Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solutions of sin(x) = cos(x) I know that the solutions to the equation $\sin(x) = \cos(x)$ are :
$ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°)
However when I try to solve it algebraically I get the following :
$$ \sin x = \cos x$$
$$ \sin^2 x = \cos^2 x$$
$$ \sin^2 = 1 - \sin^2 x$$
$$ 2\sin^2 x = 1$$
$$... | The problem is a pretty generic one. By squaring both sides of the equation earlier on, you introduced extraneous solutions. It's sort of like how in solving the equation
$$x+\sqrt{x+9} = 2$$
you can get to
$$\sqrt{x+9} = 2-x$$
and then square both sides to get
$$x+9 = 4 - 4x + x^2 \implies x^2 - 5x - 5 = 0$$
You shoul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve $\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-4x}}{x}$ without L'Hospital's Rule. I need to solve $\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-4x}}{x}$ without using L'Hospital's Rule. Using that rule I found the equation becomes $\lim_{x \to 0}(\frac{1}{\sqrt{1+2x}} - \frac{2}{\sqrt{1-4x}}) = \frac{-1}{\sqrt{1}}$.... | There is a mistake, it must be $$\frac{1}{\sqrt{1+2x}}+\frac{2}{\sqrt{1-4x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find the equation of a hyperbola given the asymptote, equation of axis and a point Given that a hyperbola has asymptote $y=0$, passes through the point $(1,1)$ and has axis $y=2x+2$, determine its equation.
The answer arrived at is $\displaystyle{4xy+3y^2+4y-11=0}$. However, I have had no success in reaching it.... | As pointed out in the comment by Jan-MagnusØkland, solving the problem requires using the property that the axis bisects the angle between the two asymptotes of the hyperbola. We are given one asymptote $y=0$ thus we can find the second one by reflecting $y=0$ across the axis $y=2x+2$. Let's rewrite it in the standard ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find two independent orthogonal eigenvectors We are given that $C$ is an $n\times n$ complex matrix, and $C$ is Hermitian ($C = C^*$). Now we define $C = A + Bi$ where $A$ and $B$ are real matrices.
Let $M$ be defined by
\begin{bmatrix}
A & -B \\
B & A
\end{bmatrix}
Given that $\lambda$ is an eigenvalue of $C$ ... | Given a Hermitian matrix
$C = C^\dagger \in M_{n \times n}(\Bbb C), \tag 1$
which has an eigenvalue $\lambda$ with associated eigenvector $z$:
$Cz = \lambda z, \tag 2$
it is well-known that
$C^\dagger = C \Longrightarrow \lambda \in \Bbb R; \tag 3$
since the entries of $C$ lie in $\Bbb C$, with real $\lambda$, we will ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find length of the arc of $y^2=x$ I am embarrassingly stuck on this example. My textbook provides the answer (picture below) and the steps, but I am unable to follow the math. I have been stuck for more then an hour.
Find the length of the arc of the parabola $y^2=x$ from (0,0) to (1,1).
Since $x=y^2$, we have $\frac{... | When you say:
"Here is where I get my $y$ value from: $$\sqrt{a^2 + y^2} = \sqrt{1+
> 2^2 y^2} \\
> = \sqrt{\frac{1}{4} + y^2}$$ Which gives me $y = \frac{1}{4} tan\theta$"
From here you want not $y = \frac{1}{4} \tan\theta$ but $y^2 = \frac{1}{4} \tan^2\theta$. Taking square roots gives $y = \frac{1}{2} \tan\theta$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation... | By false position:
Assume $x=10,y=3$, which fulfills the first equation, and let $x=10+x',y=3+y'$. Now, after simplification
$$3x'+2y'=0,\\5x'+4y'=2.$$
We easily eliminate $y'$ (using $4y'=-6x'$) and get
$$-x'=2.$$
Though this method is not essentially different from, say elimination, it can be useful for by-hand compu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 8
} |
Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the v... | Using Newton's identities
$$
\begin{aligned}
e_{1}&=p_{1}\\
2e_{2}&=e_{1}p_{1}-p_{2}\\
3e_{3}&=e_{2}p_{1}-e_{1}p_{2}+p_{3}\\
4e_{4}&=e_{3}p_{1}-e_{2}p_{2}+e_{1}p_{3}-p_{4}\\
5e_{5}&=e_{4}p_{1}-e_{3}p_{2}+e_{2}p_{3}-e_{1}p_{4}+p_{5}\\
\end{aligned}
$$
with $p_1=1,p_2=2,p_3=3,e_4=0, e_5=0$, we get $p_5 = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
Series of Beta Function Solve:
$$\int_0^\frac{\pi}{2}\frac{\sqrt{\sin{x}\cos{x}}}{\cos{x}+1}dx$$
I tried
$$\int_0^\frac{\pi}{2}\frac{\sqrt{\sin{x}\cos{x}}}{\cos{x}+1}dx=\sum_{k=0}^\infty(-1)^k\int_0^\frac{\pi}{2}(\sin{x})^\frac{1}{2}(\cos{x})^{k+\frac{1}{2}}=\sum_{k=0}^\infty(-1)^kB\left(\frac{3}{4},\frac{k}{2}+\frac{... | We are given:
$$ \int_0^{\pi/2} \frac{\sqrt{\sin x\cos x}\,dx}{1+\cos x} $$
Put $t=\tan\tfrac{x}{2}$; then
$$
\sin x = \frac{2t}{1+t^2}
\quad\quad
\cos x = \frac{1-t^2}{1+t^2}
\quad\quad
dx = \frac{2\,dt}{1+t^2}
$$
Then
$$ \sqrt{\sin x\cos x} = \frac{2t(1-t^2)}{1+t^2} $$
and
\begin{align}
1 + \cos x &= 1 + \frac{1-t^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the median in a triangle with trigonometry In a triangle $ABC$, $AB=7$, $AC=4$ and $\angle CAB=50º$. Let $M$ be the midpoint of $BC$. Determinate $AM$.
My try
I applied law of cosines $3$ times, first to find $BC$, then I let $\angle BCA=\alpha$ and $\angle ABC=130-\alpha$, and applied law of cosines in triangle $... |
\begin{align}
\triangle ADC:\quad
|AD|=4\cos50^\circ
,\quad
|CD|&=4\sin50^\circ
,\\
\triangle BCD:\quad
|BC|&=\sqrt{(4\sin 50^\circ)^2+(7-4\cos 50^\circ)^2}
\\
&=\sqrt{16\sin^2 50^\circ
+49-56\cos 50^\circ+16\cos^2 50^\circ}
\\
&=\sqrt{65-56\cos50^\circ}
,\\
\triangle BCD,\triangle BME:\quad
|ME|&=\tfrac12|CD|
=2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3189380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show with induction that $\sum_{k=1}^{n} \frac{k^{2}}{2^{k}} = 6 - \frac{n^2+4n+6}{2^{n}}$ Show with induction that
$\sum_{k=1}^{n} \frac{k^{2}}{2^{k}} = 6 - \frac{n^2+4n+6}{2^{n}}$
n = 1
$LHS = \frac{1}{2}$
$RHS = 6 - \frac{1+4+6}{2} = \frac{1}{2}$
n = p
$LHS_{p} = \frac{1^{2}}{2^{1}} + \frac{2^{2}}{2^{2}} + \frac{3^{... | I will assume the correct formula is:
$$\sum_{k=1}^n \dfrac{k^2}{2^k} = 6-\dfrac{n^2+4n+6}{2^n}$$
Assume this is true for $n=p$ and try to prove for $n=p+1$:
$$\begin{align*}\sum_{k=1}^{p+1} \dfrac{k^2}{2^k} & = \sum_{k=1}^p \dfrac{k^2}{2^k} + \dfrac{(p+1)^2}{2^{p+1}} \\ & = 6-\dfrac{p^2+4p+6}{2^p}+\dfrac{(p+1)^2}{2^{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Drawing without replacement: why is the order of draw irrelevant? I am trying to wrap my head around this problem:
Daniel randomly chooses balls from the group of $6$ red and $4$ green. What is the probability that he picks $2$ red and $3$ green if balls are drawn without replacement.
What I remember from my college ... | You can comprehend the calculation in a simpler way with smaller numbers.
Daniel randomly chooses balls from the group of $3$ red and $2$ green. What
is the probability that he picks $2$ red and $2$ green if balls are drawn
without replacement.
Indeed we have to regard the order. There are $\frac{4!}{2!\cdot 2!}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find function $f(x)$ satisfying $\int_{0}^{\infty} \frac{f(x)}{1+e^{nx}}dx=0$ I am looking for a non-trivial function $f(x)\in L_2(0,\infty)$ independent of the parameter $n$ (a natural number) satisfying the following integral equation: $$\displaystyle\int_{0}^{\infty} \frac{f(x)}{1+e^{nx}}dx=0$$
or prove that $f(x)=0... | The issue condition can be written in the form of
$$\int_0^\infty f\left(\dfrac zn\right)\dfrac{dz}{1+e^z} = 0.\tag1$$
Let the Maclaurin series of $f(x)$ exists, then
$$f(z) = \sum\limits_{k=0}^\infty a_kz^k.$$
Taking in account the integral representation of the Riemann zeta function in the form of
$$\int\limits_0^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 6,
"answer_id": 4
} |
Evaluating $\int_{-2}^{-1} \int_{-2}^{-1} \int_{-2}^{-1}\frac{x^2}{x^2+y^2+z^2} \,dx \,dy \,dz$
I am trying to solve this triple integral problem , but I am having some issues.
$$\int_{-2}^{-1} \int_{-2}^{-1} \int_{-2}^{-1}\frac{x^2}{x^2+y^2+z^2} dx dy dz$$
I tried with the 2 different approaches.
*
*I converte... | Following @kimchilover's suggestion, suppose $X,Y,Z$ are independent and identically distributed random variables having the uniform distribution on $(-2,-1)$.
So the pdf of $(X,Y,Z)$ is just $$f(x,y,z)=\mathbf1_{-2<x,y,z<-1}$$
Now,
$$E\left(\frac{X^2+Y^2+Z^2}{X^2+Y^2+Z^2}\right)=1$$
Or,
$$E\left(\frac{X^2}{X^2+Y^2+Z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$f: \mathbb{R} \to \mathbb{R},\space\space\space f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,\space\space$find $f(3)$ in terms of $f(0)$.
$f: \mathbb{R} \to \mathbb{R},\space\space\space\space f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,\space\space\space\space$
Find $f(3)$ in terms of $f(0)$.
My approach:
$$f(x)-2f(\frac{... | Making $y=2^x$ we have
$$
F(y)-2F(y-1)+F(y-2)=4^y
$$
which is a linear difference functional equation with solution
$$
F(y) = y \Phi_1(y)+\Phi_2(y)+\frac{4^{y+2}}{9}
$$
Here $\Phi_1(y),\Phi_2(y)$ are generic periodic functions with period $1$ Assuming $\Phi_1(y) = C_1,\Phi_2(y)= C_2$ we have
$$
f(x) = C_1\log_2 x + C_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding a Probability for a Gamma Distributed Random Variable Below is a problem from the book "Probability and Statistics". It is one of the Schaum's books. I am getting the
wrong answer and I believe that I am doing the integration incorrectly.
Thanks
Bob
Problem:
A random variable $X$ is gamma distributed with $\alp... | Until here everything is perfect:
$$P( X \le 1 ) = \int_0^1 \frac{x^{2}e^{-\frac{x}{2}}}{16} \, dx $$
Now when you substitute $-\frac{x}{2}=u\Rightarrow x=-2u\Rightarrow dx=-2du\,$ don't forget to add the coefficient of $du$, like this:
$$P( X \le 1 ) = \int_0^1 \frac{x^{2}e^{-\frac{x}{2}}}{16} \, dx=\int_0^{-1/2}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derive Distance of skew lines with methods from analysis It is well known, that given two skew lines
\begin{align*}
g: x &= a + \lambda b \quad \\
h: y &= c + \mu d
\end{align*}
the distance is given by
$$D = \frac{|(a - c) \cdot b \times d|}{|b \times d|}$$
See e.g. here
I thought, it would be a nice exercise to deri... | Without loss of generality, let $\ c=0,\ $ but we do not let $\ |b|=|d|=1\ $
because we would lose the valuable homogeneous property.
We recall the standard scalar
Triple product identity:
$$ \Delta := (a \cdot (b \times d))^2 = \begin{vmatrix} a\cdot a & a \cdot b & a\cdot d \\
b \cdot a & b\cdot b & b\cdot d \\
d ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$a_n=(1-\frac{1}{n})a_{n-1}+\frac{1}{n}a_{n-2}$, $\lim_{n\to \infty}a_n$ is Given $a_1,a_2,n\in \mathbb N$
$$a_n=(1-\frac{1}{n})a_{n-1}+\frac{1}{n}a_{n-2}$$
Then $\lim_{n\to \infty}a_n$ is
(A) $2(a_2-a_1)+a_1e^{-1}$
(B) $2(a_1-a_2)e^{-1}+a_2$
(C) $2(a_1-a_2)e^{-1}+a_1$
(D) $2(a_2-a_1)e^{-1}+a_1$
My attempt,
$a_1,a_2\in... | Multiplying both sides of the equation by $n$ and rearranging, you get that: $$n(a_n-a_{n-1}) = -(a_{n-1}-a_{n-2})$$
Setting $b_n = a_n-a_{n-1}, n\geq 2$ and $b_1 =a_1$, this becomes: $$b_n = \frac{-1}{n}b_{n-1} = ... = (-1)^n\frac{2}{n!}b_2 = (-1)^n\frac{2}{n!}(a_2-a_1),\text{ for } n \geq2.$$
So, $a_n = a_n-a_{n-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How do i show$\lim_{n\rightarrow \infty} \frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=\frac{4}{e}$ without using integration? Without using integration show that $$\lim_{n\rightarrow \infty} \frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=\frac{4}{e}$$. It could have been easier with integration, but I cannot proceed with t... | If we let $a_n=\frac{[(n+1)(n+2)...(2n)]}{n^n}=\frac{(2n)!}{n!n^n}$, then $\frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=(a_n)^{\frac{1}{n}}$. Then $\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{(2n)!}×\frac{n!}{(n+1)!}×\frac{n^n}{(n+1)^{n+1}}=\frac{(2n+1)(2n+2)}{(n+1)}×\frac{n^n}{(n+1)^{n+1}}=\frac{2(2n+1)}{(n+1)}×\frac{n^n}{(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $x^4+y^4+x^2+y^2+x^3y+y^3x\geq 0$ I tried 2 ways, first, take $x$ and $y$ $\geq0$ then obviously true. Take $x$ and $y$ both $\leq0$ same thing.
Now $y<0<x$ and wlog $|y|\leq x,$ then for big $x$ small $|y|$ we have $-x^4\leq x^3y$ and $-x^2\leq y^3x$ but if both $|y|$ and $x$ are big I got stuck.
So I decided to... | divide by $x^2y^2$if $x\neq0$ and $y\neq0$
$$\frac{x^2}{y^2}+\frac{y^2}{x^2}+\frac{1}{y^2}+\frac{1}{x^2}+\frac{x}{y}+\frac{y}{x}\geq 0$$
$$(\frac{x}{y}+0.5)^2+(\frac{y}{x}+0.5)^2+\frac{1}{y^2}+\frac{1}{x^2}-0.5\geq 0$$
now if $x\geq y$ then $(\frac{x}{y}+0.5)^2>1$
or if $y\geq x$ then $(\frac{y}{x}+0.5)^2>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve $2x^2-5x+2=$ $\frac{5-\sqrt{9+8x}}{4}$ Solve $2x^2-5x+2$= $\frac{5-\sqrt{9+8x}}{4}$
I simply do square both sides solve it and I get two value of x one is 2 and other is $\frac{3-√5}{2}$ but this approach it take more time so is there any approach for solving this equation.
| Taking Jerry Chang's observation (which amounts to the fact that the expression on the right-hand side is what you get when you plug the coefficients of the quadratic into the quadratic formula, just choosing the minus sign for the square root) and setting $y=2x^2-5x+2$ so that $x=2y^2-5y+2$ we can subtract one of thes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
When recreating the quadratic formula by completing the square of $ax^2+bx+c=0$ I cannot shorten the right hand side I am attempting to derive the quadratic formula by completing the square on the generic generic expression:
$$ax^2+bc+c=0$$
I'm struggling with the right hand side of the equation which, for the step I... | Here is an alternate approach which avoids working with fractions
\begin{eqnarray}
ax^2+bx+c&=&0\\\
&&\\\
\text{subtract }c&&\\\
ax^2+bx&=&-c\\\
&&\\\
\text{multiply by }4a&&\\\
4a^2x^2+4abx&=&-4ac\\\
&&\\\
\text{add }b^2&&\\\
4a^2x^2+4abx+b^2&=&b^2-4ac\\\
&&\\\
\text{factor the left side}&&\\\
(2ax+b)^2&=&b^2-4ac\\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Eliminating $\theta$ from the system $x\sin\theta-y\cos\theta=-\sin4\theta$, $x\cos\theta+y\sin\theta=\frac52-\frac32\cos4\theta$
Eliminate $\theta$ from the system of equations.
$$\begin{align}
x\sin\theta-y\cos\theta&=\phantom{\frac52\frac32}-\sin4\theta \\
x\cos\theta+y\sin\theta&=\frac52-\frac32\cos4\theta
\end{... | My believe that one of the coefficients of $x,y$ in Eliminating $\theta$ from trigonometric system will be negative has been corroborated by the current question.
Solving for $x,y$
$$\dfrac x{-\cos t(3\cos4t-5)-2\sin t\sin4t}=\dfrac y{2\cos t\sin4t-\sin t(3\cos4t-5)}=\dfrac12$$
$$4x=10\cos t-5\cos3t-\cos5t\iff x=5c-4c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I find the generator $(g)$ that generates $(29, \sqrt{-5} ± 13)$
Find the generator $(g)$ that generates $(29, \sqrt{-5} + 13)$. The ring is $\mathbb{Z}\left[\sqrt{-5}\right]$.
The fact I used was that $\text{Norm}(g)$ must divide both 29 and $\text{Norm}(\sqrt{5} + 13)$, which means solving the Pell equation ... | To show that $(29, 13+\sqrt{-5}) \subseteq (g)$ we have to find integers $x,y,z,t$ such that $g(x + y\sqrt{-5})=29$ and $g(z + t\sqrt{-5})=13+\sqrt{-5}$.
If $g=3 + 2\sqrt{-5}$ then
$\cases{3z-10t=13\\
2z+3t=1}.$
That is $z=\tfrac{49}{29}$ and $t=-\tfrac {23}{29}$, which is impossible.
Let's try $g=3 - 2\sqrt{-5}$ (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving $2x^2-3xy+y^2=0$ is transitive and anti-symmetric or symmetric and reflexive. Let $R$ be the binary relation defined on $\mathbb{R}$ by $xRy$ iff $2x^2-3xy+y^2=0$
For reflexive we get $2x^2=2x^2\implies-x=x$ which means reflexive on $xRx$
$2x^2-3xy+y^2=0$ tried going for $2y^2-yz+z^2=0$ then adding them togeth... | $\text{R}$ is a binary relation on $\mathbb{R}$ defined as $x\text{R}y\iff 2x^2-3xy+y^2=0$
Reflexive:
If $x=y$, then $2x^2-3xy+y^2=2x^2-3x^2+x^2= 0$
Hence, we have $x\text{R}x\ \ \forall \ x \in\mathbb{R}$ and the relation is Reflexive
Now using a technique called Completing the Square we have:
$$\begin{align}x\text{R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Invertibility of a matrix in portfolio optimization Let $A$ be an $n\times n$ symmetric matrix with non-negative entries. Let $\mathbf{1}$ be the column vector of dimension $n$ with all entries being $1$.
Consider the $(n+1)\times (n+1)$ matrix
$$ B=
\begin{bmatrix}
A & \mathbf{1} \\
\mathbf{1}^T & 0
\end{bmatrix}
$... | If $\det(A) \neq 0$ so that $A^{-1}$ exists
and the scalar $\alpha = \mathbf{1}^T A^{-1} \mathbf{1} \neq 0$, then we have
$$B^{-1} = \begin{bmatrix}
A^{-1} - \alpha^{-1}A^{-1}\mathbf{1}\mathbf{1}^TA^{-1} & \alpha^{-1}A^{-1}\mathbf{1} \\
\alpha^{-1}\mathbf{1}^TA^{-1} & -\alpha^{-1}
\end{bmatrix}$$
Note that
$$\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Comparing two variable expressions quickly Let us say $S_1=2xy^2+3xy$, and $S_2=3y^2+7x+7y+8$. Then, can we say that $S_1\ge S_2$ if $y\le x-2$ and $x,y\in\mathbb{N}$?
I think yes, but the usual quadratic function method is taking too much time, although I found that the discriminant of the final quadratic is $\ge0$ as... |
can we say that $S_1\ge S_2$ if $y\le x-2$ and $x,y\in\mathbb{N}$?
No, we cannot.
If $y=1$, then $S_1-S_2=-2x-18\lt 0$ for $x\ge 3$.
If $y=2$, then $$S_1-S_2=7x-34\ \begin{cases}\lt 0&\text{for $x=4$}\\\\ \gt 0&\text{for $x\ge 5$}\end{cases}$$
If $y\ge 3$, then we have $2y^2+3y-7\gt 0$, so
$$\begin{align}S_1-S_2&=x (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$ Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$
After few calculations I get that if I take $\frac{3}{2}x$ then after integral I get $\frac{3}{4}x^2+ C$ and
$$ \int_{\pi/6}^{\pi/2} \frac{3}{4}x^2+ C = \pi^2 / 6$$ so I shou... | $$(f=)\ \frac{x}{\sin\ x} = \frac{x}{x-x^3/6+\cdots }
=\frac{1}{1-x^2/6+\cdots }=1+Cx^2+\cdots ,\ C>0$$ so that
$\frac{x}{\sin\ x}$ is a positive increasing convex function on
$[{\pi/6}, {\pi/2}]$
By drawing a graph of $f$, the integral $\int\ f(x)\ dx$ is smaller than an area of
trapezoid i.e. $$ \int\ f(x)\ dx \leq \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that for all $n \in \mathbb{N}$, either 3 or 13 divides $3^n + 13n^2 + 38$ Let $a\in \{3,13\}.$ I'm having trouble with this proof. I know that
$$3^{n+1} + 13(n+1)^2 + 38 = (3^n + 13n^2 + 38) + (2\cdot 3^n + 26n + 13)$$
But I can't prove that $a \mid 2\cdot3^n + 26n + 13$. I know that 13 doesn't divide this becau... | If $n\equiv1$ or $2\pmod3$ then $3^n+13n^2+38\equiv n^2+2\equiv0 \pmod 3$
because $3^n\equiv0$ and $13\equiv1$ and $38\equiv2 \pmod 3$, so $3|3^n+13n^2+38$.
If $n\equiv0\pmod3$ then $3^n+13n^2+38\equiv 1+12\equiv0\pmod { 13}$
because $3^3=27\equiv1 $ and $13\equiv0$ and $38\equiv12 \pmod {13}$, so $13|3^n+13n^2+38$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
How many Taylor series terms are needed to accurately approximate $\sqrt{a+x}-\sqrt{a}$? Naive evaluation of $\sqrt{a + x} - \sqrt{a}$ when $|a| >> |x|$ suffers from catastrophic cancellation and loss of significance.
WolframAlpha gives the Taylor series for $\sqrt{a+x}-\sqrt{a}$ as:
$$\frac{x}{2 \sqrt{a}} - \frac{x^2}... | The Taylor series is
$$ \sqrt{a+x} - \sqrt{a} = \sum_{k=1}^\infty (-1)^{k+1} \frac{(2k)!}{(k!)^2(2k-1)} 4^{-k} a^{1/2-k} x^k$$
If $|x/a| < 1$, the absolute values of the terms decrease, since if $c_k = (2k)!/((k!)^2 (2k-1) 4^k)$,
$$ \frac{c_{k+1}}{c_k} = \frac{2k-1}{2k+2} < 1$$
Thus if $a > x > 0$ the absolute value o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prime number and square problem How many pairs of natural numbers, not bigger than 100, are such that difference between that pair is a prime number, and their product is a square of a natural number.
My attempt: I tried writing relationship such as $x-y=p$ and $xy=n^2$ but I can't seem to find any pattern to enumerate... | If $x - y$ is prime, then there can be at most one number that divides both $x$ and $y$, and it has to be that prime. Suppose the prime $p$ divides both $x, y$. Thus we can write $x = np$, and $y = (n-1)p$. Then we would have that $n(n-1)p^2$ is a square, hence so is $n(n-1)$. But $n, n-1$ are coprime, so this would me... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Can't solve $\int_{0}^{\pi} \frac{x}{1 + \cos^2x} dx$ I tried this :-
Let $$I =\int_{0}^{\pi}\frac{x}{1 + \cos^2x}dx\tag{1}$$
then $$I = \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2(\pi-x)}dx= \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2x}dx\tag{2}$$
Adding (1) and (2), we get
$$
2I = \int_{0}^\pi\frac{\pi}{1 + \cos^2x}dx\\
= \pi\in... | Hint For a slightly more theoretically demanding solution you can start out like this:
Let $$h(x) = f(x)\cdot g(x)\\ f(x) = x\\ g(x) = \frac{1}{1+\cos(x)^2}$$
Now we seek $$\int_0^{\pi}h(x)dx$$
But with Fourier analysis we know:
$$\hat h(0) = \int h(x)dx$$
And furthermore we know $${\hat {(f\cdot g)}} = \hat f * \hat ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Linear dependence of 3 vectors in $\mathbb{R}^4$ Let $a,b,c \in \mathbb{R},$ $\vec{v_1}=\begin{pmatrix}1\\4\\1\\-2 \end{pmatrix},$ $\vec{v_2}=\begin{pmatrix}-1\\a\\b\\2 \end{pmatrix},$ and $\vec{v_1}=\begin{pmatrix}1\\1\\1\\c \end{pmatrix}.$ What are the conditions on the numbers $a,b,c$ so that the three vectors are l... | The matrix
$$
\begin{pmatrix}
1 & -1 & 1 &0\\
4 & a & 1 &0\\
0& b+1 & 0 &0\\
0 & 0 & c+2 &0\\
\end{pmatrix}
$$
is not in row echelon form. Sum to the second column the first multiplied by $-4$, getting
$$
\begin{pmatrix}
1 & -1 & 1 &0\\
0 & a+4 & -3 &0\\
0& b+1 & 0 &0\\
0 & 0 & c+2 &0\\
\end{pmatrix}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Long division of $\frac{3x^3-x^2-13x-13}{x^2-x-6}$ I'm self-studying from Stroud & Booth's amazing textbook "Engineering Mathematics", and am on the "Partial Fractions" chapter. As part of an exercise I need to do long division of two polynomial equations.
The problem is, long division of polynomials was never explaine... | Polynomial long division is exactly same as long division of decimals.
With respect to your posed problem here are steps.
Step1: We have to divide $3x^3-x^2-\dots$ with $x^2$, so we first multiply the divisor $(x^2-x-6)$ by $3x$ to give $3x^3$. THen we subtract and copy the remaining terms.
Step2: Again now we have $2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate the Limit Without L'Hopital Rule Let $\lim_{x\to a}\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=L$, find the value of $\lim_{x\to a}\frac{x(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}$ for $a\neq0$.
Using L'Hopital rule I found that the answer is $a^4-aL$. My question is how to solve this pr... | Case 1: $a>0$. Since $x \to a$, we can assume that $x>0.$ Then show that $\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=0.$ Hence $L=0.$
Case 2: $a<0$. Since $x \to a$, we can assume that $x<0.$ Then show that $\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=-4x^2(x+a).$ Hence $L=-8a^3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3236774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove that hat if $a,b \ge 2$ then $ab \ge a+b$ Prove that if $a,b \ge 2$ then $a+b \le ab$
so if $a \ge 2$ and $b \ge 2$ then $a-1 \ge 1$ and $b-1 \ge 1$
$(a-1)(b-1) \ge b-1$
$(a-1)(b-1) \ge 1$
$(a-1)(b-1) - 1 \ge 0$
$ab -a -b\ge 0$
$ab \ge a + b$
Thanks in advance, is this valid?
| Let $a\equiv x+ 2,\,b\equiv y+ 2$
$$\therefore\,ab- a- b= xy+ x+ y\geqq 0$$
$$\because\,x,\,y\geqq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Calculus of $\sum_{n=1}^{\infty}( \frac{1}{2^n})$, $n$ is even I want to prove that $\sum_{n=1}^{\infty}( \frac{1}{2^n}) = \frac{1}{3}$, where $n$ can only be an even integer.
That is, how can I use linear algebra to prove that $\frac{1}{2^2} + \frac{1}{2^4} + \frac{1}{2^6} + \frac{1}{2^8} + \cdots = \frac{1}{3}$?
I he... | Use the fact that the set of all even (non-zero) integers is $2\mathbb N^*$. This gives the following sum to compute
$$\sum_{k=1}^{\infty} \dfrac{1}{2^{2k}}=\sum_{k=1}^{\infty}\dfrac{1}{4^k}=\dfrac{1}{4}\dfrac{1}{1-1/4}=\dfrac{1}{4}\dfrac{4}{3}=\dfrac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3241633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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solving a problem with generating functions This is a problem from a course of MIT.
Find the coefficients of the power series $y = 1 + 3 x + 15 x^2 + 184 x^3 + 495 x^4 + \cdots $ satisfying
$$ (27 x - 4)y^3 + 3y + 1 = 0 . $$
This is an interesting problem, but I am really cluelss (My naive approaches all failed). Can... | The given coefficients of the power series
\begin{align*}
y(x)=1+3x+15x^2+\color{blue}{84}x^3+495x^4+\cdots
\end{align*}
indicate they are $\binom{3n}{n}$ resulting in
\begin{align*}
y(x)=\sum_{n=0}^\infty \binom{3n}{n}x^n \tag{1}
\end{align*}
We prove (1) by showing it fulfils the functional equation
\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Prove that $\sum_{0\leq k\leq r} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$ where $m,n,r\in \mathbb{N} $ $\sum_{0\leq k\leq r} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$ where $m,n,r\in \mathbb{N} $.
Exam problem which stayed unproven for me. I tried ind... | Induction always works for hypergeometric sum identities.
$r=0$: $$\sum_{k=0} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$$
$r > 0$: $$\begin{eqnarray}
\textrm{LHS} &=& \binom{m+n}{n} \binom{m+n+(r-1)+1}{m+n+1} + \binom{n+r}{r} \binom{m+n+r}{n+r} \\
&=& \binom{m+n}{n} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Domain of $f(x,y) = {\sqrt{x+y-1 \over x-y+1}}$ How I can get the domain of the function
$$ f(x,y) = \sqrt{x+y-1 \over x-y+1}?$$
I know that is: $x+1 \neq y$ and $x^2 \ge (y-1)^2$
But I don't know how to get the second condition.
| We need $\dfrac{x+y-1}{x-y+1} \ge 0$.
1) $x+y-1 \ge 0$, and $x-y+1 >0$
$(x+(y-1)) \cdot (x-(y-1)) \ge 0$;
$x^2 \ge (y-1)^2$;
2) $x+y-1\le 0$, and $x -y +1 <0;$
Proceed likewise, note that multiplying
$(x+y-1) \le 0$ by $(x-y+1) <0$
changes the sign of the inequality.
$(x +(y-1))(x-(y-1)) \ge 0.$
Same result as in 1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Prove or disprove: If $a+b \leq \frac{1}{2}$, then $\frac{(1-a)(1-b)}{ab} \geq 1$ for positive $a,b$
Let $a,b$ be two positive numbers. Prove or disprove the statement:
If $a+b \leq \frac{1}{2}$, then $\dfrac{1-a}{a} \dfrac{1-b}{b} \geq 1$.
True. Assume $a+b \leq \frac{1}{2}$. Then
$$\dfrac{1-a}{a} \dfrac{1-b}{b}=\df... | You have succeeded in proving that $\dfrac{1-a}{a} \dfrac{1-b}{b} \geq 1$
if $a,b>0$ and $a+b \leq 1.$
Therefore, since $\frac12<1$, certainly $\dfrac{1-a}{a} \dfrac{1-b}{b} \geq 1$ if $a+b\leq\dfrac12.$
Here is another way to write the proof:
$\dfrac{1-a}a\dfrac{1-b}b\ge1\iff(1-a)(1-b)\ge ab\iff 1-a-b\ge0\iff1\ge a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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find all values of y such that $(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0$ One of the questions on my Algebra 1 homework was to find all values of $y$ such that $$(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0$$
This was listed under "Lecture 13: Quadratic Equations"
I can not figure out how to approach this problem? Should I expand it? I al... | $$(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0 \\ (y-2)(y+3)(x^2-6y+9)-2(y-3)(y+3)=0 \\ (y-2)(y+3)(x^2-6y+9)=2(y-3)(y+3) \\ (y-2)(x^2-6y+9)=2(y-3) \\ x^2-6y+9=\frac{2(y-3)}{y-2} \\ x^2=\frac{2(y-3)}{y-2}+6y-9 \\ x^2= \frac{2(y-3)}{y-2}+\frac{6y(y-2)}{y-2}-\frac{9(y-2)}{y-2} \\x^2= \frac{2(y-3)+6y(y-2)-9(y-2)}{y-2} \\ x^2=\frac{2y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof by induction when the first element is defined in an interval Given that, for a sequence $x_n$:
*
*$1<x_1<2$
*$x_{n+1} = 1 + x_n - \frac{1}{2}(x_n)^2$ for $n \ge 2$
Prove that, for $n\ge3$, $|x_n - \sqrt2| < \frac{1}{2^n}$
So, at first I tried to verify the base case by first calculating $x_2$.
Thus, I got $... | This problem can be solved by letting $a_n = x_n - \sqrt 2$. Rewriting the condition, we obtain $a_{n+1} + \sqrt 2 = 1 + a_n + \sqrt 2 - \frac{1}{2}(a_n + \sqrt 2)^2$, and expanding and rearranging gives $a_{n+1} = (1-\sqrt 2)a_n - \frac{1}{2} a_n^2 = (1- \sqrt 2 - \frac{a_n}{2})a_n$.
Now if $|a_n| < 3 - 2 \sqrt 2$, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3248520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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For every integer $n$, the quantity $n^2 + 2n \equiv 0\pmod 4$ or $n^2 + 2n \equiv 3\pmod 4$ I'm trying to prove this question using induction
So far I have
Base Case
Let $n = 1$, $(1^2 + 2)\equiv 3 \pmod 4 $
Claim holds for base case
Induction
Assume $n = k$ holds, that is $k^2 + 2k \equiv 0\pmod 4$ or $k^2 + 2k \equ... | $$n^2+2n=(n+1)^2-1$$
Now $n+1\equiv0,\pm1,2\pmod4$
$\implies(n+1)^2\equiv0,1\pmod4$
$(n+1)^2-1\equiv?,?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove for all positive a,b,c that $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$ Prove for all positive a,b,c $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$$
My Try
I tried taking common denominator of the expression,
$\frac{a^2b+ab^2+b^2c+c^2b+ac^2+a^2c}{abc}$
How to proceed? Is there a way to write them as per... | It is $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\geq 6$$ and now use that $$x+\frac{1}{x}\geq 2$$ for all $$x>0$$
You can also use that $$\frac{a^2 b+ab^2+b^2c+c^2b+a^2c+ac^2}{6}\geq \sqrt[6]{(abc)^6}=abc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Convergence of the sequence $x_n= \frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}$
$\left\{x_n\right\}$ is a convergent sequence where $x_n= \frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}$.
What is $\lim\limits_{n\to\infty}x_n?$
Here are my two approaches:
Using Euler's constant($\gamma$):
$$\begin{align}\lim_{n\righ... | Observe that you can rewrite $x_n$ as,
$$x_n =\frac{1}{n+1}+\frac{1}{n+1}+\dots+\frac{1}{2n}=\left(1+\frac{1}{2}+\frac{1}{3}+\dots \frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{n}\right)$$
$$=\left(1+\frac{1}{2}+\frac{1}{3}+\dots +\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Nice olympiad inequality Let's go for an olympiad inequality :
let $a,b,c>0$ then we have :
$$\sum_{cyc}\frac{ab}{a+b}\geq \frac{3\sqrt{3}}{2}\sqrt{\frac{abc}{a+b+c}}$$
My proof :
$$\sum_{cyc}\frac{ab}{a+b}=\frac{(a^2 b^2 + 3 a^2 b c + a^2 c^2 + 3 a b^2 c + 3 a b c^2 + b^2 c^2)}{(a+b)(b+c)(c+a)}=\frac{(ab + ac + ... | By your work we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=3v^4+uw^3+\frac{1}{2}(w^3-9uv^2)\sqrt{\frac{w^3}{u}}.$$
We see that $f$ increases, which says that it's enough to prove our inequality for a minimal value of $w^3$, which happens in the following cases.
*
*$w^3\rightarrow0^+$.
In this case our inequa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplification of proof for $\exists c>0$ such that $|x+y|\leq c(1+\sqrt{x^2+y^2})$ Question: Consider function $f(x,y)=|x+y|$ for $x,y\in \mathbb{R}$. Show that $\exists c>0$ such that $|x+y|\leq c(1+\sqrt{x^2+y^2})$.
My proof:
i) if $|x|,|y|\leq 1$, then $|x+y|\leq 2\leq 2(1+\sqrt{x^2+y^2})$;
ii): if $|x|\leq 1$ and ... | Notice that
$$|x|=\sqrt{x^2}\leq\sqrt{x^2+y^2}\quad\text{and}\quad|y|=\sqrt{y^2}\leq\sqrt{x^2+y^2}$$
and so
$$|x+y|\leq|x|+|y|\leq 2\sqrt{x^2+y^2}<2\left(1+\sqrt{x^2+y^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Minimize a function in two variables with constraint I have to minimize this:
\begin{align*}
\min&\quad{ (x-3)^2+(y-1)^2} \\
s.t.& \quad 2x+y \leq 2 \\
&\quad x^2 + 2y = 3\\
&\quad x, y \geq 0
\end{align*}
Can I isolate $y$ in the second constraint and substitute it in the first?
| Let $$(x-3)^2+(y-1)^2=z~~~(1)$$ Let us put $x^2=(3-2y)$ in (1), we get $$y^2-4y-6x+13-z=0.~~~(2)$$ $z$ will attain optimum value if when the line $y=2-2x$ touches the curve (2)
which is a parabola. Let us but this line in (2)
$$4x^2-6x+9-z=0.~~~~(3)$$
Now demand $B^2=4AC$.This gives $z=27/4,$ the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ \sum_{1 \le i < j \le n} a_{i} a_{j} \ge n(n-1)/2 $ , prove that $a_{1} + ... + a_{n} \ge n$ for $n \ge 2$ using AM-QM Let $a_{1}, a_{2}, ...$ be a sequence of positive real numbers. Let the following relation holds:
$$ a_{k+1} \ge \frac{k a_{k}}{a_{k}^{2} + (k-1)}, \:\: k \ge 1$$
Prove that $ S_{n} = a_{1} + a_{2} +... | Since
$$(a_1+a_2+\cdots +a_n)^2=(a_1^2+a_2^2+\cdots +a_n^2)+2\sum_{1 \le i < j \le n} a_{i} a_{j}$$
we can write
$$\sum_{1 \le i < j \le n} a_{i} a_{j}=\frac{S_n^2-(a_1^2+a_2^2+\cdots +a_n^2)}{2}$$
So, the hint $$ \sum_{1 \le i < j \le n} a_{i} a_{j} \ge \frac{n(n-1)}{2} $$
is equivalent to
$$\frac{S_n^2-(a_1^2+a_2^2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
value of $2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$
If $ x^3+bx^2+cx+1=0$ has only real root $\alpha $.
Where $(b<c)$. Then $\displaystyle 2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$ is
Plan
$$\tan^{-1}\bigg(\frac{2\csc \alpha}{1-\csc^2\alpha}\bigg)+\tan^{-1}\bigg(2\sin \alpha\sec^... | Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$2\arctan p=\begin{cases} \arctan\dfrac{2p}{1-p^2} &\mbox{if } p^2<1 \\
\pi+ \arctan\dfrac{2p}{1-p^2} & \mbox{if } p>1\\-\pi+ \arctan\dfrac{2p}{1-p^2} & \mbox{if } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find all possible integers $n$ such that $\sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.
Find all possible integers $n$ such that $m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.
Guess what? This problem is adapted from a recent competition. There have been a solution below for you to check out. I ... | We have that $$m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}} \ (m \in \mathbb N)$$
$$\iff m - \sqrt{n + 2} = \sqrt{n + \sqrt{n + 2}} \iff (m - \sqrt{n + 2})^2 = n + \sqrt{n + 2}$$
$$\iff m^2 - (2m - \sqrt{n + 2})\sqrt{n + 2} = (\sqrt{n + 2} + 1)\sqrt{n + 2} - 2$$
$$\iff m^2 + 2 = (2m + 1)\sqrt{n + 2} \iff \sqrt{n + 2} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\iint\limits_D(x^2-2xy+y^2)\,\mathrm{d}x\,\mathrm{d}y$ where $4\leqslant x^2-y^2 \leqslant 16,~-5\leqslant y-\frac{x}{2}\leqslant 1$
Evaluate
$$I=\iint\limits_D(x^2-2xy+y^2)\,\mathrm{d}x\,\mathrm{d}y$$
where $D$ is bounded by $$x^2-y^2=4,~x^2-y^2=16,~y=\frac{x}{2}-5,~y=\frac{x}{2}+1.$$
Attempt. The obvious ... | The problem with $u = x^2 - y^2, v = 2y-x$ is that it is difficult to express the objective as a function of $u,v$
Which gets me thinking about
$u = x+y\\
v = x-y\\
du dv = 2 dx dy$
This works nice with the objective and one of the functions on the boundary, but not the other.
Our transformed objective is now $\frac 12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $e^{x \cos(x)}=1+x+\frac{x^2}{2} - \frac{x^3}{3}-\frac{11x^4}{24}- \frac{x^5}{5} + \cdots$ How do we do this question using Maclaurin's Series. I tried expanding it by putting $x \cos x$ in place of $x$ in Maclaurin's expansion of $e^x$, and then using multinomial theorem to open the squares, cubes etc of $\cos x... | There's a sign error in your $x^3$ coefficient. With $\equiv$ denoting equality up to $x^5$ terms, $$\exp(x\cos x)\equiv\exp\left(x-\frac{x^3}{2}+\frac{x^5}{24}\right)\\\equiv\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\right)\left(1-\frac{x^3}{2}\right)\left(1+\frac{x^5}{24}\right)\\\equiv\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Integrating $\int^2_0 xe^{x^2}dx$ Well what I was thinking was to integrate the indefinite integral first.
$u=x^2$, $x=\sqrt u$
$du=2xdx = 2\sqrt {u} dx$
$dx= \frac{1}{2\sqrt{u}}du$
$\int xe^{x^2} dx = \int \sqrt{u}\frac{1}{2\sqrt{u}} du =\frac{1}{2}\int e^u du = \frac{1}{2}e^u =\frac{1}{2}e^{x^2} +C$
Now I can evalu... | Also, one might set
$g(x) = e^{x^2}; \tag 1$
then
$g'(x) = 2xe^{x^2}; \tag 2$
then
$\displaystyle \int_0^2 xe^{x^2} \; dx = \dfrac{1}{2} \int_0^2 g'(x) \; dx = \dfrac{1}{2}(g(2) - g(0))$
$= \dfrac{1}{2}(e^{2^2} - e^0) = \dfrac{1}{2} (e^4 - 1) = \dfrac{1}{2}(e^4 - 1) = \dfrac{1}{2}e^4 - \dfrac{1}{2}. \tag{3}$
If one wan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is the sum of series?? What is the sum of first 50 terms of the series
$$(1\times3)+(3\times5)+(5\times7)+\ldots$$
I had tried to solve this question
It seems that it is mixed arithmetic series
In which first $A.p$ is $1,3,5\ldots$
Nd other one is $3,5,7\ldots$
But i don't know how to solve it together
Please giv... | You have better go through Falling Factorials
$$
\eqalign{
& \sum\limits_{n = 0}^{49} {\left( {2n + 1} \right)\left( {2n + 3} \right)} = 4\sum\limits_{n = 0}^{49} {\left( {n + 1/2} \right)\left( {n + 3/2} \right)} = \cr
& = 4\sum\limits_{n = 0}^{49} {\left( {n + 3/2} \right)^{\,\underline {\,2\,} } } \cr}
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
quadratic equation solving mistake I'm a student who started self learning quadratic equations for a youth university program. I'm busy at trying to solve such equation:
$$
(1 - 4x)^2 + 9x + 7 = 2(x+3)(1-x) + (x+4)^2
$$
this is my current progress:
\begin{align}
(1 - 4x)^2 + 9x + 7 &= 2(x+3)(1-x)+ (x+4)^2\\
(1 - 4x)(1 ... | You have not made any mistake.
The final equation you have obtained is
$$17x^2-3x-14=0$$
$$17x^2-17x+14x-14=0$$
$$17x(x-1)+14(x-1)=0$$
$$(17x+14)(x-1)=0$$
which has the roots $1$ and $\frac{-14}{17}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Solution Of Diophantine Equations. Find all positive integers $x$ and $y$ such that $(x+y)(xy+1)=2^z$
Now as $2$ is a prime number we get $x+y=2^a$ and $xy+1=2^b$ for some natural numbers $a$ and $b$. Now as $xy+1>x+y$ for every $x, y>1$. So we get $2^b>2^a=> b>a=> 2^a|2^b$. So $x+y|xy+1$. Now let $p$ be a common prime... | You have $xy=2^k-1$ for some integer $k> 0$
Note that with $x=1$ and $y=2^k-1$ you get $$(x+y)(xy+1)=2^{2k}$$
So we always have that solution.
We also have other solutions if we can factor $2^k-1$ into two integers whose sum is a power of $2$
For example with $k=4$ we get $2^{k}-1=3(5)$ where $3+5=2^3$ so $(x+y)(xy+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int\limits_0^{1}\frac{\sqrt{1+x^2}}{1+x}dx$ Evaluate:
$I=\int\limits_0^1 \frac{\sqrt{1+x^2}}{1+x}dx$
My try:
Let $x=\tan y$ then $dx=(1+\tan^{2} y)dy$
As for the integration limits: if $x=0$ then $y=0$ and if $x=1$ then $y=\frac{π}{4}$
So:
$I=\int\limits_0^{\frac{π}{4}}\frac{1+\tan^{2} y}{(1+\tan y)\cos... | \begin{align}
\int\limits_0^{1}\frac{\sqrt{1+x^2}}{1+x}dx
= &\int\limits_0^{1}\frac{x-1}{\sqrt{1+x^2}}+\frac2{\sqrt{1+x^2}(1+x)}\ {dx}\\
=&\ \left(\sqrt{1+x^2}-\sinh^{-1}x -\sqrt2\sinh^{-1}\frac{1-x}{1+x} \right)_0^1\\
=& \ \frac{\sinh^{-1}1+1}{\sqrt2+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to integrate $\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4 x} \,dx$ So I followed the explanations made in this post and I got that:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \frac{\sqrt{2}}{2}\arctan\left(\frac{\sqrt{2}}{2}\tan\left(2x\right)\right) + C$$
But when I try to use the Leibniz-Newton formula and evaluate... | You've got many methods here to find the anti-derivative, but hey, why not one more?
Here
\begin{align}
I &= \int \frac{1}{\sin^4(x) + \cos^4(x)}\:dx = \int \frac{1}{\cos^4(x)\left(\tan^4(x) + 1\right)}\:dx \nonumber \\
&= \int \sec^2(x)\sec^2(x) \cdot\frac{1}{\tan^4(x) + 1}\:dx = \int \sec^2(x) \cdot\frac{\tan^2(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Solve the following equation: $\sqrt {\sin x - \sqrt {\cos x + \sin x} } = \cos x$ Solve the following equation:
\begin{array}{l}{\sqrt{\sin x-\sqrt{\cos x+\sin x}}=\cos x} \\ \text{my try as follows:}\\{\sin x-\sqrt{\cos x+\sin x}=\cos ^{2} x} \\ {\sin x-\cos ^{2} x=\sqrt{\cos x+\sin x}} \\ {\sin ^{2} x+\cos ^{4} x-2... | It can be seen that the radical of the form $\sqrt{x-\sqrt{x+..}}$ is undefined for $x<1$. Since $\sin{x}$ is on the interval $[-1,1]$. This means that it is only defined for $x=\arcsin(1)$. Or generalizing this, we get;
$$x=2n\pi+\frac{\pi}{2}, \tag{for n ∈ Z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Differentiating from the first principle How do I convincingly differentiate $\sin 3x$ or any trig function that comes in the form $\tan nx$, $\sin nx$, $\cos nx$ from the first principle?
I have tried expanding some forms using some trig identities but the $n$ coefficient isn't just showing up
| If you allow the trigonometric functions to be defined by their McLauren series, then you can differentiate these term-by-term.
For example:
$$\cos nx = 1 - \frac{(nx)^2}{2!} + \frac{(nx)^4}{4!}-\frac{(nx)^6}{6!}+\cdots$$
$$
\frac{d\cos nx}{dx} = - \frac{2n(nx)}{2!} + \frac{4n(nx)^3}{4!}-\frac{6n(nx)^5}{6!}+\cdots$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to choose your $u$ value properly to solve $\int\frac{1}{5+3\cos (x)} \cdot dx$. We are given the following facts:
*
*$\cos(x) = \frac{1-u^2}{1+u^2}$,
*$dx = \frac{2du}{1+u^2}$
Using 1. and 2. along with u substitution solve the following integral.
$$\int\frac{1}{5+3\cos (x)} \cdot dx$$
I re-wrote the integral... | You should know by heart the formulæ
\begin{align}\int\frac{\mathrm dx}{a^2+x^2}&=\frac 1a\,\arctan\frac xa,&\int\frac{\mathrm dx}{a^2-x^2}&=\frac 1{2a}\,\ln\biggl|\frac{a+x}{a-x}\biggr|\\
&&&=\frac 1{a}\arg\tanh\frac xa\quad\text{ if } |x|<a.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integral $\int_{0}^{1} dx \int_{- \sqrt{x- x^2 }}^{\sqrt{x- x^2}}\sqrt{1- x^2 - y^2} dy $ I am having problems with integration.
I have a function $f(x, y) = \sqrt{1- x^2 - y^2}$
And I need to integrate it on a circle $(x- \frac{1}{2} )^2 + y^2 = \frac{1}{4}$
I couldn't get far with usual coordinates:
$$
\int_{0}^{1}... | Near the end of your work, it should be $(1-\cos{\phi}^2)^{3/2}=|\sin{\phi}|$ (with the absolute value!) and therefore
$$\frac{-1}{3} \bigg( \int_{-\pi / 2}^{\pi / 2} |\sin{\phi}|^3 d\phi - \int_{-\pi / 2}^{\pi / 2} d\phi\bigg)=-\frac{2}{3} \int_{0}^{\pi / 2} \sin^3{\phi}\, d\phi +\frac{\pi}{3}=-\frac{4}{9}+\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of all the numbers in the grid.
A square containing numbers
$$
\begin{array}{|c|c|c|}
\hline 1 & 2 & 3 \\ \hline
1 & 2 & 2 \\ \hline
1 & 1 & 1 \\\hline
\end{array}
\qquad \qquad\qquad
\begin{array}{|c|c|c|c|}
\hline 1 & 2 & 3 & 4 \\ \hline
1 & 2 & 3 & 3 \\ \hline
1 & 2 & 2 & 2 \\ \hline
1 & 1 & 1 & 1 \\\hline
\e... | Another way to think of it is by summing diagonals: A $n \times n$ square has the following diagonal sums
$$
1, \\ 1 + 2 , \\ \vdots \\ 1 + \ldots + n - 1, \\ 1 + \ldots + n, \\ 1 + \ldots + n - 1, \\ \vdots \\ 1 + 2, \\ 1.
$$
Therefore, the formula for the sum is
$$
2 \sum_{k = 1}^{n - 1}\sum_{j = 1}^{k} j + \sum_{j =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Division of 11 people into 3 groups with at least 2 people in each
How many ways can 11 people be divided into three teams where each
team must have at least two members?
We are supposed to use multinomial coefficients and number of integer solutions. I have tried this
The number of ways to divide 5 people into thr... | It is ill-advised to treat the compulsory two people in each group separately. Rather, the partitions of $11$ into $3$ with each part at least $2$ should be looked at directly, which as you have worked out are
$$(7,2,2),(6,3,2),(5,4,2),(5,3,3),(4,4,3)$$
These lead to the following counts for each partition.
*
*$(7,2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solutions modulo square of a number. Let $p$ be a prime such that $$p^n\equiv 5 \mod{6}$$ for each odd positive integer $n$. Then how to find the values of $p^n$ modulo $36$, i.e. I need to find $$p^n\equiv ?\mod{36}.$$ Please help.
| If $p^n \equiv 5 \pmod 6$ for every odd $n$ then $p^1 \equiv 5 \pmod 6$.
(Which is okay; that means $p^{2k+1} \equiv (p^2)^k*p \equiv (25^k)*5 \equiv 1^k*5 \equiv 5 \pmod p$.
So $p = 5 + 6M$ for some $M$ and $p \equiv 5 + 6m\pmod {36}$ where $m \equiv M \pmod 6$
$p^2 \equiv (5+6k)^2 \equiv 25 + 60m + 36m^2 \equiv 1+ 24... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inverse Function Verification
Are functions $f(x)=\frac{7x+4}{x+6}$ and $g(x)=\frac{6x-4}{7-x}$ inverses of each other?
I'm experiencing a very strange issue with this problem. When I find the inverse of $g$ I get $f(x)$. However, when I do the same thing for $f$ I do not get $g(x)$.
\begin{align*}
f(x)&=\frac{7x+4}... | It should be
\begin{align*}
-y&=-(\frac{7x+4}{6+x}) \\ \therefore g^{-1}(x)&=\frac{7x+4}{x+6}=\ f(x)
\end{align*}
And you already have $$f^{-1}(x)=\frac{-6x+4}{x-7}=\frac{6x-4}{7-x}=g(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the problem here (all integers are irrational proof...I think so)? Let us assume $a$ is an integer which is rational which implies $a=p/q$ (where $p$ and $q$ are integers and $q$ not equal to $0$). If $p$ and $q$ are not coprime, let us simplify the fraction so this it is (I don't know how to talk like mathemat... | I think you are confusing that if $p$ is prime and $p$ divides $b^k$ then $p|b$. That is true if $p$ is prime.
Actually it's also true for a composite $a|b^k$ then $a|b$ if $a$ has no square factors. But if $a$ as any prime factors to a power greater than $1$ it need not be true.
And in fact its obviously not true ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Solve $x^{3} = 6+ 3xy - 3 ( \sqrt{2}+2 )^{{1}/{3}} , y^{3} = 9 + 3xy(\sqrt{2}+2)^{{1}/{3}} - 3(\sqrt{2}+2)^{{2}/{3}}$ Solve the system of equations for $x,y \in \mathbb{R}$
$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}} $
$ y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}$
I j... | Using Dr. Sonnhard Graubner's answer, multiply everything by $27x^3$ and you face an awful cubic equation in $x^3$ for which $\Delta=-1062882$ which means that there is only one real root.
Use the hyperbolic methid for that case and you will end with the "beautiful"
$$x^3=6 \left(a+b \cosh \left(\frac{1}{3}\cosh (c)\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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If at least one of the selected items must be vegetarian, how many different dinners could Jane create?
Problem: Jane must select $3$ different items for each dinner she will serve. The items are to be chosen from $5$ different vegetarian and $4$ different meat selections. If at least one of the selected items must be... | You are counting meals with more than one vegetarian item multiple times. Also, the order in which the courses are selected does not matter.
Method 1: Direct count.
Since there are five different vegetarian options and four different meat options, the number of ways of selecting $k$ vegetarian options and $3 - k$ mea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How can I prove that $\left\lceil \frac{4N}{3} \right\rceil \bmod 4 \ne 1$? Ok this is for Base $64$ padding.
The length of Base $64$ representation of a sequence of bytes is represented by
$\left\lceil \frac{4N}{3} \right\rceil$
The padding should add $M$ characters such that $$\frac{\left(\left\lceil \frac{4N}{3} \r... | For any $n$, we have either $n=3m$, $n=3m+1$, or $n=3m+2$. Then we just have three cases to consider:
\begin{align*}
\left\lceil \frac{4n}{3} \right\rceil
&=
\left\lceil 4m \right\rceil
= 4m \equiv 0 \pmod 4
\\
\left\lceil\frac{4n}{3} \right\rceil
&=
\left\lceil 4m+\frac{4}{3} \right\rceil
=4m+2 \equiv 2 \pmod 4
\\
\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding $\lim_{n\to \infty} n^{b-a}\frac{1^a+2^a+\cdots +n^a}{1^b+2^b+\cdots +n^b}$ For $a,b>-1$
$$\lim_{n\to \infty} n^{b-a}\frac{1^a+2^a+\cdots +n^a}{1^b+2^b+\cdots +n^b}$$
I am really confused in this one. I tried to calculate it but the answer comes out to be $1$ as I divided the numerator and denominator with $n^a... | Another similar solution using generalized harmonic numbers
$$S_a=\sum_{i=1}^n i^a=H_n^{(-a)}$$ Using asymptotics
$$S_a=n^a \left(\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12
n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-a)$$
$$n^{-a} S_a=\frac{n}{a+1}+\frac{1}{2}+\frac{a}{12
n}+O\left(\frac{1}{n^3}\right)+n^{-a}\zeta (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove $x+y+z \ge xy+yz+zx$ Given $x,y,z \ge 0$ and $x+y+z=4-xyz$ Then Prove that
$$x+y+z \ge xy+yz+zx$$
My try:
Letting $x=1-a$, $y=1-b$ and $z=1-c$ we get
$$(1-a)+(1-b)+(1-c)+(1-a)(1-b)(1-c)=4$$
$$-(a+b+c)-(a+b+c)+ab+bc+ca-abc=0$$
$$ab+bc+ca-abc=2(a+b+c)$$
Where $a, b,c \le 1$
is there a clue here?
| Suppose otherwise $x+y+z<xy+xz+yz$. Let $x=ka$, $y=kb$ and $z=kc$, such that $k>0$ and $a+b+c=ab+ac+bc$.
Thus, $$k(a+b+c)<k^2(ab+ac+bc),$$ which gives $$k>1$$ and
$$4=k(a+b+c)+k^3abc>a+b+c+abc,$$ which is a contradiction because we'll prove now that
$$a+b+c+abc\geq4.$$
Indeed, we need to prove that
$$\frac{(ab+ac+bc)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Finding constant $a$ so that differential equation becomes correct Question:
Find the value of $a$ so that the function
$$y = \sqrt{x} \ln{x}$$
is a solution to the differential equation
$$y' - \frac{a}{x} \cdot y = \frac{1}{\sqrt{x}}$$
Attempted solution:
My basic approach would be to take the derivative of y, put y a... | You forgot in the last line the following:
$$a = \frac{x(\frac{\sqrt{x}}{x} + \frac{\ln {x}}{2\sqrt{x}}) \color{blue}{-\sqrt{x}}}{\sqrt{x}\ln {x}} = \frac{1+\frac{\ln{x}}{2}\color{blue}{-1}}{\ln{x}} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all extrema of a complicated trigonometric function Problem
Find all local extrema for
$$f(x) = \frac{\sin{3x}}{1+\frac{1}{2}\cos{3x}}$$
Attempted solution
My basic approach is to take the derivative, set the derivative equal to zero and solve for x.
Taking the derivative with the quotient rule and a few cases of ... | Looks like you have a problem with the differentiation. You should have.
You dropped a factor of 3 in the right-hand term. It should be $(\frac 32 \sin 3x)(\sin 3x)$ in the first line.
You have brought it back by the time you get to.
$3\cos 3x + 1.5\cos^2 3x + \frac 12 \sin^2 3x = 0$
But then $1.5$ becomes $1$ in the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to find Sum of Sum of Partial Harmonic Series: $\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} $? How do I find the Sum of the Sum of Finitely Many Harmonic Series: $\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} $?
According to maple it is:
$\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} = \left( \left( m+1 \right) ^{2}-m-4 \r... | Let us use
$$H_k=- \sum_{m=1}^{k}(-1)^k\frac {{k \choose m}}{m},~~ {n \choose r}={n-1 \choose r-1}+{n-1 \choose r},~~ \sum_{k=m}^{n} {k \choose m}= {n+1 \choose m+1}.$$ Then
$$S_n=\sum_{k=1}^{n} H_k=- \sum_{k=1}^{n} \sum_{m=1}^{n} (-1)^m \frac{{n \choose m}}{m}= -\sum_{m=1}^{n} (-1)^{m} \frac{{n+1 \choose m+1}}{m}
$$ $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Proof Verification: $4x^2+6xy+4y^2>0$ unless both $x$ and $y$ are equal to $0$ This is a problem from Spivak's Calculus 4th ed., Chapter 1, Problem 16(b).
Using the fact that
$x^2 + 2xy + y^2 = (x + y)^2 > 0$,
show that $4x^2 + 6xy + 4y^2 >0$ unless $x$ and $y$ are both $0$.
This is my proof:
Firstly, as any non-... | More generally,
since
$\begin{array}\\
ax^2+2bxy+ay^2
&=a(x^2+y^2)+b(x^2+2xy+y^2)-b(x^2+y^2)\\
&=(a-b)(x^2+y^2)+b(x+y)^2\\
\end{array}
$
if
$a > b$ then
$ax^2+2bxy+ay^2 \ge 0$
with equality if and only if
$x = y = 0$.
Your case is
$a=4, b=3$.
You can handle the more general case of
$ax^2+2bxy+cy^2$
by looking at how
th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum value of $(\sin x+ \csc x)^2 + (\sec x + \cos x)^2$ I made it till $5+\tan^2x + \cot^2x$, but I don’t know how to proceed further.
| I don't think your function is $5+\tan^2x+\cot^2x$, actually. With $c:=\cos x,\,s:=\sin x$ your sum is $$\left(s+\frac{1}{s}\right)^2+\left(c+\frac{1}{c}\right)^2=5+\frac{1}{s^2}+\frac{1}{c^2}=5+\frac{1}{c^2s^2}=5+4\csc^22x\ge 9,$$with equality iff $\sin2x=\pm 1$, as occurs e.g. if $x=\frac{\pi}{4}$ for which we get$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
im trying to use AM-GM
$\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}}
\le ... | Jensen's inequality is $\rm{avg~}f(x_i) \le f(\rm{avg~} x)$ (for concave up functions). So put in $f(z) = 1/z$:
$$\frac{1}{n} \sum_{i} \frac{1}{x_i} \le \frac{1}{\frac{1}{n} \sum_i x_i}$$
Note that this assumes $x_i > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$.
I have tried:
$$\frac{10!}{(3!\times7!)} \times (-x + 2)^7 \times (x^2)^3 $$
But got an incorrect answer $-15360$.
| The terms in the development are of the form $x^{2a}(-x)^b2^c\propto x^3$, and the exponents must satisfy
$$\begin{cases}a+b+c=10,\\2a+b=3.\end{cases}$$
The solutions are $0,3,7$ and $1,1,8$.
Hence by the multinomial formula
$$\frac{10!}{0!3!7!}1^0(-1)^32^7+\frac{10!}{1!1!8!}1^1(-1)^12^8=-120\cdot128-90\cdot256=-38400... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to show the matrix has Rank $\le 5$
I want to show that the following matrix has Rank $\le 5$.
The matrix is
\begin{bmatrix}
2&1&1&1&0&1&1&1\\
1&2&1&1&1&0&1&1\\
1&1&2&1&1&1&0&1\\
1&1&1&2&1&1&1&0\\
0&1&1&1&2&1&1&1\\
1&0&1&1&1&2&1&1\\
1&1&0&1&1&1&2&1\\
1&1&1&0&1&1&1&2
\end{bmatrix}
I found that there is a submatri... | The given matrix is equal to $A+B$ where
$$
A=\begin{bmatrix}
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1\\
1&1&1&1&1&1&1&1
\end{bmatrix},\quad
B=\begin{bmatrix}
1&0&0&0&-1&0&0&0&\\
0&1&0&0&0&-1&0&0&\\
0&0&1&0&0&0&-1&0&\\
0&0&0&1&0&0&0&-1&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$. Let $n \in \mathbb{N}$. Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$.
I began this problem giving some values for $n$ and I found that $\gcd(n^2+3, (n+1)^2+3)=1$ for most of $n$ I tried, but if $n=6$, then $\gcd=13$.
Then I tried to prove that only for $n=... | I think that I just solve it:
Let $d=gcd(n^2+3, (n+1)^2+3).$ By the property $\gcd(a,b)=\gcd(a,b-a)$, we obtain:
$$d=\gcd(n^2+3,2n+1).$$
Since $d \mid n^2+3,$ we have $n^2+3=ds$ for some integer $s$.
Similarly, $2n+1=dr$ for some integer $r$, and by this relation we obtain: $4n^2=d^2 r^2 -2dr +1 \iff 4ds=4(n^2+3)=d^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$
Is there an elegant proof for this identity for all real $s \neq -1, -1/2$?
$$\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$$
Whe... | Partial solution to the second double sum:
Lets work on the inner sum first
\begin{align}
S_n=\sum_{k=1}^{n+1}\frac{H_{k-\frac12}}{k}=\sum_{k=1}^{n}\frac{H_{k-\frac12}}{k}+\frac{H_{n+\frac12}}{n+1}
\end{align}
by substituting $\ H_{k-\frac12}=2H_{2k}-H_k-2\ln2$, to have
\begin{align}
\sum_{k=1}^{n}\frac{H_{k-\frac12}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating $\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx$ My first instinct was to evaluate the indefinite form of the integral, which I did by substituting $x=\tan t$, therefore yielding
\begin{align}
\int \frac{x\ln x}{(1+x^2)^2} \,dx
&= \int \frac{\tan t \sec^2 t \ln\tan t}{(1+\tan^2 t)^2} \,dt
&& \text{by substi... | \begin{align}
I&=\int_0^\infty\frac{x\ln x}{(1+x^2)^2}\ dx\overset{\large x\ \mapsto\frac1x}{=}-I\\
2I&=0\\
I&=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
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Find the asymptote of the function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ We have a function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ and when $x$ goes towards $-\infty$, we have an asymptote
$y = -2x - 3/2$.
How we get this asymptote?
| Note that if $ax+b$ is an asymptote for the function $f(x)$ in $-\infty$, then$$a=\lim_{x\to -\infty}{f(x)\over x}\\b=\lim_{x\to -\infty}{f(x)-ax}$$therefore$$a{=\lim_{x\to -\infty}{\sqrt{x^3\over x-3}-x\over x}\\=\lim_{x\to -\infty}{-\sqrt{x\over x-3}-1}\\=-2}$$and $$b{=\lim_{x\to -\infty}\sqrt{x^3\over x-3}-x+2x\\=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Interesting four-sum inequality $n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right) \ge...$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \lef... | We have$$\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}{=\sum \limits_{k=1}^n (2k-1)\left\{1+{1\over k}\right\}\\=\sum \limits_{k=1}^n 2k-1+\sum \limits_{k=1}^n {2k-1\over k}\\=n^2+2n-\sum_{k=1}^{n}{1\over k}\\=n^2+n-H_n}$$where $H_n$ is the famous Harmonic Number. Similarly$$\sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}=n^2-2n-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Prove that for $\forall m \ge 2$, $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid [(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}]$.
Let $(a, b, c)$ be a Pythagorean triple. Prove that for $\forall m \ge 2$, $$\large \dfrac{1}{2}(a^4 + b^4 + c^4) \mid \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}\right]$$
Let $P_{n} = a^{2^n} + b^{2^n} + c^{... | HINT.-Given that $a, b, c$ are Pythagorean triples, we can deduce that the problem is equivalent to demonstrating that for every pair of integers $a, b$ the quotient
$$\frac{(a^{2^m}+b^{2^m})(a^2+b^2)^{2^{m-1}}+a^{2^m}b^{2^m}}{(a^2+b^2)^2-a^2b^2}$$ is always an integer for $m\ge2$ (which can be verified numerically).
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding the surface area of a region which is generated by revolving a curve around a line The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney. It is early on in the book
so I would expect / hope any integral would be easy to solve.
Problem:
Find the area of the surface gener... | The integral formula for $S$ In the post is questionable. It is good for surfaces revolving around $x$, not $y$,
The following expression should be used, instead, to integrate surfaces around $y$,
$$2\pi \int_1^3 (y+1)\sqrt{1 + \left( \frac{dy}{dx} \right)^2} \,\, dx $$
Otherwise, the integrand goes to infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve $3n^2+3n-1=0\pmod{5}$?
How do I solve $3n^2+3n-1=0\pmod{5}$?
I know the answer should be $1 \pmod 5 $ and $3 \pmod5$, but am unsure how to get there. Thanks!
| The quadratic formula still applies, as the integers modulo $5$ is a field (where division by $2$ is allowed), like the rational numbers and the real numbers (this is because $5$ is a prime number):
$$
n\equiv\frac{-3\pm\sqrt{3^2+4\cdot3}}{2\cdot3}\\
=\frac{-3\pm\sqrt{21}}{6}
$$
Now, $-3$ is the same as $2$, $21$ is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Prove $(x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $ Prove
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $$
if $x,y,z \ge 0$
I made this problem using Holders inequality, no... | Yes, you can!
A full expending gives
$$\sum_{cyc}x^6\sum_{cyc}(x+1)^2-\left(\sum_{cyc}(x^4+x^3)\right)^2=$$
$$=\sum_{cyc}(x^6y^2+x^6z^2-2x^4y^4+2x^6y-2x^4y^3-2x^4z^3+2x^6-2x^3y^3)=$$
$$=\sum_{cyc}x^2y^2(x^2-y^2)^2+2\sum_{cyc}xy(x+y)(x^2+xy+y^2)(x-y)^2+\sum_{cyc}(x^3-y^3)^2\geq0.$$
Actually, the inequality that you name... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Getting different answers for an integral: $\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$ vs $\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$ Problem:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
Using two different methods I am getting two different answers and have trouble finding why.
Method 1:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
$$\int\... | Both answers are the same, mind the constant:
$$\ln|2x+4|+\color{blue}{C_1}=\ln|2\left(x+2\right)|+\color{blue}{C_1}=\ln|x+2|+ \underbrace{\ln 2 + \color{blue}{C_1}}_{\color{purple}{C_2}} = \ln|x+2|+\color{purple}{C_2}$$
| {
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"answer_count": 3,
"answer_id": 0
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Do all unitary matrices belong to a one-parameter unitary group (for Stone's theorem). Background. Per Stone's theorem, a one-parameter unitary matrix group $U_t$ corresponds to a Hermitian matrix $H$:
$$U_t=e^{iHt}$$
Example. The group of unitary matrices
$$U_t=
\left(
\begin{matrix}
1&0&0&0 \\ 0&1&0&0 \\ 0&0&\cos t&... | It is indeed true that every unitary matrix $U$ is an element of some one-parameter subgroup. Equivalently, the map $H \mapsto \exp(iH)$ is a surjective map from the Hermitian matrices to the unitary matrices. One can say that this is a consequence of the fact that the unitary matrices form a connected, compact Lie gr... | {
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"question_score": "6",
"answer_count": 1,
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If $0 < x < y$, then $x^n < y^n$ The problem asks to prove that if $0 < x < y$ then $x^{n} < y^{n}$, where $n$ is a positive integer, so I started by assuming that $0 < x < y$. I then wrote this chain of inequalities: $x^{n} < x^{n-1}y^{1} < x^{n-2}y^{2} < x^{n-3}y^{3} < ... < x^{2}y^{n-2} < x^{1}y^{n-1} < y^{n}$. It i... | Your proof is absolutely correct. But to make the proof clearer and standard I would recommend you to write these first.
$\because$ $0<x<y$.
$\implies \frac{y}{x}>1$ .....$(i)$
Multiplying both sides by $x^n$ we get,
$x^{n-1}y^1>x^n$
Again Multiplying both sides of $(i)$ by $x^{n-1}y$ we get,
$x^{n-2}y^2>x^{n-1}y$
Con... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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What is the probability that the same set of outcomes is obtained when $n$ fair dice are rolled twice? Question
Let's say I have $n$ fair 6-sided dice. What is the probability that the same set of outcomes is obtained when $n$ fair dice are rolled twice?
Example
$n=8$:
Suppose $8$ dice are rolled. If the first roll is ... | You have multinomial distribution with $n$ trials, $k = 6$ outcomes and all $p_i = \frac{1}{6}$. So probability of getting two equal rolls is $\sum\limits_{x_1 + \ldots + x_6 = n} \left(\frac{n!}{x_1! \ldots x_6!}\cdot \frac{1}{6^n}\right)^2$. It's unlikely there is good closed form for it. For asymptotic see, for exam... | {
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"question_score": "1",
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a hint for this Taylor series$ \frac{\cos\left(2x\right)-1}{x^2}$ Compute the first three terms (nonzero) $\frac{\cos\left(2x\right)-1}{x^2}$
the first term is $\cos \left(2\right)-1$ but in the answer, the first term that I have to choose is... $-2$ or $2$ or $-1/2$
UPDATE: Use the substitution method
| Assuming that you want to find the Maclaurin series which is the Taylor series expansion about $0$, you can start from
$$\cos(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+ \dots$$
therefore
\begin{align}\cos(2x)=\sum_{n=0}^{\infty}(-1)^n\frac{(2x)^{2n}}{(2n)!}&=1-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325291",
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"source": "stackexchange",
"question_score": "1",
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General rule for integrals of irrational functions involving $r = \sqrt{a^2 + x^2}$ Browsing the List of integrals of irrational functions I found out that if
$$r=\sqrt{(a^2+x^2)}$$
then
$$\int{rdx}=\frac12(xr+a^2\ln{(x+r)})$$
$$\int{r^3dx}=\frac14xr^3+\frac38a^2xr+\frac38a^4\ln(x+r)$$
$$\int{r^5dx}=\frac16xr^5+\frac{5... | If we denote the integral by
$$I_k=\int\left(a^2+x^2\right)^{(2k+1)/2}\mathrm{d}x$$
then we can apply integration by parts to get
$$\begin{align}
I_k
&=x\left(a^2+x^2\right)^{(2k+1)/2}-(2k+1)\int x^2\left(a^2+x^2\right)^{(2k-1)/2}\mathrm{d}x\\
&=x\left(a^2+x^2\right)^{(2k+1)/2}-(2k+1)\int \left(a^2+x^2\right)^{(2k+1)/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326472",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Find sum $\sum _{k=0}^nF_kF_{n-k}$
Find sum $\sum _{k=0}^nF_kF_{n-k}$
My try
Let $$ a_n = \sum _{k=0}^nF_kF_{n-k} // \cdot x^n
\\ a_n x^n = \sum _{k=0}^nF_k x^k F_{n-k} x^{n-k} // \sum_n \\
A(x) = (F(x))^2 + \cdots + (F(x))^2 = n\cdot (F(x))^2$$
I have some doubts:
*
*I am not sure that summing $F_k x^k$ by $n$ g... | I guessed the formula, then prove it by induction.
Let $s_n = \sum_0^n F_k F_{n-k}$
From identity: $F_m F_n = F_m F_{n+1} + F_{m-1} F_n$
$s_n = (n-1) F_{n-1} - s_{n-2}$
$s_n = (n-1) F_{n-1} - (n-3) F_{n-3} + (n-5) F_{n-5} - \cdots$
Simplify the sum, so that $s_n = a F_{n} + b F_{n-1}$, and build a table:
$$\begin{matr... | {
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"answer_count": 2,
"answer_id": 1
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Finding the integral $\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}$ One may take $x= \cos t$ and get
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= -\frac{1}{4}\int \csc^4(t/2)~ dt=-\frac{1}{4} \int [\csc^2(t/2) +\csc^2(t/2) \cot^2(t/2)]~ dt.$$
$$\Rightarrow I=\frac{1}{2} \left [\cot (t/2)] +\frac{1}{3}\cot^3(t/2)\right]=\frac{(2-x)}... | Because we have the radical $\sqrt{1-x^2}=\sqrt{(1-x)(1+x)}$ (and because we know the anser), we may try the Euler substitution $$
t=\sqrt{\frac {1+x}{1-x}}\ .
$$
In
https://en.wikipedia.org/wiki/Euler_substitution
it is the third Euler substitution:
$$
\sqrt{(1+x)(1-x)}=t(1-x)\ .
$$
Then $t^2=(1+x)/(1-x)$, $t^2(1-x)=... | {
"language": "en",
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Show that $\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1$
Show that $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1 .$$
I know that $\arctan 1 = \frac{\pi}{4}$ and that the sequence ... | you can actually get a stronger result such as $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 1/11 $$
because the error in a convergent alternating series is less than the absolute value of the first missing term which in this case is $1/11.$
The alterna... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Maclaurin expansion of $\arccos(1-2x^2)$ Maclaurin expansion of $\arccos(1-2x^2)$
This is what I tried.
$f'(x)=2(1-x^2)^{-1/2} \\
f''(x)=2(1-x^2)^{-3/2}+3 \cdot 2 x^2(1-x^2)^{-5/2} \\
f^{(3)}(x)=18x(1-x^2)^{-5/2}+2\cdot 3\cdot 5x^3(1-x^2)^{-7/2} \\
f^{(4)}(x)=18(1-x^2)^{-5/2}+180x^2(1-x^2)^{-7/2}+2\cdot 3\cdot 5\cdot 7... | As David has pointed out, $\cos^{-1}(1-2x^2)$ is not differentiable at $x=0$, so it does not have a Maclaurin series in a strict sense.
But now let $x\ge0$ and $\cos^{-1}(1-2x^2)=u$. We have
$$1-2x^2=\cos u$$
$$\sqrt{\frac{1-\cos u}2}=x=\sin\frac u2$$
$$\sin^{-1}x=\frac u2$$
$$u=2\sin^{-1}x=\cos^{-1}(1-2x^2)$$
Now the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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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.