Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find $\sin A$ and $\cos A$ if $\tan A+\sec A=4 $ How to find $\sin A$ and $\cos A$ if
$$\tan A+\sec A=4 ?$$
I tried to find it by $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$, therefore
$$\tan A+\sec A=\frac{\sin A+1}{\cos A}=4,$$
which implies
$$\sin A+1=4\cos A.$$
Then what to do?
| Let $t=\tan \frac{A}{2}$ then $\tan A = \frac{2t}{1-t^2}$ and $\cos A = \frac{1-t^2}{1+t^2}$ so $$\begin{align*}\frac{1+t^2}{1-t^2} + \frac{2t}{1-t^2}&=\frac{(1+t)^2}{1-t^2} \\ & = \frac{(1+t)(1+t)}{(1-t)(1+t)} \\ & = \frac{1+t}{1-t} = 4\end{align*}$$
So, assuming $t\neq 1$ we get $t = \frac{3}{5}$. From this, you can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Subtraction of fractional exponents with different bases I wanna know if $$\left[(x^2+h)^3-(x+h)\right]^{1/2}-[x^3-x]^{1/2}=\left[(x+h)^3-(x+h)-x^3+x\right]^{1/2}.$$
| Does $a^2 - b^2 = (a-b)^2\;$?
No, Because $$(a-b)^2 = a^2 -2ab +b^2 \neq a^2 - b^2$$
Alternatively $$a^{1/2} - b^{1/2} = \sqrt a -\sqrt b \neq \sqrt{a-b}$$
Simply take $a = 4, b = 9$. Then we have $2 - 3 = -1$. Whereas $\sqrt{2-3}$ is undefined in the reals, and which will never be equal to a negative real number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some ... | One more way to prove it is by substituting out for $z$:
$$
x^2+y^2+z^2 = x^2+y^2+(1-x-y)^2=2x^2+2y^2+2xy-2x-2y+1
$$
Now, substitute $x=\hat x+a$ and $y=\hat y+b$. We get
$$
2\hat x^2+2\hat y^2+2\hat x\hat y+(4a+2b-2)\hat x+(2a+4b-2)\hat y+(2(a^2+ab+b^2-a-b)+1)
$$
We can eliminate the order 1 terms by letting $4a+2b-2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 15,
"answer_id": 6
} |
Reverse engineer the gamma question from a typo $\int_0^1(\ln1/x)^{1/3}$dx and the answer $\Gamma(4/3)$ Background: question 9.1.2.d in a newly published book Mathematics for Physical Science by Harris.
My question is what was the question supposed to be? Taking ln(x) for ln1 I'm doing something wrong or not guessing t... | Notice that:
$$\int_0^\infty \frac{3}{2}\left(-3/2u\right)^{\frac{1}{3}}e^{-u}\mathrm{du} = \int_0^\infty \frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}u^{\frac{1}{3}}e^{-u}\mathrm{du} = \\
\frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\int_0^\infty u^{\frac{4}{3}-1}e^{-u}\mathrm{du} =\frac{3}{2}\left(-\frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Let $x, y, z$ be different prime numbers with $x, y, z > 3$. Prove that if $x + z = 2y$, then $6 | (y - x)$. I have troubles to prove the following task:
Let $x, y, z$ be different prime numbers with $x, y, z > 3$. Prove that if $x + z = 2y$, then $6 | (y - x)$.
The only idea I have is that every prime number $> 3$ di... | $6| y-x$ if and only if $3|y-x$ and $2|y-x$.
1) $2|y-x$.
$y$ and $x$ are both primes larger than $2$ so $x$ and $y$ or both odd. So $y-x$ is even.
2) $3|y-x$.
$y > 3$ and $y$ is prime so $y \equiv \pm 1 \mod 3$. $x > 3$ and $x$ is prime so $x \equiv \pm 1 \mod 3$. and similarly $z \equiv \pm 1 \mod 3$.
Either $x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Use relations to show the value of $ C $ I am asked to show that $$ \lim_{n \to \infty} \frac{n!}{a_n} = \sqrt{2 \pi}, $$ where $$ a_n = \sqrt{n} \cdot \left (\frac{n}{e} \right)^n. $$
And I am only allowed to use the following relations $$ \sum_{k = 0}^{2n} \dbinom{2n}{k} \cdot 2^{-2n} = 1, \qquad \int_{- \infty}^{\in... | The following proof has two parts: in the first part I will prove that $n!\sim C\sqrt{n}\left(\frac{n}{e}\right)^n$, in the second part I will show that $C=\sqrt{2\pi}$. It is crucial to notice that
$$ m = \prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)\tag{1}$$
from which:
$$ n! = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $ab$ is a square number and $\gcd(a,b) = 1$, then $a$ and $b$ are square numbers.
Let $n, a$ and $b$ be positive integers such that $ab = n^2$. If $\gcd(a, b) = 1$, prove that there exist positive integers $c$ and $d$ such that $a = c^2$ and $b = d^2$
So far I have tried this:
Since $n^2 = ab$ we have that $n = \s... | Because $n$ is a positive integer, $n^2$ has an even number of factors, and each factor appears an even number of times in its factorization. The prime factorizations of $a$ and $b$ must together contain $2$ of each of all elements of the prime factorization of $n$. However, since $\gcd(a,b)=1$, $a$ and $b$ share no fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
if $2^n+1$ is prime, then n = $2^r$ and $(x - y) \mid(x^k - y^k$) only with natural numbers I need to prove that if $2^n + 1$ is prime, then $n = 2^r$ for a natural number $r$.
I do know how to prove it with the lemma $(x - y) \mid (x^k - y^k)$, but in order to prove it with this lemma, I need to substitute $y$ with $-... | Suppose that $p$ is odd prime divisior of $n$. Then $n=p\cdot m$,
$$ 2^n +1 = (2^m)^p + 1^p = (2^m + 1)\cdot ((2^m)^{p-1}-(2^m)^{p-2}+ \cdots + 1)$$
and each factor greater than $1$. So $2^n +1 $ isn't prime which gives a contradiction.
Some Notes:
1) We use the identity $x^{2k+1}+y^{2k+1}=(x+y)\cdot (x^{2k} - x^{2k-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
normal equations and projections (linear algebra) I have three points $(-1,7)$, $(1,7)$ and $(2,21)$ and a linear model $b = C + Dt$. It is asking to give and solve the normal equations in $\hat{x}$ to find the projection $p = A\hat{x}$ of $b$ onto the column space of $A$ (i.e. find $\hat{x}$ with least $||A\hat{x} - ... | Problem statement
Given a sequence of $m=3$ data points of the form $\left\{ x_{k}, y_{k} \right\}$, and a model function
$$
y(x) = a_{0} + a_{1} x
$$
find the least squares solution
$$
a_{LS} = \left\{
a\in\mathbb{C}^{n} \colon
\lVert
\mathbf{A} a - y
\rVert_{2}^{2}
\text{ is minimized}
\right\}
$$
Linear system
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Maximum value of $4x-9y$ Suppose xand y are real numbers and that $x^2 +9y^2 -4x +6y+4=0$ then we have to find the maximum vale of 4x-9y
I found that the equation given is that of ellipse .
I can consider a line $4x-9y=0$ which cut the ellipe .
bt by doing this its getting very comlicated .
| We'll use Lagrange's multiples:
$$f(x,y) = 4x-9y , \phi(x,y) = (x-2)^2 + (3y+1)^2 $$
$$F(x,y) = 4x-9y - \lambda(x-2)^2 - \lambda (3y+1)^2 $$
And so:
$$\frac{\partial F}{\partial x} = 4 - 2\lambda (x-2) = 0 \Rightarrow x-2 = \frac{2}{\lambda} $$
$$ \frac{\partial F}{\partial y} = -9 - 6\lambda (3y+1) = 0 \Rightarrow 3y+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find all positive integer solutions of the equation $n^5+n^4 = 7^m-1$
Find all positive integer solutions of the equation $n^5+n^4 = 7^m-1$.
Suppose $n$ is odd. Then we have $$n^5+n^4 \equiv \pm 2 \not \equiv 7^m-1 \pmod{7}.$$ Thus $n$ is even. Let $n = 2k$ for some positive integer $k$, and we have $$n^4(n+1) = 16k^... | Let's start by observing that
$$
n^5+n^4 + 1 = (n^2 + n + 1)(n^3-n+1).
$$
Now, let $d=gcd(n^2 + n + 1,n^3-n+1)$. Note that, $d\mid n^2 + n + 1 \implies d\mid n^3 - 1$. Hence, $n^3 \equiv 1\pmod{d}.$ But since $d|n^3-n+1$, we must have, $n\equiv 2\pmod{d}$. Hence, $d|7$, therefore $d$ is either $1$ or $7$.
Note that $n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability of drawing a spade on the first draw, a heart on the second draw, and an ace on the third draw My answer is
$(\frac{1}{52})(\frac{1}{51})(\frac{2}{50})$ + $2(\frac{1}{52})(\frac{12}{51})(\frac{3}{50})$ + $(\frac{12}{52})(\frac{12}{51})(\frac{4}{50})$
= $\frac{1}{204}$
But the answer in the textbook is
$(\fr... | Using the probability tree diagram:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove this inequality? Thanks
Let $x_{1},x_{2},\cdots,x_{n}\in [0,1]$, show that
$$\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^{\frac{n}{2}}\cdot\sqrt{1-x_{i}}\right]\le 1$$
Maybe it can use McLaughlin inequality to solve it?I found this inequality simaler this problem
| Claim: For all integer $n$, positive $c$ satisfying $n/2 \le c$ and $x_{1},x_{2},\cdots,x_{n}\in [0,1]$,$$\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^c\cdot\sqrt{1-x_{i}}\right]\le 1$$
Proof:
Let $f(x_1, x_2, \cdots, x_n, c)=\sum_{i=1}^{n}\left[\left(\prod_{j\neq i }x_{j}\right)^c\cdot\sqrt{1-x_{i}}\right]$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Volume of the revolution solid Let $K \subset \mathbb R^2$ be area bounded by curves $x=2$, $y=3$, $xy=2$ and $xy=4$.
What is the volume of the solid we get after rotating $K$ around $y$-axis?
I tried to express the curves in terms of $y$ and solved the integral $\displaystyle \pi\left(\int_2^3 \frac{16}{y^2}dy - \int_... | Do this in two parts. First find the section of the volume between $y=3$ and $y=2$, then the section of the volume between $y=2$ and $y=1$. For the first volume, use the integral
$$\pi\int_{2}^{3} \frac{16}{x^2}dx-\pi\int_{2}^{3} \frac{4}{x^2}dx$$
This is the difference of the volumes formed by rotating the regions bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Deriving left nullspace of matrix from $EA=R$ Let $A$ be an $m\times n$ matrix, $R$ be its row-reduced echelon form, and $E$ be the sequence of matrices $E_k\dots E_1$ used to bring $A$ to $R$, such that $EA = R$. In one of the books on linear algebra it is said that we can use the fact that $EA=R$ to find the basis fo... | Here are two detailed examples:
Find base and dimension of given subspace
Given a matrix and its reduced row echelon form, resolve the image and the kernel.
Fundamental Theorem of Linear Algebra
Given $\mathbf{A}\in\mathbb{C}^{m\times n}$, the four fundamental subspaces are
$$
\begin{align}
%
\mathbf{C}^{n} =
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Uniform convergence of $f_n(x) =\sqrt[n]{1+x^n}$ I need to decide if the following function uniformly converges:
$$f_n(x) =\sqrt[n]{1+x^n} \quad, \quad x\in[0,\infty)$$
I found the sequence pointwise converges to
$$f(x) =
\begin{cases}
1 \; , & \text{$x\in[0,1]$} \\
x \; ,& \text{$x \in(1,\infty)$}
\end{cases}$$
I tri... | Hint. For $x>1$,
$$ \sqrt[n]{1+x^n} - x = \frac{1}{x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}(1+x^n)^{\frac{1}{n}}+\cdots +x^{\frac{1}{n}}{(1+x^n)^{\frac{n-1}{n}}}} \le \frac{1}{n}$$
and for $x\le 1$,
$$ \sqrt[n]{1+x^n} - 1 = \frac{x^n}{1+(1+x^n)^{\frac{1}{n}}+\cdots +{(1+x^n)^{\frac{n-1}{n}}}} \le \frac{1}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum of $x^3+y^3+z^3$ with $x+y+z=3$ It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$.
My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that,
$f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f... | You have correctly established that $z=0$. From there you have $y=3-x$ so substitute that into $f$. As $y\le2$ then $1\le x\le2$.
$$f(x,3-x,0)=x^3+(3-x)^3=9x^2+27x+27=9(x^2+3x+3)$$
$$=9\left(x-\frac{3}{2}\right)^2+\frac{27}{4}$$
This quadratic has minimum at $x=\frac{3}{2}$ and the maximum is only limited by the domain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Infinite Sum Calculation: $\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$ Problem
Show the following equivalence: $$\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$$
Good afternoon, dear StackExchange community. I'm studying real analysis ... | $\displaystyle \begin{align}\sum_{n\ge 0} \dfrac{1}{(2n+1)^s}&= \sum_{n=1}^\infty \dfrac{1}{n^s}-\sum_{n=1}^\infty \dfrac{1}{(2n)^s} \\ &= \sum_{n=1}^\infty \dfrac{1}{n^s}(1-2^{-s}) \\ &= (1-2^{-s})\sum_{n=1}^{\infty} \dfrac{1}{n^s}\end{align}$
Putting $s=2$ gives you the result and obviously the above is true for $s\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now,
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^... | Almost finished from there:
$$(\sin x + 1)(\sin^3 x + \sin^2 x)\\ = (\sin x + 1)(\sin x + 1) \sin^2 x\\ = (\sin^2 x + \sin x) (\sin^2 x + \sin x) \\= 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Different ways to solve $\int_0^1 \sqrt{1+x^2} dx $ So I have the task to solve this integral in $4$ different ways, but have solved it only with substitution. ($x=tg(t)$, $dx=\sec^2(t)dt$) and so on. Any advice on the other $3$ ways ? Thank you :)
| IBP? $u = \sqrt{1+x^2}$ and $\mathrm{d}v = 1$ so $v = x$. Hence $$I = \int_0^1 \sqrt{1+x^2} \, \mathrm{d}x = \bigg[x\sqrt{1+x^2}\bigg]_0^1 - \int_0^1 \frac{x^2}{\sqrt{1+x^2}} \, \mathrm{d}x$$
Then $\int_0^1 \frac{x^2}{\sqrt{1+x^2}} \, \mathrm{d}x = \int_0^1 \frac{1+x^2}{\sqrt{1+x^2}} \, \mathrm{d}x - \int_0^1 \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How many real roots of $f(x)=x^{12}-x^9+x^4-x+1$ between $0$ and $1$? How can we know if $f(x) = x^{12} - x^9 + x^4 - x + 1$ has How many real roots between 0 and 1. ?
Okay so at both 1 and 0 the value of function is 1. So there can be no or even roots between
them so how can we tell can we use calculus ?
| $\begin{cases}
f(x)=x^{12}-x^9+x^4-x+1 \\
f'(x)=12x^{11}-9x^8+4x^3-1 \\
f''(x)=132x^{10}-72x^7+12x^2=12x^2(11x^8-6x^5+1)=12x^2g(x) \\
g'(x)=88x^7-30x^4=x^4(88x^3-30)
\end{cases}$
And we can stop here, because we are only interested in the signs of the derivatives.
$a=\sqrt[3]{\frac{30}{88}}$
$\begin{array}{|c|ccccc|}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$ In MAO 1991,
Find $2x+5$ if $x$ satisfies $\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$
My attempt,
I squared the the equation then I got $144x^2+1648x+4480=144x^2-1632x+4624$, which results $x=-9$, and $2x+5=-13$.
I want to ask is there another way to solve this question as my method is v... | An $x$ that solves your equation also solves the system
\begin{align}
40&&-9x&-a^2&=0\\
7&&-x&-b^2&=0\\
&&-x&-c^2&=0\\
&&a-2b&-c&=0\\
\end{align}
If we subtract the second and the third equation we get
$$-b^2+c^2+7=0.$$
It turns out that
$$-9(-b^2+c^2+7)=-(3b-4c)(-3b-4c)-(-7c^2+63).$$
Therefor one solution is $c^2-9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$ Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$
Hints are appreciated. Thanks in advance.
| Another way:
$$x^{18} + x^{24} = x^{21} (x^3 + \frac{1}{x^3})\\ = x^{21} ((x+\frac{1}{x})^3 - 3(x+\frac{1}{x}))\\ = x^{21}(3\sqrt{3} - 3\sqrt{3}) = 0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
For integer m greather than 2, $\frac{1}{m} + \frac{1}{m+2}$, the numerators and denomitors are primitive pythagorean triples $a$ and $b$ For $m = 2$, the fraction is $\frac{3}{4}$. for $m=3$, the fraction is$\frac{8}{15}$. I was wondering why numerators and demoninators of $\frac{1}{m} + \frac{1}{m+2}$ show primitive... | $$\frac{1}{m} + \frac{1}{m+2} = \frac{2m+2}{m^2 + 2m}$$
Now notice that
$$(2m+2)^2 + (m^2 + 2m)^2=8m^2+4m+4+m^4+4m^3=(m^2+2m+2)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving Ramanujan's Nested Cube Root Ramanujan's Nested Cube:
If $\alpha,\beta$ and $\gamma$ are the roots of the cubic equation$$x^3-ax^2+bx-1=0\tag{1}$$then, they satisfy$$\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}=(a+6+3t)^{1/3}\tag{2.1}$$
$$(\alpha\beta)^{1/3}+(\beta\gamma)^{1/3}+(\gamma\alpha)^{1/3}=(b+6+3t)^{1/3}\tag{... | We have $\alpha$ , $\beta$ and $\gamma$ are roots of the equation $x^3-ax^2+bx-1=0$ so
\begin{eqnarray*}
a= \sum \alpha \\
b= \sum \alpha \beta \\
\alpha \beta \gamma =1
\end{eqnarray*}
so $\sqrt[3]{\alpha \beta \gamma} =1$ and let
\begin{eqnarray*}
A= \sum \sqrt[3]{\alpha} \\
B= \sum \sqrt[3]{\alpha \beta}
\end{eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Find all solutions to $|x^2-2|x||=1$ Firstly, we have that $$\left\{
\begin{array}{rcr}
|x| & = & x, \ \text{if} \ x\geq 0 \\
|x| & = & -x, \ \text{if} \ x<0 \\
\end{array}
\right.$$
So, this means that
$$\left\{
\begin{array}{rcr}
|x^2-2x| & = & 1, \ \text{if} \ x\geq 0 \\
|x^2+2x| & = & 1, \ \tex... | The most general way is to grow a tree and then for each leaf of the tree a table. If more than a couple of layers of $|.|$ you will probably be starting to confuse yourself if you don't stick to a systematic approach.
Each branch in the tree reduce the set the variable is valid for. We need to store a pair $(expressio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$a/b < (1 + \sqrt{5})/2 \iff a^2 - ab - b^2 < 0$? For positive integers $a$ and $b$, I want to show that $a/b < (1 + \sqrt{5})/2$ if and only if $a^2 - ab - b^2 < 0$.
I had a loose proof ready to go, but I noticed a fatal flaw. Perhaps there is a way to work around this though.
My tactic was to start from $a^2 - ab - b... | If $b=0, a^2<0$ which is impossible
So, $b\ne0, b^2>0$ consequently, $$a^2-ab-b^2<0\iff\left(\dfrac ab\right)^2-\left(\dfrac ab\right)-1<0$$
Now the roots of $x^2-x-1=0$ are $$x=\dfrac{1\pm\sqrt5}2$$
We can prove if $(x-a)(x-b)<0$ with $a<b;$
$$a<x<b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Is there name for Ring, with the following Cayley tables I'm solving a Cayley table for Ring with four elements and noticed that the resulting multiplication table looks pretty interesting. It has two left multiplication 1s and two left multiplications zeros and one of those zeros is ofcourse the zero in the Additive g... | I've never heard a name for this rng, but I have seen it expressed this way:
Given the two-element semigroup $S=\langle b,c\mid b^2=cb=b, c^2=bc=c\rangle$, your rng is the semigroup rng $F_2[S]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show without multiplying out Show without multiplying out,
$$\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}=\frac{(a-b)(b-c)(a-c)}{abc}$$
So, I'm given a solution which is
\begin{align}\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}&=\frac{bc(b-c)+ac(c-a)+ab(a-b)}{abc}\\
&=\frac{bc(b-c)+a^2(b-c)-a(b^2-c^2)}{abc}\\
&=\frac{(b-c... | Between line $1$ and line $2$ you've got $ac(c-a)+ab(a-b)$ as terms in the numerator on the right-hand-side. These terms expand out as $ac^2-a^2c+a^2b-ab^2$ (we're not breaking the rules since we're not multiplying out the whole thing). Notice the two terms with $a^2$. We can rearrange these terms to $a^2(b-c)-a(b^2-c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
fine the limits :$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=?$ fine the limits-without-lhopital rule and Taylor series :
$$\lim_{x \to 0} \frac{(\sin 2x-2x\cos x)(\tan 6x+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x \tan x\sin 2x}=... | If you know, that $\enspace\displaystyle \lim\limits_{x\to 0}\frac{1}{x^2}(1-\frac{\sin x}{x})=\frac{1}{3!}\enspace$ then you can answer your question easily:
$\displaystyle \frac{(\sin(2x)-2x\cos x)(\tan(6x)+\tan(\frac{\pi}{3}-2x)-\tan(\frac{\pi}{3}+4x))}{x\sin x\tan x\sin(2x)}=$
$\displaystyle =\frac{(\sin(2x)-2x\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
Let $x^3+ax^2+bx+c=0$ are $\alpha, \beta, \gamma$. Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
My attempt,
As I know from the original equation,
$\alpha+\beta+\gamma=-a$
$\alpha\beta+\beta\gamma+\alpha\gamma=b$
$\alpha\beta\gamma=-c$
I'... | Using basic algebra:
*
*Calculating $\alpha^3 + \beta^3 + \gamma^3$ :
As, \begin{aligned}
(&\alpha + \beta + \gamma)^3 = \alpha^3 + \beta^3 + \gamma^3 + 3\alpha^2\beta + 3\alpha^2\gamma + 3\alpha\beta^2 + 3\beta^2\gamma + 3\alpha\gamma^2 + 3\beta\gamma^2 + 6\alpha\beta\gamma
\end{aligned}
We can factor this, \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How many solutions does the equation $x^2-y^2 = 3^8\cdot 5^6\cdot 13^4$ have? Given that $x$ and $y$ are positive integers. Problem: How many solutions does the equation $x^2-y^2 = 3^8\cdot 5^6\cdot 13^4$ have? Given that $x$ and $y$ are positive integers.
I tried a similar approach to the ones described here, but to ... | The product $n=3^8\cdot 5^6\cdot13^4$ has a total of $(8+1)(6+1)(4+1)=315$ factors $d$. Those can be paired up as $(d,n/d)$. For the choice $d=n/d=\sqrt n$ the two factors are equal. For the remaining $157=(315-1)/2$ pairs we must use the bigger one as $x+y$ and the smaller as $x-y$. Because all the factors $d, n/d$ ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Values of $m$ that satisfy the quartic equation $x^4-(3m+2)x^2+m^2 = 0$ which has 4 real roots in AP The question says find all values for $m$, such that the below equation has 4 real roots in Arithmetic Progression.
My approach:
\begin{gather}
x^4-(3m+2)x^2+m^2 = 0 \\
\\
\text{Let roots be }\\
\beta+d \\
\beta-d \\
\... | In case others need help in a similar question, I am putting up the solution.
\begin{gather}
x^4-(3m+2)x^2+m^2 = 0 \\
\\
\text{Let roots be }\\
\beta+d \\
\beta-d \\
\beta+3d \\
\beta-3d \\
\\
x^4+0x^2-(3m+2)x^2+0x+m^2 = 0\\
\implies 0 = (\beta-3d) + (\beta-d) + (\beta+3d) + (\beta+d) = 4\beta \\
\implies \beta = 0\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Logarithmic inequality, can't define the scope of $x$ I'm solving and getting answer that $x >1$.
\begin{align*}
\ln(x)-\ln(2-x)&>0\\
\implies \ln\left(\frac{x}{2-x}\right)&>0\\
\implies e^{\ln(x/(2-x))}&>e^0\\
\implies \frac{x}{2-x}&>1\\
\implies x &> 2-x\\
\implies 2x &> 2\\
\implies x &> 1
\end{align*}
But when I as... | You have $\dfrac x {2-x} > 1.$
You cannot go from there to $x>2-x$ unless you somehow know that $2-x$ is positive.
There are at least two ways to deal with that:
*
*First of course you rule out $x=2$ as a possible solution because it makes the denominator zero. Then consider what happens when $x>2.$ Then $2-x$ is ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$ \lim_{x \to \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $ fine limit :
$$ \lim_{x \to \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $$
such that : $$ a \in (0,2)$$
and :
$[x]: \ \ $ floor function
My Try :
$$f(x):=( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]... | $\begin{array}\\
( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]}
&=(x^{2/3} \sqrt[3]{1+4 x^{a-2}} - x^{2/3}\sqrt[3]{1+ x^{a-2} } )^{x-[x]}\\
&=x^{2(x-[x])/3} (\sqrt[3]{1+4 x^{a-2}} - \sqrt[3]{1+ x^{a-2} } )^{x-[x]}\\
&=x^{2(x-[x])/3} (1+4 x^{a-2}/3+O(x^{a-3}) - (1+ x^{a-2}/3+O(x^{a-3})) )^{x-[x]}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
how many $3\times 3$ matrices with entries from $\{0,1,2\}$.
How many $3 × 3$ matrices $M$ with entries from $\left\{0, 1, 2\right\}$ are there for which taken from
the sum of the main diagonal of $M^TM$ is $5$.
Attempt: Let $M = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$. where $a,b,c,d,e,f... | Following Lord Shark the Unknown's answer.
For five 1s and the rest zero, there are $C^9_5$ matrices.
For one 1 and one 2, there are $9\times 8$ matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
A summation question ,Is $S \to \ln 2$? I am in doubt with this question .
let $S=\dfrac{\dfrac12}{1} +\dfrac{(\dfrac12)^2}{2}+\dfrac{(\dfrac12)^3}{3}+\dfrac{(\dfrac12)^4}{4}+\dfrac{(\dfrac12)^5}{5}+...$ Is it converge to $\ln 2$ ?
I tried this
$$x=\dfrac12 \to 1+x+x^2+x^3+x^4+...\sim\dfrac{1}{1-x}\to 2$$ by integr... | Yes.
This series was already known to Jacob Bernoulli (Gourdon and Sebah http://plouffe.fr/simon/articles/log2.pdf, formula 14) and can be written
$$S=\sum_{k=0}^\infty \frac{1}{k+1}\left(\frac{1}{2}\right)^{k+1}$$
To evaluate it, we can change the identity
$$\int_0^1 x^n dx = \frac{1}{n+1}$$
into
$$\int_0^\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
completing square for a circle In the following question:
I don't understand how we can get from the original equation to the final equation using completing the square.
Any thoughts as how to get to the final equation?
| $x + y = c(x^2 + y^2 + 1)$
$x^2 + y^2 - \frac xc - \frac yc + 1 = 0$
$x^2 - \frac xc + (\frac 1{2c})^2- (\frac 1{2c})^2 + y^2 - \frac yc + (\frac y{2c})^2- (\frac 1{2c})^2+ 1 = 0$
$(x - \frac 1{2c})^2 + (y-\frac 1{2c})^2 + 1 - \frac 1{4c^2} - \frac 1{4c^2} = 0$
$(x - \frac 1{2c})^2 + (y-\frac 1{2c})^2 = \frac 1{2c^2} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $g(x) = \sqrt{1 + x^2}$ is continuous Show that $g(x) = \sqrt{1 + x^2}$ is continuous on $\mathbb{R}$ using the $\epsilon - \delta$ argument.
This is what I was thinking...
Let $p \in \mathbb{R}$. Then $g(x)-g(p) = \sqrt{1 + x^2} - \sqrt{1 + p^2}$
$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \qua... | A slightly "different" approach is to use ( to abuse to be correct ) the big brother inequality called Minkowski inequality of the form: $\sqrt{(a+b)^2 + (c+d)^2} \le \sqrt{a^2+c^2} +\sqrt{b^2+d^2}$. Use this inequality with $a = b = 1, c = x, d = p$ as in your answer, the follows it with the popular trigonometric subs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the value of this integral? $\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx=-\frac{G}{4}+\frac{1}{24}+\frac{\pi}{64}-\frac{\pi}{16}\ln2$ $$I=\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx=-\frac{G}{4}+\frac{1}{24}+\frac{\pi}{64}-\frac{\pi}{16}\ln2$$
Where G is the Catalan's contant
My try:
Integrating by parts, we have:
$$3I=\... | It is easy to check that $\int_{0}^{1}x^2\sqrt{1-x^2}\,dx=\frac{\pi}{16}$. By setting $x=\cos\theta$ the problem boils down to computing
$$ \int_{0}^{\pi/2}\cos^2(\theta)\sin^2(\theta)\log\sin^2\frac{\theta}{2}\,d\theta$$
that is simple through the Fourier series of $\log\sin$ and $\log\cos$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Differentiate function: $\ y=\sin(x)\sqrt{x+3}$ Differentiate function:
$\ y=\sin x \sqrt{x+3}$
$\frac{1}{y}dy= \frac{1}{\sqrt{x+3}}dx + \frac{1}{\sin x}\cos xdx$
$\ln(y)=\ln(\sqrt{x+3})+\ln(\sin x)$
$\frac{dy}{dx}=y \left(\frac{1}{\sqrt{x+3}}+\frac{\cos x}{\sin x} \right)$
$\frac{dy}{dx}=\sqrt{x+3}\sin x \left(\frac{1... | I assume the function is $y=\sqrt{x+3}\sin x$. With the (formal) logarithmic derivative you have
$$
\log y=\frac{1}{2}\log(x+3)+\log\sin x
$$
so
$$
\frac{y'}{y}=\frac{1}{2}\frac{1}{x+3}+\frac{\cos x}{\sin x}
$$
and therefore
$$
y'=\sqrt{x+3}\,\sin x\left(\frac{1}{2}\frac{1}{x+3}+\frac{\cos x}{\sin x}\right)=
\frac{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$
How to prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$?
I tried Eisenstein criteria on $f(x+n)$ with $n$ ranging from $-10$ to $10$. None can be applied. I tried factoring over mod $p$ for primes up to $1223$. $f... | Assume that $f(x) = f_1(x) f_2(x)$.
Note that $f(1) = - 23$, so one of $f_1(1)$, $f_2(1) = \pm 1$, and the other $\mp 23$.
The polynomial factors $\mod 5$ as $(x^4+2)(x^4 + x^2 + 2)$ ( see Jyrki's answer). Say we have $f_1(x)\equiv x^4 + 2$ and $f_2(x) \equiv x^4 + x^2 + 2 \mod 5$. Now, $f_1(1) \equiv 1^4 + 2 = 3 \mod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 5,
"answer_id": 2
} |
Find all prime $a, b, c$ such that $ab+bc+ac > abc$ Find all primes $a, b, c$ such that $ab + bc + ac > abc$
| let $a\geq b \geq c$
if $c\geq 3$ then $ab+bc+ac\leq 3ab\leq cab\implies c=2$.
We now need $ab+2(a+b)> 2ab\iff 2(a+b)>ab$
if $b\geq 4$ then $2(a+b)\leq4a \leq ab$, so $b\in \{2,3\}$
if $b=2$ then $2(a+2)>2a$.
if $b=3$ then $2a+6>3a\iff a<6$.
Hence ordered solutions are $(p,2,2)$ and $(3,3,2),(5,3,2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int_{-1}^1\frac{y}{\pi\sqrt{1-y^2}}\,dy$ The solution says that we will get this $$\left[-\frac{\sqrt{1-y^2}}{\pi} \right]^1_{-1} $$.
How do we get this definite integral. I know we are integrating an odd function over symmetric domain so the answer will be 0.
| Let $u = 1-y^2$, then $du = -2y ~dy$ so that
\begin{align}
\int_{-1}^{1} \frac{y}{\pi\sqrt{1-y^2}} dy &= \int_{y=-1}^{y=1} \frac{\frac{-1}{2}du }{\pi\sqrt{u}} = -\frac{1}{2\pi}\int_{y=-1}^{y=1} u^{\frac{-1}{2}}du = -\frac{1}{2\pi}\left(2u^{\frac{1}{2}}\right)_{y=-1}^{y=1}\\
&=-\left(\frac{\sqrt{1-y^2}}{\pi}\right)_{y=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Doubts about derivative of $|x|$ We know that the derivative of $f(x)=|x|$ is $-1$, if $x\lt 0$ and $1$, if $x\ge 0$.
However for every $x\in \mathbb R$, the following holds
$f(x)=|x|=\sqrt{x^2}\implies f'(x)=\frac{1}{2}(x^2)^{-1/2}2x=1$.
So where is my mistake?
| As you said
$$ f(x) = \sqrt{x^2} = |x| = \begin{cases} x, & \text{if } x\ge0\\ -x, & \text{if } x < 0, \end{cases} $$
then
$$\frac{1}{\sqrt{x^2}} = \frac{1}{|x|} = \begin{cases} \frac{1}{x}, & \text{if } x > 0, \\ -\frac{1}{x} , & \text{if } x < 0. \end{cases}$$
Hence
$$ f′(x)=\frac{1}{2}(x^2)^{−1/2}2x=\frac{x}{|x|} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $x^3-x^2+8=y^2$ has no integer solution
Show that $x^3-x^2+8=y^2$ has no integer solution.
I spent several hours on this problem but I couldn't figure out how to solve it. A hint from the professor was to find a proper field where to find the solutions, or use Gaussian integers, but I still can't find the p... | Here is an official solution: https://drive.google.com/file/d/1qOPYHK3p_S8tlwlewQ91X6JbDrZ-oKrc/view. Following is an English version (not literal translation, also little reordered):
If $x$ is even, say $x=2a$, we have $y^2=x^3-x^2+8=8a^3-4a^2+8$, implying $2 \mid y^2$, and so $2 \mid y$. Let $y=2b$, giving $8a^3-4a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Solving $(a-1)^{p+1}+a^p=(a+1)^{p-1}$
Suppose that $a,p$ are nonnegative integers such that $p$ is prime and $p\nmid (a-1)$. If $(a-1)^{p+1}+a^p=(a+1)^{p-1}$, find the sum of all possible values of $a$.
We can't have $p = 2$ since the equation $$(a-1)^3+a^2 = (a+1)$$ has no integer solutions. We now take two cases:
C... | I will assume $p$ is odd.
Consider the monic polynomial $f(X) = (X-1)^{p+1} + X^p - (X+1)^{p-1}$.
It has $0$ as a root, so dividing by $X$ we find $g(X) = f(X)/X$ is another monic polynomial.
The constant term of $g$ is ${p+1 \choose p}(-1)^p-{p-1 \choose p-2} = (p+1)(-1)^p-(p-1) = -2p$.
The rational root theorem gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all positive integers $n$ such that $3^n+5^n = x^3$
Find all positive integers $n$ such that $3^n+5^n = x^3$ for some positive integer $x$.
One solution is $n = 1, x = 2$.
We have $1 < 3^n+5^n \leq 8^n$, so $1 < 3^n+5^n \leq 2^{3n}$. Thus $1 < x \leq 2^n$. How can we continue from here?
| Modulo $9$, if $n>1$, the equation simplifies to $5^n\equiv x^3 \pmod 9$. Because the cubes mod 9 are $0,1,8$, and $5^n \pmod 9$ repeats with period $\phi(9)=6$ with the pattern (starting at $n=0$) $1,5,7,8,4,2,\ldots$, we have that $n$ is a multiple of $3$ (if $n>1$).
However, working modulo $7$, where $3^n+5^n$ also... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
If $f(x) =mx$, then $f(a + b) = f(a) + f(b)$ for all $a$ and $b$?
If $f(x) =mx$, then $f(a + b) = f(a) + f(b)$ for all $a$ and $b$. True or False.
Verification of work: I found a similar problem where $f(x)=y-mx+b$ and test values for $m$ and $b$ were used. $f(x)=y-mx+b$
My problem has $m$ and $x$ as the values so I ... | So if $f(x) = mx$, $f(a+b) = m(a+b) = ma + mb = f(a) + f(b)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integer solutions to $y=(2^x-1)/3$ where $x$ is odd For the equation $y=(2^x-1)/3$ there will be integer solutions for every even $x$.
Proof: When $x$ is even the equation can be written as $y=(4^z-1)/3$ where $z=x/2$.
$$4^z =1 + (4-1)\sum_{k=0}^{z-1} 4^k$$
If you expand that out you get:
$$4^z=1+(4-1)4^0+(4-1)4^1+(4-1... | If x is odd we can write x= k+ 1 for k an even integer. Then $\frac{2^x- 1}{3}= \frac{2^{k+1}- 1}{3}= \frac{2(2^k)- 1}{3}= \frac{2(2^k)- 2+ 1}{3}= \frac{2(2^k- 1)+ 1}{3}= 2\frac{2^k- 1}{3}+ \frac{1}{3}$.
You have already shown that $\frac{2^k-1}{3}$ is an integer so this is an integer plus 1/3, not an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})}$ without L'Hopital I tried:
$$\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})} =
\frac{\frac{\cos(2x)}{x-\frac{\pi}{4}}}{\frac{\sin(x-\frac{\pi}{4})}{x-\frac{\pi}{4}}}$$ and $$\begin{align}\frac{\cos(2x)}{x-\frac{\pi}... | You are trying this the hard way.
Let $x = \frac\pi 4 -y$. Then
$$
\lim_{y\to 0}\frac{\cos(2(\frac\pi 4 -y))}{\sin(- y)}
=\lim_{y\to 0}\frac{\cos(\frac\pi 2-2y)}{-\sin y}
=-\lim_{y\to 0}\frac{\sin(2y)}{\sin y}
=-\lim_{y\to 0}\frac{2\sin y \cos y}{\sin y}
=-\lim_{y\to 0}2 \cos y=-2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find all polynomials $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$ for $x\in \Bbb R$
Find all polynomials $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$ for $x\in \Bbb R$
I've tried different substitutions like put $Q(x)$ $=$ $k$ for some $k$ and getting the equation $k(x^2-6x+8)$ $=$ $(k-2)(x^2-6x)$ $<=>$ $k$=$(6x-x^2)/4$, but that doesn't gi... | maybe Hint:Some information come out from $Q(x)(x^2-6x+8) =Q(x-2)(x^2-6x)$
$$x=0 \to Q(0)(8)=Q(-2)0 \to Q(0)=0\\
x=6 \to Q(6)(8)=Q(6-2)0 \to Q(6)=0\\
x=4 \to Q(4)(0)=Q(2)(16-24) \to Q(4)=0$$ so $Q(x)$ at least contain $$Q(x)=(x-0)(x-4)(x-6)$$ or
$$Q(x)=(x-0)(x-6)(x-4)q(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is $e$ a coincidence? $e$ has many definitions and properties. The one I'm most used to is
$$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n $$
If someone asked me (and I didn't know about $e$):
Is there a constant $c$ such that the equation $\frac{d}{dx}c^x=c^x $ is true for all $x$?
Then I'd likely answer that:
I... | Just to add how these definitions of $e$ are matched together. You can call this an overkill. I use Taylor polynomials around $x=0$ and then Taylor series to show that $e$ satisfies the equation.
We first note that $c=0$ will not work (why?). It is obvious that the function $f(x)=c^x$ is infinitely many times different... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 7,
"answer_id": 6
} |
Show that $f(x)=x^2+x+4$ is irreducible over $\mathbb{Z}_{11}$ I know this can be done by evaluating $f$ at the points $0,1,...10$ to check if $f$ has a linear factor.
Is there any other shorter way?
| We start with $$x^2+x+4=0$$ multiplying with $4$ we get $$4x^2+4x+16=0$$ This can be written as $$(2x+1)^2+15=0$$ modulo $11$ this is equivalent to $$(2x+1)^2-7=0$$
So we have to solve $$(2x+1)^2\equiv 7\mod 11$$
Using the quadratic reciprocity law we can calculate the legendre symbol
$$(\frac{7}{11})=-(\frac{11}{7})=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is the set of solutions to $2\log_{\cos x}\sin x\le \log_{\sin x}\cot x$ where $0
$$2\log_{\cos x}\sin x\le \log_{\sin x}\cot x$$ where $0<x<\pi$
My process:
Since both $\sin x$ and $\cos x$ must be greater than $0$ and less than $1$, the initial set shrinks to $(0,\pi/2)$
Then after transformations I get
$2\log_{... | We need $\displaystyle 0<x<\frac{\pi}{2}$. Note that $\ln\cos x<0$ and $\ln \sin x<0$.
\begin{align}
\frac{2\ln \sin x}{\ln \cos x}&\le\frac{\ln \cos x-\ln\sin x}{\ln\sin x}\\
2(\ln\sin x)^2&\le(\ln\cos x)^2-\ln\sin x\ln \cos x\\
(\ln\cos x)^2-\ln\sin x\ln \cos x-2(\ln\sin x)^2&\ge0\\
(\ln \cos x+\ln \sin x)(\ln\cos x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Injectivity and Range of a function Let, $A=\{(x,y)\in \Bbb R^2:x+y\neq -1\}$
Define, $f:A\to \Bbb R^2$ by $f(x,y)=\left(\displaystyle \frac{x}{1+x+y},\frac{y}{1+x+y}\right)$
$(1)$Is $f$ injective on $A$?
$(2)$What is the Range of $f$?
I started with $f(x,y)=f(u,v)$, then tried to show $(x,y)=(u,v)$. But I could not do... | To continue our discussion in the comments, pick any $(x,y), (u,v) \in A$.
Suppose $f(x,y) = f(u,v)$. Further suppose $x, y, v \neq 0$. (You have to prove the cases (i) $x = 0$, (ii) $y = 0$ and (iii) $v = 0$ later on)
Then $$\frac{x}{1+x+y} = \frac{u}{1+u+v}\quad \text{and} \quad \frac{y}{1+x+y} = \frac{v}{1+u+v}$$
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\frac{2}{3}tx+\sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x}=t+\sqrt{2}+\sqrt{6}$ where $t=\sqrt{2}-\sqrt{3}-\sqrt{5}+\sqrt{6}$?
Let $t=\sqrt{2}-\sqrt{3}-\sqrt{5}+\sqrt{6}$
then fine the $x$:
$$\frac{2}{3}tx+\sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x}=t+\sqrt{2}+\sqrt{6}$$
since : $$\sqrt{2x-3} \to 2x-3 \geq 0 \to x\geq\... | if you see $d(d(\sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x})/dx)/dx$, you will see that is less than zero from $\frac{3}{2}$ to $3$. So $f = \frac{2}{3}tx + \sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x} - t - \sqrt{2} - \sqrt{6}$ is convex upward. At $x = \frac{3}{2}$ is zero, at $x = 3$ is zero too. So we have no other roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Completing the square but in different situation
Solve the equation $$x^2+4\left(\frac{x}{x-2}\right)^2=45$$
My attempt,
I decided to use completing the square method, so I change it to $$x^2+\left(\frac{2x}{x-2}\right)^2=45$$
But I never encounter this before. Normally, for example $x^2+4x=5$, we can change it to $x... | The suggested solution posted by OP provide a clever method.
\begin{align}
x^2+4\left(\frac{x}{x-2}\right)^2&=45\\
x^2+2(x)\left(\frac{2x}{x-2}\right)+\left(\frac{2x}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\
\left(x+\frac{2x}{x-2}\right)^2&=45+4\left(\frac{x^2}{x-2}\right)\\
\left(\frac{x^2}{x-2}\right)^2&=45+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof that $\log_a b \cdot \log_b a = 1$
Prove that $\log_a b \cdot \log_b a = 1$
I could be totally off here but feel that I have at least a clue.
My proof is:
Suppose that $a = b$, then $a^{1} = b$ and $b^{1} = a$ and we are done.
Suppose now that $a \neq b$. We wish to show that: $\log_b a = \frac{1}{\log_a b}$.
... | Method I
Let $\log_ab=x$ and $\log_ba=y$. Then we have
$$a^x=b \quad\text{and}\quad b^y=a$$
So,
\begin{align}
(b^y)^x&=b\\
b^{xy}&=b\\
xy&=1
\end{align}
Method II
Let $\log_ab=x$. Then we have
\begin{align}
a^x&=b\\
\log_b(a^x)&=\log_bb\\
x\log_ba&=1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving equation of type $a\cos x+b\cos y-c=0$ and $a\sin x+b\sin y-d=0$ Here's the questions
There are two equations:
$a\cos x+b\cos y-c=0 $ and $ a\sin x+b\sin y-d=0$ .
For instance
What is the value of $x$ and $y$ in following question?
$$2\cos x+3\cos y-2=0$$
$$2\sin x+3\sin y-8=0$$
| The equations can be rewritten as
$$a\cos x+b\cos y=c$$
$$a\sin x+b\sin y = d$$
Squaring both gives us
$$a^2\cos^2x+b^2\cos^2 y+2ab\cos x\cos y=c^2$$
$$a^2\sin^2 x+b^2\sin^2 y+2ab\sin x\sin y=d^2$$
Adding these together gives
$$a^2+b^2+2ab\cos(x-y)=c^2+d^2$$
Which allows us to solve for $\cos(x-y)$ in terms of given co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle What I tried:
Let $A(2,0,4)$, $B(4,1,-1)$, $C(6,7,7)$ then
$$\vec{AB}=(2,1,-5), \vec{AC}=(4,7,3), \vec{BC}=(2,6,8)$$
Then I calculated the angle between vectors:
$$\begin{aligned}
\alpha_1 &= \cos^{-1}\left(\frac{(2,1,-5)(4,7,3)}{\sqrt{2^2+1^2+(-5)^2}\sqrt{... | To find $\alpha_3=\angle ABC$, you should consider $\overrightarrow{BA}\cdot \overrightarrow{BC}$ instead of $\overrightarrow{AB}\cdot \overrightarrow{BC}$. That's why you have a negative cosine and obtained the supplementary angle of the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 1
} |
Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 $ So far I've done this:
LHS $ =\cos^{2}3x-\sin^{2}3x$
$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$
$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$
I can tell I'm going in the right direction but how should I proceed further?
EDIT ... | Expand $(\cos(x)+i\sin(x))^6=\cos(6x)+i\sin(6x) $ and take raeal part of both sides
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Function question $f(x+1)$ and $f(x)$
Given that $f(x+1)-f(x)=4x+5$, $f(0)=6$. Find $f(x)$.
My attempt,
$f(1)-6=5$,
$f(1)=11$
How to proceed then? I've never solve this kind of question before.
| This type of question is called a functional equation. First, suppose $f(x)$ is in the form
$$f(x)=ax+b$$
Then you can use this assumption to solve for $a$ and $b$. So
$$a(x+1)+b-ax-b=4x+5$$
$$a=4x+5$$
Which cannot be, since $a$ is a constant. So $f(x)$ must not be of that form. So, instead consider the quadratic form
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$x^4 -ax^3 +2x^2 -bx +1$ has real root $\implies$ $a^2+b^2 \ge 8$ it is requested to show that if the quartic polynomial $f(x) \in \mathbb{R}[x]$, defined by:
$$
f(x) = x^4 -ax^3 +2x^2 -bx +1,
$$
has a real root, then
$$
a^2 +b^2 \ge 8
$$
this question was asked by @medo, then deleted a few minutes ago. however having ... | Let $x$ be a root. Thus, $x\neq0$ and $b=\frac{x^4-ax^3+2x^2+1}{x}$ and we need to prove that
$$a^2+\frac{(x^4-ax^3+2x^2+1)^2}{x^2}\geq8$$ or
$$(x^6+x^2)a^2-2(x^7+2x^5+x^3)a+x^8+4x^6+6x^4-4x^2+1\geq0,$$
for which it's enough to prove that
$$(x^7+2x^5+x^3)^2-(x^6+x^2)(x^8+4x^6+6x^4-4x^2+1)\leq0$$ or
$$(x^2-1)^4\geq0.$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Give me some hints in calculation this limit. $$\lim_{x\to2\ \\ y\to2}\frac{x^{6}+ \tan (x^{2}-y^{2}) - y^{6}}{\sin(x^{6}-y^{6}) - x^{5}y +xy^{5} + \arctan(x^{2}y -xy^{2})}$$
I used a fact that $$\tan \alpha \sim \alpha \\ \arctan \alpha \sim \alpha \\ \sin\alpha \sim \alpha$$
Since now we have $$\lim_{x\to2\ \\ y\to... | Observe that
\begin{align*}
\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}&=\frac{\left(x^{6}+ x^{2}-y^{2} - y^{6}\right)/(x-y)}{\left(x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}\right)/(x-y)}\\
\end{align*}
$$=\frac{x^5+x^4y+x^3y^3+x^2y^3+xy^4+y^5+x+y}{x^5+x^4y+x^3y^3+x^2y^3+xy^4+y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given that $\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$ Question:
Given that $\displaystyle\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $$\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$$
Attempt:
Substituting $... | I would propose a more efficient strategy. Given that $\sum_{r=-2}^{n}r^3$ is a fourth-degree polynomial in the variable $n$, the same holds for $\sum_{r=0}^{n}r^3=\sum_{r=1}^{n}r^3$, since the two sums differ by a constant. Two different fourth-degree polynomials cannot agree on $5$ points or more: otherwise their dif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Help to understand polynomial factoring I'm following some proof, but got stuck at how the factoring works. I can follow this part:
$$\begin{align*}
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\
&= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\
\end{align*}$$
The next two steps are not so clear to me ... | $\frac{k^2 (k+1)^2 + 4(k+1)^3}{4} = (k+1)^2 \cdot \frac{k^2 + 4(k+1)}{4} = (k+1)^2 \cdot \frac{k^2 + 4k + 4}{4} = (k+1)^2 \cdot \frac{(k + 2)^2}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$ then find $B^{16}$.
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$, then find $B^{16}$.
My Method:
Given $$AB^2=BA \tag{1}$$ Post multiplying with $B^2$ we get
$$AB^4=BAB^2=B^2A$$ Hence
$$AB^4=B^2A$$ Pre Multiplying with $A$ and using $(1)$ we g... | The relation $AB^2 = BA$ allows to reduce the number of $B$s by one. Using this idea, we can show that for every even power $2n$, we have that
$$AB^{2n} = B^nA.$$
Applying this and using $A^4 = I$, we can simplify
$$B^{16} = A^4B^{16} = A^3B^8A = \ldots = B.$$
When we pull the first $A$ through, we half the power of $B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 3
} |
Calculate line integral $\oint _C d \overrightarrow{r } \times \overrightarrow{a }$ Calculate line integral:
$$\oint_C d \overrightarrow{r} \times \overrightarrow{a},$$
where $\overrightarrow{a} = -yz\overrightarrow{i} + xz \overrightarrow{j} +xy \overrightarrow{k}$ , and curve $C$ is intersection of surfaces given by... | $x^2 + y^2 + z^2 = \frac 34\\
x^2 = y\\
y^2 + y + z^2 = 1\\
(y+\frac 12)^2 + z^2 = \frac 54$
$y = \frac {\sqrt 5}{2} \cos\theta - \frac 12\\
z = \frac {\sqrt 5}{2} \cos\theta\\
x = \sqrt {\frac {\sqrt 5}{2} \cos\theta - \frac 12}$
or we could do standard spherical:
$x = \sin\phi\cos\theta\\
y = \sin\phi\sin\theta\\
z ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The perimeter is equal to the area The measurements on the sides of a rectangle are distinct integers. The perimeter and area of the rectangle are expressed by the same number. Determine this number.
Answer: 18
It could be $4*4$ = $4+4+4+4$ but the answer is 18.
Wait... Now that I noticed, the sides are different ... | Let $x$ and $y$ be the sides of the rectangle. Then the given condition implies that $xy = 2x + 2y$. Solving for $y$ in terms of $x$, we get:
$$y = \frac{2x}{x-2}.$$
It's not immediately obvious from this equation for what integer values of $x$ this gives an integer value of $y$. However, by doing a polynomial divis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate $\cos{\frac{\pi}{8}}$? I have to evaluate
$\cos{\frac{\pi}{8}}$
and I'm supposed to do so evaluating first
$\cos^2{\frac{\pi}{8}}$
(since it's an exercise to practice half-angle formulas). Solving this second formula I get to
$\cos^2{\frac{\pi}{8}} = \frac{1}{2} + \frac{1}{2\sqrt[]{2}}$
where I'm stuck... | $$\cos2\theta = 2\cos^2\theta -1$$
Put $\theta = \frac{\pi}{4}$,
$$\frac{1}{\sqrt{2}} = 2\cos^2\frac{\pi}{8} - 1$$
$$\cos^2\frac{\pi}{8} = \frac{\sqrt{2}+1}{2\sqrt{2}} = \frac{2 + \sqrt{2}}{4}$$
$$\implies \cos\frac{\pi}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x... | We have used here $$\sin { 2x } =2\sin { x } \cos { x } \\ \cos { x } =\cos ^{ 2 }{ \frac { x }{ 2 } } -\sin ^{ 2 }{ \frac { x }{ 2 } } \\ \cos ^{ 2 }{ \frac { x }{ 2 } } +\sin ^{ 2 }{ \frac { x }{ 2 } } =1\\ $$$$\frac { 2\sin { x } -\sin { 2x } }{ 2\sin { x+\sin { 2x } } } =\frac { 2\sin { x } \left( 1-\cos { x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Find $\sin(\frac{1}{2}\arccos(\frac{1}{9}))$ This question puts me in mind of the previous alike question beginning with tangent. So, tried to solve it using that method: if $\arccos(\frac{1}{9})=\alpha,$ then $\sin(\frac{\alpha}{2}).$ The formula: $$\sin(\frac{x}{2})=\pm\sqrt{\frac{1-\cos(x)}{2}}$$ doesn't work becaus... | Let $\alpha = \cos^{-1}(\frac{1}{9})$ so $\cos(\alpha)= \frac{1}{9}$, we have
\begin{eqnarray*}
\sin(\frac{1}{2} \cos^{-1}(\frac{1}{9}))= \sin(\frac{1}{2} \alpha) = \sqrt{\frac{1-\cos(\alpha)}{2}} = \sqrt{\frac{1-\frac{1}{9}}{2}}=\color{red}{\frac{2}{3}}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
| Since $z^2+\frac{1}{z^2}=7$, we obtain $z+\frac{1}{z}=3$ or $z+\frac{1}{z}=-3$.
In the first case we obtain
$$z^3+\frac{1}{z^3}=3\left(z^2+\frac{1}{z^2}-1\right)=3(7-1)=18$$
in the second case we obtain
$$z^3+\frac{1}{z^3}=-3(7-1)=-18$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
How to show that $\lim_{n\to\infty}[a_1,\cdots,a_n]$ exists if $a_k\geq 2$ for all $k$?
Consider a sequence of positive real numbers $(a_n)$. Define $[a_1]=\frac{1}{a_1}$ and recursively inductively $[a_1,\cdots,a_n]=\frac{1}{a_1+[a_2,\cdots,a_n]}$. Suppose $a_k\geq 2$ for all $k$. How to show that $\lim_{n\to\infty}... | You only need $a_i \ge 1$.
Define
$$\begin{bmatrix} p_{-1} & p_{0} \\ q_{-1} & q_{0} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. $$
Recursively, set
$$p_k = a_kp_{k-1} + p_{k - 2}, q_k = a_kq_{k-1} + q_{k - 2} \tag{1} $$
for $k \ge 1$. We will show that
$$ [a_1,\dots,a_n,a_{n+1}] = \frac{a_{n+1}p_n + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
find matrix $A$ in other basis Original matrix is:
$$A=\begin{pmatrix}15&-11&5\\20&-15&8\\8&-7&6 \end{pmatrix}$$
original basis: $$\langle \mathbf{e_1},\mathbf{e_2},\mathbf{e_3} \rangle$$
basis where I have to find matrix: $$\langle f_1,f_2,f_3 \rangle$$
Where:
$$f_1=2\mathbf{e_1}+3\mathbf{e_2}+\mathbf{e_3} \\ f_2 = 3\... | To compute the matrix in the other basis We need to compute $T(f_1), T(f_2), T(f_2)$, where $T$ is the linear map defined by $A$ in the basis $e$.
By definition of matrix of a linear map, the coordinates of $T(f_i)$ in the basis $e$ will be obtained if we multiply $A$ by the coordinates of $f_i$ in the basis $e$.
For e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
show that $x+y+z\ge 2$
Let $x,y,z\ge 0$. If
$$x^{2016}y^{2016}+y^{2016}z^{2016}+x^{2016}z^{2016}=1$$ then show
that $$x+y+z\ge 2$$
I have tried
$$(x+y+z)^2-3(xy+yz+xz)=\dfrac{1}{2}\left[ (x-y)^2+(y-z)^2+(z-x)^2\right] \ge 0$$
so we have
$$(x+y+z)^2\ge 3(xy+yz+xz)$$
but I am stuck here. How could I continue this p... | For $z=0$ and $x=y=1$ we get a value $2$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\left(\frac{x+y+z}{2}\right)^{4032}\geq\sum_{cyc}x^{2016}y^{2016},$$
for which we'll prove the following.
Let $x$, $y$ and $z$ be non-negative numbers and nutural $n\geq2$. Prove that:
$$\left(\frac{x+y+z}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation for $4\csc (x+2)-3\cot^2(x+2) = 3$ for $ 0\leq x \leq 8$ For this question, what I did was simply change all of the trigonometric functions to $\sin , \cos, \tan$ form. So the final step before I got stuck was $3(\sin^2 (x+2) + \cos^2 (x+2))=4\sin(x+2)$.
Not quite sure how to proceed from here, any h... | As $\csc^2A-\cot^2A=1$
$$3=4\csc(x+2)-3\cot^2(x+2)=4\csc(x+2)-3\{\csc^2(x+2)-1\}$$
$$\iff\csc(x+2)\{3\csc(x+2)-4\}=0$$
As for real $x+2,\csc(x+2)\ge1$ or $\le1$
$$3\csc(x+2)-4=0\iff\sin(x+2)=\dfrac34$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\i}{\mathrm{i}} \newcommand{\text}[1]{\mathrm{#1}} \newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}\newcomman... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 5
} |
Prove that $\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$ I figure out these thing when "playing" with numbers:
$$3^2-2.2^2+1^2=2=2!$$
$$4^3-3.3^3+3.2^3-1^3=6=3!$$
$$5^4-4.4^4+6.3^4-4.2^4+1^4=24=4!$$
So I go to the conjecture that:
$$\binom{n}{n}(n+1)^n-\binom{n}{n-1}n^n+\binom{n}{n-2}(n-1)^n-...=n!$$
or
$$\sum_{x=0... | We can test the case $n=1$:
\begin{align}\sum_{x=0}^1(−1)^x{1\choose 1−x}(1+1−x)^1&=(−1)^0{1\choose 1−0}(1+1−0)^1+ (−1)^1{1\choose 1−1}(1+1−1)^1\\
&=2+(-1)\\
&=1\\
&=1!\end{align}
Now we assume it holds for $n=k$, that is to say that $$\sum_{x=0}^k(−1)^x{k\choose k-x}(k+1−x)^k=k!$$
We need to prove that it holds for $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 0
} |
Simplifying the Solution to the Cubic I am trying to solve the cubic. I currently have that, for $ax^3+bx^2+cx+d=0$, a substitution to make this monic. Dividing by $a$ gives
$$x^3+Bx^2+Cx+D=0$$
where $B=\frac{b}{a}, C=\frac{c}{a}, D=\frac{d}{a}$. Then, with the substitution $x=y-\frac{B}{3}$, I got
$$y^3+\left(C-\fr... | For the calculation of the roots of the depressed cubic
$$
y^{\,3} + p\,y + q = 0
$$
where $p$ and $q$ are real or complex,
I personally adopt a method indicated in this work by A. Cauli, by which putting
$$
u = \sqrt[{3\,}]{{ - \frac{q}
{2} + \sqrt {\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}}} }}\quad v = - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + \cdots$ Attempt: I found the Fourier series for $f(x) = \begin{cases} 0,& -\pi < x < 0 \\ x/2,& 0 < x < \pi \end{cases}$
a) $a_0 = \frac{1}{2\pi}\int_0^{\pi} r\,dr = \pi/4$
$a_n = \frac{1}{2\pi}\int_0^r \frac{r\cos(nr)}{2}dr = \frac{(-1)^n - 1}{2\pi n^2}$
$b_n = \frac... | First of all your $b_k$ are wrong, they should be:
$$b_k= \frac{(-1)^{k+1}}{2k}$$
Not that it matters beacause of the following. Second of all notice that $f(x)$ is continuos at zero, which is to say $f(0^+)= f(0^-)=0$. Once you get the expansion right is not that hard, just make $x=0$, easy peasy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solution to $7^{2x-2} \equiv 4 \mod13$ I need to solve the congruence
$$ 7^{2x-2} \equiv 4 \mod13 $$
I think I need to use a primitive root to transform it into $2x-2 \equiv ??? \mod (\phi(13)=12)$ but I'm stumped on how to actually do it.
| $\mod 13$:
$$7^2 \equiv 10 \equiv -3 \\
7^4 \equiv (-3)^2 \equiv 9 \equiv -4\\
7^8 \equiv (-4)^2 \equiv 16 \equiv 3\\
7^{10} \equiv 7^2\cdot 7^8 \equiv (-3)\cdot 3\equiv -9 \equiv 4
$$
So
$$
2x-2 = 10 \\
x = 6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Two ways to calculate the primitive of $f(x) =\frac{x^2}{x^4-1}$ but two different results, why? When I search primitives of $f(x)= \dfrac{x^2}{x^4-1}$, using partial fraction decomposition the solutions are
$\boxed {F(x) =\dfrac{1}{4} \ln|x-1| + \dfrac{1}{4} \ln|x+1| + \dfrac{1}{2} \arctan{x} + C}$
But if I use an o... | Alternate forms exist. According to Wolfram Alpha:
The first one and the second one are the same because:
$$
\tanh^{-1}(x):=\frac12\log\left(\frac{1+x}{1-x}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to get the $\phi$ from $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$? $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$,
where $c=\sqrt{a^2+b^2+2ab\cos(\theta)}$, and $\displaystyle\tan(\phi)=\frac{b\sin(\theta)}{a+b\cos(\theta)}$.
I want to know how to get to this result.
I'm able to derive $c$ by taking the derivative of the e... | Use the sin addition formula $\sin(\alpha+\beta)=\sin \alpha \cos \beta + \cos \alpha \sin \beta$
\begin{eqnarray*}
a \sin x + \underbrace{b \sin(x+\theta)}_{ b\sin x \cos \theta+b \cos x \sin \theta}= \underbrace{c \sin(x+ \phi)}_{b\sin x \cos \phi+b \cos x \sin \phi} \\
(a + b \cos \theta) \color{red}{\sin x} + b \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
For a odd prime $p, \exists a,b$ such that $a^2 + ab + b^2 \equiv 0 \pmod{p} \iff \exists x,y $ such taht $x^2 + xy + y^2 = p$? We know that $p=x^2+xy+y^2$ if and only if $p \equiv 1 \pmod {3}$.
But I need $a^2+ab+b^2 \equiv 0 \pmod{p} $ if and only if $p \equiv 1 \pmod {3}$, more generalized theorem.
I think that i s... | Given any sum with a particular prime factor $p$ we can isolate the factor $p$ via Eisenstein integer multiplication.
Take your example with $10, 26$ having the sum $28×37$. How to isolate the factor $37$?
First get rid of the common factor $2^2$ which is totally useless:
$5^2+(5×13)+13^2=7×37$
Next we associate thus ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the probability of the sum of four dice being 22? Question
Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?
My Approach
I simplified it to the equation of the form:
$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\le... | In order for the sum to equal 22, either three dice equal $6$ and one equals $4$, or two dice equal $6$ and two dice equal $5$. The number of valid outcomes thus equals:
$${4 \choose 1} + {4 \choose 2} = 4 + 6 = 10$$
As such, the probability of the four dice having a sum of $22$ equals:
$$\frac{{4 \choose 1} + {4 \choo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 11,
"answer_id": 9
} |
$x^4 + y^4 \mod p$ The primes $p = 5, 13, 17, 29$ have the property that there exist fewer than $p$ values for $x^4 + y^4 \mod p$. For example, $x^4 \equiv 0$ or $1 \mod 5$ so $x^4 + y^4 \equiv 0, 1$ or $2$, but never $3$ or $4$, $\mod 5$. Are there any other primes with this property?
The question arose in connectio... | For a given prime $p$, let $Z_p$ denote the ring of integers, mod $p$, and let $f\colon Z_p^2 \to Z_p$ be defined by $f(x,y) = x^4+y^4$.
Partial result: If $p \equiv -1\;\,(\text{mod}\;4)$, then $f$ is surjective.
Proof:
Assume $p$ is a prime, with $p \equiv -1\;\,(\text{mod}\;4)$.
Then $-1$ is not a quadratic resi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
How can the sum of squares be negative?
If $a,b,c,d$ are the roots of the equation $x^4-Kx^3+Kx^2+Lx+M=0$, where $K,L,M$ are real numbers, then the mininmum value of $a^2+b^2+c^2+d^2$ is?
My answer:
$\sum a=K,\ \sum ab=K\implies$
$a^2+b^2+c^2+d^2=K^2-2K=(K-1)^2-1$. For $K=1$, $(a^2+b^2+c^2+d^2)_{min}=-1$
This matches... | Consider $K=1,$ that is, $$f(x)=x^4-x^3+x^2+Lx+M.$$ We have that
$$f''(x)=2(6x^2-3x+1).$$ Note that $$f''=0$$ has no real roots. Thus $f'=0$ has a real root and so $f=0$ has at most two real roots. In other word, at least two roots of the equation $$f(x)=0$$ are complex. Thus there is no contradiction with the fact tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$ Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$.
I completed the first part of the question by finding the roots of the first equation. I obtained $3$ roots, one of them being $-1$ and the other two complex. It is ev... | The roots of $x^2+x+1$ are $\omega$ and $\omega^2$ (The two complex cube roots of unity)
and $\omega^3=1$. Therefore
$$ \omega^{1990}+\omega^{200}+1=\omega+\omega^2+1=0 $$
and
$$ \omega^{2\times1990}+\omega^{2\times200}+1=\omega^2+\omega+1=0$$
So $\omega$ and $\omega^2$ are the common roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this ... | On the LHS you have $$\frac{a^5}{b^3c^3} + \frac{b^5}{a^3c^3} +
\frac{c^5}{a^3b^3}$$
Relabel this as $$x_1 + x_2 + x_3$$
The standard way to proceed with AM-GM is to fiddle around with products of $x_1, x_2, x_3$ until you find some $\sqrt[n_1 + n_2 + n_3]{x_1^{n_1}x_2^{n_2}x_3^{n_3}}$ that yields one of the RHS terms... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Reflection across a line Let $g$ be a line with equation $g:ax+by+c=0$ in Hesse normal form. I want to show that the reflection across $g$ is described by \begin{equation*}\binom{x}{y}\mapsto \binom{x}{y}-2(ax+by+c)\binom{a}{b}\end{equation*}
At the reflection across $g$ it holds the following for the image $P'$ of ea... | This reflection is an affine map, so we can represent it as linear map if we embed plane in $\mathbb R^3$ with $(x,y)\mapsto (x,y,1)$.
I will use the same idea as Robert Z, $P'$ is reflection of $P$ along the line $ax + by + c=0$ if two conditions are met:
(1) $\ \displaystyle\frac{P+P'}2$ lies on the line $ax + by + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determine if these functions are injective
Determine if the following functions are injective.
$$f(x) = \frac{x}{1+x^2}$$
$$g(x) = \frac{x^2}{1+x^2}$$
My answer:
$f(x) = f(y)$
$$\implies \frac{x}{1+x^2} = \frac{y}{1+y^2}$$
$$\implies x+xy^2 =y+yx^2$$
$$\implies x=y$$
Hence $f(x)$ is injective
$g(x) = g(y)$
$$\implies... | If $x + xy^2 = y+yx^2$, then $(x - y) = yx^2 - xy^2 = xy(x-y)$. This gives $(x-y)(xy-1) = 0$, hence either $x=y$ or $xy=1$. You can check that for example, $f(\frac 12) = f(2) = \frac 25$.
Hence, $f$ is not injective.
The answer to the second question is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$
Evaluate the following limit.
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
My Attempt:
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-... | Clearly, for $b<1$, $-\infty$ and for $b>1$, $\infty$ (because $\sqrt{1-a/x}-\sqrt b$ tends to the constant $1-\sqrt b$).
Then
$$\sqrt{x-a}-\sqrt{x}=\frac{x-a-x}{\sqrt{x-a}+\sqrt{x}}$$ and the limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Simplifying $\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$ I'm struggling trying to simplify $$\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$$
Here's my attempt:
All of these coefficients can be converted to base $2$ exponentials.
$$\frac{2^{4^{(x+1)}}+20^\ (2^{2^{(2x)}})}{2^{x-3}2^{3^{(x+2)}}}$$
Doing some algebra:
$$\... | You just need to get more comfortable converting bases. Then any way you want to do it will work:
$\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$
$=\frac {2^{4(x+1)} + 5*4*2^{2*2x}}{2^{x-3}2^{3(x+2)}}$
$=\frac {2^{4x + 4} + 5*2^22^{4x}}{2^{(x-3)+3x + 6}}$
$=\frac {2^{4x+4} + 5*2^{4x+2}}{2^{4x + 3}}$
$= \frac {2^{4x+2}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving solution set inequalities problems I need some guidance with solving problems of the following nature:
$$|x^2-3x-1|<3$$
Instinctively, my way of solving it is splitting it into the following equations:
$$x^2-3x-1< 3$$
$$x^2-3x-1> -3$$
Solving $x^2-3x-1< 3$
$$x^2-3x-4< 0$$
$$(x+1)\ (x-4)< 0$$
$$x < -1... | $$|x^2-3x-1|<3\Rightarrow -3<x^2-3x-1<3\\\Rightarrow -3<x^2-2(3/2)x+(3/2)^2-(3/2)^2-1<3\\\Rightarrow 1/4<(x-3/2)^2<25/4$$
Now you have two inequation:
*
*$1/4<(x-3/2)^2\Rightarrow x-3/2<-1/2\space \text{or}\space x-3/2>1/2\Rightarrow x<1\space \text{or}\space x>2$.
*$(x-3/2)^2<25/4\Rightarrow -5/2<x-3/2<5/2\Rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
arctan series multisection by roots of unity I'm trying to multisect the series for $\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots$ using the method of roots of unity, as described in the paper linked in one of the comments in How to express a power series in closed form. I wan... | The problem is that the arctangent series has already had this procedure applied once:
$$ \arctan{x} = \frac{1}{2} (-i\log{(1+ix)}-(-i\log{(1-ix)})). $$
If one applies the sifting procedure again, the answer comes out the same (each term in the sum is sifted and returns the same terms as the original, which add to give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Given $1 \le |z| \le 7$ Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$ Given $1 \le |z| \le 7$
Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$
I have taken $z=r e^{i \theta}$ $\implies$ $1 \le r \le 7$
Now $$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{r \cos \thet... | $\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{z^2+24}{4z}\right|$
The function $f(z)=\frac{z^2+24}{4z}$ is analitic on $1\leq|z|\leq7$. Therefore, by the maximum modulus theorem its maximum absolute value is attained at the boundary. The boundary are the circles $|z|=1$ and $|z|=7$.
For $|z|=1$, observe that $z^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find $\det(A^3B^2+A^2B^3)$ for $A,B$ s.t. $AB = BA$ given $\det A,\det B$. If two matrices $A$ and $B$ are such that $AB = BA$ with $\det A = 1$ and $\det B = 0$, then what is $\det(A^3B^2 + A^2B^3)$ ?
| \begin{align*}
\text{det}(A^3B^2+A^2B^3) & = \text{det}A^2 \text{det} B^2\text{det}(A+B)\\
& = (\text{det}A)^2 (\text{det} B)^2\text{det}(A+B)\\
&=0.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2367774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understa... | I was thinking of writing
\begin{align}
n^4 - n^2 + 64 &= (n^2 + A)^2 \\
n^4 - n^2 + 64 &= n^4 + 2An^2 + A^2 \\
2An^2 + n^2 + A^2 &= 64 \\
n^2(2A+1) &= 64 - A^2 \\
n^2 &= -\dfrac{A^2 - 64}{2A+1} \\
n^2 &= -\frac 12A + \frac 14 + \frac{255}{4(2A+1)} \\
4n^2 &= -2A + 1 + \frac{255}{2A+1}
& \text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
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.