Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the exact value of integration $ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$ Can you help me find the exact value for integration with the given steps?
$$ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$$
Some of my attempts as indefinite Integral
$$
\int \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}} \, dx\approx \left(\sqrt... | Hint:
$$
\begin{aligned}
& \int\frac{dx}{\sqrt{1-x}+\sqrt{x+1}+2}\\
& \stackrel{x\to\cos2\phi}=
\int\frac{\sin{2\phi}\,d\phi}{1+\frac1{\sqrt2}(\sin\phi+\cos\phi)}
=\int\frac{\sin{2\phi}\,d\phi}{1+\sin(\phi+\frac\pi4)}\\
&\stackrel{\phi\to\theta+\frac\pi4}
=\int\frac{\cos{2\theta}\,d\theta}{1+\cos\theta}=\int\frac{2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3607599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the derivative of $\mathbf{a}^TX^2\mathbf{a}$ with respect to the symmetric matrix $X$? Given a constant vector $\mathbf{a}\in{\rm I\!R}^n$ and a real symmetric matrix $X\in{\rm I\!R}^{n\times n}$, what is the derivative of $\mathbf{a}^T X^2 \mathbf{a}$ with respect to $X$?
I tried a simple example using $n=2$... | Hint
Fréchet derivative of $f(X) =\mathbf{a}^TX^2\mathbf{a}$ is given by
$$\partial_{X_0}f(h) = \mathbf{a}^TX_0 h\mathbf{a} + \mathbf{a}^Th X_0 \mathbf{a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Area of square that is inscribed in a circle that is also inscribed in a rhombus
A circle is inscribed in a rhombus whose diagonals are $17 cm$ and $27 cm$. What is the area of square inscribed on the same circle?
Solution:
Centred at the origin, one side can lie on the line $\frac{x}{(13.5)} + \frac{y}{(8.5)} = 1$.
... | Let $r$ be the circle of the radius. Then,
$$r=\frac{a_1a_2}{\sqrt{a_1^2+a_2^2}}$$
with $a_1$ and $a_2$ being the half diameters of the rhombus. Then, the area of the square is
$$ Area = 2r^2 = \frac{2\left( \frac{17}2 \cdot\frac{27}2\right)^2}{{\left(\frac{17}2\right)^2+\left(\frac{27}2\right)^2}} = \frac12\cdot\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Express the matrix of $f$ with respect to the basis $\{1,x+1,x^{2}+x\}$. Let $\textbf{P}_{2}(\textbf{R})$ be the real vector space of polynomials of degree less than or equal to $2$.
Let $f:\textbf{P}_{2}(\textbf{R})\rightarrow \textbf{P}_{2}(\textbf{R})$ be the linear map given by differentiation, i.e., $f(p(x)) = p'(... | Since we are dealing with a linear operator, we can assume the same basis for the domain and the counterdomain.
Given the basis $\mathcal{B} = \{1,x+1,x^{2}+x\}$, we have the following system of equations to solve
\begin{align*}
\begin{cases}
a_{11} + a_{12}(x+1) + a_{13}(x^{2}+x) = f(1) = 0\\\\
a_{21} + a_{22}(x+1) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Range of $f(x)=ax^2-c$ The function $f(x)=ax^2-c$ satisfies $-4\le f(1) \le -1$ and $-1\le f(2) \le 5$. Which of the following statement is true ?
(1)$-7\le f(3) \le 26$
(2)$-4\le f(3) \le 15$
(3)$-1\le f(3) \le 20$
(4)$-\frac {28}{3}\le f(3) \le \frac {35}{3}$
My approach is as follow
$f(1)=a-c$
$f(2)=4a-c$
$f(3)=9a-... | First, consider how to express $f(3)$ as a linear combination of $f(1)$ and $f(2)$ so you can effectively use their adjusted limits to determine the limits for $f(3)$. To do this, for some real constants $d$ and $e$ you have
$$\begin{equation}\begin{aligned}
9a - c & = d(a - c) + e(4a - c) \\
& = (d)a - (d)c + (4e)a - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving Proposition 4.1.8. from Terence Tao's Analysis I Let $a$ and $b$ be integers such that $ab = 0$. Then either $a = 0$ or $b = 0$ (or both $a=b=0$).
MY ATTEMPT
Let us consider that $a = m - n$ and $b = p - q$, where $m,n,p,q$ are natural nubmers.
Then we have that
\begin{align*}
ab = (m-n)(p-q) & = (mp + nq) - (m... | You can try and prove the contrapositive, namely, that $a \neq 0$ and $b \neq 0$ implies $ab \neq 0$. Use trichotomy of integers for $a$ and $b$ and Lemma 2.3.3.
Or you can use contradiction and assume that $a \neq 0$ and $b \neq 0$. Then, use trichotomy of integers and see that it contradicts Lemma 2.3.3.
Or if you pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Conjecture Prove that : $\sum_{cyc}\frac{a}{a^n+1}\leq \sum_{cyc}\frac{a}{a^2+1}\leq \frac{3}{2}$ Conjecture, Prove that :
$$\sum_{cyc}\frac{a}{a^n+1}\leq \sum_{cyc}\frac{a}{a^2+1}\leq \frac{3}{2}$$
Under the assumptions $a\geq b\geq 1\geq c>0$ such that $abc=1$ and $\frac{c}{c^n+1}\geq \frac{b}{b^n+1}\geq \frac{... | I will prove the second inequality $\sum_{cyc}\frac{a}{a^2+1}\leq \frac{3}{2}$ for you.
Note that for $x\gt0,$
$$\dfrac{x}{x^2+1}=\left(x+\dfrac1x\right)^{-1}$$ and by AM-GM inequality $$x+\dfrac1x\ge2$$ with equality occurs at $x=1.$ Hence the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Linear combinations problem The vectors $\dbinom{3}{2}$ and $\dbinom{-4}{1}$ can be written as linear combinations of $\mathbf{u}$ and $\mathbf{w}$:
\begin{align*}
\dbinom{3}{2} &= 5\mathbf{u}+8\mathbf{w} \\
\dbinom{-4}{1} &= -3\mathbf{u}+\mathbf{w} .
\end{align*}The vector $\dbinom{5}{-2}$ can be written as the lin... | Here is how I would do it.
$\pmatrix{3\\2} - 2\pmatrix{-4\\1} = \pmatrix{11\\0}$
Allowing us to find one of the princicipal component vectors of the standard basis in terms of the $\{u,w\}$ basis.
$5u + 8w - 2(-3u+w) = \pmatrix{11\\0}\\
11u + 6w = \pmatrix{11\\0}\\
\pmatrix{1\\0} = u + \frac {6}{11} w$
With that we can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3628989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$ Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$
My Attempt:
Given
$$y-3px+ayp^2=0$$
$$3px=y+ayp^2$$
$$x=\frac {1}{3} \cdot \frac {y}{p} + \frac {a}{3} \cdot yp$$
This is solvable for x. Differentiating both sides with respect to $y$
$$\frac {dx}{dy... | Hint 1:
$$\frac{ap^2-1}{p(2-ap^2)} = -\frac{1-ap^2}{p(2-ap^2)} = - \frac{(2-ap^2) - 1}{p(2-ap^2)} = - \frac{1}{p} + \frac{1}{p(2-ap^2)}$$
Hint 2: The second expression involves two simple logarithmic expressions and can be and explicitly calculated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$ Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$
My Attmept:
$$x+yp=ap^2$$
$$x=ap^2-yp$$
This is solvable for $x$ so differentiating both sides w.r.t $y$
$$\frac {dx}{dy} = 2ap\cdot \frac {dp}{dy} - y\cdot \frac {dp}{dy} -p$$
$$\frac {1}{p} + p = (2ap-y) \cdot \frac {dp}{dy}$$
$$\fr... | Yes go ahead $$y\sqrt{1+p^2}=ap\sqrt{1+p^2}-a\sinh^{-1}{p}+C \implies y(p)=ap+\frac{C-\sinh^{-1}p}{\sqrt{1+p^2}}.$$
put this in $$x(p)=ap^2-py(p)$$ to get $x(p)$.
Finally
$x(p)$ and $y(p)$ constitue the parametric solution of the ODE, where $p$ acts as a parameter only.$C$ is the integration-constant,
Note that $$\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3634280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ramanujan's rational elementary results on $A^3+B^3=C^2$. In the bottom of one of the pages of Ramanujan's notebooks (see below), he writes down the following equations: $$\left(11\frac 12\right)^3+\left(\frac 12\right)^3=39^2$$ $$\left(3\frac 17\right)^3-\left(\frac 17\right)^3=\left(5\frac 47\right)^2$$ $$\left(3-\fr... | Ramanujan could not have had a general formula with the machinery of the time. Note that all of the listed results can be rewritten to have the same denominator $A$ in each number, so after multiplying by $A^3$ we have
$$x^3+y^3=Az^2$$
where $x,y,z$ are integers.
The given solutions correspond to the following integral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3639358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Transform Collatz sequence to a strictly decreasing sequence While playing with numbers, I found that every Collatz sequence $n, T(n), T^2(n), \ldots, 1$ can be associated with a strictly decreasing sequence of integers.
The Collatz conjecture asserts that a sequence defined by repeatedly applying the Collatz function
... | It suffers from the same pitfall as other representations which relies on the fact that the sequence reaches 1.
e.g. in the Collatz tree, you pick a number, and it does not matter if it seems to rise, in the tree it is a step closer to the root.
another one is the "inverse Collatz" representation of a number:
$7 = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3639525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$.
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.
Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x... | Denote: $x^2+x=a$. Then:
$$f(a)+2f(a-4x+2)=9a-24x.$$
Plug $x=\frac12$ to get:
$$f(a)+2f(a)=9a-12 \Rightarrow f(a)=3a-4.$$
Hence:
$$f(2016)=3\cdot 2016-4=6044.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 6
} |
Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$.
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.
At first, I tried to evalua... | Let $7x=2k\pi$ where $k=\pm1,\pm2,\pm3$
like Prove that $\cos(\pi/7)$ is root of equation $8x^3-4x^2-4x+1=0$
$q_k=2\cos\dfrac{2k\pi}7; k=1,2,3$ are the roots of $$c^3+c^2-2c-1=0$$
Use Veita's formula $$\sum_{k=1}^3\dfrac1{q_k}=\dfrac{q_1q_2+q_2q_3+q_3q_1}{q_1q_2q_3}=\dfrac{-\dfrac21}{\dfrac11}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3649434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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A parabola touches the bisectors of the angles formed by lines $x+2y+3=0$ and $2x+y+3=0$ at $(1,1)$ and $(0,-2)$. Find its focus and directrix.
A parabola touches the bisectors of the angles obtained by the lines $x+2y+3=0$ and $2x+y+3=0$ at the points $(1,1)$ and $(0,-2)$. Then find its focus and the equation of the ... | Since angular bisectors are perpendicular to each other and are tangent to the parabola at $A(1,1)$ and $B(0,2)$, the point of intersection of the two given lines $(-1,-1)$ is a point on the directrix of the parabola.
The median through $P$ to the Archimedes triangle $PAB$ is known to be parallel to axis, which gives t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3651655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$ Question -
Let $a, b, c$ be positive real numbers such that $a+b+c=3 .$ Prove that
$$
a^{2}+b^{2}+c^{2} \geq \frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}
$$
My try -
i tried putting $a+2 = x, b+2=y , c+2=z$
then we get... | Here's another way. Notice that
$$\frac{2+a}{2+b} = \frac{5a+2b+2c}{2a+5b+2c} = \frac52 -\frac32\cdot\frac{7b+2c}{2a+5b+2c} $$
Also using CS inequality ($\sum $ representing cyclic sums):
$$\sum \frac{7b+2c}{2a+5b+2c} \geqslant \frac{\left( \sum (7b+2c)\right)^2}{\sum (7b+2c)(2a+5b+2c)}= \frac{81(a+b+c)^2}{18\sum a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find limit (Riemann sum) $\displaystyle a_n=\sum^{n^2}_{k=1}\sqrt{\frac{k^2-k}{n^8+k^2n^4-kn^4}}$.
I tried solving it by changing it a Riemann sum then integrating, however I couldn't manipulate the algebra to its form.
| We have:
\begin{align*}
\sum^{n^2}_{k=1}\sqrt{\frac{k^2-k}{n^8+k^2n^4-kn^4}} &= \frac{1}{n^2} \sum^{n^2}_{k=1}\sqrt{\frac{k^2-k}{n^4+k^2-k}} \\ &= \frac{1}{n^2} \sum^{n^2}_{k=1}\sqrt{1 - \frac{n^4}{n^4+k(k-1)}} \\ &= \frac{1}{n^2} \sum^{n^2}_{k=1}\sqrt{1 - \frac{1}{1+\frac{k}{n^2}\frac{k-1}{n^2}}}\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3656848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find any two integers that satisfy $19x+47y = 1$ How can I find two integers that satisfy $19x+47y = 1$? Is there some technique to finding 2 numbers? I can't find any 2 numbers in $\mathbb{Z}$ that make this work.
Thank you!
| Consider the continued fraction of $\frac{47}{19}$:
$$ \frac{47}{19}=2+\frac{1}{\frac{19}{9}}=2+\frac{1}{2+\frac{1}{9}}=[2;2,9] $$
truncate it and expand it back:
$$ [2;2] = 2+\frac{1}{2} = \frac{5}{2}. $$
We have that $\frac{5}{2}$ and $\frac{47}{19}$ are consecutive convergents of the same continued fraction, hence t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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There are two positive integers of the form $p-n^2$ such that one divides the other.
Let $p>3$ be a prime number. Consider the numbers of the form $p - n^2$ where $n = 1 , 2 , 3 , ... , \lfloor\sqrt{p}\rfloor$. Show that there exist two such numbers, $a$ and $b$, that $a | b$.
I tried show that $p - n_{max}^2 | p -... | Let $m:=\lfloor \sqrt{p}\rfloor$. If $p=m^2+1$, then obviously, we have $a\mid b$ if $a:=p-m^2$ and $b:=p-(m-1)^2$. We now suppose that $p-m^2>1$.
We shall prove that, if $a:=p-m^2$, then there exists an integer $k$, where $1\le k<m$, such that $a\mid p-k^2$. First, we note that $p< (m+1)^2-1$; therefore, $p\leq (m+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$
The expression $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$, where $0<x<1$, is equal to $x$ or $\sqrt{(1+x^2)}$ or $\frac1{\sqrt{(1+x^2)}}$ or $\frac x{\sqrt{(1+x^2)}}$? (one of these 4 is correct).
My attempt: $$0<x<1\implies 0<\arctan x<\frac{\pi}{4}\im... | Since $x>0$ and $$1+\tan^2\alpha=\frac{1}{\cos^2\alpha},$$ we obtain: $$\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}=\sqrt{\left(\frac{1}{\sqrt{1+x^2}}+x\cdot\frac{x}{\sqrt{1+x^2}}\right)^2-1}=$$
$$=\sqrt{\left(\sqrt{1+x^2}\right)^2-1}=\sqrt{x^2}=x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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If $z_1,\dotsc,z_n$ are the vertices of a regular $n$-gon, with $z_3 + z_n = Az_1 + \bar{A}z_2$, find $\lfloor |A| \rfloor$. The question goes:
The complex numbers $z_1, z_2, z_3 \ldots z_n, \sum_{i=1}^{n} z_i \neq 0$ represent the vertices of a regular polygon of $n$ sides in order, inscribed in a circle of unit radi... | Let $n \geq 4$. We are given that $z_1, \dotsc, z_n$ are the vertices of a regular $n$-gon inscribed in a circle in the complex plane, with $\sum_i z_i \neq 0$. So, letting $w := x_0 + iy_0$ denote the centre of the circle, we have $w \neq 0$.
Let $\theta \in [0,2\pi)$ such that $z_1 = w + e^{i \theta}$. Then, $$z_j = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to find the sum of this complex series? Question says:
The convergent infinite series C and S are defined as
$$
C=1+\frac{1}{4}\cos(2\theta)+\frac{1}{16}\cos(4\theta)+\frac{1}{64}\cos(6\theta)+...
$$
$$
S=\frac{1}{4}\sin(2\theta)+\frac{1}{16}\sin(4\theta)+\frac{1}{64}\sin(6\theta)+...
$$
Show that $C+iS=\frac{k}{k-... | Add the corresponding terms of C and iS
You get
$$1+\frac{1}{4}\left(\cos 2\theta +i\sin 2\theta \right)+\frac{1}{16}\left(\cos 4\theta +i\sin \theta \right)...$$
use the identity $\begin{array}{l}e^{i\theta }=\cos \theta +i\sin \theta \end{array}$
$$C+iS=\left(\frac{e^{i2\theta}}{4}\right)^0+\left(\frac{e^{i2\theta}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Investigate the convergence of the integral Investigate the convergence of the integral
$$\int_{0}^{1} \left\vert \sin\frac{1}{x} \right\vert\ \frac{1}{x^{\alpha}} dx \ \text{where $\alpha \geq 1$ }$$
I assume that it diverges, so I rated function from below
$|sin\frac{1}{x}| \ \frac{1}{x^{\alpha }} \geq \sin^2(\frac... | Using integration by substitution, we have to study the convergence of the integral
$$I_\alpha=\int_1^\infty x^{\alpha-2} \left\vert \sin x \right\vert \ dx.$$
For $k \in \mathbb N$, and $x \in [k \pi + \frac{\pi}{6}, k\pi + \frac{5\pi}{6}]$, you have
$$\left\vert \sin x \right\vert \ge \frac{1}{2}$$ hence
$$g(x) = x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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When $ab/(a+b)$ is an integer, where $a,b$ are positive integers. When $ab/(a+b)$ is an integer, where $a,b$ are positive integers?
clear;
maxn:=30;
for a in [1..maxn] do
for b in [a..maxn] do
q1:=a*b; q2:=a+b;
if q1 mod q2 eq 0 then
print a,b,q1 div q2;
end if;
end for;
end for;
the Magma code given above outputs t... | We can do way better.
Thm. We have $a+b\mid ab$ if and only if $a=\dfrac{u+v+w}{2}$ and $b=\dfrac{-u+v+w}{2}$, where $u,v,w\in\mathbb{Z}$ such that $v$ is even and $u^2+v^2=w^2.$
Since we know how to describe completely Pythagorean triples, this gives a full description of the solutions.
Proof. If $a+b\mid ab$, there ... | {
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"url": "https://math.stackexchange.com/questions/3668069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of the squares of twelve real numbers which add to $1.$ Let $a, b, c,\cdots, l$ be real numbers such that $a + b + c \dots + l = 1.$ Find the minimum value of $$a^2 + b^2 + c^2 + \dots + l^2.$$
My first glance at this problem, I would assume the minimum would be $1,$ gotten by a bunch of zeros, and then a $1.$ I'm ... | Minimum occurs when $a_1 = a_2 = a_3 = \ldots = a_{12} = \frac{1}{12}$, and the minimum such value is $\frac{1}{12}$. To prove this, consider shifting the variables by setting $b_i = a_i - \frac{1}{12}$. We are then trying to minimise
$$\left(b_1 + \frac{1}{12}\right)^2 + \left(b_2 + \frac{1}{12}\right)^2 + \ldots + \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3668926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can we prove that $\gcd((n^4) + (n+1)^4 , (n+1)^4 + (n+2)^4) = 1$? I have found through experimentation that the consecutive sums of consecutive natural numbers raised to certain powers$(1,2,4)$ are always coprime. I was looking for modular multiplicative inverses and came across this facet.
If you define a sequen... | Suppose $d$ is a common factor of $n^4+(n+1)^4$ and $(n+1)^4+(n+2)^4$. Then
$$d\bigg|(n+2)^4-n^4\\
\implies d\bigg|4(n+1)(n^2+(n+2)^2)$$
But note that $4(n+1)$ and $d$ are coprime (since both of the given numbers are odd, hence $d$ is also odd, also, if $k$ is a common factor of $d$ and $n+1$, then $k$ also divides $n^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the area of $\triangle ABC$ with $B$ and $C$ lying on the line $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$
The vertices $B$ and $C$ of a $\triangle ABC$ lie on the line, $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$ such that $BC = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1, –... | Let $AD$ be an altitude of the triangle.
Thus, $$D(2+3t,1,4t)$$ and since $$\vec{AD}\cdot\vec{(3,0,4)}=0,$$ we obtain:
$$(1+3t,2,4t-2)(3,0,4)=0,$$ which gives $$t=\frac{1}{5}.$$
Can you end it now?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$\sum(-1)^{k}k^{2}C_{n}^{k}$ I'm asked to compute $\sum^{n}_{k=1}(-1)^{k}k^{2}C_{n}^{k}$ for $n\geqslant 1$. I tried to use generating functions to solve the problem: $A(x):=\sum^{n}_{k=1}k^{2}C_{n}^{k}x^{k}$. So, I need to compute $A(-1)$.
For $n=1$ $A(-1)=-1$, for $n=2$ $A(-1)=2$, for $n=2$ $A(-1)=2$, for $n=3,4,5,6... | Computing
$$\sum_{k=0}^n {n\choose k} (-1)^k k^2$$
we write
$$\sum_{k=0}^n {n\choose n-k} (-1)^k k^2
= [z^n] (1+z)^n \sum_{k=0}^n z^k (-1)^k k^2.$$
Here the coefficient extractor controls the range and we find
$$[z^n] (1+z)^n \sum_{k\ge 0} z^k (-1)^k k^2
= [z^n] (1+z)^n \frac{z(z-1)}{(1+z)^3}
\\ = [z^n] (1+z)^{n-3} (z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there missing information to answer this question! I have two quantities $A_1$ and $A_2$, and I would like to compare them in order to know which one is bigger.
Knowing that $x$, $y$ are two constants such that $y \leq \frac{x}{2}$
$$A_1 = \sum_{z=1}^y (x-y+z)!(y-z+1)! (z+1)$$
$$A_2 = \sum_{z=1}^y (x-y+z+2)! (y-z)!... | We have $x-y\ge y.$ So, for each given $z,$ the ratio of the "$z$-term" in $A_2$ to the "$z$-term" in $A_1$ is $$\frac {((x-y)+z+2)((x-y)+z+1)}{y-z+1}\cdot \frac {z+3}{z+1}\ge \frac {(y+z+2)(y+z+1)}{y-z+1}\cdot \frac {z+3}{z+1}=$$ $$=(y+z+2)\cdot \frac {(y+1)+z}{(y+1)-z}\cdot \frac {z+3}{z+1}>$$ $$>(y+z+2)\cdot 1 \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3681811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculation of Complex Limit When trying to calculate a residue, I came across this limit:
$$L:=\lim_{z\to \pi k} \frac{z^3-2z^2}{(1-\mathrm e^{\mathrm iz})\sin(z)}\left[\frac{(3z^2-4z)(z-\pi k)^2}{z^3-2z^2}+2(z-\pi k)-\frac{(z-\pi k)^2\cos(z)}{\sin(z)}+\frac{(z-\pi k)^2\mathrm i\,\mathrm e^{\mathrm i z}}{1-\mathrm e^{... | We can calculate the limit using series expansion at $z=\pi k$ where $0\neq k\in\mathbb Z, k \text{ even}$. We recall
\begin{align*}
\cos(z)&=1+\mathcal{O}\left((z-\pi k)^2\right)\\
e^{iz}&=1+i(z-\pi k)+\mathcal{O}\left((z-\pi k)^2\right)\\
\frac{1}{\sin(z)}&=\frac{1}{z-\pi k}+\mathcal{O}\left(z-\pi k\right)\\
\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Do we reduce the problem into the known $3^x+4^x=5^x $? I want to solve the following equations: \begin{align*}&(i) \ \ \ \ \ 3^x+4^x=5^x \\ &(ii) \ \ \ \ \ (x+1)^x+(x+2)^x=(x+3)^x, \ x>-1 \\ &(iii) \ \ \ \ \ (x+1)^x+(x+2)^x=(x+4)^x, \ x>-1\end{align*}
$$$$
I have done the following:
For (i):
We have that $$3^x+4^x... | Another approach
$x=2$ is an obvious solution to the problem. So consider the function
$$f(x)=\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x-1.$$
We know $f(2)=0$. Also $f'(x)=\left(\frac{3}{5}\right)^x\ln\left(\frac{3}{5}\right)+\left(\frac{4}{5}\right)^x \ln\left(\frac{4}{5}\right) <0$ for all $x$. Should this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Recurrence relations: prove that if $a_i = a_j$, then $i=j$
let $$a_n = \frac{n^2+10}{10^4}a_{n-1}, \space a_1=99$$
Prove (or disprove) that $a_i = a_j \implies i = j$
My proof:
The above sequence is non-monotonic: $\frac {a_n}{a_{n-1}} > 1$ when $n^2 > 9900$. Therefore the least element will be $a_{99}$. Since a d... | Well, this is the same as proving that $f:\mathbb{Z^{+}}\rightarrow\mathbb{Z^{+}}$ is injective $(f(a)=f(b) \implies a=b)$
while $f(1)=99$ and
$$f(x)=\frac{x^2+10}{10^4}\cdot f(x-1)$$
$$f(x)=\frac{x^2+10}{10^4}\cdot\frac{(x-1)^2+10}{10^4}\cdot \dots \cdot\frac{(x-k)^2+10}{10^4} \cdot f(x-k-1) \tag{1}$$
for all positiv... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does $a^2 + b^2 = 2 c^2$ have any integer solution?
Does the equation $a^2 + b^2 = 2 c^2$ have any integer solution with $|a| \neq |b|$?
I think not, because of these equations for pythagorean triplets:
$$\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$$
$$x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$$
I think I... | Suppose $(x,y,z)$ is any Pythagorean triple. Then:
$$(x-y)^2+(x+y)^2=2x^2+2y^2=2z^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and $33x+56y=\frac{x}{x^2+y^2}$. Let (x,y) be a pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and $33x+56y=\frac{x}{x^2+y^2}$. If $|x|+|y|=\frac{p}{q}$ (where p and q are relatively prime), then find the value (6p – q).
I used the concept $\frac... | Both of the existing answers posted here are incorrect, so here I will start with @Aditya Dwivedi's solution and go from there.
Let $ z = x - yi $
Then, subtract the first equation times $i$ from the second equation.
\begin{align*}
&33x - 56y - 56xi - 33yi = \frac{x + yi}{x^2 + y^2} \\
\implies &33(x-yi) - 56(y + xi... | {
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"timestamp": "2023-03-29T00:00:00",
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How does $x - 4y = 12$ result in $y = \frac{1}{4}x - 3$ instead of $y = -\frac{1}{4}x - 3$? I'm learning math. That's why I'm asking a simple explanation from a math expert.
When I solve this equation it results in $y = -\frac{1}{4}x - 3$, but the right answer is $y = \frac{1}{4}x - 3$.
Here is my solution:
$$x-4y = 12... | The error is that when dividing both sides by $-4$ you went from $-x$ to $-\frac x4$; it should go to $-\frac{x}{-4}=+\frac{x}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove by induction that for all $n\in\mathbb N, (\sqrt3+i)^n+(\sqrt3-i)^n=2^{n+1}\cos(\frac{n\pi}6)$ I want to prove by induction that for all $n \in \mathbb{N}$, $$(\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n+1} \cos\left(\frac{n\pi}{6} \right)$$
I can prove the identity using direct complex number manipulation and de M... | You need to induct the two statements below simultaneously
$$(\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n+1} \cos\frac{n\pi}{6} \\
(\sqrt{3} + i)^n - (\sqrt{3} - i)^n = i 2^{n+1} \sin\frac{n\pi}{6} $$
Note
\begin{align}
&(\sqrt{3} + i)^{n+1}+ (\sqrt{3} - i)^{n+1}
\\
=& \sqrt3[ (\sqrt{3} + i)^{n}+ (\sqrt{3} - i)^{n}]+i ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3685150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$f:\mathbb N_0\to\mathbb N_0$ with $2f\left(m^2+n^2\right)=f(m)^2+f(n)^2$ and $f\left(m^2\right)\geqslant f\left(n^2\right)$ when $m\geqslant n$
Determine all $f : \mathbb N_0 \to \mathbb N_0$ that satisfy $2f\left(m^2 + n^2\right) = f(m)^2 + f(n)^2$ and $f\left(m^2\right) \geqslant f\left(n^2\right)$ when $m \geqslan... | Observe the following things :
*
*For any $m, n \in \mathbb{N}_0$, $f(m)^2 + f(n)^2 $ is an even number, so the values of $f(x)$ are all odd or all even.
*$2f(0^2+ 0^2) = 2f(0)^2$ so we have $f(0) = 0$ or $1$.
*$2f(1^2 + 0^2) = f(1)^2 + f(0)$.
First assume that $f(x)$ values are all odd. then we have $f(0 ) = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Vieta's formulas for $x^2+px+1$ and $x^2+qx+1$.
Let $\alpha$ and $\beta$ be the roots for $x^2+px+1$ and let $\gamma$ and $\delta$ be the roots
for $x^2+qx+1$. Show that $$(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)=q^2-p^2.$$
This seemed to be rather peculiar. It should be a simple application of Vi... | Note $x^2+px+1= (x-\alpha)(x-\beta)$. Then
\begin{align}
& (\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)\\
= &(\gamma^2+p\gamma +1)(\delta^2-p\delta +1)\\
= &(\gamma\delta)^2+\delta^2 +\gamma^2+1-(\gamma\delta)p^2\\
= &(\delta+\gamma)^2-2\gamma\delta+2-p^2\\
=&q^2-p^2
\end{align}
where $\gamma+\delta=-q $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ . What is the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ ?
I have been stuck on this problem with no direction. I have tried multiplying the sequence with $x$ and trying out $S-Sx$ but have gotten nowhere. Any help?
Thanks.
| Expanding my hint, you have:
$$\frac{s}{x} = \frac{\text{d}^2}{\text{d} x^2}\left[ x^2 + x^3 + x^4 + x^5 + \cdots \right] = \frac{\text{d}^2}{\text{d} x^2}\left[ 1 + x + x^2 + x^3 + x^4 + x^5 + \cdots \right] = \frac{\text{d}^2}{\text{d} x^2}\left[ \frac{1}{1 - x} \right] = \frac{2}{(1 - x)^3}$$
(you can add $1+x$ to t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691770",
"timestamp": "2023-03-29T00:00:00",
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Finding the general formula for the sequence with $d_0=1$, $d_1=-1$, and $d_k=4 d_{k-2}$
Suppose that we want to find a general formula for the terms of the sequence
$$d_k=4 d_{k-2}, \text{ where } d_0=1 \text{ and } d_1=-1$$
I have done the following:
\begin{align*}d_k=4d_{k-2}&=2^2d_{k-2} \\ &=2^2\left (2^2d_{(k-2)... | You found $d_k=2^k$ when $k$ is even and $d_k=-2^{k-1}$ when $k$ is odd.
To put this in one formula, note that $\dfrac{1+(-1)^n}2$ is $0$ when $n$ is odd and $1$ when $n$ is even,
whereas $\dfrac{1-(-1)^n}2$ is $1$ when $n$ is odd and $0$ when $n$ is even.
So you could say $d_k=2^k\dfrac{1+(-1)^k}2-2^{k-1}\dfrac{1-(-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3693082",
"timestamp": "2023-03-29T00:00:00",
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How to find the intergral $I_{A}=\int_{0}^{2\pi}\frac{\sin^2{x}}{(1+A\cos{x})^2}dx$ Let $A\in (0,1)$be give real number ,find the closed form intergral
$$I_{A}=\int_{0}^{2\pi}\dfrac{\sin^2{x}}{(1+A\cos{x})^2}dx$$
This integral comes from a physical problem，following is my try:
since
$$I_{A}=\int_{0}^{2\pi}\dfrac{\sin... | Most probably you would like to have a formula depending on $A$.
Using complex residues we need to find the residues of
$$f(z) = -\frac{(z^2-1)^2}{A^2 z \left(1 + \frac{2z}{A}+ z^2\right)^2}$$
within the unit disc.
$f(z)$ has a single pole at $z=0$ and for $0<A<1$ a pole of order $2$ at $z_A=\frac{\sqrt{1-A^2}-1}{A}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3694900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $3(x+y)(x+z)(y+z)\neq a$ cubic when $x,y,z$ are different co-primal positive integers Prove $3(x+y)(x+z)(y+z)\neq a$ cubic when $x,y,z$ are different co-primal positive integers.
I believe you can look at the prime factors of $x,y$ and $z$. As for the equation to equal a cubic, the prime factors must all be prese... | What you're asking to prove is not always true. One specific counter-example, among infinitely many, is
$$x = 19 \tag{1}\label{eq1A}$$
$$y = 324 = 2^2 \times 3^4 \tag{2}\label{eq2A}$$
$$z = 4\text{,}589 = 13 \times 353 \tag{3}\label{eq3A}$$
These are all distinct, co-primal positive integers. You next have
$$x + y = 34... | {
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"timestamp": "2023-03-29T00:00:00",
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Problem 1, Exercise 3.4, Linear Algebra, Hoffman and Kunze The fundamental question is this: Let $V$ be a n-dimensional F-vector space. Is the matrix $A \in F^{n \times n}$ of $T \in L(V,V)$ relative to $\mathcal{B}$ is unique upto row-equivalence or not. With this out, let me discuss what is my issue with the problem.... | I have figured it out. I was wrong and the right answer is:
$T = [T]_{\mathcal{B}}=\begin{bmatrix} 1&0\\0&0 \end{bmatrix}$
Let $\alpha = \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix} $ and $e = \begin{bmatrix} e_1 \\ e_2 \end{bmatrix} $. Then $\alpha = P e$ where, $P = \begin{bmatrix}1 & -i\\ i&2 \end{bmatrix} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What are the steps to factor $x^2 - 1$ into $(x+1)(x-1)$? Does $(x+1)(x-1) = x^2+1x-1x-1$? If so where are the $+1x$ and the $-1x$ when it is being factored from $x^2-1$ into $(x+1)(x-1)$?
What exactly are we dividing $x^2-1$ by to get $(x+1)(x-1)$ and how did you know what to divide it by?
| Let $y:=x-1$. Then
$$x^2-1=(y+1)^2-1=y^2+2y$$ which obviously factors as $$(y+2)y=(x+1)(x-1).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Strategy to calculate $ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right) $. I am asked to calculate the following: $$ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right). $$
I simplify this a little bit, by moving the constant multiplicator out of the derivative:
$$ \left(\frac{1}{2}\right) \frac{d}{dx} \left(... | Logarithmic differentiation can also be used to avoid long quotient rules. Take the natural logarithm of both sides of the equation then differentiate:
$$\frac{y'}{y}=2\left(\frac{1}{x-3}-\frac{1}{x}-\frac{1}{x+3}\right)$$
$$\frac{y'}{y}=-\frac{2\left(x^2-6x-9\right)}{x(x+3)(x-3)}$$
Then multiply both sides by $y$:
$$y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3707227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $\sqrt{x^2+8x+7}+\sqrt{x^2+3x+2}=\sqrt{6x^2+19x+13}$ I tried squaring both sides but doesn't seem like a good idea.
$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$
Is there a better way for solving this equation?
| I would factor both sides first so that:
$$\sqrt{(x+7)(x+1)}+\sqrt{(x+2)(x+1)}=\sqrt{(6x+13)(x+1)}$$
$$\sqrt{x+1}(\sqrt{x+7}+\sqrt{x+2})=\sqrt{x+1}\sqrt{6x+13}$$
$$\sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}$$
$$2x+9+2\sqrt{(x+2)(x+7)}=6x+13$$
$$\sqrt{(x+2)(x+7)}=2x+2$$
$$(x+2)(x+7)=4x^2+8x+4$$
$$3x^2-x-10=0$$
$$(3x+5)(x-2)=0$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that if $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$, then $V = W_{1}\oplus W_{2}$. (a) Let $W_{1}$ and $W_{2}$ be subspaces of a vector space $V$ such that $V = W_{1}\oplus W_{2}$. If $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ are bases for $W_{1}$ and $W_{2}$, respectively, show that $\mathcal{B}_{1}\ca... | I prefer denoting a basis as a list and not a set.
(a)
Let $B_{1}= (v_{i}:i\in I), B_{2}= (u_{j}:j\in J)$ (where $I, J$ are arbitrary index sets).
By hypothesis $V = W_{1} + W_{2}, W_{1}\cap W_{2}=\{0\}$. Suppose $B_{1}\cap B_{2} \neq \emptyset,$ then $\exists x\in B_{1}\cap B_{2},$ then $x\neq 0$ (since $B_{1}, B_{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3708857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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A Systematic way to solve absolute value inequalities? So, I had to solve this problem: $\left\vert \dfrac{x^2-5x+4}{x^2-4}\right\vert \leq 1$
I factored it in the form: $\left\vert \dfrac{(x-4)(x-1)}{(x-2)(x+2)} \right\vert \leq 1$.
After that I found the intervals in which the expression is positive: $x \in(-\infty, ... | I can think of two methods to solve these kinds of equations, so I'll write them both down and you can see which you prefer. The first is a lot more algebraic, the second you can argue is more geometric. It may also not fly in your class, depending on the syllabus and professor.
First, we have $-1\leq\frac{x^2-5x+4}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determining $\arg(u-\sqrt{3})$ in exact form. Let $u$ be the solution to $z^5=-9\sqrt{3}i$ so that $\frac{\pi}{2}\le arg(u) \le \pi$.
Determine $\arg(u-\sqrt{3})$ in exact form.
How would I go about completing this question?
| Let $z=r(\cos\theta+i\sin\theta)$. We know $-9i\sqrt3=9\sqrt3(\cos-\frac{\pi}{2}+i\sin-\frac{\pi}{2} )$
This means (using De Moivre's Theorem)
$$r^{5}(\cos5\theta+i\sin5\theta)=9\sqrt3(\cos-\frac{\pi}{2}+i\sin-\frac{\pi}{2})=9\sqrt3(\cos(-\frac{\pi}{2}+2k\pi)+i\sin(-\frac{\pi}{2}+2k\pi)$$
So $r^{5}=9\sqrt3$ so $r=\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Value of $\alpha$ for which $x^5+5\lambda x^4-x^3+(\lambda\alpha-4)x^2-(8\lambda+3)x+\lambda\alpha-2=0$ has roots independent of $\lambda$
Consider the equation $$x^5 + 5\lambda x^4 -x^3 + (\lambda \alpha -4)x^2 - (8\lambda +3)x + \lambda\alpha - 2 = 0$$ The value of $\alpha$ for which the roots of the equation are in... | The problem with your answer is that it's true that $2$ would be a root independent on $\lambda$ but there may be other roots which depends on $\lambda$.
Indeed with $\alpha=-\frac{64}{5}$ your equation becomes
$$\frac{1}{5}(x-2)(5(1+x+x^2)^2+\lambda(5x+8)(4+x(2+5x))$$
which may have solutions different from $2$ and $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
The number of critical points of the function $f(x,y)=(x^2+3y^2)e^{-(x^2+y^2)}$ is
The number of critical points of the function $$f(x,y)=(x^2+3y^2)e^{-(x^2+y^2)}$$ is _____.
My attempt:
$f_x=-(2x^3+(6y^2-2)x)e^{-(x^2+y^2)}$ and $f_y=-(6y^3+(2x^2-6)y)e^{-(x^2+y^2)}$
Clearly, $(0,0)$ is a critical point since at $(0,0... | Clearly, you can factor:
\begin{align}
f'x&=-2x(x^2+3y^2-1)\mathrm e^{-(x^2+y^2)} \\
f'y&=-2y(x^2+3y^2-3)\mathrm e^{-(x^2+y^2)}
\end{align}
and observe that $x^2+3y^2-1$ and $x^2+3y^2-3$ can also be $0$, but not simultaneously. Therefore you also have the solutions of the systems
$$
\begin{cases}
x=0\\x^2+3y^2-3=0
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the solution of this summation? $$S(x) = \frac{x^4}{3(0)!} + \frac{x^5}{4(1)!}+\frac{x^6}{5(2)!}+.....$$
If the first term was $$x^3$$ and the next terms were $$x^{3+i}$$ then differentiating it would have given $$x^2.e^x$$ and then it was possible to integrate it. But how to solve this one?
| Your idea is right, just let $x^3$ appear and differentiate
$$\left(\frac{S(x)}x\right)'=\left(\frac{x^3}{3\cdot0!}+\frac{x^4}{4\cdot1!}+\frac{x^5}{5\cdot2!}+\cdots\right)'=\frac{x^2}{0!}+\frac{x^3}{1!}+\frac{x^4}{2!}+\cdots=x^2e^x.$$
Then by integration,
$$\color{green}{\frac{S(x)}x=(x^2-2x+2)e^x-2}$$ (the constant of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Integrate $\int_0^{\frac{\pi}{2}} \frac{dx}{{\left(\sqrt{\sin{x}}+\sqrt{\cos{x}}\right)}^4} $ I found a challenge problem and am confused$$\int_0^{\frac{\pi}{2}} \frac{dx}{{\left(\sqrt{\sin{x}}+\sqrt{\cos{x}}\right)}^4} $$
$u=\frac{\pi}{2}-x$ is no good and square or 4th power the denominator does not help? Suggestion... | \begin{align}
\int_0^{\frac{\pi}{2}} \frac{1}{{\left(\sqrt{\sin{x}}+\sqrt{\cos{x}}\right)}^4}{\rm d}x &= \int_0^{\pi/2}\dfrac{\sec^2 x}{(\sqrt{\tan x} + 1)^4}{\rm d}x.
\end{align}
Denote the upper integral by $I$.
Put $u = \tan x$ to get
$$I = \int_0^\infty \dfrac{1}{(1 + \sqrt u)^4}{\rm d}u.$$
Put $u = t^2$ to get
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluation of integral $\int_{S^2} \frac{dS}{((x-a)^2 +y^2+z^2)^{1/2}}$, where $a>1$ and $S$ is the unit sphere. I want to evaluate$$\int_{S^2} \frac{dS}{((x-a)^2 +y^2+z^2)^{1/2}}$$
where $a >1$ and $S$ is the unit sphere.
I'm not sure how to do this using only multivariable calculus techniques. My only idea was to use... | We can use a few symmetry conditions to make our lives easier. Note that your proposed integral is equivalent to
$$\iint_{(x+a)^2+y^2+z^2=1} \frac{1}{\sqrt{x^2+y^2+z^2}}\:dS = \iint_{(x+a)^2+y^2+z^2=1} \frac{(x+a,y,z)\cdot(x+a,y,z)}{\sqrt{x^2+y^2+z^2}}\:dS$$
Thus we can use the divergence theorem
$$= \iint_{(x+a)^2+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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For real values with $abc\neq0$, if $\frac{xb+(1-x)c}{a}=\frac{xc+(1-x)a}{b}=\frac{xa+(1-x)b}{c}$, show that $x^3 = -1$ and $a=b=c.$
Let $a,b,c$ and $x$ are real numbers such that $abc \neq 0$ and $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c}.$$ Prove that $x^3=-1$ and $a=b=c.$
My attempt $:... | $$bc((b-c)x+c)=ca((c-a)x+a)=ab((a-b)x+b)$$
has a solution in $x$ iff
$$\begin{vmatrix}bc(b-c)&bc^2&1\\ca(c-a)&ca^2&1\\ab(a-b)&ab^2&1\end{vmatrix}=-abc(3abc-a^3-b^3-c^3)\\=-abc(a+b+c)(a+\omega b+\omega^2c)(a+\omega^2b+\omega c)=0$$
where $\omega$ is a cube root of unity.
There are two cases:
*
*one of the complex fact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How does the square root disappear when differentiating $y=\frac{\sqrt{2x^2}}{\cos x}$? Finding the derivative of $$y=\frac{\sqrt{2x^2}}{\cos x}$$ I am going through the steps and having trouble using the quotient rule. I have seen the final answer, and I've had no trouble using the quotient rule in the past, but this... | You have
$$\frac{\frac{4x\cos x}{2\sqrt{2x^2}}+\sqrt{2x^2}\sin x}{(\cos x)^2}$$
You correctly mentioned that this can be simplified to
$$\frac{\frac{2x\cos x}{\sqrt{2x^2}}+\sqrt{2x^2}\sin x}{(\cos x)^2}$$
To simplify this you can remove the compound fraction. To get rid of the fraction in the numerator, multiply numera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Second system of equations I've solved that system and recieved $y=\frac{1}{4}$, $x = -\frac{4}{5}$ but there are one more pair of $(x, y)$ in book and I don't know what I have to do to find its.
\begin{cases} 2x - 3xy + 4y=0 \\ x + 3xy -3x = 1 \end{cases}
In book there are two answers and the second answer is $(1, -2)... | We have
$$2x-3xy+4y\quad=x+3xy-3x-1=0\implies 4x-6xy+4y+1=0$$
$$\implies\quad x = \frac{(4 y + 1)}{(6 y - 4)} \land 3 y\ne 2\qquad y = \frac{(4 x + 1)}{(6 x - 4)} \land 3 x\ne2$$
so the solution for one variable in terms of the other yields the same results in that $1)$ either independent variable can be any of an infi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proof for the general formula for $a^n+b^n$. Based on the following observations. That is
$$a+b = (a+b)^1 \\ a^2+b^2 = (a+b)^2-2ab \\ a^3+b^3 = (a+b)^3-3ab(a+b) \\ a^4+b^4= (a+b)^4-4ab(a+b)^2+2(ab)^2\\ a^5+b^5
= (a+b)^5 -5ab(a+b)^3+5(ab)^2(a+b)\\\vdots$$
I came to make the following conjecture as general formula... | You want to express $x^n+b^n$ in terms of their sum, or (for convenience) half-sum $s:=\dfrac{a+b}2$, and geometric mean $p:=\sqrt{ab}$.
We have
$$2as=a^2+ab=a^2+p^2,$$ giving
$$a=s\pm\sqrt{s^2-p^2},b=s\mp\sqrt{s^2-p^2}.$$
Now
$$a^n+b^n=\sum_{k=0}^n\binom nk\left(s^{n-k}(s^2-p^2)^{k/2}+(-1)^ks^{n-k}(s^2-p^2)^{k/2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ for positive $a$, $b$, $c$
For $a,b,c>0$ Prove that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$
My attempt: By AM-GM we obtain $$\frac{a}{b}+\frac{a}{b}+\fr... | Suppose the inequality is true,then,
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$$$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac3{a+b+c}\ge 4\cdot \frac{1}{\sqrt[3]{abc}}$$
$$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac3{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Proofs of the Reflection Rules I couldn't find a formal proof for the rule:
when a point $(a,b)$ is reflected along $y=x$, it becomes $(b,a)$.
I tried to prove it by sketching out the situation:
However, I still don't know how to prove that $b'=b, a'=a$.
Furthermore, I just want to make sure, for the following two rul... | To find the coordinates of the reflected point $P'$, let us first find the intersection point of the line $y=x$ and the line perpendicular to that line and passing through the point $P=(a,b)$.
As we know, the equation of the line perpendicular to the line $y=x$ and passing through the point $P=(a,b)$ is$$y=-(x-a)+b.$$S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)...$ How can I prove the identity $\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)\ldots$ for $|x|<1$? I am preferably looking for a derivation rather than using the RHS. I have tried using binomial expansion, but it only seems to give the LHS back. I also tried taking the logarith... | Use $(P+Q)((P-Q)=P^2-Q^2$, repeatedly:
$$F=(1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)......=(1-x^2)(1+x^2)(1+x^4)(1+x^8).....=(1-x^4)(1+x^4)(1+x^8)(1+x^{16})....=(1-x^8)(1+x^8)(1+x^{16})...(1+x^{2^n})=(1-x^{2^{n+1}})$$
When $|x|<1$ and $n \to \infty$, then $F =1$,
$$\implies \frac{1}{1-x}=(1+x)(1+x^2)(1+x^4)(1+x^8)....., |x|<1.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Continuity of $a^x+b$ with $a, b \in \mathbb R$ Let $a,b \in \mathbb{R}$ with $a > 0$. find $a$, $b$ so the function would be continuous
$$
f(x) = \begin{cases} a^x + b, & |x|<1 \\
x, & |x| \geq 1 \end{cases}
$$
I got $b = -a^x+x$ as my answer, but I'm unsure.
| Just do the definitions. $x$, and $a^x + b$ are continuous so the the only possible point of discontinuity is as $|x| = 1$.
If $f$ is continuous at $x = 1$ then $\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-}a^x + b = a+b$ must equal $f(1) = x|_1 = 1$ which must equal $\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} x = 1$. So we mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=... | Hint:
$$\left(\dfrac12\right)^2=(\sin\alpha-\cos\alpha)^2=?$$
So, we know $\sin\alpha\cos\alpha=?$
$$\dfrac1{\sin\alpha}-\dfrac1{\cos\alpha}=\dfrac{?}{\sin\alpha\cos\alpha}=?$$
Finally use $$\left(\dfrac1{\sin\alpha}-\dfrac1{\cos\alpha}\right)^3=\dfrac1{\sin^3\alpha}-\dfrac1{\cos^3\alpha}-\dfrac3{\sin\alpha\cos\alpha}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Generalized repetitions of letters with limited amount of adjacent letters Say I have the first $x$ letters of the alphabet, and I want to generate a sequence of length $y$, such that there are no more than $z$ adjacent repeated letters. For example, if $x = 2$, $y = 3$ and $z = 2$, here are all the valid sequences:
AA... |
The following answer is based upon the Goulden-Jackson Cluster Method which provides a convenient technique to solve problems of this kind. We consider the set of words of length $y\geq 0$ built from an alphabet $$\mathcal{V}=\{A_1,A_2,\ldots A_x\}$$ and the set $B=\{A_1^{z+1},A_2^{z+1},\ldots,A_x^{z+1}\}$ of bad word... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Solve the following logarithmic equation over real numbers
Solve the equation:
$$\log_{2020} {(x^{10} + x^9 + x^8 + x^7 + x^6+ x^5+ x^2 )}=\log_2 x$$
over real numbers.
I found out that $x=2$ is a solution and I suspect is the only one, but cannot prove it.
|
Solve for $x$ in $\log_{2020} {(x^{10} + x^9 + x^8 + x^7 + x^6+ x^5+ x^2 )}=\log_2 x$.
$$\log_2{2020} =\log_x{(x^{10} + x^9 + x^8 + x^7 + x^6+ x^5+ x^2 )}$$
$x=2$ is a solution on checking.
To check if it is the only solution, let $t=\log_2 2020$ and consider
\begin{align*}
f(x)&=\ln(x^{10}+x^9+\cdots+x^2)-t\ln x\\
f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{... | One alternate method is thus: make the substitution $t=u^2$ which gives $dt = 2u\; du \iff du = dt/2 \sqrt t$. Then
$$\mathcal I := \int \frac{u^3}{(u^2+1)^3} du = \int \frac{t \sqrt t}{(t+1)^3} \cdot \frac{dt}{2 \sqrt t} = \frac 1 2\int\frac{t}{(t+1)^3}dt$$
Now let $w = t+1 \implies dw = dt$. Then
$$\mathcal I = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 6
} |
For $x≠y$ and $2005(x+y) = 1$; Show that $\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$ Problem:
Let $x$ and $y$ two real numbers such that $x≠0$ ; $y≠0$ ; $x≠y$ and $2005(x+y) = 1$
*
*Show that $$\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$$
*Calculate $l$:
$$l = \frac{y}{y-x} - \frac{y-... | According to the problem, $x, y$ $\in$ $\mathbb{R}$ and $x \neq y$.
$\bullet$ For the first part,
(I) From the given, we have that
\begin{align*}
&2005(x + y) = 1\\
\implies & 2005 \cdot \frac{(x + y)}{xy} = \frac{1}{xy}\\
\implies & 2005 \cdot \bigg( \frac{1}{x} + \frac{1}{y} \bigg) = \frac{1}{xy}
\end{align*}
Hence, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Solve second order linear ODE if particular solution of the homogenous part is known Here's this ODE: $$x(x-1)y'' -(2x-1)y' + 2y = 2x^3 -3x^2$$
and $$y_1 =x^2$$
I know that I have to consider the homogenous part of the ODE first, which is $$x(x-1)y'' -(2x-1)y' + 2y = 0$$
If one solution is already known, then the secon... | First, after considering $y_1 = c_1 x^2$ as a particular solution for the homogeneous
$$
x(x-1)y_h'' -(2x-1)y_h' + 2y_h = 0
$$
we have by constants variation (Lagrange)
$$
y_h = \left(\frac{c_2 (2 x-1)}{2 x^2}+c_3\right)x^2 = \frac{c_2}{2}(2x-1)+c_3 x^2
$$
now, using the Lagrange method again we propose for the complet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \t... | Hint: Use substitution $\space 9+4x^2 = t^2. \space$ New integral shall be the integral of rational function...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos ... | @GregMartin's hint is to compute the numerator and denominator each to $O(x^2)$, respectively as $1+\tfrac12x^2-(1-\frac14x^2)=\tfrac34x^2$ and $x^2$, so the limit is $\tfrac34$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
in the triangle ABC on the AC side, points M and N are chosen such that ABM = MBN = NBC in the triangle ABC on the AC side, points M and N are chosen such that <ABM = <MBN = <NBC It turned out that NB = BC. On the side AB, a point K was marked such that BK = BM. Prove that AK + NC> AM.
I tried to get a triangle with s... | let $BC=a$ and $\measuredangle ABC=3\beta$.
Thus, $$\measuredangle ACB=90^{\circ}-\frac{\beta}{2},$$
$$\measuredangle ANC=90^{\circ}+\frac{\beta}{2},$$
$$\measuredangle AMB=90^{\circ}+\frac{3\beta}{2}$$ and
$$\measuredangle BAC=180^{\circ}-\left(90^{\circ}+\frac{5\beta}{2}\right),$$ which says
$$90^{\circ}+\frac{5\beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
if $f'(x)\ne0, g(x)>0 \forall x \in 0$, prove that $|f(x)| < 5$
if $f(x)$ is a twice differentiable real-valued function satisfying $f'(x)\ne0, f(0) = -3, f'(0) = 4$, such that $f(x) + f''(x) = -xg(x)f'(x), \space g(x) > 0$ for all $x > 0$. Prove that $|f(x)| \le 5$
There was a hint in the problem to first prove that... | $\boxed{\textit{Solution:}}~$We consider 2 cases.
$\bullet~$When $x > 0$.
as we know that $g(x) \geqslant 0$, for $x > 0$, we have for the interval $[0, t]$ for $t > 0$.
\begin{align*}
&f(x) + f''(x) = - x g(x) f'(x)\\
\implies & f(x) f'(x) + f'(x) f''(x) = - x g(x) \{ f(x) \}^{2} \leqslant 0 \quad [\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many roots does $(x+1)\cos x = x\sin x$ have in $(-2\pi,2\pi)$? So the nonlinear equation that I need to find the number of its roots is
$$(x+1)\cos x = x\sin x \qquad \text{with } x\in (-2\pi,2\pi)$$
Using the intermediate value theorem I know that the equation has at least one root on this interval, and if I use ... | First we can manipulate the expression given:
$(x+1)\cos x= x\sin x$
divide both sides by $x\cos x$
Yielding: $(\frac{x+1}{x})= \tan x $
The Inverse tangent of both sides yields:
$\operatorname{arctan}(\frac{x+1}{x})=x$
We can state without requiring a proof, although can be proved, that:
$\frac{d}{dz} \operatorname{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sum of squares and linear sum
For which positive integer $n$ can we write $n=a_1+a_2+\dots+a_k$ (for some unfixed $k$ and positive integers $a_1,a_2,\ldots,a_k$) such that $\sum_{i=1}^k a_i^2 = \sum_{i=1}^k a_i + 2\sum_{i<j}a_ia_j$?
When $k=1$, the equation is $a_1^2=a_1$, so $a_1=1$ and $n=1$ is the only possibility... | Here we prove an auxiliary claim for WhatsUp’s answer.
Let $S$ be the set of all pairs $(a,b)$ of natural numbers such that a system
$$\cases{
c_1 + 2c_2 + 3c_3 + 4c_4 + 5c_5 + 6c_6=x\\
c_1 + 4c_2 + 9c_3 + 16c_4 + 25c_5 + 36c_6=y}$$
has a solution in non-negative integers.
The set $S$ can be constructed recursively as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\... | By setting
$$
\frac{1}{\left(x^2-4 x+13\right)^2}=\frac{A (2 x-4)+B}{x^2-4 x+13}+\frac{d}{dx}\left(\frac{C x+D}{x^2-4 x+13}\right)
$$
you get
\begin{align}
A &= 0,\\
B &= \frac{1}{18},\\
C &= \frac{1}{18},\\
D &= -\frac{1}{9}
\end{align}
so that
\begin{align}
\int\frac{1}{\left(x^2-4 x+13\right)^2}dx
&= \frac{1}{18}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 6
} |
Find the value of $\lim _{a \to \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} \cdot \tan ^{-1}\left(\frac{1}{x}\right) \,d x $ Find the value of : $$
\lim _{a \rightarrow \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} \cdot \tan ^{-1}\left(\frac{1}{x}\right) \,d x
$$ I have tried ... | Your solution attempt honestly works just fine as well (and it's also the first thing I thought of when I saw the question!). Indeed from ${x=\frac{1}{t}}$ you get
$${\int_{0}^{\infty}\frac{x^2 + ax + 1}{1+x^4}\arctan\left(\frac{1}{x}\right)dx=\int_{0}^{\infty}\frac{\left(\frac{1}{t}\right)^2 + \frac{a}{t} + 1}{1 + \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{i-3 n}$ Evaluate $\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{i-3 n}$
Here $\sum_{i=1}^{n} \frac{1}{i-3 n}=\frac{1}{1-3 n}+\frac{1}{2-3 n}+\cdots+\frac{1}{n-3 n}$.
I tried to make the sum squeezed
between Convergent Sequences, but failed by getting... | One approach is to use an expansion of harmonic numbers where $H(n)= \sum \limits_{i=1}^n \frac1n = \log_e(n) + \gamma + \frac{1}{2n}+O(n^{-2})$
You then have $\sum\limits_{i=1}^{n} \frac{1}{i-3 n} = H(2n-1)-H(3n-1) = \log_e(\frac{2n-1}{3n-1}) + \frac{n}{2(2n-1)(3n-1)}+O(n^{-2})$
and so $\lim\limits _{n \rightarrow \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rolling Dice Game, Probability of Ending on an Even Roll The game is described as follows. $A$ and $B$ take turns rolling a fair six sided die. Say $A$ rolls first. Then if $A$ rolls {1,2} they win. If not, then $B$ rolls. If $B$ rolls {3,4,5,6} then they win. This process repeats until $A$ or $B$ wins, and the game st... | The answer = 1/2
The game has to end by either A winning or B winning
Let's say A wins. He is just as likely to roll a 1 or a 2 on the last roll. Therefore in a game that A wins, probability of an even roll ending the game is 1/2, as 1(odd) and 2(even) are equally likely.
Let's say B wins. He is just as likely to roll ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to solve $\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$ The original question is:
Prove that:$$\begin{aligned}\\
\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\neq\int_0^1dy&\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\\
\end{aligned}\\$$
But I can't evaluate the integral $$\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^... | Hint:
$$\int_0^1\frac {x^2-y^2}{(x^2+y^2)^2}\, dy = \int_0^1\frac{\partial}{\partial y} \left(\frac{y}{x^2 + y^2}\right)\, dy = \frac{1}{1+x^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
System of congurences and the Chinese Remainder Theorem I have the following system of congruences:
\begin{align*}
x &\equiv 1 \pmod{3} \\
x &\equiv 4 \pmod{5} \\
x &\equiv 6 \pmod{7}
\end{align*}
I tried solving this using the Chinese remainder theorem as follows:
We have that $N = 3 \cdot 5 \cdot 7 = 105$ and $... | Here's how I would do it: using Bezout coefficients, we get $2\cdot5-3\cdot3=1$. So the solution to $\begin{cases}x\cong1\pmod3\\x\cong4\pmod5\end{cases}$ is $x=1\cdot{10}-4\cdot9=-26\pmod{15}=4\pmod{15}$.
Next we solve $\begin{cases} x\cong{4}\pmod{15}\\x\cong6\pmod7\end{cases}$.
Since $-6\cdot15+13\cdot7=1$, we get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2 $ I'm trying to calculate:
$$T = \lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2$$
Here is my attempt.
Put $x^2=\dfrac{1}{t}$ so when $x\to \infty, t \to 0$ and the limit become
\begin{align*}
T &= \lim\limits_{t \to 0} \sqrt[... | By definition of pochammer symbol $$(x^2+1)^{(n)}=(1+x^2)(2+x^2)\cdots (n+x^2)=\frac{\Gamma(x^2+1+n)}{\Gamma(x^2+1)}\sim x^{2n}\left(1+\frac{n(n+1)}{2x^2}+O(x^{-4})\right)$$ thus $$\sqrt[n]{(x^2+1)^{(n)}}-x^2 =x^2\left(1+\frac{n(n+1)}{2x^2}\right)^{\frac{1}{n}}-x^2$$ using the fractional binomial theorem we have limi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Proving that $ \sum_{k=1}^\infty \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ converges by the comparison test I would like to prove that the following series converges:$ \sum_{k=1}^\infty \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ by comparing it with a series that I already know converges. One such series could be the geometri... | $$\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}= \left(\frac{2}{3}\right)^{k} \cdot \frac{1+ \frac{k^8}{2^{k}} }{1-\left( \frac{2}{3}\right)^{k}}$$
As $\frac{1+ \frac{k^8}{2^{k}} }{1-\left( \frac{2}{3}\right)^{k}} \to 1$, then we can say, that $\exists N \in \mathbb{N}$ such that for $k>N$ holds $\frac{k^{8} + 2^{k} }{3^{k} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Greatest Common Divisor Problem: Prove that $\gcd(\frac{a^3+b^3}{a+b}, a+b) = \gcd(a+b, 3ab)$ I've been stuck in this problem for some time now. Currently what I have accomplished is, using the propriety $\gcd(a,b) = \gcd(b,a \bmod(b))$ to get in the equation
$$ \gcd(a+b, \frac{a^3+b^3}{a+b}\bmod(a+b)) $$
but I don't ... | We have
$$
\begin{pmatrix}
\frac{a^3+b^3}{a+b} \\ a+b
\end{pmatrix}
=
\begin{pmatrix}
a^2-ab+b^2 \\ a+b
\end{pmatrix}
=
\begin{pmatrix}
-1 & a+b \\ 0 & 1
\end{pmatrix}
\begin{pmatrix}
3ab \\ a+b
\end{pmatrix}
$$
The result follows because the matrix has determinant $-1$ and so has an integer inverse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in t... | Consider
$$\frac{a x^9+bx^8+1}{x^2-x-1}$$ perform long division to get
$$-1+x-2 x^2+3 x^3-5 x^4+8 x^5-13 x^6+21 x^7-(b+34) x^8+x^9
(-a+b+55)+O\left(x^{10}\right)$$
So $b=-34$ and $a=21$ and the result is
$$\frac{a x^9+bx^8+1}{x^2-x-1}=-1+x-2 x^2+3 x^3-5 x^4+8 x^5-13 x^6+21 x^7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Evaluate integral $\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}$ How can I evaluate this integral $$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\;\;?$$
My attempt:
I tried using substitution $x=\sec\theta$, $dx=\sec\theta\ \tan\theta d\theta$,
$$\int (\sec^2\theta-1)(\sec^3\theta-3\sec\theta)^{4/3}
\sec\theta\ \tan\theta d\the... | Let $x^3-3x=t\implies (3x^2-3)dx=dt$ or $(x^2-1)dx=\frac{dt}{3}$
$$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\int t^{4/3}\frac{dt}{3}$$
$$=\frac13\frac{t^{7/3}}{7/3}+C$$$$=\frac{(x^3-3x)^{7/3}}{7}+C$$
or alternatively,
$$\int (x^2-1)(x^3-3x)^{4/3}\ dx=\frac13\int (3x^2-3)(x^3-3x)^{4/3}\ dx$$
$$=\frac13\int (x^3-3x)^{4/3}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Calculus of $ \lim_{(x,y)\to (0,0)} \frac{8 x^2 y^3 }{x^9+y^3} $ By Wolfram Alpha I know that the limit
$$
\lim_{(x,y)\to (0,0)} \dfrac{8 x^2 y^3 }{x^9+y^3}=0.
$$
I have tried to prove that this limit is $0$, by using polar coordinate, the AM–GM inequality and the change of variable $ x^9= r^2 \cos^2(t) $ and $y^3= ... | As suggested in the comments, let consider the following path "near" the problematic points $y=-x^3$ for which denominator vanishes:
*
*$x=t$
*$y=-t^3+t^5$
then we have
$$\frac{8 x^2 y^3 }{x^9+y^3}
=\dfrac{8 t^2 (-t^3+t^5)^3 }{t^9+(-t^3+t^5)^3}
=\frac{8 t^2(-t^{9}+3t^{11}-3t^{13}+t^{15})}{t^9+(-t^{9}+3t^{11}-3t^{13}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $a$, $b$, $c$, $d$ are positive reals so $(a+c)(b+d) = 1$, prove the following inequality would be greater than or equal to $\frac {1}{3}$.
Let $a$, $b$, $c$, $d$ be real positive reals with $(a+c)(b+d) = 1$. Prove that $\frac {a^3}{b + c + d} + \frac {b^3}{a + c + d} + \frac {c^3}{a + b + d} + \frac {d^3}{a + b + ... | By C-S $$\sum_{cyc}\frac{a^3}{b+c+d}=\sum_{cyc}\frac{a^4}{ab+ac+ad}\geq\frac{(a^2+b^2+c^2+d^2)^2}{\sum\limits_{cyc}(ab+ac+ad)}\geq$$
$$\geq\frac{a^2+b^2+c^2+d^2}{\sum\limits_{cyc}(ab+ac+ad)}\geq\frac{1}{3},$$
where the last inequality it's $$\sum_{sym}(a-b)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I integrate $\frac1{x^2+x+1}$? I have tried this:
$$\frac1{x^2+x+1} = \frac1{\left( (x+\frac12)^2+\frac34\right)}$$
Now $u = x+\frac12$
$$\frac1{ u^2+\frac34 }$$ Now multiply by $ \frac34$
$$\frac1{ \frac43 u^2 + 1}$$
Now put the $\frac43$ outside the integral
$$\frac34 \int \frac1{u^2+1}\,du=\frac34\arctan(u)=... | Note that $$\frac{1}{u^2+3/4}=\frac{1}{\frac34((2u/\sqrt 3)^2+1)}=\frac43 \frac{1}{v^2+1}$$
where $v=2u/\sqrt 3$. Can you finish now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$ Prove that $$(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$$
First of all, I don't really know if by proving it means finding the function Sum ... | Another approach.
$$f(x)=(x-1)\log(1 - 2 x) -2x \implies f''(x)=\frac{4 x}{(1-2 x)^2}$$ Let $t=2x$ to make
$$g''(t)=\frac{2 t}{(1-t)^2}=2\sum_{n=1}^\infty n t^n\implies g'(t)=2\sum_{n=1}^\infty\frac{ n t^{n+1}}{n+1}\implies g(t)=2\sum_{n=1}^\infty\frac{ n t^{n+2}}{(n+1) (n+2)}$$ Replace $t$ by $2x$ and shift the index ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do we solve pell-like equations? I need to find all solutions $(x,y)∈Z^2$ to the Pell-like equation $x^2-21y^2= 4$
Method I used to solve above problem:-
I solved the pell-equation $x^2-21y^2= 1$ and calculated the possible solutions to the equation and further multiplied the above equation with the initial equatio... | Given
$x^2-21y^2= 4$ we can see $(5,1)$ as an easy solution where $5^2-21= 4$.
Another observation is
$$x^2-21y^2= 4\implies \frac{x^2-4}{21}=y^2=\frac{x-2}{p}\cdot\frac{x+2}{q}\quad \text{ where }\quad p|x-2\quad\land\quad q|x+2$$
The factors of $21$ are $1,3,7,21$ and trying the cofactors $(1,21)$ we get conflicting ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that by means of the transformation $w=\frac 1z$ the circle $C$ given by $|z-3|=5$ is mapped into the circle $|w+\frac{3}{16}|=\frac{5}{16}$
Show that by means of the transformation $w=\frac 1z$ the circle C given by $|z-3|=5$ is mapped into the circle $\left|w+\frac{3}{16}\right|=\frac{5}{16}$
My try:
$$\begin{... | $$|z-3|=5.$$
Let $\displaystyle w= \frac{1}{z}$ and let $w^\star$ be the conjugate of $w$.
$$\begin{aligned}
\left(\frac{1}{w} -3 \right) \left(\frac{1}{w^\star} -3 \right) &= 25 \\
(1 -3 w^\star) \left( 1 - 3w \right) &= 25 w w^\star\\
1-3(w+w^\star) + 9 w w^\star &= 25 w w^\star \\
16 w w^\star + 3(w + w^\star) -1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Calculate the minimum value of $\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$.
Given positives $a, b, c$ such that $abc = 1$, if possible, calculate the minimum value of $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \lef... | Answer to the question: $\lim_{h \to 0}6+h$
The given expression$$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right| $$
upon condition that $abc=1$
This can be rewritten as
$$\left|\frac{a^2 - \frac{1}{a}}{b - c}\right| + \left|\frac{b^2 - \frac{1}{b}}{c - a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving $\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
Proving $\displaystyle\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
My atempt:
\begin{align*}
\int_0^1 \int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx \, dy &=\int_0^1I_x(y)\,dy\\[6pt]
\text{where }I_x(y)=\int_0^1\frac{-x\ln... |
I thought it might be instructive to present an approach that uses elementary calculus tools to reduce the double integral to a single integral.
Then, after a standard use of a Taylor series expansion of $\log(1+y)$, the answer is expressed as the familiar alternating series of reciprocal squares $\sum_{n=1}^\infty\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What's the problem with differentiating $y = \sin(x^2)$ by applying the limit definition of a derivative directly? I was taking the derivative of $y = \sin(x^2)$. I know that we can solve it by applying chain rule, but i tried without any rules, just like a normal method. This is what i did:
$$\frac{\sin((x + h)^2) - \... | This is OK,
but the chain rule is easier.
Continue like this,
using $\sin(x) = x+O(x^3),
\cos(x) = 1-x^2/2+O(x^2)$
for small $x$:
$\begin{array}\\
\Delta_h(\sin(x^2))
&=\dfrac{\sin((x + h)^2) - \sin((x)^2)}{h}\\
&=\dfrac{\sin(x^2+2hx+h^2) - \sin(x^2)}{h}\\
&=\dfrac{\sin(x^2)\cos(2hx+h^2)+\cos(x^2)\sin(2hx+h^2) - \sin(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving: $\int_0^1 \int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$
Proving:$$\displaystyle\int_0^1\displaystyle\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$$
I tried using variable switching
$\ln(xy)=t$ But I did not reach any results after the calculation
\begin{align*}
k&=\displaystyl... | Starting off after your first substitution, notice that $$\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}, \text{ for } x \in(-1,1)$$
Since the domain of $x,y$ is $(0,1)$, we can write $$\frac{e^t}{(1+e^t)^2}=-\sum_{n=1}^{\infty} {(-1)}^n n e^{tn}$$
In addition, you made a slight error when calculating $dt$ I suppose. It... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all values of $a$ for which the maximum value of $f(x)=\frac{ax-1}{x^4-x^2+1}$equals $1$. Find all values of $a$ for which the maximum value of $$f(x)=\frac{ax-1}{x^4-x^2+1}$$equals $1$.
I equated $f(x)$ equal to $1$ to obtain a polynomial $x^4-x^2-ax+2=0$. Now this must have at least one repeated root.But I could... | Compute the polynomial GCD of your polynomial and its derivative.
A repeated root of a polynomial is a root of its derivative. So we want to find $a$ such that $g = \gcd(x^4 - x^2 - a x+2, 4 x^3 - 2x - a) \neq 1$.
Taking $4$ of the left and $-x$ of the right, as well as the right, we have
$$ g = \gcd(-2x^2-3ax+8, 4 x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to solve double absolute value inequality? This question comes from Spivak Calculus Chapter 1.
How can we algebraically solve $|x − 1|+|x − 2| > 1?$
I know that if we 2 absolute values and no constants, we can square both sides, but I'm pretty sure this is not the case here. My attempt was to split this into differ... | "However, after doing this, I obtained conflicting solutions and unsolvable expressions"
Those are cases with no solutions. Nothing wrong with that.
Do cases be keep track of you initial assumptions.
Case 1: $x-1 \ge 0; x-2 \ge 0$. Thus $x\ge 1$ and $x \ge 2$. This is the case that $x \ge 2$.
Okay $|x-1| + |x-2|> 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
} |
Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using ... | If we assume $z\in(-1,1)$ and start with
$$ \sum_{n\geq 0} (2n+1) z^n = \frac{1+z}{(1-z)^2} \tag{1}$$
granted by stars&bars
$$ \frac{1}{(1-z)^{m+1}}=\sum_{n\geq 0}\binom{n+m}{m}z^n\tag{2} $$
by convolution it follows that
$$ \sum_{n\geq 0}\left(\sum_{k=0}^{n}(2k+1)\right) z^n = \frac{1+z}{(1-z)^3}\tag{3} $$
and by eval... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Check the convergence of the series : Which test do we apply? Check the convergence of the following series:
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^2+1}{n^4+n}}$
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^{n^2}}{(n+1)^{n^2}}}$
For the first series do we use the comparison test? But with which sequence do we... | The first one is fine, for the second one, by $\left(1+\frac1n\right)^{n}\ge 2$, we have
$$\frac{n^{n^2}}{(n+1)^{n^2}}=\frac1{\left(1+\frac1n\right)^{n^2}}\le\frac1{2^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int\limits_{0}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$ Notice that $\frac{n \sin x}{1+n^2x^2}\to 0$ pointwise.
And we have,$$\int\limits_{0}^{\infty}\frac{n \sin x}{1+n^2x^2}dx=\int\limits_{0}^{1}\frac{n \sin x}{1+n^2x^2}dx+\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$$
Then for $\int\limits_{1}^{\infty}... | The change fo variable $u=nx$ gives
$$
\int^\infty_0\frac{\sin(u/n)}{1+u^2}\,du
$$
The integrand is dominated by $\frac{1}{1+u^2}$ which is integrable. Then, by dominated convergence
$$
\lim_{n\rightarrow\infty}\int^\infty_0\frac{n\sin x}{1+n^2x^2} =\lim_{n\rightarrow\infty}\int^\infty_0\frac{\sin(u/n)}{1+u^2}\,du=
\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
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.