Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
derive differential form $$\frac{l}{r^2}\frac{d}{d\theta}(\frac{l}{mr^2} \frac{dr}{d\theta})-\frac{l^2}{mr^3}=f(r)$$
I have to prove it is equal to
by taking $u=\dfrac{1}{r}$.
$$\frac{l}{r^2}\frac{l}{mr^2}\frac{d^2r}{d\theta^2}-\frac{l^2}{mr^3}=f(r)$$
$$\frac{d^2r}{d\theta^2}=\frac{mr^4}{l^2}f(r)+\frac{1}{r}$$
$$\fra... | We have $u=\dfrac{1}{r},$ so $r=\dfrac{1}{u}$ and $\dfrac{dr}{d\theta}=-\dfrac{1}{u^2}\dfrac{du}{d\theta}$ by the chain rule and thus substituting we have
$$\frac{l}{r^2}\frac{d}{d\theta}\left(\frac{l}{mr^2} \frac{dr}{d\theta}\right)-\frac{l^2}{mr^3}=f(r)$$
$$lu^2\frac{d}{d\theta}\left(\frac{lu^2}{m}\left(-\frac{1}{u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\frac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$
Prove
$$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$$
Proving right hand side to left hand side:
$$\begin{align}\csc^2x+2\csc x \cot x+\cot^2x
&= \frac{1}{\sin^2x}+\dfrac{2\cos x}{\sin^2x}+\dfrac{\cos^2x}{\si... | We have that, using $\sin 2x=2\sin x\cos x$ and cancelling out $2\sin x\neq0$
$$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}= \dfrac{2\sin x+2\sin x\cos x}{2\sin x-2\sin x\cos x}= \dfrac{1+\cos x}{1-\cos x}$$
then, assuming $\cos x \neq -1$ ( which holds since we need $\sin x\neq0$ for the original expression), multiply by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $a^{2n+1}+(a-1)^{n+2}$ is divisible by $a^2-a+1$ for any $a \in\mathbb{Z}$. Prove that $a^{2n+1}+(a-1)^{n+2}$ is divisible by $a^2-a+1$ for any $a \in \mathbb{Z}$.
I tried to use induction here, but it seems like a dead end. Maybe I should somehow use binomial theorem?
I need a hint
| For the divisibility, I am going to prove that $a^{2 n+1}+(a-1)^{n+2}\equiv 0 \quad \pmod{a^2-a+1}$
\begin{aligned}
& a^{2 n+1}+(a-1)^{n+2} \\
\equiv & a^{2 n+1}+\left(a^{2}\right)^{n+2} \quad\left(\bmod a^{2}-a+1\right) \\
\equiv & a^{2 n}\left(a+a^{4}\right) \quad\left(\bmod a^{2}-a+1\right)\\
\equiv & a^{2 n}\left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$10$ circles ($2$ large of radius $R$, $6$ small of radius $r$ and 2 small of radius $t$) are enclosed in a square. How we find $r$ in terms of $t$? Let us embed $2$ large intersecting circles of radius $R$ into a square as depicted by the figure below. These two circles are highlighted green. Into these $2$ circles we... |
The blue triangle is easy and has already been illustrated by the OP. Without loss of generality, suppose the square is the unit square in the coordinate plane. Then the common point of tangency of the two small circles is $(1/2,1/2)$, and the distance between this point and a center of the larger circle is simply $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Eliminating $\theta$ from $\cos^3\theta +a\cos\theta =b$ and $\sin^3\theta +a\sin\theta =c$
Eliminate $\theta$ from the equations.
$$\cos^3\theta +a\cos\theta =b$$
$$\sin^3\theta +a\sin\theta =c$$
Can anyone solve this question?
| I have now my own answer developed out of conversation with David Quinn, so I thank him. I publish this answer to demonstrate the calculation. It seems to coincide with the calculation of Macavity in the comments above. Wow what a monster of a question.
We have
$$x^3+ax=b$$
$$y^3+ay=c$$
Where $x^2+y^2=1$.
Adding and su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4265926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove $\int_{0}^{\infty} \psi^{(2)} (1+x) \ln (x) \, dx = \zeta'(2) + \zeta(2)$ I'm looking for alternate methods of proving the following result:
$$\int_{0}^{\infty} \psi^{(2)} (1+x) \ln (x) \, dx = \frac{\pi^2}{6} \left( \gamma + \ln (2\pi)-12 \ln A +1\right) = \zeta'(2) + \zeta(2)$$
Where $\psi$ is the polygamma fu... | For another solution using the complete beta function:
Using the same expansion
$$\psi^{(2)}(x) = - 2 \sum_{n=0}^{\infty} \frac{1}{(n+x)^{3}},$$
we have
$$ \begin{align} \int_{0}^{\infty} \psi^{(2)}(x+1) \ln (x) \, \mathrm dx &= -2\int_{0}^{\infty}\ln(x) \sum_{n=0}^{\infty} \frac{1}{(n+x+1)^{3}} \, \, dx \\ &= - 2 \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
the value of $\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}$ I want to compute this limit
$$\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}.$$
I tried to apply Hopital rule, but I cannot compute it.
| Various tricks can simplify the limit. Multiply by $\frac{\sin(x^2)}{x^2}$ which has a limit of one to reduce to
$$\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x^3}.$$Add $$\frac{\sin x^2 -x^2}{x^3}$$ to reduce to
$$\frac{(x^2+1) \ln(x^2+1)-x^2}{x^3}.$$Now break into pieces,
$$\frac{\ln(x^2+1)-1}{x}+\frac{\ln(x^2+1)}{x^3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What's the measure of the segment $LR$ in the figure below? For reference:In the picture shown we have an isosceles trapezoide,
The $QR$ side measures $b$ and the $PQ$ side measures $a$. find $LR$
(answer:$\frac{b(a+b)}{a-b}$ )
My progress:
I didn't get much on this question... a colleague indicated if PQRL form a har... | Showing the pencil $O (LQ; RP)$ is harmonic definitely makes the work easier. Here is an alternate approach that requires solving a quadratic equation in the end.
$EF$ is perpendicular bisector of $AD$ and $BC$. In $EFRP$, $\angle FRP = 180^\circ - \angle EPR = 180^\circ - 2\theta$
As $F$ and $T$ are points of tangenc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Maximum of a multivariable function with constraints I have to maximize the function
$$p(a,b,c)=\frac{1}{2} \left(\frac{1}{3} \sqrt{2 (\cos (a)+\cos (b)+\cos (c))+3}+1\right);$$
with the constarint that following function is a constant.
$$t(a,b,c)=\frac{1}{2} \left(\sqrt{2 (\cos (a)-\cos (b)-\cos (c))+3}+\sqrt{2 (-\cos... | Remark: In the proof, we use the expression $\color{blue}{p=\frac{1}{12} \left(t+6 +\sqrt{3(12-t^2)}\right)}$ due to @Claude Leibovici which is simpler than @Domen's $p = \frac{1}{12}(\sqrt{2 t (\sqrt{36-3
t^2}-t)+36}+6) $ in the linked page, though they are the same (easy to prove).
Consider the following optimiza... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4280771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric equation in quadratic form If $\cos^4α+k$ and $\sin^4α+k$ are the roots of $x^2+\lambda \ (2x+1 )=0$ and $\sin^2α +\ell$ and $\cos^2 α +\ell$ are the roots of $x^2+8x+4=0$ , then the sum of the possible values of $λ$
is _________.
My approach is as follow
${\sin ^2}\alpha + \ell + {\cos ^2}\alpha + ... | For a quadratic equation, $Ax^2+Bx+C=0$, the difference of its roots can be shown to satisfy
$$(x_1-x_2)^2 = (x_1+x_2)^2-4x_1x_2 = \frac{B^2-4AC}{A^2} \tag{1}$$
Since $$\cos^4 \theta - \sin^4\theta = (\cos ^2 \theta - \sin ^2\theta)(\cos ^2 \theta + \sin ^2\theta)$$ $$=\cos ^2 \theta - \sin ^2\theta$$
the difference of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the diophantine equation $(n+1)^{2}X+n^2Y=1$ where $n\in\mathbb{N}$. I having difficulty applying Euclid's algorithm to find a particular solution.
| Either $n=2k$ or $n+1=2k$ for some integer $k$.
In the first case, Euclid's algorithm gives:$$\begin{array}{c|c} (2k+1)^2&4k^2\\4k+1&-k\\1 \end{array}$$
Working backwards: \begin{eqnarray*}1&=&(4k+1) +4(-k)\\
&=& (4k+1)+4(4k^2-k(4k+1)\\ &=& (1-4k)(4k+1)+4(4k^2)\\&=&
(1-4k)((2k+1)^2-(4k^2))+4(4k^2)\\&=&
(1-4k)(2k+1)^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Zero divisors in a Ring Let's say $a, b$ are zero divisors in a Ring, (i.e., there exist some $x,y \in R$ s.t. $ax=0, by=0$).
I feel that $a$ is a zero divisor of $xy$ (as $axy=0y=0$), but is $b$ a zero divisor of $xy$?
If I take a look at $bxy$, I know I can't commute $b$ and $x$, but can $b$ be a zero divisor of $(xy... | Your intuition is right. Indeed, $axy=0$, but we cannot say that $bxy=0$.
For example, consider $M_2(\mathbb{Z})$, the $2\times2$ matrices with integer coefficients. We have that
\begin{align*}
\begin{pmatrix}
1&0\\0&0
\end{pmatrix}
\begin{pmatrix}
0&0\\1&1
\end{pmatrix}=\begin{pmatrix}
0&0\\0&0
\end{pmatrix}
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4283794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Approximate solution to a transcendental equation I'm working on a physics problem and stumbled upon the following equation:
$$h=\frac{2n\pi+\arctan\left(\frac{c}{b}\right)}{b}$$
where $n \in \mathbb{Z}$, $c \in [0,20]$ and $h \in \mathbb{R}^+$. This equation has to be solved for $b$, which can be done numerically ofco... | Let $\frac c b=x$ to make the equation
$$2 n\pi x+x \tan ^{-1}(x)=k \qquad \text{with} \qquad k=ch$$ Expand the lhs as a Taylor series around $x=0$
$$2 n\pi x+x \tan ^{-1}(x)=2 n\pi x+x^2-\frac{x^4}{3}+\frac{x^6}{5}+O\left(x^8\right)$$ Use series reversion
$$x=\frac{k}{2 \pi n}-\frac{k^2}{8 \pi ^3 n^3}+\frac{k^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4284566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Should I ignore ± sign when integrating square roots? I was solving the following integral:
$$
\int \:\frac{x^2}{\sqrt{x^2+4}}dx
$$
$$
u=\sqrt{x^2+4}
$$
$$
\:du=\frac{2x}{2\sqrt{x^2+4}}dx=\frac{x}{u}dx
$$
$$
\int \:\frac{x^2}{\sqrt{x^2+4}}dx=\int \:\frac{x^2}{u}dx=\int \:xdu
$$
Now I only need to find what x means in t... | What the $\pm$ sign in $x=\pm \sqrt{u^2-4}$ implies is that you have a choice of how to make the substitution of $u$ for $x.$ You need this choice to be available in order that your evaluation of the integral can be valid for all values of $x.$
If we suppose that $x$ can have either positive or negative values, then th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Area of a square on top of a quadrilateral
In quadrilateral $ABCD, \angle ABC + \angle DCB = 90^\circ$ and $ADEF$ is a square constructed on side $AD$ in the exterior of the quadrilateral $ABCD.$ If $BC = 10 $ cm, $AC = 9$ cm and $BD = 8$ cm, then the area of the square $ADEF$ lies between
(a) $70$ and $80$
(b) $60$ a... | Extend $\small CD$ to meet $\small AB$ at $\small E$.
Since $\small \measuredangle B+\measuredangle C=90^\circ$, we can see $\small \angle BEC=90^\circ.$
Now from Pythagorean theorem,
$\small BC^2=EB^2+EC^2\tag1$
$\small AD^2=EA^2+ED^2\tag2$
Adding both,
$\small \begin{align} BC^2+AD^2&=(EB^2+ED^2)+(EC^2+EA^2)\\ &=BD^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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 $$\fr... | Proof:
First of all, clearly we have
$$ (a - b)^2 + (b - c)^2 + (c - a)^2 \ne 0. \tag{1}$$
Now, we split into two cases:
Case 1 $~ x = y$:
We have
$$\frac{x^2-xz}{a^2-bc}=\frac{x^2-zx}{b^2-ca}=\frac{z^2-x^2}{c^2-ab} \ne 0$$
which results in
\begin{align*}
a^2 - bc &= b^2 - ca, \tag{2}\\
x(c^2 - ab) &= - (z + x)(b^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Dealing more efficiently with fractional forms in system of equations As an example, suppose we have to solve the following system of two equations and two unknowns:
$$
\begin{cases}
-\frac{10}{x}-\frac{8}{y} &= \frac{8}{3} \\
-\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}
\end{cases}
$$
My approach and solution
I opted ... | We start by getting rid of the fractions.
\begin{align*}
\begin{cases}
-\frac{10}{x}-\frac{8}{y} &= \frac{8}{3}
&\implies -30y -24x=8xy \\
-\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}
&\implies -18y+18x=-xy
\end{cases}
\\\text{We mulitply the second equation by 8 and then add}
\\
\begin{cases}
-\space \space 30y -\spa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4288341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Limits of integration for theta in double integrals I have a question that asks me to find the volume lying inside both the sphere $x^2+y^2+z^2=a^2$ and the cylinder $x^2+y^2=ax$. The worked solution in the textbook goes like this
One quarter of the required volume lies in the first octant. In polar coordinates the cyl... | In polar coordinates, $ \theta \in (0, 2\pi)$ is valid for the circle centered at the origin. But we have a circle that is centered on x-axis away from the origin and in fact the origin is a point on the circle.
The projection of the cylinder in XY-plane is circle $x^2 + y^2 = ax$.
Or, $ \displaystyle \left(x - \frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4288853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int_{\gamma} \nabla f \cdot d\mathbf{x}$ Compute $\int_{\gamma} \nabla f \cdot d\mathbf{x}$ for the following choices of $f$ and $\gamma$.
(a) $f(x,y) = x^2+y^2; \gamma:g(t) = (1+t^2, 1-t^2), -1 \le t \le2$
What I have tried:
$$\int_{-1}^2(1+t^2)^2dx + \int_{-1}^2(1-t^2)^2dy$$
Where $dx = 2t$ and $dy = -2t$
S... | For $f(x,y) = x^2+y^2$ we have
$$ \nabla f. X = 2(x^2+y^2)$$
Thus,
$\begin{array}{ccl}
\displaystyle\int_{\gamma} \nabla f. X ds &=& 2 \displaystyle\int_{\gamma} (x^2+y^2) ds\\
&=& 2 \displaystyle\int_{-1}^{2} ((1+t^2)^2+(1-t^2)^2) \sqrt{8t^2} dt\\
&=& 2 \sqrt{8} \left( - \displaystyle\int_{-1}^{0} ((1+t^2)^2+(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4292070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Multiplying power series I'm trying to understand how the expression
$$
\sum\limits_{n=0}^\infty x^n \cdot \sum\limits_{n=0}^\infty x^n
$$
results in the following expression:
$$
\sum\limits_{n=0}^\infty (1+n)x^n.
$$
Could you please walk me through it?
| A simple observation might be enlightening:
$$(1+x+x^2+x^3)(1+x+x^2+x^3)\\
=1+x+x^2+x^3+x+x^2+x^3+x^4+x^2+x^3+x^4+x^5+x^3+x^4+x^5+x^6\\
=1+2x+3x^2+4x^3+3x^4+2x^5+x^6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Infinite binomial series over a column of Pascal's triangle: $F_k(x)= \sum\limits_{n=0}^\infty \binom{2n+k}n x^{2n+k}$ Is there any closed form formula (or an equivalent) for this binomial infinite series :
$$F_k(x)= \sum_{n=0}^{\infty} \binom{2n+k}{ n } x^{2n+k} $$ in which $|x|<1$ and $k$ is a given integer?
This is... | We have
$$
S(k,x) = \sum\limits_{n = 0}^\infty {\left( \matrix{ 2n + k \cr n \cr} \right)x^{2n + k} }
= x^k \sum\limits_{n = 0}^\infty {\left( \matrix{ 2n + k \cr n \cr} \right)\left( {x^2 } \right)^n }
$$
Indicating the series coefficients as
$$
t_n = \left( \matrix{ 2n + k \cr n \cr} \right)
$$
then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove that is a perfect square. Let $x,y$ and $z$ be positive integers such that $$\frac{1}{x}-\frac{1}{y}=\frac{1}{z}.$$ Let $h=(x,y,z).$ Prove that $hxyz$ is a perfect square.
Solution: Let $z=hc, x=ha,y=hb$ so $(a,b,c)=1$. Notice that:
\begin{align*}
hxyz=h^4abc
\end{align*}
We need to see that $abc$ is an square. N... |
Why is $z=\frac{xy}{y-x}$?
Initially, $x,y,z$ satisfy
$$\frac 1 x - \frac 1 y = \frac 1 z$$
Therefore, by solving for $z$,
$$z = \dfrac{1}{\left( \dfrac 1 x - \dfrac 1 y \right)} = \dfrac{1}{\left(\dfrac{y-x}{xy} \right)} = \dfrac{xy}{y-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4297140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\arctan(\cos(\alpha)*\tan(x)) = \arccos(\dfrac{\cos(x)}{\cos(\arcsin(\sin(\alpha)*\sin(x)))})$ In trying to calculate an arc length of a right spherical triangle, I reached this $\arccos$ expression. However, I see the expression given in the manuals for the same problem is much simpler. checking against ... | Notice that $\cos(\arcsin(x)) = \sqrt{1 - x^2}$ by drawing a triangle where $\sin\theta = x$.
Similarly, $\tan(\arccos(x)) = \frac{\sqrt{1 - x^2}}{x} = \sqrt{\frac{1}{x^2} - 1}$.
Now \begin{align*}\tan(RHS) &= \tan\arccos\big(\cos x\cdot\frac{1}{\sqrt{1 - (\sin\alpha\sin x)^2}}\big) \\ &= \sqrt{\frac{1}{\big(\cos x\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4297536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Fibonacci numbers, sum of squares and divisibility Let $(F_n)$ be the Fibonacci sequence (i.e. $F_{n+2} = F_{n+1}+F_n$ with $F_{-1}=1$ and $F_0=0$).
Consider $a,b \in \mathbb{Z}_{\ge 1}$ with $a \le b$, and $D=1+a^2+b^2$, such that $a$ and $b$ divide $D$.
Question: Must the ordered pair $(a,b)$ be equal to $(F_{2n-1},F... | Yes. This uses Vieta Jumping. Note first that your $\gcd(a,b) = 1,$ so that $ab| a^2 + b^2 + 1.$ Then $a^2 + b^2 + 1 = 3ab.$ Furthermore, the only Ground Solution, in the sense of Hurwitz 1907, is $(1,1).$ Every solution derives as a finite number of jumps from $1,1.$ That all solutions are Fibonacci numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4298283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Putting 12 balls of different colors in 4 urns, 3 balls in each. 5 red, 4 green and 3 blue balls are put in 4 boxes so that every box has exactly 3 balls.
Find the probability that:
*
*At least 1 box has a ball of every color;
*Only 1 box has a ball of every color
I've tried to compute the probability of at least 1... | Line up the urns from left to right. There are
$$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
to distribute three balls to each of the four urns.
An urn with a ball of each color: There are four ways to select the urn which will receive a ball of each color, five ways to select a red ball, four ways to select ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all positive integers $x,y,z$ for which the expression is true. So I wish to find all the positive integers $x,y,z$ such that the following expression is true.
$$\frac{-1 + x + y + x y}{-1 - x - y + x y} = z$$
The problem is that the expression does not simplify any further as far as I can tell. The only constrain... | Let $x$, $y$ and $z$ be positive integers such that
$$\frac{-1+x+y+xy}{-1-x-y+xy}=z.\tag{1}$$
First, it should be clear that $x\geq2$ and $y\geq2$ because the ratio $z$ is positive, and that $z\geq2$ because the numerator is strictly greater than the denominator. If $x=2$ then $(1)$ simplifies to
$$z=\frac{1+3y}{-3+y}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to find following limit of sequence?
How to following limit
$$\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{(n-1)}{n^{2}}\right)\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right). $$
My attempt:
\begin{align*}
L_n&=\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{(n-1... | Hint:
Rewrite the second factor as
$$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}=\frac 1n\biggl(\frac 1{1+\dfrac1n}+\frac 1{1+\dfrac2n}+\dots+\frac 1{1+\dfrac nn}\biggr) $$
and observe this is a Riemann sum of the function $\dfrac1{1+x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Integrate: $\frac{x^2+x-1}{x^2-1}$ with respect to $x$ Ok so here is where I am up to:
I can find some similarity with the numerator and denominator and managed to reduce to the following:
$\int\frac{x^2+x-1}{x^2-1}dx = \int\frac{x^2-1}{x^2-1}+\frac{x}{x^2-1}dx$
$=\int1+\frac{x}{x^2-1}dx.$
I tried to reduce it further ... | $\int x^2 + x - 1\over x^2 - 1$ dx = $\int ({x^2 - 1\over x^2 - 1} + {x\over x^2 - 1})$dx
= $\int 1 dx + \int {x\over x^2 - 1}$ dx
Let $x^2 - 1$ = t $\implies$ 2x dx = dt
= x + $\int {dt\over 2t}$
= x + ${1\over 2} \int {1\over t}$ dt
since, integration of ${1\over x}$ = log x
= x + ${1\over 2}$ log t + C
Subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Number of real solutions of $\begin{array}{r} {\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} =\frac{3 x-1}{2} \end{array}$ Solve for $x \in \mathbb{R}$ $$\begin{array}{r}
{\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]}
=\frac{3 x-1}{2}
\end{array}$$ where $[x]$ denotes greatest integer less th... |
Letting $\displaystyle a=\frac{2x+1}{3}$ we get
$$\left[a\right]+\left[a+\frac{1}{2}\right]=\frac{9 a-5}{4} \tag{1}$$
The solution will involve identifying all satisfying values of $a$, and then translating these values into all satisfying values of $x$ via $\displaystyle a=\frac{2x+1}{3}$.
The problem may be attacke... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4311152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Use Induction to prove recurrence
if the above screenshot is not visible, here is the text format:
Question:
Solve the following recurrence and prove your result is correct using induction:
$a_1 = 0$
$a_n = 3(a_{n-1}) + 4^{n}$ for $n>=2$
Use induction to prove this recursive sequence.
So my approach was that, I plug ... | I assume you mean $a_0=0$ and $a_n = 3a_{n-1}+4^n$ for $n\ge 1$ (not $n\ge 2$). And by "prove this recursive sequence", you mean to find a closed formula.
Consider the generating function $f(x)=\sum_{n=0}^\infty a_nx^n$,
then by the initial condition and recursive relation, we have $$3xf(x)+\sum_{n=1}^\infty 4^nx^n = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Optimal wager on two games given probability and odds Question. Suppose I have two games I would like to wager \$100 dollars on, but that I'm not required to wager the full \$100, i.e. I can bet \$50 if I want.
The games: $A$ vs. $B$ and $C$ vs. $D$, where the probabilities of each team winning are
$$P(A) = \frac{7}{10... | If the player is infinately risk averse she should bet $x$ on $C$ and $100-x$
on $D$ where $x$ is determined by the equation
\begin{eqnarray*}
\frac{x}{2}\ast 7+x &=&\frac{100-x}{3}+100-x\rightarrow \\
x &=&\frac{160}{7}
\end{eqnarray*}
Then, if $C$ wins she will get a net profit of $\frac{160}{7}\frac{7}{2}+%
\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Reduction formula for $\int x\tan^n(x)$ So I've been trying to derive a reduction formula for the following integral
$$\int x\tan^n(x)dx$$
I tried to use integration by parts by factoring out a $\tan^2(x)$ and taking the following:
$$u=x\tan^{n-2}(x)\implies du=\tan^{n-2}(x)+(n-2)x\tan^{n-3}(x)\sec^2(x)dx\\
dv=\tan^2(x... | $$\begin{split}
I_{n+2} +I_n &= \int x \tan^n (x) (1+\tan^2 x)dx \\
&= \frac 1 {n+1}x \tan^{n+1}(x)-\frac 1 {n+1}\underbrace{\int \tan^n x dx}_{J_n} &\,\,\,\,\,\,\,\,(1)
\end{split}$$
where the last line comes by integrating by parts.
Now
$$\begin{split}
J_{n+2}+J_n &= \int \tan^n (x)(1+\tan^2x) dx\\
&= \frac 1 {n+1}\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4318915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$. I have written an attempted proof of the theorem on the title, and I need help verifying it.
I have used the following theorems to proof the theorem on the title.
Theorem 5.14)a) Let $x$ be a real number. $-|x| \leq x \leq |x|$.
Theorem 5.14)b) Let $... | We have :
$(1+z)(1+w)-1 = z+w+zw$
Thus :
$|(1+z)(1+w)-1| = |z+w+zw| \le |z| + |w| + |zw| = (1+|z|)(1+|w|) -1$
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4321116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How many four-digit positive integers are there that contain the digit $3$ and are divisible by $5$?
How many four-digit positive integers are there that contain the digit $3$ and are divisible by $5$?
The answer is:
the number of four-digit integers that are divisible by $5\;-\;$ the number of four-digit integers ... | You double counted, for example $3335$ appears in all your sum.
You can use inclusion-exclusion to adjust your count.
\begin{align}
&|A|+|B|+|C|-|A\cap B|-|A \cap C|-|B \cap C| + |A \cap B \cap C|\\
&=560 - 18 - 20-20+2\\
&=504
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$1^2 - 2^2 + 3^2 - 4^2 + \dots + 1999^2$ using closed form formula for sum of squares of first $n$ natural numbers The question is simple, the solution, not so much
Q.Find the sum of the given expression $1^2- 2^2 + 3^2 - 4^2 + \dots + 1999^2$
My idea is we know $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n + 1)(2n ... | Use the decomposition of squares into triangular numbers, $n^2=T_{n-1}+T_n$, and telescope:
$$1^2-2^2+3^2-\cdots+1999^2=1-(1+3)+(3+6)-(6+10)+\cdots+(T_{1998}+T_{1999})$$
$$=T_{1999}=\frac{1999×2000}2=1999000$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Different solutions to the same system of linear equations I want to solve a real homogeneous system of linear equations represented by this matrix:
\begin{bmatrix}0&2&2&7&1&0\\1&0&1&3&1&0\\-1&2&1&4&0&0\end{bmatrix}
The reduced echelon form is:
\begin{bmatrix}1&0&1&3&1&0\\0&1&1&7/2&1/2&0\\0&0&0&0&0&0\end{bmatrix}
Since... | It is easy to make Maple to use your choices of free variables, how? Let us rewrite what you are looking for mathematically. What you are asking can be written as a new linear system.
$$\begin{bmatrix}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
1 & 0 & 1 & 3 & 1\\
0 & 1 & 1 & \frac{7}{2} & \frac{1}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4336818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $x^2 \equiv 106234249 \bmod{12^2}$ I have the following congruence:
$$x^2 \equiv 106234249 \bmod{12^2}$$
When I tried to solve it, the equation becames in 2 equations:
$x^2 \equiv 4 \bmod{9}$
$x^2 \equiv 9 \bmod{16}$
Because $12^2 = 9 \times 16$ and, 9 and 16 are coprimes.
Then I applied the method of this vide... | When you solved $x^2 \equiv 4 \pmod 9$, the only solutions are $x \equiv ±2 \pmod 9$. This is because if $x = 3k ± 2, x^2 = 9k^2 ± 12k + 4$ and $±3k + 4$ is never a multiple of $9$.
However, solving $x^2 \equiv 9 \pmod {16}$ is different. When $x = 16k ± 3$, $x^2 \equiv ±(2 \cdot 16k \cdot 3) + 9 \equiv 9 \pmod {16}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there a polynomial $g(x)$ such that $p \circ f=g \circ p$, where $f(x)=9x+30x^3+27x^5+9x^7+{{x}^{9}}$? Let us consider a polynomial $f(x)=9x+30x^3+27x^5+9x^7+{{x}^{9}}$ over a field of characteristic $0$.
Set $p(x)=x^nh(x), n \geq 1$, where $h(x)$ can be any rational function.
I am trying to find out another polynom... | $$
\begin{array}{rcl}
p(x)&=&9x,\\
g(x)&=&\displaystyle \frac{x^9}{43046721}+\frac{x^7}{59049}+\frac{x^5}{243}+\frac{10 x^3}{27}+9 x.
\end{array}
$$
Upd. For the new version
$$
\begin{array}{rcl}
p(x)&=&x^2,\\
g(x)&=&x (x+3)^2 \left(x (x+3)^2+3\right)^2;
\end{array}
$$
$$
\begin{array}{rcl}
p(x)&=&x^4+4 x^2,\\
g(x)&=&x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving the system $x^2+y^2+x+y=12$, $xy+x+y=-7$ I've been trying to solve this system for well over the past hour.$$x^2+y^2+x+y=12$$
$$xy+x+y=-7$$
I've tried declaring $x$ using $y$ ($x=\frac{-7-y}{y+1}$) and solving from there, but I've gotten to $$y^4+3y^3-9y^2-45y+30=0$$ and I don't see how we can get $y$ from here... | Let $S = x+y$ and $P=xy$. You have $S^2 - 2P +S = 12$ and $P+S=-7$. Therefore
$$S^2 - 2P +S = S^2-2(-7-S) + S=S^2+3S+14=12$$ or $$S^2+3S+2=0.$$ The roots of this last equation are $-1,-2$. Therefore $(S,P) \in \{(-1, -6) , (-2,-5)\}$. Which implies that $x,y$ are the roots of either $u^2 +u -6 = 0$ or of $u^2 +2u -5=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Can one integral can give more than one answer one with natural log and other with tan inverse question is $$\int\frac{1}{\sin^6(x) + \cos^6(x)}\,dx$$
My method :
$$\sin^6\left(x\right)+\cos^6\left(x\right)$$
$$=\left(\sin^2\left(x\right)+\cos^2\left(x\right)\right)\left(\sin^4\left(x\right)-\cos^2\left(x\right)\sin^2\... | You made a mistake : \begin{aligned}\int{\frac{\mathrm{d}x}{1-3\cos^{2}{x}\sin^{2}{x}}}&=\int{\frac{\sec^{2}{x}}{\sec^{2}{x}-3\color{red}{\sin^{2}{x}}}\,\mathrm{d}x}\\ &=\int{\frac{\mathrm{d}y}{1+y^{2}-3\left(1-\frac{1}{1+y^{2}}\right)}}\\ &=\int{\frac{y^{2}+1}{y^{4}-y^{2}+1}\,\mathrm{d}y}\\ &=\int{\frac{1+\frac{1}{y^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find parameters $a,b$ such that $x^6-2 x^5+2 x^4+2 x^3-x^2-2 x+1-\left(x^3-x^2+a x+b\right)^2>0$ The probrem is to prove that
$$x^6-2 x^5+2 x^4+2 x^3-x^2-2 x+1>0.$$
(the minimum value is about 0.02, tested by wolframalpha.)
I use sos(sum of squares) method, my idea is to reduce the degree of the polynomial gradually.
F... | For the original problem, observe that
$$x^6 - 2x^5 + 2x^4 + 2x^3 - x^2 - 2x + 1 = (x^3 - x^2)^ 2 + (x^2 +x - 1)^2.$$
Since both squares cannot be 0 at the same time, hence it is strictly positive.
For your stated problem, $ a, b = 0 $ is still not a solution to your strict inequality because the expression does hit 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How far can I go with the integral $\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x, $ where $n\in N$? Latest Edit
By the aid of my recent post, a closed form for its definite integral is obtained as below:
$$
\int_{0}^{\frac{\pi}{2}} \sin ^{k} \theta d \theta= \frac{\sqrt{\pi} \Gamma\left(\frac{k+1}{2}\right)... | Just for your curiosity.
As said in comments, the result is not very pretty.
Using
$$I_k=\int_0^{\frac \pi 2} \sin^k(x)\,dx=\frac{\sqrt{\pi }}{2}\,\,\frac{\Gamma \left(\frac{k+1}{2}\right)}{\Gamma \left(\frac{k+2}{2}\right)}$$
$$S_n=\sum_{k=1}^{n-1} \frac{1}{2^{k+1}} \int_{0}^{\frac{\pi}{2}} \sin ^{k}( x)\, d x$$
$$S_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
A 3d line intersecting 2 other 3d lines The equation of the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and intersecting the lines $9x + y + z + 4 = 0 = 5x + y + 3z$ & $x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
My solution is as follow
$\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4}$ represent ... | To find a line parametric equation, we have to find a point on the line, and then use the cross product of the normal vectors of the the two planes given.
The two planes for the first line are
$9x + y + z = -4 $
$5 x + y + 3 z = 0 $
To find a point on both these planes, set $x = 0$ and solve the system
$ y + z = -4 $
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4346271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is this series known? I encountered the following series and was wondering if it can be expressed using a known value?
$$\sum_{a,b=1}^{\infty} \frac{(-1)^a+1}{a^3-a^2+ab^2+b^2},$$
where all terms in the sum for which $a$ is odd are interpreted as zero.
| Since $a$ can only contribute nontrivially when even, we may as well replace $a\to 2a$
$$\sum_{(a,b)\in\Bbb{Z}^+\times\Bbb{Z}^+} \frac{2}{8a^3-4a^2+(2a+1)b^2} = \sum_{a=1}^\infty \frac{2}{2a+1}\sum_{b=1}^\infty\frac{1}{4a^2\left(\frac{2a-1}{2a+1}\right)+b^2}$$
The sum on the interior is the famous result
$$\sum_{n=1}^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4346417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
show that $(x,y)\to \ln(\sqrt{x^2+y^2})$ is harmonique over $\mathbb{R}^2\backslash \{(0,0)\} $ by use of following result:
If $f$ is a holomorphic function over the open $O$ of $\mathbb{C}$ then the real part of $f$ is harmonic over the open $O$
show that $(x,y)\to \ln(\sqrt{x^2+y^2})$ is harmonique over $\mathbb{R}^2... | We may also show $u(x, y)$ harmonic by direct differentiation as follows:
$u(x, y) = \ln(x^2 + y^2)^{1/2} = \dfrac{1}{2} \ln(x^2 + y^2); \tag 1$
$u_x = \dfrac{1}{2} \dfrac{1}{x^2 + y^2} (2x) = \dfrac{x}{x^2 + y^2}; \tag 2$
$u_{xx} = \dfrac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} = \dfrac{y^2 - x^2}{(x^2 + y^2)^2}; \tag 3$
$u_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4354683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Area of a circle passing through two vertices of a parallelogram touching one edge. For reference:
Let $ABCD$ be a parallelogram, $AB = 6, BC= 10$ and $AC = 14$ ; traces a circle passing through the vertices $C$ and $D$ with $AD$ being tangent and this circle is $BC$ a secant. Calculate the area of the circle. (Answ... | Drop the perp from $A$ to $BC$ and call the foot $E$.
Let $BE=x$, then using Pythagoras' theorem, $$6^2-x^2=14^2-(10+x)^2\implies x=3.$$
Therefore in right triangle $ABE$, $\angle BAE=30^\circ$ and so is $\angle CDO$.
Let $M$ be the midpoint of side $CD$, then considering $\triangle ODM$, $OD$, or, radius of the circl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that if $a^5-a^3+a=3$, then $a^6\geq 5$ The problem, which I encountered in a highschool book, goes as following:
Prove that if $a^5-a^3+a=3,$ then $a^6\geq 5$ must hold.
Now, obviously, I have tried a lot of things such as:
$a^6=a^4-a^2+3a\Rightarrow a^4-a^2+3a\geq 5$ or multiplying that again by $a^2$ to crea... | The equation $a^5-a^3+1-3=0$ has at least one real solution.
As
$$3=a(a^4-a^2+1)=a\bigl((a^2-1/2)^2+3/4\bigr)$$
we have $a>0$ assuming $a$ is real.
Now
$$a^4-a^2=\frac3a-1,$$
hence
$$a^6=a^4-a^2+3a=\frac3a+3a-1
=3(\sqrt{a}-1/\sqrt{a})^2+6-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the system $3^x-2^{y^2}=77$, $3^{\frac{x}{3}}-2^{\frac{y^2}{2}}=7$ in $\mathbb{R}$ I had to solve the similar system $3^x-2^{y^2}=77,\; 3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7$ before, which can be solved like this:
$$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7$$
$$2^{\frac{y^2}{2... | If $u=3^{\frac x 2}$ and $v=2^{\frac {y^2}2}$ then
$$\left \{
\begin{split}
u^3-v^2&=77\\
u-v&=7
\end{split}
\right.$$
So $v=u-7$ and thus $u^3-u^2+14u-126=0$.
Solving the latter gives you 3 roots (1 real and 2 complex), and you can find $u$ and $v$, and get back to $x$ and $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $x_n = (-1)^n \frac {3n + 2} {n + 1}$ is divergent Prove that the sequence {${x_n}$} to be defined by $x_n = (-1)^n \frac {3n + 2} {n + 1}$ is divergent.
Also, you can assume without proof that $\lim{n\to\infty}$ $\frac{3n+2}{n+1} = 3 $
Proposed Solution: (by contradiction)
Given that there were a limit $x$,... | Suppose the limit exists and call it $x$. Thus, for $\epsilon = \dfrac{1}{2}$, there is a $n$ such that $|x_n - x| < \dfrac{1}{2}, |x_{n+1} - x| < \dfrac{1}{2}\implies |x_{n+1} - x_n| < 1$. If $n$ is even, then $|x_n - x_{n+1}| = \left|\dfrac{3n+2}{n+1} + \dfrac{3n+5}{n+2}\right| > 2$, contradiction. The case $n$ is od... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Describe the locus of $w$ if $w=\frac{1-z}{1+z}$ and $z=1+iy$ (i.e $z$ is a complex number that moves along the line $x=1$) So I'm trying to solve the following problem: If $z=x+iy$, express $w=\frac{1-z}{1+z}$ in the form $a+bi$ and hence find the equation of the locus of $w$ if $z$ moves along the line $x=1$.
My atte... | $\,w=\frac{1-z}{1+z} \;\;(z \ne -1)\, \iff z=\frac{1-w}{1+w} \;\; (w \ne -1)\,$, then the condition $\,\text{Re}(z) = 1\,$ translates to:
$$
\require{cancel}
2 = 2\text{Re}(z)=z + \bar z = \frac{1-w}{1+w}+\frac{1-\bar w}{1+\bar w}
= \frac{\left(1-|w|^2-\cancel{w}+\bcancel{\bar w}\right)+\left(1-|w|^2+\cancel{w}-\bcanc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum \frac{a^3}{a^2+b^2}\le \frac12 \sum \frac{b^2}{a}$
Let $a,b,c>0$. Prove that
$$ \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\le \frac12 \left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).\tag{1}$$
A idea is to cancel the denominators, but in this case Muirhead don't work because th... | Let me continue your approach. We will use the following simple observation:
Lemma. For positive $x$ and $y$ the following inequality holds
$$
\frac{x^2}{y}\ge 2x-y.
$$
Proof. It is equivalent to $(x-y)^2\ge 0$.
We need to prove that
$$
\sum_{cyc}\frac{(ab)^2}{a^3+ab^2}\ge \frac{1}{2}\sum_{cyc}\frac{2a^2-b^2}{a}.
$$
No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Six people (half are female, half are male) for seven chairs. Problem:
Suppose there are $7$ chairs in a row. There are $6$ people that are going to randomly
sit in the chairs. There are $3$ females and $3$ males. What is the probability that
the first and last chairs have females sitting in them?
Answer:
Let $p$ be th... | Here is an approach that is closest to your thought process. The three females and the three males are all indistinguishable, so we use combinations instead of permutations. Picking any one sex first gives:
$${7 \choose 3} \cdot {4 \choose 3}$$
total possibilities.
Now if two women are already sitting at the ends, then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$A+B+C+D=\pi$, and $0\leq A,B,C,D \leq \frac{\pi}{2}$. Prove that $\sin^4(A)+\sin^4(B)+\sin^4(C)+\sin^4(D)\leq 2$ I tried to simplify it from some ways.
(1).$\sin^4(A)+\sin^4(B)+\sin^4(C)+\sin^4(D)$
$=\left(\frac{1-\cos(2A)}{2}\right)^2+\left(\frac{1-\cos(2B)}{2}\right)^2+\left(\frac{1-\cos(2C)}{2}\right)^2+\left(\frac... | A bit of an ugly solution and would love a clever one.
First, let's assume wlog $D \le C \le B \le A$ and then $D+C \le \pi/2$ and we show that $\sin^4C+\sin^4D \le \sin^4(C+D)$ so we reduce the problem to three angles where we use calculus.
Noting that $\sin^2x-\sin^2y=\sin(x-y)\sin(x+y)$, one has $\sin^4(C+D)-\sin^4C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n}=\frac n4\csc\frac{2\pi}n$ I would like to evaluate the sum
$$\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n}
$$
which is said to reduce to the simple close-form $\frac n4\csc\frac{2\pi}n$. I have verified it numerically for a large number ... | Noting that
$$
\begin{aligned}
\sum_{k=1}^{n-1} k \sin (k \theta) &=-\sum_{k=1}^{n-1} \frac{d}{d \theta}(\cos (k \theta))=-\frac{d}{d \theta}\left(\sum_{k=1}^{n-1} \cos k \theta\right)
\end{aligned}
$$
Using the identity
$$\sum_{k=1}^{n-1} \cos k \theta=
\frac{1}{2}\left[-\cos (n \theta)+\cot \left(\frac{\theta}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Using Lagrange Mult. to find maximum of $2x + y^2$ with constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ I was trying to maximize the function $2x + y^2$ with the following constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ using Lagrange multipliers. I already did all the work of proving that the maximu... | $x^2+y^2+z^2=2$.
$x+y^2+z=0.$
Minimize: $2x+y^2$
$z=-(x+y^2)\implies z^2=x^2+2xy^2+y^4.$
$x^2+y^2+(x^2+2xy^2+y^4)=2\implies 2x^2+(2x+1)y^2+y^4=2$
$4x+2y^2=2\lambda\implies 2x+y^2=\lambda$
$2y(2x+1)+4y^3=2y\lambda$
$y=0$ or $(2x+1)+2y^2=\lambda\implies 1+y^2=0$
Case 1: $y=0$.
$x^2+z^2=2$ and $x+z=0$. Then $x=\pm1$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Functional analysis problem. Find the smallest value of $f(x) = \frac{x^2}{x-9}$ on the interval $(9, +\infty)$.
We should basically find the biggest $a$ such that $\frac{x^2}{x-9} \geq a$. We can multiply both sides by $x-9$ since it's positive and than we get $x^2-ax+9a \geq 0.$ I don't know how to proceed.
| Rephrased:
$y:=x-9,$ $y\in (0,\infty)$;
$f(y)=\dfrac{(y+9)^2}{y} =$
$y+18+\dfrac{81}{y}=$
$(y^{1/2}-\dfrac{9}{y^{1/2}})^2+18+18 \ge 36;$
Equality for
$y^{1/2}=\dfrac{9}{y^{1/2}}$,
$y=9.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_0^\infty\frac{x^2\ln x}{x^4+x^2+1}dx$ by the residue theorem The result should be $\frac{\pi^2}{12}$.
Edit: I have tried to reproduce the image, but limitations of MathJax required some reformatting. Here is the original image.
$$
\begin{align}
\int_0^\infty\frac{x^2\ln x\,\mathrm{d}x}{x^4+x^2+1}&=\int_0... | The Residues
$\newcommand{\Res}{\operatorname*{Res}}$
If we have $g(z)$ with a simple zero at $z=z_0$, then
$$
\begin{align}
\Res_{z=z_0}\left(\frac{f(z)}{g(z)}\right)
&=\lim_{z\to z_0}\frac{(z-z_0)f(z)}{g(z)}\tag{1a}\\
&=\frac{f(z_0)}{g'(z_0)}\tag{1b}
\end{align}
$$
Applying $(1)$, we get
$$
\begin{align}
\Res_{z=e^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4387831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati) $y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati)
(a) Find the solutions.
(b) $y(x_0)=y_0$, prove two cases:
$$0<y_0<x_0 \implies \text{solution's domain is} [x_0,\infty) $$
$$0<x_0<y_0 \implies \text{solution's domain is} [x_0,x_0+\alpha) , \alpha \in \mathbb {R}.$$
I will b... | The equation is homogeneous, so you can use $v=\frac{y}{x}$ so that $y=vx$ and $y'=v+xv'$. Then
$$v+xv'= 2 -3v +2v^2$$
$$xv' = 2(1-2v +v^2)$$
Then separating variables
$$\int \frac{dv}{(v-1)^2} = \int\frac{2dx}{x}$$
$$-\frac{1}{v-1} = \ln(x^2) -C$$
$$ v-1 = \frac{1}{C-\ln(x^2)}$$
$$ v = 1+\frac{1}{C-\ln(x^2)}$$
Then
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4389227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculating $\int \frac{dx}{\sqrt[3]{(x+1)^2(x-1)^7}}$ I am trying to calculate the integral
$$
\int \frac{dx}{\sqrt[3]{(x+1)^2(x-1)^7}}.
$$
I know that the answer is
$$
\frac3{16}(3x-5)\sqrt[3]{\frac{x+1}{(x-1)^4}}+C;
$$
this led me to the idea of introducing the new variable
$$
t^3=\frac{x+1}{(x-1)^4}.
$$
But that go... | Substitute $t^3=\frac{x+1}{x-1}$. Then, $x=\frac{t^3+1}{t^3-1}$, $dx= -\frac{6t^2}{(t^3-1)^2}dt$ and
\begin{align}
&\int \frac{1}{\sqrt[3]{(x+1)^2(x-1)^7}}dx\\
=&-\frac34 \int (t^3-1)dt = -\frac3{16}t^4 +\frac34t +C\\
= & -\frac3{16}\left(\frac{x+1}{x-1} \right)^{\frac43} +\frac34 \left(\frac{x+1}{x-1} \right)^{\frac13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How can I show that $\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$ exists, using limit def? I am trying to solve an exercise to show that this function $$f(x,y)=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$$ has a limit as $(x,y)$ approaches $(0,0)$:
$$\underset{\left(x,y\right... | A proof involving geometry:
Set $r^2:=x^2+y^2$, where $r \gt 0$.
Need to show: $(r^2+1)^{1/2}-1 \lt r$.
Consider a right triangle with the $2$
legs $r$ and $1$.
Pythagoren theorem:
The hypothenuse is given by $(r^2+1)^{1/2}.$
In any triangle:
The sum of $2$ sides is greater than the $3$rd side.
Hence
$r+1> (r^2+1)^{1/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Confusions in Holder's Inequality Holder's Inequality states that for nonnegative real numbers $a_1,...,a_n$ and $b_1,...,b_n$ we have $$\left(\sum_{i=1}^na_i\right)^p\left(\sum_{i=1}^nb_i\right)^q\ge \left(\sum_{i=1}\sqrt[p+q]{a_i^pb_i^q}\right)^{p+q}$$
Where $p$ and $q$ are positive real numbers.
Here is my problem :... | It should be $a^5 - a^2 + 3$ etc. rather than $a^5 - a^2 + 2$ etc.
Problem: Let $a, b, c $ be positive reals. Prove that
$$(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\ge (a+b+c)^3.$$
Proof: It is easy to prove that $x^5 - x^2 + 3 \ge x^3 + 2$ for all $x \ge 0$.
It suffices to prove that
$$(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3.$$
Us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $ax^2+bx+c$ is irreductible, then exists constants $k_1,k_2,k_3$ such that $ax^2+bx+c = k_1((k_2x+k_3)^2+1).$ I want to prove that:
If $ax^2+bx+c$ is irreductible, then exists constants $k_1,k_2,k_3$ such that
$$ax^2+bx+c = k_1((k_2x+k_3)^2+1)$$
We note that
\begin{align*}
&ax^2+bx+c = \frac{1}{4a}(4a^2x^2+4abx+4a... | $$
q(x) = \frac{1}{4a} ((2ax+b)^2 - (b^2 - 4ac) )\\
D = b^2 - 4ac < 0\\
r \equiv \sqrt{-D}\\
q(x) = \frac{1}{4a} ((2ax+b)^2 + r^2 )\\
= \frac{r^2}{4a} (\frac{(2ax+b)^2}{r^2} + 1 )\\
= \frac{r^2}{4a} ( (\frac{2a}{r} x + \frac{b}{r})^2 + 1)\\
k_1 = \frac{r^2}{4a}\\
k_2 = \frac{2a}{r}\\
k_3 = \frac{b}{r}\\
$$
Note that yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$ with other approaches
$$\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$$
$$x=?$$
I solved this problem as follow:
$x=0$ is not a root, we can divide numerator and denominator of each fraction by $x$:
$$\frac{3}{t+1}+\frac{2}{t-1}=3\quad\quad\text{where $t=x+\frac1x$}$$
... | We have $(x-1)^2\ge 0$ with equality if and only of $x=1$.
But
\begin{align}
(x-1)^2 \ge 0 &\iff x^2-2x+1\ge 0\\
&\iff x^2+x+1\ge 3x\\
&\iff \frac{3x}{x^2+x+1}\le 1
\end{align}
and
\begin{align}
(x-1)^2 \ge 0 &\iff x^2-2x+1\ge 0\\
&\iff x^2-x+1\ge x\\
&\iff \frac{x}{x^2-x+1}\le 1\\
&\iff \frac{2x}{x^2-x+1}\le 2
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to obtain the asymptotic behavior of this integral? Consider the integral
$$ I(x) = \int_{0}^{\infty} \exp \bigl( - s^2/2 \bigr) \cos \bigl( xs + \lambda s^3 \bigr) \, \text{d} s,$$
where $\lambda >0$ is a constant. I would like to know the asymptotic behavior of this integral in the limit $x \rightarrow \infty$. A... | This is an elaboration on Maxim's comment. Your integral is
$$
I(x) = \frac{1}{2}\int_{ - \infty }^{ + \infty } {\exp \left( { - \frac{1}{2}s^2 + i\lambda s^3 + ixs} \right)ds} .
$$
With the change of variables $s = \frac{t}{{(3\lambda )^{1/3} }} + \frac{1}{{6i\lambda }}$, we find
$$
I(x) \!=\! \frac{1}{2}\frac{1}{{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4403218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\left(\frac{a+b+c}{3}\right)^p\leq \frac{a^p+b^p+c^p}{3}$. Prove:
Let $p$ be an integer greater than $1$. Suppose $a,b,c$ be positive real numbers. Then $\left(\frac{a+b+c}{3}\right)^p\leq \frac{a^p+b^p+c^p}{3}$.
By AM-GM, I get $\frac{a+b+c}{3}\geq (abc)^{1/3}$ and $\frac{a^p+b^p+c^p}{3}\geq (a^pb^pc^p)^{1/3}$.... | To make the problem less tedious, let $x = \dfrac{a}{a+b+c}, y = \dfrac{b}{a+b+c}, z = \dfrac{c}{a+b+c}$, then you prove: $x^p+y^p+z^p \ge \dfrac{1}{3^{p-1}}$, with $x+y+z=1$. Let $u = 3x, v = 3y, w = 3z$, then you again prove: $u^p+v^p+w^p \ge 3$, with $u+v+w = 3$. Observe that by AM-GM inequality: $u^p + 1(p-1) \ge p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find an orthonormal basis and the signature of the quadratic form
Consider the quadratic form given by the matrix below (in the canonical basis)
\begin{pmatrix}
1 & 1 & -1\\
1 & 1 & 3\\
-1 & 3 & 1
\end{pmatrix}
Find an orthonormal basis of it and find its signature.
First I calculated the eigenvalues, which are $4, ... | Using SymPy:
>>> from sympy import *
>>> A = Matrix([[ 1, 1,-1],
[ 1, 1, 3],
[-1, 3, 1]])
>>> Q, D = A.diagonalize(normalize=True)
>>> D
Matrix([
[4, 0, 0],
[0, -1/2 + sqrt(17)/2, 0],
[0, 0, -sqrt(17)/2 - 1/2]])
>>> simplify... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rhombus rotates in a circle. Two vertexes of the rhombus is on the circle, rotate the rhombus $ABCD$ clockwise around the point A to the rhombus $AB'C'D'$, where the point $B'$ falls on the circle, link $B'D,C'C$.
If $B'D:CC'=4:3$, then value of $tan\angle BAD$ will be?
It seems $B',B,D'$ or $D,D',C'$ are in a straight... | If the angle of rotation is $\phi$ (clockwise) then
$ CC' = 2 \overline{AC} \sin(\dfrac{\phi}{2}) $
and
$ BD' = 2 \overline{AD} \sin(\angle BAC + \dfrac{\phi}{2} ) $
Now, $ \dfrac{\overline{AD}}{\overline{AC}} = \dfrac{1}{2} \sec(\angle BAC) $
Also, from the figure, we have $\phi = \pi - 4 \angle BAC $
Let $ x = \angl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find $k\in\mathbb{R}$ given $w = k+i$ and $z=-4+5ki$ and $\arg(w+z)$ I am working on problem 2B-11 from the book Core Pure Mathematics, Edexcel 1/AS. The question is:
The complex numbers $w$ and $z$ are given by $w = k + i$ and $z = -4 + 5ki$ where $k$ is a real constant. Given at $\arg(w + z) = \frac{2\pi}{3}$, find t... | It seems to me that one can simply say
$$
w+z=k+4+i(1+5k)\\
\arg(w+z)=\tan^{-1}\frac{1+5k}{k+4}=\frac{2\pi}{3}\\
\frac{1+5k}{k+4}=\tan\frac{2\pi}{3}=-\sqrt{3}\\
k=\frac{4\sqrt{3}-1}{5+\sqrt{3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $f(x)=\frac {x^ 2 -2x +4}{ x^ 2 +2x+4}$ for $x \in \mathbb{R}$, prove that the range of $f(x)$ is $[1/3, 3]$ One method to solve it would by putting $y = f(x)$ then multiplying the denominator with $y$ hence making a quadratic equation in x then we can just use the inequalities for $x$ being real to prove it.
For an... | Notice that $f(-x) = \frac{1}{f(x)}$, so the analysis we do for one side will apply to the other (aka take $x\geq0$ WLOG).
We have
$$\frac{x^2-2x+4}{x^2+2x+4} \leq 1$$
with equality achieved at least at $x=0$. Then rewriting the function we have
$$\frac{(x-2)^2+2x}{(x-2)^2+6x} = 1 - \frac{4x}{(x-2)^2+6x} \geq 1 - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4424161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Spivak Calculus, Ch 10, problem 32b: Can we use Leibniz's formula to calculate $f^{(k)}(x)$ if $f(x)=\frac{1}{x^2-1}$? Spivak's Calculus, Chapter 10 on Differentiation, problem 32:
*What is $f^{(k)}(x)$ if
a)$f(x) = \frac{1}{(x-1)^n}$
*b) $f(x)=\frac{1}{x^2-1}$
My question regards part $b)$.
$a)$ can be solved qui... | The calculation can be done in both ways. On the one hand we have
\begin{align*}
\color{blue}{f^{(k)}(x)}&=\left(\frac{1}{x^2-1}\right)^{(k)}\\
&=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)^{(k)}\\
&=\frac{1}{2}\left(\left((x-1)^{-1}\right)^{(k)}-\left((x+1)^{-1}\right)^{(k)}\right)\\
&=\frac{1}{2}\left((-1)(-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4426875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Elegant solutions to $ \int_{0}^{\frac{\pi}{2}} x\tan 2 x \ln (\tan x) d x $ . Letting $x\mapsto \frac{\pi}{2} -x$ converts the integral
$\displaystyle I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \tan 2 x \ln (\tan x) d x \tag*{} $
Using the identity $ \displaystyle \tan x=\frac{\sin 2 x}{1+\cos 2 x} $ , we get
$$\display... | Substitute $t=\tan^2x$, along with $\tan 2x =\frac{2\tan x}{1-\tan^2x}$\begin{align}
& I
= \frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \tan 2 x \ln (\tan x) d x = \frac\pi8 \int_0^\infty\frac{\ln t}{1-t^2}dt
=\frac\pi8\left(-\frac{\pi^2}4\right)=-\frac{\pi^3}{32}
\end{align}
where $\int_0^\infty \frac {\ln t}{1-t^2}dt=-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4427985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find $\angle{CAD}$ in equilateral triangle $\triangle{ABC}$ for $D$ inside with $\angle{ABD}=18^{\circ}, \angle{BCD}=12^{\circ}$
This problem is feasible to solve using trigonometric functions. I am looking for pure geometric solution as usual. Thanks...
|
Let $E$ be the circumcenter of $\triangle{BCD}$, so
$BE=CE=DE, \angle{BED}=2\angle{BCD}=2*12^{\circ}=24^{\circ} \\
\implies \angle{EBD}=\angle{BDE}=\dfrac{180^{\circ}-24^{\circ}}{2}=78^{\circ} \\
\implies \angle{EBC}=\angle{ECB}=78^{\circ}-42^{\circ}=36^{\circ} \\
\implies \angle{BEC}=108^{\circ}$
Find $F$ so that $BE... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
An inequality $2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd)$
Let $a, b, c, d \ge 0$ such that $a+b, a+c, a+d, b+c, b+d, c+d \le 1$. Show that
$$2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd).$$
I am trying to maximise
$$f(a,b,c,d)=(a+b+c+d) /(1-(1-a)(1-b)(1-c)(1-d) +abcd) $$ over the set of nonnegativ... | Remark: Here is a proof which is similar to my answer
in An inequality $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$
Let $x = a + b, \, y = c + d$.
We have $x, y \in [0, 1]$.
We have
\begin{align*}
\mathrm{RHS} - \mathrm{LHS} &= (x + y) + 2(ab y + cdx) - 2(ab + cd + xy)\\
&= (x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Turning a pole/ladder horizontally at a corner I've been attempting to solve the following problem:
A lane runs perpendicular to a road 64 ft wide. If it is just possible to carry a pole 125 ft long from the road into the lane, keeping it horizontal, then what is the minimum width of the lane?
At first, I tried to solv... | From the figure attached, we'll relate the maximum possible pole length $L$ to $w$, the width of the lane.
From the figure
$L = w \csc(\theta) + 64 \sec(\theta)$
for a given $w$, we want to find the minimum attainable length $L$ and this critical value will be our maximum possible pole length $L$
$\dfrac{dL}{d\theta} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find integral of $\sqrt{x}$ using Riemann sum definition Let $a > 1$ be a real number. Evaluate the definite integral
\begin{equation}
\int_{1}^{a} \sqrt{x} \,dx
\end{equation}
from the Riemann sum definition.
My approach I know a Riemann sum consists of a sigma notation with a width and function. However, I am confuse... | Using the hints and tips provided, I successfully proved the integral from the Riemann sum definition:
The function $f(x)=\sqrt{x}$ is continuous on $[1,a]$, hence integrable on $[1,a]$. For every positive integer $n$, we consider the left Riemann sum of $f$ with respect to the partition $[1,a^{1/n},a^{2/n},a^{1/n}...a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that $p^k \mid \mid (x-y)$ iff $p^k \mid \mid (x^6-y^6)$. Definition. Let $n>1$ be an integer and $p$ be a prime. We say that $p^k$ fully divides $n$ and write $p^k \mid \mid n$ if $k$ is the greatest positive integer such that $p^k \mid n$.
Let $p>6$ be a prime. Let $x$ and $y$ be two distinct integers such that ... | $(\implies)$
Let $p^k \mid \mid (x-y)$. Then $p^k \mid (x-y)$ with $k$ is the greatest positive integer such that $x-y = p^km$ for some nonzero integer $m$. Notice that we must have $p \nmid m$, because otherwise we would have $m=pc$ for some nonzero integer $c$ such that $x-y=p^{k+1}c$, which implies $p^{k+1} \mid (x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Is my construction of one-point compactification of $\mathbb{R}$ correct? (+ Clarifying questions) Could you please check if this construction makes sense and answer to questions to the parts in bold?
My construction:
The construction can be given explicitly as an inverse stereographic projection. Consider the map
$s: ... | Stereographic projection from $p = (-1,0)$ is given by
$$\sigma : S^1 \setminus \{p\} \to \mathbb R, \sigma(x,y) = \frac{y}{1+x}.$$
Your map $s$ has the property $p \notin s(\mathbb R)$ (because $s(x)$ has second coordinate $0$ only for $x = 0$, and $s(0) = (1,0)$). Thus we can regard $s$ as a map $s : \mathbb R \to S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there any closed form for $\frac{d^{2 n}( \cot z)}{d z^{2 n}}\big|_{z=\frac{\pi}{4}}$? Latest Edit
Thanks to Mr Ali Shadhar who gave a beautiful closed form of the derivative which finish the problem as
$$\boxed{S_n = \frac { \pi ^ { 2 n + 1 } } { 4 ^ { n + 1 } ( 2 n ) ! } | E _ { 2 n } | }, $$
where $E_{2n}$ is an ... | $$\begin{align}\cot(x+\pi/4)&=\sec2x-\tan2x\\&=\sum_{n\ge0}\frac{(-1)^n}{(2n)!}E_{2n}(2x)^{2n}+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}\cdot2^{2n}(2^{2n}-1)B_{2n}(2x)^{2n-1}\\&=1+\sum_{n\ge1}\frac{(-1)^n2^{2n}}{(2n)!}E_{2n}x^{2n}+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}2^{4n-1}(2^{2n}-1)B_{2n}x^{2n-1}\\\implies\cot z&=1+\sum_{n\ge1}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to factorize $2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c$? If $a,b,c$ are in AP $\implies \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are in AP $\implies
\frac{ab+bc+ca}{bc}, \frac{ab+bc+ca}{ca}, \frac{ab+bc+ca}{ab}$ are in AP $\implies \frac{bc+ca}{bc},\frac{ab+bc}{ca},\frac{bc+ca}{ab}$ are in AP $\implies \frac{bc}{bc+ca}, \f... | Given expression is $2(a+c)b^2-(a^2+c^2)b-ac(a+c)$
=$2(a+c)b^2-(a+c)^2b+2acb - ac(a+c)=(a+c)b\left[2b-(a+c) \right]+ac\left[2b-(a+c)\right]$
$=(ab+bc+ca)(2a-b-c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Determining if a die is fair or not by rolling six times and observing only the sum Suppose you have two bags and each has 6 dice. In one bag all the die are fair. In the other each die has a bias - die “1” has a probability of returning 1 of $\frac{1}{6} +\epsilon$ and $\frac{1}{6} -\frac{\epsilon}{5}$ for any of the ... | So let's look at $P(\mathrm{sum}=S \mid \mathrm{die\ biased\ toward\ } k)$.
Each roll of the biased die is equivalent to a process where we first choose between case $A$ with probability $\frac{6}{5} \epsilon$ and case $B$ with probability $1 - \frac{6}{5} \epsilon$. In case $A$, the result of the process is $k$. In ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Question from isi previous years (a) Show that $\left(\begin{array}{l}n \\ k\end{array}\right)=\sum_{m=k}^{n}\left(\begin{array}{c}m-1 \\ k-1\end{array}\right)$.
(b) Prove that
$$
\left(\begin{array}{l}
n \\
1
\end{array}\right)-\frac{1}{2}\left(\begin{array}{l}
n \\
2
\end{array}\right)+\frac{1}{3}\left(\begin{array}{... | Suppose we seek to evaluate
$$S_n = \sum_{k=1}^n \frac{(-1)^{k-1}}{k}{n\choose k}.$$
We introduce the function
$$f(z) = n! (-1)^{n-1} \frac{1}{z} \prod_{q=0}^n \frac{1}{z-q}.$$
We have for $1\le k\le n$ that
$$\mathrm{Res}_{z=k} f(z) =
n! (-1)^{n-1} \frac{1}{k}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What would be the Mean shortest distance from random points in the right angled triangle to the Hypotenuse. The problem is to find the average shortest distance of uniformly random points from the hypotenuse in a right angled rectangle.
The distance d shows the shortest distance to the hypotenuse from a random point N ... | Take $P$ to be the origin. Then the equation of the hypotenuse is
$ y = \dfrac{b}{a} x , \hspace{15pt} x \in [0, a] $
Now pick a point $ (x_1, y_1) $ such that $ y_1 \in [0, \dfrac{b}{a} x_1 ], then its perpendicular distance from the hypotenuse is
$d (x_1) = \dfrac{ \dfrac{b}{a} x_1 - y_1 } {\sqrt{1 + \left(\dfrac{b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine all solutions of the linear equation system $A \vec{x} = \vec{b} = \begin{bmatrix} 2 \\ 4 \\ 18\\ -10 \end{bmatrix}$ Could you give me your feedback ?
Determine **all** solutions of the linear equation system $A \vec{x} = \vec{b}$ for
$$A = \begin {bmatrix}
1 & -2 & 4 & 3 & 2 \\
-7 & 14 & -28 & -12 & -23 \\
... | I took these steps
*
*$(01)~ R_2: R_2 + 7 R_1$
*$(02)~ R_3: R_3 - 4 R_1$
*$(03)~ R_4: R_4 + 3 R_1$
*$(04)~$ Swap $R_2$ with $R_4$
*$(05)~ R_3: R_3 + \dfrac{1}{2} R_2$
*$(06)~$ Swap $R_3$ with $R_4$
*$(07)~ R_4: R_4 -\dfrac{4}{9} R_3$
*$(08)~ R_3: \dfrac{1}{9} R_3$
*$(09)~ R_2: R_2 - 8 R_3$
*$(10)~ R_1: R_1 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does this proof of the binomial expansion (a+b)^2 work? I was rereading Terence Tao's Analysis 1 and found this question in the section:
Exercise $2.3.4.$ Prove the identity $(a + b)^2 = a^2 + 2ab + b^2$ for all natural
numbers a, b.
Prior to this we already have proved:
$1.a\cdot b=b\cdot a\\2.a\cdot b=0\implies a=0... | Here is the modified version of your proof with few skipped steps added and properties mentioned.
\begin{align}
(x+y)^2 &=(x+y).(x+y) & \\
&=(x(x+y))+(y(x+y))&\text{(using property 4 and 1)} \\
&=(x.x +x.y )+(y.x +y.y) &\text{(using property 4 again)}\\&=x^2+(x.y+y.x)+y^2 &\text{(using associativity of addition)}\\ &=x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve for x: $\lfloor x\rfloor \{\sqrt{x}\}=1$ where $\{x\}=x-\lfloor x\rfloor$ Solve for x:
$\lfloor x\rfloor \{\sqrt{x}\}=1$
where $\{x\}=x-\lfloor x\rfloor$
My Attempt:
I took intervals of $x$ as $x\in (2,3)$ so $\sqrt x\in (1,2)$. Due to which $\lfloor x\rfloor=2$ and $\{\sqrt x\}=\sqrt x-\lfloor \sqrt x\rfloor=\sq... | We can proceed similar as you did. Let $n=\lfloor x\rfloor$. From the assumption $\lfloor x\rfloor\{\sqrt x\}=1$ one concludes $\{\sqrt x\}=\frac1n$. Since $\sqrt x$ changes its integral part at integral values, we can conclude that $\lfloor\sqrt n\rfloor=\lfloor\sqrt x\rfloor$, so that
$$\sqrt x=\lfloor\sqrt x\rfloor+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculate: $\Delta=\left|\begin{array}{ccc} b c & c a & a b \\ a(b+c) & b(c+a) & c(a+b) \\ a^{2} & b^{2} & c^{2} \end{array}\right|$ Calculate:
$$\Delta=\left|\begin{array}{ccc}
b c & c a & a b \\
a(b+c) & b(c+a) & c(a+b) \\
a^{2} & b^{2} & c^{2}
\end{array}\right|$$
Does anyone know any easy way to calculate this dete... | \begin{gathered}
\left|\begin{array}{ccc}
b c & a c & a b \\
a b+a c & a b+b c & a c+b c \\
a^{2} & b^{2} & c^{2}
\end{array}\right|=b c \cdot(a b+b c) \cdot c^{2}+a c \cdot(a c+b c) \cdot a^{2}+a b \cdot(a b+a c) \cdot b^{2}-a^{2} \cdot(a b+b c) \cdot(a b)-b^{2} \cdot(a c+b c) \cdot(b c)-c^{2} \cdot(a b+a c) \cdot(a c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solving $\tan ^{-1}(\frac{1-x}{1+x})=\frac{1}{2}\tan ^{-1}(x)$ I was solving the following equation,$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
But I missed a solution (don't know where's the mistake in my work).
Here's my work:
$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\ta... | Thank you to @Robin'sPremiumCoffee for a nice algebraic solution. Now the problem with your original method is that although:
$$\frac{1-\tan \theta}{1+\tan \theta} = \tan(\frac \pi 4 -\theta)$$
It is also equal to:
$$\frac{1-\tan \theta}{1+\tan \theta} = \tan(\frac \pi 4 -\theta)=\tan(\pi +\frac \pi 4 -\theta)$$
This i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find $\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$? By factorization:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}\tag{1}$$
$$=\lim_{x\to-\infty} \frac{x\sqrt{1+\frac{2}{x}}}{-x}$$
$$=\lim_{x\to-\infty}-\sqrt{1+\frac{2}{x}}$$
If I input $x=-\infty$, the limiting value seems to be $-1$. But according to desmos, ... | \begin{eqnarray}
\frac{\sqrt{x^2+2x}}{-x}&=&\frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}}{-x}\\
&=&\frac{|x|\sqrt{1+\frac{2}{x}}}{-x}\\
&=&\frac{-x\sqrt{1+\frac{2}{x}}}{-x} \text{ since }x<0\\
&=&\sqrt{1+\frac{2}{x}}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How many different whole numbers are factors of number $2 \times 3 \times 5 \times 7 \times 11 \times 13$? The question is:
How many different whole numbers are factors of number $2 \times 3 \times 5 \times 7 \times 11 \times 13$?
My answer to this question is $63$ but the right answer is $64$. I don't know why it is... | You can obtain the solution as well by omitting the factor 1. One unique whole number is if you multiply all 6 primes. The corresponding sequence is $2,3,5,7,11,13$. So we have one sequence. The corresponding binomial coefficient is $\binom{6}{0}=1$
Then we can remove one of the 6 factors. This can be done with all 6 p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\sum_{k=1}^\infty\frac{1}{x_k^2-1}$ where $x_1=2$ and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}$ for $n \ge 2$ Given $x_1=2$
and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}, n\geq 2$
Prove that $y_n=\sum_{k=1}^{n}\frac{1}{x_k^2-1}, n\geq 1$ converges and find its limit.
*
*To prove a convergence we can jus... | Making use of the following equality, and standard partial fraction decompositions: $$x_{k+1}^2-(x_k+1)x_{k+1}-1=0\implies(x_k+1)=\frac{x_{k+1}^2-1}{x_{k+1}}$$
$$\begin{align}y_n&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{1}{x_k+1}\right)\\&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{x_{k+1}}{x_{k+1}^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $2\arccos(\frac45)-\arcsin(\frac45)=\arctan(y)$ then find the value of $y$ If $2\arccos(\frac45)-\arcsin(\frac45)=\arctan(y)$ then find the value of $y$
My Attempt:
Using $2\arccos(x)=\arccos(2x^2-1)$, I get
$$2\arccos(\frac45)=\arccos(2\times\frac{16}{25}-1)=\arccos(\frac7{25})$$
Also,
$$\arcsin(\frac45)=\arccos(\f... | Since $$7^2 + 24^2 = 25^2$$ and $$3^2 + 4^2 = 5^2,$$ it follows that
$$\frac{24}{25} = \sqrt{ 1 - \frac{7^2}{25^2}},$$ and $$\frac{4}{5} = \sqrt{1 - \frac{3^2}{5^2}}.$$
Hence
$$\frac{7}{25} \cdot \frac{3}{5} + \sqrt{ 1 - \frac{7^2}{25^2}} \sqrt{1 - \frac{3^2}{5^2}} = \frac{21}{125} + \frac{96}{125} = \frac{117}{125}.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show $a^3+b^3+c^3+d^3 \le 27$ Given $a, b, c, d$ are real numbers and $a^2+b^2+c^2+d^2 = 9$ show that
$$a^3+b^3+c^3+d^3\le27$$
I have tried to do the following:
$$(a^2+b^2+c^2+d^2)^{\frac{3}{2}} = 27 \\ \text{after simplifying} \\ (a^3+b^3+c^3+d^3) + \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=27$$
By setting the fol... | Sorry didn't really read your solution, but this might work:
Clearly $|a|,|b|,|c|,|d|\leq 3$ and thus $a^3\leq 3a^2$ and so on... and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
what is the solution of the ordinary differential equation? would have any complete solution to the equation plus complete below?
$$
y'= \frac{4y-{3x}}{2x-y}
$$
a part of the solution:$$
y'=xv'+v
$$
$$
y=xv
$$
not complete solution
$$
xv'+v= \frac{4xv-{3x}}{2x-xv}
$$
Continuation
$$
xv'+v= \frac{2v-{3}}{1-v}
$$
Contin... | That's a homogeneous differential equation. I'd solve it like this:
divide by x $$y' = \frac{4\left(\frac{y}{x}\right) - 3}{2 - \left(\frac{y}{x}\right)}$$
Set $v = \frac{y}{x}$, so $y = xv$ and $y' = v + xv'$
Substitute: $$v + xv' = \frac{4v - 3}{2 - v}$$
$$xv' = \frac{4v - 3}{2 - v} - v$$
$$xv' = \frac{4v - 3 - 2v + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine the image of the unit circle $S^1$ by the action of the matrix $e^A$. We have:
$$e^{ \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} }$$
I need to determine the image of the unit circle $S^1$ by the action of the matrix $e^A$.
I think that I know how to calculate $e^A$:
I get the Jordan decomposition:
$$A = \be... | This address the general question of shape of image of an unit circle under the action of $e^A$. We can think of a linear dynamic system $\dot x=Ax$, the solution of which will be $x=e^{tA}$. Then what you are seeking is the ending position of points starting from the unit circle, after following linear dynamics for $t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
To prove an Inequality: $ ( x^2 +2x)e^x + (x^2-2 x)e^{-x} \ge 0$
$ \left(x^2 +2x\right)e^x + \left(x^2-2 x\right)e^{-x} \ge 0$.
I used photomath to plot its graph: $y=(x^{2}+2x))e^{x} + \frac{{x}^{2}-2x}{{e}^{x}}$
But how do I prove it without an image? Should I take the derivative of it and reason, please tell me ... | HINT:
$f(x) =(x^2 + 2 x) e^x$ has a Taylor series with all coefficients positive. So $f(x) + f(-x)$ will have a Taylor series with all coefficients $\ge 0$.
$\bf{Note}$: the function $g(x) = x e^x$ has a positive Taylor expansion at $0$, so $g(x) + g(-x) \ge 0$. We get stronger inequalities
$(x^2 +a x) e^x + (x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Is it possible to expand $\arctan x \ln(1-x^2)$ in series? I am not sure if its possible to find the Taylor series of $\arctan x \ln(1-x^2)$ so I am giving it a try:
By using the Cauchy product
$$\left(\sum_{n=1}^\infty a_n x^n\right)\left(\sum_{n=1}^\infty b_n x^n\right)=\sum_{n=1}^\infty \left(\sum_{k=1}^n a_k b_{n-k... | A bit tricky but it makes things simpler.
Rewrite
$$\tan ^{-1}(x)\log \left(1-x^2\right) = x \frac{\tan ^{-1}(x)}{x}\log \left(1-x^2\right)$$ and let $x=\sqrt t$
$$\frac{\tan ^{-1}(\sqrt t)}{\sqrt t}=\sum_{n=0}^\infty \frac{(-1)^n }{2 n+1}t^n$$
$$\log \left(1-t\right)=-\sum_{n=1}^\infty \frac {t^n} n$$
Where is see a p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Study stability of a linear time-invariant system by its 5x5 A matrix I need to study the Lyapunov stability of these matrices.
Is there a criterion that does not need to calculate the eigenvalues?
For example if the trace of a matrix is > 0 then it is unstable (like the third one).
$$
\begin{bmatrix}
-1 & 2 & 0 & -5 &... | Note that these matrices are block triangular. Thus, each matrix is stable iff its blocks on the diagonal are each (separately) stable.
For example, the second matrix can be partitioned as
$$
A =
\left[\begin{array}{cc|ccc}
1 & 2 & 0 & 0 & 0 \\
-5 & -8 & 0 & 0 & 0 \\
\hline
-1 & 3 & -12 & 1 & 1 \\
1 & -3 & 0 & -3 & 0 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum value of a trigonometric function If the minimum value of $f\left(x\right)=\left(1+\frac{1}{\sin ^6\left(x\right)}\right)\left(1+\frac{1}{\cos ^6\left(x\right)}\right),\:x\:∈\:\left(0,\:\frac{\pi }{2}\right)$ is $m$, find $\sqrt m$.
How do I differentiate this function without making the problem unnece... | Let $s\equiv\sin x$ and $c\equiv\cos x$ then
$$
\begin{align}
f(x)
&=
\left(1+{1\over s^6}\right)
\left(1+{1 \over c^6}\right)
\\
&=
{s^6+1\over s^6}{c^6+1 \over c^6}
\\
&=
{(sc)^6+s^6+c^6+1\over (sc)^6}
\\
\end{align}
$$
Now $s^6+c^6=(s^2+c^2)^3-3(sc)^2(s^2+c^2)=1-3(sc)^2$
so
$$
f(x)=1+{2\over (sc)^6}-{3\over (sc)^4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
How to solve $\sin(x) = \pm a$ for $a \not = 0$? I was solving the below equation: $\left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| = 1$ for $x \in [0, 2\pi]$.
My attempt:
$$\begin{align}&\left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| = 1\\\implies& \left|\sqrt{2... | In order to solve $\sin(x)=a$; such that $a\in[-1,1]$ and $x\in [-\pi/2,\pi/2]$, one can use the $\arcsin$ function indeed for all $a\in[-1,1]$, $\arcsin(a)=x$ where $x\in[-\pi/2,\pi/2]$, in order to find all solutions one can use the periodicity of $\sin$, note that as long as $a\in [-1,1]$ there will be an infinite a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
help me to evaluate these integrals Find the value of $$\int\frac{x}{x^2-x+1} dx$$ and $$\int\frac{1}{x^2-x+1} dx$$ This was not the original question. Original question was tougher and I have simplified that to these two integrals. I am having a hard time evaluating these two integrals but what I know is that, in thes... | Integral No. 1
$$\int \frac{dx}{x^2-x+1}= \int \frac{dx}{(x-\frac12)^2+\frac34}$$ Substitute $x-\frac12=t$ so $du = dt$. The integral becomes $$\int \frac{dt}{(t^2+\frac34)}=\frac{2}{\sqrt3}\arctan\frac{t}{\frac{\sqrt3}{2}}+c= \frac{2}{\sqrt3}\arctan\frac{2(x-\frac12)}{\sqrt3}+c = \frac{2}{\sqrt3}\arctan\frac{2x-1}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"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.