Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{... | If you are familiar with the Maclaurin series for the basic trig functions, you can solve this limit and others like it very quickly or even in your head. It also avoids differentiation mistakes from L'Hospital's.
In your case, second order expansion is enough. Taking
$$
\sin(x)\sim x-\frac{x^3}{6}\\
x\cos(x)\sim x(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Proving a trigonometric expression is identical to $2(\operatorname{cosec}^{2}{B}-1).$ I came across this trigonometric identity:
$$\frac{\operatorname{cosec}{B} - \cot{B}}{\operatorname{cosec}{B} + \cot{B}} + \frac{\operatorname{cosec}{B} + \cot{B}}{\operatorname{cosec}{B} - \cot{B}} = 2(\operatorname{cosec}^{2}{B} - ... | Get rid of cosecants and cotangent:
$$
\frac{\csc B-\cot B}{\csc B+\cot B}=
\frac{\dfrac{1}{\sin B}-\dfrac{\cos B}{\sin B}}
{\dfrac{1}{\sin B}+\dfrac{\cos B}{\sin B}}=
\frac{1-\cos B}{1+\cos B}
$$
Thus your left-hand side is
$$
\frac{1-\cos B}{1+\cos B}+\frac{1+\cos B}{1-\cos B}=
\frac{(1-\cos B)^2+(1+\cos B)^2}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Polar to cartesian form of r=sin(3θ) $$\sin(2θ) = \sinθ\cosθ + \cosθ\sinθ = 2\sinθ\cosθ $$
$$\cos(2θ) = \cosθ\cosθ - \sinθ\sinθ = \cos²θ - \sin²θ $$
$$\sin(3θ) = \sin(2θ+θ) $$
$$= \sin(2θ)\cosθ + \cos(2θ)\sinθ $$
$$= 2\sinθ\cos²θ + \cos²θ\sinθ - \sin³θ
$$ $$= 2\sinθ(1-\sin²θ) + (1-\sin²θ)\sinθ - \sin³θ
$$ $$= 2\sinθ ... | Hint
$$\sin3\theta=3\sin(\theta)\cos^2(\theta)-\sin^3(\theta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$ What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$?
My effort:
We have $x+y-1\lt \lfloor x+y\rfloor \leq x+y$. What can I do after this? P... | As by definition, $\,\lfloor x\rfloor\le x <\lfloor x\rfloor+1$, we deduce
$$\lfloor x\rfloor+\lfloor y\rfloor\le x+y<\lfloor x\rfloor+\lfloor y\rfloor+2, $$
whence
$$\lfloor x\rfloor+\lfloor y\rfloor\le \lfloor x+y\rfloor\le\lfloor x\rfloor+\lfloor y\rfloor+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}$ I want to evaluate:
$$\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}.$$
Here's what I did. We know that as soon as ${x\to 0}$
$$1 - \cos x = \frac{x^2}{2} + O(x^4)$$
Therefore
$$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$
Now... | More generally, for any integer $N\geq 1$,
$$\lim_{x\to 0}\frac{1-\prod_{n=1}^N \sqrt[n]{\cos(nx)}}{x^2}=\frac{N(N+1)}{4},$$
because
$$\prod_{n=1}^N \sqrt[n]{\cos(nx)}=\prod_{n=1}^N \sqrt[n]{1+\frac{(nx)^2}{2}+o(x^2)}=\prod_{n=1}^N \left(1+\frac{(nx)^2}{n2}+o(x^2)\right)\\
=\prod_{n=1}^N \left(1+\frac{nx^2}{2}+o(x^2)\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solve for $x$ in $\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$ I am taking physics right now and I have gotten my problem down to the following equation:
$$\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$$
I am looking for some guidance as to what to do with the square roots in order to solve for $x$. I know I can't take the square ... | \begin{align}
& \sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=.25 \\
& \sqrt{3.5^2+x^2}=.25+\sqrt{3.0^2+x^2} \\
\end{align}
Next, square both sides. You get:
\begin{align}
& 3.5^2+x^2=(.25+\sqrt{3.0^2+x^2})^2 \\
3.5^2+x^2=.0625+.50(\sqrt{3.0^2+x^2})+3.0^2+x^2
\end{align}
Simplifying further gets you: $3.1875=.50(\sqrt{3.0^2+x^2})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Question about Diophantine equations? Solve the Diophantine equation.
$6x+ 9y = 15$
We can see that $x = 1$ and $y=1$ solves the equation, so all solutions are given by $x = 1 + 3n$ and $y = 1 - 2n, n \in \mathbb{Z}$. By using:
$$x=x_0 + n \times \frac{b}{gcd(a,b)}$$ and $$y=y_0 - n \times \frac{a}{gcd(a,b)}$$
Howe... | Given $6x+9y=15\implies 2x+3y=5$
\begin{equation}
x = \frac{(5 - 3 y)}{2}\quad\land\quad y =\frac{(5 - 2 x)}{3}
\end{equation}
A pattern in a spreadsheet suggests $\quad x = (3 n + 1), n\in\mathbb{Z}\quad$
and, by substitution we get
\begin{equation}
y =\frac{(5 - 2 (3n-1)}{3}=\frac{(5 - 6n+1)}{3}=\frac{(6 - 6n)}{3}=(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\c... | Hint:
$$\sin ^{4}\theta ={\frac {3-4\cos(2\theta )+\cos(4\theta )}{8}}\!$$
$$\cos ^{4}\theta ={\frac {3+4\cos(2\theta )+\cos(4\theta )}{8}}\!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 5
} |
Conjecture $\sum_{n=1}^\infty\frac{n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}=\frac{1}{12\pi^2}$ Wolfram Alpha numerical calculation shows that the quantity
$$
\sum_{n=1}^\infty\frac{12\pi^2n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}
$$
is 1 with high accuracy. Can anybody prove the resulting conjecture:
$$
\sum_{n=1}^\infty\frac{n... | Take the function
$$
f(z) = \frac{z^2}{\sin (\pi z) \left(\cosh \left(\sqrt{3} \pi z\right)-\cos (\pi z) \right)}.
$$
Observe that $\sin (\pi z) \left(\cosh \left(\sqrt{3} \pi z\right)-\cos (\pi z) \right)=2\sin (\pi z) \sin (\pi z \omega) \sin (\pi z\omega^2) $, where $\omega=e^{2\pi i/3}$. This means that $f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
How to show this conjecture：$\frac{1}{\sqrt{a_{1}}}+\frac{1}{\sqrt{a_{2}}}+\cdots+\frac{1}{\sqrt{a_{n}}}\ge\frac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$ I have prove this inequality
$$\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}>2\sqrt{n+1}-2$$
because
$$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>\sum_{k=1}^{n}\d... | Note that
$$
\begin{align}
\frac1n\sum_{k=1}^n\frac2{\sqrt{a+kd}+\sqrt{a+(k-1)d}}
&=\frac1n\sum_{k=1}^n\frac2d\left(\sqrt{a+kd}-\sqrt{a+(k-1)d}\right)\\
&=\frac2{nd}\left(\sqrt{a+nd}-\sqrt{a}\right)\\[4pt]
&=\frac2{\sqrt{a+nd}+\sqrt{a}}\tag{1}
\end{align}
$$
Let $a_k=\sqrt{a+kd}$.
$$
\begin{align}
\frac1{n+1}\sum_{k=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $\cos^n(x) - \sin^n(x)=1$
Solve the equation $\cos^n(x) - \sin^n(x)=1,n \in \mathbb{N}-\{0\}$
If $n$ is even then $\cos^n(x) = \sin^n(x)+1$ is only possible if $\sin(x)=0$ therefore the solution is $x=k\pi, k \in \mathbb{Z}$.
I'm having problems with $n$ odd case.
UPDATE
For $n=1$ we have $\cos(x)... | It suffices to investigate the solutions over $[0,2\pi)$, as all solutions will be some integer multiple of $2\pi$ away from a solution in $[0,2\pi)$. Let $f(x) = \cos^n(x)-\sin^n(x)$, with $n$ odd and $n\ge 3$. Then
$$f'(x) = n(\cos^{n-1}(x)(-\sin x) - \sin^{n-1}(x)\cos(x)) = -n\sin(x)\cos(x)(\cos^{n-2}(x)+\sin^{n-2}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How can I evaluate: $\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$ I need to find this limit:
$$\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$$
The answer I got from using the limit laws is $\sqrt{\infty} - \sqrt{\infty}$.
How do I proceed now?
Added I took the conjugate of the function and I got a new... | If you know Differential Calculus, you may also use L'hopital's Rule to solve for the limit. Just as a recap, one instance of L'hopital's Rule states:
Given $\frac{f(x)}{g(x)}$ with some arbitrary limit to $c$ imposed on it, if all of the following constraints are met:
$\lim_{x\rightarrow{c}} f(x) = 0 $
$\lim_{x\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1941003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
2 Related Questions About Finding A Closed Form
*
*Consider the sequence defined by
$$
\begin{cases}
s_0=0\\
s_1=3\\
s_n=6s_{n-1}-9s_{n-2} & \text{if }n\ge 2
\end{cases}
.$$
Find a closed form for $s_n$.
*Consider the sequence defined by
$$
\begin{cases}
t_0=5\\
t_1=9\\
t_n=6t_{n-1}-9t_{n-2} & \text{if }n\ge 2
\... | Hint:
For you first problem you could rewrite it as a matrix equation
\begin{align}
\begin{pmatrix}
s_n\\
s_{n-1}
\end{pmatrix}
=
\begin{pmatrix}
6s_{n-1}-9s_{n-2}\\
s_{n-1}
\end{pmatrix}
=
\begin{pmatrix}
6 &-9\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
s_{n-1}\\
s_{n-2}
\end{pmatrix}
=
\begin{pmatrix}
6 &-9\\
1 & 0
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_0^{2\pi}\frac1{5-4\cos x}\ dx$ $$\int_0^{2\pi}\frac1{5-4\cos x}\ dx$$
How do I compute this integral? An online integral calculator gives an antiderivative as
$$\frac{2\arctan(3\tan\frac x2)}3$$
but then gives the definite integral as $\frac{2\pi}3$. Obviously this doesn't make sense as the antiderivative va... | A simple way:
$$ I=\int_{0}^{2\pi}\frac{dx}{5-4\cos x}=\int_{0}^{\pi}\left(\frac{1}{5-4\cos x}+\frac{1}{5+4\cos x}\right)\,dx\tag{1}$$
$$ I = 10\int_{0}^{\pi}\frac{dx}{25-16\cos^2 x}=20\int_{0}^{\pi/2}\frac{dx}{25-16\cos^2 x} \tag{2}$$
and now, by setting $x=\arctan t$,
$$ I = 20 \int_{0}^{+\infty}\frac{dt}{25(1+t^2)-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Nth power of the following matrix The matrix is
$ \begin{pmatrix}0&1\\-1&0\end{pmatrix}^n $
for $n=2 \implies \left(\begin{matrix}-1 & 0\\ 0 &-1\end{matrix}\right)$ for $n=3 \implies \begin{pmatrix}1&0\\0&1\end{pmatrix}$ for $n=4 \implies \begin{pmatrix}-1&0\\0&-1\end{pmatrix}$
so I assume that for every $n=2k $, wh... | You just have to calculate the first power:
$$A=\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}\;,\;\;A^2=\begin{pmatrix}\!-1&0\\0&-1\end{pmatrix}=-I$$
and that's all we need, since then
$$A^3=A\cdot A^2=-A\;,\;\;A^4=I\;,\;\;A^5=A;,\ldots\text{etc.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Arrangements of a,a,a,b,b,b,c,c,c in which no three consecutive letters are the same Q: How many arrangements of a,a,a,b,b,b,c,c,c are there such that
$\hspace{5mm}$ (i). no three consecutive letters are the same?
$\hspace{5mm}$ (ii). no two consecutive letters are the same?
A:(i). 1314. ${\hspace{5mm}}$ (ii). 174.
I t... | This answer is based upon a generating function of generalized Laguerre polynomials
\begin{align*}
L_k^{(\alpha)}(t)=\sum_{i=0}^k(-1)^k\binom{k+\alpha}{k-i}\frac{t^i}{i!}
\end{align*}
The Laguerre polynomials have some remarkable combinatorial properties and one of them is precisely suited to answer problems of thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
What time are the minute and hour hands of a clock perpendicular?
At noon the minute and hour hands of a clock coincide. Assuming the hands of the clock move continuously:
a) What is the first time $T_1$ when they are perpendicular?
b) What is the next time $T_2$ when they again coincide?
I believe I have solved ... | At noon the hands coincide. In $15 + \delta$ minutes, the angle between them will be $90^\circ$.
The minute hand moves at $\dfrac{360^\circ}{60\ \text{min}} = 6$ degrees per minute. The hour hand moves at $\dfrac{360^\circ}{12\ \text{hr}} = \dfrac 12$ degrees per minute. So we need to solve
\begin{align}
(15+\delta)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solving $\lim\limits_{x \to 0} \frac{\ln(1+x)}{\ln(1+4x+x^2)}$ without L'Hospital or Sandwiching Using the L'Hospital I got: $\lim\limits_{x \to 0} \frac{\ln(x+1)}{\ln(1+4x+x^2)} = \lim\limits_{x \to 0}\frac{\frac{1}{1+x}}{\frac{4+2x}{1+4x+x^2}}= \frac{1}{4}$,
I then wondered, if I could sandwich it andobserved that ... | Hint:
You can evaluate it based on this limit, $$\lim_{x\to0}\frac{\ln(x+1)}x=1,$$ by writing,
$$\lim\limits_{x \to 0} \frac{\ln(1+x)}{\ln(1+4x+x^2)}=\lim\limits_{x \to 0} \frac{\ln(1+x)}{x}\cdot\frac{4x+x^2}{\ln(1+4x+x^2)}\cdot\frac{x}{4x+x^2}=\ldots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is my proof that $x^2+y^2+z^2 ≥ xy+yz+xz$ correct? The question:
Prove that $x^2+y^2+z^2 ≥ xy+yz+xz$ for all real numbers $x$, $y$ and $z$.
This problem has been posed before, but my question is whether my proof below is correct, since it seems the other answers to this problem are different.
If $x$, $y$ and $z$ are re... | Your concerns were already addressed in the comments, so there is no need to keep underlying your actual sign mistake. But on a meta-mathematical point of view, something should have struck you, since your method would have derived a symmetric inequality from a non-symmetric one, kind of strange.
A working and fast pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any succ... | If you expand the squares, you'll see that every column is a linear combination of
$$
\begin{bmatrix}a^2 \\ b^2 \\ c^2 \\ d^2\end{bmatrix},
\begin{bmatrix}a \\ b \\ c \\ d\end{bmatrix}, \text{ and }
\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}.
$$
There are four columns that are linear combinations of three vectors. So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 2
} |
If $x*y$ is the operation $x*y = xy/(x+y+1)$, then how does $x*y*z = xyz(x+y+1)/(x+y+z+xy+xz+1)$? $$x*y = \frac{xy}{x+y+1}$$
What I got was,
$$x*(y*z) = x* \frac{yz}{y+z+1} =x\frac{\frac{yz}{y+z+1}}{x+\frac{yz}{y+z+1}+1}$$
The book's answer is
$$x*(y*z) = x*\frac{yz}{y+z+1} = \frac{xyz(y+z+1)}{x+y+z+xy+xz+1}$$
| You can go on from where you stopped:
$$
\frac{x\dfrac{yz}{y+z+1}}{x+\dfrac{yz}{y+z+1}+1}=
\frac{\dfrac{xyz}{y+z+1}}{\dfrac{(x+1)(y+z+1)+yz}{y+z+1}}=
\frac{xyz}{(x+1)(y+z+1)+yz}
$$
Thus you get
$$
x*(y*z)=\frac{xyz}{x+y+z+xy+yz+zx+1}
$$
The book's answer is wrong. Set $x=y=1$. Then
$$
y*z=\frac{1}{3}
$$
and so
$$
x*(y*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proof $\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$
$\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$ Given hint: $(n^2+3n+2) = (n+2)(n+1)$
I've tried converting the series to a geometric one but failed with that approach and don't know other methods for normal series that help determine the actu... | $$
\begin{aligned}
& S_{0} = \sum_{n=0}^{\infty} \frac{n^{0}}{4^{n}} = \sum_{n=0}^{\infty} \frac{1}{4^{n}} = \sum_{n=0}^{\infty} (1/4)^{n} = \frac{1}{1 - (1/4)} \Rightarrow \color{red}{S_{0} = \frac{4}{3}} \\ \\
& S_{1} = \sum_{n=0}^{\infty} \frac{n^{1}}{4^{n}} = \sum_{n=0}^{\infty} \frac{n + 1 - 1}{4^{n}} = \sum_{n=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
Find all $0^\circ\leq A\leq360^\circ$ with $\tan A + \tan 2A + \tan 3A = 0$ solve:
$\tan A+ \tan 2A+ \tan 3A=0$
My Attempt:
$$\tan A+\tan 2A+\tan 3A=0$$
$$\tan A+\frac {2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
$$\frac {\tan A-\tan^3A+2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
What sh... | $$\tan(a) + \tan(2a) + \tan(3a) = 0 $$
$$\tan(2a) = 2 \cdot \frac{\tan(a)}{ (1 - \tan(a)^2) }$$
$$\tan(3a) =\tan(2a + a)$$
$$= \frac{\tan(a) + \tan(2a }{ 1 - \tan(a) \cdot \tan(2a)}$$
$$= \frac{(\tan(a) + \frac{2 \cdot \tan(a }{1 - \tan(a)^2)} }{ \frac{(1 - tan(a) \cdot 2 \cdot \tan(a) }{ 1 - \tan(a)^2}}$$
$$\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Limit to infinity using $\delta$-$\epsilon$ language. I'm trying to prove:
$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2 + x - 1}{3x^2 + 2x + 1} = \frac{1}{3}$$
using $\delta$-$\epsilon$ language. Here is my attempt:
Proof: We wish to show that as $x$ gets arbitrarily large that the difference between $f(x... | It's correct, but it can be simplified.
When you are at
$$
\left|\frac{x-4}{9x^2+6x+3}\right|
$$
you can observe that it's not restrictive to assume $x>4$; next, $x-4<x$ and $9x^2+6x+3>9x^2$, so
$$
\frac{x-4}{9x^2+6x+3}<\frac{x}{9x^2}<\frac{1}{9x}
$$
Therefore we can take
$$
\delta=\max\left\{4,\frac{1}{9\varepsilon}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If the value of $xy$ , If $ x^2 + y^2 = 34$ and $x + y =8$ If $ x^2 + y^2 = 34$ and $x + y =8$ find the value of xy
So what should I be considering here the sum of two squares or..
I'm struggling here & any help to figure this out will be kindly appreciated
| Use:
$$x+y=8\Longleftrightarrow x=8-y\Longleftrightarrow x^2=\left(8-y\right)^2$$
So:
$$x^2+y^2=34\Longleftrightarrow\left(8-y\right)^2+y^2=34$$
So, for $y$ we find $y=3$ or $y=5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding features of parabolic formula I have a question that requires me to find the intercepts and the turning point of a parabola from it's formula, but I'm not really sure how to find all this information out from the equation below. I tried using completing the square but I ended up with a weird answer that I don't... | There are two intercepts. The $y$-intercept, and the $x$-intercepts.
$y$-intercept:
To find the $y$-intercept, you simply plug in $x=0$ into the equation to get the $y$ value. So thus, we have$$y=-3\cdot 0^2-5\cdot 0+2=2$$ so the $y$-intercept is $(0,2)$.
$x$-intercepts:
The $x$ intercepts are points that give $y=0$ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Is this inequality true? And can it be shown? If $x$ and $y$ are both positive, odd integers, greater than $1$, is it true that,
$(y-1) \left( \displaystyle\sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right) - \ln(4) \right) + \displaystyle\sum_{i=\frac{x+1}{2}}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right)... | By multiplying and assembling the sums
$(y-1) \left( \displaystyle\sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right) - \ln(4) \right) + \displaystyle\sum_{i=\frac{x+1}{2}}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) = $
$= \displaystyle\sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) - \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A problem regarding inequality There is a question in an algebra textbook that asks to prove $|x^2-1|\lt\frac{21}{100}$, given that $|x-1|\lt\frac{1}{10}$.
Is the question correct(solvable)? If it is, how?
| Suppose $|x - 1| < \frac{1}{10}$. Then
$$|x^2 - 1| = |(x-1)(x+1)| = |x-1|\cdot|x+1| < \frac{1}{10} |x+1|$$
We now need to show $|x+1| < \frac{21}{10}$ in order to show $|x^2 - 1| < \frac{21}{100}$.
Since $|x-1| < \frac{1}{10}$, this means
$$ -\frac{1}{10} < x-1 < \frac{1}{10}.$$
Adding 2 to each side gives us:
$$ -\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Hard problem I can not prove properly. I managed to concoct a weak proof for the below problem but I would love to see if there are any different ways to solve it.
Find the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3 .... +2^3-1^3$
the options listed are
a)$1^2$
b)$2^2$
c)$1007^2$
d)$2014^2$
I thi... | Consider:
\begin{align}
A_{2014} &= \sum_{n=1}^{1007} [(2n)^3 - (2n-1)^3] = \sum_{n=1}^{1007} ((2n) -(2n-1))((2n)^2 + (2n)(2n-1) + (2n-1)^2) \\
&= \sum_{n=1}^{1007} (12n^2 - 6n + 1) \\
&= 4087631519 = 19^{2} \cdot 29^{1} \cdot 53^{2} \cdot 139 = 29 \cdot 139 \cdot (1007)^2.
\end{align}
In comparison: $2014^{2} = 2^2 \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Probability that number is divisible by 5 Two numbers $x$ and $y$ are chosen at random without replacement from the set $${1,2,\cdots,5n}$$.then the probability that $x^4-y^4$ is divisible by $5$ is----
I could write down $x^4-y^4$ as $(x-y)(x+y)(x^2+y^2)$ and then i could easily figure out the possoble cases for first... | Note that if $x$ is not a multiple of $5$ then $x^4-1$ is divisible by $5$ (why?).
This implies that if $x$ and $y$ are not a multiple of $5$ then $x^4-y^4=(x^4-1)-(y^4-1)$ is divisible by $5$.
Can you take it from here?
P.S. Finally you will find that $x^4-y^4$ is NOT divisible by 5 iff the set $\{x,y\}$ ($5n(5n-1)/2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac... | Considering the problem geometrically, then
$$
f(x) = x^{\,2} - ax + 1
$$
is a vertical parabola, convex (upward concave), with axis $x=a/2$ and vertex in $V=(a/2,1-a^2/4)$.
Thus the parameter $a$ is then just a translation parameter.
So, from the above and with the help of the sketch it is clear that, to satisfy the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 10,
"answer_id": 3
} |
A Cauchy–Schwarz-like inequality involving dot products Let $\mathbf{a}_1,\ldots,\mathbf{a}_N$ and $\mathbf{b}_1,\ldots,\mathbf{b}_N$ be vectors in $\mathbb{R}^n$. Prove that:
\begin{equation}
\bigg(\sum_{i=1}^{N} \mathbf{a}_i^\top\mathbf{a}_i\bigg) \bigg(\sum_{i=1}^{N} \mathbf{b}_i^\top\mathbf{b}_i\bigg)
\geq
N \sum... | The inequality doesn't seem to be true. Consider the following example in $\mathbb{R}^2$
\begin{align}
\mathbf{a}_1 = 2\mathbf{e}_1,\ \ \mathbf{a}_2 = \mathbf{e}_2 \ \ \text{ and } \ \ \mathbf{b}_1 = 2\mathbf{e}_2,\ \ \mathbf{b}_2 = \mathbf{e}_1.
\end{align}
Observe
\begin{align}
(\mathbf{a}_1^T\mathbf{a}_1+\mathbf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Internal angle bisectors and right-angled triangle? Imagine a right angled triangle. Two of its internal bisectors are of length $7cm$ and $4cm$, respectively. The first internal angle bisector belongs to the angle of the side $a$ and the second one belongs to the angle of the side $b$ Calculate the length of the hypot... | Here is a sketch of the triangle and its two internal bisectors $\overline{AP}=7$ , $\overline{BQ}=4$ :
We shall need a cosine law:
$$
\cos(A) = 2\cos^2(\frac{1}{2}A)-1 \quad ; \quad
\cos(B) = 2\cos^2(\frac{1}{2}B)-1
$$
Three highly non-linear equations with three unknowns can be derived then:
$$
2\cos^2(\frac{1}{2}A)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find a line which is tangent to the curve $y=x^4-4x^3$ at 2 points How can I solve this? Should I set the two points as $a^4-4a^3$ and $b^4-4b^3$?
| Let $(t,t^4-4t^3)$ be a point of contact, then
$$\frac{dy}{dx} = \frac{\dot{x}}{\dot{y}} = 4t^3-12t^2$$
The equation of tangent is
$$(y-t^4+4t^3)=4t^2(t-3)(x-t)$$
Let $(s^4-4s^3,s)$ be the self-intersection of its tangent.
\begin{align*}
(s^4-4s^3)-(t^4-4t^3) &= 4t^2(t-3)(s-t) \\
(s^4-t^4)-4(s^3-t^3) &= 4t^2(t-3)(s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find a number $\alpha > 1$ such that the following holds
We want to find a real number $\alpha > 1$ such that
$$ \int\limits_{0}^{\infty} \frac{ dx }{(1+x^{\alpha})^{\alpha}} = 1
$$
What I got so far
First, I used Integration by parts to obtain
$$ \int\limits_{0}^{\infty} \frac{ dx }{(1+x^{\alpha})^{\alpha}} = \un... | If $\alpha = \phi\textrm{( the golden ratio )}, \textrm{ then }\alpha^2-\alpha -1=0,$
$$
\begin{aligned}
\int_0^{\infty} \frac{1}{\left(1+x^\alpha\right)^\alpha} d x & \stackrel{x\mapsto\frac{1}{x}}{=} \int_{\infty}^0 \frac{1}{\left(1+\frac{1}{x^\alpha}\right)^\alpha} \frac{d x}{-x^2} \\
& =\int_0^{\infty} \frac{x^{\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Compute the determinant of circulant matrix with entries $\cos j\theta$ compute the determinant
$
\begin{vmatrix}
\cos \theta & \cos 2\theta & \cos 3\theta &\cdots &\cos n\theta \\
\cos n\theta & \cos \theta & \cos 2\theta & \cdots & \cos (n-1)\theta \\
\cdots \\
\cos 2\theta & \cos 3\theta & \cos 4\theta & \cdots & ... | let $ A = \pmatrix{\cos t & \cos 2t &\ldots & \cos nt\\
\cos nt & \cos t & \ldots &\cos(n-1)t\\\ldots & \ldots &\ldots &\ldots\\
\cos 2t & \cos 3t &\ldots & \cos t}
$ and $\omega$ the $n$ the root of unity. that is $\omega^n = 1.$ you can verify that
$\pmatrix{1 & \omega & \ldots &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving complex series is convergent Question 1: Test the following series for convergence
$$\sum_{n=0}^\infty \left(\frac{2+i}{3-4i}\right)^{2n}$$
I got that it was convergent by doing a ratio test and getting
$$p = \frac{\sqrt{3}}{5} < 1 $$
which checks out for convergence. Wondering if that was the correct way to a... | The first series is a geometric series of the form $\sum_{n=0}^{\infty} q^n$ with $q = \left( \frac{2 + i}{3 - 4i} \right)^2$. Since
$$ \left| \frac{2 + i}{3 - 4i} \right| = \left| \frac{2 + i}{3 - 4i} \frac{3 + 4i}{3 + 4i}\right| = \left| \frac{2 + 11i}{25} \right| = \frac{\sqrt{4 + 121}}{25} = \frac{1}{\sqrt{5}} $$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $n > 1$ is an integer not of the form $6k + 3$, prove that $n^{2} + 2^{n}$ is composite. I try to divide in to $5$ case $6k, 6k+1, 6k+2, 6k+4, 6k+5$
and put them in $n^{2}$ and consider the possible last digit 0f $2^n$
and sum with those form
I can only prove $6k, 6k+2, 6k+4$ case .the last digit is even number and ... | The hint is $n > 1$. When $n = 1$, $n^2 + 2^n = 3$ which is prime. This hints that $n^2 + 2^n$ may be divisible by $3$.
We know that $6k+1$ can be written as $3(2k)+1$ and $6k+5$ can be written as $3(2k+1)+2$. Thus if $n$ is of the form $6k+1$ or $6k+5$, then $n^2 \equiv 1\pmod3$.
The next part is to prove that $2^n \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1975284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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prove that if x is a real number, then $\lfloor2x\rfloor=\lfloor x\rfloor+\lfloor x + \frac{1}{2}\rfloor$ prove that if $x$ is a real number, then $\lfloor2x\rfloor=\lfloor x\rfloor + \lfloor x + \frac{1}{2}\rfloor$
I've seen the proof in my discrete math book, but it was very confusing. I don't understand why they con... | One has to distinguish these cases since in this example "something special" happens with the floor function when the variable $x$ passes an integer multiple of ${1\over2}$. Nevertheless it can be said that the proof in your book is much too complicated.
Instead begin with the remark that both sides of the stated equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find matrix $A$. How many solutions are there? The matrices below is matrix multiplcation of $AB=C$ but someone has changed the order of the coloumns in $C$ and erased what was in $A$. What is matrix $A$? Is it the only one? I have no idea how to start. I appreciate all the help I can get.
$$\begin{bmatrix} & & & \\ ... | Preliminary: I assume it is clear that equation $AB=C'$ (where $C'$ is a column-shuffled version of $C$) is equivalent to 7 relationships between the 7 columns of $B$ and the corresponding 7 columns of $C'$:
$$A.\begin{pmatrix}2 \\-1\\ -4\\ -1\end{pmatrix}=\begin{pmatrix} . \\ .\\ . \\ . \end{pmatrix}, \ \ A.\begin{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Which number is greater, $2^\sqrt2$ or $e$?
Claim: $\color{red}{2^\sqrt2<e}$
Note: $2^\sqrt2=e^{\sqrt2\ln2}$
Different approach: We show $e^{x-1}>x^\sqrt x$ for $x>2$.
Let $f(x) = x -1 - \sqrt{x} \ln x $. We have $f'(x) = 1 - \frac{ \ln x }{2 \sqrt{x} } - \frac{1}{\sqrt{x}}$. By inspection, note that $f'(1)=0$ and si... | Another way:
since $e^x$ is increasing and $\sqrt2\approx1.414213$ we have
$$(2^{\sqrt2})^{\sqrt2}=4\lt e^{1.4}\approx4.055199\lt e^{\sqrt2}$$It follows
$$\left((2^{\sqrt2})^{\sqrt2}\right)^{\frac{1}{\sqrt2}}\lt(e^{\sqrt2})^{\frac{1}{\sqrt2}}\iff2^{\sqrt2}\lt e
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Quadratic equation with absolute values Solve $|x^{2}+x-2|+|x^{2}-x-2|=2$
My attempt:
$|x^{2}+x-2|=
x^{2}+x-2, x\in (-\infty,-2]\cup[1,\infty)$
$|x^2+x-2|=-x^{2}-x+2, x\in(-2,1)\\$
$|x^{2}-x-2|=x^{2}-x-2, x\in (-\infty,-1]\cup[2,\infty)$
$|x^{2}-x-2|=-x^{2}+x+2, x\in (-1,2)\\$
$1)$ $x\in(-\infty,-2]\cup[1,\infty)$ a... | Your work is essentially correct but, as you noted, the case 1) is not well solved. In this case the range is
$x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$, that is :
$$
\{(-\infty,-2]\cup[1,\infty)\}\cap\{(-\infty,-1]\cup[2,\infty)\}=(-\infty,-2]\cup [2,\infty)=D_1
$$
that is: $x\le -2 \mbox{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How to Solve Summation by Hand I am having trouble solving the following summation by hand.
$$ \sum_{i=0}^{n-1} 2^i (n-i) $$
Can someone guide me in the right direction, especially for the $i \cdot 2^i$ part?
| Here's a hint:
$$ \sum_{i=0}^k i 2^i = 2 + 2*2^2+ 3*2^3 + \dots + k* 2^k \\
= ( 2 + 2^2 + 2^3 + \dots + 2^k ) \\
+ (2^2+2^3+\dots + 2^k) \\
+ \dots + (2^{k-1}+2^k) + (2^k) \\
= 2(1+ \dots + 2^{k-1}) + \dots + 2^{k-1}(1+2) + 2^k \\
= 2(2^k-1)+2^2(2^{k-1}-1) + \dots + 2^{k-1}(2^2-1) \\
= (k-1) 2^{k+1} + 2^k - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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How can I solve this equation of functions? I need to find two functions continuous on $(-a, a)$, such that:
*
*$f(0) = 0$
*$f'(0) = 0$
*$g(0) = 0$
*$\lim_{x \to 0}\frac{f(x)g'(x)}{g(x)^2} \not= 0$
*The functions are not both writable as $x^n$, where n is a real number (one can be, but not both).
*The $\sin$ an... | Let
$$ f(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n} \hspace{5mm} \mbox{ and } \hspace{5mm} g(x) = \sum_{n=0}^{\infty} b_{n} \, x^{n} $$
then by the given conditions, $f(0) = 0$, $f'(0) = 0$, $g(0) = 0$, it is found that $b_{0} = 0$, $a_{0} = a_{1} = 0$ and lead to
$$ f(x) = \sum_{n=1}^{\infty} a_{n} \, x^{n} \hspace{5mm} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the 6 lines can be represented by the following equation: $(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$
Question: Figure $3$ shows six lines passing through the origin. The lines are separated by equal angles. Some
exact values of $\tan(t)$ are given in Table $1$.
$(i)$ Show that the lines ... | First you can see that there are six lines on your graph and the given equation is the product of three terms in $x^2$ and $y^2$. So each of those three terms gives two of those lines :
$$(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$$
Is equivalent to :
$$x^2-y^2=0$$
$$x^2-(7-4\sqrt{3})y^2=0$$
$$x^2-(7+4\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Using the Chinese Remainder Theorem to solve the following linear congruence: $17x \equiv 9 \pmod{276}$ The book I am following (Elementary Number Theory by David Burton) uses the Chinese Remainder Theorem to solve $17x \equiv 9 \pmod{276}$ by breaking it up into a system of three linear congruences,
$$17x \equiv 9 \pm... | First Bit - why the solution of the system is a valid answer
If we have a number $y$ such that $y\equiv9\pmod{3}$ and $y\equiv9\pmod{4}$ and $y\equiv9\pmod{23}$ then we know that $y-9\equiv0\pmod{3}$ and $y-9\equiv0\pmod{4}$ and $y-9\equiv0\pmod{23}$. In other words $y-9$ divides by $3$, $4$ and $23$. As these are all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Show that $\{a_n\}$ is an increasing sequence and bounded from above Question
Let $a_1$ = 1 and $a_{n+1} = \sqrt{1+2a_n}$. Show that $\{a_n\}$ is increasing and bounded from above.
Attempted solution
showing that the sequence is increasing:
\begin{align*}
a_n &\geq 1 \quad \forall n = {1,2,3,...} \\
\Rightarrow 1+2a_n ... | We prove by induction that the generalized sequence
$$a(1)=1,a(n+1)=\sqrt{2 c \;a(n)+1},\;c>0$$
is
(i) increasing.
The quotient of two successive terms of the sequence is
$$q(n+1)=\frac{a(n+1)}{a(n)}$$
Obviously
$$q(2)=\sqrt{2 c+1}>1$$
Now
$$q(n+1)=\sqrt{\frac{2 c a(n-1) q(n)+1}{2 c a(n-1)+1}}>\sqrt{\frac{2 c a(n-... | {
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"question_score": "2",
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If $a,b,x,y>0$ and $a^{11}+b^{11}\leq 1$ and $x^{11}+y^{11}\leq 1\;,$ Then prove that $a^{5}x^6+b^5y^6\leq 1$
If $a,b,x,y>0$ and $a^{11}+b^{11}\leq 1$ and $x^{11}+y^{11}\leq 1\;,$ Then prove that $a^{5}x^6+b^5y^6\leq 1$
$\bf{My\; Try::} $Using $\bf{A.M\geq G.M}$
$$\frac{a^{11}+a^{11}+a^{11}+a^{11}+a^{11}+x^{11}+x^{11... | Using Hölder's inequality with $p=\frac{11}{5}$ and $q=\frac{11}{6}$:
\begin{align}a^5x^6+b^5y^6 & \leq ((a^5)^p+(b^5)^p)^{\frac{1}{p}}((x^6)^q+(y^6)^q)^{\frac{1}{q}}\\ & \leq (a^{11}+b^{11})^{\frac{5}{11}}(x^{11}+y^{11})^{\frac{6}{11}}=1.\end{align}
| {
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"question_score": "4",
"answer_count": 1,
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Prove $a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)\geq 6abc$ Prove $a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)\geq 6abc$
My attempt:
$a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)-6abc\geq 0$
$\implies a^{2}+a^{2}b^{2}+b^{2}+b^{2}c^{2}+c^{2}+c^{2}a^{2}-2abc-2abc-2abc\geq 0$
$\implies (a-bc)^{2}+(b-ac)^{2}+(c-ab)^{2}\ge... | Alternatively by $AM \ge GM$,
\begin{align*}
a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2) &=
a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2 \\
& \ge 6\sqrt[6]{a^6b^6c^6} \\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is $x*y = \frac{1}{\frac{1}{x} + \frac{1}{y}}$ closed? Injective? Surjective? Let $x*y = \frac{1}{\frac{1}{x} + \frac{1}{y}}$.
*
*Is $x*y$ closed in $(0, \infty)$?
*For a given $x > 0$, is the function $f(y) = x*y, (0, \infty) \to (0,\infty)$ injective?
*For a given $x > 0$, is the function $f(y) = x*y, (0, \infty... | What I tried:
*
*Yes.
If $x,y > 0$, then $x*y > 0$.
QED
*Yes
Let $y_1,y_2 > 0$ s.t. $f(y_1) = f(y_2)$.
Then
$$\frac{1}{\frac{1}{x} + \frac{1}{y_1}} = \frac{1}{\frac{1}{x} + \frac{1}{y_2}}$$
$$\to \frac{1}{x} + \frac{1}{y_1} = \frac{1}{x} + \frac{1}{y_2}$$
$$\to \frac{1}{y_1} = \frac{1}{y_2}$$
$$\to y_1 = y_2$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the system of equations $\sqrt{x+y}+\sqrt{2x+y+2}=7, 3x+2y=23$. Solve the system of equations $\sqrt{x+y}+\sqrt{2x+y+2}=7, 3x+2y=23$.
$x=\frac{23}{3}-\frac{2}{3} y$ so when I plug it in I get
$\sqrt{\frac{23}{3}-\frac{2}{3}y+y} +\sqrt{2\times\frac{23}{3}-\frac{2}{3}y+y+2}=7$
$\sqrt{\frac{23}{3}+\frac{y}{3}} + \s... | Put $a=x+y,\; b=2x+y+2$.
Equations:
$$\begin{cases}\begin{align}
&\sqrt{a}+\sqrt{b}=7&\Rightarrow a+b+2\sqrt{ab}&=49\\
&a+b-2=23&\Rightarrow a+b&=25\end{align}\bigg\rbrace\Rightarrow 25+2\sqrt{ab}=49\Rightarrow ab=144\\
\end{cases}\\$$
$$\begin{align}
a+\frac {144}a&=25\\
a^2-25a+144&=0\\
(a-16)(a-9)&=0\\
a&=16, 9\\
b&... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why do these two summations not equate? Why is the below equation false?
(1) $\sum_{i=0}^n i^2 = \ (1/4)\sum_{i=0}^{2n} {i}^2 $
When If we let $\ j=2i $ , and substitute $\ i=(j/2)$ into the leftmost equation above, then:
(2) $\sum_{j=0}^{2n} {(j/2)}^2 $
Which gives the same sum as:
$\ (1/4)\sum_{i=0}^{2n} {i}^2$
Note... | $$\begin{align}
\sum_{i=0}^n i^2&=0^2+1^2+2^2+3^2+4^2+\cdots +n^2\qquad\qquad\qquad\qquad\qquad (1)\\
\sum_{j=0}^{2n} (j/2)^2&=\left(\frac 02\right)^2+\left(\frac 12\right)^2+\left(\frac 22\right)^2+\left(\frac 32\right)^2+\left(\frac 42\right)^2\cdots+\cdots +\left(\frac n2\right)^2+\cdots+\left(\frac {2n}2\right)^2\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Reduction Integration of $I_n=\int{\frac{x^n}{\sqrt{x^2+a^2}} dx}$ How do I derive this using reduction formula?
$$I_n=\int{\frac{x^n}{\sqrt{x^2+a^2}} dx}$$
I got $\sqrt{x^2+a^2}(x^n) - \frac{1}{n-1}(x^{n-1})\sqrt{x^2+a^2} + \frac{1}{n-1}(I_n) $ . I'm not too sure if this is correct. I would appreciate any form of help... | For starters, I would get rid of the $a$ parameter:
$$ I_n = a^n \int \frac{x^n\,dx}{\sqrt{x^2+1}} = a^n J_n \tag{1} $$
then apply integration by parts:
$$ J_n = \int x^{n-1}\frac{x\,dx}{\sqrt{1+x^2}} = x^{n-1}\sqrt{1+x^2}-(n-1)\int x^{n-2}\sqrt{1+x^2}\,dx\tag{2}$$
and notice that $\sqrt{1+x^2}=\frac{1}{\sqrt{1+x^2}}+\... | {
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"question_score": "2",
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Nontrivial rational solution of $x^3-3x^2+3x+1=y^2$ So, I'm looking for a rational solution of $$x^3-3x^2+3x+1=y^2$$ that is different from $(0,\pm 1)$. I tried to substitute $y=xt+1, t\in\mathbb Q$, getting following equation $$
x(x^2-(t^2+3)x+(-2t+3))=0$$ with the discriminant of the quadratic term $t^4+6t^2+8t-3$. T... | As hinted in the comments, we may consider the equivalent equation
$$y^2 = x^3 + 2$$
This is an elliptic curve. Finding rational points is an active domain in research and many books have been written, e.g. Silverman's Rational Points on Elliptic Curves.
Now, you already have some rational points on your curve: $(-1,-1... | {
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"source": "stackexchange",
"question_score": "2",
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Factorize $x^{2n}+1$ to evaluate $\int\tan^{1/n}(x)dx$ I've been attempting to find a formula for $\int\tan(x)^{1/n}dx, n\in\Bbb N$.
I started out by performing the substitution $$u=\tan(x)^{1/n}$$
$$dx=\frac {nu^{n-1}}{1+u^{2n}}du$$ to transform the integral into $$n\int \frac{u^n}{1+u^{2n}}du$$
This is a relatively... | I do not think that this is an answer but it is too long for a comment.
Adressing this problem, you enter in the world of the gaussian hypergeometric function since $$I_n=n\int \frac{u^n}{1+u^{2n}}\,du=\frac{n \,u^{n+1} }{n+1}\,\, _2F_1\left(1,\frac{n+1}{2 n};\frac{3n+1}{2 n};-u^{2 n}\right)$$ which, as you showed, can... | {
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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System of equations with inequality
Question Statement:-
If $$\begin{eqnarray}
x=cy+bz\\
y=az+cx\\
z=bx+ay
\end{eqnarray}$$ where $x,y,z$ are not all zero, prove that $a^2+b^2+c^2+2abc=1$
Further if at least one of $a,b,c$ is a proper fraction, prove that:-
(i) $a^2+b^2+c^2\lt 3$
(ii) $abc\gt-1$
Attempt at a soluti... | Assuming $a$ is a proper fraction, then
$$a^2<1$$
Considering quadratic in $c$,
$$c^2+2abc+a^2+b^2-1=0$$
Admitting real value of $c$,
\begin{align*}
\Delta & \ge 0 \\
4(a^2-1)(b^2-1) &\ge 0 \\
b^2-1 & \le 0
\end{align*}
Hence,
$$b^2 \le 1$$
Similarly,
$$c^2 \le 1$$
Hence,
$$a^2+b^2+c^2<3$$
Also
$$abc=\frac{1-a... | {
"language": "en",
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"source": "stackexchange",
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Value of $\sin(\pi/18)$ by nested radicals I recently asked a question about how to find value of $\sin(\pi/18)$ and I understand that there is no expression for $\sin(\pi/18)$ that uses the ordinary arithmetical operations. but I know that we have exact value by nested radicals. I want to know how we get this value ?
... | Deriving $\sin\frac{\pi}{18}$ will be simpler when we implement Half angle cosine formula
$$2\sin\frac{\theta}{2} = \sqrt{2-2\cos\theta}$$
$$2\cos\frac{\theta}{2} = \sqrt{2+2\cos\theta}$$
and cosine angle identity
$$2\cos(\pi-\theta) = -2\cos\theta$$
We can derive as follows
$2\sin(\frac{\pi}{18})$ = $\sqrt{2-2\cos(\f... | {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
... | Under the Extended Euclidean algorithm, you annotate the process of finding a GCD with how you got there. This can be used for finding multiplicative inverses:
$$
\begin{array}{c|c|c|c|l}
q & r & a & b & \text{implied equation} \\ \hline
& 41 & 1 & 0 & 41 = 1\cdot 41 + 0\cdot 17\\
& 17 & 0 & 1 & 17 = 0\cdot 41 + 1\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003116",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Calculating the probability of reaching each absorbing state in Markov Chain I'm starting with a Matrix that looks like this:
[[ 0 , 1/2, 0, 0 , 0 , 1/2],
[4/9, 0 , 0, 3/9, 2/9, 0 ],
[ 0 , 0 , 1, 0 , 0 , 0 ],
[ 0 , 0 , 0, 1 , 0 , 0 ],
[ 0 , 0 , 0, 0 , 1 , 0 ],
[ 0 , 0 , 0, 0 , 0 , 1 ]]
A... | The state transition diagram for the given problem is given by:
Consider transitions from A to D:
It could happen in several ways, as listed in the following. All these paths are mutually exclusive. Invoking the definition of of first passage time, we can obtain the required probability.
\begin{eqnarray*}
A\rightarr... | {
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"source": "stackexchange",
"question_score": "5",
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If $x$ is a real and $0 < x < 4$, then $\frac{4}{x(4-x)} \geq 1$ I got this exercise from the textbook Book of Proof, CH4 E12. I've tackle this problem in the following manner:
Suppose x is a real and $0 < x < 4$, it follows that,
\begin{align*}
&\Rightarrow 0 - 2 < x - 2 < 4 - 2 \\
&\Rightarrow 4 < (x - 2)^2 < 4\\... | Hint See that: $0\le x\le 4 $ implies $0\le 4-x\le 4$ and then,
$$0\le x(4-x) =-(x-2)^2+4\le4$$
Thus directly implies your inequality.$$\dfrac{4}{x(4 - x)} \geq 1.$$
| {
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"url": "https://math.stackexchange.com/questions/2003660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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What are the linear factors of $x^3 - 2$? The polynomial is defined as $$x^3 - 2 \in \mathbb{C}[x]$$
How do you figure out the linear factors of the term? If I evaluate the function in polar coordinates I get some weird terms that can't possibly be correct. Wolfram says that I have to multiply the real solution by the ... | See that any root that you could have of $x^3-2$ is of the style $\sqrt[3]{2}\ e^{2it\pi/3}$ with $t=0,1,2$, so it factors into
$$(x-\sqrt[3]{2})(x-\sqrt[3]{2}\ e^{2i\pi/3})(x-\sqrt[3]{2}\ e^{4i\pi/3})$$
If you want to know what is $e^{2ti\pi/3}$ use the polar coordinates:
$$e^{2it\pi/3}=\cos\left(\frac{2\pi t}{3}\righ... | {
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"url": "https://math.stackexchange.com/questions/2007159",
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"source": "stackexchange",
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image of circle after inversion Given a circle $C$ of radius $r$ and center $(x_0,y_0)$, what is the center and radius of $C$ after inversion in the unit circle?
A simpler question has been asked:
Image of a circle under conformal map $1/z$
| You have $(x-x_0)^2+(y-y_0)^2=r^2$ as the original circle. Performing the inversion means replacing $x$ by $\frac x{x^2+y^2}$ and $y$ by $\frac y{x^2+y^2}$. Abbreviating $d:=x^2+y^2$ and $f:=x_0^2+y_0^2-r^2$ one can write the circle after inversion as
\begin{align*}
\left(\frac{x}{d}-x_0\right)^2+\left(\frac{y}{d}-y_0\... | {
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Prove that $21|a^2 +b^2 \implies 441|a^2+b^2$
Prove that $$21|a^2+b^2 \implies 441|a^2+b^2$$
I can show that if $3$ divides $a^2+b^2$ than so does $9$, so the part that's left to do is to prove that $7$ and $49$ divide $a^2+b^2$ and I am stuck at this point.
| Note that
\begin{align}
1^2&\equiv 1\pmod{7},\\
2^2&\equiv 4\pmod{7},\\
3^2&\equiv 2\pmod{7},\\
4^2&\equiv 2\pmod{7},\\
5^2&\equiv 4\pmod{7},\\
6^2&\equiv 1\pmod{7}.
\end{align}
It is impossible to get 2 numbers with replacement from $\{1,2,4\}$ to get a sum that is divisible by $7$. Thus, to have $a^2+b^2\equiv 0\pmod... | {
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"source": "stackexchange",
"question_score": "2",
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the diophantine forms of the equation $a^2 +b^3 = c^5$ I just want to know the forms of the numbers a,b and c in the Diophantine equation:
$a^2 +b^3 = c^5$
| See OEIS sequence A178130
$$\eqalign{104^2 + 28^3 &= 8^5\cr
654^2 + 127^3 &= 19^5\cr
2816^2 + 32^3 &= 24^5\cr
3912^2 + 124^3 &= 28^5\cr
4096^2 + 256^3 &= 32^5\cr
6048^2 + 288^3 &= 36^5\cr
48500^2 + 275^3 &= 75^5\cr
19683^2 + 1458^3 &= 81^5\cr
65216^2 + 1008^3 &= 88^5\cr
77824^2 + 1280^3 &= 96^5\cr
\ldots}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving System of Linear Equations with LU Decomposition of $4 \times 3$ matrix The following is all confirmed to be true:
Matrix A =
$
\begin{bmatrix}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{bmatrix}
$
U =
$
\begin{bmatrix}
-1 & 2 ... | $$
\begin{align}
\mathbf{P} \mathbf{A} &= \mathbf{L} \mathbf{U} \\
% P
\left[
\begin{array}{cccc}
0 & \boxed{1} & 0 & 0 \\
\boxed{1} & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right]
% A
\left[
\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{array}
\right]
%... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding the Jordan Canonical Form of a Classical Adjoint of a Jordan Block Let $A$ be a size $n$ Jordan matrix with $0$ on its diagonal, that is
$$A = J_n(0) = [a_{ij}] = \begin{cases}
1, &j=i+1\\
0, &\text{elsewhere}
\end{cases}
$$
What is the Jordan Canonical Form of the classical adjoint of A, $\text{adj} A$?
Can w... | If you just start computing the classical adjoint for $n=2,3,4...$ you should notice a pattern as to what they look like.
$$adj\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & - 1\\ 0 & 0\end{pmatrix}$$
$$adj\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 ... | {
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"source": "stackexchange",
"question_score": "3",
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Fibonacci numbers solution to this recurrence relation Prove that the Fibonacci number's are the solutions the following recurrence relation, $$S_n=5S_{n-4}+3S_{n-5}$$
For all n greater than or equal to 5, where we have
$$S_0=0$$ $$S_1=1$$ $$S_2=1$$ $$S_3=2$$ $$S_4=3$$
Then use the formula to show that the Fibonacci nu... | hint:characteristic equation$$x^5=5x+3\\x^5-5x-3=0$$ then $$x^2-x-1|x^5-5x-3\\ \to $$divide $x^5-5x-3$ by $x^2-x-1$
$$x^5-5x-3=(x^2-x-1)(x^3+x^2+2x+3)$$
and $x^2-x-1$ is the characteristic equation of $f_n=f_{n-1}+f_{n-2}\\----\\$
Another solving :
$$f_{n+5}=f_{n+4}+f_{n+3}\\
=(f_{n+3}+f_{n+2})+f_{n+3}=\\2f_{n+3}+f_{n+... | {
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For $a$, $b$, $c$ the sides of a triangle, prove $\left| \frac{a}b-\frac{b}a+\frac{b}c-\frac{c}b+\frac{c}a-\frac{a}c\right|< 1$ The inequality I have to prove that if $a$, $b$, and $c$ are the sides of a triangle, then
$$\left| \frac{a}b-\frac{b}a+\frac{b}c-\frac{c}b+\frac{c}a-\frac{a}c\right|< 1$$
I initially though... | Hint:
Using the substitution $a = x+y, b=y+z, c = z+x$ (if $a, b, c$ are sides of a triangle then $x, y, z$ always exist and are positive) we get
$$\left| \sum_{cyc} \left( \frac{a}b-\frac{b}a \right)\right| = \left|\frac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)} \right| < 1$$
| {
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"source": "stackexchange",
"question_score": "3",
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Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$ Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that
$$\sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}. $$
Solution.
Using AM GM inequality we have
\begin{gather*}
x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\
xy+yz \geq 2 y ... | From your assumption, we obtain that
\begin{align*}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le 1.
\end{align*}
Since
\begin{align*}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} &=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z} +\frac{1}{y} + \frac{1}{z} \right)\\
&\ge\frac{1}{2}\left(\frac{2}{\sqrt{xy}} + \frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Show the coefficient of $x^3$ in the expansion of $(1+x+x^2+x^3)(1+x^2)(1+x^3)$ is the same as the number of partitions of $3$ (let a partition of $n \in \Bbb N, n>0 $ be a sum of positive integers)
How would you show that the number of different partitions of $3$ is the coefficient of
$x^3$
in the
expansion of:
$(1+x... | You can see the "situation" as I try and depict in this sketch
For what concerns the partitions of $3$ (and of $2$ and $1$) , all the necessary "pieces" are provided.
Explanation of the sketch
When you expand the first polynomial, this will provide a factor $x^k$ to multiply the rest, and $k$ is either $0$ or $1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Find all positive three - digit integers $\overline{abc}$ such that $\overline{abc}=5abc$ Number is divisible by $5$ if the last digit is $0$ or $5$. Since $c\neq 0$, $c=5$.
Using brute force (checking each combination with three digits), I get that the only solution is $175$ : $a=1,b=7,c=5$.
How to solve a problem ef... | You already know that $c=5$.
Hence $100a+10b+5=25ab$.
Hence $20a+2b+1=5ab$.
$20a+2b+1$ is odd.
Hence $5ab$ is odd.
Hence $5ab$ ends with $5$.
Hence $20a+2b+1$ ends with $5$.
Hence $2b+1$ ends with $5$.
Hence $2b$ ends with $4$.
Hence $b=2$ or $b=7$.
Since $5ab$ is odd, $b$ must be odd.
Hence $b=7$.
Hence $20a+2\cdot7+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Solve for $x$ in $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$ $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$
Note that I didn't forget any parenthesis. I copied it as it is in my book.
I tried:
$$3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0 \Leftrightarrow 3\cdot 5^{x^2} - 5 \cdot 3^{2x} = 0 \Leftrightarrow 5 \cdot 3^{x^2} = ... | $3^{x^2+1} = 3^{x^2} \cdot 3$ is correct.
Furthermore,$$ 5^{x^2} \cdot 3 = 3^{x^2} \cdot 5 \iff 5^{x^2-1} = 3^{x^2-1} \iff \Big(\frac{5}{3} \Big)^{x^2-1} = 1$$.
This happens exactly when $x^2-1=0$,or when $x = \pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Simplify $\frac {2\sqrt{x^2+x+1}-2-x} {x^2} $ In order to find the limit of the function at $0$ I need to simplify the fraction $$\frac {2\sqrt{x^2+x+1}-(2+x)} {x^2} $$
so that at $0$ it didn't look like $\frac 0 0$. I'm almost certain that it can be solved with a clever substitution of the variable but I didn't manage... | Multiplying numerator and denominator by $$2\sqrt{x^2+x+1}+2+x$$ we get
after simplification $$\frac{3x^2}{x^2(2\sqrt{x^2+x+1}+2+x)}$$ and after cancelling $$x^2$$ you will get the searched limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is $\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} $ always an integer? In a textbook I found the rather strange identity:
$$ \frac{2^4}{(5-2)(3-2)}+\frac{3^4}{(5-3)(3-2)}+\frac{5^4}{(5-3)(5-2)}= \frac{414}{6}=69 $$
just kind if out of nowhere and I wonder if it generalizes and why. Perhaps it is ... | Because
$$ \frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} = a^2 + a b + a c + b^2 + b c + c^2 \text{,} $$
the result is certainly an integer for distinct integers $a$,$b$,$c$. (It's undefined if any pair of them are equal.)
Your guess about $P$ being the product of differences of alphabetical ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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The number of paths of length $n$ that end at a specific vertex in a grid-like graph Suppose you have a graph consisting of nine vertices laid out in a $3\times3$ grid where each vertex is connected to each of its horizontal, vertical, and diagonal neighbors. So the corner vertices have degree three, the edge vertices ... | Instead of using normal $3\times 3$ matrix, write it out as a $9\times 1$ vector : $$\left[\begin{array}{c}1&2&3\\4&5&6\\7&8&9\end{array}\right] \to [1,2,3,4,5,6,7,8,9]^T$$
and use adjacency matrix for multiplication:
$$
\left[
\begin{array}{c}
0&1&0&1&1&0&0&0&0\\
1&0&1&1&1&1&0&0&0\\
0&1&0&0&1&1&0&0&0\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Can the antiderivate of $\frac{1}{(x^2+1)^2}$ be calculated without using the iteration formula? To calculate the antiderivate of $$\frac{1}{(x^2+1)^2}$$ , we can either use the iteration formula reducing the exercise to the integral of $\ \frac{1}{x^2+1}\ $ or we can use $$(\frac{x}{x^2+1})'=\frac{1-x^2}{(x^2+1)^2}$$ ... | For substitution method, see How do I solve this trigonometric substitution integral?.
For intgration by parts,
$$\int\dfrac{dx}{(1+x^2)^2}=\int\dfrac{2x}{(1+x^2)^2}\cdot\dfrac1{2x}\ dx$$
$$=\dfrac1{2x}\int\dfrac{2x}{(1+x^2)^2}dx-\int\left(\int\dfrac{2x}{(1+x^2)^2}dx\cdot \dfrac{d(1/2x)}{dx}\right)dx$$
$$=-\dfrac1{2x(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $x>0$, then find the greatest value of the expression $ \frac{x^{100}}{1+x+x^2+x^3+\cdots+x^{200}}$
If $x>0$, then find the greatest value of the expression $ \dfrac{x^{100}}{1+x+x^2+x^3+\cdots+x^{200}}$
This expression simplifies to $ \frac{(x^{100})(x-1)}{x^{200}-1}$ using sum of n terms of GP. Now one can ... | AM-GM inquelity $=>$
$$
\\1+x+\cdots+x^{2n}\geq (2n+1)(x^{0+1+\cdots+2n})^{1/{2n}}=(2n+1){x^{\frac{2n+1}{2}}}
\\f(x)=\frac{x^n}{1+x+\cdots+x^{2n}}\leq\frac{x^n}{(2n+1){x^{\frac{2n+1}{2}}}}=\frac{1}{(2n+1)\sqrt{x}}
\\
\\\max_{x>0}(f(x))=\frac{1}{(2n+1)\sqrt{x}}
$$
then $f(x)=\frac{1}{(2n+1)\sqrt{x}}\>$ if and only if $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Calculate the exact value of $\sin\frac{11\pi}{8}$ Calculate the exact value of $\sin\frac{11\pi}{8}$.
The formula $\sin^2x=\frac12(1–\cos2x)$ may be helpful.
I was thinking of using the Angle-Sum and -Difference Identity:
$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$
For instance: $\alpha=\frac{20\pi}{... | Another idea: Let be $x = 3\pi/8$
$$0 = \sin(3\pi) = \sin(8x) = 2\sin(4x)\cos(4x) = \cdots $$
$$= (-128\sin^7 x + 192\sin^5 x - 80\sin^3 x + 8\sin x)\cos x.$$
Then, $s = \sin x$ is a root of
$$0 = 128s^7 - 192s^5 + 80s^3 - 8s = 8(16s^6 - 24s^4 + 10s^2 - 1)s = 8(8s^4 - 8s^2 + 1)(2s^2 - 1)s.$$
The roots of the biquadrati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solving $a^2-b^2-a=0$ ∧ $-2ab+b={1\over 2}$ I want to solve
$$a^2-b^2-a=0 ∧ -2ab+b={1\over 2}$$
I'm a bit stuck in the first part. I thought of the solutions, $a={1±\sqrt {1+4b^2} \over 2}$. But can't figure out solutions to the second part. Am I approaching it the right way?
| Let $z=a-bi$, then $z^2-z=(a^2-b^2-a)+(-2ab+b)i$
\begin{align*}
z^2-z &= \frac{i}{2} \\
z &= \frac{1\pm \sqrt{1+2i}{}}{2} \\
a-bi &=
\frac{1}{2}
\left(
1 \pm \sqrt{\frac{\sqrt{5}+1}{2}} \pm i\sqrt{\frac{\sqrt{5}-1}{2}} \,
\right) \\
a &= \frac{1}{2} \pm \sqrt{\frac{\sqrt{5}+1}{8}} \\
b &= \mp \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to find the general formula of the recurrence sequence? Let two sequences $\{a_n\},\{b_n\}$ such that
$a_{2n-1}=a_{2n}=a_{2n-2}+\frac{1}{b_{2n-2}}$,
$b_{2n}=b_{2n+1}=b_{2n-1}+\frac{1}{a_{2n-1}}$,
$a_1=b_1=1$.
With the help of OEIS, I find a solution:
$a_{2n-1}=a_{2n}=\frac{(2n-1)!!}{(2n-2)!!}$,
$b_{2n}=b_{2n+1}=\fr... | I only manage to decouple the recurrence relations: We have
\begin{array}{rlcrl}
a_{2n} &= a_{2n-2} + 1/b_{2n-2} \quad (*) & \quad &
b_{2n+1} &= b_{2n-1} + 1/a_{2n-1} \quad (**)\\
a_{2n} &= a_{2n-1} & \quad &
b_{2n+1} &= b_{2n} \\
a_1 &= 1 & \quad &
b_1 &= 1
\end{array}
then solving equation $(**)$ for $a_{2n-1}$ in te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate $\int_{0}^\infty\frac{dx}{\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}$ I'm interested in the integral
$$
I=\int_{0}^\infty\frac{dx}{\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}.\tag{1}
$$
So far I have b... | define:
$$ Q^n(x) = \prod_{a=1}^n{1-\left(\frac{x}{a}\right)^3}
= \left(1-\left(\frac{x}{a}\right)^3\right)
\cdot\prod_{b\ne a}{1-\left(\frac{x}{b}\right)^3}
= \left(1-\left(\frac{x}{a}\right)^3\right)\cdot Q_a(x)
$$
using:
$$\begin{align*}
(a) && & 1-\left(\frac{x}{a}\right)^3 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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On the integral $\int_0^1\frac{dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{1-x\,\gamma^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{2\gamma}}$ V. Reshetnikov gave the interesting integral,
$$\int_0^1\frac{\mathrm dx}{\sqrt[4]x\ \sqrt{1-x}\ \sqrt[4]{2-x\,\sqrt3}}=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$
After some experimentation, it tur... | It seems solving $I(\beta^2;\tfrac{1}{2},\tfrac{1}{4})=\frac1N$ also solves $I(\gamma^2;\tfrac{1}{4},\tfrac{1}{4})=\frac1N$ as they involve reciprocal polynomials. So again following Nemo's lead in this answer, we employ the duplication formula,
$$\frac{1}{2}I(p^2;\tfrac{1}{2},\tfrac{1}{4})=I(1-q^2;\tfrac{1}{2},\tfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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The sequence $a_n=\frac{2n+1}{(n-1)^2}$ decreases monotonically How could we show that the sequence $a_n=\frac{2n+1}{(n-1)^2}$ decreases monotonically?
When we take the quotient $\frac{a_n}{a_{n+1}}$ we get $\frac{n^2(2n+1)}{(n-1)^2(2n+3)}$. Howcan we conclude that this quotient is $\geq 1$ ?
| Every term of the sequence is positive. So, the sequence is decreasing if, and only if, the ratio $a_{n+1}/a_{n} < 1$ for every integer $n \geq 2$.
\begin{equation*}
\frac{a_{n+1}}{a_{n}} = \frac{(2n+3)(n-1)^2}{(2n+1)n^{2}} = \frac{(2n+3)(n^{2} - 2n + 1)}{2n^{3} + n^{2}}
= \frac{2n^{3} - n^{2} - 4n + 3}{2n^{3} + n^{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that a group is abelian under given conditions Let $(G, \cdot)$ be a group with the property that there exists $a \in G$ such that $ax^3a = x$, $\forall x \in G$. Prove that $(G, \cdot)$ is abelian.
Here is my solution:
By plugging in $x := ax$ in the given condition, we get $a^3x^3a = x$. But $ax^3a$ is also equ... | There are a couple of issues with your argument, but it starts with
By plugging in $x := ax$ in the given condition, we get $\color{red}{a^3x^3a = x}$.
The equation in red is incorrect. If you replace $x$ with $ax$, then you would get $a(ax)^3a = ax$.
Replace $x$ with $axa^{-1}$ in the condition to get $a(axa^{-1})^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the minimum area of a nth-side polygon whose vertices satisfy In the Argand diagram, a square has its vertices satisfying a 4th power equation with coefficients being integers, i.e.
$$z^4+az^3+bz^2+cz+d=0, z\in \mathbb C, a,b,c,d \in \mathbb Z$$
Find the minimum area.
I suspect that the vertices statisfying $$z^4... | \begin{align*}
f(z) &= \left( z+\frac{a}{4} \right)^4+d-\frac{a^4}{256} \\
&= z^4+az^3+\frac{3a^2}{8}z^2+\frac{a^3}{16}z+d \\
\left| z_k+\frac{a}{4} \right| &=
\sqrt[4]{\left| d-\frac{a^4}{256} \right|} \\
\text{area} &= 2\sqrt{\left| d-\frac{a^4}{256} \right|}
\end{align*}
Case I: $a=0$
$$\text{area}=2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove $\sin(\cos x) \neq \cos(\sin x)$ Prove that the trigonometric equation
sin(cosx) = cos(sinx)
has no solutions.
I graphed the functions and I can see they are never equal to eachother, but how do I do a formal proof for this?
| $$\cos(\sin x)=\sin(\cos x)=\cos\left(\frac\pi2-\cos x\right)$$
$$\text{So}, \sin x=2n\pi\pm \left(\frac\pi2-\cos x\right)$$
$$\text{Taking the '+' sign,} \sin x=2n\pi+ \left(\frac\pi2-\cos x\right)\implies \sin x+\cos x=\frac{(4n+1)\pi}2$$ which can be $\cdots,-\frac{3\pi}2,\frac{\pi}2,\frac{5\pi}2,\cdots$
$$\text{Tak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
| You could solve it this way:
The second fraction, $\frac {2}{2x+6}$ can be simplified into $\frac 1{x+3}$. Thus,
$$\frac 5{x+3}+\frac 2{2x+6}=4\implies\frac 6{x+3}=4\tag1$$
And cross multiplying, we get $4(x+3)=6\implies x=x=-\frac 32$
As to where you went wrong, nothing has gone wrong yet. Use the quadratic formula a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
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Proving two infinite nested radicals from a condition Kindly please tell me how to prove the following problem:
Solve completely : $$ x^2 =y+a ; y^2=z+a ; z^2 =x+a ...(1) $$ and show that (with or without using the preceding question): $$\sqrt{11-2\sqrt{11+2\sqrt{11-...}}} = 1+4\sin 10^\circ$$ and $$\sqrt{23-2\sqrt{2... | We need re-scaling:
\begin{align*}
bx &= \color{red}{-2}\sqrt{11\color{green}{-2}\sqrt{11 \color{blue}{+2}\sqrt{11 \color{red}{-2}\sqrt{11-\ldots}}}} \\
by &= \color{green}{-2}\sqrt{11 \color{blue}{+2} \sqrt{11 \color{red}{-2} \sqrt{11 \color{green}{-2}\sqrt{11-\ldots}}}} \\
bz &= \color{blue}{+2}\sqrt{11 \color{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Principal part: $ \frac{1}{z(\exp(z)-1)^2} $ (Laurent series) I have to compute from the following, its principal part of the Laurent series around $z=0$.
$$ \frac{1}{z(\exp(z)-1)^2} $$
I have trouble with computing the series of $(\exp(z)-1)^2$. I wanted to take out the $\frac{1}{z}$ and then multiply it by the series... | It is convenient to represent $\exp(z)-1$ using the big-O notation.
\begin{align*}
\exp(z)-1&=\sum_{n=1}^\infty\frac{z^n}{n!}\\
&=z+\frac{1}{2}z^2+\frac{1}{6}z^3+\frac{1}{24}z^4+O(z^5)\tag{1}
\end{align*}
We obtain
\begin{align*}
\frac{1}{z\left(\exp(z)-1\right)^2}&=\frac{1}{z\left(z+\frac{1}{2}z^2+\frac{1}{6}z^3+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How do I prove this using mathematical induction? \begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n(n+2) = \frac{n(n+1)(2n+7)}{6}
\end{align}
Using the mathematical induction step I arrive at this :
\begin{align}
1\cdot3+2\cdot4+3\cdot5+...+n+1(n+3) = \frac{n+1(n+2)(2n+9)}{6}
\end{align}
And I don't see any other way to cont... | All you need to do is realize that
$$\frac{(n+1)(n+2)(2n+9)}6=\frac{(n+1)(n+1+1)(2(n+1)+7)}6$$
which is what you are trying to prove, but with $(n+1)$ everywhere $n$ originally was.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{a+3b}}\geq\sqrt{a+b+c+d}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3d}}+\frac{d}{\sqrt{d+3a}}\geq\sqrt{a+b+c+d}$$
I tried Holder, AM-GM and more, but without success.
| Hello Michael Rozenberg . I start my proof with a classical substitution .
Put:
$A=a$;
$AB=b$;
$AC=c$;
$AD=d$
We get :
$$\frac{1}{\sqrt{1+3B}}+\frac{B}{\sqrt{B+3C}}+\frac{C}{\sqrt{C+3D}}+\frac{D}{\sqrt{D+3}}\geq \sqrt{1+B+C+D}$$
Make a new substitution like this :
$x=B$;
$C=xh$;
$D=x\beta$
We get :
$$\frac{1}{\sqrt{1+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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Check induction proof that $\sum\limits_{i=0}^ni\left(i+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}$ claim: $\sum _{i\:=\:0}^n\:i\left(i+1\right)\:=\:\frac{n\left(n+1\right)\left(n+2\right)}{3}$
base case n = 0
$\frac{0\left(0+1\right)\left(0+2\right)}{2}$ ; $0\:=\:0$
inductive hyp (n = k):
$\sum _{i=0}^k\:i\l... | For the case $n=k+1$:
$$\sum_{i=0}^{k+1}\:i\left(i+1\right)\:=\sum_{i=0}^{k}\:i\left(i+1\right)+(k+1)(k+2)$$
Using inductive hyp we get:
$$\sum_{i=0}^{k+1}\:i\left(i+1\right)\:=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=$$
$$(k+1)(k+2)\left(\frac{k}{3}+1\right)=\frac{(k+1)(k+2)(k+3)}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How do I calculate surface of ellipse rotating around x axel? How to calculate surface of ellipse rotating around x axel when ellipse equation is :
$y^{2}+4x^{2} = 36$
I started with this formula for calculating surface of rotating object around x axel
$ \int_{a}^{b}f(x) \sqrt{1 + f'(x)^{2}} dx$
but i always end up w... | The surface area is given by
\begin{equation}
A=2\pi\int_{a}^{b}y\sqrt{1+(y^\prime)^2}\,dx
\end{equation}
with $y=2\sqrt{9-x^2}$, $a=-3,\,b=3$. Because of symmetry we can double the integral from $0$ to $3$.
Since $\sqrt{1+(y^\prime)^2}=\sqrt{\dfrac{9-3x^3}{9-x^2}}$ we have
\begin{eqnarray}
A&=&4\pi\int_{0}^{3}2\sqrt{9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to calculate the length of a parabolic arc if two end points & angle at both ends are specified? First I confess I my mathematics knowledge is not very good. So first can an expert please first confirm if this question is actually solvable? Because elsewhere on net I have read that you actually need to specify 3 po... | Without loss of generality we can suppose that the coordinate are chosen in such a way that the parabola has equation $ y=ax^2+bx$. So we have $y'=2ax+b$.
From the condition that the starting angle is $30°$ we have $y'(0)=\tan 30°=\frac{\sqrt{3}}{3}$ and this gives $b=\frac{\sqrt{3}}{3}$.
Now the problem is to find a p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the Sum of an Infinite Series Compute
$${1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.}$$
I tried creating a partition for this but no such luck.
What would the the equation to generate the sum for the nth term
| Let
$$f(x)=\sum_{n=0}^{+\infty}\frac{x^n}{2^n}=\frac{2}{2-x}$$
with a radius of convergence $R=2$
$$f'(x)=\sum_{n=0}^{+\infty}n\frac{x^{n-1}}{2^n}=\frac{2}{(2-x)^2}$$
thus, your sum is $f'(1)=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Vector Parametric Form Flat Please see the image with the question and my answer. The thing I'm having an issue with is understanding what this vector description actually is. Is it a 2-dimensional flat in $\mathbb{R}^6$?
In fact, according to the table, we have
$n-2$ Cartesian equations and $2$ free parameters in th... | The table constructs a formula between the number of free parameters $F$, number of equations $E$, and the dimension of the space $n$ as follows
$$n-F=E\tag{1}$$
however, it should be emphasized that $E$ is the number of linearly independent equations and not just the number of equations! Now, let us see how many linea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is my result to $z^2 - (3+4i)z - 1 + 7i = 0$ right? I solved it much like I would a second degree equation, with a=1, b=-(3+4i) and c=(-1+7i). Is that the correct approach?
My final result was $z = \frac{3+4i \pm i\sqrt{3+4i}}{2}$
Is that correct? There is no way for me to check that.
| Here is the standard way t o compute the square roots of a complex number: $$(x+iy)^2=3+4i\iff \begin{cases}x^2-y^2=3,\\2xy=4.\end{cases}$$
Further the square of the module of $x+iy$ is equal to the module of $3+4i$:
$$x^2+y^2=5.$$
So we first solve the linear system in $x^2$ and $y^2$:
$$\begin{cases}x^2-y^2=3,\\x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove that $\sum\limits_{cyc}\frac{a^2}{a^3+2}\leq\frac{4}{3}$ if $a, b, c, d > 0$ and $abcd=1$
Let $a$, $b$, $c$ and $d$ be positive numbers such that $abcd=1$. Prove that:
$$\frac{a^2}{a^3+2}+\frac{b^2}{b^3+2}+\frac{c^2}{c^3+2}+\frac{d^2}{d^3+2}\leq\frac{4}{3}.$$
Vasc's LCF Theorem does not help here. Also I tried ... | WLOG, assume that $a\ge b \ge c \ge d$.
Let
$$f(u) := \frac{u^2}{u^3 + 2}.$$
Fact 1: $f(u)$ is strictly increasing on $(0, \sqrt[3]{4}]$
and strictly decreasing on $[\sqrt[3]{4}, \infty)$.
We split into four cases.
Case 1: $c > \sqrt[3]{13 - 3\sqrt{17}}$
and $a < \sqrt[3]{13 + 3\sqrt{17}}$
Let
$$g(u) := \frac{\mathrm{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Prove at $(0,0)$, $f_{xy}$ and $f_{yx}$ aren't equal If $f(x, y) =xy(\frac{x^2-y^2}{x^2+y^2})$ when $x$ and $y$ are not simultaneously zero when $x=0,y=0$. Show that at $(0,0),\ f_{xy} \neq f_{yx} $
| First find $f_x$ and $f_y$.
For $(x,y) \neq (0,0)$ you have
$$f_x(x,y) = y \frac{x^2-y^2}{x^2+y^2} + xy \frac{2x(x^2+y^2) - 2x (x^2 - y^2)}{(x^2+y^2)^2} = y \frac{x^2-y^2}{x^2+y^2} + xy \frac{4xy^2}{(x^2+y^2)^2} = y \frac{x^2-y^2}{x^2+y^2} + \frac{4x^2y^3}{(x^2+y^2)^2}. \qquad (1)$$
Similarly,
$$f_y (x,y) = x \frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove an inequality with positive numbers Let $n\in\mathbb N$ and $a_{1},...,a_{n} >0$, such that $a_{1} + \ldots +a_{n} =1$. Show that
$$
\left(\frac{1}{a_{1}^{2}} + \ldots +\frac{1}{a_{n}^{2}} \right)\left(2(a_{1}^{2}+\ldots+a_{n}^{2}) - \frac{1}{n}\right)\ge n^2.
$$
I tried induction, but it does not work.
An... | Using Cauchy-Schwarz, we obtain
$$
a_1^2+\cdots+a_n^2\ge \frac{1}{n}(a_1+\cdots+a_n)=\frac{1}{n},
$$
and hence
$$
2(a_1^2+\cdots+a_n^2)-\frac{1}{n}\ge a_1^2+\cdots+a_n^2.
$$
Next, Cauchy-Schwarz again
$$
(a_1^2+\cdots+a_n^2)\left(\frac{1}{a_1^2}+\cdots+\frac{1}{a_n^2}\right)\ge (1+\cdots+1)^2=n^2,
$$
and hence
$$
\big(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.