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Inequality $\left(\frac{2x-1}{2x+1}\right)^2>0$ The inequality
$$\left(\frac{2x-1}{2x+1}\right)^2>0$$
has the same solution set as
a) $(2x-1)^2 > 0$
b) $2x-1 \neq 0$
c) $2x-1 > 0$
d) None of the above
Is this equivalent to$$\frac{2x-1}{2x+1}>0 \ ? $$
|
The answer for your question is shortly "No", because $$\left( \frac { 2x-1 }{ 2x+1 } \right) ^{ 2 }>0\quad \Rightarrow x\in R-\left\{ \pm \frac { 1 }{ 2 } \right\} \\ \\ \frac { 2x-1 }{ 2x+1 } >0\quad \Rightarrow x\in \left( -\infty ;-\frac { 1 }{ 2 } \right) \cup \left( \frac { 1 }{ 2 } ;+\infty \right) $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to show that $\int_{-\alpha}^{\alpha}\arccos\left({x\over \alpha}\right)\ln(\alpha+x)\mathrm dx=\alpha \pi \ln\left({\alpha \over 2}\right)?$ Given that:
Where $\alpha >0$
$$\int_{-\alpha}^{\alpha}\arccos\left({x\over \alpha}\right)\ln(\alpha+x)\mathrm dx=\alpha \pi \ln\left({\alpha \over 2}\right)\tag1$$
Setting $y=\arccos(x/\alpha)\implies \cos(y)={x\over \alpha}$
$dx=-\alpha \sin(y)dy$
$$\alpha\int_{0}^{\pi}y\sin(y)\ln(\alpha+\alpha\cos y)\mathrm dy\tag2$$
$$\alpha \ln\alpha \int_{0}^{\pi}y\sin y\mathrm dy+\alpha\int_{0}^{\pi}y\sin y\ln(1+\cos y)\mathrm dy\tag3$$
$$\alpha \pi \ln\alpha +\alpha\int_{0}^{\pi}y\sin y\ln(1+\cos y)\mathrm dy\tag4$$
|
$\begin{align} J= \int_{-1}^{1} \arccos x\ln(1+x)\,dx&=\Big[(x+1)(\ln(1+x)-1)\arccos x\Big]_{-1}^1+\int_{-1}^{1}\dfrac{(x+1)(\ln(1+x)-1)}{\sqrt{1-x^2}}dx\\
=&\int_{-1}^{1}\sqrt{\dfrac{1+x}{1-x}}(\ln(1+x)-1)dx\\
\end{align}$
Perform the change of variable $y=\sqrt{\dfrac{1+x}{1-x}}$,
$\begin{align}
J&=8\int_0^{+\infty} \dfrac{x^2\ln x}{(1+x^2)^2}\,dx+4(\ln 2-1)\int_0^{+\infty} \dfrac{x^2}{(1+x^2)^2}\,dx-4\int_0^{+\infty} \dfrac{x^2\ln(1+x^2)}{(1+x^2)^2}\,dx\\
&=-4\left(\Big[\dfrac{x\ln x}{1+x^2}\Big]_0^{+\infty}-\int_0^{+\infty}\left(\dfrac{\ln x}{1+x^2}+\dfrac{1}{1+x^2}\right)\,dx\right)+4(\ln 2-1)\int_0^{+\infty} \dfrac{x^2}{(1+x^2)^2}\,dx+\\
&2\left(\Big[\dfrac{x\ln(1+x^2)}{1+x^2}\Big]_0^{+\infty}-\int_0^{+\infty}\left(\dfrac{\ln(1+x^2)}{1+x^2}+\dfrac{2x^2}{(1+x^2)^2}\right)\,dx\right)\\
&=2\pi+4(\ln 2-2)\int_0^{+\infty} \dfrac{x^2}{(1+x^2)^2}\,dx-2\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}\,dx\\
&=2\pi+2(\ln 2-2)\left(\Big[-\dfrac{x}{1+x^2}\Big]_0^{+\infty}+\int_0^{+\infty}\dfrac{1}{1+x^2}\,dx\right)-2\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}\,dx\\
&=2\pi+(\ln 2-2)\pi-2\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}\,dx\\
&=\pi\ln 2-2\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}\,dx\\
\end{align}$
In the latter integral perform the change of variable $y=\arctan x$,
$\begin{align}J&=\pi\ln 2+4\int_0^{\tfrac{\pi}{2}} \ln(\cos x)\,dx\\
&=\pi\ln 2-2\pi\ln 2\\
&=-\pi\ln 2
\end{align}$
(Thanks to Gebrane for inspiration)
|
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|
$S_n=\sum_{k=1}^n\frac{1}{k}$. then $S_{2^n}$=? Let $S_n$=$\sum_{k=1}^n\frac{1}{k}$. which of the following is true?
*
*$S_{2^n}\ge\frac{n}{2}$ for every n$\ge1$.
*$S_n$ is a bounded sequence.
*$|S_{2^n}-S_{2^{n-1}}|\to0$ as n$\to\infty$.
*$\frac{S_n}{n}\to1$ as n$\to\infty$.
I have a confusion that whether $S_{2^n}$=$\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^n}$?
or
$S_{2^n}$=$1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2^n}$?
|
The second interpretation is correct: $S_{2^n}=1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2^n}$.
Hint for your questions: note that
$$S_{2^{n+1}}=S_{2^{n}}+\sum_{k=1}^{2^n}\frac{1}{2^n+k}\geq S_{2^{n}}+\sum_{k=1}^{2^n}\frac{1}{2^n+2^n}=S_{2^{n}}+\frac{1}{2}.$$
Hence $S_{2^{n+1}}-S_{2^{n}}\geq 1/2$ and 3) is false.
Show by induction that 1) holds and therefore 2) is false.
What about 4)?
P.S. 4) is false. We have that
$$S_{2^{n+1}}=S_{2^{n}}+\sum_{k=1}^{2^n}\frac{1}{2^n+k}< S_{2^{n}}+\sum_{k=1}^{2^n}\frac{1}{k}=2S_{2^{n}}.$$
Hence the sequence $(S_{2^{n}}/2^n)_n$ is positive and strictly decreasing.Therefore it converges to a non-negative limit $L$. Since $S_{1}/1=1$ it follows that $L<1$ which contradicts 4).
|
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|
How to prove an inequality I didn't know where can I start because it doesn't fit to any theories or formulas.
If $a>0$, $b>0$, $c>0$ and $a+b+c=1$, prove that:
$$\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\geq \frac{1}{2}$$
|
Note that using AM-GM,
$$\frac{a^3}{a^2+b^2} = a - \frac{ab^2}{a^2+b^2} \geqslant a - \frac{ab^2}{2ab} = a - \frac{b}2 $$
Hence $$\sum_{cyc} \frac{a^3}{a^2+b^2} \geqslant \frac{a+b+c}2 = \frac12$$
|
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|
Finding solutions to $2^x+17=y^2$
Find all positive integer solutions $(x,y)$ of the following equation:
$$2^x+17=y^2.$$
If $x = 2k$, then we can rewrite the equation as $(y - 2^k)(y + 2^k) = 17$, so the factors must be $1$ and $17$, and we must have $x = 6, y = 9$.
However, this approach doesn't work when $x$ is odd.
|
I think that there are not solutions for $x$ odd.
Indeed,
$y^2=2^x+17=2^{2k+1}+17=2 \cdot 4^k +17 \equiv 2+17 \equiv3 \mod4$, which cannot happen.
(If $x$ is of the form $x=2k$, then the equation can be easily written $(y-2^k)(y+2^k)=17$, and so we go on).
EDIT: Wrong solution!
|
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|
Show that $\int_{0}^{\pi/2}{\ln[\cos^2(x)+\sin^2(x)\tan^4(x)]\over \sin^2(x)}\mathrm dx=\pi$ Given that:
$$\int_{0}^{\pi/2}{\ln[\cos^2(x)+\sin^2(x)\tan^4(x)]\over \sin^2(x)}\mathrm dx=\pi\tag1$$
$$1-\sin^2(x)+\sin^2(x)\tan^4(x)$$
$$=1+\sin^2(x)[\tan^4(x)-1]$$
$$=1+\sin^2(x)[\tan^2(x)-1][\tan^2(x)+1]$$
$$=1+\tan^2(x)[\tan^2(x)-1]$$
$$=1-\tan^2(x)+\tan^4(x)$$
$$\int_{0}^{\pi/2}{\ln[1-\tan^2(x)+\tan^4(x)]\over \sin^2(x)}\mathrm dx\tag2$$
$t=\tan(x)\implies dt=\sec^2(x) dx$
$$\int_{0}^{\infty}{\ln(1-t^2+t^4)\over t^2}\mathrm dt\tag3$$
$u=t^2\implies du=2tdt$
$${1\over 2}\int_{0}^{\infty}{\ln(1-u+u^2)\over u^{3/2}}\mathrm du\tag4$$
|
$J=\displaystyle \int_{0}^{\infty}{\ln(1-t^2+t^4)\over t^2}\mathrm dt$
$\begin{align}
J&=\Big[-\dfrac{\ln(1-t^2+t^4)}{t}\Big]_{0}^{+\infty}+\int_{0}^{+\infty}\dfrac{-2+4t^2}{1-t^2+t^4}dt\\
&=\int_{0}^{+\infty}\dfrac{-2+4t^2}{1-t^2+t^4}dt
\end{align}$
Observe that,
$\displaystyle\int_{0}^{+\infty}\dfrac{t^2}{1-t^2+t^4}dt=\int_{0}^{+\infty}\dfrac{1}{1-t^2+t^4}dt$
(perform the change of variable $y=\dfrac{1}{t}$ )
Therefore,
$\begin{align}
J&=\int_{0}^{+\infty}\dfrac{2}{1-t^2+t^4}dt\\
&=\int_{0}^{+\infty}\dfrac{t^2}{1-t^2+t^4}dt+\int_{0}^{+\infty}\dfrac{1}{1-t^2+t^4}dt\\
&=\int_{0}^{+\infty}\left(1+\dfrac{1}{t^2}\right)\dfrac{1}{1+\left(t-\tfrac{1}{t}\right)^2}dt
\end{align}$
Perform the change of variable $y=t-\dfrac{1}{t}$,
$\begin{align}
J&=\int_{-\infty}^{+\infty}\dfrac{1}{1+t^2}dt\\
&=\boxed{\pi}
\end{align}$
|
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|
Find the value of $√(4+6\sqrt{5}i) + √(4-6\sqrt{5}i)$
Find the value of $\sqrt{(4+6\sqrt{5}i)} + \sqrt{(4-6\sqrt{5}i)}$
$\sqrt{(4+6\sqrt{5}i)} = ± (3+\sqrt{5}i)$
$\sqrt{(4-6\sqrt{5}i)} = ± (3-\sqrt{5}i)$
There are two solutions to each which implies on adding them together, we will get four combinations and four different answers.
These answers are $6,-6,-2\sqrt{5}i,2\sqrt{5}i$.
But the only answer given in my book is $6$; they have only taken the + and + combination. Why we don't take the remaining three combinations? Why aren't all four the answers?
|
I think the book means the following.
$$\left(\sqrt{4+6\sqrt5i}+\sqrt{4-6\sqrt5i}\right)^2=8+2\cdot14=36$$
and the book gets $\sqrt{4+6\sqrt5i}+\sqrt{4-6\sqrt5i}=6$.
This reasoning is wrong, of course.
|
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|
Condition for three roots The equation $$2\sin ^3 x +(2\lambda -3)\sin ^2 x -(3\lambda +2)\sin x -2\lambda=0$$ has excatly three roots in $(0,2\pi)$ then what can be the value of $\lambda$ .
I thought for three roots the differentiation of the equation should have two roots . but from that I am not getting anya graph .
|
The equation can be factored as follows:
$$
(\sin x - 2)(2\sin x + 1)(\sin x + \lambda) = 0
$$
For the lefthand side to be $0$, one of the factors needs to be $0$.
$\sin x - 2$ cannot be $0$.
$2\sin x + 1 = 0$ has two solutions ($\frac{7\pi}{6}$ and $\frac{5\pi}{3}$). The number of solutions of $\sin x + \lambda = 0$ depends on $\lambda$.
Because the total number of solutions needs to be exactly $3$, we see that $\sin x + \lambda = 0$ has either exactly one solution, this happens when $\lambda$ is $-1$, $0$ or $1$, or there have to be $2$ solutions, and exactly one of these solutions coincides with $\frac{7\pi}{6}$ or $\frac{5\pi}{3}$. This last option is not possible, because then $\lambda$ would be $\frac{1}{2}$, which means that both solutions of $\sin x + \lambda = 0$ are $\frac{7\pi}{6}$ and $\frac{5\pi}{3}$, so the total number of solutions would be 2.
So we see that the only possibilities for $\lambda$ are $-1$, $0$ and $1$.
|
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|
How to integrate $\int_a^b t \cdot \sqrt{\frac{t - a}{b - t}} \,dt$? How to integrate $\displaystyle \int_a^b t \cdot \sqrt{\frac{t - a}{b - t}} \,dt$?
I literally have no idea how to integrate this integral.
I've tried all basic methods, but it seems like a quite hard problem for a beginner.
|
Do the following substitution
$$t=a\cos^2 \theta+b \sin^2 \theta \implies \mathrm dt=(b-a) \sin 2 \theta \, \mathrm d \theta$$
Your function $\displaystyle \sqrt{\frac{t - a}{b - t}}$ will be tremendously simplified to $\tan \theta$. Hence, your integral will be simplified :
\begin{align}
\int_a^b t \cdot \sqrt{\frac{t - a}{b - t}} \, \mathrm dt&=\int_0^{{\pi}/{2}} (a\sin^2 \theta+b \cos^2 \theta) \cdot \tan \theta \,((b-a) \sin 2 \theta \, \mathrm d \theta)\\
&= ~a\int_0^{{\pi}/{2}} \frac{\sin^3 \theta}{\cos \theta}\, \cdot (b-a) \sin 2 \theta \, \mathrm d \theta+ b\int_0^{{\pi}/{2}}\frac{ \sin 2 \theta}{2} \, (b-a) \sin 2 \theta \, \mathrm d \theta\\
&=~(b-a) \left(2a \int_0^{{\pi}/{2}} {\sin^4 \theta}\, \mathrm d \theta + \frac b2 \int_0^{{\pi}/{2}}{ \sin^2 2 \theta} \, \mathrm d \theta \right)\\
&=~(b-a) \left(2a \cdot\frac{3\pi}{16} + \frac b2 \cdot\frac{\pi}{4}\right) \\
&=\frac\pi8 (b-a) (3a+b)
\end{align}
|
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|
How to find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$?
Find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$. The correct answer should be $(8\pi-16)a^2$.
This is an illustration:
First, there's the following formula to calculate surface area:
$$
\int\int_R\sqrt{Z_x^2+Z_y^2+1}\cdot dA
$$
In our case it's:
$$
\int\int_R \sqrt{\frac{x^2+y^2}{4a^2-x^2-y^2}+1}dA
$$
We can move into polar coordinates (and multiply the original operand by $r$ as the Jacobian in polar), then:
$$
\int\int_R\ r\sqrt{\frac{r^2}{4a^2-r^2}+1}drd\theta
$$
Now the only thing left is to find the bounds of $r$ and $\theta$.
The projection of the surface to $xy$ plane will be a circle with the center in $(0,a)$ with the radius $a$ so it's symmetric around $y$ axis. Then $0\le\theta\le\pi$.
Because $x^2+y^2=2ay$ then $r=2a\sin\theta$, therefore we now have the integral:
$$
\int_0^{\pi}\int_0^{2a\sin\theta}r\sqrt{\frac{r^2}{4a^2-r^2}+1}drd\theta
$$
And this is where I'm stuck. How can I integrate $dr$?
|
$$\frac{r^2}{4a^2-r^2}+1=\frac{r^2+4a^2-r^2}{4a^2-r^2}\implies r\sqrt{\frac{r^2}{4a^2-r^2}+1}=\frac r{\sqrt{4a^2-r^2}}\implies$$
$$\left.\int_0^{2a\sin\theta}\frac r{\sqrt{4a^2-r^2}}dr=-\sqrt{4a^2-r^2}\right|_0^{2a\sin\theta}=-\sqrt{4a^2-4a^2\sin^2\theta}+2a=-2a\cos\theta+2a$$
and etc. Yet I think the final answer is different from what you said it is...
|
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|
Predicting if a $3 \times 3$ matrix has a real square root Here are a few $3 \times 3$ matrices
A)$$
\begin{pmatrix} 1& 0& 0\\
0 & 1 & 0\\
0& 0 & 1\end{pmatrix}$$B)$$
\begin{pmatrix} 1& 0& 0\\
0 & 1 & 0\\
0& 0 & -1\end{pmatrix}$$C)$$
\begin{pmatrix} 1& 0& 0\\
0 & -1 & 0\\
0& 0 & -1\end{pmatrix}$$D)$$
\begin{pmatrix} -1& 0& 0\\
0 & -1 & 0\\
0& 0 & -1\end{pmatrix}$$
How can say if these matrices are the square of a $3 \times 3$ matrix with real entries or not?
|
A necessary condition for a real matrix to have a real square root is to have a non-negative determinant.
Indeed, if $A=B^2$, then $\det(A)=\det(B)^2\geqslant 0$.
Method. If $A$ is diagonalizable and has positive eigenvalue, a square root can be found taking the square root of its eigenvalues, namely if $A=P\Lambda P^{-1}$, then $B=P\Lambda^{1/2}P^{-1}$ is a square root of $A$.
|
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|
How to calculate $\int{\frac {x^2}{ ( x\cos x -\sin x ) ^2}}\,{\rm d}x$? When I want to calculate
$$\int{\frac {x^2}{ ( x\cos x -\sin x ) ^2}}\,{\rm d}x$$
I have tested with software and get
$${\frac {x\sin \left( x \right) +\cos \left( x \right) }{x\cos \left( x
\right) -\sin \left( x \right) }}$$
But I can not come to this conclusion, neither using integration by parts, nor using trigonometric identities, nor multiplying by their conjugate, Even by rational trigonometric substitution. I do not know what else to try. Could you give me any suggestions?
|
We can tackle the integral by integration by parts as below:
$$
\begin{aligned}
& \int \frac{x^{2}}{(x \cos x-\sin x)^{2}} d x \\
=& \int \frac{x}{\sin x} d\left(\frac{1}{x \cos x-\sin x}\right) \\
=& \frac{x}{\sin x(x \cos x-\sin x)} -\int \frac{1}{x \cos x-\sin x} \cdot \frac{\sin x-x \cos x}{\sin ^2x}dx\\=& \frac{x}{\sin x(x \cos x-\sin x)} +\int \csc ^{2} x d x \\
=& \frac{x}{\sin x(x \cos x-\sin x)}-\cot x+C\\=& \frac{x \sin x+\cos x}{x \cos x-\sin x}+C
\end{aligned}
$$
|
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|
Calculus Spivak Chapter 2 problem 16(c) The question asks to prove that if $\frac mn \lt \sqrt{2}$, then there is another rational number $\frac {m'}{n'}$ with $\frac mn \lt \frac {m'}{n'} \lt \sqrt{2}$.
Intuitively, it's clear that such a number exists, but I don't understand the solution to this problem. It states: let $m_1 = m + 2n$ and $n_1 = m + n$, and choose $m' = m_1 + 2n_1 = 3m + 4n$, and $n' = m_1 + n_1 = 2m + 3n$.
Apparently $\frac {(m + 2n)^2}{(m+n)^2} \gt 2$, but can someone explain why and how plugging in those equations for $m'$ and $n'$ ensures that $\frac {m'}{n'}$ lies between $ \frac mn$ and $\sqrt {2}$?
|
Actually we have the following estimates:
\begin{gather*}
\frac{m}{n}<\frac{2(m+n)}{m+2n}<\sqrt{2},
\end{gather*}
provided $m,n$ are positive integers with $\frac{m}{n}<\sqrt{2}.$
Verification: Since $0<\frac{m}{n}<\sqrt{2},$ we have $\frac{n}{m}>\frac{1}{\sqrt{2}},$ and so
\begin{gather*}
\frac{2(m+n)}{m+2n}=\frac{m}{n}\cdot\frac{2\left(1+\frac{n}{m}\right)}{2+\frac{m}{n}}>\frac{m}{n}\cdot\frac{2\left(1+\frac{1}{\sqrt{2}}\right)}{2+\sqrt{2}}=\frac{m}{n}.
\end{gather*}
In the other direction, we have
\begin{align*}
&\frac{2(m+n)}{m+2n}=\frac{2}{\frac{m+2n}{m+n}}=\frac{2}{1+\frac{n}{m+n}}\\
=&\frac{2}{1+\frac{1}{\frac{m}{n}+1}}<\frac{2}{1+\frac{1}{\sqrt{2}+1}}=\frac{2}{1+(\sqrt{2}-1)}=\sqrt{2}.
\end{align*}
Therefore, we have
\begin{gather*}
\frac{m}{n}<\frac{m'}{n'}=\frac{2(m+n)}{m+2n}<\sqrt{2},
\end{gather*}
if $0<m/n<\sqrt{2}.$
|
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|
$1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ..........$ $1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ......$
Can anyone help me out how to solve this.
My try : I was thinking about the expansion of $tan^{-1}x$. But in that series positive and negative will come alternatively.
|
$$\sum_{r=0}^\infty\dfrac1{(2r+1)}\left(\dfrac12\right)^{2r}=2\sum_{r=0}^\infty\dfrac{x^{2r+1}}{(2r+1)}$$ where $2x=1$
Now $-\ln(1-x)=\sum_{n=0}^\infty\dfrac{x^n}n$ for $-1\le x<1$
$\ln(1+x)-\ln(1-x)=?$
Alternatively, $$\sum_{r=0}^\infty\dfrac{x^{2r+1}}{(2r+1)}=\int\sum_{r=0}^\infty(x^2)^rdx$$
|
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|
Calculating sample size using Central Limit Theorem How many times do we have to throw a dice to achieve the situation, that the proportion of sixes to all numbers is between $\frac{9}{60}$ and $\frac{11}{60}$ with the probability $\leq\frac{1}{100}$?
This is my solution, that gives a wrong answer.
I know, that number of sixes in multiple dice throws is following the binomic distribution ($n$ is the sample size I need to calculate):
$$
\mu = n \cdot\frac{1}{6}\\
\sigma = \sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}
$$
Based on that, I can normalize the random variable $X$ (expressing number of sixes in $n$ throws):
$$
Z=\frac{\overline{X}-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
$$
Now I can use the Central Limit Theorem:
$$
P\left(
\frac{n\cdot\frac{1}{6}-\left(n\cdot\frac{1}{6}\cdot\frac{1}{60}\right)-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}<
Z<
\frac{n\cdot\frac{1}{6}+\left(n\cdot\frac{1}{6}\cdot\frac{1}{60}\right)-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
\right)=0.99$$
$$
P\left(Z<
\frac{n\cdot\frac{1}{6}+\left(n\cdot\frac{1}{6}\cdot\frac{1}{60}\right)-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
\right)=\frac{0.99}{2}=0.495
$$
I can look up this value in the standardized normal distribution table and use it to calculate the value of $n$.
$$
\Phi(-0.012)=0.495
$$
$$
\frac{n\cdot\frac{1}{6}+\left(n\cdot\frac{1}{6}\cdot\frac{1}{60}\right)-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
=-0.012\\
\frac{n\cdot\frac{1}{6}\cdot\frac{1}{60}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
=-0.012\\
n=0.000595
$$
And that is what seems suspicious. I don't think that i have to throw the dice $0.000595$ times to ensure that the ration of sixes is within a $\frac{1}{60}$ error margin from the mean value with the probability of $\frac{1}{100}$.
Can somebody help me to determine where I am wrong? (Sorry about my English.)
Edit based on Doug M's comment:
I made a mistake in the left tailed distribution, It should be:
$$
P\left(Z<
\frac{n\cdot\frac{1}{6}+\left(n\cdot\frac{1}{6}\cdot\frac{1}{60}\right)-n\cdot\frac{1}{6}}{\sqrt{n\cdot\frac{1}{6}\cdot\frac{5}{6}}}
\right)=0.995
$$
With this corrected the result for $n$ seems more reasonable:
$$
\Phi(2.58)=0.995
$$
$$
\frac{n\cdot\frac{1}{6}\cdot{1}{60}}{\sqrt{n\cdot\frac{5}{36}}{}}=2.58\\
n=119815,2
$$
Which means, the optimal sample size is $119816$, which is more than I expected but sounds more reasonable than my previous result.
|
for a single die
$\mu = \frac 16\\
\sigma^2 = \frac {5}{36}$
the expected number of sixes in $n$ throws
$\mu = \frac 16n\\
\sigma^2 = \frac 5{36}n\\
\sigma = \sqrt{\frac{5n}{36}}$
The average number of sixes in $n$ throws
$\mu = \frac 16\\
\sigma = \frac 1n \sqrt{\frac{5n}{36}}\\
\sigma = \sqrt{\frac{5}{36n}}$
The expected number of sixes is governed by a binomial distribution. But, for a large number of throws the binomal resembles the normal distribution.
If $99\%$ are inside of a two tailed distribution, $99.5\%$ are to the left of a 1 tailed distribution.
+normsinv(0.995) = 2.575
For how large does $n$ have to be such that $2.57\sigma < \frac {1}{60}$ ?
$60\sqrt 5\cdot 2.57 < 6\sqrt n\\
500\cdot 2.57 < n\\
n>1288$
|
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|
Taylor series expansion of $\ln(5 + x)$. If I start with the function $1/(5+x)$ I can find its Taylor series about 0 by using the familiar geometric series $1/(1 + x)$.
\begin{align}
\frac{1}{5 + x} &= \frac{1}{5 (1 + x/5)} \\
&= \frac15 \cdot \frac{1}{1 + (1/5) x} \\
&= \frac15 - \frac{x}{25} + \frac{x^2}{125} - \frac{x^3}{625} + \frac{x^4}{3125} - \frac{x^5}{15625} + \dots
\end{align}
If I take the antiderivative of both sides I get:
$$ \ln(5 + x) = \frac{x}{5} - \frac{x^2}{50} + \frac{x^3}{375} - \frac{x^4}{2500} + \frac{x^5}{15625} - \dots$$
So in particular I have $\ln(5) = 0$.
|
First the correct answer:
If you do the transformation without integration you get
$$ \ln(5 + x) = \ln \left(5 \left(1 + \frac15 x\right) \right) = \ln 5 + \ln \left(1 + \frac15 x\right). $$
Then using the series for $\ln(1 + x)$ you get
$$ \ln(5 + x) = \ln 5 + \sum_{n = 1}^\infty (-1)^{n + 1} \frac{x^n}{n \cdot 5^n}. $$
This is the same as your answer except for the $\ln 5$ in front.
It comes down to the fact that there is no difference between the derivatives of $\ln(5 + x)$ and $\ln(5 + x) - \ln 5$. When you take the integral you have to account for the constant term seperately.
|
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|
A question on the Euler prime of odd perfect numbers A number $N \in \mathbb{N}$ is said to be perfect if $\sigma(N)=2N$, where $\sigma=\sigma_{1}$ is the classical sum-of-divisors function. For example, $\sigma(6)=1+2+3+6=2\cdot{6}$, so that $6$ is perfect. (Note that $6$ is even.)
It is currently unknown whether there are any odd perfect numbers. Euler proved that an odd perfect number, if any exists, must take the form $N = q^k n^2$ where $q$ is prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. We call $q$ the Euler prime of the odd perfect number $N$.
In what follows, we denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$.
MOTIVATION
Since the non-Euler part $n^2$ is deficient but not almost perfect, we know (from this preprint) that
$$I(n^2) < \frac{2n^2 + D(n^2)}{n^2 + D(n^2)},$$
where $D(n^2)={2n^2}-\sigma(n^2)$ is the deficiency of $n^2$.
When the Descartes-Frenicle-Sorli conjecture that $k=1$ is true, we also know that we have the lower bound
$$\frac{5}{3} \leq I(n^2),$$
whereupon we obtain, from the last two inequalities, the following lower bound
$$\frac{n^2}{D(n^2)} > 2.$$
(When $k=1$, ${n^2}/D(n^2)=(q+1)/2$, so that ideally we should have ${n^2}/D(n^2) \geq 3$ since $q \geq 5$.)
From the following Answer 1 and Answer 2 to a related MSE question, we obtain improved upper bounds for $I(n^2)$ by successively computing the mediant of
$$I(n^2) = \frac{2n^2 - D(n^2)}{n^2}=I_0$$
and the upper bounds $u_i$ (for $I(n^2)$) given by
$$u_0 = \frac{2n^2 + D(n^2)}{n^2 + D(n^2)}$$
$$u_1 = \frac{\bigg(2n^2 - D(n^2)\bigg)+\bigg(2n^2 + D(n^2)\bigg)}{n^2 + \bigg(n^2 + D(n^2)\bigg)} = \frac{4n^2}{2n^2 + D(n^2)}$$
$$u_2 = \frac{\bigg(2n^2 - D(n^2)\bigg)+4n^2}{n^2 + \bigg(2n^2 + D(n^2)\bigg)}=\frac{6n^2 - D(n^2)}{3n^2 + D(n^2)}$$
$$u_3 = \frac{\bigg(2n^2 - D(n^2)\bigg)+\bigg(6n^2 - D(n^2)\bigg)}{n^2 + \bigg(3n^2 + D(n^2)\bigg)} = \frac{8n^2 - 2D(n^2)}{4n^2 + D(n^2)}$$
$$u_4 = \frac{\bigg(2n^2 - D(n^2)\bigg)+\bigg(8n^2 - 2D(n^2)\bigg)}{n^2 + \bigg(4n^2 + D(n^2)\bigg)} = \frac{10n^2 - 3D(n^2)}{5n^2 + D(n^2)}$$
$$\ldots \ldots \ldots$$
$$\ldots \ldots \ldots$$
$$\ldots \ldots \ldots$$
(That is, $u_i = \text{mediant}(u_{i-1},I_0)$.)
Note that it is easy to prove that $u_i$ (for $i=0, 1, 2, \ldots$) is a strictly decreasing sequence.
The resulting lower bounds for ${n^2}/D(n^2)$ (on assuming $k=1$) are:
$$u_0 : \frac{n^2}{D(n^2)} > 2$$
$$u_1 : \frac{n^2}{D(n^2)} > \frac{5}{2}$$
$$u_2 : \frac{n^2}{D(n^2)} > \frac{8}{3}$$
$$u_3 : \frac{n^2}{D(n^2)} > \frac{11}{4}$$
$$u_4 : \frac{n^2}{D(n^2)} > \frac{14}{5}$$
$$\ldots : \ldots$$
QUESTIONS
(1) Do the resulting lower bounds for ${n^2}/D(n^2)$ (using this procedure) have a closed form?
(2) Does the lower bound ever reach $3$, or is it possible to prove that it does not?
|
It appears that the numerators of the lower bounds form OEIS sequence A016789. Hence, the lower bounds have the closed form
$$\frac{3i + 2}{i + 1}.$$
Consequently,
$$\frac{3i + 2}{i + 1}=3-\frac{1}{i + 1},$$
so that the lower bounds never reach $3$.
|
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|
Given $A$ and $B$, find matrix $C$ such that $CA=B$ Hoffman and Kunze, Linear Algebra, ch. 1
$A=\left[\begin{matrix} 1 & -1 \\ 2 & 2 \\ 1 & 0 \end{matrix}\right], B=\left[\begin{matrix} 3 & 1 \\ -4 & 4 \end{matrix}\right]$
Is there a matrix $C$ such that $CA=B$?
I assumed $C=\left[\begin{matrix} a & b & c\\ d & e & f \end{matrix}\right]$, then computed $CA$, equated each of the four elements of $CA$ with the corresponding elements of $B$, there are many solutions and got one of them.
My question: is there any other method to find C that involves elementary matrices ?
|
We have
$$
B = C A
$$
where $C$ consists of two unknown row vectors. I am more used to having the unknows on the right hand side ($Ax=b$) so I would, like Siong Thye Goh did, look at the transposed equations:
$$
B^T = (b_1^T, b_2^T)= A^T C^T = A^T (c_1^T, c_2^T)
$$
which can be interpreted as two linear systems
$$
A^T c_1^T = b_1^T \\
A^T c_2^T = b_2^T
$$
with column vectors $c_i^T, b_i^T$, which can be solved like usual.
Example: The first system can be solved like
$$
\begin{pmatrix}
1 & 2 & 1 \\
-1 & 2 & 0
\end{pmatrix}
\begin{pmatrix}
a \\
b \\
c
\end{pmatrix}
=
\begin{pmatrix}
3 \\
1
\end{pmatrix}
\iff \\
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
-1 & 2 & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
0 & 4 & 1 & 4
\end{array}
\right]
\to
\\
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
0 & 1 & 1/4 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
0 & 1 & 1/4 & 1
\end{array}
\right]
\to \\
\left[
\begin{array}{rrr|r}
1 & 0 & 1/2 & 1 \\
0 & 1 & 1/4 & 1
\end{array}
\right]
\iff \\
(a, b, c) = (1-(1/2) c, 1-(1/4) c, c)
$$
where we used $c$ as parameter for the line (1D space) of solutions of that system.
Doing the second calculation, or doing both at the same time, we get:
$$
\left[
\begin{array}{rrr|rr}
1 & 2 & 1 & 3 & -4\\
-1 & 2 & 0 & 1 & 4
\end{array}
\right]
\to
\left[
\begin{array}{rrr|rr}
1 & 0 & 1/2 & 1 & -4\\
0 & 1 & 1/4 & 1 & 0
\end{array}
\right]
\iff \\
(a, b, c) = (1-(1/2) c, 1-(1/4) c, c) \\
(d, e, f) = (-4-(1/2) f, -(1/4) f, f)
$$
for $c, f \in \mathbb{R}$.
|
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|
Integrating rational functions. $\int\frac{x^2-3x+1}{x^4-x^2+1}\mathrm{d}x$ Is there really another way of solving this problem $$\int\frac{x^2-3x+1}{x^4-x^2+1}\mathrm{d}x$$
avoiding partial fractions decomposition? I have tried the partial fractions decomposition, but it is not serving me right.
|
We have that $$\int\frac{x^2-3x+1}{(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)}dx=\\=\int\frac{x^2-\sqrt{3}x+1}{(x^2-\sqrt3{x}+1)(x^2+\sqrt3{x}+1)}-\int\frac{(3-\sqrt{3})x}{x^4-x^2+1}=\\\int\frac{dx}{x^2+\sqrt{3}x+1}-\frac{3-\sqrt{3}}{2}\int\frac{2x}{x^4-x^2+1}dx$$
Now putting $u=x^2$ for the second integral
$$\int\frac{dx}{(x+\frac{\sqrt{3}}{2})^2+\frac{1}{4}}-\frac{3-\sqrt{3}}{2}\int\frac{du}{(u-\frac{1}{2})^2+\frac{3}{4}}$$
Those two integrals can be made into $\arctan$ form I'll leave it to you.
|
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|
To find the last digit of the number $4^{7^5}$? To find the last digit of the number $4^{7^5}$, I use the following method.
first, we know $4\equiv_{10}4,4^2\equiv_{10}6,4^3\equiv_{10}4,...,4^{2n}\equiv_{10}6,4^{2n+1}\equiv_{10}4$
Then$7^1\equiv_21,7^2\equiv_21,7^3\equiv_21,...,7^5\equiv_21$
Therefore, $4^{7^5}\equiv4^{2k+1}\equiv_{10}4$.(because of the remainder 1)
and if to find the last digit of $4^{8^3}$ instead, the answer will be 6. Because $8^n\equiv_20,n\in \mathbb{N^+}$ so $8^n=2k,k\in\mathbb{N^+}$. From above, $4^{2k}\equiv_{10}6$
Is my method correct?
|
$$4^{7^5}=4\cdot\left(4^2\right)^{\frac{7^5-1}{2}}\equiv4\cdot6(\mod10)\equiv4(\mod10)$$
|
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|
Integer solutions for $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xyz}=1$ Is there a beautiful solution for this equation over the integers?
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xyz}=1$$
|
$$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}+\frac{1}{xyz}\color{red}{=1+1=2}$$
Rewriting $1+\frac{1}{x}=\frac{x+1}{x}$ and mutliplying both sides by $xyz$ (assumed nonzero), we get $$(x+1)(y+1)(z+1)=2xyz$$
Now, $x,y,z$ are all integers, so this gives a lot of conditions. For example, if we take $z=5$, we have $$(x+1)(y+1)6=10xy$$
or $$3(x+1)(y+1)=5xy$$
Hence, either $x$ or $y$ is a multiple of $3$. Also, either $x+1$ or $y+1$ is a multiple of $5$. Assume that we have $x=3k, y+1=5m$, so we rewrite as
$$3(3k+1)5m=5(3k)(5m-1)$$
Cancelling, we get $$(3k+1)m=k(5m-1)$$
If $m=1$, then $3k+1=4k$, so $k=1$, which gives $(x,y,z)=(3,4,5)$.
If $m=2$, then $6k+2=9k$, so $k=2/3$, which gives $(x,y,z)=(2,9,5)$.
If $m=3$, then $9k+3=14k$, so $k=3/5$, which does not give an integer solution.
Etc, etc, we can find as many solutions as we want by trial and error.
|
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|
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_n(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
How can I find $P$? I am doing Gauss but it does not work?$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \sim \left(\begin{array}{cc|cc} 2&0&-8&6\\ 0&-1/2&-3/2&1 \end{array}\right)$$
What am I doing wrong? Steps would be much appreciated.
|
You need to perform simulatenous row and column operations on the left hand side while performing only column operations on the right hand side. Then, when the left side becomes diagonal, the right side will be your $P$. In your case,
$$ \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \xrightarrow[C_2 = C_2 - \frac{3}{2}C_1]{R_2 = R_2 - \frac{3}{2}R_1}
\left(\begin{array}{cc|cc} 2&0&1& -\frac{3}{2}\\ 0& -\frac{1}{2}&0&1 \end{array}\right) $$
and indeed
$$ \begin{pmatrix} 1 & 0 \\ -\frac{3}{2} & 1 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & -\frac{3}{2} \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & -\frac{1}{2} \end{pmatrix}. $$
|
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|
Is Matrix Multiplication Commutative? I am doing some self-study in math. Below is a problem that I did but I am not sure if I have the right answer. I am hoping that somebody could check it for me.
Thanks
Bob
Problem:
If $A$ and $B$ are two by two matrices, does it imply that $A * B = B * A$?
Answer:
\begin{eqnarray*}
A &=& \begin{bmatrix}
a_{1\,1} & a_{1\,2} \\
a_{2\,1} & a_{2\,2} \\
\end{bmatrix} \\
B &=& \begin{bmatrix}
b_{1\,1} & b_{1\,2} \\
b_{2\,1} & b_{2\,2} \\
\end{bmatrix} \\
A * B &=& \begin{bmatrix}
a_{1\,1} b_{1 \, 1} + a_{1 \, 2 }b_{2 \, 1} &
a_{1\,1} b_{1\,2} + a_{1\,2}b_{2\,2 } \\
a_{2\,1} b_{1\,1} + a_{2\,2}b_{2\,1} & a_{2\,1} b_{1\,2} + a_{2\,2}b_{2\,2} \\
\end{bmatrix} \\
B *A &=& \begin{bmatrix}
b_{1\,1} a_{1 \, 1} + b_{1 \, 2 }a_{2 \, 1} &
b_{1\,1} a_{1\,2} + b_{1\,2}a_{2\,2 } \\
b_{2\,1} a_{1\,1} + b_{2\,2}a_{2\,1} & b_{2\,1} a_{1\,2} + b_{2\,2}a_{2\,2} \\
\end{bmatrix} \\
\end{eqnarray*}
Assume that $A * B = B * A$ then we have:
\begin{eqnarray*}
a_{1\,1} b_{1 \, 1} + a_{1 \, 2 }b_{2 \, 1} &=&
b_{1\,1} a_{1 \, 1} + b_{1 \, 2 }a_{2 \, 1} \\
\end{eqnarray*}
If $a_{1\,1} = b_{1 \, 1} = 1$ we have:
\begin{eqnarray*}
1(1) + a_{1 \, 2 }b_{2 \, 1} &=& 1(1) + b_{1 \, 2 }a_{2 \, 1} \\
a_{1\,2} b_{2\,1} &=& b_{1\,2}a_{2\,1} \\
\end{eqnarray*}
If $a_{1\,2} = b_{2 \, 1} = b_{1\,2} = 1$ and $a_{2\,1} = 0$ we have:
\begin{eqnarray*}
1(1) &=& 1(0) \\
1 &=& 0 \\
\end{eqnarray*}
Therefore, I conclude for the set of two by two matrices the operator $*$ is
not communicative.
|
Your approach is correct, but there is an easier way:
You can prove that something is not true by giving a counterexample (the claim is that matrix multiplication is not commutative, so it is sufficient to find at least one example where commutativity fails). So, a much easier approach is:
$$\begin{pmatrix} 1 &0\\0&0\end{pmatrix} \begin{pmatrix} 1 &0\\1&0\end{pmatrix} = \begin{pmatrix} 1 &0\\0&0\end{pmatrix}$$
$$ \begin{pmatrix} 1 &0\\1&0\end{pmatrix}\begin{pmatrix} 1 &0\\0&0\end{pmatrix} = \begin{pmatrix} 1 &0\\1&0\end{pmatrix}$$
Hence, matrix multiplication is not commutative.
You can ask yourself why matrix multiplication is defined this way. This definition seems somewhat strange when one is exposed to matrix multiplication for the first time. The deeper underlying reason is that we can represent a certain type of function, called linear transformation, with matrices. Matrix multiplication is defined such that it corresponds with function composition of linear transformation. Since function composition is not commutative, it also makes sense that matrix multiplication is not commutative.
|
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|
Calculate the sum of the series $\sum_{1\leq aCalculate the sum of the series:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c}$$
My attempt:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c} = \sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}$$
Is it equal?
What's next?
From Wolfram Mathematica I know that $\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}= \frac{1}{1624}$.
|
\begin{eqnarray*}
\sum_{1 \leq a < b < c} \frac{1}{2^a 3^b 5^c} = \sum_{a=1}^{\infty} \frac{1}{2^a} \sum_{ b=a+1}^{\infty} \frac{1}{3^b} \sum_{c=b+1}^{\infty} \frac{1}{5^c}
\end{eqnarray*}
\begin{eqnarray*}
= \sum_{a=1}^{\infty} \frac{1}{2^a} \sum_{ b=a+1}^{\infty} \frac{1}{3^b} \frac{1}{5^b \times 4}
\end{eqnarray*}
\begin{eqnarray*}
= \sum_{a=1}^{\infty} \frac{1}{2^a} \frac{1}{15^a \times 14 \times 4}
\end{eqnarray*}
\begin{eqnarray*}
=\color{red}{\frac{1}{29 \times 14 \times 4}} =\frac{1}{1624}.
\end{eqnarray*}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2327827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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|
What would the solution be if I cannot simplify my solution to find more basic variables(pivots)? 4x3 matrix $$x - 5y + 4z = -3$$
$$2x - 7y + 3z = -2$$
$$-2x + y + 7z = -1$$
The furthest I was able to reach before coming to the conclusion that I could not move any further was:
Row 1 => $x - 5y + 4z = -3$
Row 2 => $9y + 15z = -7$
Row 3 => $8y + 5z = 1$
Please explain to me how I would be able to express my solution in the simplest form! Also let me know what type of solution this system would contain (independent, dependent, inconsistent)
|
Here is the reduction of the augmented matrix.
Column 1
$$
\left[
\begin{array}{rcc}
1 & 0 & 0 \\
-2 & 1 & 0 \\
2 & 0 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{rrc|ccc}
1 & -5 & 4 & 1 & 0 & 0 \\
2 & -7 & 3 & 0 & 1 & 0 \\
-2 & 1 & 7 & 0 & 0 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{crr|rcc}
\boxed{1} & -5 & 4 & 1 & 0 & 0 \\
0 & 3 & -5 & -2 & 1 & 0 \\
0 & -9 & 15 & 2 & 0 & 1 \\
\end{array}
\right]
$$
Column 2
$$
\left[
\begin{array}{ccc}
1 & \frac{5}{3} & 0 \\
0 & \frac{1}{3} & 0 \\
0 & 3 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{crr|rcc}
\boxed{1} & -5 & 4 & 1 & 0 & 0 \\
0 & 3 & -5 & -2 & 1 & 0 \\
0 & -9 & 15 & 2 & 0 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{crr|rcc}
\boxed{1} & 0 & -\frac{13}{3} & -\frac{7}{3} & \frac{5}{3} & 0 \\
0 & \boxed{1} & -\frac{5}{3} & -\frac{2}{3} & \frac{1}{3} & 0 \\
0 & 0 & 0 & -4 & 3 & 1 \\
\end{array}
\right]
$$
Least squares solution
The least squares solution to $$\mathbf{A}x = b$$ is
$$
x = \frac{1}{5827}
%
\left[
\begin{array}{r}
-257 \\ 1369 \\ -1168 \\
\end{array}
\right]
$$
The least squares residual errors is
$$
\begin{align}
\mathbf{A} x - b &= \mathbf{0} \\
%
\left[
\begin{array}{rrc}
1 & -5 & 4 \\
2 & -7 & 3 \\
-2 & 1 & 7 \\
\end{array}
\right]
\frac{1}{5827}
%
\left[
\begin{array}{r}
-257 \\ 1369 \\ -1168 \\
\end{array}
\right]
%
-
\left[
\begin{array}{r}
-3 \\ -2 \\ -1 \\
\end{array}
\right]
%
&=
\frac{1}{26}
\left[
\begin{array}{r}
20 \\ -15 \\ -5 \\
\end{array}
\right]
%
\end{align}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2327923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Let $f(x)$ be a quadratic expression which is positive for all real values of $x.$ If $g(x)=f(x)+f'(x)+f''(x)$
Let $f(x)$ be a quadratic expression which is positive for all real values of $x$. If $g(x)=f(x)+f'(x)+f''(x)$, then for any real $x$,
$$(A)\, g(x)<0,\quad (B)\,g(x)>0,\quad (C)\, g(x)=0,\quad (D)\;g(x)\ge 0.$$
My work so far.
Let $f(x)=ax^2+bx+c$ be the quadratic polynomial. Then
$b^2-4ac<0$ and
$$g(x)=ax^2+x(b+2a)+(c+b+2a).$$
Now I am stuck. I am not able to pick the right answer.
|
$f(x)=ax^2+bx+c$. $f>0 \Rightarrow a>0$.
$f(x)+f^\prime(x)+f^{\prime\prime}(x)=ax^2+(2a+b)x+c+b+2a=g(x)$
Reformulating $g$ in terms of $x+1$ gives
$$a(x+1)^2+ (2a+b)(x+1)+c+b+2a-(2ax+a)-(2a+b) = $$
$$a(x+1)^2+ (2a+b)(x+1)+c-2a(x+1)+a = $$
$$a(x+1)^2+b(x+1)+c+a=g(x)$$.
So $g(x) = f(x+1)+a$, so $g(x)$ is $f(x)$ translated by 1 to the left and by $a$ upwards.
$f>0 \Longrightarrow g>0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2328578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$
And,
$$x^2-5x+1=0$$
??
|
For a more brut'ish force alternative, note that the problem assumes $x \ne 0$ for $\sqrt {x} + 1 / \sqrt {x}$ to be defined, so that $x$ must be a root of $x^2-5x+1=0 \iff x = \frac{1}{2}\left(5 \pm \sqrt{21}\right)\,$.
By known radical denesting techniques, $\sqrt{5 \pm \sqrt{21}}=\frac{1}{2}\left(\sqrt{14} \pm \sqrt{6}\right)\,$, so:
$$\require{cancel}
\sqrt{x}=\frac{1}{2 \sqrt{2}}\left(\sqrt{14} \pm \sqrt{6}\right) = \frac{1}{2}\left(\sqrt{7}\pm\sqrt{3}\right) \;\;\implies\;\; \frac{1}{\sqrt{x}} = \frac{1}{2}\left(\sqrt{7} \mp \sqrt{3}\right)
$$
Then $\displaystyle x + \frac{1}{\sqrt{x}}=\frac{1}{2}\left(\sqrt{7}\pm\cancel{\sqrt{3}}\right) + \frac{1}{2}\left(\sqrt{7} \mp \cancel{\sqrt{3}}\right)=2 \cdot \frac{1}{2} \sqrt{7} = \sqrt{7}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2329371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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|
Polar Coordinates (Area enclosed)
Find the common area enclosed by the curves
$$r= 3 - 2 \cos \theta$$
$$r= 2$$
My attempt,
Area$$=4\pi-\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}}\frac{1}{2}(4-(3-2\cos \theta)^2)d\theta$$
$$=36.753$$
Am I correct?
|
Area:
$$\begin{align}
2\left(\int_0^{\pi/3} \frac 12\big(\overbrace{3-2\cos\theta}^{\color{red}{r_1}}\big)^2 d\theta +\int_{\pi/3}^{\pi}\frac 12\cdot (\overbrace{\;\;2\;\;}^{\color{blue}{r_2}})^2 d\theta\right)
&=2\left(\frac {11}2\bigg(\frac {\pi}3-\frac{\sqrt{3}}2\bigg)+ \frac {4\pi}3\right)\\
&=11\bigg(\frac {\pi}3-\frac{\sqrt{3}}2\bigg)+\frac {8\pi}3\\
&=\color{red}{\frac {19}3\pi-\frac {11\sqrt{3}}2}\\
&=\color{red}{10.37}\end{align}$$
|
{
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"url": "https://math.stackexchange.com/questions/2329876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove that $\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$ where $\mathcal{C}$ is the unit circle On the generalization of a recent question, I have shown, by analytic and numerical means, that
$$\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$$
where $\mathcal{C}$ is the unit circle. Thus, $z=e^{i\theta}$ and $dz=iz~d\theta$. There remains to prove it, however.
What I have done: consider the absolute value part of the integrand,
$$
\begin{align}
|1+z+z^2+\cdots+z^{2n}|^2
&=(1+z+z^2+\cdots+z^{2n})(1+z+z^2+\cdots+z^{2n})^*\\
&=(1+z+z^2+\cdots+z^{2n})(1+z^{-1}+z^{-2}+\cdots+z^{-2n})\\
&=(1+z+z^2+\cdots+z^{2n})(1+z^{-1}+z^{-2}+\cdots+z^{-2n})\frac{z^{2n}}{z^{2n}}\\
&=\left(\frac{1+z+z^2+\cdots+z^{2n}}{z^n} \right)^2\\
&=\left(\frac{1}{z^n}\cdots+\frac{1}{z}+1+z+\cdots z^n \right)^2\\
&=(1+2\cos\theta+2\cos 2\theta+\cdots+2\cos n\theta)^2\\
\end{align}
$$
We now return to the integral,
$$
\begin{align}\frac{1}{2\pi i}\int_C |1+z+z^2+\cdots z^n|^2dz
&=\frac{1}{2\pi}\int_0^{2\pi}(1+2\cos\theta+2\cos 2\theta+\cdots+2\cos n\theta)^2 (\cos\theta+i\sin\theta)~d\theta\\
&=\frac{1}{2\pi}\int_0^{2\pi}(1+2\cos\theta+2\cos 2\theta+\cdots+2\cos n\theta)^2 \cos\theta~d\theta
\end{align}$$
where we note that the sine terms integrate to zero by virtue of symmetry. This is where my trouble begins. Clearly, expanding the square becomes horrendous as $n$ increases, and even though most of the terms will integrate to zero, I haven't been able to selectively find the ones that won't.
The other thing I tried was to simplify the integrand by expressing it in terms of $\cos\theta$ only using the identity
$$\cos n\theta=2\cos (n-1)\theta\cos\theta-\cos(n-2)\theta$$
but this too unfolds as an algebraic jungle very quickly. There are various other expressions for $\cos n\theta$, but they seem equally unsuited to the task. I'll present them here insofar as you may find them more helpful than I did.
$$
\cos(nx)=\cos^n(x)\sum_{j=0,2,4}^{n\text{ or }n-1} (-1)^{n/2}\begin{pmatrix}n\\j\end{pmatrix}\cot^j(x)=\text{T}_n\{\cos(x)\}\\
\cos(nx)=2^{n-1}\prod_{j=0}^{n-1}\cos\left(x+\frac{(1-n+2j)\pi}{2n} \right)\quad n=1,2,3,\dots
$$
where $\text{T}_n$ are the Chebyshev polynomials. Any suggestions will be appreciated.
|
You can simplify things as follows. We have:
$$f(z) = \sum_{k=0}^{2n}z^k = \frac{z^{2n+1}-1}{z-1}$$
Then on the unit circle, we have:
$$\left|f(z)\right|^2 = f(z)f^*(z) = f(z)f(z^*) = f(z)f\left(z^{-1}\right) = \frac{\left(z^{2n+1}-1\right)^2}{z^{2n}(z-1)^2}$$
The integral over the unit circle is then given by the coefficient of $z^{2n-1}$ of $\dfrac{\left(z^{2n+1}-1\right)^2}{(z-1)^2}$ which is the same as the coefficient of $z^{2n-1}$ in $\dfrac{1}{(z-1)^2}$, we can get this from differentiating the geometric series, which yields the result of $2n$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2331657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
$\sin(2\arcsin(\frac{1}{3}))=? $ A broad one I have written it as $$\sin(2\arcsin(\frac{1}{3}))=2\sin(\arcsin(\frac{1}{3}))\cos(\arcsin(\frac{1}{3})).$$Then, solved for $$\cos(\arcsin(\frac{1}{3}))=\frac{\sqrt{10}}{3}$$ $$\arcsin(\frac{1}{3})=\theta$$$$\sin\theta=\frac{1}{3}\Rightarrow\cos\theta=\frac{\sqrt{10}}{3}.$$But noticed something weird (1) it didn't satisfy the Pythagorean theorem as $\sqrt{10}\approx3.1622,$ which means that the adjacent side to the angle $\theta$ is greater than the hypotenuse, (which is just 3) why this method (using right triangle to determine $\cos(\theta)$) didn't work here although the answer ($\frac{\sqrt{10}}{3}$) satisfied the range of the cosine function, which is $R(\cos\theta)=\Bbb{R}$? After it (2) used this formula$$\arcsin(x)=\arccos(\sqrt{1-x^2});$$$$\arcsin(\frac{1}{3})=\arccos(\sqrt{1-\frac{1}{9}})=\arccos(\frac{\sqrt{8}}{3});$$then, checked if it satisfies or not $D(\arccos(\theta))=[-1;1],$ satisfied, since $\frac{\sqrt{8}}{3}\approx0.942.$ Then proceeded on my problem $$2\sin(\arcsin(\frac{1}{3}))\cos(\arcsin(\frac{1}{3}))=\frac{2}{3}\cdot\frac{\sqrt{8}}{3}=\frac{2\sqrt{8}}{9}.$$ Anyway, it didn't turn out to be the right answer. Where is my mistake?
|
Your mistake is that $\cos\theta=\dfrac{2\sqrt{2}}{3}$, not $\dfrac{\sqrt{10}}{3}$.
\begin{eqnarray}
\cos\left(2\arcsin\left(\frac{1}{3}\right)\right)&=&2\sin\left(\arcsin\left(\frac{1}{3}\right)\right)\cos\left(\arcsin\left(\frac{1}{3}\right)\right)\\
&=&2\left(\frac{1}{3}\right)\cdot\frac{2\sqrt{2}}{3}\\
&=&\frac{4\sqrt{2}}{9}
\end{eqnarray}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Changing the order of double integration The function is $\int_0^8 \int_{y^{1/3}}^2 \sqrt{x^4 + 1}dxdy$ and the prompt is to change the order of integral and evaluate it.
Plotting the graphs for the limits $$y^{\frac{1}{3}} < x < 2$$
$$0 < y < 8$$
Using the graph we can identify, changing the order of the limits will give us
$$\int_0^2 \int_0^{x^3} \sqrt{x^4 + 1}dydx$$
$$\int_0^2 x^3\sqrt{x^4 + 1}dx$$
By substitution, assuming $x^4 + 1 = t$
$$4x^3 dx = dt$$
$$= \frac{1}{4} \int \sqrt{t}$$
$$= \frac{1}{2}\left(\frac{(17)^{\frac{3}{2}}}{3}\right)$$
Is this the correct way to change the order?
|
Yes, you changed your order of integration correctly. Good Job! However, your final result is wrong because you integrated incorrectly in the last steps: $$\int_{x=0}^{x=2}x^3\sqrt{x^4+1}dx=\frac{1}{4}\int_{{\color{red}{t=1}}}^{t=17}\sqrt{t}dt=\frac{1}{6}\left(17\sqrt{17}-1\right)$$
where you mistakenly treated the lower bound for your $t$ as $0$, but when $x=0$, then $t=x^4+1=0^4+1=1$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Significance of the Triangle Inequality After working through a few problems regarding the Cauchy Integral Formula, I'm still a little confused on the significance of the triangle inequality. Why do we use it and what information does it tell us?
See the following example below.
Determine $\int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx$
First, we'll create the contour $$ \begin{cases} C_1: Rt & -1 \le t \le 1 \\ C_2:R e^{i\theta} & \ \ \, 0 \le \theta \le \pi \\ C=C_1+C_2 \end{cases} $$
$$ x^2 +1 =0 \Rightarrow x=\pm i \ \ \ ; \ \ \ \ \int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx$$
$$ \Rightarrow \int_C \frac{z^2}{(z+1)^2}dz = \int_C \frac{z^2}{[(z+i)(z-i)]^2}
= \int_C \frac{z^2}{(z+i)^2 (z-i)^2} $$
$z=-i$ is outside the contour and therefore irrelevant, so choosing $z=i$ for the integration.
$$ \Rightarrow z_0=i \ \ \text{and} \ \ f(z)=\frac{z^2}{(z+i)^2} $$
$$ f'(z)=\frac{d}{dz}\left[ \frac{z^2}{(z+i)^2} \right] = \frac{2z}{(z+i)^2}-\frac{2z^2}{(z+i)^3}$$
Therefore, by the Cauchy Integration Formula,
$$ \int_C \frac{f(z)}{(z-i)^2}dz = 2\pi i f'(i) = 2\pi i \left[ \frac{2i}{(2i)^2}-\frac{2i^2}{(2i)^3} \right] = \pi \left[1 + \frac{2}{4i^2} \right] = \pi - \frac{\pi}{2} = \frac{\pi}{2} $$
The following part is where I'm a little confused. I know the work, but am not quite sure why I'm doing it or what the end result is telling us. So, by the triangle inequality,
$$ \left| z^2 +1 \right|^2 \le \left( \left|z^2 \right| +1 \right) ^2 = z^4 +2\left| z^2 \right| +1 \Rightarrow \left| \int_C \frac{z^2}{(z+1)^2}dz \right| \le \frac{R^2}{R^4 +2R +1} $$
$$ \Rightarrow \int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx \le \lim_{R \rightarrow \infty} \int_C \frac{z^2}{(z+1)^2}dz \\ \le \lim_{R \rightarrow \infty} \frac{R^2}{R^4 +2R +1} = 0 $$
$$ \therefore \int_{0}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{(x^2 + 1)^2} dx = \frac{1}{2} \left[ \frac{\pi}{2} \right] = \frac{\pi}{4} $$
Any help in deciphering the part in the work I pointed out would be very helpful, thank you for your time.
|
The reason you're confused is that you should be using the reverse triangle inequality here. What we should have is the following; $$|z^2+1|^2 \ge ||z|^2-1|^2 = (|z|^2-1)^2$$
Where the last equality is because we will consider the semicircular contour letting $R=|z| \to \infty$ (so certainly bigger than $1$).
Then we get the result; $$\left\vert \int_{C_2} \frac{z^2}{(z^2+1)^2} dz \right\vert \le \int_{C_2} \left\vert\frac{z^2}{(z^2+1)^2}\right\vert dz \le \int_0^{\pi} \frac{R^2}{(R^4-2R^2+1)} Rd\theta = \pi\cdot R \cdot \frac{R^2}{(R^4-2R^2+1)} $$.
This tends to zero as $R$ tends to infinity. So the integral around the semi circle $C_2$ disappears.
What this means, is that when considering our integral around whole contour $C$ and letting $R \to \infty $ the result we get ( which is constant from Cauchy's residue theorem) is simply the contribution from the integral over the real line. Then noting that the integrand is even we halve the result to get to integral between $0$ and $\infty$
I hope this helps clear up what is going on, and why we do it!!
|
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"timestamp": "2023-03-29T00:00:00",
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|
(Dice), population mean and variance of the odd numbers.
A fair die is rolled 100 times. Find the population mean and variance of the sum of the odd numbers that appear.
I am aware that the expected value for discrete multinomial distributions is $\Sigma \Sigma ... g(x_1 , x_2,...)f(x_1 ,x_2...)$. In this case, to find the expected value of the sum of the odd numbers that were rolled, should I not simply find the sum of the expected values for each of the individual odd numbers?
Couldn't something similar be done for variance? Or must I compute it using something like $Y= X_1 + 3X_3 +5X_5$?
|
For the mean, we find:
$$E[100X] = 100 E[X] = 100 \bigg(\frac{1}{6} \cdot 1 + \frac{1}{6} \cdot 3 + \frac{1}{6} \cdot 5 + \frac{3}{6} \cdot 0\bigg) = 100 \cdot \frac{3}{2} = 150$$
For the variance, we find:
$$Var[100X] = 100 Var[X] = 100 \cdot \frac{\big(1-\frac{3}{2}\big)^2 + \big(3-\frac{3}{2}\big)^2 + \big(5-\frac{3}{2}\big)^2 + 3 \big(0-\frac{3}{2}\big)^2}{6}$$
$$= 100 \cdot \frac{0.5^2 + 1.5^2 + 3.5^2 + 3 \cdot 1.5^2}{6} = 100 \cdot \frac{43}{12} \approx 358$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find the series of $f(x) = x^3\arctan(x^3)$ around $x=0$ Am I correct?
$$x^3\arctan(x^3)$$
$$\frac{d}{dx}\arctan(x^3) = \frac{3x^2}{x^6+1} = \sum_{n=0}^{\infty}3x^2(x^6)^n$$
$$x^3\sum_{n=0}^{\infty}3x^2(x^6)^n$$
$$f(x) = x^3\arctan(x^3) = 3x^5\sum_{n=0}^{\infty}(x^6)^n$$
|
For $X $ such that $X^2 <1$,
$$\frac {1}{1+X^2}=\sum_{n=0}^{+\infty}(-1)^nX^{2n} $$
and by integration
$$\arctan (X)=\arctan (0)+\sum_{n=0}^{+\infty}\frac {(-1)^nX^{2n+1}}{2n+1} $$
replace $X $ by $x^3$ and you can finish and get
$$f (x)=\sum_{n=0}^{+\infty}\frac {(-1)^nx^{6n+6}}{2n+1} $$
for $x $ such that $|x|<1$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Proving trigonometric identity $\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$ Show that
$$\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$$
Starting from the left hand side (LHS)
\begin{align}
\text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\
&=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\
&=\cos^4A-\cos^2A\sin^2A+\sin^4A
\end{align}
Can anyone help me to continue from here
|
HINT:
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$ can also be written as
$$a^3+b^3=(a+b)^3-3ab(a+b)$$
|
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|
If the point of minima of the function ,$f(x)=1+a^2x-x^3$ satisfy the inequality $\frac{x^2+x+1}{x^2+5x+6}<0$, If the point of minima of the function ,$f(x)=1+a^2x-x^3$ satisfy the inequality $\frac{x^2+x+1}{x^2+5x+6}<0$,then $a$ must lie in the interval:
$(A)(-3\sqrt3,3\sqrt3)$
$(B)(-2\sqrt3,-3\sqrt3)$
$(C)(2\sqrt3,3\sqrt3)$
$(D)(-3\sqrt3,-2\sqrt3)\cup (2\sqrt3,3\sqrt3)$
$$f'(x)=a^2-3x^2=0\implies x=\pm\frac{a}{\sqrt3}$$
$$f''(x)=-6x$$
So $x=-\frac{a}{\sqrt3}$ is the point of minima.It satisfies the inequality $\frac{x^2+x+1}{x^2+5x+6}<0$.
$$\frac{x^2+x+1}{x^2+5x+6}<0\implies \frac{1}{x^2+5x+6}<0$$
$$\frac{x^2+5x+6}{(x^2+5x+6)^2}<0\implies x^2+5x+6<0$$
$$(x+2)(x+3)<0\implies (2-\frac{a}{\sqrt3})(3-\frac{a}{\sqrt3})<0$$
$$\implies (\frac{a}{\sqrt3}-2)(\frac{a}{\sqrt3}-3)<0$$
$$a\in (2\sqrt3,3\sqrt3)$$
But the answer is $(-3\sqrt3,-2\sqrt3)\cup (2\sqrt3,3\sqrt3)$
|
Finding the $x$-value of the minimum point, you have assumed without warrant that $a>0$. We know that the minimum occurs at $x=\pm\frac{a}{\sqrt{3}}$, but we still don't know if $a$ is positive or negative.
Then, you find out that $f''(x)=-6x$, which must be positive at the minimum value. That means we have two cases from here. Either $a>0$ and the minimum occurs at $x=-\frac{a}{\sqrt3}$, or else $a<0$ and the minimum occurs at $x=\frac{a}{\sqrt3}$.
Your reasoning from there is fine; just apply it to both cases.
Backing away from those technical details, we can see that the solution must be symmetric, because we are only given information about $a^2$, so there is no way to tell whether $a$ will be positive or negative. Thus, the solution set for $a$ must be symmetric across the point $a=0$.
|
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|
Solve three equations with three unknowns Solve the system:
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$
The solution is:
$a=1,b=2,c=3$
How can I solve it?
|
As an alternative to the other derivations, here's my approach.
In the second equation, move everything to the left side, factor $ac+bc$ into $(a+b)c$, multiply everything by $c$, replace $a+b$ with $6-c $ from the first equation, and $abc$ with $6$ from the third. Then you have $6-11c+(6-c)c^2 = 0$, which simplifies to $c^3-6c^2+11c-6 = 0$, which has roots $1, 2,$ and $3$.
|
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|
Proving inequality $\ 1+\frac14+\frac19+\cdots+\frac1{n^2}\le 2-\frac1n$ using induction Question:
Prove $$\ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n},
\text{ for all natural } n$$
My attempt:
Base Case: $n=1$ is true:
I.H: Suppose $1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}\le 2-\frac{1}{k},$ for some natural $k.$
Now we prove true for $n = k+1$
$$ 1+\frac{1}{4}+\cdots+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\le 2-\frac{1}{k}+\frac{1}{\left(k+1\right)^2},\text{ by induction hypothesis} $$
Now how do I show that $2-\frac{1}{k}+\frac{1}{\left(k+1\right)^2}\le 2-\frac{1}{\left(k+1\right)}\text{ ?}$
Have I done everything correctly up until here?
If yes, how do I show this inequality is true?
Any help would be appreciated.
|
You may use creative telescoping for proving a much tighter inequality.
If we set $H_n^{(2)}=\sum_{k=1}^{n}\frac{1}{k^2}$, for any $n\geq 3$ we have:
$$\begin{eqnarray*} H_n^{(2)} &=& \sum_{k=1}^{n}\frac{1}{k(k+1)}+\sum_{k=1}^{n}\frac{1}{k^2(k+1)}\\&=&\left(1-\frac{1}{n+1}\right)+\frac{1}{2}+\sum_{k=1}^{n-1}\frac{1}{(k+1)^2 (k+2)}\\&=&\frac{3}{2}-\frac{1}{n+1}+\sum_{k=1}^{n-1}\frac{1}{k(k+1)(k+2)}-\sum_{k=1}^{n-1}\frac{1}{k(k+1)^2(k+2)}\\&=&\frac{5}{3}-\frac{2n+1}{2n(n+1)}-\sum_{k=1}^{n-2}\frac{1}{(k+1)(k+2)^2(k+3)}\\ &\color{red}{\leq}&\color{red}{\frac{5}{3}-\frac{1}{n}} \end{eqnarray*}$$
and the given inequality for $n\in\{1,2\}$ can be easily checked by hand.
It is interesting to point out that this approach leads to a short proof of Stirling's inequality.
As an alternative, the convexity of $\frac{1}{x^2}$ over $\mathbb{R}^+$ leads, by the Hermite-Hadamard inequality, to:
$$ H_{n}^{(2)}=\frac{\pi^2}{6}-\sum_{m\geq n+1}\frac{1}{m^2}\color{red}{\leq} \frac{\pi^2}{6}-\int_{n+1}^{+\infty}\frac{dx}{x^2}=\frac{\pi^2}{6}-\frac{1}{n+1} $$
$$ H_{n}^{(2)}\color{blue}{\geq} \frac{\pi^2}{6}-\frac{1}{n+\frac{1}{2}}. $$
|
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|
The sequence $a_1 = \frac{1}{2}, a_2 = 1, a_{n+1} = \frac{na_n+1}{a_{n-1}+n}$ is decreasing
Consider the sequence $\{a_n\}$ defined by $$a_1 = \frac{1}{2}, a_2 = 1, a_{n+1} = \frac{na_n+1}{a_{n-1}+n}, \forall n\ge 2.$$
Prove that $\{a_n\}_{n\ge 3}$ is decreasing.
I get the first $200$ values of $\{a_n\}$ and recognize this fact, but I cannot prove it.
Thank you very much.
|
From comments: it would appear that part III is not relevant for the original question. Plus, if we drop part III, the remaining induction hypotheses take effect with smaller $n.$ On the other hand, finding III is what allowed me to solve the problem; together, parts III and IV give a quite precise rate of convergence. I put in some lines of symbols to divide the parts.
Take $$ a_n = 1 + \delta_n $$
I get
$$ 1 + \delta_{n+1} = \frac{n+1+ n \delta_n}{n+1+ \delta_{n-1}} $$
The induction hypotheses become extensive, and are true for $n \geq 8:$
$$ \mbox{I:} \hspace{20mm} \delta_n > 0, $$
$$ \mbox{II:} \hspace{20mm} n \delta_n < 1, $$
$$ \mbox{III:} \hspace{20mm} n \delta_n > (n-3) \delta_{n-1} > \delta_{n-1}, $$
$$ \mbox{IV:} \hspace{20mm} n \delta_n < (n-2) \delta_{n-1}. $$
We begin with part III using only the $n \delta_n > \delta_{n-1}$,
$$ 1 = \frac{n+1}{n+1} < \frac{n \delta_n}{\delta_{n-1}}, $$
by the ``mediant'' inequality
$$ 1 = \frac{n+1}{n+1} < \frac{n+1+ n \delta_n}{n+1+ \delta_{n-1}} <\frac{n \delta_n}{\delta_{n-1}}, $$
$$ 1 = \frac{n+1}{n+1} < 1 + \delta_{n+1} < \frac{n \delta_n}{\delta_{n-1}}, $$
and $$ 0 < \delta_{n+1}. $$ This was the induction step part I.
Context: in the setting of (simple) continued fractions, with two consecutive convergents (let's say it is for a positive irrational), when the "partial quotient" $a_n$ is equal to $1,$ the next convergent $\frac{h_n}{k_n}$ is the mediant of the previous two.
$$\frac{h_n}{k_n} = \frac{a_n h_{n-1}+ h_{n-2}}{a_n k_{n-1}+ k_{n-2}} $$
Whatever $a_n$ might be, the new convergent is between the two given.
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
From part II we get
$$ 1 + \delta_{n+1} = \frac{n+1+ n \delta_n}{n+1+ \delta_{n-1}} < \frac{n+2}{n+1+ \delta_{n-1}} < \frac{n+2}{n+1} = 1 + \frac{1}{n+1}, $$
$$ \delta_{n+1} < \frac{1}{n+1}. $$
This was induction part II.
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
Back to III, we have
$$ \delta_{n-1} < \left( \frac{n}{n-3} \right) \delta_n. $$ So
$$ 1 + \delta_{n+1} > \frac{n+1+ n \delta_n}{n+1+ \left( \frac{n}{n-3} \right) \delta_n}. $$
$$ n+1 + (n+1) \delta_{n+1} + \left( \frac{n}{n-3} \right) \delta_n (1 + \delta_{n+1}) > n+1 + n \delta_n. $$
$$ (n+1) \delta_{n+1} + \left( \frac{n}{n-3} \right) \delta_n (1 + \delta_{n+1}) > n \delta_n. $$
$$ 1 + \delta_{n+1} < \frac{n+2}{n+1} $$
$$ (n+1) \delta_{n+1} + \left( \frac{n}{n-3} \right) \left( \frac{n+2}{n+1} \right) \delta_n > n \delta_n. $$
For $n \geq 7,$
$$ \frac{n^2 + 2n}{n^2 -2n-3} < 2. $$
For $n \geq 7,$
$$ (n+1) \delta_{n+1} > (n-2) \delta_n $$
This was induction part III.
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
In the other direction, part IV gives
$$ \delta_{n-1} > \left( \frac{n}{n-2} \right) \delta_n. $$
$$ 1 + \delta_{n+1} < \frac{n+1+ n \delta_n}{n+1+ \left( \frac{n}{n-2} \right) \delta_n}. $$
$$ n+1 + (n+1) \delta_{n+1} + \left( \frac{n}{n-2} \right) \delta_n (1 + \delta_{n+1}) < n+1 + n \delta_n. $$
$$ (n+1) \delta_{n+1} + \left( \frac{n}{n-2} \right) \delta_n (1 + \delta_{n+1}) < n \delta_n. $$
$$ (n+1) \delta_{n+1} + \left( \frac{n}{n-2} \right) \delta_n < n \delta_n. $$
$$ \frac{n}{n-2} > 1 $$
$$ (n+1) \delta_{n+1} < (n -1) \delta_n. $$
This was induction part IV.
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
|
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|
How to find this simplification integral involving products of roots? Consider the following integral
$$
f = \int_0^1 \frac{1}{\sqrt{-\frac{1}{2} \, t^{2} + 1} \sqrt{-t^{2} + 1}} \mathrm \,dt.
$$
If we change variable by letting $x^2=t^2/(2-t^2)$, then we have
$$
f = \int_0^1 \sqrt{2} \cdot\sqrt{\frac{1}{1-x^{4}}} \mathrm \,dx,
$$
which is a simpler form.
I read this in a book and wonder how can we come up with this sort of simplification? Is it just experience or is there systematic way to do it?
Note: The integral is the complete elliptic integral of the first kind $K(1/\sqrt{2})$.
|
$\int\frac{dt}{\sqrt{1-t^2}}=\arcsin(t)$, $\int\frac{dt}{\sqrt{1-\frac{t^2}{2}}}=\sqrt{2}\arcsin\left(\frac{t}{\sqrt{2}}\right)$, hence two candidate substitutions for simplifying things are $t=\sin\theta$ and $t=\sqrt{2}\sin\frac{\theta}{\sqrt{2}}$. Let us try the first one:
$$ I = \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\frac{1}{2}\sin^2\theta}} $$
followed by the substitution $\theta=\arctan u$:
$$ I = \int_{0}^{+\infty}\frac{du}{(1+u^2)\sqrt{1-\frac{1}{2}\cdot\frac{u^2}{1+u^2}}}=\sqrt{2}\int_{0}^{+\infty}\frac{du}{\sqrt{(u^2+1)(u^2+2)}}=\frac{\pi}{\text{AGM}(2,\sqrt{2})}$$
The AGM allows an efficient numerical evaluation (it immediately tells us that $I\geq\frac{2\pi}{2+\sqrt{2}}$, for instance) and the identity $\text{AGM}(a,b)=\text{AGM}\left(\frac{a+b}{2},\sqrt{ab}\right)$ gives many equivalent integrals. We may notice that
$$ \int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}\stackrel{x\mapsto\sqrt{z}}{=}\frac{1}{2}\int_{0}^{1}\frac{dz}{\sqrt{z(1-z^2)}}\stackrel{z\mapsto t^{-1}}{=}\frac{1}{2}\int_{1}^{+\infty}\frac{dt}{\sqrt{t(t-1)(t+1)}}$$
leads to
$$ \int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}=\frac{1}{2}\int_{0}^{+\infty}\frac{dt}{\sqrt{t(t+1)(t+2)}}=\int_{0}^{+\infty}\frac{du}{\sqrt{(u^2+1)(u^2+2)}} $$
This completely explains how to find the useful substitution by underlying the relation between the initial elliptic integral, the lemniscate constant and the AGM. We may also add a fourth actor on the scene, since by the substitution $x=w^{1/4}$ the integral $\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}$ is related with the Beta function, hence with the $\Gamma$ function. Here it is a complete summary:
$$\boxed{ \int_{0}^{1}\frac{dx}{\sqrt{1-x^4}} = \frac{\pi}{\text{AGM}(1,\sqrt{2})}=\frac{1}{\sqrt{2}}\,K\left(\frac{1}{\sqrt 2}\right)=\frac{1}{4}B\left(\frac{1}{4},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{4}\right)^2}{4\sqrt{2\pi}}. }$$
This excursion gives as a by-product an efficient numerical approach for computing $\Gamma\left(\frac{1}{4}\right)$, proving $\Gamma\left(\frac{1}{4}\right)=\frac{(2\pi)^{3/4}}{\sqrt{\text{AGM}(1,\sqrt{2})}}$.
|
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|
Proving right angle triangle A triangle $A$$B$$C$ is integer sided and has inradius $1$.prove that it is right angled. Please help.give some hints please.
|
Let the segments of tangents from the vertices to the incircle be $x,y,z$ (what is popularly known as the Ravi-substitution.
We have $(x+y), (y+z), (z+x)$ are all integers and so is $2(x+y+z)$ so $x+y+z = \frac{n}{2}, n \in \mathbb{N}$. Now its easy to see that $x,y,z$ are in the form $\frac{a}{2}, \frac{b}{2}, \frac{c}{2}$. Since $x+y, y+z, z+x \in \mathbb{N}, a,b,c$ have the same parity
We are given that $r=\frac{xyz}{x+y+z} =1 $ i.e. $abc = 4(a+b+c)$ from which its clear that $a,b,c$ are all even $\Rightarrow x,y,z \in \mathbb{N}$.
So we are looking for solutions in $\mathbb{N}$ for $xyz = x+y+z$.
WLOG $x \le y \le z$. Hence $xyz = x+y+z \le 3z \Rightarrow xy \le 3$
Now taking the cases $(x,y) = (1,1), (1,2), (1,3)$ we see that $x=1, y=2, z=3$ is the only solution.
This gives the sides as $3,4,5$ which is a Pythagorean Triplet and hence sides of a right angled triangle.
|
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|
Compute $\mathbb{E}[\text{max}(x,0) \text{max}(y,0)]$ where $(x,y)$ is jointly gaussian with given covariance and nonzero mean
Is there a closed form expression for the expectation of $g(x,y) = \text{max}(x,0) \text{max}(y,0)$ where $(x,y)$ is jointly gaussian with the bivariate gaussian distribution, when the mean is not zero and the covariance is non singular?
I tried to apply Price's theorem that can be written in the following form:
$$\frac{\partial^2 \mathbb{E}[g(x,y)]}{\partial \rho^2} = \mathbb{E} \left[ \frac{\partial^4 g(x,y)}{\partial x^2 \partial y^2} \right] = \mathbb{E}[\delta(x) \delta(y)] = \frac{1}{2 \pi \sigma_x \sigma_y\sqrt{1 - \rho^2}} \text{exp}\left(-\frac{1}{2(1 - \rho^2)} \left(\frac{\mu_x^2}{\sigma_x^2} + \frac{\mu_y^2}{\sigma_y^2} - \frac{2\rho \mu_x\mu_y}{\sigma_x \sigma_y} \right)\right)$$
Unfortunately, integrating the last equality with respect to the correlation coefficient doesn't seem to have a closed form in terms of known functions. I have posted the integral question earlier before here.
Moreover, it doesn't seem to be possible to find the expectation directly as:
$$\int_0^{\infty} \int_0^{\infty} xy f(x,y) dx dy$$
EDIT: Price's theorem can nicely handle the zero mean case with an elegant solution.
|
The calculation in question is pretty straightforward and can be done using elementary methods. Assume for simplicity that the variables have variance one and mean zero. Then the joint pdf reads:
\begin{eqnarray}
\rho(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp\left[ -\frac{1}{2} \frac{1}{(1-\rho^2)} (x^2+y^2-2 \rho x y )\right]
\end{eqnarray}
We integrate over $x$ first.
\begin{eqnarray}
I(y):=\int\limits_0^\infty x \rho(x,y) dx &=& \frac{1}{2\pi \sqrt{1-\rho^2}} \int\limits_0^\infty \exp\left[-\frac{1}{2} \frac{1}{(1-\rho^2)} ((x-\rho y)^2 + y^2(1-\rho^2))\right]\\
&=&\frac{1}{2\pi \sqrt{1-\rho^2}}e^{-\frac{1}{2} y^2} \int\limits_{-\rho y}^\infty (x+ \rho y) e^{-\frac{1}{2} \frac{1}{1-\rho^2} x^2} dx \\
&=& \frac{1}{2\pi} \sqrt{1-\rho^2} e^{-\frac{1}{2} y^2 \frac{1}{1-\rho^2}} + \frac{\rho e^{-\frac{y^2}{2}} y}{2 \sqrt{2 \pi }} + \frac{\rho e^{-\frac{y^2}{2}} y \text{erf}\left(\frac{\rho y}{\sqrt{2-2 \rho ^2}}\right)}{2 \sqrt{2 \pi }}
\end{eqnarray}
Now we multiply the result by $y$ and integrate the whole thing over $y\in(0,\infty)$. Even at the first glance it is clear that the integrals from the first two terms on the right hand side are doable it is only the last integral that might cause difficulties. However even that integral is doable and it reads:
\begin{equation}
\int\limits_0^\infty y^2 e^{-\frac{1}{2}y^2} Erf[a y] dy = \frac{1}{\sqrt{\pi}} \left[ \frac{2 a}{1+2 a^2} + \sqrt{2} \arctan(\sqrt{2} a)\right]
\end{equation}
The result was derived by differentiating with respect to the parameter $a$.
The final result is as follows:
\begin{eqnarray}
\left< max(x,0) max(y,0) \right> = \frac{\left(1-\rho ^2\right)^{3/2}}{2 \pi } + \frac{\rho }{4} + \frac{\rho \left(\sqrt{1-\rho ^2} \rho +\tan ^{-1}\left(\sqrt{\frac{\rho ^2}{1-\rho ^2}}\right)\right)}{2 \pi }
\end{eqnarray}
rho =.;
myrho[x_, y_] :=
1/(2 Pi Sqrt[1 - rho^2]) Exp[-1/
2 1/(1 - rho^2) (x^2 + y^2 - 2 rho x y)];
rho = RandomReal[{0, 1}, WorkingPrecision -> 50];
NIntegrate[x y myrho[x, y], {x, 0, Infinity}, {y, 0, Infinity},
WorkingPrecision -> 20]
(1 - rho^2)^(3/2)/(2 \[Pi]) + rho/4 + (
rho (rho Sqrt[1 - rho^2] + ArcTan[Sqrt[rho^2/(1 - rho^2)]]))/(
2 \[Pi])
|
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|
A limit involving $\sum_{k=1}^{n}\frac{2^k}{k}$ I would like to show that
$\lim_{n \to \infty} \frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{k} = 2$
I have seen proofs that $\sum_{k=1}^n \frac{2^k}{k} \sim \frac{2^{n+1}}{n}$ using the Euler-Maclaurin method but instead I tried with the squeeze theorem.
For one side I showed that
$$\frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{k} > \frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{n} = \frac{1}{2^n}\sum_{k=1}^n 2^k = \frac{2^{n+1} - 2}{2^n} = 2 - \frac{1}{2^{n-1}}$$
and the limit on this side is $2$.
Now I am having trouble finding an upper bound that converges to $2$.
Thank you.
|
Note that
$$\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} = \frac{n}{2^n}\sum_{k=0}^{n-1} \frac{2^{n-k}}{n-k} = \sum_{k=0}^{n-1} \frac{1}{2^k} + \sum_{k=1}^{n-1} \frac{k}{2^k(n-k)}.$$
The limit of the first sum on the RHS is $\sum_{k=0}^\infty 2^{-k} = 2$ and we can show that the limit of the second sum is $0$ giving the result
$$\lim_{n \to \infty}\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}= 2.$$
To show that
$$\lim_{n \to \infty} \sum_{k=1}^{n-1} \frac{k}{2^k(n-k)} = 0, $$
we have, since $2^k > k(k-1),$
$$0 < \sum_{k=1}^n \frac{k}{2^k(n-k)}= \frac{1}{2(n-1)}+ \sum_{k=2}^{n-1} \frac{k}{2^k(n-k)} \\ < \frac{1}{2(n-1)}+ \sum_{k=2}^{n-1} \frac{1}{(k-1)(n-k)} \\ = \frac{1}{n-1} \left(\frac{1}{2} + \sum_{k=2}^{n-1}\frac{1}{k-1} + \sum_{k=2}^{n-1}\frac{1}{n-k} \right) \\ \frac{1}{n-1} \left(\frac{1}{2} + 2\sum_{k=1}^{n-2}\frac{1}{k} \right)$$
Clearly the RHS converges to $0$ as $n \to \infty$ since a harmonic sum is of order $\log n$.
|
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|
convergence and divergence of $\iiint\limits_{x^2+y^2+z^2\geq 1}(x^2+y^2+z^2)^\alpha \ln(x^2+y^2+z^2)\,dx\,dy\,dz$ I want to study the divergence or convergence of this integral :
$$\iiint\limits_{x^2+y^2+z^2\geq 1}(x^2+y^2+z^2)^\alpha \ln(x^2+y^2+z^2)\,dx\,dy\,dz$$
I think that if $\alpha\geq-3$, then in the domain:
$$(x^2+y^2+z^2)^\alpha \ln(x^2+y^2+z^2)\geq(x^2+y^2+z^2)^\alpha$$
so the integral on the domain over the R.H.S. function diverges.
But know I am left with the situation when $\alpha\lt-3$. Can I say that $\alpha=-3-\epsilon$, so there exist some ball $B(0,r_\epsilon)$, such that in the domain $\mathbb{R}^3\setminus B(0,r_\epsilon)$:
$$\ln(x^2+y^2+z^2)\leq(x^2+y^2+z^2)^\frac{\epsilon}{2}$$
$$\iff(x^2+y^2+z^2)^\alpha \ln(x^2+y^2+z^2)\leq(x^2+y^2+z^2)^{-3-\frac{\epsilon}{2}}$$
so the integral on the domain over the R.H.S. function converges. Thus concluding that:
If $\alpha\geq-3$, our integral diverges, otherwise converges. Is it true?
|
The surface area of $x^2+y^2+z^2=\rho^2$ is given by $4\pi\rho^2$, hence the given integral equals
$$ \int_{1}^{+\infty} 4\pi\rho^2 \cdot \rho^{2\alpha}\cdot 2\log\rho\,d\rho $$
i.e. $\frac{8\pi}{(2\alpha+3)^2}$ as soon as $\alpha<-\frac{3}{2}$.
|
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|
MCQ The nth derivative of $f(x)=\frac{1+x}{1-x}$ Let $f(x)=\dfrac{1+x}{1-x}$ The nth derivative of f is equal to:
*
*$\dfrac{2n}{(1-x)^{n+1}} $
*$\dfrac{2(n!)}{(1-x)^{2n}} $
*$\dfrac{2(n!)}{(1-x)^{n+1}} $
by Leibniz formula
$$ {\displaystyle \left( \dfrac{1+x}{1-x}\right)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}\ (1+x)^{(k)}\ \left(\dfrac{1}{1-x}\right)^{(n-k)}}$$
using the hint
*
*$\dfrac{1+x}{1-x}=\dfrac{2-(1-x)}{1-x}=\dfrac2{1-x}-1$ and
*$\left(\dfrac{1}{x}\right)^{n}=\dfrac{(-1)^{n}n!}{x^{n+1}}$
so
$${\displaystyle \left( \dfrac{1+x}{1-x} \right)^{(n)} = \left( \dfrac{2}{1-x}-1 \right)^{(n)}=2\dfrac{ (-1)^{n}n! }{ (1-x)^{n+1} } } $$ but this result isn't apear in any proposed answers
what about the method of Lord Shark the Unknown
tell me please this way holds for any mqc question contain find the n th derivative so it's suffice to check each answer in y case i will start with first
*
*let $f_n(x)=\dfrac{2n}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2(n+1)}{(1-x)^{n+2}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{-2n(n+1)}{(1-x)^{n+2}}\neq f_{n+1}$$
*let $f_n(x)=\dfrac{2(n!)}{(1-x)^{2n}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{2(n+1)}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{-2(n!)(2n)}{(1-x)^{4n}}\neq f_{n+1}$$
*let $f_n(x)=\dfrac{2(n!)}{(1-x)^{n+1}}$ then $f_{n+1}(x)=\dfrac{2((n+1)!)}{(1-x)^{n+2}}$ do i have $f'_{n}=f_{n+1}$ let calculate $$ f'_n=\dfrac{2(n!)(n+1)}{((1-x)^{n+1})^{2}}=\dfrac{2((n+1)!)}{(1-x)^{2n+2}}\neq f_{n+1}$$
|
$$y=\dfrac{1+x}{1-x}=\dfrac{2-(1-x)}{1-x}=\dfrac2{1-x}-1$$
$$\implies\dfrac{dy}{dx}=\dfrac{2(-1)}{(1-x)^2}$$
$$\dfrac{d^2y}{dx^2}=\dfrac{2(-1)(-2)}{(1-x)^3}=\dfrac{2(-1)^22!}{(1-x)^{2+1}}$$
Can you follow the pattern?
|
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|
Common complex roots
If the equations $ax^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ have two common roots then show that $a=b=c$.
My attempts:
Observing $-2$ is a root of $x^3+3x^2+3x+2=0\implies x^3+3x^2+3x+2=(x+2)(x^2+x+1)=0$
Hence $ax^2+bx+c=0$ can have complex roots in common, comming from $(x^2+x+1)=0$
Both the roots of $(x^2+x+1)=0$ and $ax^2+bx+c=0$ are common should imply $a=b=c$ not only this but $a=b=c=1$.
Is this solution correct?
|
Yes, I just check your calculus and are all correct. Your polynomial of grade $3$ defined as $x^3+3x^2+3x+2=0$ has one real root $x_{1}=-2$ and two complex roots $x_{2,3}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$ so it can be written as $P(x)=(x+2)(x-(\frac{1}{2}-\frac{\sqrt{3}}{2}i))(x-(\frac{1}{2}+\frac{\sqrt{3}}{2}i))$. Then, remember the following: "a property of the second degree equations with real coefficients that have complex roots is that they are conjugated to each other". So, knowing this, there is no way you will have in common with the other polynomial one real root and other complex (is the only other option we have available), then: $a=b=c=1$.
|
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|
If $a + \frac{1}{a} = -1$, then the value of $(1-a+a^2)(1+a-a^2)$ is?
If $a + \frac{1}{a} = -1$ then the value of $(1-a+a^2)(1+a-a^2)$ is?
Ans. 4
What I have tried:
\begin{align}
a + \frac{1}{a} &= -1 \\
\implies a^2 + 1 &= -a \tag 1 \\
\end{align}
which means
\begin{align}
(1-a+a^2)(1+a-a^2) &=(-2a)(-2a^2) \\
&=4a^{3}
\end{align}
as $1 + a^{2} = -a$ and $1 + a = -a^{2}$ from $(1)$.
|
Just solve the quadratic $(1)$ to get $a$, then substitute $a$ back in the original expression or $4a^3$.
Also, since $a^3-1=(a-1)(a^2+a+1),$ $a^3=1$.
In more detail, from $(1)$,
$$\begin{align}a&=\frac{-1\pm\sqrt{1-4}}{2} \\
&=\frac{-1\pm i\sqrt{3}}{2}.\end{align}$$
Now
$$\begin{align}
a^2&=\frac{-1\pm i\sqrt{3}}{2}\times \frac{-1\pm i\sqrt{3}}{2} \\
&=\frac{1\mp i2\sqrt{3}-3}{4} \\
&=\frac{-1\mp i\sqrt{3}}{2}
\end{align}$$ and
$$\begin{align}
a^3&=\frac{-1\pm i\sqrt{3}}{2}\times \frac{-1\mp i\sqrt{3}}{2} \\
&=\frac{1\pm i\sqrt{3}\mp i\sqrt{3}+3}{4} \\
&=1.
\end{align}$$
|
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|
If a is a non real root of $x^7 = 1$, find the equation whose roots are $ a + a^6 , a^2 + a^5, a^3 + a^4$ If a is a non real root of $ x^7 = 1$, find the equation whose roots are $a + a^6 , a^2 + a^5, a^3 + a^4$. This is one of the questions I have encountered while preparing for pre rmo. I feel the question requires the concept of the nth roots of unity and de moivre's theorem. But i actually couldnt work it out. Any help will be appreciated. Thanks in advance.
|
WLOG let $a=e^{\frac{2i\pi}{7}}$. Then expand
$$(x-(a+a^6))(x-(a^2+a^5))(x-(a^3+a^4))$$
using the facts that $a^7=1$ and $a^6+a^5+a^4+a^3+a^2+a+1=0$.
|
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|
If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{x^2-4)^2}=2x^2-4+2x\sqrt{x^2-4}$
Now,
$a^2-ax=2x^2-4+2x\sqrt{x^2-4}-x^2-x\sqrt{x^2-4}=x^2+x\sqrt{x^2-4}-4=xa-4$
Why am I not getting the intended value?
|
Rationalizing $a$, you should get $$\frac{x+\sqrt{x^2-4}}{2}$$
|
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|
$(xyz)_b$ is divisible by $n$ if and only if $z+3y-4x$ is divisible by $n$
Determine all natural numbers $n > 1$ with the following property: there exists a base $b \geq 5$ such that any three digit-number $(xyz)_b$ is divisible by $n$ if and only if $z+3y-4x$ is divisible by $n$.
Suppose that $b$ is such a base for some $n$. Then $xb^2+yb+z \equiv z+3y-4x \pmod{n}$, so that $x(b^2+4)+y(b-3) \equiv 0 \pmod{n}$. How can we continue?
|
Immediately, we only have
$$n\mid xb^2+yb+z\iff n\mid z+3y-4x\qquad \text{for }0\le x,y,z\le b-1$$
and not necessarily a congruence in general.
Nevertheless, $1+3\cdot 1-4\cdot1=0$ implies that $n\mid 111_b=b^2+b+1$.
By a similar argument, $n\mid 104_b=b^2+4$ (note that $4$ is a valid digit), hence also $n\mid b-3$.
Then also $n\mid (b^2+b+1)-(b-3)(b+4)=13$.
The only candidates are $n=1$ and $n=13$.
$n=1$ trivially has the desired property for any $b$ (but is excluded per problem statement).
$n=13$ works as well if we pick $b=16$: As $(256x+16y+z)-(z+3y-4x)=13\cdot(20x+y)$, we even have $xyz_{13}\equiv z+3y-4x\pmod{13}$.
|
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|
Can anybody solve the following using binomial theorem? If $x,y \in\mathbb{R}$ such that $x^2 + y^2 - 6x + 8y + 24 = 0$ then what is the greatest value of $$\frac{16\cos^2(\sqrt{x^2+y^2})}{5} - \frac{24\sin(\sqrt{x^2+y^2})}{5}?$$
|
Hint. Note that $x^2 + y^2 - 6x + 8y + 24 = 0$ is equivalent to
$$(x-3)^2 + (y-4)^2 = 1$$
that is the circle $C$ centered in $(3,4)$ of radius $1$. Hence for any point $(x,y)$ on this circle its distance from the origin, that is $\sqrt{x^2+y^2}$, attains by continuity all the values in the interval $[\sqrt{3^2+4^2}-1,\sqrt{3^2+4^2}+1]=[4,6]$.
Therefore
\begin{align*}
M:&=\max_{(x,y)\in C} \left(\frac{16\cos^2(\sqrt{x^2+y^2})}{5}-\frac{24\sin(\sqrt{x^2+y^2})}{5}\right)\\
&=\frac{1}{5}\max_{t\in [4,6]} \left(16\cos^2(t)-24\sin(t)\right)\\
&=\frac{1}{5}\max_{t\in [4,6]} \left(16(1-\sin^2(t))-24\sin(t)\right)\\
&=\frac{8}{5}\max_{t\in [4,6]} \left((\sin(t)+2)(1-2\sin(t))\right).
\end{align*}
Now the quadratic polynomial $(s+2)(1-2s)$ attains its maximum value at the midpoint of its roots that is $\frac{-2+1/2}{2}=-3/4$.
Is there a value of $t\in [4,6]$ such that $\sin(t)=-3/4$? What is $M$?
P.S. I am not sure you need the binomial theorem for this problem.
|
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|
How to integrate: $\int_0^{\infty} \frac{1}{x^3+x^2+x+1}dx$ I want to evaluate $\int_0^{\infty} \frac{1}{x^3+x^2+x+1}$. The lecture only provided me with a formula for $\int_{-\infty}^{\infty}dx \frac{A}{B}$ where $A,B$ are polynomials and $B$ does not have real zeros. Unfortunately, in the given case $B$ has a zero at $z=-1$ and is not even. Is there a straight forward way to solve this in terms of complex analysis?
|
Yes there is , integrate the following function
$$f(z) = \frac{\log(z)}{z^3+z^2+z+1}$$
Around a key-hole contour. Due to the evaluation of the lines on the positive x-axis the $\log$ will cancel. What is remaining is the evaluation of the residues of the three poles.
More explanation
we have three poles $\pm i , -1$. Integrate around the following contour
$$\int_{r}^{\varepsilon}\frac{\log|x|}{x^3+x^2+x+1}dx +\int^{r}_{\varepsilon}\frac{\log|x|+2\pi i }{x^3+x^2+x+1}dx+ \int_{\gamma_r}f(z) dz +\int_{\gamma_{\varepsilon}}f(z) dz = 2\pi i \left(\mathrm{Res}(f,\pm i)+\mathrm{Res}(f,-1)\right)$$
The evaluation of the residues
$$\mathrm{Res}(f,i) = \frac{\log(i)}{3\cdot i^2+2\cdot i +1} = \frac{i\pi/2}{2i-2} $$
$$\mathrm{Res}(f,-i) = \frac{\log(-i)}{3\cdot i^2-2\cdot i +1} = \frac{3i\pi/2}{-2i-2}$$
$$\mathrm{Res}(f,-1) = \frac{\log(-1)}{3-2 +1} = \frac{\pi i}{2}$$
By summing the residues and taking $r\to \infty , \varepsilon \to 0$. Note that the complex logarithm is easily bounded hence the integrals on the smaller and bigger circles vanish and we get
$$\int^\infty_0 \frac{\log(x)}{x^3+x^2+x+1}dx -\int^ \infty_0 \frac{\log(x)+2\pi i }{x^3+x^2+x+1}dx = -\frac{\pi}{2}i$$
Hence we get
$$\int^\infty_0 \frac{\log(x)}{x^3+x^2+x+1}dx = \frac{\pi}{4}$$
|
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|
Middle binomial coefficient size How can we prove that $$\lim_{n\rightarrow\infty}\frac{\dbinom{2n}{n}^2}{\dbinom{4n}{2n}}=0?$$
From plotting this for some values of $n$ it is clear, but using an approximation like $\binom{n}{k}\approx \left(\frac{ne}{k}\right)^k$ gives a value of $1$, since both numerator and denominator are $(2e)^{2n}$.
|
$$ \binom{2n}{n}= \frac{2^n (2n-1)!!}{n!} = 4^n \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{1}$$
hence:
$$\begin{eqnarray*} \binom{2n}{n}^2\binom{4n}{2n}^{-1}&=&\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 \prod_{k=1}^{2n}\left(1-\frac{1}{2k}\right)^{-1}\\&=&\frac{1}{2n}\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right)\prod_{k=2}^{2n}\left(1-\frac{1}{2k}\right)^{-1}\\&=&\frac{1}{2\sqrt{n}}\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right)\prod_{k=2}^{2n}\left(1-\frac{1}{(2k-1)^2}\right)^{-1/2}.\tag{2}\end{eqnarray*}$$
Since $\prod_{k\geq 2}\left(1-\frac{1}{(2k-1)^2}\right)=\frac{\pi}{4}$ by the Weierstrass product for the cosine function, the product $\binom{2n}{n}^2\binom{4n}{2n}^{-1}$ behaves like $\frac{K}{\sqrt{n}}$ for large values of $n$ and the given limit equals zero.
|
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Confused by Proof by induction Im just confused by this equation and how it was used.
Assume
$$P(n) = 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}2$$
Inductive step
$$P(n+1) = 1 + 2 + 3 + 4 + \cdots + n + (n+1) = \frac{(n+1)(n+2)}2$$
Why is the inductive step not written like this?
$$P(n+1) = 1 + 2 + 3 + 4 + \cdots + (n+1) = \frac{(n+1)(n+2)}2$$
|
You assume that $P(n)$ is true and prove that $P(n+1)$ is also true from that.
So assume $1+2+3+\cdots+n=\frac {n(n+1)}2$.
Then from this we get $1+2+3+\cdots+n+(n+1)=\frac {n(n+1)}2 + (n+1)=(n+1)(\frac n2+1)=\frac {(n+1)(n+2)}2.$(Which the answer has avoided to mention here)
This is exactly what is done in that step you have stumbled upon. That's why writing the "$n$" matters in $1+2+3+\cdots+n+(n+1)$. Because you are using trueness of $P(n)$ here to show trueness of $P(n+1)$.
|
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|
Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
using L'Hopital's rule we get
$$L=\lim_{y \to 0} \frac{1}{1+\left(\frac{1+y}{1+4y}\right)^2} \times \frac{-3}{(1+4y)^2}$$
$$L=\lim_{y \to 0}\frac{-3}{(1+y)^2+(1+4y)^2}=\frac{-3}{2}$$
is this possible to do without Lhopita's rule
|
Using the formula for the tangent of a sum,
$$
\tan\left(x-\frac\pi4\right)=\frac{\tan(x)-1}{\tan(x)+1}
$$
we get
$$
\tan^{-1}(x)-\frac\pi4=\tan^{-1}\left(\frac{x-1}{x+1}\right)
$$
Thus,
$$
\begin{align}
\lim_{x\to\infty}\left(\tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac\pi4\right)x
&=\lim_{x\to\infty}\tan^{-1}\left(\frac{-3}{5+2x}\right)x\\
&=\lim_{x\to\infty}\frac{\tan^{-1}\left(\frac{-3}{5+2x}\right)}{\frac{-3}{5+2x}}\lim_{x\to\infty}\frac{-3x}{5+2x}\\
&=-\frac32
\end{align}
$$
|
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|
To find number of ways of solving $x + y + z = 12$, with $0 \leq x, y, z \leq 6$, using generating functions I am learning generating functions so I tried to solve the below question using generating functions.
Number of ways in which value of three variables add up to 12.
$x + y + z = 12$ and $0 \leq x,y,z \leq 6$.
To solve this question, we can form a generating functions as :
$$(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)^3$$ and find the coefficient of $x^{12}$.
The above function can be written as:
$$\left(\frac{1-x^7}{1-x}\right)^3$$
How do I find coefficient of $x^{12}$ in the above expression?
|
You can write $\displaystyle\left(\frac{1-x^7}{1-x}\right)^3=(1-x^7)^3(1-x)^{-3}=(1-3x^7+3x^{14}-x^{21})\sum_{n=0}^{\infty}\binom{n+2}{2}x^n,$
so the coefficient of $x^{12}$ is given by $\displaystyle\binom{14}{2}-3\binom{7}{2}=28.$
|
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|
integral $\int_{}^{}\frac{dx}{1+x^4+x^8} $ looking for help for the following integral -
$$
\int_{}^{}\frac{dx}{1+x^4+x^8}
$$
what I tried to do:
$$\int_{}^{}\frac{dx}{1+x^4+x^8} = \int_{}^{}\frac{dx}{\frac14+x^4+x^8 + \frac{3}{4}}= \int_{}^{}\frac{dx}{\left(x^4+\frac{1}{2}\right)^2 + \frac{3}{4}} $$
and now I am stuck :-(
|
I want to present another method such that we calculate 2 partial fractions incited of 4 partial fractions. By substitution $\frac{1}{x}=t$ we have $dx=-\frac{1}{t^2}dt$ and we obtain the form $J$ that is equal the form $I$ as follows:
$$
\begin{array}{l}
I=\int\,\frac{dx}{1+x^4+x^8} \\
\\
J=\int\, \frac{-\frac{1}{t^2}dt}{1+{(\frac{1}{t})}^4+{(\frac{1}{t})}^8}
=J=\int\, \frac{-\frac{1}{t^2}dt}{\frac{t^8+t^4+1}{t^8}}=
\int\,\frac{-t^6\,dt}{1+t^4+t^8}
\end{array}
$$
which results that
$$
2\,I=I+J=\int\,\frac{(1-t^6)\,dt}{1+t^4+t^8}=
\int \, \frac{(1-t^2)(1+t^4+t^8)\,dt}{(t^4+t^2+1)(t^4-t^2+1)}
=\int\, \frac{(1-t^2)\, dt}{(t^4-t^2+1)}
$$
|
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|
Find the number of five digit numbers of the form $d_1d_2d_3d_4d_5$ and satisfying $d_1Find the number of five digit numbers of the form $d_1d_2d_3d_4d_5$ and satisfying $d_1<d_2\le d_3<d_4\le d_5.$
I considered $d_1,d_2,d_3,d_4,d_5$ as 5 baskets and these have to be filled with 0 to 9 apples.I found total ways as $\binom{10+5-1}{5-1}$ but the answer given is $\binom{11}{6}.$
|
Method 1: The leading digit must be at least $1$, so $d_1, d_2, d_3, d_4, d_5 \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
There are three possibilities:
*
*All the digits are different: There are $\binom{9}{5}$ such cases since selecting five of the nine digits completely determines the number since $d_1 < d_2 < d_3 < d_4 < d_5$
*Four different digits are used: There are $\binom{9}{4}$ such cases in which $d_2 = d_3$ since the repeated digit must be the second smallest digit selected. There are also $\binom{9}{4}$ such cases in which $d_4 = d_5$ since the repeated digit must be the largest digit selected. Thus, there are $2\binom{9}{4}$ such cases.
*Three different digits are used: There are $\binom{9}{3}$ such cases since the two repeated digits must be the two largest digits selected.
Total: Since the three cases are disjoint, there are
\begin{align*}
\binom{9}{5} + 2\binom{9}{4} + \binom{9}{3} & = \binom{9}{5} + \binom{9}{4}
+ \binom{9}{4} + \binom{9}{3}\\
& = \binom{10}{5} + \binom{10}{4} && \text{by Pascal's Identity}\\
& = \binom{11}{5} && \text{by Pascal's Identity}
\end{align*}
numbers with the desired property.
Method 2: Since the leading digit must be at least $1$, $d_1, d_2, d_3, d_4, d_5 \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
We can set up a bijection between selections of five numbers from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ satisfying $d_1 < d_2 \leq d_3 < d_4 \leq d_5$ and selections of five distinct elements $x_1, x_2, x_3, x_4, x_5$ of the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ by setting $x_1 = d_1$, $x_2 = d_2$, $x_3 = d_3 + 1$, $x_4 = d_4 + 1$, and $x_5 = d_5 + 2$. For instance, the number $35579$ corresponds to the selection $\{3, 5, 6, 8, 11\}$. Hence, the number of five-digit numbers with the desired property is $\binom{11}{5}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Horizontal line(s) that intersect $f(x)=x-2+\frac{5}{x}$ in two points. Compute exactly the value(s) of $q$ for which the horizontal line $y = q$
intersects the graph of $f(x)$ in two points that are located on a distance $4$ from each other.
While I found the two lines as $y = 4$ and $y = -8$
, I do not know how to find it by any mathematical means.
|
We want to values of $x$ (of distance $4$ apart) such that $x-2 + \frac{5}{x} = q$, but this is equivalent to $x^2 - 2x + 5 = qx$ by multiplying throughout by $x$, more simply: $x^2 - (2+q)x + 5 = 0$.
Now say that the smaller point is $\alpha$, then the other one is necessarily $\alpha + 4$. Hence $2\alpha + 4= 2 +q$ and $\alpha(\alpha + 4) = 5$ using the relationship between roots and coefficients.
Isolating $\alpha$ from the first equation and plugging it into the second gives $(q-2)(q+6) = 20$ and equivalently $(q-4)(q+8) = 0.$
|
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|
Find $k$ such that $(\alpha-1)^3+(\beta-2)^3+(\gamma-3)^3=0$ holds for the roots of $x^3-6x^2+kx+k=0$
If $\alpha$, $\beta$ and $\gamma$ are the three roots of the equation $x^3-6x^2+kx+k=0$, find the values of $k$ such that $(\alpha-1)^3+(\beta-2)^3+(\gamma-3)^3=0$. For each of the possible values of $k$ solve the equation.
I conclude from the equation that $\alpha+\beta+\gamma=6$, $\alpha\beta+\beta\gamma+\gamma\alpha=k$ and $\alpha\beta\gamma=-k$ and with the condition that $(\alpha-1)^3+(\beta-2)^3+(\gamma-3)^3=0$. I found the solution from wolfram: http://www.wolframalpha.com/input/?i=Solve%5Ba%2Bb%2Bc%3D6,ab%2Bbc%2Bca%3Dk,abc%3D-k,(a-1)%5E3%2B(b-2)%5E3%2B(c-3)%5E3%3D0%5D
From the solution I can see that either $\alpha=1$ or $\beta=2$ or $\gamma=3$. I guess to calculate the term $(\alpha-1)(\beta-2)(\gamma-3)$ and prove it to be zero. Should I continue with this or are there another other method?
|
Since $\alpha-1+\beta-2+\gamma-3=0$ and $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$
we see that $$(\alpha-1)(\beta-2)(\gamma-3)=0$$ and we have three following cases:
*
*$\alpha=1$, which gives $k=\frac{5}{2}$ and our equation is
$$2x^3-12x^2+5x+5=0$$ or
$$2x^3-2x^2-10x^2+10x-5x+5=0$$ or
$$(x-1)(2x^2-10x-5)=0,$$
which gives $\left\{1,\frac{5+\sqrt{35}}{2},\frac{5+\sqrt{35}}{2}\right\}$.
*and 3. are similar.
|
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|
How do I rotate 3 groups of 4 people into teams of 3 so that each person in one group works with each person in the other groups? I have three teams of four people. I would like to create rotating groups of three, where each group has one person from each team, and those people rotate on a staggered schedule so that each person is in the group for 3 slots.
For example, if team 1 is people a, b, c, and d; team 2 is people e, f, g, and h; team 3 is people i, j, k, l, the start of the schedule might look like this
*
*a, e, i
*b, e, i
*b, f, i
*b, f, j
*c, f, j
*c, g, j
*c, g, k
*d, g, k
*d, h, k
*d, h, l
*a, h, l
*a, i, l
How do I shift the order in future to make sure that the groups change and everyone gets to work with all of the people on the other teams?
Thanks!
|
\begin{eqnarray*}
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
a&b&c&d& &e&f&g&h&& i&j&k&l\\ \hline
\color{red}{1}&0&0&0& &\color{red}{1}&0&0&0& &\color{red}{1}&0&0&0 \\ \hline
0&\color{red}{1}&0&0& &0&\color{red}{1}&0&0& &0&\color{red}{1}&0&0 \\ \hline
0&0&\color{red}{1}&0& &0&0&\color{red}{1}&0& &0&0&\color{red}{1}&0 \\ \hline
0&0&0&\color{red}{1}& &0&0&0&\color{red}{1}& &0&0&0&\color{red}{1} \\ \hline
\color{blue}{1}&0&0&0& &0&\color{blue}{1}&0&0& &0&0&\color{blue}{1}&0 \\ \hline
0&\color{blue}{1}&0&0& &0&0&\color{blue}{1}&0& &0&0&0&\color{blue}{1} \\ \hline
0&0&\color{blue}{1}&0& &0&0&0&\color{blue}{1}& &\color{blue}{1}&0&0&0 \\ \hline
0&0&0&\color{blue}{1}& &\color{blue}{1}&0&0&0& &0&\color{blue}{1}&0&0 \\ \hline
\color{red}{1}&0&0&0& &0&0&\color{red}{1}&0& &0&\color{red}{1}&0&0 \\ \hline
0&\color{red}{1}&0&0& &0&0&0&\color{red}{1}& &\color{red}{1}&0&0&0 \\ \hline
0&0&\color{red}{1}&0& &\color{red}{1}&0&0&0& &0&0&0&\color{red}{1} \\ \hline
0&0&0&\color{red}{1}& &0&\color{red}{1}&0&0& &0&0&\color{red}{1}&0 \\ \hline
\color{blue}{1}&0&0&0& &0&0&0&\color{blue}{1}& &0&0&0&\color{blue}{1} \\ \hline
0&\color{blue}{1}&0&0& &\color{blue}{1}&0&0&0& &0&0&\color{blue}{1}&0 \\ \hline
0&0&\color{blue}{1}&0& &0&\color{blue}{1}&0&0& &0&\color{blue}{1}&0&0 \\ \hline
0&0&0&\color{blue}{1}& &0&0&\color{blue}{1}&0& &\color{blue}{1}&0&0&0 \\ \hline
\end{array}
\end{eqnarray*}
|
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|
How can I solve $y''=x^{-1/3}y^{3/2}$? I am trying to solve the differential equation.
$$y''=x^{-1/3}y^{3/2}$$
I tried the solution: $y=Ax^s$
$$\implies y'=Asx^{s-1} \\ \iff y''=As(s-1)x^{s-2}$$
$$\implies As(s-1)x^{s-2}= x^{-1/3}(Ax^s)^{3/2} \\ \iff As(s-1)x^{s-2}=x^{-1/3}A^{3/2}x^{\frac{3s}{2}} \\ \iff A^{5/2}s(s-1)=x^{-1/3}x^{3s/2}x^{2-s} \\ \iff Cs(s-1)=x^{\frac{3s}{2}-\frac{1}{3}+2-s} \\ \iff Cs(s-1)=x^{\frac{1}{6}(9s-2+12-6s)}\\ \iff Cs(s-1)=x^{\frac{1}{6}(3s+10)} \\ \iff \ln(Cs^2-Cs)=\frac{1}{6}(3s+10)\ln(x) \\ \iff \ln(C)+\ln(s^2-s)=\frac{1}{6}(3s+10)\ln(x)$$
I am stuck here and I don't know if this is even the best approach. I even tried to solve it with Mathematica but I got no answer. How do I solve this equation?
|
This is in the form of the Emden-Fowler equation, $y'' = Ax^ny^m$. For $m\neq1$, the solution is
\begin{equation*}
y = Cx^{\frac{n+2}{1-m}}
\end{equation*}
for a constant $C$ that depends on $A$, $n$, and $m$. In your case, the solution is $y = Cx^{\frac{5/3}{-1/2}} = Cx^{-10/3}$. You would have gotten this constant also by requiring $\frac16(3s+10) = 0$, as mentioned by @Winther. Indeed, $y' = C\frac{-10}3x^{-13/3}$, $y'' = C\frac{-10}3\frac{-13}3x^{-16/3}$, and $x^{-1/3}y^{3/2} = x^{-1/3}A^{3/2}x^{-10/2} = A^{3/2}x^{-16/3}$ and so setting $C = \frac{-3}{10}\cdot\frac{-3}{13}\cdot A^{3/2}$ will give you the desired solution.
|
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|
Common root between a cubic and a quadratic The equation $x^2-ax+b=0$ & $x^3-px^2+qx=0$, have one common root & the second equation has two equal roots. Prove that $2(q+b) = ap$
($b\neq0 q\neq0$)
I am not able to think of any way to solve this any hint would be appreciated.
|
Counterexample
Let $a = 1, b = 0, p = 4, q = 4$. Then, $x^2 - ax + b = x(x-1)$ and $x^3 - 4x^2 + 4x = x(x-2)^2$ have one common root, $0$, and the second equation has two equal roots, $2$. However,
$$
2(q+b) = 2(4+0) = 8\neq 4 = 1\cdot 4 = ap.
$$
Perhaps you're missing some restrictions?
Edit
Given that $b,q\neq 0$, we can write the polynomials as
$$
x^2-ax+b = (x-r_1)(x-r_2),\qquad x^3-px^2+qx = x(x-r_1)^2.
$$
Now, try writing $a, b, p, q$ in terms of $r_1$ and $r_2$ and see if you can prove the statement.
|
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|
Pythagorean Triple: $\text{Area} = 2 \cdot \text{perimeter}$ Find the unique primitive Pythagorean triple whose area is equal to twice the perimeter.
So far I set the sides of the triangle to be $a, b,~\text{and}~c$ where $a$ and $b$ are the legs of the triangle and c is the hypotenuse.
I came up with 2 equations which are:
$\dfrac{ab}2 = 2(a+b+c)\;\;$ and $\;\;a^2+b^2=c^2$
but I'm not sure how to proceed and solve for $a, b, c$.
|
With Euclid's formula $F(m,k)$, we can solve $R=$area/perimeter for $k$ and test a defined range of $m$-values to see which yield integers.
$$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$
$$R=\frac{area}{perimeter}=\frac{AB}{2P} =\frac{2mk(m^2-k^2)}{2(2m^2+2mk)}=\frac{mk-k^2}{2}$$
\begin{equation}
R=\frac{mk-k^2}{2}\quad\implies k=\frac{m\pm\sqrt{m^2-8R}}{2}\\ \quad\text{for}\quad \big\lceil\sqrt{8R}\big\rceil\le m \le (2R+1)
\end{equation}
The lower limit insures that $k\in \mathbb{N}$ and the upper limit ensures that $m> k$.
$$R=2\implies \lceil\sqrt{8(2)}\rceil=4\le m \le (2(2)+1)=5 \\
m\in\{ 4\}\implies k\in\{ 2\}
\qquad\land\qquad
m\in\{ 5\}\implies k\in\{ 4,1\}
$$
$$F(4,2)=(12,16,20)\qquad\land\qquad\frac{96}{48}=2\\
F(5,4)=(\space 9,40,41)\qquad\land\qquad\frac{180}{90}=2\\
F(5,1)=(24,10,26)\qquad\land\qquad\frac{120}{60}=2\\
$$
We can see that only one of these is primitive but this formula finds non-primitives as well.
|
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|
If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then differentiating we get
$$\frac{f'(x)}{f(x)}=\sum_{k=0}^n \frac{A_k}{x+k}$$
hence
$$\frac{f'(x)}{f(x)}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$
any clue here?
|
Note that:
$$\frac1x\frac1{\binom{x+n}n}=\frac{n!}{x(x+1)(x+2)\dots(x+n)}$$
And thus,
$$\frac1x\left[\frac1{\binom{x+n}n}-\frac1{\binom{x+{n-1}}{n-1}}\right]=\frac{A_n}{x+n}$$
Which gives the general solution immediately.
|
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|
How to bound above $| \frac{n}{\sqrt{n^2+n} +n}-\frac{1}{2} |$
Find and Justify $\lim\limits_{n \rightarrow \infty} S_n$
$$S_n = \sqrt{n^2+n} - n = $$
$$\lim\limits_{n \rightarrow \infty} \sqrt{n^2+n} - n
= \lim\limits_{n \rightarrow \infty} \frac{n}{\sqrt{n^2+n} +n}
= \lim\limits_{n \rightarrow \infty} \frac{1}{\sqrt{1+1/n} + 1} = 1/2$$
$S_n$ converges to $\frac{1}{2}$ if for every $\epsilon>0 $ there is an integer $N$ s. t.
$$|S_n-\frac{1}{2}| <\epsilon,n\geq N$$
Considering
$$|S_n-\frac{1}{2}| = \left| \frac{n}{\sqrt{n^2+n} +n}-\frac{1}{2} \right|= \left|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}+2n} \right|
\leq \left|\frac{n-\sqrt{n^2+n}}{2n+2n} \right|
$$
$$ \leq
\left|\frac{n-\sqrt{n^2+n}}{4n} \right| \leq
\left|\frac{1 -\sqrt{1+1/n}}{4}\right| \leq
\frac{\sqrt{1/n}}{4} \leq \epsilon$$
let $N$ be such that
$$ \frac{\sqrt{1/N}}{4} \leq \epsilon => \frac{1}{16 \epsilon^2}<N$$
It follows for any $\epsilon>0$, if $n \geq N > \frac{1}{16 \epsilon^2}$, then
$$|S_n-\frac{1}{2}| <\epsilon$$
I am a bit uncertain if it is the right way to proceed particularly with the bound of $N$ in relation to $\epsilon$. Is my method to bound $N$ correct?
Much appreciated for your input.
|
Your method of proof for me it is correct.
Now the bound of $N$ you need is $$N=[\frac{1}{16 \epsilon^2}]+1$$
This $N$ is the least natural number with the property:
$$ \frac{\sqrt{1/n}}{4} < \epsilon, \forall n\geqslant N $$
As for answering the comment below,you proved that $N> \frac{1}{16\epsilon^2}$.
We use the double inequality $[x]\leqslant x \leqslant [x]+1$ where $[x]$ is the integer part of $x$.
Thus we have $[\frac{1}{16\epsilon^2}] \leqslant \frac{1}{16\epsilon^2} \leqslant [\frac{1}{16\epsilon^2}]+1$
You see now that the the least positive natural number greater than $\frac{1}{16\epsilon^2}$ is $N=[\frac{1}{16\epsilon^2}]+1$
Thus whenever $n \geqslant N$ we have that desired inequality for the limit.
|
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|
A nice but somewhat challenging binomial identity
When working on a problem I was faced with the following binomial identity valid for integers $m,n\geq 0$:
\begin{align*}
\color{blue}{\sum_{l=0}^m(-4)^l\binom{m}{l}\binom{2l}{l}^{-1}
\sum_{k=0}^n\frac{(-4)^k}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\binom{k+l}{l}
=\frac{1}{2n+1-2m}}\tag{1}
\end{align*}
I have troubles to prove it and so I'm kindly asking for support.
Maybe the following simpler one-dimensional identity could be useful for a proof. We have for non-negative integers $n$:
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}=\frac{4^{n}}{2n+1}\binom{2n}{n}^{-1}\tag{2}
\end{align*}
The LHS of (2) can be transformed to
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}&=\sum_{k=0}^n(-1)^k\binom{n}{k}\int_{0}^1x^{2k}dx\\
&=\int_{0}^1\sum_{k=0}^n(-1)^k\binom{n}{k}x^{2k}\,dx\\
&=\int_{0}^1(1-x^2)^n\,dx
\end{align*}
Using a well-known integral representation of reciprocals of binomial coefficients the RHS of (2) can be written as
\begin{align*}
\frac{4^{n}}{2n+1}\binom{2n}{n}^{-1}&=4^n\int_{0}^1x^n(1-x)^n\,dx
\end{align*}
and the equality of both integrals can be shown easily. From (2) we can derive a simple one-dimensional variant of (1).
We consider binomial inverse pairs and with respect to (2) we obtain
\begin{align*}
&f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}g_k \quad&\quad g_n=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k\\
&f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1} \quad&\quad\frac{1}{2n+1}=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k
\end{align*}
We conclude again with (2)
\begin{align*}
\frac{1}{2n+1}&=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k\\
&=\sum_{k=0}^n\frac{(-4)^{k}}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\\
\end{align*}
This identity looks somewhat like a one-dimensional version of (1). Maybe this information can be used to solve (1).
|
Let us complete the OP's work, started with
$$ \frac{1}{2k+1}\stackrel{\text{Binomial transform}}{\longleftrightarrow} \frac{4^k}{(2k+1)\binom{2k}{k}}\tag{$d=0$}$$
by computing first the binomial transform of $\frac{1}{2k+3}$. We have:
$$\begin{eqnarray*}\sum_{k=0}^{n}\frac{(-1)^k}{2k+3}\binom{n}{k}=\int_{0}^{1}x^2(1-x^2)^n=\frac{B\left(n+1,\tfrac{3}{2}\right)}{2}=\frac{1}{2n+3}\cdot\frac{B\left(n+1,\frac{1}{2}\right)}{2}\end{eqnarray*}$$
hence:
$$ \frac{1}{2k+3}\stackrel{\text{Binomial transform}}{\longleftrightarrow} \frac{4^k}{(2k+1)(2k+3)\binom{2k}{k}}\tag{$d=1$}$$
and in general:
$$ \frac{1}{2k+2d+1}\stackrel{\text{Binomial transform}}{\longleftrightarrow} \frac{4^k\binom{k+d}{d}\binom{2k}{k}^{-1}}{(2k+2d+1)\binom{2k+2d}{2d}}\tag{$d\geq 1$}$$
I need some time to check the above computations, but the last identity, together with creative telescoping, should be the key for proving OP's statement. Indeed, we have:
$$ \sum_{k=0}^{n}\frac{(-4)^k}{(2k+1)\binom{2k}{k}}\binom{n}{k}=\frac{1}{2n+1}\tag{$l=0$} $$
$$ \sum_{k=0}^{n}\frac{(-4)^k}{(2k+1)\binom{2k}{k}}\binom{n}{k}(k+1)=-\frac{1}{(2n+1)(2n-1)}\tag{$l=1$} $$
$$\begin{eqnarray*} \sum_{k=0}^{n}\frac{(-4)^k}{(2k+1)\binom{2k}{k}}\binom{n}{k}\binom{k+l}{l}&=&\frac{(-1)^l(2l-1)!!(2n-2l+1)!! }{(2n+1)!!}\\
&=&\frac{(-1)^l 4^{n-l} n! (2l)! (n-l)!}{(2n+1)!l! (2n-2l+1)!}\tag{$l\geq 1$} \end{eqnarray*}$$
hence the whole problem boils down to computing:
$$ \frac{4^n}{(2n+1)\binom{2n}{n}}\sum_{l=0}^{m}\frac{\binom{m}{l}}{(2n-2l+1)!\binom{n}{l}}$$
|
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|
Center of mass using vectors I encountered a physics problem in this book where I was given trees located at points $A,B,C,D,E$.
The problem tells me to move from $A$ to $B$, but only covering half the distance. From my current location, I would then move to $C$, but only covering one-third the distance. This pattern continues until, finally, I move to $E$, covering one-fifth the distance.
I was asked what would happen if the order of the trees was rearranged and the answer given was that I would get the same final point, which turned out to be the center of mass.
I'm failing to see how the procedure described in the problem is equivalent to averaging the $x$- and $y$-coordinates of the points.
I would really appreciate it if anyone could show how they are equivalent and why the order does not matter.
|
Remember that the center of mass (assuming all trees have the same mass) is $\frac{1}{5}(A + B + C + D + E)$ (treating the locations as vectors).
After the first step, we end up with the point $\frac{1}{2}(A + B)$, found by taking their midpoint. Now from here, we travel $\frac{1}{3}$ of the distance to $C$. We take the weighted average to get the new position of
$$\frac{2}{3}\left(\frac{1}{2}(A + B)\right) + \frac{1}{3}C = \frac{1}{3}A +\frac{1}{3}B + \frac{1}{3}C$$
Similarly, after the next step, we end up at
$$\frac{3}{4}\left(\frac{1}{3}A +\frac{1}{3}B + \frac{1}{3}C\right) + \frac{1}{4}D = \frac{1}{4}A + \frac{1}{4}B + \frac{1}{4}C + \frac{1}{4}D$$
Likewise, the final step brings us to
$$\frac{4}{5}\left(\frac{1}{4}A + \frac{1}{4}B + \frac{1}{4}C + \frac{1}{4}D\right) + \frac{1}{5}E = \frac{1}{5}A + \frac{1}{5}B + \frac{1}{5}C + \frac{1}{5}D + \frac{1}{5}E$$
Intuitively, each of the vectors adds less proportional weight each time, while those already used have their contributions slightly diminished each time. This has the effect of balancing out in the end to give the average.
|
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|
Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) = \frac{(1 + \sin\theta + i\cos\theta)^2}{(1 + \sin\theta)^2 + \cos^2\theta}$$
$$=\frac{(1 + \sin\theta)^2 + \cos^2\theta + 2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
thus
$$x = \frac{(1 + \sin\theta)^2 + \cos^2\theta }{(1 + \sin\theta)^2 + \cos^2\theta} $$
$$y= \frac{2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
According to the binomial theorem,
$$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k$$
we get
$$z = \frac{1}{(1 + \sin\theta)^2 + \cos^2\theta}\sum_{k=0}^n \binom{n}{k} ((1 + \sin\theta)^2 + \cos^2\theta)^{n-k}\cdot(2i(1 + \sin\theta)\cos\theta)^k$$
...and that is where I'm stuck. What do you think? Thanks for the attention.
|
The denominator is $2+2\sin\theta$, the numerator is, setting $2\alpha=\theta$,
$$
1+e^{2i\alpha}=2e^{i\alpha}\cos\alpha
$$
Then your number is
$$
\left(\frac{\cos\alpha}{1+\sin2\alpha}\right)^{n}e^{ni\alpha}
$$
|
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|
Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}\tag1$$
Moving the things in RHS to LHS:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} - \frac{1}{(x-3)} + \frac{1}{(x-4)} = 0\tag2$$
Writing everything above a common denominator:
$$\frac{1}{(x-4)(x-1)(x-2)(x-3)}\bigg[(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3)\bigg] = 0\tag3$$
Multiplying both sides with the denominator to cancel the denominator:
$$(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3) = 0\tag4$$
Multiplying the first two factors in every term:
$$(x^2-3x-2x+6)(x-4) - (x^2-3x-x+3)(x-4) - (x^2-x-2x+2)(x-4) + (x^2-2x-x+2)(x-3) = 0\tag5$$
Simplifying the first factors in every term:
$$(x^2-5x+6)(x-4) - (x^2-4x+3)(x-4) - (x^2-3x+2)(x-4) + (x^2-3x+2)(x-3) = 0\tag6$$
Multiplying factors again:
$$(x^3-4x^2-5x^2+20x+6x-24) - (x^3-4x^2-4x^2+16x+3x-12) - (x^3-4x^2-3x^2-12x+2x-8) + (x^3-3x^2-3x^2+9x+2x-6) = 0\tag7$$
Removing the parenthesis yields:
$$x^3-4x^2-5x^2+20x+6x-24 - x^3+4x^2+4x^2-16x-3x+12 - x^3+4x^2+3x^2+12x-2x+8 + x^3-3x^2-3x^2+9x+2x-6 = 0\tag8$$
Which results in:
$$28x - 10 = 0 \Rightarrow 28x = 10 \Rightarrow x = \frac{5}{14}\tag9$$
which is not correct. The correct answer is $x = \frac{5}{2}$.
|
A hint to make your life much simpler.
Always simplify your variables to start with. Here you can easily make the substitution $x-4 = y$ to get an easier equation in $y$.
Then noting that $\frac 1y - \frac 1{y+1} = \frac 1{y(y+1)}$, your can very easily see that the numerators on both sides when combining the rational expressions is just $1$ (to avoid confusion, note that I transposed the terms on each side to get a positive numerator). Taking the reciprocal on both sides and expanding, you get quadratics on both sides where the square terms immediately cancel, giving you a simple linear equation. Solve for $y$, then add $4$ to get $x$.
|
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|
Find all positive integers such that: $n^2-3|3n^3-5n^2+7$
Find all positive integers such that: $$n^2-3|3n^3-5n^2+7.$$
I did the following:
$$n^2-3|3n^3-5n^2+7,n^2-3\Rightarrow n^2-3|3n^3-5n^2+7,3n^3-9n$$
$$\Rightarrow n^2-3|5n^2-9n-7,5n^2-15\Rightarrow n^2-3|9n-8$$
Now we must have: $9n-8\geq n^2-3$.We should determine the sign of $n^2-9n+5$, but the equation: $n^2-9n+5=0$ does not have integer roots.How can we proceed?
|
Yes, $n^2-9n+5\leq0$, which gives $1\leq n\leq8$ and check it.
$n=2$ is valid.
|
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|
Zeroes of function with real exponents Let real $x \geq 0 $ and real $p > 2$.
Let $ f(x) = (x - 1)(x + 1)^{p - 1} - x^p + 1$.
Show that, for the given range of $x$ and $p$, $f(x)=0$ only for $x=0$ and $x=1$.
Since $f(x)$ is not convex, I find it difficult to show that $f(x) < 0$ for $0<x<1$ and $f(x) > 0$ for $x>1$.
EDIT (generalization): the same zeroes hold for all $p>0$ ; $p \neq 1$ , $p \neq 2$.
|
We will show that $f(x) > 0$ for $x>1$ and $f(x) < 0$ for $0<x<1$.
Let's start with the first one, $f(x) > 0$ for $x>1$.
For real $p > 2$, we can use Bernoulli's inequality (which holds for real exponents!):
$(1 + w)^r \ge 1 + r w $, for real $w > -1$ and real $r > 1$.
This gives for the first exponent in $f(x)$:
$(x+1)^{p-1} = x^{p-1} (1+1/x)^{p-1} \geq x^{p-1} (1 + \frac{p-1}{x})$
Inserting into $f(x)$ gives
$$f(x) = (x - 1)(x + 1)^{p - 1} - x^p + 1 \\\ge (x - 1)x^{p-1} (1 + \frac{p-1}{x}) - x^p + 1 \\= x^{p-1} (p-2) - x^{p-2} (p-1) + 1 = h(x)$$
with equality for $x=1$. Now note that $h(x=1) = 0$. Further,
$h'(x) = (p-2)(p-1)x^{p-3} (x-1)$
so for $x>1$, $h'(x) > 0$ everywhere. Hence $h(x) \geq 0$ with equality for $x=1$. This establishes that for $x\geq 1$, $f(x) \geq 0$ with equality for $x=1$.
For the second case $0<x<1$, we will see now that this can be reduced to the first case $x>1$, so indeed there is nothing to do.
Let $y =\sqrt x$. Then the function becomes:
$$
f(y = \sqrt x) = y^p \left[ (y - \frac1y)(y + \frac1y)^{p - 1} - y^p + \dfrac{1}{y^p} \right]
$$
We have shown above that for $y>1$, $f(y) >0$. I.e. we have that the following inequality holds:
$$
g(y) = (y - \frac1y)(y + \frac1y)^{p - 1} - y^p + \dfrac{1}{y^p} > 0 \qquad \bf\text{[ineq. 1]}
$$
Now, we want to show that for $0<y<1$, $f(y) <0$. Here, we need to show
$$
g(y) =(-y + \frac1y)(y + \frac1y)^{p-1} -[- y^p + \dfrac{1}{y^p} ]> 0
$$
Now replacing $y$ by $1/y$ we get $y > 1$ and, in the replaced variable,
$$
\tilde g(y) = (y - \frac1y)(y + \frac1y)^{p - 1} - y^p + \dfrac{1}{y^p} > 0
$$
So this is the same condition as before in ineq. 1, which has already been proven.
$\qquad \qquad \Box$
|
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|
Finding the Equation of Circle, Given is its Differential Equation I was studying for some quizzes when I stumbled upon this question.
It goes like this:
Find the equation of the circle whose differential equation is
$y'' = (1 + (y')^2)^{\frac{3}{2}}$ and which passes through the points
(0, 0) and (1,1).
My work:
The differential equation $y'' = (1 + (y')^2)^{\frac{3}{2}}$ doesn't have a dependent
variable $y,$ so we let
$$\frac{dy}{dx} = p$$
$$\frac{d^2 y}{ dx^2} = \frac{dp}{dx}$$
Now we had all we need, we can now solve for the general solution of
$y'' = (1 + (y')^2)^{\frac{3}{2}}$
$$y'' = (1 + (y')^2)^{\frac{3}{2}}$$
$$ \left(\frac{dp}{dx}\right) = (1 + p^2)^\frac{3}{2}$$
$$dp = (1 + p^2)^\frac{3}{2} dx$$
$$\frac{dp}{(1 + p^2)^\frac{3}{2}} = dx$$
$$\int \frac{dp}{(1 + p^2)^\frac{3}{2}} = \int dx$$
Remembering that $\int \frac{du}{(u^2 + a^2)^\frac{3}{2}} = \frac{u}{a^2 \sqrt{u^2 + a^2}} + c$
$$\frac{p}{\sqrt{p^2 + 1}} = x + c$$
$$\frac{y'}{\sqrt{(y')^2 + 1}} = x + c$$
$$\int \frac{y'}{\sqrt{(y')^2 + 1}} = \int x + c$$
At this point, I'm stuck, because is it possible to do this?
$$\int \frac{y'}{\sqrt{(y')^2 + 1}} = \int x + c$$
How will I go forward, and ultimately, getting the equation of the circle?
|
$$\frac{y'}{\sqrt{(y')^2 + 1}} = x + c$$
$$\frac{y'^2}{(y')^2 + 1} = (x + c)^2 \quad\to\quad
y'^2 = \frac{(x + c)^2}{(1-(x + c)^2)}$$
$$y' =\pm \frac{x + c}{\sqrt{1-(x + c)^2}}$$
$$y =\pm \int\frac{x + c}{\sqrt{1-(x + c)^2}}dx=\mp\sqrt{1-(x + c)^2}+C$$
$$(y-C)^2+(x+c)^2=1$$
I suppose that you can take it from here to find the values of $C$ and $c$ according to the specified conditions.
|
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|
Determine the Polynomial $P$ for which $16P(x^2) = P(2x)^2$ Determined the Polynomial $P$ for which $16P(x^2) = P(2x)^2$
Okay, this problem was given as an example on an Olympiad training material I'm Self-studying, however I do not understand most of the arguments in the solution. Here's the solution.
Plugging $x = 0$ in $16P(x^2) = P(2x)^2$ , we obtain $P(0) = 0$ or $16$
$(i)$, suppose $P(0) = 0$. Then $P(x) = xQ(x)$, for some Polynomial $Q$, hence
$16x^2Q(x^2) = 4x^2Q(2x)^2$, which reduces to $4Q(x^2) = Q(2x)^2$, setting $4Q(x) = R(x)$ for some Polynomial $R$, gives $16R(x^2) = R(2x)^2$ hence $P(x) = \frac {1}{4}xR(x)$, with R satisfying the same relation as $P$
I understand the above argument very well, but the argument presented below is what I do not understand at all.
$(ii)$, Suppose that $P(0) = 16$ putting $P(x) = xQ(x) + 16$ in the given relation, we obtain $4xQ(x^2) = xQ(2x)^2 + 16Q(2x)$. Hence $Q(0) = 0$, i.e. $Q(x) = xQ_1(x)$ for some Polynomial $Q_1$
Furthermore; $x^2Q(2x)^2 = x^2Q_1(2x)^2 + 8Q_1(2x)$, implying that $Q_1(0) = 0$ so $Q_1$ too is divisible by $x$, thus $Q(x) = x^2Q_1(x)$, now, suppose $x^n$ is the highest degree of $x$ dividing $Q_1$, and $Q(x) = x^nR(x)$, where $R(0) \ne 0$, then $R$ satisfies
$4x^{n-1}R(x^2) = 2^{2n}x^{n+1}R(2x)^2 + 2^{n+4}R(2x) . . (*)$ which implies that $R(0) = 0$, a contradiction, it follows that $Q(0) = 0$ and $R(x) = 16.
Please I need someone to break down $(ii)$ for me, especially the (*) part. Thanks in advance.
|
I assume you understood how the relation $4xQ(x^2)=xQ(2x)^2+16Q(2x)$ was obtained. Now you just have to substitute $Q(x)=x^nR(x)$.
In particular, $4xQ(x^2)=4x^{2n+1}R(x^2)$, $xQ(2x)^2=2^{2n}x^{2n+1}R(2x)^2$ and $16Q(2x)=2^{n+4}x^nR(2x)$.
After simplifying $x^n$ on both sides, you find $4x^{n+1}R(x^2)=2^{2n}x^{n+1}R(2x)^2+2^{n+4}R(2x)$, which is (the correct form of) (*).
|
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|
Where the object that moves along the intersection of $x^2+y^2=1$ and $y+z=1$ needs to be if we want the sum $x+2y+z$ to be max/min?
An object is moving along the curve which is derived from the intersection of the cylinder $x^2+y^2=1$ and the plane $y+z=1$. Where does the object need to be located if we want to maximize/minimize the sum $x+2y+z$? What is the min/max sum?
I'm really not sure how to determine the constraint function. The intersection of the cylinder and the plane is:
$$
x^2+(1-z)^2=1
$$
Then:
$$
z=\frac{x^2+z^2}{2}
$$
I guess we can define new function:
$$
g(x,y)=x+2y+z=x+2y+\frac{x^2+z^2}{2}=x+2y+\frac{x^2+(1-y)^2}{2}
$$
Then we can look for critical points:
$$
g_x=1+x\\
g_y=1+y
$$
Which means that $(-1,-1)$ is the critical point. I don't see how to proceed though.
|
Express $z$ and we are interested in maximum/minimum of $E = x+y+1$. By the Cauchy inequality we have: $$|x+y|\leq \sqrt{2(x^2+y^2)} = \sqrt{2}$$ So $E_{\max} = 1+\sqrt{2}$ which is reached at $x=y=1/\sqrt{2}$ and $E_{\min} = 1-\sqrt{2}$ which is reached at $x=y=-1/\sqrt{2}$
|
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|
Prove a matrix has an inverse Prove the matrix
\begin{bmatrix}
A & B \\
B^T & C
\end{bmatrix}
has the inverse
\begin{bmatrix}
D & -DBC^{-1} \\
-C^{-1}B^TD & C^{-1} + C^{-1}B^TDBC^{-1}
\end{bmatrix}
whenever C and D = $(A - BC^{-1}B^T)^{-1}$ are nonsingular.
I tried to see if their multiplication results in $I_{2}$, but I got stuck.
|
Note that
\begin{align*}
&\begin{bmatrix}
A & B \\ B^T & C
\end{bmatrix}
\begin{bmatrix}
D & -DBC^{-1} \\ -C^{-1}B^TD & C^{-1}+C^{-1}B^TDBC^{-1}
\end{bmatrix}
\\ &=
\begin{bmatrix}
AD - BC^{-1}B^TD & -ADBC^{-1} + BC^{-1} + BC^{-1}B^TDBC^{-1} \\
B^TD + -CC^{-1}B^TD & -B^TDBC^{-1}+CC^{-1}+CC^{-1}B^TDBC^{-1}
\end{bmatrix}
\\ &=
\begin{bmatrix}
(A-BC^{-1}B^T)D & (I-(A-BC^{-1}B^T)D)BC^{-1} \\
0 & I
\end{bmatrix}
\\ &=
\begin{bmatrix}
I & 0 \\
0 & I
\end{bmatrix}.
\end{align*}
If instead you choose to multiply in the reverse order, then you will have similar cancellations using $D=(A-BC^{-1}B^T)^{-1}$ and the fact that $C$ is invertible.
|
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|
Is it possible to solve for $\theta$ in this multivariable equation? Question: Given the equation $$c^2=a^2+(\frac{pq}{\sqrt{q^2\sin^2{\theta}+p^2\cos^2{\theta}}})^2-2a\frac{pq}{\sqrt{q^2\sin^2{\theta}+p^2\cos^2{\theta}}}\cos{\theta},$$
is it possible to solve for $\theta$ in terms of $a,c,p,$ and $q$?
Original Problem: I am attempting to develop an algorithm to solve an extension of a spherical codes problem. This image shows a partial cutaway of a spherical codes solution (a center sphere with as many smaller spheres packed onto it as possible, cut away to show the inner sphere).
I'd like to extend this problem to ellipsoids, so essentially to take an ellipsoid with given dimensions and pack as many spheres of another given dimension on its surface as possible. The equation above is a central part of my solution to this problem, and is derived from the law of cosines with $\theta$ as the opposite angle to side $c$ and the side $b$ substituted by the large fraction ($\frac{pq}{...}$).
Attempts: I've plugged the equation into Matlab's solve feature, Mathematica's Reduce function (both to no avail), and Mathematica's Solve function , which returned this:
$x = \arccos{\sqrt{(-((a^2 p^2 pq^2)/(a^4 p^4 - 2 a^2 c^2 p^4 +
c^4 p^4 - 4 a^2 p^2 pq^2 - 2 a^4 p^2 q^2 +
4 a^2 c^2 p^2 q^2 - 2 c^4 p^2 q^2 + 4 a^2 pq^2 q^2 +
a^4 q^4 - 2 a^2 c^2 q^4 +
c^4 q^4)) + (c^2 p^2 pq^2)/(a^4 p^4 - 2 a^2 c^2 p^4 +
c^4 p^4 - 4 a^2 p^2 pq^2 - 2 a^4 p^2 q^2...}}$ (truncated, else it would continue for 21 total lines)
However, after substituting variables into the full expression above, I determined that it too returned false values (e.g. $\arccos(n)$ where $n>1$).
I was wondering if there is a way to solve this by hand or if it is impossible to do so? Thank you in advance.
|
EDIT: After the OP has been edited, there is no hope to find a closed form solution, since you are likely to obtain at least fourth degree powers in $\cos\theta$ when squaring :(
First try to isolate the $\theta$:
$$\frac{c^2-a^2}{pq}=\frac{1-2a\cos \theta}{\sqrt{q^2 \sin^2\theta+p^2\cos^2\theta}}$$
now take squares in both sides and substitute the $\sin\theta$:
$$\left(\frac{c^2-a^2}{pq}\right)^2=\frac{1-4a\cos \theta+4a^2\cos^2\theta}{q^2 \sin^2\theta+p^2\cos^2\theta}=\frac{1-4a\cos \theta+4a^2\cos^2\theta}{q^2 +(p^2-q^2)\cos^2\theta}$$
Let us call $k$ the left hand side for brevity, multiply:
$$kq^2+k(p^2-q^2)\cos^2\theta=1-4a\cos \theta+4a^2\cos^2\theta$$
and group:
$$\left(k(p^2-q^2)-4a^2\right)\cos^2\theta + 4a\cos \theta+(kq^2-1)=0$$
Can you take it from here? It is a standard second degree equation in $\cos\theta$.
|
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|
Expression for the element $(a+u)^{-1}$ in the extension field $G = F(u)$, where $u$ is a root of $x^{2}+px +q$ over a field $F$ Let $f = x^{2}+px+q$ be an irreducible polynomial over a field $F$ and let $G = F(u)$ be the extension of $F$ by a root $u$ of $f$. Every element of $G$ can be written in the normal form $a+bu$ where $a, b \in F$. I need to find an expression (in the normal form) for thee element $(a+u)^{-1}$ where $a \in F$.
This problem has become the bane of my existence over the past few days, as I have tried to model my answer off of an example given to us in our lecture notes, which I suspect is incorrect itself. So, I suppose part of my question here is asking whether the example problem is actually incorrectly solved.
The example problem says that in the case of the extension of $\mathbb{Q}$ by a root $u$ of the irreducible polynomial $f = x^{2} + x + 1$, to find an expression for $(a+bu)^{-1}$, we have to divide $f$ by $a+bx$ with remainder by using the Extended Euclidean Algorithm. The result of this is the following sequence of steps, which I have tried myself and found to be correct:
$$x^{2} + x + 1 = \left( \frac{1}{b}x\right)(a+bx) + \left( 1 - \frac{a}{b}\right)x + 1 \\ = \left( \frac{1}{b}x\right) (a+bx) + \frac{1}{b} \left( 1 - \frac{a}{b}\right)(a+bx) + \left( \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right) \\ = \left(\frac{1}{b}x+\frac{1}{b}-\frac{a}{b^{2}} \right)(a+bx)+\left( \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right)$$
My professor then proceeds to say that $\frac{a^{2}}{b^{2}}- \frac{a}{b} + 1 \neq 0$ because then a rational number $-\frac{a}{b}$ would be a root of $f$. Also, even though technically, one additional step needs to be completed in the Euclidean Algorithm, we already have a polynomial $h \in \mathbb{Q}[x]$ such that $h(a+bx)-1$ is a multiple of $f$.
So, he multiplies the equality through by $\left( \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right)^{-1}$ to obtain
$$ \left( \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right)^{-1}(x^{2}+x + 1) = \left( \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right)^{-1}\left(\frac{1}{b}x+\frac{1}{b}-\frac{a}{b^{2}} \right)(a+bx) + 1$$
and then takes $h = -\left( \frac{a^{2}}{b^{2}} - \frac{a}{b} + 1\right)^{-1}\left(\frac{1}{b}x+\frac{1}{b}-\frac{a}{b^{2}} \right) = \frac{1}{a^{2}-ab+b^{2}}(a-b-bx) $, so that $h(u) = \frac{1}{a^{2}-ab+b^{2}}(a-b-bu)$.
However, $h(u)(a+bu) \neq 1$ like he says it is: I multiplied it out myself to get $$ h(u) (a+ bu) = \frac{a-b-bu}{a^{2}-ab+b^{2}}(a+bu) \\ = \frac{a^{2}-ab-abu + abu -b^{2}u - b^{2}u^{2}}{a^{2}-ab+b^{2}} \\ = \frac{a^{2}-ab-b^{2}u-b^{2}u^{2}}{a^{2}-ab+b^{2}} \\ = \frac{a^{2}-ab-b^{2}(u+u^{2}}{a^{2}-ab+b^{2}} \neq 1 $$.
But, I can find absolutely no error in the process by which my professor found his function $h$! What am I missing here??
I need to know how to do that example problem in order to do the real problem I am trying to tackle, and that is finding $(a+u)^{-1}$ for $f = x^{2}+px+q$ over a general field $F$ (so, $p,q \in F$ in case you were wondering). I did a similar process to what my professor did, applying the extended Euclidean algorithm to divide $f$ by $a+x$:
$$x^{2}+px+q = x(a+x) + (p-a)(a+x) + q -(p-a)a \\ = (x+(p-a))(a+x)+(q-(p-a)a) $$
Like in the example problem, here the remainder $q - (p-a)a \neq 0$, because then otherwise $-a \in F$ would be a root of $f$ when we claimed it was irreducible over $F$ (a contradiction).
So, because of this, I know that the remainder has an inverse, and I can multiply the equality through by said inverse, $(q - (p-a)a)^{-1} $ to obtain
$$(q - (p-a)a)^{-1} (x^{2}+px+q) = (q - (p-a)a)^{-1}(x+(p-a))(a+x) + 1$$
Then, I took $$h = - (q - (p-a)a)^{-1}(x+(p-a)) \\ = \frac{-x-(p-a)}{q-(p-a)a} = \frac{a - p - x}{a^{2}-pa+q} $$
Then, letting $h(u) = \frac{a-p-u}{a^{2}-pa+q}$, we should have $h(u)(a+u) = 1$, but instead I get $h(u)(a+u) = \left(\frac{a-p-u}{a^{2}-pa+q} \right)(a+u) = \frac{a^{2}-pa - up +u^{2}}{a^{2}-pa+q}$, and not $1$!
What am I doing wrong??!
|
Looks like a sign error in the very last line.
The top multiplies out to be $a^2-pa-pu-u^2$, but then $-pu-u^2=q$, so that it is equal to the denominator.
In the other example you gave,
$$ \frac{a^{2}-ab-b^{2}u-b^{2}u^{2}}{a^{2}-ab+b^{2}} \\ = \frac{a^{2}-ab-b^{2}(u+u^{2})}{a^{2}-ab+b^{2}}=\frac{a^2-ab-b^2(-1)}{a^2-ab+b^2} =1 $$.
You didn't make any mistakes. You simply failed to use the relation $u^2+u+1=0$.
|
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"url": "https://math.stackexchange.com/questions/2394866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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|
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result that simplifies to an integer; (b) tried to find $a$ and $b$ such that $(a+\sqrt{b})^3=2+\sqrt{5}$ without success.
The answer stated for the problem in the original source (a local Math Olympiad Constest) is $x=1$.
|
You can solve these types of problems by denesting each radical and then simplifying the terms that cancel out. In general, we have$$\sqrt[n]{A+B\sqrt[m]C}=a+b\sqrt[m]{C}$$From which you raise both sides to the nth power and expand using Newton's binomial theorem.
In this case, we have$$\sqrt[3]{2\pm\sqrt5}=a\pm b\sqrt5$$
Cubing both sides, and simplifying, we get the following systems of equations$$\begin{cases}a^3+15ab^2=2\\\\3a^2b+5b^3=1\end{cases}$$
Cross multiplying, and dividing both sides by $b^3$, we get a cubic in $\tfrac ab$. With a rational root at $a/b=1$. Therefore, we see that $a=b$ and$$a^3+15a^3=2\iff 16a^3=2\iff a=b=\frac 12$$Therefore, your radical simplifies into$$\sqrt[3]{2\pm\sqrt5}=\frac 12\pm\frac {\sqrt5}2$$And I will leave the rest of the problem for the OP to finish.
|
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"url": "https://math.stackexchange.com/questions/2395060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Integral $\int\cos (2nx)\,\log \sin(x)\; dx $ Please help me out wrt this question -
$$\int_0^\frac{\pi}2 \cos (2nx)\; \log \sin(x)\; dx = -\frac {\pi}{4n}$$
Here $n>1$.
I tried doing integration by parts but then how to calculate
$$\int_0^\frac{\pi}2 \cot (x) \,\sin (2nx)\, dx $$
The question is given under Improper Integrals.
Thank you in advance. I really need to sort this out
|
A different approach.
By integrating by parts, one has
$$
\begin{align}
\int_0^{\Large \frac{\pi}2} \cos (2nx) \log \sin (x)\; dx&=\left[ \frac{\sin (2nx)}{2n}\cdot \log \sin (x)\right]_0^{\Large \frac{\pi}2} -\frac{1}{2n}\int_0^{\Large \frac{\pi}2} \sin (2nx)\: \frac{\cos (x)}{\sin (x)}\; dx
\\&=\color{red}{0}-\frac{1}{2n}\int_0^{\Large \frac{\pi}2} \sin (2nx)\: \frac{\cos (x)}{\sin (x)}\; dx. \tag1
\end{align}
$$Let
$$
u_n:=\int_0^{\Large \frac{\pi}2} \sin (2nx)\: \frac{\cos (x)}{\sin (x)}\; dx,\quad n\ge1.
$$ One may observe that, for $n\ge1$,
$$
\begin{align}
u_{n+1}-u_n&=\int_0^{\Large \frac{\pi}2} \left[\frac{}{}\sin (2nx+2x)-\sin(2nx)\right]\cdot \frac{\cos (x)}{\sin (x)}\; dx
\\&=\int_0^{\Large \frac{\pi}2} \left[2\frac{}{}\sin (x)\cdot \cos(2nx+x)\right]\cdot \frac{\cos (x)}{\sin (x)}\; dx\quad \left({\small{\color{blue}{\sin p-\sin q=2 \sin \frac{p-q}{2}\cdot \cos \frac{p+q}{2}}}}\right)
\\&=\int_0^{\Large \frac{\pi}2} 2\cdot\cos(2nx+x)\cdot \cos (x)\; dx
\\&=\int_0^{\Large \frac{\pi}2} \left[\frac{}{}\cos(2nx+2x)+\cos(2nx)\right] dx\qquad \quad \left({\small{\color{blue}{2\cos a \cos b= \cos (a+b)+ \cos (a-b)}}}\right)
\\\\&=\color{red}{0} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left({\small{\color{blue}{\sin(m\cdot \pi)=0,\, m=0,1,2,\cdots }}}\right)
\end{align}
$$ giving
$$
u_{n+1}=u_n=\cdots=u_1=2\int_0^{\Large \frac{\pi}2} \cos^2 (x)\; dx=\frac \pi2, \tag2
$$ then inserting $(2)$ in $(1)$ yields
$$
\int_0^{\Large \frac{\pi}2} \cos (2nx) \log \sin (x)\; dx=-\frac {\pi}{4n},\qquad n\ge1
$$
as wanted.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Partial fractions - integration $$\int \frac{4}{(x)(x^2+4)} $$
By comparing coefficients,
$ 4A = 4 $,
$A = 1$
$1 + B = 0 $,
$B= -1 $
$xC= 0 $,
$C= 0 $
where $\int \frac{4}{x(x^2+4)}dx =\int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 4}\right)dx$.
So we obtain $\int \frac{1}{x} - \frac{x}{x^2+4} dx$.
And my final answer is
$\ln|x| - x \ln |x^2 + 4| + C$.
However my answer is wrong , the answer is - $\ln|x| - \frac{1}{2} \ln |x^2 + 4| + C$.
Where have I gone wrong?
|
Your partial fraction expansion was correct.$$\frac 4{x(x^2+4)}=\frac 1x-\frac x{x^2+4}$$However, the error lies in integrating the second term of the right-hand side. To simplify$$\int\frac x{x^2+4}\, dx$$Make a substituting $z=x^2+4$. The derivative is $dz=2x\, dx$ so $x\, dx=dz/2$. Therefore, the integral transforms into$$\begin{align*}\int\frac {x\, dx}{x^2+4} & =\frac 12\int\frac {dz}{z}\\ & =\tfrac 12\log z+C\\ & =\tfrac 12\log(x^2+4)+C\end{align*}$$Hence, we have$$\boxed{\int\frac {4\, dx}{x(x^2+4)}=\log x-\tfrac 12\log(x^2+4)+C}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Obtain the value of $\int_0^1f(x) \ dx$, where $f(x) = \begin{cases} \frac {1}{n}, & \frac{1}{n+1}
Is the function Riemann integrable? If yes, obtain the value of $\int_0^1f(x) \ dx$
$f(x) =
\begin{cases}
\frac {1}{n}, & \frac{1}{n+1}<x\le\frac{1}{n}\\
0, & x=0
\end{cases}$
My attempt
$f$ is bounded and monotonically increasing on $[0,1]$. Also, $f$ has infinite discontinuities but only one limit point. Therefore $f$ is Riemann integrable. Now, to calculate the integration
$\int_0^1f(x) \ dx=\int_{1/2}^{1}1 \ dx + \int_{1/3}^{1/2}\frac{1}{2} \ dx + \int_{1/4}^{1/3}\frac{1}{3} \ dx+...$
$=\sum_{n=1}^\infty \frac{1}{n^2}-\frac{1}{n}+\frac{1}{n+1}$
How do I proceed from here? How do I calculate these summations? I know $\sum \frac{1}{n}$ is $\log 2$, but not the other two summations.
|
Since
$$
(0,1]=\bigcup_{n=1}^\infty\left(\dfrac{1}{n+1},\dfrac{1}{n}\right]=\bigcup_{n=1}^\infty I_n,
$$
and
$$
I_n\cap I_m=\emptyset\quad \forall m,n\in \mathbb{N},
$$
we have
\begin{eqnarray}
\int_0^1f(x)\,dx&=&\int_{[0,0]}f(x)\,dx+\int_{(0,1]}f(x)\,dx\\
&=&\sum_{n=1}^\infty \int_{I_n}f(x)\,dx\\
&=&\sum_{n=1}^\infty\dfrac{1}{n}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\\
&=&\sum_{n=1}^\infty\dfrac{1}{n^2}-\sum_{n=1}^\infty\dfrac{1}{n(n+1)}\\
&=&\sum_{n=1}^\infty\dfrac{1}{n^2}-\sum_{n=1}^\infty\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\\
&=&\dfrac{\pi^2}{6}-1
\end{eqnarray}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Prove or disprove : If $a\equiv b$ mod $m$, when $a,b,m\in \mathbb{Z}$ , then $a^3\equiv b^3$ mod $m^2$. Prove or disprove : If $a\equiv b$ mod $m$, when $a,b,m\in \mathbb{Z}$ , then
$a^3\equiv b^3$ mod $m^2$
My attempt:
since $a\equiv b$ mod $m$ then $a=b+mk$ for some $k$
Now $a^3=m^3k^3+b^3+3m^2k^2b+3mkb^2$
$a^3-b^3=m^3k^3+3m^2k^2b+3mkb^2$
from here I stuck
|
For every $\color{Red}{m \notin \{ \pm 1, \pm 3 \}} $ ,
we will construct a
$\color{Red}{\text{counter-example}}$!
Remark(I):
Let $\color{Red}{m \neq \pm 3}$,
and $\color{Red}{m \neq \pm 1}$;
in this case note that $m^2 \color{Red}{\nmid} 3m$.
Let $a:=m+1$ and $b:=1$;
then one can see easily that:
$$a^3-b^3=(m+1)^3-1^3=\color{Blue}{m^3+3m^2}+\color{Red}{3m}.$$
Notice that the above $a$ and $b$ are
$\color{Red}{\text{counter-example}}$.
[
Suppose on contrary that $a^3-b^3$ is divisible by $m^2$;
on the otherhadn note that
$\color{Blue}{\text{the first two terms}}$ are divisible by $m^2$,
so $\color{Red}{\text{the third term}}$ must divisible by $m^2$,
which has an obvious contradiction with remark(I).
]
Note that your assertion is $\color{Green}{\text{true}}$ when
$\color{Green}{m \in \{ \pm 1, \pm 3 \}} $.
Remark:
Let $m \in \{ \pm 1, \pm 3 \} $;
in this case
for every $b,k \in \mathbb{Z}$
we have:
$m^2 \color{Green}{\mid} 3mkb^2.$
Notice that $a \overset{m}{\equiv} b$
implies that there is an integer $k$,
such that $a=b+mk$,
as you noted, therefor we have:
$$a^3-b^3=\color{Blue}{m^3k^3}+\color{Blue}{3m^2k^2b}+\color{Green}{3mkb^2} ,$$
$\color{Blue}{\text{the the first two terms}}$ are divisible by $m^2$,
and by the remark(II)
$\color{Green}{\text{the third term}}$
is again divisible by $m^2$,
so we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2396640",
"timestamp": "2023-03-29T00:00:00",
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|
Find the area of the triangle defined in the figure provided Find the area of the triangle defined in the figure below
This question appeared in a math olympiad contest and was considered invalid later on without any specific reason. Is it solvable? If yes, provide the answer!
|
In $\triangle ABC$,
$|BC|=a$, $|CA|=b$, $|AB|=c$,
points $D,E,F$ are midpoints,
$|D_1D_2|=|E_1E_2|=|F_1F_2|=2\,R$.
Using the power of a points $D,E$ and $F$ with respect to
the circumscribed circle with radius $R$, we have
\begin{align}
\tfrac{a}2\cdot\tfrac{a}2&=2\cdot(2\,R-2)
,\\
\tfrac{b}2\cdot\tfrac{b}2&=1\cdot(2\,R-1)
,\\
\tfrac{c}2\cdot\tfrac{c}2&=3\cdot(2\,R-3)
,\\
a&=4\sqrt{R-1}
,\\
b&=2\sqrt{2\,R-1}
,\\
c&=2\sqrt{6\,R-9}
,\\
S&=\frac{a\,b\,c}{4\,R}
=\frac{4\sqrt3}R\,\sqrt{(R-1)(2\,R-1)(2\,R-3)}\tag{1}\label{1}
,\\
S&=\tfrac14\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}
=4\sqrt{R\,(2\,R-3)}\tag{2}\label{2}
.
\end{align}
From $\eqref{1}=\eqref{2}$
we have
\begin{align}
3\,(R-1)(2\,R-1)(2\,R-3)
&=
R^3
,
\end{align}
and as it was noted
in the comments, this results in a cubic equation
\begin{align}
R^3-6\,R^2+9\,R-3&=0 \tag{3}\label{3}
.
\end{align}
Using substitution $R=2\,t+2$ and dividing the resulting equation by 2,
we get
\begin{align}
4\,t^3-3\,t-\tfrac12&=0
.
\end{align}
Using identity
\begin{align}
4\,\cos^3\phi-3\,\cos\phi&=\cos3\phi
\end{align}
for $\cos3\phi=\tfrac12$,
\begin{align}
t&=\cos\phi
=\cos\left(
\tfrac13(\arccos\tfrac12+2\pi k)
\right)
=\cos
\frac{\pi\,(1+6 k)}9
,\quad k=0,1,2
.
\end{align}
Thus there are three real solutions for \eqref{3},
\begin{align}
R_0&=2+2\cos\tfrac\pi9\approx 3.87938524157182
,\\
R_1&=2+2\cos\tfrac{7\pi}9\approx 0.467911113762044
,\\
R_2&=2+2\cos\tfrac{13\pi}9\approx 1.65270364466614
,
\end{align}
and the only valid solution that
satisfies condition $R>3$
is $R=2+2\cos\tfrac\pi9$, which gives the area
\begin{align}
S&=4\sqrt{R\,(2\,R-3)}
=4\sqrt2\sqrt{(\cos\tfrac\pi9+1)(4\,\cos\tfrac\pi9+1)}
\approx 17.1865547357625
.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$
show that there exsit infinitely many postive integers triples $(x,y,z)$
such
$$(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$$
May try it is clear $(x,y,z)=(1,1,1)$ is one solution，and
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
|
Lemma($\color{Green}{\text{Vieta's formula}}$):
Let $\alpha_1$
be the root of the quadratic polynomial equation
$aY^2+bY+c=0$;
then we have: $\alpha_2= \color{Blue}{\dfrac{-b}{a}}-\alpha_1$.
Proof:
Only notice that
$\alpha_1 + \alpha_2 = \dfrac{-b}{a}$.
$
\color{Purple}
{
\text{Let's to look at one of the}
\ \
x, y, z
\ \
\text{as the}
} $
$\color{Red}{\text{variable}}$
$
\color{Purple}
{
\text {and to look at the others as}
} $
constants.
For example let's to look at $\color{Red}{y}$ as an indeterminate,
and to look at $x,z$ as constants;
as @user399601 has been done.
$$(x+\color{Red}{y}+z)^2+2(x+\color{Red}{y}+z)=5(x\color{Red}{y}+\color{Red}{y}z+zx)
\Longrightarrow
\\
\Bigg[
\color{Red}{y^2}
+
\big(2(x+z)\big)
\color{Red}{y}
+
(x+z)^2
\Bigg]
+
\Bigg[
2
\color{Red}{y}
+(x+z)
\Bigg] =
\Bigg[
5(x+z)
\color{Red}{y}
+ 5zx
\Bigg]
\Longrightarrow
\\
\color{Red}{y^2}
+
\Big(
2(x+z)
+
2
-5(x+z)
\Big)
\color{Red}{y}
+
\Big(
(x+z)^2
+
(x+z)
-5zx
\Big)
=0
\ \ \ \ \ \ \ \ \ \ \
\Longrightarrow
\\
\color{Red}{y^2}
+
\Big(
\color{Blue}{
2
-3(x+z)
}
\Big)
\color{Red}{y}
+
\Big(
(x+z)^2
+
(x+z)
-5zx
\Big)
=0
\ \ \ \ \ \ \
\color{Green}{\star\star\star\star}
$$
Suppose that
$\color{Red}{y}$
satisfies the polynomial equation
$\color{Green}{\star\star\star\star}$
;
then by
$\color{Green}{\text{Vieta's formula}}$
;
we can see that
:
$
\Big(
\color{Blue}{
3(x+z)
-2
}
- \color{Red}{y}
\Big)
$
will satisfies
$\color{Green}{\star\star\star\star}$
.
So we proved that:
If
$(x,\color{Red}{y},z)$ satisfies
$\color{Green}{\star\star\star\star}$ ;
then
$
(
x
,
\color{Blue}{
3(x+z)
-2
}
- \color{Red}{y}
,
z
)
$
will satisfies
$\color{Green}{\star\star\star\star}$
.
[
More specially if we let $x=1$ we have the following:
If
$(1,\color{Red}{y},z)$ satisfies
$\color{Green}{\star\star\star\star}$ ;
then
$
(
x
,
\color{Blue}{
3z
+
3
-2
}
- \color{Red}{y}
,
z
)
$
will satisfies
$\color{Green}{\star\star\star\star}$
.
]
$
\color{Purple}
{ \text {This method is called}}$
$
\color{Green}
{ \text {vieta-jumping}}$
For more information you can see here:
https://math.stackexchange.com/questions/tagged/vieta-jumping
https://en.wikipedia.org/wiki/Vieta_jumping#Constant_descent_Vieta_jumping
|
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|
Integration with logarithmic expression. I'm trying to solve an integration problem from the book which is the $$\int\frac{\sqrt{9-4x^2}}{x}dx$$ using trigonometric substitution. The answer from the book is $$3\ln\left|\frac{3-\sqrt{9-4x^2}}{x}\right|+\sqrt{9-4x^2}+C.$$
I have almost the same solution where there's a $$3\ln|\csc\theta-\cot\theta|+3\cos\theta+C.$$
The problem is when the substitution comes in. I end up having
$$3\ln\frac{|3-\sqrt{9-x^2}|}{2x}+\sqrt{9-x^2}$$ and when I tried to simplify it further, it resulted to $$3\ln\left|3-\sqrt{9-x^2}\right|-3\ln|2x|+\sqrt{9-x^2}.$$ I hope you could help me to tell where i did wrong.
By the way, I set $a=3$ and $x=\frac{3}{2}\sin\theta$.
|
I tried to follow your steps. Here we are assuming that $0<x\leq 3/2$. Let $x=3/2\sin(t)$ then $$3\cos(t)=3\sqrt{1-(2x/3)^2}=\sqrt{9-4x^2}$$ and
\begin{align*}\int\frac{\sqrt{9-4x^2}}{x}dx&=\int\frac{3\cos(t)}{3/2\sin(t)}d(3/2\sin(t))\\&
=3\int\frac{\cos^2(t)}{\sin(t)}dt=3\int\frac{1-\sin^2(t)}{\sin(t)}dt\\&=3\int\frac{dt}{\sin(t)}-3\int \sin(t)dt
\\&=
3\ln\frac{1-\cos(t)}{\sin(t)}+3\cos(t)+C\\
\\&=
3\ln\frac{3-3\cos(t)}{3\sin(t)}+3\cos(t)+C\\
&=3\ln\frac{3-\sqrt{9-4x^2}}{2x}+\sqrt{9-4x^2}+C\\
&=3\ln\frac{3-\sqrt{9-4x^2}}{x}+\sqrt{9-4x^2}+C'.
\end{align*}
where $C'=C-3\ln(2)$ is an arbitrary constant.
P.S. We have that for $\sin(t)>0$,
\begin{align*}
\int\frac{dt}{\sin(t)}&=\int\frac{1+\cos(t)-\cos(t)}{\sin(t)}\,dt\\
&=\int\frac{(1+\cos(t))(1-\cos(t))}{\sin(t)(1-\cos(t)}\,dt-\int\frac{\cos(t)}{\sin(t)}\,dt\\
&=\int\frac{\sin(t)}{1-\cos(t)}\,dt-\int\frac{\cos(t)}{\sin(t)}\,dt\\
&=\ln(1-\cos(t))-\ln(\sin(t))+C=\ln\left(\frac{1-\cos(t)}{\sin(t)}\right)+C
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding the quadratic equation from its given roots.
If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c
=0$ , then form an equation whose roots are:
$\alpha+\dfrac{1}{\beta},\beta+\dfrac{1}{\alpha}$
Now, using Vieta's formula,
For new equation,
Product of roots ($P$) = $\dfrac{a^2+c^2+2ac}{ac}$
Sum of roots ($S$) = $\dfrac{2c-b}{c}$
Hence,
the required equation is:
$acx^2 - ax(2c-b)+(a+c)^2=0$ // as quadratic equation = $x^2-Sx +Px=0$
But the answer key states that the answer is:
$acx^2 +b(a+c)x
+(a+c)^2=0$
I am really doubtful of this answer. Where have I gone wrong (or is the answer in the key wrong?)?
|
For an alternative approach, note that $\alpha \beta=c/a\,$, so the two new roots are $\alpha+1/\beta= \alpha(1+a/c)$ and $\beta+1/\alpha= \beta(1+a/c)$.
If $P(x)=ax^2+bx+c\,$ has roots $\alpha,\beta\,$, then the polynomial with roots $\lambda \alpha, \lambda\beta\,$ is:
$$
P\left(\frac{x}{\lambda}\right) = \frac{a}{\lambda^2} x^2 + \frac{b}{\lambda } x+ c
$$
For $\displaystyle \lambda =1+ \frac{a}{c} = \frac{a+c}{c}$ the above gives:
$$
\frac{ac}{(a+c)^2} x^2 + \frac{bc}{a+c} x+ c = \frac{1}{(a+c)^2}\left(ac x^2 + bc(a+c)x + (a+c)^2\right)
$$
|
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|
Proving $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$. How would I show $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$?
I tried starting from the LHS, and rationalising and what-not but I can't get the result...
Also curious to how they got the LHS expression from considering the right.
|
$$\sqrt{3+\sqrt{13+4\sqrt{3}}}= \sqrt{3+\sqrt{{(2\sqrt{3}+1)}^2}} = \sqrt{4+ 2\sqrt{3}} = \sqrt{{(\sqrt{3}+1)}^2} = \sqrt{3}+1.$$
|
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|
Infinite power summation where no common ratio can be seen Problem in book states to evaluate the sum:
$$\sum_{i=0}^\infty \frac{i^2}{4^i}$$
I can come up with the series S = $\frac{1^2}{4^1} + \frac{2^2}{4^2} + \frac{3^2}{4^3} + \frac{4^2}{4^4} + \frac{5^2}{4^5} + ...$
And 4S = $ 1 + \frac{2^2}{4^1} + \frac{3^2}{4^2} + \frac{4^2}{4^3} + \frac{5^2}{4^4} + \frac{6^2}{4^5} + ...$
If I subtract S from 4S I get 3S = $ 1 + \frac{3}{4^1} + \frac{5}{4^2} + \frac{7}{4^3} + \frac{9}{4^4} + \frac{11}{4^5} + ...$
I cannot find a common ratio to plug in to $\frac{1}{1-r}$ in order to figure this out. I asked my professor for some assistance, and he said the answer should be $\frac{20}{27}$, but I cannot figure out what steps he took to get that. He didn't explain it very well, and all other resources I've checked only show how to compute a geometric sum when there is a common ratio like $\frac{1}{2}$.
How do I determine a common ratio to figure out the sum? I tried division by sequential terms, and that gave me a common ratio of $\frac{5}{16}$, but that left me with an answer of S = $\frac{16}{27}$, which isn't correct.
What am I doing wrong? Or what am I missing?
|
$S=\sum_{i=1}^{\infty}\frac{i^2}{4^i} \rightarrow 4S=\sum_{i=1}^{\infty}\frac{i^2}{4^{i-1}}=1+\sum_{i=1}^{\infty}\frac{(i+1)^2}{4^i}$
Thus we have $3S=1+\sum_{i=1}^{\infty}\frac{2i+1}{4^i}=1+\frac{1}{3}+2\sum_{i=1}^{\infty}\frac{i}{4^i}$
$T=\sum_{i=1}^{\infty}\frac{i}{4^i} \rightarrow 4T=\sum_{i=1}^{\infty}\frac{i}{4^{i-1}}=1+\sum_{i=1}^{\infty}\frac{i+1}{4^i}$
Thus we have $3T=1+\sum_{i=1}^{\infty}\frac{1}{4^i}=\frac{4}{3}$, or $T=\frac{4}{9}$. Substituting this leads to $3S=\frac{20}{9}$, or $S=\frac{20}{27}$.
|
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|
$\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$: short proof? The identity $\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$ arises from a question on probability in my textbook. A proof by induction on $n$, which exploits the fact that $\binom{a}{b}+\binom{a}{b+1}=\binom{a+1}{b+1}$, is straightforward but not enlightening.
Is it possible to find any very clever approaches? Via a combinatorial or probabilistic interpretation, for instance?
|
Here is an answer based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g.
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\binom{n+k}{k}\frac{1}{2^{n+k}}}
&=\sum_{k=0}^n\binom{2n-k}{n-k}\frac{1}{2^{2n-k}}\tag{1}\\
&=\sum_{k=0}^\infty[z^{n-k}](1+z)^{2n-k}\frac{1}{2^{2n-k}}\tag{2}\\
&=2^{-2n}[z^n](1+z)^{2n}\sum_{k=0}^\infty\left(\frac{2z}{1+z}\right)^{k}\tag{3}\\
&=2^{-2n}[z^n](1+z)^{2n}\cdot\frac{1}{1-\frac{2z}{1+z}}\tag{4}\\
&=2^{-2n}[z^n](1+z)^{2n+1}\cdot\frac{1}{1-z}\tag{5}\\
&=2^{-2n}\sum_{k=0}^n[z^k](1+z)^{2n+1}\tag{6}\\
&=2^{-2n}\sum_{k=0}^n\binom{2n+1}{k}\tag{7}\\
&=2^{-2n}\frac{1}{2}2^{2n+1}\tag{8}\\
&\color{blue}{=1}
\end{align*}
Comment:
*
*In (1) we change the order of summation $k \rightarrow n-k$.
*In (2) we apply the coefficient of operator. We also set the limit to $\infty$ without changing anything since we are adding zeros only.
*In (3) we do a rearrangement and apply the formula $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (4) we apply the geometric series expansion.
*In (5) we do some simplifications.
*In (6) we do the Cauchy multiplication with the geometric series $\frac{1}{1-x}$ and restrict the upper limit of the sum with $n$ since other terms do not contribute to $[z^n]$.
*In (7) we select the coefficient of $z^k$.
*In (8) we apply the binomial theorem.
|
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|
Express $(\sqrt{3}+i)^{6-i}$ in the form $x+iy$ I have used that $2e^{i\pi /6}=\sqrt{3}+i $ and I have reached that $(\sqrt{3}+i)^{6-i}= 2^{6-i}(-e^{\pi /6}) $ but I do not know how to pass $2^{6-i}$ to the form $x + iy$, could anyone help me? Thank you.
|
$$\left( \sqrt { 3 } +i \right) ^{ 6-i }=\left( 2{ e }^{ \frac { \pi }{ 6 } i } \right) ^{ 6-i }={ 2 }^{ 6-i }{ e }^{ \pi i }{ e }^{ \frac { \pi }{ 6 } }=64{ e }^{ \frac { \pi }{ 6 } }\left( { e }^{ i\left( \pi -\ln { 2 } \right) } \right) =\\ =64{ e }^{ \frac { \pi }{ 6 } }\left( \cos { \left( \pi -\ln { 2 } \right) } +i\sin { \left( \pi -\ln { 2 } \right) } \right) =\\ =64{ e }^{ \frac { \pi }{ 6 } }\left( -\cos { \left( \ln { 2 } \right) } +i\sin { \left( \ln { 2 } \right) } \right) =\\ =-64{ e }^{ \frac { \pi }{ 6 } }\left( \cos { \left( -\ln { 2 } \right) } +i\sin { \left( -\ln { 2 } \right) } \right) $$
|
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|
Solve in positive integers, $ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$ Solve in positive integers, $$ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$$
My attempt :
$ x^3y^3+x^6-y^6+3xy(x^2-y^2)^2$
$=x^3y^3+3x^2y^2(x^2-y^2)+3xy(x^2-y^2)^2+(x^2-y^2)^3$
$=(xy+(x^2-y^2))^3$
$=(x^2+xy-y^2)^3 = 1$
so $x^2+xy-y^2 = 1$
Please suggest how to proceed.
Thank you, AmateurMathGuy and lhf.
Please check my work on Induction for Fibonacci sequence.
$$t_{k+2}-3t_{k+1}+t_{k}=0$$
$$y_{k+2}-3y_{k+1}+y_{k}=0$$
$(t_1,y_1)=(3,1)\to(x_1,y_1)=(1,1)$
$(t_2,y_2)=(7,3)\to(x_2,y_2)=(2,3)$
$(t_3,y_3)=(18,8)\to(x_3,y_3)=(5,8)$
$(t_4,y_4)=(47,21)\to(x_4,y_4)=(13,21)$
Since $1, 2, 3, 5, 8, 13, 21$ are in Fibonacci sequence, we predict that $(x_n,y_n)=(F_{2n-1},F_{2n})$
and will prove by Induction.
$(x_n,y_n)=(F_{2n-1},F_{2n})$ is true for $n=1, 2, 3, 4$
Suppose that $(x_k,y_k)=(F_{2k-1},F_{2k})$ is true,
Since $y_{k+1}-3y_k+y_{k-1}=0$, so $y_{k+1}=3y_k-y_{k-1}=3F_{2k}-F_{2k-2}$
$=3F_{2k}-(F_{2k}-F_{2k-1})=2F_{2k}+F_{2k-1}=F_{2k}+F_{2k+1}=F_{2k+2}$
$x_{k+1}=\frac{t_{k+1}-y_{k+1}}{2}=\frac{(3t_k-t_k)-3y_{k-1}-y_{k-1}}{2}$
$=3\left(\frac{t_k-y_k}{2}\right)-\left(\frac{t_k-1-y_{k-1}}{2}\right)=3x_k-x_{k-1}$
Similarly, $x_{k+1}=F_{2k+1}$, so $(x_n,y_n)=(F_{2n-1},F_{2n})$
Answer : $(x_n,y_n)=(F_{2n-1},F_{2n})$
|
It's obvious that $(1,1)$ is one of solutions.
Let $y-1=m(x-1)$ be an equation of the line which has other solutions.
Hence, $m\in\mathbb Q$ and after substitution in $x^2+xy-y^2=1$ we obtain:
$$x=\frac{m^2-2m+2}{m^2-m-1}$$ and
$$y=\frac{2m-1}{m^2-m-1}.$$
I hope it will help.
|
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|
Constrained Optimisation Problem: Confusion with Algebra/Contradiction I completed the following constrained optimisation problem:
Maximum and minimum values of $f(x,y) = x^2 + 2xy + y^2$ on the ellipse $g(x,y) = 2x^2 + y^2 - xy - 4x = 0$.
$\nabla f = \lambda \nabla g$
$\therefore (2x + 2y, 2x + 2y) = \lambda(4x - y - 4, 2y - x)$
We have a system of 3 equations in 3 unknowns:
$2x + 2y = \lambda(4x - y - 4)$,
$2x + 2y = \lambda(2y - x)$,
$2x^2 + y^2 - xy - 4x = 0$.
$\\$
$2x + 2y = \lambda(4x - y - 4)$
$\implies \lambda = \dfrac{2x + 2y}{4x - y - 4} \forall \ 4x - y - 4 \not= 0$
$\\$
$2x + 2y = \lambda(2y - x)$
$\implies \dfrac{2x + 2y}{2y - x} \forall \ 2y - x \not= 0$
$\\$
$\therefore \dfrac{2x + 2y}{4x - y - 4} = \dfrac{2x + 2y}{2y - x} \forall \ y \not= \dfrac{x}{2}, y \not= 4x - 4$ Remember this part: This is where my confusion lies.
But if we continue with our reasoning, we get the following (correct) points of potential maxima/minima:
$(x,y) = (0,0)$, $(x,y) = (1, -1)$, $(x,y) = (2,2)$, and $(x,y) = \left(\dfrac{2}{7}, \dfrac{-6}{-7}\right)$.
And, finally, using these points we find that $f_{max} = 16$ at $(x,y) = (2,2)$, and $f_{min} = 0$ at $(x,y) = (0,0)$ and $(x,y) = (1,-1)$.
But here's the problem: We had above that $y \not= \dfrac{x}{2}$; therefore, $(x,y) = (0,0)$ would get us $0 \not= \dfrac{0}{2} \implies 0 \not= 0$, which is obviously false. Also, as we just saw, $(0,0)$ is a solution for $f_{min}$!
The algebra here is confusing me, since we cannot have $y = \dfrac{x}{2}$, but $(0,0)$ is apparently the minima. Is this not a contradiction? Did I do something wrong? Or am I supposed to discard the solution $(0,0)$ and just have $f_{min} = 0$ at $(1, -1)$? What am I misunderstanding?
I would greatly appreciate it if people could please take the time to clarify this.
|
$(x+y)^2\geq0$ for all real variables.
The equality occurs for $x=y=0$ for example.
Thus, $0$ is a minimal value.
For $x=y=2$ we'll get a value $16$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$(x+y)^2\leq16$$ or
$$16-(x+y)^2\geq0$$ or
$$16-(x+y)^2+4(2x^2+y^2-xy-4x)\geq0$$ or
$$3(x-y)^2+4(x-2)^2\geq0,$$
which is obvious.
Done!
|
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|
Sequence : $f(f(n))+f(n+1)=2n$ Does there exist a function $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ such that $f(f(n))+f(n+1)=2n$, $\forall n \in \mathbb{Z}^+$ ?
My attempt :
Substitute $n=1$, $f(f(1))+f(2)=2$ so $f(f(1))=f(2)=1$
Substitute $n=2$, $f(f(2))+f(3)=4$ so $f(1)+f(3)=4$
Substitute $n=3$, $f(f(3))+f(4)=6$
Assume that there exists $k >1$ such that $f(k) \geq k$,
then $f(f(k-1))+f(k)=2k$, we have $f(f(k-1))<k$
|
For a positive integer $n$, let $P(n)$ denote the condition that $f\big(f(n)\big)+f(n+1)=2n$. The condition $P(1)$ implies that $f(2)=1$. Therefore, $P(2)$ leads to $f(1)\in\{1,2,3\}$. Hence, there are only three functions satisfying the condition, and they are listed below.
If $f(1)=1$, then $P(2)$ gives $f(3)=3$. Then, $P(3)$ gives $f(4)=3$. By induction on the positive integer $k$, we can easily see that
$$f(2k-1)=2k-1\text{ and }f(2k)=2k-1$$
for every $k=1,2,3,\ldots$. In other words,
$$f(n)=2\,\left\lfloor\frac{n}{2}\right\rfloor+1$$
for all $n\in\mathbb{Z}_{>0}$.
If $f(1)=2$, then $P(2)$ implies $f(3)=2$. Then, $P(3)$, $P(4)$, and $P(5)$ lead to $f(4)=5$, $f(5)=4$, and $f(6)=5$. It can be proven again by induction that
$$f(3k-2)=3k-1\,,\,\,f(3k-1)=3k-2\,,\text{ and }f(3k)=3k-1$$
for all $k=1,2,3,\ldots$. In other words,
$$f(n)=3\,\left\lceil\frac{n}{3}\right\rceil-1-\left\lfloor\frac{n\mod 3}{2}\right\rfloor$$
for $n=1,2,3,\ldots$.
If $f(1)=3$, then $P(1)$ gives $f(3)=1$. Then, $P(3)$, $P(4)$, and $P(5)$ yield $f(4)=5$, $f(5)=4$, and $f(6)=5$, respectively. We claim that, for all positive integers $k$,
$$f(3k+1)=3k+2\,,\,\,f(3k+2)=3k+1\,,\text{ and }f(3k+3)=3k+2\,.$$
However, this claim can be easily verified by induction on $k$. Thus,
$$f(n)=3\,\left\lceil\frac{n}{3}\right\rceil-1-\left\lfloor\frac{n\mod 3}{2}\right\rfloor$$
for $n=4,5,6,\ldots$, with $f(1)=3$, $f(2)=1$, and $f(3)=1$.
|
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|
Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\left\lceil{\frac{1}{n^2}}\right \rceil = 5 - 4$$
Therefore, for $n \in \mathbb R$, where $n \neq 0$
$$4 < \left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil \le 5$$
$$4 < \left\lceil{4}+\frac{1}{n^2}\right \rceil \le 5$$
$$0 < \left\lceil\frac{1}{n^2}\right \rceil \le 1$$
I feel kind of without purpose once I hit point. Should I be scrapping the inequality and solving for $n$ instead?
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For any $n\in\mathbb{N}$ we have that
$$4=\dfrac{4n^2}{n^2}<\color{red}{\dfrac{4n^2+1}{n^2}}\le\dfrac{4n^2+n^2}{n^2}=\dfrac{5n^2}{n^2}=5.$$ So, we have that
$$\left\lceil \frac{4n^2+1}{n^2} \right\rceil=5$$ and
$$\left\lfloor \frac{4n^2+1}{n^2} \right\rfloor=4.$$
|
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|
Find all of the points of the form $(x, −1)$ which are $4$ units from the point $(3, 2)$
Find all points of the form $(x, −1)$, which are $4$ units from the point
$(3, 2)$.
I understand the distance formula, I think. Also, I don't know how to format for math on here yet so I apologize for that.
\begin{align}
&\text{dist} = 4 = \sqrt{(x-3)^2 + (2-(-1))^2)} \\
&\implies 16 = (x-3)^2 + (2 + 1)^2 \tag{square both sides} \\
&\implies 16 = x^2 -6x + 18 \tag{expand} \\
&\implies 0 = x^2 -6x +2 \tag{zero on the left}
\end{align}
Next should be finding the factors which would give me the answers but I have no idea what would work for this.
$$(x + \text{something})(x - \text{something}).$$
I must be doing something wrong or missing something. Please correct me.
|
Your computations are correct up to the factoring
$$ (x+\text{something})(x-\text{somethign}).$$
In general, there is no reason to expect that your polynomial will factor like a difference of squares. In fact, the presence of a linear term (the $-6x$) indicates that this won't work. Instead, you need to either factor the polynomial somehow, or solve the quadratic using some other technique. In this case, completing the square seems reasonable.
Every quadratic polynomial can be written in the form
$$ A(x-h)^2 + k, $$
where $A$, $h$, and $k$ are real numbers. Hence we have
$$x^2 - 6x + 2
=A(x-h)^2 + k
= Ax^2 - 2Ahx + Ah^2 + k. \tag{1}$$
Since two polynomials are equal to each other if and only if they have the same coefficients, we can equate the coefficients on the left with the coefficients on the right in order to obtain
$$ \begin{cases}
1 = A, \\
-6 = -2Ah, \\
2 = Ah^2 + k.
\end{cases} $$
Solving this system, we immediately have that $A = 1$. Substituting this into the second equation, we get $-6 = -2h$, which implies that $h = 3$. Finally, plugging these into the last equation, we get $2 = 3^2 + k$, which implies that $k = -7$. Putting these back into (1), we get
$$ x^2 - 6x + 2 = (x-3)^2 - 7.$$
Since we are looking for the roots of this polynomial, we set it equal to zero, then solve by extracting roots. This gives us
\begin{align}
0 = (x-3)^2 - 7
&\implies (x-3)^2 = 7 \\
&\implies x-3 = \pm \sqrt{7} \tag{extract the roots} \\
&\implies x = 3\pm \sqrt{7}.
\end{align}
Hence the two points on the line $x=-1$ that are a distance of 4 from the point $(3,2)$ are the points
$$\left( 3+\sqrt{7}, -1 \right)
\qquad\text{and}\qquad
\left( 3-\sqrt{7}, -1 \right).$$
|
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Alternative method to find this infinite sum?
Find where m is an integer:
$$S = 1+\sum_{n=1}^{\infty}\frac{n!}{(mn)!}$$
When we rewrite this sum as one plus a function of $x$ and multiply
the $n^{th}$ term of the series by $x^{mn-1}$, the resulting function
solves for the differential equation $$\frac{d^{m-1}f(x)}{dx^{m-1}} = \frac1m + \frac1{xf(x)}$$ and hence we can replace $f(x)$ with $s(x)$ and set
$s(0) = 1$. Then use that to find $s(1)$ which is $S$.
I'm looking for an alternative method because I'm having trouble dealing with this differential equation.
|
Considering $$S_m = \sum_{n=1}^{\infty}\frac{n!}{(mn)!}$$ beside the case $m=2$ for which $$S_2=\frac{1}{2} \sqrt[4]{e} \sqrt{\pi } \text{erf}\left(\frac{1}{2}\right)$$ as Patrick Stevens commented, we are left with hypergeometric functions which apparently do not simplify at all.
They show quite nice patterns
$$S_3=\frac{1}{3!} \, _1F_2\left(1;\frac{4}{3},\frac{5}{3};\frac{1}{3 ^3}\right)$$
$$S_4=\frac{1}{4!} \,
_1F_3\left(1;\frac{5}{4},\frac{6}{4},\frac{7}{4};\frac{1}{4^4}\right)$$
$$S_5=\frac{1}{5!} \,
_1F_4\left(1;\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5};\frac{1}{5^5}\right)$$
$$S_6=\frac{1}{6!} \,
_1F_5\left(1;\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6};\frac{
1}{6^6}\right)$$
$$S_7=\frac{1}{7!}\,
_1F_6\left(1;\frac{8}{7},\frac{9}{7},\frac{10}{7},\frac{11}{7},\frac{12}{7},\frac{13}{7};\frac{1}{7^7}\right)$$
In fact, it seems that the asymptotics is just $S_m\sim\frac{1}{m!}$.
|
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Solve this trigonometric equation ${\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0$
Solve the following equation. $${\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0.$$
I can't figure out the way to solve this equation. This was my attempt
\begin{array}{l}
{\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0\\
\Leftrightarrow \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - 3(\frac{{\sin x}}{{\cos x}} - \frac{{\cos x}}{{\sin x}}) = 0\\
\Leftrightarrow \frac{{{{\sin }^4}x + {{\cos }^4}x}}{{{{\cos }^2}x + {{\sin }^2}x}} - 3\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}} = 0\\
\Leftrightarrow \frac{{1 + 2{{\sin }^2}x{{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} + 3\frac{{\cos 2x}}{{\sin x\cos x}} = 0\\
\Leftrightarrow \frac{4}{{(1 - \cos 2x)(1 + \cos 2x)}} + 2 + 3\frac{{\cos 2x}}{{\sin x\cos x}} = 0\\
\Leftrightarrow \frac{4}{{1 - {{\cos }^2}2x}} + 2 + 3\frac{{\cos 2x}}{{\sin x\cos x}} = 0
\end{array}
I don't know how to solve after that. Can anyone help me?
|
Let $\tan{x}-\cot{x}=t$.
Thus, $$t^2+2-3t=0.$$
For $t=1$ we obtain
$$\tan^2x-\tan{x}-1=0,$$
which gives $x=\arctan\frac{1\pm\sqrt5}{2}+\pi k$, where $k\in\mathbb Z$.
For $t=2$ we obtain
$$\tan^2x-2\tan{x}-1=0,$$ which gives
$x=\arctan(1\pm\sqrt2)+\pi k$, where $k\in\mathbb Z$.
Actually, $\arctan(1+\sqrt2)=\frac{3\pi}{8}$ and $\arctan(1-\sqrt2)=-\frac{\pi}{8}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability distribution: finding 'a' Consider the random variable X with probability distribution,
$$
P(x) =
\begin{cases}
x^2/a & -2,-1,0,1,2 \\
0 & \text{otherwise}
\end{cases}
$$
a) Find $a$ and $E(X)$.
b) What is the probability distribution of random variable $Z = (X - E[X])^2$?
c) Using part b), compute the variance of $X$.
I think I understand parts b, c, and the second part of a, but I'm not sure how to find $a$. Maybe there's a formula I'm missing?
a) a = 10, E(X) = 0
b) $$P(Z = z) =
\begin{cases}
0 & z = 0 \\
\frac{2}{10} & z = 1 \\
\frac{8}{10} & z = 2
\end{cases}$$
alt. b) $$P(Z = z) =
\begin{cases}
\frac{2}{5} & z = 1 \\
\frac{4}{5} & z = 4 \\
0 & otherwise
\end{cases}$$
|
a) a = 10, E(X) = 0
Yes. $
{~\\{P_X(x) =
\begin{cases} 0/a^2 & : & x=0 \\ 1/a^2 &:& x\in\{-1,1\} \\ 4/a^2 & : & x\in\{-2,2\}\\ 0 & : & \text{otherwise}
\end{cases} }\\~\\ {\therefore ~a=10, \mathsf E(X)=0}}
$
b) ${P(Z = z) = \begin{cases}
\frac{1}{5} & z = 0 \\
\frac{2}{5} & z = 1,4
\end{cases}}$
No, but close. Try again.
Having found $a=10$ and $\mathsf E(X)=0$ you know know that $Z=X^2$ and can find $P_Z(z)= P_X(+\surd z)+ P_X(-\surd z)$ since: $$
P_X(x) =
\begin{cases} 0 & : & x=0 \\ 1/10 &:& x\in\{-1,1\} \\ 4/10 & : & x\in\{-2,2\}\\ 0 & : & \text{otherwise}
\end{cases}
$$
Therefore ...
c) Hint $\mathsf {Var}(X)=\mathsf E((X-\mathsf E(X))^2)$ and $Z=(X-\mathsf E(X))^2$ so...
|
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|
Find all triples $(a, b, c) \in \mathbb{R} : ab + c = bc + a = ca + b = 4$ Any help on this question would be great. Ive found that (1,1,3), (3,1,1), (1,3,1) but am confused on how to find more anymore (if there are any).Any help would be appreciated.
|
$ab + c = bc + a = ca + b=4$
By symmetry any solution for any any one of $a,b,c$ can be a solution for any other.
$ab +c = bc +a$ means $ab - bc = b(a-c) = a-c$ So either $b=1$ or $a=c$.
So by symmetry:
if $a\ne 1$ then $b=c$ and if $b\ne 1$ then $a = c = b$. So it is not possible that only one of $a,b,c$ is equal to $1$.
So either:
1) All $a,b,c$ equal $1$
2) None of $a,b,c$ equal $1$ and $a=b=c$.
3) Two of $a,b,c$ equal $1$ and the third does not.
....
1) $1*1 + 1 = 4$ is not possible.
2)we get $a^2 + a = 4$ and $a = \frac {-1 \pm \sqrt{17}}{2}$.
3) wolog $a= b = 1$ so $1 + c = c + 1 = c + 1 = 4$. $c = 3$.
So there are 5 solutions:
And $(a,b,c) = (1,1,3)$, $(1,3,1)$ , $(3,1,1), ( \frac {-1+ \sqrt{17}}2,\frac {-1+ \sqrt{17}}2,\frac {-1+ \sqrt{17}}2)$ or $ ( \frac {-1- \sqrt{17}}2,\frac {-1- \sqrt{17}}2,\frac {-1- \sqrt{17}}2)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2420692",
"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.