Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Integrate $\int\frac{x}{x^3-8}dx$ Integrate
$$\int\frac{x}{x^3-8}dx$$
I solved this integral by dividing it in partial fractions, than i came in two integrals $I_1,I_2$. Then, partioning $I_2$ into $I_3,I_4$ and partioning $I_4$ into $I_5,I_6$. But it took me so much work even though i got correct answer.
Is there any ... | Recognize
$$\frac{4x}{x^3-8}
= \frac{(x^2+2x+4) -x^2 +2(x-2)}{x^3-8}
= \frac1{x-2} - \frac{x^2}{x^3-8}+\frac2{x^2+2x+4}
$$
and integrate to obtain
$$4\int \frac x{x^3-8}dx=\ln|x-2|-\frac13\ln|x^3-8|+ \frac2{\sqrt3}\tan^{-1}\frac{x+1}{\sqrt3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove $\frac{\sin x}{x^2}$ is uniformly continuous at $N(0;r)^c$ for any $r >0$ I tried $\left \vert \frac{\sin x}{x^2} - \frac{\sin c}{c^2}\right \vert \leq \frac{1}{x^2} + \frac{1}{c^2} < \epsilon$, but it doesn't help me much with $\vert x - c \vert < \delta$. How can I prove this?
| Note that for $x,y \neq 0$,
$$\left|\frac{\sin x}{x^2} - \frac{\sin y}{y^2} \right| = \left|\frac{\sin x}{x^2} - \frac{\sin x}{xy} + \frac{\sin x}{xy}- \frac{\sin y}{y^2}\right| \\\leqslant \left|\frac{\sin x }{x}\right|\left|\frac{1}{x} - \frac{1}{y}\right| + \frac{1}{|y|}\left| \frac{\sin x}{x} - \frac{\sin y}{y}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Convergence of $\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$. My try. Examine the convergence of:
$$\sum_{n=1}^{\infty} x\frac{\sin(n^2x)}{n^2}$$
Pointwise convergence (for every $ x \in \mathbb{R}$ ):
$\displaystyle \lim_{n \to \infty} x\frac{\sin(n^2x)}{n^2} = 0$ , because $\sin(n^2x) \in [-1, 1]$
Uniform convergen... | In the comments you can see the issues with your solution. Here I present a possible approach to this problem. The pointwise convergence follows from the estimate
$$
\left| {x\sum\limits_{n = 1}^\infty {\frac{{\sin (n^2 x)}}{{n^2 }}} } \right| \le \left| x \right|\sum\limits_{n = 1}^\infty {\frac{{\left| {\sin (n^2 x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sum_{n=1}^{\infty}(\sqrt{n^2+3}-\sqrt{n^2-1})=\infty$ I want to prove $$\sum_{n=1}^{\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=\infty$$
If $\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\neq0$, I can prove it.
In fact, however, $$\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^... | Also:
\begin{align}
\sqrt{n^2+1} &= n\left(1+\frac{1}{n^2}\right)^{1/2} = n\left(1+\frac{1}{2n^2} + O(n^{-4})\right)
\\ &= n + \frac{1}{2n} + O(n^{-3})
\\
\sqrt{n^2-3} &= n - \frac{3}{2n} + O(n^{-3})\qquad\text{similarly}
\\
\sqrt{n^2+1} - \sqrt{n^2-3} &= \frac{2}{n} + O(n^{-3})
\end{align}
Thus the series diverges by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Area between $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$ and $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ I am trying to find the area between $x^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$ and $\left(x-\frac{1}{2}\right)^{2}+y^2=\frac{1}{4}$ using Double integrl in Polar coordinates.
Now as shown in th... | The polar equation of the circles are $r=\sin\theta=f(\theta)$ and $r=\cos\theta=g(\theta)$. They intersect at the pole and when $\theta=\frac{\pi}{4}$. Looking at the figure, the area is
$$A=\int_0^{\pi/4} \frac{\sin^2\theta}{2}\; d\theta + \int_{\pi/4}^{\pi/2}\frac{\cos^2 \theta}{2} = \frac{1}{2}\left( \int_0^{\pi/4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Converting equation into short Weierstrass form I have the curve $y^2 + 4y = x^3 + 3x^2 −x + 1$. I need to find a transformation of the form $X=x+a$ and $Y=y+b$
that turns this curve into the standard form of the elliptic curve: $$Y^2=X^3+AX+B.$$
I completed the square for the left side and now have $(y+2)^2= x^3 + 3x^... | You need to complete the cube in terms of $x$, this is very similar to completing the square, except that we have
$$(x+a)^3 = x^3 + 3ax^2 + 3a^2x + a^3$$
so if the coefficient of $x^2$ is $c$ then $(x + c/3)$ is the variable you should rewrite
$x^3 + cx^2 +dx + e$ in terms of so that you get
$$x^3 + cx^2 +dx + e =(x+c/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$N^2$ leaves a remainder of $1$ when divided by $24$. What are the possible remainders we can get if we divide $N$ by $12$? My solution approach is as below :-
$Rem[\frac{N^2}{24}] = 1$
$\Rightarrow Rem[\frac{N \times N}{24}] = 1$
$\Rightarrow Rem[\frac{N}{24}] \times Rem[\frac{N}{24}] = 1$
$\Rightarrow \text{This can... | Let's suppose the remainder of $N$ when divided by $12$ is some integer $r \in \{0, 1, 2, \ldots, 11\}$. That is to say, $$N = 12q + r,$$ where $q$ is the quotient. Then $$N^2 = (12q + r)^2 = 144q^2 + 24qr + r^2,$$ and we are told that this has a remainder of $1$ when divided by $24$. Since $144 = 6(24)$, the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the equation of the locus of the mid point of AB as m varies I am working through a pure maths book as a hobby. This question puzzles me.
The line y=mx intersects the curve $y=x^2-1$ at the points A and B. Find the equation of the locus of the mid point of AB as m varies.
I have said at intersection:
$mx = x^2-1 \... | The $x$-coordinate of the midpoint is the the sum of the other two divided by $2$, i.e. $$ x=\frac m2$$ And $$y=mx =\frac{m^2}{2} $$ and $$y=2x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to calculate a missing middle digit of a multiplication between 2 big numbers using modular arithmetic? What elementary number theory methods can I use for solving this type of questions?
One of these kind of problems is to find $x$ where $985242x6565 = 172195\cdot 572167$ without multiplying the numbers again.
I t... | The remainder of the division by 11 is equal to the alternating sum of the digits of the number. Because
$1 \equiv 1$ (mod 11)
$10 \equiv -1$ (mod 11)
$100 \equiv 1$ (mod 11)
$1000 \equiv -1$ (mod 11)
$10000 \equiv 1$ (mod 11)
and so on...
Thus,
$172195 \equiv (-1+7-2+1-9+5) \equiv 1$ (mod 11)
$572167 \equiv (-5+7-2+1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show $\begin{pmatrix}1&1 \\ 1&1\end{pmatrix}^n = 2^{n-1}\begin{pmatrix}1&1 \\ 1&1\end{pmatrix}$? Initial note: I'm interested in the combinatorics aspect of the following problem, not how to proof the relation in general.
The idea is to show the following relationship:
$$
\begin{pmatrix}1&1 \\ 1&1\end{pmatrix}^... | Some more aspects regarding OPs approach. When looking at the matrix $M$ in the form
\begin{align*}
M=A+B+C+D
\end{align*}
we can associate certain walks in the graph $G(M)$ given below. The graph consists of two nodes $1$ and $2$ and directed edges corresponding to the entries of $M$. We recall Theorem 1.1. from Algeb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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An Inequality concerning double integral
Prove $$ \frac{\pi}{8}\left(1-\cos\frac{2}{\pi}\right)\le \iint_D \sin (x^2)\cos (y^2) dxdy \le \frac{\pi}{8}(1-\cos 1),$$where the integrating region $D$ is enclosed by $x=0, y=0$ and $x+y=1$.
We can obtain
$$\iint_D \sin (x^2)\cos (y^2) dxdy=\int_0^1\int_0^{1-y}\sin(x^2)\cos... | We want to bound
$$
I = \iint_D \sin (x^2)\cos (y^2) dxdy
$$
where the integrating region $D$ is enclosed by $x=0, y=0$ and $x+y=1$. Since the integrand is symmetric in $x$ and $y$, we can write this as
$$
I = \frac12 \iint_E \sin (x^2)\cos (y^2) dxdy
$$
where the integrating region $E$ is enclosed by the rectangle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
taking an integral without trigonometric substitution I needed to evaluate $$\int_{-\infty}^\infty\frac{1}{x^2+a^2}dx$$
I looked around and found it's solved by trigonometric substitution $x = a\tan\theta$ and the answer is $\frac{\pi}{a}$. I understand the derivation but have been curious: supposing I didn't know the ... | Another approach is as follows:
$$\begin{align}
\int_{-\infty}^\infty\frac{1}{x^2+a^2}dx
&= 2\int_0^\infty \frac{1}{x^2+a^2}dx \\
&= \frac{2}{a^2} \int_0^\infty \frac{1}{1+(x/a)^2}dx \\
&= \frac{2}{a} \int_0^\infty \frac{1}{1+t^2}dt \\
&= \frac{2}{a} \arctan t\bigg|_0^\infty \\
&= \frac{\pi}{a}\end{align}$$
Note that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Minimal Polynomial Exercise $\newcommand{\Ker}{\operatorname{Ker}}$Let $E$ be a vector space. We are given a matrix $A$ that has a characteristic polynomial $q(t) = -(t-2)^5$. We know that $\dim\Ker(A-2I)^2 = 3$ and $\dim\Ker(A-2I)^4 = 5$. We are asked to find the minimal polynomial of $A$.
This is what I've done so fa... | Recall that $$ \dim \ker (A - \lambda I)^k - \dim \ker (A - \lambda I)^{k-1}$$ gives you the number of blocks corresponding to $\lambda$ of size at least $k$. See, for example: Jordan form, number of blocks.
So in this case you know that $\dim\ker(A-2I)^2 = 3$ and $\dim\ker(A-2I)^4 = 5$. Let us consider the three possi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the Largest integer that is smaller than this expression Largest integer that is smaller than
$\frac{2^{2^{2021}}}{(2^{2^1}-2^{2^0}+1)(2^{2^2}-2^{2^1}+1)...(2^{2^{2020}}-2^{2^{2019}}+1)}$
is...
My Progress
$\frac{1}{2^{2^{k+1}}-2^{2^k}+1}=\frac{2^{2^k}+1}{2^{3.2^k}+1}$
Reorder so
$\frac{2^{2^{2021}}}{(2^{2^1}-2^{2... | Your first step is, good.
Then use:
$$(x-1)\prod_{k=0}^n \left(x^{2^k}+1\right)=x^{2^{n+1}}-1 $$
Prove by induction.
Use it with $x=2,x=2^3.$
Then $$\begin{align}\frac1{\prod_{k=0}^{2019}\left(2^{2^{k+1}}-2^{2^k}+1\right)}&=\prod_{k=0}^{2019}\frac{2^{2^k}+1}{2^{3\cdot 2^k}+1} \\
&=(8-1)\frac{2^{2^{2020}}-1}{2^{3\cdot 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4146057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Smallest value of $b$ when $0<\left\lvert \frac{a}{b}-\frac{3}{5}\right\rvert\leq\frac{1}{150}$ Problem
For positive integers $a$ and $b$, $$0<\left\lvert \dfrac{a}{b}-\dfrac{3}{5}\right\rvert\leq\dfrac{1}{150}$$
What is the smallest possible value of $b$? (BdMO 2021 Junior P10)
My approach
If $\dfrac{a}{b}>\dfrac{3}{5... | HINT:
If $\frac{a}{b}$ satisfies the condition then
$$|5 a - 3 b| = 1$$
the explanation being that in the parallelogram on the vectors $(3,5)$ and $(a,b)$ there should be no other integer points excepts the vertices. Now, the solutions of
$$5 a - 3 b = 1$$
are $$a = 2 + 3 t \\ b = 3 + 5 t$$
while those of
$$5 a - 3 b =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
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Locus of point M such that its projections onto two fixed lines are always at the same distance. A point moves so that the distance between the feet of the perpendiculars drawn from it to the lines $ax^2+2hxy+by^2=0$ is a constant $c$. Prove that the equation of its locus is
$$4(x^2+y^2)(h^2-ab)=c^2(4h^2+(a-b)^2).$$
Wh... | Here's a non-trigonometric take on this problem.
The conic $ax^2+2hxy+by^2=0$ gives the two solutions $y=m_\pm x$ where $bm_\pm=-h\pm\sqrt{h^2-ab}$. Given $P(x_0,y_0)$, the perpendicular foot $A(x,m_+x)$ minimises $AP^2=(x-x_0)^2+(m_+x-y_0)^2$. This is minimised at $(m_+^2+1)x^*=x_0+m_+y_0$ so $$A\left(\frac{x_0+m_+y_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Does $\lim_{\epsilon \to 0^+}(\int_{1/2}^{1-ε}＋\int_{1＋ε}^{3/2})\frac {\log x} {(x-1)^2} dx$ exist? Does $$\lim_{\epsilon \to 0^+} \left(\int_{1/2}^{1-ε}＋\int_{1＋ε}^{3/2}\right)\frac{\log x}{(x-1)^2} dx$$ exist?
Hint says $$\lim_{\epsilon \to 0^+} \left( \frac{\log x-a-b(x-1)-c(x-1)^2}{(x-1)^3}\right)$$ exists only i... | Look at the Taylor expansion of $\log x$ around $x = 1$:
$$\log x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots.$$
Therefore, there exists a continuous function $g : [\frac{1}{2}, \frac{3}{2}] \to \mathbb{R}$ defined by
$$g(x) = -\frac{1}{2} + \frac{x-1}{3} - \frac{(x-1)^2}{4} + \cdots$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $\mathbf{A}_i=\lambda_i\mathbf{a}+\mu_i\mathbf{b}+\nu_i\mathbf{c}$, then $\mathbf{A}_1\cdot\left(\mathbf{A}_2\times\mathbf{A}_3\right)=$ Here is Prob. 2.43 in the book Vector Analysis & Cartesian Tensors by D. E. Bourne and P. C. Kendall, 3rd edition:
If
$$
\begin{align}
\mathbf{A}_1 &= \lambda_1 \mathbf{a} + \mu_1... | Your approach is correct, but there is a simpler proof using triple product formula:
$$\vec{u}.(\vec{v} \times \vec{w})=\det(u,v,w)\tag{1}$$
making the relationship to be proven equivalent to:
$$\det(A_1,A_2,A_3) \overset{?}{=} \begin{vmatrix} \lambda_1 & \mu_1 & \nu_1 \\ \lambda_2 & \mu_2 & \nu_2 \\ \lambda_3 & \mu_3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate the line intergal: $A=\int_C \sqrt{x^2+y^2}ds$ Calculate the line intergal:
$$A=\int_C \sqrt{x^2+y^2}ds, \, \text{with} \,\, C: x^2+y^2=ax$$
Set $\left\{\begin{align*}
x&=\dfrac{a}{2}\left( \dfrac{a}{2}+\cos t\right)\\[10pt]
y&=\dfrac{a}{2}sint
\end{align*}\right., \quad 0 \leq t \leq 2\pi$
So, $$\begin{a... | $C: x^2+y^2=ax \implies \left(x - \dfrac{a}{2}\right)^2 + y^2 = \dfrac{a^2}{4}$
So parametrize as $r(t) = \left(\dfrac{a}{2} + \dfrac{a}{2} \cos t, \dfrac{a}{2} \sin t \right), 0 \leq t \leq 2\pi$
$|r'(t)| = \sqrt{\left( - \dfrac{a}{2}\sin t\right)^2 + \left( \dfrac{a}{2}\cos t\right)^2} = \dfrac{a}{2}$
$\sqrt{x^2+y^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4160315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Collatz Conjecture: For a cycle where the maximum odd integer is $x_{max}$, does it follow that $x_{max} < 3^n$ I am working on understanding the upper limit in the case where a non-trivial cycle exists for the Collatz Conjecture.
Is the following reasoning valid for establishing that the maximum odd integer in a cycle... | The start of your reasoning is quite good, although there are some issues with your later steps. Otherwise, your arguments are basically correct.
In your (4), there are several relatively minor mistakes, but your overall upper bound still applies. Using your result of $p_n \lt 2n$, then each $p_{i+1} \gt p_{i}$ for $1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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$\frac{a_1}{\sqrt{1-a_1}}+\frac{a_2}{\sqrt{1-a_2}}+..+ \frac{a_n}{\sqrt{1-a_n}} \geq \frac{1}{\sqrt{n-1}}(\sqrt{a_1}+\sqrt{a_2}+\dots+ \sqrt{a_n})$ If $a_1,a_2,\dots ,a_n\ge 0$ such that $a_1+a_2+\dots a_n=1$ show that $\frac{a_1}{\sqrt{1-a_1}}+\frac{a_2}{\sqrt{1-a_2}}+\dots +\frac{a_n}{\sqrt{1-a_n}} \geq \frac{1}{\sqr... | Let $x \in (0,1)$, then notice that
$f(x)=\frac{x}{\sqrt{1-x}} \implies f''(x)=\frac{4-x}{4(1-x)^{5/2}}>0.$
So by Jensen's inequality, we can write that
$$S=\sum_{k=1}^n \frac{a_k}{\sqrt{1-a_k}}\ge n \frac{\sum_{k} a_k/n}{\sqrt{1-\sum a_k/n}}=\sqrt{n}\frac{\sum_k a_k}{\sqrt{n-1}}=\sqrt{n} \frac{\sqrt{\sum_k a_k}}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Let $x = (a^2+a+1)^{1/2}-(a^2-a+1)^{1/2}$ and $a$ belongs to real number find all possible value of $x$. Let $A = (-1/2, \sqrt{3}/2)$, $B = (1/2, \sqrt{3}/2)$, and $P = (a, 0)$. Then $P$ is a point on the $X$-axis and we are looking for all possible values of $d = PA - PB$.
By the Triangle Inequality, $PA - PB < AB = 1... | Since
$a^{1/2}-b^{1/2}
=(a^{1/2}-b^{1/2})\dfrac{a^{1/2}+b^{1/2}}{a^{1/2}+b^{1/2}}
=\dfrac{a-b}{a^{1/2}+b^{1/2}}
$,
$d(a)
=(a^2+a+1)^{1/2}-(a^2-a+1)^{1/2}
=\dfrac{2a}{(a^2+a+1)^{1/2}+(a^2-a+1)^{1/2}}
$.
Also,
by the generalized mean inequality,
$\sqrt{uv}
\le (\dfrac{\sqrt{u}+\sqrt{v}}{2})^2
\le \dfrac{u+v}{2}
$
or
$2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Computing $\lim_{x\to-1}\frac{2x+\sqrt{3-x}}{x^2+x}$ I have this limit as my question to solve:
$$\lim_{x\to-1}\frac{2x+\sqrt{3-x}}{x^2+x}$$
My procedure:
$$\lim_{x\to-1}\frac{(2x+\sqrt{3-x})(2x-\sqrt{3-x})}{(x^2+x)(2x-\sqrt{3-x})}$$
$$\lim_{x\to-1}\frac{4x^2+x-3}{(x^2+x)(2x-\sqrt{3-x})}$$
$$\lim_{x\to-1}\frac{(x+1)(x-... | Hint: Your factorisation is not correct.
It is easier to do it if you let $t=\sqrt{3-x}$, then $x= 3-t^2$ and $t\to 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using Monotonicity to find the value of a required variable Given $f(x) = \log_c\frac{x-2}{x+2}$, is defined $\forall \; x \in [a,b]$ and the function is monotonically decreasing.
We have to find the value (or range of values) of $c$ such that there exists $a$ and $b$, where $(2 < a < b)$, and the range of the function... | As you wrote, we have
$$\frac{a-2}{a+2} = c(a-1)$$
which is equivalent to
$$ca^2+(c-1)a-2c+2=0\tag1$$
Similarly, we have
$$cb^2+(c-1)b-2c+2=0\tag2$$
From $(1)(2)$, we want to find $c\ (0\lt c\lt 1)$ such that the equation
$$cx^2+(c-1)x-2c+2=0$$
has two distinct real roots $x$ satisfying $x\gt 2$.
So, we want to find $c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding the range of $a$ for which line $y=2x+a$ lies between circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting either
Find the range of parameter $a$ for which the variable line $y = 2x + a$ lies between the circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting or touching ei... | I would draw perpendicular to the line $2x - y + a = 0$ from centers of both circles.
From circle $C_1$ centered at $(1, 1)$,
$d_1 = \dfrac{|2 -1 + a|}{\sqrt5} \gt 1$
(where $1$ is the radius of the circle $C_1$).
Similarly from $(8, 1)$, $d_2 = \dfrac{|16 - 1 + a|}{\sqrt5} \gt 2$
(where $2$ is the radius of the circle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Methods for solving nonhomogeneous system of ODEs I have the system
$$ \mathbf{x}'=\begin{pmatrix}2&1\\1&2\end{pmatrix}\mathbf{x}+\begin{pmatrix}e^t\\t\end{pmatrix}. $$
The eigenvalues of the matrix are 1 and 3 with eigenvectors $(1,-1)^T$ and $(1,1)^T$ respectively. Thus, we have the fundamental matrix
$$
\Psi(t)=\beg... | For the first part, I will use this approach. We are given
$$\mathbf{x}'=\begin{pmatrix}2&1\\1&2\end{pmatrix}\mathbf{x}+\begin{pmatrix}e^t\\t\end{pmatrix}$$
The eigenvalues/eigenvectors are
$$\lambda_1 = 1, v_1 = \begin{pmatrix}-1 \\ 1 \end{pmatrix} \\ \lambda_2 = 3, v_2 = \begin{pmatrix}1 \\ 1 \end{pmatrix} $$
The Fun... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluating $\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x$ While I was working on computing $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$, I came across the integral:
$$I=\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x.$$
I tried $x=\sin(u)$ then $\tan(u/2)=t$ and got
$$I=... | Since you are primarily interested in that series I'll evaluate it.
$$\sum _{k=1}^{\infty }\frac{H_{2k}^{\left(2\right)}}{4^kk^2}\binom{2k}{k}$$
Consider the well-known expansion:
$$2\sum _{k=1}^{\infty }\frac{\left(H_{2k}^{\left(2\right)}-\frac{1}{4}H_k^{\left(2\right)}\right)}{4^k}\binom{2k}{k}x^{2k}=\frac{\arcsin ^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Partial Fraction Decomposition of $\frac{z+4}{(z+1+2i)(z+1-2i)}$ I have the fraction $\frac{z+4}{(z+1+2i)(z+1-2i)}$. I want to partial decompose this fraction, but I am not seeing how to do it. I know that the answer is a=$\frac{1}{2}$+$\frac{3i}{4}$ and b=$\frac{1}{2}$-$\frac{3i}{4}$, where
$$\frac{z+4}{(z+1+2i)(z+1-2... | Let $k=1+2i$, then we want to find $a$ and $b$ such that
$$\frac{z+4}{(z+k)(z+\bar{k})}=\frac{a}{z+k}+\frac{b}{z+\bar{k}}$$
Or $$z+4=(z+\bar{k})a+(z+k)b$$
Then setting $z=-\bar{k}$ and $z=-k$ we obtain $b=\frac{-\bar{k}+4}{k-\bar{k}}=\frac{-(1-2i)+4}{(1+2i)-(1-2i)}=\frac{3+2i}{4i}=\frac{1}{2}-\frac{3}{4}i$ and $a=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Root of the given polynomial $4x^3-3x-p$ Given the equation $$ 4x^3 - 3x-p=0 $$
In the question we were required to find the root of this equation in the interval $[1/2,1]$ and $-1\le p\le 1$. The answer is arrived at by substituting x as $\cos(\theta)$
and using the multiple angle formula. However assuming I am not ab... | If the solution is not rational, you can use Cardano's method to find one real root. $4x^3-3x-p=0$ is equivalent to $x^3-\frac{3}{4}x-\frac{p}{4}=0$. And now you let $t=u+v$, which only holds if $-\frac{3}{4}=-3uv$ and $-\frac{p}{4}=-(u^3+v^3)$. Therefore, $u=\frac{1}{4v}$ and $\frac{p}{4}=v^3+\frac{1}{64v^3}$. Multipl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How do I calculate the sum of sum of triangular numbers? As we know, triangular numbers are a sequence defined by $\frac{n(n+1)}{2}$. And it's first few terms are $1,3,6,10,15...$. Now I want to calculate the sum of the sum of triangular numbers. Let's define
$$a_n=\frac{n(n+1)}{2}$$
$$b_n=\sum_{x=1}^na_x$$
$$c_n=\sum_... | Hint:
$$\sum_{r=1}^n r=\frac{n(n+1)}{2}$$
$$\sum_{r=1}^n r^2=\frac {n(n+1)(2n+1)}{6}$$
$$\sum_{r=1}^n r^3=\frac {(n(n+1))^2}{4}$$
Use of these $3$ formulae is sufficient to prove the required result.
The derivation of the $3^{rd}$ formula can comes by noting:
$$(r+1)^4-r^4=4r^3+6r^2+4r+1$$
Now sum this identity over $r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
$\textbf{My attempts} :$
From the condition, we can write $$1+2n^2+3m^2\equiv 0(\mod 5)\tag 1$$
Now, since $5$ is pr... | squares are $0,1,4 \pmod 5$ so $3m^2 \equiv 0,3,2 \pmod 5,$ next $1+3m^2 \equiv 1,4,3 \pmod 5$ Finally
$$ -(1+3m^2) \equiv 4,1,2 \pmod 5 \; , \; \; $$
$$ 2 n^2 \equiv 0,2,3 \pmod 5 $$
The overlap of these two lists, $4,1,2$ and $0,2,3$ is the single possibility $2.$ That is, we need $2n^2 \equiv 2 \pmod 5$ and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove multiplication closure for the sequence of 1,4,7,10.... I am reading the following problem:
If S = ${1, 4, 7, 10, 13, 16, 19, ...}$ and $a \in S\space$ and $b\in
S \space$ then if $a = b\cdot c\space$ prove that $c \in S$
My approach:
The elements of $S$ are of the form $1 + 3n\space$ so $a = 1 + 3\cdot x\spac... | Take $c = r + 3k_1, a = 1 + 3k_2$ and $b = 1 + 3k_3$ where$k_1,k_2,k_3 $ and $r$ are integers. By Division theorem we can chose $r$ such that $r =0,1 $ or $2$. Now we know that $a = bc$ so
$$3k_2 + 1 = (r+3k_1)(1+3k_3) = r + 3k_3r+ 3k_1 + 9k_1k_3$$
So by rearranging terms we have
$$ 1 - r = 3( k_3r + k_1+3k_1k_3 - k_2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Is the determinant of a Hessenberg matrix whose elements are the Bernoulli numbers positive? The Bernoulli numbers $B_r$ are generated by
\begin{equation}\label{Bernoullu=No-dfn-Eq}
\frac{z}{e^z-1}=\sum_{r=0}^\infty B_r\frac{z^r}{r!}
=1-\frac{z}2+\sum_{r=1}^\infty B_{2r}\frac{z^{2r}}{(2r)!}, \quad \vert z\vert<2\pi.
\e... | The so-called guess $\mathfrak{D}_r>0$ can be proved to be true via mathematical induction and by the following recursive relation.
Let $H_0=1$ and
\begin{equation*}
H_r=
\begin{vmatrix}
h_{1,1} & h_{1,2} & 0 & \dotsc & 0 & 0\\
h_{2,1} & h_{2,2} & h_{2,3} & \dotsc & 0 & 0\\
h_{3,1} & h_{3,2} & h_{3,3} & \dotsc & 0 & 0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $a$ & $b$ , s.t. $a+b=0$ and $ab=-3$? I've a problem in this question :
In the polynomial identity $x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)$ , find $ab$ $? $
MY APPROACH
We have : $$x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$$
Now according to the Problem : $$x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)=(x^2+1)(x^4-x^2+1)$$ Or $$x^4+(a+... | The equation $t^2-(a+b)t+ab=0$ can be written $(t-a)(t-b)=0$ and thus has solutions $t=a$ and $t=b$.
In your case this equation becomes $t^2-3=0$ which has solutions $t=\pm\sqrt3,$ so $a,b=\pm\sqrt3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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prove that $x^6=(ax)^3$ implies $x^3=e,\: \forall x\in G$ Let be $(G,\cdot)$ a group and $e$ the identity element.
Let be $a\in G$.
If $$x^6=(ax)^3,\:\forall x\in G$$
Then
$$x^3=e,\: \forall x\in G$$
What I did:
$$x^6=(ax)^3,\:\forall x\in G$$
$x \gets a^2$ then $a^3=e$.
$$x^6=(ax)^3,\:\forall x\in G \implies x^6=(xa)^... | As @ancient mathematician hinted, for all $x\in G$
$$x^{24} = \left(x^6\right)^4 = (ax)^{12} = \dots = x^3\,,$$
so $x^{21}=e$.
What's more, since you proved $(ax)^3 = (xa)^3$, it follows that
$$x^3 = (a^{-1}xa)^3 = a^{-1}x^3a\,,$$
which implies that $\forall x\in G, x^3a = ax^3$.
Thus $x^{18}=(x^3)^6 = a^3x^9 = x^9$, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving $2x + 1 = 11$: Why, when subtracting $1$, do I only do it to a single term on the left but, if dividing by $2$, I must divide both terms? Solving $2x + 1 = 11$ (for example)
Why, when subtracting $1$, do I only do it to a single term on the left but, if dividing by $2$, I must divide both terms ?
| The key thing to realise here is that multiplication is distributive over addition. This is essentially a fancy way of saying that you can expand brackets: $(3+x)\times 5=(3 \times 5)+(x\times 5)$, and $(3+7)\times 4=(3\times 4)+(7\times 4)$, and $(a+y)\times 100=(a\times 100)+(y\times 100)$. In general, the distributi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
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Probability two blocks have exactly 2 out of 4 properties the same?
I have 120 blocks. Each block is one of 2 different materials, 3 different colors, 4 different sizes, and 5 different shapes. No two blocks have exactly the same of all four properties. I take two blocks at random. What is the probability the two bloc... | You have made a simple arithmetic error. I evaluated your expression and I got $\dfrac{5}{17}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Input of a function must give specific output, find number of positive integers that satisfy. Question: The prime factorization of an integer $n \geq 2$ is of the form $({p_1})^{a_1}(p_2)^{a_2} … (p_k)^{a_k}$ where $p_1$, $p_2$, …, $p_k$ are different prime numbers and $a_1, a_2, …, a_k$ are positive integers. Given an... | It seems like your solution is valid. However, you should state that $a < 8$ instead of $a \leq 6$ because $120/15 = 8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4199341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $x+y+z=3$, how do I maximize $xy+yz+zx-xyz$? Given non-negative numbers $x,y,z$ such that $x+y+z=3$. How do I maximize $xy+yz+zx-xyz$. I found out that the maximum is 2 and equality holds when one of the numbers is 1 and the other two sum up to 2 but had no idea how to prove it. Can someone help please?
| We can equivalently want to maximize:
$$
\begin{aligned}
(1-x)(1-y)(1-z)
&=1-\underbrace{(x+y+z)}_{=3}+(xy+yz+zx)-xyz\\
&=-2 + (xy+yz+zx)-xyz\\[3mm]
&\qquad\text{ so let us substitute }\\
X&=1-x\ ,\\
Y&=1-y\ , \\
Z&=1-z\ .\qquad\text{ These variables satisfy:}\\
X+Y+Z &= 3-(x+y+z)=0\ ,\text{ and}\\
(*)\qquad 1&\ge X,Y,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solving $\sqrt{(\log_2x)^2 - 5|\log_2x| + 6} \le 2\sqrt{5}$ My attempt:
Given,
$$\sqrt{(\log_2x)^2 - 5|\log_2x| + 6} \le 2\sqrt{5}$$
Taking $t=|\log_2x|$:
$$\sqrt{t^2-5t+6} \le 2\sqrt{5}$$
I know that squaring both sides will lead to extraneous solutions. In fact, I solved the rest of the problem by squaring sides and ... | This inequality is quite fiddly, but with care we can get to the solution. First, note that any inequality of the form $\sqrt{a}\le b$ is equivalent to $a\le b^2$ and $a\ge0$, meaning
\begin{align}
\sqrt{t^2-5t+6}\le2\sqrt{5} &\iff t^2-5t+6\le20 \quad\text{and}\quad t^2-5t+6\ge0 \\[5pt]
&\iff (t+2)(t-7)\le0\quad\text{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $k^7/7 + k^5/5 + 2k^3/3 - k/105$ is an integer I tried to prove this using induction.
Let $k=1$; then the equation gives
$$1/7 + 1/5 +2/3 – 1/105 = 105/105 = 1,$$
which is an integer. So it is true for $k=1$. Now let it be true for $n>k$. This gives
$$105|(15n^7 + 21n^5 + 70n^3 – n).$$
For $(n+1)$ we have
$$105|(... | I think you made a mistake, and it should have been
$15n^7+105n^6+336n^5+630n^4+805n^3+735n^2+419n+105$
where you typed $15n^7 + 105n^6 + 336n^5 + 630n^4 + 875n^3 + 945n^2 + 629n + 175$,
so it should have been
$105n^6 + 315n^5 + 630n^4 + 735n^3 + 735n^2 + 420n + 105$
where you typed $105n^6 + 315n^5 + 630n^4 + 805n^3 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Computing $\sum_{k = 1}^{n} \dfrac{k}{2^k} $ in different ways I am trying to compute the following sum \begin{align} S = \sum_{k = 1}^{n} \dfrac{k}{2^k} \end{align}
I have computed this sum and found that
$$ \bbox[5px,border:2px solid red]
{
S = 2 - \dfrac{n+2}{2^{n}}
}
$$
For prooving this let's rewrite the sum in t... | First, compute the value of a geometric series
$$
T_n = \sum_{k=1}^n (1/2)^k .
\tag1$$
We can multiply by $1/2$ to get
$$
\frac{1}{2} T_n = \sum_{k=1}^n (1/2)^{k+1} = \sum_{k=2}^{n+1} (1/2)^k
\tag2$$
Compute $(1) - (2)$:
$$
T_n - \frac{1}{2}T_n = \sum_{k=1}^{n} (1/2)^k- \sum_{k=2}^{n+1} (1/2)^k
\\
\frac{1}{2} T_n = (1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Using Rearrangement Inequality .
Let $a,b,c\in\mathbf R^+$, such that $a+b+c=3$. Prove that $$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
$Hint$ : Use Rearrangement Inequality
My Work :-$\\$
Without Loss of Generality let's assume that $0\le a\le b\le c$ then we can infer that $c^2\ge b^... | I'm not sure how to use Rearrangement Inequality but I can do it with tangent line method.
Using $a+b+c=3$ we can rewrite OP as: $$\underbrace{{a^2\over 3-a}-{a^2\over 2}}_{f(a)} + {b^2\over 3-b}- {b^2\over 2}+{c^2\over 3-c} -{c^2\over 2} \geq 0$$ If you calculate a tangent at point $x=1$ on $f(x)$ we get $y={1\over ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Asymptotic approximation for $\int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-a^2x^2}}\, dx$ as $a \rightarrow 0$
Asymptotic approximation for $$\int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-a^2x^2}}\, dx$$ as $a \rightarrow 0$.
In such an integral, when $a$ approaches 0, the term $a^2x^2$ is always << 1 because x is bounded from $0$ to... | Yes, with such approximation we find that as $a\to 0$,
$$\int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-a^2x^2}}\,dx=
\frac{\pi}{4}+\frac{\pi a^2}{32}+O(a^4).$$
You may also obtain a more precise approximation, by using Wallis' integral,
$$\int^1_0 \sqrt{1-x^2}\, x^{2n}\,dx=\int^{\pi/2}_0 \cos^2(t)\, \sin^{2n}(t)\,dt=\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(1+A)^n \ge (1+a_1)\cdot(1+a_2)\cdots(1+a_n) \ge (1+G)^n$ $A$ and $G$ are arithmetic mean and the geometric mean respectively of $n$ positive real numbers $a_1$,$a_2$,$\ldots$,$a_n$ . Prove that
*
*$(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$
*if $k$$\gt$$0$, $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdo... | Your approach works fine. I feel I should point out that the second result follows trivially from the first one by thinking about what happens when you multiply all the $a_i$ by $k$. Also, there are a few places you say "integer" when you mean real number.
For the missing step in your method, you can use the same appro... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Question on partial answer of the generalized Catalan's conjecture (case $n =2$) Edited question.
Let $n,m\in\Bbb N$ be integers greater or equal to $2$. If $3\leq m<n$ then there is no $m,n$ such that $m^n -n^m =2$. If $n<m$ and $m$ is prime then there is no $m,n$ such that $m^n-n^m =2$. If $m = 2$ then there is no $... | If $p$ is prime, then looking at $p^n-n^p=2$ mod $p$ shows that $n\equiv-2\pmod{p}$.
We don't need to consider $p=2$, since then $n$ must be even and then $p^n$ and $n^p$ are divisible by $4$.
$\boldsymbol{n\lt p}$
Since $n\equiv-2\pmod{p}$, if $n\lt p$, we must have that $n=p-2$
$$
\begin{align}
p^n-n^p
&=p^{p-2}-(p-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that $T(n)= 2^{n+1} - 1$ Show that $T(n)= 2^{n+1} - 1$, for the following,
$$
T(n) = \cases{ 1 & if, $n=0$ \cr
T(n-1) + 2^n & otherwise}
$$
I did it as follows,
\begin{align*}
T(n) = T(n-2) + 2^{n-1} + 2^n \\
T(n) = T(n-3) + 2^{n-2} + 2^{n-1} + 2^n\\
&\vdots\\
T(n) = T(... | Your work contains a number of errors.
First of all, your equation $$T(n-1) = T(n-2) + 2^{n-1} + 2^n$$ makes no sense, since $$T(n) = T(n-1) + 2^n \tag{1}$$ from the given definition implies $$T(n-1) = T(n-2) + 2^{n-1}, \tag{2} $$ simply by replacing $n$ with $n-1$.
What you probably intended to write is
$$T(n) = T(n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the system of differential equations and plot the curves given the initial conditions. We are given the following system of ordinary differential equations:
$$\dot{x} = x - 4y \quad \dot{y} = x - 2y -4.$$
Thus, the slope of the trajectories is given by
$$\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{x - 2y -4}{... | Mathematica:
$$\left\{x(t)\to \frac{1}{7} e^{-t/2} \left(56 e^{t/2} \sin ^2\left(\frac{\sqrt{7}
t}{2}\right)-3 \sqrt{7} \sin \left(\frac{\sqrt{7} t}{2}\right)+56 e^{t/2} \cos^2\left(\frac{\sqrt{7} t}{2}\right)+49 \cos \left(\frac{\sqrt{7}
t}{2}\right)\right), y(t) \to \frac{1}{7} e^{-t/2} \left(14 e^{t/2} \sin
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Stuck solving $\iint \frac{x^2}{(x^2+y^2)}dA$ over the region $R$, $R$ lies between $x^2+y^2=4$ and $x^2+y^2=49$ by changing to polar coordinates. I plugged in for $x=r \cos\theta$ and $y=r \sin\theta$ to change to polar coordinates
I set my bounds as $2 \leqq r\leqq 7$ and $0\leqq \Theta \leqq 2\pi$.
I simplified my i... | You have a mistake in your integral. First of all, it should have been $x^2 = r^2 \cos^2\theta$.
Also $ \int \displaystyle \sin^2\theta \ d\theta \ne \cos^3\theta$.
Write $\sin^2\theta$ as $\cfrac{1 - \cos2\theta}{2} \ $ or $ \ \cos^2\theta \ $ as $ \ \cfrac{1 + \cos2\theta}{2}$
But there is an easier way to do this. D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to compute $\int_0^\infty \frac{\log(x)}{\sqrt{x}(x+1)} \ dx$ using the Residue Theorem? I have made the following attempt (I won't put every detail, but if necessary, I will edit the question).
I consider $\Gamma_{R,\varepsilon}$ as the following path:
and consider computing $\int_\Gamma \frac{\log^2(x)}{\sqrt{x}... | On the "lower" part of the branch cut we have
$$\begin{align}
\int_{\gamma_3} \frac{\log^2(z)}{\sqrt{z}(z+1)}\,dz&=-\int_0^\infty\frac{(\log(x)+i2\pi)^2}{\sqrt{xe^{i2\pi}}(x+1)}\,dx\\\\
&=\int_0^\infty \frac{(\log(x)+i2\pi)^2}{\sqrt{x}(x+1)}\,dx
\end{align}$$
This might not be fit for purpose here.
But we also have on ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate via substitution: $x^2\sqrt{x^2+1}\;dx$ I need to integrate the following using substitution:
$$
\int x^2\sqrt{x^2+1}\;dx
$$
My textbook has a similar example:
$$
\int \sqrt{x^2+1}\;x^5\;dx
$$
They integrate by splitting $x^5$ into $x^4\cdot x$ and then substituting with $u=x^2+1$:
$$
\int \sqrt{x^2+1}\;x^4\c... | One way to proceed is:
\begin{align}
\int x^2 \sqrt{x^2+1} \, dx
&= \frac{1}{2} \int \sqrt{u} \sqrt{u-1} \, du \qquad (u = x^2 + 1) \\
&= - \frac{1}{16} \int \frac{y^8 - 2 y^4 + 1}{y^5} \, dy \qquad (y = \sqrt{u} - \sqrt{u-1}) \\
&= - \frac{y^4}{64} + \frac{1}{64 y^4} + \frac{\log(y)}{8} + C.
\end{align}
Here, to chan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Find the second real root for cubic $x^3+1-x/b=0$ A cubic of the form $$x^3+1-x/b=0$$ has has three real roots
Using the Lagrange inversion theorem one of the roots is given by
$$x = \sum_{k=0}^\infty \binom{3k}{k} \frac{b^{3k+1} }{(2)k+1} $$
How do you find the second one? I cannot find any info online. I am looking... | $x^3 - \frac {x}{b} =-1$
Let $x = 2\frac {\sqrt {3b}}{3b}\cos\theta$
$\frac {8\sqrt {3b}}{9b^2} \cos^3\theta - \frac {2\sqrt{3b}}{3b^2}\cos\theta = 1\\
\frac {2\sqrt{3b}}{9b^2}(4\cos^3\theta - 3\cos\theta) = 1\\
\cos3\theta = \frac {3b\sqrt {3b}}{2}\\
\theta = \frac 13 \arccos\left(\frac {3b\sqrt {3b}}{2}\right), \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Integration By Substitution： Why are the two results different? The sphere: $x^2+y^2+z^2\leq a^2$ is intercepted by the cylindrical surface $x^2+y^2=ax$. Calculate the intercepted volume.
Consider the intercepted volume of the upper hemisphere, and then multiply it by 2:
$$D=\{(x,y):x^2+y^2\leq ax\}$$
Now calculate $A$... | Thanks @Benjamin_Gal.
In the comment of this question, he pointed out:
$$
(\sin^2\theta)^{3/2}=|\sin^3\theta|\ne\sin^3\theta
$$
And the correction would be:
$$
\begin{eqnarray}
A=\iint\limits_{D}\sqrt{a^2-x^2-y^2}\text{d}x\mathrm{d}y&=&\int_{\color{black}{-\pi/2}}^{\pi/2}{\text{d}\theta\int^{a\cos\theta}_{0}}r\sqrt{a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $2^{2n+7} \equiv 2 (mod 3)$ for all integer and non-negative number n. Originally, I want to prove this problem:
for all integer and non-negative number $n$,
$$\frac{1}{3}\times2^{2n+7}+\frac{37}{3} \equiv 0\quad(\text{mod}\space3)$$
I tried like this:
$$\frac{1}{3}\times2^{2n+7}+\frac{37}{3} = \frac{1}{3}(2^{2n+... | Thanks Will199 and Evariste.
for all integer and non-negative number $n$,
$2^{n}\equiv (-1)^{n} \equiv -1($when n is odd$)$ or $1$(when n is even) $(mod \quad3)$
$2n+7$ is always odd because $2n$ is even and $7$ is odd and even+odd=odd.
So, $2^{2n+7}\equiv (-1)^{2n+7} \equiv -1 \equiv 2 \quad(mod \quad3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Which solution of this integral $\int{\frac{x^2+1}{x^4-x^2+1}}dx$ is correct? $$\begin{align}
\int{\frac{x^2+1}{x^4-x^2+1}}dx&=\int{\frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}}dx\\
&=\int{\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+1}}dx\\
&=\int{\frac{1}{u^2+1}}du \quad(u=x-\frac{1}{x})\\
&=\arctan(u)+C\\
&=\arctan(x-\fr... | Both are. Remember that antiderivatives may differ by a constant (hence the $+ C$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given $B=AA^T$ and $C=A^TA$, can we calculate $C$ if we know $B$ but not $A$? Consider $B=AA^T$ and $C=A^TA$ with second-order tensor $A$. With knowledge of $B$, but no knowledge of $A$, can we calculate C?
I suspect it should not be possible since different $A$ probably can result in the same $B$, but on the other han... | Permute the columns of $A$ doesn't change $B$ but may change $C$, so you can't establish a function $C=f(B)$.
The reason that this is true is that $B_{ij}$ is the dot product of the $i$th and $j$th rows whereas $C_{ij}$ is the dot product of the columns. Permutations on columns have no effect on dot product of rows.
Ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Pairs of integers $ (x,m)$ for which $\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$ hold?
Find all pairs of integers $(x,m)$ for which $$\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$$ hold.
I have used this property :
Property:
if $$a+b+c=0 \implies a^3+b^3+c^3=3abc, $$ I come up to the followin... | Let $y=\sqrt[3]{x-2}$, then we have $$2 = \sqrt[3]{m+y} + \sqrt[3]{m-y}\;\;\;\;\;\;|^3$$ $$8 = m+y +3\sqrt[3]{m^2-y^2}\cdot 2 +m-y$$ and thus $$4-m = 3\sqrt[3]{m^2-y^2} $$
so $$ 64 -48m+12m^2-m^3 = 27m^2-27y^2$$ or $$27y^2 =m^3+15m^2+48m-64$$ and finally $$\boxed{27y^2 = (m-1)(m+8)^2}$$
Since $\gcd(m-1,m+8) \mid 9$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$
$$\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$$
I tried factoring as $${(a^2-x^2)(a^2+x^2)}$$ $$\text{and}$$ $$(x^2+a^2+ax)(x^2+a^2-ax)$$ but didn't get much after this. I sew that the expression is symmetric in a,x but I don't know how to use that.
| Rewrite the integrand as
$$\frac{a^4 - x^4}{x^4 + a^2x^2 + a^4}
=-1+ \frac{\frac {3a}2(\frac 1a +\frac{a}{x^2})-\frac a2(\frac 1a -\frac{a}{x^2})}{\frac{x^2 }{a^2}+ \frac{a^2}{x^2}+1}
$$
and then integrate as follows
$$\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx
= -x + \frac{3a}2\int \frac{d(\frac xa-\frac ax)}{(\frac xa-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$ Let $a,b,c>0$:
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$
My solution:
We have: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} \\\... | Another way.
After full expanding we need to prove that:
$$\sum_{cyc}(2a^4b^2+2a^4c^2-4a^3b^3+5a^4bc+4a^3b^2c+4a^3c^2b-13a^2b^2c^2)\geq0,$$ which is true by Muirhead because
$(4,2,0)\succ(3,3,0),$ $(4,1,1)\succ(2,2,2)$ and $(3,2,1)\succ(2,2,2).$
Also, we can see it by AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Calculating value of $\delta$ for a function $y=\frac1{(1-x)^2}$ unbounded at $x\rightarrow 1$ The function $y=\frac 1{(1-x)^2}$ is unbounded at $x\rightarrow 1$.
How can I calculate the value of $\delta$ such that $y> 10^6$ if $\left | x-1 \right |< \delta$ ?
Taking $y=f(x)=10^6$, putting in ${y=\frac1{(1-x)^2}}$, we ... | Pick $\delta$ such that
$$y=\frac{1}{\delta^2}=10^6$$
then at $\delta=±0.001$, $y=10^6$.
So$$|x-1|<\delta=0.001$$
in order to have $$y=\frac{1}{(1-x)^2}>\frac{1}{\delta^2}=10^6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Let $x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$ are rational numbers. Show that $x+\frac{1}{x}$ is rational.
$x^3+\dfrac{1}{x^3}$ and $x^4+\dfrac{1}{x^4}$ are rational numbers. Prove that $x+\dfrac{1}{x}$ is rational number.
My solution:
$x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-1\right... | I constructed a possible elementary algebraic way:
Let,
$$\begin{cases}x+\frac 1x=v,\thinspace v \not \in \mathbb Q\\
x^2+\frac{1}{x^2}=u,\thinspace u\in\mathbb R\\
x^3+\frac{1}{x^3}=p,\thinspace p\in\mathbb Q\\
x^4+\frac{1}{x^4}=q,\thinspace q\in\mathbb Q\end{cases}$$
This implies,
$$pv=u+q\not\in \mathbb Q\implies u\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 1
} |
Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:
As $10 \equiv 1 \mod ... | You have shown that $10^n + 3 \cdot 4^{n+2} + 5=3(2+2^{n+4})$ modulo $9$, so it should be enough to show that $2+2^{n+4}=3k$ for some $k$. You can do this by induction.
The base case is obvious, so we will assume $2+2^{2n+4}=3k$ for some n, then:
$$
2+2^{2(n+1)+4}
=
2+4\cdot 2^{2n+4}
=
(2+4(2^{2n+4}+2-2))=
12k-6
=
3(4k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Coefficient of $x^{10}$ in $f(f(x))$ Let $f\left( x \right) = x + {x^2} + {x^4} + {x^8} + {x^{16}} + {x^{32}}+ ..$, then the coefficient of $x^{10}$ in $f(f(x))$ is _____.
My approach is as follow
$f\left( {f\left( x \right)} \right) = f\left( x \right) + {\left( {f\left( x \right)} \right)^2} + {\left( {f\left( x \rig... | To cut it short, for $x^{10}$ to appear in $f(x)^n$, you just need to have some $2^k$'s add up to 10. It doesn't necessarily require binary expansions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Why is this approach wrong?
If $\alpha$ and $\beta$ are the roots of $x^{2} - 4 x - 3$, then find $$\frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}$$
Solution
This is the approach that gives wrong answer.
\begin{array}{l}
x^{2}-4 x-3=0 \\
(x-4)^{2}=19 \\
So, {(\alpha-4)^{2}} {\&(\beta-4)^{2}}=19 \\
\text { Hence, }... | $x^2-4x-3=(x-2)^2-7$
This is the mistake in your analysis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4239918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding three unknowns from three equations Let $a$,$b$ and $c$ be three positive real numbers such that
$$\begin{cases}3a^2+3ab+b^2&=&75\\
b^2+3c^2&=&27\\c^2+ca+a^2&=&16\end{cases}$$
Find the value of $ ab+2bc+3ca$.
My attempt: I observed that $3 . 16+27=75$. Then on replacing $16$ by $c^2+ca+a^2$, $27$ by $b^2+3c^2... | There are two values of the expression.
The value of
$ ab+2bc+3ca=-24\sqrt{3}$
and
$ ab+2bc+3ca=+24\sqrt{3}$.
The respective values of $a$, $b$ and $c$ are:
$a=+\sqrt{\frac{24576\sqrt{3}}{6553}+\frac{93184}{6553}}$,
$b=+\sqrt{\frac{58995}{6553}-\frac{31104\sqrt{3}}{6553}}$,
$c=-\sqrt{\frac{10368\sqrt{3}}{6553}+\frac{39... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Differential equation, tricky How would I solve the following:
$(x^2-y^2)dy+(y^2+x^2y^2)dx=0$?
Here is what I did:
$\frac{dy}{dx}(x^2-y^2)=-(y^2+x^2y^2)$
Dividing through, to leave differential on one side:
$\frac{dy}{dx}=\frac{-(y^2+x^2y^2)}{x^2-y^2}$
I then proved this is a homogeneous differential equation.
This is ... | $$
\left(x^2 - y^2\right)dy = \left(y^2 + x^2y^2\right)dx
$$
This becomes
$$
\left(\frac{x^2}{y^2} - 1\right)dy = (1+x^2)dx
$$
$$
\frac{dx}{dy} = \frac{\frac{x^2}{y^2} - 1}{1+x^2}
$$
If we assume that $y = xt^a$ we have
$$
\frac{d}{dy} = \frac{dt}{dy}\frac{d}{dt} = \frac{1}{at^{a-1}x}\frac{d}{dt}
$$
Subbing into the or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Ben and Jordan have three coins between them. Two of them are fair but one of them has a 4/7 chance of showing heads. Now, Ben and Jordan both flip each coin once and write down the outcomes.
What is the probability they both get the same number of heads?
My approach:
4 possible outcomes: 3H, 3T, 2H&1T, 2T&1H.
$P(3H) ... | You don't need to multiply by binomial coefficient or in any other way consider different orders of coins. We can simply say that first and second coins are fair, and the last is weighted.
For example, $3H$ means simply that both fair coins and weighted came heads, thus $P(3H) = \frac{1}{2} \cdot \frac{1}{2} \cdot \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x \to 0} \frac{1-\cos(x^2+2x)}{1-\cos(x^2+x)}$ without L'Hopital Could you help me with this limit please:
$$
\lim_{x \to 0} \frac{1-\cos(x^2+2x)}{1-\cos(x^2+x)}
$$
I know that with L'Hopital is easy but I want to do it without that theorem. I have already tried converting it to:
$$
\frac{2\sin^2(x... | \begin{align}
\lim_{x\to0}\frac{1-\cos(x^2+2x)}{1-\cos(x^2+x)} &= \lim_{x\to0}\frac{2\sin^2(\frac{x^2+2x}{2})}{2\sin^2(\frac{x^2+x}{2})} \\
&= \lim_{x\to0}\frac{\sin^2(\frac{x^2+2x}{2})}{\left(\frac{x^2+2x}{2}\right)^2}\cdot\frac{\left(\frac{x^2+x}{2}\right)^2}{\sin^2(\frac{x^2+x}{2})}\cdot\frac{\left(\frac{x^2+2x}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find polynomial with the lowest degree
Find all polynomial $f\in \mathbb{Z}[x]$ with the lowest degree
satisfying the condition: there exists $g,h\in \mathbb{Z}[x]$ such that
$$\big(f(x)\big)^4+2f(x)+2=(x^4+2x^2+2)g(x)+3h(x)$$
This question in Korea Winter Program 2021, a program that selects students to participate ... | You probably are allowed to quote Eisenstein criterion here, so $p(x)=x^4+2x+2$ and $q(x)=x^4+2x^2+2$ are irreducible polynomials in $\mathbb{Z}[x]$.
If you have seen solving two polynomials $P(x,y)=Q(x,y)=0$ using resultants, then it is all routine. You can do it without resultants but it will just be messier (though... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Does Diophantine equation $1+n+n^2+\dots+n^k=2m^2$ have a solution for $n,k \geq 2$? When studying properties of perfect numbers (specifically this post), I ran into the Diophantine equation
$$
1+n+n^2+\dots+n^k=2m^2, n\geq 2, k \geq 2.
$$
Searching in range $n \leq 10^6$, $k \leq 10^2$ yields no solution. So I wonder ... | $1+n+n^2+n^3=2m^2$ can be transformed to $y^2 = x^3+2x^2+4x+8$ with $x=2n$ and $y=4m.$
The equation $y^2 = x^3+2x^2+4x+8$ is an elliptic curve which has one torsion point $(x,y)=(-2,0)$ and the rank of curve is $0$.
Since rank is $0$, so there are no rational points of infinite order on the curve.
The only integral poi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Calculate the diagonal $d$ of three equal squares inscribed into a triangle whose sides $a$, $b$ and $c$ (and interior angles) are given We have a triangle, whose sides $a$, $b$ and $c$ (and interior angles $\alpha$, $\beta$ and $\gamma$) are given. Into this triangle three equally sized squares with a side length $r$ ... | To establish the existence of such a construct and also find a way towards a solution it is useful to consider the inverse construction, starting from a central point $M$, and drawing three squares around it, with angles between the adjacent edges of adjacent squares being $\delta, \epsilon,\zeta$ and the length of the... | {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the equation $- x^4 - 2 x^3 + 14 x^2 - 2 x - 1=0$ This equation:
$- x^4 - 2 x^3 + 14 x^2 - 2 x - 1 = 0$
has 4 real solutions (I saw it on GeoGebra), but I can't calculate them analytically.
I know the solving formula of the fourth degree equations (https://it.wikipedia.org/wiki/Equazione_di_quarto_grado) :... | First note that
\begin{eqnarray*}
-x^{4}-2x^{3}+14x^{2}-2x-1=0 &\implies& x^{4}+2x^{3}-14x^{2}+2x+1=0\\
&\implies& \left(x+\frac{1}{x}\right)^{2}+2\left(x+\frac{1}{x}\right)-16=0\\
&\implies& u^{2}+2u-16=0.
\end{eqnarray*}
Then we have a quadratic equation with solutions $$u=-1\pm\sqrt{17}.$$
Therefore all the solution... | {
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"source": "stackexchange",
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Calculating the coefficients of $\left( 1 + x^1 + x^3 + \dots + x^{2k+1} \right)^n$
Let $1 \leq m \leq n(2k+1)$. Calculate the coefficient of $x^m$ in $$\left( 1 + x^1 + x^3 + \dots + x^{2k+1} \right)^n$$
So, what we actually need to compute here is the number of solutions to $a_1 \cdot 1 + a_3 \cdot 3 + \dots + a_{... | Complete rewrite, because I didn’t realize it was only the odd terms.
Your version with the $a_i$ does not count the same thing. For example, when $k=1,n=2,m=3,$ there is only one $(a_1,a_3)=(0,1)$, but the coefficient of $x^3$ in $(1+x+x^3)^2$ is two.
This is because your reformulation ignores the order of values, whi... | {
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$P_n$ is the maximum prime factor of $1 + P_1P_2P_3…P_{n-1}$ for any integer n ≥ 2, where $P_1$ = 2. Is there a $P_n =11$? The definition of a sequence ${P_n}$ is as following: $P_1$ = 2, $P_n$ is the maximum prime factor of $1 + P_1P_2P_3…P_{n-1}$ for any integer n ≥ 2. Is there one term equal to 11 in this sequence?... | You have verified the claim for $n \leqslant 3$. Now, let $n > 3$ and assume the contrary. As you rightly pointed out, we have:
$$P_1P_2\cdots P_{n-1} + 1 = 5^a \cdot 11^b$$
where $a$ is a non-negative integer and $b$ is a positive integer. Considering the equation modulo $4$, the LHS is $3 \bmod{4}$ since the first te... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I derive $~\text{opposite}\cdot\sin^{}\left(\theta_{}\right)+\text{adjacent}\cdot\cos^{}\left(\theta_{}\right)=\tan^{}\left(\theta_{}\right)$ Given the below equation .
$$ b \cos^{}\left(\theta_{} \right) = a \cdot \sin^{}\left(\theta_{} \right) $$
I have to derive the below equation .
$$ b \sin^{}\left(\the... | Of the last line .
$$ \frac{ b^2\cos^{}\left(\theta_{} \right) }{ a } + \frac{ a ^2 \sin^{}\left(\theta_{} \right) }{ b } $$
Just plug the below ones to the above formula and you can get the RHS of the problem .
$$ \sin^{}\left(\theta_{} \right) = \frac{ b }{ \sqrt{ a^2+b^2 } } $$
$$ \cos^{}\left(\... | {
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"timestamp": "2023-03-29T00:00:00",
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If $ \lim_{x\to \infty} (f(x+1)-f(x))=1$, then $ \lim_{x\to \infty} \frac{f(x)}{x}=1$? Let $f:\mathbb{R} \to \mathbb{R}$ be a fuction such that $\displaystyle \lim_{x\to \infty} (f(x+1)-f(x))=1$. Is it true then that $\displaystyle \lim_{x\to \infty} \frac{f(x)}{x}=1$?
I think it is and here is how I went about it. Let... | Let
$
f(x)=
-\sin \left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right)
\cdot
\cos\left( \lfloor x\rfloor\cdot \dfrac{\pi}{2} \right).
$
Then
\begin{align}
f(x+1)-f(x)
=&
\sin \left( \lfloor x+1\rfloor\cdot \dfrac{\pi}{2} \right)
\cdot
\cos\left( \lfloor x+1\rfloor\cdot \dfrac{\pi}{2} \right)
-
\sin \left( \lfloor x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Simplify summation over collection Consider a collection $V_t$ of vectors of dimension $n$ created as follows:
When $n= 3, t=0$
$$V_0=\left\{\begin{pmatrix}0\\0\\0\end{pmatrix}\right\}$$
When $n=3, t=1$
$$V_1=\left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\end{pmatr... | You can find a simple expression for this using the principle of inclusion exclusion to subtract the monomials where some variable has an exponent of zero.
$n=2$: $$(x_1+x_2)^t-x_1^t-x_2^t$$
$n=3$: $$(x_1+x_2+x_3)^t-(x_1+x_2)^t-(x_1+x_3)^t-(x_2+x_3)^t+x_1^t+x_2^t+x_3^t$$
etc. In general, the expression for $n$ is
$$
\s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability of One Event Less Than Probability of Second Event I am having a bit of trouble proving some cases when one probability is smaller than the other probability for all positive integers $a, b$, so some suggestions would be appreciated.
Here is the problem:
For all $a, b \in \mathbb{Z}^+$, if $P(A) = \dfrac{a... | $$P(A) < P(B)\\
\iff \frac{a^2 - a + b^2 - b}{a^2 + 2ab + b^2 - a - b}<\frac{a^2 + b^2}{a^2 + 2ab + b^2}\\
\iff \frac{a^2+b^2-(a+b)}{(a+b)^2-(a+b)}-\frac{a^2+b^2}{(a+b)^2}<0\\
\iff \frac{(a^2+b^2-(a+b))(a+b)-(a^2+b^2)((a+b)-1)}{(a+b)^2((a+b)-1)}<0\\
\iff \frac{(a^2+b^2)(a+b)-(a+b)^2-(a^2+b^2)(a+b)+(a^2+b^2)}{(a+b)^2((a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limit in two variables using epsilon-delta In a class assignment, I was asked to find the following limit:
$$\underset{(x,y)\to (0,0)}{\lim}\frac{1-\cos(x^2\cdot y)}{x^6+ y^4}$$
I computed it to be $0$ numerically and as such, I wanted to prove it using the $\varepsilon-\delta$ definition, but I have not been able to f... | We can use that
$$\frac{1-\cos(x^2\cdot y)}{x^6+ y^4}=\frac{1-\cos(x^2\cdot y)}{(x^2\cdot y)^2}\frac{(x^2\cdot y)^2}{x^6+ y^4}$$
with
$$\frac{1-\cos(x^2\cdot y)}{(x^2\cdot y)^2} \to \frac12$$
then for a proof by $\varepsilon-\delta$ it suffices to consider
$$\underset{(x,y)\to (0,0)}{\lim}\frac{(x^2\cdot y)^2}{x^6+ y^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258457",
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Solving $\log_{1/3} \left(x^3 - 1\right) > \log_{1/3} \left(|3x^2 - 3x|\right)$ I tried over and over again but I simply can't solve:
$$\log_{1/3} \left(x^3 - 1\right) > \log_{1/3} \left(|3x^2 - 3x|\right)$$
So, if someone could please explain to me how to solve this step by step, I would be very greatful. By the way, ... | We have that $\log_{\frac 13}x$ function is strictly decreasing then
$$\log_\frac{1}{3} \left(x^3 - 1\right) > \log_\frac{1}{3} \left(|3x^2 - 3x|\right) \iff x^3 - 1 <|3x^2 - 3x|$$
under the conditions
*
*$x^3 - 1>0 \iff x>1$
*$|3x^2 - 3x| \neq 0 \iff x\neq 0 \quad x \neq 1$
that is $x>1$ therefore
$$x^3 - 1 <|3x^2 ... | {
"language": "en",
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Zeros of a polynomial in a simple form Let $m\geq 3$ be a natural number and define
$$f(x)=1-x^m-(1-x)(1+x)^{m-1}.$$
Then, if $m$ is even, the only real roots of $f$ are $x=-1,0,1$ and if $m$ is odd, the roots are $x=0,1$.
My work: for the case where $m$ is even, i proved that
$$f(x)=\sum_{k=0}^{m-2}\left(\begin{pmatri... | I was able to prove that $f$ has no more positive roots. Indeed, I needed also to study the sign of this polynomial, so I end up proving a bit more. I used the following reasoning:
\begin{align*}
f(l)=&1-l^m-(1-l)\sum_{k=0}^{m-1}
\begin{pmatrix}
m-1\\
k
\end{pmatrix}l^{m-k-1}\\
=&1-l^m-\left[\su... | {
"language": "en",
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"question_score": "2",
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$10$ students ordering $10$ different combinations of $5$ dishes so that each dish appears in at least $1$ combination
In how many ways can $10$ students in the queue for sandwiches order $10$ different combinations of $5$ side dishes so that every dish is in at least $1$ combination? Every combination can consist of ... | To be honest , as i wrote in comment section , i could not clearly understand the question.Maybe ,it is because of my english. Anyway , it seems that OP understood their question and produced a solution way ,but he suffer from calculating the equation of $$x_0 +x_1 +x_2 +x_3 +x_4 +x_5 =10$$ where $x_i \geq 0 , x_0 \leq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the partial fraction: $\frac{x^2}{(x-1)^2(x-2)^2(x-3)}$ I'm trying to find the partial fraction for the following $\frac{x^2}{(x-1)^2(x-2)^2(x-3)}$ by the process of long-division, here's what I have tried:
Leaving $(x-3)$ as it is in the denominator and aiming for $(x-1)$ and $(x-2)$.
First step:
$\frac{1^2}{(x-1... | Hint: $f(x) = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x-2}+\dfrac{D}{(x-2)^2}+\dfrac{E}{x-3}$. And you have to find these constants $A,B,C,D,E$.
| {
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Prove that $|x-2y|$ is the squaring number with $x^2-4y+1=2(x-2y)(1-2y)$ For $x,y$ are positive integer, statisfying $x^2-4y+1=2(x-2y)(1-2y) \,\,\ (1)$. Prove that $|x-2y|$ is a square number.
I try to compute, from $(1)$ We have $(1) \Leftrightarrow x^2+4xy-2x-8y^2+1=0$ and $x=\dfrac{1}{4} \left(x+\sqrt{3x^2-4x+2}\ri... | Continuing from what you've already stated,
$$\begin{equation}\begin{aligned}
x^2 + 4xy - 2x - 8y^2 + 1 & = 0 \\
x^2 - 2x + 1 & = 8y^2 - 4xy \\
(x - 1)^2 & = (2^2)y(2y - x)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This shows $y \mid (x-1)^2$, so $\gcd(x, y) = 1$. Furthermore,
$$\gcd(y, 2y - x) = \gcd(y, (2y - x... | {
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"timestamp": "2023-03-29T00:00:00",
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Find all $f$ which satisfies the condition: $f:\mathbb{R} \to \mathbb{R}, f(f(x+y)-f(x-y))=xy$ $$ f:\mathbb{R} \to \mathbb{R}, f(f(x+y)-f(x-y))=xy $$
$$
P(x, 0): f(0)=0 \\
P(0, y): f(f(y)-f(-y))=-y^2\\
P(x, x): f(f(2x))=x^2 \\
P(x, -x): f(-f(2x))=-x^2 \\
P(-x, x): f(-f(-2x))=-x^2 \\
P(-x, -x): f(f(-2x))=x^2 \\
$$
I don... | Let us rewrite the problem as $f(f(a)-f(b)) = \frac{a^2-b^2}{4}$.
Suppose that there exists a solution;
Claim $1)$ If $f(u) = f(v)$ then $u = \pm v$.
Proof; As you noted (taking $a=b$) we have that $f(0) = 0$. Now
$$0 = f(0) = f(f(u)-f(v)) = \frac{u^2-v^2}{4} \Rightarrow u = \pm v.$$
Claim $2)$ $f(u) = f(-u)$.
Proof; S... | {
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"timestamp": "2023-03-29T00:00:00",
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Counting solutions to $x^2+y^2 \equiv d \pmod{p}$ for a prime $p \equiv 3 \pmod 4$ For any given $p \equiv 3 \pmod{4}$ and $d=1, 2, \dots, p-1$, we would like to show that there are always exactly $p+1$ solutions to $x^2 + y^2 \equiv d \pmod{p}$.
This conjecture comes from some numerical experiments, and certainly hold... | alright, there are $p^2$ ordered pairs. The only pair giving $x^2 + y^2 = 0$ is $(0,0)$ because $-1$ is not a square.
For each of $(p-1)/2$ residues, there are $p+1$ pairs, total is $ \frac{p^2 - 1}{2}$
Thus $0$ and residues total to $ \frac{p^2 + 1}{2}$
Finally, nonresidues total to $ \frac{p^2 - 1}{2},$ and the... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the range of the function $f(x) = \cos(x)[\sin(x) + \sqrt{\sin^2(x) + \frac{1}{2}}]$
$f(x) = \cos(x)[\sin(x) + \sqrt{\sin^2(x) + \frac{1}{2}}]$
First of all, I tried to convert $\sqrt{\sin^2(x) + \frac{1}{2}}$ into perfect square, I tried many attempts but I failed to do so.
Next, I thought of differentiating i... | $$f(x)-\cos x\sin x=\cos x\sqrt{\frac{1}{2}+\sin^2 x}\Rightarrow
f(x)^2-2f(x)\cos x\sin x=\frac{1}{2}\cos^2 x\Rightarrow\\ 4f(x)\cos x\sin x=2f(x)^2-\cos^2 x\Rightarrow 16f(x)^2\cos^2 x(1-\cos^2 x)=(2f(x)^2-\cos^2 x)^2$$
$$f(x)^2=y,\ \cos^2 x=t:$$
$$16yt(1-t)=(2y-t)^2\Rightarrow 4y^2+(16t^2-20t)y+t^2=0\Rightarrow\\ 8yy... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Determining the period of $ \frac{\sin(2x)}{\cos(3x)}$ I would like to compute the period of this function which is a fraction of two trigonometric functions. $$ \frac{\sin(2x)}{\cos(3x)}$$
Is there a theorem for this? what trick to use to easily find the period?
I started by reducing the fraction but I'm stuck on the ... | $$\frac{\sin(2x)}{\cos(3x)}=\frac{2\sin(x)\cos(x)}{4\cos^3(x)-3\cos x}=\frac{2\sin(x)}{4\cos^2(x)-3}$$
$\sin x$ and $\cos x$ has a period of $2\pi$. Therefore $\frac{2\sin(x)}{4\cos^2(x)-3}$ has a period of $2\pi$.
The fundamental period must be $2\pi/n$ where $n\in\mathbb{N}$. We have
$$\frac{2\sin(x)}{4\cos^2(x)-3}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$, then find the value of $\frac{(1+x)^3}{1+x^3}$
Let $x$ be a real number such that $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$. What is the value of $\frac{(1+x)^3}{1+x^3}$?
Of course, one way is that to solve for $x$ from the quadratic $37(1+x)^2=13(1+x^2)$ which gives the value $x... | To go a little further, let's calculate all $f(n)=\dfrac{1+x^n}{(1+x)^n}$
I take the inverse of the desired quantity because it makes it easier to have a factorizable expression on denominator.
Notice that $f(1)=1$ therefore:
$f(n)=f(n)f(1)=\dfrac{(1+x^n)(1+x)}{(1+x)^{n+1}}=\dfrac{(1+x^{n+1})+x(1+x^{n-1})}{(1+x)^{n+1}}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $a^2+b^2+c^2=x^2+y^2+z^2$ given that $a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2$ Prove $$a^2+b^2+c^2=x^2+y^2+z^2$$ given that $$a^2+x^2=b^2+y^2=c^2+z^2=\\(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2$$ where $a,b,c,x,y,z \in\mathbb R$
A friend forwarded me this problem which recent... | Your second observation is astute! Simply for interest, I would make one further observation, which is that the question geometrically asks us to prove
Given three $2D$ vectors (equivalently, complex numbers $r,s,t$) with given equality of lengths, prove the two $3D$ vectors $$\mathbf{u}=(a \quad b \quad c) , \qquad \... | {
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"question_score": "4",
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Finding the maximum value of $y=\frac{x^2+x-1}{x^2-x+1}$ without calculus The problem is:
Find the maximum value of
$$y=\frac{x^2+x-1}{x^2-x+1}$$ assuming $x\geq 0$. I am only interested in pre-calculus solutions.
Using calculus it is easy to see that the maximum is $y=\frac{5}{3}$ occuring at $x=2$.
Other forms are
... | A simple approach is to rewrite it as a quadratic for $x$
$$ (y-1)x^2 - (y+1)x + (y+1)=0$$
And require that the discriminant be not negative
$$(y+1)^2 - 4(y+1)(y-1) \ge 0$$
$$(y - \frac{5}{3})(y+1) \le 0$$
Which gives us the range for $y$. To be thorough, it's also good to check that the maximum value for $y$ does not... | {
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Finding all functions $f:\mathbb R\to\mathbb R$ which satisfy $f(xy)=yf(x)+x+f\bigl(f(y)-f(x)\bigr)$ -Closed: It has a solution in AOPS.-
Find all functions $f:\mathbb R\to\mathbb R$ which satisfy $$f(xy)=yf(x)+x+f\bigl(f(y)-f(x)\bigr)$$
for all $x,y\in\mathbb R$.
My attempt:
\begin{align}
&P(x, x): f\left(x^2\right)... | Solution from AOPS:
\begin{align}
&\text{let } P(x, y): f(xy)=yf(x)+x+f(f(y)-f(x)). \\
\ \\
&P(x, 1): f(f(1)=f(x))=-x. \Rightarrow f: \text{ Bijective.} \\
\ \\
&P(x, 0): f(-f(x)=1)=-x-1. \\
&P(0, x): x-1=f(f(x)+1). \\
&\therefore f(f(x)+1)+f(-f(x)-1)=-2. \\
&\text{Because } f \text{ is surjective, } f(-x)+f(x)=-2. \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $x - \sqrt{ \frac{2}{x} } = 3 $ then $x-\sqrt{2x} = ?$ Please find the value without calculate the value of $x$. I tried to multipy it by $x$, I tried to square it but I still can't find the solution. I tried to make some equation that my help to solve this:
$$x^{2} + \frac{2}{x}-2\sqrt{2x} =9, $$
$$x^{2} - \sqrt{2x... | Let $\sqrt{\dfrac x2}=y\implies x=2y^2$
$$\implies2y^2-\dfrac1y=3\iff2y^3-3y-1=0$$
Let $x-\sqrt{2x}=2y^2-2y=2a,y^2-y-a=0\ \ \ \ (1)$
$2y-\dfrac1y=3-2a\iff2y^2+(2a-3)y-1=0\ \ \ \ (2)$
$(1),(2)$ will represent the same equation iff $$\dfrac12=\dfrac{-1}{2a-3}=\dfrac a1$$
Clearly $2a=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4289510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Compute the series $\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}$ Compute the series $$\sum_{n=3}^{\infty} \frac{n^4-1}{n^6}.$$
How do I go about with the index notation, for example to arrange the series instead as $\sum_{n=1}^{\infty}a_n $?
I have tried to simplify the expression as:
$$\begin{align}&\sum_{n=3}^{\infty} \fr... | You have at least two options to transform your sum into a sum where the index goes from $1$ on.
One, you can use the fact that
$$\sum_{n=1}^\infty a_n = a_1 + a_2 + \cdots + a_N + \sum_{n=N+1}^\infty a_n,$$
which can be shown from the definition of the infinite sum.
Or, you you can use the fact that
$$\sum_{n=k}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The image set of the function $f(x) = \dfrac{x -1}{x + 1}$ generates $\Bbb{Q}$ multiplicatively? Define $f(x) = \dfrac{x - 1}{x+1}$.
Compute the composition $f\circ f(x) = \dfrac{\dfrac{x-1}{x+1}-1}{\dfrac{x - 1}{x+1} +1} =\dfrac{\dfrac{x - 1 -x -1}{x+1}}{\dfrac{x-1 + x + 1}{x+1}} =\dfrac{-2}{2x} = \dfrac{-1}{x}$
$$
f^... | Let $H$ be the subset of $\Bbb Q$ which can be written as a product of finitely many elements of $f(Z)$. We want to show that $H = \Bbb Q$.
We have:
*
*$0 \in H$ because $0 = f(1)$;
*$-1\in H$ because $-1 = f(0)$;
*$\frac 1 n \in H$ for any positive integer $n$ because $\frac 1 n = f(2n - 1)f(2n - 3)\cdots f(3)$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4292212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many negative real roots does the equation $x^3-x^2-3x-9=0$ have? How many negative real roots does the equation $x^3-x^2-3x-9=0$ have ?
My approach :-
f(x)= $x^3-x^2-3x-9$
Using rules of signs, there is 1 sign change , so there can be at most 1 positive real root
f(-x)= $-x^3-x^2+3x-9$
2 sign changes here, indicat... | Just another way - to check for negative roots, it is enough to show $x^3+x^2+9=3x$ is not possible for any positive $x$. Using AM-GM, $x^3+(x^2+4)+5 \geqslant x^3+2\sqrt{4x^2}+5=x^3+(4x)+5>3x$, so this isn't possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
A non standard way to a probability problem We have teams $A$ and $B$.Every team has equal chance to win. A team wins after they have won $4$ matches. Team $A$ won the first match, what is the probability, that team $B$ won the whole game.
My try:
So team $A$ needs $3$ more wins to win the game. Team $B$ needs $4$ more... | Your approach looks right, but I think you made a couple mistakes when calculating $N$. First, you erred with your initial inequality. It should be
$$(i-1)+(j-i-1)+(n-j-1)\leq 3$$
$$\implies n\leq 6$$
This result should make sense because if you have $7$ games then clearly $B$ would have $4$ wins since $A$'s $3$rd win ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $f(3)$ if $f(f(x))=x^{2}+2$ Let $a,b,f(x),x$ be positive integers such that If $a>b$ then $f(a)>f(b)$ and $f(f(x))=x^{2}+2$ . Find $f(3)$
My approach:
Replacing $x$ with $f(x)$ in the equation gives $f(f(f(x))) = f(x)^2 + 2$, but $f(f(x)) = x^2 + 2$ so $$f(x^2+2) = f(x)^2 + 2$$
how do i proceed after this. Please ... | Note that $f$ is by definition strictly monotonic. So consider: If $f(3)<3$ then we can only have $f(3)=1$ or $f(3)=2$. Else if $f(3)=3$ we have $f(3)=f(f(3)) = 3^2+2=11$, so this cannot be. Finally if $f(3)>3$ we have $f(3) < f(f(3))=11$.
$f(3)=1,2$ would imply $f(1)=f(f(3)) = 11>f(3)$.
Generally we get that $f(x)$ ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is there any formula for $S(k)=\sum_{n=1}^{\infty} \frac{1}{n^{k} k^{n}},$ where $k\in N$? I had just used the definite integral $$\int_{0}^{\frac{1}{2}} \frac{\ln (1-x)}{x} d x $$ to evaluate the infinite sum
$$ \displaystyle S=\sum_{n=1}^{\infty} \frac{1}{n^{2} 2^{n}}. $$
However, I just wonder if there is a formul... | As already pointed out by @Varun Vejalla in the comments section, the polylogarithm function is the way to go.
It´s defined by
$$\text{Li}_n\left(x\right)=\sum_{k=1}^\infty \frac{x^k}{k^n} \tag{1}$$
Letting $n=2$ and $x=\frac12$ in $(1)$ we obtain
$$\text{Li}_2\left(\frac12\right)=\sum_{k=1}^\infty \frac{1}{k^2 2^k} \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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How to derive easily : $|\sqrt{x+3} -2|=\frac {|x-1|}{\sqrt{x+3}+2}$? ( Context : a limit problem). Source : Lebossé & Hemery, Algèbre et Analyse ( Terminales CDT ), $1966$.
$|\sqrt{x+3} -2|=\frac {|x-1|}{\sqrt{x+3}+2}$
This equality is asserted in the solution of a limit problem , namely : show that the limit of $f(x)... | Regardless of the value of $x$, $\sqrt{x+3}+2>0$, so $\sqrt{x+3}+2=|\sqrt{x+3}+2|$ and $(\sqrt{x+3}+2)^{-1}$ exists.
Multiplying $(\sqrt{x+3}+2)^{-1}|\sqrt{x+3}+2|=1$, $|\sqrt{x+3}-2|$ changes into...
$$
|\sqrt{x+3}-2|=\frac{|\sqrt{x+3}-2|\cdot|\sqrt{x+3}+2|}{\sqrt{x+3}+2}=\frac{|\sqrt{x+3}^2-2^2|}{\sqrt{x+3}+2}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.