Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Reflection matrix in $\Bbb R^2$ (matrix $R$ satisfying $R^2 = 1$) So there is this problem
A square matrix $R$ is called a reflection matrix if $R^2 = I$. Here are some examples $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} , \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix... | That's a good start. For convenience, let $M = \pmatrix{a&b\\c&d}$.
If you want the entries of $M$ to be nonzero, notice that
$$
\pmatrix{a^2 + bc & ab + bd \\ ac + cd & b^2 + bc} =
\pmatrix{a^2 + bc & b(a + d) \\ c(a + d) & d^2 + bc}.
$$
So, if $ab + bd = b(a + d) = 0$ with all of $a, b$, and $d$ nonzero, we must ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$
find the :
$$\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$$
My Try :
$$\dfrac{1}{1+\dfrac{\sin 70^{\circ}}{\cos 70^{\circ}}}+\dfrac{1}{1+\dfrac{\sin 20^{\circ}}{\cos 20^{\circ}}}$$
$$\dfrac{\cos70^{\circ}}{\cos 70^{\circ}+\sin 70^{\cir... | $$\frac { 1 }{ 1+\tan { { 70 }^{ \circ } } } +\frac { 1 }{ 1+\tan { { 20 }^{ \circ } } } =\frac { 1 }{ 1+\tan { { 70 }^{ \circ } } } +\frac { 1 }{ 1+\cot { { 70 }^{ \circ } } } =\\ =\frac { 1 }{ 1+\tan { { 70 }^{ \circ } } } +\frac { 1 }{ 1+\frac { 1 }{ \tan { { 70 }^{ \circ } } } } =\frac { 1+\tan { { 70 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$\tan{A} \cdot \tan{B} \cdot \tan{C}=9$, find $\tan^2{A}+\tan^2{B}+ \tan^2{C}$ In $\triangle{ABC}$,
$$\tan{A}\cdot \tan{B}\cdot \tan{C}=9$$
$$\tan^2{A}+\tan^2{B}+ \tan^2{C}=\lambda$$
then,
$\lambda$ lies in the interval?
| We have the inequality $\displaystyle (ab+bc+ca)^2 \ge 3abc(a+b+c)$
Also in $\displaystyle \triangle ABC, \ \tan A+\tan B + \tan C = \tan A \tan B \tan C$,
Hence, $\displaystyle \left(\sum \tan A \tan B \right)^2 \ge 3 \times 9^2 \Rightarrow \sum \tan A \tan B \notin \left(- 9 \sqrt 3 , 9 \sqrt 3 \right)$
So $\displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Let $(X, Y, Z)$ be jointly continuous random variables over a truncated cone This may sound like a silly question (I may just be blanking right now), but I need to solve the following question:
Let $(X,Y,Z)$ be jointly continuous random variables with joint pdf given by
$$
f(x,y,z) = \left\lbrace
\begin{array}{cl}
c... | The domain of $\Bbb R^3$ defined by $x^2 + y^2 \leq z^2$ and $0<a\leq z \leq b$ is a truncated cone with axis $(Oz)$ and apex angle $\pi/4$. Its volume is given by
\begin{aligned}
\int_{\Bbb R^3} \mathbf{1}_{x^2 + y^2 \leq z^2}\, \mathbf{1}_{a\leq z\leq b}\, \text{d}x\,\text{d}y\,\text{d}z
&= \int_{\Bbb R} \mathbf{1}_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Indicial equation of $(x^2-1)^2y''+(x+1)y'-y=0$ Let $\alpha$ and $\beta$ with $\alpha>\beta$ be the roots of the indicial equation of $(x^2-1)^2y''+(x+1)y'-y=0$ at $x= -1$.
Then what is the value of $\alpha-4\beta$ ?
I am trying to solve this by Frobenius series solution method. Assuming trial solution of the form ... | $$(x^2-1)^2y''+(x+1)y'-y=0$$
At $x=-1$
We have,
$$\alpha(x)(x+1)^2y''+\beta(x)(x+1)y'+\gamma(x)y=0$$
With$ \begin{cases} \alpha(x)=(x-1)^2 \implies \alpha(-1)=4\\
\beta(x)=1 \implies \beta(-1)=1\\
\gamma(x)=-1 \implies \gamma(-1)=-1
\end{cases}
$
Then the indicial equation is
$$r^2+\left(\frac {\beta(-1)}{\alpha(-1)}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5... | By the generalized binomial theorem,
$$(1+t)^\alpha-1=\alpha t+\frac{\alpha(\alpha-1)}2t^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}t^3+\cdots$$ and higher order terms.
Hence your limit is essentially
$$\lim_{t\to0}\frac{\dfrac t5}{\dfrac t6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 8
} |
How to factor $2x^3 + 21x^2 +27x$ I am having trouble with factoring $2x^3 + 21x^2 +27x$. The answer is $x(x+9)(2x+3)$ but not sure how that was done. Obviously I factored out the $x$ to get $x(2x^2+21x+27)$ then from there I am lost. I tried the AC method and grouping. Can someone show the steps? Thanks!
| 1)
$2x^3 + 21x^2 +27x = x(2x^2 + 21x + 27) = x(2x(x + 9) + 3x + 27) =x(2x(x+9) + 3(x + 9)) = x(2x + 3)(x+9)$.
2)
$2x^3 + 21 x^2 + 27x = x(2x^2 + 21x + 27) = x*2*(x - a)(x+b)$ where $a,b$ are solutions to $2x^2 + 21x + 27=0$. i.e. $x = \frac {-21 \pm {21^2 - 8*27}}{4} = \frac {-21 \pm \sqrt {225}}4 = \frac {-21 \pm 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Let $f: \mathbb{R} \to \mathbb{R}$ and $\exists \ \ b \in \mathbb{R} : f(x+b)=\sqrt{f(x)-f^2(x)}+\frac{1}{2}$
Let $f: \mathbb{R} \to \mathbb{R}$ and $\exists \ \ b \in \mathbb{R} : f(x+b)=\sqrt{f(x)-f^2(x)}+\frac{1}{2}$
then find the :
$$\lim_{x \to \infty} f(x)=?$$
My Try : $f^2(x+b)+\frac{1}{4}-f(x+b)=f(x)-f^2(x... | Rewrite the starting equation like this:
$$\left(f(x+b)-\frac{1}{2}\right)^2={1\over 4}-{1\over 4}+{f(x)-f^2(x)}= {1\over 4}- \left(f(x)-{1\over 2}\right)^2$$
so we have also
$$\left(f(x)-\frac{1}{2}\right)^2= {1\over 4}- \left(f(x-b)-{1\over 2}\right)^2$$
Combining both equations we get $$\left(f(x+b)-\frac{1}{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Stationary points of $ \frac{1}{2} \|A-xx^T\|^2_F $ Stationary points of $ \dfrac{1}{2} \|A-xx^T\|^2_F $.
If $A$ is an $n \times n$ matrix and $x$ is an $n \times 1$ vector. How do I get the stationary points?
I know that I am supposed to find the gradient and equate it to $0$, but how do I do that for a vectorized equ... | Let $f(x) = \frac{1}{2} \|A-xx^T\|^2_F$ and $\partial_k f$ the partial derivative of $f$ with respect to $x_k$.
By the definition of the Frobenius norm
$$ f(x) = \frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} (a_{ij} - x_i x_j)^2 $$
We can split that double summation in clear way so that computing $\partial_k f$ becomes str... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Trig substitution for $\int \frac{x^2dx}{\sqrt{4 - x^2}}$ According to my textbook the answer is: $2\arcsin(\frac{1}{2}x) - \sin(2\arcsin(\frac{1}{2}x))$, but I'm getting something slightly different.
First I tried setting $x = 2 \sin \theta, dx = 2 \cos \theta, \theta = \arcsin(\frac{x}{2})$
$$\int \frac{4\sin^2 \thet... | I don't like your book's answer either.
$\sin(2\arcsin(\frac x{2}))$ should be simplified to $2\frac {x}{2}\sqrt {1-\frac {x^2}{4}} = \frac 12 x\sqrt {4-x^2}$
Others have pointed out the problem with the half-angle identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Number of point of intersection of $y=(x+2)$ and $y=(x+2)^2$
The solution curve of the differential equation $\displaystyle (x^2+xy+4x+2y+4)\frac{dy}{dx}-y^2=0,x>0$
Passes through the point $(1,3).$ Then
$(1)$ number of point of intersection of $y=x+2$
$(2)$ number of point of intersection of $y=(x+2)^2$
Try: $$\big... | $$y^2 \frac{dx}{dy} = (x+2)^2 + y(x+2)\\
\frac{1}{(x+2)^2}\frac{dx}{dy}-\frac{1}{y(x+2)} = \frac{1}{y^2}$$
Observe it is a linear differential equation by treating $\frac{-1}{x+2}$ as some variable $t$. Integration factor will be $\frac{1}{y}$ and solution is
$$\frac{-1}{(x+2)y} = \int \frac{dy}{y^3} = \frac{-1}{2y^2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $\int \frac{dx}{(a^2 + x^2)^2}$? I'm trying to use a trig substitution but I'm stuck. Here's what I did so far:
$$\int \frac{dx}{(a^2 + x^2)^2}$$
Let $x = a\sin \theta, dx = a\cos \theta d\theta$
$$\int \frac{a cos\theta d\theta}{(a^2 + a^2 sin^2 \theta)^2} = \int \frac{a\cos \theta d\theta}{(a^2(1+si... | $$\int\dfrac{dx}{a^2+x^2}=\dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right)+C$$
then
$$\dfrac{d}{da}\int\dfrac{dx}{a^2+x^2}=\dfrac{d}{da}\left(\dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right)+C\right)$$
and
$$\int\dfrac{-2adx}{(a^2+x^2)^2}=\dfrac{1}{2a^3}\arctan\left(\dfrac{x}{a}\right)-\dfrac{x}{2a^2(a^2+x^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Evaluating $\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}$
Evaluate:
$$\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}$$
I have been trying to solve this for $15$ minutes but sin(sin(x)) part has me stuck.
My attempt:
I tried multiplying with $x$ inside the $\sin$ as $\sin{(\frac{x\sin{x}}{x})}$. No leads.
| Note that by Taylor's expansion
*
*$x\sin(\sin x)=x^2-\frac13x^4+\frac1{10}x^6+o(x^6)$
*$\sin^2x=x^2-\frac13x^4+\frac2{45}x^6+o(x^6)$
thus
$$\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}=\frac{x^2-\frac13x^4+\frac1{10}x^6-x^2+\frac13x^4-\frac2{45}x^6+o(x^6)}{x^6}=\frac1{18}+o(1)\to\frac1{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Infinite sum of antisymmetric matrix? I have an antisymmetric matrix ${\bf A} =
\begin{bmatrix}
0 & \frac{1}{2} \\
- \frac{1}{2} & 0
\end{bmatrix}
$, and I’d like to prove that
$$\sum_{n=1}^{\infty} \frac{{\bf A}^{4n} - 2{\bf A}^{2n}}{n} =
\begin{bmatrix}
ln\frac{5}{3} & 0\\
0 & ln\frac{5}{3}
\end{bmatrix}
$$
I have ... | Note that
$${\bf A^2} =
\begin{bmatrix}
-1/4 & 0 \\
0 & -1/4
\end{bmatrix} = (-1/4) I \quad \text{and} \quad
{\bf A^4} =
\begin{bmatrix}
1/16 & 0 \\
0 & 1/16
\end{bmatrix} = (1/16) I$$
So now we can factor out the identity matrix and reduce the problem to solving
$$\sum_{n=1}^{\infty}\frac{(1/16)^n - 2(-1/4)^n}{n}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the natural solutions of $a^3-b^3=999$ I want to find the natural solutions of $a^3-b^3=999$.
I got $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$, so if we consider the equation in $\mathbb{Z}/3\mathbb{Z}$ we get
$$(a-b)\cdot(a^2+ab+b^2) \equiv0 \text{ mod }3$$
and because $\mathbb{Z}/3\mathbb{Z}$ is a domain, we get
$$a\equ... | You have $(a-b)(a^2+ab+b^2)=999$, so $a-b$ is a factor of $999$, that is
one of $1,3,9,27,37,111,333$ and $999$ and $a^2+ab+b^2$ is the complementary factor. There are now eight cases. If $a-b=1$, then
$a=b+1$ and $999=a^2+ab+b^2=3b^2+3b+1$. This is a quadratic equation;
has it any integer solution? Once this is decide... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How to calculate $\int _\frac{3}{4}^{\frac{3}{2}} \sqrt{9 - 4x^2}$ $$\int _\frac{3}{4}^{\frac{3}{2}} \sqrt{9 - 4x^2}$$
First I set $x = \frac{3}{2}\sin\theta, dx = \frac{3}{2}\cos \theta d\theta$:
$$\int \sqrt{9 - 4x^2}dx = \int \sqrt{4\left(\frac{9}{4} - x^2\right)}dx = 2\int \sqrt{\left(\frac{3}{2}\right)^2 - \left(\... | Now you use the fact that$$\sin(2\arcsin x)=2\sin(\arcsin x)\cos(\arcsin x)=2x\sqrt{1-x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Identity involving binomial coefficients I might have written this in a needlessly cumbersome way, but I want to prove that for odd positive integers $n$, $$\sum_{k\ odd}^{n}\binom{2n+1}{2k}=\begin{cases}
\binom{2^n+1}{2}, & \text{if}\ n\ \text{mod}\ 4 =1\\
\binom{2^n}{2}, & \text{if}\ n\ \text{mod}\ 4 =3
... | Hint: Use the binomial formula for $(1+1)^{2n+1}$, $(1-1)^{2n+1}$, $(1+i)^{2n+1}$ and $(1-i)^{2n+1}$:
$$A=(1+1)^{2n+1}=\binom{2n+1}{0}+\binom{2n+1}{1}+\binom{2n+1}{2}+\binom{2n+1}{3}+\cdots$$
$$B=(1-1)^{2n+1}=\binom{2n+1}{0}-\binom{2n+1}{1}+\binom{2n+1}{2}-\binom{2n+1}{3}+\cdots$$
$$C=(1+i)^{2n+1}=\binom{2n+1}{0}+\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to prove this integration equality? How to prove
\begin{align*}
&\mathrel{\phantom{=}}6\int_0^1\mathrm{d}x\int_1^{\infty}\frac{x\left( x-1 \right)}{t}\exp \left( -\frac{t}{\sqrt{x\left( 1-x \right)}} \right) \,\mathrm{d}t\\
&=-\int_1^{\infty}e^{-2x}\frac{2x^2+1}{2x^4}\sqrt{x^2-1}\,\mathrm{d}x
\end{align*}
I have n... | We can write
\begin{align}
I&=6\int_0^1\mathrm{d}x\int_1^{\infty}\frac{x\left( x-1 \right)}{t}\exp \left( -\frac{t}{\sqrt{x\left( 1-x \right)}} \right) \,\mathrm{d}t\\
&=6\int_0^1 x\left( x-1 \right)\mathrm{d}x\int_{\tfrac{1}{\sqrt{x\left( 1-x \right)}}}^{\infty}\frac{e^{-u}}{u} \,du\\
&=-12\int_0^{1/2} x\left( 1-x \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to find the shaded area How to find the shaded area crossed by semi-circle of radius 2 and quarter-circle of radius 4?
| Hint.
(This space intentionally left blank.)
Solution.
$$\begin{align}
\frac12\cdot\text{target area} &= A + B \\
&= \left( \frac12 \cdot (2s)^2\cdot\alpha - 4 C\right) + \left(\frac12 \cdot s^2 \cdot \beta - C \right) \\[4pt]
&= \frac12 s^2\left( 4 \alpha + \beta \right) - 5 C \\[4pt]
&=\frac12 s^2 \left( \frac\pi2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$ Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$
The only idea I have is that I could apply Cauchy-Schwartz but i don't see how, any hints?
| The minimal value of the parabola
$$
a^2+b^2=\left( \frac 1 2-2\,b \right) ^{2}+{b}^{2}=\frac 1 4-2\,b+5\,{b}^{2}
$$
is at its vertex $b=\dfrac{1}{5}$ and it equals exactly $\dfrac{1}{20}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Prove that $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\geq \frac{2}{3}$ Prove that for $n \in \mathbb{N}$
(a) $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\geq \frac{2}{3}$
And
(b) $\frac{1}{2} \leq \frac{1}{3n+1}+\frac{1}{3n+2}+.....\frac{1}{5n}+\frac{1}{5n+1}+....< \frac{2}{3}$
my attempt... | By C-S
$$\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\geq\frac{(1+1+...+1)^2}{n+(n+1)+...+(2n)}=\frac{(n+1)^2}{\frac{(n+1)(2n+n)}{2}}=\frac{2(n+1)}{3n}>\frac{2}{3}$$
$$\frac{1}{3n+1}+\frac{1}{3n+2}+.....\frac{1}{5n}+\frac{1}{5n+1}\geq\frac{(1+1+...+1)^2}{(3n+1)+(3n+2)+...+(5n+1)}=$$
$$=\frac{(2n+1)^2}{\frac{(2n+1)(2(3n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$ I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$:
$$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$
which gives (Wolfram Alpha)
$$D= \left[
\frac{\si... | Note that
$$D = 2\int_0^{\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$$
Let $t=\tan\frac{\theta}2$, we have
\begin{align}
D&=2\int_0^{\infty} \frac{\frac{4t^2}{(1+t^2)^2}}{(1-a\frac{1-t^2}{1+t^2})^3}\frac{2}{1+t^2}\,dt\\
&=\int_0^{\infty} \frac{16t^2}{(1-a+(1+a)t^2)^3}\,dt\\
&=\frac{\pi}{(1-a^2)^{3/2}}\tag{1}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Am I on the right track to figure out this combinatorics problem? Say, we have 4 red balls, 4 blue balls, 4 green balls, 4 yellow balls. How many ways are there to form a sequence of 10 balls such that every color of ball occurs at least twice?
My thought:
r = red b = blue g = green y = yellow * = undetermined
Every ar... | The partition by distinct colors can be either $10=4+2+2+2$ or $10=3+3+2+2$. In the first case, you also need to choose the $1$ color that is to be used 4 times, and in the second case, the $2$ colors that are to be used $3$ times. Thus, the total number of choices is
$$
\binom{4}{1}\binom{10}{4,2,2,2}+\binom{4}{2}\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Tangent substitution in trigonometric substitution Find:
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}}\;dx$$
Let $u=2x$ and with tangent substitution we have $x=\frac{3}{2}\,tan\,\theta$ and now we have $dx=\frac{3}{2}\,sec^2\,d\theta$.
Also, $\sqrt{4x^2+9}=3\,sec\,\theta$.
When $x=0$, $tan\,\theta=0$, so $\theta=0$... | $$\frac { -3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \frac { 1-u^{ 2 } }{ u^{ 2 } } du=\frac { -3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \left( \frac { 1 }{ { u }^{ 2 } } -1 \right) du=\frac { -3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \left( { u }^{ -2 }-1 \right) du=\\ =\frac { 3 }{ 16 } \int _{ 1 }^{ \frac { ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Derivative of natural log with chain rule. Is there a better way? I'm a bit stuck on taking these two derivatives:
$$h(x) = \ln(x + \sqrt{x^2-1})$$
\begin{align}
h'(x) &= \frac{1}{x + \sqrt{x^2 - 1}} \cdot \frac{d }{dx} (x + \sqrt{x^2 -1} )\\
&= \frac{1}{x + \sqrt{x^2 - 1}} \biggl(1 + \frac{2x}{2\sqrt{x^2 - 1}}\biggr)
... | You can also do it this way:
$$h(x)=\log(x+\sqrt{x^2-1})$$
$$\exp(h(x))=x+\sqrt{x^2-1}$$
Now we can differentiate both sides with respect to $x$:
$$\exp(h(x))*h'(x)=1+\frac{x}{\sqrt{x^2-1}}=\frac{\sqrt{x^2-1}+x}{\sqrt{x^2-1}}=\frac{\exp(h(x))}{\sqrt{x^2-1}}$$
So:
$$h'(x)=\frac{1}{\sqrt{x^2-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Dividing balls into boxes with capacity limits Problem:
In how many ways can you divide $13$ identical balls into $3$ different boxes
$k_1$, $k_2$, $k_3$, such that $k_1$ contains no more than $5$ balls, $k_2$ contains no more than $6$ balls and $k_3$ contains no more than $4$ balls?
My idea:
So my idea is to use the ... | To show the working behind the generating function approach:
A binomial expansion of $(1-x)^{-n}$ gives $$(1-x)^{-n} = 1 + (-n)(-x) + \frac{(-n)(-n-1)}{2!} (-n)^2 + \frac{(-n)(-n-1)(-n-2)}{3!} (-n)^3 + \cdots \\
= \sum_{i\ge 0}\frac{(-1)^i (-n)^\underline{i}}{i!}x^i \\
= \sum_{i\ge 0}\frac{n^\overline{\,i\,}}{i!}x^i \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate the surface integral of the cylinder that cuts the cone
How do I calculate this? How do I find dS? What are the boundaries of the integral? Thanks!
| HINTS
Note that the equation of the cylinder is
$$x^2+y^2=2x \implies x^2-2x+1+y^2=1\implies (x-1)^2+y^2=1$$
thus it is generate by a circle with radius $1$ and center at $(1,0)$.
For a fixed $z$, the intersection between the culynder ant the cone is given by
$$x^2+y^2=2x=z^2\implies x=\frac{z^2}2 \quad y=\sqrt{z^2-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How is $x(2x+7)+3$ equal to $(2x+1)(x+3)$? For some reason, $x(2x+7)+3$ seems like it should be equal to $(2x+7)(x+3)$ instead of $(2x+1)(x+3)$. How does $(2x+1)$ factor out of here?
The original equation was $2x^2+7x+3$.
Proof of this: https://www.desmos.com/calculator/2xpcqznkio
Notice how $x(2x+7)+3$ and $(2x+1)(x+3... | $2x+7$ is a factor of the first term, clearly. But it's not a factor of the second term. And therefore it can't possibly be a factor of the whole expression.
One could argue for the $(2x+1)$ version like this:
$$
x(2x+7)+3 = x(2x+1+6) + 3\\
= x(2x+1) + 6x + 3\\
= x(2x+1) + 3(2x+1)\\
= (x+3)(2x+1)
$$
however, this doesn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What is the general solution of $xy' - 2y = -x$? I am trying to find the general solution of $xy' - 2y = -x$.
I normalize the equation to get $y' - \frac{2}{x}y = -1$.
I get the integrating factor as $x^{-2}$, and so $y = x^{2}\int x^{-2}*(-1)\,dx $.
Solving for this integral, I get $ y = x + \frac{C}{x^2}$, but this ... | After dividing both sides by $x$ you get
$$y'(x) - \frac{2}{x}y(x) = -1$$
The integrating factor is
$$\mu(x) = e^{\int -\frac{2}{x}\ dx} = \frac{1}{x^2}$$
Then by following the procedure, you multiply both sides by $\mu(x)$:
$$\frac{1}{x^2}y'(x) - \frac{2}{x^3}y(x) = -\frac{1}{x^2}$$
Use the tricky substitution:
$$-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Secant and Tangent identity i've been stuck on this question too long
$x = \sec A + \tan A$
show $x + \frac{1}{x} = 2\cdot \sec A$
I've been using $\tan^2 \theta + 1 = \sec^2 \theta$
and $\tan\theta = \frac{\sin \theta}{\cos\theta}$
help would be much appreciated
$x=\sec A +\tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos... | Because $$x+\frac{1}{x}=\frac{1+\sin{A}}{\cos{A}}+\frac{\cos{A}}{1+\sin{A}}=$$
$$=\frac{(1+\sin{A})^2}{\cos{A}(1+\sin{A})}+\frac{\cos^2{A}}{\cos{A}(1+\sin{A})}=\frac{(1+\sin{A})^2+\cos^2A}{\cos{A}(1+\sin{A})}=$$
$$=\frac{1+2\sin{A}+\sin^2A+\cos^2A}{\cos{A}(1+\sin{A})}=\frac{1+2\sin{A}+1}{\cos{A}(1+\sin{A})}=$$
$$=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$. Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.
Attempt at a solution:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \co... | It is a typo. It must be $$\tan y=\dfrac{3\sin x\cos x}{1+3\sin^2 x}=\dfrac{3\sin 2x}{5-3\cos 2x}$$for minimizing it we should have the 1st-order derivation of $\tan y$ equal to $0$ or $$\dfrac{d\tan y}{dx}=0$$which yields to $$6\cos 2x(5-3\cos 2x)=6\sin 2x\cdot 3\sin 2x$$which yields to $\cos 2x=\dfrac{3}{5}$ and $\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $3\sin^2 x - \cos^2 x - 2 =0$
Find all the angles between $0$ and $360^\circ$ that satisfy $$3\sin^2 x - \cos^2 x - 2 =0$$
My attempt -
$3\sin^2 x - (1-\sin^2x) - 2 =0$
$ 3 \sin^2 x + \sin^2 x = 3 $
$4\sin^2 x = 3 $
$ \sin x= \frac{\sqrt{3}}{2} $
I found that $x= 60,120 $
Why is the answer for this $60,12... | $$\sin^2x=\sin^2A\iff\cos^2x=\cos^2A\iff\tan^2x=\tan^2A$$
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$\sin^2x=\sin^2A\iff\sin(x+A)\sin(x-A)=0$$
Now $\sin(x\pm A)=0\implies x=n180^\circ\mp A$ where $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $ Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $
Find the minimum value of $h(\theta)$
$h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} = 5 \sin (\theta + 53.13) + \sqrt{2} $
Minimum value -
$5\sin (\theta + 53.13) + \sq... | Hint
You must note that the range of $a\sin\theta \pm b\cos\theta$ is $\left[ -\sqrt {a^2+b^2}, \sqrt {a^2+b^2}\right]$
Hence in your case range of given function is $[-5+\sqrt 2, 5+\sqrt 2]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the determinant of the following $5\times 5$ real matrix:
Let $A\in\mathbb{R^{5\times5}}$ be the matrix: $\left(\begin{array}{l}a&a&a&a&b\\a&a&a&b&a\\a&a&b&a&a\\a&b&a&a&a\\b&a&a&a&a\end{array}\right)$
Find the determinant of $A$.
Hey everyone. What I've done so far: $det\left(\begin{array}{l}a&a&a&a&b\\a&a&a&b&a... | The eigenvalues of
$$ \left(\begin{array}{l}a&a&a&a&a\\a&a&a&a&a\\a&a&a&a&a\\a&a&a&a&a\\a&a&a&a&a\end{array}\right) $$
are $$ 5a,0,0,0,0 $$
with eigenvectors as the columns (pairwise perpendicular) of
$$
\left( \begin{array}{rrrrr}
1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 \\
1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int x \frac{\sqrt {a^2 - x^2}}{\sqrt{a^2+x^2}} dx$ How to solve $\displaystyle\int x\, \frac{\sqrt {a^2 - x^2}}{\sqrt{a^2+x^2}}\, dx$?
I have tried substituting $x = a \sin(\theta)$ and got
$$a^2\int\frac{\sin(\theta) \cos^2(\theta)}{\sqrt{1+\sin^2(\theta)}}\, d \theta.$$
I'm not sure how to proceed from her... | Note
$$\int\frac{\sin \theta \cos^2 \theta}{\sqrt{1+\sin^2 \theta}} d \theta=-\int\frac{\cos^2 \theta}{\sqrt{2-\cos^2 \theta}} d \cos\theta.$$
Let $u=\cos\theta$ and then
$$\int\frac{\cos^2 \theta}{\sqrt{2-\cos^2 \theta}} d \cos\theta=\int\frac{u^2}{\sqrt{2-u^2}}du.$$
Let $u=\sqrt{2}\sin t$ and then
$$\int\frac{u^2}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Show $\sqrt[\sqrt{i}]{i}\approx 23$. (What are the other values?) Let $\arg(z)=\theta$ and $|z|=r$. So,
\begin{equation*}
\begin{aligned}
\sqrt[\alpha]{z}
& = r^{1/\alpha}\cdot \left(\sin\left(\dfrac{\theta+2\pi k}{\alpha}\right)+i\cos\left(\dfrac{\theta+2\pi k}{\alpha}\right)\right)
\end{aligned}
\end{equation*}
where... | Using the multivalued complex logarithm we have that
$$\frac1{i^{1/2}}=\frac1{\exp(\frac12\ln i)}=\frac1{\exp(\frac12i(\pi/2+2\pi\Bbb Z))}=\frac1{\exp(i(\pi/4+\pi\Bbb Z))}\\=e^{-i\pi/4}e^{i\pi\Bbb Z}=\pm e^{-i\pi/4}=\pm\frac{1-i}{\sqrt 2}$$
Thus
$$i^{1/i^{1/2}}=\exp\left(\pm\frac{1-i}{\sqrt 2}\ln i\right)=\exp\left(\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove the identity $ \sum\limits_{i=0}^{n} (-1)^{i} (a+2i) \binom{n}{i} \binom{a+n+i}{n+1}^{-1}=(n+1) \delta_{a,0}$ Playing with hypergeometric series I come across the identity
$$
\sum_{i=0}^{n} (-1)^{i} (a+2i) \binom{n}{i} \binom{a+n+i}{n+1}^{-1}=\begin{cases} n+1, a=0, \\
0, a\neq 0 \end{cases}
$$
and $a \notin ... | Without use of integrals one can take the following path. Let
$$
S_{n} = \sum_{i=0}^{n} (-1)^{i} (a+2i) \binom{n}{i} \binom{a+n+i}{n+1}^{-1}
$$
and
$$
S_{n}^{(1)} = 2 \, \sum_{i=0}^{n} (-1)^{i} (i) \binom{n}{i} \binom{n+i}{n+1}^{-1}.
$$
Now,
\begin{align}
S_{n}^{(1)} &= 2 \, \sum_{k=0}^{n} (-1)^{k} (k) \binom{n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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A rough idea to prove $\int_{0}^{\infty}\frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}dx = \frac{\pi}{2} ?$
I have a rough idea of approach to prove the Borwein integral in $(1)$ via Complex-Analytic techniques, is it valid ?
$(1)$
$$ \int_{0}^{\infty}\frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}dx = \frac{\pi}{2}.$$
From glancing ... | Perhaps you might be interested in seeing a real method used to evaluate this integral.
We begin by enforcing a substitution of $x \mapsto 3x$. This gives
\begin{align*}
\int_0^\infty \frac{\sin x}{x} \frac{\sin (x/3)}{x/3} \, dx &= \int_0^\infty \frac{\sin (3x) \sin x}{x^2} \, dx\\
&= 3 \int_0^\infty \frac{\sin^2 x}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ How do I solve this equation I found in my textbook:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
This is what I tried:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
$\mapsto \sqrt{(x+4)(x-4)} - (x-4) = \sqrt{(x-1)(x-4)}$
Dividing both sides by $\sqrt{x-4}$
$\mapsto \sqrt{x+4} - ... | The domain gives $x\geq4$ or $x\leq-4$.
*
*$x\leq-4$.
We need to solve $$\sqrt{(4-x)(-4-x)}+\left(\sqrt{4-x}\right)^2=\sqrt{(4-x)(1-x)}$$ or
$$\sqrt{-4-x}+\sqrt{4-x}=\sqrt{1-x}$$ or
$$-4-x+4-x+2\sqrt{x^2-16}=1-x$$ or
$$2\sqrt{x^2-16}=x+1,$$ which has no real roots.
*$x\geq4.$
We have $$\sqrt{(x-4)(x+4)}=\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Problem with generalized eigenvectors in a 3x3 matrix. I have this matrix:
$$
A= \begin{pmatrix}
0 & 1 & 1 \\
0 & 1 & 0 \\
-1 & 1 & 2 \\
\end{pmatrix}
$$
I have founded the eigenvalues:
$$\lambda_{1,2,3}=1$$
So
$$\lambda=1$$$$\mu=3$$
I'm expecting to have one eigenvector plus two generalized eigenvec... | Since $A-I=\begin{pmatrix}-1&1&1 \\0&0&0\\ -1&1&1\end{pmatrix}$, the eigenspace is defined by the single equation $x=y+z$, and it has dimension $2$, being isomorphic to $\mathbf R^2$ by the isomorphism:
\begin{align}\mathbf R^2&\longrightarrow \ker(A-I)\\\begin{pmatrix} y\\z\end{pmatrix}&\longmapsto \begin{pmatrix}y+z\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2667609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to prove $e^x\ge \left(1+\frac xn\right)^n$ for any real numbers $x, n > 0$ Can someone provide a detailed proof? I saw a proof here
$$\begin{align}
\frac{e_{n+1}(x)}{e_n(x)}&=\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}\\\\
&=\left(1+\frac{-x}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right... | By the binimial theorem $$\left(1+\frac{x}{n}\right)^n=1+n\cdot\frac{x}{n}+\frac{n(n-1)}{2!}\cdot\frac{x^2}{n^2}+...+\frac{n(n-1)...(n-n+1)}{n!}\cdot\frac{x^n}{n^n}=$$
$$=1+x+\left(1-\frac{1}{n}\right)\frac{x^2}{2!}+...+\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{n-1}{n}\right)\frac{x^n}{n!}<$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Finding $\frac{3x+y}{6x-1} = ?$ $$2^{3x} = 18$$
$$2^y = 9$$
$$\frac{3x+y}{6x-1} = ?$$
Let me show my attempt:
$$2^y = 9, 2^y = 3^2, y = 1$$
$$2^{3x} = 18, x = 1$$
I think I've gone too wrong
| You can get the value of $x$ by doing :
$2^{3x} = 18 \implies \ln(2^{3x})=\ln(18) \implies 3x \cdot \ln(2)= \ln(18) \implies 3x = \frac{\ln(18)}{\ln(2)} \implies x=\frac{\ln(18)}{3 \cdot \ln(2)}$
And similarly, the value of $y$ by doing:
$2^y = 9 \implies \ln(2^y) = \ln(9) \implies y \cdot \ln(2) = \ln(9) \implies y= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Solve $x^4 - 8x^3 + 21x^2 - 20x + 5 = 0$ given that the sum of two of its roots is $4$ Here's what I tried:
Let the roots be $a$, $b$, $c$ and $d$, $a+b=4$. Then,
$$a + b + c + d = 8 \Longrightarrow 4 + c+ d = 8 \Longrightarrow a+b = c+d = 4$$
$$(a + b)(c + d) + ab + cd = 21$$
$$ab (c + d) + cd (a + b) = 20 \Longrighta... | $x^4-8x^3+21x^2-20x+5=(x^2-4x+a)(x^2-4x+b)$
$\begin{cases}a+b+16=21 \\ -4a-4b=-20\\
ab=5 \end{cases}$
So $a+b=5$ and $ab=5$.
$a$, $b$ are the roots of $t^2-5t+5=0$.
| {
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"url": "https://math.stackexchange.com/questions/2670645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Prove $\sum\limits_{n=0}^\infty\frac{5^n(3^{5^{n+1}}-5\cdot3^{5^n}+4)}{(729)^{5^n}-(243)^{5^n}-5\cdot3^{5^n}+1}=\frac12$ This problem is taken from Algerian Olympiad and asks to prove that
$$\sum_{n=0}^{\infty} \dfrac{5^n(3^{5^{n+1}} -5\cdot3^{5^n} + 4)}{(729)^{5^n} - (243)^{5^n}-5\cdot3^{5^n}+1} = \frac 12.$$
Noticing... | The series does indeed converge (by the ratio test, for example), but its sum is not $\,\dfrac{1}{2}\,$, it is rather strictly greater than $\,\dfrac{1}{2}\,$. The first two terms add up to $\dfrac{29}{59}$ $+$ $\dfrac{529555380145}{25630480435499}$ $\simeq 0.512$ $\gt \dfrac{1}{2}\,$ already, and adding more positive... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Can we calculate $ i\sqrt { i\sqrt { i\sqrt { \cdots } } }$? It might be obvious that $2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } $ equals $4.$ So what about $i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } \text{ ?} $ The answer might be $-1$, but I'm not sure as ... | Let $$x=i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } }$$ $$\implies x^2=-1i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } }$$ $\implies x^2=-x$ $\implies x^2+x=0$ $\implies x(x+1)=0\implies x=0\; \text{or} -1$ since $x$ cannot be $0$, hence $x=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 7,
"answer_id": 1
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Computing $\sum_{n=1}^ \infty n^2 \cdot \left(\frac{2}{3}\right)^n$ I've been dealing with the following series for a while now, without real progress.
$$\sum_{n=1}^ \infty n^2 \cdot \left(\frac{2}{3}\right)^n$$
After using WolframAlpha, I know it converges to $30$, but I can't see how to calculate it by myself.
Any le... | For $-1<x<1$, $\displaystyle \sum_{n=1}^{k}x^n=\frac{x(1-x^{k})}{1-x}$.
Differentiating,
\begin{align*}
\sum_{n=1}^knx^{n-1}&=\frac{(1-x)[1-(k+1)x^k]-(x-x^{k+1})(-1)}{(1-x)^2}\\
&=\frac{1-(k+1)x^k+kx^{k+1}}{(1-x)^2}\\
\end{align*}
So.
\begin{align*}
\sum_{n=1}^knx^{n}&=\frac{(1-x)[1-(k+1)x^k]-(x-x^{k+1})(-1)}{(1-x)^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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To prove an inequality for $n\ge 11$ Let
$$f(n)=3(n-3)(n-5)(n-1+\sqrt{n^2-14n+61})$$
$$g(n)=(n-4)(3n^2-19n+34+\sqrt{(3n^2-19n+34)^2-48(n-3)(n-4)^2})$$ For $n\ge 11$, I have to prove that $f(n)<g(n).$
My Try:
I tried breaking into two parts,namely to prove that $$3(n-3)(n-5)(n-1)<(n-4)(3n^2-19n+34)$$ and $$3(n-3)(n-5)\s... | For $n = 11, \cdots, 15$, the values of $f(n)$ and $g(n)$ can be computed directly to compare them.
Now suppose $n \geqslant 16$. Because$$
n^2 -14n + 61 < \left(n - \frac{19}{3}\right)^2 \Longleftrightarrow \frac{4}{3}n > \frac{188}{9} \Longleftrightarrow n > \frac{47}{3},
$$
then$$
f(n) < 3(n - 3)(n - 5)\left(n - 1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $A+B+C=180^{\circ}$, then show that: $\cos B=\sin A\sin C-\cos A\cos C$ Here is the question :
If $A+B+C = 180^{\circ}$, then show that: $\cos B = \sin A \sin C - \cos A \cos C$.
EDIT : Here is my reviewed working :
$$
\cos B=-\cos (A+C)
$$
Since $$\space A+B+C = 180^\circ, \space B =180^\circ-(A+C)$$
And
$$\begi... | Your first line should be deleted, ok. But the main mistake is at the last line. You write
$$
-\cos(B)=\cos(A+C),
$$
which is fine. But what is $\cos(A+C)$?
Also, more directly: $\cos(x)=-\cos(180-x)$ for all $x$. So $\cos(B)=-\cos(A+C)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing that $\frac 1 {(x+1)^2}=\frac 1 {(x+1)(x+2)}+\frac {1!} {(x+1)(x+2)(x+3)} + \frac {2!} {(x+1)(x+2)(x+3)(x+4)} + \cdots$
Prove that if $x>-1$ then
$$\frac 1 {(x+1)^2}=\frac 1 {(x+1)(x+2)}+\frac {1!} {(x+1)(x+2)(x+3)} + \frac {2!} {(x+1)(x+2)(x+3)(x+4)} + \cdots$$
Could I have a hint for this? I tried writing... | Another way. from the properties of the Pochhammer symbol and Beta function we can note that $$\sum_{n\geq0}\frac{n!}{\left(x+1\right)\left(x+2\right)\cdots\left(x+n+1\right)}=\sum_{n\geq0}\frac{n!}{\left(x+1\right)_{n+1}}=\sum_{n\geq0}\frac{\Gamma\left(n+1\right)\Gamma\left(x+1\right)}{\Gamma\left(x+n+2\right)}$$ $$=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2678866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Incorrect in solving $\frac{a}{b} - \frac{a}{c} = 1$ for $c$ I have this:
$$\frac{a}{b} - \frac{a}{c} = 1$$ Solve for $c$. Then,
$$\frac{a}{b} - \frac{a}{c} = 1 \cdot bc$$
$$ = ac - ab = bc$$
$$ = a(c - b) = bc$$
$$ c = \frac{bc}{a} + b$$ This is my final result.
But the correct result is:
$$c=\frac{ab}{a-b}$$
What... | To isolate $c$ we can proceed as follow
$$\frac{a}{b} - \frac{a}{c} = 1\iff \frac{a}{c}=\frac{a}{b}-1=\frac{a-b}{b}\iff\frac{c}{a}=\frac{b}{a-b}\iff c=\frac{ab}{a-b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Binomial Expansion of (1+4x) I have been asked to prove the following:
$a(n) = \frac{1}{n+1}\ \binom{2n}{n}$, given that,
$f(x) = \sum_{n = 0}^{\infty} a(n) x^n$ and $x\{f(x)\}^2$ = $f(x) -1.$
So far I have divided $x\{f(x)\}^2 = f(x) -1$ through by $x$ so I have a quadratic to work with, $\{f(x)\}^2 - \frac{f(x)}{x} ... | Notice that $f(x) = 1 + \sum_{n=1}^{\infty} a_nx^n$, so that
$$\frac{f(x)-1}{x} = \sum_{n=1}^{\infty} a_nx^{n-1} = \sum_{n=0}^{\infty} a_{n+1}x^{n}$$
With this in mind, it suffices to show that writing $f(x)^2 = \sum_{n=0}^{\infty}b_nx^n$ we have $b_n = a_{n+1}$.
Do you think you can take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$1^2-2^2+3^2-4^2+…-2016^2+2017^2=2017k$ (Solve for $k$) Question:
$1^2-2^2+3^2-4^2+…-2016^2+2017^2=2017k$
Solve for $k$
My attempt:
$$1^2-2^2+3^2-4^2+…-2016^2+2017^2\\
\begin{align}= (1-2)(1+2)+(3-4)(3+4)+…+(2015-2016)(2015+2016)+2017^2 \end{align}
$$
What should I do next?
| As mentioned by sharding4, simplify each parenthesized difference into $-1$ to achieve:
$$-(1+2+3+4+\dots+2016)+2017^2$$
Then recall the partial sum formula:
$$\sum_{k=1}^n k={n(n+1)\over 2}$$
Then apply:
$$-\left({2016\cdot2017\over 2}\right)+2017^2 = 2017k$$
Divide by $2017$:
$$\require{cancel}{-\left({2016\,\cdot\,2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to finds the smallest $n$ such that $I=A^0, A^1, A^2, \dots, A^n$ are linearly dependent?
Let $A$ be the following $2\times 2$ matrix:
$$A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$$
Find the smallest value of $n$ such that matrices $I=A^0, A^1, A^2, \dots, A^n$ are linearly dependent.
I don't quite know how to begin ... | Calculate the powers of $A$ upto $A^2$ and write them in a row:
$$\underbrace{\begin{pmatrix} \fbox{1} & 0 \\ 0 & 1 \end{pmatrix}}_{I}, \underbrace{\begin{pmatrix} \fbox{1} & 2 \\ 3 & 4 \end{pmatrix}}_{A},\underbrace{\begin{pmatrix} \fbox{7} & 10 \\ 15 & 22 \end{pmatrix}}_{A^2}$$
Use the coefficient at the position $(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solve the equation $\cos^2x+\cos^22x+\cos^23x=1$
Solve the equation: $$\cos^2x+\cos^22x+\cos^23x=1$$
IMO 1962/4
My first attempt in solving the problem is to simplify the equation and express all terms in terms of $\cos x$. Even without an extensive knowledge about trigonometric identities, the problem is solvable.... | This is a summary of the solution found in this hyperlink.
We can write the LHS as a cubic function of $\cos^2 x$. This means that there are at most three values of $x$ that satisfy the equation.
Hence, we look for three values of $x$ that satisfy the equation and produce three distinct $\cos^2 x$. Indeed, we find th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Mathematical Induction prove that $n^3+5n$ is divisible by $6$ Sorry, I know this will be a duplicate on the site but the other solution I found confusing and the method look completely different to what I was taught.
Prove that $n^3 + 5n$ is divisible by $6$ by using induction
The question is
Prove by mathematical Ind... | $$n^3+5n = n(n^2+5)$$
On the RHS, factor occur in odd, even pairs, so product is even and divisible by $2$.
$$n(n^2+5) \equiv n(n^2+2) \qquad \mod 3$$
Hence if $n \equiv 0$ (mod $3)$ then $3$ divides into the result and if $n \equiv \pm 1$ then $n^2 \equiv +1$ thus $3 | (n^2+2)$ and thus if both $2$ and $3$ are factor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Find the Fourier series of the function $f$
Define $f (\theta) = (\pi - \theta)^2/4$ for $0\leq \theta \leq 2\pi$
If $n\neq 0,$
Fourier coefficient of $f$ is:
$$\hat f(n)=\frac 1 {2\pi}\int_0^{2\pi}\frac {(\pi-\theta)^2} 4 e^{-in\theta}d\theta$$
which I have found to be equal to: $$=\frac 1 {8\pi}[\pi^2\frac i n e^... | Expand the answer in terms of exponentials:
$$
f(\theta) = \frac{\pi^2}{12} + \sum_{n=1}^\infty \frac{\cos(\theta)}{n^2} =
\frac{\pi^2}{12} + \sum_{n=1}^\infty \frac{e^{in\theta}}{2n^2} + \frac{e^{-in\theta}}{2n^2} = \frac{\pi^2}{12} + \sum_{n=-\infty}^\infty \frac{e^{in\theta}}{2n^2}
$$
which is precisely your solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why did this extraneous root creep into the solution? I was solving this equation and proceeded as follows:
$$\arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$$
$$\implies \arcsin(1-x) = \frac{π}{2} + 2\arcsin (x)$$
$$\implies \sin (\arcsin (1-x)) = \sin \left( \frac {π}{2} + 2\arcsin (x)\right)$$
$$\implies (1-x) = \cos \l... | When $x=\frac12$, you have $\arcsin(x)=\frac{\pi}{6}$ and $\arcsin(1-x)=\frac{\pi}{6}$
so $\sin (\arcsin (1-x))=\sin \left( \frac {π}{6} \right)$ and $\sin \left( \frac {π}{2} + 2\arcsin (x)\right) =\sin \left( \frac {5π}{6} \right) $
showing your third line would be a correct equality when $x=\frac12$ since $\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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If the null space of $T_2$ is a subspace of the range of $T_1$, deduce three linearly independent vectors in the null space of $T_2 \circ T_1$ Below is a question from an old A Level Further Mathematics paper. Parts 1 through 3 are straightforward, but part 4 has stumped me. It is easy to go the long way and just find ... | First note that $\ker T_1 \subseteq \ker T_2T_1$ so you already have one vector, the one in $\ker T_1$.
$$\ker T_1 = \operatorname{span}\left\{\begin{pmatrix}1\\-1\\1\\1\end{pmatrix}\right\}, \quad\ker T_2 = \operatorname{span}\left\{\begin{pmatrix}-5\\1\\0\\1\end{pmatrix}, \begin{pmatrix}8\\-5\\-2\\0\end{pmatrix}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find three rational numbers $a,b,c$ s.t. $b^2-a^2=c^2-b^2=5$. A rational numbers cannot have irrational as$\exists n \in \mathbb{Z}, \, \sqrt[n]{\frac xy}$, but the two equalities give: $b^2=\frac{(a^2+c^2)}{2} \implies b = \sqrt[2]{\frac {(a^2+c^2)}{2}}$.
To avoid this, need $4\mid a$, & $a=c$; so that if $\exists t=a... | The general solution of $b^2-a^2=5$ has
$$b=\frac12\left(t+\frac 5t\right)$$
for $t\in\Bbb Q^*$. Then
$$c^2=b^2+5=\frac{t^4+10t^2+25}{4t^2}+5=\frac{t^4+30t^2+25}{4t^2}.$$
The problem boils down to whether the genus-one curve
$$y^2=x^4+30x^2+25$$
has rational points.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why do I get that $0 = 1$ when trying to prove that $\forall n\in\mathbb{N}, \ 3 + 2\left(n + \sum_{i=1}^{2n-1}(i+1)\right) = (2n+1)^2\,?$ I was playing around with numbers and by looking at some similar observations, I found an interesting pattern. For a natural number $n$, it seemed like the following was always true... | Hint
Use that $$\sum_{i=1}^k i= \frac {k(k+1)}{2}$$
Hence $$\sum_{i=1}^{2n-1} i= \frac {2n(2n-1)}{2}$$. Hence your expression turns to
$$3+2\left(\frac {2n(2n-1)}{2} +(2n-1)+n\right)$$ $$=3+2n(2n-1)+6n-2$$ $$=4n^2+4n+1$$ $$=(2n+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2696865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the general term of the sequence $a_n = (0, 3, 1/2 , 5/3 , 2/3 , 7/5 , 3/4 , ...)$. Then find the limit of this sequence. I have an exercise in which I am stuck. The exercise is
Find the general term of the sequence
$$a_n = (0, 3, \dfrac{1}{2}, \dfrac{5}{3}, \dfrac{2}{3}, \dfrac{7}{5}, \dfrac{3}{4}, ...)$$.
The... | $$a_{2n+1}=\frac {2n+3}{2n+1} $$
$$a_{2n}=\frac {n}{n+1} $$
$$\lim a_{2n}=1=\lim a_{2n+1} \implies $$
$$\lim a_n=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2697815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to use integration by parts to solve an integral? I'm having trouble with solving this integral by parts. Here is what I have so far. Can anyone please help me out?
Solve the integral $\int x^3\sqrt{x^2-1}dx$ by parts, choosing u = $x^2$ and
$dv = x\sqrt{x^2-1} dx$
$du_1 = 2xdx$
$v_1$ = $\int x\sqrt{x^2-1}dx$ $ t =... | I would make the substitution $u=x^2-1$ first. This makes the integral
$$\frac{1}{2}\int (u-1)\sqrt{u} \; du.$$
Parts will work on this integral, but it's not the easiest way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int \cos^5 x \sin^{1/2} x \ dx $ $$I=\int \cos^5 x \sin^{1/2} x \ dx $$
My attempt:
$$\cos^5 x = \cos^4 x \cos x =(1-\sin^2x)^2 \cos x$$
\begin{align}
I & =\int (1-\sin^2x)^2 \sin^{1/2} x \cos x \ dx \\[10pt]
& =\int \sin^{1/2} x \cos x \ dx+\int \sin^2 x \cos x \, dx + \int -2\sin x \cos x\, dx \\[10pt]
& ={2\over 3... | You made a tiny mistake with the powers of $\,sin(x)$
$I=\int \cos^5 x \sin^{1/2} x \ dx$
$I =\int (1-\sin^2x)^2 \sin^{1/2} x \cos x \ dx $
let
$\,u= sinx \implies du = cos(x)dx$
$I= \int (1-u^2)^2.u^\frac12\,du \\I = \int u^\frac12 + u^{4+\frac12} - 2u^{2+\frac12}\,du\\I = \int u^\frac12+ u^\frac92 - 2u^\frac52\,du... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\left ( 1- \sqrt{2}\sin x \right )\left ( \cos 2x+ \sin 2x \right )= \frac{1}{2}$ Solve
$$\left ( 1- \sqrt{2}\sin x \right )\left ( \cos 2x+ \sin 2x \right )= \frac{1}{2}$$
Now I did not understand how can i solve that.
I have tried substituting $\cos(2x)=\cos^2(x)−\sin^2(x)$ and$\,$ $\sin(2x)=2\sin(x)\cos(x)... | You can substitute:
$$y=2x\Rightarrow x=\frac y2$$
$$\left ( 1- \sqrt{2}\sin \frac y2 \right )\left ( \cos y+ \sin y \right )= \frac{1}{2}$$
Remember that:
$$\sin\frac\theta2=\pm\sqrt{\frac{1-\cos\theta}2}$$
So you get: $\left ( 1- \sqrt{2} (\pm\sqrt{\frac{1-\cos\theta}2})\right )\left ( \cos y+ \sin y \right )= \fr... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the rational number of a, b, c, solving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+ \sqrt[3]{b}+\sqrt[3]{c}$ I try as following
let
\begin{eqnarray}
x= \sqrt[3]{a} \\
y= \sqrt[3]{b} \\
z= \sqrt[3]{c} \\
x+y+z = \sqrt[3]{\sqrt[3]{2}-1 }\\
\end{eqnarray}
We know that
\begin{equation}
x^3+y^3+z^3 = (x+y+z)^3... | A systematic approach is to apply the denesting formula
$$\sqrt[3]{\sqrt[3]{A}-B} = \sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 } $$
where $x_1$, $x_2$ and $x_3$ are the roots of the cubic equation
$$x^3 + \frac{B+2C}3x^2 - \frac{(B-C)(2B+C)}{27}x+ \frac{(B-C)^3}{729}=0$$
with $C =\sqrt[3]{B^3-A}$. Thus, to denest $\s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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4 digit numbers divisible by 11 Four digit numbers are formed using the
digits 1,2,3,4 (repetition is allowed). The
number of such four digit numbers divisible
by 11 is-
(1) 22 (2) 36 (3) 44 (4) 52
I know for a number to be divisible by 11 the sum of digits at even places must be equal to the sum of those at odd places... | Let $abcd$ be the number.
If $a=b$ then $c=d$. There are $4*4=16$ ways that can occur. (Four options for $a$ and four options for $b$).
If $a=b\pm 1$ then $c=d \mp1$. And there are $2*3*3=18$ ways this can occur. (Two choices whether $a > b$ or $b > a$ and three choices from $1,2,3,4$ that are one apart.
If $a = b\pm... | {
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$A289B$ is divisible by $90$, then what is $A+B$? It is known that $A289B$ is divisible by $90$, are with $A$ and $B$ as digits. What is $A+B$?
My approach :
If $A289B$ is divisible by $90$, then the unit-value must be a zero, $B=0$.
So now we have $A2890$, which can be written as
$$ A2890 = A\times10000 + 2890 $$
sin... | You are going to a lot of effort to avoid noting that if $A2890=A289*10$ is divisible by $90$ and $\gcd(9,10) = 1$ then $A289$ is divisible by $9$.
And $A289 = A28*10 + 9$ being divisible by $9$ means $A28*10 + 9 \equiv A28 \equiv A27 + 1 \equiv 100A + 27 + 1 \equiv A + 1 \mod 9$. And $A289\equiv 0 \mod 9$ so $A+1 \eq... | {
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"answer_id": 2
} |
Partial Derivative of arctan Given that $$f(x,y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$
Find $f_x(x,y)$
My attempt,
$$
\begin{aligned}
f_x(x,y)&=\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2}\cdot\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\\
&=\frac{1+y^2}{(1-xy)^2+(x+y)^2}\\
&=\frac{1+y^2}{1+y^2+x^2+x^2y^2}
\end{aligned}
$$
B... | Because
$$\frac{1+y^2}{1+y^2+x^2+x^2y^2}$$
and
$$\frac{1}{x^2+1}$$
are the same thing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Sum of the first integer powers of $n$ up to k Pascal's triangle has a lot of interesting patterns in it; one of which is the triangular numbers and their extensions. Mathematically:
$$\sum_{n=1}^k1=\frac{k}{1}$$
$$\sum_{n=1}^kn=\frac{k}{1}\cdot\frac{k+1}{2}$$
$$\sum_{n=1}^kn^2=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2... | We can use telescoping sums to find formula for $$\sum_1^n k^p $$ for $p\ge 1$
Note that $$ (k+1)^2 - k^2 =2k+1 \implies$$
$$ \sum_1^n \big[(k+1)^2 - k^2\big]=\sum_1^n (2k+1)\implies $$
$$ (n+1)^2 -1=2 \sum_1^n k + n\implies $$
$$\sum_1^n k = \frac {n(n+1)}{2}$$
Similarly we can find formula for $$\sum_1^n k^2 $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
System and determinant We assume that the following determinant is equal to zero:
$$\det A=\begin{vmatrix}
-x&1&1&1\\
1&-y&1&1\\
1&1&-z&1\\
1&1&1&-t\\
\end{vmatrix}=0$$
with $x$, $y$, $z$, $t$ positive integers. Then,
$$\det A=(xt-1)(yz-1)-(x+t+2)(y+z+2)$$
I have to determine positive integers $a$, $b$, $c$, $d$, of th... | Consider the system of linear equations determining $(a,b,c,d)$:
$$
\begin{cases}
\begin{align}
-xa+b+c+d&=0\\
a-yb+c+d&=0\\
a+b-zc+d&=0\\
a+b+c-td&=0
\end{align}
\end{cases}\tag{1}
$$
Subtracting pairwise the equations one obtains:
$$
a:b:c:d=\frac{1}{1+x}:\frac{1}{1+y}:\frac{1}{1+z}:\frac{1}{1+t}.\tag{2}
$$
Multiplyi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^... | Another contour integration approach:
Let's integrate the function $$f(z) = \frac{z^{2}}{\sin z}$$ around a tall rectangular contour with vertices at $z=0$, $z=\frac{\pi}{2}$, $z= \frac{\pi}{2}+ i R$, and $z= i R$.
There are no singularities inside contour, and the singularity at $z=0$ is removable.
Also, since the mag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 4
} |
Prove by induction that $4n^2 + 1 < 3\cdot 2^n$ for every $n \ge 6$ My question is about solving for $k+1$
I did the base case and tried to solve the induction step.
this is what I tried
my hypothesis is $4k^2 + 1 < 3\cdot 2^k$ is true then I need to show that it is true for $k+1$
I did right hand side by doing this
$3... | $3\cdot2\cdot2^k = 3\cdot 2^{k+1}$
Okay, we have to bring $3*2^k > 4k^2 + 1$ into this.
$3*2*2^k > 2(4k^2 + 1)$
And we want to somehow relate this to $4(k+1)^2 + 1 = 4k^2 + 8k + 5$
So $3\cdot2^{k+1} = 2*3*2^k > 2(4k^2 + 1) = 8k^2 +2 = 4k^2 + 4k^2 + 2$
So we have to show that $4k^2 + 2 \ge 8k + 5$. Which.... seems reas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
GCD of $X^3+1$ and $X^2+1$ in a field K let K be a field of characteristic $p$.
I try to find $\gcd(X^3+1,X^2+1)$
We have, $\gcd(X^3+1,X^2+1)=\gcd(X^3+1,X-1)=\gcd(X+1,X-1)=\gcd(X-1,2)$
I didn't know how to go from there, but the solution in my book says:
if $p \neq 2$, then $ \gcd(X^3+1,X^2+1)=1$ else $\gcd(X^3+1,X^2+... | Since $X^3+1=X(X^2+1)-(X-1)$, we have
$$
\gcd(X^3+1,X^2+1)=\gcd(X^2+1,X-1)
$$
Since $X^2+1=X(X-1)+(X+1)$, we have
$$
\gcd(X^2+1,X-1)=\gcd(X-1,X+1)
$$
Since $X-1=(X+1)-2$, we have
$$
\gcd(X-1,X+1)=\gcd(X+1,2)
$$
If $p=2$, then $2=0$, so the greatest common divisor is $X+1$ (the last nonzero remainder). If $p\ne2$, the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$ Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$
i am trying to solving integrating combination
since given that
$\cot (x^2+y^2)dy+xdx+ydy=0$
then $\cot (x^2+y^2)dy+d(xy)=0$ is it correct way ? and we can apply integration from here? can any one help me this problem
| $$2\cot(x^2+y^2)dy+2xdx+2ydy=0$$
$$2\cot(x^2+y^2)dy+d(x^2+y^2)=0$$
$$\dfrac{d(x^2+y^2)}{\cot(x^2+y^2)}=-2dy$$
$$-\ln\cos(x^2+y^2)=-2y+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
either inequality $\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$ is hard or I need to go back to study ASAP Prove that for every $x \in(0,\frac{\pi}{2})$, the following inequality:
$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$
holds
I don't see room to use derivatives, since it seems a little messy to calculate the ... | You need the following two inequalities:
$$\cos x = 1- \frac{x^2}{2}+ \frac{x^4}{24}-\frac{x^6}{6!} + \cdots < 1- \frac{x^2}{2}+ \frac{x^4}{24}$$
and
$$\ln (1+y) = y -\frac{y^2}{2}+ \cdots\le y$$
(you can formally prove them using derivatives, if you want).
Combining them substituting $y= - \frac{x^2}{2}+ \frac{x^4}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{... | What you have done is correct so far. Nice job! Next time, you could divide both sides of $2x^2-9x+9 = 0$ to get $x^2-\frac{9}{2}x + \frac{9}{2}=0$.
From your method, we have:
$$(x-\frac{9}{4})^2 - (\frac{9}{4})^2+\frac{9}{2} =0$$
$$(x-\frac{9}{4})^2 = \frac{81}{16} - \frac{9}{2}$$
$$(x - \frac{9}{4})^2 = \frac{9}{16}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
A Diophantine equation solved when N is not a square? In the following all variables are assumed to be integers.
It is easy to write a Diophantine equation which has solutions only when $N$ is a square. i.e.
$$N=A^2$$
It's trivial to write a Diophantine equation which has solutions if and only if $N$ is divisible by 4:... | Here's one approach, though possibly not the simplest.
A number $N$ is not a perfect square if $A^2+1 \le N \le (A+1)^2-1$ for some $A$. But how can we encode $X \le Y$? Over the real numbers, the standard trick would be to write $Y = X + Z^2$, because $Z^2$ is always nonnegative. Over the integers, that doesn't quite ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Proving expression$\geq 2$, where variables are $x,y,z$
If $x,y,z$ are distinct real number. Then prove that $$\bigg(\frac{x}{y-z}\bigg)^2+\bigg(\frac{y}{z-x}\bigg)^2+\bigg(\frac{z}{x-y}\bigg)^2\geq 2$$
Try : let $\displaystyle \frac{x}{y-z}=p\Rightarrow x=py-pz\Rightarrow x-py+pz=0$
And let $\displaystyle \frac{y}{... | Note
$$ \sum x(2x-y-z)=2(x^2+y^2+z^2-xy-yz-zx),$$
and
$$\sum (y-z)^2(2x-y-z)^2=2(x^2+y^2+z^2-xy-yz-zx)^2.$$
Therefore, according to the Cauchy-Schwarz inequality we have
$$\sum \frac{x^2}{(y-z)^2} \geqslant \frac{\left[x(2x-y-z)+y(2y-z-x)+z(2z-x-y)\right] ^2}{(y-z)^2(2x-y-z)^2+(z-x)^2(2y-z-x)^2+(x-y)^2(2z-x-y)^2}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating integral with residues I'm trying to evaluate the following integral: $$\int_0^{2\pi} \frac{d\theta}{8\cos^2 (\theta) + 1}$$ using residues. To begin, assume $z(t) = e^{i\theta}$ is a parametrization of the unit circle, for $0 \le \theta \le 2\pi$, so that $dz = ie^{i\theta} d\theta$. Making the necessary su... | First of all I suggest the use of the identity $2\cos^2(\theta)=\cos(2\theta)+1$ in order to have a second degree polynomial at the denominator (later you will have to compute just ONE residue).
$$I:=\int_0^{2\pi} \frac{d\theta}{8\cos^2 (\theta) + 1}=\int_0^{2\pi} \frac{d\theta}{4\cos(2\theta)+4 + 1}=\int_0^{2\pi} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of positive integral solutions in the given inequality
Find the number of positive integral solutions of the inequality $$3x+y+z \leq 30$$
My attempt:
Introducing a dummy variable '$a$' then the equation becomes $3x+y+z+a=30$, where $x,y,z \geq 1$ and $a\geq0$, then we have to find coefficient of $x^{30}$ in t... | The method you have used is correct (by correct, I mean that the coefficient of $t^{30}$ does indeed correspond to the number of positive integral solutions of the inequality), but you still need a way to compute the coefficient. Since what we want is the coefficient of a term in the expansion, we first impose the rest... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A series about $n!$ and Riemann zeta function Compute
$$
\sum_{n=1}^{\infty}{\left( \frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}} \right)}.
$$
By the software Mathematica, I find
$$
\sum_{n=1}^{\infty}{\left( \frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}} \right)}=-\frac{2}{3}-\frac{\zeta \left( 1/2 \right)}{\sqrt{2\pi}}.
$$
| Taking $$F\left(x\right)=\sum_{n\geq1}\frac{n^{n-1}}{n!e^{n}}x^{n}-\frac{1}{\sqrt{2\pi}}\sum_{n\geq1}\frac{x^{n}}{n^{3/2}}=-W\left(-\frac{x}{e}\right)-\frac{\mathrm{Li}_{3/2}\left(x\right)}{\sqrt{2\pi}},\,\left|x\right|<1$$ where $W\left(x\right)$ is the Lambert $W$ function and $\mathrm{Li}_{3/2}\left(x\right)$ is the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$? I'm trying to find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$ in $\mathbb{Z}[\sqrt[3]{2}]$. I know it is a unit, so there is an inverse, but I feel like I may be doing too much work in the wrong direction. Here's what I have so far:
Let $\alpha = 5+4\sqrt[3]{2}+3\sqr... | From your last line you get three simultaneous equations for $a,b,c$ which you solve
$$5a+6b+8c=1 \\4a+5b+6c=0 \\3a+4b+5c=0\\a+b+c=0\\b+3c=1\\b+2c=0\\c=1\\b=-2\\a=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Probability - Peculiar die
A peculiar die has the following properties: on any roll the probability of rolling either a $4$, a $6$, or a $1$ is $1/2$, just as it is with an ordinary die. Moreover, the probability of rolling either a $1$, a $3$, or a $2$ is again $1/2$. However, the probability of rolling a $1$ is $5/1... | Your calculations are correct but slightly more complicated than necessary.
In a way, the order in which parts (a) and (b) are presented is a hint at a simpler solution.
Letting $R_{i,j,\ldots,k}$ be the event that we roll one of the numbers in the set $\{i,j,\ldots,k\},$
then we are given that $R_{1,4,6} = R_{1,2,3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that $x^3 - 2 x^2 +\log(1+x)(x(3x+4) -2(1+x)^2 \log(1+x))$ is positive I want to show that (the following just gets rid off large brackets)
$$x^3 - 2 x^2 +x(3x+4)\log(1+x)-2(1+x)^2 \log^2(1+x)>0, \ \ \mbox{for}\ \ x\in(0,\infty).$$
My attempt: Transfer all negative terms to the other side of the inequality. Evalua... | Solve with respect to $\log(1+x)$ to obtain
\begin{equation}
\frac{3x^2 + 4x - x^2\sqrt{8x+9}}{4(1+2x+x^2)}<\log(1+x)<\frac{3x^2 + 4x + x^2\sqrt{8x+9}}{4(1+2x+x^2)}, \ \ \mbox{for}\ \ x>0. \tag{*}
\end{equation}
Start with the first inequality. Observe that both sides of the inequality go to zero when $x\to 0$. Also ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Removing squared values by dividing I am solving a simple trig equation:
$$
2\cos^2{\theta} + \cos{\theta} = 0
$$
My first approach was to move the $\cos\theta$ to the other side, and then divide by $\cos\theta$.
$$
2\cos^2{\theta} = -\cos{\theta}
$$
$$
2\cos{\theta} = -1
$$
$$
\cos{\theta} = -\frac{1}{2}
$$
$$
\theta ... | The point is only that if $\cos\theta=0$ then you cannot divide both sides by $\cos\theta;$ i.e. you cannot divide both sides by $0.$ Note, for example that it is true that $3\times0=5\times0,$ but dividing both sides by $0$ and getting $3=5$ is wrong since $3\ne5.$
So dividing both sides by $\cos\theta$ gives you all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Proof. So i'm going to prove this by induction. The first case when $n=1$ is trivial since:
$$9+16=25,$$
implying $ 5 \mid 25$.
Now we need to show is divisible when $n=k+... | HINT.-$9^n$ is congruent with $1$ or $9$ modulo $10$ and $4^m$ is congruent with $4$ or $6$ modulo $10$. The integer $9^n+4^{n+1}$always end with $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Simultaneous equations involving indices Please can someone help me solve this. I saw it in a text but I have tried to solve all to no avail.
$$3^x + 9^{2y} = 27\\2^x + 4^{-y} = \frac18 $$
Find 2$x$ + 3$y$
| \begin{align}
3^x + 9^{2y} &= 27\\
3^{4y} &= 3^3 - 3^x \\
3^{4y} &< 3^3 \\
y &< 0.75 \\
\hline
2^x + 4^{-y} &= \frac18 \\
2^{-2y} &= 2^{-3} - 2^x \\
2^{-2y} &< 2^{-3} \\
-2y &< -3 \\
y &> 1.5
\end{align}
Since we can't have both $y < 0.75$ and $y > 1.5$, there is no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Baffled with $\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}$
Calculate $$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}$$
Personal work:
$$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}=^{0 \over 0}\lim\limits_{x\to 0}{e^x-e^{\sin x}\cdot\cos x \over 1-\cos x}=^{0 \over 0}\lim\limits_{x\to 0}{{e^x-(e^... | You are exactly one application of L'Hospital away from getting the answer, because the next denominator will be (non-vanishing!) $\cos x$. Alternatively, we can see that three applications of L'Hospital's rule are necessary because $x - \sin x$ vanishes to third order at $0$.
Here's an approach using series with a de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
How to find value of floor function of a given number? I am considering numbers $n = 3m + 1$ and $p = 3q$, where $m$ and $q$ are some positive integers.
I want to find the value of $\lfloor \frac{3m + 1}{2} \rfloor$ and $\lfloor \frac{3q}{2} \rfloor$. I know somewhere the answer depends on the values of $m$ and $q$.
F... | If $m$ is odd, $3m+1$ is even and $\displaystyle \left\lfloor \frac{3m+1}{2}\right\rfloor=\frac{3m+1}{2}$.
If $m$ is even, $3m+1$ is odd and $\displaystyle \left\lfloor \frac{3m+1}{2}\right\rfloor=\frac{3m}{2}$.
If $q$ is odd, $3q$ is odd and $\displaystyle \left\lfloor \frac{3q}{2}\right\rfloor=\frac{3q-1}{2}$.
If $q$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How many solution does $1/a+1/b+1/c+1/d=1$ have? From my friend, he gives me a competition question:
"How many solution $(a,b,c,d)$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$ have where $a,b,c,d$ are positive integers? (the size of $a,b,c,d$ doesn't matter, either one can be the biggest or smallest, and t... | Above equation shown below has parametric form:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$
$a=k(3-k)/2k$
$b=k(3-k)/k$
$a=3(k-3)/2k$
$a=3(k-3)/k$
For $k=-7$ we get $(a,b,c,d)=[5,10,(15/7),(30/7)]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $\alpha$, $\beta$, $\gamma$ are roots of $x^3+p x+q=0$, then $\alpha^5+\beta^5+\gamma^5=5\alpha\beta\gamma(\alpha\beta+\beta\gamma+\gamma\alpha)$.
If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3+px+q=0$, show that $$\alpha^5+\beta^5+\gamma^5=5\alpha\beta\gamma\ (\alpha\beta+\beta\gamma+\gamma\alpha)$$
I tried... | plugging $$\gamma=-\alpha-\beta$$ in the left-hand sode we get
$$-5\,\alpha\,\beta\, \left( \beta+\alpha \right) \left( {\alpha}^{2}+
\alpha\,\beta+{\beta}^{2} \right)
$$ (after factirization)
and the right-hand side:
$$5\,\alpha\,\beta\, \left( \beta+\alpha \right) \left( {\alpha}^{2}+
\alpha\,\beta+{\beta}^{2} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Stuck calculating the derivative of $f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}$. I have to calculate the derivative of this:
$$f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}$$
But I'm stuck. This is the point where I have arrived:
$$f'(x) = \frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})+x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$$
H... | The first derivative is given by $$f'(x)= \left( \left( 1+\sqrt {5-{x}^{2}} \right) ^{-1}+{\frac {{x}^{2}}{
\left( 1+\sqrt {5-{x}^{2}} \right) ^{2}\sqrt {5-{x}^{2}}}} \right)
\left( 1+\sqrt {5-{x}^{2}} \right) {x}^{-1} \left( \ln \left( 10
\right) \right) ^{-1}
$$ and can be simplified to
$$f'(x)={\frac {\sqrt {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Figure out all positive integers n with consecutive + integers a,b,c. When $2018^n$ = $a^4$ + $b^4$ + $({b^2+c^2})^2$,
then what is the possible positive integers n be?
| Let us consider the given equation modulo $4$. Then $a,b$ are consecutive, of the shape $2k$ and $2k\pm1$ (in the right order), so $a^4+b^4$ is $0+1$ or $1+0$ modulo four. The same argument shows $b^2+c^2\equiv 0+1$ (or $1+0$) modulo four. The whole expression is thus $1+1^2=2$ modulo four.
So on the L.H.S of the given... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2750558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the polynomial if remainder is given If $f$ is a quintic polynomial which leaves remainder $1$ when divided by $(x-1)^3$, and $-1$ when divided by $(x+1)^3$ , then find the value of first derivative of $f$ at $x=2$.
My approach
Let $$ f = A(x-1)^5 + B(x-1)^4 + C(x-1)^3 +1 $$
Also
$$ f = A(x+1)^5 + D(x+1)^4 + E(x+... | $f$ is our quintic polynomial, and we have the following:
$$f=g \cdot (x-1)^3+1 \tag{1}$$
$$f = h \cdot (x+1)^3-1 \tag{2}$$
Where $g$ and $h$ are quadratic polynomials.
Now, since we require $f'(2)$, let's just differentiate these 2 descriptions of $f$, and let's note that since $f$ is quintic, $f'$ is biquadratic. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\lim _{x\to \infty} (\sqrt[3]{(x+1)(x+2)(x+3)}-x)$
Evaluate: $\lim _{x\to \infty}( \sqrt[3]{(x+1)(x+2)(x+3)}-x)$
I set $y= x-3$ for simplification and then tried to solve $y\to \infty$.
I tried to use the tayor expansion of the cubic function f(y) in the cuberoot but that didn't help.
How do I approach... | $y = x+2$:
$$L = \lim _{x\to \infty} (\sqrt[3]{(x+1)(x+2)(x+3)}-x) = 2+ \lim _{y\to \infty} (\sqrt[3]{y^3-y}-y) = 2+ \lim _{y\to \infty} \frac{\sqrt[3]{1-\frac{1}{y^2}}-1}{\frac{1}{y}}$$
$y = \frac{1}{h}$:
$$L = 2+ \lim _{h\to 0^+}\frac{\sqrt[3]{1-h^2}-1}{h} \stackrel{L'Hopital}{=}2+ \lim _{h\to 0^+}\frac{\frac{-2h}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Without using the Rule of Sarrus, prove that: Without using the Rule of Sarrus, prove that:
$$\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|=3abc-a^3-b^3-c^3$$
My Approach:
$$LHS=
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|$... | Here is a way to break it down to a factor and only one $2\times 2$-determinant containing only binomials before expanding:
$$\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|\stackrel{R_3 \mapsto R_3+R_2+R_1}{=}
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
2(a+b+c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solving a complex cubic equation I am trying to solve the following equation:
$$
z^3 + z +1=0
$$
Attempt: I tried to factor out this equation to get a polynomial term, but none of the roots of the equation is trivial.
| Substitute $x=\frac2{\sqrt3}\sinh t$ to rewrite the equation $x^3+x+1=0 $ as
$$4\sinh^3t+3\sinh t + \frac{3\sqrt3}2=0$$
Comparing with the identity $4\sinh^3t+3\sinh t = \sinh3t$
results in $\sinh 3t =- \frac{3\sqrt3}2$, or
$t = -\frac13\sinh^{-1} \frac{3\sqrt3}2$. Thus, one real root of the cubic equation is
$$x_0= -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solving nonlinear system to change variable in a PDE I want to change variables of the PDE
\begin{align*}
\left(-\partial_x^2 - \frac{1}{x}\partial_x + \frac{t^2}{x^2}\partial_t^2\right)\psi = \lambda \psi
\end{align*}
to variables
\begin{align*}
r &:= \sqrt{(x+2bt)(x-2c/t)} \\
s &:= te^{\tau}\sqrt{\frac{1-\frac{2c}{xt... | Denote
\begin{align}
A&=\sqrt{x-\frac{2c}{t}},\\
B&=\sqrt{x+2bt},
\end{align}
and your change-of-variable reads
\begin{align}
r&=BA,\\
s&=te^{\tau}\frac{A}{B}.
\end{align}
These two relations yield
\begin{align}
rs&=te^{\tau}A^2=te^{\tau}\left(x-\frac{2c}{t}\right)=e^{\tau}\left(xt-2c\right),\\
\frac{r}{s}&=\frac{1}{te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the sum $1+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots =\ ?$ Let $\phi(x)=\begin{cases}0, & 0\lt x\lt 1\\ 1, & 1\lt x\lt3 \end{cases}$
We have that the Fourier cosine series is given by $$\phi(x)=\begin{cases}0, & 0\lt x\lt1\\ \frac{4}{3}+\displaystyle\sum_{m... | In this answer, it is shown that
$$
\sum_{k=-\infty}^\infty\frac{(-1)^k}{z+k}=\pi\csc(\pi z)
$$
Therefore,
$$
\begin{align}
\sum_{k=0}^\infty(-1)^k\left(\frac1{3k+1}+\frac1{3k+2}\right)
&=\sum_{k=0}^\infty(-1)^k\left(\frac1{3k+1}-\frac1{-3(k+1)+1}\right)\\
&=\sum_{k=0}^\infty\left(\frac{(-1)^k}{3k+1}+\frac{(-1)^{-k-1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}=\frac{1}{5}$ using integral(lower-/upper- or Riemann-sums) how can I show that $$
\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}=\frac{1}{5}$$
using integral(lower-/upper- or Riemann-sums).
I tried the following:
$\lim_{n\rightarrow \infty} \frac{1^4+2^4+.... | It depends on what tools you have available.
Regardless,
$$\lim_{n \to \infty}\dfrac{1^4+2^4+\cdots+n^4}{n^5}=\lim_{n \to \infty}\dfrac{\sum_{i=1}^{n}i^4}{n^5}\tag{*}$$
Method 1: You don't know of the Fundamental Theorem of Calculus.
I assume you have knowledge of the formulas
$$\begin{align}\sum_{i=1}^{n}i &= \dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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Fractions in Questions and Answers
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