Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the number of real solutions to the system of equations $x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$ My approach is naive:
Given
$x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+... | From the system of equations, you can infer that $x,y,z \ge 0$ and consider the function $f(a) = \dfrac{2a^2}{1+a^2}=2 - \dfrac{2}{1+a^2}$. Thus if $a > b>0 \implies a^2>b^2\implies 1+a^2>1+b^2\implies \dfrac{2}{1+a^2} < \dfrac{2}{1+b^2}\implies -\dfrac{2}{1+a^2} > -\dfrac{2}{1+b^2}\implies 2-\dfrac{2}{1+a^2} > 2-\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$ Question: Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$
Attempt: When I tried solving this problem, I paired up $1$ with $\frac{1}{2018^3}$, $\frac{1}{2^3}$ with $\frac{1}{2017^3}$ etc. but it wasn't useful.
| Another method that is generally applicable:
$\zeta(2n+1) = 1 + \frac{1}{2^{2n+1}} + \frac{1}{3^{2n+1}} + \frac{1}{4^{2n+1}} + \ldots$
$< 1 + \frac{1}{2^{2n+1}} + \frac{1}{3^{2n+1}} + \left(\frac{1}{4^{2n+1}} + \frac{1}{4^{2n+1}} + \frac{1}{4^{2n+1}} + \frac{1}{4^{2n+1}}\right) + \ldots $
$= 1 + \frac{1}{2^{2n+1}} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I need the steps to solving this limit without using l´Hopital rule I've tried many ways of solving this limit without using l'Hopital and I just can't figure it out. I know the answer is $3/2 \sin (2a).$
$$\lim_{x\,\to\,0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x$$
Thank you!
| \begin{align*}
&\dfrac{\sin(a+x)\sin(a+2x)-\sin^{2}a}{x}\\
&=2\cdot\dfrac{\sin(a+x)[\sin(a+2x)-\sin a]}{2x}+\dfrac{(\sin a)[\sin(a+x)-\sin a]}{x}\\
&\rightarrow 2\sin(a+0)(\sin x)'\bigg|_{x=a}+(\sin a)(\sin x)'\bigg|_{x=a}\\
&=2(\sin a)(\cos a)+(\sin a)(\cos a).
\end{align*}
The true without L'Hopital:
\begin{align*}
&... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve system of equations: $2x^2-4y^2-\frac{3}{2}x+y=0 \land 3x^2-6y^2-2x+2y=\frac{1}{2}$ $$
\begin{cases}
2x^2-4y^2-\frac{3}{2}x+y=0 \\
3x^2-6y^2-2x+2y=\frac{1}{2}
\end{cases}
$$
I multiplied the first with $-6$ and the second with $4$ and get two easier equations:
$9x-6y=0 \land -8x+8y=2 $ and out of them I get th... | It looks like you took the two scaled equations $$-12x^2+24y^2+9x-6y = 0 \\ 12x^2-24y^2-8x+8y=2$$ and then simply lopped off the common quadratic parts to produce two linear equations that are entirely unrelated to the original system. Once you’ve scaled the two equations so that the coefficients of their squared terms... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$
Evaluate$$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$$
My attempt:
$$1+2\cos 2\theta= 1+2(1-2\sin^2\theta)=3-4\sin^2\theta$$
$$=\frac{3\sin \theta-4\sin^3\theta}{\sin \theta}=\frac{\sin 3\theta}{\sin \thet... | It's unclear if you are asking about $$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$$ or about $$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi k\cdot 3^k}{3^{100}+1}\right]$$
I will assume it is the former. At the moment, that is in the body of your question, while the latter is in the title. If i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why can't I cancel $2x-3$ from $(2x-3)(x+5)=9(2x-3)$? Why are these simplifications wrong?
$$\begin{align}
(2x-3)(x+5)=9(2x-3) &\quad\to\quad \frac{(2x-3)(x+5)}{2x-3} = \frac{9(2x-3)}{2x-3} \quad\to\quad x+5 = 9\\[4pt]
x(x+2)=x(-x+3) &\quad\to\quad \frac{x(x+2)}{x} = \frac{x(-x+3)}{x} \quad\to\quad x+2=-x+3
\end{align}... | $$(2x-3)(x+5)=9(2x-3).$$ Since you are multiplying both sides by the same constant, namely $2x-3$, then they cancel out, thus bringing forward the following equation: $$x+5=9.\tag1$$
$$x(x+2)=x(-x+3).$$ In the same fashion as before, the $x$'s cancel out, i.e. $$x+2=-x+3.$$ Now, adding $x$ to both sides yeilds the fol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the maximum value of $y\cdot{x^2}+x$ if $y^2+x^2+x+y=100$. We know that $y^2+x^2+x+y=100$. Find the maximum value of $$y\cdot x^2+x$$
I tried to simplify it and use inequalities but I failed. Is there a way to solve it without calculus?
| You Can reduce your Problem in two variables in to one: from $$y^2+y-100+x^2+x=0$$ follows $$y=-\frac{1}{2}\pm\sqrt{\frac{1}{4}+100-x^2-x}$$
so you will get
$$h(x)=\left(-1/2\pm\sqrt{\frac{1}{4}+100-x^2-x}\right)x^2+x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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On the series $\sum \limits_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )$ I managed to prove through complex analysis that
$$\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right ) = 1 -2 \log 2$$
However, I'm having a difficult time proving this result with r... | In the same spirit as achille hui,
$$S_p=\sum_{n=1}^p \left(\frac1n - \frac{1}{2n-1} - \frac{1}{2n+1}\right)$$
$$(2p+1)S_p=2 \gamma p+2 p-2 p \psi ^{(0)}\left(p+\frac{1}{2}\right)+2 p \psi ^{(0)}(p+1)+2 p
\psi ^{(0)}\left(\frac{1}{2}\right)-\psi ^{(0)}\left(p+\frac{1}{2}\right)+\psi
^{(0)}(p+1)+\gamma +\psi ^{(0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Finding points on a line that are closest
Find the points that give the shortest distance between the lines$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2-t\\-1+2t\\-1+t\end{pmatrix}\\\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}5+3s\\0\\2-s\end{pmatrix}$$
So I subtracted the second line from the first t... | $$\begin{pmatrix}-t-3s-3\\ -1+2t\\ t+s-3\end{pmatrix}\cdot \begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}=0\implies t+3s+3-2+4t+t+s-3 = 6t+4s=2$$
$$\begin{pmatrix}-t-3s-3\\ -1+2t\\ t+s-3\end{pmatrix}\cdot \begin{pmatrix}3\\ 0\\ -1\end{pmatrix}=0\implies -3t-9s+9-t-s+3=-4t-10s=-12$$
Sove above equations for $t$ and $s$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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When $2$ $6$-sided die are rolled, what is the probability that the first is odd and the difference is $0$? The probability the the first die is odd is $\frac{3}{6}=\frac{1}{2}$.
The probability that first is odd and diff is $0$ is $\frac{3}{6}\cdot \frac{1}{6} = \frac{3}{36} = \frac{1}{12}$.
Is $\frac{1}{12}$ correct?... | Set $X_1$: outcome of the first die. There are three successful outcomes: $X_1 = 1, X_1 = 3, X_1 = 5$; since the probabilities are uniform, $P(X_1= \{1,3,5\}) = \frac{1}{6}$. 'Difference equal to $0$' means the second outcome is the same; hence, we need
$P(X_2 \cap X_1) = P(X_2 = 1|X_1=1)P(X_1=1) + P(X_2 = 3|X_1=3)P(X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding an invertible matrix P and some matrix C Find an invertible matrix $P$ and a matrix $C$ of the form
$C=\begin{pmatrix}a & -b\\b & a\end{pmatrix}$
such that the given matrix $A$ has the form $A = PCP^{-1}$
$A=\begin{pmatrix}5 & -2\\1 & 3\end{pmatrix}$
The first thing i tried to do was to find the eigenvectors o... | To find $P$ and $C$, note that
$$A = PCP^{-1}\iff AP=PC$$
since A and C are similar we have that
*
*$Tr(A)=Tr(C) \implies 2a=8 \implies a=4$
*$\det(A)=\det(C) \implies a^2+b^2=17 \implies b=\pm1$
then let $P=[v_1\, v_2]$ and we have
*
*$Av_1=av_1+bv_2$
*$Av_2=-bv_1+av_2$
and with $v_1=(x,y)\quad v_2=(z,w)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Inconsistency for solving $x' = x^{1/2}$ The proposed system $x' = x^{1/2}$ can be solved easily to obtain $x(t) = \frac{1}{4} (t^2 + t c + c^2)$, where $c$ is the integration constant.
However, differentiate the newly-found $x(t)$, one gets: $x(t)' = \frac{1}{2}t+\frac{1}{4}c$. This implies that $x^{1/2} = \frac{1}{2}... | You can do it as follows:
$\frac{dx}{dt}$ = $\sqrt{x}$. Rewrite this as follows:
$\frac{dx}{\sqrt{x}}$ = dt.
Let u = $\sqrt{x}$, then du = $\frac{dx}{2\sqrt{x}}$. After this, you'll get 2du = $\frac{dx}{\sqrt{x}}$ = dt, and an indefinite integration gives u = $\frac{1}{2}$(t + c) = $\sqrt{x}$, with c $\in$ $\mathbb{R}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\lim_{x\rightarrow \frac{\pi }{3}}\frac {\sin x-\sqrt {3}\cos x}{\sin 3x}$ has two different values? Given a limit like:
$$\lim_{x\rightarrow \frac{\pi }{3}}\frac {\sin x-\sqrt {3}\cos x}{\sin 3x}$$
How did I solve it:
$$\begin{align}\lim_{x\rightarrow \frac{\pi }{3}}\frac {-2 (-\frac {1}{2}\sin x+\frac {\sqrt {3}}{2}... |
Can anyone show me if there's any error above?
Your mistake:
$$
\sin 3 (\frac{\pi}{3}-t)=\color{red}{-}\sin 3t
$$ instead of
$$
\sin\left[ 3 \left(\frac{\pi}{3}-t\right)\right]=\sin 3t.
$$ (one may recall that $\cos (3\pi)=-1$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How to find the equation of the tangent to the parabola $y = x^2$ at the point (-2, 4)? This question is from George Simmons' Calc with Analytic Geometry. This is how I solved it, but I can't find the two points that satisfy this equation:
$$
\begin{align}
\text{At Point P(-2,4):} \hspace{30pt} y &= x^2 \\
\frac{dy}{d... | The derivative of the function $f(x)$ is the slope of the tangent line at that particular value of $x$. Meaning when you differentiate $f(x)=x^2$, the tangent line at that value is at $x=-2$ so$$f'(2)=-4$$Therefore you're tangent line is actually
$$y-4=-4(x+2)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how to compute Z^n? assuming $Z$ is a complex number i.e $a + bi$. we have
$Z^2 = (a + bi)^2 = (a^2-b^2)+(2ab)i$
$Z^3 = (a+bi)^3 = (a^3-3ab^2) - (b^3-3a^2b)i$
but what would be the general formula for computing $Z^n$ where n is any real number?
I wanted to use this method but i dont know how the computation goes for co... | If, when you say "I want to use this method", you mean that you really want to use the binomial expansion, rather than using polar form, then $(a+ ib)^n$ will be a sum of monomials of the form $\frac{a^(n-i)(ib)^i}{i!}$. "i" has "period 4". That is $i^1= i$, $i^2= -1$, $i^3= -1$, and $i^4= 1$ so that further powers a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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if $d\mid n$ then $x^d-1\mid x^n-1$ proof How would you show that if $d\mid n$ then $x^d-1\mid x^n-1$ ?
My attempt :
$dq=n$ for some $q$. $$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$ in fact, $$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$
By multiplying both sid... | Let $\alpha$ be a solution to the equation $x^d = 1$. We then have that $\alpha^d$ = 1. Since $d|n$ we can write $n = d\cdot k$ for some integer $k$. Thus $$1 = 1^k = \left( \alpha^d \right)^k = \alpha^{d\cdot k} = \alpha^n$$
This shows that $\alpha$ is a solution to $x^n = 1$.
This shows that $x^d-1|x^n-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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A series of Wallis-like infinite products of fractions Would like to derive an analytic solution to the following two infinite products:
$$P_1=\frac{4}{5}\frac{8}{7}\frac{10}{11}\frac{14}{13}\frac{16}{17}...$$
$$=\prod_{n=1}^\infty \frac{3+6n+(-1)^n} {3+6n-(-1)^n}$$
and
$$P_2=\frac{8}{7}\frac{13}{14}\frac{20}{19}\frac{... | These products can be computed through standard Weierstrass products. For instance
$$ \prod_{n=1}^{2N}\frac{3+6n+(-1)^n}{3+6n-(-1)^n} = \prod_{n=1}^{N}\frac{4+12n}{2+12n}\cdot\frac{-4+12n}{-2+12n}=\frac{\sqrt{3}\,\Gamma\left(N+\tfrac{2}{3}\right)\,\Gamma\left(N+\tfrac{4}{3}\right)}{2\,\Gamma\left(N+\tfrac{5}{6}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748023",
"timestamp": "2023-03-29T00:00:00",
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Minimum length of the hypotenuse
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Prove that the minimum length of the hypotenuse is $(a^{2/3}+b^{2/3})^{3/2}$.
My Attempt
$\frac{x}{y}=\frac{a}{CM}=\frac{AN}{b}$
$$
\frac{x}{y}=\frac{AN}{b}\implies y=\frac{xb}{\sqrt{x^2-a^... | Following your solution from this point:
$$(x^2-a^2)^{3/2}=a^2b \Rightarrow x^2=(a^2b)^{2/3}+a^2 \Rightarrow x=a^{2/3}(a^{2/3}+b^{2/3})^{1/2}.$$
So:
$$y=\frac{xb}{\sqrt{x^2-a^2}}=\frac{a^{2/3}(a^{2/3}+b^{2/3})^{1/2}b}{\sqrt{((a^2b)^{2/3}+a^2)-a^2}}=(a^{2/3}+b^{2/3})^{1/2}b^{2/3}.$$
Hence:
$$x+y=a^{2/3}(a^{2/3}+b^{2/3})... | {
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"url": "https://math.stackexchange.com/questions/2748447",
"timestamp": "2023-03-29T00:00:00",
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If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$ If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$
It is very long to direct differentiate it.Can someone help me?
| Denote $a=\sin \alpha$ and $b=\cos \alpha$. Then:
$$xy=(x+y)(xa^4+yb^4) \qquad (1) \Rightarrow \\
xy=x^2a^4+y^2b^4+xy(b^4+a^4) \Rightarrow \\
xy=x^2a^2(1-b^2)+y^2b^2(1-a^2)+xy(1-2a^2b^2) \overbrace{\Rightarrow}^{:a^2b^2} \\
(x+y)^2-\frac{x^2}{b^2}-\frac{y^2}{a^2}=0 \Rightarrow F(x,y)=0.$$
Take the derivative:
$$y'=-\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the length and width of rectangle when you are given the area The area of a rectangle is $x^2 + 4x - 12$. What is the length and width of the rectangle?
The solution says the main idea is to factor $x^2 + 4x -12$.
So, since $-12 = -2 \times 6$ and $-2 + 6 = 4$, it can be written as $x^2 + 4x - 12 = (x - 2)(x + 6)... | The length is $(x+6)$ and the width is $(x-2)$.
The area is $(x+6)(x-2)=x^2+4x-12$
You may take this answer as the length is 6 units longer and the width is 2 units shorter than a give number $x$
For example, given $x=10$, you may get that the area is $10^2+4\times10-12=128$, or, the area is $(10+6)(10-2)=16\times8=128... | {
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Show that the two straight lines $x^2(\tan^2 (\theta)+\cos^2 (\theta))-2xy\tan (\theta)+y^2.\sin^2 (\theta)=0$ Show that the two straight lines $x^2(\tan^2 (\theta)+\cos^2 (\theta))-2xy\tan (\theta)+y^2.\sin^2 (\theta)=0$ make with x axis such that the difference of their tangents is $2$.
My Attempt:
$$x^2(\tan^2 (\the... | We have $$\left(\dfrac yx\right)^2\sin^2\theta-\dfrac yx(2\tan\theta)+\tan^2\theta+\cos^2\theta=0$$
If $m_1,m_2$ are the two roots
$m_1m_2=\dfrac{\tan^2\theta+\cos^2\theta}{\sin^2\theta}=\dfrac{s^2+c^4}{c^2s^2}$
$m_1+m_2=\dfrac{2s}{cs^2}=\dfrac2{cs}$
$$(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2=4\cdot\dfrac{1-(s^2+c^4)}{c^2s^2}=\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Nonhomogenous variable coefficient How would you solve this ODE:
$$(1 + x^{2})y’’ + 4xy’ + 2y = 1/(1+x^{2})$$
I have the answer for homogenous part of this ODE but do not know how to do this, variation of parameter is a mess at integral step when applying to this
Any help would be appreciated. Thank you very much :) ... | Using variation of parameters is actually not that messy.
Dividing by $(1+x^2)$, we get
$$ y'' + \frac{4x}{x^2+1}y' + \frac{2}{x^2+1} y = \frac{1}{(x^2+1)^2} $$
which has inhomogeneity of $g(x) = \frac{1}{(x^2+1)^2}$. The solutions to the homogeneous equation are $$y_1 = \frac{1}{x^2+1} \ \text{ and } \ y_2 = \frac{x}{... | {
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Show that $x^3-a$ is irreducible in $\mathbb{Z}_7$ unless $a=0$ or $\pm1$ Show that $x^3-a$ is irreducible in $\mathbb{Z}_7$ unless $a=0$ or $\pm1$
My Idea:
Suppose $x^3-a=(Ax+b)(Bx^2+cx+d).$
Then $A=B=1$ or $A=B=-1$ WLOG $A=B=1.$
Then $x^3-a=x^3+x^2(b+c)+x(bc+d)+bd\\
\Rightarrow c+b=0,d+bc=0,bd=-a.$
I can't go furthe... | It's a cubic, so you need not work so hard. Indeed, you can just check if it has a root.
By exhaustion: $x^3-a\cong 0\implies x^3\cong a$, so
$$
0^3=0=a\\
1^3=1=a\\
2^3=1=a\\
3^3=-1=a\\
(-3)^3=1=a\\
(-2)^3=-1=a\\
(-1)^3=-1=a
$$
where equality is in the the mod $7$ sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $n$ is prime. (Fermat's little theorem probably)
Let $x$ and $n$ be positive integers such that $1+x+x^2\dots x^{n-1}$ is prime. Prove $n$ is prime
My attempt:
Say the above summation equal to $p$ $$1+x+x^2\dots x^{n-1}\equiv 0\text{(mod p)}\\
{x^n-1\over x-1}\equiv0\\
\implies x^n\equiv1\text{ (as $p$ can't di... | This does not involve any Fermat's little theorem type tricks. Indeed, it follows from a very simple factorization trick.
In some sense, if $n = pq$, then we arrange the $n$ powers $x^0,...,x^{pq-1}$ in a $p \times q$ box fashion, and then collect terms.
So, write $n = pq$ for some $1 < p,q < n$ if $n$ is not prime. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Area of polygon on complex plane formed by complex roots of a polynomial
Compute the area of the polygon whose vertices are the solutions in the complex plane to the polynomial $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$.
What I did was find the complex numbers by factoring and then using Surveyor's formula. However, that requir... | Simply note that the polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Range of $k$ for which equation has positive roots
Range of $k$ for which both the roots of the equation $(k-2)x^2+(2k-8)x+3k-17=0$ are positive.
Try: if $\alpha,\beta>0$ be the roots of the equation. Then $$\alpha+\beta=\frac{8-2k}{k-2}>0\Rightarrow k\in(2,4)$$
And $$\frac{3k-17}{k-2}>0\Rightarrow k\in(-\infty,2)\cu... | Use quadratic formula. $$(k-2)x^2+(2k-8)x+3k-17=0$$ gives $$x=\frac{-(2k-8)\pm\sqrt{(2k-8)^2-4(k-2)(3k-17)}}{2(k-2)}=-\frac{k-4}{k-2}\pm\frac{\sqrt{-8k^2+60k-72}}{2(k-2)}$$ Hence $$x=-1+\frac2{k-2}\pm\frac{\sqrt{(k-6)(3-2k)}}{k-2}$$ Now we need only find when the negative root is positive since the positive one would t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Computing $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3}$ without L'Hôpital's rule.
Computing $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3}$ without L'Hopital
Say $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3} = L$
For $L$:
$$L=\lim_{x\to0}\frac{\tan x-x}{x^3}\\
L=\lim_{x\to0}\frac{\tan 2x-2x}{8x^3}\\
4L=\lim_{x\to0}\frac{\frac12\tan2x-x}{x^3... | Let $x=\arctan y.$ We have $y\to 0^+$ as $x\to 0^+.$
For $x\in (0,\pi /2)$ we have $\frac {\tan x-x}{x^3}=$ $\frac {y-\arctan y}{y^3}\frac {y^3}{x^3}=$ $\frac {(y-\arctan y)}{y^3}(\frac {\tan x}{x})^3.$
We have $\lim_{x\to 0^+}(\frac {\tan x }{x})^3=1.$
Consider $g(y)=y-\arctan y.$
We have $g'(y)-y^2=$ $(1-\frac {1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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prove this nice inequality $\left|\prod_{i=1}^{n}(a_{i}-a_{i+1})\right|\le \frac{3\sqrt{3}}{16}$ Let $n$ be odd number, and $a_{i}\ge 0$, such that
$$2(a_{1}+a_{2}+\cdots+a_{n})=n$$
Show that
$$\left|\prod_{i=1}^{n}(a_{i}-a_{i+1})\right|\le \frac{3\sqrt{3}}{16}$$
where $a_{n+1}=a_{1}$
Seeing this inequality reminds me ... | Here is a proof for $n=3$, perhaps that helps someone to solve the
general problem.
Let $a, b, c$ be non-negative real numbers with $a+b+c = \frac 32$.
Then
$$ f(a, b, c) = \vert (a-b)(b-c)(c-a) \vert \le \frac{3 \sqrt 3}{16} \, .
$$
Equality holds if and only if $(a, b, c)$ is a permutation of
$$
(0, \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Show that if $2x + 4y = 1$ where x and y are real numbers. Show that if $2x + 4y = 1$ where $$x, y \in \mathbb R $$, then $$x^2+y^2\ge \frac{1}{20}$$
I did this exercise using the Cauchy inequality, I do not know if I did it correctly, so I decided to publish it to see my mistakes. Thank you!
If $x=2$ and $y=4$, then $... | You have one big confusion here. It is always true:$$(2^2+4^2)(x^2+y^2)\ge (2x+4y)^2$$
with equality iff $\frac{x}{2}=\frac{y}{4}$. Now, since $2x+4y=1$ we get $$20(x^2+y^2)\ge 1$$
so $$x^2+y^2\ge \frac{1}{20}$$
and you are done.
Now if you are interested when eqaulity ocurres then just plug $y=2x$ in $2x+4y=1$ and yo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Tangent lines to the curve parameterized by $x = a \cos^4t$, $y = a \sin^4t$ The parametric equations of a curve are
$$x = a \cos^4t \qquad y = a \sin^4t$$
where $a$ is a positive constant.
(i) Express $\dfrac{dy}{dx}$ in terms of $t$. (3)
(ii) Show that the equation of the tangent to the curve at the point with parame... | $$x = a \cos^4t \qquad y = a \sin^4t$$
This is what I got.
$$\dfrac{dy}{dx}
=\dfrac{4a \sin^3(t) \cos(t)}{-4a \cos^3(t) \sin(t)}
=-\dfrac{\sin^2(t)}{\cos^2(t)}\tag{i}$$
\begin{align}
y-a\sin^4(t) &= -\dfrac{\sin^2(t)}{\cos^2(t)}(x-a\cos^4(t)) \\
y\cos^2(t) - a\sin^4(t)\cos^2(t) &= -x\sin^2(t) + a\cos^4(t)\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Sequence and series with specific nth term Let $a(1) = 2$ , $a(n+1) = a(n)^2-a(n) + 1$ for $n\geq 1$, Find $$\sum_{n=1}^{\infty} \frac{1}{a(n)}$$
| We first prove the following by induction:
$$
\frac{1}{a(n + 1)-1} + \sum_{k = 1}^n \frac{1}{a(k)} = 1.
$$
For $n = 1$, there stands
$$
\frac{1}{4 - 2 + 1 - 1} + \frac{1}{2} = 1,
$$
which holds. Suppose it holds for $n$. Then,
\begin{align*}
\frac{1}{a(n + 2) - 1} + \sum_{k = 1}^{n + 1} \frac{1}{a(k)} &= \frac{1}{a(n +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What are the local extreme values of $f(x,y)=x^3+x^2y-y^2-4y$? $f'_x=3x^2+2xy$,
$f'_y=x^2-2y-4$
so I have to solve the equation system of
$3x^2+2xy=0,x^2-2y-4=0$ as solutions I get that $((x=0),(y=0)),((x=-4),(y=6)),(x=1),((y=-\frac{3}{2}))$
After that:
$f''_x=6x+2y$,
$f''_y=-2$,
$f''_{xy}=2x=f''_{yx}$
so I get the d... | The stationary points are determined as the solutions for
$$
3x^2+2xy = 0\\
x^2-2y-4 = 0
$$
giving the set
$\{(-4,6),(0,-2),(1,-3/2)\}$
Now taking the Hessian
$$
H = \left(\begin{array}{cc}3x+y & x\\ x & -1\end{array}\right)
$$
To qualify the stationary points we should evaluate $H$ in such set, verifying it's eigenval... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating a determinant. $D_n$=\begin{vmatrix}
a & 0 & 0 & \cdots &0&0& n-1 \\
0 & a & 0 & \cdots &0&0& n-2\\
0 & 0 & a & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&a&1 \\
n-1 & n-2 & n-3 & \cdots... | Develop with respect to the first column. Then
$$
\begin{aligned}
D_n &= aD_{n-1} -(n-1)\cdot (n-1)\cdot a^{n-2}
\\
&=aD_{n-1}-(n-1)^2a^{n-2}\ .
\end{aligned}
$$
This recursion, together with $D_1=a$ gives for $n\ge 2$ the solution
$$
D_n= (a^2-(1^2+2^2+\dots+(n-1)^2)a^{n-2}\ .
$$
(The sum in the first factor has a cl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Question about irreducible polynomials over a finite field. If a polynomial $f(x)$ is irreducible over a finite field, does that mean the only factors are $\{1, f(x)\}$?
How would I go about proving a polynomial $f(x)$ is irreducible over a finite field? A bit of searching on StackExchange showed me this:
Irreducibili... | You don't need a CAS for this low degree polynomial
$$f(x)=x^8+x^4+x^3+x+1.$$
A pencil & paper solution follows:
First we calculate the remainders $r_k(x)$ of the monomials $x^{2k}, 0\le k\le7$, modulo $f(x)$. These are:
$$
\begin{array}{c|c}
k& r_k\\
\hline
0&1\\
1&x^2\\
2&x^4\\
3&x^6\\
4&x^4+x^3+x+1\\
5&x^6+x^5+x^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find vector $x \not= 0$ that satisfies the equation $Ax = x$. Given a matrix A:
$$A = \begin{bmatrix}
0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{3} & 0 & 1 & 0 & 0 \\
\frac{1}{3} & 0 & 0 & 0 & \frac{1}{2} \\
\frac{1}{3} & \frac{1}{2} & 0 & 0 & 0 \\
0 & \frac{1}{2} & 0 & \f... | You want an $x\in\mathrm{Null}(A-I)$. Note that each column of the matrix $A-I$ adds up to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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The shortest distance from a point to the graph of the function To compute the distance from the point (5,5) to the graph of xy=4. I choose an arbitrary point (u,v) on the graph of $xy=4$.
I get $d(u,v)=\sqrt{(u-5)^2+(v-5)^2}$ again $(u,v)$ satisfies equation of hyperbola so that $uv=4$. Now what shall i do next?
| You're after the minimum of $f(x)=(x-5)^2+\left(\frac4x-5\right)^2$, with $x\in\mathbb R$. Now,\begin{align}f'(x)&=-10-\frac{32}{x^3}+\frac{40}{x^2}+2x\\&=2\frac{x^4-5x^3+20x-16}{x^3}\\&=2\frac{(x^2-4)(x^2-5x+4)}{x^3}.\end{align}So, see what happens at the roots of $f'(x)$, which are $\pm2$, $1$ and $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving that this sequence is convergent Given $a_1 = 2$, and $ a_{n+1} = \frac{a_n+5}{4} $ for all $n > 1$ , is this sequence convergent? Give a formal proof in either case (converges or diverges).
Attempt: I do think this converges, but cannot say for sure.
$a_1 = 2$
$a_2 = \frac{7}{4}$
$a_3 = \frac{27}{16}$
$a_4 = ... | Let $x \in \mathbb R$. If $x \gt 5/3$ then we always have
$\tag 1 \frac{5}{3} \lt \frac{x + 5}{4} \lt x$
Since our sequence $a_n$ is strictly decreasing and always greater than $5/3$, it has a limit which we denote by $\alpha$.
For the sequence to converge, $(a_{n+1} - a_n)$ must converge to zero, and
$\quad a_{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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How is $2^{n-k-1} \cdot 3^k = (3/2)^k \cdot 2^{n-1}$? I'm currently reading through some litterature about difference equations, and i came across a relation i just can't seem to wrap my head around.
How come:
$2^{n-k-1}\cdot 3^k=(\frac{3}{2})^k\cdot 2^{n-1}$?
The original problem in the book is:
(1*) $y(n)=a^ny_0+\sum... | $$2^{n-k-1}\cdot 3^k=\frac{2^{n-1}}{2^k}\cdot 3^k=2^{n-1}\cdot \frac{3^k}{2^k} =\left(\frac{3}{2}\right)^k\cdot 2^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Converting $\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$ to trigonometric form
Convert complex to trig.
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$$
Let us consider
$$(3+3i)^5$$
Here $a = 3$ ,$b=3$
$$\sqrt{a^2+b^2}=\sqrt{3^2+3^2} = \sqrt{18}=3\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\bigg... | I can't see any mistakes in your answers. Your calculations, however, are sometimes very wrong, not even leading the correct answers you pull out of them.
For $z^3=(-2+2i)^3$, $r=2\sqrt{2}$ as you stated. $\theta=\tan^{-1}{(-1)}=\frac{-\pi}{4}$. The reason the correct result is $\frac{3\pi}{4}$, is due to $z$'s positi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Using de-Moivre's theorem to find the reciprocal of each number
Use de-Moivre's theorem to find the reciprocal of each number below.
$$\sqrt 3 - i$$
Given $\sqrt{3}-i$ , we need to find the reciprocal of it using de-Moivre's theorem.
$$\frac{1}{\sqrt 3-i} $$
$$= \frac{1(\cos0^c + i\sin 0^c)}{2\big(\frac{\sqrt3}{2}-... | De Moivre's Theorem states that:
$$[r(\cos \theta + i \sin \theta)]^n=r^n(\cos (n\theta)+ i \sin (n\theta))$$
To find the reciprocal, take $n=-1$.
$$z=\sqrt{3}-i\to r=2, \theta=\tan^{-1}\bigg({\frac{-1}{\sqrt{3}}}\bigg)=\frac{-\pi}{6}$$
Hence $z^{-1}=2^{-1}(\cos{(-1)(\frac{-\pi}{6})}+ i\sin{(-1)(\frac{-\pi}{6})})=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find Cauchy principal values of $\int_{-\infty}^{\infty}\frac{x}{(x^2+4)(x^2-2x+5)}\,dx$ I have been asked to find the Cauchy principal vlaues of the following problem using residues:
$\int_{-\infty}^{\infty}\frac{x}{(x^2+4)(x^2-2x+5)}\,dx$
So far I have taken $\oint_C\frac{z}{(z^2+4)(z^2-2z+5)}\,dz$
which gives me:
$2... | factor the denominator.
$\frac {z}{(z+2i)(z-2i)(z-1+2i)(z-1-2i)}$
We only care about the poles in the upper half-plane.
You have poles at $z = 2i, z = 1+2i$
at $z=2i$
$Res f(z) = \lim_\limits{z\to 2i} (z-2i)f(z) = \frac {2i}{(4i)(-1+4i)(-1)} = \frac {1+4i}{2(17)} = \frac {1}{34} + \frac {2i}{17} $
and at $z = 1+2i$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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show whether $\frac {xy}{x^2+y^2}$ is differentiable in $0$ or not? (multivariable) Q: $f(x,y)=\frac {xy}{x^2+y^2}$ if $(x,y)\not=(0.0)$, and $0$ if $(x,y)=(0,0)$.
Is $f$ differentiable at $(0,0)$?
Attempt: $$\lim_{(x,y) \to (0,0)} \frac {f(x,y)-f(0,0)}{||(x,y)-(0,0)||} = \lim_{(x,y) \to (0,0)} \frac {\frac{xy}{x^2+... | $$
f\left(\frac{1}{n},\frac{1}{n}\right)=\frac{1}{n^2}\frac{n^2}{2}=\frac{1}{2}
$$
And $\displaystyle \left(\frac{1}{n},\frac{1}{n}\right) \underset{n \rightarrow +\infty}{\rightarrow}(0,0)$, so it is not continuous at $(0,0)$. It cannot be differentiable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2779130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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USAMO 2018: Show that $2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$ Here is question 1 from USAMO 2018 Q1 (held in April):
Let $a,b,c$ be positive real numbers such that $a+b+c = 4 \sqrt[3]{abc}$.
Prove that: $$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$$
This question is on symmetric poly... | Since our inequality and condition are symmetric and homogeneous,
we can assume that $abc=1$ and $a\geq b\geq c$.
Thus, $a+b+c=4$ and we need to prove that
$$2(ab+ac+bc)\geq a^2+b^2-3c^2$$ or
$$(a+b+c)^2\geq2(a^2+b^2-c^2)$$ or
$$8\geq a^2+b^2-c^2$$ or $$8\geq a^2+b^2-(4-a-b)^2$$ or
$$8\geq a^2+b^2-(16+a^2+b^2-8a-8b+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Maximizing $3\sin^2 x + 8\sin x\cos x + 9\cos^2 x$. What went wrong?
Let $f(x) = 3\sin^2 x + 8\sin x\cos x + 9\cos^2 x$. For some $x \in \left[0,\frac{\pi}{2}\right]$, $f$ attains its maximum value, $m$. Compute $m + 100 \cos^2 x$.
What I did was rewrite the equation as $f(x)=6\cos^2x+8\sin x\cos x+3$. Then I let $\m... | $f(x)=2\sin^2x+8\sin x\cos x+8\cos^2x+1=2(\sin x+2\cos x)^2+1$
Let $\displaystyle \alpha\in\left[0,\frac{\pi}{2}\right]$ and $\displaystyle\cos\alpha=\frac{2}{\sqrt{5}}$. Then $\displaystyle\sin\alpha=\frac{1}{\sqrt{5}}$ and
$$\sin x+2\cos x=\sqrt{5}(\sin x\sin\alpha+\cos x\cos\alpha)=\sqrt{5}\cos(x-\alpha).$$
attainin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Inertia Tensor of an ellipsoid Given is the following inertia tensor of a certain mass distribution $\rho(\vec{r})$ :
$$ I_{ij} = \int dV \rho(\vec{r}) \left( \vec{r}^2 \cdot \delta_{ij} - r_ir_j \right) $$
I should compute the inertia tensor for the following ellipsoid:
$$x^2/a^2 + y^2/b^2 + z^2/c^2 \le 1$$
And I'm ... | You can use the parametrization
\begin{eqnarray}
x &=& a r \sin\theta\sin\phi \\
y &=& b r \sin\theta\cos\phi \\
z &=& c r \cos\theta
\end{eqnarray}
with $0<r<1$, $0<\theta<\pi$ and $0<\phi<2\pi$. The Jacobian for this choice of coordinates is $abc r^2\sin\theta$, so the integral becomes
\begin{eqnarray}
I_{ij} &=& \rh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding the smallest n satisfying $S_n > 10$ Let $S_n = 1 + \frac 12 + \frac 13 + \cdots + \frac 1n$, where $n \in \{ 1,2,3,\cdots\}$ Find the smallest $n$ satisfying $S_n > 10$.
Sorry, it's my first time asking and I don't know how to format this thing. I still don't see anything even after staring at this for really... | We have an asympotic expansion for $H_n$:
$$H_n = \ln n + \gamma + \frac{1}{2n} - \sum_{k=1}^{\infty} \frac{B_{2k}}{2 k n^{2k}} = \ln n + \gamma + \frac{1}{2n} -\frac{1}{12 n^2}+ \frac{1}{120 n^4} - \dots$$
Reference : Wikipedia on Harmonic number
So
$$\ln n + \gamma + \frac{1}{2n} -\frac{1}{12 n^2} < H_n < \ln n + \ga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer. QUESTION: Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer.
ATTEMPT (AND SOME SOLUTIONS): So, for a positive integer $x$, $x^2-1\ge0$. In the case $x^2-1=0$ i.e $x=1$ (since it's a positiv... | Since $xy+1\mid xy+1$ we have $$xy+1\mid (x^2-1)+(xy+1)= x(x+y)$$
Now $\gcd(xy+1,x)=1$ so $$xy+1\mid x+y\implies xy+1\leq x+y$$
So $(x-1)(y-1)\leq 0$ and thus if $x>1$ and $y>1$ we have no solution.
Ergo $x=1$ or $y=1$...
| {
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"url": "https://math.stackexchange.com/questions/2792995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $a+a =0$ for Boolean ring
Suppose $R$ is a Boolean ring. Prove that $a+a=0$ for all $a\in R$.
Also prove that $R$ is commutative. Give an example (with explanation)
of a Boolean ring.
From what I know, a Boolean ring is a ring for which $a^2=a$ for all $a\in R$.
Under addition a ring is a commutative grou... | Take $x \in R$, with $R$ boolean ring, in particular is a ring, therefore $(x+x) \in R$ and $(x+x)^{2}=(x+x)$. Thus
\begin{align*}
(x+x)^{2} &= (x+x)(x+x) \\
&=(x+x)x + (x+x)x \\
&= x^2 + x^2 + x^2 + x^2
\end{align*}
since $x^2 = x$ and $(x+x)^2 = (x+x)$ because $R$ is boolea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is there a pattern to the last digits of square numbers? I was programming and I realized that the last digit of all the integer numbers squared end in $ 0, 1, 4, 5, 6,$ or $ 9 $.
And in addition, the numbers that end in $ 1, 4, 9, 6 $ are repeated twice as many times as the numbers that end in $ 0, 5$
I checked th... | Note that any number can be written in the form $10a+(5 \pm b)$ where $0 \leq b \leq 5$. If n = $10a+(5 \pm b)$, then we can calculate $n^2$ as
$(10a)^2+2(10a)(5 \pm b) + (5 \pm b)^2=$$100a^2 + 100a \pm 20ab+(5 \pm b)^2$
$100a^2$, $100a$, and $\pm 20ab$ are all divisible by 10, so we can ignore them, and we're left wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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If $\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$ prove that $\cos\alpha=\frac{2-m^2}{m}$
If $$\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$$
prove that
$$\cos\alpha=\frac{2-m^2}{m}$$
My approach:
$$\cos^2(\alpha-3\theta)+... | $$\sin\alpha\sin3\theta+\cos\alpha\cos3\theta-m\cos^3\theta=0$$
$$\sin\alpha\cos3\theta-\cos\alpha\sin3\theta-m\sin^3\theta=0$$
$$-\dfrac{\sin\alpha}{m(\cos^3\theta\sin3\theta+\cos3\theta\sin^3\theta)}=-\dfrac{\cos\alpha}{m(\cos3\theta\cos^3\theta-\sin3\theta\sin^3\theta)}=-1$$
Using $\sin3\theta,\cos3\theta$ formula
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Divisibility of Sum of Equally Spaced Binomial Coefficients According to a numerical calculation I did for small values of $k$, it appears that the following is true.
$$4|\left[\sum_{j=1}^{n-1}\binom{3n}{3j}\right]$$ or $$\sum_{j=1}^{n-1}\binom{3n}{3j}=4p, p\in\mathbb{Z}$$
Ex.
If $n=2, \binom{6}{3}=20=4\cdot 5$
If $n... | Starting with the binomial theorem,
$$(1+x)^{3m} = \sum_{k=0}^{3m} \binom{3m}{k} x^k,$$
we see that for the cube root of unity $\omega = e^{2\pi i/3}$,
$$(1+x)^{3m} + (1+\omega x)^{3m} + (1+\omega^2 x)^{3m} = 3 \sum_{\substack{k=0 \\ k \equiv 0 \pmod{3}}}^{3m} \binom{3m}{k} x^k.$$
The terms where $k \not\equiv 0 \pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solving first order differential equation with power series So, I was told solve the equation $y' - y = x^2$ using power series. Normal methods tell me that the solution is $y = c_{0}e^{x}-x^{2}-2x-2$, and this can be verified by plugging it back in. However, I am stuck on trying to solve this with power series.
We ass... | Your approach is correct and you did no mistake
$$y=\sum_{n=0}^{\infty}\frac{a_{0}}{n!}x^{n}+\sum_{n=3}^{\infty}\frac{2}{n!}x^{n}$$
Transform the last part as an exponential and it will be aborbed by the first series excet for the three first terms
$$\sum_{n=3}^{\infty}\frac{2}{n!}x^{n}=\sum_{n=0}^{\infty}\frac{2}{n!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Area of Triangle inside a Circle in terms of angle and radius A circle $O$ is circumscribed around a triangle $ABC$, and its radius is $r$. The angles of the triangle are $\angle CAB = a, \angle ABC = b$ and $\angle ACB = c$.
The area $\triangle ABC$ is expressed by $a, b, c$ and $r$ as:
$\Large r^2 \over\Large2$$\Bigg... | The area formula you derived is a good one to know, but if you want something in terms of a sum of sines, use the trigonometric sum-product relations. Along the way you will also note that the angles of the rriangle sum to 180°. And be careful with signs or this won't come out pretty.
Our starting point, taking $S$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the limit with exponents using L'Hôpital's rule or series expansion Evaluate the limit$$\lim_{x\to 0}\dfrac{\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}
-\sqrt{ab}}{x}$$ It is known that $a>0,b>0$
My Attempt:
I could only fathom that $$\lim_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\sqrt{ab}$$
| We have that
*
*$a^x=e^{x\log a}=1+x\log a+\frac12x^2\log^2 a+o(x^2)$
*$b^x=e^{x\log b}=1+x\log b+\frac12x^2\log^2 b+o(x^2)$
then
$$\left( \frac{a^x+b^x}{2} \right)^{\frac{1}{x}}=\left(\frac{2+x\log ab+\frac12x^2(\log^2 a+\log^2 b)+o(x^2)}{2}\right)^{\frac{1}{x}}=\left(1+x\log \sqrt{ab}+\frac14x^2(\log^2 a+\log^2 b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence. As the question states:
"Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence."
I have consulted this related q... | In fact a lot of the work needed is already done. Let's recapitulate. We have $f(x)=x^3\ln\left(\sqrt{ x}\right)$ and the Taylor expansion at $a=1$ is given as
\begin{align*}
f(x)=\sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(a)(x-a)^n
\end{align*}
We obtain
\begin{align*}
f(x)&=x^3\ln\left(\sqrt{ x}\right)=\frac{1}{2}x^3\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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2014 AIME II: Cubic Question Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$, and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$. Find the sum of all possible values of $|b|$.
I assumed that the other root of p(x) would be t. From here, I used Vieta's formulas to get $-rst=b$ and $rs+st+rt=a$. Then, I assumed t... | Let $r$, $s$, and $-r-s$ be the roots of $p(x)$ (per Vieta's). Then $r^3 + ar + b = 0$ and similarly for $s$. Also, $q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0$.
Set up a similar equation for $s$:
$q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.$
Simplifying and adding the equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a fraction $\frac{m}{n}$ which satisfies the given condition Find a fraction such that all of $\frac{m}{n}$, $\frac{m+1}{n+1}$, $\frac{m+2}{n+2}$, $\frac{m+3}{n+3}$, $\frac{m+4}{n+4}$, $\frac{m+5}{n+5}$ are reducible by cancellation. Condition: $m≠n$.
What I tried was... I wrote $$\frac{m}{n}=k$$ Then, I replaced ... | Fixed $n$ a solution is given by
$$m=n+n(n+1)(n+2)(n+3)(n+4)(n+5)=n+\frac{(n+5)!}{(n-1)!}$$
indeed
$$\frac{m}{n}=\frac{n+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n}\\=1+(n+1)(n+2)(n+3)(n+4)(n+5)$$
$$\frac{m+1}{n+1}=\frac{n+1+n(n+1)(n+2)(n+3)(n+4)(n+5)}{n+1}\\=1+n(n+2)(n+3)(n+4)(n+5)$$
$$\frac{m+2}{n+2}=\frac{n+2+n(n+1)(n+2)(n+3)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How did root of the equation $x= \sqrt{(2-x)(3-x)} + \sqrt{(3-x)(5-x)} + \sqrt{(2-x)(5-x)}$ vanish while solving? The original question was to solve for $x \in \mathbb{R}$ in:
$$x= \sqrt{(2-x)(3-x)} + \sqrt{(3-x)(5-x)} + \sqrt{(2-x)(5-x)}$$
My Solution:
By domain requirements: $x \in (-\infty,2) \cup (5,\infty) $.
\be... | Your error occurs at the step where you removed the factor $(5-x)$.
It should be $|5-x|$, which leads to two cases.
If $x \le 5$, then $(5-x)$ is correct.
If $x > 5$, the correct factor is $(x-5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Number of solutions to the congruence $x^2\equiv 121\pmod {1800}$ I'm trying to find the number of solutions to this congruence:
$$x^2\equiv 121\pmod {1800}$$
I thought about writing it as a system of congruences. As $1800=3^2 \cdot 5^2 \cdot 2^3$, we get:
$x^2\equiv 121\pmod {5^2} \;,\; x^2\equiv 121\pmod {3^2} \;,\;... | No, there's not a fast way. To solve $x^2\equiv a \pmod{p^n}$ we first solve $x^2\equiv a \pmod{p}$ and then use Hensel's Lemma to "lift" the solution to $\pmod{p^2}$ and then $\pmod{p^3}$, etc. Sometimes the solutions lift uniquely, sometimes not.
In the case of your problem, there are two solutions modulo $5$ whi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The implication: $x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$ $$x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$$
The graphs/range: $\quad y_1(x)=x+\frac{1}{x}, \quad y_7(x)=x^7+\frac{1}{x^7} \quad$
and do not touch the line $\quad y=1\quad$. The relation $\quad x+\frac 1 x=1 \quad$ appears to only be true for certain ... | The given equation implies $x^2-x+1=0$; multiplying by $x+1$, we get $x^3+1=0$. This implies $x^6=1$ and therefore
$$
x^7=x
$$
Your argument is sound as well.
The key is that a solution to $x+\frac{1}{x}=1$ is a 6th root of $1$. Indeed,
$$
x^6-1=(x^3-1)(x^3+1)=(x-1)(x+1)(x^2+x+1)(x^2-x+1)
$$
Thus any root of $x^2-x+1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Someone's Inequality ? $(a + b)^2 \le 2(a^2 + b^2) $ For real $a, b$ then $(a + b)^2 \le 2(a^2 + b^2) $
This fairly trivial inequality crops up a lot in my reading on (Lebesgue) integration, is it named after someone ? It extends rather obviously for positive reals to $a^2 + b^2 \le (a + b)^2 \le 2(a^2 + b^2) $.
Proof... | That inequality probably does not have a name as it is so basic. In any case it can be viewed as a special case of Young's inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution of the functional equation $f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right) $
If
$$f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right)\text,$$
prove that
$$f(4x^3 -3x) + 3f(x) =0\text.$$
I started by substituting $x = y$, in the expression and I get
$$2f(x)=f\left(2x\sqrt{1-x^2}\right)\text.$$... | Easy way to prove that for every real number $ x $ with $ - 1 \le x \le 1 $, $ f ( x ) = 0 $ (using some steps already mentioned in the comments, but repeated here to be complete):
Let $ x = y = 0 $ in
$$ f ( x ) + f ( y ) = f \Big( x \sqrt { 1 - y ^ 2 } + y \sqrt { 1 - x ^ 2 } \Big) \tag { * } \label { * } $$
and you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Dodecahedron volume How can I derive the volume of a regular dodecahedron? A. P. Kiselev's Stereometry suggests that we cut it into a cube and congruent "tent-like" solids. I'm finding some difficulty to compute the volume of these "tent-like" solids.
| First, note that if $s$ is the side length of the dodecahedron, then the side length of the cube is $$c = \frac{s}{2} \csc \frac{\pi}{10} = s \left(\frac{1 + \sqrt{5}}{2}\right).$$
Now consider looking at the tent solid from above, it looks something like this:
where the dotted line $x$ denotes the slant height of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Exponential inequality solution I am trying to solve
$$ \displaystyle (2^{(3x-1)/(x-1)})^{1/3} < 8^{((x-3)/(3x-7))}$$
Here's how I proceeded.
$LHS = 2^{((3x-1)/(3(x-1))} , RHS = 2^{((3*(x-3))/(3x-7))}$
hence inequating the exponents ,
$$\frac {(3x-1)}{(3(x-1))} < 3\frac {(x-3)}{(3x-7)}$$
$$ \implies (3x-1)(3x-7) < ... | Be carefull when you multiply your inequality
$$\frac {(3x-1)}{(3(x-1))} < 3\frac {(x-3)}{(3x-7)}$$
Because $3x-7$ or $3(x-1)$ may be negative and change the order of the ineqaulity
You need to study the sign of $3(x-1)(3x-7)$...
A)$ \,3(x-1)(3x-7) $ is positive
$$\implies (3x-1)(3x-7) < 9(x-1)(x-3)$$
$$ \implies x<5/... | {
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"timestamp": "2023-03-29T00:00:00",
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What is $\lim_{n\to\infty} \left(\sum_{k=1}^n \frac1k\right) / \left(\sum_{k=0}^n \frac1{2k+1}\right)$? I have the following problem:
Evaluate
$$ \lim_{n\to\infty}{{1+\frac12+\frac13 +\frac14+\ldots+\frac1n}\over{1+\frac13 +\frac15+\frac17+\ldots+\frac1{2n+1}}} $$
I tried making it into two sums, and tried to make it... | One approach is as follows: it suffices to note that
$$
\sum_{k=2}^n\frac{1}{k} \leq \int_1^n \frac 1x \,dx \leq \sum_{k=1}^n\frac{1}{k}, \\
\sum_{k=2}^{n+1}\frac{1}{2k-1} \leq \int_1^{n+1} \frac 1{2x-1} \,dx \leq \sum_{k=1}^{n+1}\frac{1}{2k-1},
$$
and apply the squeeze theorem. In particular, we can use the above to ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How to solve the given integral of the type: $\int_{0}^{\infty}\tfrac{1}{(a + b x)(c + d x) (1 + p x)}dx$ Can anyone give a hint how to solve integral of the following type:
$$\int_{0}^{\infty}\dfrac{1}{(a + b x)(c + d x) (1 + p x)}dx$$
The problem is if we proceed with partial fraction, we get
$$\dfrac{1}{(a + bx)(c ... | For simplicity, write this as $$J = \dfrac{1}{bdp} \int_0^\infty \dfrac{dx}{(A+x)(B+x)(C+x)}$$
where $A = a/b$, and $B = c/d$, $C = 1/p$. We're assuming $a,b,c,d,p > 0$ and $A,B,C$ are all distinct. The partial fraction decomposition of the integrand is
$$ F(x) = \frac{1}{(A-B)(A-C)(A+x)} + \frac{1}{(B-A)(B-C)(B+x)} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Sum of digits when $99$ divides $n$ Let $n$ be a natural number such that $99\mid n$. Show that $S(n)\ge 18$.
It is clear that $9 \mid S(n)$, but I cannot apply the divisibility rule of $11$. Please help!
| If $99|n$ then $9|n$ and $11|n$.
Two well known rules.
1) If $9|n$ then $9|S(n)$.
So $S(n) = 9, 18,27,.... $ etc. So the the only way $S(n) \ge 18$ can be false is if $S(n) = 9$.
2) If $11|S(n)$ then the sum of the even positioned digits plus the sum of the odd positions digits are either equal, or off by a multip... | {
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"url": "https://math.stackexchange.com/questions/2820979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$ $a, b, c ∈ \mathbb{R}+.$
WLOG assume $a \leq b \leq c.$ I tried substitution: $x=\frac{1} {1+a^2}, y=\frac{1} {1+b^2}, z=\frac{1} {1+c^2},$ so $x \geq y \geq z$ and $(1-x)+(1... | Cauchy-Schwartz ...
\begin{eqnarray*}
\left( \frac{a}{\sqrt{1+a^2}} \frac{1}{\sqrt{1+a^2}} + \frac{b}{\sqrt{1+b^2}} \frac{1}{\sqrt{1+b^2}} + \frac{c}{\sqrt{1+c^2}} \frac{1}{\sqrt{1+c^2}} \right)^2 \\\leq \left( \frac{a^2}{1+a^2} + \frac{b^2}{1+b^2} +\frac{c^2}{1+c^2} \right) \left( \frac{1}{1+a^2} + \frac{1}{1+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
An interesting algebraic inequality involving a square root I came across the following inequality as a step in the proof of another: Let $x,\ y,\ z>0$, then this is true:
$$\sqrt{\frac{x^2 + 1}{y^2 + 1}} + \sqrt{\frac{y^2 + 1}{z^2 + 1}} + \sqrt{\frac{z^2 + 1}{x^2 + 1}}\le \frac{x}{y} + \frac{y}{z} + \frac{z}{x}.$$
I c... | Let $z=\min\{x,y,z\}$.
Thus, we need to prove that
$$\frac{x}{y}+\frac{y}{x}-2+\frac{y}{z}-\frac{y}{x}+\frac{z}{x}-1\geq\sqrt{\tfrac{x^2+1}{y^2+1}}+\sqrt{\tfrac{x^2+1}{y^2+1}}-2+\sqrt{\tfrac{y^2+1}{z^2+1}}-\sqrt{\tfrac{y^2+1}{x^2+1}}+\sqrt{\tfrac{z^2+1}{x^2+1}}-1$$ or
$$\tfrac{(x-y)^2}{xy}+\tfrac{(z-x)(z-y)}{xz}\geq\tf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Help setting up integral: $\iint_{\mathbb R^2} \frac{1}{(1+4x^2+9y^2)^2} dx dy$
Evaluate $\iint_{\mathbb R^2} \frac{1}{(1+4x^2+9y^2)^2} dx dy$
If I use polar coordinates I get
$$\int_0^{2\pi}\int_0^\infty \frac{r}{(1+4r^2\cos^2\theta+9r^2\sin^2\theta)^2} drd\theta.$$
However, I am not sure where to go from here.
| We substitute first $2x=s$, $3y=t$, so $6dx\wedge dy = ds\wedge dt$. The integral is after this easily computed.
$$
\begin{aligned}
\iint_{\mathbb R^2} \frac{1}{(1+4x^2+9y^2)^2} \; dx\; dy
&=
\frac 16\iint_{\mathbb R^2} \frac{1}{(1+s^2+t^2)^2} \;ds\; dt
\\
&=
\frac 16\int_0^\infty\int_0^{2\pi} \frac{1}{(1+r^2)^2} \; r\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluation of limits and finding the value of $\lambda$
If $f(x)= \lim_{n\to \infty}\dfrac{\left(1- \cos \left(1- \tan \left(\dfrac {\pi}{4}-x\right)\right)\right)(x+1)^n+ \lambda\sin((n - \sqrt{n^2 -8n})x)}{x^2(x+1)^n+x}$ ,$x\ne 0$ is continuous at $x=0$, then find the value of $(f(0)+ 2\lambda)$
Attempt:
I know t... | Rewrite the given function as:
$$f(x) = \lim_{n\to\infty}\frac{2\sin^2\left(\frac{\tan(x)}{1+\tan(x)}\right)(x+1)^n + \lambda \sin\left(\frac{8nx}{n+\sqrt{n^2-8n}}\right)}{x^2(x+1)^n+x}$$
We need left and right limits at $x=0$ to be equal as $n\to \infty$. First let us consider individual limits:
*
*For right hand l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find for how many values of $n$. $I_n=\int_0^1 \frac {1}{(1+x^2)^n} \, dx = \frac 14 + \frac {\pi}8$
Find for how many values of $n$. $I_n=\int_0^1 \frac {1}{(1+x^2)^n} \, dx=\frac 14 + \frac {\pi}8$
My attempt (integration by parts):
\begin{align}
I_n & = \int_0^1 \frac 1{(1+x^2)^n}\,dx = \left. \frac {x}{(1+x^2)^n}... | Hint. The sequence $I_n$ is strictly decreasing because for $x\in (0,1]$,
$$0<\frac {1}{1+x^2} <1.$$
.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^... | $$n^4-4n^2=n^2(n^2-1-3)=n\underbrace{(n-1)n(n+1)}_{\text{product of three consecutive integers}}-3n^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Choose $y\in\mathbb{R}$ so that $\sin y=\frac{u}{\sqrt{u^2+v^2}}$ and $\cos y=\frac{v}{\sqrt{u^2+v^2}}$. I know the identity $\sin^2y+\cos^2y=1$. Also, I notice that
$$\left(\frac{u}{\sqrt{u^2+v^2}}\right)^2+\left(\frac{v}{\sqrt{u^2+v^2}}\right)^2=1$$
without $u,v$ vanishing simultaneously. But I am not sure whether $\... | It is clear that $\frac v{\sqrt{u^2+v^2}}\in[-1,1]$. Therefore, since $\cos0=1$ and $\cos\pi=-1$, the intermediate value theorem implies that there's a $y\in[0,\pi]$ such that $\cos y=\frac v{\sqrt{u^2+v^2}}$. Then\begin{align}\sin^2y&=1-\cos^2y\\&=1-\frac{v^2}{u^2+v^2}\\&=\left(\frac u{\sqrt{u^2+v^2}}\right)^2,\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\dots+\frac{1}{(n+1)\sqrt{n}}<2$
For any positive integer $n$ prove by induction that:
$$ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\dots+\frac{1}{(n+1)\sqrt{n}}<2.$$
The author says that it is sufficient to prove that
$$ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+... | The stronger inequality is easier to be proved by using induction than the original one. This is another example: in order to prove
$$\dfrac{1}{1^2}+ \dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2} < 2$$ show that
$$\dfrac{1}{1^2}+ \dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2} \le 2-\dfrac{1}{n}.$$
See Induction on inequalities: $\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to show $ \sum\limits_{n=0}^{a_1} \dfrac{a_2 +n \choose a_2}{a_1+a_2 \choose a_2} $ identity? How do i show this identity:
$ \sum\limits_{n=0}^{a_1} \dfrac{a_2 +n \choose a_2}{a_1+a_2 \choose a_2} = \sum\limits_{n=0}^{a_1} \dfrac{a_1 \choose n}{a_1+a_2 \choose n} $
I don't think the partial sums match $ \dfra... |
We start with the right-hand side and obtain
\begin{align*}
\color{blue}{\sum_{n=0}^{a_1}}&\color{blue}{\binom{a_1}{n}\binom{a_1+a_2}{n}^{-1}}\\
&=\sum_{n=0}^{a_1}\frac{a_1!}{n!(a_1-n)!}\cdot\frac{n!(a_1+a_2-n)!}{(a_1+a_2)!}\\
&=\frac{a_1!}{(a_1+a_2)!}\sum_{n=0}^{a_1}\frac{(a_1+a_2-n)!}{(a_1-n)!}\\
&=\frac{a_1!a_2!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Probable a Binomial Probability Problem :) (not sure) I need to calculate the average profit in a month of a trading strategy.
My probability of a successful trade is 33%
My probability of a bad trade is 67%
The strategy has a Risk Return ratio of 1:3 meaning that I risk 1 to gain 3
So in a bad trade I loose one and in... | Expected payouts per day:
A = 3 losses = -3 units
B = 1 win / 2 losses = +1 unit
C = 2 wins / 1 loss = +5 units
D = 3 wins = +9 units
Probabilities:
$\Pr(A) = \binom{3}{0}(1/3)^0(2/3)^3=8/27$
$\Pr(B) = \binom{3}{1}(1/3)^1(2/3)^2=12/27$
$\Pr(C) = \binom{3}{2}(1/3)^2(2/3)^1=6/27$
$\Pr(D) = \binom{3}{3}(1/3)^3(2/3)^0=1/27... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Short way to evaluate $\int_{0}^{1}\ln(\frac{a-x^2}{a+x^2})\cdot\frac{dx}{x^2\sqrt{1-x^2}}$ I would like to evaluate this integral, $$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}$$
An approach:
1. Using integration by parts
$u=\ln\left(\frac{a-x^2}{a+x^2}\right)$, $du=\frac{x... | Under $x\to\sin x$, one has
$$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}=\int_{0}^{\pi/2}\ln\left(\frac{a-\sin^2x}{a+\sin^2x}\right)\cdot\frac{\mathrm dx}{\sin^2x}.$$
Let
$$I(k)=\int_{0}^{\pi/2}\ln\left(\frac{a-k\sin^2x}{a+k\sin^2x}\right)\cdot\frac{\mathrm dx}{\sin^2x}$$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Equation involving floor function and fractional part function How to solve $\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} = \{x\} + \frac{1}{3}$ , where $\lfloor \rfloor$ denotes floor function and {} denotes fractional part. I did couple of questions like this by solving for {x} and bounding it from 0 ... | Let $x = n+z$ where $n$ is an integer and $0\leq z <1,$ and break into two cases.
Case 1: If $z<1/2$ the equation becomes
$$\frac{1}{n}+\frac{1}{2n} = z+\frac{1}{3},$$
which leads to
$$\frac{1}{3} \leq \frac{3}{2n} <\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$$
which is equivalent to $1.8 < n \leq 4.5.$ So $n = 2, 3,$ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Position of Q in which the area of triangle CDQ is a third part of the area of square ABCD. Given
*
*$A(8,0)$
*$B(0,4)$
*$C(-4,-4)$
*$D(4,-8)$
*$P(2,3)$
Question
In the figure the triangle $CDQ$ is grayed out. There is a position of $Q$ in which the area of triangle $CDQ$ is a third part of the area of square ... | The side of the square is $4\sqrt{5}$, so the area of the square is $80$.
The area of $\triangle_{QCD}=\frac{1}{2}\cdot4\sqrt{5}\cdot h=\frac{80}{3}$, so the height $h=\frac{8\sqrt{5}}{3}$, which is $\frac{2}{3}$ the side of the square. Now imagine drawing a line through the midpoint of $CD$, say $R$, through $Q$, to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Equation involving inverse trigonometric function $$I=\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $$
Expression of $I$ as a rational integer equation in $x$ and $y$ is
$$
(a) 27x^2=y^2(9-8y^2)
\\(b) 27y^2=x^2(9-8x^2)
\\(c) 27x^4 - 9x^2 + 8y^2=0
\\(d) ... | Hint:
Let $\sqrt{\dfrac{3-4x^2}{x^2}}=2a$
$$2\tan^{-1}a-\tan^{-1}(2a)=\tan^{-1}a+\tan^{-1}a+\tan^{-1}(-2a)$$
$$\tan\left(\tan^{-1}a+\tan^{-1}a+\tan^{-1}(-2a)\right)=\dfrac{a+a-2a+4a^3}{1-a^2-a(-2a)-a(-2a)}$$
which is $$=\dfrac{\sqrt{1-x^2-y^2}}y$$
Now square both sides and replace the value of $a^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I need to find the real and imaginary part of this $Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\right )^{n} + \left (\frac{ \sqrt{3} - i }{2}\right )^{n}$ I have a test tomorrow and i have some troubles understanding this kind of problems, would really appreciate some help with this
$$ Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\r... | Note that $z=\left (\frac{ \sqrt{3} + i }{2}\right )^{n}=(a+bi)^n$ we have
$$|z|=\sqrt{(\sqrt{3}/2)^2+(1/2)^2}=1, \quad \theta_z=\arctan(\tfrac{b}{a})=\arctan(\tfrac{1}{\sqrt{3}})=\tfrac{\pi}{6}$$
and for $w=\left (\frac{ \sqrt{3} - i }{2}\right )^{n}=(a-bi)^n$ we have
$$|w|=\sqrt{(\sqrt{3}/2)^2+(1/2)^2}=1, \quad \th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Find the derivative of a function that contains a sum If $$f(x)=\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}$$ find $f''(-2)$.
I know that, by ratio test, the previous sum converges iff $x\in(0,6)$.
I did:
$$\begin{matrix}
f'(x)&=&\left(\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}... | In terms of finding $f''(-2)$ with
$$f(x)=\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}$$
then consider the following.
Method 1
Since
\begin{align}
f(x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \, \left(1 - \frac{x}{3}\right)^{n} = \ln\left(\frac{x}{3}\right)
\end{align}
then
\begin{align}
f'(x) &= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is
Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a... | You can try also like this. Write $$a_1 =\cos \alpha \sin \beta$$
$$a_2 =\cos \alpha \cos \beta$$
$$a_3 =\sin \alpha \sin \beta$$
$$a_4 =\sin \alpha \cos \beta$$
for some $\alpha$ and $\beta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Is $ \lim_{x\to \infty}\frac{x^2}{x+1} $ equal to $\infty$ or to $1$? What is
$ \lim_{x\to \infty}\frac{x^2}{x+1}$?
When I look at the function's graph, it shows that it goes to $\infty$, but If I solve it by hand it shows that the $\lim \rightarrow 1$
$ \lim_{x\to \infty}\frac{x^2}{x+1} =$ $ \lim_{x\to \infty}\frac{... | Writing the fraction as follows may clear it up:
$$
\frac{x^2}{x+1}
= \frac{x^2-1+1}{x+1}
= \frac{x^2-1}{x+1} + \frac{1}{x+1}
= x-1 + \frac{1}{x+1}
\to \infty
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Limit $\lim_{n\to\infty} n^2\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2\right)$ Greetings I am trying to solve $$\lim_{n\to\infty} n^2\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2\right)$$ Using binomial series is pretty easy: $$\lim_{n\to\infty}n^2\left(1+\frac{1}{2n}-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1... | Hint: multiplying numerator and denominator by $\sqrt{1+1/n}+\sqrt{1-1/n}+2$ we get
$$2 n^2 \frac{\sqrt{1-1/n^2}-1}{\sqrt{1+1/n}+\sqrt{1-1/n}+2}$$ and then do the same with $$\sqrt{1-1/n^2}+1$$
you will get
$$\frac{n^2(2(\sqrt{1-1/n^2}-1))(\sqrt{1-1/n^2}+1)}{(\sqrt{1+1/n}+\sqrt{1-1/n}+2)(\sqrt{1-1/n^2}+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the quotient and remainder Find the quotient and remainder when $x^6+x^3+1$ is divided by $x+1$
Let $f(x)=x^6+x^3+1$
Now $f(x)=(x+1).q(x) +R $ where r is remainder
Now putting $x=-1$ we get $R=f(-1)$
i.e $R=1-1+1=1$
Now $q(x)=(x^6+x^3)/(x+1)$
But what I want to know if there is another way to get the quotient ex... | $$f(x)=x^6+x^3+1= x^6 +x^5 -x^5 +x^3+1\\ = x^5(x+1) -x^5-x^4+x^4+x^3+1 \\=x^5(x+1)-x^4(x+1)+x^3(x+1)+1\\= (x+1)(x^5-x^4+x^3)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\q... | This is a summary of the arguments and comments above:
$a^2 + b^2 = 1$ and they symmetry argument that $a = b$ shows that $a = b = {1 \over \sqrt{2}}$, and thus $a + b + 1/a + 1/b = 3 \sqrt{2}$.
Calculus confirms it:
$$f(a) = a + 1/a + \sqrt{1 - a^2} + {1 \over \sqrt{1 - a^2}}$$
so
$${df(a) \over da} = -\frac{a}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 3
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Find the Determinant when $p$ and $q$ are roots of $x^2-2x+5=0$ If $p$ and $q$ are roots of $x^2-2x+5=0$ Then find value of
$$\Delta=\begin{vmatrix}
1 & 1+p^2+q^2 & 1+p^3+q^3\\
1+p^2+q^2& 1+p^4+q^4 & 1+p^5+q^5\\
1+p^3+q^3 & 1+p^5+q^5 & 1+p^6+q^6
\end{vmatrix}$$
My try: I tried to express the determinant as product o... | HINT:
Note that by Vieta's formulas, $p+q=2$ and $pq=5$ so $$1+p+q=3$$ and $$1+p^2+q^2=1+2p-5+2q-5=2(p+q)-9=-5$$ giving $$1+p^3+q^3=(1+p+q)(1+p^2+q^2)-(p^2+q^2+p+q+pq(p+q))$$ and since $x^2=2x-5$, $$1+p^3+q^3=3(-5)-(2p-5+2q-5+2+5(2))$$ or $$1+p^3+q^3=-15-(2(p+q)+2)=-15-(6)=-21$$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\left(1+\frac1 n\right)^n > 2$ I'm trying to demonstrate that $\left( 1+\frac1 n \right)^n$ is bigger than $2$. I have tried to prove that $\left( 1+\frac1 n \right)^n$ is smaller than $\left( 1+\frac1{n+1} \right)^{n+1}$ by expanding
$\left( 1+\frac1n \right)^n = \sum\limits_{i=0}^n \left( \frac{n}{k} \ri... | \begin{align}
\left(1+\frac1n\right)^n &= \sum_{k=0}^n {n \choose k} \frac1{n^k} \\
&= \sum_{k=0}^n \frac{n(n-1)\cdots(n-k+1)}{k!n^k} \\
&= \sum_{k=0}^n \frac1{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots \left(1-\frac{k-1}n\right)
\end{align}
so
\begin{align}
\left(1+\frac1{n+1}\right)^{n+1} &= \sum_{k=0}^{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluating ${\Large\int} _{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(x)+\cos(x)}{\sin^4(x)-4}dx$ This integral is giving me hard times, could anyone "prompt" a strategy about? I tried, resultless, parameterization and some change of variables.
$${\Large\int} _{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(x)+\cos(x)}{\sin^... | $$\int\dfrac{\sin x}{\sin^4 x-4}dx=-\int\dfrac{1}{(1-u^2)^2-4}du\\\int\dfrac{\cos x}{\sin^4 x-4}dx=\int\dfrac{du}{u^4-4}=0.25\int\dfrac{du}{u^2-2}-0.25\int\dfrac{du}{u^2+2}$$the first one is zero over a symmetric interval and for the second we have$$I=0.25\int_{-1}^{1}\dfrac{du}{u^2-2}-0.25\int_{-1}^{1}\dfrac{du}{u^2+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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a tough sum of binomial coefficients
Find the sum: $$\sum_{i=0}^{2}\sum_{j=0}^{2}\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\binom{4}{k-l+i+j},\space\space 0\leq k,l\leq 6$$
I know to find $\sum_{i=0}^{2}\binom{2}{i}\binom{2}{2-i}$, I need to find the coefficient of $x^2$ of $(1+x)^4$ (which is $\binom{4}{2}$). But I fa... | Computing the generating function:
$$
\begin{align}
&\sum_k\sum_l\sum_i\sum_j\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\binom{4}{k-l+i+j}x^ly^k\\
&=\sum_k\color{#C00}{\sum_l}\sum_i\sum_j\binom{2}{i}\binom{2}{j}\binom{2}{k-i-j}\color{#C00}{\binom{4}{k-l+i+j}x^{l+4-k-i-j}}x^{k+i+j-4}y^k\\
&=\color{#C00}{(1+x)^4}\sum_k\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove that $|z_{1}+z_{2}|^2\leq (1+c)|z_{1}|^2+\bigg(1+\frac{1}{c}\bigg)|z_{2}|^2$
If $z_{1},z_{2}$ are two complex numbers and $c>0.$ Then prove that
$\displaystyle |z_{1}+z_{2}|^2\leq (1+c)|z_{1}|^2+\bigg(1+\frac{1}{c}\bigg)|z_{2}|^2$
Try: put $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}.$ Then
from left side
$$(x_... | You are on the right track. Following your approach the inequality becomes
$$x^2_{1}+x^2_{2}+2x_{1}x_{2}+y^2_{1}+y^2_{2}+2y_{1}y_{2}\leq (1+c)(x_1^2+y_1^2)+\bigg(1+\frac{1}{c}\bigg)(x_2^2+y_2^2)$$
that is
$$2x_{1}x_{2}+2y_{1}y_{2}\leq c(x_1^2+y_1^2)+\frac{1}{c}(x_2^2+y_2^2).$$
Is it true that $2uv\leq cu^2+\frac{v^2}{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
find all the points on the line $y = 1 - x$ which are $2$ units from $(1, -1)$ I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by ... | Since you have the general form $2x^2-6x+1=0$
Now solve for x,
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\dfrac32\pm\dfrac12\sqrt{7},$$
Now try to plug in these values of $x$.
Edit:
$$2(x-1.5)^2-3.5=0$$
$$2(x^2+2.25-3x)=3.5$$
$$x^2-3x+2.25=1.75$$
$$x^2-3x+0.5=0$$
$$2x^2-6x+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find $\log _{24}48$ if $\log_{12}36=k$ Find $\log _{24}48$ if $\log_{12}36=k$
My method:
We have $$\frac{\log 36}{\log 12}=k$$ $\implies$
$$\frac{\log 12+\log 3}{\log 12}=k$$ $\implies$
$$\frac{\log3}{2\log 2+\log 3}=k-1$$ So
$$\log 3=(k-1)t \tag{1}$$
$$2\log 2+\log 3=t$$ $\implies$
$$\log 2=\frac{(2-k)t}{2} \tag{2}$$
... | We have
$$\log _{24}48=\frac{\log _{12}48}{\log _{12}24}=\frac{\log _{12}36+\log _{12}\frac43}{\log _{12}36+\log _{12}\frac23}=\frac{k+\log _{12}\frac43}{k+\log _{12}\frac23}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi... | Hint: multiply both sides with $\frac{\sqrt 2}{2} $: $$\underbrace{\sin\frac{\pi}{20}+\cos\frac{\pi}{20}}+\underbrace{\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}}=\frac{\sqrt 2}{2}\;\;\;/\cdot \frac{\sqrt 2}{2} $$
so we have to prove:
$$\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}=\frac{1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
} |
Is there a simpler way to determine m, n, p, such that the following holds for all reals? I am given the following equation and I am asked to find $m$, $n$ and $p$, so that the equation holds for all reals:
$$ \sin^4x + \cos^4x + m(\sin^6x + \cos^6x) + n(\sin^8x + \cos^8x) + p(\sin^{10}x + \cos^{10}x) = 1, \space \fora... | Your reasoning is correct. You may save a little algebraa by formulating it as follows. Define $\, f(t) := t^2 + m t^3 + n t^4 + p t^5. \,$ We want the equation
$\, f(t) + f(1-t) - 1 = 0\,$ to be true for all $\, t = \sin^2(x). \,$ If you expand the left side, it is a fourth degreee polynomial in $\,t.\,$ All of its c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $\min\{x \in\mathbb{N} : xn+1 \text{ is a square}\} \geq \frac{n}{4}$ then n is prime.
Let $n$ be a odd natural number greater than 3 and $k=\min\{x \in\mathbb{N} : xn+1 \text{ is a square}\}$ and $t=\min\{x \in\mathbb{N} : xn \text{ is a square}\}.$ If $k>\frac{n}{4}$ and $t>\frac{n}{4}$ then n is prime.
I proved... | If $n\in \mathbb{N}$ is odd and not a prime power, then there is always an $a$ such that $1 < a \leqslant \frac{n-1}{2}$ and $a^2 \equiv 1 \pmod{n}$. Then it follows that
$$k \leqslant \frac{a^2-1}{n} < \frac{n^2/4}{n} = \frac{n}{4}\,.$$
To see that such an $a$ exists, let $p$ be a prime dividing $n$, and write $n = p^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The value of $a+b=ab=a^2-b^2$ If $a+b=ab=a^2-b^2$ and $a$ and $b$ are real numbers then what do all of the expressions evaluate to?
*
*$(1+\sqrt5)/2$
*$(3+\sqrt5)/2$
*2
*$\sqrt5$
Not sure at all how to find the answer. I tried $a=ab/b$ and $a=(a^2-b^2)/b$ but after that I am lost.
| Any time you see $a^2-b^2$ you should think $(a+b)(a-b)$. Here if $a+b \neq 0$ you can divide it out to get $1=a-b$ Then $a=b+1, a+b=2b+1=(b+1)b$ and feed it to the quadratic formula.
If $a+b=0$ we must have $a=b=0$ and the sum is again $0$, which should have been one of the choices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2860977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many fair dice of this kind exist? I am not talking about the shape of the dice here, I am talking about another type. You will see what I mean soon.
For example, when there are 1 dice, a normal dice is a fair dice, because the probability of getting each number is the same by $ \frac{1}{6} $
When 2 normal dice are... | Add $1$ to each number in one of the dice:
$A: (0,1,2,3,4,5),(0,6,12,18,24,30)$
$B: (0,1,2,6,7,8),(0,3,12,15,24,27)$
$C: (0,1;2,9,10,11),(0,3,6,18,21,24)$
$D: (0,1,2,18,19,20),(0,3,6,9,12,15)$
$E: (0,1,4,5,8,9),(0,2,12,14,24,26)$
$F: (0,1,6,7,12,13),(0,2,4,18,20,22)$
$G: (0,1,12,13,24,25),(0,2,4,6,8,10)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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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.