Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Easy way to decompose $\frac{1}{X^{3}\cdot (X-2)^{3}}$ into partial fractions? Is there any easy way or shortcut to decompose $$\frac{1}{X^{3}\cdot (X-2)^{3}}$$ into partial fractions ? because dealing with the usual way of replacing and giving values to $X$ is too clumsy in this case , so I was wondering if there ... | You can also do this as follows. If $R(x)$ is a rational function that tends to zero at infinity, then expanding $R(x)$ around each of its singularities in the complex plane, keeping only the singular parts of the expansion and adding up all the terms from all these expansions will yield the partial fraction expansion.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Taylor series expansion of $f(x) = \sin^3 \left(\ln(1+x) \right)$. How does one use Taylor series expansion to compute $f^{(3)}(0)$ in which $f(x) = \sin^3 \left(\ln(1+x) \right)$.
| Probably easier to use a linearized expression for $\sin^3(\theta)$ :
$$\sin^3(\theta)=\frac{3}{4}\sin(\theta)-\frac{1}{4}\sin(3\theta)$$
Now, for all $x>-1$ :
$$f(x)=\sin^3(\ln(1+x))=\frac{3}{4}\sin(\ln(1+x))-\frac{1}{4}\sin(3\ln(1+x))$$
and the Taylor expansion at $x=0$ is :
\begin{align}
f(x)&=&\frac{3}{4}\sin(x-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find a common direct complement for $K$ and $L$ in $M_n$ Given the matrices $$A=\begin{pmatrix} 1 & 0 \\ -1 & -1\end{pmatrix}, \ B=\begin{pmatrix} 1 & 2 \\ 0 & -1\end{pmatrix}$$ and subspaces of $M_2$:
$$K=\{X\in M_2: tr(XA)=0\}, L=\{X\in M_2: tr(XB)=0\}$$
Find bases for $K$ and $L$ and determine one common direct c... | First, a general observation. Let $V$ be an $n$-dimensional vector space and let $\varphi \colon V \rightarrow \mathbb{F}$ be a non-zero linear functional. By the rank-nullity theorem, we know that $\dim \ker(\varphi) = n - 1$. Let us choose any vector $w \in V$ such that $\varphi(w) \neq 0$. Then
$$ V = \ker(\varphi) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Identity of an inverse hyperbolic function Recently I stumbled upon the identity: $$\operatorname{csch}^{-1}(x) = \ln \left( \frac 1 x + \frac{\sqrt{1+x^2}}{|x|}\right)$$
So i decided to prove this identity by taking the integral of the derivative of inverse $\operatorname{csch}(x)$. Although i tried expressing the int... | The usual algebraic proof: Recall that $$\sinh x = \frac{e^{x} - e^{-x}}{2}.$$ Thus, $$\operatorname{csch} x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}.$$ It follows that $y = \operatorname{csch}^{-1} x$ is the solution to the equation $$x = \frac{2}{e^y - e^{-y}}.$$ To this end, we write $$e^y - e^{-y} = 2/x,$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Intersection of 3 circles on equilateral triangle given the difference in their radii Imagine we are given three intersecting circles centered on the vertices of an equilateral triangle of side length $1$, with $(0,0)$ arbitrarily placed at the bottom left corner. The circles have radii $r$, $r+a$, and $r+b$ respectiv... | First, note that (x,y) is determined by the intersection of the circles with radii r and r+b, Given the "safe" values assumed, so we should be able to express both x and y in terms of r and b only.
Safe values values are those such that:
* 2r + a > 1
* 2r + b > 1
* r + a < 1
* r + b < 1
Since it was specifie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How do I continue solving this polynomial division? I had this equation as a bonus question on a quiz today. I got to a certain extend and was completely unsure how to continue.
*$$(10x^4-18x^3-94x^2-8x+2) \div (10x+2)$$
I figured the easiest first step would be to divide. (Actually, I assumed it was factorable, bu... | multiples of $10 x + 2$
$$
\begin{array}{rrrrrr}
x^3 \mapsto & 10 x^4 & + 2 x^3 &&& \\
-2x^2 \mapsto & & -20 x^3 & -4 x^2 && \\
-9x \mapsto & & & -90 x^2 & -18 x & \\
+1 \mapsto & & & & 10 x & + 2 & \\
&&&&& \\
& 10x^4 & - 18 x^3 & -94 x^2 & -8 x & + 2 & \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Value of the integral $\int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx$ Evaluate
$ \displaystyle \int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx$.
(I am relatively confident of my answer below but I would still like to make sure that it is correct.)
| \begin{align}
\int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx &= \sum_{n=1}^\infty \int_n^{n+1} \frac n{x^3} \, dx \\
&= -\frac 12 \sum_{n=1}^\infty \frac n{x^2} \bigg\vert_n^{n+1} \\
&= \frac 12 \sum_{n=1}^\infty \left( \frac 1n - \frac n{(n+1)^2} \right) \\
&= \frac 12 \sum_{n=1}^\infty \frac{2n+1}{n(n+1)^2} \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Triangle inequality in C Show that $\left|Im(2+ z^{c} -4z^2) \right| \leq 9.5$ When $ \left| z \right| \leq \frac {3}{2}$
$z^c$= compliment of z
$\left|Im(2+ z^{c} -4z^2) \right| \leq \left| 2+z^{c} - 4z^2 \right|$
I have tried to split it up directly i have tried to force complete the square i always get a weird valu... | Credit: @MartinBladt is right that something is wrong with the question.
Let $z = \frac{\sqrt{5}}{2}+i$
$$|z|=\sqrt{\frac54+1}=\frac32$$
$$|\Im(2+z^c+-4z^2)|=|-\Im (z)-4(2) \Re(z)\Im(z)|=1+8\frac{\sqrt{5}}{2}=1+4\sqrt{5}>9.5$$
Edit:
The following statement is not valid:
$$\left| z^{c} - 4z^2 \right|= 2^{\frac {1}{2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the sum of the digits of the four-digit number
It is known that $\overline{abcd}-\overline{dcba}=x$ and $a^3+b^3+c^3+d^3=x$. Find $a+b+c+d$.
From the given we get $999a+90b-90c-999d = a^3+b^3+c^3+d^3$. We could just test out the possibilities, but is there another way to find the sum of the digits?
| Partial Answer
As $x=999a+90b-90c-999d=9(111a+10b-10c-111d)$ then $x\equiv0\pmod{9}$. Looking at cubes modulo 9 we see:
$0^3\equiv0\pmod{9}$, $1^3\equiv1\pmod{9}$, $2^3\equiv8\pmod{9}$, $3^3\equiv0\pmod{9}$, $4^3\equiv1\pmod{9}$, $5^3\equiv8\pmod{9}$, $6^3\equiv0\pmod{9}$, $7^3\equiv1\pmod{9}$, $8^3\equiv8\pmod{9}$, $9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Factor $ a^4-3a^2-2ab+1-b^2$ In order to factor $a^4-3a^2-2ab+1-b^2$, I find that $a=1, b=-1$ makes the value of the expression 0. Thus, I assume $b=-a$.
I rewrite the expression on the assumption as:
$$a^4-3a^2-2ab+1-b^2$$
$$=a^4-3a^2+2a^2+1-a^2$$
$$=a^4-2a^2+1$$
$$=(a^2-1)^2$$
Then I insert $a+b$, which is another fo... | Here is probably the most simple method to factor it.
Note that $$a^2-b^2=(a-b)(a+b)$$So $$a^4-3a^2-2ab+1-b^2=a^4-2a^2+1-a^2-2ab-b^2=(a^2-1)^2-(a+b)^2$$
So we can factor it into $$(a^2+a+b-1)(a^2-a-b-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Write the quadratic function in the form $g(x)= a(x-h)^2 + k$ Write the quadratic function in the form $g(x)= a(x-h)^2 + k$
Then, give the vertex of its graph.
$g(x)= 2x^2-16x+35$
~
I tried, but still lost:
$(2x^2-16x)+35$
$2(x^2-4^2-16)+35$
$2(x^2-4^2-32+35$
| $2x^2-16x+35$
$=2(x^2-8x)+35$
$=2(x^2-8x+16)-2\cdot 16+35$
$=2(x-4)^2+3$
There you go. Completing the square.
The vertex is easy from this. It's just x=4 since squares are always nonnegative so to make the square part zero you let x=4. (to be more specific the vertex is (4, 3))
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $(f^{-1})''(x)$. If $f$ is strictly monotonous and differentiable for a given interval, we have
$$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$$
How much is $(f^{-1})''$?
| \begin{align*}
y &= f(x) \\
x &= f^{-1} (y) \\
f'(x) &= \frac{dy}{dx} \\
&= \frac{1}{\frac{dx}{dy}} \\
f''(x) &= \frac{d^{2}y}{dx^{2}} \\
&= \frac{d}{dy} \left( \frac{1}{\frac{dx}{dy}} \right) \times \frac{dy}{dx} \\
&= -\frac{\frac{d^{2}x}{dy^{2}}}{\left( \frac{dx}{dy} \right)^{3}} \\
\frac{d^{2}x}{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim\limits_{n\to\infty} ( \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2})$ Find $\lim\limits_{n\to\infty} ( \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2})$
Please help me find this limit .
I cannot use cauchys limit theorem on this problem . So I am stuck and clueless
| Observe that
$$0 < \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2} < n\cdot\frac{1}{(n+1)^2}.$$
Since the limit of the right side is $0,$ the squeeze theorem shows our limit is $0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve $y'=-\frac{x+y-2}{x-y+4}$
$$y'=-\frac{x+y-2}{x-y+4}$$
I have used the substation:
$x=u-1$
$y=v+3$
And got $$\frac{dv}{du}=\frac{-1-\frac{v}{u}}{1-\frac{v}{u}}$$
$z=\frac{v}{u}$
And got:
$\frac{du}{u}\frac{1-z}{z^2-2z-1}$
$$u=e^{\frac{-z+2z+1}{4}+C}$$
What should I do next?
| $$\frac { dv }{ du } =\frac { -1-\frac { v }{ u } }{ 1-\frac { v }{ u } } \\ z=\frac { v }{ u } \\ v=uz\\ z+{ u }z^{ \prime }=\frac { -1-z }{ 1-z } =\frac { 1+z }{ z-1 } \\ u{ z }^{ \prime }=\frac { 1+z }{ z-1 } -z=\frac { 1+z-{ z }^{ 2 }+z }{ z-1 } =\frac { -{ z }^{ 2 }+2z+1 }{ z-1 } =\frac { { z }^{ 2 }-2z-1 }{ 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $x^2+y^2$ if $x^2-\frac{2}{x}=3y^2$ and $y^2-\frac{11}{y}=3x^2$
Find $x^2+y^2$ if $x^2-\frac{2}{x}=3y^2$ and $y^2-\frac{11}{y}=3x^2$.
My try:
$$\frac{y^2-\frac{11}{y}}{3}-\frac{2}{x}=3y^2$$
then?
| [ EDIT #3 ] The previously posted answer remains copied below, under Alternative Solution.
Write the equations as:
$$
\begin{cases}
\begin{align}
2 &= x\,\left(x^2-3y^2\right) \\
11 &= y\,\left(y^2-3x^2\right)
\end{align}
\end{cases}
$$
Square the two and add together:
$$
\begin{align}
125 = 2^2 + 11^2 &= x^2\,\left(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Weird limit: $\lim_{n\to\infty}\left(\frac{f(1)+f(2)+...+f(n)}{n}\right)^n$ I couldn't find the right path for this limit to solve. $$\lim_{n\to\infty}\left(\dfrac{f(1)+f(2)+...+f(n)}{n}\right)^n$$
$$f:\mathbb{R}\to\mathbb{R}, f(x)=\sqrt[3]{x^3+3x^2+2x+1}-\sqrt[3]{x^3-x+1}$$
I know that we have the indeterminate $1^{\i... | As N.S.John says in the comments, we can write $$\sqrt [3]{x^3+3x^2+2x+1} =\sqrt [3]{(x+1)^3-(x+1)+1}$$ and $$\sqrt [3]{x^3-x+1} =\sqrt [3]{(x-1+1)^3-(x-1+1)+1} $$ Thus $$f (1)=\sqrt [3]{2^3-2+1}-\sqrt [3]{1^3-1+1} $$ $$f (2)=\sqrt [3]{3^3-3+1}-\sqrt[3]{2^3-2+1}$$ $$\vdots $$ $$f (n)=\sqrt [3]{(n+1)^3-(n+1)+1}-\sqrt [3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What number should be subtracted from What number should be subtracted from $4x^3+5x+3$
so that the resulting polynomial leaves remainder $-80$ when divided by $2x+5$?.
Let the required number to be subtracted be $K$.
let: $$P(x)=4x^3+5x+3-k$$
$$g(x)=2x+5=2(x+5/2)$$
Comparing $g(x)$ with $x-a $ , $a =\frac {-5}{2}$
... | $$4x^3+5x+3=2x^2\underbrace{(2x+5)}-5x\underbrace{(2x+5)}+15\underbrace{(2x+5)}+3-75$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
convergence of $\sum\frac{\sin(\sqrt{n})}{n^a}$ For $a>0$, define $x_n=\dfrac{\sin(\sqrt{n})}{n^a}$.
Prove that $\sum x_n$ converges if and only if $a>\dfrac1 2$.
I proved the convergence when $a>\dfrac1 2$, but I can't derive the other direction. Could someone give me some help? Thanks a lot.
| A straightforward way to show that this series diverges if $a \leqslant 1/2$ is to show that the Cauchy criterion is not satisfied.
If $\sqrt{n} \in [c_1,c_2] = [\pi/6 + 2k\pi, 5\pi/6 + 2k\pi]$ then $\sin \sqrt{n} \geqslant 1/2$ and, with $a \leqslant 1/2$, we have
$$\begin{align} \left|\sum_{c_1 \leqslant \sqrt{n} \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Any hints on how to show that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $ I am trying to prove that $$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $$
My attempt was to suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10}$, and show that it was absurd.
Here's what I did:
Suppose that $\frac{2 - \sqrt{... | multiplying by $10$ and $$2\sqrt{3}$$ we get $$20-10\sqrt{3}<2\sqrt{3}$$ and we get
$$20<12\sqrt{3}$$ dividing by $$4$$ and we have $$5<3\sqrt{3}$$ this is true since we have $$25<27$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Finding the number of $4$-digit numbers so no two adjacent digits are even Find the number of $4$-digit numbers with distinct digits chosen from the set $\{0,1,2,3,4,5\}$ in which no two adjacent digits are even.
| You can approach this thinking that you have two different subsets: the even numbers $\{0,2,4\}$ and the odd numbers $\{1,3,5\}$. In order to avoid adjacent even numbers you have the two following cases:
Alternating even and odd numbers, starting with an odd number:
$3 \cdot 3 \cdot 2 \cdot 2 = 36$ possible combination... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the rank of the following matrix depending on $\lambda\in\Bbb R$ Find the rank of the following matrix depending on $\lambda\in\Bbb R$.
$$A=\begin{pmatrix}
1&2&3&4\\
2&\lambda&6&7\\
3&6&8&9\\
4&7&9&10
\end{pmatrix}$$
My attempt:
$$\begin{pmatrix}
1&2&3&4\\
2&\lambda&6&7\\
3&6&8&9\\
4&7&9&10
\end{pmatrix}\sim\begin... | I will point out that you can also check the result in Wolfram Alpha.
Here is my computation - rather similar to yours.
$$\begin{pmatrix}
1 & 2 & 3 & 4\\
2 &\lambda& 6 & 7\\
3 & 6 & 8 & 9\\
4 & 7 & 9 &10
\end{pmatrix}\overset{(1)}\sim
\begin{pmatrix}
1 & 2 & 3 & 4\\
2 &\lambda& 6 & 7\\
3 & 6 & 8 & 9\\
1 & 1 & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Solve the system ${x}^{2}+ \left( y-1 \right) ^{2}=4,z^{4}+y{z}^{2}+xz+1=0. $ Find all real solutions of the system of equation
\begin{cases}
{x}^{2}+ \left( y-1 \right) ^{2}=4,\\{z}^{4}+y{z}^{2}+xz+1=0.
\end{cases}
| This system of 2 equations has infinite solutions.
Let $x=2\cos\theta$ and $y=2\sin\theta+1$ which satisfy the first equation. Then
$$z^4+yz^2+xz+1=0$$
$$z^4+(2\sin\theta+1)z^2+(2\cos\theta)z+1=0$$
$$(z^2+\sin\theta)^2+(z+\cos\theta)^2=0$$
$z=-\cos\theta$ and $z^2=-\sin\theta$, this leads us to find solutions of $\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to integrate $\int\sqrt{x^2+1}dx$ $$\int\sqrt{x^2+1}dx$$
I've been attempting this problem for days and I've made very little progress. I understand that I should substitute $x=\tan(x)$ and $dx=\sec^2(x)dx$ but beyond that I am very much lost.
| You can also do:
$$\int\sqrt{x^2+1}\,dx = \int\frac{x^2+1}{\sqrt{x^2+1}}\,dx = \int x\cdot \frac{x}{\sqrt{x^2+1}}\,dx + \int\frac{1}{\sqrt{x^2+1}}\,dx = x\sqrt{x^2+1} - \int\sqrt{x^2+1}\,dx\,+ \ln(x+\sqrt{x^2+1}) \implies \int\sqrt{x^2+1}\,dx = \frac{1}{2}\big(x\sqrt{x^2+1}+\ln(x+\sqrt{x^2+1})\big) + C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve $y'+y=|x|, y(-1)=0$. Find y(1)=? This is my try.
$y(x)e^x=\int|x|e^xdx$
$=\int_x^0(-t)e^tdt+\int_0^xte^tdt$
$=2+2(x-1)e^x+c$
$y(x)=2e^{-x}+(x-1)2+ce^{-x}$.
with $y(-1)=0$ gives $c=1-\frac{3}{e}$
but later I find which is wrong. where is my mistake?
| The homogenous equation
$y' + y = 0\\
y = C_1 e^-t$
A particular solution
Suppose $y = Ax + B$
$y' = A\\
A + Ax + B = |x|\\
A = sgn x\\
B = -A$
$y = \begin{cases} C_1 e^{-t} + x - 1 & x\ge 0\\C_2 e^{-t} - x + 1 & x< 0 \end{cases}$
$y(-1) = 0$
$C_2 e + 2 = 0\\
C_2 = -2 e^{-1}$
$y(0) = $$-2e^{-1}+1 = C_1 -1\\
C_1 = -2e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the particular solution of $\sin(2x)dx + \cos(3y)dy = 0$ for $y(\pi/2) = \pi/3$ Find the particular solution of $\sin(2x)dx + \cos(3y)dy = 0$ for $y(\pi/2) = \pi/3$, plot the graph of the solution, and determine the interval in which the solution is defined.
Because the terms are separable, we integrate the expres... | Your calculus is correct.
$$-\frac{1}{2}\cos(2x) = -\frac{1}{3}\sin(3y) + C$$
The condition $\quad y(\pi/2) = \pi/3\quad $ leads to $\quad C=\frac{1}{2}$
Note that, if the condition was $\quad y(\pi/2) = k\pi/3\quad $ the constant would be the same $\quad C=\frac{1}{2}\quad$ because $\quad \sin(3k\pi/3)=\sin(3\pi/3)=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve the following simultaneous trig equations? Equation 1
$X$ = $a_1\sin(\theta_1) + a_2\cos(\theta_2)$
Equation 2
$Y$ = $a_1\cos(\theta_1) + a_2\sin(\theta_2)$
where $X, Y, a_1,a_2$ are known.
WHAT I HAVE DONE SO FAR
Let
$sin(\theta_1) = u_1$
$cos(\theta_1) = v_1$
$sin(\theta_2) = u_2$
$cos(\theta_2) = v_2... | As I said in the comments, it looks like there is no nice closed-form solution for this. So I plugged your set of equations into Mathematica. The result is:
$$\begin{align}
\sin\theta_1=\frac{({a_1}^3-a_1{a_2}^2)X-\sqrt{-{a_1}^2 Y^2 \left({a_1}^4-2 {a_1}^2
\left({a_2}^2+X^2+Y^2\right)+\left(-{a_2}^2+X^2+Y^2\right)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Polynomials with integer coefficients – find minimal $m$ Let $m$ be the minimal positive integer such that
$$(x+4)(x+5)(x+9)p(x) - (x-4)(x-5)(x-9)q(x) = m$$
where $p(x)$ and $q(x)$ are polynomials with integer coefficients. What is the value of $m$?
My work :
By Bezout's identity, we have, $\exists p(x), q(x) \in \ma... | LEMMA: If $f(x),g(x)\in \mathbb Z[x]$ and a prime number $p$ divides every co-efficient of the product $f(x)g(x)$ then $p$ divides every co-efficient of $f(x)$ or of $g(x).$
From this we can show that if $n\in \mathbb N$ divides every co-efficient of $f(x)$ and $n$ is co-prime to some (any) co-efficient of $g(x)$ then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solution of $\lim_{x\to 0} \frac{(x \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $ using Mclaurin series The problem is
$$\lim_{x\to 0} \frac{(x \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $$
I know that
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}... $$
got this
$$\tan x = x + \frac{x^3... | Note that $$x\sqrt{1+\sin x} = x\sqrt{1+x+O(x^3)} = x(1+\frac x2-\frac{x^2}8+O(x^3)) = x+\frac{x^2}{2}-\frac{x^3}8+O(x^4)$$
and
$$\ln\sqrt{1+x^2} = \ln(1+\frac{x^2}2+O(x^4)) = \frac{x^2}{2}+O(x^4)$$
so we have
$$\lim_{x\to 0}\frac{x\sqrt{1+\sin x}-\ln\sqrt{1+x^2}-x}{\tan^3x} = \lim_{x\to 0}\frac{-\frac{x^3}8+O(x^4)}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determining if a subset of $\mathbb{C}$ is a subfield. I am trying to deduce whether or not
$$K=\{a+b\sqrt{2}|a,b\in\mathbb{Q}\text{ and }ab<\sqrt{2}\}$$
is a subfield of $\mathbb{C}$. I feel like I'm getting stuck. If I subtract two numbers $a+b\sqrt{2}$ and $c+d\sqrt{2}$, then we have to show that
$$(a-c)(b-d)=ab... | The key here is observing that, for $a,b,c,d\in\mathbb{Q}$,
$$
a+b\sqrt{2}=c+d\sqrt{2}
\quad\text{if and only if}\quad
a=c\text{ and }b=d
$$
which is easy, but should be noted before going on. One direction is obvious; for the reverse, if $b\ne d$, then
$$
\sqrt{2}=\frac{a-c}{d-b}
$$
which is a contradiction. Therefore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Number of integral quotient of $n^2$ Is there a theorem which states the number of integral quotient of $n^2$ divided by $\{1,2,3, ... n^2\}$ is $2n-1$?
Example: If $n=4$,then $16 \div \{1,2,3, ... 16\} = 16,8,5,4,3,2,1$. There are $7$ integral quotients.
| The set $\{\,\lfloor \frac {n^2}d\rfloor\mid 1\le d\le n^2\,\}$ contains the numbers $1,2,\ldots,n$ from taking $d=n,\ldots, n^2$ because $d=n$ leads to $\lfloor\frac{n^2}{n}\rfloor=n$, and because $\lfloor \frac{n^2}{d+1}\rfloor$ is either $\lfloor \frac {n^2}d\rfloor$ or $\lfloor \frac {n^2}d\rfloor-1$ for $d\ge n$ (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find all matrices $X$ such that $ABXB^tA^t=I$ Find all matrices $X$ such that:
$$ABXB^tA^t=I$$ if
$A=\begin{pmatrix}
1 &-2 &2\\
3 &-5 &6\\
-1 &2 &-1
\end{pmatrix}$ and $B=\begin{pmatrix}
-3 &-2 &-2\\
2 & 1 &1\\
6 &3 &4
\end{pmatrix}$.
So I managed to get that $AB=\begin{pmatrix}
5 &2 &4\\
17 &7 &13\\
1&1&0
\end{pmat... | Matrices multiplication is not commutative. You are allowed to multiply an equation by a non-zero matrix, but you have to mind the side from which you are applying the multiplication.
Let us say that $P=AB$ and $Q=B^TA^T$.
Your equation takes the form of $P \cdot X \cdot Q = I$.
You can multiply each side by $P^{-1}$ f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is the sum of $100$ terms of the sequence $a_{k} = \frac{1}{k} - \frac{1}{k+1}$? In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence?
The answer given ... | $$a_{100} = \left( \frac11 - \frac12 \right)+\left( \frac12 - \frac13 \right)+
\left( \frac13 - \frac14 \right)+\cdots + \left( \frac1{100} - \frac1{101} \right)
\\= \frac11 + \left( -\frac12+\frac12\right)+\left( -\frac13+\frac13\right)
+\cdots +\left( -\frac1{100}+\frac1{100}\right) -\frac1{101} = 1 - \frac1{101}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Which quaternions are solutions of $x^2+1=0$? What would the final Quaternion Solutions look like for $x^2+1=0$?
I substituted in $x = a+bi+cj+dk$ and came up with a very long +/- square root.
| The solutions have the form
$$
bi + cj + dk
$$
where $b^2 + c^2 + d^2 = 1$. They are the "pure imaginary unit quaternions".
Why? Because $x^2 = -1$, so $|x|^2 = 1$ so $a^2 + b^2 + c^2 + d^2 = 1$. Since the real part of $x^2$ is $a^2 - b^2 - c^2 - d^2 = -1$, we get that $2a^2 = 0$, so $a = 0$.
Once you settle that, it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum of positive integer $k$ so that for all positive real numbers $x$ we have: $x^6+x^5+x^4+x^2+x+4>kx^3$
Find the maximum of positive integer $k$ so that for all positive real numbers $x$ we have: $x^6+x^5+x^4+x^2+x+4>kx^3$
Since the power of the polynomial on LHS is greater than 3 I have no idea for it... | Assume $k \ge 9$. Then we have $$9=1^6+1^5+1^4+1^2+1+4>k \times 1^3 \ge 9$$
Contradiction. So we have $k<9$.
Note that $$x^6+x^5+x^4+x^3+x^2+x+4=(x^6+4)+(x^5+x)+(x^4+x^2)>8x^3$$
As follows from $\text{AM-GM}$, and because the equality condition can not be satisfied, there is no equality symbol.
So since $8$ is possib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\cos(\pi/7)$ is root of equation $8x^3-4x^2-4x+1=0$
Prove that $\cos\theta$ is root of equation $8x^3-4x^2-4x+1=0$, given $\theta=\frac{\pi}{7}$.
I put $\cos\theta$ in equation, but couldn't show the left-hand side to be zero.
| We have
$$ \cos{2\theta} = 2\cos^2{\theta}-1 \\
\cos{3\theta} = \cos{\theta}(2\cos^2{\theta}-1) - 2\sin^2{\theta}\cos{\theta} = 4\cos^3{\theta}-3\cos{\theta}, $$
so
$$ \cos^2{\theta} = \frac{1}{2}(\cos{2\theta}+1) \\
\cos^3{\theta} = \frac{1}{4}(\cos{3\theta}+3\cos{\theta}) $$
Putting these into the equation gives
$$ 8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to solve this system of equations in Lagrange Multiplier problem Find the maximum and minimum values of ${x^{2} + y^{2} + z^{2}}$ subject to the conditions ${\frac{x^{2}}{4} + \frac{y^{2}}{5} + \frac{z^{2}}{25} = 1}$ and ${x + y - z = 0}$.
Using Lagrange multiplier method, I got following equations:
$$ {2x = \frac... | I think Lagrange multipliers method is not necessary here.
We need to find a maximal and a minimal value of $x^2+y^2+(x+y)^2$, where $\frac{x^2}{4}+\frac{y^2}{5}+\frac{(x+y)^2}{25}=1.$
Let $2(x^2+xy+y^2)=k$.
Hence, the condition gives $29x^2+8xy+24y^2=100$ or
$$k(29x^2+8xy+24y^2)=200(x^2+xy+y^2),$$
which says that the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How could I know that $X^4+1$ is $(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1)$? I thought that $X^4+1$ was irreducible, but in fact, $$X^4+1=(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1).$$
In general, how can I have the intuition of such a factorisation if I don't know it ?
| We know that $$a^4+4=a^4+4a^2+4-4a^2=(a^2+2)^2-(2a)^2=(a^2+2a+2)(a^2-2a+2)$$
This is an well known identity, most easily identifiable from the difference between two squares. It is called the Sophie Germain Identity=
Putting in $x=\sqrt{2} a$, we have that $$x^2+1=(x^2+\sqrt{2} x+1)(x^2-\sqrt{2}x+1)$$
Though the first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 3
} |
show that $\lim_{x\rightarrow 0} \left(\frac{\sin(x)}{x}\right)^{1/x^2} = e^{-1/6}$ I see that this is indeterminate form, so I approach L'Hopital rule, but I can not find this limit. Please help me to find this limit.
| $$\lim _{ x\rightarrow 0 }{ { \left( \frac { \sin { x } }{ x } \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { \sin { x } }{ x } -1 \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { \sin { x-x } }{ x } \right) }^{ \frac { 1 }{ { x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Prove: $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) > 10^{2009}$
Show that the number $ 3^{{4}^{5}} {+} 4^{{5}^{6}}$ can be expressed as the product of two integers greater than $ 10^{2009}$
By Sophie Germain:
$ 3^{{4}^{5}} {+} 4^{{5}^{6}}=\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right... | $log_{10}(3^{4^5}+4^{5^6}) = 9407.1874$ then it has $9408$ digits. Since you express this number as a product of two numbers then each of those has $4704$ digits.
$log_{10}(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}) = 4703.5937$ as we can see both numbers have $4704$ digits then we conclude that :
$$3^{2^9}+3^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to solve $2\cdot \left( \frac { 1 }{ 3 } \right) ^{ 0.3-x }-4=2$ How do I solve this equation:
$$2\cdot \left( \frac { 1 }{ 3 } \right) ^{ 0.3-x }-4=2$$
I've tried it myself but ended up with x = 0.3, eventhough the correct answer is 1.3
I followed these steps:
$$ 2*(1/3)^{(0.3-x)}-4=2$$
$$2*(1/3)^{(0.3-x)}=-2... | $$2\cdot \left( \frac { 1 }{ 3 } \right) ^{ 0.3-x }-4=2\\ 2\cdot \left( \frac { 1 }{ 3 } \right) ^{ 0.3-x }=6\\ \left( \frac { 1 }{ 3 } \right) ^{ 0.3-x }=3\\ { 3 }^{ x-0.3 }=3\\ x-0.3=1\\ x=1.3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Circle tangent to a parabola this is a calculus-geometry question that I found -
A circle of radius $1$ is tangent to the parabola $y = x^2 -2x + 1$ at two points. Find the coordinates of the center of the circle.
My attempt:
Let the center of the circle be $(a, b)$. Since there are two intersection points, we have:
$... | It is simpler to translate the parabola by $1$ towards the left, solving the problem for the parabola $y=x^2$ and then re-translating the final result by one unit towards the right. In this way, we can search a circle with its center on the $y $-axis, which typically has an equation of the form
$$x^2 +(y-k)^2=1$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $a^2+b^2+c^2=1$ so $\sum\limits_{cyc}\frac{1}{(1-ab)^2}\leq\frac{27}{4}$
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that:
$$\frac{1}{(1-ab)^2}+\frac{1}{(1-ac)^2}+\frac{1}{(1-bc)^2}\leq\frac{27}{4}$$
This inequality is stronger than $\sum\limits_{cyc}\frac{1}{1-ab}\leq\frac{9}{2}$ with ... | The Buffalo Way works. (I think that Michael Rozenberg knew this.)
We only need to prove the case $a, b, c > 0$.
After homogenization and clearing the denominators, it suffices to prove that
$f(a, b,c)\ge 0$ where $f(a,b,c)$ is a homogeneous polynomial of degree $12$.
Due to symmetry, assume that $c\le b\le a$. Let $c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Why does "$x^2 - 5x + 6 = 0$", which is the same as "$(x-3)(x-2) = 0$", represent a parabola? Consider the equation $x^2 - 5x + 6 = 0$. By factorising I get $(x-3)(x-2) = 0$. Which means it represents a pair of straight lines, namely $x-2 =0 $ and $x- 3 = 0$, but when I plot $x^2 - 5x + 6 = 0$, I get a parabola, not a... |
I get $(x-3)(x-2) = 0$. which means it represent a pair of striaght
lines namely $x-2 =0 $ and $x- 3 = 0$
When we just write
$$
(x-3)(x-2) = 0
$$
and ask for $x$ we usually mean the set of solutions
$$
S = \{ x \mid (x-3)(x-2) = 0 \}
$$
Two vertical lines in two dimensions we can model as:
$$
L_1 = \{ (x,y) \mid (x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 8,
"answer_id": 3
} |
If $a^2+b^2+c^2+d^2=4$ what is the range of $a^3+b^3+c^3+d^3$? I tried to used AM-GM but could not do.Setting any one as $2$ or $-2$ and the others $0$ we get the range as $[-8,8]$ but what is the formal way to do this?
| Without resorting to higher level of math such as the use of "compactness". Observe that $a^2 \le a^2+b^2+c^2+d^2 = 4\implies a \le 2\implies a^2(a-2) \le 0$, and similarly: $b^2(b-2) \le 0, c^2(c-2) \le 0, d^2(d-2) \le 0$. Thus $a^3+b^3+c^3+d^3 \le 2(a^2+b^2+c^2+d^2) = 2\cdot 4 = 8$. And the max is $8$ which occurs wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How do you find the derivative of $x\sqrt{x}$ with nonstandard analysis? Find the derivative of $x\sqrt{x}$. I'm doing this without power rules etc. , what I know beforehand are that
$$f'(x)=st\left(\frac{f(x+\Delta x)-f(x)}{\Delta x}\right)$$ and that
$$y'=st\left(\frac{\Delta y}{\Delta x}\right).$$
The correct answe... | Let's examine your proof.
Begin with
$$
\Delta y = (x + \Delta x)^{3/2} - x^{3/2}
$$
multiply numerator and denominator by a conjugate
$$
\big((x + \Delta x)^{3/2} - x^{3/2}\big) \cdot
\frac{(x + \Delta x)^{3/2} + x^{3/2}}{(x + \Delta x)^{3/2} + x^{3/2}}
$$
this is
$$
\frac{(x - \Delta x)^{3} - x^{3}}{(x + \Delta x)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the value of $\sum\binom{6n}{2k-1}(-3)^k$
Find the value of $\sum_{k=1}^{3n}\binom{6n}{2k-1}(-3)^k$.
My working:
\begin{align}
& \sum\binom{6n}{2k-1}(-3)^k\\
=& \sum\binom{6n}{2k-1}(i\sqrt3)^{2k}\\
=& i\sqrt3\sum\binom{6n}{2k-1}(i\sqrt3)^{2k-1}\\
=& \frac{i\sqrt3[(1+i\sqrt3)^{6n}-(1-i\sqrt3)^{6n}]}{2}\\
=& \... | $\sum \binom{m}{2k-1}x^{2k-1}
+\sum \binom{m}{2k}x^{2k}
=\sum \binom{m}{k}x^k
=(1+x)^m
$.
$(1-x)^m
=\sum \binom{m}{k}(-1)^kx^k
$
so
$(1+x)^m+(1-x)^m
=\sum \binom{m}{k}x^k(1+(-1)^k)
=2\sum \binom{m}{2k}x^{2k}
$
and
$(1+x)^m-(1-x)^m
=\sum \binom{m}{k}x^k(1-(-1)^k)
=2\sum \binom{m}{2k-1}x^{2k-1}
$.
Therefore,
putting $x =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Letter counting problem using generating functions Consider 4 different letters: a, b, c and d. Imagine you have access from 4 up to 12 letters, with at least 1 letter of each. For example, if I have 5 letters, one possible combination could be one $a$, one $b$, one $c$ and two $d$'s. Considering the case where it is i... | We consider words of length $4$ up to $12$ built from a four character alphabet $\{a,b,c,d\}$ . The following holds
*
*Each word contains each of the four characters at least once
*The position of the characters don't matter: e.g. $$aabcd\equiv badca$$
*Multiplicity of characters matter, but not which character ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding an inverse matrix of an infinite matrix I tried to find the inverse matrix of
$$
\begin{pmatrix}
2 & -1 & 0 & \cdots& 0 \\
-1 & 2 & -1 & \cdots& 0\\
0 & -1 & 2 & \ddots& \vdots\\
\vdots & \vdots & \ddots & \ddots &-1\\
0 & 0 & \cdots & -1& 2\\
\end{pma... | Your examples suggest that the $(i,j)$-th entry in the upper triangular part (i.e. when $i\le j$) is $\frac{1}{n+1}i(n+1-j)$. It should be straightforward to verify this formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate integral of fractional part I have to calculate $$I=\int _0^1\left\{nx\right\}^2dx , \:\:\:\:\: n \in \mathbb N, n \ge 1$$
Where {a} is $frac(a)$.
I know that $\left\{nx\right\}^2 = (nx - [nx])^2$ so
$$I\:=\int _0^1\:\left(n^2x^2-2nx\left[nx\right]+\left[nx\right]^2\right)dx=\frac{n^2}{3}-2n\int _0^1\:x\left... | \begin{align}
\lfloor nx \rfloor = m
&\iff m \le nx < m+1 \\
&\iff \dfrac mn \le x < \dfrac{m+1}{n} \\
\end{align}
Let
\begin{align}
I_m &= \int_{x=\frac mn}^{\frac{m+1}{n}}\{nx\}^2 dx \\
&= \int_{x=\frac mn}^{\frac{m+1}{n}}(nx - m)^2 dx
& \left(\text{Let $y = x - \dfrac mn$.}\right) \\
&= n^2\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Two right triangles of equal perimeter share a side. Given an acute angle of one triangle, find the acute angles of the other.
Two right triangles shown below have equal perimeters. The hypotenuse of the orange triangle is one leg of the green triangle stacked on top of it. If the smallest angle of the orange triangle... | I'll work through a slightly-more-general problem, replacing $20^\circ$ with $\alpha$. Writing $d$ for the hypotenuse of the lower triangle, and $\beta$ for the angle shown in the upper triangle ...
... the equal perimeters condition tells us:
$$d + d \sin\alpha + d \cos\alpha = d + d\tan\beta + d\sec\beta \tag{1}$$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find series representation of a function $ y=\frac{1}{1+x+x^{2}+x^{3}} $ Let $y = \frac{1}{1+x+x^{2}+x^{3}}$.
And I want to find the series representation of this function.
I've noticed that I can rewrite this like $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{1+x}*\frac{1}{1+x^{2}}$ or I can rewite this like a sum of two fracti... | There are two possibilities here: either we use the binomial theorem on $(1+x(1+x+x^2))^{-1}$, which looks awful, or we use a trick: multiplying and dividing by $1-x$, we have
$$ \frac{1-x}{(1-x)(1+x+x^2+x^3)} = \frac{1-x}{1-x^4}, $$
because the middle terms all cancel out. Now, we can expand the denominator,
$$ \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Simplifying a quartic equation I have the following function to simplify and solve but there definitely is something wrong with my method as the initial conditions do not work with my final result so if anyone could pinpoint what I'm doing wrong, I would really appreciate it.
Solving: $\frac{(N-0.5)^4}{N^2(-N+1)^2}=Ae... | Beginning with
\begin{equation}
\frac{k^4}{\left(k^2-\frac{1}{4}\right)^2}=Ae^t
\end{equation}
we get
\begin{equation}
\frac{k^2}{\left(k^2-\frac{1}{4}\right)}=\pm\sqrt{Ae^t}
\end{equation}
Solving for $k^2$ gives
\begin{equation}
k^2=\frac{\pm\sqrt{Ae^t}}{4\left(1\pm\sqrt{Ae^t}\right)}
\end{equation}
So
\begin{equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Number of triangles with integer sides Total number of right angle triangles whose inradius is $2013$ and sides are integer
Attempt: assuming that $a,b,c>0$ are the sides of a triangle.
so form a right angle triangle with sides $a,b,c$ and right angle at $C$
$\displaystyle r=(s-c)\tan \frac{C}{2}=(s-c)=\frac{a+b-c}{2}$... | Remember the Pythagorian triple $(a,b,c)$ where
$$c=m^2+n^2, b=m^2-n^2,a=2mn. $$ We know that such triple defines a right triangle. The additional condition $$ab=2r(a+b)-2r^2 $$
implies that $$2mn(m^2-n^2)=2r(m^2-n^2+2mn)-2r^2 .$$
Now finding $m $ and $n$ answers your question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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given $I_{n}=\int^1_{0}x^n\sqrt{1-x^2}dx$, then finding value of $ \frac{I_{n}}{I_{n-2}}$ given $\displaystyle I_{n}=\int^{1}_{0}x^n\sqrt{1-x^2}dx$, then finding value of $\displaystyle \frac{I_{n}}{I_{n-2}}$
Attempt: put $x=\sin \theta$ and $dx = \cos \theta$
$\displaystyle I_{n} = \int^{\frac{\pi}{2}}_{0}\sin^{n}\the... | $I_n=\dfrac12\beta(\dfrac{n+1}{2},\dfrac32)$ so
$$I_n=\dfrac12\beta(\dfrac{n+1}{2},\dfrac32)=\dfrac12\beta(\dfrac{n-1}{2},\dfrac32)\dfrac{n-1}{n+2}=\dfrac12\beta(\dfrac{n-3}{2},\dfrac32)\dfrac{n-3}{n}\dfrac{n-1}{n+2}=I_{n-2}\dfrac{n-3}{n}\dfrac{n-1}{n+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Finding a basis for a set of matrices in a vector space. I am trying to find a basis for the following vector space:
V = {2x2 matrices A | $\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$A = A$\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$}
So far, I have augmented $\bigl( \beg... | Start with the definitions
$$
\mathbf{A} =
\left[
\begin{array}{cc}
1 & 2 \\
2 & 3 \\
\end{array}
\right],
\qquad
\mathbf{B} =
\left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right].
$$
Following the logic of @DonAntonio, compute the commutator:
$$
\left[ \mathbf{A}, \mathbf{B} \right] =
\mathbf{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Checking if two matrices are similar I have two matrices
$$ \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} $$
and $$ \begin{pmatrix}
2 & 2 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} $$
They are not diagonalizable. Share the same characteristic polynomial, the same trace, same determin... | Although using Jordan Canonical forms are probably the fastest answer, it is possible to solve this question using brute force. We want to find out if there is an invertible matrix $P$ such that
$$P \cdot
\begin{pmatrix}
2 & 1 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{pmatrix} =
\begin{pmatrix}
2 & 2 & 0\\
0 & 2 & 0\\
0 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the equation of a circle tangent to a circle and x-axis, with center on a certain line. Find the equation of the circle, which is tangent to the circle $x^2 + y^2=4$ and $x-axis$, with center on the line $x=4$.
So far I have:$$(x-h)^2 + (y-k)^2=r^2$$
Since the center is on the line $x=4$, then the center is $(4;k)... |
$$x^2+y^2=0$$
this circle have radius $=2$center$=(0,0)$
So, in $\triangle OO'A$ $$(OO')^2=(O'A)^2+(OA)^2\tag{Pythagoras theorem}$$
$$(OB+BO')^2=(O'A)^2+(OA)^2$$ $\therefore$ $$(2+k)^2=k^2+4^2$$ $$4+k^2+4k=k^2+16$$ $\therefore$ $$k=3$$
Hope it helps!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$ I am studying the book Equations and Inequalities by Herman et al, and am stuck on the following exercise:
Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$... | $$\dfrac1{k+2}\binom nk=\dfrac{k+1}{(n+1)(n+2)}\cdot\dfrac{(n+2)!}{(k+2)!\{n+2-(k+2)\}!}=\dfrac{k+1}{(n+1)(n+2)}\cdot\binom{n+2}{k+2}$$
Now $\displaystyle(k+1)\cdot\binom{n+2}{k+2}$
$\displaystyle=(k+2-1)\cdot\binom{n+2}{k+2}$
$\displaystyle=(n+2)\cdot\binom{(n+1)!}{(k+1)!\{n+1-(k+1)\}!}-\binom{n+2}{k+2}$
$\displaystyl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Determine if 3 circles intersect at a common point Given three circles centered at points $A$, $B$, and $C$ with non-zero radii of lengths $R_A$, $R_B$ and $R_C$. Where the centers of the circles form a valid triangle, and where the distance between any two centers is less than or equal to the sum of their correspondin... | If we coordinatize, say, with
$$A = (0,0) \qquad B = (c, 0) \qquad c = (b\cos A,b\sin A)$$
and take suppose circles $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ (of respective radii $r_A$, $r_B$, $r_C$) meet at a point $P = (x,y)$, then we have three equations in two unknowns $x$ and $y$:
$$\begin{align}
x^2 + y^2 &= r_A^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9$ Let $$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9,$$
$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32,$$
$$\frac{a^3}{b+c} + \frac{b^3}{c+a} + \frac{c^3}{a+b} = 122.$$
Find the value of $abc$.
Please check if my answer is correct or not.
$a+b+c + \frac... | Your solution is correct.
You find $a^2+b^2+c^2=6$ using the same trick:
$$\frac{a^3}{b+c} + \frac{b^3}{c+a} + \frac{c^3}{a+b} = 122 \iff \\
(a+b+c)\left(\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b}\right)-(a^2+b^2+c^2) = 122\Rightarrow \\
a^2+b^2+c^2=4\cdot 32-122=6.$$
Then:
$$ab+bc+ca=\frac12[(a+b+c)^2-(a^2+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all primes $p$ for which the equation $x^6 \equiv -1$ mod $p$ has a solution. I'm not sure if i could apply Wilson's theorem here.
| Since $(x^3)^2 + 1^2 = 0$ mod $p$, $p$ is $1$ mod $4$ (Fermat's two squares)
$(x^2)^3 + 1^3 = (x^2 + 1)(x^4 + x^2 + 1) = 0 $ mod $p$
$\implies p | x^2 + 1$ as $p$ has to be of the form $a^2 + b^2$ (Fermat and skipping steps)
So if we take any $p$ such $p$ is a factor of $x^2 + 1$ then the equation $x^6 + 1 = 0$ mod $p$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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then minimum number of number of roots of $p(x) = 0$ is let $p(x)=x^6+ax^5+bx^4+x^3+bx^2+ax+1.$ given that $x=1$ is a one rot of $p(x)=0$
and $-1$ is not a root. then minimum number of number of roots of $p(x) = 0$ is
Attempt: $x=0$ in not a root of $p(x)=0.$
So $\displaystyle \left(x^3+\frac{1}{x^3}\right)+a\left(x... | This answer assumes that you want to find the minimum number of the real roots of $p(x)$.
Since $x=1$ is a root of $p(x)$, we have
$$p(1)=0\iff 2a+2b+3=0\iff b=\frac{-2a-3}{2}$$
from which we can write
$$p(x)=(x-1)^2\left(x^4+(a+2)x^3+\left(a+\frac 32\right)x^2+(a+2)x+1\right)$$
So, we see that the number of the real r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $abc=1$ so $\sum\limits_{cyc}\frac{7-6a}{2+a^2}\geq1$
Let $a$, $b$ and $c$ be real numbers such that $abc=1$. Prove that:
$$\frac{7-6a}{2+a^2}+\frac{7-6b}{2+b^2}+\frac{7-6c}{2+c^2}\geq1$$
The equality occurs also for $a=b=2$ and $c=\frac{1}{4}$.
This inequality is a similar to the very many contest's inequalitie... | Clearly, that we only need to prove this inequality for case:$\ a , b, c >0$
Let $a=e^{t_1}, b= e^{t_2}, c= e^{t_3} $
$$f(t_1)+f(t_2)+f(t_3) \ge1 \ , \ t_1+t_2+t_3=0$$
$$f(t)=\dfrac{7-6e^t}{2+e^{2t}}$$
$f'(t)=\dfrac{2e^t(e^{t}-3)(3e^t+2)}{(e^{2t}+2)^2}$
Minimum of $f(t)$ is attained at $t=\ln(3)>0 \Rightarrow \min\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving for $p$ and $k$. For the question,
$$F(x)= x^{2} + 6x + 20 + k(x^{2} -3x -12)$$
I have to find the value of k and the value of $p$ given that the minimum value of $F(x)$ is $p$ and $F(-2) = p$.
What I did is: I expanded the expression of $F(x)$ and reached a term for $p= 12 - 2p$. I tried to complete the square... | A quadratic $ax^2+bx+c$ with a positive leading term $a\gt0$ attains its minimum at $-\cfrac{b}{2a}\,$.
Rewriting $F(x) = (1+k)x^2+(6-3k)x+20-12k\,$ it follows that: $$-\cfrac{6-3k}{2(1+k)}=-2 \quad \implies \quad k = \cfrac{2}{7}$$
(Since the leading term $1+k = 1 + \cfrac{2}{7}\gt 0$ this is a valid solution.)
Substi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Formula for number of ways to make $n$ units of currency if the coins we have are of value 1, 3, and 9 Suppose we live in a world where there are three forms of currency: each worth 1, 3, and 9 units. Can we find a formula for the number of ways to make $n$ units of currency, if we have an effectively unlimited supply ... | Let $f_1(n)$ be the # of ways to make $n$ units using only $1$; $f_3(n)$ be the # of ways to make $n$ units using $1$ and $3$; and $f_9(n)$ be the # of ways using $1$, $3$ and $9$.
If only $1$ and $3$ are allowed, $3$ can be used $0 \leq k \leq \lfloor \frac{n}{3} \rfloor$ times. For the remaining $n - 3k$ units, we fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating a finite sum using a double sum I want to evaluate the sum $S_n = \sum_{k=0}^n k 3^k $.
Try:
Since $ \sum_{j=0}^{k-1} 1 = k $, then
$$ S_n = \sum_{k=0}^n \sum_{j=0}^{k-1} 3^k $$
Since we are summing over $(0 \leq k \leq n )$ and $(0 \leq j \leq k -1 )$, then we are summing over $(0 \leq j < k \leq n )$. Thus... |
$$ S_n = \sum_{j=0}^n \sum_{k=j}^n 3^k = \sum_{j=1}^n \frac{ 3^{n+1} - 3^j }{2} = \frac{n 3^{n+1}}{2} - \frac{1}{2} \sum_{j=0}^n 3^j = \boxed {\frac{n 3^{n+1} }{2} - \frac{1}{2} \cdot \left( \frac{ 3^{n+1} - 1 }{2} \right) } $$
Is this a correct argument?
You have an error.
Since $j\le k-1\iff k\ge j+1$, we have
$$\s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The equations $x^3+5x^2+px+q=0$and $x^3+7x^2+px+r=0$ have 2 common roots, then find the third root of both equations
The equations $x^3+5x^2+px+q=0$and $x^3+7x^2+px+r=0$ have 2 common roots, then find the third root of both equations
From the first equation we can say, $\alpha\beta+\beta\gamma+\gamma\alpha=p/1... | If $x^3 + 5x^2 + px + q = 0$ and $x^3 + 7x^2 + px + r = 0$ have two common roots then there must exists a common monic binomial factor, $F(x)$.
So $F(x)$ must be a divisor of
$$(x^3 + 7x^2 + px + r) - (x^3 + 5x^2 + px + q) = 2x^2 - (q - r) $$
So $F(x) = x^2 - \dfrac 12(q - r)$.
So, for some $s$
\begin{align}
x^3 + 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$ let $$f(x)=\frac{1}{1+2\cos x}$$
prove that :
$$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$$
My Try :
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{4\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\... | Let $2\cos\frac{2\pi}{7}=a$, $2\cos\frac{4\pi}{7}=b$ and $2\cos\frac{6\pi}{7}=c$.
Hence,
$$a+b+c=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Regarding solutions of ordinary differential equations Got into a problem with the solution of this one:
$$
x\cos\left(\frac{y}{x}\right)(x\text{d}x+y\text{d}y)=y\sin\left(\frac{y}{x}\right)(x\text{d}y-y\text{d}x)
$$
I can identify differentials of $xy$ and $\frac{x}{y}$ and also of $\cos$ and $\sin$ functions but how ... | We have:
$x \cos (\frac{y}{x})(x dx+ydy) = y \sin(\frac{y}{x})(xdy-ydx)$
$\Rightarrow x \cos (\frac{y}{x})(x + y\frac{dy}{dx}) = y \sin(\frac{y}{x})(x\frac{dy}{dx} - y)$
Let $u = \frac{y}{x}$
Then $\frac{du}{dx} = \frac{x\frac{dy}{dx} - y}{x^2} = \frac{\frac{dy}{dx} - u}{x}$
$\Rightarrow \frac{dy}{dx} = x\frac{du}{dx} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factor $r^6 -3r^4 +3r^2 - 1 = 0$ I tried two ways to factor $r^6 -3r^4 +3r^2 - 1 = 0$
When I factor $r^4$ out of $r^6 -3r^4$:
$r^4(r^2-3)+(3r^2-1) = 0$
When I factor $r^2$ out of $r^6 + 3r^2$:
$r^2(r^4+3)-(3r^4 + 1) = 0$
For both methods, I'm stuck at an equation that isn't factorable. But Wolfram Alpha says $r^... | Let $x = r^{2}$
Then $x^{3} -3x^2+3x-1 = 0$
Notice that $x=1$ is a solution
Therefore we have $(x-1)(x^{2} +ax+1) = 0$ where $a$ is a constant we need to find
By comparing coefficients of $x$, we can see that $-a+1 = 3 \Rightarrow a = -2$
Then we have $(x-1)(x^2-2x+1)=0 \Rightarrow (x-1)(x-1)^{2}=0 \Rightarrow (x-1)^{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Finding $\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$ Finding $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$
Attempt: Assume $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx =\int \frac{2x-x^{-2}}{\bigg[x^4-x^2+2x-\frac{2}{x}-\frac{1}{x}+\frac{1}{x^2}\bigg]}dx$
could some help me how to solve,thank... | $$\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx=\frac{1}{3}\int\left(\frac{1}{x-1}+\frac{2x+1}{x^2+x-1}-\frac{3x^2+1}{x^3+x+1}\right)dx=$$
$$=\frac{1}{3}\left(\ln|x-1|+\ln|x^2+x-1|-\ln|x^3+x+1|\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$3\mid a^3+b^3+c^3$ if only if $3\mid a+b+c $
Prove the following equivalence: $3\mid (a^3 + b^3 + c^3) $ if and only if $3\mid (a + b + c) $.
My try:
I know $a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc$, but I can't seem to proceed from here.
Thanks all!
| By Fermat's Little Theorem we know that $3$ divides $x^3-x$ for any $x \in \mathbb{Z}$
Which means $(a^3-a)+(b^3-b)+(c^3-c) = (a^3+b^3+c^3)-(a+b+c)$ is divisible by $3$.
The conclusion is easy to derive now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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How to prove that $7p + 3^p -4$ is not a perfect square? How to prove that $7p + 3^p -4$ is not a perfect square?
I calculated: $\left(\frac{7p+3^p-4}{p}\right) = \left(\frac{-1}{p}\right)$. So if $p \equiv 3 \mod 4$, the result is $-1$. So in that case, $7p+3^p -4$ can't be a square. But what about the case $p \equiv... | The squares modulo $4$ are
$$(4k)^2 = 4(4k^2) \equiv 0 \pmod{4},$$
$$(4k+1)^2 = 4(4k^2 + 2k) + 1 \equiv 1 \pmod{4},$$
$$(4k + 2)^2 = 4(4k^2 + 4k + 1) \equiv 0 \pmod{4},$$
and
$$(4k + 3)^2 = 4(4k^2 + 6k + 2) + 1 \equiv 1 \pmod{4},$$
but if $p = 4 \ell + 1$, then
$$7p + 3^p - 4\equiv (-1)(1) + (-1)^{4 \ell}(-1)^1 \pmod{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can i find the sum? How can i calculate this serias?
$$\sum_{i=1}^n \frac{2^m}{2^i}\cdot i$$
I tried to do:
$$\ 2^m\cdot\left[\frac{1}{2}+\frac{2}{2^2}+...+\frac{n}{2^n}\right]$$
And i don't know how to continue..
Thank you.
| $$s(n)=\frac{1}{2}+\frac{2}{2^2}+...+\frac{n}{2^n}\quad (1)$$
multiply by $1/2$ and get
$$\frac{s(n)}{2}=\frac{1}{2^2}+\frac{2}{2^3}+...+\frac{n-1}{2^{n}}+\frac{n}{2^{n+1}}\quad (2)$$
Now make $(1)-(2)$ and get
$$\frac{s(n)}{2}=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n}}-\frac{n}{2^{n+1}}\\
s(n)=\left[\frac{1}{2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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prove an inequality with conditions Let $a,b, c \in \mathbb{R}$ such that $0\le a\le1,0\le b\le1 , 0\le c\le1.$
If $$a+b\leq c+1,
\\ a+c \leq b+1,
\\b+c\leq a +1$$
can we prove that
$a^2+b^2+c^2\le 1+2abc$ ?
| Let
$$ a+b=(c+1) k_1,a+c=(b+1) k_2,b+c=(a+1) k_3 $$
and then
$$ a=-\frac{-k_2+k_3-k_1 \left(k_2 \left(k_3+2\right)+1\right)}{k_2+k_3+k_1 \left(1-k_2
k_3\right)+2},b=-\frac{k_2-k_3-k_1 \left(\left(k_2+2\right) k_3+1\right)}{k_2+k_3+k_1
\left(1-k_2 k_3\right)+2},c= -\frac{-k_3-k_2 \left(2 k_3+1\right)+k_1 \left(1-k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Verify my proof of the irrationality of $\sqrt{6}$. Here's my proof for the irrationality of $\sqrt{6}$:
Proof:
Let $\sqrt{6}$ be a rational number. Then
$\frac{a}{b}$ = $\sqrt{6}$ , where $a$ and $b$ are coprime.
Therefore,
$a^2 = 6 b^2$.
Here $a^2$ is divisible by both $2$ and $3$. Which means, $a$ too is divisible ... | That works fine, but you don't need to do both $2$ and $3$. If:
$$a^2=6b^2$$
Then $a$ must be even, $a=2a_0$, and then: $4a_0^2=6b^2$ or $$2a_0^2=3b^2.$$
So $3b^2$ must be even, so by the fundamental theorem of arithmetic, $b$ must be even, and you are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the coefficient of $x^2y^3z^6$ in the expression $(x^2+y+z)^{10}$? How can find the coefficient of $x^2y^3z^6$ in the expression $(x^2+y+z)^{10}$?
I know this involves the binomial theorem, but I can't figure out how to solve it.
| We can apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of an expression.
We obtain
\begin{align*}
[x^2y^3z^6](x^2+y+z)^{10}&=[x^2y^3z^6]\sum_{k=0}^{10}\binom{10}{k}x^{2k}(y+z)^{10-k}\tag{1}\\
&=[y^3z^6]\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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$x\sqrt{1-\frac{y(x+1)}{(x+y+1)^2}} = y\sqrt{1-\frac{z(y+1)}{(y+z+1)^2}} = z\sqrt{1-\frac{x(z+1)}{(z+x+1)^2}}$
If $x,y,z > 0$, and $$x\sqrt{1-\frac{y(x+1)}{(x+y+1)^2}} = y\sqrt{1-\frac{z(y+1)}{(y+z+1)^2}} = z\sqrt{1-\frac{x(z+1)}{(z+x+1)^2}}$$
Prove $x=y=z$.
I try it by Mathematica but can not solve, and Maple give a... | Let $$f(x,y)=x^2\Bigg(1-\frac{(x+1)y}{(x+y+1)^2}\Bigg)$$
Then the hypothesis says that $f(x,y)=f(y,z)=f(z,x)$. Note that for $x\neq y$ and $x,y$ positive,
$$
\frac{f(y,x)-f(x,y)}{y-x}=\frac{x^3+y^3+2(x^2+y^2)+xy(2(x+y)+3)+x+y}{(x+y+1)^2} > 0 \tag{1}
$$
Similarly, if we expand $\frac{f(x_2,y)-f(x_1,y)}{x_2-x_1}$, we f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Conditional Poisson processes with Multiple conditions Let $\lbrace N(t)\rbrace_{t\geq 0}$ be a Poisson process with intensity $\lambda = 3$.
Compute $$P\left[N(6) = 2 \,|\, N(8) = 4, N(3) = 1\right].$$
I understand when there is only one condition i.e. $P\left[N(6) = 2 | N(8) = 4\right]$. Since that is $$P\left[N(6) =... | You can handle this using the definition of conditional probability, stationary increments and independent increments.
$$
\begin{align*}
P\left[N(6) = 2 \,|\, N(8) = 4, N(3) = 1\right] &= \frac{P[N(6) = 2,N(8) = 4, N(3) = 1]}{P[N(8) = 4, N(3) = 1]}\\
&=\frac{P[N(3) = 1, N(6) - N(3) = 1, N(8) - N(6) = 2]}{P[N(8) = 4, N(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many positive integer solutions does the equation $a+b+c=100$ have if we require $aHow many positive integer solutions does the equation $a+b+c=100$ have if we require $a<b<c$?
I know how to solve the problem if it was just $a+b+c=100$ but the fact it has the restriction $a<b<c$ is throwing me off.
How would I solv... | First we count the number of triples of positive integers $(a, b, c)$ with $a + b + c = 100$. There are $\binom{99}{2}$ of them, which results from an elementary application of stars and bars. Just line up a hundred dots, place a bar in between the $a$th and $(a +1)$th dot, and then place another bar $b$ dots to the ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Area inside outer curve of limacon $r=1+2\cos\theta$ I need to find the area inside the outer curve of limacon $r=1+2\cos\theta$
Here is what I tried:
$$
0=1+2\cos\theta
$$
$$
\theta=\cos^{-1}(-\frac{1}{2})
$$
$$
\theta=\frac{2}{3}\pi , \theta=\frac{8}{3}\pi
$$
$$
A = 2\int^{\frac{8}{2}\pi}_{\frac{2}{3}\pi}r^2d\theta... | You pretended that the $d\theta$ was a $dr$. Unfortunately you cannot do that.
$$2\int^{\frac{8}{2}\pi}_{\frac{2}{3}\pi}r^2d\theta\color{red}{\neq}\frac{r^3}{3}|^{\frac{8}{3}\pi}_{\frac{2}{3}\pi}$$
Now I'll let you try it. If you still need help, just ask!
| {
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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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curve on sphere from $(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ to $(-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$ Consider the set:
$$
A \equiv \left\{ (x,y,z) \in \mathbb{R}^{3} | x + y + z = 0\ \mathrm{and}\ x^{2} + y^{2} + z^{2} = 1 \right\}
$$
This is the intersection of the unit ... | When a plane intersects a sphere the curve of intersection will be a circle.
the origin is at the center of this cricle (because the plane goes through the origin, and the sphere is centered at the origin). Now you just need to parametarize the circle.
We have two points on this circle, but what we really need are ort... | {
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"url": "https://math.stackexchange.com/questions/2218352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding value of $\lfloor b_{100} \rfloor $ in recursive relation
If $\displaystyle b_{n+1} = b_{n}+\frac{1}{b_{n}}$ and $b_{1} = 1,$ then find the value of $\lfloor b_{100}\rfloor$.
My attempt: $\displaystyle b_{n+1}b_{n} = b^2_{n}+1\Rightarrow b_{n}b_{n+1}-b^2_{n} = 1$
$$\frac{1}{b_{n}b_{n+1}-b^2_{n}} = 1\Rightarr... | Note that $b_n$ is strictly increasing and for $n\geq 2$, $b_n^2\ge 2n$, because $b_2=2$ and
$$b_{n+1}^2=\left(b_n+\frac{1}{b_n}\right)^2=b_n^2+2+\frac{1}{b_n^2}\geq 2n +2=2(n+1).$$
Moreover, for $n\geq 2$,
$$\begin{align}b_n^2&=b_2^2+\sum_{k=2}^{n-1}(b_{k+1}^2-b_k^2)=4+\sum_{k=2}^{n-1}\left(2+\frac{1}{b_k^2}\right)\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\int_{0}^{\infty}{1\over x^{2}-x+1}{\mathrm dx\over \sqrt{x}}=\int_{0}^{\infty}{1\over (x^{2}-x+1)^2}{\mathrm dx\over \sqrt{x}}={\pi}$ Consider these integrals
$$\int_{0}^{\infty}{1\over x^{2}-x+1}\cdot{\mathrm dx\over \sqrt{x}}={\pi}\tag1$$
$$\int_{0}^{\infty}{1\over (x^{2}-x+1)^2}\cdot{\mathrm dx\over \sq... | Alternative computation.
$\displaystyle J=\int_{0}^{\infty}{1\over x^{2}-x+1}\cdot{\mathrm dx\over \sqrt{x}}$
Perform the change of variable $y=\sqrt{x}$,
$\begin{align} J&=2\int_{0}^{\infty}{1\over x^{4}-x^2+1}dx\\
&=2\int_{0}^{\infty}{1+x^2\over x^{6}+1}dx\\
&=2\int_{0}^{\infty}{1\over 1+x^{6}}dx+2\int_{0}^{\infty}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Singular value decomposition works only for certain orthonormal eigenvectors, not all? I'm trying to find the SVD of the following matrix:
$$A=
\begin{pmatrix}
1 & 1 \\
2 & -2 \\
2 & 2 \\
\end{pmatrix}
$$
I found the eigenvalues and vectors for $A'A$:
$$
\begin{array}{cc}
\lambda_1=10 & \lambda_2=8 \\
e_1'=(1,1) ... | The answer is yes. Since your $m\times n$ matrix $A$ is such that $m\geq n$, you first compute $\sigma_i$s and $v_i$s and than get the $u_i$s via relation
$$u_i = \frac{1}{\sigma_i}Av_i\text{.}$$
This follows from the proof of "SVD theorem": There exist unitary matrices $U$ and $V$ with the columns $u_i$ and $v_i$ (and... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Evaluating $\int_0^\infty \frac{\sin^2(xu)du}{u^2}$ How do I evaluate$$\int_0^\infty \frac{\sin^2(xu)du}{u^2}$$
I tried integration by parts, but it kept getting messier.
| Here is a slight different line of reasoning: We begin by noting that
\begin{align*}
\frac{1}{a} &= \int_{0}^{\infty} e^{-ax} \, dx, \qquad (a > 0) \\
\frac{1}{a^2} &= \int_{0}^{\infty} x e^{-ax} \, dx, \qquad (a > 0) \\
\frac{a}{a^2+b^2} &= \int_{0}^{\infty} \cos(bx) e^{-ax} \, dx, \qquad (a > 0) \\
\end{align*}
which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Find the value of $3+7+12+18+25+\ldots=$ Now, this may be a very easy problem but I came across this in an examination and I could not solve it.
Find the value of
$$3+7+12+18+25+\ldots=$$
Now here is my try
$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$
After that, I co... | I'm adding another answer because people ask how to find $a_n$ without trial and error.
We note that:
$a_2 - a_1 = 4; a_3 - a_2 = 5; a_4 - a_3 = 6$,
which leads us to conclude that $a_n$ is given by the recurrence relation:
$a_{n+1} - a_n = n+3$
Let's start by solving the homogeneous equation:
$a_{n+1} - a_n = 0$
The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Find the greatest common divisor of the polynomials $ \ f(x)=x^{3}+2 \ \ and \ \ g(x)=x- 1 \ $ over the field $ \ \ \mathbb{Z}_{3}$ Find the greatest common divisor of the polynomials $ \ f(x)=x^{3}+2 \ \ and \ \ g(x)=x- 1 \ $ over the field $ \ \ \mathbb{Z}_{3}$. $$ $$ We have $ \ f(x)=x^{3}+2 \ \ and \ \ g(x)=x-1=x... | It's easier if you write $x^3+2=x^3-1=(x-1)(x^2+x+1)$, so it becomes apparent that $x-1$ is a divisor, hence the greatest common divisor.
Anyway, your computation is correct, using $x^3+2=x^3+8=(x+2)(x^2-2x+4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$ If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$
My Attempt:
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2a.\dfrac {1}{a}$$
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2$$
How do I proceed further?
| You have $$a-\frac 1a=6$$ so squaring gives $$a^2-2+\frac{1}{a^2}=36$$ so...?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Solutions of an equation Problem:
Find all real solutions for $x$ in
$$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $$
I know that you can move all the terms to one side, then divide by $2$ and factor, but what next?
| We seek all real solutions for $2(2^x-1)x^2+\left(2^{x^2}-2\right)x=2^{x+1}-2$.
Let's rewrite our equation into a very interesting form.
\begin{align}
2(2^x-1)x^2+\left(2^{x^2}-2\right)x&=2^{x+1}-2\\
2(2^x-1)x^2+\left(2^{x^2}-2\right)x-2^{x+1}+2&=0&&\text{Set equation = 0}\\
(2^x-1)x^2+\left(2^{x^2-1}-1\right)x-2^x+1&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to turn $12x^2-8x+1$ into $(2x-1)(6x-1)$ without quadratic equation? After almost seven years I recently started again learning math and have few holes in my algebra knowledge, so I apologize for the beginner question.
My question is:
Is there any simple trick to turn $12x^2-8x+1$ into $(2x-1)(6x-1)$ without using ... | Write $x=y/12$, so the polynomial becomes
$$
12\frac{y^2}{12^2}-\frac{8y}{12}+1=
\frac{1}{12}(y^2-8y+12)=\frac{1}{12}(y-2)(y-6)
$$
The decomposition of $y^2-8y+12$ is obtained from the search for two numbers whose sum is $8$ and product is $12$.
Substituting back $y=12x$, we have $y-2=2(6x-1)$ and $y-6=6(6x-1)$, so the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 7
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Show this $|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\frac85$ Let $x\in R$ show that
$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{8}{5}$$
since
$$f(x)=f(x+\pi),$$it sufficient to show $x\in (0,\pi]$
| Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of $\sin$ on $[0, \pi]$ quickly reduces the problem to showing the inequality
$$2 \sin 1 > \frac{8}{5} .$$
From $\pi < \frac{22}{7}$ we deduce $\frac{3 \pi}{10} < \frac{66}{70} < 1$, and so
$$2 \sin 1 > 2 \sin \frac{3 \pi}{10} = 2 \cdot \frac... | {
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"url": "https://math.stackexchange.com/questions/2237952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
For $\Bbb K^{n \times n}$ matrices $A$ and $B$, $\exists p[p(A)=B] \implies \exists p[p(B)=A]$? For $\Bbb K^{n \times n}$ matrices $A$ and $B$, is it true that if there exists a matrix polynomial $p$ such that $p(A)=B$, then there exists another matrix polynomial $p$ (not necessarily the same) such that $p(B) = A$?
My ... | Take $A=\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$, and let $p(x)=x^2$. Then $p(A) = I$, but there's no polynomial $q$ with $q(I) = A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $\int_{0}^{1}\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right) dx=\int_{0}^{1}x\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right) dx=\pi$ Two integrals exhibit the same closed form
Motivated by this interesting question.
$$\int_{0}^{1}\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right)\mathrm dx=\pi\tag1$$
... | On the path of Latte,
$\displaystyle J=\int_0^1 \sqrt{\dfrac{x}{1-x}}\ln\left(\dfrac{x}{1-x}\right)dx$
Perform the change of variable $y=\sqrt{\dfrac{x}{1-x}}$ and then, use integration by parts,
$\begin{align} J&=4\int_0^{\infty} \dfrac{x^2\ln x}{(1+x^2)^2}dx\\
&=\left[-\dfrac{2x\ln x}{1+x^2}\right]_0^{\infty}+2\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Cotangent expansion to compute infinite products At the end of class my professor remarked that the infinite product expansion of cotangent,
$$\text{cot}(z)=\frac{1}{z}+\prod \frac{2z}{z^2-n^2\pi^2}$$
can be used to compute infinite products like $\prod \frac{1}{n^2+1}$,$\prod \frac{1}{n^2+a^2}$. I'm assuming I'd subs... | Let $z = \pi i a$. Then
\begin{align*}
\cot (\pi i a) &= \frac{1}{\pi i a} + \prod \frac{2\pi i a}{i^2 \pi^2 a^2 - n^2 \pi^2}\\
- i \coth (\pi a) &= - \frac{i}{\pi a} - \frac{2 i a}{\pi} \prod \frac{1}{a^2 + n^2}\\
\frac{2 a}{\pi} \prod \frac{1}{a^2 + n^2} &= \coth (\pi a) - \frac{1}{\pi a} \\
\prod \frac{1}{a^2 + n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $x$ and $y$ such that $15\sin(x+y)+7\sin x+7\sin y$ is maximized Question is: Find $x$ and $y$ such that
$$15\sin(x+y)+7\sin x+7\sin y$$
is maximized.
What I tried was
$$f(x,y)=15\sin(x+y)+7\sin x+7\sin y=15(\sin x\cos y+\cos x\sin y)+7(\sin x +\sin y)$$
$$\frac{\partial f}{\partial x}=15\cos(x+y)+7\cos x=0$$
$$\f... | Hint:
Observe that
$$\frac{\partial f}{\partial x}=0=\frac{\partial f}{\partial y}\qquad\implies\qquad\cos x=\cos y\qquad\iff\qquad x=2n\pi\pm y\;\;\text{for some }n\in\mathbb{Z}$$
If $x=2n\pi+y$ we have
\begin{align*}
\frac{\partial f}{\partial x}=0&\quad\implies &15\cos(2n\pi+2y)+7\cos(2n\pi+y)&=0\\
&\quad\iff&15\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Let a general term $T_n$ be defined as $T_n =\left(\frac{1\cdot 2\cdot 3 \cdot 4 \cdots n}{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}\right)^2$ Let a general term $T_n$ be defined as
$$T_n =\left(\frac{1\cdot 2\cdot 3 \cdot 4 \cdots n}{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}\right)^2$$
Then prove that
$\lim_{n\to\infty... | $$\dfrac{T_m}{T_{m-1}}=\left(\dfrac m{2m+1}\right)^2<\left(\dfrac m{2m}\right)^2=\dfrac14$$ for $m>0$
$$\sum_{r=1}^\infty T_r<\sum_{r=1}^\infty T_1\left(\dfrac14\right)^{r-1}=\dfrac{\dfrac19}{1-\dfrac14}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve a simple equation: $x+x\sqrt{(2x+2)}=3$
$x+x\sqrt{(2x+2)}=3$
I must solve this, but I always get to a point where I don't know what to do. The answer is 1.
Here is what I did:
$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1... | As in the comment $x-1$ is a factor of $f(x) = -2x^3 -x^2-6x+9$. After an easy long division we get $f(x) = (x-1)(-2x^2-3x-9)$. From this we can use the quadratic equation to see the other two roots are not real.
We can do the long division in the following way:
We need to divide $-2x^3-x^2-6x+9$ by $x-1$ so we can ask... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos\left (\theta -\frac{\pi}{4}\right)$ If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos \left(\theta -\frac{\pi}{4}\right).$
I could not get any idea to solve. However I tried by using $\theta =0^\circ $. Bu... | Rearrange into sines and cosines. Form a cosine addition identity.
\begin{align}
\tan(\pi\cos t)\tan(\pi\sin t) &= 1
\\
\sin(\pi\cos t)\sin(\pi\sin t) &= \cos(\pi\cos t)\cos(\pi\sin t)
\\
0 &= \cos(\pi\cos t)\cos(\pi\sin t) - \sin(\pi\cos t)\sin(\pi\sin t)
\\
0 &= \cos(\pi\cos t + \pi\sin t)
\\
0 &= \cos(\pi[\cos t + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove inequality with integral Prove the following inequality
$$\ln \frac{\pi + 2}{2} \cdot \frac{2}{\pi} < \int \limits_0^{\pi/2} \frac{\sin\ x}{x^2 + x} < \ln \frac{\pi + 2}{2}$$
I can prove that $\frac{\sin\ x}{x^2 + x} < \frac{1}{x + 1} \ \forall x \in (0, +\infty) \Rightarrow \int \limits_0^{\pi/2} \frac{\sin\ x}{... | The given inequality just depends on the convexity inequality
$$ \forall x\in\left[0,\frac{\pi}{2}\right],\qquad \frac{2}{\pi}x\leq \sin(x)\leq x.$$
We may use another convexity inequality
$$ \forall x\in\left[0,\frac{\pi}{2}\right],\qquad x-\frac{1}{6}x^3\leq \sin(x)\leq x-\left(\frac{4}{\pi^2}-\frac{8}{\pi^3}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.