Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $a,b>0$ and $a+b=2$ , prove that $a^{2b}+b^{2a}+(\frac{a-b}2)^2\leqslant2$ If $a,b>0$ and $a+b=2$ , prove that
$$
a^{2b}+b^{2a}+(\frac{a-b}2)^2\leqslant2
$$
The equality occurs if and only if $(a,b)\sim(1,1)$ or $(a,b)\sim(2,0)$ or its cyclic permutations.
My attempt:
By symmetry and the constraint, we may let $x=a-1=1-b$ , the inequality converts to
$$
(1+x)^{2(1-x)}+(1-x)^{2(1+x)}+x^2\leqslant2
$$
I tried to let $f(x):=(1+x)^{2(1-x)}+(1-x)^{2(1+x)}+x^2$ , but its derrivative is too complicated. I also tried to write $(1+x)^{1-x}$ as $e^{2(1-x)\ln(1+x)}$ and apply inequalities like $\ln x\leqslant x-1$ , $e^x\leqslant\dfrac{1}{1-x}$ and such, but it would be either too complicated or too crude.
How to solve it?
| Following your approach, WLOG, let $a=1-x, b=1+x$ for some non-negative real $x \in [0, 1)$. Then we have to show: $f = (1-x)^{2+2x}+(1+x)^{2-2x}+x^2 \leqslant 2$
Using the case of Bernoulli's inequality where exponent is in $(0,1)$, we have $(1-x)^{1+x} \leqslant (1-x)(1-x^2)$.
Again by Bernoulli's inequality, $(1+ x)^{1-x} \leqslant 1+ x(1-x) = (1+x-x^2)$
Hence $f \leqslant (1-x)^2(1-x^2)^2+(1+x-x^2)^2+x^2 = 2-x^2(1+x)(1-x)^3\leqslant 2$, with maximum when $x=0 \implies (a,b)=(1,1)$.
--
Of course, if $x=1$ is possible, then that is a maximum as well, but then one among $a, b \not >0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
What is the probability f(p) of A winning the match?
Here's my approach:
Let A denote the event "A wins the game" and B denote the event "B wins the game"
I am supposed to calculate the probability f(p)
There are two possible cases:-
i) A wins within 4 games
Possible cases: AAA,ABAA,BAAA,AABA
ii) A doesn't win within 4 games
Possible cases: AABB(A and B occur 2 times each) ABABABAB...(A and B occur k times each) AA
AABB can be arranged in 6 ways
ABABAB.... can be arranged in $(2!)^k$ ways (A and B are adjacent to one another)
Now :
f(p) = P(case i) + P(case ii)
P(case ii) = $(6 (p^2) (1-p)^2 ) \sum_{k=1}^∞ 2^k . p^k . (1-p)^k. p^2 $=$(6 (p^4) (1-p)^2 ) \sum_{k=1}^∞ 2^k . p^k . (1-p)^k$
= $(6. 2 . p^5 . (1-p)^3)/(1-2p+2p^2)$ (The denominator seems correct)
P(case i ) = $p^3 +3p^3(1-p)$
f(p) = $p^3 +3p^3(1-p) + \frac{(6. 2 . p^5 . (1-p)^3)}{(1-2p+2p^2)}$
The expected answer is not produced by solving the above equation.Where did I go wrong?
Edit:-
I finally figured out the mistake that I made in the above solution.
In P(case ii) = $(6 (p^4) (1-p)^2 ) \sum_{k=1}^∞ 2^k . p^k . (1-p)^k $, I replaced $\sum_{k=1}^∞$ with $\sum_{k=0}^∞$ . There has to be one case when B does not occur after the tie, i.e. A occurs only two times.
P(case ii) = $(6 (p^4) (1-p)^2 ) \sum_{k=0}^∞ 2^k . p^k . (1-p)^k $
= $(6 (p^4) (1-p)^2 ) .[1+2p(1-p)^1+2^2.p^2.(1-p)^2+...]$
=$(6 (p^4) (1-p)^2 ) .\frac{1}{1-2p(1-p)}$
f(p) = P(case i) + P(case ii)
=$p^3 +3p^3(1-p)+(6 (p^4) (1-p)^2 ) .\frac{1}{1-2p(1-p)}$
=$p^3[1+3(1-p)+\frac{6p(1-p)^2}{1-2p(1-p)}]$
=$p^3[1+(3-3p)+\frac{6p-12p^2+6p^3}{1-2p+2p^2)}]$
=$p^3[\frac{(1-2p+2p^2)+(3-6p+6p^2-3p+6p^2-6p^3)+(6p-12p^2+6p^3)}{1-2p+2p^2)}]$
=$p^3\frac{(4-5p+2p^2)}{1-2p+2p^2}$
| Let $q = 1 - p.$
Suppose that the match is tied 2-2 after 4 games.
Let $R$ denote the probability of A's winning the match, in that event.
Then
$$R = p^2 + 2pqR \implies R(1 - 2pq) = p^2 \implies R = \frac{p^2}{1 - 2pq}.$$
After $4$ games either A has won all $4$, A has won $3$ out of $4$ (where it is harmless to play the 4th game if A won the first three games), the match is split 2-2, or A has lost.
Therefore, A's probability of winning is
$$p^4 + 4p^3q + 6p^2q^2R = p^4 + 4p^3q + \left[6p^2q^2 \times \frac{p^2}{1 - 2pq}\right]$$
$$= p^3 \times \left\{ ~ p + 4q + \frac{6pq^2}{1 - 2pq}
~\right\}$$
$$=
p^3 \times \left\{ ~\frac{[ ~(1 - 2pq)(p+4q) ~] + 6pq^2}{1 - 2pq}
~\right\}. \tag1 $$
Then:
*
*$1 - 2pq = 1 - 2p(1 - p) = 1 - 2p + 2p^2.$
*$p + 4q = p + 4(1 - p) = 4 - 3p.$
*$6pq^2 = 6p(1 - 2p + p^2) = 6p - 12p^2 + 6p^3.$
The given answer is
$$\frac{p^3(4 - 5p + 2p^2)}{1 - 2p + 2p^2}. \tag2 $$
Comparing (1) and (2), the problem reduces to showing that
$$[(1 - 2pq)(p + 4q)] + (6p - 12p^2 + 6p^3) = (4 - 5p + 2p^2). \tag3 $$
$$(1 - 2p + 2p^2)(4 - 3p) = 4 - 11p + 14p^2 - 6p^3.$$
Therefore, the equality in (3) above is established.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
An urn has 2 white, 3 red and 5 black balls. Problem An urn has 2 white, 3 red and 5 black balls. 3 balls are randomly drawn, one at a time and without replacement.
Calculate the probability of extracting the sequence of colors (white, black, red) knowing that you have extracted a black ball.
solution
Let E be the event that refers to the indicated color sequence. The required probability is $P(E|X= 1)$ from which it follows $P(E|X= 1) = \frac{P(E⋂(X=1))}{P(X=1)} = \frac{P(E)}{P(X=1)} \overset{(question1)}{=} \frac{\frac{2}{10}\frac{5}{9}\frac{3}{8}}{\frac{5}{12}} = \frac{\frac{1}{24}}{\frac{5}{12}} = \frac{1}{10}$
question 1
If the sequence is (white, black, red) in the numerator we should not have: $\frac{3}{10}\frac{5}{9}\frac{2}{8}$ in the end the result is the same but surely the reasoning to obtain it is different.What can be the reasoning done in the solution?
And in the denominator , why $\frac{5}{12}$. $P(X=1)$ indicates the probability that a black ball has been extracted. This shouldn't be $\frac{5}{10}$
| Start by computing the probability of drawing a black ball. The probability that no black ball is drawn is $\frac{1}{2} \frac{4}{9} \frac{3}{8} = \frac{1}{12}$. Therefore, the probability that a black ball is drawn is $1 - \frac{1}{12} = \frac{11}{12}$.
Now, compute the probability of observing the sequence (white, black, red). This is straightforward: $\frac{1}{5} \frac{5}{9} \frac{3}{8} = \frac{1}{24}$.
The desired conditional probability is thus $\frac{\frac{1}{24}}{\frac{11}{12}} = \boxed{\frac{1}{22}}$.
Edit: Suppose that instead of "a black ball", we are interested in the condition that "exactly one black ball" is drawn. For this, consider that the number of permutations of three selected balls (regardless of color, and assuming they are distinguishable) is $(10) (9) (8) = 720$. To extract exactly one black ball, we must select it among $5$ choices, and choose one position out of $3$ possible. Then, for the first unselected position, we choose one of the $5$ non-black balls, and for the last position, we choose one of the $4$ remaining non-black balls. The number of ways is $(5) (3) (5) (4) = 300$. The probability of extracting exactly one black ball is thus $\frac{300}{720} = \frac{5}{12}$. And the desired conditional probability would be $\frac{\frac{1}{24}}{\frac{5}{24}} = \boxed{\frac{1}{10}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4630946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}$ is discontinuous at $0$?
How to prove that $\sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}$ is discontinuous at $0$?
As we know, $\sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}$ converges uniformly on $[\delta,\pi-\delta]$ for $\delta>0$. and not uniformly on $(0,\pi)$.
But this could not be used to show discontinuity of $\sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}=S(x)$ at $0$. Any ideas?
| To establish the discontinuity of $S(x)$ at $x = 0$, it suffices to show that $S(\pi/N) \to \infty$ as $N \to \infty$. Indeed, by grouping the terms into $N$ consecutive terms, the defining sum for $S(\pi/N)$ is recast as
\begin{align*}
S(\pi/N)
&= \sum_{r=1}^{N} \sin\left(\frac{r\pi}{N}\right) \sum_{q=0}^{\infty} \frac{(-1)^q}{\sqrt{qN + r}}.
\end{align*}
Then by the property of the alternating series, we have
$$ \frac{1}{\sqrt{r}} - \frac{1}{\sqrt{N + r}} \leq \sum_{q=0}^{\infty} \frac{(-1)^q}{\sqrt{qN + r}} \leq \frac{1}{\sqrt{r}}. $$
Together with the fact that $\sin(r\pi/N) \geq 0$ for all $ r = 1, 2, \ldots, N$, we obtain the bound
\begin{align*}
S(\pi/N)
&\geq \sum_{r=1}^{N} \sin\left(\frac{r\pi}{N}\right) \left( \frac{1}{\sqrt{r}} - \frac{1}{\sqrt{N + r}} \right) \\
&= \sqrt{N} \cdot\sum_{r=1}^{N} \sin\left(\frac{r\pi}{N}\right) \left( \frac{1}{\sqrt{r/N}} - \frac{1}{\sqrt{1 + r/N}} \right) \frac{1}{N}.
\end{align*}
Now the sum in the last line, excluding the prefactor $\sqrt{N}$, is the Riemann sum for the integral
$$ C := \int_{0}^{1} \sin(\pi x) \left( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{1+x}} \right) \, \mathrm{d}x > 0, $$
and so, it follows that
$$ S(\pi/N) \geq (C + o(1)) \sqrt{N}. $$
This shows that $S(\pi/N) \to \infty$ as $N \to \infty$, hence $S(x)$ is discontinuous at $x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $\arcsin$ of dual number: $\arcsin\left( x+y\cdot\varepsilon\right)$ where $x,y\in\mathbb{R}$ and $\varepsilon\ne0=\varepsilon^{2}$? How to calculate $\arcsin$ of dual number:
$\arcsin\left( x + y \cdot \varepsilon \right) = \dots$
where
$\begin{align*}
a, ~b &\in \mathbb{R}\\
\varepsilon^{2} &= 0\\
\varepsilon &\ne 0\\
\end{align*}$?
I recently saw the question on SE where someone asked how to get $\sin$ from a dual number, so I had the idea of reversing it, i.e. getting the $\arcsin$ from a dual number , to investigate.
For this I wanted to derive it myself, where several ideas would come to me on how to do it (via the logarithm relationship, the series expansion and calculating backwards).
With the individual methods, however, I always came up with other solutions.
So I'm wondering what's correct, what I'm doing wrong and if there is a better method for this?
I gave up on the logarithm method, since I would have to use the product of two imaginary units there, which, however, are not commutative with regard to multiplication, which would severely limit the validity of the formulas.
My calculating backwards attempt
I'm using the formula from this SE answer:
$$
\begin{align*}
\sin\left( u + v \cdot \varepsilon \right) &= \sin\left( u \right) + v \cdot \cos\left( u \right) \cdot \varepsilon\\
\sin\left( u + v \cdot \varepsilon \right) &= x + y \cdot \varepsilon\\
\\
x = \sin\left( u \right) &\wedge y = v \cdot \cos\left( u \right)\\
\arcsin\left( x \right) = u &\wedge y = v \cdot \cos\left( \arcsin\left( x \right) \right)\\
\arcsin\left( x \right) = u &\wedge y = v \cdot \cos\left( \arcsin\left( x \right) \right)\\
\arcsin\left( x \right) = u &\wedge y = v \cdot \sqrt{1 - x^{2}}\\
\arcsin\left( x \right) = u &\wedge \frac{y}{\sqrt{1 - x^{2}}} = v\\
\\
\sin\left( \arcsin\left( x \right) + \frac{y}{\sqrt{1 - x^{2}}} \cdot \varepsilon \right) &= x + y \cdot \varepsilon\\
\arcsin\left( x + y \cdot \varepsilon \right) &= \arcsin\left( x \right) + \frac{y}{\sqrt{1 - x^{2}}} \cdot \varepsilon\\
\end{align*}
$$
And this can only be true for $\left| x \right| < 1$ which I find strange since I didn't expact that $\left| x \right| = 1$ gets lost as a possible argument.
My series expansion attempt
$$
\begin{align*}
\arcsin\left( z \right) &= z + \left( \frac{1}{2} \right) \cdot \frac{z^{3}}{3} + \left( \frac{1 \cdot 3}{2 \cdot 4} \right) \cdot \frac{z^{5}}{5} + \left( \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \right) \cdot \frac{z^{7}}{7} + \cdots\\
\arcsin\left( x + y \cdot \varepsilon \right) &= \left( x + y \cdot \varepsilon \right) + \left( \frac{1}{2} \right) \cdot \frac{\left( x + y \cdot \varepsilon \right)^{3}}{3} + \left( \frac{1 \cdot 3}{2 \cdot 4} \right) \cdot \frac{\left( x + y \cdot \varepsilon \right)^{5}}{5} + \left( \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \right) \cdot \frac{\left( x + y \cdot \varepsilon \right)^{7}}{7} + \cdots\\
\arcsin\left( x + y \cdot \varepsilon \right) &= x + y \cdot \varepsilon + \left( \frac{1}{2} \right) \cdot \frac{x^{3} + 3 \cdot x^{2} \cdot y \cdot \varepsilon}{3} + \left( \frac{1 \cdot 3}{2 \cdot 4} \right) \cdot \frac{x^{5} + 5 \cdot x^{4} \cdot y \cdot \varepsilon}{5} + \left( \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \right) \cdot \frac{x^{7} + 7 \cdot x^{6} \cdot y \cdot \varepsilon}{7} + \cdots\\
\arcsin\left( x + y \cdot \varepsilon \right) &= \arcsin\left( x \right) + \left( 1 + x^{2} + x^{4} + x^{6} + \cdots \right) \cdot y \cdot \varepsilon\\
\arcsin\left( x + y \cdot \varepsilon \right) &= \arcsin\left( x \right) + \left( \sum_{k = 0}^{\infty} x^{2 \cdot k} \right) \cdot y \cdot \varepsilon\\
\arcsin\left( x + y \cdot \varepsilon \right) &= \arcsin\left( x \right) + \frac{1}{1 - x^{2}} \cdot y \cdot \varepsilon, \text{ for } \left| x \right| < 1\\
\arcsin\left( x + y \cdot \varepsilon \right) &= \arcsin\left( x \right) + \frac{y}{1 - x^{2}} \cdot \varepsilon, \text{ for } \left| x \right| < 1\\
\end{align*}
$$
But obviously something is wrong:
$$ \arcsin\left( x \right) + \frac{y}{1 - x^{2}} \cdot \varepsilon \ne \arcsin\left( x \right) + \frac{y}{\sqrt{1 - x^{2}}} \cdot \varepsilon$$
| "Automatic differentiation" is the way to go. Namely, for any real-analytic function $f\colon \Bbb R \to \Bbb R$, its natural extension to $\Bbb R\oplus \Bbb R \varepsilon$ in fact satisfies $$f(x+y\varepsilon) = f(x) + yf'(x)\varepsilon,$$for all $x,y\in \Bbb R$. To see this, write $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6} x^3+\cdots$$and replace $x\mapsto x+y\varepsilon$, noting that $$(x+y\varepsilon)^n = x^n + nx^{n-1}y\varepsilon$$for all $x,y\in \Bbb R$, to obtain $$\begin{align} f(x+y\varepsilon) &= f(0) + f'(0)(x+y\varepsilon) + \frac{f''(0)}{2}(x^2+2xy\varepsilon) + \frac{f'''(0)}{6}(x^3+3x^2y\varepsilon) + \cdots \\ &= f(0) + f'(0) x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3+\cdots \\ &\qquad + y\varepsilon\left(f'(0) + f''(0)x + \frac{f'''(0)}{2}x^2+\cdots\right) \\ &= f(x) + yf'(x)\varepsilon.\end{align}$$
Taking $f = \arcsin$, we have that $$\arcsin(x+y\varepsilon) = \arcsin(x) + \frac{y\varepsilon}{\sqrt{1-x^2}}.$$Trying to do this calculation with $\arcsin$ instead of an abstract $f$ turns out to be a distraction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int_{0}^{\pi/2}\ln^3(\sin x)\,\mathrm dx$ There are closed forms for $\int_{0}^{\pi/2}\ln(\sin x)\,\mathrm dx\,$ and $\,\int_{0}^{\pi/2}\ln^2(\sin x)\,\mathrm dx\,$ but I can’t seem to find a closed form for $$\int_{0}^{\pi/2}\ln^3(\sin x)\,\mathrm dx\;.$$
How would I calculate it?
| Noting that $$
\int_0^{\frac{\pi}{2}} \ln ^3(\sin x) d x=\left.\frac{\partial^3}{\partial a^3} I(a)\right|_{a=0}
$$
where
$$
I(a)=\int_0^{\frac{\pi}{2}} \sin ^a x d x=\frac{1}{2}B\left (\frac{a+1}{2},\frac{1}{2} \right)$$
\begin{aligned}
&\quad \frac{\partial^3}{\partial x^3}(\mathrm{~B}(x, y)) \\&=\left[\left(\psi^{(0)}(x)-\psi^{(0)}(x+y)\right)^3+3\left(\psi^{(1)}(x)-\psi^{(1)}(x+y)\right) \left(\psi^{(0)}(x)-\psi^{(0)}(x+y)\right)\\ +\psi^{(2)}(x)-\psi^{(2)}(x+y)\right] \mathrm{B}(x, y)
\end{aligned}
Putting $x=\frac{a+1}{2} $ and $y=\frac{1}{2} $ at $a=0$ in the derivative yields
\begin{aligned}
& \quad \frac{\partial^3}{\partial a^3} B\left(\frac{a+1}{2}, \frac{1}{2}\right)\\&=\frac{\pi}{8}[(\left.\psi\left(\frac{1}{2}\right)-\psi(1)\right)^3+3\left(\psi^{\prime}\left(\frac{1}{2}\right)-\psi^{\prime}(1)\right)\left(\psi\left(\frac{1}{2}\right)-\psi(1)\right) \\
&\left. \quad +\psi^{(2)}\left(\frac{1}{2}\right)-\psi^2(1)\right] \\
&=\frac{\pi}{8}\left[(-\gamma-\ln 4+\gamma)^3+3\left(\frac{\pi^2}{2}-\frac{\pi^2}{6}\right)(-\gamma-\ln \psi+\gamma)\right. +(-14 \zeta(3)+2 \zeta(3))] \\
&= \frac{\pi}{8}\left(-\ln ^3 4-\pi^2 \ln 4-12 \zeta(3)\right)
\end{aligned}
Hence $$\boxed{I=-\frac{\pi}{16}\left(\ln ^3 4+\pi^2 \ln 4+12 \zeta(3) \right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral inequality $I_2(z) = \mathcal{O}\left(\frac{1}{(1+z)^2}\right)$ I want to prove the inequality of an integral like this:
$$I_2(z) = \int_1^{+\infty} \frac{\rho(t)}{(z+t)^2} d t = \mathcal{O}\left(\frac{1}{(1+z)^2}\right)$$ , when $z \to +\infty$
Below is my current attempt：
$$\begin{aligned}
I_2(z) & =\int_1^{+\infty} \frac{\rho(t)}{(z+t)^2} d t \\
& =\frac{R(t)}{(z+t)^2}\Big|_1 ^{+\infty}+2 \int_1^{+\infty} \frac{R(t)}{(z+t)^3} d t \\
&\leq \int_{0}^{+\infty} \frac{1}{|(z+1)+t|^3} d t
\end{aligned}$$
Where，
$$\begin{gathered}
\rho(t)=\{t\}-\frac{1}{2},|\rho(t)| \leq \frac{1}{2}, \\
R(t)=\int_1^t \rho(s) d s, \quad|R(t)| \leq \frac{1}{2} .
\end{gathered}$$
The last integral with absolute value I don't know how to handle it properly.
| For $z=x+iy$ with $x > 0$ is, as you already calculated,
$$
I_2(z) = \int_1^\infty \frac{2R(t)}{(z+t)^3} \, dt
$$
and $|2 R(t)|\le 1$. Let us first assume that $y \ne 0$. Then
$$
\begin{align}
|I_2(z)| &\le \int_1^\infty \frac{1}{(x+t)^2+y^2)^{3/2}} \, dt \\
&= \left[ \frac {x+t}{y^2 \sqrt{(x+t)^2+y^2}}\right]_{t=1}^{t=\infty} \\
&= \frac{1}{y^2} \left( 1 - \frac{x+1}{\sqrt{(x+1)^2+y^2}}\right) \\
&= \frac{1}{\sqrt{(x+1)^2+y^2} \cdot (\sqrt{(x+1)^2+y^2} + (x+1))} \\
&\le \frac{1}{(x+1)^2+y^2} \\
&= \frac{1}{|z+1|^2} \, .
\end{align}
$$
By continuity it follows that the same estimates holds also if $y=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find the $n$th power of a 3-by-3 circulant matrix Consider the matrix given as $$A=\begin{bmatrix}a_0 & a_2 & a_1\\ a_1 & a_0 & a_2\\ a_2 & a_1 & a_0\end{bmatrix}$$
Write down a formual for $A^n$ for $n\in\mathbb{N}$.
$$$$
My attempt: The first that comes to mind is to diagonalize it and hence find the formual for $A^n$, but that is very messy, so I tried to do something else it goes as:
bserve that $$A=\begin{bmatrix}a_0 & a_2 & a_1\\ a_1 & a_0 & a_2\\ a_2 & a_1 & a_0\end{bmatrix}=a_0\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}+a_1\begin{bmatrix}0 & 0 & 1\\1 & 0 & 0\\0 & 1 & 0\end{bmatrix}+a_2\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{bmatrix}$$
Let $U=\begin{bmatrix}0 & 0 & 1\\1 & 0 & 0\\0 & 1 & 0\end{bmatrix}$ then we will have that
$$\begin{matrix}U^2=\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{bmatrix} & \text{ and } & U^3=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\end{matrix}$$
Hence we have $A=a_0I+a_1U+a_2U^2 = a_0U^3+a_1U+a_2U^2=(a_0U^2+a_1I+a_2U)U$
This got me thinking that there might be an easy way to solve the above problem but, I was not able to make any further progress.
Please Help and thanks in advance.
| Circulant matrices are related to Fourier transform.
Consider the DFT matrix of length $N=3$
$$
\mathbf{W}
=
\begin{pmatrix}
1 & 1 & 1 \\
1 & \Omega & \Omega^2 \\
1 & \Omega^2 & \Omega^4
\end{pmatrix}
$$
where
$\Omega
= e^{-2\pi i/N}$.
Consider now the DFT of the signal
$$
\mathbf{y}=
\mathbf{W}
\begin{pmatrix}
a_0 \\ a_1 \\ a_2
\end{pmatrix}
$$
From here, you can observe that
$$
\mathbf{WA}=
\mathbf{W}
\begin{pmatrix}
a_0 & a_2 & a_1 \\ a_1 & a_0 & a_2 \\ a_2 & a_1 & a_0
\end{pmatrix}
=
\mathrm{Diag}(\mathbf{y})
\mathbf{W}
$$
and thus
$$
\mathbf{A}
=
\mathbf{W}^{-1}
\mathrm{Diag}(\mathbf{y})
\mathbf{W}
=
\frac{1}{N} \mathbf{W}^{H}
\mathbf{D}
\mathbf{W}
$$
with $\mathbf{D}=\mathrm{Diag}(\mathbf{y})$.
Finally the $k$th power of $\mathbf{A}$ is
$$
\mathbf{A}^k
=
\frac{1}{N}
\mathbf{W}^{H} \mathbf{D}^k \mathbf{W}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove that $\prod_{i=1}^n\left(\frac{1}{x_i^2}-1\right)\ge (n^2-1)^n$ Let $n$ be a positive integer and $x_1,...,x_n$ positive reals such that $x_1+...+x_n=1$. Prove that $$\prod_{i=1}^n\left(\frac{1}{x_i^2}-1\right)\ge (n^2-1)^n$$
Now notice that if we apply Jensen's Inequality to $$f(x)=\ln\left(\frac{1}{x_i^2}-1\right)$$ We would get the exact inequality...except that $f$ is not convex on $(0,1)$. But I do know this,
Claim: If for all $k\le n$ we have $0< a_k\le 1/2$. Then, $$\prod_{i=1}^n\left(\frac{1}{a_i}-1\right)\ge \left(\frac{n}{a_1+...+a_n}-1\right)^n$$
The proof is not too hard. Just consider $$g(x)=\ln\left(\frac{1}{x}-1\right)$$
$g$ is convex on $(0,1/2)$ and you can finish with Jensen. Now what if we let $a_i=x_i^2$? Well, we would get $$\prod_{i=1}^n\left(\frac{1}{x_i^2}-1\right)\ge \left(\frac{n}{x_1^2+...+x_n^2}-1\right)^n\ge (n^2-1)^n$$
Because $x_1+...+x_n=1$ implies, $$\sum_{i=1}^n x_i^2\ge \frac{1}{n}\left(\sum_{i=1}^n x_i\right)^2=\frac{1}{n}$$
And we're done...Or are we? The argument is only valid if $x_i^2\le 1/2$ for all $i$. Well if there exists $x_i^2>1/2$ Can we still get the inequality?
By the way there is a way to solve the inequality without any use of Jensen. But I want to know if we can solve it like that.
| WLOG, assume that $x_1 \le x_2 \le \cdots \le x_n$.
We split into two cases:
Case 1: $x_n < \frac{1}{\sqrt 3}$
Note that $x \mapsto \ln(\frac{1}{x^2} - 1)$ is convex on $(0, \frac{1}{\sqrt 3})$.
We have
$$\sum_{i=1}^n \ln\left(\frac{1}{x_i^2} - 1\right) \ge n\ln\left(\frac{1}{(\frac{x_1 + x_2 + \cdots + x_n}{n})^2} - 1\right) = n\ln(n^2 - 1).$$
Case 2: $x_n \ge \frac{1}{\sqrt 3}$
Let $y_1 = x_1, y_2 = x_2, \cdots,
y_{n-2} = x_{n-2}$ and $y_{n-1} = y_n = \frac{x_{n-1} + x_n}{2}$.
Then $0 < y_i < \frac{1}{2}$ for all $i$.
Note that $x \mapsto \ln(\frac{1}{x^2} - 1)$ is convex on $(0, \frac{1}{\sqrt 3})$.
We have
\begin{align*}
\sum_{i=1}^n \ln\left(\frac{1}{y_i^2} - 1\right) \ge n\ln\left(\frac{1}{(\frac{y_1 + y_2 + \cdots + y_n}{n})^2} - 1\right) = n\ln(n^2 - 1)
\end{align*}
which results in
$$\prod_{i=1}^{n-2}\left(\frac{1}{x_i^2}-1\right)
\cdot \left(\frac{1}{(\frac{x_{n-1} + x_n}{2})^2} - 1\right)^2\ge (n^2-1)^n$$
It remains to prove that
$$\left(\frac{1}{x_{n-1}^2}-1\right)\left(\frac{1}{x_n^2}-1\right) - \left(\frac{1}{(\frac{x_{n-1} + x_n}{2})^2} - 1\right)^2 \ge 0$$
which is true since
\begin{align*}
&\left(\frac{1}{a^2}-1\right)\left(\frac{1}{b^2}-1\right) - \left(\frac{1}{(\frac{a + b}{2})^2} - 1\right)^2 \\[6pt]
={}& \frac{a^2 + 6ab + b^2 - (a^2 + 4ab + b^2)(a + b)^2}{a^2b^2(a+b)^4}(a-b)^2\\
\ge{}& 0
\end{align*}
where $a = x_{n-1}, b = x_n$ (using $a + b < 1$).
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ How do I evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ ?
Note that this is a Q&A post and I've presented my solution below.
| Letting $x^2=\sec \theta$ transform the integral into
\begin{aligned}
I & =\int \frac{\sec \theta \tan \theta d \theta}{(1-\sec \theta) \tan \theta} \\
& =\int \frac{1}{\cos \theta-1} d \theta \\
& =\int \frac{\cos \theta+1}{-\sin ^2 \theta} d \theta \\
& =-\int \cot \theta \csc \theta d \theta-\int \csc ^2 \theta d \theta \\
& =\csc \theta+\cot \theta+C \\
& =\frac{x^2+1}{\sqrt{x^4-1}}+C
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4640836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 4,
"answer_id": 1
} |
$(a+1)x^{2} +(b+1)y^{2} +(c+1)z^{2} \geq 2(xy+yz+zx)$ Prove $(a+1)x^{2} +(b+1)y^{2} +(c+1)z^{2} \geq 2(xy+yz+zx)$ for $a,b,c>0$ and $abc\geq1$.
My try: $(a+1)x^2-4xy+(b+1)y^2+(a+1)x^2-4xz+(c+1)z^2+(b+1)y^2-4yz+(c+1)z^2\geq0$
Then I had to prove that $(a+1)x^2-4xy+(b+1)y^2\geq2x^2-4xy+2y^2$, but I realized that there's no way of using $abc\geq1$.
Any help is appreciated!
| *
*Clearly we just need to focus on $ abc = 1$. Suppose $x, y, z$ are fixed. How can we minimize the $LHS$ subject to $ abc = 1$?
*If $c$ is fixed, then we have $(a+1)x^2 + (\frac{1}{ac} + 1) y^2 + (c+1) z^2$, and differentiating with respect to $a$ gives $ x^2 - \frac{1}{c a^2} y^2 = 0 $, or that $ \frac{x^2}{y^2} = \frac{b}{a}$. (Check that this yields a minimum.)
*Hence, the minimum occurs when $ x^2 : y^2 : z^2 = \frac{1}{a} : \frac{1}{b} : \frac{1}{c}$ and $ abc = 1$, which yields the solution $ a = \frac{ y^{2/3} z^{2/3} } { x^{4/3}}$ etc. Hence
$$LHS = (a+1)x^2 + (b+1) y^2 + (c+1)z^2 \geq \sum ( \frac{ y^{2/3} z^{2/3} } { x^{4/3}} + 1 ) x^2 . $$
*
*Let $ x^{1/3} = p, y^{1/3} = q, z^{1/3} = r$, then it remains to show that
$$ \sum (p^2q^2r^2 + p^6) \geq 2\sum p ^3 r^3 = RHS$$
*
*This is true by Shurs on $ p^2, q^2 r^2$, which gives us the first inequality, and the second is just AM-GM:
$$ p^6 + q^6 + r^6 + 3 p^2 q^2 r^2 + \geq \sum p^2q^2(p^2+q^2) \geq 2 \sum p^3 q^3.$$
*
*Equality holds when
*
*$ p = q = r \Rightarrow x = y = z, a = b = c = 1$
*$ p = 0 , q = r \Rightarrow x = 0, y = z$, but then $a \rightarrow \infty, b = c \rightarrow 0 $ which isn't allowed.
Notes
*
*Yes, we could get to $ x^2 : y^2 : z^2 = \frac{1}{a} : \frac{1}{b} : \frac{1}{c}$ separately, but differentiation was the fastest / easiest.
*It is not true that $ax^2 + by^2 + cz^2 \geq xy+yz+zx$. Likewise, the inequality of $ \sum (1.01 a+1) x^2 \geq 2.01\sum xy$ is also not true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
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} |
Find the maximum value of $a^2(b-c)+b^2(c-b)+c^2(1-c)$, if $0\le{a}\le{b}\le{c}\le{1}$.
How do I find the maximum value of $a^2(b-c)+b^2(c-b)+c^2(1-c)$, if $0\le{a}\le{b}\le{c}\le{1}$?
I have tried plugging in values, tried manipulating and inequality by means but that didn't worked. How do I approach this question?
| $$\underbrace{(c-b)}_{\ge0}\underbrace{(b^2-a^2)}_{\ge0}+c^2\underbrace{(1-c)}_{\ge0}\\
$$
$b^2-a^2$ is the only term dependent on $a$ and thus it attains a maximum at $a=0.$ The problem then reduces to
$$b^2(c-b)+c^2(1-c)$$ for $0\le b\le c\le1.$ Sub $b=rc.$
$$r^2c^3(1-r)+c^2(1-c)\\
c^2(r^2(1-r)c+1-c)$$
$r^2(1-r)\overset{\partial_r}{\longrightarrow}r(2-3r)$ has a maximum at $r=\frac23.$
We then maximize
$$c^2\left(1-\frac{23}{27}c\right)\overset{\partial_c}{\longrightarrow}c\left(2-\frac{23}9c\right)\\
c=\frac{18}{23}\\
b=\frac23\cdot\frac{18}{23}=\frac{12}{23}$$
The maximum is thus $$\frac{108}{529}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Fraction of prime number between consecutive powers of two A $n$-bit integer is an integer $x$ such that $2^{n-1} \le x < 2^n$. In [1] it is claimed without proof that a corollary of the prime number theorem is that:
For any $n > 1$, the fraction of $n$-bit integers that are prime is at least $\frac{1}{3n}$.
Can somebody point me to a proof of this?
It is (perhaps) worth noticing that a direct application of the prime number theorem allows to prove an asymptotic version of the above (with a slightly better constant).
$$
\begin{align*}
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg( \pi(2^n-1) - \pi(2^{n-1})\bigg) &=
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg( \pi(2^n) - \pi(2^{n-1}) \bigg) \\&=
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg( \frac{2^n}{n \ln 2} - \frac{2^{n-1}}{(n-1)\log2} \bigg)\\ & =
\lim_{n \to \infty} \frac{n \ln 2}{2^{n-1}} \bigg( \frac{2^n}{2n \ln 2 } \cdot\frac{n-2}{n-1} \bigg) = 1.
\end{align*}
$$
Therefore $\frac{\pi(2^n-1) - \pi(2^{n-1})}{2^{n-1}} \sim \frac{1}{n \ln 2}$, and for every $\varepsilon >0$ there is some $n_0$ such that, for all $n \ge n_0$, the fraction of $n$-bit integers that are prime is at least $\frac{1-\varepsilon}{n \ln 2}$.
A not-so-elegant way to prove the the claim would be that of choosing $\varepsilon =1-\frac{\ln 2}{3} \approx 0.7689$ and finding an upper bound on $n_0$ that lies in the range for which $\pi(\cdot)$ has been computed. Are such upper bounds known?
[1] Jonathan Katz, Yehuda Lindell. Introduction to Modern Cryptography (3rd edition). CRC Press. ISBN 9781351133012.
| According to
the classic article
"Explicit Bounds for Some Functions of Prime Numbers",
Barkley Rosser
American Journal of Mathematics
Vol. 63, No. 1 (Jan., 1941), pp. 211-232 (22 pages)
for $x \ge 55$,
$\dfrac{x}{\log x+2}
\lt \pi(x)
\lt \dfrac{ x}{\log x-4}
$.
Therefore,
putting $x=2^n$,
$\dfrac{ 2^n}{n\log 2+2}
\lt \pi(2^n)
\lt \dfrac{ 2^n}{n\log 2-4}
$.
Therefore
$\dfrac{ 2^n}{n\log 2+2}-\dfrac{ 2^{n-1}}{(n-1)\log 2-4}
\lt \pi(2^n)-\pi(2^{n-1})
\lt \dfrac{ 2^n}{n\log 2-4}-\dfrac{ 2^{n-1}}{(n-1)\log 2+2}
$.
Those differences are about
(more explicitness later)
$\begin{array}\\
D
&=\dfrac{ 2^n}{n\log 2}-\dfrac{ 2^{n-1}}{(n-1)\log 2}\\
&=\dfrac{ 2^n}{n\log 2}\left(1-\dfrac{\frac12}{1-\frac1{2n}}\right)\\
&=\dfrac{ 2^n}{n\log 2}\left(\dfrac{1-\frac1{2n}-\frac12}{1-\frac1{2n}}\right)\\
&=\dfrac{ 2^n}{n\log 2}\left(\dfrac{\frac12-\frac1{2n}}{1-\frac1{2n}}\right)\\
&=\dfrac{ 2^n}{2n\log 2}\left(\dfrac{1-\frac1{n}}{1-\frac1{2n}}\right)\\
&=\dfrac{ 2^n}{2n\log 2}\left(\dfrac{1-\frac1{2n}-\frac1{2n}}{1-\frac1{2n}}\right)\\
&=\dfrac{2^n}{2n\log 2}\left(1-\dfrac{\frac1{2n}}{1-\frac1{2n}}\right)\\
&=\dfrac{2^n}{2n\log 2}\left(1-\dfrac{1}{2n-1}\right)\\
&\approx\dfrac{ 2^n}{2n\log 2}\\
\end{array}
$
For the lower bound,
$\begin{array}\\
L(n)
&=\dfrac{ 2^n}{n\log 2+2}-\dfrac{ 2^{n-1}}{(n-1)\log 2-4}\\
&=\dfrac{ 2^n((n-1)\log 2-4)-2^{n-1}(n\log 2+2)}{(n\log 2+2)((n-1)\log 2-4)}\\
&=\dfrac{ 2^{n-1}(2(n-1)\log 2-8)-(n\log 2+2))}{n(n-1)\log^22-4n\log 2+2(n-1)\log 2-8}\\
&=\dfrac{2^{n-1}(n \log 2-2\log 2-10)}{n(n-1)\log^22-2n\log 2+2\log 2-8}\\
\end{array}
$
I will leave it at this for now,
and a similar expression
can be derived for the
upper bound.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Derivation of $\sin(15^\circ)$ geometrically This is my attempt:-
Let us consider a right $\triangle ABC$ such that angle $A$ is $15^\circ$ and $C$ is $75^\circ$.
On the line $AB$, let us assume a point $D$ such that $\frac{BC}{BD} =\frac{ 1}{\sqrt{3}}$ (Without Loss Of Generality). So $\angle BDC$ becomes $30^\circ$ and $\angle BCD$ becomes $60^\circ$. Then $\angle DCA$ becomes equal to $\angle BAC$, that is $15^\circ$; so $CD = DA$.
$CD$ will be $2$ times $BC$ (angle $BDC = 30^\circ$; $\sin 30^\circ$). On adding $BD$ and $AD$ we get $$AB = BC(2+\sqrt{3})$$
$$BC^2 + AB^2 = AC^2$$
$$\therefore AC = 2BC\sqrt{2 + \sqrt{3}}$$
$$\sin(15^\circ) = \frac{BC}{AC}$$
$$\sin(15^\circ) = \frac{1}{\sqrt{2 + \sqrt{3}}}$$
Rationalising the denominator 2 times we get:-
$$\sin(15^\circ) = \frac{(4-2\sqrt{3})(\sqrt{2+\sqrt{3}})}{4}$$
Further simplifying:-
$$\sin(15^\circ) = \frac{(2-\sqrt{3})(\sqrt{2+\sqrt{3}})}{2}$$
$$\sin(15^\circ) = \frac{(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}}) (\sqrt{2+\sqrt{3}})}{2}$$
$$\sin(15^\circ) = \frac{\sqrt{2-\sqrt{3}}}{2}$$
Is my answer correct? And is there any other method or way to get the value of $\sin(15^\circ)$ geometrically?
| There is another method that starts with a fun and challenging geometry problem. Not sure if you've seen it before, but it is fairly famous (I don't know if it has a specific name though):
Now, once you've solved that purely with elementary geometry, no trig, you can then get some nice trig results from that. $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively:
Let the square have side $2$. You've already shown $\triangle AOB$ is equilateral, so $\angle OAB = 60^{\circ}$. Which gives $OM = AM\tan 60^{\circ} = \sqrt 3$. Then $ON = 2 - \sqrt 3$. You can get $OD$ by Pythagoras': $OD = \sqrt{(2 - \sqrt 3)^2 + 1^2} = 2\sqrt{2 - \sqrt 3}$. So $\sin 15^{\circ} = \frac{ON}{OD} = \frac 12\sqrt{2-\sqrt 3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to evaluate $\int x \sqrt{x^2 - x}\ dx$
Problem is to integrate $ x \sqrt{x^2 - x}$.
My attempt:
I made it ready for a substitution $u = x^2 - x$
$$\begin{aligned} \int x \sqrt{x^2 -x}\ dx &= \int (2x-1)\sqrt{x^2 - x} \ dx - \int(x - 1)\sqrt{x^2 - x}\ dx\\& = \int \sqrt u\ du - \int(x - 1)\sqrt{x^2 - x}\ dx\\& = \frac23(x^2 -x)^{3/2} + C_1 - \int(x-1) \sqrt{x^2 -x}\ dx\end{aligned}$$
I don't know how to continue from here.
Alternatively I tried this:
$$\begin{aligned} \int x \sqrt{x^2 - x}\ dx &= \int x^2 \sqrt{1- \frac{1}{x}}dx\\ & \overset{1- \frac1x = t^2}{=} \int \frac{2t^2}{(1-t^2)^4}\ dt\\& \overset{t =\sin(\theta)}{=} \int \frac{2\sin^2(\theta) \cos(\theta)\ d\theta}{\cos^4(\theta)}\\& = \int 2 \tan(\theta)\tan(\theta) \sec(\theta) \ d\theta\\ & = \int 2\sqrt{\sec^2(\theta) - 1}\tan(\theta) \sec(\theta) \ d\theta\\& \overset{\sec(\theta) = u}{=} \int 2 \sqrt{\sec^2(\theta) - 1}\tan(\theta) \sec(\theta) \ d\theta\\& = \int 2 \sqrt{u^2 - 1}\ du\\& = u \sqrt{u^2 - 1} - \ln|u + \sqrt{u^2- 1}| + C\\& = \sqrt{x^2 - x} - \ln|x + \sqrt{x^2 - x}| + C\end{aligned}$$
This method is very tedious. Is there any easy way to do the original integral?
| Substitute
$$t=\sqrt{x^2-x}-x \implies x = -\frac{t^2}{2t+1} \implies dx = -\frac{2t^2+2t}{(2t+1)^2} \, dt$$
and the resulting integrand is primed for partial fraction expansion.
$$\int x \sqrt{x^2-x} \, dx = 2 \int \frac{t^4(t+1)^2}{(2t+1)^4} \, dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4646982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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Proving Generating Function holds a specific recurrence. Consider the generating function $$\dfrac{1}{1 − 2x − x^2} = \sum_{n=0}^{\infty}a_nx^n$$
Prove that for each integer $n \ge 0$,
$$a_n^2+a_{n+1}^2 = a_{2n+2}$$
Hint: Find a $2 \times 2$ matrix $A$ such that
$$A^{n+2} =\begin{bmatrix}a_n &a_{n+1}\\
a_{n+1} &a_{n+2}\end{bmatrix}$$
and consider the top left entry of the matrix product $A^{n+2}A^{n+2}$.
Looking at the hint, I think about how we used matrices and eigenvalues to find the closed form expression of recurrence relations, but I can only really do that for stuff of the form $a_n=a_{n-1}+a_{n-2}$, and I'm not sure how the matrix product fits in there.
Instead, I tried to find the generating function for the recurrence listed above as:
$$\begin{align*}
a_n^2+a_{n+1}^2 &= a_{2n+2}\\
\left(x^n\right)^2a_n^2+\left(x^n\right)^2a_{n+1}^2 &= \left(x^n\right)^2a_{2n+2}\\
\sum_{n=0}^{\infty}\left(x^n\right)^2a_n^2+\sum_{n=0}^{\infty}\left(x^n\right)^2a_{n+1}^2 &= \sum_{n=0}^{\infty}\left(x^n\right)^2a_{2n+2}\\
A(x)^2+\dfrac{A(x)^2}{x^2} &= \sum_{n=0}^{\infty}x^{2n}a_{2n+2}\\
A(x)^2+\dfrac{A(x)^2}{x^2} &= \dfrac{A(x)}{x^2}\\
x^2A(x)^2+A(x)^2&=A(x)\\
A(x)=\dfrac{1}{x^2+1}
\end{align*}$$
But that's clearly not what we wanted. What mistakes did I make, and how does the hint fit into all of this? Thanks!
| Too long for a comment
$$\frac{1}{1 − 2x − x^2}=\frac{1}{(1+\sqrt2+x)(\sqrt2-1-x)}=\frac{1}{2\sqrt2}\left(\frac{1}{1+\sqrt2+x}+\frac{1}{\sqrt2-1-x}\right)$$
$$=\frac{1}{2\sqrt2}\sum_{n=0}^\infty\left(\frac{x^n}{(\sqrt2-1)^{n+1}}+(-1)^n\frac{x^n}{(\sqrt2+1)^{n+1}}\right)$$
Then, given that $\frac{1}{(\sqrt2+1)^{n+1}}=\frac{(\sqrt2-1)^{n+1}}{(\sqrt2+1)^{n+1}(\sqrt2-1)^{n+1}}=(\sqrt2-1)^{n+1}$
$$a_n=\frac{1}{2\sqrt2}\left(\frac{1}{(\sqrt2-1)^{n+1}}+(-1)^n\frac{1}{(\sqrt2+1)^{n+1}}\right)=\frac{1}{2\sqrt2}\left((\sqrt2+1)^{n+1}+(-1)^n(\sqrt2-1)^{n+1}\right)$$
and
$$a_{n+1}=\frac{1}{2\sqrt2}\left((\sqrt2+1)^{n+2}+(-1)^{n+1}(\sqrt2-1)^{n+2}\right)$$
Then
$$a_n^2+a_{n+1}^2=\frac{1}{8}\left((\sqrt2+1)^{2n+2}+(\sqrt2-1)^{2n+2}+(\sqrt2+1)^{2n+4}+(\sqrt2-1)^{2n+4}\right)$$
$$=\frac{1}{8}\left((\sqrt2+1)^{2n+2}+(\sqrt2-1)^{2n+2}+(\sqrt2+1)^{2n+2}(3+2\sqrt2)+(\sqrt2-1)^{2n+2}(3-2\sqrt2)\right)$$
$$=\frac{1}{2\sqrt2}\left((\sqrt2+1)^{2n+3}+(\sqrt2-1)^{2n+3}\right)=a_{2n+2}$$
| {
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"url": "https://math.stackexchange.com/questions/4648320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$. Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$.
Let $[x]$ denote $\lceil{x-\frac{1}{2}}\rceil$.
Using Desmos, I got $2.46736022133$ and WolframAlpha does not give me a solution. My intuition tells me that it might be possible to find an exact value using Trapezoidal Reimann Sums but I am not really sure how to go about doing it. After my attempt, I got stuck but I was at a point where I could plug it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. Why did it come out so nicely?
My attempt:
Where $A_n$ denotes the area of the $nth$ trapezoid from the right:
$$A=\frac{h}{2}(a+b)$$
$$A_n=\frac{\frac{2}{2n-1}-\frac{2}{2n+1}}{2}(\frac{2n}{2n-1}+\frac{2n}{2n+1})$$
$$A_n=\frac{\frac{4n+2}{4n^{2}-1}-\frac{4n-2}{4n^{2}-1}}{2}\left(\frac{4n^{2}+2n}{4n^{2}-1}+\frac{4n^{2}-2n}{4n^{2}-1}\right)$$
$$A_n=\frac{2}{4n^{2}-1}\left(\frac{8n^{2}}{4n^{2}-1}\right)$$
$$A_n=\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
Then:
$$\int_{0}^{2}x[\frac{1}{x}]dx=\sum_{n=1}^{\infty}A_n=\sum_{n=1}^{\infty}\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
I do not know how to solve this infinite summation so I plugged it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. How did it get to this conclusion? Is there a more efficient way to solve this?
| An alternate solution:
$$
\begin{align}
\int_{0}^{2}x\operatorname{round}\left(\frac{1}{x}\right)dx &= \int_0^2 x\Biggl\lfloor{\frac{1}{x}+\frac{1}{2}\Biggr\rfloor}dx \tag{1}\\
&= 8\int_{1}^{\infty}\frac{\lfloor{x\rfloor}}{\left(2x-1\right)^{3}}dx \tag{2}\\
&= 8\sum_{n=1}^{\infty}n\int_{n}^{n+1}\frac{dx}{\left(2x-1\right)^{3}} \\
&= 16\sum_{n=1}^{\infty}\frac{n^{2}}{\left(1-4n^{2}\right)^{2}} \\
\end{align}
$$
where in $(1)$ we used the Desmos interpretation $\displaystyle \operatorname{round}\left(\frac{1}{x}\right) := \Biggl\lfloor{\frac{1}{x}+\frac{1}{2}\Biggr\rfloor}$ and in $(2)$ we used the mapping $\displaystyle \frac{1}{x}+\frac{1}{2} \mapsto x$.
Next, let $f$ be a piecewise smooth function on $[0,L]$. Then the Fourier sine expansion is given by
$$f\left(x\right)\ =\ \sum_{n=1}^{\infty}b_{n}\sin\left(\frac{\pi nx}{L}\right)$$
where we have the sequence
$$b_{n}=\frac{2}{L}\int_{0}^{L}f\left(x\right)\sin\left(\frac{\pi nx}{L}\right)dx.$$
Let $f(x) = \cos(x)$ and $L = \dfrac{\pi}{2}$. Then using some basic integration, we can prove that
$$b_{n}=\frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\cos\left(x\right)\sin\left(2nx\right)dx\ =\ \frac{4}{\pi}\cdot\frac{2n}{4n^{2}-1}.$$
This means
$$
\begin{align}
\cos\left(x\right) &= \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{2n}{4n^{2}-1}\sin\left(2nx\right) \\
\implies \cos^2(x) &= \frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n\cos\left(x\right)\sin\left(2nx\right)}{4n^{2}-1} \\
\implies \int_{0}^{\frac{\pi}{2}}\cos^{2}\left(x\right)dx &= \frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n}{4n^{2}-1}\int_{0}^{\frac{\pi}{2}}\cos\left(x\right)\sin\left(2nx\right)dx \\
&= \frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n}{4n^{2}-1}\cdot\frac{2n}{4n^{2}-1}. \\
&= \frac{16}{\pi}\sum_{n=1}^{\infty}\frac{n^{2}}{\left(4n^{2}-1\right)^{2}} \\
\end{align}
$$
But $\displaystyle \int_{0}^{\frac{\pi}{2}}\cos^{2}\left(x\right)dx=\frac{\pi}{4}$. By transitivity, we get
$$\frac{\pi}{4}=\frac{16}{\pi}\sum_{n=1}^{\infty}\frac{n^{2}}{\left(4n^{2}-1\right)^{2}}.$$
Therefore,
$$\int_{0}^{2}x\operatorname{round}\left(\frac{1}{x}\right)dx = \frac{\pi^{2}}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
Is there an integral that proves $\pi > 333/106$? The following integral,
$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$
is clearly positive, which proves that $\pi < 22/7$.
Is there a similar integral which proves $\pi > 333/106$?
| Another solution is given by the integral
$$
0 < \int_0^1 \frac{x^4(1-x)^8}{4(1+x^2)}dx = \pi -\frac{2419}{770} = \pi - \frac{333}{106}-\frac{1}{20405}
$$
This proves the stricter condition
$$
\pi > \frac{333}{106}+\frac{1}{20405}
$$
which implies
$$
\pi > \frac{333}{106}
$$
Similarly, for the fourth convergent (formula (6) http://www.math.ucla.edu/~vsv/resource/general/Lucas.pdf)
$$0<\int_0^1 \frac{x^{10}(1-x)^8}{4(1+x^2)}dx=\frac{3849155}{1225224}-\pi=\frac{355}{113}-\frac{5}{138450312}-\pi$$
Therefore
$$\pi<\frac{355}{113}-\frac{5}{138450312}$$
and
$$\pi<\frac{355}{113}$$
Even for the first convergent
$$0<2\int_0^1 \frac{x(1-x)^2}{(1+x^2)}dx=\pi-3$$
so
$$\pi>3$$
(See https://math.stackexchange.com/a/1618454/134791 for a proof for $3<\pi<4$ that uses this integral)
A series proof that $\pi>\frac{333}{106}$ is given by
$$\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)} \\=\pi-\frac{333}{106}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "129",
"answer_count": 6,
"answer_id": 0
} |
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| Here are two ways to calculate this sum. First is by symmetry of another sum:
$\begin{aligned} \displaystyle & \sum_{0 \le k \le n}k^2 = \sum_{0 \le k \le n}(n-k)^2 = n^2\sum_{0 \le k \le n}-2n\sum_{0 \le k \le n}k+\sum_{0 \le k \le n}k^2 \\& \implies 2n\sum_{0 \le k \le n}k = n^2(n+1) \implies \sum_{0 \le k \le n}k = \frac{1}{2}n(n+1).\end{aligned}$
The second is writing it as double sum and switching the order of summation:
$\begin{aligned}\displaystyle & \begin{aligned}\sum_{1 \le k \le n}k & = \sum_{1 \le k \le n}~\sum_{1 \le r \le k} = \sum_{1 \le r \le n} ~\sum_{r \le k \le n} = \sum_{1 \le r \le n}\bigg(\sum_{1 \le k \le n}-\sum_{1 \le k \le r-1}\bigg) \\& =\sum_{1 \le r \le n}\bigg(n-r+1\bigg) = n\sum_{1 \le k \le n}-\sum_{1 \le k \le n}k+\sum_{1 \le k \le n}\end{aligned} \\& \implies 2\sum_{1 \le k \le n}k = n^2+n \implies \sum_{1 \le k \le n}k = \frac{1}{2}n(n+1), ~ \mathbb{Q. E. D.} \end{aligned}$
Note I started using k back on the third line for convenience because r is just a dummy vairable at this point, and our sum no longer depends on k. Note that the first trick can easily be generalised:
$\begin{aligned} & \hspace{0.5in}\begin{aligned}\displaystyle \sum_{0 \le k \le n}k^{2p} &= \sum_{0 \le k \le n}(n-k)^{2p} \\& = \sum_{0 \le k \le n}~\sum_{0 \le r \le 2p}\binom{2p}{r}n^r(-1)^{2p-r}k^{2p-r}\\& = \sum_{0 \le k \le n}k^{2p}-2pn\sum_{0 \le k \le n}k^{2p-1}+\sum_{0 \le k \le n}~\sum_{2 \le r \le 2p}\binom{2p}{r}n^r(-1)^{2p-r}k^{2p-r} \end{aligned} \\& \implies \sum_{0 \le k \le n}k^{2p-1} = \frac{1}{2pn}\sum_{0 \le k \le n}~\sum_{2 \le r \le 2p}\binom{2p}{r}n^r(-1)^{2p-r}k^{2p-r}. \end{aligned}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "136",
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"answer_id": 16
} |
If $AB = I$ then $BA = I$
If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.
I do not understand anything more than the following.
*
*Elementary row operations.
*Linear dependence.
*Row reduced forms and their relations with the original matrix.
If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?
P.S.: Please avoid using the transpose and/or inverse of a matrix.
| An alternative.
Let $\textbf{A}$ be a $n \times n$ matrix, and let $\lambda_k$ be the eigenvalues, then we can write
$$
\prod_{k=1}^n \Big( \textbf{A} - \lambda_k \textbf{I} \Big) = 0
$$
So we obtain
$$
\textbf{A}^n = a_0 \textbf{I} + \sum_{k=1}^{n-1} a_k \textbf{A}^k
$$
Let us write
$$
\textbf{B} = b_0 \textbf{I} + \sum_{k=1}^{n-1} b_k \textbf{A}^k$$
It is clear that
$$
\textbf{A} \textbf{B} = \textbf{B} \textbf{A}
$$
We also find that
$$
\begin{eqnarray}
\textbf{A} \textbf{B}
&=& b_0 \textbf{A} + \sum_{k=1}^{n-1} b_k \textbf{A}^{k+1}\\
&=& b_0 \textbf{A} + \sum_{k=1}^{n-2} b_k \textbf{A}^{k+1} + b_{n-1} \textbf{A}^{n}\\
&=& b_0 \textbf{A} + \sum_{k=1}^{n-2} b_k \textbf{A}^{k+1} + b_{n-1} a_0 \textbf{I}
+ b_{n-1} \sum_{k=1}^{n-1} a_k \textbf{A}^k\\
&=& b_{n-1} a_0 \textbf{I} + \Big( b_0 + b_{n-1} a_1 \Big) \textbf{A} + \sum_{k=2}^{n-1} \Big( b_{k-1} + b_{n-1} a_k \Big) \textbf{A}^k\\
\end{eqnarray}
$$
When we set (for $a_0 \ne 0$)
$$
\begin{eqnarray}
b_{n-1} a_0 &=& 1\\
b_{k-1} + b_{n-1} a_k &=& 0
\end{eqnarray}
$$
we obtain
$$
\textbf{A} \textbf{B} = \textbf{I}
$$
So we can write
$$
\textbf{B} = -a_0^{-1} a_1 \textbf{I}
- \sum_{k=1}^{n-2} a_0^{-1} a_{k+1} \textbf{A}^k
+ a_0^{-1} \textbf{A}^{n-1}
$$
or
$$
\textbf{A}^{-1} = -a_0^{-1} \left( a_1 \textbf{I}
+ \sum_{k=1}^{n-2} a_{k+1} \textbf{A}^k
- \textbf{A}^{n-1} \right)
$$
the inverse of $\textbf{A}$ can be expressed as a linear sum of $\textbf{A}^k$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "379",
"answer_count": 34,
"answer_id": 2
} |
Explain why calculating this series could cause paradox? $$\ln2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots
= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - 2(\frac{1}{2} + \frac{1}{4} + \cdots)$$
$$= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) = 0$$
thanks.
| In this chain of 4 equations, #1, #3 and #4 are correct, and no.2 is the mistake. The equations are assertions about (limits of) some finite sums. Let $H_n = \Sigma_{j=1}^n 1/j$ and $A_n = \Sigma_{i=1}^n (-1)^{i-1} 1/i$.
The correct formula is $A_n = H_n - H_{[n/2]}$ (which is approximately $\log(n) - \log(n/2) \sim \log(2) = A_{\infty}$) , but the second equation cut off at $n$ terms is claiming $A_n = H_n - H_n$ (which is zero).
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Factoring a Cubic Polynomial I've been trying to understand how
${x^3-12x+9}$
factors to
$(x-3) (x^2+3 x-3)$
What factoring rule does this follow? The net result seems to be similar to what is attained through the sum/difference of cubes factoring pattern, but the signs are different.
Additionally, what type of problem is this, so I can make better and more relevant searches for help on future questions. Is it a cubic trinomial?
| In order to factor any cubic, you must find at least one root. You acknolwedged that $3$ is a root, thus $x=3$ and $x-3=0$. And since $x-3$ is a factor of $x^3-12x+9$, split the polynomial in accordance with $x-3$ and factor as follows:
\begin{align}
x^3-12x+9 &= x^3-3x^2+3x^2-9x-3x+9\\
&=x^2(x-3)+3x(x-3)-3(x-3)\\
&=(x-3)(x^2+3x-3)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/5646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 3
} |
How to prove an identity in radicals? (4 / (3 - sqrt(5))) ^ 2 - ((6 - 5 * sqrt(6)) / (5 - sqrt(6))) ^ 2 = 2 * sqrt(61 + 24*sqrt(5))
$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$
How to prove it is right equality?
I come up with $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$, but still can't get it to the obvious equality.
Any ideas?
| Specialize $\rm\ \ \ b=3,\ \ c = 5,\ \ d = 6 \ \Rightarrow\ a = 4\ \ $ in this simple derivation:
$\rm\quad\quad\quad\quad\displaystyle \bigg(\frac{b^2-c}{b-\sqrt{c}}\bigg)^2 - \bigg(\frac{d -5\sqrt{d}}{5-\sqrt d}\bigg)^2$
$\rm\quad\quad =\quad \:(\: b \ \: + \: \sqrt{c}\ )^{\:2} \ \ \:-\ \ \ (\:-\:\sqrt{d}\:)^{2} $
$\rm\quad\quad =\ \ 2\ (\:a + b \sqrt{c}\ )\:, \quad 2\ a\ =\ b^2+c-d $
$\rm\quad\quad =\ \ 2\:\sqrt{a^2+b^2\:c+2\:a\:b\sqrt{c} } $
NOTE $\:$ Replacing numbers by functions makes the proof both simpler and more general - similar to your recently asked question
| {
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"source": "stackexchange",
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If $\gcd(a,b) = 1$ and $a,b\mid x$ then $ab\mid x$. If $\gcd(a,b) = 1$ and $a,b\mid x$ then $ab\mid x$.
My attempt at answering the question:
\begin{align*}
x &\equiv 0 \pmod{a}\\\
&\Longrightarrow x\text{ is divisible by $a$}\\\
&\Longrightarrow x = ma\text{ for some integer $m$}\\\
\ \\\
x &\equiv 0 \pmod{b}\\\
&\Longrightarrow x\text{ is divisible by $b$}\\\
&\Longrightarrow x = mb\text{ for some integer $m$}\\\
\ \\\
x^2 &= (ma)(mb)\\\
x^2 &= (m^2)(ab)\\\
x &= \sqrt{m^2ab}\\\
x &= m\sqrt{a}\sqrt{b}
\end{align*}
Let $m$ be $k\sqrt{a}\sqrt{b}$. Then
\begin{align*}
x &= kab\\\
&\Longrightarrow x \equiv 0 \pmod{ab}
\end{align*}
Is this correct, if not can someone point me in the right direction?
| You are given that $a$ divides $x$. Therefore, you can write $x = ma$ where m is an integer. You are also given that $b$ divides $x$. This implies that $b$ divides $ma$. But $b$ and $a$ are coprime. Therefore $b$ must divide $m$. So you can write $m = kb$ where $k$ is another integer. Therfore, you have $x = kab$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
| Let $f(x)=\frac 12-x$ on the interval $[0, 1)$, and extend $f$ to be periodic on $ \mathbb{R} $.
By definition,
\begin{align*}
\hat f(0)=\int_0^1 f(x)dx=\int_0^1 \left(\frac 12-x\right)dx=0.
\end{align*}
And for $ \kappa\ne 0 $:
\begin{align*}
\hat f(\kappa)&=\int_{0}^{1}f(x)e^{-2\pi i\kappa x }dx=\int_0^1\left( \frac 12 -x \right)e^{-2\pi i\kappa x}dx=-\int_0^1xe^{-2\pi i \kappa x}dx\\
&=\frac{1}{2\pi i\kappa }\int_{0}^{1}xd(e^{-2\pi i\kappa x})=\left.\frac{1}{2\pi i\kappa}xe^{-2\pi i\kappa x}\right|_0^1+\frac{1}{2\pi i\kappa}\int_0^1 e^{-2\pi i\kappa x}dx\\
&=\frac{1}{2\pi i\kappa}.
\end{align*}
By the Parseval identity
\begin{align*}
\int_{0}^{1}|f(x)|^2dx=\sum_{k=-\infty}^{\infty}|\hat{f}(k)|^2=|\hat{f}(0)|^2+2\sum_{k=1}^{\infty}|\hat{f}(k)|^2=2\sum_{k=1}^{\infty}\frac{1}{4\pi^2 k^2}.
\end{align*}
On the other hand,
\begin{align*}
\int_{0}^{1}|f(x)|^2dx&=\int_{0}^{1}\left( \frac{1}{2}-x \right)^2 dx=\frac 14-\frac 12+\frac 13=\frac 1{12}.
\end{align*}
Hence, we have $$ \sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}. $$
Remark: This is an exercise(Chapter 8.13 on page 254 ) in Folland's book.
| {
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"answer_count": 48,
"answer_id": 27
} |
Probability and Integrals Suppose $f(x,y) = c$ for $0\lt y\lt x\lt 1$ and $0$ outside. What is $P(X+Y \leq 1)$? What is $P(X^2+Y^2 \leq 1)$?
So
\begin{equation*}
P(X+Y \leq 1) = \int_{0}^{1} \int_{0}^{1-x} 2 \ dy \ dx?
\end{equation*}
Likewise,
\begin{equation*}
$P(X^2+Y^2 \leq 1) = \int_{0}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2 \ dy \ dx$?
\end{equation*}
This is assuming that $c=2$.
| Since the joint density is constant on the support region, you can get your answers by considering areas. The answer in each case is the area that corresponds to the event in question divided by the total area of the support region.
Thus, to calculate $P(X + Y \leq 1)$, you want the area of the lower triangular region below (i.e., the region for which $x + y \leq 1$), divided by the total area of the triangle.
This is $$\frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}.$$
Similarly, to calculate $P(X^2 + Y^2 \leq 1)$, you want the area of the circle sector divided by the total area of the triangle.
This is $$\frac{\frac{\pi}{8}}{\frac{1}{2}} = \frac{\pi}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/8526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Expected number of neighbors Given a row of 16 houses where 10 are red and 6 are blue, what is the expected number of neigbors of a different color?
| The chances of a particular neighbour pair being the same colour is
$$ \frac{{14 \choose 8} + {14 \choose 10}}{{16 \choose 6}} = \frac{4004}{8008} = \frac{1}{2}$$
Hence the answer is $\displaystyle 7.5$
This is happening because $\displaystyle {14 \choose 8}, {14 \choose 9}, {14 \choose 10}$ are in arithmetic progession: The number of ways of being same colour is $\displaystyle {14 \choose 8} + {14 \choose 10}$ and the number of ways of being different is $\displaystyle 2{14 \choose 9}$. The probability is $\displaystyle \frac{1}{2}$ if these two are equal.
Interestingly, $\displaystyle {n \choose r}, {n \choose r+1}, {n \choose r+2}$ are in arithmetic progression if and only if $\displaystyle n+2$ is a perfect square and $\displaystyle r$ is given by $\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$ (see the end of the answer for a proof).
So for instance, the whole bunch of problems:
15 red, 10 blue
21 red, 15 blue
etc give rise to this neat probability of being $\displaystyle \frac{1}{2}$.
Proof that n+2 is a perfect square
$\displaystyle {n \choose r}, {n \choose r+1}, {n \choose r+2}$ are in arithmetic progression iff
$\displaystyle 2{n \choose r+1} = {n \choose r} + {n \choose r+2}$
i.e
$\displaystyle 2 = \frac{r+1}{n-r} + \frac{n-r-1}{r+2}$
Doing some manipulations gives us
$\displaystyle (n-2r-2)^2 = n+2$
Hence
$\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$
Which has an integer solution iff $\displaystyle n+2$ is a perfect square.
| {
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Prove by induction $T(n) = 2T(\frac{n}{2}) + 2$ I'm stuck with this induction proof:
So far, given:
$\begin{align*}
T(1) & = 2 \\
T(n) & = 2T(n/2)+2 \\
& = 2(2T(n/[2^2])+2) + 2 \\
& = [2^2]T(n/[2^2]) + [2^2] + 2 \\
& = [2^2](2T(n/[2^2])+2) + [2^2] + 2 \\
& = [2^3]T(n/[2^3]) + [2^3] + [2^2] + 2 \\
& = [2^3]T(n/[2^3]) + 2\{[2^2] + [2^1] + 1\} \\
& \vdots \\
& = [2^k]T(n/[2^k]) + 2\{2^{k} - 1\}
\end{align*}$
How then do I show this to be correct (the proof). So far I have:
Let $(n/[2^k]) = 1$
$\Rightarrow n = 2^k$
So, $T(n) = nT(1) + 2(n - 1)$
$T(n) = 4n - 2$ //This is where I'm stuck.
Proof (by induction):
When $n = 1$, $T(1) = 2$.
Assume $T(k)$ is true [$T(n) = 4n - 2$] //This is where I am stuck.
| HINT $\: $ From the first few values we guess $\rm\ T(2^n)\ =\ 2^{n+2}-2\ $ and induction confirms it:
$$\rm T(2^{n+1})\ =\ 2\ T(2^n) + 2 \ =\ 2\ (2^{n+2}-2) +\ 2\ =\ 2^{n+3} - 2$$
One can extend $\rm\:T\:$ to $\:\mathbb N\:$ by defining $\rm\ T(2k+1) = 2\ T(k+1)-2 $ and now one easily proves by induction that $\rm\ T(k) = 4\:k-2\ $ since
$\rm\quad\quad\quad\quad\quad\quad\quad T(2k+1)\ =\ 2\ T(k+1)-2\ =\ 2\ (4k+2)-2\ =\ 4(2k+1)-2 $
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\ T(2k)\ =\ 2\ \ \ \ T(k)\ \ +\ \ \: 2\ =\ 2\ (4k-2) + 2\ =\ 4 (2k) - 2 $
| {
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Can I find the limit of $\sum_\limits{i=1}^{n^2}\frac{1}{\sqrt{n+i}}$ with the squeeze theorem? Problem: Calculate limit of $\frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+3}} + \cdots + \frac{1}{\sqrt{n+n^2}}$ as $n$ approaches infinity.
Solution: Denote the above some as $X$, then we can bound it:
$$ \infty\longleftarrow\frac{1}{\sqrt{n+n^2}} \lt X \lt \frac{n^2}{\sqrt{n+1}} \lt \frac{n^2}{\sqrt{n}} = \sqrt{\frac{n^4}{n}}\longrightarrow \infty.$$
So, from the Squeeze Principle, $\lim X = \infty$. Am I doing the right thing?
| The term $\frac{1}{\sqrt{n+n^2}}$ goes to zero as $n \rightarrow \infty$, so that side doesn't work. You should think about how many terms you are adding.
| {
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How to derive the following series expansion for $\ln \left( \frac{x+1}{x-1} \right)$? How to prove :
$$ \ln \biggl(\frac{x+1}{x-1}\biggr) = 2\biggl[\frac{1}{x} + \frac{1}{3x^3} + \frac{1}{5x^5} + \cdots \biggr]$$ where $|x| \gt 1$
I am not able to get how to proceed on this one ?
| If $|x|>1$ then $\frac{1}{|x|}<1$, so:
$$\ln\left(\frac{x+1}{x-1}\right)=\ln\left(\frac{x(1+\frac{1}{x})}{x(1-\frac{1}{x})}\right)=\ln\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right)=\ln(1+t)-\ln(1-t)$$
With $t=\displaystyle\frac{1}{x}<1$, then $|t|<1$
Expanding :
$$\ln(1+t)=t-\frac{t^{^2}}{2}+\frac{t^{^3}}{3}-\frac{t^{^4}}{4}+...$$
$$\ln(1-t)=-t-\frac{t^{^2}}{2}-\frac{t^{^3}}{3}-\frac{t^{^4}}{4}-...$$
Then:
$$\ln(1+t)-\ln(1-t)=\left(t-\frac{t^{^2}}{2}+\frac{t^{^3}}{3}-\frac{t^{^4}}{4}+...\right)-\left(-t-\frac{t^{^2}}{2}-\frac{t^{^3}}{3}-\frac{t^{^4}}{4}-...\right)=$$
$$2\left(t+\frac{t^{^3}}{3}+\frac{t^{5}}{5}...\right)$$
Now $t=\displaystyle\frac{1}{x}$;
$$\ln\left(\frac{x+1}{x-1}\right)=2\left(\frac{1}{x}+\frac{1}{3x^{^3}}+\frac{1}{5x^{^5}}+\frac{1}{7x^{^7}}...\right)$$
| {
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Proof of the formula $1+x+x^2+x^3+ \cdots +x^n =\frac{x^{n+1}-1}{x-1}$
Possible Duplicate:
Value of $\sum x^n$
Proof to the formula
$$1+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1}.$$
| Observe that
\begin{eqnarray}
x^{n+1} - 1 = x^{n+1} + (x^{n} - x^{n}) + \cdots + (x - x) - 1 = (x^{n} + x^{n-1} + \cdots + x + 1)(x - 1).
\end{eqnarray}
| {
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Show that $\gcd(7^{79}+5,7^{78}+3) = 4$ How can I prove that $\gcd(7^{79}+5,7^{78}+3) = 4$ ? This was a question on a past exam, so the naive euclidean algorithm doesn't seem to suffice.
I'm not really sure where to start with this.
Note: This is exam prep, not homework.
| Since $\gcd(a,b)=\gcd(a-b,b)$, $$\begin{align}
\gcd(7^{79}+5,7^{78}+3)
&=\gcd(7\cdot 7^{78}+5-7^{78}-3,7^{78}+3)
\\
&=\gcd(6\cdot 7^{78}+2,7^{78}+3)
\\
&=\gcd(6\cdot 7^{78}+2-7^{78}-3,7^{78}+3)
\\
&=\gcd(5\cdot 7^{78}-1,7^{78}+3)
\\
&\vdots
\\
&=\gcd(7^{78}-13,7^{78}+3)
\\
&=\gcd(7^{78}-13,7^{78}+3-7^{78}+13)
\\
&=\gcd(7^{78}-13,16)
\\
\end{align}$$
From there, I'd determine the remainder when $7^{78}$ is divided by 16 and use that to see how $7^{78}-13$ compares to 16.
| {
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Algebra Problem The expression $x^2-4x+5$ is a factor of $ax^3+bx^2+25$. Express the sum a+b as an integer.
Please give an explanation of how the answer
| If $x^2-4x+5$ is a factor of $ax^3+bx^2+25$, then $ax^3+bx^2+25=(x^2-4x+5)(\text{something})$. Since $ax^3+\cdots$ is a polynomial of degree 3 and $x^2-\cdots$ is a polynomial of degree 2, the "something" must be a polynomial of degree 1:
$$ax^3+bx^2+0x+25=(x^2-4x+5)(\underline{\;\;\;\;\;\;}x+\underline{\;\;\;\;\;\;})$$
Try to fill in the two blanks based the terms on the left side that don't have $a$ and $b$ in them (for example, how will $+25$ end up in the product?), then finish the multiplication to find the values of $a$ and $b$.
| {
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Derivative of a rational function I've found the following derivative in my Calculus book and I can't get my my head around the algebra involved. Can anybody help me?
Thanks.
| \begin{align*}
\frac{d}{dx}\left(-\frac{x^2+1}{(x^2-1)^2}\right) &= -\frac{d}{dx}\left(\frac{x^2+1}{(x^2-1)^2} \right)&\quad&\mbox{(1)}\\
&= -\left(\frac{(x^2-1)^2(x^2+1)' - (x^2+1)\left((x^2-1)^2\right)'}{\left((x^2-1)^2\right)^2}\right)&&\mbox{(2)}\\
&= - \frac{(x^2-1)^2(2x) - (x^2+1)\left(2(x^2-1)(x^2-1)'\right)}{(x^2-1)^4}&&\mbox{(3)}\\
&= -\frac{2x(x^2-1)^2 - (x^2+1)(2(x^2-1)2x)}{(x^2-1)^4}&&\mbox{(4)}\\
&= -\frac{2x(x^2-1)^2 - 4x(x^2+1)(x^2-1)}{(x^2-1)^4}&&\mbox{(5)}\\
&= - \frac{2x(x^2-1)\left((x^2-1) - 2(x^2+1)\right)}{(x^2-1)^4}&&\mbox{(6)}\\
&= - \frac{2x(x^2-1)\left(x^2-1-2x^2-2\right)}{(x^2-1)^4}&&\mbox{(7)}\\
&= - \frac{2x(x^2-1)(-x^2-3)}{(x^2-1)^4}&&\mbox{(8)}\\
&= - \frac{2x(-x^2-3)}{(x^2-1)^3}&&\mbox{(9)}\\
&= -\frac{-2x(x^2+3)}{(x^2-1)^3}&&\mbox{(10)}\\
&= -(-2)\frac{x(x^2+3)}{(x^2-1)^3}&&\mbox{(11)}\\
&= 2\frac{x(x^2+3)}{(x^2-1)^3}.&&\mbox{(12)}
\end{align*}
Notes.
*
*Pull out the minus sign fromt he derivative.
*Use the Quotient Rule.
*Do the derivatives in the numerator, using the Chain Rule for $(x^2-1)^2$.
*Finish the derivative.
*Do some of the algebra in the numerator. Notice that both summands in the numerator have a factor of $2x(x^2-1)$.
*Factor out $2x(x^2-1)$ from both summands in the numerator.
*Do the operations in the other factor.
*Do the algebra in the numerator.
*Cancel the $x^2-1$ in the numerator with one in the denominator.
*Pull out the minus sign from $(-x^2-3)$.
*Pull out the $-2$ from the fraction.
*Simplify $-(-2)$ to $2$, and rejoice for your answer matches the one in the book.
| {
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Factoring $a^{10}+a^5+1$ I'm very interested to know how I can factorise $a^{10} +a^5 +1$ in two factors with integer coefficients. I've tried a lot but I don't have any idea how do that.
| Another hint: can you factor $a^{15}-1$? In more than one way?
Six years later, for the record, let's spell out what I meant by this. What I had intended was to observe
$$a^{15}-1 = (a^3)^5 - 1 = (a^3-1)(a^{12} + a^9 + a^6 + a^3 + 1)
= (a-1)(a^2+a+1)(a^{12} + a^9 + a^6 + a^3 + 1)$$
and also
$$a^{15}-1 = (a^5)^3 - 1 = (a^5-1)(a^{10} + a^5 + 1) = (a-1) (a^4+a^3+a^2+a+1) (a^{10}+a^5+1)$$
Thus we have
$$(a^2+a+1) (a^{12}+a^9 + a^6 + a^3 + 1) = (a^4 + a^3 + a^2 + a + 1)(a^{10} + a^5 + 1)$$
and so
$$a^{10} + a^5 + 1 = (a^2 + a + 1) \times {a^{12} + a^9 + a^6 + a^3 + 1 \over a^4 + a^3 + a^2 + a + 1}$$
Now, since $a^2 + a + 1 = (a^3-1)/(a-1)$, the roots of $a^2 + a + 1$ are just the two complex cube roots of unity; it's easy to check that $a^{10} + a^5 + 1$ has those same roots. So that quotient is in fact a polynomial. We can find out what polynomial it is by long division, and we get the answer:
$$a^{10} + a^5 + 1 = (a^2 + a + 1) (a^8 - a^7 + a^5 - a^4 + a^3 - a + 1).$$
| {
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How can I find $\int\frac1{\sqrt[4]{1+x^4}}\mathrm dx$? My question is, how can I evaluate the following integral?
$$\int\frac1{\sqrt[4]{1+x^4}}\mathrm dx$$
Thanks.
| Integral
$$\begin{align*}
\int\frac{1}{(1+x^4)^{1/4}}dx&=\int \frac{1}{x(1+1/x^4)^{1/4}}dx\\
&=\int \frac{x^4}{x^5(1+1/x^4)^{1/4}}dx.
\end{align*}$$
Substitution: $z^4=(1+1/x^4)$
$4z^3 dz=-4\frac{1}{x^5}dx$
Therefore,
$$\begin{align*}
\int\frac{1}{(1+x^4)^{1/4}}dx&=\int \frac{x^4}{x^5(1+1/x^4)^{1/4}}dx\\
&=-\int\frac{z^2}{(z^4-1)}dz\\
&=-\frac{1}{2}(\frac{1}{z^2-1}+\frac{1}{z^2+1})dz\\
&=-\frac{1}{2}\ln\left|\frac{1-z}{1+z}\right|+\arctan z+ C
\end{align*}$$
where $z=(1+1/x^4)^{1/4}$.
| {
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Where is the mistake in this incorrect proof in Eisenstein integers? The Diophantine equation is $n^2 + n + 1 = m^3$ my attempt to solve it shows there is no solutions to this equation, but in fact there are four. I could not find my mistake so I hope someone could point it out please.
Factoring in the Eisenstein integers $\mathbb{Z}[\omega]$ with $\omega^2 + \omega + 1 = 0$: $n^2 + n + 1 = (n - \omega)(n + 1 + \omega)$.
If $d$ is a common divisor then $d$ also divides their difference $1 + 2 \omega$ which is a prime.
If the greatest common divisor is $1$ then both numbers are cubes, but that is impossible because $(a + b \omega)^3 = (a^3 - 3 a b^2 + b^3) + 3 (a^2 b - a b^2)\omega$ and the $\omega$ term of the factors are not multiples of $3$.
If the greatest common divisor is $1 + 2 \omega$ then $(n - \omega)(n + 1 + \omega) = -3 \frac{n - \omega}{1+2 \omega}\frac{n + 1 + \omega}{1 + 2 \omega}$ so either $n - \omega$ or $n + 1 + \omega$ is $1 + 2 \omega$ times a cube, but that would imply $2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 = (a-2b)(a+b)(2a-b) = \pm 1$ which is impossible.
Edit: Does he make the same mistake here Link ?
| From
$$\mbox{cube}= -3 \frac{n - \omega}{1+2 \omega}\frac{n + 1 + \omega}{1 + 2 \omega}$$ doesn't the inference
"so either $n−\omega\ $ or $n+1+\omega\ $ is $1+2\omega\ $ times a cube"
also need $-3$ to be prime? This is not true in the Eisenstein integers.
| {
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How can I can solve integrals of rational functions of polynomials in $x$? I would like to know a way to solve integrals such as this one:
$$\int \frac{x}{3x - 4}dx$$
Also, I assume similar integrals where x is squared are solved in a similar manner. (If the answer is yes then don't also show me how to solve this second one, I want to see if I can do it myself. :) )
$$\int \frac{x^2}{x^2 - 1}dx$$
(I haven't included the conditions that x must meet in order for those to be valid expressions.)
| Whenever you are trying to integrate a rational function, the first step is to do the division so that the numerator is of degree strictly smaller than the numerator (this is what Eugene Bulkin and J.M. are saying in the comments). For example, for
$$\int \frac{x}{3x-4}\,dx$$
you should do the division of $x$ by $3x-4$ with remainder. This is
$$x = \frac{1}{3}(3x-4) + \frac{4}{3}$$
which means that
$$\frac{x}{3x-4} = \frac{1}{3} + \frac{4/3}{3x-4}.$$
So the integral can be rewritten as
$$\int \frac{x}{3x-4}\,dx = \int\left(\frac{1}{3} + \frac{4/3}{3x-4}\right)\,dx = \int\frac{1}{3}\,dx + \frac{4}{3}\int \frac{1}{3x-4}\,dx.$$
The first integral is immediate. The second integral yields to a change of variable $u=3x-4$. We get
$$\begin{align*}
\int\frac{x}{3x-4}\,dx &= \int\frac{1}{3}\,dx + \frac{4}{3}\int\frac{1}{3x-4}\,dx\\
&= \frac{1}{3}x + \frac{4}{9}\int\frac{du}{u}\\
&= \frac{1}{3}x + \frac{4}{9}\ln|u| + C\\
&= \frac{1}{3}x + \frac{4}{9}\ln|3x-4| + C.
\end{align*}$$
In general, if you have a denominator of degree $1$, by doing the long division you can always express it as a polynomial plus a rational function of the form
$$\frac{k}{ax+b}$$
with $k$, $a$, and $b$ constants. The polynomial is easy to integrate, and the fraction can be integrated with a change of variable.
The same is true for your second integral. Doing the long division gives, as you note, that
$$\int \frac{x^2}{x^2-1}\,dx = \int\left(1 + \frac{1}{x^2-1}\right)\,dx = \int\,dx + \int\frac{1}{x^2-1}\,dx.$$
The first integral is easy. The second is as well, using partial fractions:
$$\frac{1}{x^2-1} = \frac{1}{(x+1)(x-1)} = \frac{1/2}{x-1} - \frac{1/2}{x+1}$$
so:
$$\int\frac{1}{x^2-1}\,dx = \frac{1}{2}\int\frac{dx}{x-1} - \frac{1}{2}\int\frac{dx}{x+1} = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1|+C.$$
See also some of the comments in this answer on solving integrals by partial fractions.
| {
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Simpler way to compute a definite integral without resorting to partial fractions? I found the method of partial fractions very laborious to solve this definite integral :
$$\int_0^\infty \frac{\sqrt[3]{x}}{1 + x^2}\,dx$$
Is there a simpler way to do this ?
| By using techniques of complex analysis ($\text{Residue Theory}$) one can actually show that $$\int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx} = \frac{\pi}{b \sin(\pi{a}/b)}, \qquad 0 < a <b$$
You can obtain the value of your $\text{Integral}$ by putting $a=\frac{4}{3}$ and $b=2$.
Set $$I = \int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx}$$ and integrate $$f(z) = \frac{z^{a-1}}{1+z^{b}} = \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}}$$
Simple pole at $z_{1} = e^{\pi{i}/b}$ and hence $$\text{Res} \Biggl[\frac{z^{a-1}}{1+z^{b}}, e^{\pi{i}/b}\Biggr] = \frac{z^{a-1}}{bz^{b-1}}\Biggl|_{z =e^{\pi i / b}} = -\frac{1}{b}e^{\pi i a/b}$$
Integrate along $\gamma_{1}$, and let $R \to \infty$ and let $ \epsilon \to 0^{+}$. This gives,
\begin{align*}
\int\limits_{\gamma_{1}} f(z) \ dz & = \int\limits_{\gamma_{1}} \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}} \ dz \\ &= \int\limits_{\epsilon}^{R} \frac{x^{a-1}}{1+x^{b}} \to \int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ dx =I
\end{align*}
Integrate along $\gamma_{2}$, and let $R \to \infty$. This gives $0 < a < b$ and $$\Biggl|\int\limits_{\gamma_{2}} f(z) dz \Biggr| \leq \frac{R^{a-1}}{R^{b}-1} \cdot \frac{2\pi R}{b} \sim \frac{2 \pi}{b R^{b-a}} \to 0$$
Integrate along $\gamma_{3}$ and let $R \to \infty$ and $\epsilon \to 0^{+}$. This gives
\begin{align*}
\int\limits_{\gamma_{3}} f(z) \ dz &= \int\limits_{\gamma_{3}} \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}} = \Biggl[\begin{array}{c} z=x e^{2\pi i/b} \\ dz=e^{2\pi i/b} \ dx \end{array}\Biggr] \\ &= \int\limits_{R}^{\epsilon} \frac{x^{a-1}e^{2\pi i(a-1)/b}}{1+x^{b}} \cdot e^{2\pi i b} \ dx \to \int\limits_{\infty}^{0} \frac{x^{a-1}e^{2\pi i(a-1)/b}}{1+x^{b}} \cdot e^{2\pi i b} \ dx \\ &= -e^{2\pi ia/b}I
\end{align*}
Integrate along $\gamma_{4}$ and let $\epsilon \to 0^{+}$. This gives $0 < a <b$, $$\Biggl|\int\limits_{\gamma_{4}} f(z) \ dz \Biggr| \leq \frac{\epsilon^{a-1}}{1-\epsilon^{b}} \cdot \frac{2\pi\epsilon}{b} \sim \frac{2\pi\epsilon}{b} \to 0$$
Using the $\text{Residue Theorem}$ and letting $R \to \infty$ and $\epsilon \to 0^{+}$, we obtain that $$ I + 0 - e^{2\pi a/b}I + 0 = 2\pi i \cdot \Bigl(-\frac{1}{b} e^{\pi ia/b}\Bigr)$$ This yields, $$(e^{-\pi i a/b} - e^{\pi i a./b})I= -\frac{2\pi i}{b}$$ and hence solving for $I$, we have $$I= \frac{2\pi i}{b \cdot (e^{\pi ia/b} - e^{-\pi i a/b})}=\frac{\pi}{b \sin(\pi a/b)}$$
| {
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A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$? I am currently trying to show that $\int_{-\infty}^\infty \cos(x^2) \, \mathrm dx = \sqrt{\frac{\pi}{2}}$ and the last integral I have to evaluate is
$$\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx.$$
Now of course I'm familiar with wolframalpha, however the way it solves this integral seems very awkward and also not elegant to me, even though the function to me looks quite simple. So, is there a simpler way to solve this integral or is the way described on wolframalpha already (one of) the simplest approach(es)? I ask this because often wolframalpha doesn't see tricks (occurred to me when I wanted to find a formula for the n-th derivative of some function) which a human eye might see.
Thanks for any answers in advance.
| $\int_{-a}^a\dfrac{x^2}{x^4+1}~dx$
$=\int_{-a}^a\dfrac{1}{x^2+\dfrac{1}{x^2}}~dx$
$=\dfrac{1}{2}\int_{-a}^a\dfrac{1+\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}~dx+\dfrac{1}{2}\int_{-a}^a\dfrac{1-\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}}~dx$
$=\dfrac{1}{2}\int_{-a}^a\dfrac{1+\dfrac{1}{x^2}}{\left(x-\dfrac{1}{x}\right)^2+2}~dx+\dfrac{1}{2}\int_{-a}^a\dfrac{1-\dfrac{1}{x^2}}{\left(x+\dfrac{1}{x}\right)^2-2}~dx$
$=\dfrac{1}{2}\int_{-a}^a\dfrac{d\left(x-\dfrac{1}{x}\right)}{\left(x-\dfrac{1}{x}\right)^2+2}+\dfrac{1}{2}\int_{-a}^a\dfrac{d\left(x+\dfrac{1}{x}\right)}{\left(x+\dfrac{1}{x}\right)^2-2}$
$=\dfrac{1}{2\sqrt2}\left[\tan^{-1}\dfrac{x-\dfrac{1}{x}}{\sqrt2}\right]_{-a}^a-\dfrac{1}{2\sqrt2}\left[\tanh^{-1}\dfrac{x+\dfrac{1}{x}}{\sqrt2}\right]_{-a}^a$
$=\dfrac{1}{\sqrt2}\tan^{-1}\dfrac{a-\dfrac{1}{a}}{\sqrt2}-\dfrac{1}{\sqrt2}\tanh^{-1}\dfrac{a+\dfrac{1}{a}}{\sqrt2}$
| {
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Where are these additional solutions coming from? Solve for $x$: $2\sin(2x)-\sqrt{2} = 0$ in interval $[0,2\pi)$
Step $1$: Add $\sqrt{2}$ and divide by $2$ to get $\sin(2x) = \dfrac{\sqrt{2}}{2}$
Step $2$: Set $2x$ equal to the angles where $\sin(x) = \dfrac{\sqrt{2}}{2}$:
$2x = \dfrac{\pi}{4}$ and $2x = \dfrac{3\pi}{4}$
Step $3$: Solve for $x$ by dividing by $2$: $x = \dfrac{\pi}{8}$ and $x = \dfrac{3\pi}{8}$
My textbook also lists $\dfrac{9\pi}{8}$ and $\dfrac{11\pi}{8}$ as additional solutions, anyone know where they may have came from? thanks
| Your step two is correct except for a minor omission. More properly,
$2x = \displaystyle \frac{\pi}{4} + 2k\pi$, $k$ integer, and
$2x = \displaystyle \frac{3\pi}{4} + 2k\pi$, $k$ integer
Dividing these two expressions by 2 yield
$x = \displaystyle \frac{\pi}{8} + k\pi$, $k$ integer, and
$x = \displaystyle \frac{3\pi}{8} + k\pi$, $k$ integer
While $k=0$ gives the solutions you have, $k=1$ gives solutions that are ALSO in the interval $[0,2\pi]$. That's why your textbook has two additional solutions.
| {
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"url": "https://math.stackexchange.com/questions/38583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $\sqrt{3}\tan\theta=2\sin\theta$ I am trying to solve $\sqrt{3}\tan\theta=2\sin\theta$ on the interval $[-\pi,\pi]$.
$$\sqrt{3}\tan\theta=2\sin\theta \Rightarrow \sqrt{3}=\frac{2\sin\theta}{\tan\theta}$$
$$\Rightarrow \sqrt{3}=2\sin\theta \cdot \frac{\cos\theta}{\sin\theta} \Rightarrow 3 = 4\cos^2\theta$$
I get $\displaystyle \cos \theta = \pm{\frac{\sqrt{3}}{2}}$; the cosine of $30^{\circ}$ and $150^{\circ}$ so arrived at the solutions $\displaystyle -\frac{5}{6}\pi,-\frac{1}{6}\pi,\frac{1}{6}\pi,\frac{5}{6}\pi$.
Looking in the back of the book (and checking with Wolfram), the answer is $\displaystyle -\pi,-\frac{1}{6}\pi,0,\frac{1}{6}\pi,\pi$.
Where am I going wrong please?
| $$\sqrt{3}\tan\theta=2\sin\theta$$
$$\sqrt{3}\frac{\sin\theta}{\cos\theta}=2\sin\theta$$
$$\sqrt{3}\sin\theta-2\sin\theta\cos\theta=0$$
$$\sin\theta(\sqrt3-2\cos\theta)=0$$
$$\sin\theta=0, \ \ \cos\theta=\frac{\sqrt3}{2}$$
$$\theta=k\pi , \ \ \ \theta=2k\pi\pm \frac{\pi}{6}$$
For given interval, $\theta\in[-\pi, \pi]$, we get
$$\color{blue}{\theta=-\pi, -\frac{\pi}{6}, 0, \frac{\pi}{6}, \pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/40077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
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Define term using Stirling Numbers I am trying to solve the following exercise - quite unsuccessful yet.
Let a(m,n) be defined as
$$ \sum\limits_{n=0}^m a(m,n) \prod\limits_{i=1}^n (x+i-1) = x^m $$
Express a(m,n) using S(m,n) while S(m,n) are the Stirling numbers of the second kind which count the number of ways to partition a set of n elements into k nonempty subsets.
Hint: use the following identity : $$x^m = \sum\limits_{n=0}^m S(m,n) \cdot x \cdot (x-1) \cdots (x-n + 1) $$
First I rewrote the "hint"-identity as
$$ x^m = \sum\limits_{n=0}^m S(m,n) \prod\limits_{i=1}^n (x+1-i)$$
and got
$$ m = 0 \rightarrow a(0,0) = x^0 = S(0,0) $$
$$ m = 1 \rightarrow a(1,0) + a(1,1) \cdot x = S(1,0) + S(1,1) \cdot x $$
and m = 2
$$ a(2,0) + a(2,1) \cdot x + a(2,2) \cdot x \cdot (x+1) = S(2,0) \cdot x + S(2,1) \cdot x + S(2,2) \cdot x \cdot (x-1)$$
and both compared for m = 3
$$ \begin{array}{llll} a(3,0) & + a(3,1) \cdot x & + a(3,2) \cdot x \cdot (x+1) & + a(3,3) \cdot x \cdot (x+1) \cdot (x+2) \\ \underbrace{S(3,0) \cdot x}_{\text{always 0}} & +S(3,1) \cdot x & +S(3,2) \cdot x \cdot (x-1) & +S(3,3) \cdot x \cdot (x-1) \cdot (x-2) \end{array}
$$
Replacing x with -x in the "hint"-identity as recommended by user9325 results in
$$ \begin{array}{llll} a(3,0) & + a(3,1) \cdot x & + a(3,2) \cdot x \cdot (x+1) & + a(3,3) \cdot x \cdot (x+1) \cdot (x+2) \\ S(3,0) & +S(3,1) \cdot (-x) & +S(3,2) \cdot (-x) \cdot (-x-1) & +S(3,3) \cdot (-x)(-x-1)(-x-2) \end{array} $$
Multiplying each summand of the already modified identity by $(-1)^{(n+1)}$ gets
$$ \begin{array}{llll} a(3,0) & + a(3,1) \cdot x & + a(3,2) \cdot x \cdot (x+1) & + a(3,3) \cdot x \cdot (x+1) \cdot (x+2) \\ S(3,0) & +S(3,1) \cdot x & +S(3,2) \cdot x \cdot (x+1) & +S(3,3) \cdot x\cdot(x+1)\cdot(x+2) \end{array} $$
Is this correct? How do I put this altogether?
| The shortest way to find the answer is to replace $x$ by $-x$ in one of the identities and then compare them.
| {
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The sum of $(-1)^n \frac{\ln n}{n}$ I'm stuck trying to show that $$\sum_{n=2}^{\infty} (-1)^n \frac{\ln n}{n}=\gamma \ln 2- \frac{1}{2}(\ln 2)^2$$
This is a problem in Calculus by Simmons. It's in the end of chapter review and it's associated with the section about the alternating series test. There's a hint: refer to an equation from a previous section on the integral test. Specifically:
$$L=\lim_{n\to\infty} F(n)=\lim_{n\to\infty}\left[a_1+a_2+\cdots+a_n-\int_1^n\! f(x)\,\mathrm{d}x\right]$$
Here, $\{a_n\}$ is a decreasing sequence of positive numbers and $f(x)$ is a decreasing function such that $f(n)=a_n$, and $\gamma$ is this limit in the case that $a_n=\frac{ 1}{n}$.
New users can't answer their own questions inside of 8 hours, so I'm editing my question to reflect the answer.
Ok, I got it.
Following the hint in the book
$$L=\lim_{n\to\infty}\left[\frac{\ln 2}{2}+\frac{\ln 3}{3}+\cdots+\frac{\ln n}{n}-\int_2^n\! \frac{\ln x}{x}\,\mathrm{d}x\right]$$
$$=\lim\left[\frac{ \ln 2}{2}+\cdots+\frac{ \ln n}{n}-\left.\frac{ \ln^2x}{2}\right|_2^n\right]$$
The partial sum for the positive series is:
$$\left(\frac{\ln^2n}{2}-\frac{\ln^2}{2}\right)+L+o(1)$$
Returning to the original, alternating series:
$$-S_{2n}=\frac{ -\ln 2}{2}+\frac{ \ln 3}{3}-\frac{\ln 4}{4}+\frac{\ln 5}{5}-\cdots$$
$$=\frac{\ln 2}{2}+\frac{\ln 3}{3}+\cdots+\frac{\ln 2n}{2n}-2\left(\frac{\ln 2}{2}+\frac{\ln 4}{4}+\cdots+\frac{\ln 2n}{2n}\right)$$
Consider the partial sum in parentheses
$$ \frac{\ln 2}{2}+\frac{\ln 4}{4}+\cdots+\frac{\ln 2n}{2n}=\frac{\ln 2}{2}+\frac{\ln 2 +\ln 2}{4}+\frac{\ln 2+\ln 3}{6}+\cdots+\frac{\ln 2+\ln n}{2n}$$
$$=\frac{1}{2}\left(\ln 2\left(1+\frac{ 1}{2}+\cdots+\frac{ 1}{n}\right)+\left(\frac{ \ln 2}{2}+\frac{ \ln 3}{3}+\cdots+\frac{ \ln n}{n}\right)\right)$$
Now, plug that back in
$$-S_{2n}=\left(\frac{\ln 2}{2}+\cdots+\frac{ \ln 2n}{2n}\right)-\ln 2\left(1+\frac{ 1}{2}+\cdots+\frac{ 1}{n}\right)-\left(\frac{ \ln 2}{2}+\frac{ \ln 3}{3}+\cdots+\frac{ \ln n}{n}\right)$$
$$=\frac{ \ln^2(2n)}{2}-\frac{ \ln^2 2}{2}+L+o(1)-\ln 2\left(\ln n +\gamma+o(1)\right)-\left(\frac{ \ln^2 n}{2}-\frac{ \ln^2 2}{2}+L+o(1)\right)$$
$$=\frac{ (\ln 2 +\ln n)^2}{2}-(\ln 2)(\ln n)-\gamma\ln 2-\frac{ \ln^2 n}{2}+o(1)$$
$$=\frac{ \ln^2 2}{2}+(\ln 2)(\ln n)+\frac{ \ln^2 n}{2}-(\ln 2)(\ln n)-\gamma \ln 2 - \frac{ \ln^2 n}{2}+o(1)$$
$$-S_{2n}\to\frac{ \ln^2}{2}-\gamma\ln 2$$
Which gives the desired result
$$\sum_2^{\infty}(-1)^n \frac{ \ln n}{n}=\gamma\ln 2 -\frac{ \ln^2 2}{2}$$
| Consider
$$P=\frac{\ln 2}{2}+\frac{\ln 3}{3}+\frac{\ln 4}{4}+...+\frac{\ln 2n}{2n}$$
$$Q=\frac{2\ln 2}{2}+\frac{2\ln 4}{4}+\frac{2\ln 6}{6}+...+\frac{2\ln 2n}{2n}=\left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n} \right)\ln 2+\frac{\ln 2}{2}+\frac{\ln 3}{3}+\frac{\ln 4}{4}+...+\frac{\ln n}{n}$$
$$Q-P=\sum\limits_{j=2}^{2n}{\frac{\left( -1 \right)^{j}}{j}\ln j}=\left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n} \right)\ln 2-\left( \frac{\ln \left( n+1 \right)}{n+1}+\frac{\ln \left( n+2 \right)}{n+2}+\frac{\ln \left( n+3 \right)}{n+3}+...+\frac{\ln 2n}{2n} \right)$$
$$=\left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n \right)\ln 2+\ln 2\ln n-\sum\limits_{j=1}^{n}{\frac{\ln n}{n+j}}-\sum\limits_{j=1}^{n}{\frac{1}{n}\cdot \frac{\ln \left( 1+\frac{j}{n} \right)}{1+\frac{j}{n}}}$$
$$\sum\limits_{j=2}^{+\infty }{\frac{\left( -1 \right)^{j}}{j}\ln j}=\gamma \ln 2-\int\limits_{0}^{1}{\frac{\ln \left( 1+x \right)}{1+x}dx}=\gamma \ln 2-\frac{\ln ^{2}2}{2}$$
Done
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $\sqrt{x+5} = x - 1$ I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:
$$\sqrt{x+5} = x - 1$$
So I used this logic:
$$
\begin{align}
\sqrt{x+5} &= x - 1 \\
x + 5 &= (x-1)^2 \\
x + 5 &= (x-1)(x-1) \\
x + 5 &= x^2 - 2x + 1 \\
0 &= x^2 - 3x - 4 \\
0 &= (x-4)(x+1) \\
\end{align}
$$
Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.
But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:
$$
\begin{align}
\sqrt{4 + 5} &= 4 - 1 \\
\sqrt{9} &= 3 \\
3 &= 3
\end{align}
$$
So using 4 works as expected, but when using $-1$:
$$
\begin{align}
\sqrt{-1 + 5} &= -1 - 1 \\
\sqrt{4} &= -2 \\
2 &\ne -2
\end{align}
$$
At what stage am I going wrong?
And according the WolframAlpha, the solution is $x = 4$.
| At no stage...
Here is what happens, you squared the equation
$$\sqrt{x+5} =x-1 \,.$$
But then, if the two sides of the equation have the same absolute value, but opposite signs they are not equal in this equation but they become equal at the next step.
This means that the equation $x+5=(x-1)^2$ possibly has more solutions. So the answers you got at the end are no necessarily the solution, they are just the possible solutions.
You have to check which one works.
+1 for showing your work :)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of eigenvectors versus eigenvectors of limit Consider the matrix
$$
A=\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & g
\end{pmatrix},
$$
where $g$ is a real parameter. If I set $g=0$ and calculate the normalized eigenvectors of $A|_{g=0}$ with Mathematica, I find that they are
$$
v_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\
v_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix},\
v_3 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.
$$
If instead I calculate the eigenvectors of $A$ leaving $g$ as an unknown and then take their limit as $g\to 0$, I find
$$
u_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\
u_2 = \frac{1}{\sqrt{6}}\begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix},\
u_3 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.
$$
My question is, why are these two sets of eigenvectors different?
| Both results are correct. $u_1$ and $u_2$ correspond to the same eigenvalue $-1$, and $\left( \matrix{-1\cr 0\cr 1\cr} \right) = \frac{1}{2} \left(\matrix{-1 \cr 1 \cr 0\cr} \right) + \frac{1}{2} \left(\matrix{-1 \cr -1 \cr 2\cr} \right)$, so $u_1$ and both versions of $u_2$ span the same vector space. Any nonzero vector in this space is an eigenvector for eigenvalue $-1$.
| {
"language": "en",
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Avoiding matching first digit of $a^n$ with $b^n$ For any given pairs of positive integers $a$ and $b$, is it possible that the first digit of $a^n$ never matches the first digit of $b^n$ for any positive integer $n$?
(If $a=2$ and $b=5$ the only possible matching first digit is "3".)
Edit: I meant to say $a=2$ in the parenthetical comment above. I had previously written $a=3$, which does not work as several comments indicate.
| I'm not answering the question, but justifying the claim in the last sentence of the question: when $a=2$ and $b=5$ the only shared leading digit of $a^n$ and $b^n$ is 3.
Suppose $2^n$ and $5^n$ share leading digit $d$ and have $r+1$ and $s+1$ digits, resp. For $n>3$ we have $d10^r < 2^n < (d+1)10^r$ and $d10^s < 5^n < (d+1)10^s$. Therefore $d^2 10^{r+s} < 10^n < (d+1)^2 10^{r+s}$ so $d^2 < 10^{n-r-s} < (d+1)^2$. Since $d$ is a digit, $d \geq 1$ and $(d+1) \leq 10$; these restrictions force $n-r-s = 1$, so $d^2 < 10 < (d+1)^2$ meaning only $d=3$ works. And since $2^5 = 32$ and $5^5 = 3125$ it does in fact occur as a leading digit.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluating $\int_{0}^{1} \frac{dx}{1+{}_2F_{1}\left(\frac{1}{n},x;\frac{1}{n};\frac{1}{n}\right)}$ On a lark (as a followup to this question), I was playing around with Wolfram alpha, and it seems that
$$\int_{0}^{1} \frac{dx}{1+{}_2F_{1}\left(\frac{1}{n},x;\frac{1}{n};\frac{1}{n}\right)} = \frac{\log\left(\frac{2n}{2n-1}\right)}{\log\left(\frac{n}{n-1}\right)}$$
Could someone take a stab at proving this?
| First note that $${}_2F_{1}\left(\frac{1}{n},x;\frac{1}{n};\frac{1}{n}\right) = 1 + \frac{x}{n} + \frac{x(x+1)}{2! n^2} + \frac{x(x+1)(x+2)}{3! n^3} + \frac{x(x+1)(x+2)(x+3)}{4! n^4} + \cdots$$
$$ = 1 + (-x) \frac{-1}{n} + \frac{(-x)(-x-1)}{2!} \left( \frac{-1}{n} \right)^2 + \frac{(-x)(-x-1)(-x-2)}{3!} \left( \frac{-1}{n} \right)^3 + \cdots$$
$$ = \left(1 - \frac1{n} \right)^{-x}$$
Hence,
$${}_2F_{1}\left(\frac{1}{n},x;\frac{1}{n};\frac{1}{n}\right) = \left(\frac{n}{n-1} \right)^x$$
Hence,
$$\int_{0}^{1} \frac{dx}{1+{}_2F_{1}\left(\frac{1}{n},x;\frac{1}{n};\frac{1}{n}\right)} = \int_{0}^{1} \frac{dx}{1+\left(\frac{n}{n-1} \right)^x} = \int_{0}^{1} \frac{dx}{1+e^{kx}}$$ where $e^k = \frac{n}{n-1}$
Hence, we are interested in evaluating an integral of the form $$\int_0^1 \frac1{1+e^{kx}} dx$$
Let $t = 1 + e^{kx}$. We get $dt = k e^{kx} dx = k (t-1) dx$
$$\int_0^1 \frac1{1+e^{kx}} dx = \int_2^{1+e^k} \frac1{k} \frac{dt}{t(t-1)} = \frac1{k} \int_2^{1+e^k} \left(\frac{dt}{t-1} - \frac{dt}{t} \right) = \frac1{k} \left( k - \log(e^k+1) + \log(2) \right)$$
In our case, $e^k = \frac{n}{n-1}$ and hence the integral becomes
$$1 - \frac{\log(2n-1) - \log(n-1)}{\log(n) - \log(n-1)} + \frac{\log(2)}{\log(n) - \log(n-1)} = \frac{\log \left( \frac{2n}{2n-1} \right)}{\log \left(\frac{n}{n-1} \right)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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How to find remainder modulo $n$, when $n$ is a large number I am doing RSA questions and I really could use help! Can someone show me a simple way to find $25^9 \pmod{33}$?
| For variety, let's invoke the Chinese remainder theorem
$$ 25^9 \equiv 1^9 \equiv 1 \pmod 3 $$
$$ 25^9 \equiv 3^9 \pmod{11}$$
That latter one isn't so bad to compute by repeated squaring:
$$ 3^9 \equiv 3 \cdot 3^8 \equiv 3 \cdot 9^4 \equiv 3 \cdot 81^2 \equiv 3 \cdot 4^2 \equiv 3 \cdot 5 \equiv 4 \pmod{11}$$
but I'm even lazier. Since $3^{10} \equiv 1 \pmod{11}$ (because $11$ is prime):
$$ 3^9 \equiv 3^{-1} \equiv 1/3 \equiv 12/3 \equiv 4 \pmod{11} $$
The extended Euclidean algorithm (or inspection) lets us compute
$$ 4 \cdot 3 + (-1) \cdot 11 = 1 $$
and therefore
$$ 25^9 \equiv 1 \cdot (-1 \cdot 11) + 4 \cdot (4 \cdot 3)
\equiv -11 + 48 \equiv -11 + 15 \equiv 4 \pmod{33} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/44533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Anyone know if this integral has an analytic solution? I've come across the following integral:
$$\int_{-\pi}^{\pi}\left[\frac{1}{A-R \cos(2\theta-\phi)}\right]^{\frac{N-1}{2}}d\theta$$
I know how to approximate this integral using the Laplace method, just wondering if:
a) Does this integral have an exact answer?
b) Is there a better approximation than Laplace method for this integral? If so, under what conditions will it be better?
My thinking is that it will be a hypergeometric function (mainly because every hard integral I've come across turns out to be one of these). Conditions (if needed) are $A>R>0$, and $N$ is an integer.
| From Maple... $K$ and $E$ are elliptic integrals.
$$
\int_{0}^{\pi} \sqrt{\frac{1}{2 - \operatorname{cos} (2 t)}} d t = \frac{2 \sqrt{3} K \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{3}
$$ $$
\int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)} d t = \frac{\pi \sqrt{3}}{3}
$$ $$
\int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)}^{\frac{3}{2}} d t = \frac{2 \sqrt{3} E \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{3}
$$ $$
\int_{0}^{\pi} \bigl(2 - \operatorname{cos} (2 t)\bigr)^{(-2)} d t = \frac{2 \pi \sqrt{3}}{9}
$$ $$
\int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)}^{\frac{5}{2}} d t = \frac{-2\sqrt{3} K \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{27} + \frac{16 \sqrt{3} E \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{27}
$$ $$\int_{0}^{\pi} \bigl(2 - \operatorname{cos} (2 t)\bigr)^{(-3)} d t = \frac{\pi \sqrt{3}}{6}
$$ $$
\int_{0}^{\pi} \frac{1}{2 - \operatorname{cos} (2 t)}^{\frac{7}{2}} d t =
\frac{-32\sqrt{3} K \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{405} + \frac{202 \sqrt{3} E \Bigl(\frac{\sqrt{3} \sqrt{2}}{3}\Bigr)}{405}
$$ $$
\int_{0}^{\pi} \bigl(2 - \operatorname{cos} (2 t)\bigr)^{(-4)} d t = \frac{11 \pi \sqrt{3}}{81}
$$
This suggests a reduction formula, together with the first two.
| {
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Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accordingly, $a^2 +b^2 + c^2 \geq ab + ac +bc~$
Multiplying both sides by $(a + b + c)$:
$(a^2 +b^2 + c^2)(a+b +c) \geq (ab + ac +bc)(a + b + c)~~$ and simplifying this, $ a^3 + b^3 + c^3 + \Sigma a^2 b \geq 3abc + \Sigma a^2 b $
Therefore, it follows that $a^3 + b^3 + c^3 \geq 3abc~$, and letting $a^3 = x~$, $b^3 = y~$, $c^3 = z~$: $x + y + z\geq 3\sqrt[3]{xyz}$
Cubing both sides, $(x + y + z)^3 \geq 27 xyz~$ which was to be proven.
I was wondering if there are alternative approaches to solve this problem (possibly using higher-level maths), and is my proof entirely correct?
| I know a nice proof. It goes like this:
Let $x,y,z>0$. You know that $\frac{x+y}{2} \geq \sqrt{xy}$. This can be generalized for four numbers
$$\frac{a+b+c+d}{4}=\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2}\geq \sqrt[4]{abcd}.$$
Now pick $a=x,b=y,c=z,d=\sqrt[3]{xyz}$ and you'll get your inequality.
For $x,y,z$ not positive the inequality may not hold. Check $x=-1, y=-2, z=-3$.
| {
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Determinant of a special matrix The question is related to the eigenvalue problem. Using MAPLE, I calculated the following determinants:
$$\begin{align}
\begin{vmatrix}
-\lambda & 1 \\
1 & -\lambda \\
\end{vmatrix}&=\lambda^2-1\\
\begin{vmatrix}
-\lambda & 1 & 0 \\
1& -\lambda & 1 \\
0& 1 & -\lambda \\
\end{vmatrix}&=-\lambda^3+2\lambda\\
\begin{vmatrix}
-\lambda & 1 &0 &1 \\
1& -\lambda & 1 &0 \\
0& 1& -\lambda & 1 \\
1& 0& 1& -\lambda \\
\end{vmatrix}&=\lambda^4-4\lambda^2\\
\begin{vmatrix}
-\lambda &1 &0 &1 &0 \\
1& -\lambda &1 &0 &1 \\
0& 1& -\lambda &1 &0 \\
1& 0& 1& -\lambda &1 \\
0& 1& 0& 1& -\lambda \\
\end{vmatrix}&=-\lambda^5+6\lambda^3\\
\begin{vmatrix}
-\lambda &1 &0 &1 &0 &1 \\
1& -\lambda &1 &0 &1 &0 \\
0& 1& -\lambda &1 &0 &1 \\
1& 0& 1& -\lambda &1 &0 \\
0& 1& 0& 1& -\lambda &1 \\
1& 0& 1& 0&1 & -\lambda \\
\end{vmatrix}&=-9\lambda^4+\lambda^6\\
\end{align}
$$
But I have no idea how to calculate the determinants quickly by hand. Here is my question:
What is the determinant in the $n$ by $n$ case?
| Based on the given results, you should be able to guess what the eigenvalues are. Can you write down an explicit set of eigenvectors for those eigenvalues? Does your construction generalize?
| {
"language": "en",
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"source": "stackexchange",
"question_score": "15",
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Which integers are the areas of squares with vertices in the 3d integer lattice? For which integers $n$ does there exist a square of area $n$ with vertices in the 3d integer lattice $\mathbb{Z}^3$?
A sufficient condition is that $n$ is a square or the sum of two squares, and I have verified that the condition is also necessary when $n < 10^5$.
Edit: This question was posed by James Tanton on Twitter. I thought it was very interesting, so I took the liberty of posting it here.
Edit 2: I have extended the search to $n < 10^6$ without finding any counterexamples.
| With Qiaochu's observation, one can find a general parametric solution to this problem, starting from:
$a^2 + b^2 + c^2 = d^2 + e^2 + F^2$
$a d + b e + c F = 0$
Start by forgetting about integers for now, and dividing both equations by $c^2$, and considering $a, b, d, e, F$ as rational. In other words, in effect take $c = 1$.
Then plugging $- F = a d + b e$ into the first gives a result equivalent to:
$(a^2 + 1) d^2 + 2 a b d e + (b^2 + 1) e^2 = a^2 + b^2 + 1$
Multiplying throughout by $a^2 + 1$, this can be expressed in the form:
$((a^2 + 1) d + a b e)^2 = (a^2 + 1 - e^2) (a^2 + b^2 + 1)$
Letting:
$(a^2 + 1) d + a b e = (a^2 + b^2 + 1) f$
this becomes:
$a^2 + 1 - e^2 = (a^2 + b^2 + 1) f^2$
or equivalently:
$(a^2 + 1) (1 - f^2) = e^2 + (b f)^2$
It isn't hard to prove that this implies the existence of rational $g, h$ with:
$1 - f^2 = g^2 + h^2$
whence by composition:
$e^2 + (b f)^2 = (a g + h)^2 + (a h - g)^2$
This in turn implies the existence of rational $u, v$ with:
$u^2 + v^2 = 1$
such that again by composition:
$e = u (a g + h) + v (a h - g)$
$b f = v (a g + h) - u (a h - g)$
Now $f^2 + g^2 + h^2 = 1$ has general solution:
$f = \frac{p^2 + q^2 - 1}{p^2 + q^2 + 1}$
$g = \frac{2 p}{p^2 + q^2 + 1}$
$h = \frac{2 q}{p^2 + q^2 + 1}$
and $u^2 + v^2 = 1$ of course has general solution:
$u = \frac{r^2 - 1}{r^2 + 1}$
$v = \frac{2 r}{r^2 + 1}$
So plugging these two solutions into the preceding equations for $e$ and $b f$ expresses $b$ and $e$ in terms of $a, p, q, r$.
Also plugging these two solutions, and $b$ and $e$ just obtained, into the equation (near
the start) where $f$ was introduced, expresses $d$ in terms of $a, p, q, r$.
Finally, $F$ follows from $- F = a d + b e$.
Then simply homogenize to obtain equations giving all integer solutions.
I later realized that the above, although correct, is suboptimal.
Solving $a^2 _+ b^2 + 1 = d^2 + e^2 + (a d - b e)^2$, as we did, by expressing $b, d, e$ in terms of $a$ and parameters should require only 2 parameters instead of our 3.
This mystery can be resolved by observing that $u^2 + v^2 (= 1)$ can be absorbed by composition into $g^2 + h^2$. So with a pair of new parameters $G, H$, with $1 - f^2 = G^2 + H^2$ we can conclude:
$e = a G + H$
$b f = a H - G$
and $b, d, e, F$ can each now be expressed in terms of $a, G, H$
| {
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"source": "stackexchange",
"question_score": "15",
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I need explanation for this solution for the proof. (Perfect square ends with 0,1,4,5,6,9) Give a proof to the sentence: "The final decimal digit of a perfect square is 0, 1, 4, 5, 6 or 9."
Solution:
A integer $n$ can be expressed as $10a+b$, where $a$ and $b$ are positive integers and $b$ is 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9.
Here $a$ is the integer obtained by subtracting the final decimal digit of $n$ and dividing by 10. (so $a=(n-b)/10$)
Next note that $(10a + b)^2 = 100a^2+20ab+b^2=10(10a^2+2b)+b^2$ so the final decimal digit of $n^2$ is same as the final decimal digit of $b^2$.
I understand until this point but not below:
Furthermore, note that the final decimal digit of $b^2$ is the same as final decimal digit of $(10-b)^2 = 100 - 20ab + b^2$. (how did you get this equation?) consequently we cab reduce our proof to the consideration of fix cases.
final digit of:
1) $n$ is 1 or 9 is 1
2) $n$ is 2 o 8 is 4
3) $n$ is 3 or 7 is 9
4) $n$ is 4 or 6 is 6
5) $n$ is 5 is 5
6) $n$ is 0 is 0
THANKS!
| Well, it should say "note that the final decimal digit of $b^{2}$ is the same as the final decimal digit of $(10-b)^{2} = 100-20b+b^{2}$." Since $100-20b+b^{2}=10(10-2b)+b^{2}$, the only number affecting the units digit of $(10-b)^{2}$ is $b^{2}$. Thus, the units digit of $b^{2}$ is the same as the units digit of $(10-b)^{2}$.
Note that this whole part of the proof is completely unnecessary. They already showed that since any integer can be written in the form $10a+b$ and since $$(10a+b)^{2}=100a^{2}+20ab+b^{2}=10(10a^{2}+2ab)+b^{2}$$ the only part that matters in determining the units digit is $b^{2}$. From here, you could simply check all integer values of $b$ from 0 to 9, but they instead slightly simplified the number of cases needed by showing that $b^{2}$ and $(10-b)^{2}$ have the same units digit.
| {
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"source": "stackexchange",
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Showing $\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}$ How can one show :
$$\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}?$$
| HINT: The number $\alpha=\sqrt{65+18\sqrt{13}}$ is one of the four solutions of the equation $(\alpha^2-65)^2-4212=0$. Then, show that the LHS is also a solution of the same equation. The other three solutions must be disregarded somehow.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/55120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
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} |
Get number of elements of a square matrix given a vector that has upper right elements of that matrix Suppose I have some vectors:
*
*$[1,2,1]$ its length is $3$ wich represents a matrix like $\begin{pmatrix}
1 & 2\\
-1 & 1
\end{pmatrix}$ the complete matrix would have $4$ elements
*$[1,2,1,3,4,1]$ its length is $6$ wich represents $\begin{pmatrix}
1& 2& 3\\
-1& 1 & 4\\
-1& -1 &1
\end{pmatrix}$ the complete matrix would have $9$ elements
*$[1,2,1,3,4,1,5,6,7,1]$ its length is $10$ wich represents $\begin{pmatrix}
1 & 2& 3& 5\\
-1 &1 & 4& 6\\
-1 &-1 &1 & 7\\
-1 &-1 &-1 &1
\end{pmatrix}$ the complete matrix would have $16$ elements
*$[1,2,1,3,4,1,5,6,7,1,8,9,10,11,1]$ its length is $15$ wich represents
$\begin{pmatrix}
1 &2 &3 &5 &8 \\
-1 &1 &4 &6 &9 \\
-1 &-1 &1 &7 &10 \\
-1 &-1 &-1 &1 &11 \\
-1 &-1 &-1 &-1 &-1
\end{pmatrix}$ the complete matrix would have $25$ elements
How do I find a general formula to get all elements of a square matrix from the number of elements in a vector?
So
*
*$3\to 4$
*$6\to 9$
*$10\to 16$
*$15\to 25$
*etc...
also: How would you know the matrix size $2\times 2$, $3\times 3$,...$n\times n$?
| Notice that all of your matrices are $n^2$ in size, where $n$ is the number of rows (or columns), and all of your vectors are 1+2+3+...+n = $n(n+1)/2$ in size. Your function takes numbers $n(n+1)/2$ to $n^2$, where $n$ is an integer. If you want to know what $n$ is from the length of your vector, just solve the quadratic equation: e.g. if your vector is 15 in length, solve n(n+1)/2 = 15. You get n = 5 or n = -6, and you should obviously disregard the latter because it's negative. Then you can square 5 to get the size of your matrix.
Alternatively: 3, 6, 10, 15, ... are the numbers of elements in the upper right of your matrix, including the diagonal. These elements (marked with a star below) form a 'triangle' shape in your matrix, of side length 2, 3, 4, 5 respectively:
$\underbrace{\begin{pmatrix} *&*&*&*&* \\ &*&*&*&* \\ &&*&*&* \\ &&&*&* \\ &&&&* \end{pmatrix}}_{\text{length }5}$
The "missing" elements in the lower left ('marked' as a blank space above) form another slightly smaller 'triangle' shape of size 1, 2, 3, 4 etc. respectively - that is, the lower left has 1, 3, 6, 10, ... elements in it.
These are obviously the same sequence, just shifted along by 1: that is, the upper right of a matrix of side length 10 (with the diagonal) will have as many elements as the lower left of a matrix of side length 11 (without the diagonal). So to get the total number of elements in your matrix, just add the number in the upper right to the number in the lower left: 3+1, 6+3, 10+6, 15+10, ... In recurrence relation speak, if $n_i$ is the number of elements in your vector, the number of elements in your matrix is $n_{i-1} + n_i$.
| {
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Complex number + trigo : $-1 + \tan(3)i$ , find modulus and argument I have $-1 + \tan(3)i$ and must find its modulus and its argument. I tried to solve it by myself for hours, and then I looked at the answer, but I am still confused with a part of the solution.
Here is the provided solution:
$$\begin{align}
z &= -1 + \tan(3)i \\
&= -1 + \frac{\sin(3)}{\cos(3)}i \\
&= \frac1{\left|\cos(3)\right|} ( \cos(3) + i(-1)\sin(3)) \\
&= \frac1{\left|\cos(3)\right|} e^{-3i} \\
&= \frac1{\left|\cos(3)\right|} e^{(2\pi-3)i}
\end{align}$$
I don't understand how we get to $$ \frac1{\left|\cos(3)\right|}(\cos(3) + i(-1)\sin(3)) $$
How did they get this modulus $1/|\cos(3)|$, and the $-1$ in the imaginary part? How did they reorder the previous expression to obtain this?
I also don't see why they developed the last equality. They put $2\pi-3$ instead of $-3$; OK, it is the same, but what was the aim of a such development?
Thanks!
| Let $z = -1 + \tan(3) \ i$. In the complex plane, this would be the point $(-1, \tan(3))$, which has length
$$|z| = \sqrt{(-1)^2 + \tan^2(3)} = \sqrt{1 + \frac{\sin^2(3)}{\cos^2(3)}} = \sqrt{\frac{1}{\cos^2(3)}} \sqrt{\cos^2(3) + \sin^2(3)} = \frac{1}{|\cos(3)|}$$
For the last equality, we used $\sin^2(x) + \cos^2(x) = 1$. Now we want to write
$$z = |z| \ e^{\omega i} = |z| \ (\cos(\omega) + i \sin(\omega))$$
for some $\omega$. It turns out this can be done easily by writing
$$z = \frac{1}{\cos(3)}(-\cos(3) + i \sin(3)) = \frac{1}{|\cos(3)|}(\cos(-3) + i \sin(-3)) = \frac{1}{|\cos(3)|} e^{-3i}$$
Since $-3 \notin [0, 2\pi)$ they decided to add $2 \pi$ to the angle, so that it is inside this interval.
| {
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Simplify using the Tangent Difference identity I solved a problem to the point that I know the answer is
$$\frac{2nr}{\tan\left( \frac{(n-2)\pi}{2n} \right)}$$
The question tells me that the answer is going to be
$$2rn\tan\left(\frac{\pi}{n}\right)$$
Wolfram Alpha tells me the two are equal, so at least I'm on the right track. [link]
I was able to simplify
$$\frac{2nr}{\tan\left( \frac{(n-2)\pi}{2n} \right)}$$ to $$\frac{2nr}{\tan\left( \frac{\pi}{2} - \frac{\pi}{n} \right)}$$
and then attempted to use the tangent difference identify,
$$\tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta}$$
to simplify further.
But I am unable to find $\tan\alpha$ due to the fact that $\tan\left(\frac{\pi}{2}\right)$ is undefined. How do I work around this?
(Sorry if I went overboard with my LaTeX, I was really excited to be able to use it.)
| First, as you note:
$$\tan\left(\frac{(n-2)\pi}{2n}\right) = \tan\left(\frac{\pi}{2}-\frac{\pi}{n}\right) =- \tan\left(\frac{\pi}{n}-\frac{\pi}{2}\right).$$
Now remember that $\sin\left(a - \frac{\pi}{2}\right) = -\cos(a)$ and $\cos\left(a - \frac{\pi}{2}\right) = \sin(a)$. So:
$$\begin{align*}
\tan\left(\frac{(n-2)\pi}{2n}\right) &= -\tan\left(\frac{\pi}{n}-\frac{\pi}{2}\right)\\
&= -\frac{\sin\left(\frac{\pi}{n}-\frac{\pi}{2}\right)}{\cos\left(\frac{\pi}{n}-\frac{\pi}{2}\right)}\\
&= -\frac{-\cos\frac{\pi}{n}}{\sin\frac{\pi}{n}}\\
&= \cot\frac{\pi}{n}.
\end{align*}$$
Therefore,
$$\frac{2nr}{\tan\left(\frac{(n-2)\pi}{2n}\right)} = \frac{2nr}{\cot\frac{\pi}{n}} = 2nr\tan\frac{\pi}{n}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Triples of Numbers I have a question:
How many triples $(a,b,c)$ are there such that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$$ and $a <b<c$? They have to be positive integers. Also find those triples.
I know that all of them have to be $\geq 2$. So do I just fix a number and count the other pairs?
If I choose $a = 3$ then I count the other pairs $(b,c)$? If I choose a very large $a$ then it seems that no triples will satisfy the condition since the sum will be too small.
| Well, $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} < 1$, so we must have $a=2$. So we really just need $\frac{1}{b} + \frac{1}{c} = \frac{1}{2}$.
Since $\frac{1}{4} + \frac{1}{5} < \frac{1}{2}$, $b = 3$. That leaves $c = 6$.
I think a nice way to think about it is to view the number $1$ as $\frac{1}{1}$. We can decompose $\frac{1}{n}$ into $\frac{1}{n+1} + \frac{1}{n^2+n}$, then decompose one of those to get an expression for $\frac{1}{n}$ as the sum of three harmonic numbers. In this case, we see that $\frac{1}{1} = \frac{1}{2} + \frac{1}{2} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$.
See the Leibniz Harmonic Triangle.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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finding the coefficient of $x^{14}$ in the expression: $\frac{5x^2-x^4}{(1-x)^3}$ I have a homework question which requires me to find the coefficient of $x^{14}$ in the expression: $\dfrac{5x^2-x^4}{(1-x)^3}$
I have not figured out a way to do this (I believe this is because my algebra is weak).
This is a question in a combinatorics class and has to do with Generating Functions if that is any help.
Help is greatly appreciated. Thanks, Jason
| You are after:
\begin{align}
[x^{14}] \frac{5 x^2 - x^4}{(1 - x)^3}
&= 5 [x^{12}] (1 - x)^{-3} - [x^{10}] (1 - x)^{-3} \\
&= 5 (-1)^{12} \binom{-3}{12} - (-1)^{10} \binom{-3}{10} \\
&= 5 \binom{12 + 3 - 1}{3 - 1} - \binom{10 + 3 - 1}{3 - 1} \\
&= 389
\end{align}
Here I used:
$$
\binom{-n}{k} = (-1)^k \binom{k + n - 1}{n - 1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/61879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Orthogonal projection of a point onto a line How would I go about solving the following problem?
Find an orthogonal projection of a point T$(-4,5)$ onto a line $\frac{x}{3}+\frac{y}{-5}=1$.
| The slope of the line $r$, with equation $\frac{x}{3}+\frac{y}{-5}=1$, is $m_{r}=\frac{5}{3}$ (because $\frac{x}{3}+\frac{y}{-5}=1$ is equivalent to $y=\frac{5}{3}x-5$). The slope of the line $s$ orthogonal to $r$ is $m_{s}=-\frac{3}{5}$ (because $m_{r}m_{s}=-1$). Hence the equation of $s$ is of the form
$$y=-\frac{3}{5}x+b_{s}.$$
Since $T(-4,5)$ is a point of $s$, we have
$$5=-\frac{3}{5}\left( -4\right) +b_{s},$$
which means that $b_{s}=\frac{13}{5}$. So the equation of $s$ is
$$y=-\frac{3}{5}x+\frac{13}{5}.$$
The coordinates of the orthogonal projection of $T$ onto $r$ are the solutions of the system
$$\left\{
\begin{array}{c}
y=\frac{5}{3}x-5 \\
y=-\frac{3}{5}x+\frac{13}{5},
\end{array}
\right. $$
which are $(x,y)=\left( \frac{57}{17},\frac{10}{17}\right) $.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the integral $I(u,v,w)=\iint_{(0,\infty)^2}\sinh(upq) e^{-vq^2 - wp^2}pq(q^2-p^2)^{-1}dpdq$ I found in an article "Imperfect Bose Gas with Hard-Sphere Interaction", Phys. Rev. 105, 776–784 (1957) the following integral, but I don't know how to solve it. Any hints?
$$\int_0^\infty {\int_0^\infty {\mathrm dp\mathrm dq\frac{\sinh(upq)}{q^2 - p^2}pq} } e^{-vq^2 - wp^2} = \frac{\pi}{4}\frac{u(w - v)}{\left[(w + v)^2-u^2 \right]\left(4wv-u^2\right)^{1/2}}$$
for $u,v,w > 0$.
| I will assume that $u, v, w > 0$ and $4vw > u^2$. Let $$
I := \int_{0}^{\infty} \int_{0}^{\infty} \frac{\sinh (upq)}{p^2 - q^2} \, pq \, e^{-vp^2} e^{-wq^2} \; dpdq.
$$ By polar coordinate transform, we obtain $$
\begin{eqnarray*}
I & = & \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{\sinh (r^2 u \cos \theta \sin \theta)}{r^2 \cos^2 \theta - r^2 \sin^2 \theta} \, r^2 \cos\theta \sin \theta \, e^{-vr^2 \cos^2 \theta} e^{-w r^2 \sin^2 \theta} \; r dr d\theta \\
& = & \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\tan \theta}{1 - \tan^2 \theta} \int_{0}^{\infty} \sinh(r^2 u \sin \theta \cos\theta) \, e^{-r^2 (v \cos^2 \theta + w \sin^2 \theta)} \; d(r^2) d\theta \\
& = & \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\tan \theta}{1 - \tan^2 \theta} \left( \frac{u \cos\theta \sin\theta}{(v \cos^2 \theta + w \sin^2 \theta)^2 - u^2 \cos^2 \theta \sin^2 \theta} \right) d\theta \\
& = & \frac{1}{4} \int_{-\infty}^{\infty} \frac{u t^2}{(1-t^2)\left( (v + w t^2)^2 - u^2 t^2 \right)}\; dt. \qquad (\text{where} \ t = \tan \theta)
\end{eqnarray*}
$$ Now the last integral can be attacked by standard contour integration techinque. In particular, let $$ f(z) = \frac{u}{4} \frac{z^2}{(1-z^2)\left( (v + w z^2)^2 - u^2 z^2 \right)} $$ be the integrand. Then considering appropriate upper-semicircular contour with vanishing dents at $\pm 1$, we obtain $$ \begin{align*} I = & \pi i \Bigg[ \mathrm{Res} \left\{ f, 1 \right\} + \mathrm{Res} \left\{ f, -1 \right\} \Bigg] \\ & + 2\pi i \Bigg[ \mathrm{Res} \left\{ f, \frac{u+i\sqrt{4vw-u^2}}{2w} \right\} + \mathrm{Res} \left\{ f, \frac{-u+i\sqrt{4vw-u^2}}{2w} \right\} \Bigg], \end{align*}$$
which yields the desired formula. (A tip : $\mathrm{Res} \{ f, 1 \} + \mathrm{Res} \{ f, -1 \} = 0$ because $\pm 1$ are simple poles of an even function $f$.)
p.s. While posting my solution, Andrew gave a nice solution.
| {
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How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$? How can the following inequation be proven?
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
| $a^2+b^2+c^2-ab-bc-ca\\
=(a, b, c)\begin{pmatrix}1&-1/2&-1/2\\-1/2& 1&-1/2\\-1/2&-1/2&1\end{pmatrix}\left(\begin{array}{c}a\\b\\c\end{array}\right)$
$=(a, b, c)A\begin{pmatrix}a\\b\\c\end{pmatrix}$
It is sufficient to prove $A$ is a positive semi-definite.
It follows from that the minor determinant of $A$ is one of $1,\dfrac{3}{4},0$
| {
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Evaluating $ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $ I would like to evaluate: $$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $$
$$ \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}=\frac{\sqrt{1-x}+\sqrt{1+x}-2}{2(\sqrt{1-x^2}-1)} $$
The substitution $ x \rightarrow \sin(x) $ or $ \cos(x) $ can only simplify the denominator, and $ x \rightarrow \sqrt{1+x}$ or $ \sqrt{1-x} $ is also useless...
Can you help me find a useful substitution?
$$ x=\cos(2t) $$
$$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx=-\int {\frac{\sqrt{2}\sin(t)\cos(t)}{\sqrt{2}+\sin(t)+\cos(t)}}\mathrm dt $$
$$ u=\tan(t/2) $$
$$ -4\sqrt{2}\int \frac{u(1-u^2)}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})}\mathrm du $$
But now it looks even more complicated... ?
| $$\newcommand{\ct}[0]{\color{grey}{\text{constant}}}
\newcommand{\b}[1]{\left(#1\right)}
\begin{align}
\int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}dx&=\int\frac{-2\sin(2t)}{2+\sqrt2\sin t+\sqrt2\cos t}dt\tag{1}\\
&=-2\int\frac{\sin(2t)}{2+2\sin\b{t+\frac{\pi}4}}dt\\
&=-2\int\frac{\sin(2t)}{2+2\cos\b{\frac{\pi}4-t}}dt\\
&=-\frac12\int\sin(2t)\sec^2\b{\frac{\pi}8-\frac t2}dt\\
&=\int\sin(2\b{\pi/4-2u})\sec^2(u)du\tag{2}\\
&=\int\cos(4u)\sec^2(u)du\\
&=\cos(4u)\tan(u)+4\int\sin(4u)\tan(u)du\tag{3}\\
\int\sin(4u)\tan(u)du&=\int(2\cos(2u)-\cos(4u)-1)du\tag{4}\\
&=\sin(2u)-\frac14\sin(4u)-u+\ct\\
&=\sin\b{\frac{\pi}4-t}-\frac14\cos(2t)+\frac t2+\ct\\
&=\frac1{\sqrt2}\b{\sin(t)-\cos(t)}-\frac14\cos(2t)+\frac t2+\ct\\
&=\frac12\b{\sqrt{1-x}-\sqrt{1+x}}-\frac14x+\frac 14\arccos(x)+\ct\\
\cos(4u)\tan(u)&=\sin(2t)\tan\b{\pi/8-t/2}\\&=\sin(2t)\sqrt{\frac{1-\cos(\pi/4-t)}{1+\cos(\pi/4-t)}}\\&=\sin(2t)\sqrt{\frac{1-\sin(t+\pi/4)}{1+\sin(t+\pi/4)}}\\&=\sqrt{1-x^2}\sqrt{\frac{\sqrt2-\sqrt{1-x}-\sqrt{1+x}}{\sqrt2+\sqrt{1-x}+\sqrt{1+x}}}
\end{align}$$
So:
$$\large\int \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}dx=\sqrt{1-x^2}\sqrt{\frac{\sqrt2-\sqrt{1-x}-\sqrt{1+x}}{\sqrt2+\sqrt{1-x}+\sqrt{1+x}}}+2\sqrt{1-x}-2\sqrt{1+x}-x+\arccos(x)+\color{grey}{\rm constant}$$
$(1)$ let $x=\cos(2t)\mid dx=-2\sin(2t)$
$(2)$ let $u=\pi/8-t/2\iff t=\pi/4-2u\mid dt=-2du$
$(3)$ use integration by parts
$(4)$ use $$\begin{align}\sin(4u)\tan(u)&=2\sin(2u)\cos(2u)\tan(u)\\&=4\sin^2(u)\cos(2u)\\&=2\cos(2u)(1-\cos(2u))\\&=2\cos(2u)-2\cos^2(2u)\\&=2\cos(2u)-\cos(4u)-1\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/67172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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A question about limit $\lim_{n\to\infty} {n \choose n/2}/2^n$ My question is: What is the result of this limit:
$\displaystyle \lim_{n \to +\infty} \frac{{n \choose n/2}}{2^n}=$ ?
| As $\lim_{n\to+\infty} \frac{1}{2^n} \binom{n}{n/2} = \lim_{n\to+\infty} \frac{1}{4^n} \binom{2n}{n}$, and since for $r(n) = \frac{1}{4^n} \binom{2n}{n}$ the ratio
$$
r(n) >0 \land \frac{r(n+1)}{r(n)} = \frac{2n+1}{2(n+1)} < 1 \qquad \forall n \ge 1
$$
it follows that $\lim_{n \to \infty} r(n) = 0$ since $r(n)$ monotonically decreases as $n$ grows.
Added: To address the gap in my demonstration pointed out by Henning:
$$
\begin{eqnarray}
r(n) = \frac{1}{4^n} \binom{2n}{n} &=& r(0) \prod_{k=0}^{n-1} \frac{r(k+1)}{r(k)} = \prod_{k=0}^{n-1} \frac{2k+1}{2(k+1)} \\
&=& \prod_{k=0}^{n-1} \left( 1 - \frac{1}{2(k+1)} \right) \le \prod_{k=0}^{n-1} \left( 1 + \frac{1}{2(k+1)} \right)^{-1} \\
&\le& \frac{1}{\sum_{k=0}^{n-1} \frac{1}{2(k+1)}} = \frac{2}{H_n}
\end{eqnarray}
$$
Thus $0 < r(n) \le \frac{2}{H_n}$ and $\lim_{n\to \infty} r(n) = 0$ follows from $\lim_{n\to\infty} \frac{2}{H_n} = 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A 10 digit positive number and ordered triplets A 10 digit positive number is said to be a “LearnHub”number if its digits are all distinct and it is a multiple of 11111. How many “LearnHub” numbers are there?
Find the number of ordered triplets (a, b, c) of positive integers for which LCM (a, b) =1000, LCM (b, c) = 2000 and LCM (c, a) = 2000.
How do i do this ? I have worked a lot on them.
| $1000 = 2^3 * 5^3$
$2000 = 2^4 * 5^3$
Since LCM of $a$ and $b$ is $1000$, they at the most contain $2$ three times and $5$ three times.
Also the LCM of terms containing $C$ is $2000$, the term $c$ should contain $2$ four times, as $a$ and $b$ can contain $2$ only three times.
so $c$ should either be
$2^4 = 16$
$2^4 * 5^1 = 80$
$2^4 * 5^2 = 400$
or $2^4 * 5^3 = 2000$.
Either $a$ or $b$ should have $2^3$. Hence fixing one as $2^3$ other can take values from $2^0$ to $2^3$.
Hence total combination= $2*4 - 1$ (the one combination is $2^3,2^3$ which we took twice)$= 7$
2 of the three should have power of $5$ as $5^3$, whereas the third can take 4 values in terms of powers of $5$ $(5^0$ to $5^3)$. Hence total no of such arrangement = $(3!/2! *4) -2$(this is for $5^3,5^3,5^3$ which we took thrice) = $12 -2 =10$
Hence ordered triplets will be $10*7 = 70$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/74267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral solutions of $x^2+y^2+1=z^2$ I am interested in integral solutions of $$x^2+y^2+1=z^2.$$ Is there a complete theory comparable to the one for $x^2+y^2=z^2?$
| For the equation: $qX^2+Y^2=Z^2+j$
In the case where a square: $a=\sqrt{\frac{j}{q}}$
Using equation Pell: $p^2-(q+1)s^2=1$
Then the solution can be written:
$X=2s(s\pm{p})L\pm{ap^2}+2aps\pm{a(q+1)s^2}=bL+af$
$Y=(p^2\pm2ps+(1-q)s^2)L\pm{ap^2}+2aps\pm{a(q+1)s^2}=cL+af$
$Z=(p^2\pm2ps+(q+1)s^2)L\pm{ap^2}+2a(q+1)ps\pm{a(q+1)s^2}=fL+at$
$L$ - any integer number given by us.
The most interesting thing is that these numbers are solutions of equations:
$qb^2+c^2=f^2$
$t^2-(q+1)f^2=\pm{q}$
If we use the equation Pell: $p^2-(q+1)s^2=k$
And substituting the solutions in the upper formula, we have solutions of the following equations.
$qb^2+c^2=f^2$
where: $c-b=k$
$t^2-(q+1)f^2=\pm{qk^2}$
True, I use this formula in reverse order. Find solutions of Pell's equation is much more complicated than the simple equations like Pythagorean triples. So find them and then have solutions of Pell's equation. The most interesting thing is that the solution of Pell related to Pythagorean triples.
| {
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"url": "https://math.stackexchange.com/questions/74931",
"timestamp": "2023-03-29T00:00:00",
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Need help finding: $\frac{d}{dx}\frac{\sec{x}}{1+\tan{x}}$ So I am trying to find:
$$\frac{d}{dx}\frac{\sec{x}}{1+\tan{x}}$$
And tried doing:
$$\frac{(1+\tan{x})(\tan{x}\times\sec{x})-(\sec^{2}{x})(\sec{x})}{(1+\tan{x})^{2}}$$
Because of the Quotient Rule. Then I did some simplifying:
$$\frac{(1+\tan{x})(\tan{x}\times\sec{x})-(\sec^3{x})}{(1+\tan{x})^{2}}$$
Further simplification (crossed out the $(1+\tan{x})^{2})$:
$$\frac{\tan{x}\times\sec{x}-\sec^{3}{x}}{1+\tan{x}}$$
Then I got:
$$\frac{\sec{x}\times(\tan{x}-\sec^{2}{x})}{1+\tan{x}}$$
But Wolfram Alpha says differently. Where did I go wrong? Thanks.
Update:
So I tried regrouping:
$$\frac{(1+\tan{x})\tan{x}\times(\sec{x}-(\sec^3{x}))}{(1+\tan{x})^{2}}$$
Factored out a $\sec{x}$:
$$\frac{(1+\tan{x})\tan{x}\times\sec{x}(1-(\sec^2{x}))}{(1+\tan{x})^{2}}$$
Which then gives:
$$\frac{(1+\tan{x})\tan{x}\times\sec{x}\times-\tan^{2}{x}}{(1+\tan{x})^{2}}$$
Which then I said:
$$-\frac{\tan^{3}{x}\times\sec{x}}{(1+\tan{x})}$$
Which still isn't right. Sorry, if I made another obvious mistake.
| Decided to use product rule:
$$\frac{d}{dx}\frac{\sec{x}}{(1+\tan{x})}$$
$$\frac{(1+\tan{x})(\tan{x}\times\sec{x})-(\sec^3{x})}{(1+\tan{x})^{2}}$$
I just worked on the top for a while:
$$(1+\frac{\sin{x}}{\cos{x}})(\frac{\sin{x}}{\cos{x}}\times\frac{1}{\cos{x}})-(\frac{1}{\cos{x}^{3}})$$
$$(\frac{\sin{x}}{\cos{x}}+\frac{\sin^2{x}}{\cos^2{x}})(\frac{1}{\cos{x}})-(\frac{1}{\cos{x}^{3}})$$
$$(\frac{\sin{x}}{\cos^2{x}}+\frac{\sin^2{x}}{\cos^3{x}})-(\frac{1}{\cos{x}^{3}})$$
$$\frac{\sin{x}}{\cos^2{x}}+\frac{\sin^2{x}}{\cos^3{x}}-\frac{1}{\cos{x}^{3}}$$
$$\frac{\sin{x}}{\cos^2{x}}+\frac{\sin^2{x}-1}{\cos^3{x}}$$
$$\frac{\sin{x}}{\cos^2{x}}+\frac{-\cos^2{x}}{\cos^3{x}}$$
$$\frac{\sin{x}}{\cos^2{x}}-\frac{1}{\cos{x}}$$
$$\frac{\sin{x}}{\cos^2{x}}-\frac{1}{\cos{x}}*\frac{\cos{x}}{\cos{x}}$$
$$\frac{\sin{x}}{\cos^2{x}}-\frac{\cos{x}}{\cos^2{x}}$$
$$\frac{\sin{x}-cos{x}}{\cos^2{x}}$$
$$\sec^2{x}\times(\sin{x}-\cos{x})$$
$$\sin{x}sec^2{x}-\cos{x}\sec^2{x}$$
$$\sin{x}\times\sec^2{x}-\cos{x}\times\sec^2{x}$$
$$\sin{x}\times\sec^2{x}-\frac{1}{\cos{x}}$$
$$\sin{x}\times\sec^2{x}-\sec{x}$$
$$\sin{x}\times\frac{1}{\cos^2{x}}-\sec{x}$$
$$\frac{\sin{x}}{\cos^2{x}}-\sec{x}$$
$$\frac{\sin{x}}{cos{x}}\frac{1}{cos{x}}-\sec{x}$$
$$\tan{x}\sec{x}-\sec{x}$$
$$(\tan{x}-1)\times\sec{x}$$
Put it all over the original denominator:
$$\frac{(\tan{x}-1)\times\sec{x}}{(1+\tan{x})^{2}}$$
So that is what I did, there probably should be an easier way though...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Interscholastic Mathematics League Senior B #12 Compute the product of the nonreal roots of the equation $x^4+4x^3+6x^2+1004x+1001=0$.
So here is what I have done so far. I got two of the roots to be zero and 4 since $x^3(x+4)+(6x^2+1004x+1001)=0$. But for the other two roots by using the quadratic formula I get $$\frac{-1004 \pm \sqrt{1004^2-24024}}{12}.$$ This contest was already completed but I couldn't solve that quadratic equation during the competition. Can someone please post a easier way to solve this problem?
| We are solving the equation
$$x^4+4x^3+6x^2+1004x+1001=0.$$
Note that
$$(x+1)^4=x^4+4x^3+6x^2+4x+1.$$
Interesting! So we are looking at
$$(x+1)^4 +1000x+1000=0.$$
Let $y=x+1$. Our equation can be rewritten as $y^4+1000y=0$. The real roots are $y=0$ and $y=-10$, and the other two roots are non-real. Now it's essentially over.
Added: For completeness, note that two of the roots are $x=-1$ and $x=-11$. The product of the roots of the original equation is $1001$. So the product of the two non-real roots is $1001/11$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Formula for completing the square? My math teacher said that this was the formula for completing the square.
Original function: $$ax^2 + bx + c$$
Completed square: $$a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$$
However, using this formula I'm not getting the same answers that I would get just by determining the stuff myself. Is this correct?
| For the polynomial $4x^2+4x+5$ that you mention, I would not use the formula, since it is fairly clear that $4x^2 +4x$ is "almost" $(2x+1)^2$. In fact, $(2x+1)^2=4x^2+4x+1$, so $4x^2+4x+5=(2x+1)^2-1+5=(2x+1)^2+4$.
In general, suppose that $a \ne 0$, and we want to deal with $ax^2+bx+c$. Multiply the expression by $4a$, and to keep things unchanged, divide by $4a$.
We get
$$ax^2+bx+c=\frac{1}{4a}(4a^2x^2 +4abx +4ac).$$
But
$4a^2x+4abx$ is almost the square of $2ax+b$. In fact, $4a^2x^2+4abx=(2ax+b)^2-b^2$. It follows that
$$4a^2x^2+4abx+4ac=(2ax+b)^2-(b^2-4ac),$$
so
$$ax^2+bx+c=\frac{1}{4a}\left((2ax+b)^2-(b^2-4ac)\right).$$
The formula is useful as is, and more pleasant to work with than the formula of the post. We can transform it to look like that formula by multiplying the top and bottom of the front by $a$, and using the fact that $\frac{1}{4a^2}(2ax+b)^2=\left(x+\frac{b}{2a}\right)^2$.
Comment: If we want to derive the Quadratic Formula, we don't need to bother with dividing by $4a$, for $ax^2+bx+c=0$ iff $4a^2+4abx+4ac=0$. Complete the square like above. We get
$$ax^2+bx+c=0 \qquad\text{if and only if}\quad (2ax+b)^2=b^2-4ac,$$
and we are a couple of easy steps away from the Quadratic Formula.
Important: One should not try to remember a formula for completing the square. What one needs to understand is the process, the idea. Students, particularly those blessed (?) with good memories, find that throughout high school they can achieve easy success by memorizing formulas. Finding out what's really going on may in the short term look like more work, but it will last.
| {
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"timestamp": "2023-03-29T00:00:00",
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Irreducible polynomial which is reducible modulo every prime
How to show that $x^4+1$ is irreducible in $\mathbb Z[x]$ but it is reducible modulo every prime $p$?
For example I know that $x^4+1=(x+1)^4\bmod 2$. Also $\bmod 3$ we have that $0,1,2$ are not solutions of $x^4+1=0$ then if it is reducible the factors are of degree $2$. This gives that $x^4+1=(x^2+ax+b)(x^2+cx+d)$ and solving this system of equations $\bmod 3$ gives that $x^4+1=(x^2+x+2) (x^2+2x+2) \pmod 3$. But is there a simpler method to factor $x^4+1$ modulo a prime $p$?
| For every odd prime $p$ we have $8\mid p^2-1$. The multiplicative group of the finite field $F=GF(p^2)$ is cyclic of order $p^2-1$. Putting these two bits together tells us that there is a primitive root $u$ of order $8$ in $F$. We must have $u^4=-1$, because $-1$ is the only element of multiplicative order two. Because $F$ is a quadratic extension of $\mathbf{Z}/p\mathbf{Z}$, the minimal polynomial of $u$ is of degree $\le 2$. That minimal polynomial is then a factor of
$$x^4+1=(x-u)(x-u^3)(x-u^5)(x-u^7)=(x-u)(x-u^3)(x+u)(x+u^3).$$
====================
Edit: Here's an idea for finding the factorization. I split it into cases according to the residue class of $p$ modulo 8. Assume first that $p\equiv 1\pmod 4$ (or $p$ equivalent to $1$ or $5$ modulo 8). In that case all we need is a square root $i$ of $-1$ modulo $p$. IIRC there is an algorithm for finding two integers $x,y$ such that $p=x^2+y^2$, and then $i=x*y^{-1}$ is the desired square root in the prime field $F_p=GF(p)$. A factorization is then
$$
x^4+1=(x^2+i)(x^2-i).
$$
Observe that if $p\equiv1\pmod8$ then both quadratic factors will split further.
If $p\equiv 3\pmod 8$, then $u$ is not in the prime field, and its conjugate is $u^p=u^3$. Therefore the minimal polynomial is
$$
m(x)=(x-u)(x-u^p)=(x-u)(x-u^3)=x^2-[u+u^3]x + u^4= x^2-ax-1,
$$
where $a$ is some unknown element of the prime field. Because $u^5=-u$ and $u^7=-u^3$, the other factor of $x^4+1$ must be $m(-x)=x^2+ax-1$. We need to find the coefficient $a$.
Let's multiply
$$
(x^2-ax-1)(x^2+ax-1)=(x^2-1)^2-a^2x^2=x^4-(2+a^2)x^2+1.
$$
We see that we have found the factorization, if we can find $a=\sqrt{-2}$. It is well known that when $p\equiv 3\pmod 8$, then $-2$ is a quadratic residue modulo $p$ confirming our finding.
In the last case $p\equiv 7\pmod8$ the minimal polynomial of $u$ over $F_p$ is
$$
m(x)=(x-u)(x-u^p)=(x-u)(x-u^7)=x^2-[u+u^7]x+u^8=x^2-bx+1
$$
for some $b\in F_p$. Again the other factor is $m(-x)$, and a similar calculation shows that we need $b=\sqrt{2}$. Again this fits together with the known fact that in this case $2$ is a quadratic residue modulo $p$.
==================
Edit(2): TonyK described the following methods for finding the square roots. They depend on the fact that if $p$ is an odd prime, and $gcd(a,p)=1$, then $a^{(p-1)/2}\equiv\pm1\pmod p$. Here we have the plus sign, if and only if $a$ is a quadratic residue (=QR) modulo $p$.
If $p\equiv 3\pmod8$, then we know that $2$ is not a QR modulo $p$. Therefore
$2^{(p-1)/2}\equiv -1\pmod p$. Hence $2^{(p+1)/2}\equiv -2\pmod p$. But here $(p+1)/2$ is an even integer, so writing $z=2^{(p+1)/4}$ we get $z^2\equiv 2^{(p+1)/2}\equiv -2$, and we have found a square root of $-2$.
Similarly, if $p\equiv 7\pmod 8$, we know that $2$ is a quadratic residue modulo $p$. This time $2^{(p+1)/2}\equiv 2$, and the same calculation shows that $z=2^{(p+1)/4}$ is a square root of $2$ in $F_p$.
If $p\equiv 5\pmod 8$, then again $2$ is not a QR modulo $p$, so $2^{(p-1)/2}\equiv -1\pmod p$ and $(p-1)/2$ is even. Thus $z=2^{(p-1)/4}$ is a square root of $-1$. If $p\equiv 1\pmod 8$, then we cannot use $2$ (but could use any non-QR in its place, or the method mentioned earlier).
| {
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Solving the equation $\frac{2x+3}{x+1} = \frac{2x+2}{x-1}$ I'm a college student and I've got a problem solving this:
$$\frac{2x+3}{x+1} = \frac{2x+2}{x-1}.$$
Since neither the numerators nor the denominators are equal, I figured that $(2x+3)(x-1) = (x+1)(2x+2)$ would provide me with a suitable $x$.
However calculating this I found that $x = -\frac13$ while my answer sheet states that $x = -\frac23$. What am I doing wrong?
| There’s nothing wrong with your basic approach, but you seem to have solved the equation $(2x+3)(x-1)=(x+1)(2x+2)$ incorrectly. After multiplying out both sides, you should have $$2x^2+x-3=2x^2+4x+2\;,$$ which simplifies progressively to $$\begin{align*}
x-3&=4x+2\;,\\
-3&=3x+2\;,\\
-5&=3x\;,\text{ and}\\
x&=-\frac53.
\end{align*}$$
Substituting this into the original equation yields $$\begin{align*}
\frac{2\left(-\frac53\right)+3}{-\frac53+1} &\stackrel{?}= \frac{2\left(-\frac53\right)+2}{-\frac53-1}\\
\frac{-\frac13}{-\frac23} &\stackrel{?}= \frac{-\frac43}{-\frac83}\\
\frac12&=\frac12.
\end{align*}$$
In other words, both you and the answer sheet are wrong.
| {
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Summation Identity: $\sum_{i=1}^ni^3 = \left( \frac{n(n+1)}{2} \right)^2$ I have to prove:
$$\sum\limits_{i = 1}^n i^3 = \Bigg( \frac{n(n+1)}{2}\Bigg)^2$$
Using the following:
$$n^3 = 6 {n \choose 3} + 6 {n \choose 2} + n \quad \forall n \in \mathbb{N}$$
My work is that first I substitute $n^3$ for $6 {n \choose 3} + 6 {n \choose 2} + n$. Then I go and invoke the sum over that (I am assuming this is how it works). That is,
$$\sum\limits^n_{i=1} \bigg(6 {i \choose 3} + 6 {i \choose 2} + i \bigg)$$
$$6 \sum\limits^n_{i=1} {i \choose 3} + 6 \sum\limits^n_{i=1} {i \choose 2} + \sum\limits^n_{i=1} {i \choose 1}$$
The summation identity is:
$$\sum\limits^n_{i=0} {i \choose k} = {n+1 \choose k+1}$$
invoking it over the sums yields:
$$6{n+1 \choose 4} + 6{n+1 \choose 3} + {n+1 \choose 2}$$
I think these are the right steps, but simplification seems a bit difficult to get the ending result.
| Suppose we have arrived at the expression
$$6{n+1 \choose 4} + 6{n+1 \choose 3} + {n+1 \choose 2},$$
and we remain in a combinatorial mood, not an algebraic one. Then we might work a little harder and give a bijective argument.
From a group of $n+1$ boys and $n+1$ girls, we can choose $2$ boys and $2$ girls (a pair of pairs) in
$$\binom{n+1}{2}^2$$
ways. We will count the number of pairs of pairs in another way.
Let the boys be called $b_0,b_1,\dots,b_n$, and the girls $g_0, g_1,\dots,g_n$. If $2$ boys and $2$ girls are chosen they either (a) share no number or (b) share $1$ number or (c) share $2$ numbers.
(a) The pairs of pairs that share no number can be chosen as follows. Choose $4$ numbers from the set $\{0,1,\dots,n\}$. Then choose $2$ of the $4$ numbers, and select the boys with these $2$ numbers, and the girls with the remaining $2$ numbers. The choosing of $4$ numbers can be done in $\binom{n+1}{4}$ ways, and the choosing of $2$ from $4$ can be done in $\binom{4}{2}=6$ ways, for a total of
$$6\binom{n+1}{4}.$$
(b) The pairs of pairs that share exactly one number can be chosen as follows. Choose $3$ numbers from $\{0,1,\dots,n\}$. Choose $1$ of these $3$ numbers to be the "duplicated" number (boy and girl), then select the boy who has $1$ of the remaining numbers, and the girl with the other. The choosing of the $3$ numbers can be done in $\binom{n+1}{3}$ ways. For each of these, we can choose the duplicated number in $3$ ways, and decide which of the remaining numbers will be a boy number in $2$ ways, for a total of
$$(3)(2)\binom{n+1}{3}.$$
(c) Finally, we count the pairs of pairs that share two numbers.
All we need to do is to choose these $2$ numbers, and the rest is determined, so the number of type (c) pairs of pairs is
$$\binom{n+1}{2}.$$
Add up.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
} |
Show that the value of a definite integral is unity $$ \int_2^4\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(3+x)}}dx=1$$
| We show that $I=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}\,dx$ equals $1$, where $f(x)=\sqrt{\log(9-x)}$.
Proof. By making a substitution $y=6-x$, we get
$$I=\int_4^2 -\frac{f(6-y)}{f(y)+f(6-y)}\,dy=\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}\,dy.$$
Therefore
$$\begin{align*}2I&=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}dy\\
&=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-x)}{f(x)+f(6-x)}dx=\int_2^4 1\,dx =2.\end{align*}$$
Edit. In the last part, $y$ is a dummy variable and can be changed to $x$ or any other variable you like.
Edit2. If you don't like the same $x$ being used, you could use another variable, say $s$,
so that with two substitutions
$$\begin{align*}2I&=\int_2^4 \frac{f(x)}{f(x)+f(6-x)}dx+\int_2^4 \frac{f(6-y)}{f(y)+f(6-y)}dy\\
&=\int_2^4 \frac{f(s)}{f(s)+f(6-s)}ds+\int_2^4 \frac{f(6-s)}{f(s)+f(6-s)}ds=\int_2^4 1\,ds =2.\end{align*}$$
| {
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} |
Square Roots of Complex Number $3-4i$
What I did
$z^2=3-4i$
$(a+bi)^2 = 3-4i$
$a^2-b^2+2abi = 3-4i$
Then got 2 simultaneous equations
$a^2-b^2=3$ and $2ab=-4$
Solve for $a^2$ in 1st equation: $a^2=3+b^2$
Subbed into 2nd equation to power of 2
$(3+b^2)b^2=4$
$b^4+3b^2-4=0$
Is there a better way than below? Solving power 4 equation then cubic?
Then solved solved power 4 equation ...
$(b-1)(Ab^2+Bb^2+Cb+D)=b^4+3b^2-4$
$(b-1)(b^3+b^2+4b+4)=b^4+3b^2-4$
then solved for cubic equation, getting ...
$(b-1)(b+1)(b^2+4)=0$
So
$b=\pm 1 \text{ or } 2i$
What did the book do now?
I did: (subbing into $a^2=3+b^2$)
When $b=1, a^2=3+1^2 \Longrightarrow a = 2$
When $b=-1, a^2=3+(-1)^2 \Longrightarrow a=2$
When $b=2i, a^2=3+(2i)^2=3+4(-1)=-1, a=i$
So I will have 3 equations
*
*$2+i$ // why is this not in the book?
*$2-i$
*$i-2$
UPDATE (In response to @David Mitra)
Ok. I let $x=a^2$ and $b=b^2$ giving the quadratic equation: $x=3+y$ and $xy=4$. Then (after subbing) $(y-1)(y+4)=0$
Then when $y=1, x=4$. $y=-4, x=-1$
Then $b=\pm 1 or \pm 2i$ and $a=\pm 2i or \pm i$
Finally testing equations:
$(2+i)^2 = 3+4i$ (rej)
$(2-i)^2=5$ what to do?
$(i-2)^2=3$
$(-2-i)^2=3$
... 8 equations (but 4 unique)
| You can look at this as a problem in the arithmetic of the Gaussian Integers, $\mathbb Z[i]$. I’ll make use of the fact that this ring is a Unique Factorization Domain, and I’ll also use the fact that every prime $p\equiv1\pmod4$ is writable as the sum of two squares, $p=m^2+n^2$, equivalently, $p=(m+ni)(m-ni)$, and that $m\pm ni$ are primes in $\mathbb Z[i]$.
For the prime $5$, the first where these facts become operative, we have $5=1^2+2^2=(1+2i)(1-2i)$, and certainly the two factors are not related by a unit, so that they generate different prime ideals of the G-Integers. What about $3+4i$? We have $(3+4i)(3-4i)=25=(1+2i)^2(1-2i)^2$, a product of primes in the UFD $\mathbb Z[i]$. Up to units, you see that it must be that $3+4i$ is equal to $(1+2i)^2$ or $(1-2i)^2$. You check, and you see that you don’t need to worry about units, and the first possibility is the right one.
| {
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"answer_count": 5,
"answer_id": 3
} |
How to calculate the determinant of all-ones matrix minus the identity? How do I calculate the determinant of the following $n\times n$ matrices
$$\begin {bmatrix}
0 & 1 & \ldots & 1 \\
1 & 0 & \ldots & 1 \\
\vdots & \vdots & \ddots & \vdots \\
1 & 1 & ... & 0
\end {bmatrix}$$
and the same matrix but one of columns replaced only with $1$s?
In the above matrix all off-diagonal elements are $1$ and diagonal elements are $0$.
| $$D_n(a,b)=
\begin{vmatrix}
a & b & b & b \\
b & a & b & b \\
b & b & a & b \\
b & b & b & a
\end{vmatrix}$$
($n\times n$-matrix).
$$D_n(a,b)=
\begin{vmatrix}
a & b & b & b \\
b & a & b & b \\
b & b & a & b \\
b & b & b & a
\end{vmatrix}$$
$$=[a+(n-1)b]
\begin{vmatrix}
1 & 1 & 1 & 1 \\
b & a & b & b \\
b & b & a & b \\
b & b & b & a
\end{vmatrix}$$
$$=[a+(n-1)b]
\begin{vmatrix}
1 & 1 & 1 & 1 \\
0 & a-b & 0 & 0 \\
0 & 0 & a-b & 0 \\
0 & 0 & 0 & a-b
\end{vmatrix}$$
$$=[a+(n-1)b](a-b)^{n-1}
$$
(In the first step we added the remaining rows to the first row and then "pulled out" constant out of the determinant. Then we subtracted $b$-multiple of the first row from each of the remaining rows.)
You're asking about $D_n(0,1)=(-1)^{n-1}(n-1)$.
If you replace one column by 1's, you can use this result to get the following. (I've computed it for $n=4$, but I guess you can generalize this for arbitrary $n$.)
$$
\begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
\end{vmatrix}
=
\begin{vmatrix}
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
\end{vmatrix}
+
\begin{vmatrix}
1 & 0 & 0 & 0 \\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
\end{vmatrix}=
\begin{vmatrix}
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
\end{vmatrix}
+
\begin{vmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{vmatrix}
$$
Note that both these determinants are of the type you already handled in the first part.
| {
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"answer_count": 8,
"answer_id": 6
} |
how to solve $\pm y \equiv 2x+1 \pmod {13}$ with Chinese remainder theorem or iterative method? how to solve $\pm y \equiv 2x+1 \pmod {13}$ with Chinese remainder theorem or iterative method?
It comes from solving $x^2+x+1 \equiv 0 \pmod {13}$ (* ) and background is following:
13 is prime. (* ) holds under Euclidean lemma if and only if $4(x^2+x+1) \equiv \pmod {13}$
or if and only if $(2x+1)^2 \equiv -3 \pmod {13}$. So if $p=13$, so by Euler's criterion
$[ \frac{-3}{13} ] \equiv (-3)^{\frac{13-1}{2}} = (-3)^6 = 9^3 \equiv (-4)^3 =-64 \equiv 1 \pmod{13} $. Hence equation $y^2 \equiv -3 \pmod{13}$ has two incongruent solution( lemma 4.1.3) $\pm y$ so solutions of the equations $\pm y \equiv 2x+1 \pmod{13}$ are solutions of the equation (* )
So my most important question is how you change equation $\pm y \equiv 2x+1 \pmod{13}$ to the form $ax\equiv b \pmod{13} $ in other words to the form where you can use either Chinese remainder theorem or iterative method to solve $\pm y \equiv 2x+1 \pmod{13}$ and finally (* )?
Finally just because of curiosity. Is $[\frac{-3}{7}]\equiv (-3)^3 = -27 \equiv -1 \pmod{7}$? So is mod(-27,7)=1 or -1?
| To solve $x^2 + x + 1 \equiv 0 \pmod{13}$, you can use the usual quadratic formula, interpreted appropriately. The solutions are given by
$$x = \frac{-1 \pm\sqrt{-3}}{2},$$
where "$\sqrt{-3}$" means an integer $y$ such that $y^2\equiv -3\pmod{13}$, and "$\frac{1}{2}$" means an integer $z$ such that $2z\equiv 1\pmod{13}$.
The latter is easy: $z\equiv 7\pmod{13}$ does the job.
For the latter, you need $-3$ to be a quadratic residue modulo $13$; it is, because $3\equiv 4^2\pmod{13}$, and $-1\equiv 5^2\pmod{13}$, so $4\times 5 = 20\equiv 7\pmod{13}$ does the trick; indeed, $7^2 = 49 \equiv -3\pmod{13}$.
So the two solutions to $x^2+x+1\equiv 0\pmod{13}$ are:
$$x= (2)^{-1}(-1+\sqrt{-3}) \equiv 7(-1+7) = 7(6) = 42 \equiv 3\pmod{13}$$
and
$$x= (2)^{-1}(-1-\sqrt{-3}) \equiv 7(-1-7) = 7(-8) = -56 \equiv 9\pmod{13}.$$
You can verify this easily:
$$\begin{align*}
3^2 + 3 + 1 &= 9+3+1 = 13\equiv 0\pmod{13},\\
9^2+9+1&=81+9+1 = 91=13\times 7\equiv 0\pmod{13}.
\end{align*}$$
If you want to go your route, you are trying to find values of $x$ such that $(2x+1)^2\equiv -3\pmod{13}$. That means finding the two square roots of $-3$ modulo $13$; as above, they are $7$ and $-7\equiv 6$, so you are looking for the values of $x$ such that $2x+1\equiv 7\pmod{13}$ and $2x+1\equiv 6\pmod{13}$. These translate to $2x\equiv 6\pmod{13}$ and $2x\equiv 5\pmod{13}$. We can solve these by multiplying both sides by $7$ (the multiplicative inverse of $2$), so we get
$$x\equiv 7(6) = 42\equiv 3\pmod{13}\qquad\text{and}\qquad x\equiv 7(5) = 35 = 9\pmod{13},$$
the same answers as above.
I don't know why you are trying to find $\pm y\equiv 2x+1\pmod{13}$. This does not give you what you want.
| {
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"answer_count": 2,
"answer_id": 0
} |
Probability of rolling a die I roll a die until it comes up $6$ and add up the numbers of spots I see.
For example, I roll $4,1,3,5,6$ and record the number $4+1+3+5+6=19$. Call this sum $S$.
Find the standard deviation of $S$.
I have been looking for an easy way to do this because I know I can use the definitions here to calculate the variance of $S$ and then take the square root of it. But I am sure there is an easier way to do this.
| Let $Y$ be the number of rolls before a $6$ is rolled. Let $X$ be the sum of the dice rolled before a $6$. Straightforward calculation yields
$$
\mathsf{E}(X|Y=n)=n\;\mathsf{E}(X|Y=1)=3n
$$
and
$$
\mathsf{Var}(X|Y=n)=n\;\mathsf{Var}(X|Y=1)=2n
$$
and
$$
\mathsf{P}(Y=n)=\left(\frac{5}{6}\right)^n\frac{1}{6}
$$
Using the Law of Total Variance, we get
$$
\begin{align}
\mathsf{Var}(X)
&=\mathsf{E}(\mathsf{Var}(X|Y))+\mathsf{Var}(\mathsf{E}(X|Y))\\
&=\sum_{n=0}^\infty\;2n\left(\frac{5}{6}\right)^n\frac{1}{6}+\sum_{n=0}^\infty\;(3n)^2\left(\frac{5}{6}\right)^n\frac{1}{6}-\left(\sum_{n=0}^\infty\;3n\left(\frac{5}{6}\right)^n\frac{1}{6}\right)^2\\
&=\frac{2}{6}\frac{\frac{5}{6}}{(1-\frac{5}{6})^2}+\frac{3^2}{6}\left(\frac{2\left(\frac{5}{6}\right)^2}{(1-\frac{5}{6})^3}+\frac{\frac{5}{6}}{(1-\frac{5}{6})^2}\right)-\left(\frac{3}{6}\frac{\frac{5}{6}}{(1-\frac{5}{6})^2}\right)^2\\
&=10+495-225\\
&=280
\end{align}
$$
Therefore, the standard deviation is $\sqrt{280}$.
Afterword:
Although not requested in the question, the expected value of $S$ is simple to compute by the linearity of expectation. Since the probability of rolling a $6$ is $\frac{1}{6}$, the mean number of rolls is $6$. Since each non-$6$ roll has a mean of $3$ and on average there will be $5$ non-$6$ rolls, we get $\mathsf{E}(S)=5\cdot3+6=21$.
We can also compute this using the set-up for the variance above. Since $S=6+X$,
$$
\begin{align}
\mathsf{E}(S)
&=6+\mathsf{E}(X)\\
&=6+\mathsf{E}(\mathsf{E}(X|Y))\\
&=6+\sum_{n=0}^\infty3n\left(\frac{5}{6}\right)^n\frac{1}{6}\\
&=6+\frac{3}{6}\frac{\frac{5}{6}}{(1-\frac{5}{6})^2}\\
&=6+15\\
&=21
\end{align}
$$
| {
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"answer_count": 3,
"answer_id": 1
} |
Finding a matrix How might I find matrix $M\in M_2(\mathbb C)$ such that $M^t A =M^{-1}$
where $A=\left[
\begin{array}{cc}
a & a \\
a & a+1 \\
\end{array}
\right]\in M_2(\mathbb N)$ without using the brute force method of writing $M=\left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right]$? (I don't quite know how to solve the resulting system even in the brute force case anyway.)
Thanks in advance.
| Clearly $a$ is nonzero, or else $M^\top A$ is non-invertible and it cannot be equal to the invertible matrix $M^{-1}$. Now,
$$
\begin{align}
A=\begin{pmatrix}a&a\\a&a+1\end{pmatrix}
&=\begin{pmatrix}a&a\\a&a\end{pmatrix}+\begin{pmatrix}0&0\\0&1\end{pmatrix}\\
&=\begin{pmatrix}\sqrt{a}\\ \sqrt{a}\end{pmatrix}\begin{pmatrix}\sqrt{a}&\sqrt{a}\end{pmatrix}+\begin{pmatrix}0\\1\end{pmatrix}\begin{pmatrix}0&1\end{pmatrix}\\
&=\begin{pmatrix}\sqrt{a}&0\\ \sqrt{a}&1\end{pmatrix}
\begin{pmatrix}\sqrt{a}&\sqrt{a}\\0&1\end{pmatrix} = (M^{-1})^\top M^{-1}.
\end{align}
$$
So you may set $M=\begin{pmatrix}\sqrt{a}&\sqrt{a}\\0&1\end{pmatrix}^{-1}=\begin{pmatrix}\frac{1}{\sqrt{a}}&-1\\0&1\end{pmatrix}$.
| {
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"answer_count": 3,
"answer_id": 2
} |
How to simplify $\frac{(3x^{3/2}y^3)} {(x^2y^{-1/2})}^{-2}$? I can't figure this one out on my own either
$$\frac{(3x^{3/2}y^3)} {(x^2y^{-1/2})}^{-2}$$
I am a little confused on all the small rules at play here but I know that a negative exponent will flip a fraction so I square the top and then flip it. But before that I should work in the parentheses first since that is the order of operataions. I am not sure how to cancel out each of the numbers though I am a little confused what a negative fraction in the numerator does to a larger positive exponent in the denominator.
| You have:
$$\begin{align*}
\frac{(3x^{3/2}y^3)^{-2}}{(x^2y^{-1/2})} &= \frac{1}{(3x^{3/2}y^3)^2}\times \frac{1}{x^2} \times y^{1/2}\\
&= \frac{1}{(3^2)(x^{3/2})^2(y^3)^2}\times \frac{1}{x^2}\times\frac{y^{1/2}}{1}\\
&= \frac{1}{9x^3y^6}\times\frac{1}{x^2}\times \frac{y^{1/2}}{1}\\
&= \frac{1}{9x^3x^2}\times\frac{y^{1/2}}{y^6}\\
&= \frac{1}{9x^5}\times \frac{1}{y^{11/2}}\\
&= \frac{1}{9x^5y^{11/2}}.
\end{align*}$$
| {
"language": "en",
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"answer_id": 0
} |
Systems of equations finding right triangles I need help setting up the equation for the question, "Find all right triangles for which the perimeter is $24$ units and the area is $24$ square units."
I know that the area is $A = \frac12 b h$ and perimeter is $P = a + b + h$.
Using this, would the system look something like?
$a + b + c = 24$
$\frac12b c = 24$
(I'm using $c$ for consistency)
Then continue on using substitution?
| You might be more comfortable using $a$ and $b$ for the two "legs" of the triangle, and $c$ for the hypotenuse.
Then, exactly as you did, we obtain
$$a+b+c=24 \qquad \text{and}\qquad \frac{1}{2}ab=24.\qquad\qquad(1)$$
By the Pythagorean Theorem, we have
$$a^2+b^2=c^2.\qquad\qquad(2)$$
Now what? One approach (but not the only one) is to start by writing $c=24-(a+b)$, square both sides, and use Equation $(2)$ to eliminate $c$ and obtain a simple equation that involves only $a$ and $b$. We have
$$c^2=(24-(a+b))^2=24^2-48(a+b)+a^2+2ab+b^2=a^2+b^2.$$
There is a fair bit of cancellation. Note that $2ab=96$. So we get
$$48(a+b)=24^2+96$$
and therefore $a+b=14$.
Now we could write $b=14-a$ and substitute into $ab=48$ to get a quadratic equation in $a$. But the following is I think prettier. From $(a+b)^2=196$ and $4ab=192$ we conclude by subtraction that $(a-b)^2=4$, so $a-b=\pm 2$. We can decide now that without loss of generality $a>b$. So $a-b=2$. From $a+b=14$, by adding and dividing by $2$, we find that $a=8$ and $b=6$. And of course $c=24-(a+b)=10$.
| {
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} |
Geometric inequality: $2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$ Suppose $a$, $b$, and $c$ are the lengths of the sides of a triangle, and $R$ and $r$ are its circumradius and inradius respectively. How can one prove the following inequality? $$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$
| We know that
$$\begin{align*}
a+b+c&=2s,\\
ab+bc+ca&=s^2+r^2+4rR,
\end{align*}$$
where $s$ is semiperimeter, $r$ is inradius and $R$ is circumradius;
see e.g., Manfrino, Ortega, Delgado - Inequalities: A Mathematical Olympiad Approach, Section 2.5.
Using these inequalities we can rewrite the LHS as
$$
\begin{align*}
2(r^2+4rR) &= 2(ab+bc+ca-s^2) \\
&= 2(ab+bc+ca)-\frac{(a+b+c)^2}2 \\
&= ab+bc+ca-\frac{a^2+b^2+c^2}2.
\end{align*}$$
After this the inequality we want to prove becomes
$$ab+bc+ca\le a^2+b^2+c^2,$$
which follows easily from AM-GM inequality. (Simply add the inequalities $ab\le \frac{a^2+b^2}2$, $ac\le \frac{a^2+c^2}2$ and $bc\le \frac{b^2+c^2}2$. ) See also this question for more proofs of the last inequality.
For the sake of completeness, let me copy here the proofs of these equalities as they are given in the book I mentioned above. (I hope that such a short excerpt still qualifies as a fair use. I tried to find a proof of them online, but I did not succeed.)
$$
\begin{align*}
a + b + c &= 2s, \tag{2.5}\\
ab + bc + ca &= s^2 + r^2 + 4rR, \tag{2.6}\\
abc &= 4Rrs. \tag{2.7}
\end{align*}
$$
The first is the definition of $s$ and the third follows from the fact that the area of
the triangle is $\frac{abc}{4R}=rs$. Using Heron’s formula for the area of a triangle, we have
the relationship $s(s - a)(s - b)(s - c) = r^2s^2$, hence
$$s^3 - (a + b + c)s^2 + (ab + bc + ca)s - abc = r^2s.$$
If we substitute $(2.5)$ and $(2.7)$ in this equality, after simplifying we get that
$$ ab + bc + ca = s^2 + r^2 + 4Rr.$$
| {
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"answer_count": 3,
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} |
Conditional Normal $X,Y,Z$ are standard normal R.V.
What is the value of $\operatorname{E}[X|X+Y+Z=1]$ and $\operatorname{Var}(X|X+Y+Z=1)$?
I think the first one should be $1/3$ by symmetry but don't know how to tackle the second part.
Any thought on this? Thanks a lot!
| If $A$ and $B$ are jointly normal random variables, then the conditional density of $A$ given $B = b$ is a normal density with mean and variance as follows:
$$E[A\mid B = b] = \mu_A + \left.\left.\frac{\text{cov}(A,B)}{\sigma_B^2}
\right(b - \mu_B\right), ~~
\text{var}(A\mid B = b) = \frac{\sigma_A^2\sigma_B^2
- (\text{cov}(A,B))^2}{\sigma_B^2}$$
where ${cov}(A,B)$ is the covariance
of $A$ and $B$.
If $X$, $Y$, $Z$ are jointly normal, then $X$ and $X+Y+Z$ are also
jointly normal with
$$\begin{align*}
E[X+Y+Z] &= \mu_X + \mu_Y + \mu_Z,\\
\text{var}(X+Y+Z) &= \sigma_X^2 + \sigma_Y^2+\sigma_Z^2
+2\text{cov}(X,Y)+2\text{cov}(X,Z)+2\text{cov}(Y,Z)\\
\text{cov}(X,X+Y+Z) &= \sigma_X^2 +
\text{cov}(X,Y)+\text{cov}(X,Z)
\end{align*}$$
and plugging and chugging in the result of the previous
paragraph gives the desired answers for
$E[X\mid X+Y+Z=1]$ and $\text{var}(X\mid X+Y+Z=1)$.
Of course, considerable simplifications occur in
the special case when $X$, $Y$, and $Z$ are independent standard normal
random variables. We have $E[X+Y+Z]=0$, $\text{var}(X+Y+Z)=3$,
$\text{cov}(X,X+Y+Z)=1$, and so
$$E[X\mid X+Y+Z=1] = \left.\left.\frac{1}{3}\right(1-0\right) = \frac{1}{3},
~\text{var}(X\mid X+Y+Z=1) = \frac{1\times 3 - 1^2}{3} = \frac{2}{3}$$
exactly as found by others as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/96192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a sum of appropriate values of $\cos$ and $\sin$ to determine the value of a series The task is to find a sum of multiple values $\cos$ and $\sin$ to determine the value of
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}$$
Since I had no clue how to approach this I consulted Wolfram|Alpha which returned this result:
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \frac{\sin(1)}{2} + \cos(1) - \cos(0)$$
So I wrote down the partial sums of the given series and $\sin(1)$ and $\cos(1)$:
$$
\qquad\qquad\quad\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \quad - \frac{1}{4!} + \frac{2}{6!} - \frac{3}{8!} \cdots
$$
$$
\;\;\sin(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n+1}}{(2n+1)!} = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} \cdots
$$
$$
\cos(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n}}{(2n \qquad)!} = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} \cdots
$$
Looking at the numbers I can see that Wolfram|Alpha's result is correct: $\frac{1}{2}1 - \frac{1}{2!} = 0$ and $\frac{1}{2}\frac{-1}{3!} + \frac{1}{4!} = \frac{1}{4!}$, so the $\cos(1)$-series is shifted by $1$ since there is no $1$ at the beginning of the given series, so it needs to be subtracted from $\cos(1)$: $-cos(0)=-1$. But how do I get here without Wolfram|Alpha?
| Of course you know the series for $\cos(1)$ and $\sin(1)$, so you want to express this
using those.
I prefer to start the sum at $n=0$: since the $n=0$ term is $0$, this is harmless.
Note that $\frac{n}{(2n+2)!} = \frac{n}{(2n+1)!(2n+2)}$. Now
$\frac{n}{2n+2} = \frac{1}{2} - \frac{1}{2n+2}$. So
$$\sum_{n=0}^\infty (-1)^n \frac{n}{(2n+2)!} = \frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!} - \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+2)!}$$
Now note that $ \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!} = \sin(1)$ and
$$\sum_{n=0}^\infty (-1)^n \frac{1}{(2n+2)!} = \frac{1}{2!} - \frac{1}{4!} + \frac{1}{6!} - \ldots = 1 - \cos(1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How does one prove $\int_0^\infty \prod_{k=1}^\infty \operatorname{\rm sinc}\left( \frac{t}{2^{k+1}} \right) \mathrm{d} t = 2 \pi$ Looking into the distribution of a Fabius random variable:
$$
X := \sum_{k=1}^\infty 2^{-k} u_k
$$
where $u_k$ are i.i.d. uniform variables on a unit interval, I encountered the following expression for its probability density:
$$
f_X(x) = \frac{1}{\pi} \int_0^\infty \left( \prod_{k=1}^\infty \operatorname{\rm sinc}\left( \frac{t}{2^{k+1}} \right) \right) \cos \left( t \left( x- \frac{1}{2} \right) \right) \mathrm{d} t
$$
It seems, numerically, that $f\left(\frac{1}{2} \right) = 2$, but my several attempts to prove this were not successful.
Any ideas how to approach this are much appreciated.
| Since $u_k$ and $1-u_k$ are equal in distribution, it follows that
$$
X \stackrel{d}{=} \sum_{k=1}^\infty 2^{-k} \left(1-u_k\right) = \sum_{k=1}^\infty 2^{-k} - X = 1 - X
$$
Since $u_k \geq 0$, it follows that $\mathbb{P}\left(0 \leqslant X \leqslant 1\right) = 1$.
Therefore $f_X(x) = 0$ for $x < 0$ or $x>1$. It also shows that $f_X(x) = f_X(1-x)$ for $0<x<1$.
Notice that
$$ \begin{eqnarray}
f_X^\prime(x) &=& \frac{1}{\pi} \int_0^\infty \left(- t \sin\left( t \left(x-\frac{1}{2} \right) \right) \right) \operatorname{\rm sinc}\left( \frac{t}{4} \right) \prod_{k=2}^\infty \operatorname{\rm sinc}\left( \frac{t}{2^{k+1}} \right) \mathrm{d} t
\end{eqnarray}
$$
Now using $t \cdot \operatorname{\rm sinc}\left( \frac{t}{4} \right) = 4 \sin\left( \frac{t}{4} \right)$, and
$$
\sin\left( t \left(x-\frac{1}{2} \right) \right) \sin\left(\frac{t}{4} \right) = \frac{1}{2} \left[
\cos\left( \frac{t}{2} \left( (2 x-1) -\frac{1}{2} \right) \right) -
\cos\left( \frac{t}{2} \left( 2 x -\frac{1}{2} \right) \right)
\right]
$$
which gives
$$
\begin{eqnarray}
f_X^\prime(x) &=& 4 f_X(2 x) - 4 f_X(2x-1)
\end{eqnarray}
$$
Now let $0 < z \leqslant \frac{1}{2}$. Then $f_X^\prime(z) = 4 f_X(2z)$ and thus
$$
f_X(z) - f_X(0) = \int_0^{z} 4 f_X(2x) \mathrm{d} x = 2 \left(F_X(2z) - F_X(0)\right)
$$
Clearly $F(0) = 0$, $F(1) = 1$ and $F_X\left(\frac{1}{2} \right) = \frac{1}{2}$, since the distribution is symmetric about $x=\frac{1}{2}$, thus
$$
f_X\left(\frac{1}{2}\right) = f_X\left(0\right) + 2 \qquad
f_X\left(\frac{1}{4}\right) = f_X\left(0\right) + 1
$$
If I now assume that $f_X(0) = 0$, the result follows.
Added A missing proof of $f_X(0)=0$, as well as alternative proof of $f_X\left(\frac{1}{2}\right) = 2$ results from writing:
$$
X \stackrel{d}{=} \frac{u_1}{2} + \frac{1}{2} \sum_{k=2}^\infty 2^{-k+1} u_k \stackrel{d}{=} \frac{u_1+X}{2}
$$
This equality in distribution implies the following integral equation for $f_X$:
$$
f_X(x) = 2 \int_0^1 f_X(2x -u) \mathrm{d} u
$$
Substituting $x=0$ we get $f_X(0) = 2 \int_0^1 f_X(-u) \mathrm{d} u =0$, since $f_X(x)=0$ for $x<0$. Incidentally, using $x=\frac{1}{2}$ gives the desired identity as well:
$$
f_X\left(\frac{1}{2}\right) = 2 \int_0^1 f_X(1-u) \mathrm{d} u = 2 \int_0^1 f_X(u) \mathrm{d} u = 2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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For what values of $b$ does this function lack extrema? I need to find the range of values of the parameter $b$ for which the function below has no extrema.
$$ \frac{b}{3}\,8^x + (2) 4^x + (b+3) 2^x + b \ln(2)$$
In the beginning I thought it'd be easy, Very simple, just finding a range when derivative is not zero. But I haven't succeeded. Could anyone please help me out with this problem? I'd be very grateful.
$\ln(x)$ is the natural logarithm.
| Let $f(b,x) = \frac{b}{3}8^x + 2 \cdot 4^x + (b+3) 2^x + b \log(2)$. What you want is the set of values for $b$ such that $\frac{\partial}{\partial x}f(b,x)\neq 0$ for all $x\in\mathbb{R}$. Computing the partial derivative with respect to $x$ gives
$\begin{eqnarray*}
\frac{\partial}{\partial x}f(b,x) &=& \frac{\partial}{\partial x}\left(\frac{b}{3}8^x + 2 \cdot 4^x + (b+3) 2^x + b \log(2)\right)\\
&=& \frac{b}{3}\frac{\partial}{\partial x}8^x + 2 \frac{\partial}{\partial x}4^x + (b+3) \frac{\partial}{\partial x}2^x + b \log(2)\frac{\partial}{\partial x}1\\
&=&\frac{b}{3}\ln(8)8^x+2\ln(4)4^x+(b+3)\ln(2)2^x\\
&=&\ln(2)2^x(b(4^x+1)+2^{x+2}+3)
\end{eqnarray*}$
and setting this equal to $0$ gives us $b(4^x+1)+2^{x+2}+3 = 0$, which clearly cannot happen for $b\geq 0$. However, for $-4\leq b<0$ we can set $x=-1$ to get
$$b(4^{-1}+1)+2^{-1+2}+3= \frac{5b}{4}+5\geq 0$$
while for any $b<0$ the $4^x$ term is eventually much larger than the others so for sufficiently large values of $x$,
$$b(4^x+1)+2^{x+2}+3\approx b4^x<0$$
hence we must have some value of $x$ between $-1$ and $\infty$ such that $b(4^x+1)+2^{x+2}+3 = 0$ when $-4\leq b<0$. If $b< -4$ we get
$$b(4^x+1)+2^{x+2}+3< -4(4^x+1)+2^{x+2}+3=-(2^{x+1}-1)^2\leq 0$$
and so we cannot have $b(4^x+1)+2^{x+2}+3 = 0$.
This shows that $f(b,x)$ has no extrema (treated as a function in $x$) for $b\geq 0$ or $b<-4$, and has extrema when $-4\leq b<0$.
EDIT: As Andre points out, the case $b=-4$ is special and I should have treated it differently. In this case we have only $1$ root for $-4(4^x+1)+2^{x+2}+3$, and this root is an extremum (which is otherwise not the case) so it reflects an inflection point of $f(b,x)$, rather than an extremum. So in fact there are no extrema for $b=-4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given a die, what is the probability that the second roll of a die will be less than the first roll? If you are given a die and asked to roll it twice. What is the probability that the value of the second roll will be less than the value of the first roll?
| If the:
first roll is a 6 odds are: 5/6
first roll is a 5 odds are: 4/6
first roll is a 4 odds are: 3/6
first roll is a 3 odds are: 2/6
first roll is a 2 odds are: 1/6
first roll is a 1 odds are: 0/6
Therefore the total odds are the average of all those roll possibilities so:
$$
\frac{\frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6} + \frac{0}{6}}{6} = \frac{\frac{15}{6}}{6} = \frac{15}{36} = \frac{5}{12} = \frac{2.5}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/100615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "97",
"answer_count": 10,
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} |
Which is the fastest way to find the remainder when $2^{400}$ is divided by $400$?
Which is the fastest way to find the remainder when $2^{400}$ is divided by $400$?
My approach is to break up $400$ as $16 \times 25$ and then apply CRT, I was wondering if there is any other approach that gives the result faster than using CRT.
Thanks,
| As a relatively fast but purely mechanical method (not a whole lot of thought involved), I'd first note that the exponent $$400_{10}=110010000_2=2^4+2^7+2^8$$ so that $$2^{400}=2^{(2^4+2^7+2^8)}=2^{2^4}\cdot2^{2^7}\cdot2^{2^8}.$$
Now, by successively squaring:
$$\begin{align}
2^{2^1}&\equiv 4\mod 400
\\
2^{2^2}\equiv 4^2&\equiv 16\mod 400\qquad(*)
\\
2^{2^3}\equiv 16^2\equiv 256&\equiv-144\mod 400
\\
2^{2^4}\equiv (-144)^2\equiv 336&\equiv-64\mod 400
\\
2^{2^5}\equiv (-64)^2&\equiv 96\mod 400
\\
2^{2^6}\equiv 96^2&\equiv 16\equiv2^{2^2}\mod 400\qquad(*)
\\
2^{2^7}\equiv2^{2^3}&\equiv-144\mod 400
\\
2^{2^8}\equiv2^{2^3}&\equiv-64\mod 400
\end{align}$$
(Because we get the same result on the two $(*)$ lines, we now have a repeating pattern.)
So, $$\begin{align}
2^{400}&\equiv2^{2^4}\cdot2^{2^7}\cdot2^{2^8}\mod400
\\
&\equiv(-64)(-144)(-64)\mod400
\\
&\equiv176\mod400.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/100982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
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What's the prime period of $\frac{\sin 2x + \cos 2x}{\sin 2x - \cos 2x}$ What's the prime period of the following function?
$$\frac{\sin 2x + \cos 2x}{\sin 2x - \cos 2x}$$
| $$\begin{align*}
\frac{\sin 2x+\cos 2x}{\sin 2x-\cos 2x}&=\frac{\sin 2x+\cos 2x}{\sin 2x-\cos 2x}\cdot\frac{\sin 2x+\cos 2x}{\sin 2x+\cos 2x}\\\\
&=\frac{(\sin 2x+\cos 2x)^2}{\sin^2 2x-\cos^2 2x}\\\\
&=-\frac{\sin^2 2x+2\sin 2x\cos 2x+\cos^2 2x}{\cos 4x}\\\\
&=-\frac{1+\sin 4x}{\cos 4x}\\\\
&=-\sec 4x-\tan 4x\;.
\end{align*}$$
The first term has primitive period $\dfrac{2\pi}4=\dfrac{\pi}2$, and the second has primitive period $\dfrac{\pi}4$. Can you finish it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/101581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Explanation of the Fibonacci sequence appearing in the result of 1 divided by 89? What's the explanation of the Fibonacci sequence appearing in the result of 1/89, as demonstrated by http://www.goldennumber.net/Number89.htm and shown below? If you wish, also explain the relation to the number 109 too.
1 / 89 =
0 / (10 ^ 1 ) +
1 / (10 ^ 2 ) +
1 / (10 ^ 3 ) +
2 / (10 ^ 4 ) +
3 / (10 ^ 5 ) +
5 / (10 ^ 6 ) +
8 / (10 ^ 7 ) +
13 / (10 ^ 8 ) +
...
0.011235955... =
0.0 +
0.01 +
0.001 +
0.0002 +
0.00003 +
0.000005 +
0.0000008 +
0.00000013 +
...
(This question was inspired by What is special about the numbers 9801, 998001, 99980001 ..?.)
| Consider the generating function
$$f(x) = \sum_{k=0}^{\infty} F_k x^k$$
where $F_k$ are the Fibonacci numbers. The Fibonacci recurrence $F_k = F_{k-1} + F_{k-2}$ gives
$$f(x) = x + \sum_{k=2}^{\infty} (F_{k-1} + F_{k-2}) x^k = x + \sum_{k=1}^{\infty} F_k x^{k+1} + \sum_{k=0}^{\infty} F_k x^{k+2} = x + (x + x^2) f(x).$$
It follows that $(1 - x - x^2) f(x) = x$, so
$$f(x) = \frac{x}{1 - x - x^2}.$$
Substituting $x = \frac{1}{10}$, we conclude that
$$\sum_{k=0}^{\infty} \frac{F_k}{10^k} = \frac{10}{89}.$$
Similarly, substituting $x = - \frac{1}{10}$, we conclude that
$$\sum_{k=0}^{\infty} (-1)^k \frac{F_k}{10^k} = - \frac{10}{109}.$$
Generating functions are a very powerful method for understanding many sequences in combinatorics and other areas of mathematics. In this example we can use the generating function to go even further: via partial fraction decomposition we can quickly deduce Binet's formula
$$F_k = \frac{\phi^k - \varphi^k}{\phi - \varphi}$$
for the Fibonacci numbers, where $\phi, \varphi$ are the two roots of $x^2 = x + 1$, and this idea generalizes to other sequences defined by a linear recurrence.
A standard reference on generating functions is Wilf's generatingfunctionology.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{16}{27} \left( \frac a{b+c} + \frac b{a+c} +\frac c{a+b} \right) ^3 + \left( \frac{abc}{(a+b)(b+c)(a+c)}\right)^{\frac{1}{3}} \geq \frac52$ I don't quite remember where this problem is from. I came across is sometime last summer, when I was in an olympiad-problem mood and I decided to improve my inequality skills.
Suppose $a,b,c > 0$. Then we want to show that
$$\frac{16}{27} \left( \frac{a}{b+c} + \frac{b}{a+c} +\frac{c}{a+b} \right) ^3 + \left( \frac{abc}{(a+b)(b+c)(a+c)}\right)^{\frac{1}{3}} \geq \frac{5}{2}$$
I think that there are many things to notice. Firstly, it's homogenous. The left part is tantalizingly close to Nesbitt's inequality. The right part seems to demand AM-GM attention.
| First, make the substitutions
$$
x=
\frac{a}{b+c},
\quad
y=
\frac{b}{a+c},
\quad
z=
\frac{c}{a+b}.
$$
The strategy will be to reduce the problem to an inequality in the single variable $t=(xyz)^{1/3}$. Note that $xy+yz+xz+2xyz=1$, and the inequality to be proved is
$$
\frac{16}{27}\left(x+y+z\right)^3+(xyz)^{1/3}\geq\frac{5}{2}.
$$
Now
$$
\frac{(x+y+z)^2}{3}\geq xy+yz+xz=1-2xyz
$$
and also $x+y+z\geq3/2$ by Nesbitt's inequality. Therefore,
$$
\frac{16}{27}(x+y+z)^3=\frac{16}{9}\cdot(x+y+z)\cdot\frac{(x+y+z)^2}{3}\geq\frac{8}{3}(1-2xyz),
$$
and it is sufficient to prove the inequality
$$
\frac{8}{3}(1-2xyz)+(xyz)^{1/3}\geq\frac{5}{2}.
$$
Now $xyz\leq1/8$, because AM-GM gives
$8abc\leq (a+b)(b+c)(a+c)$ by grouping pairs on the right-hand side (e.g., $2abc\leq a^2b+bc^2$). Thus by setting $t=(xyz)^{1/3}$, we are reduced to proving that the polynomial
$$
f(t):=
8\left(\frac{1-2t^3}{3}\right)+t-\frac{5}{2}
=
\frac{1}{6}+t-\frac{16}{3}t^3.
$$
is nonnegative for $t\in[0,1/2]$. Since $f(0)>0$ and $f(1/2)=0$, we can show that $f(c)>0$ whenever $c$ is a critical point of $f$. But $f'(t)=1-16t^2$, which has $c=1/4$ as its only zero in $[0,1/2]$. As $f(1/4)=4/12>0$, we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Perfect Squares ending in 576 I want to find out perfect squares ending in 576, after the number 576.
Here is my derivation to arrive at such a number. Let the perfect square ending in $576$ be $1000k+576$. Every perfect square can be expressed as a the sum of a certain number of consecutive odd numbers. For eg: $2^2 = 1+3$, $3^2 = 1+3+5$, $4^2 = 1+3+5+7$, and so on..
Hence I can write my required perfect square ending in 576 as -
$$(1+3+5+7+\cdots+49) + \underbrace{(51+53+55+57+\cdots)}_{(n\text{ summands})}$$
Therefore,
$$(1+3+5+7+\cdots+49) + \underbrace{(51+53+55+57+\cdots)}_{
(n \text{ summands})} = 1000k +576.$$
Since $(1+3+5+7+ ....49) = 576$, the equation reduces to
$$\underbrace{(51+53+55+57+\cdots)}_{n\text{ summands})} = 1000k$$
Using formula for Arithmetic Progression starting with 51 and a common difference of 2,
$$\begin{align*}
\frac{n}{2}\left(2(51) + (n-1)2\right) &= 1000k\\
n(n+50) &= 1000k
\end{align*}$$
Put $n = 100$, $100\times 150 = 1000k$, hence $k = 15$.
Put $k = 15$ in the perfect square term $1000k+576$ we get the number $15576$.
But $15576$ is not a perfect square.
What is flawed in my derivation? Kindly help.
| The flaw is
Since, (1+3+5+7+ ....49) = 576
The sum is actually 625. You need to stop a term earlier.
| {
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"timestamp": "2023-03-29T00:00:00",
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Method for determining irreducibles and factorising in $\mathbb Z[\sqrt{d}]$ I know that $\mathbb Z[\sqrt{7}]$ is a UFD, and I can write the equation $(2 + \sqrt{7})(3 - 2\sqrt{7}) = (5 - 2\sqrt{7})(18 + 7\sqrt{7}) $. So clearly these are not all irreducibles. How do I determine which of these aren't irreducible? How do I factor those into irreducibles?
It seems difficult when dealing with $\mathbb Z [\sqrt{d}] $ with $d$ positive, since the norm map isn't as nice (it's not always positive).
Thanks
| An irreducible of $\mathbb{Z}[\sqrt{7}]$ must be a prime of $\mathbb{Z}$, or must lie above a prime of $\mathbb{Z}$.
Since $\mathbb{Z}[\sqrt{7}]$ is the splitting field of $x^2-7$, you want to determine how $x^2-7$ factors modulo $p$, for $p$ prime. In other words, for which primes is $7$ a square modulo $p$?
If $p=2$ or $p=7$, $x^2-7$ factors as a perfect square; so $2$ and $7$ ramify; that is, there is a unique irreducible (up to units) that divides $2$ and that divides $7$. It is not hard to find $3+\sqrt{7}$ as a divisor of $2$: $(3+\sqrt{7})(3-\sqrt{7}) = 2$. Note that $N(3+\sqrt{7}) = 2$ is a prime, so that means that $3+\sqrt{7}$ is necessarily irreducible. And of course, $\sqrt{7}$ divides $7$, and is irreducible.
For $p\neq2, 7$, we have: for $p\equiv 1\pmod{4}$,
$$\left(\frac{7}{p}\right) = \left(\frac{p}{7}\right) = \left\{\begin{array}{ll}
1 & \text{if }p\equiv 1,2,4\pmod{7}\\
-1 & \text{if }p\equiv 3,5,6\pmod{7}.
\end{array}\right.$$
For $p\equiv 3\pmod{4}$,
$$\left(\frac{7}{p}\right) = -\left(\frac{p}{7}\right) = \left\{\begin{array}{ll}
1 & \text{if }p \equiv 3,5,6\pmod{7}\\
-1 &\text{if }p\equiv 1,2,4\pmod{7}
\end{array}\right.$$
So a rational prime $p$ splits into a square (times a unit) if $p=2$ or $p=7$; factors into two distinct irreducibles in $\mathbb{Z}[\sqrt{7}]$ if and only if $p\equiv 1, 3, 9, 19, 25, 27\pmod{28}$; and remains irreducible if $p\equiv 5, 11, 13, 15, 17, 23$.
The units are precisely the elements of norm $\pm 1$; in fact, it cannot be $-1$, since that would require $a^2 -7b^2 =-1$, which has no solutions modulo $7$. You can find, through the "usual" methods, that every unit is of the form $\pm(8+3\sqrt{7})^k.$
| {
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Correlation between variables I asked this question on stats SE but did not find a suitable answer so far. Maybe someone can help.
Given n random variables x1,...,xn (one-dimensional).
The following is known (corr() = Pearson correlation):
corr(x1,x2) = a
corr(x2,x3) = a
The actual values of the random variables and their covariances are unkown though. Only some of their correlations are known.
From this, is it possible to calculate
corr(x3,x1) = ?
or give an estimate of the lowest possible correlation coefficient
corr(x3,x1) > a
More generally:
Given set of correlations
corr(x_i, x_i+1) with i=[1..c], c<n
is it possible to either directly calculate
corr(x_1, x_c+1)
or give a lower bound a of the coefficient with
corr(x_1, x_c+1) > a
| I find it most intuitive to use the cholesky-decomposition of some correlation-matrix to look at such questions. The cholesky-decomposition provides a lower triangular matrix which always has (given the variables $\small x_1,x_2,x_3 $) the form
$\qquad \small \begin{array} {r|lll}
x_1: & 1 & . & . & \\
x_2: & a_1 & a_2 & . \\
x_3: & b_1 & b_2 & b_3 \\
\end{array} $
which can be continued to more rows/columns and where the dots mean (systematical) zeroes.
The squares of the entries of one row sum up to 1 , and the correlations are the sum of the products of the entries along two rows, say for $\small corr(x_1,x_2)=1 \cdot a_1 $ or $\small corr(x_2,x_3)=a_1 \cdot b_1 + a_2 \cdot b_2 $
If we now want to know the possible range for the correlation $\small corr(x_2,x_3) $ given $\small corr(x_1,x_2)=a $ and $\small corr(x_1,x_3)=b $ then we know immediately that a,b must be the entries in the first column:
$\qquad \small \begin{array} {r|lll}
x_1: & 1 & . & . & \\
x_2: & a & a_2 & . \\
x_3: & b & b_2 & b_3 \\
\end{array} $
and by the rule of sum-of-squares = 1 we get
$\qquad \small \begin{array} {r|lll}
x_1^*: & 1 & . & . & \\
x_2^*: & a^2 & 1-a^2 & . \\
x_3^*: & b^2 & b_2^2 & 1-b^2-b_2^2 \\
\end{array} $
Here all except the entry $\small b_2$ are fixed or determined by the choice of $\small b_2$, which is also limited to the obvious interval $\small 0 \le b_2^2 \le 1-b^2$.
Let's for simpliness assume a and b are positive values.
Then it is also obvious, that we get the possible range for the correlation $\small corr(x_2,x_3) $ if we set $\small x_2 $
*
*to its maximum, that is $\small b_2^2 = 1-b^2, b_2=\sqrt{1-b^2} b_3=0$
$\qquad \small \begin{array} {r|lll} x_1: & 1 & . & . &
\\ x_2: & a & \sqrt{1-a^2} & . \\ x_3: & b & \sqrt{1-b^2} & 0 \\
\end{array} $
and $\small corr(x_2,x_3)=a \cdot b + \sqrt{1-a^2}\cdot \sqrt{1-b^2} $
If a=b we have then $\small corr(x_2,x_3)=a^2 + (1-a^2) = 1 $
*to some mean value, (which, when we allow only positive values for all entries
is also its minimum) that is $\small b_2^2 = 0, b_3^2=1-b^2,b_3=\sqrt{1-b^2}$ and
$\qquad \small \begin{array} {r|lll}
x_1: & 1 & . & . & \\
x_2: & a & \sqrt{1-a^2} & . \\
x_3: & b & 0 & \sqrt{1-b^2} \\
\end{array} $
and $\small corr(x_2,x_3)=a \cdot b + 0 $
If a=b we have then $\small corr(x_2,x_3)=a^2 + 0 $
*to its minimum (possibly negative, and then not minimal in its absolute value), that is $\small b_2^2 = 1-b^2, b_2=-\sqrt{1-b^2} ,\qquad b_3=0$
$\qquad \small \begin{array} {r|lll} x_1: & 1 & . & . &
\\ x_2: & a & +\sqrt{1-a^2} & . \\ x_3: & b & - \sqrt{1-b^2} & 0 \\
\end{array} $
and $\small corr(x_2,x_3)=a \cdot b - \sqrt{1-a^2}\cdot \sqrt{1-b^2} < a\cdot b $
If a=b then we get $\small corr(x_2,x_3)=a \cdot a - \sqrt{1-a^2}\cdot \sqrt{1-a^2} = 2a^2-1 < a^2 $ which might also come out to be zero or even negative.
Completely similarly this can be done if more variables in the correlation-matrix are existent, because only the number of rows/columns in the cholesky-factor increases accordingly.
(Remark: for simpliness of the exposition of the principle of that calculations I did not attempt a more exact case-distinction)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/106101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
$f(x) = x^\frac{1}{3}$ is continuous at any $p \neq 0$. Trying to show $f(x) = x^\frac{1}{3}$ is continuous at any $p \neq 0$. The case with 0 is easy, this one has me stuck. I should use that $a^3 - b^3 = (a-b)(a^2+ab+b^2).$ I started off by $$|x^\frac{1}{3} - p^\frac{1}{3}| = |x^\frac{1}{3} - p^\frac{1}{3}| \frac{|x^\frac{2}{3}+(xp)^\frac{1}{3}+p^\frac{2}{3}|}{|x^\frac{2}{3}+(xp)^\frac{1}{3}+p^\frac{2}{3}|} = \frac{|x-p|}{|x^\frac{2}{3}+(xp)^\frac{1}{3}+p^\frac{2}{3}|}.$$ The numerator is exactly what I want it to be. The denominator is giving me headache because $(xp)^\frac{1}{3}$ can be negative, so i can't do $\leq \frac{|x-p|}{|p^\frac{2}{3}|}$ and get the $\delta$ easily. I have already shown that either $x^\frac{2}{3}+(xp)^\frac{1}{3} = x^\frac{1}{3}(x^\frac{1}{3} + p^\frac{1}{3}) \geq 0$ or $p^\frac{2}{3}+(xp)^\frac{1}{3} = p^\frac{1}{3}(x^\frac{1}{3} + p^\frac{1}{3}) \geq 0$ (or both). The first case allows me to proceed to $\leq \frac{|x-p|}{|p^\frac{2}{3}|}$, but the second leaves me with $\leq \frac{|x-p|}{|x^\frac{2}{3}|}$, so I can't do anything about the variable quantity. Is there an easier way to solve the problem? I would like not to go case-by-case, but that might be the option.
| If you make sure that $|x-p|<|p|$ (as part of how you choose $\delta$), then $x$ and $p$ have the same sign, and all terms in the denominator are positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/107732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How can I make the following 2 fractions integers? Let $m,n$ be integers. I want to find the possible values of $m,n$ such that $4(m+n)\over (2m+n)^2+3n^2$ and $4n\over (2m+n)^2+3n^2$ are both integers too. Would someone please help? Of course letting $(2m+n)^2+3n^2=4$ gives some good values, but is this all the $m,n$ I can get?
Added: I can see that the problem can be reduced to asking for $4k\over (2m+n)^2+3n^2$ to be an integer for both $k=m,n$
| We find all integer values of $m$ and $n$ such that $\frac{4n}{(2m+n)^2+3n^2}$ is an integer. Once this is done, your problem that imposes additional conditions is easily solved.
If $3n^2 > 4n$, then the bottom has absolute value greater than the absolute value of the top. Thus all but $n=0$, $n=\pm 1$ are immediately ruled out.
If $n=0$, we are looking at $\frac{0}{4m^2}$, which is an integer for all non-zero $m$.
If $n=1$, then we are looking at $\frac{4}{4m^2+4m+4}$, or equivalently at $\frac{1}{m^2+m+1}$. This is an integer only in the cases $m^2+m+1=\pm 1$. The equation $m^2+m+1=1$ has the solutions $m=0$ and $m=-1$. The equation $m^2+m+1=-1$ has no real solutions, let alone integer solutions.
By a similar argument, or by symmetry, the case $n=-1$ gives the solutions $m=0$ and $m=-1$.
So for integer values of $n$ and $n$, $\frac{4n}{(2m+n)^2+3n^2}$ is an integer precisely in the following cases:
(i) $n=0$, $m$ arbitrary non-zero; (ii) $n=1$, $m=0$ or $m=-1$; and (iii) $n=-1$, $m=0$ or $m=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Telescoping Series I have a question about a particular formula that is supposed to be used to simplify difficult summations into telescoping series. The formula is as follows.
$$\sum k(k+1) = \sum \frac{1}{3} \Big(k(k+1)(k+2) - (k-1)k(k+1)\Big) $$
So now here is the question. How would this be used if one were to sum $k^3$? Thank you very much in advance and I sincerely apologize for any MathJax errors in advance.
| We have
$$k(k+1)(k+2)= \frac{1}{4}(k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)),$$
which gives a telescoping sum to yield $$\sum_{k=1}^n k(k+1)(k+2)=\frac{1}{4}n(n+1)(n+2)(n+3).$$
Similarly, from the identity you wrote we get $$\sum_{k=1}^n k(k+1)=\frac{1}{3}n(n+1)(n+2),$$ and from the identity $$k=\frac{1}{2}(k(k+1)-(k-1)k)$$ we get $$\sum_{k=1}^n k = \frac{1}{2}n(n+1).$$
Finally, since $$k^3=k(k+1)(k+2) - 3k(k+1) +k,$$ you can combine these results to get
$$\sum_{k=1}^n k^3 = \frac{1}{4}n(n+1)(n+2)(n+3) -3\cdot\frac{1}{3}n(n+1)(n+2)+\frac{1}{2}n(n+1).$$ This can be written in a nicer form.
Here's a somewhat easier way: $$k^3=(k-1)k(k+1) + k,$$ so a similar telescoping yields
$$\sum_{k=1}^n k^3=\frac{1}{4}(n-1)n(n+1)(n+2)+\frac{1}{2}n(n+1).$$ (Of course both simplify to the same answer Peter shows.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a differential equation by power series. Here's a question I'm struggling with and I'd like help on it. Please, this is not a homework problem.
I want to find the power series solutions about the origin of two linearly independent solutions of $$w''-zw=0.$$
Also, how do I show that these solutions are analytic?
thanks.
| $w(x)=a_0+a_1x+\frac{a_2x^{2}}{2!}+\frac{a_3x^{3}}{3!}+...$
$w''(x)=a_2+a_3x+\frac{a_4x^{2}}{2!}+\frac{a_5x^{3}}{3!}+...$
$w''-xw=0$
$(a_2+a_3x+\frac{a_4x^{2}}{2!}+\frac{a_5x^{3}}{3!}+...)-x(a_0+a_1x+\frac{a_2x^{2}}{2!}+\frac{a_3x^{3}}{3!}+...)=0$
$a_2+(a_3-a_0)x+(\frac{a_4}{2!}-a_1)x^{2}+(\frac{a_5}{3!}-\frac{a_2}{2!})x^{3}+....=0$
$a_2=0$
$a_3-a_0=0$
$\frac{a_4}{2!}-a_1=0$
$\frac{a_5}{3!}-\frac{a_2}{2!}=0$
$[n>2]$
$\frac{a_n}{(n-2)!}-\frac{a_{n-3}}{(n-3)!}=0$
$a_n=(n-2)a_{n-3}$
$a_0=c_1$
$a_1=c_2$
$a_2=0$
$a_3=a_0=c_1$
$a_4=2a_1=2c_2$
$a_5=3a_2=0$
if $n>=0$ then $a_{3n+2}=0$
$w(x)=c_1+c_2x+\frac{c_1x^{3}}{3!}+\frac{2c_2x^{4}}{4!}+\frac{4c_1x^{6}}{6!}+\frac{2.5c_2x^{7}}{7!}+\frac{4.7c_1x^{9}}{9!}+\frac{2.5.8c_2x^{10}}{10!}+......$
$w(x)=c_1(1+\frac{x^{3}}{3!}+\frac{4x^{6}}{6!}+\frac{4.7x^{9}}{9!}+.....)+c_2(x+\frac{2x^{4}}{4!}+\frac{2.5x^{7}}{7!}+\frac{2.5.8x^{10}}{10!}+.....)$
$w(x)=c_1(1+\sum_{k=1}^\infty \frac{1.4.7...(3k-2)x^{3k}}{(3k)!})+c_2(x+\sum_{k=1}^\infty \frac{2.5.8...(3k-1)x^{3k+1}}{(3k+1)!})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Easy way to solve this non-linear second degree DY $\sqrt{y}\;y''=1$? I prooceeded by integrating both sides $$y'=\int y^{-\frac{1}{2}} dx=\cdots$$
so I got $(y')^{2}+C y' - \frac{1}{2} \sqrt{y} = 0$ but I am thinking that I am proceeding the wrong or the hard way. Some easy to solve this kind of 2nd-degree DYs?
Page 633 on the book I have been reading earlier.
| $$y'=\int y^{-\frac{1}{2}} dx=...$$
$$y''= y^{-\frac{1}{2}} $$
$$y'y''= y'y^{-\frac{1}{2}} $$
$$\int y'y'' dx=\int y^{-\frac{1}{2}} y'dx$$
$$\frac {y'^{2}}{2} = 2y^{\frac{1}{2}} +k $$
$$y'^{2} = 4y^{\frac{1}{2}} +2k=4y^{\frac{1}{2}} +c $$
$$y' = \sqrt{4y^{\frac{1}{2}} +c} $$
$$\frac{y'}{\sqrt{4y^{\frac{1}{2}} +c}} = 1 $$
$$\int \frac{y'}{\sqrt{4y^{\frac{1}{2}} +c}} dx=\int dx$$
If you select $$ u^{2}= 4y^{\frac{1}{2}} +c $$
$$ y= \frac{(u^{2}-c)^{2}}{16} $$
$$ y'= uu'\frac{(u^{2}-c)}{4} $$
$$\int \frac{u'u\frac{(u^{2}-c)}{4}}{u} dx=\int dx$$
$$\int \frac{u'u\frac{(u^{2}-c)}{4}}{u} dx=x+c_1$$
$$\int \frac{u'u^{2}}{4} dx -\int \frac{u'c}{4} dx=x+c_1$$
$$\frac{u^{3}}{12} -\frac{cu}{4} =x+c_1$$
After solving cubic equation you must put $$ u= \sqrt{4y^{\frac{1}{2}} +c} $$
then you must find y depend on X
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/110339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In how many ways can a batsman score a century in all $6$'s and $4$'s? A batsman scored a century in all $6$'s and $4$'s. In how many ways can he do this?
The given answer is $8$, but there is no explanation, how are they doing it?
As century is $100$ runs a very loose translation of this problem would be "In how many ways we can get a sum of $100$ by using only $4$'s and $6$'s?"
| If you want exactly $100$ (and not $102$ or $104$, etc)
Then these are the possibilities
$10\cdot 6+10\cdot 4=100$,
$12\cdot 6+7\cdot 4=100$,
$14\cdot 6+4\cdot 4=100$,
$16\cdot 6+1\cdot 4=100$,
$8\cdot 6+13\cdot 4=100$,
$6\cdot 6+16\cdot 4=100$,
$4\cdot 6+19\cdot 4=100$,
$2\cdot 6+22\cdot 4=100$
And if you allow that no $6$'s required and he can make $100$ with only $4$'s then this is the $9$th possibility,
$0\cdot 6+25\cdot 4=100$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/110579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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