Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculate the value of the series $\,\sum_{n=1}^\infty\frac{1}{2n(2n+1)(2n+2)}$ Calculate the infinite sum
$$\dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{1}{4\cdot 5\cdot 6}+\dfrac{1}{6\cdot 7\cdot 8}+\cdots$$
I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum.
Is th... | Hint. First observe that
$$
\frac{1}{2i(2i+1)(2i+2)}=\frac{1}{2}\left(\frac{1}{2i(2i+1)}-\frac{1}{(2i+1)(2i+2)}\right)=\frac{1}{2}\left(\frac{1}{2i}-\frac{2}{2i+1}+\frac{1}{2i+2}\right)
$$
Then
$$
\sum_{i=1}^n\frac{1}{2i(2i+1)(2i+2)}=\sum_{i=1}^n\left(\frac{1}{2i}-\frac{1}{2i+1}\right)-\frac{1}{4}+\frac{1}{2(2n+2)}\\=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$
Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$
I get the answer $\frac{-1}{ \sqrt2}$
By solving like this
\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\
2\sin^{-1}(- x) &= \frac\pi2\\
\sin^{-1}(- x) &= \frac\pi4\\
-x &= \sin\left(\frac\pi4\right)\end{align}
Thus $x =\frac{-1}{\... | Okay so expand the double angle to get $\cos^2(\sin^{-1}(-x))-\sin^2(\sin^{-1}(-x)=0$.
This should give you $(1-(-x)^2)-(-x)^2=0$.
Finally you have $1-2x^2=0$.
Solutions are $\displaystyle \boxed{\pm \frac{1}{\sqrt{2}}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Green's Theorem in the plane When calculating $\oint _{ C }^{ }{ \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2 } dx+\left( \frac { { x }^{ 2 } }{ 2 } +{ y }^{ 4 } \right) dy } $ where $C$ Is the boundary of the region $D=\left\{ \left( x,y \right) \in \mathbb{R}^{ 2 }: 1\le { x }^{ 2 }+{ y }^{ 2 }\le 4,x\ge 0,y\ge 0 \right\}... | Let $P(x,y)= \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2 }$ and $Q(x,y)= \frac { { x }^{ 2 } }{ 2 } +{ y }^{ 4 }$.
Then
$$
\begin{align*}
\oint _{ C }^{ }{ \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2 } dx+\left( \frac { { x }^{ 2 } }{ 2 } +{ y }^{ 4 } \right) dy }
&= \iint_D ( Q_x-P_y ) dA \\
&= \iint_D x+y \: dA \\
&= \int_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find this limit. Compute the value of the limit :
$$
\lim_{x\to\infty}{\frac{1-\cos x\cos2x\cos3x}{\sin^2x}}
$$
I've tried simplifying the expression to
$$
\lim_{x\to\infty}\frac{-8\cos^6x+10\cos^4x-3\cos^2x+1}{\sin^2x}
$$
But I don't know what to do after this.
| The limit at $\infty$ does not exist: consider the sequence
$$
a_n=\frac{\pi}{3}+2n\pi
$$
Then the numerator evaluated at $a_n$ is
$$
1-\cos\frac{\pi}{3}\cos\frac{2\pi}{3}\cos\pi=1-\frac{1}{4}=\frac{3}{4}
$$
and the denominator is $\frac{3}{4}$, so the quotient is $1$.
On the other hand, for the sequence
$$
b_n=\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
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Find the minimum and maximum distance between the ellipsoid of ecuation $x^2+y^2+2z^2=6$ and the point $P=(4,2,0)$ I've been asked to find, if it exists, the minimum and maximum distance between the ellipsoid of ecuation $x^2+y^2+2z^2=6$ and the point $P=(4,2,0)$
To start, I've tried to find the points of the ellipsoid... | Since by C-S $$2x+y\leq\sqrt{(x^2+y^2)(2^2+1^2)}=\sqrt{5(x^2+y^2)},$$
we obtain
$$(x-4)^2+(y-2)^2+z^2=x^2+y^2+z^2-4(2x+y)+20\geq $$
$$\geq x^2+y^2+z^2-4\sqrt{5(x^2+y^2)}+20=6-2z^2+z^2-4\sqrt{5(6-2z^2)}+20=$$
$$=26-z^2-4\sqrt{5(6-2z^2)}\geq26-4\sqrt{30}.$$
It's obvious that the equality occurs, which gives the minimal v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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If 25 divides $a^5 + b^5 + c^5 + d^5 + e^5$ then prove that 5 divides $abcde$. I solved the following problem and giving the outline of my solution. I'm curious to know if a shorter and more elegant solution exists.
Problem
Let $a,b,c,d,e \in \mathbb{N}$ such that $25 | (a^5 + b^5 + c^5 + d^5 + e^5)$. Prove that $5 | a... | Since $5^1\binom{5}{1}=25$, we have $(n+5)^5\equiv n^5\pmod{25}$. Thus, we can generate the table
$$
\begin{array}{c|c}
n\pmod{5}&n^5\pmod{25}\\
0&0\\
1&1\\
2&7\\
3&-7\\
4&-1\\
\end{array}
$$
Since there are no solutions to $7x+y=25$ with $|x|+|y|\le5$, we must be looking for $7x+y=0$ with $|x|+|y|\le5$, which has only... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Can we identify the limit of this arithmetic/geometric mean like iteration? Let $a_0 = 1$ and $b_0 = x \ge 1$. Let
$$
a_{n+1} = (a_n+\sqrt{a_n b_n})/2, \qquad b_{n+1} = (b_n + \sqrt{a_{n+1} b_n})/2.
$$
Numeric computation suggests that regardless of the choice of $x$, $a_n$ and $b_n$ always converge to the same value. ... | We have $1\le a_0\le b_0\le x$.
We can prove that $1\le a_n\le a_{n+1}\le b_{n+1}\le b_n\le x$ by mathematical induction.
$\displaystyle a_1=\frac{1+\sqrt{x}}{2}$ and $\displaystyle b_1=\frac{x+\sqrt{a_1x}}{2}$.
Obviously, $a_1\ge 1$. Hence $\displaystyle b_1-a_1=\frac{(x-1)+(\sqrt{a_1}-1)\sqrt{x}}{2}\ge0$
Note that
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Integrating the Fourier series to find the Fourier series of $\frac{1}{2}x^2$
The function $\phi(x) = x$ on the interval $[-l,l]$ has the Fourier series
$$x = \frac{2 l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi x}{l} \right) = \frac{2 l}{\pi}\left(\sin\left(\frac{\pi x}{l} \right) - \frac{1}{2... | What has been given is
$$\frac{x^2}{2} = -\frac{2 l^2}{\pi^2}\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m^2}\left(\cos\left(\frac{m \pi l}{l}\right) - 1\right).$$
Now consider:
$$\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$
and
\begin{align}
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} &= \frac{1}{1^2} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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What are the diagonals in the matrix called? How correctly to designate these diagonals? They are highlighted in different colors.
\begin{pmatrix}\color{green}1&\color{orange}2&\color{red}3&\color{blue}4\\
\color{orange}5&\color{red}6&\color{blue}7&\color{green}8\\
\color{red}9&\color{blue}{10}&\color{green}{11}&\color... | A cyclic generator matrix ( a generator for the cyclic group of order 4 ):
$$ {\bf C} =\left[\begin{array}{cccc}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0\end{array}\right]$$
if we multiply to the left of
$${\bf M} = \left[\begin{array}{cccc}0&0&0&\color{blue} 4\\0&0&\color{blue}3&0\\0&\color{blue}2&0&0\\\color{blue}1&0&0&0\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Finding values of $t$ I have this equation -
$3t^{\frac{1}{2}} - \frac{2}{5} t^{\frac{-3}{2}} = 0 $
I'm struggling on how to find the values of $t$
First, I power both sides by $2$ to make the power be a whole number .
I get
$3t - \frac{2}{5} t^{-3} = 0 $
From here I'm stunned and stuck . Can I get a hint ! Thank... | The equation
$$
3t^{\frac{1}{2}} - \frac{2}{5} t^{\frac{-3}{2}} = 0
$$
is equivalent to
$$
t^{\frac{1}{2}} = \frac{2}{15} t^{\frac{-3}{2}} \;.
$$
Since $t$ cannot be zero$\color{green}{^*}$, you are allowed to multiply both sides by $t^{\frac{3}{2}}$:
$$
t^{2} = \frac{2}{15} \;.
$$
So, the solution is
$$
t = \pm\sqrt{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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PA = LU Decomposition with Row Exchange I am not sure how to deal with the L with we do row exchange in PA = LU decomposition. Here's my example:
$
A =
\left[ {\begin{array}{ccc}
1 & 1 & 1\\
0 & 0 & 1\\
2 & 3 & 4\\
\end{array} } \right]
$
Step 1: Swap row 1 and row 3
$
U = \left[ \begin{array}{ccc}
... | When you check you answer you have to check it by PA = LU.
What I did first was put A in an upper triangle.
$
A =
\left[ {\begin{array}{ccc}
1 & 1 & 1\\
0 & 0 & 1\\
2 & 3 & 4\\
\end{array} } \right]
$
Step 1: $r_3$-2$r_1$→$r_3$
$
= \left[ \begin{array}{ccc}
1 & 1 & 1\\
0 & 0 & 1\\
0 & 1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
continuity of a function series I need help finding for which $x$ the function:
$$\sum_{n=1}^{\infty}\frac{x+n(-1)^n}{x^2+n^2}$$
is continuous.
I first need to show that the series converges uniformly, the thing is I don't know how to deal with the $(-1)^n$. I tried comparing to the series $\sum_{n=1}^{\infty}\frac{x+... | $f(x) = \sum_{n=1}^{\infty}\frac{x+n(-1)^n}{x^2+n^2}$
We need to show that for any $\epsilon$ there exists a $\delta$ such that for all $x,y$ in the domain, when $|x-y|< \delta,$ then $|f(x) - f(y)| < \epsilon$.
The first thing we are going to need is some upper bound on $f(x)-f(y)$
$f(x) -f(y) =$$ \sum_{n=1}^{\infty}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $5^n +5 <5^{n+1}$ $∀n∈N$ using induction Prove $5^n +5 <5^{n+1}$ $∀n∈N$
Base Case: $n=1$
$\implies 5^1 +5 <25$
$\implies 10<25$ ; holds true
Induction hypothesis: Suppose $5^k +5 < 5^{k+1}$ is true for k∈N
Then;
$\implies 5^{k+1} +5 < 5^{k+2}$
$\implies 5\cdot 5^k +5 < 25*5^k$
I don't know how to proceed after t... | $5^{k+1} + 5 < 5^{k+2} = 5(5^{k} + 1 < 5^{k+1})$. As $5^{k} + 1 < 5^{k} + 5 < 5^{k+1}$(from the inductive hypothesis), you have your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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The perimeter is equal to the area The measurements on the sides of a rectangle are distinct integers. The perimeter and area of the rectangle are expressed by the same number. Determine this number.
Answer: 18
It could be $4*4$ = $4+4+4+4$ but the answer is 18.
Wait... Now that I noticed, the sides are different ... | Let's see, let $x$ be the length of the one side and $y$ be the length of one of the adjacent sides. Then the perimeter of the rectangle is $P(x,y)=2x+2y$ and the area is $A(x,y)=xy$. So we need $P=A$ i.e. $2x+2y=xy$. Thta is, $$2x+2y=xy \Leftrightarrow 2x-xy+2y+4=4 \Leftrightarrow (x-2)(y-2)=4.$$ Note that $4=2 \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $... | Inspired/ Trying to make sense of Jaideep's Answer ...
\begin{eqnarray*}
\tan(5 \theta) =\frac{5 \tan(\theta)-10 \tan^3(\theta)+\tan^5(\theta)}{1-10\tan^2(\theta) +5\tan^4(\theta)}
\end{eqnarray*}
Let $\theta=\frac{\pi}{5}$ and $x=tan(\frac{\pi}{5})$. We have the quintic
\begin{eqnarray*}
5x-10x^3+x^5=0
\end{eqnarray*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
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Find the magnitude of the vertex angle of an isosceles triangle of the given area $A$ Find the magnitude of the vertex angle $\alpha$ of an isosceles triangle with the given area $A$ such that the radius $r$ of the circle inscribed into the triangle is maximal.
My attempt:
| Suppose that the base of our trangle is $2a$ then the coresponding height is $A/a$ and the length of the side is $b=\sqrt{a^2+A^2/a^2}$, and finally the semi perimeter of the triangle is
$$p= a+\sqrt{a^2+\frac{A^2}{a^2}}$$
Now because $pr=A$ the maximal value of $r$ corresponds to the smallest value of $p$. So, we nee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Finding the volume of the solid generated by revolving the given curve. The objective is to find the volume of the solid generated by revolving the curve $y=\dfrac{a^3}{a^2+x^2}$ about its asymptote.
Observing the given function yields that $y\ne0$, hence $y=0$ is the asymptote to the given curve. Thus, the volume of t... | This seems like a trig substitution problem to me, albeit a not-so-straightforward one.
Let $x = a \tan\theta$. Then $dx = a \sec^2\theta\,d\theta$, and $a^2 + x^2 = a^2 \sec^2 \theta$. Since $\theta = \tan^{-1}\left(\frac{x}{a}\right)$, $\theta = 0$ when $x=0$, and $\theta\to\frac{\pi}{2}$ as $x\to\infty$. Theref... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Expand $\frac{z}{z^4+9}$ To Taylor Series
expand $$\frac{z}{z^4+9}$$ to taylor series
$$\frac{z}{z^4+9}=\frac{z}{9}\frac{1}{1--\frac{z^4}{9}}$$
Can we write $$\frac{z}{9}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z^4}{9}\right)^n=\sum_{n=0}^{\infty}(-1)^n\frac{z^{4n+1}}{9^{n+1}}$$?
| Your expansion as Taylor series around $0$ (i.e. as Maclaurin series) is fine. But we also have to state the validity of the series representation of $f(z)$
\begin{align*}
f(z)=\frac{z}{z^4+9}=\frac{z}{9}\cdot\frac{1}{1-\left(-\frac{z^4}{9}\right)}=\sum_{n=0}^\infty (-1)^n\frac{z^{4n+1}}{9^{n+1}}
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find all orthogonal $3\times 3$ matrices of the form... Find all orthogonal $3\times 3$ matrices of the form
\begin{bmatrix}a&b&0\\c&d&1\\e&f&0\end{bmatrix}
Using the fact that $A^TA$ = $I_n$, I set that all up and ended up with the following system of equations:
$$\left\{\begin{array}{l}a^2 + e^2 = 1\\
ab + ef = 0\\
... | There are few possibilities in fact. First of all, since $c^2+d^2+1^2=1$, $c=d=0$.
Now, you know that $a^2+b^2=1$, that $e^2+f^2=1$, and that $ab+ef=0$. This means that the matrix $\left(\begin{smallmatrix}a&b\\e&f\end{smallmatrix}\right)$ is orthogonal. Therefore, there is some $\theta\in\mathbb R$ such that$$\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $a^2+b^2+2(a+b)$ minimum if $ab=2$ Let $a,b\in R$,and such
$$ab=2$$
Find the minimum of the $a^2+b^2+2(a+b)$.
I have used $a=\dfrac{2}{b}$, then
$$a^2+b^2+2(a+b)=\dfrac{4}{b^2}+b^2+\dfrac{4}{b}+2b=f'(b)$$
Let $$f'(b)=0,\,b=-\sqrt{2}$$
So $$a^2+b^2+2(a+b)\ge 4-4\sqrt{2}$$
I wanted to know if there is other way to ... | AM-GM inequality is (at least usually) only applied to positive real numbers.
Therefore I first consider the region $a>0,b>0$.
You can also write
$$
\begin{aligned}
f(a,b)&=a^2+b^2+2(a+b)\\
&=(a+b)^2+2(a+b)-2ab\\
&=s^2+2s-4,
\end{aligned}
$$
where $s=a+b$.
By the AM-GM inequality $s\ge 2\sqrt2$ with equality only when ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Evaluate an indefinite integral Find the value of
$$\int{\frac{x^2e^x}{(x+2)^2}} dx$$
My Attempt: I tried to arrange the numerator as follows:
$$ e^xx^2 = e^x(x+2-2)^2 $$ but that didn't help.
Any guidance on this problem will be very helpful.
| Another approach, let $u \mapsto x+2$ and expand
\begin{align*}
\int \frac{x^2e^x}{(x+2)^2}dx &= e^{-2} \int \frac{(u-2)^2 e^u}{u^2} \,du \\
&= e^{-2} \int \left( \frac{4e^u}{u^2} - \frac{4e^u}{u} + e^u \right) \, du \\
&= e^{-2} \left( \int e^u \,du -4 \int \frac{e^u}{u} - \frac{e^u}{u^2}\, du \right) \\
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find the largest constant $k$ such that $\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$
Find the largest constant $k$ such that $$\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$$
My attempt,
By A.M-G.M, $$(a+b)^2+(a+b+4c)^2=(a+b)^2+(a+2c+b+2c)^2$$
$$\geq (2\sqrt{ab})^2+(2\sqrt{2ac}+2\sqrt{2bc})^2$$
$$=4ab+8ac+8bc+16c\sqrt{ab}$... | I think you mean that $a$, $b$ and $c$ are positives.
For positive variables your solution is true, I think.
Another way:
Let $a+b=4xc$.
Hence, by AM-GM $$\frac{((a+b)^2+(a+b+4c)^2)(a+b+c)}{abc}\geq\frac{((a+b)^2+(a+b+4c)^2)(a+b+c)}{\left(\frac{a+b}{2}\right)^2c}=$$
$$=\frac{(16x^2+(4x+4)^2)(4x+1)}{4x^2}=\frac{4(2x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Rate of convergence of a sequence I was just playing around rediscovering calculus (trying to figure out the derivative of $\arcsin x$) when I stumbled upon this rather untrivial problem.
$\large \int _0^1 \frac{dx}{\sqrt{1-x^2}} = \frac{\pi}{2}$
From this formula we can apply the definition of definite integral and ge... | Since the integrand is unbounded at $x = 1$, it is not Riemann integrable and the integral is improper:
$$\int_0^1 \frac{dx}{\sqrt{1 - x^2}} = \lim_{c \to 1}\int_0^c \frac{dx}{\sqrt{1 - x^2}} = \frac{\pi}{2}.$$
In general, there is no guarantee that a Riemann sum converges to the improper integral. Fortunately, the int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Geometric series with complex numbers In a geometric series
$a_{1}=1-\sqrt{3}i$ and $a_{2}=2$
Prove that for every natural number $n$, the numbers at the $3n+2$ location in the series are real numbers.
I have started by finding the ratio of the series, which is:
$q=1-\sqrt{3}i$
Now I am stuck. The elements of the seri... | The geometric progression has ratio
$$\frac{a_2}{a_1}=\frac{2}{1-i \sqrt{3}}=\frac{1}{2} \left(1+i \sqrt{3}\right)$$
Therefore the sequence can be written as
$$\{a_n\}=\left\{\left(1-i \sqrt{3}\right) \left(\frac{1}{2} \left(1+i \sqrt{3}\right)\right)^{n-1}\right\};\;n\geq 1$$
Observing that $\frac{1}{2} \left(1+i \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the Roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$ Once I came across the following problem: find the roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$.
Here it is how I proceeded:
\begin{align*}
(x+1)(x+3)(x+5)(x+7) + 15 & = [(x+1)(x+7)][(x+3)(x+5)] + 15\\
& = (x^2 + 8x + 7)(x^2 + 8x + 15) + 15\\
& = (x^2 + 8x + 7)[(x^2 + 8x +... | Hint: let $x+4=y$ then the equation writes as:
$$0 = (y-3)(y-1)(y+1)(y+3)+15=(y^2-1)(y^2-9)+15=y^4-10y^2+24$$
The latter is a biquadratic with solutions $y^2 \in \{4, 6\}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Laplace transform of $x^{3/2}$ Solve the Laplace transform of $x^{3/2}$
$L[x^{3/2}]=L[x (x^{1/2})]=-\frac{d}{dp}L[x^{1/2}]=-\frac{d}{dp}L[x(x^{-1/2})]=-\frac{d}{dp}(-\frac{d}{dp})\sqrt{\frac{\pi}{p}}=\frac{1}{2p}\frac{1}{2p}\sqrt{\frac{\pi}{p}}=\frac{1}{4p^2}\sqrt{\frac{\pi}{p}}$
The correct answer is $\frac{3}{4p^2}\... | Here's the correct way to do the integral.
\begin{align*}
\int_0 ^\infty e^{-st} t^{3/2} \ dt &= \int_0 ^\infty e^{-u} \left(\frac{u}{s} \right)^{3/2} \ \frac{du}{s} \\
&= \frac{1}{s^{5/2}} \int_0 ^\infty e^{-u} u^{\frac{5}{2}-1} \ du \\
&= \frac{1}{s^{5/2}} \Gamma \left( \frac{5}{2} \right) \\
&= \frac{1}{s^{5/2}} \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $x-\sqrt{7x}$ given that $x - \sqrt{\frac7x}=8$
If $ x - \sqrt{\frac{7}{x}}=8$ then $x-\sqrt{7x}=\text{?}$
I used some ways, but couldn't get the right form :) by the way, the answer is $1$.
Thanks in advance.
| \begin{eqnarray*}
x-\sqrt{\frac{7}{x}}=8 \\
x-8 =\sqrt{\frac{7}{x}}
\end{eqnarray*}
Now square both sides and multiply by $x$
\begin{eqnarray*}
x^2-16x+64=\frac{7}{x} \\
x^3-16x^2+64x-7=0 \\
(x-7)(x^2-9x+1)=0
\end{eqnarray*}
Now assume $x$ does not equal $7$ and taking the positive square root will give
\begin{eqnarray... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2347703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$ I figure out these thing when "playing" with numbers:
$$3^2-2.2^2+1^2=2=2!$$
$$4^3-3.3^3+3.2^3-1^3=6=3!$$
$$5^4-4.4^4+6.3^4-4.2^4+1^4=24=4!$$
So I go to the conjecture that:
$$\binom{n}{n}(n+1)^n-\binom{n}{n-1}n^n+\binom{n}{n-2}(n-1)^n-...=n!$$
or
$$\sum_{x=0... | Here is a combinatorial proof: consider the set $F$ of functions $\{ 1, 2, \ldots, n \} \to \{ 1, 2, \ldots, n, n + 1 \}$, and for $k = 1, \ldots, n$, let $S_k$ be the set of such functions such that $k$ is not in the range. Then, by inclusion-exclusion:
$$\left|F \setminus \bigcup_{k=1}^n S_k \right| = |F| - \sum_{1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 3
} |
Finding a Closed Form for $\int_0^\infty A\sin(\frac{2\pi}{T}x) \exp(-bx)dx$ I'm struggling in finding a closed form for,
$$ \int_0^\infty Ae^{-bx}\sin\left(\frac{2\pi}{T}x\right) \,dx $$
My Attempt: Let $H = \int Ae^{-bx}\sin\left(\frac{2\pi}{T}x\right)\,dx$. First I integrate by parts, splitting $f=e^{-bx}$ and $d... | Even without Laplace transform, consider the two antiderivatives
$$I=\int \sin(ax) e^{-bx}\,dx \qquad \text{and} \qquad J=\int \cos(ax) e^{-bx}\,dx$$ $$J+iI=\int e^{iax} e^{-bx}\,dx=\int e^{-(b-ia)x}\,dx=-\frac{e^{ -(b-ia)x}}{b-ia}\tag 1$$
$$J-iI=\int e^{-iax} e^{-bx}\,dx=\int e^{-(b+ia)x}\,dx=-\frac{e^{-(b+ia)x}}{b+ia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$ Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$.
I completed the first part of the question by finding the roots of the first equation. I obtained $3$ roots, one of them being $-1$ and the other two complex. It is ev... | $$x^3+2x^2+2x+1=x^3+x^2+x^2+x+x+1=(x+1)(x^2+x+1).$$
We see that $-1$ it's not common root and
$$x^{1990}+x^{200}+1=x^{1990}-x+x^{200}-x^2+x^2+x+1=$$
$$=x((x^3)^{663}-1)+x^2((x^3)^{66}-1)+x^2+x+1\equiv0(\mod(x^2+x+1)),$$
which says that roots of $x^2+x+1$ are all common roots and we get the answer:
$$\left\{-\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Matrix and sequences Given the following matrix : $$
A= \begin{bmatrix}
1 & 1 \\
1 & 0 \\
\end{bmatrix}
$$ and the sequences $a_n$, $b_n$, $ c_n$, $d_n$ such that $ A^n$= \begin{bmatrix}
a_n& b_n \\
c_n & d_n \\
\end{bmatrix}
I need to find the pair $(x,y)$ s... | Note that
$$\begin{pmatrix} a_{n+1}&b_{n+1}\\c_{n+1}&d_{n+1} \end{pmatrix} = \begin{pmatrix} a_n&b_n\\c_n&d_n \end{pmatrix} \begin{pmatrix} 1&1\\1&0 \end{pmatrix} = \begin{pmatrix} a_n + b_n & a_n \\ c_n + d_n & c_n \end{pmatrix} $$
and
$$\begin{pmatrix} a_{n+2}&b_{n+2}\\c_{n+2}&d_{n+2} \end{pmatrix} = \begin{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If a and b are two distinct real values such that $F (x)= x^2+ax+b $ And given that $F(a)=F(b)$ ; find the value of $F(2)$
If $a$ and$ b$ are two distinct real values such that
$$F (x)= x^2+ax+b $$
And given that $F(a)=F(b)$ ; find the value of $F(2)$
My try:
$$F(a)=a^2+a^2+b= 2a^2+b,\quad
F(b)=b^2+ab+b $$
$F... | Due to the symmetry of parabolas of that form, you can see $\frac{|a+b|}{2}$ is a minimum or maximum.
This gives you $f'\left(\frac{|a+b|}{2}\right)=|a+b|+a=0$ which implies $b=0\lor 2a+b=0$.
$b=0 \implies F(b)=0 \land F(a)=2a^2$
But $F(a)=F(b)\implies a=0$ which leads to a contradiction (as $a$ and $b$ are distinct).
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$ , and $a \leq b \leq c $?
What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$, and $a \leq b \leq c$?
My try:
The prime factorization of $1000$ is $2^3\cdot 5^3$
$a\cdot b \cdo... | For all $a,b,c$ you got 100 but that didn't take into account $a \le b \le c$.
So count $a=b=c$ that is $a=b=c=10$ you counted that once. The remaining $99$ were overcounted.
Consider $a=b, c\ne a$. Then $a=b= 2^j5^k; c=2^{3-2j}3^{3-2j}$ so there are $3$ such cases ($j = 0,1; k = 0,1$ but not $j=k=3-2j=3-2k= 1$). Yo... | {
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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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proof that $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ Let $n$ be a positive integer number.
How do we show that the irreducible polynomial $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ for all $n$?
| Because $$n^4+22n^3+71n^2+218n+384=n(n+1)(n+2)(n+3)+16n^3+60n^2+212n+384=$$
$$=n(n+1)(n+2)(n+3)+384+16(n^3-n)+60n^2+228n=$$
$$=n(n+1)(n+2)(n+3)+384+16(n-1)n(n+1)+60n(n+1)+168n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How to calculate $\lim_{x\to 0^+} \frac{x^x- (\sin x)^x}{x^3}$ As I asked, I don't know how to deal with $x^x- (\sin x)^x$.
Please give me a hint. Thanks!
| Note that
$$\frac{x^x - (\sin x)^x}{x^3} = x^x\frac{1 - \left(\frac{\sin x}{x}\right)^x}{x^3}$$
We can find the Taylor expansion
$$\left(\frac{ \sin x}{x} \right)^x = \exp\left(x \log \left(\frac{\sin x}{x} \right) \right) = 1 - \frac{x^3}{6} + O(x^5),$$
using
$$\frac{\sin x}{x} = 1 - \frac{x^2}{6} + O(x^4)\\ \log (1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Checking divisibility of $C^{20}_7-C^{20}_8 +C^{20}_9 -C^{20}_{10}+\dots-C^{20}_{20}$ Let $N=C^{20}_7-C^{20}_8 +C^{20}_9 -C^{20}_{10}+\dots-C^{20}_{20}$. Prove that it is divisible by $3,4,7,19$.
| Hint. Since by the binomial theorem
$$0=(1-1)^{20}=\sum_{n=0}^{20}\binom{20}{n}(-1)^n=\sum_{n=0}^{6}\binom{20}{n}(-1)^n+\sum_{n=7}^{20}\binom{20}{n}(-1)^n=\sum_{n=0}^{6}\binom{20}{n}(-1)^n-N$$
it follows that
\begin{align*}
N&=\sum_{n=0}^6\binom{20}{n}(-1)^n\\
&=1-20+190-\frac{20\cdot 19\cdot 18}{3!}+\frac{20\cdot 19... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358397",
"timestamp": "2023-03-29T00:00:00",
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How does the trigonometric identity $1 + \cot^2\theta = \csc^2\theta$ derive from the identity $\sin^2\theta + \cos^2\theta = 1$? I would like to understand how would the original identity of $$ \sin^2 \theta + \cos^2 \theta = 1$$ derives into
$$ 1 + \cot^2 \theta = \csc^2 \theta $$
This is my working:
a) $$ \frac{\... | It is taken from Pythagoras theorem,
Let
Know that
let
c= Hypothenuse
a= Opposite
b=Adjacent
length of the triangle can be found by
$$a^2+b^2=c^2$$
Now divide everything by b
We have
$$\frac{a^2}{b^2}+1=\frac{c^2}{b^2}$$
Notice that
$$tan \theta=\frac{a}{b}$$
$$sec \theta=\frac{1}{\cos \theta}=\frac{c}{b}$$
There ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Find $\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)$
Find $$\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)$$
$$\lim\limits_{x \rightarrow \infty} \sqrt{x} (e^{-\frac{1}{x}}-1)
= \lim\limits_{x \rightarrow \infty} \frac{e^{-\frac{1}{x}}-1}{x^{-0.5}}
= \lim\limits_{x \rightarrow \i... | Yes your way is correct,but you can do it simply as $$\lim _{ x\rightarrow \infty } \sqrt { x } \left( e^{ -\frac { 1 }{ x } }-1 \right) =\lim _{ x\rightarrow \infty } \sqrt { x } \frac { \left( e^{ -\frac { 1 }{ x } }-1 \right) }{ -\frac { 1 }{ x } } \left( -\frac { 1 }{ x } \right) =\lim _{ x\rightarrow \inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Lengthy integration problem $$ \int\frac{x^3+3x+2}{(x^2+1)^2(x+1)} \, dx$$
I managed to solve the problem using partial fraction decomposition.
But that approach is pretty long as it creates five variables. Is there any other shorter method to solve this problem(other than partial fractions)?
I also tried trigonometri... | HINT:
$$x^3+3x+2=x^3+x+2x+2=x(x^2+1)+2(x+1)$$
$$\dfrac{x^3+3x+2}{(x^2+1)^2(x+1)}=\dfrac x{(x^2+1)(x+1)}+\dfrac2{(x^2+1)^2}$$
For the second set $x=\tan y$
For the first
Method$\#1:$
write the numerator as $$x\cdot\dfrac{(x^2+1)-(x^2-1)}2$$
Method$\#2:$
set $x=\tan y$ to find $$\int\dfrac{\sin y}{\sin y+\cos y}dx$$
Now... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the limit $ \lim_{x \to (\frac{1}{2})^{-}} \frac{\ln(1 - 2x)}{\tan \pi x} $ Question
$$
\lim_{x \to (\frac{1}{2})^{-}} \frac{\ln(1 - 2x)}{\tan \pi x}
$$
I'm not sure how to go about this limit, I've tried to apply L'Hopital's rule
(as shown).
It seems that the form is going to be forever indeterminate? Unless I... | You were doing fine:
Using the first derivatives gives
$$ \lim_{x \to (\frac{1}{2})^{-}}
\frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x}$$
Now simplify a bit first:
$$\frac{\frac{-2}{1 - 2x}}{\pi \sec^2 \pi x} = \frac{2\cos^2\pi x}{\pi \left(2x-1\right)}$$
Applying l'Hôpital to the fraction in this form gives the eas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Given $1 \le |z| \le 7$ Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$ Given $1 \le |z| \le 7$
Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$
I have taken $z=r e^{i \theta}$ $\implies$ $1 \le r \le 7$
Now $$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{r \cos \thet... | We want to find the minimum and maximum of
$\left| \dfrac{z}{4}+\dfrac{6}{z}\right|$
When $1\le |z|\le 7$
Let $z=x+iy$ and plug it in the given expression
$$\left| \dfrac{x+iy}{4}+\dfrac{6}{x+iy}\right|=\left| \dfrac{x+iy}{4}+\dfrac{6(x-iy}{x^2+y^2}\right|=\left|\frac{6 x}{x^2+y^2}+\frac{x}{4}+i\left(\frac{y}{4}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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On understanding the use of binomial theorem to find asymptotes of a real valued function. I have the function
$$y=\frac{x^3+2x+9}{\sqrt{4x^2+3x+2}}$$
Wolfram Alpha says that it has a non linear asymptote at
$y=\frac{x^2}{2} - \frac{3x}{16}+ \frac{251}{256}$.
I have been told to expand using the binomial theorem to o... | hint
Let $$f (x)=(x^3+2x+9)(4x^2+3x+2)^\frac{-1}{2}$$
$$g (x)=f (1/x)=$$
$$(1+\frac {9x}{2}+\frac {1}{2x^2})(1+\frac {3x}{4}+\frac {x^2}{2})^\frac {-1}{2} $$
expand $g (x) $ near $x=0$ using binomial formula.
$$(1+\frac {3x}{4}+\frac {x^2}{2})^\frac {-1}{2} $$
$$=1-\frac {3x}{8}-\frac {x^2}{4}+\frac {27x^2}{112}+x^2\ep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4... | Let $\arctan\frac{1+x}{4+x}=\frac{\pi}{4}+y$.
Hence, $y\rightarrow0$ and $$\frac{1+x}{4+x}=\tan\left(\frac{\pi}{4}+y\right)$$
or $$x=\frac{1-4\tan\left(\frac{\pi}{4}+y\right)}{\tan\left(\frac{\pi}{4}+y\right)-1}$$ and we need to calculate
$$\lim_{y\rightarrow0}\frac{y\left(1-4\tan\left(\frac{\pi}{4}+y\right)\right)}{\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I solve this limit without L'Hopital rule? I have found this interesting limit and I'm trying to solve it without use L'Hopital's Rule.
$$\lim\limits_{x\rightarrow 0}\frac{\sinh^{-1}(\sinh(x))-\sinh^{-1}(\sin(x))}{\sinh(x)-\sin(x)}$$
I solved it with L'Hopital's rule and I found that the solution is $1$. But if... | You can use equivalents and expansion in power series.
Let's begin with the denominator:
$$\sinh x-\sin x=x+\frac{x^3}{3!}+o(x^3)-\Bigl(x-\frac{x^3}{3!}+o(x^3)\Bigr)=\frac{x^3}3+o(x^3)\sim_0\frac{x^3}3.$$
Now for the numerator:
First, by definition, $\;\operatorname{arsinh}(\sinh(x))=x$.
Next,
$$\operatorname{arsinh}... | {
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"url": "https://math.stackexchange.com/questions/2366608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve $\frac{3}{2}x-\frac{4}{5}=\sqrt{\sin(30^\circ)+\sin\frac{7\pi}{4}}$ I did it as: $$\Bigl(\frac{3}{2}x-\frac{4}{5}\Bigr)^2=\Biggl(\sqrt{\frac{1}{2}-\frac{\sqrt{2}}{2}}\Biggr)^2.$$ But after, got big numbers $$225x^2-63-\sqrt{2}=0.$$ It would be very good, if someone showed the correct way of solving this. I... | $$\frac{3}{2}x-\frac{4}{5}=\sqrt{\frac{1}{2}-\frac{1}{\sqrt2}},$$
which is impossible because $\frac{1}{2}-\frac{1}{\sqrt2}<0$.
The answer is $\oslash$
| {
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"url": "https://math.stackexchange.com/questions/2367995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Logarithmic series as $\sum_{n=3}^{\infty}(-1)^n\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right)$ Inspired by this question, I've designed the following series.
$$
\begin{align}
S&=\sum_{n=3}^{\infty}\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right) \tag1... | Noting that
$$ 2n^2+n-6=(n+2)(2n-3),2n^2+n-10=(n-2)(2n+5)$$
one has
\begin{eqnarray}
S&=&\sum_{n=3}^{\infty}\ln\!\left(1+\frac1{2n}\right) \!\ln \!\left(\frac{2n^2+n-6}{2n^2+n-10}\!\right)\\
&=&\sum_{n=3}^{\infty}\ln\!\left(\frac{2n+1}{2n}\right) \!\ln \!\left(\frac{(n+2)(2n-3)}{(n-2)(2n+5)}\!\right)\\
&=&-\sum_{n=3}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Inequality in sum of fractions
Is there a constant $C\ge 0$ such that for any $a,b,c,d>0$ the inequality holds:
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge C+\frac{a+b}{b+c}+\frac{b+c}{c+d}+\frac{c+d}{d+a}
$$
My attempt:
$a=b=c=d$ gives $1\ge C$.
Now, take $b=c=d$; this leads to inequality with two vari... | For $C=1$ it's wrong. Try $a=10$, $b=15$, $c=10$ and $d=11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$?
$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$
I managed to get the answer as $1$ by standard methods of so... | With integrals:
$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)=\lim_{n\rightarrow\infty} \frac{1}{n}\biggl(\frac{1}{n} + \frac{2}{n} + \frac{3}{n}+\dots+\frac{n}{n}\biggr)= \int_{0}^1 x dx =\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 11,
"answer_id": 4
} |
Intersection of rectangular and polar equations Find the points of intersection for the following relations:
$x^2+y^2-20 = 0$ and $\theta = -3\pi / 4$
So the first one is obviously a circle, and the second one, converting into rectangular form, is y/x = 1, and thus y=x. Substituting y as x in the first equation, I then... | How about $x=\sqrt{20}\cos \phi$, and $y=\sqrt{20}\sin \phi$. Putting $\phi = -3 \frac{\pi}{4}$ and $\phi = -3 \frac{\pi}{4}+\pi$ gives you the answer that you mentioned $(\sqrt{10}$,$\sqrt{10})$ and $(-\sqrt{10}$,$-\sqrt{10})$.
Also, for the other question $(3,0)$ cannot be the answer as it does not satisfy the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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There exists rational number $a_n$, $(x^2+\frac{1}{2}x+1)\mid(x^{2n}+a_nx^n+1)$ Let $n\in \mathbb{N}$. Prove that there exists a rational number $a_n$ for which $$(x^2+\frac{1}{2}x+1)\mid(x^{2n}+a_nx^n+1)$$
My attempt :
I try $n=2$,
$(x^2+\frac{1}{2}x+1)(x^2-\frac{1}{2}x+1)=x^4+\frac{7}{4}x^2+1$
$a_2 = \frac{7}{4}$
| We define $$p_n(x)=x^{2n}+a_n\,x^n+1,$$ so we have to prove
$$\left.x^2+\frac12\,x+1\right|p_n(x).$$
This is clearly true for $n=1$ with $a_1=1/2$, and it was shown for $n=2$ with $a_2=7/4$. Induction step: we assume it's true for $n=k-1$ and $n=k$. Then, as can be seen by expanding the products,
$$p_{k+1}(x)=(x^{2k}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Finding the minimum value of $\cot^2A + \cot^2B+ \cot^2C$ where $A$, $B$ and $C$ are angles of a triangle. The question is:
If $A+B+C= \pi$, where $A>0$, $B>0$, $C>0$, then find the minimum value of $$\cot^2A+\cot^2B +\cot^2C.$$
My solution:
$(\cot A + \cot B + \cot C)^2\ge0$ // square of a real number
$\implies ... | Since $\cot^2$ is a convex function, we can use Jensen.
$$\sum_{cyc}\cot^2\alpha\geq3\cot^2\frac{\alpha+\beta+\gamma}{3}=1.$$
The equality occurs for $\alpha=\beta=\gamma=\frac{\pi}{3},$ which says that $1$ is a minimal value.
Actually, $(\cot^2x)''=\frac{2(2+\cos2x)}{\sin^4x}>0$.
Also, we can use the following way.
Le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find the maximum positive integer that divides $n^7+n^6-n^5-n^4$
Find the maximum positive integer that divides all the numbers of the form $$n^7+n^6-n^5-n^4 \ \ \ \mbox{with} \ n\in\mathbb{N}-\left\{0\right\}.$$
My attempt
I can factor the polynomial
$n^7+n^6-n^5-n^4=n^4(n-1)(n+1)^2\ \ \ \forall n\in\mathbb{N}.$
If... | Newton's interpolation formula
gives
$$
n^7+n^6-n^5-n^4
=144 \binom{n}{2}+ 2160 \binom{n}{3}+ 9696 \binom{n}{4}+ 18480 \binom{n}{5}+ 15840 \binom{n}{6}+ 5040 \binom{n}{7}
$$
The gcd of the coefficients in the binomial representation is $48$ and so $n^7+n^6-n^5-n^4$ is always a multiple of $48$.
$48$ is largest factor ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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If $ x \in \left(0,\frac{\pi}{2}\right)$. Then value of $x$ in $ \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
If $\displaystyle x \in \left(0,\frac{\pi}{2}\right)$ then find a value of $x$ in $\displaystyle \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
$\bf{Attempt:}$ From $$\frac{3}{\sqrt{2}\cos x}-\frac{\sqrt{2}}{\s... | To take advantage of $\sqrt2$ in simplification restructure the equation to:
$$\frac{3 \cdot \dfrac{1}{\sqrt{2}}}{\cos x}-\frac{2 \cdot \dfrac{1}{\sqrt{2}}}{\sin x} = 1,$$
and so by inspection it is obviously $ x= \pi/4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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What are the steps for deriving a complicated generalization of a partial sum of a taylor series? I looked at the Taylor series for $$-\frac{x}{x-2}$$ and found it to be $$ \sum_{k=1}^{\infty}\frac{x^k}{2^k}$$
but then I also found that this series' partial sum is a bit more complicated in the form of
$$\frac{x 2^{-k}(... | The Taylor series has a strict formula. Suppose $f(x)$ has any derivative at $x_0$, then
$$f(x)=\sum _{n=0}^{\infty } \dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
where $f^{(n)}$ are the derivatives of the function, with the convention that $f^{(0)}=f,\;f^{(1)}=f',\ldots$
If $x_0=0$ then the formula simplifies in the MacLaurin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Finding the value of a given trigonometric series.
Find the value of $\tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\tan^2\dfrac{4\pi}{16}+\tan^2\dfrac{5\pi}{16}+\tan^2\dfrac{6\pi}{16}+\tan^2\dfrac{7\pi}{16}.$
My attempts:
I converted the given series to a simpler form:
$\tan^2\dfrac{\pi}{16}+... | Use the following identity:
$$ \cot x - \tan x = 2 \cot 2x$$
Square this
$$ \cot^2 x + \tan^2 x = 2 + 4 \cot^2 2x$$
Use this again with your idea of grouping terms as $\tan^2 x + \cot^2x$, and you will see that all you need to know is the value of $$\cot \frac{4\pi}{16} = \cot \frac{\pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find $\cos2\theta+\cos2\phi$, given $\sin\theta + \sin\phi = a$ and $\cos\theta+\cos\phi = b$
If
$$\sin\theta + \sin\phi = a \quad\text{and}\quad \cos\theta+\cos\phi = b$$
then find the value of $$\cos2\theta+\cos2\phi$$
My attempt:
Squaring both sides of the second given equation:
$$\cos^2\theta+ \cos^2\phi + 2... | $$\cos2\theta+\cos2\phi=2\cos(\theta+\phi)\cos(\theta-\phi),$$
$$a^2+b^2=2+2\cos(\theta-\phi)$$ and
$$b^2-a^2=\cos2\theta+\cos2\phi+2\cos(\theta+\phi).$$
Thus, $$\cos2\theta+\cos2\phi=2\cdot\frac{b^2-a^2-(\cos2\theta+\cos2\phi)}{2}\cdot\frac{a^2+b^2-2}{2},$$
which gives
$$\cos2\theta+\cos2\phi=\frac{(a^2+b^2-2)(b^2-a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find three numbers such that the sum of all three is a square and the sum of any two is a square It seems like it should be a well known problem. Ive read that Diophantine himself first posed it. I couldnt find a solution when i researched for one.
It asks to find ordered triples $(x,y,z) $ such that
$$x+y+z=a^2 $$
... | For the General case of formula there. https://artofproblemsolving.com/community/c3046h1172008_combinations_of_numbers_in_squares
The system of equations:
$$\left\{\begin{aligned}&a+b=x^2\\&a+c=y^2\\&b+c=z^2\\&a+b+c=q^2\end{aligned}\right.$$
Solutions have the form:
$$a=4t((2t-p)k^2+2(2t-p)^2k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculation of a partial derivative I have to find the $\frac{\partial}{\partial x}\left( f(x,y)\right)=\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$.
1) Thinking of that as $\frac{\partial}{\partial x}\left( (x^2+y^2)^{-\frac{1}{2}}\right)$ and $\frac{\partial}{\partial a}x^a = ax^{a-1}$ I get $\f... | In your calculations you should use these 3 formulas:
*
*$\frac{\partial}{\partial x}\frac{1}{a(x,y)} = \frac{-1}{a^2(x,y)} \cdot \frac{\partial}{\partial x} a(x,y)$
*$\frac{\partial}{\partial x}\sqrt{b(x,y)} = \frac{1}{2\sqrt{b(x,y)}} \cdot \frac{\partial}{\partial x} b(x,y)$
*$\frac{\partial}{\partial x}(x^2+c(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there a geometric method to show $\sin x \sim x- \frac{x^3}{6}$ I found a geometric method to show $$\text{when}\; x\to 0 \space , \space \cos x\sim 1-\frac{x^2}{2}$$ like below :
Suppose $R_{circle}=1 \to \overline{AB} =2$ in $\Delta AMB$ we have $$\overline{AM}^2=\overline{AB}\cdot\overline{AH} \tag{*}$$and
$$\o... | Now a full answer (assuming we have a result slightly stronger than yours, that $\cos x = 1-\frac{x^2}{2}+o(x^2)$, which is the true meaning of $\cos x\sim 1-\frac{x^2}{2}$, but which you haven't proven - you haven't come up with a bound for your error.)
Let $D$ be the midpoint of the arc in $MA$. The area of the quadr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 3
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How to change the appearence of the correct answer of $\cos55^\circ\cdot\cos65^\circ\cdot\cos175^\circ$ I represented the problem in the following view and solved it: $$\begin{align}-\sin35^\circ\cdot\sin25^\circ\cdot\sin85^\circ\cdot\sin45^\circ&=A\cdot\sin45^\circ\\ -\frac{1}{2}(\cos20^\circ-\cos70^\circ)\cdot\frac{1... | \begin{align}
A &= - \sqrt{\left(\frac{\sqrt{6}+\sqrt{2}}{16}\right)^2}\\
&=- \sqrt{\left(\frac{\sqrt{3}+1}{8\sqrt{2}}\right)^2}\\
&=- \frac18\sqrt{\left(\frac{3+1+2\sqrt3}{2}\right)}\\
&=- \frac18\sqrt{2+\sqrt{3}}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find the determinant of the following $4 \times 4$ matrix
Use a cofactor expansion across a row or column to find the determinant of the following matrix
$$B=\begin{pmatrix}1 &c&0&0\\0&1&c&0\\0&0&1&c\\c&0&0&1\end{pmatrix}$$
Clearly indicate the steps you take.
I have tried
$$\det B = 1 \det \begin{pmatrix}1 &c&0\\0&1... | Viewing $\rm B$ as a block matrix,
$$\det \left[\begin{array}{ccc|c} 1 & c & 0 & 0\\ 0 & 1 & c & 0\\ 0 & 0 & 1 & c\\ \hline c & 0 & 0 & 1\end{array}\right] = \det \left( \begin{bmatrix} 1 & c & 0\\ 0 & 1 & c\\ 0 & 0 & 1\end{bmatrix} - \begin{bmatrix} 0\\ 0\\ c\end{bmatrix} \begin{bmatrix} c\\ 0\\ 0\end{bmatrix}^\top \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2}$ then $\sum\limits_na_n$ diverges
Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that, for every $n\ge1$,
$$\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2} \tag 2$$ Prove that $x_n=a_1 + a_2 + .. + a_n$ diverges.
It is clear t... | You may notice that
$$ \frac{a_{n+1}}{a_n} \geq \left(1-\frac{1}{n}\right)\left(1+\frac{1}{n^2-n-1}\right)^{-1} \tag{1}$$
hence:
$$ \frac{a_{N+1}}{a_2}\geq \frac{1}{N}\prod_{n=2}^{N}\left(1+\frac{1}{n^2-n-1}\right)^{-1} \tag{2}$$
but the infinite product $\prod_{n\geq 2}\left(1+\frac{1}{n^2-n-1}\right)^{-1}$ is converg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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By using the definition of limit only, prove that $\lim_{x\rightarrow 0} \frac1{3x+1} = 1$
By using the definition of a limit only, prove that
$\lim_{x\rightarrow 0} \dfrac{1}{3x+1} = 1$
We need to find $$0<\left|x\right|<\delta\quad\implies\quad\left|\dfrac{1}{3x+1}-1\right|<\epsilon.$$
I have simplified $\left|\dfr... | If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $\frac{1}{3x+1}$is unbounded.
Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < \frac14$. Hence $\delta < \frac14$.
if $-\frac14 < x < \frac14$, $$-\frac34+1 < 3x+1< \frac34+1$$.
$$\frac14 < 3x+1< \frac74$$
$$\left|\frac{1}{3x+4}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality : $\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}\geq \frac{3}{5}$
Let $a$, $b$ and $c$ be positive real numbers. Prove that:
$$\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}+\frac{\sqrt{b^3a}}{2\sqrt{c^3b}+3ca}+\frac{\sqrt{c^3b}}{2\sqrt{a^3c}+3ab}\geq \frac{3}{5}$$
My attempt :
Substitute $a=x^2, b=y^2, c=z^2$, ... | Let $a = x^2, b = y^2, c=z^2$, where $x, y, z > 0$. By C-S, we have
$$ \sum_{cyc} \frac{x^3 z}{2 y^3 x + 3y^2z^2} \sum_{cyc} \frac{2yx + 3z^2}{zx} \geq \left( \sum_{cyc} \frac{x}{y} \right)^2.$$
But
$$ \sum_{cyc} \frac{2yx + 3z^2}{zx} = 5 \sum_{cyc} \frac{x}{y}.$$
Combining them we get
$$ \sum_{cyc} \frac{x^3 z}{2 y^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Taylor Series of $\frac{1}{3z+1}$ at $a=-2$
Find the Taylor Series of $\frac{1}{3z+1}$ at $a=-2$
When we are asked to find the taylor series at a point $p$ the taylor series should be at the form $\sum a_n(z-p)^n$?
So
$$\frac{1}{3z+1}=\frac{1}{2z-1+z+2}=\frac{1}{2z-1}\frac{1}{1+\frac{z+2}{2z-1}}=\frac{1}{2z-1}\frac{... | No, your expansion is not a power series centered at $-2$.
Let $w=z+2$ then for $|3w/5|<1$, or $|z+2|<5/3$,
\begin{align*}\frac{1}{3z+1}&=\frac{1}{3(w-2)+1}=\frac{1}{3w-5}=\frac{-1/5}{1-3w/5}=
-\frac{1}{5}\sum_{n=0}^{\infty}\left(\frac{3w}{5}\right)^n\\
&=-\frac{1}{5}\sum_{n=0}^{\infty}\left(\frac{3(z+2)}{5}\right)^n.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Taylor expansion of $\cos^2(\frac{iz}{2})$
Expand $\cos^2(\frac{iz}{2})$ around $a=0$
We know that $$\cos t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}$$
So $$\cos^2t=[\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}]^2=\sum_{n=0}^{\infty}(-1)^{2n}\frac{t^{4n}}{{4n^2!}}$$
We have $t=\frac{iz}{2}$
$$\sum_{n=0}^{\infty}... | If $f(x) = \sum_{n=0}^{\infty} a_n x^n$ then
$$ f^2(x) = \left( \sum_{n=0}^{\infty} a_n x^n \right) \left( \sum_{m=0}^{\infty} a_m x^m \right) = \sum_{k = 0}^{\infty} \left( \sum_{l = 0}^k a_l a_{k - l} \right) x^k \neq \sum_{n=0}^{\infty} a_n^2 x^{2n}.$$
In your case,
$$ \cos^2(x) = \left( \sum_{n=0}^{\infty} \frac{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Polynomial $x^{8} + x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1$ is reducible over $\mathbb{Q}$? Clearly, there are no roots, but how can I find factors of higher degree?
| To see how to get the answer, know the factorization of a finite geometric series:
$$ \begin{array}{ll} 1+x+\cdots+x^8 & \displaystyle=\frac{x^9-1}{x-1} \\[5pt] & \displaystyle =\frac{(x^3)^3-1}{x-1} \\[5pt] & \displaystyle =\frac{(x^3-1)\big((x^3)^2+x^3+1\big)}{x-1} \\[5pt] & = (x^2+x+1)(x^6+x^3+1) \end{array} $$
As C... | {
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"url": "https://math.stackexchange.com/questions/2397116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right)<1 $ $$\ f(x)=\log _{x-2}\left(\frac{2x+3}{7-x}\right) $$
Which is the solution of$\ f(x)<1 $ ? I did:
$$\ \frac{2x+3}{7-x}>0 $$
$\ x<7 $ results in $\ x> \frac{-3}{2} $ whereas $\ x>7 $ results in $\ x< \frac{-3}{2} $ so x=($\ \frac{-3}{2} $,7) and also
$\ x-2>0 $ so $\ ... | The domain gives $2<x<7$, $x\neq3$ and since $\frac{2x+3}{7-x}\neq x-2$,
we get the answer without any cases immediately by the intervals method:
$$(2,3).$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/2398066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $1^n+2^n+ \ldots +n^n=k!$ over positive integers If $k,n \in \mathbb{N}^*$, solve the following equation
$1^n+2^n+ \ldots +n^n=k!$, where $k! $ denotes $1 \cdot 2 \cdot 3 \cdots k$.
| This is a supplement to the answer by knm to show that $k\ge n+2$ (for large enough $n$).
Let $S_n=1^n+\cdots+n^n$. First, we show that $S_{n+1}/S_n>2(n+1)$. To do that, it suffices to show that $(r+1)^{r+1}/r^r>2(n+1)$ for all $r=0,\ldots,n$. Since the LHS is an increasing function, it has it's maximum at $r=n$, so we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I sti... | There's actually a general formula for these kinds of expressions. Namely$$\sqrt{X\pm Y}=\sqrt{\frac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}2}$$
Where $X,Y$ are real numbers. Simply substituting $X=6$ and $Y=\sqrt{20}$ gives the proper denesting. The proof of this is quite simple. Assume that$$X\pm Y=\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Compute the determinant The following problem is taken from here exercise $2:$
Question: Evaluate the determinant:
\begin{vmatrix}
0 & x & x & \dots & x \\
y & 0 & x & \dots & x \\
y & y & 0 & \dots & x \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y & y & y & \dots & 0
\end{vmatrix}
My attempt:
I tried to use... | If you continue subtracting the $(k+1)$th row from the $k$th, you end up with
$$ \Delta_n = \begin{vmatrix} -y & x \\
& -y & x \\
&& -y & x \\
&&&\ddots & \ddots \\
&&&& -y & x \\
y & y & y & \cdots & y & 0 \end{vmatrix}, $$
where blank spaces are zeros. One can also pull out a factor of $xy$ now, but there's not a lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the quadratic equation from its given roots.
If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c
=0$ , then form an equation whose roots are:
$\alpha+\dfrac{1}{\beta},\beta+\dfrac{1}{\alpha}$
Now, using Vieta's formula,
For new equation,
Product of roots ($P$) = $\dfrac{a^2+c^2+2ac}{ac}$
Su... | $$\alpha +\beta =-\frac { b }{ a } \\ \alpha \beta =\frac { c }{ a } \\ a\left( x-\left( \alpha +\frac { 1 }{ \beta } \right) \right) \left( x-\left( \beta +{ \frac { 1 }{ \alpha } } \right) \right) =0\\ a\left( x-\frac { \alpha \beta +1 }{ \beta } \right) \left( x-\frac { \beta \alpha +1 }{ \alpha } \right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
proof that the pattern exists for all n Explain the pattern
$$(\sqrt2-1)^1= \sqrt2 - \sqrt1$$
$$(\sqrt2-1)^2 = \sqrt9 - \sqrt8$$
$$ (\sqrt2-1)^3 = \sqrt{50} - \sqrt{49} $$
that is $(\sqrt2-1)^n$ is equal to difference of two consecutive numbers one of which are squares.
| We will prove a stronger result that $(\sqrt{2}-1)^{2n}=\sqrt{x_n^2}-\sqrt{2y_n^2}$ for some $x_n, y_n$ where $x_n^2-2y_n^2=1$, and $(\sqrt{2}-1)^{2n-1}=\sqrt{2z_n^2}-\sqrt{w_n^2}$ for some $z_n, w_n$ where $2z_n^2-w_n^2=1$ (here $\{x_n\}, \{y_n\}, \{z_n\}, \{w_n\}$ are all increasing sequences of positive integers suc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all positive integers $n > 1$ such that the polynomial $P(x)$ belongs to the ideal generated by the polynomial $x^2 +x +1$ in $\Bbb Z_n[x]$
Find all positive integers $n > 1$ such that the polynomial $x^4 + 3x^3 + x^2 + 6x + 10$ belong to the ideal generated by the polynomial $x^2 + x + 1$ in $\Bbb Z_n[x]$.
My a... | Because $x^4+3x^3+x^2+6x+10$ is in the ideal,
$p(x)=x^4+3x^3+x^2+6x+10-(x^2 + x + 1)( x^2 + 2x -2)=6x+12 $ is also in the ideal.
So $6x+12=q(x)*(x^2+x+1)=q(x)x^2+(x+1)q(x)$.
As $deg(6x+12)=1$, $q(x)x^2=0$. Also $deg q(x)=1$, which leads to $6x+12=0$.
Therefore, n=2,3 or 6.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Simplify $\frac1{\sqrt{x^2+1}}-\frac{x^2}{(x^2+1)^{3/2}}$ I want to know why
$$\frac1{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}}$$
can be simplified into
$$\frac1{(x^2+1)^{3/2}}$$
I tried to simplify by rewriting radicals and fractions. I was hoping to see a clever trick (e.g. adding a clever zero, multiplying by a cle... | Hint:$$term_a = \frac{1}{\sqrt{x^2+1}} - \frac{x^2}{(x^2+1)^{3/2}}=\\
\frac{1}{\sqrt{x^2+1}} - \frac{x^2}{(\sqrt{x^2+1})^{3}}=\\
\frac{x^2+1}{(\sqrt{x^2+1})^{3}} - \frac{x^2}{(\sqrt{x^2+1})^{3}}=\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$ To convert it into a number with base $9$
$$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7 = 331$$
But answer is $481$?????
| Here is a layout of the conversion algorithm (successive Euclidean divisions)
$$\begin{array}{ccrcrc*{10}{c}}
&&44&&4 \\
9&\Bigl)&397&\Bigl)&44&\Bigl)& \color{red}{\mathbf 4}\\
&&\underline{36}\phantom{7}&&\underline{36}\\
&&37&&\color{red}{\mathbf 8} \\
&&\underline{36} \\
&&\color{red}{\mathbf 1}
\end{array}$$
This i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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How can I prove the following sequence converges Given the following n-th term sequence:
$$a_{n} = \sqrt[n]{1^2+2^2+...+n^2}$$
You're asked to evaluate the limit of the given sequence, justifying your operations.
What strategy should I take on this? I have considered taking some inequality in order to, eventually, be ... | $$1 = \sqrt[n]{1^2} \leq \sqrt[n]{1^2 + 2^2 + \ldots + n^2} \leq \sqrt[n]{n^2 + n^2 + \ldots + n^2} = \sqrt[n]{n^3} \xrightarrow{n\to\infty}1$$
By the squeeze theorem, the sequence converges to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Reduction formula for Integral I am trying to find a reduction formula for $$\int\frac{\sin ^n(x)}{x} dx .$$
Some numerical calculations seem to indicate a reduction formula should exist, but the usual tricks (break up the $\sin ^n(x)$ etc) don't seem to help.
Any pointers/solutions would be great.
| This is not an answer, but can lead you to the answer
$\text{Si}=\int\frac{\sin (x)}{x} \,dx;\;\text{Ci}=\int\frac{\cos (x)}{x} \,dx$
Some solutions of
$$\int\frac{\sin ^n(x)}{x} \,dx$$
for $n=1..10$
$$
\begin{array}{l|l}
n & integral\\
\hline
1 & \text{Si}(x) \\
2 & \frac{\log (x)}{2}-\frac{\text{Ci}(2 x)}{2} \\
3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Mathematical Induction - Is this correct? The question is :
Prove that
$$3 + 6 + 12 + 24 + ... + 3 \cdot 2^{n-1} = 3 \cdot 2^n - 3$$
for all integers $n ≥ 1$
My Solution (I'll skip the basis step and go straight to the inductive step):
II.
Assume that
$$3 + 6 + 12 + 24 + ... + 3 \cdot 2^{k-1} = 3\cdot 2^k - 3$$
Cons... | This question has been answered by this user (and others) via the comment. In particular quoting the comment:
$2^k+2^k=2×2^k=2^{k+1}$. This is the part where you made mistake. Otherwise it's fine, and you have actually proved it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Volume of a solid bounded by 6 planes I need the volume of a figure delimited by planes
$$ 1 \le x + y + z \le 2 \\
0 \le x - y - z \le 3 \\
-1 \le 2x + y - z \le 4 $$
That should hold whenever this sum here :
$$ x \begin{pmatrix}
1 \\ 1 \\ 2
\end{pmatrix} +
y\begin{pmatrix}
1 \\-1 \\ 1
\end{pmatri... | Apart from a typo $\bigl((2-0)$ should be replaced by $(3-0)\bigr)$ this is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Show identity for $e^x$ I want to show that, using the definition $e^x := \lim_{n \to \infty} (1+\frac{x}{n})^n$, the following identity holds:
$$ e^{x+y}=e^xe^y$$
This is what I tried:
\begin{align}
e^xe^y &=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n\cdot\left(1+\frac{y}{n}\right)^n \\
&= \lim_{n \to \infty}\lef... | For $x, y \geq 0$ we have:
$$\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right) = 1 + \frac{x+y}{n} + \frac{xy}{n^2}\geq 1+\frac{x+y}{n}$$
For $n \in \mathbb{N}$, using the identity $u^n-v^n = (u-v)\sum_{k=0}^{n-1}u^{n-1-k}v^k$ we have:
\begin{align}\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n - \left(1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.
I started with our conergence definition, i.e. $\lvert a_n -... | For one, we have
$$
\frac{\sqrt{n^2+2}}{4n+1} \le \frac{\sqrt{n^2+2}}{4n} = \sqrt{\frac{1}{16}+\frac{1}{8n}} \le \frac{1}{4}+ \sqrt{\frac{1}{8n}},
$$
and we also have
$$
\frac{\sqrt{n^2+2}}{4n+1} \ge \frac{n}{4n+1} = \frac{1}{4 + 1/n}.
$$
As $n\to\infty$, both of these tend to $1/4$. By the squeeze theorem, we obtain t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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How to find the closed form of $\sum_{k=2}^{\infty}{\lambda(k)-1\over k}?$ Given that:
$$\sum_{k=2}^{\infty}{\lambda(k)-1\over k}\tag1$$
Where $\lambda(k)$ is Dirichlet Lambda Function
We are seeking to determine the closed form $(1)$ and came close to estimates it to $1-{\frac12}\left(\gamma+\ln{\pi}\right)$.
where $\... | Another approach is to use the ordinary generating function of the Riemann zeta function, namely $$\sum_{n=2}^{\infty} \zeta(n) x^{n-1} = - \gamma - \psi (1-x), \quad |x| <1. $$
(This is just the Maclaurin series of the digamma function $\psi(1-x)$.)
Integrating both sides of the generating function, we get
$$\sum_{n=2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How do I find the closed form of this integral $\int_0^2\frac{\ln x}{x^3-2x+4}dx$? How do I find the closed form of this integral:
$$I=\int_0^2\frac{\ln x}{x^3-2x+4}dx$$
First, I have a partial fraction of it:
$$\frac{1}{x^3-2x+4}=\frac{1}{(x+2)(x^2-2x+2)}=\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+2}$$
$$A=\frac{1}{(x^3-2x+4)'}... | Just for your curiosity.
If you enjoy special functions of complex arguments, the antiderivative can be computed.
$$\frac 1 {x^3-2x+4}=\frac{\frac1{10}}{
x+2}-\frac{\frac{1}{20}-\frac{3 i}{20}}{x-(1-i)}-\frac{\frac{1}{20}+\frac{3 i}{20}}{x-(1+i)}$$ which makes that we are left with integrals $$I_a=\int \frac{\log(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluating $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ I was evaluating
$$\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$$
My work:
I see that in the denominator, there is the radical $\sqrt{x^2 + 3}$. This reminds me of the trigonometric substitution $\sqrt{u^2 + a^2}$ and letting
$ u = a \tan \theta$. With ... | Hint:
$$\dfrac{\sec t}{\tan^4t}=\dfrac{\cos^3t}{\sin^2t}=\dfrac{(1-\sin^2t)\cos t}{\sin^2t}$$
Set $\sin t=u$
OR directly, $u=\dfrac{\tan t}{\sec t}=\dfrac x{\sqrt{x^2+3}}$ assuming $0\le t<\dfrac\pi2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove that for any integer $n, n^2+4$ is not divisible by $7$. The question tells you to use the Division Theorem, here is my attempt:
Every integer can be expressed in the form $7q+r$ where $r$ is one of $0,1,2,3,4,5$ or $6$ and $q$ is an integer $\geq0$.
$n=7q+r$
$n^2=(7q+r)^2=49q^2+14rq+r^2$
$n^2=7(7q^2+2rq)+r^2$
$n... | Here is a proof that seems a bit roundabout, but it relates simpler methods to Fermat's Little Theorem.
First note that clearly $n^2+4$ will not be a multiple of $7$ if $n$ is one. Now for other values of $n$, raise $n^2+4$ to the power of $7-1=6$:
$(n^2+4)^6=n^{12}+(6)(4)n^{10}+(15)(4^2)n^8+...+4^6$
$\equiv n^{12}+3n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
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Finding generating function for a given sequence The task is to find the generating function for the sequence $$d_n = (\sum_{k=0}^n \binom{n}{k})*(\sum_{k=0}^n \frac{(k-1)^2}{2^k})$$
let's call this function $D(x)$.
I marked $a_n = \frac{1}{2^n}$; $b_n = \frac{-2n}{2^n}$; $c_n = \frac{n^2}{2^n}$ with respective generat... | I think you had to compute the sums first and then multiply. I mean, the sequence is:
$$d_n = \left(\sum_{k=0}^n \binom{n}{k}\right)\cdot\left(\sum_{k=0}^n \frac{(k-1)^2}{2^k}\right)$$
Which means
$$d_n=2^n\cdot\left(\sum _{k=0}^n \frac{k^2}{2^k}-2\sum _{k=0}^n \frac{k}{2^k}+\sum _{k=0}^n \frac{1}{2^k}\right)=\\=2^n\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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domain and range of function $y= 1+\sum_{n=2}^\infty x^n$ My friend give me a question to find domain and range of
$y= 1+\sum_{n=2}^\infty x^n$
there is no more description about the problem, so I think the domain of that function is all $\mathbb{R} $ and range of that function is $ \mathbb{R}$ too
but he told me th... | The expression
$$ 1 + x + x^2 + x^3 + \dotsb = \sum_{k=0}^{\infty} x^k $$
is a geometric series. We say that a series converges if the corresponding sequence of partial sums converges. That is
$$ \sum_{k=0}^{\infty} x^k \text{ converges}\iff \lim_{n\to\infty} \sum_{k=0}^{n} x^k =: S_n < \infty.$$
But note that
$$ S_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove an equation with summation and binomial coefficients Good day everybody,
I am looking to prove this:
$$\delta_{a,b}=\left(\frac{1}{2}\right)^{2\left(a-b\right)}\sum_{\gamma=0}^{a-b}\left(-1\right)^{a-b-\gamma}\frac{b+\gamma}{b}\left(\begin{array}{c}
2a\\
a-b-\gamma
\end{array}\right)\left(\begin{array}{c}
2b-1+\g... | We write $n$ instead of $\gamma$, focus on the essentials and skip the constant $(-1)^{a-b}2^{-2(a-b)}$.
We obtain for integers $a\geq b>0$
\begin{align*}
\color{blue}{\sum_{n=0}^{a-b}}&\color{blue}{(-1)^n\frac{b+n}{b}\binom{2a}{a-b-n}\binom{2b-1+n}{n}}\tag{1}\\
&=\sum_{n=0}^{a-b}\frac{b+n}{b}\binom{2a}{a-b-n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Some formula related with factor of (a+b+c+d) I am looking for some math formula
For example
\begin{align}
& a^2 -b^2 = (a+b)(a-b) \\
&a^3 +b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca)
\end{align}
First one related with factor a+b and the second one related with factor a+b+c
then
How about some formula relat... | It will be probably
$$a^4+b^4+c^4+d^4-b^2a^2-c^2a^2-d^2a^2-b^2c^2-b^2d^2-d^2c^2+16abcd=(a+b+c+d)(a^3+b^3+c^3+d^3-ba^2-ca^2-da^2-ac^2-bc^2-dc^2-ab^2-cb^2-db^2-ad^2-bd^2-cd^2+4bca+4bda+4cad+4bcd)$$
OR
$$\sum a^4 -\sum a^2b^2 +16abcd=(\sum a)(\sum a^3-\sum ab^2 +4\sum abc)$$
$$$$
$$\sum \text{ represents cyclic summa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution of $\sqrt{5-2\sin x}\geq 6\sin x-1$
Solve the following inequality. $$\sqrt{5-2\sin x}\geq 6\sin x-1.$$
My tries:
As $5-2\sin x>0$ hence we do not need to worry about the domain.
Case-1: $6\sin x-1\leq0\implies \sin x\leq\dfrac{1}{6}\implies -1\leq\sin x\leq\dfrac{1}{6}\tag*{}$
Case-2:$6\sin x-1>0\implies \d... | Hint:)
Other way that may be helpful. Let $y=5-2\sin x$, from $\sqrt{5-2\sin x}\geq 6\sin x-1$ you find
$$3y-\sqrt{y}-14\geq0$$
gives
$(\sqrt{y}-2)(\sqrt{y}+\dfrac73)\geq0$. You can discuss about conditions and find that $\sin x\leq\dfrac12$. This concludes
$$\color{blue}{\left[2k\pi+\dfrac{3\pi}{2}-\dfrac{2\pi}{3},2k... | {
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"url": "https://math.stackexchange.com/questions/2429305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{3}{abc} \ge a + b + c$, prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge a + b + c$ Let $a$, $b$ and $c$ be positive numbers such that $\frac{3}{abc} \ge a + b + c$. Prove that: $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge a + b + c.$$
Any way I use - I get stuck after 2 or 3 steps...
| The condition gives
$$1\geq\sqrt{\frac{(a+b+c)abc}{3}}.$$
Thus,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\sqrt{\frac{(a+b+c)abc}{3}}=$$
$$=\sqrt{\frac{(ab+ac+bc)^2(a+b+c)}{3abc}}\geq\sqrt{\frac{3abc(a+b+c)(a+b+c)}{3abc}}=a+b+c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solving system of equations (3 unknowns, 3 equations) So, I've been trying to solve this question but to no avail. The system of equations are as follows:
1) $x+ \frac{1}{y}=4$
2) $y+ \frac{1}{z}=1$
3) $z + \frac{1}{x}=\frac{7}{3}$
Attempt:
Using equation 1, we can rewrite it as $\frac{1}{y}=4-x \equiv y=\frac{1}{4-x}$... | Eq. (2) implies $z=1/(1-y)$, and Eq. (1) says $y = 1/(4-x)$. Consequently, Eq. (3) becomes
$$\frac{1}{x}+\frac{4-x}{3-x}=\frac{7}{3} \implies 4x^2-12x+9 = (2x-3)^2 = 0 \implies x = \frac{3}{2}.$$
Consequently, $y = 1/(4-x) = 2/5$, and $z = 1/(1-y) = 5/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\lim_{x\to 0^+} \Biggr(\tan^{-1}{\Big(\dfrac{b}{x}\Big)} - \tan^{-1}{\Big(\dfrac{a}{x}\Big)}\Biggr)= 0$? For $$f(x) = \tan^{-1}{\Big(\frac{b}{x}\Big)} - \tan^{-1}{\Big(\frac{a}{x}\Big)}$$
where $a$ and $b$ are differently valued constants, why is $\lim_{x\to0^+}= 0$? I understand that separately both expression... | You can use the MTV to see it. For a differentiable function $g$ holds
$g(x_1)-g(x_2)=g'(\xi)(x_1-x_2)$ for some $\xi\in (x_1,x_2)$.
Now define $g(y)=\arctan(y)$ and you get
$$
f(x)=g\left(\frac{b}{x}\right)-g\left(\frac{a}{x}\right)=g'(\xi)\left(\frac{b}x-\frac{a}x\right)=\frac1{1+\xi^2}\cdot\frac{b-a}{x}
$$
where $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Two methods of finding $\int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\sin x}$ Two methods of finding $$I=\int_{0}^{\pi/2} \frac{dx}{1+\sin x}$$
Method $1.$ I used substitution, $\tan \left(\frac{x}{2}\right)=t$ we get
$$I=2\int_{0}^{1}\frac{dt}{(t+1)^2}=\left. \frac{-2}{t+1}\right|_{0}^{1} =1$$
Method $2.$ we have $$I=\int_0^{... | \begin{align}
\tan x - \sec x & = \frac{\sin x -1}{\cos x} = \frac{(\sin x - 1)(-\sin x -1)}{(\cos x)(-\sin x - 1)} = \frac{1-\sin^2 x}{(\cos x)(-\sin x - 1)} \\[10pt]
& = \frac{\cos^2 x}{(\cos x)(-\sin x - 1)} = \frac{-\cos x}{1+\sin x} = \frac 0 2 \text{ if } x = \frac \pi 2.
\end{align}
Or you can use the identity
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
} |
$a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$. Prove $(a-c)(b-c)<0$
$a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$.
Show that $(a-c)(b-c)<0$.
This is a question presented in the "Olimpiadas do Ceará 1987" a math contest held in Brazil. Sorry if this a duplicate.
Given the assumptions,... | If $a<b$, then
$$
c^2 = a^2+b^2-ab = a^2+b(b-a) > a^2 \\
c^2 = a^2+b^2-ab = b^2-a(b-a) < b^2
$$
which means $a<c<b$ and $(a-c)(b-c)<0$.
If $a>b$, then
$$
c^2 = a^2+b^2-ab = a^2-b(a-b) < a^2 \\
c^2 = a^2+b^2-ab = b^2+a(a-b) > b^2
$$
which means $b<c<a$ and $(a-c)(b-c)<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
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Integrating modular function $f(x,y)=|xy|$ in the area of the circle $x^2+y^2=a^2$ I need to solve the integral $$\iint_\omega |xy|\,dx\,dy$$ where $\omega:x^2+y^2=a^2$ I created the following integral and I have no idea how can I integrate modular function:
$$\int_{-a}^a dy\int_{-\sqrt{a^2-y^2}}^\sqrt{a^2-y^2} |xy|\,d... | Why not convert to polar coordinates?
With $x=r\cos(\theta)$ and $y=r\sin(\theta)$. The area element becomes $rdrd\theta$. Then the integral over the area of the circle becomes
\begin{align}
I &=\int_0^a\int_0^{2\pi}r^2|\cos(\theta)\sin(\theta)|rdrd\theta=\int_0^ar^3dr\int_0^{2\pi}|\cos(\theta)\sin(\theta)|d\theta\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the limit of $ \frac{\cos (3x) - 1}{ \sin (2x) \tan (3x) } $ without L'Hospital technique I would like to find the limit as $x$ goes to zero for the following function, but without L'Hospital technique.
$$ f(x) = \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)} $$
This limit will go to zero (I had tried using calcul... | Just another way using the standard Taylor expansions.
$$\cos(3x)=1-\frac{9 x^2}{2}+\frac{27 x^4}{8}+O\left(x^6\right)$$
$$\sin(2x)=2 x-\frac{4 x^3}{3}+O\left(x^5\right)$$
$$\tan(3x)=3 x+9 x^3+O\left(x^5\right)$$
makes
$$\cos(3x)-1=-\frac{9 x^2}{2}+\frac{27 x^4}{8}+O\left(x^6\right)$$
$$\sin(2x)\tan(3x)=6 x^2+14 x^4+O\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Help converting spherical to cartesian?
Convert the following equation from spherical to coordinate:
sin2(ϕ) + cos2(ϕ)⁄4 = 1/ρ2;
I have converted the right side of the equation to 1/x2+y2+z2, but the fraction on the left side of the equation is throwing me off. Any advice/help/solutions?
| $x = \rho\cos\theta\cos\phi\\
y = \rho\sin\theta\cos\phi\\
z = \rho\sin\phi$
$\sin^2\phi + \frac 14 \cos^2\phi = \frac 1{\rho^2}\\
\rho^2\sin^2\phi + \frac 14 \rho^2 \cos^2\phi = 1\\
z^2 + \frac 14 \rho^2 \cos^2\phi = 1$
$x^2+y^2 = \rho^2 \cos^2\phi$
$z^2 + \frac 14 x^2 + \frac 14 y^2 = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality on tangent and secant function
Let $\{\alpha, \beta, \gamma, \delta\} \subset \left (0,\frac {\pi}{2}\right)$ and $ \alpha + \beta+\gamma+\delta = {\pi}$. Prove that $\sqrt {2} \left(\tan \alpha +\tan \beta +\tan\gamma+\tan \delta\right)\ge \sec \alpha +\sec \beta +\sec \gamma +\sec \delta$
| Let $f(x)=\sqrt2\tan{x}-\frac{1}{\cos{x}}$.
Thus, $$f''(x)=\frac{(1+\sqrt2-\sin{x})(\sin{x}-\sqrt2+1)}{\cos^3x}>0$$
for $x\in\left[\frac{\pi}{4},\frac{\pi}{2}\right)$.
Thus, by Vasc's RCF Theorem it's enough to prove our inequality for
$\beta=\gamma=\alpha$ and $\delta=\pi-3\alpha$, where $\alpha\in\left[\frac{\pi}{4},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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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.