Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Show that this lengthy integral is equal to $\int_{0}^{\infty}{1\over x^8+x^4+1}\cdot{1\over x^3+1}\mathrm dx={5\pi\over 12\sqrt{3}}$ $$\displaystyle\int_{0}^{\infty}{1\over x^8+x^4+1}\cdot{1\over x^3+1}\mathrm dx={5\pi\over 12\sqrt{3}}$$
$x^8+x^4+1=(x^2+x-1)(x^2-x-1)(x^2+x+1)(x^2-x+1)$
$x^3+1=(x+1)(x^2-x+1)$
${A\over x+1}+{Bx+C\over x^2-x+1}={1\over x^3+1}$
${Ax+B\over x^2+x-1}+{Cx+D\over x^2-x-1}+{Ex+F\over x^2+x+1}+{Gx+H\over x^2-x+1}={1\over x^8+x^4+1}$
it would be a nightmare trying to decomposition of fraction.
I would like some help please?
| If you really love partial fractions you may note that $$I=\int_{0}^{1}\frac{1-x^{3}+x^{6}}{1+x^{4}+x^{8}}\mathrm dx=\frac{1}{4\sqrt{3}}\int_{0}^{1}\frac{1}{x^{2}+\sqrt{3}x+1}\mathrm dx+\frac{1}{4}\int_{0}^{1}\frac{1}{x^{2}+x+1}\mathrm dx
$$ $$+\frac{3}{4}\int_{0}^{1}\frac{1}{x^{2}-x+1}\mathrm dx-\frac{1}{4\sqrt{3}}\int_{0}^{1}\frac{1}{x^{2}-\sqrt{3}x+1}\mathrm dx
$$ $$=\frac{1}{2\sqrt{3}}\int_{\sqrt{3}}^{2+\sqrt{3}}\frac{1}{x^{2}+1}\mathrm dx+\frac{1}{2\sqrt{3}}\int_{1/\sqrt{3}}^{\sqrt{3}}\frac{1}{x^{2}+1}\mathrm dx
$$ $$+\frac{\sqrt{3}}{2}\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{1}{x^{2}+1}\mathrm dx-\frac{1}{2\sqrt{3}}\int_{-\sqrt{3}}^{2-\sqrt{3}}\frac{1}{x^{2}+1}\mathrm dx
$$ so we have $4$ elementary integrals hence $$I=\color{red}{\frac{5\pi}{12\sqrt{3}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Help to prove that $\int_{0}^{\infty}{\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}\mathrm dx={5\over 6}$ I am trying to prove that
$$\displaystyle \int_{0}^{\infty}{\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}\mathrm dx={5\over 6}$$
$u=(x^2+1)^{1/2}$ then $du=(x^2+1)^{-1/2}dx$
$$\int_{1}^{\infty}{u^2+(u^2-1)\sqrt{u^2+1}\over u^2\sqrt{u^2+1}}du$$
$v=(u^2+1)^{1/2}$ then $dv=(u^2+1)^{-1/2}du$
$$\int_{1}^{\infty}{v^3+v^2-2v-1\over v^2-1}dv$$
$\int_{1}^{\infty}{v^2\over v^2-1}-{1\over v^2-1}dv$ -$\ln{(v^2-1)}|_{1}^{\infty}$
I am sure I when wrong somewhere, but I can figured it out.
Any help?
| We have $$I=\int_{0}^{\infty}\frac{\sqrt{x^{2}+1}+x^{2}\sqrt{x^{2}+2}}{\sqrt{\left(x^{2}+1\right)\left(x^{2}+2\right)}}\frac{1}{\left(x^{2}+1\right)^{2}}dx=\int_{0}^{\infty}\frac{x^{2}}{\left(x^{2}+1\right)^{5/2}}dx+\int_{0}^{\infty}\frac{1}{\left(x^{2}+1\right)^{2}\sqrt{x^{2}+2}}dx=I_{1}+I_{2}.
$$ Now recalling that the Beta function has the representation $$\int_{0}^{\infty}\frac{u^{m}}{\left(u+1\right)^{m+n+2}}du=B\left(m+1,n+1\right)
$$ we get $$I_{1}=\frac{1}{2}\int_{0}^{\infty}\frac{u^{1/2}}{\left(u+1\right)^{5/2}}du=\frac{B\left(\frac{3}{2},1\right)}{2}=\frac{1}{3}.
$$ If you prefer, you can also compute $I_{1}
$ using the substitution $x=\tan\left(v\right)
$. For $I_{2}
$ we get $$I_{2}\stackrel{x=\sqrt{2}\tan\left(v\right)}{=}\int_{0}^{\pi/2}\frac{\left(1-\sin^{2}\left(v\right)\right)\cos\left(v\right)}{\left(\sin^{2}\left(v\right)+1\right)^{2}}dv
$$ $$\stackrel{z=\sin\left(v\right)}{=}\int_{0}^{1}\frac{1-z^{2}}{\left(z^{2}+1\right)^{2}}dz=\int_{0}^{1}\frac{2}{\left(z^{2}+1\right)^{2}}dz-\int_{0}^{1}\frac{1}{z^{2}+1}dz
$$ and note that $$\int_{0}^{1}\frac{2}{\left(z^{2}+1\right)^{2}}dz\stackrel{z=\tan\left(w\right)}{=}2\int_{0}^{\pi/4}\cos^{2}\left(w\right)dw=\frac{1}{2}+\frac{\pi}{4}
$$ hence $$I=\frac{1}{3}+\frac{\pi}{4}+\frac{1}{2}-\frac{\pi}{4}=\color{red}{\frac{5}{6}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
For $abc=1$ prove that $\sum\limits_\text{cyc}\frac{1}{a+3}\geq\sum\limits_\text{cyc}\frac{a}{a^2+3}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}\geq\frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}$$
I tried TL, BW, the Vasc's Theorems and more, but without success.
I proved this inequality!
I proved also the hardest version: $\sum\limits_{cyc}\frac{1}{a+4}\geq\sum\limits_{cyc}\frac{a}{a^2+4}$.
Thanks all!
| Another way.
Since for any $a>0$ we have $$\frac{1}{a+3}-\frac{a}{a^2+3}+\frac{9}{64}\geq\frac{27}{64\left(a^{\frac{8}{3}}+a^{\frac{4}{3}}+1\right)}$$ and for positives $a$, $b$ and $c$ such that $abc=1$ we have $$\sum_{cyc}\frac{1}{a^2+a+1}\geq1,$$ we obtain:
$$\sum_{cyc}\left(\frac{1}{a+3}-\frac{a}{a^2+3}\right)=\sum_{cyc}\left(\frac{1}{a+3}-\frac{a}{a^2+3}+\frac{9}{64}\right)-\frac{27}{64}\geq$$
$$\geq\frac{27}{64}\left(\sum_{cyc}\frac{1}{a^{\frac{8}{3}}+a^{\frac{4}{3}}+1}-1\right)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Question about inverse function $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ This was from iranian university entrance exam .Suppose $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ find $f^{-1}(x)+f^{-1}(\frac{1}{x}),x \neq 0$.
It is easy to find $f^{-1}$ and solve this like below ...
$$y=\frac{1}{2}(x+\sqrt{x^2+4}) \\(2y-x)^2=(\sqrt{x^2+4})^2\\4y^2+x^2-4xy=x^2+4\\4y^2-4=4xy\\x=\frac{y^2-1}{y}=y-\frac{1}{y} \\ \to f^{-1}(x)=x-\frac{1}{x}\\
f^{-1}(x)+f^{-1}(\frac{1}{x})=x-\frac{1}{x}+(\frac{1}{x}-\frac{1}{\frac{1}{x}})=0\\$$ And now my question is ...
is there an other method to solve this question ?
I was thinking about $f(x)f(-x)=1$ but I can't go anymore ...
Any hint ,or other Idea ?
Thanks in advanced.
| $f(2\sinh\theta) = \frac{1}{2}\left(2\sinh\theta+2\cosh\theta\right) = e^\theta $ implies that the range of $f$ is $\mathbb{R}^+$ and
$$ f(x) = \exp\text{arcsinh }\frac{x}{2} \to \\
\qquad f^{-1}(x)=2\sinh\ln x=x-\frac{1}{x}\tag{1}$$
so $ f^{-1}(x)+f^{-1}\left(\frac{1}{x}\right) = \color{red}{0}$ for any $x\in\mathbb{R}^+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx$ $$\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx$$
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$$z=\alpha(x)=e^{ix}$$
$$\sin(x)=\frac{z^2-1}{2zi}$$
$$2+\sin(x)=\frac{4zi+z^2-1}{2zi}$$
$$\frac{1}{2+\sin(x)}=\frac{2zi}{4zi+z^2-1}$$
$$\int_\alpha \frac{2zi}{4zi+z^2-1} \ \ \frac{1}{zi}$$
$$4zi+z^2-1=0$$
$$z_1=\frac{-4i+2\sqrt{3}i}{2}$$
$$z_2=\frac{-4i-2\sqrt{3}i}{2}$$
$$\rvert z_1 \rvert <1$$
$$\rvert z_2 \rvert >1$$
Using Residue theorem:
$$\int_\alpha \frac{2}{4zi+z^2-1}=2 \pi i \ \lim_{z\rightarrow-2i+\sqrt{3}i} \ \frac{2}{z+2i+\sqrt{3}i}=\frac{2 \pi}{\sqrt{3}}$$
Is it correct?
Thanks!
| \begin{align}
\int_0^{2\pi} \frac{1}{2+\sin(x)} dx&=\int_0^{2\pi} \frac{1}{1+\sin^2(x/2)+\cos^2(x/2)+2\sin(x/2)\cos(x/2)} dx\\
&=\int_0^{2\pi} \frac{1}{1+(\sin(x/2)+\cos(x/2))^2} dx\\
&=\int_0^{2\pi} \frac{1}{1+2\sin^2(x/2+\pi/4)} dx\\
\end{align}
From here you should be able to continue recalling the behaviour of $\arctan$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Find indefinite integral $\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$ $$\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$$
$$\frac {1}{x^2\sqrt{1-x^2}}+2\int \frac{1}{x^3\sqrt{1-x^2}}\, \mathrm{d}x$$
I try to integrate by parts method, but its doesnt want to be solved. I try to substitue $x=\sin u \mathrm{d}x=\cos u$ but failed somewhere
| Let $u=\arcsin x$ and $dv=\frac{dx}{x^2}$. Then we will get (see Michael Burr answer. He used partial fractions and mine is trigo substitution)
$$\int \frac{\arcsin x}{x^2}dx=-\frac{\arcsin x}{x}+\int\frac{dx}{x\sqrt{1-x^2}}.$$
Let $x=\sin \theta$. Then $dx=\cos\theta\ d\theta$ and $$\sqrt{1-x^2}=\cos\theta.$$ Thus,
\begin{align}
\int \frac{\arcsin x}{x^2}dx&=-\frac{\arcsin x}{x}+\int\frac{\cos\theta\ d\theta}{\sin\theta\cos\theta}\\
&=-\frac{\arcsin x}{x}+\int \csc\theta\ d\theta\\
&=-\frac{\arcsin x}{x}-\ln|\csc\theta+\cot\theta|+C\\
&=-\frac{\arcsin x}{x}-\ln\left| \frac{1+\sqrt{1-x^2}}{x}\right|+C\\
&=-\frac{\arcsin x}{x}-\ln\left|1+\sqrt{1-x^2}\right|+\ln|x|+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the maximum value of the expression ${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$.
Find the maximum value of the expression :
$${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$$
where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.
Taking $x=y=z=\frac {1}{3}$, I get the expression as $\frac {3x}{1+x^2}$, which is equal to $\frac {1}{1+{\frac{1}{9}}}$ or $\frac {9}{10}$.
How can I actually solve the problem without making unnecessary assumptions ?
| We use the method of Lagrange Multipliers.
First, we define the function to maximize as $f(x,y,z)=\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ and our constraint as $g(x,y,z)=x+y+z-1$.
Thus, the Lagrange function is:
$$\mathcal{L}(x,y,z,\lambda)=f(x,y,z)+\lambda\cdot g(x,y,z)$$
$$\mathcal{L}(x,y,z,\lambda)=\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}+\lambda(x+y+z-1) \tag{1}$$
Where $\lambda$ is some scalar. This function's partial derivatives must equal all equal to zero:
$$\frac{\partial \mathcal{L}}{\partial x}=\frac{1-x^2}{(x^2+1)^2}+\lambda=0$$
$$\frac{\partial \mathcal{L}}{\partial y}=\frac{1-y^2}{(y^2+1)^2}+\lambda=0$$
$$\frac{\partial \mathcal{L}}{\partial z}=\frac{1-z^2}{(z^2+1)^2}+\lambda=0$$
$$\frac{\partial \mathcal{L}}{\partial \lambda}=x+y+z-1=0$$
Solving this system of equations will yield values for $x,y,z,\lambda$ which will either minimize or maximize $f(x,y,z)$.
After some tedious calculation, we notice that the solution which maximizes the value of $f(x,y,z)$ is:
$x=y=z=\frac{1}{3}$ and $\lambda=-0.72$, as you have correctly predicted.
Thus, the maximum value of $f(x,y,z)=\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}=\frac{9}{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Prove that the elements of the sequence are integers A sequence is given by the following recursion. Prove that the elements of this sequence are integers!
$a_0=1$
$a_1=41$
$a_{n+2}=3a_n+\sqrt{8(a_n^2+a_{n+1}^2)}$
| Sequence is Integral
Suppose
$$
a_{n+2}=3a_n+b_n\tag{1}
$$
where
$$
b_n^2=8\!\left(a_n^2+a_{n+1}^2\right)\tag{2}
$$
Then
$$
\begin{align}
b_{n+1}^2
&=8\!\left(\color{#C00}{a_{n+2}^2}+\color{#090}{a_{n+1}^2}\right)\\
&=8\!\left(\color{#C00}{9a_n^2+6a_nb_n+b_n^2}+\color{#090}{\frac{b_n^2}8-a_n^2}\right)\\
&=64a_n^2+48a_nb_n+9b_n^2\\[6pt]
&=(8a_n+3b_n)^2\tag{3}
\end{align}
$$
Therefore,
$$
b_{n+1}=8a_n+3b_n\tag{4}
$$
We can continue the integer sequences using $(1)$ and $(4)$:
$$
\begin{bmatrix}a_{n+1}\\a_{n+2}\\b_{n+1}\end{bmatrix}
=
\begin{bmatrix}0&1&0\\3&0&1\\8&0&3\end{bmatrix}
\begin{bmatrix}a_n\\a_{n+1}\\b_n\end{bmatrix}\tag{5}
$$
That is,
$$
\begin{bmatrix}a_n\\a_{n+1}\\b_n\end{bmatrix}
=
\begin{bmatrix}0&1&0\\3&0&1\\8&0&3\end{bmatrix}^{\,\large n}
\begin{bmatrix}1\\41\\116\end{bmatrix}\tag{6}
$$
Closed Form for the Sequence
Using the Jordan Normal Form of the matrix in $(6)$, we get
$$
\begin{bmatrix}a_n\\a_{n+1}\\b_n\end{bmatrix}
=
\begin{bmatrix}
-1&\frac{-1-\sqrt3}8&\frac{-1+\sqrt3}8\\
1&\frac{1-\sqrt3}8&\frac{1+\sqrt3}8\\
2&1&1
\end{bmatrix}
\begin{bmatrix}
-1&0&0\vphantom{\frac{\sqrt3}8}\\
0&2-\sqrt3&0\vphantom{\frac{\sqrt3}8}\\
0&0&2+\sqrt3
\end{bmatrix}^{\,\large n}
\begin{bmatrix}
\frac{22}3\\
\frac{152-84\sqrt3}3\\
\frac{152+84\sqrt3}3
\end{bmatrix}\tag{7}
$$
from which, we can extract
$$
a_n=\frac{25+17\sqrt3}6\left(2+\sqrt3\right)^n-\frac{22}3(-1)^n+\frac{25-17\sqrt3}6\left(2-\sqrt3\right)^n\tag{8}
$$
It is not immediately evident that $(8)$ is always an integer; however, since we have $\left(x-2-\sqrt3\right)\left(x-2+\sqrt3\right)(x+1)=x^3-3x^2-3x+1$, equation $(8)$ implies that
$$
a_n=3a_{n-1}+3a_{n-2}-a_{n-3}\tag{9}
$$
which does imply that $a_n$ is always an integer since $a_0=1$, $a_1=41$, and $a_2=119$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Generating function for counting compositions of $n$ with parts from a given set I was given the next question:
$A_n$ is marked to be the number of all of the sequences from a subset of the
naturals to $\{1,2,3,4,5,6\}$ such that their sum is $n$ (a natural number) ** which means that the order matters
For example: for $A_5$ some part of the sequences are <1,1,1,1,1> , <1,2,2> , <2,2,1>
I was asked to find the generating function that suits this problem, and what I mean by this is finding the function $f$ such that $f$= $A_0$ + $A_1x$ + $A_2x^2$+...
I'd really like some help with this question, Thank you very much!
p.s.:
$A_0=0$
|
We observe: $$\frac{x}{1-x}=x^1+x^2+x^3+\ldots$$
carries as exponents the possible solutions $1,2,3,\ldots$ of one $x_i$ in
$$x_1+x_2+x_3+\ldots+x_k=n\qquad k\geq 1, n\geq k$$
Therefore the number of compositions of $n$ with $k$ variables $x_1,x_2,\ldots,x_k$ is the coefficient of $x^n$ of
$$\left(\frac{x}{1-x}\right)^k$$
Since we want to restrict the solutions to be elements from $\{1,2,3,4,5,6\}$ we take
\begin{align*}
\frac{x-x^7}{1-x}=x^1+x^2+x^3+x^4+x^5+x^6
\end{align*}
Since the number of variables is at least $1$ up to $n$ for $A_n$ we obtain as generating function
\begin{align*}
f(x)&=\sum_{n=1}^\infty A_nx^n\\
&=\sum_{n=1}^\infty \left(\frac{x-x^7}{1-x}\right)^n\\
&=\frac{x-x^7}{1-x}\cdot\frac{1}{1-\frac{x-x^7}{1-x}}\\
&=x+2x^2+4x^3+8x^4+\color{blue}{16}x^5+32x^6+63x^7+\cdots
\end{align*}
whereby the last line was obtained with some help of Wolfram Alpha.
$$ $$
Example: We find e.g. $A_5=16$ which represents the $\color{blue}{16}$ compositions of $5$
\begin{align*}
&<1,1,1,1,1,1>,\\
&<1,1,1,2>,<1,1,2,1>,<1,2,1,1>,<2,1,1,1>,\\
&<1,1,3>,<1,3,1>,<3,1,1>\\
&<1,2,2>,<2,1,2>,<2,2,1>\\
&<1,4>,<4,1>\\
&<2,3>,<3,2>\\
&<5>
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$ I was asked to find the following integral:
$$\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$
What I tried to replace $\sqrt{1-\frac{1}{x^2}}$ with $u$ so that: $$du=\frac{dx}{x^3\sqrt{1-\frac{1}{x^2}}} \Rightarrow du*x=\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$
And:$$x=\sqrt{\frac{1}{1-u^2}}$$
And we can replace:
$$\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}=\int\frac{du}{\sqrt{1-u^2}}=\arctan(u)+C=\arctan(\sqrt{1-\frac{1}{x^2}})+C$$
The problem is, when that result is derived we don't get the original expression. I just can't find my mistake, so some help would be appreciated.
| HINT:
$$\dfrac1{x^2\sqrt{1-\dfrac1{x^2}}}=\dfrac{|x|}{x^2\sqrt{x^2-1}}$$
Set $\sqrt{x^2-1}=u\implies x^2=u^2+1$ and $x\ dx=u\ du$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proving that $19\mid 5^{2n+1}+3^{n+2} \cdot 2^{n-1}$ How can I prove that $$5^{2n+1}+3^{n+2} \cdot 2^{n-1} $$ can be divided by 19 for any nonnegative n? What modulo should I choose?
| For $n=0$, the formula says that $\left.19\middle|\frac{19}2\right.$, which is false. So consider $n\ge1$:
$$
\begin{align}
5^{2n+1}+3^{n+2}2^{n-1}
&=125\cdot25^{n-1}+27\cdot6^{n-1}\\
&\equiv11\cdot6^{n-1}+8\cdot6^{n-1}&\pmod{19}\\
&=19\cdot6^{n-1}\\
&\equiv0&\pmod{19}
\end{align}
$$
Since
$$
\begin{align}
125&\equiv11&\pmod{19}\\
25&\equiv6&\pmod{19}\\
27&\equiv8&\pmod{19}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Solving two Cubic Equation on their Roots.
Let $x^{3}+ax+10=0$ and $x^{3}+bx^{2}+50=0$ have two roots in common. Let $P$ be the product of these common roots. Find the numerical value of $P^{3}$, not involving $a,b$.
My attempts:
Let roots of $x^{3}+ax+10=0$ be $\alpha,\beta,\gamma\implies$
\begin{align*}
&\alpha+\beta+\gamma =0 \\
&\alpha\beta+\alpha\gamma+\beta\gamma =a \\
&\alpha\beta\gamma = -10
\end{align*}
and of $x^{3}+bx^{2}+50=0$ be $\alpha,\beta,\gamma'\implies$
\begin{align*}
&\alpha+\beta+\gamma' =-b \\
&\alpha\beta+\alpha\gamma'+\beta\gamma' =0 \\
&\alpha\beta\gamma' = -50
\end{align*}
Few important equations:
*
*$\dfrac{\alpha\beta\gamma'}{\alpha\beta\gamma}=\dfrac{-50}{-10}=5$ $\implies \gamma'=5\gamma $
*$\gamma-\gamma'=b \because$ Substracting first eq
*$(\alpha+\beta)(\gamma-\gamma')=a$ Substracting second eq. from above
*$\alpha+\beta=\dfrac{a}{b} \ \ \because(2),(3)$ squaring gives: $(\alpha+\beta)^{2}=\dfrac{a^{2}}{b^{2}}$
*$\alpha\beta\gamma-\alpha\beta\gamma'=40\implies\alpha\beta(\gamma-\gamma')=40\implies\alpha\beta=\dfrac{40}{b}$
Also $(\alpha+\beta+\gamma')^{2} =b^{2}$
$\implies \alpha^2+\beta^2+\gamma'^2=b^2 \implies \alpha^2+\beta^2=b^2-\gamma'^2=b^2-\dfrac{25b^{2}}{16}=\dfrac{-9b^2}{16}\because (2),(1)$
Now, squaring first eq. of first set of eq.$\implies \alpha^2+\beta^2+\gamma^2+2a =0 \implies \alpha^2+\beta^2=\dfrac{-b^2}{16}-2a\because (2),(1)$
Equating this with previous equations, $\implies \alpha^2+\beta^2=\dfrac{-9b^2}{16}=\dfrac{-b^2}{16}-2a\implies a^2=\dfrac{b^4}{16}\rightarrow(7)$
Also, $\alpha^2 +\beta^2 +2\alpha\beta=\dfrac{a^2}{b^2} \because(4)$. And we calculated very thing in terms of $b$, putting all these,$\implies \dfrac{-9b^2}{16} +2\alpha\beta=\dfrac{b^4}{16b^2}\because(7)\implies \dfrac{80}{b}=\dfrac{10b^2}{16}\ \because (5)\implies b^3=128\rightarrow(6)$
Now, $(eq.5)^3\implies (\alpha\beta)^3=\dfrac{64000}{b^3}=\dfrac{64000}{128}=500=P^3\because(6)$
Is this correct, and if, then what are easiest/shortest method besides my GIANT method.
| I thought that there was an impossibility that such polynomials exist. But in fact this is not the case, as evidenced by the edit @dxiv has done in his answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Perfect numbers adding up to 10 In aliquot sequences, am I right in saying that all perfect numbers when you add up the digits the result always add up to $10$ ( apart from the perfect no $6$?) if so, is this just a coincidence?
Example: "$496\mapsto 4+9+6=19\mapsto 1+9=10$".
| An even number is perfect if and only if it is of the form $N=2^n\cdot(2^{n+1}-1)$ and $2^{n+1}-1$ is prime (Euler).
This implies that $n+1$ is also prime (although this is not a sufficient condition for $2^{n+1}-1$ to be prime).
On the other hand, if $S(n)$ is the sum of the digits of $n$, then $S(n)\equiv n\pmod 9$. So we need to see if $2^n\cdot (2^{n+1}-1)\equiv 1\pmod 9$.
We have that $2$ is a primitive root modulo $9$, that is, the first power of $2$ that is $1$ in $\Bbb Z_9$ is $2^6$. Now write $n=6q+r$. Since $n+1$ must be prime, if we assume that $n+1$ is not $2$ or $3$, $r+1$ must be $1$ or $5$, so $r$ is $0$ or $4$. Putting all together we get
$$N\equiv2^r(2^{r+1}-1)=2^{2r+1}-2^r\equiv1\pmod 9$$
in both cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $x^2+xy+y^2+ \sqrt{3}y + 1=0$ for $xy$. I have a problem on my textbook:
$x,y \in\mathbb{R}$ and $x^2+xy+y^2+ \sqrt{3}y + 1=0$ is given. Find the value of $xy$.
I couldn't find.
| You can rewrite this equation as :
$$\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+y\sqrt{3}+1=0$$
which can be transformed further into :
$$\left(x+\frac{y}{2}\right)^2+\frac{3}{4}\left(y^2+\frac{4y}{\sqrt3}+\frac{4}{3}\right)=0$$
Finally, you should notice that :
$$y^2+\frac{4y}{\sqrt3}+\frac{4}{3}=\left(y+\frac{2}{\sqrt3}\right)^2$$
When the sum of two positive numbers is zero, each of them must be $0$.
Hence :
$$x+\frac{y}{2}=0\qquad\mathrm{and}\qquad y+\frac{2}{\sqrt3}=0$$
which leads to :
$$x=\frac{1}{\sqrt3}\qquad\mathrm\qquad y=-\frac{2}{\sqrt3}$$
Finally :
$$\boxed{xy=-\frac{2}{3}}$$
| {
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"url": "https://math.stackexchange.com/questions/2104259",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Known that $a + b + c = 0$ and $a^3 + b^3 + c^3 = 27$. Find ABC!
Known that
$$a + b + c = 0$$
$$a^3 + b^3 + c^3 = 27$$
What is the value of $abc?$
A.) 1
B.) 0
C.) 7
D.) 8
E.) 10
My Work:
$$a + b = -c$$
$$a + c = -b$$
$$b + c = -a$$
Then
$$(a + b + c)(a + b + c)(a + b + c) = 0$$
Expanded into:
$$a^3 + b^3 + c^3 + 3a^2b+3b^2c+3a^2c+3ab^2 + 3bc^2+ 3ac^2+6abc = 0$$
Putting the Values:
$$27+3a^2(-a)+3b^2(-b)+3c^2(-c)+6abc = 0$$
$$27 -3(a^3 + b^3 + c^3) = -6abc$$
$$27-3(27) = -6abc$$
$$-54=-6abc$$
$$abc = 9$$
$9$ wasn't an option in the question. Am I missing something?
| if $a+b+c=0$
then ,
$a^3+b^3+c^3=3abc $ and $ a^3+b^3+c^3=27$
equating both equations
$3abc=27$
therefore $abc=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Different ways to come up with $1+2+3+\cdots +n=\frac{n(n+1)}{2}$ I am trying to compile a list of the different ways to come up with the closed form for the sum in the title. So far I have the famous Gauss story, the argument by counting doubletons, and using generating functions. Is there any other (significantly) different way to discover the closed form?
To clarify, Gauss's method was to take $1+2+3+\dots+n$, write it backwards, and add it up.
| Here is one with the use of Indeterminant Coefficients:
Starting with $$1+2+3+4+5+\cdots+n\tag1$$Assume it to be equal to $A+Bn+Cn^2+Dn^3+\cdots\&\text c$. Hence$$1+2+3+4+5+\cdots+n=A+Bn+Cn^2+Dn^3+En^4+\cdots\&\text c\tag2$$
Therefore, we also have$$1+2+3+4+\cdots+n+1=A+B(n+1)+C(n+1)^2+D(n+1)^3+\cdots\&\text c\tag3$$
Subtracting $(2)$ from $(3)$, we get$$n+1=B+C(2n+1)\tag4$$And equating coefficients, we have$$2C=1\\B+C=1\\\implies C=B=\dfrac 12,\ A=0$$
Therefore$$1+2+3+4+\cdots n=0+\dfrac 12n+\dfrac 12n^2=\dfrac {n^2+n}2\tag5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Write the quadratic function in the form $g(x)= a(x-h)^2 + k$ Write the quadratic function in the form $g(x)= a(x-h)^2 + k$
Then, give the vertex of its graph.
$g(x)= 2x^2-16x+35$
~
I tried, but still lost:
$(2x^2-16x)+35$
$2(x^2-4^2-16)+35$
$2(x^2-4^2-32+35$
| It may be prudent to write out all of the steps without any shortcuts.
$$\begin{align}
g(x)&=2x^2 - 16x + 35\\
g(x)&=2(x^2 - 8x) + 35\\
g(x)&=2(x^2 - 8x) + 35\\
g(x)&=2(x^2 - 4x -4x) + 35\\
g(x)&=2(x^2 - 4x -4x + 0) + 35\\
g(x)&=2(x^2 - 4x -4x + 16 - 16) + 35\\
g(x)&=2(x^2 - 4x -4x + 16) -32 + 35\\
g(x)&=2(x(x - 4) -4(x - 4)) +3\\
g(x)&=2(x - 4)(x - 4) +3\\
g(x)&=2(x - 4)^2 +3\\
\end{align}$$
You know that the vertex will occur when $x-h = 0$. Maybe draw a table of values, or draw a graph, or both to determine the coordinates of the vertex.
| {
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"url": "https://math.stackexchange.com/questions/2106909",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Does $\int_{0}^{\pi/ n}{\sin^3(x)\over [1+\cos^2(x)]^2}\mathrm dx=F(n)$ have a closed form?
$$\int_{0}^{\pi/ n}{\sin^3(x)\over [1+\cos^2(x)]^2}\mathrm dx=F(n)\tag1$$
For $n\ge 1$
I have calculated out for:
$F(1)=1$
$F(2)={1\over 2}$
$F(3)={1\over 10}$
The general for $F(n)$ I wasn't able to find...
What is the closed form for $(1)?$
An attempt
Using $\cos^2(x)={1+\cos(2x)\over 2}$ then $1+\cos^2(x)={3+\cos(2x)\over 2}$
Calling upon a subsitition $u=3+\cos(2x)$ then $du=-2\sin(2x)dx$
Then $(1)$ becomes:
$$-\int_{1}^{\cos(2\pi/n)}{\sin^2(x)\over u^2\cos(x)}\mathrm du\tag2$$
$2+2\cos^2(x)=u$ then $\sin^2(x)={4-u\over 2}$
$\cos(x)=\sqrt{u-2\over 2}$
$${\sqrt{2}\over 2}\int_{1}^{\cos(\pi/n)}{u-4\over u^2\sqrt{u-2}}\mathrm du\tag3$$
$${\sqrt{2}\over 2}\int_{1}^{\cos{(\pi/n)}}{1\over u\sqrt{u-2}}\mathrm du+{2\sqrt{2}}\int_{1}^{\cos(\pi/n)}{1\over u^2\sqrt{u-2}}\mathrm du\tag4$$
From Standard integral table
$$\int{dx\over x\sqrt{ax+b}}={2\over \sqrt{-b}}\arctan\sqrt{ax+b\over -b}$$
$$\int{dx\over x^2\sqrt{ax+b}}=-{\sqrt{ax+b}\over bx}-{a\over 2b}\int{dx\over x\sqrt{ax+b}}$$
Substitute it in,
$${\sqrt{2}\over 2}\cdot{2\over\sqrt{2}}\arctan\sqrt{u-2\over 2}-2\sqrt{2}\left[-{\sqrt{u-2\over -2u}}+{1\over 4}\left({2\over\sqrt{2}}\cdot\arctan{\sqrt{u-2\over2}}\right)\right]\tag5$$
Break down to
$$\arctan\sqrt{u-2\over 2}-\sqrt{2}\cdot{\sqrt{u-2}\over u}-\arctan\sqrt{u-2\over 2}\tag6$$
$$=-\sqrt{2}\cdot{\sqrt{u-2}\over u}\tag7$$
I don't thing I am on the right track, any help?
| $$ \int_{0}^{\pi/ n}{\sin^3(x)\over [1+\cos^2(x)]^2}\mathrm dx $$
$$ = \int_{0}^{\pi/n}\frac {(1-\cos^2 x)(\sin x) }{(1+\cos^2 x)^2} dx $$
Put $ \cos(x) = t$ , then, $\sin(x)dx = dt$
Rewriting the integral,
$$-\int_{1}^{\cos\frac{\pi}{n}} \frac{1-t^2}{(1+t^2)^2} $$
$$ = F(n) = \frac {1}{2}-\frac{\cos\frac {\pi}{n}}{\cos^2(\frac {\pi}{n})+1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the sum to n term of the series. $1+3x+5x^2+7x^3................, X\ne1$ Here, $a=1, d=2, b=1, r=x$
\begin{align}
S_n&= \frac{ab}{1-r}+\frac{bdr(1-r{^n}^{-1})}{(1-r)^2}-\frac{[a+(n-1)d]br^n}{1-r}\\
S_n&=\frac 1 {1-x}+\frac{ 2x(1-x{^n}^{-1})} {(1-x)^2}-\frac{[1+(n-1)(2)]x^n} {1- x}\\
&= \frac 1 {1-x}+\frac{2x}{(1-x)^2}-\frac {2x.x{^n}^{-1}}{(1-x)^2}-\frac{[1+2n-2]x^n}{(1-x)}\\
&= \frac 1 {1-x}+\frac{2x}{(1-x)^2}-\frac {2x^n}{(1-x)^2}-\frac{[2n-1]x^n}{(1-x)}\\
\end{align}
Is it correct. I have not got the answer, please show me how to move to this answer without skipping any line
$\frac {1-3x} {(1-x)^2}+\frac {2x^n}{(1-x)^2}-\frac{(2n-1)x^n}{(1-x)}$
| Another way
\begin{align*}
1+3x+5x^2+7x^3+\cdots+(2n-1)x^{n-1}&=(1+2x+3x^2+4x^3+\cdots+nx^{n-1})\\&+x(1+2x+3x^2+\cdots+(n-1)x^{n-2})\\
&=\frac{d}{dx}(x+x^2+x^3+\cdots+x^n)\\&+x\frac{d}{dx}(x+x^2+x^3+\cdots+x^{n-1})\\
&=\frac{d}{dx}\left(\frac{x^{n+1}-x}{x-1}\right)+x\frac{d}{dx}\left(\frac{x^{n}-x}{x-1}\right)\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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find the root : $x^8-24x^4+256x^2+144=0$ find the root of $\mathbb{R} $:
$$x^8-24x^4+256x^2+144=0$$
my try:
$$t=x^2\\t^4-24t^2+256t+144=0\\t^4-(8×3)t^2+(8×2^5)t+(8×18)=0$$
now ?!!
thank you very much!
| The equation can be factored as
$$(x^2-8x+36)(x^2+8x+4)=0$$
So here are the roots:
$$x_1 = -2(2+\sqrt{3})$$
$$x_2 = -2(2-\sqrt{3})$$
$$x_3 = 4-2\sqrt{5}i$$
$$x_4 = 4+2\sqrt{5}i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})...$ Is there any well-known value (Or Approximation) for this?
$$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})...$$
we know that it converges as $$\sum_{i=2}^{\infty}\frac{(-1)^{i+1}}{i}=ln2-1$$
So there is a trivial upper bound $\frac{2}{e}$ for it. Is there any better result? In addition is there any similar result for
$$(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})...$$
or
$$(1+\frac{1}{2})(1+\frac{1}{4})(1+\frac{1}{8})(1+\frac{1}{16})...$$
| $$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})(1-\frac{1}{6})(1+\frac{1}{7})... $$
Except the first term, consider consecutive couples of terms :
$(1+\frac{1}{3})(1-\frac{1}{4})= (\frac{4}{3})(\frac{3}{4})= 1$
$(1+\frac{1}{5})(1-\frac{1}{6})= (\frac{6}{5})(\frac{5}{6})= 1$
$(1+\frac{1}{7})(1-\frac{1}{8})= (\frac{8}{7})(\frac{7}{8})= 1$
And so on
All couples $=1$ . So, only the first term remains : $(1-\frac{1}{2})=\frac{1}{2}$
Finally
$$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})(1-\frac{1}{6})(1+\frac{1}{7})... = \frac{1}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing pointwise and uniform convergence of $\frac{nx^2}{1+n^2x}$ Suppose we have:
$$f_n: \mathbb{R}^{+} \rightarrow \mathbb{R} $$ $$ f_n(x)=\frac{nx^2}{1+n^2x}$$
I started of by showing the pointwise convergence.
For $x=1$, we have:
$$f_n(1)=\frac{n}{1+n^2} \leq \frac{1}{n}$$ thus if $f=0$, then $f_n$ converges to $f$.
For $x < 1$ we have:
$$x^2 < x \implies nx^2 < n^2x \implies lim_{n \rightarrow \infty} = 0$$
Thus if $f=0$, then $f_n$ converges to $f$.
For $x>1$, we have:
$$f_n(x)=\frac{nx^2}{1+n^2x}=\frac{n}{\frac{1}{x^2}+\frac{n^2}{x}} \leq \frac{n}{ \frac{n^2}{x} }=\frac{nx}{n^2}=\frac{x}{n}$$, thus if $f=0$, then $f_n$ converges to $f$.
In all cases $f_n$ converges to $0$. So it pointwise converges to $0$.
Now I want to study the uniform convergence of the function. And this is where I am stuck. $f_n(x) \leq \frac{1}{n}$ if $x \leq 1$, and otherwise we have $f_n(x) \leq \frac{x}{n}$.
So if I put $a_n= \frac{x+1}{n}$, we have then $||f_n -f||_{\infty} \leq a_n$, and $lim_{n \rightarrow \infty}=0$, thus $f_n$ convergence uniformely. Am I correct?
| uniform convergence: you can rewrite $f_n$ as follows :
$$ \begin{array}[rcl] ~f_n(x) &=& \frac{nx^2}{1+n^2x} \\
&=& \frac{x^2}{\frac{1}{n^2}+x}\frac{1}{n} \end{array}$$
If $\bar{x} = n^2 \in \mathbb{R}^+$, we have
$$ f_n(\bar{x}) = \frac{n^4}{\frac{1}{n^2}+n^2}\frac{1}{n} = \frac{n^3}{\frac{1}{n^2}+n^2} \rightarrow +\infty $$
Hence, the sequence cannot converge uniformly for $x \in \mathbb{R}^+$.
| {
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"url": "https://math.stackexchange.com/questions/2111591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Algebraic Fractions grade 7 I am a student and I am having difficulty with answering this question. I keep getting the answer wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.
Write as a single fraction in its simplest Form.
$$\frac{3}{2x+1}+\frac{8}{2x^2-7x-4}$$
I factorised both fractions:
$$\frac{3}{(2x+1)} \qquad \frac{8}{(2x+1)(x+4)}$$
And got rid of $(2x+1)$. I don't know what to do next.
Kind Regards
| Getting $\frac{1}{2x+1}$ common, we have
$\frac{1}{2x+1} \left[3 + \frac{8}{x-4} \right]$
$\frac{1}{2x+1} \left[\frac{3(x-4) + 8}{x-4}\right]$
$\frac{1}{2x+1} \left[\frac{3x - 12 + 8}{x-4}\right]$
$\frac{1}{2x+1} . \frac{3x - 4}{x-4}$
| {
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"url": "https://math.stackexchange.com/questions/2113579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Writing elements of field extension in terms of minimal polynomial
Let $\alpha \in \mathbb{C}$ be a root of the irreducible polynomial $f(x)=x^3+x+1$. Write $\alpha^{-1}$ and $(\alpha+1)^{-1}$ in terms of $\{1,\alpha,\alpha^2\}$.
So I was able to find a similar question here and was able to solve some of the problems i was given where $\alpha$ had a positive power but I am confused on how to divide the polynomials when it is instead raised to a negative power? Thanks in advance!
| Let us assume that $\alpha^{-1} = a \cdot 1 + b \cdot \alpha + c \cdot \alpha^2$ and try and find the coefficients $a,b,c \in \mathbb{C}$. Multiplying both sides of the equation by $\alpha$ we get
$$ 1 = a \cdot \alpha + b \cdot \alpha^2 + c \cdot \alpha^3. $$
Since $\alpha^3 + \alpha + 1 = 0$ we have $\alpha^3 = -\alpha - 1$ and we get
$$ 1 = (-c) \cdot 1 + (a - c) \cdot \alpha + b \cdot \alpha^2 $$
or
$$ (-c -1) \cdot 1 + (a - c) \cdot \alpha + b \cdot \alpha^2 = 0. $$
Since $1,\alpha,\alpha^2$ are linearly independent, we must have $-c - 1 = a - c = b = 0$ so $a = c = -1$ and $b = 0$ leading to
$$ \alpha^{-1} = -1 - \alpha^2. $$
This can be verified directly by multiplying both sides by $\alpha$:
$$ 1 =^{?} (-1 - \alpha^2)(\alpha) = -\alpha -\alpha^3 = -(\alpha + \alpha^3) = -(-1) = 1. $$
| {
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Problem in showing $\lim_{(x,y)\to (0,0)} \frac{12x^3y^5+4x^4y^4}{x^6+4y^8}=0$ using polar coordinates I'm trying to show that
$$\lim_{(x,y)\to (0,0)} \frac{12x^3y^5+4x^4y^4}{x^6+4y^8}=0$$
I've used polar coordinates but when I do this I get the possibility of $\frac{0}{0}$ if $\cos(\theta)\to 0$ as $r\to 0$. So it must be that I need some sort of bound to make sense of this limit. But I'm not sure how to proceed.
| If we wish to use polar coordinates $(r,\theta)$, then we can write
$$\begin{align}
\frac{12x^3y^5+4x^4y^4}{x^6+4y^8}&=r^2\left(\frac{12\cos^3(\theta)\sin^5(\theta)+4\cos^4(\theta)\sin^4(\theta)}{\cos^6(\theta)+4r^2\sin^8(\theta)}\right)\\\\
&=r\left(3\sin(\theta)+\cos(\theta)\right)\left(\frac{4r\cos^3(\theta)\sin^4(\theta)}{\cos^6(\theta)+4r^2\sin^8(\theta)}\right)\tag1
\end{align}$$
Let $ g(r,\theta)=\frac{4r\cos^3(\theta)\sin^4(\theta)}{\cos^6(\theta)+4r^2\sin^8(\theta)}$. Denote $\sin(\theta)$ by $s$ and $\cos(\theta)$ by $c$.
We will view $g(r,\theta)$ as a function of $\theta\in \mathbb{R}$, which is differentiable and $2\pi$-periodic. Therefore the extrema occur at points for which $\frac{\partial g(r,\theta)}{\partial \theta}=0$.
Then, taking the partial derivative with respect to $\theta$, and , we have
$$\begin{align}
\frac{\partial g(r,\theta)}{\partial \theta}&=4rs^3c^2\,\left(\frac{(4c^2-3s^2)(c^6+4r^2s^8)-s^2c^2(32r^2s^6-6c^4)}{(c^6+4r^2s^8)^2}\right)\\\\
&= 4rs^3c^2\,\left(\frac{(3+c^2)(c^6-4r^2s^8)}{(c^6+4r^2s^8)^2}\right)\tag 2
\end{align}$$
We see that $\frac{\partial g(r,\theta)}{\partial \theta}=0$ when $\sin(\theta)=0$ or $\cos(\theta)=0$ or $\cos^6(\theta)=4r^2\sin^8(\theta)$.
When $\sin(\theta)=0$ or $\cos(\theta)=0$, $g(r,\theta)=0$. When $\cos(\theta)^6=4r^2\sin^8(\theta)$,
$$g(r,\theta)=\text{sgn}(\cos(\theta))\tag 3$$
Finally, using $(3)$ in $(1)$ reveals
$$\left|\frac{12x^3y^5+4x^4y^4}{x^6+4y^8}\right|\le r|3\sin(\theta)+\cos(\theta)|$$
whereupon applying the squeeze theorem yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{(x,y)\to (0,0)}\frac{12x^3y^5+4x^4y^4}{x^6+4y^8}=0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Factorise $\,a(a - 4)(a^2 - 4a - 1) - 20\,$ I am making an attempt to factorise the above equation.
To do this i am expanding the already factored part of the above equation to obtain ->
$\,f(a) = a^4 - 8a^3 + 15a^2 + 4a - 20\,$
Now i do some random guesses to find that f(2) = 0, hence i know that (a-2) will divide f(a) so i perform long division to get f(a) = (a-2)( Something ) and so on... until i get - > $\,f(a) = (a - 2)^2(a + 1)(a - 5)\,$
I feel that i am approaching the problem in an incorrect and long way. Please let me know if i can get the factorised form in a cleaner and faster way.
PS: Another linked question is to factorise $\,f(x) = (x + 1)(x + 3)(x + 5)(x + 7) + 15\,$
| $1)$ Call $x=a(a-4)$ then
$$x(x-1)-20=x^2-x-20=(x+4)(x-5)$$
then
$$(a^2-4a+4)(a^2-4a-5)=(a-2)^2(a+1)(a-5)$$
$2)$ For
$$(x+1)(x+3)(x+5)(x+7)+15=[(x+3)(x+5)][(x+1)(x+7)]+15=(x^2+8x+15)(x^2+8x+7)+15$$
Call $y=x^2+8x+7$ then:
$$(y+8)y+15=y^2+8y+15=(y+3)(y+5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question on the proof of divergence of harmonic series. The harmonic series, $\sum_{k=1}^{\infty}\frac{1}{k}$, diverge. Let the partial sum be denoted as $s_n = \sum_{k=1}^{n}\frac{1}{k}$. If $s_n$ was convergent, then it would have to be bounded and its subsequence $s_{2^n}$ must be also bounded. But, $$s_{2^n} = (1) + (\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})+\dots+(\frac{1}{2^{n-1}+1}+\dots+\frac{1}{2^n})>1+\frac{n}{2}$$ as $n \rightarrow \infty$. I can see that each term in parenthesis are greater than $\frac{1}{2}$. But the book doesn't show how $(\frac{1}{2^{n-1}+1}+\dots+\frac{1}{2^n}) > \frac{1}{2}$ for each $n$. Is this supposed to be intuitively obvious or is there some way to prove this rigorously?
| $$\left(\frac{1}{2^{n-1}+1}+\dots+\frac{1}{2^n}\right) > \underbrace{\left(\frac{1}{2^n}+\dots+\frac{1}{2^n}\right)}_{2^{n-1} \text{times}} = \frac{2^{n-1}}{2^n} = \frac{1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit $\lim_{x\to 0} \frac{\tan ^3 x - \sin ^3 x}{x^5}$ without l'Hôpital's rule. I need to solve $$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}$$
I did like this:
$\lim \limits_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5} = \lim \limits_{x\to 0} \dfrac{\tan ^3 x}{x^5} - \dfrac{\sin ^3 x}{x^5}$
$=\dfrac 1{x^2} - \dfrac 1{x^2} =0$
But it's wrong. Where I have gone wrong and how to do it?
| HINT:
$\lim_{x\to0}\left(\left(\dfrac{\tan x}x\right)^3\cdot\dfrac1{x^2}-\left(\dfrac{\sin x}x\right)^3\cdot\dfrac1{x^2}\right)$ is of the form $\infty-\infty$
See List of indeterminate forms
Use
$$\dfrac1{\cos^3x}\cdot\left(\dfrac{\sin x}x\right)^3\cdot\dfrac{1-\cos x}{x^2}\cdot(1+\cos x+\cos^2x)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $a^b = b^a$ and $a=2b$ then find the value of $a^2+b^2$ If $a^b = b^a$ and $a=2b$ then find the value of $a^2+b^2$
My Attempt,
$$a^b = b^a$$
$$a^b =b^{2b}$$
$$a^b =b^b.b^b$$.
Now, what should I do further?
| Continue from your solution -
$a = (b^{2b})^{\frac 1b}$
$a = b^2$
Also a = 2b.
So we have,
$b^2 = 2b$
Assuming b is real positive number.
b = 2.
a = 4.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
How to prove that $3^{3n+2}+2^{4n+1}$ is a multiple of 11 How can I prove that following is a multiple of 11 with the induction method?
$$3^{3n+2}+2^{4n+1}$$
| Base case, $n=0$: $\quad 3^{0+2}+2^{0+1} = 9+2 = 11 \quad \checkmark$
Induction hypothesis: $\quad 3^{3n+2}+2^{4n+1} = 11k$
$\begin{align} 3^{3(n+1)+2}+2^{4(n+1)+1} &= 27\cdot3^{3n+2}+16\cdot 2^{4n+1} \\
&= 22\cdot3^{3n+2}+5\cdot3^{3n+2}+11\cdot 2^{4n+1}+5\cdot 2^{4n+1}\\
&= 22\cdot3^{3n+2}+11\cdot 2^{4n+1}+5(3^{3n+2}+ 2^{4n+1})\\
&= 11(2\cdot3^{3n+2}+ 2^{4n+1})+5(11k)\\
&= 11(2\cdot3^{3n+2}+ 2^{4n+1}+5k)\quad \text{as required}\\
\end{align}$
(without induction):
$3^3 \equiv 27 \equiv 5 \bmod 11$ and $2^4 \equiv 16 \equiv 5\bmod 11$
So $3^{3n+2}+2^{4n+1} \equiv 9\cdot 5^n+2\cdot 5^n \equiv 11\cdot 5^n \equiv 0 \bmod 11$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can both $n+3\; \text{and}\; n^2+3$ both be cubic numbers at same time?
Can both $n+3\; \text{and}\; n^2+3$ both be cubic number at same time? Where $n$ is an integer number. Not necessarily positive.
I tried writing $x^3 = n+3$ and expressing $n^2+3$ in terms of $x$. I found $x^6 -6x^3+12$ but this doesn't help. How do I prove this?
| Your question is asking if there is $x,a \in \mathbb{Z}$ such that $$x^6-6x^3+12=a^3$$
Note that if we have $x \ge 2, x \le -2$, then we have that $$(x^2-1)^3 = x^6-3x^4+3x^2-1 < x^6-6x^3+12=a^3$$
And also $$(x^2+1)^3 =x^6+3x^4+3x^2+1 > x^6-6x^3+12=a^3$$
This give us $x^2-1<a<x^2+1$, thus forcing $a$ to be $x^2$. However, as $x^3 \neq 2$ for any integer $x$, this is a contradiction.
The only cases left are $x=-1, 0,1 $, which can be manually checked to be never be cubes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega\neq 1$, find ... Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}$$
I've been having trouble with this unit, need help on solving this problem.
| The sum of the first and fourth terms is
\begin{align*}
\frac{\omega}{1 - \omega^2} + \frac{\omega^4}{1 - \omega^3} &= \frac{\omega (1 - \omega^3) + \omega^4 (1 - \omega^2)}{(1 - \omega^2)(1 - \omega^3)} \\
&= \frac{\omega - \omega^4 + \omega^4 - \omega^6}{(1 - \omega^2)(1 - \omega^3)} \\
&= \frac{\omega - \omega^4 + \omega^4 - \omega}{(1 - \omega^2)(1 - \omega^3)} \\
&= 0,
\end{align*}and the sum of the second and third terms is
\begin{align*}
\frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} &= \frac{\omega^2 (1 - \omega) + \omega^3 (1 - \omega^4)}{(1 - \omega^4)(1 - \omega)} \\
&= \frac{\omega^2 - \omega^3 + \omega^3 - \omega^7}{(1 - \omega^4)(1 - \omega)} \\
&= \frac{\omega^2 - \omega^3 + \omega^3 - \omega^2}{(1 - \omega^4)(1 - \omega)} \\
&= 0.
\end{align*}Therefore, the sum of all four terms is $\boxed{0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Can we prove a arc as circle
In this question ,according to me
The argument term shows a part of circle(major arc) and we have to prove it a circle , how it can be possible.
| Let $z=x + iy $. We then get $$\arg (\frac {z-z_1}{z-z_2}) =\pi/4$$ $$\Rightarrow \arg (x-10+i (y-6)) -\arg (x-4) +i (y-6)) =\pi/4 $$ $$\Rightarrow \arctan (\frac {y-6}{x-10}) - \arctan (\frac {y-6}{x-4}) =\pi/4$$ Using the relevant formula, we have, $$ \frac {y-6}{x-10} -\frac {y-6}{x-4} = 1+ \frac {y-6}{x-10}\frac {y-6}{x-10}$$ $$\Rightarrow (x-7)^2 + (y-9)^2 = 18$$ $$\Rightarrow |z -7 -9i|^2 =18 $$ $$\boxed {|z-7-9i| = 3\sqrt{2}}$$ Hope it helps.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Square Root of Wallis Product How can we prove or disprove this:
$$\begin{align}
\prod_{n=1}^{\infty}\frac {2n}{2n-1}=\frac {2\cdot 4\cdot 6\cdot 8\cdots}{1\cdot 3\cdot 5\cdot 7\cdots }
&=\sqrt2\cdot \sqrt{\frac{2\cdot 4\cdot 4\cdot 6\cdot 6\cdot8\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdots}}\\
&=\sqrt{\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdots}}\\
&=\sqrt{\frac{\pi}2}&&\text{(using Wallis Product)}
\end{align}$$
According to Wolframlpha, the product diverges, and the partial product is given by $$\prod_{n=1}^m\frac {2n}{2n-1}=\sqrt{\pi}\frac {\Gamma(m+1)}{\Gamma(m+\frac 12)}$$
The main objective of this question is ascertain why the Wallis product cannot be used in deriving the solution.
| Every terms count
\begin{align*}
\frac{2 \cdot 4}{1\cdot 3} &= \sqrt{5} \times
\sqrt{\frac{2\cdot 2 \cdot 4 \cdot 4}{1\cdot 3 \cdot 3 \cdot 5}} \\
\frac{2 \cdot 4 \cdot 6}{1\cdot 3 \cdot 5} &= \sqrt{7} \times
\sqrt{\frac{2\cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6}
{1\cdot 3 \cdot 3 \cdot 5 \cdot 5\cdot 7}} \\
\frac{(2n))!!}{(2n-1)!!} &= \sqrt{2n+1} \times
\sqrt{\prod_{k=1}^{n} \frac{4k^2}{4k^2-1}} \\
& \sim \sqrt{n\pi}
\end{align*}
which blows up!!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Summation problems involving $k c^{-k}$ I am trying to find a better way of solving problems involving summations of the form: $\sum_{k=0}^\infty k\cdot c^{-k}$ and $\sum_{k=0}^\infty k^2 \cdot c^{-k}$
For example, finding $\sum_{k=0}^\infty k\cdot 2^{-k}$, I used the following approach:
$S = \frac{1}{2} + 2 \cdot \frac{1}{2^2} + 3 \cdot \frac{1}{2^3} + ....$
$S = \frac{1}{2}[1+2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2^2} + .....]$
$S = \frac{1}{2}[1+\sum_{k=0}^\infty (k+1)\cdot 2^{-k}]$
$\sum_{k=0}^\infty (k+1)\cdot 2^{-k} = \sum_{k=0}^\infty k\cdot 2^{-k} + \sum_{k=0}^\infty 2^{-k} = S + \sum_{k=0}^\infty 2^{-k}$
Therefore, $S = 1 + \sum_{k=0}^\infty 2^{-k} = 1 + \frac{1}{1-\frac{1}{2}} = 3$
I may have made a few calculation mistakes (I would appreciate if you could point that out), but my main concern is a simpler way to approach this kind of problems. Can I use this method to evaluate problems of the form $\sum_{k=0}^\infty k^2 \cdot c^{-k}$ ? I tried, but failed.
I would appreciate if someone could point out the most efficient way to solve this
| Note that we can start with the sum (where I replaced $c$ with $x$)
$$\frac{x}{x-1}=\sum_{k=0}^\infty x^{-k}$$
Differentiating and multiplying by $x$, we get that
$$\frac{x}{(x-1)^2 }= \sum_{k=0}^{\infty} kx^{-k}$$
Doing this again
$$\frac{(x+1)x}{(x-1)^3 }= \sum_{k=0}^{\infty} k^2x^{-k}$$
If you continue this process your denominator will not change, and your numerator will be an $n$th degree polynomial, where $n$ is the power of $k$ on the RHS
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the inequality $\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ for $a,b,c>0$ As in the title. Prove the inequality $$\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ for $a,b,c>0$.
Thsi inequality can be proved in a pretty straightforward manner utilizing the Muirhead's inequality, yet I ought to prove it using the rearrangement inequality. I can't however figure out any suitable sequences and permutations of them.
| multiplying with $$a^3b^3c^3$$ then our inequality is equivalent to
$$a^8+b^8+c^8\geq a^2b^3c^3+a^3b^2c^3+a^3b^3c^2$$ now we use the wellknown inequality
$$x^2+y^2+z^2\geq xy+yz+zx$$ thus we have
$$(a^4)^2+(b^4)^2+(c^4)^2\geq (ab)^4+(bc)^4+(ca)^4$$ and now again
$$((ab)^2)^2+((bc)^2)^2+((ca)^2)^2\geq a^2b^3c^3+a^3b^2c^3+a^3b^3c^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The matrix representation of a reflection operator across the plane $x+2y+3z=0$ Let $T:\mathbb{R}^3\rightarrow \mathbb{R}^3$ be the reflection across the plane $x+2y+3z=0$, find the matrix of this linear operator $T$ in respect to the basis $B=\left\{v_1,v_2,v_3\right\}$, where we have:
$v_1=\begin{bmatrix}1\\ 1\\ -1\end{bmatrix}$ $v_2=\begin{bmatrix}-1\\ 2\\ -1\end{bmatrix}$ $v_3=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$
First of what does it mean that $T$ is a reflection across the given plane? What will hapen to let's say the vector $(1,0,0)$ if it's passed through $T$? In my textbook, there was a short example that mentioned a reflection operator, which transformed a given vector $(a,b,c)$ to the vector $(a,b,-c)$ So that would be the reflection across the standard plane $x+y+z=0$, but here, will it act as if given a vector $(a,b,c)$ you get out a vector $(a,2b,-3c)$?
EDIT:
after reading the comments, I got the matrix representation as:
$T=\begin{bmatrix}1&-1&-1\\ 1&2&-2\\ -1&-1&-3\end{bmatrix}$
Now how can I find the representation of this operator in respect to the standard basis vectors $(e_1,e_2,e_3)$?
I read about finding the change of basis matrix so I calculated $T^{-1}$:
$T^{-1}=\begin{bmatrix}\frac{4}{7}&\frac{1}{7}&-\frac{2}{7}\\ -\frac{5}{14}&\frac{2}{7}&-\frac{1}{14}\\ -\frac{1}{14}&-\frac{1}{7}&-\frac{3}{14}\end{bmatrix}$
Is this it or is there more to it?
| As noted above, vectors $v_1,v_2 \in p$ (plane),and $v_3$ is $\perp$ to $v_1,v_2$.
Since $v_1,v_2$ are in the plane their reflectection will be the same vector $v_1,v_2$ and since $v_3$ is the normal vector for this plane it's reflection will be the same vector just in the opposite direction ( $-v_3$).
So your reflection matrix in the base $B=\{v_1,v_2,v_3\}$ will look like this:
$R=\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{bmatrix}_B$
To get the matrix representation in the standard base you can use the change of basis matrix $T$.
Notice that $T^{-1}$ is the matrix which will change $B_{std} \rightarrow B$ and $T$ will change $B\rightarrow B_{std}$.
The matrices will look like this:
$T^{-1}=\begin{bmatrix}
\frac{4}{7}& \frac{1}{7} & \frac{-2}{7}\\
& & &\\
\frac{-5}{14} & \frac{2}{7} & \frac{-1}{14}\\
& & &\\
\frac{1}{14} & \frac{1}{7} & \frac{3}{14}
\end{bmatrix}
, T=\begin{bmatrix}
1 & -1 & 1\\
1 & 2 & 2 \\
-1 & -1 & 3
\end{bmatrix}$
And finally $R'=TRT^{-1}=\begin{bmatrix}
\frac{6}{7} & \frac{-2}{7} & \frac{-3}{7}\\
&&&\\
\frac{-2}{7} & \frac{3}{7} & \frac{-6}{7}\\
&&&\\
\frac{-3}{7} & \frac{-6}{7} & \frac{-2}{7}
\end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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How many non-negative integer solution is there for the equation $ x_1 + x_2 + x_3 + x_4 + 8*x_5 = 20 $? I am trying to figure out this problem and the way I think I might need to solve this is to account for the first 4 $x_i$ terms and then add on the possibilities for the $8x_5$ but I am not 100% sure that this is what I need to do. Any suggestions would be appreciated.
| Partitioning for $x_5$ $$x_1 + x_2 + x_3 + x_4 + 8*x_5 = 20\\ \to
8x_5 \leq 20 \to x_5\leq\frac{20}{8} \to x_5=0,1,2\\
\begin{cases}x_5=0 \to x_1 + x_2 + x_3 + x_4 + 8*0= 20\\\\x_5=1 \to x_1 + x_2 + x_3 + x_4 + 8*1= 20\\\\x_5=2 \to x_1 + x_2 + x_3 + x_4 + 8*2= 20\end{cases}\\$$
WE know non-negative integer solution is there for the equation $x_1+x_2+..=+x_k=n$ is $\color{red}{\left(\begin{array}{c}n+k-1\\ k-1\end{array}\right)
}$ so
$$
\begin{cases}x_5=0 \to x_1 + x_2 + x_3 + x_4 = 20&\left(\begin{array}{c}20+4-1\\ 4-1\end{array}\right)
\\\\x_5=1 \to x_1 + x_2 + x_3 + x_4 = 12&\left(\begin{array}{c}12+4-1\\ 4-1\end{array}\right)
\\\\x_5=2 \to x_1 + x_2 + x_3 + x_4 = 4 &\left(\begin{array}{c}4+4-1\\ 4-1\end{array}\right)
\end{cases}$$
final answer is :$$\left(\begin{array}{c}20+4-1\\ 4-1\end{array}\right)+\left(\begin{array}{c}12+4-1\\ 4-1\end{array}\right)+\left(\begin{array}{c}4+4-1\\ 4-1\end{array}\right)=\\\left(\begin{array}{c}21\\ 3\end{array}\right)+\left(\begin{array}{c}15\\ 3\end{array}\right)+\left(\begin{array}{c}7\\ 3\end{array}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this differential equation involving a polynomial function? I am struggling with this question please help..
Suppose $f(x)$ is a polynomial function as well as continuous in $\mathbb{R} \to \mathbb{R}$. Given that $f(2x)=f'(x) f''(x)$, then find $f(3)$.
| if f is a polynomial of the degree $n$ ($n>1$), then $f'$ and $f''$ will have the degrees $n-1$ and $n-2$. So, we'll have:
$$n=n-1+n-2\Rightarrow n=3$$. So our polynomial will be something like this:
$$f(x)=ax^3+bx^2+cx+d$$. By $f(2x)=f'(x)f''(x)$, we will have:
$$8ax^3+4bx^2+2cx+d=(3ax^2+2bx+c)*(6ax+2b) \Rightarrow 8ax^3+4bx^2+2cx+d= 18a^2x^3+(6ab+12ab)x^2+(6ac+4b^2)x+2bc \Rightarrow (8a-18a^2)x^3+(4b-18ab)x^2+(2c-6ac-4b^2)x+(d-2bc)=0 $$. Thus, we will have the following equations:
$$8a-18a^2=0 \Rightarrow a=\frac{8}{18}=\frac{4}{9}$$ Note that $a$ cannot be zero due to the degree of f.
$$4b-18ab=0\Rightarrow b=0$$
$$2c-6ac-4b^2=0\Rightarrow c=0$$
$$d-2bc=0\Rightarrow d=0$$. Thus,
$$f(x)=\frac{4}{9}x^3\Rightarrow f(3)=\frac{4}{9}*3^3=12$$
if $n<2$, then $f''=0$;hence, $f(2x)=0\Rightarrow f(3)=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the second derivative of $f(x)=\frac{x^2}{10+x}$ How do you find the second derivative of $f(x)=\frac{x^2}{10+x}$?
I get $f'(x)={20x+x^2}/{(10+x)^2}$
Then get stuck here:
$(10+x)^2 (20+2x) - (20x+x^2) \frac {2(10+x)}{((10+x)^2)^2}$
| Just do it:
$f(x) = \frac {x^2}{10 - x}$
$f'(x) = \frac {2x(10 - x) - x^2(-1)}{(10-x)^2} = \frac{-2x^2 + 20x +x^2}{(10-x)^2} = \frac{20x - x^2}{(10 - x)^2}=\frac {x(20-x)}{(10-x)^2}$
$f''(x) =\frac{[x(-1)+ (20-x)](10-x)^2 - x(20 -x)2(10-x)(-1)}{(10-x)^4}$
$=\frac{[x(-1)+ (20-x)](10-x)^{\not 2} - x(20 -x)2\not{(10-x)}(-1)}{(10-x)^{\not 43}}$
$=\frac{[-x+ (20-x)](10-x) + 2x(20 -x)}{(10-x)^3}$
$=\frac{2(10-x)^2 + 2x(20-x)}{(10-x)^3}=\frac{200 - 40x + 2x^2 + 40x - 2x^2}{(10-x)^3}=\frac{200}{(10-x)^3}$
Or you can generalize first:
$(\frac fg)'' =$
$((\frac fg)')' = $
$(\frac{f'g - fg'}{g^2})'=$
$(\frac{f'}g)' - (\frac{fg'}{g^2}) =$
$\frac{f''g - f'g'}{g^2} - \frac{(f'g'+fg'')g^2 - fg'*2gg'}{g^4} =$
$\frac{f''}g - \frac{f'g'}{g^2} - \frac{f'g'}{g^2} - \frac{fg''}{g^2} + \frac{2fg'^2}{g^3}=$
$\frac{f''}g - 2\frac{f'g'}{g^2} - \frac{fg''}{g^2} + \frac{2fg'^2}{g^3}=$
$\frac{2}{10-x} - 2\frac{-2x}{(10-x)^2} - 0 + \frac{2x^2}{(10-x)^3}=$
$\frac {2(10-x)^2 + 2x(10 - x) + 2x^2}{(10-x)^3}=$
$\frac{2[(10-x)+x]^2}{(10-x)^3 }=\frac {2*10^2}{(10-x)^3}= \frac {200}{(10-x)^2}$
But I can't see that $(\frac fg)'' = \frac{f''}g - 2\frac{f'g'}{g^2} - \frac{fg''}{g^2} + \frac{2fg'^2}{g^3}$ is anything worth memorizing.... but still....
| {
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"url": "https://math.stackexchange.com/questions/2130806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$
Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=3$. Prove that:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}.$$
This inequality we can prove by BW with computer.
I am looking for an human proof, which we can use during competition.
For example, we need to prove that
$$\sum_{cyc}\left(\frac{a^3}{a+b}-\frac{a^2}{2}\right)\geq\frac{3}{2}-\frac{a^2+b^2+c^2}{2}$$ or
$$\sum_{cyc}\frac{a^2(a-b)}{a+b}\geq\sqrt[3]{3(a^3+b^3+c^3)^2}-a^2-b^2-c^2$$ or
$$\sum_{cyc}\frac{a^2(a-b)}{a+b}\geq\frac{3(a^3+b^3+c^3)^2-(a^2+b^2+c^2)^3}{\sqrt[3]{9(a^3+b^3+c^3)^4}+(a^2+b^2+c^2)\sqrt[3]{3(a^3+b^3+c^3)^2}+(a^2+b^2+c^2)^2}$$
and since $\sqrt[3]{3(a^3+b^3+c^3)^2}\geq a^2+b^2+c^2$ or $3(a^3+b^3+c^3)^2\geq(a^2+b^2+c^2)^3$,
it remains to prove that
$$\sum_{cyc}\frac{a^2(a-b)}{a+b}\geq\frac{3(a^3+b^3+c^3)^2-(a^2+b^2+c^2)^3}{3(a^2+b^2+c^2)^2}$$
and from here I don't see what to do.
| I propose a simple alternative of the third step of my proof in order to avoïd Lagrange multipliers.
New Step 3. (version without Lagrange multipliers.)
I suppose here that $\max (a,b,c)\geq \sqrt{2}.$
without loss of generality I suppose that $a\geq \sqrt{2}.$
We also have $a\leq 3^{1/3}$ and $b,c \leq (3-\sqrt{2}^3)^{1/3}.$
Thus $\max(a+b,a+c,a+c)\leq 3^{1/3}+(3-\sqrt{2}^3)^{1/3}<2$
Hence
$$a^3(a+b)+b^3(b+c)+c^3(a+c)< 2(a^3+b^3+c^3)=6.$$
c.q.f.d
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
every linear transformation is associated with a matrix and vice versa I'm quite confused with this theorem
Can someone explain me step by step how do I find a linear transformation from a given matrix and finding a matrix from a given linear transformation please?
Examples will be great.
Thanks alot.
| Let
$$A=\begin{pmatrix} 1 & 0 & 9 \\ 0 & -2 & 1 \\ 0 & 0 & 3 \end{pmatrix}\in\Bbb R^{3\times 3}$$
be a matrix. The linear map associated to it is the map
\begin{align*}
F_A:\Bbb R^3 &\longrightarrow \Bbb R^3 \\
\begin{pmatrix} a\\b\\c\end{pmatrix} &\longmapsto
\begin{pmatrix} 1 & 0 & 9 \\ 0 & -2 & 1 \\ 0 & 0 & 3 \end{pmatrix}
\cdot\begin{pmatrix} a\\b\\c\end{pmatrix}
=
\begin{pmatrix} a+9c\\-2b+c\\3c\end{pmatrix}.
\end{align*}
Conversely, assume you have any linear map $F:\Bbb R^3\to \Bbb R^3$. Let
\begin{align*}
e_1 &= \begin{pmatrix} 1\\0\\0\end{pmatrix}, &
e_2 &= \begin{pmatrix} 0\\1\\0\end{pmatrix}, &
e_3 &= \begin{pmatrix} 0\\0\\1\end{pmatrix}.
\end{align*}
Let
$$
\begin{pmatrix} A_{1i}\\A_{2i}\\A_{3i}\end{pmatrix} := F(e_i),
$$
this defines a Matrix $A\in\Bbb R^{3\times 3}$ with entries $A_{ij}$.
If you apply this to our $F_A$ above, you see how we get the matrix $A$ back. All of this generalizes as explained in the other answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that: $\cos^2 20° + \cos^2 40° +\cos^2 80° = \sin^2 20° + \sin^2 40° + \sin^2 80°$ Prove that: $\cos^2 20° + \cos^2 40° +\cos^2 80° = \sin^2 20° + \sin^2 40° + \sin^2 80°$
My Attempt:
$$L.H.S=\cos^2 20° + \cos^2 40° + \cos^2 80°$$
$$=\dfrac {1+\cos 40}{2}+\dfrac {1+\cos 80}{2} + \dfrac {1+\cos 160°}{2}$$
$$=\dfrac {3+\cos 40°+\cos 80°+\cos 160°}{2}$$
I.could not solve further from here..
| I shall prove the generalization of $$\cos x+\cos(x+120^\circ)+\cos(x+240^\circ)=0$$ (here $x=40^\circ$)
$$S=\sum_{r=0}^{n-1}\cos\left(x+\dfrac{360^\circ r}n\right)=0$$
Method $\#1:$
$S=$ real part of $\sum_{r=0}^{n-1} \exp i\left(x+\dfrac{360^\circ r}n\right)$
Now $\displaystyle \sum_{r=0}^{n-1} \exp i\left(x+\dfrac{360^\circ r} n \right) = \frac{e^{i360^\circ}-1}{\exp i\left(x+\dfrac{360^\circ(-n)}n\right)-1}=0$
Now equate the real & the imaginary parts.
Method $\#2:$
Using multiple angle formula cosine:
$$\cos(nx)=2^{n-1}\cos^nx-n2^{n-3}\cos^{n-2}x+\cdots$$
Now if $\cos nx=\cos ny, nx=360^\circ m\pm ny$ where $m$ is any integer
$x= y+\dfrac{360^\circ m}n$ where $m\equiv0,1,2,\ldots, n-1\pmod n$
So, the roots of $$2^{n-1}\cos^nx-n2^{n-3}\cos^{n-2}x+\cdots-\cos ny=0$$ are
$\cos\left(y+\dfrac{360^\circ m}n\right)$ where $m\equiv0,1,2,\ldots n-1\pmod n$
Using Vieta's formula,
$$\sum_{r=0}^{n-1}\cos\left(y+\dfrac{360^\circ m}n\right)=\dfrac0{2^{n-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the indicated derivative Find the indicated derivative
$\frac {d}{dt}$ $\frac {(6t-5)^6}{t+9}$
I'm stuck after getting to this part
$\frac {(6t-5)^6-36(6t-5)^5(t+9)}{(t+9)^2}$
How do they get to the answer
$\frac {(6t-5)^5(30t+329)}{(t+9)^2}$
| \begin{align*}
\frac{d}{dt}\left(\frac{(6t-5)^6}{t+9} \right) & = \frac{6((6t-5)^5.6)(t+9)-(6t-5)^6}{(t+9)^2}\\
& = \frac{(6t-5)^5(36(t+9)-6t+5)}{(t+9)^2}\\
& = \frac{(6t-5)^5(30t+329)}{(t+9)^2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Show via direct proof that $k(k+1)(k+2)$ is divisible by $6$. How do I show via direct proof that $k(k+1)(k+2)$ is divisible by $6$. I showed it was divisible by $2$ because at least one of the multiples is even but could not figure out how to show it is divisible by $3$. I tried making $k$ even or odd and substituting $2q$ or $2q+1$ but have not made much progress. Does anyone have any tips as to what direction I should take? Thanks!
| If $k = 6a+b$,
where $0 \le b \le 5$,
then
$\begin{array}\\
k(k+1)(k+2)
&= k(k^2+3k+2)\\
&= k^3+3k^2+2k\\
&= (6a+b)^3+3(6a+b)^2+2(6a+b)\\
&= (6a)^3+3((6a)^2b+6ab^2)+b^3+3(36a^2+12ab+b^2)+12a+2b\\
&= 6(6^2a^3+18a^2b+ab^218a^2+6ab+2a)+b^3+3b^2+2b\\
&= 6(6^2a^3+18a^2b+ab^218a^2+6ab+2a)+b(b+1)(b+2)\\
\end{array}
$
By directly computing
$b(b+1)(b+2)$
for $0 \le b \le 5$,
the values are all divisible by $6$
(they are
$0, 6, 24, 60, 120, 210$).
Therefore,
the product is
always divisible by $6$,
being the sum
of two terms
each of which
is divisible by $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that: $ \frac{x^2}{2(x + 1)} < x − \ln(x + 1) < \frac{x^2}{ 2} , x > 0$ Show that: $ \frac{x^2}{2(x + 1)} < x − \ln(x + 1) < \frac{x^2}{ 2} , x > 0$
One method is considering a function $f(x) = x − \ln(1 + x) − \frac{x^2}{2}$ and showing its derivative $f'(x) < 0$.
Similarly, considering another function $g(x) = x − \ln(1 + x) − \frac{x^2}{ 2(x + 1)}$ and showing its $g'(x) > 0$.
Is there some other method that is not lengthy?
| You can use limits for the natural log (which can basically be seen after Taylor expansion, proof is here) to see that
$\ln(x+1) > x - x^2/2$ and $(x+1)\ln(x+1) < x + x^2/2$
The first inequality immediately gives
$x-\ln(x+1) < x^2/2$ and
and the second one gives
$(x+1) (x-\ln(x+1)) > x + x^2 - (x + x^2/2) = x^2/2$
so we have the two required limits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Odd powers of trig integrands such as in $\int \sin^7 \theta\cos^5\theta\, d\theta$ I am doing as many trig integrals exercices as I can to develop the skills and thought patterns. I have a rather simple question about the result I get when computing integrals such as $\int \sin^7 \theta\cos^5\theta\, d\theta$, I use the basic substitution technique
$$\int \sin^7 \theta\cos^5\theta\, d\theta\,=\,\int u^7\theta\left(1-u^2\theta\right)^2\, du\,:\,u=\sin\theta \quad du=\cos\theta\, d\theta$$
wich gives me
$$\frac 18 \sin^8\theta - \frac 15 \sin^{10}\theta + \frac 1{12} \sin^{12}\theta +C$$
I check my answers with $Mathematica$ which gives,
$$\frac {-5\cos{2\theta}}{1024}+\frac {5\cos{4\theta}}{8192}+\frac {5\cos{6\theta}}{6144} - \frac {\cos{8\theta}}{4096} - \frac {\cos{10\theta}}{10240}+\frac {\cos{12\theta}}{24 576}$$
this is far from the result I got by substitution so I thought I was doing something wrong and I tried to achieve the same result but it is a bit painful, so I tried a definite integral as a quicker way to convince myself. Sure enough, I get the same answer as $Mathematica$ when I compute definite integrals. For example,
$$\int_0^\frac \pi 2 \sin^7 \theta\cos^5\theta\, d\theta\,=\,\frac1{120}$$
$$\frac 18 \sin^8\frac \pi 2- \frac 15 \sin^{10}\frac \pi2+ \frac 1{12} \sin^{12}\frac \pi2\,=\,\frac 1{120}$$
Does this have to do with the way $Mathematica$ does computations or am I doing something wrong? Also, is it useful to get it down to the same form as does $Mathematica$ or is the simple u-substitution result always reliable and correct?
| Actually, both of your solutions are correct. They are just of a different form.
Using the following identities:
$$\cos(2\theta)\equiv \cos^2 \theta-\sin^2 \theta \tag{1}$$
$$\sin(2\theta)\equiv 2\sin \theta \cos \theta \tag{2}$$
$$\sin^2 \theta+\cos^2 \theta\equiv 1 \tag{3}$$
We can obtain the result you've obtained from Mathematica:
$$\frac 18 \sin^8\theta - \frac 15 \sin^{10}\theta + \frac 1{12} \sin^{12}\theta +C$$
To do so, replace each of the $\cos$ terms on the result you've obtained from Mathematica with powers of $\sin \theta$.
$$\frac {-5\cos(2\theta)}{1024}+\frac {5\cos(4\theta)}{8192}+\frac {5\cos(6\theta)}{6144} - \frac {\cos(8\theta)}{4096} - \frac {\cos(10\theta)}{10240}+\frac {\cos(12\theta)}{24 576}$$
For the $\cos(4\theta), \cos(6\theta),\cdots ,\cos(12\theta)$ terms apply all the 3 above identities:
This is an example for $\cos(4\theta)$:
$$\cos(4\theta)=\cos^2(2\theta)-\sin^2(2\theta)=(\cos^2 \theta-\sin^2\theta)^2-(2\sin \theta\cos\theta)^2$$
$$\therefore \cos(4\theta)=\sin^4\theta+\cos^4\theta-6\sin^2 \theta\cos^2 \theta $$
$$\therefore \cos(4\theta)=\sin^4\theta+(1-\sin^2\theta)^2-6\sin^2\theta\cdot (1-\sin^2\theta)$$
$$\therefore \cos(4\theta)=8\sin^4\theta-8\sin^2 \theta+1$$
Now apply this process to the other $\cos$ terms.
If you would not like to do this method by hand (since it is extremely tedious), you can check this using Wolfram|Alpha. Do not be alarmed that it suggests that no solutions exist on this specific input. That is because the arbitrary constant of integration $C$ was not considered (It was assumed that $C=0$). If you let $C=-\frac{462}{122880}$, you will see that they are identical and apply for all $\theta \in \mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the max value of $a^2+b^2$ if $x^4+ax^3+3x^2+bx+1\ge0$ Let $f(x)=x^4+ax^3+3x^2+bx+1$, where $a,b \in \mathbb{R}$. If $f(x) \ge 0$ for all $x \in \mathbb{R}$, what is the maximum possible value of $a^2+b^2$?
I don't know how to proceed. Hints or help will be appreciated.
| Following mathlove's strategy, we can do a little better . . .
If $f(x) = x^4 + ax^3 + 3x^2 + bx + 1$, where
\begin{align*}
a&=0\\[8pt]
b&=\sqrt{6+{\small{\frac{14}{9}}}\sqrt{21}}
\end{align*}
then $f(x) \ge 0$ for all $x \in \mathbb{R}$, and
$$a^2 + b^2= 6+{\small{\frac{14}{9}}}\sqrt{21} \approx 13.12845108$$
I don't know if this is best possible, however it is best possible for the case $a=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Problem with $\int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}} \text{dx}$ I wrote
\begin{eqnarray}
I
&=& \int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}} \text{dx}\\
&=&
\int \frac{\sin^{3}x}{\sin{x}\cos{x}}\text{dx}+\int \frac{\cos^{3}x}{\sin{x}\cos{x}}\text{dx}\\
&=&
\int \frac{\sin^{2}x}{\cos^2{x}}\cos{x}\text{dx}+\int \frac{\cos^{2}x}{\sin^2{x}}\sin{x}\text{dx}\\
&=&
\int \frac{\sin^{2}x}{1-\sin^2{x}}\cos{x}\text{dx}+\int \frac{\cos^{2}x}{1-\cos^2{x}}\sin{x}\text{dx}\\
&=&
\int\frac{u^2}{1-u^2}du-\int\frac{m^2}{1-m^2}dm\\
&=&
\color{blue}{\int\frac{u^2}{1-u^2}du-\int\frac{u^2}{1-u^2}du}\\
&=&
0
\end{eqnarray}
and Other
Let $x=\dfrac{\pi}{2}-t$ so
$$I=\int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}} \text{dx}=-\int \frac{\sin^{3}t+\cos^{3}t}{\sin{t}\cos{t}} \text{dt}=-I$$
but correct answer is
$$
I=\ln\left|\dfrac{1+\tan\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}\right|+\ln\left|\tan\dfrac{x}{2}\right|-\sin x+\cos x+C
$$
Question.1 Where is wrong.?
Question.2 what conditions guarantee that our changing variable in indefinite integrals doesn't change our final solutions!.
| The step that is wrong is precisely the one highlighted in blue. You have $u=\cos(x)$ and $m=\sin(x)$, how can you then happily set $m$ to $u$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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The roots of the equation $x^2+3x-1=0$ are also the roots of $x^4+ax^2+bx+c=0$
The roots of the equation $x^2+3x-1=0$ are also the roots of quartic equation $x^4+ax^2+bx+c=0$. Find $a+b+4c$.
This problem is from yesterday's Bangladesh National Math Olympiad 2017. I tried this using Vieta Root Jumping but no luck. After the contest my friend laughed at me "One doesn't simply try a 10 point problem with Vieta Root Jumping".
How to solve this problem?
| $$(x^2+3x-1)(x^2-3x+3)=x^4-7x^2+12x-3$$ has the same real roots as $x^2+3x-1$.
In this case $a=-7,b=12,c=-3$ so $$a+b+4c=-7.$$
Moreover $$(x^2+3x-1)(x^2-3x+d)=x^4+(d-10)x^2+(3d+3)x-d$$ for $$d>\frac{9}{4}$$ has the same real roots as $x^2+3x-1$. In this case $a=d-10,b=3d+3,c=-d$ so $$a+b+4c=-7.$$
So if the statement of the problem says the same real roots then it seems to be that $a+b+4c=-7$ in all cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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If $m^2 +n^2 - m$ is a multiple of $mn$ and $p$ divides $m$ then $p^2$ divides $m$ Given natural numbers $m,n$ such that $m^2 + n^2 - m \equiv 0 \pmod{mn}$ and $p$ a prime dividing $m$, then I want to show that $p^2$ divides $m$.
I have tried multiple approaches: Euclidean division gives that $m = qp^2 + r$ for some $q, r \in \mathbb{N}$ where $0 \leq r < p^2$ and filling this in to show that $r = 0$. I also tried using that $p$ divides $m$ and $mn$ divides $m^2 + n^2 - m$ and again filling this in to find that $p^2$ has to divide $m$, but with no succes.
Any hints would be appreciated.
$\textbf{EDIT}$ Based on the given hint I have the following solution:
Since $p$ divides $m$, we have that $p$ divides $mn$ and therefore $m^2 + n^2 - m$. Hence, there is some $k \in \mathbb{Z}$ such that $kp = m^2 + n^2 -m$ and therefore we have that $kp - m^2 + m = n^2$, so that $p$ divides $n^2$. Since $p$ is prime, it must divide $n$. Since $p$ divides both $m,n$ we have that $p^2$ divides $mn$ and therefore $m^2 + n^2 - m$. Also $p^2$ divides $m^2, n^2$ hence it must divide $m$.
| $p$ divides $m\implies p$ divides both $mn$ and $m^2-m=m(m-1)$.
$p$ divides $mn$ and $mn$ divides $m^2+n^2-m\implies p$ divides $m^2+n^2-m$.
$p$ divides both $m^2-m$ and $m^2+n^2-m\implies p$ divides their substraction $m^2+n^2-m-(m^2-m) = n^2$.
$p$ is prime and $p$ divides $n^2=n\cdot n\implies p$ divides $n$.
Conclusion:$p$ divides both $m^2$ and $n^2\implies p$ divides $m^2+n^2-(m^2+n^2-m) =m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Let $A = \mathbb{R} \times \mathbb{R}$, and the operation $* : A \times A \rightarrow A$ on $A$. Check if associative, commutative, identity, inverse. I have this exercise:
on the set $A = \mathbb{R} \times \mathbb{R}$ is defined the operation $* : A \times A \rightarrow A$ such that
$\forall (a,x), (b,y) \in A \quad (a,x)*(b,y) = \left ( \frac{2}{5}ab , \frac{3}{4} + y + x \right )$
*
*Check if the operation is associative;
*Check if the operation is commutative;
*Check if exists the identity element of the algebric structure $(A, *)$;
*Check if exists the inverse of $(2,-3)$ in $(A,*)$.
This is what I have done for the first two points, and I think that is right:
*
*Associativity:
I have to check if the following equation holds
$\left [ (a,x) * (b,y) \right ] * (c,z) \overset{?}{=} (a,x) * \left [ (b,y) * (c,z) \right ]$
hence,
LHS:
in the following I have assigned $m = \frac{2}{5}ab, n = \frac{3}{4}+y+x$
$\begin{array}{lcl}\left [ (a,x) * (b,y) \right ] * (c,z) & = & \left ( \frac{2}{5} ab, \frac{3}{4} +y+x\right ) * (c,z) \\ & = & (m,n)*(c,z) \\ & = & \left [ \frac{2}{5}mc, \frac{3}{4} + z +n \right ] \\ & = & \left [ \frac{2}{5} \left ( \frac{2}{5}ab \right )c, \frac{3}{4} + z + \left ( \frac{3}{4} +y +x \right )\right ] \\ & = & \left [ \frac{4}{25}abc, \frac{3}{2} +z+y+x\right ] \end{array}$
RHS:
in the following I have assigned $g = \frac{2}{5}bc, h = \frac{3}{4}+z+y$
$\begin{array}{lcl}(a*x)*\left [ (b,y) * (c,z) \right ] & = & (a,x) * \left [ \frac{2}{5}bc, \frac{3}{4}+z+y \right ] \\ & = & (a,x) * (g,h) \\ & = & \left [ \frac{2}{5} ag, \frac{3}{4} + h + x \right ] \\ & = & \left [ \frac{2}{5}a \left ( \frac{2}{5}bc \right ), \frac{3}{4}+\left ( \frac{3}{4} + z + y \right ) +x\right ] \\ & = & \left [ \frac{4}{25}abc, \frac{3}{2} +z+y+x \right ] \end{array}$
the equation holds and the operation $*$ is associative.
*Commutative element:
I have to check if the following equation holds
$(a,x) * (b,y) \overset{?}{=} (b,y) * (a,x)$
LHS:
$(a,x) * (b,y) = \left ( \frac{2}{5}ab, \frac{3}{4} + y + x \right )$
RHS:
$(b,y) * (a,x) = \left ( \frac{2}{5}ba, \frac{3}{4}+x+y \right )$
the equation holds and the operation $*$ is commutative.
*Identity element:
here, I have some problems, I am not sure on what to do, I have tried this but without success:
maybe, we have to check the following equation?
$(a,x)*(I_b, I_y) \overset{?}{=} (a,x)$
hence,
$\begin{array}{lcl}(a,x)*(I_b, I_y) & = & \left [ \frac{2}{5}aI_b, \frac{3}{4} + I_y + x \right ] \end{array}$
i.e. we should find a value to $(I_b,I_y)$ such that $\left [ \frac{2}{5}aI_b, \frac{3}{4} + I_y + x \right ] = (a,x)$
and therefore I have problems in point 4.
I don't know!
Please, can you help me?
Many thanks really!
| Supposing $(a,x)$ is an identity element immediately tells us that $\left(\frac{2}{5}(1)(a), \frac{3}{4} + x + 1\right) =(1,1) \ast (a,x) = (1,1)$. Now you can see what $a,x$ specifically have to be in order for $(a,x)$ to qualify as an identity element: $a = \frac{5}{2}$ and $x = -\frac{3}{4}$. But you still need to check that $\left(\frac{5}{2},-\frac{3}{4}\right)$ is truly an identity element.
| {
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"url": "https://math.stackexchange.com/questions/2146430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does "$x^2 - 5x + 6 = 0$", which is the same as "$(x-3)(x-2) = 0$", represent a parabola? Consider the equation $x^2 - 5x + 6 = 0$. By factorising I get $(x-3)(x-2) = 0$. Which means it represents a pair of straight lines, namely $x-2 =0 $ and $x- 3 = 0$, but when I plot $x^2 - 5x + 6 = 0$, I get a parabola, not a pair of straight lines. Why?
Plotting: x^2 - 5x + 6 = 0
at Wolfram Alpha, I get the result:
| As has been pointed out $x^2-5x+6=0\quad$ is not the equation of the parabola.
The equation of the parabola is:
$y_p = x^2-5x+6 = (x-2)(x-3)$
The equations of the two straight lines within the equation are is $y_1 = x-2$ and $ y_2 = x - 3$ and both lines can be plotted as a function of the same variable $x$. But if we multiple the two lines together we get:
$y_1y_2 = (x-2)(x-3)$
and $y_2= y_1 - 1$ so
$y_1(y_1-1) = (x-2)(x-3)$
$y_1^2 -y_1 = (x-2)(x-3)$
and thus $y_p = y_1^2 - y_1$ which isn't terribly useful.
But as $x$ goes to infinity the constant term 6 in the equation becomes negligible. Thus the parabola approaches the curve $y_3 = x^2-5x$ as $x$ approaches $+\infty$ or as $x$ approaches $-\infty$.
Now if we translate the axis to the point ($\dfrac{5}{2}$, $-\dfrac{1}{4}$), we can define:
$y' = y + \dfrac{1}{4}$ and $x' = x - \dfrac{5}{2}$
so that
$y' = x'^2$
and the quandary would seem to disappear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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If $a+b+c=1$ then $\sum\limits_{cyc}\frac{a}{\sqrt[3]{a+b}}\leq\frac{31}{27}$
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$ and $a+b+c=1$.
Prove that:
$$\frac{a}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c}{\sqrt[3]{c+a}}\leq\frac{31}{27}$$
The equality occurs for $(a,b,c)=\left(\frac{19}{27},\frac{8}{27},0\right)$.
This inequality is similar to the following inequality, which was proposed by Walther Janous.
For all non-negatives $x$, $y$ and $z$ such that $xy+xz+yz\neq0$ prove that:
$$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq\frac{5}{4}\sqrt{x+y+z}$$
My proof:
By Cauchy-Schwarz $$\left(\sum_{cyc}\frac {x}{\sqrt {x+y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}.$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}$$ or
$$\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$$
$$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$$ or
$$\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$$
$$+\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0,$$ which is obvious.
If we want to use a similar way for the starting inequality, we need to use Holder, which gives very big numbers.
Maybe there is another reasoning?
Thank you!
| Here is my personal take on the problem;
We can use the substitutions $b+c = 1-a$, $c+a = 1-b$, $c = 1-a-b$ to reduce the problem to the following inequality in arbitrary $a,b$;
$$\dfrac{a}{\sqrt[3]{a+b}} + \dfrac{b}{\sqrt[3]{1-a}} + \dfrac{1-a-b}{\sqrt[3]{1-b}} \leq \dfrac{31}{27}$$
Define $f(a,b) = \dfrac{a}{\sqrt[3]{a+b}} + \dfrac{b}{\sqrt[3]{1-a}} + \dfrac{1-a-b}{\sqrt[3]{1-b}}$. I suspect that using the first and second derivative tests to find local maxima-minima could solve the problem from here, as $\dfrac{\partial f}{\partial a}$ and $\dfrac{\partial f}{\partial b}$ are, while messy, not difficult to calculate.
In fact, this can be generalized quite a bit. This method can be generalized as follows;
Consider $\sum_{cyc} g(x_1,\dots,x_n)$ for some well-defined function $g$, where the cyclic behavior occurs over a set of variables $x_1,\dots,x_k$ for $k > n$ and we restrict these $x$ by stating that $\sum_{j=1}^k x_j = A$ for a fixed constant $A$. Then we can write $x_k = A - \sum_{j=1}^{k-1} x_j$ and use this as the basis for a substitution to create a function $f(x_1,\dots,x_{k-1}) = \sum_{cyc} g(x_1,\dots,x_n)$ with restrictions on the $x_j$ removed and using the substitution previously given for $x_k$. If $g$ is differentiable, then so will $f$ be, and so local extremes can be calculated. If $g$ is twice differentiable, then so is $f$ and we can use the second derivative test to test whether these values are true extrema or saddle points.
| {
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"url": "https://math.stackexchange.com/questions/2148909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Integral $\int\frac{(1 - x)^2 e^x}{(1 + x^2)^2}dx$ need help integrating this have tried to do this but keep getting stuck due to the whole square in the fraction. Do keep in mind that I have recently started integration and only know the basic methods.
$$\int\frac{(1 - x)^2 \mathrm e^x}{(1 + x^2)^2} \; \mathrm d x$$
| $$\int\frac{(1 - x)^2 \ e^x}{(1 + x^2)^2} \; d x\\
\int\frac{(1 + x^2 - 2x) \ e^x}{(1 + x^2)^2} \; d x\\
\int\frac{e^x}{(1 + x^2)} \; d x +\int\frac{- 2x \ e^x}{(1 + x^2)^2} \; d x
$$
Tackling the integral on the left first.
$$\int\frac{e^x}{(1 + x^2)} \; d x$$
Integration by parts
$u = (1+x^2)^{-1} ;du = -2x (1+x^2)^{-2} dx\\
dv = e^x dx ; v = e^x$
$$\frac {e^x}{(1+x^2)} + \int \frac {2x e^x}{(1+x^2)^2} dx$$
And this remaining integral cancels.
$$\frac {e^x}{(1+x^2)} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving a Triangle relationship $1$" />
Given $\triangle ABC$, we have$$c=a\cos B+b\cos A\tag{1}$$
Where $a=BC$, $b=AC$ and $c=AB$.
Question: How do you prove the relation?
I started with$$\begin{align*} & CD=a\cos B\\ & CD=b\cos A\end{align*}$$
So adding them up gives$$2CD=a\cos B+b\cos A$$However, what's confusing is how $2CD=c$ for the equation $(1)$ to hold.
| Using the Law of cosines : $c=a\cos B+b\cos A \Leftrightarrow c=a \frac {a^2 + c^2 - b^2}{2 ac}+b\frac {b^2 + c^2 -a^2}{2bc} \Leftrightarrow 2c^2=2c^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solutions to $8^n = a^3+b^3+c^3-3abc$ Let $n$ be a positive integer. What is the number of solutions to the equation
$$8^n = a^3+b^3+c^3-3abc$$
with integers $a\geq b\geq c\geq 0$?
We have the factoring
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
meaning that both factors must be powers of $2$.
| Note $$a^2+b^2+c^2-ab-ac-bc=(1/2)((a-b)^2+(a-c)^2+(b-c)^2)$$ Now suppose $a+b+c$ is even. Then all three variables are even, or else exactly one of them is even. If all three are even, then we can divide the original equation through by 8, getting a smaller solution. Repeat this enough times, and we get to a solution with exactly one even variable; let it be $a$. Then $a^2+b^2+c^2-ab-ac-bc$ is odd, but it's a power of 2, so it must be 1, so $(a-b)^2+(a-c)^2+(b-c)^2=2$. The only way to get three squares to sum to $2$ is for one of them to be zero and the others to be $1$, so $a-b=\pm1$, $a-c=\pm1$, $b=c$. Taking the plus sign, we get $3a-2=2^r$ for some $r$, so $a=(2^r+2)/3$, $b=c=(2^r-1)/3$. Then $$a^3+b^3+c^3-3abc=(1/27)((2^r+2)^3+2(2^r-1)^3-3(2^r+2)(2^r-1)^2)=2^r$$ so we need to take $r$ to be a multiple of $3$. For example, $r=6$ leads to $a=22$, $b=c=21$, $a^3+b^3+c^3-3abc=64=8^2$.
The reader may enjoy working through the case with the minus sign.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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An inequality concerning Lagrange's identity
Does the following inequality still hold
$$(a^2_{1}+b^2_{2}+b^2_{3})(a^2_{2}+b^2_{3}+b^2_{1})(a^2_{3}+b^2_{1}+b^2_{2})\ge (b^2_{1}+b^2_{2}+b^2_{3})(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2 $$
$$+\dfrac{1}{2}(b_{1}a_{2}b_{3}-b_{1}b_{2}a_{3})^2+\dfrac{1}{2}(b_{1}b_{2}a_{3}-a_{1}b_{2}b_{3})^2+\dfrac{1}{2}(a_{1}b_{2}b_{3}-b_{1}a_{2}b_{3})^2\tag{*}$$
for $a_{i},b_{i}\in \mathbb R,i=1,2,3$?
we know Lagrange's identity
$$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})=(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2+\sum_{i=1}^{2}\sum_{j=i+1}^{3}(a_{i}b_{j}-a_{j}b_{i})^2$$
then we have Cauchy-Schwarz inequality
$$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})\ge (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2$$
| We have the following identity.
$$(a^2+y^2+z^2)(b^2+x^2+z^2)(c^2+x^2+y^2)-(x^2+y^2+z^2)(ax+by+cz)^2-$$
$$-\frac{1}{2}(ayz-bxz)^2-\frac{1}{2}(ayz-cxy)^2-\frac{1}{2}(bxz-cxy)^2=$$
$$=(x^2+y^2)(x^2+z^2)(y^2+z^2)+a^2b^2c^2+a^2b^2(x^2+y^2)+a^2c^2(x^2+z^2)+b^2c^2(y^2+z^2)-$$
$$-2abxy(x^2+y^2)-2acxz(x^2+z^2)-2bcyz(y^2+z^2)-xyz(abz+acy+bcx).$$
I hope it can help.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find Minima and Maxima of $ y = \frac{x^2-3x+2}{x^2+2x+1}$ $$ y = \frac{x^2-3x+2}{x^2+2x+1}$$
I guess I made some mistakes cause after taking the first derivative and simlifying I have
$$y = \frac{2x^3-4x^2+5}{(x+1)^2}$$
but then numerator has complex roots. which should not be, IMO
| Hint. Write it as:
$$ y = \frac{x^2+2x+1-5(x+1)+6}{(x+1)^2}=1 -\frac{5}{x+1}+\frac{6}{(x+1)^2}$$
The latter is a quadratic in $w=\frac{1}{x+1}\,$ with easy to determine extrema: $\;y=1-5w+6w^2\,$.
| {
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"url": "https://math.stackexchange.com/questions/2154273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation of three lines forming a equilateral triangle.
Prove that if general equation $$\cos \left(3\alpha\right)\left(x^3 -3xy^2\right) + \sin3\alpha\left(y^3 - 3x^2y\right) + 3a\left(x^2 + y^2\right) - 4a^3 = 0$$ represents three lines then they form a equilateral triangle of area $3a^2\sqrt 3$.
This equation looks like an good candidate to solve in polar coordinates than in cartesian coordinates.
Converting it into polar coordinates by parts
\begin{align}
\left(x^3 -3xy^2\right) &= r^3 \left(\cos^3 \theta - 3\cos \theta \sin^2 \theta\right) =r^3 \left(4\cos^3 \theta - 3\cos \theta\right) = r^3\cos 3\theta \tag1\label1 \\
\left(y^3 - 3x^2y\right) &= r^3\left(4\sin^3 \theta - 3\sin\theta\right) = r^3\left(\cos\left(3\pi/2 - 3\theta\right)\right) = -r^3\sin 3\theta\tag2\label2
\end{align}
Using $\eqref1$, $\eqref2$
\begin{align}
r^3\left(\cos 3\alpha\cos 3\theta - \sin3\alpha\sin 3\theta\right) + 3ar^2 - 4a^3 &= 0 \\
r^3\cos \left(3\alpha+ 3\theta\right) + 3ar^2 - 4a^3 &= 0
\end{align}
On dividing by $a^3$ and substituting $z = r/a$
$$z^3\cos \left(3\alpha+ 3\theta\right) + 3z^2 - 4 = 0.$$
Now this happens very often to me, I am not able to solve this cubic for $z$.I did try to find the factor by putting $\,z = \sin\left(3\alpha + 3\theta\right),\,\tan\left(3\alpha + 3\theta\right),\,\ldots$
What should I do now ? how to solve this cubic ?
| The given equation always represents three lines. When we rotate the axes by an angle $\alpha$, in polar coordinates, $r$ remains unchanged but $\theta$ changes to $\theta - \alpha$. Thus the combined equation becomes
$$r^3\cos \left(3\alpha+ 3(\theta-\alpha)\right) + 3ar^2 - 4a^3 = 0$$
This is equivalent to rotating the curve clockwise by $\alpha$ about the origin and thus the shape of the curve does not change. Hence it is enough to prove that the above equation represents three lines.
We have
\begin{align*}
r^3\cos3\theta +3ar^2 - 4a^3 &= 0\\
r^3(4\cos^3 \theta -3\cos\theta) + 3ar^2 - 4a^3 &= 0\\
4x^3 -3x(x^2+y^2)+3a(x^2+y^2) - 4a^3 &= 0\\
x^3 - 3xy^2 + 3ax^2 + 3ay^2 - 4a^3 &= 0
\end{align*}
where in the last two steps, we have converted the equation back to Cartesian. Clearly $x=a$ is a factor and dividing this out we get
$$(x-a)(x^2 - 3y^2 + 4ax + 4a^2) = 0$$
and hence
$$(x-a)((x+2a)^2 - 3y^2) = 0$$
Thus
$$(x-a)(x+2a-\sqrt{3}y)(x+2a +\sqrt{3}y) = 0$$
Now it is easy to see that the circum center of this triangle is the origin and the circum radius is $2a$. Hence the area is $3\sqrt{3}a^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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For $a^3+b^3+c^3=3$ prove that $a^4b+b^4c+c^4a\leq3$
Let $a$, $b$ and $c$ be non-negative numbers such that $a^3+b^3+c^3=3$. Prove that:
$$a^4b+b^4c+c^4a\leq3$$
This inequality similar to the following.
Let $a$, $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2=3$. Prove that:
$$a^3b+b^3c+c^3a\leq3,$$
which follows from the following identity.
$$(a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)=\frac{1}{2}\sum_{cyc}(a^2-b^2-ab-ac+2bc)^2.$$
I tried Rearrangement.
Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, $$a^4b+b^4c+c^4a=a^3\cdot ab+b^3\cdot bc+c^3\cdot ca\leq x^3\cdot xy+y^3\cdot xz+z^3\cdot yz=$$
$$=y(x^4+y^2xz+z^4)$$
and I don't see what is the rest.
Thank you!
| The Buffalo Way works. However it is not nice.
Let me describe Michael Rozenberg's solution. We have
\begin{align}
\sum_{\mathrm{cyc}} a^4b &\le \frac{1}{3}\sum_{\mathrm{cyc}} (a^{9/2}b^{3/2} + a^{9/2}b^{3/2} + a^3)\\
&= \frac{2}{3} \sum_{\mathrm{cyc}} a^{9/2}b^{3/2} + \frac{1}{3} (a^3+b^3+c^3)\\
&\le \frac{2}{9} (a^3+b^3+c^3)^2 + \frac{1}{3} (a^3+b^3+c^3)\\
&= 3
\end{align}
where we have used Vasc's inequality $(x^2+y^2+z^2)^2\ge 3(x^3y+y^3z+z^3x)$
to obtain $\sum_{\mathrm{cyc}} a^{9/2}b^{3/2} \le \frac{1}{3}(a^3+b^3+c^3)^2$.
This inequality can be used to prove the following inequality:
Let $x,y,z>0$ and $x^7+y^7+z^7=3$. Prove that
$\frac{x^4}{y^3}+\frac{y^4}{z^3}+\frac{z^4}{x^3}\ge 3$.
Proof: Using AM-GM, we have $7\frac{x^4}{y^3} + 9 x^{28/3} y^{7/3} \ge 16 x^7$
which results in $$7 \sum_{\mathrm{cyc}} \frac{x^4}{y^3} + 9 \sum_{\mathrm{cyc}} x^{28/3} y^{7/3} \ge 16 (x^7+y^7+z^7). $$
It suffices to prove that $\sum_{\mathrm{cyc}} x^{28/3} y^{7/3} \le 3$.
Let $a = x^{7/3}, \ b = y^{7/3}, \ c = z^{7/3}$. The condition becomes $a^3+b^3+c^3 = 3$.
We need to prove that $a^4b+b^4c+c^4a \le 3$. We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area under curve, infinite rectangles I'm trying to calculate the area under the curve of $ y=x^2 $ between $ x=1 $ and $x = 3$ and above $y=0$ using the sum of infinitely many rectangles.
So far I've tackled it by first defining the width of every rectangle to be $\Delta x = \frac{3-1}{n} = \frac{2}{n}$.
After that I proceeded by stating that the area of every rectangle under the curve combined is
$\sum_{i=1}^n (\frac{2i}{n})^2*(\frac{2}{n})$
I'm told that $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ which I then used in my calculations:
$\sum_{i=1}^n (\frac{2}{n})^2*i^2*(\frac{2}{n}) = \frac{8}{n^3}*(\frac{n(n+1)(2n+1)}{6})$
Simplifying to:
$8 * \frac{(n+1)(2n+1)}{6n^2} = \frac{8n^2+12n+4}{3n^2}$
Proceeding by calculating the limit as $x$ approaches infinity:
$\lim _{x\to \infty }\left(\frac{\left(8n^2+12n+4\right)}{3n^2}\right) = \frac{8}{3}$.
However, the answer should be $\frac{26}{3}$ which is indeed very close to what I have, but I have no idea where the problem lies. Any ideas?
| Following up on my comment, you need
$$
R_n = \sum_{i=1}^n \left(1 + \frac{2i}{n}\right)^2 \cdot \frac{2}{n}
$$
Expand the square and collect the sums of powers:
\begin{align*}
R_n &= \sum_{i=1}^n \left(1 + \frac{4i}{n} + \frac{4i^2}{n^2}\right)\cdot \frac{2}{n} \\
&= \frac{2}{n} \sum_{i=1}^n 1
+ \frac{8}{n^2} \sum_{i=1}^n i
+ \frac{8}{n^3} \sum_{i=1}^n i^2 \\
&= \frac{2}{n} \cdot n + \frac{8}{n^2} \cdot \frac{n(n+1)}{2}
+ \frac{8}{n^3} \frac{n(n+1)(2n+1)}{6}
\end{align*}
As $n\to \infty$, this tends to $2 + 4 + \frac{8}{3} = \frac{26}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2156646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric-substitutions for $\int \frac{x}{\sqrt{x^2+x+1}}\,dx$ Working integrals involving trigonometric substitutions, I evaluate
$$\int \frac{x}{\sqrt{x^2+x+1}}\,dx$$
and I am not sure what I am doing wrong.
What I do is change the denominator to $x^2+x+1=(x+1)^2-1$ then let $v=x+1$ and so $dv=dx$ such that I get
$$\int \frac {v-1}{\sqrt{v^2-1}}\,dv$$
Then, I set $v=a\sec\theta$ such that $dv=a\sec\theta\tan\theta\,d\theta$ and $a=1$ and I get,
$$\int \frac {(\sec\theta-1)}{\sqrt{\sec^2\theta-1}}\,\sec\theta\tan\theta\,d\theta$$
$$=\;\int \sec^2\theta-\sec\theta\,d\theta$$
$$\tan\theta - \ln|\sec\theta+\tan\theta|+C$$
Getting back in terms of the initial variable $x$ I get,
$$=\;\sqrt{v^2-1}-\ln|v+\sqrt{v^2-1}|+C$$
$$=\; \sqrt{x^2+x+1}-\ln|x+1+\sqrt{x^2+x+1}|+C$$
Which is incorrect. What I should get is
$$=\; \sqrt{x^2+x+1}-\frac 12\ln|x+\frac 12+\sqrt{x^2+x+1}|+C$$
The fact that my answer is so close and has the correct form leads me to believe that my substitutions might be correct and that I am making silly arithmetic mistakes somewhere.
All help is appreciated.
| $$\int\frac{xdx}{\sqrt{x^{2}+x+1}}$$
$$x^{2}+x+1=(x+\frac{1}{2})^{2}+\frac{3}{4}=\frac{3}{4}\left[ (\frac{2x+1}{\sqrt{3}})^{2}+1 \right]$$
$$\frac{2x+1}{\sqrt{3}}=z\Rightarrow dz=\frac{2}{\sqrt{3}}dx,x=\frac{\sqrt{3}z-1}{2}$$
$$
\int\frac{xdx}{\sqrt{x^{2}+x+1}}=\frac{\sqrt{3}}{4}\int\frac{(\sqrt{3}z-1)dz}{\sqrt{z^{2}+1}}$$
$$=\frac{3}{8}\int\frac{2z}{\sqrt{z^{2}+1}}-\frac{\sqrt{3}}{4}\int\frac{dz}{\sqrt{z^{2}+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Floor Function. Let x and y be rational numbers let x and y be rational numbers.
A. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor
= \left \lfloor x+y \right \rfloor$
B. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor
\le \left \lfloor x+y \right \rfloor$
C. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor
\ge \left \lfloor x+y \right \rfloor$
D. None of the above.
Can anyone please explain why the answer is B? Thank You!
| Let $\left \lfloor x \right \rfloor = x_{0} + x_{1}, x_{0} \in \mathbb{Z}, 0\leq x_{1}<1$
Similarily we define $\left \lfloor y\right \rfloor$
Then $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor = x_{0} + y_{0}$
Now there are two cases to consider:
(i) $x_{1} + y_{1} < 1$
and
(ii) $x_{1} + y_{1} \geq 1$
Case (i):
$\left \lfloor x + y \right \rfloor = \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor = x_{0} + y_{0}$
Case (ii):
$\left \lfloor x + y \right \rfloor = x_{0} + y_{0} + 1 > \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor $
Therefore $\forall x,y \in \mathbb{Q}$, $\left \lfloor x + y \right \rfloor \geq \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor$ as required
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
If $x \not \equiv 1 \pmod 3$, when is $x^2 + x +1$ not a prime? If $x \not\equiv1 \pmod 3$, when is $x^2 + x + 1$ not a prime?
I am especially interested in an example that is not prime or even better, an explanation why the frequency of such primes goes down as $x$ gets larger.
| Suppose $x^2 + x + 1$ is a prime $p$ other than $p = 3$. Then $4p - 3$ is a square; e.g., for $p = 31$, $4p - 3 = 124 - 3 = (11)(11)$. Conversely if $4p - 3$ is a square for $p$ a prime, there is a divisor $x$ of $p - 1$ such that $p = x^2 + x + 1$. The complement of $x$, $\frac{p - 1}{x}$, is $x + 1$. With $p = 31$, $x = 5$ and $\frac{p - 1}{x} = 6$. Notice that since $x(x + 1) = p - 1$, $x$ cannot be congruent to 1 mod 3, unless $p = 3$. Thus, an answer is that $x^2 + x + 1$ is not a prime exactly when $4(x^2 + x + 1) - 3$ is not a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find 4x3 matrix B such that AB = I I got stuck with this problem.
\begin{matrix}
1 & 1 & 1 & 1\\
0 & 1 & 1 & 0\\
0 & 0 & 1 & 1\\
\end{matrix}
Consider the $3\times 4$ matrix $\bf A$ (above). Do the columns of $\bf A$ span $\mathbb R^3$?
Prove your answer. Also, Find a $4\times 3$ matrix $\bf B$, such that $\bf AB = I_3$
--
I know that the columns of $\bf A$ span $\mathbb R^3$ as there more columns than rows. But I cannot understand how to find matrix $\bf B$ because I cannot implement "super-augmented" matrix and do Gauss-Jordan elimination. Looks like I need to do something with 4th column of $\bf A$ and 4th row of $\bf B$. What do you think?
Thanks!
| B is a 4*3 matrix. we will find the columns of B using the fact that the columns of AB are linear combinations of columns of A with column elements of B.
$\\1\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}+0\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}+0\begin{bmatrix}
1\\
1\\
1
\end{bmatrix}+0\begin{bmatrix}
1\\
0\\
1
\end{bmatrix}=\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$ So the first column of B is given by $ \begin{bmatrix}
1\\
0\\
0\\
0
\end{bmatrix}$
similarly $ \\-1\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}+1\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}+0\begin{bmatrix}
1\\
1\\
1
\end{bmatrix}+0\begin{bmatrix}
1\\
0\\
1
\end{bmatrix}=\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}$ so the second column of B is given by $\begin{bmatrix}
-1\\
1\\
0\\
0
\end{bmatrix}$
$\\0\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}-1\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}+1\begin{bmatrix}
1\\
1\\
1
\end{bmatrix}+0\begin{bmatrix}
1\\
0\\
1
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}$ So the third column of B is given by $\begin{bmatrix}
0\\
-1\\
1\\
0
\end{bmatrix}$
$B=\begin{bmatrix}
1 &-1 &0 \\
0& 1 &-1 \\
0& 0 &1 \\
0& 0 & 0
\end{bmatrix}$.
Note that the matrix B is not unique.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Show that the tangent intercept the the ellipse in thease positions I have been stuck on this question for a couple of hours and can not figure it out, it states:
Consider the tangent line to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}= 1$ at the point $(p,q)$ in the first quadrant. Show that the tangent line has $x$-intercept $\dfrac{a^2}{p}$ and $y$-intercept $\dfrac{b^2}{q}$.
How I tried to solve this problem was to start differentiating the function with respect of $x$.
$$\frac{d}{dx} \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} \right) = 0 \implies \frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 \implies y'= -\frac{xb^2}{ya^2}$$
Now I want to know what I can substitute $y$ for, so I solve it in the first equation.
$$\frac{x^2}{a^2} + \frac{y^2}{b^2}= 1 \implies \ldots \implies y = \frac{\sqrt{a^2b^2-x^2b^2}}{a}$$
I substitute $y$ to this in the second equation.
$$y'= -\frac{2xb^2}{2a^2 \left( \dfrac{\sqrt{a^2b^2-x^2b^2}}{a} \right)} \implies \ldots \implies y' = -\frac{xb\sqrt{a^2-x^2}}{a(a^2-x^2)}$$
So now I know that the tangent line have the equation $y=-\dfrac{xb\sqrt{a^2-x^2}}{a(a^2-x^2)} + m$ where $m$ is the intersection in the $y$-axis. I know that I have a point $(p,q)$ witch are constants, so I solve $m$.
$$m = q + \frac{pb\sqrt{a^2-p^2}}{a(a^2-p^2)}$$
Now I have the equation for the tangent line I was after.
$$y = -\frac{xb\sqrt{a^2-x^2}}{a(a^2-x^2)} + q + \frac{pb\sqrt{a^2-p^2}}{a(a^2-p^2)}$$
But, here is the problem. If I solve $y(p)$, I get that $y(p) = q$, which make sense. But that was not really what I was after.
Have I misinterpreted the question or can any one of you spot my error? Personally I do not know what else I can do except what I have already done.
| As $y'=-\dfrac{b^2x}{a^2y}$
So, the equation of the tangent at $x_1,y_1$
$$\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1\iff\dfrac x{\dfrac{a^2}{x_1}}+\dfrac b{\dfrac{b^2}{y_1}}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Double integral over area bounded by ellipse Evaluate the double integral of $\int \int(x+y)^{2} dxdy$ over the area bounded by the ellipse $(\frac{x}{a})^2+(\frac{y}{b})^2=1$
Please check answer = $\frac{ab\pi}{4}(a^{2}+b^{2})$
| Let,
$$x=ar\cos(\theta)$$
$$y=br\sin(\theta)$$
With $\theta \in [0,2\pi]$ and $r \in [0,1]$. Then the Jacobian corresponding to this transformation is $abr$. And we have the double integral,
$$\int_{0}^{2\pi} \int_{0}^{1} (ar\cos(\theta)+br\sin(\theta))^2 ab r dr d\theta$$
$$=ab \int_{0}^{2 \pi} \int_{0}^{1} \left(a^2r^2\cos^2(\theta)+b^2r^2 \sin (\theta)+\frac{ab}{2}r^2 \sin(2\theta) \right) r dr d\theta$$
Noticing that cosine is $2\pi$ periodic so we quickly see we can drop the $\sin(2\theta)$ term then separate the double integral as follows.
$$=ab \int_{0}^{1} r^3 dr \int_{0}^{2\pi} \left(a^2\cos^2(\theta)+b^2\sin^2(\theta) \right) d\theta$$
$$=\frac{ab}{4} \int_{0}^{2\pi} \left(\frac{1}{2}a^2(1+\cos(2\theta)+\frac{1}{2}b^2(1-\cos(2\theta) \right) d\theta$$
By $2\pi$ periodicity of sine we will see we can drop the $\cos(2\theta)$ terms to get,
$$=\frac{ab}{4} \int_{0}^{2\pi} \left(\frac{a^2}{2}+\frac{b^2}{2} \right) d\theta$$
$$=\frac{\pi ab}{4}(a^2+b^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Efficient way to solve modular arithmetic equations I want to find all integer solutions ($x$ value) the following equation:
$((2(x+2))\mod 3+(x-3)\mod 2+3)\mod 5 = 3$
I have simplified the above equation as
$((2x+4))\mod 3+(x-3)\mod 2+3)\mod 5 = 3$
$((2x+3+1))\mod 3+(x-3)\mod 2+3)\mod 5 = 3$
$((2x+1) \mod 3 + (x-1)\mod 2 +3)\mod 5$ ==> I used the fact that $(k+3)\mod 3 = k\mod 3 = (k-3)\mod 3$
I can only solve it by such reasoning methods. Is there any specific fixed way or algorithm to solve such expressions?
| You can continue simplifying : $(2x+1)\mod 3+(x-1)\mod 2\equiv 0\mod 5$
So $2x+1=3p+r$ and $x-1=2s+u$ and $u+r=5k$ but since $r\in\{0,1,2\}$ and $u\in\{0,1\}$ the only possible multiple of $5$ is $0$.
Thus $u=r=0$.
$\begin{cases}
2x+1\equiv 0\mod 3\iff 2x\equiv -1\equiv 2\mod 3\iff x\equiv 1\mod 3\\
x-1\equiv 0\mod 2\iff x\equiv 1\mod 2
\end{cases}$
So finally $\quad x=6k+1$.
The method is to try if possible to arrive at a system of $x\equiv a_i\mod p_i,\ i=1..n$ then you can apply chinese theorem.
When having a look at your equation :
$f(x)=(3(2(x\mod 2)+2 - (x\mod 4)))\mod 4 + (x\mod 8) = 9$
we first notice there are three modulo used $2,4$ and $8$, so if you are to find an $x$, the Chinese remainder theorem states that $x=lcm(2,4,8)a+t=8a+t$ with $t\in\{0,1,2,3,4,5,6,7\}$.
The success of the method resides in the fact that only $8$ values for $t$ need to be plugged and tested in the equation.
Note: since $(x\mod 2,4,8)=(t\mod 2,4,8)$ just replace $x$ by $t$ in the equation and try all values.
$\begin{array}{c|ccc|c}
t & t\%2 & t\%4 & t\%8 & f(t) \\
\hline
0 & 0 & 0 & 0 & 2 \\
1 & 1 & 1 & 1 & 2 \\
2 & 0 & 2 & 2 & 2 \\
3 & 1 & 3 & 3 & 6 \\
4 & 0 & 0 & 4 & 6 \\
5 & 1 & 1 & 5 & 6 \\
6 & 0 & 2 & 6 & 6 \\
7 & 1 & 3 & 7 & 10
\end{array}$
And you see that your equation has no solution, but solving it was a very automated process.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the limit without L'hopital rule $\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=$? Find the limit without L'hopital rule
$$\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=?$$
My Try:
$$1-\cot( \frac{π}{4}x)=1-\frac{1}{\tan( \frac{π}{4}x)}=\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}\\\sin (πx)=\sin (\pi-\pi x)=-\sin \pi(x-1)\\\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=\lim_{ x \to 1}\frac{\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}}{-\sin \pi(x-1)}\\u=x-1⇒x=u+1\\\lim_{ u \to 0}\frac{\frac{\tan( \frac{π}{4}(u+1))-1)}{\tan( \frac{π}{4}(u+1))}}{-\sin \pi u}$$
now ?
| \begin{align*}
lim_{ u \to 0}\frac{\frac{\tan( \frac{π}{4}(u+1))-1)}{\tan( \frac{\pi}{4}(u+1))}}{-\sin \pi u} &= lim_{ u \to 0}\frac{\frac{1+\tan\frac{\pi}{4} u}{1-\tan\frac{\pi}{4} u}-1}{-\sin \pi u} \\
&= lim_{ u \to 0}\frac{2\tan\frac{\pi}{4} u}{-\sin \pi u}\\
&= lim_{ u \to 0}\frac{\frac{\frac{\pi}{2}\tan\frac{\pi}{4} u}{\frac{\pi}{4}u}}{-\frac{\sin \pi u}{u}}\\
&= -\frac{1}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
Compute a partial fractions decomposition. How do I compute a partial fractions decomposition of
$$\dfrac{x+1}{x^2-x}$$
I've tried,
Since, $x^2-x$ is a quadratic expression so, I wrote the partial fraction decomposition of this as
$$\dfrac{Ax+B}{x^2-x}$$
But the answer came incorrect. Thanks
| One of the biggest mistakes people make is not factoring the deonimator completly! You have made the same mistake.
Notice how:$$\dfrac{x+1}{x^2-x} = \dfrac{x+1}{x(x-1)}$$
Now node the $x$ and $x-1$. Both are linear. Therefore the simplest form is juts a constant, lets call it $A$, and $B$.
So you have:
$$\frac{A}{x}+\frac{B}{x-1} = \frac{x+1}{x(x-1)}$$
Now let's solve.
$$\frac{A(x-1)+Bx}{x(x-1)} = \frac{x+1}{x(x-1)}$$
$$\frac{Ax-A+Bx}{x(x-1)} = \frac{x+1}{x(x-1)}$$
$$\frac{(A-B)x-A}{x(x-1)} = \frac{x+1}{x(x-1)}$$
Now equate the equation. From the numerator $x+1$, the degree $1$ term is just the coefficient on $x$, which is $1$, and the degree $0$ term is $+1$. So we have:
$$A-B=1$$
$$-A=1$$
Therefore, $A=-1$, B=2$
Thus, the $p.f.d$ is:
$$\frac{-1}{x}+\frac{2}{x-1} = \frac{x+1}{x(x-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Proving an inequality by mathematical induction I'm trying to solve a problem with inequalities using mathematical induction but I am stuck halfway through the process.
The problem: Use mathematical induction to establish the inequality -
$(1 + \frac{1}{2})^n \ge 1 + \frac{n}{2}$ for n $\in \mathbb{N}$
Steps
1) $n = 1$, $(1 + \frac{1}{2})^1 \ge 1 + \frac{1}{2}$ is TRUE
2) $n = k$, assume that $(1 + \frac{1}{2})^k \ge 1 + \frac{k}{2}$ for n $\in \mathbb{N}$
3) Show the statement is true for $k + 1$
$(1 + \frac{1}{2})^{k+1}$ = $(1 + \frac{1}{2})^k * (1 + \frac{1}{2})$
$\ge$ $(1 + \frac{k}{2}) * (1 + \frac{1}{2})$ - using the assumption in step $2$
My question is, how do I continue this problem? Or did I go wrong somewhere? I just can't figure out what the next step is.
| Continue with:
$(1 + \frac{k}{2}) * (1 + \frac{1}{2}) =$
$1 + \frac{k}{2} + \frac{1}{2} + \frac{k}{4} >$
$1 + \frac{k}{2} + \frac{1}{2}=$
$1 + \frac{k+1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
The digits of a 3-digit integer are reversed to form a new integer of greater value. Find the new integer if their product is 91,567. The digits of a 3-digit integer are reversed to form a new integer of greater value. The product of this new integer and the original integer is 91,567. What is the new integer?
| Let the original $3$ digit integer be $I$.
Let $I$ = $100a + 10b + c$, with $0 \leq a,b,c \leq 9$ and $c\geq a$
Then the new integer, $J = 100c + 10b + a$
We are given that $IJ = 91567 = (100a+10b+c)(100c+10b+a)$
So let's factorise $91567 = 7 \cdot 103 \cdot 127$
Then, since $I$ and $J$ are both integers, one of them is equal to $127$, and the other is equal to $7\cdot 103$. Since we are given that $J > I$, it follows that $J = 7 \cdot 103 = 721$ and $I = 127$
Now we must solve the following equations:
$721 = 100c + 10b + a$
$127 = 100a + 10b + c$
We can deduce that $a = 1$
Then we get $72 = 10c +b$
and $27 = 10b + c$
This implies:
$c = 27-10b, 72 = 270 -100b + b$
Then we get $b =2$ and $c = 7$
So the new integer is $721$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding maximum of $x+y$ Let x and y be real numbers satisfying $9x^{2} + 16y^{2} = 1$. Then $x + y$ is maximum when
a. $y = \frac{9x}{16}$
b. $y = −\frac{9x}{16}$
c. $y = \frac{4x}{3}$
d. $y = −\frac{4x}{3}$
| We can solve for $y$
$$
y = \pm \frac{\sqrt{1 - 9 x^2}}{4}
$$
on the ellipse:
$$
\left(\frac{x}{1/3}\right)^2 + \left(\frac{y}{1/4}\right)^2 = 1
$$
Then
$$
f(x) = x + y(x) \le f_+(x) = x + \frac{\sqrt{1 - 9 x^2}}{4}
$$
and for an extremum:
$$
f_+'(x) = 1 + \frac{1}{8 \sqrt{1-9x^2}}(-18 x) = 0 \iff \\
1 = \frac{9x}{4 \sqrt{1-9x^2}} \iff \\
4 \sqrt{1-9x^2} = 9 x \iff \\
16 (1 - 9x^2) = 81 x^2 \iff \\
16 = 225 x^2 \iff \\
x^2 = \frac{16}{225} \iff \\
x = \pm \frac{4}{15}
$$
and thus picking the positive solution for $x$:
$$
y = \frac{\sqrt{1 - 9\cdot 16/225}}{4} = \frac{9}{4\cdot 15} = \frac{3}{20} \Rightarrow \\
\frac{y}{x} = \frac{3}{20} / \frac{4}{15} = \frac{45}{80} = \frac{9}{16}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 2
} |
Integrate $\int \cos^n x dx$ I know how to solve it with reduction formula, but is there concrete answer for this integral (without other integrals like in reduction formula)?
WolframAlpha gives me expression with $F_1(\frac{1}{2}, \frac{n+1}{2};\frac{n+3}{2}; cos^2(x))$, but I don't know really what it is. I want to know if there is a simpler solution for this without other integrals like in reduction formula and difficult things like $F_1(\frac{1}{2}, \frac{n+1}{2};\frac{n+3}{2}; cos^2(x))$?
| if n is odd
$\int \cos^{2k+1} x dx\\
\int \cos x (\cos^2 x)^k dx\\
\int \cos x (1-\sin^2 x)^k dx\\
u = \sin x,\ du = cos x\ dx\\
\int (1-u^2)^k du\\
\int \sum_\limits{i=0}^k {k\choose i} (-1)^i u^{2i} du\\
\sum_\limits{i=0}^k {k\choose i} (-1)^i \frac {sin^{2i+1} x}{2i+1}$
if n is even...$n = (2^p)k$
where k is odd.
$\int \cos^{2^pk} x dx\\
\int \frac 12 (1+\cos^{2^{p-1}k} 2x)\ dx\\
\int \frac 12 (1+\frac 12(1+\cos^{2^{p-2}k} 2x)\ dx\\
\int \frac 34 + \frac 14 \cos^{2^{p-2}k} 4x\ dx\\
\int 1-\frac 1{2^p} + \frac 1{2^p}\cos^k (2^p x)\ dx\\
(1-\frac 1{2^p})x + \frac 1{2^{2p}}\sum_\limits{i=0}^k {k\choose i} (-1)^i \frac {sin^{2i+1} 2^px}{2i+1}\\
$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Sum of three variables given two equations Given $$x^2+y^2+z^2=121$$
$$x\sqrt{11} + 4y + z\sqrt{22}=77$$
Find $$ \frac{\sqrt{11} + 4 + \sqrt{22}}{x+y+z} $$
I tried to plug in something for z at first, since x and y should have unique values for every value of z, but that didn't seem to work.
The answer is 7/11, which is clearly the second equation divided by the first but I don't understand how or why that would lead the final expression.
| $x^2+y^2+z^2=121$
Is the equation of a sphere centered at the origin of radius 11.
$x\sqrt{11} + 4y + z\sqrt{22}=77$ is the equation of a plane
Origin is $\frac {77}{\sqrt {11 + 4^2 +22}} = 11$ units from the plane!
The plane is tangent at to the sphere.
the point of tangency $(x,y,z) = \frac {11}{7}\cdot(\sqrt{11} , 4, \sqrt {22})$
$x+y+z = \frac {11}{7} (\sqrt {11} + 4 + \sqrt{22})$
$\frac {\sqrt {11} + 4 + \sqrt{22}}{x+y+z} = \frac {7}{11}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim\limits_{x\rightarrow 0} \frac{(\cos{x})^2}{\sin{(x^2)}}-\frac{1}{x^2}$
Find
$$\lim_{x\rightarrow 0} \frac{(\cos{x})^2}{\sin{(x^2)}}-\frac{1}{x^2}$$
I suppose I have to use L'Hopital's rule here but how?
Edit: I cannot use Taylor yet.
| If you can only use L'Hospital, rewrite $$\frac{(\cos{x})^2}{\sin{(x^2)}}-\frac{1}{x^2}=\frac{x^2(\cos{x})^2-\sin{(x^2)}}{x^2\sin{(x^2)}}=\frac uv$$ Now, you will have a funny time since you must apply the rule four times.
I give below the successive derivatives of $v=x^2\sin{(x^2)}$
$$v'=2 x \sin \left(x^2\right)+2 x^3 \cos \left(x^2\right)$$
$$v''=2 \sin \left(x^2\right)+10 x^2 \cos \left(x^2\right)-4 x^4 \sin \left(x^2\right)$$
$$v'''=24 x \cos \left(x^2\right)-8 x^5 \cos \left(x^2\right)-36 x^3 \sin \left(x^2\right)$$
$$v''''=-156 x^2 \sin \left(x^2\right)+\color{red}{24 \cos \left(x^2\right)}+16 x^6 \sin
\left(x^2\right)-112 x^4 \cos \left(x^2\right)$$ where, for first time, you see a term which is not going to $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the smallest natural number n such that the decimal expansion of $9^n$ has $< n$ digits? We can see that,
*
*$9^1=9$
*$9^2=81$
*$9^3=729$
*$9^4=6561$
if there exists such n then what is the prove.
Thank you.
| The base 10 log shows us how many digits there are in the decimal expansion. Notice how $log_{10}100=2$
$log_{10}1000=3$
$log_{10}500=2.7$
The number of digits in $x$ is given by the formula:
$\text{digits}=\lfloor{log_{10}x}\rfloor+1$
Where $\lfloor{z}\rfloor$ is the number $z$ rounded down to the nearest integer.
So for your question you can use this formula:
$\lfloor{log_{10}9^n}\rfloor+1 < n$
Use laws of logs to simplify:
$\lfloor{nlog_{10}9}\rfloor < n-1$
From the definition of the floor function clearly
$\lfloor{nlog_{10}9}\rfloor \leq nlog_{10}9 < n-1$
Therefore we can get a lower bound for $n$ by solving
$nlog_{10}9 < n-1$
$1 < n(1-log_{10}9)$
$21.85 < n$
Therefore you only have to search for $n \geq 22$
You can see that $9^{22}=9.85\times 10^{20}$ which clearly has 21 digits
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area of the intersection of two cones I would like to find an expression for the area of the intersection of the two cones shown in the following figure:
The axes of both cones are situated on the plane YZ, and the angles are respectively $\alpha_1$ and $\alpha_2$, while the angle of the intersection is $\theta$, as indicated in the picture. Obviously, I'm interested in the case when $\alpha_1+\alpha_2<\theta$, since otherwise there would be no intersection between the cones.
My idea is to do it by integration on the sphere, but how can I find the limits I need for the integral?
| Your question can be answered with spherical geometry. In the notation of the question, and referring to the diagram, the circle of angular radius $\alpha_{1}$ centered at $p_{1}$ and the circle of angular radius $\alpha_{2}$ centered at $p_{2}$ meet at $p_{3}$ and $p_{4}$. The segments shown are great circle arcs.
*
*The area sought is the area of the spherical sector $p_{1} p_{3} p_{4}$ plus the area of the sector $p_{2} p_{4} p_{3}$ minus the area of the spherical quadrilateral $p_{1} p_{3} p_{2} p_{4}$.
*If $\phi_{1} = \angle p_{4} p_{1} p_{3}$ is the "apex" angle of the spherical triangle $\triangle p_{1} p_{3} p_{4}$, the area of the spherical sector $p_{1} p_{3} p_{4}$ is $\phi_{1}(1 - \cos \alpha_{1})$ by Archimedes' hat box theorem.
Similarly, if $\phi_{2} = \angle p_{3} p_{2} p_{4}$, the area of the spherical sector $p_{2} p_{4} p_{3}$ is $\phi_{2}(1 - \cos \alpha_{2})$.
*If $\psi_{1} = \angle p_{1} p_{3} p_{4}$ and $\psi_{2} = \angle p_{2} p_{4} p_{3}$ are the "base" angles of the spherical triangles, the spherical quadrilateral $p_{1} p_{3} p_{2} p_{4}$ has area $\phi_{1} + \phi_{2} + 2(\psi_{1} + \psi_{2}) - 2\pi$.
Putting everything together, your "digon" has area
\begin{multline*}
\phi_{1}(1 - \cos\alpha_{1}) + \phi_{2}(1 - \cos\alpha_{2}) - \bigl[\phi_{1} + \phi_{2} + 2(\psi_{1} + \psi_{2}) - 2\pi\bigr] \\
= 2\pi - 2(\psi_{1} + \psi_{2}) - \phi_{1} \cos\alpha_{1} - \phi_{2} \cos\alpha_{2}.
\end{multline*}
If $A$, $B$, $C$ are points on the unit sphere, the normalized cross products
$$
n_{B} = \frac{A \times (B - A)}{\|A \times (B - A)\|},\qquad
n_{C} = \frac{A \times (C - A)}{\|A \times (C - A)\|}
$$
are perpendicular to the planes through $O$, $A$, $B$ and $O$, $A$, $C$ respectively, so
$$
\angle CAB = \arccos(n_{B} \cdot n_{C}).
\tag{1}
$$
To express this completely in terms of $\alpha_{1}$, $\alpha_{2}$, and $\theta$, use Cartesian coordinates with
\begin{align*}
p_{1} &= (0, 0, 1), \\
p_{2} &= (\sin\theta, 0, \cos\theta), \\
p_{3} &= (a, -b, \cos\alpha_{1}), \\
p_{4} &= (a, \phantom{-}b, \cos\alpha_{1}).
\end{align*}
The condition $p_{2} \cdot p_{3} = \cos\alpha_{2}$ gives
$$
a = \frac{\cos\alpha_{2} - \cos\alpha_{1} \cos\theta}{\sin\theta},
$$
and then
$$
b = \sqrt{\sin^{2}\alpha_{1} - a^{2}}.
$$
Equation (1) now gives the angles $\phi_{i} = \angle p_{3} p_{i} p_{4}$ and $\psi_{i} = p_{3} p_{4} p_{i}$ in terms of $\alpha_{1}$, $\alpha_{2}$, and $\theta$.
(I haven't tried to substitute everything and simplify.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $abc=1$ so $\sum\limits_{cyc}\frac{7-6a}{2+a^2}\geq1$
Let $a$, $b$ and $c$ be real numbers such that $abc=1$. Prove that:
$$\frac{7-6a}{2+a^2}+\frac{7-6b}{2+b^2}+\frac{7-6c}{2+c^2}\geq1$$
The equality occurs also for $a=b=2$ and $c=\frac{1}{4}$.
This inequality is a similar to the very many contest's inequalities, but nothing helps.
At least, I don't see how we can prove it.
An example of my trying.
We need to prove that
$$\sum_{cyc}\frac{7-6a}{2+a^2}\geq1$$ or
$$\sum_{cyc}\left(\frac{7-6a}{2+a^2}+1\right)\geq4$$ or
$$\sum_{cyc}\frac{(a-3)^2}{2+a^2}\geq4.$$
By C-S $$\sum_{cyc}\frac{(a-3)^2}{2+a^2}=\sum_{cyc}\frac{(a-3)^2(a+k)^2}{(2+a^2)(a+k)}\geq\frac{\left(\sum\limits_{cyc}(a-3)(a+k)\right)^2}{\sum\limits_{cyc}(2+a^2)(a+k)^2}$$
Now we'll find a value of $k$, for which the equality in the last inequality occurs for $a=b=2$ and $c=\frac{1}{2}$.
Since in all equality case we have
$$\frac{a-3}{(2+a^2)(a+k)}=\frac{b-3}{(2+b^2)(b+k)}=\frac{c-3}{(2+c^2)(a+k)},$$
we obtain:
$$\frac{2-3}{(2+2^2)(2+k)}=\frac{\frac{1}{4}-3}{(2+\left(\frac{1}{4}\right)^2)(\frac{1}{4}+k)},$$
which gives $k=-\frac{9}{4}$.
Thus, it remains to prove that
$$\left(\sum\limits_{cyc}(a-3)(4a-9)\right)^2\geq4\sum_{cyc}(2+a^2)(4a-9)^2,$$
which is wrong for $a=4$ and $b=c=\frac{1}{2}$.
Any hint?
Thank you!
| Michael Rozenberg's solution without UVW:
Michael Rozenberg gave a nice application of Cauchy-Bunyakovsky-Schwarz inequality.
Then it suffices to prove that, for all $a, b, c > 0$ with $a b c = 1$,
$$9(a + b + c - 1)^2 - 8(a^2 + b^2 + c^2) - 12 \ge 0.$$
Denote LHS by $f(a, b, c)$. WLOG, assume $c\ge 1$. We have
$$f(a, b, c) - f(\sqrt{ab}, \sqrt{ab}, c)
= (\sqrt{a} - \sqrt{b})^2(a + 2\sqrt{ab} + b + 18c - 18) \ge 0.$$
Also, we have
$$f(\sqrt{ab}, \sqrt{ab}, c)
= \frac{(\sqrt{c} + 5)(\sqrt{c} + 1)(\sqrt{c} - 2)^2(\sqrt{c} - 1)^2}{c} \ge 0.$$
We are done.
| {
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"url": "https://math.stackexchange.com/questions/2183109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find all points $(x,y)$ on the graph of $f(x)$ with tangent lines passing through a certain point Find all points $(x,y)$ on the graph of $f(x) = x^2$ with tangent lines passing through the point $(3,8)$.
My attempt:
$f'(x) = x^2$
I substituted a random point into $f'(x)$ to get the gradient of the tangent, so $f'(0) = 0$.
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
$$ m = \frac{8-0}{3-0} = 8/3$$
$$y-y_1 = m(x-x_1)$$
Now I substituted the point $(3,8)$:
$$ y - 8 = \frac{8}{3}(x-3)$$
$$y = \frac{8}{3}x$$
$$ y = x^2$$
Solving simultaneously we get $x = 0$ and $y = 0$ or $x = 8/3 $ and $ y = 64/9$
So we have the points $(0;0)$ and $(\frac{8}{3} ; \frac{64}{9})$. However, I am not so confident with my answer. Is this correct?
| Here is an alternative method:
A line with gradient $m$ passing through the point $(3,8)$ has equation $$y-8=m(x-3)$$
Solving this simultaneously with the curve $y=x^2$ leads to the quadratic equation $$x^2-mx+3m-8=0$$
This must have double roots so the discriminant is zero, so $$m^2-12m+32=0\implies m=4,8$$
Then $x=-\frac{b}{2a}=\frac m2=2,4$
So the points are $$(2,4),(4,16)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Change of basis for a polynomial So I am given the following.
$B=\{1+t+t^2,2-t^2,t+t^2\}$ and $C=\{1+t,1+t^2,1-t+t^2\}$ are two bases for $\mathbb P_2$. I have to find
a) The standard basis for $\mathbb P_2$ is $S=\{1,t,t^2\}$. I am told to find the matrices $P_{S\leftarrow B}$ and $P_{S\leftarrow C}$
b) The change of basis from $P_{C\leftarrow B}$
c) If $[p(t)]_{B}=\begin{bmatrix}
1 \\
-2 \\
3
\end{bmatrix}
$, what is $[p(t)]_{C}$ ?
d) If $[q(t)]_{B}=\begin{bmatrix}
3 \\
2 \\
-1
\end{bmatrix}
$, what is $q(t)$ ?
Work for part $a)$ :
$P_{S\leftarrow B}=\begin{bmatrix}
1 & 2 & 0 \\
1 & 0 & 1 \\
1 & -1 & 1
\end{bmatrix}
$
and
$P_{S\leftarrow C}=\begin{bmatrix}
1 & 1 & 1 \\
1 & 0 & -1 \\
0 & 1 & 1
\end{bmatrix}
$
I think that's okay. Since I am changing the basis back to the standard one, nothing special needs to be done.
Work for part $b)$ :
If both of these bases span $\mathbb P_2$ then I should be able to write the bases as a linear combination of each other I think.
$i)$ $1+t+t^2 = a_{1} (1+t) + a_{2} (1+t^2) + a_{3} (1-t+t^2)$
$ii)$ $2-t^2$ = $b_{1} (1+t) + b_{2} (1+t^2) + b_{3} (1-t+t^2)$
$iii)$ $t+t^2$ = $c_{1} (1+t) + c_{2} (1+t^2) + c_{3} (1-t+t^2)$
If I solve the coefficients, I get:
$a_{1}=0 , a_{2}=2 , a_{3}=-1$
$b_{1}=-3 , b_{2}=2 , b_{3}=-3$
$c_{1}=-1 , c_{2}=3 , c_{3}=-2$
So the matrix would just be:
$P_{C\leftarrow B}=\begin{bmatrix}
0 & 3 & -1 \\
2 & 2 & 3 \\
-1 & -3 & -2
\end{bmatrix}
$
Is that right? I was told something with inverses but I am not sure how that is related.
I have no idea what part $c)$ and $d)$ is asking though. Can someone guide me through what it's even asking? Thanks!
Also in general, what is a change of basis? How is it useful?
| $P_{B\to C} = P_{B\to S}P_{S\to C}$
$P_{S\to C} = P_{C\to S}^{-1}$
$P_{B\to C} = \begin{bmatrix}
1 & 2 & 0 \\
1 & 0 & 1 \\
1 & -1 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0 & -1 \\
-1 & 1 & 2 \\
1 & -1 & -1
\end{bmatrix} = \begin{bmatrix}
0 & 3 & -1 \\
2 & -4 & 3 \\
1 & 3 & 2
\end{bmatrix}$
$[p(t)]_{B}= \begin{bmatrix}
1 \\
-2 \\
3
\end{bmatrix}$
$[p(t)]_{C} = P_{B\to C} [p(t)]_B =
\begin{bmatrix}
0 & 3 & -1 \\
2 & -4 & 3 \\
1 & 3 & 2
\end{bmatrix}\begin{bmatrix}
1 \\
-2 \\
3
\end{bmatrix}=\begin{bmatrix}
9 \\
19 \\
-13
\end{bmatrix}$
or,
$p(t) = (1+t+t^2) - 2(2t-t^2) + 3(t+t^2) = -3 +4t + 6t^2 = -9(1-t) + 19(1+t^2) - 13(1-t+t^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the easiest way to do partial fraction? $$\frac{2a^4x}{x^4+4a^4}=\frac{A}{x-a(1+i)}+\frac{B}{x-a(1-i)}+\frac{C}{x+a(1+i)}+\frac{D}{x+a(1-i)}$$
I am trying to calculate this, but it is too complicated. I am thinking that maybe I am using wrong method. What is the easiest method to solve this?
| Just for completeness, I'll do this using the method I mentioned in the comment above (even though this is an old question). Use $u=x^2\implies du=2xdx$ to get $$\int\dfrac{2a^4x}{x^4+4a^4}dx=\int\dfrac{a^4}{u^2+4a^4}du$$
Now, use the substitution $u=2a^2\tan(t)\implies du=2a^2\sec^2(t)dt$ to get $$\int\dfrac{a^4}{u^2+4a^4}du=\int\dfrac{a^4}{(2a^2\tan(t))^2+4a^4}2a^2\sec^2(t)dt=\int\dfrac{a^4}{4a^4(\tan^2(t)+1)}2a^2\sec^2(t)dt=\frac{1}{2}a^2\int 1dt=\frac{1}{2}a^2t+C=\frac{1}{2}a^2\arctan\Bigl(\frac{u}{2a^2}\Bigr)+C=\boxed{\frac{1}{2}a^2\arctan\Bigl(\frac{x^2}{2a^2}\Bigr)+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Ramanujan's Proof for Morley's Hypergeometric Identity
Introduction:
Ramanujan's Proof of Morley's Identity:
If $\text{Re}(x+y+n+1)>0$, then$$_2F_1(-x,-y;n+1)=\frac {\Gamma(n+1)\Gamma(x+y+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)}\tag{8.1}$$
Assume $x,n\in\mathbb{Z^+}$. Expanding $(1+u)^{y+n}$ and $(1+1/u)^x$ in their formal binomial series and taking their product, we find that, if $a_n$ is the coefficient of $u^n$,$$a_n=\sum\limits_{k=0}^{\infty}\binom{y+n}{k+n}\binom xk=\frac {\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(y+1)}\sum\limits_{k=0}^\infty\frac {(-x)_k(-y)_k}{(n+1)_k(1)_k}\tag{8.2}$$
On the other hand, expanding $(1+u)^{x+y+n}$ in its binomial series and dividing by $u^x$, we find that$$a_n=\binom{x+y+n}{x+n}=\frac {\Gamma(x+y+n+1)}{\Gamma(x+n+1)\Gamma(y+1)}\tag{8.3}$$
Comparing $(8.2)$ and $(8.3)$, we deduce $(8.1)$.
Questions:
I understand most of the proof, but there are still "trippy" spots that I'm not sure about.
*
*How do you find the coefficient of $u^x$ in $(1+u)^{y+n}\left(1+\frac 1u\right)^x$?
*How was $(8.3)$ obtained (Specifically, the binomial portion)?
*How is$$\frac {\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(y+1)}\sum\limits_{k=0}^{\infty}\frac {(-x)_k(-y)_k}{(n+1)_k(1)_k}=\sum\limits_{k=0}^{\infty}\binom{y+n}{k+n}\binom xk$$
I'm just not exactly sure how the coefficient summations are obtained. A detailed explanation would be awesome!
For $3$, I tried on my own, but I must've gone wrong somewhere because I ended up getting$$\frac {\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(y+1)}\sum\limits_{k=0}^{\infty}\frac 1{(n+1)_k(1)_k(x+1)_{-k}(y+1)_{-k}}$$
Which I don't think is right because that would imply that $(-x)_k(-y)_k=(x+1)_{-k}(y+1)_{-k}$.
| We use the coefficient of operator $[u^n]$ to denote the coefficient of $u^n$ in a series. This way we can write e.g.
\begin{align*}
[u^k](1+u)^n=\binom{n}{k}
\end{align*}
Ad question 1:
We obtain
\begin{align*}
a_n&=[u^n]\left(1+\frac{1}{u}\right)^x\left(1+u\right)^{y+n}\\
&=[u^n]\sum_{k=0}^\infty\binom{x}{k}u^{-k}\sum_{l=0}^\infty\binom{y+n}{l}u^l\tag{1}\\
&=\sum_{k=0}^\infty\binom{x}{k}[u^{n+k}]\sum_{l=0}^\infty\binom{y+n}{l}u^l\tag{2}\\
&=\sum_{k=0}^\infty\binom{x}{k}\binom{y+n}{n+k}\tag{3}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we use the binomial series expansion.
*In (2) we use the linearity of the coefficient of operator and apply the rule
\begin{align*}
[u^{p+q}]A(u)=[u^p]u^{-q}A(u)
\end{align*}
*In (3) we select the coefficient of $u^{n+k}$.
Ad question 2:
We obtain
\begin{align*}
a_n&=[u^n]\left(1+\frac{1}{u}\right)^x\left(1+u\right)^{y+n}\\
&=[u^n]u^{-x}\left(1+u\right)^{x+y+n}\\
&=[u^{n+x}]\left(1+u\right)^{x+y+n}\\
&=\binom{x+y+n}{x+n}
\end{align*}
and the claim follows.
Ad question 3:
Since
\begin{align*}
(-x)_k&=(-x)(-x+1)(-x+2)\cdots(-x+k-1)\\
&=(-1)^kx(x-1)(x-2)\cdots(x-k+1)\\
&=(-1)^k\frac{x!}{(x-k)!}\\
(n+1)_k&=(n+1)(n+2)\cdots(n+k)\\
&=\frac{(n+k)!}{n!}\\
(1)_k&=k!
\end{align*}
we obtain with $x!=\Gamma(x+1)$
\begin{align*}
\frac{\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(y+1)}&\sum_{k=0}^\infty\frac{(-x)_k(-y)_k}{(n+1)_k(1)_k}\\
&=\frac{(y+n)!}{n!y!}\sum_{k=0}^\infty(-1)^k\frac{x!}{(x-k)!}(-1)^k\frac{y!}{(y-k)!}\cdot\frac{n!}{(n+k)!}\cdot\frac{1}{k!}\\
&=\sum_{k=0}^\infty\frac{(y+n)!}{(y-k)!(n+k)!}\cdot\frac{x!}{(x-k)!k!}\\
&=\sum_{k=0}^\infty\binom{y+n}{k+n}\binom{x}{k}
\end{align*}
and the claim follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A infinite and alternating square root of 2 I want to show
$$
A=\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+...}}}}
$$
converges to a finite value. It's easy to see that
$$
\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}
$$
converges to 2 and obviously the value of A is bounded by 2. But it's not easy to see if A converges to a finite value or not. Any suggestion, idea, or comment is welcome, thanks!
| Clearly we have $$A=\sqrt{2-\sqrt{2+A}}$$
Now $A>0$ and $\sqrt{2+A}\le2\iff A\le2$ in fact $A<2$
WLOG $A=2\cos4y$ where $0<4y<90^\circ$
$\sqrt{2+A}=\sqrt{2(1+\cos4y)}=2\cos2y$ as $\cos2y>0$
$\sqrt{2-\sqrt{2+A}}=\sqrt{2-2\cos2y}=2\sin y$ as $\sin y>0$
So, we have $2\cos4y=2\sin y=2\cos(90^\circ-y)$
$\implies4y=360^\circ m\pm(90^\circ-y)$ where $m$ is any integer
$+\implies y=72^\circ m+18^\circ\implies4y=288^\circ m+72^\circ\equiv72^\circ,0,288^\circ,216^\circ,144^\circ\pmod{360^\circ}$
$-\implies4y=480^\circ m-120^\circ\equiv0,120^\circ,240^\circ\pmod{360^\circ}$
But $0<4y<90^\circ\implies4y=72^\circ$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to find max integer value of $6\sin(x)-8\cos(x)$ without using derivative I have $6\sin(x)-8\cos(x)$ and want to find its maximum value. If we use derivative and assuming that it is equal to 0, we get $6\cos(x)+8\sin(x)=0$ which implies that $\tan(x)=-\cfrac{3}{4}$ and $x = 143^\circ$ and we can conclude that the maximum value is equal to $10$ . How can I find the maximum integer value of $6\sin(x)-8\cos(x)$ without using derivative?
| Suppose,
$$a\cos(\theta)-b\sin (\theta)=R \cos (\theta+y)$$
Expanding out the right hand side using the trigonometric angle addition theorem we have,
$$R\cos (\theta)\cos(y)-R\sin (\theta)\sin(y)$$
Than all we need to make sure is that,
$$a=R\cos (y)$$
$$b=R \sin (y)$$
Squaring both equations and adding gives,
$$R^2=a^2+b^2$$
But we need to chose $y$ so both the previous conditions are satisfied.
In the case $a=6$ and $b=8$ then $R^2=6^2+8^2=100$. Let us chose $R=10$.
Then we need to satisfy,
$$6=10\cos (y)$$
$$8=10 \sin (y)$$
Dividing the second equation by the third first $\tan (y)=\frac{8}{6}=\frac{4}{3}$. It is easy to check that $\arctan (\frac{4}{3}) \in [0,\frac{\pi}{2}]$ solves both equations, so it is a valid one. And we have,
$$6\cos (\theta)-8 \sin (\theta)=10 \cos( \theta+\arctan (\frac{4}{3}))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial fractioning an exponential function $\sum\limits_{n=1}^{\infty}\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}$ I saw this and immediately thought it was a telescoping series, so I tried to partial fraction it. $$\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}=\frac{A}{3^{n+1}-2^{n+1}}+\frac{B}{3^n-2^n}$$
\begin{align}
6^n &= A(3^n-2^n)+B(3^{n+1}-2^{n+1})\\
6^n &=3^n(A+3B)+2^n(-A-2B)
\end{align}
This is where I got stuck, how can the sum of these 2 be $6^n$ so I figured one way it could be true was if
$$A+3B=\frac{2^n}{2}$$
$$-A-2B=\frac{3^n}{2}$$
Because then I will just be doing $\frac{6^n}{2}+\frac{6^n}{2}=6^n$
Solving this system gave me
\begin{align}
A&=-\frac{3^{n+1}+2^{n+1}}{2}\\ B&=\frac{3^n+2^n}{2}
\end{align}
As terrible as it looks, I got that it did telescope to $\frac{5}{2}$ however it actually came out to $2$ with the partial fraction of $$\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$$
So 2 questions,
Why am I wrong? and
What would be the correct approach to get this answer?
| Hint: write your term in th form $$\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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} |
Last three digits of $\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$ Find the last three digits of $$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$$
I don't know how to continue my work:
$$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}=\sqrt{2^{2(2016)}+2\cdot (3\cdot 2)^{2016}+3^{2(2016)}}=\sqrt{2^{4032}+2\cdot 3^{2016}\cdot 2^{2016}+3^{4032}}=\sqrt{2^{4032}+2^{2017}\cdot 3^{2016}+3^{4032}}=?$$
| If you follow the hint from @user35508, lets make a congruence.
$$2^{2016}+3^{2016} \text{ mód } 1000$$
Thanks to the Euler's totient function, we know that $a^{\phi(b)}\equiv 1 \text{ mod } b$ if mcd(a,b)=1. The factorization of $1000 = (5*2)^3$ so its trivial that $mcd(2,3,1000)=1$.
$$\phi(1000)=\phi((5*2)^3)=\phi(5^3)\cdot\phi(2^3)=(5^3-5^2)\cdot(2^3-2^2)=400$$
Finally, lets divide $2016/400 = 5\cdot 400 + 16$ and apply it to the first formula.
$$2^{5*400+16}+3^{5*400+16}\text{ mod } 1000 = (2^5)^{400}\cdot{2^{16}}+(3^5)^{400}\cdot{3^{16}}\text{ mod } 1000 = 2^{16}+3^{16}\text{ mod } 1000$$
The remainder is $43112257\text{ mod } 1000 = 257$ that is also the three last digits.
| {
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"timestamp": "2023-03-29T00:00:00",
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When, where and **how often** do you find polynomials of higher degrees than two in mathematical, pure/applied, research? A formula for solving a polynomial of degree three, see this link; $ax^3+bx^2+cx+d=0$, is
$$\begin{align}
x\quad&=\quad
\sqrt[3]{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right)
+
\sqrt{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right) ^2
+
\left(
\frac{c}{3a} - \frac{b^2}{9a^2}
\right) ^3
}
}\\
&+\quad
\sqrt[3]{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right)
-
\sqrt{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right) ^2
+
\left(
\frac{c}{3a} - \frac{b^2}{9a^2}
\right) ^3
}
}
\;-\;\frac{b}{3a}
\end{align}$$
Unlike quadratic, cubic, and quartic polynomials, the general quintic
cannot be solved algebraically in terms of a finite number of
additions, subtractions, multiplications, divisions, and root
extractions, as rigorously demonstrated by Abel (Abel's impossibility
theorem) and Galois. However, certain classes of quintic equations can
be solved [...] Source: http://mathworld.wolfram.com/QuinticEquation.html
At levels of $5^{\text{th}}$ degree polynomials, things are starting to look really serious in my eyes. My question is:
If it is possible to not answer subjectively: When, where and how often do you find polynomials of higher degrees than two in mathematical, pure/applied, research?
| Vaughan Jones won the Fields Medal for finding a polynomial, which is usually not of degree two.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Combinatorics and T-tetromino In how many ways can I fill a grid $4 \times N$ with only T-tetrominoes?
my idea is that it's possible to fill it if and only if $N$ is a multiple of $4$, in that case the solution is:
\begin{equation}
\bigg \{
\begin{array}{rl}
2^{N/4} & N \text{ is multiple of } 4,\\
0 & \text{otherwhise}.\\
\end{array}
\end{equation}
Is that correct? It seems too easy!
| At first we can show that if $N$ is not divisible by $4$ the answer is $0$.
Let's consider $4$ types of cells identified by diagonals:
$$\begin{array}{|c|c|c|c|c}
\hline
1 & 2 & 3 & 4 & 1 & \\\hline
4 & 1 & 2 & 3 & 4 & \\\hline
3 & 4 & 1 & 2 & 3 & \\\hline
2 & 3 & 4 & 1 & 2 & \\\hline
\end{array}
\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}$$
It is easy to see that any $4 \times N$ field has equal number of cell of each type. Each T-tetramino takes $2$ cells of one type (let's call it main type), $1$ cell of each of two other types, the $0$ cells of the remaning type. This means that filling has the same number of tetraminoes for each of $4$ main types. Therefore total number of cells is divisible by $16$ and then $N$ is divisible by $4$.
Let's call a filling solid if each inner vertical line (crossing $4$ lines) contains inner point of some tetramino. Each filling can be splitted into a set of solid fillings. If $f(N)$ is the number of all fillings of field $4 \times N$ and $g(N)$ is the number of solid fillings of the same field, then $$f(n) = \sum_{K = 1}^{N/4}g(4K)f(N - 4K).$$
Now let's compute $g(N)$. Upper left corner cell can be covered in two ways:
$$\begin{array}{|c|c|c|c}
\hline
1 & 1 & 1 & ~ & \\\hline
& 1 & & & \\\hline
& & & & \\\hline
& & & & \\\hline
\end{array}
\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}\qquad
\begin{array}{|c|c|c|c}
\hline
1 & & ~ & \\\hline
1 & 1 & & \\\hline
1 & & & \\\hline
& & & \\\hline
\end{array}\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}$$
That implies
$$\begin{array}{|c|c|c|c}
\hline
1 & 1 & 1 & & \\\hline
2 & 1 & & & \\\hline
2 & 2 & 3 & & \\\hline
2 & 3 & 3 & 3 & \\\hline
\end{array}
\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}\qquad
\begin{array}{|c|c|c|c}
\hline
1 & 3 & 3 & 3 & \\\hline
1 & 1 & 3 & & \\\hline
1 & 2 & & & \\\hline
2 & 2 & 2 & & \\\hline
\end{array}\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}$$
These two cases are symmetrical reflection of each other, so we can consider any of them and multiply result by $2$. If $N = 4$ we can place one more tetramino to fininish filling:
$$\begin{array}{|c|c|c|c|}
\hline
1 & 1 & 1 & 4 \\\hline
2 & 1 & 4 & 4 \\\hline
2 & 2 & 3 & 4 \\\hline
2 & 3 & 3 & 3 \\\hline
\end{array}$$
If $N > 4$ the only way to continue is the following:
$$\begin{array}{|c|c|c|c}
\hline
1 & 1 & 1 & 4 & 7 & 7 & 7 & \\\hline
2 & 1 & 4 & 4 & 4 & 7 & & \\\hline
2 & 2 & 3 & 5 & 5 & 5 & 6 & \\\hline
2 & 3 & 3 & 3 & 5 & 6 & 6 & 6 \\\hline
\end{array}
\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}$$
This is absolutely the same situation: for $N = 8$ we can finish filling and for $N > 8$ we have exactly $1$ way to continue. Therefore $g(N) = 2$.
$f(4) = 2$ and we have
$$f(N) = \sum_{K = 1}^{N / 4}g(4K)f(N - 4K) = 2\sum_{K = 1}^{N / 4}f(N - 4K)\\
= 2f(N - 4) + 2\sum_{K = 1}^{N / 4 - 1}f(N - 4 - 4K) = 3f(N - 4).$$
Therefore we have $f(N) = [4 \mid N]2 \cdot 3^{N / 4 - 1}$ for $N \ge 1$ and $f(0) = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a closed formula for a generating function involving catalan numbers For $n$ $\ge$ $1$, let the Catalan number $C_n$ be defined to be the number of ways of partitioning a convex $(n+2)$-gon into $n$ triangles by using diagonals that do not cross one another (except perhaps at their ends). By convention set $C_0$ $=$ $1$.
Find a recurrence relation expressing $C_n$ in terms of the Catalan numbers of smaller index.
I have found this to be $C_{n+1}$ $=$ $\frac{4n+2}{n+2}C_n.$
I want to use this to find a closed formula for the generating function $$g(x) = \sum_0^{\infty}C_nx^n.$$
Please help me..
| Going directly from your recursion is not to difficult, but first you need to correct the recursion -- your index is off by 1. The right relation is
$$
C_k = C_{k-1} \frac{4k+2}{k+2} + [k = 0]
$$
where $[k = 0]$ stands for a function of $k$ which is $1$ when $k=0$ and zero otherwise.
We define (for this derivation) $C_{n} = 0$ for all negative $n$.
Break up
$$ \frac{4k+2}{k+2} = 4 - \frac{8k+8 -6}{k+2} =4-\frac{6}{k+2}$$
to get (R)
$$
C_k = 4 C_{k-1} -6 \frac{C_{k-1}}{k+2} + [k = 0]
$$
Multiply (R) by $z^k$ for all values of $k$ and sum those equations to get a generating function relation:
$$
\sum_{k=0}^\infty C_k z^k= 4 \sum_{k=0}^\infty C_{k-1}z^{k} -6\sum_{k=0}^\infty \frac{1}{k+2}C_{k-1}z^{k}+1
$$
Here, we have used the fact that $[k=0]$ contributes 1 to the sum when $k$ is zero, and nothing else.
Now shift the summation indices on the right by using $m=k-1$ so that only $C_d$ appears, where $d$ is a "dummy" (summed over) index, to get (S):
$$
C(z) \equiv \sum_{k=0}^\infty C_k z^k= 4 \sum_{m=0}^\infty C_{m}z^{m+1} -6\sum_{m=0}^\infty \frac{1}{m+1+2}C_{k-1}z^{m+1}+1
\\=4 z C(z) - 6\sum_{m=0}^\infty \frac{1}{m+3}C_{k-1}z^{m+1}+1
$$
The $(m+3)$ in the denominator would disappear if we had a $z^{m+3}$ multiplying it, and differentiated that term. So multiply (S) by $z^2$ and differentiate with respect to $z$:
$$\frac{d}{dz} \left( z^2 C(z) \right) =
4 \frac{d}{dz} \left( z^3 C(z) \right) + z^2 C(z) + \frac{d}{dz}(z^2)
$$
Use the multiplication rule for derivatives to simplify this and get (T):
$$
z^2 \frac{dC(z)}{dz} + 2z C(z) = 4z^3 \frac{dC(z)}{dz} +12 z^2 C(z) - 6 z^2 C(z) + 2 z\\
(4z^3-z^2) \frac{dC(z)}{dz} + (6 z^2 -2 z) C(z) + 2 z = 0
$$
Solve differential equation (T) to get
$$
C(z) = \frac{1-2z+\kappa \sqrt{1-4z}}{2z^2}
$$
The behavior of this function near $z=0$ is
$$
\frac{1+\kappa}{2z^2} - \frac{1+\kappa}{z} -\kappa + \cdots
$$
and since the coefficients for negative exponents of $z$ need to be zero, this determines that $\kappa = -1$.
$\kappa$ can be determined by noting that the constant term of the generating function must be $C_0 = 1$, and the which forces $\kappa = 1$.
So the generating function for $C_k$ is
$$
C(z) = \frac{1-2z-\sqrt{1-4z}}{2z^2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x,y$ and $z$ when: $x+y=\sqrt{4z-1}, \ y+z=\sqrt{4x-1}\ \mathrm{and} \ z+x=\sqrt{4y-1}$ How to Solve for $x,y$ and $z$ when: $x+y=\sqrt{4z-1}, \ y+z=\sqrt{4x-1}\ \mathrm{and} \ z+x=\sqrt{4y-1}$.
I've added the equations and get: $2(x+y+z)=\sqrt{4x-1}+\sqrt{4y-1}+\sqrt{4z-1}$ and after this how I proceed further?
Even if I subtract the first two equations then I'm getting:
\begin{align*}
x-z&=\sqrt{4z-1}-\sqrt{4x-1}\\
\Rightarrow\left(x+\sqrt{4x-1}\right)^2&=\left(z+\sqrt{4z-1}\right)^2\\
\Rightarrow\left(x^2-z^2\right)+4(x-z)&=2z\sqrt{4z-1}-2x\sqrt{4x-1}
\end{align*}
after this also I can't understand how to help with this.
| Hint: $\;x,y,z \ge \frac{1}{4}$ for the square roots to be defined. Assume WLOG that $x \ge y \ge z\,$, then $\sqrt{4z-1}=x+y \ge z+z = 2z \implies 4z-1 \ge 4z^2\iff (2z-1)^2 \le 0 \,$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to solve an equation of the form $f(x)=f(a)$ for a fixed real a. I got stuck on this question: find all solutions $x$ for $a\in R$:
$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{(a^2-a+1)^3}{a^2(a-1)^2}$$
I see that if we simplify we get:
$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{[(x-{\frac 12})^2+{\frac 34}]^3}{[(x-{\frac 12})^2-{\frac 14}]^2}$$
From the expression $(x-{\frac 12})^2$, I see that if $x=x_1$ is a solution, then $x=1-x_1$ is also a solution. But in the solution to this exercise, it was stated that $x=\frac{1}{x_1}$ must also be a solution, and I don't see how.
[EDIT]
Ok, thx for the help guys. What do you think of this solution (doesn't involve any above precalculus math, and needs no long calculations)?
From the above we know that if $x_1=a$ is a solution, then $x_2=1-a$ is also a solution.
Also, from here:
$$\require{cancel}\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{\cancel{x^3}(x+{\frac 1x}-1)^3}{\cancel{x^3}(x+{\frac 1x}-2)}$$
in the expression $x+{\frac 1x}$ we see that if $x=x_1$ is a solution, then $x=\frac{1}{x_1}$ is also a solution, so $x_3=\frac{1}{a}$.
With these two rules we can now keep generating roots until we have 6 total.
If $x=x_2$ is a solution, then $x=\frac{1}{x_2}$ is also a solution, so $x_4=\frac{1}{1-a}$.
If $x=x_3$ is a solution, then $x=1-x_3$ is also a solution, so $x_5=\frac{a-1}{a}$.
Finally, if $x=x_5$ is a solution, then $x=\frac{1}{x_5}$ is also a solution, so $x_6=\frac{a}{a-1}$
The 6 obtained values are distinct, so they cover all the roots.
[EDIT2]
I guess this is answered. No sure whose particular answer to actually select as the right one since they're all correct, so I'll just leave it like this.
| $$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{(a^2-a+1)^3}{a^2(a-1)^2}$$
Now multiply both sides by $x^2(x-1)^2$ :
$$(x^2-x+1)^3=x^2(x-1)^2\frac{(a^2-a+1)^3}{a^2(a-1)^2}
\\ \implies (x^2-x+1)^3-x^2(x-1)^2\frac{(a^2-a+1)^3}{a^2(a-1)^2}=0$$
Without expanding, it can be written as :
$$x^6+a_1x^5+a_2x^4+a_3x^3+a_4x^2+a_5x+1=0$$
Since, product of roots of this equation is $1$, each solution's reciprocal will also be a solution to this equation.
| {
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"source": "stackexchange",
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Identity of $\frac{n!}{x(x+1)(x+2)...(x+n)}$
$$\frac{n!}{x(x+1)(x+2)\dots(x+n)}=\sum_{k=0}^n{n\choose k}\frac{(-1)^k}{x+k}$$
This is the identity. I tried proving it by induction but it got really complicated and couldn't solve it.
| A proof by induction is possible, using the identity $$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}.$$ To this end, let $$P_n(x) = \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{x+k}.$$ Then $$\begin{align*} P_{n+1}(x) &= \sum_{k=0}^{n+1} \left(\binom{n}{k-1} + \binom{n}{k}\right) \frac{(-1)^k}{x+k} \\
&= \sum_{m=0}^n \binom{n}{m} \frac{(-1)^{m+1}}{x+1+m} + \sum_{m=0}^n \binom{n}{m} \frac{(-1)^m}{x+m} \\
&= -P_n (x+1) + P_n (x).
\end{align*}$$
Now let $Q_n(x) = \frac{n!}{x(x+1)\ldots(x+n)}$. Then if $P_n(x) = Q_n(x)$ for some positive integer $n$, we have
\begin{align}
P_{n+1}(x) &= -Q_n(x+1) + Q_n(x) \\
&= -\frac{n!}{(x+1)(x+2)\ldots(x+n+1)} + \frac{n!}{x(x+1)\ldots(x+n)} \\
&= \frac{n!}{(x+1)(x+2)\ldots(x+n)} \left( \frac{1}{x} - \frac{1}{x+n+1} \right) \\
&= \frac{n! (n+1)}{x(x+1)\ldots(x+n+1)} \\
&= Q_{n+1}(x),
\end{align} completing the induction step.
| {
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Solve for $x$ in $\frac{x^2+12x+36}{x^2+4x-12}=2$ $\frac{x^2+12x+36}{x^2+4x-12}=2$
I factored top and bottom to $\frac{(x+6)(x+6)}{(x+6)(x-2)}=2$
and eliminated the common factor (x+6) for $\frac{x+6}{x-2}=2$
then $x+6=2(x-2)$ and $x+6=2x-4$ and $6+4=2x-x$ and finally $x = 10$
Plugging this back into the original fraction proves true
My instructor however in her exam prep test says
$\frac{x^2+12x+36}{x^2+4x-12}=2$
$$x^2+12x+36 = 2(x^2+4x-12)$$
$$x^2+12x+36 = 2x^2+8x-24$$
$$0 =2x^2-x^2 +8x -12x -24 -36$$
$$0 =x^2-4x -60$$
which she then factors to $(x-10)(x+6)$ saying the answer to the problem is $x=10$ or $x=-6$
But if you plug -6 back into the original fraction you get $0=2$. Where is the problem here? Thanks
| In short, you are correct.
Slightly longer, the expression
$$\frac{x^2+12x+36}{x^2+4x-12}$$
is not defined for $x = {-6}$ because $x^2 + 4x -12 = 0$ when $x = -6$ and division by zero is undefined. Hence a priori to even talk about this fraction we assume that $x\neq -6$ and $x\neq2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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How do we show that $S=\sum_{k=1}^{\infty}{k^3\over (4k^2-1)(9k^2-1)(16k^2-1)}?$ Let:
$$S=\sum_{k=1}^{\infty}{k^3\over (4k^2-1)(9k^2-1)(16k^2-1)}\tag1$$
How can we show that $$S={1\over 420}+{1\over 7}\ln\left[2^{1/12}\left({4\over 3\sqrt{3}}\right)^{1/5}\right]?\tag2$$
An attempt:
Apply decomposition of partial fraction to
$${k^3\over (4k^2-1)(9k^2-1)(16k^2-1)}={A\over 4k^2-1}+{B\over 9k^2-1}+{C\over 16k^2-1}\tag3$$
$$k^3=A(9k^2-1)(16k^2-1)+B(4k^2-1)(16k^2-1)+C(4k^2-1)(9k^2-1)$$
$(1)$ becomes
$$\sum_{k=1}^{\infty}\left({32\over 15(4k^2-1)}-{32\over 35(9k^2-1)}-{43\over 21(16k^2-1)}\right)\tag4$$
Further simplify to
$${1\over 2}\sum_{k=1}^{\infty}\left[{32\over 15}\color{red}{\left({1\over 2k-1}-{1\over 2k+1}\right)}-{32\over 35}\left({1\over 3k-1}-{1\over 3k+1}\right)-{43\over 21}\left({1\over 4k-1}-{1\over 4k+1}\right)\right]\tag5$$
The red part telescope
$$\sum_{k=1}^{N}\left({1\over 2k-1}-{1\over 2k+1}\right)={1\over 2N+1}\tag6$$
So $(5)$ simplify to
$${1\over 2}\sum_{k=1}^{\infty}\left[{32\over 35}\left({1\over 3k-1}-{1\over 3k+1}\right)-{43\over 21}\left({1\over 4k-1}-{1\over 4k+1}\right)\right]\tag7$$
I got this far, but I can't evaluate any further.
| By computing the residues of $f(z)=\frac{z^3}{(4z^2-1)(9z^2-1)(16z^2-1)}$ at $\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{1}{4}$ we get
$$ f(z)=\frac{1}{120}\left(\frac{1}{z-\frac{1}{2}}+\frac{1}{z+\frac{1}{2}}\right)-\frac{1}{70}\left(\frac{1}{z-\frac{1}{3}}+\frac{1}{z+\frac{1}{3}}\right)+\frac{1}{168}\left(\frac{1}{z-\frac{1}{4}}+\frac{1}{z+\frac{1}{4}}\right)\tag{A}$$
where the sum of residues is $0=\frac{1}{120}-\frac{1}{70}+\frac{1}{168}$ and
$$ \sum_{n\geq 1}\left(\frac{1}{n-\frac{1}{m}}+\frac{1}{n+\frac{1}{m}}-\frac{2}{n}\right)=-2\gamma-\psi\left(1-\frac{1}{m}\right)-\psi\left(1+\frac{1}{m}\right). \tag{B}$$
It follows that the sum of the given series just depends on values of the digamma function at rational points, that can be computed through Gauss' digamma theorem. In particular:
$$ \sum_{n\geq 1}\frac{n^3}{(4n^2-1)(9n^2-1)(16n^2-1)}=\frac{1}{420}\left[1+29 \log(2)-18 \log(3)\right].\tag{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Value of n for which f(n) = $\,n^2 + 9n + 30\,$ is a perfect square. I attempted this by setting $f(n) = \,m^2.\,$
So $\,n^2 + 9n + 30 = m ^2\,$.
Then $\,9(n + 10/3) = (m + n)(m - n)\,$.
So $m = 10/3$ and $n = -17/3$ which is incorrect.
| Let's assume that $n$ is positive. Observe that
$$(n+4)^2=n^2+8n+16<n^2+9n+30<n^2+12n+36=(n+6)^2$$ and the only possibility is $n^2+9n+30=(n+5)^2$.
Now, let's assume that $m=-n$ is positive and $m^2-9m+30$ is perfect square. Also observe that$$(m-5)^2=m^2-10m+25<m^2-9m+30$$
and when $m$ is large enough, $m^2-9m+30<m^2-8m+16=(m-4)^2$ and you can check the case $m^2-9m+30 \ge (m-4)^2$ manually.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How many positive integer solutions does the equation $a+b+c=100$ have if we require $aHow many positive integer solutions does the equation $a+b+c=100$ have if we require $a<b<c$?
I know how to solve the problem if it was just $a+b+c=100$ but the fact it has the restriction $a<b<c$ is throwing me off.
How would I solve this?
| $a<b<c$ can be translated by $b=a+x$ and $c=b+y=a+(x+y)$, where $a,x,y>0$.
$$a + b + c = 100\iff a+a+x+a+x+y=100 \iff3a+2x+y=100$$
Let $a_k$ the number of solutions of the diophantine equation $3a+2x+y=k$, where $a,x,y>0$.
$$\sum_{d=0}^\infty a_k t^k=\sum_{k=0}^\infty(\sum_{3a+2x+y=k\\a,x,y>0}t^k)=(\sum_{k=1}^\infty t^{3k})(\sum_{k=1}^\infty t^{2k})(\sum_{k=1}^\infty t^{k})=\frac{t^6}{(1-t^3)(1-t^2)(1-t)}$$
We will calculate explecitely $a_k$ now :
$$\frac{t^6}{(1-t^3)(1-t^2)(1-t)}\\=\frac{t + 2}{9 (t^2 + t + 1))} - \frac{89}{72 (t - 1)} + \frac{1}{8 (t + 1)} - \frac{3}{4 (t - 1)^2} -\frac{1}{6 (t - 1)^3}-1\\=\sum_{n=1}^∞ t^n\frac{1}{72} (47 + 9 (-1)^n + 8 e^{-(2 i n π)/3} + 8 e^{(2 i n π)/3} + 6 n^2-36n)\\=\sum_{n=1}^∞ t^n\frac{1}{72} (47 + 9 (-1)^n + 16\cos(2\pi \frac{n}{3}) + 6 n^2-36n)$$
So : $$a_{100}=\frac{1}{72} (47+9 -8 +60000-3600)=784$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Trig - Difference formulas to solve $\sqrt3\sin \theta- \cos\theta = 1$ My exam review states that I need to utilize the difference formula for sine to solve the equation on the interval $0 \leq \theta < 2\pi $
$$\sqrt3\sin \theta- \cos\theta = 1$$
I know that: $\sin \frac\pi3 = \frac{\sqrt3}{2}$
and $\cos\frac\pi3 = \frac12 $, so I divide each term by 2 and rewrite the equation:
$$\frac{\sqrt3\sin\theta}{2} - \frac{\cos\theta}{2} = \frac12$$
from here, I apply the difference formula for sine:
$$\sin(\alpha-\beta) = (\sin\alpha \cdot \cos\beta) - (\cos\alpha \cdot \sin\beta)$$
(this is the step that I believe I'm doing incorrectly. I've tried plugging in the corresponding numerical values for sin/cos into the equation instead, but I was still unable to go past this step.)
$$\sin(\alpha-\beta) = \left(\frac {\pi}{3} \cdot \frac{\pi}{3}\right) - \left(\frac{\pi}{6} \cdot \frac{\pi}{6}\right) = \frac12$$
| $$\sqrt3\sin \theta- \cos \theta = 1$$
$$2(\frac{\sqrt3}{2} \sin \theta- \frac{1}{2}\cos \theta)=1$$
$$(\frac{\sqrt3}{2} \sin \theta- \frac{1}{2}\cos \theta)=\frac12$$
$$ (\cos \dfrac{\pi}{6}\sin \theta -\sin \dfrac{\pi}{6} \cos \theta) = \sin \dfrac{\pi}{6} $$
$$\sin\left(\theta-\dfrac\pi6\right)=\sin\dfrac\pi6$$
$$\left(\theta-\dfrac\pi6\right)=\dfrac\pi6,\, \pi-\dfrac\pi6 .... $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find $\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}})$ $$\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}})$$
First I don't know how to solve this problem since there is $\sin (\frac{1}{\sqrt[3]{x-1}})$ in there. But I manage to do this:
$$\lim\limits_{x \to 1} (x-1)^2 \sin (\frac{1}{\sqrt[3]{x-1}}) =
(\lim\limits_{x \to 1} (x-1)^2) (\lim\limits_{x \to 1}\sin (\frac{1}{\sqrt[3]{x-1}}))
$$
$$= ((1-1)^2) (\lim\limits_{x \to 1}\sin (\frac{1}{\sqrt[3]{x-1}}))$$
$$= (0) (\lim\limits_{x \to 1}\sin (\frac{1}{\sqrt[3]{x-1}}))$$
$$= 0$$
But I'm pretty sure that I'm wrong here.
| Well, according what I read. Using, the squeeze theorem meaning that:
$$-1 \le \sin (\frac{1}{\sqrt[3]{x-1}}) \le 1$$
$$-(x-1)^2 \le \sin (\frac{1}{\sqrt[3]{x-1}}) \le (x-1)^2$$
$$-(x-1)^2 \le (x-1)^2\sin (\frac{1}{\sqrt[3]{x-1}}) \le (x-1)^2$$
$$\lim\limits_{x \to 1} -(x-1)^2 \le \lim\limits_{x \to 1} (x-1)^2\sin (\frac{1}{\sqrt[3]{x-1}}) \le \lim\limits_{x \to 1} (x-1)^2$$
Then I got this:
$$\lim\limits_{x \to 1} -(x-1)^2 = -(1-1)^2 = 0$$
And this:
$$\lim\limits_{x \to 1} (x-1)^2 = (1-1)^2 = 0$$
Meaning that:
$$0 \le \lim\limits_{x \to 1} (x-1)^2\sin (\frac{1}{\sqrt[3]{x-1}}) \le 0$$
So, I can conclude that:
$$\lim\limits_{x \to 1} (x-1)^2\sin (\frac{1}{\sqrt[3]{x-1}}) = 0$$
Is this the correct way?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Subsets and Splits
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