Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the values of a and b so that $x^4+x^3+8x^2+ax+b$ is exactly divisible by $x^2+1$ I have been trying this question for a long time but I am not getting it. So please help me and try to make it as fast as possible
| Long division gives
$$
\frac{x^4+x^3+8x^2+ax+b}{x^2+1}=x^2+x+7+\frac{(a-1)x+b-7}{x^2+1}
$$
Therefore, the remainder is zero iff $(a-1)x+b-7$ is the zero polynomial, that is, iff
$$
.\begin{cases}
a-1=0\\
b-7=0\\
\end{cases}
$$
Hence, for $a=1$ and $b=7$, we have that $x^4+x^3+8x^2+ax+b$ is divisible by $x^2+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Determinant of a matrix and chech whether it is non negative definite or not
Let $V = \{ f : [0,1] \to \mathbb R | f$ is a polynomial of degree less than or equal to n $\}$. Let $f_j(x) = x^j$ for $0\leq j \leq n$ and let $A$ be the $(n+1) \times (n+1)$ given by $a_{ij} = \int_0^1 f_i(x)f_j(x) dx$. Then which of the... | Hint: for any real vector $v$, express $v^T A v$ as an integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to easily solve this trigonometric equation? Given equation:
$$\frac{\sin(x) + \sin(5x) - \sin(3x)}{\cos(x) + \cos(5x) - \cos(3x)} = \tan(3x),$$
what is the easiest way to solve it? I know it can be solved by expanding each $\sin(nx)$ and $\cos(nx)$ terms, but is there an easier way?
| $$\frac { 2\sin { \left( \frac { x+5x }{ 2 } \right) \cos { \left( \frac { x-5x }{ 2 } \right) -\sin { \left( 3x \right) } } } }{ 2\cos { \left( \frac { x+5x }{ 2 } \right) \cos { \left( \frac { x-5x }{ 2 } \right) } -\cos { \left( 3x \right) } } } =\frac { \sin { \left( 3x \right) \left( 2\cos { 2x-1 } \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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New series formula for $\arctan(x)$? Ln(x)? I discovered this equation, but have no idea if it has been previously discovered. Please help determine if it has been previously developed. Or please prove that the equation is not correct.
$$\sum_{n=0}^\infty \frac{x^{2n+1}}{(x^2+1)^{n+1}}\cdot\frac{(2n)!!}{(2n+1)!!}=\arc... | Evidently from marty cohen's answer, the formula for $\arctan$ was previously known, and as shown below, the result for $\log(x)$ is not too hard to derive. That you have discovered them yourself is no less impressive however.
Substituting $x\mapsto4x^2$ in $(2)$ from this answer, we get
$$
\sum_{k=0}^\infty\frac{4^kx... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
| $$\dfrac{3\sqrt3-8}{1-2\sqrt3}=\dfrac{8-3\sqrt3}{2\sqrt3-1}$$
Now $$\dfrac{3\sqrt3-4}{7-2\sqrt3}-\dfrac{8-3\sqrt3}{2\sqrt3-1}=\dfrac{(3\sqrt3-4)(2\sqrt3-1)-(7-2\sqrt3)(8-3\sqrt3)}{(7-2\sqrt3)(2\sqrt3-1)}$$
$$=\dfrac{26\sqrt3-52}{(7-2\sqrt3)(2\sqrt3-1)}$$
Now the denominator $>0,$ and $26\sqrt3-52=26(\sqrt3-2)<0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \... | $$\int_{-1}^{1}\tan^{-1}(\sqrt{x+2})dx=\int_{1}^{3}\tan^{-1}\sqrt{x}\,dx$$
if$f$ be differentiable, increasing and one-to-one on $[a,b]$ such that $f(a)=\alpha$ and $f(b)=\beta$ then
$$\int_{a}^{b}f(x)dx+\int_{\alpha}^{\beta}f^{-1}(x)dx=b\beta-a\alpha$$
therfore
$$\int_{1}^{3}\tan^{-1}\sqrt{x}\,dx+\int_{\large\frac{\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 0
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Solutions of $\sin^2\theta = \frac{x^2+y^2}{2xy} $ If $x$ and $y$ are real, then the equation
$$\sin^2\theta = \frac{x^2+y^2}{2xy}$$
has a solution:
*
*for all $x$ and $y$
*for no $x$ and $y$
*only when $x \neq y \neq 0$
*only when $x = y \neq 0$
| Suppose that $x$ and $y$ are not $0$.
We know that $$x^2 + y^2 \geq 2xy. \quad (1)$$
Two cases:
*
*$x>0$ and $y<0$ (or $x<0$ and $y>0$), then $\frac{x^2+y^2}{2xy} < 0$, and you cannot have $\sin^2(\theta) < 0$
*$x>0$ and $y>0$ (or $x<0$ and $y<0$). Then $(1)$ becomes
$$\frac{x^2+y^2}{2xy} \geq 1.$$
The only soluti... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solving $2x^4+x^3-11x^2+x+2 = 0$ I am having no idea how I can solve this problem.
I need help!
Here's the problem
$2x^4+x^3-11x^2+x+2 = 0$
I am learning Quadratic Expressions and this is what I need to solve, and I can't understand how :C
| $2x^4+x^3-11x^2+x+2=0$
Note that the coefficients: $2,1,-11,1,2$ are symmetrical.
$2(x^4+1)+(x^3+x)-11x^2=0$
$2(x^4+4x^3+6x^2+4x+1)-7(x^3+x)-23x^2=0$
$2(x^4+4x^3+6x^2+4x+1)-7(x^3+2x^2+x)-9x^2=0$
$2(x+1)^4-7(x+1)^2x-9x^2=0$
$2\left(\frac{(x+1)^2}{x}\right)^2-7\left(\frac{(x+1)^2}{x}\right)-9=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Inequality involving ArcTan How to prove that for $x\in[0, +\infty]$ the following inequality is true: $$\arctan x\geq\frac{3 x}{1+2\sqrt{1+x^2}}?$$
I don't have idea from where to start, so any hint is welcome. Thanks in advance.
| Let's say $x=\tan \theta$, so that:
$$\theta \geq \frac{3\tan \theta}{1+2\sqrt{1+\tan^2 \theta}}$$
$\sqrt{1+\tan^2\theta}=\sec\theta$ because the range of $\arctan x$ always has $\cos \theta > 0$:
$$\theta \geq \frac{3\tan \theta}{1+2\sec\theta}$$
Multiply both the numerator and denominator by $\cos \theta$:
$$\theta \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Converting $\cos\phi$ into $\frac{1−t^2}{1+t^2}$, given that $t = \tan\frac{\phi}{2}$
I have to figure out the working to convert $\cos\phi$ into $\dfrac{1−t^2}{1+t^2}$, given that $t = \tan\dfrac{\phi}{2}$.
It would be amazing if someone could help I've been trying to do it for hours.
| If $t = \tan \frac \phi 2$, then by drawing a right triangle, one can see that $$\sin \frac \phi 2 = \frac{t}{\sqrt{t^2+1}} \quad \text{ and } \quad \cos \frac \phi 2 = \frac{1}{\sqrt{t^2+1}}$$
So, by using $\cos (2x) = 1 - 2 \sin^2x$ and letting $x = \frac \phi 2$, we get
$$\cos \phi = 1 - 2 \cdot \left(\frac{t}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many subsets contain no consecutive elements? How many subsets of $\{1,2,...,n\}$ have no two consecutive numbers ?
Here is the solution :
The subsets are interpreted as $n$-words from the alphabet $\{0,1\}$. Let $a_n$ be the number of words with no consecutive ones. Then, a word can start from $0$ and proceed in $... | Choose the first value in the set:
$$\sum_{q=1}^n z^q = z \sum_{q=0}^{n-1} z^q = z\frac{1-z^n}{1-z}.$$
Choose some number of gaps that are at least two:
$$\sum_{p=0}^{n-1} \left(\frac{z^2}{1-z}\right)^p
= \frac{1-z^{2n}/(1-z)^n}{1-z^2/(1-z)}.$$
Sum the contributions that end in at most $n$ and extract the
coef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
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construct polynomial from other polynomials If I have a polynomial, P, with root $a$ and a polynomial, Q, with root $b$, is there a way to construct polynomial R such that $a+b$ is a root of R?
Here's a concrete example. a = $\sqrt2$. $P(x) = x^2 -2$ and $P(a) = 0$.
b = $\sqrt3$. $Q(x) = x^2 - 3$ and $Q(b) = 0$.
From ... | Gaussian elimination is the key. The ring
$$ \mathbb{Q}[x,y]/(P(x),Q(y)) $$
is a vector space over $\mathbb{Q}$ with dimension $\partial P\cdot \partial Q$, and a base given by $x^i y^j$ for $i\in[0,\partial P-1],j\in[0,\partial Q-1]$. By representing $1,(x+y),(x+y)^2,\ldots,(x+y)^{\partial P\cdot\partial Q}$ in such a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Find $\lim\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}$ The question arise in connection with this problem
Prove that
$$\lim_{n\rightarrow \infty}\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+... | Note that the sequence you're interested in can be written as $a_n=\dfrac1{\sqrt{n}}\left(\sqrt{4n-1}+\sum\limits_{k=1}^{2n-1}\sqrt{2k-1}-\sqrt{2k}\right).$ Applying the Stolz-Cesàro theorem, we have \begin{align}\lim_{n\to\infty} a_n&=\lim_{n\to\infty}\frac{\sqrt{4n+3}-\sqrt{4n-1}+\sqrt{4n-1}-\sqrt{4n}+\sqrt{4n+1}-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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What's the mistake on my answer for this inequality $ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} $ Good evening to everyone! I have the following inequality: $$ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} $$. I don't know what's wrong with my answer: $$ \... | ALthough in fact $a>b$ does not necessarily mean that $a^2 > b^2$ even if $a$ and $b$ are both positive, the real place you went wrong is in solving $2x^2-4 = 0$. The answer to that is $x = \pm \sqrt{2}$ not $1 \pm \sqrt{2}$.
The $-\sqrt{2}$ zero is spurious; it was introduced by the quaring operation, and for $x=-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the power series representation of $e^{-x^2}$ I know that the Maclaurin expansion of $e^x$ is $$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
But i'm not sure how to find the Maclaurin series here
I tried this
$$
f'_{(0)}=-2xe^{-x^2}=0
$$
And that follows to every derivative that follows, so how can I get a power series... | That is easier than what you think.
General expansion:
$$e^A = 1 + A + \frac{1}{2}A^2 + \frac{1}{6}A^3 + \cdots$$
So if $A = -x^2$ you get
$$e^{-x^2} = 1 - x^2 + \frac{1}{2}(-x^2)^2 + \frac{1}{6}(-x^2)^3 + \cdots$$
So
$$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$$
To find the general series expansion... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$x^3 +y^2 +z =100z+10y+x$ What is the largest and smallest integer that satisfies this equation. $x^3 + y^2 +z=\overline{zyx}$, where $\overline{zyx}$ denotes the sequence of the digits.
$x^3 +y^2 +z =100z+10y+x$,where $x,y,z>0$
The maximum value of $x$, $y$, $z$ individually can only be $9$.
$\text{Maximum value}= 9^3... | $$x^3+y^2+z=100z+10y+x$$
$$x(x+1)(x-1)=y(10-y)+99z$$
No needs to check $x=1$ and $y$ have symmetrical sense (e.g. if $y=1$ satisfies so as $y=9$).
Also, $9\leq y(10-y) \leq 25$ hence $$9(1+11z)\leq x(x+1)(x-1) \leq 25+99z$$
$$108\leq x(x+1)(x-1) \leq 916$$
$$\because \quad 3\times 4\times 5 < 108 < 4\times 5\times 6 \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Summing $3+7+14+24+37...$ up to $10$ terms
What is $3+7+14+24+37...$ up to $10$ terms?
I only see that difference between two consecutive terms is in AP ie $7-3=4,14-7=7,24-14=10$ and so on. Any ideas on how to do it?
| $x=1 \phantom{n} 2 \phantom{n} 3 \phantom{n} 4$
$y=3 \phantom{n} 7 \phantom{n} 14 \phantom{n} 24$
$y=m_2(x-x_1)(x-x_2)+m_1(x-x_1)+y_1$
$m_1$ is slope
$m_1=(y_2-y_1)/(x_2-x_1)$
$m_1=(7-3)/(2-1)$
$m_1=4$
$m_2$ changes in slope
$m_2=((y_3-y_2)-(y_2-y_1))/(x_3-x_1)$
$m_2=((14-7)-(7-3))/(3-1)$
$m_2=3/2$
$y=3/2*(x-2)(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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The value of $x^2+y^2+z^2+w^2$ Let$x,y,z,w$ satisfy
$$\frac{x^2}{2^2 - 1^2} +\frac{y^2}{2^2 - 3^2} +\frac{z^2}{2^2 - 5^2} +\frac{w^2}{2^2 - 7^2} =1$$
$$\frac{x^2}{4^2 - 1^2} +\frac{y^2}{4^2 - 3^2} +\frac{z^2}{4^2 - 5^2} +\frac{w^2}{4^2 - 7^2} =1$$
$$\frac{x^2}{6^2 - 1^2} +\frac{y^2}{6^2 - 3^2} +\frac{z^2}{6^2 - 5^2} +\... | I multiplied the equations out to get:
$$\begin{pmatrix} -4725 & 2835 & 675 & 315 \\
2079 & 4455 & -3465 & -945 \\
-3861 & -5005 & -12285 & 10395 \\
32175 & 36855 & 51975 & 135135 \\
\end{pmatrix}
\begin{pmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{pmatrix} =
\begin{pmatrix} -14175 \\
31185 \\
-135135 \\
2027025 \\
\end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Factoring $x^4-11x^2y^2+y^4$ I am brushing up on my precalculus and was wondering how to factor the expression
$$
x^4-11x^2y^2+y^4
$$
Thanks for any help!
| $$
\begin{align}
x^4-11x^2y^2+y^4
&=\left(x^2-y^2\right)^2-(3xy)^2\\
&=\left(x^2-3xy-y^2\right)\left(x^2+3xy-y^2\right)\\
&=\left(x-\tfrac{3-\sqrt{13}}2y\right)\left(x-\tfrac{3+\sqrt{13}}2y\right)\left(x+\tfrac{3-\sqrt{13}}2y\right)\left(x+\tfrac{3+\sqrt{13}}2y\right)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Solving $\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$ Does anyone have some tips for me how to go about the problem in the image?
$$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$$
I know it's supposed to be simple, but I can't figure out why the solution is:
$90^{\circ}+ 720^{\circ}k$ where $k... |
Notice:
$$\cos\left(\frac{\theta}{2}\right)+\sin\left(\frac{\theta}{2}\right)=\frac{\sqrt{2}\cos\left(\frac{\theta}{2}\right)}{\sqrt{2}}+\frac{\sqrt{2}\sin\left(\frac{\theta}{2}\right)}{\sqrt{2}}=\sqrt{2}\left[\frac{\cos\left(\frac{\theta}{2}\right)}{\sqrt{2}}+\frac{\sin\left(\frac{\theta}{2}\right)}{\sqrt{2}}\right]=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$
$\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{... | Just differentiate the $f(x)$ wrt $x$, you will se that $f'(x)=0$ which means $f(x)=constant=p$. SO you can find $p$ by just putting $x=\frac{\pi}{4}$. So
$$p=\int_{\frac{1}{e}}^{1} \frac{tdt}{t^2+1}+\int_{\frac{1}{e}}^{1} \frac{dt}{t(t^2+1)}=\int_{\frac{1}{e}}^{1}\frac{dt}{t}=0-(-1)=1$$
SO $f(x)=1$ which is the final... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$
Find the number of solutions of the equation
$$\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,$$
... | Alt. solution: Write the fractional part of $x$ in binary:
$$
x = n + 0.b_1 b_2 b_3 \ldots = n + \sum_{i=1}^\infty \frac{b_i}{2^i},
$$
where $n \in \mathbb{Z}$ and $b_i \in \{0,1\}$.
Also give your function $\mathbb{R} \to \mathbb{R}$ a name:
$$
f(x) = \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem
What is the remainder when $10^{35}$ is divided by $... | Do you know Fermat little theorem?
for prime $p, n^p \equiv n \mod p\\
n^{p-1}\equiv 1 \mod p$
$10^35 = (10^6)^5(10^5)\\
10^35 \equiv (10^5) \mod 7\\
10^35 \equiv 5 \mod 7$
Alternative:
You could say, $10^{35} -1 = (10^7 - 1)(10^{28} + 10^{21} + 10^{14} + 10^7 + 1)$
And the remainder of the product equals the product o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Find the determinant using colum or row operations I find problem in simplification. When I tried to simplify I ended up doing the regular process of finding the determinant value. The matrix is $\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ ab & bc & ca \end{pmatrix}$.
| $$\det\left(\begin{array}{ccc}1&1&1\\a&b&c\\ab&bc&ca\end {array}\right)=abc\det\left(\begin{array}{ccc}\frac{1}{a}&\frac{1}{b}&\frac{1}{c}\\1&1&1\\b&c&a\end {array}\right)=abc\det\left(\begin{array}{ccc}\frac{1}{a}-\frac{1}{b}&\frac{1}{b}&\frac{1}{c}-\frac{1}{b}\\0&1&0\\b-c&c&a-c\end {array}\right)=$$
$$abc\det\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
number of non differentiable points in $g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$
If $\displaystyle f(x) = \lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}\;,n\in \mathbb{N}$
and $\displaystyle g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x)... | Hint: using the formula $\tan ^{ 2 }{ \frac { x }{ 2 } = } \frac { 1-\cos { x } }{ \sin { x } } $ we can transform $g(x)$ function so :$$g\left( x \right) =\tan \left( \frac { 1 }{ 2 } \arcsin \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } } \right) \right) =\\ =\pm \sqrt { \frac { 1-\cos { \left( \arcsin \left( \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Generalized Harmonic Number Summation $ \sum_{n=1}^{\infty} {2^{-n}}{(H_{n}^{(2)})^2}$
Prove That $$ \sum_{n=1}^{\infty} \dfrac{(H_{n}^{(2)})^2}{2^n} = \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2 $$
Notation : $ \displaystyle H_{n}^{(2)} = \sum_{r=1}^{n} \dfrac{... | second approach suggested by Cornel Ioan Valean using summation by parts and lets start with the following sum:
with ${N \in \mathbb{N}_{\ \geq\ 1}}$
\begin{align}
\sum_{n=1}^N\frac{\left(H_{n-1}^{(2)}\right)^2}{2^n}=\sum_{n=1}^N\frac{\left(H_n^{(2)}\right)^2}{2^n}-2\sum_{n=1}^N\frac{H_n^{(2)}}{n^22^n}+\sum_{n=1}^N\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to identify this sequence? Interested by this question, $j$ being a positive integer,
I tried to work the asymptotics of
$$S^{(j)}_n=\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+j)}=\frac{\, _2F_1\left(j,-n;j+1;-\frac{1}{n}\right)}{j}$$
I quickly noticed (not a proof) that the asymptotics write
$$S^{(j)... | Notice
$$S_n^{(j)}
= \int_0^1 \left(1+\frac{x}{n}\right)^n x^{j-1} dx
= \int_0^1 e^{n\log\left(1+\frac{x}{n}\right)} x^{j-1} dx
= \int_0^1 e^{x - \frac{x^2}{2n} + \frac{x^3}{3n^2} + O(n^{-3})} x^{j-1}dx\\
= \int_0^1 e^x \left[1 - \frac{x^2}{2n} + \frac{8x^3+3x^4}{24n^2} + O(n^{-3})\right]x^{j-1} dx
$$
Compare with exp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Quadratic Functional equations. Suppose $f$ is a quadratic ploynomial, with leading cofficient $1$, such that $$f(f(x) +x) = f(x)(x^2+786x+439)$$ For all real number $x$. What is the value of $f(3)$?
| Suppose $f(x) = x^2 + bx + c$
Then
$$f(f(x)+x) = (f(x)+x)^2 + b(f(x)+x) + c = f(x)^2 + 2xf(x) + x^2 + bf(x) + bx + c = f(x)^2 + 2xf(x) + bf(x) + f(x) = f(x)(f(x) + 2x + b+ 1)$$
Equating factors yields
$$f(x) + 2x + b+ 1 = x^2 + 786 x + 439$$
Then letting $x=3$ we have
$f(3) = 2799-b$
Finding $b$ is a trivial matter by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Factor polynomial over $C, Q, R$, if one comlex root is given The polynomial is: $P(x)=x^6+x^4-x^3+x^2+1$. I need to factor it over $C, Q, R$ if one complex root is $\sqrt[3]{1}$. Also find all fields in which $P$ is reducible.
Now, I know how to find one factor of P using given complex root, and I end up with $(x^2+x... | A start: To factor the quartic, it may be useful to find its roots.
Rewrite the quartic as $x^2(x^2+\frac{1}{x^2}-\left(x+\frac{1}{x}\right)+1)$.
Let $t=x+\frac{1}{x}$. Then to solve our quartic we first solve
$$t^2-2-t+1=0.$$
Another way: Our quartic is $\frac{x^5+1}{x+1}$. So its roots are certain $10$-th roots of un... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Indefinite Integral - How to do questions with square roots?
$$\int \frac{dx}{x^4 \sqrt{a^2 + x^2}}$$
In the above question, my first step would be to try and get out of the square root, so I would take $ t^2 = a^2 + x^2 $. But that gets me nowhere. How would you solve this, and if you are going to take a substituti... | You can use trigonometric functions. Substitute $x=a*tan(u)$ and $dx=a*\frac{du}{{cos}^{2}u}$ this gives:
$$\int \dfrac{dx}{x^4 \sqrt{a^2 + x^2}} =\int \dfrac{du}{a^4\cos(u){\tan}^{4}(u)}$$
Working further on the solution of Battani and by simplifying:
$${ \sin{ \left( \arctan { \frac { x }{ a } } \right) } }=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$ for all $x,y$ positive.
Let's look at
$$\begin{split} &(x-y)^2(x+y)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff & 3x^3+3y^3\geq +3xy^2+3yx^2\\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3... |
Is this proof valid?
Your proof is valid and correct. However, it would be better if you prove that an equivalent statement is true, from an inequality, instead of doing the opposite.
$$\begin{split} & x^3+y^3\geq \frac{1}{4}(x+y)^3 \\
\iff & 4x^3+4y^3\geq (x+y)^3 \\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3-y^3 \\
\iff & 3x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find $a + b + c$, given that $(a+1)^{1/2} - a + (b+2)^{1/2} \cdot 2 - b + (c+3)^{1/2} \cdot 3 - c = \frac{19}{2}.$
Let $a,b,c$ be real number such that $$(a+1)^{1/2} - a + (b+2)^{1/2} \cdot 2 - b + (c+3)^{1/2} \cdot 3 - c = \frac{19}{2}.$$
Find $a + b + c$.
The answer is: $-\frac{5}{2}$. Please give me some clues o... | Using CS-inequality: let $x = a+b+c \implies \left(x+\frac{19}{2}\right)^2 = \left(1\sqrt{a+1}+2\sqrt{b+2}+3\sqrt{c+3}\right)^2\leq 14(x+6)\implies \left(x+\frac{5}{2}\right)^2 \leq 0\implies x = -\dfrac{5}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int \frac{\sqrt{64x^2-256}}{x}\,dx$ QUESTION
Evaluate $$\int \frac{\sqrt{64x^2-256}}{x}\,dx$$
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
MY ATTEMPT
*
*Typed
$\newcommand{\dd}{\... | $$\dfrac{\sqrt{64x^2-256}}x=8x\cdot\dfrac{\sqrt{x^2-4}}{x^2}$$
Let $\sqrt{x^2-4}=y\implies x^2-4=y^2\implies x\ dx= y\ dy$
$$\int\dfrac{\sqrt{64x^2-256}}xdx=8\int\dfrac{y^2dy}{y^2+4}=8\int dy-32\int\dfrac{dy}{y^2+4}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Convergence of the series $\sum \frac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$
To prove that nature of the following series : $$\sum \dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$$
they use in solution manual :
My questions:
*
*I don't know how to achieve ( * ) could someone complete my... | Fix $N>0
$ and $N
$ even (you can also take $N
$ odd, the proof is essentially the same). We have $$\sum_{n=1}^{N}\frac{\left(-1\right)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+\left(-1\right)^{n}}=\sum_{n=1}^{N/2}\frac{1}{\left(2n\right)^{\frac{2}{3}}+\left(2n\right)^{\frac{1}{3}}+1}-\sum_{n=1}^{N/2-1}\frac{1}{\left(2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Linear combination issue I have 4 vectors:
$u_1 = \begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix} $, $\; u_2 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{pmatrix} $, $\; u_3 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{p... | You need three linearly independent vectors to form a basis for three dimensions.
As $u_2$ and $u_3$ are in fact identical and the same, the redundant and duplicated one is unnecessary and also not needed.
Reapply your process with only three column vectors: $u_1, u_2, u_4$, and $v$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the orthocentre of a trinagle. Now, I know this has been asked here but my question is something else so please bear with me.
Question:- If the vertices of a triangle are represented by $z_1, z_2, z_3$ respectively then show that the orthocentre of the $\triangle ABC$ is
$$\dfrac{(a \sec A)z_1+(b \sec B)z_2+(... |
$$\dfrac{AH}{HL}=\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)}$$
Because the two triangles have the common height $BL$.
$$\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)}=\dfrac{\dfrac{1}{2}AB.BH\sin{\alpha}}{\dfrac{1}{2}BL.BH \sin{\beta}}$$
$$\dfrac{ar(\triangle ABH)}{ar(\triangle HBL)}=\dfrac{\dfrac 12 AB\cdot NH}{\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Differentiate and simplify. $m(x) = \frac{x}{\sqrt{4x-3}}$ My work so far is:
\begin{align}
m'(x) &= \frac{(1)(\sqrt{4x-3})-(x)(1/2)(4x-3)^{-1/2}(4)}{(\sqrt{4x-3})^2} \\
&= \frac{\sqrt{4x-3} - 2x(4x-3)^{1/2}}{4x-3}
\end{align}
and now I'm stuck on how to simplify further
| One method for differentiating $m(x)$ is treating $m(x)=\frac{f(x)}{g(x)}$ with
$f(x)=x$ and $g(x)=\sqrt{4x-3}$ and using the quotient rule for differentiation.
Then $m'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}=\frac{\sqrt{4x-3}-(x\frac{2}{\sqrt{4x-3}})}{4x-3}$.
Simplyfying we obtain the desired result $m'(x)=\frac{2x-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Circle Problem:Which of the following are true If the circle $x^2+y^2+2gx+2fy+c=0$ cuts the three circles $x^2+y^2−5=0$, $x^2+y^2−8x−6y+10=0$ and $x^2+y^2−4x+2y−2=0$ at the extremities of their diameters, then which of the following are true
?
$c=-5$
$fg=147/25$
$g+2f=c+2$
$4f=3g$
Can't think of a simple method.Help pl... | The center of $x^2+y^2−5=0, x^2+y^2−8x−6y+10=0$ and $x^2+y^2−4x+2y−2=0$ is $(0,0), (4,3)$ and $(2,-1)$ respectively.
So, $$(x^2+y^2+2gx+2fy+c)-(x^2+y^2-5)=0$$
$$(x^2+y^2+2gx+2fy+c)-(x^2+y^2−8x−6y+10)=0$$
$$(x^2+y^2+2gx+2fy+c)-(x^2+y^2−4x+2y−2)=0$$
passes through $(0,0), (4,3)$ and $(2,-1)$ respectively.
(For example, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Problem 1, Ch. 6 in Piskunov's, Differential and Integral calculus
Find the curvature of the curve at indicated points
$b^2x^2+a^2y^2=a^2b^2$ at $(0,b)$ and $(a,0)$
My attempt
$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$
Differentiating the implicit equ... | I solved it as follows:
$\begin{align}
y'&=-\frac{b^2}{a^2}\cdot\frac{x}{y}\\
y''&=-\frac{b^2}{a^2}\cdot{\frac{y-xy'}{y^2}}\\
&= -\frac{b^2}{a^2}\cdot{\frac{y-x\cdot{\left(-\frac{b^2}{a^2}\cdot\frac{x}{y}\right)}}{y^2}}\\
&=-\frac{b^2}{a^2}\cdot{\frac{a^{2}y^{2}+b^{2}x^{2}}{a^{2}y^{3}}}\\
&=-\frac{b^2}{a^2}\cdot{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Help solving this recurrence relation I wanted to resolve the determinant of the next (nxn) matrix via recurrence relations:
$$
\begin{vmatrix}
a & 1 & 0 & 0 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0\\
1 & a & 1 & 0 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0 \\
0 & 1 & a & 1 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & a & 1 & 0 &.... ... | To solve for the constants we just evaluate the expression for $D_n$ at $n=1$ and $n=2$, as well as use the information that $D_1=a$ and $D_2=a^2-1$ to get
$$C_1\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)+C_2\left(\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\right)=a$$
$$C_1\left(\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\right)^2+C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculation of the squared Euclidean norm Just a simple question. I'm reading an article and come across an equation that I cannot replicate as done in the original.
\begin{align}
&\|\mathbf{x} - \mathbf{\alpha} \|^2 - \|\mathbf{x} - \mathbf{\beta}\|^2\\
&= \|\mathbf{x}\| \|\mathbf{x}\| - 2 \|\mathbf{\mathbf{\alpha}}\|... | Your transition from the first line to the second is incorrect. We should have
$$
\|x - \alpha\|^2 - \|x - \beta\|^2 = \\
(x - \alpha)^T(x - \alpha) - (x - \beta)^T(x - \beta) = \\
\|x\|^2 + \|\alpha\|^2 - \|x\|^2 - \|\beta\|^2 - x^T\alpha - \alpha^Tx + x^T\beta + \beta^Tx =\\
\|\alpha\|^2 - \|\beta\|^2 - 2\alpha^Tx +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove $\prod\limits_{i=1}^{n}(x-4i+2)(x-4i+1)>\prod\limits_{i=1}^{n}(x-4i+3)(x-4i)$ for all $x\in\mathbb{R}$? I would like to prove that for $n\in\mathbb{N}$ we have $f_n(x):=\prod\limits_{r=1}^{n}(x-4r+2)(x-4r+1)>\prod\limits_{r=1}^{n}(x-4r+3)(x-4r)=:g_n(x)$ for all $x\in\mathbb{R}$ (actually it would suffice f... | $\begin{array}\\
(x-4i+2)(x-4i+1)-(x-4i+3)(x-4i)
&=((x-4i)^2+3(x-4i)+2)-((x-4i)^2+3(x-4i))\\
&=2\\
\end{array}
$
so each individual term is greater,
so their product is greater.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the maximum value that the quantity $2m+7n$ can have
Find the maximum value that the quantity $2m+7n$ can have such that there exist distinct positive integers $x_i$ $(1 \leq i \leq m)$, $y_j$ $(1 \leq j \leq n)$ such that the $x_i$'s are even, the $y_j$'s are odd, and $\displaystyle \sum_{i = 1}^m x_i+\sum_{j=1}... | Alternatively, rewrite $m(m+1)+n^2\leq 1986$ as $(2m+1)^2+(2n)^2\leq 4\cdot 1986+1=7945$. By the Cauchy-Schwarz Inequality,
$$\begin{align}
(2m+7n)+1=(2m+1)+\frac{7}{2}(2n)
&\leq \sqrt{1^2+\left(\frac{7}{2}\right)^2}\,\sqrt{(2m+1)^2+(2n)^2}
\\
&\leq\sqrt{\frac{53}{4}}\,\sqrt{7945}<325\,.
\end{align}$$
Thus, $$2m+7n<32... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $n = 3k + 5l$ for $n > 7$
Show that for every n greater than $7$, there are non-negative integers $k$ and $l$ such that $$n = 3k+ 5l.$$
So induction seems like a possibility.
$n = 3k + 5l$ and so $n + 1 = 3k + 5l + 1$.
What can be done in such a case? Can I get a hint?
| By strong induction: We can write
\begin{align*}
8 & = 1 \cdot 3 + 1 \cdot 5\\
9 & = 3 \cdot 3 + 0 \cdot 5\\
10 & = 0 \cdot 3 + 2 \cdot 5
\end{align*}
Let $n \geq 10$. Assume that we can write each integer $m$ such that $8 \leq m \leq n$ in the form $m = 3k + 5l$ for some non-negative integers $k, l$. We wish to sho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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What approach should I use to solve integrals like this? $$\int{\sqrt{1-{x^3}}}dx$$
I tried with $t=x^3$ but then I have the $3x^2$ dt that I can't get rid of.
| For any real number of $x$ ,
When $|x|\leq1$ ,
$\int\sqrt{1-x^3}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{3n}}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{3n+1}}{4^n(n!)^2(1-2n)(3n+1)}+C$
When $|x|\geq1$ ,
$\int\sqrt{1-x^3}~dx$
$=i\int\sqrt{x^3-1}~dx$
$=i\int x^\frac{3}{2}\sqrt{1-\dfrac{1}{x^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integral $\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$ $$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$
$$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$
$$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$
$$t= \tan {\frac{x}{2}}$$
On solving ,
$$\frac{1}{\tan x + \cot x +... | $\displaystyle\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}=$
$\displaystyle\int\frac{\sin x\cos x}{1+\sin x+\cos x}\,dx=\int\frac{\sin x\cos x}{1+\sin x+\cos x}\cdot\frac{1-(\sin x+\cos x)}{1-(\sin x+\cos x)}\,dx$
$=\displaystyle\int\frac{\sin x\cos x(1-(\sin x+\cos x))}{1-(\sin x+\cos x)^2}\,dx=\int\frac{\sin x\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Positive semidefinite ordering for covariance matrices Suppose that X and Z are matrices with the same number of rows. Let
$$ D = \left[\begin{array}{cc} X' X & X'Z \\ Z'X & Z'Z \end{array} \right]^{-1} - \left[\begin{array}{cc} (X' X)^{-1} & 0 \\ 0 & 0 \end{array} \right],$$ where all inverses are assumed to exist and... | $U = \left(\begin{matrix} X^{T}X & X^TZ \\ Z^TX & Z^TZ \end{matrix}\right) = \left( \begin{matrix} X^T \\ Z^T\end{matrix}\right) \left(\begin{matrix} X & Z \end{matrix} \right) $ is clearly positive semidefinite and since we are given it is invertible it must be positive definite.
Consequently for any invertible matrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$ I am working on the following exercise. Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$.
I have a couple of ideas for going about this exercise.
$1)$ By moving $1$ to the other side of the equation we obtain:
$y^4-1=x^4-x^3+x^2-x \rightar... | One has $$1-x+x^{2}-x^{3}+x^{4}=y^{4}\iff(2x^2+1)^2=4y^4+4(x^3+x)-3$$ Notice that necessarily $y$ must be odd and $x^3+x$ is even when $x$ is even or odd. Hence one has $$(2x^2+1)^2\equiv 4y^4-3 \pmod 8$$ The solutions of this congruence are
$$(x,y)=(x,1),(x,3),(x,5),(x,7)\text { with } x=0,1,2,3,4,5,6,7$$ we can veri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Fundamental group of $S^2 - \{N,S\}$ I would like to calculate the fundamental group $\Pi_1$ of $S^2 - \{N,S\}$, where $N,S$ are the north respectively south pole of $S^2$, but do not see yet how to do this.
| The map \begin{align}\phi:S^2\setminus \{N,S\}&\to S^1\times (-1,1)\\ \begin{pmatrix}x\\ y\\ z\end{pmatrix}&\mapsto \frac{1}{\sqrt{x^2+y^2}}\begin{pmatrix}x\\ y\\ z\sqrt{x^2+y^2}\end{pmatrix}\end{align} is a diffeomorphism.
Its inverse is $$\begin{pmatrix}a\\ b\\ c\end{pmatrix}\mapsto \begin{pmatrix}a\sqrt{1-c^2}\\ b\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac... | $$\dfrac{\sqrt{5-2\sqrt6}}{1+\sqrt6}=\dfrac{\sqrt3-\sqrt2}{1+\sqrt3\cdot\sqrt2}$$
$$\implies\arctan\dfrac{\sqrt{5-2\sqrt6}}{1+\sqrt6}=\arctan\sqrt3-\arctan\sqrt2$$
$$\arctan\sqrt3=\dfrac\pi3$$ and
$$\arctan\sqrt2=\text{arccot}\dfrac1{\sqrt2}=\dfrac\pi2-\arctan\dfrac1{\sqrt2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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If 'x' is real then $\frac{x^2-x+c}{x^2+x+2c}$ can take all real values if? the question asks for the interval in which c lies so that $\frac{x^2-x+c}{x^2+x+2c}$ gives all real values for all x belongs to R.
how to proceed in this problem?
| Let $y = \frac{x^2-x+c}{x^2+x+2c} $ be a value taken by the given fraction. This gives
\begin{equation*}
x^2(y-1) +x(y+1) +c(2y-1) = 0
\end{equation*}
should have a real root. Thus
\begin{equation*}
(y+1)^2 -4c(y-1)(2y-1) \geq 0
\end{equation*}
and hence
\begin{equation*}
y^2(1-8c) + 2y(1+6c)+(1-4c) \geq 0
\end{equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the least squares solution to $A\vec{x} = \vec{b}$ using the transpose of $A$
Find the least squares solution to $A\vec{x} = \vec{b}$ using the transpose of $A$. Let $A = \begin{bmatrix}1&1&0\\1&0&-1\\0&1&1\\-1&1&-1\end{bmatrix}$ and $\vec{b} = \begin{bmatrix}2\\5\\6\\6\end{bmatrix}$
The formula says $x = (A^TA)... | $$A^T A = \left(
\begin{array}{ccc}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3 \\
\end{array}
\right)$$
$$(A^T A )^{-1} = \left(
\begin{array}{ccc}
\frac{1}{3} & 0 & 0 \\
0 & \frac{1}{3} & 0 \\
0 & 0 & \frac{1}{3} \\
\end{array}
\right)$$
$$A^T b = (1,14,-5)$$
$$(A^T A )^{-1} A^T b = \left(\dfrac{1}{3},\dfrac{14}{3},-\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Computing $\lim_{n \to \infty} \left(\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\cdots+\frac{n}{(n+n)^2}\right)$
Calculate $$\lim_{n \to \infty} \left(\dfrac{n}{(n+1)^2}+\dfrac{n}{(n+2)^2}+\cdots+\dfrac{n}{(n+n)^2}\right).$$
I tried turning this into a Riemann sum, but didn't see how since we get $\dfrac{1}{n} \cdot \dfrac{... | I thought it might be instructive to present a solution that relies on the Euler-Maclaurin Summation Formula (EMSF). To that end, we proceed.
Note that we can write
$$\sum_{k=1}^{n}\frac{n}{(n+k)^2}=\sum_{k=n+1}^{2n}\frac{n}{k^2}$$
From the EMSF, we have
$$\begin{align}
\sum_{k=n+1}^{2n}\frac{n}{k^2}&=n\left(\int_n^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Can anyone please help me to find the generalized equation for $x$,$y$ and $z$? $$
A= P(9+2\cos(x)+2\cos(y)+2\cos(z))
$$
$$
B= P(9+2\cos(x-2\pi/3)+2\cos(y-2\pi/3)+2\cos(z-2\pi/3))
$$
$$
C= P(9+2\cos(x+2\pi/3)+2\cos(y+2\pi/3)+2\cos(z+2\pi/3))
$$
| Observation: replacing $A,B,C$ with $A/p$, etc., we have
\begin{align}
A&= (9+2\cos(x)+2\cos(y)+2\cos(z))\\
B&= (9+2\cos(x-2\pi/3)+2\cos(y-2\pi/3)+2\cos(z-2\pi/3))\\
C&= (9+2\cos(x+2\pi/3)+2\cos(y+2\pi/3)+2\cos(z+2\pi/3))
\end{align}
Summing, we get
\begin{align}
A+B+C&= 27 + 2[\cos(x) + \cos(x-2\pi/3) + \cos(x + 2\pi/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1880441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $x^{x^{x^{16}}}=16$ calculate the value of $x^{x^{x^{12}}}$ I have done the following but I'not satisfied.
if $x^a=a$ then by substitution follows that $x^a=x^{x^a}=x^{x^{x^a}}$ etc.
So $x^{x^{x^{16}}}=16$ is equivalent to $x^{16}=16$, and $x=2^{\frac{1}{4}}$
Substituting $x$ to $x^{x^{x^{12}}}$ returns $2$.
| It's obvious that if $x=\sqrt[16]{16}$, then $x^{16} = 16$, and $x^{x^{16}} =16$ and $x^{x^{x^{16}}} = 16$, so $x=\sqrt[16]{16}$ solves your equation.
So then $x^{12} = {16^{0.75}} = 8$, and $x^{x^{12}} = x^8 = \sqrt{16} = 4$, and $x^{x^{x^{12}}} = x^4 = \sqrt[4]{16} = 2$. Hence $x^{x^{x^{12}}} = 2$
Ok hang on, I've ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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The integral $\int_0^1 x^n \log(2-x) \, dx$ In this question I had stuck on the integral:
$$\int_0^1 x^n \log(2-x) \, {\rm d}x$$
Claude Leibovici says that $\displaystyle \int_0^1 x^n \log(2-x) \, {\rm d}x=\frac{\, _2F_1\left(1,n+2;n+3;\frac{1}{2}\right)}{2 (n+1) (n+2)}=a_n+b_n \log 2$ where $a_n, b_n$ are sequences of... | We can easily develop a recursive formula for the integral of interest. To that end we proceed.
Let $I_n(\alpha)=\int_0^1 x^n \log(\alpha -x)\,dx$. Integrating by parts with $u=x^n$ and $v=(x-\alpha)\log(\alpha -x)-x$ yields
$$\begin{align}
I_n(\alpha)&=(1-\alpha)\log(\alpha-1)-1-n\int_0^1 x^{n-1}\left((x-\alpha)\log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proofs involving summations I am working on the following proof.
Prove that $\displaystyle \sum_{k=1}^{n} \frac{2k^4+1}{k+2} \geq \frac{n(n+1)(2n+1)}{6}$.
My first question is if the above is equivalent to saying "Prove that:$\displaystyle \sum_{k=1}^{n} \frac{2k^4+1}{k+2} \geq \displaystyle \sum_{n=1}^{t} \frac{n(n+1... | I think what you are asking is $\frac{2k^4 + 1}{k + 2} \ge \frac{k(k+1)(2k+1)}{6}$.
$\frac{k(k+1)(2k+1)}{6}= \frac{k(k+1)(k+2)(2k+1)}{6(k+2)}=\frac{\frac 16(2k^4 + 7k^3 +7k^2 + 2k)}{k+2}$
So your question is true if and only if $2k^4 + 1 \ge \frac 16(2k^4 + 7k^3 +7k^2 + 2k)$
Which is true if and only if $ 10k^4 + 6 \ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Contour inegration involving a confluent hypergeometric function with a pole at zero I have the following integral
\begin{equation}
\int_0^1 t^{-5/4} {_1}F_1(-1/4;1/2;t) dt
\end{equation}
I know that it only has a pole at zero since the confluent hypergeometric function is analytic in the entire complex plane but I don... | This integration does not have a convergent result.
$$
\int t^{-5/4} {_1}F_1(-1/4;1/2;t) dt
= \int t^{-5/4} \sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k} \frac{t^k}{\Gamma(k+1)} dt \\=\sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k} \frac{\int t^{k-\frac{5}{4}}dt}{\Gamma(k+1)} =\sum_{k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square
Prove that there exist infinitely many positive integers $n$ such that $\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square. Obviously, $1$ is the least integer having this property. Find the next two least integers with this property.
The given condition is equivalen... | From the equation $(2n+1)(2n+2) = 12p^2$, we know that there are two possibilities:
*
*$2n+2$ is of the form $4x^2$, $2n+1$ is of the form $3y^2$.
*$2n+2$ is of the form $12x^2$, $2n+1$ is of the form $y^2$.
This follows from $2n+1, 2n+2$ being relatively prime. Thus, solutions correspond to integer solutions of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
How to find $\sqrt[3]{0.5964}$ using logarithms? \begin{align*}
\sqrt[3]{0.5964} &= \left(0.5964\right)^{\frac{1}{3}}\\
\log \sqrt[3]{0.5964} &= \frac{1}{3} \log 0.5964\\
&= \frac{1}{3} \cdot\overline{1}.7755\\
&= \frac{1}{3}\cdot\left(\overline{3} +2.7755\right)\\
&= \overline{1}.9252
... | My guess is that $$\overline{1}.7755=-1+0.7755=-3+2.7755$$ Then divide by $3$ to get $$-1+0.9252=\overline{1}.9252$$ This is a way of keeping a positive mantissa.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A question about substitute equivalent form into limit: $\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$ When I had the calculus class about the limit, one of my classmate felt confused about this limit:
$$\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$$
What he thought that since $x^2 > x$ and $x^2 >... | Consider the general case $$A=\sqrt{x^2+ax+b}-\sqrt{x^2+cx+d}=x\left(\sqrt{1+\frac a x+\frac b {x^2}}-\sqrt{1+\frac c x+\frac d {x^2}}\right)$$ and use the fact that, for small $y$, using the generalized binomial theorem or Taylor series, $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ For the first radica... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to prove the solution of these inequalities is empty? Prove: There does not exist 4 unit vectors $\mathbf{v}_1$, $\mathbf{v}_2$, $\mathbf{v}_3$, $\mathbf{v}_4$ in $\mathbb{R}^3$ such that
$$
\left \{ \
\begin{array}{ll}
\dfrac{4}{3} < \left \|\mathbf{v}_i + \mathbf{v}_j \right \| _2 ^2 < \dfrac{8}{3} \, , \\
\dfra... | Without loss of generality, take $$\mathbf{v}_1=(0,0,1)$$
Let $\displaystyle \mathbf{v}_{k}=\left( \sqrt{1-z_k^2}\cos \theta_{k},\sqrt{1-z_k^2}\sin \theta_{k},z_{k} \right)$ where $k=2,3,4$
$$|\mathbf{v}_1-\mathbf{v}_k|^2=2(1-z_k)$$
$$\frac{4}{3}< 2(1-z_k) < \frac{8}{3}$$
$$-\frac{1}{3} < z_{k} < \frac{1}{3}$$
To minim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given $\tan a = \frac{1}{7}$ and $\sin b = \frac{1}{\sqrt{10}}$, show $a+2b = \frac{\pi}{4}$.
Given
$$\tan a = \frac{1}{7} \qquad\text{and}\qquad \sin b = \frac{1}{\sqrt{10}}$$
$$a,b \in (0,\frac{\pi}{2})$$
Show that $$a+2b=\frac{\pi}{4}$$
Does exist any faster method of proving that, other than expanding $\sin... | Just for fun... Adding a picture for the first time, so bear with me.
$\sin b = 1/\sqrt{10} \implies \cos b = 3/\sqrt{10}$ and $\sin 2b = 3/5$.
Now consider an isosceles right triangle $ABC$ as in the picture, with $AB = AC = 4$. Pick $D$ on $AC$ so that $AD = 3$, and $DE$ perpendicular to $BC$. $BD = 5$ by the Pyth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$ Could it be possible to find the solution for the following series?
$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Thanks in advance!
| You have $$\sum_{n=0}^\infty \left(a^n\sum_{m=0}^nb^m\right)\text{.}$$ Before determining what this converges to, it is worth establishing for which values of $(a,b)$ it converges at all. To that end, consider the Ratio Test:
$$\left\lvert\frac{a^{n+1}\sum_{m=0}^{n+1}b^m}{a^n\sum_{m=0}^nb^m}\right\rvert=\left\lvert a\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Problem solving recurrence equation for $a_n = a_{n-1} + 2a_{n-2}, \,\,\, a_0=a_1=1$ I want to try the generating function $A(x)$ and a formula for $a_n$ for the following sequence:
$$a_n = a_{n-1} + 2a_{n-2}, \,\,\, a_0 = a_1 = 1$$
So I multiplied the above equation on both sides by $x^n$ and summed them for each $n \... | I corrected the mistake in the second line of the calculations and got the following:
$$\Longrightarrow \sum_{n=2}^{\infty}a_{n}x^{n} = \sum_{n=2}^{\infty}a_{n-1}x^{n} + 2\sum_{n=2}^{\infty}a_{n-2}x^{n}$$
$$\Longrightarrow \sum_{n=2}^{\infty}a_{n}x^{n} = x\sum_{n=1}^{\infty}a_{n}x^{n} + 2x^2\sum_{n=0}^{\infty}a_{n}x^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How do I integrate this. How can I find the integral of
$$\int\frac{x+2}{(x^2+2x+2)\sqrt{x+1}} dx $$
So that the answer=$\sqrt{2}\tan^{-1}(\frac{x}{\sqrt{2(x+1)}})+c$
| First of all, just substitute $u=\sqrt{x+1}$ and $\text{d}u=\frac{1}{2\sqrt{x+1}}\space\text{d}x$, then we get:
$$\text{I}=\int\frac{x+2}{(x^2+2x+2)\sqrt{x+1}}\space\text{d}x=2\int\frac{u^2+1}{u^4+1}\space\text{d}u$$
Now, use partial fractions:
$$\frac{u^2+1}{u^4+1}=\frac{1}{2(u^2+u\sqrt{2}+1)}-\frac{1}{2(u\sqrt{2}-u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $ABC$ is right-angled Prove that if $\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$, then $ABC$ is right-angled.
I only found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$, but I have no idea what to do next.
Thank you in advance for your answers!
| $$\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$$
Use $\cos^2{A} =\frac{1+\cos2A}2$. Then
$$\cos2A+\cos2B+\cos2C=-1$$
Use $\cos2A+\cos2B+\cos2C=-1-4\cos A\cos B\cos C$
Then $$-1-4\cos A\cos B\cos C=-1$$
$$4\cos A\cos B\cos C=0$$
Then $\angle A$ or $\angle B$ or $\angle C$ is $\frac{\pi}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find $\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$
Find
$$\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$$
My attempt :-
$$3(x-1)+2\sqrt{2x^2 - 7x - 4}$$
$$\implies 3x - 3 + 2\sqrt{2x^2 - 7x - 4} + 2x^2 - 2x^2 -7x + 7x - 1 +1 $$
$$\implies 2x^2 -7x - 4 + 2\sqrt{2x^2 - 7x - 4} + (- 2x^2 + 10x + 1)... | $$\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}=$$
$$=\sqrt{\frac{3(x-1)+2\sqrt{2x^2 - 7x - 4}}2}=$$
$$=\sqrt\frac{\color{red}{3x-3}+2\sqrt{(2x+1)(x-4)}}2=$$
$$=\sqrt{\frac{\color{red}{2x+1}+2\sqrt{(2x+1)(x-4)}+\color{red}{x-4}}2}=$$
$$=\sqrt{\frac{(\sqrt{2x+1}+\sqrt{x-4})^2}2}=$$
$$=\frac{\sqrt{2x+1}+\sqrt{x-4}}{\sqrt2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $7$ Divides $\binom{2^n}{2}+1$ then $n =3k+2$ for some positive integer $k$ It is a straightforward question.
If $$ 7 \text{ }\Bigg | \binom{2^n}{2}+1$$ then $n=3k+2$ for some
positive integer $k$.
This is just curiosity no motivation just rummaging through some old question in a notebook. A simple counter exam... | $$\binom{2^n}2+1=\frac{2^n(2^n-1)}2+1=2^{2n-1}-2^{n-1}+1$$
Now, let's apply mod $7$. Note that $2^3\equiv 1\pmod 7$.
If $n=3k$ we have $4-4+1\neq 0$.
If $n=3k+1$ we have $2-1+1\neq 0$.
This proves your statement, but we can check easily if the "iff" statement is true:
If $n=3k+1$ we kave $1-2+1=0$.
So $7$ divides $\bin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this a new method for finding powers? Playing with a pencil and paper notebook I noticed the following:
$x=1$
$x^3=1$
$x=2$
$x^3=8$
$x=3$
$x^3=27$
$x=4$
$x^3=64$
$64-27 = 37$
$27-8 = 19$
$8-1 = 7$
$19-7=12$
$37-19=18$
$18-12=6$
I noticed a pattern for first 1..10 (in the above example I just compute the firs... | For a little bit more, see the answer "General method for indefinite summation" which explains how exactly this representation using forward differences allows you to easily find the formula for indefinite summation of powers. Applied to your case you get:
0, 1, 8, 27
1, 7, 19
6, 12
6
and hence:
$n^3 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
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Expressing products as the sum of two squares Express $(1+x^2)(1+y^2)(1+z^2)$ as the sum of two squares.
I have no clue where to start, I know that we should not post just for solution but a hint will definitely help.
| By Lagrange's identity
$$ (a^2+b^2)(c^2+d^2) = (ad-bc)^2 + (ac+bd)^2 \tag{1}$$
it follows that numbers that are a sum of two squares form a semigroup.
A straightforward proof of $(1)$ is to consider that $a^2+b^2$ is the norm of $a+bi$ in $\mathbb{C}=\mathbb{R}[i]$.
The norm is multiplicative and $(a-bi)(c+di)=(ac+bd)+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $a\equiv b\pmod p\Rightarrow a^p \equiv b^p \pmod {p^2}$ Suppose $p \nmid b$ and $p \nmid a$. Let $a^p \equiv b^p (\text{mod}(p))$. Prove that $a^p \equiv b^p (\text{mod}(p^2))$.
Below is my proof. I'm interested in two things.
1) Verifing this proof
2) Finding a simpler proof (preferably one that doesn't ... | Simpler proof:
By Fermat $p \ | \ a-b$, then $a \equiv b$ in modulo $p$
\begin{align*}
\frac{a^p-b^p}{a-b} &= (a^{p-1} + a^{p-2}b + \cdots b^{p-2}a + b^{p-1})\\
&\equiv (b^{p-1} + b^{p-2}b + \cdots b^{p-2}b + b^{p-1})\\
&\equiv p \ b^{b-1} \equiv 0
\end{align*}
in modulo $p$
Since $\frac{a^p-b^p}{a-b}$ and $a-b$ are d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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why exist $a_{i},a_{j},a_{k}(1\le iAssmue that postive real numbers $a_{1},a_{2},a_{3},a_{4}$ such
$$2(a^4_{1}+a^4_{2}+a^4_{3}+a^4_{4})<33a_{1}a_{2}a_{3}a_{4}$$
show that
There exist $a_{i},a_{j},a_{k}(1\le i<j<k\le 4)$ are side lengths of a triangle
if $a,b,c$ is trangle three side ,then we have
$$\sum a^4<2\sum a... | Let $a_4\geq a_3\geq a_2\geq a_1$ and $a_4=a$, $a_3=b$, $a_2=c$, $a_1=d$ and $c=xd$.
Hence, $x\geq1$.
Let $a\geq b+c$, $b\geq c+d$ and $f(a)=2(a^4+b^4+c^4+d^4)-33abcd$.
Hence, $f'(a)=8a^3-33bcd\geq8(b+c)^3-33bcd>0$, which says that
$$f(a)\geq f(b+c)=2(b+c)^4+2b^4+2c^4+2d^4-33(b+c)bcd$$
Let $g(b)=2(b+c)^4+2b^4+2c^4+2d^... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $z^n + 1/{z^n}$ is a rational number Knowing that $z + \frac{1}{z} = 3$ ($z \in \mathbb{R}$), prove that $z^n + \frac{1}{z^n} \in \mathbb{Q}$ ($n \in \mathbb{N}$ and $n \geq 1$).
I've tried to find a general formula for $z^n + \frac{1}{z^n}$.
$$\left(z + \frac{1}{z}\right)^n = \binom{n}{0}z^n + \binom{n}{1}z... | Here is an uglier solution:
$$z+\frac{1}{z}=3 \Rightarrow z^2-3z+1=0 \Rightarrow z=\frac{3\pm \sqrt{8}}{2}$$
Note that $\frac{3-\sqrt{8}}{2}=\frac{1}{\frac{3+\sqrt{8}}{2}}$, thus $\{ z, \frac{1}{z} \} = \{ \frac{3+ \sqrt{8}}{2}, \frac{3- \sqrt{8}}{2} \}$.
Then
$$z^n+\frac{1}{z^n}=\left( \frac{3 + \sqrt{8}}{2} \right)^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove each element of the following sequence is a perfect square? Sequence $\{a_n\}$ satisfies the following formula:
$a_{n+2}=14a_{n+1}-a_n+12$,
and $a_1=1, a_2=1$.
It is easy to check that $a_3=25$ and $a_4=361$.
The question is how to prove each element of the sequence $\{a_n\}$ is a perfect square?
| As Umberto P. commented, it is sufficient to prove by induction that
$$a_{n+2}=b_n^2\tag1$$
where $b_{n+2}=4b_{n+1}-b_{n}$ with $b_0=1, b_1=5$.
The base case : $a_2=1=b_0^2$ and $a_3=25=b_1^2$
Suppose that $a_{n+2}=b_n^2$ and $a_{n+1}=b_{n-1}^2$.
Then, using that
$$b_n^2-4b_nb_{n-1}+b_{n-1}^2=6\tag2$$
we get
$$\begin... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for?
$$x^4+y^4$$
Any help would be appreciated.Thanks in advance!
| $$(x+y)^2=3^2$$
$$x^2+2xy+y^2=9$$
$$x^2+y^2=9-2xy$$
square both sides
$$x^4+2x^2y^2+y^4=(9-2xy)^2$$
$$x^4+y^4=(9-2xy)^2-2x^2y^2=\color{red}{-9}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a and b such that $x^2 + 8x + 7 = (x + a)(x + b)$ for all real values of $x$. Find $a$ and $b$ such that $x^2 + 8x + 7 = (x + a)(x + b)$ for all real values of $x$.
That is what the question states, I think you should factor
Any Ideas on how to begin?
| By factorization,
$$x^2+8x+7 \equiv (x+7)(x+1) \equiv (x+a)(x+b)$$
$$\therefore \quad (a,b)=(1,7) \quad \text{or} \quad (7,1)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is solution to inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ equal to interval $[0, \frac{3 - \sqrt{5}}{6})$? Given inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ we can easily determine, that it's domain is $D = [0, 1]$. Because each term is real, we can take square of the inequality, which ... | The number $\sqrt{1-x}-\sqrt{x}$ can be negative and, in this case, the inequality doesn't hold. Squaring both sides can be done only if both sides are nonnegative. Thus your inequality becomes
$$
\begin{cases}
0\le x\le 1 \\[8px]
\sqrt{1-x}-\sqrt{x}\ge0 \\[4px]
1-x-2\sqrt{x(1-x)}+x>\dfrac{1}{3}
\end{cases}
$$
The seco... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Doubt about asymptotic analysis $\frac{1}{x^{a}*(4+9x)^{b+1}}\sim \frac{1}{9^{b+1}x^{a+b+1}}$ I don't understand why is correct
$$a,b\in \mathbb{R}$$
$$x \mapsto \infty $$
$$\frac{1}{x^{a}*(4+9x)^{b+1}}\sim \frac{1}{9^{b+1}x^{a+b+1}}$$
I would write
$$x \mapsto \infty $$
$$\frac{1}{(4+9x)^{b+1}}\sim \frac{1}{9x^{b+1}}$... | The problem lies in
$$\frac{1}{(4+9x)^{b+1}}\sim \frac{1}{9x^{b+1}}.$$
It should be
$$\frac{1}{(4+9x)^{b+1}}\sim \frac{1}{(9x)^{b+1}}$$
because
$$\lim_{x\to\infty}\frac{(9x)^{b+1}}{(4+9x)^{b+1}}=1.$$
| {
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can we express $x=f(y)$, if $y=\exp(x/1.45)(1-\exp(-x))$? Given $y=\exp(x/1.45)(1-\exp(-x))$, I need to express $x = f(y)$.
I tried taking log of both side but didnt get any fruitful expression.
Is it possible? If yes, how?
| Here are the details of the Lagrange Inversion computation. We have
$$y = \exp(bx)(1-\exp(-x)).$$
where $b=20/29.$ Putting $\exp(x) = t$ so that $x = \log t$ we get
$$y = t^b (1-1/t).$$
We seek to compute
$$[y^n] t(y) =
\frac{1}{2\pi i}
\int_{|y|=\epsilon} \frac{1}{y^{n+1}} t(y) dy.$$
Using the definition we put with $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$ Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$
I find out the answer $104$ but my friend find the answer $96$ using c... | Let $x=\sqrt7$ and $y=\sqrt5+\sqrt6$ and $z=\sqrt5-\sqrt6$. It becomes:
$$(y+x)(y-x)(x+z)(x-z)$$
$$=(y^2-x^2)(x^2-z^2)$$
$$=-x^4+x^2(y^2+z^2)-y^2z^2$$
$$=-7^2+7((\sqrt5+\sqrt6)^2+(\sqrt5-\sqrt6)^2)-(\sqrt5+\sqrt6)^2(\sqrt5-\sqrt6)^2$$
$$=-49+7(5+2\sqrt5\sqrt6+6+5-2\sqrt5\sqrt6+6)-(5-6)^2$$
$$=-49+7\times22-1$$
$$=104$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculus 2: Partial fractions problem. Finding the value of a constant I encountered the following problem.
Let $f(x)$ be a quadratic function such that $f(0) = -6$ and
$$\int \frac{f(x)}{x^2(x-3)^8} dx $$
is a rational function.
Determine the value of $f'(0)$
Here's what I tried. I decomposed the fraction integrand b... | Consider that your partial fraction decomposition is equal to $\frac {A_1}{x}+\frac {B_1}{x-3}+h(x),$ where $h(x)$ is the sum of the other terms. The anti-derivative of each of those other terms is a rational function. But $\int (\frac {A_1}{ x}+\frac {B_1}{x-3})\;dx=A_1\ln |x|+B_1\ln |x-3|$ is not a rational function... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve $ 1 - \sqrt{1 - 8\cdot(\log_{1/4}{x})^2} < 3\cdot \log_{1/4}x $ My answer: $2^\frac{-1}{\sqrt{2}} < x < 1$
Textbook answer: $2^\frac{-12}{17} < x < 1$
The only difference between my resolution and the Textbook one is that I solved by saying that
$$\text{(I) }\sqrt{1 - 8\cdot (\log_{1/4}{x})^2} > 1 -3\cdot \log_{... | Initially we have $$\sqrt { 1-8\cdot (\log _{ 1/4 }{ x } )^{ 2 } } <1-3\cdot \log _{ 1/4 } x\\ \log _{ 1/4 }{ x } =t\\ \\ 1-8t^{ 2 }>0\\ \left( t-\frac { 1 }{ 2\sqrt { 2 } } \right) \left( t+\frac { 1 }{ 2\sqrt { 2 } } \right) <0\\ t\in \left( -\frac { 1 }{ 2\sqrt { 2 } } ,\frac { 1 }{ 2\sqrt { 2 } } \right) \\ ... | {
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"timestamp": "2023-03-29T00:00:00",
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Suppose that $q=\frac{2^n+1}{3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$ I am curious for a starting point to prove the following claim
Suppose that $q={2^n+1 \above 1.5pt 3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$
For example take $n=61$. Then $q ={2^{61}+1 \above 1.5pt ... | Most probably yes, because:
$$\binom{2^n}{2}-1=\frac{2^n(2^n-1)}{2}-1=2^{n-1}(2^n-1)-1=$$
$$=2^{n-1}(2^n+1 - 2)-1 = 2^{n-1}(2^n+1) - 2^n-1=(2^n+1)(2^{n-1}-1)=$$
$$=3q(2^{n-1}-1)$$
Additionally:
$$2 \equiv -1 \pmod{3}$$
$$2^n \equiv (-1)^n \pmod{3}$$
$$2^n + 1 \equiv (-1)^n +1 \pmod{3}$$
So $n$ is odd, further leading t... | {
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"url": "https://math.stackexchange.com/questions/1912522",
"timestamp": "2023-03-29T00:00:00",
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Find integral $\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x$ I have to find definite integration $$\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x$$
I tried to subtitute $x = \sin u$ and $x = e^u$
but got no result . Please help me in proceeding.
| The first step is to turn the one integral into two as follows:
\begin{align}
\int_{0}^{1} \frac{\ln(x)}{1-x^2} \, dx = \frac{1}{2} \, \left[ \int_{0}^{1} \frac{\ln(x)}{1-x} \, dx + \int_{0}^{1} \frac{\ln(x)}{1+x} \, dx \right]
\end{align}
then using the integrals
\begin{align}
\int_{0}^{1} \frac{\ln(x)}{1-x} \, dx &= ... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the value of $ \sum\limits_{k=0}^{n-1}\binom {n-k-1}{j-1} \binom {r+k}{j+k}$?
What is the value of $$ \sum_{k=0}^{n-1}\binom {n-k-1}{j-1} \binom {r+k}{j+k}$$ where $r \ge j \ge 1$?
I know that
$$ \sum_{k=0}^{n}\binom {n-k}{m}\binom{r+k}{s} = \binom {n+r+1}{m+s+1} \text{ where }n,m \ge0,\text{ and }s\ge r\ge... | $$
\begin{align}
\sum_{k=0}^{n-1}\binom{n-k-1}{j-1}\binom{r+k}{j+k}
&=\sum_{k=0}^{n-1}\binom{n-k-1}{j-1}\binom{r+k}{r-j}
\end{align}
$$
which looks like $\binom{n+r}{r}$, but we are missing some $-j\le k\lt0$.
For example, let $n=6$, $j=3$, and $r=4$
$$
\begin{align}
\sum_{k=0}^{n-j}\binom{n-k-1}{j-1}\binom{r+k}{r-j}
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve quadratic fraction I would like to simplify the fraction
$$\frac{x^2-2x}{x^2+x-6}$$
I know from Mathematica that it should equal $\frac3{3x}$ but how do I get there?
| The numerator factors as $x^{2}-2x=x(x-2)$.
By the quadratic formula, the denominator has roots $2$ and $-3$. So the denominator factors as $x^{2}+x-6=(x-2)(x+3)$.
Hence
$\frac{x^{2}-2x}{x^{2}+x-6}=\frac{x(x-2)}{(x-2)(x+3)}=\frac{x}{x+3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$ If $a,b,c$
are the sides of a triangle and $p,q,r$ are positive real numbers then prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$
After modification. I have to prove $(a^2 p^2+b^2 q^2 + c^2 r^2) \ge pr(a^2+c^2-b^2)+qr(b^2+c^2-a^2)+pq(a^2+b^2-c^2)$
| Note that substitution $p'=p-t, q'=q-t, r'=r-t$, where $t$ is any real number leads to an equivalent form
$$a^2(p'-q')(p'-r')+b^2(q'-p')(q'-r')+c^2(r'-p')(r'-q')\ge 0.$$
This, as you proved already, is equivalent to
$$(a^2 p'^2+b^2 q'^2 + c^2 r'^2) \ge p'r'(a^2+c^2-b^2)+q'r'(b^2+c^2-a^2)+p'q'(a^2+b^2-c^2).$$
However,... | {
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$a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Let $a,b,c$ and $d$ be real numbers such that $a^4+b^4+c^4+d^4=16$. Then $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Why does this hold?
| First, none of them can be in $]-\infty,-2[$ or in $]2,+\infty[$, otherwise it would mean $a^4+b^4+c^4+d^4 > 16$
If some number (let us say $a$) is in $]-2,0[\cup]0,2[$. Then, since $a^4+b^4+c^4+d^4 = 16$ , we have $|b|,|c|,|d|<2$ and $m=\max \{|a|,|b|,|c|,|d|\} < 2$. However, since $a^5+b^5+c^5+d^5\leq m(a^4+b^4+c^4+d... | {
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Expanding $(2^b-1)\cdot(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})$ Can someone please explain to me how the following works (primarily interested in an explanation of the second step when $2^b$ is expanded?
I understand how each series cancels out to equal $2^n-1$ at the end.
$$\begin{align*}
xy&=(2^b-1) \cdot (1 + 2^b + 2^{2... | You can perhaps understand this better if you set $t=2^b$, so your product becomes
$$
(t-1)(1+t+t^2+\dots+t^{a-1})
$$
Distribute:
$$
t(1+t+t^2+\dots+t^{a-1})-(1+t+t^2+\dots+t^{a-1})
$$
Push $t$ in the first term inside the parenthesis:
$$
(t+t^2+t^3+\dots+t^{a})-(1+t+t^2+\dots+t^{a-1})
$$
Regroup:
$$
\underbrace{(t+t^2... | {
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Finding singularities and residues I am trying to solve this question.
$$\frac{(\pi- z)(z^4 -3z^2)}{\sin^{2}z}$$
The above is the given function and I am suppose to find the singular points and their corresponding residues.
My approach was as follows.
$\sin^2z = 0$ means $z = n\pi$ for $n \in \mathbb{N}$
But $z = 0$ is... | Let $z=w+n\pi$ with $n\in\mathbb{Z}$. Then $\sin^2(z)=\sin^2(w+n\pi)=((-1)^n\sin(w))^2=\sin^2(w)$. Therefore
$$f(z):=\frac{(\pi- z)(z^4 -3z^2)}{\sin^{2}z}=\frac{(\pi- (w+n\pi))(w+n\pi)^2((w+n\pi)^2 -3)}{\sin^{2}w}\\=\frac{(\pi- (w+n\pi))(w+n\pi)^2((w+n\pi)^2 -3)}{w^2(1+O(w^2))}\\
=\frac{(\pi- (w+n\pi))(w+n\pi)^2((w+n\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find all integers $x$ that can't be written as $x=a+b+c$, where $a$ divides $b$ and $b$ divides $c$ with $aFind all integers $x$ that can't be written as $x=a+b+c$, where $a$ divides $b$ and $b$ divides $c$ with $a<b<c$
This implies that $a$ divides $c$.
I can also rewrite $c=zb$ and $b=ya$ so $c=zya$ with $z\neq0$ and... | There is no solution for $x=1,2,3,4,5,6,8,12,24$ and solutions for all other $x$.
You should probably consider the case when $a=1$ first. There is a solution to $x-1=z(1+y)$ when $x-1$ is non-prime and bigger than $4$ (since $1+y>2$.)
On to the general equation.
If $X$ has an odd factor $d>5$, then $d-1=2\left(1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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$f(x+2xy)=f(x)+2f(xy)$; find $f(1992)$ given $f(1991)=a$
A function $f$ satisfies
$$f(x+2xy) = f(x) + 2f(xy)$$
for all $x,y\in\mathbb{R}$. If $f(1991)=a$, what is $f(1992)$? Give an answer dependent on $a$.
I have seen that $f(x)=x$ is solution, but then $f(1992)$ isn't dependent on $a$...
| For any $x \neq 0$, let $y = \frac{1}{2x}$. We then see that
$$ f(x + 1) = f(x) + 2f\left( \frac{1}{2} \right). $$
Thus by induction, we can prove that
$$
f(n) = f(1) + 2(n-1) f\left( \frac{1}{2} \right)
$$
for all natural numbers $n \geq 1$.
In particular,
$$
f(3) = f(1) + 4f\left( \frac{1}{2} \right).
$$
But taki... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Equal slicing of my spherical cake Recently I baked a spherical cake ($3$ cm radius) and invited over a few friends, $6$ of them, for dinner. When done with main course, I thought of serving this spherical cake and to avoid uninvited disagreements over the size of the shares, I took my egg slicer with parallel wedges(a... | Let me recast Lovsovs' answer in another perspective.
Consider at first to split an hemi-sphere into $n$ parts of equal volume.
Let designate as $h(k,n)$ the relative position of the $k$-th cut along the radius of the emisphere starting from the base circle
(corresponding to $k=0$).
$$
0 \leqslant h(k,n) \leqslant 1\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to show that the harmonic function $H(n) = 1 + \frac{1}{2} + · · · + \frac{1}{n} = O(\log n)$ using simple inequalities on fractions? How can I prove that $H(n) = 1 + \frac{1}{2} + · · · + \frac{1}{n} = O(\log n)$ using simple inequalities on fractions?
| For any $n\geq 2$ we have that
$$ \frac{1}{n}<\frac{1}{n}+\frac{1}{3n^3}+\frac{1}{5n^5}+\ldots=\text{arctanh}\frac{1}{n} = \frac{1}{2}\log\frac{n+1}{n-1}\tag{1}$$
hence we may exploit a telescopic sum:
$$\begin{eqnarray*} H_n = 1+\sum_{k=2}^{n}\frac{1}{k} &<& 1+\frac{1}{2}\sum_{k=2}^{n}\left(\log(k+1)-\log(k-1)\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Is there a quick way to find the remainder when this determinant is divided by $5$?
Find the remainder when the determinant $\begin{vmatrix} { 2014 }^{ 2014 } & { 2015 }^{ 2015 } & { 2016 }^{ 2016 } \\ { 2017 }^{ 2017 } & { 2018 }^{ 2018 } & { 2019 }^{ 2019 } \\ { 2020 }^{ 2020 } & { 2021 }^{ 2021 } & { 2022 }^{ 2022 ... | Note that, modulo $5$,
\begin{align}
2014 &\equiv-1, & 2015 & \equiv 0, & 2016 &\equiv 1,\\
2017 & \equiv2, & 2018 &\equiv-2, & 2019 &\equiv-1, \\
2020 &\equiv 0, & 2021 &\equiv 1, & 2022 & \equiv2.
\end{align}
Furthermore $2$ and $-2$ have order $4$ modulo $5$, so the determinant is congruent to
$$\begin{vmatrix}
1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove or disprove $\sum\limits_{cyc}\frac{x^4+y^4}{x+y}\le 3\frac{x^4+y^4+z^4}{x+y+z}$
Let $x,y,z\ge 0$. Prove or disprove
$$\dfrac{x^4+y^4}{x+y}+\dfrac{z^4+y^4}{z+y}+\dfrac{z^4+x^4}{x+z}\le 3\dfrac{x^4+y^4+z^4}{x+y+z}$$
This is what I tried. Without loss of generality, let $x+y+z=1$, then
$$\Longleftrightarrow \su... | Something different than brute force would be the usage of Schur's inequality and your transformation of $x^4+y^4=(x+y)(x^3+y^3)-xy(x+y)^2+2x^2y^2$
$$\sum_{cyc}\frac{x^4+y^4}{x+y}\le 3\frac{x^4+y^4+z^4}{x+y+z}$$
$$(x+y+z)\sum_{cyc}\frac{x^4+y^4}{x+y}\le 3(x^4+y^4+z^4)$$
$$\sum_{cyc}(x^4+y^4+\frac{(x^4+y^4)z}{x+y})\le 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find the square roots of $z = 5-12i$ I am asked the following question:
Find the square roots of $z = 5-12i$
I know that this problem can be easily solved by doing the following:
$$
z_k^2 = 5-12i\\
(a+bi)^2 = 5-12i\\
(a^2-b^2) + i(2ab) = 5-12i\\
\\
\begin{cases}
a^2 - b^2 = 5\\
2ab = -12
\end{cases} \quad \Rig... | All you need to do is to convert your $z$ to polar form:
$$z=5-12i=\sqrt{12^2+5^2}e^{i(\arctan(-\frac{12}{5})+k\pi)}=13e^{i(-1.176+k\pi)}$$
Hence, one answer of $z^{\frac{1}{2}}$ is
$$z_1=13^{\frac{1}{2}}e^{i\frac{-1.176}{2}}=\sqrt{13}e^{i(-0.588)}$$
Converting this to Cartesian will give:
$$z_1=\sqrt{13}\left(\cos(-0.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find a derivable function $f$ for which $f(x) - f(x-1) =\alpha ( f(x-1) - f(x-2) )$ Find a derivable function $f$ for which $f(x) - f(x-1) =\alpha ( f(x-1) - f(x-2) )$.
My initial conditions would be:
$$\begin{align*}
f(0) &= 0\\
f(1) &= \beta
\end{align*}$$
and $\alpha < 1$
and my domain $[0,+\infty[$
Basically, if lo... | Starting from your equation
$$
f(x) - f(x - 1) = \alpha \left( {f(x - 1) - f(x - 2)} \right)
$$
put
$$
g(x) = f(x) - f(x - 1)
$$
from which you get the general solution for $g(x)$
$$
g(x) = \alpha g(x - 1)\quad \Rightarrow \quad g(x) = \alpha ^{\,x} + c
$$
with $c$ being a constant (actually, any periodic function of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the smallest real value of $x$ I don't know why my answer is different from the answer sheet.
Find the smallest real value of $x$ that satisfies the equation: $(x+5)(x^2-x-11)=x+5$
Here is what I did. This equation can be rewritten as $(x+5)(x^2-x-12)=(x+5)(x-4)(x+3)=0$, which give $$(x+5)=0$$ or $$(x-4)=0$$ or... | If $x \neq -5$ then divide by $x+5$ and have
$$x^2 - x - 11 = 1$$
$$x^2 - x - 12 =0$$
Solve it:
$$x = \frac{1\pm\sqrt{1 + 48}}{2}$$
Hence you get two solutions:
$$x_1 = +4 ~~~~~~~ x_2 = -3$$
So.....
In any case
$x = -5$ solves the equation too.
So.....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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