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How to find the sum: $\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) $ $$S_n=\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) = \frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{r}\right)$$
I fail to see how this:
$$S_n = \frac{1}{r-s}\left( \frac{1}{s+1}-\frac{1}{r-1}+\frac{1}{s+2}-\frac{1}{r+2}+\cdots\right)$$
converges to the right answer. Can someone please give a detailed explanation?
$r,s$ are integers and $r>s>0$
|
Let $r=s+k$. Then:
$$\begin{align}S_n&=\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) = \\
&=\frac1{r-s}(\ \boxed{\color{black}{\frac1{s+1}}}-\color{red}{\frac1{s+k+1}}+\color{black}{\boxed{\frac1{s+2}}}-\color{blue}{\frac1{s+k+2}}\color{black}{+\cdots +\boxed{\frac1{s+k}}}-\color{green}{\frac1{s+2k}}+\\
&\ \ \ \ +\color{red}{\frac1{s+k+1}}-\color{#FF00FF}{\frac1{s+2k+1}}+\color{blue}{\frac1{s+k+2}}-\color{#6495ED}{\frac1{s+2k+2}}+\cdots +\color{green}{\frac1{s+2k}}-\color{red}{\frac1{s+3k}}+\\
&\ \ \ \ +\color{#FF00FF}{\frac1{s+2k+1}}\color{black}{-\frac1{s+3k+1}}+\color{#6495ED}{\frac1{s+2k+2}}\color{black}{-\frac1{s+3k+2}+\cdots +}\color{red}{\frac1{s+3k}}\color{black}{-\frac1{s+4k}+\cdots})=\\
&=\frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{s+k}\right)=\\
&=\frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{r}\right).\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3229113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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|
Simplifying $\frac{1}{x^4+4x^2}$ I'm trying to solve this problem on my own and it involves simplifying the expression in the title.
In the solutions it says it's this:
$$\frac{1}{x^4+4x^2} = \frac{1}{4}\Biggl[\frac{1}{x^2}-\frac{1}{x^2+4}\Biggr]$$
But I can't for the life of me figure out where the 1/4 came from. Or exactly how they got to this answer...
|
We can write the fraction as,
$$\frac{1}{x^4 +4x^2} = \frac{1}{4}\cdot \frac{4}{x^2(x^2+4)}$$
The fraction $\frac{4}{x^2(x^2+4)}$ can be written as..
\begin{align}
\frac{1}{4}\cdot \frac{4}{x^2(x^2+4)}=\frac{1}{4}\Biggl[\frac{(x^2+4)-x^2}{x^2(x^2+4)}\Biggr]\\
=\frac{1}{4}\Biggl[\frac{1}{x^2}-\frac{1}{x^2+4}\Biggr]
\end{align}
Understand?
|
{
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"url": "https://math.stackexchange.com/questions/3231405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to prove De Moivre's theorem inductively It is given that $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ where $n\in Z^+$.
I can show it works for $n=1$ but I am stuck in showing it inductively. I have got as far as below but are stuck in the rearranging:
$$(\cos\theta + i\sin\theta)^{k+1}= (\cos k\theta + i\sin k \theta)(\cos\theta + i\sin\theta)$$
How can I rearrange the left to look like the right?
|
$$(\cos \theta + i\sin \theta)^k = \cos k\theta + i\sin k\theta$$
$$\text{So for } (\cos \theta + i\sin \theta)^{k+1} = (\cos \theta + i\sin \theta)^k \cdot (\cos \theta + i\sin \theta) = (\cos k\theta + i\sin k\theta)\cdot (\cos \theta + i\sin \theta)$$
$$=\cos k\theta\cos\theta +i\sin\theta\cos k\theta +i\sin k\theta\cos\theta +i^2\sin{k\theta}\sin\theta$$
$$=(\cos{k\theta}\cos\theta - \sin k\theta\sin\theta)+i(\sin\theta\cos k\theta +\sin k\theta\cos\theta)$$
$$=\cos(k+1)\theta + i\sin(k+1)\theta$$
Hence, this proves that $(\cos\theta +i\sin\theta)^n = \cos n\theta + i\sin n\theta$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3231547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Eliminate $\theta$ from from $\frac{x-k\sin \theta \cot \alpha}{k \cos \theta}=\frac{y-k\cos \theta \tan \alpha}{k \sin \theta}=\frac{z}{c}$ I am stuck with the following problem :
Eliminate $\theta$ from
$$\frac{x-k\sin \theta \cot \alpha}{k \cos \theta}=\frac{y-k\cos \theta \tan \alpha}{k \sin \theta}=\frac{z}{c}$$
to get the following result:
$$k^2(c^2-z^2)^2=c^2\left(\,(cx \tan \alpha-yz)^2+(cy \cot \alpha-zx)^2\,\right)$$
I will be grateful if someone explains the problem which was actually the part of another problem . Thanks in advance for your time.
|
Write your line as
$$
\frac{x-k\sin\theta\cot\alpha}{k\cos\theta}=\frac{z}{c}\quad\text{and}\quad\frac{y-k\cos\theta\tan\alpha}{k\sin\theta}=\frac{z}{c}.
$$
Manipulate the two equations into
\begin{align*}
cx/k&=c\cot\alpha\sin\theta+z\cos\theta\\
cy/k&=z\sin\theta+c\tan\alpha\cos\theta
\end{align*}
so solving for $\sin\theta,\cos\theta$:
\begin{align*}
k(c^2-z^2)\sin\theta &=c\tan\alpha\cdot cx-z\cdot cy\\
k(c^2-z^2)\cos\theta &=c\cot\alpha\cdot cy-z\cdot cx
\end{align*}
Now remember $\sin^2\theta+\cos^2\theta=1$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Pythagorean Quadruples and Stereographic Projection I am trying to solve Diophantine equation $a^2+b^2+c^2=d^2$ by transforming this equation, assuming $d \neq 0$, into a sphere $ (\frac{a}{d})^2 + (\frac{b}{d})^2 +(\frac{c}{d})^2 = x^2 + y^2 +z^2 =1$ and using stereographic projection of this sphere from $N = (0,0,1)$ onto $ {x,y}$ plane.
Each point $P^{'}=(x_{0},y_{0},0)$ from $x,y$ plane is a projection of a point $P=(x,y,z)$ on the sphere, where $x=\frac{2x_{0}}{1+x_{0}^2+y_{0}^2}$, $y=\frac{2y_{0}}{1+x_{0}^2+y_{0}^2}$ and $z=\frac{x_{0}^2+y_{0}^2-1}{1+x_{0}^2+y_{0}^2}$. Using $P^{'} \in \mathbb{Q^2} \iff P \in \mathbb{Q^3}$ we know that for each rational point from the plane we can obtain a solution. Therefore for $x_{0}=\frac{m}{n}$ and $y_{0}=\frac{p}{q}$ such that $m,n,p,q \in \mathbb{N}$ and $GCD(m,n)=GCD(p,q)=1$ we get $$x=\frac{m^2q^2+p^2n^2-n^2q^2}{n^2p^2+m^2q^2+n^2q^2}, y=\frac{2mnq^2}{n^2p^2+m^2q^2+n^2q^2}, z=\frac{2pn^2q}{n^2p^2+m^2q^2+n^2q^2}$$
Then multiplying by $n^2p^2+m^2q^2+n^2q^2$: $$a=m^2q^2+p^2n^2-n^2q^2, b=2mnq^2, c=2pn^2q, d=n^2p^2+m^2q^2+n^2q^2$$
I want to find all primitive quadruples, so $GCD(a,b,c,d)$ must be equal to one, however, at this point I am not sure how to continue. I've tried to compute $GCD(2pn^2q,2mnq^2)=2nq \ GCD(pn,mq)$, unsuccessfully.
|
You are on the right track; indeed you have $\gcd(b,c)=2nq\gcd(np,mq)$, and also
$$\gcd(a,d)=\gcd(d,d-a)=\gcd(n^2p^2+m^2q^2+n^2q^2,2n^2q^2),$$
which shows that $\gcd(a,b,c,d)$ divides $2n^2q^2$. It follows that
\begin{eqnarray*}
\gcd(a,b,c,d)&=&\gcd(d,d-a,\gcd(b,c))=\gcd(d,2n^2q^2,2nq\gcd(np,mq)),
\end{eqnarray*}
where in turn
$$\gcd(2n^2q^2,2nq\gcd(np,mq)),=2nq\gcd(nq,\gcd(np,mq))=2nq\gcd(nq,np,mq).$$
By assumption $\gcd(m,n)=\gcd(p,q)=1$ and so
$$\gcd(nq,np,mq)=\gcd(n,mq)=\gcd(n,q),$$
which means that
$$\gcd(a,b,c,d)=\gcd(d,2nq\gcd(n,q))=\gcd(n^2p^2+m^2q^2,2nq\gcd(n,q)).$$
Now if $r$ is an odd prime dividing $\gcd(a,b,c,d)$ then it divides $2nq\gcd(n,q)$ and hence it divides either $n$ or $q$, or both. If $r$ divides only $n$, then it does not divide $n^2p^2+m^2q^2$ a contradiction. Similarly, if $r$ divides only $q$ then it does not divides $n^2p^2+m^2q^2$, a contradiction. This shows that the odd prime factors of $\gcd(a,b,c,d)$ divide both $n$ and $q$, and hence $\gcd(n,q)$.
If $\gcd(a,b,c,d)$ is even then $n^2p^2+m^2q^2$ is even, meaning that either $n$ and $q$ are even, or $m$ and $p$ are even.
It is clear that $\gcd(n,q)$ divides $a$, $b$, $c$ and $d$ and therefore also $\gcd(a,b,c,d)$. It follows that $\gcd(a,b,c,d)=1$ if and only if $\gcd(n,q)=1$ and $2\nmid\gcd(m,p)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Linear Algebra - How to find the axis of a parabola? Given the parabola $x^2+y^2-2xy+4x=0$ defined in $\mathbb{R}^2$, how can I find the axis?
The matrices associated to the curve are:
$$A=\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix}, B=\begin{bmatrix}2 \\ 0\end{bmatrix}, C=\begin{bmatrix}1 & -1 & 2\\ -1 & 1 & 0\\ 2 & 0 & 0\end{bmatrix}$$
My textbook says that we need to compute the characteristic polynomial $P_A(\lambda)=\lambda(\lambda-2)$ of the matrix A and then find the eigenspaces associated to the respective eigenvalues, but I can't understand why we would do that.
|
Equation of a parabola can be written as $L_1^2= 4 A L_2$, provided $L_1$ and $ L_2$ are equations of lines which are non-parallel.
Further if $L_1$ and $L_2$ are perpendicular and normalized, then $4A$ is the length of
latus rectum. Equation of axis is $L_1=0$, Equation of tangent at vertex is $L_2=0$. The focus is given by $(L_2=A, L_1=0)$. The equation the directrix is $L_2=-A$. The equation of the latus rectum is $L_2=A$. A line is called normalized if written as $\frac{ax+by+c}{\sqrt{a^2+b^2}}.$
So in this problem as done by Santos above we have $(x-y+1)^2=(1-x-y)$. We re-write it as:
$$ \left ( \frac {x-y+1}{\sqrt{2}}\right)^2=\frac{1}{\sqrt{2}} \left(\frac{1-x-y}{\sqrt{2}}\right)$$
So the equation of axis is $$\frac{x-y+1}{\sqrt{2}}=0.$$ Tangent at vertex is $$\frac{1-x-y}{\sqrt{2}}=0.$$
The directrix is $$\frac{1-x-y}{\sqrt{2}}=- ~\frac{1}{4\sqrt{2}}.$$
The length of the latus rectum of the parabola is $1/\sqrt{2}$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of powers of $3\times3$ matrix Consider the matrix
$$A = \begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$$
What is $\lim_{n→\infty}$$A^n$ ?
A)$\begin{bmatrix} 0 & 0 & 0\\ 0& 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
B)$\begin{bmatrix} \frac{1}{4} &\frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\end{bmatrix}$
C)$\begin{bmatrix} \frac{1}{2} &\frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\end{bmatrix}$
D)$\begin{bmatrix} 0 &\frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}$
E) The limit exists, but it is none of the above
The given answer is D). How does one arrive at this result?
|
If you are in $1$, you have same probability to stay there or to pass to $2$, but no way to get back from there. Thus you are finally drifting to $2$.
States $2$ and $3$ are symmetrical: at long they will tend to be equally populated, independently of the starting conditions.
Therefore also starting from $1$ you will at long be split between $2$ and $3$.
Thus the answer is D).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3238042",
"timestamp": "2023-03-29T00:00:00",
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|
Compute convolution $f*g(x)$ such that : $f,g\in L^{p}$ Question :
Compute convolution : $f*g(x)$
$f(x)=\begin{cases}3x^2 & \text{ if } |x|\leq4 \\0 & \text{ otherwise}\end{cases}$
$g(x)=\begin{cases}1 &\text{ if }|x|\leq 2 \\0 & \text{ otherwise}\end{cases}$
My try :
\begin{align}
f*g(x)&=\int f(x-y)g(y)dy\\
&=\int_{[-4,4]}3(x-y)^{2}1_{[-2,2]}(x-y)dy\\
&=\int_{[-4,4]∩[x-2,x+2]}3(x-y)^{2}dy
\end{align}
Now, how I find or discussed with this?
$[-4,4]\cap[x-2,x+2]=?$
Please, give me ideas and method to approach it.
|
Your bounds aren't quite right. We have $$f(x-y) = 3(x-y)^2\cdot\mathsf 1_{[-4,4]}(x-y) $$
and $g(y) = \mathsf 1_{[-2,2]}(y)$, so we have the following inequalities for $y$:
\begin{align}
x-4&\leqslant y\leqslant x+4\\
-2&\leqslant y\leqslant 2.
\end{align}
For $-6\leqslant x\leqslant -2$ we have
$$
f\star g(x) = \int_{-2}^{x+4} 3(x-y)^2\ \mathsf dy = (x+6)(x^2+12).
$$
For $-2\leqslant x\leqslant 2$ we have
$$
f\star g(x) = \int_{-2}^2 3(x-y)^2\ \mathsf dy = 12x^2+16.
$$
For $2\leqslant x\leqslant 6$ we have
$$
f\star g(x) = \int_{x-4}^{2} 3(x-y)^2\ \mathsf dy = (-x+10)(x^2-8x+28).
$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Does dividing a common factor out from numerator and denominator of a rational function create a new function with different domain? Suppose a function is defined as $ f(x) =\frac{x^2 - 9}{x-3}. $
If we divide the common factor $ x-3 $ from both the numerator and denominator :-
$$ \frac{x^2 - 9}{x - 3} \\ = \frac{(x+3)(x-3)}{x-3} \\ = x+3 $$
1) Is $ x+3 $ a different function from the original $ \frac{x^2 - 9}{x - 3} $ ?
2) Since $ \frac{x^2 - 9}{x - 3} $ does not have 3 in its domain, when we simplify it to become $x + 3$, does it now have 3 in its domain ?
|
$x+3$ is different from $\frac{x^2-9}{x-3}$, but only in the domain. All the simplification does is to add $x=3$ to the domain.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why do we take the positive square root only from $\sqrt{a^2-x^2}$ when integrating using trig substitutions? $$\int\frac{\mathrm dx}{x^2 \sqrt{16-x^2}}$$
when substituting $\,x=4\sin\theta ,\;\mathrm dx=4\cos\theta\, \mathrm d\theta,\,$ it becomes
$$\int\frac{4\cos\theta\, \mathrm d\theta}{4^2\sin^2\theta\sqrt{16-4^2\sin^2\theta}}$$
then
$$\frac{1}{4}\int\frac{\cos\theta \,\mathrm d\theta}{\sin^2\theta\times4\sqrt{1-\sin^2\theta}}$$
The question is why did we take the positive root of 16 and took it out? as $\sqrt{16} \;is \pm 4 $
|
With $x=4\sin\theta$ your integral becomes $\int\frac{4\cos\theta d\theta}{64\sin^2\theta|\cos\theta|}=\frac{1}{16}\int\frac{\operatorname{sgn}(\cos\theta)d\theta}{\sin^2\theta}$ with $\operatorname{sgn}y:=\frac{y}{|y|}$. Luckily, choosing $\theta$ to obtain $\sin\theta=\frac{x}{4}$ lets us arbitrarily choose whether $\cos\theta$ is $\sqrt{1-\frac{x^2}{16}}$ or $-\sqrt{1-\frac{x^2}{16}}$, so we can choose a constant $\operatorname{sgn}(\cos\theta)$ factor. Since $\sqrt{16-x^2}$ is real only if $|x|\le4$, the most convenient choice is $\theta=\arcsin\frac{x}{4}\in\left[-\frac{\pi}{2},\,\frac{\pi}{2}\right]$ so, on this half-period of $\theta$, $\theta$ is a monotonic function of $x$ with $\cos\theta\ge0$.
|
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"timestamp": "2023-03-29T00:00:00",
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Computing the matrix powers of a non-diagonalizable matrix Define
\begin{equation}
A = \begin{pmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{pmatrix}.
\end{equation}
Note that the sum of the dimensions of the eigenspaces of $A$ is only two. $A$ is thus not diagonalizable. How can we compute $A^n$?
|
Note that your matrix $A$ has the generalized eigenvectors
\begin{equation}v_1=\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}, v_2=\begin{pmatrix}0 \\ 2 \\ -2\end{pmatrix}, v_3=\begin{pmatrix}0\\ 0 \\ 1\end{pmatrix}.\end{equation}
Thus, by Jordan decomposition, $A=\big(v_1,v_2,v_3\big)J\big(v_1,v_2,v_3\big)^{-1}$, where
\begin{equation}J=\begin{pmatrix}\frac12 & 1 & 0\\0 & \frac12 & 0\\0 & 0 & 1\end{pmatrix}.\end{equation}
The problem of calculating $A^n$ is thus reduced to calculating $J^n$. Let $a_{ij}^{(n)}$ denote the entry of $J^n$ in the $i$-th row and $j$-th column.
The product of an arbitrary $3\times3$-matrix with $J$ is given by:
\begin{equation}
\begin{pmatrix}
a&b&c\\d&e&f\\g&h&i
\end{pmatrix} J =
\begin{pmatrix}
\frac a2&a+\frac b2&c\\\frac d2&d+\frac e2&f\\\frac g2&g+\frac h2&i
\end{pmatrix}.
\end{equation}
We can deduce that, for all $n\in\Bbb N$:
\begin{align}
a_{11}^{(n)}&=a_{22}^{(n)}=\frac1{2^n}, \\a_{21}^{(n)}&=a_{31}^{(n)}=0,\\
a_{13}^{(n)}&=a_{23}^{(n)}=a_{32}^{(n)}=0, \\
a_{33}^{(n)}&=1,\\
a_{12}^{(n+1)}&=a_{11}^{(n)}+\frac{a_{12}^{(n)}}2=\frac1{2^n}+\frac{a_{12}^{(n)}}2.
\end{align}
Thus, all $a_{ij}^{(n)}$ are explicitly known except for $a_{12}^{(n)}$. Note that, by the last equation, \begin{equation}a_{12}^{(n+1)}=2^{-n}+\frac{a_{12}^{(n)}}2 = 2^{-n}+2^{-n}+\frac{a_{12}^{(n-1)}}4 = \dots = (n+1)\cdot2^{-n}.\end{equation}
Thus,
\begin{equation}J^n=\begin{pmatrix}2^{-n}&n\cdot 2^{1-n} & 0\\0 & 2^{-n} & 0\\0 & 0 & 1\end{pmatrix}.\end{equation}
And by some calculations, we find that
\begin{equation}
A^n=\big(v_1,v_2,v_3\big)J^n\big(v_1,v_2,v_3\big)^{-1}=
\begin{pmatrix}
2^{-n} & n\cdot 2^{-n-1} - 2^{-n-1} + \frac12 & {1-\frac{n+1}{2^n}\over2}\\
0 & {2^{-n}+1\over2} & {1-2^{-n}\over2} \\
0 & {1-2^{-n}\over2} & {2^{-n}+1\over2}
\end{pmatrix}.
\end{equation}
|
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|
How to find probability of having at least 2 out of 3 students selected to sit together? I have the problem. 25 students are seated in a circle at the campfire night. 3 students are selected (the probability of being selected for each student is the same) and asked to join a game. Find the probability of having at least 2 out of 3 students selected to sit together.
I tried.
*
*There are $ \binom{25}{3} $ ways to select.
*If 3 students selected to sit together, then we have $ 25 $ ways.
*If there are exact 2 students selected to sit together, then we have $ 25 \times 21 $ ways.
*Probability we need to find is $ \dfrac{25 +25 \times 21 }{ \binom{25}{3}} =\dfrac{11}{46}.$
Is my solution true?
|
Yes, your solution is correct.
Refer to the diagram:
$\hspace{4cm}$
You considered two cases:
Case 1: triple students (order does not matter):
$$(\color{red}1,2,3),(\color{red}2,3,4),(\color{red}3,4,5),...,(\color{red}{23},24,25),(\color{red}{24},25,1),(\color{red}{25},1,2) \Rightarrow 25$$
Case 2: double students (order does not matter):
$$(\color{red}1,2,4),(1,2,5),...,(1,2,24) \Rightarrow 24-4+1=21\\
(\color{red}2,3,5),(2,3,6),...,(2,3,25)\Rightarrow 25-5+1=21\\
\vdots\\
(\color{red}{25},1,3),(25,1,4),...,(25,1,23)\Rightarrow 23-3+1=21$$
Hence the result.
Alternatively, select any student, say $1$. Now we will consider the probability of selecting neighboring one or two students. Refer to the diagram (order matters):
$\hspace{5cm}$
Hence:
$$\mathbb P(1,(2 \text{ or } 25))+P(1,3,(2 \text{ or } 4 \text{ or } 25))+P(1,24,(2 \text{ or } 23 \text{ or } 25))+\\
P(1,\color{red}4,(2 \text{ or } 3 \text{ or } 5 \text{ or } 25))+\cdots+P(1,\color{red}{23},(2 \text{ or } 22 \text{ or } 24 \text{ or } 25))=\\
\frac2{24}+\frac{3}{24\cdot 23}+\frac{3}{24\cdot 23}+\frac{4}{24\cdot 23}+\cdots+\frac{4}{24\cdot 23}=\\
\frac{2\cdot 23+3+3+4\cdot (23-4+1)}{23\cdot 24}=\frac{132}{23\cdot 24}=\frac{11}{46}.$$
|
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Modular Arithmetic. Integer Solutions How can you show that $|a^2 -10b^2|=2$ has no integer solutions for a and b using modular arithmetic?
Thank you.
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This is a case where the principal form also represents $-1,$ as $3^2 - 10 = -1.$ So, if $a^2 - 10 b^2 = n,$ we find
$$ (3a+10b)^2 - 10 (a+3b)^2 = -n $$
Meanwhile, $x^2 - 10 y^2$ represents primes $p \equiv 1, 9, 31, 39 \pmod {40}$
The other class of this discriminant $2x^2 - 5 y^2,$ represents $2$ and $5,$ then primes $p \equiv 3, 13, 27, 37 \pmod {40}$
If $2a^2 - 5 b^2 = k,$ we find
$$ 2(3a+5b)^2 - 5 (2a+3b)^2 = -k $$
|
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|
studying the series $\sum_\limits{n=0}^\infty \tan (\frac{n}{1+n^3})$ studying the series $\sum_\limits{n=0}^\infty \tan (\frac{n}{1+n^3})$.
The series has got a sense for each natural number.
$1+n^3>n, \forall n \in N \Rightarrow \frac{n}{1+n^3}<1,\forall n \in N\Rightarrow \tan \frac{n}{1+n^3} $ tends to 0 as $n \in N$ increases.
So all the elements are positive and tends to 0.
$$\tan \frac{n}{1+n^3} \sim \frac{\sin\frac{n}{1+n^3} }{\cos\frac{n}{1+n^3}} \sim \frac{\frac{n}{1+n^3} }{\cos\frac{n}{1+n^3}} \sim \frac{n}{1+n^3} <\frac{n}{n^3}= \frac{1}{n^2}$$
$\sum \frac{1}{n^2}$ converges $\Rightarrow $ the original series converges.
Is it right?
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From $$x\leq\tan x\leq \frac{4x}{\pi} \quad \left(0\leq x \leq \frac{\pi}{4}\right)$$
We get $$0\leq \frac{n}{1+n^3}\leq \tan \left(\frac{n}{1+n^3}\right) \leq \frac{4n}{\pi(1+n^3)} \leq \frac{4}{\pi n^2}$$
So original series converges by comparison test.
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If $x + y + z = 2$, then show $\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$, added a second question(Problem 2). Problem number 1:
The problem is that $x, y, z$ are proper fractions, and each one of them is greater than zero.
Given $x + y + z = 2$, prove
$$\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$$
I have tried to solve this using AM $\geq$ GM inequality.
Attempt :
$$\frac{\dfrac{1-x}{x} + \dfrac{1-y}{y} + \dfrac{1-z}{z}}{3}\geq\left(\frac{(1-x)(1-y)(1-z)}{xyz}\right)^{1/3}$$
What should I do to calculate the value of $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$?
Doubt:
*
*Let $a, b, c, d$ all be positive numbers.
*Let $a > c$ and $b > d$.
*Can we say that $\dfrac {a}{b} > \dfrac{c}{d}$?
Problem number 2:
$x,y,z$ are unequal positive quantities, prove that
$(1+x^3)(1+y^3)(1+z^3) > (1+xyz)^3$
My attempt:
$ \frac{x^3+y^3+z^3}{3} > xyz$
$ => (1+x^3)+(1+y^3)+(1+z^3) > 3(1+xyz)$
now by cubing the both sides we get,
$ => \frac{(1+x^3)+(1+y^3)+(1+z^3)}{27} > (1+xyz)^3$ ---(eqn. 1)
$ \frac{(1+x^3)+(1+y^3)+(1+z^3)}{3} > {\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)}^{\frac{1}{3}}$
$=> \frac{(1+x^3)+(1+y^3)+(1+z^3)}{27} > {\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)}$ ---(eqn. 2)
So, now if we do : $\frac{(eqn. 1)}{(eqn. 2)}$
then the result will be :
${\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)} > (1+xyz)^3$
and this is the correct result for all $x,y,z$ greater than zero.
But if we do $\frac{(eqn. 2)}{(eqn. 1)}$ then the result will be :
${\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)} < (1+xyz)^3$
So, my question is, how to prove it correctly without doing the division operation, cause when we divide the two equations then we get two separate results.
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The first inequality:
We need to prove that:
$$\frac{8xyz}{\prod\limits_{cyc}(x+y+z-2x)}\geq8$$ or
$$xyz\geq\prod_{cyc}(y+z-x)$$ or
$$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.
The second inequality it's just Holder for three sequences:
$$\prod_{cyc}(1+x^3)\geq\left(\sqrt[3]{1\cdot1\cdot1}+\sqrt[3]{x^3y^3z^3}\right)^3=(1+xyz)^3.$$
|
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|
Find a matrix $C \in M_{3\times 3} (\mathbb C)$ such that $A_4=C^{T}A_3C$
Let $$A_3=\begin{bmatrix} -1 & 2 & -1 \\ 2 & 0 & 0 \\ -1 & 0 & -1 \end{bmatrix},$$ $$A_4=\begin{bmatrix} 4 & 4 & 4 \\ 4 & 0 & 2 \\ 4 & 2 & 5 \end{bmatrix}.$$
a) Show that $A_3$ and $A_4$ are congruent over $\mathbb C$, but not over $\mathbb R$.
b) Find a matrix $C \in M_{3\times3} (\mathbb C)$ such that $A_4=C^{T}A_3C$.
My try:
a) $$\det A_3=4, \det A_4=-32 \Rightarrow \det A_3 \cdot \det A_4=-128$$ When the matrices are congruent over $K$ then the product of determinants is equal to $c^2$ for some $c\in K$. But $\sqrt{-128} \notin \mathbb R$, so $A_3$ and $A_4$ are not congruent over $\mathbb R$.$\det A_3 \neq 0$, $\det A_4 \neq 0$ so $\operatorname{rank} A_3= \operatorname{rank} A_4 =3$ and $A_3$ is congruent with $A_4$ over $\mathbb C$.I think this is a good way, but I have a problem with b):The matrix $ C $ is determined by finding a base perpendicular to the bilinear form $ h $ where $A_3=G(h;st)$:$h(\epsilon_1, \epsilon_1)=-1, h(\epsilon_2, \epsilon_2)=0,h(\epsilon_3, \epsilon_3)=-1$ so let $\alpha_1=\epsilon_1$ because it is an isotropic vector. We're looking for one $\alpha_2$ such that $h(\alpha_1,\alpha_2) = 0$ and $h(\alpha_2,\alpha_2)\neq0$$h(\alpha_1,\alpha_2) = 0 \Leftrightarrow x_{1}=2x_2-x_3$ so let $\alpha_2=(0,1,2)$ and then $h(\alpha_2, \alpha _2)=-4\neq 0$. We're looking for one $\alpha_3$ such that $h(\alpha_1,\alpha_3) = 0,h(\alpha_2,\alpha_3) = 0$ and $h(\alpha_3,\alpha_3)\neq0$.
$h(\alpha_1,\alpha_3) = 0,h(\alpha_2,\alpha_3) = 0 \Leftrightarrow -a+2b-c=0, -2c=0$ so let $\alpha_3=(2,1,0)$ then $h(\alpha_3,\alpha_3) = 4 \neq0$. So we have $A=\left\{ \alpha _1, \alpha _2, \alpha _3\right\} $ and $C=M(id)_A^{st}$.
Unfortunately it is not a correct answer because when I multiply $C^{T}, A_3, C$ I don't get $A_4$. Can you tell me where did I make a mistake?
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Note that if you write $A_{4}=C^{T}A_{3}C$, then you are essentially applying simultaneous row and column operations. $C^{T}$ represent some operations applied to the rows of $A_{3}$ and $C$ represent the same operations applied to the columns. Now you can start doing calculations
$$A_{3}=\left(\begin{matrix}-1&2&-1\\2&0&0\\-1&0&-1\end{matrix}\right)\stackrel{\sqrt{2}iR_{3}\rightarrow R_{3}\\\sqrt{2}iC_{3}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}-1&2&-\sqrt{2}i\\2&0&0\\-\sqrt{2}i&0&2\end{matrix}\right)\stackrel{R_{1}+\frac{1}{2}R_{2}\rightarrow R_{1}\\C_{1}+\frac{1}{2}C_{2}\rightarrow C_{1}}{\longrightarrow}\left(\begin{matrix}1&2&-\sqrt{2}i\\2&0&0\\-\sqrt{2}i&0&2\end{matrix}\right)\stackrel{R_{3}+\frac{1}{2}\left(1+\sqrt{2}i\right)R_{2}\rightarrow R_{3}\\C_{3}+\frac{1}{2}\left(1+\sqrt{2}i\right)C_{2}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}1&2&1\\2&0&0\\1&0&2\end{matrix}\right)\stackrel{R_{1}+R_{3}\rightarrow R_{3}\\C_{1}+C_{3}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}1&2&2\\2&0&2\\2&2&5\end{matrix}\right)\stackrel{2R_{1}\rightarrow R_{1}\\2C_{1}\rightarrow C_{1}}{\longrightarrow}\left(\begin{matrix}4&4&4\\4&0&2\\4&2&5\end{matrix}\right)$$
Applying the same operations to the identity matrix, we get
$$I=\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right)\stackrel{\sqrt{2}iC_{3}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}1&0&0\\0&1&0\\0&0&\sqrt{2}i\end{matrix}\right)\stackrel{C_{1}+\frac{1}{2}C_{2}\rightarrow C_{1}}{\longrightarrow}\left(\begin{matrix}1&0&0\\\frac{1}{2}&1&0\\0&0&\sqrt{2}i\end{matrix}\right)\stackrel{C_{3}+\frac{1}{2}\left(1+\sqrt{2}i\right)C_{2}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}1&0&0\\\frac{1}{2}&1&\frac{1}{2}\left(1+\sqrt{2}i\right)\\0&0&\sqrt{2}i\end{matrix}\right)\stackrel{C_{1}+C_{3}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}1&0&1\\\frac{1}{2}&1&\frac{1}{2}\left(2+\sqrt{2}i\right)\\0&0&\sqrt{2}i\end{matrix}\right)\stackrel{2C_{1}\rightarrow C_{1}}{\longrightarrow}\left(\begin{matrix}2&0&1\\1&1&\frac{1}{2}\left(2+\sqrt{2}i\right)\\0&0&\sqrt{2}i\end{matrix}\right)$$
The end result is
$$C=\left(\begin{matrix}2&0&1\\1&1&\frac{1}{2}\left(2+\sqrt{2}i\right)\\0&0&\sqrt{2}i\end{matrix}\right)$$
Indeed, according to Wolfram Alpha this is correct.
|
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|
$\frac{x}{4} = \frac{y}{x} = \frac{7}{y}$ Could someone please explain to me what's wrong here?
$\cfrac{x}{4} = \cfrac{y}{x} = \cfrac{7}{y}$
So $$\cfrac{x^2}{4} = y \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ y = \sqrt{7x}$$
Hence $$\cfrac{x^2}{4} =\sqrt{7x}$$
With a solution of $x = 2.57$
But $$\cfrac{2.57^2}{4} \neq \sqrt{7 (2.57)}$$
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Hint $ $ Multiplyimg $\,\overbrace{ \dfrac{\color{#c00}x}{4} = \dfrac{\color{#0a0}y}{\color{#c00}x} = \dfrac{7}{\color{#0a0}y}}^{\Large a}\,\Rightarrow\,a^{\large 3} = \dfrac{7}4$
hence $\, x = 4a,\ \ y = xa = 4a^2,\ \ 7 = ya = 4a^3\ \ \checkmark$
Remark $ $ This telescoping product view works generally
$\,\overbrace{ \dfrac{\color{#c00}{x_1}}{x_0} = \dfrac{\color{#0a0}{x_2}}{\color{#c00}{x_1}} = \dfrac{x_3}{\color{#0a0}{x_2}}= \,\cdots\, = \dfrac{x_n}{x_{n-1}} }^{\Large a}\ \Rightarrow\ a^{\large n} = \dfrac{x_n}{x_0}$
so $\, x_1 = a x_0,\ x_2 = a x_1 = a^2 x_0,\,\ldots,\, x_n = a^n x_0$
See here for many more examples of multiplicative telescopy.
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Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$.
$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b}$$
I have not come up with any ideas to solve the problem yet. I will probably in the near future but right now, I can't.
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By rearrangement-inequality
$$\frac{a^{\,2}}{b+ c}+ \frac{b^{\,2}}{c+ a}+ \frac{c^{\,2}}{a+ b}\geqq \frac{a^{\,2}}{a+ b}+ \frac{b^{\,2}}{b+ c}+ \frac{c^{\,2}}{c+ a}$$
We have to prove
$$\frac{a^{\,2}}{a+ b}+ \frac{b^{\,2}}{b+ c}+ \frac{c^{\,2}}{c+ a}\geqq \frac{1}{2}(\,\sqrt{\frac{a^{\,2}+ b^{\,2}}{2}}+ \sqrt{\frac{b^{\,2}+ c^{\,2}}{2}}+ \sqrt{\frac{c^{\,2}+ a^{\,2}}{2}}\,)$$
By a.m.-g.m.-inequality
$$\sqrt{\frac{a^{\,2}+ b^{\,2}}{2}}= \frac{2\sqrt{2(\,a^{\,2}+ b^{\,2}\,)(\,a+ b\,)^{\,2}}}{4(\,a+ b\,)}\leqq \frac{(\,a+ b\,)^{\,2}+ 2(\,a^{\,2}+ b^{\,2}\,)}{4(\,a+ b\,)}$$
We need to have
$$\frac{4\,a^{\,2}}{a+ b}+ \frac{4\,b^{\,2}}{b+ c}+ \frac{4\,c^{\,2}}{c+ a}\geqq \frac{1}{2}\times \sum\limits_{cyc}\frac{(\,a+ b\,)^{\,2}+ 2(\,a^{\,2}+ b^{\,2}\,)}{(\,a+ b\,)}$$
$$\because\,\frac{a^{\,2}- 2\,ab+ b^{\,2}}{a+ b}+ \frac{b^{\,2}- 2\,bc+ c^{\,2}}{b+ c}+ \frac{c^{\,2}- 2\,ca+ a^{\,2}}{c+ a}\geqq 0$$
So it holds !
$$\therefore\,\frac{a^{\,2}}{b+ c}+ \frac{b^{\,2}}{c+ a}+ \frac{c^{\,2}}{a+ b}\geqq \frac{1}{2}(\,\sqrt{\frac{a^{\,2}+ b^{\,2}}{2}}+ \sqrt{\frac{b^{\,2}+ c^{\,2}}{2}}+ \sqrt{\frac{c^{\,2}+ a^{\,2}}{2}}\,)= \frac{1}{2\sqrt{2}}$$
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Discrepancy in finding the coefficient of a polynomial After asking a question here let us consider on the following case which is simpler, this might help me to make some progress:
suppose we have the following polynomial:
$$f(x)=(x^0+x^1+x^2+x^3)^3$$ expanding this gives:
$$f(x)=1 + 3 x + 6 x^2 + 10 x^3 + 12 x^4 + 12 x^5 + 10 x^6 + 6 x^7 +
3 x^8 + x^9$$
Aim is to find the coefficients for a given power. Now let us use multinomial theorem. Let us now call:
$$a=x^0$$
$$b=x^1$$
$$c=x^2$$
$$d=x^3$$
then the polynomial becomes:
$$f(a,b,c,d)=(a+b+c+d)^3$$
which has the following expansion:
$$f(a,b,c,d)=a^3 + 3 a^2 b + 3 a b^2 + b^3 + 3 a^2 c + 6 a b c + 3 b^2 c +
3 a c^2 + 3 b c^2 + c^3 + 3 a^2 d + 6 a b d + 3 b^2 d + 6 a c d +
6 b c d + 3 c^2 d + 3 a d^2 + 3 b d^2 + 3 c d^2 + d^3$$
surely upon substituting we recover the original polynomial or its expansion. Now as a matter of example say we want to determine the coefficient of $a^2d$, we can see that the coefficient is 3, we can derive this as:
$$a^2d: \frac{3!}{2!1!}=3$$
which is the write coefficient, now we have
$$3a^2d=3(x^0)^2(x^3)=3x^3$$
which is not the coefficient of the expanded original polynomial, the expanded original polynomial has coefficient equal to 10, now question is what am I missing here? why there is discrepancy in here?
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$$f(x)=\frac{(1-x^4)^3}{(1-x)^3}$$
We have that $$\frac{1}{(1-x)^3}=\sum_{k=0}^{\infty}\binom{k+2}{2}x^k.$$
And $(1-x^4)^3=1-3x^4+3x^8-x^{12}.$
So $f(x)=\sum_{k=0}^{\infty} a_kx^k$ where:
$$a_k = \binom{k+2}{2}-3\binom{k-2}{2}+3\binom{k-6}{2}-\binom{k-10}{2}.$$
More general, if $$\begin{align}f_{n,m}(x)&=\left(1+x+x^2+\cdots+x^n\right)^m\\&=\frac{\left(1-x^{n+1}\right)^m}{(1-x)^m}\\&=\left(1-x^{n+1}\right)^m\sum_{k=0}^{\infty}\binom{k+m-1}{m-1}x^k
\end{align}$$
From this we get that $$a_k=\sum_{j=0}^{m}\left(-1\right)^j\binom{m}{j}\binom{k+m-1-j(n+1)}{m-1}$$
When $n=m,$ this gives:
$$a_k=\sum_{j=0}^{n}\left(-1\right)^j\binom{n}{j}\binom{k-(n+1)(j-1)-2}{n-1}$$
There is no simpler form, I strongly suspect.
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|
Solve integer equation $2^m.m^2=9n^2-12n+19$
Problem: Find $m,n\in \mathbb{N}^*$ satisfies: $2^m.m^2=9n^2-12n+19$.
This is my attempt:
We have $9n^2-12n+19\equiv 1 \pmod 3$, so: $2^{m}m^2\equiv 1 \pmod 3\tag{1}$
In addition, we have: $$m^2\equiv 0\text{ or }1 \pmod 3 $$
So $(1)\implies m\equiv \pm 1\pmod 3$.
Suppose that: $m=3k+1(k\in \mathbb{N})$.
We have: $$2^{m}m^2=2^{3k+1}(3k+1)^2=8^{k}(3k+1)^22\equiv 2^{k+1}(3k+1)^{2}\pmod 3\\ \implies k\equiv 1\pmod 2$$
I can only come here!!
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The case left open by the above answer is for $m$ odd. If $m$ is odd, then $m^2\times 2^m = 2A^2$ for some integer $A$; namely $A = m\times 2^{\frac{m-1}{2}}$. Thus we are left with the equation
$$2A^2 = (3n-2)^2+15$$
Note that the LHS of the equation is either 2 or 0 mod 3, whereas the RHS is always 1.
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|
Double Integration Problem $\int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$
Compute
$$I = \int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$$
Here are my steps:
$$\begin{split}
I &=\int_{0}^{1} \left(\int_0^1 \frac{dy}{1+y(x^2-x)}\right)dx\\
&=\int_{0}^{1} \left[\frac{\ln(1+y(x^2-x))}{x^2-x}\right]_0^1dx\\
&=\int_{0}^{1} \left[\frac{\ln(1+(1)(x^2-x))}{x^2-x}
-\frac{\ln(1+(0)(x^2-x))}{x^2-x}\right]dx\\
&=\int_{0}^{1} \frac{\ln(1+x^2-x)}{x^2-x}dx
\end{split}
$$
And here I can't find any substitution to solve this integral.
Can anyone help me?
By the way, I also used Simpson's 3/8 method to find the approximation and got $1.063$.
But I want to find it using Calculus.
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Here is an alternate way that avoids your integral.
Writing the integrand in terms of a geometric sum we have:
$$\frac{1}{1 + y(x^2 - x)} = \sum_{n = 0}^\infty y^n (x - x^2)^n, \quad |x|, |y| < 1,$$
then
\begin{align}
I &= \int_0^1 \int_0^1 \frac{1}{1 + y(x^2 - x)} \, dy dx\\
&= \sum_{n = 0}^\infty \int_0^1 y^n \, dy \int_0^1 x^n (1 - x)^n \, dx\\
&= \sum_{n = 0}^\infty \frac{1}{n + 1} \cdot \operatorname{B}(n + 1, n + 1)\\
&= \sum_{n = 0}^\infty \frac{1}{n + 1} \cdot \frac{(n!)^2}{(2n + 1)!}\\
&= \sum_{n = 0}^\infty \frac{1}{(n + 1)(2n + 1) \binom{2n}{n}}.\tag1
\end{align}
Here $\operatorname{B}(x,x)$ denoted the Beta function while $\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$ is the central binomial coefficient.
To find the sum in (1), the following (reasonably?) well known Maclaurin series can be recalled (for a proof, see here)
$$\frac{\sin^{-1} x}{\sqrt{1 - x^2}} = \sum_{n = 0}^\infty \frac{2^{2n} x^{2n + 1}}{(2n + 1) \binom{2n}{n}}, \quad |x| < 1.$$
Enforcing a substitution of $x \mapsto \sqrt{x}$, after rearranging one has
$$\frac{\sin^{-1} \sqrt{x}}{\sqrt{x} \sqrt{1 - x}} = \sum_{n = 0}^\infty \frac{2^{2n} x^n}{(2n + 1) \binom{2n}{n}}, \quad |x| < 1.$$
Integrating both sides with respect to $x$ from $0$ to $\frac{1}{4}$ leads to
$$\sum_{n = 0}^\infty \frac{1}{(n + 1)(2n + 1) \binom{2n}{n}} = 4 \int_0^{\frac{1}{4}} \frac{\sin^{-1} \sqrt{x}}{\sqrt{x} \sqrt{1 - x}} \, dx.$$
Making a substitution of $x = \sin^2 u$ in the integral readily leads to
\begin{align}
\sum_{n = 0}^\infty \frac{1}{(n + 1)(2n + 1) \binom{2n}{n}} &= 8 \int^{\frac{\pi}{6}}_0 u \, du = \frac{\pi^2}{9},
\end{align}
from which it follows that:
$$\int_0^1 \int_0^1 \frac{1}{1 + y(x^2 - x)} \, dy dx = \frac{\pi^2}{9}.$$
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Find the sum of $\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$ $$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
The question is divided into three parts:
1. Determine its radius of convergence
2. By using the power series of $\frac{1}{1-x}$, show that for all x $\in$ ]-1,1[ , we get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
3. Find the value of the sum S(x)=$\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$
I was able to solve 1 and 2 but I found some difficulty in 3. I will write how I solved the first two questions and if possible I would like to know if I have made a mistake.*
1. Using the ratio test with $U_n=\frac{1}{n(n+2)}$
$$\frac{U_{n+1}}{U_n}=\frac{n(n+2)}{(n+1)(n+3)}=\frac{n^2+2n}{n^2+4n+3}$$
$$L=\lim\limits_{n \to \infty}(\frac{n^2+2n}{n^2+4n+3})=\lim\limits_{n\to \infty}(\frac{\frac{1}{n^2}(1+\frac{2}{n})}{\frac{1}{n^2}(1+\frac{4}{n}+\frac{3}{n^2})})=1$$
Radius of converge R=1/L $\rightarrow$ R = 1
2.
$$\frac{1}{1-x}=-\frac{dy}{dx}\ln(1-x)$$
and since it is known that $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ for $x \in]-1,1[$ (too lazy to write proof) we can take the integral of $x^n$ to get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
$$-\int_{0}^{x} x^n = -\frac {x^{n+1}}{n+1} \rightarrow -\sum_{n=0}^\infty\frac {x^{n+1}}{n+1}= -\sum_{n=1}^\infty\frac {x^n}{n}$$
3.
$$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
I took
$$U_n=\frac{1}{n(n+2)}=\frac{A}{n}+\frac{B}{n+2}$$
I got
$$U_n=\frac{1}{n(n+2)}=\frac{1}{2n}+\frac{-1}{2(n+2)}$$
$\therefore$
$$S(x)=\sum_{n=1}^\infty (\frac{1}{2n}+\frac{-1}{2(n+2)})x^n$$
First, I started to solve $\sum_{n=1}^\infty (\frac{1}{2n})x^n$, and by using the proof of question n.2 I got: $$\sum_{n=1}^\infty (\frac{1}{2n})x^n=\frac{1}{2}\sum_{n=1}^\infty (\frac{1}{n})x^n=\frac{1}{2}(-ln(1-x))=ln(\frac{1}{\sqrt{1-x}})$$
However, I was not able to solve the second part: $$\sum_{n=1}^{\infty}\frac{-1}{2(n+2)}x^n$$
How can I find the sum of the second part to find S(x)? I appreciate the help and I would be glad to know if I have made any mistakes too.
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Another approach is to start with the geometric series $\sum_{n=1} x^{n-1}=\frac{1}{1-x}$ (assuming you are in appropriate interval for convergence etc.).
\begin{align*}
\frac{1}{1-x} & = \sum_{n=1}x^{n-1}\\
\int \frac{1}{1-x} \, dx& =\sum_{n=1}\frac{x^n}{n}\\
-\ln(1-x)+c & = \sum_{n=1}\frac{x^n}{n}\\
-x\ln(1-x)+xc & = \sum_{n=1}\frac{x^{n+1}}{n}\\
\int -x\ln(1-x) \, dx+c\int x \, dx & = \sum_{n=1}\frac{x^{n+2}}{n(n+2)}\\
\int -x\ln(1-x) \, dx+c\int x \, dx & = x^2\left[\sum_{n=1}\frac{x^{n}}{n(n+2)}\right]\\
\end{align*}
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|
Simplifying $\sum_{cyc}\tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)$. I get $0$, but the answer is $\pi$. So the question is
$$ \tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)+\tan^{-1}\left(\sqrt{\frac{y(x+y+z)}{xz}}\right)+\tan^{-1}\left(\sqrt{\frac{z(x+y+z)}{yx}}\right) =\ ? $$
So my take on the question is to rewrite it as
$$ \tan^{-1}\left(x\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(y\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(z\sqrt{\frac{(x+y+z)}{yzx}}\right) $$
Then say $$\frac{x+y+z}{yzx}= a^2.$$
We get
$$ \tan^{-1}\left( \frac{a((x+y+z)-a^2xyz)}{1-a^2(xy+yz+zx)}\right)$$
And since $ (x+y+z) = a^2xyz $ , this is just equal to $\tan^{-1}(0)= 0 $ but the answer given is $\pi.$
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Hint:-
$$ \tan^{-1}a + \tan^{-1}b + \tan^{-1}c= \pi$$
Only and only if
$$a+b+c=abc$$
|
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|
Easier method of finding the equation of the circle circumscribing the triangle formed by 3 lines? The equation of the circle circumscribing the triangle formed by the lines $y = 0, y = x$ and $2x + 3y = 10$ is?
I know this can be done by solving two equations at a time and finding the vertex. Then forming 3 different equations to solve for the centre (abscissa, ordinate) and the radius( which are 3 variables).
I would like to know if there's a shorter method to find the equation of the circle for any 3 random lines given. (It's easier in this particular case as one of the coordinates is $(0,0)$ which makes $c=0$)
|
Easy to see that $(0,0)$, $(2,2)$ and $(5,0)$ are the vertices of the triangle.
Let $M\left(\frac{5}{2},b\right)$ be the center of the circle.
Thus, $$\left(\frac{5}{2}\right)^2+b^2=\left(\frac{1}{2}\right)^2+(b-2)^2,$$ which gives $$b=-\frac{1}{2}$$ and $$M\left(\frac{5}{2},-\frac{1}{2}\right).$$
Now, for the radius of the circle we obtain:
$$\sqrt{\frac{25}{4}+\frac{1}{4}}=\sqrt{\frac{13}{2}},$$ which gives the answer:
$$\left(x-\frac{5}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=\frac{13}{2}.$$
|
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|
How to tackle this squaring of inequality problem
If the roots of quadratic equation $$x^2 − 2ax + a^2 + a – 3 = 0$$
are real and less than $3$, find the range of $a$.
The roots are $a \pm \sqrt {3 – a}$
For the roots to be real, we must have a < 3.
Also, for the roots to be less than 3, we must have $\pm \sqrt {3 – a } \lt 3 – a $
If squaring both sides is allowable, I will get $(a – 2)(a – 3) > 0$. Then the problem is solved.
The question is:- how to convince others that the squaring of both sides of $\pm \sqrt {3 - a } \lt 3 - a $ is allowable?
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Since $x_1,x_2<3$ we have $3-x_i>0$ so their product is positive:$$ 0<9-3(x_1+x_2)+x_1x_2$$
thus $$0<9-6a+a^2+a-3 = a^2-5a+6=(a-3)(a-2)$$
So $a\in (-\infty ,2)\cup (3,\infty)$. But since the discriminat must be $\geq 0$ we get $a\leq 3$ so we have $a\in (-\infty ,2)$
Now let us prove that all $a<2$ are good.
We have $$x_1= a-\sqrt{3-a}\leq a<3$$ We are left if $x_2= a+\sqrt{3-a}$ is smaller than $3$ if $a<2$ i.e. $$\sqrt{3-a} <3-a$$ which is true since $3-a>1$
|
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|
How to evaluate $\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$
How to evaluate $$\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$$
I tried to integrate by parts, but no way so far, help me, thanks.
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From here , we have $\ \displaystyle \ 3\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=0$
or $\ I=\displaystyle\int_{0}^{1}\frac{\arctan x}{x} \ln{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\ dx=\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx$
using $\ \displaystyle\arctan x\ln(1+x^2)=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}} {2n+1}x^{2n+1}$ ( proved here) , we get
\begin{align}
I&=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1x^{2n}\ dx\\
&=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}\\
&=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\
&=-2\Im\sum_{n=1}^\infty\frac{i^nH_n}{n^2}+\frac{\pi^3}{16}
\end{align}
using the generating function with $x=i$
$$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$
we get $\qquad\displaystyle\Im\sum_{n=1}^\infty\frac{i^nHn}{n^2}=-\frac{\pi}{16}\ln^22-\frac12G\ln2-\Im\operatorname{Li}_3(1-i)$
Plugging this result, we get $\quad\boxed{\displaystyle I=\frac{\pi^3}{16}+\frac{\pi}{8}\ln^22+G\ln2+2\Im\operatorname{Li}_3(1-i)}$
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|
Taylor expansion of $\sqrt{n-k}$ I am reading a paper which casually assumes the asymptotic $\sqrt{n-k} \simeq \sqrt{n}-\frac{k}{2\sqrt{n}}$. This expression is what Wolfram calls Taylor expansion at infinity and from what I understand we work with $\sqrt{x-k}=\sqrt{x}\cdot \sqrt{1-k/x}$ and then we proceed on doing the Taylor expansion. My problem is that we also cannot do Taylor expansion near zero because said function is not differentiable there.
Another point is: do the Taylor expansion near point $1$ for $\sqrt{x}$ so that
$$\sqrt{x}\sqrt{1-k/x} = \sqrt{x} - \frac{k}{2\sqrt{x}} - \frac{k^2}{8x^{3/2}}\frac{1}{\xi^{3/2}},$$ where $\xi$ as in Legendre residual form (between $1-k/x$ and $1$). We can then deduce that $\sqrt{n-k}\leq \sqrt{n}-k/2\sqrt{n},$ but what about the lower bound? Is there a way to fix an inequality relating the remainder and the second term of the above expansion?
My real question, is how to use the aforementioned fact in order to get a result of the form $$\sqrt{x-1/2}\geq \sqrt{x}\left(1-\frac{4}{x}\right).$$ The first two terms of $\sqrt{x-1/2}$ in the above expansion should be $\sqrt{x}-\frac{1}{4\sqrt{x}},$ so the remainder term should be $\geq -\frac{15}{4\sqrt{x}}$ or something.
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If you want to show that $\sqrt{x-1/2}\geq \sqrt{x}\left(1-\frac{4}{x}\right),$
the following argument is simpler than bounding Taylor series terms:
(a) for $\frac12\le x<4$: $\sqrt{x-1/2}\geq0\gt \sqrt{x}\left(1-\frac{4}{x}\right);$
(b) for $x>\frac{32}{15}, \frac{15}2>\frac{16}x,$ so $x-\frac12>x-8+\frac{16}x=x(1-\frac4x)^2, $ so $\sqrt{x-1/2}>\sqrt x(1-\frac4x);$
so we have shown $\sqrt{x-1/2}>\sqrt x\left(1-\frac4x\right)$ for all $x$ where $\sqrt{x-1/2}$ is defined.
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|
Ordinary Differential Equation getting two different answers Solve $y(y^2-2x^2)dx+x(2y^2-x^2)dy=0$ and find a particular curve passing through $(1,2)$
My attempt:
1st Solution: Rewrite as
$y(y^2dx+x2ydy)-x(x^2dy+y2xdx)=0$
$\implies xy^2d(xy^2)-x^2yd(x^2y)=0$ (multiplying $xy$)
$\implies (xy^2)^2-(x^2y)^2 = c\,\,\,$ Or $x^2y^2(y^2-x^2) = c$
$\implies x^2y^2(y^2-x^2) = 12 $
2nd Solution: Rewrite as
$\frac{dy}{dx}=\frac{2\frac yx-(\frac yx)^3}{2(\frac yx)^2-1}$
Let $y=tx\implies \frac{dy}{dx}=t+x\frac{dt}{dx}=\frac{2t-t^3}{2t^2-1}$
$\implies x\frac{dt}{dx}=\frac{3(t-t^3)}{2t^2-1}$
$\implies \int\frac{2t^2-1}{t^3-t}dt+\int\frac 3x dx=0$
$\implies \int \left(\frac 2t+\frac{1}{(t-1)}+\frac{1}{(t+1)}\right)dt+\int\frac 6x dx = 0 $
$\implies \ln|t^2(t^2-1)x^6| = ln c $
$\implies |t^2(t^2-1)x^6| = c $
$\implies |x^2y^2(y^2-x^2)| = 12 $
$\implies x^2y^2(y^2-x^2) = \pm 12 $
Where I am doing wrong in second or in first solution!
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In your second solution, you work out your constant too soon. It is the case that if $|x| = c$, then you can conclude that $ x = c$ because the arbitrary constant can absorb the $\pm$. In the first equation, $c$ nonzero. In the second, it's allowed to be negative.
So I think you should have written
$$x^2y^2(y^2-x^2) = c$$
and THEN worked out what $c$ is.
You'd have the same trouble if solved a DE and got $(y+x)^2 =c$ and then worked out that $c =4$ and then wrote $y+x = \pm 2$, when you should rather have done this:
$$(y+x)^2 = c$$
$$y+x = c$$
$$y+x = 2$$
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Evaluate $(\sqrt{3}+i)^{14}+(\sqrt{3}-i)^{14}$ Evaluate $(\sqrt{3}+i)^{14}+(\sqrt{3}-i)^{14}$
I tried by using the De'moivers theorem but I didn't get proper value I get a mess value 4..but I am not sure about the answer can anyone please tell me
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First of all, try to factorise by $2^{14}$ :).
$(\sqrt{3} + i)^{14} + (\sqrt{3} - i)^{14} = 2^{14}(\dfrac{\sqrt{3}}{2} + \dfrac{i}{2})^{14} + 2^{14}(\dfrac{\sqrt{3}}{2} - \dfrac{i}{2})^{14} = 2^{14}\left((\dfrac{\sqrt{3}}{2} + \dfrac{i}{2})^{14} + (\dfrac{\sqrt{3}}{2} - \dfrac{i}{2})^{14}\right)$
Now, use the fact that: $\dfrac{\sqrt{3}}{2} + \dfrac{i}{2} = \cos(\pi/6) + i\sin(\pi/6) = \exp(i\pi/6)$.
And similarly: $\dfrac{\sqrt{3}}{2} - \dfrac{i}{2} = \cos(-\pi/6) + i\sin(-\pi/6) = \exp(-i\pi/6)$.
I think you'll be able to finish from there, by using De Moivre's theorem :)
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$\sqrt{2} x^2 - \sqrt{3} x +k=0$ with solutions $\sin\theta$ and $\cos\theta$, find k
If the equation $\sqrt{2} x^2 - \sqrt{3} x +k=0$ with $k$ a constant has two solutions $\sin\theta$ and $\cos\theta$ $(0\leq\theta\leq\frac{\pi}{2})$, then $k=$……
My approach is suggested below but I am not sure how to continue.
Since $\sin\theta$ and $\cos\theta$ are two solutions of the equation,
Then we have,
$\sqrt{2} \sin^2\theta - \sqrt{3} \sin\theta +k=0$ .....Equation (1)
$\sqrt{2} \cos^2\theta - \sqrt{3} \cos\theta +k=0$ .....Equation (2)
Add (2) to (1),
$\sqrt{2} (\sin^2\theta + \cos^2\theta) - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$
$\sqrt{2} - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$
The answer key provided is $\frac{\sqrt{2}}{4}$. I think I am probably on the right track here but not sure how I should proceed with $\sin\theta$ and $\cos\theta$ next. Please help.
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$\sin\theta+\cos\theta=\dfrac{\sqrt3}{\sqrt2}$ and $\sin\theta\cos\theta=\dfrac k{\sqrt2}$.
$(\sin\theta+\cos\theta)^2-2\sin\theta\cos\theta=1$
$\dfrac32-\sqrt2 k=1$
$k=\dfrac 1{2\sqrt2}$
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Prove that $x\mathcal{R}y\iff x^2-y^2=2(y-x)$ is an equivalence relation Prove that $$x\mathcal{R}y\iff x^2-y^2=2(y-x)$$ is an equivalence relation.
Reflexive. For all $x$ we have $x^2-x^2=2(x-x)$, so $x\mathcal{R}x$.
Symmetric. For all $x,y$ we have \begin{align}x\mathcal{R}y&\implies x^2-y^2=2(y-x)\\&\implies(-1)(x^2-y^2)=(-1)2(y-x)\\&\implies y^2-x^2=2(x-y)\\&\implies y\mathcal{R}x.\end{align}
Transitive. For all $x,y,z$ we have $$\begin{cases}x\mathcal{R}y\\y\mathcal{R}z\end{cases}\implies\begin{cases}x^2-y^2=2(y-x)\\y^2-z^2=2(z-y)\end{cases}\implies x^2-y^2+y^2-z^2=2(y-x)+2(z-y),$$ so $$x^2-z^2=2(y-x+z-y)=2(z-x)\implies x\mathcal{R}z.$$
Is it correct?
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Your proof is correct and well-presented. Good job!
Mostly posting this so this question can be finally considered to have an answer. Made it Community Wiki since I have nothing of substance to add.
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Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$ Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$
Start by squaring both sides
$$\frac{-4x^2-x+14}{(x-1)^2}<0$$
Factoring and multiplied both sides with -1
$$\frac{(4x-7)(x+2)}{(x-1)^2}>0$$
I got
$$(-\infty,-2)\cup \left(\frac{7}{4},\infty\right)$$
Since $x\leq2$ then
$$(-\infty,-2)\cup \left(\frac{7}{4},2\right]$$
But the answer should be $(-\infty,1)\cup \left(\frac{7}{4},2\right]$. Did I missed something?
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For $\dfrac{3\sqrt{2-x}}{x-1}$ to be defined, $x\le2$ and $x\ne1$.
If $x<1,$ then the expression is negative (i.e., $<0$), so of course it is $< 2$.
If $x>1,$ then, as you showed, the inequality holds when $x>\dfrac74$.
Therefore, the solution set is $x<1$ or $\dfrac74<x\le2$.
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Find remainder of division of $x^3$ by $x^2-x+1$ I am stuck at my exam practice here.
The remainder of the division of $x^3$ by $x^2-x+1$ is ..... and that of $x^{2007}$ by $x^2-x+1$ is .....
I tried the polynomial remainder theorem but I am not sure if I did it correctly.
By factor theorem definition, provided by Wikipedia,
the remainder of the division of a polynomial $f(x)$ by a linear polynomial $x-r$ is equal to $f(r)$.
So I attempted to find $r$ by factorizing $x^2-x+1$ first but I got the complex form $x=\frac{1\pm\sqrt{3}i}{2}=r$.
$f(r)$ is then $(\frac{1+\sqrt{3}i}{2})^3$ or $(\frac{1-\sqrt{3}i}{2})^3$ which do not sound right.
However, the answer key provided is $-1$ for the first question and also $-1$ for the second one. Please help.
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Since $$x^3+1 = (x+1)(x^2-x+1)$$ so $$x^3 = (x+1)(x^2-x+1)-1$$ the answer is $-1$.
Similarly for \begin{eqnarray}x^{3n}+1 &=& (x^3+1)\underbrace{\Big((x^3)^{n-1}-(x^3)^{n-2}+...-(x^3)+1\Big)}_{q(x)}\\
&=& (x+1)(x^2-x+1)q(x)\\
\end{eqnarray}
so the answer is again $-1$.
|
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Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ . Given three positive numbers $a,\,b,\,c$ . Prove that $$(\!abc+ a+ b+ c\!)^{3}\geqslant 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!).$$
My own problem is given a solution, and I'm looking forward to seeing a nicer one(s), thank you !
Solution. Without loss of generality, we can suppose $(a- 1)(b- 1)\geqslant 0 \rightarrow 1+ ab\geqslant a+ b$. Then
\begin{align*} (abc+ a+ b+ c)^{3}&= \left ( c(1+ ab)+ a+ b \right )^{2}\left ( c(1+ ab)+ a+ b \right )\\& \geqslant 4\,c(1+ ab)(a+ b)\left ( c(a+ b)+ (a+ b) \right )\\&= 2c(a+ b)^{2}(2+ 2ab)(1+ c)\\& \end{align*}
Again, by using a.m.-g.m.-inequality, we have $$2c\left(a+ b\right)^{2}\left( 2+ 2ab\right)\left( 1+ c\right)\geqslant 8abc\left(1+ a\right)\left(1+ b\right)\left(1+ c\right)$$
That is q.e.d!
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After replacing $a\rightarrow\frac{1}{a},$ $b\rightarrow\frac{1}{b}$ and $c\rightarrow\frac{1}{c}$ we need to prove that
$$(1+ab+ac+bc)^3\geq8abc(1+a)(1+b)(1+c).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$ and $abc=w^3$.
Thus, we need to prove that:
$$(1+3v^2)^3\geq8w^3(1+w^3+3v^2+3u)$$ or
$$(1+3v^2)\geq8(w^3+w^6+3v^3w^2+3uw^3)$$ and since by AM-GM $v\geq w$ and $v^4\geq uw^3,$ it's enough to prove that:
$$(1+3v^2)^3\geq8(v^3+v^6+3v^5+3v^4)$$ or
$$(1+3v^2)^3\geq8v^3(1+v)^3$$ or
$$1+3v^2\geq2v(1+v)$$ or
$$(1-v)^2\geq0$$ and we are done!
Also, $uvw$ kills this inequality, but it's not so nice.
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|
Is the complexity of $\binom{2n}{n} = O(2^n)$? How to find the complexity of $f(n)=\binom{2n}{n}$?
We know that $f(n)=\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$.
Is this $O(n^2)$? What concerns me is $n!$
|
First off, you know that:
$\begin{align*}
2^{2 n}
&= (1 + 1)^{2 n} \\
&= \sum_{0 \le k \le 2 n} \binom{2 n}{k} \\
&\ge \binom{2 n}{n}
\end{align*}$
so that $\binom{2 n}{n} = O(2^{2 n})$.
More precise estimates are from Stirling's approximation, in the variant given by Robbins ("A Remark on Stirling's Formula", AMM 62:1 (1955), 26-29):
$\begin{align*}
&n!
= \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \cdot e^{r(n)} \\
&\frac{1}{12 n + 1} < r(n) < \frac{1}{12 n}
\end{align*}$
We have:
$\begin{align*}
\binom{2 n}{n}
&= \frac{(2 n)!}{(n!)^2} \\
&= \frac{1}{\sqrt{\pi n}}\cdot 2^{2 n} \cdot e^{r(2 n) - 2 r(n)} \\
\end{align*}$
This means that:
$\begin{align*}
\binom{2 n}{n}
= \Theta\left(2^{2 n} \cdot n^{-1/2}\right)
\end{align*}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Using triangle inequality to find $\lim_{(x,y)\to(0,0)}\frac{x^3-x^2y}{x^2+y^2+xy}$ This is an exercise from my textbook where the problem is to find the limit of the function $\frac{x^3-x^2y}{x^2+y^2+xy}$ when $(x,y) \to (0,0)$.
So after changing to polar coordinates and simplifying I get the equivalent function $$r\cdot\frac{\cos^3(x)-\cos^2(x)\sin(x)}{1+\frac{1}{2}\sin(2x)}$$
In my textbook the limit of this function is solved by estimating the function using the triangle inequality (I am guessing thats $|x+y|\leq |x|+|y|$) :
$$\left|\frac{x^3-x^2y}{x^2+y^2+xy}\right|\leq r\cdot \frac{|\cos^3(x)|+\cos^2(x)|\sin(x)|}{1+\frac{1}{2}\sin(2x)} \leq r\cdot\frac{1+1}{1-\frac{1}{2}} $$
My question is how exactly was the triangle inequality used in this problem?
|
Because by the triangle inequality we obtain:
$$\left|\frac{x^3-x^2y}{x^2+xy+y^2}\right|=\frac{x^2|x-y|}{x^2+xy+y^2}\leq\frac{x^2(|x|+|y|)}{x^2+xy+y^2}\leq$$
$$\leq\frac{x^2(|x|+|y|)}{x^2-|xy|+y^2}\leq\frac{x^2(|x|+|y|)}{\left(\frac{|x|+|y|}{2}\right)^2}=\frac{4x^2}{|x|+|y|}=4|x|\cdot\frac{|x|}{|x|+|y|}\leq4|x|\rightarrow0.$$
|
{
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|
Finding general solution of $3{\times}3$ system I am given the following:
$$
x'=
\begin{bmatrix}
2 &0 &0 \\
-7&9 &7 \\
0&0 &2
\end{bmatrix}
x
$$
Solving $\det(A-\lambda I)$, I get $\lambda = 2,2,9$. Solving $\det(A-2\lambda)$, I get
\begin{bmatrix}
0&0 &0 \\
-7&7 &7 \\
0&0 &0
\end{bmatrix}
So we have geom multi. = alg. multi, so our matrix is complete.
Take $v_1 =
\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}
$
Similarly, solve $\det(A-9\lambda)$ to get
$
\begin{bmatrix}
-7&0 &0 \\
-7&0 &7 \\
0&0 &-7
\end{bmatrix}
$
So take $v_2 =
\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
$
So the general solution should be $x(t) = C_1e^{2t}\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}
+C_2e^{2t} \begin{bmatrix}
1\\
1\\
0
\end{bmatrix}
+ C_3e^{9t}\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}$
However, according to the back of the book solution, this is incorrect. What am I missing here? Thank you.
|
For $\lambda=2$, $(A-2I)v=0$ gives $-x+y+z=0$.
Letting $y=k_1$ and $z=k_2$ you get $x=k_1+k_2$.
Then $v=\begin{pmatrix}k_1+k_2\\k_1\\k_2\end{pmatrix}=k_1\begin{pmatrix}1\\1\\0\end{pmatrix}=k_2\begin{pmatrix}1\\0\\1\end{pmatrix}$.
$\implies v_1=\begin{pmatrix}1\\1\\0\end{pmatrix}$ and $v_2=\begin{pmatrix}1\\0\\1\end{pmatrix}$ are two corresponding eigenvectors.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $ \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx ~~ ?$ How to evaluate $$ \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx ~~ ?$$
I have tried to evaluate, $$ I = \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx = 5\int_{0}^2 x^{27} (x-1)^{17} dx - 3\int_{0}^2 x^{26} (x-1)^{17} dx$$ by parts but I am getting very lengthy expressions for both the integrals . Please help.
|
$$\frac{d}{dx} x^{a+1} (x-1)^{b+1} = x^a (x-1)^b \left\{(a+b+2)x - (a+1)\right\}$$
So
\begin{align*}
\int x^{26} (x-1)^{17} \left\{(26+17+2)x - (26+1)\right\} dx
&= \int x^{26} (x-1)^{17} (45x - 27) dx \\
&= 9\int x^{26} (x-1)^{17} (5x - 3) dx \\
&= x^{27} (x-1)^{18}
\end{align*}
Thus $$\int x^{26} (x-1)^{17} (5x - 3) dx = \frac{1}{9} x^{27} (x-1)^{18}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))dx$
Evaluate
$$\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))\mathrm dx$$
I tried the following ideas:
$u=a+b\tan(x)$, $du=b\sec^2(x)dx$
$$\frac{1}{b}\int \cos^4(x)\cos(u)du$$
using $1+\tan^2(x)=\sec^2(x)$
$$b^3\int \frac{\cos(u)du}{(u-a)^4}$$
I am stuck at this point. Any help?
|
$$I=\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))\mathrm dx\overset{x\to -x}=\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a-b\tan(x))\mathrm dx$$
Summing up the two integrals from above gives us:
$$ 2I= 2\cos a \int_{-\pi/2}^{\pi/2} \cos^2 x \cos(b\tan x) dx\Rightarrow I=2\cos a\int_0^{\pi/2}\cos^2 x\cos(b\tan x) dx$$
$$\overset{\tan x=t}=2\cos a\int_0^\infty \frac{\cos(b t)}{(1+t^2)^2}dt\overset{bt=x}=2b^3 \cos a\int_0^\infty \frac{\cos x}{(b^2+x^2)^2}dx$$
Consider the following integral, found here: $$I(b)=\int_0^\infty \frac{\cos x}{b^2+x^2}dx=\frac{\pi}{2b e^{b}}\Rightarrow I'(b)=-2b\int_0^\infty \frac{\cos x}{(b^2+x^2)^2}dx$$
$$\Rightarrow I= -2b^3\cos a\cdot \frac{1}{2b}I'(b)= \frac{\pi}{2}\cos a\frac{b+1}{ e^b}$$
|
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|
How can I formally prove that $3\mid 2^n+1\iff n=2m+1,m \in \mathbb N$ How can I formally prove that two to the power of some n all plus one is divisible by three when n is odd (1,3,5,7,...)? Or, another words,$$2^n+1=3k \Leftrightarrow (n=2m+1,m \in \mathbb N)\bigwedge(k \ge 1,k \in \mathbb N)$$It is actually true, for example:$$2^1+1=2+1=3=3\cdot1\\2^3+1=8+1=9=3\cdot3\\2^5+1=32+1=33=3\cdot11\\etc.$$
The Little Fermat's theorem does NOT look quite convenient for this case. Should I consider the group $1+\mathbb Z/3\mathbb Z$? Considering $2^{2\cdot m+1}+1=2\cdot2^{2m}+1=2\cdot4^m+1=3\cdot(???)?$ It just has no sense.
|
1) Let $n$ be odd.
Polynomial.$y:= x^n+1^n$, has a zero at $x=-1$.
$(x-(-1)) =(x+1)$ is a factor.
$x^n+1^n= (x+1)p_{n-1}(x)$, where $p_{n-1}$ is a polynomial of degree $n-1$.
$x=2$:
$2^n +1= (2+1)p_{n-1}(2)$, and we are done.
2) Le n be even.
$y= x^n+1^n$;
Is $x=-1$ a zero, i.e $(x+1)$ a factor?
|
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|
Prime factor inversion Define a function $f(n)$ for $n \in \mathbb{N}$ to "invert" the prime
factorization of $n$ in the following sense. Let me start with an example.
If $n= 3564 = 2^2 \cdot 3^4 \cdot 11^1$,
then $f(n) = 2^2 \cdot 4^3 \cdot 1^{11} = 256$,
inverting the base primes and their exponents.
In general, if the prime
factorization of $n$ is $p_1^{m_1} \cdot \cdots \cdot p_k^{m_k}$,
then $f(n) = m_1^{p_1} \cdot \cdots \cdot m_k^{p_k}$.
Continuing the above example:
\begin{eqnarray}
f(n) &= f(3564 = 2^2 \cdot 3^4 \cdot 11^1) & = 2^2 \cdot 4^3 \cdot 1^{11} = 256\\
f^2(n)&=f(256=2^8) &= 8^2 =64 \\
f^3(n)&=f(64 = 2^6) &= 6^2 = 36 \\
f^4(n)&=f(36 = 2^2 \cdot 3^2) &= 2^2 \cdot 2^3 = 32\\
f^5(n)&=f(32 = 2^5) &= 5^2 = 25\\
f^6(n)&=f(25 = 5^2) &= 2^5 = 32
\end{eqnarray}
and we have fallen into a $2$-cycle. My questions concern the iterated behavior of $f(n)$.
*
*Q1. If $n= \Pi p_i$, $p_i$ primes, then $f(n) = 1$.
If $n= \Pi p_i^{p_i}$, $p_i$ primes, then $f(n) = n$, a fixed point.
Are any other $n$ fixed points or $n$ that eventually lead to $1$-cycles?
*Q2. If $n= \Pi p_i^{q_i} \cdot q_j^{p_j}$, all of $p_i,p_j,q_i,q_j$ primes, then $f^2(n)=n$, a $2$-cycle.
Can you see a characterization of those $n$ that eventually fall into a $2$-cycle?
*Q3. Are there any $n$ that eventually fall into $k$-cycles
for $k \ge 3$?
|
Your function is OEIS sequence A008477, where I find the comment
For any n, the sequence n, a(n), a(a(n)), a(a(a(n))), ... is eventually periodic with period <= 2 [Farrokhi]. - N. J. A. Sloane, Apr 25 2009
The reference is to
M. Farrokhi, The Prime Exponentiation of an Integer: Problem 11315, Amer. Math. Monthly, 116 (2009), 470.
|
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|
find the $\cos 40(2\cos 80-1)=?$ find the $\cos 40(2\cos 80-1)=?$
My try :
$\cos 40(2\cos 80-1)=2\cos 40 \cos 80 -\cos40=2\cos40 (2\cos ^240-1)-\cos 40\\=4\cos^340-3\cos 40=\cos 3(40)=\cos (120)=-1/2$
I do not want to use the formula $4\cos^3x -3\cos x=\cos 3x$ .
Now How to solve ?
|
$$ \cos(40°)\cdot(2\cdot \cos(80°)-1)\\
= \cos(40°)\cdot(4\cos^2(40°)-3)\\
= 4\cos^3(40°)-3\cos(40°)\\
= \cos(120°)\\
= -\frac{1}{2} $$
Just look how simple the solution is. Why would you want to not use this approach.
Unless you want to prove the triple angle identity from scratch.
Or
$$ \cos(40°)\cdot(2\cdot \cos(80°)-1)\\
= 2\cdot \cos(40°) \cdot \cos(80°) - \cos(40°)\\
= \frac{2\cdot \sin(40°)\cdot \cos(40°) \cdot \cos(80°)}{\sin(40 °)} - \cos(40°)\\
= \frac{\sin(20°)}{2\sin(40°)} - \cos(40°)\\
= \frac{1}{4\cos(20°)} - \cos(40°)\\
= \frac{1-4\cos(40°)\cos(20°)}{4\cos(20°)}\\ $$
And this mess is so unnecessary that I let you continue if you wish so.
|
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|
Recurrence relation for $\int_{0}^{\infty} \frac{1}{(1+x^2/a)^n}dx$
Let$$I_{n,a} = \int_{0}^{\infty} \frac{1}{(1+x^2/a)^n}dx$$ where $a>0$.
Show that $$I_{n+1,a} = \frac{2n-1}{2n}I_{n,a}$$
I have tried integrating by parts but it didn't work for me, and I don't know what else can I try. Can anyone please help?
|
Just in case you're interested you can solve this for any $n \in \mathbb{R}^+$, $n \geq 1$ using the Beta and by extension the Gamma Function.
Here we will address your integral:
\begin{equation}
I(a,n) = \int_0^\infty \frac{1}{\left(1 + \frac{x^2}{a}\right)^n}\:dx
\end{equation}
We begin by making the substitution $x = \sqrt{a}t$:
\begin{equation}
I(a,n) = \int_0^\infty \frac{1}{\left(1 + \frac{at^2}{a}\right)^n} \cdot \sqrt{a}\:dt = \sqrt{a}\int_0^\infty \frac{1}{\left(1 + t^2\right)^n}\:dt
\end{equation}
We now let $t = \tan(\theta)$:
\begin{align}
I(a,n) &= \sqrt{a}\int_0^{\frac{\pi}{2}} \frac{1}{\left(1 + \tan^2(\theta)\right)^n} \cdot \sec^2(\theta)\:d\theta \nonumber \\
&= \sqrt{a}\int_0^{\frac{\pi}{2}} \cos^{2n - 2}(\theta)\:d\theta
\end{align}
Now from here you could form your recurrence relationship by IBP twice, but alternatively you can call upon one of the common definitions of the Beta Function:
\begin{equation}
B(a,b) = 2\int_0^{\frac{\pi}{2}} \cos^{2a - 1}(x)\sin^{2b - 1}(x)\:dx
\end{equation}
For $I(a,n)$ we observe that (1) $2a - 1 = 2n - 2$ and so $a = \frac{2n - 1}{2}$ and (2) $2b - 1 = 0$ and so $b = \frac{1}{2}$.We also observe that we need to scale the integral by $\frac{1}{2}$:
\begin{equation}
I(a,n) = \sqrt{a}\int_0^{\frac{\pi}{2}} \cos^{2n - 2}(\theta)\:d\theta = \frac{\sqrt{a}}{2}B\left( \frac{2n - 1}{2}, \frac{1}{2} \right)
\end{equation}
Using the relationship between the Beta and Gamma Function, this reduces to:
\begin{equation}
I(a,n) = \frac{\sqrt{a}}{2}\cdot \frac{\Gamma\left(\frac{2n - 1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{2n - 1}{2} + \frac{1}{2} \right)} = \frac{\sqrt{a}\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{2n - 1}{2}\right)}{\Gamma(n)}
\end{equation}
We can also see that this satisfies the recurrence relationship you sought.
|
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|
What is the radius of convergence for the Taylor expansion of $\frac{e^x\sin(x)}{x^2+25}$ about $x=0$? I would like to know how I could go about finding the radius of convergence for the Maclaurin series of $\frac{e^x\sin(x)}{x^2+25}$.
I am familiar with how to find the radius of convergence of more simple series such as the Maclaurin series of $\frac{1}{1-x}$, but I don't even know where I should start for something more complex like $\frac{e^x\sin(x)}{x^2+25}$.
|
$$\frac{e^x\sin x }{x^2+25}= e^x \cdot \sin x \cdot f(x)$$
where
$$f(x) = \frac{1}{25} \frac{1}{1 + \left( \frac{x}{5} \right)^2}$$
Since $e^x$ and $\sin x$ have Maclaurin series with infinite radius of convergence, we need look only at $f(x)$ which has radius of convergence $r=5$.
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots, \quad (r=1)$$
$$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots, \quad (r=1)$$
$$\frac{1}{1+\left(\frac{x}{5}\right)^2} = 1 - \left(\frac{x}{5}\right)^2 + \left(\frac{x}{5}\right)^4 - \left(\frac{x}{5}\right)^6 + \cdots, \quad (r=5)$$
The radius of convergence of the product is determined by the intersection of the circles of convergence, which is a circle of radius 5 centered at the origin: $r=5.$
|
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|
Evaluate $\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx$
How to prove $\ \displaystyle\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx=\frac{3\pi^3}{64}+\frac{\pi}{16}\ln^22-\Im\operatorname{Li}_3(1+i)$
Where $\ \displaystyle\operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}\ $
is the trilogarithm.
I managed to prove the above equality using integral manipulation, but I would like to see different approaches.
|
Using $\displaystyle\int_0^1\frac{u\ln^nx}{1-ux}\ dx=(-1)^nn!\operatorname{Li}_{n+1}(u)$ .
which can be found in Cornel's book, (Almost) impossible integrals, sums and series.
and with $n=1$, we get $\displaystyle\int_0^1\frac{u\ln x}{1-ux}\ dx=-\operatorname{Li}_2(u)$
Divide both sides by $1+u^2$ then integrate from $u=0$ to $u=1$, we get
\begin{align}
\int_0^1\int_0^1\frac{u\ln x}{(1-ux)(1+u^2)}\ dx\ du=-\int_0^1\frac{\operatorname{Li}_2(u)}{1+u^2}\ du=I
\end{align}
Lets evaluate the double integral:
\begin{align}
I&=\int_0^1\ln x\left(\int_0^1\frac{u}{(1-ux)(1+u^2)}\ du\right)\ dx\\
&=\int_0^1\ln x\left(\frac{\ln2}{2}\frac{1}{1+x^2}-\frac{\ln(1-x)}{1+x^2}-\frac{\pi}{4}\frac{x}{1+x^2}\right)\ dx\\
&=\frac{\ln2}{2}\int_0^1\frac{\ln x}{1+x^2}\ dx-\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx-\frac{\pi}{4}\int_0^1\frac{x\ln x}{1+x^2}\ dx\\
&=-\frac12G\ln2-\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx-\frac{\pi}{4}\left(\frac{\pi^2}{48}\right)\tag{1}
\end{align}
And by applying IBP to $\displaystyle\int_0^1\frac{\operatorname{Li}_2(u)}{1+u^2}\ du\ $, we get
$$I=-\int_0^1\frac{\operatorname{Li}_2(u)}{1+u^2}\ du=-\frac{\pi^3}{24}-\int_0^1\frac{\arctan u\ln(1-u)}{u}\ du\tag{2}$$
From $(1)$ and $(2)$, we get
$$\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx=\frac{3\pi^3}{64}-\frac12G\ln2+\int_0^1\frac{\arctan u\ln(1-u)}{u}\ du$$
substituting $\ \displaystyle\int_0^1 \frac{\arctan u\ln(1-u)}{u}\ du=\frac{\pi}{16}\ln^22+\frac12G\ln2-\text{Im}\operatorname{Li}_3(1+i)\ $ ( proved here)
gives: $$\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx=\frac{3\pi^3}{64}+\frac{\pi}{16}\ln^22-\Im\operatorname{Li}_3(1+i)$$
|
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|
Integrating quadratics in denominator I'm following a book on Calculus that introduces partial fraction expansion. They discuss common outcomes of the partial fraction expansion, for example that we are left with an integral of the form:
$$
\int \frac{dx}{x^2+bx+c}
$$
And then we can use complete the square and $u$-substitution:
$$
x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right) = u^2+\alpha^2
$$
where $u=x+\frac{b}{2}$ and $\alpha=\frac{1}{2}\sqrt{4c-b^2}$. The book says: "... this is possible because $4c-b^2>0$."
Eagerly I tried an example, using the quadratic $x^2-8x+1$.
Then let $a=1, b=-8, c=1$ and:
$$
x^2+bx+c = \left(x+\frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right) = u^2 + \alpha^2
$$
where $u = x+b/2 = x-4$ but we run into a problem:
$$\alpha=\frac{1}{2}\sqrt{4c-b^2} = \frac{1}{2}\sqrt{(4)(1) - (-8)^2} = \frac{1}{2}\sqrt{4-64}$$
So $\alpha$ doesn't satisfy $4c-b^2>0$. Maybe I'm missing something obvious? Or is the book missing a caveat that this method doesn't always work. Because in the book they make it sound like "... this is possible because $4c-b^2>0$." is always true.
|
The answer is that this method doesn't always work. In this particular case, we can use the following: Since
$$x^2-8x+1=0 \quad \Leftrightarrow \quad x=\frac{8\pm \sqrt{60}}{2}=4\pm \sqrt{15}$$
this polynomial can be factored as $$\big(x-(4+\sqrt{15})\big)\big(x-(4-\sqrt{15})\big)$$
use partial fraction like this
$$\frac{1}{x^2-8x+1}=\frac{1}{(x-4-\sqrt{15})(x-4+\sqrt{15})}=\frac{A}{x-4-\sqrt{15}}+\frac{B}{x-4+\sqrt{15}}$$
and proceed. Or maybe, since $x^2-8x+1=(x-4)^2-15$ just set $x-4=\sqrt{15}\sec \theta$ (trigonometric substitution).
|
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"url": "https://math.stackexchange.com/questions/3281352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Designate set $ \lbrace n \in\mathbb N : 2^{n-1} | n! \rbrace $ Actually I found out that if $n = 2^k$, $k\in\mathbb Z_{+}$ we can say that it's true.
Here's the proof:
Let $n = 2^k$, $k\in\mathbb Z_{+}$
Then $\nu _{2}(n!) = \nu _{2}((2^k)!) = \lfloor \frac{2^k}{2} \rfloor + \lfloor \frac{2^k}{4} \rfloor +\dots+ \lfloor \frac{2^k}{2^p} \rfloor $
where $2^p \leq n < 2^{p+1}$.
Further we have:
$\nu _{2}(n!) = 2^{k-1} + 2^{k-2} +\dots+2+1 = 2^k - 1$ and $ \nu _{2}(2^{n-1}) = 2^k - 1 $. Since $\nu _{2}(2^{n-1}) \leq \nu _{2}(n!) $ we can say that $\nu _{p}(2^{n-1}) \leq \nu _{p}(n!) $ and this part implies that $2^{n-1} \mid n!$.
My problem is that I can't prove it's false for numbers $n = 2^k*a $ where $a$ is a non-negative odd number and $a > 1$.
|
The power of $2$ in $n$ is given by $\left\lfloor\frac n2\right\rfloor+\left\lfloor\frac n4\right\rfloor+\cdots\left\lfloor\frac n{2^{p}}\right\rfloor$ where $2^p\leq n<2^{p+1}$
For the given condition to be satisfied, we have
$$n-1\leq \left\lfloor\frac n2\right\rfloor+\left\lfloor\frac n4\right\rfloor+\cdots+\left\lfloor\frac n{2^{p}}\right\rfloor$$
$$\left\lfloor\frac n2\right\rfloor+\left\lfloor\frac n4\right\rfloor+\cdots+\left\lfloor\frac n{2^{p}}\right\rfloor\leq\frac n2+\frac n4+\cdots+\frac n{2^p}=n-\frac{n}{2^p}$$
However, as $n\geq2^p$, $$n-\frac{n}{2^p}\leq n-1$$
As equality must be satisfied everywhere, we have that $\frac n{2^k}$ is an integer for $1\leq k\leq p$, that is, $n=2^p$.
|
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|
Prove that the diophantine equation $2x^2-5y^2=7$ has no integer solutions. My attempt: I rewrote it as $2x^2=5y^2+7. 2x^2$ is always even, so in order for the RHS to be even, this means that $5y^2$ must be odd since an odd number plus $7$ is even.
If I evaluate when y is odd, so if $y=2k+1$ for some integer $k$, I get: $2x^2=20k^2+20k+12$. This is the same as $x^2=10k^2+10k+6$, which implies that $x^2$ is congruent $6$ (mod $10$).
Here, I arrive at an issue because if $x=4$, then I get that $x^2$ is congruent to $6$ (mod $10$), but I am supposed to show that the equation does not have a $6$ (mod $10$) congruency.
|
Modulo $7$, $2x^2-5y^2=7$ would mean $2x^2+2y^2\equiv0$ or $x^2+y^2\equiv0$ or $x^2\equiv-y^2$.
Now $x^2, y^2\equiv 0, 1, 2, $ or $4 \pmod 7$, so the only solution would be $x^2\equiv y^2\equiv0\pmod7$.
But this means $7|x,y$, which means $49|2x^2-5y^2=7,$ a contradiction.
|
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|
A finite product : $\prod_{k=0}^{n-1}(1-\frac{1}{n-1+k})$ Find the maximum and minimum of the following products :
$A)$ $\prod_{k=0}^{n-1}(1-\frac{1}{n-1+k})$
$B)$ $\prod_{k=0}^{n-1}(1-\frac{1}{n+1-k})$
My idea is :
$n-1+k>k$ then : $\frac{1}{n-1+k}<\frac{1}{k}$
We obtain :
$\prod_{k=0}^{n-1}(1-\frac{1}{k})$
But I don't have ideas to complete my work , and is my attempt correct ?
|
HINT: (Assuming $ n \geq 2 $)
$\prod_{k=0}^{n-1}(1-\frac{1}{n+k-1}) = \prod_{k=0}^{n-1}\frac{n+k-2}{n+k-1} = \frac{n-2}{n+(n-1)-1} = \frac{1}{2} \cdot \frac{n-2}{n-1}$
|
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|
The integral $\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}}dx =\frac{\pi}{2}\ln (1+\sqrt{2})$ At Mathematica the numerical value of the integral
$$\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}} dx$$
equals 1.3844.., which is nothing but $\frac{\pi}{2}\ln (1+\sqrt{2})=z$. Also, one of its transformed forms is evaluated to be $z$ by Mathematica. The question is: How to do it by hand?
|
Here we will address your integral:
\begin{equation}
I = \int_0^\infty \frac{\cot^{-1}\left(\sqrt{1 + x^2} \right)}{\sqrt{1 + x^2}}\:dx
\end{equation}
We first let $x = \tan(s)$:
\begin{align}
I &= \int_0^\frac{\pi}{2} \frac{\cot^{-1}\left(\sqrt{1 + \tan^2(s)} \right)}{\sqrt{1 + \tan^2(s)}}\cdot \sec^2(s)\:ds = \int_0^\frac{\pi}{2} \sec(s)\cot^{-1}\left(\sec(s)\right)\:ds \nonumber \\
&= \int_0^\frac{\pi}{2} \frac{\arctan\left(\cos(s)\right)}{\cos(s)}\:ds
\end{align}
Here I will now employ Feynman's Trick and introduce the following function:
\begin{equation}
J(t) = \int_0^\frac{\pi}{2} \frac{\arctan(t\cos(s))}{\cos(s)}\:ds
\end{equation}
Here $I = J(1)$ and $J(0) = 0$. ,By Leibniz's Integral Rule we now differentiate under the curve with respect to $t$:
\begin{align}
J'(t) &= \int_0^\frac{\pi}{2} \frac{1}{t^2\cos^2(s) + 1}\cdot \cos(s) \cdot \frac{1}{\cos(s)}\:ds = \int_0^\frac{\pi}{2}\frac{1}{t^2\cos^2(s) + 1}\:ds \nonumber \\
&= \left[ \frac{1}{\sqrt{t^2 + 1}}\arctan\left(\frac{\tan(s)}{\sqrt{t^2 + 1}}\right) \right]_0^\frac{\pi}{2} = \frac{\pi}{2}\frac{1}{\sqrt{t^2 + 1}}
\end{align}
Thus,
\begin{equation}
J(t) = \frac{\pi}{2} \int \frac{1}{\sqrt{t^2 + 1}}\:dt = \frac{\pi}{2}\sinh^{-1}(t) + C
\end{equation}
Where $C$ is the constant of integration. To resolve $C$ we use our condition $J(0) = 0$:
\begin{equation}
J(0) = 0 = \frac{\pi}{2}\sinh^{-1}(0) + C \rightarrow C = 0
\end{equation}
Thus,
\begin{equation}
J(t) = \frac{\pi}{2}\sinh^{-1}(t)
\end{equation}
We now may resolve $I$
\begin{equation}
I = J(1) = \frac{\pi}{2}\sinh^{-1}(1)
\end{equation}
Noting that
\begin{equation}
\sinh^{-1}(t) = \ln\left|t + \sqrt{t^2 + 1}\right|
\end{equation}
We see that another representation for $I$ is given by:
\begin{equation}
I = \frac{\pi}{2}\ln(1 + \sqrt{2})
\end{equation}
|
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|
Solve $x^{3} = 6+ 3xy - 3 ( \sqrt{2}+2 )^{{1}/{3}} , y^{3} = 9 + 3xy(\sqrt{2}+2)^{{1}/{3}} - 3(\sqrt{2}+2)^{{2}/{3}}$ Solve the system of equations for $x,y \in \mathbb{R}$
$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}} $
$ y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}$
I just rearranged between those equations and get $ \frac{y^{3}-9}{x^{3} -6} = (\sqrt{2}+2)^{\frac{1}{3}}$ then I don't know how to deal with it.
Please give me a hint or relevant theorem to solve the equation.
Thank you, and I appreciate any help.
Furthermore I get an idea
how about we subtract two equation and get
$y^{3}-x^{3} = 3 + 3xy((\sqrt{2}+2)^{\frac{1}{3}} -1) - (3(\sqrt{2}+2)^{\frac{2}{3}} - 3(\sqrt{2}+2)^{\frac{1}{3}})$
$(y-x)(x^2+xy+y^2)= 3[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)]$
$y-x = 3$ and $x^2 +xy+y^2 =[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $
or,
$y-x = [1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $ and $x^2 +xy+y^2 = 3$
Am I on the right track?
|
Write the equations as $$
x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}\\
\frac{(xy)^{3}}{x^3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}
$$
Now let $z = xy$ which gives you
$$
f(z) = z^3 - (9 + 3z(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}})(6+ 3z - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}) = 0
$$
You can analyse $f(z)$ easily and find that it has only one root at roughly $z = 14.485$.
From here, the original two equations give $x$ and $y$.
|
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|
$\int_{0}^{\infty }{{{x}^{n}}\sin \left( {{x}^{1/4}} \right)\exp \left( -{{x}^{1/4}} \right)dx}=0$ I'd like to show that for all positive integers $n$ we have
$$I\left( n \right)=\int_{0}^{\infty }{{{x}^{n}}\sin \left( {{x}^{1/4}} \right)\exp \left( -{{x}^{1/4}} \right)dx}=0.$$
This is true after some computer experiments, besides after setting $x={{u}^{4}}$ we get
$$I\left( n \right)=4\int_{0}^{\infty }{{{u}^{4n+3}}\sin \left( u \right)\exp \left( -u \right)du}.$$
But how to proceed? Integration by parts maybe?
|
Let
$$I_n = \int_0^\infty x^n \sin (\sqrt[4]{x}) \exp (\sqrt[4]{x}) \, dx, \qquad n \in \mathbb{N}.$$
After enforcing a substitution of $x \mapsto \sqrt[4]{x}$ one has
$$I_n = 4 \int_0^\infty x^{4n + 3} e^{-x} \sin x \, dx.$$
The following useful property for the Laplace transform will now be employed to evalaute the integral:
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$
Noting that
$$\mathcal{L} \{\ x^{4n + 3} \sin x \}(t) = \frac{(4n + 3)!}{(1 + t^2)^{2n + 2}} \sin \left [4(n + 1) \tan^{-1} \left (\frac{1}{t} \right ) \right ],$$
and
$$\mathcal{L}^{-1} \{e^{-x} \} (t)= \delta (t - 1),$$
where $\delta (x)$ is the Dirac delta function,
then
\begin{align}
I_n &= 4\int_0^\infty x^{4n + 3} \sin x \cdot e^{-x} \, dx\\
&= 4\int_0^\infty \mathcal{L} \{x^{4n + 3} \sin x\} (t) \cdot \mathcal{L}^{-1} \{e^{-x} \} (t) \, dt\\
&= 4(4n + 3)! \int_0^\infty \frac{1}{(1 + t^2)^{2n + 2}} \sin \left [4(n + 1) \tan^{-1} \left (\frac{1}{t} \right ) \right ] \cdot \delta (t - 1) \, dt\\
&= \frac{4(4n + 3)!}{2^{2n + 2}} \sin [4(n + 1) \tan^{-1} (1)]\\
&= \frac{(4n + 3)!}{4^n} \sin ((n + 1)\pi)\\
&= 0,
\end{align}
as required to show.
A direct approach
Here is an approach where the results of the above Laplace transform and its inverse are not quoted in advance.
From
$$I_n = 4 \int_0^\infty e^{-x} x^{4n + 3} \sin x \, dx,$$
we re-write this as
$$I_n = -4 \operatorname{Im} \int_0^\infty x^{4n + 3} e^{-(1 + i)x} \, dx.$$
Integrating by parts $(4n + 3)$ times gives
\begin{align}
I_n &=-4 \, \operatorname{Im} \left [\frac{(-1)^{4n + 3} (4n + 3)!}{(1 + i)^{4n + 3}} \right ] \int_0^\infty e^{-(1 + i)x} \, dx\\
&= 4 \, \operatorname{Im} \left [\frac{(-1)^{4n + 4} (4n + 3)!}{(1 + i)^{4n + 4}} \right ]\\
&= 4(4n + 3)! \operatorname{Im} \left [\frac{1}{(1 + i)^{4n + 4}} \right ]\\
&= 0,
\end{align}
where the last line is due to the fact that
$$\frac{1}{(1 + i)^{4n + 4}} = \frac{1}{2^{2n + 2}} \left [\cos ((n + 1)\pi) + i \sin ((n + 1)\pi) \right ].$$
|
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"url": "https://math.stackexchange.com/questions/3289346",
"timestamp": "2023-03-29T00:00:00",
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|
Show $\int_{\max\{0,1-c/x\}}^1(1-v)\,dv=\frac{1}{2}\min\{c/x,1\}^2$ Let $c,x\in [0,\infty)$. Using $-\max\{a,b\}=\min\{-a,-b\}$ we get
\begin{align}
\int_{\max\{0,1-c/x\}}^1(1-v)\,dv&= 1-\max\{0,1-c/x\}-\left(\frac{1}{2}-\frac{1}{2}\max\{0,1-c/x\}^2\right)\\
&=\frac{1}{2}-\max\{0,1-c/x\}+\frac{1}{2}\max\{0,1-c/x\}^2\\
&= \frac{1}{2}+\min\{0,c/x-1\}+\frac{1}{2}\min\{0,c/x-1\}^2 \\
&=\min\left\{\frac{1}{2},c/x-\frac{1}{2}\right\}+\frac{1}{2}\min\{0,c/x-1\}^2
\end{align}
How do I get to $\frac{1}{2}\min\{c/x,1\}^2$?
|
\begin{align*}
&=\frac{1}{2}\left(1+2\min(0,c/x-1)+\min(0,c/x-1)^2\right) \\
&=\frac{1}{2}\left(1+\min(0,c/x-1)\right)^2 \\
&=\frac{1}{2}\min(1,c/x)^2
\end{align*}
|
{
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"url": "https://math.stackexchange.com/questions/3291908",
"timestamp": "2023-03-29T00:00:00",
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|
Radius of convergence of $\sum_{n=0}^\infty a_n x^n$, with $a_{n+2} = \frac{n(n+1) a_{n+1} - a_n}{(n+2)(n+1)}, a_2 = -a_0/2$ Problem
Find the radius of convergence of the power series
$$
\sum_{n=0}^\infty a_n x^n
$$
where $a_n$'s are defined by the following recurrence relation
$$
\begin{aligned}
a_{n+2} &= \frac{n(n+1) a_{n+1} - a_n}{(n+2)(n+1)}, n\ge 1 \\[8pt]
a_2 &= -a_0/2
\end{aligned}
$$
with arbitrary $a_0, a_1$.
Try1
I have tried to directly applying the ratio test, by dividing the above recurrence relation by $a_{n+1}$,
$$
r_{n+1} = \frac{n}{n+2} - \frac{1}{(n+2)(n+1)} \frac{1}{r_n}
$$
where $r_n := a_{n+1}/a_n$. We can observe that
$$
r_n r_{n+1} = \frac{n}{n+2} r_n - \frac{1}{(n+2)(n+1)}
$$
but I cannot proceed to find the expression for
$$
\lim_{n \to \infty} \vert r_n \vert
$$
Try2
Since the recurrence relation depends on the arbitrary choice pf $a_0, a_1$, let us proceed by letting $a_0:= 0$. We have
$$
\begin{aligned}
a_2 &= a_0 = 0 \\[7pt]
a_3 &= -\frac{1}{6} a_1 \\[7pt]
a_4 &= \frac{2}{4} a_3 = -a_1/12 \\[7pt]
a_5 &= \frac{3}{5} a_4 - \frac{1}{5\cdot 4}a_3 = -\frac{1}{24} a_1 \\[7pt]
\end{aligned}
$$
where I cannot find any simple rules. Likewise, by letting $a_1:= 0$, we have
$$
\begin{aligned}
a_2 &= -a_0/2 \\[7pt]
a_3 &= \frac{1}{3} a_2 = -\frac{1}{6}a_0 \\[7pt]
a_4 &= \frac{2}{4} a_3 - \frac{1}{4\cdot 3}a_2 = -\frac{1}{24} a_0\\[7pt]
a_5 &= \frac{3}{5} a_4 - \frac{1}{5\cdot 4}a_3 = -\frac{1}{60} a_0 \\[7pt]
\end{aligned}
$$
where again I have failed to find any rules. So, I cannot find the interval that the following composition is valid.
$$
\sum_{n=0}^\infty a_n x^n = a_0 \left[ 1 - x^2/2 - x^3/6 - x^4/24 - x^5/60 + \cdots \right] + a_1 \left[ x - x^3/6 - x^4/12 - x^5/24 + \cdots\right]
$$
Any help will be appreciated.
|
The trick is to notice $a_{n+2}=-\frac{f_n}{(n+2)!}(a_0+a_1)$ for $n\ge1$ with $f_1=f_2=1,\,f_n=nf_{n-1}-f_{n-2}$. Asymptotically $f_n\sim nf_{n-1}$, so $\frac{f_n}{(n+2)!}$ is a sequence in which the ratio of consecutive terms $\to1$. Thus the radius of convergence is also $1$.
|
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|
Given three positive numbers $a,b,c$. Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{b+1}}\geqq3\sqrt[3]{\frac{4\,abc}{3\,abc+1}}$ .
Ji Chen. Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{cyc}\sqrt{\frac{a+ b}{b+ 1}}\geqq 3\sqrt[3]{\frac{4\,abc}{3\,abc+ 1}}$$
Of course, we've to solve it by $uvw$, before that, I tried to use Holder-inequality with integer polynomials but without a high probability of success for me against this particular problem ...
I found here: https://artofproblemsolving.com/community/c6h538065p3209975, something obvious
|
By AM-GM $$\sum_{cyc}\sqrt{\frac{a+b}{b+1}}\geq3\sqrt[6]{\prod\limits_{cyc}\frac{a+b}{a+1}}.$$
Thus, it's enough to prove that
$$(a+b)(a+c)(b+c)(3abc+1)^2\geq16a^2b^2c^2(a+1)(b+1)(c+1).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$(9uv^2-w^3)(3w^3+1)^2\geq16w^6(w^3+3v^2+3u+1)$$ and since by AM-GM $uv^2\geq w^3,$ it's enough to prove that
$$uv^2(3w^3+1)^2\geq2w^6(w^3+3v^2+3u+1),$$ which is true by AM-GM.
Can you end it now?
|
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|
Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if someone has a more straightforward solution.
So applying the sum formula for sine and doing simple algebra we have:
$$\lim_{x\to 0} \frac{\cos{2} \,(\sin{x^2}-\sin{x})}{x} - \frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
The first limit is easy to evaluate and is equal to $-\cos{2}$. However, the second limit is harder, as it follows:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
I came across a solution by using the following sum-to-product identity:
$$\cos{A}-\cos{B}=-2\sin{\Big(\frac{A+B}{2}\Big)} \sin{\Big(\frac{A-B}{2}\Big)}$$
Setting $A=x^2$ and $B=x$, we have that
$$\cos{x^2}-\cos{x}=-2\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}$$
This is my only point of concern whether I applied the identity correctly. The rest of it flows more easily:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} = \lim_{x\to 0} \frac{-2\sin{2}\,\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}}{x}$$
$$= -2 \sin{2} \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} \lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}\Big)} $$
The first limit can be solved as it follows:
$$\lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} = \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)} \Big(\frac{x^2+x}{2}\Big)}{x \Big(\frac{x^2+x}{2}\Big)} = 1 \cdot \lim_{x \to 0} \frac{x^2 + x}{2x} = \frac{1}{2} $$
The second limit is equal to zero
$$\lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}}\Big)=0$$
|
The second limit is not really hard: leaving out the $\sin2$ factor, we have
$$
\lim_{x\to0}\frac{\cos(x^2)-\cos x}{x}=
\lim_{x\to0}\frac{\cos(x^2)-1+1-\cos x}{x}
$$
It's easy to show that
$$
\lim_{x\to0}\frac{1-\cos x}{x}=0
$$
so we remain with
$$
\lim_{x\to0}\frac{\cos(x^2)-1}{x}=\lim_{x\to0}\frac{\cos(x^2)-1}{x^2}x
$$
which is zero for the same reason.
|
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|
Find all extrema of a complicated trigonometric function Problem
Find all local extrema for
$$f(x) = \frac{\sin{3x}}{1+\frac{1}{2}\cos{3x}}$$
Attempted solution
My basic approach is to take the derivative, set the derivative equal to zero and solve for x.
Taking the derivative with the quotient rule and a few cases of the chain rule for the trigonometric functions with a final application of the Pythagorean identity:
$$f'(x) = \frac{(1+\frac{1}{2}\cos{3x})(3\cos{3x})+1.5\sin{3x}\sin {3x}}{(1+\frac{1}{2}\cos{3x})^2} = \frac{3\cos 3x+1.5\cos^2 3x + 1.5\sin^2 3x}{(1+\frac{1}{2}\cos 3x)^2} = \frac{3 \cos 3x + 1}{(1+\frac{1}{2}\cos 3x)^2}$$
Putting it equal to zero and solving for x:
$$3\cos 3x + 1 = 0 \Rightarrow x = \frac{\arccos{\Big(-\frac{1}{3}\Big)}}{3} = \frac{\pi}{6} + \frac{2\pi n}{3}$$
...however the expected answer is $\pm\frac{2\pi}{9} + \frac{2\pi n}{3}$
So I must have gone wrong somewhere.
|
This function has period $\frac{2\pi}3$ and it is an odd function, so we need to determine its variations only on $\bigl[0,\frac\pi 3\bigr]$.
Now simplifying the derivative, you get
$$f'(x)=\frac{3\bigl(\frac12+\cos 3x\bigr)}{\bigl(1+\frac{1}{2}\cos{3x}\bigr)^2},$$
which has the sign of $\;\frac12+\cos 3x$, so we have to solve the inequation
$$\cos 3x>-\tfrac 12\quad\text{on}\quad \bigl[0,\tfrac\pi 3\bigr].$$
Can you continue?
|
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|
Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$.
Given non-negatives $x, y, z$ such that $x + y + z = 4$. Calculate the maximum value of $$\large x^3y + y^3z + z^3x$$
As an assumption, the maximum value is $27$, occured when $(x, y, z) = (0, 1, 3)$.
I have a guess about a working-in-process prove. Let $y$ be the median of $x, y, z$.
$$\iff (zx - yz)(y^2 - z^2) \ge 0 \iff y^2zx - y^3z - z^3x + yz^3 \ge 0$$
$$\iff x^3y + y^3z + z^3x \le x^3y + y^2zx + yz^3 = y(x^3 + xyz + z^3)$$
And further down the line is what I haven't accomplished yet.
|
I like the following way.
Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Thus, by Rearrangement and AM-GM we obtain:
$$x^3y+y^3z+z^3x=x^2\cdot xy+y^2\cdot yz+z^2\cdot zx\leq a^2\cdot ab+b^2\cdot ac+c^2\cdot bc=$$
$$=b(a^3+c^3+abc)\leq b(a+c)^3=27\cdot b\left(\frac{a+c}{3}\right)^3\leq27\left(\frac{b+3\cdot\frac{a+c}{3}}{4}\right)^4=27.$$
|
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|
Matrices commuting with a given $3\times 3$ complex matrix.
Let $A$ be a $3\times 3$ complex matrix. Let $C(A)$ be the vector space of complex matrices that commute with $A$. Show that the complex dimension of $C(A)$ is at least $3$.
I know that this kind of questions has been asked many times on this site. And there is an explicit formula for the dimension of $C(A)$ given by Frobenius viewing the matrices $B$ that commute with $A$ as endomorphisms of $\mathbb C[\lambda]$-module.
But I am looking for a more elementary way to show the lower bound of the dimension of $C(A)$ is $3$.
For example, I have already found that $\operatorname{Span}\{ I, A \}$ is a two-dimensional subspace of $C(A)$ for $A\notin\operatorname{Span}\{I\}$, where $I$ is the identity matrix. But how to find another matrix that is linearly independent of $\operatorname{Span}\{I, A\}$? Thanks.
|
By change of basis we can assume that $A$ is in Jordan normal form.
Case 1: $A = \begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}$.
Then the matrices $\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}, \begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}, \begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$ are linearly independent and commute with $A$.
Case 2: $A = \begin{pmatrix}a&1&0\\0&a&0\\0&0&b\end{pmatrix}$.
Then the matrices $\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}, \begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}, \begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$ are linearly independent and commute with $A$.
Case 3: $A = \begin{pmatrix}a&1&0\\0&a&1\\0&0&a\end{pmatrix}$.
Then the matrices $\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}, \begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}, \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$ are linearly independent and commute with $A$.
More generally, for each Jordan block of size $k\times k$ we can quickly find $k$ independent commuting matrices $A_i$ by setting the entries on the $(i-1)$-th upper off-diagonal to $1$ and all other entries to $0$. Then for any matrix $A$ in Jordan form, you can construct commuting matrices whose nonzero entries are an off-diagonal of ones spanning the location of a single Jordan block of $A$.
|
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|
How can one integrate $\int\frac{1}{(x+1)^4(x^2+1)} dx$?
How can one integrate $\displaystyle\int\frac{1}{(x+1)^4(x^2+1)}\ dx$?
Attempt:
I tried partial fraction decomposition (PFD) and got lost. The method of u-substitution didn't work for me either.
What else can I do? Can one calculate the integral without PFD?
|
We use a variant of the Heaviside method. Shift by one and consider
$$\frac{1}{z^4(z^2-2z+2)}\text{.}$$
Develop $1/(z^2-2z+2)$ in series about $z=0$, keeping the remainder exactly as you go:
$$\frac{1}{z^2-2z+2}=\frac{1}{2}+\frac{z}{2}+\frac{z^2}{4}-\frac{z^4}{4(z^2-2z+2)}\text{.}$$
Then
$$\frac{1}{z^4(z^2-2z+2)}=\frac{2 + 2z + z^2}{4z^4}-\frac{1}{4(z^2-2z+2)}\text{.}$$
Can you take it from here?
|
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|
How to show the matrix has Rank $\le 5$
I want to show that the following matrix has Rank $\le 5$.
The matrix is
\begin{bmatrix}
2&1&1&1&0&1&1&1\\
1&2&1&1&1&0&1&1\\
1&1&2&1&1&1&0&1\\
1&1&1&2&1&1&1&0\\
0&1&1&1&2&1&1&1\\
1&0&1&1&1&2&1&1\\
1&1&0&1&1&1&2&1\\
1&1&1&0&1&1&1&2
\end{bmatrix}
I found that there is a submatrix in the matrix which has rank $ =4$ given by
$[2,1,1,1],[1,2,1,1],[1,1,2,1],[1,1,1,2]$.
I need to show the given matrix has at least 3 zero rows in order to show that Rank $\le 5$..
But I dont know how to show it.
Can someone help.
|
Let
$$
A = \begin{bmatrix} 2 & 1 & 1 & 1 \\ 1 & 2 & 1 &1 \\ 1 &1&2&1\\1&1&1&2\end{bmatrix}, \quad B = \begin{bmatrix} 1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1 \end{bmatrix} = vv^{T}
$$
where $v = [1, 1, 1 ,1]^{T}$. It is easy to see that $rank(B) = 1$. Now our original matrix is
$$
X = \begin{bmatrix} A & 2B - A\\ 2B -A & A\end{bmatrix}
$$
and we have
$$
\begin{bmatrix} I & I \\0&I \end{bmatrix}X\begin{bmatrix} I&0\\I&I \end{bmatrix} = \begin{bmatrix} 4C&2C \\ 2C & A\end{bmatrix}
$$
so the rank of $X$ is same as the rank of the above matrix. (Rank of matrix is preserved by multiplying invertible matricies) Since all 4 colums and 4 rows are the same, Gaussian elimination (elementary row and column operation) gives a matrix with same rank
$$
\begin{bmatrix} 0_{3\times 3} & 0_{3 \times 5} \\ 0_{5\times 3} & C \end{bmatrix}
$$
where
$$
C = \begin{bmatrix} 4&2&2&2&2 \\ 2&2&1&1&1\\2&1&2&1&1\\2&1&1&2&1\\2&1&1&1&2\end{bmatrix}
$$
which is invertible by Gaussian elemination again.
|
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|
an interesting inequality with Muirhead If $x,y,z>0$ I have to prove that
$\sum\limits_{cyc}^{} \frac { x(x^3 yz+x^2-x y^3 z-yz) }{(1+x y^2)(1+xyz)} \ge 0$ holds. My approach is that from Muirhead's inequality the inequality is true since $(4,1,1)≻(2,3,1)$ and $(2,1)≻(1,1)$. Am I right?
|
We need to prove that
$$\sum_{cyc}\frac{x(x^2(1+xyz)-yz(1+xy^2)}{(1+xy^2)(1+xyz)}\geq0$$ or
$$\sum_{cyc}\frac{x^3}{1+xy^2}\geq\frac{3xyz}{1+xyz}.$$
Now, by C-S
$$\sum_{cyc}\frac{x^3}{1+xy^2}=\sum_{cyc}\frac{x^4}{x+x^2y^2}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(x+x^2y^2)}.$$
Id est, it's enough to prove that
$$(1+xyz)(x^2+y^2+z^2)\geq3xyz\sum_{cyc}(x+x^2y^2),$$ which is true because
$$(x^2+y^2+z^2)^2\geq3xyz(x+y+z)$$ and
$$xyz(x^2+y^2+z^2)^2\geq3xyz(x^2y^2+x^2z^2+y^2z^2).$$
Indeed, we can say that both last inequalities are true by Muirhead.
|
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|
Prove equality - Necessity and sufficiency I'm trying to figure out an example from a book with math problems. So here is an example (the equality should be proven)
$$ \sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}} = 4 $$
The author is suggesting to have $ \sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}} = x $, so after cubing both parts of the equation we get:
$$ 38 + \sqrt{1445} + 38 - \sqrt{1445} + 3\sqrt[3]{(38 + \sqrt{1445})(38 - \sqrt{1445})} \cdot x = x^3 $$ or $ x^3 + 3x - 76 = 0. $ If we substitute $4$ in the equation we'll figure out that $4$ is a root. So this part is very clear for me, but what I don't understand is the fact that the $4$ is a root of the equation is insufficient to prove the problem. The author is trying to show that the $4$ is a single root of the equation and only after that he concludes that the problem is proved.
So why is the author doing so, why proving that $4$ is a root of the equation doesn't prove the whole problem?
|
Because the substitution $\sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}}=x$ in the equation
$$38 + \sqrt{1445} + 38 - \sqrt{1445}+3 \sqrt[3]{(38 + \sqrt{1445})(38 - \sqrt{1445})}\left( \sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}}\right)=x^3$$
is not an equivalent transformation of the last equation.
We can get another roots except a value $\sqrt[3]{38 + \sqrt{1445}} + \sqrt[3]{38 - \sqrt{1445}}$, which we need to find.
For example, if we'll try to make the similar thing with $$1+1=x,$$ so we obtain:
$$1+1+3\cdot1\cdot1(1+1)=x^3$$ and after a similar substitution $1+1=x$ in the last equation we obtain:
$$2+3x=x^3$$ or
$$x^3-3x-2=0$$ or
$$x^3-2x^2+2x^2-4x+x-2=0$$ or
$$(x-2)(x+1)^2=0,$$ which gives also $$x=-1.$$
Id est, we need to prove that the equation $$x^3+3x-76=0$$ has an unique real root, that Robert Z made.
We can use also the way, which is based on on the identity:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
The factor $$a^2+b^2+c^2-ab-ac-bc=\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=$$
$$=\frac{1}{2}\sum_{cyc}(a^2+b^2-2ab)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$
The equality occurs for $a=b=c$ only.
In our case $a=\sqrt[3]{38 + \sqrt{1445}},$ $b=\sqrt[3]{38 - \sqrt{1445}}$ and $c=-x$ and we see that $a\neq b$, which gives $\sum\limits_{cyc}(a^2-ab)\neq0,$ which says that in our case the transform from $$a+b+c=0$$ to
$$(a+b+c)\sum\limits_{cyc}(a^2-ab)=0$$ is an equivalent transformation and we don't get an extraneous root.
After this transform we obtain:
$$a^3+b^3+c^3-3abc=0$$ or
$$38 + \sqrt{1445} + 38 -\sqrt{1445}-x^3-3 \sqrt[3]{38 + \sqrt{1445}}\cdot\sqrt[3]{38 - \sqrt{1445}}\cdot(-x)=0$$ or
$$x^3+3x-76=0$$ and since $4$ is a root of the equation, we see that this root is an unique because our transformations was equivalent.
|
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Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$. Let $n \in \mathbb{N}$. Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$.
I began this problem giving some values for $n$ and I found that $\gcd(n^2+3, (n+1)^2+3)=1$ for most of $n$ I tried, but if $n=6$, then $\gcd=13$.
Then I tried to prove that only for $n=6$, $\gcd \neq 1$ but I couldn't.
Can someone help me with this problem?
|
$$\gcd(n^2+3,(n+1)^2+3)=\gcd(n^2+3,2n+1)=\gcd(4n^2+12,2n+1)=$$
$$=\gcd(4n^2+4n+1-4n+11,2n+1)=\gcd(4n-11,2n+1)=$$
$$\gcd(4n+2-13,2n+1)=\gcd(13,2n+1).$$
Can you end it now?
|
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|
Natural number solutions of $x^3+y^3=p^2$, $x$ and $y$ are integers, $p$ is prime number. Natural number solutions of $x^3+y^3=p^2$, $x$ and $y$ are integers, $p$ is prime number.
I have found $(1,2,3)$ is a solution and there seem to be no other solution.
Can anyone prove it?
|
Hint: Since $p$ is a prime number, we must have : $x+y = p, x ^2 - xy + y^2 = p$ or $x + y = 1, x^2 - xy + y^2 = p^2$ or $x+y = p^2, x^2-xy+y^2 = 1$. Either case is quite simple enough to handle. But for the last case which is the case the OP pointed out as a missing case in the comment.We claim that this case can't happen.For if it were, then since $p^2 > 0$, $x,y$ can't be both negative. Thus if both are $>0$, then let's say if $x > y$, then $x^2-xy+y^2 = x(x-y)+y^2 \ge x+y^2 > x+y \implies 1 > p^2$, contradiction. If $xy < 0$, then $x^2-xy+y^2 > x^2+y^2 > x+y^2 > x+y \implies x^2 -xy+y^2 > x+y \implies 1 > p^2$ contradiction again. Thus this case is ruled out.
|
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Evaluating $\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx$ My first instinct was to evaluate the indefinite form of the integral, which I did by substituting $x=\tan t$, therefore yielding
\begin{align}
\int \frac{x\ln x}{(1+x^2)^2} \,dx
&= \int \frac{\tan t \sec^2 t \ln\tan t}{(1+\tan^2 t)^2} \,dt
&& \text{by substitution} \\
&= \int \tan t \cos^2 t \ln \tan t \,dt \\
&= \int \sin t \cos t \ln \tan t \,dt \\
&= -\frac{1}{2} \cos^2 t \ln \tan t + \frac{1}{2} \int \cot t \, dt
&& \text{by parts} \\
&= -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t + k
\end{align}
I run into a wall when I introduce the limits of the integral, since I get
\begin{align}
\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx
&= \bigg[ -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t \bigg] ^\frac{\pi}{2} _0
\end{align}
I'm not too sure how to evaluate the limit of the final equation as $t \rightarrow 0$. I feel like the solution is something very trivial, but I can't quite put my finger on what I'm forgetting.
|
Using $x\mapsto\frac1x$ for $x>1$,$$\int_{0}^{\infty}\frac{x^{a}\ln^{b}x}{\left(1+x^{2}\right)^{c}}dx=\int_{0}^{1}\left(\frac{x^{a}\ln^{b}x}{\left(1+x^{2}\right)^{c}}+\left(-1\right)^{b}\frac{x^{2c-a-2}\ln^{b}x}{\left(1+x^{2}\right)^{c}}\right)dx$$vanishes provided $b$ is odd with $c=a+1$, as in your problem viz. $a=b=1,\,c=2$.
|
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|
Calculate $\int \frac{dx}{1-\sin^4x}$
Calculate $$\int \frac{dx}{1-\sin^4x}$$
My try:
\begin{align}
\int \frac{dx}{1-\sin^4x}&=\int \frac{(1-\sin^2x)+(1+\sin^2x)dx}{1-\sin^4x}
\\&=\int \frac{dx}{1-\sin^2x}+\int \frac{dx}{1+\sin^2x}
\\&=\tan x+\int \frac{dx}{1+\sin^2x}
\end{align}
How to deal with the second one?
|
For the 2nd integral, rewrite the integrand,
$$\frac{1}{1+\sin^2x}=\frac{\sec^2x}{2\sec^2x-1}=\frac{(\tan x)’}{2\tan^2x+1}$$
Then, integrate,
$$\int \frac{d\tan x}{2\tan^2x+1}
=\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan x) + C$$
|
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|
$\epsilon - N$ proof of $\sqrt{4n^2+n} - 2n \rightarrow \frac{1}{4}$ I have the following proof for $\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$ and was wondering if it was correct. Note that $\sqrt{4n^2+n} - 2n = \frac{n}{\sqrt{4n^2+n} + 2n}$.
$$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| \\
= \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\
= \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \leq \left|\frac{n}{4(4n)^2}\right| = \left|\frac{n}{64n^2}\right| \\
= \left|\frac{1}{64n}\right| < \epsilon \\
\implies n>\frac{1}{64\epsilon}$$
|
Like the other commenters have mentioned, the implication is in the wrong direction. Since you want $|\frac{1}{64n}| < \epsilon$, you are required to have $n > \frac{1}{64 \epsilon}$. In other words, $n > \frac{1}{64 \epsilon}$ implies $|\frac{1}{64n}| < \epsilon$. What you have written is the other way around.
|
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|
Help differentating $f(x) = \sqrt\frac{x^2-1}{x^2+1}$ The equation I'm trying to differentiate is, $ f(x) = \sqrt\frac{x^2-1}{x^2+1}$ and I know the answer is meant to be
$$=\frac{\frac{x\sqrt {x^2+1}}{\sqrt {x^2-1}}-\frac{x\sqrt {x^2-1}}{\sqrt {x^2+1}}}{x^2+1}$$
But when I do the working out I get this
$$=\frac{(x^2-1)^\frac{1}{2}}{(x^2+1)^\frac{1}{2}}$$
$$=\frac{\frac{1}{2}(x^2-1)^\frac{-1}{2}\cdot2x\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot\frac{1}{2}(x^2+1)^\frac{-1}{2}\cdot2x}{x^2+1}$$
simplify
$$=\frac{x(x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot x(x^2+1)^\frac{-1}{2}}{x^2+1}$$
$$=\frac{\frac{\sqrt {x^2+1}}{x\sqrt {x^2-1}}-\frac{\sqrt {x^2-1}}{x\sqrt {x^2+1}}}{x^2+1}$$
As you can see two of my $x$'s are in the wrong location, and I just can't figure out what I'm doing wrong. Any help as to what steps I'm doing wrong or missing would be much appreciated.
|
You have erroneously transposed the two lone $x$'s in the second-last line into $1/x$'s in the last. (That is, they incongrously turn from being mulitplied with the radicals to dividing them.)
$$=\frac{\color{blue}x(x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot \color{blue}x(x^2+1)^\frac{-1}{2}}{x^2+1}$$
$$=\frac{\frac{\sqrt {x^2+1}}{\color{red}x\sqrt {x^2-1}}-\frac{\sqrt {x^2-1}}{\color{red}x\sqrt {x^2+1}}}{x^2+1}$$
|
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|
Evaluating $\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$ Problem:
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$$
$$\lim_{x \to -\infty} \sqrt{\frac{1}{x^6}}=0$$ so...
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}=\frac{1}{0}$$
The answer is $-1$ and I know how to get that answer. Where is the mistake in this method though?
|
$\frac 1 {-\sqrt {\frac 1 {x^{6}}} \sqrt {x^{6}+4}}$ is nothing but $\frac 1 {- \sqrt {\frac 4 {x^{6}}+1}}$ so the limit is $\frac 1 {-1}=-1$.
|
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|
Sum of products of combinatorials In the proof of some proposition, it appears that the following statement should hold:
$$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\beta} =2^{2n}. $$
However, using the definition of combinatorials does not help:
$$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\beta} = \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} \frac{(n+1)!}{(n+1-r)!r!} \frac{n!}{(n-\beta)! \beta!}.$$
I suspect that this has something to do with the binomial expansion. Any ideas?
|
Let the sum be $S$. By using the definition $C^n_r=C^n_{n-r}$, we’ll change the sum into:
$\quad S \\= \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^{n}_\beta \\ = \sum_{0 \le \beta < r \le n+1} C^{n+1}_r C^{n}_\beta \\ = \sum_{0 \le \beta < r \le n+1} C^{n+1}_{n-r+1} C^{n}_{n-\beta}$
Let $s=n-r+1$ and $\alpha=n-\beta$, then by the inequality $0 \le \beta < r \le n+1$, we’ll get $$0 \le s \le \alpha \le n+1$$
Substitute back to the sum and changing the variables:
$\quad\sum_{0 \le \beta < r \le n+1} C^{n+1}_{n-r+1} C^{n}_{n-\beta} \\ = \sum_{0 \le s \le \alpha \le n+1} C^{n+1}_s C^n_\alpha \\= \sum_{0 \le r \le \beta \le n+1} C^{n+1}_r C^n_\beta$
$\quad S+S \\= \sum_{0 \le \beta < r \le n+1} C^{n+1}_r C^{n}_\beta + \sum_{0 \le r \le \beta \le n+1} C^{n+1}_r C^n_\beta \\= \sum_{0 \le r, \beta \le n+1} C^{n+1}_r C^{n}_\beta \\= \left(\sum_{r=0}^{n+1} C^{n+1}_r \right)\left(\sum_{\beta=0}^n C^n_\beta\right) \\= 2^{n+1}2^{n} \\= 2^{2n+1}$
Therefore, the sum $S = \dfrac{2^{2n+1}}{2} = 2^{2n}$
|
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|
If $a+b+c = 4, a^2+b^2+c^2=7, a^3+b^3+c^3=28$ find $a^4+b^4+c^4$ and $a^5+b^5+c^5$ I have tried to solve it but cannot find any approach which would lead me to the answer
|
Hint
Use Newton’s identities and Vieta's formulas to find $a,b,c$ as roots of a degree 3 polynomial.
|
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|
Computing a circle from a given set with a tangent line condition I was asked to compute the circles' equations from the set
$$x^2 + y^2 -3x + (k-6)y + (9-3k)=0$$
that fulfill the following request: the circles must be tangent to the line
$$x+y-3=0$$
I started to compute
$$
\begin{cases}
x^2 + y^2 -3x + (k-6)y + (9-3k)=0\\
x+y-3=0
\end{cases}
$$
$$
\begin{cases}
x^2 + y^2 -3x + (k-6)y + (9-3k)=0\\
x=3-y
\end{cases}
$$
obtaining
$$
\begin{cases}
2y^2 +(k-3)y +(9-3k)=0\\
x=3-y
\end{cases}
$$
Now, in order to obtain a tangent line to the circle, I must put the condition $\Delta=0$, yielding
$$(k-3)^2-8(9-3k)=0$$
and so
$$k_1=3, \quad k_2=-24$$
So I have the circles, by replacing $k$:
$$x^2 + y^2 -3x -3y =0$$
and
$$x^2 + y^2 -3x -30y + 81=0$$
while the solutions reported were $$x^2+y^2-3x-5y+8=0$$ and $$x^2+y^2-3x-13y+32=0$$
Could anyone tell me where I am getting wrong?
Thanks in advance.
|
Once you spotted the obvious solution
$k=3$, you can factor
$$(k-3)^2-8(9-3k)=(k-3)\left[k-3+24\right]=(k-3)(k+21)$$
and so the second solution is $21$, not $24$.
There is another mistake earlier when you develop $x^2=(y-3)^2$, it looks like you implicitly wrote $(y-3)^2=y^2+3^2$, forgetting the rectangular term.
|
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|
How can we not use Muirhead's Inequality for proving the following inequality? There was a question in the problem set in my math team training homework:
Show that $∀a, b, c ∈ \mathbb{R}_{≥0}$ s.t. $a + b + c = 1, 7(ab + bc + ca) ≤ 2 + 9abc.$
I used Muirhead's inequality to do the question (you can try out yourself):
By Muirhead's inequality, $$\begin{align}7(ab+bc+ca)&=7(a+b+c)(ab+bc+ca)\\&=21abc+6\big(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\big)+\sum_{sym}a^2b\\&\le21abc+6\big(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\big)+\sum_{sym}a^3\\&=2(a+b+c)^3+9abc\\&=2+9abc\end{align}$$As $(3,0,0)$ majorizes $(2,1,0)$.
Is above proof correct? Also, can we find a proof not using Muirhead's inequality?
Any help is appreciated!
|
Schur's inequality:
$$a^3+b^3+c^3+3abc \ge a^2(b+c)+b^2(c+a)+c^2(a+b) \iff \\
(a+b+c)^3+9abc\ge 4(a+b+c)(ab+bc+ca) \Rightarrow \\
1+9abc\ge 4(ab+bc+ca) \quad (1)$$
Rearrangement:
$$a+b+c=1 \Rightarrow a^2+b^2+c^2=1-2(ab+bc+ca)\ge ab+bc+ca \Rightarrow \\
1\ge 3(ab+bc+ca) \quad (2)$$
Now add $(1)$ and $(2)$.
|
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|
What is the general solution of this equation :$2^x 3^y+1=7^z$ with $x, y , z$ are integers? I have got these triplet solution $(x,y,z)=(1,1,1),(4,1,2)$ for this equation:
$$2^x 3^y+1=7^z$$
with $x, y , z$ are integers, But i can't get general solution of it, I have attempted to use Gausse theorem for the solution of $ ax+by= c $, with $a, b, c$ are integers but my problem i can't transfer the titled equation to that of Gausse as a linear form, any way ?
|
Clearly $z>0$, since $2^x3^y+1>1$, so $7^z-1\in \mathbb{Z}$ so $x,y\in \mathbb{N}$.
If $y\geq 2$ then $1\equiv _9 7^z$. Since ord$_9(7) = 3$ we have $3\mid z$ so $z=3t$.
Now we can write: $$2^x3^y = (7^3-1)\Big((7^3)^{t-1}+\ldots+7^3+1\Big)$$
Since $7^3-1 = 19\cdot 9\cdot 2$ we see this is impossible, so $y\leq 1$ or $y=1$.
Can you finish?
Added:
Now we have: $$2^{x-1} = 7^{z-1}+\ldots+7^2+7+1$$
Say $x>1$, then $z$ is even so $z=2s$ and $$2^x3 = (7^s-1)(7^s+1)$$
Since factors on right differ for $2$ at most one is divisible by $4$.
*
*$7^s-1 = 1$ and $7^s+1 = 2^{x}3$...
*$7^s-1 = 2$ and $7^s+1 = 2^{x-1}3$...
*$7^s-1 = 3$ and $7^s+1 = 2^{x}$...
*$7^s-1 = 6$ and $7^s+1 = 2^{x-1}$...
*$7^s+1 = 1$ and $7^s-1 = 2^{x}3$...
*$7^s+1 = 2$ and $7^s-1 = 2^{x-1}3$...
*$7^s+1 = 3$ and $7^s-1 = 2^{x}$...
*$7^s+1 = 6$ and $7^s-1 = 2^{x-1}$...
|
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|
tangent inequality in triangle Let $a$, $b$ and $c$ be the measures of angles of a triangle (in radians).
It is asked to prove that
$$\tan^2\left(\dfrac{\pi-a}{4}\right)+\tan^2\left(\dfrac{\pi-b}{4}\right)+\tan^2\left(\dfrac{\pi-c}{4}\right) \ge 1$$
When does equality occur ?
My try :
Letting $u:= \tan\left(\dfrac{\pi-a}{4}\right)$ and $v:= \tan\left(\dfrac{\pi-b}{4}\right)$ the inequality reduces to proving
$$u^2+v^2+\dfrac{(1-uv)^2}{(u+v)^2} \ge 1\quad\quad (*)$$ ($$u,v\in (0,1)$$)
( using $a+b+c=\pi$ and the formula for $\tan(x+y)$ and the fact that $\tan\left(\dfrac{\pi}{2}-x\right)=\dfrac{1}{\tan x}$ )
I'm having trouble in proving that last inequality.
Any suggestions are welcome.
Thanks.
Edit : is the following reasoning to prove the inequality (*) sound ?
(*) is obvious when $u^2 + v^2 \ge 1$ so we only have to deal with the case $u^2 + v^2 \le 1$ which we assume true in what follows.
(*) $\iff (u^2 + v^2)(u + v)^2 \ge (u+v)^2+(1-uv)^2$
Setting $x:= u^2 + v^2$ and $a:=uv$ we get
(*) $\iff x^2+(2a-1)x \ge a^2+4a-1$
Now some calculus $(x^2+(2a-1)x)' = 2x + 2a-1$
the function $\phi:x\mapsto x^2+(2a-1)x$ then has a minimum at $\dfrac 12 - a$ which is $\phi\left(\dfrac 12 - a\right) = -a^2+a-\dfrac 14$
It then suffices to have $-a^2+a-\dfrac 14 \ge a^2+4a-1$
This last ineq is equivalent to
$8a^2+12a-3 \le 0$
which, in turn, is equivalent to
$a \in \left[\dfrac{-6-\sqrt{60}}{8}, \dfrac{-6+\sqrt{60}}{8}\right]$
Recall that we're working under the assumption $u^2 +v^2 \le 1$, that yields in particular that $uv \le \dfrac 12$.
since $ \dfrac{-6-\sqrt{60}}{8} \le 0 \le a := uv \le \dfrac 12 \le \dfrac{-6+\sqrt{60}}{8}$, we're done.
Thanks for taking time to check the correctness of the above proof.
|
Let $\dfrac{\pi-A}4=x$ etc.
$\implies4(x+y+z)=3\pi-\pi\iff x+y+z=\dfrac\pi2$
Now as $\tan x,\tan y,\tan z$ are real,
$$(\tan x-\tan y)^2+(\tan y-\tan z)^2+(\tan z-\tan x)^2\ge0$$
$$\implies\tan^2x+\tan^2y+\tan^2z\ge\tan x\tan y+\tan y\tan z+\tan z\tan x$$
Finally $$\tan(x+y)=\tan\left(\dfrac\pi2-z\right)$$
$$\iff\dfrac{\tan x+\tan y}{1-\tan x\tan y}=\dfrac1{\tan z}$$
Simplify
|
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|
Finding the integral $\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}$ One may take $x= \cos t$ and get
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= -\frac{1}{4}\int \csc^4(t/2)~ dt=-\frac{1}{4} \int [\csc^2(t/2) +\csc^2(t/2) \cot^2(t/2)]~ dt.$$
$$\Rightarrow I=\frac{1}{2} \left [\cot (t/2)] +\frac{1}{3}\cot^3(t/2)\right]=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
The question is: How to obtain this integral by other methods not using the trigonometric substitution.
|
Let us use $$1-x=\frac{1}{u} \Rightarrow x=1-\frac{1}{u} \Rightarrow dx=\frac{du}{u^2}.$$ Then
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= \int \frac{u du}{\sqrt{2u-1}}.$$
Next use $$2u-1=v^2 \Rightarrow du =v dv.$$ Then
$$I=\frac{1}{2} \int (v^2+1) dv= \frac{v}{2}[\frac{v^2}{3}+1]=\frac{\sqrt{2u-1}}{3} (u+1)=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
|
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|
Show that $\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1$
Show that $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1 .$$
I know that $\arctan 1 = \frac{\pi}{4}$ and that the sequence being subtracted is a partial sum of its Taylor series. I believe you use the alternating series test to explain, but all I get from it is that the series will converge on $[-1,1]$.
|
Simple answers above. The alternative is as follows:
*
*We know that $$1-\frac13+\frac15-\frac17+\frac19=\frac{263}{315}$$ which can be calculated by hand.
*We also know that $$\frac\pi4<\frac{3.2}4=0.8\quad\text{whereas}\quad\frac{263}{315}>\frac{252}{315}=0.8$$ so the inequality is equivalent to $$\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)-\frac{\pi}{4} < 0.1\iff\frac\pi4>\frac{463}{630}\iff\pi>\frac{926}{315}$$
*This is true, since $\pi>3$ and $$\frac{926}{315}<\frac{945}{315}=3.$$
|
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|
Locus of mid point of $AB$
If the family of lines $tx+3y-6=0.$ where $t$ is variable intersect the lines $x-2y+3=0$ and $x-y+1=0$ at point $A$ and $B.$ Then locus of mid point of $AB$ is
what i try
Intersection of line $tx+3y-6=0$ and $x-2y+3=0$ is
$\displaystyle A\bigg(\frac{3}{3+2t},\frac{6+3t}{3+2t}\bigg)$
and Intersection of $tx+3y-6=0$ and $x-y+1=0$ is
$\displaystyle B\bigg(\frac{3}{3+t},\frac{6+t}{3+t}\bigg)$
Let locus of mid point of $AB$ is at $M(h,k)$
Then
$\displaystyle h=\frac{1}{2}\bigg(\frac{3}{3+2t}+\frac{3}{3+t}\bigg)$ and $\displaystyle k = \frac{1}{2}\bigg(\frac{6+3t}{3+2t}+\frac{6+t}{3+t}\bigg)$
How do i eliminate $t$ in an easy way
or please help me is any other easier ways to solve it thanks
|
Let $M(x,y).$
Thus, $$y-x=\frac{1}{2}\left(\frac{3t+3}{2t+3}+1\right)$$ or
$$t=\frac{6x-6y+6}{4y-4x-5}.$$
Can you end it now?
|
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|
Convergence of recurrence sequence Let $a_1=1$ define $a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})$ . Show that the sequence converges to $\sqrt{2}$. The idea is to show the sequence is bounded and monotone . But how should I do it?
|
*
*$a_2=\frac{3}{2}>\sqrt{2}$
*Show if $a_n>\sqrt{2}, \sqrt{2}<a_{n+1}<a_n$
$a_{n+1}-\sqrt{2}=\frac{1}{2}(a_n-\sqrt{2}+\frac{2-\sqrt{2}a_n}{a_n})=(a_n-\sqrt{2})\frac{1-\frac{\sqrt{2}}{a_n}}{2}$
so $0<a_{n+1}-\sqrt{2}<\frac{a_n-\sqrt{2}}2$
So $\lim_{n\to\infty}a_n=\sqrt{2}$
|
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Find value for $k$ such that $(x^2-k)$ is a factor for $f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$ Find value for $k$ such that $(x^2-k)$ is a factor for, $$f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$$
My Try
Since $x^2-k=0$ when we substitute $x=\pm k$ to $f(x)$ it should be equal to $0.$
But this gives a polynomial of k where it has $\sqrt{k}$ terms as well. Is my approach correct or is there a simpler way? Please any hint would be highly appreciated.
|
Use that $$\frac{f(x)}{x^2-k}=2\,{k}^{2}+3\,kx+2\,{x}^{2}-3\,k-4\,x-5+{\frac {2\,{k}^{3}x+2\,{k}^{3}
+{k}^{2}x-3\,{k}^{2}-7\,kx-5\,k-6\,x+6}{{x}^{2}-k}}
$$
|
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|
How to factorize $\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$? How to factorize and simplify the following?
$$\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$$
I've tried everything I know. Trying to factorize the numerator first then denominator, but I get no where. Usual identities like $(x+y)^2=x^2+2xy+y^2$ don't work either, and neither does long division. I'm pretty stuck.
The answer from wolfram is
$(2x-1)/((2x+1)(x-2))$.
But I can't get there.
|
Just above the line that says GCD it shows the quotients, which you would want because of wishing to reduce the fraction
$$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) $$
$$ \left( 4 x^{4} - 17 x^{2} + 4 \right) $$
$$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) = \left( 4 x^{4} - 17 x^{2} + 4 \right) \cdot \color{magenta}{ \left( 0 \right) } + \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) $$
$$ \left( 4 x^{4} - 17 x^{2} + 4 \right) = \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( - 6 x^{2} - 9 x + 6 \right) $$
$$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) = \left( - 6 x^{2} - 9 x + 6 \right) \cdot \color{magenta}{ \left( \frac{ - 2 x + 1 }{ 3 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 0 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 0 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( x - 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ - 2 x + 1 }{ 3 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 2 x + 1 }{ 3 } \right) }{ \left( \frac{ - 2 x^{2} + 3 x + 2 }{ 3 } \right) } $$
$$ \left( 2 x - 1 \right) \left( \frac{ x - 1 }{ 3 } \right) - \left( 2 x^{2} - 3 x - 2 \right) \left( \frac{ 1}{3 } \right) = \left( 1 \right) $$
$$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) = \left( 2 x - 1 \right) \cdot \color{magenta}{ \left( 2 x^{2} + 3 x - 2 \right) } + \left( 0 \right) $$
$$ \left( 4 x^{4} - 17 x^{2} + 4 \right) = \left( 2 x^{2} - 3 x - 2 \right) \cdot \color{magenta}{ \left( 2 x^{2} + 3 x - 2 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( 2 x^{2} + 3 x - 2 \right) } $$
$$ \left( 4 x^{3} + 4 x^{2} - 7 x + 2 \right) \left( \frac{ x - 1 }{ 3 } \right) - \left( 4 x^{4} - 17 x^{2} + 4 \right) \left( \frac{ 1}{3 } \right) = \left( 2 x^{2} + 3 x - 2 \right) $$
|
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|
Evaluate $\tan 195^{\circ}$ without using the calculator How to evaluate $\tan 195^{\circ}$ without using the calculator, and how to give the answer in the form $a+b \sqrt{3}$, where $a$ and $b$ are integers?
|
The Tangent Identities:
$$\tan (2\theta) = \frac{2\tan\theta}{1-\tan^2\theta} $$
$$\tan (\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta} $$
$$\tan (\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1+\tan\alpha\tan\beta} $$
$$\tan 195^\circ = \tan (180^\circ + 15^\circ)$$
$$\tan(180^\circ) = 0$$
$$\frac{0 + \tan(15^\circ)}{1 - 0 (\tan(15^\circ))} = \tan(15^\circ)$$
$$\tan(15^\circ) = \tan(45^\circ - 30^\circ)$$
$$\tan (45^\circ - 30^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1+\tan(45^\circ)\tan(30^\circ)} $$
$$\tan(45^\circ) = 1$$
$$\tan(30^\circ) = \frac{1}{\sqrt3}$$
$$\frac{1 - \frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}$$
|
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|
Proving the equation using binomial theorem I want to prove this theorem using Binomial theorem and I've got trouble in understanding 3rd step if anyone knows why please explain :) Prove that sum:
$\sum_{r=0}^{k}\binom{m}{r}\binom{n}{k-r}=\binom{m+n}{k}$
1st step:
$(1+y)^{m+n}=(1+y)^m(1+y)^n$
2nd step: we use the binomial theorem formula and evaluate that sum
$\sum_{m+n}^{r=0}\binom{m+n}{r}y^r=\sum_{r=0}^{m}\binom{m}{r}y^r \sum_{r=0}^{n}\binom{n}{r}y^r$
3step: equating the coefficient of $y^k$
$\binom{m+n}{k}=\binom{m}{0}\binom{n}{k}+\binom{m}{1}\binom{n}{k-1}+...+\binom{m}{k}\binom{n}{k-k}$
Why and how is the 3rd step allowed? Thank you
|
By the equation of 2nd step,
Left:
$$\binom{m+n}{0}y^{0}+\binom{m+n}{1}y^{1}+\cdots+\binom{m+n}{m+n-1}y^{m+n-1}+\binom{m+n}{m+n}y^{m+n}$$
The coefficient of $y^{k}$ is $\binom{m+n}{k}$
Right:
$$(\binom{m}{0}y^{0}+\cdots+\binom{m}{m}y^{m})(\binom{n}{0}y^{0}+\cdots+\binom{n}{n}y^{n}) $$
The term $y^{k}$ come from
$$\binom{m}{0}y^{0}\binom{n}{k}y^{k}+\binom{m}{1}y^{1}\binom{n}{k-1}y^{k-1}+ \cdots +\binom{m}{k-1}y^{k-1}\binom{n}{1}y^{1}+\binom{m}{k}y^{k}\binom{n}{0}y^{0}$$
So the coefficient of $y^{k}$ is $\binom{m}{0}\binom{n}{k} + \binom{m}{1}\binom{n}{k-1}+ \cdots +\binom{m}{k-1}\binom{n}{1}+\binom{m}{k}\binom{n}{0}$
By the equality of the equation of 2nd step, the coefficient of $y^{k}$ of left and right might be the same, so $$\binom{m+n}{k} = \binom{m}{0}\binom{n}{k} + \binom{m}{1}\binom{n}{k-1}+ \cdots +\binom{m}{k-1}\binom{n}{1}+\binom{m}{k}\binom{n}{0}$$
|
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|
Inequality $\left(1-\frac{x(1-x)+y(1-y)}{1-x+1-y}\right)^2+(1-\frac{x+y}{2})^2\geq (1-x)^2+(1-y)^2$ it's a little problem found by myself
let $x,y\neq 1$ .be real numbers then we have :
$$\left(1-\frac{x(1-x)+y(1-y)}{1-x+1-y}\right)^2+\left(1-\frac{x+y}{2}\right)^2\geq (1-x)^2+(1-y)^2$$
I tried Jensen's inequality and Am-Gm but without success .
I also tried Karamata's inequality but it fails.
I have a ugly proof using derivative and it's too long to be explain here .
I'm looking for a contest proof .
Thanks a lot to share your knowledge and your time.
|
For $x+y\neq2$ by AM-GM we obtain:$$\left(1-\frac{x(1-x)+y(1-y)}{1-x+1-y}\right)^2+\left(1-\frac{x+y}{2}\right)^2=$$
$$=\left(\frac{(1-x)^2+(1-y)^2}{2-x-y}\right)^2+\left(\frac{2-x-y}{2}\right)^2\geq$$
$$\geq2\sqrt{\left(\frac{(1-x)^2+(1-y)^2}{2-x-y}\right)^2\left(\frac{2-x-y}{2}\right)^2}= (1-x)^2+(1-y)^2$$
|
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|
Volume of a solid created by an extended tetrahedron
Every edge of a tetrahedron with length $p$ is extended through the vertices by $p$.
Now all 12 points create a new solid $J$ of which I seek the volume dependent on the volume of the tetrahedron in the centre.
With some help the solution becomes clear:
The whole Volume of all pyramids is:
$$V_P=4\cdot\frac{1}{6}\cdot\left(\frac{a}{\sqrt{2}}\right)^3 +4\cdot\frac{1}{6}\cdot\left(\sqrt{2}a\right)^3 =\frac{3}{\sqrt{2}}a^3 \tag{1}$$
The side of the large cube is
$$s=\frac{3}{\sqrt{2}}a\tag{2}$$
and the Volume is respectively
$$V_C=\left(\frac{3}{\sqrt{2}}a\right)^3=\frac{27\sqrt{2}}{4}\cdot a^3 \tag{3}$$
The last step is subtracting the Volume of the pyramids.
$$V_J=V_C-V_P= \frac{27\sqrt{2}a^3}{4}-\frac{3}{\sqrt{2}}a^3=\frac{21\sqrt{2}a^3}{4} \tag{4}$$
Thx for the help.
|
There are only three types of faces for the new solid. They are either equilateral triangles of side lenghth $p$ that are $\dfrac{7}{4} h$ away from the center, where $h = p \sqrt{\dfrac{2}{3}} $, or equilateral triangles of side $2p$ that are $ \frac{5}{4} h $ from the centroid, or rectangles of sides $ p $ and $ 2 p $ that are $ \dfrac{3 \sqrt{2}}{4 } p $ from the center.
$V = (1/3) \left(4 (\dfrac{\sqrt{3}}{4}) ( p^2 h (\dfrac{7}{4}) + 5 p^2 h ) + 6 (2 p^2)( \dfrac{3\sqrt{2}}{4} p ) \right) = \sqrt{3} ( \dfrac{9}{4} p^2 h ) + 3 \sqrt{2} p^3 $
$V = \sqrt{2} p^3 ( \dfrac{21}{4} ) = 63 V_0$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\lim_{x\to -\infty} \frac{4x^3+1}{2x^3 + \sqrt{16x^6+1}}$ In finding this limit:
$$\lim_{x\to -\infty} \frac{4x^3+1}{2x^3 + \sqrt{16x^6+1}}$$
I've been told to divide all the terms by $-x^3$ (as opposed to $x^3$ if we take the limit as $x \to \infty$), and go from there. Dividing by a negative $x^3$ doesn't make sense to me, because we will be plugging in negative numbers approaching $-\infty$ anyways. Why double up?
Is there a different way to think about/solve the limit?
|
You can do it either way. If you divide by $x^3$ you have to remember that $x^3$ is a negative number and $x^3 =-\sqrt{x^6}$. you get
$\frac {4+\frac 1{x^3}}{2 +\frac {\sqrt{16x^6 + 1}}{x^3}}=$
$\frac {4+\frac 1{x^3}}{2+\frac {\sqrt{16x^6+1}}{-\sqrt{x^6}} }=$
$\frac {4+\frac 1{x^3}}{2-\sqrt{16+\frac 1{x^6}}}$
It's easier to avoid mistakes if you divide by $-x^3$ and get:
$\frac {-4-\frac 1{x^3}}{-2+\sqrt{\frac {16x^3+1}{(-x)^6}}}=\frac {-4-\frac 1{x^3}}{-2+\sqrt{16+\frac 1{x^6}}}$
|
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|
Eigenvalue of $A-aI_3$ Question: Let $A=\begin{pmatrix} a+1 & 1 & 1 \\ 1 & a+1 & 1 \\ 1 & 1 & a+1\end{pmatrix}$. Show that $A-aI_3$ has eigenvalue of 3. Also find eigenvector.
My thinking:
I know that we have to apply characteristic polynomial $|A- \lambda I|$ to find the eigenvalue. I don't understand the part $A-aI_3$. What should I suppose to do here with this $A-aI_3$ ? I am kinda confused. Your kind suggestion will be appreciated.
|
Alternative answer: $A-aI_3=\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}$ clearly has eigenvalue $0$
with independent eigenvectors $\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}.$
The sum of the eigenvalues is the trace of the matrix, which is $3$, so the other eigenvalue is $3$.
An eigenvector corresponding to the eigenvalue $3$ is in the kernel of $\begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{pmatrix}$ , i.e. $(-2,1,1)\times(1,-2,1)=(-3,-3,-3)$.
|
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|
Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Note: $x$ and $y$ are obtuse angles.
My attempt that is not simple is as follows.
Expand both known constraints, so we have
\begin{align}
\cos x \cos y &=4/15\\
\sin x \sin y &=-1/15
\end{align}
Eliminate $x$ using $\sin^2 x +\cos^ 2 x=1$, we have
$$
225 \sin^4 y -210 \sin^2 y +1=0
$$
with its solution
$\sin^2 y = \frac{7\pm4\sqrt3}{15}$.
Then, $\cos^2 y = \frac{4(2\mp\sqrt3)}{15}$.
\begin{align}
\sin^2(2y) &= 4\cos^2 y\sin^2 y\\
&= 4 \times \frac{4(2\mp\sqrt3)}{15}\times \frac{7\pm4\sqrt3}{15} \\
\sin 2 y & = - \frac{4}{15}\sqrt{(2\mp\sqrt3)(7\pm4\sqrt3)}
\end{align}
$\sin 2y$ must be negative.
Edit
Thank you for your effort to answer my question. However, the existing answers seem to be more complicated than my attempt above.
By the way, I am confused in deciding which the correct pair among $(2\mp\sqrt3)(7\pm4\sqrt3)$ is.
|
As $90<x,y<180$ and $\cos(x+y)>0$
$$270<x+y<360\implies\sin(x+y)=-\sqrt{1-(1/3)^2}$$
Again, $-90<x-y<90^\circ\implies\sin(x-y)=\pm\sqrt{1-(1/5)^2}$
Finally $$\sin2y=\sin(x+y+(x-y))=?$$
|
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|
Problems with understanding a given Sum Identity In a textbook I found the following equation without any explaination:
\begin{align}
\sum_{\substack{j, k=1,\\ j\neq k}}^n\frac{1}{(x-x_j)(x-x_k)} - \frac{3}{2}\left( \sum_{j=1}^n \frac{1}{x-x_j}\right)^2 \\
= -\frac{1}{2}\sum_{j=1}^n \left( \frac{1}{x-x_j}\right)^2 - \left( \sum_{j=1}^n \frac{1}{x-x_j} \right)^2
\end{align}
I was wondering why this is true and tried to proof it but somehow failed. Can anyone give a hint?
|
Consider the following expression
\begin{eqnarray*}
\left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right)^2 = \left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right) \left( \sum_{k=1}^{n} \frac{1}{x-x_k} \right).
\end{eqnarray*}
Now when you calculate the product we have either $j \neq k$ or $j=k$, so we have
\begin{eqnarray*}
\left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right)^2 = \sum_{\substack{j, k=1,\\ j\neq k}}^n \frac{1}{x-x_j} + \sum_{j=1}^{n} \frac{1}{(x-x_j)^2} .
\end{eqnarray*}
So the first term in the stated formula needs a minus sign!
|
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|
Odd or even function. Determine whether the following function is odd or even :
$$f(x) =(\cos{x} + \sin{x} +1)(\cos{x} + \sin{x} -1)$$
My turn :
$$f(x) = (\cos{x}+ \sin{x})^2 -1= \sin{2x}$$
Then f is an odd function
Another turn :
$$f(-x) = (\cos{x}- \sin{x} +1)(\cos{x} -\sin{x} -1)$$
But the last form seems neither odd nor even. My question is why i could not reach the same another using the last form ?
|
While it seems that everything in the original post is correct, you want to show that:
$f(x) = -f(-x)$
$(\cos x + \sin x + 1)(\cos x +\sin x - 1) = -(\cos (-x) +\sin (-x) + 1)(\cos (-x) + \sin (-x) -1)\\
(\cos x + \sin x + 1)(\cos x +\sin x - 1) = -(\cos x -\sin x + 1)(\cos x - \sin x -1)\\
(\cos x + \sin x)^2 - 1 = -(\cos x - \sin x)^2 + 1\\
2\sin x\cos x = 2\sin x\cos x$
It is not obvious that
$f(x) = (\cos x + \sin x + 1)(\cos x +\sin x - 1)$ because neither of the factors are odd nor even. Nonetheless, it is.
|
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|
Eigen Decomposition Check I am following the wiki entry on eigen dicomposition with the following matrix:
$$A = \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}$$
I wish to find a diagonalizing matrix T.S.
$$T^{-1}AT=\Lambda$$
$$AT=T\Lambda$$
where,
$$\Lambda = \begin{pmatrix}
x & 0 \\
0 & y \\
\end{pmatrix}$$
$$T = \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}$$
I am following with the wiki:
$$A\begin{pmatrix}a \\c \\\end{pmatrix}=x\begin{pmatrix}a \\c \\\end{pmatrix}$$
$$A\begin{pmatrix}b \\d \\\end{pmatrix}=y\begin{pmatrix}b \\d \\\end{pmatrix}$$
and the eigenvalues are:
$$\left|A-I\lambda\right| = \left|\begin{pmatrix}-\lambda & 1 \\1 & -\lambda \\\end{pmatrix}\right|=(\lambda^2-1)=(\lambda-1)(\lambda+1)$$
Hence, $\lambda_{1,2}=\pm 1$
Now I am finding the eigenvectors with:
$$A\begin{pmatrix}a \\c \\\end{pmatrix}=\begin{pmatrix}0 & 1\\1 & 0 \\\end{pmatrix}\begin{pmatrix}a\\c \\\end{pmatrix}=\begin{pmatrix}c \\a \\\end{pmatrix}=1\begin{pmatrix}a \\c \\\end{pmatrix}$$
Hence, $a=c$. And,
$$A\begin{pmatrix}b \\d \\\end{pmatrix}=\begin{pmatrix}0 & 1\\1 & 0 \\\end{pmatrix}\begin{pmatrix}b\\d \\\end{pmatrix}=\begin{pmatrix}d \\b \\\end{pmatrix}=-1\begin{pmatrix}b \\d \\\end{pmatrix}$$
Hence, $b=-d$.
So I take $T$ to be:
$$T = \begin{pmatrix}
a & b \\
a & -b \\
\end{pmatrix}=\begin{pmatrix}
1 & 1 \\
1 & -1 \\
\end{pmatrix}$$
with:
$$T^{-1} = \frac{1}{det(T)}\begin{pmatrix}
1 & -1 \\
-1 & -1 \\
\end{pmatrix}=\frac{1}{-1-1}\begin{pmatrix}
1 & -1 \\
-1 & -1 \\
\end{pmatrix}=\frac{1}{-2}\begin{pmatrix}
1 & -1 \\
-1 & -1 \\
\end{pmatrix}=\frac{1}{2}\begin{pmatrix}
-1 & 1 \\
1 & 1 \\
\end{pmatrix}$$
Now, when I try to calculate $T^{-1}AT$ I get:
$$\Lambda =\frac{1}{2}\begin{pmatrix}
-1 & 1 \\
1 & 1 \\
\end{pmatrix}\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}\begin{pmatrix}
1 & 1 \\
1 & -1 \\
\end{pmatrix}=\frac{1}{2}\begin{pmatrix}
1 & -1 \\
1 & 1 \\
\end{pmatrix}\begin{pmatrix}
1 & 1 \\
1 & -1 \\
\end{pmatrix}=\frac{1}{2}\begin{pmatrix}
0 & -2 \\
2 & 0 \\
\end{pmatrix}=\begin{pmatrix}
0 & -1 \\
1 & 0 \\
\end{pmatrix}$$
Where have I gone wrong here?
|
Your answer is correct except that you forgot to switch the elements on the diagonal of $T$ when finding its inverse.
|
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|
Solve $xu_x+yu_y+zu_z=4u$ \begin{cases}
xu_x+yu_y+zu_z=4u\\
u(x,y,1)=xy\\
\end{cases}
Using Lagrange method we get:
$$\frac{x}{y}=c_1, \frac{y}{z}=c_2, \frac{z}{\sqrt[4]{u}}=c_3$$
So the general solution is $$u=\frac{z^4}{c_3}$$?
|
Converting to spherical coordinates, we get
$$ru_r = 4u \implies u = f(\theta,\phi)r^4$$
Then plugging in our boundary condition at $r\cos\theta = 1$ and $xy=r^2\sin^2\theta\sin\phi\cos\phi$, we can get
$$f(\theta,\phi)r^4 = r^2\sin^2\theta\sin\phi\cos\phi\cdot(1)=r^2\sin^2\theta\sin\phi\cos\phi\cdot (r^2\cos^2\theta)$$
$$\implies f(\theta,\phi) = \cos^2\theta\sin^2\theta\sin\phi\cos\phi$$
by canceling out the $r^4$ on both sides. In other words when we convert back to Cartesian we get the solution
$$u(x,y,z) = xyz^2$$
|
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|
Finding $\displaystyle \lim_{x \to 2}\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}$ I came across this question.
Evaluate the limit $$ \lim_{x \to 2}\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}$$
I tried rationalizing the denominator, substitution, yet nothing seems to cancel out with the denominator. I don't think we are supposed to use squeeze theorem or L'Hopital rule for this.
Can someone give me a hint in the right direction?
|
Using a small trick that I enjoy.
Let $x=t+2$ to make
$$y=\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}=\frac{\sqrt{4 t+9}-\sqrt{t^3+6 t^2+12 t+9}}{\sqrt{t+4}-\sqrt{t^3+6 t^2+10 t+4}}$$ and now use the binomial expansion or Taylor series around $t=0$.
We have
$$\sqrt{4 t+9}=3+\frac{2 t}{3}-\frac{2 t^2}{27}+O\left(t^3\right)$$
$$\sqrt{t^3+6 t^2+12 t+9}=3+2 t+\frac{t^2}{3}+O\left(t^3\right)$$
$$\sqrt{t+4}=2+\frac{t}{4}-\frac{t^2}{64}+O\left(t^3\right)$$
$$\sqrt{t^3+6 t^2+10 t+4}=2+\frac{5 t}{2}-\frac{t^2}{16}+O\left(t^3\right)$$
So
$$y=\frac{-\frac{4 t}{3}-\frac{11 t^2}{27}+O\left(t^3\right) } {-\frac{9 t}{4}+\frac{3 t^2}{64}+O\left(t^3\right) }$$ Now, using the long division
$$y=\frac{16}{27}+\frac{47 }{243}t+O\left(t^2\right)$$ which shows the limit and also how it is approached.
|
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|
Problem with approximating function I want to approximate the expression $$ \frac{a z +1/2 b z^2 + c z^3}{a + b z + d z^2}.$$
This should be approximately equal to $z - \frac{b}{2a}z^2.$ But no matter the approach, I do not get this answer. Anyone who sees what I should have done?
Edit: some users had a fair point, in this case $z$ is very small
|
$$\begin{array}{rcl}\dfrac{az + \dfrac{1}{2} bz^2 + cz^3}{a + bz + dz^2} &=& \dfrac{az + bz^2 +dz^3-\dfrac{1}{2}bz^2-(d-c)z^3}{a +bz+dz^2}\\
&=&\dfrac{az + bz^2 +dz^3}{a+bz+dz^2}-\dfrac{\dfrac{1}{2}bz^2 +(d-c)z^3}{a+bz+dz^2}\\
&=&z -\dfrac{\dfrac{1}2b\left(z^2 +\dfrac{2(d-c)}{b}z^3\right)}{a+bz+dz^2}\\
&=& z - \dfrac{\dfrac{b}{2a}\left(az^2 +\dfrac{2a(d-c)}{b}z^3\right)}{a+bz+dz^2}\\
&=& z - \dfrac{b}{2a}\cdot \dfrac{az^2+bz^3+dz^4-dz^4+\dfrac{2a(d-c)-b^2}{b}z^3}{a+bz+dz^2}\\
&=&z-\dfrac{b}{2a}z^2 +\dfrac{b}{2a}\cdot\dfrac{dz^4-\dfrac{2a(d-c)-b^2}{b}z^3}{a+bz+dz^2}\\
\end{array}$$
This should be approximately equal to $z-\dfrac{b}{2a}z^2$ when $z$ is sufficiently small.
|
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|
Formulas for Sequences Removing Multiples of 2, 3, and 5 First off, I am a programmer so please excuse if some of the terms I use are not the correct mathematical terms. I was working on devising a function to improve one of my prime number generation algorithms. With this in mind, I first set out to find the formulas for a sequence removing multiples of 2 and 3:
\begin{array}{c|c}
x&y\\ \hline
0&5\\ \hline
1&7\\ \hline
2&11\\ \hline
3&13\\ \hline
4&17\\ \hline
5&19\\ \hline
6&23\\ \hline
7&25\\ \hline
8&29\\ \hline
\vdots&\vdots
\end{array}
The equations that I came up for for this sequence are as follows:
$$y = 3x + 5 - x \bmod 2$$
$$x = \left\lfloor\frac{y - 5 + [y \bmod 3 \neq 0]}{3}\right\rfloor$$
After this, I tried to do the same for a sequence removing multiples of 2, 3, and 5:
\begin{array}{c|c}
x&y\\ \hline
0&7\\ \hline
1&11\\ \hline
2&13\\ \hline
3&17\\ \hline
4&19\\ \hline
5&23\\ \hline
6&29\\ \hline
7&31\\ \hline
8&37\\ \hline
9&41\\ \hline
10&43\\ \hline
11&47\\ \hline
12&49\\ \hline
13&53\\ \hline
\vdots&\vdots
\end{array}
While I think I found an equation to get $y$ from a value of $x$, I cannot find a way to get the value of $x$ from a given value $y$.
$$y = 4x + 7 - 2\left\lfloor\frac{1}{8}x\right\rfloor - 2\left[\{2, 3, 6\} \ \text{contains}\ (x \bmod 8)\right] - 4\left[\{4, 5, 7\} \ \text{contains}\ (x \bmod 8)\right]$$
$$x =\ ?$$
I am wondering if an equation that produces the corresponding value of $x$ for a given value of $y$ for the aforementioned sequence exists, and if indeed it does, what the equation is.
|
my way is with an 8 value addition sequence 2, 6, 4, 2, 4, 2, 4, 6. and Repeat.
Start @ -1
(-1) [+ 2] = 1 [+6] = 7 [+4] = 11
thus (-1) 1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, 79, 83...
and so on
Repeating this addition sequence from start position of -1
[2, 6, 4, 2, 4, 2, 4, 6]{2, 6, 4, 2, 4, 2, 4, 6}[2, 6, 4, 2, 4, 2, 4, 6]
Generates a continuous sequence without any multiples of 2, 3 or 5
|
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Prove that $\frac{x^3 - x + 10}{x^2-9}$ is surjective I'm looking for an elegant way to prove that this function is surjective. One way to solve this would be to set $\frac{x^3-x+10}{x^2-9} = y$ and solve for $x$ in terms of $y$ but this would require using the cubic formula to solve. Is there any other nice way?
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One can also approach the problem by way of Descartes' Rule of Signs.
render $\frac{x^3-x+10}{x^2-9} = y$ as $x^3-x^2y-x+(10+9y)=0$ for $|x|\ne 3$ and let
\begin{eqnarray}
f(x)&=&x^3-x^2y-x+(10+9y)\\
f(-x)&=&-x^3-x^2y+x+(10+9y)
\end{eqnarray}
Then for $y$ in the interval $\left(\infty, -\frac{10}{9}\right)$ Descartes' signs for $f(x)$ are $+\,+\,-\,-$, so there is one positive $x$ which maps to $y$.
Clearly, $x=0$ maps to $y=-\frac{10}{9}$.
In the interval $\left(-\frac{10}{9},0\right)$ Descartes' signs for $f(-x)$ are $-\,+\,+\,+$ so there is one negative $x$ which maps to $y$.
For $y=0$ a root of $x^3-x+10$ maps to $y$.
For $y$ in the interval $(0,\infty)$ Descartes' signs for $f(-x)$ are $-\,-\,+\,+$ so one negative value of $x$ maps to $y$.
Thus for each value of $y$ in $(-\infty,\infty)$ there is an $x$ which maps to $y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3361354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Equation of circle touching three circles, two of which are intersecting Find the equation of the circle which is tangentially touching three given circles: $x^2+y^2=49$, $x^2+(y-3.5)^2=49/4$, and $y^2+(x-3.5)^2=49/4$.
By tangentially i mean, it touches the smaller two circle externally and the larger one internally.
The problem would have been much easier had the three given circles been tangent to each other but the smaller two of them intersect, making finding the radius of the circle in question much difficult for me.
Well, I should write the general equation of the circle and equate sum of radius with the distance between the centre with the three given equations.
I am not getting the correct answer with this approach.
Please help.
|
Let $a$ be the radius of the inscribed circle and apply the cosine rule to the triangle formed by the centers of the large circle, one of the small circles and the inscribed circle,
$$(a+3.5)^2=(7-a)^2+3.5^2-2\cdot 3.5 \cdot (7-a)\cos 135^\circ$$
which yields
$$a= \frac{7(\sqrt{2}+1)}{3\sqrt{2}+1}$$
Therefore, the equation of the inscribed circle is
$$\left( x+ \frac{7-a}{\sqrt{2}} \right)^2
+\left( y+ \frac{7-a}{\sqrt{2}} \right)^2=a^2$$
Similarly, let $b$ be the radius of the smaller inscribed circle. With the cosine rule,
$$(b+3.5)^2=(7-b)^2+3.5^2-2\cdot 3.5 \cdot (7-b)\cos 45^\circ$$
which yields
$$b= \frac{7(\sqrt{2}-1)}{3\sqrt{2}-1}$$
Therefore, the equation of the smaller inscribed circle is
$$\left( x- \frac{7-b}{\sqrt{2}} \right)^2
+\left( y- \frac{7-b}{\sqrt{2}} \right)^2=b^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3362007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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|
If $a_{k}=2^{2^k}+2^{-2^k}$ then evaluate $\prod_{k=1}^\infty\left(1-\frac{1}{a_{k}}\right)$ If $$a_{k}=2^{2^k}+2^{-2^k}$$ then evaluate $$\prod_{k=1}^\infty\left(1-\frac{1}{a_{k}}\right)$$
I tried using Sophie-Germaine Identity about factorisation for $x^4+4$ but it did not work
|
Note that
$$1 - \frac{1}{a_k} = \frac{1 - 2^{-2^k} + 4^{-2^k}}{1 + 4^{-2^k}} = \frac{1 + 8^{-2^k}}{(1 + 2^{-2^k})(1 + 4^{-2^k})}$$
and for $|x| < 1$, we have
$$\prod_{k=1}^{\infty} (1 + x^{2^k}) = \lim_{n \to \infty} \prod_{k=1}^n (1 + x^{2^k}) = \lim_{n \to \infty} \frac{1 - x^{2^{n+1}}}{1 - x^2} = \frac{1}{1-x^2}$$
which gives
$$\prod_{k=1}^\infty \left(1 - \frac{1}{a_k}\right) = \frac{\prod_{k=1}^\infty (1 + 8^{-2^k})}{\prod_{k=1}^\infty (1 + 2^{-2^k})\prod_{k=1}^\infty (1 + 4^{-2^k})} = \frac{\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{16}\right)}{1 - \frac{1}{64}} =
\frac{5}{7}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3362348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Integration of a rational function involving a quadratic
Evaluate the integral: $$\int\frac{3x}{(1-4x-2x^2)^2}\ dx$$
Here is my work:
*
*Complete the square on the denominator: $$(1-4x-2x^2)=(1-2(2x+x^2+1-1))=(3-2(x+1)^2)$$
*Insert back into denominator. Use substitution $u=x+1$; $x=u-1$; $dx=du$. $$\int\frac{3(u-1)}{(3-2u^2)^2}\ du$$
*Split the integral into two: $$\int\frac{3u}{(3-2u^2)^2}\ du\ -\int\frac3{(3-2u^2)^2}\ du$$
*Integrate the first integral: $$\int\frac{3u}{(3-2u^2)^2}\ du$$
$v=(3-2u^2)$
$\frac{dv}{du}=-4u$
$\frac{-dv}4=u\ du$
$$\frac{-1}4\int\frac{3}{v^2}\ dv=\frac{-3}{4}\int\frac{1}{v^2}\ dv=\frac1{4v}$$$$=\frac{1}{4(3-2u^2)}=\frac1{4(3-2(x+1)^2)}$$
*
*Integrate the second integral: $$-\int\frac3{(3-2u^2)^2}\ du$$
Use sine trig substitution: $u=\frac{\sqrt3}{\sqrt2}\sin\theta$; $\frac{\sqrt2}{\sqrt3}du=\cos\theta\ d\theta$
$$-\frac{\sqrt2}{\sqrt3}\int\frac{3\cos\theta}{(3-2(\frac{\sqrt3}{\sqrt2}\sin\theta)^2)^2}\ d\theta=-\frac{\sqrt2}{\sqrt3}\int\frac{3\cos\theta}{(3-2(\frac{3}{2}\sin^2\theta))^2}\ d\theta$$
$$=-\frac{\sqrt2}{\sqrt3}\int\frac{3\cos\theta}{(3-3\sin^2\theta))^2}\ d\theta=-\frac{\sqrt2}{3\sqrt3}\int\frac{\cos\theta}{(\cos^2\theta))^2}\ d\theta=-\frac{\sqrt2}{3\sqrt3}\int\frac{1}{(\cos^3\theta)}\ d\theta$$$$=-\frac{\sqrt2}{3\sqrt3}\int\sec^3\theta\ d\theta=\frac{\sqrt2}{3\sqrt3}\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|$$
Besides for the last resubstitution which I still have to do, I also seem to have a different leading fraction for my second integral than the answer key. Where did I go wrong?
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Here is how to carry out the second integral. With the substitution $u=\frac{\sqrt3}{\sqrt2}\sin\theta$,
$$I=-\int\frac3{(3-2u^2)^2}\ du$$
$$=-\frac{\sqrt3}{\sqrt2}\int\frac{3\cos\theta}{(3-2(\frac{3}{2}\sin^2\theta))^2}\ d\theta$$
$$=-\frac{\sqrt3}{\sqrt2}\int\frac{3\cos\theta}{(3-3\sin^2\theta)^2}\ d\theta
=-\frac{\sqrt3}{3\sqrt2}\int\frac{\cos\theta}{\cos^4\theta}\ d\theta$$
$$=-\frac{1}{\sqrt6}\int\frac{1}{\cos^3\theta}\ d\theta =-\frac{1}{\sqrt6}\int\sec^3\theta\ d\theta$$
$$=-\frac{1}{2\sqrt6}(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)+C$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3362838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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.