Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed... | Here's an alternative approach, motivated by the appearance of even numbers in the summand.
Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$
we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$
Now take $a_k=k\binom{n}{k}$ to obtain
\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
How to prove that $1-\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} \geq \frac{2^{N-1}+1}{2^N}$? The question is in the title. Numerical computation suggests the result is true, but I don't know how to prove it rigorously.
| I think I got it, but my answer isn't very slick. Improvements are very welcome.
Rearranging, it suffices to show that
$$\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} \leq \frac{1}{2} - \frac{1}{2^N}.$$
Now, I can verify this inequality by hand for $N \in \{1,2\}$. Let $N > 2$. First observe that
$$\frac{2^{n-1}-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Using residue theorem to calculate integral $\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}$ - where is my mistake? I am to calculate:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}
$$
We can set $\gamma(t)=e^{it}$ for $t \in [0, 2\pi]$ and then $z = e^{it}$, $\dfrac{dz}{iz}=dt$, $\sin t =\dfrac{1}{2i}(z-\frac{1}{z})$ so th... | $$
\operatorname{Res}\left(f,\frac{-i}{3}\right) = \lim_{z \to \frac{-i}{3}} \frac{1}{\color {red}3(z+3i)}=\frac{1}{8i}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve for $a_{n}$ where $a_{n} = 4a_{n-1} + 2^{2n-1}$, $a_0=1$ and $a_1=6$ . $a_{n} = 4a_{n-1} + 2^{2n-1}$, $a_0=1$ and $a_1=6$.
So I am trying to find $a_n$ by using the generating function let's call it $A(x)$.
the equation then is written as (if I doing this correctly) :
$A(x)-ao = 4xA(x) + \sum_{k=1}2^{2n-1}*x^k$
n... | Note, you didn't provide the boundary value for $a_1$.
Do you specifically want to use a GF? If you set
$$
b_n = \frac{a_n}{2^{2n-1}}
$$
to get (since $2(n-1)-1 = 2n -3$)
$$
b_n = b_{n-1} + 1
$$
and,
$$
a_n = 2^{n-2}a_1 + 2^{n-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Projection of triangle on coordinate axes? A triangle in the $xy$-plane is such that when projected onto the $x$-axis, $y$-axis and the line $y=x$ the results are line segments whose end points are $(1,0)$ and $(5,0)$, $(0,8)$ and $(0,13)$ and $(5,5)$ and $(7.5,7.5) $ respectively. If the area of triangle is $\Delta$,
... | Considering the inequalities, which should hold for every triangle point, namely $\begin{cases}1\le x\le 5\\8\le y\le 13\\10\le x+y\le 15\end{cases}$, we have the following figure containing all the triangle:
The border of the figure consists of $6$ straight line segments, each segment should contain a triangle vertex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$ Let's say P is equal to $x^4 + x^2 +1$ and $Q$ is equal to $x^3 + x$.
For $x <0$, $P$ is positive and $Q$ is negative. Hence, in this region, $P>Q$.
For $x=0$, $P>Q$.
Also, for $x = 1$, $P>Q$.
For $x > 1$, I factored out $P$ as $x^2(x^2+1) + 1$ and $Q$ as $x(x^2+1)$... | For $x\lt1$ we have $x(x^2+1)\lt(x^2+1)$, since $x^2+1\gt0$, and therefore
$$x^3+x=x(x^2+1)\lt x^2+1\le x^4+x^2+1$$
For $x\ge1$, we have $x^3+x\le x(x^3+x)$, and therefore
$$x^3+x\le x(x^3+x)=x^4+x^2\lt x^4+x^2+1$$
Thus $x^3+x\lt x^4+x^2+1$ for all $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 3
} |
Turning a fraction with repeating decimals into a mixed number: why doesn't this work? Problem:
Turn $\frac{0.\overline{48}}{0.\overline{15}} $ into a mixed number.
My solution:
$0.\overline{15}$ goes into $0.\overline{48}$ 3 times, with a remainder of $0.\overline{48} - 3 x 0.\overline{15} = 0.\overline{48}-0.\overlin... | When you want to compute, say,
$$
\frac{25}{7}
$$
you say "$7$ goes into $25$ three times, with a remainder of $4$."
But does that mean that
$$
\frac{25}{7} = 3 + 4?
$$
Not at all! It means that
$$
\frac{25}{7} = 3 + \frac{4}{7}.
$$
By analogy, in your case, you have
$$
\frac{0.\overline{48}}{0.\overline{15}}
$$
is $3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Construct a Reed Solomon code: find the parity check matrix I am trying to solve the following exercise, but I need a check/opinion on how to solve it.
Construct a Reed-Solomon code with dimensions $[12,7]$ over $\mathbb{F}_{13}$ and find a parity check matrix for the code $C$. Hint: $2$ is a primitive element of $\ma... | As an alternative to @rschweib's answer and possibly requiring little computation you have good look-up tables, a cyclic Reed-Solomon code whose generator polynomial has $2, 2^2, 2^3, 2^4, 2^5$ as roots has parity check matrix
$$H = \left[\begin{matrix}
1&2&2^2&2^3&\quad \cdots&2^{11}\\
1&2^2&(2^2)^2&(2^2)^3&\quad \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $(D^2-1)y=e^x(1+x)^2$ I did like this:
$$\text{Let,} y=e^{mx} \text{ be a trial solution of } (D^2-1)y=0$$
$$\therefore \text{The auxiliary equation is } m^2-1=0$$
$$\therefore m=\pm1\\
\text{C.F.} = c_1e^x+c_2e^{-x}$$
$$\begin{align}
\text{P.I.}& =\frac{1}{D^2-1}e^x(1+x)^2\\
& =e^x\frac{1}{(D+1)^2-1}(1+2x+x^2)... | It is the same. Your coefficient for $e^x$ is $c_1-\frac 18$, which you can rename it to say $c_3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving trigonometric equations: $\cos(3x)+ \cos(x)=0$ Recently I worked on a problem where I had to solve $$\cos(3x)+\cos(x)=0$$
When I tried calculating it by evaluating $x$, $x=2nπ+\dfrac{\pi}{4}$, $x=2n\pi-\dfrac{\pi}{4}$, or $x=2n\pi+\dfrac{\pi}{2}$ was the solution I reached.
Unfortunately, it was apparently wro... | So, we have ${\cos(2x + x) + \cos(x) = 0}$. Using the compound trig formulae:
$${\Rightarrow \cos(2x)\cos(x) - \sin(2x)\sin(x) + \cos(x) = 0}$$
Using the double angle identity for ${\sin(2x)}$, and ${\cos(2x)}$ gives
$${(2\cos^2(x) - 1)\cos(x) - 2\cos(x)\sin^2(x) + \cos(x)= 0}$$
$${\Rightarrow 2\cos^3(x) - 2\cos(x)\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Prove that $\sum_nF_{2n+1}x^n=\frac{1-x}{1-3x+x^2}$
Prove that $$\sum_nF_{2n+1}x^n=\frac{1-x}{1-3x+x^2}$$
This is a problem of generating functions. I know that $\sum_nF_{2n}x^=\frac{x}{1-3x+x^2}$ and I am guessing this is a good point to start. But I don't really know where to go from here. Could someone help me?
| $$\sum\limits_{n=0}^\infty F_{2n}x^n=\dfrac{x}{1-3x+x^2}$$
$$=F_0+\sum_{n=0}^\infty F_{2n+2}x^{n+1}=x\sum_{n=0}^\infty F_{2n+2}x^n.$$
$$\therefore \sum_{n=0}^\infty F_{2n+2}x^n=\dfrac{\sum\limits_{n=0}^\infty F_{2n}x^n}x=\dfrac{1}{1-3x+x^2}.$$
$$\therefore\sum_{n=0}^\infty F_{2n+1}x^n=\sum_{n=0}^\infty F_{2n+2}x^n-\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int_0^{\infty} \frac{e^{\frac{2}{1+x^2}} \cos{\left(\frac{2x}{1+x^2}\right)}}{x^2+1} \mathop{dx}$ A challenge problem $$\int_0^{\infty} \frac{e^{\frac{2}{1+x^2}} \cos{\left(\frac{2x}{1+x^2}\right)}}{x^2+1} \mathop{dx}$$
Someone said that differentiation in the integral should be used, I dont know how? I try... | Expanding on my comment, the integrand is the real part of the complex function $$f(z) = \frac{\exp \left(\frac{2}{1-iz} \right)}{1+z^{2}}. $$
And by letting $z= x+iy$, we find that the real part of $\frac{2}{1-iz}$ is $$\frac{2(y+1)}{x^{2}+(y+1)^{2}}.$$
Therefore, in the upper half of the complex plane, the magnitude ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
For real values with $abc\neq0$, if $\frac{xb+(1-x)c}{a}=\frac{xc+(1-x)a}{b}=\frac{xa+(1-x)b}{c}$, show that $x^3 = -1$ and $a=b=c.$
Let $a,b,c$ and $x$ are real numbers such that $abc \neq 0$ and $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c}.$$ Prove that $x^3=-1$ and $a=b=c.$
My attempt $:... | A Comment and the full scenario:
$a,b,c,x$ are real and $abc \ne 0$
Case (1) : $a+b+c \ne 0$
(i): $a \ne b \ne c:$
We get $x^3=-1\implies x=-1$ (from OP's work if multiply three expressions of $x$) without $x$ becoming indeterminate ($0/0$).
(i): $a=b=c$ : After OP's second-step we can write two equations for $a,b,c$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $ 3a+2b+c=7$ then find minimum value of $ a^2+b^2+c^2$ Question:- If $ 3a+2b+c=7$ then find the minimum value of $ a^2+b^2+c^2 $.
I used vectors to solve this problem.
Let $$α=3\hat{i}+2\hat{j}+\hat{k}$$
$$β=a\hat{i}+b\hat{j}+c\hat{k}$$
Using Cauchy-Schwarz inequality
we have, $|α.β|\le |α| |β|$
$=|3a+2b+c|\le\sqrt{... | There is a variant of the Cauchy-Schwarz inequality and its name is by a certain community of problem solvers as Titu’s lemma:
$a^2+b^2+c^2 = \dfrac{(3a)^2}{9}+\dfrac{(2b)^2}{4}+\dfrac{c^2}{1} \ge \dfrac{(3a+2b+c)^2}{9+4+1} = \dfrac{49}{14} = \dfrac{7}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+... | Let $x^2+y^2=2uxy$.
Thus, $u\geq1$ and we need to prove that:
$$x^2+y^2-\sqrt[3]{(x^3+y^3)^2}>6\left(\sqrt[3]{x^3+y^3}-x\right)\left(\sqrt[3]{x^3+y^3}-y\right)$$ or
$$x^2+y^2-6xy+6(x+y)\sqrt[3]{x^3+y^3}-7\sqrt[3]{(x^3+y^3)^2}>0$$ or
$$(x^2+y^2-6xy)^3+216(x+y)^3(x^3+y^3)-343(x^3+y^3)^2+$$
$$+126(x^2+y^2-6xy)(x+y)(x^3+y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Determinant of a Toeplitz matrix How can I calculate the determinant of the following Toeplitz matrix?
\begin{bmatrix}
1&2&3&4&5&6&7&8&9&10\\
2&1&2&3&4&5&6&7&8&9 \\
3&2&1&2&3&4&5&6&7&8 \\
4&3&2&1&2&3&4&5&6&7 \\
5&4&3&2&1&2&3&4&5&6 \\
6&5&4&3&2&1&2&3&4&5 \\
7&6&5&4&3&2&1&2&3&4 \\
8&7&6&5&4&3&2&1&2&3 \\
9&8&7&6&5&4&3&2&1... | We define the following $n \times n$ (symmetric) Toeplitz matrix
$${\rm A}_n := \begin{bmatrix}
1 & 2 & 3 & \dots & n-1 & n \\
2 & 1 & 2 & \dots & n-2 & n-1 \\
3 & 2 & 1 & \dots & n-3 & n-2 \\
\vdots & \vdots & \vdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
There exists $c$ such that $\int_a^{(a+b)/2}f(x)dx=(b-a)/4(f(a)+f((a+b)/2))-(b-a)^3/96f''(c)$ I am struggling to understand a statement from a solution to a problem concerning integrals.
The hypotesis of the problem is that $f:[a,b]\to\mathbb{R}$ is twice differentiable with continuous second derivative and (but I don'... | We have that
$$\int_a^\frac{a+b}{2}f(x)\mathop{}\!\mathrm{d}x=\int_{a}^\frac{a+b}{2}f(x)\left(x-\frac{3a+b}{4}\right)^\prime \mathop{}\!\mathrm{d}x=$$
$$=f(x)\left(x-\frac{3a+b}{4}\right)\Bigg|_a^\frac{a+b}{2}-\int_a^\frac{a+b}{2}f'(x)\left(x-\frac{3a+b}{4}\right)\mathop{}\!\mathrm{d}x=$$
$$=\frac{b-a}{4}\left(f\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximize $\boxed{\mathbf{x}+\mathbf{y}}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$ Maximize $\mathbf{x}+\mathbf{y}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$
My approach
$\frac{x^{2}}{1 / 2}+\frac{y^{2}}{1 / 3} \leq 1$
Let $z=x+y$
$\mathrm{Now}, 4 \mathrm{x}+6 \mathrm{y} \frac{d y}{d x}=0 \Ri... | Alternatively, the objective function $z=x+y$ will achieve its maximum when the contour line $y=-x+z$ will touch the ellipse $2x^2+3y^2=1$ from above. So, the slope of the tangent line must be equated to the slope of the contour line:
$$4x_0^2+6y_0^2y'=0\Rightarrow y'=-\frac{2x_0}{3y_0}=-1\Rightarrow x_0=\frac32y_0$$
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Number of Real Solutions of $\frac{7^{1+\cos(\pi x)}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}=1$
Find the number of real solutions of the equation $$\frac{7^{1+\cos(\pi x)}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}=1\,.$$
By hit and trial i got the solution at $x=\pm 1$ but i am not able to solve it as it involves power of 7
| Note that $1+\cos(\pi x)\geq 0$ for all $x\in\mathbb{R}$. Therefore,
$$1-\frac{7^{1+\cos(\pi x)}}{3}\leq 1-\frac{1}{3}=\frac{2}{3}\,.$$
Thus, if $x$ is a real solution of the required equation, then
$$3^{x^2-1}+3^{2\big(1-|x|\big)}=3\,\left(3^{x^2-2}+9^{\frac12-|x|}\right)=3\,\left(1-\frac{7^{1+\cos(\pi x)}}{3}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many positive integers less than or equal to $500$ have exactly $3$ positive divisors?
How many positive integers less than or equal to $500$ have exactly $3$ positive divisors?
My approach was to use the following inequality $N \leqslant 500$, where $N$ is the number of positive integers that have exactly $3$ po... | For $n$ to have exactly $3$ positive divisors, $1$ must be a factor, $n$ must be a factor and also we need another factor say $p$.
We need $\frac{n}{p}=p$ and $p$ is a prime. Prime is needed or additional divisors exists.
Hence $n$ must be a prime square.
Since $\sqrt{500}\approx 22.3$, the numbers are $2^2, 3^2, 5^2, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
For which $k$ does $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ hold? By generalizing this (1) and this (2) questions and performing some research
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to find a maximal $k$ for which the following inequality is true for any positives $a$, $b$ and $c$.
$$\frac{9uv^2}{w^3}+k-3\geq \left(2+\frac{k}{3}\right)\frac{3u}{w},$$
which says that it's enough to show it for a minimal value of $v^2$.
Now, $a$, $b$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Intuition for weighted average. Why $\frac{w_1}{w_1 + w_2}x_1 + \frac{w_2}{w_1 + w_2}x_2 = \frac{\sum_i w_ix_i}{\sum_i w_i}$? I know $\dfrac{w_1}{w_1 + w_2}x_1 + \dfrac{w_2}{w_1 + w_2}x_2 = \dfrac{\sum_i w_ix_i}{\sum_i w_i}$, because $\sum_i w_i$ is common denominator. I'm not asking about this algebra.
It's intuitive ... | Here is an example from statistics.
The table shows the sales of sugar (in kilograms) during $10$ days:
$$\begin{array}{c|c|c}
\text{Sales of sugar (in kg)}, x & \text{Number of days}, f & \text{Percentage of days}, P(x)\\
\hline
0&1&0.1\\
1&3&0.3\\
2&4&0.4\\
3&2&0.2\\
\hline
&10&1
\end{array}$$
On $3$ days (or during ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Solve the inequality $|3x-5| - |2x+3| >0$. In order to solve the inequality $|3x-5| - |2x+3| >0$, I added $|2x+3|$ to both sides of the given inequality to get $$|3x-5| > |2x+3|$$ Then assuming that both $3x-5$ and $2x+3$ are positive for certain values of $x$, $$3x-5 > 2x+3$$ implies $$x>8$$ If $3x-5$ is positive and ... | Needless to explicit the absolute values in function of the intervals where $x$ lives:
$$|3x-5| > |2x+3| \iff (3x-5)^2>(2x+3)^2\iff 5x^2-42x+16>0,$$
so it comes down to a quadratic inequality.
The reduced discriminant is $\Delta'=21^2-80=361=19^2$, and the quadratic is positive outside the interval of the roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
If $a_{n+1}=2a_n −n^2+n$ Define a sequence $a_n$ that satisfy the recurrence relation as described above, with $a_1 = 3$ If
$$a_{n+1}=2a_n −n^2+n$$
Define a sequence $a_n$
that satisfy the recurrence relation as described above, with $a_1 = 3$
Find the value of $$\dfrac{ |a_{20} - a_{15} | }{18133} $$
Attempt
First ev... | As an alternative, you can use more elementary methods for linear difference equations. In fact, the general solution will be of the form
$$
a_n = a_n^h + a_n^*
$$
where $a_n^h$ is the general solution of the homogeneous equation, i.e. $a_n^h = c 2^n$, and $a_n^*$ is a particular solution of the full equation. If you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{... | Letting
$v = u^2+1$,
$dv = 2u\, du, u^2 = v-1$
so
$\begin{array}\\
\int\dfrac{u^3}{(u^2+1)^3}du
&=\int\dfrac{(v-1)}{2v^3}dv\\
&=\frac12\int(v^{-2}-v^{-3}dv\\
&=\frac12\left(\dfrac{v^{-1}}{-1}-\dfrac{v^{-2}}{-2}\right)\\
&=\frac12\left(-v^{-1}+\frac12 v^{-2}\right)\\
&=\frac12\left(-\dfrac1{u^2+1}+\dfrac1{2(u^2+1)^2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 8
} |
Calculate the Covariance of X and Y for a Uniformly Distributed Quadrilateral
Exercise 8.15: Let $(X,Y)$ be a uniformly distriubted random point on the quadrilateral $D$ with vertices $(0,0), (2,0), (1,1),$ and $(0,1)$. What is the Covariance of $X$ and $Y$?
After numerous attempt I calculate Cov($X,Y) = \frac{83}{32... | Thanks to the comment from YJT, I realize that it was a computational error. The integrals were set up correctly, but I forgot to add the integrals for the squares into my expectation, i.e., while computing by hand I simply dropped a term.
Correction:
\begin{equation*}
E[X] = \int_0^1\int_0^1\frac{2}{3}xdxdy + \int_1^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Geometric sequence problem including sum of the numbers Numbers: $a,b,c,d$ generate geometric sequence and $a+b+c+d=-40. $ Find these numbers if $a^2+b^2+c^2+d^2=3280$
I tried this problem and I have system of equations which I can't solve. I think there should be different way to handle this problem.
| Let $r$ be the common difference of the GP. Then using GP sum formula, we have,
\begin{align}
a+b+c+d=-40&\implies a\left(\dfrac{r^4-1}{r-1}\right)=-40\tag1\\
a^2+b^2+c^2+d^2=3280&\implies a^2\left(\dfrac{r^8-1}{r^2-1}\right)=3280\tag2
\end{align}
Now, divide the second equation from the square of the first equation to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \t... | The OP correctly identified the power rule for exponents
$$(b^n)^m= b^{n\times m}=b^{nm}.$$
Therefore for $b=\sec\theta,n=2,$ and $m=3/2$,
$$27\ ( \sec^2 \theta ) ^\frac{3}{2}=27(\sec\theta)^{2\times\frac32}=27(\sec^3\theta).$$
To finish the problem, the OP can apply the substitution $x = \frac{3}{2} \tan \theta \impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
Small-angle approximation of $ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} $ I need to show the following:
$$ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} \approx 1-\frac{x^2}{6} $$ when $ x $ is small.
I think this problem is trickier than most other questions like it because in the original source there is c... | Suppose for simplicity you had two polynomials $P(z) = 1+ z + z^2 + 4z^4 + 7z^5$ and $Q(z) = 1 + 2z + 3z^2 + 4z^3$, and I asked you to calculate the product $P(z)Q(z)$... but not the entire thing. Suppose I only want the terms up to quadratic term; i.e if we write
\begin{align}
P(z)Q(z) &= a_0 + a_1z + a_2z^2 + a_3z^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Let $f(x) = 2x^2 + nx - 6$ and $g(x) = mx^2 + 2x - 4$? The functions are combined to form the new functions $h(x) = f(x) - g(x)$. Points $(2, 4)$ and $(-3, 17)$ are given to satisfy the new function. Determine the values of $m$ and $n$.
I know these formulas but I don't know what is $m$ and $n$ values are.
$$h(x)=(2-m)... | Just continue... But be careful of arithmetic
$h(x) = f(x) - g(x) = (2-m)x^2 + (n-2)x -2$
$h(2) = 4(2-m) + 2(n-2)-2 = -4m + 2n +2 = 4$.
so simplify $-4m + 2n = 2$ and $-2m + n = 1$
$h(-3) = 9(2-m)-3(n-2) -2 = -9m -3n+22 = 17$
so simplify $-9m -3n = -5$ so $3m +n = \frac 53$
Now just solve for $n$ and $m$.
Simple subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\... | $\begin{aligned} I&=\int \frac{d x}{\left[(x-2)^{2}+9\right]^{2}} \\ &=\int \frac{d y}{\left(y^{2}+9\right)^{2}}, \text { where } y=x-2 \\ &=-\frac{1}{2} \int \frac{1}{y} d\left(\frac{1}{y^{2}+9}\right) \quad \text{(By IBP)}\\ &--\frac{1}{2 y\left(y^{2}+9\right)}-\frac{1}{2} \int\left(\frac{1}{y^{2}} \cdot \frac{1}{y^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 8
} |
How to evaluate $\int \frac{\cos x}{a-\cos x} \mathrm{d}x $ in a more elegant way? I am trying to evaluate
$$\int \frac{\cos x}{a-\cos x} \mathrm{d}x \quad (1)$$
Since there is a ratio of trigonometric functions, I tried to reduce the problem to a polynomial ratio by using Weierstrass substitutions:
$$t = \tan\left(\f... | Let $|a| > 1$. Observe $$f_a(x) = \frac{\cos x}{a - \cos x} = \frac{(\cos x - a) + a}{a - \cos x} = -1 + \frac{a}{a - \cos x} = -1 + \frac{a}{a - (1 - 2 \sin^2 \frac{x}{2})}$$ where we have employed the half-angle identity $$\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2}.$$ Consequently
$$\begin{align} f_a(x)
&= -1 + a \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the matrix $A^{15}$. Let $I=
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}$ and $O=\begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}$.
1.Let $A=\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}$ and $ B=\begin{pmatrix}
x & 3 \\
3 & 6 \\
\end{pmatrix}$. Find the value of $x$ which satisfies $AB=BA$.
$AB=\begin{pm... | If you recognize $x^2-x+1$ as a cyclotomic polynomial, then
$$x^{15} + 1 = (x^3+1)a(x) = (x^2-x+1)b(x)$$ gives $A^{15}+I=0$.
The systematic way, which does not need insights, is to use polynomial division:
$$
x^{15}=(x^2-x+1)q(x)-1
$$
where $q(x)=x^{13} + x^{12} - x^{10} - x^9 + x^7 + x^6 - x^4 - x^3 + x + 1$ is not re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Strange Cube Root Offense in an Inequality I don't know how to tackle the unusual cube root present in this inequality-
$1.$For real numbers $a,b,c > 0$ and $n\le3$ prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c}\right)\ge 3+n$$
Here is another question with the same lesser side (an... | A proof for $n=3$ of the second one.
By AM-GM $$\sum_{cyc}\frac{a^2}{b}+\frac{9}{a^2+b^2+c^2}\geq2\sqrt{\sum_{cyc}\frac{a^2}{b}\cdot\frac{9}{a^2+b^2+c^2}}.$$
Thus, it's enough to prove that:
$$\sum_{cyc}\frac{a^2}{b}\geq a^2+b^2+c^2$$ or
$$\sum_{cyc}\frac{a^2}{b}\sum_{cyc}ab\geq\sum_{cyc}a^2\sum_{cyc}a$$ or
$$\sum_{cyc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating an improper integral - issues taking the cubic root of a negative number Problem:
Evaluate the following integral.
$$ \int_{-1}^{-1} \frac{dx}{x^\frac{2}{3}} $$
Answer:
This integral includes the point $x = 0$ which results in a division by $0$. To get around this difficulty, we break the integral into two i... | Cube root of a real number $p$ is the unique real number $q$ such that $q^3=p$. Therefore, $x^{\frac{1}{3}} \to -1$ as $x \to -1$ because $(-1)^3=-1$ and so $(-1)^{\frac{1}{3}}=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the value of $1 -\omega^h + \omega^{2h} -...+(-1)^{n-1} \omega^{(n-1)h}$ when $\omega$ is a root of unity? I'm reading Ahlfors' complex analysis book. One of the problems in the book says as follows
What is the value of $1 -\omega^h + \omega^{2h} -...+(-1)^{n-1} \omega^{(n-1)h}$?
where $h$ is some integer and... | A bit late of an answer, but I think there is a simpler form for the even case (provided I haven't made any mistakes).
Startig with the form $\frac{2}{1+\omega^h}$, we begin with the usual fare of multiplying on top and bottom by the conjugate, leaving us with
$$\frac{2(1+\bar{\omega}^h)}{1+\omega^h + \bar{\omega}^h + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding coefficients in expansions Show that the coefficient of $x^{−12}$ in the expansion of
$$\left(x^4−\frac{1}{x^2}\right)^5\left(x−\frac{1}{x}\right)^6$$
is $−15$, and calculate the coefficient of $x^2$. Hence, or otherwise, calculate the coefficients of $x^4$ and $x^{38}$ in the expansion of $$(x^2−1)^{11}(x^4+x^... | \begin{align*}
(x^2 - 1)^{11}(x^4 + x^2 + 1)^5
&=(x^2 - 1)^{5}(x^4 + x^2 + 1)^5(x^2 - 1)^{6} \\
&=(x^6 - 1)^5(x^2 - 1)^6 \\
&=x^{16}\left(x^4 - \frac{1}{x^2}\right)^5\left(x - \frac{1}{x}\right)^6
\end{align*}
Thus, the coefficient of $x^4$ in the new equation is the same as the coefficient of $x^{-12}$ in the origina... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2 $ I'm trying to calculate:
$$T = \lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2$$
Here is my attempt.
Put $x^2=\dfrac{1}{t}$ so when $x\to \infty, t \to 0$ and the limit become
\begin{align*}
T &= \lim\limits_{t \to 0} \sqrt[... | $$\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2=\lim\limits_{x \to \infty}x^2\left[ e^{\frac{1}{n}\ln \left(1+ \frac{1}{x^2} \right)\cdots \left(1+ \frac{n}{x^2} \right)}-1 \right] =\\=\lim\limits_{x \to \infty}\frac{1}{n}x^2\left[ \ln \left(1+ \frac{1}{x^2} \right)+\cdots+\ln \left(1+ \frac{n}{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Show that for $p \geq 1$, $[\Gamma(\frac{p+1}{2})]^{1/p}=O(\sqrt{p})$ as $p \rightarrow \infty$ Actually it is Exercise 2.5.1 in High Dimensional Probability by Vershynin.
I got stuck at showing that for $p \geq 1$,
$$\left[\Gamma\left(\frac{p+1}{2}\right)\right]^{1/p}=O(\sqrt{p}) \ \ \ \text{as }p\rightarrow\infty$$
I... | $$
\begin{align}
\Gamma\!\left(\frac{p+1}2\right)
&=\Gamma\!\left(1+\left\{\frac{p-1}2\right\}\right)\prod_{k=1}^{\left\lfloor\frac{p-1}2\right\rfloor}\left(k+\left\{\frac{p-1}2\right\}\right)\tag1\\
&\le1\cdot\left[\frac1{\left\lfloor\scriptstyle{\frac{p-1}2}\right\rfloor}\sum_{k=1}^{\left\lfloor\frac{p-1}2\right\rflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find limits of $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor Expansion. Find the limit $\displaystyle \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor expansion.
My Try
$\displaystyle =\lim _{x \to 0} \frac {\cos x - \frac{\sin x}{x... | While searching more easy way, let me suggest one possible way to solve main difficult part of suggested limit. I change denumerator to $x^3$, for simplicity, as it's equivalent $x^2\sin x$
Suppose we know existence of limit. Then
$$L=\lim_{x\to0}\frac{x-\sin x}{x^3} = \lim_{x\to0}\frac{x-3\sin \frac{x}{3}+4 \sin^3 \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in t... | Put $A = x^2 - x - 1$, $B = ax^9 + bx^8 + 1$ now we want $B/A$ to divide without remainder. We can subtract multiples of $A$ from $B$ to kill off high order terms and see what the remainder would be:
B
- a*x^7*A
= (a + b)*x^8 + a*x^7 + 1
- (a + b)*x^6*A
(2*a + b)*x^7 + (a + b)*x^6 + 1
- (2*a + b)*x^5*A
(3*a + 2*b)*x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Evaluate integral $\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}$ How can I evaluate this integral $$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\;\;?$$
My attempt:
I tried using substitution $x=\sec\theta$, $dx=\sec\theta\ \tan\theta d\theta$,
$$\int (\sec^2\theta-1)(\sec^3\theta-3\sec\theta)^{4/3}
\sec\theta\ \tan\theta d\the... | $$t=x^{3}-3x\\
I=\frac{1}{3}\int t^{\frac{4}{3}}dt=\frac{1}{7}t^{2}\sqrt[3]{t}+c=\frac{1}{7}(x^{3}-3x)^{2}\sqrt[3]{x^{3}-3x}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
How to prove $\int_0^\pi\biggl |\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx \le c(k+1)$ for some constant $c$? Let $|a_i|\le1$, prove that $$\int_0^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr|\,dx \le c(k+1)$$ for some constant $c$.
I tried to solve ... | On the other forum, I found the answer to this question.The website link is here, https://www.zhihu.com/question/410940277/answer/1400850373.
I copied the proof roughly here.
$$\int_0^\pi \biggl|\sum_{i=0}^k \frac{a_i\sin(i+\frac{1}{2})x}{2\sin \frac{1}{2}x}\biggr| \, dx = \int_0^\frac{\pi}{k+1} \biggl|\sum_{i=0}^k \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a$, $b$, $c$, $d$ are positive reals so $(a+c)(b+d) = 1$, prove the following inequality would be greater than or equal to $\frac {1}{3}$.
Let $a$, $b$, $c$, $d$ be real positive reals with $(a+c)(b+d) = 1$. Prove that $\frac {a^3}{b + c + d} + \frac {b^3}{a + c + d} + \frac {c^3}{a + b + d} + \frac {d^3}{a + b + ... | Also, we can use Holder here:
$$\sum_{cyc}\frac{a^3}{b+c+d}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c+d)}=\frac{(a+b+c+d)^2}{12}\geq\frac{\left(2\sqrt{(a+c)(b+d)}\right)^2}{12}=\frac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int \left(\frac{1}{3x}-2\sec^2\left(\frac x2\right)-e^{-2x+3}\right)dx$ Given that $\displaystyle f(x)=\frac{1}{3x}-2\sec^2\left(\frac{x}{2}\right)-e^{-2x+3}$, evaluate $\int f(x) \mathrm{d}x$.
Attempt: $$\begin{aligned} \int f(x) \mathrm{d}x&=\int \left[ \frac{x}{3}-2\sec^2\left(\frac{x}{2}\right)-e^{-2x+... | Instead of evaluating
$$\int \left(\frac{1}{3x}-2\sec^2\left(\frac x2\right)-e^{-2x+3}\right)dx$$
You seem to have evaluated
$$\int \left(\frac{x}{3}-2\sec^2\left(\frac x2\right)-e^{-2x+3}\right)dx$$
Your evaluation for the incorrect integrand is correct though, and to get the actual answer, you only need to change th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The triangle that has the longest possible smallest side of a triangle inscribed in a unit square is equilateral The points $A$, $B$ and $C$ lie on the sides of a square of side $1$ cm and no two points lie on the same side. Show that the length of at least one side of the triangle $ABC$ must be less than or equal to $... | Here is an entirely analytical attempt. Let $(x_1,0), (1,y),(x_2,1)$ be the vertices of the triangle inscribed in the unit square $[0,1]^2$. Since only one of them can be in a vertex of the square, we will allow only $x_1 \in [0,1\rangle$ and $x_2,y \in \langle 0,1\rangle$. Also we can assume $x_1 \le x_2$. The length ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Maximum value of $|z|$ given $\lvert z-\frac 4z \rvert = 8$? The question is $$ \left|z-\frac 4z \right| = 8$$ Find the max value of $ |z|$
You know how the triangle inequality is:
$$ \bigg| | z_1 | - | z_2| \bigg| \leqq | z_1 \pm z_2 | \leqq | z_1 | + | z_2 | $$
The solutions used only the left hand side inequality, a... | Redo:
We have three potential inequalities
$|z| - |\frac z4| \le ||z| - |\frac z4|| \le |z -\frac z 4|=8$ or
*
*$|z| -|\frac z4| \le 8$.
$-|z| + |frac z4| \le ||z| - |\frac z4|| \le |z -\frac z 4|=8$ or
*$|\frac z4| -|z| \le 8$.
$8=|z-\frac z4|\le |z| + |\frac z4|$ or
*$|z| +|\frac z4| \ge 8$.
All three of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Laurent Series of $\frac{1}{(z-1)(z-2)}$ for $|z-1|>1$ I am calculating Laurent series of
$$
f(z)=\frac{1}{(z-1)(z-2)}
$$
which converges when $|z-1|>1$.
I started as
$$
\frac{1}{(z-1)(z-2)}=\frac{1}{z-1}\cdot\frac{1}{(z-1)-1}=\frac{1}{(z-1)^2-(z-1)}.
$$
Is this even the right way to transform function to some kind of ... | $$f(z) = \frac{1}{(z-1)(z-2)}$$
We write $f(z)$ in its partial fraction expansion, and then we expand $f(z)$ in powers of $\frac{1}{z-1}$:
$$\begin{aligned}
f(z) &= \frac{1}{z-2}-\frac{1}{z-1} \\
&= \frac{1}{(z-1)-1} - \frac{1}{z-1} \\
&= \frac{1}{z-1} \frac{1}{1-\frac{1}{z-1}}- \frac{1}{z-1}\\
&= \frac{1}{z-1} \left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving $\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
Proving $\displaystyle\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
My atempt:
\begin{align*}
\int_0^1 \int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx \, dy &=\int_0^1I_x(y)\,dy\\[6pt]
\text{where }I_x(y)=\int_0^1\frac{-x\ln... | \begin{align}
\int _0^1\int _0^1-\frac{x\ln \left(xy\right)}{1-x^2y^2}\:dx\:dy
&=\int _0^1\int _0^1\left(\underbrace{-\frac{x\ln \left(x\right)}{1-x^2y^2}-\frac{x\ln \left(y\right)}{1-x^2y^2}}_{t=x^2}\right)\:dx\:dy
\\[3mm]
&=\int _0^1\int _0^1\left(-\frac{1}{4}\underbrace{\frac{\ln \left(t\right)}{1-ty^2}}_{K}-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find the determinant of this $6\times 6$ X-matrix? This question was asked in my quiz and i was unable to solve it, so I am asking it here.
Find the value of determinant of this particular matrix .
$$\begin{pmatrix}1&0&0&0&0&2\\0&1&0&0&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{pmatrix}$$
I ... | Permuting the rows and columns, we obtain a block diagonal matrix.
$$\det \begin{bmatrix} \color{red}{1} & 0 & 0 & 0 & 0 & \color{red}{2}\\ 0 & \color{orange}{1} & 0 & 0 & \color{orange}{2} & 0\\ 0 & 0 & \color{magenta}{1} & \color{magenta}{2} & 0 & 0\\ 0 & 0 & \color{magenta}{2} & \color{magenta}{1} & 0 & 0\\ 0 & \col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How can we take the derivative of this function: $y = \frac{x}{x^2+1}$ from first principles (using the limit definition of the derivative)? I was taking the derivative of the function: $y = \frac{x}{x^2+1}$. I know that we can solve it by the quotient rule. But I tried using the limit definition of differentiation. Th... | Factor out $h$ from the numerator and the denominator. You can then cancel the factored term:
$$\frac{1-x^2-xh}{x^4+x^2h^2+2x^3h+2x^2+2xh+h^2+1}$$
Then take the limit as $h$ tends to $0$ and you get:
$$\frac{1-x^2}{x^4+2x^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $\int_0^{\infty} \frac{\ln^2(x^3+1)}{x^3+1} dx = \frac{\sqrt{3} \pi}{18} \left(9\ln^2(3)+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\ldots$ Prove $$\int_0^{\infty} \frac{\ln^2{(x^3+1)}}{x^3+1} \; \mathrm{d}x = \frac{\sqrt{3} \pi}{18} \left(9\ln^2{(3)}+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\frac{\pi... | Consider the parameterized integral $I(a)$ where the integral in question is equal to $I''(1)$:
$$I(a)=\int_0^{\infty} \frac{1}{\left(x^3+1\right)^a} \; \mathrm{d}x$$
First, let $x^3 \to x$:
\begin{align}
I(a) &= \frac{1}{3} \int_0^{\infty} \frac{t^{-\frac{2}{3}}}{(t+1)^a} \notag \\
& \; \; = \frac{\Gamma\left(\frac{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find limit of $f(x)$ as $x$ tends to $0$ I need some help answering this question:
$$f(x) = \frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}$$
find the limit of $f(x)$ as $x$ tends to $0$ by writing $f(x)$ as a quotient of two powers series.
I have so far:
$$\frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} ... | We have that $$ \lim_{x \to 0} \frac{\sinh(x)}{\cosh(x)} - \frac{1}{x} = \lim_{x \to 0} \frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{x(x + \frac{x^3}{3!}+\cdots)}$$ $$= \lim_{x \to 0} \frac{(x+\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{(x^2 + \frac{x^4}{3!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
What is the range of $\vec{z}^{ \mathrm{ T } }A\vec{z} $? Let A be a 3 by 3 matrix
$$\begin{pmatrix}
1 & -2 & -1\\
-2 & 1 & 1 \\
-1 & 1 & 4
\end{pmatrix}$$
Then we have a real-number vector $\vec{ z }= \left(
\begin{array}{c}
z_1 \\
z_2 \\
z_3
\end{array}
\right)$ such that
$$\vec{z}^{ \mathrm{ T } }\v... | If we eliminate $z_3$ by replacing it with $1-z_1-z_2$, you want to find the minimum and maximum of
$$\{z^TAz + b^Tz + c : zQz+q^Tz = 0\}$$
with
$$A=\begin{pmatrix}7 & 2 \\ 2 & 3\end{pmatrix}, \; b=\begin{pmatrix}-10\\-6\end{pmatrix}, \; c=4, \; Q=\begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}, \; q=\begin{pmatrix}-2\\ -2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Plese help me to express this Jacobian determinant $\frac{ \partial (x,y) }{ \partial (u,v) }$ only using u and v We define u and v as
$$u=\frac{2x}{x+y+1}$$
$$v=\frac{2y}{x+y+1}$$
I'm trying to get Jacobian determinant $\frac{ \partial (x,y) }{ \partial (u,v) }$ and express it only using $u$ and $v$.
What I have tri... | Calculate $\frac{\partial(u,v)}{\partial(x,y)}$ first
$$\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{2}{x+y+1} - \frac{2x}{(x+y+1)^2} & \frac{-2x}{(x+y+1)^2} \\ \frac{-2y}{(x+y+1)^2} & \frac{2}{x+y+1}-\frac{2y}{(x+y+1)^2} \\ \end{vmatrix} = \frac{4}{(x+y+1)^3}$$
Then notice that
$$u+v = 2-\frac{2}{x+y+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $V_n(a)$ counts sign changes in the sequence $\cos a, \cos2a,\cos3a,\ldots,\cos na,$ show that $\lim_{n\to\infty}\frac{V_n(a)}n=\frac{a}\pi$
Let $0\leq\alpha\leq \pi $. $V_n (\alpha) $ denote the number of sign changes in the sequence $\cos\alpha,\cos2\alpha,\cos3\alpha,\ldots,\cos n\alpha $. Then prove that $$\lim... | HINT: let $ b_n\equiv n a \pmod {2\pi}$ indicate the angle formed with the $x$- axis in the $n^{th}$ term of the sequence. Assume that $b$ is uniformly distributed in the range between $0$ and $2\pi$.
Now firstly consider the case in which $0<b_n<\pi/2$ or $3\pi/2<b_n<2\pi$. In the next step, a change of sign will occu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
What I Tried :
I used some clever ways to get $x + y + z = 26xyz$ , but I suppose we have some solution as a number .
All all $3$ to get :-
$$2(x + y + z) = 5xy + 6yz + 7zx$$
Or,... | Assuming $x,y,z \neq 0$, we have that,
$$x + y = 5xy \iff \frac1x+\frac1y =5$$
$$y + z = 6yz \iff \frac1y+\frac1z =6$$
$$z + x = 7zx \iff \frac1z+\frac1x =7$$
then solve for $1/x$, $1/y$, $1/z$.
The case $x=0 \lor y=0 \lor z=0$ is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding $a$ such that the complex solutions of $z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0$ form a parallelogram in the complex plane Find all values of the real number $a$ so that the four complex roots of
$$z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0$$ form the vertices of a parallelogram in the complex pl... | Expanding on @AlexeyBurdin's answer (in case anyone wants to see how to do it by hand), since $x=3/2$ we have $y^2+z^2=27/2-11a$ and$$6a^2+9a-9=27/2-3(y^2+z^2)=33a-27\implies a\in\{1,\,3\}.$$Each such $a$ works viz.$$1=x^4-(y^2+z^2)x^2+y^2z^2=81/16-(27/2-11a)9/4+y^2z^2,$$which we can simplify to obtain $y^2z^2$. We the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $(a-b^2)b>0$, then $\sqrt[3]{a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}+\sqrt[3]{a-\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}$ is rational From Hardy´s "A course of pure mathematics" 10th edition, problem 31 miscellaneous problems of chapter I.
If $(a-b^2)b>0$, then
$$
\sqrt[3]{a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}... | No proof, but making the start. From the hint we have $$\left( \alpha + \beta\sqrt{\frac{a-b^3}{3b}} \right)^3 = a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}} \\ =\frac{\beta \left(\beta^2(a-b^3)+9\alpha^2b\right)}{3b} \sqrt{\frac{a-b^3}{3b}} + \frac{\alpha\left(\beta^2(a-b^3) + \alpha^2b\right)}{b}$$
so by comparison we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$
$a,b,c$ are three real numbers such that $a+b+c=5$ and $a^2+b^2+c^2=11$,
what's the maximum value of $abc$?
I thought of a way, $ab+bc+ca$ is not hard to find, $a,b,c$ satisfy the cubic equation $x^3 - 5 x^2 + 7 x - abc = 0$ , then use the discriminant of the ... | Plugging in $c=5-a-b$ into the quadratic and simplifying shows that
$$b^2+(a-5)b+(a^2-5a+7)=0.\tag{1}$$
Because $b+c=5-a$ we see that $b$ and $c$ are precisely the roots of this quadratic, so
$$abc=a(a^2-5a+7)=a^3-5a^2+7a.\tag{2}$$
Then the quadratic $(1)$ has two real roots so its discriminant is positive, i.e.
$$0\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve $x^5 - 1 = y^2$ for integer values of $(x,y)$ .
Solve $x^5 - 1 = y^2$ for integer values of $(x,y)$ .
What I Tried :- I see that $x^5 = y^2 + 1$ , from here I can conclude that $x$ has to be positive , because if $x \leq 0$ , then $x^5 \leq 0$ , but $y^2 + 1 > 0$.
Also I thought that maybe $(x^\frac{5}{2} + 1)... | So
$$x^5=y^2+1=(y+i)(y-i).$$
Considering this modulo $4$ gives $y$ even and $x$ odd.
The gcd of $y\pm i$ in the Gaussian integers divides $2$ and $y^2+1$ (which is odd)
so it is $1$. As the Gaussian integers has unique factorisation
and has four units,
both $y\pm i$ are fifth powers, so
$$y+i=(a+bi)^5=(a^5-10a^3b^2+5ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Proving continuous differentiability over the domain I have the following problem from Tao.
Let $f: \mathbb R^2 \to\mathbb R$ be the function defined by $f(x,y)=\dfrac {xy^3}{x^2+y^2}$ when $(x,y) \neq (0,0)$ and $f(0,0):=0$. Show that $f$ is continuously differerentiable.
I have $\dfrac{\partial{f(x,y)}}{\partial{x}... | Let $f(x,y)$ be given by
$$f(x,y)=\begin{cases}\frac{xy^3}{x^2+y^2}&,(x,y)\ne (0,0)\\\\0&,(x,y)=(0,0)\end{cases}$$
For $(x,y)\ne(0,0)$, we have
$$\begin{align}
\frac{\partial f(x,y)}{\partial x}&=\frac{y^3(y^2-x^2)}{(x^2+y^2)^2}\\\\
\frac{\partial f(x,y)}{\partial y}&=\frac{xy^2(3x^2+y^2)}{(x^2+y^2)^2}
\end{align}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Diophantine equation:$x^5+x^4+1=p^y$
find all triplets $(x,y,p)$ satisfying $x^5+x^4+1=p^y$ where x, y are positive integers and p is a prime.
My attempt:I didn't know how to start. So I tried finding some triplets. Interestingly, $(1,1,3)$ satisfies the given equation, but I am not able to find any more.
Next, I tri... | For anybody like me who is unable to view the solution at Diophantus Era begins! that Math Lover's question comment linked to because they're not a member of Brilliant and doesn't want to join, here is a solution.
First, apart from $x = 1$, which leads to the solution of $(1, 1, 3)$ you've already found, then both of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find $\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$? I tried to solve it, but the answer I got was different from the answer given.
Answer given:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)} = \frac{n(2n+1)}{4(2n-1)(2n+3)}$$
My working:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$= \sum_{r=1}^{n} \Biggl... | Substituting $n=1$, the "correct answer" gives a wrong result as mentioned in the comments.
Indeed $$\sum_{r=1}^{1}\frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{1}{1\times3\times5}=\frac{1}{15}$$ but $$\frac{n(2n+1)}{4(2n-1)(2n+3)}\big|_{n=1}=\frac{3}{4\times1\times5}=\frac{3}{20}\neq \frac{1}{15}.$$
Also you should have:
$$\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How does one go about solving $arg(\frac{z-2i}{z-6}) = \frac{1}{2}\pi$ This should give $$\frac{z-2i}{z-6} = bi$$
but solving that gives me $$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$
and substituting $z$ for $x + yi$ gives me $x = \frac{-2b +6b^2}{1+b^2}$ and $y=\frac{-6b +2}{1+b^2}$
And I have no clue how to continue now... | As you said solving gets you here:
$$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$
and finally, substituting gets u here:
$$x = \frac{-2b +6b^2}{1+b^2}$$ and $$y=\frac{-6b +2}{1+b^2}$$
Just divide above two equations you get
$$x = \frac{-2b + 6b^2}{1 + b^2} = -\frac{2 - 6b}{1 + b^2}\cdot b = -by$$
$$\implies x+by=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3826976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Strange series that converges I'm trying to evaluate this series
$$S= \sum_{n=2}^{\infty} a_n \frac{\ln (n)}{n}$$
But I have some conditions on $a_n$ making the problem hard. Namely, $a_n=3$ for $n = 2 \mod 4$ and $a_n = -1$ otherwise.
Albeit 2 mod 4 =2 is just a number, a friend in the comments suggested that the 2[mo... | As I pointed out in the comments, this is Question B-4 from the 2017 William Lowell Putnam competition. For convenience, I am providing the first of two solutions provided from the link in my comment. I take no credit--only minor changes to phrasing were made.
The key insight is to define an auxiliary telescoping ser... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Epsilon Delta Proof Verification for $14 + \frac{12}{x^2}$ Can someone check if my steps are completely justified:
Question: Prove $\lim\limits_{x \to \infty} 14 + \frac{12}{x^2} = 14$
For all $\epsilon > 0 \ \exists \ N>0$ such that $|14+\frac{12}{x^2} -14| < \epsilon$ for all $x > N$
Let $N = \frac{2 \sqrt{3}}{\sqrt... | Write your proof clearer:
Proof: Let $\varepsilon>0$ and define $N := \sqrt{\frac{12}{\varepsilon}}$. Then, for all $x > N$ we have $x^2 > \frac{12}\varepsilon$, and so
$$\bigg|\Big( 14 + \frac{12}{x^2} \Big) - 14 \bigg| = \frac{12}{x^2} < \varepsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a^2+b^2+c^2+d^2=4$ then $(a+2)(b+2)\geq cd$ Let $a,b,c,d$ be real numbers with $a^2+b^2+c^2+d^2=4$. Prove that $(a+2)(b+2)\geq cd$.
My approach: I have considered an expression $$\begin{aligned}(a+2)(b+2)-cd=&4+2(a+b)+(ab-cd)\\=&(a^2+b^2+c^2+d^2)+2a+2b+(ab-cd)\end{aligned}$$ I was trying to write it as the sum of s... | Using the AM-GM inequality, we have
$$(a+2)(b+2) = \frac{4-a^2-b^2-c^2-d^2}{2}+\frac{(a+b+2)^2}{2}+\frac{c^2+d^2}{2}$$
$$\geqslant \frac{c^2+d^2}{2} \geqslant cd.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I prove that $y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}>0$ when $x>0$ and $1I would like to prove that
$$y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}>0$$
for all real numbers $x > 0$ and $1 < y < 1.5$. This seems true when plotted on WolframAlpha, but I don't know how to pro... | Let $f(y)=y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}.$
Thus, $$f''(y)=-\frac{6xy^2}{(1+x^2)^2}+2x^3\left(\frac{y}{(1+y^2)^2}\right)'=$$
$$=-\frac{6xy^2}{(1+x^2)^2}+2x^3\left(\frac{1}{(1+y^2)^2}-\frac{4y^2}{(1+y^2)^3}\right)=$$
$$=-\frac{6xy^2}{(1+x^2)^2}+\frac{2x^3(1-3y^2)}{(1+y^2)^3}<0,$$
which says... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is the solution to $\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$ $$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$$
I've already tried this many times but still not get the answer, this was one of my solution which I can't go further anymore.
$\frac{{x}^{2}}{(... | The domain gives $x>-1$ and we need to solve $$\frac{x^2}{(\sqrt{x+1}-1)^2}<\frac{x^2+3x+18}{x+1}$$ or
$$x^2(x+1)<(x+2-2\sqrt{x+1})(x^2+3x+18)$$ or
$$2(x+3)^2>(x^2+3x+18)\sqrt{x+1}$$
or $$4(x+3)^4>(x^2+3x+18)^2(x+1)$$ or after factoring
$$x^2(3-x)(x^2+6x+21)>0,$$ which gives $$(-1,3)\setminus\{0\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Can a triangle ABC be made if $\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$
I would like to know the simplest approach to find out whether a triangle ABC will be made if $$\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$$
The counterpart questions for sine and tangent can be handled as follows:
*
*If $\df... | Strating from @Teresa Lisbon's answer, the exact results are
$$k=\frac{31}{126} \left(2 \cos \left(\frac{1}{3} \left(2 \pi n-\cos
^{-1}\left(-\frac{17884}{29791}\right)\right)\right)-1\right)\qquad (n=0,1,2)$$ and this gives angles (in degrees) $a=76.358$, $b=69.281$, $c=34.361$.
Using algebra, the problem is very ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Understanding convergence of $a_n := \frac{2n^3 +n^2 +3}{n^3-4}$ We proved by definition that the sequence $a_n := \frac{2n^3 +n^2 +3}{n^3-4}$ converges.
Let $\epsilon > 0$. Choose $N := \lceil \frac{24}{\epsilon} \rceil + 2 $.
Then for all $n \geq N$ it holds, that
$$|a_n - 2| = \big |\frac{2n^3+n^2+3}{n^3-4} - \frac{... | We have $$-\frac{2n^{3}-8}{n^3-4}=-\frac{2(n^3-4)}{n^3-4}=-2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
question from Euclid 2011 about proving that $\frac{(a^3+b^3)c}{a^2+b^2}+\frac{(b^3+c^3)a}{b^2+c^2}+\frac{(c^3+a^3)}{c^2+a^2}\ge 1$ I just did the following question:
If $a, b, c$ positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}$ prove that $\frac{(a^3+b^3)c}{a^2+b^2}+\frac{(b^3+c^3)a}... | SOS helps:
$$\sum_{cyc}\frac{(a^3+b^3)c}{a^2+b^2}-1=\sum_{cyc}\frac{(a^3+b^3)c}{a^2+b^2}-ab-ac-bc=$$
$$=\sum_{cyc}\left(\frac{(a^3+b^3)c}{a^2+b^2}-\frac{c(a+b)}{2}\right)=$$
$$=\sum_{cyc}\frac{c(a+b)}{2}\left(\frac{2(a^2-ab+b^2)}{a^2+b^2}-1\right)=\sum_{cyc}\frac{c(a+b)(a-b)^2}{2(a^2+b^2)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Directional derivatives at the origin and conditions for differentiability
Let $f:\mathbb{R^2} \to \mathbb{R}$ $$f(x) = \left\{
\begin{array}{ll}
\frac{xy^2}{x^2+y^4}, & (x,y)\ne0 \\
0, & (x,y) =0 \\
\end{array}
\right.$$ Find the directional derivatives at the origin $D_af(0,0)$ for every direction $a=(a... | We are requested to find the directional derivatives at the origin that is for $a\cdot b \neq 0$
$$\lim_{(ah,bh)\to(0,0)} \frac{\frac{ab^2h^3}{a^2h^2+b^4h^4}-0}{h} =\lim_{(ah,bh)\to(0,0)} \frac{ab^2h^3}{a^2h^3+b^4h^5}=\frac{b^2}a$$
with $f_x=f_y=0$.
For differentiability just note that for $x=y^2$
$$\lim_{(x,y)\to(0,0)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Induction proof of a known harmonic sum I want to prove that $$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} \leq 1$$ only by induction!
I check for the first one, $\frac12 \leq 1 $ correct.
Then I assume for $n=k$ : $$\frac12 + \frac14 + \dots + \frac{1}{2^k} \leq 1$$
And Try and prove for $n=k+1$
$$\frac12 + \fr... | $\frac12 + \frac14 + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \leq 1 \iff \frac14 + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \leq \frac12 \iff \frac12 + \dots + \frac{1}{2^{k-1}} + \frac{1}{2^{k}} \leq 1$.
The last step is multiplying by $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Use mathematical induction to prove that $\sum_{j=1}^n \frac{1}{(j^2-1)}$ = $\frac{3}{4}$ - $\frac{2n+1}{2n(n+1}$ for all n $\geq$ 2 so I know how to do induction and all I'm just struggling with the algebra for proving the case for n = k+1. My proof is below:
For the case n=2 we have
$\sum_{j=1}^2 \frac{1}{j^2-1}$ = $... | \begin{align}
\frac34 - \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} &= \frac34 - \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} \\
\end{align}
Your goal is to simplify $- \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} $ and show that it is equal to $-\frac{2(k+1)+1}{2(k+1)(k+2)}$
\begin{align}
- \frac{2k+1}{2k(k+1)} + \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the value of the following limit So I have given the limit:
$$\lim_{x\to0} \frac{2\sin\left ( e^{-\frac{x^2}{2}} -\cos x \right)}{(\arctan(\sinh(x^2))^2}$$
I have been struggling for hours with it. Since i got the undefined form when i put $x=0$ i tried out with L'Hopital method and I come to this point:
$$\l... | If you are aware of Taylor expansions then:
$$
e^{\frac{-x^2}{2}} =_{x \rightarrow 0} 1-\frac{x^2}{2} + \frac{x^4}{8} + o(x^4)
$$
$$
\cos(x) =_{x \rightarrow 0} 1 - \frac{x^2}{2} + \frac{x^4}{4!} + o(x^4)
$$
Then
$$
e^{\frac{-x^2}{2}} - \cos(x) =_{x \rightarrow 0} \frac{x^4}{12} + o(x^4)
$$
Now because $sin(x) =_{x \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the height of a trapezoid
A trapezoid $ABCD$ is given $(AB$ $||$ $CD)$ with side lengths $AB=18,BC=\sqrt{74},CD=5$ and $AD=\sqrt{61}.$ Find the sines of $\measuredangle A$ and $\measuredangle B.$
(I have only studied trig functions of acute angles)
Since we are given the four sides of the trapezoid, it is enoug... | First Solution
$PC=AD=\sqrt{61}$, $BC=\sqrt{74}$, $BP=13$
$cos \angle{B}=\frac{74+169-61}{26\sqrt{74}}=\frac{7}{\sqrt{74}}$
$sin \angle{B}=\frac{5}{\sqrt{74}}$
$cos \angle{A}= cos \angle{P}=\frac{61+169-74}{26\sqrt{61}}=\frac{6}{\sqrt{61}}$
$sin\angle{A}=\frac{5}{\sqrt{61}}$
Second Solution
Let $AD_1=x, BC_1=13-x, CC_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the area bounded by the curve $x^4+y^4=x^2+y^2$ I am stuck with this problem which deals with evaluating an Area
The problem reads :
Find the area bounded by the curve $x^4+y^4=x^2+y^2$.
I tried factorizing the expression and expressing $y$ in terms of $x$, not able to proceed with that idea. Someone please help... | Use polar co-ordinates $x=r \cos t, y=r \sin t$, then this curve is $$r^2=\frac{1}{\sin^4 t+ \cos^4 t}=\ $$, the area the area is
$$A=4 ~ \frac{1}{2}\int_{0}^{\pi/2} r^2 dt=2 \int_{0}^{\pi/2} \frac{1}{\cos^4 t+\sin^4 t}dt =2 \int_{0}^{\pi/2}\frac{\sec^4 t}{1+\tan^4 t} dt $$
Let $z=\tan t \implies dz=\sec^2 t dt$
$$=2\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A probability question about choosing 6 shoes are chosen randomly from the 20 distinct pairs of shoes drawer. $\mathbf{Question:}$
A drawer has 20 distinct pairs of shoes. 6 shoes are chosen randomly from the drawer. The drawer contain only one my favorite pair.
(a) what is the probability that my favorite pair is chos... | Not all of your answers are correct.
(a) If you choose both shoes from your favorite pair, you must also choose four shoes from the remaining $40 - 2 = 38$ shoes. Thus, the number of favorable cases is
$$\binom{2}{2}\binom{38}{4}$$
You failed to account for the fact that you are choosing six shoes, not two. Selecting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}$ for $a,b,c\in\mathbb{R}^+$ with $abc=1$ Suppose that $a,b,c$ are positive reals such that $abc=1$. Prove that $$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}.$$
Hint: Use Titu's lemma.
My approach: I am trying to use Titu's lemma direct... | Notice that $$\sum_{cyc}\frac{a^3}{b+c}=\sum_{cyc}\frac{a^4}{a(b+c)}\geqslant \frac{(a^2+b^2+c^2)^2}{2(ab+bc+ca)}\geqslant \frac{a^2+b^2+c^2}{2}\geqslant \frac{3\sqrt[3]{abc}}{2}=\frac32 $$ Where we first used Titu's Lemma, then the well-known $a^2+b^2+c^2\geqslant ab+bc+ca$, and finally AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
What does the geometric series $ \sum\limits_{n=0}^{\infty} n \cdot \left( \frac{1}{4} \right) ^{n}$ converge to? I am trying to find the value to which the geometric series
$$
\sum\limits_{n=0}^{\infty} n \cdot \left( \frac{1}{4} \right) ^{n},
$$
converges. Now I know how to find a sum when we have a common ratio, but... | Hint
$$S_n=1\cdot\left( \frac{1}{4} \right) ^{1}+2\cdot\left( \frac{1}{4} \right) ^{2}+...+n\cdot\left( \frac{1}{4} \right) ^{n}$$
Now, multiply by $1/4$
$$\frac{1}{4}S_n=1\cdot\left( \frac{1}{4} \right) ^{2}+2\cdot\left( \frac{1}{4} \right) ^{3}+...+n\cdot\left( \frac{1}{4} \right) ^{n+1}$$
Now subtract,
$$S_n-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find $E[B 1_{\{ B\ge \frac{n}{2} \} }]$ where $B$ is Binomial $(n,\frac{1}{2})$. How to find the following expectation
$$E[B 1_{\{ B\ge \frac{n}{2} \} }]$$
where $B$ is Binomial random variable with $(n,\frac{1}{2})$.
Here is what I did. The pmf in this case is given by
\begin{align}
E[B 1_{\{ B\ge \frac{n}{2} \} }]=... | Note that we have
$$
E[B 1_{\{ B\ge \frac{n}{2} \} }]=\frac n2 - E[B1_{\{B< \frac n2\}}].
$$
Then we evaluate $E[B1_{\{B< \frac n2\}}]$. We have
$$
E[B1_{\{B< \frac n2\}}]=\sum_{k<\frac n2} k\binom nk \frac 1{2^n}=\sum_{k<\frac n2} n \binom{n-1}{k-1} \frac1{2^n}.
$$
The second equality is due to $k\binom nk = n \bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there an identity to combine a sum of more than two sines; eg $\sin(a)+\sin(b)+\sin(c)+\sin(d)$? I get these trigonometric product to sum formulas like:
$$\sin(a)+\sin(b)=2\sin\frac12(a+b)\cos\frac12(a-b)$$
And that's useful, but I'm not too sure what to do if I need to turn a product into a sum if there's more than... | The following formula works:
$$
\begin{align}
\sin(a)+sin(b)+sin(c)+sin(d) = 4*\sin\left(\frac{a+b+c+d}{4}\right)*\cos\left(\frac{a-b+c-d}{4}\right) \\
*\cos\left(\frac{a+b-c-d}{4}\right)*\cos\left(\frac{a-b-c+d}{4}\right)- \\
4*\cos\left(\frac{a+b+c+d}{4}\right)*\sin\left(\frac{a-b+c-d}{4}\right) \\
*\sin\left(\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3848224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $ \prod_{n=1}^{\infty}\Bigg\{ \left(1+\frac{1}{n} \right)^x\left(1-\frac{x}{n}\right)\Bigg\}$ I have to discuss the convergence of the product
$ \prod_{n=1}^{\infty}\Bigg\{ \left(1+\frac{1}{n} \right)^x\left(1-\frac{x}{n}\right)\Bigg\}$
Here is my solution:
Based on the binomial formula, we have
$ \left... | Yes, this is mostly right, except that it should be $\int \limits_1^\infty \frac{1}{n^2}\, \mathrm{d}n$. And I'd use some simpler test for that, $1/n^2\le1/n(n-1)$, or condensation test, but that's a matter of taste, naturally.
Looking at https://en.wikipedia.org/wiki/Gamma_function#Euler's_definition_as_an_infinite_pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3848469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$? From the Pre-Regional Mathematics Olympiad, 2019:
If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, ... | Using a bit of Galois theory: first, note that the intermediate fields between $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ are $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, and $\mathbb{Q}(\sqrt{6})$, none of which contains $\sqrt{2} + \sqrt{3} + \sqrt{6}$. Therefore, $\sqrt{2} + \sqrt{3} + \sqrt{6}$ is a prim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3851795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is $\tan \alpha$, if $(a+2)\sin\alpha +(2a - 1)\cos\alpha =2a + 1$? I tried the following:
$$\begin{aligned}a\sin\alpha +2\sin\alpha + 2a\cos\alpha - \cos\alpha &= 2a+1\\
a(\sin\alpha +2\cos\alpha)+(2\sin\alpha-\cos\alpha)&=2a+1\end{aligned}$$
Therefore,
$$\sin\alpha +2 \cos\alpha=2$$
$$2\sin\alpha - \cos\alpha=1$... | The problem is that $a$ is a constant, not a variable. Therefore equating by parts does not work here: if we separate $2a - 1$ into $2(a-1) + 1$, $2(a-2) + 3$ and so on, we end up with a different set of solutions each time. If you try substituting $\sin \alpha$ and $\cos \alpha$ back into the original question, the LH... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to prove this inequality: $\sum_{cyc} \frac {1}{\alpha + \log_a {b}} \le \frac {2}{\alpha}$? This is another Olympiad problem my teacher gave us to train.
$$\sum_{cyc} \frac {1}{\alpha + \log_a{b}} \le \frac {2}{\alpha}$$ with $\alpha \in (0, 2]$ and $a, b, c \in (0, 1)$ or $ a, b, c \in(1, \infty)$. After working ... | Starting from @Robert Z's answer, consider that we look for the maximum value of function
$$f(x,y,z)=\frac {1}{\alpha + x}+\frac {1}{\alpha + y}+\frac {1}{\alpha + z}$$ Using the constraint $xyz=1$, we look at the maximum of function
$$g(x,y)=\frac{2 \alpha +x^2 y (2 \alpha +y)+x \left(2 \alpha y^2+3 \alpha ^2
y+1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3855023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why does this equality Dirichlet series hold? Following on from my question here, I have hit a second roadblock.
I am working (very slowly!) through a paper here that demonstrates Riemann's analytic continuation of the zeta function $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$ to the complex plane (except for the pole at... | For $\Re s >2$ so everything is absolutely convergent and then extending by analytic continuation wherever RHS converges:
$\frac{1}{s-1} \sum_{n=1}^\infty (\frac{n}{(n+1)^s}-\frac{n-s}{n^s})=\frac{1}{s-1} \sum_{n=1}^\infty (\frac{n+1}{(n+1)^s}-\frac{n}{n^s}-\frac{1}{(n+1)^s}+\frac{s}{n^s})=$
(by telescoping the first t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$. My attempt.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$
$$=\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k}\binom{2k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}$$
The fi... | It seemed something was missing, so with the right tools the proof isn't difficult.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$
Consider:
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k}H_k\binom{2k}{k}=\frac{2}{\sqrt{1-x}}\ln \left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)$$
$$\sum _{k=1}^{\infty }\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Prove $\forall z\in\mathbb C-\{-1\},\ \left|(z-1)/(z+1)\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$ I'm trying to prove
$$\forall z\in\mathbb C-\{-1\},\ \left|\frac{z-1}{z+1}\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$$
thus showing that the solutions to $\left|(z-1)/(z+1)\right|=\sqrt2$ form the circle of center $-3$ and ... | For $r>0, a,b \in \mathbb{C}, a\neq b\;$ the equation $\left|\frac{z-a}{z-b}\right|=r$ defines a hyperbolic pencil of Apollonian circles. Their centers lie on the line $AB,$ where $A(a), B(b).$
To find the Apollonian circle (its center and radius) in the particular case
$$\left|\frac{z-1}{z+1}\right|=\sqrt2,$$
it suff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int \frac{1-x^2}{(1+x^2)\sqrt {1+x^4}} dx$ $$\int \frac{\frac{1}{x^2}-1}{(x+\frac 1x)\sqrt{\frac{1}{x^2} + x^2}}dx$$
Let $x+\frac 1x = t$
$$-\int \frac{dt}{t\sqrt {t^2-2}}$$
Let $\sqrt{t^2-2} =u$
$$-\int \frac{du}{t^2}$$
$$-\int \frac{du}{u^2+2}$$
$$-\frac{1}{\sqrt 2} \arctan (\frac{u}{\sqrt 2})$$
$$-\frac{1... | You need to take care of the sign in your approach. The result you derived is valid only for $x>0$. The full result valid for all domain is, instead
$$\int \frac{1-x^2}{(1+x^2)\sqrt {1+x^4}} dx
=-\frac{\text{sgn}(x)}{\sqrt 2} \arctan \frac{\sqrt { x^2+\frac{1}{x^2}}}{\sqrt 2}
=-\frac{1}{\sqrt 2} \arctan \frac{ \sqrt {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3867040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
From 7 men & 4 women, 4 are to be selected to form a committee so that at least a woman is there on the committee. In how many ways can it be done?
From $7$ men & $4$ women, $4$ people are to be selected to form a committee so that at least a woman is there on the committee. In how many ways can it be done?
I was try... | There are 7 men and 4 women, and we need 4 members on the committee such that at least there is one woman on it. In that regard, we can make the following combinations.
We can have 4 women and 0 men. Or we can have 3 women and 1 man. Or we have 2 women and 2 men. Or we can have 1 woman and 3 men. We have to stop at thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3867924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve $\cos\theta-3\cos2\theta+\cos3\theta=\sin\theta-3\sin2\theta+\sin3\theta$ My attempt:
\begin{align*}
\cos\theta-3\cos2\theta+\cos3\theta&=\sin\theta-3\sin2\theta+\sin3\theta\\
\cos\theta-3\cos2\theta+4\cos^3\theta-3\cos\theta&=\sin\theta-3\sin2\theta+3\sin\theta-4\sin^3\theta\\
-2\cos\theta-3\cos2\theta+4\cos^3\t... | It is a false identity. Click here to see the graphs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2}$ Please help me find:
$$
\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2}
$$
I cannot use L'Hospital's rule.
I tried to eliminate $x-3$, but I have no idea what to do next.
$$
\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cd... | Let $x=3+u$ with $u\to 0$ then use binomial formula to expand $(3+u)^{15}$.
You can ignore terms in $u^k$ with $k\ge 3$ because when divided by $u^2$ they converge to $0$.
$\require{cancel}\begin{align}f(x)&=\dfrac{(3+u)^{15}-3^{15}-15\cdot3^{14}u}{u^2}\\\\&=\dfrac{\bigg(\cancel{3^{15}}+\cancel{15\cdot3^{14}u}+\binom{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Application of the Cauchy-Schwarz Inequality Need to prove the following:
$(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c)^2\leq \frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}c^2$
using the Cauchy-Schwarz inequality however direct application yields $(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c)^2\leq \frac{7}{18}(a^2+b^2+c^2)$ which i... | Because by C-S:
$$\frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}c^2=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)\left(\frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}{c^2}\right)\geq\left(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given $\cos x + 3\sin x = \sqrt{10} \cos(x-71.6)$, find the second solution in the interval $0 < \theta < 90$ $$\cos x + 3\sin x = \sqrt{10} \cos(x-71.6)$$
I've proven the above true. I must now solve this:
$$\cos 2\theta + 3\sin 2\theta = 2, 0<\theta<90$$
Here's what I've done so far:
$$\alpha = 2\theta - 71.6$$
$$\co... | $\cos 2x + 3\sin 2x = \sqrt 10\cos (2x - \arctan 3)=2\\
\cos (2x - \arctan 3) = \frac {2}{\sqrt 10}\\
2x - \arctan 3 = \arccos\frac {2}{\sqrt 10}\\
x = \frac 12 (\arccos \frac {2}{\sqrt {10}} + \arctan 3)$
Regarding the second solution.
Since $\cos x = \cos -x$
Then $\alpha = -\arccos \frac {2}{\sqrt 10}$ will point to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $. If $\overrightarrow a = \hat j + \sqrt 3 \hat k;\overrightarrow b = - \hat j + \sqrt 3 \hat k;\overrightarrow c = 2\sqrt 3 \hat k$ form a triangle then find the internal angle of the triangle between $\overrightarrow a \& \o... | The best way to find the angle is tp find the lengths of the side vectors here they are $a=2,b=2, c=\sqrt{12}$, so the required angle is $C$, then by cosine law
$$\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{2+2-12}{8}=-\frac{1}{2} \implies C=\frac{2\pi}{3}.$$
If you want to it by dot product of $\vec a$ and $\vec b$, the angl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Understanding the solution to $4^{3x+6}+3^{2x+3}=4^{3x+4}+2\times3^{2x+4}$ For the last 5 hours I tried to solve the equation below (without luck).
$$4^{3x+6}+3^{2x+3}=4^{3x+4}+2\times3^{2x+4}$$
A friend of mine then told me that the solution is
$$\frac{\ln(16/3)}{\ln(3/8)}$$
Can anyone help me understanding why this i... | $$4^{3x+6}+3^{2x+3}=4^{3x+4}+2 \times 3^{2x+4}$$
$$\Longleftrightarrow 16(4^{3x+4})+3^{2x+3}=4^{3x+4}+6(3^{2x+3}) $$
$$\Longleftrightarrow 15(4^{3x+4})=5(3^{2x+3}) $$
$$\Longleftrightarrow 3(4^{3x+4})=3^{2x+3} $$
$$\Longleftrightarrow \ln(3)+(3x+4)\ln(4)=(2x+3)\ln(3)$$
$$\Longleftrightarrow x(3\ln(4)-2\ln(3))=2\ln(3)-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Conditional Expectation of Square of a Gaussian Consider two Gaussian random variables $X\sim N(0,\sigma_X^2)$ and $Y\sim N(0,\sigma_Y^2)$ with known $E[XY] = \sigma_{XY}$. The question is how to find $E[X^2\mid Y]$ assuming $X$ and $Y$ are jointly Gaussian.
My approach: I thought of the following steps:
*
*Find the ... | Your approach is right if the distribution of $X$ and $Y$ is jointly Gaussian, rather than only each by itself being Gaussian. For example, suppose one had $Y = \begin{cases} +X & \text{if } |X|>c, \\ -X & \text{if } |X|<c. \end{cases}\quad$ Then it can be shown that $X$ and $Y$ are both Gaussian if $X$ is Gaussian, bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.