Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding the infinite Sum of a series: $\sum\frac1{n(n+1)(n+2)}$ Find the infinite Sum of the series with general term $\frac{1}{n(n+1)(n+2)}$.
I decomposed the fraction upto this $1/(2n)-1/(n+1)+1/(2n+4)$. But I find no link about cancelling terms. So how should I find the infinite Sum?
Please don't say that this is o... | Notice, use partial fractions in form of difference of two terms as follows $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$$$=\sum_{n=1}^{\infty}\frac 12\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)$$
$$=\small \frac 12\lim_{n\to \infty}\left(\left(\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}\right)+\left(\frac{1}{2\cdot 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluate the sum $\sum_{k=1}^\infty \big(\frac{1}{36k^2-1}+\frac{2}{(36k^2-1)^2}\big)$ I have to evaluate the series:
$$\sum_{k=1}^\infty \left(\frac{1}{36k^2-1}+\frac{2}{(36k^2-1)^2}\right).$$
I tried using the identity $\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$ but I got stuck.
| A more general approach, proving the identity:
$$ \sum_{k\geq 1}\frac{k^2+x^2}{(k^2-x^2)^2}=-\frac{1}{2x^2}+\frac{\pi^2}{2\sin^2(\pi x)}\tag{1}$$
for any $x\in(-1,1)$ is to check that both the LHS and the RHS of $(1)$ are meromorphic functions with the same-behaving singularities at any $x\in\mathbb{Z}\setminus\{0\}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the solutions of the diophantine equation $(x^2-y^2)(z^2-w^2)=2xyzw$ Let $x,y,z,w$ be postive integers. Find all solutions of:
$$(x^2-y^2)(z^2-w^2)=2xyzw$$
This gives:
$$\left(\dfrac{x}{y}-\dfrac{y}{x}\right)\left(\dfrac{z}{w}-\dfrac{w}{z}\right)=2$$
$$\left(p-\dfrac{1}{p}\right)\left(q-\dfrac{1}{q}\right)=2$$
| There are NO solutions in rationals or integers.
The equation
\begin{equation*}
\left( p-\frac{1}{p} \right) \left( q-\frac{1}{q} \right) =2
\end{equation*}
is equivalent to the quadratic
\begin{equation*}
p^2 + \frac{2q}{1-q^2} p - 1=0
\end{equation*}
For this to have rational solutions the discriminant must be a rati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find an equation of the tangent line to $y = \cos(x)+3\sin(x)$ at $x=\pi/3$
Find an equation of the tangent line to
$$y = \cos(x)+3\sin(x)$$ at $x=\pi/3$.
This is what I have done...
Find $y$, $y= \cos(\pi/3) + 3\sin(\pi/3)$
this equals $1 + \sqrt 3/2$
Next
Find $f'(x) = \sin(\pi/3) + 3\cos(\pi/3)$
this equals $3+... | While your strategy is correct, you incorrectly evaluated the sine and cosine functions at $x = \pi/3$ and incorrectly took the derivative.
Let $f(x) = \cos x + 3\sin x$. Since
$$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$
and
$$\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$
when we evaluate $f(x)$ at $\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$.
The term inside the bracket is divisible by 6 since we have assumed t... | Continuing from your start, let $k^3+6k^2+5k=6p$ for some $p\in\mathbb Z$.
Now,
$$(k+1)(k+2)(k+6)=(k^3+6k^2+5k)+3k^2+15k+12=6p+3k^2+15k+12=6p+3(k^2+5k+4)=6p+3(k+1)(k+4)$$
Case 1: $k$ is even.
Then, $k+4$ is also even. We can thus write $k+4=2q$ for some $q\in\mathbb Z$.
Hence,
$$6p+3(k+1)(k+4)=6p+3(k+1)(2q)=6p+6q(k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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The number and amount of dividers - a power of two For a positive integer $n$ is known that the sum of all divisors of that number is a power of $2$. Prove that the number of these divisors is also a power of $2$.
My work so far:
Several of these numbers I found
Let $\tau(n) -$ the number of divisors of $n$
1) $n=3; 1+... | The first two divisor functions of a number with prime factorization $n=\prod_ip_i^{a_i}$ are
$$
\sigma_0(n)=\prod_i(a_i+1)
$$
and
$$
\sigma_1(n)=\prod_i\left(1+p_i+\cdots+p_i^{a_i}\right)\;.
$$
Since $\sigma_0$ is clearly a power of $2$ if and only if the $a_i+1$ are, we need to show that $a_i+1$ is a power of $2$ if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Solving differential equation - how to find inhomogeneous solution Given the following
$\ y'= \frac{3y^2-x^2}{2xy}$
I need to tell if the equation is linear, which I think it is because:
$\ y'= \frac{3y^2-x^2}{2xy} = \frac{3y}{2x}-\frac{x}{2y}$
Now I need to solve the equation with separation of variables which is only... | $$y'(x)=\frac{3y(x)^2-x^2}{2xy(x)}\Longleftrightarrow$$
$$y'(x)-\frac{3y(x)}{2x}=-\frac{x}{2y(x)}\Longleftrightarrow$$
$$2y(x)y'(x)-\frac{3y(x)^2}{2x}=-x\Longleftrightarrow$$
Let $r(x)=y(x)^2$, which gives $r'(x)=2y(x)y'(x)$:
$$r'(x)-\frac{3r(x)}{x}=-x\Longleftrightarrow$$
Let $v(x)=\exp\left[\int-\frac{3}{x}\space\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Line $mx + ny = 3$ is normal to the hyperbola $x^2 – y^2 = 1$ If the line $mx + ny = 3$ is normal to the hyperbola $x^2 – y^2 = 1$, then evaluate $\frac{1}{m^2}+\frac{1}{n^2}$.
I compared given equation of normal to equation of normal at parametric point i.e
$x\sin\theta+y=2\tan\theta$ and obtained $\frac{1}{m^2}-\frac... | There must be a flaw in the text, expression $\frac{1}{m^2}+\frac{1}{n^2}$, should have been $\frac{1}{m^2}-\frac{1}{n^2}$.
Let us see why:
First of all, I confirm the expression given by @Mathematician :
$$\frac{1}{m^2}-\frac{1}{n^2}=\frac{4}{9} \ \ \ (1)$$
Proof below.
In fact, as the question is formulated, the only... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve integral $ \int \frac{dx}{( \sin x + 2 \cos x )^3}$ Alright, so this one has been giving me nightmares for a couple of days now... I've tried the substitutes for $u = \tan \frac{x}{2}$ where you can substitute
sin(x), cos(x) and dx accordingly and pretty much every possibility and combination that came to mind w... | As what randomgirl did by auxiliary angle, I am going to complete the solution for reference.
Letting $\alpha = \arctan (\frac{1}{2})$ changes the denominator into $$
\sin x+2 \cos x=\sqrt{5} \cos (x-\alpha)
$$
which convert the integral into
$$
\begin{aligned}
I &=\int \frac{d x}{(\sin x+2 \cos x)^{3}} \\
&=\int \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $p^n+12^2=m^2$ Find all triples $(m,n,p) \in \mathbb{N}^3$, with $p$ prime, which satisfy $$p^n+12^2=m^2$$
| I have rewritten it as
$$ p^n = (m+12)(m-12) $$
Then there exists $\ 0 \le a \le n\ $ such that $\ p^a=m-12$. It follows that
$ p^n = p^a (p^a+24) $ and then $$ p^a(p^{n-2a}-1)= 24 = 2^3 \cdot 3 $$
We have the following cases:
*
*$\; p^a=1$ and then $a=0$ and $p^n=25$, which yields $p=5$, $n=2$ and $m=13$
*$\; p^a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Taylor series of $\ln{\sqrt[4]{\frac{x-2}{5-x}}}$ to $o((x-x_0)^n)$ when $x_0 = 3$ Well I have tried to get it as
$$f(x) = f(x_0) + \frac{f'(x_0)(x-x_0)}{1!} + \frac{f''(x_0)(x-x_0)^2}{2!} + ... + o((x-x_0)^n)$$
and got wrong results:
First: $$f'(x) = \frac{3}{4(x-2)(5-x)}$$
Second: $$f''(x) = \frac{3(2x-7)}{4(x-2)^2(... | First of all, set $x-3=t$ so the computations will possibly be easier; then note that $x=t+3$ and that
$$
\ln{\sqrt[4]{\frac{x-2}{5-x}}}=
\frac{1}{4}\ln(1+t)-\frac{1}{4}\ln(2-t)=
\frac{1}{4}\ln(1+t)-\frac{\ln{2}}{4}-
\frac{1}{4}\ln\left(1-\frac{t}{2}\right)
$$
Now just use the series expansion of $\ln(1+z)$ around $z=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the laurent series for $\frac{1}{z(z-2)^2}$ centered at z=2 and specify the region in which it converges. My attempt:
$$\frac{1}{z(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{A}{z}+\frac{B}{z-2}+\frac{C}{(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{(1/4)}{z}+\frac{(-1/4)}{z-2}+\frac{(1/2)}{(z-2)^2}$$
This is where I get stu... | HINT:
Write $\frac1z$ as
$$\frac1z=\frac{1/2}{1+(z-2)/2}$$
Then, recall the sum of a geometric series.
SPOILER ALERT: Scroll over the highlighted area to reveal the solution
$$\begin{align}\frac1z&=\frac{1}{2+(z-2)}\\\\&=\frac{1/2}{1+(z-2)/2}\\\\&=\frac12 \sum_{n=0}^\infty (-1)^n\left(\frac{z-2}{2}\right)^n\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What is the fast way to evaluate the following integral: $\int{\frac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$? I am trying to evaluate the following integral:
$$\int{\dfrac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$$
I tried the trigonometric substitution: $u = \tan(x)$. Generally, The whole integral needs two substitutions: $u = \tan(x)$... | For clarifying the sign of the antiderivative, I decide to integrate it by cases:
A. When $x>0,$
$$
\begin{aligned}
I &=\int \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} d x \\
&=-\frac{1}{2} \int \sqrt{1+\frac{1}{x^{2}}} d\left(1+\frac{1}{x^{2}}\right) \\
&=-\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{\frac{3}{2}}+C\\&=-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the first few Legendre polynomials without using Rodrigues' formula If a polynomial is given by $$y=\color{red}{a_0\left[1-\frac{l(l+1)}{2!}x^2+\frac{l(l+1)(l-2)(l+3)}{4!}x^4-\cdots\right]}+\color{blue}{a_1\left[x-\frac{(l-1)(l+2)}{3!}x^3+\frac{(l-1)(l+2)(l-3)(l+4)}{5!}x^5-\cdots\right]}\tag{1}$$ where $l$ is a co... |
We consider for even $l$ the polynomial (OPs red part)
\begin{align*}
a_0^{(l)}\left[1-\frac{l(l+1)}{2!}x^2+\frac{l(l+1)(l-2)(l+3)}{4!}x^4-\cdots\right]\tag{1}
\end{align*}
and for odd $l$ the polynomial (OPs blue part)
\begin{align*}
a_1^{(l)}\left[x-\frac{(l-1)(l+2)}{3!}x^3+\frac{(l-1)(l+2)(l-3)(l+4)}{5!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Inequality using Cauchy-Schwarz Let $a,b,c\in\mathbb{R}^+$, prove that $$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\leq \sqrt{\frac{3}{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)} $$
Hi everyone, I've been trying to do this exercise but any method that I tried has failed. First I tried to use... | I finally understood what tong_nor meant so I want to write it in details.
So treat this answer as a more detailed version of the tong_nor answer.
The Cauchy-Schwarz inequality gives us that:
(1) $(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2) \geq (a_1b_1 + a_2b_2 + a_3b_3)^2$
If we put here $a_i = \sqrt {a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Evaluation of $\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}dx$
Evaluation of $$\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}dx$$
$\bf{My\; Try::}$ We can write $$x\cos x+\sin x= \sqrt{1+x^2}\left\{\frac{x}{\sqrt{1+x^2}}\cdot \cos x+\frac{1}{\sqrt{1+x^2}}\cdot \sin x\right\}$$
So we get $$(x\cos x+\sin x) = \sqrt{1+x... | Your integral becomes $$I=\int { \frac { \left( 2+{ x }^{ 2 } \right) }{ \left( 1+{ x }^{ 2 } \right) } { \left( \sec { \left( x-\cot ^{ -1 }{ x } \right) } \right) }^{ 4 } } dx$$
On substituting $$\left( x-\cot ^{ -1 }{ x } \right) =t$$
diffrentiating $$dt = \frac { \left( 2+{ x }^{ 2 } \right) }{ \left( 1+{ x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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"answer_id": 0
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How do you calculate the area of the intersection between a rectangle and a doughnut? I'm dealing with an engineering problem, involving concentric pipes, with air flowing through the outer pipe (doughnut).
I need a cross-beam to support the inner pipe, so I need to calculate how much of the outer doughnut's area will ... | The equations for the upper semicircles of the small pipe and large pipe are $f(x) = \sqrt{r^2 - x^2}$ and $g(x) = \sqrt{R^2 - x^2}$ respectively. The area of the cross-beam is then found by integrating $g(x) - f(x)$ over the interval $[-\frac{w}{2}, \frac{w}{2}]$:
\begin{align*}
\int_{-w/2}^{w/2} (g(x) - f(x))\,dx &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is
A. 0;
B. 3;
C. 5;
D. 1.
I don't know how to solve this.
| If $x \ge 0$ the $x^5 + 2x^3 + x^2 + 2 \ge 2$ so there are no solutions for $x \ge 0$.
If $x = -humongouseffinglargenumber$ then $x^5 + 2x^3 + x^2+2 < 0$ so somewhere between $-humongouseffinglargenumber$ and $0$ there must be an $x$ where the term is $0$. Let's try to narrow that range down a little.
As we are only l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1700615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 3
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Prove that $1+\frac12+\frac13+\cdots+\frac1{2^n} \ge 1+\frac{n}{2}$ Is it true that $1+\dfrac12+\dfrac13+ \dots +\dfrac{1}{2^n} \geq 1+\dfrac{n}{2}$?
If so, a proof would be great!
Thank you!
| We have
$$1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\cdots$$
$$\geq 1+\frac{1}{2} + \frac{1}{4} + \frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\cdots$$
$$=1+\frac{1}{2} + \frac{1}{2} + \frac{1}{2}+\cdots$$
$$=1+\frac{n}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Given the following relation between the angles of a triangle, find the number of possible triangles. If $A,B,C$ are the interior angles of triangle $\Delta ABC$ such that
$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2=9$, then the number of possible triangles is
$(A)0\hspace{1cm}(B)1\hspace{1cm}(C)3\hspace{1cm}(D)... | $$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2=9$$
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B+2\cos B\cos C+2\cos C\cos A+\sin^2A+\sin^2B+\sin^2C+2\sin A\sin B+2\sin B\sin C+2\sin C\sin A=9$$
$$3+2(\cos A\cos B+\cos B\cos C+\cos C\cos A+\sin A\sin B+\sin B\sin C+\sin C\sin A)=9$$
$$\cos A\cos B+\cos B\cos C+\cos C\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The graph of the equation $x+y=x^3+y^3$ is the union of The graph of the equation $x+y=x^3+y^3$ is the union of
$(A)$line and an ellipse$(B)$line and a parabola$(C)$line and hyperbola$(D)$line and a point
I tried to factorize the given equation.
$x^3-x+y^3-y=0$
$x(x^2-1)+y(y^2-1)=0$
The answer given is a line and an e... | $$x+y=x^3+y^3=(x+y)(x^2-xy+y^2)$$$$\therefore (x+y)(x^2-xy+y^2-1)=0$$Hopefully you can conclude from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding a General Solution to a Nonhomogeneous Matrix Equation I've come across a question in which I've been asked to find the general solution to the matrix equation:
$\begin{bmatrix}1 & -2 & 1\\-2 & 4 &-2\\1 & -2 & 1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\-16\\8\end{bmatrix}$
by first ... | Notice the column structure of the target matrix:
$$
\mathbf{A}=
%
\left[
\begin{array}{rrr}
1 & -2 & 1 \\
-2 & 4 & -2 \\
1 & -2 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{rrr}
c_{1} & -2 c_{1} & c_{1}
\end{array}
\right].
$$
We have one essential column. The rank plus nullity theorem reveals there will... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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There are $5$ women, $3$ men. How many ways to form a committee of $3$ with at least $1$ member of the opposite sex? I have looked through several topics for similar solutions and I have attempted an answer to the question. Unfortunately, the sample question itself does not have an answer.
From $5$ women and $3$ men, ... | One more formulation. From the number of all possible committees, substract the number of committees with only men and the number of committees with only women. That is :
$$
\binom{5+3}{3}-\binom{5}{3}-\binom{3}{3}=56-10-1=45
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Is there a simple way of proving that $\lfloor\sqrt{n}\rfloor+\lfloor\sqrt{4n+1}\rfloor = \lfloor\frac{3}{2} \lfloor \sqrt{4n+1} \rfloor\rfloor$? It appears that $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor = \left\lfloor\frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor\right\rfloor$$ for a... | Following on from your comment, suppose that
$$\left\lfloor\sqrt{n}\right\rfloor<\left\lfloor\sqrt{n+\frac14}\right\rfloor\;,$$
and let $m=\left\lfloor\sqrt{n+\frac14}\right\rfloor$. Then
$$\sqrt{n}<m\le\sqrt{n+\frac14}\;,$$
so
$$n<m^2\le n+\frac14\;.$$
Since $n$ and $m^2$ are integers, this is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1709105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Complex numbers- elementary question How can I calculate the argument of the complex number $z= (\frac{1}{2}+ \frac{i\sqrt{3}}{2}) \cdot (1+i)$?
I always get $\tan^{-1}(-2-\sqrt{3})$, but the book answer is $7 \pi/12$.
| I) Tedious: $z= (\frac{1}{2}+ \frac{3^{1/2}i}{2}) * (1+i)= (1/2 - \sqrt{3}/2) + (1/2 + \sqrt 3/2) i = $. Arg($z$) = $\arctan (1+ \sqrt{3})/(1-\sqrt{3}) = -5\pi/12 = 7\pi/12$.
(To be fair, I made data entry errors the first six times I tried to enter this into a calculator.)
II) Easier (albeit it notation intimidatin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplifying the integral $\int\frac{dx}{(3 + 2\sin x - \cos x)}$ by an easy approach $$I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}$$
If $$\tan\left(\frac{x}{2}\right)=u$$
or $$x=2\cdot\tan^{-1}(u)$$
Then,
$$\sin{x}=\dfrac{2u}{1+u^2}$$
$$\cos{x}=\dfrac{1-u^2}{1+u^2}$$
$$dx=\dfrac{2}{1+u^2}$$
Substitute $$\tan\l... | With $a=\tan^{-1}\frac1{\sqrt5}$ and $t=\frac12(x+a)$
\begin{align}
\int\frac{dx}{3 + 2\sin x - \cos x}
& = \int \frac{dx}{3-\sqrt5\cos(x+a)}\\
&= \int \frac{2d(\tan^2t)}{(3-\sqrt5)+(3+\sqrt5)\tan^2t}=\tan^{-1} \frac{2\tan t}{3-\sqrt5} +C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find $\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$$
What I have done is to take the $\ln$
$$e^{\lim_{x \to 0}\ln\left(\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}\right)}$$
$$y={\lim_{x \to 0}{\frac{1}{x^2}}\cdot \ln\f... | We can proceed using the expansion $\cos(x)=1-\frac12 x^2+O(x^4)$.
First note that
$$\frac{\cos(x)}{\cos(2x)}=1+\frac32 x^2+O(x^4)$$
so that the limit of interest becomes
$$\begin{align}
\lim_{x\to \infty}\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}&=\lim_{x\to \infty}\left(\left(1+\frac{3/2}{1/x^2}\right)^{1/x^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find polynomials $f(x),g(x)$ such that $f(x)p(x) + g(x)q(x) = 2x^2 + 6x +2$ Let $$p(x) = x^4 + 7x^3 + 14x^2 + 7x+1 $$$$q(x) = x^4 +10x^3 + 23x^2 + 10x+1$$
Find polynomials $f(x),g(x)$ with rational coefficients such that $$f(x)p(x) + g(x)q(x) = 2x^2 + 6x +2$$
I totally have no idea to solve this problem... Please help... | You can use Grobner basis to solve this,
$f1=x^4+7x^3+14x^2+7x+1;$,
$f2=x^4+10x^3+23x^2+10x+1;$
$f3=(f2-f1)/3=x^2+3x^2+x$
$f4=-3(f2-(x+7)f3)=x^2+3x+1$
Since,
$2x^2+6x+2 = 2*f4$
Now just replace it as $f4->f3->f2$ till everything is in terms of $f1$ and $f2$,
Answer: $2/3(-(x+4)f2+(x+7)f1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim_{x \to 3} \sqrt{x+1} = 2$ Prove that $\displaystyle\lim_{x \to 3} \sqrt{x+1} = 2$
Attempt:
$0 < |x - 3| < \delta \Rightarrow |\sqrt{x+1} - 2| < \epsilon$
Well $|\sqrt{x+1} - 2| = |(\sqrt{x+1} - 2) \cdot \displaystyle\frac{\sqrt{x+1} + 2}{\sqrt{x+1}+2}| = |\frac{x-3}{\sqrt{x+1} + 2}| = |x-3| \cdot \fra... | Your derivation is correct (I believe, it looks right but I didn't check every detail), but you are going for too much. To prove the limit statement, you don't need to identify specifically the largest $\delta$ that works for each $\epsilon$. You just need to prove there is some positive $\delta$ that will work. And yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
tricky system of trigonometric equations I am not very fresh in math, but I need to solve this system:
\begin{gather}
A\sin(x-y)+B\sin(z-y)=C\\
A\cos(x-y)+B\cos(z-y)=D
\end{gather}
where $A,B,C,D$ and $x$ are given.
I tried to expand and combine the bracket terms and I suppose that there are some tricky substitutions t... | Let $x-y=U, z-y=V$. Then
$$A\sin U+B \sin V=C$$
$$A\cos U+B \cos V=D$$
Square both equations:
\begin{align}
A^2\sin^2 U + B^2 \sin^2 V + 2 AB \sin U \sin V&=C^2\\
A^2\cos^2 U + B^2 \cos^2 V + 2AB \cos U \cos V &=D^2
\end{align}
Add to get
\begin{align}
A^2(\sin^2 U + \cos^2 U) + B^2 (\sin^2 V + cos^2 V) + 2 AB (\sin U... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$ While I know that $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^{2}}{6}$$
But trying to evaluate this has left me stumped
$$\sum_{n=1}^{\infty} \frac{1}{n^2+1}$$
I evaluated it through wolfram alpha, it gave me $\frac{1}{2}(\pi\coth(\pi)-1)$.
What would be a good way t... | This sum is harder, and perhaps less natural, to find than the sum for $\sum \frac{1}{n^2}$. We use Fourier series of $e^{x}$ with respect to $\{e^{i nx} / 2\pi\}$and apply Parseval's theorem.
Note that we can use this method of evaluation to find $\sum \frac{1}{n^2}$, too! We do the same for the (simpler) function $f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 5
} |
Find the roots of quadratic polynomial given one root of another quadratic polynomial? if $a,b,c$ are Real numbers and $1$ is a root of equation $ax^2+bx+c=0$ then curve $y = 4ax^2+3bx+2c$ , (a is not zero) intersects $x$ axis at how many points?
I get a relation $a+b+c = 0$ I tried substituting this in given polynomia... | If $1$ is a root of $ax^2 + bx + c = 0$ Then
\begin{align}
ax^2 + bx + c
&= a(x-1)(x - \lambda) \\
&= ax^2 - a(1 + \lambda)x + a\lambda \\
\end{align}
So $b = -a(1+\lambda)$ and $c = a\lambda$.
The descriminant of $4ax^2 + 3bx + 2c$ is
\begin{align}
9b^2 - 32ac
&= 9a^2(1+\lambda)^2 -32a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Steps to solve $\lim_{n \to \infty} (\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} ) = 4 $ $$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} \right) = 4 $$
I think $\sqrt{n^4+1} - \sqrt{n^4-1}$ is approaching to zero, but it is not correct. What steps can evaluate above limit t... | Using jim's answer, by Taylor $$\sqrt{n^4+1}=n^2+\frac{1}{2 n^2}-\frac{1}{8 n^6}+O\left(\frac{1}{n^{9}}\right)$$ $$\sqrt{n^4-1}=n^2-\frac{1}{2 n^2}-\frac{1}{8 n^6}+O\left(\frac{1}{n^{9}}\right)$$ $$\sqrt{n^4+1}-\sqrt{n^4-1}=\frac{1}{n^2}+O\left(\frac{1}{n^{9}}\right)$$ Now, multiplying by $(2n+1)^2$, you get for the wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Establish the inequality $\frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} \geq (1-\frac{2\Omega}{N})^{\sqrt{N}}$ Establish the inequaity
$$ \frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} \geq (1-\frac{2\Omega}{N})^{\sqrt{N}}$$ where $N > \Omega > \sqrt{N}$ and $N$ is a perfect-square.
My attempt:
$$ \fra... | According to the comments after the question, we assume $\sqrt{N}<Q<\frac{N}{2}$.
As you started,
$$
\dfrac{\binom{N-\sqrt{N}}{\Omega}}{\binom{N}{\Omega}}
= \prod_{k=0}^{\Omega-1} \frac{N-\sqrt{N}-k}{N-k}
= \prod_{k=0}^{\Omega-1} \left(1-\dfrac{\sqrt{N}}{N-k}\right) >
\\
> \left(1-\frac{\sqrt{N}}{N-Q}\right)^{\Omega}
>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Summation $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+....$ I came across a question today...
Q. The sum $\dfrac{3}{1^2}+\dfrac{5}{1^2+2^2}+\dfrac{7}{1^2+2^2+3^2}+....$ upto $11$ terms is?
Okay, I think it can be written as
$$\sum_{r=1}^{11}\dfrac{2r+1}{1^2+2^2+...+r^2}$$
I can't see how to simplify it.
An... | After the help provided by @lab bhattacharjee, I would like to complete the answer.
$$6\sum_{r=1}^{11}\left(\dfrac{r+1-r}{r(r+1)}\right)$$
$$=6\sum_{r=1}^{11}\left(\dfrac{1}{r}-\dfrac{1}{r+1}\right)$$
$$=6\left[ \left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Unable to derive reason/formula for permutation problem What is the probability of $n$ preceding $1$ and $n$ preceding $2$ when we randomly
select a permutation of ${1, 2, . . . , n}$ where $n ≥ 4$?
I wrote out examples of n! when n equals some number like $3$ and $4$, and I see that the answer should be $\frac{1}{3}$,... | The result holds for all $n \geq 3$.
What matters here is that of the $3! = 6$ permutations of $1, 2, n$, two of them have $n$ appearing before both $1$ and $2$. The placement of the other $n - 3$ numbers in the sequence has no effect on whether $n$ appears before both $1$ and $2$. Hence, the probability that in a p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How can I solve this using prime factors? I'm stuck with this problem: $2^x \cdot 3^3 \cdot 26^y = 39^z$ for $x, y, z \in \mathbb{N}$.
I know that there isn't a natural solution for the equation, but I need to "prove" it using prime factors. I'm stuck here:
$$
2^x \cdot 3^3 \cdot (2\cdot 13)^y = (3\cdot 13)^z \\
\impli... | Assuming "natural solutions" means positive integers:
$2 \times 27 \times 26 = 702 = 2^2 \times 3^3 \times 13$ while $39 = 3 \times 13$. Therefore $x = y = z = 1$ is not a solution. In fact, no solution will work because the prime factorization of $2^x 3^3 26^y$ will always have that pesky oddly even prime while the fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integration by Parts with a logarithm Im not sure how i go about integrating a function being:
$$\int^{\sqrt{2}}_{1} r^{3}\log({r^2}) dr$$
I assume you use parts but it keeps making me go round in circles.
Any help would be appreciated.
My attempt was:
$$u=log(r^{2}) \rightarrow u'=\frac{2}{r},$$ $$v=\frac{r^4}{4} \rig... | $$\int_{1}^{\sqrt{2}}r^3\ln(r^2)\space\text{d}r=$$
Integrate by parts, $\int f\space\text{d}g=fg-\int g\space\text{d}f$ where:
$$f=\ln(r),\text{d}g=r^3\space\text{d}r,\text{d}f=\frac{1}{r}\space\text{d}r,g=\frac{r^4}{4}$$
$$\left[\frac{r^4\ln(r)}{2}\right]_{1}^{\sqrt{2}}-\frac{1}{2}\int_{1}^{\sqrt{2}}r^3\space\text{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the sum of series $\sum\limits_{n=2}^{+\infty}\frac{(-1)^n}{n^2+n-2}$. Using the power series $\sum\limits_{n=2}^{+\infty}\frac{1}{n^2+n-2}x^{n}$, the first derivative gives $\sum\limits_{n=2}^{+\infty}\frac{n}{n^2+n-2}x^{n-1}$ where the sum doesn't exist.
Integration of function $\frac{1}{n^2+n-2}x^n$ given $\fra... | Decompose $n^2+n-2$ as $(n+2)(n-1)$ and then do: $$\frac{1}{n^2+n-2}=\frac{A}{n+2}+\frac{B}{n-1} \Rightarrow B=-A=\frac{1}{3}$$ And from this you can make the sum (just putting 1/3 out of the same as it is a common factor): $$\frac{1}{3}\left(\sum_{n=2}^{\infty}(-1)^n \left(\frac{1}{n-1} - \frac{1}{n+2}\right)\right) $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simple differential equation $\frac{x}{y}=\frac{y'}{x+1}$ $$\frac{x}{y}=\frac{y'}{x+1}$$
The solution to this is very easy, we just multiply both sides by
$$(x+1)y\ \mathrm dx$$
Then we have
$$(x^2+x)\ \mathrm dx=y\ \mathrm dy$$
Then we integrate both sides to get solution
$$\frac{x^3}{3}+ \frac{x^2}{2}=\frac{y^2}{2... | From the form of the DE given at the start it is implied that $x=−1$ and $y=0$ are excluded from the domain and range respectively. If you had an an initial condition this has more impact on your question.
Suppose you had the initial condition that $y\left(\frac{3}{2}\right)=\frac{3}{2}$. This initial condition will re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Median BM of triangle ABC two results Given
*
*Calculate the measure of the median $\overline{BM}$ of ABC triangle, given A (-6.1); B (-5,7) and C (2,5)
I get this result:
$Xm = \frac{Xc - Xa}{2} + Xa$
$Xm = \frac{2-(-6)}{2} + (-6) = 4 - 6 = -2$
$Ym = \frac{(Yc - Ya)}{2} + Ya$
$Ym = \frac{5-1}{2} + 1 = 2 + 1 = 3$... | First, both you and someone else seem to be using $B(-5,7)$, not $B(5,7)$.
The "someone else" appears to be computing the midpoint between some other point $B(-5,7)$ and $C(2,5)$, and you are computing the midpoint between $A(-6,1)$ and $C(2,5)$.
If you are really trying to compute the length $BM$ where $M$ is the mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Define the image of the function $f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$ Question
Define the range of the function:
$$f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$$
Answer attempt
I assume that the domain for $f(x)$ is in ${\rm I\!R}$.
If we draw a triangle with the sides: $1, x$ and $\sqrt{1+x^2}$, with the ... | Since $-1\le 2x/(1+x^2)\le 1$ for every $x\in\mathbb{R}$, the domain of $f$ is indeed the whole real axis. Moreover
$$
\lim_{x\to-\infty}f(x)=-\pi,\qquad
\lim_{x\to\infty}f(x)=\pi
$$
Let's compute the derivative:
$$
f'(x)=\frac{2}{1+x^2}+\frac{1}{\sqrt{1-\frac{4x^2}{(1+x^2)^2}}}
\frac{2(1+x^2)-4x^2}{(1+x^2)^2}
$$
A sim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute $\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$ I have to compute $$\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$$ I know that $\sinh t$ can be represented as a series1, but for t... | \begin{align*}
\sqrt{t} \sinh \sqrt{t} &= \sum_{n=1}^{\infty} \frac{t^{n}}{(2n-1)!} \\
(\sqrt{t} \sinh \sqrt{t})' &=\sum_{n=1}^{\infty} \frac{nt^{n-1}}{(2n-1)!} \\
t(\sqrt{t} \sinh \sqrt{t})' &=\sum_{n=1}^{\infty} \frac{nt^{n}}{(2n-1)!} \\
[t(\sqrt{t} \sinh \sqrt{t})']' &=
\sum_{n=1}^{\infty} \frac{n^{2}t^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Prove $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$ Prove that $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$
$$\frac{2\sec\... | It is possible using the Rule $ P = n p, Q = n q $, $n$ any real number
and recognizing
$$ \frac{1-\cos\theta}{\sin\theta}= \frac{\sin\theta}{1+\cos\theta}\tag{0} ,$$
$$ \frac {u}{v} = \frac {p}{q} = \frac {P}{Q}=\frac {u \pm P}{v \pm Q}, $$ and entirely avoid all trigonometric calculations.
Multiply numerator and d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding $\int \frac{dx}{a+b \cos x}$ without Weierstrass substitution. I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ without using Weierstrass substitution, which is the usual technique.
When $a,b=1$ we can just multiply the numerator and denominator... | Another way to get to the same point as C. Dubussy got to is the following:
\begin{align*}
\frac{1}{a + b \cos x} &= \frac{1}{a \left (\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} \right ) + b \left (\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} \right )}\\
&= \frac{1}{(a - b) \sin^2 \frac{x}{2} + (a + b) \cos^2 \frac{x}{2}}\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
} |
Equation arrangement. How to arrange $x=\frac{1}{2} \ln({1+\frac{10}{x}})$ to $x=\frac{10}{e^{2x}-1}$ ?
Can anyone give me step by step explanation? Thanks in advance.
| $x=\frac{1}{2} \ln(1+ \frac{10}{x}) \implies 2x= \ln(1+ \frac{10}{x}) \implies
e^{2x} = 1+ \frac{10}{x} \implies e^{2x} - 1 =\frac{10}{x} \implies x= \frac{10}{ e^{2x} - 1}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
| The partial sum decomposition of the term of your serie is : $\frac{\frac{1}{3}}{n}-\frac{\frac{1}{3}}{n+3}$. You recognise telescoping series : $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}=\sum_{n=1}^{\infty} \frac{\frac{1}{3}}{n}-\frac{\frac{1}{3}}{n+3}=\frac{\frac{1}{3}}{1}+\frac{\frac{1}{3}}{2}+\frac{\frac{1}{3}}{3}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Solve the equation $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$ Solve the following equation: $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$
Unfortunately I have no idea.
| First notice that $x$ must be in $[-1,1]$. Then rewrite $$\sqrt{1-x}-2x\sqrt{1-x^2} =2x^2-1$$ and square : $$(1-x)+4x^2(1-x^2)-4x(1-x)\sqrt{1+x}=4x^4-4x^2+1.$$ This can be simplified $$4x(x-1)\sqrt{1+x}=8x^4-8x^2+x=x(8x^3-8x+1).$$ Symplify by $x$ and square another time to get $$16(x+1)(x-1)^2=64x^6+64x^2+1+16x^3-16x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Branches of $\log(z^2)$ Suppose $z=x+iy$, then $r=\sqrt{x^2+y^2}$ and $\theta_p=\tan^{-1}\left(\frac y x\right)$, so we have
$$z=re^{i\theta},\theta=\theta_p+2n\pi$$
Now, $z^2=x^2-y^2+i(2xy)$, so $r'=x^2+y^2$ and $\theta_p'=\tan^{-1}\left(\frac{2xy}{x^2-y^2}\right)$, so we have
$$z^2=(x^2+y^2)e^{i\theta'},\theta'=\thet... | $$\begin{align*}&\color{red}{\frac{\partial}{\partial x}\left(\arctan\frac{2xy}{x^2-y^2}\right)}=\frac{2y(x^2-y^2)-4x^2y}{(x^2-y^2)^2}\frac1{1+\frac{4x^2y^2}{(x^2-y^2)^2}}=\frac{-2y}{(x^2+y^2)}\\
&\color{green}{\frac{\partial}{\partial x}\left(2\arctan\frac yx\right)}=\frac{-2\frac y{x^2}}{1+\frac{y^2}{x^2}}=\frac{-2y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Examine convergence of $\sum_{n=1}^{\infty}(\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2})$ How to examine convergence of $\sum_{n=1}^{\infty}(\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2})$ for $a, b, c> 0$ using Taylor's theorem?
| You have, from Taylor expansion of $e^u$ around $0$ to second order,
$$\begin{align}
\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2} &=
e^{\frac{1}{n}\ln a} - \frac{1}{2}\left(e^{\frac{1}{n}\ln c}+e^{\frac{1}{n}\ln b}\right)\\ &=
1+\frac{1}{n}\ln a + \frac{1}{2n^2}\ln ^2 a \\&\qquad- \frac{1}{2}\left(2+\frac{1}{n}\ln ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Maclaurin Expansion for $e^{e^{z}}$ at $z=0$ I need to find terms up to degree $5$ of $e^{e^{z}}$ at $z=0$.
I tried letting $\omega = e^{z} \approx 1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\cdots$, and then substituting these first few terms into the Taylor series expansion for $e^{z}$ as follows:
$e^{\omega} = \sum_{n... | Your approach works, but the Taylor series for the outer function should not necessarily be centered at $0$. Rather, it should be centered at the constant term of the Taylor series of the inner function.
In your example, the outer series should be centered at $x=1$. Expanding $e^x$ in a Taylor series around this poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Convergence of $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ I need to prove $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges. The root test is inconclusive, so I check in W.A., $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges by comparison test, but I don't know what seri... | Hint $$\frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} = \frac{\sqrt{4n + 3}-\sqrt{4n + 1}}{\sqrt{4n + 3}\sqrt{4n + 1}}=\frac{2}{\sqrt{4n + 3}\sqrt{4n + 1}(\sqrt{4n + 3}+\sqrt{4n + 1})}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Epsilon delta proof with fractions, given what delta must be Choose $\delta = \min (1, \frac{\epsilon}{10})$
is the following statement true?
$$0 < |x − 1| < δ\text{ implies that }\left|\frac{x^2+3x}{x^2+1} − 2\right| < ε$$
Okay so this is what I have so far
$$\left|\frac{x^2+3x}{x^2+1} − 2\right| = |x-1|\frac{|-x+2|}... | As $|x-1|<1$ for all possible $\delta$ values then $$|x-2|< 2\ \text{and} \ |x^2+1|>1$$ So then we have
$$|x-1|\frac{|-x+2|}{|x^2+1|}< \frac{2|x-1|}{1} = 2|x-1|$$
If $\epsilon\geq10$ then remembering $|x-1|<1$ as $\delta =1$ in this case $$|x-1|\frac{|-x+2|}{|x^2+1|}<2\cdot 1<10\leq\epsilon$$ as desired. If $\epsilon<1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to plot correct best fit line? I'm working on a software where I'm plotting graphs and finding best fit line.
I have used Least-Square Method and linear regression technique with y = mx + c
My problem is that when most of the X values of graph are equal (not all) at that time best fit line is not proper but when th... | The problem has horrific conditioning.
What is the slope$-$intercept equation for a vertical line?
The value for the slope is swinging between $\pm \infty$. This mean a very small perturbation in your data can cause an enormous change in the results.
The trial function is
$$
y(x) = a_{0} + a_{1} x.
$$
The data will ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Bag of 24 distinct objects, four colors, six objects per color, select three (probability). Could someone confirm my combinatorics solutions for this question?
Question:
A bag holds 24 different objects, of which 6 are orange, 6 are white, 6 are yellow, and 6 are red. If a juggler selects three objects to juggle, what... |
What is the probability that all three objects are the same color.
Your answer is correct.
Alternatively, we choose a ball. The probability that the second ball is the same color as the first is $\frac{5}{23}$. The probability that the third ball is the same color as the first two is $\frac{4}{22}$. Hence, the pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}}-\frac{1}{S_n}$ for partials sums of a divergent series Let $(a_n)$ be a sequence of non-negative numbers such that $a_1 > 0$ and $\sum a_n$ diverges. Let $S_n = \sum_{k=1}^n a_k$. Prove that, for all $n \geq 2$,
$$\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}}-\frac{1}{S_n}$$
... | You can prove this result by induction. First, in the case $n=2$,
\begin{align*}
\frac{a_2}{S^2_{2}} \leq \frac{1}{S_{1}}-\frac{1}{S_{2}},
\end{align*}
which follows that $\frac{1}{S_{1}}-\frac{1}{S_{2}}= \frac{S_2-S_1}{S_1S_{2}}=\frac{a_2}{S_1S_{2}}$ and $S_2 \geq S_1$.
Suppose that \begin{align*}
\frac{a_n}{S^2_{n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$
Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that
$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$$
I tried various methods.... | The original post was to show that
$$\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge
\sqrt{a}+\sqrt{b}+\sqrt{c}+3.
$$
Observe that
\begin{align}
\frac{a}{b}+\frac{c}{a}&\ge2\sqrt{\frac{c}{b}}=2c\sqrt{a},\\
\frac{b}{a}+\frac{c}{b}&\ge2\sqrt{\frac{c}{a}}=2c\sqrt{b},\\
\frac{c}{a}+\frac{c}{b}&\ge2\sqrt{\frac{c^2}{ab}}=2c\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's ... | Notice that $\tan x = \sin x \sec x$; so we'll have:
$$\lim_{x \to 0} \frac{-\sin x}{2\sec^2x\sec x \tan x} = \lim_{x \to 0} \frac{-\sin x}{2\sec^2x\sec^2 x \sin x} = \lim_{x \to 0} \frac{-\sin x}{2\sec^4x\sin x} = \lim_{x \to 0} \frac{-1}{2\sec^4x} = \cdots$$
Hope you can go from here. :x
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Without using modular equivalence, show that: $\gcd(4n^2+1,24)=1$ Without using modular equivalence, show that: $\gcd(4n^2+1,24)=1$
Let $d=\gcd(4n^2+1,24)$ then we have:
$$d|24n^2+6,24n^2\ \Rightarrow\ d|6\ \Rightarrow\ d|6n^2,4n^2+1\ \Rightarrow\ d|12n^2,12n^2+3\ \Rightarrow\ d|3\ \Rightarrow\ d=1\ or\ 3$$
Using modul... | Theorem. $4n^2 + 1$ is not a multiple of $3$.
Proof. Since $n^2 = (-n)^2$, we need only prove this for
$n \in \mathbb W = \{0, 1, 2, \dots\}$.
Let $T = \{k\in \mathbb W : 3 | 4k^2 + 1\}$.
We need to show that $T$ is the empty set.
If we assume that $T$ is not the empty set, then it must contain a smallest member, say $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum of the function $f(x_1,x_2) = (x_1 + x_2 + 3)^2 + (3x_1 + x_2 - 1)^2 + (2x_1 + x_2 + 1)^2$ Find the minimum of the function $f(x_1,x_2) = (x_1 + x_2 + 3)^2 + (3x_1 + x_2 - 1)^2 + (2x_1 + x_2 + 1)^2$.
We've been instructed to solve this by writing $f(x_1,x_2)$ as $||Ax - b||^2$, and $x = [x_1,x_2]^T$.
I... | $$
\begin{align}
&\nabla\left[(x_1+x_2+3)^2+(3x_1+x_2-1)^2+(2x_1+x_2+1)^2\right]\\
&=2(x_1+x_2+3)\begin{bmatrix}1\\1\end{bmatrix}
+2(3x_1+x_2-1)\begin{bmatrix}3\\1\end{bmatrix}
+2(2x_1+x_2+1)\begin{bmatrix}2\\1\end{bmatrix}\\
&=\begin{bmatrix}28x_1+12x_2+4\\12x_1+6x_2+6\end{bmatrix}
\end{align}
$$
This vanishes when $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solution to exponential congruence Is there a clever solution to the congruence without going through all the values of x up to 58?$$2^x \equiv 43\pmod{59}$$
Can I somehow use the fact that $2^4 \equiv -43\pmod{59}$ ?
| We know that $2$ is a quadratic non-residue of $59$ from a theorem that states that $2$ is a quadratic residue of $p$, where $p$ is a prime if and only if $\frac{p^2-1}{8}$ is even. Obviously $\frac{59^2-1}{8}$ is odd.
Also, there is another theorem stating that $a$ is a quadratic residue of $p$ (prime) if and only if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Finding out a limit using Taylor series. So the limit is the following:
$$\lim_{x \to 0}{\frac{x^2-\frac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}}$$
Expansions for $\sin(x)$ and $\cos(x)$ are given:
$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} + o(x^{2n})$$
$$\cos x = 1-\frac{x^2}{... | Another way is to note that $\frac{\sin x^{10}}{x^{10}} \to_x 1$, so you get your $x^{10}$ in the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$a,b$ are positive integers . Prove that if $k=\frac{a^2+b^2}{a+b+1}$ is an integer then $k=5$. This is the original question that was posted and answered. For the question the OP intended to ask, please see the edit history.
$a,b$ are positive integers. Prove that if $k=\frac{a^2+b^2}{a+b+1}$ is an integer then $k=5$... | In this case, if $k=5,$ then we would have $70$ as the sum of two integer squares, which it is not.
$$ a^2 + b^2 = 5a + 5b + 5, $$
$$ 4a^2 + 4 b^2 = 20a + 20 b + 20, $$
$$ 4 a^2 - 20a + 4 b^2 - 20 b = 20, $$
$$ 4 a^2 - 20a + 25 + 4 b^2 - 20 b + 25 = 20 + 25 + 25 = 70, $$
$$ (2a-5)^2 + (2b-5)^2 = 70. $$
However, $70 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A Chinese Exam Question which is.....quite hard Let $f(x)=x^2-2x-3$, and $x_n$ be some sequence.
$x_1=2$, $x_n =$ the $x$ coordinate of the point of intersection of the $x$ axis and the line joining $P(4,5)$ and $Q_n(x_n, f(x_n))$.
Find an expression for $x_n$
Find $\lim \limits_{n\to\infty} Q_n(x_n,f(x_n))$.
| I suppose you mean $x_1=2$ and $x_{n+1}=$ the $x$ coordinate of the point of intersection of the $x$ axis and the line joining $P(4,5)$ and $Q_n(x_n,f(x_n))$.
Now the equation of the line joining $Q_n(x_n,f(x_n))$ and $P(4,5)$ is
$$\frac{y-5}{x-4}=\frac{f(x_n)-5}{x_n-4}=\frac{x_n^2-2x_n-3-5}{x_n-4}$$
Setting $y=0$ to f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How can we show that $ \sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4 ?$ We proposed this sum, but we are lacking in knowledge of this area of maths and we would ask if any of the authors would be willing to show us step by step how to go about proving this... | Generally the following sums may help:
$$
\sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}}=4 \left(\frac{1}{4-x}+\frac{\sqrt{x} \arcsin\left(\frac{\sqrt{x}}{2}\right)}{(4-x)^{3/2}}\right)
$$
$$
\sum_{n=0}^\infty\frac{nx^n}{\binom{2n}{n}}=\frac{\partial}{\partial x}\sum_{n=0}^\infty\frac{x^{n+1}}{\binom{2n}{n}}-\sum_{n=0}^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
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Let $f(x)=x^5+x^2+1$ with $x_1,x_2,x_3,x_4,x_5$ as zeros and ... Let $f(x)=x^5+x^2+1$ with $x_1,x_2,x_3,x_4,x_5$ as zeros and $g(x)=x^2-2.$ Show that $$g (x_1)g (x_2)g (x_3)g (x_4)g (x_5)-30g(x_1x_2x_3x_4x_5)=7$$.
I found this question in a local question paper. And I have no idea how to solve it...
Please Help.
| It is known that $$x^2+1+x^5=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\tag1$$
Comparing the constants of each side, we have $-1=x_1x_2x_3x_4x_5$.
Now, puting in $-x$ in $\text{(1)}$, we have $$x^2+1-x^5=-(x+x_1)(x+x_2)(x+x_3)(x+x_4)(x+x_5)\tag 2$$Multipling $\text{(1)}$ and $\text{(2)}$, we have $$h(x)=(x^2+1)^2-x^{10}=(x_1^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of $\lim_{n\rightarrow \infty}\frac{1}{n^{2m}}\left[(n^2+1^2)^{m}(n^2+2^2)^m(n^2+3^2)^m..............(2n^2)^{m}\right]^{\frac{1}{n}}$
Evaluation of $$\lim_{n\rightarrow \infty}\frac{1}{n^{2m}}\left[(n^2+1^2)^{m}(n^2+2^2)^m(n^2+3^2)^m..............(2n^2)^{m}\right]^{\frac{1}{n}}$$
$\bf{My\; Try::}$ Let $$L=... | If we set
$$ \color{blue}{L}=\lim_{n\to +\infty}\left[\left(1+\frac{1^2}{n^2}\right)\cdot\left(1+\frac{2^2}{n^2}\right)\cdot\ldots\cdot\left(1+\frac{n^2}{n^2}\right)\right]^{1/n}\tag{1}$$
the original limit is just $L^m$. But:
$$ \log L = \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left[1+\left(\frac{k}{n}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Under certain conditions $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}\Rightarrow \{a,b,c\}=\{a',b',c'\}$ Let $a,b,c,a',b',c'\in \mathbb{Z}_{\geq 1}$ be such that
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<1,\quad \frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}<1.
$$
Suppose
$$
\frac{1}{a}+\frac{1}{b... | For the more general case when $a\neq b\neq c$ and $a'\neq b'\neq c'$, the answer is also no.
In particular, let $\frac1a +\frac1b = \frac{1}{a'}$, and $\frac{1}{c} = \frac{1}{b'} +\frac{1}{c'}$. Then $a' = \frac{ab}{a + b}$, and $c = \frac{b' c' }{b'+ c'}$. We'd want these to be integers.
A simple way of making $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$A^2+B^2=AB$ and $BA-AB$ is non-singular The question is:
Are there square matrices $A,B$ over $\mathbb{C}$ s.t. $A^2+B^2=AB$ and $BA-AB$ is non-singular?
From $A^2+B^2=AB$ one could obtain $A^3+B^3=0$. Can we get something from this?
Edit:
$$A^2+B^2=AB\implies\\
A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\
... | We consider the equations
(1) $A^2+B^2=AB$,
(2) $A^2+B^2=2AB$.
A couple $(A,B)$, solution of (2), is simultaneously triangularizable, and, moreover, when $n=2$, $AB=BA$.
Eq. (1) has not the second property: indeed $$A=\begin{pmatrix}0&-1\\1&-1\end{pmatrix},\ B=\begin{pmatrix}\dfrac{1-i\sqrt{3}}{2}&0\\\dfrac{3+i\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
How to Find $ \lim\limits_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$. Can someone help me with this limit? I'm working on it for hours and cant figure it out.
$$ \lim_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$$
I started transforming to the form
$ \lim_{x\to 0} e^{ {\frac{\ln \left(\fra... | In the same spirit as other answers, consider $$A=\left(\frac {\tan (x) }{x} \right)^{\frac{1}{x^2}}$$ Take logarithms $$\log(A)=\frac{1}{x^2}\log\left(\frac {\tan (x) }{x} \right)$$ Now, consider Taylor series $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ $$\frac {\tan (x) }{x}=1+\frac{x^2}{3}+\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 7
} |
If $p>5$ is prime, $2p+1$ is a prime, $\frac{4p+1}{3}$ is prime, $8p+1$ is prime, Then $p \equiv 29 (mod \; 30)$ Assume that $p>5$ is prime, $2p+1$ is a prime, $\frac{4p+1}{3}$ is prime, $8p+1$ is prime. Then I want to prove that $p \equiv 29 (mod \; 30).$
First of all I have to show that $4p+1$ is a multiple of 3 if ... | We have $p\equiv \pm 1\pmod{3}$. But if $p\equiv 1\pmod{3}$ then $2p+1$ is divisible by $3$, so $p\equiv -1\pmod{3}$, and therefore $p\equiv 5\pmod{6}$.
Also, $p$ is congruent to one of $1,2,3,4$ modulo $5$. Since $2p+1$ is prime, we cannot have $p\equiv 2\pmod{5}$. Since $\frac{4p+1}{3}$ is prime, $p$ cannot be congru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $a^2b+b^2c+c^2a \leqslant 3$ for $a,b,c >0$ with $a^ab^bc^c=1$
Let $a,b,c >0$ and $a^ab^bc^c=1$. Prove that
$$a^2b+b^2c+c^2a \leqslant 3.$$
I don't even know what to do with the condition $a^ab^bc^c=1$. At first I think $x^x>1$, but I was wrong. This inequality is true, following by the verification from M... | We shall prove the following inequalities:
If $x+y+z=3$ and $x,y,z>0$, then
$$\tag{1}x^2y+y^2z+z^2x+xyz\le 4\label{1},$$
$$\tag{2}3x^xy^yz^z+xyz\ge 4\label{2},$$
i.e.,
$$\tag{*}x^2y+y^2z+z^2x\le 3x^xy^yz^z\label{*}.$$
Accept $\eqref{*}$ for a moment. Homogenizing the inequality by substitution
$$x=\frac{3a}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 0
} |
Evaluation of $\cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$
Evaluation of $$\lim_{n\rightarrow \infty}\left(\tan \frac{\pi}{2n}\cdot \tan \frac{2\pi}{2n}\cdot \tan \frac{\pi}{3n}\cdot ...............\tan \frac{(n-1)\pi}{2n}\right)^{\frac{1}{n}} = $$ wi... | First we have
\begin{align}
\lim_{n\to \infty}\frac1{n}\ln{\left(\tan\frac{\pi}{2n}.\tan\frac{2\pi}{2n}\cdots\tan\frac{(n-1)\pi}{2n}\right)}
&=\lim_{n\to \infty}\frac1{n}\sum\limits_{k=0}^{n-1}\ln{\left(\tan\frac{k\pi}{2n}\right)}
\\
&=\frac{2}{\pi}\int_0^{\pi/2}\ln{\tan{x}}\:dx
\\
&=\frac{2}{\pi}\left(\int_0^{\pi/2}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Splitting fractions with a linear denominator: $\frac{2x-1}{x+2}$ How can $$\frac{2x-1}{x+2}$$ be split to give $$A-\frac{B}{x+2}$$
where $A$ and $B$ are integers?
The solution is $$2-\frac{5}{x+2}.$$
| $\dfrac{2x-1}{x+2}:$
\begin{array}{ccccccc}
&&&& 2 \\
&&&& -- & -- & --\\
x & + & 2 & | & 2x & - & 1 \\
&&&&2x & + & 4 \\
&&&& -- & -- & --\\
&&&&&&-5
\end{array}
So $\dfrac{2x-1}{x+2} = 2 + \dfrac{-5}{x+2} = 2 - \dfrac{5}{x+2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to differentiate this fraction $\frac{2}{x^2+3^3}$? $\frac{2}{(x^2+3)^3}$.
I have ${dy}/{dx}$ x 2 x ${x^2+3^3}$ - 2 x ${dy}/{dx}$ x ${x^2+3^3}$ over $({x^2+3)^6}$
And then simplifying to $-12x^5 + 36x^2$ over $({x^2+3)^6}$
I'm not sure if this is right.
| You can use the quotient rule or the product rule. Let's use the quotient rule for this:
$$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'g-g'f}{g^2}$$
In your case we have: $f(x)=2 \implies f'(x)=0$ and $g(x)=x^2+3^3 \implies g'(x)=2x$
It follows:
$$\implies \frac{f'g-g'f}{g^2}=\frac{0\cdot (x^2+3^3)-2x\cdot2}{(x^2+3^3)^2}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Trigonometric identities involved $\sin(\alpha)$, $\cos(\alpha)$, $\tan(\alpha)$ $\alpha$ is valid between $-90<\alpha<90$ Degrees
Show that,
$$ \frac{\tan^6(\alpha)-\tan^4(\alpha)+2}{\tan^6(\alpha)-2\tan^2(\alpha)+4 }\cdot\cos^2(\alpha) =
\frac{\sin^6(\alpha)+\sin^4(\alpha)-2}{\sin^6(\alpha)-2\sin^2(\alpha)-4 } $$
... | So, for the LH fraction:
$$
{{x^{\,6} - x^{\,4} + 2} \over {x^{\,6} - 2x^{\,2} + 4}} = {{y^{\,3} - y^{\,2} + 2} \over {y^{\,3} - 2y + 4}}
$$
then by Ruffini's method
$$
= {{\left( {y + 1} \right)\left( {y^{\,2} - 2y^{\,2} + 1} \right)} \over {\left( {y + 2} \right)\left( {y^{\,2} - 2y^{\,2} + 1} \right)}} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\cos^4x - \sin^4x - \cos^2x + \sin^2x$ is always $0$? So I have a small problem here where I have to prove the following :
$$\cos^4x - \sin^4x - \cos^2x + \sin^2x = 0 $$
I know that the 2nd part is always $1$, so I need to prove that the first part also equals $1$. So how should I prove it ?
Edit : Sorr... |
Notice, your statement isn't true. Now we know that:
$$\cos^2(x)+\sin^2(x)=1$$
So, we can see that:
$$\cos^4(x)+\sin^4(x)-\cos^2(x)-\sin^2(x)=\cos^4(x)+\sin^4(x)-1$$
Now, prove that $\cos^4(x)+\sin^4(x)=\frac{\cos(4x)+3}{4}$:
$$\cos^4(x)+\sin^4(x)=\frac{\cos(4x)+3}{4}\Longleftrightarrow$$
$$4\left(\cos^4(x)+\sin^4(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
What is the limit of $\lim_{n\to \infty}\frac{1-2+3-4+\cdots+(-2n)}{\sqrt{n^2+1}}$? Find the $$\lim_{n\to \infty}\frac{1-2+3-4+\cdots+(-2n)}{\sqrt{n^2+1}}.$$
I thought to apply squeeze theorem so
$$\frac{1-2+3-4+\cdots+(-2n)}{\sqrt{n^2+1}}\leq \frac{1+2+3+4+\cdots+(2n)}{\sqrt{n^2+1}}=\frac{n(2n+1)}{\sqrt{n^2+1}}$$ But... | You can do the following. For any $k$ we have $$1-2+3-4 + ... +(-2k) = (1-2) + (3-4) + \cdots +(2k-1 - 2k) =-k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Problem solving limit infinity/infinity. I cannot solve this limit:
$$\lim_{x\rightarrow\infty}
\frac {(3x^2-4) \left( \sqrt [3]{2x^2+1}+1
\right)^2}{ (2x-1) \left( 4-\sqrt {8x^3-2}
\right)x^{3/2}}$$
I make this:
$${\frac { \left( 3\,{x}^{2}-4 \right) \left( \sqrt [3]{2\,{x}^{2}+1}+1
\right) ^{2}}{ \left( 2\,x-1 ... | The total power of $x$ in the numerator is $2 + 4/3 = 10/3.$ The total power of $x$ in the denominator is $1 + 3/2 + 3/2 = 4.$ The denominator wins: The limit must be $0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_1^N \frac{-3N+6t-3}{t^3(N-t+1)^4}dt$ when $N=3$ or $N=5$ Let the Cauchy product $$(\zeta(3))^2=\sum_{n=1}^\infty c_n,$$
where $$c_n=\sum_{k=1}^n\frac{1}{k^3(n-k+1)^3},$$
and $\zeta(3)$ is the Apèry constant.
Taking $f(x)=\frac{1}{x^3(N-x+1)^3}$ in Abel's identity after tedious computations using the div... | $$\frac{1}{n(N-n)} = \frac{1}{N}\left(\frac{1}{n}+\frac{1}{N-n}\right) \tag{1}$$
gives, by squaring:
$$\frac{1}{n^2(N-n)^2} = \frac{1}{N^2}\left(\frac{1}{n^2}+\frac{1}{(N-n)^2}+\frac{\frac{2}{N}}{n}+\frac{\frac{2}{N}}{N-n}\right)\tag{2} $$
as well as:
$$ \frac{1}{n^3 (N-n)^3} = \frac{1}{N^3}\left(\frac{1}{n^3}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to write this cubic equation For an imaginary number $i=\sqrt{-1}$ ,the cubic equation $24x^3+21x^2-72x-7=0$ can be represented in the form $$\frac{((x+i)^3-(x-i)^3)}{((x+i)^3+(x-i)^3)}\frac{((2+i)^4+(2-i)^4)}{((2+i)^4-(2-i)^4)}=1$$
How can we write the cubic $141x^3+72x^2-141x-8=0$ in a similar form?
| HINT:
By applying Componendo and Dividendo repeatedly,
we find $$\dfrac{x^3-3x}{-7}=\dfrac{3x^2-1}{24}\iff24x^3+21x^2-72x-7=0$$
Now,
$$14x1^3+72x^2-141x-8=0\iff\dfrac{x^3-x}8=\dfrac{1-9x^2}{141}$$
Again, $(ax+i)^3=a^2x^3+i(3a^2x^2)+3ax(-1)-i$
Comparing with $x^3-x+i(1-9x^2)$
we need, $$\dfrac{a^3}1=\dfrac{3a}1\iff a^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the closed form for $\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{x^{\pi}}{1+x^{\pi}}\cdot\frac{1}{1+x^ e}dx $? On my previou page Jack D'Aurizio offered a concise elegant prove of Vladimir Reshetnikov's identity and a closed form for it.
(1)
$$\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{1}{1+x^{\pi}}dx=\int_{0}^{... | Through the substitution $x\mapsto\frac{1}{x}$ we have:
$$ I(a,b) = \int_{0}^{+\infty}\frac{x^a}{1+x^a}\cdot\frac{1}{1+x^b}\cdot\frac{dx}{1+x^2} = \int_{0}^{+\infty}\frac{1}{1+x^a}\cdot\frac{x^b}{1+x^b}\cdot\frac{dx}{1+x^2} = I(b,a) $$
hence, by your previous question:
$$\begin{eqnarray*} I(a,b) &=& \frac{1}{2}\int_{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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For which $\lambda$ do we have solutions I'm trying to find for what values of $\lambda$ the following matrix has either no solutions, infinitely many or unique solutions.
$$A=\begin{pmatrix} 1 & 1 & \lambda & 1 \\ 4 & \lambda ^2 & -8 & 4 \\ \lambda & 2 & 4 & \lambda + 1\end{pmatrix}$$
I know that in general a system... | Because of the way your question is phrased, I will assume that $A$ is the augmented matrix of a system of 3 equations on 3 variable
If we perform row reduction:
\begin{align}
\begin{bmatrix} 1 & 1 & \lambda & 1 \\ 4 & \lambda ^2 & -8 & 4 \\ \lambda & 2 & 4 & \lambda + 1\end{bmatrix}
&\xrightarrow[R3-\lambda R1]{R2-4R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Shortest distance between two lines in 3-dimensional space Can someone explain to me how to solve this question?
Find the shortest distance between the lines
$L_1 = \left\{t \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} : t \in \mathbb{R}\right\}$ and $L_2 = \left\{s \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} + \begin{bmatrix} 1... | The Euclidean distance between the two lines is
$$\left\|s \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} - t \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} + \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}\right\| = \left\|\begin{bmatrix} 1 & -1\\ 2 & -1\\ 3 & -1\end{bmatrix} \begin{bmatrix} s\\ t\end{bmatrix} - \begin{bmatrix} -1\\ 0\\ 0\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Factors of the numbers of the form $a^2+nb^2$
Let $N=a^2+nb^2$ with $\gcd(a,b) =1$ and $n \in \mathbb{Z^+}$. If $N=xy$ where $x$ and $y$ are relatively prime numbers, in what condition can $x$ and $y$ be also written in the same form as $N$ (i.e, $c^2+nd^2$) ? How can we prove it?
The above statement is true for sum... | This is true when $a^2 + n b^2$ is of class number one ( well, the form class number of the discriminant $-4n$). I refer to the number of (equivalence classes of) primitive binary quadratic forms with that discriminant, $-4n.$ Even in this case, we need to be cautious when there are also imprimitive forms of that discr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $(x^2y+y^2z+z^2x)\cdot \left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2} +\frac{1}{(z+x)^2}\right) \geqslant \frac94$ $x,y,z > 0$ and $x+y+z=3$, prove
$$(x^2y+y^2z+z^2x)\cdot \left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2} +\frac{1}{(z+x)^2}\right) \geqslant \frac94$$
My immediate thought is that this inequality is similar to ... | A full expanding gives
$$\sum\limits_{cyc}(4x^6y+5x^5y^2+x^5z^2+3x^4y^3-x^4z^3+$$
$$+2x^5yz+9x^4y^2z-7x^4z^2y-2x^3y^3z-14x^3y^2z^2)\geq0$$
and since by Muirhead and AM-GM
$$\sum\limits_{cyc}(x^5y^2+x^5z^2)\geq\sum\limits_{cyc}(x^4y^3+x^4z^3)$$
$$6x^4y^3+2y^4z^3+5z^4x^3\geq13x^3y^2z^2$$
$$19x^6y+2y^6z+10z^6x\geq31x^4z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
write down the expression for $\sin (15°)$ using the double angle formula. show that $\sin 15^\circ=\frac {\sqrt3 -1}{2\sqrt2}$ using $\cos2A=1-2\sin^2A$
However I got $\sin 15^\circ= \sqrt{\frac {2-\sqrt 3}{4}}$ instead.
| Your form, $\sqrt{\frac{2-\sqrt3}{4}}$, is already close:
$$\begin{align*}
\sin15^\circ &=\sqrt{\frac{2-\sqrt3}{4}}\\
&= \sqrt{\frac{4-2\sqrt3}{8}}\\
&= \sqrt{\frac{1-2\sqrt3+3}{8}}\\
&= \sqrt{\frac{(1-\sqrt3)^2}{8}}\\
&= \frac{\sqrt3 -1 }{2\sqrt2}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding K such that a line is perpendicular or parallel to a given plane? Known information:
Plane $P = 3x -5y + 2z =1 (R^3)$
Parametrics:
x = 5 + t
y = 2 - 2t
z = -1 - 6t
I've solved for $t$, $t = -2$. Point of intersection: $(3,6,11)$
The part causing problems is, given vectors:
A = (2, -7, -6)
B = (-4, 3, k)
... | Your plane is described by
$$
(3, -5, 2) \cdot (x,y,z) = 1 \iff \\
\frac{1}{\lVert (3,-5,2) \rVert} (3, -5, 2) \cdot (x,y,z) =
\frac{1}{\lVert (3,-5,2) \rVert} \\
\frac{1}{\sqrt{38}} (3, -5, 2) \cdot (x,y,z) =
\frac{1}{\sqrt{38}} \iff \\
n \cdot u = d
$$
where $n$ is a unit normal vector and $d$ is the signed distan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Can anyone help me with this finite sum? I have to calculate the sum
$\displaystyle\sum_{k=1}^n \displaystyle\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1}$
We can re-write the sum as follows
$\displaystyle\sum_{k=1}^n \displaystyle\frac{3^k-1+1}{(3^{k+1}-1)(3^k-1)}$
And then we obtain
$\displaystyle\sum_{k=1}^n \displaystyle\frac... | For such problem, we usually look for some telescoping structure, note that
$$\frac{1}{3^{k} - 1} - \frac{1}{3^{k + 1} - 1} = \frac{2 \times 3^k}{3^{k + 1} - 1)(3^k - 1)}.$$
So based on the decomposition you already found, the summation can be written as
$$\frac{1}{2}\sum_{k = 1}^n \left(\frac{1}{3^{k} - 1} - \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Evaluating the inverse trigonometric limit $\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}$
$$
\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}
$$
I was doing some questions on limits, I saw one in which there is $\arc... | Set $t=\arccos(2x\sqrt{1-x^2})$, so $2x\sqrt{1-x^2}=\cos t$ and, by definition, $0\le t\le\pi$. Then
$$
\sin t=\sqrt{1-\cos^2t}=\sqrt{1-4x^2+4x^4}=|2x^2-1|
$$
and so $t=\arcsin|2x^2-1|$. Thus your limit (from the right) can be written
$$
\lim_{x\to(1/\sqrt2)^+}\sqrt{2}\frac{\arcsin|2x^2-1|}{\sqrt{2}x-1}=
\lim_{x\to(1/\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factor proofs problem The coolness of an integer is equal to the integer divided by the total number of factors that it has. For example, $48$ has $10$ factors therefore, coolness $(48) = \frac { 48 }{ 10 } =\quad 4.8$
1. Provide an explanation for why coolness$(xy)$ cannot be equal to an integer if both $x$ and $y... |
1. Provide an explanation for why coolness$(xy)$ cannot be equal to an integer if both $x$ and $y$ are different prime numbers.
You are correct in your thinking for this part:
As $x$ and $y$ are distinct primes, $xy$ has $4$ factors: $1, x, y, xy$. This means that $xy$ must be divisible by $4$. However, as $2$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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proof that $\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$ diverges, by comparsion I need to prove that
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$$
diverges by comparsion.
The way I did was to use
$$\frac{n}{n^2+2n+1}>\frac{n}{n^2+2n^2+n^2} = \frac{n}{4n^2} = \frac{1}{4n}$$
which diverges. Can I do that? Because $$2n+1<2n^2+n... | $$\frac{n}{n^2+2n+1} = \frac{n}{(n+1)^2} = \frac{n + 1 - 1}{(n+1)^2} = \frac{1}{(n+1)} - \frac{1}{(n+1)^2}$$
Which diverges because $\frac{1}{n+1}$ does so and $\frac{-1}{(n+1)^2}$ doesn't.
| {
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"url": "https://math.stackexchange.com/questions/1801693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $nx_n=(n+2)x_{n-1} + 1$ by the telescoping method I am trying to solve this recurrence relation from a book "Problem solving through Problems" by Loren c. Larson (5.3.14 (b)) using the telescoping method.
$$x_0=0\qquad nx_n=(n+2)x_{n-1} + 1\ (n > 0)$$
Here some additional terms I computed: $x_1=1, x_2 = 5/2, x_... | Dividing both parts of the expression by $n(n+1)(n+2)$, one gets:
$$\frac{x_n}{(n+2)(n+1)}=\frac{x_{n-1}}{(n+1)n} + \frac{1}{(n+2)(n+1)n}$$
Let us define $$y_n = \frac{x_n}{(n+2)(n+1)}$$
Then the new recurrence is easier to solve:
$$y_n = y_{n-1} + \frac{1}{(n+2)(n+1)n}\qquad y_0 = 0$$
One knows that there exists $(A,B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of discontinuous values We have to find the number of values of $x$ at which the function $$ f(x) = \frac{2x^5-8x^2+11}{x^4+4x^3+8x^2+8x+4}$$ is discontinuous.
I thought that since both numerator and denominator are polynomials, and so are continuous, that the function would be continuous except when the denomi... | Note $(x+1)^4=x^4+4x^3+6x^2+4x+1$.
Then, $$x^4+4x^3+8x^2+8x+4=(x+1)^4+2x^2+4x+3=(x+1)^4+2(x+1)^2+1$$
Thus we conclude that it is never discontinuous because $$x^4+4x^3+8x^2+8x+4=(x^2+2x+2)^2>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral ${\large\int}_0^\infty\big(2J_0(2x)^2+2Y_0(2x)^2-J_0(x)^2-Y_0(x)^2\big)\,dx$ I'm interested in the following definite integral:
$$\int_0^\infty\big(2J_0(2x)^2-J_0(x)^2+2Y_0(2x)^2-Y_0(x)^2\big)\,dx,\tag1$$
where $J_\nu$ and $Y_\nu$ are the Bessel functions of the first and the second kind.
Mathematica evaluates... | Somewhat similar to this answer, we can use Ramanujan's master theorem to show that $$\int_{0}^{\infty}\big(2J_0(2x)^2-J_0(x)^2\big)\,dx = \frac{\ln 2}{\pi}. $$
As $x \to \infty$, $$\big(2J_0(2x)^2-J_0(x)^2\big) \sim \frac{\sin (4x)-\sin (2x)}{\pi x}. \tag{1}$$ So the integral does indeed converge (but not absolutely)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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If $a,b>0$ and $a+b=1\;,$ Then minumum value of $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ is If $a,b>0$ and $a+b=1\;,$ Then minumum value of $\displaystyle \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2$ is
$\bf{My\; Try::}$ Let $a=\sin^2 \theta$ and $b=\cos^2 \theta\;,$ Then We have to minimize
$$\displaystyl... | The following is not true
$$\sin^4 \theta+\csc^4 \theta+\cos^4 \theta+\sec^4 \theta+4=1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4.$$
For example, let $\theta=\pi/4$. Then $\text{LHS}=1/4+4+1/4+4+4=12.5$, while $\text{RHS}=1-1/2+4+4=8.5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Linear Regression: linear or reciprocal function? The problem is given below:
Simultaneous values of time $t$ and output $y$ from a specific sensor has been measured and is tabulated below $$\begin{array}{cc}
t & y \\
\hline
1 & 17 \\
2 & 15 \\
3 & 11 \\
4 & 10\\
5&8\\
6&7\\
7&7
\end{array} $$
Determine whether t... | Linear function
$$
y_{1}(t) = \beta_{0} + \beta_{1} t
$$
Linear system:
$$
\begin{align}
\mathbf{A} \beta &= y \\
%
\left[
\begin{array}{cc}
1 & 1 \\
1 & 2 \\
1 & 3 \\
1 & 4 \\
1 & 5 \\
1 & 6 \\
1 & 7 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
\beta_{0} \\
\beta_{1}
\end{array}
\right]
%
&=
%
\left[ \beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $3^n | 2^{3^n} + 1$ Question:
Proof by induction that $3^n | 2^{3^n} + 1$.
Attempt: $$ 2^{3^{n+1}} + 1 = 2^{3^n} 2^3 + 1 = 2^{3^n} 2^3 + 1 + 2^3 - 2^3 =
2^3( 2^{3^n} + 1 ) + 1 -2^3$$
And the first is $3^n |$ but second I don't know how to proof that.
| Hint: you should be looking at $2^{3^{n+1}}$, not $2^{3^n+1}$. Assume, by induction that $2^{3^n} + 1 = 3^nd$, i.e., $2^{3^n} = 3^nd - 1$, and see what you can get from $2^{3^{n+1}} + 1 = 2^{3^n\cdot 3} + 1 = (2^{3^n})^3 + 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a+b+c=0$ what is the value of $\frac{a^2}{2a^2 +bc }+\frac{b^2}{2b^2 +ca }+\frac{c^2}{2c^2 +ab }$ Let $s=\frac{a^2}{2a^2 +bc }+\frac{b^2}{2b^2 +ca }+\frac{c^2}{2c^2 +ab }$.
If we use inequality $\frac{x^2}{a}+\frac{y^2}{b} \ge \frac{(x+y)^2}{(a+b)}$ we get $s \ge 0$ as $a+b+c=0$.
Again $s \le \frac{a^2}{bc }+\frac... | From your observation, we have
$$\sum_{cyc}\frac{a^2}{2a^2+bc}-1=\frac{abc(a+b+c)(ab+bc+ca-a^2-b^2-c^2)}{(2a^2+bc)(2b^2+ac)(2c^2+ab)}$$
Since $a+b+c=0$, we get the value of $S=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Trouble with a Probability (Integral) Computation from *All of Statistics* From All of Statistics pg. 36:
Suppose that $X$ and $Y$ are independent and both have the same density $f(x) = 2x$ for $0 \le x \le 1$ and $f(x) = 0$ otherwise. Let us find $\mathbb{P}(X + Y \le 1)$. Using independence, the joint density is
$$... | Question 1: The integral is over the region $x+y \leq 1$. You are also constrained to $0 \leq x \leq 1$ and $0 \leq y \leq 1$ because of the densities of $X$ and $Y$. If you draw the $1 \times 1$ box where $0 \leq x \leq 1$ and $0 \leq y \leq 1$, then draw the line $x+y=1$ and shade the part below the line where $x+y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.