Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Help to use change of variables to solve the double integral I'm trying to evaluate:
$$\iint_D \left(\sqrt{a^2-x^2-y^2}-\sqrt{x^2+y^2}~\right)dxdy$$
where $D_{xy}$ is the disk $x^2+y^2\le a^2$.
The exercise is to use change of variables to solve this integral.
My solution
I chose $\varphi (r,\theta)=(ra\cos\theta,r... | The answer is zero; the answer in the book is incorrect.
To verify, let's show that the following integrals (from your derivation above) are equal:
$$ A = \int_0^1 r\sqrt{1 - r^2} dr$$
$$ B = \int_0^1 r^2 dr $$
Let's change variable in $A$:
$$\quad t^2 = 1 - r^2 \Rightarrow 2tdt = -2rdr\Rightarrow rdr = -tdt.$$
Hence,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve the systems of equations for real numbers $x$, $y$, $z$ Solve the system of equations in real solutions:
$x+2y+z$ $=$ $5(x+y)(y+z)$,
$y+2z+x$ $=$ $7(y+z)(z+x)$,
$z+2x+y$ $=$ $6(z+x)(x+y)$,
I substituded $a=x+y$, $b=y+z$ and $c=z+x$ and got:
$5ab-a-b$ $=$ $0$,
$7bc-b-c$ $=$ $0$,
$6ac-a-c$ $=$ $0$, and got to the ... | Considering $$5ab-a-b=0\tag 1$$ $$7bc-b-c=0\tag 2$$ $$6ac-a-c=0\tag 3$$ extract $b=\frac{a}{5 a-1}$ from $(1)$ (assuming $a\neq \frac 15$), $c=\frac{a}{6 a-1}$ from $(3)$ (assuming $a\neq \frac 16$). Replace in $(2)$ and simplify to get $$\frac{2 (1-2 a) a}{30 a^2-11 a+1}=0$$
Lord Shark's solution is better
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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A proof that $\frac{\log^k(1)+\log^k(2)+\dotsb +\log^k(n)}{1^k+2^k+\dotsb +n^k} \to 0$ Let $\left(a_n\right)$ be the following sequence:
$$a_n =\frac{\log^k\left(1\right)+\log^k\left(2\right)+\dotsb +\log^k\left(n\right)}{1^k+2^k+\dotsb +n^k},$$ for a fixed $k \in \mathbb{N}$. Prove that $a_n \to 0$. There are many pr... | Noting
\begin{eqnarray}
0&<&\frac{\log^k(1)+\dots+\log^k(n)}{1^k+\dots+n^k}=\frac{\frac{1}{n}\sum_{i=1}^n(\frac{\log i}{n})^k}{\frac1n\sum_{i=1}^n(\frac{i}{n})^k}\\
&\le& \frac{\frac{1}{n}\sum_{i=1}^n(\frac{\log n}{n})^k}{\frac1n\sum_{i=1}^n(\frac{i}{n})^k}=\frac{(\frac{\log n}{n})^k}{\frac1n\sum_{i=1}(\frac{i}{n})^k},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the exact value of a radical How do I show that $\sqrt{97 +56\sqrt3}$ reduces to $7 +4\sqrt3?$. Without knowing intitially that it reduces to that value.
| Wikipedia describes a simple method to write
$$
\sqrt{a+b \sqrt{c}\ } = \sqrt{d}+\sqrt{e}
$$
with
$$
d=\frac{a + \sqrt {a^2-b^2c}}{2},
\qquad
e=\frac{a - \sqrt {a^2-b^2c}}{2}
$$
This works iff $a^2 - b^2c$ is a square.
For $\sqrt{97 +56\sqrt3}$ we have $a^2 - b^2c=1$ and so
$$
\sqrt{97 +56\sqrt3} = \sqrt{49}+\sqrt{48} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Prove that $\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$ is a closed and bounded set.
Prove that the following set $$V=\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$$ is closed and bounded.
The set $V$ is closed. In fact, for any continuous function $f:\mathbb R^n\rightarrow \mathbb R$, the set $\{\... | Note that $(a-b)^2 \ge 0$, so $ab \le \frac{a^2+b^2}{2}$. Thus, since $x^2 - xy + y^2 - 1 = 0$ we have
$x^2 + y^2 = 1 + xy \le 1 + \frac{x^2+y^2}{2}$
and so
$\frac{1}{2} (x^2 + y^2) \le 1$
and finally
$x^2 + y^2 \le 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the condition that the diagonals of a parallelogram formed by $ax+by+c=0$. Find the condition that the.diagonals of a parallelogram formed by $ax+by+c=0$, $ax+by+c'=0$, $a'x+b'y+c=0$ and $a'x+b'y+c'=0$ are at right angles.
My Attempt:
The equation of diagonal passing through the point of intersection of $ax+by+c=0... | $ax+by+c=0$ ...(1)
$a'x+b'y+c=0$ ...(2)
$ax+by+c'= 0$ ...(3)
$a'x+b'y+c'=0 $...(4)
Let P be the intersection of (1) and (2)
Equation of a line passing through P is given by
$ ax+by+c +K(a'x+b'y+c) = 0$ ...(5)
Let R be the intersection of (3) and (4) for which $ ax+by=-c'$ and $a'x+b'y=-c'$
Substituting in (5) $(c-c')+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Using Ptolemy's Theorem to find length In $\triangle ABC$ we have $AB=7, AC=8, BC=9$. Point $D$ is the midpoint of the arc $BC$ of the circumcircle of $\triangle ABC$. Compute $\displaystyle\frac{AD}{BD}$, $BD$, and $CD$.
This is what I have so far but I am unsure if what I have is correct:
Since arc$BD$ = arc$CD$, $B... | $AD$ is the angle bisector of $\widehat{BAC}$, hence if we call $X$ the intersection of $AD$ and $BC$, we have $BX=\frac{7}{15}\cdot 9$ and $CX=\frac{8}{15}\cdot 9$ by the bisector theorem. By Stewart's theorem the length of $AX$ is given by
$$ AX^2 = \frac{7\cdot 8}{(7+8)^2}((7+8)^2-9^2) $$
hence $AX=\frac{8}{5}\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , ... | $$-1<\frac{3x+1}{x-2}<1$$
Multiplying expression by $x-2$,
$-x+2<3x+1<x-2$
$-x+2<3x+1$ and $3x+1<x-2$
$1<4x$ and $2x<-3$
$\frac 14<x$ and $x<\frac{-3}{2}$
We have either $x<2$ or $x>2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simple inequality $\frac{4x+1}{2x-3}>2$
$$\frac{4x+1}{2x-3}>2$$
I have started with looking at positive/negative situations and there are 2 or that both expressions are positive or both are negative.
If both a positive we can solve for $$\frac{4x+1}{2x-3}>2$$
$${4x+1}>2(2x-3)$$
$${4x+1}>4x-6$$
$${0x}>-7$$
Or both a... | Let's use equivalent but simpler inequalities
$$
\frac{4x+1}{2x-3}>2 \quad \Leftrightarrow \quad
\frac{4x+1}{2x-3}-2>0 \quad \Leftrightarrow \quad
\frac{7}{2x-3}>0
$$
Therefore we must have
$$
2x-3>0 \quad \Leftrightarrow \quad x>{3\over2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove expression is not prime $$
(n + 4)^4 + 4
$$
If n is natural number, how to prove that above expression is not prime?
I am stuck here
$$
(n+4)^2 \cdot (n+4)^2 + 2 . 2
$$
$$
\left(\left(n^2+4^2\right) \cdot 2\right)\left(\left(n^2+4^2\right) \cdot 2\right)
$$
| Hint: $(x+4)^4+4=0$ has roots $x = -4 \pm \sqrt{\pm 2i}=-4 \pm (1 \pm i)\,$, so the expression factors as: $$(x+4)^4+4=(x^2 + 6 x + 10) (x^2 + 10 x + 26)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some ... | It is enough to prove the result for $x,y,z$ positive since
$$\frac{|x|+|y|+|z|}{3} \geq \frac{x+y+z}{3}$$
One can use the generalized AM-GM inequality: If
$$M_p = \left(\frac{x^p+y^p+z^p}{3}\right)^{\frac{1}{p}}$$ then for $p < q$, $M_p \leq M_q$ with equality holding if and only if $x=y=z$. Here, using $M_2 \geq M_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 15,
"answer_id": 2
} |
Solve $\sin(\frac{\pi}{5})$ analytically How do I solve $\sin(\frac{\pi}{5})$ analytically? I looked up here and in the first step I have to
Show that: $\cos \left(\frac{\pi }{5}\right)-\sin
\left(\frac{\pi }{10}\right)=\frac{1}{2}$
My question is, why do I have to do that?
| By repeated application of angle sum formulas we may get,
$$\sin (5x)=\sin^5 x+5 \cos^4 x\sin x-10 \sin^3 x \cos^2 x$$
Let $x=\frac{\pi}{5}$ and let $\sin (\frac{\pi}{5})=u$ then we have,
$$0=u^5+5(1-u^2)^2 u-10(1-u^2)u^3$$
It is safe to say $\frac{\sqrt{2}}{2}>u>0$. So that we may divide by $u$ to get.
$$0=u^4+5(1-u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $A= 2 \cdot \pi/7$ then show that $ \sec A+ \sec 2A+ \sec 4A=-4$ If $A= 2 \times \pi/7$ then how to show,
$$\sec A+ \sec 2A+ \sec 4A=-4$$ I have tried using formula for $\cos 2A$ but I failed.
| $$\sec\frac{2\pi}{7}+\sec\frac{4\pi}{7}+\sec\frac{8\pi}{7}=\frac{1}{\cos\frac{2\pi}{7}}+\frac{1}{\cos\frac{4\pi}{7}}+\frac{1}{\cos\frac{8\pi}{7}}=$$
$$=\frac{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{8\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Given: $\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$, find $f(x)$.
$$\int f(x)\sin x \cos x dx = \log(f(x)){1\over 2( b^2 - a^2)}+C$$
On differentiating, I get,
$$f(x)\sin x\cos x = {f^\prime(x)\over f(x)}{1\over 2( b^2 - a^2)}$$
$$\sin 2x (b^2 - a^2) = {f^\prime( x)\over (f(x))^2} $$
On integr... | On differentiating you get indeed
$$
f(x)\sin x\cos x=\frac{f'(x)}{f(x)}\frac{1}{2(b^2-a^2)}
$$
so the differential equation
$$
\frac{f'(x)}{f(x)^2}=(b^2-a^2)\sin2x
$$
Integrating it you get
$$
-\frac{1}{f(x)}=-\frac{1}{2}(b^2-a^2)\cos2x+c
$$
hence
$$
f(x)=\frac{2}{(b^2-a^2)\cos2x-2c}
$$
You can expand $\cos2x=\cos^2x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the Maximum value of $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ if $x$, $y$ and $z$ are positive real numbers such that $x+y+z=4$ Find the maximum value of $$S=\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$$
I tried as follows.
The given expression can be rewritten as
$$S=\s... | No, this approach is not correct, since
$$S=\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
does not necessarily hold. For example, consider $x=\frac12$, $y = \frac32$ and $z=2$, then $S=\left(\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \right) \approx 2.420$, but $\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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A combinatoric problem
The question is to find which of the following option is correct regarding $$\left(\frac{\ \ 2^{10}}{11}\right)^{11}$$
$A)$ strictly larger than $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2\binom{10}{5}$
$B)$ strictly larger than $\binom{10}{1}^2 \binom{10}{2}^2\binom{10}{3}^2\... | Is this AM-GM? We have
$$\binom{10}1^2\binom{10}2^2\binom{10}3^2\binom{10}4^2\binom{10}5
=\prod_{k=0}^{10}\binom{10}k<\left(\frac1{11}\sum_{k=0}^{10}
\binom{10}k\right)^{11}=\left(\frac{2^{10}}{11}\right)^{11}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of exponential quadratic and linear Is there any closed form of the integral below?
$$
y = \int_0^\infty \exp\left[-\frac{(\mathbf{b}-\mathbf{Ax})^2}{2\sigma^2}-\lambda \sum_{i=0}^{N-1} x_i\right]\ \mathrm{d}\mathbf{x}
$$
where $\mathbf{x}$ is an $N\times1$ vector, $\mathbf{A}$ is an $M\times N$ matrix, and $\... | With a vector $\mathbf{1}$ with all components equal to 1, write is as
$$
y = \int_0^\infty \exp\left[-\frac{(\mathbf{b}-\mathbf{Ax})^2}{2\sigma^2}-\lambda \mathbf{x}\mathbf{1}\right]\ \mathrm{d}\mathbf{x}
$$
and further, for $M = N$,
$$
y = \int_0^\infty \exp\left[-\frac{1}{2\sigma^2}(\mathbf{b}^2-2 \mathbf{bAx} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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For which primes $p$ are there integers $a,b,q$ so that $pq^2=a^4+4b^4$? There is a right triangle with rational sides and are $p$ iff there exists a solution.
$$pq^2=a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$$
Without loss of generality, choose $a$ odd and $a,b$ coprime, then $a^2+2b^2+2ab$ and $a^2+2b^2-2ab$ are also cop... | Unless I am getting really stupid in my old age, a rational triangle with area $p$ means that $p$ will be a Congruent Number. The first prime examples are $p=5$ and $p=7$. But a solution of $pq^2=a^4+4b^4$ must have $p \equiv 1 (\bmod 4)$, as the OP states. Thus the "iff" statement in the first sentence must be wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the radius of convergence of $\sum_{n=1}^{\infty} \left(\frac{5^{n} + (-1)^{n}}{n^{3}}\right)(x-2)^{n}$ Find the radius of convergence of the following:
$$\sum_{n=1}^{\infty} \left(\frac{5^{n} + (-1)^{n}}{n^{3}}\right)(x-2)^{n}$$
My attempt:
I used Ratio Test and managed to get until
$$\lim_{n\to\infty} \bigg|\fr... | L'Hospital's rule is not necessary here, one may just write, as $n \to \infty$,
$$
\left|\frac{5^{n+1} + (-1)^{n+1}}{5^{n} + (-1)^{n}} \cdot\left(\frac{n}{n+1}\right)^{3} (x-2)\right|=5\left|\frac{1 + \frac{(-1)^{n+1}}{5^{n+1}}}{1 + \frac{(-1)^{n}}{5^{n}}} \cdot\frac{1}{\left(1+\frac1n\right)^{3}} \right|\cdot|x-2| \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate limit of sums Evaluate $$\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n + \sqrt{k^2+n}}$$
I saw this as a Riemann Sum and tried to rewrite as
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(\frac{k}{n})^2 + \frac{1}{n}}}$$
but I can't say this is equal to $$\int_0^1 \frac{1}{1 + \sqrt{x^2 + 1}}... | A general approach for such "almost" Riemann sums is to evaluate as a double limit which, in this case, can be justified by uniform convergence of the inner limit:
$$\begin{align} \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/n}} &= \lim_{n \to \infty} \lim_{m \to \infty} \frac{1}{n} \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Minimal polynomial problems Let $V = \mathbb R^4, \phi \in$ End(V)
$$D(\phi)= \begin{pmatrix}
1 & -3 & 1 &-2 \\
2 & 1 & 1 & 1 \\
-2 & 4 & -2 & 3 \\
-2 & -2 & -1 & -2
\end{pmatrix}$$
The characteristic polynomial is $$h_\phi(x) = (x^2+1)(x+1)^2$$
After Cayley-Hamilton i know that the minimal polynomial divides the ch... | I have made a mistake when calculating.
The minimal polynomial is indeed $g(x)=(x^2+1) \cdot (x+1)^2$
I still have another question.
Let $p_i^{e^i}$ be a primary factor of the characteristic polynomial.
Determining $V_i = \operatorname{Ker}(p_i^{e^i}(\phi))$, I get
$$V_1 = \left<\begin{pmatrix}1 \\ 0 \\ -1 \\ 0\end{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding a sum of $1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$ I need someone to find a mistake in my soliution or maybe to solf it much more easily... I have got a sum $$1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$$ and need to evaluate it. So here's my... | To apply a discrete Fourier transform to Taylor series of $-\log(1-x)$ is a good idea.
Since
$$ -\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n} $$
we have
$$ \sum_{n\geq 0}\frac{x^{3n+1}}{3n+1} = \int_{0}^{x}\frac{dt}{1-t^3}\\= -\frac{1}{3}\log(1-x)+\frac{1}{18} \left(-\sqrt{3} \pi +6 \sqrt{3} \arctan\left(\frac{1+2 x}{\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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To prove $\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$ if $a,b,c$ are non zero positive reals prove $$\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$$ I have used A.M G.M inequality as follows:
$$a^3+b^3+c^3 \ge 3abc$$ adding $abc$ both sides we get
$$a^3+b^3+abc \ge 4abc-c^3=c(4ab-c^2)$$ so
$$\frac{1}{a^3+b... | The best proof is use $a^3+b^3 \geqslant ab(a+b).$ I have another proof
Because
$$(2c^2+ab)(a^3+abc+b^3)-abc\left[2(a^2+b^2+c^2)+ab+bc+ca\right]$$
$$=\frac{c(ab+bc-2ca)^2}2+\frac{c(3a^2+4ab+2ac+3c^2)(a-b)^2+b(a^2+5b^2)(c-a)^2}{4} \geqslant 0.$$
Thefore
$$\frac{abc}{a^3+abc+b^3} \leqslant \frac{2c^2+ab}{2(a^2+b^2+c^2)+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Integral $ \int_a^\infty \frac{dr}{r^5 \sqrt{r-a}}$
$$ \int_a^\infty \frac{dr}{r^5 \sqrt{r-a}}$$
By taking $r-a=t^2$ we are getting
$$ \int_0^\infty \frac{2dt}{(a+t^2)^5}$$
Now I am stuck after this. Thanks.
| Let $t = \sqrt{a}\tan(\theta), dt = \sqrt{a}\sec^2(\theta)$. Then your integral becomes
$$2\int \frac{\sqrt{a}\sec^2(\theta)}{(a+a\tan^2(\theta))^5} d \theta = \frac{2\sqrt a}{a^5} \int \cos^8(\theta)\, d\theta$$
$$ = \frac{2}{a^{9/2}} \int \left( \frac{1+\cos(2\theta)}{2}\right)^4 d\theta$$$$ = \frac{1}{8a^{9/2}}\int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $a,b,c$ are positive, prove that $\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a} \geq \frac{9}{a+b+c}$ If $a,b,c$ are positive real numbers, prove that
$$\frac{2}{a+b}+\frac{2}{b+c}+ \frac{2}{c+a}≥ \frac{9}{a+b+c}$$
| Let $a+b+c = s$. Then we have to prove
$$\dfrac{1}{s-a} + \dfrac{1}{s-b} + \dfrac{1}{s-c} \geq \dfrac{9}{2s},$$
or, equivalently,
$$\dfrac{3}{\dfrac{1}{s-a} + \dfrac{1}{s-b} + \dfrac{1}{s-c}} \leq \dfrac{2s}{3}.$$
Note that the LHS is the harmonic mean of $s-a,s-b,s-c$ and the RHS is the arithmetic mean of the same ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Matrix determinant as Dickson polynomial $\frac{x^{n+1}-y^{n+1}}{x-y}$
Given matrix $$
A=\begin{bmatrix}
x+y&xy&0& .&.&. &0\\
1&x+y&xy&0& .&.&0 \\
0&1&x+y&xy&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&0 \\
.&.&.&.&.&.&xy \\
0&.&.&.&0&1&x+y
\end{bmatrix}
$$
prove by induction that $$|A|=\frac{x^{n+1}-y^{n+1}}{x-y}$$ $x \neq... | You did the base cases $n=1,2$, then you do the inductive step using strong induction; suppose your claim is true for any $m \leq n$, then using the recursive formula, for $n \geq 3$:
\begin{align}
D_{n+1} &=(x+y)D_n-xyD_{n-1} =\\
&=(x+y)\frac{x^{n+1}-y^{n+1}}{x-y} -xy\frac{x^n-y^n}{x-y} = \\
&= \frac{1}{x-y}(x^{n+2}-... | {
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"url": "https://math.stackexchange.com/questions/2262488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Given point $P=(x,y)$ and a line $l$, what is a general formula for the reflection of $P$ in $l$ I am doing Isometries i.e. Rotations, translations and reflections at the moment. General equations for 2 of them are "easy"
Translation: $f(x,y)=(x+a,y+b)$
Rotation about origin through the angle $\alpha$ is just $f(r,\th... | This is fun. If point P is given and a line $y=m x + c$, then the mirror point Q is
$$ \begin{pmatrix} x_Q \\ y_Q \end{pmatrix} =\begin{bmatrix} \frac{1-m^2}{1+m^2} & \frac{2 m}{1+m^2} \\ \frac{2 m}{1+m^2} & \frac{m^2-1}{1+m^2} \end{bmatrix} \begin{pmatrix} x_P \\ y_P \end{pmatrix} + \begin{pmatrix} -\frac{2 c m}{1+m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a, b, c$ are positive and $a+b+c=1$, prove that $8abc\le\ (1-a)(1-b)(1-c)\le\frac{8}{27}$ If $a, b, c$ are positive and $a+b+c=1$, prove that $$8abc\le\ (1-a)(1-b)(1-c)\le\dfrac{8}{27}$$
I have solved $8abc\le\ (1-a)(1-b)(1-c)$ (by expanding $(1-a)(1-b)(1-c)$)
but do not get how to show that $(1-a)(1-b)(1-c)\le\fra... | Hint: $a,b,c \ge 0$ and $a+b+c=1$ implies $a, b, c \le 1\,$ so $1-a, 1-b, 1-c \ge 0$, then by AM-GM:
$$\sqrt[3]{(1-a)(1-b)(1-c)} \le \cfrac{(1-a)+(1-b)+(1-c)}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Continued Cosine Product. Is there a way to evaluate,
$$
\large \cos x \cdot \cos \frac{x}{2} \cdot \cos \frac{x}{4} ... \cdot \cos \frac{x}{2^{n-1}} \tag*{(1)}
$$
I asked this to one of my teachers and what he told is something like this,
Multiply and divide the last term of $(1)$ with $\boxed{\sin \frac{x}{2^{n-1... | By the sine duplication formula $\sin(2z)=2\sin(z)\cos(z)$ we have $\cos(z)=\frac{1}{2}\cdot\frac{\sin(2z)}{\sin(z)}$.
In particular
$$ \prod_{k=0}^{n-1}\cos\left(\frac{x}{2^k}\right)=\frac{1}{2^n}\prod_{k=0}^{n-1}\frac{\sin\frac{x}{2^{k-1}}}{\sin\frac{x}{2^k}}=\frac{\sin(2x)}{2^n\sin\frac{x}{2^{n-1}}} \tag{1} $$
is a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
null space of a matrix for a matrix with a 0 column I'm not a mathematician. Math is a hobby and currently I'm learning bits and pieces of linear algebra when I have time. Confused about a null space for the following matrix
\begin{bmatrix}
0 & 0 \\
0 & 2
\end{bmatrix}
How does one find a null space for this matrix?... | We find the null space by inspection. (Systematic methods are demonstrated in the linked pages below)
Given
$$
\mathbf{A} =
\left(
\begin{array}{cc}
0 & 0 \\
0 & 2 \\
\end{array}
\right)
$$
we note that
$$
\mathbf{A}
\left(
\begin{array}{cc}
1 \\
0 \\
\end{array}
\right)
=
\left(
\begin{array}{cc}
0 \\
0 \\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the coefficient in an expansion? I'm reviewing for my combinatorics final and have completely forgotten how to find the coefficient in the expansion of a polynomial. Here's an example I'm struggling with
Find the coefficient of $x^{11}$ in the expansion of
$$(x+x^2+x^3+x^4+x^5)^7(1+x+x^2+x^3+\dots)^4$$
So f... | Hint Pulling out the factor of $x^7$ we see that this is the same as finding the coefficient of $x^4$ in the expansion of $$(1 + x + x^2 + x^3 + x^4)^7 (1 + x + x^2 + x^3 + x^4 + \cdots)^4 .$$ Since no term of degree $\geq 5$ can contribute, this is the same as the coefficient of $x^4$ in
$$(1 + x + x^2 + x^3 + x^4)^7 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proof by induction $1\cdot 3 \cdot 5\cdots (2n-1) / (2 \cdot 4\cdot 6\cdots 2n) \leq 1/ \sqrt{n+1}, \forall n \geq 1, n\in \mathbb N$ Prove by induction for
$$ \frac{\left(1\right)\left(3\right)\left(5\right)...\left(2n-1\right)}{\left(2\right)\left(4\right)\left(6\right)...\left(2n\right)}\le\frac{1}{\sqrt{n+1}}\quad ... | There's no need to use factorials here, you can do it with basic arithmetic. The base case is
$$\frac{1}{2} \leq \frac{1}{\sqrt{2}},$$
which you can square both sides to find that it's true.
Then we note that if $f(n)$ is the fraction on the left of the inequality, then $f(n+1) = f(n)\frac{2n+1}{2n+2}$.
So we assume fo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $2^n + 4^n \leq 5^n$ I'm trying to prove by induction that $2^n + 4^n \leq 5^n$. Through some value plugging I've established that the induction must start from $n = 2$ because $2^2 + 4^2 \leq 5^2 \equiv 20 \leq 25$; for $n = 1$ it doesn't hold since $2 + 4 \geq 5$.
Now I assume that $2^k + 4^... | By Karamata for all $n \geq2$ we obtain:
$$2^n+4^n<5^n+1^n=5^n+1.$$
Thus, $2^n+4^n\leq5^n$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
$a+b+c+d+e=79$ with constraints How many non-negative integer solutions are there to $a+b+c+d+e=79$ with the constraints $a\ge7$, $b\le34$ and $3\le c\le41$?
I get that for $a\ge7$ you do $79-7=72$, $\binom{72+5-1}{5-1}=\binom{76}4$. For $b\ge35$ I think it's $\binom{47}4$ and I'm not too sure what it is for $3\le c\le... |
Here is an answer based upon generating functions.
*
*$a\geq 7$ can be encoded as
\begin{align*}
z^7+z^8+z^9+\cdots=z^7\left(1+z+z^2+\cdots\right)=\frac{z^7}{1-z}\tag{1}
\end{align*}
*$b\leq 34$ can be encoded as
\begin{align*}
1+z+z^2+\cdots+z^{34}=\frac{1-z^{35}}{1-z}\tag{2}
\end{align*}
*$3\leq c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3$ I have to prove that $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3 $$
is always true for real numbers $a, b, c>0$ with $abc=1$.
Using the AM-GM inequality I got as far as $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq \frac{b}... | By AM-GM$$\sum_{cyc}\frac{1+ab}{1+a}\geq3\sqrt[3]{\prod_{cyc}\frac{1+ab}{1+a}}=3\sqrt[3]{\prod_{cyc}\frac{1+\frac{1}{c}}{1+a}}=$$
$$=3\sqrt[3]{\prod_{cyc}\frac{1+c}{1+a}}=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$
If $\cos^2\theta+\cos\theta = 1$, express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$.
We have \begin{align*}\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\the... | Note that $\cos^2\theta+\cos\theta =1$ gives $(1-\sin^2\theta)+\cos\theta = 1$, so $\cos\theta = \sin^2\theta$. Therefore, we have \begin{align*}\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2 &= \cos^4\theta+\cos^3\theta+\cos^2\theta+\cos\theta-2\\&=(1-\sin^2\theta)^2+\cos\theta(1-\sin^2\theta)+\sin^2\theta-2\\&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the number of non-negative integral solutions of $x+y+z≤ 20$
Find the number of non-negative integral solutions of $x+y+z≤ 20$.
Conventional method :
$ x+y+z=20$, then no. of ways= $^{22}C_2$ (using beggar's method)
$x+y+z=19$, then no. of ways= $^{21}C_2$
$x+y+z=18$, then no. of ways= $^{20}C_2$
...
...
...
...... | If $x+y+z \leq 20$ Then the number of solutions will be the same as solving for the number of solutions to $w + x + y + z = 20$.
One approach to compute this is to look at the coefficient of $x^{20}$ in the generating function $f(x) = (1 + x + x^2 + ... + x^{20})^{4}$.
Use the identity $\displaystyle \sum_{n=0}^{k}x^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How can we tackle this integral $\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=-1?$ Something is wrong with this integral (in terms of splitting them out)
$$\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=\color{blue}{-1}\tag1$$
My try:
Splitting the integral
$$\int_{0... | You have to split as
$$
I=\int_0^1\frac{x^2/2+x+\log(1-x)}{x^3\sqrt{1-x^2}}+\int_0^1\frac{3x^2/2-3x+3\log(1+x)}{x^3\sqrt{1-x^2}}=\color{red}{I_1}+\color{blue}{I_2}
$$
Now we can calculate
$$
\color{red}{I_1}=-\sum_{n=3}^{\infty}\frac{1}{n}\int_0^1\frac{x^{n-3}}{\sqrt{1-x^2}}=-\frac{\sqrt{\pi}}{2}\sum_{n=3}^{\infty}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2271011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n\left[\sqrt{4i/n}\right]$ $$
\lim_{n \to \infty}
\sum_{i = 1}^{n}{1 \over n}
\,\left\lfloor\,\sqrt{\,{4i \over n}\,}\,\right\rfloor\quad
\mbox{where}\ \left\lfloor\,x\,\right\rfloor\
\mbox{is the greatest integer function.}
$$
I approached the problem this way-
$$
\lim_... | The Riemann integral is $$\lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left(a + \frac{b-a}{n} k\right) = \int_{x=a}^b f(x) \, dx,$$ and with $a = 0$, $b = 1$,and $f(x) = \lfloor 2 \sqrt{x} \rfloor$, we obtain $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \lfloor 2 \sqrt{k/n} \rfloor = \int_{x=0}^1 \lfloor 2 \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as... If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as
a) $\frac{1+t^2}{1-t^2}$
b) $\frac{2t}{1+t^2}$
c) $\frac{1-t^2}{1+t^2}$
d) $\frac{2t}{1-t^2}$
Attempt: I tried using the half angle formula but it just leaves me with an expression in terms of $... | In my opinion the simplest solution is as follows:
$\tan(\frac{x}{2})=t\implies\cos(\frac{x}{2})=\frac{1}{\sqrt{t^2+1}}$;
this can be seen by constructing a triangle with opposite side of angle $\frac{x}{2}$ equal to $t$, and adjacent side equal to $1$.
Now, it is known $\cos(x)=2\cos^2(\frac{x}{2})-1$.
$\cos(\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Matrices and Divisibility Let $p$ be an odd prime number and $T_p$ be the following set of $2$ x $2$ matrices
$$T_p=\{A=\left(\begin{array}{cc} a & b\\ c & a \end{array}\right); a,b,c \in\{0,1,2,3...,p-1\}\}$$
Q.1) The no. of $A$ in $T_p$ such that $det(A)$ is not divisible by p.
(A) $2p^2$
(B) $p^3-5p$
(C) $p^3-3p$
(... | The answers are
Q1: $ \ p^3-p^2$,
Q2: $ \ (p-1)^2$.
To obtain these answers, let us solve an auxiliary problem first:
How many combinations of $a,b,c\in[0,p-1]$ result in $\det A$ divisible by $p$?
Example $1$: $ \ p=3. \ $ We have only $p^2=9$ combinations with $\det A$ divisible by $p$, namely:
$$
a=0, \quad b=0, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
showing $ 1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0 $ How to show that the following infinite series
$$
1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0?
$$
The above series is of the form $\sum_{n \ge 1} \frac{f(n)}{n}$,... | Your series is
$$\begin{align}
S &=\sum_{k=0}^\infty \left( \frac{1}{4k+1}-\frac{3}{4k+2} + \frac{1}{4k+3} + \frac{1}{4k+4}\right)\\
&=\sum_{k=0}^\infty \int_0^1 (x^{4k}-3x^{4k+1}+x^{4k+2}+x^{4k+3})dx\\
&=\int_0^1 \left( \sum_{k=0}^\infty \left(x^{4k}-3x^{4k+1}+x^{4k+2}+x^{4k+3}\right) \right)dx\\
&=\int_0^1 \frac{1-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \frac{2x}{1-x^2} \right) $ where $|x| < \frac {1}{\sqrt {3}}$, then what is $y$? If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \dfrac{2x}{1-x^2} \right)$ where $|x| < \dfrac {1}{\sqrt {3}}$ then find the value of $y$.
. . .
Let $\tan^{-1} y= A$
$$y=\tan A$$
$$\tan^{-1} x=B$$
... | We use the identity $$\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}.$$
Letting $a=\tan^{-1}(x)$ and $b=\tan^{-1}\left(\frac{2x}{1-x^2}\right)$, we have $$y=\tan(a+b)=\frac{x+\frac{2x}{1-x^2}}{1-\frac{2x^2}{1-x^2}}=\frac{3x-x^3}{1-3x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ . Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ .
...
Let $\cot^{-1} x=z$
$$x=\cot z$$
Then,
$$\sin \cot^{-1} (x)$$
$$=\sin z$$
$$=\dfrac {1}{\csc z}$$
$$=\dfrac {1}{\sqrt {1+\cot^2 z}}$$
$$=\dfrac {1}{\sqrt {1+x^2}}$$
| Let $y=\frac{1}{\sqrt{1+x^2}}$ and let $\tan^{-1}y=w$
Then, $$\begin{align}
\ \cos{\tan^{-1}y} & =\cos{w} \\
\ \text{Now since $1 \ge y \gt 0$ we have $\cos{w} \gt 0$}\\
\ &= \frac{1}{\sec{w}}\\
\ &= \frac{1}{\sqrt{1+\tan^2{w}}}\\
\ &= \frac{1}{\sqrt{1+y^2}}\
\end{align}$$
And then substitute $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove: $\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\ge 2\sqrt{3}$ Let $a,b,c$ be the lengths of the sides of triangle $ABC$ opposite $A,B,C$, respectively, and let $m_a,m_b,m_c$ be the lengths of the corresponding angle medians. Prove:
$$\frac{a}{m_a}+\frac{b}{m_b}+\frac{c}{m_c}\ge 2\sqrt{3}.$$
Source: I thought about it... | By AM-GM $$\sum_{cyc}\frac{a}{m_a}=\frac{2a}{\sqrt{2b^2+2c^2-a^2}}=\sum_{cyc}\frac{4\sqrt3a^2}{2\sqrt{3a^2(2b^2+2c^2-a^2)}}\geq$$
$$\geq\sum_{cyc}\frac{4\sqrt3a^2}{3a^2+2b^2+2c^2-a^2}=2\sqrt3\sum_{cyc}\frac{a^2}{a^2+b^2+c^2}=2\sqrt3.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Function defined by its own integral at a given point. If $y=f(x)$ is a differentiable function such that
$$f(x)=x+ \int_0^1x^2zf(z)dz + \int_0^1xz^2f(z)dz$$ Then what is the value of $f\left(\frac{-9}{4}\right)$
The options were
(A)$\frac{-4}{9}$ (B)$\frac{4}{9}$
(C)$-1$ (D)$0$
First I tried differentiating with re... | Let $\displaystyle a=\int_0^1zf(z)dz$ and $\displaystyle b=\int_0^1z^2f(z)dz$. Then
$$f(x)=ax^2+(b+1)x$$
We have
\begin{align*}
\int_0^1zf(z)dz&=\int_0^1[az^3+(b+1)z^2]dz\\
a&=\left[\frac{az^4}{4}+\frac{(b+1)z^3}{3}\right]_0^1\\
a&=\frac{a}{4}+\frac{b+1}{3}\\
\frac{3a}{4}&=\frac{b+1}{3}\\
b+1&=\frac{9a}{4}
\end{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve for $x$ in $\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$ Solve for $x$ in $$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
i have done in a lengthy way:
By inspection we observe that $x=41$ and $x=-24$ are the solutions
we have
$$\sqrt[4]{57-x}=5-\sqrt[4]{x+40}$$ squaring both sides we get
$$\sqrt{57-x}=25+\sqrt{x+40}-10 \sqrt[4]{x+40}$$... | Since you found two roots you may try to argue that there are no more roots.
One way to do this is study the function $f(x)=(57-x)^{1/4}+(x+40)^{1/4}$.
Now,
we can calculate $$f''(x)= \dfrac{3}{16} \left (-\dfrac{1}{(57 - x)^{7/4}} - \dfrac{1}{(40 + x)^{7/4}} \right )<0.$$
Therefore $f$ cannot have three roots in $(-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How can I find maximum and minimum modul of this complex? Let be given the complex $z$ such that $$\left |z + \dfrac{1}{z}\right |=3.$$ Find maximum and minimum modul of complex $z$.
I tried. Put $z=x+i y$. From $$\left |z + \dfrac{1}{z}\right |=3.$$ We have $z^2 + 3z|$ = 3|z|.$ Threrefore
$$1 + 2 x^2 + x^4 - 2 y^2 +... | Hint. By using samjoe's comment $z\bar{z}=|z|^2$, we have that the given equation is equivalent to
$$|z|^2+2\cos(\theta)+\frac{1}{|z|^2}=\left |z + \dfrac{1}{z}\right |^2=3^2=9, \;\;|z|^2+\frac{1}{|z|^2}=9-2\cos(\theta)$$
where $z=|z|(\cos(\theta)+i\sin(\theta))$. Note that
$$7=9-2\cos(0)\leq|z|^2+\frac{1}{|z|^2}\leq 9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the following triple summation
Evaluate $$\sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \sum_{i=0}^{\infty} \frac{1}{3^i3^j3^k}$$ for $i\neq j\neq k$
My Attempt
Evaluate the summation with no restriction on $i,j,k$. Let this be $a_0$
When $i= j=k$. Let this be $a_1$.
When $i\neq j=k$, let this be $a_2$.
The... | Let's rearrange the sum:
$$ S := \sum_{i,j,k \ge 0, \ i \neq j \neq k} 3^{-i} 3^{-j} 3^{-k} = \sum_{j \ge 0} \left( 3^{-j} \sum_{i \ge 0, \ i \neq j} 3^{-i} \sum_{k \ge 0, \ k \neq j} 3^{-k} \right). $$
The two inner sums are equal
$$ \sum_{i \ge 0, \ i \neq j} 3^{-i} = \left( \sum_{i \ge 0} 3^{-i} \right) - 3^j = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplifying $\lim_{n\to \infty} \frac{\sqrt[n]{2}-1}{\sqrt[n]{8}-1}$ I was attempting to go through a worked solution from my lecturer but I can't really understand this step:
$\lim_{n\to \infty} \frac{\sqrt[n]{2}-1}{\sqrt[n]{8}-1} = \lim_{n\to \infty} \frac{1}{\sqrt[n]{4}+\sqrt[n]{2}+1}$
What was actually even done he... | Hint:
$$\frac{\sqrt[n]{2} - 1}{\sqrt[n]{8} - 1}=\frac{\sqrt[n]{2} - 1}{\sqrt[n]{8} - 1}\cdot\frac{\sqrt[n]{4} + \sqrt[n]{2} + 1}{\sqrt[n]{4} + \sqrt[n]{2} + 1}=\frac{\sqrt[n]{8} - 1}{\bigg(\sqrt[n]{8} - 1\bigg)\bigg(\sqrt[n]{4} + \sqrt[n]{2} + 1\bigg)}$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Functions and Sequences Problem
The function $F(k)$ is defined for positive integers as $F(1) = 1$,
$F(2) = 1$, $F(3) = -1$ and $F(2k) = F(k)$, $F(2k + 1) = F(k)$ for $k \geq
2$. Then $$F(1) + F(2) + \dotsb + F(63)$$ equals
$\begin{array}{lr}
(\text{A}) & 1 \\
(\text{B}) & -1 \\
(\text{C}) & -32 \\
(\text{D})... | We have $F(1)=1, F(2)=1, F(3)=-1$ and $F(2k)=F(k)$ and $F(2k+1)=F(k)$.
We have to evaluate $S=\sum_{i=1}^{63}F(i)$
Notice that $S=F(1)+F(2)+F(3) + \sum_{i=2}^{31}[F(2k) + F(2k+1)]$
$\Rightarrow S=F(1)+F(2)+F(3) + \sum_{i=2}^{31}[2F(k)]$
$\Rightarrow S=1+2\sum_{i=2}^{31}F(k)$
$\Rightarrow S=1+2[F(2)+F(3)+\sum_{i=2}^{15... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Is it true that for $n \ge5$, ${{3n} \choose {2n}} > \frac{6^n}{n}$ For $n \ge 5, \frac{9n^2 - 9n + 2}{4n^2-2n} > 2$
$9n^2 -9n +2 -8n^2+4n = n^2 -5n +2 \ge 5^2-25+2$
Here's the basis step:
$${{9} \choose {6}} = 84 > \frac{6^3}{3} = 72$$
Here's the inductive step:
$${{3n} \choose {2n}} = {{3(n-1)}\choose {2(n-1)}}\... | I will show that
$0.850
\lt \dfrac{\binom{3n}{2n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.
More precise bounds,
of the form
$1+O(\frac{c}{n})$,
can be easily gotten
by the method below.
Since
$n! \approx \sqrt{2\pi n}(n/e)^n$,
$\begin{array}\\
\binom{an}{bn}
&=\dfrac{(an)!}{(bn)!((a-b)n!}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find the inverse Laplace transform of the following fraction? Please help me to find the inverse Laplace transform of the following:
$$\frac{s^2+6s+9}{(s-1)(s-2)(s-3)}$$
Do I have to divide the fractions and use partial fractions decomposition?
Like this:
$$\frac{s^2}{(s-1)(s-2)(s-3)}=\frac{A}{s-1}+\frac{B}{s-2}... | We perform partial fraction decomposition on the expression first:
\begin{align*}
\frac{s^2 + 6s+9}{(s-1)(s-2)(s-3)} &= \frac{A}{s-1} + \frac{B}{s-2} + \frac{C}{s-3} \\
&= \frac{A(s-2)(s-3) + B(s-1)(s-3) + C(s-1)(s-2)}{(s-1)(s-2)(s-3)} \\
s^2 + 6s+9 &= A(s^2-5s+6) + B(s^2-4s+3) + C(s^2-3s+2) \\
&= (A+B+C)s^2 + (-5A-4B-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Evaluate $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$ How do you evaluate $$\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$$ using the identity $$\sin(\pi z)=\pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)?$$
I assume I'll have to express $1+\frac{1}{n^2}+\frac{1}{n^4}$ as $\... | Observing that
$$
\left(1+\frac1{n^2}+\frac1{n^4}\right)
=\left(1-\frac{a^2}{n^2}\right)\left(1-\frac{b^2}{n^2}\right)
$$
where
$$
a^2+b^2=-1\\
a^2b^2=1
$$
find $a$ and $b$ by a direct computation.
We get
\begin{align*}
\prod_{n\ge1}\left(1+\frac1{n^2}+\frac1{n^4}\right)
=&\prod_{n\ge1}\left(1-\frac{a^2}{n^2}\right)\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Minimum value of $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$ Given $a,b,c \in \mathbb{R^+}$ such that $a+b+c=12$
Find Minimum value of $$S=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$$
My Try: By Cauchy Schwarz Inequality we have
$... | I think it's correct.
By C-S we obtain:
$$\sum_{cyc}\left(a+\frac{1}{b}\right)^2=\frac{1}{3}\sum_{cyc}1^2\sum_{cyc}\left(a+\frac{1}{b}\right)^2\geq$$
$$\geq\frac{1}{3}\left(\sum_{cyc}\left(a+\frac{1}{b}\right)\right)^2=\frac{1}{3}\left(12+\frac{1}{12}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)^2\geq$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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equation involving absolute value function has $4$ solutions
If the equation $|x^2-5|x|+k|-\lambda x+7 \lambda = 0$ has exactly $4$ solution, Then $(\lambda,k)$ is
Attempt:$x^2-5|x|+k = \lambda x-7 \lambda$
$\star$ For $x>0,$ then $x^2-5x+k=\lambda x- 7 \lambda.$
$\star$ For $x\leq 0,$ then $x^2+5x+k=\lambda x- 7 \l... | First, notice that if $\lambda=k=0$ then the only solutions are $0,\,-5$ and $5$.
Suppose $\lambda=0$ and $0<k<\dfrac{25}{4}$.
Then
\begin{eqnarray}
{\Big\vert}\, |x|^2-5|x|+k\, {\Big\vert}&=&0\\
{\Huge\vert}\left(|x|-\frac{5}{2}\right)^2+k-\frac{25}{4}{\Huge\vert}&=&0\\
\left(|x|-\frac{5}{2}\right)^2&=&\frac{25}{4}-k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2288586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $xyz$, given that the value of $x^2+y^2+z^2$, $x+y+z=x^3+y^3+z^3=7$
Given that $$x^2+y^2+z^2=49$$ $$x+y+z=x^3+y^3+z^3=7$$
Find $xyz$.
My attempt,
I've used a old school way to try to solve it, but I guess it doesn't work.
I expanded $(x+y+z)^3=x^3+y^3+z^3+3(x^2y+xy^2+x^2z+xz^2+yz^2)+6xyz$
Since I know substit... | Ok, so you've got $$xy+yz+zx=0$$
Now, we know that $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\
\implies7-3xyz=7(49-0)\\
\implies xyz=-\dfrac{7\times48}{3}=-112$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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solve the set of simultaneous congruences using Chinese Remainder Theorem 2x= 1 (mod 5), 3x= 9 (mod 6), 4x = 1 (mod 7), 5x = 9 (mod 11) Solve the set of simultaneous congruences using Chinese Remainder Theorem
$\begin{cases}
2x \equiv 1 \pmod{5} \\
3x \equiv 9 \pmod{6} \\
4x \equiv 1 \pmod{7} \\
5x \equiv 9 \pmod{11} \... | $$ \begin{cases}
2x \equiv 1 \pmod{5} \\
3x \equiv 3 \pmod{6} \\
4x \equiv 1 \pmod{7} \\
5x \equiv 9 \pmod{11} \\
\end{cases} \implies \begin{cases}
x \equiv 3 \pmod{5} \\
x \equiv 1 \pmod{2} \\
x \equiv 2 \pmod{7} \\
x \equiv 4 \pmod{11} \\
\end{cases} \implies x \equiv 653 \pmod{770} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Given that $a\cos\theta+b\sin\theta+A\cos 2\theta+B\sin 2\theta \leq 1$ for all $\theta$, prove that $a^2+b^2\leq 2$ and $A^2+B^2 \leq 1$ I tried to solve this question but things got too much complicated and hence my efforts were completely futile.
Let $a,A,b,B \in \mathbb{R}$ and $$F(\theta)= 1- a\cos \theta - b\sin... | We have
$$0 \leq F(\theta) + F(\theta + \pi) = 2 - 2(A \cos 2\theta + B \sin 2\theta).$$
Let $u$ be the vector $(A,B)$. Pick $\theta$ so that $v = (\cos 2\theta, \sin 2\theta)$ is a unit vector in the same direction as $u$. Then
$$\sqrt{A^2 + B^2} = |u| = u \cdot v = A \cos 2\theta + B \sin 2\theta \leq 1.$$
For the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int_{|z|=1} \frac{1}{z^2 -\frac{3}{2}z + 1} dz$ Evaluate :
$$\int_{|z|=1} \frac{1}{z^2 -\frac{3}{2}z + 1} dz$$
Using residue method :
$$z=\frac{3}{4} \pm i \frac{\sqrt{7}}{4}$$
The problem is the two roots on the boundary $|z|=1$
|
As stated in the comments, the integral $\oint_{|z|=1}\frac{1}{z^2-\frac32 z+1}\,dz$ is not defined.
However, we can define the Cauchy Principal Value of the integral as the limit
$$\begin{align}
\text{PV}\left(\oint_{|z|=1}\frac{1}{z^2-\frac32 z+1}\,dz\right)&=\lim_{\epsilon \to 0}\int_{C_\epsilon} \frac{1}{z^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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The circumference of the circle is $C$, what is the area of circle in terms of $C$? The circumference of the circle is $C$, what is the area of circle in terms of $C$?
a). $\dfrac {C^2}{4\pi }$
b). $2\pi C$
c). $\dfrac {4}{3} \pi C^2$
d). $2\pi C^2$
My Attempt:
$$\textrm {Circumference}=2\pi r$$
$$C=2\pi r$$
$$\dfrac {... | Your attempt is correct, you should just cancel out the $\pi$ in the expression:
$$\require{cancel} \dfrac {\pi C^2}{4\pi^2} = \dfrac {\bcancel{\pi} C^2}{4 \bcancel{\pi} \pi} = \dfrac {C^2}{4\pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Intersection of two parabolas Given $a>0$ and $b>0$, I want to find the points of intersection of the two parabolas
\begin{align}
y&=1-ax^2 \\x&=1-by^2
\end{align}
Clearly I can just eliminate one of the variables, and I'll get a quartic equation, whose general solutions will be an enormous mess (according to Mathe... | Re-writing the second equation to be in terms of $y$, we have the two equations $y = 1 − a x^2$ and $y = \sqrt{\frac{1 − x}{b}}$.
Set them equal to each other, giving $1 − a x^2 = \sqrt{\frac{1 − x}{b}}$.
Square both sides and expand to get $a^2 x^4 − 2 a x^2 + 1 = \frac{1}{b} − \frac{1}{b} x$.
Move everything to the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Solutions of $\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$
:Solutions of:
$$\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$$
My try :
$$\lfloor 2x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac12 \rfloor $$
$$\lfloor 3x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac13 \rfloor... | Just do it in 6 cases:
$x = [x]+\{x\}$
Case 1: $\{x\} < 1/4$
Then $$[2x]\times[3x] + [4x] = 2[x]3[x] + 4[x] = 0$$ so $[x] = 0$ of $[x] = -\frac 32$. So $[x]$ is an integer $[x] = 0$ and $0 \le x < \frac 14$.
Case 2: $1/4 \le \{x\} < 1/3$
Then $$[2x]\times[3x] + [4x] = 2[x]3[x] + 4[x]+ 1 = 6[x]^2 + 4[x] + 1 = 0.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Constructing Series : Alternating between Rational Number and Irrational Number In the range of $[0,1] \in \Bbb R$, I'd like to construct a series which is strictly increasing and rational number and irrational numbers are alternatively appearing, like $0< \sqrt{1\over 3} < 2/3 < \sqrt{1 \over 2} < ... $
I want to make... | You may use numbers of the form $\frac{n}{n+1}$ such as $\frac{2}{3}$ for rational numbers and their geometric mean $\sqrt{\frac{n}{n+1}\frac{n+1}{n+2}}=\sqrt{\frac{n}{n+2}}$ in between.
This way, your sequence would start
$$\frac{1}{2}<\sqrt{\frac{1}{3}}<\frac{2}{3}<\sqrt{\frac{1}{2}}<\frac{3}{4}<\sqrt{\frac{3}{5}}<\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2299742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find values of constants a and b such that the given improper integral converges Find values of constants a and b such that:$$\int_{3}^\infty \left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right) dx=k$$ then by partial fractions we get:
$$\lim_{N \to \infty} \int_{3}^N \left(\frac{2}{3}\frac{1}{x}+\frac{a-\frac{2}{3}}{x+3}-\... | Since $\frac{ax+2}{x^2+3x}=\frac{a}x+O\!\left(\frac1{x^2}\right)$ and $\frac{b}{3x-2}=\frac{b}{3x}+O\!\left(\frac1{x^2}\right)$, it should be that $a=\frac{b}3$.
$$
\begin{align}
&\int_3^\infty\left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right)\,\mathrm{d}x\\
&=\int_3^\infty\left(\frac{\frac23}{x}+\frac{a-\frac23}{x+3}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integer solutions of equation $x^3+10x-1=y^3+6y$? I want to find all integer solution of the following equation in integers$$x^3+10x-1=y^3+6y$$ I tried to factor it somehow, but failed. Also $x$ and $y$ are have different parity. How can one solve this?
| Comparing terms $6y$ and $10x$ tells us that $x$ can not be greater than $y$. We'll find out that, let $y-x=z$ and equation becomes
$$x^3+10x-1=(x+z)^3+6(x+z)$$ $$-3zx^2-(3z^2-4)x-(z^3+6z+1)=0 $$
Discriminant of the obtained quadratic in $y$ must be non-negative, that is
$$(3z^2-4)^2-4(3z)(z^3+6z+1)\ge0,$$which gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find Volume V bounded by surface $\sqrt[3]{ x^2}+\sqrt[3]{ y^2}+\sqrt[3]{ z^2}=1$
I try this substitution $x={u}^3$,$y={v}^3$ and $z={k}^3$ ,and I get sphere with radius 1
| Let $x=r\sin^3\alpha\cos^3\beta$, $y=r\sin^3\alpha\sin^3\beta$
and $z=r\cos^3\alpha$.
Thus, $J=9r^2\sin^5\alpha\cos^2\alpha\sin^2\beta\cos^2\beta$ and the volume it's
$$9\int_{0}^1r^2dr\int_{0}^{\frac{\pi}{2}}\sin^5\alpha\cos^2\alpha d\alpha\int_{0}^{2\pi}\sin^2\beta\cos^2\beta d\beta=\frac{4\pi}{35}$$
About Jacobian... | {
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Solving $Т(n)=\frac{1}{\frac{1}{T(n-1)}+n^2}$ and $T(1) = 1$ $$T(n) = \frac1{\frac1{T(n-1)}+n^2}$$
I did find $T(2) = \frac{1}{1+n^2}$, but I don't know how to proceed.
How do I go on from here? How can I find the solution? Thanks. :)
| $$\frac{1}{t_{n}}=\frac{1}{t_{n-1}}+n^2$$ and the rest is smooth I think:
$$\frac{1}{t_2}=\frac{1}{t_1}+2^2$$
$$\frac{1}{t_3}=\frac{1}{t_2}+3^2...$$
$$\frac{1}{t_n}=\frac{1}{t_{n-1}}+n^2.$$
A summing of these equalities gives:
$$\frac{1}{t_n}-\frac{1}{t_1}=2^2+3^2+...+n^2$$ or
$$\frac{1}{t_n}=1^2+2^2+3^2+...+n^2$$ or
$... | {
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Inclusion exclusion distribution problem Prompt: How many ways are there to give 20 different presents to 4 different children, so that no child gets exactly 6 presents. All presents are different.
Here's what I tried doing
Since there are 20 presents to be distributed among 4 children,
$$C_1 + C_2 + C_3 + C_4 = 20$$... | For future reference here is a verification of the result by
N.F. Taussig, which I upvoted. This is the case where it is possible
for a child not to receive any presents. The combinatorial species
here is
$$\mathfrak{S}_{=4}(\mathfrak{P}_{\ne 6}(\mathcal{Z})).$$
We get the generating function
$$G(z) = \... | {
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How to parametrise intersection of two cylinders I have a question that asks me to verify Stokes Theorem on
$S:=$portion of the cylinder $ y^2+z^2=4$ with $ z>0$ and $x^2+y^2\leq1$
How would I parametrise the curve and surface so that I can apply Stokes Theorem. I have plotted this surface and can imagine what it look... | Based on the fact that $z = \sqrt{4-y^2}$ :
$$
\sigma(r,\theta) = (x,y,z) = (r\cos\theta, r\sin\theta, \sqrt{4-(r\sin\theta)^2}) \\
r \in [0,1],\ \theta \in [0,2\pi[
$$
According to WA, (And my understanding of the surface), this seems right.
The cross product is painful, I agree...
$$
\sigma_r = (\cos\theta, \sin\the... | {
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Integrating an expression by parts and by substitution giving two different solutions? So I'm given this example in my notes:
$$\int x(x+3)^5dx$$
And when I integrate by parts and by substitution I end up getting two different answers, but I don't know why.
Integrating by parts:
If I let $u = x$ and $dv = (x + 3)^5$, ... | In fact, the two solutions are equivalent (if you put the forgotten $C$ on the first integral):
\begin{align*}
\frac{x}{6}(x+3)^6 - \frac{1}{42}(x+3)^7 + C & = \frac{(x+3)-3}{6}(x+3)^6 - \frac{1}{42}(x+3)^7 + C \\
& = \frac{1}{6} (x+3)^7 - \frac{1}{2} (x+3)^6 - \frac{1}{42} (x+3)^7 + C \\
& = \frac{1}{7} (x+3)^7 - \fra... | {
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Evaluate $\int_0^{2\pi} \frac{\cos 2x}{1-2a\cos x+a^2} $ with $a^2<1$ How do I compute the integration for $a^2<1$,
$$\int_0^{2\pi} \dfrac{\cos 2x}{1-2a\cos x+a^2}dx=? $$
I think that:
$$\cos2x =\dfrac{e^{i2x}+e^{-2ix}}{2},
\qquad\cos x =\dfrac{e^{ix}+e^{-ix}}{2}$$
But I cannot. Please help me.
| One can factorise the denominator:
$$ 1-2a\cos{x}+a^2 = (e^{ix}-a)(e^{-ix}-a). $$
Then using partial fractions gives
$$ \frac{1}{2}\frac{e^{2ix}+e^{-2ix}}{(e^{ix}-a)(e^{-ix}-a)} = -\frac{a^2+1}{2 a^2} + \frac{a^4+1}{2 a (a^2-1)}\left( -\frac{1}{e^{i x}-a}-\frac{1}{1-a e^{i x}} \right)-\frac{e^{i x}+e^{-i x}}{2 a} $$
No... | {
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Prove that $\sum_{cyc}\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}\ge \frac{\sqrt{181}}{5}$ For $a,b,c>0$ satisfy $ab+bc+ca\ge \frac{4}{3}$. Prove that $$\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}+\sqrt{b^2+\frac{1}{\left(c+1\right)^2}}+\sqrt{c^2+\frac{1}{\left(a+1\right)^2}}\ge \frac{\sqrt{181}}{5}$$
My try: By Minkowski:
$... | Let $a=\frac{2x}{3}$, $b=\frac{2y}{3}$ and $c=\frac{2z}{3}$.
Hence, the condition gives $xy+xz+yz\geq3$ and we need to prove that
$$\sum_{cyc}\sqrt{\frac{4x^2}{9}+\frac{9}{(2y+3)^2}}\geq\frac{\sqrt{181}}{5}.$$
Now, by C-S
$$\sum_{cyc}\sqrt{\frac{4x^2}{9}+\frac{9}{(2y+3)^2}}=\frac{15}{\sqrt{181}}\sum_{cyc}\sqrt{\left(\f... | {
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Recursive relationship for incomplete beta function Consider the incomplete Beta function $I_x(a, b)$
$$ I_x(a, b) =
\dfrac{B(x; a, b)}{B(a, b)} =
\dfrac{\int_0^x t^{a-1} \left( 1-t \right)^{b-1} dt}{\int_0^1 t^{a-1} \left( 1-t \right)^{b-1} dt}.
$$
How can I prove the recursive property given in Wikipedia
$$
I_x(a+1,... | Using the Beta function property $B(a+1,b)=aB(a.b)/(a+b)$ compute the derivative
$$I_x'(a+1, b) = \frac{x^a (1-x)^{b-1}}{B(a+1,b)} =
\frac{x^a (1-x)^{b-1}}{B(a,b)a/(b+a)}=
(a+b)\frac{x^a (1-x)^{b-1}}{B(a,b)a}$$
Now let $$
f(a,b,x) = I_x(a+1, b) - I_x(a, b) + \dfrac{x^a \left( 1-x \right)^b}{a \, B\left( a, b \right)} ... | {
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Inequality "A la Rozenberg" Hello I want to solve this
The inequality is equivalent to this :
$$\frac{a^2}{b^2}cos(arctan(\sqrt{\frac{b}{a}}))^2+\cos(\arctan(\sqrt{\frac{c}{b}}))^2+\frac{c^2}{b^2}\cos(\arctan(\sqrt{\frac{a}{c}}))^2\geq \frac{3}{2b^2}$$
We put :
$\sqrt{\frac{b}{a}}=\frac{x+y}{1-xy}$
$\sqrt{\frac{c}{b}}=... | To start we study the following function :
$$f(x)=\frac{x^3}{x^2+1}+\frac{1}{h^2+1}+\frac{h}{y^2+1}-\frac{3^{\frac{2}{3}}}{2}(x^3+h^3+1)^{\frac{1}{3}}$$
The derivative of function is :
$$f'(x)=\frac{1}{2}x^2[\frac{-3^{\frac{2}{3}}}{(h^3+x^3+1)^{\frac{2}{3}}}-\frac{(4x^2)}{(x^2+1)^2}+\frac{6}{(x^2+1)}]$$
So we study thi... | {
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Prove that for any $n \neq 4$, if for any $k < n$ we have that $a_k$ is relatively prime to $a_n$, then $a_n$ is prime
Consider the sequence $a_n = |n(n+1)-19|$ for $n = 0,1,2,\ldots$. Prove that for any $n \neq 4$, if for any $k < n$ we have that $a_k$ is relatively prime to $a_n$, then $a_n$ is prime.
The first few... | If $a_n$ is not prime and $n > 4$, then $a_n>1$ and there exists a prime natural number $p\leq n$ such that $p\mid a_n$. Then, $p$ divides both $a_n$ and $a_{n-p}$. (In fact, one can show that $p=n$ implies $n=19$.)
| {
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Finding the sum of digits of $9 \cdot 99 \cdot 9999 \cdot ... \cdot (10^{2^n}-1)$ Find the sum of the digits (in terms of $n$) of $$9 \cdot 99 \cdot 9999 \cdot 99999999 \cdot ...
\cdot (10^{2^n} - 1)$$
Where every term has two times more digits than the previous term.
I tried to factorise as $$(10-1)(10^2-1)(10^4-1).... | Let us assume that $N$ is a number with $2^k-1$ decimal digits, whose last digit is $\geq 1$.
Let $S(N)$ be the sum of digits of $N$. Let us study the sum of digits of
$$ N\cdot(10^{2^k}-1) = N\cdot 10^{2^k}- N = N\cdot 10^{2^k} - 10^{2^k} + (10^{2^k}-1-N)+1. $$
We have:
$$ S(N\cdot(10^{2^k}-1)) = S(N)-1+\left(9\cdot(... | {
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Is this an ellipse?
Is this parameterisation an ellipse:
\begin{align}x(t) &= \frac{2 \cos(t)}{1 + a \sin(t)}\\
y(t) &= \frac{2 \sin(t)}{1 + a \sin(t)}\end{align}
where $a$ is a real positive parameter.
I tried to do it the naive way but couldn't find a definitive answer.
Plotting our curve with the help of Geog... | $$\begin{align}
x&=\frac {2\cos t}{1+a\sin t}\tag{1}\\
y&=\frac {2\sin t}{1+a\sin t}\tag{2}\\
(2)/(1):\hspace{3cm}\\
\frac yx&=\tan t\tag{3}\\
(1)^2+(2)^2:\hspace{3cm}\\
x^2+y^2&=\frac 4{(1+a\sin t)^2}\\
&=\frac {4(x^2+y^2)}{\big(\sqrt{x^2+y^2}+ay\big)^2}
&&\scriptsize \bigg(\sin t=\frac y{\sqrt{x^2+y^2}}\bigg)\\
(x^2+... | {
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"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 0
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Smith Normal Form I would like to put this matrix below into Smith Normal Form over $\mathbb{Q}[x]: $
$$\left(
\begin{array}{ccc}
7 & x & 0 & -x \\
0 & x-3 & 0 & 3\\
0 & 0 & x-4 & 0 \\
x-6 & -1 & 0 & x+1
\end{array}
\right)$$
but I am stuck here:
$$\left(
\begin{array}{ccc}
7 & 0 & 0 & -x \\
0 & x & 0 & 3\\
0 & 0 & x-... | Here are the first few steps as per my suggestion in the comments:
\begin{align*}
\left(\begin{array}{rrrr}
x & 7 & 0 & -x \\
x - 3 & 0 & 0 & 3 \\
0 & 0 & x - 4 & 0 \\
-1 & x - 6 & 0 & x + 2
\end{array}\right) &\leadsto
\left(\begin{array}{rrrr}
-1 & x - 6 & 0 & x + 2 \\
x - 3 & 0 & 0 & 3 \\
0 & 0 & x - 4 & 0 \\
x & 7 ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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How do you integrate $\frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)}$?
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)} \,dx$$
I tried by dividing the terms in both the numerator and denominator by $\cos^5x$ but still cant find my way.
| $$I =\int_{0}^{\frac{\pi}{2}} \frac{\cos^5x}{\sin^5x + \cos^5x} dx\tag{1}$$
Let $u =\frac{\pi}{2}-x$
$$I = -\int_{\frac{\pi}{2}}^{0} \frac{\sin^{5}u}{\sin^5u+\cos^5u}du\tag{2} =\int_{0}^{\frac{\pi}{2}} \frac{\sin^5x}{\sin^5x + \cos^5x} dx$$
Then $$(1)+(2)\implies 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^5x+\cos^5x}{\sin^5... | {
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"source": "stackexchange",
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$e^i.e_j=\delta_{ij}$ not true in a two dimensional coordinate system with dual. Suppose vector space $\mathbb{R}^2$ with standard basis $E=\{e_1,e_2\}$, vector space $V$ with basis $E^\prime=\{e^\prime_1,e^\prime_2\}$, $V^*$ (dual of vector space $V$) with basis $E^*=\{e^1,e^2\}$ and a linear transformation $A^\prime... | I get
$$
\left[\begin{matrix} e^\prime_1 \\ e^\prime_2\end{matrix} \right]=
\left[ \begin{matrix}
\cos\beta & \sin\beta \\
\cos\gamma & \sin\gamma
\end{matrix} \right]
\left[\begin{matrix} e_1\\ e_2\end{matrix} \right]
$$
At least that makes
$$
\left[\begin{matrix} e^\prime_1 & e^\prime_2\end{matrix} \right]
\left[\b... | {
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"question_score": "4",
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Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 $ So far I've done this:
LHS $ =\cos^{2}3x-\sin^{2}3x$
$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$
$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$
I can tell I'm going in the right direction but how should I proceed further?
EDIT ... | Another way:
$$\cos(3\cdot2x)=4\cos^3(2x)-3\cos2x$$
Now use $\cos2x=2\cos^2x-1$
| {
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Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$
$αβ=\... | By the formula for the roots of a quadratic equation, the squared difference between them is $$\left(\frac{\pm\sqrt{b^2-ac}}{a}\right)^2=\frac{b^2-ac}{a^2}$$ (factor $4$ omitted) and is invariant by translation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 4
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Prove by this inequality to with $x^2+y^2+z^2=1$ Let $x,y,z>0,x^2+y^2+z^2=1$ show that
$$\sum_{cyc}\sqrt{\dfrac{x^2+y}{2}}\ge\sqrt{2}$$
Here's what I have done. The expression $x^2+y\ge x^2+y^2$. it is is equivalent to
$$\sum_{cyc}\sqrt{1-x^2}\ge 2$$ shows that the function is neither concave or convex. So I don't thi... | We need to prove that
$$\sum_{cyc}\sqrt{\frac{x^2+y^2}{2}}\geq\sqrt2$$ or
$$\sum_{cyc}\sqrt{x^2+y^2}\geq2,$$
which is true because
$$\sum_{cyc}\sqrt{x^2+y^2}=\sqrt{\sum_{cyc}(x^2+y^2+2\sqrt{(x^2+y^2)(x^2+z^2)})}\geq$$
$$\geq\sqrt{\sum_{cyc}(x^2+y^2+2x^2)}=2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\int \frac{1}{6+(x+4)^2} dx$. $\int \frac{1}{6+(x+4)^2} dx = 6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx$
Now $u=\frac{(x+4)}{\sqrt{6}}$ and $du=\frac{1}{\sqrt{6}}dx$.
$6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx=\frac{6}{\sqrt{6}}\int\ \frac{1}{1+u^2}=\frac{6}{\sqrt{6}}$ arctan$(u)$=$\frac{6}{\sqrt{6}}$ arctan$(\frac{... | Start U-substitution:
$u$ = $x$+4 and $dx$= 1$dx$ $$\\$$
Making the integral:
$$\int \frac{1}{6+(u)^2} du$$ $$\\$$
Now do another U-substitution:
$u$ = $\sqrt6v$ and $du$= $\sqrt6 dv$ $$\\$$
Making the integral:
$$\int \frac{1}{\sqrt6+(v^2+1)} dv$$ $$\\$$
Take out the constant:
$$\frac{1}{\sqrt6}\int \frac{1}{(v^2+1)... | {
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convergence of $\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^\alpha dx $ For which value $\alpha$ does the following integral converges:
$\int_{0}^{+ \infty} (\sqrt{x^2 + 2x + 2} - x - 1)^\alpha dx $
I tried to use: $ \sqrt{x^2 + 2x + 2} - 1 = \frac{x^2 + 2x + 1}{\sqrt{x^2 + 2x + 2} + 1} $
I wanted to simplify i... | $$\sqrt{x^2+2x+2}-x-1=\frac{x^2+2x+2-(x^2+2x+1)}{\sqrt{x^2+2x+2}+x+1}=\frac{1}{\sqrt{(x+1)^2+1}+x+1}\sim \frac{1}{2(x+1)}$$
So our integral converges if
$$\int_0^\infty\frac{dx}{(2(x+1))^\alpha}$$
converges, do you know when this integral converges? You can compute the anti derivative to assure yourself.
| {
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If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
| HINT:
You could further find $$z+\frac{1}{z}=3$$
and note that $$a^3+b^3=(a+b)(a^2-ab+b^2)$$
| {
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Why a $20$ digit number starting with eleven $1$'s cannot be a perfect square? Why a $20$ digit number starting with eleven $1$'s cannot be a perfect square?
I haven't been able to figure out which properties of perfect squares disallows such a number from being a perfect square.
| Let $x$ be a positive integer with $2n$ digits when written in base $b>1$, where the $n+1$ first digits are 1's, and where we will assume that $b-1=d^2$ is a perfect square. Note that this holds in the decimal system ($b=10$), as $10-1=9=3^2$. It also holds in the binary system ($b=2$), and with respect to several othe... | {
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$a-b \nmid a^2+b^2$ if $a>b+2$ and $\gcd(a,b)=1$ Any two co-prime number $a,b$ with $a>b+2$ we have $a^2+b^2$ is not divisible by $a-b$, $a,b \in \mathbb{N}$.
But how to prove this?
| If $a-b \mid a^2+b^2$ then $\gcd(a^2+b^2,a-b)=a-b \ge 3$. Now
$$
\gcd(a^2+b^2,a-b)=\gcd(a^2+b^2-(a-b)^2,a-b)=\gcd(2ab,a-b).
$$
Since $\gcd(a,b)=1$ then every prime dividing $a$ or $b$ cannot divide $a-b$. Hence
$$
\gcd(a^2+b^2,a-b) \in \{1,2\}.
$$
In particular, your conjecture is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a positive point of maximum
The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a
positive point of maximum lies in the interval
$(a_1,a_2)\cup(a_3,a_4)$.find the value of $a_2+11a_3+70a_4$
I found $f'(x)=3x^2+6(a-7)x+3(a^2-9)=0$ and L... | The point of maximum is the least root of $f'(x)=0$. In order for it to be positive, both roots must be. So we need, after writing the equation as
$$
x^2+2(a-7)x+(a^2-9)=0,
$$
that the following conditions are satisfied:
\begin{cases}
-2(a-7)>0 & \text{sum of roots is positive} \\
a^2-9>0 & \text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If the roots of an equation are $a,b,c$ then find the equation having roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$. Actually I have come to know a technique of solving this kind of problem but it's not exactly producing when used in a certain problem.
Say we have an equation
$$2x^3+3x^2-x+1=0$$
the roots of this ... | Hint: let $y=(1-x)/x \iff x=1 / (y+1)\,$, then:
$$
P(x)=P\left(\frac{1}{y+1}\right)=\frac{1}{(y+1)^3}\big(1+3(y+1)^2+(y+1)^3\big)
$$
The polynomial in $y$ is therefore:
$$
1+3(y+1)^2+(y+1)^3=y^3 + 6 y^2 + 9 y + 5
$$
This can be verified in WA by calculating the resultant[ x^3+3x+1, 1-x-xy, x ].
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplify $5(3 \sin(x) + \sqrt3 \cos(x))$ So Wolfram tells me I can reach $10 \sqrt3 \sin(x + π/6)$ from this expression, but I cant grasp how to do it. Any help is appreciated.
| We know
$$\sin (x + a) = \sin x \cos a + \sin a \cos x$$
While we have
$$15 \sin x + 5 \sqrt 3 \cos x \tag {*}$$
Therefore if we can find an numbers $a, C$ such that $15 = C \cos a$ and $5\sqrt3 = C\sin a$, we can rewrite $(*)$ as
$$C \cos a \sin x + C \sin a \cos x = C\sin (x+a)$$
To find appropriate $C$ and $a$, ... | {
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"url": "https://math.stackexchange.com/questions/2340143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Definite integral of exponential of square root Does anyone have a clue as to how integrals of the kind
$$
I(a) = \int_{-\infty}^{\infty} dt \frac{e^{-\sqrt{t^2 + a^2}}}{\sqrt{t^2 + a^2}}
$$
or
$$
I_2(a) = \int_{-\infty}^{\infty} dt e^{-\sqrt{t^2 + a^2}}
$$
can be solved? I realize there is a problem as the square roo... | For this one $$
I(a) = \int_{-\infty}^{\infty} dt \frac{e^{-\sqrt{t^2 + a^2}}}{\sqrt{t^2 + a^2}}
$$
Note that $(\ln |t+ \sqrt {t^2 + a^2} |)' = \frac {1}{\sqrt {t^2 +a^2}}$
Next notice that $(e^{-\sqrt{t^2 + a^2}})' = -e^{-\sqrt{t^2 + a^2}} * \sqrt{t^2 + a^2}$
so $(\ln |t+ \sqrt {t^2 + a^2} |* e^{-\sqrt{t^2 + a^2}})' ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Simplify $x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0$ to $x^2 +45x-8=0$ Simplify ($x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0$) to ($x^2 +45x-8=0$)
These two equations have the same solutions.
If given the first equation how would you go about simplifying such that you end up with the quadratic form ?
I tried cubing the first expr... | Cubing is not a mess. Look here:
$$x^{2/3}+3 x^{1/3}=2$$
cubing both sides
$$\left(x^{2/3}+3 x^{1/3}\right)^3=8$$
Expand:
$$27 x^{4/3}+9 x^{5/3}+x^2+27 x=8$$
Collect $x$ between (among?) the first two terms:
$$9 \left(x^{2/3}+3 x^{1/3}\right) x+x^2+27 x=8$$
Inside the parenthesis there is the left side of the given equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Characteristic polynomial of a $7 \times 7$ matrix whose entries are $5$ Avoiding too many steps, what is the characteristic polynomial of the following $7 \times 7$ matrix? And why?
\begin{pmatrix}5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\end{pmatrix}
| It is easy to see, that $v=(1,1,1,1,1,1,1)^T$ is an eigenvector of that matrix. By calculation the corresponding eigenvalue is $35$ (just calculate $Av$).
since the rank of the matrix is $1$ and it has the eigenvalues with their multiplicities as zeros it has to be of the form $p(t) = a t^6 (t-35)$ with $a\neq 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
what is $\sqrt{(i^4)}$ We know that,
$(a^m)^n = a^{mn} = (a^{n})^m$
So ,
*
*$\sqrt {i^4} = (i^{1/2})^4 = \left(\pm\frac{1+ i}{\sqrt2} \right)^4 = -1$
*$\sqrt {i^4} = (i^{2}) = -1$
*$\sqrt {i^4} = \sqrt1 = 1$
I think only no $3$ is right. I must have violated some rule in $1$ & $2$. Please let me know.
Maybe $(... | $$\sqrt{i^4}=i^2 = -1\tag1$$
Or
$$\sqrt{i^4}= \sqrt{(\sqrt{-1})^4} = (\sqrt{-1})^2 = -1 = i^2 \tag2$$
Or
$$\sqrt{i^4} = \sqrt{i^2\cdot i^2} = \sqrt{i^2}\cdot \sqrt{i^2}= i\cdot i = i^2 = -1\tag3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$? In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?
Alright, so if d is the ... | By Stewart's theorem the squared length of the median from $A$ is given by $\frac{2b^2+2c^2-a^2}{4}$.
In our case we know that $c=3, b=4$ and $2b^2+2c^2-a^2 = 4a^2$, hence
$$ a^2 = \frac{2}{5}(b^2+c^2) = 10 $$
and $a=\color{red}{\sqrt{10}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
second derivative of a function equals a power of said function Is it possible to find an expression for $f(x)$ so that:
$$f''(x) = (f(x))^{-2}$$
and
$$f(0) = 1$$
I suspect that no such expression exists but i might be wrong.
Thanks in advance.
| One can take the hard path and let
$$f(x) = 1 + a_{1} x + a_{2} x^2 + a_{3} x^3 + a_{4} x^4 + \cdots,$$
since $f(0) = 1 = a_{0}$, such that
\begin{align}
1 &= f^2 \, f'' = (1 + a_{1} x + a_{2} x^2 + a_{3} x^3 + a_{4} x^4 + \cdots)^2 \cdot (2 a_{2} + 6 a_{3} x + 12 a_{4} x^2 + \cdots) \\
&= (1 + 2 a_{1} x + (2a_{2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
| If you just want the value of expression $(a+b)^2 + (b+c)^2 + (a+c)^2$than set the vale of $a=0$, $b=1$ and $c=1$ (these satisfy the required conditions ) and you are going to get the answer.
$$(1)^2 + (2)^2 + (1)^2 = 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
Where was my mistake (integration by trig-substitution problem)? I am attempting to solve the problem
$$\int \frac{dx}{x^2+x+1}$$
First, I complete the square, then factor out a $\frac {3}{4}$:
$$\int \frac{dx}{\frac{3}{4}(\frac{4}{3}(x+\frac{1}{2})^2+1)}$$
Let $u = \sqrt{\frac{4}{3}}(x+\frac{1}{2})$
$$\frac{du}{dx} = ... | Your only mistake appears to be a failure to notice that $\displaystyle \frac{4}{3} \frac{\sqrt{3}}{2} \tan^{-1} \left( \sqrt{\frac 4 3} \left(x+\frac 1 2 \right)\right) $ is exactly the same thing as $ \displaystyle \frac 2 {\sqrt 3} \tan^{-1} \left( \frac{2x+1}{\sqrt 3} \right).$
First you have
$$
\frac 4 3 \cdot \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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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.