Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Simple factor of equation I have this polynomial: $5z^4-12z^3+30z^2-12z+5$
How do I factor it to get the following?: $(5z^2-2z+1)(z^2-2z+5)$
Can someone show me the procedure to perform whenever I encounter with a case like this? Thank you.
| You should exploit the symmetry: write the polynomial as
$$
z^2\left(5z^2+\frac{5}{z^2}-12z-\frac{12}{z}-30\right)
$$
and observe that
$$
z^2+\frac{1}{z^2}=\left(z+\frac{1}{z}\right)^{\!2}-2
$$
so you can rewrite the expression as
$$
z^2\left(5\left(z+\frac{1}{z}\right)^{\!2}-12\left(z+\frac{1}{z}\right)-40\right)
$$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving that $\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$ I am trying to prove that
$$I=\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$$
where $\beta(s)$ is the Dirichlet Beta function and $G$ is the Catalan's constant. I managed to derive the following series... | The generalization of the main integral follows easily by employing the same ideas used in this post and the previous one.
Let $n$ be a natural number. Then, we have
$$\int_0^1 \frac{\log^{2n}(x)\operatorname{arctanh}(x)}{1+x^2}\textrm{d}x$$
$$=\lim_{s\to0}\frac{d^{2n}}{ds^{2n}}\left(\frac{\pi}{16}\cot \left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 0
} |
Prove: $2^k$ is the sum of two perfect squares If $k$ is a nonnegative integer, prove that $2^k$ can be represented as a sum of two perfect squares in exactly one way. (For example, the unique representation of $10$ is $3^2+1^2$; we do not count $1^2+3^2$ as different.)
I understand that $2^{2n}=0+2^{2n}$ and $2^{2n+1}... | Suppose
$2^{2n+1}$
is the smallest odd power of two
that can represented in
more than one way
as the sum of two squares.
Then
$2^{2n+1}
=(2^{n}+a)^2+(2^{n}-b)^2
$
where
$1 \le a < 2^n$
and
$1 \le b < 2^n$.
Then
$\begin{array}\\
2^{2n+1}
&=(2^{n}+a)^2+(2^{n}-b)^2
\qquad(0)\\
&=2^{2n}+a2^{n+1}+a^2+2^{2n}-b2^{n+1}+b^2\\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
When is matrix $A$ diagonalizable? I got the following matrix:
$$ A =
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & 0 \\
1 & 2 & 1 \\
\end{pmatrix}
$$
I need to answer when this matrix is diagonalizable.
Its characteristic polynomial is $ t(t-a)(t-1) $. So its 3 eigenvalues are 0, 1 and ... | 1) If $a\ne 0, 1,\;$ then A is diagonalizable since it has 3 distinct eigenvalues.
2) If $a=0$, then A is diagonalizable $\iff$ $\text{nullity}(A-0I)=\text{nullity}(A)=2 \iff \text{rank}(A)=1$
$\hspace{2.3 in}\iff\text{rank}\begin{pmatrix} 0&0&0\\b&0&0\\1&2&1\end{pmatrix}=1\iff b=0$
3) If $a=1$, then A is diagonalizabl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Three questions about the form $X^2 \pm 3Y^2 = Z^3$ and a related lemma In Ribenboim’s Fermat’s Last Theorem for Amateurs, he gives the following lemma [Lemma 4.7, pp. 30–31].
Lemma. Let $E$ be the set of all triples $(u, v, s)$ such that $s$ is odd, $\gcd(u,v) = 1$ and $s^3 = u^2 + 3v^2$. Let $F$ be the set of all pai... | The formula quoted in Ribenboim has the general form,
$$u^2+dv^2 = (p^3 - 3 d p q^2)^2 + d(3 p^2 q - d q^3)^2 = (p^2+dq^2)^3\tag1$$
Assume $\gcd(u,v)=1$. For $d=3$, apparently it is integrally complete.
But we cannot necessarily extend the same conclusion to general $d$. For example, for $d=11$, the formula above beco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integer solutions for $n$ for $|{\sqrt{n} - \sqrt{2011}}| < 1$ $$|{\sqrt{n} - \sqrt{2011}}| < 1$$
What is the number of positive integer $n$ values, which satisfy the above inequality.
My effort:
$
({\sqrt{n} - \sqrt{2011}})^2 < 1 \\n + 2011 -2\sqrt{2011n} < 1\\ n+2010<2\sqrt{2011n}\\ n^2+2 \times 2010 \times n +2010^... | $\sqrt{n}\in(\sqrt{2011}-1,\sqrt{2011}+1)\iff n\in ( (\sqrt{2011}-1)^2,(\sqrt{2011}+1)^2)=(2012-2\sqrt{2011},2012+2\sqrt{2011})$
There are $2\lfloor2\sqrt{2011} \rfloor + 1$ integers in this range.
We approximate $\sqrt{2011}$ as $44.8$ and we find the answer is $2\cdot 89 + 1=179$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove the function Prove
$\forall n\in\mathbb{N}: n\ge 1 \rightarrow 2^n\le 2^{n+1}-2^{n-1}-1.$
I did that $2^n \le\ 2^{n+2} - 2^{n}$ and then $2^n < 2^{n+2} - 2^{n-1}$
but have no idea how to add $-1$ in the function and let the $2^n \le\ 2^{n+2} - 2^{n} - 1$
| $2^{n+1}-2^{n-1}-1=2^{n-1}*(4-1)-1=2^{n-1}*3 -1 \ge 2^{n-1}*3-2^{n-1}=2^{n-1}*2=2^n $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $abcd=1$,where $a,b,c,d$ are positive reals,then find the minimum value of $a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd$. If $abcd=1$,where $a,b,c,d$ are positive reals,then find the minimum value of $a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd$.
Let $E=a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd=(a+b+c+d)^2-(ab+ac+ad+bc+bd+cd)$
I do not know h... | with AM-GM, we get
a²+b²+c²+d² ≥ 4√((abcd)²) = 4
ab+ac+ad+bc+bd+cd ≥ 6√((abcd)³) = 6
minimum value of a²+b²+c²+d²+ab+ac+ad+bc+bd+cd is 4+6 = 10 when a=b=c=d=1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The only natural number $x$ for which $x+\sqrt{-2}$ is a cube in $\mathbb{Z}[\sqrt{-2}]$ is $x=5$ Let $A = \mathbb{Z}[\sqrt{-2}]= \{a+b\sqrt{-2} \ : a, b \in \mathbb{Z}\}$. Show that the only natural number $x$ for which $x+\sqrt{-2}$ is a cube in $A$ is $x=5$.
So I have to show that there exists $c,d \in \mathbb{Z}$... | Directly
$$(a+b\sqrt{-2})^2(a+b\sqrt{-2})=(a^2-2b^2+2ab\sqrt{-2})(a+b\sqrt{-2})=$$
$$=(a^3-6ab^2)+(3a^2b-2b^2)\sqrt{-2}$$
We get that it must be
$$3a^2b-2b^2=1\iff b(3a^2-2b)=1\iff b,\,3a^2-2b=\pm 1$$
But $b=-1\implies 3a^2-2(-1)=-1\implies3a^2=-3$ , impossible, so it must
$$\;b=1\implies3a^2-2=1\iff 3a^2=3\implies a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Converging sequence $a_{{n+1}}=6\, \left( a_{{n}}+1 \right) ^{-1}$ I know the sequence is converging. But I find it difficult proving it, by induction. So far I have drawn a diagram and calculate the five first numbers. From the diagram I can se that the sequence can be split into two sequences, one that is increasing ... | (At the end,
I have added my form
of the explicit solution.)
$a_{1}=1, a_{n+1}=\dfrac{6}{a_{{n}}+1},1\leq n
$
If it has a limit $L$,
then
$L = \dfrac{6}{L+1}$
or
$L^2+L-6 = 0
$
or
$(L+3)(L-2) = 0
$.
Since
$a_n > 0$
for all $n$,
the only possibility
is $L=2$.
To get a sequence
whose terms should
go to $0$,
let
$a_n = b_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove $1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$ The task is to prove the following non-equality by hand:
$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$
Wolframalpha s... | Putting $x=\cos(\frac{2\pi}{7})$ you have the polynomial $1+x-4x^2-8x^3\not= 0$ then $8x^3+4x^2-x-1\not=0$ from that
$$8x^3+4x^2-x-1=4x^2(2x+1)-x-1-x+x=4x^2(2x+1)-(2x+1)+x=(2x+1)(4x^2-1)+x=(2x+1)^2(2x-1)+x$$
Since $(2x+1)^2>0$ and $x>\frac{1}{2}$ since $\frac{2\pi}{7}<\frac{\pi}{3}$ we have that $(2x-1)>0$ and a sum of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac... | we have after squaring $$\sin(x)^2+\cos(x)^2+\sin(2x)=\frac{1}{9}$$ or $$\sin(2x)=-\frac{8}{9}$$
the answer is $$c_1\in \mathbb{Z}\land \left(x=2 \pi c_1+2 \tan ^{-1}\left(\frac{1}{10}
\left(9-\sqrt{161}\right)\right)\lor x=2 \pi c_1+2 \tan
^{-1}\left(\frac{1}{10} \left(9+\sqrt{161}\right)\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 2
} |
integrate $\int \frac{\tan^4x}{4}\cos^3x$
$$\int \frac{\tan^4x}{4}\cos^3x$$
$$\int \frac{\tan^4x}{4}\cos^3x=\frac{1}{4}\int \frac{\sin^4x}{\cos^4x}\cos^3x=\frac{1}{4}\int\frac{\sin^4x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot\sin^2x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot(1-\cos^2x)}{\cos x}=\frac{1}{4}\int \frac... | More easy way $tan(x).cos(x)=sin(x)$ so integral becomes $sin^3(x).tan(x)=sin^4(x)/cos(x)$ Now $sinx=u$ so integral becomes $$\int\frac{u^4}{1-u^2}$$ which on simplification becomes $$\int \frac{2}{(1/u^2)(1/u^2-1)}du$$ now let $1/u=t$ thus the common $ u^2$ cancels to give $\int\frac{1}{1-t^2}dt$ whose integral is $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$
In $\triangle ABC$, if
$$\cos A \cos B \cos C=\frac{1}{3}$$
then can we find value of
$$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$
Please give some hint. I am not sure if $\tan A \tan B+\tan ... | Hint: For a triangle ABC
$A+B=\pi-C$
and
$1-2\cos A \cos B \cos C=\cos^2A+\cos^2B+\cos^2C$
Edit $1$:
$A+B=\pi-C$
Apply $\cos$ on both sides
Divide each term by $\cos A.\cos B$
We get $\tan A.\tan B=1+\frac{\cos C}{\cos A. \cos B}$
Similarly write $2$ more equations and add three equations.
Now use $1-2\cos A \cos B \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Last Digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
Given $x$ and $p$. Find the last digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
I need a general formula. I can find that the sum is equal to
$\dfrac{x^{p+1}-1}{x-1}$
But how to find the last digit.
P.S: $x\leq 999999$ and $p \leq 10^{15}$
| We'll work it out mod 5 and mod 2. We'll assume $p > 2$.
Mod 2, we have $x^i \equiv x$ for all $i$, except for $x^0$. Therefore the parity of the sum is just the same as the opposite of that of $x$ [unless $p=2$].
Mod 5: if $x \equiv 0 \pmod{5}$ then the result is certain to be $1$ mod $5$. Otherwise, $x^4 \equiv 1 \pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to solve $a = \cos x - b\sin x$ where $a$ and $b$ are real numbers? I found this equation when solving a physics problem related to finding an angle when entering a river, that has a known current, and trying to get to a specific point on the other side. I'm not sure how to solve this equation explicitly for the a... | Method 1:
*
*$a=\cos x - b\sin x$
*$\cos x = a+b\sin x$
*$\cos^2x = a^2+2ab\sin x+b^2\sin^2x$
*$1-\sin^2x = a^2+2ab\sin x+b^2\sin^2x$
*$(a^2-1)+2ab\sin x+(b^2+1)\sin^2x=0$
Then quadratic.
Method 2:
$$\sin(\alpha-x)=\sin\alpha\cos x-\cos\alpha\sin x$$
$$\frac{\sin(\alpha-x)}{\sin\alpha}=\cos x-\frac1{\tan\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find sum of power series. Having a small mistake. Find the sum of the series. My answer is $-\frac{3}{4}$, but it should be $\frac{3}{4}$. Where did i make a mistake?
$$ \sum_{n=1}^{\infty} \frac{n}{3^n} $$
$$ \frac{d}{dx} (\frac{1}{1-x}) = \sum_{n=0}^{\infty} \frac{d}{dx} (x^n) $$
$$ \frac{-1}{(1-x)^2} = \sum_{n=0}^{\... | Since $(1-x)'=-1$,
$$
\frac{d}{dx}\frac{1}{1-x}=-\frac{(1-x)'}{(1-x)^2}=\frac{1}{(1-x)^2}
$$
by chain rule. Now the correct answer will appear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by 4 I'm studding D.M Burton & want to solve: Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by $4$. . Please help me by giving your solution to it. I'm new comer to number theory so please don't use theorems above Theory of Congruence.
| Another way is to use the "paired element" approach... Since $a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$, we have
$$1^5+99^5+2^5+98^5+\dots+ 50^5+100^5\\
=100(1-99+99^2-99^3+99^4)+100(2^4-\dots)+\dots+100(49^4-\dots)+2^5\cdot(25)^5+100^5\\
\equiv 0\pmod 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find all $x,y$ so that $\dfrac{x+y+2}{xy-1}$ is an integer. I am trying to find the integers $x,y$ so that
$\dfrac{x+y+2}{xy-1}$ is an integer.
What I have done:
I suppose there exists $t$ such that $$t=\dfrac{x+y+2}{xy-1}$$ where $xy\neq 1$ then consider the following scenarios:
$$x=y$$ $$x>y>0$$ $$x>0>y$$ ,etc.
Th... | Wlog $|x|\le |y|$.
$x=0$ leads to $\frac{y+2}{-1}$, which is always an integer.
$x=1$ leads to $\frac{y+3}{y-1}=1+\frac4{y-1}$ which is an integer iff $y-1$ is a (positive or negative) divisor of $4$, so $y\in\{-3,-1,2,3,5\}$ ($y=0$ is excluded by $|y|\ge |x|$)
$x=-1$ leads to $\frac{y+1}{-y-1}=-1$, always an integer
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
solve $\sin 2x + \sin x = 0 $ using addition formula
$\sin 2x + \sin x = 0 $
Using the addition formula, I know that
$\sin 2x = 2\sin x \cos x$
=> $2\sin x \cos x + \sin x = 0$
=> $\sin x(2\cos x + 1) = 0$
=> $\sin x = 0$ and $\cos x = -\frac{1}2 $
I know that $\sin x = 0$ in first and second quadrant so $x = 0$ and ... | We know that $\cos 60^\circ = \frac{1}{2}$.
The cosine of an angle is defined to be the $x$-coordinate of the point where the terminal side of an angle in standard position (initial side on the positive $x$-axis and vertex at the origin) intersects the unit circle. Therefore, the cosine function is positive if the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find the Min of P(x,y) Find the Minimum of the following function :
$$P(x,y) = \frac{(x-y)}{(x^4+y^4+6)}.$$
This is a math problem I found in an internet math competition but it is really complex to me !!!
| Hint: Armed with your guess, it is not hard to complete the square to get
$$\frac{x-y}{x^4+y^4+6}+\frac14= \frac{x^4+y^4+4x-4y+6}{4(x^4+y^4+6)}=\frac{(x^2-1)^2+(y^2-1)^2+2(x+1)^2+2(y-1)^2}{4(x^4+y^4+6)} \geqslant 0$$
with equality possible iff $x=-1, y=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these
The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+... | $$a^2+b^2+c^2=ac+ab\sqrt3$$ The above equation can be re-written as
$$\frac{a^2}{4}-ac+c^2+\frac{3a^2}{4}-ab\sqrt3+b^2=0$$
which is
$$(\frac{a}{2}-c)^2+(\frac{\sqrt3 a}{2}-b)^2 = 0$$
which implies that
$\frac{a}{2} = c$ and $\frac{\sqrt3 a}{2}= b$
.Based on this it can be concluded that the ratio of sides is 1:$\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Proof of $\sum_{j=1}^{p-1} \lfloor jq/p \rfloor = \frac{1}{2}(q-1)(p-1)$ Involving Pairing of Summands I've seen the proof of the identity
$$\sum_{j=1}^{p-1} \lfloor jq/p \rfloor = \frac{1}{2}(q-1)(p-1)$$
where $p$ and $q$ are coprime positive integers. This involves counting the remainders $r_{j}/p$ of the summands $... | Your procedure is correct, in fact your steps can be reformulated as follows:
$$
\eqalign{
& \sum\limits_{j = 1}^{p - 1} {\left\lfloor {j\,q/p} \right\rfloor } = {1 \over 2}\left( {\sum\limits_{j = 1}^{p - 1} {\left\lfloor {j\,q/p} \right\rfloor + \sum\limits_{j = 1}^{p - 1} {\left\lfloor {\left( {p - j} \right)\,q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of tan function This is a question from an old tutorial for a basic mathematical analysis module.
Show that $$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = e^2$$
My tutor has already gone through this in class but I am still confused. Is there anything wrong with the following reasoning?
Since $\... | Solution
Applying $\textbf{L'Hospital's Rule}$, we have $$\begin{align*}\lim_{x \to +\infty} \left[x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]&=\lim_{x \to +\infty}\dfrac{\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\&=\lim_{x \to +\infty}\dfrac{1}{\sin\left(\dfrac{\pi}{4}+\dfrac{1}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Calculating eigenvalues and eigenvectors Question: Calculate the eigenvectors and eigenvalues of the following matrix $$\begin{pmatrix}3&-3\\0&-2\\ \end{pmatrix}$$
My attempt:
I have calculated the eigenvalues to be $\lambda = 3$ and $\lambda =-2$ and I have managed to get the eigen vector for $\lambda = 3$ to be \begi... | The eigenvalues of a triangular matrix are simply the diagonal entries, so your eigenvalues $3$ and $-2$ are correct. As
$$\begin{pmatrix}
3 & -3 \\
0 & -2 \\
\end{pmatrix}
\begin{pmatrix}
1 \\ 0
\end{pmatrix}
= \begin{pmatrix}
3 \\ 0
\end{pmatrix}
= 3
\begin{pmatrix}
1 \\ 0
\end{pmatrix},$$
the vector $\begin{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Is the following solution correct? Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$
My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$
$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) =... | A suggested simplification. You should always look for simplifications to the algebra if they're easy to find.
Let $y = x^2 + 1$
Then you're solving $\sqrt y + \frac{8}{\sqrt y} = \sqrt{y + 8}$
$\frac{y + 8}{\sqrt y} = \sqrt{y + 8}$
$(y+8)^2 = y(y+8)$
$8(y+8) = 0$
From this, it should be obvious that $y = -8$ is the on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Show that $\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}$ Show that $$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}.$$
Any solutions or hints are greatly appreciated.
I know I can rewrite the integral as $$\int_{-\infty}^\infty {(x-1)(x-2)\over {(x^2+1)(x^2+9)}}dx.$$ ... | Well, $\int_{-\infty}^{+\infty}\frac{-3x}{x^4+10x^2+9}\,dx = 0$, hence the problem boils down to computing:
$$ \int_{-\infty}^{+\infty}\frac{(x^2+2)}{(x^2+1)(x^2+9)}\,dx=\frac{1}{8}\int_{-\infty}^{+\infty}\frac{1}{x^2+1}\,dx+\frac{7}{8}\int_{-\infty}^{+\infty}\frac{1}{x^2+9} $$
that is trivially equal to $\left(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Extract coefficients for a formal power series using Lagrange Inversion Formula Given $f(x)$ is a formal power series that satisfies $f(0) = 0$
$(f(x))^{3} + 2(f(x))^{2} + f(x) - x = 0$
I know that the Lagrange inversion formula states given f(u) & $\varphi(u)$ are formal power series with respect to u, and $\varphi(0)... | Suppose we have
$$f(z)^3 + 2f(z)^2 + f(z) = z$$
and we seek $[z^n] f(z).$
While we wait for a contribution from LIF experts in the meantime we
can use Poor Man's Lagrange Inversion which is the Cauchy Residue
Theorem. We have
$$[z^n] f(z)
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} f(z) \; dz.$$
Now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
n tends to infinity $\displaystyle\lim_{n\to\infty}
{\left(
\frac{\sqrt {n^2+n} - 1}{n}
\right)}^{
\left(2 \sqrt{n^2+n} - 1
\right)}$
Sorry for the bad format of the question I don't know much latex
I tried rationalization of the power and the base but was not able to get the answer.
| When we see both the base and exponent as variable then the best approach is to take logs. Thus if $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{\sqrt{n^{2} + n} - 1}{n}\right)^{(2\sqrt{n^{2} + n} - 1)}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{\sqrt{n^{2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Show that $\sum_{k=1}^{\infty}\frac{F_{2^{k-1}}}{L_{2^k}-2}=\frac{15-\sqrt{5}}{10}$
Show that$$\sum_{k=1}^{\infty}\frac{F_{2^{k-1}}}{L_{2^k}-2}=\frac{15-\sqrt{5}}{10},$$
where $F_n$ is a Fibonacci number and $L_n$ is a Lucas number.$^1$
Motivation: For example, when calculating Millin series
$$
\sum_{n=1}^{\infty} ... | Using
$$F_{2N}=F_NL_N$$
$$L_{2N}=L_N^2-2(-1)^N$$
we have
$$F_{2^{n+1}}=F_{2^n}L_{2^n}$$
$$L_{2^{n+1}}=L_{2^n}^2-2$$
and so
$$\begin{align}\frac 32-\frac 12\cdot\frac{F_{2^n}}{L_{2^n}-2}+\frac{F_{2^n}}{L_{2^{n+1}}-2}&=\frac 32-\frac 12\cdot\frac{F_{2^n}}{L_{2^n}-2}+\frac{F_{2^n}}{L_{2^n}^2-4}\\\\&=\frac 32-\frac{F_{2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Is $\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\cdots$ an irrational number? Obviously:
$$\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\cdots=0.1111\dots=\frac{1}{9}$$
is a rational number.
Now, if we make terms with demoninators in the form:
$$q_n=\sum_{k=0}^{n} 10^k$$
Then the sum will be:
$$\sum_{n=1}^{\infty}\frac{1}{q... | This is not the answer to the question of irrationality. This a piece of information about an interesting closed form of the series.
$$q_n=\sum_{k=0}^n 10^k = \frac{10^{n+1}-1}{9}$$
$q_0=1\quad;\quad q_1=11\quad;\quad q_2=111\quad...$
$\sum_{n=0}^m \frac{1}{q_n}=\frac{1}{1}+\frac{1}{11}+\frac{1}{111}+... \quad$ ($m+1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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Contour Integral of $\int\limits_0^{2\pi}\frac{d\theta}{1+a\cos\theta}$ for $a^2<1$ (textbook wrong?) My book is telling me that the answer is $\frac{2\pi}{\sqrt{1-a^2}}$. I'm getting an extra a on the numerator. Could somebody verify if I'm wrong, or if it's my book (it has been wrong numerous times).
| Contour Integration
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}\theta}{1+a\cos(\theta)}
&=\oint\frac{2\,\mathrm{d}z}{iz\left(2+az+\frac az\right)}\tag{1}\\
&=-\frac{2i}a\oint\frac{\mathrm{d}z}{z^2+\frac2az+1}\tag{2}\\
&=-\frac{2i}a\oint\frac{\mathrm{d}z}{\left(z+\frac1a+\sqrt{\frac1{a^2}-1}\right)\left(z+\frac1a-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find continuities with square root? I don't understand how to find $$\frac{4-x^2}{3-\sqrt{x^2+5}}$$
The book says to multiply the equation by $\frac{3 + \sqrt {x^2+5}}{3 + \sqrt {x^2+5}}$. I don't understand where that comes from. It says the multiplication simplifies to "$3 + \sqrt {x^2+5}$" - I don't see how t... | For this simplification, you need to know that $(a+b)(a-b)=a^2-b^2$. It is important to recognize this, as this is something you need very often. You also need this when simplifying expressions with square roots in the denominator. In this case, this gives us the following simplification: $(3+\sqrt{x^2+5})(3-\sqrt{x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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finding the smallest number $n$ such that $n!=n(n+1)(n+2)(n+3)$ What is the smallest number $n$ such that $n!=n(n+1)(n+2)(n+3)$?
How will I solve this type of problems?
| Okay... Is there only one solution? Well, yeah... because $n!$ increases by a factor of $(n+1)!/n! = n+1$ as $n$ increases by one, while $n(n+1)(n+2)(n+3)$ only increases by $(n+1)(n+2)(n+3)(n+4)/n(n+1)(n+2)(n+3) = (n+4)/n = 1 + 4/n$. As $n!$ increases faster for $n > 2$ there is at most one solution.
(Unless there i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving system of non-linear equations. So I'm trying to find the stationary points for $$f(x,y,z) = 4x^2 + y^2 +2z^2 -8xyz$$
Setting the partial derivatives to zero leads to:
$$x-yz=0 \\ y-4xz=0\\z-2xy=0$$
Substiting $z=2xy$ into the first two equations to get $y(1-8x^2)=0$ and $x(1-2y^2)=0$. Which have the solution $... | You have two conditions:
*
*$y=0$ or $x=\frac{1}{\sqrt{8}}$ or $x=-\frac{1}{\sqrt{8}}$
*$x=0$ or $y=\frac{1}{\sqrt{2}}$ or $y=-\frac{1}{\sqrt{2}}$
Both of these must be true. That means that one of the conditions gives the value for $x$, while the other must give the value for $y$. For example, if $y=0$ then the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the sum using The question is as follows:
Find the sum:
$1\cdot2 + 2\cdot3 + ... + (n-1)n$
What I have tried so far:
We can write $(n-1)n$ as $\frac{(n+1)!}{(n-1)!}$ which we can also write as $2\cdot\dbinom{n+1}{2}$
I believe it is possible to use the binomial theorem here, setting $a = b = 1$ in $(a+b)^n$. ... | $$1\cdot2 + 2\cdot3 + ... + (n-1)n+\color{red}{n(n+1)-n(n+1)}=$$
$$=1(1+1)+2(2+1)+...+(n-1)n+\color{red}{n(n+1)-n(n+1)}=$$
$$=1^2+2^2+...+n^2+1+2+...+n-n(n+1)=$$
$$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}-n(n+1)=$$
$$=\frac{n(n+1)(n+2)}{3}-n(n+1)=\frac{n(n^2-1)}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the minimum $k$
Find the minimum $k$, which $\exists a,b,c>0$, satisfies
$$ \frac{kabc}{a+b+c}\geq (a+b)^2+(a+b+4c)^2$$
My Progress
With the help of Mathematica, I found that when $k=100$, we can take $a=1,b=1,c=1/2$. And I'm pretty sure that $k=100$ is the answer, but I couldn't prove it.
| Effectively, you want to show
$$\frac{(a+b)^2+(a+b+4c)^2}{abc}(a+b+c) \geqslant 100$$
and you already have a case of equality.
Using homogeneity, we may set $a+b+c=5$, to equivalently show
$$(5-c)^2+(5+3c)^2 \geqslant 20 abc$$
Now $a+b = 5-c$, so for any $c$, we have $ab$ maximized when $a=b$. Thus it is enough to sho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Solving a 6th degree polynomial equation I have a polynomial equation that arose from a problem I was solving. The equation is as follows:
$$-x^6+x^5+2x^4-2x^3+x^2+2x-1=0 .$$
I need to find $x$, and specifically there should be a real value where $\sqrt3<x<\sqrt{2+\sqrt2}$, in accordance to the problem I am solving. ... | Using a complicated computer algebra system (Mathematica 10.4), we can get all the roots to this equation as radicals. Two roots are complex and the rest are all real (surprisingly).
Module[{roots},
roots = Solve[-x^6 + x^5 + 2 x^4 - 2 x^3 + x^2 + 2 x - 1 == 0, x];
Transpose[{
N[x /. roots],
FullSimplify[E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$ - I keep getting imaginary numbers $$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$$
My attempt
$\sqrt{x+938^2} + \sqrt{x + 140^2} = 1116$
$(\sqrt{x+938^2} + \sqrt{x + 140^2})^2 = (1116)^2$
$x+938^2 + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} + x + 140^2 = 1116^2$
$2x... | You received good answers and solutions. However, it seems that you made a few mistakes in your calculations. The line $$2x + 2\sqrt{x+938^2}\sqrt{x + 140^2} = 1116^2 - 938^2 - 140^2$$ is correct. What is not correct is in the next line since $$(x+938^2)(x + 140^2)=x^2+(938^2+140^2)x+938^2\times 140^2$$ I do not know... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How can we see that $ \sum_{n=0}^{\infty}\frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}}=(\pi-1)(\pi-3) $? I wonder will it help me so prove it if I was to decompose it into partial fractions?
Mathematica approves of the identity; it is converges. can anyone help me to prove it?
$$
\sum_{n=0}^{\infty}\frac{2^n(1-n)^3}{(n... | Hint. One may observe that, for $n\geq 0$, using the Euler beta function,
$$
\frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}}=-\frac{(n-1)^3}{n+1}\int_0^1(2x(1-x))^ndx. \tag1
$$ Then, one may write
$$\begin{align}
&\sum_0^\infty\frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}}
\\\\&=-\sum_0^\infty\frac{(n-1)^3}{n+1}\int_0^1(2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the infinite simple continued fractions for ... Find the infinite simple continued fractions for $\sqrt{2};\sqrt{5};\sqrt{6};\sqrt{7};\sqrt{8}$.
*
*I have solved similar equations for continued fractions but only using a fraction, if someone could please demonstrate how to do this to ANY of these values I will ... | Just an example: $\sqrt{7}$. Since $4<7<9$, $\left\lfloor \sqrt{7}\right\rfloor = 2$, so:
$$ \sqrt{7} = 2+(\sqrt{7}-2) = \color{blue}{2}+\frac{1}{\frac{\sqrt{7}+2}{3}}\tag{1}.$$
Since $2+\sqrt{7}\in (4,5)$, $\left\lfloor\frac{\sqrt{7}+2}{3}\right\rfloor =1$, so:
$$ \frac{\sqrt{7}+2}{3} = 1+\frac{\sqrt{7}-1}{3} = 1+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can I apply Chinese remainder theorem here?
A number when divided by a divisor leaves $27$ remainder. Twice the number when divided by the same divisor leaves a remainder $3$. Find the divisor.
My attempt:
Let, the number be=$n$ and the divisor be=$d$.
Thus, we have $n\equiv 27\pmod d$ and $2n\equiv 3\pmod d$.
Thus w... | Let the number be $n$, and the divisor be $d$.
$\frac{n}{d} = X + \frac{27}{d}$ and also $\frac{2n}{d} = Y + \frac{3}{d}$
Where $X$ and $Y$ are different or same positive integers, comparing the two
$$2X + \frac{54}{d} = Y + \frac{3}{d}$$
$$2X = Y + \frac{3}{d} - \frac{54}{d}$$
$$2X = Y - \frac{51}{d}$$
Now the value o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Roll 6 dice, find the number of outcomes with 3 distinct numbers. Suppose you roll six dice, how many outcomes are there with 3 distinct numbers.
My attempt:
First there are ${6 \choose 3}$ ways to choose these 3 distinct numbers.
We consider 3 cases;
Case 1: 3 pars of repeated numbers e.g. $223344$. There are ${3\choo... | This is correct, except you forgot to include the factor $\binom31$ for case $3$ in the final result. We can double-check the result using inclusion-exclusion: There are $\binom63$ ways to choose the $3$ numbers, and by inclusion-exclusion there are
$$\sum_{k=0}^3(-1)^k\binom3k(3-k)^6=3^6-3\cdot2^6+3\cdot1^6=540$$
ways... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show $3\cos 2x + 1 = 4\cos^2 x - 2\sin^2 x$ Show $3\cos 2x + 1 = 4\cos^2 x - 2\sin^2 x$
Using the formula $\cos 2x = \cos x - \sin^2 x$
I can say $3\cos 2x + 1 = 3(\cos^2 x - \sin^2 x) + 1$
$\Rightarrow 3\cos x^2 - 3\sin^2 x + 1$
But from there I don't see how I can get the answer.
| You are very close. All you need to do is use the identity $\sin^2{x} + \cos^2{x} = 1$:
$\begin{eqnarray}3\cos{2x} + 1 & = & 3(\cos^2{x}-\sin^2{x}) + 1 \\
& = & 3(\cos^2{x}-\sin^2{x}) + \sin^2{x} + \cos^2{x} \\
& = & 4\cos^2{x} - 2\sin^2{x}
\end{eqnarray}$
QED.
Alternatively, if you don't feel comfortable turning the $... | {
"language": "en",
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Finding $a, b, c$ values of a polynomial from a graph. I was doing my homework and I am now stuck on question number 7 which is:
The diagram shows the curve with the equation $y = (x + a)(x - b)^2$ where $a$ and $b$ are positive integers.
(i) Write down the values of $a$ and $b$, and also of $c$, given that the curve c... | From the graph it is evident that the roots are $-2$ and $1$ ($1$ being a repeated root),
So the equation becomes:
$$(x+2)(x-1)^2=0$$
Comparing it with $$(x+a)(x-b)^2=0$$ We find $a=2,b=1$
Also for finding the point $(0,c)$ we can plug in $x=0$ in the equation $y=(x+2)(x-1)^2$ and we get $c=2$
| {
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Finding lim${_{n \rightarrow \infty}}\left( \frac{n^3}{2^n} \right)$ For a class of mine we were given a midterm review; however, I just cannot figure out how to finish this one:
Find the limit $$\lim_{n \rightarrow \infty}\left( \dfrac{n^3}{2^n} \right)$$
My attempt so far:
Let $s_n=\dfrac{n^3}{2^n}$.
Note that $2^... | Using your first idea, what about using the fact that $$(1+1)^n = \sum_{k=1}^n \binom{n}{k} \geq \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24} \geq \frac{(n-3)^4}{24}$$and
concluding by the squeeze theorem? (as $\frac{24n^3}{(n-3)^4} \xrightarrow[n\to\infty]{} 0$).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the double integral.
To find:
$$\iint_R(x^2-xy)dA$$
enclosed by
$$y=x, y=3x-x^2,$$
$$x=3x-x^2,$$
$$x=0, x=2,$$
$$y=0, y=2,$$
$$R=((x,y): 0\le x\le 2), (x \le y \le 3x-x^2)$$
$$\int_0^2\int_x^{3x-x^2}(x^2-xy)\,dy\,dx$$
$$\int_0^2 \left. \left(x^2y-x\frac{y^2}{2}\right) \right|_x^{3x-x^2}\,dx$$
where to fro... | Given your domain, let's first make a small correction:
\begin{align}
R &= \left\{(x,y) \in \mathbb{R}^2: 0 \leq x\leq 2, x \leq y \leq 3x-x^2 \wedge 2 \right\} \\
&= \left\{(x,y) \in \mathbb{R}^2: 0 \leq x\leq 1, x \leq y \leq 3x-x^2 \right\} \\
&\qquad\cup \left\{(x,y) \in \mathbb{R}^2: 1 < x \leq 2, x \leq y \leq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving that a function maintain a certain equation I need to prove that the function $$g(x,y,z)=f(\frac{1}{y}-\frac{1}{x},xye^{\frac{-z^2}{2}})$$ maintain the equation $$x^2g_x+y^2g_y=-\frac{x+y}{z}g_z$$ while $y\neq0$ and $f(u,v)$ differentiable for all $(u,v)$
How do i prove it without knowing what is $f$?
| You start by writing $g_x$ as function of $f_u$ and $f_v$, thus
$$g_x = \frac{ \partial g}{\partial x} = \frac{ \partial u}{\partial x} \frac{\partial f}{\partial u} + \frac{ \partial v}{\partial x} \frac{\partial f}{\partial v} = \frac{ \partial \frac{1}{y} -\frac{1}{x}}{\partial x} f_u + \frac{ \partial xy \exp\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of $1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=\frac{F_nF_{n+1}^2+(-1)^{n+1}[F_{n-1}+(-1)^{n+1}]}{2}$ Fibonacci series
$F_0=0$, $F_1=1$; $F_{n+1}=F_n+F_{n-1}$
This is a well known identity
$1^2+1^2+2^2+3^2+5^2+\cdots +F_n^2=F_nF_{n+1}$
I was curious and look every websites for a closed form of
$1^3+1^3+2^3+3^3+5^3+\cdots ... | For a (somewhat complicated) combinatorial proof see Identity 5 in Benjamin, Carnes, Cloitre. Recounting the Sums of Cubes of Fibonacci
Numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Derivation of Spherical Law of Cosines I am trying to get a derivation of the spherical law of cosines. The Wikipedia page [https://en.wikipedia.org/wiki/Spherical_law_of_cosines ] contains a proof that I don't understand because there are not enough intermediate steps shown.
The Wikipedia page says that for unit vect... | It looks like you just follow the angle definitions given and distribute.
You have:
$$
\begin{align}
\cos(a) = \mathbf{u}\cdot\mathbf{v}
\end{align}
$$
\begin{align}
\cos(b) = \mathbf{u}\cdot\mathbf{w}
\end{align}
$$
\begin{align}
\cos(c) = \mathbf{v}\cdot\mathbf{w}
\end{align}
$$
$$
\begin{align}
\mathbf{t}_a =
{{{\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Factor $6x^2 −7x−5=0$ I'm trying to factor
$$6x^2 −7x−5=0$$
but I have no clue about how to do it. I would be able to factor this:
$$x^2-14x+40=0$$
$$a+b=-14$$
$$ab=40$$
But $6x^2 −7x−5=0$ looks like it's not following the rules because of the coefficient of $x$. Any hints?
| Consider the product
\begin{align*}
(4x + 7)(3x - 8) & = 4x(3x - 8) + 7(3x - 8)\\
& = \color{blue}{12}x^2 \color{green}{- 32}x + \color{green}{21}x \color{blue}{-56}\\
& = \color{blue}{12}x^2 \color{green}{- 11}x \color{blue}{- 56}
\end{align*}
Observe that the product of the quadrati... | {
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"url": "https://math.stackexchange.com/questions/1776775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 0
} |
Prove: $\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$
Prove the trigonometric identity
$$\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$$
I've searched high and low on the net and cannot find identities where there is $+$ or $- 1$'s in the equation. Any help is appreciated.
Edit af... | Substitute $\frac1{\cos{x}}\;$ for $\sec{x}\;$ in the identity, and it will shout its truth to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I show that $\ln(2) = \sum \limits_{n=1}^\infty \frac{1}{n2^n}$? My task is this:
Show that $$\ln(2) = \sum \limits_{n=1}^\infty \frac{1}{n2^n}$$
My work so far:
If we approximate $\ln(x)$ around $x = 1$, we get:
$\ln(x) = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + ...$
Substituting $x ... | Note that
\begin{align}
\frac{1}{1-x}&=1+x+x^2+\cdots\qquad(-1<x<1).
\end{align}
Thus
\begin{align}
-\ln(1-x)&=\int\frac{1}{1-x}{\rm d}x\\
&=x+\frac{1}{2}x^2+\frac{1}{3}x^3+\cdots\\
&=\sum_{n=1}^\infty\frac{1}{n}x^n.
\end{align}
By taking $x=\frac{1}{2}$ into the above equation, we have
$$\sum_{n=1}^\infty\frac{1}{n2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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is $y = \sqrt{x^2 + 1}− x$ a injective (one-to-one) function? I have a function $y = \sqrt{x^2 + 1}− x$ and I need to prove if it's a Injective function (one-to-one). The function f is injective if and only if for all a and b in A, if f(a) = f(b), then a = b
$\sqrt{a^2 + 1} − a = \sqrt{b^2 + 1} − b$
$\sqrt{a^2 + 1} +... | Hint:
$$y=\sqrt{x^2+1}-x=\frac1{\sqrt{x^2+1}+x}$$
Then $$f(a)=f(b) \Leftrightarrow \frac1{\sqrt{a^2+1}+a}=\frac1{\sqrt{b^2+1}+b}\Leftrightarrow$$
$$\Leftrightarrow \sqrt{a^2+1}+a=\sqrt{b^2+1}+b \Leftrightarrow a=b$$
$f(t)=\sqrt{t^2+1}+t -$ strictly increasing
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
sum of all non real roots of the equation in a bi-quadratic equation
Consider the equation $8x^4-16x^3+16x^2-8x+a=0\;\left(a\in \mathbb{R}\right)\;,$ Then the sum of
all non real roots of the equation can be
$\bf{OPTIONS::}\;\; (a)\;\; 1\;\;\;\;\;\; (b)\;\; 2\;\;\;\;\;\; (c)\; \displaystyle \frac{1}{2}\;\; \;\;\;\; ... |
Now How can i solve it after that
Noting that $f'(1/2)=0$ is a key.
We have
$$f'(x)=8(4x^3-6x^2+4x-1)=8(2x-1)\left(2\left(x-\frac 12\right)^2+\frac 12\right)$$
So, $f(x)$ is decreasing for $x\lt 1/2$ and is increasing for $x\gt 1/2$.
By the way,
$$f\left(\frac 12+s\right)=8s^4+4s^2-\frac 32+a$$
and so $f(1/2-\alpha)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solution to a 2D recurrence equation I am seeking an explicit solution to this 2D recurrence equation:
\begin{eqnarray}
f(0,b) & = & b\\
f(a,0) & = & a\\
f(a,b) & = & f(a-1,b) - f(a,b-1)
\end{eqnarray}
So, for example, for $a=3$,
$$f(3,0)=3 \;,$$
$$f(3,1)=f(2,1)-f(3,0)=-2-3=-5 \;,$$
etc. Here is $f(3,b)$ for $b=0,\ldot... | Consider the generating function $g_n(x) = \sum_{k=0}^n f(k,n-k) x^k$. Thus $g_0(x) = f(0,0) = 0$ and
$$g_n(x) = n + n x^n + \sum_{k=1}^{n-1} (f(k-1,n-k) - f(k,n-1-k)) x^k
= 2n-1 + x^n + (x-1) g_{n-1}(x)$$
which has solution (according to Maple)
$$ g_n(x) = -2\,{\frac {n+1}{x-2}}-{\frac {-x+4}{ \left( x-2 \right) ^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Indicate on an Argand Diagram the region of the complex plane in which $ 0 \leq \arg (z+1) \leq \frac{2\pi}{3} $
Question: Indicate on an Argand Diagram the region of the complex plane in which $$ 0 \leq \arg (z+1) \leq \frac{2\pi}{3} $$
I've tried this
Consider $$ 0 \leq \arg (z+1) \leq \frac{2\pi}{3} $$
Split it... |
You switched the sign of the inequality somewhere. It should be $y\ge-\sqrt3(x+1)$. So you plot the two lines $y=0$ and $y=-\sqrt3(x+1)$ and mark the sector between them as shown.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+... | Your long division is wrong, because:
$$\frac{x^4+1}{x^3+x^2}=\frac{x^4+1}{x^2(x+1)}=\frac{1}{x^2}+x+\frac{2}{x+1}-\frac{1}{x}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Find $\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})}$ Find the value of: $$I=\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})}$$
First, I rewrote the limits as the function goes from $\frac{\pi}{4}$ to $\frac{9\pi}{4}$. Now the integral with the reduced limits has the value $\frac{I}{8}... | \begin{align*}
I&=\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})} \\
&= 8\int_{-\pi}^{\pi} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})} \tag{1}\\
&= 8\int_{-\pi}^{\pi}\frac{dx}{\left(1+2^{\cos (-x)}\right)\left(1+2^{\sin (-x)}\right)} \tag{2}\\
&=8\int_{-\pi}^{\pi} \frac{dx}{\left(1+2^{\cos x}\right)\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Derivatives using matrices good $$\left|\begin{matrix}
(1+x)^{a_1b_1} & (1+x)^{a_1b_2} & (1+x)^{a_1b_3} \\
(1+x)^{a_2b_1} & (1+x)^{a_2b_2} & (1+x)^{a_2b_3} \\
(1+x)^{a_3b_1} & (1+x)^{a_3b_2} & (1+x)^{a_3b_3} \\
\end{matrix}\right |$$
Find the coefficient of $x$ in the above determinant.
Try using differentiability.
| Denote the determinant by $f(x)$. Then the $x$ coefficient is given by $f'(0)$. Now the derivative of a determinant equals the sum of determinants of the matrices obtained by taking the derivative of each row separately. Hence
$$
\begin{align}
f'(0)
&=
\frac{\mathrm{d}}{\mathrm{d}x}
\left|
\begin{matrix}
(1+x)^{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $ \int \frac{1}{1 + x^3}dx $ $$ \int \frac{1}{1 + x^3}dx $$
Attempt:
I added and subtracted $x^3$ in the numerator but after a little solving I can't get through.
| $$\begin{align}
\int\frac{1}{1+x^3}\,dx
&=\int\frac{1-x^2+x^2}{1+x^3}\,dx\\
&=\int\frac{1-x^2}{1+x^3}\,dx+\int\frac{x^2}{1+x^3}\,dx\\
&=\int\frac{1-x}{1-x+x^2}\,dx+\frac13\ln\left(1+x^3\right)+C\\
&=\int\frac{1-(u+1/2)}{1-(u+1/2)+(u+1/2)^2}\,du+\frac13\ln\left(1+x^3\right)+C\\
&=\int\frac{1/2-u}{3/4+u^2}\,du+\frac13\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
What is a sumset? Say we have sets $X = \{0,2,3\}$ and $Y = \{1,2,5\}$.
Is the sumset defined to be
$X + Y = \{0 + 1,2 +2,3+5\} = \{1,4,8\}$
or summing every element pairwise
$X + Y = \{1,2,5,3,4,7,8\} $
?
| $$ A + B = \{a+b : a \in A, b \in B\}. $$
If $X = \{0,2,3\}$ and $Y = \{1,2,5\}$.
$$X+Y=\{0+1;0+2;0+5;2+1;2+2;2+5;3+1;3+2;3+5\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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538.com's Puzzle of the Overflowing Martini Glass - How to compute the minor and major axis of an elliptical cross-section of a cone FiveThirtyEight.com Riddler Puzzle / May 13
The puzzle goes like this;
"It’s Friday. You’ve kicked your feet up and have drunk enough of your martini that, when the conical glass () is up... | Diagram below shows side and aerial views of upright and tilted martini glass.
For the upright glass, the shape of the martini liquid is an inverted right circular cone, with height $p\sin\theta$ and circular base of area $A_1$. Volume of martini is given by:
$$\begin{align}
V&=\frac 13\cdot A_1 \cdot p\cos\theta\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 1
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If $x^4 + 3\cos(ax^2 + bx +c) = 2(x^2-2) $ has two solutions with $a,b,c \in (2,5)$, then find the maximum value of $\frac{ac}{b^2} $ The answer given is 1. i tried like this $3\cos(ax^2 + bx +c) = -x^4 +2x^2-4 = -(x^2 -1)^2 -3 $. The maximum value of $-x^4 +2x^2-4$ is $-3$ so $3\cos(ax^2 + bx +c) =-3$ and the two va... | You are on the right track. The next step is to utilize the condition $a, b, c \in (2, 5)$.
Since $\cos(a + b + c) = \cos(a - b + c)$ and $a + b + c > a - b + c$, we have
$$a + b + c = a - b + c + 2k\pi$$
for some positive integer $k$, implying $b = k\pi$. Since $b \in (2, 5)$, it can be seen that
$k = 1$, and $b = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792012",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$ $a,b,c >0$ and $a+b+c=3$, prove
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$$
I try to app... | Partial Hint :
With your work one can show that :
$$\left(\frac{a+1}{3-x-a+a} \right)\left(\frac{3-x-a+1}{3-x-a+x}\right)\left(\frac{x+1}{x+a} \right) \geqslant 1 $$
With $1\leq a\leq 2$ and $x\in[2-a,1]$
I cannot prove it but using Bernoulli's inequality we have :
$$ \left(\frac{a+1}{x+a} \right)^{\frac25}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 2,
"answer_id": 1
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How to prove $\sum_{n=1}^{\infty} \frac{3}{\sqrt[3]{n^2+2}}$ diverges? $$\sum_{n=1}^{\infty} \frac{3}{\sqrt[3]{n^2+2}}$$
It seems clear to me that this series diverges because the dominant temr is $1/n^{2/3}$, a p-series with $p < 1$
However I need to prove divergence using something like the comparison test, integral ... | Note that for all positive integers $n$,
$$n^2 + 2 < n^2 + 4n + 4 = (n+2)^2.$$ Therefore, $$\frac{3}{\sqrt[3]{n^2 + 2}} > \frac{3}{\sqrt[3]{(n+2)^2}} > \frac{1}{(n+2)^{2/3}},$$ hence $$\sum_{n=1}^\infty \frac{3}{\sqrt[3]{n^2+2}} > \sum_{n=1}^\infty \frac{1}{(n+2)^{2/3}} = - 1^{2/3} - \frac{1}{2^{2/}} + \sum_{n=1}^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding the area of a circle that is formed by cutting a sphere. Say I have a sphere $x^2+y^2+z^2=a^2$ and a plane $x+y+z=b.$ How do I find the surface area of the circle cut by the sphere on the plane? I think I would use the surface integral and for graphs the surface "element" is $\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}$ whe... | By symmetry the center of the circle is $c=\left(\frac b3,\frac b3,\frac b3\right)$, as noted in the comments. By the Pythagoras's theorem the radius of the circle is
$$r=\sqrt{a^2-|c|^2}=\sqrt{a^2-\frac{1}{3}b^2}$$
whenever the expression is defined, otherwise there is no intersection between the sphere and the plane,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$. Problem :
Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$.
My approach :
Since :
$(a^3+b^3)=(a+b)^3-3ab(a+b)$
$\Rightarrow z^3+\frac{1}{z^3}=(z+\frac{1}{z})^3-3(... | You are on the right path; you have shown that
$$\left(z+\tfrac{1}{z}\right)^3=z^3+\tfrac{1}{z^3}+3\left(z+\tfrac{1}{z}\right).$$
Taking norms it follows from the triangle inequality that
$$\left|\left(z+\tfrac{1}{z}\right)^3\right|=\left|z^3+\tfrac{1}{z^3}+3\left(z+\tfrac{1}{z}\right)\right|\leq\left|z^3+\tfrac{1}{z^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Limit of ($\sqrt{x^2+8x}-\sqrt{x^2+7x}$) as $x$ approaches infinity I've been stuck on this one problem for 3 days now, I don't know how to proceed. Any help would be appreciated.
The problem is asking for the $$\lim_{x\to\infty} (\sqrt{x^2+8x}-\sqrt{x^2+7x}) $$
Every time I attempt this problem, I can never get rid o... | You want
$\lim_{x\to\infty} (\sqrt{x^2+8x}-\sqrt{x^2+7x})
$.
For reals $a$ and $b$,
$\begin{array}\\
\sqrt{x^2+ax}-\sqrt{x^2+bx}
&=(\sqrt{x^2+ax}-\sqrt{x^2+bx})\dfrac{\sqrt{x^2+ax}+\sqrt{x^2+bx}}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{(\sqrt{x^2+ax}-\sqrt{x^2+bx})(\sqrt{x^2+ax}+\sqrt{x^2+bx})}{\sqrt{x^2+ax}+\sqrt{x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Generalized eigenvector space with $\lambda_1=\lambda_2=\lambda_3=2$ Let be $$M:=\begin{pmatrix}
1 & 1 & 1 \\
-1 & 3 & 1 \\
0 & 0 & 2
\end{pmatrix}$$
The characteristic polynomial would then be $\chi_M(\lambda)=\lambda^3-6\lambda^2+12\lambda-8$, and his zeros points are $\lambda_1=\lambda_2=\lambda_3=2$. Now we have t... | Let $$v_3=\begin{pmatrix}-1\\0\\0\end{pmatrix}.$$ Then $$(M-2I)v_3 = \begin{pmatrix}-1&1&1\\-1&1&1\\0&0&0\end{pmatrix}\begin{pmatrix}-1\\0\\0\end{pmatrix} = \begin{bmatrix}1\\1\\0\end{bmatrix}=v_1, $$
so
$$(M-2I)^2v_3 = (M-2I)(M-2I)v_3=(M-2I)v_1=0, $$
and hence $v_3$ is a generalized eigenvector of $M$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the range of $a$ for $a^2 - bc - 8a + 7 = 0$ and $b^2 + c^2 + bc - 6 a + b = 0$. Let $a, b, c$ be real numbers such that
\begin{cases}
a^2 - bc - 8a + 7 = 0\\
b^2 + c^2 + bc - 6 a + b = 0\\
\end{cases}
Find the range of $a$.
By adding the two equations I have $$(a - 7)^2 + (b + \frac 12)^2 + c^2 = \frac {169}4,$$
... | This is the "official" solution which I found later:
\begin{cases}
a^2 - bc - 8a + 7 = 0\\
b^2 + c^2 + bc - 6 a + \mathbf6= 0\\
\end{cases}
is equivalent to
\begin{cases}
bc=a^2 - 8a + 7\\
b+c=\pm(a-1)\\
\end{cases}
So $b,c$ are the roots of the equation
$$x^2\pm(a-1)x+(a^2-8a+7)=0$$
A necessary and sufficient conditio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Prove $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32+ \frac{27}{16}\frac{(y-z)^2}{(x+y+z)^2}$
$x,y,z >0$, prove
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32+ \frac{27}{16}\frac{(y-z)^2}{(x+y+z)^2}$$
This inequality is easier than the other one. Previously, I learned from Jack D'Aurizi... | Here's a proof.
Following the task description, we want to establish
$$
\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2}
$$
We will follow two paths for separate cases.
Path 1:
Clearing denominators, we obtain
$$
(x-y)^2 (x+y) + (y-z)^2 (y+z)+ (z-x)^2 (z+x) \geq \frac{27}{8} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How should I try to evaluate the integral $\int_a^b \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ I've tried to evaluate $\displaystyle\int_{-r}^r \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ on my own, but I have encountered a problem I cannot get around.
The indefinite integral $\sqrt{\frac{r^2}{r^2-x^2}} \sqrt{r^2-x^2} \tan ^{... | $$\displaystyle\int_{-r}^r \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx=2r\int_{0}^r \frac{1}{\sqrt{r^2-x^2}}dx=2r\,\left. {{\sin }^{-1}}\left( \frac{x}{r} \right) \right|_{0}^{r}=r\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\fr... | Hint. From what you have proved, one may deduce that
$$
\frac{\pi}{2}\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}=\int_{0}^{\large \frac{\pi}{2}}\sum^{\infty}_{n=0}\frac{1}{4^n}\sin^{2n}xdx=\int_{0}^{\large \frac{\pi}{2}}\frac{4}{4-\sin^2 x}dx
$$ the latter integral being easy to evaluate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
How does $\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$ relate to $\sqrt{x^2+y^2+z^2}$? Another possible 'mean' for three positive real numbers $x,y,z$ is made of pairwise quadratic means:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}$$
By QM-AM inequality it is greater than or equal to arithmetic m... | Never mind, it's easy to prove that
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}$$
Square the RHS:
$$(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2=2(x^2+y^2+z^2)+2 \sum_{cyc} \sqrt{(x^2+y^2)(y^2+z^2)}$$
Using AM-GM we obtain:
$$\sqrt{(x^2+y^2)(y^2+z^2)} \leq \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Compute the minimum value of $ \underline{A}\ \underline{B}\ \underline{C} - (A^2 + B^2 + C^2). $ Let $\underline{A}\ \underline{B}\ \underline{C}$ represent a three-digit base 10 number whose digits are $A$, $B$, and $C$ with $A \ge 1$. Compute the minimum value of $$ \underline{A}\ \underline{B}\ \underline{C} - (A^2... | Your attempts are good; you want to minimize the expression
$$100a-a^2+10b-b^2+c-c^2,$$
which is the sum of three quadratics. It is minimal when the three quadratics
$$100a-a^2,\qquad 10b-b^2,\qquad c-c^2,$$
are minimal. These are parabolas with their maxima at $50$, $5$ and $\tfrac{1}{2}$, respectively. They are minim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int\frac{\sin x}{\sqrt{1-\sin x}}dx=?$ Calculate this integral $\displaystyle\int\dfrac{\sin x}{\sqrt{1-\sin x}}dx=?$
Effort;
$1-\sin x=t^2\Rightarrow \sin x=1-t^2\Rightarrow \cos x=\sqrt{2t^2-t^4}$
$1-\sin x=t^2\Rightarrow-\cos x dx=2tdt\Rightarrow dx=\frac{2t}{\sqrt{t^4-2t^2}}dt$
$\displaystyle\int\frac{1-t^2}{t}... | Slightly different attempt to get rid of the trig functions (assuming a domain where no sign issues cause trouble):
$$\begin{array}{rl}
\displaystyle \frac{\sin x}{\sqrt{1-\sin x}}
& \displaystyle
= \frac{\sin x\sqrt{1+\sin x}}{\sqrt{1-\sin x}\sqrt{1+\sin x}}\\[7 pt]
& \displaystyle
= \frac{\sin x\sqrt{1+\sin x}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
derivative with square root I have been trying to figure this equation for some time now, but have come up empty. I have tried multiple ways on solving it. Whether by using the Quotient Rule or some other method, I can't seem to figure it out. Any help would be appreciated.
Find the derivative of the function
$
y = \f... | For differentiating $\sqrt x$ one should rewrite it as $x^{\frac{1}{2}}$ and use the power rule. Given $y=\frac{x^2+8x+3}{\sqrt x}$ we have:
$$y'=\frac{\left(\sqrt x\right)\left(2x+8\right)-\left(\frac{1}{2}x^{-\frac{1}{2}}\right)\left(x^2+8x+3\right)}{\left(\sqrt x\right)^2}
\\=x^{-\frac{3}{2}}\left(2x^2+8x-\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Integration by Parts on $\sqrt{1+x^2}dx$ I have been asked to show that $$\int\sqrt{1+x^2}dx=\frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\int\frac{1}{{\sqrt{1+x^2}}}dx$$
I'm aware of how to do this with trig substitution, but the question is specifically regarding integration by parts, and the only examples I can find keep usi... | Start with integration by parts
$$v=x, \ dv=dx$$
$$u=\sqrt{1+x^2}, \ du=\frac{x}{\sqrt{1+x^2}}dx$$
Therefore
\begin{align*}\int \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{(1+x^2)-1}{\sqrt{1+x^2}}dx\\
&=x\sqrt{1+x^2}-\int\frac{1+x^2}{\sqrt{1+x^2}}dx+\int\frac{dx}{\sqrt{1+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Problem solving a word problem using a generating function
How many ways are there to hand out 24 cookies to 3 children so that they each get an even number, and they each get at least 2 and no more than 10? Use generating functions.
So the first couple steps are easy.
The coefficient is $x^{24}$
$g(x) = x^6(1+x^2+x... | Your expression $g(x) = x^6 (1+x^2+\cdots+x^8)^3$ is correct, but the geometric series formula needs to be used correctly. Recall the geometric series formula for the sum of 5 terms: $a+ar+\cdots+ar^4 = \frac{a(1-r^5)}{1-r}$. In our case, $r=x^2$ and $a=1$. So $1+x^2+\cdots+x^8$ simplifies to $\frac{1-(x^2)^5}{1-x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I tho... | Hint: This is equivalent to asking what $$1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$$ is modulo $10$. But $m=m-10$ mod $10$, so modulo 10 the above is the same as
$$1^n+2^n+3^n+4^n+5^n+(-4)^n+(-3)^n+(-2)^n+(-1)^n.$$
What does this equal if $n$ is odd versus even?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
expansion of a generating function I found this formula in a book
$$\sqrt{1-4x}=1-2\sum_{n=1}^{\infty}\frac{1}{n} {{2n-2}\choose {n-1}} x^n$$
How can I prove that?
| There are quite a few ways to prove it.
You could note that $\frac1n\binom{2n-2}{n-1}$ is $C_{n-1}$, the $(n-1)$-st Catalan number, so that the formula can be written
$$\sqrt{1-4x}=1-2\sum_{n\ge 0}\frac1{n+1}\binom{2n}nx^{n+1}=1-2\sum_{n\ge 0}C_nx^{n+1}\;.$$
Now you can simply manipulate the known generating function ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
proof of $\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}$ for large $n$ I'm trying to conclude this question in which I just need to prove that:
$$0<\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}$$
and then I'll have that, for large $n$:
$$\begin{align}
\left\lvert\cos\left(\sqrt... | Well, as $x$ is positive, by monotonicity of the square root
$$\sqrt{x^2+\frac1n}+x \geq \sqrt{x^2+\frac1n}+\geq \sqrt{\frac1n}=\frac1{\sqrt{n}}$$
Thus
$$\frac{\frac{1}{n}}{\sqrt{x^2+\frac1n}+x}\leq \frac{\frac{1}{n}}{\sqrt{x^2+\frac1n}}\leq \frac{\frac{1}{n}}{\frac1{\sqrt{n}}}=\frac{1}{\sqrt{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Positive integers $a$ and $b$ are such that $ab+1$ is a factor of $a^2+b^2$. Prove that $\frac{a^2+b^2}{ab+1}$ is a perfect square. IMO 1988 question No. 6
I have a confusion in the question.The question is as follows:
$a$ and $b$ are positive integers and $ab+1$ is a factor of $a^2+b^2$. Prove that $\displaystyle\fra... | Let us assume $(a^2 + b^2)/(ab + 1) = k \in \mathbb{N}$. We then have $a^2 - kab + b^2 = k$. Let us assume that $k$ is not an integer square, which implies $k \ge 2$. Now, we observe the minimal pair $(a, b)$ such that $a^2 - kab + b^2 = k$ holds. We may assume, without loss of generality, that $a \ge b$. For $a = b$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
number of real triplets $(x,y,z)$ in system of equations
Total number of real triplets of $(x,y,z)$ in $x^3+y^3+z^3=x+y+z$ and $x^2+y^2+z^2=xyz$
$\bf{My\; Try::}$ Let $x+y+z=a$ and $xy+yz+zx=b\;,$
Then $(x+y+z)^2-2(xy+yz+zx)=xyz\implies a^2-2b=xyz$
And using $\displaystyle x^3+y^3+z^3-3xyz=(x+y+z)\left[x^2+y^2+z^2-(x... | Assume that $x+y+z=a$ and $xyz=b$. We have:
$$ e_2 = xy+xz+yz = \frac{a^2-b}{2} $$
and we may consider that:
$$ p(t) = (t-x)(t-y)(t-z) = t^3 - e_1 t^2 + e_2 t^2 - e_3 = t^3 - at^2 + \frac{a^2-b}{2}t - b. $$
On the other hand, for any $w\in\{x,y,z\}$ we have $p(w)=0$ or
$$ w^3 = e_1 w^2 - e_2 w + e_3,\tag{1} $$
hence by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integral of $\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx$
Find the integral of the following:
$$\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx$$
Do set $u=x^3+3x^2+1$?
So, $du=(3x^2+6x)dx$?
And, $x^2+2x={u-1-x^2\over x}$?
So then,
$$\int{({u-1-x^2\over x})\over \sqrt u} du$$
This seems very complicated, is there any easier w... | $du = (3x^2+6x)dx = 3(x^2+2x)dx$, so your answer changes as $\frac{1}{3}\int \frac{du}{\sqrt{u}} = \frac{2}{3}\sqrt{u}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n}{\left(n+1\right)!} = 1-\frac{1}{\left(n+1\right)!}$
So I proved the base case where $n=1$ and got $\frac{1}{2}$
Then since $n=k$ implies $... | You have your base case.
Explicitly state your inductive hypothesis.
Suppose:
$\sum_\limits{n=1}^k\frac{n}{(n+1)!} = 1-\frac{1}{(k+1)!}$
We will show that:
$\sum_\limits{n=1}^{k+1}\frac{n}{(n+1)!} = 1-\frac{1}{(k+2)!}$
$\sum_\limits{n=1}^k\frac{n}{(n+1)!}+\frac{(k+1)}{(k+2)!}$
$1-\frac{1}{(k+1)!}+\frac{(k+1)}{(k+2)!}$(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\f... | Put $x=\dfrac1y$
$$I=\int_\infty^0\dfrac{y^4\cdot -\ln(y)}{y(y^2+1)^2}\cdot-\dfrac{dy}{y^2}=\int_\infty^0\dfrac{y\ln y}{(1+y^2)^2}dy=-\int_0^\infty\dfrac{y\ln y}{(1+y^2)^2}dy=-I$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
first orderr non linear ODE I came along this first order non linear ODE, and cannot solve it.
$$\frac{dv}{dt}=\frac{-b}{(vt)^2}+k$$ (where b and k are constants)
The question asked to express v as a function of t.
Thank you very much, any help is welcome!
| Assume $b\neq0$ and $k\neq0$ for the key case:
Hint:
Approach $1$:
$\dfrac{dv}{dt}=-\dfrac{b}{(vt)^2}+k$
$\dfrac{dv}{dt}-k=-\dfrac{b}{v^2t^2}$
$\dfrac{d(v-kt)}{dt}=-\dfrac{b}{v^2t^2}$
Let $x=v-kt$ ,
Then $\dfrac{dx}{dt}=-\dfrac{b}{(x+kt)^2t^2}$
$\dfrac{dt}{dx}=-\dfrac{k^2t^4+2kxt^3+x^2t^2}{b}$
This belongs to a "Chini-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral solution and odd prime Here is my question:
Let $p\neq 7$ be an odd prime. Suppose that only one of the two equations,
$$
x^2 + 7y^2 = p, \quad x^2 - 7y^2 = p
$$
has integral solution $(x,y)$. Prove that $p \equiv 3\;(\mathrm{mod}\; 4)$.
What I have done is that in modulo $7$, $x^2$ can be $1, 2$ or $4$.... | First we show that there are integers $x$ and $y$ such that $x^2+7y^2=p$, and $p$ is a prime not equal to $7$, then $-7$ is a quadratic residue of $p$.
If $x^2+7y^2=p$, then $y$ is not divisible by $p$. For suppose to the contrary that $p$ divides $y$. Then $p$ divides $x$, so $p^2$ divides $x^2+7y^2$, which is impossi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $x$ and $y$ are non-negative integers for which $(xy-7)^2=x^2+y^2$. Find the sum of all possible values of $x$. I am not able to reach to the answer. I have used discriminant as $x$ and $y$ are both integers but it didn't give any hint to reach to answer. I am not able to understand how should I deal with these type... | We have $$(xy)^2-12xy+49=(x+y)^2\iff(xy-6)^2+13=(x+y)^2$$
hence $$13=(x+y+xy-6)(x+y-xy+6)$$ It follows the only possibility (after discard$-13$ and $-1$)
$$x+y+xy-6=13\\x+y-xy+6=1$$ This implies clearly $$x+y=7$$ which gives the candidates $(0,7),(1,6),(2,5),(3,4)$ and symmetrics and it is verified that the only to be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Integrating $\int\frac{x^3}{\sqrt{9-x^2}}dx$ via trig substitution What I have done so far:
Substituting $$x=3\sin(t)\Rightarrow dx=3\cos(t)dt$$ converting our integral to
$$I=\int\frac{x^3}{\sqrt{9-x^2}}dx=\int \frac{27\sin^3(t) dt}{3\sqrt{\cos^2(t)}}3\cos(t)dt\\
\Rightarrow \frac{I}{27}=\int \sin^3(t)dt=-\frac{\sin^... | A nice trick to do in this situation is to write
$$
\sin^3t = (1-\cos^2t)\sin t.
$$
You can now do a secondary $u$-substitution where you can take $u = \cos t$ and $du = -\sin tdt$.
Edit
Where you left off you had
$$
\frac{I}{27} \;\; =\;\; \int \sin^3t dt \;\; =\;\; \int (1-\cos^2t)\sin t dt
$$
If we now take the $u$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826494",
"timestamp": "2023-03-29T00:00:00",
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Probability that at least 2 edges of $\Gamma_{n,N}$ shall have a point in common In the classic paper of Erdos,Renyi On the evolution of random graphs[page 7] ,it is argued that the probability that at least 2 edges of $\Gamma_{n,N}$ shall have a point in common is given by $1-\frac{{n\choose 2N}(2N)!}{2^N N!{{n \choos... | An important condition on $N$ in that paper is that $N = o(n^{1/2}).$ This allows us to more easily deal with the binomial coefficient terms. Namely, since $N = o(n^{1/2})$, we have that
$$
{n \choose 2N} = (1+o(1)) \frac{n^{2N}}{(2N)!},
$$
and
$$
{{n \choose 2} \choose N} = (1+o(1)) \frac{ {n \choose 2}^N}{N!}.
$$
But... | {
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"url": "https://math.stackexchange.com/questions/1827559",
"timestamp": "2023-03-29T00:00:00",
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Prove that $2+2\sqrt{12n^2+1}$ is perfect square Problem:
Let $n \in \mathbb{N}$ such that $2+2\sqrt{12n^2+1}$ is the integer. Prove that $2+2\sqrt{12n^2+1}$ is perfect square.
I tried to found $n$ such that $\sqrt{12n^2+1}$ is integer, i.e. $12n^2 + 1 = k^2$. It is a Pell equation, and it has solution: $(n_0,k_0) =... | Let $$12n^2+1=4k^2+4k+1 \Rightarrow 3n^2=k^2+k=k(k+1)$$
This implies that $k=3a^2, k+1=b^2$ or $k=b^2, k+1=3a^2$. However, the second case is impossible as $$b^2 =3a^2-1 \equiv -1 \pmod 3$$
Thus, we have that $k+1$ is a perfect square.
Note $$2+2\sqrt{12n^2+1}=4k+4=(2b)^2$$ Our proof is done.
| {
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"url": "https://math.stackexchange.com/questions/1828275",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\... | Both integrands are even, so the integrals can be extended to negative infinity. The integrands decay quickly enough for the contour to be closed with a semicircle at infinity in the upper half plane without chaning the integral. Then the integrals can be determined with the residue theorem.
Multiplying the denominator... | {
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"url": "https://math.stackexchange.com/questions/1829298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Prove that $\frac{2^a+1}{2^b-1}$ is not an integer
Let $a$ and $b$ be positive integers with $a>b>2$. Prove that $\frac{2^a+1}{2^b-1}$ is not an integer.
This is equivalent to showing there always exists some power of a prime $p$ such that $2^a+1 \not \equiv 0 \pmod{p^a}$ but $2^b-1 \equiv 0 \pmod{p^a}$. How do we pr... | Assume that $2^b-1$ is a divisor of $2^a+1$ and write $a$ as $kb+r$ with $0\leq r<b$. Then:
$$2^a+1\equiv (2^b)^k\cdot 2^r+1 \equiv 2^r+1 \pmod{(2^b-1)} $$
but since $r<b$ and $b>2$, $2^r+1$ is too small to be $\equiv 0\pmod{2^b-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Given: 2 lines containing the diameter of a circle and a point lying on this circle; Find: the equation of this circle The lines $ y = \frac{4}{3}x - \frac{5}{3} $ and $ y = \frac{-4}{3}x - \frac{13}{3} $ each contain diameters of a circle. and the point $ (-5, 0) $ is also on that circle.
Find the equation of this ci... | Prerequisites
Given
*
*We have 2 lines with given equations (see below).
*
*$ y = \frac{4}{3}x - \frac{5}{3} $
*$ y = \frac{-4}{3}x - \frac{13}{3} $
*These lines each contain a diameter of the circle-in-question.
*We have a point with the given coordinate (see below).
*
*$ P = (-5, 0) $
*This point is o... | {
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"url": "https://math.stackexchange.com/questions/1832557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the values of a and b so that $x^4+x^3+8x^2+ax+b$ is exactly divisible by $x^2+1$ I have been trying this question for a long time but I am not getting it. So please help me and try to make it as fast as possible
| Method 1.
Subtracting $x^2(x^2+1)$ from $x^4+x^3+8x^2+a x+b$ gives $x^3+7 x^2 +a x+b.$
Subtracting $x(x^2+1)$ from $x^3+7 x^2+a x+b$ gives $7 x^2+(a-1)x+b.$
Subtracting $7(x^2+1)$ from $7 x^2+(a-1)x+b$ gives $(a-1)x+(b-7).$
Observe that $(a-1)x+(b-7)=0$ for every $x$ if and only if $a=1$ and $b=7.$
Method 2.
$f(x)=x^4+... | {
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"url": "https://math.stackexchange.com/questions/1834138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + 1/β$ and $β + 1/α$. Quadratic equation question, as specified in the title.
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + \frac{1}{β}$ and $β + \frac{1}{α}$.
I gather that ... | As $\alpha\beta=\dfrac71,$
let $y=\dfrac{\alpha\beta+1}\alpha=\dfrac{7+1}\alpha\iff\alpha=\dfrac8y$
Put the value of $\alpha$ in $$x^2-6x+7=0$$ and rearrange.
| {
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"url": "https://math.stackexchange.com/questions/1837156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$
Prove that $\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$.
I tried using the partial fraction decomposition $a_j = \frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}$, but I don't see... | Hint. From what you wrote one may observe that
$$
\frac{1}{j (j+1)(j+2)}=\frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}=\frac12\left(\frac1j - \frac1{j+1}\right)+ \frac12\left(\frac1{j+2} - \frac1{j+1}\right)
$$ then noticing that terms telescope.
Remark. One may also write
$$
\frac{1}{j (j+1)(j+2)}=\frac12\frac1{j(j... | {
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"url": "https://math.stackexchange.com/questions/1837431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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.