Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers
Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers
Clearly $k$ is a rational number, so I let $k=\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \neq 0$, this gives ... | Starting from $\frac{6}{k-1}=n \in \mathbb N$, we can generate all possible rational $k=\frac{n+6}{n}$ without regard to the second condition.
Considering the second condition, $\frac{12}{k+1}=m \in \mathbb N$, we substitute to get $m=\frac{12}{\frac{n+6}{n}+1}=\frac{12n}{2n+6}=\frac{6n}{n+3}$. We need to find positive... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My sol... | Consider the partial term
$$T_n=\left(1+\frac{H_n}{n^2}\right)^n\implies \log(T_n)=n \log\left(1+\frac{H_n}{n^2}\right)$$
For large $n$
$$H_n=\log (n)+\gamma +\frac{1}{2 n}-\frac{1}{12 n^2}+\frac{1}{120
n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log\left(1+\frac{H_n}{n^2}\right)=\frac{\log (n)+\gamma }{n^2}+\frac{1}{2 n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
Prove that the sum $2^0+2+2^2+\cdots+2^{p-1}(2^p-1)$ is $2^{p-1}(2^p-1)$ Attempt: In an attempt to find the sum of the series given that $2^{p-1}$ is prime,
$$1+2+2^2+\cdots+2^{p-2}+2^{p-1}+(2^p-1)+2(2^p-1)+2^2(2^p-1)+\cdots+2^{p-1}(2^p-1)$$
A teacher grouped elements as follows,
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cd... | Your grouping method is correct. Your teacher seems to have made a mistake.
From your sum you get
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$
which is equal to
$$(2^p)+2(2^p)+2^2(2^p)+\cdots+2^{p-2}(2^p)+2^{p-1}(2^p)=$$
$$2^p(2+2^2+...+2^{p-1})$$
Now, you only have to solve
$$2+2^2+..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\lim_{x\to-1}\frac{2x^2+5x+3}{2-\sqrt{2+\sqrt{3-x}}}$
What is the value of the limit:
$$\lim_{x\to-1}\frac{2x^2+5x+3}{2-\sqrt{2+\sqrt{3-x}}}$$
$1)8\quad\quad\quad\quad\quad2)12\quad\quad\quad\quad\quad3)16\quad\quad\quad\quad\quad4)24$
I evaluated the limit by L'Hopital rule:
$$\lim_{x\to-1}\dfrac{4x+5}{\... | HINT
What about multiplying by the conjugate twice?
\begin{align*}
\frac{1}{2 - \sqrt{2+\sqrt{3-x}}} = \frac{2 + \sqrt{2+\sqrt{3-x}}}{2 - \sqrt{3-x}}
\end{align*}
Similarly, we have
\begin{align*}
\frac{1}{2-\sqrt{3-x}} = \frac{2 + \sqrt{3-x}}{1 + x}
\end{align*}
Finally, notice that
\begin{align*}
2x^{2} + 5x + 3 & =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given the number $N=3^{12}-1$. Find how many even and odd divisors of $N$ exist, which divide $N$ but don't divide $3^k-1$, for $1\le k\le 11$ Given the number $N=3^{12}-1$. Find how many even and odd divisors of $N$ exist, which divide $N$ but don't divide $3^k-1$, for $1\le k\le 11$
Initially I tried working out how ... | Let $d$ be a divisor of $3^{12}-1$. Note $d\neq3$. Call $d$ "bad" if $d$ divides some $3^k-1$ for $k\in\{1,2,\ldots,11\}$.
If $d$ divides $3^k-1$, then mod $d$: $3^k\equiv1\implies3^{12}\equiv3^{12-k}\implies1\equiv3^{12-k}$. So if $d$ is bad, then $d$ divides some $3^k-1$ for $k\in\{1,2,\ldots,6\}$.
If $d$ divides $3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4075126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int x^4\sqrt{x^2-3} \, dx$ Integrate $\int x^4\sqrt{x^2-3} \, dx$
I tried substituting $\sqrt{x^2-3}=t$ than by squaring both sides and by simplifying i got $x\,dx=t\,dt$ and $x^2=t^2+3$
Now after substituting to integral i have $\int (t^2+3)^2\cdot t\frac{t\,dt}{x}$, can't get rid of $x$ :(
Don't know how... | Utilize the recursive relation
$$\int x^n\sqrt{x^2-3}\ dx=I_n=\frac{x^{n-1}}{n+2}(x^2-3)^{3/2}+\frac{3(n-1)}{n+2}I_{n-2}
$$
to evaluate the integral systematically
\begin{align}
\int x^4\sqrt{x^2-3}\ dx=
\left(\frac16 x^3 +\frac38 x \right)(x^2-3)^{3/2}+\frac98I_0
\end{align}
where
$$I_0= \int \sqrt{x^2-3}\ dx=\frac x2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find all natural solutions that satisfy $2^ + 3^ = ^2$ It looks like an easy question but I couldn't find a way to solve it. I found (0,1,2),(3,0,3),(4,2,5) by trial and error and I'm kinda sure they are the only answers but I'm not sure how to prove it.
| First consider the cases $x=0$ and $y=0$. If $x=0$, then $3^y=z^2-1=(z-1)(z+1)$ and hence $z-1$ and $z+1$ are two powers of $3$ with distance $2$; that immediately implies $z+1=3, z-1=1$ and so $z=2, y=1$. If $y=0$, then $2^x=z^2-1=(z-1)(z+1)$ and hence $z-1$ and $z+1$ are two powers of $2$ with distance $2$; that imme... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Problem understanding integration by parts from the book An example from my book (from unrelated subject) has the following steps:
$$ -2 \int \sqrt{9-y^2} dy = - y\sqrt{9-y^2} + 9\sin^{-1}\frac{y}{3} \ + C$$
I am trying to understand the steps in between from my books example when they use the integral forumla (secon... | As mentioned by @Deepak you can use substitution and find the answer that way.
Next it should be
$$-2\int \sqrt{9-y^2}dy=-y\sqrt{9-y^2}\color{red}{-}9\sin^{-1}\big(\frac{y}{3}\big)+C$$
Here is the integration by parts method
First note that $\frac{d}{dy}\big(\sqrt{9-y^2}\big)=\color{red}{-}\frac{y}{\sqrt{9-y^2}},$ then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Different values of $A^n$ using Cayley-Hamilton Theorem And Direct Multiplication Let's say there's a matrix A
$$
A = \begin{pmatrix}
3 & -4\\
1 & -1\\
\end{pmatrix}
$$
Now I want to find $A^n$
I tried the following two methods but get different answers. Any solution ?
By Direct Multiplication:
$$
A^2= \beg... | $(A-I)^{2}=0$ does not imply that $A=I$.
[For example, $M=\left[\begin{array}{llll}0 & 1 \\ 0 & 0 \end{array}\right]$ is a matrix whose square is $0$. Bur $M$ itself is not $0$].
C-H Theorem gives $A^{2}=2A-I$. A simple induction argument gives $A^{n}=nA-(n-1)I$ for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\{a_n\}$ converges.
Suppose $a_1,a_2>0$ and
$a_{n+2}=2+\dfrac{1}{a_{n+1}^2}+\dfrac{1}{a_n^2}(n\ge 1)$. Prove
$\{a_n\}$ converges.
First, we may show $\{a_n\}$ is bounded for $n\ge 3$, since $$2 \le a_{n+2}\le 2+\frac{1}{2^2}+\frac{1}{2^2}=\frac{5}{2},~~~~~~ \forall n \ge 1.$$
But how to go on?
| Note that $2<a_n<3$ for $n\ge5$. Since, for $n\ge5$,
\begin{eqnarray}
&&|a_{n+3}-a_{n+2}|\\
&=&\bigg|\frac{1}{a_{n+2}^2}-\frac{1}{a_{n}^2}\bigg|=\frac{(a_{n+2}+a_{n})|a_{n+2}-a_{n}|}{a_{n+2}^2a_{n}^2}\\
&\le&\frac{6}{2^2\cdot 2^2}|a_{n+2}-a_{n}|\le\frac12|a_{n+2}-a_{n}|\\
&\le&\cdots\le\frac{1}{2^{n-2}}|a_5-a_3|
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4094333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
$ax^2 + 2hxy + by^2 +2gx + 2fy+c = 0$ represents a point when $\Delta = 0$ and $h^2 - ab < 0$. How can I show that $ax^2 + 2hxy + by^2 +2gx + 2fy+c = 0$ represents a point when $\Delta = 0$ and $h^2 - ab < 0$.
I know I can do it by completing the square. But it turns out to be too tedious. Can anyone help me ?
| We can write $$y=\frac{hx+f\pm \sqrt{(h^2-ab)x^2+2(hf-bg)x+(f^2-ac)}}{b}~~~~~(1)$$
For a pair of lines, $y$ needs to be real for all real values of $x$. inside square root we should have a perfect square of like $(px+q)^2>0$. For a point it should be like $-(rx+s)^2$ ($y$ non real excepting one value of $x=-s/r$. For ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding variance from a beta distribution Suppose $[X\mid \theta]\sim\operatorname{Uniform}([0, \theta])$ and $B$ is is a beta distributed random variable. Additionally, $\max(X_1,…,X_n)/θ\sim \operatorname{Beta}(n+1,1)$
Show that $\hat{θ}= \frac{n+2}{n+1} \max(X_1,…,X_n)$ and find the variance of $\hat\theta$ given $... | I will assume you meant $B\sim\operatorname{Beta}(n+1,1).$
Recall that $\operatorname{var}(B) = \operatorname E\big(B^2\big) - \big( \operatorname E(B) \big)^2$.
\begin{align}
& \operatorname E(B^2) = \int_0^1 x^2\cdot f_B(x)\,dx \\[8pt]
= {} & \int_0^1 x^2\cdot \frac{\Gamma(n+2)}{\Gamma(n+1)\Gamma(1)} \cdot x^n\,dx \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$ The problem
Given that $a,b>0$ and $$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$$
Find the value of $$\log _{a b}\left(\frac{1}{a}+\frac{1}{b}\right)$$
My attempt
We have from the given condition
$$2+\frac{\log a}{\log 2}=3+\frac{\log b}{\log 3}=\frac{\log (a+b)}{\log 6}... | $$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k$$
$$2+\log _{2} a=k \implies a=2^{k-2}$$
$$3+\log _{3} b=k \implies b=3^{k-3}$$
So, now, we have
$$\frac{\log \left(2^{k-2}+3^{k-3}\right)}{\log (6)}=k\tag 1$$ and we need to compute
$$\log _{a b}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{\log \left(2^{2-k}+3^{3-k}\right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4103740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the intersection of $x= y^2 + y - 2$ and $y = -x^2 - \frac32x + 1 $ Goodmorning, I am struggling in finding the points of intersection of the following parabolas:
$$\begin{align}
x &= y^2 + y - 2 \\[4pt]
y &= -x^2 - \frac32x + 1
\end{align}$$
I know that these two can be solved either algebraically or with matr... | $$x = y^2 + y - 2 \tag 1$$
$$y = -x^2 - \frac32x + 1 \tag 2$$
Inspection :
$x=0$ in Eq.(2) gives $y=1$ and $y=1$ in Eq.(1) gives $x=0$ . Thus a first solution is
$$(x=0\:;\:y=1)$$
$y=0$ in Eq.(1) gives $x=-2$ and $x=-2$ in Eq.(2) gives $y=0$ . Thus a second solution is
$$(x=-2\:;\:y=0)$$
Full solving :
Put $y$ from Eq.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$ \int \frac{x-4}{\sqrt{x^2-4x+5}}\, dx$ I'm trying to solve this irrational integral $$ \int \frac{x-4}{\sqrt{x^2-4x+5}}\, dx$$
doing the substitution
$$ x= \frac{5-t^2}{2 \cdot (2+t)}$$
according to the rule.
So the integral becomes:
$$ \int \frac{1}{2} \cdot \frac{t^2+8t+11}{t^2+4t+5}\, dt =\frac{1}{2} \int 1+\fr... | To check your results, here's an other approach
$$\int \frac{2x-4-4}{2\sqrt{(x-2)^2+1}}dx=$$
$$\sqrt{x^2-4x+5}-2\int \frac{dx}{\sqrt{(x-2)^2+1}}$$
put
$$x-2=\sinh(t)=\frac{e^t-e^{-t}}{2}$$
the last integrale becomes
$$\int dt=t+C$$
but
$$e^{2t}-2(x-2)e^t-1=0$$
gives
$$e^t=(x-2)+\sqrt{x^2-4x+5}$$
So, the final result is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$a^4+b^4+c^4+d^4=4abcd$ , prove that $a=b=c=d$.
If $a,b,c$ and $d$ are positive real numbers satisfying the expression: $$a^4+b^4+c^4+d^4=4abcd$$ then, prove that $a=b=c=d$.
Approach:
$$a^4+b^4+c^4+d^4=4abcd$$
$$a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4=4abcd-2a^2b^2-2c^2d^2$$
$$(a^2-b^2)^2 +(c^2-d^2)^2 = -2(ab-cd)^2$$
$$(a^2... | Using AM-GM with $a^4, b^4, c^4,d^4$, we get
$$
\frac{a^4 + b^4 + c^4 + d^4}{4} \ge \sqrt[4]{a^4 b^4 c^4 d^4}
$$
This means that
$$
a^4 + b^4 + c^4 + d^4 \ge 4abcd
$$
With equality if and only if $a^4 = b^4 = c^4 = d^4 \implies a = b = c =d$
Alternatively, we can just transform your proof into an AM-GM one:
We use AM-G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x $ I am trying to evaluate this antiderivative $$
\int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x
$$
What i have done:
$$
\begin{split}
I
&= \int \frac{1}{2+\sqrt{x+1}+\sqrt{3-x}} \cdot d x \\
&= \int \frac{2+\sqrt{x+1}-\sqrt{3-x}}{4+x+1+4 \sqrt{x+1}-3+x} \cdot d x\\... | hint
$$x+1=(x-1)+2$$
$$3-x=2-(x-1)$$
put
$$x-1=2\cos(2t)$$
to get
$$x+1=4\cos^2(t)$$
and
$$3-x=4\sin^2(t)$$
the integrale becomes
$$\int\frac{-2\sin(2t)dt}{1+\sin(t)+\cos(t)}$$
which can be computed using the substitution
$$u=\tan(\frac t2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find $\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}$? Problem was to find $$\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}.$$ My partial progress i tried to motivate such that upper term in binomial terms gets constant rather than variable , so i thought of changing it to form $\binom{n}{n-r}\binom{n-r}{n... | Applying
$$\binom{n}{k}\binom{n-k}{m-k}=\binom{n}{m}\binom{m}{k}$$
with $m=2r$ and $k=r$, we have
$$\binom{n}{r}\binom{n-r}{r} = \binom{n}{2r}\binom{2r}{r}.$$
Now snake oil yields
\begin{align}
\sum_{n\ge 0}\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r} z^n
&=\sum_{n\ge 0}\sum_{r\ge 0} \binom{n}{2r}\binom{2r}{r} 2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Solve the equation $\frac{x^2-10x+15}{x^2-6x+15}=\frac{4x}{x^2-12x+15}$ Solve the equation $$\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}.$$
First we have $$x^2-6x+15\ne0$$ which is true for every $x$ ($D_1=k^2-ac=9-15<0$) and $$x^2-12x+15\ne0\Rightarrow x\ne6\pm\sqrt{21}.$$ Now $$(x^2-10x+15)(x^2-12x+15)=4x(x^... | let $p=x^2+15,q=x$ then we have to solve $$\frac{p-10q}{p-6q}=\frac{4q}{p-12q}$$ $$\iff (p-10q)(p-12q)-4q(p-6q)=0$$ $$\iff (p - 8 q) (p - 18 q)=0$$ provided that $p\neq 6q,p\neq 12q$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Using polar coordintaes to evaluate $\int \int {\sqrt {\frac {1-x^2-y^2} {1+x^2+y^2}} }\ dx \ dy$ I want evaluate the following integral using polar coordinates.
$$ \int \int {\sqrt {\frac {1-x^2-y^2} {1+x^2+y^2}} }\ dx \ dy $$ over the positive quadrant of the circle $$x^2+y^2=1$$
I used the substitution $x=r\cos\the... | Substitute $u=1+r^2$ and $\mathrm du=2r\,\mathrm dr$, so that
$$\sqrt{\frac{1-r^2}{1+r^2}}r\,\mathrm dr=\sqrt{\frac{2-(1+r^2)}{1+r^2}}r\,\mathrm dr=\frac12\sqrt{\frac{2-u}u}\,\mathrm du$$
Then take $v=\sqrt u$ (so that $v^2=u$) and $2v\,\mathrm dv=\mathrm du$:
$$\frac12\sqrt{\frac{2-u}u}\,\mathrm du=\frac12\sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$ Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$
And Prove it by induction.
Attempt:
$$A(x)= \sum_{k=0}^n (k+1)(n-k+1)$$
$$\sum_{n=0}^\infty(\sum_{k=0}^n (k+1)(n-k+1))x^n$$
The series of differences is :$\{(n+1),2(n),3(n... | Note that
\begin{align}
\sum_{n \ge 0} \sum_{k=0}^n a_k a_{n-k} x^n
&=\sum_{k \ge 0} a_k x^k \sum_{n\ge k} a_{n-k} x^{n-k} \tag1\\
&=\sum_{k \ge 0} a_k x^k \sum_{n\ge 0} a_n x^n \tag2\\
&=\left(\sum_{n \ge 0} a_n x^n\right)^2
\end{align}
Equality $(1)$ arises from interchanging the order of summation.
Equality $(2)$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4124684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
An immediate alternative to a trigonometry problem (high school) I have a right triangle, $AH\perp BC$, where $\cos \beta=\sqrt 5/5$ and $\overline{AH}+\overline{CH}+\overline{HB}=7$.
I have to found the area.
My steps (or solution):
$$\mathcal A(\triangle ABC)=\frac 12\overline{AB}\cdot \overline{BC} \sin \beta$$
wit... | HINT
Let $a = \overline{AH}$, $b = \overline{CH}$ and $c = \overline{HB}$. Then we have that
\begin{align*}
\begin{cases}
a = \tan(\beta)c\\\\
a = \tan(\pi/2 - \beta)b
\end{cases} \Rightarrow a(1 + \tan(\beta) + \cot(\beta)) = 7
\end{align*}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Finding the probability density function of the $n$th largest random variable. Let $X_1,...,X_{25}$ be independent Unif $[0,1]$ random variables. Let $Y$ be the $13$th largest of the $25$ random variables. Find the probability density function of $Y$.
I already know the answer for this one, but the answer that was prov... | First note that the $13$-largest of $25$ is also the $13$-smallest, i.e. $13$th from the bottom:
$$
\begin{array}{rcccl}
& i & & 26-i \\
\hline
\text{smallest} \rightarrow & 1 & & 25 \\ \text{2nd-smallest} \rightarrow & 2 & & 24 \\ \text{3rd-smallest} \rightarrow & 3 & & 23 \\ & 4 & & 22 \\ & \vdots & & \vdots \\
& 11 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $A$ and $B$ are positive-definite matrices with $A>B$, must $A^k > B^k$ for $k>0$? Assume $A$ and $B$ are positive-definite matrices. If $A>B$, can we conclude that $A^k > B^k$ for any positive scalar $k$?
Note that $A>B$ means $A-B$ is a positive-definite matrix, not an element-wise comparison. I've tried using Cay... | No. In $\mathbb{R}^{2 \times 2}$, take $B=I$ and
$$ A = \left( \begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array} \right) $$
$A$ is positive-definite since
$$ (\begin{array}{cc} x & y \end{array})
\left( \begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array} \right)
\left( \begin{array}{c} x \\ y \end{array} \right) = 2 x^2 + 2 y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
This question comes from counting the number of digits of $10^n$ in terms of the number of digits of $... | We can first do the steps that are obvious: rearranging. This would give us $\left \lfloor{\log_{10}(2^n)}\right \rfloor+\left \lfloor{\log_{10}(5^n)}\right \rfloor=n-1$. Using logrithm properties, we can simplify that to $\left \lfloor{n\log_{10}(2)}\right \rfloor+\left \lfloor{n\log_{10}(5)}\right \rfloor=n-1$.
Let's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find definite integral as power series I need to find the following indefinite integral as power series and then compute the definite integral
$$\int_0^{1/2}\sqrt{1+x^3}dx$$
with the accuracy of $10^{-4}$.
My attempt:
Using the binomial theorem, we can express $\sqrt{1+x^3}$ as
$$\sum_{k=0}^\infty{\frac{1}{2}\choose k}... | Use the well-known Taylor series for the function $\sqrt{1+x}$ and substitute $x\to x^3$ to get
$$\sqrt{1+x^3}=\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}{4^n (n!)^2 (2n-1)} x^{3n}.$$
This series converges for $\lvert x\rvert <1$. Now integrating the series yields
$$\int_0^{1/2}\sum^{\infty}_{n=0} \frac{(-1)^{n-1}(2n)!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Local maximum uniqueness of the logarithm of the Student-t distribution The negative logarithm of the Student-t distribution partial density function is
$$f(\nu,x) := -\ln\Gamma\left(\frac{\nu+1}{2}\right)
+\ln\Gamma\left(\frac{\nu}{2}\right)
+\frac{1}{2}\ln(\pi\nu)
+\frac{\nu+1}{2}\... | Just some thoughts (I will add proofs in the future, if possible.)
For convenience, we replace $x^2$ with $y$.
We have
$$
\frac{\partial f}{\partial v}
= \frac{1 - y}{2v + 2y} + \frac12 \ln \left(1 + \frac{y}{v}\right) - \frac12 \psi\left(\frac{v + 1}{2}\right) + \frac12\psi\left(\frac{v}{2}\right)
$$
where $\psi(\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4139399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Let P and Q be two points and let R be the point on PQ whose distance from P is twice its distance from Q.Verify that $w= \frac{1}{3}u + \frac{2}{3}v$ Let $P$ and $Q$ be two points in space and let $R$ be the point on $PQ$ whose distance from $P$ is twice its distance from $Q$. If $U=OP$, $v=OQ$ and $W=OR$. Verify that... | If you want to use your initial Cartesian coordinate approach:
Let $\vec{u} = OP = (u_x, u_y)$, $\vec{v} = OQ = (v_x, v_y)$, and $\vec{w} = OR = (w_x, w_y)$.
Let's assume $\vec{w} = \frac{1}{3} \vec{u} + \frac{2}{3} \vec{v}$, i.e.
$$\left\lbrace \begin{aligned}
w_x &= \frac{1}{3} u_x + \frac{2}{3} v_x = \frac{u_x + 2 v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Showing $\tan70° = \tan20° + 2\tan50°$ Q. Prove that $\tan 70° = \tan 20° + 2\tan 50°$.
My approach:
$ LHS = \tan 70° = \dfrac1{\cot 70°} = \dfrac1{\tan 20°}$
$ \begin{align}
RHS
&= \tan 20° + 2\tan 50° \\
&= \tan 20° + 2\tan (20+30)° \\
&= \tan 20° + \dfrac{2(\tan 20° + 1/√3)}{1 - \tan 20°/√3} \\
&=\dfrac{2 + 3√3 \t... | This will answer the question you asked in bold.
In fact, the two expressions you got involving $\tan 20^\circ$ are equal, not in the sense that algebraic manipulation will produce identical expressions, but rather depending upon a special fact regarding $\tan 20^\circ$. You want to show for $x=\tan 20^\circ$, $$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Finding distance from Local maximum of $f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant
What is the distance of the local maximum of the function
$f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant?
$1)1\qquad\qquad2)\sqrt2\qquad\qquad3)2\qquad\qquad3)2\sqrt2$
It is a question from timed exam so the fastest answ... | This is an arguably faster way of getting to the $x$-coordinate of the local maximum. $$f(x)=2+(x-2)+\sqrt{4-(x-2)^2} $$ Let $x-2 = 2\sin t$: $$f(t)= 2+2(\sin t +\cos t)= 2+2\sqrt 2\sin\left( t+\frac{\pi}{4} \right) $$ For a local maximum, $t=\frac{\pi}{4}$ (e.g.) for which $x=2+\sqrt 2$. And then you can do the same t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Indefinite integration of $\int\frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$
Integrate $$\int\dfrac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$$
My Attempt:
Using, $$\tan A-\tan B=\dfrac{\sin(A-B)}{\cos A\cdot \cos B}$$
The given integral can be transformed as
$$\int\dfrac{\tan(x-1)}{\cos(x-2)}\,\textrm dx - \i... | Little bit of trigonometry
$$\begin{align}
\frac{\tan(x-1)}{\cos(x-2)}
&= \frac{\tan(x-1)}{\cos((x-1) - 1)} \\
&= \frac{\tan(x-1)}{\cos(x-1)\cos 1 + \sin(x-1)\sin 1} \\
&= \frac{\tan(x-1)\sec(x-1)}{\cos 1 + \tan(x-1)\sin 1} \\
&= \frac{d\left(\sec(x-1)\right)}{\cos1 + \sin 1 \sqrt{\sec^2(x-1) - 1}} \\
&= \frac{du}{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Proving $\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ for positive reals $a$, $b$, $c$ Question $5$ of BMO1 $2008$:
For positive real numbers $\;a,\;b,\;c,\;$ prove that $$\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
I noticed that the right side can be grouped, but did not get further.... | Note that
$$
(a+b+c)(a+b-c)(a-b+c)(-a+b+c)=((a+b)^2-c^2)(c^2-(a-b)^2)=
\\
=(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)=(2ab)^2-(a^2+b^2-c^2)^2.
$$
Can you continue now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How can I expand on this part of this proof concerning magnetic fields? I am supposed to prove:
$$B_y=\frac{\mu_0}{4\pi}Iaz\int_{0}^{2\pi} \frac{\sin\phi}{(a^2+y^2+z^2-2ay\sin\phi)^{3/2}}d\phi=\frac{\mu_0Ia^2}{4r^3}\biggl(\frac{3yz}{r^2}\biggl)$$
and
$$B_z=\frac{\mu_0}{4\pi}Iaz\int_{0}^{2\pi} \frac{a-y\sin\phi}{(a^2+... | First note that:
*
*$r = \sqrt{z^2+y^2} \gg a \implies z^2+y^2 \gg a^2 \implies a^2+y^2+z^2 \approx y^2+z^2$
*$r \gt y \;\land\; r \gg a \implies r^2 \gg ay \implies r^2 \gg ay \sin \phi$
Then:
$$
\require{cancel}
\begin{align}
(a^2+y^2+z^2-2ay\sin\phi)^{-3/2} &\approx (y^2+z^2-2ay\sin\phi)^{-3/2}
\\ &= \left(r^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\frac{a}{b}+\left(\frac{3b+c}{3c+b}\right)^2+\left(\frac{2c+a}{2a+c}\right)^3\ge 3$ for positive $a$, $b$, $c$
Let $a,b,c$ be positive real number. Prove that
$$\dfrac{a}{b}+\left(\dfrac{3b+c}{3c+b}\right)^2+\left(\dfrac{2c+a}{2a+c}\right)^3\ge 3$$
It is not a symmetric inequality. I know that the equality o... | Using the AM-GM inequality,
$$\left(\frac{3b+c}{3c+b}\right)^2+1\ge2\left(\frac{3b+c}{3c+b}\right) $$
$$\left(\frac{2c+a}{2a+c}\right)^3+1+1\ge3\left(\frac{2c+a}{2a+c}\right)\;\;\;\;\;$$
$$\therefore\;\;\frac{a}{b}+\left(\frac{3b+c}{3c+b}\right)^2+\left(\frac{2c+a}{2a+c}\right)^3+3\ge \frac{a}{b}+2\left(\frac{3b+c}{3c+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Right Tangential Trapezoid with an incircle: Find the radius of the incircle and the area of the trapezoid The incircle of the trapezoid $ABCD$ that has a right angle $A$ with the center of $O$ is tangent with the point $E$ on $CD$ in a way that $CE=3$ and $ED=12$.
Find the angle of $COD$, the radius of the incircle an... | Suppose the $AD$ and $BC$ are the bases of the trapezoid $ABCD$:
According to the data given, $CE = 3$ and $DE = 12$
Suppose $\angle ADC = 2\alpha$. Since $\angle BAD = 90^\circ$, $\angle ADC + \angle DCB = 180^\circ = 2\alpha + \angle DCB$:
$$\therefore \, \angle DCB = 180^\circ - 2\alpha = 2(90^\circ - \alpha) \tag1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the number of 5-digit number divisible by 6 which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
Find the number of $5$-digit number divisible by $6$ which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
I started by considering the following cases:
*
*Unit ... | Case $1$:
$1,2,3,4,5$ are chosen.
For the number to be divisible by $2$, last digit is $2$ or $4$, i.e. last digit can be selected in 2 ways and the other digits can arrange in $4!$. Thus, the number of ways =$4! \cdot 2=48$.
Case $2$:
$0,1,2,4,5$ are chosen.
a) if last digit is $0$: number of ways =$4!=24$
b) if last ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving $\sum_{k=1}^{n}\cos\frac{2\pi k}{n}=0$ I want to prove that the below equation can be held.
$$\sum_{ k=1 }^{ n } \cos\left(\frac{ 2 \pi k }{ n } \right) =0, \qquad n>1 $$
Firstly I tried to check the equation with small values of $n$
$$ \text{As } n=2 $$
$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 2 } \ri... | Your sum is the sum of the real parts of the $n$th roots of the unity. That is, they are the roots of the polynomial $z^n-1$. The sum of the roots is equal to the coefficient of $z^{n-1}$ by Vieta's formulas, hence it is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Given the distance measure for two points and the point $p = (0,2)$, which of the following points have the same distance as $p$ from the origin? Given the distance measure dist:
$$\mathrm{dist}(x,y) =
\sqrt{(x_1-y_1,x_2-y_2)\begin{pmatrix}
3 & 0\\
0 & 4
\end{pmatrix} (x_1-y_1,x_2-y_2)^{T} } $$
for two dimensional p... | For two points $x = (x_1,x_2)$ and $y =(y_1,y_2)$ we have,
$$\begin{split}
\mathrm{dist}(x,y) =&
\sqrt{(x_1-y_1,x_2-y_2)\begin{pmatrix}
3 & 0\\
0 & 4
\end{pmatrix} (x_1-y_1,x_2-y_2)^{T} } \\
=& \sqrt{(x_1-y_1,x_2-y_2)\begin{pmatrix}
3 & 0\\
0 & 4
\end{pmatrix} \begin{pmatrix}
x_1-y_1 \\
x_2-y_2
\end{pmatrix} }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $y' = (1+\frac{y-1}{2x})^2$ Solve $y' = (1+\frac{y-1}{2x})^2$
My first thought was to expand to see if I can get a linear form:
$$y' = 1 + \frac{y-1}{x} + (\frac{y-1}{2x})^2 = 1 + \frac{y-1}{x} + \frac{y^2-2y+1}{4x^2}$$
$$y' = 1 + \frac{y}{x} -\frac{1}{x} + \frac{y^2}{4x^2} + \frac{-2y}{4x^2} + \frac{1}{4x^2} ... | Hint
If you consider $$u=1+\dfrac{y-1}{2x}$$ then we get $$y'=2xu'+2u-2.$$ Thus the original equation becomes
$$2xu'+2u-2=u^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric Simplification Question I was working through a trigonometry problem, and was having some difficulty so I decided to look at the solution. Here are the steps:
$$\frac{\sin(2x+50^\circ)+\sin(150^\circ)}{\sin(2x+50^\circ)-\sin(150^\circ)}=\frac{\cos(50^\circ)-\cos(2x+50^\circ)}{\cos(50^\circ)+\cos(2x+50^\ci... | This is similar to Siong's answer, but written differently using ratios:
\begin{align}
\frac{a+b}{a-b} - 1 &= \frac{c-d}{c+d} -1 \\[2ex]
\require{cancel} \frac{2b}{a-b} &= -\frac{2d}{c+d} \\[2ex]
\frac{a-b}{b} &= -\frac{c+d}{d} \\[2ex]
\frac{a}{b} - 1 &= -\frac{c}{d} -1 \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find value of $|(\overrightarrow{a} \times\overrightarrow{b}) \times\overrightarrow{c}|$
let $\overrightarrow{a} = 2 \hat {i}+ \hat {j}-2 \hat {k}$ and $\overrightarrow{b}= \hat {i} + \hat{j}$. $\:$ if $\overrightarrow{c}$ is a vector such that $\overrightarrow{a} \cdot \overrightarrow{c}+2|\overrightarrow{c}|=0$ an... | The simple answer is that, the question is erroneous and $\overrightarrow{c}$ simply does not exist.
Note that from your calculations, $\overrightarrow{a}\cdot \overrightarrow{c}=-2|\overrightarrow{c}|$, and $|\overrightarrow{c}|=1$. Since $|\overrightarrow{a}|=3$, this means that if $\phi$ is the angle between these t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve this integral: $\int_0^\infty{\frac{ln x}{x^2+2x+4}}\cdot dx$ I have been trying to solve this integral:- $$\int_0^\infty{\frac{\ln x}{x^2+2x+4}}\cdot dx$$ But I have been unsuccessful so far. I have tried integration by parts ( taking $1/(x^2+2x+4)$ as the second function does not work because it ends of ... | The Path
To evaluate
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\int_0^\infty{\frac{\log(x)}{x^2+2x+4}}\,\mathrm{d}x\tag1
\end{align}
$$
we will evaluate
$$
\int_\gamma{\frac{\log(x)^2}{x^2+2x+4}}\,\mathrm{d}x\tag2
$$
where
$$
\begin{align}
\gamma
&=\left[\epsilon+i\epsilon^2,R+i\epsilon^2\right]\cup\sqrt{R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A series representation for $e^x$ I want to show that for $x \notin \{2,3,4,\dots\}$ $$e^x = \frac{2+x}{2-x} + \sum_{k=2}^{\infty} \frac{-x^{k+1}}{k!(k-x)(k+1-x)}$$
It is pretty clear that why $x \notin \{2,3,\dots\}$. I proved that the series converges for all other $x$ by using the ratio test. My initial thoughts wer... |
Can someone suggest some other ways to do it?
Yes, one can prove the identity directly, without computing the derivative. We start with a partial fraction decomposition:
$$
\sum_{k=2}^{\infty} \frac{-x^{k+1}}{k!(k-x)(k+1-x)}
= \sum_{k=2}^{\infty}\frac{-x^{k+1}}{k!} \left( \frac{1}{k-x} - \frac{1}{k+1-x}\right) \, .
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(n+1)a\leq a^{n+1}+n, \forall a,n\in\mathbb{N}$. We can start from the fact that:
\begin{align*}
0\leq a^n + a^{n-1} +\ldots + a^2 + a-n,\forall a,n\in\mathbb{N}.
\end{align*}
The above is true, since if $a>1$, then $a^n>n, \forall n \in\mathbb{N}$. Also if $a = 1$, then we will have $1 + 1 + \ldots + 1... | Define $f(x)=x^{n+1}-(n+1)x+n$. Note that $f(1)=0$ and $f'(x)=(n+1)(x^n-1)$. Therefore, $f'(x)$ takes negative values in $(0,1)$ and positive values in $(1,\infty)$, as desired. So $f(x)$ takes its minimum value (which is zero) at $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to represent Hadamard gate as product of Rx and Ry gates? As simple as question asks - how to represent Hadamard gate as product of Rx and Ry gates assuming that values for rotation of Rx and Ry gate can be different?
| We have
$$R_x(\pi)R_y(\pi/2)=\left(
\begin{array}{cc}
0 & -i \\
-i & 0 \\
\end{array}
\right)\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
$$=\left(
\begin{array}{cc}
-\frac{i}{\sqrt{2}} & -\frac{i}{\sqrt{2}} \\
-\frac{i}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is $y^2+f(y)b+c$ a quadratic equation? The solution to the question:
Let $x, y, z \in R$ such that $x+y+z=6$ and $x y+y z+z x=7$. Then find the range of values of $x, y$, and $z$.
given in book is as follows:
$x, y, z \in R$
$x+y+z=6$ and $x y+y z+z x=7$
$\Rightarrow y(6-y-z)+y z+z(6-y-z)=7$
$\Rightarrow \quad-y^{2}... | Just go back and recall the derivation of the quadratic formula, it never uses the fact that $a$, $b$ and $c$ are constant, you can use it any time. Just as an example, consider:
$$\begin{align*}
2x+4&= 20\\
4+2x-20&= 0\\
(1)\color{blue}{2}^2+(x)\color{blue}{2}+(-20)&= 0\\
\color{blue}{2}&= \dfrac{-x\pm\sqrt{x^2-4(1)(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality... | Here is a proof following Momo answer.
$$\dfrac{1}{a^2+b^2+ab}+\dfrac{1}{b^2+c^2+bc}+\dfrac{1}{a^2+c^2+ac}\ge \dfrac{10.5}{a^2+b^2+c^2+2.5(ab+bc+ca)}$$ where all variables are nonnegative or positive. This is the inequality with $r=2.5$ also if the inequality in Momo answer is true for $r\ge 0$ it is also true for eve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Order on three real numbers given they are roots of a cubic, their sum and sum of pairwise product
Let $a,b,c$ be three real numbers which are roots of a cubic polynomial and satisfy $a+b+c=6$ and $ab+bc+ca=9.$ Suppose $a<b<c.$ Show that $$0<a<1 <b<3<c<4$$
Source: ISI Bmath Entrance 18-Jul-2021
Let the polynomial be ... | Solution using the derivative:
$P(x)$ is a polynomial, hence its roots are separated by the roots of the first derivative $P'(x).$ $$P'(x)=3x^2-12x+9=3(x-1)(x-3),$$ thus $$a<1<b<3<c.$$
To complete the proof, we have to show that $a>0$ and $c<4.$
*
*Let us prove $a>0$ by contradiction.
*
*Case $a=0:$
Here, the poly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why and when can I just peacefully substitute into $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$ without checking range conditions? This is an example question in my book:
To solve for $x$:
$$\tan^{-1}\frac{x-1}{x+2}+\tan^{-1}\frac{x+1}{x+2}=\frac{\pi}{4}$$
and it is solved by a direct formula given by
$$\tan^{-1... | We have,
$\mathsf{tan^{-1}\left(\dfrac{x-1}{x+2}\right)+tan^{-1}\left(\dfrac{x+1}{x+2}\right)=\dfrac{\pi}{4}}$
$\mathsf{\implies\,tan^{-1}\left(\dfrac{x+1}{x+2}\right)=\dfrac{\pi}{4}-tan^{-1}\left(\dfrac{x-1}{x+2}\right)}$
$\mathsf{\implies\,tan\left\{tan^{-1}\left(\dfrac{x+1}{x+2}\right)\right\}=tan\left\{\dfrac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4205737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modul... | It still involves cases, but here's a somewhat slicker proof.
By Euler's theorem, $a^{\varphi(n)}\equiv 1$ mod $n$, where $\varphi$ is Euler's totient function, which count the number of positive integers less than $n$ coprime to $n$.
As $\varphi(9)=6$, this implies that the residue of
$$2^n+5^n+56$$
mod $9$ depends on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 2
} |
Finding $a$ such that $f(x)=a\,|x-b|$ is satisfied by $(2,1)$ and $(10,3)$
I was given the following equation: $$f(x)=a\,|x-b|$$ and the information that both $(2,1)$ and $(10,3)$ are solutions for the equation. The question asked to solve for $a$.
This is what I did: $$a=\frac{1}{2-b}\\[10pt]a=\frac{-1}{2-b}\\[10pt]... | Your answers are correct, but there is a more efficient way of finding them using what we know about the vertex in this context.
First, note that $f(x) = a|x-b|$ has two transformations, a shift to the right by $b$ and a vertical stretch by $a$. So, we know that the vertex has coordinates $(b,0)$. Since the vertex must... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Calculate the limit $\lim_{n\rightarrow\infty}\left(\frac{\prod_1^n(1 +r/n^2)}{e^{1/2}}\right)^n$ How to calculate the limit
$$\lim_{n\rightarrow\infty}\left(\frac{\prod\limits_{r=1}^n\left(1 +\frac{r}{n^2}\right)}{e^{1/2}}\right)^n?$$
I tried using the sandwich theorem by taking logarithm of the limit and using the ex... | Hint:
If we take
$\begin{align}
&P(x) := \prod_{r=1}^n\left(x +\frac{r}{n^2}\right) \\[1mm]
&= x^n + \frac{\sum_{r_1=1}^n r_1}{n^2}x^{n-1} + \frac{\sum_{r_1=1}^n\sum_{r_2=1}^n r_1r_2}{n^4}x^{n-2} + \dots + \frac{\sum_{r_1=1}^n\sum_{r_2=1}^n\dots\sum_{r_n=1}^n r_1r_2\cdot...\cdot r_n}{n^{2n}} \\[1mm]
&= x^n + \frac{\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4209348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Extrema of $f(x,y)=(1-x^2-y^2)\cdot xy$ subject to $x^2+y^2\leq 1$ I want to determine the extrema of $f(x,y)=(1-x^2-y^2)\cdot xy=xy-x^3y-xy^3$ subject to $x^2+y^2\leq 1$.
We use Lagrange Multipliers to check the critical points on the circle $x^2+y^2=1$.
To check the critical points inside the circle, $x^2+y^2<1$, we... | Let $x=r\cos t$ and $y=r\cos t$, where $0\leq r\leq1$ and $t\in[0,2\pi)$.
Thus, by AM-GM $$(1-x^2-y^2)xy=(1-r^2)r^2\sin t\cos t=\frac{1}{2}(1-r^2)r^2\sin2t\leq$$
$$\leq\frac{1}{2}(1-r^2)r^2\leq\frac{1}{2}\left(\frac{1-r^2+r^2}{2}\right)^2=\frac{1}{8}.$$
The equality occurs for $1-r^2=r^2,$ which gives $r=\frac{1}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4211762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$a^a\cdot{b^b}\ge \bigl(\frac{a+b}{2}\bigl)^{a+b}\ge{a^b}\cdot{b^a}$ If $a$ and $b$ are positive rational numbers, prove that
$$a^a\cdot b^b\ge \left(\frac{a+b}{2}\right)^{a+b} \ge a^b \cdot{b^a}$$
My try:
consider $\frac{a}{b}$ and $\frac{b}{a}$ be two positive numbers with associated weights $b$ and $a$.
Then $\displ... | The second inequality.
We need to prove that:
$$\ln\frac{a+b}{2}\geq\frac{b}{a+b}\ln{a}+\frac{a}{a+b}\ln{b}.$$
Now, since $\frac{b}{a+b}+\frac{a}{a+b}=1$ and $\ln$ is a concave function, by Jensen and AM-GM we obtain:
$$\frac{b}{a+b}\ln{a}+\frac{a}{a+b}\ln{b}\leq\ln\left(\frac{ba}{a+b}+\frac{ab}{a+b}\right)\leq\ln\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solving $2x^3-5ix^2+3x+4i=0$, need $(21x_1-1)^{-2}+(21x_2-1)^{-2}+(21x_3-1)^{-2}$ I have this polynomial from Problemas selectos (Lumbreras editors):
$$2x^3-5ix^2+3x+4i=0$$
I make $x=it$. then:
$$2(it)^3-5i(it)^2+3it+4i=0$$
$$i(-2it^3+5it^2+3it+4)=0$$
$$2t^3-5t^2-3t-4=0$$
$$8t^3-20t^2-12t-16=0$$
$$(2t)^3-3(2t)^2(5/3)+(... | Starting from @mjw's answer, you face a cubic equation with only one real root in $y$ since $\Delta=-5447$.
Using the hyperbolic method for this root, we then have
$$y_1=\frac{5}{6}+\frac{\sqrt{43}}{3} \cosh \left(\frac{1}{3} \cosh
^{-1}\left(\frac{476}{43 \sqrt{43}}\right)\right)=3.17174\cdots$$
Now, deflate the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The Diophantine equation $x^5-2y^2=1$ I'm trying to solve the Diophantine equation $x^5-2y^2=1$.
Here's my progress so far. We can write the Diophantine equation as
$$\frac{x-1}{2}\cdot(x^4+x^3+x^2+x+1)=y^2.$$
If $x\not\equiv1\pmod{5}$, then $\gcd(\frac{x-1}{2},x^4+x^3+x^2+x+1)=1$, so both $\frac{x-1}{2}$ and $x^4+x^3+... | Also not a super elementary argument, but the only non-trivial result I use here is that $K=\Bbb Q(\sqrt{-2})$ has class number $1$, i.e. $\Bbb Z[\sqrt{-2}]$ is a PID.
Let $x^5-2y^2=1$ with $x,y\in\Bbb Z$. Note that $x$ is odd. We can factor the equation as $$x^5=(1-\sqrt{-2}y)(1+\sqrt{-2}y).$$ Let $d$ be a gcd of $(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding foot of perpendicular of centre of ellipse on its variable tangent
Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2\over a^2} +{y^2\over b^2} = 1$$
on the chord joining the points whose eccentric angles differ by $π/2.$
Ref:Locus of foot of perpendicular of origin from tangent ... | Here's one method to simplify things. First note that :
$$u^2+v^2=1/2$$
Now, consider the equation you found:
$$\frac{ax}{u}= \frac{by}{v}= \frac{a^2b^2 (u^2 +v^2)}{b^2u^2 +a^2v^2}$$
Then:
$$axv=byu\Rightarrow a^2v^2=b^2u^2y^2/x^2\Rightarrow b^2u^2+a^2v^2=b^2u^2\left(\frac{x^2+y^2}{x^2}\right)$$
Plugging this back into... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Geometry Problem from Olympiad book Given the base and the vertical angle of a triangle show that its area is greatest when it's isoceles
I am stuck on how to proceed and which theorem to use here
Thanks in advance
| Denote $S = \dfrac{1}{2}ab\sin C$. Thus: $S \le \dfrac{1}{8}(a+b)^2\sin C$, with equality occur at $a = b$ which shows $\triangle ABC$ isosceles. To show that $S$ reaches a maximum value, you need to show: $(a+b)^2$ also reaches a maximum in terms of $c$. To this end, use the law of cosines: $c^2 = a^2+b^2 - 2ab\cos C=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$ To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$
My Attempt:
First Method: we know that $\cot^{-1}x = \tan^{-1}\frac{1}{x}$ for $x>0$ and $\tan^{-1}x+\tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}, xy<1$
Now $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18$ = $\tan^{-1}\fr... | The statement is equivalent to
$$\DeclareMathOperator{\arccot}{arccot}
\arccot7+\arccot8=\arccot3-\arccot18
$$
The right-hand side is positive because the arccotangent is decreasing, so it is in the interval $(0,\pi/2)$ and we have to ensure that the same holds for the left-hand side, but this is easy, because $7>1$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Roots of polynomials with repeated roots Let $a$ and $b$ be real numbers. Consider the cubic equation $$x^3+2bx^2-ax^2-b^2=0$$
(i) Show that if $x=1$ is a solution of the cubic then $$ -1+\sqrt{2}\leq b\leq1+\sqrt{2} $$
(ii) Show that there is no value of $b$ for which $x=1$ is a repeated root of the cubic
I need some ... | If $x=1$ is a root of $P(x) = x^3 + 2bx^2 - a^2x - b^2$, then
\begin{align}
P(1) &= 0 \\
1+2b-a^2-b^2 &= 0 \\
a^2 &= 1 + 2b - b^2
\end{align}
So
\begin{align}
P(x) &= x^3 + 2bx^2 + (b^2-2b-1)x - b^2 \\
&= (x-1)(x^2+(2b+1)x + b^2)
\end{align}
If $x=1$ is a double root of $P(x)$, then
\begin{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to change $x^3\prod_{n=1}^{\infty}\left(1-\frac{x^4} {2\pi^2n^2}-\frac{x^6}{2\pi^3n^3}\right)$ to equal $x\sin{(x^2)}$? Consider a function $$f(x)=x^3\prod_{n=1}^{\infty}\left(1-\frac{x^4} {2\pi^2n^2}-\frac{x^6}{2\pi^3n^3}\right)$$
This function has its zeros when $x=0$ or $x^2=k\pi$.
This function is ve... | Note :
Consider the Weierstass factorization of $\sin z$ :
$$\sin z = z \cdot \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{\pi^2 n^2}\right).$$
let $z = x^2$ then
$$\sin(x^2 ) = x^2 \cdot \prod_{n=1}^{\infty} \left( 1 - \frac{x^4}{\pi^2 n^2} \right)$$
and
$$x \sin(x^2 ) = x^3 \cdot \prod_{n=1}^{\infty} \left( 1 - \frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve in $\mathbf{N}$ the equation $9x^2+p=y^2$ In these days, I have been trying to solve this problem:
Let $p \in \mathbf{N}$ a positive large integer ($> 10^9$). Find all $x, y \in\mathbf{N}$ such that:
$$9x^2+p=y^2$$
The first approach that I have tried is the following. We know that:
$$(t+n)^2-t^2=t^2+2\cdot t \... | If we let $\quad A^2+B^2=C^2
\qquad A=3x\quad B=\sqrt{p}\quad C=y\qquad$
we can use Euclid's formula for Pythagorean triples to find infinite solutions based on the value of $(x)$. We start with the formula
$$ \qquad A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2\qquad$$
and solve the $A$-function for $(k).\quad$
Then we tes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Finding the square root of a radical expression I have this (not necessarily optimum) method for finding the value of sin $15^{\circ}$.
(Meant to be an application of Angle Bisector Theorem).
In a equilateral triangle with side length 1, draw the altitude to get two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles
Then, in... | Presumably you mean
$\sqrt{6-\color{blue}{3}\sqrt3}=(3-\sqrt3)/\sqrt2.$
Suppose we are given two positive rational numbers $a,b$ and we seek $x,y$ such that
$\sqrt{a-\sqrt{b}}=\sqrt{x}-\sqrt{y}$
Then we will have the conjugate relation
$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$
And upon multiplication
$\color{blue}{\sqrt{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A simple general expression for a definite integral Let
$$I_k = \int_0^{\frac{\pi}{2}} \sin x \cos^{2k} x \sqrt{1 + \sin x} \, dx,$$
where $k = 0,1,2,\ldots$. I wish to find a simple general expression for $I_k$ in terms of $k$. Simple here is the operative word.
Making a tangent half-angle substitution of $t = \tan \f... | I think that we could simplify the problem letting
$$\sqrt{1+\sin(x)}=t \implies x=-\sin ^{-1}\left(1-t^2\right)\implies dx=\frac{2 t}{\sqrt{1-\left(1-t^2\right)^2}}\,dt$$ This makes
$$I_k=2\int_1^{\sqrt 2}\left(t^2-1\right) t^{2 k+1} \left(2-t^2\right)^{k-\frac{1}{2}}\,dt$$
$$\color{blue}{I_k=\frac{\sqrt{\pi }\, \Gamm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ How to find $\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$? Here is my attempt.
Put $f(x)=\frac{x}{\log x}(x^{\frac{1}{x}}-1)$ for $x>1$. Then
\begin{aligned}
\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}... | As an alternative
$$\frac{n}{\log n}(n^{\frac{1}{n}}-1)=\frac{n^{\frac{1}{n}}-1}{\log \left(n^{\frac{1}{n}}\right)}$$
with $x+1=n^{\frac{1}{n}} \to 1$ that is as $x \to 0$
$$\lim_{x\to 0}\frac{x}{\log (1+x)}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Inductively prove $1 + \frac12 + \frac14 + \cdots + \frac{1}{2^n} = 2 - \frac{1}{2^n}$. Inductively prove that the formula holds for all $n\in\Bbb{N}$:
$$1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}=2-\frac{1}{2^n}.$$
What I have so far:
base: n = 1: $$1+\frac{1}{2}=2-\frac{1}{2}=1.5$$
inductionstep: n = k: $$1+\frac... | The formula
$$
1+\dots+\frac{1}{2^n}=2-\frac{1}{2^n}
$$
is true for $n=0$ (in which case the LHS of the equation actually only has one term in it). Therefore, it is better to choose $n=0$ as your base case. Now suppose it is true when $n=k$, that is
$$
1+\dots+\frac{1}{2^k}=2-\frac{1}{2^k} \, .
$$
Then, for $n=k+1$, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find the quadratic equation from given relatioship the quadratic equation whose roots are a and b where $a^2 +b^2=5$ and
$3(a^5+b^5)=11(a^3+b^3)$
What I Tried
$a^2 +b^2=5$ $(a+b)^2-2ab=5$ $(\text{sum of roots})^2 -2(\text{products of roots})=5$
$3(a^5+b^5)=11(a^3+b^3)$ $a^3(3a^2-11)=b^3(11-3b^2)$
| There is no single answer here. It turns out there are three quadratics with real coefficients and two more with complex coefficients, the roots of which satisfy the given conditions.
The quadratic equation with roots $a$ and $b$ is $x^2-sx+p$, where $s=a+b$ and $p=ab$. Some careful algebra tells us
$$a^2+b^2=s^2-2p,\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the maximum value of $\frac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$? Let $\theta\in\left(0,\frac{\pi}{2}\right)$, then find the maximum value of $\dfrac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$
I found the derivative, which is equal to $\dfrac{(\cos\theta−\sin\theta)(\sin\theta+\c... | Let $$f(\theta)=\frac{2\sin\theta\cos\theta}{(\sin\theta+1)(\cos\theta+1)}=\frac{2\sin\theta\cos\theta}{1+\sin\theta+\cos\theta+\sin\theta\cos\theta}$$
Taking $\sin2\theta=t,\ t\in(0,1],$ $$f(t)=\frac t{1+\sqrt{1+t}+\frac t2}=\frac{2t}{2+2\sqrt{1+t}+t}=\frac2{\left(\frac 1t+\sqrt{\frac 1t+1}\right)^2}$$
In order to max... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How was the closed form of this alternating sum of squares calculated? I am reading through this answer at socratic.org.
The question is to find the closed form of the sum
$$1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-6^{2}+\ldots.$$
I understand that, if the terms were added, the sum would be
$$
\sum_{n=1}^{N} n^{2}=1^{2}+2^{2}+\ld... | It is indeed not true that $\sum_{n=1}^Nn^2=\frac{N(N+1)}{2}$ for each positive integer $N$. This is easily verified by plugging in any value $N\geq2$. Of course it is true that $\sum_{n=1}^Nn=\frac{N(N+1)}{2}$. And you are indeed correct when you say that
$$\sum_{n=1}^Nn^2=\frac{N(N+1)(2N+1)}{6}.$$
Do note that if $N$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Simple but interesting problem about the binomial coefficient from Olympiad
"Let's define $a_n=\sum\limits_{k=0}^{\lfloor n/2 \rfloor} {n-k \choose k}\left(-\frac{1}{4}\right)^k$. Evaluate $a_{1997}$."
This problem is from the final round of an old South Korean Mathematical Olympiad (1997 KMO).
I think this problem... | Snake oil:
\begin{align}
\sum_{n=0}^\infty a_n z^n &= \sum_{n=0}^\infty \left(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n-k}{k}\left(-\frac{1}{4}\right)^k \right) z^n \\
&= \sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{n=2k}^\infty \binom{n-k}{k} z^n \\
&= \sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k z^{2k}\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
The equation $x^k = -1$ over finite fields Let $k$ be a positive integer and let $F_q$ be a finite field with $q$ elements, and consider the equation $X^k + 1 = 0$. When does this equation has a solution in $F_q$? For example if $k =2$ and $q$ is a prime congruent to $3 \pmod{4}$ the equation has no solution in $F_q$.... | If $q$ is even then $x = 1$ is a solution, so assume $q$ is odd. Write $k = 2^a b$ where $b$ is odd. We can solve $x^k=-1$ iff we can solve $y^{2^a} = -1$: if $x^k = -1$ then $(x^b)^{2^a} = -1$, and conversely if $y^{2^a} = -1$ then $y^k = (-1)^b = -1$. Now $y^{2^a} = -1$ if and only if $y$ has order $2^{a+1}$, and bec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show $f\left(\frac{1}{2}\right) \le \frac{M}{4} + \int_0^1 f(x)\, dx$ Let $f$ be a differentiable function such that $f'(x)$ is continuous and $|f'(x)| \le M$ for all $x \in [0,1]$. Show that
\begin{align}
f\left(\frac{1}{2}\right) \le \frac{M}{4} + \int_0^1 f(x)\, dx
\end{align}
I have no idea how to begin prov... | Let $f$ be as in the hypothesis. We know that $|f'|\leq M$. Therefore, it follows that for $x,s\in[0,1]$ with $s>x$
$$
-(s-x)M \leq \int_{x}^{s}f'(t) dx \leq (s-x)M.
$$
Indeed, by the Fundamental Theorem of Calculus, and by letting $s=\frac{1}{2}$, we find that
$$
f\left(\frac{1}{2}\right) - f(x) \leq \left(\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Difficult limit: $ \lim_{n \rightarrow \infty} n^{3/2} \int _0^1 \frac{x^2}{(x^2+1)^n}dx $ I found this question online
Find the limit:$$\lim_{n \rightarrow \infty} n^{3/2} \int _0^1 \frac{x^2}{(x^2+1)^n}dx $$
I've been told that I need to use the gamma function by converting $n^{3/2}=n\sqrt{n}$ and doing $u$ sub $u=... | Using the inequalities $x^2-\frac12 x^4\le \log(1+x^2)\le x^2$, we find that
$$\int_0^1 x^2e^{-nx^2}e^{-nx^4/2}\,dx\le \int_0^1 \frac{x^2}{(x^2+1)^n}\,dx\le \int_0^1 x^2e^{-nx^2}\,dx\tag1$$
Then, enforcing the substitution $x\mapsto x/\sqrt n$ in $(1)$ reveals
$$\frac1{n^{3/2}}\int_0^{\sqrt n} x^2e^{-x^2}e^{-x^4/2n}\,d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the square root of a polynomial with radical I have two positive integers, $a$ and $b$, such that
$$a^2=28b\sqrt{8b^2+1}+80b^2+5, \tag{$\star$}$$
and I’d like to find $a$ in terms of $b$ (including $\sqrt{8b^2+1}$, if necessary/appropriate), with a range like $$6b + \sqrt{8b^2+1} ≤ a ≤ 2b^2+ 3\sqrt{8b^2+1}$$ be... | Suppose that $a$ and $b$ are integers such that
$$a^2=28b\sqrt{8b^2+1}+80b^2+5.$$
Then in particular $\sqrt{8b^2+1}$ is an integer, and so $8b^2+1=c^2$ for some integer $c$. Then
$$a^2=28bc+10c^2-5,$$
and a bit of rearranging shows that
$$(10c^2-(a^2+5))^2=(28bc)^2=28^2b^2c^2=98(8b^2)c^2=98(c^2-1)c^2.$$
Expanding the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of roots of the equation $ax^2+ bx + c = 0$ in $(1,2)$
Let $a, b, c \in R, a \ne 0$ such that $a$ and $4a + 3b + 2c$ have the same sign. Then the
number of roots of the equation $ax^2+ bx + c = 0$ that lie(s) in $(1,2)$ is(are)?
I began by writing that $4a^2+3ab+2ac>0$. I tried finding the sign of $f(1)\cdot f... | One can say that if the number of the roots lying in $(1,2)$ is $2$, then the axis of symmetry of the parabola $y=ax^2+bx+c$ lies in $(1,2)$.
Therefore, one can say that if the axis of symmetry of the parabola $y=ax^2+bx+c$ does not lie in $(1,2)$, then the number of the roots lying in $(1,2)$ is not $2$.
One has
$$0\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\int_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$ Prove that $$\int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$$
On simplifying by parts we get:
$$\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$$
Thus if we prove that$$\max\left(\displaystyle\int\limi... | $$\int_0^1 \frac{x^{n+1}}{x+1}\le \frac{1}{2}\int_0^1\frac{x^{n+1}+x^{n}}{x+1}=\frac{1}{2(n+1)}$$ here we used when $x\in [0,1] $ $x^{n+1}\le x^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Points of discontinuity of $\left[\frac{f(x)}3\right], f(x)=\lim_{n\to\infty}\ln(\sqrt{e^{\cos x}\sqrt{e^{3\cos x}...\sqrt{e^{(2n+1)\cos x}}}})$
Let $f(x)=\lim_{n\to\infty}\ln\left(\sqrt{e^{\cos x}\sqrt{e^{3\cos x}\sqrt{e^{5\cos x}...\sqrt{e^{(2n+1)\cos x}}}}}\right)$ and if $g(x)=\left[\frac{f(x)}3\right]$, then the ... | Hint:
Let $r = e^{\cos(x)}$. Then verify by induction or otherwise that
$$ \sqrt{r \sqrt{r^3 \sqrt{r^5 \ldots \sqrt{r ^{2n+1} }}} } = r^{\frac12+ \frac{3}{2^2} + \ldots+ \frac{2n+1}{2^{n+1}} }$$
Now evaluate $S(n) = \sum_{k=0}^{n}\frac{2k+1}{2^{k+1}}$ using this and geometric series.
Then evaluate $f(x) = \lim_{n\to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x$ I am trying evaluating this
$$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\ \text{d}x.$$
For $k=1$, there has
$$J(1)=\frac{\pi^4}{96}.$$
Maybe $J(k)$ doesn't ... | Since you have a relation with another integral, here are 2 Maclaurin series representations for a general anti derivative which you can then use to solve for J(x). Here is a Demo Graph and Integral Demo:
$$\int \frac{\arctan^3(x)}{k^2x^2+1}dx=\int \sum_{n=0}^\infty\frac{\left(\frac{d^n}{dx^n} \frac{\arctan^3(x)}{k^2x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Is there a unique partial fraction decomposition for every rational polynomial Consider this
$$({2x-3})/({x^2-1})(2x+3)$$
Here can i do decomposition as
$$Ax+B/(x^2-1)+C/(2x+3)$$
Instead of $$A/(x-1)+B/(x+1)+C/(2x+3)$$
And if not then why?
| Note tha in your case there are three unknowns in RHS. If you simplify yowill get
$A(x+1)(2x+3)+B(x-1)(2x+3)+C(x^2-1)=0x^2+2x-3$. Simplify this and equate the cpefficients of the same powers of $x$ on both sides, you get three eqns. to find $A,B,C$.
The process of partial fractions is interesting and requires some rule... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4249051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Replacing 8 red balls with blue balls You have a bag with 8 red balls in it.
Every turn you randomly select a ball from the bag. If it is red you replace it with a blue ball, and if it is blue you replace it with a red ball.
What is the expected number of turns you need to make before all 8 balls in the bag are blue?
I... | I am not finding anything wrong in the equations.
Putting it in a Jordan- Gauss calculator, I get an answer of $\dfrac{a}{b}= \dfrac{32768}{105}$
Thus $a+b= 32873$
The calculator is here
I strongly advise you to check the computations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4249185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to find the smallest parameter for trigonometric equation? Find the smallest value of parameter $\alpha$ such that equation $${\mathrm{sin}}^{2}x \times \mathrm{cos}2x + \alpha ({\mathrm{cos}}^{4}x - {\mathrm{sin}}^{4}x) = -10(2\alpha + 1{)}^{2}$$ has at least one real solution.
simplifying the expression I got $... | To find solutions to the trigonometric equation that are real, we solve the equation with respect to $sin(x)^{2}$; we get two roots:
$sin(x)^{2}=-\frac{9|2\alpha+1|}{4}-\frac{2\alpha-1}{4}$,
$sin(x)^{2}=+\frac{9|2\alpha+1|}{4}-\frac{2\alpha-1}{4}$.
We impose that
$0≤sin(x)^{2}≤1$.
We then write the system:
$0≤-\frac{9|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression:
a) $$P=\frac{4a}{b+c-a} + \frac{9b}{c+a-b} + \frac{16c}{a+b-c}$$
b) $$P=\frac{a^3}{2a+bc} + \frac{b^3}{2b+ac} + \... | For the second problem, notice that $ 2a+bc = (a+b+c)a+bc = (a+b)(a+c)$. This allows us to normalize the inequality and (mostly) forget the condition. WTS
$$ \sum \frac{ a^3(b+c) } { (a+b)(b+c)(c+a) } \geq \frac{1}{2} $$
Approach 1: It is almost immediately obvious that by AM-GM / Muirhead (since there are no $a^4$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4252731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
From a group of three biologists, two physicists and one mathematician, a committee of two people is to be randomly selected. From a group of three biologists, two physicists and one mathematician, a committee of two people is to be randomly selected. Denote by $X$ the random variable representing the number of biologi... | No that is not correct. You are supposed to find probability mass function $ \small f_X(x), f_Y(y), f_{XY} (x, y)$. Below is a tabular representation of what you need to come up with,
$ \small \begin{array}{|c|c|c|}
\hline
& Y=0 & Y=1 & Y=2 & f_X\\
\hline
X=0 & & \\
\hline
X=1 & & \\
\hline
X=2 & & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve $ \sin(x) - \cos(x) -3\sin(2x) +3\cos(2x) +\sin(3x) -\cos(3x) =0$ Hello friends please help me to solve this equation.
$ \sin(x) - \cos(x) -3\sin(2x) +3\cos(2x) +\sin(3x) -\cos(3x) =0.$
Here is what I have done so far
$ \Rightarrow \sqrt2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x ) - 3\sqrt2(\frac... | We use $$\sin A+\sin B=2\sin \left( \frac {A+B}{2}\right) \cos \left(\frac {A-B}{2}\right)$$ and $$\cos A+\cos B=2\cos \left( \frac {A+B}{2} \right) \cos \left( \frac {A-B}{2} \right)$$
Hence, $$(\sin x+\sin 3x)-3\sin 2x=2\sin 2x \cos x-3\sin 2x=\sin 2x(2\cos x-3)$$
Moreover, $$(\cos x+\cos 3x)-3\cos 2x=2\cos 2x \cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $. Simply bringing it to a common denominator does not lead me to success How can I solve this equation?
$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $$
Simply bringing it to a common denominator does not lead me to success
What I tried
| We have that for $(4+x)(4+2x)(4+3x)\neq0 $
$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 \iff 18 x^3 + 165 x^2 + 378 x + 255=0$$
$$\iff 6 x^3 + 35 x^2 + 126 x + 85=0$$
and by rational root theorem we can find that $x=-\frac 5 3$ is a root and then we obtain
$$6 x^3 + 35 x^2 + 126 x + 85=(3 x + 5) (2 x^2 + 15 x + 17)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $A=(x+a)(x+b)(x+c)-x^3-1$ is divisible by $9$. Problem: Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $$A=(x+a)(x+b)(x+c)-x^3-1$$ is divisible by $9$.
We can consider $0\le a,b,c<9... | The $0 \leq a,b,c<9$ step is fine, after all if $a,b,c$ satisfy the equation then so do $a \%9,b\%9$ and $c\%9$. (Where $\%9$ denotes the remainder upon division by $9$).
I don't think, however, that the step that realizes that $x^3+1$ can only be $0,1,2$ and reflects this on the RHS of the modulo, is a good step (eve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Making $6$ digits numbers by the digits $2,3,3,5,5,5,5$
How many $6$ digits number we can generate by the digits $2,3,3,5,5,5,5$ ?
To solve this problem I considered three cases because we have $7$ digits and looking for six digit number.
*
*Numbers without $2\Rightarrow\quad\dfrac{6!}{4!2!}=15$
*Numbers without ... | Once you have arranged the first six digits, there is only one choice for the seventh digit. Thus, the number of arranging all seven digits also gives the total number of ways of arranging the first six of those seven digits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the derivative with respect to $x^2$? I would like to find the derivative of this function with respect to x^2.
$-0.5\ln({2\pi x^{2}}) - \frac{1}{2 x^{2}}$
The answer I got is:
$-(0.5)(\frac{1}{2\pi x^{2}})(2\pi)(2) - (0.5)(-1)(\frac{1}{x^{4}})(2)$
$=-\frac{1}{x^{2}} + \frac{1}{x^{4}}$
Is it correct?
| If in some problems you are having difficulties then you can use chain rule.
y = $-0.5\ln({2\pi x^{2}}) - \frac{1}{2 x^{2}}$ = $-0.5ln(2{\pi}) - ln(x) - \frac{x^{-2}}{2} $
To find: $\frac{dy}{dx^2}$
$\frac{dy}{dx}\frac{dx}{dx^2}$ = $\frac{dy}{dx}\left({\frac{dx^2}{dx}}\right)^{-1}$ = $(2x)^{-1}\frac{dy}{dx}$
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge ab + bc + ca$ For all positive $a,b,c $ satisfying $a+b+c = 3$,Prove:
$$
\sum_{cyc} \sqrt[3]{a} \ge \sum_{cyc} ab
$$
This is a hard problem and I tried it myself, but it's really hard without using advanced techniques(e.g. EV theorem or HCF theorem).
Is there any ... | I think, TL method does not help here.
Another way.
Let $a=x^3$, $b=y^3$ and $c=z^3$.
Thus, $x^3+y^3+z^3=3$ and we need to prove that:
$$x+y+z\geq x^3y^3+x^3z^3+y^3z^3$$ or
$$(x+y+z)^3(x^3+y^3+z^3)^5\geq243(x^3y^3+x^3z^3+y^3z^3)^3.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(w^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the relationship between the length of the circumradius and the inradius in $ \triangle ABC $? For reference: In a right angle triangle ABC, an interior bisector BD is traced, where I is or incenter, $\measuredangle B = 90 ^ o$ and $3BI = 4ID$. Find the relationship between the circumraio and inraio lenght of $... | T.Poncelet:$BA+BC = AC+2r\\
R = \frac{AC}{2}\\
\frac{BA}{BI}=\frac{DA}{DI}\\
\frac{BC}{BI}=\frac{DC}{DI}\\
\therefore BA+BC = AC+2r\\
\frac{DA \cdot BI}{DI}+\frac{DC \cdot BI}{DI} = 2R+2r\\
\frac{4}{3}\cdot (DA+DC) = 2R+2r\\
\frac{4}{3}\cdot (AC) = 2R+2r\\
\frac{4}{3}\cdot (2R) = 2R+2r\\
\boxed{R =3r}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show $\frac{n^2}{2^{\sqrt{\log(n)}}} \geq \frac{n}{2}. $ So we know that $\frac{n^2}{2} \geq \frac{n}{2}$, but I'm stuck proving that $\frac{n^2}{2} \geq \frac{n^2}{2^{\sqrt{\log n}}}\geq \frac{n}{2}$. Am I missing something?
| Notice that the expression $\frac{x^2}{2^\sqrt{\ln(x)}}$ is well-defined if and only if $x\geq1$. Division by $x$ then shows that your inequality is equivalent to
$$\frac{x}{2^\sqrt{\ln(x)}}\geq\frac{1}{2}$$
We can prove this inequality with calculus:
Consider the function $f:(1,\infty)\to\mathbb{R}$ defined by
$$f(x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4264808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int \frac{1}{a+b\sec(x)}dx$ I tried to find a closed-form expression for the integral
$$\int\frac{1}{a+b\sec(x)}\>dx$$ and, afterwards, set $a=0$, $b=1$ to recover the result for the simplified integral
$$\int\cos(x)\>dx=\sin(x)+C$$
The above integral was inspired by Américo Tavares’s rather illuminating s... | To derive the result valid for $a\to 0$, integrate as follows
\begin{align}
\int \frac{1}{a+b\sec x}dx
=& \frac1a \int \left(1- \frac{b}{b+a\cos x}\right)dx\\
=& \frac1a \bigg(x-2b\int \frac{d(\tan\frac x2)}{(b+a)+(b-a)\tan^2\frac x2}\bigg)\\
=& \frac1a \bigg( x- \frac{2b}{\sqrt{b^2-a^2}}\tan^{-1}\frac{\tan\frac x2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
solve the equation $1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$ im trying to solve the equation $$(E)\quad 1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$$
attempt : because $1$ isnt a solution we have $$\begin{aligned}
1+2 z+2 z^{2}+\cdots+2 z^{n-1}+z^{n} &=2\left(z^{0}+z+z^{2}+\cdots+z^{n-1}\right)-1+z^{n} \\
&=2 \frac{1-z^{n}}{1-z}-1+z^{... | $$\begin{align}
1 + 2z + 2z^2 + \cdots + 2z^{n-1} + z^n
&= (1 + z + \cdots + z^{n-1}) + (z + z^2 + \cdots + z^n) \\
&= (1+z)(1 + z + z^2 + \cdots + z^{n-1}) \\
&= \frac{(z+1)(z^n - 1)}{z - 1}.
\end{align}$$
Thus the roots are $$z \in \{ e^{2\pi i k/n} : k \in \{1, 2, \ldots, n-1\} \} \cup \{ -1 \}.$$ When $n$ is even... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Isn't my book doing this math about differentiation wrongly? Problem:
Differentiate with respect to $x$: $\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$
My attempt:
Let,
$$y=\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$$
Both $e^x$ and $\frac{x-1}{x+1}$ are positive: $e^x$ can never be negative, and you can take the square root o... | Giving a restriction of the domain is an issue of this problem, because that quantity is defined only for $x<-1\lor x>1$. In my opinion, if this were part of a "gauntlet of exercises" in some textbook, then an indication of the domain should be specified, if the purpose of the exercise lies elsewhere: at the very least... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Prove that: $\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge\frac{6}{(k+1)\sqrt[3]{(a+b)(b+c)(c+a)}}$ Problem: Let $a,b,c>0; k\ge1$. Prove that: $$\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge\frac{6}{(k+1)\sqrt[3]{(a+b)(b+c)(c+a)}}$$
My attempt: Using C-S inequality: $$\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge... | I just get a part by AM-GM. Maybe stronger tools can help to make up the rest $1<k<2$
For $k=1$: the inequality holds obviously by AM-GM
For $k\ge2$: Using C-S inequality: $$k(a+b+c)\sum{\frac{1}{ka+b}}=3+(k-1)\sum{\frac{b}{ka+b}}+k\sum{\frac{c}{ka+b}}\ge\frac{6k}{k+1}+\frac{k(a+b+c)^2}{(k+1)(ab+bc+ca)}$$Also by AM-GM ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is this function of partitions one-to-one? Suppose we have a set of integers $H=\{1,2, ...n\}$. Let $A$ a set of partitions of H into $n/2$ pairs $\{\{x_1,y_1\},\{x_2,y_2\}, ...,\{x_{n/2},y_{n/2}\}\}$ and
function $f:A \rightarrow Z^n$ where $f(\{\{x_1,y_1\},\{x_2,y_2\}, ...,\{x_{n/2},y_{n/2}\}\})=\cup \{x_i*y_i\}$.
Fo... | No. For $n$ large enough, $f$ will not be injective.
It's easier to consider the case of addition, where we have
$$1 + (-1) = 0, \qquad 2 + (-3) = -1, \qquad 3 + (-2) = 1 \\
1 + (-2) = -1, \qquad 2 + (-1) = 1, \qquad 3 + (-3) = 0$$
That is, partition $\{-3,-2,-1, 1, 2, 3\}$ into three parts of size $2$ in at least two... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}$, prove that $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
Problem: Let $a, b, c, x, y, z$ be real numbers
such that $abc \ne 0$, $ ~ a + b + c \ne 0$, $~ (a^2 - bc)(b^2 - ca)(c^2 - ab)\ne 0$, $xyz\ne0$, $x+y+z\ne0$, $(x-y)^2+(y-z)^2+(z-x)^2\ne0$
and $$\fr... | Here is a solution that necessitates to have had a previous introduction
*
*to conic curves (maybe it is the case) and to
*barycentric coordinates (less likely, unless you have prepared some Mathematics Olympiads) ; here or here are a nice introduction to barycentric coordinates (b.c. in short). One of the most ele... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Eliminate $\theta$ from $4x=5\cos\theta -\cos 5\theta$ and $4y=5\sin\theta -\sin 5\theta$ Eliminate $\theta$ from the equations.
$$4x=5\cos\theta -\cos 5\theta$$
$$4y=5\sin\theta -\sin 5\theta$$
Alternative forms are
$$x=5\cos^3\theta-4\cos^5\theta$$
$$y=5\sin^3\theta-4\sin^5\theta$$
and $$x=\cos^3\theta(3-2\cos2\theta... | Let $c=\cos2\theta$.
$$4x=5\cos\theta -\cos 5\theta\tag1$$
$$4y=5\sin\theta -\sin 5\theta\tag2$$
Squaring and adding ,$$16(x^2+y^2)=25+1-10\cos4\theta$$ $$8(x^2+y^2)=13-5(2c^2-1)$$ $$5c^2=9-4(x^2+y^2)=k$$
Squaring and subtracting, $$16(x^2-y^2)=25\cos2\theta+\cos10\theta-10\cos6\theta$$ $$16(x^2-y^2)=25c+(16c^5-20c^3+5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
General form for $\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$ I'm wondering if there is a general form for the following sum:
$$\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$$ for $m \in \mathbb{N}$
I have obtained the following closed-forms for these speci... | I didn't want to rain on mathstackuser12's parade, but now that the bounty has been awarded, here's a simpler approach using the theory of the Barnes G function. (Mathstackuser12 seems to have reinvented a lot of it.) As that answer notes, use the log representation of arccoth. Sum in pairs up to $2N,$ where $N$ will ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Expected Value of |H-T| A coin is flipped $n$ times, with $H$ heads and $T$ tails. What is the expected value of $|H-T|$?
Here's where I'm at so far:
There are $2^n$ total different flip sequences. Each is equally likely. Thus the expected value is the sum of all the $|H-T|$ values for each flip divided by $2^n$. We ca... | Let $a_n = \operatorname{E}[|H-T|] = \operatorname{E}[|2H - n|]$ for $H \sim \operatorname{Binomial}(n,1/2)$ and $T = n - H$. Then we first observe from your sequence that $a_{2m} = a_{2m-1}$ for each positive integer $m$. So let's aim to prove this property first. Note
$$\begin{align}
a_{2m} &= \frac{1}{2^{2m}} \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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.