Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Differential of normal distribution Let
$$f(x)=\frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}}$$
(Normal distribution curve) Where $\sigma$ is constant. Is my derivative correct and can it be simplified further?
$$\begin{align} f'(x)
&=\frac d{dx}\left(\frac{\exp\left(-\frac{x^2}{2\sigma^2}\right)}{\... | Here is one approach:
$f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-x^2/2\sigma^2}$. Taking the log of both sides we get:
$ln\left(f(x)\right) = -\frac{x^2}{2 \sigma^2} + ln \left( \frac{1}{\sigma \sqrt{2 \pi}} \right)$.
Let's differentiate both sides to get:
$\frac{f'(x)}{f(x)} = -\frac{x}{\sigma^2}$, implying $f'(x) = -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/461139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Need help with logarithmic differentitation I have the expression
$$y = \sqrt{x^2(x+1)(x+2)}.$$
I have tried looking at videos but I still cannot arrive at the correct answer and don't know how to get there.
By the way, the correct answer is
$$y' = \frac{4x^2+9x+4}{2\sqrt{(x+1)(x+2)}}.$$
Please, help.
| $$y = \sqrt{x^2 (x+1)(x+2)}$$
$$\ln{y} = \ln{x} + \frac12 \ln{(x+1)} + \frac12 \ln{(x+2)}$$
$$\frac{y'}{y} = \frac{1}{x} + \frac12 \frac{1}{x+1} + \frac12 \frac{1}{x+2}$$
$$y' = x \sqrt{(x+1)(x+2)} \left ( \frac{1}{x} + \frac12 \frac{1}{x+1} + \frac12 \frac{1}{x+2}\right) = x \sqrt{(x+1) (x+2)} \left ( \frac{4 x^2+9 x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Does this integral have a closed form expression? I have been working with two related integrals. The first one yields a simple expression, but I can't seem to find a simple expression for the second.
Integral 1
$$ \begin{align*}
I_x &= \int_z^{1-y+z} \sqrt{\frac{1}{1-x-y+z} + \frac{1}{x-z}} \, dx\\
&= \int_z^{1-y+z} \... | First of all, regarding the limits of integration $a=\max{(0,x+y-1)}$ and $b=\min{(x,y)}$, you do not have to break up your analysis into four separate cases to accommodate the two possible values taken by the min/max functions. These functions can be represented in terms of elementary functions as follows:
$$\max{(a,b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_{0}^{\infty}\frac{1}{x}\big (\frac{\sinh ax}{\sinh x}-ae^{-2x}\big )dx$ Some time ago, stumbled out of an integral:
$$\int_{0}^{\infty}\frac{1}x{}\left (\frac{\sinh ax}{\sinh x}-ae^{-2x}\right )dx=\ln\frac{\pi\cos\frac{a\pi}{2}}{\Gamma^2(\frac{a+1}{2})};\left | a \right |<1$$
I have no idea where to ... | To be specific, let $I(a)$ denote the integral and differentiate the integral with respect to $a$. Then by referring to the contents and notations in my blog posting,
\begin{align*}
I'(a)
&= \int_{0}^{\infty} \left( \frac{x \cosh ax}{\sinh x} - e^{-2x} \right) \, \frac{dx}{x} \\
&= \int_{0}^{\infty} \left( \frac{x e^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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interesting Integral , alternative solution. Show the following relation:
$$\int_{0}^{\infty} \frac{x^{29}}{(5x^2+49)^{17}} \,\mathrm dx = \frac{14!}{2\cdot 49^2 \cdot 5^{15 }\cdot 16!}.$$
I came across this intgeral on a physics forum and solved by (1) making a substitution (2) finding a recursive formula (with in... | The given integral is \begin{align}
I=&\int_{0}^{\infty}\frac{x^{29}}{(5x^2+49)^{17}}dx\\
\ =&\frac{1}{2\cdot 5^{15}}\int_{0}^{\infty}\frac{z^{14}}{(z+49)^{17}}dz\quad(\mbox{substituting}\ 5x^2=z)\\
\ =&\frac{1}{2\cdot 5^{15}}\int_{0}^{\pi/2}\frac{(49)^{14}\tan^{28}\theta\cdot2\cdot49\cdot\tan \theta\sec^2\theta}{49^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Find the limit: $\lim\limits_{x\to1}\dfrac{x^{1/5}-1}{x^{1/3}-1}$ Find the limit of $$\lim_{x\to 1}\frac{x^{1/5}-1}{x^{1/3}-1}$$
How should I approach it? I tried to use L'Hopital's Rule but it's just keep giving me 0/0.
| Hint: $$\frac{x^{\frac{1}{5}}-1}{x^{\frac{1}{3}}-1}=\frac{(x^{\frac{1}{15}})^3-1}{(x^{\frac{1}{15}})^5-1}=\frac{((x^{\frac{1}{15}})-1)(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1)}{((x^{\frac{1}{15}})-1)((x^{\frac{1}{15}})^4+(x^{\frac{1}{15}})^3+(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1)}$$
This is equal to:
$$\frac{(x^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $ The question is to simplify $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
My friend has given me ... | Use $$(a+b)(a-b)=a^2-b^2.$$
With $a=\sqrt6+\sqrt7$, $b=\sqrt5$ you see that the product of first two numbers is $(\sqrt6+\sqrt7)^2-5=8+2\sqrt{42}$. With $a=\sqrt5$, $b=\sqrt7-\sqrt6$ you get that
the product of last two is $5-(\sqrt7-\sqrt6)^2=-8+2\sqrt{42}$. One last application of this rule tells you that the answer ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/465103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 2
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Factorise: $2a^4 + a^2b^2 + ab^3 + b^4$ Factorize : $$2a^4 + a^2b^2 + ab^3 + b^4$$
Here is what I did:
$$a^4+b^4+2a^2b^2+a^4-a^2b^2+ab^3+b^4$$
$$(a^2+b^2)^2+a^2(a^2-b^2)+b^3(a+b)$$
$$(a^2+b^2)^2+a^2(a+b)(a-b)+b^3(a+b)$$
$$(a^2+b^2)^2+(a+b)((a^2(a-b)) +b^3)$$
$$(a^2+b^2)^2+(a+b)(a^3-a^2b+b^3)$$
At this point I don't... | Hint
We look for a factorization on the form
$$2a^4 + a^2b^2 + ab^3 + b^4=(2a^2+\alpha ab+b^2)(a^2+\beta ab+b^2)$$
We find $\alpha=2,\beta=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/465136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Trisect a quadrilateral into a $9$-grid; the middle has $1/9$ the area Trisect sides of a quadrilateral and connect the points to have nine quadrilaterals, as can be seen in the figure. Prove that the middle quadrilateral area is one ninth of the whole area.
| This is most easily done using vectors. Let the points $A, B, C, D$ be represented by the vectors $a, b, c, d$. The area $[ABCD]$ is equal to $\frac{1}{2}(a-c) \times (b-d) $.
If you are unfamiliar with this, consider triangulation using the origin, and sum up the 4 triangle areas, to get
$$\begin{align} [ABCD] = & \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/465986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
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Sum of a sequence I need guidance for the following question.
Using the fact that $\sum_1^{\infty}\frac{(-1)^{n+1}}{n}=\log2$, $\sum_1^{\infty}\frac{(-1)^{n}}{n(n+1)}$ equals
$1.$ $1-2\log2$
$2.$ $1+2\log2$
$3.$ $(\log2)^2$
$4.$ $-(\log2)^2$
The given sequence gives us $1-\frac12+\frac13-\frac14+\cdots=log2$, but I a... | $
\begin{array}{l}
\frac{1}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}} \\
\Rightarrow \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{{n\left( {n + 1} \right)}}} = \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} }}{n}} - \sum\limits_{n = 1}^{ + \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/468406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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The Laurent series of the digamma function at the negative integers To find the Laurent series of $\psi(z)$ at $z= 0$, I would first find the Taylor series of $\psi(z+1)$ at $z=0$ and then use the functional equation of the digamma function.
Specifically,
$$\begin{align} \psi(z + 1) = \frac{1}{z} + \psi(z) &= \psi(1)... | We will use the Laurent expansion around $x=0$
$$\psi_0(x)=-\frac{1}{x}-\gamma-\sum_{k\geq 1}\zeta(k+1)(-x)^k$$
Look the proof here
Then
$$\tag{1}\psi_0(x+N)=-\frac{1}{x+N}-\gamma-\sum_{k\geq 1}(-1)^k\zeta(k+1)(x+N)^k$$
Now we use that
$$\tag{2} \psi_0(x+N)=\psi_0(x)+\sum_{k=0}^{N-1}\frac{1}{x+k}$$
Now we look at the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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How prove this inequality $\sum\limits_{1\le ilet $n\ge 2,n\in Z$,and $x_{1},x_{2},\cdots,x_{n}\in[0,1]$,
show that
$$\sum_{1\le i<j\le n}ix_{i}x_{j}\le\dfrac{n-1}{3}\sum_{k=1}^{n}kx_{k}$$
This problem is (2013,8.16) chia west compition
my idea:
let
\begin{align*}
&\dfrac{n(2n+1)(n+1)}{6}=n^2+(n-1)^2+\cdots+1\\
&\ge(... | For more convenient notation, let $L_n := \sum_{1 \leq i<j\leq n} i x_i x_j$ and $R_n := \frac{n-1}{3} \sum_{1 \leq i \leq n} i x_i$. We need to prove that $L_n \leq R_n$.
We proceed by induction, the case $n = 1$ being trivial. Suppose that the claim holds for $n$, we will prove it for $n+1$. It will suffice to show t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Logarithm integral with golden ratio $\int^1_0 \frac{\log(1+\phi x^2)}{1+x}\, dx$ How to evaluate the following integral
$$ \int^1_0 \frac{\log(1+\phi x^2)}{1+x}\, dx$$
where
$$\phi =\frac{\sqrt{5}+1}{2}$$
Is there a closed form for the integral ?
My first attempt would involve relating the integral to dilogarithms b... | Let's replace $\phi$ with a positive parameter $a$:
$$
\mathcal{I}(a) = \int_0^1 \frac{\log(1+a x^2)}{1+x} \mathrm{d}x \stackrel{x^2=t}{=} \int_0^1 \frac{\log(1+ a t)}{1+\sqrt{t}} \frac{\mathrm{d}t}{2 \sqrt{t}}
$$
Differentiating under the integral sign:
$$\begin{eqnarray}
\mathcal{I}^\prime(a) &=& \int_0^1 \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/471672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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Minima problem? This is a question in my textbook which I can't solve. Any help would be appreciated, thanks.
"A piece of wire 10 metres long is cut into two portions. One piece is bent to form a circle, and the other piece to form a square. Find the circumference of the circle if the sum of areas of the circle and squ... | Given Total length of wire $10$ meter, Now Let $x$ meter cut from it and form a Circle and
remaining $(10-x)$ form a Square.
So Here Radius of Square is $\displaystyle \frac{x}{2\pi}$ and Length of square is $\displaystyle \frac{(10-x)}{4}$
Now Area of Circle $\displaystyle \pi\cdot \frac{x^2}{4\pi^2} = \frac{x^2}{4\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/474861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Trigonometric Identities for $\sin nx$ and $\cos nx $ These are generalizations of simple trigonometric identities for $\sin 2x$ and $\cos 2x$, but in general how can we prove them?
$$\sin nx =\sum_{k=1}^{\left\lceil\frac{n}{2}\right\rceil}(-1)^{k-1}\binom{n}{2k-1}\sin^{2k-1}x\cos^{n-2k+1}x,$$
$$\cos nx =\sum_{k=0}^{\... | Have a look at http://en.wikipedia.org/wiki/Chebyshev_polynomials. Once you get the first two identities, you'll easily get the last two. As pbs suggested, you can use the Binomial Theorem. The reason is :
$$\cos(n\theta) + i\sin(n\theta) = e^{in\theta} = (e^{i\theta})^{n} = (\cos(\theta)+i\sin(\theta))^{n}$$
The idea ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Numbers of solutions $e^x=-x^2+2x+5 $ Prove that $e^x=-x^2+2x+5 $ have exactly two solutions.
Is it enoguht that Vertex of the parabola is over $y=e^x$ and arms of it looks down
| This is a little out of the blue, but...
Since $e^x > 0$ for all $x \in \mathbb{R}$, there can only be two points where RH intersects $e^x$ since $e^x = -x^2+2x+5 \implies x^2-2 x+e^x = 5$.
If we rewrite $x^2-2 x+e^x = 5$ as $x^2-2 x+e^x = y$, observe that its minima is at $x^2-2 x+e^x = W(\frac{e}{2}) (W(\frac{e}{2})+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the number of solutions I am trying to compare my answer with a friend's and we are both confident in our answers. But the problem is, they are different. So the problem goes:
Suppose I have the equation
$$x+y+z+w = 14$$
where $x,y,z,w \in \mbox{Z}^+$ (i.e. integers that can be zero or positive) such that $x,y... | Your friend is counting the solutions where $ 1 \le w,x,y,z \le 6$; you are counting the solutions where $ 0 \le w,x,y,z \le 6$.
Since the problem statement says $x, y, z, w \in Z^+$, I think your friend has the proper interpretation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/479731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\max\{x,y\}=\frac {(x-y)}{2} + \frac {(x+y)}{2}$ Problem: prove $\max\{x,y\}=\frac {(x-y)}{2} + \frac {(x+y)}{2}$
$x$ and $y$ are max elements in two sets.
Here is what i have thought of so far as a concrete problem:
$\max \{4,5\}=d(4,5)= 1$
I know that the $$\begin{align} \max\{4,5\} &= \frac{4-5}{2}+ \frac{4... | The problem is wrong as $\frac{x+y}2+\frac{x-y}2$ is always equal to $x$
where as max$(x,y)=x$ iff $x\ge y$
In fact max$\displaystyle(x,y)=\frac{x+y}2+\frac{|x-y|}2$
If $x\ge y,$
max$(x,y)=x$ and $|x-y|=x-y\implies \frac{x+y}2+\frac{|x-y|}2=\frac{x+y+(x-y)}2=x$
If $x<y,$
max$(x,y)=y$ and $|x-y|=-(x-y)\implies \frac{... | {
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"url": "https://math.stackexchange.com/questions/481038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that roots are real I am stuck with this equation, I need to prove the roots are real when $a, b, c \in R$
The equation is $$(a+b-c)x^2+ 2(a+b)x + (a+b+c) = 0$$
If someone could tell me the right way to go about this, so I can attempt it.
Thank you
EDIT: I have made an error in the question. I have now corrected... | $$ax^2 +bx^2 + cx^2 + 2ax+2bx + a+b+c = 0$$
or
$$x^2(a+b-c) + 2x(a+b) +a+b+c = 0$$
Let us now find the expression for the discriminant,
$$\Delta = 4(a+b)^2-4(a+b-c)(a+b+c)$$
If we prove that, $\Delta\geq 0$, we done.
$$4(a+b)^2-4(a+b+c)(a+b-c)=$$
$$=4a^2+8ab+4b^2-4(a^2+2ab+b^2-c^2)=$$
$$=4a^2+8ab+4b^2-4a^2-8ab-4b^2+4c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve system equation $\left\{ \begin{array}{l} xy - x - y = 1\\ 4{x^3} - 12{x^2} + 9x = - {y^3} + 6y + 7 \end{array} \right.$ Solve system equation : $$\left\{ \begin{array}{l} xy - x - y = 1\\ 4{x^3} - 12{x^2} + 9x = - {y^3} + 6y + 7 \end{array} \right. ,\quad (x,y\in\mathbb{R}).$$
My solution begin with : Set $z... | Using the straightforward approach, I find $x=\frac {y+1}{y-1} $ from the first equation and substitute it in the second one $$\frac {{y}^{6}-3\,{y}^{5}-3\,{y}^{4}+11\,{y}^{3}-6\,{y}^{2}+32}{
\left( y-1 \right) ^{3}}=0.
$$ The keypoint is the next step $$y^6-3y^5-3y^4+11y^3-6y^2+32 =(y^2-y+2)(y^2-y-4)^2.$$ The rest ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factoring $a^3-b^3$ I recently got interested in mathematics after having slacked through it in highschool. Therefore I picked up the book "Algebra" by I.M. Gelfand and A.Shen
At problem 113, the reader is asked to factor $a^3-b^3.$
The given solution is:
$$a^3-b^3 = a^3 - a^2b + a^2b -ab^2 + ab^2 -b^3 = a^2(a-b) + ab... | The second equality is obtained by first grouping terms in the middle expression and then factoring the grouped terms:
$$\begin{align}
a^3-a^2b+a^2b-ab^2+ab^2-b^3&=(a^3-a^2b)+(a^2b-ab^2)+(ab^2-b^3)\cr
&=a^2(a-b)+ab(a-b)+b^2(a-b)\cr
\end{align}$$
If the OP is wondering where the middle expression came from in the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/484281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How to find the inverse of 70 (mod 27) The question pertains to decrypting a Hill Cipher, but I am stuck on the part where I find the inverse of $70 \pmod{ 27}$. Does the problem lie in $70$ being larger than $27$?
I've tried Gauss's Method:
$\frac{1}{70} = \frac{1}{16} ^{\times2}_{\times2} = \frac{2}{32} = \frac{2}{5}... | We have $70\equiv16\pmod{27}\equiv2^4$
As $(2,27)=1,$
so using Carmichael function $\lambda(27)=\phi(27)=3^{3-1}(3-1)=18$ where $\phi$ is the Totient function
$\implies 2^{18}\equiv1\pmod{27}$
$\implies (2^a)^{-1}\equiv2^{18-a} \pmod{27}$
$\implies (2^4)^{-1}\equiv2^{14} \pmod{27}$
Now, $2^5=32\equiv5\pmod{27}\implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/484990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 4
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Does the derivative of $x^{-1}$ and of $x^3-x$ equal $-\frac{1}{x^{2}}$ and $3x^2-1$? I want to check my answers concerning the derivative of the following functions: $\displaystyle f(x)= \frac{1}{x}$ and of $\displaystyle j(x)= x^3-x$
$$\displaystyle f(x)= \frac{1}{x}$$
$$\begin{align}
f'(x) & = \lim_{h \to 0} \fra... | Your title is wrong but your limits in the actual post are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/488787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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How prove this is an equilateral triangle in $\Delta ABC$,$AB=c,AC=b,BC=a$and such
$$ab^2\cos{A}=bc^2\cos{B}=ca^2\cos{C}$$
show that
$\Delta ABC$ is an equilateral triangle
this problem I have solution,But not nice, and I think this problem have more nice methods,Thank you everyone.
my solution:
$$ab^2\cdot\dfrac{b^... | Hint:divide all the three by abc and then proceed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Find min $P$: $P=\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b)$ Let $a,b,c\geq 0$ and $a+b+c=1$. Know that never have two numbers both zero. Find min $P$:
$$P=\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b)$$
| The question is tagged integral-inequality. Therefore, I assumed that an integer solution is sought for.
At least one of two integer variables must be unequal to zero
(= greater equal one):
$$a + b \geq 1$$
$$a + c \geq 1$$
$$b + c \geq 1$$
With $a+b+c = 1$ we get:
$$c = 1 - (a+b) \leq 0$$
Because $c \geq 0$, it follow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate limits $ \lim_{x\to+\infty} \frac{3x-1}{x^2+1}$ and $\lim_{x\to-\infty} \frac{3x^3-4}{2x^2+1}$ I want to calculate the following limits
$$\begin{matrix}
\lim_{x\to+\infty} \frac{3x-1}{x^2+1} & \text{(1)} \\
\lim_{x\to-\infty} \frac{3x^3-4}{2x^2+1} & \text{(2)}
\end{matrix}$$
In both cases we have indeterminat... | I think we can change the limit $\to0$ to even avoid L'Hosiptal's Rule as follows:
Putting $\frac1x=h$
$$\lim_{x\to\infty}\frac{3x-1}{x^2+1}=\lim_{h\to0}\frac{\frac3h-1}{\frac1{h^2}+1}=\lim_{h\to0}\frac{(3-h)h}{1+h^2}=0$$
$$\lim_{x\to\infty}\frac{3x^3-4}{2x^2+1}=\lim_{h\to0}\frac{\frac3{h^3}-4}{\frac2{h^2}+1}=\lim_{h\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find $(a,b)$ if $x^2-bx+a = 0, x^2-ax+b = 0$ both have distinct positive integers roots
If $x^2-bx+a = 0$ and $x^2-ax+b = 0$ both have distinct positive integers roots, then what is $(a,b)$?
My Try:
$$\displaystyle x^2-ax+b = 0\Rightarrow x = \frac{a\pm \sqrt{a^2-4b}}{2}$$
So here $a^2-4b$ is a perfect square.
Simila... | Suppose $x^2-ax+b$ and $x^2-bx+a$ have positive integer roots. Then
$$x^2-ax+b=(x-c)(x-d)\qquad\text{and}\qquad x^2-bx+a=(x-e)(x-f),$$
for some positive integers $c$, $d$, $e$ and $f$. As the roots are must be distinct, we may assume without loss of generality that $c>d$ and $e>f$. Now compare coefficients.
Hint 1:
I... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Can an integer of the form $4n+3$ written as a sum of two squares? Let $u$ be an integer of the form $4n+3$, where $n$ is a positive integer. Can we find integers $a$ and $b$ such that $u = a^2 + b^2$? If not, how to establish this for a fact?
| Lemma 1: $a$ is odd $\Longrightarrow$ $a^2\equiv 1(\operatorname{mod} 4)$.
Proof: $a^2-1=(a-1)(a+1)$. Since $a$ is odd, both $a-1$ and $a+1$ are even, so that $a^2-1$ is divisible by $4$. $\blacksquare$
Lemma 2: $a$ is even $\Longrightarrow$ $a^2\equiv 0(\operatorname{mod} 4)$.
Proof: Trivial. $\blacksquare$
Now, suppo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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a - b > 0 algebra correction Simple Algebra
$\frac{1}{2}-\frac{1}{5} > 0$
$\frac{1}{2} > \frac{1}{5}$
$5 >2$
Looks correct, but where am I wrong in this,
$\frac{1}{2}-\frac{1}{5} > 0$
$-\frac{1}{5} > -\frac{1}{2}$
$\frac{1}{5} > \frac{1}{2}$
$ 2 > 5$
| $-\frac{1}{5} > -\frac{1}{2}$ implies that 1/2>1/5 not the other way around
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Explicit Formula for a Recurrence Relation for {2, 5, 9, 14, ...} (Chartrand Ex 6.46[b])
Consider the sequence $a_1 = 2, a_2 = 5, a_3 = 9, a_4 = 14,$ etc...
(a) The recurrence relation is: $a_1 = 2$ and $a_n = a_{n - 1} + (n + 1) \; \forall \;n \in [\mathbb{Z \geq 2}]$.
(b) Conjecture an explicit formula for $a_n$... | Let $2\leq n$, then you can write
$$a_{2}-a_{1}=2+1$$
$$a_{3}-a_{2}=3+1$$
$$...$$
$$a_{n}-a_{n-1}=n+1.$$
Now addition of all this equality implies that
$$a_{n}-a_{1}=(2+3+..+n)+(n-1),$$
$$a_{n}=(2+3+..+n)+(n+1)-1,$$
we know that $1+2+..+n=\frac{n^{2}+n}{2}$, so
$$a_{n}=\frac{(n+1)(n+2)}{2}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Limit $ \sqrt{2\sqrt{2\sqrt{2 \cdots}}}$
Find the limit of the sequence $$\left\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots\right\}$$
Another way to write this sequence is $$\left\{2^{\frac{1}{2}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}}2^{\frac{1}{8}... | An alternative proof (of sorts):
If the sequence has a limit, then it will be a fixed point of $f(x)=\sqrt{2x}$, and hence a solution to $x=\sqrt{2x}$. The solutions are clearly $x=0, 2$, and you should be able to show that 0 is out because all of the terms in the sequence are $\gt 1$ and hence the limit is $\geq 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a closed form of $\sum_{i=1}^n i^3$ I'm trying to compute the general formula for $\sum_{i=1}^ni^3$. My math instructor said that we should do this by starting with a grid of $n^2$ squares like so:
$$
\begin{matrix}
1^2 & 2^2 & 3^2 & ... & (n-2)^2 & (n-1)^2 & n^2 \\
2^2 & 3^2 & 4^2 ... | Try this link http://mathschallenge.net/library/number/sum_of_cubes It gives full explanation of how to compute it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/506784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Solve $X^2\equiv-1\pmod{13}$ I've been asked to solve the following equation:
$$X^2 \equiv-1\pmod{13}.$$
I am not sure what to do.
| You can try all possible solutions: $0,1,…,12$, though only half of them need to be tested.
For a more sophisticated solution, note that $13=4+9=2^2+3^2$ and so $3^2 \equiv -2^2 \bmod 13$. Since $2\cdot 7 \equiv 1 \bmod 13$, we conclude that $ -1 \equiv -2^2 \cdot 7^2 \equiv 3^2 \cdot 7^2 \equiv (3\cdot 7)^2 \equiv 8^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq 27$ How can I prove $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq 27$, given that $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$ and $x+y+z=1$.
I've already tried using that: $\frac{1}{x} +\frac{1}{y} +\frac{1}{z}\geq 9$ But I can't seem to manipulate tha... | You may also try this :
Apply AM-HM inequality on the set $\{x^2,y^2,z^2\}$ :
$$\frac {x^2+y^2+z^2}{3} \geq \frac {3}{\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}$$
$$\implies \big(x^2+y^2+z^2\big)\big({\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}\big) \geq 9$$
$$\implies {\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Three Boolean Algebra Proofs - I just don't get it! I'm having a very difficult time proving the following 3 expressions:
$$\begin{align*}
&x\cdot y\cdot z+x'\cdot z=y\cdot z+x'\cdot z\\
&x\cdot y+y\cdot z+x'\cdot z=x\cdot y+x'\cdot z\\
&(x'\cdot z'+x'\cdot y+x'\cdot z+x\cdot y)'=x\cdot y'
\end{align*}$$
I need to sho... | Essentially, we manipulate the equation in whatever way possible, trying to make the left-hand side look like the right-hand side. There's no real magic method here, we just keep trying until we succeed.
In the second example, we can do:
$\small
\begin{align*}
x\cdot y+y\cdot z+x'\cdot z &= x\cdot y + 1 \cdot (y\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$ let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}
{5}\cdots (1)$$
I find sometimes,and I find this same problem:
let $a,b,c$ are real... | Let $f(x) = \dfrac{1}{2x^2-6x+9}$. First let us first quickly sketch some properties of $f$ that we'll need. $f(x)=\dfrac{1}{2(x-\frac32)^2+\frac92}$, so its maximum is $f(\frac32)=\frac29$. $f~'(x)=\dfrac{6-4x}{(2x^2-6x+9)^2}$, and $f~''(x)=12\dfrac{2x^2-6x+3}{(2x^2-6x+9)^3}$. Setting $f~''(x)=0$ gives the inflection ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$ Question is to check if $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$
we have $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\prod \limits_{n=2}^{\infty}(\frac{n^2-1}{n^2})=\prod \limits_{n=2}^{\infty}\frac{n+1}{n}\frac{n-1}{n}=(\frac{3}{2}.\frac{1}{2})(\frac{4}... | Obviously, the product of any positive numbers that less than 1 is still less than one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/513053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Probability that the sum of two elements in a set is divisible by 3. Select 2 integers from the set $\{1,2,3,4,5,6,7,8\}$. What is the probability that their sum is divisible by 3? I am assuming this is without replacement although this is not explicitly stated.
My answer: There are $\binom{8}{2}$ ways of selecting tw... | Numbers can be classified as $3n$, $3n+1$, $3n+2$. Now the sum will be divisible by $3$ only if the pair is $(3n,3n)$ or $(3n+1,3n+2)$.
Digits of type: $3n = (3,6)$; type $3n+1= (1,4,7)$; type $3n+2 = (2,5,8)$. So the number of possible would be $1 + 3\cdot 3 = 10$. $\text{Total probability} = 10/28$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/514062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How is $a_n=(1+1/n)^n$ monotonically increasing and bounded by $3$? I was reading about how completeness is required for limits. And I came across this:
the sequence $a_n=(1+1/n)^n$ is monotonically increasing and bounded by 3 and so we expect it to converge, but that it does not converge within $\mathbb{Q}$. More gen... | From Rudin's PMA Theorem $3.31$, by the Binomial theorem,
$$ t_n=\left( 1 + \frac{1}{n} \right)^n $$
$$= 1 + 1 + \underbrace{\frac{1}{2!}\left(1-\frac{1}{n}\right)}_{\text{term } 2} + \underbrace{\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}_{\text{term } 3} + \ldots + \frac{1}{n!}\left(1-\frac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515317",
"timestamp": "2023-03-29T00:00:00",
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What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots? What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots?
$(x + a)(x + 1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$
This is of the form $ax^2 + bx + c$. Applying the quadratic fo... | (x + a)(x + 1991) + 1 = 0
(x + a)(x + 1991) = -1
if x is integer, then, there are two possible pairs.
I) x+a= 1 and x+1991 = -1
II) x+a = -1, x+1991 = 1.
if we eleminate x in both system euqations, we'll get a = 1989 and a= 1993.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the $3$ angles of triangle $ABC$ We have a non obtuse triangle $ABC$.
With $$\bf\dfrac{1}{2}\cos(2A)+\sqrt{2}\cos(B)+\sqrt{2}\cos(C)=\dfrac{3}{2}$$
Find the $3$ angles $A,B,C$.
| Given a non obtuse triangle $ABC$,
find the corresponding angles $\alpha,\beta,\gamma$
such that
\begin{align}
\tfrac12\cos2\alpha+\sqrt2\cos\beta+\sqrt2\cos\gamma
&=\tfrac32
\tag{1}\label{1}
\\
\text{or }\quad
\cos^2\alpha+\sqrt2\cos\beta+\sqrt2\cos\gamma
-2&=0
.
\tag{1a}\label{1a}
\end{align}
Using a known identi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Simplifying an inverse trig function? I am trying to figure out how to simplify this expression but I am not quite sure how these inverses work. What sort of approach should I take for this equation?
$$\tan\left(2\cos^{-1}\left(\dfrac{x}{5}\right)\right)$$
| As @Spencer notes, there exists a triangle with hypotenuse 5 and adjacent $x$. Pythagoras tells us that we must therefore have an opposite of $\sqrt{5^2 - x^2} = \sqrt{25-x^2}$
If
$$\cos(\theta) = x/5$$
then
$$\tan(\theta) = \frac{\sqrt{25-x^2}}{x}$$
Given
$$\tan(2\theta) = \frac{2\tan(\theta)}{1 - tan²(\theta)}$$... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the greatest common divisor (gcd) of $f(x) = x^2 + 1$ and $g(x) = x^6 + x^3 + x + 1$ Find the greatest common divisor (gcd) of $f(x) = x^2 + 1$ and $g(x) = x^6 + x^3 + x + 1$.
Since $x^6 + x^3 + x + 1 = (x^2 + 1)(x^4 - x^2 + x + 1)$, $\mathrm{gcd}[f(x),g(x)] = x^2 + 1$.
My question is how could I JUSTIFY that the ... | Clearly $x^{2}+1$ divides both $f$ and $g$, as you have shown. On the other hand, since $g=x^{2}+1$, any polynomial $h$ dividing both $f$ and $g$ must divide $g=x^{2}+1$. Thus, by definition, $x^{2}+1$ is the greatest common divisor.
This example was easy since $g$ divides $f$. In other cases, you could use the Euclide... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of irrational roots of the equation $(x-1)(x-2)(3x-2)(3x+1)=21$? The number of irrational roots of the equation $(x-1)(x-2)(3x-2)(3x+1)=21$ is
(A)0
(B)2
(C)3
(d)4
Actually im a 10 class student i don't know any of it,but my elder brother(IIT Coaching) cannot solve them,he told me post these questions on this sit... | Hint :
$$ (3x-2)(x-1) = 3x^2-5x+2 $$
and
$$ (3x+1)(x-2) = 3x^2 -5x-2 $$
$$ (x-1)(x-2)(3x-2)(3x+1) = (3x^2-5x-2)(3x^2-5x+2) = 21 $$
Now put
$$ 3x^2-5x = t $$
$$ (t-2)(t+2) = 21 $$
Now, solve
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$ I'm having problem with showing that:
$$\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$$
I would need some help in the right direction
| From the Article $240,$ Ex$-5$ of Plane Trigonometry(by Loney),
$$\arctan x+\arctan y=\begin{cases} \arctan\frac{x+y}{1-xy} &\mbox{if } xy<1\\ \pi+\arctan\frac{x+y}{1-xy} & \mbox{if } xy>1\end{cases} $$
$$\implies 2\arctan x=\arctan \frac{2x}{1-x^2}\text{ if }x^2<1$$
Here $\displaystyle x=\frac23$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/523625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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How to find the orthogonal trajectories of the family of all the circles through the points $(1,1)$ and $(-1,-1)$? I'm trying to find the orthogonal trajectories of the family of circles through the points $(1,1)$ and $(-1, -1)$. Now such a family can be given by an equation of the form $$ x^2 + y^2 + 2g(x-y) - 2 = 0, ... | Let $u = y+x$, $v = y-x$, we have
$$\begin{align}
u' = y'+1 = & \frac{2x^2-2y^2}{x^2 - 2xy - y^2+2} = \frac{-2uv}{x^2-2xy-y^2+2}\\
v' = y'-1 = & \frac{4xy-4}{x^2 -2xy- y^2+22} = \frac{u^2 - v^2-4}{x^2-2xy-y^2+2}\\
\end{align}$$
From this, we get
$$u^2 \left( \frac{v^2}{u} + u +\frac{4}{u} \right)' = u^2\left(\frac{v^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Ellipse: product of the distance from foci to a tangent is a constant I am supposed to determine what is the result of said product. Given $P(x_0,y_0)$, I need to calculate the distance from the foci of an ellipse to the tangent line that passes through $P$, and then multiply the distances.
In essence it is quite simpl... | Here I go:
$$
\frac {\frac{x_0^2c^2}{a^4}-1} {\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}
$$
Simplify
$$
\frac {\frac{x_0^2c^2-a^4}{a^4}} {\frac{x_0^2b^4 + y_0^2a^4}{a^4b^4}}
$$
Then
$$
\frac{(x_0^2c^2-a^4)(a^4b^4)}{a^4(x_0^2b^4 + y_0^2a^4)} \\
$$
Quick cancellation
$$
\frac{(x_0^2c^2-a^4)(b^4)}{x_0^2b^4 + y_0^2a... | {
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"url": "https://math.stackexchange.com/questions/526502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factorise $y^2 -3yz -10z^2$ How do I solve this question? I have looked at the problem several times. However, I cannot find a viable solution. I believe that it is a perfect square trinomial problem.
| $$y^2 - 3 y z - 10 z^2 = \begin{bmatrix} y\\ z\end{bmatrix}^T \begin{bmatrix} 1 & -\frac 32 + t\\ -\frac 32 - t & -10\end{bmatrix} \begin{bmatrix} y\\ z\end{bmatrix}$$
We want to find a $t$ that makes the matrix above rank-$1$. Computing the determinant,
$$\det \begin{bmatrix} 1 & -\frac 32 + t\\ -\frac 32 - t & -10\en... | {
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"url": "https://math.stackexchange.com/questions/526684",
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"question_score": "3",
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} |
How to find the center of mass of half a torus? Consider a halved solid torus (half a donut). The radius of the torus are $R_1$ and $R_2$. I need to find its center of mass. The hint they give is that the center of mass of a homogeneous solid object $\Omega \subset \Bbb R^3$ is calculated as
$$\overline{x}=\int_{\Omeg... | I found it easiest to use cylindrical coordinates to set up the integrals needed for the center of mass. Before we do so, however, I set my coordinate system up as follows. I have positive $x$ coming out of the screen, positive $y$ going to the right, and positive $z$ up. In cylindrical coordinates $(r,\phi,z)$:
$$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/531608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Integral $ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ Question
$ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ = ?
we tried to bound it from both sides using $x$ and $(x-1)$, which yield nice estimation ($\frac {24}\pi$) - ($\frac {36}\pi$) but not a precise one.
we also trie... | Here is another solution using Riemann-Stieltjes integral. Integrating by parts,
\begin{align*}
\int_{0}^{6} \lfloor x \rfloor \sin \left(\frac{6x}{\pi}\right) \, dx
&= \int_{0^{+}}^{6} \lfloor x \rfloor \sin \left(\frac{6x}{\pi}\right) \, dx \\
&= \left[ - \lfloor x \rfloor \frac{\pi}{6} \cos \left(\frac{6x}{\pi}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/531962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Logically speaking, why can variables be substituted? Suppose that
$$a^2+a+1=b$$
Suppose also that $a=5/4$. What makes it valid to substitute $5/4$ into the first equation? Is it because equality is transitive?
| The reason why you can make such a substitution is that $a$ and $\dfrac{5}{4}$ are precisely the same, just written differently. Here's a long-winded way of making the substitution you describe.
\begin{align*}
a &= \frac{5}{4} & \\
a^2 &= \left(\frac{5}{4}\right)^2 &\text{squaring both sides}\\
a^2 + a &= \left(\frac{5... | {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
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Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$ Prove that $1<\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.
By using the Mathematical induction. Suppose the statement holds for $n=k$.
Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{... | You are almost done. Prove that
$$\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}-\frac{1}{k+1}$$ is positive.
To do this it is enough to show that $\frac{1}{3k+2}+\frac{1}{3k+4} \gt \frac{2}{3k+3}$. The left side can be written as $\frac{6k+6}{(3k+2)(3k+4}$. So we want to show that $(3k+3)^2\gt (3k+2)(3k+4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/534117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Matrix multiplication - Express a column as a linear combination Let $A = \begin{bmatrix}
3 & -2 & 7\\
6 & 5 & 4\\
0 & 4 & 9
\end{bmatrix} $ and $B = \begin{bmatrix}
6 & -2 & 4\\
0 & 1 & 3\\
7 & 7 & 5
\end{bmatrix} $
Express the third column matrix of $AB$ as a linear combination of the column matrices of $A$
I do... | Not quite: we need to add entries. So the third column of matrix $AB$ is given by:
\begin{bmatrix}
3(4) -2(3) + 7(5)\\
6(4) + 5(3) + 4(5)\\
0(4) + 4(3) + 9(5)
\end{bmatrix}
So the third column represented as a linear combination of columns of $A$ is given by:
$$4 \begin{bmatrix}
3\\
6\\
0
\end{bmatrix} + 3 \begin{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$? I am trying to show that
$$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$$
This question stems from the underlying homework probl... | In this answer, it is derived that
$$
\begin{align}
\pi\cot(\pi z)
&=\sum_{k=-\infty}^\infty\frac{1}{z+k}\\
&=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\tag{1}
\end{align}
$$
To get the alternating series, note that
$$
\begin{align}
\frac1z+2z\sum_{k=1}^\infty\frac{(-1)^k}{z^2-k^2}
&=\sum_{k=-\infty}^\infty\frac{(-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/535116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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If $a, b, c$ are integers with $a^2 + b^2 = c^2$, then $a$ and $b$ cannot both be odd If $a, b, c$ are integers with $a^2 + b^2 = c^2$, it's true that $a$ and $b$ cannot both be odd.
But how can we prove it
| We can prove this by contradiction.
$\text{Let }\;\; a = 2m+1, b = 2n + 1 $
So, the LHS is:
$$
\begin{align}
a^2 + b^2 &= 4(m^2+n^2) + 4(m+n) + 2 \\\\
&= 2\left( 2(m^2+n^2) + 2(m+n)+1 \right) \\\\
&= 2\times\text{an odd number}
\end{align}
$$
This can not be a perfect square, as it doesn't have an ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How prove this$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+\sum_{j=0}^{n-1}\binom{m-1+j}{j}x^my^j=1$ let $m,n$ be positive numbers,and $x,y>0$ such $x+y=1$,show that
$$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+\sum_{j=0}^{n-1}\binom{m-1+j}{j}x^jy^m=1$$
My try:
$$\sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i=\sum_{i=0}^{m-1}\binom{n-1... | Suppose we seek to show that
$$\sum_{q=0}^{m-1} {n-1+q\choose q} x^n (1-x)^q
+ \sum_{q=0}^{n-1} {m-1+q\choose q} x^q (1-x)^m = 1$$
where $n,m\ge 1.$
We will evaluate the second term by a contour integral and show that
is equal to one minus the first term which is the desired result.
Introduce the Iverson bracket
$$[[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Proving an inequality using the Cauchy-Schwarz inequality Consider four real numbers $a_1, a_2, a_3, a_4$ such that $\sum a_i^3 = 10$. Prove that
$$\sum a_i^4 \geq \sqrt[3]{2500}$$
Applying the Cauchy Schwarz inequality with $a_i^2$ and $a_i$, we get
$$\left(\sum a_i^3\right)^2 \leq \left(\sum a_i^4\right)\left(\sum a... | Using the following relations from Cauchy-Schwarz (which has the advantage of being applicable for all reals):
$$\left(\sum a_i^4\right)\left(\sum 1\right)\ge \left(\sum a_i^2\right)^2 \tag{1}$$
$$\left(\sum a_i^4\right)\left(\sum a_i^2\right)\ge \left(\sum a_i^3\right)^2 = 100\tag{2}$$
Noting both LHS and RHS are posi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
| Using the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ and denoting $(a+b)$ by $A$, we have (from the assumption)
\begin{align*}
&a^3+b^3+3ab=1\\
&(a+b)(a^2-ab+b^2)+3ab=1\\
&A(A^2-3ab)+3ab=1\\
&(A^3-1)-3abA+3ab=0\\
&(A-1)(A^2+A+1)-3ab(A-1)=0\\
&(A-1)(A^2+A+1-3ab)=0,\\
\end{align*}
where we use the identities $a^2+b^2=(a+b)^2-2a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How can this equation be solved? I have no idea how to solve this equation.
$$x^2y^2+324y^2+64x^2-36xy^2-16x^2y+144xy = 0 $$
Sorry $x,y \in \mathbb{+Z}$
| Consider the equation as a quadratic in $y$, we obtain
$$ y^2 (x^2-36x+324)+y (144x-16x^2) + 64x^2 = 0 $$
First, check that if the coefficient of $y^2$ is 0, then we must have $ x = 18$, which gives $y=8$ (1 solution).
Otherwise, using the quadratic equation, we get that (for $x\neq 18$),
$$y = \frac { 8(x^2-9x \pm 3\s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it... | Consider the polynomial $S_m(x)$, satisfying $S_m(\sin^2 \theta)=\sin^2(m\theta)$.
These are known as spread polynomials, and may easily be derived from the Chebyshev polynomials $T_m(x)$, via $$1-2S_m(\sin^2(\theta)=1-2\sin^2(m\theta)=\cos(m(2\theta))=T_m(\cos(2\theta))=T_m(1-2\sin^2 \theta)$$ so $1-2S_m(x)=T_m(1-2x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
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$a^2-b^2=bc$ and $b^2-c^2=ac \Rightarrow a^2-c^2=ab$ Some weeks ago our math teacher asked the following question and gave us a week to solve it:
If $a^2-b^2=bc$ and $b^2-c^2=ac ,$ Prove $a^2-c^2=ab$, Where $a,b,c$ are non-zero real numbers.
This seemed really easy at the first, but when i tried to prove it i just ... | To begin with, note that by adding the two given equations together, we can immediately conclude that $$a^2-c^2=(a+b)c.\tag{$\star$}$$
Now, multiply both sides of $(\star)$ by $a-b,$ giving us $$(a^2-c^2)(a-b)=(a^2-b^2)c\\a^3-a^2b-ac^2+bc^2=bc^2\\a^3-a^2b-ac^2=0\\a(a^2-ab-c^2)=0,$$ so since $a\ne 0,$ we can conclude th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/545965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solutions to equation involving trigonometry and geometric series $$\cos^2 x + \cos ^3 x +\dots = 1+ \cos x$$
I want to find values of $x$ between $0$ and $ 180$ degrees for which the above equation holds true.
Attempt at a solution: left side is a converging geometric progression, for which $a_1$ is $\cos^2 x$ and $q... | As mentioned, the LHS is a geometric series, with first term $ \cos^2 x $ and common ratio $ \cos x $, so it evaluates to $ \dfrac {\cos x}{1 - \cos x} $ and our equation becomes $$ \dfrac {\cos^2 x}{1-\cos x} = 1 + \cos x \implies \cos^2 x = 1 - \cos^2 x \implies \cos^2 x = \dfrac {1}{2}. $$Finally, take the square ro... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability task (Find probability that the chosen ball is white.) I have this task in my book:
First box contains $10$ balls, from which $8$ are white. Second box contains $20$ from which $4$ are white. From each box one ball is chosen. Then from previously chosen two balls, one is chosen. Find probability that the ch... | Denote that event that eventually a white ball is drawn by $W$.
Denote that event that a white ball is drawn from the $i$-th box
by $W_{i}$ ($i=1,2$).
Denote that event that not a white ball is drawn from the $i$-th
box by $A_{i}$ ($i=1,2$).
Then $P\left(W\right)=P\left(W|W_{1}\cap W_{2}\right)P\left(W_{1}\cap W_{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/549329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that $\sqrt{3} +\sqrt{7}$ is rational/irrational I took $\sqrt{3}+\sqrt{7}$ and squared it. This resulted in a new value of $10+2\sqrt{21}$.
Now, we can say that $10$ is rational because we can divide it with $1$ and as for $2\sqrt{21}$, we divide by $2$ and get $\sqrt{21}$.
How do I prove $\sqrt{21}$ to be ra... | @Doorknob: I think he means that he took the square: $(\sqrt{3} + \sqrt{7})^2 = 10 + 2\sqrt{21}$.
If you want to prove that $\sqrt{21}$ is rational/irrational, do as follows. Suppose $a,b \in \mathbb{N}$ are such that $\frac{a}{b} = \sqrt{21}$ and that $\gcd(|a|,|b|) = 1$. I.e., $\frac{a}{b}$ can not be simplified. The... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For what $(m, n)$, does $1+x+x^2 +\dots+x^m | 1 + x^n + x^{2n}+\dots+x^{mn}$? For what $(m, n)$, does $1+x+x^2 +\dots+x^m | 1 + x^n + x^{2n}+\dots+x^{mn}$?
Well,
$$\sum_{i = 0}^{m} x^i = \frac{x^{m+1} - 1}{x - 1}$$
and,
$$\sum_{i = 0}^m x^{in} = \frac{x^{n(m+1)} - 1}{x-1}$$
Notice that $x^{m+1} - 1|(x^{m+1})^n - 1$, ... | The number $x^{m}-1$ has a unique factor for every number $b \mid m$, say $A_b$.
The number $x^{mn}-1$ has a unique factor for every $b \mid mn$.
The proposed divisor is then the product of $A_b$, where $b\mid m$, except b=1
The proposed quotient is the product of $A_b$, where $b \mid mn$, but not $b \mid n$
So we seek... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Associativity of a composition $x ∘ y = xy+\sqrt{(x^2-1)(y^2-1)}$ For many hours I had been stucked at this problem. For the following composition
$x ∘ y = xy+\sqrt{(x^2-1)(y^2-1)}$
I have to demonstrate that this composition is associative($(x ∘ y)∘z=x∘(y∘z)$
and that $x ∘ y = xy-\sqrt{(x^2-1)(y^2-1)}$ it's not assoc... | For the second operation, as was pointed out in the comments, it should be easy to just find a counterexample.
For the first one, there's a rather easy way using hyperbolic functions. Assuming that $x,y$ are greater than or equal to $1$, we can find such nonnegative real numbers $a$ and $b$ that $x = \cosh a$ and $y = ... | {
"language": "en",
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"source": "stackexchange",
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How to maximize the function I have a triangle $T=ABC$. I want to calculate $\max (a-b)$, where the the angle $ABC = \beta$, and $|AB|=c$ is fixed (pre-known). My guess is $c\times\cos (\beta)$, but I want to prove it.
Let $A$,$B$,$C$ denote the vertices of $T$, and $|AB|=c$,$|AC|=b$, and $|BC|=a$.
| We can use the cosine formula
$$b^2 = a^2 + c^2 - 2ac\cos \beta \Rightarrow b = \sqrt{a^2 + c^2 - 2ac\cos \beta}\ .$$
Write $f(a) = a-b = a- \sqrt{a^2 + c^2 - 2ac\cos \beta}$, then
$$f'(a) = 1- \frac{a-c\cos\beta}{\sqrt{a^2 + c^2 - 2ac\cos \beta}}\ .$$
As $a^2 + c^2 - 2ac\cos \beta = (a-c\cos\beta)^2 - c^2 \cos^2\bet... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Express $\cos2\theta$ in terms of $\cos$ and $\sin$ (De Moivre's Theorem) Use De Moivre's to express $\cos2\theta$ in terms of powers of $\sin$ and $\cos$.
What I have is:
$\cos2\theta + i\sin2\theta\\
= (\cos\theta + i \sin\theta)^2\\
= \cos^2\theta + 2 \cos\theta ~i \sin\theta + (i \sin)^2\theta\\
= \cos^2\theta + i(... | Yes, indeed! Since the sine and cosine functions are real-valued functions on the reals, then since $$\cos2\theta+i\sin 2\theta=\cos^2\theta-\sin^2\theta+i(2\sin\theta\cos\theta),$$ we have: $$\cos2\theta=\cos^2\theta-\sin^2\theta\\\sin 2\theta=2\sin\theta\cos\theta$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the limit $ \lim_{x\to 0}\frac{(1-3x)^\frac{1}{3} -(1-2x)^\frac{1}{2}}{1-\cos(\pi x)}$ I cannot find this limit:
$$
\lim_{x\to 0}\frac{(1-3x)^\frac{1}{3} -(1-2x)^\frac{1}{2}}{1-\cos(\pi x)}.
$$
Please, help me.
Upd: I need to solve it without L'Hôpital's Rule and Taylor expansion.
| Your limit is
$$\lim_{x\to 0} \frac{(1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} } }{1-\cos\pi x }$$
$$=\lim_{x\to 0} \frac{(1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} } }{1-\cos\pi x }\cdot \frac{1+\cos(\pi x)}{1+\cos(\pi x)}$$
$$=\lim_{x\to 0} \frac{\left((1-3x)^\frac{1}{3}-(1-2x)^{\frac{1}{2} }\right)(1+\cos(\pi x)) }{1-\cos... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the limit of this function Find
$$\lim_{n\to\infty}\left(1+x^n+\left(\frac {x^2}2\right)^n\right)^{\frac{1}{n}}$$
$(x\ge0)$.
Actually I could not find it when $x\ge1$, but when $0\leq x\lt1$ I found that the limit is $1$. Help me please.
| $$\begin{align} \lim_{n\to\infty} \left(1+x^n+2^{-n}x^{2n}\right)^{1/n} \end{align}$$
Let's consider the roots of the quadratic: $x^n=\frac{-1\pm\sqrt{1-2^{2-n}}}{2^{1-n}}$
Thus, we have
$$\begin{align} &\lim_{n\to\infty} \left(1+x^n+2^{-n}x^{2n}\right)^{1/n} \\
=&\lim_{n\to\infty} (2^{-n})^{1/n}\left(x^n+\frac{1+\sqrt... | {
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If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this?
Prove that if $n$ is a natural number and $(3n+1)$ & $(4n+1)$ are both perfect squares, then $56$ will divide $n$.
Clearly we have to show that $7$ and $8$ both will divide $n$.
I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd per... | For $7 | n$, consider adding $4n+1$ and $3n+1$.
Observe that we get $7n + 2 = x^2 + y^2$, where we can let $4n+1 = x^2$, and $3n+1 = y^2$. Now, we see that mod $7$, the only quadratic residues are $0$, $1$, $2$, $4$. The only way to get two squares summing to 2 mod 7 is either if both are congruent to $1$ mod $7$, or i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral $\int_0^\infty\frac{\ln\left(\sqrt{x+1\vphantom{x^0}}-1\right)\,\ln\left(\sqrt{x^{-1}+1}+1\right)}{(x+1)^{3/2}}dx$ Another integral similar to my previous question:
$$\int_0^\infty\frac{\ln\left(\sqrt{x+1\vphantom{x^0}}-1\right)\,\ln\left(\sqrt{x^{-1}+1}+1\right)}{(x+1)^{3/2}}dx$$
Can someone suggest how to ev... | To evaluate
$$
\int^{\infty}_{0}\frac{\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)}{(x+1)^{3/2}}dx.\tag 1
$$
I will evaluate first
$$
I(x)=\int\frac{\log\left(\sqrt{x+1}-1\right)\log\left(\sqrt{1/x+1}+1\right)}{(x+1)^{3/2}}dx\tag 2
$$
and then use limits to find the answer. The evaluation is based on Ma... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then to prove $x=\sqrt{3}+\sqrt{2}$
If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then we have to prove $x=\sqrt{3}+\sqrt{2}$
The question would have been simple if it asked us to prove the other way round.
We can multiply by $x^3$ and solve the quadratic to get $x^3$ but that would be unnecessa... | $x^3+\frac{1}{x^3}=18\sqrt{3}\Rightarrow x^6-18\sqrt{3}x^3+1=0 $
we asssume $y=x^3$
$y^2-18\sqrt{3}y+1=0\Rightarrow y=\frac{18\sqrt{3}\pm\sqrt{968}}{2}=\frac{18\sqrt{3}\pm22\sqrt{2}}{2}=9\sqrt{3}\pm11\sqrt{2}$
$x^3-(9\sqrt{3}\pm11\sqrt{2})=0$
let $x=a\sqrt{3}+b\sqrt{2}$ now,
$x^3=3\sqrt{3}a^3+2\sqrt{2}b^3+3.3a^2b\sqr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Ring theory: Ideals being equal Question: Prove directly, without gcd computations, the following equalities of ideals.
(i) $(5, 7) = (1)$ in $\Bbb{Z}$ (of course (1) = $\Bbb Z$).
(ii) $(15, 9) = (3)$ in $Z$.
(iii) $(X^3 −1,X^2 −1) = (X −1) \text{ in } \Bbb{Q} [X]$
Attempted solution:
Try to show that $(1)$ is a sub... | For (i), note that $3 \times 7 - 4 \times 5 = 1$, whence $1 \in (5, 7)$, so that $(5, 7) = Z$;
for (ii), note that $2 \times 15 - 3 \times 9 = 3$, whence $3 \in (15, 9)$, whence $(3) \subset (15, 9)$, and clearly $(15, 9) \subset (3)$, so that $(15, 9) = (3)$;
for (iii), observe that $(x^3 - 1) - x(x^2 -1) = x - 1$, wh... | {
"language": "en",
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Math induction sum of even numbers I need to prove by induction this thing:
$2+4+6+........+2n = n(n+1)$
so, this thing is composed by sum of pair numbers, so its what I do, but I'm stucked.
$2+4+6+\cdots+2n = n(n+1)$
$(2+4+6+\cdots+2n)+(2n+2) = n(n+1) + (2n+2) $
$n(n+1)+(2n+2) = n(n+1)+(2n+2) $
$n^2 + 3n + 2$
$n(n+2+1... | Suppose that
$$
2+4+6+\cdots+2n=n(n+1)
$$
add $2n+2$ to both sides:
$$
2+4+6+\cdots+2n+(2n+2)=n(n+1)+(2n+2)
$$
To finish the induction, you want the right side to be $(n+1)(n+2)$. Is it?
$$
n(n+1)+(2n+2)=n^2+n+2n+2=n^2+3n+2=(n+1)(n+2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/583884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove using mathematical induction that $2^{3n}-1$ is divisible by $7$ So, i wanna prove $2^{3n}-1$ is divisible by $7$, so i made this:
$2^{3n}-1 = 7\cdot k$ -> for some $k$ value
$2^{3n+1} = 1+2\cdot1 - 2\cdot1 $
$2^{3n+1} - 1-2\cdot1 + 2\cdot1 $
$2^{3n}\cdot2 - 1-2\cdot1 + 2\cdot1$
$2(2^{3n}-1) -1 +2$
$2\cdot7k+1$ -... | If $\displaystyle f(m)=2^{3m}-1$
$\displaystyle f(m+1)=2^{3(m+1)}-1$ and not $2^{3m+1}-1$
So, $2^{3(m+1)}-1=2^3\cdot2^{3m}-1=2^3(2^{3m}-1)-1+2^3$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maximizing slope of a secant line Two points on the curve $$ y=\frac{x^3}{1+x^4}$$ have opposite $x$-values, $x$ and $-x$. Find the points making the slope of the line joining them greatest.
Wouldn't the maximum slope of the secant line be with the max/min of the curve?
So $x=3^{1/4}$ and $x=-3^{1/4}$?
| The slope of the line will be
$$slope = \frac{y(x) - y(-x)}{2x} = \frac{2x^3}{2x(1+x^4)} = \frac{x^2}{1+x^4}.$$
Then you take the derivatives of this with respect to $x$ to find the maximum:
$$slope' = \frac{(1+x^4)(2x) - x^2(4x^3)}{(1+x^4)^2} = \frac{2x(1-x^4)}{(1+x^4)^2}.$$
This is zero at $x = 0, -1, and +1$. You'l... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help Solving Equation I've been trying to solve this equation for a few hours, but somehow I'm stuck somewhere and I could use some help.
$\frac{1} {x^2} = -\frac {2(x-a)} {a^3} + \frac {1} {a^2}$
I know from the book that there are two solutions (this is the equation of a tangent line to $x^2$). One solution is $x = a... | As lcm$(x^2,a^3,a^2)=x^2a^3$ multiplying either sides by that and rearranging we get
$$a^3-ax^2+2x^2(x-a)=0$$
$$2x^2(x-a)-a(x^2-a^2)=0$$
$$(x-a)\{2x^2-a(x+a)\}=0$$
If $\displaystyle x-a\ne0,2x^2-ax-a^2=0$
Now, $\displaystyle2x^2-ax-a^2=2x^2-2ax+ax-a^2=2x(x-a)+a(x-a)=(x-a)(2x+a)$
But we have already assumed $x-a\ne0$
| {
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"timestamp": "2023-03-29T00:00:00",
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probability:picking 3 different numbers from 1-3n. its known that their sum is devided by 3. whats the probability that each of them is devided by 3 we pick 3 different numbers from 1,2,3,...3n , which their sum is devided by 3 without remainder,what is the probability that each of the number is devider by 3 without re... | Let $a$ be the probability that each number is divisible by $3$, and let $b$ be the probability the sum is divisible by $3$. We want $\frac{a}{b}$.
To get a non-zero probability, we need $n\ge 3$. There are $\binom{n}{3}$ ways to choose $3$ distinct numbers divisible by $3$.
Now we need to find the number of ways to ... | {
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"source": "stackexchange",
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logarithms equation Hello I have following problem: solve equation $\log{(x-5)^2}+\log{(x+6)^2}=2$
and I rewrited this equation as
$2\log{(x-5)}+2\log{(x+6)}=\log{100} \implies 2(\log{(x-5)(x+6))=\log{100}} \implies \log{x^2+x-30}=\log10 \implies x^2+x-40=0 $
and I solved this equation, but I obtained only two soluti... | $$\log{(x-5)^2}+\log{(x+6)^2}=2$$
$$\log{((x-5)(x+6))^2}=2$$
$$((x-5)(x+6))^2=10^2=100$$
$$(x-5)(x+6)=\pm10$$Solve from here
| {
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"source": "stackexchange",
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struggle simplifying $\sqrt{9+\sqrt{5}}$ I need to simplify $\sqrt{9+\sqrt{5}}$
I already do this (proven it) $\sqrt{9-4\sqrt{5}}=2-
\sqrt{5}$
But I couldn't when apply to $\sqrt{9+\sqrt{5}}=\sqrt{9-4\sqrt{5}+5\sqrt{5}}=\sqrt{(2+\sqrt{5})^2+5\sqrt{5}}$
PLEASE help me out
| Let $\sqrt{9+\sqrt{5}}=A+B \sqrt{5}$. Square each side: $9+\sqrt{5} = A^2 + 2AB \sqrt{5} + 5 B^2$. Now we get two equations and two unknowns and solve for $A$ and $B$...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know... | Observe that
$$\sum_{n=0}^\infty \frac{n^2}{2^n}=2\sum_{n=0}^\infty \frac{n^2}{2^{n+1}}=2E[X^2]$$
where $X$ is a geometric random variable with support $\mathbb{Z}_{\geq 0}$ and success probability $p=1/2.$ Thus,
$$2E[X^2]=2\left[\text{Var}(X)+(E[X])^2\right]=2\left[\frac{1-p}{p^2}+\left(\frac{1-p}{p}\right)^2\right]=... | {
"language": "en",
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"source": "stackexchange",
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Factorization of $ x^2 +xy+5x+m+5 $ I want to factorize $ x^2 +xy+5x+m+5 $ . For what value of m , $ x^2 +xy+5x+m+5 $ can be resolved into linear factors ?
My try :
$ x^2 +xy+5x+m+5 $ = $ x^2 +(5+y)x+(m+5) $
To get the linear factors , we must have the determinant of this eqn is >= 0 .
D = $ (5+y)^2-4(m+5)$ . Th... | Since now you want to have the discriminant greater than $0$ which in itself is a quadratic polynomial ie $y^2+10y+(5-4m)$ . We see that this will be always greater than $0$ . If its discriminant is always less than $0$ . Because if a quadratic has real roots than it will be positive and negative .
The determinant of ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value problem
Find the minimum value of $(x+y)(y+z)$ where $x,y,z$ are positive real numbers satisfying the condition $$xyz(x+y+z)=1$$
Hint?
| Just apply AM-GM.
$(x+y)(y+z)=xy+y^2+xz+yz=y(x+y+z)+zx=\frac y{xyz}+zx=zx+\frac1{zx}\geq 2$
Equality holds if and only if $zx=1$.
Substituting $xz=1$, we have the condition that $y(x+y+\frac{1}{x}) = 1$. As such, $(x+y)(y+ \frac{1}{x}) = xy + \frac{y}{x} + y^2 + 1 = 1 + 1 = 2 $.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $4(a^3 + b^3) \ge (a + b)^3$ and $9(a^3 + b^3 + c^3) \ge (a + b + c)^3$
Let $a$, $b$ and $c$ be positive real numbers.
$(\mathrm{i})$ Prove that $4(a^3 + b^3) \ge (a + b)^3$.
$(\mathrm{ii})$Prove that $9(a^3 + b^3 + c^3) \ge (a + b + c)^3.$
For the first one I tried expanding to get $a^3 + b^3 \ge a^2b+ab... | (i) By AM-GM, $a^3+a^3+b^3\ge3a^2b$ and $a^3+b^3+b^3\ge3ab^2$.
Adding these inequalities gives $3a^3 + 3b^3 \ge 3a^2b + 3ab^2$, so $4(a^3+b^3)\ge(a+b)^3$.
(ii) Again by AM-GM, $a^3 + b^3 + c^3 \ge 3abc$.
Add two times that to the six cyclic versions of $a^3 + a^3 + b^3 \ge 3a^2b$ to get $8(a^3 + b^3 + c^3)\ge 3a^2b +... | {
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Find the limit: $\displaystyle\lim_{x\to \infty} (\sqrt{x^2+x}-\sqrt{x^2-x})$.
Find the following limit: $$ \lim_{x\to \infty} (\sqrt{x^2+x}-\sqrt{x^2-x} )$$
I tried to simplify using conjugation. This gave me the following: $$ \lim_{x\to \infty} \frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} $$
When I plug in the $\infty$, I'... | For $x\ge 1$ we have
$$
\sqrt{x^2+x}-\sqrt{x^2-x}=\frac{(x^2+x)-(x^2-x)}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}.
$$
It follows that
$$
\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})=\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Jordan Normal form: problems with finding correct generalized eigenvectors I have been tasked to find the Jordan Normal Form for the matrix $A$ shown below.
\begin{align*}
A = \begin{pmatrix}
2 & 2 & 0 & -1 \\
0 & 0 & 0 & 1 \\
1 & 5 & 2 & -1 \\
0 & -4 & 0 & 4
\end{pmatrix}
\end{align*}
In itself, finding a Jordan Norm... | JNF of $A$ should be $Q^{-1}AQ$, where $Q$ is the matrix having a Jordan basis as its columns.
The basis which you found is not a Jordan basis, so it is not a disjoint union of Jordan
chains.
Different method of getting JNF:
Theorem.
Let $\lambda$ be an eigenvalue of a matrix $A$ and let $J$ be the JNF of $A$. Then
th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Standard form of the equation of a circle involving two points.
For parts (a), (b), and (c), consider the points $\mathrm A=(-1,3)$ and $\mathrm B=(5,3)$ in the $xy$-plane. Distance is measured in meters.
c) Find the standard form of the equation of a circle that has the line segment $\overline{\mathrm{AB}}$ as the di... | The center of the circle is the midpoint of the diameter: $(2, 3).$
The radius has length $3 m$ because the diameter has length $6 m$.
Then the equation for the circle is straightforward:
$$(x-2)^2 + (y-3)^2 = 9 m^2.$$
The $x$ coordinate of the center is $\frac{1}{2} \cdot (5 + (-1)) = 2 m.$
Actually, your $y$ coordina... | {
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Integrating With Trig Subsitution Can someone please explain how to calculate the integral of
$\frac{\sqrt{1 + x^2}}{x}$ using trig substitution?
| In all the following, I intentionnally forget the constant to add to every primitive, in order to simplify the notations. All equalities with primitives are to be understood "up to a constant".
Using trigonometric functions, and substitution $x=\tan u$ (then $\mathrm{d}x=(1+\tan^2u)\,\mathrm{d}u = \frac{\mathrm{d}u}{co... | {
"language": "en",
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How find this inequality $\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$ let $a,b,c,d,e\in R$,and such
$$a^2+b^2+c^2+d^2+e^2=1$$
find this value
$$A=\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$$
I use computer have this
$$A=\dfrac{2}{\sqrt{10}}$$
then equal holds if we su... | This is a comment that’s too long to fit in the usual format. I prove below
the non-optimal bound $A\leq \sqrt{\frac{5+\sqrt{5}}{10}} \approx 0.8$. Let us denote
$$
x_1=x_6=a, x_2=b, x_3=c, x_4=d, x_5=e
$$
Then, we have
$$
\big(\frac{5+\sqrt{5}}{2}\big)
\bigg(\sum_{k=1}^5 x_k^2\bigg)-
\bigg(\sum_{k=1}^5 (x_k-x_{k+1})^2... | {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$
As is the question in the title, I am wishing to find all positive integers $n$ such that $3^n + 5^n$ is divisible by $n^2 - 1$.
I have so far shown both expressions are divisible by $8$ for odd $n\ge 3$ so trivially a solution is $n=3$. I'm n... | Here are a few necessary conditions on $n$ summarizing my statements from the chat:
*
*$n$ is odd.
*any prime factor $p\mid n^2-1$ is of the form $p \equiv
1,2,4,8 \mod 15$.
*$n \equiv 3,93 \mod 120$
proofs:
1)
It is clear that $3 \nmid 3^n+5^n$ and $5\nmid 3^n+5^n$. Thus $3\mid n$.
Write $n = 3m$, then $$... | {
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"timestamp": "2023-03-29T00:00:00",
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theory of equations finding roots from given polynomial
If the equation $x^4-4x^3+ax^2+bx+1=0$ has four positive roots then $a=\,?$ and $b=\,?$
$\textbf{A.}\,6,-4$
$\textbf{B.}\,-6,4$
$\textbf{C.}\,6,4$
$\textbf{D.}\,-6,-4$
we can replace options and check answers .. are there any other shortcuts we can use
| $$(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$
$$=x^4-x^3\left(\sum_{4\ge i\ge1} x_i\right)+x^2\left(\sum_{4\ge i>j\ge1}x_ix_j\right)-x\left(\sum_{4\ge i>j>k\ge1}x_ix_jx_k\right)+x_1x_2x_3x_4 $$
If $x_i>0,1\le i\le4$ the coefficient of $x^2$ must be $>0$ and that of $x$ must be $<0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all postive integer numbers $x,y$,such $x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$ Find all postive integer $x$ and $y$ such that
$x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$
My try: since
$$(x+y)^2-2xy=x^2+y^2$$
I know this well know reslut:
$$xy|(x^2+y^2+1),\Longrightarrow \dfrac{x^2+y^2+1}{xy... | Note that since $(x+y+1)(x+y-1)=(x^2+y^2-1)+2xy$,
$$
x+y-1\mid x^2+y^2-1\iff x+y-1\mid2xy\tag{1}
$$
Since
$$
x+y+1\mid2xy\tag{2}
$$
and $(x+y-1,x+y+1)\mid2$, $(1)$ and $(2)$ imply
$$
(x+y)^2-1\mid4xy\tag{3}
$$
However, because $4xy\le(x+y)^2$, $(3)$ implies $(x+y)^2-1=4xy$, that is,
$$
(x-y)^2=1\tag{4}
$$
Wlog, assume ... | {
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"timestamp": "2023-03-29T00:00:00",
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Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(... | There are 14 possible answers up to order for the 4-term analog of this problem.
2,3,7,42
2,3,8,24
2,3,9,18
2,3,10,15
2,3,12,12
2,4,5,20
2,4,6,12
2,4,8,8
2,5,5,10
2,6,6,6
3,3,4,12
3,3,6,6
3,4,4,6
4,4,4,4
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$ Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$
how to do it? I think it should be $3^6$ since $(3x^2)^6=3^6x^8$. (this is false)
Is this true?
| Notice
$$\frac{1}{(1+x)^2} = 1 -2x+3x^2-4x^3+5x^4-6x^5+7x^6-8x^7 + 9x^8 + O(x^9)$$
So the coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$ is the same as the
one in
$$\left[\frac{1}{(1+x)^2} + \big(8x^7 - 9x^8\big)\right]^6
= \frac{1}{(1+x)^{12}} + \binom{6}{1}\frac{8x^7-9x^8}{(1+x)^{10}} + O(x^9)
$$
Above ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Density of the sum of $n$ uniform(0,1) distributed random variables I am working on the following problem:
Let $X_1, X_2, \ldots, X_n, \ldots$ be iid. random variables, each of Uniform$(0,1)$ distribution. Denote by $f_n(x)$ the density of the random variable $S_n := \sum_{k = 1}^n X_k$. Then
\begin{align*}
f_n(x) ... | $\int f_{n}\left(y\right)f_{1}\left(x-y\right)dy=\int\frac{1}{\left(n-1\right)!}\sum_{k=0}^{\left\lfloor y\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left(y-k\right)^{n-1}1_{\left(0,1\right)}\left(x-y\right)dy$
$=\frac{1}{\left(n-1\right)!}\sum_{k=0}^{\left\lfloor x-1\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $a,b,c\in(0;+\infty)$, prove that $\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$. If $a,b,c\in(0;+\infty)$ and $$\frac{c}{1+a+b}+\frac{a}{1+b+c}+\frac{b}{1+c+a}\ge\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ca}{1+c+a}$$Prove that $$\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\... | Here are some thought I have, but no proof (yet):
Both the condition and the inequality to be proven have equality when $(a,b,c)=(x,x,x)$ and when $(a,b,c)=(1,1,4)$ (or permutations). We know that the inequalities of means only have equality when the term we apply them to are equal. Thus, we can never use AM-GM on $a$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$3$ never divides $n^2+1$ Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.
Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.
If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.
And that is where I am stuck. I try ... | If $3$ divides $n^2+1$ then it must have solution modulo $3$. But clearly
$0^2+1\equiv 1 \pmod 3$
$1^2+1 \equiv 2 \pmod 3$
$2^2+1 \equiv 5 \equiv 2 \pmod 3$
Otherwise put $n=3k,3k+1,3k+2$ and see that $3$ never divides it
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 2
} |
proving the inequality $\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$ If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$
and also show that equality hold if $a=b=c$.
$\bf{My\; Try}::$ Here we have to prove $4\tria... | We know
$$S=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}.$$
So, all we need is to prove the following :
$$(-a+b+c)(a-b+c)(a+b-c)\le abc.$$
This is a well known inequality. Proof is here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Prove that if $10|A$ then $100|A$. For : $A=a^2+ab+b^2$ with $a,b \in \mathbb{N}$.
We known that $10|A$ prove that $100|A$.
| We show that $4 \mid A$ and $25 \mid A$.
Suppose $a^2+ab+b^2 \equiv 0 \bmod 2$. As $a \equiv 1 \equiv b$ is not a solution, at least one of them is $\equiv 0$, forcing the other to be $\equiv 0$. So $2\mid a,b$, showing $4\mid a^2+ab+b^2 = A$.
Suppose $a^2+ab+b^2 \equiv 0 \bmod 5$. Then $(a+3b)^2 \equiv a^2+ab+4b^2\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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