Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Show every prime $p\equiv 1,3$ (mod 8) can be written as $p=x^2+2y^2$
Let $K=\mathbb{Q}(\sqrt{-2})$. Show that $\mathcal{O}(K)$ is a principal ideal domain. Deduce that every prime $p\equiv 1, 3$ (mod 8) can be written as $p = x^2 + 2y^2$ with $x, y \in \mathbb{Z}$.
As $−2$ is squarefree $6\equiv 1$ (mod 4) we have ... | An elementary proof starting from scratch. If $p\equiv 1\pmod{8}$, in $\mathbb{F}_p^*$ there is an element with order $8$, which we may call $\alpha$. Since $\alpha^4+1=0$ we have $(\alpha+\alpha^{-1})^2=0 $ and $-2$ is a quadratic residue $\pmod{p}$. If $p\equiv 3\pmod{8}$ we may consider the degree of the splitting f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find center of rational circle When a circle is defined by three points on its perimeter, it's possible to find its center $(h,k)$ using:
$$
h=\frac{\left| \begin{matrix}x_1^2+y_1^2 & y_1 & 1 \\x_2^2+y_2^2 & y_2 & 1 \\ x_3^2+y_3^2 & y_3 & 1\end{matrix} \right|}{2*\left| \begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x... | If you multiply a column of a determinant by $c$ the determinant is also multiplied by $c$. So if we say that the points are $(\frac{x_k}{w}, \frac{y_k}{w})$ the values would become $$
h=\frac{\frac{1}{w^3} \left|\begin{matrix}x_1^2+y_1^2 & y_1 & 1 \\x_2^2+y_2^2 & y_2 & 1 \\ x_3^2+y_3^2 & y_3 & 1\end{matrix} \right|}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Partial fraction problem in $\int \frac{dx}{x^2(x^2+1)}$ I'm stuck with this integral:
$$\int \frac{dx}{x^2(x^2+1)}$$
I know that it can be solved in a simple way adding $x^2-x^2$ in the numerator, but I have troubles with partial fraction decomposition:
$$\frac{1}{x^2(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+... | How did you get powers of $x^4$ in your expansion for the numerator?
$$\begin{align}\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+1}&=\frac{Ax(x^2+1)+B(x^2+1)+(Cx+D)x^2}{x^2(x^2+1)}\\
&=\frac{(A+C)x^3+(B+D)x^2+Ax+B}{x^2(x^2+1)}
\end{align}$$
Yielding equations:
$$\begin{align}B&=1\\A&=0\\B+D&=0\\A+C&=0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If a,b, and c is real, and following the equation, solve for
If $a,b,c$ are real and aren't 0, that complete the following equations
$a^2 + a = b^2$
$b^2 + b = c^2$
$c^2 + c = a^2$
Solve for $(2a+c)(2b+a)(2c+b)$,
What I've done:
*
*Adding all of the equation of it and substact with $(a^2 + b^2 + c^2)$ of both si... | Adding all equations we get
$$a+b+c=0$$ and now we can eliminate the other variables, we obtain: $$3a^4-3a^2+a=0$$ and so on.
With $$c=-a-b$$ we get
$$b^2+b=(a+b)^2$$ and
$$(a+b)^2-(a+b)=a^2$$
From the equation $$b^2+b=a^2+b^2+2ab$$ we get
$$a^2+2ab-b=0$$ and $$b=\frac{a^2}{1-2a}$$ Can you proceed?
And we get $a$ from
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Proof. So i'm going to prove this by induction. The first case when $n=1$ is trivial since:
$$9+16=25,$$
implying $ 5 \mid 25$.
Now we need to show is divisible when $n=k+... | By Induction:
$n=0$: $9^0+4^1=5$ , divisible by $5$.
Hypothesis:
Assume : $9^n+4^{ n+1}$ is divisible by $5$.
Step:
$9^{n+1}+4^{n+2} =9 \cdot 9^{n} +4 \cdot 4^{n+1}=$
$(5+4)9^n +4 \cdot 4^{n+1} =$
$ 5 \cdot 9^n +4(9^n+4^{n+1})=$
$ 5 \cdot 9^n +4\cdot 5c= 5(9^n+4c)$,
where $c \in \mathbb{Z^+}$, since $(9^n+4^{n+1})$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find all integer $t$ such that $\frac{t^2+9}{t+9}$ is an integer Find all integer $$ t \geq 0 $$ such that $$ \frac{t^2+9}{t+9} =k $$ k is integer.
How to solve this?
I've tried to solve this as an equation with parameter
$$
t^2-kt-9k+9=0 \\
D= k^2-36k+36\\
t_1=\frac{k-\sqrt{k^2-36k+36}}{2} \\
t_2=\frac{k+\sqrt{k^2-... | Your discriminant is wrong. You should get:
$$D=k^2-4(9-9k)=k^2\color{red}+36k\color{red}-36$$
The easier approach is to note that $\frac{t^2-81}{t+9}=t-9$ is an integer, so $$\frac{t^2+9}{t+9}=\frac{t^2-81}{t+9}+\frac{90}{t+9}$$ is an integer if and only if $\frac{90}{t+9}$ is an integer.
If you must use the discrim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $a_n = a_{n-1}a_{n-3} + a_{n-2}^2$, then is it always divisible by $a_{n-4}$? The Somos Sequences are neat because they are integral yet contain a division operation.
If we consider the numerator $a_n = a_{n-1}a_{n-3} + a_{n-2}^2$ of the Somos-4 sequence, $a_1 = a_2 = a_3 = 1$, then is it always divisible by $a_{n-... | No. Working out the first few terms of the recurrence $a_n = a_{n-1} a_{n-3} + a_{n-2}^2$, we get
$$
\begin{array}{c|c}
n & a_n \\ \hline
1 & 1\\
2 & 1\\
3 & 1\\
4 & 2\\
5 & 3\\
6 & 7\\
7 & 23\\
8 & 118\\
9 & 1355\\
\end{array}
$$
from which we see $a_{5} = 3$ and $a_9 = 1355$, but $3 \nmid 1355$.
In fact for $n \leq 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x\to 0} \frac {1-(\cos 2x)^3(\cos 5x)^5(\cos 7x)^7(\sec 4x)^9(\sec 6x) ^{11}}{x^2}$
Evaluate $$\lim_{x\to 0} \frac {1-(\cos 2x)^3(\cos 5x)^5(\cos 7x)^7(\sec 4x)^9(\sec 6x) ^{11}}{x^2}$$
Now I can apply L'Hospital rule twice but believe it or not it would seriously be a very tedious task to do. I also ... | This is what one calls a crafted problem especially designed to intimidate students. One of the easiest approaches is to use the Taylor series expansion (presented in another answer).
However the problem can be solved in step by step manner using the standard limits $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2},\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find the highest power of prime which divides a factorial. Find the highest power of $10$ that divides $50!$.
I know by De-Polignac's formula ,the highest power of $5$ that divides $50!$ is $12$ & the highest power of $2$ that divides $50!$ is $47$.
I am able to find this because $5$ & $2$ are primes but here $10$ is n... | As you found:
The highest power of $2$ that divides $50!$ is $\left\lfloor \frac{50}{2} \right \rfloor + \left\lfloor \frac{50}{2^2} \right \rfloor + \left\lfloor \frac{50}{2^3} \right \rfloor + \cdots$ which evaluates as $ 25+12+6+3+1$ $= 47$, each term being half-rounded-down of the previous one.
The highest power o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Showing that $\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$ As the title suggests, I would like to show that: $$\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$$ for every positive integer $n$.
How can I go about doing that? I understand the that there are alrea... | Integrating by parts
\begin{align*}
\int_0^{2\pi}\cos^{2n}\theta d\theta&=\Big[\cos^{2n-1}\theta\sin\theta\Big]_0^{2\pi}-\int_0^{2\pi}\sin\theta(2n-1)\cos^{2n-2}\theta(-\sin\theta) d\theta\\
&=(2n-1)\int_0^{2\pi}(\cos^{2n-2}\theta-\cos^{2n}\theta)d\theta\\
&=(2n-1)\int_0^{2\pi}(\cos^{2n-2}\theta-\cos^{2n}\theta)d\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int \frac {dx}{\sin^2 x + \tan^2x}$
I am trying to find the following integral:
$$\int \frac {dx}{\sin^2 x + \tan^2x}$$
I have tried the common thing to do when encountering rational functions that contains rational functions and converting everything in terms of $\tan \frac{x}{2}$ then substituting it... | $$\dfrac1{\sin^2x+\tan^2x}=\dfrac{\cos^2x}{\sin^2x(1+\cos^2x)}$$
Method$\#1:$
$$=\dfrac{\cos^2x+1-1}{\sin^2x(1+\cos^2x)}=\csc^2x-\dfrac1{(1-\cos^2x)(1+\cos^2x)}$$
Now $$\dfrac2{(1-\cos^2x)(1+\cos^2x)}=\dfrac1{1-\cos^2x}+\dfrac1{1+\cos^2x}$$
Finally $$\dfrac1{1+\cos^2x}=\dfrac{\sec^2x}{2+\tan^2x}$$
Method$\#2:$
$$\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integral values of $k$ for which polynomial has integral roots. How integral values of $k$ exists such that all roots of the polynomial:
$$f(x)=x^3-(k-3)x^2-11x+(4k-8)$$ are also integers.
Could someone please provide me some direction to proceed in this. First I thought first root would be obtained by hit and trial bu... | Hint:
Since $$k ={x^3+3x^2-11x-8\over x^2-4}\implies x^2-4 \mid x^3+3x^2-11x-8$$
Since also $$x^2-4\mid (x+3)(x^2-4) = x^3+3x^2 -4x-12$$
we get
$$ x^2-4\mid ( x^3+3x^2-11x-8)-( x^3+3x^2-4x-12)= -7x+4$$
thus $$x-2\mid -7x+4\implies x-2 \mid -7x+4+7(x-2) = -10$$
Finally we have $x-2\in \{-10,-5,-2,-1,1,2,5,10\}$ so $x\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Taylor series $\sqrt{1-2x+x^2+o(x^3)}$ I need to solve this asymptotic expansion for $x\to0$
$$\sqrt{1-2x+x^2+o(x^3)}$$
This is the expansion to use:
$$\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(x^3)$$
let:
$$t = -2x+x^2+o(x^3)$$
My question is, should I stop to the first order or keep going?
How can I ... | You are being told to work in the ring $\Bbb Q[[x]]/(x^3)$, and to find there a square root of $1-2x+x^2$. I think that the answer is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Precise definition of Multivariable limit I have encountered this limit: $\lim_{(x,y) to (0,0)} \frac{x^2+3y-7}{2+x^3-5y^2}$. We can substitute and find the limit to be $\frac{-7}{2}$since it is not a difficult limit, however I am trying to prove it rigorously and am getting stuck. I’m starting out with for all $\epsil... | Note that
$$
\delta > \sqrt{x^2+y^2} \implies \delta > \lvert x\rvert,\, \delta>\lvert y\rvert
$$
Then $\delta > \sqrt{x^2+y^2}$ implies that
$$
\lvert 7x^3+2x^2-35y^2+6y \rvert < 7\delta^3 + 2\delta^2 +35\delta^2+6\delta = 7\delta^3 + 37\delta^2+6\delta
$$
Now, suppose without loss of generality that $\delta < 1/5$ (I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Invariant Factors of a complete tripartite graph Let $K_{4,4,12}$ be a complete tripartite graph on 20 vertices. Considering its adjacency matrix with a convenient labeling,its nonzero eigenvalues are -4,-8,12. Now the invariant factors are 4,4,24. How did they find the invariant factors in detail, please?
| The version of the paper I found says that the invariant factors are $4,4,24$; we get $1,9,36$ for the second example $K_{2,9,9}$.
In any case, here is how you get there. These are not the invariant factors of the adjacency matrix (there's a result about those elsewhere in the paper) but of the quotient matrix, which i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limits of a product $N$ is a positive integer.
I want to calculate this limit, but i couldn't get anywhere when i tried.
$$P[n]=\left(1+\frac {1}{n^2}\right)\left(1+\frac {2}{n^2}\right)\cdots\left(1+\frac {n-1}{n^2}\right)$$ as $n\to \infty$
I tried to apply $\ln()$ at both sides to transform into a sum etc..
tried to... | $$\left(1+\frac {j}{n^2}\right)\left(1+\frac {n-j}{n^2}\right)=1+\frac{1}{n}+\frac{j(n-j)}{n^4}$$
By AM-GM we have
$$\sqrt{j(n-j)}\leq \frac{n}{2} $$
and hence
$$1+\frac{1}{n} \leq \left(1+\frac {j}{n^2}\right)\left(1+\frac {n-j}{n^2}\right) \leq 1+\frac{1}{n}+\frac{1}{4n^2}$$
Therefore
$$\left(1+\frac{1}{n}\right)^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Proving Hermite's Identity in a different approach To prove
$$S=\left [x \right]+\left [x+\frac{1}{n} \right]+\left [x +\frac{2}{n}\right]+\cdots+\left [x +\frac{n-1}{n}\right]=\left [nx \right]$$
using and starting with $$x-1 \lt \left [x \right]\le x \tag{1}$$ we have
$$x+\frac{1}{n}-1 \lt \left [x +\frac{1}{n}\righ... | Here is a proof I came up with
about a year ago:
For all $x$ which are real numbers, prove that $\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$
The basic idea is that
you have to see
how far $x$ is from
$\lfloor x \rfloor$
and,
if $d = x-\lfloor x \rfloor$,
what $\lfloor nd \rfloor$ is.
This allows you d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Without using the Rule of Sarrus, prove that: Without using the Rule of Sarrus, prove that:
$$\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|=3abc-a^3-b^3-c^3$$
My Approach:
$$LHS=
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|$... | \begin{align*}
\left|
\begin{matrix}
c&(a-b)&a \\
a&(b-c)&b \\
b&(c-a)&c \\
\end{matrix}\right|&= c\left|\begin{matrix}
(b-c)&b \\
(c-a)&c \\
\end{matrix}\right|-a\left|\begin{matrix}(a-b)&a\\(c-a)&c \\\end{matrix}\right|+b\left|\begin{matrix}(a-b)&a\\(b-c)&b \\\end{matrix}\right|\\
&=c[bc-c^2-bc+ab]-a[ac-bc-ac+a^2]+b[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
British Maths Olympiad (BMO) 2002 Round 1 Question 3 Proof without Cauchy-Schwarz? The question states:
Let $x,y,z$ be positive real numbers such that $x^2 + y^2 + z^2 = 1$
Prove that
$x^2yz + xy^2z + xyz^2 ≤ 1/3$
I have a proof of this relying on the fact that:
$x^2/3 +y^2/3 + z^2/3 \geq (x+y+z)^3/9 $ (A corollary of... | Here is my approach.
Using AM-GM,
$\begin{align} x^2+y^2+z^2 &\ge 3(xyz)^{2/3} \\ \implies 1 &\ge 3(xyz)^{2/3} \\ \implies (xyz)^{2/3} &\le \dfrac{1}{3} \\ \implies xyz \le \dfrac{1}{3\sqrt{3}} \tag 1 \end{align}$
Again, using AM-GM,
$\begin{align} \dfrac{x^3+y^3+z^3}{3} &\ge 3xyz \\ \implies x^3+y^3+z^3-3xyz &\ge 0 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Evaluating $\int \sqrt{\frac{\cos x - \cos^3 x}{1-\cos^3 x}}dx$
Evaluate $\int \sqrt{\frac{\cos x - \cos^3 x}{1-\cos^3 x}}dx$
My attempt:
$I=\int \sqrt{\frac{\cos x - \cos^3 x}{1-\cos^3 x}}dx=\int \sqrt{\frac{\cos x(1-\cos^2 x)}{1-\cos^3 x}}dx=\int \sqrt{\frac{\cos x\sin^2 x}{1-\cos^3 x}}dx=\int \sin x\sqrt{\frac{\co... | Hint: Substitute $u=z^{3/2}$. You should then be able to recognise the integral you get.
How could you have come up with this yourself? Well note that you have $\sqrt z dz$ on top (which is the same up to a constant as $d(z^{3/2})$), and you have $1-z^3$ on the denominator, whereas you're probably more used to seeing ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Calculate residue in every pole of $\frac{z^2}{(z-2)^2(\cos{z}-1)^3}$ The problem is as simple as the title suggests. Of course, the formula for the residue of an order $n$ pole involving $n-1$ derivatives could be applied, but the computation will be extremely long, as there are poles of order $6$ in the function:
$\... | Let
$$f(z) = \frac {z^2} {(z - 2)^2} =
1 + \frac 4 {z - 2} + \frac 4 {(z - 2)^2}, \\
g(z) = \frac 1 {(\cos z - 1)^3}.$$
To find the expansion of $g$, take the product of the series of $1/g$ and of $g$ and equate the coefficients:
$$(a z^6 + b z^8 + c z^{10} + O(z^{12}))
(A z^{-6} + B z^{-4} + C z^{-2} + O(1)) = \\
a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find a matrix $N$ s.t. $N^2=H$ Given the below matrix, find a matrix $N$ s.t. $N^2=H$
$$H=\begin{bmatrix}2&i&2&1+i\\-i&2&-2i&1-i\\2&2i&5&2+2i\\1-i&1+i&2-2i&3\end{bmatrix}$$
I solved for the characteristic polynomial, which is $(x-1)^3(x-9)$
Any help will be appreciated.
| Your matrix is diagonalizable. More precisely, if$$M=\begin{pmatrix}1+i & -1-i & -2 & -i \\ 1-i & 0 & 0 & 1 \\ 2+2 i & 0 & 1 & 0 \\ 2 & 1 & 0 & 0\end{pmatrix},$$then$$M^{-1}.H.M=\begin{pmatrix}9&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.$$So, an answer to your question is$$M.\begin{pmatrix}3&0&0&0\\0&1&0&0\\0&0&1&0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Express the given polar equation in simplest rectangular form: $r = 4 + 3cosθ$ Express the given polar equation in simplest rectangular form: $r = 4 + 3cosθ$
My attempt
Multiplying r on both sides, we get
$r^2=4r+3 rcos\theta$
since, $x=rcos\theta$ and $y=rsin\theta$
and $r^2=x^2+y^2$
we get $x^2+y^2=4r+3x$
can anyone ... | $$r^2 - 4 r = 3 x$$
$$x^2 +y^2 - 4\sqrt{x^2 +y^2} = 3 x$$
which represents a cardoid.. it is better to leave it that way.
if you try to get rid of radical sign you end up with another equation representing another symmetrical cardoid, not really what you wanted to represent at the outset:
$$x^2 +y^2 + 4\sqrt{x^2 +y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Without using a calculator, is $\sqrt[8]{8!}$ or $\sqrt[9]{9!}$ greater? Which is greater between $$\sqrt[8]{8!}$$ and $$\sqrt[9]{9!}$$?
I want to know if my proof is correct...
\begin{align}
\sqrt[8]{8!} &< \sqrt[9]{9!} \\
(8!)^{(1/8)} &< (9!)^{(1/9)} \\
(8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\
(8!)^{(9/72)} - (9!)^{8... | I would compare it by using logarithms:
$$\ln a = \frac{1}{8}(\ln1 + ...+\ln8)$$
$$\ln b = \frac{1}{9}(\ln1 + ...+\ln8+\ln9)=\frac{8}{9} \ln a + \frac{1}{9}\ln9$$
$$\ln b-\ln a=\frac{1}{9}(\ln9-\ln a)=\frac{1}{9} \ln\frac{9}{a}$$
$a$ is obviously less than 8 so:
$$\frac{9}{a} \gt 1$$
$$\ln\frac{9}{a}\gt0$$
$$\ln b - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 10,
"answer_id": 1
} |
What is $\sqrt[4]{-1}$ So, we all know that $\sqrt{-1}=i$, that $\sqrt{-2}=\sqrt2i$, and so on. And that, for example, $\sqrt[3]{-27}=-3$. But I was wondering; what would $\sqrt[4]{-1}$ be? Since all $n^{th}$ roots of $-1$ that are odd result in $-1$ (Because $-1^{1,3,5,7,9,\dots}=-1$), I wondered what variation of $i$... | We will use Euler's Identity, $$e^{i\pi} + 1 = 0,\tag1$$ or in particular, $$\begin{align} e^{i\theta} &= \cos\theta + i\sin\theta \\ &= \text{cis}\,\theta.\end{align}\tag2$$
From $(1)$, we get that $e^{i\pi} = -1$, thus $$\begin{align} \left(e^{(i\pi)/2}\right)^{1/2} &= e^{(i\pi)/4} \\ &= \sqrt [4] {-1}.\end{align}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How can one prove that $\sqrt{ 2} \cdot \sqrt{ 3} = \sqrt{ 6}$? I have already proved that $\sqrt{2}\cdot \sqrt{2}=2$ so I hope I can now use $\sqrt{3}\cdot \sqrt{3}=3$ and the same for 6.
The exercise comes from Stillwell: Mathematics and its History. The other exercises have not been complicated so probably the solut... | Let $\sqrt{2}\sqrt{3} = x$.
Substitute $\sqrt{2} = \dfrac{2}{\sqrt{2}}$ and $\sqrt{3} = \dfrac{3}{\sqrt{3}}$. Then we should have: $$\left(\frac{2}{\sqrt{2}}\right)\left(\frac{3}{\sqrt{3}}\right) = \sqrt{6}.\tag1$$ Note that the $LHS$ (Left Hand Side of the equation) is equal to $\dfrac{6}{\sqrt{2}\sqrt{3}}$. It follow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant? Why is the difference between these two functions a constant?
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$
Since the denominators are equal and the numerators differ in degree I would never have th... | I'm not sure anyone is speaking to your observation that the two numerators have different degrees.
Let's flip this around the other way:
\begin{align*}
\frac{x-2}{x^2-x+1} + 2 &= \frac{x-2}{x^2-x+1} + 2\frac{x^2-x+1}{x^2-x+1} \\
&= \frac{2x^2 -x}{x^2-x+1} \text{.}
\end{align*}
That is, we started with a thing havi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 16,
"answer_id": 7
} |
How to get the expected value of these numbers?
So for the E(x) of (a) I got 1 $\cdot$ ($\frac{1}{5} $) + 3 $\cdot$ ($\frac{2}{5} $)+ 9 $\cdot$ ($\frac{2}{5} $) = 9.8
Why is the answer showing E(x)= 5?
Also To compute the Variance, according to the formula, E($X^2$) - E$(X)^2$ shouldn't the answer be $\frac{181}{5} ... |
So for the E(x) of (a) I got $1 \cdot \tfrac 15 + 3 \cdot \tfrac25+ 9 \cdot\tfrac25 = 9.8$
Why is the answer showing E(x)= 5?
Because $\tfrac {1+ 6+18}5 =\tfrac {25}5$ not $\tfrac {49}5$ (...how did you even?)
Also To compute the Variance, according to the formula, E($X^2$) - E$(X)^2$ shouldn't the answer be $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Conceptual Idea of Functional Reduction I have a doubt regarding functional reduction(In mathematics, reduction refers to the rewriting of an expression into a simpler form).
I have a function, let me call it as $f(x)$ and I have rewritten it to a simpler form let me call it as $g(x)$. Can we say both $f(x)$ and $g(x)... | Equal functions agree on domain and values.
When one proves identities, one only requires the (restricted) functions to agree on their mutual domain, the intersection of their domains. If we do not make this relaxation, many identities become false. For instance, "$1 = \frac{\sin z}{\sin z}$" is undefined on the righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the value of $\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $?
What is the value of
$\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $ ?
The question is indeed intriguing. We could start by condensing it using $\sin \theta = \sin (1... |
The product of the chords in this figure is
$(2\cos \frac {2\pi}{5})(2\cos \frac {\pi}{5})(2\cos 0)(2 \cos -\frac {\pi}{5})(2\cos -\frac {2\pi}{5}) = 2^5\prod_\limits {n=-2}^2 \cos \frac {n\pi}{5}$
If we map this figure to the complex plane the product of those lengths = $|(1+e^{\frac {\pi i}{5}})(1+e^{\frac {3\pi i}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
computing probabilities of an negative binomial random variable Would appreciate if you could check if I answered the questions correctly:
$X$ is a Negative Binomial random variable with the parameters $\frac{1}{2}$ and $r =1,2,3,\ldots$.
$Y$ is a random variable that is defined as: $Y =
\begin{cases}
X, & x \le r+1 ... | Your description in (1) and the fact that you quoted $\mathbb{E}[X] = r/p$ for (2), I take it that you adopt the total number of trials for the definition of $X$ the Negative Binomial.
\begin{align*}
\Pr\bigl\{ Y = r+1 \bigr\} &= \overbrace{\Pr\bigl\{X = r+1 \bigr\} }^{Y=X} + \overbrace{ \Pr\bigl\{ X = r+2\bigr\} }^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve the integration How to find $$ \frac{\int_0^{\pi}x^3\log(\sin x)\,dx}{\int_0^{\pi} x^2 \log(\sqrt{2} \sin x)\,dx} $$
I couldn’t resolve it by using integration by parts.
| $\newcommand{\intd}{\,\mathrm{d}}$
Alternatively, and quite similarly I might add, this can be done using the substitution $u=a+b-x$ where $a$ and $b$ are the bounds of the integration. It will be important to note that $\sin(\pi-x)=\sin x$, looking at the numerator we have $I_1=\int_0^{\pi}x^3\log(\sin x)\,\mathrm{d}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Integration of $\frac{1}{1+x^2+x^4+\cdots +x^{2m}}$ Please calculate $$A_m=\int _{-\infty}^{\infty}\frac{dx}{1+x^2+x^4+\cdots +x^{2m}}$$
My idea is $$A_m=\int \frac{1-x^2}{1-x^{2m+2}}\: dx$$ so I tried to use residue theorem to upper half circle with radius $R$.
But, I cannot calculate it.
| Let $\gamma$ be the upper half circle $C$ with radius $R$ centered at $0$ and the segment from $A=(-R,0)$ to $B=(R,0)$. Clearly, inside $\gamma$, the function $f(z)=\frac{z^2-1}{z^{2m+2}-1}$ has $m$ poles $z=\omega_k\equiv=e^{\frac{\pi i k}{m+1}}=\omega_1^k$, $k=1,2,\cdots,m$ and on $AB$, $f(x)$ has two removable singu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2770635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $a\in\mathbb{R}, a\neq 0$ such that the parabolas of the functions $f$ and $g$ are tangent.
Find $a\in\mathbb{R}, a\neq 0$ such that the parabolas of the functions $f$ and $g$ are tangent.
$f_a(x)=ax^2-(a+2)x-1$.
$g_a(x)=x^2-x-a$.
At first what I tried is find their derivative and just make them equal, but it's ... | $f_a(x)
=ax^2-(a+2)x-1
$,
$g_a(x)=x^2-x-a
$.
Their tangents are
$f_a'(x)
=2ax-(a+2)
$,
$g_a'(x)=2x-1
$.
The tangents have
the same slope when
$2ax-(a+2)
=2x-1
$
or
$2x(a-1) = a+1
$
or
$x = \frac{a+1}{2(a-1)}$
(if $a \ne 1$).
If $a=1$ then
$f_a=g_a$
so the curves are identical.
They intersect when
$ax^2-(a+2)x-1
=x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Property of Bernstein polynomial $\sum_{k=0}^n k^2 B_{n,k} (x) $ $= n(n-1)x^2+nx$ Actually, I have to show:
$$ \sum_{k=0}^n B_{n,k} (x) =1 $$
$$ \sum_{k=0}^n \frac{k}{n} B_{n,k} (x) =x $$
$$ \sum_{k=0}^n \frac{k^2}{n^2} B_{n,k} (x) =x^2+ \frac{1}{n}x(1-x) $$
So the first two claims were easy, but I failed to show the l... | Evaluating the identity in the title we seek
$$\sum_{k=0}^n k^2 {n\choose k} x^k (1-x)^{n-k}.$$
This is a polyonomial of degree $n$ in $x$ from which we may
extract coefficients with $0\le q\le n$:
$$[x^q] \sum_{k=0}^n k^2 {n\choose k} x^k (1-x)^{n-k}
= \sum_{k=0}^q k^2 {n\choose k} [x^{q-k}] (1-x)^{n-k}
\\ = \sum_{k=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show that the equation $x^4 + 3 y^4 = 131$ has no integer solutions. I have seen the result demonstrated by considering the equation$\pmod{5}$, but was wondering if this proof, working in $\mathbb{Z}_3$, is also valid, since it seems to require less working:
$$\forall x \in \mathbb{Z}_3, x = \{0,1,2\} .$$
So, for any $... | I imagine if you've actually seen this done more than once via $\mod 5$ it is not because $\mod 5$ is easy, but because it is interesting.
By Fermat's little Theorem $a^4 \equiv 0, 1\mod 5$ and determined by whether $5$ divides $a$ or not. So $x^4 + 3y^4\equiv 0, 1, 3,4\mod 5$ and which value will uniquely determine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $(y_n)$ has a finite limit when $n \rightarrow \infty$. Let a real sequence $(a_n)$ defined by:
$x_1$ = 1; $x_n = \displaystyle\frac{2n}{(n-1)^2} \displaystyle\sum_{i=1}^{n-1} x_i$, $n=2,3,4, \dots$
For each integers $n$ let
$y_n = x_{n+1} - x_n$
Prove that $(y_n)$ has a finite limit when $n \rightarrow \inf... | Not really an answer, just a clarification: $(x_n)_n$ indeed does not converge.
We have:
\begin{align}
x_{n+1} &= \frac{2(n+1)}{n^2}\left(\frac{2n}{(n-1)^2} + 1\right)\left(\frac{2(n-1)}{(n-2)^2} + 1\right)\cdots\left(\frac{2\cdot 2}{1^2} + 1\right)\\
&= \frac{2(n+1)}{n^2}\cdot\frac{2n + (n-1)^2}{(n-1)^2}\frac{2(n-1) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Writing $\frac{1}{x+5}$ as $\frac{A}{x}+\frac{B}{5}$ Is this possible? (in this case specifically, but also in general for any fraction with addition/subtraction in the denominator).
I thought this would be easy, but upon trying it and thinking some more
*
*theres 2 unknowns and one equation, so we probably can't id... | Note that
$\frac{A}{x}+\frac{B}{5}=\frac{5A+Bx}{5x}=\frac{1}{x+5}\implies 5Ax+25A+Bx^2+5Bx=5x$$
then for $B=0$
$$5Ax+25A=5x \implies A=\frac{x}{x+5}$$
and for $A=0$
$$Bx^2+5Bx=5x\implies B=\frac{5}{x+5}$$
which are trivial solutions and then it is not possible obtain the equality for $A,B$ constants.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate the determinant of this $5 \times 5$ matrix
Calculate the determinant of the matrix $$A=\begin{pmatrix} \sin\alpha
& \cos\alpha & a\sin\alpha & b\cos\alpha & ab \\
-\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2 \\ 0
& 0 & 1 ... | If you calculate along the first column,
\begin{align}
|A|&= \sin\alpha \begin{vmatrix}
\sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2 \\
0 & 1 & a^2 &
b^2 \\ 0 & 0 & a &
b \\ 0 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove the equality: $\det\left[\begin{smallmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{smallmatrix}\right] = 4(a+b)(b+c)(c+a)$
i have to prove:
$$\begin{vmatrix}
-2a &a+b &a+c \\
b+a&-2b &b+c \\
c+a&c+b &-2c
\end{vmatrix} = 4(a+b)(b+c)(c+a)$$
I have tried many calculations between the rows an... | First of all, note that the determinant is going to be a polynomial in a, b, and c of degree 3. Secondly, note that the polynomial is cyclic.
Substitute $a = -b$, to get,
$$\begin{vmatrix}
-2a &0 &c+a \\
0&2a &c-a \\
c+a&c-a &-2c
\end{vmatrix} = -2a(-4ac - (c-a)^2 + (c+a)^2) = 0$$
This shows that $a+b$ is facto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inverse element in field extension $\mathbb{Q}(\sqrt{2},\sqrt{3})$ Determine $(2\sqrt{2}-3\sqrt{3}+5)^{-1}$ in $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$, which is of degree $4$ over $\mathbb{Q}$ and for $a= \sqrt{2} + \sqrt{3}$ follows $a^4-10a^2+1=0$.
I did calculate inverse elements in field e... | You can carry out the "rationalise the denominator" in two steps.
First multiply $\frac{1}{5+2\sqrt{2}-3\sqrt{3}}$ by $\frac{5+2\sqrt{2}+3\sqrt{3}}{5+2\sqrt{2}+3\sqrt{3}}$ and get $\frac{5+2\sqrt{2}+3\sqrt{3}}{6+20\sqrt{2}}$.
Now multiply by by $\frac{6-20\sqrt{2}}{6-20\sqrt{2}}$ and get
$\frac{(5+2\sqrt{2}+3\sqrt{3})(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving a fourth degree equation Well, I want to find the roots of the equation:
$$x^4+6x^3-9x^2-162x-243=0\tag1$$
And Wolfram Alpha tells me that I can factor this and get:
$$(x^2-3x-9)(x^2+9x+27)=0\tag2$$
Then it will be easy to solve... But my question is, how can I go from $(1)$ to $(2)$?
| To solve $x^4+6x^3−9x^2−162^x−243=0$,
first try and factor, if possible: $(x^2−3x−9)(x^2+9x+27)=0$
So since this factors into to quadratics, we can use quadratic formula and some other simplification methods such as completing the square.
The roots of $x^2-3x-9$ are the same as the roots of $x^2-x-1$, multiplied by thr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
The sum of angles in a triangle is greater than 180? So basically the question goes:
Let $ABC$ be a triangle and that $\cos A=-\frac 3 5$ and $\cos B=\frac 5 {13}$, find $\cos C$.
What I did was I get the value of $A$ and $B$ by calculating the arccosine of $-3/5$ and $5/13$ and the sum of both of these angles was alr... | I suspect you're meant to calculate $\cos C$ by using the cosine of sums rule, not by computing inverse cosines.
With that in mind, and on the off chance that the minus sign on $-3/5$ is an error, and the cosine of angle $A$ is really meant to be $3/5$, I'll point out
$$
\arccos 3/5 \approx 53.13 \\
\arccos 5/13 \appr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show $\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$ is equivalent to $1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$ for $\lvert x\rvert < 1$ I have been asked to show that $$\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$$ is equivalent to writing $$1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$$
From here I just ... | We have $$\frac 1 {1 - \frac {x}{3-x} + \frac x{4-x}}=\frac {x^2-7x+12}{x^2-8x+12}=1+\frac {x} {x^2-8x+12}=1+\frac 1 2 \left(\frac 1 {2-x} - \frac 3 {6-x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\tan\left(\sum_{r=1}^{\infty} \arctan\left(\frac{4}{4r^2 +3}\right)\right)$
$$\tan\left(\sum_{r=1}^{\infty} \arctan\left(\dfrac{4}{4r^2
+3}\right)\right)= ? $$
I wrote it in the form:
$$\tan\left(\sum_{r=1}^{\infty} \arctan\left(\dfrac{\dfrac43}{\dfrac{4r^2}{3}
+1}\right)\right)$$
and tried to use:
$$... | You are on the right track. Just note that
$$\frac{4}{4r^2+3} =\frac{1}{r^2+3/4}=\frac{1}{1+r^2-(1/2)^2}=\frac{(r+1/2)-(r-1/2)}{1+(r+1/2)(r-1/2)}$$
and therefore
$$\arctan\left(\frac{4}{4r^2+3}\right)=\arctan\left(r+\frac{1}{2}\right)- \arctan\left(r-\frac{1}{2}\right)\\=\arctan\left((r+1)-\frac{1}{2}\right)- \arctan\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A polynomial inequality in three variables I have a polynomial $P(A,B,C)$ where $A,B,C \in \mathbb{R}$ and $A>0,B>0$.
$$P(A,B,C)=C^2+A^2-2CA+4AB-2B+C-A$$
When will
*
*$P(A,B,C)>0$
*$P(A,B,C)=0$
*$P(A,B,C)<0$
I have tried factorizing in various ways but none of them is very helpful. Any ideas here?
My attempt:
$... | Let $x=C-A$, then $P(A,B,C)=x^2+x+2B(2A-1) \equiv f(x)$.
$$\Delta=1^2-4(2B)(2A-1)=1-8B(2A-1)$$
*
*$\Delta < 0 \implies f(x)>0$,
$$1<8B(2A-1) \implies 2A-1> \frac{1}{8B} \implies A> \frac{8B+1}{16B}$$
*
*$\Delta \ge 0 \land f(x) < 0$,
$$\frac{-1-\sqrt{1-8B(2A-1)}}{2} < C-A < \frac{-1+\sqrt{1-8B(2A-1)}}{2}$$ wher... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to show that this integral is correct? How can one show that
$$\int_0^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha} \tan(x) \tan\left(\frac{x}{2}\right)\right] \sqrt{\sin(x) \tan \left(\frac{x}{2}\right)} \, \mathrm dx=-\frac{\ln(\alpha)}{\sqrt{2}}$$
assume $\alpha\ge1$.
I can't see how to simplify $... | Use properties of trig functions to get
\begin{equation*}
\sin(x)\tan\left(\frac{x}{2}\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)=2\sin^2\left(\frac{x}{2}\right).
\end{equation*}
Now we can simplify the integral to
\begin{equation*}
\int_0^{\pi/2} \frac{1}{\sqrt{2}}\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Integrate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$
Integrate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$ where $b>a$
My Attempt
$$
\begin{align}
&\int \frac{1}{\sqrt{(x-a)(b-x)}}dx=\int\frac{dx}{\sqrt{-x^2+(a+b)x-ab}}\\
&=\int\frac{dx}{\sqrt{-(x^2-2.\frac{a+b}{2}x+ab+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4})}}\\
&=\int\frac{dx}{\sqrt{... | Hint:
From your answer if $\sin^{-1}\sqrt{\dfrac{x-a}{b-a}}=y$
$\sin y=\sqrt{\dfrac{x-a}{b-a}}$
$\implies\cos2y=1-2\sin^2y=\dfrac{a+b-2x}{b-a}$
$2y=\cos^{-1}\dfrac{a+b-2x}{b-a}=\dfrac\pi2-\sin^{-1}\dfrac{a+b-2x}{b-a}=\dfrac\pi2+\sin^{-1}\dfrac{2x-a-b}{b-a}$
So, you have reached to the right answer.
See also: Proof for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Local truncation error of Crank-Nicolson for PDE $u_t+au_x = 0$
Exercise 4:
The Crank-Nicolson scheme for $u_t + a u_x = 0$ is given by
$$ \frac{U_{j,n+1}-U_{j,n}}{\Delta t} + \frac{a}{2}\frac{D_xU_{j,n}}{2\Delta x} + \frac{a}{2}\frac{D_xU_{j,n+1}}{2\Delta x} = 0 .$$
Show that the LTE is given by
$$ \mathcal{L... | Let us write the Crank-Nicolson method for the linear advection equation $u_t + au_x = 0$ by averaging forward and backward Euler time-integration and by using centered spatial differences:
$$
\frac{u_{j}^{n+1} - u_j^n}{\Delta t} = -\frac{a}{2}\left(\frac{u_{j+1}^n - u_{j-1}^n}{2\Delta x} + \frac{u_{j+1}^{n+1} - u_{j-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Necessary condition for a special polynomial of degree $5$ to have a special quadratic factor I am trying to solve the problem no. 41(i) on page $33$, Higher Algebra by Barnard and Child.
The problem statement is:
If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$, prove that
$$(a^2-c^2)(a^2-c^2+bc)=a^2b^2.$$
By ... | Denote $f_1(t) = at^4 - 3at^2 - bt + a$, $f_2(t) = at^3 - 2at - (b - c)$, then $f_1(p) = f_2(p) = 0$. Therefore, defining\begin{align*}
f_3(t) &= t f_2(t) - f_1(t) = at^2 + ct - a,\\
f_4(t) &= a(tf_3(t) - f_2(t)) - c f_3(t) = (a^2 - c^2)t + ab,\\
f_5(t) &= (a^2 - c^2)^2 f_3(t) - a (f_4(t))^2 + 2a^2b f_4(t) - c(a^2 - c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving the system $5(\sin x + \sin y) = 1$ and $5(\sin 2x + \sin 2y) = 1$ To find the general solution $(x,y)$ satisfying the system of equations
\begin{align} 5(\sin x + \sin y) &= 1 \\ 5(\sin 2x + \sin 2y) &= 1
\end{align}
I applied $\sin C + \sin D$ rule and then divided these two equations, then I am stuck at
$$\c... | First,$$
\begin{cases}
\sin x + \sin y = \dfrac{1}{5}\\
\sin 2x + \sin 2y = \dfrac{1}{5}
\end{cases} \Longrightarrow \begin{cases}
\sin \dfrac{x + y}{2} \cos \dfrac{x - y}{2} = \dfrac{1}{10} & (1)\\
\sin(x + y) \cos(x - y) = \dfrac{1}{10} & (2)
\end{cases}.
$$
Denote $u = \cos \dfrac{x - y}{2}$, $v = \sin \dfrac{x + y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
trigonometric limit with integral: $\lim_{\alpha\to 0}\int^{\alpha}_{0}\frac{dx}{\sqrt{\cos x -\cos \alpha}}$ I've got following limit to calulate:
$$\lim_{\alpha\to 0}\int^{\alpha}_{0}\frac{dx}{\sqrt{\cos x -\cos \alpha}}$$
Is there any way to bound it by some "more convenient" limit?
| Assume $a>0$ and rewrite
$$
\cos x-\cos a=2\sin\Big(\frac{a+x}{2}\Big)\sin\Big(\frac{a-x}{2}\Big).
$$
Estimate: $t-\frac{t^3}{6}\le\sin t\le t$, $t\ge 0$ gives for $0\le x\le a$
$$
\underbrace{\frac{a^2-x^2}{2}\Big(1-\frac{(a+x)^2}{24}\Big)\Big(1-\frac{(a-x)^2}{24}\Big)}_{\ge\ \frac{a^2-x^2}{2}\big(1-\frac{a^2}{6}\big)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove the theorem of Nicomachus by induction Prove the theorem of Nicomachus(AD.100) by induction:
$$
1^3 = 1,\ 2^3 = 3+ 5,\ 3^3 = 7 + 9 + 11,\ 4^3 = 13 + 15 + 17 + 19,\ ...
$$
My approach: from looking at the above pattern you can tell there is something of the following sort:
$$ 2^{n-1} + q = n ^3,$$
where $q$ ... | If induction is not mandatory,
let the $r$th term$(T_r)$ of $1,3,7,13$ be $$a(r-1)(r-2)+b(r-1)+c$$
$r=1\implies 1=c$
$r=2\implies3=b+c\iff b=2$
$r=3\implies7=2a+2b+c\iff a=1$
$$T_r=(r-1)(r-2)+2(r-1)+1=r^2-r+1$$
Sum of $n$ term $$\sum_{m=\frac{n^2-n+1}2-1}^{\frac{n^2+n-1}2-1}(2m+1)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to factor $16x^2-y^2+8y-16$ so I get $(4x+y-4)(4x-y+4)$? I tried completing the square, making it equal to zero, and trying to think of it as a circle, but I still can't prove that these two are equivalent. I tried to factor this in Wolfram Alpha, and it spits out that answer. But I don't know how else to get to it... | I f you want to be systematic, add in a variable $z$ to make this homogeneous,
$$ 16 x^2 - y^2 - 16 z^2 + 8 y z $$
with result (below)
$$ 16 x^2 - y^2 - 16 z^2 + 8 y z = 16 x^2 - (y-4z)^2 = (4x+y-4z)(4x-y+4z) $$
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-book... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3... | Observe that $$(X+1/X)^2=a+2$$. This gives you $X+1/X$.
Now note that $$(X+1/X)^3=X^3+1/X^3+3(X+1/X)$$ and obtain $X^3+1/X^3$. Finally $$(X^2+1/X^2)(X^3+1/X^3)=X^5+1/X^5+(X+1/X)$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
} |
Solutions of $x^4+2tx^2+1=0$
I am looking for all the solutions of $x^4+2tx^2+1=0$
Is there any "quick" way to figure out all the complex-solutions $\pi_i(t)$ where $i=1,...4$ ?
I got $\pi_1(t)=\sqrt{-t+\sqrt{t^2-1}}\sqrt{t-\sqrt{t^2-1}}\sqrt{-t+\sqrt{1-t^2}}$ but apparently I am too "stupid" to verify my calculati... | $$x^4+2tx^2+1=0$$
Substitute $y=x^2$
$$y^2+2ty+1=0$$
Now use the formula $$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
So, here $a=1,b=2t,c=1$
$$y=\frac{-t\pm \sqrt{4t^2-4(1)(1)}}{2(1)}$$
$$y=\frac{-t\pm \sqrt{4t^2-4}}{2}$$
$$y=-t\pm \sqrt{t^2-1}$$
Substitute back $y=x^2$
$$x^2=-t\pm \sqrt{t^2-1}$$
$$x=\pm\sqrt{-t\pm \sqrt{t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
2013 AIME I #5: Funky Real Root The real root of the equation $8x^3-3x^2-3x-1=0$ can be written in the form $\frac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$.
I noticed that there was a $3x^2-3x-1$ present and I decided to add $(x+1)^3$ to get rid of this. When I did thi... | What I will do for most cubic equations:
Step 1:
If the equation $ax^3+bx^2+cx+d=0$ has $b=0$ then skip this step. Otherwise, let $y=x+\dfrac{b}{3a}$.
In this case:
$$8x^3-3x^2-3x-1=0$$
Let $y=x+\dfrac{b}{3a}=x-\dfrac{1}{8}\Rightarrow x=y+\dfrac{1}{8}$.
\begin{equation}\begin{aligned}
8x^3-3x^2-3x-1=0 &\Leftrightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
[Calc II]: Shell Method - Dealing with offsets So I'm told to use the shell method for a triangle bounded by $(1,1)$, $(1,2)$ and $(2,1)$. The function values, I've found, are then $y = 3 - x$, $y = 1$ and $x = 1$. I'm rotating it around the $y$-axis. I'm having trouble finding my values for $h(x)$ and $ρ(x)$ though. T... | Since we are rotating around the $y$-axis, the method of shells says that the volume is
$$ V = \int_a^b A(x)\, dx $$
where $A(x)$ is the cross-sectional area given by
$$ A(x) = 2\pi\left(\textrm{radius}\right)\left(\textrm{height}\right). $$
Since we are rotating around $x=0$ and the triangle is on the first quadrant,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Partial fraction of $\frac{s}{(s^2+2s+2)(s^2-2s+2)}$ I am trying to find the partial fraction of:
$$\frac{s}{(s^2+2s+2)(s^2-2s+2)}$$
I started off with:
$$\frac{A(s^2-2s+2)}{s^2+2s+2} + \frac{B(s^2+2s+2)}{s^2-2s+2}$$
After that I get the following equations:
$A+B = 0$; $-2A+2B =1$
Giving: $A=-B$, and secondly $4B=1$, ... | Since the factors in the denominator can't be factored over in the real numbers further, the best you can do is$$\frac {s}{(s^2+2s+2)(s^2-2s+2)}=\frac {As+B}{s^2+2s+2}+\frac {Cs+D}{s^2-2s+2}$$
Multiply both sides by the denominator to clear the fractions to get the equation$$s=(As+B)(s^2-2s+2)+(Cs+D)(s^2+2s+2)$$
Now le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Can these definite integrals be expressed in closed form? In my research, I recently came across the integral
$$
I(a,b) = \int_{\theta_1}^{\theta_2} \frac{\sin^2 \theta}{(a \sin \theta - b \cos \theta)^2}\,\mathrm{d}\theta,
$$
where the range of integration $[\theta_1, \theta_2]$ is defined slightly differently in the ... | $a\sin\theta + b\cos\theta = \sqrt {a^2+b^2}\cos (\theta -\arctan \frac {a}{b})$
$\frac {1}{a^2+b^2} \int \sin^2\theta\sec^2(\theta - \arctan \frac {a}{b}) \ d\theta\\
\phi = \theta - \arctan \frac {a}{b}\\
\frac {1}{a^2+b^2} \int \sin^2(\phi+\arctan \frac {a}{b})\sec^2\phi \ d\theta\\
\frac {1}{a^2+b^2} \int (\sin\phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding $\lim\limits_{n→∞}\left(\frac{\sqrt{n^2+n}-1}n\right)^{2\sqrt{n^2+n}-1}$ I have the limit
$$\lim_{n\to\infty}\left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}.$$
I simplified the limit by using $(f(x)-1)\cdot g(x)$, where $f(x)= \sqrt{n^2+n}-1$ and $g(x) = 2\sqrt{n^2+n}- 1$ since it is in the form of $1^... | First of all, note that the $-1$ in the exponent is irrelevant, i.e.,
$$\lim_{n\to\infty}\left(\sqrt{n^2+n}-1\over n \right)^{2\sqrt{n^2+n}-1}
=\lim_{n\to\infty}\left(\left(\sqrt{n^2+n}-1\over n \right)^{2\sqrt{n^2+n}}\over\sqrt{1+{1\over n}}-{1\over n}\right)
=\lim_{n\to\infty}\left(\sqrt{n^2+n}-1\over n \right)^{2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove identity: $\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $ $$\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $$
I am having a problem proving this identity. I write tan like $\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}$ and the squared one in the same way.
I ... | In your last formula
*
*Your nominator is $\sin x$ due to $\sin 2x = 2\sin x \cos x$, written for $\frac{x}{2}$.
*Your denominator is $1$, due to $\sin^2 x + \cos^2 x = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Number of $10$ digit numbers such that every digit that appears appears exactly twice I need to find the number of $10$-digit numbers in which every digit that appears appears exactly twice.
I have attempted to solve this but am not sure if I am not overcounting something.
The only problem is that a $0$ cannot stand o... | Since the leading digit cannot be $0$, there are $9$ ways to pick the leading digit. There are $9$ ways to pick the other location for the leading digit. There are $\binom{9}{4}$ ways to select which four of the other nine digits appear in the number. There are $\binom{8}{2}$ ways to select the locations of the smal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Characteristic Polynom of a Matrix for the complete Graph, I need to derive a formula for the characteristic polynom of a complete graph. Here are some example of the adjacency Matrices:
$$
K_2=
\begin{pmatrix}{}
0 & 1 \\
1 & 0 \\
\end{pmatrix}
$$
$$
\chi(K_2)=(-1 + x) (1 + x)
$$
$$
K_3=
\begin{pmatrix}
0 & 1 & 1 \\... | If you add the identity then you get a rank-one matrix whose characteristic polynomial is $$\chi(K_n + I) = (-1)^n x^{n-1} (x - n).$$ The factor $(x-n)$ coming from the lone eigenvalue $n$ with eigenvector $(1,1,...1).$ When passing from $K_n+I$ to $K_n$ you substitute $x \mapsto x+1$ and get the formula you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use sum and product of roots to prove type of roots. Suppose $α$ and $β$ are the roots of $x^2-px+q$. Then
$(a)$ Find the quadratic equation whose roots are $α^2+β^{-2}$ and $β^2+α^{-2}$.
$(b)$ Prove that if $(p,q) \in \mathbb{R}$ then this equation will have equal roots if $p = 0$ and $p^2=4q$.
I have tried to solve p... | For part a we have that
$(a)$ Find the quadratic equation whose roots are $α^2+β^{-2}$ and $β^2+α^{-2}$
*
*$p'=α^2+β^{-2}+β^2+α^{-2}=α^2+β^2+\frac{α^2+β^2}{α^2\beta^2}=p^2-2q+\frac{p^2-2q}{q^2}$
*$q'=(α^2+β^{-2})(β^2+α^{-2})=α^2β^2+2+\frac{1}{α^2\beta^2}=q^2+2+\frac1{q^2}$
then
$$x^2-(p^2-2q+\frac{p^2-2q}{q})x+q^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\sum_n(2n - \sqrt{n^2+1}-\sqrt{n^2-1})$? How to compute $\sum_n (2n - \sqrt{n^2+1}-\sqrt{n^2-1})$?
I tried two ways:
1. \begin{align*}
(2n - \sqrt{n^2+1}-\sqrt{n^2-1})
&= n - \sqrt{n^2+1} + n -\sqrt{n^2-1} \\
&= \frac{1}{n+\sqrt{n^2-1}}-\frac{1}{n-\sqrt{n^2+1}},
\end{align*}
but I don't know how to do l... | Let us assume that you need to compute the infinite summation with
$$a_n=2n - \sqrt{n^2+1}-\sqrt{n^2-1}$$ For large values of $n$, rewrite $$a_n=n\left(2- \sqrt{1+\frac 1{n^2}}- \sqrt{1-\frac 1{n^2}}\right)$$ and use the binomial expansion or Taylor series to get
$$a_n=\frac{1}{4 n^3}+\frac{5}{64 n^7}+O\left(\frac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric version of second difference I am reading an article
Hall P, Turlach BA. On the estimation of a convex set with corners. IEEE Transactions on Pattern Analysis and Machine Intelligence. 1999 Mar;21(3):225-34
and stuck at an identity used there (p. 228). It can be written as follows:
$$
p(\theta_i)\sin(\the... | $\def\peq{\mathrel{\phantom{=}}{}}$Since $p'''$ is bounded, then\begin{align*}
p(θ_i) &= p(θ_j) + (θ_i - θ_j) p'(θ_j) + \frac{1}{2} (θ_i - θ_j)^2 p''(θ_j) + O((θ_i - θ_j)^3), \\
p(θ_k) &= p(θ_j) + (θ_k - θ_j) p'(θ_j) + \frac{1}{2} (θ_k - θ_j)^2 p''(θ_j) + O((θ_k - θ_j)^3).
\end{align*}
Also,\begin{align*}
\sin(θ_i - θ_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Division of $f(x):= x^6+x^4+x^3-x^2-1$ by $g(x):=x^4+x^3-1$ seems not so correct Let's say $f(x):= x^6+x^4+x^3-x^2-1$ and $g(x):=x^4+x^3-1$. Given that degree $(f(x))$ $>$ degree $(g(x))$, we can divide $f$ by $g$. The issue here is, continuous division of polynomial (starting from the higher powers) we have learnt in ... | The division is correct indeed we have that
$$(x^4+x^3-1)(x^2-x+2)=x^6-x^5+2x^4+x^5-x^4+2x^3-x^2+x-2=$$
$$=x^6+x^4+2x^3-x^2+x-2=(x^6+x^4+x^3-x^2-1)+(x^3+x-1)$$
and therefore
$$x^6+x^4+x^3-x^2-1=(x^4+x^3-1)(x^2-x+2)+(-x^3-x+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Short notation for a column vector where each element is the mean of each row of a given matrix I have a matrix $A$ defined as:
$$
A =
\begin{bmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,m} \\
a_{2,1} & a_{2,2} & \ldots & a_{1,m} \\
\vdots & \vdots & \ddots & a_{1,m} \\
a_{n,1} & a_{n,2} & \ldots & a_{n,m}
\end{bmatrix}... | I think
$$b = \dfrac{1}{m}\begin{bmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,m} \\
a_{2,1} & a_{2,2} & \ldots & a_{1,m} \\
\vdots & \vdots & \ddots & a_{1,m} \\
a_{n,1} & a_{n,2} & \ldots & a_{n,m}
\end{bmatrix} \underbrace{\begin{bmatrix} 1\\1\\ \vdots \\ 1\\1\end{bmatrix}}_{m\times1}=\dfrac{1}{m}A\,\vec{1}$$
is the mo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to compute integral on calculator? So i have this integral tending from 2(on top) to 1(on bottom ) of
$$\int_1^2x\ln(x^2 +3)x\,dx$$
If I turn it into $x^2\ln(x^2 +3)\,dx$, with integration by parts I get
$$\frac{16}{3}\,\ln(2) + \frac{49}{9}.$$
I was wondering, how is it possible that my TI nspire CAS does NOT reco... | By parts
$$\begin{align}\int x^2\log(x^2+3)\,dx&=\frac{x^3}3\log(x^2+3)-\int\frac{2x^4}{3(x^2+3)}dx\\
&=\frac{x^3}3\log(x^2+3)-\frac23\int\left(x^2-3+\frac9{x^2+3}\right)dx\\
&=\frac{x^3}3\log(x^2+3)-\frac{2x^3}{9}+2x-6\int\frac1{(\sqrt3t)^2+3}d(\sqrt3t)\\
&=\frac{x^3}3\log(x^2+3)-\frac{2x^3}{9}+2x-\frac63\sqrt3\arctan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A geometric inequality for a triangle ABC I have to prove that:
$ \frac {a^2}{w_a^2} + \frac {b^2}{ w_b^2} +\frac {c^2}{ w_c^2} \ge 4,$
for the sides $a,b,c$ of the triangle $ABC,$ $ w_a, w_b, w_c $ the angle bisectors and $s$ its semiperemeter. We have for the angle bisectors of the triangle $ABC$:
$ w_a = \frac {2}{b... | Let $m_a$, $m_b$ and $m_c$ be medians of the triangle to sides $a$, $b$ and $c$ respectively.
Thus, by C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{a^2}{w_a^2}\geq\sum_{cyc}\frac{a^2}{m_a^2}=4\sum_{cyc}\frac{a^2}{2b^2+2c^2-a^2}=4\sum_{cyc}\frac{a^4}{2a^2b^2+2a^2c^2-a^4}\geq$$
$$\geq\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the acute angle between the surfaces $xy^2z = 3x + z^2 $ and $3x^2-y^2+2z=1$ at the point $(1,-2,1)$ Find the acute angle between the surfaces $xy^2z = 3x + z^2 $ and $3x^2-y^2+2z=1$ at the point $(1,-2,1)$
Angle between curves at a point is given by the angle between their tangent planes at the point.
$$f(x,y,z):... | Did you go ahead and take arccos$\left(\frac{-3}{7\sqrt{6}}\right)$? I get about 100$^o$. So your answer is no. The acute angle is 180 - 100 or about 80$^o$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that the substitution $x=X-1$ and $y=Y+3$ turns $\frac{dy}{dx}=...$ into a homogenous equation. $$\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$$
After substitution of the given variable I get $$\frac{dY}{dX} = \frac{4-\frac{Y}{X}}{2+\frac{Y}{X}}$$ which seems to give a homogenous equation. (There is no answer given for th... | Note that LHS integral should give
$$ \int \frac{2+v}{(v+4)(1-v)}dv = -\frac35 \ln(1-v) - \frac25 \ln(4+v) $$
After fixing the sign error, you can transform your answer to match the one given.
$$ 3\ln\left(1-\frac{y-3}{x+1}\right) + 2\ln\left(4+\frac{y-3}{x+1}\right) + 5\ln(x+1) = -5c $$
$$ 3\ln \left(\frac{x-y+4}{x+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2827007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to get Jordan Form above the diagonal and not below it? Let $ M=
\begin{bmatrix}
0 & 0 & 0 \\
1 & 1 &1 \\
-1 & -1 & -1\\
\end{bmatrix}$ $\in M_{3}(\mathbb R^{3})$
I found that $\chi_{M}=X^{3}$ and $\mu_{M}=X^{2}$, since $\dim_{\mathbb R}\operatorname{Im}(M-0\cdot\operatorname{Id})=1$
So $$\ker(M-0\c... | You must construct the Jordan basis more carefully.
You already determined that the Jordan form consists of one $2 \times 2$ block and one $1 \times 1$ block, i.e. $\dim \ker M = 2$ and $\dim \ker M^2 = 3$.
Write a table like this $$
\begin{array}{ccc}
\ker M^2 \,\dot-\,\ker M^1 & f_2\\
\ker M^1 \,\dot-\,\ker M^0 & f_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2827100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Expected number of correct guess in a game Question is as follow.
There is a bag. In the bag, there are $a$ red cubes and $b$ blue cubes. Assume that she knows exactly how many cubes for each of the colors before the draw. Mary is going to draw all the cubes one by one out of the bag randomly. For every turn, she will... | We find the following recurrence:
$$E(a,b) = \begin{cases}
\frac{bE(a,b-1) + a(E(a-1,b)+1)}{a+b}&\text{if } a\ge b,\\
E(b,a)&\text{if }b>a,\\
0&\text{if }a=b=0.
\end{cases}$$
These cases are pretty obvious: the first says "we gain one point if we guess right", the second says "we can pick the cube that is more common a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Problem related to quadratic equation If $x$ is a real number then what is the greatest value of
$$ f(x)=2(a-x)\left(x+\sqrt{x^2+b^2}\right)\;\;? $$
I cannot figure out how to convert this into a quadratic equation to find out the greatest value of $f(x).$
From what I've read, the answer is given by $a^2+b^2$.
| let $\sqrt{x^2+b^2}+x=y,$ then $\displaystyle \sqrt{x^2+b^2}-x = \frac{b^2}{y}$ and $\displaystyle 2x=y-\frac{b^2}{y}$
$\displaystyle f(x) = (2a-2x)(x+\sqrt{b^2+y^2}) = (2a-y-\frac{b^2}{y})\cdot y = (2ay-y^2+b^2) $
$f(y)= (a^2+b^2)-(y-b)^2\leq (a^2+b^2)$ equality hold when $y=b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Proof that squares are divisible by 3 when their sum is In this proof, they write $3|a^2+b^2 \implies 3|a$, $3|b$. I tried using the same proof used to prove $3|a^2 \implies 3|a$, where $3$ being prime and writing $a^2 = a\cdot a$ suggests that $a$ is divisible by $3$. I'm not sure how to prove the $3|a^2+b^2$ case, th... | In $\mathbb{Z}/3$, $a^2=1$, or $a^2=0$, $b^2=1$ or $0$ implies that $a^2+b^2=0$ if and only if $a^2=b^2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Grid walk with restrictions on some points
Given a $10\times10$ grid (each axis enumeration is $0,1,2,..,9$) a grid walk is started at point $(0,0)$ and finished at point $(9,9)$. Each move can be either to the right or up. How many paths are possible if points $(3,3), (6,3), (6,6)$ must not be crossed?
I know that ... | If a path passes through $(3,3)$ and then $(6,3)$, it must travel only horizontally from $(3,3)$ to $(6,3)$. This gives us
$$\binom{3+3}{3}\cdot 1\cdot \binom{9+9-6-3}{9-3}$$
total paths, where the first binomial coefficient is the number of walks from $(0,0)$ to $(3,3)$, the $1$ is the number of walks from $(3,3)$ to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\pi = 48\tan^{-1}\frac{1}{18} + 32 \tan^{-1}\frac{1}{57} - 20\tan^{-1}\frac{1}{239}$ The below equation represents $\pi$ to some decimals using tangent inverse. I need to prove that the left hand side of the equation equals the right hand side.
$$
\pi = 48\tan^{-1}\frac{1}{18} + 32 \tan^{-1}\frac{1}{57} - 20\t... | To prove given formula, we'll use identity (see here)
$$
\arctan\frac{a_1}{b_1} + \arctan\frac{a_2}{b_2} = \arctan\frac{a_1 b_2+a_2b_1}{b_1b_2-a_1a_2}.
\tag{1}
$$
Denote $$\arctan\frac{1}{b} = f(b).\tag{2}$$
Then given formula will have form
$$
f(1) = 12 f(18) + 8 f(57) - 5 f(239).\tag{3}
$$
The proof will be based on ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Use the Gauss elimination method in order to find $u$ in $\mathbb{R}^2$ such that $T(u)=\left(\begin{matrix} 3 & 1& 4 \end{matrix}\right)^T$ Consider the linear map $T:\mathbb{R}^2->\mathbb{R}^3$ with
$$T\left(\begin{matrix} x \\ y \end{matrix}\right)=\left(\begin{matrix} x-2y \\ 2x+y \\ -5x+8y \end{matrix}\right)$$ ... | You just what to eliminate in the first two columns.
You are fine through here.
$\begin{array}{cc|c} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & -2 & 19 \end{array}$
But then then you hit the problem. The vector you have been asked to find is not in the column space of the matrix. Your problem has no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ I am just wondering how to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ (This one is taken from Wo... | We need to prove that
$$(\sqrt[4]3-1)^2(5+3\sqrt3+2\sqrt[4]3(2+\sqrt3))=2$$ or
$$2(5+3\sqrt3+2\sqrt[4]3(2+\sqrt3))=(1+\sqrt[4]3+\sqrt3+\sqrt[4]{27})^2,$$
which is obvious after full expanding.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Express the following complex numbers in the form of $a+bi$ I'm struggling with the following exercises:
I tried to use the reasoning as follows:
$$(a+bi)^n=(re^{\theta i})^n=r^ne^{\theta in}=r^n(\cos(\theta n)+i\sin(\theta n))$$
So for the first one I did:
$$2^{1/6}(\cos(-\frac{\pi}{3} \cdot \frac{1}{6})+i\sin(-\fra... | We can calculate the norm of $z$ in this way:
$|z|^6=|z^6|=|1-\sqrt{3}i|=2$ so you have that
$|z|= 2^\frac{1}{6}$
Now you can define $y:= \frac{z}{2^\frac{1}{6}}$ and then
*
*$|y|=1$
*$y^6=\frac{1}{2}-\frac{\sqrt{3}}{2}i$
You can write $y$ in trigonometric form:
$y=cos(\theta)+i sin(\theta)$
and so
$\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solutions of $x^2 -xy + y^2 \equiv 0 \pmod{43}$
$x^2 -xy + y^2 \equiv 0 \pmod{43}$
We could transform this into
$\frac{x^3 + y^3}{x+y} \equiv 0 \pmod{43}$
and thus
$x^3 + y^3 \equiv 0 \pmod{43}$
From what I could manage so far, I know that if $x = 3k$ , $y= 3k+1$, as $43 = 3k+1$.
Also there are no solutions for $... | We have $x \equiv -y$ iff $x \equiv 0$ iff $y \equiv 0$.
Otherwise, as you have argued, $x^2 - xy + y^2 \equiv 0$ iff $x^3 + y^3 \equiv 0$ and this happens iff $(xy^{-1})^3 \equiv -1$.
So the solutions are $x=yu$, where $u$ is an element of order $6$ mod $43$, that is, $u=7$ or $u=37$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find solution to this ODE with powerseries I am working on solving this ODE
$$ (1+x^2)y'' + xy'-y =0 $$
with $1. y(0)=0,y'(0)=1 \\
2. y(0)=1, y'(0)=1$
with $$ y= \sum_{n=0}^{ \infty} a_n x^n , y'= \sum_{n=1}^{ \infty}na_n x^{n-1}, y''=\sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}$$
plugged into the equation fol... | For (1), you should get $a_n = 0$ for all $n \ge 2$, i.e. $y(x) = x$.
For (2), the solution turns out to be $y(x) = x + \sqrt{1+x^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$xy\frac{dy}{dx} = \sqrt{ x^2- y^2 - x^2y^2 -1}$ Solve the differential equation -
$$xy\frac{dy}{dx} = \sqrt{ x^2- y^2 - x^2y^2 -1}$$
I tried various methods such as putting $y=vx$, squarring both sides and then manipulating the expression but couldnt solve it. Any hint will be helpful. Thanks
| $$xy\frac{dy}{dx} = \sqrt{ x^2- y^2 - x^2y^2 -1}$$
Let $X=x^2$ and $Y=y^2$
$$X\frac{dY}{dX} = \sqrt{ X- Y - XY -1}$$
Let $u(X)=\sqrt{ X- Y - XY -1}$
$$Y=\frac{X-1-u^2}{X+1}$$
$$X\;Y' =X\left(\frac{1}{X+1}-\frac{2u\,u'}{X+1}-\frac{X}{(X+1)^2}+\frac{1}{(X+1)^2} +\frac{u^2}{(X+1)^2}\right)=u$$
$$u\,u'+\frac{X+1}{2X}u-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
integral $\int_{0}^{1}\left( \left\lfloor{\frac{2}{x}} \right\rfloor-2 \left\lfloor{\frac{1}{x}} \right\rfloor \right)dx$ Evaluate $$\int_{0}^{1}\left( \left\lfloor{\frac{2}{x}} \right\rfloor-2 \left\lfloor{\frac{1}{x}} \right\rfloor \right)dx$$
My Attempt
$$I_{1}=\int_{0}^{1}\left\lfloor{\frac{2}{x}} \right\rfloor dx$... | $$\begin{align*}\int_0^1 \left( \left\lfloor \dfrac{2}{x} \right\rfloor - 2\left\lfloor \dfrac{1}{x} \right\rfloor \right)dx & = \sum_{n=1}^\infty \left[\int_{\tfrac{1}{n+1}}^{\tfrac{2}{2n+1}}\left( \left\lfloor \dfrac{2}{x} \right\rfloor - 2\left\lfloor \dfrac{1}{x} \right\rfloor \right)dx + \int_{\tfrac{2}{2n+1}}^{\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find all $z$ for which $z^2+2z+2$ is real positive. Find all $z$ for which $z^2+2z+2$ is real positive.
My Attempt :
Let $z=x+iy, (x,y\in R)$:
$$z^2+2z+2= (x^2-y^2+2ixy) +2(x+iy)+2$$
$$=(x^2-y^2+2x+2)+i(2xy+2y)$$
| You want the real part to be positive and the imaginary part to be $0$
$$ (x^2-y^2+2x+2)+i(2xy+2y)$$
$$2xy+2y=0 \implies 2y(x+1)=0$$
Thus $y=0$ or $x=-1$
For $y=0$ we get $x^2+2x+2=(x+1)^2+1 >0$
For $ x=-1 $ we need $1-y^2>0$ which implies $|y|<1$
Thus the solution is either $ y=0$ or ($|y|<1$ and $x=-1$ )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 5
} |
If $\sin x +\sin 2x + \sin 3x = \sin y\:$ and $\:\cos x + \cos 2x + \cos 3x =\cos y$, If $\sin x +\sin 2x + \sin 3x = \sin y\:$ and $\:\cos x + \cos 2x + \cos 3x =\cos y$, then $x$ is equal to
(a) $y$
(b) $y/2$
(c) $2y$
(d) $y/6$
I expanded the first equation to reach $2\sin x(2+\cos x-2\sin x)= \sin y$, but I doubt i... | If $z = e^{ix} = \cos x + i \sin x$
Then we have $z + z^2 + z^3 = e^{iy} = w$ (say)
Divide by $z^2$ to see that
$$z + \frac{1}{z} + 1 = \frac{w}{z^2}$$
The left side is real and thus
$$w = az^2$$
Since $|w| = |z| = 1$ we must have that $|a| = 1$
Thus $$w = \pm z^2$$
This gives rise to two equations:
$$z + z^3 = 0$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
$\lim_{n \rightarrow \infty} ( 1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})$ Evaluate
$\lim_{n \rightarrow \infty} \left( 1 - \frac{2}{2\cdot3}\right) \left( 1 - \frac{2}{3\cdot4}\right)\ldots\left( 1 - \frac{2}{(n+1)(n+2)}\right)$
My attempts : i know that $1 - \frac {2}{k(k+1)} = \fr... | Hint:
\begin{eqnarray*}
\frac{ 1\times 4}{ 3\times 2}\times \frac{ 2\times 5}{ 4\times 3}\times \frac{ 3 \times 6}{ 5 \times 4}\times \cdots
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Are $1+p^3+p^6$ and $1+p^4+p^8$ coprime? What are the primes $p$ for which $1+p^3+p^6$ and $1+p^4+p^8$ are coprime?
I know it is true for $p=2$ and $p=3$ and not true for any $p \equiv 1 \mod 6$. I conjecture that it true for all primes $p \equiv 5 \mod 6$.
Any counterexample $> 10^8$.
This is relevant to OEIS sequenc... | Let $p$ be an integer.
Suppose $\gcd(1+p^3+p^6,1+p^4+p^8) = u > 1$.
\begin{align*}
\text{Then}\;\;&1+p^3+p^6\equiv 0\;(\text{mod}\;u)\\[4pt]
\implies\;&(p^3-1)(p^6+p^3+1)\equiv 0\;(\text{mod}\;u)\\[4pt]
\implies\;&p^9-1\equiv 0\;(\text{mod}\;u)\\[4pt]
\implies\;&p^9\equiv 1\;(\text{mod}\;u)\\[10pt]
\text{Similarly}\;\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
How do you evaluate limit of $\frac{\sqrt{a+x} - \sqrt a}{x\sqrt a(\sqrt{a+x})}$ when $x$ tends to $0$? I tried rationalization method where we multiply both the numerator and denominator with appropriate opposite factor of numerator. But I could only get $\frac{1}{\sqrt {a^2+ax}(2 \sqrt {a + \sqrt x)}}$. But the final... | Easier solution: Write $y=\sqrt{x+a}$ and $b=\sqrt{a}$, then
\begin{align}\lim_{x \to 0}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a}\sqrt{a+x}} &= \lim_{y \to b}\frac{y-b}{by(y^2-b^2)}\\&=\lim_{y \to b}\frac{1}{by(y+b)}\\
&= \frac{1}{2b^3}=\frac{1}{2a\sqrt{a}}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the sum of all numbers greater than $10000$ formed by using all digits $0,1,2,4,5$ Find the sum of all numbers greater than $10,000$ formed by using all digits $0,1,2,4,5$ and no digit being repeated in any number.
I could find that the number of such numbers are $96$ but I could not find their sum.
| Let's keep placeholders for where we will keep the digits :
[] [] [] [] []
Now, in the first box i.e: ten thousands place, we can put any number in the following set $ \{1,2,4,5 \}$ and after that have 4! ways of filling in the remaining numbers. So, there are 4! numbers with 1 as first number, 4! numbers with 2 as fir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Range of $a$ in Trigonometric equation
If the inequality $\sin^2 x+a\cos x+a^2>1+\cos x$ hold for any $x\in \mathbb{R}.$ Then range of $a$ is
Try: $1-\cos^2 x+a\cos x+a^2-1-\cos x>0$
$$\cos^2 x+(1-a)\cos x-a^2<0$$
$$4\cos^2 x+4(1-a)\cos x-4a^2<0$$
$$\bigg(2\cos x+(1-a)\bigg)^2-4a^2-(1-a)^2<0$$
Could some help me how... | Given task allows suitable algebraic approach, because it is equivalent to the system of
\begin{cases}
\cos x =t\\
t\in[-1,1]\\
t^2+(1-a)t-a^2<0,\tag1
\end{cases}
wherein the parabola $P(t,a) = t^2+(1-a)t-a^2$ has the positive discriminant
$$D=(1-a)^2+4a^2.\tag2$$
The roots of $P(t,a)$ are
$$t_1 = \dfrac{a-1 - \sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What is the probability that exactly one box remains empty Five identical(indistinguishable) balls are to be randomly distributed into $4$ distinct boxes. What is the probability that exactly one box remains empty?
we can choose the empty box in $4$ ways.
We should fit the $5$ balls now into $3$ distinct boxes in $^7C... | I couldn't see how the previous answer came up with a solution so I did it differently.
$0$ boxes empty has $1,1,1,2$ box contents times $\frac{5!}{2!}$ ball arrangements $= \frac{4!}{3!}\cdot \frac{5!}{2!} = 240$
$1$ box empty has $1,1,3$ and $1,2,2$ box contents and $\frac{5!}{3!}$ and $\frac{5!}{2!\cdot 2!}$ ball ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$ is strictly increasing I want to show that the function $f:\mathbb N \to \mathbb Q$ defined by$$f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$$
is strictly increasing. Could I have a hint on how to go about this? I have tried playing around with $f(n+1)-f(n)$ but it's all so messy ... | $$
\begin{align}
&\sum_{k=1}^{n+1}\frac{n+1}{k^2+(n+1)^2}-\sum_{k=1}^n\frac{n}{k^2+n^2}\tag1\\
&=\frac1{2n+2}+\sum_{k=1}^n\left(\frac{n+1}{k^2+(n+1)^2}-\frac{n}{k^2+n^2}\right)\tag2\\
&=\frac1{2n+2}-\sum_{k=1}^n\frac{n(n+1)-k^2}{\left(k^2+(n+1)^2\right)\left(k^2+n^2\right)}\tag3\\
&=\frac1{2n+2}-\frac1{n(n+1)}\sum_{k=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Spherical coordinates in surface integrals I am stuck on the following problem
Evaluate :
$$I=\iint x^2 y^2 z dS $$
where S is the positive side of lower half of the sphere $x^2 + y^2 + z^2 = a^2$
I tried using spherical coordinates and their jacobians but cannot seem to find the answer which is $$I= \frac {2\pi a^7} ... | Substitute $z=-\sqrt{a^2-x^2-y^2}$ and $dS=\sqrt{1+z_x^2+z_y^2}dA$. which turn $$I=-a\iint_{R} x^2y^2dA.$$ Use polar co-ordinate.$x=r\cos\theta, y=r\sin \theta$. $dxdy=rdrd\theta$. $r$ varies from $0$ to $a$ and $\theta$ varies from $0$ to $2\pi$. Which turn $I=-a\int_{0}^{2\pi}\sin^2\theta \cos^2\theta d\theta \int_{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int\frac{(1+x) \lfloor(1-x+x^2)(1+x+x^2)+x^2\rfloor}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$ If $$F(x) = \int\frac{(1+x) \lfloor(1-x+x^2)(1+x+x^2)+x^2\rfloor}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$$then find the value of $\lfloor F(99)- F(3)\rfloor$
My attempt : I could only reduce the portion of the numerator within $\l... | After some factorisation, what we're left to figure out is:
$$ I= \left \lfloor \int_3^{99} \frac{\lfloor (1+x^2)^2 \rfloor}{(1+x)(1+x^2)^2} dx \right \rfloor $$
But first, for $g(x) = (1+x^2)^2$, let's study $ \lfloor g(x) \rfloor$ and
$ \frac{\lfloor g(x) \rfloor}{g(x)}$.
Notice that for some positive integer $n$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $2+2\cos(2\pi\theta) \leq 4\exp(-2\|\theta \|^2)$. Let $\theta \in \mathbb{R}$, and let $\|\theta\|$ denote the distance to the integer nearest to $\theta$. I want to prove $$2+2\cos(2\pi\theta) \leq 4\exp(-2\|\theta \|^2).$$ From the double angle formula we arrive at the equivalent $$|\cos(\pi \theta) | \leq... | In fact, because both sides of $|\cos\pi\theta|\le e^{-\Vert\theta\Vert^2}$ are symmetric around $\frac12$ ($f(\theta)=f(1-\theta)$), we only need to check $0\le e^{-\theta^2}-\cos\pi\theta$ for $\theta\in[0,1/2]$. This can be rewritten as
$$e^{-\theta^2}\ge\cos\pi\theta\tag1$$
The idea of the proof of $(1)$ is to find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determine all possible integers $x$ and $y$ s.t. $3x + 7y \equiv 14 \pmod{28}$ and $x + 3y \equiv 8 \pmod{28}$. Determine all possible ints $x$ and $y$ that satisfy $3x + 7y \equiv 14 \pmod {28}$ and $x + 3y \equiv 8 \pmod {28}$. The answer should be in terms of $x\equiv r$ and $y \equiv s$ where $r$ and $s$ are remain... | We have
\begin{eqnarray*}
3x+7y \equiv 14 \pmod{28} \\
x+3y \equiv 8 \pmod{28}.
\end{eqnarray*}
Multiply the second equation by $3$ and then subtract the first
\begin{eqnarray*}
3x+9y \equiv 24 \pmod{28} \\
2y \equiv 10 \pmod{28}.
\end{eqnarray*}
So
\begin{eqnarray*}
y &\equiv 5 \pmod{28} \\
x & \equiv 21 \pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Roots of $f$ and $f'$ for $1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$ I have this question from an admission exams. Given that $$f(x)=1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$$ find $S(f(x))-S(f'(x))$ where $S$ denotes the sum of the real roots for $f(x)$ respectively ... | We have
$$f(x) = a_{101}x^{101} + a_{100}x^{100} + \cdots$$
and we know that $a_{101} = -\frac1{101!}$. Vieta's formulas give
$$5151 = \text{ sum of roots of } f = -\frac{a_{100}}{a_{101}}$$
so $a_{100} = \frac{5151}{101!}$.
The derivative is
$$f'(x) = 101a_{101}x^{100} + 100a_{100}x^{99} + \cdots$$
so Vieta's formul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
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.