Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\c... | alternative is to use $u=\sqrt{9-x^{2}}$
$$
\begin{aligned}
\int{\frac{x^{3}}{\sqrt{9-x^{2}}}\ dx}&=\int{(u^{2}-9)\ du}\\
&=\frac{u^{3}}{3}-9u+c\\
&=\frac{\left(\sqrt{9-x^{2}}\right)^{3}}{3}-9\sqrt{9-x^{2}}+c
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Simplify homogeneous polynomial of degree 3 with all coefficients equal 1 I want to simplify this homogeneous polynomials:
$$x^3 + y^3 + z^3 + x^2 y + x y^2 + x^2 z + x z^2 + z^2 y + z y^2 + x y z$$ to a simpler form.
For example, the homogeneous polynomials of degree 2 can be simplified as follows:
$$x^2 + y^2 + z^2... | Maybe $$(x^2+y^2+z^2)(x+y+z)+xyz$$ or $$\frac{x^5y+y^5z+z^5x-x^5z-y^5x-z^5y}{(x-y)(x-z)(y-z)},$$ where $\prod\limits_{cyc}(x-y)\neq0?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $a^2 + b^2 + c^2 \geqslant ab + bc + ca$
Let $a, b, c \in \mathbb{R}$. Show that $a^2 + b^2 + c^2 \geqslant ab + bc + ca$.
My reasoning went as follows and I would like to know if it's correct.
$a^2 + b^2 + c^2 \geqslant ab + bc + ca$
$\Leftrightarrow a^2 + b^2 + c^2 -ab - bc - ca \geqslant 0$
$\Leftrightarro... | Yes, you can.
But I think it's better to release this way and the fact $2a^2=a^2+a^2$ by the cyclic sum:
$$a^2+b^2+c^2-ab-ac-bc=\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=$$
$$=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$
Another way:
$$a^2+b^2+c^2-ab-ac-bc=a^2-(b+c)a+b^2-bc+c^2=$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$ I tried doing it with CS-Engel to get $$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c}
$$
I thought that maybe proof that $$
\frac{1}{a+b+c+a b+b c+a c} \geq \frac{1}{3(1+a b c)}
$$ or $$
3+3 a b c \geq a+b+... | As in Michael's solution, this uses the same idea of "+1" to split up the fraction.
Written up this way, it might seem like a more natural approach.
WTS
$$ \sum \frac{ 1 + abc } { a (1+b)} \geq 3 $$
$$\sum \frac{ 1 + abc + a + ab } { a (1+b) } \geq 6$$
$$ \sum \frac{ ab ( 1 + c) + (1+a ) } { a (1+b) } \geq 6$$
$$ \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Unable to prove an assertion with induction I need to prove:
$ \displaystyle \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5n + 6)} $ with mathematical induction for all $n \in \mathbb{N}$.
After successfully proving it for n = 1, I try to prove it in the Induction-Step for n + 1:
$ \displaystyle... | Your proof by induction is complete as soon as you compute the difference $$\frac{1}{5s} - \frac{1}{5(s + 5)}=\frac{(s+5)-s}{5s(s + 5)}=\frac{1}{s(s + 5)}.$$
There is also a direct proof: the sum is telescopic
$$ \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)}=\frac{1}{5}\sum_{k=1}^n\left(\frac{1}{5k + 1}-\frac{1}{5(k+1) + 1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Would my method of solving this symmetric difference set be correct and is my answer correct? Problem: If A = {1, 2, 4, 6, 7} and B = {1, 2, 4, 6} and C = {1, 3, 4, 7}, what is A ⊕ B ⊕ C?
1) A − B = {1, 2, 4, 6, 7} − {1, 2, 4, 6} = {1, 2, 4, 6, 7} − {1, 2, 4, 6} = {7}
2) B − A = {1, 2, 4, 6} − {1, 2, 4, 6, 7} = {1, 2... | Your answer and working appear to be correct. A noteowrthy fact: for a family of sets $A_1,\cdots,A_n$, we can show that
$$\bigoplus_{i=1}^n A_i = \{ \text{all elements belonging to an odd number of the } A_n \}$$
You can use this fact to quickly verify your answer in a case like this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1+\sin^4 x}{\rm d}x$ Someone gives a solution as follows:
\begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1+\sin^4 x}{\rm d}x&=2\int_0^{\frac{\pi}{2}}\frac{\sin^2 x}{1+\sin^4 x}{\rm d}x\\ &=2\int_0^{\frac{\pi}{2}}\frac{\cos^2 x}{1+\cos... | $$I=2\int_{0}^{\pi/2}\frac{\sin^2 x}{1+\sin^4 x} dx=2\int_{0}^{\pi/2} \frac{\tan^2 t \sec^2 x}{(1+\tan^2x)^2+\tan^4 x}dx=2\int_{0}^{\infty} \frac{t^2 dt}{2t^4+2t^2+1}$$
$$\implies I=2\int_{0}^{\infty} \frac{dt}{2t^2+1/t^2+2}=\frac{1}{\sqrt{2}}\int_{0}^{\infty}
\frac{(\sqrt{2}-1/t^2)+(\sqrt{2}+1/t^2)}{2t^2+1/t^2+2}dt$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What method should I use to solve this system of polynomial equations? What method should I use to solve this system of polynomial equations?
$$\begin{cases} 3x^3 - 3y^3 + z^3 - xyz - 3 &= 0\\
3y^3 - x^3 - z^3 - xyz + 5 &= 0\\
x^3 - y^3 + z^3 - xyz - 2 &= 0\end{cases}$$
I've run ... | Hint. By summing the first two equations we obtain
$$xyz=x^3+1.$$
Then solve the linear system with respect to $y^3$ and $z^3$:
$$\begin{cases}
3x^3 - 3y^3 + z^3 - (x^3+1) - 3 = 0\\
x^3-y^3+z^3-(x^3+1)-2=0
\end{cases}$$
and we find
$$y^3=x^3-\frac{1}{2},\quad z^3=x^3+\frac{5}{2}.$$
Going back to the first equation cube... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
if $\omega$ is a primitive cube root of unity then $-\omega$ is a primitive sixth root of unity. I was given a statement that if $\omega$ is a primitive cube root of unity then $-\omega$ is a primitive sixth root of unity.
The roots of $x^n−1$ in $\mathbb C$ which are not also roots of $x^m −1$ for some $1 ≤ m ≤ n$ are... | Yes, that is correct.
Another way, without computing anything is this:
Let $\zeta$ be a primitive 3rd root of unity. Then $\zeta^3=1,\zeta\neq 1$.
Now a 6th root of unity $\omega$ must satisfy $\omega^6=1$. Primitive means that $\omega^2\neq 1$ and $\omega^3\neq 1$. (It has order 6 and not a strict divisor of 6).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the points on the curve $x^4+y^4+3xy=2$ closest and farthest to the origin I need to find the points on the curve $x^4+y^4+3xy=2$ that are closest and farthest to the origin.
I believe this might be a Lagrange multiplier problem, but I am not sure. I was thinking that maybe minimizing/maximizing the function woul... | I saw people recommending to use lagrange multiplier, but I know none of that. Instead, I tried this.
Firstly, you want to maximise/minimise (sqrt) $x^2+y^2$ with $x^4+y^4+3xy-2=0$. Consider $x^2+y^2=t$ and the intersection point $(u, v)$. i.e.
$$u^2+v^2=t, u^4+v^4+3uv-2=0$$
If at the point $(u, v)$, we can get both of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Proving existence of a limit I try to prove that the folowing limit does not exist.
$$\displaystyle\lim_{(x,y)\to(0,0)} \frac{x\sin (ax^2+by^2)}{\sqrt{x^2+y^2}}, a,b>0, a\neq b$$
I make the assumption that this limit exist. So, I try to write the limit $$\displaystyle\lim_{(x,y)\to(0,0)} \frac{x}{\sqrt{x^2+y^2}}$$ as a... | As Peter Foreman's question comment indicates, the limit does exist and is equal to $0$. One way to see this is with
$$\begin{equation}\begin{aligned}
\lim_{(x,y)\to(0,0)} \frac{x\sin (ax^2+by^2)}{\sqrt{x^2+y^2}} & = \lim_{(x,y)\to(0,0)} \frac{\sin(ax^2+by^2)(ax^2+by^2)(x)}{(ax^2+by^2)\sqrt{x^2+y^2}} \\
& = \lim_{(x,y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
For $n\ge 3$ determine all real solutions of the system of $n$ equations.
Question: For $n\ge 3$ determine all real solutions of the system of $n$ equations: $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n=\frac{1}{x_i}\\ \cdots \\ x_2+\cdots+x_{n-1}+x_n=\frac{1}{x_1}.$$
... | I get the same result.
Here's my work.
I abhore "..."s,
so I'll write the equations like this:
For $i=1$ to $n$,
$\sum_{k=1, k\ne i}^n x_k
=\dfrac1{x_i}
$.
Filling in the missing term,
$\sum_{k=1}^n x_k
=\dfrac1{x_i}+x_i
$.
Letting
$S = \sum_{k=1}^n x_k$,
we have
$S
= \dfrac1{x_i}+x_i
$
so
$x_i^2-Sx_i+1 = 0$
or
$x_i
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find stability function of Runge Kutta Method starting from a 3x4 Butcher Tableau Given the scalar IVP $\dot{y}(t) = f(t, y(t))$ , $y(0)=y_0$,
and the butcher Tableau
$$
\begin{array}
{c|cccc}
0 & 0 & 0 & 0\\
\frac{1}{2} & \frac{1}{2} & 0 & 0\\
1& -1& 2& 0\\
\hline
& \frac{1}{6} &\frac{2}{3} &\frac{1}{6}
\end{array}
... | $$k_1=f(t_j, y_j) = \lambda y_j$$
$$k_2= f(t_j + \frac{1}{2}, y_j + \frac{1}{2}hk_1) = \lambda(y_j + \frac{1}{2}hk_1)$$
$$k_3= f(t_j + 1, y_j - hk_1 + 2hk_2) = \lambda (y_j - hk_1 + 2hk_2)$$
$$y_{j+1}=y_j +\frac{h}{6}(k_1 + 4k_2 + k_3)$$
Thus
$$y_{j+1}=y_j +\frac{h}{6}(\lambda y_j + 4\lambda(y_j + \frac{1}{2}h(\lambda... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Different solutions with different results for an inequality Find m such that the following inequality:
$$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$
is always true for $\forall x \in R$.
1st solution:
1st case
$$4x-2m-\frac{1}{2} > -x^2 + 2x +\frac{1}{2} -m$$
$$<=>x^2+2x-m-1>0$$
$$\Leftrightarrow 1^2... | Your two solutions are mathematically correct, but you when you reach $m<-2$, you have calculated for which values of $m$, we have $m+2>0$ and you have not shown that $\forall m>-2, (x+1)^2>m+2, \forall x \in R$. This explain the error in the first method. For the second method, you have shown that the inequality:
$$(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to use prove this $p^4\equiv p\pmod {13}$
Let a prime number $p$, and $n$ a positive integer such $$p\mid n^4+n^3+2n^2-4n+3.$$
Show that $$p^4\equiv p\pmod {13}.$$
A friend of mine suggested that I might be able to use the results problem.
| We will have modulo $13$ all the time. Notice that $$N:= n^4+n^3+2n^2-4n+3 \equiv (n^2-6n-4)^2 \equiv (n-3)^4$$
Clearly $13\mid N\iff 13\mid n-3$, so we assume $p\ne 13$ and thus $13$ does not divide $n-3$. Say prime divisor $p$ of $N$ is good if $p^4{\equiv}p$.
Suppose the statment is not true, so there exists a prim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
An A level question about partial differentiation
The equation of a curve is $2x^4+xy^3+y^4=10$. Show that $$\frac{dy}{dx}=-\frac{8x^3+y^3}{3xy^2+4y^3}.$$
I understand that you are to work out:
\begin{align}
\frac{dz}{dx} = 8x^3 + y^3\\
\frac{dz}{dy} = 3xy^2 + 4y^3
\end{align}
and therefore, $\dfrac{dy}{dx}$.
My ans... | $$2x^4+xy^3+y^4=10$$
D.w.r.t.$x$
we get $$ 8x^
3+y^3+3xy^2y'+4y^3 y'=0 \implies \frac{dy}{dx}=y'= -\frac{8x^3+y^3}{3xy^2+4y^3}$$
next you have actually found partial derivatives $\frac{\partial z }{\partial x}$ and
$\frac{\partial z }{\partial y}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How does $\frac {1} {a^n}$ compare to $\frac {1} {b^n}$ when $a>b$ and $n>0$? To put it briefly, my question is : suppose $a>b$ and $n>0$, how does $\frac {1} {a^n}$ compare to $\frac {1} {b^n}$ ?
I have considered various cases without arriving at finding a general rule.
In view of deriving the order relation betw... | If $n>0$ is an integer, this is $n\in\mathbb{Z}^+$, and $a>b$ are any two real numbers, then we have even nine different cases:
*
*$0<b<a$;
*$b<0<a$, $|b|<|a|$ and $n$ is even;
*$b<0<a$, $|b|<|a|$ and $n$ is odd;
*$b<0<a$, $|b|=|a|$ and $n$ is even;
*$b<0<a$, $|b|=|a|$ and $n$ is odd;
*$b<0<a$, $|b|>|a|$ and $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trouble with $4\times4$ matrix determinant $$
\begin{vmatrix}
1 & -6 & 7 & 5 \\
0 & 0 & 3 & 0 \\
3 & -2 & -8 & 6\\
2 & 0 & 5 & 4\\
\end{vmatrix}
$$
Clearly I want to expand along the second row yielding:
$((-1)^5)3$ times the following matrix
$$
\begin{vmatrix}
1 & -6 & 5 \\
3 & -2 ... | Your computations are almost correct. But it turns out that $-36+10=-26$, not $-16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Question about the basis of an intersection of subspaces
Let $ U=\left\langle\left(\begin{array}{c}-5 \\ 2 \\ 0\end{array}\right),\left(\begin{array}{c}-2 \\ 2 \\ -1\end{array}\right)\right\rangle $ and $ W=\left\langle\left(\begin{array}{c}-3 \\ 0 \\ 1\end{array}\right),\left(\begin{array}{c}-1 \\ -2 \\ -2\end{array}... | Hint :
You write the equation :
$$\lambda_{1}\begin{pmatrix}-5 \\ 2 \\ 0\end{pmatrix}+\lambda_{2}\begin{pmatrix}-2 \\ 2 \\ -1\end{pmatrix}
= \gamma_{1}\begin{pmatrix}-3 \\ 0 \\ 1\end{pmatrix} +\gamma_{2}\begin{pmatrix}-1 \\ -2 \\ -2\end{pmatrix}$$
And after you get a very simple linear system
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Efficient computation of $\sum_{i=1}^{i=\left \lfloor {\sqrt{N}} \right \rfloor}\left \lfloor \frac{N}{i^{2}} \right \rfloor$ I have tried to find a closed form but did not succeed but is there an efficient way to calculate the following expression
$\sum_{i=1}^{i=\left \lfloor {\sqrt{N}} \right \rfloor}\left \lfloor \f... | One approach to improving the efficiency of calculating this would be to take
$$\sum_{i=1}^{\infty}\left\lfloor\frac{N}{i^2}\right\rfloor$$ and ask how many times is the summand a $1$? How many times is it a $2$? And so on. Keep reading to the end, and this reduces the computation from $O(n^{1/2})$ time to $O(n^{1/3})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum of squared Fresnel sine integral I'm trying to find the following sum:
$$ \sum_{n=0}^{\infty} \frac{S\left(\sqrt{2n}\right)^2}{n^3}$$
where $S(n)$ is the fresnel sine integral, however, I think I made a mistake somewhere.
To start, I considered using parseval's identity:
$$ 2\pi\sum_{n=-\infty}^{\infty} |c_n|^2 = \... | With help from reddit user GamblingTheory the solution is as follows:
let
$$ f(x) = -\sqrt{2\pi} * \left( \sqrt{\frac{ix}{2}} + \sqrt{-\frac{ix}{2}} - \sqrt{\pi}\right)$$
thus,
$$ c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) * e^{-inx} \space dx = \frac{S({\sqrt{2n}})}{n^{3/2}}$$
Using Parseval's Identity:
$$\sum_{-\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to make sense of a weird Summation? I have been toying with some sums, as I saw that $123456789\div987654312 = \frac{1}{8}$. Which led me to wonder if this can be generalized to other bases. I.E. in base 4 it would be $(1*4^{2}+2*4^{1}+3*4^{0})\div(3*4^{2}+1*4^{1}+2*4^{0})=\frac{1}{2}$. So this led me to the genera... | (This has hung around unanswered for more than long enough.)
I’ll rewrite the equation like this
$$\frac{\sum\limits_{i=0}^{b-1}(b-1-i)b^i}{b+2+\sum\limits_{i=2}^{b-2}(i+1)b^i}=\frac1{b-2}$$
and (as MPW suggested in the comments) get rid of the fractions to produce the following equation, which is equivalent for $b\ne ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3623567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Finding det(A) with standard basis vectors For example, let $e_1 = [1 \quad 0]$ and $e_2 = [0 \quad 1] $be standard basis vectors. A is a $2 \times 2$ matrix. $Ae_1 = [-3\quad 7]$ and $Ae2 = [3 \quad 5] $
How do I find the $\det(A)$?
| Let $a,b,c,d \in \mathbb{R}$, then
\begin{align}
&A = \begin{bmatrix}
a&b \\ c&d
\end{bmatrix}\begin{bmatrix}
1 \\ 0
\end{bmatrix} = \begin{bmatrix}
-3 \\ 7
\end{bmatrix} \implies a = -3, \quad \text{and} \quad c = 7 \\
&A = \begin{bmatrix}
a&b \\ c&d
\end{bmatrix}\begin{bmatrix}
0 \\ 1
\end{bmatrix} = \begin{bmatrix}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Explicit formula for orthonormalized vectors (using Gram-Schmidt) For $n\in\mathbb{N}$, let
$$
B(n):=\{b_1,b_2,\ldots,b_n\}:=\left\{\begin{pmatrix}1\\2\\3\\4\\\vdots\\n\end{pmatrix},\begin{pmatrix}2\\-1\\0\\0\\\vdots\\0\end{pmatrix},\begin{pmatrix}3\\0\\-1\\0\\\vdots\\0\end{pmatrix},\begin{pmatrix}4\\0\\0\\-1\\\vdots\\... | It seems that
$$v_k=(1,2,3,\ldots,k-1,-x_k,0,\ldots,0)$$
works for constants $x_k$ that are easy to work out by dotting with $b_1$. Then normalize each $v_k$. If $v_k$ is normal to $v_1$ then it is normal to $v_h$ for all $h\gt k$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\lim\limits_{n\to\infty}\frac1{n^4}·\left[1·\sum\limits_{k=1}^nk+2·\sum\limits_{k=1}^{n-1}k+3·\sum\limits_{k=1}^{n-2}k+\cdots+n·1\right]$
Find the value of$$L=\lim_{n\to\infty}\frac1{n^4}\cdot\left[1\cdot\sum_{k=1}^nk+2\cdot\sum_{k=1}^{n-1}k+3\cdot\sum_{k=1}^{n-2}k+\cdots+n\cdot1\right].$$
My attempt is a... | Let $$S_n=1\cdot\sum_{k=1}^nk+2\cdot\sum_{k=1}^{n-1}k+3\cdot\sum_{k=1}^{n-2}k+\cdots+n\cdot1.$$
$$S_1=1=\frac{1\cdot2\cdot3\cdot4}{24}.$$
Now, by the assumption of the induction we obtain:
$$S_{n+1}=S_{n}+(n+1)+2\cdot n+3\cdot(n-1)+...+(n-1)\cdot3+n\cdot2+(n+1)\cdot1=$$
$$=\frac{n(n+1)(n+2)(n+3)}{24}+\sum_{k=1}^{n+1}(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626127",
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"source": "stackexchange",
"question_score": "3",
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Shortest distance from circle to a line
Let $C$ be a circle with center $(2, 1)$ and radius $2$. Find the shortest distance from the line $3y=4x+20$.
This should be very simple, but I seem to end up with no real solutions.
The shortest distance would be from the center of the circle perpendicular to the line right?
S... | The perpendicular line to the given line has slope $=\frac{-1}{4/3}=-\frac{3}{4}$
Also the perpendicular should pass through the centre of the circle so that the shortest distance between the given line and the given circle is the distance between the given line and the point of intersection of the perpendicular with t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Find linear mapping in basis Let A:
\begin{bmatrix}0&0&0\\0&0&1\\1&2&3\end{bmatrix}
be a matrix associated with the linear mapping T in the basis B={(1,1,1), (0,2,2), (0,0,3)}.
Find the standard matrix for T.
If I'm not misunderstanding:
We want to find the transformation T in the standard basis. I.e. what the transfor... | Here's how to start thinking about it. I'll use $\mathcal B$ to denote the coordinate representation in the basis $\mathcal B = \left\{(1,1,1),(0,2,2),(0,0,3) \right\}$ and $\mathcal S$ to denote the coordinate representation in the standard basis. The transformation matrix for the basis $\mathcal B$ is given by
$$A =... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a line integral of a vector field I'm working on the following:
For $k \in \mathbb{N}$ let the oriented piecewise $C^1$ curve $\gamma_k$ be parametrised by
$$\gamma_k(\theta):=\left ((k+1)\cos\left(\frac{\theta}{k+1}\right)−\cos(\theta),(k+1)\sin\left (\frac{\theta}{k+1}\right)−\sin(\theta) \right )$$
with $ \t... | This is a hefty computation and it needs to be done step-by-step. First notice that in evaluating $g(\gamma(\theta))$ you need to compute $x^2 + y^2$:
\begin{eqnarray*}
x^2 + y^2 & = & (k+1)^2\cos^2\left (\frac{\theta}{k+1}\right ) - 2(k+1)\cos\theta\cos\left (\frac{\theta}{k+1}\right ) + \cos^2\theta \\
&& + (k+1)^2\... | {
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"source": "stackexchange",
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Showing that $\sum_{n=1}^\infty n^{-1}\left(1+\frac{1}{2}+...\frac{1}{n}\right)^{-1}$ is divergent
How to show that $\sum_{n=1}^\infty \frac{1}{n\left(1+\frac{1}{2}+...\frac{1}{n}\right)}$ diverges?
I used Ratio test for this problem and this is the result: $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1-\frac{... | First, We need a lemma.
Lemma:
For $n>1$,
$$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n-1} < n$$
This could be checked easily with induction.
To prove the question we use Cauchy Condensation Test, and see that its equal to show $$\sum_{k=1}^{\infty}1+\frac{1}{2}+\frac{1}{3}+ ... +\frac{1}{2^k}$$ is divergent.
It can b... | {
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"source": "stackexchange",
"question_score": "2",
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How to integrate with a function in the limit $\int ^a _0 (\int_0 ^ \sqrt {a^2 - x^2} (a^2 - x^2 - y^2)dy)dx = 303$
I'm trying to solve the above question where the integral has a function as the limit.
Seeing the similarity between the limit and the variable integration, I substituted the $\sqrt {a^2 - x^2}$ with $u$... | As a general rule, whenever possible, try to draw a picture of the region over which you are integrating. In this case, the limits of integration describe a quarter circle in the first quadrant of radius $a$, centered at the origin. This suggests a change of variable to polar coordinates, which turns the integral into
... | {
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"source": "stackexchange",
"question_score": "1",
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Two nested radicals with similar result Sorry for my last post it's my bad .So I ask to this (the true ^^)nested radical :
$$\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{\cdots }}}}}}}}=\frac{1+\sqrt{5}+\sqrt{30-6\sqrt{5}}}{4}$$
The period is $4$ and the related equation is :
$$\sqrt{2+\sqrt{2-\sqrt{2... | We can solve the equation $\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+x}}}}=x$ as follows.
Since $2+x\ge 0$, we have $x\ge -2$. Since $\sqrt{2+x}\le 2$, $x\le 2$. So we can put $x=2\cos 32\alpha$ for some $0\le\alpha\le \tfrac{\pi}{32}$. Then
$$\sqrt{2+x}=\sqrt{2+2\cos 32\alpha}=2\cos 16\alpha$$
$$\sqrt{2-2\cos 16\alpha}=2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$.
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.
Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x... | First (observation):
Note that we can determine $f(0), f(2)$ easily:
$$
x=0 \qquad \rightarrow \qquad f(0)+2f(2)=0;\\
x=1 \qquad \rightarrow \qquad f(2)+2f(0)=-6;
$$
so
$$
f(0)=-4,\quad f(2)=2.
$$
Same way we can determine $f(6), f(20)$ (substituting $x=-3, x=4$).
Same way we can determine $f(56), f(30)$ (substituting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644316",
"timestamp": "2023-03-29T00:00:00",
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Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$ Any idea how ot approach
$$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$
I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted here.
here is how I came across it;
using the identity
$... | $$I=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y^2}{4}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy+\frac{7}{32} \zeta (3) \log (2)+\frac{ \text{Li}_4\left(\frac{1}{2}\right)}{8}-\frac{157 \pi ^4}{46080}-\frac{ 11\log ^4(2)}{48}+\frac{19}{384} \pi ^2 \log ^2(2)$$
Put $$y=\frac{1-x}{1+x}$$
$$\int_0^1\frac{y}{1+y^2}\ln\le... | {
"language": "en",
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Find $a,b \in \mathbb{N}$ with $\mathrm{lcm}(a,b)=12\gcd(a,b)\,$ and $\,a\bmod b = 5$
Find $a,b \in\mathbb{N}$ such that:
Remainder of $a$ divided by $b$ equals $5$
$\mathrm{lcm}(a,b)=12\gcd(a,b)$
I don't know how or where to start. The most similar problem I found was Find $a, b \in \mathbb{N}$ that verify $a + b =... | Remember $lcm(a,b) = \frac {ab}{\gcd(a,b)}$.
So $\frac {ab}{\gcd(a,b)}= 12\gcd(a,b)$ and
$\frac {a}{\gcd(a,b)}\frac {b}{\gcd(a,b)}= 12$.
Let $a' = \frac a{\gcd(a,b)}; b' =\frac b{\gcd(a,b)} = 12$ so
$(a',b') = \{(1,12), (2,6), (3,4),(4,3), (6,2), (12,1)\}$.
But $\gcd(a',b') = 1$ so $(a',b') \ne (2,6)$ or $(6,2)$.
Now ... | {
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How can I approach this inequality? Let $a, b$ and $c$ be three non-zero positive numbers. Show that:
$$\sqrt{\frac{2a}{a + b}} + \sqrt{\frac{2b}{b + c}} + \sqrt{\frac{2c}{a + c}} \leq 3$$
I know the triangular inequality would help here, but I don't know how to approach it.
I started by $a+b≥a$ then that gives $\frac{... | It's not a clean solution but it works. Consider the change of variable $0<x:=\frac{b}{a}, 0<y:=\frac{c}{b}, 0 < x :=\frac{a}{c} $. Observe that $xyz=1$. Now the inequality reads
$$f(x,y,z)=\sqrt{\frac{2}{1+x}}+\sqrt{\frac{2}{1+y}}+\sqrt{\frac{2}{1+z}}\leq 3$$
That means we have to find de maxima of $f$ subject to res... | {
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Prove that $ a^2+b^2+c^2 \le a^3 +b^3 +c^3 $ If $ a,b,c $ are three positive real numbers and $ abc=1 $ then prove that $a^2+b^2+c^2 \le a^3 +b^3 +c^3 $
I got $a^2+b^2+c^2\ge 3$ which can be proved $ a^2 +b^2+c^2\ge a+b+c $. From here how can I proceed to the results? Please help me to proceed. Thanks in advance.
| This is the same as proving that
$$(a^2+b^2+c^2)(abc)^{1/3}\le a^3+b^3+c^3$$
for all positive $a$, $b$, $c$. This is AM/GM. We get
$$a^{7/3}b^{1/3}c^{1/3}\le\frac{7a^3+b^3+c^3}9$$
applying AM/GM to seven copies of $a^{3}$ and one each of $b^{3}$ and $c^{3}$. Cyclicly
permute and add.
| {
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"source": "stackexchange",
"question_score": "6",
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Solving a first order differential equation by finding an integrating factor : Edit Problem:
Solve the following differential equations by first finding an integrating factor.
$$ (y^2(x+1) + y ) \, dx + ( 2xy + 1 ) \, dy = 0 $$
Answer:
\begin{align*}
M_y &= 2(x+1)y + 1 = 2xy + 2y + 1 \\
N_x &= 2y \\
\frac{1}{N(x,y)} \l... | Everything looks good up to the last formula, you missed there one factor in the product rule for $N_x$:
$$
N_x=(2xy+1)_xe^x+(2xy+1)(e^x)_x=(2y)e^x+(2xy+1)e^x,
$$
restoring the needed equality $M_y=N_x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to improve $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$ I have proved this inequality $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$.
Using $\left|\sin(nx)\right|\leq n\left|\sin(x)\right|$ on $[0,\frac{\pi}{2n}]$ and $\... | Alternative solution:
When $n = 2, 3, 4$, the inequality is verified directly.
In the following, assume that $n\ge 5$.
Let
$$I_n = \int_0^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x.$$
We have
\begin{align}
I_n &= \underbrace{\int_0^{\pi/n} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4... | {
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"url": "https://math.stackexchange.com/questions/3664974",
"timestamp": "2023-03-29T00:00:00",
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how is this solved? $\lim_{x→0} \frac{1}{x^2}\left(\left(\tan(x+\frac{\pi}{4})\right)^{\frac{1}{x}}−e^2\right)$ I know how to solve the trigonometric part of the limit, i.e $\left(\tan(x+\frac{\pi}{4})\right)^{1/x}$, which I think is $e^2$, I do not however, know how to carry forward the question with that.
Also, I am ... | $$\frac{\left(\tan\left(x+\frac{\pi}{4}\right)\right)^{\frac{1}{x}}-e^2}{x^2}=\frac{\left(\frac{1+\tan{x}}{1-\tan{x}}\right)^{\frac{1}{x}}-e^2}{x^2}=$$
$$=\frac{\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)^{\frac{1}{x}}-e^2}{x^2}=\frac{\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)^{\frac{1-\tan{x}}{2\tan{x}}\cdot\frac{2\tan{x... | {
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Prove that $x^2+y^2-\sqrt{2}xy < 4$ provided $1\le x,y \le \frac{5}{2}$. If possible I would like an elegant solution to the following problem:
Let $x,y\in\mathbb{R}$ such that $1\le x \le \frac{5}{2}$ and $1\le y \le \frac{5}{2}$. Prove that
$x^2+y^2-\sqrt{2}xy < 4$
I'm aware that you can use Lagrange multipliers bu... | Hint. Make the substitution $$x=t+u,\\y=\sqrt 2 u.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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diophantine question parity and sign If I have an equation say $$3(1+x+x^2)(1+y+y^2)=4x^2y^2-1 \quad (1)$$ and I rewrite it as
$$4+3(x+y+xy+x^2+y^2+xy^2+x^2y)-x^2y^2=0$$ and I find a positive integer solution where both x and y are even, can I show that no odd positive integer solutions exist?
| Since the equation is symmetric in $x$ and $y$, we only need to consider the case $x\le y$.
Dividing the both sides of
$$3(1+x+x^2)(1+y+y^2)=4x^2y^2-1\tag1$$
by $x^2y^2$ gives
$$3\bigg(1+\frac 1x+\frac{1}{x^2}\bigg)\bigg(1+\frac 1y+\frac{1}{y^2}\bigg)=4-\frac{1}{x^2y^2}\tag2$$
Suppose here that $x\ge 8$.
Then, we have
... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim_{n\to\infty} \left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}$
Find the following limit without using the L'Hopital rule:
$$\lim_{n\to\infty} \left(\dfrac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}$$
Answer: $e^{-1}$
My attempt: Since the limit is of the form $1^{\infty}$, I decided to use the... | $$a_n=\left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}\implies\log(a_n)=\left({2\sqrt{n^2+n}-1}\right)\log\left(\frac{\sqrt{n^2+n}-1}{n}\right)$$ Now, using Taylor expansions
$$\log\left(\frac{\sqrt{n^2+n}-1}{n}\right)=-\frac{1}{2 n}-\frac{1}{4 n^2}-\frac{1}{24
n^3}+O\left(\frac{1}{n^4}\right)$$
$${2\sqrt{n^2... | {
"language": "en",
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If $(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) = a + bi$, prove that $a^2 + b^2 = (4\cos x \cos\frac x2)^2$ Could anyone help me with this? I'm stuck.
If $$(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) = a + bi$$ prove that $$a^2 + b^2 = \left(4\cos x \cos\frac x2\right)^2$$
For reference, $\operator... | Hint:
$$1+\text{cis}2y=1+\cos2y+i\sin2y=2\cos y(\cos y+i\sin y)$$
$$|1+\text{cis}2y|=|2\cos y|\cdot|\cos y+i\sin y|=2|\cos y|$$
$$|1+\text{cis}2y|^2=4|\cos y|^2=4\cos^2y$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cevianas AD, BE and CF of ABC compete at P. Show $\frac{S_{DEF}}{2S_{ABC}}=\frac{PD .PE.PF}{PA.PB.PC}$ In an ABC triangle, the cevianas AD, BE and CF compete in P. Show that
$\frac{S_{DEF}}{2S_{ABC}}=\frac{PD .PE.PF}{PA.PB.PC}$
Using areas relation, I found
$\frac{3S_{ABC}-\overbrace{(S_{PAB}+S_{PAC}+S_{PBC})}^{S_{ABC}... |
Let $a=\frac{a_1}{a_2}$, $b=\frac{b_1}{b_2}$, $c=\frac{c_1}{c_2}$. Per the Ceva's theorem, $abc = 1$. Evaluate
\begin{align}
\frac{AP}{PD} &= \frac{S_{ABE}}{S_{DBE}}
= \frac{\frac{b_2}{b_1+b_2}S_{ABC} }{ \frac{a_1}{a_1+a_2}\frac{b_1}{b_1+b_2}S_{ABC} }
= \frac1b(1 + \frac1a )=c(1+a)\\
\end{align}
Likewise, $\frac{... | {
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Evaluation of $\int \:\frac{1}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}dx$ using trig substitution Recently I came accross this integral :
$$ \int \:\frac{1}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}dx $$
I would evaluate it like this, first start by the substitution:
$$ x=\cos(2u) $$
$$ dx=-2\sin(2u)du $$... | Another way (more complex than @Quanto's one)
$$I=\int\dfrac{dx}{\sqrt[3]{\left(1-x\right)^4\left(x+1\right)^2}}=\int\dfrac{dx}{\left(x-1\right)^\frac{4}{3}\left(x+1\right)^\frac{2}{3}}$$
$$u=\dfrac{1}{\sqrt[3]{x-1}}\implies I=-3\int\dfrac{du}{\left(\frac{1}{u^3}+2\right)^\frac{2}{3}}$$
$$v=\sqrt[3]{2u^3+1}\implies I=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I find the conditional entropy? I was given the joint pmf of 3 variables :
$$P_{X,Y,Z}(x,y,z) =
\begin{cases}
{\frac{3}{16}} &\quad\text{if }(x,y,z)\in \{001,111\}\\
{\frac{1}{8}} &\quad\text{if }(x,y,z)\in \{000,010,100,110\}\\
{\frac{1}{16}} &\quad\text{if } (x,y,z)\in \{011,101\}
... | Note that simply $P_{X,Y|Z}(x,y|z=0)=\frac{P_{X,Y,Z}(x,y,z=0)}{p(z=0)}$, which is equal to $\frac{1}{4}$, in the given distribution. Hence
$$H(X,Y|Z=0)=2.$$
| {
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"answer_id": 0
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If $x$ and $y$ are positive integers, and $p$ is a prime, find all triples ($x$, $y$, $p$) such that $x^5 + x^4 + 1 = p^y$ If $x$ and $y$ are positive integers, and $p$ is a prime, find all triples ($x$, $y$, $p$) such that $x^5 + x^4 + 1 = p^y$. (Titu Andreescu)
My attempt:
I factorised the LHS to get $(x^3 - x +1)(x^... | Note that $x^5 + x^4 + 1 = (x^3 - x + 1)(x^2 + x + 1)$, and
$$x^3 - x + 1 \geq x^2 + x + 1 \iff x^3 - x^2 - 2x \geq 0 \iff x(x^2-x-2) \geq 0 \iff x \geq 2.$$
Hence, for $x \geq 2$, we know if $x^3 - x + 1 = p^m$ and $x^2 + x + 1 = p^n$, then $x^2 + x + 1 \mid x^3 - x + 1$. We have
$$\begin{aligned}
x^2 + x + 1 &\mid ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show inequality $a\sqrt{b - 1} + b\sqrt{a - 1} \le ab$ Given numbers $a$ and $b$; $a, b \ge 1.$
I'm trying to prove
$$a\sqrt{b-1} +b\sqrt{a - 1} \le ab.$$
Also conditions for turning it to equality.
I tried to use AM-GM to the $(a - 1)(b - 1)$, which less than ab, but got nothing.
Applying AM-GM to $a\sqrt{b-1}$ and $... | Set $x=a-1,y=b-1.$
$$\begin{aligned}LHS&=(x+1)\sqrt y +(y+1)\sqrt x\\\\ &=\frac{(x+1)(y+1)}{2}\cdot \left[\frac{2}{\sqrt y + \frac{1}{\sqrt y}}+\frac{2}{\sqrt x + \frac{1}{\sqrt x}}\right]\\\\&\leq(x+1)(y+1)=RHS,\end{aligned}$$
because $t+{1\over t} \geq 2\;$ for all $t>0.$
Equality occurs when $t=1$ or, in terms of $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate limit of a sequence $ \lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}} $ I need a help with evaluating a limit of a sequence
$$ \lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}}. $$
The problem is th... | Divide by $3^{2n}$, numerator becomes $\frac{2}{3} - (-1)^n (\frac{2}{3})^n$, so the second term converges to 0. In he denominator you also get term that converges to $0$ - another term that converges to $0$, so in total the expression has limit of the type $\frac{1-0}{0-0}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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For angles $A$ and $B$ in a triangle, is $\cos\frac B2-\cos \frac A2=\cos B-\cos A$ enough to conclude that $A=B$? Brief enquiry:
$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$
Optionally
$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$
Is above equality sufficient to prove that it implies $A=B$?
Deta... | $$\cos\dfrac A2-\cos\dfrac B2=2\left(\cos^2\dfrac A2-\cos^2\dfrac B2\right)$$
What if $\cos\dfrac A2-\cos\dfrac B2=0?$
else
$$\dfrac12=\cos\dfrac A2+\cos\dfrac B2$$
which is clearly possible as $0<\cos\dfrac A2,\cos\dfrac B2<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Galois group of $x^6-2x^4+2x^2-2$ over $\mathbb{Q}$
Find Galois group of $x^6-2x^4+2x^2-2$ over $\mathbb{Q}$ and describe an extension corresponding to any of it's proper subgroups of maximal order.
I know that the roots are $$\sqrt{\frac{1}{3}\left(2 - \frac{2}{\sqrt[3]{17 + 3\sqrt{33}}} + \sqrt[3]{17 + 3 \sqrt{33}}... | Hint: Write $x^6-2x^4+2x^2-2=g(x^2)$ and start by considering the spitting field of $g$ and its Galois group.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Inequality involving AM-GM but its wierd Let a, b, c be positive real numbers. Prove that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \frac{3\cdot\sqrt[3]{abc}}{a+b+c} \geq 4$
Ohk now i know using AM-GM that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \cdot \sqrt[3] {\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 3 \c... | Your steps are right, but they are don't give a proof.
You proved that $$A=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq3$$ and $$B=\frac{3\sqrt[3]{abc}}{a+b+c}\leq1.$$
We need prove that $$A+B\geq4,$$ which is impossible by your work.
For example, for $A=3.1$ and $B=0.8$ we obtain $$A+B\geq4$$ is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limitations of Arithmetic-Geometric Inequality? I have to find the range of function $f(x) = x+(1/x) +1$, where $x$ is positive. Now I did it with two ways which we can see below, in equations $(1)$ and $(2)$, by using the AM-GM inequality.
$$\frac 1 3 \left( x + \frac 1 x + 1 \right) \ge \sqrt[3]{x \cdot \frac 1 x \c... | It is clear that $f(x)$ is unlimited from above. For a lover bound, the following is useful:
For $x>0\;$ is $x+\frac{1}{x}\geq 2,$ with equality at $x=1.$
Therefore, the range of the function $f(x)=x+\frac{1}{x}+1\;$ is $\;[3, \infty).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Tried to apply the ratio test to determine the convergence interval, but get the limit as a constant. So here is the series: $\Sigma_{n=1}^{\infty} \frac{x^{2n}}{1+x^{4n}}$
$$\left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x(1+x^{4n})}{1+x^{4n+4}} \right| \overset{\text{ n } \rightarrow \... | Without ratio test...
For $\vert x \vert \lt 1$ you have
$$0 \le \left\vert \frac{x^{2n}}{1+x^{4n}} \right\vert \lt \vert x \vert^{2n}$$ hence the series converges as $\sum x^{2n}$ converges.
For $\vert x \vert \gt 1$ you have
$$0 \le \left\vert \frac{x^{2n}}{1+x^{4n}} \right\vert \lt 1/\vert x \vert^{2n}$$ hence the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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1st order linear differential equation $y'+\frac{xy}{1+x^2} =x$ Can anyone help me with this one task. I need to resolve 1st order linear equation of this equation.
$$y'+\frac{xy}{1+x^2} = x.$$
I stopped when this result came out
$$e^{\ln|y|}=e^{-\frac{1}{2}\ln|1+x^2|}\cdot e^C.$$
I try solve this by wolfram
$$y=\frac... | We multiply the ODE by the integrating factor $e^{\int{\frac{x}{x^2+1}}dx}=e^{\frac{ln(1+x^2)}{2}}=\sqrt{1+x^2}$ to obtain $\frac{d}{dx}(y\sqrt{1+x^2})=x\sqrt{1+x^2}$. Then integrating gives $y\sqrt{1+x^2}=\int x\sqrt{1+x^2}dx= \frac{(1+x^2)^{\frac{3}{2}}}{3}+C$ and finally we have $y(x)=\frac{C}{\sqrt{1+x^2}}+\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A tangent to ellipse $E_1$ intersects ellipse $E_2$ at $P$ and $Q$. Prove that the tangents to $E_2$ at $P$ and $Q$ are perpendicular to each other. A tangent to the ellipse $x^2 + 4y^2 = 4$ meets the ellipse $x^2 + 2y^2 = 6$ at P and Q. Prove that tangents at $P$ and $Q$ of ellipse $x^2 + 2y^2 = 6$ are perpendicular t... | Let $A=(2\cos \theta, \sin \theta)$ be the point of tangency on $E_1$. The slope of this tangent line will be
$$t_A=-\frac{\cot \theta}{2}.$$
The equation of the tangent line $\ell$ at $A$ on ellipse $E_1$ is given by:
$$\ell: x\cos \theta+2y \sin \theta=2. \tag{1}$$
Let $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ be the points o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Parametrization of Parabola After seeing recent curve I wonder if a parametrization of red curve
of equation $ \sqrt{x}+ \sqrt{y}=1 $ can be found for extended domain/range. Parametrization $ ( x= \cos^4 t, y=\sin^4 t \; )$ is bounded $\pm1$ for $(x,y).$
This interesting curve is a parabola, intersection of a cone to... | One approach is to convert the equation to the standard bivariate polynomial form for a conic section. Start with the given equation $\sqrt{x}+\sqrt{y}=1$ and square both sides:
$x+2\sqrt{xy}+y=1$
$2\sqrt{xy}=1-(x+y)$, square again:
$4xy=1-2(x+y)+(x+y)^2$
Using the quarter-square multiplication formula $4xy=(x+y)^2-(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $\pi<\alpha<\frac{3\pi}{2}$ what is the value of $\sqrt{4\sin^4\alpha + \sin^2 2\alpha} + 4\cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$? I solved part of it this way:
$$\sqrt{4\sin^4\alpha + \sin^2 2\alpha} + 4\cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$$
$$= \sqrt{4\sin^4\alpha + 4\sin^2\alpha \cos... | Now, since $\pi<\alpha<\frac{3\pi}{2}$, it's $$-2\sin\alpha+2\sin\alpha+2=2$$ because for $x<0$ we have:
$$\sqrt{x^2}=-x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find solution set of $\dfrac{8^x+27^x}{12^x+18^x}=\dfrac{14}{12}$ What I've done is factoring it.
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
This looks like it can be factored more but it doesn't work from my attempts.
| Rearrange the equation
$$6(8^x+27^x)-7(12^x+18^x)=0$$
and factorize
$$(2^x+3^x)(3\cdot 3^x - 2\cdot 2^x)(2\cdot 3^x - 3\cdot 2^x)=0
$$
which leads to the solutions $x=\pm1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding $\lim_{n\to \infty} \sum_{r=1}^n \frac{6n}{9n^2-r^2}$ This is a question from an entrance exam for a college here in India.
We have to find $$ \lim_{n\to\infty}\sum_{r=1}^n \frac{6n}{9n^2-r^2}$$
So far I have tried:
$$ S(n)= \sum_{r=1}^n \frac{6n}{9n^2-r^2}$$
$$ S(n)= \sum_{r=1}^n \Bigl(\frac{1}{3n+r} + \frac{1... | $S(n) = \underbrace{\sum\limits_{1 \le k \le n} \frac{1}{3n - k }}_{A} + \underbrace{\sum\limits_{1 \le k \le n} \frac{1}{3n + k}}_{B}$
Let's denote $H_{n} = \sum\limits_{1 \le k \le n} \frac{1}{k}$ - harmonic series (we assume, that it's a known series).
Than A = substitution $\{ k = 3n - k \} \rightarrow \sum\limits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find t... | Notice that
\begin{align*}
f(x) = \frac{x+2}{x^{2} + 2x + 1} = \frac{(x+1) + 1}{(x+1)^{2}} = \frac{1}{x+1} + \frac{1}{(x+1)^{2}}
\end{align*}
If we set $y = 1/(x+1)$ and consider $c\in\textbf{R}$, we arrive at the following equation
\begin{align*}
y + y^{2} = c \Longleftrightarrow y^{2} + y - c = 0\Longleftrightarrow y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed form for $\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x$ I am looking for a closed form for:
$$\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x.$$
I am assuming integration by parts multiple times but I can't get anywhere with it. Any help/ hint would be greatly appreciated.... | A conjectural evaluation for the remaining alternating sum is given in equation (19) of https://arxiv.org/abs/1908.04770, namely
$$
\sum_{n=1}^{\infty} \frac{(-1)^n H_{n-1}}{n^5}
= \frac{1}{13} \left( \frac{1}{3}\operatorname{Li}_6\left(-\frac{1}{8}\right) - 162 \operatorname{Li}_6\left(-\frac{1}{2}\right) - 126 \opera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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verifying the result $\int_{B^4}e^{x^2+y^2-z^2-w^2}\,dx\,dy\,dz\,dw=\pi^2(\sinh(1)+1-\frac{1}{e})$ the problem is:
$$I=\int_{B^4} e^{x^2+y^2-z^2-w^2} \,dx\,dy\,dz\,dw $$
where $B^4$ is the unit 4 ball, explicitly: $$ B^4=\{(x,y,z,w)\in\Bbb{R}^4 \mid x^2+y^2+z^2+w^2\leq 1\}$$
here is my solution:
using fubini's theorem ... | could you verify using two sets of polar coordinates e.g.:
$$r_1=\sqrt{x^2+y^2}$$
$$r_2=\sqrt{z^2+w^2}$$
$$x=r_1\cos\theta_1,y=r_1\sin\theta_1,z=r_2\cos\theta_2,w=r_2\sin\theta_2$$
$$dxdydzdx=r_1r_2dr_1dr_2d\theta_1d\theta_2$$
and we know that $r_1^2+r_2^2\le1$
and our integral would become:
$$\iiiint_{\Omega^4}e^{r_1^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the image of an absolute value function I am trying to find the image of the following function:
$$f:\mathbb{R}\rightarrow\mathbb{R}, \\f(x)=|1-x^2|.$$
I know trivially that the image is in ${[0,\infty[}$, but I'm looking for a way to algebraically solve it.
So I have tried taking a $y\in im(f)$
$\Rightarrow$ t... | Please note that there are usually three ways to find the range of a function. I show how they work by finding the range of the given function through these ways.
First Way
We can find the range of a function by starting from the known functions, of which we know the range, and then trying to transform them into the gi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Power series approximation for $\ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ to calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $ Problem
Approximate $f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ and then calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $
My attempt
Let
$$f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)}) ... | Summing $$S=\sum_{n=1}^{\infty} \frac{1}{n(2n+1)}=2\sum_{k=1}^{\infty} \left( \frac{1}{2n}-\frac{1}{2n+1}\right)=2 [1/2-1/3+ 1/4-1/5_....]$$ $$\implies S=2(1-\ln 2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Writing a system in Polar form and finding the equilibrium points
Consider the system
\begin{align}
\dot{x}&= -y+x(1-2x^2-3y^2)\\
\dot{y}&=x+y(1-2x^2-3y^2)
\end{align}
(a)Find the equilibrium points and determine their stability
(b)Rewrite the equations in polar coordinates
My Attempt:
(a) As far as I know, an equili... | We have the system
$$\begin{align}
\dot{x}&= -y+x(1-2x^2-3y^2)\\
\dot{y}&=x+y(1-2x^2-3y^2)
\end{align}$$
We can solve $x' = y' = 0$ and find a single critical point of
$$(x, y) = (0, 0)$$
The Jacobian is
$$J(x, y) = \begin{bmatrix}
\dfrac{\partial x'}{\partial x}& \dfrac{\partial x'}{\partial y}\\
\dfrac{\partial y'}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove Stirling's formula given that for $I_n = \int_0^{\pi/2} \sin^n\theta \, d\theta$ we have $I_{2n+1}/I_{2n} \rightarrow 1$ For $I_n = \int_0^{\pi/2} \sin^n\theta \, d\theta$ it is possible to show (using integration by parts and $\sin^2\theta = 1-\cos^2\theta$) that: $nI_n = (n-1)I_{n-2}$
We can then repeatedly app... | Note that you know a little bit more from the linked question - that $n!/(n^{n+1/2} e^{-n})$ decreases to $C$.
Using this and the upper bound you have, note that $$ \frac{I_{2n+1}}{I_{2n}} \le \frac{2^{4n + 1} C^4 n^{4n + 2} e^{-4n} e^{1/n}}{\pi \cdot C(2n)^{2n+1/2}e^{-2n} \cdot C(2n+1)^{2n + 3/2} e^{-2n - 1}} \\ = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ How can I evaluate this integral $$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x } \mathop{dx}$$
I tried using the half angle formula
$$\sin x=\dfrac{2\tan\dfrac x2}{1+\tan^2\dfrac x2}, \cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$$
substituted and simplif... | Bioche's rules say you should use the substitution
$$t=\tan x,\qquad\mathrm dx=\frac{\mathrm dt}{1+t^2},$$
to obtain the integral of a rational function in $t$, which you can compute via partial fractions decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Parametrization the curve of intersection of sphere $x^2+y^2+z^2=5$ and cylinder $x^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$. I've seen similar posts here but none of the answers helped me. I am trying to parametrize a curve of intersection of a (top half $z>0$) sphere $x^2+y^2+z^2=5$ and cylinder $x^... | Use the fact that the right-circular cylinder of radius $r$ centered at $(h, k)$ in the $xy$-plane can be parametrized by $\mathbf G(t, z) = \langle h + r \cos t, k + r \sin t, z \rangle$ by the Pythagorean Identity, i.e., $\cos^2 t + \sin^2 t = 1.$
Considering that we are in the intersection of the right-circular cyli... | {
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"timestamp": "2023-03-29T00:00:00",
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Geometric representation of range and kernel of a linear transformation Let $T$ be a linear operator on $\Bbb{R}^3$ defined by $TX$ = $AX$ where $X$ is a $3$$\times$$1$ column vector and $A$ is
$$\begin{bmatrix}1 & 3 & 4\\3 & 4 & 7\\-2 & 2 & 0\end{bmatrix}$$
I need to show that the range and kernel of $T$ are a plane a... | First, X is not a "3 x 3 column vector"! "3 x 3" would mean a matrix with three rows and three columns. X is a 3 dimensional column vector.
The most direct way (not necessarily the simplest) to solve this is to actually find the kernel and range.
A vector $v= \begin{bmatrix}a \\ b \\ c \end{bmatrix}$ is in the range ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$ Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$
My attempt:
Let $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}=\sqrt{x}+\sqrt{y}$
Square both sides: $a+b+\sqrt{\left(2ab+b^2\right)}=x+2\sqrt{xy}+y$
Rearrange: $\sqrt{\left(2ab+b^2\right)}-2\sqrt{xy}=x+y-a-b$
That's where my ... | After taking squares, you can proceed as follows
$$(a+b)+\sqrt{\left(2ab+b^2\right)}=(x+y)+2\sqrt{xy}$$
Compare corresponding (conjugate) parts on both the sides of above equation, we get
$$x+y=a+b\tag 1$$
$$2\sqrt{xy}=\sqrt{2ab+b^2}\iff 4xy=2ab+b^2\tag 2$$
$$x-y=\pm\sqrt{(x+y)^2-4xy}=\pm\sqrt{(a+b)^2-(2ab+b^2)}=\pm a\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Partition of $\Bbb R^2$ arising from matrix multiplication. Describe the partition of $\Bbb R^2$ arising from the action, by matrix multiplication, of the subgroup $H=\left\{\begin{pmatrix}
1 & 0\\
a & 1
\end{pmatrix}:a\in\Bbb R\right\}$ of $GL_2(\Bbb R)$.
First of all I don't think I really understand the ques... | Let's first prove that indeed $H\le\operatorname{GL}_2(\mathbb{R})$, and that we really have a $H$-action on $\mathbb{R}^2$.
*
*Subgroup - closure:
$\space\begin{pmatrix}
1 & 0 \\
a & 1 \\
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
b & 1 \\
\end{pmatrix}=
\begin{pmatrix}
1 & 0 \\
a+b & 1 \\
\end{pmatrix}\in H$;
*Sub... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Limit of a fraction involving square and third roots. $$\lim_{n\to\infty} \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$
Is it a good idea to substitute the numerator and denominator using that $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$, since I’ll end up having to multi... | Instead, write $\sqrt{n^2+1}= |n|\sqrt{1+\frac{1}{n^2}}$, and similar for other expression in the numerator. For the denominator, multiply both numerator and denominator. Then in the denominator you will have $n^2$, and in the numerator the term
$$
(n^3+1)^{\frac{2}{3}} + ((n^3+1)(n^3+n^2+1))^{\frac{2}{3}} + (n^3+n^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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If $a=7!$, $b=_{13}P_k$, $\frac{ab}{\operatorname{lcm}(a,b)}=120$, find the $k$
Question:
$a=7!$, $b=_{13}P_k$, $\dfrac{ab}{\operatorname{lcm}(a,b)}=120$, then find the $k$.
My attempts:
$$\frac{ab}{\operatorname{lcm}(a,b)}=\gcd{(a,b)}=120$$
$\gcd{(7!, _{13}P_k)}=120 \Longrightarrow \begin{cases} \dfrac{13!}{(13-k)!×... | This was based on the original post
The problem as stated is incorrect or there is a typo. If $\frac{ab}{\gcd(a,b)}=120$, then that would mean $\text{lcm}(a,b)=120$. But $120=\text{lcm}(a,b) \geq a=7!$. This is not possible.
After the original post got modified:
Perhaps, the $\gcd(a,b)=120$. In which case we can do the... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $|z| < 1$, then $\displaystyle\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots$ Show that if $|z| < 1$, then
\begin{align*}
\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots
\end{align*}
converges
MY ATTEMPT
Le... | Your approach is correct! It can be verified in the following way too.
Consider the partial sums $$S_n=\sum_{j=0}^{n}\frac{2^jz^{2^j}}{1+z^{2^j}}$$
Then $$\sum_{m=0}^{\infty}\frac{2^mz^{2^m}}{1+z^{2^m}}=\lim_{n\rightarrow\infty}S_n$$
Observe (prove by induction on $n$) that $$S_n+\frac{z}{z-1}=\frac{2^{n+1}z^{2^{n+1}}}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\lim\limits_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$ Evaluate $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$$
My attempt: $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right) = \lim_{x \to 1} \frac{x+2}{x^2+x+1}=1$$
According to the answer key, this limit does not exist... | $$\frac1{x-1}-\frac3{1-x^3}=\frac1{x-1}+\frac3{x^3-1}\to\infty$$
as the terms do not cancel each other (they have the same sign).
Assuming a typo,
$$\frac1{x-1}+\frac3{1-x^3}=\frac1{x-1}-\frac3{x^3-1}=\frac{x^2+x-2}{x^3-1}=\frac{x+2}{x^2+x+1}\to1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Polynomial $f(x) = ax^2 – bx + c $ (where $a$, $b$ & $c$ are positive integers) Let the polynomial $f(x) = ax^2 – bx + c $ (where $a$, $b$ & $c$ are positive integers). If $f(p) = f(q) = 0$, where $ 0 < p < q < 1$, then find the minimum possible value of $a$.
The vertex is $-\frac{-b}{2a}=\frac{b}{2a}>0$ and lies betwe... | Let the quadratic be $a(x-u)^2-v$, where $u=(p+q)/2$ and $-v=f(u)\lt0$.
Then $f(0)=au^2-v\ge1$ so $au^2\gt1$ and likewise $a(1-u)^2\gt1$. Either $u\le\frac12$ or $1-u\le\frac12$ so $\frac a4\gt1$, and $a\ge5$.
An example is $f(x)=5x^2-5x+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find dV/dA in terms of r. The volume of a sphere, $V$ cm³, of radius $r$ is given by the formula $V = \frac{4}{3} \pi r^3$. The surface area of a sphere $A$ cm² of radius $r$ cm is given by the formula $A=4\pi r^2$. Find $dV/dA$ in terms of $r$.
Here's my workings to the question:
$$V= \frac{\frac{4}{3}\pi r^{3}}{4\pi ... | Given $V= \frac{4\pi}3 r^3$ and $A=4\pi r^2$, we have
$$V= \frac{4\pi}3\left(\frac{A}{4\pi}\right)^{3/2}= \frac{A^{3/2}}{3\sqrt{4\pi}}
$$
Then,
$$\frac{dV}{dA}= \frac12\frac{\sqrt A}{\sqrt{4\pi}}= \frac r2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{... | Substitute $u=\tan\theta\implies du=\sec^2\theta \ d\theta$
$$\int \frac{u^3}{(u^2+1)^3}du=\int \frac{\tan^3\theta}{(\tan^2\theta+1)^3}\sec^2\theta\ d \theta$$
$$=\int \frac{\tan^3\theta\sec^2\theta}{\sec^6\theta}\ d\theta$$
$$=\int\sin^3\theta\cos\theta d\theta$$
$$=\int\sin^3\theta\ d(\sin\theta)$$
$$=\frac{\sin^4\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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When the function equation $f(x)f(y)=axy+b$ is solvable
Assume $a,b$ are constants. The question is whether there is a continuous function $f$ defined on $\mathbb R$ or $\mathbb C$ so that
$$
f(x)f(y)=axy+b
$$
Of course, such a function $f$ exists if $b=0$ by taking $$f(x)=\sqrt{a}x\,.$$ Likewise if $a=0$ then $f$ e... | If $x$ and $y$ are independent variables, you only have to set $y = x$, to get $f(x)^2 = ax^2 + b$ and therefore $f(x) = \sqrt{ax^2 + b}$. Now,
$$f(x)f(y) = \sqrt{(ax^2 + b)(ay^2 + b)} = \sqrt{a^2x^2y^2 + abx^2 + aby^2 + b^2}$$
Needs to be equal to $axy + b$, i.e. taking the squares of both expressions
$$a^2x^2y^2 + ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$
My attempt :
\begin{align*}
f(x)&=\dfrac{5}{9\cos^2x-6\sin x\cos x+\sin^2x-6\cos^2x}\\
&= \dfrac{5}{(3\cos x+\sin x)^2-6\cos^2x}
\end{align*}
The problem is if I'm going to use
$$-1\leqslant\sin x... | this $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ can be simplified more neatly, if you use:
$\sin^2(\theta)+ \cos^2(\theta)=1$
$\cos(2\theta)=2\cos^2(\theta)-1$
$\sin(2\theta)=2\sin(\theta)\cos(\theta)$
And the simplified denominator would be in the form:
$$ a\cos(2x)+b\sin(2x)+c$$
Then you can use $$-\sqrt{a^2+b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Rotational Volume of $y^2 = \frac{x^3}{2a-x}$ around asymptote $x = 2a$ My approach is to use a shell method over the range $[0, 2a]$. One cylinder will be $C = 2\pi xy \ dx$. Let us only work with the positive quadrant and multiply by two for ease:
https://www.desmos.com/calculator/pvuquz8orz
then the volume is: $$V =... | You're on the right track. As long as $x>0$ then
$$ 4\pi\int_0^{2a} (2a-x)x \sqrt{\frac{x}{2a-x}}\:dx = 4\pi\int_0^{2a} x \sqrt{x(2a-x)}\:dx$$
Then use the substitution $x=2a\sin^2\theta$:
$$= 64\pi a^3 \int_0^{\frac{\pi}{2}} \sin^4\theta\cos^2\theta\:d\theta$$
Denote $I$ as just the integral part without the constants... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the greatest integer less than $3^\sqrt{3}$ without using a calculator and prove the answer is correct. Find the greatest integer less than $3^\sqrt{3}$ without using a calculator and prove the answer is correct.
I'm puzzled on how to solve this problem, any help is appreciated. There was hints about turning the e... | Since $3 = \frac{48}{16} <\frac{49}{16}$, you have $\sqrt{3}<\frac{7}{4}.$ So you might try to take $y=\frac{7}{4}.$ It's easy to calculate $3^7 = 2187$ which is close to $2401 = 7^4.$ So $3^7<7^4$ and you have $3^{7/4}<7.$ So your answer is $6$ or less.
Try $x=5/3$ to get the lower bound.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\... | Since you selected $$x - 2 = 3 \tan \theta$$ as your substitution, it follows that $$\tan \theta = \frac{x-2}{3},$$ and by considering a right triangle with legs $3$ and $x-2$ with hypotenuse $\sqrt{3^2 + (x-2)^2}$ via the Pythagorean theorem, we obtain $$\sin \theta = \frac{x-2}{\sqrt{3^2 + (x-2)^2}}, \\ \cos \theta =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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How to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}}\ dx$? I am trying to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}} dx$
This is apparently a binomial integral of the form $\int x^m (a+bx^k)^ndx$. Therefore, we can use Euler's substitutions in order to evaluate it. Since $\dfrac{m+1}{k} = \dfrac{5+1}{2} = 3 \in \mathbb{Z}$ w... | Take $x^5 \times (1 + x^2)^{\frac{2}{3}} = (x^{15} \times (1+x^2)^2)^{\frac{1}{3}}$ or some other form of that sort and integrate.
Thus, you'll be integrating $(x^{19} + 2x^{17} + x^{15})^{\frac{1}{3}}$
Applying substitution method here may work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Check and comment my proof of $a+b \geq 2 \sqrt{ab}$ I want to prove
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 1
$$
First question:
Isn't this wrong? Shouldn't it be
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b\geq0 \tag 2
$$
Or
$$
a+b > 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 3
$$
... | I would just want to point out that in the third equation and the proof,
\begin{align} \text{for }~ a ,~ b > 0 , \ ( \sqrt{a} - \sqrt{b} )^2 \end{align}
is non-negative, since both of them could be equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to solve polynomial rational relations for $y$ (e.g $\sqrt{4-3y-y^2} = x(y+4)$)? From time to time, I struggle to solve polynomial relations for $y$.
A trivial example is :
$$ \frac{y}{x} = x \iff y = x^2$$
Easy.
But consider this relation:
$$ \sqrt{4-3y-y^2} = x(y+4)$$
No matter how much I mess around it, seems i... | For $-4<y\le 1, x>0$
We have $$\sqrt{(y+4)(1-y)}=x (y+4) \implies (1-y)(y+4)=x^2(y+4)^2 \implies \frac{1-y}{y+4}=x^2$$
$$\implies y(x)=\frac{1-4x^2}{1+x^2}.$$
See the graph of $y(x)$ below:
| {
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"url": "https://math.stackexchange.com/questions/3764098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving for positive reals: $abcd=1$, $a+b+c+d=28$, $ac+bc+cd+da+ac+bd=82/3$ $$a,b,c,d \in \mathbb{R}^{+}$$
$$ a+b+c+d=28$$
$$ ab+bc+cd+da+ac+bd=\frac{82}{3} $$
$$ abcd = 1 $$
One can also look for the roots of polynomial
$$\begin{align}
f(x) &= (x-a)(x-b)(x-c)(x-d) \\[4pt]
&= x^4 - 28x^3 + \frac{82}{3}x^2 - (abc+abd+a... | The hint.
By your work $$\frac{x^4-28x^3+\frac{82}{3}x^2+1}{x}-\frac{244}{27}=\frac{(3x-1)^3(x-27)}{27x}$$ because by Rolle $$\left(\frac{x^4-28x^3+\frac{82}{3}x^2+1}{x}\right)'=\frac{(3x-1)^2(x^2-18x-3)}{3x^2}$$ has three positive roots and one of them must be $\frac{1}{3},$ which is also an element of $\{a,b,c,d\}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
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Help finding a centre of a circle I have been trying to work out this problem for hours but I cannot get it. A circle has its centre on the positive $x$-axis at $(a,0)$. The radius of this circle is $4$. Lines $y=2x$ and $y=-2x$ are tangent to the circle. I have to find the $x$-coordinate $a$.
So I get that the equatio... | Why not just solve the quadratic? Collect like terms in $x$ as follows. Move everything to one side to get $$(x-a)^2 + 4x^2 - 16 = 0.$$ Then expand the square:
$$\begin{align}
0 &= x^2 - 2ax + a^2 + 4x^2 - 16 \\
&= 5x^2 - 2ax + a^2 - 16.
\end{align}$$
Then using the quadratic formula gives you $$x = \frac{2a \pm \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Maximum of function abs Let function $f(x)=|2x^3-15x+m-5|+9x$ for $x\in\left[0,3\right]$ and $m\in R$. Given that $\max f(x) =60$ with $x\in\left[0,3\right]$, find $m$.
I know how to solve this kind of problem for $g(x)=|2x^3−15x+m−5|$. However, the $+9x$ is confusing me.
| Case 1:
Let $2x^3-15x+m-5\gt0$ at the point where maximum occurs.
Then our $f(x)$ becomes $2x^3-6x+m-5$
note that $f(x)$ decreases for $(0,1)$ and then increases for $x\gt1$ so the maximum of $f(x)$ is at $x=3$(as $x\in\left[0,3\right]$)
Plugging in the $f(3)=60$ we get $m = 29$.
We can put in $x=3$ and $m = 29$ in $2... | {
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"url": "https://math.stackexchange.com/questions/3766297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\prod_{n\ge0}^{ }\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma(c)\Gamma(d) }{ \Gamma(a)\Gamma(b)}$ How to prove the following identity?
$$\prod_{n\ge0}^{ }\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}\tag{$a,b,c,d \in \mathbb R$}=\frac{\Gamma(c... | You can finish the proof by using the asymptotics $$
\mathop {\lim }\limits_{N \to + \infty } \frac{{(N + a)!}}{{N^{N + a + 1/2} e^{ - N} \sqrt {2\pi } }} = 1,
$$ which can be derived from Stirling's formula ($a$ is an arbitrary fixed complex number).
| {
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"url": "https://math.stackexchange.com/questions/3766409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$
Evaluate:
$$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$$
I could find the integral by setting it equal to $$\frac{ax+b}{(1+x^3)^{1/2}}$$
and differentiating both sides w.r.t.$x$ as
$$\frac{2-x^3}{(1+x^3)^{3/2}}=\frac{a(1+x^3)^{3/2}-(1/2)(ax+b)3x^2(1+x^3)^{-1/2}}{(1+x^3)}$$$$=\f... | $$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx=\int \frac{2x^{-3}-1}{(x^{-2}+x)^{3/2}} dx$$
Now substitute $t=x^{-2}+x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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Show that if $a$ and $b$ have the same sign then $|a + b| = |a| + |b| $, and if $a$ and $b$ have opposite signs then $|a+b| < |a| + |b|$ I'm considering the different cases for $a$ and $b$
Case 1) $ a\geq 0$ and $ b\geq 0$
Given both terms are positive, $ a + b \geq 0 $
$$
|a+b| = a + b = |a| + |b|\\
$$
Case 2) $ a< 0$... | I'll suggest some techniques for being briefer, which helps you know if you've covered all bases.
Since $|a+b|-|a|-|b|$ is unchanged when multiplying $a,\,b$ by $-1$, for the first problem we can assume without loss of generality that their common sign means they're $\ge0$, reducing the equation to the trivial $a+b=a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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I need a hash function that will give all values from $0$ to $2^n - 1$. Will $f(x) = \frac{x^2 + x}{2} \; \text{mod} \; 2^n$ do the job? I need a hash function that will give all values from $0$ to $2^n - 1$. I want to know if this function does what is needed?
$$f(x) = \frac{x^2 + x}{2} \; \text{mod} \; 2^n$$
the val... | Consider if there are any $x, y \in [0, 2^n - 1]$ where $x \neq y$ and $f(x) = f(y)$. If so, then there's a $k \in \mathbb{Z}$ where
$$\begin{equation}\begin{aligned}
\frac{x^2 + x}{2} - \frac{y^2 + y}{2} & = k\left(2^n\right) \\
x^2 + x - (y^2 + y) & = k\left(2^{n+1}\right) \\
x^2 - y^2 + x - y & = k\left(2^{n+1}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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$f\left( x \right) = {x^3} + x$, then $\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $
If $f\left( x \right) = {x^3} + x$, then
$$\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $$
is________.
My approach is as follows:
$$g = {f^... | After the substitution $u=2x$ in the second integral, your expresion becomes
$$\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx}= \int\limits_1^2 {f\left( x \right)dx} + \int\limits_{f(1)=2}^{f(2)=10} {{f^{ - 1}}\left( {u} \right)du}$$
Now, you can use an integration rule for t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Finding $\lim_{n\to\infty}\frac1{n^3}\sum_{k=1}^{n-1}\frac{\sin\frac{(2k-1)\pi}{2n}}{\cos^2\frac{(k-1)\pi}{2n}\cos^2\frac{k\pi}{2n}}$
For all $n\ge 1$, let
$$
a_{n}=\sum_{k=1}^{n-1} \frac{\sin \left(\frac{(2 k-1) \pi}{2 n}\right)}{\cos ^{2}\left(\frac{(k-1) \pi}{2 n}\right) \cos ^{2}\left(\frac{k \pi}{2 n}\right)}
$$
... | This is based on one of the official solutions for the problem.
We first write $a_n$ as a telescoping sum. Notice that
$$
\frac{1}{AB} = \left(\frac{1}{A}-\frac{1}{B}\right)\cdot \frac{1}{B-A}\,.
$$
It follows that the summand of $a_n$ can be written as
$$
\left(\frac{1}{\cos ^{2}\left(\frac{(k-1) \pi}{2 n}\right)}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to evaluate $\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2}$? Note: Similar questions have been asked here and here, but this is quite different.
I am trying to evaluate
$$\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2} \quad (1)$$
I re-wrote the fraction as $$ \frac{2n+1}{2n(n+1)^2} = \frac1{2(n+1)} \cdot \frac{2n+1}{n(... | As has been already pointed out in the comments,
$$\sum_{n = 1}^\infty \frac{1}{(n+1)^2} = \frac{1}{4} + \frac 19 + ... = \frac{\pi^2}{6}-1$$
Therefore, your solution would be
$$\frac{1}{2}\left( \sum_{n=1}^\infty \frac 1n - \frac {1}{n+1} + \sum_{n=1}^\infty\frac{1}{(n+1)^2}\right) = \frac 12 \left( 1 + \frac{\pi^2}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Need help with finishing integration I have the following integral: $$y=\int \frac{1}{1-2\sqrt{x}} \, dx$$
I first got $u=2\sqrt{x}$ which gives us $x=\frac{u^2}{4}$. Plugging this in I got:
$$y=\int \frac{1}{1-2\sqrt{\frac{u^2}{4}}} \, du=\int \frac{1}{1-2(\frac{u}{2})} \, du=\int \frac{1}{1-u} \, du$$
After this I a... | We have $$ \int \frac{1}{1-2\sqrt{x}}dx$$
Substitute $u = 2\sqrt{x}$ so that $x = \frac{u^2}{4}$ and $dx = \frac{u}{2}du$. We get $$ \int \frac{u}{2(1-u)}du $$
Now, it can be done from here, but since you're learning, make another substitution which might help you seeing "typical" integral here.
Let $t = 1-u$ so that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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What is the integer part of the following fraction: $\frac{2012^{2013}+2013^{2014}}{2012^{2012}+2013^{2013}}$
What is the integer part of the following fraction: $\dfrac{2012^{2013}+2013^{2014}}{2012^{2012}+2013^{2013}}$.
This is a competition problem for 7th grade students.
The answer to this question is $2012$.
Is ... | Observe that for any $a,b > 0,$ we have $\dfrac{a^{a+1}+b^{b+1}}{a^a + b^b} = \dfrac{a(a^a + b^b)+(b-a)b^{b}}{a^a + b^b} = a + (b-a)\dfrac{b^b}{a^a+b^b}.$ Clearly, $b^b < a^a + b^b.$ So in order for $\dfrac{b^b}{a^a+b^b}$ to be a fraction, we need $\dfrac{b^b}{a^a+b^b} < \frac{1}{b-a}.$ Here we have $a = 2012, b = 2013... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding the asymptotes of the polar curve $ r=\frac{ 5\cos^2 \theta+3 }{ 5\cos^2 \theta-1 }$ Find all asymptotes ( their equations in polar or cartesian form) for the polar curve
$$ r(\theta)=\dfrac{ 5\cos^2 \theta+3 }{ 5\cos^2 \theta-1 }$$
When denominator goes to zero we have one set, but what is the other set?
| You are right, values are given by
$$5\cos^2 \theta-1=0 \implies \cos \theta = \pm\frac1{\sqrt 5}\implies \theta_{1,2}=\pm \arccos \left(\frac1{\sqrt 5}\right),\: \theta_{3,4}=\pm \arccos \left(-\frac1{\sqrt 5}\right)$$
To find the asymptotes equations let consider for any direction $\theta_0$ (wlog assume $ \theta_0=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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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.