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 $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} , \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Find a $2 \times 2$ reflection matrix with all entries nonzero. Can you find all $2 \times 2$ reflection matrices? That is, find necessary and suffcient conditions
on $a, b, c, d$ so that ${\begin{pmatrix} a & b \\ c & d \end{pmatrix}}^2 = I $
This is what I have done so far
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ = $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
So we have
$\begin{pmatrix} a^2 + bc & ab + bd \\ ca + dc & cb + d^2 \end{pmatrix}$ = $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
So what do I say now?
| 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 have $d = -a$; that takes care of two out of the four entries in $M^2$.
Now $M = \pmatrix{a&b\\c&-d}$ and
$$
M^2 = \pmatrix{a^2 + bc & 0 \\ 0 & a^2 + bc}.
$$
In order to have $M^2 = I$, you simply need $a, b, c$ such that $a^2 + bc = 1$. There are lots of solutions here, I'm sure you can find one.
As far as "conditions on $a, b, c, d$ so that $M^2 = I$", we already have four equations putting restrictions on the values of $a, b, c, d$ (that various entries should be $1$ or $0$). I'm not sure it gets much better than that. You can separate into cases (whether $a = -d$, whether $b$ and $c$ are $0$), but when $a = -d$, the only other requirement is that $a^2 + bc = 1$, which has many solutions. You could solve for $b$ or $c$ and use only two variables to write $M$, but that's about it.
As a sidenote, the lone requirement that $M^2 = I$ doesn't make a very good definition for a reflection matrix, in the usual sense (but that's not your fault!). Reflections preserve length, so in addition, we'd want $a^2 + c^2 = 1 = b^2 + d^2$ (since $M(e_1) = \begin{bmatrix}a \\ b\end{bmatrix}$ has length $a^2 + b^2$, likewise for $M(e_2)$). Further, reflections should reverse orientation, so that we should require $\det M = ad - bc = -1$, as well.
If you follow this trail, you'll see that conventional reflection matrices come in exactly one form:
$$
M = \pmatrix{
a & \pm \sqrt{1 - a^2}\\
\mp \sqrt{1 - a^2} & -a},
$$
with $a \in [-1, 1]$ (which leads nicely to their parameterization using angles and formatting with trig functions; an angle completely determines a line through the origin, which completely determines the reflection).
| {
"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 :
$$\dfrac{1}{1+\dfrac{\sin 70^{\circ}}{\cos 70^{\circ}}}+\dfrac{1}{1+\dfrac{\sin 20^{\circ}}{\cos 20^{\circ}}}$$
$$\dfrac{\cos70^{\circ}}{\cos 70^{\circ}+\sin 70^{\circ}}+\dfrac{\cos20^{\circ}}{\cos 20^{\circ}+\sin 20^{\circ}}$$
now what do i do ?
| $$\frac { 1 }{ 1+\tan { { 70 }^{ \circ } } } +\frac { 1 }{ 1+\tan { { 20 }^{ \circ } } } =\frac { 1 }{ 1+\tan { { 70 }^{ \circ } } } +\frac { 1 }{ 1+\cot { { 70 }^{ \circ } } } =\\ =\frac { 1 }{ 1+\tan { { 70 }^{ \circ } } } +\frac { 1 }{ 1+\frac { 1 }{ \tan { { 70 }^{ \circ } } } } =\frac { 1+\tan { { 70 }^{ \circ } } }{ 1+\tan { { 70 }^{ \circ } } } =1$$
| {
"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 $\displaystyle \sum \tan^2 A = \left(\sum \tan A \right)^2 - 2 \sum \tan A \tan B \notin (81-18 \sqrt 3, 81+18 \sqrt 3)$
| {
"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
} |
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
$$
f(x,y,z) = \left\lbrace
\begin{array}{cl}
cz^{-3} & x^2+y^2 \leq z^2 \text{ and } a\leq z \leq b \\
0 & \text{otherwise,}
\end{array}\right.
$$
where $a$ and $b$ are constants satisfying $0<a<b<\infty$.
(a) Find $c$.
(b) Find the marginal joint pdf of $(X,Y)$.
(c) Find $E[X]$ and $E[Y]$.
I am specifically looking for guidance for part (a). I tried solving this myself with the limits of $z$: $a$ to $b$, $y$: $-\sqrt{z^2 - x^2}$ to $\sqrt{z^2 - x^2}$, and $x$: $-\sqrt{z^2 - y^2}$ to $\sqrt{z^2 - y^2}$.
Now, my integrals are looking really messy, so I am thinking I do not have the right limits. I was just wondering if anyone could help me? Thanks a lot!
Also some guidance for (c): When I solve for the expectations, do I solve using the joint pdf $f(x, y, z)$ or do I use the marginal pdfs?
| 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}_{a\leq z\leq b} \int_{\Bbb R^2} \mathbf{1}_{x^2 + y^2 \leq z^2}\, \text{d}x\,\text{d}y\,\text{d}z \\
&= \int_{\Bbb R}\left( \mathbf{1}_{a\leq z\leq b} \int_{0}^{+\infty} \mathbf{1}_{r \leq |z|}\, r\text{d}r \int_{0}^{2\pi}\text{d}\theta\right) \text{d}z \\
&= {\pi} \int_{a}^{b} |z|^2 \,\text{d}z \\
&= \frac{\pi}{3}\left(b^3 - a^3\right) .
\end{aligned}
This computation can be adapted to the integral of the pdf
\begin{aligned}
\int_{\Bbb R^3} c z^{-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
&= {\pi}c \int_{a}^{b} z^{-3}\, |z|^2 \,\text{d}z \\
&= {\pi}c\ln\left(b/a\right) \, ,
\end{aligned}
which must be equal to one.
The marginal pdf $g$ of $(X,Y)$ is obtained by integrating the pdf with respect to $z$:
\begin{aligned}
g(x,y) &= \left\lbrace\begin{array}{ll}
0 &\text {if}\quad \sqrt {x^2+y^2}>b\\
c \int_\sqrt{x^2+y^2}^b z^{-3} \,\text{d}z &\text {if}\quad b>\sqrt{x^2+y^2}>a\\
c \int_a^b z^{-3} \,\text{d}z &\text {if}\quad a>\sqrt{x^2+y^2}
\end{array}\right.\\
&= c\int_{\max\left\lbrace a,\sqrt{x^2 + y^2}\right\rbrace}^{\max\left\lbrace b,\sqrt{x^2 + y^2}\right\rbrace} z^{-3} \,\text{d}z \\
&= -\frac{c}{2} \left({\max\left\lbrace b,\sqrt{x^2 + y^2}\right\rbrace}^{-2} - {\max\left\lbrace a,\sqrt{x^2 + y^2}\right\rbrace}^{-2} \right) .
\end{aligned}
The marginal pdfs $h$ of $X$ and $Y$ are identical, since $x$ and $y$ play symmetric roles in $g(x,y)$. They write $h(x) = \int_{\Bbb R} g(x,y)\,\text{d}y$. Then, $E[X] = \int_{\Bbb R} x\, h(x) \,\text{d}x = E[Y]$.
| {
"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
} |
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 $y=\sum_{m=0}^{\infty} c_m(x+1)^{m+\rho}$. Then I get the indicial equation as $$\rho^2-1=0.$$ So $\alpha=1$ and $\beta=-1$. Thus the value of $\alpha-4\beta$ is $5$.
But the answer is $2$. The question appears in GATE 2017. So where i am wrong. Any help is highly appreciated.
| $$(x^2-1)^2y''+(x+1)y'-y=0$$
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
$$r^2+\left(\frac {\beta(-1)}{\alpha(-1)}-1\right)r+\frac {\gamma(-1)}{\alpha(-1)}=0$$
$$r^2-\frac {3}{4}r-\frac {1}{4}=0$$
$$(r-1)(r+\frac {1}{4})=0 \implies r=1, r=-\frac 1 4$$
$$\text{And }\alpha -4\beta=1-4\frac {-1} 4=2$$
| {
"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:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{1}{6} - 1) (x^\frac{1}{6} + 1) } $$
$$\lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{2}{6} - 1) } $$
| 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. $x = \frac {-21 \pm {21^2 - 8*27}}{4} = \frac {-21 \pm \sqrt {225}}4 = \frac {-21 \pm 15}4 = -9, -\frac 32$ so
$2x^3 + 21 x^2 + 27x = x*2*(x + \frac 32)(x + 9) = x(2x +3)(x+9)$.
3)
$2x^3 + 21x^2 + 27x = (ax + b)(cx + d)(ex + f)$ and
$ace = 2$. Wolog, $a = 2$ and $c, e = 1$.
So $2x^3 + 21x^2 + 27x = (2x + b)(x + d)(x + f)$
$2f + 2d + b = 21$
$2df + bf + bd = 27$
$bdf = 0$
$b$ is odd so $b\ne 0$. Wolog $d= 0$.
$2f + b = 21$
$bf = 27$
So $b=3; f= 9$
So $x(2x + 3)(x + 9)$.
| {
"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 : $f^2(x+b)+\frac{1}{4}-f(x+b)=f(x)-f^2(x)$
and let $b=0$ then $f^2(x)+\frac{1}{4}-f(x)=f(x)-f^2(x)$ so $b \neq 0$
now what do i do ?
| 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 $$\left(f(x+b)-\frac{1}{2}\right)^2 = \left(f(x-b)-{1\over 2}\right)^2$$
and so $$\left|f(x+b)-\frac{1}{2}\right| = \left|f(x-b)-{1\over 2}\right|$$
But $f(x)\geq 1/2$ for all $x$ so we have $$f(x+b)-\frac{1}{2} = f(x-b)-{1\over 2}$$
which means that $f$ is periodic with period $2b$ and thus the limit doesn't exist unless $f$ is constant.
| {
"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
} |
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 equation? I can solve this when $n = 1$, but not for a general case.
| 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 straightforward as follows
$$\begin{align}
f(x) &= \frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} (a_{ij} - x_i x_j)^2 \\
&= \frac{1}{2} \left( \overbrace{\sum_{i=1 \\ i\neq k}^{n} \sum_{j=1 \\ j \neq k}^{n} (a_{ij} - x_i x_j)^2}^{\text{has no $x_k$}} + \overbrace{\sum_{i=1 \\ i\neq k}^{n} (a_{ik} - x_i x_k)^2}^{\text{has only one $x_k$}} + \overbrace{\sum_{j=1 \\ j\neq k}^{n} (a_{kj} - x_k x_j)^2}^{\text{also has only one $x_k$}} + \overbrace{(a_{kk} - x_k^2)^2}^{\text{has two $x_k$s}} \right)
\end{align}$$
Therefore
$$\begin{align}
\partial_k f(x) &= \frac{1}{2} \left( 0 + \sum_{i=1 \\ i\neq k}^{n} 2(a_{ik} - x_i x_k)(-x_i) + \sum_{j=1 \\ j\neq k}^{n} 2(a_{kj} - x_k x_j)(-x_j) + 2(a_{kk} - x_k^2)(-2 x_k) \right) \\
&= - \sum_{i=1 \\ i\neq k}^{n} (a_{ik} - x_i x_k) x_i - \sum_{j=1 \\ j\neq k}^{n} (a_{kj} - x_k x_j) x_j - (a_{kk} - x_k^2) 2 x_k \\
&= - \sum_{i=1}^{n} (a_{ik} - x_i x_k) x_i - \sum_{j=1}^{n} (a_{kj} - x_k x_j) x_j \\
&= - \sum_{i=1}^{n} \left( (a_{ik} - x_k x_i) x_i + (a_{ki} - x_k x_i) x_i \right) \\
&= - \sum_{i=1}^{n} (a_{ik} - x_k x_i + a_{ki} - x_k x_i) x_i \\
&= - \sum_{i=1}^{n} (a_{ik} + a_{ki} - 2 x_k x_i) x_i \\
\end{align}$$
Since $\nabla f(x) = \left[ \begin{matrix} \partial_1 f(x) & \cdots & \partial_n f(x) \end{matrix} \right]^T$ it follows that $\nabla f(x) = - (A + A^T - 2x x^T)x $
Then $\nabla f(x) = 0$ implies $x = 0$ or $A + A^T - 2x x^T = 0$.
Now suppose that $x \neq 0$. So $x x^T = \dfrac{1}{2}(A + A^T)$. That is, $x x^T$ is equal to the symmetric part of $A$. This is only feasible if $A$ meets certains conditions which I will left for you to verify. Namely if $A$ have non-negative diagonal and for all $i, j \in 1 \dots n $ it holds that
$$ \underbrace{\sqrt{a_{ii} a_{jj}}}_{\text{geometric mean}} = \underbrace{\frac{a_{ij} + a_{ji}}{2}}_{\text{arithmetic mean}} $$
(Which particularly I think is a nice relation)
So if such conditions are meet, then $x = \left[ \begin{matrix} \sqrt{a_{11}} \cdots \sqrt{a_{nn}} \end{matrix} \right]^T$.
Thus the stationary points of $f(x)$ are $x=0$, and $x = \left[ \begin{matrix} \sqrt{a_{11}} \cdots \sqrt{a_{nn}} \end{matrix} \right]^T$ if $A$'s additional conditions are met.
| {
"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 $x = 2 \sin \theta, dx = 2 \cos \theta, \theta = \arcsin(\frac{x}{2})$
$$\int \frac{4\sin^2 \theta}{\sqrt{4(1-\sin^2 \theta})}*2\cos \theta d\theta $$
$$ \int \frac{4\sin^2 \theta}{2\cos \theta}2cos\theta = \int4\sin^2\theta{d\theta}$$
Then I let $u = 2\theta$
$$4 \int \frac{\sin 2\theta + 1}{2} = 4 [\frac{1}{2}\theta + \frac{1}{4} \int \sin udu]$$
$$4[\frac{1}{2}\theta - \frac{\cos u}{4}] = 4 [\frac{\arcsin(\frac{x}{2})}{2} - \cos(2\arcsin(\frac{x}{2}))]$$
Yielding:
$$2\arcsin(\frac{x}{2}) - \frac{\cos(2\arcsin(\frac{x}{2}))}{4}$$
What's wrong with this?
| 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: $$\bigg[(x+2)^2+y(x+2)\bigg]\frac{dy}{dx}=y^2$$
$$y^2\frac{dx}{dy}=(x+2)^2+y(x+2)$$
Could some help me to solve it , thanks
| $$y^2 \frac{dx}{dy} = (x+2)^2 + y(x+2)\\
\frac{1}{(x+2)^2}\frac{dx}{dy}-\frac{1}{y(x+2)} = \frac{1}{y^2}$$
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
$$\frac{-1}{(x+2)y} = \int \frac{dy}{y^3} = \frac{-1}{2y^2}-c\\
2y=(x+2)(cy^2+1)$$
I think you can proceed.
| {
"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 $x = a\sin \theta, dx = a\cos \theta d\theta$
$$\int \frac{a cos\theta d\theta}{(a^2 + a^2 sin^2 \theta)^2} = \int \frac{a\cos \theta d\theta}{(a^2(1+sin^2\theta))^2} $$
$$\int \frac{a\cos \theta d\theta}{(a^2 \cos^2\theta)^2} =\int \frac{d\theta}{a^3cos^3\theta} $$
I don't know what to do anymore
| $$\int\dfrac{dx}{a^2+x^2}=\dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right)+C$$
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 solved for ${\bf A}^{2n}$ easily, and I recognise the Taylor series of the natural log in the summation. I am however unable to link the two together, and I can never get a perfect $\frac {5}{3}$ in the final matrix. Thank you!
| 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
$$\sum_{n=1}^{\infty}\frac{(1/16)^n - 2(-1/4)^n}{n}$$
We can, as you realized, use the taylor series of the natural log summation, namely,
$$-\ln(1-x)=\ln(\frac{1}{1-x})=\sum_{n=1}^{\infty}\frac{x^n}{n}$$
And from this we can directly solve the problem.
$$\left(\sum_{n=1}^{\infty}\frac{(1/16)^n}{n}\right) -2\left(\sum_{n=1}^{\infty}\frac{(-1/4)^n}{n}\right)$$
$$=\ln\left(\frac{1}{1-1/16}\right)-2\ln\left(\frac{1}{1+1/4}\right)$$
$$=\ln\left(\frac{16}{15}\right) -\ln\left(\frac{16}{25}\right)=\ln\left(\frac{5}{3}\right)$$
Thus
$$\sum_{n=1}^{\infty} \frac{{\bf A}^{4n} - 2{\bf A}^{2n}}{n} = \ln\left(\frac{5}{3}\right)I=
\begin{bmatrix}
\ln\frac{5}{3} & 0\\
0 & \ln\frac{5}{3}
\end{bmatrix}$$
| {
"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
$$a\equiv b \text{ mod } 3 \text{ or } a^2+ab+b^2\equiv0 \text{ mod } 3.$$
Besides, the prime factorization of $999=3^3\cdot37$, but I don't know how to go on.
I would appreciate any hints.
| 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 decided, seven more cases to go!
| {
"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
} |
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 $x = \frac{3}{2}\sin\theta, dx = \frac{3}{2}\cos \theta d\theta$:
$$\int \sqrt{9 - 4x^2}dx = \int \sqrt{4\left(\frac{9}{4} - x^2\right)}dx = 2\int \sqrt{\left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\sin^2\theta}*\frac{3}{2}\cos \theta$$
$$3\int \frac{3}{2}\cos^2\theta d\theta = \frac{9}{2}\int \frac{\cos 2\theta + 1}{2}$$
$$\frac{9}{2}\left(\int \frac{1}{2} + \int \cos 2\theta \right) = \frac{9}{2}\left(\frac{1}{2}\theta + \frac{1}{4}\sin2\theta\right) = \frac{9}{2}\left(\frac{1}{2}\arcsin\left(\frac{2}{3}x\right) + \frac{1}{4}\sin\left(2\arcsin\left(\frac{2}{3}x\right)\right)\right)$$
The problem is evaluating the definite integral, it's getting way too messy. How do I go about this?
| 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
} |
Identity involving binomial coefficients I might have written this in a needlessly cumbersome way, but I want to prove that for odd positive integers $n$, $$\sum_{k\ odd}^{n}\binom{2n+1}{2k}=\begin{cases}
\binom{2^n+1}{2}, & \text{if}\ n\ \text{mod}\ 4 =1\\
\binom{2^n}{2}, & \text{if}\ n\ \text{mod}\ 4 =3
\end{cases}$$
I have tested these identities and they should hold in general. Thank you very much in advance!
Edit:
I found the following formula in the wikipedia article for binomial coefficients, under multisections of sums (https://en.wikipedia.org/wiki/Binomial_coefficient#Multisections_of_sums) which works nicely.$$\binom{n}{2}+\binom{n}{6}+\binom{n}{10}+\cdots=\frac{1}{2}(2^{n-1}-2^{\frac{n}{2}}\cos\frac{n\pi}{4})$$
It would be great if someone can provide a reference or proof for this. Or if there is a much faster way of getting at the same result then please ignore this entirely.
| Hint: Use the binomial formula for $(1+1)^{2n+1}$, $(1-1)^{2n+1}$, $(1+i)^{2n+1}$ and $(1-i)^{2n+1}$:
$$A=(1+1)^{2n+1}=\binom{2n+1}{0}+\binom{2n+1}{1}+\binom{2n+1}{2}+\binom{2n+1}{3}+\cdots$$
$$B=(1-1)^{2n+1}=\binom{2n+1}{0}-\binom{2n+1}{1}+\binom{2n+1}{2}-\binom{2n+1}{3}+\cdots$$
$$C=(1+i)^{2n+1}=\binom{2n+1}{0}+\binom{2n+1}{1}i-\binom{2n+1}{2}-\binom{2n+1}{3}i+\cdots$$
$$D=(1-i)^{2n+1}=\binom{2n+1}{0}-\binom{2n+1}{1}i-\binom{2n+1}{2}+\binom{2n+1}{3}i+\cdots$$
so the requested sum is:
$$\binom{2n+1}{2}+\binom{2n+1}{6}+\cdots=\frac{1}{4}(A+B-C-D)$$
All it takes is to calculate $\frac{1}{4}(A+B-C-D)=\frac{1}{4}\left(2^{2n+1}-(1+i)^{2n+1}-(1-i)^{2n+1}\right)$. That can be done for the cases $n=4k+1$ and $n=4k+3$ separately.
For example, for $n=4k+1$, we have:
$$\frac{1}{4}\left(2^{2n+1}-(1+i)^{8k+3}-(1-i)^{8k+3}\right)=\frac{1}{4}\left(2^{2n+1}-2^{4k}\left((1+i)^3-(1-i)^3\right)\right)=\frac{1}{4}\left(2^{2n+1}+2^{4k+2}\right)=\frac{1}{2}(2^{2n}+2^n)=\binom{2^n+1}{2}$$
Above, we used the fact that $(1+i)^8=(1-i)^8=2^4$, and also $(1+i)^3+(1-i)^3=-4$.
The case $n=4k+3$ is similar.
| {
"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 no idea how to deal with the LHS.Any help will be grateful.
| 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 \right)\,dx\int_{\tfrac{1}{\sqrt{x\left( 1-x \right)}}}^{\infty}\frac{e^{-u}}{u} \,du\\
\end{align}
We take advantage of the symmetry with respect to $x=1/2$ for the las expression.
Now, with $y=\tfrac{1}{\sqrt{x\left( 1-x \right)}}$, one has $1-2x=\sqrt{1-\frac{4}{y^2}}$ and
$dx=-\frac{2}{y^2\sqrt{y^2-4}}dy$, and thus
\begin{align}
I&=-24\int_2^\infty \frac{dy}{y^4\sqrt{y^2-4}}\int_{y}^{\infty}\frac{e^{-u}}{u} \,du\\
&=-\frac{3}{2}\int_1^\infty \frac{dz}{z^4\sqrt{z^2-1}}\int_{2z}^{\infty}\frac{e^{-u}}{u} \,du
\end{align}
As
\begin{equation}
\int \frac{1}{v^4\sqrt{v^2-1}}dv=\frac{1}{3}\frac{\left( 2v^2+1 \right)\sqrt{v^2-1}}{v^3}
\end{equation}
the last integral can be calculated by parts:
\begin{equation}
I=-\frac{1}{2}\int_1^\infty \frac{\left( 2z^2+1 \right)\sqrt{z^2-1}}{z^4}e^{-2z}\,dz
\end{equation}
| {
"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.
$$\begin{align}
\frac12\cdot\text{target area} &= A + B \\
&= \left( \frac12 \cdot (2s)^2\cdot\alpha - 4 C\right) + \left(\frac12 \cdot s^2 \cdot \beta - C \right) \\[4pt]
&= \frac12 s^2\left( 4 \alpha + \beta \right) - 5 C \\[4pt]
&=\frac12 s^2 \left( \frac\pi2 + 3\alpha \right)-5 \cdot \frac{1}{20}(2s)^2 \\[4pt]
&= \frac12 s^2 \left( \frac\pi2 + 3\operatorname{atan}\frac{1}{2} - 2 \right)
\end{align}$$
So, the target area, with $s = 2$, is
$$2 \pi + 12 \operatorname{atan}\frac{1}{2} - 8 $$
| {
"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
} |
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 :
$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\\
>\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2n}+....+\frac{1}{2n}\\
=\frac{n+1}{2n}>\frac{1}{2}$
Where i am missing and how to processed second question
| By C-S
$$\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\geq\frac{(1+1+...+1)^2}{n+(n+1)+...+(2n)}=\frac{(n+1)^2}{\frac{(n+1)(2n+n)}{2}}=\frac{2(n+1)}{3n}>\frac{2}{3}$$
$$\frac{1}{3n+1}+\frac{1}{3n+2}+.....\frac{1}{5n}+\frac{1}{5n+1}\geq\frac{(1+1+...+1)^2}{(3n+1)+(3n+2)+...+(5n+1)}=$$
$$=\frac{(2n+1)^2}{\frac{(2n+1)(2(3n+1)+2n)}{2}}=\frac{2n+1}{4n+1}>\frac{1}{2}.$$
$$\frac{1}{3n+1}+\frac{1}{3n+2}+.....\frac{1}{5n}+\frac{1}{5n+1}<\int\limits_{3n}^{5n+1}\frac{1}{x}dx=\ln\frac{5n+1}{3n}<\frac{2}{3}$$ for all $n\geq2$.
For $n=1$ we see that $\frac{1}{4}+\frac{1}{5}+\frac{1}{6}<\frac{2}{3}$ is true.
| {
"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)
$$D= \left[
\frac{\sin \theta(\cos \theta - a)}{2(a^2-1)(a \cos\theta-1)^2}
+\frac{\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{(a^2-1)^{3/2}}\right]_0^{2\pi}
.$$
which can be expressed as
$$D= \left[
\frac{(a^2-1)^{1/2}\sin \theta(\cos \theta - a)+2 (a \cos\theta-1)^2\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{2(a^2-1)^{3/2}(a \cos\theta-1)^2}
\right]_0^{2\pi}
.$$
This expression involves discontinuities and complex numbers which is beyond my present abilities to handle.
| 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}
\end{align}
$(1):$ for all $a,b>0$,
\begin{align}
\int_0^{\infty} \frac{16t^2}{(a+bt^2)^3}\,dt&=\frac{1}{(ab)^{3/2}}\int_0^{\infty} \frac{16t^2}{(1+t^2)^3}\,dt\\
&=\frac{1}{(ab)^{3/2}}\int_0^{\infty} {8s^{1/2}}{(1+s)^{-3}}\,ds\tag{$s=t^2$}\\
&=\frac{\pi}{(ab)^{3/2}}\tag{2}
\end{align}
$(2):$ For $x,y>0$, Euler integral gives
$$\int^\infty_0 \frac {t^{x-1}}{(1+t)^{x+y }}dt =\frac {\Gamma (x)\Gamma (y)}{\Gamma (x+y)}$$
and for $x=y=\frac32$, we have
$$\frac {\Gamma^2 (\frac32)}{\Gamma (3)}=\frac\pi8$$
| {
"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
} |
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 arrangments should in this form: rrbbggyy**
Case 1: two * are same
$\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}10!}{4!2!2!2!}$
Case 2: two * are different
$\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}10!}{3!3!2!2!}$
Result = Case 1 + Case 2
Any suggestions would be appreciated!
| 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
$$
\binom{4}{1}\binom{10}{4,2,2,2}+\binom{4}{2}\binom{10}{3,3,2,2}
$$
| {
"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 $\theta=0$; when $x=3\sqrt{3}/2$, $tan\,\theta=\sqrt{3}$ so $\theta=\pi/3$
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}}\,dx=\int^{\pi/3}_0\frac{\frac{27}{8}\,tan^3\,\theta}{27\,sec^3\,\theta}\,\frac{3}{2}\,sec^2\,\theta\,d\theta$$
$$=\frac{3}{16}\int^{\pi/3}_0\frac{1-cos^2\,\theta}{cos^2\,\theta}sin\,\theta\,d\theta$$
With u substitution $u=cos\,\theta$ and $du=-sin\,\theta\,d\theta$. When $\theta=0$, $u=1$; when $\theta=\pi/3$, $u=\frac{1}{2}$
Therefore:
$$\frac{3}{16}\int^{\pi/3}_0\frac{1-cos^2\,\theta}{cos^2\,\theta}sin\,\theta\,d\theta=\frac{-3}{16}\int^{\frac{1}{2}}_1\frac{1-u^2}{u^2}du$$
My question lies in the next part:
$$=\frac{3}{16}\int^{\frac{1}{2}}_1(1-u^{-2})du$$
Why does $\frac{-3}{16}$ become positive? Also what did they do to get $1-u^{-2}$?
| $$\frac { -3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \frac { 1-u^{ 2 } }{ u^{ 2 } } du=\frac { -3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \left( \frac { 1 }{ { u }^{ 2 } } -1 \right) du=\frac { -3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \left( { u }^{ -2 }-1 \right) du=\\ =\frac { 3 }{ 16 } \int _{ 1 }^{ \frac { 3 }{ 2 } } \left( 1-{ u }^{ -2 } \right) du$$
| {
"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)
\end{align}
I'm a bit stuck on how to simplify from here? Was there a simpler way somewhere?
Also this one is giving me problems. I am thinking of changing the log forms first?:
$$\ln \frac{(2y+1)^5}{\sqrt{y^2 + 1}}= \ln(2y+1)^5 - \ln \sqrt{y^2 +1}=5 \ln(2y+1) - \frac{1}{2}\,\ln (y^2 +1)$$
so:
$$G'y = \frac{5}{2y+1} \cdot 2 - \frac{2y}{2(y^2 + 1)} = \frac{10}{2y+1} - \frac{y}{2 (y^2 +1)}$$
Is that right?
| 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 following theorem:
"There are $C(n+r-1,r)$ r-combinations from a set with $n$ elements when repetition of elements is allowed." But I'm not sure.
Any great hints would be appreciated.
| To show the working behind the generating function approach:
A binomial expansion of $(1-x)^{-n}$ gives $$(1-x)^{-n} = 1 + (-n)(-x) + \frac{(-n)(-n-1)}{2!} (-n)^2 + \frac{(-n)(-n-1)(-n-2)}{3!} (-n)^3 + \cdots \\
= \sum_{i\ge 0}\frac{(-1)^i (-n)^\underline{i}}{i!}x^i \\
= \sum_{i\ge 0}\frac{n^\overline{\,i\,}}{i!}x^i \\
= \sum_{i \ge 0}\binom{n+i-1}{i}x^i \\
= \sum_{i \ge 0}\binom{n+i-1}{n-1}x^i$$
Then $$[x^k] \frac{\sum_{j\ge 0} a_j x^j}{(1-x)^n} = \sum_{j\ge 0} [x^k] \frac{a_j x^j}{(1-x)^n} = \sum_{j\ge 0} a_j [x^{k-j}] (1-x)^{-n} = \sum_{j\ge 0} a_j \binom{n+k-j-1}{n-1}$$
Now we can apply that to your problem as outlined in Manthanein's first answer: by considering the three boxes we're looking for the coefficient of $x^{13}$ in $$\frac {(1-x^6)(1-x^7)(1-x^5)}{(1-x)^3} = \frac{1-x^5-x^6-x^7+x^{11}+x^{12}+x^{13}-x^{18}}{(1-x)^3}$$
So applying the lemma with $n=3$, $k=13$ we get $$\binom{15-0}{2}-\binom{15-5}{2}-\binom{15-6}{2}-\cdots-\binom{15-18}{2}$$
The last term is zero, and the rest give $$\frac{15\times14 - 10\times9 -9\times8 - 8\times7 + 4\times3 + 3\times2 + 2\times1}{2}\\
= \frac{210-90-72-56+12+6+2}{2} = 6$$
| {
"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
$$x^2+y^2=2x=z^2\implies x=\frac{z^2}2 \quad y=\sqrt{z^2-\frac{z^4}4}$$
The integrand function becomes
$$x^4-y^4+y^2z^2-z^2x^2+1=x^4-y^4+(y^2-x^2)(y^2+x^2)+1=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 $(2x+1)(x+3)$ overlap, but $(2x+7)(x+3)$ does not!
| $2x+7$ 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't say much about how one would find such a thing in the first place.
If you want to find the factoring on your own, without knowing the answer beforehand, you could look at the original quadratic equation, and then use that for any quadratic expression $ax^2 + bx + c$ we have
$$
ax^2 + bx + c = a(x-r_1)(x-r_2)
$$
where $r_1, r_2$ are the two roots of the equation $ax^2 + bx + c = 0$, which you find using the standard quadratic formula:
$$
r_1 = \frac{-b+\sqrt{b^2-4ac}}{2a},\quad r_2 = \frac{-b-\sqrt{b^2-4ac}}{2a}
$$
Or, you could complete the square, then use difference of squares (following basically the proof of the quadratic formula above) to get
$$
2x^2 + 7x + 3 = \frac{16x^2 + 56x + 24}8\\
= \frac{16x^2 + 56x + 49 - 49 + 24}{8}\\
= \frac{(4x + 7)^2 - 25}{8}\\
= \frac{(4x + 7+5)(4x+7-5)}{8}\\
= \frac{4x+12}{4}\cdot\frac{4x+2}{2}\\
= (x+3)(2x+1)
$$
| {
"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
} |
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 is not the answer I get on wolfram, which is $Cx^2 +x$. I'd like to know where I'm going wrong.
| 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:
$$-\frac{2}{x^3} = \frac{d}{dx} \frac{1}{x^2}$$
To get
$$\frac{1}{x^2} y'(x) + \frac{d}{dx} \left(\frac{1}{x^2}\right) y(x) = -\frac{1}{x^2}$$
Now in the central term, we apply the reverse product rule:
$$f g' = (fg)' - f'g$$
This will help you to get rid of the first term. Now integrate what remains with respect to $x$:
$$\int \frac{d}{dx}\left(\frac{y(x)}{x^2}\right) \ dx = \int -\frac{1}{x^2}\ dx$$
Easily:
$$\frac{y(x)}{x^2} = \frac{1}{x} + C$$
Hence
$$\color{red}{y(x) = x + Cx^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
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
$x=\sec A +\tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}$
| Because $$x+\frac{1}{x}=\frac{1+\sin{A}}{\cos{A}}+\frac{\cos{A}}{1+\sin{A}}=$$
$$=\frac{(1+\sin{A})^2}{\cos{A}(1+\sin{A})}+\frac{\cos^2{A}}{\cos{A}(1+\sin{A})}=\frac{(1+\sin{A})^2+\cos^2A}{\cos{A}(1+\sin{A})}=$$
$$=\frac{1+2\sin{A}+\sin^2A+\cos^2A}{\cos{A}(1+\sin{A})}=\frac{1+2\sin{A}+1}{\cos{A}(1+\sin{A})}=$$
$$=\frac{2+2\sin{A}}{\cos{A}(1+\sin{A})}=\frac{2}{\cos{A}}=2\sec{A}.$$
| {
"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
} |
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:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \cos x - 3 \sin x \tan y) \sin x$$
$$\tan y + 3 \sin^2 x + \tan y = 3 \sin x \cos x$$
$$\tan y = \frac{3 \sin x \cos x}{1 + 3 \sin^2 x}$$
I have also tried substituting $0$, $30$, $45$, $60$, $90$ to the values of $x$.
| 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 $\sin 2x=\dfrac{4}{5}$. Substituting these values in the expression of $\tan y$ we attain the maximum:$$\max_{0\le x\le\dfrac{\pi}{2}} \tan y=\dfrac{3}{4}$$
Here is a sketch of $\tan y$ respect to $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
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 $60,120,240,300$ ? How do I find 240 and 300?
| $$\sin^2x=\sin^2A\iff\cos^2x=\cos^2A\iff\tan^2x=\tan^2A$$
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
} |
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) + \sqrt{2} = -5 $
Therefore min value is = $ -5/5 - \sqrt{2} $
Why am I wrong ? And how should I do this question..
| 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: $det\left(\begin{array}{l}a&a&a&a&b\\a&a&a&b&a\\a&a&b&a&a\\a&b&a&a&a\\b&a&a&a&a\end{array}\right)=det\left(\begin{array}{l}b&a&a&a&a\\a&b&a&a&a\\a&a&b&a&a\\a&a&a&b&a\\a&a&a&a&b\end{array}\right)$ (since switching two pairs of rows does not change the determinant)
$= det\left(\begin{array}{l}b-a&0&0&0&a-b\\0&b-a&0&0&a-b\\0&0&b-a&0&a-b\\0&0&0&b-a&a-b\\a&a&a&a&b\end{array}\right)$ (since adding a multiple of one row to another does not change the determinant) for all $1\le i\le 4 \rightarrow R_i-R_5$
Now I am quite stuck. I wanted to obtain a triangular matrix so I can calculate its determinant by the diagonal entries, but I don't know what to do with the last row. I've tried some column operations as well, but have had no success.
Would be happy to get your help, thank you :)
| 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
$$
\left( \begin{array}{rrrrr}
1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 \\
1 & 0 & 2 & -1 & -1 \\
1 & 0 & 0 & 3 & -1 \\
1 & 0 & 0 & 0 & 4 \\
\end{array}
\right).
$$
After adding $(b-a)I,$ the eigenvalues of
$$ \left(\begin{array}{l}b&a&a&a&a\\a&b&a&a&a\\a&a&b&a&a\\a&a&a&b&a\\a&a&a&a&b\end{array}\right) $$
are $$ b+ 4a, \; b-a, \; b-a, \; b-a, \; b-a $$
| {
"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 here.
| 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
$$\int\frac{u^2}{\sqrt{2-u^2}}du=\int\frac{2\sin^2t}{\sqrt2\cos t}\sqrt2\cos tdt=\int(1-\cos(2t))dt=\cdots. $$
which you can handle easily.
| {
"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
} |
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 $k=0,1,\dots,\alpha-1$.
Using this formula, we want to calculate $\sqrt[\sqrt{i}]{i}$. Note that we will be using this formula twice. Once to calculate $\alpha=\sqrt{i}$ and twice to calculate $\sqrt[\alpha]{i}$.
For $\sqrt{i}$, $r=|i|=1$ and $\theta=arg(z)=\pi/2$. By the formula, for $k=0,1$
$$
\alpha=\sqrt{i}
= 1^{1/2}\cdot \left(\sin\left(\dfrac{\pi/2+2\pi k}{2}\right)+i\cos\left(\dfrac{\pi/2+2\pi k}{2}\right)\right)
$$
For $k=0$,
$$
\alpha_0=\sqrt{i}
= \sin\left(\dfrac{\pi}{4}\right)+i\cos\left(\dfrac{\pi}{4}\right) =\dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}
$$
For $k=1$,
$$
\alpha_1=\sqrt{i}
= \sin\left(\dfrac{\pi/2+2\pi}{2}\right)+i\cos\left(\dfrac{\pi/2+2\pi}{2}\right)
= \sin\left(\dfrac{3\pi}{4}\right)+i\cos\left(\dfrac{3\pi}{4}\right) = \dfrac{\sqrt{2}}{2} - i \dfrac{\sqrt{2}}{2}
$$
For the time being, we will let $\alpha$ act as either $\alpha_0$ or $\alpha_2$ in our computation of $\sqrt[\alpha]{i}$.
\null
For $\sqrt[\alpha]{i}$, $r=|i|=1$ and $\theta=arg(z)=\pi/2$. By the formula, for $k=?$
$$
\alpha=\sqrt{i}
= 1^{1/2}\cdot \left(\sin\left(\dfrac{\pi/2+2\pi k}{\alpha}\right)+i\cos\left(\dfrac{\pi/2+2\pi k}{\alpha}\right)\right)
$$
My question is how am I suppose to use this formula if my $\alpha$'s are complex numbers?
| 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
$$i^{1/i^{1/2}}=\exp\left(\pm\frac{1-i}{\sqrt 2}\ln i\right)=\exp\left(\pm\frac{1-i}{\sqrt 2}i(\pi/2+2\pi\Bbb Z)\right)\\=\exp\left(\pm\frac{1+i}{\sqrt 2}(\pi/2+2\pi\Bbb Z)\right)$$
But Im not sure how to simplify this further.
| {
"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 $a \notin \{-1, -2,-3, \ldots \}.$
My attempt to prove it by rewritting \begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz
\end{align*}
and reduce it to an integral
like as here was unsuccessful.
| 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}{k} \binom{n+k}{n+1}^{-1} \\
&= 2 \, \sum_{k=0}^{n} \frac{(-1)^k \, n! (n+1)! (k-1)!}{(k-1)! (n-k)! (n+k)!} \\
&= 2 \, (n+1) \, \sum_{k=0}^{n} \frac{(-1)^{k}}{(n+1)_{-k} \, (n+1)_{k}} \\
&= 2 \, (n+1) \, \sum_{k=0}^{n} \frac{(-n)_{k}}{(n+1)_{k}} = 2 \, (n+1) \, {}_{2}F_{1}(-n,1; n+1; 1) \\
&= 2 \, (n+1) \, \frac{\Gamma(n+1) \Gamma(2n)}{\Gamma(2n+1) \Gamma(n)} = (n+1).
\end{align}
\begin{align}
S_{n} &= \sum_{k=0}^{n} \frac{(-1)^{k} (a+2k) n! (n+1)! (a+k-1)!}{k! (n-k)! (a+n+k)!} \\
&= B(a, n+1) \, \sum_{k=0}^{n} \frac{(a+2k) (-n)_{k} (a)_{k}}{k! (a+n+1)_{k}} \\
&= B(a,n+1) \, \left[ a \, {}_{2}F_{1}(-n, a; a+n+1; 1) + 2 \, \sum_{k=1}^{n} \frac{(-n)_{k} (a)_{k}}{(k-1)! (a+n+1)_{k}} \right] \\
&= B(a,n+1) \, \left[a \, {}_{2}F_{1}(-n, a; a+n+1; 1) + \frac{2 (a) (-n)}{(a+n+1)} \, {}_{2}F_{1}(1-n, a+1; a+n+2; 1) \right] \\
&= a \, B(a, n+1) \, \left[ \frac{\Gamma(a+n+1) \Gamma(2n+1)}{\Gamma(2n+a+1) \Gamma(n+1)} - \frac{\Gamma(a+n+1) \Gamma(2n+1)}{\Gamma(2n+a+1) \Gamma(n+1)} \right] \\
&= 0.
\end{align}
From these results it can be stated that
$$\sum_{i=0}^{n} (-1)^{i} (a+2i) \binom{n}{i} \binom{a+n+i}{n+1}^{-1} = (n+1) \delta_{a,0}$$
| {
"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
} |
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 at it seems one can consider an Indented Contour $\Gamma_{R}$, and define in $(1.2)$
$(1.2)$
$$\Gamma_{R}^{1}(t) = t \, \, \, \, \text{if} \, \, \, -R \leq t \leq -1/R $$
$$\Gamma_{R}^{2}(t) = e^{it}/R \, \, \, \, \text{if} \, \, \, \pi \leq t \leq 2 \pi$$
$$\Gamma_{R}^{3}(t) = e^{it}/R \, \, \, \, \text{if} \, \, \, \pi \leq t \leq 2 \pi$$
$$ \Gamma_{R}^{4}(t) = Re^{it} \, \, \, \, \text{if} \, \, \, 0 \leq t \leq \pi$$
Our original integral $\Psi(x) = \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}$ can be defined on $\Gamma_{R}$ as $\Psi(z)=\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}$ so our original integral now becomes
$$I = \oint_{\Gamma_{R}}\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}dz; $$
Then it seems via Cauchy's Theorem one has in $(1.3)$
$(1.3)$
$$\, \, \, \, \, \, \lim_{R \rightarrow \infty} \bigg(\oint_{\Gamma_{R}^{1}} \Psi(x)dx + \oint_{\Gamma_{R}^{2}} \Psi(z)dz + \oint_{\Gamma_{R}^{3}}\Psi(x)dx + \oint_{\Gamma_{R}^{4}} \Psi(z)dz \bigg) = 0 $$
To finish the argument it seems one would have to invoke Jordan's Lemma on the last integral in $(1.3)$, then parametrize the integral over $\Gamma_{R}^{4}$ then in $(1.4)$ we have
$(1.4)$
$$\lim_{R \rightarrow \infty} \oint_{\Gamma_{R}}\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}dz \rightarrow \int_{0}^{\pi}\frac{e^{ire^{i\theta}}}{re^{i\theta}}\frac{e^{ire^{i\theta}}}{re^{i\theta}/3}d\theta = \frac{\pi}{2}$$
| 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^2} \, dx - 4 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx, \tag1
\end{align*}
since $\sin (3x) = 3 \sin x - 4 \sin^3 x$.
For the first of these integrals, as is well-known
$$\frac{\pi}{2} = \int_0^\infty \frac{\sin x}{x} \, dx,$$
(this is just the Dirichlet integral) enforcing a substitution of $x \mapsto 2x$ gives
\begin{align*}
\frac{\pi}{2} &= \int_0^\infty \frac{\sin (2x)}{x} \, dx = \int_0^\infty \frac{2 \sin x \cos x}{x} \, dx\ = \int_0^\infty \frac{(\sin^2 x)'}{x} \, dx\\
&= \left [\frac{\sin^2 x}{x} \right ]_0^\infty + \int_0^\infty \frac{\sin^2 x}{x^2} \, dx = \int_0^\infty \frac{\sin^2 x}{x^2} \, dx.
\end{align*}
And for the second of the integrals, as we just found above
$$\frac{\pi}{2} = \int_0^\infty \frac{\sin^2 x}{x^2}dx.$$
Enforcing a substitution of $x \mapsto 2x$ gives
$$\frac{\pi}{2} = \frac{1}{2} \int_0^\infty \frac{\sin^2 (2x)}{x^2} \, dx.$$
Since
$$\sin^2 (2x) = 4 \cos^2 x \sin^2 x = 4(1 - \sin^2 x)\sin^2 x = 4 \sin^2 x - 4 \sin^4 x,$$
we have
$$\frac{\pi}{2} = 2 \int_0^\infty \frac{\sin^2 x}{x^2} \, dx - 2 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx = 2 \cdot \frac{\pi}{2} - 2 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx,$$
or
$$\int_0^\infty \frac{\sin^4 x}{x^2}dx = \frac{\pi}{4}.$$
So returning to (1) we have
$$\int_0^\infty \frac{\sin x}{x} \frac{\sin (x/3)}{x/3} \, dx = 3 \cdot \frac{\pi}{2} - 4 \cdot \frac{\pi}{4} = \frac{\pi}{2},$$
as required to show.
| {
"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
} |
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ 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} - \sqrt{x-4} = \sqrt{x-1} $
Squaring both Sides
$\mapsto x - 2 \sqrt{x^2 - 16}= -1$
$\mapsto \frac{x+1}{2} = \sqrt{x^2 - 16}$
Squaring both sides
$ \mapsto \frac{x^2 + 2x + 1}{4} = x^2 - 16$
$\mapsto 3x^2 -2x - 65 = 0$
Solving the quadratic equation
$$x = 5 OR \frac{-13}{3}$$
Checking in the initial equation we can see that $5$ is a valid root. But the second value that is given in the key to the book is $4$. How do I obtain $4$?
| 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 $$\sqrt{(x-4)(x+4)}=\left(\sqrt{x-4}\right)^2+\sqrt{(x-4)(x-1)},$$ which gives
$$x=4$$ or
$$\sqrt{x+4}=\sqrt{x-4}+\sqrt{x-1},$$ which is
$$x+4=x-4+x-1+2\sqrt{x^2-5x+4}$$ or
$$2\sqrt{x^2-5x+4}=9-x,$$ which gives $$4\leq x\leq9$$ and
$$4(x^2-5x+4)=(9-x)^2$$ or
$$3x^2-2x-65=0,$$ which gives $$x=5$$ and the answer:
$$\{4,5\}.$$
| {
"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
} |
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 eigenvectors. But, proceding I have some troubles:
$$
A= \begin{pmatrix}
-1 & 1 & 1 \\
0 & 0 & 0 \\
-1 & 1 & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y\\
z\\
\end{pmatrix}
= \begin{pmatrix}
0 \\
0 \\
0 \\
\end{pmatrix}
$$
Which cleary brings to two equal equations:
$$-x+y+z=0$$
I don't know how I should proceed. I can find a solution by trying some values but I don't like this method.
Which is the best and secure method to solve this problem? Thank you very much.
| 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\\ y\\z\end{pmatrix}\end{align}
You can take as a basis of the eigenspace, say the image of the canonical basis of $\mathbf R^2$, i.e.
$$v_1=\begin{pmatrix}1 \\1\\0\end{pmatrix},\quad v_2=\begin{pmatrix}1 \\0\\1\end{pmatrix}.$$
To obtain a Jordan basis, you need to complete it with a vector $v_3$ such that $(A-I)v_3=v_2$ In such a basis the matrix of the endomorphism associated to $A$ will be
$$J=\begin{pmatrix}1&0&0 \\0&1&1\\ 0&0&1\end{pmatrix} \quad\text{since }\enspace Au_1=u_1,\;Au_2=u_2,\;Au_3=u_3+u_2.$$
Now to solve the vector equation $(A-I)v_3=v_2$, we have to solve the linear system
$$\begin{cases}1=-x+y+z, \\0=0,\\1=-x+y+z,\end{cases}$$
which has an obvious solution: $x=0,\;y=1,\;z=0$, in other words $u_3=e_2$.
As a conclusion, $A$ has a Jprdan normal form in basis
$$\mathcal B=(e_1+e_2,\, e_1+e_3,\, e_2).$$
| {
"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
} |
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
$$\begin{align}
\frac{e_{n+1}(x)}{e_n(x)}&=\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}\\\\
&=\left(1+\frac{-x}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right) \tag 1\\\\
&\ge \left(1+\frac{-x}{n+x}\right)\left(1+\frac xn\right)\tag 2\\\\
&=1
\end{align}$$
where in going from (1) to (2) we used Bernoulli's Inequality. Note
that (2) is valid whenever $n>−x$ or $x>−n$. Since $e_n(x)$
monotonically increases and is bounded above by $e^x$, then $$e^x\ge
\left(1+\frac xn\right)^n \tag 3$$ for all $n\ge 1$.
But I don't know how do we get $(1)$.
| By the binimial theorem $$\left(1+\frac{x}{n}\right)^n=1+n\cdot\frac{x}{n}+\frac{n(n-1)}{2!}\cdot\frac{x^2}{n^2}+...+\frac{n(n-1)...(n-n+1)}{n!}\cdot\frac{x^n}{n^n}=$$
$$=1+x+\left(1-\frac{1}{n}\right)\frac{x^2}{2!}+...+\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{n-1}{n}\right)\frac{x^n}{n!}<$$
$$<1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}<1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+...=e^x.$$
| {
"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:
$2^y = 9 \implies \ln(2^y) = \ln(9) \implies y \cdot \ln(2) = \ln(9) \implies y= \frac{\ln(9)}{\ln(2)}$
And then simply substitute
$$\frac{3x+y}{6x-1} = \frac{3 \cdot\frac{\ln(18)}{3 \cdot \ln(2)}+\frac{\ln(9)}{\ln(2)}}{6 \cdot \frac{\ln(18)}{3 \cdot \ln(2)}-1}=\frac{\frac{\ln(18)+\ln(9)}{\ln(2)}}{\frac{2 \ln(18)-\ln(2)}{\ln(2)}} = \frac{\ln(18) + \ln(9)}{2\cdot \ln(18) - \ln(2)} = \frac{\ln(18 \cdot 9)}{\ln(\frac{18^2}{2})} = \frac{\ln(162)}{\ln(162)} = 1.$$
| {
"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
} |
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,
$$a + b + c + d = 8 \Longrightarrow 4 + c+ d = 8 \Longrightarrow a+b = c+d = 4$$
$$(a + b)(c + d) + ab + cd = 21$$
$$ab (c + d) + cd (a + b) = 20 \Longrightarrow 4ab + 4cd = 20 \Longrightarrow ab + cd = 5$$
$$abcd = 5$$
I can't figure out how to proceed.
| $x^4-8x^3+21x^2-20x+5=(x^2-4x+a)(x^2-4x+b)$
$\begin{cases}a+b+16=21 \\ -4a-4b=-20\\
ab=5 \end{cases}$
So $a+b=5$ and $ab=5$.
$a$, $b$ are the roots of $t^2-5t+5=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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 that $729=3^6$ and $243 = 3^5$, I tried simplifying the general terms by setting $x=3^{5^n}$ but it seems to give no simplifications.
Thanks in advance for any advice / ideas.
| 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 terms will only increase the sum further.
To show that all terms are in fact positive, it is enough to note that with $\,x=3^{5^n} \ge 3\,$ both:
*
*numerator $\,5^n(x^5-5x+4) = 5^n(x-1)^2(x^3+2x^2+3x+4) \ge 0$;
*denominator $\,x^6-x^5-5x+1 = u^6 + 11 u^5 + 50 u^4 + 120 u^3 + 160 u^2 + 107 u + 23\ge 0\,$ where $\,u=x-2 \ge 0\,$.
| {
"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 $i$ is not a real number. Can anyone help?
| 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
} |
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 leads would be greatly appreciated!
| 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}\\
&=\frac{x-(k+1)x^{k+1}+kx^{k+2}}{(1-x)^2}\\
\end{align*}
Then differentiate again and multiply by $x$. Put $x=\frac{2}{3}$. Take $k\to\infty$.
| {
"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
} |
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 $$3(n-3)(n-5)\sqrt{n^2-14n+61}<(n-4)\sqrt{(3n^2-19n+34)^2-48(n-3)(n-4)^2}$$ but couldn't succeed since first inequality is not true.
Kindly help.
| 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$$
f(n) < 3(n - 3)(n - 5)\left(n - 1 + n - \frac{19}{3}\right) = 6n^3 - 70n^2 + 266n - 330. \tag{1}
$$
Also,\begin{align*}
&\mathrel{\phantom{=}} (3n^2 - 19n + 34)^2 - 48(n - 3)(n - 4)^2\\
&= \left(3n^2 - 27n + \frac{182}{3}\right)^2 + 64n - \frac{1984}{9} > \left(3n^2 - 27n + \frac{182}{3}\right)^2,
\end{align*}
then\begin{align*}
g(n) &= (n - 4)\left(3n^2 - 19n + 34 + \sqrt{(3n^2 - 19n + 34)^2 - 48(n - 3)(n - 4)^2}\,\right)\\
&> (n - 4)\left(3n^2 - 19n + 34 + 3n^2 - 27n + \frac{182}{3}\right)\\
&= (n - 4)\left(6n^2 - 46n + \frac{284}{3}\right) = 6n^3 - 70n^2 + \frac{836}{3} n - \frac{1136}{3}. \tag{2}
\end{align*}
Combining (1) and (2),$$
g(n) - f(n) > \frac{38}{3} n - \frac{146}{3} > 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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
$$\begin{align}
-\cos B &=\cos (180+B) \\
-\cos B &=\cos(180+(180-(A+C)) \\
-\cos B &=\cos(360-(A+C)) \\
-\cos B &=\cos(A+C) \\
-\cos B &=\cos A \cos C - \sin A \sin C \\
\cos B &= \sin A \sin C - \cos A \cos C \\
\cos B &= -\cos (A+C)
\end{align}$$
Can someone confirm that my working is correct?
Thanks!
| 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",
"answer_count": 2,
"answer_id": 1
} |
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 the RHS as a power series and coming up with a differential equation but it doesn't really help as I end up with an intractable integral.
| Another way. from the properties of the Pochhammer symbol and Beta function we can note that $$\sum_{n\geq0}\frac{n!}{\left(x+1\right)\left(x+2\right)\cdots\left(x+n+1\right)}=\sum_{n\geq0}\frac{n!}{\left(x+1\right)_{n+1}}=\sum_{n\geq0}\frac{\Gamma\left(n+1\right)\Gamma\left(x+1\right)}{\Gamma\left(x+n+2\right)}$$ $$=\sum_{n\geq0}\int_{0}^{1}u^{n}\left(1-u\right)^{x}du=\int_{0}^{1}\left(1-u\right)^{x-1}du=\frac{1}{x}.$$
| {
"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 I development wrong in this equation ?
| 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
} |
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, $\{f(x)\}^2 - \frac{f(x)}{x} -\frac{1}{x}$.
from there I have rearranged the equation to,
$\left(f(x)^2 -\frac{1}{2x}\right)^2 = \frac{1}{4x^2} - \frac{1}{x}$ and therefore $f(x) = \frac{1}{2x} \pm \sqrt{\frac{1}{4x^2}- \frac{1}{x} }$.
I then took $\frac{1}{2x^2}$ out as a common factor to give me $\frac{1}{2x^2}\left(1 \pm \sqrt {\frac{1}{2x} -2 }\right) $.
I then took the square root and have rearranged that to $(1-4x)^{1/2}$ which I have expanded to $$ 1 + \sum_{k=1}^{\infty} \frac{-2(2k-2)!}{k!(k-1)!} x^k.$$
I have no idea if this is correct or if I'm even going in the right direction but any help at all is greatly appreciated.
| 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?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$1^2-2^2+3^2-4^2+…-2016^2+2017^2=2017k$ (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 $2017$:
$$\require{cancel}{-\left({2016\,\cdot\,2017\over 2}\right)+2017^2 \over 2017} = k$$
$${\cancelto{-1008}{{-\left({2016\,\cdot\,2017\over 2}\right) \over 2017}} + \cancelto{2017}{{2017^2\over 2017}}} = k$$
$$-1008+2017=k$$
$$k=1009$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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 to answer this question since so far we have only defined linear independance for vectors, and these are matrices...
| 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 $(1,1)$ in the matrix $I$ to annihilate the coefficients at $(1,1)$ in $A$ and $A^2$:
$$\underbrace{\begin{pmatrix} 0 & 2 \\ 3 & 3 \end{pmatrix}}_{A-I},\underbrace{\begin{pmatrix} 0 & 10 \\ 15 & 15 \end{pmatrix}}_{A^2-7I}$$
Now notice that $A^2-7I = 5(A-I)$. Therefore:
$$0 = A^2-7I - 5(A-I) = A^2-5A-2I$$
Since $A$ is obviously not a multiple of $I$, we conclude that the smallest $n$ such that $\{I, A, \ldots, A^n\}$ is lineraly dependent is $n = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
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.
$$
\begin{align}
\cos^22x&=(\cos^2x-\sin^2x)^2\\
&=\cos^4x+\sin^4x-2\sin^2\cos^2x\\
&=\cos^4+(1-\cos^2x)^2-2(1-\cos^2)\cos^2x\\
&=\cos^4+1-2\cos^2x+\cos^4x-2\cos^2x+2\cos^4x\\
&=4\cos^4x-4\cos^2x+1
\end{align}
$$
Without knowledge of other trigonometric identities, $\cos3x$ can be derived using only Ptolemy's identities. However for the sake of brevity, let $\cos 3x=4\cos^3x-3\cos x$:
$$
\begin{align}
\cos^23x&=(4\cos^3x-3\cos x)^2\\
&=16\cos^6x+4\cos^2x-24\cos^4x
\end{align}
$$
Therefore, the original equation can be written as:
$$\cos^2x+4\cos^4x-4\cos^2x+1+16\cos^6x+4\cos^2x-24\cos^4x-1=0$$
$$-20\cos^4x+6\cos^2x+16\cos^6x=0$$
Letting $y=\cos x$, we now have a polynomial equation:
$$-20y^4+6y^2+16y^6=0$$
$$y^2(-20y^2+6y+16y^4)=0\Rightarrow y^2=0 \Rightarrow x=\cos^{-1}0=\bbox[yellow,10px]{90^\circ}$$
From one of the factors above, we let $z=y^2$, and we have the quadratic equation:
$$16z^2-20z+6=0\Rightarrow 8z^2-10z+3=0$$
$$(8z-6)(z-\frac12)=0\Rightarrow z=\frac34 \& \ z=\frac12$$
Since $z=y^2$ and $y=\cos x$ we have:
$$\biggl( y\rightarrow\pm\frac{\sqrt{3}}{2}, y\rightarrow\pm\frac{\sqrt{2}}2 \biggr)\Rightarrow \biggl(x\rightarrow\cos^{-1}\pm\frac{\sqrt{3}}{2},x\rightarrow\cos^{-1}\pm\frac{\sqrt{2}}2\biggr)$$
And thus the complete set of solution is:
$$\bbox[yellow, 5px]{90^\circ, 30^\circ, 150^\circ, 45^\circ, 135^\circ}$$
As I do not have the copy of the answers, I still hope you can verify the accuracy of my solution.
But more importantly...
Seeing the values of $x$, is there a more intuitive and simpler way of finding $x$ that does away with the lengthy computation?
| 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 that
$$\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{6}$$
all satisfy the equation, and produce three different values for $\cos^2 x$, namely $0, \frac{1}{2}, \frac{3}{4}$.
Lastly, we solve the resulting equations
$$\cos^2 x = 0$$
$$\cos^2 x = \frac{1}{2}$$
$$\cos^2 x = \frac{3}{4}$$
separately. We conclude that our solutions are:
$$x=\frac{(2k+1)\pi}{2}, \frac{(2k+1)\pi}{4}, \frac{(6k+1)\pi}{6}, \frac{(6k+5)\pi}{6}, \forall k \in \mathbb{Z}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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 Induction $(n^3+5n)$ is divisible by $6$
Here is what I have done
Assume $n=k$ is true
$\sum_{1}^{k} k =\dfrac{(k^3+5k)}{6}$
Now assume $n=k+1$ is true
$\sum_{1}^{k+1} k+1 =\dfrac{(k+1)^3+5(k+1)}{6}$
Then now
$\dfrac{(k+1)^3+5(k+1)}{6}=\sum_{1}^{k} k + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + \dfrac{(6k+6)}{6}$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+11k+6)}{6}$
But the other side doesnt equate (LHS)
$\dfrac{(k^3+3k^2+3k+1)+5k+5}{6}$
$\dfrac{(k^3+3k^2+8k+6)}{6}\ne\dfrac{(k^3+11k+6)}{6}$
I hope my method was clear enough so you can see where I went wrong. It would be much more use to me if you solved it as I learn better from looking at solutions and then applying them to other questions.
| $$n^3+5n = n(n^2+5)$$
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 factors, then so is $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
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: $$=\frac 1 {8\pi}[\pi^2\frac i n e^{-2\pi i n}+\frac {2\pi}{n^2}e^{-2\pi i n}-\frac {2i}{n^3}e^{-2\pi i n}-\pi^2\frac i n+\frac {2\pi}{n^2}+\frac {2i}{n^3}]\tag {*} $$
Also $e^{-2\pi i n}=\cos{(-2\pi n)}+i\sin{(-2\pi n)}=1$ for any $n\in \{\cdots ,-3,-2,-1,1,2,3,\cdots\}$
This in turn leads me to get $\frac 1 {2n^2}$ in $(*)$
And when $n=0,$
$$\hat f(0)=\frac 1 {2\pi}\int_0^{2\pi}\frac {(\pi-\theta)^2} 4 d\theta=\frac 1 {2\pi}\frac {(\theta-\pi)^3} {12}|_0^{2\pi}=\frac{\pi^2}{12}$$
So, how do I combine these two results to write in the Fourier series?
It seems that I pretty much there since the result is $$f(\theta) \thicksim \frac{\pi^2}{12}+ \sum_{n=1}^\infty \frac {\cos (n\theta)}{n^2}$$
| 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 solution. Your problem was you were comparing the vector in two different bases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why did this extraneous root creep into the solution? I was solving this equation and proceeded as follows:
$$\arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$$
$$\implies \arcsin(1-x) = \frac{π}{2} + 2\arcsin (x)$$
$$\implies \sin (\arcsin (1-x)) = \sin \left( \frac {π}{2} + 2\arcsin (x)\right)$$
$$\implies (1-x) = \cos \left(2\arcsin(x)\right)$$
$$\require{cancel}\implies \cancel{1}-x =\cancel{1}- 2 \left(\sin\arcsin (x)\right)^2$$
$$\implies -x = -2x^2$$
$$\therefore x = 0$$
Or $$ x=\frac{1}{2}$$
However, $x=\frac{1}{2}$ doesn't satisfy the original equation.
I understand that extraneous roots do creep in while solving inverse trig problems, but I wonder why it crept in here.
If possible (if it doesn't make the question too broad), I'd like to know the general causes for the occurrence of extraneous roots in inverse trig equations, too.
| 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 $\sin \left( \frac {π}{6} \right)=\sin \left( \frac {5π}{6} \right) =\frac12$
but your second line would not be a correct equality when $x=\frac12$ since $\frac {π}{6}\not = \frac {5π}{6}$
and it is this use of $\sin$ which creates an equality which was not in the original expression.
$\arcsin (1-x) - 2\arcsin (x) = \frac{π}{2} \Rightarrow x=0 \text{ or } x=\frac12$ is a correct statement, but checking shows $x=0 \Rightarrow \arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$ while $x=\frac12 \not \Rightarrow \arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
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 the basis for the null space, but what is the hence way?
(Further Mathematics Paper 1, Nov 1997)
The linear transformations $T_1:\mathbb{R}^4\rightarrow\mathbb{R}^4$, $T_2:\mathbb{R}^4\rightarrow\mathbb{R}^4$, $T_3:\mathbb{R}^4\rightarrow\mathbb{R}^4$ are represented by the matrices $\mathbf{M}_1$, $\mathbf{M}_2$ and $\mathbf{M}_2\mathbf{M}_1$ respectively, where
$$\mathbf{M}_1 = \begin{pmatrix}1&4&-5&8\\0&-4&1&-5\\-1&-3&0&-2\\0&1&1&0\end{pmatrix}\text{ and }\mathbf{M}_2 = \begin{pmatrix}1&2&-1&3\\1&0&4&5\\3&2&7&13\\1&4&-6&1\end{pmatrix}$$
*
*Show that the rank of $\mathbf{M}_1$ is equal to 3.
*Write down a basis for $R_1$, the range space of $T_1$, and find a basis for the null space of $T_1$.
*Find a basis for $K_2$, the null space of $T_2$, and hence show that $K_2$ is a subspace of $R_1$.
*Hence, or otherwise, find three linearly independent vectors in the null space of $T_3$.
Answers to part 1–3
Basis for $R_1 = \left\{\begin{pmatrix}1\\0\\-1\\0\end{pmatrix},\begin{pmatrix}4\\-4\\-3\\1\end{pmatrix},\begin{pmatrix}-5\\1\\0\\1\end{pmatrix}\right\}$
Basis for the null space of $T_1 = \left\{\begin{pmatrix}1\\-1\\1\\1\end{pmatrix}\right\}$
Basis for $K_2 = \left\{\begin{pmatrix}-8\\5\\2\\0\end{pmatrix},\begin{pmatrix}-5\\1\\0\\1\end{pmatrix}\right\}$
Part 4
Three linearly independent vectors in the null space of $T_3$ are
*
*$\begin{pmatrix}1\\-1\\1\\1\end{pmatrix}$ (from the basis for the null space of $T_1$)
*$\begin{pmatrix}4\\-4\\-3\\1\end{pmatrix}$ (this is just a coincidence, as far as I can tell)
*???
| First note that $\ker T_1 \subseteq \ker T_2T_1$ so you already have one vector, the one in $\ker T_1$.
$$\ker T_1 = \operatorname{span}\left\{\begin{pmatrix}1\\-1\\1\\1\end{pmatrix}\right\}, \quad\ker T_2 = \operatorname{span}\left\{\begin{pmatrix}-5\\1\\0\\1\end{pmatrix}, \begin{pmatrix}8\\-5\\-2\\0\end{pmatrix}\right\}$$
Now notice that the basis vectors for $\ker T_2$ are in fact the third and the fourth columns of the matrix $M_1$, respectively. Therefore
$$\begin{pmatrix}-5\\1\\0\\1\end{pmatrix} = T_1e_3 = T_1\begin{pmatrix}0\\0\\1\\0\end{pmatrix}$$
$$\begin{pmatrix}8\\-5\\-2\\0\end{pmatrix} = T_1e_4 = T_1\begin{pmatrix}0\\0\\0\\1\end{pmatrix}$$
So
$$\left\{\begin{pmatrix}1\\-1\\1\\1\end{pmatrix}, \begin{pmatrix}0\\0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\0\\1\end{pmatrix}\right\}$$
are three vectors in $\ker T_2T_1$ which are clearly linearly independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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 $\exists t=a/2, \, b = \sqrt[2]{\frac {(a^2+c^2)}{2}} => 2t$.
As an example, $a=c=8, t=4, b=8$.
I am giving an idea for a particular case only, & am unable to pursue.
| 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2694706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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: $$3 + 2\left(n + \sum_{i=1}^{2n-1}(i+1)\right) = (2n+1)^2.$$
How must we be able to prove this, though?
My Attempt:
Assume that the equation is true for all natural $n$, then we can similarly write the same for $n+1$.
$$\begin{align} 3 + 2\left(n+1+\sum_{i=1}^{2(n+1)-1}(i+1)\right) &= \big(2(n+1)+1\big)^2 \\ \Leftrightarrow 3 + 2\left(n+1+\sum_{i=1}^{2n+1}(i+1)\right) &= 3 + 2n + 2\left(1+\sum_{i=1}^{2n+1}(i+1)\right) \\ &= (2n+3)^2 \\ \Leftrightarrow (2n+3)^2 - (2n + 3) &= (2n+3)(2n + 3 - 1) \\ &= 2(2n+3)(n+1) \\ &= 2\left(1+\sum_{i=1}^{2n+1}(i+1)\right) \\ \Leftrightarrow (2n+3)(n+1) &= 1+\sum_{i=1}^{2n+1}(i+1) \\ &= \sum_{i=0}^{2n+1}(i+1) \\ &= \sum_{i=1}^{2n+1}i.\end{align}$$
I know that, $$\sum_{i=1}^ki = \frac{k(k+1)}{2}.\tag{The Triangular Numbers}$$ Therefore, by substituting $k = 2n+1$, we get $$\begin{align} \sum_{i=1}^{2n+1}i &= \frac{(2n+1)(2n+2)}{2} \\ &= \frac{2(n+1)(2n+1)}{2} \\ &= (n+1)(2n+1).\end{align}$$ It follows, then, that $(2n+3)(n+1) = (2n+1)(n+1)$ which means that $2n+3 = 2n+1$ and absurdly we get that $n=n+1$ or $0=1$.
I obviously did something wrong, but I don't know where. Can someone help me out please?
Thank you in advance.
| 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
} |
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
$$a_n = (0, 3, \dfrac{1}{2}, \dfrac{5}{3}, \dfrac{2}{3}, \dfrac{7}{5}, \dfrac{3}{4}, ...)$$.
Then find the limit of this sequence.
So far, I have tried like this(Idk if its right)
First thing, I have written sequence as like this
$$a_n = (0, 3, \dfrac{1}{2}, \dfrac{5}{3}, \dfrac{2}{3}, \dfrac{7}{5}, \dfrac{3}{4}, ...)$$
$$indexes \rightarrow 1, 2, 3,\hspace{0.5mm} 4,\hspace{1mm}5, 6, \hspace{0.5mm}7, 8, \hspace{2.2cm}$$
Now, I can get/ divide indexes in odd and even.
For Odd indexes:
$$a_n = (0, \dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}, ...)$$
$$indexes \rightarrow \hspace{4mm}1,\hspace{2mm} 2, \hspace{1mm}3,\hspace{2.5mm} 4,..\hspace{2cm}$$
For Even indexes:
$$a_n = (3, \dfrac{5}{3}, \dfrac{7}{5}, ...)$$
$$indexes \rightarrow \hspace{4mm}1,\hspace{2mm} 2, \hspace{1mm}3,..\hspace{2cm}$$
So Now, I know the sequence of odd and even indexes(I think):
$$Odd: \dfrac{n}{n+1}\ and\ even\ \dfrac{n+2}{n}\ \forall n : n=2(k+1)$$
I have arrived here. I don't know how to proceed. Thanks For all help.
| $$a_{2n+1}=\frac {2n+3}{2n+1} $$
$$a_{2n}=\frac {n}{n+1} $$
$$\lim a_{2n}=1=\lim a_{2n+1} \implies $$
$$\lim a_n=1$$
| {
"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
} |
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 = x^2-1$ $dt = 2x dx$
$v_1$ = $\frac{1}{2}\int \sqrt{t}dt$
$v_1$ = $\frac{1}{2}(\frac{2}{3}t^\frac{3}{2}) + c$
$v_1$ = $\frac{1}{3} (x^2-1)^\frac{3}{2} + c$
$uv - \int vdu$
= $\frac{1}{3} (x^2-1)^\frac{3}{2}x^2 - \int 2x\frac{1}{3} (x^2-1)^\frac{3}{2}dx$
$u_2 = 2x$ $dv_2 = \frac{1}{3} (x^2-1)^\frac{3}{2}dx$
$du_2 = 2dx$
| 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",
"answer_count": 3,
"answer_id": 1
} |
$\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} \sin^{3\over 2} x+{1\over 3} \sin^3x+\cos^2 x+c
\end{align}
True ?
| 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
$\,u= sinx \implies du = cos(x)dx$
$I= \int (1-u^2)^2.u^\frac12\,du \\I = \int u^\frac12 + u^{4+\frac12} - 2u^{2+\frac12}\,du\\I = \int u^\frac12+ u^\frac92 - 2u^\frac52\,du\\I = \frac23u^\frac32 + \frac{2}{11}u^\frac{11}{2}-\frac47u^\frac72 +C \\I= \frac23(sin(x))^\frac32+\frac{2}{11}(sin(x))^\frac{11}{2}- \frac47(sin(x))^\frac74 +C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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$\,$ $\sin(2x)=2\sin(x)\cos(x)$,
the equation is now $(1−\sqrt2\sin(x))(\cos^2(x)−\sin^2(x)+2\sin(x)\cos(x))=\frac12$
Help Required
Thanks
| 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: $\left ( 1- \sqrt{2} (\pm\sqrt{\frac{1-\cos\theta}2})\right )\left ( \cos y+ \sin y \right )= \frac{1}{2}$, which simplifies to:
$$\left ( 1 \pm\sqrt{{1-\cos\theta}}\right )\left ( \cos y+ \sin y \right )= \frac{1}{2}$$
Solving for the first case:$\left ( 1 -\sqrt{{1-\cos\theta}}\right )\left ( \cos y+ \sin y \right )= \frac{1}{2}$
$$\begin{align}
\cos y-(\cos y) \sqrt{1-\cos (y)}+\sin y-(\sin y) \sqrt{1-\cos (y)}&=\frac{1}{2}\\
-(\cos y) \sqrt{1-\cos (y)}-(\sin y) \sqrt{1-\cos (y)}&=\frac{1}{2}-\cos y-\sin y\\
\left(-(\cos y) \sqrt{1-\cos (y)}-(\sin y) \sqrt{1-\cos (y)}\right)^2&=\left(\frac{1}{2}-\cos y-\sin y\right)^2\\
\end{align}$$
We then evaluate each side:
$$\begin{align}
\left(-(\cos y) \sqrt{1-\cos (y)}-(\sin y) \sqrt{1-\cos (y)}\right)^2&=\cos ^2 y-\cos ^3 y+2 (\cos y) (\sin y)-2 \left(\cos ^2 y\right) (\sin y)+\sin ^2 y-(\cos y) \left(\sin ^2 y\right)\\
\left(\frac{1}{2}-\cos y-\sin y\right)^2&=y \sin ^2-y \sin +y \cos ^2-y \cos +2 (y \sin ) (y \cos )+\frac{1}{4}\\
\end{align}$$
Continuing the computation, we get:
$$\begin{align}
-\frac{1}{4}+\cos y-\cos ^3 y+\sin y-2 \left(\cos ^2 y\right) (\sin y)-(\cos y) \left(\sin ^2 y\right)&=0\\
-1+4 (\cos y)-4 \left(\cos ^3 y\right)+4 (\sin y)-8 \left(\cos ^2 y\right) (\sin y)-4 (\cos y) \left(\sin ^2 y\right)&=0\\
-1-4 (\sin y)+8 \left(\sin ^3 y\right)&=0\\
(2 (\sin y)+1) \left(-1-2 (\sin y)+4 \left(\sin ^2 y\right)\right)&=0\\
\end{align}$$
We get the correct answers as:
$$y=\begin{cases}
2 \pi n+\frac{7 \pi }{6}\qquad{n \in \mathbb{Z}}\\
2 \pi n+\frac{3 \pi }{10}\\
2 \pi n+\frac{11 \pi }{10}\\
2 \pi n+\frac{19 \pi }{10}\\
\end{cases}$$
And since $x=\frac y2$, then the first set of solutions for $x$ will be:
$$x=\begin{cases}
2 \pi n+\frac{7 \pi }{12}\qquad{n \in \mathbb{Z}}\\
2 \pi n+\frac{3 \pi }{20}\\
2 \pi n+\frac{11 \pi }{20}\\
2 \pi n+\frac{19 \pi }{20}\\
\end{cases}$$
The second case:$\left ( 1 +\sqrt{{1-\cos\theta}}\right )\left ( \cos y+ \sin y \right )= \frac{1}{2}$ returns the following answers:
$$y=\begin{cases}
2 \pi n-\frac{ \pi }{6}\qquad{n \in \mathbb{Z}}\\
2 \pi n+\frac{7 \pi }{10}\\
\end{cases}$$
However,the second answer for $x$ does not hold for the original equation. And thus, our final solution set contains:
$$\therefore x=\begin{cases}
2 \pi n-\frac{ \pi }{12}\qquad{n \in \mathbb{Z}}\\
2 \pi n+\frac{7 \pi }{12}\\
2 \pi n+\frac{3 \pi }{20}\\
2 \pi n+\frac{11 \pi }{20}\\
2 \pi n+\frac{19 \pi }{20}\\
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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)^3-3(x+y)(x+z)(y+z)
\end{equation}
that turns out to be
\begin{equation}
(x+y+z-1)(x+y+z)(x+y+z-1)=3(x+y)(x+z)(y+z)
\end{equation}
plug in $x+y+z$,
we will get
\begin{equation}
\sqrt[3]{2}-1 - \sqrt[3]{\sqrt[3]{2}-1 } = 3(x+y)(x+z)(y+z)
\end{equation}
From now I stuck to solve for the rational numbers of $a,b,c$. Could anyone show me the way how to continue to solve it?
| 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 $\sqrt[3]{\sqrt[3]{2}-1}$, solve
$$x^3 -\frac13 x^2-\frac2{27}x + \frac8{729}=0\implies
(x_1, x_2, x_3)= (\frac19,-\frac29,\frac49)$$
to obtain
\begin{align}
\sqrt[3]{\sqrt[3]{2}-1}&=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}
\end{align}
The general formula given above can be used as well to denest other nested cubic radicals, for instance
$$\sqrt[3]{21\sqrt[3]{6}-17}=\sqrt[3]{18}+\sqrt[3]{4}-\sqrt[3]{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
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. But how do I use this to get the answer?
| 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 $a = b\pm 2$ then $c = d\mp 2$ and there $ 2*2*2 = 8$ ways.
And if $a = b\pm 4$ then either $a = 1;b=4;c=4;d=1$ or $a=4;b=1;c=1;d=4$. $2$ ways.
So $16 + 18 + 8 + 2 = 44$ ways.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$A289B$ 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
$$ A2890 = A\times10000 + 2890 $$
since it is divisible by $90$, the sum of the remainders of $A \times 1000$ and $2890$ being divided by 90, must be divisible by 90.
We have $mod(2890, 90) = 10$.
Now we may find $A$ such that $mod(A \times 10000, 90) = 80$.
$$mod(10000, 90) = 10 \implies 10000 = (90 \times 111 + 10) $$
so $A$ must be $8$, since
$$ 8 \times 10000 + 2890 = 8 \times (90 \times 111 + 10) + (90 \times 32 + 10) = 90 \times (888+32) + 90 $$
is divided by 90.
$$A+B = 8$$
Is this the best approach..? Thanks.
| 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 $A+1 \equiv 0 \mod 9$ so $A = 8$.
So $A+B = 8+0= 8$.
That's assuming you are not aware that if a number is divisible by $9$ then the sum of it's digits is divisible by $9$. Which can be proven the exact same way.
$n = \sum_{k=0}n a_k 10^k = \sum_{k=0} a_k (10^k - 1) + \sum_{k=0}a_k$ and as $(10^k - 1) = 9*(\underbrace{1111....111}_{k}) \equiv 0 \mod 9$, we may conclude that $n \mod 9 \equiv \sum_{k=0}a_k \mod 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2709305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"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,
$$
\begin{aligned}
f_x(x,y)&=\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2}\cdot\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\\
&=\frac{1+y^2}{(1-xy)^2+(x+y)^2}\\
&=\frac{1+y^2}{1+y^2+x^2+x^2y^2}
\end{aligned}
$$
But the given answer is $\frac{1}{x^2+1}$.
How?
| 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:
$$\sum_{n=1}^k1=\frac{k}{1}$$
$$\sum_{n=1}^kn=\frac{k}{1}\cdot\frac{k+1}{2}$$
$$\sum_{n=1}^kn^2=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}$$
At first, we could guess that the next summation is:
$$\sum_{n=1}^kn^3 ?=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}\cdot\frac{3k+1}{4}$$
Yet this is off. However, it is off geometrically. Notice:
$$\left(\sum_{n=1}^kn^3\right)-\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2k+1}{3}\cdot\frac{3k+1}{4}=error$$
$$k=1, r=0$$
$$k=2, r=0.25$$
$$k=3, r=1$$
$$k=4, r=2.5$$
$$k=5, r=5$$
$$k=6, r=8.75$$
...
Consider the ratios of the errors:
$$er(k)=\frac{r(k+1)}{r(k)}$$
$$k=1, r=udf$$
$$k=2, r=4$$
$$k=3, r=2.5$$
$$k=4, r=2$$
$$k=5, r=1.75$$
$$k=6, r=1.6$$
Then, rewriting the error as a function of n starting at k = 5:
$$1.75=2.5-\frac{1.5}{2}$$
$$1.6=2.5-\frac{1.5}{2}-\frac{1.5}{10}$$
$$1.5=2.5-\frac{1.5}{2}-\frac{1.5}{10}-\frac{1.5}{15}$$
$$1.42857=2.5-\frac{1.5}{2}-\frac{1.5}{10}-\frac{1.5}{15}-\frac{1.5}{21}$$
The denominators in the series are from pascals triangle: (3rd columns, or dependent again on the triangular numbers)
Then the total formula for equating the two is:
$$\left(\frac{k}{1}\cdot\frac{\left(k+1\right)}{2}\cdot\frac{\left(2k+1\right)}{3}\cdot\frac{\left(3k+1\right)}{4}\right)-\left(\sum_{n=1}^kn^3\right)+\frac{1}{24}\left(k-1\right)k\left(k+1\right)=0$$
Super interesting!
At least, I thought it was interesting how this the error is related back to the previous power's formula. Am I missing something obvious? Any input is greatly appreciated. (I'm not smart, so in the likely event I missed something obvious try not to be too harsh)
Update:
For the next power (4), I found the formula with trial and error:
$$\left(\frac{k}{1}\cdot\frac{\left(k+1\right)}{2}\cdot\frac{\left(2k+1\right)}{3}\cdot\frac{\left(3k+1\right)}{4}\cdot\frac{\left(4k+1\right)}{5}\right)+\frac{1}{24}\left(k-1\right)k\left(k+1\right)+\frac{1}{12}\left(k-1\right)k\left(k+1\right)k$$
Any ideas on power (5) and so on? I'll continue to try and generalize it.
| 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 $$\sum_1^n k^2 $$
by using $$(k+1)^3 - k^3 =3k^2+ 3k+1$$
and so forth.
| {
"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 the system $Ar=0$, with $r=[a;b;c;d]$ with $a≠b≠c≠d $. The system has not the trivial solution, but what do I have to think?
| 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:
$$
a:b:c:d=\frac{1}{1+x}:\frac{1}{1+y}:\frac{1}{1+z}:\frac{1}{1+t}.\tag{2}
$$
Multiplying the RHS with $(1+x)(1+y)(1+z)(1+t)$ results in the integer solution:
$$
\begin{cases}
a=(1+y)(1+z)(1+t),\\
b=(1+x)(1+z)(1+t),\\
c=(1+x)(1+y)(1+t),\\
d=(1+x)(1+y)(1+z).
\end{cases}\tag{3}
$$
One can easily check that substitution of the values from (3) into any of equations (1) results in the expression for $\det A$, i.e. $0$.
As seen from (2) the condition $a\ne b\ne c\ne d$ implies $x\ne y\ne z\ne t$. Though it is not immediately obvious, by brute force one finds that six such solutions indeed exist:
$$
(1,2,6,41),\\
(1,2,7,23),\\
(1,2,8,17),\\
(1,2,9,14),\\
(1,3,4,19),\\
(1,3,5,11).
$$
The corresponding vectors $(a,b,c,d)$ can be computed from (3). For example for $(1,3,5,11)$ it is:
$$
(4\times6\times12,\,2\times6\times12,\,2\times4\times12,\,2\times4\times6)\sim(6,3,2,1).
$$
| {
"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:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)d\left(\frac{\sin x}{1+\cos x}\right)$$
$$=\int_{0}^{\frac{\pi}{2}}x^2d\log\left(\frac{\sin x}{1+\cos x}\right)=x^2\log\left(\frac{\sin x}{1+\cos x}\right)|_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{\sin x}{1+\cos x}\right)dx$$
$$=0+2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{1+\cos x}{\sin x}\right)dx=2\int_{0}^{\frac{\pi}{2}}x\log\left(1+\cos x\right)dx-2\int_{0}^{\frac{\pi}{2}}x\log\left(\sin x\right)dx$$
$$=2\int_{0}^{\frac{\pi}{2}}x\log\cot \left(\frac{x}{2}\right)dx=8\int_{0}^{\frac{\pi}{4}}x\log\cot xdx$$
but I can't proceed next step,help me,thanks.
| 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 magnitude of $\frac{1}{\sin z}$ decays exponentially to zero as $\Im(z) \to \infty$, the integral vanishes on the top of the contour as $R \to \infty$.
Therefore, integrating around the contour and then letting $R \to \infty$, we get $$\int_{0}^{\pi/2} \frac{x^{2}}{\sin x} \, \mathrm dx + \int_{0}^{\infty} \frac{(\pi /2 + it)^{2}}{\cosh t} \, i \, \mathrm dt + \int_{\infty}^{0} \frac{(it)^{2}}{i \sinh t} \, i \, \mathrm dt = 0. $$
And equating the real parts on both sides the equation, we get $$\int_{0}^{\pi/2} \frac{x^{2}}{\sin x} \, \mathrm dx = \pi \int_{0}^{\infty} \frac{t}{\cosh t} \, \mathrm dt - \int_{0}^{\infty} \frac{t^{2}}{\sinh t} \, \mathrm dt. $$
An integral representation of the Dirichlet beta function is $$\beta(s) = \frac{1}{2 \Gamma(s)}\int_{0}^{\infty} \frac{t^{s-1}}{\cosh t} \, \mathrm dt, \quad \Re(s) >0.$$
And an integral representation of the Dirichlet lambda function is $$\lambda(s) = \frac{1}{2 \Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{\sinh t} \, \mathrm dt, \quad \Re(s) >1.$$
Therefore, $$\begin{align} \int_{0}^{\pi/2} \frac{x^{2}}{\sin x} \, \mathrm dx &= 2 \pi \Gamma(2) \beta(2) - 2 \Gamma(3) \lambda(3) \\ &= 2 \pi \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} -4 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{3}} \\ &= 2 \pi G - 4 \left(1-2^{-3} \right) \zeta(3) \\ &= 2 \pi G - \frac{7}{2} \zeta(3). \end{align}$$
The same approach also shows that $$ \int_{0}^{\pi/2} \frac{x}{\sin x} \, \mathrm dx = \int_{0}^{\infty} \frac{t}{\cosh t} \, \mathrm dt =2 \Gamma(2) \beta(2) =2G $$ and
$$\begin{align} \int_{0}^{\pi/2} \frac{x^{3}}{\sin x} \, \mathrm dx &= \frac{3 \pi^{2}}{4} \int_{0}^{\infty} \frac{t}{\cosh t} \, \mathrm dt - \int_{0}^{\infty} \frac{t^{3}}{\cosh t} \, \mathrm dt \\ &= \frac{3 \pi^{2}}{2} \Gamma(2) \beta(2) - 2 \Gamma(4) \beta(4) \\ &= \frac{3 \pi^{2}G}{2} - 12 \beta(4). \end{align}$$
| {
"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
$3\cdot2\cdot2^k = 3\cdot 2^{k+1}$ but I am not able to do left hand side
please help
| $3\cdot2\cdot2^k = 3\cdot 2^{k+1}$
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 reasonable.
$k \ge 6$ so $4k^2 > 4*6*k$ and:
$3\cdot2^{k+1} = 2*3*2^k > 2(4k^2 + 1) = 8k^2 +2 = 4k^2 + 4k^2 + 2\ge 4k^2 + 24k + 2 = 4k^2 + 8k + 16k + 2$.
So now we just need to show $16k + 2 > 5$. That's..... well, that's.....
Again $k \ge 6$ so $16k \ge 16*6$.
$3\cdot2^{k+1} = 2*3*2^k > 2(4k^2 + 1) = 8k^2 +2 = 4k^2 + 4k^2 + 2\ge 4k^2 + 24k + 2 = 4k^2 + 8k + 16k + 2\ge 4k^2 + 8k + (16*6 + 2) > 4k^2 + 8k +4 + 1 = 4(k+1)^2 + 1$.
And that's the induction step.
| {
"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 $\gcd(X^3+1,X^2+1)=X+1$
Could you please help me understand this solution? Why it is $X+1$ and not $X-1$ ? how does $p$ play in this ?
I know that the $p$ is the minimal integer that verifies: $\forall x\in K : px=0_K$
Many thanks.
| 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 element $2$ is invertible, so we can conclude that the greatest common divisor is $1$.
| {
"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 $\lim_{x\to 0}$ of $\frac{2\ln(\cos{x})}{x^2}$
(which, I think, is necessary in order to make usage of derivatives).
Any hints? I've already tried cross multiplying but it doesn't lead anywhere, unless I missed something.
| 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 $y= - \frac{x^2}{2}+ \frac{x^4}{24}$ you get
$$\ln \cos x < \ln \left(1- \frac{x^2}{2}+ \frac{x^4}{24} \right) \le - \frac{x^2}{2}+ \frac{x^4}{24}$$
Multiply both sides by $\frac{2}{x^2}$ to get
$$\frac{2 \ln ( \cos x)}{x^2} < -1+\frac{x^2}{12}$$
| {
"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 $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{9/2}{2}\right)^2+\frac{9}{2}\right)$$
What am I missing and what is the best way to factorize when we have coefficients?(Is this the right term, ($a x^2$)?)
| 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:
$$(x-\frac{9}{4})^2 - (\frac{9}{4})^2+\frac{9}{2} =0$$
$$(x-\frac{9}{4})^2 = \frac{81}{16} - \frac{9}{2}$$
$$(x - \frac{9}{4})^2 = \frac{9}{16}.$$
The next step is very important. When taking the square root, there are always two solutions — one positive and one negative.
$$x - \frac{9}{4} = \color{red}{±} \color{black}{\frac{3}{4}}$$
From here you can find the two values of $x$, and then factorise.
| {
"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:
$$N = 4A$$
Also it is fairly easy to write a Diophantine which has solutions if and only if $N$ is not divisible by 4:
$$(N-4A-1)(N-4A-2)(N-4A-3)=0$$
But how about a Diophantine equation which has solutions if and only if $N$ is not a square number?
(sum, product and minus only can be used).
| 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 work, but we know that $X \le Y$ if and only if there are four integers $P,Q,R,S$ such that $Y = X + P^2 + Q^2 + R^2 + S^2$, by using Lagrange's four-square theorem.
So we get
$$
N = A^2+1 + B^2 + C^2 + D^2 + E^2 \text{ and } A^2+2A = N + F^2 + G^2 + H^2 + I^2
$$
but we probably want to write this as a single equation. To do this, just take
$$
(N - A^2-1-B^2-C^2-D^2-E^2)^2 + (A^2+2A-N-F^2-G^2-H^2-I^2)^2 = 0.
$$
Over the integers (or even the reals), $X^2+Y^2=0$ only if $X=Y=0$, giving us what we wanted.
Actually, the above only characterizes positive nonsquares: if $N$ is negative, there's no value of $A$ we can choose. But we can multiply the equation above by an equation representing $N\le-1$ easily, fixing this issue.
| {
"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 $\displaystyle \frac{y}{z-x}=q\Rightarrow y=qz-qx\Rightarrow qx+y-qz=0$
let $\displaystyle \frac{z}{x-y}=r\Rightarrow z=xr-yr\Rightarrow xr-yr-z=0$
So $$\begin{bmatrix}
1& -p &p \\
q& 1 & -q\\
r& -r & -1
\end{bmatrix}\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}$$
Which is $AX=B$
So for solution of equation , $|A|=pq+qr+rp+1=0$
Now using $(p+q+r)^2=p^2+q^2+r^2+2(pq+qr+rp)$
So $$p^2+q^2+r^2\geq 2$$
What i have used above is right or wrong. If it is right then i did not understand when will the equality hold
If it is wrong then could someone explain me a right solution, thanks
| 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 substitution, our integral becomes
$\int_C \frac{dz/(iz)}{2z^2 + 2z^{-2} + 5}$ , where $C$ is the circle $|z|=1$. Evaluating this integral yields
\begin{align}
\int_C \frac{dz/(iz)}{2z^2 + 2z^{-2} + 5} & = \frac{1}{i} \int_C \frac{dz}{2z^3 + 2z^{-1} + 5z} = \frac{1}{i}\int_C \frac{zdz}{2z^4 + 5z^2 + 2}
\end{align}
To solve for singularities, let $u = z^2$ and substitute this into the denominator to get $2u^2 +5u+2=0$. Solving for $u$ with the quadratic formula yields $u = -2,-1/2$. Thus, $f(z)$ has singularities at $z_0 = i\sqrt(2), z_1 = -i\sqrt(2), z_2 = -i\sqrt(2)/2,$ and $z_3 = i\sqrt(2)/2$. Only $z_2$ and $z_3$ are within the unit circle, so we need only consider those. Thus, the integral above is just $$\int_0^{2\pi} \frac{d\theta}{8\cos^2 (\theta) + 1} = \frac{1}{i}(2\pi i) \big(\text{Res}(f; -i\sqrt{2}/2) + \text{Res}(f; i\sqrt{2}/2)\big).$$ Since both singularities are simple poles, and that $P(a) \neq 0$ and $Q(a) = 0$ for either point, we know that $$\text{Res}(f; a) = \frac{P(a)}{Q'(a)} = \frac{z}{8z^3 + 10z}.$$ However, for when I evaluate this quantity I get the wrong answer. Where did I go wrong?
| 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).
$$I:=\int_0^{2\pi} \frac{d\theta}{8\cos^2 (\theta) + 1}=\int_0^{2\pi} \frac{d\theta}{4\cos(2\theta)+4 + 1}=\int_0^{2\pi} \frac{d t}{4\cos(t)+5}$$
where $t=2\theta$, and we used the fact that the $\frac{1}{4\cos(t)+5}$ has period $2\pi$.
Hence, by letting $z = e^{i t}$, we get
$$\begin{align}I&=\int_{|z|=1} \frac{1}{2(z+1/z)+5}\cdot \frac{dz}{iz}=\frac{1}{i}\int_{|z|=1} \frac{dz}{2z^2+5z+2}\\&=2\pi\mbox{Res}\left(\frac{1}{2z^2+5z+2},-\frac{1}{2}\right)=2\pi\left(\frac{1}{4(-1/2)+5}\right)=\frac{2\pi}{3}.\end{align}$$
Note that your procedure is correct. At the final step you should get
$$\begin{align}I&= 2\pi \left(\text{Res}\left(\frac{z}{2z^4 + 5z^2 + 2}, -\frac{i}{\sqrt{2}}\right) + \text{Res}\left(\frac{z}{2z^4 + 5z^2 + 2}, \frac{i}{\sqrt{2}}\right)\right)\\&=2\pi\left(\frac{1}{8(-1/2)+10}+\frac{1}{8(-1/2)+10}\right)=\frac{2\pi}{3}.\end{align}$$
| {
"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 the expansion of
$$(t^3 +t^6 +\cdots)\cdot(t^1+t^2+\cdots)^2\cdot(1+t+t^2+\cdots).$$
I am not sure this is the correct method. Also, I have doubts about the conditions of variables, i.e., whether they are correct or not.
| 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 restriction $|t| < 1$, and evaluate each multiplicands as an infinite geometric series.
This gives the product $$f(t) = (t^3 +t^6 +\cdots)\cdot(t^1+t^2+\cdots)^2\cdot(1+t+t^2+\cdots) = \frac{t^3}{1 - t^3}
\left(\frac{t}{1 - t}\right)^2 \frac{1}{1 - t} = \frac{t^5}{(1 - t^3)(1 - t)^3} = \frac{t^5}{(1 - t)^4(1 + t + t^2)}$$ which can be expanded as a Taylor series at $t = 0$, with the coefficient of the term $t^{30}$ in the expansion giving the answer you desire. However, taking the 30th derivative of this rational to compute this coefficient is undesirable. As an improvement, you can take a partial fraction decomposition of the form: $$f(t) = \frac{t^5}{(1 - t)^4(1 + t + t^2)} = \frac{At + B}{1 + t + t^2} + \frac{C}{(1 - t)^4} + \frac{D}{(1 - t)^3} + \frac{E}{(1 - t)^2} + \frac{F}{1 - t}$$ and solve for the constants. It then turns out that the $n$-th derivatives of each of these fractions (edit: except the first one) has a concise closed form in terms of $n$, and hence you can efficiently solve for the coefficient of $t^{30}$ by letting $n = 30$.
Edit 2: Very late edit, but it turns out that $A = 0$, so the first fraction can be further decomposed using partial fractions such that $$\frac{1}{1 + t + t^2} = \frac{G}{e^{2 \pi i / 3} - t} + \frac{H}{e^{4 \pi i / 3} - t},$$ which gives $$f^{(n)}(t) = \frac{BGn!}{(e^{2 \pi i / 3} - t)^{n + 1}} + \frac{BHn!}{(e^{4 \pi i / 3} - t)^{n + 1}} + \frac{C(n + 3)!/3!}{(1 - t)^{n + 4}} + \frac{D(n + 2)!/2!}{(1 - t)^{n + 3}} + \frac{E(n + 1)!}{(1 - t)^{n + 2}} + \frac{Fn!}{1 - t}.$$ The $n$-the coefficient of the $n$-th term ($n \in \mathbb{N}$) in the Taylor expansion of $f$ about $t = 0$ is $$a_n = \frac{f^{(n)}(0)}{n!} = \frac{BG}{(e^{2 \pi i / 3})^{n + 1}} + \frac{BH}{(e^{4 \pi i / 3})^{n + 1}} + \frac{C(n + 3)(n + 2)(n + 1)}{6} + \frac{D(n + 2)(n + 1)}{2} + E(n + 1) + F,$$ which can be evaluated by hand. Since it has been over a month since this question was asked, I will reveal that $B = 1/9$, $C = 1/3$, $D = -4/3$, $E = 17/9$, $F = -1$, $G = (e^{-2 \pi i / 3})/(e^{2 \pi i / 3} - 1) = i/\sqrt{3}$, and $H = -G$. It is easily checked that $a_{30} = 1215$ using the expression above.
Remark: This final closed form solution can be used to achieve a solution to the problem, where the inequality has any non-negative integer on the right side. If the number on the right side is $n$, the number of solutions is $a_n$.
| {
"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 Polylogarithm function, we obtain, differentiating both sides,that $$\sum_{n\geq1}\left(\frac{n^{n}}{n!e^{n}}-\frac{1}{\sqrt{2\pi n}}\right)x^{n-1}=-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}-\frac{\mathrm{Li}_{1/2}\left(x\right)}{x\sqrt{2\pi}}$$ so $$\sum_{n\geq1}\left(\frac{n^{n}}{n!e^{n}}-\frac{1}{\sqrt{2\pi n}}\right)=\lim_{x\rightarrow1^{-}}\left(-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}-\frac{\mathrm{Li}_{1/2}\left(x\right)}{x\sqrt{2\pi}}\right).$$ Now, we know that $$\mathrm{Li}_{v}\left(z\right)=\left(\Gamma\left(1-v\right)\left(1-z\right)^{v-1}+\zeta\left(v\right)\right)\left(1+O\left(\left|1-z\right|\right)\right),v\neq1,\,z\rightarrow1$$ and now we claim $$-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}\sim\frac{1}{\sqrt{2\left(1-x\right)}}-\frac{2}{3}$$ as $x\rightarrow1^{-}$. This is true because, since $$W\left(z\right)\sim-1+\sqrt{2ze+2}-\frac{2}{3}e\left(z+\frac{1}{e}\right)$$ as $z\rightarrow-1/e$, we have $$-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}\sim\frac{1-\sqrt{2\left(1-x\right)}+\frac{2}{3}\left(1-x\right)}{x\sqrt{2\left(1-x\right)}-\frac{2}{3}\left(1-x\right)x}$$ $$=\frac{1}{x}\left(-1+\frac{1}{\sqrt{2\left(1-x\right)}}\left(\frac{1}{1-\sqrt{2-2x}/3}\right)\right)=\frac{1}{x}\left(-1+\frac{1}{\sqrt{2\left(1-x\right)}}\sum_{k\geq0}\left(\frac{\sqrt{2-2x}}{3}\right)^{k}\right)$$ $$=\frac{1}{x}\left(-\frac{2}{3}+\frac{1}{\sqrt{2\left(1-x\right)}}+O\left(\sqrt{1-x}\right)\right)$$ then the claim.
| {
"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 $\alpha = 5+4\sqrt[3]{2}+3\sqrt[3]{4}$ and $\alpha^{-1} = a+b\sqrt[3]{2}+c\sqrt[3]{4}$ for some $a,b,c \in \mathbb{Z}$.
$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(5+4\sqrt[3]{2}+3\sqrt[3]{4})=1$
$=5a+6b+8c+4a\sqrt[3]{2}+5b\sqrt[3]{2}+6c\sqrt[3]{2}+3a\sqrt[3]{4}+4b\sqrt[3]{4}+5c\sqrt[3]{4}=1$
$=a(5+4\sqrt[3]{2}+3\sqrt[3]{4})+b(6+5\sqrt[3]{2}+4\sqrt[3]{4})+c(8+6\sqrt[3]{2}+5\sqrt[3]{4})=1$
and trying to solve for a,b, and c, but I don't know how?
Edit: Regrouping to $(5a+6b+8c)+(4a+5b+6c)\sqrt[3]{2}+(3a+4b+5c)\sqrt[3]{4}=1$
| 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 $5/16$, not $1/6$ as one would expect of an ordinary fair die.
From what you know about this peculiar die, compute the following:
a. The probability of rolling either a $5$, a $3$, or a $2$;
b. The probability of rolling a $5$.
My answer
Given $P(1) = \frac{5}{16}$, then
$$P(2) + P(3) + P(1) = P(2) + P(3) + \frac{5}{16} = \frac{1}{2} \iff P(2) + P(3) = \frac{3}{16}$$
Similarly, using $P(4) + P(6) + P(1) = \frac{1}{2}$, I find
$$P(6) + P(4) = \frac{3}{16}$$
b. $P(5) = 1 - \left(\frac{5}{16} + \frac{3}{16} + \frac{3}{16}\right) = \frac{5}{16}$
a. $P(5) + P(3) + P(2) = \frac{5}{16} + \frac{3}{16} = \frac{1}{2}$
Are my calculations correct?
| 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 $R_{1,4,6} = R_{1,2,3} = \frac12$ and that
$R_1 = \frac{5}{16}.$
For part (a), simply observe that $\{2,3,5\}$ and $\{1,4,6\}$ are disjoint and that $\{2,3,5\} \cup \{1,4,6\}$ is the entire sample space.
Therefore
$$P(R_{2,3,5}) = 1 - P(R_{1,4,6}) = 1 - \tfrac12 = \tfrac12.$$
Now for part (b) you know that
$$P(R_1) + P(R_{2,3}) = P(R_{1,2,3}) = \tfrac12$$
and
$$P(R_{2,3}) + P(R_5) = P(R_{2,3,5}) = \tfrac12,$$
so
$$P(R_{2,3}) + P(R_5) = P(R_1) + P(R_{2,3}),$$
and canceling $P(R_{2,3})$ on each side you have
$$P(R_5) = P(R_1) = \tfrac{5}{16}.$$
| {
"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. Evaluate the both sides at different limits of $x$, and see that both sides converge to $0$ for $x \to 0$ and increase infinitely for $x \to \infty$. Take the derivative of both sides with respect to $x$ to demonstrate that the left side increases faster than the right side. Burn out at that point because showing that the last point holds I need to engage into a similar task as proving the original inequality.
| 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 it is easy to check the LHS (containing the fraction term) is increasing starting from zero and then is decreasing in $x$, while the RHS (containing $\log$) is increasing in $x$ and goes to infinity when $x\to \infty$. Therefore, the inequality holds if the RHS is increasing faster than the LHS, or formally (taking derivatives, respectively):
$$\frac{-10 x^2 - 2 x^3 + 2\sqrt{9 + 8 x} + x (-9 + \sqrt{9 + 8 x})}{2(1+x)^3\sqrt{9+8x}} < \frac{1}{1+x}.$$
Proving the last inequality holds should not be a problem. Hence, the first inequality of $(*)$ is true.
Do similar steps to show that the second inequality in $(*)$ holds.
| {
"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 $\cos\theta$.
$$
2\cos^2{\theta} = -\cos{\theta}
$$
$$
2\cos{\theta} = -1
$$
$$
\cos{\theta} = -\frac{1}{2}
$$
$$
\theta = \frac{2\pi}{3}, \frac{4\pi}{3}
$$
But apparently this approach causes me to miss some of the solution. The correct approach seems to be to extract $\cos\theta$ and solve like a quadratic:
$$
\cos{\theta}(2\cos{\theta}+1) = 0
$$
$$
\cos{\theta} = 0, -\frac{1}{2}
$$
$$
\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{2\pi}{3}, \frac{4\pi}{3}
$$
I can kind of see that since $\cos\theta$ is squared, $\cos\theta$ could be $\pm$, but I'm not sure what it means for dividing by $\cos\theta$. Is it possible to come to the correct solution by dividing, or do I need to steer clear of that when it comes to squared values?
| 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 solutions for which $\cos\theta\ne0,$ but then you also have to check whether there are any solutions for which $\cos\theta=0.$
If $AB=0,$ then either $A=0$ or $B=0.$ You have $A=\cos\theta$ and $B=2\cos\theta+1.$
| {
"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 $n=k+1$. We will use this later on, but $$9^k+4^{k+1}=5k.$$
The
n,
$$\begin{align}9^{k+1}+4^{k+2}
&= 9 \cdot 9^k+4 \cdot 4^{k+1} \\
&= 9 \cdot 9^k+(9-5) \cdot 4^{k+1} \\
&=9 \cdot 9^k+9 \cdot 4^{k+1} -5 \cdot 4^{k+1} \\
&= 9(9^k + 4^{k+1})-5 \cdot 4^{k+1} \\
&=\underbrace{9(5k)-5 \cdot 4^{k+1}}_{\text{This is where the $5k$ comes in}} \\
&= 5(9k-4^{k+1}),\end{align}$$
thus, the original expression a multiple of $5$.
Is my induction correct?
Edit: I see several answers that took a different approach, all is welcome it really helps me see it in a different way. Thank You!
| 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
} |
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,
"answer_id": 0
} |
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:
$$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}=^{0 \over 0}\lim\limits_{x\to 0}{e^x-e^{\sin x}\cdot\cos x \over 1-\cos x}=^{0 \over 0}\lim\limits_{x\to 0}{{e^x-(e^{\sin x}\cdot\cos x-\sin x}\cdot e^{\sin x}\over \sin x})=\cdots$$
This gets to nowhere. Also, I substituted $t=e^{\sin x}$ but I could not replace the $e^x$.
| 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 detailed tracking of the error terms. We only need to keep terms up to the cubics, so we have
$$\frac{1 + x + x^2 / 2 + x^3/6 + O(x^4) - \big(1 + \sin x + \sin^2 x / 2 + \sin^3 x / 6 + O(\sin^4 x)\big)}{x - \big(x - x^3/6 + O(x^5)\big)}$$
which is equal to
$$\frac{x + x^2/2 + x^3 / 6 + O(x^4) - (x - x^3/6 + O(x^5) + \frac 1 2(x^2 + O(x^4)) + x^3/6 + O(x^4)}{x^3 / 6 + O(x^5)}$$
which in turn simplifies as
$$\frac{x^3 / 6 + O(x^4)}{x^3/6 + O(x^5)} \to 1.$$
| {
"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 $q$.
For a particular number I can write but my mind is not working to find the generalised value.
Kindly help.
| 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 $q$ is even, $3q$ is even and $\displaystyle \left\lfloor \frac{3q}{2}\right\rfloor=\frac{3q}{2}$.
| {
"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
} |
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 they are not necessarily distinct)"
I want to ask if there is any solution shorter than mine? I think mine is too long, and maybe yields a wrong answer.
My solution:
WLOG, let $a\leq b\leq c\leq d$
$$1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq \frac{4}{a}$$
$$a\leq4$$
Because $a=1$ yields no solution, so consider $a=2,3,4$
Case 1:$a=2$, then $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}$
Do that again: $\frac{1}{2}=\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq\frac{3}{b}$, so $b\leq 6$.
Let $b=6$, then $\frac{1}{c}+\frac{1}{d}=\frac{1}{3}$, going to $$(c-3)(d-3)=9$$ $$(c,d)=(4,12),(6,6)$$ so in the case have: $(a,b,c,d)=(2,6,4,12),(2,6,6,6)$ then eliminate some case not satisfy $a\leq b\leq c\leq d$
Then going through when $b=5$,$b=4$,$b=3$... yields $184$ distinct solutions.
Case 2: Following the same procedure as Case 1... yields $18$ solutions.
Case 3: As above... yields only a solution which is $(4,4,4,4)$
Conclude it, the equation has $203$ solutions.
That is my solution, I wrote it using one and a half piece of A4 paper, I have recently tried $abcd=abc+abd+acd+bcd$ but don't know how to continue, or should I use Vieta theorem?
---After first edit---
According to Robert Z, I had miscount quadruplet $(3,4,4,6)$ which add up the count to $215$ solutions.
-- After last edit --
Seems like there is no faster solution, I will close this question and marked as solved. Thanks to everyone who spend effort to my question.
| 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 to solve using Vieta's formula, but I was getting a sign error.
From the equation $x^3+px+q=0$,
$$\begin{align}
\sum \alpha\phantom{\beta} &= \phantom{-}0 \\
\sum \alpha\beta &= \phantom{-}p \\
\prod α \phantom{\beta} &= -q
\end{align}$$
L.H.S.
$$\begin{align}
\alpha^5+\beta^5+\gamma^5&=(\alpha^4+\beta^4+\gamma^4)(\alpha+\beta+\gamma)-\alpha^4(\beta+\gamma)-\beta^4(\gamma+\alpha)-\gamma^4(\alpha+\beta) \\
\text{(with $\sum \alpha = 0$)}\quad
&=-\alpha\beta\gamma\left(\begin{array}{c}
\phantom{+}\left(
\dfrac{\alpha^3}{\beta}
+\dfrac{\alpha^3}{\gamma}
+\dfrac{\beta^3}{\gamma}
+\dfrac{\beta^3}{\alpha}
+\dfrac{\gamma^3}{\alpha}
+\dfrac{\gamma^3}{\beta}\right)\\
+\left(
\dfrac{\alpha^3}{\alpha}
-\dfrac{\alpha^3}{\alpha}
+\dfrac{\beta^3}{\beta}
-\dfrac{\beta^3}{\beta}
+\dfrac{\gamma^3}{\gamma}
-\dfrac{\gamma^3}{\gamma}\right)\end{array}\right) \\
&=-\alpha\beta\gamma\left(\left(\alpha^3+\beta^3+\gamma^3\right)
\left(\dfrac {1}{\alpha}+\dfrac {1}{\beta}+\dfrac {1}{\gamma}\right)
-\left(\alpha^2+\beta^2+\gamma^2\right)\right) \tag{1}
\end{align}$$
Putting $\sum \alpha = 0$ in the formula
$$\begin{align}
\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma &=\left(\alpha+\beta+\gamma\right)\left(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha\right) \\
\alpha^3+\beta^3+\gamma^3&=3\alpha\beta\gamma \tag{2}
\end{align}$$
Putting (2) in (1)
$$\begin{align}
&\phantom{=}\;-\prod\alpha \cdot\left(3\cdot \prod \alpha \cdot
\dfrac{\sum\alpha\beta}{\prod\alpha}-\left(\left(\sum\alpha\right)^2-2\cdot\sum\alpha\beta\right)\right) \\
&=-\prod\alpha\cdot\left(3\cdot\sum\alpha\beta+2\cdot\sum\alpha\beta\right) \\
&=-5\cdot\sum\alpha\beta\cdot \prod\alpha
\end{align}$$
There is a negative sign in my solution. Please point out where I have gone wrong, or share your solution.
| 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:
$$5\,\alpha\,\beta\, \left( \beta+\alpha \right) \left( {\alpha}^{2}+
\alpha\,\beta+{\beta}^{2} \right)
$$
| {
"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:
$$f'(x) = \frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})+x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$$
How can I simplify? I didn't include all the passages.
| 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
$$f'(x)={\frac {\sqrt {5-{x}^{2}}+5}{x\ln \left( 10 \right) \left( 1+\sqrt {
5-{x}^{2}} \right) \sqrt {5-{x}^{2}}}}
$$
| {
"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 equation we have exactly the first power, since $1094\equiv 2$ modulo four. Sad enough, there is no solution, we compute for instance in sage:
sage: R.<x> = PolynomialRing(ZZ)
sage: a, b, c = x-1, x, x+1
sage: factor( a^4 + b^4 + (b^2+c^2)^2 - 1094 )
2 * (3*x^4 + 2*x^3 + 7*x^2 - 546)
and there is no linear factor.
| {
"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
$$ f = A(x+1)^5 + D(x+1)^4 + E(x+1)^3 -1$$
Is this method correct? Also I want to discuss other methods to solve these kind of problems.
| $f$ 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. Therefore, we get
$$f'=(x-1)^2(g' \cdot (x-1)+3g) \tag{3}$$
$$f'=(x+1)^2(h' \cdot (x+1)+3h) \tag{4}$$
From $(3),(4)$ we see that $1$ and $-1$ are double roots of $f'(x)$. Therefore, necessarily
$$f'(x)=K(x-1)^2(x+1)^2=K(x^2-1)^2=K(x^4 - 2x^2 +1) \tag{5}$$
$$\implies f(x)=K\left(\frac{x^5}{5}-\frac{2x^3}{3}+x\right)+C$$
$$(1) \implies 1=f(1)=K\left(\frac{1}{5}-\frac{2}{3}+1\right)+C=\frac{8K}{15}+C \tag{6}$$
$$(2) \implies -1=f(-1)=K\left(-\frac{1}{5}+\frac{2}{3}-1\right)+C=-\frac{8K}{15}+C \tag{7}$$
From $(6),(7)$, we get $$C=0,\ K = \frac{15}{8} \implies f'(x)=\frac{15}{8}(x^2-1)^2, \ \ f(x)=\frac{1}{8}(3x^5 - 10x^3 + 15x)$$
And from $(5)$, we get $$f'(2)=\frac{15\cdot 9}{8}$$
| {
"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 this problem then?
| $y = x+2$:
$$L = \lim _{x\to \infty} (\sqrt[3]{(x+1)(x+2)(x+3)}-x) = 2+ \lim _{y\to \infty} (\sqrt[3]{y^3-y}-y) = 2+ \lim _{y\to \infty} \frac{\sqrt[3]{1-\frac{1}{y^2}}-1}{\frac{1}{y}}$$
$y = \frac{1}{h}$:
$$L = 2+ \lim _{h\to 0^+}\frac{\sqrt[3]{1-h^2}-1}{h} \stackrel{L'Hopital}{=}2+ \lim _{h\to 0^+}\frac{\frac{-2h}{\sqrt[3]{(1-h^2)^2}}}{1} = 2+0 = 2$$
| {
"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:
$$LHS=
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|$$
$$C_1\to C_1+C_2$$
$$=
\left|
\begin{matrix}
(c+a)&(a-b)&a \\
(a+b)&(b-c)&b \\
(b+c)&(c-a)&c \\
\end{matrix}\right|$$
$$C_1\to C_1-C_3$$
$$=
\left|
\begin{matrix}
c&(a-b)&a \\
a&(b-c)&b \\
b&(c-a)&c \\
\end{matrix}\right|$$
How do I complete the rest?
| Here is a way to break it down to a factor and only one $2\times 2$-determinant containing only binomials before expanding:
$$\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|\stackrel{R_3 \mapsto R_3+R_2+R_1}{=}
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
2(a+b+c)&0&(a+b+c) \\
\end{matrix}\right| =
(a+b+c)\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
2&0&1 \\
\end{matrix}\right|
\stackrel{C_1 \mapsto C_1-2C_3}{=}
(a+b+c)\left|
\begin{matrix}
(b+c-2a)&(a-b)&a \\
(c+a-2b)&(b-c)&b \\
0&0&1 \\
\end{matrix}\right|
\stackrel{C_1 \mapsto C_1+C_2}{=}
(a+b+c)\left|
\begin{matrix}
(c-a)&(a-b)\\
(a-b)&(b-c)
\end{matrix}\right| =$$ $$=
(a+b+c)(ab+ac+bc-a^2-b^2-c^2) = 3abc - (a^3 + b^3 + c^3)
$$
| {
"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
$$x_0= -\frac2{\sqrt3}\sinh \left( \frac13\sinh^{-1} \frac{3\sqrt3}2\right) \approx -0.6823
$$
Then, factorize the cubic equation as
$$x^3+x+1=(x-x_0)(x^2+x_0 x-\frac1{x_0})$$
and the quadratic factor gives a pair of complex roots
$$x_{1,2}=-\frac{x_0}2 \pm \frac i2 \sqrt{1-\frac3{x_0}}$$
| {
"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}}{1+\frac{2bt}{x}}}
\end{align*}
$\tau, b,c$ are complex constants, which can be used to give some freedom in choosing the variable transformation.
Using the chain rule, I found that I needed to solve for $x,t$ in terms of $r$ and $s$ in order to define the operator in $(r,s)$ space. I plugged this into Mathematica, and got an indecipherable stream of characters. How can I solve for $(x,t)$ in terms of $(r,s)$?
| 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^{\tau}}B^2=\frac{1}{te^{\tau}}\left(x+2bt\right)=e^{-\tau}\left(\frac{x}{t}+2b\right),
\end{align}
or equivalently,
\begin{align}
xt&=e^{-\tau}rs+2c,\\
\frac{x}{t}&=e^{\tau}\frac{r}{s}-2b.
\end{align}
These last relations would lead to
\begin{align}
x^2&=xt\cdot\frac{x}{t}=\left(e^{-\tau}rs+2c\right)\left(e^{\tau}\frac{r}{s}-2b\right),\\
t^2&=xt\cdot\frac{t}{x}=\frac{e^{-\tau}rs+2c}{e^{\tau}r/s-2b}.
\end{align}
Therefore,
\begin{align}
x&=\sqrt{\left(e^{-\tau}rs+2c\right)\left(e^{\tau}\frac{r}{s}-2b\right)},\\
t&=\sqrt{\frac{e^{-\tau}rs+2c}{e^{\tau}r/s-2b}}.
\end{align}
By the way, I am interested in how you come up with this change-of-variable? This is quite technical to me!
| {
"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 $$\phi(x)=\begin{cases}0, & 0\lt x\lt1\\ \frac{4}{3}+\displaystyle\sum_{m=1}^{\infty}\frac{-2\sin\frac{m\pi}{3}}{m\pi}\cos\frac{m\pi x}{3}, & 1\lt x\lt3 \end{cases}$$
Put $x=0$ to find the sum
$\displaystyle 1+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots$
I tried the following
$$\phi(0)=\frac{4}{3}+\sum_{m=1}^{\infty}\frac{-2\sin\frac{m\pi}{3}}{m\pi}\\=\frac43-2\frac{\sin\frac{\pi}{3}}{\pi}-\frac{\sin\frac{2\pi}{3}}{\pi}-2\frac{\sin\pi}{3\pi}-\frac{\sin\frac{4\pi}{3}}{2\pi}-\cdots\\=\frac{4}{3}-\frac{\sqrt3}{\pi}-\frac{\sqrt3}{2\pi}-0+\frac{\sqrt3}{4\pi}\dots=\frac{4}{3}-\frac{\sqrt3}{\pi}(1+\frac{1}{2}-\frac{1}{4}\dots)=\ ? $$
And I'm stuck here,
What can I do here?
I greatly appreciate any assistance you may provide.
| In this answer, it is shown that
$$
\sum_{k=-\infty}^\infty\frac{(-1)^k}{z+k}=\pi\csc(\pi z)
$$
Therefore,
$$
\begin{align}
\sum_{k=0}^\infty(-1)^k\left(\frac1{3k+1}+\frac1{3k+2}\right)
&=\sum_{k=0}^\infty(-1)^k\left(\frac1{3k+1}-\frac1{-3(k+1)+1}\right)\\
&=\sum_{k=0}^\infty\left(\frac{(-1)^k}{3k+1}+\frac{(-1)^{-k-1}}{3(-k-1)+1}\right)\\
&=\sum_{k=-\infty}^\infty\frac{(-1)^k}{3k+1}\\
&=\frac13\sum_{k=-\infty}^\infty\frac{(-1)^k}{k+\frac13}\\
&=\frac\pi3\csc\left(\frac\pi3\right)\\[6pt]
&=\frac{2\pi}{3\sqrt3}
\end{align}
$$
| {
"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:
$\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}= \lim_{N\rightarrow\infty}\frac{1}{N^5} \lim_{N \rightarrow\infty} \sum_{n=1}^Nn^4=\frac{\frac{N^5}{5}}{N^5}=\frac{1}{5}$
But it feels like I am missing some steps here, especially why $\lim_{N \rightarrow \infty}\sum_{n=1}^N n^4=\frac{N^5}{5}$
How can I improve my try?
Thank you in advance!
| 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
$$\begin{align}\sum_{i=1}^{n}i &= \dfrac{n(n+1)}{2}\tag{1} \\
\sum_{i=1}^{n}i^2 &= \dfrac{n(n+1)(2n+1)}{6} \tag{2}\\
\sum_{i=1}^{n}i^3&=\left[\dfrac{n(n+1)}{2} \right]^2\tag{3}\end{align}$$
Observe the formula
$$\sum_{i=1}^{n}[(i+1)^5-i^5] = 2^5-1^5+3^5-2^5+\cdots+(n+1)^5-n^5=(n+1)^5-1\text{.}$$
From some FOILing, we find that
$$(i+1)^5-i^5=5i^4+10i^3+10i^2+5i+1$$
so thus
$$\sum_{i=1}^{n}[5i^4+10i^3+10i^2+5i+1]=5\sum_{i=1}^{n}i^4+10\sum_{i=1}^{n}i^3+10\sum_{i=1}^{n}i^2+5\sum_{i=1}^{n}i+\sum_{i=1}^{n}1$$
We have $\sum_{i=1}^{n}1 = n$, so inputting formulas $(1), (2), (3)$ into this equation, we obtain
$$5\sum_{i=1}^{n}i^4+\dfrac{n(15n^3+50n^2+60n+31)}{6}=(n+1)^5-1$$
and solving for $\sum_{i=1}^{n}i^4$, we obtain
$$\sum_{i=1}^{n}i^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
Inputting this into $(*)$, we obtain
$$\lim_{n \to \infty}\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}=\lim_{n \to \infty}\dfrac{6n^5+15n^4+10n^3-n}{30n^5}=\dfrac{6}{30}=\dfrac{1}{5}\text{.}$$
Method 2: You do know the Fundamental Theorem of Calculus.
Recall that $\Delta x = \dfrac{b-a}{n}$, where $(a, b)$ is the interval of integration. Since there are no constants that stick out in equation $(*)$, let's arbitrarily set $b - a = 1$. From this, we obtain $\Delta x = \dfrac{1}{n}$, and the remaining terms correspond with $f(x_i)$:
$$\lim_{n \to \infty}\dfrac{\sum_{i=1}^{n}i^4}{n^5}=\lim_{n \to \infty}\sum_{i=1}^{n}\dfrac{i^4}{n^5} =\lim_{n \to \infty}\sum_{i=1}^{n}\dfrac{i^4}{n^4}\cdot \dfrac{1}{n} = \lim_{n \to \infty}\sum_{i=1}^{n}\left(\dfrac{i}{n}\right)^4\Delta x$$
Thus, $f(x_i) = (i/n)^4$, so $x_i$ is probably $i/n$. The easiest way to obtain this formula for $x_i$ is to recall that
$$x_i = a + i\Delta x = a + \dfrac{i}{n}$$
so it follows that $a = 0$ since $x_i = i/n$. Lastly, since $b - a = 1$, it follows that $b = a + 1 = 1$. Hence, we have
$$\lim_{n \to \infty}\sum_{i=1}^{n}\left(\dfrac{i}{n}\right)^4\Delta x = \int_{0}^{1}x^4 \text{ d}x = \left.\dfrac{x^5}{5}\right|^{1}_{0}=\dfrac{1^5}{5}-\dfrac{0^5}{5}=\dfrac{1}{5}\text{.}$$
| {
"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
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.