Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $\sin^2x+\sin^2y<1 \forall x,y \in R$, then prove that $\sin^{-1}(\tan x\cdot\tan y)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ If $\sin^2x+\sin^2y<1 \forall x,y \in R$, then prove that $\sin^{-1}(\tan x\cdot\tan y)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
My attempt is as follows:-
$$f(x)=\tan x\tan y$$
$... | Using this, $$-\dfrac\pi2\le\sin^{-1}z\le\dfrac\pi2$$ for $-1\le z\le1\iff z^2\le1$
$$\sin^2y<1-\sin^2x=\cos^2x\implies\dfrac{\sin^2y}{\cos^2x}<1$$
$$\text{Similarly, }\dfrac{\sin^2x}{\cos^2y}<1$$
$$\implies\dfrac{\sin^2x\sin^2y}{\cos^2x\cos^2y}=\dfrac{\sin^2x}{\cos^2y}\cdot\dfrac{\sin^2y}{\cos^2x}<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Suppose that $x^5$ and $20x+\frac {19}x$ are rational numbers. Then $x$ is also rational Let $x\neq0$ be a real number such that $x^5$ and $20x+\frac {19}x$ are rational. How can we prove that $x$ is also rational? (This was a question from the RMO 2019 in India.)
My attempt: Let $a,b,c,d$ be integers such that $20x+\f... | It seems that one answer has been posted online. The argument is elementary and straightforward. I will repeat it here.
Let $r=20x+\frac{19}{x}\in\mathbb{Q}$. It follows that
$$
20x^2-rx+19=0
$$
and the quadratic formula implies that
$$
x=r_1\pm\sqrt{r_2},\quad
r_1=\frac{r}{40},\quad
r_2={\frac{r^2-4\cdot 20\cdot 19... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
How to determine a limit without L'Hospital's rule
How to solve this type of limit without L'Hospital rule.
$$\lim_{x\to a}\frac{a^x-x^a}{x-a}$$
| To make life easier, let $x=a+y$
$$\lim_{x\to a}\frac{a^x-x^a}{x-a}=\lim_{y\to 0}\frac{a^{a+y}-(a+y)^a}{y}$$ Factor out $a^a$ to make
$$a^{a+y}-(a+y)^a=a^a\left(a^y-\left(1+\frac{y}{a}\right)^a \right)$$
Now
$$a^y=e^{y \log(a)}=1+y \log (a)+\frac{1}{2} y^2 \log ^2(a)+O\left(y^3\right)$$
$$b=\left(1+\frac{y}{a}\right)^a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solve Exponential Equation with square Solve equality with square
$2^{2x}=7\cdot 2^{x+\sqrt{x-1}}+8\cdot 2^{2\sqrt{x-1}}$
$x-1\ge0 \\\sqrt{x-1}=t\ge0\Rightarrow x-1=t^2\Rightarrow x=t^2+1\\2^{2(t^2+1)}=7\cdot2^{t^2+1+t}+2^{3+2t}$
It looks very complicated and I don't know how to move it.
Is there a better way to appr... | Hint
If $$2^x=a,2^{\sqrt{x-1}}=b$$
Now for real $y,2^y>0$
we have $$0=a^2-7ab-8b^2=(a-8b)(a+b)$$
So, $$2^x=2^{3+\sqrt{x-1}}$$
Set $\sqrt{x-1}=p\ge0\implies x=p^2+1$
Like Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$
as $2\ne\pm1$
$$\implies p^2+1=3+p\iff(p-2)(p+1)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How many solutions does the equation $x_1+x_2+x_3+x_4+x_5 = 8$ have? $x$$1$+$x$$2$+$x$$3$+$x$$4$+$x$$5$ = 8
where $x$$i$'s can take values $\{0,1,2,3\}$.
| In this case, it's easy to check by hand that the only solutions are
\begin{align*}
0+0+2+3+3&=8\\
0+1+1+3+3&=8\\
0+1+2+2+3&=8\\
0+2+2+2+2&=8\\
1+1+1+2+3&=8\\
1+1+2+2+2&=8,
\end{align*}
up to permutations of the terms in the different sums. Thus the number of solutions is
$$\frac{5!}{2!\,2!}+\frac{5... | {
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If $n\geq 3$, $9^n \equiv a (\mod 100)$ and $9^{n+1} \equiv b (\mod 100)$, then $a+b=90$. I noticed a pattern in the powers of 9 modulo 100.
$9^1 \equiv 9 \pmod{100}$
$9^2 \equiv 81 \pmod{100}$
$9^3 \equiv 29 \pmod{100}$
$9^4 \equiv 61 \pmod{100}$
.
.
.
and conjectured the following:
If $n\geq 3$ is an odd integer wher... | $$a+b\equiv9^n(1+9)\pmod{100}$$
Now $9^n\equiv(-1)^n\pmod{10}\implies10\cdot9^n\equiv10\cdot(-1)^n\pmod{10\cdot10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiation under integration $\int_0^{\frac{\pi}{2}}\frac{\ln(1+\cos\alpha\cos x)}{\cos x}dx$
Question: Discuss the method of differentiation under the sign of integration. Hence evaluate following integrals: $$(i)\int_0^{\infty}\frac{\ln(1+a^2x^2)}{1+b^2x^2}dx\quad(ii)\int_0^{\frac{\pi}{2}}\frac{\ln(1+\cos\alpha... | Using Nyssa's comment I tried like that:
\begin{align*}
I(a) &= \int_0^{\infty}\frac{\ln(1+a^2x^2)}{1+b^2x^2}dx \\
I'(a)&=\frac{d}{da}\int_0^{\infty}\frac{\ln(1+a^2x^2)}{1+b^2x^2}dx\\
&=\int_0^{\infty}\frac{\partial}{\partial a}\left(\frac{\ln(1+a^2x^2)}{1+b^2x^2}\right)dx\\
&=\int_0^{\infty}\frac{2ax^2}{\left(1+x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446735",
"timestamp": "2023-03-29T00:00:00",
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$p_1^2 + p_2^2 = q_1^2 + q_2^2 = r_1^2 + r_2^2 = 2$, $p_1^2 + q_1^2 + r_1^2 = 3$, positive distinct rational solution I would like to know if equations
$p_1^2 + p_2^2 = 2$
$q_1^2 + q_2^2 = 2$
$r_1^2 + r_2^2 = 2$
$p_1^2 + q_1^2 + r_1^2 = 3$
have a solution where $p_1,p_2,q_1,q_2,r_1,r_2$ are pairwise distinct and positi... |
Yes, the possible solutions in the rationals are $x,y = \pm1, \pm1$, so
$$p = [1,1]$$
$$q = [-1,1]$$
$$r = [-1,-1]$$
is a solution
Edit:
Based on comments.
We seek a solution $x,y$ such that $x^2 +y^2=2$ where $x,y \in \mathbb{Q}$. To fulfill this (and without loss of generality within the specified domain) we require... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve differential equations $y' = \frac{x-y^2}{2y(x+y^2)}$ Help me!
I'm can try solve this:
$$y' = \frac{x-y^2}{2y(x+y^2)}$$
I'm try to change variable
$$ t = y^2$$
but what is next?
am I need to solve this?
$$t'= \frac{x-t}{x+t}$$
| $$ 2y(x+y^2)dy=(x-y^2)dx $$ $\Rightarrow $
$$ xdx-y^2dx-2yxdy-2y^3dy=0$$ We can make a replacement (1)
$$ 1=2\alpha=\alpha+1+\alpha-1=3\alpha+\alpha-1 $$
$ \alpha=\frac{1}{2}$ $\Rightarrow $ $ y=\sqrt{z}$
$$\frac{dy}{dx}=\frac{dz}{2\sqrt{z}dz} $$ $\Rightarrow $
$$(x-z)dx-(x+z)dz=0 $$
$$\frac{x-z}{x+z}=\frac{dz}{dx} $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all $x$ such that determinant is zero . $$\begin{vmatrix}x-a & a-b & b-c \\ c-a & x-b & a-c \\ b-a & c-b & x-c \end{vmatrix} =0 $$
I began expanding but quickly gave up, i couldn't factor it.Adding or subtracting rows/columns got me nowhere because i made no zeroes.
Any idea would be helpful. Hopefuly there is ... | Without constraints or additional information, your determinant is:
$$-a^3+a^2 (b+c+x)-a \left(b^2+2 b (c-x)-c^2+2 c x+x^2\right)+b^3-b^2 (c+x)+b \left(c^2+2 c x-x^2\right)-c^3+c^2 x-c x^2+x^3$$
and brute force is the only way forward. The three solutions (using the cubic equation) are:
$$\{x\to a-b+c\} \\
\left\{x\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve a diophantine equation The questions is: Solve in the positive integers $$3^n=m^4+m+1.$$
I can prove that $n$ is odd, and $m\equiv 4(\text{mod }9)$, but i dont know why $n$ and $m$ are $1$.
| Probably not of much use but some elementary observations about the numbers $m^4+m+1$ are as follows.
There is precisely one solution modulo $3^n$ of the equation
$$m^4+m+1\equiv 0\pmod {3^n}$$
The solution for $3^{n+1}$ can be obtained from the solution for $3^n$ as follows:-
If $m^4+m+1\equiv 3^nd\pmod {3^{n+1}}$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Prove that in any triangle
Prove that in any triangle with side lengths $a,b,c$ inradius $r$, and
circumradius $R$, we have
$\frac{a}{b+c}+ \frac{b}{c+a}+ \frac{c}{a+b}+ \frac{r}{R} >
\frac{5}{3}$
I think it is $\frac{a}{b+c}+ \frac{b}{c+a}+ \frac{c}{a+b}+ \frac{r}{R} \leq 2 $.
with $a+b+c=2s,\ ab+b+ca=(4R+r)r+s^2... | In the standard notation we need to prove that
$$\sum_{cyc}\frac{a}{b+c}+\frac{\frac{2S}{a+b+c}}{\frac{abc}{4S}}\leq2$$ or
$$\sum_{cyc}\frac{a}{b+c}+\frac{16S^2}{2abc(a+b+c)}\leq2$$ or
$$\sum_{cyc}\frac{a}{b+c}+\frac{\sum\limits_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)}{2abc}\leq2$$ or
$$\sum_{cyc}(a^5b+a^5c-2a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to calculate the determinant of a fully/negative symmetric matrix? I have two similar questions.
*
*How to calculate the determinant of the following $n \times n$ matrix?
$$\begin{vmatrix}
a & b & \dots & b & b\\
-b & a & \dots & b & b\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
-b & -b & ... & a ... | Matrix 1
Consider the determinant polynomials
$$
\begin{align}
p_n(x)
&=
\det\begin{bmatrix}
1&x&x&\cdots&x\\
-x&1&x&\cdots&x\\
-x&-x&1&\cdots&x\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-x&-x&-x&\cdots&1
\end{bmatrix}\tag{1a}\\[6pt]
&=
\det\begin{bmatrix}
1-x&0&0&\cdots&1+x\\
-x&1&x&\cdots&x\\
-x&-x&1&\cdots&x\\
\vdots&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to find singular solutions and determine their types for system of equations.
Given equations $x^\prime=x(1-x-2y)$ and $y^\prime=y(1-y-4x)$ find all the singular solutions in the upper quadrant, $x\geq 0, y\geq 0$ and determine the type and stability.
So for singular solutions I believe I want points such that:
$... | We want to simultaneously solve
$$x(1-x-2y) = 0 \\y (1-y-4x)=0$$
Clearly, $(x, y) = (0, 0)$.
When $x = 0$, we have $y(1 - y) = 0 \implies y = 0, y = 1$.
When $y = 0$, we have $x(1 - x) = 0 \implies x = 0, x = 1$.
Then we have
$$(1-x-2y) = 0 \\ (1-y-4x)=0$$
This gives us $x = \dfrac{1}{7}, y = \dfrac{3}{7}$.
So, our cri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Inequality with Fibonacci numbers $ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} \: \text{arccot} F_{2k-1} > \text{arccot} \: 2$
Prove that $$ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} \: \text{arccot} F_{2k-1} > \text{arccot} \: 2$$ holds for all $n \in \mathbb{N}$.
(The Fibonacci sequence, defined by the recurrence $F_1 = F_2 =... | Show that
$$\text{arccot} F_{2n+1} = \text{arccot} L_{2n} + \text{arccot} L_{2n+2}, \quad (1)$$ where $L_n$ is the Lucas series with $L_0 = 2, L_1 = 1 , L_3 = 3, L_{n+2} = L_{n+1} + L_n$.
Hence,
$$ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} \: \text{arccot} F_{2k-1} = \text{arccot} L_0 + \text{arccot} L_{4n+2} > \text{arcc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve in Z. $y^2+y=x^4+x^3+x^2+x$ Solve in Z. $$y^2+y=x^4+x^3+x^2+x$$
In my attempt to solve this, I used the fact that the left side is even.I write the eqyation in the following form: $$(x+1)(x^2+1)=0 \pmod 2$$. This clearly means x must be an odd number, as a result $$4|y(y+1)$$. This is as far as I could go.
Quest... | Hint: Bound between 2 squares
For all but finitely many values of $x$, we can bound $(2y+1)^2$ between 2 consecutive perfect squares, hence it is never a perfect square.
$(2x^2 + x)^ 2 < (2y+1) ^2 = 4y^2 + 4y + 1 = 4x^4 + 4x^3 + 4x^2 + 4x + 1 < 4x^4 + 4x^3 + 5x^2 + 2x + 1 = (2x^2 + x + 1)^2 $
Hence, we only need to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Split into partial fractions $\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}$ This is from "Calculus Made Easy", Exercises 10, Question 15 (page 147). I've worked this one over and over and still haven't made progress.
This is my initial setup:
$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+x+1)} + \fra... | I'll expand all the things from the comment.
Method 1:
By expanding we get
$$
\begin{cases}
A+B=0&&\hbox{ for }x^4\\\
2A+3B+C=0&&\hbox{ for }x^3\\\
3 A + 3 B + 3 C + D - 3=0&&\hbox{ for }x^2\\\
2 A + 2 B + 3 C + 2 D + E - 2=0&&\hbox{ for }x\\\
A + 2 C + 2 E - 1=0&&\hbox{ for }1
\end{cases}
$$
which boils down to $A = 1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How do I solve recurrence relation without characteristic equation? Question:
Solve the recurrence relation
$\ a_n = 3a_{n-1} - 2a_{n-2} + 1 $, for all $\ n \ge 2$
$\ a_0 = 2 $
$\ a_1 = 3 $
Write $\ a_n $ in terms of n
I tried to solve this by finding the characteristic equation, $\ r^2 - 3r + 2 - 1 = 0 $ which is $\... | A general way to solve this is given by generating functions. Define:
$\begin{equation*}
A(z)
= \sum_{n \ge 0} a_n z^n
\end{equation*}$
Take the recursion, shift so there are no subtractions in indices, multiply by $z^n$ and sum over $n \ge 0$. Recognize the resulting sums, use initial values:
$\begin{align*}
\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Given $2018 \times 4$ grids and tint them with red and blue. So that each row and each column...
Given $2018 \times 4$ grids and tint them with red and blue. So that each row and each column has the same number of red and blue grids, respectively. Suppose there're $M$ ways to tint the grids with the mentioned requirem... | This is only a response to if your way of thinking about the problem is correct.
Let's consider this a slightly different way:
$$\hat{v_1} = \begin{pmatrix}R \\ R \\ B \\ B\end{pmatrix}, \hat{v_2} = \begin{pmatrix}R \\ B \\ R \\ B\end{pmatrix}, \hat{v_3} = \begin{pmatrix}R \\ B \\ B \\ R\end{pmatrix}, \hat{v_4} = \begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$. Or stated in other terms, prove that
$$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$
| Without needing to know the numerical value of $\pi^2/6$, you can use the infinite series representation of the golden ration and some very crude bounds to arrive at the conclusion. Recall $$\varphi = \sum^\infty_{n=0} \frac{(-1)^n}{F_n F_{n+1}}$$ where $F_0 = 1, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}, \,\,\, n \ge 2$ i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Understanding the equation $|x+1|=x^2 -1$ I want to understand the equation
$$|x+1|=x^2 -1$$
$$\Leftrightarrow x^2 - |x+1| - 1 = 0$$
Case $1$:
$$x+1 \geq 0 \Rightarrow x^2 - x-2 = 0$$
$$x_{1,2} = \frac{1}{2} \big( 1\pm \sqrt{1-4\cdot(-2)} \big) = \frac{1}{2}(1\pm3) \Rightarrow x_1 = 2, x_2 = -1$$
Case $2$:
$$x+1 < 0 \... | For $x+1 \geq 0$, $|x+1| = x+1$, and hence the equation reads as:
$$x+1-x^2 +1 = 0 \Rightarrow x^2 -x - 2=0.$$
For $x+1 < 0$, $|x+1| = -(x+1)$, and hence the equation reads as:
$$-(x+1)-x^2 +1 = 0 \Rightarrow x^2 +x=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $ \arctan x + 2 \arctan(\sqrt{1 + x^2} - x) = \frac{\pi}{2} $ I want to prove that: $$\arctan x + 2 \arctan(\sqrt{1 + x^2} - x) = \frac{\pi}{2}, \forall x \in \mathbb{R} $$
I know that $$ \arctan x + arctan \frac{1}{x} = \frac{\pi}{2} $$
But that doesn't seem to be helping. How should I proceed?
| We have
\begin{eqnarray*}
\tan^{-1}(A) + \tan^{-1}(B) + \tan^{-1}(C) = \tan^{-1} \left( \frac{A+B+C-ABC}{1-AB-BC-CA} \right).
\end{eqnarray*}
If the RHS is to give $ \pi/2$ then we require the denominator to be zero, so
\begin{eqnarray*}
1-2x( \sqrt{1+x^2} -x) -(\sqrt{1+x^2} -x)^2=0
\end{eqnarray*}
Which is easi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number? Let $a,b,c,d$ be natural numbers such that $a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number?
Now assume that $a+b+c+d = p > 2$ for some choice of $a,b,c,d$. Notice that we cannot have $a=b=c=... | Another solution, whose approach is distinct from the rest.
Suppose that for some prime $ p \geq 4$, $ a + b + c + d = p$.
Then $ a+ b \equiv - (c+d) \pmod{p}$,
$ a^2 + 2ab + b^2 \equiv c^2 + 2cd + d^2 \pmod{p}$,
$ab \equiv cd \pmod{p}$.
$a^2 - 2ab + b^2 \equiv c^2 - 2cd + d^2 \pmod{p}$
WLOG $ a - b \equiv c - d \pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Limit of $\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$ I tried using symbolab to get the limit of
$$\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$$
but it couldn't solve it. With WolframAlpha I got $100$ for the limit.
Can someone show me how it's done "per ... | Are you sure you got $200$ and not $100$? Is there a typo here or in your input to WolframAlpha?
We have, either applying binomial theorem, or L'Hospital, or the fundamental exponential limit $\frac{e^x-1}{x}$ that
$\frac{(1+x^5)^{10}-1}{10x^5}\to1$
$\frac{\sqrt{1+x^3}-1}{\frac{1}{2}x^3}\to1$
$\frac{\sqrt[5]{1+x^2}-1}{... | {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
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What's wrong in my calculation of $\int \frac{\sin x}{1 + \sin x} dx$? I have the following function:
$$f: \bigg ( - \dfrac{\pi}{2}, \dfrac{\pi}{2} \bigg ) \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{\sin x}{1 + \sin x}$$
And I have to find $\displaystyle\int f(x) dx $. This is what I did:
$$\int \dfrac{\sin x}{1... | Required answer is $E$. Observe that
$$\frac{1}{\cos x}-\tan x=\frac{1-\sin x}{\cos x}$$
$$=\frac{(\cos\frac{x}{2}-\sin\frac{x}{2})^2}{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}$$
$$=\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}$$
$$=\frac{1-\tan\frac{x}{2}}{1+\tan\frac{x}{2}}$$
$$=\frac{2}{1+\tan\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Evaluate $\int\frac{dx}{(1+\sqrt{x})(x-x^2)}$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}
$$
Set $\sqrt{x}=\cos2a\implies\dfrac{dx}{2\sqrt{x}}=-2\sin2a.da$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}=\int\frac{-4\sin2a\cos2a.da}{2\cos^2a.\cos^22a.\sin^22a}\\
=\int\frac{-2.\sec^2a.da}{\sin2a\cos2... | You might have better luck with $y=\sqrt{x}$, provided you work with partial fractions. Something similar is achieved by continuing your current approach with $t=\tan a$, which amounts to starting with $\sqrt{x}=\frac{1-t^2}{1+t^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Prove that $f(x) = \sqrt{x^2 + x}$ is uniform continuity at the interval $[0, \infty)$. Prove that $f(x) = \sqrt{x^2 + x}$ is uniform continuity at the interval $[0, \infty)$.
There is a hint to the exercise:
Prove that $\forall x,y \geq 1: \frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}} \leq 2$
How does the hint help me?
| $f$ is clearly uniformly continuous on $[0,1]$.
To show uniform continuity on $[1, +\infty\rangle$ notice that for $x \ge 1$ we have
$$f'(x) = \frac{2x+1}{2\sqrt{x(x+1)}} \le \frac{2\cdot 1+1}{2\sqrt{1(1+1)}}= \frac{3}{2\sqrt2}$$
so for $x,y \ge 0$ the mean value theorem gives that there exists $\theta$ between $x$ an... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this quadratic equation (with $x$ represented by a fraction containing a square root)? If $x-\frac{4}{5} = \pm \frac{\sqrt{31}}{5}$, how can I find the values of a and b in the following equation?
$$5x^2+ax+b=0?$$
I've tried substituting $(\frac{4}{5} \pm \frac{\sqrt{31}}{5})$ for $x$ and got totally confu... | So, I have: $$x=\frac{-a\pm\sqrt{b^2-20a}}{10}=-\frac{a}{10}\pm\frac{\sqrt{b^2-20a}}{5}$$
Now $x=\frac{4}{5}\pm\frac{\sqrt{31}}{5}$, so: $-\frac{a}{10}=\frac{4}{5}$ and $\frac{\sqrt{b^2-20a}}{5}=\frac{\sqrt{31}}{5}$
From the first $a=-8$ and substituing in the second: $b=-3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Theta Functions and Partitions I am reading some papers by Ramanujan on congruence properties of the partition function. At one point he says that he will be using "theta functions" and introduces the following:
It can be shewn that
$$
\begin{align}
&\dfrac{(1-x^5)(1-x^{10})(1-x^{15})\dots}{(1-x^{1/5})(1-x^{2/5})(1... | The general Ramanujan theta function
is defined by
$$ f(a,b) := 1 + (a+b) + ab(a^2+b^2) + (ab)^3(a^3+b^3) + \dots. \tag{1} $$
which factors according to the Jacobi triple product as
$$ f(a,b) = (-a;ab)_\infty(-b;ab)_\infty(ab;ab)_\infty. \tag{2} $$
An important special case is the single variable theta function
$$ f(-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$
If real numbers $a$, $b$, $c$, $d$ satisfy
$$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$
then find $(a,b,c,d)$.
What I try:
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, pl... | Calculalte discriminant twice and the discriminat tells the value is only one.
$2a^2-2(b+1)a+b^2+(b-c)^2+c^2+1-1/4=0 $
$D/4=(b+1)^2-2{2b^2-2bc+2c^2+3/4}\geq0$
We get,
$3b^2-2(1+2c)b+4c^2+1/2=0 $
$D/4=(1+2c)^2-3(4c^2+1/2)\geq0 $
$-8c^2+4c-1/2\geq0 $
$16(c-1/4)^2\leq0 $
$c=1/4 $
$3b^2-3b+3/4<=0 $
$(2b-1)^2\leq0 $
$b=1/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Evaluating $\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$ without l'Hopital's rule or Taylor series Can anyone please help me find this limit without l'Hopital's rule, I already used it to evaluate the limit, but I didn't know how to calculate it without l'Hopital's rule.
$$\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$$
An... |
Result 1: $\displaystyle\lim_{x\to0}\dfrac{x^2 - \sin^2x}{x^4} = \frac{1}{3}$
Proof. Note that $\sin x = x - \frac{x^3}{3!} + o(x^5).$
Thus, $\sin^2x = x^2 - 2x\frac{x^3}{3!} + o(x^5).$
This, gives $x^2 - \sin^2 x = \frac{x^4}{3} + o(x^5),$ and the result follows.
Result 2: $\displaystyle\lim_{x\to0} \dfrac{\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a closed formula for the following recursive function How can I express a closed-form formula for the following equation?
$$f(n)=f(n-1)+\frac{C}{f(n-1)} $$
Where $C>0$ and $f(0)=\sqrt{C}$.
| This is the best I can do. It seems like
\begin{align}
\frac{f(n)-f(n-1)}{n-(n-1)} = \frac{C}{f(n-1)}.
\end{align}
Let us solve the differential equation
\begin{align}
f' = \frac{C}{f} \ \ \implies \ \ \frac{d}{dx}(f)^2 = C
\end{align}
which means $f^2 = Cx$ or $f(x) = \sqrt{C}\sqrt{x}$. Observe
\begin{align}
\sqrt{C}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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sum of coefficient of all even power of $x$
The sum of all Coefficient of even power of $x$ in
$(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n}),n\in \mathbb{N}$
what i try
for $n=1,$ we have
$(1-x+x^2)(1+x+x^2)=(1+1)-(1)+(1+1)=3$
for $n=2,$ we have
$(1-x+x^2-x^3+x^4)(1+x+x^2+x^3+x^4)$
$=(1+1+1)-(1+1)+(1+1+1)... | These are two standard geometric series,
one with ratio $-x$,
the other with ratio $x$.
Their product is
$(1-(-x)^{2n+1})(1-x^{2n+1})/((1+x)(1-x))\\
=(1+x^{2n+1})(1-x^{2n+1})/(1-x^2)\\
=(1-x^{4n+2})/(1-x^2)
$
At $x=1$, applying Hoppy,
we get
$(4n+2)x^{4n+1}/(2x)
=(2n+1)x^{4n}
\to 2n+1
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\begin{cases} x^3-y^3=19(x-y) \\ x^3+y^3=7(x+y) \end{cases}$
I should solve the following system: $$\begin{cases} x^3-y^3=19(x-y) \\
x^3+y^3=7(x+y) \end{cases}$$
by reducing the system to a system of second degree.
We can factor:
$$\begin{cases} (x-y)(x^2+xy+y^2)=19(x-y) \\
(x+y)(x^2-xy+y^2)=7(x+y) \end{cases}... | After considering of cases $x=y$ or $x=-y$ use
$$7(x^2+xy+y^2)=19(x^2-xy+y^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Interval Length Needed For Interpolation Let $f(x)=\sqrt{x}$ defined on $[1,2]$, What is the length needed between the sampling points such that the approximation error by interpolation polynomial of order $2$ will not exceed $5*10^{-8}$
We know that $|f(x)-P_n(x)|=|\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)...(x-x_n)|$
So to ... | If the quadratic interpolation is used for $|x-x_i|\le\frac12h$, then the maximum of
$|(x-x_{i-1})(x-x_i)(x-x_{i-1})|$ is at $x=x_i\pm \frac12h$, so that the maximum of the error bound is in fact $\frac{3/8}{3!}⋅\frac38⋅h^3=3⋅2^{-7}⋅h^3$, and to get this smaller than the prescribed error $5\cdot10^{-8}$ gives
$$
h^3\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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show that the following recursive series is bounded and monotonic I need to show that the following recursive series $$a_1=1, \quad a_{n+1}=\frac{a_n+3}{5}$$ is bounded and monotonic.
At first, I want to show that the series is bounded from below with $\frac{3}{4}$. By induction:
For $n=1$ we have $a_1=1>\frac{3}{4}$. ... | You can solve the $a_n$. Actually, for $a_{n+1}=\frac{a_n+3}5$
$$\therefore a_{n+1}+k=\frac{a_n+3+5k}5 $$
Make $k=3+5k$, so $k=-\frac 34$, now
$$a_{n+1}-\frac 34=\frac{a_n+3-\frac{15}4}5=\frac{a_n-\frac 34}5 $$
Let $b_n=a_n-\frac 34$, now $b_1=\frac 14$, and
$$b_{n+1}=\frac{b_n}5$$
So we know $a_n-\frac 34=b_n=\frac 1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that : $a,b,c\in\mathbb{Z}$ then : $abc\equiv 0\pmod{3} \implies a^{3}+b^{3}+c^{3}\equiv 0\pmod{9}$ Prove that :
$a,b,c\in\mathbb{Z}$ then :
$$abc\equiv 0\pmod{3} \implies a^{3}+b^{3}+c^{3}\equiv 0\pmod{9}$$
My try :
Let $a\not\equiv 0\pmod{3}$
Then :
$$a\equiv 1,2\pmod{3}$$
So :
$$a^{3}\equiv ±1\pmod{9}$$
... | As pointed in the comments , the equivalence is false but it's converse is always true . $$a^3+b^3+c^3 \equiv 0 \mod9\implies abc\equiv 0\mod3$$
Since a cube can be written as $x^3\equiv 0,1,8 \mod 9$ , the only way to add these residues to be divisible by $9$ is $(0,0,0)$ and $(0,1,8)$ and their permutations.
In eit... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can a non-right triangle be solved with only two sides if it is inside a parallelogram? In this triangle
I am given 2 side lengths for one triangle and two side lengths for the parallelogram. I am asked to find the length of m (FE) and n (DE)
I am given the lenghts:
*
*h (AC) = 9
*k (AF) = 15
*f (AB) = 16
I don'... | I don't believe there's enough information. Let's let
$$A=(0,0)$$
$$B=(16,0)$$
$$C=(9\cos\theta,9\sin\theta)\text{ for some }0<\theta<90^{\circ}$$
$$D=B+C=(16+9\cos\theta,9\sin\theta)$$
$$F=\left(\sqrt{15^2-9^2\sin^2\theta},9\sin\theta\right)$$
Then we have a parallelogram $ABDC$ with a point $F$ on $CD$, such that $|A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation. A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation.
My attempt is as fol... | hint...let the centre of the circle be $(p, 2-p)$ so you can use the distances to form a quadratic in $p$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$? How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$?
\begin{align*}
2\cos x(\sin x)^2+(\cos x)^2 & = \frac{1}{2}(\cos x- \cos 9x) +(\cos 4x)^2\\
4\cos x(1-(\cos x)^2)+2(\cos x)^2 & = \cos x + \cos 9x +(2(\cos 4x)^2-1)+1\\
\cos x(... | Using Euler's formula $e^{ix} = \cos x+i\sin x$ we can simplify this a bit:
\begin{align}
\sin 2x\sin x + \cos^2x &= \frac1{2i}\left(e^{2ix}-e^{-2ix}\right)\frac1{2i}\left(e^{ix}-e^{-ix} \right) + \frac14\left(e^{ix} + e^{-ix}\right)^2\\
&= \frac14\left(-e^{3ix} + e^{ix} +e^{-ix}-e^{-3ix} + e^{2ix} + 2 + e^{-2ix} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove that $\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$ I need to prove that: $$\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$$
I am new to inequalities so I only tried C-S and AM-GM but none of those work.
Any hints on how to proceed here?.
| $(9x^4 + y^4)(x^2 + y^2)\ \ge\ (3x^3 + y^3)^2\ \ge\ |6x^3y^3|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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System of equations modulo $n$ Suppose we have the following system of equations module $n$
\begin{align*}
2x-y & \equiv 1\mod{n}\\
4x+3y & \equiv 2 \mod{n}
\end{align*}
Determine the integers $n$ for which the system has solutions.
I have posted my solution as an answer below.
| Let $2x-y=1+an$ and $4x+3y=2+bn$ where $a,b$ are integers
$2+bn=4x+3(2x-1-an)\iff5(2x-1)=(3a+2b)n$
As $(3,2)=1, $ any integer can be expressed in terms of $3a+2b$
Let us find $a,b$ such that $3a+2b=5c\iff3(a-c)=2(c-b)$
$\dfrac{2(c-b)}3=a-c$ which is an integer
$\implies3|2(c-b)\iff3|(c-b),c-b=3d$ (say for some integer ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Search for a Rationalizing Factor Just something that I've been thinking about lately, and can't figure how to generalize efficiently:
Say we have a surd of the form $\frac{1}{\sqrt2+\sqrt3+\sqrt5+\sqrt7}$, or something along those lines. How can I find the rationalizing factor, without brute force? Does there exist on... | You have for example
$$(a+b\sqrt 2+c\sqrt 3 +d\sqrt 6 )(a-b\sqrt 2 +c\sqrt 3 -d\sqrt 6)(a+b\sqrt 2-c\sqrt 3 -d\sqrt 6 )(a-b\sqrt 2-c\sqrt 3+d\sqrt 6)=a^4 - 4 a^2 b^2 - 6 a^2 c^2 - 12 a^2 d^2 + 48 a b c d + 4 b^4 - 12 b^2 c^2 - 24 b^2 d^2 + 9 c^4 - 36 c^2 d^2 + 36 d^4$$
Note that the second term has the signs on the te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Logarithm Subtraction and Division with Same Bases I'm rusty on logarithms. What is the approach to a problem like this? Any hints would be appreciated.
I'm thinking the subtraction on the numerator and denominator can become division since the bases are the same?
| The identities $\log_a b + \log_a c \equiv \log_a(bc)$ and
$b\log_a c \equiv \log_a c^b$ will be needed here. You can then do
\begin{align*}
\frac{\log_2 24 - \frac 12 \log_2 72}
{\log_3 18 - \frac 13 \log_3 72}
&=
\frac{\log_2 24 - \log_2 \sqrt{72}}
{\log_3 18 - \log_3 \sqrt[3]{72}} \\
&=
\frac{\log_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Is $x^{14}+x^7+1$ irreducible over $Q[x]$? Is $x^{14}+x^7+1$ irreducible over $\mathbb{Q}[x]$?
I think it is, but I'm not able to justify it using any existing criterion (e.g. Eisenstein). Any help?
Indeed, this is a question I encounter on a linear algebra one. The original question gives a $8 \times 8$ real matrix sa... | $x^{14}+x^7+1=(x^7-\omega)(x^7-\omega^2)$
$$=(x^7-\omega^7)(x^7-\omega^{14}) ----(1)$$
Factorizing each factors further we get
$$(x^7-\omega^7)=(x-\omega)(x^6+x^5\omega +x^4\omega^2 +x^3 +x^2\omega +x \omega^2+1)$$
$$=((1+x^3+x^6)+(x^2+x^5)\omega +(x+x^4)\omega^2 ---(2)$$
Similarly $(x^7-\omega^{14})=(x-\omega^2)((1+x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why the wrong answer in integral calculus? Calculate the volume of the solid obtained by rotating the region about the $x$-axis, and bounded by $y=x^2+1, y=3-x^2$.
My calculations:
$V=\pi \int_a^b f^2(x) dx; \; V = V_1 + V_2 = \pi \int_{-1}^1 (3-x^2)^2 dx - \pi \int_{-1}^1 (x^2+1)^2 dx;$
$V_1 = \pi \int_{-1}^1 (3-x^2... | I think it's worth noting that your method, while completely correct (both in result and in detail as far as anyone has found), is far more laborious (and hence more error-prone) than it needs to be.
In this case, the "washer method" is easier.
For the washer method, we note that a given value of $x$ the disk inside t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $\frac{b^{4} - 24 a^{4}}{ab} \ge 3b^{2} + 6ab - 28a^{2} $ I made up a simple problem. I wonder if there are other ways to solve this problem? or maybe whether or not the inequality seems very obvious?
If $a,b$ are positive real numbers and $b$ is at least twice of $a$, prove that
$$ \frac{b^{4} - 24 a^{4}}{ab} ... | Your inequality equivalent to
$b^4 -24a^4 \geq ab(3b^2 +6ab-28a^2)$
We have: $LHS-RHS = (b-2a)^3 (3a+b) \geq 0$ (because: $b$ is at least twice of $a$)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I proved that $\sqrt{z}$ is a continuous function. My proof is not smart. Please show me a smarter proof. Let $U:=\{z \in \mathbb{C} - \{0\} | -\pi < \arg z < \pi\}$.
For $z \in U$, the equation $w^2=z$ has two solutions $w=\pm(u + i v)$, where $u>0$.
Let $f : U \to \mathbb{C}$ be a function such that $f(z):=u+iv$, whe... | It's easier to note that if $z = x + iy = re^{i\theta}$ then $\sqrt{z} =
r^{1 \over 2}\cos{\theta \over 2} + i r^{1 \over 2}\sin{\theta \over 2}$, where
$r^{1 \over 2} = \sqrt{x^2 + y^2}$ and $\theta$ (suitably defined) are both continuous functions of $x$ and $y$. Here you can use branches of $\theta = \arctan({y \ov... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3
If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.
I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.
*
*If $x$ ... | Your argument is fine and well expressed. More succinctly you could paraphrase parts of your argument as follows.
If precisely one of $x,y$ is divisible by $3$ then $x^2-y^2$ is not divisible by $3$.
If both of $x,y$ are divisible by $3$ then $x^2-y^2$ is divisible by $9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Coefficients that make a quartic trivariate polynomial non-negative Let
$$f(x,y,z)=a_1 x^4+a_2 x^3+a_3 x^2+a_4 x^2y+a_5 x^2z+a_6 y^2+a_7 z^2+a_8 xy+a_9 xz+a_{10} yz+a_{11}x+a_{12}y+a_{13}z$$
be a polynomial of degree $4$. Can we determine coefficients $a_1, \dots, a_{13}$ such that $f(x,y,z) \ge 0$ for all $x,y,z\in\m... | We can use the Newton polytope. A positive polynomial should have a convex hull defined by even degree monomials. Our polynomial has a polytope representation as can be observed in the following plot.
In black the convex hull monomials which are
$$
\left[
\begin{array}{ccc}
x & y & z\\
4 & 0 & 0 \\
0 & 2 & 0 \\
0 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_3^7x^3dx$ using $\sum_{j=1}^Nj^3=\left(\frac12{N(N+1)}\right)^2$ Question:
$$\int_3^7x^3dx$$
I know how to solve it using the Power Rule, but I wanted to know how to solve it using:
$$1^3+2^3+\cdots+N^3=\sum_{j=1}^Nj^3=\left({N(N+1)\over2}\right)^2$$
So I'm just confused on how we use that formula to ... | Recall $$\int_a^bf(x)dx=\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x$$
where $\Delta x=\frac{b-a}{n}$ and $x_i=a+i\Delta x$ (I'm using right-endpoints in our Riemann-sum). Thus $$\int_3^7x^3dx=\lim_{n\to\infty}\sum_{i=1}^n\left(3+\frac{4i}{n}\right)^3\cdot\frac{4}{n}$$ $$=\lim_{n\to\infty}\frac{4}{n}\sum_{i=1}^n\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate definite integral using the definition (Riemann Sum) I would like to know how to use the Riemann Sum (With Unequal width) to calculate $\displaystyle f(x) = \sqrt{x}$ from $[1,2]$
I know how to calculate it when the interval starts with 0, for example [0,1], I will do the following way:
*
*$\displaystyle c... | Let $$c_i=\left(1+ \frac{(\sqrt2-1)i}{n}\right)^2$$
Hence,
\begin{align}\Delta x_i&=\left(1+ \frac{(\sqrt2-1)i}{n}\right)^2-\left(1+ \frac{(\sqrt2-1)(i-1)}{n}\right)^2\\
&=\left( 2+\frac{(\sqrt2-1)(2i-1)}{n}\right)\left(\frac{\sqrt{2}-1}{n} \right)\end{align}
and the integral is
\begin{align}
\lim_{n \to \infty} \sum_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that two random variables are independent distributed normally Let $M$ and $N$ are independent random variables distributed Uniform$[0, 1]$.
Define $(M_n)_{n\geq 1}$ and $(N_n)_{n\geq 1}$ which are two independent sequences of iid random variables distributed uniformly over $[−1, 1]$. Let $Z = \inf \{ n ≥ 1, 0 < ... | $M, N$ are uniformly distributed on the unit disk, so $f_{M,N}(m,n) = \pi^{-1}$ in the unit disk and zero outside. We have new variables $X,Y$ expressed in terms of $M,N.$ Recall the Change of Variables theorem, which states that if we have some bijective function $g : (x,y) \ \mapsto \ (m, n)$ with continuous partial ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Conjecture about a continued fraction
Conjecture:
$$\large 2^{n-1}+\frac{1}{2+\cfrac{1}{2^{n}-1+\cfrac{1}{2+\cfrac{1}{2^{n}-1+\cfrac{1}{2+\ddots}}}}}=\frac{1+\sqrt{3a_n}}{2}\tag*{[1]}$$ such that $a_n=4a_{n-1}+1$ and $a_0=0$.$\quad(n\geqslant 1)$
Ex. If $n=1$, then $a_n=4a_0+1=4\times 0 + 1 = 1$. $$\therefore 1+\f... | Claim:
For all $n\ge1$, we have $[2^{n-1},2,2^n-1,2,2^n-1,\cdots]=(1+\sqrt{3a_n})/2$ with $a_n=4a_{n-1}+1$ and $a_0=0$.
Proof:
As $a_n=4a_{n-1}+1=4^2a_{n-2}+4^1+4^0=\cdots=4^ka_{n-k}+\sum\limits_{i=0}^{k-1}4^i$, choosing $k=n$ gives $$a_n=4^na_0+\sum_{i=0}^{n-1}4^i=\frac{4^n-1}3\implies\frac{1+\sqrt{3a_n}}2=\frac{1+\sq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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limit of $\lim_{x\to 7}(\frac{x}{7})^{(\frac{x^2-18x+80}{x-7})}$ $$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{x^2-18x+80}{x-7}\right)}$$
It is $1^{\infty}$
$$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}$$
I tried to take
$$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^... | Hint:
\begin{align}
\lim_{x \to 7} (x-8)(x-10)\frac{\ln x - \ln 7}{x-7} = 3 \lim_{x \to 7} \frac{\ln x - \ln 7}{x-7}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int{\frac{1}{(x^{3} \pm 1)^2}}$ I need to solve the following integrals:
$$\int{\frac{1}{(x^3+1)^2}}dx$$
and
$$\int{\frac{1}{(x^3-1)^2}}dx$$
My first thought was to use a trigonometric substitution but the $x^3$ messed it up. Can you guys suggest another method?
| $$I=\int{\frac{1}{(x^3+1)^2}}dx$$
$$I=-\int \frac {1}{3x^2}\color {red}{{\frac{-3x^2}{(x^3+1)^2}}}dx$$
The function in red is a derivative.
$$I=-\int \frac {1}{3x^2}\color {red}{ \left ({\frac{1}{x^3+1}} \right )'}dx$$
The integral is now of the form:
$$ \color {blue}{I=\int f(x) g'(x)dx}$$
Integrate by part
$$ \colo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$ $$\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$$
An approach I can think about is to expand $\cos$ using taylor series, is there another approach?
| Taylor expand
$$\frac x{1-x^2}=x+x^3+x^5+O(x^7),\>\>\>\>\>\cos t = 1-\frac{t^2}2+\frac{t^4}{24}+O(t^6)$$
to get
$$\cos\frac{x}{1-x^2}=1-\frac{1}2x^2-\frac{23}{24}x^4+O(x^6)$$
Thus,
$$\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$$
$$=\lim _{x\to 0}\frac{1-\frac{x^2}{2}-(1-\frac{x^2}2-\frac{23}{24}x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3542674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of integral $\lim_{n \to \infty} n\int_0^{\frac{\pi}{4}} \tan^n x\,dx$ Compute
$$\lim_{n \to \infty}n \int_0^{\frac{\pi}{4}}\tan^n x\,dx$$
I tried to define a recurrence with $I_n=\int_0^{\frac{\pi}{4}} \tan^n x$ :
$I_0 = \frac{\pi}{4}, I_1=\ln\sqrt{2}$
and $I_{n}=\frac{1}{n-1}-I_{n-2}$, but I can't complete it.
| Define $f : \mathbb{R} \to \mathbb{R}, f(x)=\dfrac{1}{1+x^2}$. Now, integrating by parts:
$$
\begin{aligned}
\int_0^1 (n+1)x^nf(x)\,dx &= \left[x^{n+1}f(x)\right]_0^1-\int_0^1x^{n+1}f'(x)\,dx \\
&= f(1)+2\int_0^1 \frac{x^{n+2}}{(1+x^2)^2}dx
\end{aligned}
$$
Using $1\leq 1+x^2\leq 2$ over $[0,1]$ we get
$$\frac{1}{2(n+3... | {
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"timestamp": "2023-03-29T00:00:00",
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How many integer solutions are there to $x+y+z=8$
How many integer solutions are there to $x+y+z=8$ When $x,y,z>0$? When $x,y,z\geq -3$?
So I know there is a formula for computing the number of nonnegative solutions
${8+3-1 \choose 3-1}={10\choose 2}$
So I then just subtracted cases where one or two integers are $0$.... | Your answer to the question of how many solutions the equation $$x + y + z = 8 \tag{1} $$ has in the positive integers is incorrect. As a sanity check, observe that there must be fewer solutions to the equation in the positive integers than there are in the nonnegative integers since we are not allowed to substitute $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$
Question: Show that
$$\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$$
From Wolfram alpha, it seems that the equality above is indeed correct.
But I do not know how to prove it.
Any hint is appreciated.
| Famously, $\displaystyle \int_0^{\pi/2} \cos^{2n}{x}\,\mathrm{d}x = \frac{\pi}{2^{2n+1}}\binom{2n}{n}$ (e.g. see here); and $\displaystyle \frac{1}{1+n} = \int_0^1 y^n \, \mathrm{d} y$, thus:
$\displaystyle \begin{aligned} \sum_{n \ge 0} \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} & = \frac{1}{\pi} \sum_{n \ge 0}\in... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Square root property proof Can anyone provide a link to a proof of the following square root property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$. Could not find it online anywhere.
| By the definition of square-root, $x=\sqrt{y}\hspace{.15cm}$ means that $x^2=y$.
Assuming that $a,b\in\mathbb{R}$ with $a\geq 0$ and $b>0$, then we can simply compute:
\begin{align*}
\left(\frac{\sqrt{a}}{\sqrt{b}}\right)^2&=\frac{\sqrt{a}}{\sqrt{b}}\cdot \frac{\sqrt{a}}{\sqrt{b}}
=\frac{\sqrt{a}\cdot \sqrt{a}}{\sqrt... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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calculating total number of allowable paths from $(0,0)$ to $(5,5)$ I'm looking at paths starting from $(0,0)$ with the following allowable steps :
1) from $(x,y)$ to $(x,y+1)$
2) from $(x,y)$ to $(x+1,y)$
3) from $(x,y)$ to $(x+2,y+1)$
how can I determine the total number of paths from $(0,0)$ to $(5,5)$?
and then g... | We algebraically encode the base steps
\begin{align*}
&(1,0)\qquad\to\qquad x\\
&(0,1)\qquad\to\qquad y\\
&(2,1)\qquad\to\qquad x^2y
\end{align*}
so that each step can be encoded as $(x+y+x^2y)$. Denoting the coefficient of $x^n$ with the coefficient of operator $[x^n]$ we want to find
\begin{align*}
[x^5y^5]\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question: Using the Cauchy-Schwarz Inequality to compare between 2 expressions Use the Cauchy-Schwarz Inequality to determine whether $a^2+b^2+c^2$ is bigger than/smaller than/equal to $ab+bc+ac$, where $a,b,c$ are integers and $a<b<c$.
Cauchy-Schwarz Inequality:
$$(\sum_{i=1}^{n}a_ib_i)^2 \leq {\left(\sum_{i=1}^{n}a... | We should set $a_1=b_3=a, a_2=b_1= b$ and $a_3=b_2=c$ in the Cauchy-Schwarz inequality, to get:
$$(ab+bc+ca)^2\leq (a^2+b^2+c^2)(b^2+c^2+a^2)=(a^2+b^2+c^2)^2$$
and therefore:
$$a^2+b^2+c^2\geq |ab+bc+ca|\geq ab+bc+ca$$
Of course, we don't need any restriction over $a,b,c$ (they don't have to be integers or ordered, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Help with trig sub integral I've gone over my work twice, and while I dropped a negative the first time, I believe that my work is sound. Webwork is not accepting it, however.
Here is the problem and my solution. Sorry for bad handwriting
| Let $x=2\sin{t}$.
Thus, $$\int\frac{x^2}{\sqrt{4-x^2}}dx=\int\frac{x^2-4+4}{\sqrt{4-x^2}}dx=4\int\frac{1}{\sqrt{4-x^2}}dx-\int\sqrt{4-x^2}dx=$$
$$=4\arcsin\frac{x}{2}-4\int\cos^2tdt=4\arcsin\frac{x}{2}-2\int(1+\cos2t)dt.$$
Can you end it now?
Also, let $x-2=v$.
Thus, we can use the similar way: $$\int\frac{x^2}{\sqrt{4... | {
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"timestamp": "2023-03-29T00:00:00",
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find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$. Let $A$be a $5 \times 5$ matrix whose characteristic polynomial is given by $(\lambda -2)^3(\lambda+2)^2.$ If $A$ is diagonalizable,find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$
Solution:
Characteristic polynomial of $A$,$Ch_{A}(\lambda)=(\lambda -2)^3(... | Since $A$ is diagonalizable, and the characteristic polynomial is $A$ is $c_A(x)=(x-2)^3(x+2)^2$, it follows that $A=PDP^{-1}$, where $P$ is an invertible $5\times5$ matrix, and
$$D=\begin{pmatrix}2&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&-2&0\\0&0&0&0&-2\end{pmatrix}.$$
It follows that $A^{-1}=PD^{-1}P^{-1}$. Note that
$... | {
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"timestamp": "2023-03-29T00:00:00",
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Tough multivariable inequality: Minimize $a^2 + b^3 + c^4$ given $a + b^2 + c^3 = \frac{325}{9}$ Let $a,$ $b,$ and $c$ be positive real numbers such that $a + b^2 + c^3 = \frac{325}{9}.$ Find the minimum value of
$a^2 + b^3 + c^4.$
| Note that $(a-2)^2+(c^2+2c+3)(c-3)^2+\frac{1}{27}(3b+4)(3b-8)^2\ge 0$ for all $a, b, c\ge 0$, i.e.
$a^2 + b^3 + c^4 + \frac{1093}{27} - 4(a + b^2 + c^3) \ge 0$ for all $a,b,c\ge 0$.
Thus, under the condition $a, b, c\ge 0;\ a+b^2+c^3=\frac{325}{9}$, we have $a^2 + b^3 + c^4 \ge 4\cdot \frac{325}{9} - \frac{1093}{27} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On the alternating quadratic Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$ My question is:
Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$.
$$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2... | Using your integral representation, the sum equals to: $$\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}= -2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx$$ $$\small=-2 C^2+2 \pi C \log (2)-4 \pi \Im(\text{Li}_3(1+i))+3 \text{Li}_4\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Which one is larger? I want to prove that for positive integer $n$,
$$\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n \geq 1-\frac{1}{n}$$
but I am stuck on how to proceed. Can someone help me?
| Consider
$$a_n=\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n=\left(1-\frac{1}{n^2}\right)^n\implies \log(a_n)=n \log\left(1-\frac{1}{n^2}\right)$$ So, by Taylor
$$\log(a_n)=n\left(-\frac{1}{n^2}-\frac{1}{2 n^4}-\frac{1}{3
n^6}+O\left(\frac{1}{n^8}\right)\right)=-\frac{1}{n}-\frac{1}{2 n^3}-\frac{1}{3
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find $\lim\limits_{x\rightarrow 0^+}\frac{1}{\ln x}\sum\limits_{n=1}^{\infty}\frac{x}{(1+x)^n-(1-x)^n}$
Find $$\lim\limits_{x\rightarrow 0^+}\dfrac{1}{\ln x}\sum_{n=1}^{\infty}\dfrac{x}{(1+x)^n-(1-x)^n}.$$
Consider $$f(x):=(1+x)^n,$$
By Lagrange MVT, we can obtain
$$\frac{2x}{f(x)-f(-x)}=\frac{1}{f'(\xi)}, -x\gtrles... | Well, for $0 < x < 1$, we have,
$S = \sum_{n=1}^{\infty}\frac{x}{(1+x)^n - (1-x)^n}
= \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{(1+x)^{n-1} + \cdots + (1-x)^{n-1}}
\geq (\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n(1+x)^{n-1}}
= \frac{1}{1+x}\sum_{n=1}^{\infty}\frac{1}{n(1+x)^n}).$
Using power series for natural logarithm,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given a solution $u = f(x-c_1t)+g(x-c_2t)$ to $Au_{tt}+2Bu_{tx}+Cu_{xx}=0$ what equation should $c_1$ and $c_2$ satisfy We find a solution to the equation $$Au_{tt}+2Bu_{tx}+Cu_{xx}=0$$ as $$u = f(x-c_1t)+g(x-c_2t)$$ with aribitrary $f,g,$ and real $c_1 < c_2.$
*
*What equation should satisfy $c_1$ and $c_2$
*When... | Calling
$$
M = \left(
\begin{array}{cc}
A & B \\
B & C \\
\end{array}
\right)
$$
we have
$$
\left(
\begin{array}{c}
\partial t \\
\partial x \\
\end{array}
\right)^{\dagger}M\left(
\begin{array}{c}
\partial t \\
\partial x \\
\end{array}
\right)u=0
$$
or
$$
\left(
\begin{array}{c}
\partial t \\
\partial x \\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate of: ${\prod_{n=1}^{\infty}\left[1+\frac{1}{\sum_{j=1}^{n}F_j^2}\right]^{(-1)^n+1}}$ How do we evaluate this infinite product with a sum within it?
$$\large{\prod_{n=1}^{\infty}\left[1+\frac{1}{\sum_{j=1}^{n}F_j^2}\right]^{(-1)^n+1}}$$
Where $F_j$ is the Fibonacci number
If I open the product, it does not help ... | We use Cassini's identity:
$$F_{n-1}F_{n+1}-F_n^2=(-1)^n\Rightarrow F_{2n-1}F_{2n+1}-F_{2n}^2=1$$
Then
$$\prod_{n=1}^\infty \left( 1+\dfrac{1}{F_nF_{n+1}}\right)^{(-1)^n+1}=\left( \prod_{n=1}^\infty \left( 1+\dfrac{1}{F_{2n}F_{2n+1}}\right) \right)^2$$
Let $P_n$ be
$$P_n=\prod_{k=1}^n \left( 1+\dfra... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $g(n,j)=\sum _{k=1}^n \frac{k^j (-1)^{n-k} \binom{n}{k}}{\frac{1}{2} n (n+1)-k}$ Denote $g(n,j)=\sum _{k=1}^n \frac{k^j (-1)^{n-k} \binom{n}{k}}{\frac{1}{2} n (n+1)-k}$, then how can we show that:
*
*$g(n,1)=\frac{n!}{\prod _{k=1}^n \left(\frac{1}{2} n (n+1)-k\right)}$
*$g(n,n)=\frac{\left(\frac{1}{2} n (n+... | We seek to evaluate
$$G_{n,j} = \sum_{k=1}^n \frac{k^j (-1)^{n-k} {n\choose k}}
{\frac{1}{2}n(n+1)-k}.$$
With this in mind we introduce the function
$$F_n(z) = n! \frac{z^{j-1}}{\frac{1}{2}n(n+1)-z}
\prod_{q=1}^n \frac{1}{z-q}.$$
This has the property that the residue at $z=k$ where $1\le k\le n$ is
the desired sum t... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the CDF if $P(X=1)=\frac{1}{2}$ and $P(a
The Problem: Let $X$ be a random variable that satisfies $P(X=1)=\frac{1}{2}$ and $P(a<X<b|X\ne1)=\frac{b-a}{2}$ for $0\leq a<b\leq2$. Find the cumulative distribution function of $X$.
My Thoughts: We have that
$$P(0<X<2)=P(0<X<2|X\ne1)P(X\ne1)+P(0<X<2|X=1)P(X=1)=1.$$
There... | Your end result is okay.
Let me add a route that makes things less error prone.
Let $Y$ have uniform distribution on interval $[0,2]$ and let $Z$ be a random variable degenerated at $1$ in the sence that $P(Z=1)=1$.
Then we find:
$$F_{X}\left(x\right)=P\left(X=1\right)P\left(X\leq x\mid X=1\right)+P\left(X\neq1\right... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Which expression about power series is correct? I found something weird about power series.
In my textbook,
$$ \frac{1}{1-x} = 1+x+x^2+x^3+\cdots=\sum_{n=0}^\infty x^n \quad |x|<1 $$
and $ \frac{1}{2+x} $ could be expressed using above equation.
$$ \frac{1}{2+x} = \frac{1}{2 \left(1 + \frac{x}{2} \right)} = \frac{1}{2... | You are just comparing two series developed around different points:
- one around $x=0$;
- the other around $x=-1$.
and remember that the radius of convergence is the distance from that point to the next irregularity ($x=-2$ in this case).
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that $Cov(AX) = A^T Cov(X) A$ For simplicity, assume random variable $X$ has mean $0$. I want to prove that $$Cov(AX) = A^T Cov(X) A$$ where A is a scalar matrix.
I tried to prove it from the definition of covariance:
$$Cov(AX) = E(AX)^T(AX) = E(X^TA^TAX)$$
I am wondering how do I bring $A$ out of the expectation.... | Are you sure the statement you’re trying to prove isn’t $ACov(X)A^T$? Also, I believe in your definition of the covariance matrix you were applying the transpose operator to the wrong parentheses.
$$Cov(AX)=E[(AX)(AX)^T]=E(AXX^TA^T)=AE(XX^T)A^T=ACov(X)A^T$$
The statement you were trying to prove is actually incorrect. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim_{x \to \infty} \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$ $$\underset{x\to \infty}{\lim} \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$$
My Attempt:
$$L = \underset{t\to 0}{\lim} \frac{\left( \left( \frac{t+1}{t-1} \right)^{\frac 1t} - e^2\right)} {t^2}$$
I Now have a $\frac 00$... | $$ \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$$
Start with
$$y=\left( \frac{x+1}{x-1} \right)^x\implies \log(y)=x\log\left( \frac{x+1}{x-1} \right)=x\log\left(1+ \frac{2}{x-1} \right)$$ Now, by Taylor
$$\log(y)=x\left(\frac{2}{x}+\frac{2}{3 x^3}+\frac{2}{5 x^5}+O\left(\frac{1}{x^7}\right)\right)={2}+\frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
A closed form for $\sum_{k=0}^n \frac{ (-1)^k {n \choose k}^2}{k+1}$ Mathematica gives $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{k+1}= ~_2F_1[-n,-n;2;-1],$$ where $~_2F_1$ that is Gauss hypergeometric function. Here the question is: Can one find a simpler closed form for this summation. Recently, the absolute summa... | One way to look at this is to take:
$\begin{align*}
\sum_k \frac{(-1)^k}{k + 1} \binom{n}{k}^2
&= \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k} \binom{n}{n - k} \\
&= [z^n] \left(
\sum_k \frac{1}{k + 1} \binom{n}{k} z^k
\right)
\cdot (-1)^n \left(
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Simplify Geometric series Can this equation simplify to the property of a sum of a geometric series, such as $ \frac{1}{1-r} $
$$\sum_{y=1}^{\infty}y^2q^{y}p$$
I understand that
$$\sum_{y=1}^{\infty}yq^{y} = q \sum_{y=0}^{\infty}(y-1)q^{y-1} = q \frac{d}{dq} \sum_{y=0}^{\infty}q^{y} = q \frac{d}{dq}\frac{1}{1-q} $$
| \begin{align}
\sum_y y^2 q^y p
&= p \sum_y (y(y-1)+y) q^y \\
&= p q^2 \sum_y y(y-1) q^{y-2} + p q \sum_y y q^{y-1} \\
&= p q^2 \frac{d^2}{dq^2} \sum_y q^y + p q \frac{d}{dq} \sum_y q^y \\
&= p q^2 \frac{d^2}{dq^2} \frac{1}{1-q} + p q \frac{d}{dq} \frac{1}{1-q} \\
&= p q^2 \frac{2}{(1-q)^3} + p q \frac{1}{(1-q)^2} \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Rewriting linear combinations of complex numbers into real numbers for probabilities Below is from Norris' Markov Chains. From the matrix decomposition of $P^n$, it is clear that $p_{11}^{(n)}=a+b(\frac{i}{2})^n+c(-\frac{i}{2})^n$ for some complex numbers $a,b,c$ that depends on the entries of $U, U^{-1}$. But how are ... | The disciplined approach here is to check this for 3 consecutive values, say $n= 0,1,2$. That is enough to uniquely specify and solve for $a$, $b$ and $c$ (using a Vandermonde matrix, shown at the end).
Based on limitting values you should be able to see $a =\alpha$. As far as the non-real eigenvalues and the real ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac{1}{1+x+x^4}dx$
Find $$\int \frac{{\rm d}x}{1+x+x^4}.$$
WA gives a result, which introduces complex numbers. But according to the algebraic fundamental theorem, any rational fraction can be integrated over the real number field.
How to do this?
| \begin{align}
\int \frac{{\rm d}x}{1+x+x^4}
\end{align}
It looks like the main problem is factorization
of the quartic, so here it is one more variant:
for
\begin{align}
a&=\sqrt{\frac1{\sqrt3}\,\cos\Big(\tfrac13\,\arctan(\tfrac19\,\sqrt{687})\Big)}
\\
&\approx 0.72713608449
,
\end{align}
\begin{align}
x^4+x+1&=
\Big... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Show that the function $f(x)=\cos x+\cos\left(\frac{\sqrt{3}}{2}x\right)$ is not periodic. Question: Show that the function $f(x)=\cos x+\cos\left(\frac{\sqrt{3}}{2}x\right)$ is not periodic.
My approach: Let us assume for the sake of contradiction that $T>0$ be the period of $f$. Then we must have $\forall a\in\mathb... | $$f(x)=\cos x+\cos\frac{\sqrt{3} x}{2}$$
Let $f_1(x)$ be periodic with a period of $T_1$ and $f_2(x)$ be periodic with a period of $T_2$. Then $f(x)=f_1(x)+f_2(x)$ only if $T_1/T_2$ is rational. Then the period is given by $T=LCM(T_1,T_2).$
Here $T_1=2\pi$ and $T_2=\frac{4\pi}{\sqrt{3}}$ This means $T_1/T_2=\frac{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all solutions to $\tan 2x = \cos \frac x2$ on the interval $[0,2\pi]$ I tried to solve this equation for almost 24 hours, using all the trigonometry identities I can think of, to no avail.
$$\tan 2x = \cos \frac x2$$
So far I've gotten to this point,
$$\cos^2 2x (\cos x + 3) = 2$$
but I've not made much of a progr... | $\tan 2x = \cos \frac{x}{2} \iff \frac {\sin 2x}{\cos 2x} = \pm \sqrt{\frac {1+\cos x}{2}} \iff \frac{\sin^2 2x}{\cos^2 2x} = \frac{1 + \cos x}{2} \iff 2\sin^2 2x=(1 + \cos x)\cos^2 2x \\ \text{condition:} \quad x \neq \frac{\pi}{4} + \frac{k\pi}{2}, k \in \mathbb{Z}$
$ \iff 8\sin^2x \cos^2x = (1 + \cos x)(\cos^2x - \s... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ I want to prove that $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ converges. My idea is that, for any integer $k \ge 1 $, we have
\begin{align*}
\frac{k!}{k^k}
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\cdot\frac{k}{k} \\
&= \frac{1}{k}\cd... | Here is another elementary solution. Using the inequality $x(a-x) \leq \frac{a^2}{4}$ for $0 \leq x \leq a$, we get
$$ \frac{k!}{k^k} = \frac{(k-1)!}{k^{k-1}} = \prod_{j=1}^{k-1} \frac{\sqrt{j(k-j)}}{k} \leq \frac{1}{2^{k-1}}. $$
Therefore the series converges by the Direct Comparison Test.
| {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
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For non-negative reals such that $a+b+c\geq x+y+z$, $ab+bc+ca\geq xy+yz+zx$, and $abc\geq xyz$, show $a^k+b^k+c^k\geq x^k+y^k+z^k$ for $0
Let $a$, $b$, $c$, $x$, $y$, $z$ be non-negative real numbers such that
$$a+b+c \geq x+y+z,$$
$$ab+bc+ca \geq xy+yz+zx,$$
$$ abc \geq xyz$$
Show that
$$a^k+b^k+c^k \geq x^k+y^k... | Let $a$, $b$ and $c$ be different numbers, $a+b+c=u$, $ab+ac+bc=v$ and $abc=w$.
Now, by $u$, $v$ and $w$ we can get any permutations of $a$, $b$ and $c$, but since $a^k+b^k+c^k$ is a symmetric expression, we can think that $a^k+b^k+c^k=f(u,v,w)$ and we need to prove that
$f$ increases as a function of $u$, as a functio... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Conclusion from the general triangle in a rectangle to an equilateral triangle
Let for the areas
$|\Delta ABP| =: F_B$,
$|\Delta PCQ| =: F_C$,
$|\Delta AQD| =: F_D$ and
$|\Delta APQ| =: F_\Delta$.
I know $~~~ F_\Delta = \sqrt{(F_B+F_C+F_D)^2-4 F_B F_D}~~~$ (see here);
is there an easy way to show, that $$F_B+F_D=F... | To proved the equality $F_B + F_D = F_C$ when $\triangle APQ$ is equilateral, you don't need that formula of $F_\Delta$. It can be proved directly using a little bit of trigonometry.
When $\triangle APQ$ is equilateral, let $s$ be its side and let $\alpha, \beta, \gamma$ be angles illustrated below.
The three angles c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdot \ldots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$
Evaluate $$\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$$
Let $$y=x + \frac{1 \cdot 3 \cdot}{2!} x^2 + \frac{1 \cdot 3 \cdot 5}{3!} x^3+\ldots$$ be the given expression.(replaci... | The hint:
It's a Taylor expansion for $f(x)=\frac{1}{\sqrt{1-2x}}-1.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Verify the integral $\int_1^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}}-u} \,du$ I'm stuck solving the integral
$$\int_1^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}}-u} \,du$$
This is what I got so far
\begin{align}
\int_{1}^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}} - u} \,du &= \int_{1}^z \frac{1 + \sqrt{u^2 + 1}}{u - u\cdot(1 +... | $$\frac{1}{2} \ln(3 - 2\sqrt{2}) - \frac{1}{2}\ln(z^2 - 2\sqrt{z^2 + 1} + 2)$$
$$=\frac{1}{2}{ln(\sqrt{2}-1)^2}-\frac{1}{2}{ln(\sqrt{z^2+1}-1)^2}$$
$$=\frac{1}{2}({ln \frac{(\sqrt{2}-1)^2}{(\sqrt{z^2+1}-1)^2})}$$
$$={ln \frac{(\sqrt{2}-1)}{(\sqrt{z^2+1}-1)}}$$
$$={ln \frac{(\sqrt{z^2+1}+1)}{z^2(\sqrt{2}+1)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I solve this limit without derivatives or series expansions? I have tried most of the common tricks but I still don't see what to do
I did get it to that form since I know that limit(ln(x+1)/x)=1, but the problem is that I can't get rid of x in the denominator.
| Let $n>1$ then from binomial theorem we know that $$(1+x)^{1/n}=1+\frac{x}{n}-\frac{n-1}{2n^2}x^2+\frac{(n-1)(2n-1)}{6n^3}x^3-\dots$$ for $0<x<1$. Clearly the series on right is alternating (ignoring the first term). You can also see that the terms of the series decrease in absolute value as $$\frac{(n-1)(2n-1)\dots((m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Range of $f(x) = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $ Here's what I did :
$$\text{Let }\quad y = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $$
$$\text{Let } \sin^2x=t$$
$$\Rightarrow\ (y-1)t^2-(y+1)t+2y+1=0 $$
$$\text{Since } t= \sin x\text{ is real,} $$
$$\text{Discriminant} \geqslant 0 $$
$... | Introducing $t = \sin{x}, t\in [-1, 1]$, gives a new function $y(t) = \frac{t^2+t-1}{t^2-t+2}$, and a task is to find a $\min_{t\in [-1, 1]}y(t)$ and $\max_{t\in [-1, 1]}y(t)$.
1) $y'(t) = \frac{(2t+1)(t^2-t+2) - (2t-1)(t^2-t+2)}{(t^2-t+2)^2} = \frac{-2t^2+8t}{(t^2-t+2)^2} = 0 \Leftrightarrow \left\{\begin{aligned}-2t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding Maximum area of the isosceles triangle $A = \{(x , y ,z): 1 \leq x , y, z \leq 6$ and $x , y ,z$ are the sides of a isosceles triangle$\}$
How can I find the maximum area of the triangle whose sides $a , b , c$ , $(a , b , c) \in A$?
Can anyone please give me some hint?
| Let us first find out the set of sample points $S = \{(x , y ,z) : 1 \leq x , y , z \leq 6$ and $ x , y, z $ are the sides of an isosceles triangle $\} $.\
So $S= \{(2,2,1),(2,2,3),(3,3,1),(3,3,2),(3,3,4),(3,3,5),(4,4,1),\\
(4,4,2),(4,4,3),(4,4,5),(4,4,6),(5,5,1),(5,5,2),(5,5,3)\\
(5,5,4)(5,5,6)(6,6,1)(6,6,2)(6,6,3)(6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the differential equation $ x^4y^3=xy' + y$
Using the change of the dependent variable $z = y^{−2}$, solve the differential equation:
$$xy' + y = x^4y^3$$.
My attempt:
$$xy' + y = x^4y^3 \tag 1$$
Now dividing $(1)$ by $y^2$
$$\frac{xy'}{y^2} + \frac{1}{y} = x^4y$$
Now put $z= \frac{1}{y^2}$,
$$\frac{dz}{dx} = ... | It would have been easier to divide by $y^3$ initially instead of $y^2$. Also, instead of just replacing $\frac{1}{y^2}$ with $z$ in your first term, if you used your $y'=\frac{-y^3dz}{2dx}$ instead, you would then get
$$\begin{equation}\begin{aligned}
-\frac{xyz'}{2} +\frac{1}{y} & = x^4y \\
-\frac{xz'}{2} + \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Roots of the equation $(1-4x)^4+32x^4=\frac{1}{27}$ find all real roots of the equation $(1-4x)^4+32x^4=\dfrac{1}{27}$
i try to use binomial expansion and am-gm inequality but i don't know how to do next.
| If $x<0$ so
$$(1-4x)^4+32x^4>1>\frac{1}{27}.$$
If $x>\frac{1}{4}$ so
$$(1-4x)^4+32x^4>32\left(\frac{1}{4}\right)^4>\frac{1}{27}.$$
Id est, it's enough to solve our equation for $x\in\left[0,\frac{1}{4}\right]$.
But in this case by Holder we obtain:
$$(1-4x)^4+32x^4=\frac{1}{27}(1+2)^3((1-4x)^4+2(2x)^4)\geq\frac{1}{27}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
What does this double summation with mod evaluate to? Let $X = \{1,2,3,\dots\}$ and $Y = \{0,1\}$.
Define $f:X\times Y \rightarrow \mathbb{R}$ by
$$
f(x,y) =
\begin{cases}
-2^{-x} &\text{ if } \mod(x, 2) = y \\
(y+1) 2^{-x} &\text{ if } \mod(x,2) = y+1 \\
0 &\text{ otherwise}
\end{cases}
$$
I've been able to calculat... | Note you have
$$\begin{equation}\begin{aligned}
\sum_{x}f(x,0) & = 2^{-1} - 2^{-2} + 2^{-3} - 2^{-4} + \ldots \\
& = \frac{1}{2}\sum_{i=0}^{\infty}\frac{1}{2}\left(\right)^{i} \\
& = \frac{\frac{1}{2}}{1 - \left(-\frac{1}{2}\right)} \\
& = \frac{\left(\frac{1}{2}\right)}{\left(\frac{3}{2}\right)} \\
& = \frac{1}{3}
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3595255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$e^{-x^4\sin^2x}\in L_2(R)$ proving $\int_{-\infty}^{\infty} e^{-2x^4\sin^2x}dx < \infty$ I am trying to prove that $e^{-x^4\sin^2x}\in L_2(R)$. It means that $\int_{-\infty}^{\infty} e^{-2x^4\sin^2x}dx < \infty$. So, i tried to use some usual methods, but $\sin(x)$ has no limit of infinity and I can't pick up an inte... | We have
\begin{align*}
& \int_{ - \infty }^{ + \infty } {e^{ - 2x^4 \sin ^2 x} dx} = 2\int_0^{ + \infty } {e^{ - 2x^4 \sin ^2 x} dx}
\\ &
= 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx} + 2\sum\limits_{n = 1}^\infty {\int_{\left( {n - \frac{1}{2}} \right)\pi }^{\left( {n + \frac{1}{2}} \right)\pi } {e^{ - 2x^4 \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3596480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Diophantine equation from "Solving mathematical problems" by Terence Tao
Find all integers $n$ such that the equation $\frac{1}{a} + \frac{1}{b} = \frac{n}{a+b}$ is satisfied for some non-zero values of $a$ and $b$ (with $a + b \neq 0$).
I'm reading "Solving mathematical problems" by Terence Tao and I'm a bit stuck ... | $$a^2 +a\left[b(2-n)\right] + b^2 = 0$$
is a quadratic equation in $a$, so we apply the quadratic formula. \begin{align*}
a &= \frac{-\left[b(2-n)\right] \pm \sqrt{\left[b(2-n)\right]^2 - 4(1)(b^2)}}{2} \\
&= \frac{-b(2-n)}{2} \pm \frac{1}{2}\sqrt{b^2( 4 - 4n + n^2) - 4b^2} \\
&= \frac{b}{2}(n-2) \pm \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding integer $x$ which its plus one becomes perfect square and its half+1 is also perfect square I am tring to find "integer $x$ which its plus one becomes perfect square and its half+1 is also perfect square". Make this statement into equation I have
\begin{align}
x + 1 = a^2, \qquad \frac{x}{2} + 1 = b^2
\end{ali... | From the second equation, we have: $$x=2b^2-2$$
Substituing, we have:
$$2(b^2-1)-(a^2-1)=0$$
And so:
$$2b^2-a^2=1 \leftrightarrow a^2-2b^2=-1$$
This is a simply Pell equation with $d=2$. The first solution is $(a_1,b_1)\rightarrow(1,1)$ and in general:
$$a_{n} = 6 a_{n-1} - a_{n-2}, \: \: b_{n} = 6 b_{n-1} - b_{n-2}, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I get from $x^4+2x^3y+3x^2y^2+2xy^3+y^4$ to $(x^2+xy+y^2)^2$? I was doing an example $$(x+y)^4+x^4+y^4$$ and I need to factor it. I've tried and couldn't really do much, so I checked if there was anything to help, and I came across a post asking about the same thing. But my question is how do I know that $$x^4+2... | Note,
$$\begin{array}
& & x^4+2x^3y+3x^2y^2+2xy^3+y^4 \\
& =( x^4+ 2x^2y^2+y^4 )+ 2x^3y+x^2y^2+2xy^3\\
&=( x^2+y^2)^2+ (2x^3y+2xy^3) + x^2y^2 \\
&=( x^2+y^2)^2+ 2xy(x^2+y^2) + (xy)^2 \\
&=(x^2+xy+y^2)^2
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Bounding the integral of $\ln(1+e^{-x})$ from $0$ to $n$ using given double inequality Let $f(x) := \ln(1+e^{-x})$ and
$$u_n := \sum_{k=1}^n \frac{(-1)^{k-1}}{k^2}.$$
Given that for $x>0$
$$ \tag{1} \sum_{k=1}^n \frac{(-1)^{k-1}}{k}e^{-kx} - \frac{e^{-(n+1)x}}{n+1} \le f(x) \le \sum_{k=1}^n \frac{(-1)^{k-1}}{k}e^{-kx}... | It is possible to give a closed formula for the integral. We have
\begin{align}
\int_0^n \ln(1+e^{-x}) dx = \int_{-n}^0 \ln(1+e^x) dx = -\left. \mathrm{Li}_2(-e^{x}) \right|_{x=-n}^0 &= \mathrm{Li}_2(-e^{-n}) - \mathrm{Li}_2(-1) \\& = \mathrm{Li}_2(-e^{-n}) + \frac{\pi^2}{12},
\end{align}
since it is well-known that $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\big|\frac{\cos\theta_1\cos\theta_0}{\cos^2\theta_2}+\frac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\big|$ If $$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{\cos\theta_0}{\cos\theta_2}+\dfrac{\sin\theta_0}{\sin\theta_2}=1,$$ where $\theta_1$ and $\theta_0$ do not diffe... | Hint:
Observe that $\theta_1,\theta_0$ are the roots of
$$\cos x\sin\theta_2+\sin x\cos\theta_2-\sin\theta_2\cos\theta_2=0$$
$$\iff\cos x\sin\theta_2=\sin\theta_2\cos\theta_2(\sin\theta_2-\sin x)$$
Now squaring both sides and replacing $\cos^2 x$ with $1-\sin^2x,$ we find
$$(1-\sin^2x)\sin^2\theta_2=\cos^2\theta_2(\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to
My attempt is as follows:-
Squaring both equations and adding
$$9+16+24\sin(P+Q)=37$$
$$\sin(P+Q)=\dfrac{1}{2}$$
either $... | A solution using complex numbers geometry :
The 2 relationships can be grouped into a single one by adding the second one to $i$ times the first one, giving :
$$3e^{iP}+4ie^{-iQ}=1+6i \ \ \iff \ \ \underbrace{3e^{iP}}_A+\underbrace{4e^{i(\pi/2-Q)}}_B=\underbrace{1+6i}_C \tag{1}$$
This defining relationship between poi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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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.