Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find $f(x)$ in the composite function $f(f(x)) = 6x-8$ where f(x) is a linear function Hi just want to know how to solve this.
find $f(x)$ in the composite function $f\bigl(f(x)\bigr) = 6x-8$
| If you consider a linear function $f(x)=mx+b$, we have $f(f(x))=m(mx+b)+b$.
This is equal to $m^2x+mb+b$.
We have that $m^2=6$, so $m=\sqrt{6}$.
Therefore, $\sqrt{6}b+b=8$,
We have $b(\sqrt{6}+1)=8$, so $b=\displaystyle \frac{8}{\sqrt{6}-1}=\frac{8\sqrt{6}+8}{5}$
Therefore, $\displaystyle f(x)=x\sqrt{6}+\frac{8\sqrt{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\int_0^7 g(r)~dr$ from $ g(r)=\begin{cases} \sqrt{9-(r-3)^2}, & 0\le r<3 \\ r+2, & 3\le r<5 \\ \sqrt{4-(r-5)^2}, & 5 \le r \le 7\end{cases} $ I have some piece-wise integration that I for some reason am not getting.
$$
g(r) =
\begin{cases}
\sqrt{9-(r-3)^2} && 0 \le r \lt 3 \\
r+2 && 3 \le r \lt 5 \\
\sqrt{4-(... | Note: I just add this as a complement to @projectilemotion's answer.
We can still the integral into three parts as
$$\int_0^7 g(r)\; dr=\int_0^3 g(r)\; dr+\int_3^5 g(r)\; dr+\int_5^7 g(r)\; dr$$
*
*For the first one, we have $y=g(r)=\sqrt{9-(r-3)^2}$, so $y^2+(r-3)^2=9$. So the region bounded by the graph of the fu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help to understand polynomial factoring I'm following some proof, but got stuck at how the factoring works. I can follow this part:
$$\begin{align*}
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\
&= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\
\end{align*}$$
The next two steps are not so clear to me ... | \begin{align}
k^2(k+1)^2 + 4(k+1)^3&=k^2(k+1)^2 + 4(k+1)(k+1)^2\\
&=(k+1)^2[k^2 + 4(k+1)]\\
&=(k+1)^2(k^2 + 4k+4)\\
&=(k+1)^2(k+2)^2\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$ then find $B^{16}$.
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$, then find $B^{16}$.
My Method:
Given $$AB^2=BA \tag{1}$$ Post multiplying with $B^2$ we get
$$AB^4=BAB^2=B^2A$$ Hence
$$AB^4=B^2A$$ Pre Multiplying with $A$ and using $(1)$ we g... | My Method:
Given $$AB^2=BA $$ Pre multiplying with $A^3$ we get
$$A^4B^2=A^3BA$$
$$B^2=A^3BA \tag{1}$$ Using $(1)$ we get
$$B^{16}=A^3B^8A \tag{2}$$
$$B^{8}=A^3B^4A \tag{3}$$
$$B^{4}=A^3B^2A \tag{4}$$
$$B^{2}=A^3BA \tag{5}$$
Now combining $(5)$ in $(4)$ , $(4)$ in $(3)$ , $(3)$ in $(2)$ and
we get
$$B^{16}=A^{12}BA^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 1
} |
Solve the following quadratic equation: $4\tan^2(\theta)x^4-4x^2+1=0$ $$4\tan^2(\theta)x^4-4x^2+1=0$$
It can be seen that as $\theta$ goes to $0$, the leftmost term disappears and $x=\pm \frac{1}{2}$.
But when I try to solve this using the quadratic formula I get:
$$x=\pm \sqrt{\frac{1\pm \sqrt{1-\tan^2(\theta)}}{2\tan... | Basically, your question can be reformulated to
What is $$\lim_{\theta\to 0}\sqrt{\frac{1\pm \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)}}$$
To calculate that, first of all, since $\sqrt{}$ is a continuous function, all you need to do is to caluclate
$$\lim_{\theta\to 0}\frac{1\pm \sqrt{1-\tan^2(\theta)}}{2\tan^2(\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding minima and maxima to $f(x,y) = x^2 + x(y^2 - 1)$ in the area $x^2 + y^2 \leq 1$ I'm asked to find minima and maxima on the function
$$f(x,y) = x^2 + x(y^2 - 1)$$
in the area
$$x^2 + y^2 \leq 1.$$
My solution:
$$\nabla (f) = (2x + y^2 - 1, 2xy)$$
Finding stationary points
$$2xy = 0$$
$$2x + y^2 - 1 = 0$$
gets... | HINT: the searched maximum is given by $2$ and will be attained for $$x=-1,y=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
In an equilateral triangle, prove that $|BQ| + |PQ| + |CP| > 2l$ I am trying to solve the following problem:
Let $ABC$ be an equilateral triangle with side $l$.
If $P$ and $Q$ are points respectively in sides $AB$ and $AC$, different from the triangle vertices, prove that $$|BQ| + |PQ| + |CP| > 2l$$
I can see that, a... | Let $AQ=x$, $AP=y$ and $l=1$.
Thus, $$PQ=\sqrt{x^2-xy+y^2},$$
$$PC=\sqrt{y^2-y+1}$$ and
$$BQ=\sqrt{x^2-x+1}$$ and we need to prove that
$$\sqrt{x^2-xy+y^2}+\sqrt{x^2-x+1}+\sqrt{y^2-y+1}\geq2.$$
Now, by Minkowwski
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}+\sqrt{\left(y-\frac{1}{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x... | All you've done is fine.
Now: since we want $\frac{x}{2}$ angles to appear, use $\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1+\cos x=\cos 0+\cos x=2(\cos\frac{x}{2})^2$ (a formula for the sum of cosines), and you're done.
you cant use half-angle or triple angle or any of those formulas.
I don't understand - how to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
| Hint. Use the identity
$$z^{m+n}+\frac{1}{z^{m+n}}=\left( z^m+\frac{1}{z^{m}}\right)\left(z^{n}+\dfrac{1}{z^{n}}\right)-\left(z^{m-n}+\frac{1}{z^{m-n}}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Proof $13 \mid (k\cdot 2^n+1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$?
Proof $13 \mid (k \cdot 2^n + 1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$
Hint: for $k$ odd: $2^n \equiv-k' \pmod p$ and $kk' \equiv1 \pmod p$
My thoughts:
$13\mid(k-3) \Rightarrow k=13a+3$
and
$12|(n-2) \Rightarrow n=12b+2$
s... | By Fermat we have $2^{12} \equiv 1 \pmod{13}$ so $2^{12b}=1+13c$.
So
\begin{eqnarray*}
k2^n+1=(3+13a)(4+4 \times 13 c)+1=13(1+4a+12c+52ac)
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If the roots of an equation are $a,b,c$ then find the equation having roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$. Actually I have come to know a technique of solving this kind of problem but it's not exactly producing when used in a certain problem.
Say we have an equation
$$2x^3+3x^2-x+1=0$$
the roots of this ... |
A much much simpler way to solve this math is; If we pick that, $f(x)= 2x^3+3x^2-x+1$ and the values of x are a,b and c then f(1/2x) will denote an equation (if we just write f(1/2x)=0 ) which has the roots 1/2a, 1/2b, 1/2c.
This is not true. Suppose we have $f(x) = (x - 1)^2$. This has the root 1. And $f(1/2x) = (1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Simplify $5(3 \sin(x) + \sqrt3 \cos(x))$ So Wolfram tells me I can reach $10 \sqrt3 \sin(x + π/6)$ from this expression, but I cant grasp how to do it. Any help is appreciated.
| Given these identities:
$$\frac{1}{2}\sqrt{3} = \cos(\frac{\pi}{6})$$
$$\frac{1}{2} = \sin(\frac{\pi}{6})$$
$$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)$$
We get:
$$5(3\sin(x) + \sqrt{3}\cos(x)) = $$
$$5\sqrt{3}(\sqrt{3}\sin(x) + \cos(x)) =$$
$$10\sqrt{3}(\cos(\frac{\pi}{6})\sin(x) + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
parametric spiral arc length I am attempting to calculate the arc length of a spiral. Given an Archamedean spiral of the parametric form:
$x(t) = \sqrt{t}\cos\left(\omega\sqrt{t} \right)$ and $y(t) = \sqrt{t}\sin\left(\omega\sqrt{t} \right)$, the arc length $L$ is calculated as
\begin{align}\label{parametricArcLength}
... | Parametric curves are just screaming out to be solved in the complex plane. Consider that
$$z=x+iy=\sqrt{t}~e^{i\omega\sqrt{t}}$$
The arc length is given by
$$s=\int |\dot z|dt$$
Thus,
$$
\dot z=\left(\frac{1}{2\sqrt{t}}+\frac{i\omega}{2} \right)e^{i\omega\sqrt{t}}\\
|\dot z|=\frac{1}{2}\sqrt{\frac{1}{t}+\omega^2}
$$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A limit involving square roots Transcribed from photo
$$\require{cancel}
\lim_{n\to\infty}\sqrt{n+\tfrac12}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)\\[12pt]
$$
$$
\lim_{n\to\infty}\sqrt{n+\tfrac12}\lim_{n\to\infty}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)\frac{\sqrt{2n+1}+\sqrt{2n+3}}{\sqrt{2n+1}+\sqrt{2n+3}}\\[12pt]
$$
$$
\lim_... | $$\sqrt{n+\frac{1}{2}}\cdot (\sqrt{2n+1}-\sqrt{2n+3})
=\sqrt{n+\frac{1}{2}}\cdot\frac{-2}{\sqrt{2n+1}+\sqrt{2n+3}}\\
=\sqrt{n}\cdot \sqrt{1+\frac{1}{2n}}\cdot\frac{-2}{\sqrt{2n}\cdot (\sqrt{2n+1}+\sqrt{2n+3})}\\
=\sqrt{1+\frac{1}{2n}}\cdot\frac{-\sqrt{2}}{\sqrt{1+\frac{1}{2n}}+\sqrt{1+\frac{3}{2n}}}$$
Thus, the require... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve an indefinite integral Find the value of
$$ \int\frac{\sqrt{1+x^2}}{1-x^2}dx $$
My Attempt: I tried to arrange the numerator as follows,
$$ \sqrt{1+x^2} = \sqrt{1-x^2 + 2x^2} $$
but that didn't help me.
Any help will be appreciated.
| HINT: Make the substitution $1+x^2 \to u$ or $x \to \sqrt{u-1}$. Then you can transform the integral into
$$\int -\frac{\sqrt u}{u}\cdot\frac{1}{2\sqrt{u-1}}du$$
$$\int -\frac{1}{\sqrt u}\cdot\frac{1}{2\sqrt{u-1}}du$$
$$-\frac{1}{2}\int \frac{1}{\sqrt{u^2-u}}du$$
Then complete the square in the denominator:
$$-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find $\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)$ I did it as follows: $$\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)=\tan\Bigg(\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)\Bigg)=\frac{\sqrt{2}-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}\sqrt{2}}=\frac{\sqrt{2}}{4}.$$ But there is no ... |
The angle you are looking for is the red angle:
$$\arctan\sqrt{2}-\arctan\frac{1}{\sqrt{2}} = \arctan\left(\frac{\sqrt{2}-\frac{1}{\sqrt{2}}}{1+1}\right)=\color{red}{\arctan\frac{1}{\sqrt{8}}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$?
If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$.
So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third derivative of $a$ is $0$, but the derivatives keep getting longer an... | Use $$x^2+x+1=\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)$$
By this hint we obtain:
$$\left(\frac{1}{x^2+x+1}\right)^{(36)}_{x=0}=\left(\frac{1}{\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)}\right)^{(36)}_{x=0}=$$
$$=\left(\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 6,
"answer_id": 2
} |
Prove that it is impossible to find any positive integers a, b, c such that $(2a+b)(2b+a) = 2^c$ Prove that it is impossible to find any positive integers a, b, c such that $(2a+b)(2b+a) = 2^c$.
This problem has been driving me crazy. Thanks for helping.
| We have $2a+b=2^k$ and $2b+a=2^m$, where $k$ amd $m$ are non-negative integer numbers.
Thus, $a=\frac{2^{k+1}-2^m}{3}$ and $b=\frac{2^{m+1}-2^k}{3}$, which gives
$k+1>m$ and $m+1>k$ and from here $k+1\geq m+1$ and $m+1\geq k+1$, which gives $k=m$, $a=\frac{2^k}{3}$, which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$? In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?
Alright, so if d is the ... | Using $BD=DC=a $, we can write
$$\begin{align} 3^2 &=a^2+(2a)^2-2 \cdot a \cdot 2a \cos(\alpha) \\
&=5a^2-4a^2\cos (\alpha) \end{align}$$
with $\alpha= \angle ADB$.
Also,
$$\begin{align} 4^2&=a^2+(2a)^2-2 \cdot a \cdot 2a\cos (\pi-\alpha) \\
&=5a^2+4a^2\cos (\alpha). \end{align}$$
By addition, we get
$$9+16=25=10a^2. $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Simplifying the Solution to the Cubic I am trying to solve the cubic. I currently have that, for $ax^3+bx^2+cx+d=0$, a substitution to make this monic. Dividing by $a$ gives
$$x^3+Bx^2+Cx+D=0$$
where $B=\frac{b}{a}, C=\frac{c}{a}, D=\frac{d}{a}$. Then, with the substitution $x=y-\frac{B}{3}$, I got
$$y^3+\left(C-\fr... | You can't. This is as simple as you can get, unless you want to re-obtain Cardano's formula, which is basically what you got.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Convert vector differential equation's order
Here is a 2nd order vector differential equation: $$\overrightarrow{Y}''= \begin{pmatrix}a & b \\c & d \end{pmatrix} \overrightarrow{Y}$$ Don't work it out, but write it as a vector differential eqn in $1$st order in higher dimensions.
I am not sure where to begin.
How doe... | Let
$$
\vec{x} =
\begin{pmatrix}
x_1 \\
x_2 \\
\end{pmatrix} =
\begin{pmatrix}
y_1' \\
y_2' \\
\end{pmatrix} = \vec{y}'.
$$
Then
$$
\vec{x}' =
\begin{pmatrix}
x_1' \\
x_2' \\
\end{pmatrix}
=
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}
\begin{pmatrix}
y_1\\
y_2 \\
\end{pmatrix} =
\begin{pmatrix}
a &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the value of the expression $\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$? This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.
What is the value of
$\sin\frac{... | $\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}$
$\sin \frac{8\pi}{7} = \sin \frac{-6 \pi}{7}$
-> $\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{-6\pi}{7}$
$\sin a \sin (b-c) + \sin b \sin (c-a)+ \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculate $ \lim_{x\to 0}\frac{ \ln(x+\cos(x))-xe^{-x} }{x^{3}}$ Calculate $\displaystyle \lim\limits_{x\to 0}\dfrac{ \ln\left(x+\cos(x)\right)-xe^{-x} }{x^{3}}$
My attempts
by l’hôpital
*
*$\left( \ln\left(x+\cos(x)\right)-xe^{-x} \right)'=\dfrac{(x+\cos(x))' }{(x+\cos(x))} -(e^{-x}-xe^{-x} )=\dfrac{(1-\si... | L’Hôpital's rule seems to be hard to use here.
This is an alternative method by using Taylor expansion at $0$:
\begin{align*}
\ln\left(x+\cos(x)\right)-xe^{-x}&=
\ln\left(1+x-\frac{x^2}{2}+o(x^3)\right)-x\left(1-x+\frac{x^2}{2}+o(x^2)\right)\\
&=\left(x-\frac{x^2}{2}\right)-\frac{1}{2}\left(x-\frac{x^2}{2}\right)^2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this ... | This is the same as
$$a^8+b^8+c^8> a^2b^3c^3+a^3b^2c^3+a^3b^3c^2.$$
You can get this by using AM/GM to get an upper bound for each term
on the right.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the maximum and minimum value of $S= \log_{2}a+\log_{2}b+\log_{2}c$. Let $a, b, c \geq 1$ and $a+b+c=4$. Find the maximum and minimum value of $S= \log_{2}a+\log_{2}b+\log_{2}c$.
I found the maximum, it is easy to prove $S_\max = 3\log_{2}\frac{4}{3}$. I think the minimum is $1$ when there are two number are $1$ a... | The max/min of $S$ will occur on the same instance as the max/min of $2^S = abc$. By $AM-GM$ the max occurs at $a= b=c = 4/3$ as you figured.
To figure out the min, fix $c$ and and set $b = (4-c) -a = K -a$.
So $abc = aK - a^2$ the derivative is $K - 2a$ which has a maximum at $a = \frac K2$ (which means $b = a = \f... | {
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Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$
Evaluate the following limit.
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
My Attempt:
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-... | An often good trick is to make the substitution $t=1/x$. However, in this case $t=1/\sqrt{x}$ seems better, because we get
$$
\lim_{t\to0^+}\frac{\sqrt{1-at^2}-\sqrt{b}}{t}
$$
The numerator has limit $1-\sqrt{b}$, therefore we see that, if $0\le b<1$ the limit is $\infty$, whereas for $b>1$ the limit is $-\infty$.
If $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$? Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$, if $x, y$ are reals greater than $1$, and $a, b, c$ are positive reals?
A proof with all the math to go from one to the other would be nice.
| Multiply the inequality by $4a^2$ then add & subtract $(a^2+b^2-c^2)y^2$
\begin{eqnarray*}
\underbrace{4a^4 x^2 +4a^2(a^2+b^2-c^2)xy +\color{red}{(a^2+b^2-c^2)^2y^2}}_{} -\color{red}{(a^2+b^2-c^2)^2y^2}+4a^2b^2 y^2>0
\end{eqnarray*}
Now complete the square
\begin{eqnarray*}
\left(2a^2x+(a^2+b^2-c^2)y\right)^2+(4a^2b^2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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In how many ways the sum of 5 thrown dice is 25? What I thought about is looking for the number of solutions to
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=25$$ such that $1\leq x_{i}\leq6$
for every $i$.
Now I know that the number of solutions to this such that
$0\leq x_{i}$ for every $i$ is $${5+25-1 \choose 5-1}={29 \choose 4}$... | In addition to my other answer using generating functions, I want to offer a more elementary solution. First note that
$$ 5 + 5 + 5 + 5 + 5 = 25 $$
and this is the only way to make $25$ without using a $6$.
Now make one of the dice a $6$ (5 possible ways) to get the equation
$$ a + b + c + d = 19 $$
where $1 \le a, b, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understa... | Clearly $n^4$ is a square, since $(n^2)^2=n^4$. The next smaller square is $(n^2-1)^2 = n^4-2n^2+1$, which is clearly less than the given expression. So if $n^4-n^2+64$ is a square, it needs to be not less than $n^4$, that is, we need $64\ge n^2 \implies |n|\le8$ (meaning the number of solutions finite and small)
We ca... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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perfect powers of the form $\frac{x^2-1}{y^2-1}$
Is there a natural number $k>1$ such that there are infinitely many
pairs $(x,y)$ of natural numbers such that $\frac{x^2-1}{y^2-1}$ is a
power of $k$?
In particular, is $k=2$ or $k=10$ good in this sense? (are there finitely or infinitely many solutions (in natur... | Plugging in the fraction $\dfrac{x^2-1}{y^2-1}$ values $x=2^{2 n + 1} - 1;\;y=2^n$ we get
$$\dfrac{\left(2^{2 n+1}-1\right)^2-1}{2^{2 n}-1}=2^{2n+2}$$
so there are infinite pairs $(x,\;y)$ which give $2^k$
I think there are also infinite solution to $10^k$ but I still can't find a closed form
Hope this helps
| {
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Solving trigonometric equations
If we have $$\sin (A) +\cos (A) + \csc (A) + \sec (A) +\tan (A) +\cot (A)= 7$$ and
$$\sin(2A) =a-b\sqrt{7}= 2\sin(A)\cos(A).$$
What values can $a$ and $b$ take?
| Let $A=x$ and $\sin{x}+\cos{x}=t$.
Hence, $|t|\leq\sqrt2$, $\sin{x}\cos{x}=\frac{t^2-1}{2}$ and we need to solve
$$t+\frac{t}{\frac{t^2-1}{2}}+\frac{1}{\frac{t^2-1}{2}}=7$$ or
$$t+\frac{2}{t-1}=7$$ or
$$t^2-8t+9=0$$ or
$$(t-4)^2=7,$$ which gives
$t=4+\sqrt7$, which is impossible, or $t=4-\sqrt7$,
which gives $\sin2x=t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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generating function formula Suppose $\lambda$ is a young tableau, with one color boxes.
Prove the generating function,
$F(x) = \Sigma_{\lambda} x^{i} = \prod_{n \geq 1} \frac{1}{1-x^n} $, where $i$ is the number of boxes in the young tableau.
attempt:
$\Sigma_{\lambda} x^{i} = \prod_{n \geq 1} \frac{1}{1-x^n} = \... | Suppose we have a Young tableaux with shape
\begin{eqnarray*}
(\underbrace{k,\cdots,k}_{a_k \text{times}}, \underbrace{k-1,\cdots,k-1}_{a_{k-1} \text{times}},\cdots,\underbrace{2,\cdots,2}_{a_2 \text{times}}\underbrace{1,\cdots,1}_{a_1 \text{times}})
\end{eqnarray*}
This will give a contribution of $x^n$ to the genera... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Series Solution to the ODE $(x-1)y'' - xy' + y = 0$ with I.C. $y(0) = -3$ and $y'(0)=4$ \begin{align*}
(x-1)y'' - xy' + y = 0 &\iff (x-1)\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2} - x\sum_{n=1}^{\infty} nc_n x^{n-1} + \sum_{n=0}^{\infty} c_n x^n = 0 \\
&\iff \sum_{n=2}^{\infty} n(n-1)c_n x^{n-1} - \sum_{n=2}^{\infty} n(n-1)... | This ODE belongs to the class which is dealt with in here Yet another example of linear second order ODEs being reduced to hypergeometric functions. .
Using that approach we get the solution:
\begin{eqnarray}
y(x) &=& \exp(x/2) \sqrt{x-1} \left( C_1 W_{1/2,1}(x-1) + C_2 M_{1/2,1}(x-1) \right)\\
&=& (x-1)^2 \left( C_1 U... | {
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"timestamp": "2023-03-29T00:00:00",
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$x^2 + y^2 = 10 , \,\, \frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ I couldn't solve the above equation. Below, I describe my attempt at solving it.
$$x^2 + y^2 = 10 \tag{1}$$
$$\frac{1}{x} + \frac{1}{y} = \frac{4}{3} \tag{2}$$
Make the denominator common in the RHS of $(2)$.
$$\frac{y + x}{xy} = \frac{4}{3} \tag{2.1}$$
Mu... | The right approach is via symmetric functions: set $s=x+y$, $p=xy$. Then
*
*$x^2+y^2=s^2-2p,\:$ so we have the relation: $\;s^2-2p=10 \tag{1}$
*$\dfrac 1x+\dfrac 1y=\dfrac sp=\dfrac43,\:$ whence a second relation: $\;3s=4p\tag{2}$.
Relation $(1)$, taking relation $(2)$ into account, yields a quadratic equation... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the probability of rolling at least two 6's with 3 Dice and 2 Rolls? Question: What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls? (with your first roll you keep dice only if they are 6's and roll the remainder for your second roll).
(6*6*6 = 216 = outcomes when rolling 3 di... | Your approach and your calculations are correct. Here is a variation based upon generating functions. We encode the roll of three dice with
\begin{align*}
(5+t)^3
\end{align*}
marking an occurrence of $6$ with $t$ and collecting all other five possibilities with $5$. The probability to get $j$ sixes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there an error in this problem? $\sin^4(\frac{23\pi}{12})-\cos^4(\frac{13\pi}{12})$
$\sin^4(\frac{23\pi}{12})-\cos^4(\frac{13\pi}{12})$
If there were written $\frac{23\pi}{12}$ instead of $\frac{13\pi}{12}$ it could be solved as $$\begin{align}[\sin^2(\frac{23\pi}{12})+\cos^2(\frac{23\pi}{12})][\sin^2(\frac{23\pi}... | $\frac{23}{12} \pi = 2\pi - \frac{\pi}{12}$, and $\frac{13}{12} \pi = \pi + \frac{\pi}{12}$. So you have:
$$\sin^4 \frac{\pi}{12} - \cos^4 \frac{\pi}{12}$$
which surprisingly gives the same answer you reached.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that $\sqrt{m-1}$ is an integer
Let $k$ be a fixed odd positive integer. Let $m,n$ be positive integers such that $(2+\sqrt{3})^k=m+n\sqrt{3}$. Prove that $\sqrt{m-1}$ is an integer.
Let $(2+\sqrt{3})^n = a_n+b_n\sqrt{3}$. From $$a_{n+1}+b_{n+1}\sqrt{3} = (a_n+b_n\sqrt{3})(2+\sqrt{3}) = (2a_n+3b_n)+(a_n+2b_n)\s... | Since $k$ is odd $m-1$ is not multiple of $3$. This is because it is a sum of multiples of $3$ and a $2^{k}=2\cdot 4^{(k-1)/2}=2\mod{3}$.
Now, $m^2-3n^2=1$. So,
$$(m-1)(m+1)=3n^2$$
Since $m-1$ is not divisible by $3$, the $3$ divides the factor $m+1$.
The only common divisor of $m-1$ and $m+1$ can be $2=(m+1)-(m-1)$. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Infinite square root muliplication $(x=3\sqrt{y\sqrt{3\sqrt{y}....}})$ I have this problem which says
$$
x=3\sqrt{y\sqrt{3\sqrt{y\cdots}}}\\
y=3\sqrt{x\sqrt{3\sqrt{x\cdots}}}
$$
What is $x+y$
I tried
$$
x^2=9y\sqrt{x}\\
x^3=81y^2\\
y^3=81x^2\\
x^3+y^3=81(x^2+y^2)\\
(x+y)(x^2-xy+y^2)=81(x^2+y^2)
$$
I don't know what to... | From second equality we obtain
$$x^2=9x\sqrt{x},$$ which gives $x=0$ or $x=81$. The rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If I select 9 random numbers from 1 to 10, what is the probability that their sum is less than 20?
If I select 9 random numbers from 1 to 10, what is the probability that their sum is less than 20?
Is there someway to figure this out without manually figuring out each possible scenario in which the 9 digits don't sum... | A generating function approach yields the result without too much computation.
If we choose 9 integers from 1 to 10, there are $10^9$ possible outcomes, all of which we assume are equally likely. We want to count the outcomes in which the total is less than or equal to 19, i.e. the number of solutions in integers to
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$ Find $\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$
$$I=\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$$
$$I=\int e^{x^4}(1+x^2+2x^4)e^{x^2}xdx$$Let $x^2=t$
$I=\int e^{t^2}(1+t+2t^2)e^t\frac{dt}{2}=\frac{1}{2}\int e^{t^2+t}(1+t+2t^2)dt$
I am stuck here.
| Hint:
As $\dfrac{d(x^4+x^2)}{dx}=4x^3+2x$
$$\int e^{x^4+x^2}(2x+2x^3+4x^5)dx=\int[e^{x^4+x^2}x^2(2x+4x^3)+e^{x^4+x^2}(2x)dx$$
which is clearly of the from $$\int e^{f(x)}[g'(x)+g(x)f'(x)]dx=e^{f(x)}g(x)+K$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\frac{1}{{1 - 100 + 500}} + \frac{{{2^2}}}{{{2^2} - 200 + 500}} + \ldots + \frac{{{{99}^2}}}{{{{99}^2} - 9900 + 500}} $ I need a hint to solve this problem : Find the sum of
$$ \frac{1}{{1 - 100 + 500}} + \frac{{{2^2}}}{{{2^2} - 200 + 500}} + \frac{{{3^2}}}{{{3^2} - 300 + 500}} + + \ldots + \frac{{{{99}^... | Hint:
$\frac{N^2}{N^2-100N+500} = \frac{N^2}{N^2-100N+2500-2500+500} = \frac{N^2}{(N-50)^2-(10\sqrt{20})^2)} $
$ a = (50-10\sqrt{20})$
$S = \frac{AN+B}{N-a} + \frac{C}{N+a}$
Simplifying this
$S = \sum_{N=1}^{99} 1+ a^2\frac{1}{N^2-a^2} = \sum_{N=1}^{99}1+ a^2\frac{1}{(\frac{N}{a})^2-1}$
$S= 99+(50-10\sqrt{20})^2*(-6.6... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrix to power 30 using eigen values issue
$$ \text{If matrix } A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \text{ then find } A^{30}.$$
I tried to approach through diagonalization using eigen values method.
I got eigen values as $-1, 1, 1$
As per diagonalization $ A = P*D*P^{-1}.$ So $ A^{... | Since $A$ is not diagonalizable we can use Jordan form
\begin{align}
A^{30} &=
\begin{pmatrix}
0 & 2 & 0 \\
-1 & 1 & 1 \\
1 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 1 & 1
\end{pmatrix}^{30}
\begin{pmatrix}
1/4 & -1/2 & 1/2 \\
1/2 & 0 & 0 \\
-1/4 & 1/2 & 1/2
\end{pmatrix} \\
&=
\begin{pmatrix... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$
Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$
Also it is a question of S.L. Loney's Plane Trignonometry
What I've tried by now:
\be... | Let $\theta = 2 \phi$, then the thing to be proven is:
Prove that $$\frac{1 + \sin(2\phi) - \cos(2\phi)}{1 + \sin(2\phi) + \cos(2\phi)} = \tan(\phi)$$
Then use:
$$\sin(2\phi) = 2 \sin \phi \cos \phi$$
$$\cos(2\phi) = \cos^2 \phi - \sin^2 \phi$$
and:
$$\sin^2 \phi + \cos^2 \phi = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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The derivative of the logistic function The logistic function is $\frac{1}{1+e^{-x}}$, and its derivative is $f(x)*(1-f(x))$. In the following page on Wikipedia, it shows the following equation:
$$f(x) = \frac{1}{1+e^{-x}} = \frac{e^x}{1+e^x}$$
which means $$f'(x) = e^x (1+e^x) - e^x \frac{e^x}{(1+e^x)^2} = \frac{e^x}{... | Just start from the answer and work backwards.
$$\begin{aligned}
f(x)(1-f(x)) &= \frac{e^x}{1+e^x} \left( 1 - \frac{e^x}{1+e^x} \right)\\
&= \frac{e^x}{1+e^x} \left( \frac{1+e^x}{1+e^x} - \frac{e^x}{1+e^x} \right)\\
&= \frac{e^x}{1+e^x} \left( \frac{1}{1+e^x} \right) = \frac{e^x}{(1+e^x)^2} = f'(x)
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Condition of existence of a triangle It's easy to prove that the triangle inequality holds for any triangle with the lengths of sides $a$, $b$ and $c$.
But how can one prove that if the triangle inequality holds for any given positives $a$, $b$ and $c$ then a triangle (geometric figure) with the lengths of the sides ... | Let $B(0,0)$ and $C(a,0)$.
Hence, $BC=a$ and we need to prove that there exists $A(x,y)$ such that $AB=c$ and $AC=b$.
You can write equations of two circles and prove that there are intersection points.
For example $x^2+y^2=c^2$ and $(x-a)^2+y^2=b^2$.
Thus, $-2ax+a^2+c^2=b^2$ or $x=\frac{a^2+c^2-b^2}{2a}$ and
$$y^2=c^... | {
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Integrating factor $(3y^2-x) + 2y(y^2-3)y'=0$ is a function
Show that the differential equation $(3y^2-x) + 2y(y^2-3)y' = 0$ admits an integrating factor which is a function of $(x+y^2)$. Hence solve the equation.
I know how to solve this by using Integrating Factor of an exact equation
But question specifically aski... | $(3y^2-x)+2y(y^2-3)y'=0$
$2y\dfrac{dy}{dx}=\dfrac{x-3y^2}{y^2-3}$
$\dfrac{d(y^2)}{dx}=\dfrac{x-3y^2}{y^2-3}$
$\dfrac{du}{dx}=\dfrac{x-3u}{u-3}$ $(\text{Let}~u=y^2)$
Let $\begin{cases}x=t+a\\u=v+b\end{cases}$ ,
Then $\dfrac{dv}{dt}=\dfrac{t+a-3v-3b}{v+b-3}$
Take $a=9$ and $b=3$ , the ODE becomes
$\dfrac{dv}{dt}=\dfrac{t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality : $\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}$
Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq
1+\sqrt[3]{3}.$$
From Micheal Rozenberg's answer :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqr... | $$x+y+z=p,\ xy+yz+zx=q,\ xyz=r$$
we can assume : $x+y+z=1$
$x^3+y^3+z^3=1-3q+3r,\ x^2+y^2+z^2=1-2q\ \ \ \left( 0 < q \le \dfrac{1}{3}\right)$
$$\left( \dfrac{1-3q}{r}+3\right)^{1/3}+\left( \dfrac{q}{1-2q}\right)^{1/2}\ge 1+\sqrt[3]3$$
We have
$$q^2=(xy+yz+zx)^2\ge 3xyz(x+y+z) =3r.$$
So: $LHS \ge \sqrt[3]3\cdot\left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove the inequality $\sum_{cyc}\frac{a}{1+\left(b+c\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$
Let $a>0$, $b>0$ and $c>0$ such that $a+b+c=3$. Prove the inequality
$$\frac{a}{1+\left(b+c\right)^2}+\frac{b}{1+\left(c+a\right)^2}+\frac{c}{1+\left(a+b\right)^2}\le \frac{3\left(a^2+b^2+c^2\right... | We need to prove that
$$\sum_{cyc}\left(\frac{a}{1+(b+c)^2}-a\right)\leq\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}-3$$ or
$$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}\geq\frac{36abc}{a^2+b^2+c^2+12abc},$$
which seems better, but I did not find something nice here.
By the way, our inequality is obviously true after homogenizati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Argument for why $a^2 + 1$ is never divisible by a $3 \mod 4$ integer How do you show that $$a^2 + 1$$ is never divisible by a $3 \mod 4$ integer (which is equivalent to showing that it has no $3 \mod 4$ prime factor) for any non-negative integer $a$ by analysing the arithmetic series representation of $a^2$, $1 + 3 + ... | Not using the series:
*
*Case 1: $a$ is even, say, $a=2k$. Then $a^2+1 = 4k^2 + 1$.
*Case 2: $a$ is odd, say, $a=2k+1$. Then $a^2+1 = 4k^2+4k+2$.
Using the series $a^2=1+3+5+...+(2a-5)+(2a-3)+(2a-1)$.
*
*Note that consecutive terms add to multiples of $4$.
*Therefore, $a^2$ is either $(1+3)+(5+7)+...+((2a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{x \to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}}$ Evaluate $$L=\lim_{x \to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}} \tag{1}$$
I have used L Hopital's Rule we get
$$L=\lim_{x \to 0} 2\frac{ \sqrt{1-\sqrt{1-x^2}} \sqrt{1-x^2}}{x}$$
$\implies$
$$L=2 \lim_{x \to 0}\frac{ \sqrt{1-\sqrt{1-x^2}}}{x}$$ and from $(1)$ we get
... | Alternatively:
$$L=\lim_\limits{x\to 0} \frac{sgn(x) \sqrt{x^2}}{\sqrt{1-\sqrt{1-x^2}}}=$$
$$\lim_\limits{x\to 0} sgn(x) \sqrt{\lim_\limits{x\to 0} \frac{x^2}{\sqrt{1-\sqrt{1-x^2}}}} =(LR)= $$
$$\lim_\limits{x\to 0} sgn(x) \sqrt{\lim_\limits{x\to 0} 2\sqrt{1-x^2}}=$$
$$\lim_\limits{x\to 0} sgn(x) \sqrt{2}.$$
Hence, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the volume of a part of sphere from $z=0.5r$ in spherical coordinates?
Given sphere $x^2+y^2+z^2=a^2$ find the volume above $z=0.5a, a>0$. The solution must use spherical coordinates.
It looks like the radius is not constant as it depends on the angle $\phi$ so $0.5a\le r\le\frac{0.5a}{\cos\phi}$
I think... | Using spherical coordinates, the radius $r$ should be $\frac{a}{2 \cos \phi}<r<a$ and indeed the angle $0<\phi<\pi/3$. So you get
$$
2 \pi \int_0^{\pi/3}\sin\phi\bigg[\frac{r^3}{3}\bigg]_{\frac{a}{2 \cos \phi}}^{a} \rm d \phi
= 2 \pi \int_0^{\pi/3}\frac{a^3\sin\phi}{3} \rm d \phi -2 \pi \int_0^{\pi/3} \frac{a^3\sin\p... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Give a sequence such that root test works while ratio test fails
Question: Give a sequence $(a_n)_{n=1}^\infty$ with $a_n>0$ such that root test works while ratio test does not work, that is,
$$\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}} \text{ exists}$$
while
$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} \t... | Your example is right. You can construct a similar exemple such that the series $\sum\limits_{n=1}^{\infty} a_n$ converges as follows:
$$ a_n = \begin{cases}
\frac{1}{2^{n+1}} & \text{if $n$ is odd,} \\
\frac{1}{2^{n}} & \text{if $n$ is even}.
\end{cases}
$$
Then for the odd subsequence $(a_{n_j})_{j=1}^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating the limit $\lim_{x\to1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)$ In trying to evaluate the following limit:
$$\lim_{x\to1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)$$
I am getting the indefinite form of:
$$\frac{1}{\mbox{undefined}}-\frac{3}{\mbox{undefined}}$$
What would be the best solution to evaluating... | Hint. Note that for $x\not=1$,
$$\frac{1}{1-x}-\frac{3}{1-x^3}=\frac{x^2+x+1-3}{(1-x)(x^2+x+1)}
=\frac{(x+2)(x-1)}{(1-x)(x^2+x+1)}=-\frac{x+2}{x^2+x+1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate $\binom{2}{2}\binom{10}{3} + \binom{3}{2}\binom{9}{3} + \binom{4}{2}\binom{8}{3} + \ldots + \binom{9}{2}\binom{3}{3}$ I really can't understand how to approach exercise 41.
I tried by comparing the 5th term of binomial expansion and something like that, but getting 12C5, I think the answer should be 13... | Consider the number of non-negative integral solutions to $X_1 + X_2 + X_3 .. + X_7 = 7$. This should equal $\binom{7+ 7 - 1}{7 - 1} = \binom{13}{6}$.
These solutions can also be counted as follows:
Case 1: Let $X_1 + X_2 + X_3 = 0$ and $X_4 + X_5 + X_6 + X_7 = 7$. The combined number of solutions for these are $\binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Let $\frac13\sin a-\frac17\cos a=\frac{1}{2017}.$ ... Let $a$ in $[0, \pi]$ such that $\frac13\sin a-\frac17\cos a=\frac{1}{2017}.$ If $\lvert\tan{a}\rvert=\frac mn$, where m and n are relatively prime, find m+n.
My work:
$\frac13\sin a-\frac17\cos a=\frac{1}{2017}$From the trig identity $a \cos{t} + b \sin{t} = \sqrt... | As $\tan=\frac \sin\cos$, we may assume $\sin a= mu$, $\cos a=\pm nu$. Thus
$$\frac m3u\mp\frac n7u =\frac1{2017}$$
or
$$ u=\frac{21}{2017\cdot (7m\mp 3n)}.$$
Additionally,
$$ 1=\sin^2a+\cos^2a=(m^2+n^2)u^2$$
so that elimination gives
$$ 2017^2\cdot(7m\mp 3n)^2=21^2\cdot (m^2+n^2).$$
The left hand side is either odd or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400613",
"timestamp": "2023-03-29T00:00:00",
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Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able ... | Denote:
$$a+b+c=0; ab+ac+bc=k; abc=t$$
Then $a,b,c$ are the roots of:
$$x^3+kx+t=0$$
Note:
$$a^3+ka+t=0 \Rightarrow a^4+ka^2+ta=0,$$
$$b^3+kb+t=0 \Rightarrow b^4+kb^2+tb=0,$$
$$c^3+kc+t=0 \Rightarrow c^4+kc^2+tc=0.$$
Add and multiply by $2$:
$$2(a^4+b^4+c^4)=-2k(a^2+b^2+c^2)-t(a+b+c)=-2k((a+b+c)^2-2k)-0=(2k)^2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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find a and b such that the limit will exist and find the limit I am given this question $$\lim_{x\rightarrow 0} \frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}$$
And asked to find the $a$ and $b$ such that the limit exists and also to compute the limit
My solution:
Let $t$=$\sqrt{1+x}$.
Then the Maclaurin polynomial is : $$\sqrt{... | Simplified computation:
Let $\sqrt{1+x^2}=t+1$, so that
$$\lim_{x\to0}\frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}=\lim_{t\to0}\frac{e\cdot e^t-a-bt(2+t)}{t^2(2+t)^2}\\
=\lim_{t\to0}\frac{e+et+\dfrac{et^2}2+\dfrac{et^3}{3!}+\cdots-a-2bt-bt^2}{4t^2}.$$
Then it is obvious that we need
$$a=e,b=\frac e2$$ and the third coefficient... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Complex number $i^{{i^{i^{.^{.^.}}}}}$ If $A+iB=i^{{i^{i^{.^{.^.}}}}}$
Principal values only being considered,
Prove that
(a)tan $ \frac {\pi}{2} $A= $\frac{B}{A}$
(b) $A^2 + B^2 = e^{-\pi B}$
I tried the concept A+iB= $y=i^y$
$i= e^{ \frac{i\pi}{2}}$
$\ln(A+iB)=i \frac{\pi}{2}$(A+iB)
After this step not able to pro... | Whatever the value of $x = i^{i^{i^\cdots}}$ (if it exists), it should satisfy the equation
$$
x = i^x.
$$
But what is the definition of $i^x$ ? Since it is ambiguous, we must loosen the constraint even further: Whatever the value of $x$, it should satisfy
$$
x = e^{x \log i} \text{ for } \textit{some} \text{ value of ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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$\sin(ωt) + \sin(ωt+\Delta\theta) + \sin(ωt+2\,\Delta\theta) + \cdots + \sin(ωt+\overline{n-1}\,\Delta\theta) ={}$? How will we add this series?
Any simple process?
This is one of the equation derived from the equation to find intesity of ray at any particular point in Fraunhofer diffraction.
I think this can be solve... | The hint.
Thy to multiply it by $2\sin\frac{\Delta\theta}{2}$ and use $$2\sin\alpha\sin\beta=\cos(\alpha-\beta)-\cos(\alpha+\beta).$$
\begin{align}
& \sin\omega t + \sin(\omega t+∆θ) + \sin(ωt+2∆θ) + \cdots + \sin(ωt+(n-1)∆θ) \\[10pt]
= {} & \frac{2\sin\frac{\Delta\theta}{2}\sin\omega t +2\sin\frac{\Delta\theta}{2} \si... | {
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Solve $\lfloor \frac{2x-1}{3} \rfloor + \lfloor \frac{4x+1}{6} \rfloor=5x-4$ I came across this problem and it doesn't seem so tricky, but I didn't do it.
Solve the equation
$$\lfloor {\frac{2x-1}{3}}\rfloor + \lfloor {\frac{4x+1}{6}}\rfloor=5x-4.$$
My thoughts so far:
Trying to use the inequality $k-1 < \lfloor k\... | $$\lfloor{\frac{2x-1}{3}}\rfloor + \lfloor{\frac{4x+1}{6}}\rfloor=5x-4.$$
As you have been mentioned;
$5x-4$ must be an integer,
so $x=\frac{a}{5},$
where $a$ is from $\mathbb Z,$
then replace in the original equation, to get
$$\lfloor\frac{4a-10}{30}\rfloor + \lfloor\frac{4a+5}{30}\rfloor=a-4.$$
Notice that by E... | {
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"source": "stackexchange",
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Find polynomial : $P(x)=x^3+ax^2+bx+c$ Find all polynomials in $\mathbb{Q}[x]$ is of the form $P(x)=x^3+ax^2+bx+c$ which has $a, b, c$ as its roots.
Is my answer below correct?
| By Vieta's formulas,
$$a+b+c=-a,\\ab+bc+ca=b,\\abc=-c.$$
Then if $c\ne 0$,
$$ab=-1$$
$$c=-2a-b$$
$$-1-(a+b)(2a+b)=-2a^2+2-b^2=b$$ and
$$2a^2b^2=(2-b-b^2)b^2=1.$$
The last equation has no rational roots and the only remaining option is $c=0$, giving
$$2a+b=0,ab=b$$
with the two solutions $a=b=0$, and $a=1,b=-2$.
| {
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If $(x^2-5x+4)(y^2+y+1)<2y$ for all real values of $y$ then what interval does $x$ belong to? Okay so I'm stuck on this one for a long time :
If $(x^2-5x+4)(y^2+y+1)<2y$ for all real values of $y$ then what interval does $x$ belong to?
My Try: Since $y$ can take any real value, I tried multiplying the first two term... | First, notice that $y^2+y+1$ is positive for all real values of $y$.
This means that we can divide by $y^2+y+1$ on both sides of the inequality, without having to worry about negative values that would change the direction of the inequality.
Divide by $y^2+y+1$ on both sides:
$$x^2 -5x + 4 < \frac{2y}{y^2+y+1}$$
We ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$ or $a^3-b^3$ will be divisible by $7$. This is a very interesting word problem that I came across in an old textbook of mine:
Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$... | There's always brute force:
If $m \equiv 0 \mod 7$ then $m^3 \equiv 0 \mod 7$.
If $m \equiv \pm 1,\pm 2, \pm 3$ then $m^3 \equiv \pm 1, \pm 8, \pm 27 \equiv \pm 1, \pm 1, \mp 1 \mod 7$.
So either i) $a^3 \equiv 0 \mod 7$ and $7|a^3$, ii) $b^3 \equiv 0 \mod 7$, and $7|b^3$, iii) $a^3 \equiv b^3 \mod 7$ and $7|a^3 - b^3... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Let $h(4x-1)=2x+7$. For what value of $x$ is $h(x)=x$? Let $h(4x-1)=2x+7$. For what value of $x$ is $h(x)=x$?
If $h(a)=a$, then $4x-1=2x+7$ which implies $x=4$. So $a=15$ when I substitute $x=4$ into both linear equations. Is the value of $x$ $15$?
| First, we find an expression for $h(x)$. From our definition of $h$, we have $h(4y-1) = 2y+7$. So, if we let $x=4y-1$, so that $y = (x+1)/4$, we have[h(x) = 2\cdot\frac{x+1}{4} + 7 = \frac{x+1}{2} + 7.]Setting this equal to $x$ gives[x =\frac{x+1}{2} + 7.]Multiplying both sides by 2 gives $2x = x+1 + 14$, so $x = \boxe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\frac{1}{x}<\frac{x}{2}<\frac{2}{x}$. First I consider the case $\frac{1}{x}<\frac{x}{2}$, manipulating to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x}<0\Leftrightarrow x < -\sqrt{2}.$ Did this by subtracting $\frac{x}{2}$ from both sides and writing everything with a common denominator.
Second case to consider is $\fra... | $\dfrac1x<\dfrac2x$ is only possible for positive $x$.
Then we may multiply by $2x$ and take the square roots,
$$\sqrt2<x<2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a$ is a root of $x^2+ x + 1$, simplify $1 + a + a^2 +\dots+ a^{2017}.$
If $a$ is a root of $x^2 + x + 1$ simplify $$1 + a + a^2 + a^3 + \cdots + a^{2017}.$$
my solution initially starts with the idea that $1 + a + a^{2} = 0$ since $a$ is one of the root then using the idea i grouped $$1 + a + a^2 + a^3 + \cdots +... | I haven't looked through your working, but here's a much easier method:
If $a^2+a+1=0$, then $(a-1)(a^2+a+1)=0$, that is, $a^3-1=0$, ie, $a^3=1$.
On the other hand, $1+a+\dots+a^{2017}=\frac{a^{2018}-1}{a-1}$. Since $a^3=1$, it follows that $a^{2018}=a^2$. So, your sum simplifies to $\frac{a^2}{a-1}$. <-- wrong! Read o... | {
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If $z$ is a complex number and $Re(z)>1$ then prove that $|\frac{1}{z}-\frac{1}{2}|<\frac{1}{2}$. If $z$ is a complex number and $Re(z)>1$ then prove that $|\frac{1}{z}-\frac{1}{2}|<\frac{1}{2}$.
I tried replacing $z=a+bi$ but it makes problem too long.
| You can use the theory of Moebius transformations/inversive geometry to see the behavior of the map $z\mapsto\frac{1}{z}$. This will tell you that the line $x=1$ will map to a circle. However, we can do this more explicitly as follows:
The line $x=1$ has points $1+yi$ on it. Now, the transformation $z\mapsto\frac{1... | {
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"url": "https://math.stackexchange.com/questions/2409966",
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"source": "stackexchange",
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Interesting way to evaluate $ \int \cos^3 x\ dx$ I have read these days a nice way for integrating $\cos^3x$:
First differentiate:
$$f=\cos^3(x)$$$$f'=-3\cos^2(x)\sin(x)$$$$f''=6\cos(x)\sin^2(x)-3\cos^3(x)$$$$f''=6\cos(x)(1-\cos^2(x))-3f$$$$f''=6\cos(x)-6\cos^3(x)-3f$$$$f''=6\cos(x)-9f$$
Then integrate:
$$f'= 6\sin(x) ... | Use that $\cos 3x = 4\cos^3 x - 3\cos x$ to get:
$$\int \cos^3 x \,dx = \frac{1}{4}\int\left(\cos 3x +3\cos x\right)\,dx$$
and the right side is easy to compute as $\frac{1}{12}\sin 3x +\frac{3}{4}\sin x$.
My original answer included a second approach, but I had the wrong formula for $dx$ in the substitution.
Use the ... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that inequality $\sum _{cyc}\frac{a^2+b^2}{a+b}\le \frac{3\left(a^2+b^2+c^2\right)}{a+b+c}$
Let $a>0$,$b>0$ and $c>0$. Prove that:
$$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$
$$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+... | Now, by C-S $$\sum_{cyc}\frac{1}{a+b}\geq\frac{9}{\sum\limits_{cyc}(a+b)}=\frac{9}{2(a+b+c)}.$$
Thus, it remains to prove that
$$a^2+b^2+c^2+\frac{9abc}{a+b+c}\geq2(ab+ac+bc)$$ or
$$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$
which is Schur.
Done!
| {
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"url": "https://math.stackexchange.com/questions/2413136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For what value of $k$ is one root of the equation
For which value of $k$ is one root of the equation $x^2+3x-6=k(x-1)^2$ double the other?
My Attempt:
$$x^2+3x-6=k(x-1)^2$$
$$x^2+3x-6=k(x^2-2x+1)$$
$$x^2+3x-6=kx^2-2kx+k$$
$$(1-k)x^2+(3+2k)x-(6+k)=0$$
| Now, $k\neq1$, $\Delta\geq0$ and since $x_1=2x_2$, we obtain:
$$3x_2=\frac{2k+3}{k-1}$$ and
$$2x_2^2=\frac{k+6}{k-1},$$
which gives
$$\frac{2(2k+3)^2}{9(k-1)^2}=\frac{k+6}{k-1}$$ or
$$k^2+21k-72=0,$$
which gives $k=3$ or $k=-24.$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The flux crossing a unit ball I am doing sample test provided by my professor. One of the question is:
Find the flux integral of
$$ \overrightarrow{F}=\left(x^3, y^3, z^3\right) $$
across the surface $\Sigma:x^2+y^2+z^2=1$, and oriented by normal vectors pointing away from the origin.
I do it by this way:
Let $\overrig... | HINT: Divergence (Gauss') Theorem would come pretty handy and reduce the amount of calculations.
Namely using it we have:
$$\iint_\Sigma \vec{F} \cdot \vec{n} \ d\sigma = \iiint_V \nabla \cdot F \ dV = 3\iiint_V x^2 + y^2 + z^2 \ dV$$
Now using the spherical coordinates we have:
$$\iint_\Sigma \vec{F} \cdot \vec{n} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A.P.: Sum of numbers in a particular group Question: A series of odd positive integers are divided into the following groups $(1), (3,5), (7,9,11),\dotsm$ . Prove that the sum of numbers in the $n^{\text{th}}$ group is $n^3$.
My attempt:
It is observed that the number of terms in the $n^{\text{th}}$ group is equal to $... | To form $n$ groups you need $1+2+3+\cdots+n=\frac{n(n+1)}{2}$ terms. To form $n-1$ groups you need $1+2+3+\cdots+n-1=\frac{n(n-1)}{2}$ terms. To get the sum of the terms in the $n$-th group you sum all the terms from the $1$st group to the $n$th and subtract the sum from the $1$st group to the $n-1$th which is
$$S_{n(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factorization of $18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc$ I solved a following problem in my mathematics homework today:
Factorize $18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc$.
That's easy. I solved by the following way and it was same as sample answer of the book:
$\ \ \ \ 18(ab^2 + bc... | Without loss of generality, assume $a, b=ax, c=axy$. Then:
$$18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc=$$
$$18a^3x(x + x^2y^2 + y) - 12a^3x(1 + x^2y + xy^2) - 19a^3x^2y=$$
$$a^3x[18x+18x^2y^2+18y-12-12x^2y-12xy^2-19xy]=$$
$$a^3x[6x(3-2xy)+6y(3-2xy)-(3-2xy)(4+9xy)]=$$
$$a^3x(3-2xy)(6x+6y-4-9xy)=$$
$$a^3x(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Induction: Bounds on sum of inverses of first $n$ square roots I am trying to show by induction that $2(\sqrt{n}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$.
For the upper bound, I have that if $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$, then $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{... | Note: I can see that there are already fine answers using induction!
An ad in favor of graph paper. The outcome of the standard sum/integral comparison is
$$ 2 \sqrt {n+1} - 2 \; < \; 1 + \frac{1}{\sqrt 2 } + \frac{1}{\sqrt 3 } + \cdots + \frac{1}{\sqrt n } \; < \; 2 \sqrt n - 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find $\int \frac{\sqrt{2-3x}\,dx}{x^2+1}$ Find $$\int \frac{\sqrt{2-3x}\,dx}{x^2+1}$$
I used substitution $x=\frac{2}{3} \sin^2 y$ So we get $dx=\frac{4}{3} \sin y \cos y dy$
hence
$$I= \frac{36\sqrt{2}}{3} \int \frac{\sin y \cos^2 y\, dy}{4 \sin^4 y+9}$$
Now if again i use substitution $\cos y=t$ we get
$$I=-12\sqrt{2... | Let $u=\sqrt{2-3x}$, then $1+x^2 = \frac{1}{9}\left((2-u^2)^2+9\right)$ and $x=\frac{1}{3}\left(2-u^2\right)$ so $dx=-\frac{2}{3}u\,du$ so you want:
$$-6\int\frac{u^2\,du}{(2-u^2)^2+9}$$
Now attempting partial fractions, we get a factoring:
$$\begin{align}(2-u^2)^2+9 &= u^4-4u^2+13\\&=\left(u^2+\left(\sqrt{4+2\sqrt{13}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Euler squared sum of order 2 : $\sum \limits_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2}$ Let $\mathcal{H}_n$ denote the $n$ - th harmonic number. What techniques would one use to prove that
$$\sum \limits_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2} = \zeta^2(3) + \frac{19 \zeta(... | Since
$$\left(\mathcal{H}_n^{(2)}\right)^2=\mathcal{H}_n^{(4)}+2\sum_{k=1}^{n}\frac{\mathcal{H}_{k-1}^{(2)}}{k^2}$$
we can rewrite the sum in terms of Multiple zeta functions:
\begin{align*}\sum_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2}
&=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n^{(4)}}{n^2}+
2\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find maximum of $f(x)=3\cos{2x}+4\sin{2x}.$ Find maximum of $f(x)=3\cos{2x}+4\sin{2x}.$ The shortest way to do this is to only consider $2x\in[0,2\pi)$, set $t=2x$ and note that
*
*Max$(\cos{t})=1$ for $t=0.$
*Max$(\sin{t})=1$ for $t=\frac{\pi}{2}.$
Max of these two functions added is when $t$ equals the angle ex... | We can express $3\cos 2x + 4\sin 2x$ in the form $R\cos(2x-\alpha)$, where $R>0$ and $\alpha$ is acute.
Use the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$ to find the values of $R$ and $\alpha$.
\begin{eqnarray*}
R\cos(2x-\alpha) &=& R(\cos 2x\cos\alpha + \sin 2x\sin \alpha) \\ \\
&=& (R\cos\alpha)\cos 2x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can one find $\lim_{x \to \pm\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$ without graphing? I'm trying to find both of these:
$$\lim_{x \to +\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$$
$$\lim_{x \to -\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$$
I know that they end up being $2$ and $-2$ by graphing, but what process can I use to find the... | First recall that $\sqrt{x^2} = x$ if $x\ge0$ and $\sqrt{x^2} = -x$ (which is positive) if $x<0.$
Then treat the two cases, $x>0$ and $x<0,$ separately:
\begin{align}
\frac{2x+1}{\sqrt{x^2+x+1}} & = \frac{2 + \frac 1 x}{\frac 1 x \sqrt{x^2 + x + 1}} \\[10pt]
& = \frac{2+\frac 1 x}{\frac 1 {\sqrt{x^2}}\sqrt{x^2 + x+1}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Another way of expanding Taylor series
I have to show that if $$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\ldots, $$ then prove that $$ f(x)=f(a)+2\left[ \frac{(x-a)}{2}f'\left(\frac{x+a}{2}\right)+ \frac{(x-a)^3}{3!\times 8}f'''\left(\frac{x+a}{2}\right)+ \frac{(x-a)^5}{32\times 5!}f^{(v)}\left(\frac{x+a}{2}\rig... | We write the Taylor expansion stated in OPs first line as
\begin{align*}
f(x)=\sum_{j=0}^\infty\frac{(x-a)^j}{j!}f^{(j)}(a)\tag{1}
\end{align*}
and the other expansion as
\begin{align*}
f(x)=f(a)+2\sum_{j=0}^\infty \frac{1}{(2j+1)!}\left(\frac{x-a}{2}\right)^{2j+1}f^{(2j+1)}\left(\frac{x+a}{2}\right)\tag{2}
\end{align... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\int\frac{x}{\sqrt{3-2x-x^2}} dx$. I was looking for the integral of
$$\int\frac{x}{\sqrt{3-2x-x^2}} dx$$
My work:
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(3)+(-2x-x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(2x+x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(1+2x+x^2) +1}} dx $$
$$\int \frac{x}{... | $$\int \frac{x+1}{\sqrt{3-2x-x^2}}dx=-\frac{1}{2}\int\frac{-2-2x}{\sqrt{3-2x-x^2}}dx=-\frac{1}{2}\int\frac{g'(x)}{\sqrt{g(x)}}dx$$
$$=-\frac{1}{2}\int g'(x)[g(x)]^{-1/2}dx=-(3-2x-x^2)^{1/2}=-\sqrt{3-2x-x^2}+C$$
This is the chain rule:
$$\int g'(x)f(g(x))dx=F(g(x))+C$$
where $F'(x)=f(x)$. In this case $f=x^{-1/2}$ and $... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for any integer $n, n^2+4$ is not divisible by $7$. The question tells you to use the Division Theorem, here is my attempt:
Every integer can be expressed in the form $7q+r$ where $r$ is one of $0,1,2,3,4,5$ or $6$ and $q$ is an integer $\geq0$.
$n=7q+r$
$n^2=(7q+r)^2=49q^2+14rq+r^2$
$n^2=7(7q^2+2rq)+r^2$
$n... | since we have $$n\equiv 0,1,2,3,4,5,6\mod 7$$ we get $$n^2\equiv 0,1,2,4\mod 7$$
therefore $$n^2+4\equiv 1,4,5,6\mod 7$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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for $abc = 1$ and $a \le b \le c$ prove that $(a+1)(c+1)>3$ This inequality has been given to me by my teacher to keep me occupied and after hours of fumbling around with it, and later trying to google it. I found nothing at all.
For any three positive real numbers $a$, $b$ and $c$, where $abc = 1$ and $a\le b \le c$, ... | First simplify the expression :
$ac + a + c + 1 \gt 3 \iff ac + a+c \gt 2 \iff 1/b + a+ c \gt 2 \iff 1 + ab+bc \gt 2b $
We should prove $1+ab+bc \gt 2b$ . Use the condition $a\le b\le c $ :
$a\le b\le c \iff 2a \le a+b \le a+c \to a+b\le a+c$
From these we have :
$a+b\le a+c \iff ab + b^2 \le ab + bc \iff ab + b^2 +1 \... | {
"language": "en",
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How to split $\frac{x^3}{(x^2+5)^2}$ into partial fractions?
Split $\frac{x^3}{(x^2+5)^2}$ into partial fractions.
I have tried letting $$\frac{x^3}{(x^2+5)^2}=\frac{ax+b}{x^2+5}+\frac{cx^3+dx^2+ex+f}{(x^2+5)^2},$$ but this yields a system of only four simultaneous equations.
I have also tried looking at $$\frac{x^3... | Note that, by adding and subtracting $5x$ at the numerator, we obtain
$$\frac{x^3}{(x^2+5)^2}=\frac{x^3+5x-5x}{(x^2+5)^2}=\frac{x(x^2+5)-5x}{(x^2+5)^2}=\frac{x}{x^2+5}-\frac{5x}{(x^2+5)^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ The title is quite clear. I am required to prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ using the epsilon-delta definition of the limit.
Given any $\varepsilon \gt 0$, there exists a $\delta =$
Such that $0 \lt \lvert x-1 \rvert \lt \delta \Rightarrow \lvert\f... | So you simplified the expression
\begin{align}
\left | \frac{x+2}{x^2+1} -\frac{3}{2} \right | = \left | \frac{(x-1)(3x+1)}{2x^2+2} \right |.
\end{align}
The good news is we have an $|x-1|$ appearing, but there is also a $|3x+1|$ and $|2x^2+2|,$ which are expressions we don't necessarily want. The thing to do here is ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove the sequence $\frac{n^2}{n^2-n-5}$ converges I know the basic procedure for proving convergence of sequences, but I'm having difficulty reducing this sequence. I believe the sequence converges to 1, so I've set up the following two approaches:
1:
$\left|\frac{n^2}{n^2 - n - 5} - 1\right| = \left|\frac{n^2}{n^2 - ... | For $n\to \infty$ we have $\frac{n^2}{n^2-n-5}=\frac{1}{1-\frac{1}{n}-\frac{5}{n^2}}$ (dividing numerator and denominator by $n^2$). All terms on the denominator go to 0 except 1 so the limit is 1.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation. If $x^2+y^2\leq 1$ then maximum of $ax+by$
Here what I have done so far.
Let $ax+by=k$ . Thus $by=k-ax$.
So we can have that $$b^2x^2+(k-ax)^2 \leq b^2$$
$$b^2x^2+k^2-2akx +a^2x^2-b^2\leq 0 $$
By re-writing as a... | Let $x=r \cos \theta$ and $y=r \sin \theta$ with $r \leq 1$ so that $x^2+y^2 \leq 1$. Then,
$$ax+by=r(a\cos \theta+b\sin \theta)$$
But one may show that we may write,
$$a \cos \theta+b \sin \theta=\sqrt{a^2+b^2} \cos (\theta+\phi)$$
For some $\phi$.
Hence $ax+by \leq (1)(\sqrt{a^2+b^2})$.
| {
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"url": "https://math.stackexchange.com/questions/2426188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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How to integrate $\int \frac {e^y}{y} dy$? The question is to evaluate $$\iint_R \frac {x}{y} e^y dx dy$$ where R is the region bounded by $0 \leq x \leq 1$ and $x^2 \leq y \leq x$.
So i write it as $$\int_0^1 \int_{x^2}^{x} \frac{x}{y} e^y dy dx$$.
The thing is, how do i evaluate $I=\int_{x^2}^{x} \frac{1}{y} e^y dy$?... |
$$\int_0^1 x \left(\int_{x^2}^{x} \frac{1}{y} e^y dy \right) dx$$
You can integrate by parts. Let $u(x)=\int_{x^2}^{x} \frac{1}{y} e^y dy$ and $dv=xdx$. Then we have,
$$du=\left(\frac{e^x}{x}-(2x)\frac{e^{x^2}}{x^2}\right) dx$$
$$=\left(\frac{e^x}{x}-2\frac{e^{x^2}}{x} \right) dx$$
We may let $v=\frac{x^2}{2}$. Then ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $B$ is added to $A$, under what condition does the resultant vector have a magnitude equal to $A+B$? If $B$ is added to $A$, under what condition does the resultant vector have a magnitude equal to $A+B$? Under what conditions is the resultant vector equal to zero?
My Attempt:
Let $\theta $ be the angle between $\ve... | Though the obvious fact that $\vec B=-\vec A$, we can infer this from your previous expression:
$$A^2+B^2+2AB\cos\theta=0$$
If $A=0$, then trivially $B^2=0 \to B=0$, and viceversa, and the angle can be any value.
If $A,B\gt0$, this leads to:
$$
\cos\theta=-{A^2+B^2 \over 2AB}
$$
Knowing that $\cos(\cdot)$ must be bound... | {
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"url": "https://math.stackexchange.com/questions/2427421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A floor function equation I want to solve below equation analytically $$\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor=1$$ but I have no idea to start .
Implicit: I solve it by graphing and the solution was $x\in[\frac {1}{4},1)$
I am thankful for any idea in advance .
I try to use $x=n+p... | If $x<0$ then $a=\lfloor \frac{2x+1}3\rfloor$ and $b=\lfloor \frac{\lfloor 4x\rfloor+2}3\rfloor$ are both inferior or equal to $0$, so the sum cannot be $1$.
Now since $a$ is smaller than $b\quad$ [ $(\lfloor 4x\rfloor+2)-(2x+1)\ge 4x-1+2+2x-1\ge 2x\ge 0$ ]
The only way to get $a+b=1$ with $0\le a\le b\quad$ integers... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Eavaluating the roots of quadratic equation If $b>a , c>0$
Determine the intervals that the roots of the equation
$(x-a)(x-b) -c =0$ belong to
My work is to get the values of the roots in terms of a , b and c using the general form but i couldn't determine those intervals
| First solve for $x$ the quadratic equation $(x-a)(x-b)-c=0$ by the general form we get that $x_1 =\frac{1}{2} \left(\sqrt{a^2-2 a b+b^2+4 c}+a+b\right)$ and $x_2=\frac{1}{2} \left(-\sqrt{a^2-2 a b+b^2+4 c}+a+b\right)$
substituting instead of $b$ the value $a$ we get that $x_1 >\frac{1}{2} \left(2 a+2 \sqrt{c}\right) $ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integration changing limits - does the question have an error? I am asking about changing the limits of integration.
I have the following integral to evaluate -
$$\int_2^{3}\frac{1}{(x^2-1)^{\frac{3}{2}}}dx$$ using the substitution $x = sec \theta$.
The problem states
Use the substitution to change the limits into t... | The book is indeed incorrect.
$$arccos(\frac{1}{3}) \neq \frac{\pi}{3}$$
Note if it was the case that $arccos(\frac{1}{3}) = \frac{\pi}{3}$ as stated in the solutions,
then the integral would total 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Binomial series $\sum^{\infty}_{n=0} {{n+k}\choose{k}} a^{n}$ Does someone know how to derive this sum?
$$\sum^{\infty}_{n=0} {{n+k}\choose{k}} a^{n}$$ where $|a|<1$ and $k$ is given integer.
| $\binom{n+k}{k} = [x^k]:(1+x)^{n+k}$
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{n+k}{k} a^k &=& [x^k]: \sum_{n=0}^{\infty} a^k (1+x)^{n+k} \\
&=& [x^k]: (1+x)^{k} \frac{1}{1-a(1+x)} \\
&=& [x^k]: (1+x)^{k} \frac{1}{1-a} \frac{1}{1-\frac{ax}{1-a}}\\
&=& \frac{1}{1-a}[x^k]: \left( \sum_{j=0}^{k} \binom{k}{j} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $ \forall n \ge 4$, $n^{3} + n < 3^{n}$
Prove that $\forall n \ge 4$, $n^{3} + n < 3^{n}$
My attempt: Base case is trivial.
Suppose $ \ n \ge 4$, $n^{3} + n < 3^{n}$. Then,
$$ (n+1)^{3} + (n+1) = n^3 + n + 3n^2 + 3n + 1 +1 < 3^{n} + 3n^2 + 3n + 2 \\< 3\cdot 3^{n} + 3n^2 + 3n +2.$$
Not sure how to get rid of $3... | You proved correctly that $(n+1)^3+(n+1)<3^n+3n^2+3n+2$. So, prove now that$$(\forall n\geqslant4):3n^2+3n+2\leqslant2\times3^n.$$This is easy to do using induction again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$, show that $x+y=0$
For $\{x,y\}\subset \Bbb R$, $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1.$
Prove that $x+y=0.$
Problem presented in a book, as being from Norway Math Olympiad 1985. No answer was presented. My developments are not leading to a productive direction. Sorry if this is... | Since
$$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1 / \cdot (x-\sqrt{x^2+1})$$
we get
$$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x$$ so
$$\sqrt{x^2+1}-\sqrt{y^2+1}=x+y\;/^2$$
and thus $$-\sqrt{y^2+1}\cdot \sqrt{x^2+1}+1= xy$$
so $$x^2y^2 +x^2+y^2+1=x^2y^2-2xy+1$$
and finaly $(x+y)^2=0$ so $x+y=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Calculating the limit $\lim\limits_{x \to 0^+} \frac{\sqrt{\sin x}-\sin\sqrt{ x}}{x\sqrt{x}}$
Calculate $$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}$$
without use Taylor serie and L'Hôpital.
$$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}\cdot\dfrac{\sqrt{\sin x... | note that
$$
\begin{split}
\lim_{x\to0^{+}}
\frac{\sin x-\sin^2\sqrt{x}}{x^2}&\overset{t=\sqrt{x}}{=}
\lim_{t\to0^{+}}
\frac{\sin(t^2)-\sin^2t}{t^4}\\
&=\lim_{t\to0^{+}}
\frac{(t^2-\frac{t^6}{6}+\frac{t^{10}}{120}-\ldots)-(t^2-\frac{t^4}{3}+\frac{2t^{6}}{45}+\frac{t^8}{315}-\ldots}{t^4}\\
&=\lim_{t\to0^{+}}
\frac{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Contractive Principle to Prove Convergence How do I use the contraction principle to show ${a_{n}}$ converges?
$a_{1}=1 \:$ and$\: a_{n+1}= 1+ \frac{1}{1+a_{n}}.$
I know I need to show that $\left | a_{n+2}-a_{n+1} \right |=k\left | a_{n+1}-a_{n} \right |$ with $k \in (0,1)$
So far I have:
$\left | a_{n+2}-a_{n+1} \rig... | Show that $f(x) = 1 + {1\over 1 +x}$ maps $[1,\infty) \to [1, \infty)$.
Show that $|f'(x)| \le {1 \over 4} < 1$ for $x \ge 1$.
Show that $|f(x)-f(y)| \le {1 \over 4} |x-y|$ for $x,y \ge 1$.
Note that $a_{n+1} = f(a_n)$ and the contraction mapping theorem shows that there is a unique $a$ such that $a_n \to a$. Since $a=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that for $n \ge 2$ the follow inequality holds $\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}$. As already shown above I need to prove that $$\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}\qquad \forall n \ge 2.$$
What I've come up with is the following:
$\underline{n=2:}\qquad$ $$\frac {16}{3} \lt 6$$
$\underline{n=k:}\... | Note that $$\frac{\frac{4^{n+1}}{n+2}}{\frac{4^n}{n+1}}=4\frac{n+1}{n+2}$$ and that $$\frac{\frac{(2(n+1))!}{(n+1)!^2}}{\frac{(2n)!}{n!^2}}=\frac{(2n+2)!}{(2n)!}\cdot\frac{n!^2}{(n+1)!^2}=\frac{(2n+2)(2n+1)}{(n+1)^2}=2\frac{2n+1}{n+1}. $$So, when is it true that$$4\frac{n+1}{n+2}\leqslant2\frac{2n+1}{n+1}?$$ Well, the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that two of the straight lines represented by $x^3+bx^2y+cxy^2+y^3 = 0$ will be at right angles if $b+c$=-2.
Prove that two of the straight lines represented by $x^3+bx^2y+cxy^2+y^3 = 0$ will be at right angles if $b+c=-2$.
I divided whole equation by $y^3$ to get a cubic in $\frac{x}{y}$. I guess product of tw... | Let $y=mx$.
Thus, me need
$$m^3+bm^2+cm+1=0$$ with $m_1m_2=-1$ and since $m_1m_2m_3=-1$,
we obtain $m_3=1$ and $$1+b+c+1=0$$ or
$$b+c=-2.$$
Id est, we proved a bit of more:
Two of the straight lines represented by $$x^3+bx^2y+cxy^2+y^3=0$$ will be at right angles iff $b+c=-2$.
Because if $b+c=-2$ we see that one ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $\left(\sum_{r=1}^n r\right)^2=\sum_{r=1}^n r^3$ without expanding to closed form It is well know that $$\left(\sum_{r=1}^n r\right)^2=\sum_{r=1}^n r^3$$
i.e.
$$(1+2+3+\cdots+n)(1+2+3+\cdots+n)=1^3+2^3+3^3+\cdots+n^3$$
and this is usually proven by showing that the closed form for the sum of cubes is $\frac ... | $$
\underbrace{(1+2+3+\cdots+n)(1+2+3+\cdots+n) = 1^3+2^3+3^3+\cdots+n^3}_{\Large\text{This will be our induction hypothesis.}}
$$
\begin{align}
& \Big( \underbrace{1+2+3+\cdots+n}_{\Large A} +\underbrace{\Big(n+1\Big)}_{\Large B}~~\Big)^2 \\[10pt]
= {} & (A+B)^2 = A^2+2AB+B^2 \\[10pt]
= {} & \overbrace{(1+2+3+\cdots+n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Find all the points that lie $ 3$ units from each of the points $ (2,0,0), (0,2,0), \text{ and }(0,0,2)$
I calculated the result $$ \{(x,y,z) \in\mathbb{R^3}: x=y=z\}.$$
I'm wondering whether I did this problem correctly and if I did how to draw the set of solutions.
I used the euclidean distance formula with the squ... | The set of all points equidistqnt from $A=(2,0,0)$ and $B=(0,2,0)$:
\begin{align}
\sqrt{(x-2)^2+y^2+z^2}&=\sqrt{x^2+(y-2)^2+z^2} \\
-4x+4 &= -4y+4 \\
x&=y
\end{align}
Similarly, the set of all points equidistant from $A=(2,0,0)$ and $C=(0,0,2)$ is described by $x=z$.
Hence the set of all points equidistant fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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