Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proving bounds of an integral $\displaystyle\frac{1}{\sqrt{3}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq 1$
I found $f'(x)=\displaystyle\frac{-3x^2}{2(x^3 + 4)^{\frac{3}{2}}}.$
Set $f'(x) = 0$. Got $x = 0$ as a local max.
$f(0) = 1/\sqrt{5}$
$f(2) = 1/\sqrt{12}$
So $1/\sqrt{3} \leq 1/\sqrt{12} \leq$ the ... | Hint:
$$\int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{(2)^3 +4}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{(0)^3 +4}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the area bounded by $y = 2 {x} - {x}^2 $ and straight line $ y = - {x}$ $$
y =\ 2\ {x} - {x}^2
$$
$$
y =\ -{x}
$$
According to me , the area
$$
\int_{0}^{2}{2x\ -\ { x} ^2}\, dx \ + \int_{2}^{3}{\ {x} ^2\ -\ 2{x} }\, dx \\
$$
Which gives the area $ \frac{8}{3}$
But the answer is $ \frac{9}{2}$
| To find the points of intersection between the curve $y = 2x - x^2$ and $y = -x$, equate the two expressions.
\begin{align*}
2x - x^2 & = -x\\
3x - x^2 & = 0\\
x(3 - x) & = 0
\end{align*}
which has solutions $x = 0$ and $x = 3$. These are the limits of integration.
Since
\begin{align*}
y & = 2x - x^2\\
& = -x^2 + 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Why isn't this approach in solving $x^2+x+1=0$ valid? There is this question in which the real roots of the quadratic equation have to be found:
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x \neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 \neq 0$
Therefore, it is legal to divide each term by $x$:
$x + 1 ... | Your error lies in your last line, where you go from $x^3 = 1$ to $x = 1$. There are actually three solutions to $x^3 = 1$. They are as follows
$$\begin{align}
x_1 &= 1, & \text{or} \\
x_2 &= -\frac{1}{2} + \frac{\sqrt 3}{2} i, & \text{or} \\
x_3 &= -\frac{1}{2} - \frac{\sqrt 3}{2} i
\end{align}$$
Only solutions $x_2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
$\ \lim_{x\to0} \frac{x}{\cos(\frac{\pi}{2}-x)} $ without L'Hospital's rule
How do I find the following?$$\ \lim_{x\to0} \frac{x}{\cos(\frac{\pi}{2}-x)} $$
I have tried to use trig identities :
$$ \frac{x}{\cos(\frac{\pi}{2}-x)} = \frac{x}{\cos\frac{\pi}{2}\cos x+\sin\frac{\pi}{2}\sin x} = $$
Can't really see anythi... | As an alternative by $f(x)=\cos\left(\frac{\pi}{2}-x\right)$ from the definition of derivative
$$\lim_{x\to0} \frac{x}{\cos\left(\frac{\pi}{2}-x\right)}=\lim_{x\to0} \frac{x-0}{\cos\left(\frac{\pi}{2}-x\right)-\cos\left(\frac{\pi}{2}\right)}=\frac1{f'(0)}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $\sin^{3}x+\cos^{3}x=1$
Solve for $x\\ \sin^{3}x+\cos^{3}x=1$
$\sin^{3}x+\cos^{3}x=1\\(\sin x+\cos x)(\sin^{2}x-\sin x\cdot\cos x+\cos^{2}x)=1\\(\sin x+\cos x)(1-\sin x\cdot\cos x)=1$
What should I do next?
| Let use by $t= \tan (x/2)$
*
*$\sin x=\frac{2t}{1+t^2}$
*$\cos x=\frac{1-t^2}{1+t^2}$
to obtain
$$2t^6-8t^3+6t^2=0$$
$$\iff t^2(t^4-4t+3)=t^2(t-1)^2(t^2+2t+3)=0$$
and since $t^2+2t+3>0$ the solutions are
*
*$t=0 \implies \frac x 2=k\pi\implies x=2k\pi$
*$t=1 \implies \frac x 2=\frac{\pi}4+k\pi \implies x=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Solve $7^x+x^4+47=y^2$
Solve $$7^x+x^4+47=y^2$$ where $x, y \in \mathbb{N}^*$
If $x$ is odd then the left term is congruent with $3$ mod $4$ so it couldn't be a perfect square, so we deduce that $x=2a$ and the relation becomes $$49^a+16a^4+47=y^2$$ and it is easy to see that the left term is divisible by $16$ so we ... | If $x$ is even and let $x=2a$, then $(7^a)^2<7^x+x^4+47<(7^a+1)^2=7^x+2\times7^a+1$ if $(2a)^4+47<2\times7^a+1$, which is true for $a \ge 4$. Therefore, it is enough to consider only $x=2, 4$ and $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Is discrete uniform distribution fully characterized by first two moments? Let $U$ be the discrete uniform on $\{0,1,\dots,n\}$. If a discrete random variable $X$ satisfies:
$$
\begin{align}
\mathbb{E}[X] &= \mathbb{E}[U] = n/2 \\
\mathrm{Var}(X) &= \mathrm{Var}(U) = ((n+1)^2-1)/12,
\end{align}
$$
is this sufficient to... | No. For $n=3$: consider the distribution of $X$ given by $p=(1/5,2/5,1/10,3/10)$. (There are many other possible choices.)
You have
$$
\mathbb{E}[X] = \sum_{k=0}^3 k p_k = 0\cdot \frac{1}{5} + 1\cdot\frac{2}{5}+2\cdot\frac{1}{10}+3\cdot\frac{3}{10} = \frac{3}{2} = \mathbb{E}[U_3]
$$
and
$$
\mathbb{E}[X^2] = \sum_{k=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving integer equation $a^3+b^3=a^2+72ab+b^2$
Find all pairs of positive integers $(a;b)$ that satisfy
$$a^3+b^3=a^2+72ab+b^2.$$
I have already solved this by letting $S=a+b$, $P=ab$. Then I have $S^3-S^2=(3S+70)P$, which will result in $3S+70$ dividing $5110S$, or $357700$ is divisible by $3S+70$. The result is ... | Let $a=x+y, b=x-y$. Then:
$$a^3+b^3=a^2+72ab+b^2 \Rightarrow \\
(x+y)^3+(x-y)^3=(x+y)^2+72(x+y)(x-y)+(x-y)^2 \Rightarrow \\
2x^3+6xy^2=2x^2+72x^2 -72y^2+2y^2 \Rightarrow \\
y^2=\frac{(37-x)x^2}{35+3x}\ge 0.$$
Note that:
$$x=\frac{a+b}{2}>0 \ \ \text{and} \ \ 37-x\ge0 \Rightarrow 0<x\le 37.$$
One can quickly check (if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Calculate limit of $\frac{1}{n}\cdot (1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})$ How to prove that
$$\lim_{n\to\infty}(\frac{1}{n}\cdot(1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})) = +\infty$$
using only basic limit operations and theorems?
| You have
$$\begin{align}
\frac{1}{n}\sum_{k=1}^n \sqrt{k}
&\geq \frac{1}{n}\sum_{k=\lfloor n/2\rfloor}^n \sqrt{k}
\geq \frac{1}{n}\sum_{k=\lfloor n/2\rfloor}^n \sqrt{\frac{n}{2}} \\
&= \frac{n-\lfloor n/2\rfloor+1}{n}\cdot \sqrt{\frac{n}{2}}
\geq \frac{n}{2n}\cdot \sqrt{\frac{n}{2}} \\
&= \frac{\sqrt{n}}{2\sqrt{2}} \xr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Radius and angular derivatives expressed in Cartesian coordinates Given a point with Cartesian coordinates $(x,y)$ and with Cartesian velocity $(\dot{x},\dot{y})$, I would like to express its radius $r$, its angle $\theta$, its radius velocity $\dot{r}$, and its angular velocity $\dot{\theta}$.
First, I know that $x = ... | Following the method provided by Rodney Dunning in another answer:
*
*Taking the time-derivative of $r^2 = x^2 + y^2$, we get
$$2 r \dot{r} = 2 x \dot{x} + 2 y \dot{y} \text{,}$$
and we may write
$$\boxed{\dot{r} = \frac{x \dot{x} + y \dot{y}}{r}} \text{.}$$
*Taking the time-derivative of $\tan\theta = \frac{y}{x}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the range of $ (\arcsin x)^2 + (\arccos x)^2 $ without using derivatives? I am expected to find the range of the following function:
$$
(\arcsin x)^2 + (\arccos x)^2
$$
I individually added up the range of $(\arcsin x)^2$ $(\arccos x)^2$
But that gave wrong answer ?
What to do with this problem?
| Let me write $f(x) = \arcsin^2 x + \arccos^2 x$. The natural domain of $f$ is the intersection of natural domains of $\arcsin$ and $\arccos$: $$\mathcal D(f) = \mathcal D(\arcsin) \cap \mathcal D(\arccos) = [-1,1]\cap [-1,1] = [-1,1].$$
From the identity $\sin x = \cos(\frac\pi 2 - x)$, it follows that $\arcsin x + \ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose t... | To prove $$\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$$
We take common denominator and prove the numerator is a multiple of $6.$
The numerator factors as $$n(2n-1)(n-1)$$
One of $n$ or $n-1$ is even so the product is multiple of $2$
The remainder of n in dividing by $3$ is either $0$ or $1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 0
} |
$\int_{-\infty}^{\infty}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\zeta(2)$ Without Feynman Integration How do I find $$\int_{-\infty}^{\infty}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\zeta(2)$$
without Feynman integration? I saw this video, which gives $$\int_{0}^{1}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\... | Notice that $\frac{\arctan x}{x} = \int_{0}^{1} \frac{dy}{1+x^2y^2}$. So
\begin{align*}
\int_{-\infty}^{\infty} \frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}} \, dx
&= \int_{-\infty}^{\infty} \int_{0}^{1} \frac{1}{(x^2+1)(1+(x^2+2)y^2)} \, dydx.
\end{align*}
Notice that
$$
\frac{1}{(x^2+1)(1+(x^2+2)y^2)}
= \frac{1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{3}{5} + \frac{4}{5}i$ is not a root of unity I want to prove that $z=\frac{3}{5} + \frac{4}{5}i$ is not a root of unity, although its absolute value is 1.
When transformed to the geometric representation: $$z=\cos{\left(\arctan{\frac{4}{3}}\right)} + i\sin{\left(\arctan{\frac{4}{3}}\right)}$$ Accordin... | From here, we know that
If $x$ is a rational multiple of $\pi$, then $2\cos(x)$ is an algebraic integer.
So if $x=\arctan(4/3)$ is a rational multiple of $\pi$, then $2\cos(x)=6/5$ is an algebraic integer, which is also a rational number, and hence an integer. This is a contradiction. Therefore $\frac{3+4i}5$ is not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition. Problem
Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition.
Note:
The problem asks us to prove that, no matter $x \to +\infty$ or $x \to -\infty$, the limit is $\infty$,which may be $+\infty$ or $-\infty.... | Your proof is correct.
You may consider shorter proofs by using some simplifications.
Since $x\to \infty $ we assume $x>1$
$$ \frac {x^3+1}{x^2+1} \ge\frac {x^3}{2x^2}=x/2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Limit of $\frac{y^4\sin(x)}{x^2+y^4}$ when $(x,y) \to (0,0)$ I have to find the following limit when $x,y$ tend towards $0$. I think the limit doesn't exist (thanks to Wolfram Alpha), but all I can find no matter what path I use (I've tried $y=x², x=y^2, x=0, y=0$) is that the limit equals $0$ (because of the $sin(0)$)... | Let $|x|,|y| <1$, $x,y$ real.
$0\le \left | \dfrac{y^4 \sin x }{x^2+y^4} \right | \le $
$\dfrac{y^4|x|}{x^4 + y^4} \le$
$\dfrac{(y^4+x^4)|x|}{x^4+ y^4}= |x| \le \sqrt{x^2+y^2}.$
Choose $\delta =\epsilon.$
Used : $|\sin x| \le |x|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the zeros of $f(z)=z^3-\sin^3z$ I want to find the zeros of $f(z)$, $$f(z)=z^3-\sin^3z$$
My attempt
$f(z)=0$
$z^3-(z-z^3/3!+z^5/5!-\dots)^3=0$
$z^3-z^3(1-z^2/3!+z^4/5!-\dots)^3=0$
$z^3[1-(1-z^2/3!+z^4/5!-\dots)^3]=0$
So $z=0$ is a zero of order $3$.
I don't feel good about this answer. Please give me some hints i... | Hint: The root at zero is more than fourth order. Also notice that
$z^3 - \sin^3 z \approx 0$
when $z= 5.3341 - 2.4614 i$. That should lead you to some more roots.
You can compute the lower order terms of the Taylor Series from the algebra in your "My Attempt" section.
$$z^3 - \sin^3z = z^3 - (z-z^3/3!+ z^5/5!-z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the volume of the solid obtained by rotating the region bounded by: Q: "Find the volume of the solid obtained by rotating the region bounded by:
$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$x + y = 4$, $ $ $ $$x = 5 - (y-1)^2$
$ $ $ $ $ $ $ $ $ $ $ $ $ $ About the $x-axis$."
I have attempted this problem for a while no... | I think shells will be easier than washer on this one.
Sketch your region. We have a line intersecting a parabola, but the parabola is sideways to the normal orientation.
shells $V = 2\pi \int_0^3 y((5-(y-1)^2)-(4-y))\ dy$
$2\pi \int_0^3 3y^2-y^3\ dy\\
2\pi (y^3 - \frac 14 y^4|_0^3) = 2\pi(27 -\frac {81}{4}) = \frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $7^n+2$ is divisible by $3$ for all $n ∈ \mathbb{N}$
Use mathematical induction to prove that $7^{n} +2$
is divisible by $3$ for all $n ∈ \mathbb{N}$.
I've tried to do it as follow.
If $n = 1$ then $9/3 = 3$.
Assume it is true when $n = p$. Therefore $7^{p} +2= 3k $ where $k ∈ \mathbb{N} $. Consider no... | Show that $7^n\equiv 1 (\text{mod} 3)$
$(6 + 1)(6 + 1) = 36 + 12 + 1$
$(6 + 1)(6 + 1)(6 + 1) = 216 + 72 + 6 + 36 + 12 + 1$
So all $(6 + 1)^n$ involve the sum of a series of terms which are multiples of $6$ (divisible by $3$) except for the last term which is $1$
Hence $7^n\equiv 1 (\text{mod} 3)$
and $7^n + 2$ is alway... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Induction: How to prove that $ab^n+cn+d$ is divisible by $m$.
If $a+d$, $(b-1)c$, $ab-a+c$ are divisible by $m$, prove that
$ab^n+cn+d$ is also divisible by $m$.
I want to prove this by induction. For proving $ab^{k+1}+c(k+1)+d$ is divisible by $m$, i want to prove that $ab^k(b-1)+c$ is divisible by $m$ and then ad... | Without induction. Let's do some tagging 1st
$$m \mid a+d \tag{1}$$
$$m \mid (b-1)c \tag{2}$$
$$m \mid ab-a+c \tag{3}$$
then
$$m \mid ab^n+cn+d \iff m \mid a\left(b^n-1\right)+cn+\color{red}{a+d} \overset{(1)}{\iff}\\
m \mid a\left(b^n-1\right)+cn \iff \\
m \mid a(b-1)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \iff \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How big does $r$ need to be, to ensure that $x^4+y^4 >r^2$ for all $x^2+y^2 =r^2$? Consider the circle $x^2+y^2 = r^2$ for some fixed $r>0$. How big does $r$ need to be, to ensure that $$x^4+y^4 >r^4$$ for all $x^2+y^2 =r^2$? I know that $x^2<x^4$ whenever $|x| >1$, but I'm not sure how to use that here.
| $x^4+y^4=r^4-2r^4\cos^2{\theta}\sin^2{\theta}=r^4(1-\frac{1}{2}\sin^2{2\theta})$
The minimum value of the term in parentheses is $\frac{1}{2}$
So $s^4<\frac{1}{2}r^4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Maximise $(x+1)\sqrt{1-x^2}$ without calculus Problem
Maximise $f:[-1,1]\rightarrow \mathbb{R}$, with $f(x)=(1+x)\sqrt{1-x^2}$
With calculus, this problem would be easily solved by setting $f'(x)=0$ and obtaining $x=\frac{1}{2}$, then checking that $f''(\frac{1}{2})<0$ to obtain the final answer of $f(\frac{1}{2})=\fra... | By AM-GM
$$(1+x)\sqrt{1-x^2}=\frac{1}{\sqrt3}\sqrt{(1+x)^3(3-3x)}\leq$$
$$\leq\frac{1}{\sqrt3}\sqrt{\left(\frac{3(1+x)+3-3x}{4}\right)^4}=\frac{3\sqrt3}{4}.$$
The equality occurs for $1+x=3-3x,$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Verify proof that $x_n = \sum_{k=1}^n {k \over 2^k}$ is bounded and find its supremum and infinum The problem I'm solving states:
Let $n\in \mathbb N$ and $x_n$ be a sequence:
$$
x_n = \sum_{k=1}^n {k \over 2^k}
$$
Prove $x_n$ is bounded and find $\sup\{x_n\}$ and $\inf\{x_n\}$
Let $S_n$ denote the sum, then:
$$... | It's may be easier to do it this way: since $(x_n)_n$ is strictly increasing, $\inf x_n=x_1=1/2$ and $\sup x_n=\lim x_n$, which you can calculate by noting that $f(x)=\sum_{n=0}x^n=\frac{1}{1-x};\ |x|<1.$
Now, $f$ may be differentiated inside its interval of convergence, so $x\sum_{n=1}nx^{n-1}=\frac{x}{(1-x)^{2}}.$ T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The rth term in $(1+x)^{1/x} = e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ Let
$$(1+x)^{\frac {1}{x}} = e.G(x)$$
Taking logarithm on both sides,
$$\frac {1}{x} \log {(1+x)} = 1 + \log {G(x)}$$
Putting in the Taylor expansion for $\log {(1+x)}$ we have,
$$\frac {1}{x}(x - \frac {1}{2}x^2 + \frac... | \begin{align}
\dfrac1e(1+x)^{1/x}
&= \exp\left(\dfrac{\ln(1+x)-x}{x}\right) \\
&= \exp\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right) \\
&= 1+\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)+\dfrac12\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)^2 \\
& +\dfrac16\left(-\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded
Let $n \in \mathbb N$ and:
$$
x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1}
$$
Prove $x_n$ is bounded sequence.
Start with $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{... | This may be weaker than the other ways (that are nice, including OP's), but seemed like an easy way to see it; note that $n \le \sqrt[3]{n^3+1} \le n+1$, and $n-1 \le \sqrt{n^2-1} \le n$ [can be verified by cubing & squaring, respectively]
So $\sqrt[3]{n^3+1} - \sqrt{n^2-1} \ge n - n = 0$ and $\sqrt[3]{n^3+1} - \sqrt{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Evaluate $\lim\limits_{n \to \infty}\frac{n}{\sqrt[n]{n!}}$. Solution
Notice that
$$(\forall x \in \mathbb{R})~~e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots.$$
Let $x=n$ where $n\in \mathbb{N_+}$. Then we obtain
$$e^n=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots>\frac{n^n}{n!}.$$
Thus, we obtain
$$e>\frac{n}{... | For the denominator, rewrite it as $ e^{\frac{\sum_{k=1}^{n} \log k}{n}}$ then approximate the sum of logs with the integral ($n \log n -n) $ to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
How can partial fractions be used for deductions?
Find partial fractions of the expression,$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}$
. Hence deduce that; $\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)}+\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)}+\frac{(d-p)(d-q)(... | You already got the correct partial fraction decomposition
$$
\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}
=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)} \cdot \frac{1}{x-a} \\
+ \frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \cdot \frac{1}{x-b} \\
+ \frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \cdot \frac{1}{x-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$, with $p$ being a prime number greater than $3$.
For $p \equiv 1 \mod 4$, there exists an integer $j$ such that $p\mid j^2+1$ (since $-1$ is a quadratic residue of $p$), therefore $R... | Let $p$ be a prime, with $p\equiv 3\;(\text{mod}\;4)$.
Claim:
$$\prod_{k=1}^p (1+k^2)\equiv 4\;(\text{mod}\;p)$$
Proof:
\begin{align*}
\prod_{k=1}^p (1+k^2)\;(\text{mod}\;p)
&\equiv
\prod_{k=1}^{p-1} (1+k^2)\;(\text{mod}\;p)
\\[4pt]
&\equiv
\left(\prod_{k=1}^{\frac{p-1}{2}}(1+k^2)\right)^{\!\!2}\;(\text{mod}\;p)
\\[4p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Proving the given inequalities Q: Prove the given inequalities for positive a,b,c:$(i) \left[\frac{bc+ca+ab}{a+b+c}\right]^{a+b+c}>\sqrt{(bc)^a.(ca)^b.(ab)^c}$$(ii) \left(\frac{a+b+c}{3} \right)^{a+b+c}<a^ab^bc^c<\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c} $I know that G.M$\le$A.M and somehow i guess it must be used... | $a)$We have by the weighted AM-GM inequality: $\sqrt{(ab)^{\frac{c}{a+b+c}}\cdot (bc)^{\frac{a}{a+b+c}}\cdot (ca)^{\frac{b}{a+b+c}}}\le \sqrt{\dfrac{3abc}{a+b+c}}\le \dfrac{ab+bc+ca}{a+b+c}\iff(ab+bc+ca)^2 \ge 3abc(a+b+c)\iff (ab)^2+(bc)^2+(ca)^2 \ge abc(a+b+c) $ which is true.
$b)$ Consider $f(x) = x\ln x \implies f'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Shortest distance between a point and a parabola. Find the shortest distance from the point $(1, 0)$ to the parabola $y^2 = 4x$.
As-
Let $D^2=(x-1)^2 + y^2$. Now reduce this to one variable by putting $y^2 = 4x$ to get $D^2 = (x-1)^2+ 4x$. We are working with $D^2$ to make the calculations easier. Clearly $$\frac{d... | Let $\left(\frac{y^2}{4},y\right)$ be a point on our parabola.
Thus, we need to find a minimal value of the following expression.
$$\sqrt{\left(\frac{y^2}{4}-1\right)^2+y^2},$$ which is equal to $1$ for $y=0$.
We'll prove that it's a minimal value.
Indeed, we need to prove that:
$$\left(\frac{y^2}{4}-1\right)^2+y^2\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum the first $n$ terms of the series $1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots$ The question
Sum the first $n$ terms of the series:
$$ 1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots. $$
This was asked under the heading using method of difference and the answer ... | Rewrite the formula as
$$\sum_2^{n+1}k^2(k+1)(k-1)=\sum_2^{n+1}k^4-\sum_2^{n+1}k^2.$$
It is a well known formula that
$$1+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.$$
Due to Fermat, we also have
$$1+2^4+3^4+\cdots+n^4=\frac15n^5+\frac12n^4+\frac13n^3-\frac1{30}n.$$
Just substitute in your values to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is the smallest integer greater than 1 such that $\frac12$ of it is a perfect square and $\frac15$ of it is a perfect fifth power?
What is the smallest integer greater than 1 such that $\frac12$ of it is a perfect square and $\frac15$ of it is a perfect fifth power?
I have tried multiplying every perfect square ... | Here's a very unsophisticated approach: Let $n$ be the smallest such integer. Then there exist integers $a$ and $b$ such that $n=5a^5$ and $n=2b^2$. It follows that $a$ is a multiple of $2$, say $a=2a_1$, and $b$ is a multiple of $5$, say $b=5b_1$. Then
$$n=2^5\cdot5\cdot a_1^5\qquad\text{ and }\qquad n=2\cdot5^2\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 6,
"answer_id": 2
} |
Derivation of the tangent half angle identity
I'm having trouble proceeding from
$$\frac{\sin(\theta)}{1+\cos(\theta)}$$
to
$$\tan\left(\frac{\theta}{2}\right)$$
Context:
Consider the function $f$ defined for all $(x,y)$ such that $y \neq 0$, with the rule $$f(x,y) = \frac{y}{\sqrt{x^2+y^2}+x}$$
Show that $... | Let $L=\frac{\sin \theta}{1+\cos \theta}$ and $R=\tan(\theta/2)$.
Left
\begin{align}
\frac{\textrm{d}L}{\textrm{d}\theta} &= \frac{\cos\theta(1+\cos\theta)-\sin\theta (-\sin\theta)}{(1+\cos \theta)^2}\\
&=\frac{1}{1+\cos\theta}
\end{align}
Again
\begin{align}
\frac{\textrm{d}^2L}{\textrm{d}\theta^2} &= -(1+\cos\theta)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
How many integers $a,b,c$, both positive and negative, such that $P=a^b b^c c^a$ is a prime number?
How many integers $a,b,c$, both positive and negative, such that $P=a^b b^c c^a$ is a prime number ?
If $a,b,c$ are positive, then two of $a,b,c$ equal to $1$. Assume that $b=c=1$, then $a$ is any prime number.
WLOG, i... | WLOG assume $b,c<0$, $a>0$. Then $a^b$ and $b^c$ will be fractional unless $a=1$, $b=-1$. So we need to find $1\cdot (-1)^c \cdot c=p$ for $c<0$. Since $c$ is negative, it must also be odd so that $(-1)^c$ will be negative and the entire product will be positive. Hence, $c$ can be the negative of any odd prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find $\cos(\alpha+\beta)$ if $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$
If $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$, then prove that $\cos(\alpha+\beta)=\dfrac{a^2-b^2}{a^2+b^2}$
My Attempt
$$
b\sin x=c-a\cos x\implies b^2(1-\cos^2x)=c^2+a^2\cos^2x-2ac\cos x\\
(a^2+... | To begin with, notice that
\begin{align*}
& a\cos(x) + b\sin(x) = c \Longleftrightarrow \frac{a}{\sqrt{a^{2}+b^{2}}}\cos(x) + \frac{b}{\sqrt{a^{2}+b^{2}}}\sin(x) = \frac{c}{\sqrt{a^{2}+b^{2}}}\Longleftrightarrow\\
& \sin(\theta + x) = \frac{c}{\sqrt{a^{2}+b^{2}}}\quad\text{where}\quad \sin(\theta) = \frac{a}{\sqrt{a^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Determinant 4x4 This looks very simple but I guess I made a mistake somwhere because the result should be 17
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 1 & -2 & 0 & 1 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$
First I added negative 1st row to 3rd
$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 0 & -4 & -4 & 3... | You start with
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 1 & -2 & 0 & 1 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$
and then, to the third row, you subtract the first, giving
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$
then duplicate 2nd row to get
$$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Determining what powers come out after polynomial multiplication Is there a quick method to determine what powers come out after polynomial multiplication? Specifically, I'm working with raising a polynomial by an integral power, so the binomial/multinomial theorem would be useful (though I have no idea how to use it).... | So for this case, the powers you can get are all the sums you can make using three terms (due to the power outside of the parenthesis) and the numbers 1,2, and 5 (the powers of the X's).
So since we allow for a given number to be repeated, there should be 10 different powers in the expansion (choose 3 terms from 3 ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding an integrating factor and solving: $(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$ I am trying to find an integrating factor and solve the following differential equation:
$$(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$$
These are my steps:
$$(2x \sin(x + y) + \cos(x + y)) + \cos(x + y)dy/dx = 0$... | $$(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$$
$$2x \sin(x + y)dx +( \cos(x + y))(dx +dy) = 0$$
Substitute $v=x+y$
$$2x \sin(v)dx + \cos(v)dv = 0$$
It's not exact. Multiply by $\mu=e^{x^2}$ as integrating factor
$$2xe^{x^2} \sin(v)dx + e^{x^2}\cos(v)dv = 0$$
The diffrential is exact..
$$\boxed{e^{x^2} \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
length of tangent to a curve passing through another point Let $A$ be a point on the curve
$\mathcal{C} : x^2+y^2-2x-4=0$
If the tangent line to $\mathcal{C}$ at $A$ passes through $P(4,3)$, then what is the length of AP?
Please, include a general method of approaching similar kind of questions.
| Welcome to MSE.Hint: first you have to find the center (o) of the curve which is a circle with equation:
$x^2-2x+y^2-4=(x-1)^2+(y-0)^2-5=0r^2$ or$(x-1)^2+(y-0)^2=5=r^2$
Where r is the radius of circle and O(-1, 0) is its center. In right triangle OAP we can write:
$AP^2=OP^2-r^2$
$OP^2=(4-1)^2+(3-0)^2=3^2+3^2=18$
$AP^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Inequality in difference of square roots I think this inequality is true and I'm trying to prove it:
For real, non-negative $a$ and $b$
$$\lvert \sqrt{a} - \sqrt{b}\rvert \le \lvert \sqrt{a - b} \rvert$$
Closest thing I've found is equation
$$\sqrt{a} + \sqrt{b} = \sqrt{a + b + \sqrt{4ab}}$$
But this breaks if I set $... | This is not true if $a < b$. If $a < b $ then $a -b < 0$ and $\sqrt{a-b}$ does not exist and $|\sqrt{a-b}|$ does not exist. And so it is not true that $|\sqrt{a} - \sqrt{b}| \le |\sqrt{a-b}|$.
I think what you meant was to prove $|\sqrt{a} -\sqrt{b}| \le \sqrt{|a - b|}$ which is true.
Notice: $\min (a,b) \le \sqrt{ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}$ without L'Hopital I have the following limit question, where different indices of roots appear in the numerator and the denominator
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}.$$
As we not allowed to use L'Hopital, I want to learn how we can pro... | A possible purely algebraic way can use
$$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$$
Using this formula note that
*
*$\sqrt[4]{x-2} - 1 = \frac{\left(\sqrt[4]{x-2}\right)^4 - 1^4 }{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1} = \frac{x-3 }{\left(\sqrt[4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Show that $2 < (1+\frac{1}{n})^{n}< 3$ without using log or binommial coefficient $2 <(1+\frac{1}{n})^{n}< 3$
Is it possible to show the inequality without using binommial coefficients thus only by induction? The leftinequality can be shown using bernoulli inequality.
| I have found the answer to my problem in a different forum:
We show that $b_n > b_{n+1}$ with $b_n:= (1+\frac{1}{n})^{n+1}$
Proof:
To show: $\frac{b_n}{b_{n+1}}\geq 1$
$\frac{b_n}{b_{n+1}} = \frac{\left(1+\frac{1}{n}\right)^{n+1}}{\left(1+\frac{1}{n+1}\right)^{n+2}} = \frac{1}{1+\frac{1}{n}} \cdot \left(\frac{1+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\sum_{(a,b,c)\in T}\frac{2^a}{3^b 5^c}$, for $T$ the set of all positive integer triples $(a,b,c)$ forming a triangle
Let $T$ be the set of all triples $(a,b,c)$ of positive integers for which there exist triangles with side lengths $a$, $b$, $c$. Express
$$\sum\limits_{(a,b,c)\,\in\,T}\;\frac{2^a}{3^b ... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\bi... | Given $$\sin x+\sin^2x+\sin^3x=1$$$$\sin x+\sin^3x=1-\sin^2x$$$$(\sin x+\sin^3x)^2=(1-\sin^2x)^2$$
$$\sin^2x+\sin^6x+2\sin^4x=\cos^4x$$
$$1-\cos^2x+(1-\cos^2x)^3+2(1-\cos^2x)^2=\cos^4x$$
$$1-\cos^2x+1-3\cos^2x+3\cos^4x-\cos^6x+1-4\cos^2x+2\cos^4x=\cos^4x$$$$\cos^6x-4\cos^4x+8\cos^2x=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
General way of solve $ax^2+by+c=0$ For example ,the diophantine equation
$$x^2+1=25y$$
we can solve this by finding particular solution $(x,y)=(7,2)$
and using this , we can get general solution.
My question is
"To solve $ax^2+by+c=0$ $(a, b, c \in \mathbb{Z})$ we must find particular solution or
not?"
| Since x and y have a difference like $d$,we can have a single variable quadratic equation by transformation.
If $x=y ±d$ we have:
$a(y ±d)^2+by+c=0$
$ay^2 +(b ±2ad)y ±ad^2+c=0$
$\Delta=(b ±2ad)^2-4a(c ±ad^2)>=0$
Limitation of value of $\Delta$ gives us a short range of numbers to try for solution. For example we solve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Recurrence Relation - # of binary strings with given property Let $a_n$ be the number of binary strings of length $n$ with the property that each entry is adjacent to at least one entry of the same type.
ex: $11000111$ is a valid string but $11011000$ is not valid
$\textbf{(a) Find $a_1,a_2,a_3,a_4,a_5,a_6,a_7$}$
If s... | You can take this as a sequence of stretches of at least two equal symbols. In ordinary generating function terms a sequence of two or more would be represented by:
$\begin{align*}
z^2 + z^3 + \dotsb
&= \frac{z^2}{1 - z}
\end{align*}$
A sequence of the above in turn would be:
$\begin{align*}
1 + \frac{z^2}{1 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given big rectangle of size $x, y$, count sum of areas of smaller rectangles. Let's say we have two integers $x$ and $y$ that describe one rectangle, if this rectangle is splitten in exactly $x\cdot y$ squares, each of size $1\cdot 1$, count the sum of areas of all rectangles that can be formed from those squares.
For ... | I assume, in your example, you mean there is one rectangle of size $(2, 2)$, because there are none $(4, 4)$.
In general, we have $(x - a + 1) \times (y - b + 1)$ rectangles of size $(a, b)$: we need to choose how many empty columns we leave before the rectangle $(0, 1, 2, \dots \text{ or } x-a)$, and how many rows we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Eigenvalue of a given operator If $u_0$ is a positive radial symmetric nontrival solution of
$$
-\frac{1}{2}\frac{d^2u}{dx^2}+\lambda u -u^3=0
$$
Then how to show $-3\lambda$ is a eigenvalue of
$$
Lu=-\frac{1}{2}\frac{d^2u}{dx^2}+\lambda u -3u_0^2 u
$$
and the corresponding eigenfunction is $u_0^2$ ?
What I try:
\beg... | Here is my try. Let $v_0=\frac{du_0}{dx}$. From
$$-\frac12\frac{d^2u_0}{dx^2}+\lambda u_0-u_0^3=0\implies \frac{d^2u_0}{dx^2}=2\lambda u_0-2u_0^4,$$
we have as you computed
$$Lu_0^2=-v_0^2-u_0\frac{d^2u_0}{dx^2}+\lambda u_0^2-3u_0^4=-v_0^2-u_0(2\lambda u_0-2u_0^4)+\lambda u_0^2-3u_0^4.$$
So $Lu_0^2=-v_0^2-\lambda u_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can area of rectangle be greater than the square of its diagonal?
Q: A wall, rectangular in shape, has a perimeter of 72 m. If the length of its diagonal is 18 m, what is the area of the wall ?
The answer given to me is area of 486 m2. This is the explanation given to me
Is it possible to have a rectangle of diago... | No, use the Pythagorean Theorem.
$$c^2 = a^2+b^2$$
$c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).
Recall for any real number, its square mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "50",
"answer_count": 12,
"answer_id": 3
} |
Proof of a Well-Known Fibonacci Identity Involving Cubes of Fibonacci Numbers The following is due to Lucas in 1876:
$$F_{n + 1}^3 + F_n^3 - F_{n - 1}^3 = F_{3n}$$
I am unable to locate an elementary proof of this identity, and am unable to reproduce it myself. Would anyone mind sharing a proof or a source?
| This is probably not the most elegant way, but you can use the explicit formula for the Fibonacci numbers.
$$
F_n = \frac{1}{\sqrt{5}} \tau^n + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^n
$$
Here $\tau = \frac{1}{2}+\frac{\sqrt{5}}{2}, -1/\tau = \frac{1}{2}-\frac{\sqrt{5}}{2}$. This formula for $F_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Attempt at sequence proof $\frac{n+3}{n^2 -3}$ converges to $0$
Prove convergence of the following sequence: $$\frac{n+3}{n^2 -3} \rightarrow 0$$
Proof discussion:
Notice that since whenever $n>3$, we have $n^2 -3 >0$, we also know that $n+3 >0$, so $\frac{n+3}{n^2 -3}>0$. This means we can drop the absolute value s... | Alternatively, decompose
$$\frac{n+3}{n^2-3}=\frac1{n-\sqrt3}+(3-\sqrt3)\frac1{n^2-3}$$
and the two terms are of the form $\dfrac1\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is $\lim_{x \to 3} (3^{x-2}-3)/(x-3)(x+5)$ without l'Hôpital's rule? I'm trying to solve the limit $\lim_{x \to 3} \frac{3^{x-2}-3}{(x-3)(x+5)}$
but I don't know how to proceed: $\lim_{x \to 3} \frac{1}{x+5}$ $\lim_{x \to 3} \frac{3^{x-2}-3}{x-3}$ = $1\over8$ $\lim_{x \to 3} \frac{\frac{1}{9}(3^{x}-27)}{x-3}$
Any... | We have that with $y=x-3 \to 0$
$$\lim_{x \to 3} \frac{3^{x-2}-3}{(x-3)(x+5)}=\lim_{y \to 0} \frac{3\cdot 3^{y}-3}{y(y+8)}=\lim_{y \to 0} \frac{3}{y+8}\frac{3^{y}-1}{y}$$
then recall that by standard limits
$$\frac{3^{y}-1}{y}=\frac{e^{y\log 3}-1}{y\log 3}\cdot \log 3 \to \log 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Show that $P(x,y)=0$ is an ellipse if $b^2-4ac<0$. I tried the following:
I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax^2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $\Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^... | For i.
Let $Y=\Delta_x(y)$. Then, note that $Y=\Delta_x(y)$ is the equation of a parabola. Moreover, the coefficient of $y^2$ is $b^2-4ac$ which is negative, so the parabola opens down. So, only one of the three cases happens.
For ii.
for example it can be a part of a hyperbola too [the normal one $\pi/2$ rotated] sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate and Simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$ I am trying to evaluate and simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$.
I am getting $\frac{11}{10}$ but the answer is $\frac{3-4\sqrt{3}}{10}$
My Process:
$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$
$\cos[\cos^{-1}(\frac{3}{5})] + \co... | By compound-angle formula,
\begin{eqnarray*}
\cos(\cos^{-1}(\frac{3}{5})+\frac{\pi}{3}) & = & \cos\left(\cos^{-1}(\frac{3}{5})\right)\cos\frac{\pi}{3}-\sin\left(\cos^{-1}(\frac{3}{5})\right)\sin(\frac{\pi}{3})\\
& = & \frac{3}{5}\cdot\frac{1}{2}-\sin\left(\cos^{-1}(\frac{3}{5})\right)\cdot\frac{\sqrt{3}}{2}.
\end{eqn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Classifying singularities of $f(z) = \frac{1}{(z-2)^2}+e^{\frac{1}{3-z}}$
Classify all the singularities of $f(z) = \frac{1}{(z-2)^2}+e^{\frac{1}{3-z}}$
I know that the singularities of $f$ are exactly $z=2,z=3$. I just want to check that if my solution for the classification is correct:
$z=2$: since $1/(3-z)$ is ana... | Your reasoning is correct.
$e^{-\frac{1}{z-3}}$ is analytic at $z=2$, so it can be expanded in a Taylor series about that point in the region $0<|z-2|<1$:
\begin{align}e^{-\frac{1}{z-3}}=e+e(z-2)+\frac{3e}{2}(z-2)^2+\cdots\end{align}
Therefore, the Laurent series of $f(z)$ about $z=2$ in the region $0<|z-2|<1$ is give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove the limit lim n→∞ $ {(2n^7 + 3n^5 + 4n) \over (4n^7 − 7n^2 + 5)} $ = 1/2 Im trying to use the formal definition of a limit to prove
$\lim \limits_{x \to ∞} $$ {(2n^7 + 3n^5 + 4n)\over(4n^7 − 7n^2 + 5)}$ = 1/2
I understand this problem is done backwards so I set up the equation
$ {(2n^7 + 3n^5 + 4n)\over(4n^7 − 7n... | Note that\begin{align}\left\lvert\frac{2n^7+3n^5+4n}{4n^7-7n^2+5}-\frac12\right\rvert&=\frac{\lvert6n^5+7n^2+8n-5\rvert}{\lvert8n^7-14n^2+10\rvert}\\&\leqslant\frac{26n^5}{8n^7-14n^2-10}\\&=\frac{26}{8n^2-14n^{-3}-10n^{-5}}\\&\leqslant\frac{26}{8n^2-24},\end{align}if $n>1$. Now, use the fact that\begin{align}\frac{26}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A Fibonacci convolution
A Fibonacci convolution. Recall that $$F(x)=\sum_{n=0}^\infty F_n x^n =\frac{x}{1-x-x^2} =\frac{1}{\sqrt{5}} \left(\frac{1}{1-\Phi x} -\frac{1}{1-\bar{\Phi}x}\right).$$
(a) Prove that $\displaystyle \sum_{n=0}^\infty F_{n+1} x^n =\frac{1}{1-x-x^2}$.
(b) Prove that $\displaystyle \sum_{n=0}^\... | Since you have already mastered a.) to c.) we can conveniently use the results to prove d.).
We obtain
\begin{align*}
\color{blue}{5F(x)^2}&=\sum_{n=0}^{\infty}(n+1)\left(\Phi^n+\bar{\Phi}^n\right)x^n-2\sum_{n=0}^\infty F_{n+1}x^n\tag{1}\\
&=\sum_{n=0}^\infty(n+1)\left(2F_{n+1}-F_n\right)x^n-2\sum_{n=0}^\infty F_{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3016349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
limit of $\sqrt{x^6}$ as $x$ approaches $-\infty$ I need to solve this limit:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}$$
The answer is $-3$, but I got 3 instead. This is my process:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6(9-\frac {5}{x^2})}... | To check or avoid confusion with sign let $y=-x \to \infty$ then
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}=\lim_{y \to \infty}{\frac {\sqrt{9y^6+5y}}{-y^3-2y^2+1}}=-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Does $z^4+2z^2+z=0$ have complex roots? Does $z^4+2z^2+z=0$ have complex roots? How to find them? Besides $z=0$, I got the equation $re^{3i\theta}+2re^{i\theta}=e^{i(\pi+2k\pi)}$, $k\in \mathbb Z$. How to find the complex roots?
| Enough to check the discriminant for $ax^3+bx^2+cx+d=0$:
$$
\Delta_3=-27 a^2 d^2+18 a b c d-4 a c^3-4 b^3 d+b^2 c^2
$$
$$
\Delta_3=\begin{cases} >0 & \text{3 distinct real roots}\\
<0 & \text{1 real, 2 conjugate complex roots}\\
=0 & \text{3 real roots with duplicates}\\
\end{cases}
$$
In your case, it's $a=1$, $b=0$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3018093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Values of $x$ satisfying $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$
For what values of $x$ between $0$ and $\pi$ does the inequality $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$ hold?
My Attempt
$$
\sin x\cos x\cdot(\cos^2x-\sin^2x)=\frac{1}{2}\cdot\sin2x\cdot\cos2x=\frac{1}{4}\cdot\sin4x>0\implies\sin4x>0\\
x\in(0,\pi)\im... | The given solution is wrong; you are correct. At $x=\frac{7\pi}8\in\left(\frac{3\pi}4,\pi\right)$, we have that $$\frac14\sin4x=\frac14\sin\frac{7\pi}2=-\frac14<0$$ which is a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Express a parametrization of a curve $C$ as a function of $X$ and $Y$ Express the curve $C$ with parametrization $\{(\cos 3t, \sin 2t) : t \in \mathbb{R} \}$ as a function of $X$ and $Y$.
Here's what I have (I don't know if this is helpful)
I let $X= \cos 3t$ and $Y= \sin 2t$. So I have $X= \cos^3 t -3\sin^2 t \cos t$ ... | Answer:
$$
X^4 - X^2 + 4 Y^6 - 6 Y^4 + \frac{9}{4} Y^2 = 0
$$
How I got it:
We can replace the trigonometric parametrization $(\cos(u), \sin(u))$ of the unit circle with a rational one: $\left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right)$. So we equate $\cos(u) = x = \frac{1-t^2}{1+t^2}$ and $\sin(u) = y =\frac{1-t^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
square root system of equations
I have a system of equations as follows that I need to solve for $x$:
$$
\sqrt{1 - x^2} + \sqrt{4 - x^2} = z\\
\sqrt{4 - y^2} + \sqrt{9 - y^2} = z
$$
Originally, I was trying to put $x$ in terms of $z,$ but I’m at a loss how I would solve this system. Trying to solve it just makes it... | Both the $x$ and $y$ equations have the form
$$\sqrt{a^2-h^2} + \sqrt{b^2 - h^2} = c$$
For everything to be real, we need $\sqrt{|a^2-b^2|} \le c \le a + b$.
Assume this is satisfied. For positive solution of $h$, this equation has a geometric interpretation. It is the equation for the height $h$ of a triangle of side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$
Find the number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$
\begin{align}
2\cos(\pi\sqrt{x-4})&.\cos(\pi\sqrt{x})=2\\\implies\cos\Big[\pi(\sqrt{x-4}+\sqrt{x})\Big]&+\cos\Big[\pi(\sqrt{x-4}-\sqrt{x})\Big]=2\\
\implie... | More simply we need as a necessary condition
*
*$\pi\sqrt{x-4}=k_1\pi$
*$\pi\sqrt{x}=k_2\pi$
that is
*
*$\sqrt{x-4}=k_1 \implies x=k_1^2+4$
*$\sqrt{x}=k_2\implies x=k_2^2$
that is
$$k_1^2+4=k_2^2 \iff k_2^2-k_1^2=4$$
and the only possible solutions is $k_1=0 \implies x=4$ which works by inspection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Upper bound for a ratio of two least common multiples Let $\text{lcm}(x)$ be the least common multiple of $\{1,2,3,\dots,x\}$
Let $x\#$ be the the primorial for $x$.
It occurs to me that for $x \ge 10$:
$$\frac{\text{lcm}(x^2+x)}{\text{lcm}(x^2)} < 2^x\frac{(x^2+x)\#}{(x^2)\#}$$
Am I right?
Here is my thinking:
(1) $\... | HINT $1$
Note that
$$r(x)={\large \dfrac {\mathrm{lcm}(x^2+x)}{\mathrm{lcm}(x^2)}\cdot{\dfrac {(x^2)\#}{(x^2+x)\#}} = \prod_{p_k<x+1} p_k^{[\log_{p_k}(x^2+x)]-[\log_{p_k}(x^2)]}}.$$
Calculations using Wolfram Alpha of function
$$\bar r(x) = \left(\dfrac45\right)^x\prod_{p_k<x+1} p_k^{[\log_{p_k}(x^2+x)]-[\log_{p_k}(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3026176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
I have tried to solve this question and did pretty well until I reached the end, so I was wondering if I... | $$(a+b-c)^2 = 2(c-a)(c-b)$$ is an equivalent statement of the Pythagorean formula. It is a well known fact that either a or b must be even. So from there, identify that $b^2=(c-a)(c+a) \text{ and } a^2=(c-b)(c+b)$. So it is unavoidable that either $(c-a)$ or $(c-b)$ must be even. This covers off the right side of the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Number Theory: Prove that $\gcd(a,b) \le \sqrt{a+b}$
For positive integers $a$ and $b$, we know $\dfrac{a+1}{b} + \dfrac{b+1}{a}$ is also
a positive integer. Prove that $\gcd(a,b) \le \sqrt{a+b}$.
Using Bézout's lemma, we know that $\gcd(a, b) = sa + tb$. I want to prove that $(sa+tb)^2 \le a+b$. We know $ab\,|\,a(... | Hagen von Eitzen does not seem to want to explain his answer further, so I will attempt to do so here:
Suppose that $d|a$ and $d|b$. Then we have the following:
\begin{align*}
\frac{a+b}{d^2} &= \frac{a^2+a}{d^2}+\frac{b^2+b}{d^2}-\frac{a^2}{d^2}-\frac{b^2}{d^2} \\ &=\left(\frac{a+1}{b}+\frac{b+1}{a}\right)\cdot\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The series $\frac{1}{2}+\frac{2}{5}+\frac{3}{11}+\frac{4}{23}+...$ Consider the expression $\frac{1}{2}+\frac{2}{5}+\frac{3}{11}+\frac{4}{23}+...$
Denote the numerator and the denominator of the $j^\text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1\qqua... | If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)\Rightarrow d_n+1=(d_1+1)2^{n-1}=3\cdot2^{n-1}$.
So 50th term is $\frac{50}{3\cdot2^{49}-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Seeking Methods to solve $F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$ I'm looking for different methods to solve the following integral.
$$ F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$$
For $\alpha > 0$
Here the method I took was to employ integration by parts and then call to special ... | Answer 2.0:
We know that for $|x|<1$,
$$\arcsin x=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)}x^{2k+1}$$
Where $$(1/2)_k=\frac{\Gamma(1/2+k)}{\Gamma(1/2)}$$
Hence we may begin with
$$F(a)=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)}\int_0^1x^{2k+a+1}\mathrm dx=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)(2k+2+a)}$$
Then we play with the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{n\to\infty} \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8})\ldots \cos(\frac{\pi}{2^n}) $ I already know that $$ a_n = \cos\left(\frac{\pi}{2^{n+1}}\right) = \overbrace{\frac{\sqrt{2+\sqrt{2+\ldots + \sqrt{2}}}}{2}}^{n\text{ roots}}$$
Also I know that $$\lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^n}\right) = 2
\tex... | If $x_n=\cos(\frac{\pi}{4}) \cos(\frac{\pi}{8})\ldots \cos(\frac{\pi}{2^n}) $ then $\ $
$$x_n\sin (\frac{\pi}{2^n})= \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8}) \ldots
\cos(\frac{\pi}{2^n}) \sin (\frac{\pi}{2^n}) $$
$$=\frac{1}{2^1} \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8}) \ldots
\cos(\frac{\pi}{2^{n-1}}) \sin (\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
$\lim_{n \to \infty}(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})=e^{\frac{1}{2}}$. Here is the beginning of a proof:
Suppose $0<k \leq n$,
$1+\frac{1}{n}<(1+\frac{k}{n^2})(1+\frac{n+1-k}{n^2})=1+\frac{n+1}{n^2}+\frac{k(n+1-k)}{n^4}\leq 1+\frac{1}{n}+\frac{1}{n^2}+\frac{(n+1)^2}{4n^4}$.
I'm confused by the sec... | As an alternative, we have that
$$\left(1+\frac{1}{n^2}\right)\left(1+\frac{2}{n^2}\right)\ldots\left(1+\frac{n}{n^2}\right)=\prod_{k=1}^{n}\left(1+\frac{k}{n^2}\right)=e^{\sum_{k=1}^{n} \log\left(1+\frac{k}{n^2}\right) }$$
and
$$\sum_{k=1}^{n} \log\left(1+\frac{k}{n^2}\right)=\sum_{k=1}^{n}\left(\frac{k}{n^2}+k^2O\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Arranging 3 girls and 9 boys
Three girls A, B and C, and nine boys are lined up in a row. In how many ways this can be done if B must lie between A and C, and A, B must be separated by exactly 4 boys.
I have used the following approach. First select 4 boys ${9 \choose 4}$ and then combine them in between A and B. Aft... | As pointed out in the comments by Barry Cipra, you made two errors. They are:
*
*$A$ can be to the left or right of $B$, so you have to double your answer.
*The number of ways to distribute $5$ identical objects to $3$ compartments is $\binom{5 + 3 - 1}{3 - 1}$.
With these corrections, your answer becomes
$$2\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3041775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int \frac{x^3}{\sqrt {x^2+1}}dx$ Evaluate $$\int \frac{x^3}{\sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=\sqrt {u-1}$ and $dx=\frac{1}{2\sqrt{u-1}}du$.
Therefore, $$\int \frac{(u-1)\sqrt{u-1}}{\sqrt u}\frac{1}{2\sqrt {u-1}}du$$
$$=\frac{1}{2}\int {\sqrt u}-{\frac {1}{\sqrt u}du}$$
$$={\frac{2}{3}}{{(x^... | Method 1:
Another approach using trigonometric substitutions
\begin{equation}
I =\int \frac{x^3}{\sqrt {x^2+1}}dx
\end{equation}
Here let $x = \tan(\theta)$ we arrive at:
\begin{align}
I &=\int \frac{\tan^3(\theta)}{\sqrt {\tan^2(\theta)+1}}\sec^2(\theta)\:d\theta = \int \frac{\tan^3(\theta)}{\sec(\theta)}\sec^2(\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
exponential equation has non positive roots
Find real values of $a$ for which the equation
$4^x-(a-3)\cdot 2^x+(a+4)=0$
has non positive roots
Try: Let $2^x=y\in (0,1].$ Then equation convert into
$y^2-(a-3)y+(a+4)=0$
For real roots its discriminant $\geq 0$
So $$(a-3)^2-4(a+4)\geq 0$$
$$a^2-10a-7\geq 0$$
$$a\in \big... | Not only the equation in $y$ must have real roots, but these roots have to be $\le 1$, since it's asked that the equation in $x$ has non-positive roots, so that $y=2^x\le 2^0=1$.
To test whether the roots are less than $1$, we can place $1$ w.r.t. the roots $y_1,y_2$ of $\;p(y)=y^2-(a-3)y+a+4$. The standard method is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
An exercise on the calculation of a function of operator The operator is given by
$$A=\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 4
\end{pmatrix}$$
I have to write down the operator $$B=\tan(\frac{\pi} {4}A)$$
I calculate $$\mathcal{R} (z) =\frac{1}{z\mathbb{1}-A}=\begin{pmatrix}
\frac{1}{z-1} & 0 & 0\\
\frac{1}{(... | Your calculation seems plausible. Let's try it the other way: the Taylor expansion of $\tan(\pi x/4)$ at $x=1$ is
$$
\tan(\pi x/4)=1+\frac{\pi}{2}(x-1)+\ldots
$$
Then for the Jordan block
$$
J=\begin{bmatrix}1 & 0\\1 & 1\end{bmatrix}
$$
we have
$$
\tan(\pi J/4)=I+\frac{\pi}{2}\begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}=\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do I solve $\lim \limits_{x \to \frac{π}{3}} \frac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$ Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.
But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.
Put $t= x- \dfrac{π}{3}$
$\lim \limits_{x... | $$\lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}=\lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}\cdot\frac{3t/2}{3t/2}$$
$$=\frac{2}{3} \lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ t}\ \ \ \ \ \ \ (1)$$
Now, $1-\cos t=2\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Proof that $x^{(n+4)} \bmod 10 = x^n \bmod 10\,$ for $\,n\ge 1$ While solving a programming challenge in which one should efficiently compute the last digit of $a^b$, I noticed that apparently the following holds (for $n > 0$)
$x^{(n+4)} \mod 10 = x^n \mod 10$
How can this be proven?
| Hint
$$ x^{n} \left( x^4 - 1 \right) \equiv x^{n} \left( x^2 - 1 \right) \left( x^2 + 1 \right)\equiv \left( x - 1 \right) x \left(x+1\right) x^{n-1} \left( x^2 + 1 \right) \pmod{2}\\
x^{n} \left( x^4 - 1 \right) \equiv x^{n} \left( x^2 - 1 \right) \left( x^2 + 1 \right)\equiv x^{n} \left( x^2 - 1 \right) \left( x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 7
} |
Trigonometric parametric system I have a very specific system of two trigonometric equations
$$\left( 3A^2\sin x \cos x - A \sin x \right) + \left( 3B^2\sin y \cos y - B \sin y \right) = AB \sin (x+y)$$
$$\left( 3A^2\cos^2 x - A \cos x \right) + \left( 3B^2\cos^2 y - B \cos y \right) = AB \cos (x+y) - \frac{3}{2}\left(... | For the first equation,
$$\left( 3A^2\sin x \cos x - A \sin x \right) + \left( 3B^2\sin y \cos y - B \sin y \right) = AB \sin (x+y)$$
set $A=0$,
you get
$$ 3B^2\sin y \cos y - B \sin y = 0$$
Let $\sin y=\alpha$
$$3B^2\alpha\sqrt{1-\alpha^2}-B\alpha=0$$
$$9B^4\alpha^2(1-\alpha^2)=B^2\alpha^2$$
Now we can see one of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral between 2 spheres
Let D be the set of all points $(x, y, z)$ satisfying
$ 1 \leq x^2 + y^2 + z^2 \leq 2 $ and $ z \geq 0 $
, find $\int_{D}x^2 $
how do i solve this question through triple integral and spherical co-ordinates ?
should i calculate $ \int_{r=1}^{r=\sqrt{2}}\int_{\theta = 0}^{\theta = 2\pi... | First, notice that, due to symmetry
$$\iiint\limits_{1<r<\sqrt 2,z>0}x^2 dxdydz=\frac{1}{2}\iiint\limits_{1<r<\sqrt 2}x^2 dxdydz$$
Now
$$\iiint\limits_{1<r<\sqrt 2}x^2 dxdydz = \iiint\limits_{1<r<\sqrt 2}y^2 dxdydz = \iiint\limits_{1<r<\sqrt 2}z^2 dxdydz$$
Thus,
$$\iiint\limits_{1<r<\sqrt 2,z>0}x^2 dxdydz = \frac{1}{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $x$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$ Find all $x \in (-1, +\infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.
What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:
$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} =... | Hint:
$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$
$$\iff1=\left(\dfrac{2x+4}{x^2+4x+5}\right)^2+\left(\dfrac{x^2+4x+3}{x^2+4x+5}\right)^2$$
We have
$$\left(\dfrac{2x+4}{x^2+4x+5}\right)^x+\left(\dfrac{x^2+4x+3}{x^2+4x+5}\right)^x=1$$
WLOG $\dfrac{2x+4}{x^2+4x+5}=\cos t,\dfrac{x^2+4x+3}{x^2+4x+5}=\sin t$ with $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
If $\lim\limits_{x\to\alpha}\frac{x-2}{x^3-2x+m}=-\infty$, then what are the possible values for $\alpha$ and $m$?
If $\lim\limits_{x\to\alpha}\dfrac{x-2}{x^3-2x+m}=-\infty$, then what are the possible values for $\alpha$ and $m$?
A student I'm tutoring came to me with this problem. I believe the limit is not one-sid... | Using the given condition we can see that $(x^3-2x+m)/(x-2)$ is negative and tends to $0$ as $x\to a$ (replaced $\alpha$ with $a$ to simplify typing). Note that the given condition also excludes $a=\pm\infty$ and hence we assume $a\in\mathbb {R} $. Then by multiplication with $(x-2)$ we get $$\lim_{x\to a} (x^3-2x+m)=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Factors in a different base $\ 2b^2\!+\!9b\!+\!7\,\mid\, 7b^2\!+\!9b\!+\!2$
Two numbers $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number, then what is the value of $B$?
Solution:
But base cannot be negative. Could someone please explain where I am going wro... | $$2B^2+9B+7\mid 7B^2+9B+2$$
Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.
Then $[2,9,7]_B \mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$
Writing this out "subtraction-style", we get
$\left.\begin{array}{c}
& 7 & 9 & 2 \\
-& 2 & 9 & 7 \\
\hline
\phantom{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})\cdots(1+\frac{1}{a_n})$, where $a_1=1$, $a_n=n(1+a_{n-1})$
Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\cdots(1+\frac{1}{a_n})$,
where $a_1 = 1$, $a_n = n(1+a_{n-1})$
\begin{align*}
&\lim_{n\to\infty} \left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\rig... | From where you stuck you can proceed as
\begin{align}
\frac{a_n+1}{n!}&=\frac{a_n}{n!}+\frac{1}{n!}=\frac{1}{(n-1)!}\frac{a_n}{n}+\frac{1}{n!}=\frac{1}{(n-1)!}(a_{n-1}+1)+\frac{1}{n!}=\\
&=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}=\ldots=
1+\frac{1}{1!}+\ldots+\frac{1}{n!}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $A=A^2$ is then $A^T A = A$? I know that for a matrix $A$:
If $A^TA = A$ then $A=A^2$
but is it if and only if? I mean:
is this true that "If $A=A^2$ then $A^TA = A$"?
| The answer is no.
Consider $A = \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}$. We have
$$A^2 = \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} = A$$
but $$A^TA = \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} 1 & -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluating an improper integral $\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$ I tried to solve the integral: $$\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$$
using $ x = \sqrt{\tan(u)}$ and $dx = \frac{ \sec^2(u)}{2\sqrt{\tan(u)}} du,$
but I ended up with an even worse looking integral $$ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\... | We first reduce the power 2 in the denominator by integration by parts.
$$
\displaystyle \begin{aligned}
\int \frac{x^{2}}{\left(x^{4}+1\right)^{2}} d x &=-\frac{1}{4} \int \frac{1}{x} d\left(\frac{1}{x^{4}+1}\right) \\
&\stackrel{ IBP }=-\frac{1}{4 x\left(x^{4}+1\right)}-\frac{1}{4} \int \frac{1}{x^{2}\left(x^{4}+1\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\le... | applying summation by parts
\begin{align}
\sum_{n=1}^N\frac{H_n}{n(n+1)}&=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^N\frac{H_n}{n+1}\\
&=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N+1}\frac{H_{n-1}}{n}\\
&=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N+1}\frac{H_{n}}{n}+\sum_{n=1}^{N+1}\frac{1}{n^2}\\
&=\sum_{n=1}^N\frac{H_n}{n}-\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Prove the identity for $\tan3\theta$
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$
Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\c... | Alternatively, to do it using what you've done so far, rather than applying the double-angle formula twice:
Note that $\frac{\sin\theta}{\cos^3\theta} = \tan\theta\sec^2\theta = \tan\theta(1 + \tan^2\theta) = \tan\theta + \tan^3\theta$. Alternatively, to do it using what you've done so far, rather than applying the do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Choosing an integer that is most likely to be the sum of die rolls Choose any positive integer $n$. Then roll an unbiased six-sided die as many times as needed until the sum of the results is either exactly equal to $n$ (you win) or greater than $n$ (you lose).
If you were to play this game; what number $n$ would you s... | Denote by $p_{k}$ the probability that a sum of $k$ is observed. Clearly
$$p_{1}=\frac{1}{6}$$
For $p_{2}$, there is a $\frac{1}{6}$ chance of rolling a $2$ and a $\frac{1}{6}$ chance of rolling a $1$, followed by another one. This can be written as
$$p_{2}=\frac{1}{6}+\frac{p_{1}}{6}=\frac{1}{6}(1+p_{1})$$
For $p_{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I create a function which describes the length between a parabola and a point on the y axis?
This is what I'm thinking:
The distance of the line segment is between points (0,y) and$ (x,y)$ and the y coordinate of the parabola is $4-x^2
so using he distance formula it is:
$\sqrt{(x-0)^{2} +((4-x^2)-y)^{2}} = ... | Distance form point $(x,y)$ on the graph and $(0,2)$ can be parametrized by;
$L(x) = \sqrt{x^2 + (|y-2|)^2} = \sqrt{x^2 + (2-x^2)^2} = \sqrt{4 - 3 x^2 + x^4}$ (assuming $y>2$)
To find the minimum use the given point $\sqrt{3/2}$ and put it into $L(X)$
$L(\sqrt{3/2}) = \sqrt{4 - 3 (\sqrt{3/2})^2 + (\sqrt{3/2})^4} = \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $a, b \in \mathbb{R^+}$ such that $a-b=10$, find smallest value of constant $K$ for which $\sqrt{(x^2 + ax)}- \sqrt {(x^2 +bx)}0$. Let $a, b$ are positive real numbers such that $a - b = 10$ , then the smallest value of the constant $K$ for which $\sqrt{(x^2 + ax)} - \sqrt {(x^2 +bx)} < K$ for all $x>0$.
My try:
U... | Hint.
$$
\sqrt{(x^2 + ax)} - \sqrt {(x^2 +bx)} = x\left(\sqrt{\frac{a}{x}+1}-\sqrt{\frac{b}{x}+1}\right)
$$
and for large $x$
$$
\sqrt{\frac{a}{x}+1}-\sqrt{\frac{b}{x}+1} = \frac{a-b}{2 x}+\frac{1}{8} \left(\frac{1}{x}\right)^2
\left(b^2-a^2\right)+O\left(\left(\frac{1}{x}\right)^3\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3075896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int\frac{\cos^2(x)-x^2\sin(x)}{(x+\cos(x))^2}dx$ I had to integrate the following integral:
\begin{equation}
\int\frac{\cos^2(x)-x^2\sin(x)}{(x+\cos(x))^2}dx
\end{equation}
but I can't find a suitable substitution to find a solution. Nothing I try works out and only seems to make it more complicated. Does a... | \begin{equation}
\int\frac{\cos^2 x-x^2\sin x }{(x+\cos x)^2}dx
\end{equation}
Divide both numerator and denominator by $x^2\cos^2 x$
\begin{equation}
\int\frac{\cos^2 x-x^2\sin x }{(x+\cos x)^2}dx
=\int\frac{\dfrac{1}{x^2}-\dfrac{\sin x}{\cos^2 x}}{(\dfrac{1}{\cos x}+\dfrac{1}{x})^2}dx
=\frac{1}{\dfrac{1}{\cos x}+\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3078486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How do I solve for $a$ and $b$ in $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \d... |
Since the limit exists, $2x$ and $x(a−b)$ must cancel out so $a−b=2$.
You wanted to say $2+a-b=0 \Rightarrow \color{red}{a-b=-2}$.
And you cannot go from the difference of logarithms (line $2$) to the difference of arguments (line $3$), because it is not a factor, but a sum $(2+\cdots)$, which is multiplied by $x$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Finding all roots of $x^6 + 2 x^5 + 2 x^4 + 3 x^2 + 6 x + 6$, knowing that one root is $x=-1+i$.
Find all roots of
$$V(x) = x^6 + 2 x^5 + 2 x^4 + 3 x^2 + 6 x + 6$$
knowing that one root is $x=-1+i$.
Sorry for the picture. I found two roots of the polynomial and also found an equation which can help me to find all... | One factor is $$x^2+2x+2=0$$ which can be solved by the quadratic formula. Solving $$x^4+3=0$$ we get
$$\frac{1}{2}\sqrt{2}\cdot3^{1/4}(1+i),\frac{1}{2}\sqrt{2}\cdot3^{1/4}(1-i),\frac{1}{2}\sqrt{2}\cdot3^{1/4}(-1+i),\frac{1}{2}\sqrt{2}\cdot3^{1/4}(-1-i)$$
Substituting $$x^2=t$$ then we get $$t^2+(\sqrt{3})^2=0$$ By the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $a, b\in\mathbb{R}$ so that $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$ is finite.
Find $a, b\in\mathbb{R}$ so that
$$\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$$
is finite.
I tried to use polynomial division, but the computations get tedious really fast. Any suggestions?
| First of all, you need the numerator to be zero at $x=1$, which implies $a+b+1=0$. When you substitute, say, $a=-b-1$, the numerator becomes $-bx^5(x-1)-(x^6-1)$ which, when factoring out $(x-1)$ gives $(x-1)(-bx^5-x^5-x^4-x^3-x^2-x-1)$. After canceling $x-1$ we still have $x-1$ in the denominator, and for the limit to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving a strict inequality in the limit I want to prove that
$$ \lim_{k \to \infty} \left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k} \right) < e .$$
Using the $AM-GM$ inequality we arrive at
$$\left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k... | You can just ignore the first term and carry on with your method from $i=2$.
Then you get $\prod\limits_{i=2}^ka_i\le\cdots\le\left(1+\dfrac 1{2(k-1)}\right)^{k-1}\le e^{0.5}$
And then $\prod\limits_{i=1}^\infty a_i=\left(1+\dfrac 12\right)\prod\limits_{i=2}^\infty a_i\le 1.5\,e^{0.5}<e$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Determine Fourier series expansion for $f(\theta)=\cos^4\!\theta$ Q: The function $f(\theta) = \cos^4\! \theta$ is a nice smooth function, so will have a Fourier series expansion. That is, it will have an expansion as a sum of functions $\cos j \theta$ and $\sin j \theta$ with real coefficients. Determine what the expa... | Use this ^.^
$\displaystyle \begin{array}{{>{\displaystyle}l}}
For\ odd\ n:\ cos^{n}( x) =\left(\frac{e^{ix} +e^{-ix}}{2^{n}}\right)^{n} =\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\binom{n}{k} \cdotp \frac{cos( x( n-2k))}{2^{n-1}}\\
\\
For\ even\ n:\ cos^{n}( x) =\left(\frac{e^{ix} +e^{-ix}}{2^{n}}\right)^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the numerical value of this expression If $x$ is a complex number such that $x^2+x+1=0$, then the numerical value of $(x+\frac{1}{x})^2+(x^2+\frac{1}{x^2})^2+(x^3+\frac{1}{x^3})^2+\ldots+(x^{27}+\frac{1}{x^{27}})^{2}$ is equal to?
A) 52 . B) 56 . C) 54. D)58 . E)None of these
Where is this question from? I... | Thank you for all your answers, I just find the other solution to solve this, so I want to post it here.
Since $x^2+x+1=0$, so we can find
$x+\frac{1}{x}=-1$
and $x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2=(-1)^2-2=-1$. Let's keep going
$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2})-(x+\frac{1}{x})=(-1)(-1)-(-1)=2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
When does the first repetition in $\;\lfloor x\rfloor, \lfloor x/2 \rfloor, \lfloor x/3\rfloor, \lfloor x/4\rfloor, \dots\;$ appear? Let $\lfloor x\rfloor$ denote the floor of $x$.
When does the first repetition in $\lfloor x\rfloor$, $\lfloor x/2\rfloor$, $\lfloor x/3\rfloor$, $\lfloor x/4\rfloor$, ... approximately ... | $\color{brown}{\textbf{Preliminary Notes.}}$
$\dfrac xN > \dfrac x{N+1},$ so the required equality
$$\left\lfloor\dfrac xN\right\rfloor = \left\lfloor\dfrac x{N+1}\right\rfloor,\tag1$$
where $\lfloor a\rfloor = \mathrm{floor}\,(a),$
has solutions iff
$$\left\{\dfrac xN\right\} > \left\{\dfrac x{N+1}\right\}.\tag2$$
Ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3083192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 0
} |
Cannot find angle for trigonometry problem
A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $\frac{13}{4}$. What are the angles of the triangle?
Here is the problem as I understand it, given that:
1. $\me... | $$\dfrac4{13}\sin(45^\circ -\alpha+\alpha)=2\sin\alpha\sin(45^\circ-\alpha)$$
$$=\cos(45^\circ-2\alpha)-\cos45^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find the value this sum converges to?$\sum_{n=2}^{\infty}\frac{n+1}{n(2n-1)(2n+1)} $ How to find the value this sum converges to?$$\sum_{n=2}^{\infty}\frac{n+1}{n(2n-1)(2n+1)} $$
I've tried separating it like
$$\sum_{n=2}^{\infty}\frac{1/2}{(2n+1)}+\frac{3/2}{(2n-1)}-\frac{1}{n}$$
and writing some terms and I ge... | Here is a way to find those two sums @Kavi Rama Murthy refers to in his answer. They are found by first converting them to double integrals.
Let
$$S = \sum_{n = 2}^\infty \frac{n + 1}{n(2n - 1)(2n + 1)} = \frac{3}{4} \sum_{n = 2}^\infty \frac{1}{n (2n - 1)} - \frac{1}{4} \sum_{n = 1}^\infty \frac{1}{n(2n + 1)} + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the set of the real numbers $x$ satisfying the inequalities $|x+4|<|2x-1|$ and $|x|+|x+1|<3$
Find the set of the real numbers $x$ satisfying the given inequality and sketch them on the real line:
1) $|x+4|<|2x-1|$
*
*If $x<-4$, we have $x+4<0$ and $2x-1<0$, Then $|x+4|=-x-4$ and $|2x-1|=1-2x$
hence, in this c... | Hint: $$|x+4| < |2x-1| \iff (x+4)^2 < (2x-1)^2 \iff x^2 -4x-5 >0 \iff (x-2)^2 > 9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.