Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
minimal polynomial of a matrix with some unknown entries Question is to prove that :
characteristic and minimal polynomial of $ \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right) $ is $x^3-ax^2-bx-c$.
what i have done so far is :
characteristic polynomial of a matrix $A$ is given by $\... | You don't need to factor anything. Your characteristic polynomial, call it $f(x) = x^3 - a x^2 - b x - c,$ satisfies $f(A) = 0,$ meaning $$ A^3 - a A^2 - b A - c I = 0 $$ as matrices. This is Hamilton-Cayley. Actually, given your question, you should carefully check that with actual matrices. What are $A^2$ and $A^3?$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How to find $x + y + z$?
Q. If $x^{1/3} + y^{1/3} + z^{1/3} = 0$, then
(A) $x + y + z = 3 xyz$
(B) $x + y + z = 0 $
(C) $( x + y + z)^3= 27 xyz$
(D)$ x^3 + y^3 + z^3 = 0$
What I've done:
$(x^{1/3} + y^{1/3} + z^{1/3})^3 = 0^3$
$=> (x^{1/3} + y^{1/3})^3 + z+3(x^{1/3} + y^{1/3})z(x^{1/3} + y^{1/3}+z^{1/3})=0$
$=> (x^{... | use this
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
so if $a+b+c=0$,then we have
$$a^3+b^3+c^3=3abc$$
then if $$x^{\frac{1}{3}}+y^{\frac{1}{3}}+z^{\frac{1}{3}}=0$$
$$\Longrightarrow x+y+z=3(xyz)^{\frac{1}{3}}$$
so
$$(x+y+z)^3=27xyz$$
\begin{align*}&a^3+b^3+c^3-3abc \\
&=(a^3+3a^2b+3ab^2+b^3+c^3)-(3ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$. Then $\lfloor k \rfloor =$ If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$, where $n\in \mathbb{N}$. Then $\lfloor k \rfloor = $
$\underline{\bf{My\; Try}}::$ We can write the expression $n\cdot (n+1)\cdot(n+2)\cdot (n+3) = (n^2+3n).(n^2+3n+2)$
$ = (n^2+3n)^2+2\cdo... | $$k<\sqrt{n.(n+1).(n+2).(n+3)+1}=n^2+3n+1$$
$$n.(n+1).(n+2).(n+3)=n^4+6n^3+11n^2+6n>(n^2+3n)^2$$
so $\lfloor k \rfloor =n^2+3n $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/508512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How find the maximum of $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{z}{1+z^2}$ let $x,y,z$ are postive numbers,and such
$$xy+zx+yz=1$$
find the maxum of
$$\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{z}{1+z^2}$$
my try
note
$$x^2+1=x^2+xy+yz+xz=(x+y)(x+z)$$
$$y^2+1=(y+x)(y+z)$$
$$z^2+1=(z+x)(z+y)$$
$$\dfrac{1}{1+x^2}+\dfrac{1}{... | Due to symmetry of the formula, you can suppose x=y=t. Hence
$$t^2+2tz=1$$
$$z=\frac{1-t^2}{2t}$$
$$f(t)=\frac{2}{1+t^2}+\frac{2t(1-t^2)}{4t^2+(1-t^2)^2}=2\frac{1+t+t^2-t^3}{(1+t^2)^2}$$
$$f'(t)=2\frac{(-3t^2+2t+1)(1+t^2)-4t(-t^3+t^2+t+1)}{(1+t^2)^3}$$
$$f'(t)=2\frac{t^4-2t^3-6t^2-2t+1}{(1+t^2)^3}$$
$P(t)=t^4-2t^3-6t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$ by Induction Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$
To solve it I used induction but it is leading me nowhere my attempt was as follows:
Lets assume the inequality i... | Using induction we first show this is true for $n=2$:
$\frac{1}{2+1}+\frac{1}{2+2}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}\gt\frac{13}{24}$
Therefore it is indeed true for $n=2$.
Now lets assume it is true for some $n=k$, therefore:
$S_k=\frac{1}{k+1}+\frac{1}{k+2}+...+\frac{1}{2k}\gt\frac{13}{24}$
Finally w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 1
} |
Help with partial fraction $\int_{4}^{5}\frac{x^3-3x^2-9}{x^3-3x^2}dx$ I can't figure out why I keep getting this answer wrong.
First, simplifying the answer:
$$\int_{4}^{5}1-\frac{9}{x^2(x-3)}dx$$
Setting up the partial fraction:
$$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-3}$$
Expanding...
$$\frac{A(x^2)(x-3)+B(x)(x-3)+... | Recall that the original integrand was $$1 - \frac{9}{x^2(x-3)}$$ Did you forget to account for the $1$, which would integrate to $x$, and/or forget that you are subtracting the quotient from $1$?
So you should end with $$\int_{4}^{5} 1 - \Big(\frac{1}{x}+\frac{3}{x^2}-\frac{1}{x-3}\Big)dx$$
Also, use the power rule fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How find this equation $\prod\left(x+\frac{1}{2x}-1\right)=\prod\left(1-\frac{zx}{y}\right)$ let $x,y,z\in(0,1)$, find the pairs of $(x,y,z)$ such
$$\left(x+\dfrac{1}{2x}-1\right)\left(y+\dfrac{1}{2y}-1\right)\left(z+\dfrac{1}{2z}-1\right)=\left(1-\dfrac{xy}{z}\right)\left(1-\dfrac{yz}{x}\right)\left(1-\dfrac{zx}{y}\ri... | My approach lacks something at the end.
$$x+\dfrac{1}{2x}-1 = \frac{2x^2}{2x}-\frac{2x}{2x}+\dfrac{1}{2x} = \frac{2x^2 -2x +1}{2x}$$
$$1 - \frac{xy}{z} = \frac{z-xy}{z}$$
$$
\begin{align}
\left(x+\dfrac{1}{2x}-1\right)\left(y+\dfrac{1}{2y}-1\right)\left(z+\dfrac{1}{2z}-1\right)&=\left(1-\dfrac{xy}{z}\right)\left(1-\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\sin25^° \sin35^° \sin85^°$ Trigo problem :
Find the value of $\sin25^° \sin35^° \sin85^°$.
My approach :
Using $2\sin A\sin B = \cos(A-B) -\cos(A+B)$
$$
\begin{align}
& \phantom{={}}[\cos10^{°} -\cos60^°] \sin85^° \\
& = \frac{1}{2}[2\cos10^°\sin85^° -2\cos60^° \sin85^°] \\
& = \frac{1}{2}[2\cos10... | \begin{align}
& \phantom{={}} \sin25^{°}\sin35^{°}\sin85^{°} \\
& =\sin85^{°}\sin35^{°}\sin25^{°} \\
& =\frac{1}{2}(2\sin85^{°}\sin35^{°})\sin25^{°} \\
& =\frac{1}{2}[ \cos50^{°} -\cos120^{°}] \sin25^{°} \\
& = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +2\cos60^{°} \sin25^{°}] \\
& = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +2 \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
boolean logic simplify To prove:
$(X+Y)(X'+Z) = XZ + X'Y$
I try to simply $(X+Y)(X'+Z)$ to $XZ + X'Y + YZ$
then I have no idea how to simply further.
Thanks in advance!!
| $$\begin{align} (X + Y)(X' + Z) & = (X+Y)X' + (X+Y)Z \\ \\ & = \underbrace{X'X}_{= 0} + X'Y + XZ + YZ \\ \\ & = X'Y + XZ + YZ \\ \\ & = X'Y + XZ + \underbrace{(X+ X')}_{= 1}YZ \\ \\ & = X'Y + XZ + XYZ + X'YZ \\ \\ & = X'Y\underbrace{(1 + Z)}_{= 1} + XZ\underbrace{(1 + Y)}_{= 1} \\ \\ &= X'Y + XZ \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many even numbers of four digits can be formed with the digits 0,1,2,3,4,5 and 6 no digit being used more? My attempt to solve this problem is:
First digit cannot be zero, so the number of choices only $6 (1,2,3,4,5,6)$
The last digit can be pick from $0,2,4,6$, so the number of choices only 4
Second digit can be o... | For an even no. last digit needs to be even = 4 choices
Total 4 digit even numbers = (Total even numbers) - (Total 3 digit even numbers)
Total even numbers: $4\cdot 6\cdot 5\cdot 4 = 480$
Total 3-digit even numbers: $3\cdot 5\cdot 4\cdot 1 = 60$
Total 4-digit even numbers: $480 - 60 = 420$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Inequality....(RMO $1994$...question $8$) If $a$, $b$, $c$ are positive real numbers such that $a+b+c = 1$, prove that
$$(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)\text{.}$$
| Note that by A.M - G.M we have $$ (b+c)+(c+a) \geq 2 \cdot \sqrt{(b+c)\cdot(c+a)}$$ This says $2c+b+a =1+c\geq 2 \cdot \sqrt{(b+c)\cdot (c+a)}$. Similary get inequalities for $(1+b)$ and $(1+c)$ multiply and get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Find the matrix representation of the following linear transformation $T: M_{2x2} (R) \rightarrow P_{2} (R)$ defined by $T \begin{pmatrix} a & b \\ c & d \end{pmatrix}= (a+b) + (2d)x + bx^2$.
Let $\beta = \left \{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix... | Let $V$ a vector space and $\vec{v}$. The $\vec{v}$ representation with respect to a basis $B=\left\langle { \vec{\beta}_1, \cdots, \vec{\beta}_n } \right\rangle$ of $V$ is a colmun vector $$Rep_B(\vec{v})=\left( {\begin{array}{*{20}{c}}
{c_1} \\
\vdots \\
{c_n} \\
\end{array}} \right)
$$
such that $\vec{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve this system of equation? I need to solve the following system of $(x,y)$:
\begin{cases}
3y^3+3x\sqrt{1-x}=5\sqrt{1-x}-2y\\
x^2-y^2\sqrt{1-x}=\sqrt{2y+5}-\sqrt{1-x}
\end{cases}
| Let's resurrect this old question. Two solutions to
\begin{cases}
3y^3+3xz=5z-2y\\
x^2-y^2z=\sqrt{2y+5}-z\end{cases}
are $(x,\,y)= (-3,\,2)$ for $z=\sqrt{1-x}$ and,
$$x = \text{Real root}(7 - x - 3 x^2 + 7 x^3 - 8 x^4 + 2 x^5 - x^6 + x^7=0)$$
$$y = \text{Real root}(2 + y + 2 y^2 + 4 y^3 - 3 y^5 + y^7=0)$$
for $z=-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/513327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$. Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$.
I know that $p$ must, $13, 43, 61, 101,103$.
| Try multiplying both sides by $11 \cdot 5 \cdot 6$, to get $$121 = 11 \cdot 6 + 11 \cdot 5 = 2 \cdot 5 \cdot 6 = 60$$ in $\mathbb{Z}/p\mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Show that the sequence ${a_n}$ converges where $a_n = \sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{n}}}}$ for $n\geq 1$. The original question was to determine whether the sequence converges, but I have checked for extremely high values of $n$ and it seems as though it does converge. This lead me to wonder if there was an "eas... | By repeatedly rationalizing the numerators and using that $a_n\geq1$ for all $n$,
$$
a_{n+1}-a_n=\frac{\sqrt{2+\sqrt{3+\cdots\sqrt{n+1}}}-\sqrt{2+\sqrt{3+\cdots\sqrt{n}}}}{a_{n+1}+a_n}\\
\leq\frac{\sqrt{2+\sqrt{3+\cdots\sqrt{n+1}}}-\sqrt{2+\sqrt{3+\cdots\sqrt{n}}}}{2}\\
=\frac12\,\frac{\sqrt{3+\sqrt{4+\cdots\sqrt{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/516077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
} |
by completing the square find in terms of k the roots of the equation $x^2 + 2kx-7=0$ By completing the square find in terms of $k$ the roots of the equation $$x^2 + 2kx-7=0$$
prove for all real values of $k$, the roots are real
| $$x^2+2kx-7=x^2+2\cdot x\cdot k+k^2-k^2-7=0$$
$$(x+k)^2=k^2+7$$
$$x+k=\pm\sqrt{k^2+7}$$
$$x=-k\pm\sqrt{k^2+7}$$
because $k^2+7\geq 0$ for all real $k$ all roots are real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/516667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The sum of three square roots bounded below by $\sqrt{82}$
Let $a,b,c >0$ and $a+b+c \le 1$. Prove: $$\sqrt{a^2+1/a^2}+\sqrt{b^2+1/b^2}+\sqrt{c^2+1/c^2} \ge \sqrt{82}$$
Progress: I tried to have 3 vectors $(a,1/a)$, $(b,1/b)$ and $(c,1/c)$, played around with the vector inequalities but no success.
| In the first proof only condition $a+b+c=1$ is used. What if $a+b+c<1$? Also a simpler alternative way to show $\frac{1}{a}+ \frac{1}{b}+\frac{1}{c}\geq 9$ when $a+b+c\leq 1$ is the Cauchy-Schwarz inequality:
$9=\|( \sqrt{a}, \sqrt{b}, \sqrt{c})\cdot(\frac{1}{\sqrt{c}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}})\|^2\leq (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/516955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Dervation of $\sum_{k=0}^n(r+1)^k= \frac{(r+1)^{n+1}-1}{r}$ How can one derive the following identity?
$\sum_{k=0}^n(r+1)^k= \frac{(r+1)^{n+1}-1}{r}$
I have playing around with binomial coefficients and index shiftings but wasn't able to get anywhere.
| For $r\neq 1$, the sum of the first n terms of a geometric series is:
$a + ar + a r^2 + a r^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r}$,
where a is the first term of the series, and r is the common ratio. We can derive this formula as follows:
\begin{align}
&\text{Let }s = a + ar + ar^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/517066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How prove this $a+b\le 1+\sqrt{2}$ let $0<c\le b\le 1\le a$, and such $a^2+b^2+c^2=3$, show that
$a+b\le 1+\sqrt{2}$
My try: let $ c^2=3-(a^2+b^2)\le b$
| Set $a^2+b^2=3-c^2=k^2$, then we want to maximize
$$a+b=b+\sqrt{k^2-b^2}$$
Since it is increasing in $k$, and $k^2=3-c^2\le 3$, we achieve maximum at $k=\sqrt{3}$, $c=0$ (!)
Set $b=\sqrt{3}\sin\alpha \le 1$. Therefore
$$0\le \sin(\alpha)\le\frac{1}{\sqrt{3}}<\frac{1}{\sqrt{2}}\implies 0^{\circ}\le\alpha<45^{\circ}$$
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/518990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Correctness of proof of division in ceiling function My professor asked me to present a proof to my fellow students tomorrow that
$$\left\lceil\frac{n}{2^k}\right\rceil = \left\lceil\frac{\left\lceil\frac{n}{2^{k-1}}\right\rceil}{2}\right\rceil$$
I tried to prove it as follows, but I am not sure if my proof is correct.... | Your reasoning is a bit shaky in a few places. Let me see if I can help you a bit with the rigor.
To further simplify, let's say $x=\left\lceil\frac{n}{2^{k+1}}\right\rceil$ and $y=\left\lceil\frac{n}{2^k}\right\rceil,$ so we must prove that $x=\left\lceil\frac{y}2\right\rceil$. Since $\frac{n}{2^k}\le y,$ then $\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/521610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
17539 decimal to binary not getting the same result I'm trying to convert 17539 to binary. My math says its 110000010010001, but online calculators like this and this say it equals to 100010010000011. Who is making something wrong.
| Note that
$$\begin{align}
17539 &= \color{red}1\cdot 2^{14}+ \color{red}0\cdot2^{13} + \color{red}0\cdot2^{12} +\color{red}0\cdot2^{11}+ \color{red}1\cdot 2^{10} \\ &+ \color{red}0\cdot2^{9} + \color{red}0\cdot2^{8} + \color{red}1\cdot 2^7 + \color{red}0\cdot2^{6} + \color{red}0\cdot2^{5} \\&+ \color{red}0\cdot2^{4} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/522730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How do I get the equation for this parabola in standard form? How do I get the equation for this parabola in standard form?
$ y = f(x)= 2x^2+3x-2$
| If you mean vertex form, you can complete the square.
$$\begin{align}f(x) &= 2x^2 + 3x - 2 \\
&= 2\left(x^2+\frac{3}{2}x\right) -2 \\
&= 2\left(x^2+\frac{3}{2}x+\frac{9}{16}-\frac{9}{16}\right) -2\\
&= 2\left(x^2 +\frac{3}{2}x+\frac{9}{16}\right) -2\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/523901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Con... | This is a partial solution.
Let us put, for $0\leq t\leq 1$,
$$F(t) = \int_0^1 \frac{\log x \log(1-tx^4)}{1+x^2} dx$$
Then
$$F'(t) = -\int_0^1 \frac{x^4\log x}{(1+x^2)(1-tx^4)} dx = -\int_0^1 \frac{x^4\log x}{1+x^2} \sum_{n=0}^\infty t^nx^{4n} dx$$
$$=-\sum_{n=0}^\infty t^{n} C_{4(n+1)}$$
where $$C_m = \int_0^1 \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 0
} |
Finding the Maximum and Minimum Distance by Lagrange's Method of Multipliers Q. Find the maximum and minimum distance of a point from origin such that the point lies in the curve $3x^2+4xy+6y^2=140$
I am unable to solve these three equations simultaneously for $(x,y)$
$2x+\lambda(6x+4y)=0$
$2y+\lambda(4x+12y)=0$
$3x^2+... | Another related approach that can be taken makes use of the symmetry of the geometric figure and the function to be extremized. The conic section, although rotated, still retains symmetry about the origin, that is, if the point $ \ (x,y) \ $ lies on the curve, $ \ (-x,-y) \ $ does as well. The distance-from-the-origi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the determinant of this matrix I have the following matrix:
$
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix}
$
My approach is to rref the matrix so that i can find the determinant by multiplying along the diagonals.
I attempted to do an rref and ended u... | To find the determinant you could just do it brute force by "Expansion by Minors" using the first row:
$$\begin{align}
\det \begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix} = a\det
\begin{bmatrix}
a & 1 & 1 \\
1 & a & 1 \\
1 & 1 & a \\
\end{bmatrix} \\- 1\det\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Finding the limit of $x \sin\frac{\pi}{x}$ How can I find the limit of the following
$x\rightarrow\infty$
$x \sin\frac{\pi}{x}$
I did
$\dfrac{\sin\frac{\pi}{x}}{\frac{1}{x}}$
I took the derivative using l hospital and got.
$\dfrac{-1x^{-2} \cos \dfrac{\pi}{x}}{-1x^{-2}}$
Simplying I get
$\cos \frac{\pi}{x}$ but I am s... | For problems like this you can expand sine in terms of the following taylor series:
$$ sin (p) = p - \frac{p^3}{3!} + \frac{p^5}{5!} - \frac{p^7}{7!} + ...$$
Substituting $p = \frac{\pi}{x}$, we get:
$sin(\frac{\pi}{x}) = \frac{\pi}{x} - \frac{\pi^3}{x^3 3!} + \frac{\pi^5}{x^5 3!} - ...$
We are evaluating $f(x) = x \sp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )}$ is decreasing in $y > 1 $. I am interested in the function
$f(y) = \frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )},$
for values of $c \in (0,1)$, and $y > 1$, and have been trying to show that the function is decreasing.
I have tried differentiating the functi... | EDIT
An easier proof:
Note
\begin{eqnarray}
f(y) &=& \frac{(1-y^{-1})^2}{(1-y^{-1-c})(1-y^{-1+c})} \\
&=& \frac{(y^{1/2}-y^{-1/2})^2}{(y^{\frac{1+c}{2}}-y^{-\frac{1+c}{2}})(y^{\frac{1-c}{2}}-y^{-\frac{1-c}{2}})} \\
&=& \left(\frac{\sinh x}{\sinh \ (1+c)x}\right)^2,
\end{eqnarray}
where $x = \log\sqrt{y}$. Thus, it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/529727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?
Could you show me the proof?
| I'll borrow newzad's caclulations and we'll prove that that last equation can't be a square number. We have:
$$a^2 - b^2 = n^2(k^4 - 6k^2 + 1)$$
Because $n^2$ is a square of some natural number $n$ we need to prove that $k^4 - 6k^2 + 1$ is a square number in order RHS to be also a square. First make a substitution $k^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/530003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the limit of $\sum\limits_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$
$$\lim_{n\rightarrow\infty}\sum_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$$
Note that $\forall x\ge 0, \sqrt{x}-1\le\sqrt{1+x}-1\le x$
Then
$$\sum_{k=1}^n\left(\sqrt{\frac{k}{n^2}}-1\right)\le S_n\le \sum_{k=1}^n\frac{k}{n^2}=\frac{1}{2}-... | We have $$S_n=\sum_{1,n} \sqrt{1+k/n^2}-1=\sum_{1,n}\frac{k/n^2}{\sqrt{1+k/n^2}+1}$$
hence
$$\frac{n(n+1)}{2n^2 (\sqrt{1+1/n}+1)}=\sum_{1,n}\frac{k/n^2}{\sqrt{1+n/n^2}+1}\leq S_n\leq \sum_{1,n}\frac{k/n^2}{\sqrt{1}+1}=\frac{n(n+1)}{4n^2}$$
and so $\lim S_n =\frac{1}{4}$ by squeezing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/532404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Use the epsilon-delta definition of limits to evaluate the limit $\lim_{x\to1}\frac{x^2-x-2}{2x-3}$ The Problem: Evaluate the given limit and prove your conclusion using only definitions from the epsilon-delta limit definition: $$\lim_{x \rightarrow1}\frac{x^2-x-2}{2x-3}$$
First I evaluated the limit and found the limi... | Note that $\frac{x^2-x-2}{2x-3} -2 = \frac{1}{2}(1-\frac{5}{2x-3})(x-1)$, so we have $|\frac{x^2-x-2}{2x-3} -2 | = \frac{1}{2}|1-\frac{5}{2x-3}||x-1|$.
The $x \mapsto |1-\frac{5}{2x-3}|$ part looks like
We see that we need to 'stay 'away' from $x=\frac{3}{2}$. If we choose $1-\frac{1}{5} < x < 1+\frac{1}{5}$, we see ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $a,b \in \Bbb R$, prove that $|ab| \le (a^2+b^2)/2$ So far I have the first case when $a=b$:
\begin{align*}
|ab|
&= |b^2|\\
&=|b|^2\\
&=\frac{2|b|^2}2\\
&=\frac{b^2+b^2}2\\
&=\frac{a^2+b^2}2
\end{align*}
Case 2: $a>b$
Case 3 $a<b$
I've been stuck on this problem for a few hours now and don't know how to proceed. Am ... | If $ab\ge0$ then $|ab|=ab\le ab+\dfrac{(a-b)^2}2=\dfrac{a^2+b^2}2$.
If $ab\le0$ then $|ab|=-ab\le-ab+\dfrac{(a+b)^2}2=\dfrac{a^2+b^2}2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/535194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Matrix Equation $A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}$ How can I solve in $\mathcal{M}_{2}(\mathbb{Z})$ the equation $$A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}?$$
I try to use $$A^2-Tr(A)A+detA\cdot I_2=O_2$$ but I don't still obtain anything.
thanks.
| Let $t=\operatorname{trace}(A)$ and $d=\det(A)$. By Cayley-Hamilton theorem, $A^2 = tA - dI$ and hence
$$
\pmatrix{-7&-9\\ 3&2} = A^3 - 3A = tA^2 - (d+3)A = (t^2-d-3)A - tdI.\tag{1}
$$
Taking traces on both sides, we get $-5 = (t^2-d-3)t - 2td$. Hence $t$ divides $5$, i.e. $t=1,-1,5,-5$, and we can express $d$ in terms... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Prove : $\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$ $a;b;c>0$ such that $a^2+b^2+c^2=\frac{5}{3}$. Prove :
$\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$
| Hint: $(a+b-c)^2 \ge 0 \implies (a^2+b^2+c^2) \ge 2(bc + ca - ab)$
Can you relate both sides to the question at hand?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/538234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrals of trignometric functions Question is to Prove that :
$$\int_0 ^{2\pi} \frac{d\theta}{a+b\sin \theta}=\frac{2\pi}{\sqrt{a^2-b^2}} \text{for}\ a>b>0$$
using residue theory.
What i have done so far is :
I transformed functio of $\theta$ as function of complex entity $z$ with $z=e^{i\theta}$.
Then, we have : $d... | Consider your roots:
$$z_{\pm} = -i \left [ \frac{a}{b} \pm \sqrt{\left ( \frac{a}{b}\right)^2-1}\right]$$
Note that, when $a>b$, $a/b > 1$. Therefore $|z_+| > 1$. Note also that
$$|z_-| = \frac{a}{b} - \sqrt{\left ( \frac{a}{b}\right)^2-1} = \frac{1}{\frac{a}{b} + \sqrt{\left ( \frac{a}{b}\right)^2-1}} = \frac{1}{|z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is $\int \frac1{1+(a\tan x)^2}dx$? What is $\int \frac1{1+(a\tan x)^2} \mathrm dx$?
This is a difficult integral. If you can, please give a step-by-step solution - I would be delighted.
| It's not that hard if you take the suggestion from Clayton:
$u=a \tan{x}$; $x = \arctan{(u/a)}$; $dx=a\, du/(u^2+a^2)$. Then the integral is
$$a \int \frac{du}{(u^2+a^2)(u^2+1)} $$
Assume $a \ne 1$. Use partial fractions:
$$\frac{1}{u^2+a^2}-\frac{1}{u^2+1} = \frac{1-a^2}{(u^2+a^2)(u^2+1)}$$
so the integral becomes
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding element in binary representation of $GF(2^6)$ I got the following task:
Let $F = GF(2^6
)$ be K[x] modulo the primitive polynomial $h(x) = 1 +x
^2 +x
^3 +x
^5 +x
^6$
, and let $\alpha$ be the
class of x. I have a table with the binary representation of the elements of $GF(2^6
) = \{0; 1; \alpha^1; \alpha^2
; ..... | You say you have $\alpha^{13}$, so you can substitute suitable values into $\alpha^{13} = b_0 + b_1 \alpha + \ldots + b_5 \alpha^5$. Then $$\begin{eqnarray}\alpha^{26} & = & (\alpha^{13})^2 \\
& = &(b_0 + b_1 \alpha + \ldots + b_5 \alpha^5)^2 \\
& = &b_0 + b_1\alpha^2 + \ldots + b_5 \alpha^{10}
\end{eqnarray}$$
where t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How is $\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$ How is $\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$?
(Prove by using algebraic manipulation not by calculation)
I've tried to come up with something myself but I can't find a solution, I must be missing something.
| $$\begin{align}\sqrt{\frac{\sqrt{3}+2}{4}} &= \frac{\sqrt{2}(\sqrt{3}+1)}{4}\\
{\frac{\sqrt{3}+2}{4}} &= \frac{{2}(\sqrt{3}+1)^2}{16}\\
{\frac{\sqrt{3}+2}{4}} &= \frac{4+2\sqrt{3}}{8}\\
{\frac{\sqrt{3}+2}{4}} &= \frac{\sqrt{3} + 2}{4}\tag*{$\blacksquare$}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/540233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Derivative of trigonometric function How i can find the derivative of this trigonometric function
$csc^4(8x^4-5)$
i tried to do it my self and i got to this
$ 4[csc(8x^4-5)]^3 * [-csc(8x^4-5)cotan(8x^4-5)] $
The answer in the book is $ -128 x^3csc^4(8x^4-5)cot an(8x^4-5)$
| $$\csc^4(8x^4-5)=\frac{1}{\sin^4(8x^4-5)}=-\frac{(\sin^4(8x^4-5))'}{\sin^8(8x^4-5)}=-\frac{4\sin^3(8x^4-5)\cdot (\sin(8x^4-5))'}{\sin^8(8x^4-5)}$$
$$=-\frac{4\sin^3(8x^4-5)\cdot \cos(8x^4-5)(8x^4-5)'}{\sin^8(8x^4-5)}=-\frac{4\sin^3(8x^4-5)\cdot \cos(8x^4-5)\cdot 32x^3}{\sin^8(8x^4-5)}$$
$$=-\frac{128 x^3\cdot \cos(8x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Error analysis of exponential function By definition: $$ e^x = \lim_{n \rightarrow \infty} ( 1 + \frac{x}{n} ) ^ n$$
I am interesting in calculating the error $$\left | e^x - \left( 1 + \frac{x}{n} \right) ^ n \right|$$ for some fixed $n \in \mathbb{N}$.
What I've done so far:
First note:
$$e^x = \sum_{k = 0}^\infty \f... | I take a simple-minded approach: whenever practical, write the Taylor formula with Lagrange form of remainder. Since $(1+x/n)^n=\exp(n \ln(1+x/n))$, I begin with the logarithm:
$$
\ln(1+x/n) = \frac{x}{n}+\frac{(x/n)^2}{2}\frac{-1}{(1+\xi)^2}
\tag{1}$$
where $\xi$ is between $0$ and $x/n$. Formula (1) is Taylor's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Conjecture $\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi$ $$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$
The equality numerically holds up to at least $10^4$ decimal digit... | Here is another approach for evaluating the integral (3).
Using @achille hui's transformation, we can write
$$
I=\int_0^1\frac{dx}{\sqrt[3]{x}\sqrt{1-x}\sqrt[6]{9-x}}= \frac{4\pi^2 2^{\frac{1}{3}}}{\Gamma\left(\frac{1}{3}\right)^3}\int_0^\frac{1}{\sqrt{2}}\frac{1}{\sqrt{1+x^6}}dx
$$
Using the substitution $x=\sqrt{t}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/541135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "91",
"answer_count": 2,
"answer_id": 1
} |
How to derive this second derivative using the quotient rule? If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $
What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $
My Steps:
\begin{align*}
{d^2y \over dx^2}
&= {-4... | First, make sure you know what the quotient rule is - which you can derive from the product rule say by taking the derivative of $g(x)/f(x)$.
Then, simplify. Looking at the product rule, expect to have to cancel a factor of $x^2+12$ from the top and bottom.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/541696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Polar coordinates differential equation I have the following ODE:
$$\dot x=-y(x^2+y^2), \dot y=x(x^2+y^2)$$
I want to sketch the phase portrait (manually) and I want to find the flow $\phi_t$, the orbit $O(x_0)$ and the limit set $\omega(x_0)$
I start by taking polar coordinates and change the system to $\dot r=-r^3\si... | We are given:
$$\dot x=-y(x^2+y^2), \dot y=x(x^2+y^2)$$
Recall, when we are doing polar coordinates, we have
$$x = r \cos \theta, y = r \sin \theta, x^2+y^2 = r^2$$
When we differentiate this, we have:
$$2 x x' + 2 y y' = 2 r r'$$
This gives us:
$$rr' = x(-y(x^2+y^2)) + y(x(x^2+y^2)) = 0 \rightarrow r' = 0$$
To find t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/541889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 1,
"answer_id": 0
} |
How find the minimum of the $abc$ let $a,b,c>2$ be prime numbers,such
$$a|b^5+1,b|c^5+1,c|a^5+1$$
Find the minimum of $abc$
My try:
I think use
$$a^{p-1}=1(modp)$$
But I don't,Thank you for your help.
| Note that $a, b, c$ are pairwise relatively prime and thus pairwise distinct.
If $\min(a, b, c) \geq 7$, WLOG assume that $a=\max(a, b, c) \geq 13$, so $a>b \pm 1$. Now $b^5 \equiv -1 \pmod{a}, b^{10} \equiv 1 \pmod{a}, b \not \equiv \pm 1 \pmod{a}, b^2 \not \equiv 1 \pmod{a}$. Thus the order of $b \pmod{a}$ is $10$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate $a^3+b^3+c^3+d^3$ for the real roots of $x^4+2x^3-3x^2-3x+2$ $a,b,c,d \in \mathbb{R}$ are the real roots of $x^4+2x^3-3x^2-3x+2$.
Calculate $a^3+b^3+c^3+d^3$.
With approximation i found out, that $a^3+b^3+c^3+d^3 = -17$, but how can I proof that without calculating the roots exactly?
Cheers
| As biquadratic equation has exactly $4$ roots, all of $a,b,c,d$ are real
Clearly, $abcd\ne0$
and $\displaystyle a^4+2a^3-3a^2-3a+2=0\implies a^3+2a^2-3a-3=-\frac2a$
Similarly, for $b,c,d$
Summing we get $\sum a^3+2\sum a^2-3\sum a-3\sum 1=-2\sum \frac 1a$
Now, we need
$\sum a$
$\displaystyle\sum\frac1a=\frac{\sum abc}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove that $C_n < 4n^2$ for all n greater than or equal to 1 $C_1 = 0$, $C_n = C_{\lfloor n/2\rfloor} + n^2$ for all $n \ge 1$
Prove that $C_n < 4n^2$ for all $n \ge 1$.
I don't know how to even approach this. I remember something about inductive proofs...but i really don't understand that, could you please explain tha... | Why not unroll the recursion to get a precise answer?
Let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
be the binary representation of $n.$
This yields the exact formula
$$C_n = \sum_{k=0}^{\lfloor \log_2 n \rfloor - 1}
\left(\sum_{j=k}^{\lfloor \log_2 n \rfloor} d_j 2^{j-k}\right)^2$$
where $n\ge 2$ and zer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/546068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Trig equations solution Solving the following for x:
$$
\frac{3\cos(2x)+5\cos(x)-1}{\sqrt{-\cot(x)}}=0
$$
The solution says that the answer is $x=-\frac{\pi}{3}+2\pi k$ where $k$ is an integer.
I am not sure why there is the minus sign in front. Can someone please help me out?
| First off, there is a domain issue with the function on the left-hand side of the equation, since the denominator is undefined wherever $ \ \cot x \ \ge \ 0 \ . $ Since the tangent function is positive in the first and third quadrants, so is the cotangent function; also, $ \ \cot x \ = \ 0 \ $ wherever $ \ \cos x \ = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/547532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find $f$ if $f(f(x))=\frac{x+1}{x+2}$ let $f:\mathbb R\to \mathbb R$,and such
$$f(f(x))=\dfrac{x+1}{x+2}$$
Find the $f(x)$
My try
I found $f(x)=\dfrac{1}{x+1}$
because when $f(x)=\dfrac{1}{x+1}$,then
$$f(f(x))=f\left(\dfrac{1}{x+1}\right)=\dfrac{1}{\dfrac{1}{x+1}+1}=\dfrac{x+1}{x+2}$$
so $f(x)=\dfrac{1}... | Suppose $f(x)=\frac{ax+b}{x+c}$. Then
$$
f(f(x))= \frac{(a^2+b)x+b(x+c)}{(a+c)x + b+c^2}=\frac{x+1}{x+2}
$$
yields the equations
$$
\frac{a^2+b}{a+c}=1, \frac{b(a+c)}{a+c}=b=1, \frac{b+c^2}{a+c}=2, \frac{a^2+1}{a+c}=1$$
from which we find
$$c=a^2-a+1$$
and hence
$$a^4-2a^3+a^2-2a=a(a^3-2a^2+a-2)=0$$
This last equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/549897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 6,
"answer_id": 4
} |
Evaluation of $\int^1_0 \cos^2\frac{(m+n)\pi x}{2}\sin^2\frac{(n-m)\pi x}{2}dx$. Just wondering whether the following integration is something special. To be more specific, is it equal to some constant real number, please? I found a integration table involving sine and cosine but it requires that sine and cosine have t... | First notice that $$\cos\frac{(m+n)\pi x}{2}\sin \frac{(n-m)\pi x}{2} = \frac{1}{2} \left( \sin n\pi x -\sin m\pi x\right)$$
From this, we deduce that the given integral formula is equal to
$$\frac{1}{4}\int_0^1 {\left( \sin n\pi x -\sin m\pi x\right)}^2dx \\= \frac{1}{4}\int_0^1 { \sin^2 n\pi x +\sin^2 m\pi x - \sin n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Function range problem $(sin^2\theta +\sin\theta -1)/(sin^2\theta -\sin\theta+ 2)$ .
It is asked to find the range of this function.
Here i was assumed a variable
$z=sin\theta$ so that i get
$-1<=z<=1 $
and then i obtained a quadratic equation as :-
$(y-1)z^2 -(y+1)z + 2y +1 =0$ from here i need some help or if ther... | Since $z^2 - z + 2 = (z - 1/2)^2 + 7/4$, the given function's domain is $\mathbb{R}$.
$$f(z) = \frac{z^2 + z - 1}{z^2 - z + 2} = 1 + \frac{2z- 3}{z^2 - z + 2}$$
Let us differentiate the function.
$$f'(z) = \frac{(2z - 3)' (z^2 - z + 2) - (2z - 3) (z^2 - z + 2)'}{(z^2 - z + 2)^2} \\= \frac{2z^2 - 2z + 4 - (2z - 3)(2z - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve for $\sin^2(x) = 3\cos^2(x)$ I am trying to solve the following equation for x. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm going wrong? Notice that this is "sin squared x" and 3 * "cos squared x"
$\sin^2x = 3\cos^2x$ //Just rewriting the eq... | we know that $\sin^2x=3\cos^2x$ so$$1-\cos^2x=3\cos^2x \Rightarrow 4\cos^2x=1 \Rightarrow \cos^2x=\frac{1}{4} \Rightarrow \cos x=\pm \frac{1}{2}$$
and now we calculate values of $x$
$$\mbox{if} \qquad \cos x=\frac{1}{2},\mbox{ then}\qquad x=\arccos\left(\frac{1}{2}\right)=\frac{\pi}{3},\frac{5\pi}{3}$$
$$\mbox{if} \qqu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$ are linearly dependent.
Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$
let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that
$\sin(x + \alpha)$, $\sin(x +... | As Mohamed pointed out in his answer, we can use the angle addition formula for $\sin$, viz.:
$\sin (x + \alpha) = \cos \alpha \sin x + \sin \alpha \cos x, \tag{1}$
$\sin (x + \beta) = \cos \beta \sin x + \sin \beta \cos x, \tag{2}$
$\sin (x + \gamma) = \cos \gamma \sin x + \sin \gamma \cos x, \tag{3}$
to see that all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/562034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Integral $\int_0^1\frac{\ln x}{\left(1+x\right)\left(1+x^{-\left(2+\sqrt3\right)}\right)}dx$ There is a curious known integral:
$$\int_0^1\frac{\ln\left(1+x^{2+\sqrt{3\vphantom{\large3}}}\right)}{1+x}dx=\frac{\pi^2}{12}\left(1-\sqrt{3\vphantom{\large3}}\right)+\ln \left(1+\sqrt{3\vphantom{\large3}}\right)\ln2.$$
If we ... | Here is a partial progress report. I am basically repeating Jim Belk's analysis from the previous answer.
Set $F(a) = \int_{x=0}^1 \frac{\log(1+x^a)}{1+x} dx$. Then
$$F(a) = \int_{x=0}^1 \int_{y=0}^{x^a} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)}$$
so
$$F(a) + F(a^{-1}) = \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/563204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 2
} |
How to prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$? Question:
If $a,b,c$ are nonnegative real numbers such that $a+b+c=3,$ then
$$(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$$
My try: I found the equality holds only if $(a,b,c)=(2,0,1)$ or all of its permutations.
But I can't prove this inequality it. ... | Without loss of generality, assume $a$ is smallest of $a, b, c$. Also, let $$f(a,b,c)=(a^2+bc^4)(b^2+ca^4)(c^2+ab^4)$$and firstly, if $a\le c\le b$, then$$f(a,b,c)-f(a,c,b)=(b^3-a^3) (c^3 - a^3) (b ^3- c^3) (a b c - 1)<0$$therefore we can assume $a\le b\le c$. Now, we will prove $$f(0,b,a+c)\ge f(a,b,c)$$which is, afte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/565217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 2,
"answer_id": 1
} |
Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$. Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$.
How to prove inequality
$$
ab^2+bc^2+ca^2\le 4.\tag{*}
$$
In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality
$$
27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**}
$$
$\color{gray}... | Let $a=3x, b=3y, c=3z$ and $f(x,y,z)=x^2y+y^2 z +z^2 x$
Without loss of generality let's assume that $x= \max { (x,y,z)}$
Then $f(x+\frac{z}{2}, y+\frac{z}{2}, 0) - f(x,y,z)=yz(x-y)+\frac{xz}{2}(x-z) + \frac{z^3}{8} + \frac{z^2y}{4} \ge 0$
And $f(x,y,0)= x^2y = 4 \frac{x}{2} \frac{x}{2} y \le 4\left( \frac{x+y}{3} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/566768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
If $x\le c\le y$ and $|x-y|<\delta$, then $|\frac{g(x)-g(y)}{x-y}-g'(c)|<\epsilon$ Alright, so here goes the question:
Consider a function $g$ that is differentiable at some point $c$ on an interval $[a,b]$.
Prove that: $\forall$$\epsilon$ $>0$ $\exists$$\delta$ $> 0$ such that when $a\leq x \leq c \leq y \leq b$, and... | By the definition of the derivative, for every $\varepsilon > 0$, there is a $\delta > 0$ such that for all $z \in [a,b]$ with $0 < \lvert z-c\rvert < \delta$, we have
$$\left\lvert\frac{g(z)-g(c)}{z-c} -g'(c)\right\rvert < \varepsilon.\tag{1}$$
Now if $c < y < c+\delta$ and $c-\delta < x < c$, we have by $(1)$
$$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/569455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
On the equation $(1-x)^2/x + (1-y)^2/y + (1-z)^2/z + 4 = 0$ The problem is to solve the equation,
$$\frac{(1-x)^2}{x} + \frac{(1-y)^2}{y} + \frac{(1-z)^2}{z} + 4 = 0\tag{1}$$
in the rationals. Treating this as an equation in $z$, easy solutions would involve $z = \pm 1$, $z = \pm x, \pm y$. More complicated ones would ... | Consider the problem written as
\begin{equation*}
\frac{(1-f)^2}{f}+\frac{(1-g)^2}{g}+\frac{(1-h)^2}{h}+4=0
\end{equation*}
which gives the quadratic in $h$
\begin{equation*}
h^2+\frac{f^2g+f(g-1)^2+g}{f\,g}+1=0
\end{equation*}
As Tito stated, for this to have a rational solution (assuming $f,g$ rational) the discrimin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/569536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $ \sum\limits_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}$ The question was: Evaluate, ${\textstyle {\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}}.$
And I go, since $\frac{n}{n^{4}+n^{2}+1}\sim\frac{1}{n^{3}}$ and we know that ${\displaystyle \sum_{n=}^{\infty}\frac{1}{n^{3}}}$ converges. so ${\displays... | $\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
How to prove $\sum_{n=1}^\infty\operatorname{arccot}\frac{\sqrt[2^n]2+\cos\frac\pi{2^n}}{\sin\frac\pi{2^n}}=\operatorname{arccot}\frac{\ln2}\pi$? How can I prove the following identity?
$$\sum_{n=1}^\infty\operatorname{arccot}\frac{\sqrt[2^n]2+\cos\frac\pi{2^n}}{\sin\frac\pi{2^n}}=\operatorname{arccot}\frac{\ln2}\pi$$
| Rewrite the sum as
$$\sum_{n=1}^{\infty} \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}+\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$
Now let
$$a_n = \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$
Then one may show (nontrivia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/577557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 1,
"answer_id": 0
} |
Proving for every odd number $x$, $x^2$ is always congruent to $1$ or $9$ modulo $24$ A problem I have been presented with asks the following:
Prove for every odd number $x$, $ x^2$ is always congruent to $1$ or $9$ modulo $24$.
This seems odd and non-intuitive to me. Of course, it must be true other wise they wouldn't... | Every number is equivalent mod 24 to one of 0, 1, 2, 3 ,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, or 23.
But our number $n$ is odd, so it is equivalent mod 24 to one of 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23.
Then:
$$\begin{array}{r|r|l}
n & n^2 & n^2\pmod{24} \\\hline
1 & 1 & 1 \\
3 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
How find this Fibonacci sequence sum $\sum_{k=0}^{\infty}\frac{1}{F_{2^k}}$ let sequence $\{F_{n}\}$ such $$F_{1}=1,F_{2}=1,F_{m+1}=F_{m}+F_{m-1},m\ge 2$$
Find this value
$$I=\sum_{k=0}^{\infty}\dfrac{1}{F_{2^k}}$$
My try: I know this
$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{\sqrt{5}+1}{2}\right)^n-\left(\dfr... | let
$$a=\dfrac{1+\sqrt{5}}{2}\Longrightarrow \dfrac{\sqrt{5}-1}{2}=a^{-1}$$
so
$$I=\sum_{n=0}^{\infty}\dfrac{\sqrt{5}}{a^{2^n}-a^{-2^n}}$$
so let $$a^{2^n}=x$$
then
$$\dfrac{1}{a^{2^n}-a^{-2^n}}=\dfrac{x}{x^2-1}=\dfrac{1}{x-1}-\dfrac{1}{x^2-1}=\dfrac{1}{a^{2^n}-1}-\dfrac{1}{a^{2^{n+1}}-1}$$
so
\begin{align*}I&=\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
new $\arctan$ series working for any $x$? Using Log series, we can write:
$$
\frac{1}{2 i} \left( \text{Log} \left( 1 + e^{\text{i2t}} \right) - \text{Log} \left( 1 + e^{\text{-i2t}} \right) \right)
=
\frac {1} {2 i}\left ( \sum _ {k = 1}^{\infty}\frac {(-1)^{k + 1}} {k} e^{\text{i2t}^k} - \sum _ {k = 1}^{\infty}\frac ... | I do not know whether your series for $\arctan$ is new (it may be), but when evaluating its advantages, you should consider alternatives other than the classical McLaurin series. For example, the identity $\tan 2\theta=\frac{2\tan \theta}{1-\tan^2\theta}$ can be turned into $$\arctan x = 2\arctan \frac{x}{1+\sqrt{1+x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/579007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral involving logarithm and cosine For $a,b>0$ I would like to compute the integral $$I=\int_0^{2\pi} -\log{\sqrt{a^2+b^2-2ab\cos{t}}}~dt.$$
Numerical computations suggest that $$I=\min\{-\log{a},-\log{b}\}.$$
How can I prove this? I tried to find the antiderivative using Mathematica, but the result looks awfull. ... | We have
$$I=\dfrac12\int_0^{2\pi} -\log(a^2+b^2-2ab\cos{t})dt=\int_0^{\pi} -\log(a^2+b^2-2ab\cos{t})dt$$
We then have
$$I=-2\pi\log(b) -\int_0^{\pi} \log((a/b)^2+1-2(a/b)\cos{t})dt$$
Let us call $a/b$ as $a$ from now on. Hence, we want to evaluate the integral
$$I(a) = \displaystyle \int_0^{\pi} \ln \left(1-2a \cos(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/579354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If the roots of $ax^3 - x^2 + bx - 1$ are real and positive, prove $b \geq \sqrt{3}$ Let
$$p(x) = ax^3 - x^2 + bx - 1$$
such that $a, b \in \mathbb{R}$ and the roots of $p(x)$ are real and positive.
Prove that
$$b \geq \sqrt{3}$$
Let $\alpha, \beta, \gamma \in \mathbb{R^{+}}$ be the roots of $p(x)$
$$\alpha + \beta +... | Everything you have derived is true, but it's not enough to finish the problem. Here is one way to do it:
We have $\sigma_2^2 = \alpha^2\beta^2 + \beta^2\gamma^2 +\alpha^2\gamma^2 + 2(\alpha+\beta+\gamma)(\alpha\beta\gamma) = \alpha^2\beta^2 + \beta^2\gamma^2 +\alpha^2\gamma^2 + 2\sigma_1\sigma_3$. Now we have that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ let $a,b,c\in \mathbb{R}$, if such
$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$
Does this problem has nice methods?
My idea:let
$$(ca-b^2)(... | One approach uses factoring. Define rational functions
$$ R_1 := \frac1{b c-a a} + \frac1{c a-b b} + \frac1{a b-c c}, \tag{1}$$
$$ R_2 := \frac{a}{(b c-a a)^2} + \frac{b}{(c a-b b)^2} + \frac{c}{(a b-c c)^2}, \tag{2}$$
and polynomials
$$ P_1:=a+b+c,\qquad \qquad P_2:=a b+b c+c a,\\ P_3:=P_2\!-\!a a\!-\!b b\!-\!c c,\qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculating limit of function To find limit of $\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4} $.
I differentiated it using L Hospital's rule. I got
$$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}$$ I divided and multiplied by $\sin x$.
Since $\lim_{x\to 0}\frac{\sin x}{x} = 1$, thus I got
$\frac{1-\cos x}{4x^2}$.... | Without L'Hôpital
The expansions $\sin(x) = x - \frac{x^3}{6} + O(x^5)$ and $\cos(u) = 1 - \frac{u^2}{2} + \frac{u^4}{24} + O(u^6)$ combine joyfully to give
$$
\cos(\sin x) - \cos(x) = \left(1 - \frac{x^2}{2} + \frac{5x^4}{24}\right) - \left(1-\frac{x^2}{2} + \frac{x^4}{24}\right) + O(x^5)
$$
so finally,
$$
\lim_{x\to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Can $a^2+b+2$ and $b^2+4a$ both be perfect squares? Are there any positive integers $a$ and $b$ so that $a^2+b+2$ and $b^2+4a$ are both perfect squares?
| No.
$b^2+4a$ is a perfect square $>b^2$, and $b^2+4a \not =(b+1)^2$ (fails by checking parity), so $b^2+4a \geq (b+2)^2$ so $a \geq b+1$
$a^2+b+2$ is a perfect square $>a^2$, so $a^2+b+2 \geq (a+1)^2$ so $b \geq 2a-1$.
Combining, $b \geq 2a-1 \geq 2(b+1)-1=2b+1>b$, a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/583550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Convex Quadrilateral: $ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D $ Problem
Let $ABCD$ be a convex quadrilateral with no right angles. Show that $$ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot... | Multiply the original identity by $\tan A \tan B \tan C \tan D$
$$
\tan A + \tan B + \tan C + \tan D = \tan B \tan C \tan D + \tan A \tan C \tan D + \tan A \tan B \tan D + \tan A \tan B \tan C.
$$
then find $\tan D$ from it
$$
\tan D = \frac{\tan A \tan B \tan C - \tan A - \tan B - \tan C}{1 - \tan B \tan C - \tan A \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
For how many integers $a$ is $\frac{2^{10} \cdot 3 ^8 \cdot 5^6}{a^4}$ an integer? In Mathleague $11316$ Target #$4$, the question is:
For how many integers $a$ is $$\frac{2^{10} \cdot 3 ^8 \cdot 5^6}{a^4}$$ an integer?
| Clearly the number
$$ q =\frac{2^{10} 3^8 5^6 }{a^4} = \left( \frac{2^{5} 3^4 5^3 }{a^2} \right)^2
$$
is a perfect square, and so
$$
\sqrt{q} = \frac{2^{5} 3^4 5^3 }{a^2}
$$ is rational. Now in order for this to be an integer, the denominator must divide the numerator. Therefore $a^2$ must divide $2^53^45^3$. Now th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\c... | I am not sure how you can continue either (the second term in the denominator can be expressed as $1-\sin(2 x)/2 - \sin^2(2x)/4,$ but I am not aware of any double angle formula for $\sin x + \cos x.$ The simplest approach to your integral is to use the feared $u = \tan \frac{x}2$ substitution, which reduces the integra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/595038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 3
} |
Show that $\frac{\sqrt[n]a}{\sqrt[n]{ab}+\sqrt[n]a+1}+\frac{\sqrt[n]b}{\sqrt[n]{bc}+\sqrt[n]b+1}+\frac{\sqrt[n]c}{\sqrt[n]{ac}+\sqrt[n]c+1}=1$ If $$\sqrt[n]{{abc}} = 1,$$
Prove that $$\frac{\sqrt[n]a}{\sqrt[n]{ab}+\sqrt[n]a+1}+\frac{\sqrt[n]b}{\sqrt[n]{bc}+\sqrt[n]b+1}+\frac{\sqrt[n]c}{\sqrt[n]{ac}+\sqrt[n]c+1}=1.$$
| Let $x = \sqrt[n]a, y = \sqrt[n]b, z = \sqrt[n]c$. Then you have $xyz = 1$ and
$$\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1} = \frac{x}{xy+x+1}+\frac{xy}{1+xy+x}+\frac{1}{x+1+xy}$$
where we multiplied the second term's numerator and denominator by $x$ and the third term's by $xy$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/596854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric inequality to prove Prove the following inequality for each $x$:
$$|\sin x+7\cos x| \leq \sqrt{50}$$
My remark: It seems to me that the solution uses the fact that $1^2+7^2=50$.
Thanks in advance!
| One of the ways is to use the addition formula: $R \sin (a+x) = (R\cos a) \sin x + (R\sin a) \cos x$. We let $R \cos a = 1$ and $R \sin a = 7$. Hence $50 = 1^2 + 7^2 = R^2 (\cos^2 + \sin^2) = R^2$ and $R = \sqrt{50}$.
$|\sin x + 7\cos x| = |\sqrt{50} \sin (a+x)| \leq \sqrt{50}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/598041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta$ If $n$ is a positive integer find
$$
\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta
$$
I know that I have to use contour integral with a circle of radius 1 centered at the... | I don't know if it will be too late for the answer. What you need is to add $i\sin(n\theta)$ to the integral. Clearly
$$ \int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\sin(n\theta)}{3+2\cos\theta} \operatorname{d}\theta=0. $$
Noting for $z=e^{i\theta}$, ones
$$ 1+2\cos\theta=|1+z|^2-1=(1+z)(1+\frac1z)-1=1+z+\frac1z, $$
and
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/599834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Limit $ \lim_{n \to \infty} \sum_{k=1}^n\frac 1 { \sqrt{n^2+k} } $ (1)
$$ \lim_{n \to \infty} \sum_{k=1}^n \frac 1 { \sqrt{n^2+k} } $$
(2)
$$ \lim_{n\to\infty} \frac {1+\sqrt[n]2 + \sqrt[n]3 + ... \sqrt[n]n} {n} $$
The answers should both be 1.. any hints?
| For the first problem, note that if $1\le k\le n$, then
$$n^2\lt n^2+k\le n^2+n\lt\left(n+\frac{1}{2}\right)^2.$$
Thus
$$\frac{1}{n+\frac{1}{2}}\le \frac{1}{\sqrt{n^2+k}}\lt \frac{1}{n}.$$
Our sum is therefore between $\frac{n}{n+\frac{1}{2}}$ and $1$.
Squeeze.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/600510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric Integral : $\int\frac{1}{\sin x+ 3\cos x}dx$ I would appreciate if somebody could help me with the following problem
Q: How to integrate this integral
$$\int\frac{1}{\sin x+ 3\cos x}dx$$
| \begin{align*}
\int \frac 1 {\sin x +3\cos x}\ dx &= \int \frac 1 {2\sin \frac x 2\cos \frac x 2 +3\cos^2 \frac x 2 - 3 \sin^2 \frac x 2}\ dx \\
&= \int \frac 1 {\cos^2 \frac x 2} \cdot \frac 1 {3 + 2\tan \frac x 2 - 3 \tan^2 \frac x 2}\ dx
\end{align*}
Switching $y=\tan \frac x 2 \implies\frac {dy}{dx} = \frac 1 2 \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/600754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$... | Let $x_1 = \sqrt{2}$ and define $x_{n+1} = \sqrt{2 x_n}$, then it suffices to show
that $\lim_n x_n = 2$. In order to achieve this goal show that the sequence
$x_n$ is monotonically increasing and bounded above (I will leave this for you to do).
Then the limit exists so let $x = \lim x_n$. Then, using $x_{n+1} = \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/601045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Limit of series $4\left( \frac {1}{8}+\frac {1}{12}\right) +6\left( \frac {1}{24}+\frac {1}{36}\right) +\ldots$ How to find this serie
$4\left( \dfrac {1}{8}+\dfrac {1}{12}\right) +6\left( \dfrac {1}{24}+\dfrac {1}{36}\right) +8\left( \dfrac {1}{64}+\dfrac {1}{96}\right) +\ldots $
I think it's telescopic, isn't it?
| The $n$th term appears to be $$2(n+1)\left(\frac1{2^{n+1}(n+1)}+ \frac1{2^{n}3(n+1)}\right).$$
Canceling factors of $2(n+1)$, we can rewrite the $n$th term as
$$\left(\frac{1}{2^n} + \frac{1}{2^{n-1}{3}}\right) = \left(\frac{3}{2^n3} + \frac{2}{2^{n}{3}}\right) = \left(\frac{5}{2^n3}\right)$$
So the series is $$\sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/604012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How find this $I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4})^{-\frac{1}{2}}dS$ Find this Surface integral
$$I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}dS$$
where
$$\Sigma:\dfrac{x^2}{a^2}+\dfra... | Let $\vec{r} = (x,y,z)$ and $\varphi(\vec{r}) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}$, we have
$$\vec{\nabla}\varphi(\vec{r}) = (\frac{2x}{a^2}, \frac{2y}{b^2},\frac{2z}{c^2})
\quad\implies\quad
\begin{cases}
\vec{r}\cdot \vec{\nabla}\varphi(\vec{r}) &= 2 \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/604085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$? Which is bigger, $a = 1+3\sqrt2$ or $b = 3\sqrt3$?
To find out result, I am doing: $(3\sqrt3)^2-(1+3\sqrt2)^2=8-2\sqrt{18}$.
But how can I find if the value of $8-2\sqrt{18}$ is positive or negative?
Thank you.
| HINT:
Note that $(3\sqrt3)^2-(1+3\sqrt2)^2=27-1-6\sqrt2-18=8-6\sqrt2$. Next note that $\sqrt2>\frac86$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/607061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
How find this equation solve this equation
$$\sqrt{\sqrt{3}-\sqrt{\sqrt{3}+x}}=x$$
My try: since
$$\sqrt{3}-x^2=\sqrt{\sqrt{3}+x}$$
then
$$(x^2-\sqrt{3})^2=x+\sqrt{3}$$
| Let $y = \sqrt{\sqrt3 + x}$, then the equation becomes: $\sqrt3 - y = x^2$, and $y^2 = \sqrt3 + x$.
So:
$$\begin{align*}
y^2 - x & = y + x^2 \\
\Rightarrow (y + x)(y - x - 1) & = 0
\end{align*}$$
And we can go from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/608875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that the sum of the series Show that the sum of the series is greater than 24
$$\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{9}+\sqrt{11}} +\cdots+\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
I see that
$\frac{1}{\sqrt{3}+\sqrt{1}}\cdot \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-\sqrt{1}}=\frac{\sqrt... | This was not far from being telescopic. So let us make this telescopic.
$$4S> \frac{2}{\sqrt{1}+\sqrt{3}}+ \frac{2}{\sqrt{3}+\sqrt{5}}+\frac{2}{\sqrt{5}+\sqrt{7}} +...+ \frac{2}{\sqrt{9997}+\sqrt{9999}}+ \frac{2}{\sqrt{9999}+\sqrt{10001}}$$
$$
=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\ldots+\sqrt{9999}-\sqrt{9997}+\sqrt{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find the solution of the equation Find all real solutions of this equation :
$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
| The function
$$f(x)=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
is non-negative and decreasing on its domain, $-2\le x\le2$. Since $f(2)=\sqrt2\lt2$, the equation $x=f(x)$ has exactly one solution. That solution will be a root of the polynomial of degree $8$ that comes from repeated squaring:
$$((x^2-2)^2-2)^2=2+x$$
L.F.'s comm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Probability that $5$ divides $a^2+b^2$ is $\frac9{25}$
Two positive integers $a$ and $b$ are randomly selected with replacement, then prove that the probability of $(a^2 +b^2)/5$ being positive integer is $9/25$.
I found out the pattern of the last digits of $a^2$ , $b^2$ but then the sample space isn't finite so I ... | $1^2 mod 5 \equiv 1$
$2^2 mod 5 \equiv 4$
$3^2 mod 5 \equiv 4$
$4^2 mod 5 \equiv 1$
$5^2 mod 5 \equiv 0$
We want $a^2+b^2 mod 5 \equiv 0$,
That happens for (1,3) (1,2) (4,3) (4,2) (3,1) (2,1) (3,4) (2,4) and (5,5)
where (x,y) represents (a mod 5, b mod 5)
5*5 ways to express (x,y)
So the probability is 9/25
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Absolute value of cubic polynomial roots lower than 1 Assume we have a cubic polynomial
$ x^3 +bx^2+xc+d=0 $, with $b,c,d$ real numbers.
Let $x_1, x_2, x_3 $ be the roots, either real or complex.
What is the relation of the coefficients $b,c$ and $d$ in order to have the roots inside the unit sphere, that means
$ |x_i... | What you're trying to determine is whether the conditions on $b,c,d$ that make $x^3 + bx^2 + cx + d$ a Schur polynomial. As mentioned before, a sufficient (but perhaps unnecessary) condition is that
$$
1>b>c>d
$$
Is true. For the precise conditions, one may apply either the Jury test or Bistritz test
Or, apply the Rou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Trouble solving $\int\sqrt{1-x^2} \, dx$ I am trying to learn how to solve integrals and I've got the hang out of a lot of examples, but I haven't got the slightest idea how to solve this example, this is how far I've got:
$$
\int\sqrt{1 - x^2} \, dx = x\sqrt{1-x^2} - 2\int\frac{x^2}{\sqrt{1-x^2}} \, dx
$$
Can you plea... | Your idea of integrating by parts is good (but you did it wrong):
\begin{align}
\int\sqrt{1 - x^2} \, dx
&= x\sqrt{1-x^2} - \int x\frac{-2x}{2\sqrt{1-x^2}} \, dx \\
&= x\sqrt{1-x^2} + \int\frac{x^2}{\sqrt{1-x^2}}\,dx\\
&= x\sqrt{1-x^2} + \int\frac{x^2-1+1}{\sqrt{1-x^2}}\,dx\\
&= x\sqrt{1-x^2} - \int\sqrt{1-x^2}\,dx+\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
calculating mod 7 Problem: Calculate $10^{10^{10}} \pmod 7$
According to Fermat's little theorem: $a^{p-1}\equiv1 \pmod p$, so $10^6\equiv1 \pmod 7$ and $10^n\equiv4 \pmod 6$, $n$ being any integer, why can we write $10^{10^{10}}\equiv10^4 \pmod 7$?
Similarly, if it's $10^n\equiv1 \pmod 3$, n being any integer, then wo... | Let me see if I can make it clear
Starting with $10^6 \equiv 1 \mod 7$, we have
$$ \begin{align}
10^{12} &= \left(10^6\right)^2 \equiv 1^2 = 1 \mod 7\\
10^{18} &= \left(10^6\right)^3 \equiv 1^3 = 1 \mod 7\\
10^{24} &= \left(10^6\right)^4 \equiv 1^4 = 1 \mod 7\\
\cdots&
\end{align} $$
This tells us how to find $10^n \mo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
equation $x^2+ax+6a = 0$ has integer roots, Then integer values of $a$ is If the equation $x^2+ax+6a = 0$ has integer roots, Then integer values of $a$ is
$\bf{My\; Try}::$ Let $\bf{\alpha,\beta}$ be two roots of given equation $x^2+ax+6a = 0$
So $\bf{\alpha+\beta = -a}$ and $\bf{\alpha \cdot \beta = 6a}$ and $\bf{\alp... | What you have done so far is absolutely correct. However, for the sake of completeness I will repeat the beginning of your solution. Using Vieta's equations, we know that $$\alpha+\beta=-a$$ and $$\alpha\beta=6a$$
By substituting the first equation into the second, we get:
$$ \alpha\beta=-6\alpha-6\beta$$
Which simplif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
How to prove $(n!)^4\le2^{n^2+n}$? This may sound like a newbie but
question is to show that
$$(n!)^4\le2^{n^2+n} for \quad n=1,2,3...$$
I know it is true for n=1. $(1!)^4\le2^2$
and
assume it is true for $1<m\le n$ for all $\quad m\in N$
we have to show for m=n+1.
$((n+1)!)^2\le^? 2^{(n+1)^2+n+1}$
$((n+1)!... | I would rather write $$2^{n^2+n}=4^{\frac{n(n+1)}2}=4^14^24^3\cdots 4^n$$
Then looking at the factors individually, it suffices to how that for each $n\geqslant 4$, that $n^4\leqslant 4^n$. Observe the last inequality is false for $n=3$; but your inequality is. Taking $\log$s, this is equivalent to showing that $$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$3$ never divides $n^2+1$ Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.
Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.
If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.
And that is where I am stuck. I try ... | Since $n-1,n,n+1$ are three successive integers so one of them must be divisible by 3 hence there product must be divisible by $3$ i.e. $3|(n-1)n(n+1) \implies 3|n^3-n$ , now
$3|n^2+1 \implies 3|n(n^2+1) \implies 3|n^3+n \implies 3|n^3+n-(n^3-n) \implies 3|2n$ , since 3 does
not divide $2$ so $3|n \implies 3|n^2 \imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 7
} |
Solving $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=c$ Question:
let $a,b,c$ be positive constants. Find $u=u(x,y)$ if is satisfies the partial differential equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
and the boundary condition
$$u=0,\dfrac{x^2}{a^2}+\df... | The shape that your boundary conditions describe is in fact an ellipse. So if you switch to an elliptic coordinate system your pde should separate out nicely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Proving that if $a^2+b^2=c^2$, then $a+b\ge c$. Hello, I'm trying to prove this statement.
Let a,b & c be three positive real numbers and if $a^2+b^2=c^2$ then $a+b\ge c$
Any help, please?
| Suppose Not, which means that $a + b < c $
Square both sides
$(a + b)^2 < c^2 $
$a^2 + b^2 + 2ab < c^2 $
But by assumption $a^2 + b^2 = c^2$
Therefore,
$a^2 + b^2 + 2ab < a^2 + b^2 $
Which means
$2ab < 0 $ where a & b are positive numbers
Contradiction as no 2 positive multiplication is less than 0
therefore,
$a+b≥c $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
integer $m$ for which $x^3-mx^2-mx-(m^2+1)=0\Rightarrow m^2+m(x^2+x)-x^3+1=0$ has integer roots Calculation of all integer values of $m$ for which the equation
$p(x)=x^3-mx^2-mx-(m^2+1)$ has integer roots.
$\bf{My\; Try}::$ Given $x^3-mx^2-mx-(m^2+1)=0\Rightarrow m^2+m(x^2+x)-x^3+1=0$
for integer roots(i.e $x\in \mathb... | $\frac{-(x^2+x) \pm \sqrt{(x^2+x)^2+4(x^3-1)}}{2} = n$ where $n$ is an integer.
$\Rightarrow 4n^2 + (x^2+x)^2 + 4n(x^2+x) = 4(x^3-1) + (x^2+x)^2$
$\Rightarrow n^2+ n(x^2+x) = x^3-1$
$\Rightarrow n(n+x^2+x) = (x-1)(x^2+x+1)$
Hint:
For what integer values of n and x are the LHS and RHS equal?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How find this $\left(\frac{1}{x^2+a^2}\right)^{(n)}$ Prove that
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
my try:
since
$$\dfrac{1}{x^2+a^2}=\dfrac{1}{2ai}\left(\dfrac{1}{x-ai}-\dfrac{1}{x+ai}\right),i=\sqrt{-1}$$
so
$$\left(\dfrac{1}{x... | First, you can change variables to reduce to the case $a=1$. Second, the integral is the arctangent, so you are asking for the $(n+1)$th derivative of $\arctan(x)$. Maple says:
$$
\left(\frac{d}{dx}\right)^n\arctan x =
\frac{1}{2}\,{2}^{n}
G^{1, 3}_{3, 3}\left({x}^{2}\, \Big\vert\,^{0, 0, 1/2}_{0, (n-1)/2, n/2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find max: $\frac{1}{a^3+2b^3+6}+\frac{1}{b^3+2c^3+6}+\frac{1}{c^3+2a^3+6}$ For $a,b,c>0$ and $abc=1$. Find max:
$\frac{1}{a^3+2b^3+6}+\frac{1}{b^3+2c^3+6}+\frac{1}{c^3+2a^3+6}$
I used AM-GM for $a^3+2b^3$ but I don't know how to continue ...
| From the AM-GM Inequality,
$a^3+b^3+1\ge 3ab$,
$b^3+1+1\ge 3b$
$\Rightarrow a^3+2b^3+6\ge 3(ab+b+1)$
Similarly, $b^3+2c^3+6\ge 3(bc+c+1)$,$c^3+2a^3+6\ge 3(ca+a+1)$.
We have $\frac 1{a^3+2b^3+6}+\frac 1{b^3+2c^3+6}+$$\frac 1{c^3+2a^3+6}\le \frac 13(\frac 1{ab+b+1}+\frac 1{bc+c+1}+\frac 1{ca+a+1})$
$=\frac 13(\frac 1{ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Describing a linear map geometrically I have the following linear map $\mathbb{R}^2\to\mathbb{R}^2:$
$$\begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}9y-5x\\7y-4x\end{pmatrix}$$
I am asked to describe this geometrically.
I can see what is happening: the upper left quadrant is being squeezed into the upper righ... | Suppose you have the map
$$
T_2
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
9y + x \\
y
\end{pmatrix}.
$$
Then this is called a horizontal shear. There is more information here: Shear mapping. Vertical shear is similarly defined. Now consider three additional maps
$$
T_1
\begin{pmatrix}
x \\
y
\end{pmatrix}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Inequality $\frac{a + \sqrt{ab} + \sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.$ Someone can to help me with a hint in the following problem:
Show that for any $a,b,c>0$,
$$\frac{a + \sqrt{ab} + \sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.$$
I have t... | By AM-GM inequality:
$\sqrt{ab}\le\sqrt[3]{ab.\frac{a+b}{2}}$
We prove that: $$\sqrt[3]{\frac{2}{a+b}.\frac{3}{a+b+c}}\left(a+\sqrt[3]{ab.\frac{a+b}{2}}+\sqrt[3]{abc}\right)\le3\sqrt[3]{a}$$
Also by AM-GM: $$\sqrt[3]{\frac{2a}{a+b}.\frac{3a}{a+b+c}}\le\frac{1+\frac{2a}{a+b}+\frac{3a}{a+b+c}}{3}$$
$$\sqrt[3]{\frac{3b}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/627543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
| Your sum is as follow:
$2(2^9+2^8+\cdots +2+1)$
which can be proved simply is equal to $2 \frac{2^{10}-1}{2-1}$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Let $n=x^2+y^2$; $n=2^{2k}m$ or $n=2^{2k+1}m$ with $m$ odd. Prove that $2^{k}$ divides both $x$ and $y$. Let $n=x^2+y^2$ where x,y are integers, be one of the forms $n=2^{2k}m$ respectively $n=2^{2k+1}m$ with m odd. Prove that $2^{k}$ divides both x and y.
| Let $(x,y)=2^rs$ where $s$ is odd
and $\displaystyle\frac xX=\frac yY=2^rs\implies (X,Y)=1$
$\displaystyle\implies x^2+y^2=2^{2r}s^2(X^2+Y^2)$
Now as $(X,Y)=1,$
Case $1:$ Either both are odd :
As $(2c+1)^2=4c^2+4c+1\equiv1\pmod4, X^2+Y^2=2\pmod4$ i.e., divisible by $2,$ but not by $4$
Case $2:$ $X,Y$ are of opposite ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $ \lim_{n \rightarrow \infty } \frac{n+6}{n^2-6} = 0 $. My attempt:
We prove that $ \lim\limits_{n \rightarrow \infty } \dfrac{n+6}{n^2-6} =0$.
It is sufficient to show that for any $ \epsilon \in\textbf{R}^+ $, there exists an $ K \in \textbf{R}$ such that for any $ n > K $
$ \displaystyle \left| \frac{... | Your proof is correct. Just to be pedantic, you can define $K$ directly as $14/\varepsilon$, since you precedently assumed that $n>4$.
As you've already noted in the comments, the actual way to solve limits without going through all those $\delta$-$\epsilon$ arguments is to use limits properties, such as the preservat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solving $2\cos(x) = \sin(x)$ How would you solve equations of the form $ a \sin (x+b) = \sin (x)$?
Eg. $ 2 \cos(x) = \sin(x) $
I realy have no idea how I would solve this kind of equations.
| I would do a tan half angle substitution with $t = \tan \frac{x}{2}$ which leads to
$$ \cos(x) = \cos\left( 2 \tan^{-1} t \right) = \frac{1-t^2}{1+t^2} \\
\sin(x) =\sin \left( 2 \tan^{-1} t \right) =\frac{2 t}{1+t^2} $$
which transforms your problem to
$$ \frac{1-t^2}{1+t^2} a \sin(b) + \frac{2 t}{1+t^2} \left(a \cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
What is the integrating factor of $2xy'+x^{2}e^{1-x^2}y=2$ I know how to start solving it, its dividing everything by $2x$, but I can't solve
$$\int \dfrac{x^2e^{1-x^2}}{2x}\,dx$$
| Starting from where you left off (divide both sides by $2x$), we have
$$\dfrac{dy}{dx} + \dfrac{1}{2}xe^{1 - x^2}y = \dfrac{1}{x}$$
Since (by the standard form of linear equation) $p(x) = \dfrac{1}{2}xe^{1 - x^2}$, the integrating factor is
$$\begin{aligned}
\mu(x) = e^{\int p(x)\,dx} = e^{\int \frac{1}{2}xe^{1 - x^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.