Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Why can't I cancel $2x-3$ from $(2x-3)(x+5)=9(2x-3)$? Why are these simplifications wrong?
$$\begin{align}
(2x-3)(x+5)=9(2x-3) &\quad\to\quad \frac{(2x-3)(x+5)}{2x-3} = \frac{9(2x-3)}{2x-3} \quad\to\quad x+5 = 9\\[4pt]
x(x+2)=x(-x+3) &\quad\to\quad \frac{x(x+2)}{x} = \frac{x(-x+3)}{x} \quad\to\quad x+2=-x+3
\end{align}... | Observe that the solution set of $x^2=x$ is $\{0,1\}$ whereas the solution set of $x=1$ is $\{1\}$. Hence dividing by $x$ is not an equivalence transformation. Instead of dividing by $x$ factor it out: $x^2-x=0\iff x(x-1)=0$ to get $x=0$ or $x=1$.
In your first case we will get $(2x-3)(x+5-9)=0\iff 2x-3=0$ or $x-4=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find k with little oh notation? For which values of $k$ are the following true (as $x \to 0$)?
(a) $\sqrt{1+x^2} = 1 + o(x^k)$
(b) $\sqrt[3]{1+x^2} = 1 + o(x^k)$
(c) $1 - \cos(x^2) = o(x^k)$
(d) $1 - \cos^2 x = o(x^k)$
How would you find $k$ for problems like these?
| The expression
\begin{align*}
f(x)=g(x)+o(x^k)\qquad\qquad x\rightarrow 0
\end{align*}
is just another notation for
\begin{align*}
\lim_{x\rightarrow 0}\frac{f(x)-g(x)}{x^k}=0\tag{1}
\end{align*}
In case $\sqrt{1+x^2} = 1 + o(x^k)$ we take the Taylor series expansion
\begin{align*}
\sqrt{1+x^2}&=\sum_{n=0}^\infty\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove if $x\gt3$ then $1\ge\frac{3}{x(x-2)}$. I tried to prove it by contradiction.
Suppose it is not true that $1\ge\frac{3}{x(x-2)}$, so $1\lt\frac{3}{x(x-2)}$. Then $\frac{3}{x(x-2)}-1\gt0$. Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.
$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$
${3-(x(x-... | Note that if $x>3\implies x-2>1$ then $\dfrac1x<\dfrac13$ and $\dfrac1{x-2}<1$ so $$\frac3{x(x-2)}<\frac3{3(3-2)}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding points on a line that are closest
Find the points that give the shortest distance between the lines$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2-t\\-1+2t\\-1+t\end{pmatrix}\\\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}5+3s\\0\\2-s\end{pmatrix}$$
So I subtracted the second line from the first t... | Two lines in $\mathbb R^3$ either intersect (distance = 0) or not intersect (distance = the distance between two parallel planes on which the two lines lie accordingly). Classify it first, then use $u\times v$, where $u, v$ indicate the direction of the two lines, to find the normal for the plane. Hope that helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\pi (r^2-x^2)$ We are told to integrate $\pi (r^2-x^2)$ from $-r$ to $r$
$$V=\int_{-r}^{r}\pi(r^2-x^2)dx=2\pi\int_0^r(r^2-x^2)dx=2\pi\left[r^2x-\frac{x^3}{3}\right]_0^r=2\pi\left(r^3-\frac{r^3}{3}\right)=\frac{4}{3}\pi r^3$$
why does the integral of $r^2$ equal $r^2 x$?
| We have $$\begin{align}V&=\int_{-r}^r\pi(r^2-x^2)\,dx=\int_{-r}^r(\pi r^2-\pi x^2)\,dx=\int_{-r}^r\pi \color{red}r^2\,d\color{blue}x\end{align}-\int_{-r}^r\pi \color{blue}x^2\,d\color{blue}x$$ and since we are integrating with respect to $x$ (in blue) we treat $r$ (in red) as a constant.
Hence $$\begin{align}V&=[\pi \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Showing $\cos^{10}x\le 1-x^2$ for $x$ in $[0,0.5]$ I have to prove this inequality:
$$\cos^{10}x\le 1-x^2 \quad \text{for all}\quad x \in [0,0.5]$$
Is there any easy and/or elegant way to do this? I can do this with Taylor, but it's really a mess. :(
Thanks in advance
| Note that
$$\cos(x)^{10}\le 1-x^2\iff 10\log\cos x\le\log (1-x^2)$$
and
*
*$10\log\cos x\le 10 \log
(1-\frac12x^2+\frac1{24}x^4)\le-5x^2+\frac5{12}x^4\quad$ by $\log(1-x)<x$
*$\log (1-x^2)\ge-x^2-x^4$ to be proved
and since
$$-5x^2+\frac5{12}x^4\le -x^2-x^4 \iff4x^2-\frac{17}{12}x^4\ge 0\iff x^2(4-\frac{17}{12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Integrating function of log $$
\int_0^4 \ln\left(\sqrt{(x-2)^2+4} + (x-2)\right)~{\rm d}x
$$
In the above given question, I am unable to integrate the function inside log. I am substituting $(x-2) = t$ and then reducing the question into a simpler form with only $t$ as a variable, but i am getting a function of $t$ ins... | First, make the substitution you've already mentioned:
$$\int_0^4 \ln\left(\sqrt{(x-2)^2+4} + (x - 2)\right)~{\rm d}x$$
$$t=x-2$$
$${\rm d}t={\rm d}x$$
$$\int_{-2}^2 \ln\left(\sqrt{t^2+4} + t\right)~{\rm d}t$$
Now, we solve using integration by parts where $u = \ln\left(\sqrt{t^2+4} + t\right)$ and ${\rm d}v = {\rm d}t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the equation of the tangent to the parabola $y = x^2$ at the point (-2, 4)? This question is from George Simmons' Calc with Analytic Geometry. This is how I solved it, but I can't find the two points that satisfy this equation:
$$
\begin{align}
\text{At Point P(-2,4):} \hspace{30pt} y &= x^2 \\
\frac{dy}{d... | You need a point and a slope. The point is $(2,4)$ and the slope is $y'(2)=4.$
Thus the equation is $$ y-4=4(x-2)$$ or $$ y=4x-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Help to solve a system equation: $x-y+xy=-4$; $xy(x-y)=-21$. I need to solve a system equation. Here's how it looks:
$x-y+xy=-4$
$xy(x-y)=-21$
I tried to substitute $x-y$ with $w$ and $xy$ with $t$ to simplify everything. After that I got this system equation:
$w+t=-4$
$tw=-21$
I solved this new system equation and got... | \begin{align}
x-y+xy&=-4
,\\
xy(x-y)&=-21
.
\end{align}
We can consider $u,v$
\begin{align}
u&=x-y
,\\
v&=xy
\end{align}
as two roots of the quadratic equation,
\begin{align}
z^2+az+b&=0
\tag{1}\label{1}
,\\
a&=-(u+v)=-(x-y+xy)=4
,\\
b&=uv=xy(x-y)=-21
,
\end{align}
so \eqref{1} is
\begin{align}
z^2+4z-21&=0
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Suppose $ x+y+z=0 $. Show that $ \frac{x^5+y^5+z^5}{5}=\frac{x^2+y^2+z^2}{2}\times\frac{x^3+y^3+z^3}{3} $. How to show that they are equal? All I can come up with is using symmetric polynomials to express them, or using some substitution to simplify this identity since it is symmetric and homogeneous but they are still... | First we can use the identities
$$ (x+y+z)^2=x^2+y^2+z^2+2(yz+zx+xy) $$
and
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-yz-zx-xy)$$
As $x+y+z=0$, we have $x^2+y^2+z^2=-2(yz+zx+xy)$ and $x^3+y^3+z^3=3xyz$.
So, R.H.S is $-xyz(yz+zx+xy)$.
From the above results, we have
\begin{align*}
(x^3+y^3+z^3)(x^2+y^2+z^2)&=x^5+y^5+z^5+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solving $x^2+y^2=9$, $\arctan\frac{y+2}{x+2} + \arctan\frac{y-2}{x+4} =2\arctan\frac{y}{x}$ without graphing? The system of equations:
$$\begin{align}
x^2 + y^2 &= 9 \\[6pt]
\operatorname{arctan}\frac{y+2}{x+2} + \operatorname{arctan}\frac{y-2}{x+4} &=2\,\operatorname{arctan} \frac{y}{x}
\end{align}$$
I tried to in... | Essentially, the equations say that the origin is at distance $3$ from $(x,y)$ and on the angular bisector of the triangle with vertices $(x,y)$, $(x+2,y+2)$, $(x+4,y-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimum length of the hypotenuse
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Prove that the minimum length of the hypotenuse is $(a^{2/3}+b^{2/3})^{3/2}$.
My Attempt
$\frac{x}{y}=\frac{a}{CM}=\frac{AN}{b}$
$$
\frac{x}{y}=\frac{AN}{b}\implies y=\frac{xb}{\sqrt{x^2-a^... | Let $\angle PCB=\angle ABN=\theta$. Then $\displaystyle \cos \theta =\frac{a}{x}$ and $\displaystyle \sin \theta=\frac{b}{y}$.
Then we have to minimize $$x+y=\frac{a}{\cos \theta}+\frac{b}{\sin \theta}$$
So using Holder,s Inequality
$$(\cos^2\theta+\sin^2\theta)\cdot \bigg(\frac{a}{\cos \theta}+\frac{b}{\sin \theta}\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Partial Fractions Decomposition of $\frac{25s}{(s^2+16)(s-3)(s+3)}$ So this is the problem..
$$
\frac{25s}{(s^2+16)(s-3)(s+3)}
$$
So what I did was...
$$
\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {A}{s^2+16}+\frac {B}{s-3}+\frac{C}{s+3}
$$
then...
$$\begin{align}
25s &= A(s-3)(s+3)+B(s^2+16)(s+3)+C(s^2+16)(s-3) \\
25s &=... | You must write
$$\frac{25s}{(s^2+16)(s-3)(s+3)}=\frac{As+B}{s^2+16}+\frac{C}{s-3}+\frac{D}{s+3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
" Let $A$ be a symmetric $2 \times 2$ matrix with the property $A^{-1} = A$. Find all possible trace values of $\operatorname{tr}A$" I need some help solving this.
I have tried:
$$
\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
=\frac{1}{\operatorname{det}A}\cdot \begin{bmatrix}
d & -b \\
-... | You have $A^{-1} = A \implies A^2 = I$. So we just calculate that for the matrix you have $$\begin{pmatrix} 1 & 0 \\ 0& 1 \end{pmatrix} = \begin{pmatrix} a & b \\ c& d \end{pmatrix}\cdot\begin{pmatrix} a & b \\ c& d \end{pmatrix} = \begin{pmatrix} a^2 + bc & (a+d)\cdot b \\ (a+d)\cdot c & bc + d^2 \end{pmatrix}$$
There... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Divisibility of $n^2+9$ by $n+3$ How would one find all integers $n$ such that $n+3 \vert n^2 +9$? I assume it is important that $(n+3)^2 - 6n = n+3$, but I am struggling to see how you can find all $n$, and confirm an upper bound such that there are no more such $n$.
| Alternatively:
$n+3|n^2+9 \iff$
$n+3|n^2 + 9 - n(n+3)\iff$
$n+3|-3n + 9\iff$
$n+3|-3n + 9 + 3(n+3)\iff$
$n+3|18$
$n+3 = \pm 1; \pm 2;\pm 3;\pm 6;\pm 9;\pm 18$
$-21,-12,-9,-6,-5,-4,-2,-1,0,3,6,15$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
British Maths Olympiad (BMO) 2006 Round 1 Question 5, alternate solution possible? The question states
For positive real numbers $a,b,c$ prove that
$(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b)$
After some algebraic wrangling we can get to the point where:
$(a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4 ≥ 2... | We need to prove that
$$(a^2+b^2)^2\geq\sum_{cyc}(2a^2b^2-a^4)$$ or
$$c^4-2(a^2+b^2)c^2+2(a^4+b^4)\geq0$$ or
$$(c^2-a^2-b^2)^2+(a^2-b^2)^2\geq0.$$
Yes, you can prove this inequality by the dividing.
Indeed, if $c^2(a^2+b^2)=0$ then the inequality is obvious.
Let $c^2(a^2+b^2)\neq0$.
Thus, by AM-GM and Cauchy-Schwarz
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Edge length of an equilateral triangle if distances from a point $P$ to its vertices is given A point $P$ is located inside an equilateral triangle and is at a distance of 5, 12, and 13 from its vertices. Compute the edge length of the triangle.
The answer is $\sqrt{169 + 60\sqrt(3)}$.
If $s$ is the edge length of the ... |
Let $ABC$ be an equilateral triangle. $P$ is a point inside $\triangle ABC$ such that $PA=5$, $PB=12$ and $PC=13$. Rotate $C$ and $P$ about $A$ through $60^\circ$ to $B$ and a point $X$. Rotate $A$ and $P$ about $B$ through $60^\circ$ to $C$ and a point $Y$. Rotate $B$ and $P$ through $60^\circ$ to $A$ and a point $Z$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Uniform Convergence of $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^2}{(1+x^2)^n}$ in $\mathbb{R}$
Uniform convergence of $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^2}{(1+x^2)^n}$ in $\mathbb{R}$
Using Dirichlet Test, it can be shown that the it uniformly converges for $\mathbb{R}$ \ $\{0\}$.
In $x=0$ there is obviously a point... | Note
\begin{eqnarray}
S_N(x)&=&\sum_{n=1}^{N}(-1)^{n-1}\frac{x^2}{(1+x^2)^n}=\frac{x^2}{1+x^2}=\sum_{n=1}^{N}(-1)^{n-1}\frac{1}{(1+x^2)^{n-1}}\\
&=&\frac{x^2}{1+x^2}\frac{1-\frac{(-1)^N}{(1+x^2)^{N}}}{1+\frac{1}{1+x^2}}=x^2\frac{1-\frac{(-1)^N}{(1+x^2)^{N}}}{x^2+2}\\
\end{eqnarray}
and hence
\begin{eqnarray}
\left|S_N(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that if $2x + 4y = 1$ where x and y are real numbers. Show that if $2x + 4y = 1$ where $$x, y \in \mathbb R $$, then $$x^2+y^2\ge \frac{1}{20}$$
I did this exercise using the Cauchy inequality, I do not know if I did it correctly, so I decided to publish it to see my mistakes. Thank you!
If $x=2$ and $y=4$, then $... | The distance from the circle's center ($ x^2+y^2 = \frac{1}{20}$) to the line $2x + 4y = 1$ is given by
$$
d = \frac{1}{\sqrt{2^2+4^2}} = \sqrt{\frac{1}{20}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Commutator Group of $\operatorname{GL}_2(\mathbb{R})$ is $\operatorname{SL}_2(\mathbb{R})$
Let $\operatorname{GL}_2(\mathbb{R})$ be the general linear group of $2\times2$ matrices and $\operatorname{SL}_2(\mathbb{R})$ the special linear group of $2 \times 2$ matrices. Show that the commutator subgroup of $\operatornam... | If $x\in\mathbb R$, then$$\begin{pmatrix}1&x\\0&1\end{pmatrix}=\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}1&x\\0&1\end{pmatrix}\begin{pmatrix}2&0\\0&1\end{pmatrix}^{-1}\begin{pmatrix}1&x\\0&1\end{pmatrix}^{-1}$$and therefore $\left(\begin{smallmatrix}1&x\\0&1\end{smallmatrix}\right)$ is a product of commutators... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Finding all complex entries? Find all complex triples $(x, y, z) $such that the following matrix is diagonalizable
$A = \begin{bmatrix}1&x&y \\ 0&2 & z \\0&0&1\end{bmatrix}$
my attempts :
matrix A is an upper triangular matrix.
so the the eigenvalues of A are diagonal entries $1,2,1$
This implies that $A$ is diagonal... | Diagonalizable means the minimal polynomial is squarefree. The full characteristic polynomial is $(\lambda - 1)^2 (\lambda - 2).$
Diagonalizable if and only if
$$ (A - I)(A - 2I) = 0. $$
This gives a restriction on the triple $x,y,z.$ And, when we do have $y = xz,$ we have
$$
\left(
\begin{array}{ccc}
1&-x&-xz \\
0&1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Result comparison of a trigonometric equation The Problem: Solve the equation: $$ \cos x=\cos3x+ 2\sin2x
\\$$
The Result: $$ x=k\frac{\pi}{2},k\in \mathbb{Z}$$
My solution:$$ \cos x=4\cos^3x-3\cos{x}+4\sin{x}\cos{x}\\ 4\cos^3x-4\cos{x}+4\sin{x}\cos{x}=0\\ 4\cos{x}(\cos^2{x}+\sin{x}-1)=0\\
\cos{x}(1-\sin^2{x}+\sin{x}-1... | My resolution:
$$\cos x= \cos \left(3x\right)+2\sin \left(2x\right)$$
$$-\cos \left(3x\right)-2\sin \left(2x\right)+\cos \left(x\right)=0$$
Using the following identity
$$-\cos \left(p\right)+\cos \left(q\right)=2\sin \left(\frac{p+q}{2}\right)\sin \left(\frac{p-q}{2}\right)$$
I obtaining,
$$-2\sin \left(2x\right)+2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$.
If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$.
My attempt:
Consider $\frac{1}{1 - z}$. Let $z = x + iy$, we know that $|z| < 1 \implies x, y < 1$. $$\frac{1}{1 - z} = \frac{1}{1 - x - iy} = \frac{1 - x + iy}{(1 - ... | You need to use more than just $x,y<1$ for example with $x=y=\frac{3}{4}$ you obtain $\Re\left(\frac{1}{1-(x+iy)} \right)=\frac{2}{5} <\frac{1}{2}$.
Hint:
Using your computations:
$$ \Re\left(\frac{1}{1-z}\right)=\frac{1-x}{(1-x)^2+y^2}=\frac{1-x}{x^2+y^2-2x+1}$$
but $x^2+y^2<1$ so:
$$\frac{1-x}{x^2+y^2-2x+1}>\frac{1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Calculating a determinant. $D_n$=\begin{vmatrix}
a & 0 & 0 & \cdots &0&0& n-1 \\
0 & a & 0 & \cdots &0&0& n-2\\
0 & 0 & a & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&a&1 \\
n-1 & n-2 & n-3 & \cdots... | To find other values of $x$ besides $0$ for which
$$
\det \begin{bmatrix}
x & 0 & 0 & \cdots & n-1 \\
0 & x & 0 & \cdots & n-2 \\
0 & 0 & x & \cdots & n-3 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n-1 & n-2 & n-3 & \cdots & x
\end{bmatrix} = 0
$$
consider taking $\frac{n-1}{x}$ times the first row, plus $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Sums of Nilpotent Matrices Let $A =$ diag$(a_1,a_2,…,a_n)$, where the sum of all $a_i$’s is zero.
Show that A is a sum of nilpotent matrices.
My idea:
$\begin{bmatrix}
2 & 0 \\
0 & -2
\end{bmatrix}$ = $\begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}$ + $\begin{bmatrix}
1 & -1 \\
1 & -1
\end{bmatrix}$
Where $\begin{bmatr... | The diagonal matrices of the form $\operatorname{diag}(0,\ldots,0,1,0,\ldots,0,-1)$ form a basis of the space of the matrices with null trace. In fact, a matrix with null trace is $\operatorname{diag}(a_1,\ldots,a_n)$ with $a_1+a_2+\cdots+a_n=0$, which is equal to$$a_1\operatorname{diag}(1,0,0,\ldots,0,-1)+a_2\operator... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding all values in $\mathbb R$ for quadratic absolute value equation The question:
Determine the solution set (in $\mathbb R$) for the equation $|x^2+2x+2| = |x^2-3x-4|$
So far, I have determined that for this to be true, $|x^2-3x-4|$ must be greater or equal to $0$, giving $(x-4)(x+1)\ge0$. To find the solution s... | Since $|x| = x$ or $-x$, $$|x^2+2x+2| = |x^2-3x-4|$$ if and only if
$$x^2+2x+2 = x^2-3x-4$$ or
$$-(x^2+2x+2) = x^2-3x-4$$ or
$$x^2+2x+2 = -(x^2-3x-4)$$ or
$$-(x^2+2x+2) = -(x^2-3x-4)$$
Some cases are equivalent (can you guess which and why?), so the computations are easier than they look.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the convergence domain of the series $\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}$
Exercise: Find the convergence domain of the series $$\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}$$
To solve the series I used the comparison test then the Leibniz criteria:
$$\sum_\limits{n=1}^{\infty}\f... | We have
$$
\sum_{k=0}^{\infty}(-1)^k y^k = \lim_{n\rightarrow \infty}\frac{y^n+1}{y+1}
$$
here $\vert y\vert < 1 \Rightarrow$ convergence
Now
$$
\sum_{k=0}^{\infty}(-1)^k \frac{y^k}{k} = \frac{1}{y}\int_0^y \sum_{k=0}^{\infty}(-1)^k y^k dy = \frac{1}{y}\int_0^y\frac{dy}{y+1} = \frac{\ln(y+1)}{y}
$$
for $y \ne 0$ and $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Continuity of functions of two variables Find the points where the function $f(x,y)$ is continuous where
$$f(x,y)=\left\{ \begin{array}{ll}
\frac{x^2\sin^2y}{x^2+2y^2} & \mbox{if $(x,y)\not=(0,0)$};\\
0 & \mbox{if $(x,y)=(0,0)$}.\end{array} \right.$$
What I attempted: Here $f(x,y)$ is continuous at all the points $(x... | Here is the answer for the second question: $\frac {x^{2}+\sin ^{2} (y)} {2x^{2}+y^{2}}$ is $\frac 1 2$ when $y=0$ and $\frac {\sin ^{2} (y)} {y^{2}}$ when $x=0$. The last expression approaches 1 as $y \to 0$. Hence the functioin does not even have a limit as $(x,y) \to 0$ and the function is not continuous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2774564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Series 1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2 + 1/7... This is the exercise 2.7.2 e) of the book "Understanding Analysis 2nd edition" from Stephen Abbott, and asks to decide wether this series converges or diverges:
$$ 1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\d... | $${1\over2n-1}-{1\over(2n)^2}={4n^2-2n+1\over4n^2(2n-1)}\gt{4n^2-4n+1\over4n^2(2n-1)}={2n-1\over4n^2}\ge{1\over4n}$$
so
$$\left(1-{1\over4}\right)+\left({1\over3}-{1\over16}\right)+\left({1\over5}-{1\over36}\right)+\cdots\gt{1\over4}+{1\over8}+{1\over12}+\cdots={1\over4}\left(1+{1\over2}+{1\over3}+\cdots\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Show $x^2$ + $y^2$ + $z^2$ = $2xyz$ has no non-trivial solutions using infinite descent. I want to show that the equation
$x^2$ + $y^2$ + $z^2$ = $2xyz$ has no non-trivial solutions using infinite descent.
This question has been solved with infinite descent by showing that $x,y,z$ are all even and we can infinitely fi... | $x^2+y^2+z^2=2xyz...........................(1)$
in this equation $RHS$ is even .then, $LHS$ is also even.
w.l.o.g, $z$ is even $\exists z_1 $ such that $z= 2 z_1$ and $x,y$ are in same parity.
substituting $z= 2 z_1$ in given equation
$x^2+y^2+4z_1^2=4xyz_1 ........................(2)$
$x^2+y^2=4xyz_1-4z_1^2$
$RHS$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2779723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inequality $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1$ Show that $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1,\:\forall n\in\mathbb{N}$$
This is a 9th grade problem.
I was trying to take the greatest numerator, which is the last numerator of the last fraction. But there are only $2n+1$ terms. Right?
After... | There are $2n+1$ terms in the sum, you just need to pair up the terms
symmetrically from both ends, take average and compare with the term in the middle.
$$\begin{align}\sum_{k=n+1}^{3n+1} \frac{1}{k}
&= \sum_{k=-n}^n\frac{1}{2n+1+k}
= \frac12\sum_{k=-n}^n\left(\frac{1}{2n+1+k} + \frac{1}{2n+1-k}\right)\\
&= \sum_{k=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
USAMO 2018: Show that $2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$ Here is question 1 from USAMO 2018 Q1 (held in April):
Let $a,b,c$ be positive real numbers such that $a+b+c = 4 \sqrt[3]{abc}$.
Prove that: $$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$$
This question is on symmetric poly... | $ 2(ab+bc+ca)$+$ 4\min( a^2,b^2,c^2)$=$ 4(ab+bc+ca)+ 4\min(a^2,b^2,c^2)-2(ab+bc+ca)$
Now, suppose $a≤b≤c$ without losing generality.
Hence,
$ 2(ab+bc+ca)$+$ 4\min( a^2,b^2,c^2)$
=$ 4(ab+bc+ca)+ 4a^2-2(ab+bc+ca)$
= $4((a+b+c)a+bc)-2(ab+bc+ca)$
=$4(4a(abc)^{1/3})+bc)-2(ab+bc+ca)≥ 16(abc)^{2/3}-2(ab+bc+ca)
=(a+b+c)^2-2(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ ?
I checked various trigonometric identities, but I am unable to derive $\sin(x)$ based on the given information.
For instance:
$\sin(2x) = 2 \sin(x) \cos(x)$
| $$\sin(2x)=\frac{24}{25}$$
The identities you want are:
$$\sin(2x)=2\sin(x)\cos(x) \space [I]$$
and
$$\sin^2(x)+\cos^2(x)=1\space[II]$$
Square $[I]$ to get:
$$4\sin^2(x)\cos^2(x)=\frac{576}{625}\to\sin^2x(1-\sin^2x)=\frac{144}{625}$$
$$\to\sin^4(x)-\sin^2x+\frac{144}{625}=0$$
Let $\sin^2(x)=\sigma$. Your equation becom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Maximizing $3\sin^2 x + 8\sin x\cos x + 9\cos^2 x$. What went wrong?
Let $f(x) = 3\sin^2 x + 8\sin x\cos x + 9\cos^2 x$. For some $x \in \left[0,\frac{\pi}{2}\right]$, $f$ attains its maximum value, $m$. Compute $m + 100 \cos^2 x$.
What I did was rewrite the equation as $f(x)=6\cos^2x+8\sin x\cos x+3$. Then I let $\m... | Let's ignore the condition that $x\in[0,\pi/2]$ for the moment.
We are maximising $3u^2+8uv+9v^2$ subject to the constraint $u^2+v^2=1$.
For a given $a$, $3u^2+8uv+9v^2=a$ is soluble under this constraint
iff $3u^2+8uv+9v^2=a(u^2+v^2)$ is. This means that $Q(u,v)=(3-a)u^2+8uv+(9-a)v^2=0$. If this has a solution $\ne(0,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the number of councils with at least two girls There are 12 people, 7 girls and 5 boys.
You are supposed to choose a council of 4 people so that there are at least 2 girls.
I know you can solve this by taking the total amount of choices and subtracting the scenarios with one or no girls.
12C4 - (5C4) - (7C1)(5C... | As lulu suggests in the comments, a way to count the number of councils with at least two girls is to add the number of ways of selecting a council with exactly two, exactly three, or exactly four girls.
The number of ways of selecting exactly $k$ girls and $4 - k$ boys from seven girls and five boys is $$\binom{7}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Without-loss-of-generality question Going through solutions of IMO'09. Bumped into a without-loss-of-generality assumption that I can't comprehend.
Here's the statement of the problem:
Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that
$$
\frac{1}{(2a+b+c)^2}+\frac{1}... | Suppose that we know that if $a+b+c=1$ the homogenized inequality is true. Then if we have arbitrary $A$, $B$, $C$, with $A+B+C=x$, we can define $a=A/x$ and so on. When we plug this into the inequality, the $x$'s drop out, so that $A$, $B$, and $C$ also satisfy the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute limit of double exponent expression. Compute $$\large\lim_{n\to \infty}\left(\left(\prod_{i=1}^{n+1}{i^{i^p}}\right)^{1/(n+1)^{p+1}} - \left(\prod_{i=1}^{n}{i^{i^p}}\right)^{1/n^{p+1}}\right),$$
where $p$ is some nonnegative real number.
I found this problem while working through the Problems in Real Analysis: ... | The calculations got a bit messy so there could be errors, let me know if you find some.
Let the sequence $a_n$ be defined by $a_n = \prod\limits_{i=1}^n i^{i^p}$.
Then by Abel's summation formula we have $$\log \left(a_n^{1/n^{p+1}}\right) = \frac{1}{n^{p+1}}\sum\limits_{i=1}^ni^p\log i = \frac{1}{n^{p+1}}\left(n^pS(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer. QUESTION: Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer.
ATTEMPT (AND SOME SOLUTIONS): So, for a positive integer $x$, $x^2-1\ge0$. In the case $x^2-1=0$ i.e $x=1$ (since it's a positiv... | $x+1=xy+1 ⇒ y=1$ then x can get any value greater than 0.
B: $x-1=xy+1 ⇒ x(y-1)=-2 ⇒ x=2, y=0$ are only possible solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the sum of $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$, $\cos\frac{5π}{7}$ by first finding a polynomial with those roots
Without using tables, find the value of $$\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$$
This is a very common high school trigonometric problem, and the usual way to solve this i... | Consider numbers: $$\omega_k=e^{i\frac{(2k-1)\pi}{7} }=\cos\left(\frac{(2k-1)\pi}{7}\right) + i \sin\left(\frac{(2k-1)\pi k}{7}\right), \qquad k=1,...,7.$$
Easy to see (De Moivre's formula) that $$w_k^7=e^{i(2k-1)\pi}=e^{i\pi}=-1, \qquad k=1,...,7.$$
So, all $\omega_k$ are solutions of equation
$$
\omega^7-1 = 0.\tag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Partitions of numbers Let $p_{2}(n)$ denote the number of partitions of the positive integer n into at most 2 parts. Prove for each $n \ge 4$ that $p_{2}(n) + p_{2}(n-3) = n$
I understand that I need to split this into odd and even cases - I would just like my reasoning validating. Any comments/feedback on my proof or ... | It is simpler to have a "compact view" of the arguments. From the post, the formula
$$p_2(n)=\left[\frac n2\right]+1$$
is known, and was shown. Now we can cover all cases in one breath:
$$
\begin{aligned}
p_2(n)+p_2(n-3)
&=
\left[\frac n2\right]+1
+
\left[\frac {n-3}2\right]+1
\\
&=
\left[\frac n2\right]
+
\left[\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is there a pattern to the last digits of square numbers? I was programming and I realized that the last digit of all the integer numbers squared end in $ 0, 1, 4, 5, 6,$ or $ 9 $.
And in addition, the numbers that end in $ 1, 4, 9, 6 $ are repeated twice as many times as the numbers that end in $ 0, 5$
I checked th... | These numbers are the squares modulo 10. Notice that the square of the number $10n+k$ is
$$ (10n+k)^2 = 10(10n^2+2nk)+k^2, $$
so the last digit of the square is determined by only the last digit of the original number. In particular, we find
$$ 0^2=0 \quad 1^2=1 \quad 2^2 = 4 \quad 3^2 = 9 \quad 4^2 = 10+6 \\
5^2 = 20+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 7,
"answer_id": 3
} |
Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question
Three randomly chosen non negative integers $x, y \text{ and } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?
My Approach
Calculating Sample space -:
Number of possible solution for... | This is in fact a composition problem. We want to find all compositions of $x$ and $y$ such that $z$ takes values in $0,2,4,6,8,10$.
Then we count All 2-Compositions of $0,2,4,6,8,10$, that is how can we add 2 numbers to get these values.
Let $C(n,k,a,b)$ be all k-compositions of n using the numbers from a to b.
We wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
If $\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$ prove that $\cos\alpha=\frac{2-m^2}{m}$
If $$\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$$
prove that
$$\cos\alpha=\frac{2-m^2}{m}$$
My approach:
$$\cos^2(\alpha-3\theta)+... | An approach using factoring expressions is the following: define
$ \quad x := e^{i\alpha}, \quad y := e^{i\theta}, \quad $ and
$$ a := \frac{\cos(\alpha -3\theta)}{\cos^3 \theta},
\, b := \frac{\sin(\alpha -3\theta)}{\sin^3 \theta},
\, c := \cos \alpha - \frac{2-a^2}a,
\, d := \cos \alpha - \frac{2-b^2}b, \, e := ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Divisibility of Sum of Equally Spaced Binomial Coefficients According to a numerical calculation I did for small values of $k$, it appears that the following is true.
$$4|\left[\sum_{j=1}^{n-1}\binom{3n}{3j}\right]$$ or $$\sum_{j=1}^{n-1}\binom{3n}{3j}=4p, p\in\mathbb{Z}$$
Ex.
If $n=2, \binom{6}{3}=20=4\cdot 5$
If $n... | I know this is an old question but I wanted to add my two cents. What you wronte has a closed for. $\sum_{3j=0}^n\binom{n}{3j} = \frac{2^n+m}{3}$ where m = 2, 1, −1, −2, −1, 1, when n is congruent to
0, 1, 2, 3, 4, 5 (mod 6). according to https://math.hmc.edu/benjamin/wp-content/uploads/sites/5/2019/06/Sums-of-Evenl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Cominatorics (sticks and plus, generating function) If I should distribute 25 identical cookies to 10 children each children should get at least 1 cookie and maxium of 4 cookies. The problem should be solved using both inclusion-exclusion and generating function.
My solution with in-ex (Wrong):
$$x_{1} + x_{2} + x_{3}... | You have $[ x^{10} ]\cdot\frac{(1-x^4)^{10}}{(1-x)^{10}}$, where $[ x^{k} ]$ denote the coefficient of operator.
$\frac{1}{(1-x)^{10}}$ can be written as $\sum_{j=0}^\infty \binom{-10}{j}(-x)^j$ (Binomial series).
$$[ x^{10} ]\cdot(1-x^4)^{10}\cdot \sum_{j=0}^\infty \binom{-10}{j}(-x)^j$$
Now we use that $\binom{-r}{s}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Dividing concentric super ellipses to equal area slices How the concentric super ellipses as shown in the figure can be divided into parts containing equal area such that the total area of the superellipse A = A1 + A2 + ... An where n = 60 in the shown figure given its semi-major axis a and semi-minor axis b. The origi... | Following the same idea presented here Dividing an ellipse into equal area, it suffices consider the affinity by $X=ax$ and $Y=by$ such that
$$x^2+y^2=1 \to \frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$$
and divide the unit circle in 60 equals parts by the lines
*
*$y=0$
*$y=\pm\frac{\sqrt 3}{3}x$
*$y=\pm{\sqrt 3}x$
*$x=0$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral. Help $\int\frac{2x+3}{x^2-4}dx=\int\frac{2x+3}{(x+2)(x-2)}dx$
So...
$A=\frac{1}{4}$
$B=\frac{7}{4}$
$\frac{1}{4}\log_{10}|x+2|+\frac{7}{4}\log_{10}|x-2|+c$
The result should be: $\log_{10}|x^2-4|+\frac{3}{4}\log_{10}|\frac{x-2}{x+2}|+c$
What did I do wrong?
| Your computation is correct (apart from using $\log_{10}$ instead of $\log$, with implied base $e$, or $\ln$).
Probably the book chose a different approach:
$$
\frac{2x+3}{x^2-4}=\frac{2x}{x^2-4}+\frac{3}{x^2-4}=
\frac{2x}{x^2-4}+\frac{3}{4}\frac{1}{x-2}-\frac{3}{4}\frac{1}{x+2}
$$
so integration gives
$$
\log\lvert x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Taylor series of $\sqrt{1+x}$ I have to give the Taylor series of $\sqrt{1+x}$ with development point $0$.
But I am not quite sure what to do.
Normally you have to give the Taylor series up to a certain order.
Well $T_f(x,0)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$
I calculated the derivative of $f(x)=\sqrt{1+x}$ up... | Almost correct:
$$\begin{align*}
f^{(0)}(x) & = (1+x)^{\frac{1}{2}} \\[1ex]
f^{(1)}(x) & = \frac{1}{2} \cdot (1+x)^{-\frac{1}{2}} \\[1ex]
f^{(2)}(x) & = \frac{1}{2} \cdot \left(-\frac{1}{2} \right) \cdot (1+x)^{-\frac{3}{2}} \\[1ex]
\vdots \\
f^{(n)}(x) & = \frac{1}{2} \cdot \left(-\frac{1}{2} \right) \cdot \ldots \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
closed formula for Sum of this series Is there any closed form of this series?
$$\frac{x!}{1!} + \frac{(x+1)!}{2!} + \frac{(x+2)!}{3!} +\cdots + \frac{(x+n-1)!}{n!}$$
I tried to manipulate the expansion of $(1+x)^n$ but can't seem to get the term $(1+x)!$ from $x!$. Need help on this one.
| By the Pascal formula we get:
$$\begin{align}
\frac{x!}{1!}+\dots+\frac{(x+n-1)!}{n!}&=(x-1)!\left(\binom{x}{x-1}+\dots+\binom{x+n-1}{x-1}\right)\\\\
&=(x-1)!\left(\binom{x}{x}+\binom{x}{x-1}+\dots+\binom{x+n-1}{x-1}-1\right)\\\\
&=(x-1)!\left(\binom{x+n}{x}-1\right)\\\\
&=x\cdot\frac{(x+n)!}{n!}-(x-1)!
\end{align}$$
W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Exponential equation with double radical I'm trying ti solve this exponential equation:
$(\sqrt{2+\sqrt{3}})^x+(\sqrt{2-\sqrt{3}})^x=2^x$.
Here my try:
$\sqrt{2+\sqrt{3}}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ and
$\sqrt{2-\sqrt{3}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$
So i get this relation:
$\sqrt{\frac{3}{2}}+\sqr... | Hint:
$$\dfrac{\sqrt{2-\sqrt3}}2=\dfrac{\sqrt3-1}{2\sqrt2}=\cos(45^\circ+30^\circ)=\cos75^\circ$$
Similarly, $$\dfrac{\sqrt{2+\sqrt3}}2=\sin75^\circ$$
Now for $0<A<90^\circ,(\cos A)^x,(\sin A)^x$ is deceasing fucntion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding $\frac1{1+1^2+1^4}+\frac2{1+2^2+2^4}+\cdots+\frac n{1+n^2+n^4}$.
Find an expression for
$$\frac1{1+1^2+1^4}+\frac2{1+2^2+2^4}+\cdots+\frac n{1+n^2+n^4}.$$
This was given in the chapter for APs. However, I do not see how this relates to them. I tried using telescopic sums, but I am not proficient in them and w... | A telescoping series:
$\sum_{n=1}^{k} f(n) - f(n+1) = f(1) - f(k+1)$
How do we get our series into that form?
$(n^4 + n^2 + 1) = (n^2+n +1)(n^2-n+1)\\
\frac {n}{(n^4 + n^2 + 1)} = \frac {1}{2(n^2 - n + 1)} - \frac {1}{2(n^2 + n + 1)}$
Here is the really tricky bit:
$\frac {1}{(n^2 + n + 1)} = \frac {1}{(n+1)^2 - (n+1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence. As the question states:
"Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence."
I have consulted this related q... | Write $x=1+t$, so you need to find the Taylor series of
$$
\frac{1}{2}(1+t)^3\ln(1+t)
$$
at $t=0$. Since
$$
\ln(1+t)=\sum_{n>0}\frac{(-1)^{n+1}t^n}{n}
$$
your Taylor series is
$$
\frac{1}{2}\sum_{n>0}\frac{(-1)^{n+1}t^n}{n}+
\frac{3}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+1}}{n}+
\frac{3}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Which integers can be written as $x^2+2y^2-3z^2\ $?
For which integers $n$ has the diophantine equation $$x^2+2y^2-3z^2=n$$ solutions ?
These theorems
https://en.wikipedia.org/wiki/15_and_290_theorems
do not apply because the given quadratic form is not positive (or negative) definite. It seems that the quadratic for... | The jpegs did not come out well.... in Modern Elementary Theory of Numbers by Leonard Eugene Dickson, (1939) , on the top of page 161, homework exercise 2 is this: given that $C$ is a positive integer, $$ x^2 + 2 y^2 - C z^2 $$ is universal if and only if $C$ is odd and every prime factor of $C$ is $\; \equiv 1 \; \mb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $\frac 1{a+b+c}= \frac 1a+ \frac 1b+ \frac 1c$, then show for odd $n$, $ \frac 1{a^n+b^n+c^n}= \frac 1{a^n}+ \frac 1{b^n}+ \frac 1{c^n}$ The stated duplicate question does not address solution in terms of $p, q, r,$ i.e. not in terms of the Viete's formula. My approach is based on the viete's formula by finding the... | The polynomial is $$f(x)=x^3+px^2+qx+r=(x-a)(x-b)(x-c)$$
You have proven that $r=pq$ from the condition that is given.
$$(x+p)(x^2+q)=(x-a)(x-b)(x-c)$$
$$(x+p)(x-i\sqrt{q)}(x+i\sqrt{q})=(x-a)(x-b)(x-c)$$
WLOG, if $a=-p$, then $b=-c$ and $\frac1{b}=-\frac1{c}$.
If $n$ is odd, then we have $b^n=(-1)^nc^n=-c^n$ and $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Summation of product of integers with consecutive integers as first $m$ factors It is well known that
$$\sum_{r=1}^n r^\overline{m}=\frac {n^{\overline{m+1}}}{m+1}\\
\text{i.e.}
\sum_{r=1}^n \scriptsize(r+1)(r+2)\cdots (r+m-1)=\frac{n(n+1)(n+2)\cdots (n+m)}{m+1}\\
\text{which can also be written as }\\
\qquad \scripts... |
The assumption is correct (and in fact can be generalized) since we have
\begin{align*}
\sum_{r=1}^nr^{\overline{m+1}}&=\frac{n^{\overline{m+2}}}{m+2}\\
&=\frac{n^{\overline{m+1}}}{m+1}\cdot\frac{m+1}{m+2}(n+m+1)\\
&=\frac{m+1}{m+2}(n+m+1)\sum_{r=1}^n r^{\overline{m}}\tag{1}
\end{align*}
as well as
\begin{align*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int_0^2\frac {\arctan{x}}{x^2+2x+2}dx$
Compute $I=\int_0^2\frac {\arctan{x}}{x^2+2x+2}dx$
My 2 attempts:
First:
We observe that $\frac {1}{x^2+2x+2}=\frac {1}{(x+1)^2+1}$ and $\frac 1{(x+1)^2+1}=(\arctan{(x+1)})^{'}.$ Then:
$$\int_0^2\frac{\arctan{x}}{x^2+2x+2}dx=\int_0^2\arctan{x}(\arctan{(x+1)})^{'}dx=\le... | Some thoughts here:
\begin{align}
I&=\int\frac{\arctan x dx}{(x+1)^2+1}\\
&=\int\frac{\arctan (u-1) du}{u^2+1}\\
&=\int\frac{\arctan(\tan w-1)\sec^2w dw}{\sec^2w}\\
&=\int\arctan(\tan w-1)dw
\end{align}
Using similar substitutions, consider
$$J=\int\frac{\text{arccot} x dx}{(x+1)^2+1}=\int\text{arccot}(\tan w-1)dw$$
No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Gauss' law and a half-cylinder The question is:
A half cylinder with the square part on the $xy$-plane, and the length $h$ parallel to the $x$-axis. The position of the center of the square part on the $xy$-plane is $(x,y)=(0,0)$.
$S_1$ is the curved portion of the half-cylinder $z=(r^2-y^2)^{1/2}$ of length $h$... | The lower and upper boundaries for $y$ in $S_1$ should be $-r$ and $r$ instead of $0$ and $r$. I was thinking in polar coordinates instead of in Cartesian coordinates...
I'll edit the original post accordingly
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Proof of divergence of harmonic series with Cauchy criteria I have a proof where the divergence harmonic series is shown via the Cauchy criteria.
$\epsilon := 1/2$ and $m := 2n$.
$$
a_m - a_n = (1 + \frac{1}{2} + \ldots + \frac{1}{m}) - (1+\frac{1}{2} + \ldots + \frac{1}{n}) \\
= (1 + \frac{1}{2} + \ldots + \frac{1}{n}... | Observe that if $1 \le k \le n \implies n+k \le 2n \implies \dfrac{1}{n+k} \ge \dfrac{1}{2n}$. Let $k$ varies from $1$ to $n$ and add them up. Also the number of inequalities is $2n-(n+1)+1 = n$ which explains the numerator $n$ of the fraction $\dfrac{n}{2n}$ that is $\dfrac{1}{2}$ after simplifying .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2809812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A couple of big-O questions about a problem from number theory I am going through official solutions for IMO'09 problems. Currently stuck at solution 2 for N7. I am copying parts that confuse me below, but if anyone is interested in the rest, this is a link to full pdf.
Now, here are my questions:
*
*Letting $s_... | $\def\peq{\mathrel{\phantom{=}}{}}$First, there is a typo in the definition of $\{γ_n\}$. It should be$$
2γ_1 = 1, \quad γ_n = \sum_{j = 1}^{n - 1} γ_j γ_{n - j}. \quad (n \geqslant 2)
$$
For the first question, there is some abuse of notations in the official answer. In fact, since $0 < γ_n < 1$ for all $n \geqslant 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tricky question on polynomials
For any real numbers $x$ and $y$ satisfying $x^2y + 6y = xy^3 +5x^2 +2x$, it is known that $$(x^2 + 2xy + 3y^2) \, f(x,y) = (4x^2 + 5xy + 6y^2) \, g(x,y)$$
Given that $g(0,0) = 6$, find the value of $f(0,0)$.
I have tried expressing $f(x,y)$ in terms of $g(x,y)$. But seems that some t... | The given equation implies $$\frac{y}{x} = \frac{y^2 + 5x + 2}{x^2 + 6}$$ so we can rewrite the expression for $f/g$ as $$\frac{4 + 5(y/x) + 6(y/x)^2}{1+2(y/x) + 3(y/x)^2} = \frac{4(x^2 + 6)^2 + 5(y^2 + 5x+2)(x^2 + 6) + 6(y^2 + 5x+2)^2}{(x^2+6)^2 + 2(y^2 + 5x+2)(x^2 + 6) + 3(y^2 + 5x+2)^2}$$
Assuming $f$ and $g$ are su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is this proof of the following integrals fine $\int_{0}^{1} \frac{\ln(1+x)}{x} dx$? $$\int_{0}^{1} \frac{\ln(1+x)}{x} dx = \int_{0}^{1} \frac1x\cdot (x-x^2/2 + x^3/3 -x^4/4 \ldots)dx = \int_{0}^{1}(1 - x/2 + x^2/3 - x^4/4 \dots)dx = 1-\frac1{2^2} + \frac1{3^3} - \frac{1}{4^4} \dots = \frac{\pi^2}{12}$$
Also,
$$\int_{0... | Hint: integrate by parts.
Let $u = \ln(x)\implies u' = \dfrac1x$ and $v' = \dfrac1{1 + x}\implies v = \ln(1 + x)$. Therefore,
$$\int\dfrac{\ln(x)}{1 + x}\,\mathrm dx = \ln(x)\ln(1 + x) - \int\dfrac{\ln(1 + x)}x\,\mathrm dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that if ${x_1, x_2, x_3}$ are roots of ${x^3 + px + q = 0}$ then ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ How to prove that ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ holds in case ${x_1, x_2, x_3}$ are roots of the polynomial?
I've tried the following approach:
If $x_1$, $x_2$ and $x_3$ are roots then
$$(x-x_1)(x-x_2)(x-x... | Every symmetric polynomial can be expressed in terms of the elementary symmetric polynomials, in this case $s_1=x_1+x_2+x_3$, $s_2=x_1x_2+x_2x_3+x_3x_1$ and $s_3=x_1x_2x_3$. Since $x_1^3+x_2^3+x^3$ is homogeneous, we can find $a$, $b$ and $c$ such that
$$
x_1^3+x_2^3+x_3^3=as_1^3+bs_1s_2+cs_3
$$
*
*For $x_1=1$, $x_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Lagrange multiplier when decisions variables are not in the same set
Find the maximum of $2x+y$ over the constraint set $$S = \left\{ (x,y) \in \mathbb R^2 : 2x^2 + y^2 \leq 1, \; x \leq 0 \right\}$$
I want to use Lagrange multipliers to find the optimal solution. However, Lagrange requires $\vec x \in A$. In our cas... | You can solve this problem introducing convenient slack variables in order to eliminate inequalities. The inequalities
$$
2x^2+y^2 \le 1 \rightarrow 2x^2+y^2-1+\epsilon_1^2 = 0\\
x \le 0 \rightarrow x+\epsilon_2^2 = 0
$$
and the lagrangian thus is
$$
L(x,y,\lambda_1,\lambda_2,\epsilon_1,\epsilon_2) = 2x+y+\lambda_1(2x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A urn contains blue balls and red balls. I need to find probabiltiy of drawing more blue balls than red balls
A urn contains 5 identical blue balls and 4 identical red balls. Taking 5 balls at random from the urn what is the probability that the number of blue balls be greater than the number of red balls?
My first ... | Hint
There are $31$ permutations for selecting $5$ balls:
$$\left(
\begin{array}{ccccc}
\color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} \\
\color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{blue}{B} & \color{red}{R} \\
\color{blue}{B} & \color{blue}{B} & \color{blue}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$ $a, b, c ∈ \mathbb{R}+.$
WLOG assume $a \leq b \leq c.$ I tried substitution: $x=\frac{1} {1+a^2}, y=\frac{1} {1+b^2}, z=\frac{1} {1+c^2},$ so $x \geq y \geq z$ and $(1-x)+(1... | Continuing from where you stopped.
Multiply both sides by $xyz$ to clear out the denominators. Then multiple the first six terms by $x+y+z=1$, while the RHS with $(x+y+z)^2=1$ to get:
$$\sum_{\text{sym}}x^4y + \sum_{\text{sym}}x^3y^2 \ge 2\sum_{\text{cyc}}x^3yz + 2\sum_{\text{cyc}}x^2y^2z$$
This inequality is true, as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Help with finding Limit What is the limit of $$\lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{\frac{3n^3+4}{4n^4-1}}$$
Does any one can help, I am not sure how to solve this.
| $$
\begin{aligned}
\lim _{n\to \infty }\left(\frac{n!}{n^n}\right)^{\frac{3n^3+4}{4n^4-1}} &= \lim _{n\to \infty}\exp\left(\frac{3n^3+4}{4n^4-1}\ln\left(\frac{n!}{n^n}\right)\right)\\ &\approx
\lim _{n\to \infty }\exp \left(\frac{3n^3+4}{4n^4-1}\ln \left(\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{n^n}\right)\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Rearrangement of polynomial under square root In this page, the following equation
$$\sqrt{(x + c)^2 + y^2} + \sqrt{(x - c)^2 + y^2} = 2a \label{a} \tag{1}$$
is suggested to be arranged in this way:
$$\sqrt{(x + c)^2 + y^2} = 2a - \sqrt{(x - c)^2 + y^2}$$
and then it is squared in both sides. What if instead the square... | I think the following way is better.
Since $a>0$ we obtain that our equation is equal to
$$\left(\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}\right)^2=4a^2$$ or
$$\sqrt{(x^2+y^2+z^2)^2-4x^2c^2}=2a^2-x^2-y^2-c^2$$ and since $a>c$ and by the triangle inequality
$$\sqrt{x^2+y^2}<\frac{d_1+d_2}{2}=a,$$ we obtain:
$$2a^2-x^2-y^2-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^... | Wihout induction, it is simply a consequence of lil' Fermat:
As $3$ is prime, Fermat asserts that for any $n$, $\;n^3\equiv n\mod 3$, so
$$n^4-n^2=n^3\cdot n-n^2\equiv n^2-n^2\mod 3. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Solve the recurrence $T(n) = 2 T(n/2) + 2$ T(2) = 1
T(1) = 0
Ans is (3/2)* n - 2
My solution is :
T(n) = 2 T(n/2) + 2
T(n) = 4 T(n/4) + 4
T(n) = 8 T(n/8) + 6
T(n)=(2^k)T(n/2^k) + 2k
where k = log(n) ..... in base 2
as n/(2^k) = 1 for T(1)
I don't how to solve this type of question to get in terms of n only. Can an... | I am a bit confused because as pointed out in the comments, the intial conditions don't exactly make sense.
However, just taking T(2) = 1, we can obtain our result:
Since, $T(n) = 2*T(n/2) + 2 => T(n) = 2*(2*T(n/4) + 2) + 2 => T(n) = 2^2*T(n/2^2) + 2^2 + 2$
Proceeding further we get $T(n) = 2^2 * (2*T(n/2^3)+2) + 2^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
What about the definite integral $\int_0^1\int_0^1\frac{(\operatorname{arctanh}(xy))(\arctan(xy))}{\log(xy)}\,dx\,dy$? I would like to know if it is possible to justify the calculation of next definite integral
$$\int_0^1\int_0^1\frac{(\operatorname{arctanh}(xy))(\arctan(xy))}{\log(xy)}\,dx\,dy.$$
Question.(Corrected,... | As already pointed out in the comments, we have
$$ I \equiv \int \limits_0^1 \int \limits_0^1 \frac{\arctan(x y) \operatorname{artanh} (x y)}{\log(x y)} \, \mathrm{d} x \, \mathrm{d} y = - \int \limits_0^1 \arctan(t) \operatorname{artanh}(t) \, \mathrm{d} t \, . $$
Using
$$ \int \limits_0^x \operatorname{artanh} (t) \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What is the range of convergence of $\sum_{n=0}^{\infty} {(-1)}^n\binom{1/2}{n}\frac{1}{2n+3}.$ I was fiddling with the integral
$$\int_0^1 x^2\sqrt{1-x^2} \ dx $$
and I expanded the term under square root using a binomial series. Integrating, I got the result
$$\sum_{n=0}^{\infty} {(-1)}^n\binom{1/2}{n}\frac{x^{2n+3}}... | Since
$\begin{array}\\
\dfrac{\binom{\frac12}{n+1}}{\binom{\frac12}{n}}
&=\dfrac{\frac{\prod_{k=0}^{n+1}(\frac12-k)}{(n+1)!}}{\frac{\prod_{k=0}^{n}(\frac12-k)}{n!}}\\
&=\dfrac{\frac12-(n+1)}{(n+1)}\\
&=-(1-\dfrac{1}{2(n+1)})\\
\end{array}
$
$(-1)^n\binom{\frac12}{n}$
decreases,
as does
$(-1)^n\binom{\frac12}{n}\dfrac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$a+b\sqrt{3}=\sqrt{21-12\sqrt{3}}, a,b \in \mathbb {Z}$ Find a+b So far I've reasoned that $\mathbf{a}$ and $\mathbf{b}$ can't be both negative, because $\sqrt{21-12\sqrt{3}}$ cannot be negative.
Also $\mathbf{a}$ and $\mathbf{b}$ can't be both positive, because $\sqrt{21-12\sqrt{3}}$ is from 0 to 1, thus there is no... | Write
$$\left(a+b\sqrt{3}\right)^2=a^2+3b^2+2ab\sqrt{3}=21-12\sqrt{3}$$
This means
$$\begin{cases}
a^2+3b^2&=21\\
2ab&=-12
\end{cases}$$
Now substitute $a=-6/b$. You get the following biquadratic
$$b^4-7b^2+12=0$$
Has two solution for $b^2$, $3$ to be discarded because $b$ is an integer and $4$ which means that
$$b=\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving $\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$
Prove for every $a, b, c \in\mathbb{R^+}$, given that $abc=1$ :
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$$
I tried using AM-GM or AM-HM but I can't f... | Let us show stronger, with an intermediate step:
$$
\begin{aligned}
\left(a+\frac{1}{b}\right)^2 +
\left(b+\frac{1}{c}\right)^2 +
\left(c+\frac{1}{a}\right)^2
&\geq 4(a+b+c)
\\
&\geq 3(a+b+c+1)
\ .
\end{aligned}
$$
The last inequality, compactly written $a+b+c\ge3\sqrt[3]{abc}=3$ is simple, so let us show the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Check the proof that $f(x) = \cos({x})\cdot\cos({\sqrt{3}x})$ is not periodic
I have to prove that $f(x) = \cos({x})\cdot\cos({\sqrt{3}x})$ is not periodic
If the function is periodic then:
$$
f(x) = f(x+T)\\
\cos(x)\cdot\cos(\sqrt{3}x) = \cos(x+T)\cdot\cos(\sqrt{3}(x+T))
$$
Consider the function at $0$:
$$
\cos(T)\c... | As an alternative, we have that
$$\cos({x})\cdot \cos({\sqrt{3}x})=\frac12 \cos(x(\sqrt3-1))+\frac12\cos(x(\sqrt 3+1))$$
and $\cos(x(\sqrt3-1))$ has period $\frac{2\pi}{\sqrt 3-1} $ while $\cos(x(\sqrt3+1)+$ has period $\frac{2\pi}{\sqrt 3+1}$
and $\not\exists k\in\mathbb{Z}$ such that
$$\frac{2\pi}{\sqrt 3-1}=k\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Number of different sums with k numbers from {1, 5, 10, 50}
Say we have $k$ numbers, each of which belongs to the set $S = \{1, 5, 10, 50\}$
How many different sums can be created by adding these numbers?
If $k = 1$, the are four different sums.
Also, if $k = 2$, there are ten:
$$\begin{align} 1 + 1 = 2 \quad 1 +... | At least for small values of $k$, this answer will be tractable. Your answer is the number of terms in the expansion of $(x+x^{5}+x^{10}+x^{50})^k$ which can be found using a computer algebra system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Is there a way to solve this rather than just declaring a matrix $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ and then trying to solve a system of cubic equations? My attempt... | The matrix $C=A^3$ satisfies its characteristic equation, that is: $det(A^3-xI)=0$, which is $(4-x)(-2-x)+9=0$.
See https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem.
This gives $C^2-2C+I=0$, thus $C$ has the eigenvalue 1. The eigenvalues of $A^3$ are the cubes of the eigenvalues of $A$. If the eigenvalues... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Domain of Trigonometric Inequality The problem: Find values of $x$ on the interval $[0,\pi]$ for which $\cos(x)< \sin(2x)$
My answer came out to be $0.52< x<1.57$ or $2.62< x\leq π$. However, the textbook's answer is slightly different: $0.52< x<1.57$ or $2.62< x<π$. At $x=π$, $\cos(x)< \sin(2x)$, and $x$ can be $[0,π]... | $\cos(x)-2\sin(x)\cos(x)<0$ $\leftrightarrow$
$\cos(x)[1-2\sin(x)]<0$
*
*$\cos(x)<0$ if and only if $ \frac{\pi}{2}+2k\pi<x<\frac{3\pi}{2}+2k\pi$
while
*$\sin(x)>\frac{1}{2}$ if and only if
$\frac{\pi}{6}+2k\pi<x<\frac{5\pi}{6}+2k\pi$
So the solution is
$\frac{\pi}{6}+2k\pi<x<\frac{\pi}{2}+2k\pi$
or
$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solution of polynomial Find the set of values of $k$ for which the equation $x^4+kx^3+11x^2+kx+1=0$ has four distinct positive root.
Attempt:
$x^4+kx^3+11x^2+kx+1=0$
$x^2+kx+11+{k\over x}+{1\over x^2}=0$
$x^2 + {1\over x^2} +k(x+{1\over x})+11=0$
$(x + {1\over x})^2 +k(x+{1\over x})+13=0$
I don't know how to proceed a... | The reciprocal quartic should be transformed as:
$$\left( x+\frac{1}{x} \right)^2+k\left( x+\frac{1}{x} \right)+\color{red}{9}=0$$
Now,
\begin{align}
x+\frac{1}{x} &= \frac{-k \color{red}{\pm} \sqrt{k^2-36}}{2} \\
x &=
\frac{-k\color{red}{\pm} \sqrt{k^2-36}}{2(2)} \color{blue}{\pm}
\frac{1}{2} \sqrt{
\left( \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Range of $f(x)= \frac{\tan{x}}{\tan{3x}} $
Prove that for the values of $x$ where the following $f(x)$ is defined, $f(x)$ does not lie between $\frac{1}{3}$ and $3$. $$f(x)=\frac{\tan{x}}{\tan{3x}}$$
My Attempt:
I wrote down, $$\tan{3x}=\frac{3\tan{x}-\tan^3{x}}{1-3\tan^2{x}}$$
This reduced $f(x)$ to,
$$f(x)=\frac{1-... | Note that $\tan(x)^2 \ge 0$ and
$$f(x)= \frac{1-3\tan(x)^2}{3-\tan(x)^2} = 3+\frac{8}{\tan(x)^2-3}.$$
Clearly, for $\tan(x)^2 > 3$, $\frac{8}{\tan(x)^2-3} > 0$, and $f(x) >3$.
For $0 \le \tan(x)^2 < 3$, we have $-3 \le \tan(x)^2-3 < 0 \implies \frac{8}{\tan(x)^2-3}\le-\frac{8}{3} \implies f(x) = 3+\frac{8}{\tan(x)^2-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find all complex numbers satisfying $z\cdot\bar{z}=41$, for which $|z-9|+|z-9i|$ has the minimum value My first attempt was to express $z$ as $x+iy$ and minimize the expression $\sqrt{(x-9)^2+y^2}+\sqrt{x^2+(y-9)^2}$ where $x^2+y^2=41$.
That said, it seems to me that using the geometric interpretation could be easier.... | Hint: Show that $$\sqrt{(8x-9)^2+y^2}+\sqrt{x^2+(y-9)^2}\geq 9\sqrt{2}$$ and the equal sign holds if $$x=4,y=5$$
ok we will prove the inequality above:
squaring all we get
$$2\sqrt{(8x-9)^2+y^2}\sqrt{(x^2+(y-9)^2}\geq 162-(8x-9)^2-(y-9)^2-x^2-y^2$$,
now we use that $x^2+y^2=41$:
we get
$$2\sqrt{(8x-9)^2+41-x^2}\sqrt{41... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What does is mean $Y=\min \{X_1,X_2\}$? $Y=\min \{X_1,X_2\}$ and $Z=\max \{X_1,X_2\}$?
Let's determine $X_1$ distribution to be $\operatorname{Bin}(2,\frac{1}{2})$ and the distribution of $X_2$ to be $U(1,2,3)$. Also $X_,X_2$ are independent.
After some calculations:
*
*$P(X_1=0)=\frac{1}{4}$
*$P(X_1=1)=\frac{1}{2... | Assume that $X_1$ and $X_2$ are independent. The joint probability function of $X_1,X_2$ is given by
$$
\begin{array}{c|lcr}
X_2/X_1 & 0 & 1 & 2 \\
\hline
1 & 1/12 & 1/6 & 1/12 \\
2 & 1/12 & 1/6 & 1/12 \\
3 & 1/12 & 1/6 & 1/12
\end{array}
$$
The table for $Y=\min\{X_1,X_2\}$ is
$$
\begin{array}{c|lcr}
X_2/X_1 & 0 & 1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Exercise about the form of elements of $\mathbb{Q}(u)$
Let $u \in \mathbb{C}$ be a root of $x^3-x^2+x+2$. Consider $f = (u^2+u+1)(u^2-u)$ and $g= (u-1)^{-1}$ in $\mathbb{Q}(u)$ and express them in the form $au^2+bu+c$ with coefficients in $\mathbb{Q}$.
We have $f = (u^2+u+1)(u-1)u=(u^3-1)u=u^4-u = u^4 -u(u^3-u^2+u+2)... | Here is a solution using Euclidean division.
For $f$, we get
$$
f=(u^2+u+1)(u^2-u) = (u + 1)(u^3-u^2+u+2) + (-4 u - 2) = -4 u - 2
$$
For $g$, we get
$$
0 = u^3-u^2+u+2 = (u^2 + 1)(u-1) + 3
$$
and so the inverse of $u-1$ is $-(u^2 + 1)/3$.
Note that this works for $g$ only because $u-1$ has degree $1$ and so the remaind... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trying to understand $2$'s complement Using $n$-bit integers we can usually represent anything from $0$ to $2^n-1$. If we're interpreting a number with $2$'s complement, then the first bit is the sign bit, and we represent $-x$ with $2^n-x$. So for example, normally:
$n=4$:
\begin{array}{|c|c|}
\hline
\text{Decimal} & ... | It's the latter – arithmetic mod $2^n$ and everything works itself out naturally. And yes, we can't represent $-15$ with $4$ bits because it's outside the scope.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the quotient and remainder Find the quotient and remainder when $x^6+x^3+1$ is divided by $x+1$
Let $f(x)=x^6+x^3+1$
Now $f(x)=(x+1).q(x) +R $ where r is remainder
Now putting $x=-1$ we get $R=f(-1)$
i.e $R=1-1+1=1$
Now $q(x)=(x^6+x^3)/(x+1)$
But what I want to know if there is another way to get the quotient ex... | $$x^6+x^3+1=(x+1)(x^5-x^4+x^3)+1.$$
You may apply so-call long division method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\q... | We have: $\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2=(a+b)^2+2(a+b)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2=1+2ab+4+2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\dfrac{1+2ab}{(ab)^2}= 5+2ab+\dfrac{2}{ab}+\dfrac{1}{(ab)^2}+\dfrac{2}{ab}= 5+2ab+\dfrac{4}{ab}+\dfrac{1}{(ab)^2}= 1+\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 5
} |
Write $\omega$ in terms of the standard basis $dx^i ∧dx^j$ at each point.
At each point $p ∈ \mathbb R^3$, define a bilinear function $\omega_p$
on $T_p(\mathbb R^3)$ by
$$\omega_p(\textbf{a},\textbf{b})=\omega_p(\begin{bmatrix}
a^1 \\
a^2 \\
a^3 \\ \end{bmatrix}, \begin{bmatrix}
b^1 \\
... | The general formula for $\omega_p=a_{12}dx_1\wedge dx_2+a_{13}dx_1\wedge dx_3+a_{23}dx_2\wedge dx_3$ is $$\omega_p(a,b)=a_{12}\det \begin{pmatrix}a^1&a^2\\b^1&b^2\end{pmatrix}+a_{13}\det\begin{pmatrix}a^1&a^3\\b^1&b^3\end{pmatrix}+a_{23}\det\begin{pmatrix}a^2&a^3\\b^2&b^3\end{pmatrix}.$$
Choose a special $\omega_p=p^3d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find out the subsequential limits of a sequence
Let $$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$ and find the upper and lower limits of the sequence $\{x_n\}_{n=1}^\infty$.
We put $n=1, 2, 3 \dots $ then $x_1=-9, x_2 = \frac{15}{2}, \dots $ after some stage we see that limit superior is $x_2$ and inferi... | Here is a less rigorous but quicker way to look at the problem.
We have
$$x_n=(-1)^n \left(2+\frac{3^n}{n!}+\frac{4}{n^2}\right)$$
Let's look at
$$\lvert x_n\rvert=2+\frac{3^n}{n!}+\frac{4}{n^2}$$
We know that
$\dfrac4{n^2}$ is decreasing and $2$ is constant.
The first few terms of $\dfrac{3^n}{n!}$ are
$$\dfrac{3}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $(1+i)^{(1-2i)}$ Find all the values of $(1+i)^{(1-2i)}$ and show that there are small values as we wish (else from $0$) and big values as we wish
\begin{align}
(1+i)^{(1-2i)}&=e^{\ln(1+i)^{(1-2i)}}
=e^{(1-2i)\ln(1+i)}
\\&=e^{(\ln\sqrt{2}+i(\frac{\pi}{2}+2\pi k))(1-2i)}
\\&=e^{\ln\sqrt{2}-2\ln\sqrt{2}*i+i(\fr... | Hint. Note that $|e^z|=|e^{\text{Re}(z)+i\text{Im}(z)}|=e^{\text{Re}(z)}$, hence here we are interested in (see your third step where the correct argument of $(1+i)$ is $\pi/4+2\pi k$)
$$\text{Re}(\ln\sqrt{2}+i(\frac{\pi}{4}+2\pi k))(1-2i))=\ln\sqrt{2}+\pi\left(4k+\frac{1}{2}\right)$$
where $k\in\mathbb{Z}$. Now consid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is $(2^2-1)(3^2-1)(4^2-1)....(300^2-1)$ divisible by $7^{95}$? Is $(2^2-1)(3^2-1)(4^2-1)....(300^2-1)$ divisible by $7^{95}$?
And what about $7^{100}$?
This is taken out of one of the TAU entry tests.
I seem to always have a struggle with these exercises, probably because I don't enough experience.
What I've tried is t... | Note that $x^2-1=(x-1)(x+1)$. Then
$$
(2^2-1)(3^2-1)\cdots (300^2-1)=(2-1)(2+1)(3-1)(3+1)\cdots(300-1)(300+1).
$$
Since $49=7^2 < 300 < 7^3=343$, we will check the number of multiples of $7$ and $49$. Since $7\cdot 43=301$ and $7\cdot 44=308$, the number of multiples of $7$ less than $300$ and $302$ is $42$ and $43$, r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi... | HINT: $$\left[\left(\sin\frac{\pi}{20}+\cos\frac{\pi}{20}\right)+\left(\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}\right)\right]^2$$ is equal to $$2+2\sin\frac{\pi}{20}\cos\frac{\pi}{20}-2\sin\frac{3\pi}{20}\cos\frac{3\pi}{20}+2\left(\sin\frac{\pi}{20}+\cos\frac{\pi}{20}\right)\left(\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 3
} |
My solution to inhomogeneous $\frac{d^2y}{dx^2} + y = \sin{x}$ does not conform to my book's solution! I need help with the solution of this particular equation:
$$\frac{d^2y}{dx^2} + y = \sin{x}$$
Due to me having to go to work, I cannot display all my work in mathjax, my shift starts in 5 min...but my solution is: ... | $$y'' + y = \sin{x}$$
$$y''\cos(x) + y\cos(x) = \frac 12 \sin{(2x)}$$
$$y''\cos(x) -y'\sin(x)+y'\sin(x)+ y\cos(x) = \frac 12 \sin{(2x)}$$
$$(y'\cos(x))'+(y\sin(x))'= \frac 12 \sin{(2x)}$$
First integration
$$(y'\cos(x))+(y\sin(x))= -\frac 14 \cos{(2x)}+K_1$$
$$(\frac y {\cos(x)})'= -\frac 14 \frac {\cos{(2x)}+K_1}{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is $\log(n+1)-\log(n)$? What is gap $\log(n+1)-\log(n)$ between log of consecutive integers? That is what precision of logarithms determines integers correctly?
| Just added for your curiosity.
In the same spirit as in other answers, instead of Taylor series, you could consider Padé approximants and get things such as
$$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{2}{2 n+1}$$
$$\log(n+1)-\log n=\log\left(1+\frac1n\right) \approx \frac{6 n+3}{6 n^2+6n+1}$$
$$\log(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
How to get the shaded region of the rectangle? I have this problem:
So my development was:
Denote side of rectangle with: $2a, 2b$.
So, $4ab= 64, ab = 16$
Denote shaded region with $S$
Denote area of triangle $DGH = A_1$ and triangle $FBE = A_2$.
So, $A_1 + A_2 + S = 64$
$S = 64 - A_1 - A_2$
The triangles $A_1, A_2$ ... | Hint:
*
*there are eight small red triangles all with the same area
*six of them are shaded
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
How to determine if a set of five $2\times2$ matrices is independent $$S=\bigg\{\left[\begin{matrix}1&2\\2&1\end{matrix}\right], \left[\begin{matrix}2&1\\-1&2\end{matrix}\right], \left[\begin{matrix}0&1\\1&2\end{matrix}\right],\left[\begin{matrix}1&0\\1&1\end{matrix}\right],
\left[\begin{matrix}1&4\\0&3\end{matrix}\ri... | Stretch out the matrices to complete the rows of the following matrix
$$\newcommand{\adj}{\operatorname{adj}}
M(v)=\begin{bmatrix}
v_1&1&2&2&1\\
v_2&2&1&-1&2\\
v_3&0&1&1&2\\
v_4&1&0&1&1\\
v_5&1&4&0&3
\end{bmatrix}\tag1
$$
Note that the top row of the adjugate of $M(v)$
$$
\begin{bmatrix}
-14&-14&-14&28&14
\end{bmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Solve: $2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1\Bigl) = 0$ The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^x\Bigl(2^x-1... | Here, you forgot red:
$$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1\cdot \color{red}{2}}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2\color{red}{2^2}}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}\color{red}{2^{99}}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Given a matrix $A$ find $A^n$. $A=$$
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} $
Find $A^n$.
My input:
$A^2= \begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} = \begin{bmatrix}
1 & 4\\
0 & 1
\end{bmatrix} $
$A^3 ... | What you did was the smart approach. That is, you computed the first few terms of the sequence $(A^n)_{n\in\mathbb N}$ and you noticed a patern. The only thing that remains to be done is to prove that the pattern is real, but that's easy. Obviously,$$A^1=A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$$and$$A^n=\begin{pmatrix}1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
A special Vandermonde system with integer coefficients Consider the Vandermonde system
$$\left(\begin{matrix} 1&1&1&\dots&\dots&1\\
1&2&3&4&\dots&n\\
1&2^2&3^2&4^2&\dots&n^{2}\\
1&2^3&3^3&4^3&\dots&n^{3}\\
\vdots&&&&&\vdots\\
1&2^{n-1}&3^{n-1}&4^{n-1}&\dots&n^{n-1}\\
\end{matrix}
\right)~
\le... | With help from OEIS A127717 we conjecture that
for $0\le p\lt n$
$$\sum_{k=1}^n k^p (-1)^{k+1}
\sum_{q=k}^n {q-1\choose k-1} q = (-1)^p.$$
This is
$$\sum_{q=1}^n q \sum_{k=1}^q {q-1\choose k-1}
(-1)^{k+1} k^p
= \sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\choose k}
(-1)^{k} (k+1)^p
\\ = \sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\cho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The product of three consecutive integers is ...? Odd? Divisible by $4$? by $5$? by $6$? by $12$? If i have the product of three consecutive integers:
$n(n+1)(n+2)$, so the result is:
$A)$ Odd
$B)$ Divisible by $4$
$C)$ Divisible by $5$
$D)$ Divisible by $6$
$E)$ Divisible by $12$
My thought was:
$i)$ If we have three ... | One formal proof is by case analysis.
Let $n$ be an integer.
Then $n$ is congruent to one of $0,1,2\;$mod $3$.
*
*If $n\equiv 0\;(\text{mod}\;3)$, then $n$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\\[4pt]$
*If $n\equiv 1\;(\text{mod}\;3)$, then $n+2$ is a multiple of $3$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
compute the summation $\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$ compute the summation
$$\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$$
My attempts : i take $a_n =\frac{2n-1}{2\cdot4\cdots(2n)}$
Now
\begin{align}
& = \frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n}... | You made a mistake after this line
$$\sum_ {n=1}^{\infty}(\frac{2n}{2.4.6........2n} -\frac{1}{2.4.6........2n})$$
You simplified numerator and denominator by $2n$ but then
$$\sum_ {n=1}^{\infty}(\frac{1}{2.4.6........2(n-1)} -\frac{1}{2.4.6........2n})$$
You have at the denominator
$$2^nn!$$
So that
$$\sum_ {n=1}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\iint_D x \sin (y -x^2) \,dA$. Let $D$ be the region, in the first quadrant of the $x,y$ plane, bounded by the curves $y = x^2$, $y = x^2+1$, $x+y=1$ and $x+y=2$. Using an appropriate change of variables, compute the integral
$$\iint_D x \sin (y -x^2) \,dA.$$
I've been reviewing for an upcoming test and this ... | I'm not sure Kishores answer is precisely the same, but I try to be more illustrative and short. First of all we transform the integral by defining \begin{align}
u &= y-x^2 \\
v &= x^2
\end{align}
which leads to
$$
\int_{\rm D} x\sin(y-x^2) \, {\rm d}x \, {\rm d}y
=\frac{1}{2}\int_{\rm D} \sin u \, {\rm d}u \, {\rm d}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.