Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
The base of an isosceles triangle; given leg and radius of circumcircle
An isosceles triangle $\triangle ABC$ is given with leg $AC=5$ and $R=\dfrac{25}{6}$ of the circumcircle. Find the base of the triangle.
This was my first sketch. The triangles $AHC$ and $OMC$ are similar, so $\dfrac{CH}{CM}=\dfrac{AC}{CO} \Left... | Since in the standard notation $$\frac{abc}{4S}=R,$$ we obtain:
$$\frac{5\cdot5\cdot c}{4\cdot\frac{c\sqrt{25-\frac{c^2}{4}}}{2}}=\frac{25}{6}$$ or
$$\sqrt{25-\frac{c^2}{4}}=3,$$
which gives $$c=8.$$
We'll prove that $$\frac{abc}{4S}=R.$$
Indeed, let $\Phi$ be our circle and $CO\cap\Phi=\{C,D\}$.
Thus, $$\Delta ACH\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $a^2 + b^2 + c^2$ is divisible by $16$, then show that$ a^3 + b^3 + c^3$ is divisible by $64$. Where $a, b, c \in \mathbb{Z}$. If $a^2 + b^2 + c^2$ is divisible by $16$, then show that $a^3 + b^3 + c^3$ is divisible by $64$; where $a, b, c \in \mathbb{Z}$.
I began by proving that if $(a^3+b^3+c^3) -(a^2+b^2+c^2)$ is... | We may write:
$$16\big|a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)$$
⇒ $$4\big| a+b+c$$
Also:
$$8\big| ab+bc+ac$$
Since the phrases are symmetric if one is a multiple of 4 others will also be, Hence a, b and c are multiple of 4 so the sum of their cubes are a multiple $4^3=64$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$ From Mr. Michael Rozenberg solution:
For $a,b,c>0$$,$ prove that$:$
$$(a+b+c)^2\prod_{cyc}(a+b)\geq4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab,$$
I found two SOS proof:
1) $$\text{LHS-RHS}={\frac { \left( a-b \right) ^{2}\cdot \text{M}+ab \left( {a}^{... | We write the inequality as
$$\frac{(a+b+c)^2}{ab+bc+ca} \geqslant \frac{\displaystyle 4 \sum (a^2b+ab^2)}{(a+b)(b+c)(c+a)},$$
equivalent to
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+2 \geqslant \frac{4[(a+b)(b+c)(c+a)-2abc]}{(a+b)(b+c)(c+a)},$$
or
$$\frac{a^2+b^2+c^2}{ab+bc+ca} + \frac{8abc}{(a+b)(b+c)(c+a)} \geqslant 2.$$
Which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integer solution of $a+b+c=15$ with restrictions on $a, b$ and $c$ I'm checking this problem:
Find the integer solutions of $a+b+c=15$ if $a$ is multiple of 3, $b$ is less than 10 and $c$ is multiple of 2. With $a, b, c ≥ 0$
We can make a series of polynomials with combinatorics. The polynomials would be:
$$P_a(x)=1+x... | Um.... why do work when simply counting them is magnitudes of effort easier.
$a = 3k; k=0... 5$. $c$ is even so $b$ is odd if and only if $a$ and $k$ are even. So $b$ can be any even/odd number $\le \min (15-a, 10)$ and $c = 15-a -b$.
Just count them. If $a=0$ then $b=1,3,...,9$ and $c= 15-b$ and there are $5$ suc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3685534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$ x\cos y + y \cos x = \pi$ , how do I find $y''(0)$ in easiest way possible? I considered brute force differentiating but that is very hard. I also tried expanding the cosines and '$y$' as a Taylor series but I don't think that helps much either
| $x\cos y + y\cos x = \pi$
What is $y(0)$?
$0\cos (y(0)) + y(0)\cos 0 = \pi\
y(0) = \pi\
What is $y'$?
$\cos y -y\sin x + (-x\sin y + \cos x)y' = 0\\
y'= \frac {y\sin x - \cos y}{\cos x - x\sin y} = \frac {u}{v}\\
$
And $y'(0)$?
$u(0) = \pi \sin 0 - \cos \pi = 1\\
v(0) = \cos 0 - 0\sin \pi = 1\\
y'(0) = 1$
$y'' = \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Double integral over $x^2+y^2 \le 1$ I am trying to calculate the double integral
$\displaystyle \iint_{x^2+y^2\leq 1} (\sin x+y+3)\,dA$
Here are my attempts so far:
1) I used polar coordinates
$ x= r \sin(\theta)$
$y= r \cos (\theta)$
where $\theta \in [0,2 \pi]$ and $r \in [0,1]$ which gives
$\displaystyle \int_0^... | There is no reason not to split this into three separate integrals:
$$\iint_{x^2+y^2\leq 1} \sin x\;dA+\iint_{x^2+y^2\leq 1}y\;dA+\iint_{x^2+y^2\leq 1}3\;dA$$
You can do the third, I think. For the first, note that $\sin x$ is antisymmetric about the $y$-axis; and for the second, note that $y$ is antisymmetric about th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve equation: $\log_2 \left(1+ \frac{1}{a}\right) + \log_2 \left(1 +\frac{1}{b}\right)+ \log_2 \left(1 + \frac{1}{c}\right) = 2$ $$
\log_2 \left(1 + \frac{1}{a}\right) + \log_2 \left(1 + \frac{1}{b}\right)+ \log_2 \left(1 + \frac{1}{c}\right) = 2 \quad \text{where $a$, $b$, $c \in N$.}
$$
Apparently, the answer is ... | The problem, as you have written it, has no solution.
Simplifying the LHS, we get $$(a^2+1)(b^2+1)(c^2+1) = 4abc$$
But, by the AM-GM inequality, we get $x^2+1\ge2x$, which gives
$$(a^2+1)(b^2+1)(c^2+1) \ge 8abc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Ahmed integral revisited $\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx$ How to prove
$$\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} dx=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{... | To evaluate that integral we can use Feynman's trick:
$$I=\int _0^1\frac{\arctan \left(\sqrt{x^2+4}\right)}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx$$
$$I\left(a\right)=\int _0^1\frac{\arctan \left(a\sqrt{x^2+4}\right)}{\left(x^2+2\right)\sqrt{x^2+4}}\:dx$$
$$I'\left(a\right)=\int _0^1\frac{1}{\left(x^2+2\right)\left(a^2x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3693547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
${\lim_{x\rightarrow \infty}}(\sqrt{x^2+2x+3} - \sqrt{x^2+3})^{x}$ $${\lim_{x\rightarrow \infty}}(\sqrt{x^2+2x+3} - \sqrt{x^2+3})^{x}$$
I tried taking log both sides (on paper).
After taking log how do I proceed? You get $\infty$ * $(\infty-\infty$). But $\infty-\infty$ could be any number. How can I take this as 0? E... | $$A=\left(\sqrt{x^2+2x+3} - \sqrt{x^2+3}\right)^{x}$$ For what is inside the parenthesis, use Taylor
$$\sqrt{x^2+2x+3}=x+1+\frac{1}{x}-\frac{1}{x^2}+O\left(\frac{1}{x^3}\right)$$
$$\sqrt{x^2+3}=x+\frac{3}{2 x}+O\left(\frac{1}{x^3}\right)$$
$$\sqrt{x^2+2x+3} - \sqrt{x^2+3}=1-\frac{1}{2 x}-\frac{1}{x^2}+O\left(\frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
First trigonometric differential equation
Show that:
$$\tan(x) \frac {dy}{dx}-y=\sin^2(x)+2\sec(x)$$ where $y=\sin^2(x)-2\cos(x)$
I get:
$\frac {dy}{dx}=2sin(x)cos(x)+2sin(x)$
=$tan(x)(2sin(x)cos(x)+2sin(x))-sin^2(x)-2cos(x)$
From here I go into many directions but not towards the RHS. Guidance is much apprecia... | You have a little sign mistake here:
$$E=\tan(x)(2\sin(x)\cos(x)+2\sin(x))-\sin^2(x)\color{red}{+2\cos(x)}$$
$$E=2\sin^2(x)+2\dfrac {\sin^2(x)}{\cos x}-\sin^2(x)+2\cos(x)$$
$$E=\sin^2(x)+2\dfrac {\sin^2(x)}{\cos x}+2\cos(x)$$
$$E=\sin^2(x)+2\dfrac {(\sin^2(x)+\cos^2(x))}{\cos x}$$
$$E=\sin^2(x)+\dfrac 2{\cos x}=\sin^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
All prime divisors of $\frac{x^m+1}{x+1}$ are of the form $2km+1$. Let $m$ be an odd prime and $x$ be the product of all primes of the form $2km+1$. Then all prime divisors of $\frac{x^m+1}{x+1}$ are of the form $2km+1$.
What I know is that $\frac{x^m+1}{x+1}$ is an integer.
Here is the link to the answer which prompte... | We have $2 \nmid \frac{x^m+1}{x+1}$. Let odd prime $p$ divide $\frac{x^m+1}{x+1}$ :
Case $1$ : $m \mid (p-1)$
We clearly have $p=mq+1$ for some $q \in \mathbb{N}$. As $p$ is an odd prime, $mq+1$ is odd, and thus, $mq$ is even. Moreover, $m$ is an odd prime, thus, $q=2k$ for some $k \in \mathbb{N}$. Substituting:
$$p=2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is this series convergent? $1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + ...$ Is this series convergent? How to prove?
$$1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{8} \ (8 \times 1/8) + \frac{1}{16} + ...$$
It's equal to $\{1 + 1 + 1 + 1 + ...\}$, wh... | The most direct way to see divergence is by noticing that
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\leq 1+\frac{1}{2}+\frac{1}{2}+\cdots$$
where the series on the right hand side is what you have. Since the LHS diverges as the harmonic series, so must the RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit Using squeeze theorem Find the limit of the following function as $x \rightarrow 0$
$$
\frac{|x|}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right)
$$
$\lim _{x \rightarrow 0} \frac{|x|}{\sqrt{\left(x^{4}+4 x^{2}+77\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right)$
My approach,
applying s... | Other approach:
Let $f(x)=\frac{|x|}{\sqrt{\left(x^{4}+4 x^{2}+7\right)}} \sin \left(\frac{1}{3 \sqrt{x}}\right)$
We note that $f(x)=-f(-x)$, hence $f$ is odd, we can study it for $x > 0$.
Remembering that
$$
\lim_{x \to 0} \frac{\sin x}{x} = 1
$$
we get
$$
\begin{split}
\lim_{x \to 0^+} \frac{x}{\sqrt{\left(x^{4}+4 x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the percentage of the area of each circle that overlaps.
The diagram shows two overlapping circles with centres $X$ and $Y$. Both circles have radius $r$ cm and the distance between the centres, $XY$, is $1.5r$ cm.
I got the answer $18.2$% (to $3$s.f), but the answer key states otherwise ($14.4$%).
My Steps are:... | Let's assume that $\text{r}=1$ and we will work through the solution and then the OP can generalize it.
Well, we know that the equation of a circle is given by:
$$\left(x-\text{a}\right)^2+\left(\text{y}-\text{b}\right)^2=\text{r}^2\tag1$$
Where $\left(\text{a},\text{b}\right)$ are the center coordinates of the circle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$ Reducing this whole expression i finally came to this
$$\sum \limits_{cyc}\left(ab^4+a^4b+a^2b^2c\right)\geq \sum \limits_{cyc}\left(a^3b^2+a^2b^3+a^3bc\right)$$
Here I am stuck. I ca... | We need to prove that:
$$(a+b+c)^2\prod_{cyc}(a+b)\geq4(ab+ac+bc)\sum_{cyc}(a^2b+a^2c)$$ or
$$\sum_{sym}(a^4b-a^3b^2-a^3bc+a^2b^2c)\geq0.$$
Now, let $a\geq b\geq c$.
Thus, $$\sum_{sym}(a^4b-a^3b^2-a^3bc+a^2b^2c)=\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4-abc(a^2-2ab+b^2))=$$
$$=\sum_{cyc}(a-b)^2(ab(a+b)-abc)=\sum_{cyc}(a-b)^2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\int \frac{x^2\,dx}{(a-bx^2)^2}$ How do I integrate $\int \frac{x^2\,dx}{(a-bx^2)^2}$
I've tried substitution and partial fraction decomposition, but I'm not getting anywhere.
| This is almost similar to the other integral you posted. The trig substitution $x=\frac{\sqrt a}{\sqrt b} \sin t\implies dx = \frac{\sqrt a}{\sqrt b} \cos t dt $ works here too, after which you get $$ \frac{1}{b\sqrt{ab}}\int\frac{\sin^2t}{\cos^3 t} dt =\frac{}1{b\sqrt{ab}}\int\tan^2t\ \sec t \ dt $$
Now substitute $u=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Integrate $\int_0^{2\pi}\frac{\ln(a + b\cos x)}{c + d\cos x} dx$, Residue theorem I have recently been given a challenge problem in my Complex Analysis class. Suppose $a > b > 0$ and $c > d > 0$. Evaluate
$$\int_0^{2\pi} \frac{\ln(a + b\cos x)}{c + d\cos x} dx$$
using the Residue Theorem. Unfortunately, I don't even kn... | Choose $r, s \in (0, 1)$ so that
$$\frac{b}{a} = \frac{2r}{1+r^2}, \qquad \frac{d}{c} = \frac{2s}{1+s^2}.$$
Then
$$ \left| (1 - re^{i\theta})(1 - re^{-i\theta}) \right| = 1 + r^2 - 2r\cos\theta. $$
From this, we get
\begin{align*}
\log(1 + r^2 - 2r\cos\theta)
&= \log\left|1 - re^{i\theta}\right| + \log\left|1 - re^{-i\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find Transition matrix given two basis Consider the ordered bases $B=(\begin{bmatrix}2 & -1\\0 & −1\end{bmatrix},\begin{bmatrix}3 & -3\\0& −1\end{bmatrix}, \begin{bmatrix}-2 & -3\\0 & 2\end{bmatrix})$ and $C=(\begin{bmatrix}-2 & 2\\0& 2\end{bmatrix},\begin{bmatrix}2 & -3\\0& −3\end{bmatrix},\begin{bmatrix}0 & -2\\0& 0\... | I believe the following approach may be what you are looking for. Consider adding the missing basis matrix to each set, unrolling into $4 \times 4$ matrices, then taking the appropriate products.
The procedure I am suggesting is adding $\begin{bmatrix}
0 & 0 \\
1 & 0 \\
\end{bmatrix}$ to each set:
$$
B=
\begin{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
solving simultaneous linear congruences and simplifying the mod I wanted to solve the following:
$x\equiv2 (\text{mod 3})$
$x\equiv5 (\text{mod 6})$
$x\equiv7 (\text{mod 8})$
I rewrote the first congruence as $x=2+3a$ and in (mod 6), this can be written $2+3a\equiv5(\text{mod 6})$, which has solution $a\equiv1 (\text{m... | Chinese remainder thereom says we can solve a system uniquely for the product of the moduli if the moduli are relatively prime. In this case the are not relatively prime. But, if the equations are consistent, we can solve them uniquely for the least common multiple of the moduli... if the equations are consistent.
We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number Theory And Vieta Jumping $\textbf{Question:}$Find all positive integers $a, b$ such that the expression $$\frac{a^2+b^2+1}{ab-1}$$ is an integer.
$$$$As the expression is symmetric in $a, b$, so let $a \geq b$. It is easy to check the cases when $a=b+k$ where $0 \leq k \leq 2$, so let $a \geq b+3$. Suppose $a, b... | There are infinitely many solutions with $k=3$ and $k=6.$ For 3 the generating point is $(2,2)$ For $k=6$ the points are $(2,1)$ and $(1,2)$
More solutions can be found in either case by Vieta jumping.
Ruling out larger $k$ is by inequalities.
See how, as soon as $k \geq 7,$ the arc of the hyperbola passes through ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find inverse element of $1+2\alpha$ in $\mathbb{F}_9$
Let $$\mathbb{F}_9 = \frac{\mathbb{F}_3[x]}{(x^2+1)}$$ and consider $\alpha = \bar{x}$. Compute $(1+2 \alpha)^{-1}$
I think I should use the extended Euclidean algorithm: so I divide $x^2 +1 $ by $(1+2x)$:
$$x^2 + 1 = (1+2x)(2x+2)+2$$
$$(2x+2)(1+2x) + 2(x^2+1) = ... | $(2x+2)(2x+1)=2(x+1)(2x+1)=2(2x^2+3x+1).$
Since we are in $\mathbb{F}_3,$ then $3x=0$ and $4x^2=x^2$ so
$(2x+2)(2x+1)=2(2x^2+1)=4x^2+2=x^2+2=(x^2+1)+1,$
reducing modulo $(x^2+1)$ we get $1,$ so yeah, you are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$A^2 + B^2 + C^2 = D^2 + E^2 + F^2$ and $A^2 + F^2 = B^2 + E^2 = C^2 + D^2$, distinct positive integers I am looking for parametric formula to this system of equations:
$A^2 + B^2 + C^2 = D^2 + E^2 + F^2$
$A^2 + F^2 = B^2 + E^2 = C^2 + D^2$
for distinct positive integers $A,B,C,D,E,F$. According to my computations, sol... | Since the equations are all homogeneous of degree $2$, we can multiply a solution by a constant to produce another solution.
We may assume we have a primitive solution, i.e. $\gcd(A,B,C,D,E,F)=1$.
Let's start with $A^2 + F^2 = B^2 + E^2 = C^2 + D^2$. Call this common value $x$. Note that if $x \equiv 0 \mod 4$, all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3725386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric inequalities in a triangle What is the proof of
$~~\cos^2 A+\cos^2 B+\cos^2 C \leq 1 ~~$
in an acute triangle ?
This will be of help in finding the answer (if such exists) to finding the minimal T in any ∆ ABC when
$$T \geq \sin^k A+ \sin^k B+ \sin^k C~ ,~~~~ k \geq 3$$
| Let's first assume that $\cos^2A +\cos^2B+ \cos^2C=1$
Let's proceed to solve this
$\cos^2A +\cos^2B+ \cos^2C=1$
$\Rightarrow \cos^2A + \cos^2B - (1−\cos^2C)=0$
$\Rightarrow \cos^2A + cos^2B − sin^2C=0$
$\Rightarrow \cos^2A+(cos(B+C)cos(B−C))=0$
$\Rightarrow \cos^2A+(cos(π−A)cos(B−C))=0$ as $A+B+C = π$
$\Rightarrow \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
For how many natural numbers(<=100) is $1111^n +2222^n+3333^n+4444^n$ divisible by 10? For how many natural numbers (0 not included) $n \leq 100$ is $1111^n +2222^n+3333^n+4444^n$ divisible by 10?
I factored out $1111^n$ and got $1111^n(1+2^n+3^n+4^n)$. So $1+2^n+3^n+4^n$ must be divisible by 10. I figured out that thi... | If you divide $1^n$, $2^n$, $3^n$, and $4^n$ by 10, each goes through a cycle of remainders:
$1: 1, 1, 1, 1$
$2: 2, 4, 8, 6$
$3: 3, 9, 7, 1$
$4: 4, 6, 4, 6$
So $1^n+2^n +3^n + 4^n$ goes through the cycle of remainders $0, 0, 0, 4$, and thus will be divisible by $10$ whenever $n$ is not divisible by $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$ Evaluate $$\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$$
I tried substitutions like $u=\frac{\pi}{4}-x$, and trig identities like... | Rewriting the integral where you left off:
$$4\int_0^{\frac{\pi}{4}} \frac{\left(\sin{(5x)}\cos{x}-\cos{(5x)}\sin{x}\right) \left(\sin{(5x)}\cos{x}+\cos{(5x)}\sin{x}\right)}{\sin^2{(2x)}} \;dx$$
$$=4\int_0^{\frac{\pi}{4}} \frac{\sin{(4x)}\sin{(6x)}}{\sin^2{(2x)}} \; dx$$
$$=8\int_0^{\frac{\pi}{4}} \frac{\cos{(2x)}\sin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$ (I need a review of my resolution please :) ) Find the limit:
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$$
This is what I have, i'm not sure about my answer (I'm ... | Your answer is correct.
The limit can be split into parts if each of them exists as follows
$$\lim_{x \to 1} \cos(\pi \cdot x) \cdot \sqrt{\frac{(x-1)^2}{(x^2-1)}} \cdot \frac{1-x}{x^2+x-1}$$
$$=\lim_{x \to 1} \cos(\pi \cdot x) \cdot \lim_{x \to 1}\sqrt{\frac{x-1}{x+1}} \cdot \lim_{x \to 1}\frac{1-x}{x^2+x-1}$$
$$=(-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculating $ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$ I need to find the limit of:
$$ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$$
I've tried throught "rationalization" and completing squares... This is my work so far (i... | Disclaimer : This method is probably overkill, you can make do with simpler arguments on limits.
What you want to do is find estimates of the numerator and denominator.
The main tool here is the following estimate for $y \rightarrow 0$ : $\sqrt{1+y} = 1 + O(y) $, where $O(y^r)$ means : something that is comparable to o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\det ((A + B + C) (A^3 + B^3 + C^3-3ABC))\geq 0 $
Suppose that A, B and C are 2x2 matrices that switch between each
other. Prove that
$$\det ((A + B + C) (A^3 + B^3 + C^3-3ABC))\geq 0. $$
I did
$$A^3+B^3+C^3-3ABC=\frac12(A+B+C)((A-B)^2+(A-C)^2+(B-C)^2)$$
So, this determinant is equivalent to
$$\frac14[\de... | Based on Aryaman Maithani's idea:
Presumably the matrices are real. Let $X=A-B,\,Y=B-C$ and $Z=C-A$. Then $X,Y,Z$ commute and $X+Y+Z=0$. Let $\omega$ be a primitive cube root of unity. Then
\begin{aligned}
\det(X^2+Y^2+Z^2)
&=\det(X^2+Y^2+(-X-Y)^2)\\
&=4\det(X^2+Y^2+XY)\\
&=4\det\left[(X-\omega Y)(X-\bar{\omega}Y)\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $a+b+c=k$ and $a^2+b^2+c^2 =2k$ what is the maximum value of $k$? $a,b,c$ are real numbers and they satisfy the following equations.
$a+b+c=k$
$a^2+b^2+c^2=2k$
Find the maximum value of $k$.
I tried substituting for k in the second equation from the first and got
$a^2+b^2+c^2=2(a+b+c)$
Rearranging the terms I got
$a... | Since $c=k-(a+b)$ both inequalities are fulfilled iff the set of points $(a;b)$ such that
$$ a^2 + b^2 + (k-(a+b))^2 = 2k $$
is non-empty. This equation can be written as
$$ 2a^2+2ab+2b^2-2ka-2kb+(k^2-2k) = 0 $$
or, by letting $a=A+\frac{k}{3},b=B+\frac{k}{3}$, as
$$ 2A^2+2AB+2B^2 = 2k-\frac{k^2}{3}.$$
The matrix $\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $2x^3-9x^2-10x+13$ for $x=3+\sqrt{5}$. Is there an efficient way to tell that this reduces to $1$? I was helping my sibling with a math problem from a past year paper for a competitive exam. It requires us to evaluate this cubic expression for a given value of $x$ which has an $a+b$ form where $b$ is a squar... | if $5 = (x-3)^2$ then $-10x = -2x(x-3)^2 = -2x^3+12x^2-18x$
Substituting this into $2x^3 - 9x^2 - 10x + 13$
we have
$$\begin{align}
2x^3 - 9x^2 +(-2x^3+12x^2-18x) + 13 &= 3x^2-18x+13\\
&= 3x^2-18x+27-14\\
&= 3(x^2-6x+9)-14\\
&= 3(x-3)^2-14\\
&= 3\cdot 5-14\\
&= 15-14\\
&= 1\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?
| Well you got
$\lim_{x\to 0}\frac{\sec^2x-\cos x}{3x^2}$
$\frac{1}{3}\lim_{x\to 0}\frac{\frac{1}{\cos^2x}-\cos x}{x^2}$
$\frac{1}{3}\lim_{x\to 0}\frac{1-\cos^3x}{x^2\cos^2x}$
$\frac{1}{3}\lim_{x\to 0}\frac{(1-\cos x)(1+\cos x+\cos^2x)}{x^2\cos^2x}$
$\frac{1}{3}\lim_{x\to 0}\frac{(1+\cos x+\cos^2x)}{\cos^2x}\lim_{x\to 0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Bound for $\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}$. In the book Complex analysis by Bak J. & Newman J., chapter 11, talks about Sums Involving Binomial coefficients and find a bound $\frac{16}{9}\sqrt{3}$ for $|(z-1)^2(z+1)|$ in the "Example 3" on the unit circle, my way was using lagrange multipliers in this fo... | The first term in the asymptotic expansion follows from the analysis given by 'Asymptotics of a Family of Binomial Sums' by R. Noble. Putting this in a form for comparison one gets
$$ \sum_{k=0}^n (-1)^k\binom{3n}{k} \binom{n}{k} \sim (-4)^n \frac{2^{1/4}}{\sqrt{\pi n}} \cos(n \tan^{-1}(10\sqrt{2}/23) + \tan^{-1}((1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solution verification: $ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $ This limit is not too difficult but I was just wondering if my work/solution looked good?
Thanks so much for your input!!
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $$
$$ 2 x - 6 = 2 x \left( 1 - \frac 6 { 2... | Your solution is ok, but a bit verbose.
To remedy to that, I personally suggest working around zero by setting $x=3+u$ with $u\to 0$, I find it triggers natural reflexes more. Also for presentation purposes, I prefer working on the expression and then make use of $\to$ to specify the limit rather than carrying the $\ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Modified central binomial coefficients generating function Given $n \in \mathbb{N}$, I would like to find the ordinary generating function of the sequence $a_k = \binom{2n-2k}{n-k}$.
If
\begin{align}
A(x) = \sum_{k = 0}^\infty a_kx^k,
\end{align}
then I find that
\begin{align}
A(x) &= \sum_{k = 0}^n \binom{2n-2k}{n-k}x... | We consider $a_{n,k}=\binom{2n-2k}{n-k}$ with $n,k\geq 0$ non-negative integers.
*
*Horizontal GF: First of all we note that
\begin{align*}
A_n(x)=\sum_{k=0}^na_{n,k}x^k=\sum_{k=0}^n\binom{2n-2k}{n-k}x^k\qquad\qquad n\geq 0
\end{align*}
is a polynomial in $x$ and as such a perfect ordinary generating function, a so-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Closed form sought for $a_1 = a_2 = 1, a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i $ where $n>2$ I've been working through a problem that I've got as far as getting a recursive answer to. I was hoping to turn this into more of a "closed form" answer, but haven't really gotten anywhere. I'm hoping that someone can help w... | This might help by converting it into an order 2 recurrence:
$$
a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i
$$
$$
a_{n-1} = 1 + \frac{2}{n-1} \sum_{i=1}^{n-3} a_i
$$
therefore
$$
\sum_{i=1}^{n-3} a_i = \frac{(a_{n-1}-1)(n-1)}{2}
$$
and
$$
\sum_{i=1}^{n-2} a_i = a_{n-2} + \sum_{i=1}^{n-3} a_i
$$
$$
a_n = 1 + \frac{2}{n} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Is the inequality true for all $n\geq 2$? Let $x,y,z>0$. I am wondering if the following inequality is true?
$$\sum_{cyc}\frac{x^n}{y^2+yz+z^2}\geq\frac{x^{2n-2}+y^{2n-2}+z^{2n-2}}{x^n+y^n+z^n},\qquad n\geq 2$$
If not, is it known for which $n$ it is true?
$\displaystyle(1)\qquad\sum_{cyc}\dfrac{x^2}{y^2+yz+z^2}\overse... | It's wrong for any $n\geq5$.
For $n=5$ try $x=1.1$ and $y=z=1$.
For $n=4$ and $n=3$ it's true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Perturbation Integral How to show that for $f(x,m)$ = $\frac{mx^2}{2} + \frac{x^4}{4}$,
the integral $Z(\lambda)$ = $\frac{1}{\sqrt \lambda}$$\int_{-\infty}^{\infty} dz e^{-f(z)/\lambda}$ is $\sqrt{\frac{m}{2\lambda}}$$e^{\frac{m^2}{8\lambda}}$
$K_{1/4}(\frac{m^2}{8\lambda})$ for $m>0$?
$m$ here is a mass term, $\lam... | Start with
\begin{eqnarray}
A &=& \frac{1}{\sqrt{\lambda}} \exp\left[-\frac{1}{\lambda}\left(\frac{m}{2} x^2 +\frac{1}{4}x^4\right)\right]
\\
&=& \frac{1}{\sqrt{\lambda}} \exp\left[\frac{m^2}{8 \lambda}\right]\exp\left\{-\frac{m^2}{8 \lambda}\left[2 \left(\frac{x^2}{m}+1 \right)^2 -1 \right]\right\} \, .
\end{eqnarray}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating Sum at bounds I have to find an expression in terms of n using standard results for $$\sum_{r=n+1}^{2n} r(r+1)$$
And have found the general equation
$$\sum_{r=n+1}^{2n} r(r+1) = \frac{2n^3+6n^2+4n}{6}$$
However evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n+1)^3+6(n+1)^2+4(n+1)}{6}$$
does n... | Another approach: it's clear that the result is a polynomial of $n$ of degree $3$, let $$\sum\limits_{r=n+1}^{2n}(r^2+r)=An^3+Bn^2+Cn+D=P(n)$$
thus
\begin{align*}P(n)-P(n-1)&=\sum\limits_{r=n+1}^{2n}(r^2+r)-\sum\limits_{r=n}^{2n-2}(r^2+r)\\
&=-(n^2+n)+((2n-1)^2+(2n-1))+((2n)^2+2n)\\
&=7n^2-n\\
&\equiv A(3n^2-3n+1)+B(2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to decouple this system of 2nd order partial differential equations? Solving a problem I found this system of equations:
$$ (\partial _t ^2 + \partial _x ^2 + \partial_y ^2 + \partial_z ^2)a_1(\vec{x},t) - 4gB(x\partial_y - y\partial_x)a_{\color{Red}2}(\vec{x},t) - 4g^2B^2(x^2 + y^2)a_1 - \lambda a_1 = 0 $$
$$ (\p... | Here's a linear algebra approach. I'll rearrange this slightly to make the next steps a bit clearer. I'm going to use $\Delta$ to represent $\partial_t^2 + \partial_x^2 + \partial_y^2 + \partial_z^2$. If it were $-\Delta_t^2$, I'd use the standard $\square$ d'Alembert notation instead. I'll also write $x\partial_y - y\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Problem with proving inequalities Question:
Prove that if $x,y,z$ are positive real numbers such that $x+y+z=a$ then $(a-x)(a-y)(a-z)>\frac8{27}a^3$ is not true.
My Approach:
$$\frac{a-x}{2}=\frac{y+z}2$$
$$\frac{a-y}{2}=\frac{x+z}2$$
$$\frac{a-z}{2}=\frac{x+y}2$$
Using $AM>GM$ we get $$\frac{x+y+z}{3}>\root 3 \of {x... | Using AM-GM identity,
$\implies$ $[(a-x)(a-y)(a-z)]^{1/3} \le \frac{(a-x)+(a-y)+(a-z)}{3}$$
Then we end up with our desired result
$$\implies (a-x)(a-y)(a-z) \le \frac{8a^3}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers? $\textbf{Question:}$Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers?
that i... | Observe that
*
*Clearly $ a > 1, b > 1 , c > 1$.
*WLOG $ a < b < c$
*$ 0 \equiv c^2 \equiv c(c+a+b) \equiv (a+b+c)(a+b+c) \pmod{a+b}$ (and similar expressions)
*$\gcd(a+b, b+c ) \mid \gcd( c^2, a^2) = 1$ so $\gcd(a+b, b+c) = 1$. (and similar expressions)
Hence
$$ (a+b)(b+c)(c+a) \mid (a+b+c)^2 \Rightarrow (a+b)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\int_0^{\pi/2} \frac{\sin x}{\sin^{2n+1}x +\cos^{2n+1}x} dx$? I have an exercise to evalute the following integral for all $n\geq 1 $
$$I(n)=\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n+1} x+\cos^{2n+1} x}dx$$
I attempted to find the closed form for the integral above in the following manner, where I ... | This integral appear in the Jozsef Wildt International mathematical competition proposed by Ovidui Furdui and Alina Sintamarian which I solved in the following way , couples of months back.
For all $n\geq 2$ we shall show that
$$I(n)=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n-1}x+\cos^{2n-1}x}dx=\frac{\pi}{2n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Using partial information to factor $x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ I wish to find exact expressions for all roots of $p(x)=x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ By observing that for the roots $x_0 \pm iy_0, x_0 \approx -0.15883609808599033632, y_0 \approx 0.27511219196092896700,$ we have that $x_0$ is the unique real ro... | Just for the fun, starting from @Michael Rozenberg's answer, we need to solve the two cubic equations
$$x^3+\frac{3}{2} \left(1-i \sqrt{3}\right) x^2-2 \left(1+i \sqrt{3}\right) x-1=0$$
$$x^3+\frac{3}{2} \left(1+i \sqrt{3}\right) x^2-2 \left(1-i \sqrt{3}\right) x-1=0$$
Prepare yourself for nasty solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Convergence of $\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$ Hello I am a high school student from germany and I am starting to study math this october. I am trying to prepare myself for the analysis class which I will attend so I got some analysis problems from my older cousin who also studied maths. But I am... | You cannot prove
$$\frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} \leq 1
$$
because the left-hand side tends to $+\infty$ for $n \to \infty$.
You should compare your series with $\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ instead:
$$
\frac{2}{\sqrt{n^3+1}+\sqrt{n^3-1}}=\frac{2}{n^{3/2}} \cdot \frac{1}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve $x^2+3y = u^2$ and $y^2+3x=v^2$ in positive integers. The question is from the pg - 59 from ' An Introduction to Diophantine Equations ' by Titu Andreescu , Dorin Andrica , Ion Cucurezeanu.
Example 1 : Solve in positive system of equations in positive integers
$$\begin{cases} x^2+3y = u^2 \\ y^2 + 3x = v^2 \end... | $\begin{cases} x^2+3y = u^2 \\ y^2 + 3x = v^2 \end{cases}$ -----(1)
"OP" gave numerical solution to equation (1) as:
$(v,u,x,y)=(2,2,1,1)=(13,17,16,11)$
There is another numerical solution & is given below:
$(v,u,x,y)=[(10),(25/4),(13/4),(19/2)]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
On Resolving a Fuss on Squares and Fractions over a few Inequalities Firstly, only AM-GM and C-S are to be sought.
$1.$Let $a, b, c, d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Show that
$${a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge 4$$
In my textbook, this problem is credited to Michael Roze... | The fourth problem it's just Minkowski (triangle inequality), but also, after squaring of the both sides, we can use C-S:
$$\sqrt{(a_i^2+b_i^2)(a_j^2+b_j^2)}\geq a_ia_j+b_ib_j$$ and we obtain an identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
Initially, this seemed like one could work it out with $AM-GM$, but it doesn't see... | The mean of the squares $1^2, \ldots, n^2$ is
$$ f(n) = \frac{1}{n} \sum_{i=1}^n i^2 = \frac{2n^2+3n+1}{6}$$
It is an integer if and only if $n \equiv 1$ or $5 \mod 6$.
The first $n > 1$ for which it is a square is $337$, where $f(337) = 38025 = 195^2$. There are infinitely many. See OEIS sequence A084231.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
2010 USAMO #5:Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for integers $m$ and $n$, then $m - n$ is divisible by $p$. Let $q = \frac{3p-5}{2}$ where $p$ is an odd prime, and let $S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7} + \cdots + \frac{1}{q(q+1)(q+2)}
$
Prove that if $\frac{1}{p}-2S_q = \frac{... | With the help of @user10354138 's hints, I think I got the solution.I will be grateful if someone proof reads it.
Note that $$2S_q = 2\sum_{x=1}^{\frac{q+1}{3}} \frac{1}{(3x-1)(3x)(3x+1)} = \sum_{x=1}^{\frac{p-1}{2}} \left[\frac{1}{3x(3x-1)}-\frac{1}{3x(3x+1)}\right]\\
=\sum_{x=1}^{\frac{p-1}{2}} \left[ \frac{1}{3x-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Probability of exactly $2$ sixes in $3$ dice rolls where $2$ dice have $6$ on $2$ faces? Three dice are rolled. One is fair and the other two have 6 on two faces.
Find the probability of rolling exactly 2 sixes.
My textbook gives an answer of $\frac{20}{147}$ but I get an answer of: $$\frac{1}{6}\frac{2}{6}\frac{4}{6}+... | Your solution seems correct, indeed also by the naif definition of probability we obtain
$$p=\frac{\text{#favorauble cases}}{\text{#total cases}}=\frac{8+8+20}{6^3}=\frac16$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$6$-digit permutation problem How many $6$-digit different numbers greater than $400000$ can be formed from the digits $2$, $2$, $5$, $6$, $7$, $7$, $8$?
Note: $22$ and $77$ are allowed in the number. For instance, $622775$ is allowed.
My approach: $5$ ways to choose the first digit, $6$ ways to choose the second, $5$ ... | We apply generating function to the problem. First, we find how many 6-digit number can be formed from 2, 2, 5, 6, 7, 7, 8 starting with any digit. Notice that 2, 2, 7, 7, can appeal at most 2, which corresponds to $(1 + x + \frac{x^{2}}{2!})^{2}$. Also, 5, 6, 8 can appeal at most 1, which corresponds to $(1 + x)^{3}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove $\sum_{b=0}^{n-1}b\binom{n+1-b}{2} = \binom{n+2}{4}$
An equilateral triangle of side $n$ is divided into $n^2$ equilateral
triangles of side $1$ such that each side of the congruent triangles
is parallel to the original triangle. Find the number of
parallelograms that can be formed by the segments.
The answer t... | Marcus M has given a very nice combinatorial proof, which in general I prefer. Here’s a computational proof, in case you’re more comfortable with that approach.
$$\begin{align*}
\sum_{b=0}^{n-1}b\binom{n-1-b}2&\overset{(0)}=\sum_{k=0}^{n-1}(n-1-k)\binom{k}2\\
&=n\sum_{k=0}^{n-1}\binom{k}2-\sum_{k=0}^{n-1}(k+1)\binom{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving : $5^{x^2+6x+8}$ = 1
Solve for $x$:
$$5^{x^2+6x+8} = 1$$
So, I took the natural logarithm on both sides,
$$(x^2+6x+8)\ln(5) = \ln(1)$$
then I divide both sides by $\ln(5)$ to set the polynomial to zero because we know $\ln(1) = 0$.
I will be left with:
$$x^2+6x+8 = 0$$
Factoring this will give:
$$(x+2)(x+4) =... | Another way to reach the same equation consists in noticing that the exponential function $a^{x}$ is injective:
\begin{align*}
5^{x^{2} + 6x + 8} = 1 = 5^{0} \Longleftrightarrow x^{2} + 6x + 8 = 0 \Longleftrightarrow \ldots
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
What is $\lim_{N\to\infty}\frac{-2}{\pi}\sum_{n=1}^N \frac{(-1)^n}{n} \sin(n\frac{N\pi}{N+1})$? Here is what I have so far:
$$\lim_{N\to\infty} f_N \left(\frac{N\pi}{N+1}\right)$$
$$f_N (x) = \frac{-2}{\pi}\sum_{n=1}^N \frac{(-1)^n}{n} \sin(nx)$$
$$\implies \lim_{N\to\infty} f_N \left( \frac{N\pi}{N+1}\right)=\lim_{N\t... | Since
$$
\sin \left( {n\frac{{N\pi }}{{N + 1}}} \right) = \sin \left( {\pi n - \frac{{\pi n}}{{N + 1}}} \right) = ( - 1)^{n+1} \sin \left( {\frac{{\pi n}}{{N + 1}}} \right),
$$
we have
$$
\frac{2}{{\pi (N + 1)}}\sum\limits_{n = 1}^N {\frac{{N + 1}}{n}\sin \left( {\frac{{\pi n}}{{N + 1}}} \right)} \to \frac{2}{\pi }\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
how to find the geometric centroid of a trapezoid? I'm doing a project involving a trapezoidal prism
I need to find the center of mass. To do this, I need to first find the geometric centroid of the trapezoid.
I have found the formulas online. According to Wolfram Mathworld,
$$\begin{align}
\bar{x} &= \frac{b}{2} + \f... | If you choose a coordinate system where $b$ is on the positive $x$ axis, between origin $(0, 0)$ and $(b, 0)$, the four vertices of the trapezoid are
$$(0, 0), \quad (b, 0), \quad (f+a, h), \quad (f, h)$$
Since trapezoids are simple polygons, we can use the shoelace formula for the area of a 2D polygon,
$$A = \displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Better methods to approximate $2^{2\over 3}$ Recently while solving a problem on thermodynamics I ended up with $2^{2\over 3}$ .
Now the problem was on a test where no calculators were allowed and answer was required upto $2$ decimal digits.
I then resorted to binomial theorem for help (for $x\lt 1$) $$\left. \begin{ar... | Perhaps more elaboration on how to use Newton's method, since I personally find it very easy to use to find the first few digits of $n$th roots using only basic calculations.
From the binomial expansion, we know that
$$(x+\Delta x)^3=x^3+3x^2\Delta x+\mathcal O((\Delta x)^2)$$
Our goal is essentially finding the $\Delt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Need help with $\arccos$ equation I have the equation
$$ \cos(2x + \frac{\pi}{9}) = 0.5$$
I know that in order to solve for $x\in \Bbb R$, I need to use
$$\arccos(0.5) = 2x + \frac{\pi}{9} $$
This yields
$$ 2x + \frac{\pi}{9} =
\begin{cases}
\frac{\pi}{3} + 2k\pi, & \text{Positive angle} \\
2 \pi - \frac{\pi}{3}+ 2k\... | Remember that $\;\cos x=\alpha\implies x=\pm\arccos x\;$. Besides this, we only need basic trigonometry to solve that equation:
$$\cos t=\frac12\iff t=\pm\frac\pi3+2k\pi\;,\;\;k\in\Bbb Z\implies$$
puting $\;t=2x+\frac\pi9\;$ we get
$$2x+\frac\pi9=\pm\frac\pi3+2k\pi\implies\begin{cases}2x=\cfrac{2\pi}9+2k\pi,&\text{(pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + ... | You are right that $(x-1)^{98}\equiv1\pmod{(x-2)}$. But that implies
$$(x-1)^{100}\equiv(x-1)^2=x(x-2)+1\equiv1\pmod{x-2}.$$
More naively, as
$$x-1\equiv1\pmod{x-2}$$
then
$$(x-1)^{100}\equiv1^{100}=1\pmod{x-2}.$$
Similarly,
$$x-2\equiv-1\pmod{x-1}$$
and
$$(x-2)^{200}\equiv(-1)^{200}=1\pmod{x-1}.$$
So $(x-1)^{100}+(x-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Given $\sinh x$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$ Given $\sinh x = 8/14$, find the exact value of $\cosh x, \cos 2x$ and $\tan 2x$.
I have been getting two answers which has made me confused. I keep getting $\sqrt{65/7}$ or $\sqrt{4/7}.$
That's what I got: $$\cosh^2 x - \sinh^2 x = 1$$ $$\cosh^... | $$\sinh x =\frac{e^x-e^{-x}}{2}= \frac{4}{7}$$
We want
$$\cosh x = \frac{e^x+e^{-x}}{2} = q$$
Adding these:
$$e^x = \frac{4}{7}+q$$
Subtracting the first equation from the second:
$$e^{-x} = q - \frac{4}{7}$$
So $$\frac{4}{7}+q = \frac{1}{q-\frac{4}{7}}$$
$$q^2 - \left(\frac{4}{7}\right)^2 =1$$
$$q^2 =1+\frac{16}{49}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Elegant way of finding the least perimeter of triangle A triangle $ABC$ has positive integer sides, $\angle A = 2\angle B$ and $\angle C > \pi/2$ , then the minimum length of the perimeter of $ABC$ is?
We have $\angle A = 2\angle B$
$\Rightarrow \sin A=\sin 2B=2 \sin B \cos B $
$\sin C=\sin(\pi-3B)=\sin(3B)=3\sin B-4\s... | First of all, it is consider very bad style to just throw random equations around. Write down in English what exactly you are doing --- is it an assumption you make? A given condition? Some logical deduction from earlier? And every sentence should start with an English word not an equation, unless you absolutely hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Homogenous Equation with Complex Vector Solution: Converting to Real Functions Solving the system
$$
\begin{array}{l}\frac{d x}{d t}=6 x-y \\ \frac{d y}{d t}=5 x+4 y\end{array}
$$
we get $\lambda_{1}=5+2 i, \lambda_{2}=5-2 i$ eigenvalues. So eigenvectors and corresponding solutions is:
$$
\mathbf{K}_{1}=\left(\begin{ar... | We have
$$\lambda_1 = 5 + 2 i, v_1 = \begin{pmatrix} 1 \\ 1 - 2 i \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} +i \begin{pmatrix} 0 \\ - 2 \end{pmatrix}$$
We form $x_1(t) = e^{\lambda_1 t}v_1$, while expanding using Euler's Formula and this approach
$$x_1(t) = e^{5t}(\cos 2 t + i \sin 2t) \left(\begin{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does $\lim_{n\to \infty} \sum_{k=1}^n\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$ exist? Let $x \in \mathbb{R}.$ Is is true that the following limit exists : $$\lim_{n \to \infty} \sum_{k=1}^n\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$$ What is the value of this limit?
I tried the Integral test for convergence, but nothing ca... | We can write
$$
\sum\limits_{k = 1}^n {\log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)} = - \frac{{x^2 }}{{2n}}\sum\limits_{k = 1}^n {\sin ^2 k} + \sum\limits_{k = 1}^n {\left[ {\frac{{x^2 \sin ^2 k}}{{2n}} + \log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)} \right]} .
$$
Here
$$
\sum\limits_{k = 1}^n {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Systems of polynomial equations involving sums of equal powers Given the following system of polynomial equations:
$$
\left\{\begin{array}{lclclcr}
x & + & y & + & z & = & 1
\\
x^{2} & + & y^{2} & + & z^{2} & = & 14
\\
x^{3} & + & y^{3} & + & z^{3} & = & 36
\end{array}\right.
$$
What is $x^{5} + y^{5} + z^{5}\ {\large ... | Let $p=x+y+z$, $q=xy+yz+zx$, $r=xyz$. So we have
$$\begin{cases}
p=1\\p^2-2q=14\\p^3-3pq+3r=36
\end{cases}$$
$$\begin{cases}
p = 1\\q = -\frac{13}{2}\\ r = \frac{31}{6}
\end{cases}$$
By consecutive eliminating the highest powers terms we get
$$x^5+y^5+z^5-(x+y+z)^5+5(xy+yz+xz)(x+y+z)^3-5(xy+yz+xz)^2(x+y+z)-5(x+y+z)^2xy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The number of three digit numbers $abc$, which satisfy $a≤b>c$ is My approach in this question is as follows,
⇒When $a=1$
$-> 1-1-(0) / 1-2-(0,1) / 1-3-(0,1,2) ... / 1-9-(0,1...,8)$ ⇒Total numbers = $1+2+3+...+9$
⇒When $a=2$
$-> 2-2-(0,1) / 2-3-(0,1,2) ... / 2-9-(0,1...,8)$ ⇒Total numbers = $2+3+...+9$
⇒When $a=3$
$->... | ⇒When $a=1$
$-> 1-1-(0) / 1-2-(0,1) / 1-3-(0,1,2) ... / 1-9-(0,1...,8)$ ⇒Total numbers = $1+2+3+...+9$
⇒When $a=2$
$-> 2-2-(0,1) / 2-3-(0,1,2) ... / 2-9-(0,1...,8)$ ⇒Total numbers = $2+3+...+9$
⇒When $a=3$
$-> 3-3-(0,1,2) ... / 3-9-(0,1...,8)$ ⇒Total numbers = $3+...+9$
So on counting, we get total numbers as,
$$(1+2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Minimizing a function by finding its critical points Let $f_n(x)$ equal:
$$(2^n+2) \left(2x+1-\sqrt{2x^2+2x} \right)^n-x^n-(x+1)^n-\left(3x+1-2\sqrt{2x^2+2x}\right)^n-\left(3x+2-2\sqrt{2x^2+2x}\right)^n$$
Mathematica suggests that this function has two critical points on $(0,\infty)$, namely $x_1=1$ and $x_2=1/\sqrt{2}... | Some thoughts
It suffices to prove that $f_n(x) \ge 0$ for $x > 0$.
With the substitution $x = \frac{2}{u^2+2u-1}$ for $u > \sqrt{2} - 1$ (correspondingly, $u = \sqrt{2 + \frac{2}{x}} - 1$), we have
$$f_n(x) = \frac{1}{(u^2+2u-1)^n}
[(2^n+2)(u^2+1)^n - 2^n - (u+1)^{2n} - (u-1)^{2n} - 2^n u^{2n}].$$
It suffices to prove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Nested functions Assuming that we have some function $L(x)$ such that $L(x) = x - \frac{x^2}{4}.$ Now, define $a_n$ as $$L \Bigl( L \Bigl( L \Bigl( \cdots L \Bigl( \frac{17}{n} \Bigr) \cdots \Bigr) \Bigr) \Bigr),$$ where we have $n$ iterations of $L.$ My question here is, what value does $n \cdot a_n$ approach as $n$ a... | I'd like to post the solution I came up with.
Note that $0 < L(x) < x$ for $0 < x < 2.$ Assuming $n$ is sufficiently large, i.e. $n \ge 9,$ we have that $0 < a_n < \frac{17}{n} < 2.$
From $L(x) = x - \frac{x^2}{2},$ we can write
$$\frac{1}{L(x)} = \frac{1}{x - \frac{x^2}{2}} = \frac{2}{2x - x^2} = \frac{2}{x(2 - x)} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Proof that if $x,y>0$ and $x+y=1$, then $(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$ For positive reals $x$ and $y$ such that $x+y=1$, prove that $$(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$$
I have tried using Jensen’s inequality but it won’t cover all the possible choices for $x$ and $y$ since the concavity varies. I am t... | $$(2x)^{\frac 1 x}=\frac{1}{\left( \frac{1}{2x}\right) ^{\frac{1}{x}}}\leq \frac{1}{1+\frac{1}{x}\left( \frac{1}{2x}-1\right)}=\frac{2x^2}{2x^2-2x+1}$$ by Bernoulli’s inequality. The same holds for $y$ and one immediately computes that if $x+y=1$, $$\frac{2x^2}{2x^2-2x+1}+\frac{2y^2}{2y^2-2y+1}=2$$ and the result follo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Knowing that $ \sin(2x) \equiv 2 \sin(x)\cos(x) $ show that $\cos(2x) \equiv \cos^2x - \sin^2x $ Apparently one option is to differentiate the identity $ \sin(2x) \equiv 2 \sin(x)\cos(x) $ to get the identity $\cos(2x) \equiv \cos^2x - \sin^2x $. Which is surprising as I didn't realize that differentiating an identity ... | $$\cos^2 (2x) = 1 - \sin^2 (2x) = 1 - 4 \sin^2 x \cos^2 x = 1 - 4 \sin^2 x (1 - \sin^2 x)$$
$$ = 4 \sin^4 x - 4 \sin^2 x +1$$
Now if we let $u = \sin x$, we have $4u^4 - 4u^2 + 1 = (2u^2-1)^2$. Thus:
$$\cos^2 (2x) = (2 \sin^2 x - 1)^2$$
$$\cos^2 (2x) = (2 \sin^2 x - (\sin^2 x +\cos^2 x) )^2$$
and now you are very close... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show that $x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded, monotone, and find its limit Prove that $x_1 = 0, x_2 = 0, x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded and monotonic. Then find its limit.
My attempt at boundedness:
(Using induction) For the base case we have $0 \leq x_... | For boundedness we use Strong Induction, it is trivial that the sequence is positive.
We want to show that for all $n \in \mathbb{N}$ we have $x_{n} < 2$
*
*For k = 1 we have: $x_{1} = 0 < 2$
*Let $n \in \mathbb{N}$ and suppose that for all $k \leq n$ we have: $x_{k} < 2$
*We have: $x_{n-1} < 2$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$ Here is a new challenging problem:
Show that
$$I=\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx=2\ln(2)G-\frac{\pi}{8}\ln^2(2)-\frac{5\pi^3}{32}+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
My attempt:
With Weierstrass substitution we have
... | Many ways to go are possible!
A simple way would be to exploit the known result,
$$\int_0^1 \frac{\arctan(x)}{x}\log\left(\frac{1+x^2}{(1-x)^2}\right)=\frac{\pi^3}{16},\tag 1$$
since with the Weierstrass subs the main integral reduces to
$$\mathcal{I}=2\int_0^1\frac{\arctan(x)}{x}\log\left(\frac{1-x^2}{1+x^2}\right)\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Simplify $\sum_{k = 0}^n \left[ \binom{m + n + k}{k} 2^{n + 1 - k} - \binom{m + n + k + 1}{k} 2^{n - k} \right]$. This is Exercise 6 from page 44 of Analysis I by Amann and Escher.
Exercise:
Simplify the sum
\begin{align*}
S(m, n) := \sum_{k = 0}^n \left[ \binom{m + n + k}{k} 2^{n + 1 - k} - \binom{m + n + k + 1}{k} 2^... | Starting with
$$
\sum_{k = 0}^n \Bigg[ 2^{n - k} \Big[ \binom{m + n + k}{k} - \binom{m + n + k}{k - 1} \Big] \Bigg],
$$
and using the hint with $\ell=m+n+k$ and $j=k$, we get
$$
\sum_{k = 0}^n \Bigg[ 2^{n - k} \Big[ \binom{m+n+k+1}{k} - 2\binom{m+n+k}{k - 1} \Big] \Bigg]=\sum_{k = 0}^n\left(2^{n-k}\binom{m+n+k+1}{k}-2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Ramanujan's $\sqrt{\frac{\pi e}{2}}$ formula The following identity is due to Ramanujan:
$$\DeclareMathOperator{\k}{\vphantom{\sum}\vcenter{\LARGE K}} \sqrt{\frac{\pi e}{2}}=\frac{1}{1+\k_{n=1}^\infty \frac{n}{1}}+\sum_{n=0}^\infty\frac{1}{(2n+1)!!}$$
or
$$\sqrt{\frac{\pi e}{2}}=\cfrac{1}{1+\cfrac{1}{1+\cfrac{2}{1+\cfr... | As pointed out by the linked page, it suffices to prove
$$ 1+\dfrac{1}{1+\dfrac{2}{1+\dfrac{3}{1+\ddots}}} = \sqrt{\frac{2}{\pi e}} \frac{1}{\operatorname{erfc}(1/\sqrt{2})}. \tag{1} $$
To this end, we will resort to the standard theory of continued fraction. Define $(p_n)$ and $(q_n)$ by the following relation:
$$
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$ f $ is differentiable in $ (0,0). $ Definition: Let $V\subseteq{\mathbb{R}^{m}}$ an open set, $a\in V$ y $f\colon V\to\mathbb{R}^{n}$ a function. We will say that $f$ is differentiable in $a,$ if exists a linear transformation $f'(a)\colon\mathbb{R}^{m}\to\mathbb{R}^{n}$
such that
\begin{equation}
f(a+h)=f(a)+f'(a)(... | A somewhat different approach:
In order to be differentiable, a function must be continuous and have a continuous derivative (or have a derivative with an essential singularity). Continuity requires that the limit as you approach the point be the same, regardless of direction of your approach.
Suppose we approach alon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Change this integral $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$. It is given that $f(a+b+1 -x) = f(x)$ where $a$ and $b$ are positive real numbers then $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$ is equal to
*
*$\int_{a-1}^{b-1} f(x) dx$
*$\int_{a+1}^{b+1} f(x+1) dx$
*$\int_... | $$I=\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx=I_1+I_2$$
$$I_1=\frac{1}{a+b}\int_{a}^{b}xf(x) dx$$
$$I_2=\frac{1}{a+b} \int_{a}^{b} x f(x+1) dx = \frac{1}{a+b} \int_{a+1}^{b+1} (u-1)f(u) du$$
N0w apply the property:
$$\int_{p}^{q} g(x) dx=\int_{p}^{q} g(p+q-x) dx$$
Then $$I_2=\frac{1}{a+b} \int_{a+1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is this proof of $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ incomplete? So, for any angle $\alpha$ :
$$\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = \dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha} = \dfrac{\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}{\dfrac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}= ... | Here's a simple way of proving the identity:
$$\frac{1-\cos x}{\sin x}=\frac{1-(1-2\sin^2\frac{x}{2})}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2} $$
as required. I hope that was helpful:)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Factor $x^8 + x^2 + 13$ into irreducible polynomials in $\Bbb F_{23} [x]$. I want to factor $x^8+ x^2 + 13$ into irreducible polynomials in $\Bbb F_{23} [x]$. I am trying using the method given in this link but not able to find its factors. Any help would be appreciated.
Thanks.
Factor $X^4 + 3$ into irreducible facto... | I'd start by viewing this as $y^4+y+13$, where $y=x^2$. Does that factor? Search for linear factors by testing numbers as roots. It takes a little patience (or a calculator) but $y=8$ is a root. So you have $$(y-8)\left(y^3+8y^2-5y+7\right)$$
Does this factor further? Only if the cubic has a root. Checking for roots, i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the lowest value of $m$ if $m>2$ and $m^3-3m^2+2m$ is divisible by $79$ and $83$?
$m^3-3m^2+2m$ is divisible by $79$ and $83$ where $m>2$. Find the
lowest value of $m$
$m^3-3m^2+2m$ is the product of three consecutive integers. Both $79$ and $83$ are prime numbers. The product of three consecutive positive in... | The brute force method: as $m^3−3m^2+2m=m(m−1)(m−2)$, and $79,83$ are prime, you can just solve the following nine congruences: $m\equiv\alpha\pmod{79}$, $m\equiv\beta\pmod{83}$, where $\alpha,\beta\in\{0,1,2\}$. This is possible as per Chinese Remainder Theorem, and the smallest of the nine $m$'s you will get (greater... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\lim_{n\to\infty} a_n$, where $a_{n+1} = \sqrt{1+\frac12 a_n}$? Note: A similar question (same recursive function) has been asked here, but none of the answers is relevant to my question.
I am trying to evaluate $\lim_{n\to\infty} a_n$. The sequence $a_n$ is given by the recursive function $$a_{n+1} ... | Let $L = \frac{1+\sqrt{17}}{4}$. Then $L^2 = 1 + \frac{1}{2}L$ and $\sqrt{1 + \frac{1}{2}L} = L$.
Let us use the mathematical induction to prove: $0 < a_n < a_{n+1} < L$ for all $n \ge 2$.
For $n = 2$, it is easy to verify it.
Assume that the inequality is true for $n = k$ ($k\ge 2$), i.e., $0 < a_k < a_{k+1} < L$.
Let... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)? I am trying to evaluate the following limit:
$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$
Approach #1
$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sq... | For
$x\ne 2$,$$\sqrt{x^2-4}=\sqrt{x+2}\sqrt{x-2}$$
and
$$\sqrt{x-2}=\sqrt{\sqrt x+\sqrt2}\sqrt{\sqrt x-\sqrt2}$$
so that by simplification
$$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}
=\frac{\sqrt{x+2}\sqrt{\sqrt x+\sqrt2}+\sqrt{\sqrt x-\sqrt2}}{\sqrt{\sqrt x+\sqrt2}}.$$
The limit is $\sqrt{2+2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Counting solutions to $x_1 + x_2 + x_3 + x_4 = 12$ with at least one $x_i\ge 5$.
Count the non-negative integral solutions of $x_1 + x_2 + x_3 + x_4 = 12$ with at least one $x_i\ge 5$.
I got really confused about that problem and really would like to know about methods to solve this.
I have tried to substitute the va... | We can first get $4$ numbers adding to $7 = ^{(7+4-1)}C_{(4-1)} = 120$. As $5$ can be added to any of the $4$ numbers, multiply the answer by $4$.
Then we need to subtract duplicate arrangements -
$\{7,0,0,0\}$ arrangements that make arrangements of $\{7,5,0,0\}$ by adding $5$ are already covered in $\{2,5,0,0\}$ arra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $x_{n+1}=x_n(2-ax_n)$ converges and find the limit
Let $a>0$ and $x_0\in I=\left (\frac{1}{2a}, \frac{3}{2a}\right )$. Show that the sequence $(x_n)$, $n\geq 0$, $$x_{n+1}=x_n(2-ax_n), \quad n \geq 0$$ converges. Which is the limit? Hint: Consider $\phi (x)=x(2-ax)$ and show that $\phi (I)\subset \left [\fra... |
P1. $\phi(x)$ has a maximum at $x_m=\frac{1}{a}$ and $\phi(x_m)=x_m$. Thus $\phi(x)\leq \frac{1}{a}$.
P2. $\phi(x)$ is ascending on $\left(-\infty,\frac{1}{a}\right]$ and descending on $\left(\frac{1}{a},\infty\right)$.
Since $\phi'(x)=2-2ax$
P3. For $x\in\left[\frac{1}{2a},\frac{3}{2a}\right] \Rightarrow
\phi(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$ If $a+b+c=1$ and $a,b,c>0$ then prove that $$\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$$
My try : We have to prove: $$ab+bc+ca+36abc(ab+bc+ca)\ge 21abc$$ or after homogenising we get :
$$\sum a^4b+3\sum a^3b^2+6\sum a^2b^2c\ge 14\sum a^3bc$$
I don't know what to do nex... | Now, we need to prove that:
$$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2)+4\sum_{cyc}(a^3b^2+a^3c^2-2a^3bc)-6\sum_{cyc}(a^3bc-a^2b^2c)\geq0$$ or
$$\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4)+$$
$$+4\sum_{cyc}(c^3a^2+c^3b^2-2c^3bc)-3abc\sum_{cyc}(a^2+b^2-2ab)\geq0$$ or
$$\sum_{cyc}(a-b)^2(ab(a+b)+4c^3-3abc)\geq0,$$ which is true by AM-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Finding $\cos ( 2 \sin^{-1}( \frac{5}{ 13} )) $ The following problem is from the $8$th edition of the book Calculus, by James Stewart. It is problem number $9$ in section $6.6$.
Problem:
Find an exact value for the expression:
$$ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } $$
Answer:
\begin{align*}
... | Alternative solution:
Let $sin^{-1}(\frac{5}{13})=x$, then $\sin x=\frac{5}{13},\space \cos x=\frac{12}{13}$
$\cos 2x=1-2\sin^2x=1-2\times(\frac{5}{13})^2=\frac{119}{169}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving $Q=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3}$ For $a,b,c\geqslant 0.$ Prove: $$\text{Q}=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3},$$
where $k={\frac {27}{8}}+\frac{9\sqrt... | Another way
Let $p = a + b + c = 3, q = ab+bc+ca, r = abc$.
Since $(a-b)^2(b-c)^2(c-a)^2 \ge 0$ that is $-4p^3r+p^2q^2+18pqr-4q^3-27r^2 \ge 0$,
we have
\begin{align}
r &\ge -\frac{2}{27}p^3 + \frac{1}{3}pq - \frac{2}{27}\sqrt{(p^2 - 3q)^3}\\
&= q - 2 - 2\sqrt{(1-q/3)^3}.\tag{1}
\end{align}
We have to prove that
$$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sequence and polynomial power relationship I recently encountered this relationship between polynomial powers and a certain associated sequence and I am seeking any help or idea that might answer why the relationship is true.
Let $P(x)$ be a polynomial, say for instance $P(x)=1+3x+2x^2$. Consider the consecutive powers... | The sequence you encountered is used to construct a particular extension of the $a-b$-based triangle. In particular, your sequence generates the $1-3-2$ triangle. In general, for $a_0a_1\ldots a_{r-1}$-based triangle, the entry in the $m^{th}$ row and $n^{th}$ column which is exactly the term $a_{mn}$ in your recurren... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the power series of $\frac{3x+4}{x+1}$ around $x=1$. I'm trying to find the power series of
$$
\frac{3x+4}{x+1}
$$
around $x=1$.
My idea was to use the equation
$$
\left(\sum_{n\ge0}a_n (x-x_0)^n\right)\left(\sum_{k\ge0}b_k (x-x_0)^k\right) = \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-x_0)^n \tag{1}
$$
t... | Correction to your answer
Numerator:
$$
3x+4=7+3(x-1)\tag1
$$
Denominator:
$$
\begin{align}
\frac1{x+1}
&=\frac1{2+(x-1)}\tag2\\[6pt]
&=\frac12\frac1{1+\frac{x-1}2}\tag3\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{2^{k+1}}(x-1)^k\tag4
\end{align}
$$
Multiplying the series:
$$
\begin{align}
&(7+3(x-1))\left(\sum_{k=0}^\infty\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
how can i find $ \frac{1}{2\pi}\left ( \frac{\pi^{3}}{1!3}-\frac{\pi^{5}}{3!5}+\frac{\pi^{7}}{5!7}-... \right ) $ Consider a sequence $ s_{n} = \frac{1}{2\pi}\left ( \frac{\pi^{3}}{1!3}-\frac{\pi^{5}}{3!5}+\frac{\pi^{7}}{5!7}-...+\frac{\left ( -1 \right )^{n-1}\pi^{2n+1}}{\left ( 2n-1 \right ) ! \left ( 2n+1 \right )} ... | Consider
$$f(x)=\frac{x^3}{1!3}-\frac{x^5}{3!5}+\frac{x^7}{5!7}-\cdots$$
(an infinite series). We want to find $f(\pi)/(2\pi)$. Then
$$f'(x)=\frac{x^2}{1!}-\frac{x^4}{3!}+\frac{x^6}{5!}-\cdots
=x\left(\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)$$
which is a series you should recognise. Now integrate to f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Showing that $\sqrt{2} \in \mathbb{Q}[\varepsilon^{2}]$ If we work in the field of complex numbers and let $\varepsilon$ be a primitive 16th root of unity and set $b=\frac{\varepsilon}{\sqrt{2}}$ and $A=\mathbb{Q}[\varepsilon]$ where $\mathbb{Q}$ represents the rationals, and also set $f(X)=X^{8}+16 \in \mathbb{Q}[X]$ ... | $\epsilon$ is a root of $p(x)=x^8+1$ as it is a primitive root of $x^{16}-1 = (x^8-1)(x^8+1)$.
Therefore $\epsilon^2$ is a root of
$$q(x) = x^4+1= x^2\left(x^2+\frac{1}{x^2}\right)= x^2\left(x+\frac{1}{x} - \sqrt 2\right)\left(x+\frac{1}{x} + \sqrt 2\right)$$
As $\epsilon^4 \neq 0$, $\epsilon^2$ is a zero of
$$r(x) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3818411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$; find my mistake solve :$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$
My try:
$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}\\\frac{y'}{y^{2}}+\frac{2}{y(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\\left(t\frac{1-x}{1+x}\right)'=\left(x-1\right)\frac{1-x}{1+x}\\\left(t\frac... | Your 5th and 4th lines are not equivalent. You're missing a -ve sign in the second term on LHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3818821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving infinite nested square roots of 2 converging to finite nested radical Can anyone explain to solve the identity posted by my friend $$2\cos12°= \sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}} }}$$ which is an infinite nested square roots of 2. (Pattern $++--$ repeating infinitely)
Converging to finite nested radical of ... | If the value of the radical is $x$, then we have $$x=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-x}}}}\tag1$$ Repeated squaring gives
$$\left(\left(\left(x^2-2\right)^2-2\right)^2-2\right)^2=2-x\tag2$$
Now, $(2)$ has $8$ solutions, and notice for all choices of the first three signs in $(1)$, repeated squaring gives $(2)$. Thus,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove $(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^{-1}$ is an algebraic integer Let $b = (4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^{-1}$, then $1 = b(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})$, and $1 - 4b = b(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})$.
Therefore $1 - 12b + 48b^2 - 64b^3 = (1 - 4b)^3 = b^3(3 \cdot 3^{1/3} + 2 \cdot 3... | Your mistake occurs at this step
$$(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^3=(81 + 18 \cdot 3^{2/3} \cdot 3^{2/3} + 12 \cdot 3^{1/3} \cdot 3^{4/3} + 72)$$
Here you have forgotten the binomial coefficients $\binom{3}{1}$ and $\binom{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3823666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sum_{i=1}^n (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$ using induction. My textbook has the following question:
Prove the follwing statement using induction for all natural numbers $n$
$$(1- \frac{1}{4})+(1- \frac{1}{9})+.......+(1- \frac{1}{(n+1)^2})=\frac{n+2}{2n+2}$$
So, I check both the sides for $n=1$. ... | The formula you are trying to prove is wrong. The actual correct formula is
$$ \prod_{i=1}^n \left(1-\frac{1}{(i+1)^2}\right)=\frac{n+2}{2n+2}=\frac{n+2}{2(n+1)} $$
For $n=1$, this is easy to see.
$$\left(1-\frac{1}{(1+1)^2}\right)=\frac{3}{4}=\frac{1+2}{2\cdot 1+2}$$
For the induction step:
$$\prod_{i=1}^{n+1} \left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3825456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Rate of convergence for a sequence (Preferably without Taylor series) I am trying to solve the following problem:
Knowing that the sequence $(a_{n})$ with:
$$a_{n+1}=\frac{1}{2}(a_{n}+\frac{3}{a_{n}})$$
converges to $\sqrt{3}$, find it's rate of convergence.
After doing some searching, I found this formula from wikiped... | We have that
$$\frac{a_{n+1}-\sqrt 3}{a_{n+1}+\sqrt 3}=\frac{a_n^2-2\sqrt 3a_n+3}{a_n^2+2\sqrt 3a_n+3}=\left(\frac{a_{n}-\sqrt 3}{a_{n}+\sqrt 3}\right)^2$$
therefore by induction with $a_0=a>0$ we have that
$$\frac{a_{n}-\sqrt 3}{a_{n}+\sqrt 3}=\left(\frac{a-\sqrt 3}{a+\sqrt 3}\right)^{2^{n}}$$
and therefore
$$a_n=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $\frac{\sin (10^\circ) \sin (30^\circ)}{\sin 40^\circ \sin (80^\circ-x^\circ)} = \frac{\sin 20^\circ}{\sin x}$ The context to this is trivial I think, I was solving a geometry problem using the trigonometric version of Ceva, I got here and I was stuck, I tried using the sum-difference, product to sums, sums to pr... | You can use
\begin{equation}
\sin a\sin b\sin c=\frac{1}{4}(\sin(a+b-c)+\sin(a-b+c)+\sin(-a+b+c)-\sin(a+b+c))
\end{equation}
With this you get
\begin{equation}
\sin(40-x)+\sin(-20+x)+\sin(20+x)-\sin(40+x) \\
=
\sin(-20+x)+\sin(60-x)+\sin(100-x)-\sin(140-x)
\end{equation}
Then you can use $\sin(140 - x) = \sin(180 - 140... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given $x^5-x^3+x-2=0$, find $\lfloor x^6\rfloor$.
If $\alpha$ is a real root of the equation $x^5-x^3+x-2=0$, find the value of $\lfloor\alpha^6\rfloor$, where $\lfloor x\rfloor$ is the least positive integer not exceeding $x$.
My approach is to bound the value of $\alpha^6=\alpha^4-\alpha^2+2\alpha$.
First I proved ... | Since $f'(x)=5x^4-3x^2+1$, which is always greater than $0$, $f$ is strictly increasing. Since, furthermore, $\lim_{x\to\pm\infty}f(x)=\pm\infty$, $f$ has one and only one real root. On the other hand, $f(1)=-1<0<24=f(2)$. So, the only real root $\alpha$ belongs to $(1,2)$ and $\lfloor\alpha\rfloor=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3832120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing $\lim_{\Delta\rightarrow 0}\int_{-\Delta}^{\Delta}\frac{\mathrm{d}\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\frac{\pi}{r^2}$ While I'm working on the proof of Bertrand's Theorem, I stuck at a limits calculation. I want to prove:
$$
\lim_{\Delta\rightarrow 0}\int_{-\Delta}^{\Delta}\frac{\mathrm{d}\omega}{(r+... | Let
$\omega=\Delta \sin(x)$
$$I=\int\dfrac{d\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\int\dfrac{dx}{(r+\Delta \sin (x))^2}$$ Now, tanent half-angle substitution
$$I=\int \frac{2 \left(t^2+1\right)}{\left(r t^2+2 \Delta t+r\right)^2}\,dt$$
Write
$$r t^2+2 \Delta t+r=r (t-a)(t-b)$$ to make
$$I=\frac 2 {r^2} \int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What's the rotation matrix needed to align a rhombic dodecahedron such that six of its contact points line up to form a hexagon?
I want to know what the rotation matrix is I need to make this mesh have a hexagonal shape from a top-side orthographic viewpoint. The point where it has at least something resembling this i... | If the eight threefold vertices are $$(\pm 0.5, \pm 0.5, \pm 0.5),$$ and the six fourfold vertices are $$(\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1),$$ then a rotation that maps $v_1 = (0.5, 0.5, 0.5)$ to $v_3 = (0, 0, \frac{\sqrt{3}}{2})$ will work. A rotation about the $z$-axis by $\pi/4$ radians transforms $v_1 = (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
probability that first $2$ drawn balls are red
A bag contain $9$ red and $12$ blue balls. If $4$ balls are selected randomly without replacement, then find the probability that the first $2$ balls are red.
What I tried: Let $A$ be the event the first drawn ball is red and $B$ be the event that the second drawn ball i... | What happens after the first two balls are drawn has no effect on whether the first two balls are red, so we only need to consider what happens on the first two draws.
The probability that the first ball is red is $9/21$. If the first ball is red, the probability that the second ball is also red is $8/20$. Therefore,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Minimum value of $f(x,y,z) = x^z + y^z - (xy)^{\frac{z}{4}}, x > 0, y > 0, z > 0$ $$f(x,y,z) = x^{z}+y^{z}-(xy)^{\frac{z}{4}}$$
for all real positive numbers x, y, z
Does anyone have a clue to find the minimum value of $f(x,y,z)$?
I honestly don't know where to start the solution, I just come up with $AM \geq GM$
$\fra... | An alternative approach: set $a = x^{z/4}$ and $b = y^{z/4}$. Then we want to minimize $a^4 + b^4 - ab$. Due to symmetry, this quantity achieves the minimum value when $a = b$ so we have $2a^4 - a^2$. The derivative is equal to $8a^3 - 2a$ and its equal to $0$ when $4a^2 - 1 = 0$ (since $a > 0$). And we find $a = 1/2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Computation of a complex series I got stuck in a computation for hours and don't know where I did wrong. Let $N$ be a positive integer, I want to show the identity,
\begin{equation*}
\sum_{n=1}^{N}e^{inx} = e^{ix}\frac{e^{iNx}-1}{e^{ix}-1} = e^{i\frac{1}{2}(N+1)x}\frac{\sin (\frac{1}{2}Nx)}{\sin (\frac{1}{2}x)}
\end{e... | You didn't do anything wrong, just forgot to multiply by $e^{i\frac{1}{2}(N+1)x}$.
$$\begin{align*}
\frac{\sin [\frac{1}{2}Nx]}{\sin [\frac{1}{2}x]} &= \frac{e^{\frac{1}{2}Nix} - e^{-\frac{1}{2}Nix}}{e^{\frac{1}{2}ix}- e^{-\frac{1}{2}ix}}\\
& = \frac{e^{\frac{1}{2}(N+1)ix} - e^{-\frac{1}{2}(N-1)ix}}{e^{ix}- 1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ z,w\in\mathbb{C},|z|=|w|=R\gt0$. Show that $\left(\frac{z+w}{R^2+zw}\right)^2+\left(\vcenter{\frac{z-w}{R^2-zw}}\right)^2\ge\frac1{R^2}$ Let $ z, w \in \mathbb{C} $ be such that $ |z| = |w| = R > 0 $. Show that
$ \left(\frac{z + w}{R^2 + zw}\right)^2 + \left(\frac{z - w}{R^2 - zw}\right)^2 \geq \frac{1}{R^2} $
Wel... |
From the picture, if $\arg{(w)}=a$ and $\arg{(z)}=b$, then
$$
\begin{aligned}
|z+w|&=2R\left|\cos{\left(\frac{a-b}{2}\right)}\right|\\
|z-w|&=2R\left|\sin{\left(\frac{a-b}{2}\right)}\right|\\
\\
|R^{2}+zw|&=2R^{2}\left|\cos{\left(\frac{a+b}{2}\right)}\right|\\
|R^{2}-zw|&=2R^{2}\left|\sin{\left(\frac{a+b}{2}\right)}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.