Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
A generalization of arithmetic and geometric means using elementary symmetric polynomials Let $a_1, a_2, \ldots, a_n$ be positive real numbers. A while ago I noticed that if you form the polynomial
$$
P(x) = (x - a_1)(x-a_2) \cdots (x-a_n)
$$
then:
*
*The arithmetic mean of $a_1, \ldots, a_n$ is the positive number... | Denote, $(\overline{a}) = (a_1,\cdots,a_n)$ and $\displaystyle P(x) = \prod\limits_{k=1}^{n}(x - a_k) = x^n +\sum\limits_{k=1}^{n} (-1)^k\binom{n}{k}u_k(\overline{a})x^{n-k}$
where, $\displaystyle u_k(\overline{a}) = \dfrac{\sum\limits_{1\le j_1<\cdots< j_k \le n}a_{j_1}\cdots a_{j_k}}{\binom{n}{k}} = m_{n-k}^{k}$ (in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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"answer_id": 0
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Calculate $\sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} \times \ldots $ $$
\sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} \times \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}}}} \times\ldots$$
I already know a way to calculat... | Well probably the easiest way to derive this result is to employ the same method as it's creator, François Viète. He started with a circle of radius 1 and inscribed a square, whose area we will denote $A_0 = 2$. Now let $A_k$ be the area of the inscribed regular polygon with $2^{k+2}$ sides. Since $\lim_{k\to\infty}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$? How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?
I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
| Let's square both of them:
\begin{align}
\sqrt{\frac{1}{2}}^2 &= \frac{1}{2}, \text{ while} \\
\left(\frac{\sqrt{2}}{2}\right)^2
&=\frac{\sqrt{2}^2}{2^2} \\
&=\frac{2}{4} \\
&= \frac{1}{2}
\end{align}
So they're both positive numbers, and their squares are the same, so they must be the same.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$ $a,b,c \in \mathbb{R}$ and $a+b+c=0$.
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$
I think that $2^{a}.2^{b}.2^{c}=1$, but i don't know what to do next
| As it was said, you can take $x=2^a$, $y=2^b$ and $z=2^c$ to obtain:
$$
x^3+y^3+z^3\ge x+y+z
$$
With $x,y,z>0$ and $xyz=1$. It is equivalent to:
$$
x^2(x-1)+y^2(y-1)+z^2(z-1)\ge 0 \iff \frac{x^2(x-1)+y^2(y-1)+z^2(z-1)}{3}\ge 0
$$
Since $x^2$ and $x-1$ ar equally ordered, we might apply Chebychevs inequality to obtain:
... | {
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"timestamp": "2023-03-29T00:00:00",
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Removing the root squares from this expression? I would like to understand how to remove the root squares from this expression:
$$x = \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}$$
How to do it?
| Its all about rationalization,
\begin{align}
\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}} &= \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}
\cdot \frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} - \sqrt{5}}
\\[10pt] &=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}
\\[10pt] &=\frac{\sqrt{2} + \sq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Formal proof of limit By using $\epsilon-\delta$ definition, prove that $$\lim_{x\rightarrow1}\frac{3}{x^2}=3$$
Here is what I have tried.
$$|\frac{3}{x^2}-3|=3|\frac{1}{x^2}-1|=3|\frac{1-x^2}{x^2}|=3\frac{|x-1||x+1|}{x^2}$$
So I have problem with how to elimate the denominator.
| Given $\epsilon > 0$, choose $\delta = \text{min}\left(\frac{1}{2},\frac{\epsilon}{30}\right) \to |1|-|x| < |1-x|=|x-1| < \delta < \dfrac{1}{2} \to |x| > \dfrac{1}{2}$, and $|x+1| = |x-1+2| < |x-1|+|2| < \delta + 2 < \dfrac{1}{2} + 2 = \dfrac{5}{2} \to \left|\dfrac{3}{x^2} - 3\right| < 3\cdot |x-1|\cdot \dfrac{5}{2}\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cauchy P.V. Of an improper inegral the poles are $x=+1,-1,i,-i$
we should take only the upper have of axis
so we should take residue of $1$ and $i$? right
in this problem the book took only $x= i$.
I don't know why !! please help
| Denoting by $a$ and $b$ the relevant poles ($\Im a>0, \; \Im b>0 $), we have:
$$a= (1+i)/\sqrt{2}, \quad b= (-1+i)/\sqrt{2}$$
$$\bar{a}= (1-i)/\sqrt{2}, \quad \bar{b}= (-1-i)/\sqrt{2}$$
$$ a - \bar{a} = i\sqrt{2}, \quad a - b = \sqrt{2}, \quad a - \bar{b} = (1+i)\sqrt{2}$$
$$ b - \bar{b} = i\sqrt{2}, \quad b - a = -\sq... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Complex numbers problem: $ |\frac 1z - \frac 14 | = \frac 14 $ How do you go about solving the following equation $ |\frac 1z - \frac 14 | = \frac 14 $ where $ z = a + bi $.
A hint is provided, and apprently the equation can be simplified to $ | {z-4\over z} | = 1 $ (don't understand how they did this).
| We have:
$\left\lvert\dfrac1z-\dfrac14\right\rvert=\dfrac14$. Combine fractions:
$\left\lvert\dfrac{4-z}{4z}\right\rvert=\dfrac14$. Get rid of fractions (multiply by $\lvert4z\rvert$):
$\lvert4-z\rvert=\lvert z\rvert$. Let $z=a+bi$:
$\lvert4-a-bi\rvert=\lvert a+bi\rvert$.
$\sqrt{(4-a)^2+b^2}=\sqrt{a^2+b^2}$. Get rid of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How prove $\bigl(\frac{\sin x}{ x}\bigr)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$ How prove $\left(\frac{\sin x}{ x}\right)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$.
Can this be proved with simple way?
| Set $y = 2\log\left(\frac{x}{\sin x}\right)$ and $z=\log\left(\frac{\tan x}{x}\right)$.
Exploting Weierstrass products, for every $x\in I=\left(0,\frac{\pi}{2}\right)$ we have:
$$ z-y = \sum_{n=1}^{+\infty}\sum_{m=1}^{+\infty}\frac{x^{2m}}{m\cdot \pi^{2m}}\left(\frac{4^m}{(2n-1)^{2m}}-\frac{3}{n^{2m}}\right), $$
so, g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Series acceleration with Fourier-Bessel series coefficients I was investigating methods for series acceleration and I found this identity:
$$e=\sum _{n=0}^{\infty } \left(\sqrt{\frac{\pi }{2}} (2 n+1)\right) I_{n+\frac{1}{2}}(1)$$
where $I$ is the modified Bessel function of the first kind.
Could you explain me where ... | good work Jack but the Integral it is very difficult to resolve anyway i try to get the series of the Sinh(x)/x using series Bessel Fourier and we get a different seriesthan you give it above i think the following is it faster
$$\frac{\sinh (x)}{x}=\sum _{n=0}^{\infty } I_{n+\frac{1}{2}}(1) \left(\pi 2^{n-\frac{7}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Integration of $x/\sqrt{x^2-4x}$ How can one integrate: $$\int \frac{x}{\sqrt{x^2-4x}}\,dx$$
I tried setting $x^2-4x$ as a $t$ (and changing $dx$ concordantly) but it didn't work....
| $t=x^2-4x$ and $\frac 1 2 \,dt=(x-2)\,dx$, so
$$
\int\frac x {\sqrt{x^2-4x}} \,dx = \int \frac{x-2}{\sqrt{x^2-4x}}\,dx+\int\frac2{\sqrt{x^2-4x}}\,dx.
$$
Use the substitution to evaluate the first integral.
Then: $\underbrace{x^2-4x = (x^2-4x+4)}_{\text{completing the square}}-4=(x-2)^2-4=4\left(\dfrac{x-2}{2}\right)^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Justify the solution of $x^2-x+\arctan{x}=0$ It is obvious that a solution of $x^2-x+\arctan{x}=0$ is $x=0$, but I would like you to show me how this can be derived more conceptually than by plugging in $0$, and how can one prove that such solution is unique.
| like you conjectured, $x = 0$ is the only solution. i will first show that there are no positive solutions.
$$\arctan x = \int_0^x {dt \over 1 + t^2} = \int_0^x \left( 1 - t^2 + {t^4 \over 1 + t^2} \right) dt = x -{1 \over 3}x^3 + \int_0^x {t^4 \over 1 + t^4}dt$$
for $x > 0, \arctan x > x - {1 \over 3}x^3,$ therefore
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Prove convergence of series Let $$\displaystyle a_n=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{\frac{1}{3}}+(-1)^{\frac{n(n+1)}{2}}}$$
so I divide it into four series $4k, 4k+1, 4k+2, 4k+3$ and I pair for instance $4k, 4k+1$ and $4k+2, 4k+3$ and prove that these two series is convergent and conclude that since
both series ... | Consider the series $\displaystyle\sum_{m=1}^{\infty} a_m=\sum_{n\in A}(-1)^n\frac{1}{n^{1/3}-1}=\frac{1}{2^{1/3}-1}-\frac{1}{5^{1/3}-1}+\frac{1}{6^{1/3}-1}-\frac{1}{9^{1/3}-1}+\cdots$ $\hspace{.8 in}$and $\displaystyle\sum_{m=1}^{\infty}b_m=\sum_{n\in B}(-1)^n\frac{1}{n^{1/3}+1}=-\frac{1}{3^{1/3}+1}+\frac{1}{4^{1/3}+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Formalize a proof without words of the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$ This website gives the following proof without words for the identity $(1 + 2 + \cdots + n)^2 = 1^3 + 2^3 + \cdots + n^3$.
I find it interesting but have trouble seeing the proof behind it. Could anyone could give me ... | Method 1 (By $k^3=k\cdot k^2=k$ pieces of squares of side $k$)
$\displaystyle \quad (1^{3}+2^{3}+3^{3}+\ldots+n^{3} )cm^3
\\= 1 \cdot (1^{2} cm^2\times 1 cm)+2 \cdot (2^{2}cm^2\times 1 cm)+3 \cdot (3^{2} cm^2\times 1cm)+\ldots+n \cdot (n^{2}cm^2\times 1 cm)
\\=(1cm+2cm+3cm+\cdots+ncm)^2\times 1cm$
Method 2(By Inductio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to solve $\tan x =\sin(x+45^{\circ})$? How do I solve $\tan x = \sin(x +45^{\circ})$?
This is how far I have come: $\sqrt{2}\sin x = \sin x\cdot\cos x + \cos^2 x$
| Set $x=y+45^\circ$ to get $\tan(y+45^\circ)=\sin(45^\circ+45^\circ+y)=\cos y$
Clearly, $y=0$ is a solution
Now $-1\le\cos y\le1\implies-1\le\tan(y+45^\circ)\le1$
If we consider $-180^\circ<y\le180^\circ,-45^\circ\le y+45^\circ\le45^\circ\iff-90^\circ\le y\le0\implies\cos y\ge0$
$\implies0\le\tan(y+45^\circ)\le1\implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Ways to prove $ \int_0^1 \frac{\ln^2(1+x)}{x}dx = \frac{\zeta(3)}{4}$? I am wondering if we can show in a simple way that
$$
I=\int_0^1 \frac{\ln^2(1+x)}{x}dx = \int_1^2 \frac{\ln^2(t)}{t-1}dt = \frac{\zeta(3)}{4}.
$$
Because the end result is very simple, I suspect that there might be a fast way to prove this.
Can you... | Here is a particularly efficient way to get to your Euler sum.
In this post I show that
$$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n}.$$
Replacing $x $ with $-x$ gives
$$\ln^2 (1 + x) = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1} x^n}{n}.$$
So if we replace the term $\ln^2 (1 + x)$ with its above Ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 3
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$e^{\pi\sqrt N}$ is very close to an integer for some smallish $N$s. What about $\pi^{e\sqrt N}$? Heegner numbers (1, 2, 3, 7, 11, 19, 43, 67, 163 - let's use symbol $H_n$) are know for peculiar property that $e^{\pi\sqrt{H_n}}$ are almost integers:
$$e^{\pi \sqrt{19}} \approx 96^3+744-0.22$$
$$e^{\pi \sqrt{43}} \app... | Up to 100000, the 10 best $N$ such that $e^{\pi\sqrt{N}}$ is almost an integer. The error $\delta$ is given such that the nearest integer is at $10^{\delta}$ from the result.
$$
\begin{array}{c|c}
N & \delta \\\hline
163 & -12.12\\
4\cdot163 & -9.79\\
9\cdot163 & -8.01\\
58 & -6.75\\
16\cdot163 & -6.51\\
67 & -5.87\\
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Two-dimensional limit, is my approach correct? The limit is
$$\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}$$
As usual, I tried checking along particular paths, namely the axes and the curves $y=mx^n$ for various values of $n$, but to no avail; all the limits evaluate to $0$. I resorted to converting to polar coordinates, g... | $$ (x^2 - y)^2 \geq 0, $$
$$ x^4 - 2 x^2 y + y^2 \geq 0, $$
$$ x^4 + y^2 \geq 2 x^2 y. $$
$$ (x^2 + y)^2 \geq 0, $$
$$ x^4 + 2 x^2 y + y^2 \geq 0, $$
$$ x^4 + y^2 \geq -2 x^2 y. $$
$$ x^4 + y^2 \geq 2 x^2 |y|. $$
$$ \color{magenta}{ \left| \frac{x^2 y}{x^4 + y^2} \right| \leq \frac{1}{2} } $$
Your numerator... | {
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"timestamp": "2023-03-29T00:00:00",
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How can I show that $\lim_{n\to\infty} 2^n \left( \frac{n}{n+1} \right ) ^{n^2} = 0$? How to calculate limit of the following expression:
$$2^n \left( \frac{n}{n+1} \right ) ^{n^2} $$
I know that limit of this sequence is equal to zero, but how to show that?
| $$\left(\frac{n}{n+1}\right)^{n^2}=\left(\frac{n+1}{n}\right)^{-n^2}=\left(1+\frac{1}{n}\right)^{-n^2}=\exp\left\{-n\cdot\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}} \right\}$$
therefore
$$2^n\left(\frac{n}{n+1}\right)^{n^2}=\exp\{n\ln 2\}\cdot \exp\left\{-n\cdot\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the matrix of linear transformation What is the orthogonal projection on the line of equation $x = y$ of the point
$\begin{pmatrix} 3 \\ -1 \end{pmatrix}$? Assume this is a linear transformation.
The matrix for this linear transformation is $\begin{bmatrix} \frac 12 & \frac 12 \\ \frac 12& \frac 12 \end{bmatr... | When your quote says "the perpendicular line from $(1,0)$ to the line of equation $x=y$", what is meant is the line through $(1,0)$ that is perpendicular to $x=y$.
It looks like you understood it as perpendicular to the vector $(1,0)$, but that is not how an orthogonal projection works.
| {
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"timestamp": "2023-03-29T00:00:00",
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Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on I am noticing this pattern:
\begin{align}
1+2&=3\\
4+5+6&=7+8\\
9+10+11+12&=13+14+15 \\
16+17+18+19+20&=21+22+23+24 \\
&\vdots
\end{align}
Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire... | $n^2 = \sum_{k=0}^{n-1} n = \sum_{k=0}^{n-1} ((n(n+2)-k)-(n(n+1)-k))$,
since $(n(n+2)-k)-(n(n+1)-k) = n$.
So we have
$n^2+\sum_{k=0}^{n-1} (n(n+1)-k) = \sum_{k=0}^{n-1} (n(n+2)-k)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Calculate limit on series with nested sum I want to calculate the limit of following series:
$$\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{1}{3^k} \cdot \frac{1}{2^{n-k}}$$
As far I could simply the series to:
$$\sum_{n=0}^{\infty} (\sum_{k=0}^{n} (\frac{1}{3})^k) \cdot (\sum_{k=0}^{n} 2^{n-k})$$
which would then allow me... | Consider writing the sum as $\sum_{n=0}^{\infty}\dfrac{1}{2^n}\sum_{k=0}^{n}\dfrac{2^k}{3^k}=\sum_{n=0}^{\infty}\dfrac{1}{2^n}\dfrac{1-\dfrac{2^{n+1}}{3^{n+1}}}{1-\dfrac{2}{3}}=\sum_{n=0}^{\infty}\dfrac{1}{2^n}\dfrac{3^{n+1}-2^{n+1}}{3^n}=3\sum_{n=0}^{\infty}\dfrac{1}{2^n}-2\sum_{n=0}^{\infty}\dfrac{1}{3^n}=3(2)-2(\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this limits $\lim_{n\to\infty}\frac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)}{n^2}=\frac{1}{8}$ Interesting Question:
Let denote by $v_{p}(a)$ the exponent of the prime number $p$ in the prime factorization of $a$,
show that
$$\lim_{n\to\infty}\dfrac{v_{5}(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\c... | This seemed like a fun problem and wanted to prove it for general prime p. I use Legendre's formula directly which is,
$$v_p(n!) = \frac{n-s_p(n)}{p-1}$$
$s_p(n)$ being the sum of digits of $n$ when written in base $p$.
$$\frac{1}{n^2}v_p(1^1\cdot 2^2\cdot 3^3\cdot 4^4\cdots\cdot n^n)=\frac{1}{n^2}v_p\left(\dfrac{(n!)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?
| Use the remainder theorem for polynomials.
$$
f(a) = 0 \implies f(x) = (x-a)p(x)
$$
Where $p(x)$ is reduced a degree. To find the $2$ you would trail solutions to find a root using a change of sign.
$\textbf{update}$
Since we have found one root we can assume that the we have an equation of the form
$$
(x+2)\left(ax... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Closed form solution for $\sum_{n=1}^\infty\frac{1}{1+\frac{n^2}{1+\frac{1}{\stackrel{\ddots}{1+\frac{1}{1+n^2}}}}}$. Let
$$
\text{S}_k = \sum_{n=1}^\infty\cfrac{1}{1+\cfrac{n^2}{1+\cfrac{1}{\ddots1+\cfrac{1}{1+n^2}}}},\quad\text{$k$ rows in the continued fraction}
$$
So for example, the terms of the sum $\text{S}_6$ a... | This answer contains computer-assisted algebra bashing. Pencil wielding mathematicians be warned.
Let
$$f_k(n)=\frac{1}{1+\frac{n^2}{1+\frac{1}{\frac{^\ddots_1}{1+n^2}}}}$$
where there are $2k$ horizontal fraction bars. Then, we have
$$f_2(n)=\frac{1}{1+\frac{n^2}{1+\frac{1}{1+\frac{1}{1+n^2}}}}=\frac{\frac{1}2}{n^2+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 2,
"answer_id": 1
} |
Complex Solution I was looking at a simple complex problem and I came by this:
Solve: $z^2 = 3 - 4i$
Let
$$z= x + yi$$
Rewrite as
$$(x + yi)^2 = 3 - 4i$$
Expand
$$x^2 +2xyi + y^2i^2 = 3 - 4i$$
Simplify
$$x^2 - y^2 + 2xyi = 3 - 4i$$
real/imaginary parts
$$x^2 - y^2 = 3$$
and
$$2xy = -4$$
$$xy = -2$$
Therefore $y = -\fra... | Lets say you said let:$$z=x+yi$$but you assumed $x$ and $y$ themselves could also be complex numbers. Then, in effect, you are stating that:$$x=a+bi$$$$y=c+di$$where $a,b,c$ and $d$ are real numbers. This effectively means that you are saying:$$z=(a+bi)+(c+di)i=(a-d)+(b+c)i$$so you will be computing these extra compone... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find minimum number of sums of $3$ out of $5$ whose vanishing implies all five to be zero Problem:
let $a,b,c,d,e$ be real numbers, now there are $\left(\binom{5}{3}=10\right)$ numbers
$$a+b+c,a+b+d,a+b+e,a+c+d,a+c+e,a+d+e,b+c+d,b+c+e,b+d+e,c+d+e$$
Question1:($\textbf{Jérémy Blanc has solve it}$)
Find the least k ... | Answer: $k=7$
$(i)$ If $a=-2$ and $b=c=d=e=1$, then $\binom{4}{2}=6$ of the numbers are zero (all sums involving $a$). Hence, $k\ge 7$.
$(ii)$ In order to show that $k\le 7$, we assume that at least $7$ sums are zero and show that this implies that $a=b=c=d=e=0$.
In your $7$ sums you have $21$ letters, each one belongi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\tan{\frac{2\pi}{15}}=1$ From this geometry problem, I can not find geometry solution.
However the answer is $X=\frac{2\pi}{15}$ by geometry method.
Then I get the identity $$\left(\sqrt{3}\sec{\frac{\pi}{5}}+\tan{\frac{\pi}{30}}\right)\ta... | We need $\displaystyle\sqrt3\sec\frac\pi5+\tan\frac\pi{30}=\cot\frac{2\pi}{15}=\tan\dfrac{11\pi}{30}$ as $\dfrac{2\pi}{15}+\dfrac{11\pi}{30}=\dfrac\pi2$
$\iff\displaystyle\sqrt3\sec\dfrac\pi5=\tan\dfrac{11\pi}{30}-\tan\dfrac\pi{30}$
$\iff\displaystyle\dfrac{\sqrt3}{\cos\dfrac\pi5}=\frac{\sin\left(\dfrac{11\pi}{30}-\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Question about equivalent equations I am solving equations with radicals using this substitution:
$f(x)^{\frac{1}{3}}+g(x)^{\frac{1}{3}}=h(x)$
$f(x)+g(x)+3(f(x)g(x))^{\frac{1}{3}}(f(x)^{\frac{1}{3}}+g(x)^{\frac{1}{3}})=h(x)^{{3}}$
Now I substitute the expression $f(x)^{\frac{1}{3}}+g(x)^{\frac{1}{3}}$ with $h(x)$. Why... | To make things clear, let's denote $a = f(x), b = g(x), c = h(x)$, then
$$a^{1/3}+b^{1/3} = c$$
$$(a^{1/3}+b^{1/3})^3 = a + 3a^{2/3}b^{1/3} + 3a^{1/3}b^{2/3} + b = a+3a^{1/3}b^{1/3}(a^{1/3} + b^{1/3}) + b =$$
$$= a+3a^{1/3}b^{1/3}c + b = a+3(ab)^{1/3}c + b =c^3.$$
Therefore
$$(a+3(ab)^{1/3}c + b)^{1/3} = c = a^{1/3} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Boolean algebra Simplification of "xy + x'z + yz" I'd like to simplify the following expression "xy + x'z + yz":
xy + x'z + yz = xy + z(x' +y)
= (xy + z)(xy + x' + y)
= (xy + z)(y(x + 1) + x')
= (xy + z) ( y + x')
What do I do after the la... | By either drawing a Karnaugh map or recognizing that this is just the Consensus Theorem, observe that:
\begin{align*}
xy + x'z + yz
&= xy + x'z + (1)yz \\
&= xy + x'z + (x + x')yz \\
&= xy + x'z + (xyz + x'yz) \\
&= (xy + xyz) + (x'z + x'yz) \\
&= xy(1 + z) + x'z(1 + y) \\
&= xy(1) + x'z(1) \\
&= xy + x'z
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determining conic section equation given foci and sum of distance to each point Disclaimer: This title was hard to formulate. Edits welcome.
Problem:
Given foci $$F_1 = (1,0)$$ $$F_2 = (3,0)$$
of a conic section, find the equation for all points $P$ that satisfy $$|PF_1| + |PF_2| = 6$$
My attempt:
I tried going about i... | It's not extremely difficult, just take care not to have a square root everywhere:
$$
\sqrt{(x-1)^2+y^2}+\sqrt{(x-3)^2+y^2}=6 \\
\sqrt{(x-1)^2+y^2} = 6 - \sqrt{(x-3)^2+y^2} \\
(x-1)^2+y^2 = 36 + (x-3)^2+y^2 - 12\sqrt{(x-3)^2+y^2} \\
x^2-2x+1+y^2-36-(x^2-6x+9)-y^2 = -12\sqrt{(x-3)^2+y^2} \\
-12\sqrt{(x-3)^2+y^2} = 4x-44... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n>0\ \forall\ n\ge1$.
Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.
I can't show that $a_n > 0\ \forall n\ge1$. I tried using induction but it wouldn... | $$a_{n}=\left(\sqrt{n}-\sqrt{n-1}\right)-\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{1}{\sqrt{n}+\sqrt{n-1}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Closed form of a recursive relation A sequence $\langle a_n\rangle$ is defined recursively by $a_1=0$, $a_2=1$ and for $n\ge 3$,
$$a_n=\frac 12 na_{n-1}+\frac 12n(n-1)a_{n-2}+(-1)^n\left(1-\frac n2\right).$$
Find a closed form expression for
$$f_n=a_n+2\binom n1a_{n-1}+3\binom n2a_{n-2}+\cdots+(n-1)\binom n{n-2}a_2+n\b... | Let
$$ f(z) = \sum_{n\geq 0} b_n z^n.\tag{1}$$
Since:
$$ \sum_{n\geq 0}\left(1-\frac{n}{2}\right)(-1)^n z^n = \frac{2+3z}{2(1+z)^2}$$
the recursion on $b_n$ gives:
$$ f(z) = \frac{1}{2}z\,f(z)+z^2\, f(z)+\frac{2+3z}{2(1+z)^2},$$
hence:
$$ f(z) = \frac{2+3z}{(1-z)^2(2-3z^2)}.\tag{2}$$
We want to find:
$$ f_n = n b_1 + (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Demonstration of the inequality of Cauchy-Schwarz After the demonstration of the inequality of Cauchy-Schwarz make by my professor, I still don't understand some steps of the demonstration.
To prove this inequality, my professor use the induction princile.
First, verify $P(1)$.
\begin{align}
\big(\sum^1{a_kb_k}\big)^2 ... | Actually, this expression is Cauchy-Schwarz's inequality itself. I'll give a proof and I hope you can follow. We have that $$(A_1+tB_1)^2 + (A_2+tB_2)^2 \geq 0$$ for all $t \in \Bbb R$, for being a sum of squares. This is equivalent to: $$(A_1^2+A_2^2)+2(A_1B_1+A_2B_2)t + (B_1^2+B_2^2)t^2 \geq 0.$$
This a polynomial in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
If gcd$(a, 4) = 2$ and gcd$(b, 4) =2$, then gcd$(a + b, 4) = 4$ If gcd$(a, 4) = 2$ and gcd$(b, 4) =2$, then gcd$(a + b, 4) = 4$
can someone help me solve this.
| $$\gcd(a, 4) = 2 \rightarrow a=4k+2\\\gcd(b, 4) = 2 \rightarrow b=4q+2\\\gcd(a+b, 4) = \gcd((4k+2+4q+2,4)=\gcd(4k+4q+4,4)\\\gcd(4(q+k+1),4)=4 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\sum_1^n 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} $ converge or not? how to check if this converge? $$\sum_{n=1}^\infty a_n$$
$$a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}$$
what i did is to show that:
$$a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n} - 2\sqrt{n+1} = -2({\sqrt{n+1... | $$
\begin{align}
&2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\\[16pt]
&=(\sqrt{n}-\sqrt{n-1})-(\sqrt{n+1}-\sqrt{n})\\[9pt]
&=\frac1{\sqrt{n}+\sqrt{n-1}}-\frac1{\sqrt{n+1}+\sqrt{n}}\\
&=\frac{\sqrt{n+1}-\sqrt{n-1}}{(\sqrt{n}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})}\\
&=\frac2{(\sqrt{n}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}+\sqrt{n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Problem with Lagrange multipliers I am asked to find local extrema of $f(x,y,z)=ax+by$ ($a,b$ non-zero and fixed) defined on $\{(x,y,z)\colon (x,y)\neq 0\}$ subject to
$$\left (R-\sqrt{x^2+y^2}\right)^2 + z^2 - r^2 = 0.$$
(here $0<r<R$ are fixed). Okay, let us define the auxiliary function by:
$$F(x,y,z) = ax+by - \lam... | First we calculate $F_x'$, $F_y'$, and $F_z'$:
$$F_x'=a-\lambda\left(2\left(R-\sqrt{x^2+y^2}\right)\times\frac{-2x}{2\sqrt{x^2+y^2}}\right)=a+\frac{2\lambda x\left(R-\sqrt{x^2+y^2}\right)}{\sqrt{x^2+y^2}}$$
$$F_y'=b-\lambda\left(2\left(R-\sqrt{x^2+y^2}\right)\times\frac{-2y}{2\sqrt{x^2+y^2}}\right)=b+\frac{2\lambda y\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is $\lim_{n\to\infty}\left(\frac{2}{5}\right)^{1/n}$? What is $$\lim\limits_{n\to\infty}\left(\frac{2}{5n}\right)^{\frac{1}{n}}$$ This is clearly equal to $\lim\limits_{n\to\infty}\left(\frac{2}{5}\right)^{\frac{1}{n}}\left(\frac{1}{n}\right)^{\frac{1}{n}}$. We know that $\lim\limits_{n\to\infty}\left(\frac{1}{n}\... | We have $$\left(\frac{2}{5n}\right)^{\frac{1}{n}} = e^{\frac{1}{n} \log \left(\frac{2}{5n}\right)}$$ and $$\frac{1}{n} \log \left(\frac{2}{5n}\right) = \frac{\log2 - \log5 - \log n}{n} \to 0$$ Thus $$\lim\limits_{n\to\infty}\left(\frac{2}{5n}\right)^{\frac{1}{n}} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
All automorphisms of splitting fields Let $M \le \mathbb{C} $ be the splitting field of polynomial $ f(x)
\in \mathbb{Q}[x] $. Find all automorphisms of field $ M $ in cases:
1) $ f(x) = x^6 - 1 $
2) $ f(x) = x^{2011} - 1 $
1) In the first case I found all complex roots of the 6th degree of 1. It is $$ 1, \frac{1}{2... | For $(2)$, we have that $2011$ is a prime number taking the thousand and eleventh root of unity, that is $\xi \neq 1$. Then $L = \mathbb{Q}[\xi]$ and as
$$\xi^{2010}+\xi^{2009}+\ldots +\xi^2 + \xi + 1 = 0$$
and $p(x) = x^{2010}+x^{2009}+\ldots +x^2 + x + 1$ is irreducible* over $\mathbb{Q}$ then $[L:\mathbb{Q}] = 201... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Study the convergence of $\int_1^\infty \frac{x\ln x}{x^4-1} dx$
Study the convergence of $\int_1^\infty \frac{x\ln x}{x^4-1} dx$
So first we have two potentially problematic points which are $1,\infty$
We split the integral to $$\int_1^2 \frac{x\ln x}{x^4-1} dx + \int_2^\infty \frac{x\ln x}{x^4-1} dx$$
Now first I t... | If the problem asks "Study the convergence", you're not required to find a closed form. Once you know the function is "nice" at $1$, you're done with that part: $I_1$ exists. Next look at $I_2$, and again don't worry about finding a formula for it, just whether or not it converges.
By the way, the integral from $1$ to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $a^2+b^2+c^2=1$ then prove the following. If $a^2+b^2+c^2=1$, prove that $\frac{-1}{2}\le\ ab+bc+ca\le 1$.
I was able to prove that $ ab+bc+ca\le 1$. But I am unable to gain an equation to prove that
$ \frac{-1}{2}\le\ ab+bc+ca$ .
Thanks in advance !
| (a - b)^2 + (b - c)^2 + (a - c)^2 = 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 2 - 2(ab + bc + ac). This is how we prove that ab + bc + ca <= 1.
Now:
If ab + bc + ac < -1/2, then 2(ab + bc + ac) < -1.
(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2(a + b)(c) = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 1 + 2(ab + ac + bc) < 1 + (-1) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Understanding how to estimate $\pi(x)$ based on Paul Erdos's proof of Bertrand's Postulate I am reading the 4th Edition of Proofs from the Book. I am not clear on how the proof behind Bertrand's postulate leads to the following statement on page 10 (of my edition):
From (2) one can derive with the same methods that $\... | We can put (1) into (2) and we get$$4^{n}\leq\left(2n\right)^{1+\sqrt{2n}}\underset{\sqrt{2n}<p\leq\frac{2}{3}n}{\prod}p\underset{n<p\leq2n}{\prod}p\leq\left(2n\right)^{1+\sqrt{2n}}4^{\frac{2}{3}n-1}\underset{n<p\leq2n}{\prod}p\,\Longrightarrow$$
$$\frac{2^{2n-\frac{4}{3}n-\sqrt{2n}+1}}{n^{1+\sqrt{2n}}}\leq\underset{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ an... | A nice solution can be obtained by modifying the "exact" solution. The "exact" solution is
$$
f(x) = \frac{8-\pi}8 + \sqrt{x(1-x)}
$$
which has an arc length of $\frac{8+\pi}4$.
As such, I propose a solution of the form
$$
f(x) = \sqrt{x(1-x)}(1+g(x))
$$
where the "exact" solution uses $g(x)=(8-\pi)/(8\sqrt{x(1-x)})$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 12,
"answer_id": 1
} |
Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta
I am having difficulty finding the $\delta$ value.
Here is what I have done so far:
What I want to show:
$$\forall \epsilon > 0, \exists \delta > 0 \; such \; that \; 0<\mid x -2 \mid < \d... | Here's a very easy and brief way to do it:
Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-2|<\delta$, then $|(x^4-2x^3+x+3)-5|<\epsilon$.
Now,
$$|(x^4-2x^2+x+3)-5|=|(x-2)(x^3+1)|.
$$ If $|x-2|<1$, that is, $1<x<3$, then $x^3+1<3^3+1=28$, and so
$$
|(x^4-2x^3+x+3)-5|=|x-2|(x^3+1)<28|x-2|.
$$ So if we take ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Existence of solution to Congruence relation $(x^2-2)(x^2-6)(x^2-3) \equiv 0\pmod p$ I'm taking the final exam in "Number Theory" tomorrow and stuck with:
Prove that $\,\,\forall p\in\mathbb{Z}_p\,$ the congruence relation: $$(x^2-2)(x^2-6)(x^2-3) \equiv 0\pmod p$$ has a solution.
Here $p$ is an arbitrary prime numb... | For $p=2,$
$$(x^2-2)(x^2-3)(x^2-6)\equiv x^2(x^2-1)x^2\pmod2$$
But $x^2(x^2-1)x^2$ is divisible by $x(x-1)$ which being a product of two consecutive integers is always divisible by $2$
$$(x^2-2)(x^2-3)(x^2-6)\equiv0\pmod2 $$ for all integer $x$
For $p=3,$
$$(x^2-2)(x^2-3)(x^2-6)\equiv (x^2+1)(x^2)x^2\pmod3$$
But $3\nm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What are all values of $x$ in $\mathbb{R}$ that satisfy $4 < |x+2| + |x-1| < 5$? I am having some problems getting started with this problem, as I never had to deal with an inequality that was between two values with absolute values. Any help is appreciated. The problem is find all values of $x$ in $\mathbb{R}$ that sa... | If $4 < |x+2| + |x-1| < 5$ then $\ \pm (x+2) \pm (x-1) < 5$. Note that for an opposite choice of sign the inequality is never satisfied since $(x+2) - (x-1)=3$. So the inequality implies
$$ 4 < (x+2) + (x-1) < 5$$
or
$$4 < -(x+2) -(x-1) < 5$$
The inequality $ 4 < (x+2) + (x-1) < 5$ is equivalent to $4 < 2x+1<5 \iff 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is $\int \frac{1}{\sqrt{25y^2-10y-3}}dy$ $= \int \dfrac{1}{\sqrt{(5y-1)^2-4}}dy$
$=\int \dfrac{1}{\sqrt{u^2-4}}\dfrac{du}{5}, \quad U$ substitution
$=\int \dfrac{1}{10\cos(\theta)} 2\cos(\theta) d\theta, \quad$ Trig substitution
$= \dfrac{1}{5} \theta$
$= \dfrac{\cos^{-1}\frac{\sqrt{(5y-1)^2-4}}{2}}{5} +C$
Whe... | The problem is in the 3rd line. $$\sqrt{4\sin^2\theta-4} \neq 2\cos \theta$$
Try $x=2\cosh\theta$ or $x=2\sec\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculation of determinant of an arrowhead matrix Is there any easier way to make sure the determinant of the following $n \times n$ matrix is $n$?
$$\begin{vmatrix}
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 \\
1 & 0 & 0 & 1 & \cdots & 0 \\
\vdots &... | We know that for $n=2$ the statement is true. We'll prove the rest by induction.
We can compute the determinant based on the last row. The result is the sum of two determinants
$\ \ \ \begin{vmatrix}
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 0 & 0 & \cdots & 0 \\
1 & 0 & 1 & 0 & \cdots & 0 \\
1 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Alternative matrix representation for translation The ''usual" way to write translation for $\textbf{v}\in \mathbb{R}^2$ is with the following $3\times3$ matrix
$$ \left( {\begin{array}{ccc}
1 & 0 & x_{0} \\
0 & 1 & y_{0} \\
0 & 0 & 1 \\
\end{array} } \right) \cdot \left( {\begin{array}{c}
x \\
y \\
... | I want to answer my own question in response to the answer given above. In the answer Marc van Leeuwen says that ''matrix representing linear transformation should not depend on the input"
Why exactly? What properties of linear transformations prevent that?If we find a transformation that depends on the input but sati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Does $\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0$? $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0$$
My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =):
(Sorry for the small font, but you can zoom in to see better with Firefox)
$$
\b... | \begin{align*}
\left(\frac{\sin x}x\right)^{\frac1{x^2}}
&=\left[1+\left(\frac{\sin x}x-1\right)\right]^{\frac1{x^2}}\\
&=\left[1+\left(\frac1{x/(\sin x-x)}\right)\right]^{\frac1{x^2}}\\
&=\left[\underbrace{\left[1+\left(\frac1{x/(\sin x-x)}\right)\right]^{[x/(\sin x-x)]}}_{=:A(x)}\right]^{\frac{\sin x-x}{x^3}}
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Area of an isosceles triangle where the tangents of some angles are in geometric progression In $\triangle ABC$, $AB=BC$ and $\overline{BD}$ is an altitude. Point $E$ is on the extension of $\overline{AC}$ such that $\overline{BE}=10$. The values of $\tan CBE$, $\tan DBE$, and $\tan ABE$ form a geometric progression, a... | Here is a picture:
You know that $\triangle ABC$ is isosceles, and $D$ is the midpoint of the base across from $B$. Further, $\triangle BDE$ is a right triangle.
Let $\theta = \angle DBE$ and $\phi = \angle DBC$, so $\tan \angle CBE = \tan (\theta - \phi)$ and $\tan \angle ABE = \tan(\theta+\phi)$. Applying sum and di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$
I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$... | Use the linearisation formulae. The expression can be rewritten as:
$$2^2\Bigl(2\sin\frac{\pi}{16}\,\sin\frac{7\pi}{16}\Bigr)\Bigl(2\sin\frac{3\pi}{16}\,\sin\frac{5\pi}{16}\Bigr)=2^2\cos\frac{3\pi}{8}\cos\frac{\pi}{8}=2\cos\frac{\pi}{4}=\sqrt2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
limit of $ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $ Hello I am trying to find the limit of
$ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $
I've tried applying L'H rule but it ends up getting really messy.
The answer is $ \frac {1}{e} $ so I assume it must simplify into something which I ca... | $$\lim_{x\rightarrow \infty }\frac{x(x+1)^{x+1}}{(x+2)^{x+2}}=\\\lim_{x\rightarrow \infty }\frac{(x+1)^{x+1}}{(x+2)^{x+1}}*\frac{x}{x+2}=\\\lim_{x\rightarrow \infty }(\frac{x+1}{x+2})^{x+1}\frac{x}{x+2}=\\(1-\frac{1}{x+2})^{x+1}* \lim_{x\rightarrow \infty }\frac{x}{x+2}=\\e^{-1}*\lim_{x\rightarrow \infty }\frac{x}{x+2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Jordan matrices with the only eigenvalue $1$. So, I need to list all of the Jordan matrices of a $4x4$ matrix with the only eigenvalue $1$. If the only eigenvalue is 1, then there can't be any other value on the diagonal, correct? So am I wrong in thinking that only Jordan matrix would be this? $$ J_{1,n} =\begin{bmatr... | Read egreg's comment and complete. Some of the possibilities are:
$$\begin{align}&4=4\;\;\rightarrow&\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1\end{pmatrix}\\{}\\
&4=3+1\;\;\rightarrow&\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\\{}\\
&4=2+2\;\;\rightarrow&\begin{pmatrix}1&1&0&0\\0&1&0&0\\0&0&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve $\int_{0}^{t}\cos x\sin (t-x)dx$ I want to solve the following integral $$\int_{0}^{t}\cos x\sin (t-x)dx$$
What I tried to do is integration by parts twice, but it just brought me straight back to $$\int_{0}^{t}\cos x\sin (t-x)dx=\int_{0}^{t}\cos x\sin (t-x)dx$$ Not very informative indeed.
Is there any way to do... | We have
\begin{align}
\int_0^t\cos\left(x\right)\sin\left(t-x\right)\:dx,\tag{1}
\end{align}
and since $\sin\left(t-x\right)=\sin\left(t\right)\cos\left(x\right)-\sin\left(x\right)\cos\left(t\right)$,we may re-write this integral as
\begin{align}
&\int_0^t \cos\left(x\right)\left[\sin\left(t\right)\cos\left(x\right)-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
$\int \frac{x^3+2}{(x-1)^2}dx$ In order to integrate
$$\int \frac{x^3+2}{(x-1)^2}dx$$
I did:
$$\frac{x^3+2}{(x-1)^2} = x+2+3\frac{x}{(x-1)^2}\implies$$
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+3\frac{x}{(x-1)^2}dx$$
But I'm having trouble integrating the last part:
$$\int \frac{x}{(x-1)^2}dx$$
Wolfram alpra said me ... | Here's a way to find the partial fraction expansion:
\begin{align}
\frac{x}{(x-1)^2} &= \frac{A}{x-1} + \frac{B}{(x-1)^2}\\
(x-1)^2\frac{x}{(x-1)^2} &= (x-1)^2\frac{A}{x-1} + (x-1)^2\frac{B}{(x-1)^2}\\
x &= A(x-1) + B
\end{align}
You can turn this into two linear equations and solve, or you can do the following:
*
*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1139221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
System of 4 tedious nonlinear equations: $ (a+k)(b+k)(c+k)(d+k) = $ constant for $1 \le k \le 4$ It is given that
$$(a+1)(b+1)(c+1)(d+1)=15$$$$(a+2)(b+2)(c+2)(d+2)=45$$$$(a+3)(b+3)(c+3)(d+3)=133$$$$(a+4)(b+4)(c+4)(d+4)=339$$ How do I find the value of $(a+5)(b+5)(c+5)(d+5)$. I could think only of opening each expressio... | First, by expanding the four equations, we note that the system of equations is equivalent to
$$ \begin{bmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 4 & 2 & 4 & 4 & 8 & 2 & 4 & 4 & 8 & 4 & 8 & 8 \\
1 & 3 & 3 & 9 & 3 & 9 & 9 & 27 & 3 & 9 & 9 & 27 & 9 & 27 & 27 \\
1 & 4 & 4 & 16 & 4 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 2
} |
Finding $\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
Find the limit: $\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
My attempt:
$\begin {align}\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}} &=
\lim _{n\to \infty}\frac{\sqrt{n+n^4\sqrt{1+\frac{... | $$\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}\cdot\frac{\frac1{\sqrt n}}{\frac1{\sqrt n}}=\frac{\sqrt{1+\sqrt{1+\frac1{n^2}}}}{\sqrt{3+\frac1n}}\xrightarrow[n\to\infty]{}\sqrt\frac23$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Solve: $ab+bc+ca\mid (a+b+c)^2$ I couldn't make any progress on this problem, can anyone help?
I found it's the same as:
Find all integers $a,b,c$ such that $ab+bc+ca$ divides $a^2+b^2+c^2$.
I found a solution $a=-b=1$, and $c$ any integer.
Any more solutions?
| Since $(a+b+c)^2 = 2(ab+ac+bc)+(a^2+b^2+c^2)$, if
$$ (a+b+c)^2 = k(ab+ac+bc), \tag{1}$$
then $k\geq 2$, and for $k=2$ we have only the trivial solution $(a,b,c)=(0,0,0)$.
Assuming $k=3$, we have:
$$ a^2+b^2+c^2 = ab+ac+bc \tag{2}$$
and by the Cauchy-Schwarz inequality $(2)$ holds only for $a=b=c$.
Assuming $k=4$ we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int... | $$
\begin{aligned}
\text { Let } y &=\sqrt{\frac{x}{x+1}}, \text { then } y^{2}=\frac{x}{x+1}=1-\frac{1}{x+1} \text{ and } \quad 2 y d y =\frac{1}{(x+1)^{2}} d x=\left(1-y^{2}\right)^{2} d x \\
I &=\int y \frac{2 y}{\left(1-y^{2}\right)^{2}} d y \\
&=2 \int\left(\frac{y}{1-y^{2}}\right)^{2} d y \\
&=2 \int\left[\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
From $\frac{1-\cos x}{\sin x}$ to $\tan\frac{x}{2}$ How can I write $\frac{1-\cos x}{\sin x}$ as $\tan\frac{x}{2}$? I wrote $\sin x$ as $2\sin\frac{x}{2} \cos\frac{x}{2}$ also used the double angle identity for $\cos$ but wasn't able to make much progress
| Replacing 1 with $\cos\frac2{x}{2}$ + $\sin\frac{x}{2}$, $\cos$$x$ with $\cos\frac2{x}{2}- \sin\frac{x}{2}$ and $\sin$$x$ with $2\sin\frac{x}{2}$$\cos\frac{x}{2}$ the numerator and denominator simplify to $2\sin^2\frac{x}{2}$ and $2\cos\frac{x}{2}\sin\frac{x}{2}$ and hence the fraction simplifies to $\tan\frac{x}{2}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How can I prove that $2/9$x$ and $y$ are real numbers.
Given that $1<x^2-xy+y^2<2$, how can I show that $\frac 29<x^4+y^4<8$ ?
Then can I use that to prove that for any natural number $n>3$
$$x^{2^n}+y^{2^n}>\frac 2{3^{2^n}} \text{?}$$
| 1) to show $x^4+y^4 < 8$, you already have a good solution by squaring $x^2+y^2<2+xy$.
2) By AM-GM and Cauchy Schwarz, we get
$$\left(x^4+y^4+\frac{x^4+y^4}2\right)(3) \ge (x^4+y^4+x^2y^2)(1^2+1^2+(-1)^2) \ge (x^2+y^2-xy)^2 > 1$$
hence $x^4+y^4 > \frac29$
For the more general case, you could use Power Means in additio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1145506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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System of Exponential Equations in $x$ and $y$ Three of the elements in the solution set of the simultaneous system $x^{x+y} = y^4, y^{x+y} = x$ are ordered pairs of integers $(x, y)$. Find these ordered pairs.
I found the trivial solution at $(1,1)$, but was unable to find the other two solutions. Could I get help w... | $$1) \quad x^{x+y} = y^4$$
$$2) \quad y^{x+y}=x$$
$$x^{x+y}=(y^{x+y})^{x+y}=y^{(x+y)^2}=y^4\implies (x+y)^2=4\implies x+y =\pm 2$$
Else $(x+y)^2$ need not equal $4$ if $y=-1,0,1$ as Meelo has commented(but $(x+y)^2$ need be even for $y=-1$ case.)
Clearly they have designed this so$x=y=0$ and $(-1,-1)$ solutions are i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Method of characteristics Given the equation
$y^2 u_x + x u_y = \sin(u^2)$ with
initial condition $u(x,0) = x$,
determine the values of $u_x$, $ u_y$, $u_{xx}$, $u_{yy}$, $u_{xy}$, $u_{yx}$ on the $x$-axis.
I tried the following:
$$\begin{cases}
\frac{dx}{dt} = y^2 \\\\
\frac{dy}{dt} = x \\\\
\frac{dz}{dt} = \sin(z^2)... | Hint:
$$\begin{align}
\frac{dx}{dt} &= y^{2} \ \ \ (1)\\
\frac{dy}{dt} &= x \ \ \ \ \ (2)\\\\
(2) \implies \frac{d}{dt} \bigg( \frac{dy}{dt} \bigg) &= \frac{d^{2}y}{dt^{2}} \\
&= \frac{dx}{dt} \\
&= y^{2} \\
&= (1)
\end{align}$$
EDIT:
We have
$$y'' = y^{2}$$
Multiply through by $y'$ and integrating
$$\begin{align}
y' y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(n+1)^{n-1}How would one prove that $$(n+1)^{n-1}<n^n \ \forall n>1$$
I have tried several methods such as induction.
| My try to prove $\forall n>1 : (n+1)^{n-1} < n^n$:
$$(n+1)^{n-1} < n^n \Leftrightarrow (n+1)^{n} < n^n (n+1) \Leftrightarrow \left(\frac{n+1}{n}\right)^n < n+1$$
Now apply induction on this result:
Base case: $n = 2 \Rightarrow \left(\frac{3}{2}\right)^2 < 3 \Rightarrow 2.25 < 3$ OK
Induction hypothesis: assume $\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
} |
Proving $\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right) = \frac35$ $$\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right)$$
Can someone help me to solve it?
result of online calculator: 3/5
| Hint:
$$\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}} = \frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\cdot\frac{\sqrt{x+1}+\sqrt{x-2}}{\sqrt{x+2}+\sqrt{x-3}}\cdot\frac{\sqrt{x+2}+\sqrt{x-3}}{\sqrt{x+1}+\sqrt{x-2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Number of solutions of a variable matrix Given matrix:
$$\left(\begin{array}{ccc|c}
c & c & 1-c & 1\\
c & c^{2} & 1-c^{2} & 1\\
2c & c+c^{2} & 2-2c & c+2
\end{array}\right)$$
row reduced to:
$$\left(\begin{array}{ccc|c}
c & 1 & 0 & 1\\
0 & c-1 & 1-c & 0\\
0 & 0 & c+1 & 1
\end{array}\right)$$
question asks to find... | let us look at the cases separately.
in case $c = 0,$ you have the system $\pmatrix{0&0&1&|&1\\0&0&1&|&1\\0&0&2&|&2}$ which is and has infinitely many solutions.
in case $c = 1,$ you have the system $\pmatrix{1&1&0&|&1\\1&1&0&|&1\\2&2&0&|&3}$ which has no solutions.
in case $c \ne 0, c \neq 1$ we can reduce
$\pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Induction exercise check-up Prove by induction on $n$ that $13$ divides $2^{4n+2} + 3^{n+2}$ for all natural $n$.
For base case it is divisble by 13, and
$2^{4n+6} + 3^{n+3}$ must be divisble too.
$16 * 2^{4n+2}+ 3* 3^{n+2}$
If we denote $2^{4n+2} + 3^{n+2}$ as some $13y$,
we have $3*13y + 13*2^{4n+2}$
Didn't I make ... | Your proof isn't wrong per se, but the wording is a bit confusing (and I would probably end up taking a point or two off if I were grading it). Here's how I would write it:
For $n=0$, $$2^{4\cdot0+2}+3^{0+2} = 2^2 + 3^2 = 13.$$
Now assume that $13|(2^{4n+2}+3^{n+2})$ for some integer $n\geqslant0$. Then $2^{4n+2}+3^{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Identity with Harmonic and Catalan numbers Can anyone help me with this.
Prove that
$$2\log \left(\sum_{n=0}^{\infty}\binom{2n}{n}\frac{x^n}{n+1}\right)=\sum_{n=1}^{\infty}\binom{2n}{n}\left(H_{2n-1}-H_n\right)\frac{x^n}{n}$$
Where $H_n=\sum_{k=1}^{n}\frac{1}{k}$.
The left side is equal to $$2\log(C(x))=2\log\left(\fr... | We start with the key identity:
$$\sum\limits_{j=1}^{n-k} \frac{(-1)^{j-1}}{j}\binom{n}{k+j} = \binom{n}{k}(H_n - H_k)$$
which can be proved elementarily by induction on $n$ or otherwise.
Thus we have for our special case: $\displaystyle \binom{2n}{n}(H_{2n} - H_n) = \sum\limits_{j=1}^{n} (-1)^{j-1}\frac{1}{j}\binom{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
What is the coefficient of $x^{10}$ in $\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2$? I did (parcially) the following exercise:
There are $10$ identical gift boxes. Each one must be colored with a unique color and there are the colors red, blue, green and yellow. It's po... | This is mechanized in Maple:
coeftayl(((-x^3+1)/(1-x)*((-x^4+1)/(1-x)))/(1-x)^2, x = 0, 10);
$$ 102. $$
See coeftayl for info.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Non vanishing of an infinite product I need to prove that the infinite product
$$\prod_n \left(1-\frac{1} {(a^n+1)^2} \right)^{\frac{a^n}{n}} $$
with $a$ an integer $\geq 3$,
converges to a real number $L$ such that $0<L<1$.
It's immediate to see that $L<1$, as we have $\left(1-\frac{1} {(a^n+1)^2} \right)^{\frac{a^n}{... | Since $a\ge3$,
$$
\frac{1} {(a^n+1)^2}\le\frac{1}{16}\quad\forall n\ge1.
$$
Let $C>0$ be such that $\log(1-x)\ge-C\,x$ if $0<x\le1/16$. For any $N>1$
$$\begin{align}
\log\Biggl(\prod_{n=1}^N \Bigl(1-\frac{1}{(a^n+1)^2} \Bigr)^{\frac{a^n}{n}}\Biggr)&=\sum_{n=1}^N\frac{a^n}{n}\log\Bigl(1-\frac{1}{(a^n+1)^2} \Bigr)\\
&\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove: $\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Let $a;b;c>0$. Prove that :
$\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
I think:
$\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ac}+\frac{a+b}{c^2+ab}\leq \frac{1}{a}+\f... | WLOG,you can assume $a\ge b\ge c>0$, then
$$\frac{(b-c)(a-b)}{b^3+abc}\le\frac{(a-c)(b-c)}{c^3+abc}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving the following set of real numbers is a field Show that the following set $A$ of real numbers under addition and multipication is a field:
$A = {a + b\sqrt{2} : a,b \ \text{rational}}$
I am not sure if I am right but here is what I have thus far:
Closure: Let $a_{1} + b_{1}\sqrt{2}$, $a_{2} + b_{2}\sqrt{2}\in \m... | Hints:
After realizing that $\;0=0+0\cdot\sqrt2\;,\;\;1=1+0\cdot\sqrt2\;$, we see all the axioms of a field that are inherited to subsets are fulfilled in $\;A\;$ since the operations used are exactly the same as the ones in $\;\Bbb R\;$ .
The only thing thus that is left to show is closedness of operations and existen... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find all integers $x$, $y$, and $z$ such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ Characterize all positive integers $x$, $y$, and $z$ such that:
$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$$
For example, $\dfrac{1}{x+1} + \dfrac{1}{x(x+1)} = \dfrac{1}{x}$.
| ${1\over x}+{1\over y}={1\over z}$ if and only if it has the form
$${1\over a(a+b)d}+{1\over b(a+b)d}={1\over abd}\quad\text{with}\quad \gcd(a,b)=1$$
(all variables being understood as positive integers).
The "if" part is trivial to verify. The "only if" part comes by setting $x=ga$ and $y=gb$ with $g=\gcd(x,y)$, whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Integrating $1/{(1-x)}$ Given $y = 1 - x = -1 \cdot (x-1) $ with $x \in (0,1)$ then
\begin{equation} \int \frac{1}{y} \: \mathrm{d}x = \int \frac{1}{1-x} \: \mathrm{d}x = \ln(1-x) + C
\end{equation}
but also
\begin{equation} \int \frac{1}{y} \: \mathrm{d}x = - \int \frac{1}{x-1} \: \mathrm{d}x = -\ln(x-1) + C
\end{equa... | Your first attempt is not correct:
Given the integral $$\int \frac{1}{1-x} \,dx$$ let's put $1-x = u,$ and so $\,du = (1-x)'\,dx = -dx \iff dx = -du.$
Upon substitution, $$\int \frac{1}{1-x} \,dx = -\int \frac{1}{u} \,du = -\ln|u| + C = -\ln|1-x| + C = -\ln|x-1| + C$$
Since $x \in (0, 1)$, we can write the answer as $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Geometric proof of $\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$ It is well-known that $$\sin{20^\circ}\sin{40^\circ}\sin{80^\circ}=\frac{\sqrt{3}}{8}$$
It follows that $$\frac{\sin{60^\circ}}{\sin{40^\circ}}=4\sin{20^\circ}\sin{80^\circ}$$
But how to prove this by geometry?
Thank you.
| Another Trigonometric & algebraic proof:
If $\sin3x=\sin3A,3\sin x-4\sin^3x=\sin3A\iff4\sin^3x-3\sin x+\sin3A=0$
$\implies\prod_{r=1}^3\sin x_r=-\dfrac{\sin3A}4$
Again $\sin3x=\sin3A\implies3x=n180^\circ+(-1)^n3A$ where $n$ is any integer
$\implies x=n60^\circ+(-1)^nA$
If we choose even $n=2m,x=120^\circ m+A$
$\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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If $n$ is an odd integer prove that $n - 2^k$ is divisible by $3$ So let $n$ be a odd integer. Show that $n - 2^k$ is divisible by $3$ if $k$ is SOME SPECIFIC positive integer. $k \ge 0$. So there only has to exist one. For example:
$$7 - 2^2 = 3$$ is divisible by $3$
The approach is modular arithemetic, but it is har... | So if $n$ is an odd integer, you have to prove that there exists some integer exponent $k$ such that $3|(n - 2^k)$? Trouble is, it depends on what $n$ is. There are either infinitely many solutions or none at all.
Let's look at this modulo 6: there are three possibilities: $n \equiv 1, 3 \textrm{ or } 5 \pmod 6$. For t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Tough inequality in positive reals numbers. Let $a, b, c$ be positive real. prove that
$$\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\geq\left(1+\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
Thanks
| If the equation out this
$$(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})\geq2(1+\frac{a+b+c}{\sqrt[3]{abc}})$$
We can rewrite the equation:
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$$
writing
$\frac{a+b}{c}=\frac{a+b+c}{c}-1$
$\frac{b+c}{a}=\frac{a+b+c}{a}-1$
$\frac{c+a}{b}=\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Prove formally that $\frac {n^2 + 2}{3n^3 - 5n}\to 0$ as $n \to \infty$. I'm reviewing some Sequences notes from a Mathematics Analysis course I'm taking. I'm finding the beginning of the formal proof below confusing. Some clarity on the following questions would be much appreciated.
Questions
*
*Why is $n\ge3$ noted... | If $n \geqslant 3$ then $n^3 > 5n$ which implies $3n^3 -5n > 2n^3$. This is not true if $n = 2$.
Therefore,
$$\frac{n^2+2}{3n^3 -5n} < \frac{n^2+2}{2n^3}.$$
We also have $2 < 2n^2$ for $n \geqslant 2$ which implies $n^2 + 2 < n^2 +2n^2 = 3n^2$.
Hence if $n \geqslant 3$,
$$\frac{n^2+2}{3n^3 -5n} < \frac{3n^2}{2n^3} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Closed form of $ \int _{ 0 }^{ \pi /2 }{ x\sqrt { \tan { x } } \log { (\cos { x } ) }\ dx }$ Does there exists a closed form of$$ \displaystyle \int _{ 0 }^{ \pi /2 }{ x\sqrt { \tan { x } } \log { (\cos { x } ) }\ dx }$$
If exists can someone find a way to tackle this integral and provide a closed-form of it. Many sim... | Substitute $t = \sqrt{\tan x}$
\begin{align}
I=\int_{0}^{\pi/2}x\sqrt{\tan{x}}\ln({\cos x})\ dx=J(1)\\
\end{align}
where $J(a)= -\int_0^\infty \frac{t^2 \ln(1+t^4) \tan^{-1}(a^2t^2)}{1+t^4}dt$
\begin{align}
J’(a)=&-\int_0^\infty \frac{2at^2 \ln(1+t^4)}{(1+t^4)(1+a^4t^4)}dt\\
=& \ \frac{2a}{1-a^4}\int_0^\infty
\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational I've been struggling to show that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational. I would like to restructure it to prove it, but I can't find anything besides $\sqrt{50} =5 \sqrt{2}$. Could anybody give ... | use this well known identity
$$(a+b)^3=a^3+b^3+3ab(a+b)$$,so we Let
$$x=\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}$$
then we have
$$x^3=14+3\sqrt[3]{49-50}\cdot x=-3x+14$$
so $$x=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
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Prove that $ab \leq \frac14$ and $ (1+1/a)(1+1/b)\ge 9$ when $a+b=1, a \ge 0, b \ge 0$ Our teacher gave us some identities firstly and said we could use one of them to prove it. The identities are: $$\frac{(a^2+b^2)}{2}≥\left(\frac{(a+b)}{2}\right)^2$$ $$(x+y)^2≥2xy$$ and $$\frac{(x+y)}{2} \ge \sqrt{xy}$$
Here's what ... | If the inequality$$\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2$$is used, then we can deduce that $a^2+b^2\ge2ab$ must hold. This can be seen from squaring the terms on the LHS and simplifying the resulting expression.
Now, $$ab\le\frac12(a^2+b^2)=\frac{(a+b)^2-2ab}{2}=\frac12-ab$$This implies that $$2ab\le\frac12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Positive numbers inequality Let $ x_1, x_2, \ldots, x_n \; \in \; \mathbb{R}_{+} $ so that $ x_1 \geq x_2 \geq \cdots \geq x_n $
We also know that:
$$ \frac{1}{2^{x_1}} + \frac{1}{2^{x_2}} + \cdots + \frac{1}{2^{x_n}} = 1 $$
Prove the following inequality:
$$ \frac{1}{3^{x_1}} + \frac{2}{3^{x_2}} + \cdots + \frac{n}{3... | Following Michael Hardy's comment, the problem is equivalent to showing
$$\sum_{k=1}^n k\cdot p_k^{\log_2 3}\ge \frac{n+1}{2n}$$
if $0<p_1\le \cdots\le p_n$ and $p_1+\cdots+p_n=1$. This can be proven as follows:
\begin{align*}
&\sum_{k=1}^n k\cdot {p_k}^{\log_2 3}\\
&\ge \sum_{k=1}^n \frac{n+1}{2}\cdot {p_k}^{\log_2 3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find the derivative of the function. y = $\sqrt{7x+\sqrt{7x+\sqrt{7x}}} $ This question is really tricky. I am wondering if I am right?
| \begin{array}{l}
f(x) = \sqrt {7x + \sqrt {7x + \sqrt {7x} } } \\
{f^2}(x) = 7x + \sqrt {7x + \sqrt {7x} } \\
{f^2}(x) - 7x = \sqrt {7x + \sqrt {7x} } \\
{\left( {{f^2}(x) - 7x} \right)^2} = 7x + \sqrt {7x}
\end{array}
Now I'm doing calculus using chain rule:
\begin{array}{l}
2\left( {{f^2}(x) - 7x} \right)\left( {2f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Can anyone tell me if this is correct? Suppose that the temperature of a metal plate is given by $T(x; y) = x^2 +2x+y^2$, for points $(x, y)$ on the elliptical plate defined by $x^2 + 4y^2 <= 24$.
Find the maximum and minimum temperatures on the plate.
This is what i have done so far.
Finding critical point:
$T(x)=2x+2... | For minimum, point $(-1,0)$ should be correct.
Although I didn't understand your approach for finding maximum, maximum temperature will occur at the boundaries.
$x^2 + 4y^2 = 24$
$y^2 = \frac{24-x^2}{4}$
Substituting this value into the equation for temperature yields $T(x)= \frac{3x^2+8x+24}{4}$
As $x$ ranges from $-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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For a random variable $X$ such that $P(aI've worked on the following problem and have a solution (included below), but I would like to know if there are any other solutions to this problem, particularly more elegant solutions that apply well known inequalities that I've overlooked.
QUESTION: Suppose we have a random va... | For any number $x$ such that $0<a\le x\le b$, we have $x-a\ge0$ and $b-x\ge0$, so
$$ (x-a)(b-x)\ge0,$$
which rearranges into the equivalent form
$$ {ab\over x} + x\le a+b,\tag{1}$$
which is the inequality you obtained earlier. You can slickify the rest of your proof as follows. Let $X$ be a random variable with $0<a\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right)$ $$\sum\limits_{n=1}^{\infty} \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right) = \;?$$
I have been trying to see if it can be written as sum of two telescope terms but it looks tricky. Any help ?
| $$\sum\limits_{n=1}^k \left(\frac{1}{4 n-3}+\frac{1}{4 n-1}-\frac{1}{2 n}\right) =1-\frac12+\frac13-\frac14+\cdots-\frac{1}{2k} + \sum\limits_{n=k}^{2k-1} \frac{1}{2n+1}$$
Can you evaluate the limits of the two series on the right-hand side?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Why is $\sqrt{3}=[1;1,2,1,2,\dots]$?
Why is $\sqrt{3}=[1;1,2,1,2,\dots]$ ?
$\displaystyle[1;1,2,1,2,\dots]=1+\frac{1}{[1;2,1,2,\dots]}=1+\frac{1}{1+\frac{1}{2+\frac{1}{[1;2,1,2,\dots]}}}$
If I set $x=[1;2,1,2,\dots]$ then;
$\frac{x+1}{x}=\frac{5x+2}{3x+1}$
with solutions $x_{1,2}=\frac{1\pm\sqrt{3}}{2}$
what did go w... | But then $\sqrt{3} = 1+1/x_1$. That's because $$\frac{1}{x_1}=\frac{2}{1+\sqrt 3} = \frac{2(\sqrt{3}-1)}{2} = \sqrt{3}-1$$
The other approach is to define:
$$y = 1+ \dfrac{1}{1+\dfrac1{1+y}}$$
Then you get $y=1+\frac{y+1}{y+2}=\frac{2y+3}{y+2}$ which reduces to $y^2=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determining convergence of $\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}$ I have the following infinite series:
$$\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}$$
I want to examine its convergence. First thing that came to my mind was "unfolding"... | We have
$$\sum_{n=1}^{\infty} \sqrt{\sum_{k=3}^{\infty} \frac{1}{n^k k!}}$$
We can use the Cauchy condensation test to study the convergence of
$$\sum_{n=1}^{\infty} 2^n \sqrt{\sum_{k=3}^{\infty} \frac{1}{2^{nk} k!}} = \sum_{n=1}^{\infty} 2^n \sqrt{\frac{1}{2^{2n}} \sum_{k=3}^{\infty} \frac{1}{2^{n(k-2)} k!}}$$
$$\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solving a radical equation for real roots I'm attempting to solve the derivative of my function $f(x)$ for real roots.
$$
\\ \begin{align*}
\\ f(x) &= 3x^2 + 3\arcsin{x}
\\ f^{\prime}(x) &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x \sqrt{1-x^2} + 3
\\ 0 &= \sqrt{36x^2 - 36x^4} ... | $0(-1)=(-4x^4+4x^2-1)(-1) \\
0=4x^4-4x^2+1 \\
0=(2x^2-1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Eigenvectors of a complex matrix Given the following matrix
$\begin{pmatrix}
0 & 1-i & 0\\
1+i & 0 &1-i\\
0& 1+i &0\\
\end{pmatrix}$
I have found the Eigenvalues $0, 2,-2$. But I have no idea how to calculate the corresponding Eigenvectors and I failed with Gaussian method. What could you recommend?
Thanks in advance!... | For the eigenvalue $-2$:
\begin{align}
\begin{bmatrix}
2 & 1-i & 0\\
1+i & 2 &1-i\\
0& 1+i &2\\
\end{bmatrix}
&\to
\begin{bmatrix}
1 & (1-i)/2 & 0\\
1+i & 2 &1-i\\
0& 1+i &2\\
\end{bmatrix} && R_1\gets \tfrac{1}{2}R_1
\\[6px]&\to
\begin{bmatrix}
1 & (1-i)/2 & 0\\
0 & 1 &1-i\\
0& 1+i &2\\
\end{bmatrix} && R_2\gets R_2-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1203497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
How to solve $\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$? Here is my question
$$\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$$
I have tried it by substituting $x$ = $\frac{1}{t}$. I got the answer $0$ but the correct answer is $\pi log(2)$. Any suggestion would be appreciated.
| Approach 1:
\begin{align}
I&=\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx\\
&=\int_0^{1} \frac{\log(x+\frac{1}{x})}{1+x^2}dx+\int_1^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx\\
&=2\int_0^{1} \frac{\log(x+\frac{1}{x})}{1+x^2}dx\\
&=2\int_2^{\infty} \frac{\log(u)}{u\sqrt{u^2-4}}du\\
\end{align}
using $u=x+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pic... | You are probably familiar with "tree diagrams" to solve these types of problems. For the first pick you have a (2/5) chance to remove an even, and a (3/5) chance to remove an odd.
When you remove an odd first, there are only (2/4) odds left. Meanwhile, if you remove an even first, there are now (3/4) odds left.
Finally... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Integral of rational function with trigonometric functions $$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4}
$$
I saw this problem online and it looked like an interesting/difficult problem to try and tackle. My attempt so far is to use tangent half-angle substitution.
Let $t= \tan^2 (\frac{x}{2})$, then $dt= \frac{... | Multiplying top and bottom of the original integral by $\sec^2x$ yields
$$\int\frac{\sec^2xdx}{(1+\sqrt{\tan x})^4}$$
$$u=\tan x,du=\sec^2xdx$$
$$\int\frac{du}{(1+\sqrt u)^4}$$
$$u=t^2,du=2tdt$$
$$\int\frac{2tdt}{(1+t)^4}=\int\frac{(2t+2-2)dt}{(1+t)^4}=\int\frac{2dt}{(1+t)^3}-\int\frac{2dt}{(1+t)^4}=$$
$$\frac{2}{3(1+t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Showing Function is Continuous Let $f: \mathbb{R} \backslash \{2\} \to \mathbb{R}$ be the function given by $f(x) = \frac{2x^2+x-10}{3x-6}$.
Let $g: \mathbb{R} \to \mathbb{R}$ given by:
$$g(x)=\begin{cases}{f(x)} & \text{if } x \neq 2 \\ 3 &\text{if } x = 2 \end{cases}$$
Prove that $g$ is continuous for all $x \in \ma... | You just have to calculate the 2 limits :
$$l_1 = \lim_{x \rightarrow 2; x > 2} g(x) \text{ and } l_2 = \lim_{x \rightarrow 2; x < 2} g(x)$$
and show that $l_1 = l_2 = g(2) = 3$. But $$l_1 = \lim_{x \rightarrow 2; x > 2} g(x) = \lim_{x \rightarrow 2; x > 2} f(x) = \lim_{x \rightarrow 2; x > 2} \frac{2x^2 + x - 10}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Differential Equations (how to proceed) $$y^2+\frac{dy}{dx}=4$$ $y=0$, when $x=\ln 2$
$$\frac{dy}{dx}=-y^2+4$$
$$\frac{\frac{dy}{dx}}{-y^2+4}=1$$
$$\int \frac{\frac{dy}{dx}}{-y^2+4}dx=\int 1dx$$
$$-\frac{1}{4}\ln (-y+2)+\frac{1}{4}\ln (y+2)=x+c_1$$
How to proceed to solve y?
| $$ \begin{align}x - \ln 2 = \int_{\ln 2}^x dx &= \int_0^y\frac{dy}{4-y^2}\\& = \frac 14\int_0^y \left(\frac{1}{2-y} + \frac 1{2+y}\right)\\&=\frac 14 \left(\ln(2+y) - \ln(2-y)\right)\\&=\frac 14\ln\left(\frac{2+y}{2-y}\right)\end{align}$$
that is $$\frac{2 + y}{2-y}=\frac{e^{4x}}{16}\to \frac{y}{2} = \frac{e^{4x}-16}{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplify $\left(\sqrt{\left(\sqrt{2} - \frac{3}{2}\right)^2} - \sqrt[3]{\left(1 - \sqrt{2}\right)^3}\right)^2$ I was trying to solve this square root problem, but I seem not to understand some basics.
Here is the problem.
$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^... | Nicely put question.
You are right about the absolute value missing somewhere. Indeed, we have:
$$\sqrt{x^2} = |x|.$$
In your case, we have
$$\sqrt{\left(\sqrt{2}-\frac{3}{2}\right)^2}=\left|\sqrt{2}-\frac{3}{2}\right|.$$
But $\sqrt{2}-\frac{3}{2}$ is negative, so the absolute value "chooses" the positive version of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Does $a \leq b + c $ imply $a^2 \leq (b+c)^2 + (b-c)^2$? Givens
$$
a \leq b + c
$$
or
$$
a^2 \leq b^2 + c^2+2bc
$$
Can we prove that?:
$$
a^2 \leq (b+c)^2 + (b-c)^2 = 2b^2 + 2c^2
$$
| $$(b-c)^2\geq0$$
therefore $$b^2+c^2\geq 2bc$$
Hence $$2b^2+2c^2\geq b^2+c^2+2bc\geq a^2$$
But the problem is that the very first inequality $b+c\geq a$ doesn't tell us if $a$ is non-negative. It is possible that $b+c\geq a$ but $a$ is very negative like $a=-\sqrt{(b^2+c^2)+1}$, while $b,c\geq0$. Then $b+c\geq0\geq a$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Error in approximating the sum I am watching one of the online probability courses and in one of the lectures, the professor simplifies the sum:
$$A = \sum_{j=0}^{N}\frac{j^k}{N^k} \cdot \frac{1}{N+1}$$
in the following way: $A \approx \int_{0}^{1}x^kdx=\frac{1}{k+1}$
The problem is that in my opinion the simplificati... | SalmonKiller gave a good point and I shall not repeat. However, one could notice that $$A_{k,N} = \sum_{j=1}^{N}\frac{j^k}{N^k} \cdot \frac{1}{N+1}=\frac{ H_N^{(-k)}}{(N+1)N^k}$$ where appear the harmonic numbers. Expanding as series for large values of $N$, we then have $$A_{k,N}=\left(\frac{1}{k+1}+\frac{\frac{1}{2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Let $p$=prime and $\sqrt{x}+\sqrt{y}<\sqrt{2p}$ Let $p$ be a fixed odd prime. Let $x,y\in \mathbb{Z}_+$ such that $\sqrt{x}+\sqrt{y}<\sqrt{2p}$. Prove that $$\sqrt{x}+\sqrt{y}\le \sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}.$$
Any ideas at all? This seems extremely difficult to do using elementary methods.
Note: It is fro... | Let's suppose that the problem is not true and there exists $x,y$ such that
$$\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}<\sqrt{x}+\sqrt{y}<\sqrt{2p}\qquad (1)$$
First, note that if $x+y\le p$ then $$\sqrt{x}+\sqrt{y}\le\sqrt{x}+\sqrt{p-x}\le\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}$$
Since $\sqrt{x}+\sqrt{p-x}$ is conca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1214870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.