Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
prove $\int_0^\infty \frac{\log^2(x)}{x^2+1}\mathrm dx=\frac{\pi^3}{8}$ with real methods Context: I looked up "complex residue" on google images, and saw this integral. I, being unfamiliar with the use of contour integration, decided to try proving the result without complex analysis. Seeing as I was stuck, I decided ... | \begin{align}
J&=\int_0^\infty\frac{\ln^2 x}{x^2+1}\mathrm dx\\
K&=\int_0^\infty\frac{\ln(x)}{x^2+1}\mathrm dx\\
K^2&=\int_0^\infty\frac{\ln x}{x^2+1}\mathrm dx\int_0^\infty\frac{\ln y}{y^2+1}\mathrm dy\\
&=\int_0^\infty \int_0^\infty\frac{\ln x\ln y}{(x^2+1)(y^2+1)}\mathrm dx\mathrm dy\\
&=\frac{1}{2}\int_0^\infty \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Integral $\int\frac{1}{1+x^3}dx$
Calculate$$\int\frac{1}{1+x^3}dx$$
After calculating the partial fractions I got:
$$\frac{1}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2-x}{x^2-x+1}dx=\frac{1}{3}\ln(x+1)+\frac{1}{3}\int\frac{2-x}{x^2-x+1}dx$$
I have no idea on how to proceed. Am I missing a substitution or something... | To integrate $\frac{2- x}{x^2- x+ 1}$, complete the square in the denominator: $\int \frac{2- x}{x^2- x+ 1}dx= \int\frac{2- x}{x^2- x+ \frac{1}{4}-\frac{1}{4}+ 1}dx= \int\frac{2-x}{\left(x- \frac{1}{2}\right)+ \frac{3}{4}}dx$.
Now let $u= x- \frac{1}{2}$ so that $du= dx$ and $x= u+ \frac{1}{2}$ and $2- x= \frac{3}{2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$ solve $$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$$
my attempt
$$t = \frac{1}{2}\tan(u)$$
$$dt = \frac{1}{2}\sec^2(u)du\\$$
$$\begin{align}
\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt&=\int_{0}^{1} \frac{\tan^2(u)}{4} \sqrt{1+\tan^2(u)}\frac{1}{2}\sec^2(u)du\\
&=\frac{1}{8}\int_{0}^{1} \tan^2(u)\se... | To compute the integral in question first consider that $$\cosh^2(x)-\sinh^2(x)=1,$$ so $$\cosh^2(x)=1+\sinh^2(x).$$
Making the substitution $t=\frac{\sinh(x)}{2},$ then $4t^2=\sinh^2(x),$ and we get $dt=\frac{\cosh(x)}{2}dx$
$$\int t^2 \sqrt{(1+4t^2)}dt=\frac{1}{8}\int{\sinh^2(x)}\sqrt{1+\sinh^2(x)}\cosh(x)dx$$
$$=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove $\sqrt{b} - \sqrt{a} < \sqrt{b-a}$
Prove that if $0 < a < b$ then
$$\sqrt{b} - \sqrt{a} < \sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2\sqrt{ab} < b-a$
subtract $b$ from both sides $a-2\sqrt{ab} < -a$
add $a$ to both sides $2a-2\sqrt{ab} < 0$
than add $2 \sqrt{ab}$ to both sides ... | The reasoning of your argument is basically correct but you can't argue from the conclusion to be the beginning without clarifying why.
(particularly you need to state what is known and why from what is being speculated and what required conditions are necessary for the speculation. You are stating things that you don'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
maximum and minimum of $x^2+y^4$ for real $x,y$ If $y^2(y^2-6)+x^2-8x+24=0$ then maximum and minimum value of $x^2+y^4$ is
what i try
$y^4-6y^2+9+x^2-8x+16=1$
$(x-4)^2+(y^2-3)^2=1\cdots (1)$
How i find maximum and minimum of $x^2+y^4$ from $(1)$ relation
help me to solve it please
| By C-S $$0=y^2(y^2-6)+x^2-8x+24=$$
$$=x^2+y^4-8x-6y^2+24\geq$$
$$\geq x^2+y^4-\sqrt{(8^2+6^2)(x^2+y^4)}+24=$$
$$=x^2+y^4-10\sqrt{x^2+y^4}+25-1=\left(\sqrt{x^2+y^4}-5\right)^2-1,$$
which gives
$$4\leq\sqrt{x^2+y^4}\leq6$$ and
$$16\leq x^2+y^4\leq36.$$
The equality occurs for $(|x|,y^2)||(8,6),$ which says that $36$ is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
equating coefficients in algebraic expansion If $\displaystyle \bigg(\frac{1+x}{1-x}\bigg)^n=1+b_{1}x+b_{2}x^2+b_{3}x^3+\cdots +\infty,$ then value of
$(1)\; \displaystyle \frac{3b_{3}-b_{1}}{b_{2}}$
$(2)\; \displaystyle \frac{2b_{4}-b_{2}}{b_{3}}$
$(3)\; \displaystyle \frac{3b_{6}-2b_{4}}{b_{5}}$
$(4)\; \displaystyle... | Here is a slightly different method to find the coefficients. It is convenient to use the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^j]\left(\frac{1+x}{1-x}\right)^n}
&=[x^j](1-x)^{-n}(1+x)^n\\
&=[x^j]\sum_{k=0}^\infty\binom{-n}{k}(-x)^k(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can these bounds in terms of the abundancy index and deficiency functions be improved for deficient-perfect numbers? Let
$$\sigma(x) = \sum_{e \mid x}{e}$$
denote the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$, and the deficiency of $x$ by $D(x)=2x-\sigma(x)$. ... | ILLUSTRATING VIA A TOY EXAMPLE
Let $M$ be an odd perfect number given in the so-called Eulerian form
$$M = p^k m^2$$
(i.e. $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$).
It is known that the non-Euler part $m^2$ is deficient-perfect if and only if the Descartes-Frenicle-Sorli conj... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How does $d_1$ equal $\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$? I'm going over the proof of the midpoint formula and the solution in my textbook solves its first distance as follows
$$d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}$$
$$=\frac{1}{2}\sqrt... | You forgot that the $\frac{1}{2}$ had to be squared when you were factoring it out:
$$
d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}=
\sqrt{\left(\frac{x_1+x_2}{2}-\frac{2x_1}{2}\right)^2 + \left(\frac{y_1+y_2}{2}-\frac{2y_1}{2}\right)^2}=\\
\sqrt{\left(\frac{x_1+x_2-2x_1}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$ efficiently with combinatorics? To find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$,
I used factorization on $(1+x+\frac{x^2}{2})$ to obtain $\frac{((x+(1+i))(x+(1-i)))}{2}$, then simplified the question to finding the coefficient of $x^6... | You are looking for weak compositions of $6$ with $10$ parts of at most $2$. The nonzero entries can be $(2,2,2), (2,2,1,1), (2,1,1,1,1)$ or $(1,1,1,1,1,1)$ You can fill those out to $10$ with zeros. Now sum up multinomial coefficients and you are there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} = n^2$, then find the value of $\frac{m^2 - n^2}{n^2}$ I am a beginner at trigonometry, I want to know the answer to this question.
If $m^2\cos{\frac{2\pi}{15}}\cos{\frac{4\pi}{15}}\cos{\frac{8\pi}{15}}\cos{\frac{14\pi}{15}} =... | You can use Morrie's law:
*
*$2^n\prod_{k=0}^{n-1}\cos(2^k a) = \frac{\sin(2^na)}{\sin a}$
To do so, note that
*
*$\cos\left(\frac{14}{15}\pi \right) = -\cos\left(\frac{1}{15}\pi\right)$ and
*$\sin\left(\pi + \frac{1}{15}\pi \right) = -\sin\left(\frac{1}{15}\pi \right) $
It follows
$$\cos{\frac{2\pi}{15}}\cos{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ .
I think that must I use from $\dfrac{a^2}{2}+\dfrac{b^2}{2} \geq ab$ but no result please help me .!
| A slightly different approach using $\frac{a^2}2+\frac{b^2}2\ge ab$ :
$$\begin{align*}
\frac{a^2}{2}+\frac{b^3}{3}+\frac{c^6}{6}+\frac{abc}3&=\frac13 a^2+\frac13 abc+\left(\frac16 a^2+ \frac16 b^3\right)+\left(\frac16 b^3+\frac16 c^6\right)\\&\ge\frac13a^2+\frac13abc+\frac13ab^{3/2}+\frac13b^{3/2}c^3 \\
&\ge\frac23a^{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Characters of $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ From the Cayley table:
\begin{align*}
\begin{array}{c | c c c c }
& (0,0) & (0,1) & (1,0) & (1,1)\\
\hline
(0,0) & (0,0) & (0,1) & (1,0) & (1,1)\\
(0,1) & (0,1) & (0,0) & (1,1) & (1,0)\\
(1,0) & (1,0) & (1,1) & (0,0) & (0,1)\\
(1,1) & (1,1) & (1,0) & ... | Since $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is abelian, all characters are one-dimensional so they take on values $\pm 1$. So we have the same character table as the Klein-4 group:
\begin{align*}
\begin{array}{c | c c c c }
& (0,0) & (0,1) & (1,0) & (1,1)\\
\hline
\chi_{(0,0)} & 1 & 1 & 1 & 1\\
\chi_{(0,1)} & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simultaneous congruences $3x \equiv 2 \pmod{5}$, $3x \equiv 4 \pmod{7}$, $3x \equiv 6 \pmod{11}$ I am stuck in a simultaneous linear congruence problem:
\begin{cases}
3x \equiv 2 \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \\[4px]
3x \equiv 6 \pmod{11}
\end{cases}
Using the Chinese remainder theorem,
I started with the 'hig... | This is an old-fashioned solution.
$$3x \equiv 2 \mod 5 \qquad 3x \equiv 4 \mod 7 \qquad 3x \equiv 6 \mod{11}$$
\begin{array}{|r|rrr|}
\hline
385 & 5 & 7 & 11 \\
\hline
77 & 2 & 0 & 0 \\
55 & 0 & 6 & 0 \\
35 & 0 & 0 & 2 \\
\hline
231 & 1 & 0 & 0 \\
-55 & 0 & 1 & 0 \\
-175 & 0 & 0 & 1 \\
\hl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is $f$ in the vector space of cubic spline functions? Let $S_{X,3}$ be the vector space of cubic spline functions on $[-1,1]$ in respect to the points $$X=\left \{x_0=-1, x_1=-\frac{1}{2}, x_2=0, x_3=\frac{1}{2}, x_4\right \}$$ I want to check if the function $$f(x)=\left ||x|^3-\left |x+\frac{1}{3}\right |^3\right |$$... | For $x\in(-1/3,0)$ we have
$$f(x)=\left|x^3+\left(x+\frac{1}{3}\right)^3\right|$$
which has strictly positive slope, because $x\mapsto x^3$ is monotonous with zero derivative only at $x=0$.
On $(-1/3,0)$ the first term goes from negative to zero, and the second term goes from zero to positive.
Hence the sum goes from n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Symmetric matrix factorization of a $2 \times 2$ symmetric matrices Let S be a symmetric matrix such that
$$S= \begin{bmatrix}
A & B\\
B & C
\end{bmatrix}, \text{where A, B, C $\in$ $\mathbb{Z_P},$}$$
where p is prime.
I am tasked to get the solution(s) of the symmetric matrix factorization of $2 \times 2$ symmetric m... | First, if $S$ is diagonal then this is fairly straightforward. We distinguish three cases:
*
*If $S=0$ then there are $(p-1)^2(p+1)$ pairs $(X,Y)$ of nonzero matrices such that $S\equiv XY\pmod{p}$, and $2p^4-1$ such pairs $(X,Y)$ with either $X=0$ or $Y=0$.
*If $B=0$ and $A=C\neq0$ then $S=AI$, and hence the solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $a,b$ at $f(x)=\frac{x^2+x-12}{x^2-ax+b}$ An High school question:
Given :$$f(x)=\frac{x^2+x-12}{x^2-ax+b}$$ it's given that $x=3$ is a vertical asymptote find $a$ and $b$.
I tried:
Since $x=3$ is a vertical asymptote then $3^2-3a+b=0$, but now what
| Note that $x^2+x-12$ also has $3$ as a root, so for an asymptote to exist we require $3$ to be a repeated root. Hence $x^2-ax+b=(x-3)^2\implies a=6,\,b=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Proving that two determinants are equal without expanding them So I need to prove that
$$
\begin{vmatrix}
\sin^2(\alpha) & \cos(2\alpha) & \cos^2(\alpha) \\
\sin^2(\beta) & \cos(2\beta) & \cos^2(\beta) \\
\sin^2(\gamma) & \cos(2\gamma) & \cos^2(\gamma) \\
\end{vmatrix}
$$
$$
= \begin{vmatrix}
... | HINT: Relate the third column to the first two using the trigonometric identity
$$\sin(\theta+\delta)=\sin\theta\cos\delta+\cos\theta\sin\delta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $x = a \cos t^3 , y = b \sin t^3$ then what is $d^3y/dx^3$? If $ x = a \cos t^3 $, $ y = b \sin t^3 $, then what is $ \frac{d^3y}{dx^3} $?
I tried doing this problem by dividing $ \frac{d^3y}{dt^3} $ by $ \frac{d^3x}{dt^3} $ and got $ \frac{b}{a} $.
However my book says the third derivative doesn't exist. Why is ... | We can in fact obviate the parameterisation altogether. Since $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $\frac{x}{a^2}+\frac{y}{b^2}\frac{dy}{dx}=0$ so $\frac{dy}{dx}=-\frac{b^2x}{a^2y}$ and $$\frac{d^2y}{dx^2}=-\frac{b^2}{a^2}\left(\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}\right)=-\frac{b^4x^2}{a^4y^3}-\frac{b^2}{a^2y},\\\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Understanding a proof from the APMO 1998 on inequalities. I was having trouble with proving the following inequality.The question was from the book Secrets to Inequalities by Pham Kim Hung.
$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{xyz} }$
I know that the AM-GM inequality must be applied but I... | It is by AM-GM: $$\frac{x}{y}+\frac{x}{y}+\frac{y}{z}\geq 3\sqrt[3]{\frac{x^2y}{y^2z}}=3\sqrt[3]{\frac{x^3}{xyz}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to find the jacobian of the following? I am stuck with the following problem that says :
If $u_r=\frac{x_r}{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}$ where $r=1,2,3,\cdot \cdot \cdot ,n$, then prove that the jacobian of $u_1,u_2,\cdot \cdot, u_n$ with respect to $x_1,x_2,\cdot \cdot, x_n$ is $(1-x_1^2... | The partial derivatives seems easy to find due to symmetry:-
Applying chain rule we get $$ u_{r,i}=\frac{x_ix_r}{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}$$ if i
!= r
$$u_{i,i}=\frac{1}{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}-\frac {x_i^2}{{\sqrt{1-x_1^2-x_2^2-x_3^2 \cdot \cdot \cdot-x_n^2}}}$$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
To determine a constant in an ODE
Let $w(r)$ be a function of $r$, we have the following ODE:
$$r^{n-1}w'+\frac{1}{2}r^nw=a$$
for a constant $a$.
Assume the equation holds for all positive integer $n$. The book claims that if assuming $\lim\limits_{r\to \infty}w= 0$ and $\lim\limits_{r\to \infty}w'= 0$, we have ... | Assume $a=1$ and $n=3$, then you have
\begin{align}
w'+\frac{r}{2}w= \frac{1}{r^2}.
\end{align}
Let us assume $w(1) = 1$, then we see that
\begin{align}
w(r) = \exp\left( \frac{1-r^2}{4}\right)+\exp\left(\frac{-r^2}{4} \right) \int^r_1 t^{-2} \exp\left(\frac{t^2}{4} \right)\ dt
\end{align}
and
\begin{align}
w'(r) = -\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is my integral in fully reduced form?
I have to integrate this:
$$\int_0^1 \frac{x-4}{x^2-5x+6}\,dx$$
Now
$$\int_0^1 \frac{x-4}{(x-3)(x-2)}\, dx$$
and by using partial fractions we get
$$\frac{x-4}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$$
$$x-4 = A(x-2) + B(x-3)$$
$$= Ax - 2A + Bx - 3B$$
$$x-4 = (A+B)x - 2A - ... | First, $\log 1= 0$. And there is a mistake when you compute the definite integral:
$$
-\log 2 +\log 3 -2\log2 = -3\log 2 +\log 3 = -\log 8+\log3 = -\log\frac 83.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Amidst $7$ prime numbers, difference of the largest and the smallest prime number is $d$. What is the highest possible value of $d$?
Let $a, b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ be $7$ distinct prime numbers. Among $a+b, b+c$ and $c+a$, only one of the three numbers is equal to $800$. If the difference of the larges... | Say $a+b=800$.
If $c\equiv 0\pmod 3$ then $c=3$ and then $a+b+c=803$ which is not a prime
If $c\equiv 1\pmod 3$ then $800+c\equiv 0\pmod 3$ so $800+c =3$ impossibile.
If $c\equiv -1\pmod 3$ then $800-c\equiv 0\pmod 3$ so $800-c =3\implies c=797$ so $a+b+c =1597$ and thus $\boxed{d= 1597-3 = 1594}$
Such a prime numb... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Disprove or prove using delta-epsilon definition of limit that $\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$ I want to prove if the following limit exists, using epsilon-delta definition, or prove it doesn't exist:$$\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$$
My attempt:
First I proved some direction... | Use polar coordinates $x=r\cos\theta$, $y=r\sin\theta$. Because of the domain we have $\cos\theta\ne\pm\sin\theta$. Substituting into your formula gives
$$\frac{x^3-y^3}{x^2-y^2}=r\,\frac{\cos^3\theta-\sin^3\theta}{\cos^2\theta-\sin^2\theta}\ .$$
Now it is not hard to prove that
$$\frac{\cos^3\theta-\sin^3\theta}{\co... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding out the remainder of $\frac{11^\text{10}-1}{100}$ using modulus
If $11^\text{10}-1$ is divided by $100$, then solve for '$x$' of the below term $$11^\text{10}-1 = x \pmod{100}$$
Whatever I tried:
$11^\text{2} \equiv 21 \pmod{100}$.....(1)
$(11^\text{2})^\text{2} \equiv (21)^\text{2} \pmod{100}$
$11^\text{4} \... | $$11^{10}=(10+1)^{10}=10^{10}+10×10^9+\frac {(10×9)}{2}×10^9+\cdots+(10×10)+1$$(using binomial expansion ). Now note that every term except last one is a multiple of $100$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that if $x$, $y$, and $z$ are real numbers such that $x^2(y-z)+y^2(z-x)+z^2(x-y)=0,$ then at least two of them are equal
Prove that if $x$, $y$, and $z$ are real numbers such that $x^2(y-z)+y^2(z-x)+z^2(x-y)=0,$ then at least two of them are equal
This question was asked in a past international math exam. My id... | Factoring is hard and I'm not good at it.
But if:
$x^2(y-z)+y^2(z-x) + z^2(x-y) = 0$
$x^2(y-z)+y^2(z-x)= z^2(y-x)$
If $y =x$ then $y=x$. But if $y\ne x$ then $y-x \ne 0$.
$\frac {x^2(y-z) + y^2(z-x)}{y-x} = z^2$
So $\frac {xy*y - x^2z + y^2z - xy*y}{y-x} =$
$ -xy + \frac{y^2z -x^2z}{y-x} = $
$-xy + z\frac{(y-x)(y+x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3127617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove/show this actually defines a homomoprhism
We define the homomorphism $f: \text{SL}_2(\mathbb Z / 2 \mathbb Z) \to \text{SL}_2(\mathbb Z / 2 \mathbb Z)$ that maps the generators to:
$ \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
$ \begin{pmatrix} 0 & 1 \... | Note that$$\mathrm{SL}_2(\mathbb{Z}_2)=\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&1\\1&0\end{bmatrix},\begin{bmatrix}1&0\\1&1\end{bmatrix},\begin{bmatrix}1&1\\0&1\end{bmatrix},\begin{bmatrix}0&1\\1&1\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix}\right\}.$$Here, the first element is $\operatorname{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For matrices, under what conditions can we write $AB = BC$? If $A$ and $B$ are real matrices, when does there exist a matrix $C$ such that $AB = BC$? I understand that there is the case where $A$ and $B$ commute so $AB =BA$, but is there a more general rule (ie. necessary and sufficient conditions)?
Thanks!
| One could interpret the question as: given a matrix $B \in M^{m\times n}$, describe the subspace of $A \in M^{m \times m}$ for which there exists $C \in M^{n \times n}$ such that $AB = BC$.
If $B$ has rank $r$ and $P \in \text{GL}_n$ and $Q \in \text{GL}_m$ are such that $B = P \begin{pmatrix}I_r & 0_{r \times n-r} \\ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve for $x$ : $\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$? I want to solve the following equation for $x$ :
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$$
My approach:
Let the given eq.:
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9 \tag {i}$$
On rearranging, we get:
$$\sqrt{x-6} \, +... | You're fine so far. Now from $(ii)$, you have
$$18 \sqrt{x-1} + 2 \sqrt {x^2-36} = 78-x.$$
You'll end up needing to square both sides a couple of more times but you'll be able to clear the radicals. Then you need to confirm that none of your potential solutions are spurious.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$ \lim_{(x,y)\rightarrow (0,0)} \frac{x^{5}y^{3}}{x^{6}+y^{4}}. $ Does it exist or not? I have this limit:
$$ \lim_{(x,y)\rightarrow (0,0)} \frac{x^{5}y^{3}}{x^{6}+y^{4}}. $$
I think that this limit does not exist (and wolfram|alpha agrees with me). But I can't find a way to prove it. I chose some paths and the limit ... | The first limit actually exists and is zero. You can show that by definition.
Namely,
$$\left(\frac{x^5y^3}{x^6+y^4}\right)^2 = \frac{|x^5 y^3|}{x^6+y^4} \cdot \frac{|x^5y^3|}{x^6+y^4}$$
Because
$$x^6+y^4\ge x^6 \implies \frac{1}{x^6+y^4} \le \frac{1}{x^6}$$
$$x^6+y^4\ge y^4 \implies \frac{1}{x^6+y^4} \le \frac{1}{y^4}... | {
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how to find the point obtained by reflecting over a line? Given the point $A=(-2,6)$ and the line $y=2x$,what are the coordinates of the point B obtained by reflecting A over the line $y=2x$ ?
Can someone teach me how to solve this question please?
| Note that AB has a gradient of $-\frac{1}{2}$, and that AB, which is $y=-\frac{1}{2}x+c$ intersects $y=2x$ at the midpoint of AB.
Sub $(-2,6)$ into $y=-\frac{1}{2}x+c$, and solving, $c=5$.
Equate the two equations.
$-\frac{1}{2}x+5=2x$
Coordinates of midpoint of AB: $x=2$, $y=4$
Coordinates of B: $x=2+2+2=6$, $y=4-(6-... | {
"language": "en",
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Find $\max\{y-x\}$
If $x+y+z=3, $ and $x^2+y^2+z^2=9$ , find $\max\{y-x\}$.
I tried to do this geometrically, $x+y+z=3$ is a plane in $\Bbb{R}^3$ and $x^2+y^2+z^2=9$ is a ball with radius 3 and center of origin . So the candidate points for $y-x$ are on the intersection of the plane and the ball. But now I am confu... | Our conditions give $$x^2+y^2+z^2=(x+y+z)^2$$ or
$$xy+xz+yz=0.$$
Now, let $y-x=t$.
Thus, $y=x+t$, $z=3-x-y=3-2x-t$ and we obtain that the equation
$$x(x+t)+(3-2x-t)(x+x+t)=0$$
has real roots $x$, which says that $\Delta\geq0.$
We obtain:
$$3x^2+3(t-2)x+t^2-3t=0,$$ which gives
$$9(t-2)^2-12(t^2-3t)\geq0$$ or
$$-2\sqrt3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving the inequality $\angle A+\angle COP < 90^\circ$ in $\triangle ABC$
In an acute angled $\triangle ABC$, $AP \perp BC$and $O$ is its circumcenter. If $\angle C \ge \angle B + 30^\circ$, then prove that $$\angle A + \angle COP < 90^\circ$$
My Attempt:
Extending the line $AP$ to the circumferential point $D$, I ... | Well, afterall this is an IMO 2001 geometry problem and you can find two solutions, along with a solutions of the other problems from that year, here:
https://sms.math.nus.edu.sg/Simo/IMO_Problems/01.pdf
| {
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Finding integer solutions out of a a | b Determine all positive integer values of (n) such that
$$ { n \choose 0 } + { n \choose 1 } + { n \choose 2 } + { n \choose 3 } \ \bigg| \ 2 ^ { 2008 } $$
What is the sum of all these values?
CURRENT PROGRESS:
I was able to find out that this is equivalent to $(n+1)(n^2 - n + 6... | Set $n+1=m$
$n^2-n+6=(m-1)^2-(m-1)+6=m^2-3m+8$
$g=(m, m^2-3m+8)=(m,8)$
Check all the four possible cases
For example,
If $g=4$
$\dfrac{n+1}4\cdot\dfrac{ n^2-n+6}4$ will divide $3\cdot2^{2008-4}$
where the two factors are relatively prime
i.e. one of the factors must be odd $1$ or $3$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Integrate $\int \frac{\sqrt{x^2-1}}{x^4}dx$ I am trying to integrate $\int \frac{\sqrt{x^2-1}}{x^4}dx$ via trig substitution. I decided to substitute $x = \sec\theta$ into the square root and $dx = \sec\theta \tan\theta\,d\theta$.
$$\int \frac{\sqrt{\sec^2 \theta-1^2}}{\sec^4\theta} \,dx
= \int \frac{\sqrt{\tan^2\theta... | Alternative solution (without trigonometry). Note that by integration by parts
$$\begin{align}
\int \frac{\sqrt{x^2-1}}{x^4}dx&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{3}\int \frac{1}{x^2\sqrt{x^2-1}}dx\\
&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{6}\int \frac{D(1-1/x^2)}{\sqrt{1-1/x^2}}dx\\
&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137719",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $I=\int_{0}^{\infty}\frac{(x^2+x+1)dx}{(x^3+x+1)\sqrt{x}}$ Evaluate $$I=\int_{0}^{\infty}\frac{(x^2+x+1)dx}{(x^3+x+1)\sqrt{x}}$$
My try:
Letting $\sqrt{x}=t$ we get
$$I=2\int_{0}^{\infty} \frac{t^4+t^2+1}{(t^6+t^2+1)}\,dt$$
Now
$$t^4+t^2+1=(t^2+1)^2-(t^2+1)+1$$
$$t^6+t^2+1=(t^2+1)^3-2(t^2+1)^2+(t^2+1)+1$$
Any ... | According to Wolfram Alpha, the answer is
$$\frac{\pi}{\sqrt{186}}\left(\sqrt{40-14\sqrt[3]{\frac{2}{47-3\sqrt{93}}}-\sqrt[3]{4\left(47-3\sqrt{93}\right)}}+\sqrt{80+14\sqrt[3]{\frac{2}{47-3\sqrt{93}}}+\sqrt[3]{4\left(47-3\sqrt{93}\right)}+36\sqrt{\frac{186}{40-14\sqrt[3]{\frac{2}{47-3\sqrt{93}}}-\sqrt[3]{4\left(47-3\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
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Find the coefficient of $x^{13}$ in the convolution of two generating functions
Four thiefs have stolen a collection of 13 identical diamonds. After the
theft, they decided how to distribute them.
3 of them have special requests:
*
*One of them doesn't want more than 2 diamonds ($\leq2$).
*The other one... | It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain
\begin{align*}
\color{blue}{[x^{13}]\frac{x^3}{(1-x)^2(1-x^2)}}
&=[x^{10}]\frac{1}{(1-x)^2(1-x^2)}\tag{1}\\
&=[x^{10}]\sum_{j=0}^5 x^{2j}\sum_{k=0}^\infty \binom{-2}{k}(-x)^k\tag{2}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many permutations are there of M, M, A, A, A, T, T, E, I, K, so that no two consecutive letters are the same? How many permutations are there of
$$ M, M, A, A, A, T, T, E, I, K $$
so that there are no two consecutive letters are the same?
I would use the Inclusion-exclusion principle where
$$ A_{i} = \{ \text{on} ... | As you observed, there are $10$ letters, of which $3$ are $A$s, $2$ are $M$s, $2$ are $T$s, $1$ is an $E$, $1$ is an $I$, and $1$ is a $K$.
If there were no restrictions, we would choose three of the ten positions for the $A$s, two of the remaining seven positions for the $M$s, two of the remaining five positions for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Is $T_n$ an unbiased estimator of $\theta$? Prove your answer.
Let $(x_1, x_2, \dots, x_n)$ be an observed sample from a distribution with probability density function given by $$f(x) =\begin{cases}\displaystyle\frac{1}{5\theta+8}&\text{if }0 \leq x \leq 5 \theta + 8\\0&\text{otherwise. }\end{cases}$$
Where $\theta \i... | The computation in (a) is not strictly necessary, because $$\Pr[T_n > \theta] = 0,$$ whereas $\Pr[T_n < \theta] > 0$; therefore, $\operatorname{E}[T_n] < \theta$ hence is trivially biased.
However, once the computation is performed, (b) immediately follows, and (c) is also correct.
| {
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Equilateral triangle with vertices on 3 concentric circles Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of the equilateral triangle?
My idea is to set a point at the midd... | While the posted geometric solutions are much simpler, it is possible to do this with algebra and coordinate geometry.
Centering the circles at the origin, we get the equations that you provided: $$x^2+y^2=1$$
$$x^2+y^2=4$$
$$x^2+y^2=9$$
Let's choose an arbitrary point on the smallest circle, say $(0, 1)$ for simplicit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\sqrt 7 + 2 \sqrt 5 < 2 + \sqrt{35}$ How can I prove that $\sqrt 7 + 2 \sqrt 5 < 2 + \sqrt{35}$?
I can't convert the right side to get the $\sqrt 5$.
| Since $2^2 <7$. we get
$2=\sqrt{2^2} <√7$ since $√$ is strictly increasing.
Also since $1 < 5$ we get $1 <√5.$
Finally :
$2(√5-1) < √7(√5-1)$; (Why?)
$√7 +2√5 < 2 +\sqrt{35}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence behavior of a rational iterative procedure of the form $\frac{1}{2}(z+1/z)$. I am trying to prove the following result.
First, we have an auxiliary sequence satisfying
$E_{n+1}=\dfrac{E_n^2}{(1+\sqrt{1-E_n^2})^2}$ with $E_0<1$.
It is called the Landen transformation.
Let $S_0=1+r\exp(i\theta)$ with $r\le ... | Suppose that
$|w_n|<F_n$, then we set
\begin{equation*}
w_n=r\exp(i \theta), \quad r\le F_n.
\end{equation*}
Thus,
\begin{align*}
w_{n+1}=w_n\frac{w_n+E_n}{1+E_nw_n}=
r\exp(i \theta)\frac{r\exp(i \theta)+E_n}{1+E_nr\exp(i \theta)}
\end{align*}
Then
\begin{align*}
|w_{n+1}|&=r\left|\frac{r\exp(i \theta)+E_n}{1+E_nr\e... | {
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"timestamp": "2023-03-29T00:00:00",
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Area between curves. I have to calculate area bounded by curves :
$(x^3+y^3)^2=x^2+y^2 $ for $ x,y \ge 0 $.
I tried to use polar coordinates, but I have :
$r^4(\cos^6\alpha +2\sin^3\alpha\cos^3\alpha + \sin^6\alpha)=1$
| By your work: $$r=\frac{1}{\sqrt{\sin^3\alpha+\cos^3\alpha}}.$$ Since it's symmetric in respect to $y=x$, we obtain that the needed area it's
$$2\int\limits_0^{\frac{\pi}{4}}d\alpha\int\limits_0^{\frac{1}{\sqrt{\sin^3\alpha+\cos^3\alpha}}}rdr=\int_0^{\frac{\pi}{4}}\frac{1}{\sin^3\alpha+\cos^3\alpha}d\alpha.$$
Can you e... | {
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Prove without induction that $2×7^n+3×5^n-5$ is divisible by $24$. I proved this by induction. But I want to show it using modular arithmetic. I tried for sometime as follows
$$2×7^n-2+3×5^n-3\\
2(7^n-1)+3(5^n-1)\\
2×6a+3×4b\\
12(a+b)$$
In this way I just proved that it is divisible by 12 but it is not enough. Am I mis... | Yes, it can be done by another method. Note that $7^2=2\times24+1$ and that $5^2=24+1$ and that therefore$$7^n\equiv\begin{cases}7\pmod{24}&\text{ if $n$ is odd}\\1\pmod{24}&\text{ otherwise}\end{cases}$$and$$5^n\equiv\begin{cases}5\pmod{24}&\text{ if $n$ is odd}\\1\pmod{24}&\text{ otherwise.}\end{cases}$$So:
*
*if ... | {
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"timestamp": "2023-03-29T00:00:00",
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About the fact that $\frac{a^2}{b} + \frac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$ There's a math competition I participated yesterday (19/3/2019).
In these kinds of competitions, there will always be at least one problem about inequalities.
Now this year's problem about inequality is very easy. I am more interest... | Squaring the given inequality and factorizing we get
$${\frac { \left( {a}^{2}+18\,ba+{b}^{2} \right) \left( a-b \right) ^{4
}}{{b}^{2}{a}^{2}}}
\geq 0$$ which is true.
| {
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GCD of cubic polynomials I would appreciate some help finding $GCD(a^3-3ab^2, b^3-3ba^2)$; $a,b \in \mathbb{Z}$. So far I've got here: if $GCD(a,b)=d$ then $\exists \alpha, \beta$ so that $GCD(\alpha, \beta)=1$ and $\alpha d=a$, $\beta d=b$.
Therefore we know that $GCD(a^3-3ab^2, b^3-3ba^2)=d^3 GCD(\alpha^3-3\alpha \b... | I'll write $\alpha=m,\beta=n$ for the ease of typing
If $d(\ge1)$ divides $m^3-3mn^2,n^3-3m^2n$
$d$ will divide $n(m^3-3mn^2)+3m(n^3-3m^2n)=-8m^3n$
and $3n(m^3-3mn^2)+m(n^3-3m^2n)=-8mn^3$
Consequently, $d$ must divide $(-8m^3n,-8mn^3)=8mn(m^2,n^2)=8mn$
So, $d$ will divide $8$
As $(m,n)=1,$ both $m,n$ cannot be even
If ... | {
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Quotient rule and implicit differentiation Find $\frac{dy}{dx}$ for $x^2=\frac{x-y}{x+y}$.
I have solved this in two ways.
First, I multiplicated the whole equation by $x+y$ and then I calculated the implicit derivative. I got the following solution:
$\frac{1-3x^2-2xy}{x^2+1}$
So far so good. When I calculated the i... | The two solutions that you found are equal. So where is no contradiction.
Note that solving $\quad x^2=\frac{x-y}{x+y}\quad$ for $y$ gives $\quad y=x\frac{1-x^2}{1+x^2}$
Your first solution :
$$\frac{1-3x^2-2xy}{x^2+1} =\frac{1-3x^2-2x\left(x\frac{1-x^2}{1+x^2} \right)}{x^2+1} = \frac{1-4x^2-x^4}{(1+x^2)^2}$$
Your seco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing $\int_{a}^{b} \left\lfloor x \right\rfloor dx + \int_{a}^{b} \left\lfloor -x \right\rfloor dx=a-b$ I want to show
$$\int_{a}^{b} \left\lfloor x \right\rfloor dx + \int_{a}^{b} \left\lfloor -x \right\rfloor dx=a-b$$
I know that
\begin{equation}
\left\lfloor -x \right\rfloor = \begin{cases} -\left\lfloor x \... | What happens at the integers does not matter since they have content 0. Suppose
$$n < x < n + 1.$$
then $\lfloor x \rfloor = n$, and $-n > x > -(n+1)$, so $\lfloor -x \rfloor = -(n+1)$. Adding gives $\lfloor x\rfloor + \lfloor -x \rfloor = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\frac{1}{\sqrt{x^2+x+1}}$
How to integrate $$\frac{1}{\sqrt{x^2+x+1}}$$
I tried to solve this integral as follows
$\displaystyle
\int \frac{1}{\sqrt{x^2+x+1}} \ dx=
\int \frac{1}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}} \ dx=
\int \frac{1}{\sqrt{(\frac{2x+1}{2})^2+\frac{3}{4}}} \ dx=
\int \frac... | You have got mixed up with the $\arcsin$. Note that $$\int\frac{1}{\sqrt{\color{red}{1-g^2}}}\, dg = \color{red}{\arcsin g},$$
but
$$\int\frac{1}{\sqrt{\color{blue}{g^2-1}}}\, dg = \color{blue}{\ln\left(\left|g+\sqrt{g^2 -1}\right|\right)}.$$
You can verify this by differentiating the right-hand side, or try to derive ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
For which $a>0$ series is convergent?
For which $a>0$ series $$\sum { \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^a } $$ $(n \in \mathbb N)$ is convergent?
My try:From Taylor theorem I know that:$$a_{n}={ \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{... | It is much simpler to use equivalents, which can be found with Taylor's formula:
*
*$2(1-\cos x)=x^2-\dfrac{x^4}{12}+o(x^4)$,
*$\sin(\sin x)=\sin\Bigl(x-\dfrac{x^3}6+o(x^3)\Bigr)=\Bigl(x-\dfrac{x^3}6\Bigr)-\frac16\Bigl(x-\dfrac{x^3}6\Bigr)^{\!3}+o(x^3)=x-\dfrac{x^3}3+o(x^3)$, so
$$x\sin(\sin x)=x^2-\frac{x^4}3+o(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
induction proof: number of compositions of a positive number $n$ is $2^{n-1}$ So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
... | What you know: $\color{blue}{2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}}$
What you want to show: $\color{green}{2^0+2^1+2^2+...+(2^{k-1}+1)=2^{k}}$
Use what you know:
$\color{green}{2^0+2^1+2^2+...+2^{k-2} + (2^{k-1}+1)} =$
$\color{blue}{2^0+2^1+2^2+...+2^{k-2}} + (\color{green}{2^{k-1}} + \color{blue}1) =$
$\color{blue}{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Integers $n$ satsifying $\frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}}$
If $\displaystyle \frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}},n\in \mathbb{Z}$, then number of $n$ satisfies given equation ,is
What I tried:
Let $\displaystyle \frac{\pi}{n}=x$ and equation is $\sin 5x=\sin 3x$
$\di... | No need to go with sum-to-product formulas. From $\sin5x=\sin3x$ you get either
$$
5x=3x+2k\pi
$$
or
$$
5x=\pi-3x+2k\pi
$$
The former yields $x=k\pi$, so $1=kn$, whence $n=\pm1$.
The latter yields
$$
\frac{\pi}{n}=\frac{1}{8}(2k+1)\pi
$$
that is, $(2k+1)n=8$. Hence $n=\pm8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Does it follow that a+b+c=x+y+z=m+n+p If
$$abc=xyz=mnp$$
$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=m^4+n^4+p^4-2m^2n^2-2n^2p^2-2p^2m^2$
$$\frac{x}{m}=\frac{n}{b}=\frac{c}{z}$$
Then $a+b+c=x+y+z=m+n+p$?
$a,b,c$ positive and are lenghts of triangles,
$x,y... | Hint: Note that
$$
a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a + b + c)(a + b - c)(a - b + c)(a - b - c).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3166291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2$ Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$
Hence or otherwise prove:
$$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$
| To minimize $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ , we need to have
$$
\begin{align}
0
&=\delta\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\\
&=\left(\frac1{b+c}-\frac{b}{(c+a)^2}-\frac{c}{(a+b)^2}\right)\delta a\\
&+\left(\frac1{c+a}-\frac{c}{(a+b)^2}-\frac{a}{(b+c)^2}\right)\delta b\\
&+\left(\frac1{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Let $y = \sqrt{x + 3} \implies 3 = y^2 - x$.
$$\large \begin{align}
&13x + 2(3x + 2)\sqrt{x + 3} +42\\
= &14(x + 3) + (6x + 4)y - x\\
= &14y^2 + [6(x + 3) - 14]y - x\\
= &14y(y - 1) - (y^2 - x - 9)y^3... | I have no idea where you are trying to go with your approach, but I would suggest squaring both sides of
$$13x+42=-2(3x+2)\sqrt{x+3}.$$
This yields a cubic equation in $x$ which can be solved by standard methods; it turns out that all roots are rational so the rational root theorem yields them all. It remains to check ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that the equation $x^5+x^4=1$ has a unique solution. Show that the equation $x^5 + x^4 = 1 $ has a unique solution.
My Attempt:
Let $f(x)=x^5 + x^4 -1 $
Since $f(0)=-1$ and $f(1)=1$, by the continuity of the function $f(x)$ has at least one solution in $(0,1)$.
| Let $f(x) = x^5+x^4-1$.
Then we have $f'(x)=5x^4+4x^3=x^3(5x+4)$
There are two turning points, one is at the location when $x=-\frac45$ and the corresponding $f$ value is $-\frac{4^5}{5^5}+\frac{4^4}{5^4}-1 =\frac{-4^5+5(4^4)}{5^5}-1=\frac{4^4-5^5}{5^5}<0$ and the other is at the location when $x$ takes value $0$ with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding null space and image Let $T(f(x)) = g(x) = \int_{-1}^{1}f(t)(t-x)^2dt$ be a linear transformation from $V$ to $V$ where $V$ is the vector space of continuous functions on $[-1 , 1]$ . Find $\text{Nul}(T)$ and $\text{Im}(T)$ . I really have no idea about the solution . Can $x$ be considered constant in the integ... | As other have pointed out the image of $T$ is a subspace of the $2^{nd}$ degree polynomials
To find the image let us only look at how $T$ operates on polynomials
$$ f(t) = \sum_{k=1}^n a_kt^k$$
\begin{align*}
\int_{-1}^1 f(t) (t-x)^2 dt &= \int_{-1}^1f(t)(t^2 + x^2 -2tx) \;dt \\
&= \int_{-1}^1 \sum_{k = 1}^n a_k t^{k+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3173719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What is the easiest way to get: $2+ \sqrt{-121} = (2+ \sqrt{-1})^3$ I was reading the book Seventeen equations have changed the world.
At some point, while the book was talking about complex numbers, I see this equation:
$2+ \sqrt{-121} = (2+ \sqrt{-1})^3$
Even if it's easy to proof the truth of this equivalence (it is... | Not sure how you approached it before.
If you want to utlise the binomial expansion
$$
(x+y)^n = \sum_{k=0}^n\left(\matrix{n\\k}\right)x^ky^{n-k}
$$
then we have
$$
\begin{align}
(x+y)^3 &= \left(\matrix{3\\0}\right)2^3 +\left(\matrix{3\\1}\right)2^2\sqrt{-1} + \left(\matrix{3\\2}\right)2^1(\sqrt{-1})^2 + \left(\matri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Given $\cos 3\theta = 4(\cos\theta)^3 - 3\cos\theta$, solve $4x^3 - 9x - 1 = 0$, correct to 3 decimal places. I am trying to solve the following problem:
Given $\cos 3\theta = 4(\cos\theta)^3 - 3\cos\theta$, solve the equation $4x^3 - 9x - 1 = 0$, correct to 3 decimal places.
Assuming that $x = \cos 3\theta$, you ca... | As you received as answers shows that
$$x=\sqrt{3} \cos \left(\frac{1}{3} \cos ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)\right)$$ what you want to evaluate to a given accuracy assuming that your calculator does not have trigonoetric functions at all.
Let us write
$$x=\sqrt{3} \cos \left(\frac{1}{3} \cos ^{-1}(\epsilon )\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Roots of a Complex Number expression If $z_1$, $z_2$, $z_3$, $z_4$ are roots of the equation $z^4+z^3+z^2+z+1=0$,
then what is the least value of $\lfloor mod(z_1 + z_2)\rfloor + 1$?
($\lfloor.\rfloor$ denotes Greatest Integer Function)
| $$z^4+z^3+z^2+z+1=\frac{z^5-1}{z-1}=0$$
$$\therefore z^5=1\text{ and } z\ne1$$
These solutions can be plotted on an Argand diagram which more easily shows that the solution is given by the two complex conjugates with the smallest magnitude real parts - $e^{\frac{2i\pi}5}$ and $e^{\frac{-2i\pi}5}$. Adding these values g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3178747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the last three digits of $p$ if the equations $x^6 + px^3 + q = 0$ and $x^2 + 5x - 10^{2013} = 0$ have common roots.
Find the last three digits of $p$ if the equations $x^6 + px^3 + q = 0$ and
$x^2 + 5x - 10^{2013} = 0$ have common roots.
Let $a,b $ be the solutions of second equation, then by Vieta we have ... | Clearly $5|x$; when $x_1+x_2=5$, that means $x_1$ is $10^u+5$ and $x_2=10^v$. Hence $x_1^3=10 k_1+125$ and $x_2^3=10^{3v}$. Now in equation $x^6+px^3+q$ we have $p=x_1^3+x_2^3$ if $x_1$ and $x_2$ are common roots. The result is that $p=10^t+125$, that is the last three digits of $p$ is $125$.Generally $p=10^t+5^n$ in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$
Find $f^{(22)}(0)$ for $f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$
I know that I should use Taylor's theorem and create power series. However I don't have idea how I can find $a_{n}$ such that $f(x)=\sum_{n=1}^{+\infty} a_{n}x^{n}$. I know only that $f(x)=\frac{x}{(x-1)(... | By way of a contrast, here's an intuitve method that lets you see the bigger picture rather than getting bogged down in technical details,
$$f(x)=\frac{x}{x^{3}-x^{2}+2x-2}$$
$$=\frac{1}{1-x}\times \frac{x}{2+x^2}$$
$$=\frac{1}{1-x}\times \frac{x}{2(1+(\frac{x}{\sqrt{2}})^2}$$
$$=\frac{1}{1-x}\times \frac{1}{\sqrt{2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3184465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\tan(\sin(x))=x-\frac{x^{3}}{6}+o(x^{3})$
Prove that $\tan(\sin(x))=x-\frac{x^{3}}{6}+o(x^{3})$
I know that when I calculate derivative then I get true answer. However I want to know why the way that I will present soon does not work. My try:$$\tan(x)=x+r_{1}(x), \quad r_{1}(x)=o(x)$$ $$\tan(\sin(x))=\si... | Your algebraic manipulation is correct, but your desired result is wrong (as mentioned in another answer). The problem with your approach is that although it is correct it does not lead you to anything useful. The equation $$\tan(\sin x) =x-\frac{x^3}{6}+o(x^3)+o(\sin x) \tag{1}$$ although correct is much less useful c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What causes the extraneous root when intersecting parabola $y^2=4ax$ with circle $x^2+y^2=9a^2/4$? If I solve the parabola: $y^2=4ax$ and the circle: $x^2+y^2=\frac{9a^2}{4}$ I get a quadratic in $x$; ie:
$$4x^2+16ax-9a^2=0$$
which has roots $x=\frac{a}{2},-\frac{9a}{2}$. But if we see the graphs of these two, both t... | The full solution is this $$\begin{cases}x^2+y^2=\frac94a^2\\ y^2=4ax\end{cases}\iff \begin{cases}x^2+4ax-\frac94a^2=0\\ y^2=4ax\end{cases}\iff\begin{cases}x=\frac a2\\ y^2=2a^2\end{cases}\lor\begin{cases}x=-\frac 92a\\ y^2=-18a^2\end{cases}$$
Now the system must be "discussed" in terms of the parameter, in the sense t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Minimizing lengths of cevians in an isosceles right triangle Consider isosceles right triangle $ABC$ with $BC$ as the hypotenuse and $AB=AC=6$. $E$ is on $BC$ and $F$ is on $AB$ such that $AE+EF+FC$ is minimized. Compute $EF$.
My thought process:
I reflected triangle $ABC$ across $BC$ to get a square $ABA'C$. Then, I m... |
Let $|AB|=|AC|=6=b$.
It is known that the minimal path would be
the path of the light ray from the point $C$
to point $A$, reflected at points $F$ and $E$.
Consider points $C_1,A_1$ and $E_1$
as reflection of the points $C,A$ and $E$
with respect to $AB$.
Let
$\angle A_1CC_1=\angle CFG=\angle GFE=\phi$,
$\angle FEH=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3189411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\lim_{n\to\infty}\left(\frac{\sin(2\sqrt 1)}{n\sqrt 1\cos\sqrt 1} +\cdots+\frac{\sin(2\sqrt n)}{n\sqrt n\cos\sqrt n}\right)$ Using the trigonometric identity of $\sin 2\alpha = 2\sin \alpha \cos \alpha$, I rewrote the expression to:
$$\lim_{n\to\infty}\left(\frac{\sin(2\sqrt 1)}{n\sqrt 1\cos\sqrt 1} + \cdo... | The inequality
$$\sum_{k=1}^{n}{\frac{1}{\sqrt k}} \le 1+\int_{1}^{n}{\frac{1}{\sqrt x}}\,\mathrm{d}x \leq 2\sqrt{n},$$
yields
$$\left|\frac{2\sin(\sqrt 1)}{n\sqrt 1} + ...+\frac{2\sin(\sqrt n)}{n\sqrt n}\right| \leq 4\frac{1}{\sqrt n} \to 0.$$
Note that $\sin(\cdot)$ can change signs, so absolute value is needed in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Complex number cannot arive at $\frac{9}{2}-\frac{9}{2}i$ with problem $\frac{4+i}{i}+\frac{3-4i}{1-i}$ I am asked to evaluate: $\frac{4+i}{i}+\frac{3-4i}{1-i}$
The provided solution is: $\frac{9}{2}-\frac{9}{2}i$
I arrived at a divide by zero error which must be incorrect. My working:
$\frac{4+i}{i}$, complex conjugat... | You have made a mistake in the second last step of your simplification.
$$1 - i^2 = 1 - (-1)=1 + 1 = 2$$
Then,
$$-4i + 1 + \frac{7 - i}{2}
= 1 + \frac{7}{2} + i(-4 - \frac{1}{2})
= \frac{9}{2} - \frac{9}{2}i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How can I find the P matrix in $P^{-1}AP $= D? I have the following exercise:
Let a matrix A = $\begin{pmatrix}
4 & 0 & 1\\
2 & 2 & 1\\
0 & 4 & 1
\end{pmatrix}$
1) Determine its eigenvalues and their multiplicity.
2) Give a basis of the eigen spaces associated with each of the
distinct eigenvalues of A.
3) fin... | A quick look at your calculations, I think your last eigen vector is wrong. The last eigen vector is (1, 1, 1) and not (-1, 1, 1). I think that would solve your problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a remainder when $(x^5+1)^{100} + (x^5-1)^{100}$ is divided by $x^4+x^2+1$ The question is:
Find a remainder when $f(x)=(x^5+1)^{100} + (x^5-1)^{100}$ is divided by $x^4+x^2+1$
I first began by decomposing $$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$
and using $$x^3-1=(x-1)(x^2+x+1)\\x^3+1=(x+1)(x^2-x+1)$$
and could get rem... |
Therefore, $ $ with $\,n = 3^{50}\!+1$
$$\begin{align} &f(x)\,\equiv\ \ \ n\,\color{#c00}x\quad(\mathrm{mod}\ \ x^2\!+\!x\!+\!1)\\[.2em]
&f(x)\,\equiv -n\,\color{#0a0}x\quad(\mathrm{mod}\ \ x^2\!-\!x\!+\!1)\end{align}\qquad\qquad$$
Hint: make the following substitutions $$\begin{align}\color{#c00}x &\equiv -(x^2\!+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Show that $c^2 + a^2d=abc$ for a monic quartic polynomial I apologize in advance for asking a homework question, but I have genuinely no idea on how to approach part b). The question is as follows:
Consider the polynomial equation $\rm{P}(x)=x^4 + ax^3 + bx^2 + cx+d$, with a root $ki$. In addition, $a,b,c,d\in Z$ and ... | We know that $ki$ is a solution to $P(x)=0$ so in particular,
$$\begin{split}
(ki)^4+a(ki)^3+b(ki)^2+c(ki)+d&=0 \\
\Leftrightarrow k^4-ak^3i-bk^2+cki+d&=0 \\
\Leftrightarrow k^4-bk^2+d+ki(c-ak^2)&=0
\end{split}$$
Using part (a), this gives $k^4-bk^2+d=0$. Now add $bk^2$ on both sides and multiply through with $a^2$ to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3194881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ .
Given three non-negatve numbers $a, b, c$. Prove that:
$$1+ a^{2}+ b^{2}+ c^{2}+ 4abc\geqq a+ b+ c+ ab+ bc+ ca$$
Let $t= a+ b+ c$, we have to prove
$$\left(\!\frac{1}{t^{3}}- \frac{1}{t^{2}}+ \frac{1}{t}\!\right)\sum ... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$1+9u^2-6v^2+4w^3\geq3u+3v^2,$$ which is a linear inequality of $w^3$.
Id est, it's enough to prove our inequality for the extreme value of $w^3$, which happens in the following cases.
*
*$w^3=0$.
Let $c=0$.
Thus, we need to prove that
$$1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving polynominals equations (relationship of roots)
The roots of $x^3-4x^2+x+6$ are $\alpha$, $\beta$, and $\omega$.
Find (evaluate):
$$\frac{\alpha+\beta}{\omega}+\frac{\alpha+\omega}{\beta}+\frac{\beta+\omega}{\alpha}$$
So far I have found:
$$\alpha+\beta+\omega=\frac{-b}{a} = 4 \\
\alpha\beta+\beta\omega+\... | $$\frac{\alpha + \beta}{\omega} + \frac{\beta + \omega}{\alpha} + \frac{\alpha + \omega}{\beta}$$
$$= \frac{\alpha + \beta + \omega - \omega}{\omega} + \frac{\beta + \omega + \alpha - \alpha}{\alpha} + \frac{\alpha + \omega + \beta - \beta}{\beta}$$
$$ = (\alpha + \beta + \omega) \left(\frac{1}{\alpha} + \frac{1}{\beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Given $(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$ Find $\frac{a_7}{a_{13}}$ Given $$(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$$ Find Value of $\frac{a_7}{a_{13}}$
My try:
I assumed $A=2x^2$,$B=3x$ and $C=4$
Then we have the following cases to collect coefficient of $x^7$:
Case $1.$ $A^3 \times B^1 \times C^6$
Case $2.$ $A^2 ... | This approach is from the idea of Pascal's triangle. We write
$$
(2x^2+3x+4)^n = \sum_r a_{n,r} x^r
$$
Then
$$
(2x^2+3x+4)^{n+1}=(\sum_r a_{n,r} x^r )(2x^2+3x+4) = \sum_r (2a_{n,r-2}+3a_{n,r-1}+4a_{n,r})x^r.$$
Thus, by comparing coefficients, we obtain a recurrence relation
$$
a_{n+1,r}=2a_{n,r-2}+3a_{n,r-1}+4a_{n,r}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$
If $$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}= \frac{3\sqrt{3}}{4}$$ for $a,b,c,d>0$
Then
Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$ is
Try: us... | With thanks to Michael for pointing this beautiful topic.
For proving
$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2,$$
I have used the following result discovered by me:
$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2+$$
$$+9\sum_{1\leq i<j\leq4}{(xy-1)^2}+3\sum_{1\leq i<j<k\leq4}{(xyz-1)^2}+(xyzt-1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Minimizing $\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2$ While solving a problem I came across this task, minimizing
\begin{align}
\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2.
\end{align}
One can eas... | Knowing the answer it's not that difficult to get it another way. Let $y = \cos 2x$. We have $\sin^2 x = \frac{1-y}{2}$, $\cos^2 x =\frac{1+y}{2}$. Then
\begin{align} \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 + \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 &= \left(\frac{1-y}{2} + \frac{2}{1-y}\right)^2 + \left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find the limit without using Lhopital this question came out on my analysis exam: Evaluate
$$\lim_{x\to 0}\left(\frac{5^{x^2}+7^{x^2}}{5^x+7^x}\right)^{\frac{1}{x}} $$
I did it using L'hopital rule but is there another way to do this?
| We call your limit A.
By using the exp, you get :
$$A=\lim_{x \rightarrow 0} \exp(\frac{1}{x}\ln\frac{e^{x^2\ln{5}}+e^{x^2\ln{7}}}{e^{x\ln{5}}+e^{x\ln{7}}})$$
Which gives you the following limit, as PierreCarre mentionned.
$$\lim_{x \rightarrow 0} \frac{1}{5}\Big(\frac{1+a^{x^2}}{1+a^x}\Big)^{\frac{1}{x}}$$
$$=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\fr... | Note that $$4+\frac1x-\frac1{x^2}=4+\frac1x-\left(\frac1x\right)^2,$$ so making the substitution $y=\frac1x,$ we obtain the quadratic $$-y^2+y+4=0,$$ which should be more familiar. Solve for $y,$ and since neither solution for $y$ should be equal to $0,$ use $x=\frac1y$ to solve for $x.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$? I need to remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$, and find out what $f(7)$ equals. I am not sure what I've done wrong, but I'm getting 33, which the website I'm using tells me is wrong.
Please help me figure out what I... | You probably want to remove it at $7$ and not at $f(7)$. Write $$ f(x) = {(x-7)(x-4)\over x-7 } = x-4$$
So $$\lim_{x\to 7} f(x) =3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Limit of $n(n\pi + \frac{\pi}{2} - x_n)$, where $x_n$ is the solution of $x = \tan(x)$ in the interval $(n \pi, n\pi + \frac{\pi}{2})$ Let $x_n$ denote the solution to $x = \tan(x)$ in the interval $(n \pi,n \pi +\frac{\pi}{2}).$ Find
$$ \lim_{n \to \infty}n \left(n\pi + \frac{\pi}{2} - x_n\right)$$
| $x_n\in(n\pi,n\pi+\frac{\pi}{2})\implies n\pi+\frac{\pi}{2}-x_n\in(0,\frac{\pi}{2})$ and $\frac{1}{n\pi+\frac{\pi}{2}}<\frac{1}{x_n}<\frac{1}{n\pi}$.
\begin{align*}
\tan\left(n\pi+\frac{\pi}{2}-x_n\right)
&=\tan\left(\frac{\pi}{2} - x_n\right)\\
&=\frac{1}{\tan x_n}\\
&=\frac{1}{x_n}.
\end{align*}
So
$$n\pi+\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to solve this D.E $y''(\frac{x}{2})+y'(\frac{x}{2})+y(x)=x$ I know how to slove $y''(x)+y'(x)+y(x)=x$
But I couldn't solve this
$$y''(\frac{x}{2})+y'(\frac{x}{2})+y(x)=x$$
any hint to help me?
Thanls
| Here is an incomplete attempt that I might be able to rectify later.
Write the equation as:
$$y''(x)+y'(x)+y(2x) = 2x$$ and let $y(x) = f(x)+x-1.$ Then we get an equation for $f:$
$$f''(x)+f'(x)+f(2x) = 0.$$
Now assume $x\neq 0$ and By Taylor's theorem:
$$f(2x) = f(x)+xf'(x)+\frac{x^2}{2}f''(x)+x^2h(2x)$$
with $h(t)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Indefinite integral through factorization So I have
$$\int \frac{1}{(x^2+1)^2}dx$$
And the professor does some magic
I'm confused. what's with the derivative?
I solved the integral via substitution but I'm curious how this works, so I can learn it
thanks!
| It is possible to shorten this magic as follows
\begin{eqnarray*} \int \frac{1}{(x^2+1)^2}dx
& = & \int \frac{1+x^2 - x^2}{(x^2+1)^2}dx \\
& = & \int \frac{1}{x^2+1}dx - \frac{1}{2}\int x\frac{2x}{(x^2+1)^2}dx \\
& = & \arctan x - \frac{1}{2}\left(- \frac{x}{x^2+1} + \int \frac{1}{x^2+1}dx\right) \\
& = & \arctan x + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3218154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Number of possible integer values of $x$ for which $\frac{x^3+2x^2+9}{x^2+4x+5}$ is integer How many integer numbers, $x$, verify that the following
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5}
\end{equation*}
is an integer?
I managed to do:
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5} = x-2 + \frac{3x+19}{x^2+4x+5}
... | Here is another method that works well when Diophantine quadratics $ax^2+bx+c=0$ are involved, it consists in saying that since a solution should exists then $\exists \delta\in\mathbb Z\mid b^2-4ac=\delta^2$.
In this problem we'd like $\quad\dfrac{3x+19}{x^2+4x+5}=n\quad$ to be an integer.
Thus applying the method to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Count the number of permutations of certain cycles type Suppose we have $n$ elements, assume there is a permutations over $k$ elements among the $n$ elements so $n-k$ are fixed. Let that the permutation over the k elements is represented by permutation cycles so the length of all permutation cycles $=k$.
As an example:... | Let's take your example of a cycle structure corresponding to the partition $3^1 + 2^2 + 1^8 \vdash 15$. I think the easiest way to handle the $a_i$ cycles of a given length $i$ is to select the lot (e.g. for the two two-cycles select four elements) and then repeatedly force the lower unselected element to be in the ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The odds in rolling two dice together. Two dice are rolled. I want to see the odds of the following:
$1.$ A sum of $5$.
$2.$ A sum of $8$ or $10$.
$3.$ A sum less than $6$.
$4.$ Not a sum of $7$.
Solution: $1.$ A sum of $5$. $5 = 1+4,2+3,3+2,4+1$. So the odds is $4/36 = 1/9.$
$2.$ A sum of $8$ or $10$. We can express ... | As stated in the comments, you have correctly calculated the probabilities.
When two dice are rolled, there are $6 \cdot 6 = 36$ possible sums. Of these, four outcomes yield a sum of $5$, while the remaining $36 - 4 = 32$ do not. Assuming each of the $36$ possible sums are equally likely to occur, the odds for a sum o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can re... | Let $C\leq B$ and $B+C=\frac{3\pi}{4}$.
i) When $\frac{3\pi}{8}\leq B<\frac{\pi}{2}$, then
$\frac{\pi}{4}<C\leq \frac{3\pi}{8}$.
Define $$f=\tan\ B\tan\ ( \frac{3\pi}{4}-B)$$ The range of $f$ is
$\{ \tan^2\frac{3\pi}{8} \leq t<\infty\}$, by (1) continuity, (2)
considering $B\rightarrow \frac{\pi}{2},\ B\rightarrow
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Compute polynomial $p(x)$ if $x^5=1,\, x\neq 1$ [reducing mod $\textit{simpler}$ multiples] The following question was asked on a high school test, where the students were given a few minutes per question, at most:
Given that,
$$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$
and,
$$Q(x)=x^4+x^3+x^2+x+1$$
what is the re... | This may be accessible to a high school student:
$x^{104}+x^{93}+x^{82}+x^{71}+1$
$ = (x^{104}-x^4)+(x^{93}-x^3)+(x^{82}-x^2)+(x^{71}-x)+(x^4+x^3+x^2+x+1)$
$=x^4(x^{100}-1)+x^3(x^{90}-1)+x^2(x^{80}-1)+ x(x^{70}-1)+(x^4+x^3+x^2+x+1)$
We know that $(x^n-1)|(x^{mn}-1), m,n \in \mathbb{N}$ so $x^5-1$ divides $x^{100}-1, x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "63",
"answer_count": 6,
"answer_id": 3
} |
Evaluating $\int_0^1 \frac{3x}{\sqrt{4-3x}} dx$
So this is the integral I must evaluate:
$$\int_0^1 \frac{3x}{\sqrt{4-3x}} dx$$
I have this already evaluated but I don't understand one of the steps in its transformation.
I understand how integrals are evaluated, but I don't understand some of the steps when it is bei... | Addressing question 1:
\begin{align*}
& - \left( \frac {4-3x-4}{\sqrt{4-3x}}\right) \\
= & - \left( \frac {4-3x}{\sqrt{4-3x}} - \frac{4}{\sqrt{4 - 3x}}\right) \\
= & - \frac {(\sqrt{4-3x})^2}{\sqrt{4-3x}} - \left(- \frac{4}{\sqrt{4 - 3x}}\right) \\
= & - \sqrt {4-3x} + \frac {4}{\sqrt{4-3x}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $(\sum_{k=1}^{7} \tan^2(\frac{k\pi}{16})) - \tan^2(\frac{4\pi}{16})$ The question is:
Evaluate
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
The given answer:$34$
What I've tried:
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\f... |
The summation is 35 and subtract -1 because tan pi/4 is 1
Finally the result is 34
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$ if $3\leq aI don't really know if I should use brute force or some kind of theorem, it comes on a calculus past exam and it says:
suppose: $3≤a<b≤8$
prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$
| Note: This answer was finished after farruhota's post and has a similar argument.
$\quad \frac{b-a}{6} \le \sqrt{1+b} - \sqrt{1+a} \; \text{ iff }$
$\quad\quad\quad 1 \le 6 \frac{\sqrt{1+b} - \sqrt{1+a}}{b-a} = \frac{6}{ \sqrt{1+b} + \sqrt{1+a}}$
Ok, the denominator, $\sqrt{1+b} + \sqrt{1+a}$, of the RHS of the last i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Fourier Series Representation for piecewise function I've been posed the following question:
$$
f(x)=
\begin{cases}
1-x^2, & 0 \leqslant|x|<1,\\
0, & 1\leqslant|x|<2\\
\end{cases}
$$
I'm trying to find the Fourier series representation but I am having trouble with $ a_n$ and $b_n$ coefficients.
For reference, I obta... | As you said $a_0=\dfrac43$ and $b_n=0$. Also integrating by parts shows
$$a_n=\dfrac{1}{2}\int_{-2}^{2} (1-x^2)\cos\dfrac{m\pi x}{2}\ dx
= \dfrac{1}{2}
\left[
(1-x^2)\left(\dfrac{2}{m\pi}\right)\sin\dfrac{m\pi x}{2}
-2x\left(\dfrac{2}{m\pi}\right)^2\cos\dfrac{m\pi x}{2}
+2\left(\dfrac{2}{m\pi}\right)^3\sin\dfrac{m\pi x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$
Solve the following system of equations. $$\large \left\{ \begin{aligned} x^2 + y^2 &= 2x^2y^2\\ (x + y)(1 + xy) &= 4x^2y^2 \end{aligned} \right.$$
From the system of equations, we have that
$$\left\{ \begin{align*} (x + y)^2 \le 2(x^2 + y^2)... |
My simple solution is $(\!x^{2}+ y^{2}- 2x^{2}y^{2}\!)+ \left (\!(\!x+ y\!)(\!1+ x^{2}y^{2}\!)\!\right )= (\!x+ y- 2xy\!)(\!3xy+ x+ y+ 1\!)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$ I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions.
other approaches would be appreciated.
| knowing that fact that
$$2\sum_{n=1}^\infty f(2n)=\sum_{n=1}^\infty f(n)(1+(-1)^n)$$
then
\begin{align}
2\sum_{n=1}^\infty (-1)^nx^{2n}H_{2n}&=2\sum_{n=1}^\infty (i)^{2n}x^{2n}H_{2n}\\
&=\sum_{n=1}^\infty (ix)^nH_{n}+\sum_{n=1}^\infty (-ix)^nH_{n}\\
&=-\frac{\ln(1-ix)}{1-ix}-\frac{\ln(1+ix)}{1+ix}\\
&=-\frac{\ln(1-ix)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Use real integral to calculate $\int_R \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$ Problem :
Evaluate $$\int_{-\infty}^{\infty} \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$$
Use only real integral.
What I did :
$$\int_{-\infty}^{\infty} \frac{x^2 \cos (\pi x)}{(x^2 + 1)(x^2 + 2)}dx$$
$$=2\int_{0}^{\infty} \frac{... | Just A Bit of a Generalization
We may actually evaluate the integral
$$J(a,b;t)=2\int_0^\infty \frac{x^2\cos(tx)dx}{(x^2+a^2)(x^2+b^2)}$$
for some $a, b>0$, $a\ne b$. We see that
$$J(a,b;t)=\frac{2a^2}{a^2-b^2}f(a;t)+\frac{2b^2}{b^2-a^2}f(b;t)$$
where $$f(q;t)=\int_0^\infty \frac{\cos(tx)}{x^2+q^2}dx\, .$$
We take th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}dx$
Calculate $$\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}dx$$
I tried to do this task in several ways, but none of them proved to be effective. For example:
$$\int ^{4\pi} _{-4\pi} \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x... | Hint:
$$ \frac{(\sin x)^2-(\sin x)^4}{1-(\sin x)^4}= 1-\frac{1-\sin^2 x}{1-(\sin x)^4}=1-\frac{1}{1+\sin^2x}=1-\frac{\sec^2x}{1+2\tan^2x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Difference in my and wolfram's integration. Calculate
$$\int \frac{\sin ^3(x)+1}{\cos (x)+1} \, dx$$
Let $$u = \tan(x/2)$$
$\int \frac{\sin ^3(x)+1}{\cos (x)+1} \, dx = \int \frac{2\left(\frac{8u^3}{(u^2+1)^3}+1 \right)}{(u^2+1)\left( \frac{1-u^2}{u^2+1} + 1 \right)} \, du = \int \frac{8u^3}{(u^2+1)^3 } + 1 \, du$
$$ ... | There's no error in your solution. Both results (yours and Wolfram's) are equal up to a constant (which for an indefinite integral is arbitrary). You can check that using the equalities
$$ \cos (5\alpha) = \cos^5 \alpha - 10\cos^3\alpha \sin\alpha + 5\cos\alpha \sin^4\alpha = 16\cos^5\alpha -20\cos^3\alpha + 5\cos\alph... | {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$. This question is taken from book: Exercises and Problems in Calculus, by
John M. Erdman, available online, from chapter 1.1, question $4$.
Request help, as not clear if my approach is correct.
(4) Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$.
Ha... | Your solution is not correct. But contains good ideas.
You write $\sqrt{2}i\leq x\leq \sqrt{11}$.
You can not do that, because you can not order complex numbers! So that is a mistake.
Also it is not clear, how you get your three conditions, which are also the wrong intervals.
It is $x^2+2>0$ for all $x\in\mathbb{R}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$ Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?
From the problem I know that $f(4)=51$.
Using long division, I found that remainder of $\frac{ax^3-ax^... | $f(x)=ax^3-ax^2+bx+4$
Since $f(4) = 51$,
$51
=a(64-16)+4b+4
=48a+4b+4
$
so
$12a+b
=47/4
$.
Since $f(i) = 0$,
$0
=a(-i+1)+ib+4
=i(b-a)+a+4
$
so
$ a+4 = 0,
a=-4,
b-a = 0,
b=a=-4
$.
Therefore
$f(x) = -4x^3+4x^2-4x+4
$.
But this does not satisfy
$f(4) = 51$.
Therefore the problem is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How to solve $x^2+y^2 \equiv 8\pmod 9$? How to solve $x^2+y^2 \equiv 8 \pmod 9$?
I know the Chinese Remainder Theorem, but how do I apply it here?
| If we consider each possible residue of $x^2\mod9$, we see that they are
$0^2\equiv 0$
$1^2\equiv 1$
$2^2\equiv 4$
$3^2\equiv 0$
$4^2\equiv 7$
$5^2\equiv 7$
$6^2\equiv 0$
$7^2\equiv 4$
$8^2\equiv 1$
Obviously (in a minor abuse of notation) the solutions have to be $x\in\{4,5\},y\in\{1,8\}$ (or vice versa) (where number... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ I have just started learning sums. I need to evaluate the following sum:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)}$$
$$a_{1} = \frac{1}{2}, a_{2} = \frac{1}{6}, a_{3} = \frac{1}{12}, a_{4} = \frac{1}{20}, a_{5} = \frac{1}{30}, a_{6} = \frac{1}{42}, ...$$
I tried splitting it... | $\require{cancel}$Note that$$S_n=1-\cancel{\frac12}+\cancel{\frac12}-\cdots-\cancel{\frac1n}+\cancel{\frac1n}-\frac1{n+1}=1-\frac1{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to prove the equality $|\sqrt{x}-\sqrt{a}| = \frac{|x-a|}{\sqrt{x}+\sqrt{a}}$ How to prove this equality?
$|\sqrt{x}-\sqrt{a}| = \frac{|x-a|}{\sqrt{x}+\sqrt{a}}$
related to this post
| Assuming both $a$ and $x$ are positive. Two cases (1) $x \gt a$ and (2) $x\le a$.
Case(1) becomes $\sqrt{x}-\sqrt{a}=\frac{x-a}{\sqrt{x}+\sqrt{a}}$. Clear the denominator and get $x-a=x-a$. Since $x\gt a$, $x-a=|x-a|$
Case(2) becomes $\sqrt{a}-\sqrt{x}=\frac{a-x}{\sqrt{x}+\sqrt{a}}$. Clear the denominator and get $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256861",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"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.