Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Given $\frac{dS}{dt}=-\frac{SI}{S + I + R}$ and $R = -(SI + S + I)/ \frac{dS}{dt}$ Determine $\frac{d^2S}{dt^2}$ I am given a system of equations:
\begin{align}
\frac{dS}{dt} &= -\frac{SI}{S + I + R}\\
\frac{dI}{dt} &= \frac{SI}{S + I + R} -I \\
\frac{dR}{dt} &= I - R\\
\end{align}
I am to subsitute $R = -\left(SI + S ... | Try differentiating $\frac{dS}{dt}$ directly.
$\begin{align}\frac{d^2S}{dt^2} &= -\frac{(S+I+R)\frac{d(SI)}{dt}- SI\frac{d(S+I+R)}{dt}}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[I\frac{d(S)}{dt}+S\frac{d(I)}{dt}\right]- SI\frac{d(S+I+R)}{dt}}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[-I\frac{SI}{S + I + R}+S\frac{SI}{S + I + R} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$ For $n$ normal number, the book solved it like this:
If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.
If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .
But I didn't solve it like this.
My logic i... | Problem: Find $n$ where $(2n+1)2^{4n+5} \equiv 3 \bmod{7}$
My approach:
Starting at $n=0,1,2...$, all $\bmod 7$, knowing that $2^3\equiv 1$, we can see that $2^{4n+5}\equiv 2^{n+2}$ cycles through $\{4, 1, 2\}$. Since $4^{-1}\equiv 2$, the implied $3$-cycle of $required$ values for $(2n+1)$ to satisfy is thus $\{2\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$.
Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$
Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:
\begin{align}
n = 2 \Longri... | All the other answers are for the case where there are infinite x's. However, there are finite ($n$) x's in this problem!
Let's first denote the function as $f_n(x)=x^{x^{\cdots^x}}$ with $n$ x's in the exponent. For example, $f_1(x)=x^x, f_0(x)=x.$
Take the logarithm of both sides, we get $\ln f_n(x)=f_{n-1}(x)\ln x$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3992667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Two Lebesgue's integrals I need to calculate these two integrals:
$$\lim_{n\rightarrow\infty}\int_\mathbb{R}\frac{nx^2+1}{nx^4+n^2}d\lambda(x)$$
$$\int_0^1\sum_{n=1}^\infty n\cdot\chi_{(\frac{1}{n+1},\frac{1}{n})}d\lambda$$
I did the second one like this:
$$\int_0^1\sum_{n=1}^\infty n\cdot\chi_{(\frac{1}{n+1},\frac{1}{... | You have for $\vert x \vert \ge 1$
$$\begin{aligned}
\left\vert \frac{nx^2+1}{nx^4+n^2} \right\vert &\le \frac{x^2+1}{x^4+n} \le \frac{x^2+1}{x^4+1}
\end{aligned}$$
and for $\vert x \vert \le 1$
$$\left\vert \frac{nx^2+1}{nx^4+n^2} \right\vert \le \frac{n+1}{n^2} \le 2.$$
Therefore $f_n(x) = \frac{nx^2+1}{nx^4+n^2} $ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3994198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Failing a basic integration exercise; where did I go wrong? (This is a basic calculus exercise gone wrong where I need some feedback to get forward.)
I've attempted to calculate an integral by first integrating it by parts and then by substituting. The result I got is not correct though. Can I get a hint about where st... | The mistake is
$$v'=\sin\sqrt{x}\,dx \to v=\int \sin\sqrt{x}\,dx\neq -\cos\sqrt{x}+k, \quad k\in \Bbb R$$
$$\int \sin\sqrt{x}\,dx=2\left(-\sqrt{x}\cos \left(\sqrt{x}\right)+\sin \left(\sqrt{x}\right)\right)+k',\quad k'\in \Bbb R$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3994724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Integral of a polynomial fraction: $\int \frac{4x^2-3x+2}{4x^2-4x+3}dx$ I have initially used long division to obtain: $$1+\frac{x-1}{4x^2-4x+3}$$
Now, I'm having difficulty solving $$\int \frac{x-1}{4x^2-4x+3}dx$$
Even if I do the $u$-substitution i.e. $u=4x^2-4x+3$ hence, $du=8x-4$ and $dx=\frac{1}{8x-4}du$, I can't ... | Hints:
$$2\frac{x-1}{4x^2-4x+3}=\frac{2x-2}{(2x-1)^2+2}=\frac{2x-1}{(2x-1)^2+2}-\frac1{(2x-1)^2+2}.$$
In the first term, the numerator is the derivative of the denominator (to a coefficient). In the second, we recognize the derivative of the arc tangent, modulo simple transformations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve:$(1-y^2 +\frac{y^4}{x^2})p^2- (2\frac{y}{x})p + \frac{y^2}{x^2}=0$ I have the differential equation: $(1-y^2 +\frac{y^4}{x^2})p^2- (2\frac{y}{x})p + \frac{y^2}{x^2}=0$ where $p=\frac{dy}{dx}$
I have solved this up until $\frac{dy}{dx}$, but I am not able to reduce it to an exact differential equation.
Steps unti... | $$xdy-ydx=-y\sqrt{x^2 - y^2}dy$$
$$yx'-x=y\sqrt{x^2 - y^2}$$
$$\left(\dfrac xy \right)'=\sqrt{\dfrac {x^2}{y^2} -1}$$
$$w'=\sqrt{w^2 -1}$$
Where $w=\dfrac xy$.Then integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Help solving $\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$ Reading an article online I came across this integral I've been trying to solve since yesterday morning and I literally tried every integration method I know and I still can't solve it. I am a beginner to integrals so help will be appreciated!
I tried substitution, in... | Try this.
\begin{align}
\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}
&={\displaystyle\int}\left(\dfrac{\sqrt{9x^2+9x+2}}{8x^2+9x+2}-\dfrac{x}{8x^2+9x+2}\right)\mathrm{d}x \\
{\displaystyle\int}\dfrac{\sqrt{9x^2+9x+2}}{8x^2+9x+2}\,\mathrm{d}x
&={\displaystyle\int}\dfrac{\sqrt{\left(3x+\frac{3}{2}\right)^2-\frac{1}{4}}}{8x^2+9x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Prove this formula $\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}= \sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)$ I wish to use geometric power series of $\dfrac{1}{1-x}$ to prove this formula by Euler
$$\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=r\sin(x)+r^2\sin(2x)+r^{3}\sin(3x)...=1+\sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)$$
In this ... | Use infinite GP:
$$S=\sum_{n=1}^{\infty} r^n \sin nx=\Im \sum_{n=1}^{\infty} r^ne^{inx}=\Im \frac{re^{ix}}{1-re^{ix}}$$
$$\implies S=\Im \frac{re^{ix}(1-re^{-ix})}{(1-re^{ix})(1-re^{-ix})}=\frac{r\sin x}{1-2r\cos x+r^2}$$
This result has been check with Mathematica numerically. For instance for $r=1/2,x=\pi/3$, we get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $\left \lfloor \frac{1+\lfloor na+1/a\rfloor}{a} \right \rfloor=n$ If $a \geq \frac{1+\sqrt{5}}{2}$ and $n \in \Bbb W$, prove that
$$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=n.$$
I could prove only when $a$ is an integer, that is $a \geq 2$. If $a \in \Bbb Z$ we hav... | We wish to show that $\exists\epsilon\in[0,1)$ such that $$1+\left\lfloor\frac1a+na\right\rfloor=a(n+\epsilon)=na+\epsilon a\implies\left\lfloor\frac1a+\{na\}\right\rfloor=\{na\}+b$$ where $b=\epsilon a-1\in[-1,a-1)$. As the LHS is either $0$ or $1$, either $b=-\{na\}$ or $1-\{na\}$ so we will always find an $\epsilon\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that the algebraic expression is greater than zero Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.
It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.
My idea was to use AM-GM: $x^8 + x^2 \geq 2\s... | I considered parsing the polynomial as $ \ (x^8 - x^5) \ + \ (x^2 -x + 1) \ $ . Restricting techniques to ones that would likely appear in junior high school and high school curricula, we can show that $ \ x^2 - x + 1 \ $ is never equal to zero since its discriminant is $ \ -3 \ $ . So its value can never change sign... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
What's the distance between two circumcenters This problem was came to a Facebook post of mathigon ......the problem seems trickier than I expected...can you help me to find the distance??
|
Let $\angle CAB = \angle ABC = \angle BAD = \angle DBA = \theta$
Area of the $\bigtriangleup CAD = \frac{1}{2} 1^2 \sin 2 \theta = \frac{1}{2} \sin 2 \theta$
Area of the sector $CAD$ (consider the right side of the circle on the left) $ = \frac{1}{2} 1^2 2\theta = \frac{1}{2} (2 \theta)$
So the common area between the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is it possible to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ without using trigonometric substitution? The normal approach to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ is using the substitution $x=2\tan\theta$. But I wonder is is possible to do it without using trigonometric substitution? I tried this approach:
$$\int\frac{dx... | Knowing the answer, as i mentioned it in the comments, the substitution in
the answer of J.G. has best chances to work, and it works in a line.
Alternatively...
Starting from the last expression in the OP, we may use the Euler substitution
$$
t = \sqrt{\frac{u+4}u}=\sqrt{1+ \frac 4u}\ .
$$
Then we have formally success... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
inequality:$(a+c)(b+d)(ac+bd)+4\ge 3(a+b+c+d)$
If $a,b,c,d>0$ such that $abcd=1$ prove that $$(a+c)(b+d)(ac+bd)+4\ge 3(a+b+c+d)$$
My first idea was to find counter examples and also check the bahaviour of inequality in the edge cases.Easy to see the equality occurs when $a=b=c=d=1$ However there is a big problem when... | WLOG assume that $bd\ge 1$ and $ac\le 1$. The inequality rewrites as
$$(a+c)((b+d)(ac+bd)-3)+4\ge 3(b+d).$$
AM-GM and $bd\ge 1$ yield
$$(b+d)(ac+bd)-3\ge 2\sqrt{bd}\cdot 2\sqrt{acbd}-3 \ge 2\cdot 2-3=1>0,$$
hence by AM-GM
$$(a+c)((b+d)(ac+bd)-3) \ge 2\sqrt{ac}((b+d)(ac+bd)-3) = 2(b+d)((ac)^{3/2}+(bd)^{1/2})-6(bd)^{-1/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Inequality with $abc=1$ Let $a;b;c$ be positive real numbers and $abc=1$
Find maximum value of: $P=\dfrac{1}{\sqrt{a^2+2b^2}}+\dfrac{1}{\sqrt{b^2+2c^2}}+\dfrac{1}{\sqrt{c^2+2a^2}}$
I tried to use Cauchy-Schwarz: $a^2+2b^2 \geq \dfrac{1}{3}(a+2b)^2$
Then: $P \leq \dfrac{\sqrt{3}}{a+2b}+\dfrac{\sqrt{3}}{b+2c}+\dfrac{\sqr... | There is no maximum .
Note that when $a\to 0^+,b\to 0^+,c\to \infty$ we get $P\to \infty$ .This is because the $\sqrt{a^2+2b^2}\to 0$
Note that we chose $c\to \infty$ because $c=\frac{1}{ab}$
Now even if you are looking for the minimum there is no minimum .
Note again that $a\to 0^+,b\to \infty,c\to \infty$ we see ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Finding eigenvectors of matrix of matrices
Let $A$ be a matrix with $\lambda_1,...,\lambda_n$ eigenvalues, and $v_1,...,v_n$ the corresponding eigvenvectors. Let $B= \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix}$. It's known that the eigenvalues of $B$ are $\{\pm \lambda_1,..., \pm \lambda_n\}$. Find the eigenvectors of ... | Let's look at $u_i =\begin{pmatrix} v_i \\ -v_i \end{pmatrix}$:
$B \begin{pmatrix} v_i \\ -v_i \end{pmatrix} = \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix} \begin{pmatrix} v_i \\ -v_i \end{pmatrix}= \begin{pmatrix} A(-v_i) \\ Av_i \end{pmatrix} = \begin{pmatrix} -\lambda_iv_i \\ \lambda_iv_i \end{pmatrix} = -\lambda_i\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show $\log\left(\frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}}<\log\left(\frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}$ if $x\geq 5$ and $1\leq y\leq x-2$ Assume that all logarithms are natural.
Let $x$ and $y$ be integers that satisfy $x \geq 5$ and $1 \leq y \leq x-2$. I am trying to show that $$\log\left( \frac{x-y}{x}\rig... | Try working with the inequalities on $x$ and $y$. We are given $x\in[5,\infty)$ and $y\in[1,x-2]$. First, we find $x-y\in[2,x-1]$ and $x+y\in[6,2x-2]$. For a fraction, we can find the minimum by maximizing the denominator and minimizing the numerator, for instance $\frac{x-y}{x+y}\in[\frac{2}{2x-2},\frac{x-1}{6}]$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to proceed $\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $
$$I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $$
I have taken the second term of denominator (in square root) as $(x - 2)t$. But cannot go further. Please suggest how to proceed or any alternate method to solve it.
| Note that $7x-10-x^2 = \frac94-(\frac72-x)^2$. So, substitute $\frac72-x=\frac32\sin t$ to integrate as follows
\begin{align}
& \int \frac{dx}{(x-2)(1+\sqrt{7x-10-x^2})}\\
= & - \int \frac{2\cos t}{(1-\sin t)(2+3\cos t)}dt\\
= & - \int \left ( \frac{\cos t}{1-\sin t}-\frac {3\sin t}{2+3\cos t} - \frac {3}{2+3\cos t} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Eliminating $x$, $y$, $z$ from $\frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$ and $ax+by+cz=0$ Here is a Math Tripos problem I cannot solve.
Eliminate $x$, $y$, $z$ from the equations
$$\frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$$ and
$$ax+by+cz=0$$
I dont know if there is a general (p... | Let's denote $t$ be equal to
$$t = \frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$$
We have:
\begin{align}
t &=\frac{(y^2-yx-yz)+(z^2-zx-zy)-(x^2-xy-xz)}{b+c-a} \\
&=\frac{(y^2-2yz+z^2)-x^2}{b+c-a} \\
&=\frac{(y-z)^2-x^2}{b+c-a} \\
&=-\frac{(z+x-y)(x+y-z)}{b+c-a} \\
\end{align}
Hence,
\begin{align}
t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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The center of the circumcircle lies on a side of a triangle Consider a triangle $ABC$. Let the angle bisector of angle $A$ be $AP,P\in BC$. $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on the segment $AC$. Find $AB$.
$$AB=\dfrac{144\sqrt5}{5}$$
By Triangle-Angle-Bisector Theorem $$\dfrac... | Observe that $\angle OPA=\angle PAO=\frac\alpha2=\angle BAP\implies OP\parallel AB$. Thus $$\frac{CO}{OA}=\frac{CP}{PB}\iff \frac dR=\frac{20}{16}=\frac54$$ Also, Power of a point yields $$\begin{align*}\text{Pow}(C)_{(APB)}=\lvert d^2-R^2\rvert&=20\cdot 36\\\iff \left\lvert\left(\frac54R\right)^2-R^2\right\rvert&=720\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Estimations of some new recurrence sequences
Problem 1. Given the recursion $a_{n+ 1}= \sqrt{a_{n}^{2}+ a_{n}}$ with $a_{1}= 1.$ Prove that
$$a_{n}\sim\frac{n}{2}+ \frac{1}{2n}\,{\rm as}\,n\rightarrow\infty$$
For problem 1, I only can prove $a_{n}\sim\left ( n+ 1 \right )/2$ by using the Laurent series for the increa... | We may use the approach in [1].
For Problem 1:
First, clearly $\lim_{n\to \infty} a_n = \infty$.
Second, we have
\begin{align}
a_{n+1} - a_n &= \sqrt{a_n^2 + a_n} - a_n\\
&= a_n\left(\sqrt{1 + \frac{1}{a_n}} - 1\right)\\
&= a_n\left(1 + \frac{1}{2} \frac{1}{a_n}
+ o\left(\frac{1}{a_n}\right) - 1\right)\\
&\to \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
proving that $89 \mid 2^{44}-1$ i tried to prove that $2^{44} \equiv 1\pmod{89}$.
I noticed that by Fermat's little theorem $2^{88} \equiv 2^{44}\cdot 2^{44} \equiv 1\pmod{89}$
which means that $2^{44}$ is the inverse of itself $\rightarrow 2^{44} \equiv 1 \pmod{89}$ or $2^{44}\equiv 88 \pmod{89}$.
how can I rule out t... | $2^{44} - 1 = (2^{22})^2 - 1 = (2^{22}+1)(2^{22} - 1) = (2^{22}+1)(2^{11} + 1)(2^{11} - 1) = (2^{22}+1)(2^{11} + 1)\times 2047 = (2^{22}+1)(2^{11} + 1)\times 23 \times 89$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
A simpler way to prove that $\frac 32-x-\frac{1}{x+1}$ is negative for $x\geq1$, instead of computing derivative? I have this function
$$f(x)=\frac 32-x-\frac{1}{x+1}$$
I want to prove that $f(x)$ is negative for $x\geq1$. We can easily prove this by calculating the first derivative. Is there a simpler way to prove tha... | \begin{align} \left(-x-\frac{1}{x+1}\leq -\frac{3}{2}\right)\quad \text{ and }\quad x\geq1\\
\\
\iff\quad \left(x+\frac{1}{x+1}\geq \frac{3}{2}\right)\quad \text{ and }\quad x\geq1 \\
\\
\iff \quad \left(x(x+1) + 1 \geq \frac{3}{2}(x+1)\right)\quad \text{ and }\quad x\geq1\\
\\
\iff \quad \left(x^2 - \frac{1}{2}x -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ABCD$ is a trapezoid $ABCD$ is a cyclic quadrilateral inscribed of circle $O$. The tangent line of circle $O$ at $B$ intercepts $CD$ at $K$. The tangent line of circle $O$ at $C$ intercepts $AB$ at $M$.
if $AB = BM$ and $CD=CK$, prove that $ABCD$ is a trapezoid.
I was going down the route of:
$\frac{BM}{\s... | It's enough to prove that $BD=AC$.
Observe that, $\triangle BCK\sim \triangle DBK$ and $\triangle CBM\sim \triangle ACM$.
Hence, $\frac{BK}{DK}=\frac{CK}{BK}$
$\Rightarrow BK=CK\sqrt{2}$
Thereafter, $\frac{BD}{BC}=\frac{BK}{CK}=\sqrt{2}$
Similarly yield $\frac{AC}{BC}=\sqrt{2}$ and thus $BD=BC\sqrt{2}=AC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4027066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Computing the value of $\zeta(6)$ and $\zeta(8)$ I know the closed form formula for calculating the values of Zeta function at even integers. I was able to derive it from the coefficients of the power series of $-\dfrac{\pi x}{2}\cot(\pi x)$. Now, I am looking at the way with which Euler was able to derive the values o... | The generalisation you are looking for is given by the functions
$$f_n \colon \mathbb{R} \to \mathbb{R}, \, f_n (x) = \prod \limits_{k=0}^{n-1} \operatorname{sinc}\left(\mathrm{e}^{\mathrm{i} \pi \frac{k}{n}} x\right) \, , $$
for $n \in \mathbb{N}$ ($\operatorname{sinc}(z) = \frac{\sin(z)}{z}$ with $\operatorname{sinc}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that the set $a+b\sqrt{2}$ where a and b are rational without zero is a group under multiplication We are given $G=\{a+b\sqrt{2}: a,b\in\mathbb{Q}\}$ and asked to show that $G-\{0+0\sqrt{2}\}$ is a group under multiplication.
Let us first specify that multiplication takes place as expected:
$$(a+b\sqrt{2})(c+d\sq... | Bear in mind the multiplication in $G$ is the same as the multiplication of the reals so whatever is true about multiplication of $\mathbb R$ will still be true of the multiplication of $G$.
Multiplication is associative. We don't have to prove that.
We do have to prove that mulitplicaition is closed on $G$. That is:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Obtain the coefficient of $x^2$ in the expansion of $1+\frac{6}{2x+1}+\frac{5}{2-3x}$ Hello so this is a 2 part question and I managed to express that praction as a partial fraction which was equaled to $$1+\frac{6}{2x+1}+\frac{5}{2-3x}$$ I will add my work below I tried lot to Obtain the coefficient of $x^2$ but cant ... | This is how to do it the hard way.
Let's suppose that, for $x$ close enough to $0$,
$$1 + \frac{6}{2x+1} + \frac{5}{2-3x} = a_0 + a_1x + a_2x^2 + \dots$$
$$(2x+1)(2-3x)\left(1 + \frac{6}{2x+1} + \frac{5}{2-3x}\right) = 19 - 7x - 6x^2 + \dots$$
$$(2x+1)(2-3x)(a_0 + a_1x + a_2x^2)
= 2a_0 + (a_0+2a_1)x + (-6 a_0 + a_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does $\sum{\frac{\sin (n+1/n)}{ \ln n}} \sim \sum{\frac{\sin n}{\ln n}}$? As was mentioned in the title, can we say that both the series
$$
\sum{\frac{\sin (n+1/n)}{\ln n}} \quad \text{ and } \quad \sum{\frac{\sin n}{\ln n}}
$$
converges or diverges simultaneously? I mean, it looks very intuitive since
$$
\sin(n+1/n) \... | Since $\sin(a+b) = \sin a \cos b + \cos a \sin b$, by a Taylor expansion around $0$ we get
$$\begin{align*}
\sin\!\left(n+\frac{1}{n}\right) &= \sin n \cos \frac{1}{n} + \cos n \sin \frac{1}{n} \\
&= \sin n\left(1-\frac{1}{n^2} + o\!\left(\frac{1}{n^2}\right) \right)
+ \cos n\left(\frac{1}{n} + o\!\left(\frac{1}{n^2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4042989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Remainder When Divided By 70 : Remainder Theorem Problem If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?
I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{... | Another way, where there is no need to compute the numeric values of the four cubes.
Note that $16+19=17+18=35=70/2$, hence
\begin{align*}
16^3+17^3+18^3+19^3
&=16^3+(35-18)^3 +18^3+(35-16)^3\\
&\equiv 16^3+35^3-18^3+18^3+35^3-16^3\\
&=2\cdot 35^3\equiv 0\pmod{70}.
\end{align*}
In particular, when we expand the cube $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Need help with De'Moivre's Theorem, or maybe that is the wrong approach to this problem.
Let ($a_{n}$) and ($b_{n}$) be the sequences of real numbers such that $(2+i)^n=a_{n}+b_{n}i$ for all integers $n\geq 0$, where $i=\sqrt{-1}$. What is $\sum_{n=0}^{\infty}\frac{a_{n}b_{n}}{7^n}$?
I need some help with the questio... | Let $\theta$ be the acute angle satisfying $\tan \theta = \frac{1}{2}$,
$$a_n = \sqrt{5^n} \cos(n \theta)$$
$$b_n = \sqrt{5^n}\sin (n\theta)$$
$$\frac{a_nb_n}{7^n} = \left( \frac57\right)^n\sin(n\theta)\cos(n\theta)=\frac12 \left( \frac57\right)^n \sin (2n\theta)= \frac12 \Im\left[\left(\frac57e^{2i\theta}\right)^n\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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floor function inequality $n+ \left \lfloor \tfrac{n}{3} \right \rfloor \le x$ I have the following problem which has two parts:
Given any $x>0$, I define a function $A$ by
\begin{equation}
A(x)=\max \{ n \in \mathbb{N} : n+ \left \lfloor \tfrac{n}{3} \right \rfloor \le x\}
\end{equation}
I want to compute $A(x)$ and
... | Since $n\in\mathbb{N}$, it follows by contradiction that
$$n+\left\lfloor{\frac{n}{3}}\right\rfloor\leq x \iff n+\left\lfloor{\frac{n}{3}}\right\rfloor\leq \lfloor x\rfloor$$
Suppose that it is possible to make the above equation an equality. Writing $n=3k+r,0\leq r\leq 2$ we note that
$$\lfloor x\rfloor=4k+r$$
Thus th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving a Polynomial Equation by Factoring. $n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.
| Here it is an approach based on the Binomial Theorem:
\begin{align*}
n^{3} + 12n^{2} + 48n + 64 = {3\choose 0}n^{3}4^{0} + {3\choose 1}n^{2}4^{1} + {3\choose 2}n4^{2} + {3\choose 3}n^{0}4^{3} = (n+4)^{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 2
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Binomial coefficients identity : $\sum_{k=1}^{n-m+1} k\binom{n-k+1}{m}=\binom{n+2}{m+2}$ For any positive integer m&n.$n\ge m$ , let $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) = {}^n{C_m}$. Prove that $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
{n - 1}... | Here is a combinatorial approach. Consider the set
$$\{ 1,2,3,4,\ldots,n-1,n,n+1,n+2\}$$
We will pick $m+2$ distinct numbers. This can be done randomly in $\binom{n+2}{m+2}$ ways.
Another way to do the same would be prioritizing the second smallest element.
If second smallest number is $2$, we can choose $1$ as the sma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4061296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the intergal of $\int {\dfrac{1}{(x^2+9)^2}}\,dx$ with recursive formula? I used partial integration and I got this:
${\dfrac{x}{(x^2+9)^2}} + 4 $$\int {\dfrac{x^2}{(x^2+9)^3}}\,dx$
According to the recursice formula I should continue like this:
${\dfrac{x}{(x^2+9)^2}} + 4 $$\int {\dfrac{x^2 +9-9}{(x^2+9)^3}}... | Integrate as follows
\begin{align}
\int {\dfrac{1}{(x^2+9)^2}}\,dx
&= \frac19 \int {\dfrac{1}{x^2+9}}\,dx - \frac19\int {\dfrac{x^2}{(x^2+9)^2}}\,dx\\
&= \frac19 \int {\dfrac{1}{x^2+9}}\,dx + \frac1{18}\int x\>d({\dfrac{1}{x^2+9}})\\
&= \frac1{18} \frac x{x^2+9} + \frac1{18}\int {\dfrac{1}{x^2+9}}\,dx\\
&= \frac1{18... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find a value of $\;\lim\limits_{n\rightarrow\infty}\frac{a_{n}}{n}$
(If you're good at it, you can do it by heart.)
Let $a_n$ be the average value of lengths of all the diagonal lines of regular n-gon whose side is 1.
Find a value of
$$\lim_{n\rightarrow\infty}\frac{a_{n}}{n}$$
Source: Fujino_Yusui
After messing arou... | Inscribe the n-gon in a circle. When calculating the length of a diagonal, instead of using the cosine theorem, you can bisect the central angle subtended on the diagonal. The bisector, the radia leading to the end points of the diagonal, and the diagonal will create two right triangles, from which you can get that the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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How can I prove this for $a+b+c=0$ We wish to prove that
$$\left(\frac{a-b}{c}+\ \frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9,$$ for $a+b+c=0$ and $$abc \neq 0 , a\neq b , b\neq c, c \neq a.$$
I tried working with only $2$ fractions at a time, switching some of the variable... | You can expand directly, by noticing that the first parantheses is
$$\frac{ab(a-b) + bc(b-c) + ca(c-a)}{abc} = -\frac{(a-b)(b-c)(c-a)}{abc}$$
So the expression becomes
$$\frac{a^3+b^3+c^3 + 3abc - a^2b - a^2c - b^2a -b^2c - c^2a - c^2b}{abc}$$
Collecting $a^3, -a^2b, - a^2c $ and similarly this becomes
$$\frac{a^2(a - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Möbius transformation mapping $\mathbb{D}$ into itself. I want to show that a Möbius transformation $f(z) = \frac{az + b}{cz + d}$ for $a,b,c,d \in \mathbb{C}$ maps the unit disk $\mathbb{D}$ into itself if and only if the coefficients satisfie
\begin{equation*}\label{ineq}
|\overline{b}d - \overline{a}c| + |ad - bc| \... | I thought to share my solution if someone is interested:
\begin{align*}
|f(z) - w|^2 &= \left|\frac{az + b}{cz + d} - \frac{b \overline{d} - a\overline{c}}{|d|^2 - |c|^2} \right|^2 \leq r^2 = \frac{|ad - bc|^2}{(|d|^2 - |c|^2)^2} \\
&\iff \left|\frac{az + b}{cz + d}(|d|^2 - |c|^2) - b \overline{d} - a\overline... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Fastest way to find derivative of the function at the given point
What is derivative of $f(x)=\left(\dfrac{\sqrt[3]{x^2+2x}}{x^2-x}\right)^3$ at the point $x=2$? $$1)-\frac34\qquad\qquad2)-\frac54\qquad\qquad3)-\frac52\qquad\qquad4)-\frac{15}4$$
This is a problem from an timed Exam, so I am looking for the fastest w... | Perhaps slightly...but very slightly...easier: using that $\;(x^2-x)^3=x^3(x-1)^3\;$ and also $\;x=2\implies (x-1)^n=1\;$
$$f(x)=\frac{x^2+2x}{(x^2-x)^3}=\frac{x+2}{x^2(x-1)^3}=\frac1{x(x-1)^3}+\frac2{x^2(x-1)^3}\implies$$
$$f'(x)=\left(\frac1{x(x-1)^3}\right)'+2\left(\frac1{x^2(x-1)^3}\right)'=$$
$$=-\frac{(x-1)+3x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Mathematical Induction: Can I assume that $P(k-1)$ is also true in the induction hypothesis? Suppose that $y \in \mathbb{R}$, $y \neq 0$ and $y + \frac{1}{y} \in \mathbb{Z}$. Prove using mathematical induction that $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.
Basis step: The statement is true for $n=1$ s... | For $n=2$ and $n=1$ the statement is correct.
Because, $$\left(y^2+\frac{1}{y^2}\right)=\left(y+\frac 1y\right)^2-2$$
Then ,suppose that $n=k$ and $n=k-1$ are also correct.....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Can we obtain a sufficient constraint over the parameters $A$, $B$, and $C$ in the given equation? I have this equation
$$ A\;\big(\cos (x-y)+\cos x+\cos y\big)+B\;\big(\sin (x-y)-\sin x+\sin y\big)+C=0, $$
where $-\pi\leq x,y \leq \pi$, and $A$, $B$, and $C$ are some real constants.
Then, according to the ranges of t... | Using Cauchy-Schwarz inequality repeatedly: $C^2 = (A\cos(x-y)+B\sin(x-y) +A\cos x-B\sin x + A\cos y + B\sin y)^2 \le 3(m^2+p^2+q^2), m^2 \le A^2+B^2, n^2 \le A^2+B^2, p^2 \le A^2+B^2\implies C^2 \le 3(A^2+B^2+A^2+B^2+A^2+B^2)= 9(A^2+B^2). $, Thus the necessary condition is: $C^2\le 9A^2+9B^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4079869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Floor Function Quotient Fact So I was perusing wikipedia for some floor function facts for a problem I am working on and I saw this:
Let $x,y\in \mathbb{R}$
$$\lfloor x\rfloor +\lfloor y\rfloor \leq \lfloor x+y\rfloor \leq \lfloor x\rfloor +\lfloor y\rfloor +1$$
However when I adapted my problem to this
Let $n,m,k\in \... |
To prove : for $A,B \in \Bbb{R} $
$\lfloor(A + B)\rfloor = \lfloor A\rfloor
+ \lfloor B\rfloor + \lfloor \{A\} + \{B\}\rfloor.$
By definition:
$A = \lfloor A\rfloor + \{A\}.$
$B = \lfloor B\rfloor + \{B\}.$
Therefore,
$$(A + B) = \lfloor A\rfloor + \lfloor B\rfloor
+ \{A\} + \{B\}. \tag1$$
Also, since $0 \leq \{A\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Help with $\int \frac{1}{(1-t^2)t^2} \, dt$ $$\int \frac{1}{(1-t^2)t^2} \, dt$$
By using partial fractions I get:
$$\frac{1}{(1-t)(1+t)t^2} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t+1} + \frac{D}{1-t}$$
$$1 = t^3 (-A-C+D) + t^2 (-B+C+D) + At+ B$$
$$-A-C+D = 0$$
$$-B+C+D=0$$
$$A=0$$
$$B=1$$
So, $A=0, B=1, C=\frac{1}{2}... | I agree that$$\int\frac{1}{t^2(1-t^2)}dt=\int\left(\frac{1}{t^2}+\frac{1}{1-t^2}\right)dt=-\frac1t+\frac12\ln\left|\frac{1+t}{1-t}\right|+K,$$where the locally constant function $K$ can change values either side of $t=\pm1$, so its values can be different for $t<-1,\,t\in(-1,\,1),\,t>1$. (I've called it $K$ rather than... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solve definite integral $\int_{\frac{\sqrt2}2}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}}$ by variable substitution I have this integral
$$\int_{\sqrt{2}/2}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}}$$
I have to solve it by substituting x.
I believe that the easiest way would be to substitute with $t = \sqrt{x^2+1}$. The other way... | Let $$ \mathscr{I} = \int \dfrac{dx}{x\sqrt{x^2+1}}$$
Put $ x = \dfrac1t \implies dx = -\dfrac{dt}{t^2}$
$$ \begin{align} \mathscr{I} &\ = -\int t\dfrac{1}{\sqrt{\frac{1}{t^2}+1}} \dfrac{dt}{t^2} \\ &\ =- \int\dfrac{dt}{\sqrt{t^2+1}} \\ &\ = - \ln \left| t + \sqrt{t^2+1} \right| + C
\end{align} $$
Now, substitute... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Integrating $\int \frac{r^3}{\sqrt{16 + r^2}} dr$ I need to calculate: $$\int \frac{r^3}{\sqrt{16 + r^2}} dr$$
I am currently in the midst of learning AP Calculus BC and the lesson from which this problem came from goes over integration by parts.
Basically, it utilizes that : $$\frac{d}{dx} \left[ f(x)g(x) \right] = f'... | The fastest way to solve this integral (IMO), does not use IBP. We let
\begin{align*}
u &= \sqrt{16 + r^2} &&\implies \boxed{r^2 = u^2 - 16}\\
\frac{du}{dr} &= \frac{2r}{2\sqrt{16+r^2}}&&\implies\boxed{du = {\frac{r}{\sqrt{16+r^2}}}\,dr.}
\end{align*}
Making this substitution:
\begin{align*}\int \frac{r^{3}}{\sqrt{16+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is $\sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}} = 3$ This is a problem from SASMO Grade 8 (Secondary 2) Sample Questions.
Solve for $x$
$\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = 3$
Answer: $x=6$
I have tried this on a calculator: the more $x$ we add, the closer we are to $3$. But how can we prove it, and is there a way to f... | Let $x= \sqrt{6+ \sqrt{6+\sqrt{6+ \sqrt{\cdot\cdot\cdot}}}}$
then $x^2= 6+ \sqrt{6+ \sqrt{6+ \sqrt{\cdot\cdot\cdot}}}= 6+ x$. Solve $x^2-x- 6= 0$.
Clearly this has to be positive. What is the positive root of $x^2- x- 6= 0$?
(There is only one positive root and this clearly has to be positive.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4084931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Calculating $\cos^{-1}{\frac{3}{\sqrt10}} + \cos^{-1}{\frac{2}{\sqrt5}}$ $$\cos^{-1}{\frac{3}{\sqrt{10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}= ?$$
Let $\cos^{-1}{\frac{3}{\sqrt{10}}}=\alpha,
\cos^{-1}{\frac{2}{\sqrt 5}}=\beta$ then, $\cos\alpha=\frac{3}{\sqrt{10}}, \cos\beta=\frac{2}{\sqrt5}$
Therefore $$\cos\alpha=\frac{3... | Hint : apply this formula:
$$\cos^{-1} x +\cos^{-1}y=\cos^{-1}[xy-\sqrt{(1-x^2)(1-y^2)}]$$
Put $x=\frac2{\sqrt {10}}$ and $y=\frac 2{\sqrt 5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How should I check that $6^{\log_{10} x} + 8^{\log_{10} x} = x $ does not have other solutions? The question is to solve the equation $6^{\log_{10} x} + 8^{\log_{10} x} = x $
I know one of the solutions is $x=100$ using Pythagorean triples but I can't show that this is the only solution.
I'm looking for an idea without... | \begin{align*}
6^{\log_{10} x} + 8^{\log_{10} x}
&= 6^{\log_6(x)/ \log_6 10} + 8^{\log_8(x)/ \log_8 10} \\
&= x^{1/\log_6 10} + x^{1/\log_8 10} \text{.}
\end{align*}
We have
\begin{align*}
6^{\log_{10} x} + 8^{\log_{10} x}
&= x \qquad \text{when} \\
\frac{6^{\log_{10} x} + 8^{\log_{10} x}}{x} &= 1 \q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Identities similar to $\frac{a^3 + b^3 + c^3}{3} \cdot \frac{a^7 + b^7 + c^7}{7} = \left( \frac{a^5 + b^5 + c^5}{5} \right)^2$ This question was inspired by the post here, which asks for a proof of the following fact:
If $a+b+c = 0$ then show that
$$\frac{a^3 + b^3 + c^3}{3} \cdot \frac{a^7 + b^7 + c^7}{7} = \left( \f... | The first step is to pin down the parity of $m$ and $n$. Since $m+n$ is even, we must have $m\equiv n\pmod{2}$.
Considering the case $(a,b,c)=(x,-x,0)$, we have $$x^{m+n}\cdot\frac{(1+(-1)^m)(1+(-1)^n)}{mn}=x^{m+n}\cdot\frac{4\left(1+(-1)^{\frac{m+n}{2}}\right)^2}{(m+n)^2}$$ Canceling $x$ and considering when these v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Let $f(x)=(x+1)(x+2)..(x+100)$ and $g(x)=f(x)f’’(x)-(f’(x))^2$. Find number of roots $n$ of $g(x)=0$ After relevant simplification ie. Log on both sides and then differentiating, the expression received is
$$\frac{f’(x)}{f(x)} =\sum_{k=1}^{100} \frac{1}{x+k}$$
Differentiating again will give
$$\frac{g(x)}{(f(x))^2}=-\s... | $$\frac{f’(x)}{f(x)} =\sum_{k=1}^{100} \frac{1}{x+k}$$
D.w.r.t. $x$ both sides
$$\frac{f(x)f''(x)-f'^2(x)}{f^2(x)}=-\sum_{k=1}^{100} \frac{1}{(x+k)^2}<0 \ne 0.$$
So $$g(x)=f(x)f''(x)-f'^2(x)=\sum_{k=1}^{n}\frac{f^2(x)}{(x+k)^2}$$
Then $$-g(x)=(x+2)^2(x+3)^2(x+4)^2....(x+n)^2+ (x+1)^2(x+3)^2(x+4)^2....(x+n)^2+......>0 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integral involving the modified Bessel function of the second kind K_0 The following integral
$$
\int_0^{+\infty} K_0(\alpha\sqrt{x^2+z^2})\, dx, \alpha>0,
$$
can be computed according to Gradshteyn-Ryzhik 6.596 (3) taking $\nu=0$ and $\mu=-1/2$. Its value is $\frac{\pi}{2\sqrt{\alpha}} e^{-\alpha|z|}$.
My question is:... | Assume that $\alpha>\beta$ and $|\arg z|<\frac{\pi}{4}$. Then using http://dlmf.nist.gov/10.32.E10 and interchanging the order of integrations,
\begin{align*}
& \int_0^{ + \infty } {\cosh (\beta x)K_0 (\alpha \sqrt {x^2 + z^2 } )dx}
\\ &
= \frac{1}{2}\int_0^{ + \infty } {\exp \left( { - t - \alpha ^2 \frac{{z^2 }}{{... | {
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"url": "https://math.stackexchange.com/questions/4108140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the least value of $\frac{1}{x}+\frac{3}{y}+\frac{5}{z}$ Given that $x$,$y$ and $z$ are positive numbers and $x+y+z=1$ , Find the least value of $\frac{1}{x}+\frac{3}{y}+\frac{5}{z}$.
My approach:
We write,
$$\frac{1}{x}+\frac{3}{y}+\frac{5}{z}=\frac{x+y+z}{x}+\frac{(3)(x+y+z)}{y}+\frac{(5)(x+y+z)}{z}$$
This furth... | One can show that $\sum \limits_{I=1}^{n}\dfrac{a_i^2}{b_i}\geq \dfrac{(a_1+\dots+a_n)^2}{b_1+\dots+b_n}.$ using this inequality it follows that $$\frac{1}{x}+\frac{3}{y}+\frac{5}{z}\geq (1+\sqrt{3}+\sqrt{5})^2$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Extraneous solutions without performing squaring operation
Solve in $[0, 2\pi]$ $$\sec x+\tan x=2\cos x$$
My mind boggled while solving the trigonometric equation in two different ways:
Method $1.$
Assuming $x\ne \frac{\pi}{2},\frac{3\pi}{2}$
We have:
$$\sec x+\tan x=2\cos x$$
$\implies$
$$\sec x-\tan x=\frac{1}{2\co... | Note that this is not limited to trigonometric equations.
You obtained $\sec(x)-\tan(x)=\frac 1{2\cos(x)}$ by taking the inverse of both sides of the equation.
Let examine the very simple case $x=2\iff \dfrac 1x=\dfrac 12$
Notice that when summing these two equations $x+\dfrac 1x=\dfrac 52$ you now end up with two solu... | {
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"url": "https://math.stackexchange.com/questions/4111434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving $7^x\bmod {29} = 23 $ I have $$7^x\bmod {29} = 23 $$
It is possible to get $x$ by trying out different numbers but that will not be possible if $x$ is actually big.
Are there any other solutions for this equation?
Kind regards
| I don't think there is anything but trial and error.
But look. $7^2 \equiv 23 \equiv -6 \pmod {29}$
$7^{2x} \equiv 36 \equiv 7\pmod {29}$.
So $7^{2x-1} \equiv 1 \pmod {29}$.
We know by FLT the $7^{28}\equiv 1$ so for the least power to which $7$ is a multiple of $1$ must be a divisor of $28$.
So $2x-1$ is a multiple o... | {
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"url": "https://math.stackexchange.com/questions/4122288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
determine the coefficient of $x^2yz$ in the expansion of $(2x-y+z+1)^7$
determine the coefficient of $x^2yz$ in the expansion of $(2x-y+z+1)^7$
I found this question in my textbook , i seem very easy in the beginning but my answer is wrong according to answerkey.
I said that the coefficient is equal to $\frac{7!}{2! ... | A small crosscheck. We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
\begin{align*}
[x^k](ax+z)^n=\binom{n}{k}a^kz^{n-k}\tag{1}
\end{align*}
We obtain by successively applying (1)
\begin{align*}
\color{blue}{[x^2yz]}&\color{blue}{(2x-y+z+1)^7... | {
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"url": "https://math.stackexchange.com/questions/4123460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $x^4-2x^3-x^2+2x+1=0$? How to solve $x^4-2x^3-x^2+2x+1=0$?
Answer given is: $$\frac{1+\sqrt5}{2}$$
I tried solving it by taking common factors:
$$x^3(x-2)-x(x-2)+1=0 $$ $$x(x-2)(x^2-1)+1=0 $$ $$(x+1)(x)(x-1)(x-2)+1=0$$
But it's not leading me anywhere.
| $x^4-2x^3-x^2+2x+1 = 0$
You should split the middle terms and try to look for a common factor.
$x^4-x^3-x^2-x^3+x^2+x-x^2+x+1 = 0$
$x^2(x^2-x-1)-x(x^2-x-1)-1(x^2-x-1) = 0$
$(x^2-x-1)^2 = 0$
Take the square root:
$x^2-x-1 = ±0$
$x^2-x-1 = 0 \lor x^2-x-1 = -0$
Add $\frac{5}{4}$:
$x^2-x+\frac{1}{4} = \frac{5}{4} \lor x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$
Evaluate: $$\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$$
This looks like an unusual hockey stick sum. Here are my attempts:
Method 1:
The sum is equivalent to
$$S=\sum_{r=0}^n 2^{n-r} \binom{n+r}{n}=\sum_{r=0}^n 2^{r} \binom{2n-r}{n-r}$$
and I could evaluate neither of these.
... | In seeking to evaluate
$$S_n = \sum_{r=0}^n 2^{n-r} {n+r\choose r}$$
we find that it is
$$[z^n] \frac{1}{1-2z} \frac{1}{(1-z)^{n+1}}
= \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{1-2z} \frac{1}{(1-z)^{n+1}}.$$
We will use the fact that residues sum to zero, which requires the
residue at $z=1/2$ and the residue at $z=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 8,
"answer_id": 5
} |
Find the $\max$ and the $\min$ of $f(x,y)=xy$ subject to $g(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$, using Lagrange multipliers
Find the $\max$ and the $\min$ of $$f(x,y)=xy$$ subject to the constraint $$g(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$$ using Lagrange multipliers.
My Attempt:
| As you wrote, you want to solve the system$$\left\{\begin{array}{l}y+\frac{2\lambda}{a^2}x=0\\x+\frac{2\lambda}{b^2}y=0\\\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.\end{array}\right.$$So, $x$ is both equal to $-\frac{2\lambda}{b^2}y$ and to $-\frac{a^2}{2\lambda}y$. So, since there is no solution with $x=y=0$, you must have $-\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-{\sqrt{4n^2+n+9}}$ Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9}$.
My first thoughts are the that I can remove a factor of $2n$ from each root, and obtain
$$ \lim_{n\to\infty}2n\sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-2n\sqrt{1+\frac... | The expression seems designed for sheer cussedness if one attempts many of the common methods, since the indices of the roots differ so the "conjugate-factor" method is not much of an option. (And one should avoid applying LHR to expressions with sums or differences of radicals unless one enjoys suffering...)
An appro... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Evaluating $\int_0^B \frac{1}{A^2 + (B^2-x^2)^2}\,dx$ I am trying to solve the following integral:
$$\int_0^B \frac{1}{A^2 + (B^2-x^2)^2}\,dx$$
However the polynomial have imaginary roots:
$$-(-i A + B^2)^{1/2},\; (-i A + B^2)^{1/2},\; -(i A + B^2)^{1/2},\;(i A + B^2)^{1/2}$$
I am only interested on the solution from $... | Let $p=\sqrt{1+\frac{A^2}{B^4}}$ and substitute $x=Bt$
$$I=\int_0^B \dfrac{1}{A^2 + (B^2-x^2)^2} dx
= \frac1{B^3} \int_0^1 \dfrac{1}{t^4-2t^2+p^2} dt
$$
Decompose the integrand as
\begin{align}
\frac{1}{t^4-2t^2+p^2} = \frac1{2ps}\left( \frac{t+s}{t^2+st+p}- \frac{t-s}{t^2-st+p}\right)
\end{align}
with $s=\sqrt{2(p+1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Global maxima and minima of $f(x,y)=x^2 + y^2 + \beta xy + x + 2y$ I am self-learning basic optimization theory and algorithms from An Introduction to Optimization by Chong and Zak. I would like someone to verify my solution to this problem, on finding the minimizer/maximizer of a function of two variables, or any tips... | Using elementary algebra, we have
$$f(x,y)=x^2+x(\beta y+1)+2y+y^2$$
Applying well known method,
$$ax^2+bx+c=a(x-m)^2+n$$
$$m=-\frac{b}{2a}, n=-\frac{\Delta}{4a}$$
In this case, we have
$$\begin{align}m:=-\frac{\beta y+1}{2}, ~n:&=\frac 14 \left(y^2(4-\beta ^2) + 2y(4-\beta)-1\right)&\end{align}$$
which gives,
$$f(x,y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the area of the square with vertex at $C$ The figure below shows $6$ squares connected edge by edge. The top three are unit squares.If $$ar(\Delta ABD)=\frac{3}{2} \times ar(\Delta ABC)$$ Find the area of the square with vertex at $C$.
My try:
I actually attacked this using Analytical geometry:
I assumed the foll... | On your solution, you have some mistakes in the sign. First note that $|b| = 3$. With $F$ as origin, and $|FG| = a$ and $|DM| = x$,
Coordinates of $C$ is $(-3 + a, -a)$ and of $D$ is $(-3 + a - x, -3 + x)$. Now using determinant for area and solving, you get $ \displaystyle a = \frac{1}{2} \ $ so side length of the squ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Here why is it wrong to differentiate both sides and put $x=2$ to find $g'(2)$? Here why is it wrong to differentiate both sides and put $x=2$ to find $g'(2)$?
If $\displaystyle I = \int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \,\mathrm dx = g(x) + c$, then $\left\lfloor \dfrac{1}{g'(2)} \right\rfloor = \cdots$,
(where $\lf... | $$
\int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \,\mathrm dx = g(x) + c
$$
$$
\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} = g'(x)
$$
$$
\frac 1 {3\sqrt{14}} = g'(2)
$$
$$
\frac 1 {g'(2)} = 3\sqrt{14}, \text{ which is between 11 and 12}
$$
$$
\left\lfloor \frac 1 {g'(2)} \right\rfloor = 11
$$
Addendum :
$$\begin{align}
11=\sqrt{121}&<\... | {
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"url": "https://math.stackexchange.com/questions/4138124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$
Prove that $$\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$$
for $a,b \in \mathbb{R}^+$. Establish when the equality holds.
My approach was to use AM-GM however trying both sides individually assert to $\geqslant 2ab.$ RHS is baby AM-GM if considered.... | Here is a trick, using only AM-GM, twice.
$$a^2+b^2\geq 2ab \Rightarrow \\
\frac{a^3}{b}+\color{red}{a^2}+\frac{b^3}{a}+\color{red}{b^2} \geq
\frac{a^3}{b}+\color{red}{ab}+\frac{b^3}{a}+\color{red}{ab}\geq\\
2\sqrt{\frac{a^3}{b}\cdot ab} + 2\sqrt{\frac{b^3}{a}\cdot ab}=2a^2+2b^2$$
Or
$$\frac{a^3}{b}+\color{blue}{a^2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4139952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Does $\sum_{n \geq 1}f_n(x)$ converges uniformly on $[0, +\infty)$? Let $f_n(x) = \frac{x^2}{1 + nx^2}\sin(\frac{nx}{x + 1})$.
Is it true that $$ \sum_{n \geq 1} \frac{x^2}{1 + nx^2}\sin(\frac{nx}{x + 1})$$
converges uniformly on $[0, +\infty)$
I want to say 'Yes' and use Abel-Dirichlet test with:
$$a_n(x) = \sin(\frac... | The series converges uniformly on $[0,\infty)$ by the Dirichlet test.
Factoring the terms of the series as
$$\sum_{n=1}^\infty \underbrace{\frac{x^2}{1+nx^2}}_{b_n(x)} \underbrace{\sin \frac{nx}{x+1}}_{a_n(x)}$$
is not helpful for applying the Dirichlet test. The requirement that $b_n(x) \to 0$ as $n \to \infty$ both... | {
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"url": "https://math.stackexchange.com/questions/4143124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove convergence of MacLaurin expansion of$ (1+x)^{1/2}$ at $|x|=1$. I know that the Maclaurin expansion of $(1+x)^{1/2}$ has a domain of convergence of $|x|\leq1$. It is easy to prove using the ratio test that this expansion is convergent when $|x|<1$, however, I don't know how to prove convergence when $|x|=1$. Does... | If you carefully write it out, you will get (for $c$ either $3/2$ or $1/2$):
$$f(\pm1)\,-\,c\,\,=\,\,-\frac{2^{-2}}{1(2)}\,\pm\,\frac{2^{-3}(3)}{1(2)(3)}\,-\frac{2^{-4}(3)(5)}{1(2)(3)(4)}\,\pm\,\frac{2^{-5}(3)(5)(7)}{1(2)(3)(4)(5)}\,-\pm\dots.$$
We want to show absolute convergence of this series for either choice $x=+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find all integers n such that $(\frac{n^3-1}{5})$ is prime
Find all integers n such that $(\frac{n^3-1}{5})$ is prime?
My Approach:
I wrote all the prime which i will get after dividing $(n^3-1)$ by $5$.
$n^3-1=10,15,25,35,55,...,215$
which lead me to $n^3=11,16,26,...,216$, then I obtained $n=6$
My doubt is that h... | First, since you need the fraction to be a prime, you have $n^3-1 = 5p$ with $p$ a prime number. Now, $n^3-1=(n-1)(n^2+n+1)$. So this leads to the idea that one of those factors is $5$ and the other $p$.
*
*If $n-1=5$, $n=6$ so $n^2+n+1= 36+6+1= 43$.
*If $n-1\neq 5$, $n^2+n+1=5$, so $n(n+1)=4$, and you can check in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Logarithmic integral $ \int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}\ \mathrm{d}x $ I found this integral weeks ago.
$$ \int_0^1 \dfrac{x\ln(x)\ln(1+x)}{1+x^2}\ \mathrm{d}x $$
I tried to solve this integral using various series representation and ended up with a complicated double series which I have asked here. How can I solv... | Let $K= \int_0^1 \frac{\ln x\ln(1-x)}{1+x} dx$ and note
\begin{align}
I&= \int_0^1 \dfrac{x\ln x\ln(1-x^2)}{1+x^2} dx
\overset{x^2\to x}
=\frac14K \\
J &= \int_0^1 \dfrac{x\ln x\ln\frac{1-x}{1+x}}{1+x^2} dx
\overset{x\to \frac{1-x}{1+x}}=\int_0^1 \dfrac{\ln x\ln\frac{1-x}{1+x}}{1+x} dx -J\\
&= \frac12K-\frac12 \int_0^1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve system of equations $3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$, $xy+yz+zx = 1$ Find all $x,y,z>0$ such that
$$3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$$ $$xy+yz+zx = 1$$
The only solution should be $x=\frac{1}{3}$, $y = \frac{1}{2}$, $z=1$.
There is a way to do it with $x = \tan \a... | Hint
Note that $xy+yz+zx=1$, then we have:
$$\frac{4}{3}=\frac{x+\frac{1}{x}}{y+\frac{1}{y}}=\frac{x+\frac{xy+yz+zx}{x}}{y+\frac{xy+yz+zx}{y}} = \frac{\frac{(x+y)(x+z)}{x}}{\frac{(y+z)(y+x)}{y}} = \frac{y(x+z)}{x(z+y)}=\frac{yz+yx}{xy+xz}$$
Do similiarly with $\frac{x+\frac{1}{x}}{z+\frac{1}{z}}$, you can imply the res... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.
The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.
Let $a,b,c$ be there positive integers. So, $a+b+c=20$
Total number of solutions would be $^{19}C_2=171$
For... | It is a bad question as they have not described how the sides have been chosen, but their answer almost reveals what they intended:
*
*There are $171$ compositions of $20$ into three positive integer parts (any order). Of these,
*
*$144$ compositions have the three parts distinct
*$27$ compositions have two of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Unique ways to write $n$ as sum of three distinct nonnegative integers up to the order of the summands How many ways are there to express a natural number, $n$, as the sum of three whole numbers, $a,b,c$, where $a,b,c$ are allowed to be 0 but are unique?
For example: $n=9$ there are only seven ways: $1+2+6, 1+3+5, 2+3+... | Suppose $S$ is the set of ordered triplets $\{x,y,z\}$ such that $x+y+z=n$. As you know the size of $S$ is $n+2 \choose 2$. There are three types of triplets in $S$.
*
*$x,y,z$ are all distinct.
*Exactly two of $x,y,z$ are the same.
*$x=y=z$.
Let $t_1,t_2,t_3$ be the number of triplets of type 1, 2 and 3 respectiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4156787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Fourier expansion of $4x(1-x)$ Im triyng expand $4x(1-x)$ in a Fourier sine series in the interval $(-1\leq x \leq 1)$
where:
$f(x) = 4x(1-x), (0\leq x \leq 1)$
$f(x) = 4x(1+x), (-1\leq x \leq 0)$
But on my calculations the coeficient $b_n$ is going to zero, where is supose to be $b_n=\frac{32}{n^3\pi^3}$ if n is odd a... | It would take forever to typeset all the IBP, but we have
\begin{align}
b_n &= \int_{-1}^0 (4x+4x^2)\sin(n\pi x)dx+\int_{0}^1 (4x-4x^2)\sin(n\pi x)dx\\
&= \int_{-1}^0 4x\sin(n\pi x)dx+\int_{-1}^0 4x^2\sin(n\pi x)dx+\int_{0}^1 4x\sin(n\pi x)dx-\int_{0}^1 4x^2\sin(n\pi x)dx\\
&= \int_{-1}^1 4x\sin(n\pi x)dx+\int_{-1}^0 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4159495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $ \int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx$
Find $\displaystyle \int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx$
I couldn't find my answer so I looked up the solution which is as follows
$$\displaystyle \begin{align}&\int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx\\&= \int \frac{2\cos x+1}{(2+\cos x)^2}\... | Using the "universal substitution":
$$\begin{align}
\int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}dx
&= \int\frac{(\tan^2(x/2) + 1)\left(2+ \frac{\tan^2(x/2) + 1}{1-\tan^2(x/2)}\right)}{(1-\tan^2(x/2))\left(1+\frac{\tan^2(x/2) + 1}{1-\tan^2(x/2)}\right)^2} \\
&= \begin{bmatrix}u = \tan(x/2) \\ du = \frac{1}{2}\sec^2(x/2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
High school integration: $-\frac{1}{2} \int \sec \theta d\theta$ Original question was: $\int \frac{1}{(x+3)^2 -4}$
Substitution: $x=2\sin \theta -3$
This is a follow up on my last thread because I cannot seem to get the answer.
Anyway, I got to:
$$-\frac{1}{2} \int \sec \theta d\theta$$
And had used the substitution $... | A much easier approach is to start by applying the difference of squares identity to the denominator, factorising and doing partial fractions. You'll get to the "intended" answer much quicker. Avoid trigonometric substitutions where possible.
So the integrand becomes $$\begin{align}\frac 1{(x+3)^2 - 2^2} &= \frac 1{(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$.
Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$.
What I Tried:- Given this expr... | First, substitute $x=m+1$ and move the constant term to the left side:
$$2x^3+16n^3-36n^2+27n=439$$
Then, note that $n$ must be odd. We can replace $n=2y+1$ and divide out by $2$ and get:
$$x^3+64y^3+24y^2+3y=216$$
Some algebraic manipulation (multiplying both sides by $8$, adding $1$ to both sides, factoring, and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\alpha,\beta$ are roots of the equation $3x^2-(m-2)x+(m-5)=0$ such that $\alpha^5+\beta^5=33$. Find the value of $m$.
$\alpha,\beta$ are roots of the equation $3x^2-(m-2)x+(m-5)=0$ such that $\alpha^5+\beta^5=33$. Find the value of $m$.
$$\alpha+\beta=\frac{m-2}3$$
Squaring and cubing it one by one and then multiply... | Observe,
$$\alpha+\beta=\frac{m-2}{3}\;\;\;\text{and}\;\;\; \alpha\beta=\frac{m-5}{3}\implies \alpha+\beta=\alpha\beta+1\implies (\alpha-1)(\beta-1)=0$$
Therefore, either $\alpha=1$ or $\beta=1$. Since $\alpha^5+\beta^5=33,$ the roots of the quadratic must be $1$ and $2$. Plugging this into one of the previous equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that $u-v=\sqrt5$ given that $u-v>0$ given that $u=b+b^4$ and $v=b^2+b^3$ We're also given that $b$ is a root of $z^5-1=0$ $b^4+b^3+b^2+b+1=0$
If $u=b+b^4$ and $v=b^2+b^3$, show that
i) $u+v=uv=-1$
ii) $u-v=\sqrt5$ given that $u-v>0$
I managed to do part i):
Plugging in $u$ and $v$: $(b+b^4)+(b^2+b^3)=-1$ (using $... | One approach: the quadratic equation $(x-u)(x-v) = 0$ which has roots $u$ and $v$ expands to $x^2 - (u+v)x + uv = 0$. Once you've computed $u+v = uv = -1$, you know that this is the quadratic equation $x^2 + x - 1 = 0$, which you can just solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Inconsistent solution for the factorized equation When I try to solve this equation as follows:
\begin{aligned}-k_1^2 \, \cos^2\theta \, - 2 \left( k_1k_5 + k_4 \right) \cos\theta+k_4^2-k_5^2+1 \geqslant 0 \end{aligned}
Noting that $k_1$, $k_4$, $k_5$ are real constants. I assumed:
$$
\left\{ \matrix{
x=\cos\theta ... | [A Quadratic equation][1]
The values of x that satisfy the equation are called solutions of the equation, and roots or zeros of the expression on its left-hand side. A quadratic equation has at most two solutions. If there is no real solution, there are two complex solutions. If there is only one solution, one says tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$? If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$ ?
$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)5$
First I tried plugging in some values for $\theta$ like $0,\frac{\pi}4,\frac{\pi}3,...$ but neithe... | Alternative:
We have $$2\sin\theta=\sqrt3 -\cos \theta$$
Squaring, $$4-4\cos^2 \theta=3+\cos^2\theta-2\sqrt 3 \cos\theta$$
So, $$5\cos^2 \theta-2\sqrt 3 \cos\theta-1=0$$
Or $$\cos \theta=\frac {2\sqrt 3 \pm 4\sqrt 2}{10}$$
Since further conditions haven't been given, both values would be valid, and from here $\tan \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4177501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$.
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.
My a... | Let $u=x^2+x+1,$ $v=x^2-x+1$ so that $uv=x^4+x^2+1.$ Let $P$ be the given polynomial.
$$P=uq_1-x+1\tag1$$
$$P=vq_2+3x+5\tag2$$
$$P=uvq_3+R\tag3$$
There are polynomials $\lambda$ and $\mu$ such that
$$\lambda u+\mu v=1\tag4$$
Suitable values are $\lambda={-x+1\over2}$ and $\mu= {x+1\over2} .$
By (1), (2) and (4),
$$P=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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How many real solutions does this equation have ? $4^x\times9^\frac{1}{x} + 9^x\times4^\frac{1}{x} + 6^{x+\frac{1}{x}} = 108$ How many real solutions does this equation have ?
$4^x\times9^\frac{1}{x} + 9^x\times4^\frac{1}{x} + 6^{x+\frac{1}{x}} = 108$
I am pretty sure it has something to do with prime factorization of ... | You obtained the form $$a^2+b^2+ab=108$$
Dividing both sides with $ab$, we obtain the following equation:
$$(\frac 23)^{x-\frac 1x}+\frac {1}{(\frac 23)^{x-\frac 1x}}+1=\frac {108}{6^{x+\frac 1x}}$$
Note that, from the original equation, it can be easily shown that there can never be any negative solutions, since all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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An elementary way to show that the determinant is non zero
Show that the determinant of the matrix
\begin{pmatrix} a && -c && -b \\
b && a - 2c && -c -2b \\ c && b && a -2c
\end{pmatrix} is non zero for all integers $a,b,c$ where $abc \ne 0$
There is an interesting way to do this by using integral domains. It is e... | This might be equivalent to what you did, but I'll go ahead and write this as an answer anyway.
Let $X = \begin{bmatrix}0 & 0 & -1 \\ 1 & 0 & -2 \\ 0 & 1 & 0\end{bmatrix}$. Then, $X^2 = \begin{bmatrix}0 & -1 & 0 \\ 0 & -2 & -1 \\ 1 & 0 & -2\end{bmatrix}$, and so, $$Y := \begin{bmatrix}a & -c & -b \\ b & a - 2c & -c -2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$ Let $k>0~$ be fixed. Find
$$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$$
over all $c, b, a\geq0~\text{satisfying}~ab+bc+ac=a+b+c>0~.$
==================
In case we have 2 positive numbers $a+b=ab$ results that $a, b > 1$ and $a= \dfrac{b}{b-1}$,
so the mi... | Solution for $k\ge 24$:
Let $k\ge 24$ be a real number (fixed).
Let $f(a, b, c) = \sqrt{ka + 1} + \sqrt{kb + 1} + \sqrt{kc + 1}$.
Let us prove that $f(a, b, c) \ge f(2, 2, 0) = 2\sqrt{2k + 1} + 1$
for all $a, b, c\ge 0$ with $ab + bc + ca = a + b + c > 0$.
WLOG, assume that $c = \min(a, b, c)$.
From $ab + bc + ca = a +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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$x^4+6x^2+1$ is reducible modulo $p$ for every prime $p$. Let $f(x)=x^4+6x^2+1 \in \mathbb{Z}[X]$. We have to prove that $f$ is reducible modulo $p$ for all prime $p$.
I am not sure how to proceed. Please give me some hints so that I can try solving the problem.
Edit I:
I have tried with the following primes.
*
*For... | Here is a bit of a different approach.
For $p=2$, we can factor
$$
\begin{align}
x^4+6x^2+1
&\equiv(x-1)^4&\pmod2
\end{align}
$$
For the odd primes, where $p\in\{1,3,5,7\}\pmod8$, we will use the results from this answer, to show that $-1$ is a quadratic residue mod $p$ iff $p\equiv1\pmod4$, and this answer, to show th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Alternative proofs of convergence of geometric series The usual proof for the convergence of a geometric series of ratio $C: |C|\in [0,1)$ makes use of the formula $$\sum_{0\leq k \leq n} C^k = \frac{1-C^{n+1}}{1-C}.$$
I'm looking for alternative ways to prove it. The motivation for this is that, if someone who never s... | If $c = \dfrac1{1+b}$ where $b > 0$
then,
since,
for $k \ge 2$,
$(1+b)^k
\ge 1+bk+b^2k(k-1)/2
\gt b^2k(k-1)/2$
(readily proved by induction),
so
$c^k
= \dfrac1{(1+b)^k}
\le \dfrac{2}{b^2k(k-1)}
$
so
$\begin{array}\\
\sum_{k=2}^n c^k
&\le \sum_{k=2}^n\dfrac{2}{b^2k(k-1)}\\
&= \dfrac{2}{b^2}\sum_{k=2}^n\dfrac1{k(k-1)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $ab... | I think your answer is correct. Assume that $a>2$, since $bc\geq b+c$, you get $$abc\geq ab+ac\geq a+b + a +c>a+b+c+2.$$
All $a,b,c$ must be less than or equal to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
Find $a^2 + b^2 + c^2 + d^2$.
$$\begin{align}\dfrac{a^2}{2^2 - 1^2} + \dfrac{b^2}{4^2 - 1^2} + \dfrac{c^2}{6^2 - 1^2} + \dfrac{d^2}{8^2 - 1^2} &= \dfrac{a^2}{2^2 - 3^2} + \dfrac{b^2}{4^2 - 3^2} + \dfrac{c^2}{6^2 - 3^2} + \dfrac{d^2}{8^2 - 3^2}
\\&= \dfrac{a^2}{2^2 - 5^2} + \dfrac{b^2}{4^2 - 5^2} + \dfrac{c^2}{6^2 - 5^... | We write down the following polynomial:
\begin{eqnarray}
f(x) &=& (x - 2^2)(x - 4^2)(x - 6^2)(x - 8^2)\\
&+& a^2(x - 4^2)(x - 6^2)(x - 8^2)\\
&+& b^2(x - 2^2)(x - 6^2)(x - 8^2)\\
&+& c^2(x - 2^2)(x - 4^2)(x - 8^2)\\
&+& d^2(x - 2^2)(x - 4^2)(x - 6^2).
\end{eqnarray}
The given conditions then imply that $f(1^2) = f(3^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Constructing a polynomial with integer coefficient's sharing roots with another two polynomials
Let $i$ be a root of the equation $ x
^2 + 1 = 0 $and let $ω$ be a root of the
equation $x^2 + x + 1 = 0$. Construct a polynomial $f(x) = \sum_k a_k x^k$ with $a_k \in \mathbb{Z}$ such that $f(i+w)=0$. Source
After some l... | Multiplying $\,\alpha=\omega+i\,$ by $\,i\,$ and using that $\,i^2=-1\,$:
$$
\begin{align}
\begin{cases}
\alpha &= \omega+i
\\ i \,\alpha &= i \omega - 1
\end{cases}
\end{align}
$$
Eliminating $\,i\,$ between the two equations, for example by taking $\,i=\alpha-\omega\,$ from the first equation and substituting into th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
For $|x|\gt1,\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac2{x^{2^k}+x^{-2^k}}\right)=f(x)$
For $|x|\gt1,$ $$\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac2{x^{2^k}+x^{-2^k}}\right)=f(x)$$
Show that
*
*A) $\int_2^5f(x)dx=3+\ln16$
*B) $\lim_{x\to\infty}f(x)=1$
*C) $f(x)=0$ has one solution
*D) $f(x)$ is decrea... | Let $P$ denote the product, without the limit.
Note that:
$$\left(1+\frac {2}{x^{2^k}+\frac {1}{x^{2^k}}}\right)=\frac {\left(x^{2^{k-1}}+\frac {1}{x^{2^{k-1}}}\right)^2}{x^{2^k}+\frac {1}{x^{2^k}}}$$
Using this leads to a partial telescopic product. I shall skip the cancellation steps, and write directly the result as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the longest side of a triangle when the sides are in a sequence with a common difference as 2 The Question:
Sides of a triangle form an arithmetic sequence with a common difference of 2. The area of triangle is $24 cm^2$. If the length of the greatest side is $P$, find the length of $P$.
My thoughts on this stat... | Building on Jaap and Math Lover's comments
Your solution is perfectly correct but it is advisable not to try this in olympiads as you have limited time over here and you can't devote this much time to one comment otherwise you won't be able to do the other questions. Instead let $x-2,x,x+2$ be the sides of the triangle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\alpha\ne1,\alpha^6=1$ and $\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x,$ then find the value of $|x|$. The following question is taken from the practice set of JEE exam.
If $\alpha\ne1,\alpha^6=1$ and $\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x,$ then find the value of $|x|$.
$$\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x\\\implies {^6}C_... | I'll do the case where $\alpha$ is a primitive $6$-th root of unity.
Then $\alpha$ satisfies $\alpha^2-\alpha+1=0$,$\;$hence
\begin{align*}
x&=\sum_{i=1}^6 {\small{\binom{6}{i}}}\alpha^{i-1}
\\[4pt]
&=
\alpha^5+6\alpha^4+15\alpha^3+20\alpha^2+15\alpha+6
\\[4pt]
&=
(\alpha^3+7\alpha^2+21\alpha+34)(\alpha^2-\alpha+1)+(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Upper limit for $(1+1/2)(1+1/3)\dots(1+1/n)$ I am trying to find an upper limit for the product $(1+1/2)(1+1/3)\dots(1+1/n)$. I tried using AM-GM inequality as follows:
\begin{align}
(1+1/2)(1+1/3)\dots(1+1/n) \leq \left(\frac{1}{n-1}\left( -1 + n + 1/2 + 1/3 + 1/4 + \dots + 1/n\right)\right)^{n-1}
\end{align}
From Wik... | The product is equal to
$$\prod_{i=2}^n \left(1+\frac{1}{i}\right)=\prod_{i=2}^n \frac{i+1}{i}=\frac{n+1}{2}$$
:D
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given $3$ points on a unit circle, figure out something about them. Question: Given three points $(a, b), (c, d)$ and $(x, y)$ on the unit circle in a rectangular coordinate plane, find the maximum possible value of the expression $(ax + by - c)^2 + (bx - ay + d)^2 + (cx + dy + a)^2 + (dx - cy - b)^2. $
Answer: We will... | Here is another way I would approach it. Given all the three points are on the unit circle, their coordinates can be written as,
$(a, b) \mapsto (\cos \theta_1, \sin \theta_1); \ (c, d) \mapsto (\cos\theta_2, \sin\theta_2)$;
$(x, y) \mapsto (\cos\theta_3, \sin\theta_3)$
So, $ \ (ax + by - c)^2 + (bx - ay + d)^2 + (cx +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4198297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate surface of a triangle It's quite a simple question solvable with many methods but the $2$ of them I picked don't agree.
First of I have chosen the triangle with corners: $(3,0,0) \quad (0,3,0) \quad (0,0,6)$
Thus, this one here:
First Method is by evaluating the surface Integral: $A = \int_{S}1\,\mathrm{dS} ... | Given $V=6$ and $S=3$
The length of the triangle base is base is
$B=\sqrt{S^2+S^2}=\sqrt{3^2+3^2}=3\sqrt{2}. \quad$
The distance from the back corner to the base is
$Z=\sqrt{S^2-\bigg(\dfrac{B}{2}\bigg)^2}=\sqrt{3^2-\bigg(\dfrac{3\sqrt{2}}{2}\bigg)^2}
=\dfrac{3\sqrt{2}}{2}.\quad$
The altitude of the triangle is
$H=\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
A non-inductive proof regarding a property of natural numbers Here is the question:
For each natural number $n$, prove that
$$n(n + 1)(n +2) (n+3)$$
is never a perfect square.
My approach: Here,we notice that two of the factors given in the product are bound to be odd while
the remaining two are bound to be even. WLO... | $$(n^2+3n)^2<n(n+1)(n+2)(n+3)=n(n+3)(n+1)(n+2)=(n^2+3n)(n^2+3n+2)=(n^2+3n)^2+2(n^2+3n)<(n^2+3n)^2+2(n^2+3n)+1=(n^2+3n+1)^2$$
$n(n+1)(n+2)(n+3)$ is between two consecutive perfect squares, it is not a perfect square
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find segment ratio when three points form three lines dividing area of equilateral triangle evenly $\triangle{ABC}$ is equilateral. Points $E,F,G$ are inside it and form four triangles so that $S_1=S_2=S_3=S_4$. Find $\dfrac{AF}{FG}$.
I saw this math problem on social media but did not see a strict solution. I had som... | Join G to C , we have:
$\frac{A_{GFC}}{A_{AFC}}=\frac{FG}{FA}=\frac{EG}{GB}$
$A_{GFC}=A_{GFE}+A_{GEC}$
Putting this in first relation we get:
$\frac{FG}{FA}=\frac{EG}{GB}=1+\frac{EG}{EB=EG+GB}=1+\frac{FG}{GA=FA+FG}$
Rearranging we get:
$\frac{FG}{FA}-1=\frac{FG}{FA+FG}$
Or: $FG^2-FA^2=FA\times FG$
Let $FG=x$ and $FA=y$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Comparing $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the values
I want to compare $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the actual value of square roots.
I tried to apply square on both side but it still carries the root terms.
Any trick I was missing here?
| Note that $8>\sqrt{11}$. Therefore\begin{align}\sqrt5+\sqrt6+\sqrt{11}>8&\iff\sqrt5+\sqrt6>8-\sqrt{11}\text{ ($3$ square roots)}\\&\iff11+2\sqrt{30}>75-16\sqrt{11}\text{ (only $2$ square roots)}\\&\iff\sqrt{30}>32-8\sqrt{11}\\&\iff30>1728-512\sqrt{11}\text{ (a single square root)}\\&\iff512\sqrt{11}>1698\\&\iff256\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Integral of exponential within a region Are there methods to compute the following integral for $a \leq b$? Here $x\in\mathbb{R}$
$$
\int\limits_{a \leq -\frac{x^2}{2} \leq b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\,dx
$$
Substitution
The error function is
$$
\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} dt
... | We can use $\operatorname{erf}$ in place of $\Phi$. By definition,
$$
\frac{d}{dx}\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}e^{-x^2}
\tag1$$
Now start with
$$
V = \int\limits_{a \leq -\frac{x^2}{2} \leq b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\,dx
$$
where $a < b \le 0$.
Divide into two equal parts, $x>0$ and $x<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to calculate $\mathbb{E}\left(\varPhi^{2}\left(\frac{W_{s}}{\sqrt{1-s}}\right)\right)$? How can we calculate $$\mathbb{E}\left(\varPhi^{2}\left(\frac{W_{s}}{\sqrt{1-s}}\right)\right)$$ where $W$ denotes a Wiener process, and $\varPhi$ is the CDF of the standard normal random variable... If $\varphi$ denotes the PDF... | Let $$M_t = \Phi \left( \frac{W_t}{\sqrt{1-t}} \right) = g(t, W_t)$$ where $g(t,x) =\Phi \left( \frac{x}{\sqrt{1-t}} \right)$. We claim that $M_t$ is a martingale. First, observe that $\Phi^\prime (s) =\frac{1}{\sqrt{2 \pi}} \exp \left( \frac{s^2}{2} \right)$ and $\Phi^{\prime \prime}(s) = - \frac{s}{\sqrt{2 \pi}}\exp ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$f(z)$ is holomorphic on $|z| < 1$, with $|f(z)| ≤ 3$ and $f(\frac{1}{2}) = 2$. Show that $f(z) \neq 0$ when $|z| < \frac{1}{8}.$ This is an old Schwarz lemma problem from the August 2020 UMD qualifying exam for analysis, which is posted here. The precise wording from the test is:
Suppose $f(z)$ is a holomorphic functi... | We know that the hyperbolic unit disk metric, $d(z, w) := |\frac{z - w}{1 - \bar{w}z}|$, satisfies $d(g(z), g(w)) \le d(z, w)$ for analytic maps $g : \mathbb{D} \to \mathbb{D}$. Note that $|\frac{f(z)}{3}| \le 1$ when $|z| < 1$, and the maximum principle implies we can assume $|\frac{f(z)}{3}| < 1$, otherwise $|\frac{f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Proving that the reduction formula for $I_{m,n}=\int\frac{x^m}{(a^2-x^2)^n}dx$ is $I_{m,n}=a^2I_{m-2,n}-I_{m-2,n-1}$
So I'm trying to prove the reduction formula for
$$I_{m,n}=\int\frac{x^m}{(a^2-x^2)^n}dx$$
which is listed as
$$I_{m,n}=a^2I_{m-2,n}-I_{m-2,n-1}$$
I tried integrating by parts by taking $u=(a^2-x^2)^{-... | Integrate both sides of the equation below
\begin{align}
\frac{x^m}{(a^2-x^2)^n}
= \frac{x^{m-2}(a^2+x^2-a^2)}{(a^2-x^2)^n}
= \frac{a^2 x^{m-2}}{(a^2-x^2)^{n}}- \frac{x^{m-2}}{(a^2-x^2)^{n-1}}
\end{align}
to arrive at
$$I_{m,n}=a^2I_{m-2,n}\> \>-I_{m-2,n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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