^n ds}  = \mathcal{O}\left( {\frac{{e^{ - x} }}{x}} \right).
$$
Thus,
\begin{align*}
D_n (x) & = n!\zeta (n + 1) - x^n e^{ - x}  + \mathcal{O}(x^{n - 1} e^{ - x} ) + \mathcal{O}(x^n e^{ - 2x} ) \\ & = n!\zeta (n + 1) - x^n e^{ - x}  + \mathcal{O}(x^{n - 1} e^{ - x} ).
\end{align*}
Addendum. It can be shown that for any positive integer $n$,
$$
D_n (x) = n!\zeta (n + 1) - n!x^n \sum\limits_{j = 0}^n {\frac{1}{{(n - j)!}}\frac{1}{x^j}\operatorname{Li}_{j + 1} (e^{ - x} )}
$$
where $\operatorname{Li}_s$ is the polylogarithm.

Not sure, but it seems useful to make the substitution $xu = t$:
\begin{align}
D_n(x) & = \int_0^1 \frac{x^{n+1} u^n}{e^{xu} - 1} \ du \\
& = x^{n+1} \int_0^1 e^{-xu} \frac{u^n}{1 - e^{-xu}} \ du \\
& = x^{n+1} \int_0^1 e^{-xu} u^n \sum_{m=0}^\infty e^{-m x u} \ du \\
& = x^{n+1} \sum_{m=1}^\infty \int_0^1 u^n e^{-xu m} \ du. \\
\end{align}
The first term $\Gamma(n+1) - \Gamma(n+1,x)$ can be written in terms of asymptotic formulae well known for the Gamma function. Though this expansion seems to do better than the first two terms you have there...


Trouble with Discrete Math proof This seemingly easy proof is giving me some trouble..

For every number n $\in$ $Z$, if $n>n^2+1$ then $n\leq0$.

I find that proving the conditional statement P implies Q is false. For example $n=-2$, it is false that $-2>(-2)^2+1$.

But proving the contrapositive is true (not Q implies not P). If $n>0$ then $n\leq n^2 +1$

Can someone point out where I might be going wrong? Thanks!

The statement given is false. Note that $n^2+1$ is strictly positive so if $n>n^2+1$ then $$n>n^2+1>0$$


Integrating $\int _0^1\sqrt{9\sin^2(x)+9\cos^2(x)+16}\ \mathrm dx$ What I'm trying to do is find the arc length of $r(x) = \langle 3\cos(x),4x,3\sin(x)\rangle$ when $0\le x\le 1$.
After I derived the vector and found the magnitude of that derived vector, I came up with $$\int_0^1\sqrt{9\sin^2(x)+9\cos^2(x)+16}\ \mathrm dx$$
as the solution. Now I have to integrate it and I'm not so sure where to begin because I haven't seen a problem like this before.

Hint
$\sin^2 x+\cos^2 x = 1$ due to Pythagoras, so the question is
$$\int_0^1 \sqrt{25}\ \mathrm dx = ?$$

Since there has been a discussion if the integral was actually correct (wich it was), here is the long answer:
$$r(x) = \langle 3\cos x, 4x, 3\sin x\rangle\\
\nabla r(x) = \langle -3\sin x, 4, 3\cos x\rangle$$
Now the arc length is defined by
$$L_a^b (r) := \int_a^b \| \nabla r(x) \|_2\ \mathrm dx$$
Note that $\|x\|_2 = \sqrt{x_1^2 + x_2^2 + \ldots}$ so
$$\|\nabla r(x)\|_2 = \sqrt{(-3\sin x)^2 + 4^2 + (3\cos x)^2} = \sqrt{9 (\sin^2 x + \cos^2 x) + 16} = \sqrt{25} = 5$$
Where we used Pythagoras ($\sin^2 x +\cos^2 x = 1$ for all $x$). Thus
$$L_0^1 (r) = \int_0^1 5\ \mathrm dx = 5$$


A question regarding $\lim\limits_{n\to\infty} n(n+1)(n+2)\dots(n+r)$. We know that $\lim\limits_{n\to\infty} \frac{n(n+1)(n+2)\dots(n+r)}{n^{r+1}}$, where $r$ is a finite positive integer, is $1$.

Hence, $\lim\limits_{n\to\infty} n(n+1)(n+2)\dots(n+r)=\lim\limits_{n\to\infty}{n^{r+1}}$.

This should imply that $\lim\limits_{n\to\infty} \left[n(n+1)(n+2)\dots(n+r)\right]-{n^{r+1}}=0$

However, $$\left[n(n+1)(n+2)\dots(n+r)\right]-{n^{r+1}}=n^r(1+2+\dots+n)+n^{r-1}(1.2+1.3+\dots+(r-1)r)+\dots+r!$$

Shouldn't $\lim\limits_{n\to\infty}n^r(1+2+\dots+n)+n^{r-1}(1.2+1.3+\dots+(r-1)r)+\dots+r!$ tend to infinity?

Thanks in advance!

This is false: "This should imply that $\lim\limits_{n\to\infty} \left[n(n+1)(n+2)\dots(n+r)\right]-{n^{r+1}}=0$"

We know $\lim_{n=1}^\infty \frac{n+1}{n}=1$, yet we know $n+1\neq n$.  All the limit tells you is that the terms with the highest power are the same, it doesn't tell you anything about the terms with lower powers.


discrete math: n pennies among k children with each child having atleast 2 pennies This problem is posed in Lovasz's Discrete math book chapter 3 and I understand the correct answer which is $$ \binom{n-k-1}{k-1} $$

However, why is my approach not right ? Here it goes.
We have n pennies. We group them into 2 each  and hence have
$$
\lfloor{\frac{n}{2}}\rfloor
$$
units. Lets call them U.
Now we have to distribute U units among k-1 children which will lead to

$$ \binom{\lfloor{\frac{n-2}{2}}\rfloor-2}{k-1} $$
ways. I subtracted 2 because 2 pennies will goto the first child.

Assume you have n pennies, and you want to distribute them among k children, with each child having at least 2 pennies. You can go as follow:

Distribute 2 pennies to k child.

Since the pennies are identical, the order you distribute the pennies to the children does not matter.

Then there is only one way to do that.

Now you left with n-2K pennies.

And the number of distributing the remaining pennies is:

$(n-2k)+k-1 \choose k-1 $ = $ n-k-1 \choose k-1$.
[By the previous theorem in the textbook].

So the total number is:

$ (1) $ $ n-k-1 \choose k-1$ = $ n-k-1 \choose k-1$.

You can see that your answer is wrong when different $n$ give the same answer.

What you should do is give to each child $2$ pennies.

You are left with $n-2k$ pennies and you want to give them to $k$ children.

Can you take it from here?

Recall that the number of ways to give $m$ pennies to $r$ children is
$$\binom{m+r-1}{r-1}.$$


Sum $\sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+2}$ I have to first prove whether the following series is convergent, and then find its limit if it exists:

$\sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+2}$

So, I have proved that it is convergent, but I'm having trouble in finding the limit. Can someone help me out?

If the problem is due to
$$
\frac{1}
{n} - \frac{1}
{{n + 2}}
$$
write it as
$$
a_n  = \left( {\frac{1}
{n} - \frac{1}
{{n + 1}}} \right) + \left( {\frac{1}
{{n + 1}} - \frac{1}
{{n + 2}}} \right)
$$
At this point both the terms in the brackets telescoping. So I thin now it is easy.


Covariance between Wiener process and stochastic integral I want to calculate $cov($$w_t$ , $\int_0^{t}s^{n}dw_s)$ .
I have tried integration by parts:
$$\int_0^{t}s^{n}dw_s = t^{n}w_{t} - n\int_{0}^{t}s^{n-1}w_sds$$ Further, I think to use this formula $n$ times, but I can't do it on this step. My idea was to get the equation at the end that will contain next components (maybe with some coefficients): $\int_0^{t}sdw_s$, $\int_0^tw_sds$, $w_t$ and the component that if I count a mathematical expactation from it I will get the

$cov($$w_t$, $\int_0^{t}s^{n}dw_s)$ (the mathemetical expectation for the first three components I know). And from this equation I would count the covariance, but I can't understand how I can continue. I'll be happy for any idea.

The most clever way is, as @Did suggested, to use Itô's isometry. However, you can also use the integration by parts formula:

$$\int_0^t s^n \, dW_s = t^n W_t - n \int_0^t s^{n-1} W_s \, ds.$$

Multiplying both sides by $W_t$ and taking the expectation we get

$$\begin{align*} \mathbb{E} \left( W_t \cdot \int_0^t s^n \, dW_s \right) &= t^n \underbrace{\mathbb{E}(W_t^2)}_{t} - n \int_0^t s^{n-1} \underbrace{\mathbb{E}(W_s W_t)}_{\min\{s,t\}=s}\,d s \\ &= t^{n+1} - n \int_0^t s^n \, ds =\frac{1}{n+1} t^{n+1}.\end{align*}$$

The solution is
$$cov(w_t, \int_0^ts^ndw_s)=E(w_t - Ew_t)(\int_0^ts^ndw_s- E\int_0^ts^ndw_s)=E(w_t\int_0^ts^ndw_s-w_tE\int_0^ts^ndw_s)=\int_0^ts^nds-Ew_tE\int_0^ts^ndw_s=\int_0^ts^nds=\frac{t^{n+1}}{n+1}$$


Showing $f$ is a polynomial given sequence Suppose $\{p_k \}_{k=1}^{\infty}$ is a sequence of polynomials and that $f: \mathbb{C} \to \mathbb{C}$ is an entire function such that $p_k(z) \to f(z)$ as $k \to \infty$ with the convergence uniform on each closed disk $D_R=\{z \in \mathbb{C}: \left |z \right |\le R \}$.

A) Show that if there is a finite constant $M$ such that degree$(p_k) \le M$ for each $k$, then $f$ must be a polynomial.

B) Give an example to show that if there is no such constant $M$,then $f$ might not be a polynomial.

My approach for $A$,

Suppose $k$=n where $n<+ \infty$ , then $$ \left |f^{(n)}(z) \right |\le \frac{n!}{R^n}M $$

This implies $f^{(n)}$ is constant by Lioville's theorem and therefore $$f^{(n)}(z)=a$$ where $a$ is some finite constant. Therefore integrating both sides we can conclude we get a polynomial of degree $n$.

For part B, I have no clue.

For b, take partial sums for the Maclaurin series of any entire (non-polynomial) function.

For example,
$$
p_n(z) = 1 + z + \frac{z^2}{2!} + \cdots + \frac{z^n}{n!}
$$
converges locally uniformly on $\mathbb{C}$  to $e^z$. (This follows from general theory of convergence for power series. If you don't have that, you can for example use the Weierstrass $M$-test to show uniform convergence on every closed disc.)


A concrete dam of height 128 feet was built in gorge. One side $AB$ of the gorge slopes at an angle of $60^o$, the other side $CD$ at $45^o$. I was wandering across the internet when a wild question appears. It goes like this:

  A concrete dam of height 128 feet was built in gorge. One side $AB$ of the gorge slopes at an angle of $60^o$, the other side $CD$ at $45^o$.
  The bases of the dam are horizontal and rectangular in shape. The lower base is 1215 feet by 152 feet, and the upper base is 32 feet wide.
  How many cubic yards of concrete were required?

My work

I imagine that dam to look like this:

Delving deeper into the picture, it becomes like this:

Positioning the dam so it is easier to get the volume of this dam, it looks like this:

Notice that the it is easier to get the volume of the dam if we separate the shapes that makes up the dam, the red and the green part...

The green part's volume is pretty straightforward. We'll just use the formula $$V = BH$$ where $V$ is the volume of the figure, $B$ is the
area of the base and $H$ is the height of the figure. Looking at the green part of the figure, the area of it's base is a trapezoid. The area would
be $$A = \frac{1}{2}h(b_1 + b_2)$$ where $h$ is the height of plane, $b_1$ is the length of the upper base, and $b_2$ is the length of the lower
base. With that in mind, the volume of the green part would be...

$$V=BH$$
$$V = \left( \frac{1}{2}h(b_1 + b_2)\right)(H)$$
$$V = \left( \frac{1}{2}(128 \space feet)(1013.1 \space feet + 1215 \space feet)\right)(32 \space feet)$$
$$V = 4563148 \space ft^3$$

The red part is pretty tricky, but solvable. Imagining the red part as a one part cut in halves, it becomes easier to get its volume. Getting the
volume of the whole red part below:

$$V=BH$$
$$V = \left( \frac{1}{2}h(b_1 + b_2)\right)(H)$$
$$V = \left( \frac{1}{2}(128 \space feet)(1013.1 \space feet + 1215 \space feet)\right)(120 \space feet)$$
$$V = 17111808 \space ft^3$$

Now getting the volume of the red part we see, it is $\frac{V}{2}$ or $8555904\space ft^3$

The total volume of the dam would be $4563148 \space ft^3 + 8555904 \space ft^3$ or $13119052.8 \space ft^3$

It's not yet the final answer because it must be converted to cubic yards. Now converting from cubic feet to yards:

$$13119052.8 \space ft^3\left(\frac{0.3048 \space meters}{1 \space feet}\right)^3 = 371490.18 \space m^3 $$
$$371490 \space m^3 \left( \frac{1.0936 \space yards}{1 \space meter}\right)^3 = 485873.9 \space cubic \space yards$$

Is my solution, ladies and gentlemen, correct?


Is orthogonality of column vectors preserved after right-multiplication by unitary matrix? $\mathbf V$ is an $n \times (n-1)$ matrix with mutually orthogonal columns.
$\mathbf Q$ is a unitary matrix of size $(n-1) \times (n-1)$. Is there a concise algebraic proof that the columns of $\mathbf {VQ}$ are mutually orthogonal?

In a proof of the Spectral Theorem, it is claimed that the columns of $\mathbf {VQ}$ are orthogonal by construction. I would like to find a way to verify that this is generally true.

Let $Q(i)$ be the $i$-th column of $Q$. Think of the product $VQ$ as
$$R = VQ = \begin{bmatrix} VQ(1) & VQ(2) &VQ(3) &\cdots&VQ(n) \end{bmatrix}$$.

We want to show that $\langle R(i), R(j) \rangle = 0,\forall i \neq j$,
\begin{align*}
&\langle R(i), R(j) \rangle\\
=&Q(i)^TV^T V Q(j) \\
=&Q(i)^TIQ(j)\\
=&Q(i)^TQ(j) \\
=& 0
\end{align*}
Since the columns of $V$ are mutually orthogonal (actually they need to be mutually orthonormal),  $V^TV = I \in R^{n \times n}$. $Q$ is an unitary matrix, therefore its columns are also mutually orthogonal.

A unitary matrix preserves the scalar product, hence preserves the orthogonality of two vectors.


Prove that $f'(0)$ exists and $f'(0) = b/(a - 1)$ Problem:
  If $f(x)$ is continous at $x=0$, and $\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|>1$, prove that $f'(0)$ exists and $f'(0)=\dfrac{b}{a-1}$.

This approach is definitely wrong:

\begin{align}
b&=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\
&=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\
&=af'(0)-f'(0)\\
&=(a-1)f'(0)
\end{align}

I will show you a case why this approach is wrong:

  \[f(x)= \begin{cases}
1,&x\neq0\\
0,&x=0
\end{cases}\]
  $\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0$
  but $\lim_{x\to0}\dfrac{f(3x)}{x}=\infty$,$\lim_{x\to0}\dfrac{f(x)}{x}=\infty$

Does anyone know how to prove it? Thanks in advance!

This is a tricky question and the solution is somewhat non-obvious. We know that $$\lim_{x \to 0}\frac{f(ax) - f(x)}{x} = b$$ and hence $$f(ax) - f(x) = bx + xg(x)$$ where $g(x) \to 0$ as $x \to 0$. Replacing $x$ by $x/a$ we get $$f(x) - f(x/a) = bx/a + (x/a)g(x/a)$$ Replacing $x$ by $x/a^{k - 1}$ we get $$f(x/a^{k - 1}) - f(x/a^{k}) = bx/a^{k} + (x/a^{k})g(x/a^{k})$$ Adding such equations for $k = 1, 2, \ldots, n$ we get $$f(x) - f(x/a^{n}) = bx\sum_{k = 1}^{n}\frac{1}{a^{k}} + x\sum_{k = 1}^{n}\frac{g(x/a^{k})}{a^{k}}$$ Letting $n \to \infty$ and using sum of infinite GP (remember it converges because $|a| > 1$) and noting that $f$ is continuous at $x = 0$, we get $$f(x) - f(0) = \frac{bx}{a - 1} + x\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$ Dividing by $x$ and letting $x \to 0$ we get $$f'(0) = \lim_{x \to 0}\frac{f(x) - f(0)}{x} = \frac{b}{a - 1} + \lim_{x \to 0}\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$

The sum $$\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$ tends to $0$ as $x \to 0$ because $g(x) \to 0$. The proof is not difficult but perhaps not too obvious. Here is one way to do it. Since $g(x)\to 0$ as $x \to 0$, it follows that for any $\epsilon > 0$ there is a $\delta > 0$ such that $|g(x)| < \epsilon$ for all $x$ with $0 <|x| < \delta$. Since $|a| > 1$ it follows that $|x/a^{k}| < \delta$ if $|x| < \delta$ and therefore $|g(x/a^{k})| < \epsilon$. Thus if $0 < |x| < \delta$ we have $$\left|\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}\right| < \sum_{k = 1}^{\infty}\frac{\epsilon}{|a|^{k}} = \frac{\epsilon}{|a| - 1}$$ and thus the sum tends to $0$ as $x \to 0$.

Hence $f'(0) = b/(a - 1)$.

BTW the result in question holds even if $0 < |a| < 1$. Let $c = 1/a$ so that $|c| > 1$. Now we have $$\lim_{x \to 0}\frac{f(ax) - f(x)}{x} = b$$ implies that $$\lim_{t \to 0}\frac{f(ct) - f(t)}{t} = -bc$$ (just put $ax = t$). Hence by what we have proved above it follows that $$f'(0) = \frac{-bc}{c - 1} = \frac{b}{a - 1}$$ Note that if $a = 1$ then $b = 0$ trivially and we can't say anything about $f'(0)$. And if $a = -1$ then $f(x) = |x|$ provides a counter-example. If $a = 0$ then the result holds trivially by definition of derivative. Hence the result in question holds if and only if $|a| \neq 1$.

Hint: You're very close.

Write the expression as $$a\frac{f(ax)-f(0)}{ax}-\frac{f(x)-f(0)}{x}$$ Note that $x\to 0$ if and only if $ax\to 0$ (since $a\neq 0$).

Can you see it from this?


Prove the following identity for the Apéry constant Perhaps this kind of integral is well knonw, or can be easily deduced from other. I don't know it but I would like to see the computation of this to refresh the computation of iterated integrals. I was inspired to change the factors of those integrals in the Wikipedia Page for Apéry 's constant. Firht I tried modify those integrals writting some factors $\sqrt{x}$. After with factors $1-x^2$ and computing with the online calculator of  Wolfram Alpha one find with the code

  $\displaystyle{\int_0^1\int_0^1\int_0^1\int_0^1 (1-y^2)(1-t^2)/(1-xyz)
\,dx\,dy\,dz\,dt}$

an integral involving the cited constant.

I am assuming that these families were well known , because was easy to find those identities by similarity, when one modify such integrands as I've said. See also this MathWorld Page for the Apéry's constant, truly those formulas and identites are more complicated than my question, thus the kind of integral that

  Question. Show us how to prove
  $$\int_0^1\int_0^1\int_0^1\int_0^1\frac{(1-y^2)(1-t^2)}{1-xyz}dxdydzdt=\frac{1}{36}(24\zeta(3)+9-2\pi^2).$$
   Thanks in advance.

Since the idea was easy, see comments for other example, it is easy that it was in the literature.

Also you are welcome to answer the following optional question,

  Question (Optional) Can you find a integral now involving at least a factor $\sqrt{\text{variable}}$ of previous variables $x,y,z,\ldots$, to get an identity, as the previous, for $\zeta(3)$?

Because the $t$ integral is trivial, we will not bother with it and just calculate

$$
I=\int_{[0,1]^3} dx dy dz  \frac{1-y^2}{1-xyz}
$$

the first integration is standard an yields (let's take $z$)
$$
I=\int_{[0,1]^2} dx dy (1-y^2)\frac{\log(1-xy)}{xy}
$$

now, using one of the standard properties of Polylogarithms, namely $z \text{Li}_{\nu}'(z)=\text{Li}_{\nu-1}$, this becomes

$$
I=-\int_{[0,1]} dy (1-y^2)\frac{\text{Li}_2(y)}{y}
$$

Using the above mentioned property again this yields

$$
I=\text{Li}_3(1)+\int_{[0,1]} dy y\text{Li}_2(y)
$$

the last integral is easily tackeld using integration by parts with $y=u'$ and $v=\text{Li}_2(y)$. After putting back in the limits of integration this becomes

$$
I=\text{Li}_3(1)+\frac{1}{2}\text{Li}_2(1)+\frac{3}{8}
$$

which is equal to

  $$
I=\zeta(3)-\frac{\pi^2}{12}+\frac{3}{8}
$$

which gives us the announced result if we take a factor $2/3$ into account which stems from the integral over $t$

EDIT:  I know this isn't as slick as the answer using polylogs but I always liked this approach to any integration with $\frac{1}{1-xyz}$ from [0,1].  Also I should note I got this approach from the book "Proofs from the Book"  in the section "Three Times $\frac{\pi^2}{6}$", I think is the name.  They use this approach to show $\zeta(2)=\frac{\pi^2}{6}$.

The t portion of the integral can be done separately using Fubini's theorem so noting that $\int_0^1(1-t^2)dt=t-\frac{t^3}{3}|_0^1=\frac{2}{3} $ this will be a prefactor that I will insert at the end.

So the remaining integral is:

$\int_0^1 \int_0^1 \int_0^1 \frac{1-y^2}{1-xyz} dxdydz$.  We can rewrite this into two integrals:  $\int_0^1 \int_0^1 \int_0^1 \frac{1}{1-xyz} dxdydz-\int_0^1 \int_0^1 \int_0^1 \frac{y^2}{1-xyz} dxdydz$.  Now since we're working on $[0,1]$ we can use the following geometric series $\frac{1}{1-xyz}=\sum_{n=0}^\infty(xyz)^n$.  So the first integral becomes $\int_0^1 \int_0^1 \int_0^1 \sum_{n=0}^\infty (xyz)^n dxdydz$, exchanging the integrals and the sum and breaking up into three integrals we have, $\sum_{n=0}^\infty \int_0^1 x^ndx \int_0^1 y^n dy \int_0^1 z^n dz$ this is simply, $\sum_{n=0}^\infty (\frac{x^{n+1}}{n+1}|_0^1)(\frac{y^{n+1}}{n+1}|_0^1)(\frac{z^{n+1}}{n+1}|_0^1)$ = $\sum_{n=0}^\infty \frac{1}{(n+1)^3}$ or reindexing we have $\zeta(3)$.  The second integral can be tackled in the same way as the first changing everything into sums just as before we have now -$\sum_{n=0}^\infty \int_0^1 x^ndx \int_0^1 y^{n+2} dy \int_0^1 z^n dz$ or $-\sum_{n=0}^\infty (\frac{x^{n+1}}{n+1}|_0^1)(\frac{y^{n+3}}{n+3}|_0^1)(\frac{z^{n+1}}{n+1}|_0^1)=-\sum_{n=0}^\infty \frac{1}{(n+3)(n+1)^2}=-\sum_{n=1}^\infty \frac{1}{(n+2)n^2}$.  Using partial fractions this becomes: $\sum_{n=1}^\infty -\frac{1}{2n^2}-\frac{1}{4(n+2)}+\frac{1}{4n}=-\frac{1}{2} \zeta(2) - \frac{1}{4}(\sum_{n=1}^\infty\frac{1}{(n+2)}-\frac{1}{n})=-\frac{\pi^2}{12}+\frac{3}{8}$.  Where I used that I know what the zeta function is at 2 (it's $\frac{\pi^2}{6}$) and also that the last set of sums telescopes to $\frac{3}{2}$.  So grouping all this together we get $\zeta(3)-\frac{\pi^2}{12}+\frac{3}{8}$.  Multiplying this by $\frac{2}{3}$ from the t integral gives the desired result.


Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$. Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.

I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides which would give me a simple quadratic to solve for $x$, but I don't know how to simplify $\frac{(2^{x^2}-1)}{(2^x-1)}$. Am I missing a simple trick here, or am I on a completely wrong path to solving this?

As per dxiv's hint: by elementary methods, the equation is equivalent to

$$f(x)\stackrel{\text{def}}{=}(2^x−1)(x^2-1)+(2^{x^2-1}−1)x=0$$
Note that, for any $y$, $2^y-1$ and $y$ have the same sign (positive, negative, or zero); write $\mathrm{sgn}\,y$ for the sign of $y$.
Then

$$\mathrm{sgn}[(2^x-1)(x^2-1)]=\mathrm{sgn}(2^x-1)\mathrm{sgn}(x^2-1)=\mathrm{sgn}(x)\mathrm{sgn}(x^2-1)$$
and likewise
$$\mathrm{sgn}[(2^{x^2-1}-1)x]=\mathrm{sgn}(x^2-1)\mathrm{sgn}(x)$$

Now, if $\mathrm{sgn}(y)=\mathrm{sgn}(z)$, then $\mathrm{sgn}(y+z)=\mathrm{sgn}(y)=\mathrm{sgn}(z)$. Consequently
$$\mathrm{sgn}[f(x)]=\mathrm{sgn}(x)\mathrm{sgn}(x^2-1)=\mathrm{sgn}(x)\mathrm{sgn}(x+1)\mathrm{sgn}(x-1)\text{.}$$
By the zero product property and the fact that $\mathrm{sgn}(y)=0$ iff $y=0$, we find that $f(x)=0$ iff $x\in\{-1,0,1\}$.

$\begin{align*} 2\left (2^{x}-1 \right )x^{2}+\left(2^{x^{2}}-2\right)x&=2^{x}.2-2\\ 2\left(2^{x}-1\right)x^{2}-\left(2^{x^{2}}-2\right)x&=2(2^{x}-1) \end{align*}$

Let $2^{x}-1=a$ and $2^{x^{2}}-2=b$ and equation becomes:
$2ax^{2} + bx=2a$ or $2ax^{2}+bx-2a=0$. It's a quadratic equation. The quadratic equation has real solution if and only if $b^{2}-4ac>0$.


Find eigenvalues of P if PP=P I've been given that matrix $P$ is an $n\times n$ matrix such that $PP=P$. Is it correct to say that, pre-multiplying by $P^{-1}$:

$P^{-1}PP = P^{-1}P$

Therefore $P = I$? Where $I$ is the identity matrix. That would mean the eigenvalues can only be one and zero, but I'm not sure if $P$ can only be $I$. Is that really the only matrix that can satisfy the equation?

You are assuming that $P$ is invertable but it need not be. However your conclusion about the eigenvalues is correct. If $\lambda$ is an eigenvalue of $P$ with corresponding eigenvector $v$ then $PPv = P\lambda v = \lambda^2v$ but also $PPv = Pv = \lambda v$. Thus $\lambda^2 = \lambda$ from which follows $\lambda=0$ or $\lambda=1$.

In general matrices $P$ with $PP=P$ are called projections. And in fact they are in one-to-one correspondes with linear subspaces of $\mathbb{R}^n$.

Is that really the only matrix that can satisfy the equation?

Note: We have the determinant
$$
\lvert P^2 \rvert = \lvert P \rvert^2 = \lvert P \lvert \Rightarrow
\lvert P \rvert \in \{0, 1\}
$$
so it is either not invertible or some rotation.


Let $R=\{(x,y): x=y^2\}$ be a relation defined in $\mathbb{Z}$. Is it reflexive, symmetric, transitive or antisymmetric? Let $R=\{(x,y): x=y^2\}$ be a relation defined in $\mathbb{Z}$. Is it reflexive, symmetric, transitive or antisymmetric).

I'm having most trouble determining if this relation is symmetric, how can I tell?

The relation is not reflexive: $(2,2)\notin R$
The relation is not symmetric: $(4,2)\in R$, but $(2,4)\notin R$
The relation is not transitive: $(16,4)\in R$, $(4,2)\in R$, but $(16,2)\notin R$

Now, let's see if the relation is antisymmetric. Suppose $(x,y)\in R$ and $(y,x)\in R$. Then $x=y^2$ and $y=x^2$, which implies $x=x^4$. Can you now end proving the relation is antisymmetric?

None. $(4,2) \in R, (2,4) \notin R, (2,2) \notin R$. It is anti-symmetric.


Have I totally misunderstood the Banach–Tarski paradox? I was just reading about the Banach–Tarski paradox, and after trying to wrap my head around it for a while, it occurred to me that it is basically saying that for any set A of infinite size, it is possible to divide it into two sets B and C such that there exists some mapping of B onto A and C onto A.

This seems to be such a blatantly obvious, intuitively self-evident fact, that I am sure I must be missing something. It wouldn't be such a big deal if it was really that simple, which means that I don't actually understand it.

Where have I gone wrong? Is this not a correct interpretation of the paradox? Or is there something else I have missed, some assumption I made that I shouldn't have?

That's not what the paradox says. It says that you can take the unit ball in $\mathbb{R}^3$, divide it in certain disjoint subsets, then you can rotate and translate these subsets to obtain two unit balls. You need at least $5$ weird subsets if you want to do this 'explicitly'. The weird thing about this construction is that it seems that you somehow doubled the volume of the ball simply by cutting it into several parts.

The simple explanation is that there was absolutely no reason to expect that the volume should be preserved under the construction, as some of the disjoint subsets are not measurable, i.e. have no volume.

A first step to understanding the paradox is showing that it is impossible to define a meaningful measure on all subsets of $\mathbb{R}$ that is translation-invariant and such that the measure of an interval $[a,b]$ is $b-a$ (and a bunch of other desired properties). You can look up Vitali sets as an easy example of non-measurable sets. These certain subsets in the paradox are also going to be very wild, much like the Vitali sets.

Edit: To avoid any confusion. I just want to remark that the Banach-Tarski paradox is in fact not a paradox. Mathematically speaking this construction of "the doubling of the ball" is possible.

The Banach-Tarski paradox is the theorem in ZFC that a sphere can be partitioned into finitely many subets, then those subsets can be rearranged into 2 copies of the original sphere using only translation and rotation. Actually according to ZFC, it can be partitioned into only 5 parts.

Some people don't think you can just take on blind faith the axiom of choice. Indeed, some mathematicians work in the formal system of ZF and the axiom of choice is not a theorem of ZF. ZFC is ZF with the assumption of axiom of choice. It turns out that the Banach-Tarski paradox cannot be derived in ZF. According to https://www.emis.de/journals/RSMT/61-4/393.pdf, the axiom of determinacy implies Lebesgue measurability and according to https://brilliant.org/wiki/axiom-of-determinacy/, the axiom of determinacy is consistent with ZF.


Theorem 6.20 rudin Integration How does he do the algebra? (page 134 Rudin, chapter 6 ,theorem 6.20)

$\left| \frac {F(t)- F(s)}{t-s} -f(x_o) \right| = \left| \frac{1}{t-s} \int_s^t[f(u) - f(x_o)]du \right|< \epsilon $

also, how does he conclude that $F'(x_o) = f(x_0)$

: here is the theorem(and the proof by rudin)

Let $f \in \Re$ on $[a,b]$. For $  a \leq x \leq b$, put: $F(x)  = \int_a^x f(t)dt$, Then $F$ is continuous on $[a,b]$; furthermore, if $f$ is continuous at a point $x_0$  of $[a,b]$, then $F$ is differentiable at $x_o$ and $F'(x_0) = f(x_0)$.

(i will omit the proof of continuity of $F$ on $[a,b]$)

Suppose $f$ is continuous at $x_0$. Given $\epsilon > 0 $ choose $\delta > 0$ such that:

$\vert f(t)- f(x_o) \vert < \epsilon $

if $\vert t- x_0 \vert < \delta$, and  $a \leq t \leq b $.Hence, if

$x_0 - \delta < s \leq x_0 \leq t < x_0 + \delta$ $\enspace$ with:  $a\ \leq s < t \leq b$

we have by theorem 6.12(d)

$\left| \frac{F(t) - F(s)}{t-s} - f(x_0) \right| = \left| \frac{1}{t-s} \int_s^t [f(u) - f(x_0)]du \right| < \epsilon$

it follows that $F'(x_0) = f(x_0)$

You forgot to subtract $f(x_0)$ in the absolute value. First, note that $$F(t)-F(s)=\int_a^tf(u)du-\int_a^sf(u)du=\int_s^tf(u)du. $$ Also, note that since $f(x_0)$ is a constant  $$f(x_0)=\frac{1}{t-s} \int_s^t f(x_0)du .$$
Combining the two we find $$\left| \frac{F(t)-F(s)}{t-s}-f(x_0) \right|=\left|\frac{1}{t-s} \int_s^t [f(u)-f(x_0)]du \right|=\frac{1}{|t-s|} \left| \int_s^t [f(u)-f(x_0)]du \right|. $$ The final step is using the "triangle inequality for integrals" (Theorem 6.12(d)); you need to remember that $\left|f(u)-f(x_0)\right| < \varepsilon$ .

I'm not exactly sure of the context, but I can provide some information that I believe is relevant. There appears to be an error in what you have written.

Let $f\in L^{1}(-\infty, \infty)$. Define the function $F(t)=\int_{-\infty}^{t}f(y)dy$. Now let $x_{0}\in [s,t]$
\begin{eqnarray*}
\left| \frac{F(t)-F(s)}{t-s}-f(x_{0})\right| &=& \left| \frac{1}{t-s} \int_{s}^{t}f(y)dy-f(x_{0})\right|
\\
&=& \left| \frac{1}{t-s}\int_{s}^{t}\left(f(y)-f(x_{0})\right)dy\right|
\\
&\leq & \frac{1}{t-s}\int_{s}^{t}\left|f(y)-f(x_{0})\right|dy
\end{eqnarray*}

Now if $x_{0}$ is a Lebesgue point, the right hand side converges to $0$ as the length of the interval goes to $0$. More explicitly, for every $\epsilon>0$, there is $\delta$ such that if $t-s<\delta$, the right hand side is $<\epsilon$. Now, by Lebesgue's theorem, almost every point in $(-\infty,\infty)$ is a Lebesgue point. Therefore, the limit on the left hand side converges to $f(x_{0})$ for almost all $x_{0}\in(-\infty,\infty)$. By definition, this limit is $F^{\prime}(x_{0})$.

This is really a proof that the Radon-Nikodym derivative with respect to the Lebesgue measure agrees with the usual definition of derivative as the limit of the difference quotient.


Find $\lim_{n\to \infty}n^3\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)$ I have to show that$$\lim_{n\to \infty}n^3\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)=\frac{1}{6}.$$ I have now this rewriting:

$$\lim_{n\to \infty}n^2-n^3\sin\left(\frac{1}{n}\right)=\lim_{n\to \infty}-n^2\left(\sin\left(\frac1n\right)n-1\right).$$
Maple says it is $1/6.$ But how can I show it formally?

By elementary means:

If existence of a limit $L$ is granted,
$$n^3\left(\frac1n-\sin\frac3{3n}\right)=\frac{(3n)^3}{27}\left(\frac3{3n}-3\sin\frac1{3n}+4\sin^3\frac1{3n}\right),$$
and
$$L=\lim_{n\to\infty}n^3\left(\frac1n-\sin\frac3{3n}\right)=\lim_{n/3\to\infty}\frac{n^3}{27}\left(\frac3{n}-3\sin\frac1{n}+4\sin^3\frac1{n}\right)=\frac L9+\frac4{27}$$ giving $$L=\frac16.$$

Existence can be proven by

$$x-\frac{x^3}6\le\sin x\le x-\frac{x^3}6+\frac{x^5}{120}$$

which is easy to establish by five integrations of

$$0<\sin x<1,$$ but this is not elementary and completely solves the question by squeezing.


What is the actual definition of a function? I am learning precalculus and my book defines the following:

  A function $f$ from a set $A$ to a set $B$ is a rule that assigns to every element $a$ in $A$ one and only one value in $B$.

Well, I am thinking, a rule isn't something that I've seen defined mathematically. So what is a function, really? Is it a subset of $A\times B$ or something?

Indeed, a function is a subset of $A \times B$ with the following very important property: for every $x \in A$, there exist a unique $y \in B$ such that $(x,y) \in A \times B$. Intuitively, this tells you that a function cannot take an element $x$ into several distinct values $y$ - that wouldn't be a function anymore, but somehing called "binary relation" (that you need not worry about).

Nevertheless, for many practical purposes, thinking of a function as a rule, or a correspondence between $x$ values and $y$ values, will suffice. Still, keep the correct definition somewhere in the back of your mind, readily available if necessary.

A function is a rule which relates the values of one variable quantity to the values of another variable quantity, and does so in such a way that the value of the second variable quantity is uniquely determined by (i.e. is a function of) the value of the first variable quantity.

This is clearly a very general definition. Such rules are general enough to serve as mathematical models for a large number of natural relationships between variable physical quantities.

We name functions by using symbols like "f," "g," "h," etc. Just as your name signifies all of the many things that make "you," a symbol like "f" serves as a shorthand for what may be a long or complicated rule expressing a particular relationship between variable quantities. We name the variable quantities with symbols like "x," "y," "t," etc. We then write  y = f(x)  to show that  f  is a rule which assigns a unique value of the variable quantity  y  to values of the variable quantity  x.

Notice that the definition distinguishes between a first variable quantity and a second variable quantity. It is the second variable which is a function of the first, and when we write  y = f(x)  we mean that the value of  y  depends (in a way specified by the rule  f ) on the value of  x, but not necessarily vice versa. For example, what you wear may depend on the weather, but the reverse is certainly not true. We make this distinction among variables by calling  x  the input to the function and  y  the output, or by calling  x  the independent variable and  y  the dependent variable.


$\forall a,b,c \in \mathbb{N}\!:\ a\mid b \Rightarrow ac\mid bc. $ Divisibility Proof Could someone verify this proof?

Given the following definition of divisibility $(\forall{a,b \in \mathbb{N}})[a \mid b \equiv (\exists{c \in \mathbb{N}})(ac=b)]$.

Statement: $(\forall{a,b,c \in \mathbb{N}})[(a\mid b) \rightarrow (ac\mid bc)] $ .

Proof:

Let $a,b \in \mathbb{N}$ be arbitrary.  Assuming that $a\mid b$, we can conclude that $\exists{d \in \mathbb{N}}$ such that $ad = b$.  By Algebra we can deduce the following sequence of statements.  If $(ad)=b$, then $(ad)c=bc$.  Therefore $(ac)d=bc$.  Hence by the definition of divisibility we know that $ac\mid bc$.

Yes, your proof is correct. You might find it instructive to view the proof from a more conceptual viewpoint. Generally scaling a polynomial $f(x)$ by an invertible (so cancellable) element $c$ does not change its solution set since $c f(d) = 0 \iff f(d) = 0$. In particular the  solvability of $f(x) = 0$ is not altered - which is what we are concerned with for divisibility: $\,a\mid b\iff ax = b\,$ is solvable.

Let's consider the less trivial case of quadratic $f(x)$. The formula $g(a,b,c)$ for the solutions of a quadratic $ax^2+bx+c$ remains unaltered by nonzero coefficient scalings $g(ad,bd,cd) = g(a,b,c)$. As for solvability, scaling the polynomial by $d\neq 0$ multiplies the discriminant $b^2-4ac$ by $d^2$ so it does not affect the solvability criterion - that the discriminant be a perfect square.

But we don't need to know these explicit (linear / quadratic) formulas  to deduce this. Rather, we can deduce it at a conceptually higher-level, simply from the cancellability of $\,c,\,$ i.e the invertibility of  scaling transformations $x\to cx$.

Remark $\ $ Below are some general remarks about such equational inferences.

Since the proof  employs only the associative and commutativite axioms for multiplication it remains true in any commutative semigroup. The proof employs only the inference rules of first-order equational logic, namely that equality is an equivalence relation (reflexive, symmetric, transitive) and the rules of substitution: $p(x) = q(x) \rightarrow p(a) = q(a)$ and replacement: $a = b \rightarrow p(a) = p(b)$.

By a famous theorem of Birkhoff, for any algebraic structure that can be defined by axioms - like comutativity, associativity, etc - that are all universally quantified equations, said inference rules are complete, i.e. an equation is provable from the axioms using said inference rules iff it is true in every structure that satisfies the axioms. However, finding such a purely equational proof can be highly difficult. For example, consider the famous Jacobson theorem that rings satisfying the axiom $x^n = x \;$ are commutative $xy = yx$.  It is  notoriously difficult for $n>2$ to find purely equational proofs of this theorem. So ingenuity will be essential in general (e.g. exploiting innate structure such as the above equivalence under invertible scalings).


direct proof of $U_1\cup U_2$ is subspace of $V$ $\implies$ $U_1\subset U_2$ or $U_2\subset U_1$ I'm trying to prove the following by direct proof (I've already got a proof by contradiction).

  The union of two subspace of $V$ is a subspace of $V$ implies one of the subspaces is contained in the other.

Here's how far I've gotten:  Suppose $U_1\cup U_2$ is a subspace of $V$. Let $u_1 \in U_1$ and $u_2 \in U_2$.  Then, by closure of vector addition in $U_1\cup U_2$, $u_1 + u_2 \in U_1\cup U_2$.  This implies that $u_1 + u_2 \in U_1$ or $u_1 + u_2 \in U_2$.  WLOG, suppose $u_1 + u_2 \in U_1$.  Then $u_2 = (u_1 + u_2) - u_1 \in U_1$.

At this point, I'd like to be able to say that because $u_2$ was chosen arbitrarily, this implies that $U_2 \subset U_1$.  But I also made the choice that $u_1 + u_2 \in U_1$, so for some other choice of $u_2$ it could be the case that $u_1 + u_2 \in U_2$ which would imply $u_1 \in U_2$.

Is there any way to complete this proof directly or do I have to stick with the proof by contradiction for this one?

You don't need to use contradiction really, but mere inspection of cases. Just prove that if $U_1$, say, is not contained in $U_2$ i.e. if there's a $u_1\in U_1\smallsetminus U_2$, then $U_2\subset U_1$.

The idea is OK, go by contradiction.

Suppose  neither inclusion holds, but $U_1 \cup U_2$ is a subspace.
So we have $u_1 \in U_1 ,u_1 \notin U_2$ and $u_2 \in U_2, u_2 \notin U_1$ to witness the non-inclusions.

But $u_1 + u_2 \in U_1 \cup U_2$, say in $(u_1 + u_2) \in U_i$.
But then $(u_1 + u_2 ) - u_i \in U_i$, while it equals $u_j$ where $j \neq i$. Contradiction, as $u_j \notin U_i$ when $j \neq i$.


Do there exist any numbers such that any in-order combination of digits is prime (including the original number)? I was inspired by this question, in particular, where removing any digit of a prime yielded another prime. My question is, are there any numbers that this holds true for, and will continue holding true for if digits continue to be removed?

I have a strong feeling that this is false. However, I'm very interested in seeing a proof for the falsehood of this, or maybe even a proof that one does exist but is not computable within reason.

If you're asking what I think you're asking, it shouldn't be hard to work through it. First, since you eventually get down to one digit, only prime numbers can be digits; 2, 3, 5, and 7. But 2 and 5 are no good, since any 2-digit number ending in 2 or 5 will be composite, so we just have 3 and 7. But we can't use either digit twice, since if we remove all the other digits we'll get 33 or 77, both composite. So Matthew's 37 (and 73) is the only answer of more than one digit.

If you choose to make believe that 1 is a prime, then you also get 13 (and 31) and 17 (and 71), and you have to check whether the numbers 137, 173, 317, 371, 713, and 731 are all prime.

Found this when I went looking for numbers where any combination of their digits result in a prime, since earlier I happened to stumble upon this property of 3 digit numbers comprised of two 1's and one 3, those being 113, 131, and 311, which are all prime, and all satisfy the "take any 1 digit and a prime will remain" condition, though they all start with a crutch since they're comprised of only 2 unique digits. I also happened to find something interesting while testing the combinations of the digits in the number provided in the other post you linked, there are two of the six combinations which were not prime, 791 and 917, both of which had a 7 and a 3 digit number comprised of two 1's and one 3 as factors, 791 = 7 * 113 and 917 = 7 * 131, it's oddly coincidental how the example managed to connect to my number at all, but I thought it was really interesting. Also I'm 9 years late lmao


A property of positive convergent series Let $(a_k)_{k\geq 1}$ be a positive sequence such that the series $\sum_{k\geq 1} a_k$ is convergent to $L$. Then, by C-S inequality, for any $n\geq 1$,
$$f(n):=\sum_{k=1}^n \frac{1}{a_k}\geq \frac{n^2}{\sum_{k=1}^n a_k}> \frac{n^2}{L}.$$
Is it true that eventually $f(n)>n^2\ln(n)$ (i.e. there is $n_0$ such that the inequality holds for all $n\geq n_0$).

In mathematical jargon "eventually" would mean the bound holds for all sufficiently large $n$ i.e.
$$\text{there exists $N$ such that $f(n) >n^2\ln n$ for all $n\geq N$.}$$

This is not the case, in fact we can have $f(n)\leq n^2\omega(n)$ infinitely often for any given function $\omega(n)\to\infty$ (in particular for $\omega(n)=\ln n$). This is the best possible because $\liminf f(n)/n^2=+\infty$.

If you instead ask whether $f(n)$ ever exceeds $n^2\ln n$, then the answer is true (even excepting $n=1$). I give two proofs below.

Sequence with $f(n)\leq n^2\omega(n)$ infinitely often given a function $\omega(n)\to\infty$

Let $1\leq n_1\leq n_2\leq \cdots$ be a sequence to be chosen later. Define
$$a_n=\sum_{k\text{ such that }n_k\geq n} \frac{1}{n_k2^k}.$$

Note $\sum_{k\geq 1}a_k$ converges (to $2$), and for each $k\geq 1$,

$$f(n_k)=f(n_{k-1})+\sum_{i=n_{k-1}+1}^{n_k} \frac{1}{a_i} = f(n_{k-1})+2^kn_k(n_k-n_{k-1}).$$

If $n_1,\dots,n_{k-1}$ are fixed, the right-hand side is $O(n_k^2)$. So if $\omega(n_k)\to\infty$ as $n_k\to\infty$, we can always pick $n_k$ large enough such that $f(n_k)\leq (n_k)^2\omega(n_k)$.

Proof of $\liminf f(n)/n^2=+\infty$

Your Cauchy-Schwarz argument similarly gives, for even $n$,

$$f(n)/(n/2)^2>(1/(n/2)^2)\sum_{k=n/2+1}^{n}\frac{1}{a_k} \geq \tfrac 1 4 \left(\sum_{k=n/2+1}^{\infty}a_k\right)^{-1}$$

and the right-hand-side converges to $+\infty$ as $n\to\infty$. Of course odd $n$ obey a similar bound but it is more fiddly to write down.

Grouping proof that $\sup_{n\geq 2} f(n)/n^2\ln n = +\infty$

There must be some $d\geq 1$ such that
$$\sum_{k=2^{d-1}+1}^{2^d}a_k \leq \frac 1{4C \ln(2^d)}$$
since the sum over $d$ of the right-hand-side diverges.

Your Cauchy-Schwarz argument gives
$$f(2^d)>\sum_{k=2^{d-1}+1}^{2^d}\frac{1}{a_k} \geq 4C\ln(2^d)(2^d-2^{d-1})^2=C\ln(2^d)(2^d)^2$$

so $f(n)>Cn^2\ln n$ for $n=2^d$.

Majorization proof that $\sup_{n\geq 2} f(n)/n^2\ln n = +\infty$

Fix a constant C. Define $(r_n)_{n\geq 1}$ to be the unique sequence with
$$\sum_{k=1}^n r_k = R(n) := Cn^2\max(1, \ln n).$$

We can estimate $r_n=O(n\ln n)$ so $\sum 1/r_k$ diverges. Pick $N$ large enough that $\sum_{k=1}^N 1/r_k>L$. Given any sequence $a_1,\dots,a_N>0$ such that $f(n)\leq R(n)$ for $1\leq n\leq N$, we can reduce arbitrary elements of $a_n$ if necessary to increase $f(N)$ to $R(N)$. Then $1/a_1,\dots,1/a_N$ majorizes $r_1,\dots,r_N$: assuming w.l.o.g. that $a_n$ are in descending order,
$$f(n)\leq R(n) \text{ for all $1\leq n<N$, and }f(N)=R(N)$$
(This is majorization "from above", which is equivalent to majorization "from below" because the total sums match.)
Since $1/x$ is a convex function for $x>0$, by Karamata's inequality we have
$$a_1+\dots+a_N\geq \frac{1}{r_1}+\dots+\frac{1}{r_N}$$
which gives $\sum_{k=1}^Na_k >L.$


You roll a die 8 times. What is the probability of getting an odd number at least twice and an even number at least three times? My initial idea was to do something such as $3^5 * 6^3$ for the possible cases. However, I believe this is not appropriate for this question, since simply multiplying is usually used in cases in which order matters (e.g. combination of a locker), and here order does not matter.
I'd imagine that maybe we should use something like Bernoulli trials or Combinations, but I'm not sure.

Note: I'm not saying this the the answer to your question. I'm showing you the binomial distribution and a process for adding desired
terms. You need to decide which terms are
needed to answer your question.
The probability of an Odd result on any one roll is $p=1/2.$ The number $X$ of odd results in $n = 8$ rolls is
distributed $X \sim \mathsf{Binom}(n = 8, p = 1/2).$
Specifically, the probability that $X = k$ is
$$P(X = k) = {8\choose k}(1/2)^8,$$
for $k = 0,1,\dots,8.$
Suppose you want the probability $P(2 \le X \le 5).$ That's
$$P(2 \le X \le 5) = P(X \le 5) - P(X \le 1)\\
= \sum_{k=2}^5 {8\choose k}(1/2)^8 = 0.8203,$$
to four places. To get this probability, you have four terms to add.
In R statistical software this can be computed
as follows, where dbinom is a binomial PDF and pbinom is a binomial CDF.
k = 2:5; dbinom(k, 8, 1/2)
[1] 0.1093750 0.2187500 0.2734375 0.2187500
sum(dbinom(k, 8, 1/2))  # add 4 PDF terms
[1] 0.8203125

pbinom(1:5, 8, 1/2)
[1] 0.03515625 0.14453125 0.36328125 0.63671875 0.85546875
diff(pbinom(c(1,5), 8, 1/2))  # subtract 2 CDF values
[1] 0.8203125

In the figure below, we are adding the heights of the four
maroon bars.

R code for figure above:
plot(0:8, dbinom(0:8, 8, 1/2), type="h", lwd=2,
     col="blue", ylab="PDF", xlab="k",
     main="PDF of BINOM(8, .5)")
 abline(h = 0, col="green2")
 lines(2:5, dbinom(2:5, 8, 1/2), type="h",
       lwd=2, col="maroon")

The probability of getting  an odd number at least 2 times = the probability of getting an odd number exactly 2 times + probability of getting odd number 3 times
$n$=8.
Hint: Probability of getting an odd number exactly twice:
$8*(\frac{1}{2})^2$*$\frac{1}{2}$.


Rate of surface increase of a sphere related to its volume increase I'm trying to solve problems in the Thompson's classic "Calculus made easy" (not so easy by the way).

One of this problems is to find the rate increase of the surface of a sphere which volume is increasing at $4$ square feet/second rate.

I've found a solution which unfortunatly doesn't agree with author's one.

My solution:

$r$ = radius

$v$ (sphere volume) = $\frac{4}{3}\pi r^3$

$s$ (sphere surface) = $4\pi r^2$

I "link" the volume with the surface

$$v = \frac{4}{3}\pi r^3 = 4\pi r^2 \frac{r}{3}= s \frac{r}{3}\ \Longrightarrow s = \frac{3v}{r}.$$

Now I apply the derivatives chaine rule

$$d\left[\frac{s}{t}\right] = d\left[\frac{s}{v}\right]\cdot d\left[\frac{v}{t}\right]
=d\left[\frac{3v}{r}\right]\cdot d\left[\frac{v}{t}\right]
= \frac{3}{r}\cdot 4
= \frac{12}{r}.$$

The author finds $\frac{8}{r}$, but I suspect a typographical error as an other site has found also my result.

Thank you very much to say me if I've made a mistake somewhere.

Using the natural notation, we have that since $S=4\pi r^2$, by the Chain Rule we get
$$\frac{dS}{dt}=8\pi r\frac{dr}{dt}\tag{1}$$
and similarly that
$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.$$
Since $\frac{dV}{dt}=4$ (cubic feet per second), we have $\frac{dr}{dt}=\frac{4}{4\pi r^2}$.
Substitute in (1). We get $\frac{dS}{dt}=\frac{8}{r}$.

Remark: Since we have nice formulas for $S$ and $V$, it is best to differentiate immediately. That is often the case in "related rates" problems.

Note: The answer given in Calculus Made Easy-Third Edition is 8/r.
While the answer given for this problem in the 1998 edition (with Martin Gardner's additions) is 8*(pi^.5). I believe 8/r to be correct.


Help to understand a comment in Hoffman and Kunze's linear algebra book I'm reading Hoffman and Kunze's Linear Algebra and on page 52, the authors said:

  Let $P$ be the $n \times n$ matrix whose $i,j$ entry is the scalar $P_{ij}$, and let $X$ and $X'$ be the coordinate matrices of the vector $\alpha$ in the ordered bases $\mathscr{B}$ and $\mathscr{B}'$. Then we may reformulate (2-15) as
  $$
X = PX'. \tag{2-16}
$$
  Since $\mathscr{B}$ and $\mathscr{B}'$ are linearly independent sets, $X=0$ if and only if $X'=0$. Thus from (2-16) and Theorem 7 of Chapter 1, it follows that $P$ is invertible. Hence
  $$
X' = P^{-1}X. \tag{2-17}
$$

I didn't find any mention of this result (highlighted in bold) in the book. How can I prove this fact?

$\mathcal{B} = \{b_1, b_2, \dots, b_n\}$ and $\mathcal{B}' = \{b_1', b_2', \dots, b_n'\}$ are bases, in particular linearly independent sets. Given a vector $\alpha$ in the vector space, the coordinate matrix of $\alpha$ with respect to $\mathcal{B}$ has coordinates corresponding to the unique coefficients
$$
\alpha = x_1b_1 + x_2b_2 + \dots + x_nb_n,
$$
and its coordinate matrix with respect to $\mathcal{B}'$ has coordinates corresponding to the unique coefficients
$$
\alpha = x_1'b_1' + x_2'b_2' + \dots + x_n'b_n'.
$$

Suppose $X = 0$, then clearly $\alpha = 0$. However, since $\mathcal{B}'$ is linearly independent, we must have $X'=0,$ since $X'\neq 0$ would imply there exists a nontrivial linear combination of the basis vectors in $\mathcal{B}'$ equal to the zero vector, contradicting the linear independence of a basis. Similarly if we assume $X' = 0$.

It is actually related to the matrix of basis change.

P is actually the matrix represents the basis change from the basis of X' to that of X.
Now we know that the basis change matrix is always invertible because of the fact that B and B ' are independent  .
It is easy to see.
You can proove  by inequality of the ranks of matrices
Or from the scratch that are required to proove the inequalities .

Hope that it is of some use.


Transform $dx/dt = 1 + r + x^2$ into normal form $dX/dt = R +- X^2$ (saddle node bifurcation) I am trying to find a substitution $X(x,r)=?$ and $R(x,r)=?$ to allow me to transform

$dx/dt = 1 + r + x^2$

into

$dX/dt = R +- X^2$   (normal form for saddle node bifurcation)

I'm sure it's a simple substitution, but I just can't see it for the life of me.

Thanks!

The solution was in fact to complete the square which gives:

$dx/dt = 1 - (r/2) + (x+r/2)^2$

then let X=x+r/2, which means dX/dt = dx/dt, and then let R=1-(r/2). Substituting then gives:

$dX/dt = R + X^2$

which is the normal form for a saddle node

We are working with the equation:
$$\frac{dx}{dt}=r+1+x^2$$
I believe the answer is connected to the total derivative of the substitution function. For example we can simplify:
$$\frac{dX(x,r)}{dt}=R(x,r)\pm X(x,r)^2$$
Into
$$\frac{\partial X}{\partial x}\frac{dx}{dt}+\frac{\partial X}{\partial r}\frac{dr}{dt}=R(x,r)\pm X(x,r)^2$$
Therefore, we match the left side to the first equation and state that:
$$\frac{\partial X}{\partial r}=0\rightarrow X(x,r)=X(x)$$
Therefore we rearrange the equation to:
$$\frac{dX}{dx}\frac{dx}{dt}=r+1+x^2\rightarrow\frac{dx}{dt}=\frac{R(x,r)}{X'(x)}\pm\frac{X(x)^2}{X'(x)}$$
By matching expressions, we obtain two ODEs for the functions:
$$\frac{X(x)^2}{X'(x)}=\pm(1+x^2)\text{ }\&\text{ }R(x,r)=r\cdot\frac{dX}{dx}$$
With the solution of:
$$X(x(t))=\frac{1}{C\mp\tan^{-1}x(t)}\text{ and }R(x,r)=r\cdot\pm\frac{1}{\left(C\mp\tan^{-1}x(t)\right)^2\left(1+x(t)^2\right)}$$


Why is $\infty^0$ an indeterminate form? I can't seem to find a situation where $\lim_{x\to a}f(x)=\infty$ and $\lim_{x\to a}g(x)=0$, but $\lim_{x\to a} f(x)^{g(x)}\neq 1$.  This makes me think that, even though this form ($\infty^0$) is often stated as an indeterminate form, it is not actually indeterminate (since the limit is always 1).

Moreover, I cant find a question asking this on stackexchange, and the calculus book I use neglects to provide an explanation for why it is considered indeterminate.

I have tried to prove that the limit is always 1, but it has been a few years since my real analysis days, and I am getting stuck in the proof.  I'll continue to look for a proof, but in the mean time, I figured I would ask the question, and see if anyone can provide an example I've missed.

You aren't trying hard enough. \begin{align}&\lim_{x\to\infty} x^{1/\ln x}=\lim_{x\to\infty} e=e\\ &\lim_{x\to\infty} x^{(\ln x)^{-1/2}}=\infty\end{align}

$\infty^0$ is directly related to $\infty\cdot0$ by taking the logarithm.


Find values of $x$, such as $\log_3 \sqrt{x+3}−\log_3(9−x^2) < 0$ The Function is $$f(x) = \log_3\sqrt{(x+3)}−\log_3(9−x^2)$$
and I need to figure out arguments for which $$ f(x) < 0 $$

So I calculated the domain of function which is   $ D: (-3;3)$

However I am still unable to solve $f(x) < 0$

I simplified $  \log_3\sqrt{(x+3)}−\log_3(9−x^2) < 0$    to   $\log_3\frac{\sqrt{(x+3)}}{(9-x^2)} < 0$

which gets me to $$ \frac{\sqrt{(x+3)}}{9-x^2} < 1$$

But the solutions of the above equation are complex numbers and I definitely should get them as my result. So, what I am doing wrong here?

Would apprecite every answer

First of all, we need $9-x^2>0\iff-3<x<3$

As $\log_ax$ is increasing function for $a>1$

We need $$\sqrt{x+3}<9-x^2$$

Let $\sqrt{x+3}=y>0\implies x=y^2-3$

$$\implies0<9-(y^2-3)^2-y=-y+6y^2-y^4=-y(1-6y+y^3)$$

As $y>0,$ we need $$y^3-6y+1<0$$


prove that $\lim\limits_{x\to 1}\frac{x^{1/m}-1}{x^{1/n}-1}=\frac{n}{m}$ I'm trying to find a way to prove this:

EDIT: without using LHopital theorem.
$$\lim_{x\rightarrow 1}\frac{x^{1/m}-1}{x^{1/n}-1}=\frac{n}{m}.$$
Honestly, I didn't come with any good idea.

We know that $\lim_{x\rightarrow 1}x^{1/m}$ is $1$.

I'd love your help with this.

Thank you.

HINT $\ $ If you change variables $\rm\ z = x^{1/n} $ then the limit reduces to a very simple first derivative calculation. See also some of my prior posts for  further examples of limits that may be calculated simply as first derivatives.


Length of chord on ellipse Suppose I have an ellipse centered at the origin, preferably expressed in its matrix form, and I want to know the chord length of a segment that passes through the origin with the endpoints at the ellipse, of which I know the angle theta. How can I compute it? Thank you!

If your chord makes an angle $\phi$ with the $x$-axis, then its length is
$$
\frac{2ab}{\sqrt{  {b^2}\cos^2{\phi} + {a^2}\sin^2{\phi}  }}
$$
See this answer for a bit more detail.

The angle $\theta$ used in apurv's answer is not the angle between the chord and the $x$-axis. That's probably the reason for his question, where he asked you what you mean by "theta".

The answer from Bubba assumes the chord passes through the center, so it's actually a diameter.  If you start with parametric equations [x,y]=R*[cos(phi),sin(phi)], substitute them into the equation for an ellipse, and solve for R, you get half of Bubba's answer.


then there exist $m,n$ such $(m!)\cdot (n!)\equiv r\pmod p$ let $p$is give odd prime number,show that: for any postive integer $r$,then there exists postive ineteger $m,n$ such that
$$(m!)\cdot (n!)\equiv r\pmod p$$
I try use this $$(m)!\equiv (p+1)(p+2)\cdots (p+m) \pmod p$$
$$(n)!\equiv (p+1)(p+2)\cdots (p+n) \pmod p$$
so
$$(m!)\cdot (n!)\equiv (p+1)(p+2)\cdots (p+m) (p+1)(p+2)\cdots (p+n)\pmod p$$
But I can't it

I am a tenth grader, so I find this a somewhat hard problem. Hence, I'd be doing it the other way around.
If $\exists x : x \leq m , p \mid x$ or $\exists x : x\leq n, p \mid x \implies r = 0 \implies m,n < p$ for $r \neq 0$
The only easiest thing we can do is to prove that $\exists r :m! \times n! \equiv r (\mod{p})$ \
Since $m < p, m \equiv m \pmod{p}$.
If $m! > p$, let $m! = px + a \implies ,m! \equiv a \pmod{p} (\implies a = m! - px)$
Also, if $n < p$, $n \equiv n \pmod{p}$.
If $n! > p$, let $n! = py + b \implies n! \equiv \pmod{p}(\implies b = n! - px)$
If we take $r = a\times b$, $r = (m! - px)(n! - py) = p(-m!y -n!x + xy) + m!n! = p(-pxy -ay -pxy -bx + xy) +m!n! = p(xy(1- 2p) -(ax + by)) + m!n! $
Since the term before the plus sign is divisible by $p$, we can say $r \equiv m!n! \pmod{p}$
Or, if you take $m!n!$, you get $p^2xy + pxb + pya + ab = pq + r (r = ab)$. Thus, $m!n! \equiv ab \pmod{p} \implies m!n! \equiv r \pmod{p}$
This necessarily means that for most $r$'s there exists some $m,n$ satisfying $m!n! \equiv r \pmod{p}$. Why I say most $r$'s is because there could be exceptions (as JMP pointed out in the comments). Also, you can't all of a sudden generalise this. As in the comments, simple algebra won't work here.
This solution might be faulty.. please notify me if it is so.


Finding closure/interior of subset of function space Consider the subset $$A=\left\{f\in C(\Bbb R): |f(x)|< \frac{1}{1+|x|} \, \text{for all } x\in \Bbb R\right\}\subset \left\{f\in C(\Bbb R): \lim_{|x|\to \infty}f(x)=0 \right\}=X.$$

where $X$ is equipped with the operator norm. What is the closure and interior of $A$?

  My guess is that the closure of $A$ is the set $$B=\left\{f\in C(\Bbb R): |f(x)|\leqslant \frac{1}{1+|x|} \, \text{for all } x\in \Bbb R\right\}$$
  and the interior is the set
  $$C=\left\{f\in C(\Bbb R): \exists\epsilon>0 \text{ such that }|f(x)|< \frac{1}{1+|x|}-\epsilon \, \text{for all } x\in \Bbb R\right\}.$$

I'm fairly certain I got the closure right, but I'm unsure about the inerior.

Certainly $C$ is open and a subset of $A$, therefore, $C\subset \operatorname{int}A$. Does $\operatorname{int A}\subset C$ also hold?

Are my guesses correct?

According to your definition of $C$, $A\cap C =\emptyset$ ( if $f\in A$, $|f|$ eventually becomes smaller than $\epsilon$.)

Perhaps it makes more sense to consider the set:

$
C=\left\{ f\in A\,\,\, |\,\,\, \exists M>0 \,\,\,s.t.\,\,\, f(x)=0\,\,\, \forall |x| \geq M \right\}.
$


Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose the row that they chose. This bothers me. As I should be able to do it as I see fit. I will post my work and someone point out the problem in my work. The matrix is as follows:

$$A = \left(
        \begin{matrix}
        5&-7&2&2\\
        0&3&0&-4\\
        -5&-8&0&3\\
        0&5&0&-6\\
        \end{matrix} \right)
$$

I decided to expand across row one and cross out columns as I found the minors. For the first minor obtaining:
$$
        \begin{pmatrix}
        3 & 0 & -4 \\
        -8 & 0 & 3 \\
        5 & 0 & -6 \\
        \end{pmatrix}
$$

M1 being row one column one we attain $-1^2 = 1$. This is to be multiplied by the determinate of the minor. Now finding the determinant I did:

3 times $$
        \begin{pmatrix}
        0 & 3 \\
        0 & -6  \\
        \end{pmatrix}
$$
giving $3(0-0)= 0$
then:

0 times $$
        \begin{pmatrix}
        -8 & 3\\
         5 & -6\\
        \end{pmatrix}
$$

giving 0(48-15)=0

Then:
4 times
$$
        \begin{pmatrix}
        -8 & 0 \\
         5 & 0 \\
        \end{pmatrix}
$$
giving $4(0-0)=0$
adding the determinants we get $0+0+0=0$
So det M1 $= 0(1) = 0$

M2--> M(1,2)---> $-1^1+2= -1^3 = -1$

$$
        \begin{pmatrix}
        0 & 0 & -4 \\
        -5 & 0 & 3 \\
        0 & 0 & -6 \\
        \end{pmatrix}
$$

o*
$$
        \begin{pmatrix}
        0 & 3 \\
        0 & -6 \\
        \end{pmatrix}
$$
giving $0(0-0)=0$

obviously the next matrix will look the same as the top term in column two is a zero so the determinant for that will be $0$. Now finally

4 times
$$
        \begin{pmatrix}
        -8 & 0 \\
        5 & 0 \\
       \end{pmatrix}
$$
giving 4(0-0)= 0

So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3

M3 --> $-1^4 = 1$

$$
        \begin{pmatrix}
        0 & 3 & -4 \\
        -5 & -8 & 3 \\
        0 & 5 & -6 \\
        \end{pmatrix}
$$

for the determinant:

0 times
$$
        \begin{pmatrix}
        -8 & 3 \\
         5 & -6 \\
        \end{pmatrix}
$$
which gives $0(48-15)=0$

-3 times
$$
        \begin{pmatrix}
        -5& 3 \\
        0 & -6 \\
        \end{pmatrix}
$$

which gives $-3(30-0)= -90$

it is redundant to go on from here because after the final computation for this minor I get -100 and as a result get det M3 = -190 and get determinant of zeros for the following determinant of M4.
which gives: $0(5)+ 0(-7) + (-90)(2) + (0)(2)$ giving
Det Ax $= -380.$ The book says its $20$ and when I did it in a calculator it got 20 but the problem is that both the book and calculator expand across the row with the most zeros but theoretically speaking NO MATTER WHICH row or column you choose to expand across you should get the same answer. So what is it? Is my computation wrong or is my assumption that you can expand across any row or column wrong? Isn't it only important if the determinant doesn't equal zero? or does the exact value matter in more advanced cases?

Here is how you should write it down in practice.
$$
    A =
        \begin{pmatrix}
        5  & -7 & 2 & 2  \\
        0  & 3  & 0 & -4 \\
        -5 & -8 & 0 & 3  \\
        0  & 5  & 0 & -6 \\
        \end{pmatrix}
$$
so that
$$
    \det A =
        \begin{vmatrix}
        5  & -7 & 2 & 2  \\
        0  & 3  & 0 & -4 \\
        -5 & -8 & 0 & 3  \\
        0  & 5  & 0 & -6 \\
        \end{vmatrix} \\
    = 2 \cdot
        \begin{vmatrix}
        0  & 3  & -4 \\
        -5 & -8 & 3  \\
        0  & 5  & -6 \\
        \end{vmatrix} \\
    = 2 \cdot -(-5) \cdot
        \begin{vmatrix}
        3  & -4 \\
        5  & -6 \\
        \end{vmatrix} \\
    = 2 \cdot 5 \cdot (-18 -(-20)) \\
    = 20.
$$
I expanded along the third column in the first case and then along the first column, because these are the ones with the most zeros, so it saves a lot of computations. I am assuming that's what's in your book.
If you want to expand along other columns or rows, just keep track of the appropriate minors without making any computation mistakes (I guess that's the hard part ; the only trick is to make it slowly and be careful). So here we go (along the first row):
$$
    \det A =
        \begin{vmatrix}
        5  & -7 & 2 & 2  \\
        0  & 3  & 0 & -4 \\
        -5 & -8 & 0 & 3  \\
        0  & 5  & 0 & -6 \\
        \end{vmatrix}
    =
5         \begin{vmatrix}
        3  & 0 & -4 \\
        -8 & 0 & 3  \\
        5  & 0 & -6 \\
        \end{vmatrix}
-(-7)   \begin{vmatrix}
        0  & 0 & -4 \\
        -5 & 0 & 3  \\
        0  & 0 & -6 \\
        \end{vmatrix}
 + 2
        \begin{vmatrix}
        0  & 3  & -4 \\
        -5 & -8 & 3  \\
        0  & 5  & -6 \\
        \end{vmatrix}
-2         \begin{vmatrix}
        0  & 3  & 0 \\
        -5 & -8 & 0 \\
        0  & 5  & 0 \\
        \end{vmatrix}.
$$
I don't know how you defined the determinant, but in any definition you chose it should be obvious that a determinant of a matrix with a column of zeros is zero. (If it is not clear to you, feel free to tell me your definition and I will happily answer in the comments.) So the only non-zero term in this sum is
$$
2
        \begin{vmatrix}
        0  & 3  & -4 \\
        -5 & -8 & 3  \\
        0  & 5  & -6 \\
        \end{vmatrix}.
$$
Note that this is the same determinant as when I expanded along the third column. So you can finish this computation by looking at what I did before.
A useful trick to remember the signs in the Laplace expansion (that's the name of the trick of expanding along a row or a column) is the following matrix :
$$
\begin{vmatrix}
+ & - & + & - \\
- & + & - & + \\
+ & - & + & - \\
- & + & - & +
\end{vmatrix}
$$
It works for any determinant size, just make sure that the coordinate of the matrix in the top left is a $+$ sign.
Hope that helps,

Write out the determinant. Alongside write the first  3 columns"Downward diagonal multiplication" gives ( O + O + 2OO + O ).  "Upward diagonal multiplication" gives -(O +  O + O + 18O) . det = 2O.


Integral of $\cos(3x) \cos(4x) \cos(5x)$ The integral $$\int_0^{\pi/8}\cos(3x)\cos(4x)\cos(5x) \,dx$$ is equal to $k/24$. Find the constant $k$.

So far, I assume that the best way to solve this question is to solve the integral and compare the answer to find $k$.

I have thought about rewriting the integral by using the identity
$\cos(2x) = \cos^2(x) - \sin^2(x)$ or $\cos(2x) = 1 - 2\sin^2(x)$,
but that results in a very long and confusing function.
I don't see any possible substitutions.

Please help!

Hint
$$\cos 5x\cdot\cos 3x\cdot\cos 4x
\\=\cos 5x\cdot\frac{(\cos 7x+\cos x)}{2}
\\=\frac{\cos 5x\cdot\cos 7x+\cos 5x\cdot\cos x}{2}
\\=\frac{\cos 13x+\cos 2x+\cos 6x+\cos 4x}{4}$$


Algebra 101: Making subject I am trying to figure out how the upper equation can be equivalent to the one at the bottom.

My best guess is that something is being multiplied or divided from both sides but cannot figure out what.

I proceed as follows:

I first expand everything and cancel out similar terms to ensure that I have L+W on one side and I get the following:

$$ rL(\rho-1)>= -{\sigma}{\tau}(L+W)$$

So isolating L+W gets a very different result.

I'd really appreciate if anyone can guide me in the right direction.

Continuing from where you left off, plus assuming $1 \gt \rho \implies 1 - \rho \gt 0$, $L + W \gt 0$ and $1-\frac{\sigma\tau}{r(1-\rho)} \gt 0$, gives

$$\begin{equation}\begin{aligned}
rL(\rho-1) & \ge -\sigma\tau(L+W) \\
-\frac{rL(1 - \rho)}{L+W} & \ge -\sigma\tau \\
-\frac{L}{L+W} & \ge -\frac{\sigma\tau}{r(1-\rho)} \\
1-\frac{L}{L+W} & \ge 1-\frac{\sigma\tau}{r(1-\rho)} \\
\frac{W}{L+W} & \ge 1-\frac{\sigma\tau}{r(1-\rho)} \\
W & \ge (L + W)\left(1-\frac{\sigma\tau}{r(1-\rho)}\right) \\
\frac{W}{1-\frac{\sigma\tau}{r(1-\rho)}} & \ge L+W \\
L + W & \le \frac{W}{1-(\sigma\tau)/(r(1-\rho))}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$


Suppose that $g: A \rightarrow B$, $h: B \rightarrow C$. Prove that if $g$ is onto and $h$ is not one-to-one, then $h \circ g$ is not one to one. How can I start to approach this question? I know that if $g$ and $h$ are one-to-one, then $h \circ g$ is one-to-one, and also if $h$ and $g$ are both onto, and $h \circ g$ is onto. But this question didn't fit into both cases...

Since $h$ is not one to one, there is $b_1, b_2 \in B$ distinct such that $h(b_1) = h(b_2)$. Since $g$ is onto, there is $a_1, a_2 \in A$ such that $g(a_1) = b_1$ and $g(a_2) = b_2$.
Then notice that $h(g(a_1)) = h(b_1) = h(b_2) = h(g(a_2))$. So then $h\circ g$ is not one to one.

I'd start with considering that is $h$ is not one to one considering the $x\ne y$ so that $h(x) = h(y)$ and see what happens.
$g$ is onto so there are $a,\alpha \in A$ so that $g(a) = x$ and $g(\alpha) = y$.  And.... well, seems the obvious thing to try is to  $h(g(a)) = h(x)$ while $h(g(\alpha) = h(y) = h(x)$.
So if $a \ne \alpha$ then $h\circ g$ is not one-to-one.  And clearly $a \ne \alpha$ because if $a = \alpha$ then $g(a) = g(\alpha)$ but $g(a) = x\ne y = g(\alpha)$.
So that's that.


Is $a$ < $b$ if $f(x)$ is differentiable on $(a,b)$ and continuous in $[a,b]$? $f(x)$ is differentiable on $(a,b)$ and continuous in $[a,b]$.

In theory $a$ can be equal to $b$. But then if $f(x)$ is defined on one point it's not differentiable because by definition derivative is the rate of change between 2 points on an interval.

By conventional notation we write the left-most coordinate of an interval adjacent to the left bracket and the right-most coordinate of the interval adjacent to the right bracket which in such case would mean that $a$ is before $b$.

So can we claim that $a<b$?

Or maybe if given $(a,b)$ the convention pre-supposes that $a \neq b$?

In my opinion (and these three words imply that this is just some very subjective thinking) it depends on how you define differentiability. If your definition of differentiability requires $f$ to be defined on an open interval (an interval of the form $(a,b)$ with $a<b$), then you are right, because then if $f$ is defined only on a set $[a,a]=\{a\}$ with one point, talking about differentiability of $f$ doesn't make any sense (because for this case it is not defined, and - as you have pointed out already - cannot tell anything about the rate of change, cause $f$ cannot change anyhow since there is only one value at which $f$ is defined).

However, it is sometimes convenient to define differentiability for those functions by just defining them to be differentiable (From a logical point of view this also makes sense and guides you sooner or later to the question how statements about the empty set are handled, see the comments above). This is, at first glance, pure convention. That way, one can replace a statement like "Let $f:[a,b]\to\mathbb R$ $\color{red}{\text{for }a<b}$ be continuous and differentiable in $(a,b)$ such that ..." by something like "Let $f:[a,b]\to\mathbb R$ be continuous and differentiable in $(a,b)$ such that ...". The second statement is easier to read because there are no redundant information. But as you can see, it is only a matter of taste.

Let's take a closer look at the definition of differentiability. What does
\begin{align*}
\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\quad\text{exists}
\end{align*}
really mean? It means that for each sequence of numbers $(x_n)$ contained in the domain of definition of $f$ and so that $x_n\neq x_0$ for each $n$, the limit
\begin{align*}
\lim_{n\to\infty}\frac{f(x_n)-f(x_0)}{x_n-x_0}
\end{align*}
exists and is the same, no matter how the sequence $(x_n)$ actually looks like. Thus, $f$ is differentiable at $x_0$, if this works for each such sequence. But if there is no such sequence, then there is nothing to check, there is nothing that can go wrong, so, well, then we say, $f$ is indeed differentiable. But, again, this is our choice. We also could say, that $f$ is not differentiable in this case, and we are again talking about empty sets and how to treat them.

From a porgrammers point of view: Imagine that you would like to write a program that checks whether all elements of a set of objects (a list) $L$ satisfy a given property $P$, say. Thus, your function yields true, if all elements of $L$ satisfy $P$, otherwise false. You probably know that functions (depends on the language) have to return something, in any case! So we would probably write our function like this:

bool function check(List L, Property P) {
   for each (x in L) {
       if (!P(x))
           return false;
   }
   return true;
}

In your case, $P(x)=f$ is differentiable at $x$, and $L=(a,b)$. In case $L=\emptyset$, check(L, P) returns true. Try to rewrite this function so that it returns false if $L=\emptyset$. You would need a couple more lines. In summary, it is just handy to say that any property for elements of the empty set is true.
Again (see comments) from a logical point of view, this is the only reasonable way to go. If the statement "Each element of the empty set is red" was false, then its negation "There is one element of the empty set which is not red" had to be true, which is clearly impossible.

To avoid all those discussions, my former Analysis Prof. just defined intervals to contain at least two elements. But again, just definition/convention. To cite my former Algebra Prof.: "When it comes to mathematical definitions there is no true or false, just convenient or inconvenient."

Note that if $a=b$ then $(a,b)$ is an empty set thus "f has a derivative on $(a,b)$" is a true statement.

Thus, by convention of noting lower bound first we can only claim $a \leq b$.

EDIT:
About the derivative on empty set.
In logic, a statement is either true or its logical negation is true (which is the same as saying that the statement is false).

Now, the logical negation of "f has a derivative $\forall x \in \emptyset$" is "for atleast one $x \in \emptyset$ $f(x)$ has no derivative".

Since there are no points in the empty set, one can't find a point for which $f(x)$ has no derivative.  Thus logical negation is false, making our statement true.

Conclusion: "f has a derivative $\forall x \in \emptyset$".


Find the maximum of $f(x) =\int_0^x\sqrt{(x^2-x)^2+y^4}\, dy, x\in[0,1].$ Find the maximum of $$f(x) =\int_0^x\sqrt{(x^2-x)^2+y^4}\, dy, x\in[0,1].$$

I am generally familiar with the process of how to find the maximum of a function on a given interval, however I really don't even know where to start with this one, any hints would be greatly appreciated!

Mathematica 12.0 answers

Maximize[{Integrate[Sqrt[(x^2 - x)^2 + y^4], {y, 0, x}],  x >= 0 && x <= 1}, x]

$$\left\{\frac{\Gamma \left(\frac{3}{4}\right)}{4 \Gamma \left(\frac{7}{4}\right)},\{x\to 1\}\right\} $$

First, we have to get the derivarive of $ f $.

So, let’s use the Fundamental Theorem of Calculus in order to find it:

$$f’(x) = \sqrt{(x^2 - x)^2 + x^4} \cdot 1
= |x| \sqrt{2x^2-2x+1}$$

Since the interval is positive, it’s just $x $ times that square root.

Therefore, we must avaluate at $f’=0$ or where the function “does not exist”. However, the function is well behaved at the whole real line, since the roots are complex conjugates.
$$f’(x) = x \sqrt{2x^2 -2x +1} = 0.$$
Therefore, $ x = 0, f(0)=0$.

Now, do not forget the boundaries of the interval: 0, 1. Since, we have already gotten the result above:

$f(1) =$ something that I will finish further, I’m kinda busy right now. But that’s the idea.


Maximum distance between a line segment and a polygon Given a (non-convex) Polygon $P$ and a straight line segment $S$ in $\mathbb{R}^2$, where $S$  and $P$ are disjoint, I am looking for an efficient algorithm to find a point $s$ on $S$ which has the maximum distance from $P$ among all points on $S$ (and the actual distance, which is then trivial, of course).

To be clear, the distance $dist(s,P)$ between one point $s$ and $P$ is the minimum distance between $s$ and all points in $P$ (which I can calculate by a standard point-to-polygon algorithm). So another way to write this formally is:

"Determine $max(min(|p-s| : p \in P) : s \in S)$ (and the related points)".

Note the point of maximum distance can be somewhere in the middle of $S$:

I have checked some standard resources (including this site) and books from computational geometry, but had no luck so far. If $S$ is described as a parametric curve $c(t)$ where $t \in [0,1]$, this is the problem of finding the global maximum of $dist(c(t),P)$ in $[0,1]$, so it would even help me if someone has an idea how to split $[0,1]$ into subintervals where each subinterval contains only a local maximum. That would allow me to apply a standard numerical maximum search algorithm to each of the subintervals.

Any ideas?

Hint:

If I am right, the Euclidean distance map to a polygon is made of planar and conic patches (along edges and around vertices respectively). Their construction is analogue to that of the Voronoi diagram of the polygon. (But strangely, the case of the Voronoi diagram on the outside of a filled polygon seems to never have been addressed.)

The case of a convex polygon is easy. That of a concave one, much less.

If you project the line segment S onto this map, you will traverse line segments and hyperbolas. As the concavity is downward, the searched maximum will occur either at the intersection of patches or at the segment endpoints.

Could be solved with a version of Fortune's sweepline algorithm, adapted to polygons, if that exists...

Edit: Since the poster clarified himself, the below is irrelevant to his new formulation of the problem. I'll leave it since it solves the old formulation: $\max_{s \in S}{\max_{p \in P}{d(s,p)}}$.

Measure all distances between the polygon's vertices and the line segment's endpoints, pick the maximum. The algorithm works in $O(V)$, there cannot be a faster one since you need to at least iterate over the vertices to get the correct answer.


36 Teams split into 4 groups of 9. There are 9 events and 9 rounds. A teams must face all other teams in the other groups. Okay here is my problem. I am not sure if this is possible or not as the 8 hours i have spent on this havn't been the best.

There are 4 groups of 9 teams. So we have teams A1 to A9, B1 to B9, C1 to C9, D1 to D9.
The are 9 rounds of 9 different events. Each event can hold 1 team from each group. How do i make it so that each team faces every other team that isn't in their group exactly once and no team twice. And every team goes to each event once and no event twice.

Some background, this is for a school event where our school has 4 different houses. I have been put in charge of setting together the timetable for the day and the way the teams mvoe around the events. There is no way i am able to just have 2 houses face each other at an event as 4 teams from different houses has to compete at each event. Principle wants 9 teams for each house to compete.

Is this possible? If this isnt possible, would it be possible if the number of teams went down to 8 with the number of events go down to 8 and rounds go down to 8 also?

Thanks

Initially, there are $9^4$ to organize an event. But every time you do it, you're eliminating not only one possibility out of the $9^4$. Indeed, if I picked up $\{A_1,B_1,C_1,D_1\}$ for the first event, I could not pick later any order in which $\{A_1, B_1\}$ appeared, which is $9^2-1$. On top of that, there are $\binom 4 2=6$ ways of choosing a pair out of the first event, so the previous number gets multiplied by $6$.

So you organized your first event and now, out of the $9^4$ options, there are only $9^4-6\times(9^2-1)$ left. For how many events can you do that? Simply calculate $\frac {9^4}{6\times(9^2-1)}$ and take the integer part. This actually already answers your second question.


The derivative of $\sin(\alpha x)/ \sin(x)$ is continuous at $0$? Let $\alpha$ be a real number. Define $f = \sin(\alpha x)/ \sin(x)$ if $x \neq 0$ and $\alpha$ at $x = 0.$ How can I prove the derivative of $\sin(\alpha x)/ \sin(x)$ is continuous at $0$? I took the derivative and tried to apply L'Hopital's rule but I kept getting $0/0$ and could not make the argument work... Any explanation is appreciated.

Both $\operatorname{sinc}(x)$ and its reciprocal are analytic, non-vanishing functions in a neigbourhood of zero, so it is
$$ \alpha\operatorname{sinc}(\alpha x)\frac{1}{\operatorname{sinc}(x)}. $$

An elementary proof without l'Hôpital nor Landau's big O notation:
for all $x\neq 0$,
\begin{align*}
f'(x)&=\frac{\alpha\cos(\alpha x)\sin(x)-\sin(\alpha x)\cos(x)}{\sin^2(x)}\\
%
&=\left(\frac{(x-0)}{\sin(x)-\sin(0)}\right)^2\frac{\alpha\cos(\alpha x)\sin(x)-\sin(\alpha x)\cos(x)}{(x-0)^2}.
\end{align*}
By continuity of the square and inverse functions, we have:
\begin{align*}
\lim_{x\to0}\left(\frac{(x-0)}{\sin(x)-\sin(0)}\right)^2&=\left(\frac{1}{\lim_{x\to0}\frac{\sin(x)-\sin(0)}{(x-0)}}\right)^2=\left(\frac{1}{\cos(0)}\right)^2=1.
\end{align*}
Let us write:
\begin{align*}
\alpha\cos(\alpha x)\sin(x)-\sin(\alpha x)\cos(x)&=\alpha\big(\cos(\alpha x)-\cos(0)\big)\sin(x)+\alpha\cos(0)\sin(x)\\
&\quad-\sin(\alpha x)\big(\cos(x)-\cos(0)\big)-\sin(\alpha x)\cos(0)
\end{align*}
Notice first that
\begin{align*}
\lim_{x\to0}\frac{\alpha\big(\cos(\alpha x)-\cos(0)\big)\sin(x)}{(x-0)^2}&=\alpha\lim_{x\to0}\frac{\cos(\alpha x)-\cos(0)}{x-0}\frac{\sin(x)-\sin(0)}{x-0}=\alpha(-\alpha\sin(0))\cos(0)=0,\\
%
\lim_{x\to0}\frac{\sin(\alpha x)\big(\cos(x)-\cos(0)\big)}{(x-0)^2}&=\lim_{x\to0}\frac{\sin(\alpha x)-\sin(0)}{x-0}\frac{\cos(x)-\cos(0)}{x-0}=\alpha\cos(0)(-\sin(0))=0
\end{align*}
by definition of the derivative at 0. Next,
\begin{align*}
\lim_{x\to0}\frac{\alpha\cos(0)\sin(x)-\sin(\alpha x)\cos(0)}{(x-0)^2}&=\cos(0)\lim_{x\to0}\frac{\alpha\sin(x)-\sin(\alpha x)}{(x-0)^2}\\
%
&=\cos(0)\lim_{x\to0}\frac{\left[\frac{\alpha\big(\sin(x)-\sin(0)\big)}{x-0}-\alpha\cos(0)\right]-\left[\frac{\sin(\alpha x)-\sin(0)}{x-0}-\alpha\cos(0)\right]}{x-0}\\
%
&=\cos(0)\big(-\alpha\sin(0)+\alpha\sin(0)\big)\\
%
&=0
\end{align*}
by definition of the second derivative at 0. As the limit of products is the products of the limits (provided that all limits exist in $\mathbb{R}$), we therefore have established the following:
\begin{align*}
\lim_{x\to0}f'(x)&=0.
\end{align*}
On the other hand,
\begin{align*}
\lim_{x\to0}\frac{f(x)-f(0)}{x-0}&=\lim_{x\to0}\frac{\frac{\sin(\alpha x)}{\sin(x)}-\alpha}{x-0}\\
%
&=\lim_{x\to0}\frac{x-0}{\sin(x)}\frac{\sin(\alpha x)-\alpha\sin(x)}{(x-0)^2}\\
%
&=\lim_{x\to0}\frac{x-0}{\sin(x)-\sin(0)}\frac{\sin(\alpha x)-\alpha \sin(x)}{(x-0)^2}
\end{align*}
is proved to be 0 as well using the above computations. This shows that $f'(0)=0$ and thus $f'$ is continuous at $x=0$.
Remark. The above proof works as well for any function $f(x)=\frac{g(\alpha x)}{g(x)}$ where $g$ is smooth and satisfies $g(0)=0$ and $g'(0)=1$.


random shortest grid walk Let $W$ be a random shortest grid walk from $(0,0) \to (20,20)$. Compute exactly

a) the expected area above or to the left of $W$,

b) the expected number of turns of $W$.

Please help, give some hints!

Hint for B: you might try to look the star and bars theorem. If you write the paths like $x^{p_1}y^{p_2}\ldots z^{p_{i+1}}$(x is to go up and y to go right) where z is x or y depending in the parity of i, you would have i changes of direction. So you are interested in how many ways you can write 20 as sum of $p_1+p_3+\ldots+p_{2k+1}+\ldots p_{t}$ and $p_2+p_4+\ldots+p_{2k}+\ldots p_{t'}$ where t and t' ara i or i+1 depending in the parity of i such that $p_j\geq 1$ for every j.

You are in Pak's course right?

Consider 11 positions between Rs, each Up take place will have 0, 1, 2...10 added area.


If a prime $p\mid ab$, then $p\mid a$ or $p\mid b\ $ [Euclid's Lemma] If a prime number $p$ is a divisor of a product $ab$, $p$ has to be a divisor of $b$ or $a$. How can I demonstrate this theorem? I demonstrated this theorem on one way using Bezout's theorem in an answer below but are there other proofs??

The set $\,S\,$ of naturals $\,n\,$ such that $\,\color{#c00}{p\mid nb}\,$ is $\rm\overbrace{closed\ under\ subtraction}^{\overbrace{\large p\mid nb,kb\,\Rightarrow\, p\mid nb-kb\, =\, (n-k)b}^{\LARGE n,\,k\ \in\ S\ \ \ \Longrightarrow\ \ \ n-k\ \in\ S\ \ }}$ and $\, a,p\in S\,$ therefore by the Lemma its least positive element $\,\color{#0a0}{d\mid a,p}.\,$ Since $\,\color{#a0f}{d\mid p\ \ \rm prime},\,$ either $\,\color{#a0f}{d=p}\,$ (so $\ \color{#a0f}{p = }\color{#0a0}{d\mid a}),\,$ or $\,\color{#a0f}{d=1}\in S\ $ (so $\ \color{#c00}{p\mid d b = b}),\ $ i.e. $\,\ p\mid \color{}a\,$ or $\ p\mid \color{}b.\ \ $ QED

Lemma $\ \ $ Let $\,\rm S\ne\emptyset \,$ be a set of integers $>0$ closed under subtraction $> 0,\,$ i.e. for all $\rm\,n,m\in S, \,$ $\rm\ n > m\ \Rightarrow\  n-m\, \in\, S.\,$  Then the least $\rm\:\ell\in S\,$ divides every element of $\,\rm S.$
Proof ${\bf\ 1}\,\  $ If not there is a least nonmultiple $\rm\,n\in S,\,$ contra  $\rm\,n-\ell \in S\,$ is a nonmultiple of $\rm\,\ell.$
Proof ${\bf\ 2}\,\rm\,\ \   S\,$ closed under subtraction $\rm\,\Rightarrow\,S\,$ closed under remainder (mod), when it is $\ne 0,$ since mod is simply repeated subtraction, i.e. $\rm\ a\ mod\ b\, =\, a - k b\, =\, a\!-\!b\!-\!b\!-\cdots\! -\!b.\,$ Thus $\rm\,n\in S\,$ $\Rightarrow$ $\rm\, (n\ mod\ \ell) = 0,\,$ else it is in $\,\rm S\,$ and smaller than $\rm\,\ell,\,$ contra minimality of $\rm\,\ell.$

Remark $\ $ In a nutshell, two applications of induction yield the following inferences
$\begin{eqnarray}\rm S\ closed\ under\ {\bf subtraction} &\Rightarrow\:&\rm S\ closed\ under\ {\bf mod} = remainder = repeated\ subtraction \\
&\Rightarrow\:&\rm S\ closed\ under\ {\bf gcd} = repeated\ mod\ (Euclid's\ algorithm) \end{eqnarray}$
Interpreted constructively, this yields the extended Euclidean algorithm for the gcd.
If fractions are known then $\,S\,$ can be interpreted as the set of all possible denominators of $\,b/p,\,$ since $\,ap = nb \iff a/n = b/p.\,$ See various posts on denominators ideals for more.
The Lemma describes a fundamental property of integer arithmetic whose essence will become clearer when one studies ideals of rings, viz. a Euclidean domain  is a PID, i.e. in any domain $D$ enjoying a Euclidean algorithm for division with remainder, all ideals $I$ are principal, i.e. $\, I = dD\,$ where $d$ is the least nonzero element of the ideal (= gcd of all elements). Here a subset $\,I\subset D$ is an ideal if is closed under subtraction and scaling by elements of $D$. The proof is exactly as in the 2nd proof above - by Euclidean descent using remainder (mod) via the division algorithm, e.g. see here and here and here.
See here for a few alternative proofs.
See here for a precise comparison of various forms of Euclid's Lemma
Variations of the above Theorem yield analogous "direct proofs" of the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations of integers), e.g. see this answer. Various forms of such direct proofs were given by Klappauf, Lindemann, and Zermelo.

$ab$ contains p as it's prime factor. That's what $p|ab $ means. Since $p$ is prime, it can't be expressed as product of two numbers except $1$ and $p$.  So either $a $ should contain it or $ b$.


Probability of $x$ non-zero multinomial coefficients? Let $\prod_{k=1}^K p_k^{c_k}$ be the joint probability distribution of some $n$-long memoryless random process (thus, $\sum_{k=1}^Kp_k=1$), where each $c_k$ tells how many times the $k$-th element appeared. Thus, the vector $c=(c_1,...,c_K)$ follows the multinomial distribution with $\sum_{k=1}^Kc_k=n$.

What is the probability that $x\in \mathbb{N}$ entries in the vector $c$ are non-zero?

Or analogously, what is the probability that $y$-entries are equal to zero?

Happened to come across this. Think I solved that for this article (for a uniform distribution of the p_k):
https://www.tandfonline.com/doi/abs/10.1080/00031305.2018.1444673?src=recsys&journalCode=utas20


How do I prove that if x is odd then x^16 is odd? I know how to show $x^2$ is odd but unsure about x^16

Also assuming an odd number = 2k+1

Just apply your proof four times.
For example $x^2$ is odd gives $(x^2)^2=x^4$ is odd... and so on and so forth

If $x=2y+1$ then
$$
x^n=\sum_{i=0}^n \binom{n}{i}2^iy^i=2\left(\sum_{i=1}^n \binom{n}{i}2^{i-1}y^i\right)+1,
$$
which is odd.


Existence of upper triangular matrix proof I am reading Linear ALgebra Done Right and Thm $5.27$ says:

  Suppose $\textsf V$ is a finite-dimensional complex vector space.
  Then a square matrix $\textsf T$ has an upper-triangular matrix with respect to some basis of $\textsf V$.

The proof given is based on induction and I am not getting much of a feel for it. Is there another proof of this theorem, I cant find any? Hopefully more intuitive.

My understanding so far is that suppose $\textsf T$,of $\dim (\textsf{V})$, has null space as $\{0\}$. Then I can select any linearly independent set of vectors and make that my basis. So I choose vectors that satisfy the upper triangular matrix template such as $(x,0,0,0,0,.....),(y,z,0,0,0....),\dots$

Any vectors of this form are linearly independent and thus I select this as my basis.(This would imply upper-triangular matrices are not unique).

But what about the case where null space has dimension greater than zero. How is upper-triangular matrix possible here?

Bonus Theorem. If $k$ is a field then the following are equivalent: (i) every square matrix over $k$ has an eigenvalue (in $k$) (ii) every square matrix is similar to an upper-triangular matrix (iii) the characteristic polynomial of every square matrix splits over $k$.

Note: As I suspected when I  wrote this, (i) is also equivalent to saying $k$ is algebraically closed. See here for that.

Anyway, (ii) implies (iii) is trivial, as is (iii) implies (i). For (i) implies (ii): Assume (i).

First note that if $B=(b_1,\dots,b_n)$ is a basis then

    Lemma 1. The matrix for $T$ wrt the basis  $B$ is upper triangular if and only if $Tb_k$ is a linear combination of $b_1,\dots,b_k$ for every $k$.

So it's enough to prove this:

    Lemma 2. There exists a non--zero  $v\in V$ such that if  $E$ is the span of $v$ then there is a subspace $W$ with $V=E\oplus W$ and $TW\subset W$.

Proof that Lemma 2 implies the result: By induction on the dimension there is a basis $b_1,\dots,b_{n-1}$ for $W$ such that  $Tb_k$ is a linear   combination of $b_1,\dots,b_k$ for $1\le k\le n-1$. Let $b_n=v$.

Edit: This is seeming easier. At first I was wondering how to get $v$ as in Lemma 2. But no, we just concentrate on $W$ and then $v$ comes along for free. It's clear that Lemma 2 follows from Lemma 3:

    Lemma 3. There is a subspace $W\subset V$ of codimension $1$ such that $TW\subset W$.

(Given $W$ as in Lemma 3, let $v$ be any non-zero element of $V\setminus W$ and Lemma 2 follows.)

And Lemma 3 is finally something we can prove:

Let $\Lambda$ be an eigenvector of the adjoint  $T^*:V^*\to V^*$, and let $W$ be the kernel of $\Lambda$. Then $W$ certainly has codimension $1$, and if $x\in W$ then $$\Lambda(Tx)=(T^*\Lambda)x=\lambda\Lambda x=0;$$hence $TW\subset W$.

I find a theorem in my textbook. Hopefully this is a little bit more intuitive.
Theorem : Each complex matrix A is similar to a Jordan matrix.
proof: Suppose ${\left (\lambda -\lambda_{1} \right )}^{k_{1}},{\left (\lambda -\lambda_{2} \right )}^{k_{2}},\cdots ,{\left (\lambda -\lambda_{s} \right )}^{k_{s}}$ are the primary elementary of matrix A. Each primary elementary correspond to $J_{i}$
$$J_{i}=\begin{pmatrix} \lambda _i& 0& \cdots & 0 & 0\\ 1& \lambda _i & \cdots & 0 & 0\\ 0 & 1& \cdots & 0 &0 \\ \vdots &\vdots & & \vdots &\vdots \\ 0 & 0 & \cdots & 1& \lambda _i \end{pmatrix}_{k_{i}}$$
And then you get the jordan matrix J.
$$J=\begin{pmatrix} J_{1}& & & \\ & J_{2} & & \\ & & \ddots & \\ & & & J_{s} \end{pmatrix}$$
Because A and J have the same primary elementary, so A is similar to J ( A has an upper-triangular matrix with respect to some basis of V).


Does commutativity imply Associativity? Does commutativity imply associativity? I'm asking this because I was trying to think of structures that are commutative but non-associative but couldn't come up with any. Are there any such examples?

NOTE: I wasn't sure how to tag this so feel free to retag it.

Consider the operation $(x,y) \mapsto xy+1$ on the integers.

For our example the carrier set will be the set ${\Bbb Z}^{\Bbb Z}$ of all functions from $\Bbb Z$ to $\Bbb Z$.

For any two function $f, g \in {\Bbb Z}^{\Bbb Z}$ we can form the pointwise sum, $f + g: n \mapsto f(n) + g(n)$.

We have the (left) bilateral shift operator $T$ on ${\Bbb Z}^{\Bbb Z}$,

$\tag 1 T: (a_k)_{k=-\infty}^\infty \mapsto (a_{k+1})_{k=-\infty}^\infty$

We define a binary operation $\oplus$ on the set ${\Bbb Z}^{\Bbb Z}$ as follows,

$\tag 2 (f,g) \mapsto f \oplus g = T \circ (f+g)$

This binary operation is commutative but not associative; it satisfies

$\tag 3 f \oplus g =  h \oplus k \; \text{ iff } \; f + g = h + k$


Indefinite Integral of Modulo Function By defining the modulo function with the floor function:

$$ x \bmod y = x - y \left\lfloor \frac xy \right\rfloor $$

Is it possible to find the antiderivative of the Modulo function?
I attempted at integrating with the and approximate indefinite integral of the floor function, but completely got lost due to my limited knowledge in calculus.

Can someone explain to me how this will or will not work?

Between $0$ and $y$ it is easy enough to integrate; you get $\frac12x^2+C$.

Since the integrand is periodic, the integral in the other periods is just a translated version of the integral in the base period, with the constants of integration chosen such that neighboring periods fit together. The translation is excatly what the modulo operation itself does, so we'll get something like
$$ \frac12(x\bmod y)^2 + \frac{y^2}2\left\lfloor \frac xy\right\rfloor + C $$
Here $\frac{y^2}2$ is the integral of one period (and therefore the amount we need to raise the side-translated next period to fit), and $\lfloor x/y\rfloor$ counts how many periods to the right of the base period we are.

In order to integrate a function containing an expression of floor, you must understand that the floor function is defined within calculus to be an extension of the floor function. That is to say:

$$\int f(x)dx = F(x) + floor(b) + c$$

Where b is any expression of x. Now, from this definition I can determine your integral to be the following:

$$\int x - y(x)\lfloor \frac x{y(x)} \rfloor dx = \frac 12x^2 - Y(x)\lfloor \frac x{y(x)} \rfloor + \lfloor b \rfloor + c$$

I made the y variable a function of x. I did that to avoid a possible 3 dimensional integral and to avoid the potential for it being interpreted as a constant. Now, the integrals that will be useful (since these are a range), will be the one's that have $c=0$ and b having a value that produces a continuous graph. It is likely to be some manipulation of the $\lfloor
\frac x{y(x)}\rfloor$ function, and I will also say that the floor could be to a power or added upon itself. The point is that any functions of x have to be wrapped inside floor(), otherwise it is not an extension of the C-constant of integration.

I hope this helps you in solving your integral.


limit of $|n^t\sin n|$ It is known that $\{\sin n : n\in\mathbb{N}\}$ is dense in $[-1,1]$, hence $\lim_{n\to\infty}\sin n$ doesn't exist and also $\lim_{n\to\infty} n^t\sin n$ doesn't exist for all $t>0$ (the reason is that the density implies that inequalities $\sin n>\frac{1}{2}$ and $\sin n<-\frac{1}{2}$ are satisfied infinitely many times, so there are subsequences tending to $+\infty$ and $-\infty$).

What about $\lim_{n\to\infty} |n^t\sin n|$ ?

The above argument shows that the limit - if exists - is infinite

I dont't think it does converge, but I don't know how to prove it.

The question is strongly connected with irrationality measure of $\pi$. That is such number $\mu$, that for every numbers $\lambda, \nu$ with $\lambda < \mu < \nu$:

there exist infinitely many distinct rational numbers $p/q$ for which
$$
\left| \frac{p}{q} - \pi \right| < \frac{1}{q^\lambda} \;\;\Longleftrightarrow\;\; \lvert p - \pi q \rvert < q^{1-\lambda}
$$
for each rational $p/q$ with sufficiently large denominator
$$
\left| \frac{p}{q} - \pi \right| > \frac{1}{q^\nu} \;\;\Longleftrightarrow\;\; \lvert p - \pi q \rvert > q^{1-\nu}
$$

Exact value of $\mu$ is not known, but it's true that $2\leqslant \mu \leqslant 7.6063$. Note, that by Dirichlet's theorem, the first item is always true for $\lambda = 2$, disregarding the irrationality measure.

Returning to question: there is watershed for the parameter. If $t > \mu-1$, then the limit is $+\infty$; if $t < \mu-1$, then it doesn't exist.  The remaining case $t = \mu-1$ depends on the behavior of diophantine approximations of $\pi$.

First case: $t < \mu-1$.

If $t < \mu-1$ and $p/q$ satisfies the first inequality, then
$$
\lvert \sin p \rvert = \lvert \sin (p-\pi q) \rvert \leqslant \lvert p - \pi q \rvert \leqslant q^{-t} \sim \pi^t p^{-t} = O(p^{-t}) \;\text{ as }\; p\rightarrow \infty
$$
hence there is bounded subsequence of $\{ n^t \sin n \}_{n=1}^\infty$ and it cannot have infinite limit.

Second case: $t > \mu-1$.

Take $\varepsilon > 0$, such that $t-\varepsilon > \mu-1$. Given $n\in \mathbb N$, choose $m \in \mathbb N$, such that $\lvert n -\pi m \rvert \leqslant \frac{\pi}{2}$. When $n$ is sufficiently large,
$$
\lvert \sin n\rvert = \lvert \sin(n-\pi m) \rvert \geqslant \tfrac{2}{\pi} \lvert n - \pi m \rvert = \tfrac{2}{\pi} \lvert n - \pi m \rvert \geqslant \tfrac{2}{\pi} m^{-t+\varepsilon} \sim 2\pi^{t-\varepsilon-1} n^{-t+\varepsilon},
$$
hence $\lvert n^t \sin n \rvert \geqslant C n^\varepsilon \rightarrow +\infty$.

The remaining case: $t = \mu-1$.

There are two alternatives:
$(A)$ whether exist $C > 0$ and infinitely many rational solutions $p/q$ of
$$ \left| \frac{p}{q} - \pi \right| \leqslant \frac{C}{q^\mu} \;\;\Longleftrightarrow\;\; \lvert p - \pi q \rvert \leqslant C q^{1-\mu} = C q^{-t}$$
or $(B)$ the converse. As I already mentioned, if $\mu = 2$, then $(A)$ holds.

Suppose $(A)$ is true. Then by the same argument, as in the first case, $\nexists\lim\limits_{n\rightarrow\infty} \lvert n^t \sin n \rvert$.

Oppositely, we have $(B)$, which, in fact, is equivalent that for every sequence of distinct rationals $\{p_n / q_n \}_{n=1}^\infty$ the sequence
$\{ q_n^t \lvert p_n - \pi q_n \rvert \}_{n=1}^\infty$
is unbounded. Combining this with the second case solution yields $\lvert n^t \sin n \rvert \rightarrow +\infty$.

However, it's probably open problem which of $(A)$ or $(B)$ holds, along with the very value of $\mu$.

Consider the following two cases:

$t<0$
$t>0$

Note that $t=0$ reduces to $\lim_{n\rightarrow \infty}|\sin n|$.

For case 1, since $\lim_{n\rightarrow \infty}n^t=0$, and $\sin n$ is bounded by $-1\leq \sin n\leq 1$. It follows that $\lim_{n\rightarrow \infty}|n^t\sin n|=|0\cdot\lim_{n\rightarrow \infty}\sin n|=0$ (I'm abusing notation slightly here, technically you need to bound the absolute value with the 'worse case' (that being $\sin n=\pm 1$) and then use that bound to prove the limit).

For case 2, It's easy to find a subsequence that grows without bound, but I'm not sure how to get a bounded subsequence the converges to $0$ fast enough to prove that it doesn't diverge to infinity, but I strongly suspect that such a sequence exists. Edit: Nvm, see quartermind's answer (learn something new everyday).


Which of the following numbers can be orders of a permutation $\sigma$ of $11$ symbols Which of the following numbers can be orders of a permutation $\sigma$ of $11$ symbols, such that $\sigma$ does not fix any symbols?

$1. \;18$

$2.\; 30$

$3.\;15$

$4.\; 28$

could any one just give me hints?

Think about the orders of the permutations (1 2) (3 4 5) and (1 2) (3 4 5 6) and see if you can think of a rule that gives the order of a permutation written in disjoint cycle notation. Now if $\sigma$ fixes no points, what does it tell you about the permutation written in disjoint cycle notation? From this you should be able to tell which orders are possible.