Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If a number is rational, then it has a periodic decimal expression? I have proved the following:
If the decimal expansion of a number is periodic, then it is a rational number.
Now I am trying to prove the converse. For this, I am taking the rational numbers smaller than $1$, that is $\frac{m}{n}$ with $n>m$ becaus... | EDIT: I'd like to elaborate on something I glossed over in this answer originally - namely, why we can assume that "...we necessarily hit a repeat remainder." The added explanation is included at the bottom of the answer (to keep it readable).
I think you're on the right track. But the idea here is, for $\frac{p}{q}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2347979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding the global minimum of $x^{n}+x^{n-1}+...+1$ for even $n.$ Inspired by this question, I am curious if there is an asymptotic of the global minimum of the function: $$f_n(x) = x^{2n}+x^{2n-1}+...+x^2+x+1.$$
In the referred question, I showed that $$f_n(x) = x^{2n}+x^{2n-1}+...+x^2+x+1 =x^{2n-2}(x+\dfrac{1}{2})^2+... | Claim. The minimum of $f_n$ tends to $\frac12$ when $n \to \infty$.
Let $a_n$ denote the minimum of $f_n(x) = \frac{x^{2n+1}-1}{x-1}$ for $x \in \mathbb{R}$.
For fixed $t>0$ and $n > t$ we also have $$a_n \leq f_n\left(-1 + \frac{t}n\right) = \frac{-1 + (-1 + \frac{t}{n})^{1 + 2 n}}{-2 + \frac{t}{n}} = \frac{1 + (1-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2350273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Is it possible to evaluate analytically the following nontrivial triple integral? In a physical mathematical problem, I came across a nontrivial triple integral below obtained upon 3D inverse Fourier transformation. It would be great if someone here could provide with some hints that could help to evaluate analytically... | Well, for the first integral we have:
$$\mathscr{I}:=\int_0^{2\pi}\int_0^\infty\int_0^\pi\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)\cdot\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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find a,b,c where $\sqrt{5^{1/3}-4^{1/3}}=\frac{1}{3}(a^{1/3}+b^{1/3}-c^{1/3})$ Is it ramanujan problems? $$\sqrt{5^{1/3}-4^{1/3}}=\frac{1}{3}(a^{1/3}+b^{1/3}-c^{1/3})$$
find $a,b,c$
Any helps would be appreciated.
| Note that
\begin{eqnarray*}
\left( \sqrt[3]{\alpha}+\sqrt[3]{\alpha^2 \beta }-\sqrt[3]{\beta^2}\right) ^2 &=& \color{blue}{\sqrt[3]{\alpha^2}}+\color{red}{\alpha\sqrt[3]{\alpha \beta^2}} +\beta\sqrt[3]{\beta}+2 \alpha\sqrt[3]{\beta} \color{red}{-2 \sqrt[3]{\alpha\beta^2}} \color{blue}{-2 \beta\sqrt[3]{\alpha^2}} \\
&... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does this integral have a closed-form answer? Does this integral have a closed-form solution?
$$\int\limits_{0}^{\infty}\dfrac{x^4\left(2x^{22}+\left(11\sqrt{5}+11\right)x^{17}-20\right)}{2\left(x^{22}+1\right)\sqrt{4\left(\sqrt{x^{22}+1}+x^{10}+(\sqrt{5}+1)x^{5}\right )+6+2\sqrt{5}}}dx$$
EDIT
Sorry for the lack of con... | After much effort, I asked my teacher to solve this. Here's his solution:
You need to notice that $$4(\sqrt{x^{22}+1}+x^{10}+(\sqrt{5}+1)x^{5})+6+2\sqrt{5}=4\left (\sqrt{x^{22}+1}+x^{10}+(\sqrt{5}+1)x^{5}+\frac{6+2\sqrt{5}}{4} \right )=4\left (\sqrt{x^{22}+1}+(x^{5}+\varphi)^{2} \right )$$
where $\varphi=\frac{\sqrt{5}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this ... | The inequality is equivalent to
$$a^8 + b^8 + c^8 > a^2b^3c^3 + a^3b^2c^3+ a^3b^3c^2.$$
Now apply Muirhead's inequality (i.e. $[8,0,0]\geq [3,3,2]$) and note that the equality holds iff $a=b=c$.
| {
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Find the maximum and minimum value of $S= \log_{2}a+\log_{2}b+\log_{2}c$. Let $a, b, c \geq 1$ and $a+b+c=4$. Find the maximum and minimum value of $S= \log_{2}a+\log_{2}b+\log_{2}c$.
I found the maximum, it is easy to prove $S_\max = 3\log_{2}\frac{4}{3}$. I think the minimum is $1$ when there are two number are $1$ a... | Let $f(x)=\log_2{x}$ and $a\geq b\geq c$. Hence, $f$ is a concave function.
Also we have: $a\leq2$ and $a+b=4-c\leq3$.
Thus, $(2,1,1)\succ(a,b,c)$ and by Karamata
$$\sum_{cyc}\log_2a\geq\log_22+\log_21+\log_21=1.$$
The equality occurs for $a=2$ and $b=c=1$, which says that $1$ is a minimal value.
In another hand, by A... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving equations $a+b+c=5,a^2+b^2+c^2=11,a^3+b^3+c^3=27$
Let $a, b, c $ be real numbers such that $a<b<c$ and satisfying
$$a+b+c=5;$$
$$a^2+b^2+c^2=11;$$
$$a^3+b^3+c^3=27.$$
Prove that $0<a<1<b<2<c<3$.
My understanding :
$(a+b+c)^2 - (a^2+b^2+c^2) = 14$
so, $ab+bc+ca = 7$
$a^2+b^2+c^2-ab-bc-ca = 4$
$(a+b+c)(... | This is a classical exercise about symmetric functions. By Newton's identities
$$ e_1 = p_1 = 5, $$
$$ 2e_2 = p_1^2 - p_2 = 14, $$
$$ 3e_3 = e_2 p_1 - e_1 p_2 + p_3 = 7 $$
hence $a,b,c$ are roots of the polynomial
$$ p(z)=(z-a)(z-b)(z-c) = z^3-e_1 z^2+e_2 z-e_3 = z^3-5z^2+7z-\frac{7}{3} $$
and the claim follows from th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\sqrt{9x^2-16}>3x+1$ I'm trying to solve the following inequality:
$$\sqrt{9x^2-16}>3x+1$$
Here's my attempt:
$\sqrt{9x^2-16}>3x+1$
$\longrightarrow 9x^2-16>9x^2+6x+1$
$\longrightarrow -16>6x+1$
$\longrightarrow x<-\frac{17}{6}$
Now, I need to check the constraints:
$9x^2-16 > 0$
$\longrightarrow (3x)^2 > 4^2$
$\lo... | at first it must be $|x|\geq \frac{4}{3}$ if $$x\le -\frac{1}{3}$$ our inequality is true. let $$x>-\frac{1}{3}$$ we can square it:
$$9x^2-16>9x^2+6x+1$$ otr $$x<-\frac{17}{6}$$ together with the conditions $$-\infty<x<-\frac{17}{6}$$ and $$-\frac{1}{3}<x$$ or $$-\frac{1}{3}\geq x$$ and $$x\geq \frac{4}{3}$$ or $$x\le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the equation of straight lines through the point $(\dfrac {1}{\sqrt {3}}, 1)$ whose perpendicular distance from the origin is unity. Find the equation of straight lines through the point $\left(\dfrac {1}{\sqrt {3}}, 1\right)$ whose perpendicular distance from the origin is unity.
My Attempt:
Let the equation of l... | Using the normal form of straight line, let the equation of the line be
$$x \cos \alpha + y \sin \alpha = 1$$
Now since the line passes through $\left ( \frac{1}{\sqrt{3}}, 1 \right )$, we get
$$\frac{\cos \alpha}{\sqrt{3}} + {\sin \alpha} = 1 \implies \alpha = \frac{\pi}{6}
\ \ or \ \ \frac{\pi}{2} $$
Hence we get t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\frac {1}{\sqrt{5}}[(\frac {1}{x+r_+}) - (\frac {1}{x+r_-}) = \frac {1}{\sqrt{5}x}[(\frac {1}{1-r_{+}x}) - (\frac {1}{1-r_{-}x})] $ I need to manipulate this equation:
$$
\frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right)
$$
to show that
$$ \frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac ... | From your last step:
$$\frac{1}{\sqrt{5}}\left(\frac{r_{+}}{1 - r_{+}x} - \frac{r_{-}}{1 - r_{-}x}\right)$$
$$= \frac{1}{\sqrt{5}x}\left(\frac{r_{+}x}{1 - r_{+}x} - \frac{r_{-}x}{1 - r_{-}x}\right)$$
$$= \frac{1}{\sqrt{5}x}\left(\left[\frac{1}{1 - r_{+}x} - 1\right] - \left[\frac{1}{1 - r_{-}x} - 1\right]\right)$$
$$=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx$ via residue calculus. Let $\Gamma_R$ be the semicircle of radius $R$ in the upper half plane. Then,
\begin{align}
\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx
&= \lim_{R\to \infty}\int_{\Gamma_R}\frac{1}{(1+z^2)^{n+1}}dz \\
&= 2\pi i \operat... | Your answer is actually
$$\pi\frac{(2n)!}{2^{2n}(n!)^2}.$$
(Check the derivative.)
Then use
$$ (2n)!=(1 \cdot 3 \cdots (2n-1))\cdot 2^n \cdot (n!)$$ to get
$$\pi\frac{(2n)!}{2^{2n}(n!)^2}=\pi\frac{1 \cdot 3 \cdots (2n-1)}{2^n \cdot (n!)}=\pi\frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to prove that $2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1$ When I'm reading Computer Systems: A Programmer's Perspective, I met the sum of binary numbers and failed to prove it:
$$ 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1 $$
This might be preliminary knowledge, I'm not good at mathematics, any bod... | Exploiting the binary representation,
$$111\cdots111_2+1_2=1000\cdots000_2$$ because the carry propagates.
This is exactly
$$2^0+2^1+2^2+\cdots2^{n-2}+2^{n-1}+2^n+1=2^{n+1}.$$
The proof generalizes to other bases. Let $a:=b-1$, then
$$a\cdot111\cdots111_b+1_b= aaa\cdots aaa_b+1_b=1000\cdots000_b,$$ which is a rewrite ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 6
} |
Radical problem: $\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$ What is the value of
$$\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$$
| With $x=\sqrt{7+2\sqrt{12}}$ and $y=\sqrt{7-2\sqrt{12}}$, we have $x^2+y^2=14$ and $xy=\sqrt{7^2-2^2\cdot12}=1$, so $(x+y)^2=14+2\cdot1=16$ and thus $x+y=4,$ since $x,y$ are positive.
In general, we'd have $$\sqrt{a+b\sqrt{c}}+\sqrt{a-b\sqrt{c}}=\sqrt{2a+2\sqrt{a^2-b^2c}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363879",
"timestamp": "2023-03-29T00:00:00",
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Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$? Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$, if $x, y$ are reals greater than $1$, and $a, b, c$ are positive reals?
A proof with all the math to go from one to the other would be nice.
| since you added a clause about $x,y > 1,$ some two minutes after first posting the question, the conclusion you require is false. In fact, take any $a,b$ (not both zero) at all and then take $c = 0,$ then for $x,y > 1,$
$$ a^2 x^2 + (a^2 + b^2) xy + b^2 y^2 > 0 $$
however
$$ (a^2 + b^2)^2 - 4 a^2 b^2 = (a^2 - b^2)^2 ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $\ln{\frac{n+1}{n}}\le\frac{2}{n+1},\forall n\in\mathbb{N}^+$? I found an inequality: $$\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1},\forall n\in\mathbb{N}^+.$$
I tried induction. It is obvious if $k=1$, when $n=k$, $\ln\sqrt{\left(\frac{k+1}{k}\right)^{k+1}}\le 1$, but bogged down for $n=k+1$:
$$\ln\sqrt... | Integrate $\frac{1}{x}$ between $n$ and $n+1$: $\ln \frac{n+1}{n} = \ln (n+1) - \ln (n) = \displaystyle\int_n^{n+1} \frac{1}{x} \operatorname{d}x$. As $x\to \frac{1}{x}$ is decreasing, you get $\ln \frac{n+1}{n}\leq \displaystyle\int_n^{n+1} \frac{1}{n} \operatorname{d}x = \frac{1}{n}$.
It is moreover clear that $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Quadratic equations having a common root. I have a cubic polynomial $f(x)=x^3+px^2+qx+72$ which is divisible by both $x^2+ax+b$ and $x^2+bx+a$ (where a,b,p,q are constants and a$ \neq $b).I have to find the sum of the squares of the the roots of the cubic polynomial.
I tried to attempt it like this.
Since the quadratic... | The roots are 1, 8, 9. From here you can simply sum the cubes.
Now for the solution:
As you noted your two quadratics have a common root. Let this root be called $y$. Then $y^2 + ay + b = 0$ and $y^2 + by + a = 0$. This is a system of two equations. Solving we get $(a-b)y = (a-b)$. Since $a \neq b$ we conclude $y=1$.
N... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understa... | Following the author's solution, since $n^4-n^2+64>(n^2-1)^2$ any $n^4-n^2+64$ which equals a square can be written as $(n^2+k)^2$ for $k$ a non-negative integer (i.e. $k>-1$ in the $(n^2-1)^2$). That means $$n^4-n^2+64 = n^4+2kn^2+k^2$$
and $$n^2=\frac{64-k^2}{(2k+1)}.$$ We need to check for $0\leq k\leq 8$ (since $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Arithmetic-geometric mean, prove that $c_n = 4me^{-\ell 2^n+\epsilon_n}$ Let $a$ and $b$ reals with $a > b > 0$. Let $(a_n)$ and $(b_n)$ with $a_0 = a$, $b_0 = b$ and
$$a_{n+1} = \tfrac{a_n+b_n}{2} \quad\text{;}\quad b_{n+1} = \sqrt{a_nb_n}$$
We know that $\lim_{n \rightarrow +\infty} a_n = \lim_{n \rightarrow +\infty}... | Using $c_{n+1} = \frac{c_n^2}{2(a_n+b_n)} = \frac{c_n^2}{4a_{n+1}}$, we get.
\begin{align*}
u_{n+1} & = -2^{-n-1} \ln(c_{n+1}) \\
& = -2^{-n-1} (\ln(c_n)-\ln(4a_{n+1})) \\
& = u_n+2^{-n-1} \ln(4a_{n+1})
\end{align*}
$\ln(4a_{n+1}) = \ln(4m)+\underbrace{\ln(\frac{4a_{n+1}}{4m})}_{\alpha_n}$. The sequence $(\alpha_n)$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Representing polynomials as quadratic forms I have the following cubic equation
$$x^3-6x^2+11x-6=(x-1)(x-2)(x-3)=0$$
whose solutions are $x=1,2,3$.
What if the above equation were represented by quadratic form? Let $\mathbf x = \begin{bmatrix} x^2 & x & 1\end{bmatrix}^T$. Can the cubic equation above be represented as ... | The simplest solution is to use the 3 homogeneous coordinates $\pmatrix{ x^2 & x & 1}$ and a 3×3 symmetric matrix
$$A x^3 + B x^2 + C x + D = \pmatrix{x^2 \\x \\1}^\top
\begin{bmatrix}
0 & \frac{A}{2} & 0 \\
\frac{A}{2} & B & \frac{C}{2} \\
0 & \frac{C}{2} & D
\end{bmatrix} \pmatrix{x^2 \\x \\1} $$
But the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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An equation for prime numbers $\frac{p-1}{2}(2^p-1)+1=7k^2$
If $2^p-1$ is a Mersenne prime, and $k$ is an integer, then solve
$$\frac{p-1}{2}(2^p-1)+1=7k^2$$
$$$$
If I take modulo $p$ I get $1 \equiv 14k^2 \pmod{p}$.
If I take modulo $q=2^p-1$ I get $1 \equiv 7k^2 \pmod{q}$.
Can I use these to solve the equation?... | This is only a partial answer, but I hope it may help a little. Your evaluations modulo $p$ and $q$ are of course correct, but the argument can be carried a little further. You already know that the case $p=3$ is impossible, so I will assume in what follows that $p > 3$.
It is easily verified that $2^p - 1 \equiv 1 \pm... | {
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"url": "https://math.stackexchange.com/questions/2372659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Problems while solving the differential equation. $$x^2\frac{d^2y}{dx^2}+x^2\frac{dy}{dx}-2y=0$$
$x=0$ is a regular singular point.
$$y=\sum_{n=0}^\infty c_nx^{n+r}$$
$$\frac{dy}{dx}=(n+r)\sum_{n=0}^\infty c_nx^{n+r-1}$$
$$\frac{d^2y}{dx^2}=(n+r)(n+r-1)\sum_{n=0}^\infty c_nx^{n+r-2}$$
$$(n+r)(n+r-1)\sum_{n=0}^\infty c_... | In your equations they are some terms with $n$ in them.
These terms should be in the $\displaystyle{\sum_n}$ , not outside.
Nevertheless, your result is correct which is well. But one can go further.
Case $r=2$ :
$$c_n=\frac{-(n+1)}{n^2+3n}c_{n-1}$$
$$c_n=c_0\prod_{k=1}^n\left(\frac{-(k+1)}{k^2+3k}\right)=
(-1)^n c_0\... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving $(1 + i) z - 2 \overline{z} = -11 + 25i$ How do we find the complex number $z$ that satisfies
$$(1 + i) z - 2 \overline{z} = -11 + 25i.$$
How should I start? Am I supposed to express $z$ as $x+yi$?
| Since both, left hand side and right hand side of
\begin{align*}
(1 + i) z - 2 \overline{z} = -11 + 25i\tag{1}
\end{align*}
represent a complex number, we can equate real and imaginary part of them. We set $z=x+iy$ with real part $x$ and imaginary part $y$.
We obtain
\begin{align*}
\Re\left((1+i)(x+iy)-2(x-iy)\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve for four values of x where x lies in complex plane. I came across this problem and have to find four roots of this equation. Direct multiplication was of no help and factorizing it is a mess.
I tried to factor it using $a^2+b^2=(a+ib)(a-ib) $
but was of no help.
$$x^2 + (\frac{ax}{x+a})^2= 3a^2$$
Can anyone thro... | Okay I tried to solve this question using two techniques, one is by factorizing and then substitution as below:
$$x^2 +(\frac{ax}{x+a})^2 = 3a^2$$
We can write $ a^2 + b^2$ as $$a^2 + b^2= (a-b)^2 +2ab$$ so $x^2 +(\frac{ax}{x+a})^2$ can be written as $$(x -\frac{ax}{x+a})^2 + 2 * x * \frac{ax}{x+a}$$ $$\implies (\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$. Question:
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$.
My attempt:
Proof by cont... | $$n^2\equiv(0,1,-2,4)mod9.$$ It's all!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Expanding $(x+1)(x+2)(x+3)(x-1)(x-2)(x-3)$ Is there any trick to multiplying this?: $(x+1)(x+2)(x+3)(x-1)(x-2)(x-3)$?
The brackets are to be eliminated and the result is to be simplified as much as possible.
I have started with this: \begin{align}&(x+1)(x+2)(x+3)(x-1)(x-2)(x-3)\\&= (x^3+2x+3x+x+2+x+3+3x+2x+6)(x^3-2x-3x... | $(x+a)(x-a)=x^2-a^2$ so collect these terms to obtain a third degree polynomial in $x^2$. Solve that as normal.
${(x+1)(x+2)(x+3)(x-1)(x-2)(x-3) \\\qquad=~ (x^2-1)(x^2-4)(x^2-9) \\\qquad=~ x^{2\cdot 3}-(\phantom{1+4+9})x^{2\cdot 2}+(\phantom{4\cdot 9+4+9})x^2-1\cdot 4\cdot 9 \\ \qquad\quad\ddots}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What are the real and imaginary parts of $(x+e^{ix})^{0.5}$? Finding the real and imaginary parts of
$(x+e^{ix})^{n}$ is easy if n is an integer greater than 1. But what if n is a fraction? Suppose I have a function like $(x+e^{ix})^{0.5}$. How do you find the real and imaginary parts of this?
Edit: x is a real n... | If $n$ is not an integer, $(x + e^{ix})^{n}$ is multi-valued.
First write $x + e^{ix} = x + \cos x + i\sin x = e^{i\theta} \sqrt{x^{2} + 2x \cos x + 1}$ for some real $\theta$ (unique up to an additive integer multiple of $2\pi$).
If $n$ is real, then
\begin{align*}
(x + e^{ix})^{n}
&= (x^{2} + 2x\cos x + 1)^{n/2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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$\lim\limits_{n \rightarrow \infty} \sqrt{n^2+n} -n$?
Calculate
$\displaystyle\lim_{n \to \infty}
\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right)$.
$\displaystyle\lim_{n \to \infty}\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right) =
\infty - \infty$ We have an indeterminate form
So I proceeded to factorize $$\sqrt{n^2+n... | Using you approach $$\sqrt{n^2+n} -n = \sqrt{ \frac{n^2(n+1)}{n}}-n =n \left[ \sqrt{\frac{n+1}{n}}-1 \right]=n \left[ \sqrt{1+\frac{1}{n}}-1 \right]$$ Now, use the generalized binomial theorem or Taylor series to get, for small values of $x$, $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ $$\sqrt{1+x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How many 5 letter different words can be formed using the letters AAASSSBB I saw that these sort of questions have been asked before, but not exactly this kind of problem: creating 5 DIFFERENT words, not just all possible combinations.
The words don't need to have any meanings.
I'd appreciate an explanation in addition... | $$
\begin{array}{c|c|c}
\text{Letters} & \text{Calculation} & \text{# Possibilities}\\
\hline\\
AAASS & \displaystyle\binom{5}{3} & 10 \\\\
AAASB & \displaystyle\binom{5}{3} \cdot 2 & 20\\\\
AAABB & \displaystyle\binom{5}{3} & 10\\\\
AASSS & \displaystyle\binom{5}{2} & 10\\\\
AASSB & \displaystyle\binom{5}{2} \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
My attempt:
*
*$x^2 + y^2 + z^2 = 1$
*$x+y+z=0$
$$(2) \implies z = -(x+y)$$
$$(1) \implies x^2+y^2+(x+y)^2 = 1$$
$$2x^2 + 2y^2 + 2xy = 1$$
This is the curve i... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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How can i find the least number n that can be presented as product of a*b Which the least number n can we imagine in product n = a∙b like k ways? Products a∙b and b∙a is one of the way, where all numbers is natural (1≤ k ≤50)
I tried to loop from 1 to 1000000 and with each number perform the following:
looping from 1 t... | Suppose $$n=2^{n_1}\cdot 3^{n_2}\dots \cdot p_r^{n_r}$$ where $p_r$ is the $r^{th}$ prime. The number of divisors of $n$ $$d(n)=(n_1+1)(n_2+1)\dots (n_r+1)$$
The number of factorisations is half of this unless $n$ is a square in which case it is half $d(n)+1$.
To find a number having $k$ factorisations, reverse this pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Continuity- function Let $f : R \to R $ be a function which is continuous at $0$ and $f(0)=1$ . Also assume that f satisfies the following relation for all $x$ $$f(x)-f\left(\frac{x}{2}\right)=\frac{3x^2}{4}+x$$ Find $f(3)$
[I have tried it solve. I will be grateful indeed if someone can review the following solution a... | hint
$$f (3)-f (\frac {3}{2})=\frac {3^3}{2^2}+3$$
$$f (\frac {3}{2})-f (\frac {3}{2^2})=\frac {3^3}{2^4}+\frac {3}{2} $$
...
$$f (\frac {3}{2^n})-f (\frac {3}{2^{n+1}})=\frac {3^3}{2^{2n+2}}+\frac {3}{2^n} $$
sum and $n\to +\infty $.
You will find
$$f (3)=f (0)+\frac {27}{4}\frac {1}{1-\frac {1}{4}}+3\frac {1}{1-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the Diophantine equation $x^6 + 3x^3 + 1 = y^4$
Find all pairs $(x, y)$ of integers, such that $x^6 + 3x^3 + 1 = y^4$.
My solution:
Claim: The pair $(0, 1)$ are the only solutions.
Proof. Suppose there exists other solutions for $y \gt 1$ and $x \gt 0$, then I shall show that such pairs are impossible if $x$ a... | $x^6+3x^3+1-y^4=0$ is a quadratic equation.
Thus, there is an integer $n$ for which $$3^2-4(1-y^4)=n^2$$ or
$$n^2-4y^4=5$$ or
$$(n-2y^2)(n+2y^2)=5$$
and we have four cases only.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Trigonometry limit's proof: $\lim_{x\to0}\frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x}=\frac{k(k+1)}{2}$ How to prove that $$\lim_{x\to0}\frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x}=\frac{k(k+1)}{2}$$
I tried to split up the fraction and multiple-divide every new fraction with its $x$ factor but didn't work out.
ex: $$\lim_... | You are so close. Note that
\begin{align*}
\frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x}
&= \frac{\sin(x)}{x}+\frac{\sin(2x)}{x}+\cdots+\frac{\sin(kx)}{x} \\
&= \frac{\sin(x)}{x}+2\frac{\sin(2x)}{2x}+\cdots+k\frac{\sin(kx)}{kx} \\
&\to 1 + 2 + \cdots + k \\
&= \frac{k(k+1)}{2}
\end{align*}
as $x\to0$.
I suspect you may no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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Proof that $n!$ is divisible by $(n+1)^2$ for $n=xy+x+y$ I noticed the other day that for $n>8, n\in\mathbb{N}$, the factorial $n!$ seems to be divisible by $(n+1)^2$ when $n$ can be written in the form $xy+x+y$ (where $x,y\geq1$ and $\in\mathbb{N}$). Some examples:
*
*$n=14=2 \times 4+2+4$ (i.e. $x=2$, $y=4$), and ... | $\color{Green}{\text{Question}}$ :
Find all integers $n$;
such that $\color{Green}{(n+1)! \mid n!} $ .
$\color{Blue}{\star \ \ \ \ \text{condition}}$ :
Now suppose that $n+1$ can be written
in the form $n+1=ab$;
with $2 < a,b$.
Also assume that one of the following occures:
*
*If $a= b$; then $5 \leq a=b$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\... | Make the problem simpler using $x=y+1$ and later $\pi y=z$ $$\lim_{x\to 1} \dfrac {1+\cos (\pi x)}{\tan^2 (\pi x)}=\lim_{y\to 0} \dfrac {1-\cos (\pi y)}{\tan^2 (\pi y)}=\lim_{z\to 0} \dfrac {1-\cos (z)}{\tan^2 (z)}$$ and consider either equivalents $$\cos(z)\sim 1-\frac{z^2}2 \qquad \text{and}\qquad\tan(z)\sim z \impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for every real number $x$, if $|x − 3| > 3$ then $x^2 > 6x$. This is Velleman's exercise 3.5.10:
Prove that for every real number $x$, if $|x − 3| > 3$ then $x^2 > 6x$.
And here's my proof of it:
Proof. Suppose $|x − 3| > 3$. We now consider two cases:
Case 1. $x - 3 \ge 0$. Then $|x − 3| = x - 3$, so we hav... | Your proof is incomplete: you should show in BOTH cases that $x^2>6x$. So in case 2, when $x<0$, at the end, it suffices to sya that "and therefore $x^2>3x>6x$".
However, there is a much shorter way that I warmly recommend. Since $|x − 3|$ and $3$ are both non-negative, the inequality $|x − 3| > 3$ is equivalent to $|x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to find linear dependency relationship
The Question I am attempting to answer gives vectors of,
(2,1,4,3) (2,3,2,3) (4,11,-1,6) and asks for an explicit dependency relationship.
I have found them to be linearly dependent. (found the determinant = 0)
How do i then find a dependency relationship?
Looking at t... | Look for the relationship between the three vectors:
$$
xu+yv+zw=
x
\left[\begin{matrix}
2\\
1\\
4\\
3
\end{matrix}\right]
+
y
\left[\begin{matrix}
2\\
3\\
2\\
3
\end{matrix}\right]
+
z
\left[\begin{matrix}
4\\
11\\
-1\\
6
\end{matrix}\right]
=
\left[\begin{matrix}
2 & 2 & 4\\
1 & 3 & 11\\
4 & 2 & -1\\
3 & 3 & 6\\
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able ... | A systematic way doing this is using Newton's identifites.
Let $p_k = a^k + b^k + c^k$ for $k = 1, 2, 3, 4$ and
$$\begin{align}
s_1 &= a + b + c\\
s_2 &= ab+bc+ca\\
s_3 &= abc
\end{align}$$
be the elementary symmetric polynomials associated with $a, b, c$.
Newton's identities tell us:
$$\require{cancel}\begin{array}{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
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find a and b such that the limit will exist and find the limit I am given this question $$\lim_{x\rightarrow 0} \frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}$$
And asked to find the $a$ and $b$ such that the limit exists and also to compute the limit
My solution:
Let $t$=$\sqrt{1+x}$.
Then the Maclaurin polynomial is : $$\sqrt{... | Hint. When you write
$$
e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}=1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{O}(x^6)
$$ it is wrong, since it is not right that, as $x \to 0$,
$$
\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)^m=\mathcal{O}(x^6),\qquad m=2,3,\cdots.
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$, $\frac{P(x)}{x^3+x}$ remainder Given the polynomial:
$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$
What is the remainder of $\frac{P(X)}{x^3+x}$?
I don't think the long division is efficient the way to go, and the remainder theorem doesn't seem to be applicable here as $x^... | The remainder is
$$\frac{x+2}{x^2+1}\:+\:\frac 1x\qquad\text{or}\quad2x^2+2x+1\;\;\text{respectively}\\[3ex]$$
obtained by taking a lazy approach: Feeding the command
apart((x**100+x**50-2*x**4-x**3+x+1)/(x**3+x))
into http://live.sympy.org/ ,
thus asking Python to go through the long division in short time,
yields
$$... | {
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"url": "https://math.stackexchange.com/questions/2403263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Is this true $n!\leq(\frac{5n+7}{12})^n,n∈N$? Is the following inequality true?
For all $n\in \Bbb N$ prove that:
$$n!\leq\left(\frac{5n+7}{12}\right)^n.$$
I know the answer,but I want to see other people how to prove the problem.
In my proof I used $\frac{5n+7}{12}=\frac{\frac{n+1}2+\frac{n+2}3}2\geq \sqrt{\frac{(n+... | By AM-GM
$$\frac{1\cdot n+2(n-1)+...+n\cdot1}{n}\geq\sqrt[n]{(n!)^2}$$ or
$$\left(\sqrt{\frac{(n+1)(n+2)}{6}}\right)^n\geq n!.$$
Thus, it remains to prove that
$$\frac{5n+7}{12}\geq\sqrt{\frac{(n+1)(n+2)}{6}},$$
which is $$(n-1)^2\geq0.$$
Done!
$$1\cdot n+2(n-1)+...+n\cdot1=\sum_{k=1}^nk(n-k+1)=$$
$$=(n+1)\sum_{k=1}^nk... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Polynomials$ P(x^2)=(x+1)P(x)$ Find all polynomials $P(x) \in\mathbb{R}[x]$ satisfying $$ P(x^2)=(x+1)P(x)$$
Please check my work :
Let $r$ be root of $P(x)$ so $P(r)=0$
Substitute $x=r$, $ P(r^2)=(r+1)P(r)$ so $P(r^2)=0$, i.e., if $r$ is root then $r^2$ is also root.
Substitute $x=-1$, so $P(1)=0$, i.e., if $r$ is roo... | Suppose $P(x)$ has a root other than $1$ and $-1$ (say $r$). Then $r^2$ is also a root of the polynomial, i.e. $r^4$ is also a root of the polynomial which leads $P$ has infinitely many roots, which is not possible.
Then only roots $P$ can have are $1$ and $-1$. It is easy to see that $1$ is a root of $P$.
$c\neq 0$
Le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a... | Follows directly from Hölder's inequality on a finite measure space (with dual exponents $(2,2)$): $1 = (a+b)^2 \le (a^2+b^2)(1^2+1^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
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Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$ or $a^3-b^3$ will be divisible by $7$. This is a very interesting word problem that I came across in an old textbook of mine:
Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$... | If $a$ or $b$ is a multiple of 7 then it's obvious, else $a^6\equiv b^6 \equiv 1 \pmod 7$ and $(a^3+b^3)(a^3-b^3)=a^6-b^6\equiv 0 \pmod 7$. Since 7 is prime it divides $a^3+b^3$ or $a^3-b^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
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Solve $\frac{1}{x}<\frac{x}{2}<\frac{2}{x}$. First I consider the case $\frac{1}{x}<\frac{x}{2}$, manipulating to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x}<0\Leftrightarrow x < -\sqrt{2}.$ Did this by subtracting $\frac{x}{2}$ from both sides and writing everything with a common denominator.
Second case to consider is $\fra... | If $x>0$, multiply by $x$ to get
$$1<\frac{x^2}{2}<2\iff 2<x^2 <4\implies \sqrt 2<x<2$$
If $x< 0$ multiply by $x$ and
$$1>\frac{x^2}{2}>2\iff 2>x^2>4\implies \text{false}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simple logarithmic differentiation Given the equation $\frac{dy}{dx} = \frac{-x}{y}$
How can you solve for $\frac{d^2y}{d^2x}$ using logarithmic differentiation?
Here is my work:
$ln(\frac{dy}{dx}) = ln(\frac{-x}{y})$
$ln(\frac{dy}{dx}) = ln(-x) - ln(y)$
(Take derivative of both sides)
$(\frac{d^2y}{d^2x})/(\frac{dy}{d... | That equation can be rewritten with another notation like y'y = -x
simply claculate the dervative of that expression
$$y"y+y'²= -1 $$
where y'= dy/dx and y" = d²y/dx²
Rewriting this last equation with your notation
$$d²y/dx²=-(x²+y²)/y³$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If the constant term of binomial expansion $ (2x-\frac{1}{x})^{n}$ is$-160$ ,then $n$ equal to? As I know the expansion binomial expression of $ (2x-\frac{1}{x})^{n}$ .But I don't know that for which $-160$ in $n$th term of given expression.
Please help me to solve this.
| As usual, the binomial expansion helps:
$$
\left(2x - \frac 1x\right)^n = \sum_{k=0}^n (-1)^k\binom nk\frac{1}{x^k}(2x)^{n-k} = \sum_{k=0}^n \binom nk (-1)^k2^{n-k}x^{n-2k}
$$
Now, when does the constant term appear? Precisely when $n=2k$, as one sees above. Also, if $n$ is odd, then you see that the constant term is z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ without the quadratic formula and without calculus I'm asked to prove that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ in an algebra textbook.
The formula for the real roots of a second degree polynomial is not introduced yet. And the book is written wit... | To pursue your ideas:
"And here maybe prove that there is no $x$ such that $(x+1)^2=x$ ???"
If $x > 0$ then $x + 1 > x$ and $x + 1 > 1$ so $(x+1)^2 = (x+1)(x+1) > x *1 = x$.
If $x = 0$ then $(x+1)^2 = 1 \ne 0 = x$
If $x < 0$ then $(x+1)^2 \ge 0 > x$.
"With this method I have trouble proving the case x<0"
If $x > 0$ the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 14,
"answer_id": 4
} |
$|x^2-5x+2|\leq 4$
Solve the following inequality.
$$|x^2-5x+2|\leq 4.$$
I know how to interpret $|x-a|$ as the distance between $x$ and $a$ along the $x$-axis, but how does one interpret an absolute value of $|ax^2+bx+c|$?
For the inequality $|x^2-5x+2|\leq 4$ I factored the LHS and got $$\left|\left(x+\frac{\sqrt... | Hints:
*
*$$|x^2-5x+2| \le 4 \iff -4 \le x^2-5x+2 \le4 \iff \frac{1}{4} \le x^2-5x+\frac{25}{4} \le \frac{33}{4}.$$
*$$x^2-5x +\frac{25}{4}=\left(x-\frac{5}{2}\right)^2.$$
*$$a \le y^2 \le b \iff \sqrt{a} \le y \le \sqrt{b} \quad \text{or} \quad -\sqrt{b} \le y \le -\sqrt{a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding the minimum value of $a^2+b^2+c^2$
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.
I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, wh... | Notice that by Inequality of arithmetic and geometric means we know that:
$$
2ab \leq a^2+b^2;
\\
2ac \leq a^2+c^2;
\\
2bc \leq b^2+c^2;
$$
so we can conclude that:
$$
2\left(ab+ac+bc\right)
\leq
2\left(a^2+b^2+c^2\right)
\Longrightarrow
\\
\ \
\left(ab+ac+bc\right)
\leq
\ \
\left(a^2+b^2+c^2\right)
\Lon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Find all integers $n>1$ such that$\frac{2^n+1}{n+1}$ is an integer. Find all integers $n>1$ such that $\frac{2^n+1}{n+1}$ is an integer.
Second attempt : I'd like to know if it's correct or not. Thank you.
Since $\frac{2^n+1}{n+1}$ for all integers $n>1$ is equivalent to $\frac{2^{n-1} \;+1}{n}$ for all integers $n>... | HINT: The following procedure is a sort of descent proving the impossibility of solution.
It is obvious that the denominator cannot be even so $n$ should be even and the quotient (supposing it is integer) should be odd.
$$2^{2n}+1=(2n+1)(2k+1)\iff2^{2n}=4kn+2(k+n)\iff2^{2n-1}=2kn+k+n$$ It follows that $k$ and $n$ have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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Question about parametric divisibility relations Let $a,b,c,d$ be pairwise relatively prime positive integers such that the divisibility relations
$$
(kc-d) \mid \bigl((2k^2+1)b^2-2kab-(k^2-1)a^2\bigr)
$$
and
$$
(c-kd) \mid \bigl((k^2+2)b^2-2kab+(k^2-1)a^2\bigr)
$$
hold for every nonzero integer $k$.
QUESTION: Can... | \begin{cases}
\begin{align}
(2k^2+1)b^2-2kab-(k^2-1)a^2 &= (kc-d)q, \\
(k^2+2)b^2-2kab+(k^2-1)a^2 &= (c-kd)r.
\end{align}
\end{cases}
\begin{align}
a &= ((k+1)s+(k-1)p)((k-1)p^2+2(k^3-k^2-1)ps+(k+1)(k^2+2)s^2), \\[0.5em]
b &= ((k+1)s+(k-1)p)(-(k-1)^2p^2+2k(k-1)ps+k^2(k^2-1)s^2, \\[0.5em]
c &= (2k^4-5k^3+7k^2-5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Induction: Bounds on sum of inverses of first $n$ square roots I am trying to show by induction that $2(\sqrt{n}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$.
For the upper bound, I have that if $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$, then $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{... | $$\begin{array}{cl}
& 1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}} \\
>& \dfrac2{1+\sqrt2}+\dfrac{2}{\sqrt{2}+\sqrt3}+\cdots+\dfrac{2}{\sqrt{n}+\sqrt{n+1}} \\
=& \dfrac{2(\sqrt2-1)}{(\sqrt2+1)(\sqrt2-1)}+\dfrac{2(\sqrt3-\sqrt2)}{(\sqrt{3}+\sqrt2)(\sqrt3-\sqrt2)}+\cdots+\dfrac{2(\sqrt{n+1}-\sqrt n)}{(\sqrt{n+1}+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Finding $\int\frac{x}{\sqrt{3-2x-x^2}} dx$. I was looking for the integral of
$$\int\frac{x}{\sqrt{3-2x-x^2}} dx$$
My work:
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(3)+(-2x-x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(2x+x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(1+2x+x^2) +1}} dx $$
$$\int \frac{x}{... | Hint: You have done till here
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(4)-(x+1)^2}} dx$$
now let $x+1=2\sin t$ and simplify!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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How to calculate the determinant of this n by n matrix? Find the determinant of this n by n matrix.
$$
\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
where, $$ k=n-1 $$.
I am new... | You should just factorize:\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix} = \begin{equation}\begin{pmatrix}
-1 & x_1& \cdots &x_k\\
0 &1 &\cdots &0\\
\vdots &&1\\
0&&&1
\end{pmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Show $\sqrt[4]{2}$ on horizontal axis It is very basic question ,to show $\sqrt[2]{2}$ on $x$ axis .we do like below
but,how can we show $\sqrt[4]{2}$ on $x$ axis ?(like we do for $\sqrt2)$
I am thankful if you guide me .
$\bf{Remark}:$ may be a question for many people .
| The following steps will work . . .
\begin{align*}
1.\;\,&\text{Construct a segment of length$\,\sqrt{2}$.}\\[4pt]
2.\;\,&\text{Construct a segment $AB$ of length$\,\sqrt{2} + {\small{\frac{1}{4}}}$.}\\[4pt]
3.\;\,&\text{Construct a circle with $AB$ as a diameter.}\\[4pt]
4.\;\,&\text{Construct a point $P$ on $AB$ such... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Partial Differentiation Ex 2.2 Eng Mathematics by H.K. Das Test for continuity when
$f(x,y) = \frac{x^3\,y^3}{x^3+y^3}$ when $x\neq0, y\neq0$ and f(x,y)=0 when $x=0,y=0$.
| for $$x>0,y>0$$ we have $$x^3+y^3\geq 2(xy)^{3/2}$$ thus we obtain
$$\frac{x^3y^3}{x^3+y^3}\le \frac{x^3y^3}{2(xy)^{3/2}}=\frac{1}{2}(xy)^{3/2}->0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
let : $\forall n \in \mathbb{N}$ : $x_n=\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}$ then : $\displaystyle{\sum_{i=1}^{20}}\dfrac{1}{x_i}=?$ let : $\forall n \in \mathbb{N}$ : $x_n=\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}$
then :
$$\sum_{i=1}^{20}\dfrac{1}{x_i}=?$$
my try :
$$\dfrac{1}{\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{13}}+.... | I notice that:
$$\begin{align}\frac{1}{\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}} &= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{(1+(1+1/n)^2)-(1+(1-1/n)^2)}\\
&= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{(1+1/n)^2-(1-1/n)^2}\\
&= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{4/n}\\
&= \frac14\left(n\sqrt{1+(1+1/n)^2}-n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to integrate $\int \frac {e^y}{y} dy$? The question is to evaluate $$\iint_R \frac {x}{y} e^y dx dy$$ where R is the region bounded by $0 \leq x \leq 1$ and $x^2 \leq y \leq x$.
So i write it as $$\int_0^1 \int_{x^2}^{x} \frac{x}{y} e^y dy dx$$.
The thing is, how do i evaluate $I=\int_{x^2}^{x} \frac{1}{y} e^y dy$?... | There is a theorem that says when you are given double integrals, you must flip them. Well, not always, but in this case you want to do that.
Reparametrize your region $R$ as $R = \{(x,y) | \sqrt{y} \le x \le y, 0 \le x \le 1\}.$
Then by Fubini, we can write
$$\int_0^1 \int_{x^2}^x \frac{x}{y}e^y dydx = \int_0^1 \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
A floor function equation I want to solve below equation analytically $$\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor=1$$ but I have no idea to start .
Implicit: I solve it by graphing and the solution was $x\in[\frac {1}{4},1)$
I am thankful for any idea in advance .
I try to use $x=n+p... | Use the fact that for each $x$ we have:
$$x-1<[x]\leq x$$
So $$ \frac{2x+1}{3}+ \frac{\lfloor4x\rfloor+2}{3}-2<\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor \leq \frac{2x+1}{3}+ \frac{\lfloor4x\rfloor+2}{3}$$
So $$ \frac{2x+1}{3}+ \frac{4x+1}{3}-2<1 \leq \frac{2x+1}{3}+ \frac{4x+2}{3}$$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
For which $k \in \mathbb{R}$ the vector $(1,k^2-k,k) \in Im(f)$ where $f$ is a linear map? Let $f:\mathbb{R^3}\to \mathbb{R^3} $ the map defined by $A_h$
$$
A_h= \begin{pmatrix}
2 & 1 & -1 \\
1 & 2 & 1 \\
-1 & 1 & h \\
\end{pmatrix}
$$
where $h \in \mathbb{R}$. $f(x,y,z)=(x-y-2z,... | $$a(1,2,1) + b(-1,1,2) = (1,k^2-k,k)$$
Separating components:
$$\begin{cases}
a - b &=& 1 &(1) \\
2a + b &=& k^2-k &(2) \\
a+2b &=& k &(3)
\end{cases}$$
$(1)+(3)$ gives:
$$2a+b=k+1$$
Comparing it with $(2)$ gives:
$$k^2-k=k+1$$
Solving the quadratic equation gives:
$$k=\dfrac{2\pm\sqrt{8}}{2}=1\pm\sqrt2$$
We're basical... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to solve $\cos({r^{-4}}\cos 4\theta - {4\theta}) = 0$ for $\theta$ and $r$? Problem Statement
How does one solve the trigonometric equation
$$\cos({r^{-4}}\cos 4\theta - {4\theta}) = 0$$ for $\theta$ and $r$?
My Try
Taking the inverse cosines of both sides, I get
$${r^{-4}}\cos 4\theta - {4\theta} = \cos^{-1}(0... | Notice that, from the equation
$$\cos 4\theta = {r^4}\left(4\theta + \frac{\pi}{2} + k\pi\right),$$
and the equation
$$r = (x^2 + y^2)^{1/2}$$
I get that $r \geq 0$, so that
$$\cos 4\theta \geq 0.$$
This last inequality implies that
$$-\frac{\pi}{2} \leq 4\theta \leq \frac{\pi}{2}$$
which implies that
$$-\frac{\pi}{8} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \forall n \ge 4$, $n^{3} + n < 3^{n}$
Prove that $\forall n \ge 4$, $n^{3} + n < 3^{n}$
My attempt: Base case is trivial.
Suppose $ \ n \ge 4$, $n^{3} + n < 3^{n}$. Then,
$$ (n+1)^{3} + (n+1) = n^3 + n + 3n^2 + 3n + 1 +1 < 3^{n} + 3n^2 + 3n + 2 \\< 3\cdot 3^{n} + 3n^2 + 3n +2.$$
Not sure how to get rid of $3... | Hint:
$$ 3n^2+3n+2<n^3+n<3^n. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Expectation of $Y := \min\{\frac{a}2, X\}$ if $X$ is uniform on $(0,a)$ The density of X is given by
$
f(x)=
\begin{cases}
\frac{1}{a}&\, 0 \leq x < a\\
0&\, otherwise\\
\end{cases}
$
Y := min{$\frac{a}{2}$, X} so Y belongs to [0, $\frac{a}{2}$]
Based on this solution, $E(Y) = \frac{3a}{8}$. Why is $E(Y) = \frac{3a}... | A different approach would be:
$$EY=E\left[Y|X<\frac{1}{2}a\right]+E\left[Y|X\ge\frac{1}{2}a\right]= \int_0^{\frac{1}{2}a}x\frac{1}{a}dx= 0.125a+ 0.5a=\frac{3}{8}a$$
It relays on the properties of conditional expectations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $y' = y^{2} + 1$ if $y(0) = 1$ The solution to the initial-value problem $y' = y^{2} + 1$ with $y(0)=1$ is $y = \tan(x + \frac{\pi}{4})$. I would like to show that this is the correct solution in a way that is analogous to my solution to differential equation $y' = y - 12$.
Solution to Differential Equation
If ... | The equation can be written as
$$\frac {y'}{y^2+1}=1 $$
and after integration
$$\arctan (y)=x+C $$
for $x=0$, it gives
$$\arctan (1)=0+C=\frac {\pi}{4} $$
thus
$$y=\tan (x+\frac {\pi}{4}) .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$\arctan{x}=\arccos{2x}.$ First, let's determine some domains. By definition, it follows that $\arccos:[-1,1]\rightarrow[0,\pi]$ and $\arctan:\mathbb{R}\rightarrow\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Since the domain of $\arctan$ is the entire reals, the restricting factor here should be the domain of arccos. Th... | Since $x \in \left[\frac{1}{2}, \frac{1}{2}\right]$ we have $\arctan x \in \left[-\arctan\frac{1}{2}, \arctan\frac{1}{2}\right] \subseteq \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $\cos(\arctan x) \ge 0$.
So $\cos(\arctan x) = 2x$ implies $x \geq 0$, which discards the negative solution $x_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculating the limit $\lim\limits_{x \to 0^+} \frac{\sqrt{\sin x}-\sin\sqrt{ x}}{x\sqrt{x}}$
Calculate $$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}$$
without use Taylor serie and L'Hôpital.
$$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}\cdot\dfrac{\sqrt{\sin x... | $$\mathrm L =\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}= \lim\limits_{x \to 0^+} \dfrac{(\sqrt{\sin x}- \sqrt{x}) -(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}} \\= \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^\prime} - \overbrace{\lim\limits_{x \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Computing $\int_0^1 \sin(3\pi x)\sin(2\pi x) \mathrm{d}x$
How do I show
$$\int_0^1 \sin(3\pi x)\sin(2\pi x) \mathrm{d}x=0.$$
I tried using
$$\sin (x) \sin (y) = \cos ( x \pm y ) \mp \cos (x) \cos (y) $$
Then the integral becomes
$$\int_0^1\cos(3\pi x)\cos(2\pi x) \mathrm{d}x$$
Which doesn't help me. I also trie... | Using integration by parts twice,
$\begin{align}J&=\int_0^1 \sin\left(3\pi x\right)\sin\left(2\pi x\right)\,dx\\
&=\left[-\frac{1}{2\pi}\sin\left(3\pi x\right)\cos\left(2\pi x\right)\right]_0^1+\frac{1}{2\pi}\int_0^1 \sin\left(3\pi x\right)\cos\left(2\pi x\right)\, dx\\
&=\frac{1}{2\pi}\int_0^1 \sin\left(3\pi x\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Summation involving double factorials How can we evaluate exact infinite sum of the following?
$${ \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!}}$$
where
$k!! = \begin{cases} 2\cdot4\cdot6 . . . k, & \text{if $k$ is even} \\ 1\cdot3\cdot5 . . . k, & \text{if $k$ is odd} \end{cases}$
What approach do we need to solve such... | Starting with
My attempt:
\begin{align}
S = \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!}
&= \sum_{r=1}^{\infty}\frac{(2r)!!\cdot (2r+2)!!}
{(2r+3)!!(2r+2)!!} \\
&= \sum_{r=1}^{\infty}\frac{2^r\cdot (r)!\cdot 2^{r+1}\cdot (r+1)!}{(2r+3)!} \\
&= \sum_{r=1}^{\infty}\frac{2^{2r+1}\cdot (r)!\cdot (r+1)!}{(2r+3)!}
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the range of function We have the function \begin{equation*}f(x)=-\frac{(x-3)^2}{x+1}\end{equation*}
I want to determine the domain and the range of the function.
The root of the deniminator is $x=-1$. Therefore, the domain is $D_f=\mathbb{R}\setminus\{-1\}=(-\infty, -1)\cup (-1,+\infty )$.
Does it holds that \... | $$-\frac{(x-3)^2}{x+1}=\frac{-x^2+6x-9}{x+1}=\frac{-x^2+6x+7-16}{x+1}=7-x-\frac{16}{x+1}.$$
Thus, for $x<-1$ by AM-GM we obtain
$$f(x)=8-x-1-\frac{16}{x+1}\geq8+2\sqrt{(-x-1)\left(-\frac{16}{x+1}\right)}=16$$
and for $x>-1$ by AM-GM again we obtain:
$$f(x)=8-x-1-\frac{16}{x+1}\leq8-2\sqrt{(x+1)\cdot\frac{16}{x+1}}=0.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Finding all values such that column vector is a linear combination Question:
For which values(s) of $ \ a$ is the column $ \ c = \begin{bmatrix} a\\a^2\\0\\a+1\end{bmatrix}$ a linear combination of the columns of,
$ \ x = \begin{bmatrix} 1\\1\\1\\1\end{bmatrix}$, $\ y = \begin{bmatrix} 1\\0\\-1\\0\end{bmatrix}$, $ \ z=... | $\require{cancel}$
Using your four obtained equations, you know that $r=s$ and so $2r=a.$ You then have $t=a^2-\frac{a}{2}.$ Inserting all that into the last equation, you get
$$\cancel{\frac{a}{2}}+a^2-\cancel{\frac{a}{2}}=a+1,$$
the solutions of which are $\frac{1\pm\sqrt{5}}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determine the derivative of $\arctan$ function Find $f'(x)$ for the function:
$f(x)= \arctan(\frac{a+x}{1-ax}))$ , $a\in R$
So this is what I've done:
$f(x) = \arctan x$
$f'(x) = \frac{1}{1+x^2}$
$x= \frac{a+x}{1-ax}$
$f'(x) = \frac{1}{1+\frac{(a+x)^2}{(1-ax)^2}}$
$f'(x) = \frac{(1-ax)^2}{(1-ax)^2+(a+x)^2}$
Is this co... | $$\arctan\left(\frac{x+a}{1-ax}\right)=\arctan x+\arctan a$$
(give or take a multiple of $\pi$), so
$$\frac d{dx}\arctan\left(\frac{x+a}{1-ax}\right)=\frac1{1+x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ Proposition
For any positive numbers $a$, $b$, and $c$,
\begin{equation*}
\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}
\geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c} .
\end{equation*}
I am requesting an elementary, alg... | By C-S and Holder we obtain:
$$\sum_{cyc}\frac{a^3}{b^2}=\sum_{cyc}\frac{a^5}{a^2b^2}\geq\frac{\left(a^{\frac{5}{2}}+a^{\frac{5}{2}}+a^{\frac{5}{2}}\right)^2}{\sum\limits_{cyc}a^2b^2}=$$
$$=\frac{\left(a^{\frac{5}{2}}+a^{\frac{5}{2}}+a^{\frac{5}{2}}\right)^2(a+b+c)}{(a+b+c)\sum\limits_{cyc}a^2b^2}\geq\frac{(a^2+b^2+c^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Is the limit $\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$ zero? I have this limit:$$\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$$
At first sight seems that limit equals 0. But WolframAlpha says that there is no limit. I tried to prove it. I considered cases $y = kx$, and so on. I never got to find ... | $$
\begin{align}
(x-y)^2+y^2
&=x^2-2xy+2y^2\\
&=(2-\phi)x^2+(\phi-1)x^2-2xy+\phi y^2+(2-\phi)y^2\\
&=(2-\phi)\left(x^2+y^2\right)+\left(\sqrt{\phi-1}\,x+\sqrt{\phi}\,y\right)^2\\
&\ge(2-\phi)\left(x^2+y^2\right)\\
&=\frac1{\phi^2}\left(x^2+y^2\right)
\end{align}
$$
Therefore,
$$
\begin{align}
\left|\frac{x+2y}{(x-y)^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Find $\operatorname{ord}_{17} (a)$ for all $a = 1, \ldots, 16$ The $\operatorname{ord}_{m} (a)$ is defined as the least positive integer $x$ satisfying $a^x \equiv 1 \mod m$.
To find $\operatorname{ord}_{17} (a)$ for all $a = 1, ..., 16,$ I know I can go through each numbers from 1 to 16 one by one (and have gotten ... | So here are some tricks which might help to reduce the work.
If $a\neq \pm 1$ and the order of $a$ is even, then $-a(\equiv 17-a)$ will have the same order as $a$.
The numbers for which $a^r\equiv 1 \bmod m$ form a group. Therefore $a$ and $a^{-1}$ have the same order. Here you have $18\equiv 1 \bmod 17$ and $18=2\time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Equation of a straight line passing through passing through a point and equally inclined to two other lines
Find the equation of the straight line passing through the point $(4,5)$ and equally inclined to the lines $3x= 4y+7$ and $5y=12x+6$.
I know that the equation of the bisector is given by:
$\dfrac{a_1x+b_1y+c_1}... | You have TWO lines which form equal angles with the given straight lines
$3x-4y-7=0;\;12x-5y+6=0$
$\dfrac{3x-4y-7}{\sqrt{3^2+4^2}}=\pm\dfrac{12x-5y+6}{\sqrt{12^2+5^2}}$
$\dfrac{3x-4y-7}{5}=\pm\dfrac{12x-5y+6}{13}$
$13(3x-4y-7)=\pm 5(12x-5y+6)$
$99 x-77 y-61=0;\;21 x+27 y+121=0$
$y=\dfrac{9 x}{7}-\dfrac{61}{77};\;y=-\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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$f(x)=?$ if we have $f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$ $f(x)=?$
If we have $$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$$ to fractions are very similar. I don't have an idea to find $f(x)$. Can someone show me a clue ?
| Let $y = \frac x{x^2+x+1}$ and $z = \frac x{x^2-x+1}$. Then we have
\begin{align}
y(x^2+x+1) = x &\implies y(x^2-x+1)=x -2xy\\
&\implies y=z(1-2y)\\
&\implies z=\frac y{1-2y}\\
\end{align}
which gives us $f(y) = \frac{y}{1-2y}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $\alpha,a,b$ are integers and $b\neq-1$, then prove that, if $\alpha$ satisfies the equation $x^2+ax+b+1=0$, $a^2+b^2$ must be composite.
Let $\alpha,a,b$ be integers such that $b\neq-1$. Assume that $\alpha$ satisfies the equation $x^2+ax+b+1=0$. Prove that the integer $a^2+b^2$ must be composite.
$\alpha=\frac{-... | Let the roots be $r,s$.
By hypothesis, at least one of $r,s$ is an integer.
By Vieta's formulas
\begin{align*}
r + s &= -a\\[4pt]
rs &= b + 1\\[4pt]
\end{align*}
From the first of the above equations, since one of $r,s$ is an integer, and $a$ is an integer, $r,s$ must both be integers.
From the second of the above e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Beta Gamma functions How to convert this integral into Beta Gamma function?
Any hints will be appreciated.
$$\int_0^{b} x \sqrt{b^3-x^3} dx$$ Thanks
| Set $x =bt$ so that, $dx =bdt$
$$I:=\int_0^{b} x \sqrt{b^3-x^3} dx = \int_0^{1} b^2t \sqrt{b^3-b^3t^3} dt = b^{7/2}\int_0^{1} t (1-t^3)^{1/2} dt$$
Setting $u=t^3~~$ i.e $~~t= u^{1/3}~~$ yields $dt =\frac{1}{3}u^{-2/3}du~~~$ therefore,
$$I= b^{7/2}\int_0^{1} t (1-t^3)^{1/2} dt = \frac{1}{3}b^{7/2}\int_0^{1} u^{-1/3} (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find a formula for the sum up to $n$ terms of the sum $1+11+111+11111...$ using $1+2x+3x^2+4x^3...$ I'm trying to find the formula for the sum up to $n$ terms of the sum $1+11+111+11111...$ using $1+2x+3x^2+4x^3...=\frac{1-x^n}{(1-x)^2}-\frac{nx^n}{1-x}$. It's easy to find the formula if we expressed the sum as ... | $1 + 11 + 111 + .... = n + (n-1)10 + (n-2)10^2 + ....+2*10^{n-2} + 10^{n-1}$
$= 10^{n-1}(n*10^{-n} + (n-1)*10^{-n+1} + (n-2)10^{-n+2} + .....+2*10 +1)$
$= 10^{n-1}(1 + 2*\frac 1{10} + ..... + (n-2)*\frac 1{10} + (n-1))$
$= 10^{n-1}[\frac{1-\frac{1}{10}^n}{(1-\frac{1}{10})^2}-\frac{n\frac{1}{10}^n}{1-\frac{1}{10}}]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Solving $x_{n} - 3x_{n-1} = -8$ with $n\geq 1$ and $x_0 = 2$ I tried two methods which gave different answers:
Method 1:
$$x_{n} - 3x_{n-1} = -8 \\ x_n = 3(3x_{n-2} - 8) - 8 \\ = 3^2 x_{n-2} -8 ( 1+3) \\ = 3^3 x_{n-3} - 8(1+3+3^2) \\ = 3^n x_{0} - 8(1+3+3^2 + \ldots + 3^{n-1}) \\ = 2\times 3^n - 8\left(\frac{3^n - 1}... | Here's another method. Convert the recurrence to a generalized Fibonacci form as follows:
$$
x_n=Ax_{n-1}+B,\quad x_{_0} ~\text{given}\\
x_{n-1}=Ax_{n-2}+B\\
x_n-x_{n-1}=Ax_{n-1}-Ax_{n-2}\\
x_n=(A+1)x_{n-1}-Ax_{n-2},\quad x_{_1}=Ax_{_0}+B
$$
The characteristic roots for this equation are always
$$\alpha,\beta=A,1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\iint_D x dxdy$ using polar coordinates Using polar coordinates, I want to calculate $\iint_D x dxdy$, where $D$ is the disk with center $(2,3)$ and radius $2$.
$$$$
I have done the following:
We have $D=\{(x,y)\mid (x-2)^2+(y-3)^2\leq 4\}$.
We use $(x,y)=(r\cos \theta, r\sin \theta)$.
From the inequali... | Hint: Use substitution
\begin{cases}
x=2+r\cos\theta,\\
y=3+r\sin\theta,\\
\end{cases}
then
$$\iint_D xdxdy=\int_0^{2\pi}\int_0^2r(2+r\cos\theta)\,dr\,d\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Transform term for induction proof Could you help me to mathematically show that these two terms are the same (they are). This is the last (but probably the most important :( ) step of an induction proof.
$$First: \frac{(n+1)(n+2)(2(n+1)+7)}{6}$$
$$Second: \frac{n(n+1)(2n+7)+6(n+1)(n+3)}{6}$$
Thank you! :)
| A (hopefully) more elegant variant:
Once you've factored out $\dfrac{n+1}6$, there remains
*
*$n(2n+7)+6(n+3)=n\bigl(2(n+3)+1\bigr)+6(n+3)=\color{red}{2(n+3)^2+n}$.
*$(n+2)\bigl(2(n+1)+7\bigr)=\bigl((n+3)-1\bigr)\bigl(2(n+3)+3\bigr)=2(n+3)^2+(3-2)(n+3)-3=\color{red}{2(n+3)^2+n}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2467922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find $\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$?
Find
$$\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$$
My attempt: ON THE basis of This post
$$\lim_{x\to1}\tan\frac{\pi x}{4} =1,\quad \lim_{x\to1}\tan\frac{\pi x}{2}=\infty$$
$$\implies\lim_{x\to1}\left(\tan\fr... | $$\lim_{x\rightarrow1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}} = \lim_{x \to 1}e^ {{\tan\frac{\pi x}{2}}\ln(\tan\frac{\pi x}{4})}$$
Now we will check
\begin{align*}
\lim_{x \to 1} {{\tan\frac{\pi x}{2}}\ln\left(\tan\frac{\pi x}{4}\right)}
&\stackrel{\text{(L'Hôpital's rule)}}{=}
\lim_{x \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $$
I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in.
Is there a... | The theory for this type of questions are Newton's Identities.
For a cubic polynomial
$$\begin{align}ax^3+bx^2+cx+d\end{align}\\$$
Vieta's formulas for the roots $\alpha_1, \alpha_2, \alpha_3$ are:
$$\begin{align}a(\alpha_1 + \alpha_2 + \alpha_3) + b=0\tag{V}\\
a(\alpha_1\alpha_2 + \alpha_2\alpha_3 + \alpha_3\alpha_1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$\lim_{n\to \infty} F(k)=\frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{3})}$ Find F(5) and F(6) Find the value of F(5) & F(6).It is given that
$$F(k)= \lim_{n\to \infty} \frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{... | For $f$ increasing function we have that $ \int \limits_{a-1}^{b} f(s)ds \leq \sum \limits_{s=a}^{b} f(s) < \int \limits_{a}^{b+1} f(s) ds$
So $ \lim \limits_{n \to \infty} \frac{\int \limits_{0}^{n} s^k ds}{\sum \limits_{s=1}^{n} s^2 \sum \limits_{s=1}^{n} s^3}< F(k) <\lim \limits_{n \to \infty} \frac{\int \limits_{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Stokes' Theorem verified problem
For the given vector field
$$\vec{H(r)} = rcos( \phi - \frac{\pi}{4} ) \vec{ a_{r} } +sin \phi \vec{ a_{ \phi }} $$
a) Calculate line integral of $\vec{H(r)} $ over the close path $ \Gamma $ with corners at ABCD on xy-plane shown in Figure.
b) Confirm the result of the line integral... | Solution:
$ \int_{r=2}^1 r.cos(\phi-\frac{ \pi }{4}) .dr |_{\phi = \frac{ \pi }{4}} + \int_{\phi=\frac{ \pi }{4}}^\frac{ -5\pi }{4} sin\phi.d\phi.r |_{r=1} + \int_{\phi=\frac{ 3\pi }{4}}^\frac{ 9\pi }{4} sin\phi.d\phi.r |_{r=2} $
$=-\frac{ 3 }{2} - \sqrt{2}$ Line integral boundaries:
B $\Rightarrow $ C $\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find $n\in\mathbb N$ so that: $\sqrt{1+5^n+6^n+11^n}\in\mathbb N$ Find $n\in\mathbb N$ so that: $\sqrt{1+5^n+6^n+11^n}\in\mathbb N$
My attempt led me to have $\quad n=2k+1:\quad k\in\mathbb N$
The expression under square root is odd, so the square root's value is also odd.
I assumed $\sqrt{1+5^n+6^n+11^n}=2a+1:\quad a\... | You could do also like this (for kids in elementary school).
$5^a$ always ends with $5$
$6^b$ always ends with $6$
$11^c$ always ends with $1$
So $1+5^a+6^b+11^c$ always end with 3 and there for it can not be perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Prove convergence of the series Help please to prove the convergence: $$\sum_{n=1}^{\infty}\left(\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n\right)\cos \frac{\pi n(n-1)}2$$
It can be proved with Dirichlet's test, but there are come problems with $$\left(\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n\right)$$ monotone. Other steps... | Let $\frac{1}{n}=x$.
Hence,
$$\lim_{n\rightarrow\infty}\left(\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n\right)=\lim_{x\rightarrow0}\left(\frac{1}{\sin{x}}-\frac{\cos{x}}{x}\right)=$$
$$\lim_{x\rightarrow0}\frac{x-\frac{1}{2}\sin2x}{x\sin{x}}=\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin{x}+x\cos{x}}=$$
$$=\lim_{x\rightarrow0}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Calculate a triple integral I want to draw the space $D=\{x,y,z)\mid z\geq 0, x^2+y^2\leq 1, x^2+y^2+z^2\leq 4\}$ and calculate the integral $\iiint_D x^2\,dx\,dy\,dz$.
How can we draw that space?
About the integral I have done the following:
We have that $x^2+y^2+z^2\leq 4\Rightarrow z^2\leq 4-x^2-y^2 \Rightarrow -... | $D$ is a cylinder inside a hemisphere. Or, a cylinder with a spherical cap.
Your set-up for the integral is correct.
You could proceed with a trig substitution $y = \sqrt {4-x^2} \sin \theta$
$\int \int x^2 (4-x^2) \cos^2 \theta \ d\theta\ dx$
Or you could convert to polar coordinates (which is also a trig substituti... | {
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"url": "https://math.stackexchange.com/questions/2475940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Given $x\odot y=x+y+xy$, prove $x\odot x\odot\ldots\odot x= (1+x)^n -1$
Given $x\odot y=x+y+xy$, prove that
$$\underbrace{x\odot x\odot\ldots\odot x}_{n\text{ times}}= (1+x)^n -1$$ for all $n\in \Bbb{N}$ and $x \in\Bbb R\setminus\{-1\}$.
I have tried to use the binomial theorem on $(1+x)^n$ but was unable to simpl... | By induction:
$$x_1 = x$$
$$x_n = x \odot x_{n-1} = x+x_{n-1} + xx_{n-1} \Rightarrow \\
x_n = (1+x)x_{n-1} + x.$$
Then:
$$x_n = (1+x)\left[(1+x)x_{n-2} + x\right] + x = \\
= (1+x)^2x_{n-2} + (1+x)x + x = \\
= (1+x)^3x_{n-3} + (1+x)^2x + (1+x)x + x \Rightarrow\\
x_n = (1+x)^{h}x_{n-h}+x\sum_{k=0}^{h-1} (1+x)^k.$$
For $h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the general solution to non homogeneous linear equation The question:
State the general solutuon of:
$$5x_1-2x_2+4x_3=5$$
My attempt: I tried following closely an example provided for homogeneous equations without any luck:
$$
x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=
\begin{bmatrix}2x_2-4x_3+5\\x_2\\x_3\end{bm... | Consider the pattern:
For
$$5x_1-2x_2+4x_3=5$$
then $x_{3} = \frac{1}{4} \, (5 - 5 \, x_{1} + 2 \, x_{2})$
and
\begin{align}
x &=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=
\begin{bmatrix} x_{1} \\ x_2 \\ \frac{1}{4} \, (5 - 5 \, x_{1} + 2 \, x_{2}) \end{bmatrix}
= \frac{1}{4} \, \begin{bmatrix} x_{1} \\ x_2 \\ 5 - 5 \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a,b,c,a^2+b^2+c^2$ are primes, then $a$ or $b$ or $c$ is equal to $3$ Given primes $a,b,c$ such that $a^2+b^2+c^2$ is prime, then $3\in\{a,b,c\}$.
Tested for $a,b,c<500$.
| If $p\ge 5 $ is a prime then it is always of the form $6n\pm 1 ~, n \ge 1$.
Let $a,b,c$ be primes $\ge 5$. Then we've ;
$$(6n_1\pm 1)^2+(6n_2 \pm 1)^2+(6n_2 \pm 1)^2 = \text{prime}$$
$$36(n_1^2+n_2^2+n_3^2)+12(\pm n_1\pm n_2 \pm n_3)+3 =\text{prime}$$
But in the above equation, LHS is clearly divisible by $3$ and henc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
$4y^2+y=3x^2+x$ implies that $x-y$ is a perfect square. I need a help to prove the following statement. (Sorry for my bad english).
If $x,y\in\mathbb{Z}$ are solutions of $4y^2+y=3x^2+x$, then $x-y$ is a perfect square.
I've tried to rewrite the equation as $\cfrac{y^2}{x-y}=3(x+y)+1 $ and conclude that $x-y\mid y^2$... | The equation
$$
4Y^2 + Y = 3X^2 + X
$$.
Let me proof why $X - Y = c$.
where $c$ is a perfect square, note: the values $X$ and $Y$ here are the points $( X, Y )$ on the graph of the equation.
$4Y^2+Y = 3X^2+X$.
make $Y$ the subject of the formula there.
$Y = \frac{-1 \pm \sqrt{48X^2+16X+1}}{8}$.
Now say $X-Y = c$.
So th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
How to solve a xor equation? I have the following equations I need to solve for A and B:
2A ⊕ B = 7
A ⊕ B = 14
Note:
All variables are unsigned 32-bit integers
⊕ is the XOR operator for 32-bit integers.
| XOR is the same as addition in the finite field $F_2$. Thus if we assume a solution exists with $A<8$ and $B<16$ (reasonable given the magnitudes of the right-hand sides) you can rewrite your equations as
\begin{align*}
\left[\begin{array}{cccc}0 & 1 & 0 & 0\\0 & 0& 1 & 0\\0 & 0& 0& 1\\0 & 0& 0&0\end{array}\right]\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $k$ such that $f(k)$ is Minimum Find $k$ such that
$$f(k)=\int_{0}^{4} |4x-x^2-k|dx$$ is Minimum
I splitted the Modulus in to two cases:
$1.$ if $4x-x^2-k \ge 0$ Then
$$f(k)=\int_{0}^{4} (4x-x^2-k)dx=\frac{32}{3}-4k$$
$2.$ if $4x-x^2-k \lt 0$ Then
$$f(k)=4k-\frac{32}{3}$$
But How to minimize $f(k)$ which i... | We need to separate it into three cases :
*
*Case 1 : If $k\ge 4$, then we have $4x-x^2-k=-(x-2)^2+4-k\le 0$, so $$f(k)=\int_0^4 (x^2-4x+k)dx=4k-\frac{32}{3}$$So, $f(k)\ge f(4)=\frac{16}{3}$ for $k\ge 4$.
*Case 2 : If $0\lt k\lt 4$, then $0\lt 2-\sqrt{4-k}\lt 2+\sqrt{4-k}\lt 4$ where $x=2\pm\sqrt{4-k}$ are roots o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Non unique factorization in $\mathbb{Z}_5$
Prove $3X^3+4X^2+3=(X+2)^2(3X+2)=(X+2)(X+4)(3X+1)$ in $\mathbb{Z}_5$.
I see that $2$ is a root (since $-24+16+3=-5$) , then $3X^3+4X^2+3=(X+2)(3X^2)$, and also $-4$ (since -$64\cdot3+16\cdot4+3=-125$) and although $x=-1/3$ is not in $\mathbb{Z}_5$, because it also works as a... | No contradiction.
In a UFD (such as $\mathbb{Z}_5[x]$), the factorization is unique, only up to unit factors.
Note that $2$ and $3$ are units in $\mathbb{Z}_5$, hence are also units in $\mathbb{Z}_5[x]$.
Also, we have $6=1$.
Then since
$$x+4 = 6x+4 = 2(3x+2)$$
and
$$3x+1 = 3x + 6 = 3(x+2)$$
the factors match up, on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Factorising $(4 + 3i)z^2 + 26iz + (-4+3i)$? Quadratic formula attempt included. I'm trying to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$.
I tried to use the quadratic formula to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$, where $b = 26i$, $a = (4 + 3i)$, $c = (-4+3i)$. This gives us the roots $z = \dfrac{-4i - 12}{25}$ and... | The solutions of the equation $(4+3i)z^2+26iz+(3i-4)=0$ are $z=-3-4i$ and $z=\dfrac{-3-4i}{25}$. Try using the Factor Theorem now, it should work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove or disprove : $\sqrt{1 + \sqrt{4+\sqrt{16+\sqrt{256...}}}} = \sqrt{2+\sqrt{5}}$ Edit: My initial question was regarding the expressions:
$\sqrt{1 + \sqrt{4+\sqrt{16+\sqrt{64... + \sqrt{4^{n}}}}}}$
And in general:
$\sqrt{1 + \sqrt{k+\sqrt{k^2+\sqrt{k^3... + \sqrt{k^{n}}}}}}$
And their limits, however, my working o... | Let $\phi_n:=\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}$ ($n$ square roots); note that $\phi_n\to\phi$, the golden ratio.
Let $a_{n,r}:=\sqrt{k^{2^{r-n+1}}+\sqrt{\cdots+\sqrt{k^{2^r}}}}$.
For induction in $n$, assume that $a_{n,r}=k^{2^{r-n}}\phi_n$.
Then $$a_{n+1,r}=\sqrt{k^{2^{r-n}}+a_{n,r}}=\sqrt{k^{2^{r-n}}+k^{2^{r-n}}\phi_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"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.