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Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$ Determine the number of ordered triple $(x, y, z)$ of integer numbers (negatives and positives) satisfying $|x| + |y| + |z| \le 6$
I know that final answer is 377, but how?
Edit:
Drawing from David K's answer:
One way to count the ways is to first do the ways for which $x > 0,$ $y > 0,$ and $z > 0.$
That's the number of ways to put $6$ or fewer indistinguishable balls into $3$ numbered bins under the constraint that every bin must have at least one ball,
which is the number of ways to put $3$ or fewer indistinguishable balls in $3$ numbered bins without that constraint, which is the number of ways of putting exactly $3$ indistinguishable balls in $4$ numbered bins.
Multiply by $8$ to take into account all the cases where $x < 0$ or $y < 0$ or $z < 0.$
$x,y,z >0 => x + y + z <= 3, n = Cr(3+4-1,3) * (8) = 160$
Now let $x = 0,$ $y > 0,$ and $z > 0.$
Count the number of ways to put up to $6$ balls in $2$ bins if each bin must contain at least one ball. Multiply by four to account for all the cases were $y < 0$ or $z < 0.$ Multiply that result by $3$ to account for the fact that we could have chosen $y= 0$ or $z=0$ instead of $x = 0.$
$x=0, x,y>0 => y+z <=6, n = Cr(4+3-1, 4) * 3 * 14 = 180$
Now let $x = y = 0$ and $z > 0.$ There are $6$ ways for that to happen. Multiply by $2$ to account for $z < 0,$ then by $3$ to account for the other choices of which variables are zeros.
$x,y = 0, z>0 => z<=6, n = 6 * 2 * 3 = 36$
$x,y,z=0, n = 1$
Sum of all of them is : 160 + 180 + 36 + 1 = 377
|
Let $a_{n,m}$ be the number of $n$-tuples of integers so that
$$
\sum_{k=1}^n|x_k|\le m\tag1
$$
Pretty simply, we have
$$
a_{1,m}=2\binom{m}{1}+\binom{m}{0}\tag2
$$
Furthermore, we have the recurrence
$$
\begin{align}
a_{n+1,m}
&=\overbrace{\quad a_{n,m}\quad\vphantom{\sum_1^m}}^\text{$0$ in position $n+1$}+\overbrace{2\sum_{k=1}^ma_{n,m-k}}^\text{$\pm k$ in position $n+1$}\\
&=a_{n,m}+2\sum_{k=0}^{m-1}a_{n,k}\tag3
\end{align}
$$
Thus,
$$
a_{2,m}=4\binom{m}{2}+4\binom{m}{1}+\binom{m}{0}\tag4
$$
and
$$
a_{3,m}=8\binom{m}{3}+12\binom{m}{2}+6\binom{m}{1}+\binom{m}{0}\tag5
$$
In general, we get by induction, using $(2)$ and $(3)$ and $\sum\limits_{k=0}^{n-1}\binom{k}{j}=\binom{n}{j+1}$,
$$
a_{n,m}=\sum_{k=0}^n2^k\binom{n}{k}\binom{m}{k}\tag6
$$
Plugging in $n=3$ and $m=6$ gives
$$
\bbox[5px,border:2px solid #C0A000]{a_{3,6}=377}\tag7
$$
I just came across this answer which proves
$$
\sum_{k=0}^n\binom{n}{k}\binom{m+k}{n}=\sum_{k=0}^n2^k\binom{n}{k}\binom{m}{k}
$$
|
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|
Prove: if $c^2+8 \equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$. I want to show:
If $c^2+8 \equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.
I have calculated that $c^3-7c^2-8c \equiv -7c^2-16c \equiv 56- 16c \equiv 8(7-2c) \equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?
|
We have that
$$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) \equiv 2c - 7 \mod p.$$
This proves the claim.
|
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|
Find limit supremum from 3 sequence theorem Find limit $$\limsup_{n\rightarrow \infty} \sqrt[n]{\left| \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^p\right |} $$
dependency on value of $p$
I think that $p$ doesn't matter there (by doesn't matter I mean that I can compute limit without knowledge about p and on the end power result to p) because it is just const value so I can write that as
$$\limsup_{n\rightarrow \infty} \left(\sqrt[n]{\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|} \right)^p$$ and just calculate
$$\limsup_{n\rightarrow \infty} \sqrt[n]{\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|} $$
I suspect that this sequence has normal $lim$ so probably
$$\limsup_{n\rightarrow \infty} \sqrt[n]{\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|} = \lim_{n\rightarrow \infty} \sqrt[n]{\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|} $$
Ok now I want to use three sequence theorem:
$$ ... \le \sqrt[n]{\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|} \le \sqrt[n]{2} $$
But I don't know how to estimate that from below...
|
Hint
Compose Taylor series for large values of $n$ to get
$$2-2 \cos\left(\frac{1}{n}\right)-\frac 1n{\sin \left(\sin \left(\frac{1}{n}\right)\right)}=\frac{1}{4 n^4}-\frac{7}{72 n^6}+O\left(\frac{1}{n^8}\right)$$
|
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|
Finding place of the nine digits The nine digits 1, 2, 3, ... .., 9 are placed in the nine triangles of the attached figure in such a way that the digits around each circle add up as indicated. Calculate the value of N.
|
The total of the digits from $1$ to $9$ is $45$. Four digits on the left side of the diagram add to $16$; four on the right add to $25$. That leaves $N = 45 - 16 - 25 = 4$.
EDIT: Let's find the rest.
With $4$ assigned, the additional four digits in the total $32$ must add up to $28$. The only possibility is $5,6,8,9$; correspondingly, the four digits not adjacent to $32$ are $1,2,3,7$.
The total of $25$ is made up of two digits each from these two sets of four. No choice without $7$ is large enough; considering whether $1,2$ or $3$ is included quickly reveals two possibilities: $25 = 1+7+8+9$ or $3+7+6+9$. These correspond to $16 = 2+3+5+6$ and $1+2+5+8$.
The total of $20$ comes from four digits: two from $1,2,3,7$ and two from $5,6,8,9$, but also two each from the quadruples making up $25$ and $16$. If the correct way of forming $25$ and $16$ is $(1+7)+(8+9)$ and $(2+3)+(5+6)$, then $20$ must be made with one digit chosen from each of the four parenthesized groups. None of the sixteen possibilities work. So it must be $25 = (3+7)+(6+9)$ and $16 = (1+2)+(5+8)$. Only one possibility works: $20 = 7+6+2+5$.
Thus the unique solution has the digits $2,5,6,7$ from left to right in the top half and $1,8,4,9,3$ in the bottom half.
|
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|
Find the angle in an isosceles triangle Let triangle $\Delta ABC$ have $AB=AC$. Then we draw the angle bisector from $B$ to $AC$ intersecting at $D$. Find the angle $\angle BAC$ if $BC=AD+BD$.
My attempts:
I know that the answer is 100° but I couldn't prove that if you extended $AD$ to a point $E$ so it is equal to $BC$, then the angle $\angle DCE$ is the same as $\angle ACB$.
|
In the standard notation we obtain:
$$\frac{AD}{DC}=\frac{AB}{BC}=\frac{c}{a}$$ and $$AD+DC=AC=b,$$ which gives
$$AD=\frac{bc}{a+c},$$ $$DC=\frac{ab}{a+c}$$ and
$$BD^2=AB\cdot BC-AD\cdot DC=ac-\frac{b^2ac}{(a+c)^2}=ab-\frac{ab^3}{(a+b)^2},$$
which gives $$BD=\frac{a\sqrt{b(a+2b)}}{a+b}.$$
Id est, by the given we obtain:
$$a=\frac{b^2}{a+b}+\frac{a\sqrt{b(a+2b)}}{a+b}$$ or
$$a\sqrt{b(a+2b)}=a^2+ab-b^2.$$
Now, $\sin\frac{\alpha}{2}=\frac{\frac{a}{2}}{b}=\frac{a}{2b}.$
Let $\sin\frac{\alpha}{2}=x.$
Thus, $a=2xb$ and we obtain:
$$2x\sqrt{2(x+1)}=4x^2+2x-1$$ or
$$8x^2(x+1)=(4x^2+2x-1)^2,$$ where $x>0$ and $4x^2+2x-1>0.$
We obtain:
$$16x^4+8x^3-12x^2-4x+1=0$$ or
$$8x^3(2x+1)-12x^2-6x+2x+1=0$$ or
$$(2x+1)(8x^3-6x+1)=0$$ or
$$3x-4x^3=\frac{1}{2}$$ or
$$\sin\frac{3\alpha}{2}=\frac{1}{2},$$ which gives
$$\frac{3\alpha}{2}=30^{\circ},$$ which is impossible because $4x^2+2x-1>0,$ or
$$\frac{3\alpha}{2}=150^{\circ},$$ which gives $\alpha=100^{\circ}$ and $\beta=\gamma=40^{\circ}.$
The fact that $$BD^2=AB\cdot BD-AD\cdot DC$$ we can prove by the following reasoning.
Let $\Phi$ be a circumcircle of $\Delta ABC$ and $AD\cap\Phi=\{A,E\}$.
Thus, $\measuredangle BAC=\measuredangle BEC$ and $\measuredangle ABD=\measuredangle EBD,$ which gives $\Delta ABD\sim\Delta EBC.$
Hence, $$\frac{AB}{BE}=\frac{BD}{BC}$$ or
$$AB\cdot BC=BD\cdot BE$$ or
$$AB\cdot BC=BD(BD+DE)$$ or
$$BD^2=AB\cdot BC-BD\cdot DE$$ or
$$BD^2=AB\cdot BC-AD\cdot DC.$$
|
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|
Solution of this Diophantine Equation
If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.
My attempt:
$x^2-2y^2=1$
$\implies (x+\sqrt{2}y)(x-\sqrt{2}y)=1$
$\implies (x+\sqrt{2}y)=1$ and $(x-\sqrt{2}y)=1$
$\implies x=1$ and $y=0$
Clearly $x$ and $y$ are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?
|
What about
\begin{align*}&x^2-2y^2=1\tag{1}\\\iff & x^2-1=(x+1)(x-1)=2y^2\end{align*}
Since $2\mid (x+1)(x-1)$, we conclude that both $(x+1)$ and $(x-1)$ have to be even, and hence $$4\mid 2y^2\implies 2\mid y^2\implies 2\mid y$$ and since $y$ is prime, $\color{red}{y=2}$. Can you end it now?
From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus,
the only solution is $\color{blue}{(3, 2)}$.
Addendum
The problem with your method is that for $a,b\in\mathbb R$
$$a·b=1\not\Rightarrow a=1\;\text{ and }\;b=1$$
In fact, this only works if $$a·b=0\implies a=0\;\text{ or }\;b=0$$
|
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|
Prove independence of two random variables. Let two random variables $\xi_1$ and $\xi_2$ be given. They are independent and have a standard normal distribution. Proove that $\frac{\xi_1^2 - \xi_2^2}{\sqrt{\xi_1^2+\xi_2^2}}$ and $\frac{2\xi_1\xi_2}{\sqrt{\xi_1^2+\xi_2^2}}$ are independent.
I was given a little hint that I need to think about the angles and may be it's necessary to go to the polar coordinate system.
Please help to prove it or give me a small hint)
|
according to
Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$
they are independent.
but if you want to continue your way:
$X=r\cos(\theta) \hspace{.5cm}
Y=r\sin(\theta) \hspace{.5cm}
X,Y \sim normal(0,1) \Leftrightarrow \theta \sim Uniform(0,2\pi) \hspace{.5cm} r^2\sim \chi^2_2$.
$X$ and $Y$ are independent $\Leftrightarrow $ $\theta$ and $r$ are independent.
also $\sin(\theta) \sim \cos(\theta) \sim \sin(2\theta) \sim 2\sin(\theta) \cos(\theta)
\sim \cos(2\theta) \sim \cos(2\theta) \sim f $ that $ f(z) =\frac{1}{\pi \sqrt(1-z^2)} I_{[-1,1]}(z) $ since
$z=\sin(\theta) \Rightarrow f(z)=|\frac{d}{dz} \sin^{-1}(z)| f_{\theta}(\sin^{-1}(z)) + |\frac{d}{dz} (\pi-\sin^{-1}(z))| f_{\theta}(\pi -\sin^{-1}(z)) =\frac{1}{\sqrt(1-z^2)} \frac{1}{2\pi} +\frac{1}{\sqrt(1-z^2)} \frac{1}{2\pi} =\frac{1}{\pi\sqrt(1-z^2)} $
similar for others.
$\frac{2XY}{\sqrt(X^2+Y^2)}=\frac{2r^2 \cos(\theta) \sin(\theta)}{r}=2r \cos(\theta) \sin(\theta) =r \sin(2\theta) \sim r \sin(\theta) $
$\frac{X^2-Y^2}{\sqrt(X^2+Y^2)}=\frac{r^2(\cos^2(\theta)-\sin^2(\theta))}{r}$
$=r\cos(2\theta)\sim r\cos(\theta)$
note $X=r \sin(\theta) $ and $Y=r\cos(\theta)$ are independent.
|
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|
Proving difference between two functions is small when $x$ is small... I'm really struggling to prove the following claim and I was wondering whether anyone could help me.
Claim:$$ 0<|x| \leq 10^{-4} \implies \bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} \leq 10^{-12}$$
My attempt at a proof: I have already proven the following inequality,
$$\bigg{|} \frac{e^x-1}{x} - \sum_{k=1}^n \frac{x^{k-1}}{k!} \bigg{|} \leq \frac{|x|^ne^{|x|}}{(n+1)!}$$
So let $n=3$, then
\begin{align*}
\bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} &= \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \cdot \left( \frac{e^x-1}{x} - \sum_{k=1}^3 \frac{x^{k-1}}{k!} \right) \bigg{|} \\
&= \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \bigg{|} \frac{e^x-1}{x} - \sum_{k=1}^3 \frac{x^{k-1}}{k!} \bigg{|} \\
&\leq \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \frac{|x|^3e^{|x|}}{(3+1)!} \\
&\leq \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \frac{(10^{-4})^3 e^{10^{-4}}}{4!} \\
&= 10^{-12} \cdot \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} \cdot \frac{ e^{10^{-4}}}{4!}
\end{align*} Hence, it suffices to show that for $0 < |x| \leq 10^{-4}$,
\begin{align*}
f(x) := \bigg{|} \frac{x}{(e^x-1)(\sum_{k=1}^3 \frac{x^{k-1}}{k!})} \bigg{|} < \frac{4!}{e^{10^{-4}}}
\end{align*}
Since $f(x)$ is monotonically decreasing on the intervals $[-10^{-4}, 0)$ and $(0, 10^{-4}]$, we know that for $0 <|x| \leq 10^{-4}$,
\begin{align*}
f(x) \leq f \left( -10^{-4} \right) \approx 1.0001 < \frac{4!}{e^{10^{-4}}}
\end{align*} Therefore, for $0 < |x| \leq 10^{-4}$,
\begin{align*}
\bigg{|} \frac{x}{e^x-1} - \frac{1}{\sum_{k=1}^3 \frac{x^{k-1}}{k!}} \bigg{|} \leq 10^{-12}
\end{align*} $\square$
My big problem here is that I don't actually know how to prove my claim that $f(x)$ is monotonically decreasing. Does anyone know how to prove this?
Or even better, does anyone know of a more direct proof that avoids this problem altogether?
Thanks!
|
Write $|\frac{x}{e^x-1} - \frac{x}{\sum_{k=1}^3 \frac{x^k}{k!}}|
= |x||\frac{\sum_{k=1}^3 \frac{x^k}{k!} - (e^x-1)}{(e^x-1)\sum_{k=1}^3 \frac{x^k}{k!}}| $
You can use the $|x|<10^-4$ bound given, and also instead an easier inequality for the numerator using the series sum for $e^x$. Then it is clear that the denominator is obviously increasing in $x$.
So I guess this comes under "direct proof". I think your function $f(x)$ is probably decreasing though.
|
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What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$
They all are positive terms so arithmetic mean is greater than equal to geometric mean.
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq 3( \sec^6 x \csc^6 x \sec^6 x\csc^6 x)^\frac{1}{3} $$
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x \geq 3( \sec x \csc)^4 $$
$$
\sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq
\frac{3 * 2^4}{\sin ^4 2x} $$
Clearly least value is 48, but something is wrong here, as the answer is 80, if I use other methods.
|
You want to find the least value of $f(x)=\sec^6(x)+\csc^6(x)+\sec^6(x)\csc^6(x)$. You found that $f(x) \geq g(x)=3(\sec(x)\csc(x))^4$. In addition, the minimum value of $g(x)$ is $48$. Therefore, you can conclude that $f(x) \geq 48$ for all $x$. But why would you expect there to exist some $x$ such that $f(x)=48$, when $g(x)$ was simply a lower bound?
This is like saying find the least value of $x^2+4$. Well, $x^2+4 \geq (4x^2)^{1/2} = 4|x|$, whose minimum value is $0$. But clearly $x^2+4$ has a minimum value of $4$. The problem is that the lower bound is not tight.
|
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|
Find all values of $c$ for the Mean Value Theorem's $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ I need to find all values of $c$ for the $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ such that $f'(c) = -10$. I have already found that $f'(c) = \frac{-130}{13}=-10$.
I thought I could find the values of $c$ by differentiating $f(x)$ which becomes $6x^2-12x-90$, and applying the Mean Value Theorem $f'(c) = \frac{f(b)-f(a)}{b - a}$ again.
$f(b) = 6(-5)^2-12(-5)-90 = 150+60-90 = 120$
$f(a) = 6(8)^2-12(8)-90 = 384-186 = 198$
$f'(c) = \frac{120-198}{-5-8} = \frac{78}{13} = 6$
Then I took $6x^2-12x-90$ again, replaced $x$ with $c$, and set it equal to $6$ before solving for $c$.
$6c^2-12c-90 = 6 \to 6(c^2-2c-16) = 0$
$\frac{2 \pm \sqrt{2^2-4(1)(-17)}}{2(1)} = \frac{2 \pm \sqrt{74}}{2}$ or $c = \frac{2 - \sqrt{74}}{2}, \frac{2 + \sqrt{74}}{2}$
But this answer is wrong.
How can I find the values for $c$ here?
|
You got the wrong answer because you set $f'(c)=\frac{f'(b)-f'(a)}{b-a}$ instead of $\frac{f(b)-f(a)}{b-a}.$
Otherwise your approach was correct.
|
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Any hints on how to compute this integral? Could anyone please give me a hint on how to compute the following integral?
$$\int \sqrt{\frac{x-2}{x^7}} \, \mathrm d x$$
I'm not required to use hyperbolic/ inverse trigonometric functions.
|
Write $y(x):=\sqrt{\frac {x-2} {x^7}}$.
Note that $$y'(x)= \frac {7-3x}{x^8} \frac 1 y $$
Hence $$\frac d {dx} x^n y=n x^{n-1} y + x^{n-8} \frac {7-3x} y.$$
Do the ansatz $$F(x)=\sum_{n=0}^k a_nx^ny ~~~~\text{ and } ~~~~F'(x)=y(x). $$
We get
$$y\sum_{n=1}^{k}a_n nx^{n-1} +\frac 1 y\sum_{n=0}^{k} a_n x^{n-8} ({7-3x})=y $$
Thus $$\sum_{n=0}^{k} a_n x^{n-8} ({7-3x})=y^2(1-\sum_{n=1}^{k}a_n nx^{n-1}).$$
Now insert $y^2$
and get
$$({7-3x})\sum_{n=0}^{k} a_n x^{n-1} = (x-2 )(1-\sum_{n=1}^{k}a_n nx^{n-1}) $$
or equivalently
$$2-x+({7-3x})\sum_{n=0}^{k} a_n x^{n-1}+ (x-2)\sum_{n=1}^{k}a_n nx^{n-1}=0. $$
We solve this recursively.
The lowest order is $x^{-1}$. There we have $7a_0 x^{-1}=0$, so $a_0=0$.
In constant order: $2+7a_1-3a_0-2a_1=0$, so $a_1=-\frac 2 5$
In order $x$: $(-1+7a_2 -3 a_1 +a_1-4 a_2)x=0$ , so $a_2= \frac 1 3 (1+2
a_1)=\frac 1 {15}$
in order $x^2$: $(7a_3-3a_2+2a_2-6 a_3)x^2=0$, so $a_3=a_2=\frac 1 {15}$.
in order $x^3$: $(7a_4-3a_3+3a_3 - 8a_4)x^4=0$, so $a_4=0$
in order $x^n$ for $n>3$: $(7a_{n+1} - 3 a_n +n a_n - 2(n+1) a_{n+1})x^n$, so $a_{n+1}=\frac {n-3}{7-2(n+1)} a_n=0$.
In conclusion it follows that
$$\int y(x)= const + F(x)= const+ \frac 1 {15} (-6x+ x^2+x^3) y $$
|
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"url": "https://math.stackexchange.com/questions/3170871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Why does $\sin(x) - \sin(y)=2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$? Why does this equality hold?
$\sin x - \sin y = 2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$.
My professor was saying that since
(i) $\sin(A+B)=\sin A \cos B+ \sin B \cos A$
and
(ii) $\sin(A-B) = \sin A \cos B - \sin B \cos A$
we just let $A=\frac{x+y}{2}$ and $B=\frac{x-y}{2}$. But I tried to write this out and could not figure it out. Any help would be appreciated
|
Following your notation, let $A=\dfrac{x+y}{2}$ and $B=\dfrac{x-y}{2}$.
Note that $A+B=x$ and $A-B=y$.
Now, $\sin x=\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin y=\sin(A-B)=\sin A\cos B - \cos A\sin B$ from your professor's advice.
To get the LHS, $\sin x-\sin y = 2\cos A\sin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3171404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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|
What are the integer coeffcients of a cubic polynomial having two particular properties? Let $f(x) = x^3 + a x^2 + b x + c$ and $g(x) = x^3 + b x^2 + c x + a\,$ where $a, b, c$ are integers and $c\neq 0\,$. Suppose that the following conditions hold:
*
*$f(1)=0$
*The roots of $g(x)$ are squares of the roots of $f(x)$.
I'd like to find $a, b$ and $c$.
I tried solving equations made using condition 1. and relation between the roots, but couldn't solve.
The equation which I got in $c$ is $c^4 + c^2 +3 c-1=0$ (edit: eqn is wrong).
Also I was able to express $a$ and $b$ in terms of $c$. But the equation isn't solvable by hand.
|
Let $u$, $v$ and $w$ be the roots of $f$, so that $u^2$, $v^2$ and $w^2$ are the roots of $g$. Then comparing the coefficients of
$$(x-u)(x-v)(x-w)=f(x)=x^3+ax^2+bx+c,$$
$$(x-u^2)(x-v^2)(x-w^2)=g(x)=x^3+bx^2+cx+a,$$
yields the equations
\begin{eqnarray*}
a&=&-u-v-w&=&-u^2v^2w^2,\\
b&=&uv+uw+vw&=&-u^2-v^2-w^2,\\
c&=&-uvw&=&u^2v^2+u^2w^2+v^2w^2.
\end{eqnarray*}
This immediately shows that $a=-c^2$, and the identities
\begin{eqnarray*}
u^2+v^2+w^2&=&(u+v+w)^2-2(uv+uw+vw),\\
u^2v^2+u^2w^2+v^2w^2&=&uvw(u+v+w)-(uv+uw+vw)^2,
\end{eqnarray*}
show that $-b=a^2-2b$ and $c=ac-b^2$, respectively, hence $b=a^2=c^4$ and so
$$f(x)=x^3-c^2x^2+c^4x+c,$$
for some $c$. Then $f(1)=1$ implies that
$$c^4-c^2+c+1=0,$$
which has the clear root $c=-1$. Then $a=-1$ and $b=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3173125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Solving $\tan (2x) = 6\cos^2(x) - 4\sin(x)\cos(x) - 2\sin^2(x)$ I need to solve the following trigonometric equation:
$$\tan (2x) = 6\cos ^2(x) - 4\sin (x)\cos (x) - 2\sin ^2(x)$$
My attempt:
$$\frac{\sin(2x)}{\cos(2x)} = 3(\cos(2x)+1) - 2\sin(2x) -2(1-\cos^2(x))$$
$$\frac{\sin(2x)}{\cos(2x)} = 3\cos(2x)+3 - 2sin(2x) - 2(1-\frac{\cos(2x)+1}{2})$$
$$\frac{\sin(2x)}{\cos(2x)} = 3\cos(2x)+3-2\sin(2x)-2+\cos(2x)+1$$
$$\frac{\sin(2x)}{\cos(2x)} = 4\cos(2x) +2 - 2\sin(2x)$$
Adding $2\sin(2x)$ to each side and then multiplying by $\cos(2x)$
$$\sin(2x)(1+2\cos(2x)) = 4\cos^2(2x)+2\cos(2x)$$
$$\sin(2x)(1+2\cos(2x)) +1= 4\cos^2(2x)+2\cos(2x)+1$$
$$\sin(2x)(1+2\cos(2x))+1 = (2\cos(2x)+1)^2$$
$$1 = (2\cos(2x)+1)) (2\cos(2x)+1 - \sin(2x))$$
After that I got stuck.
|
You find correctly that
$$
\tan2x=4\cos2x-2\sin2x+2
$$
Multiplying by $\cos2x$ leads to
$$
4\cos^22x-2\sin2x\cos2x+2\cos2x-\sin2x=0
$$
that can be rewritten as
$$
2\cos2x(2\cos2x-\sin2x)+(2\cos2x-\sin2x)=0
$$
so
$$
(2\cos2x+1)(2\cos2x-\sin2x)=0
$$
This splits into
$$
\cos2x=-\frac{1}{2}\qquad\text{or}\qquad \sin2x=2\cos2x
$$
and both are elementary.
|
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"url": "https://math.stackexchange.com/questions/3176171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(e^{(1+\frac{k}{n})^2} - \frac{3e^{(1 + \frac{3k}{n})}}{2\sqrt{1 + \frac{3k}{n}}}\right).$ Find
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(e^{(1+\frac{k}{n})^2} - \frac{3e^{(1 + \frac{3k}{n})}}{2\sqrt{1 + \frac{3k}{n}}}\right)$$
Choices:
|
Let $u=\sqrt{1+3x}$, then $\frac{du}{dx}=\frac{3}{2u}$
\begin{align}
\frac32 \int_0^1 \frac{\exp(1+3x)}{\sqrt{1+3x}} \, dx&= \frac32 \int_1^2\frac{\exp(u^2)}{u}\cdot \frac{2u}{3}\, du = \int_1^2 \exp(u^2)\,du
\end{align}
Also, let $v=1+x$,
$$\int_0^1 \exp((1+x)^2) \, dx = \int_1^2 \exp(v^2) \, dv$$
Hence, the two integral cancels out.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3176671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Calculate the limit of the following recurrent series Find the limit of the series : $\ q_{n+1,}=q_n + \frac{2}{q_n}$ with $q_0=1,n\ge 0$.
|
I will show that
$q_n$ is unbounded
and that
$\frac12+\sqrt{2n-\frac74}
\lt q_n
<3+2\sqrt{ 2 n+\frac14}
$.
Moreover,
if,
as these inequalities make plausible,
$\lim_{n \to \infty} \dfrac{q_n}{\sqrt{n}}
$
exists and this limit is $c$,
then $c = 2$,
so that
$\lim_{n \to \infty} \dfrac{q_n}{\sqrt{n}}
=2
$.
Here we go.
$q_n > 1$ for $n > 1$
and $q_n$ is increasing.
If $q_n$ is bounded,
there is an $L$
such that $q_n \le L$
for all $n$.
Writing
$q_{n+1}=q_n + \frac{2}{q_n}
$
as
$q_{n+1}-q_n
= \frac{2}{q_n}
$
and summing
$\begin{array}\\
q_{n+1}-q_1
&=\sum_{k=1}^n (q_{k+1}-q_k)\\
&=\sum_{k=1}^n \frac{2}{q_k}\\
&\ge\sum_{k=1}^n \frac{2}{L}\\
&\ge \dfrac{2n}{L}\\
\end{array}
$
which is unbounded,
a contradiction.
Note that this can be used,
as Will Jagy wrote,
to show that
if
$q_{n+1} = q_n+f(q_n)$
where
$f(x) > 0$
and
$f'(x) < 0$
then
$q_n$ is unbounded.-
This can be used
to get explicit bounds.
$\begin{array}\\
q_{n+1}-q_1
&=\sum_{k=1}^n (q_{k+1}-q_k)\\
&=\sum_{k=1}^n \frac{2}{q_k}\\
&\ge\sum_{k=1}^n \frac{2}{q_n}\\
&\ge \dfrac{2n}{q_n}\\
&\gt \dfrac{2n}{q_{n+1}}\\
\end{array}
$
so,
letting
$\frac{q_1}{2} = a$,
$\begin{array}\\
q_n > \dfrac{2n-2}{q_n}+2a\\
\text{or}\\
q_n^2-2aq_n > 2n-2\\
\text{or}\\
q_n^2-2aq_n+a^2 > 2n-2+a^2\\
\text{or}\\
(q_n-a)^2 > 2n-2+a+a^2\\
\text{or}\\
q_n
\gt a+\sqrt{2n-2+a^2}\\
\end{array}
$
Therefore
$q_n
\gt a+\sqrt{2n-2+a^2}\\
= \frac12+\sqrt{2n-\frac74}\\
$.
With this lower bound for $q_n$,
we can now get an upper bound.
Let $2-a^2 = b$.
$\begin{array}\\
q_{n+1}-q_1
&=\sum_{k=1}^n \frac{2}{q_k}\\
&=\frac{2}{q_1}+\sum_{k=2}^n \frac{2}{q_k}\\
&\lt\frac{2}{q_1}+2\sum_{k=2}^n \frac{1}{a+\sqrt{2n-b}}\\
&\lt\frac{2}{q_1}+2\sum_{k=2}^n \int_{k-1}^k\frac{dx}{a+\sqrt{2x-b}}\\
&\lt\frac{2}{q_1}+2\int_{1}^n\frac{dx}{a+\sqrt{2x-b}}\\
&=\frac{2}{q_1}++2(\sqrt{ 2 x-b} - a \log(\sqrt{ 2 x-b} + a))|_{1}^n\\
&=\frac{2}{q_1}+2(\sqrt{2 n-b} - a \log(\sqrt{2 n-b} + a))\\
&\qquad-2(\sqrt{a^2+2}-\log(\sqrt{a^2+2}+a))\\
&=\frac{2}{q_1}+2(\sqrt{a^2 + 2 (n+1)} - a \log(\sqrt{a^2 + 2( n+1)} + a))\\
&\qquad-2(\sqrt{2-b}-\log(\sqrt{2-b}+a))\\
\text{or}\\
q_{n}-q_1
&<\frac{2}{q_1}+2(\sqrt{a^2 + 2n} - a \log(\sqrt{a^2 + 2n} + a))\\
&\qquad-2(\sqrt{a^2+2}-\log(\sqrt{a^2+2}+a))\\
&<\frac{2}{q_1}+2\sqrt{a^2+ 2 n}\\
\end{array}
$
Therefore
$q_n
\lt 2a+\frac{2}{q_1}+2\sqrt{a^2+ 2 n}
= 3+2\sqrt{ 2 n+\frac14}
$.
Assuming a little more,
if
$q_n \approx c\sqrt{n}+O(1)$,
then
$c\sqrt{n+1}-c\sqrt{n}+O(1)
=\dfrac{2}{c\sqrt{n}+O(1)}
=\dfrac{2}{c\sqrt{n}}+O(\dfrac1{n})
$
so, since
$\begin{array}\\
\sqrt{n+1}-\sqrt{n}
&=\dfrac1{2\sqrt{n+d}}
\qquad 0 \le d \le 1, mvt\\
&=\dfrac1{2\sqrt{n}\sqrt{1+d/n}}\\
&=\dfrac1{2\sqrt{n}}(1+O(\dfrac1{n}))\\
\end{array}
$
$\dfrac{c}{2\sqrt{n}}+O(\dfrac1{n^{3/2}})
=\dfrac{2}{c\sqrt{n}}+O(1/n)
$
so
$\dfrac{c}{2}=\dfrac{2}{c}$
so that $c = 2$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Distributing $60$ identical balls into $4$ boxes if each box gets at least $4$ balls, but no box gets $20$ or more balls How many different ways can the balls be placed if each box gets at least $4$ balls each, but no box gets $20$ or more balls?
I was thinking about finding all the possible ways which every box gets at least $4$ balls which would be $47\choose44$ ways $\rightarrow$ $16215$ ways. Then subtracting that number by the number of possible ways that a box could get 20 or more balls ($16215$ - $20$ or more balls possibilities). The thing is I don't know how to get the number of ways if a box gets $20$ or more balls. Any tips?
|
What you have done thus far is correct.
As you observed, after distributing four balls to each box, we are left with $60 - 4 \cdot 4 = 44$ balls to distribute. If we let $x_i, 1 \leq i \leq 4$, represent the number of additional balls we distribute to the $i$th box, then
$$x_1 + x_2 + x_3 + x_4 = 44 \tag{1}$$
Equation 1 is an equation in the nonnegative integers. A particular solution of equation 1 corresponds to the placement of $4 - 1 = 3$ addition signs in a row of $44$ ones. The number of such solutions is
$$\binom{44 + 4 - 1}{4 - 1} = \binom{47}{3} = \binom{47}{44}$$
since we must choose which three of the $47$ positions required for $44$ ones and three addition signs will be filled with addition signs or, equivalently, which $44$ positions will be filled with ones.
From these, we must subtract those cases in which a box receives at least $20$ balls. Since four balls have already been placed in each box, that means we must subtract those cases in which at least $20 - 4 = 16$ additional balls are placed in a box. Since $3 \cdot 16 = 48 > 44$, at most two boxes could have at least $16$ additional balls placed in them.
There are four ways to select a box that will receive at least $16$ additional balls. Suppose it is the first box. Then $x_1' = x_1 - 16$ is a nonnegative integer. Substituting $x_1' + 16$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 16 + x_2 + x_3 + x_4 & = 44\\
x_1' + x_2 + x_3 + x_4 & = 28 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{28 + 4 - 1}{4 - 1} = \binom{31}{3} = \binom{31}{28}$$
solutions. Therefore, there are
$$\binom{4}{1}\binom{31}{3}$$
solutions in which at least $16$ additional balls are placed in one of the boxes.
However, if we subtract this amount from the total, we will have subtracted too much since we will have subtracted each case in which two boxes contain at least $20$ balls twice, once for each way we could have designated one of the boxes as the box that receives at least $16$ additional balls. We only wish to subtract such cases once, so we must add them back.
Choose which two of the four boxes receive at least $16$ additional balls. Suppose they are boxes 1 and 2. Let $x_1' = x_1 - 16$; let $x_2' = x_2 - 16$. Then $x_1'$ and $x_2'$ are nonnegative integers. Substituting $x_1' + 16$ for $x_1$ and $x_2' + 16$ for $x_2$ in equation 1 yields
\begin{align*}
x_1' + 16 + x_2' + 16 + x_3 + x_4 & = 44\\
x_1' + x_2' + x_3 + x_4 & = 12 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{12 + 4 - 1}{4 - 1} = \binom{15}{3} = \binom{15}{12}$$
solutions. Hence, there are
$$\binom{4}{2}\binom{15}{3}$$
solutions in which at least $16$ additional balls are placed in two of the boxes.
By the Inclusion-Exclusion Principle, the number of distributions of $60$ indistinguishable balls to four boxes in which each box receives at least four balls and no box receives at least $20$ balls is
$$\binom{47}{3} - \binom{4}{1}\binom{31}{3} + \binom{4}{2}\binom{16}{3}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3182428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Laurent series of $\sin z/(1 - \cos z)$ I have trouble solving this exercise: find the first three terms of the Laurent series of $\sin z/(1 - \cos z)$ centered at $z=0$.
I have found the first two. I proved that at $z=0$ we have a first order pole and the first one I calculated the residue. I also thought that the second term is zero because this function is odd. Now I have problems with the third. Someone can help me?
|
You know that laurent series of $sin(z) = z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7})$.
Then, the laurent series of $ 1-cos(z)= \frac{z^2}{2}-\frac{z^4}{24}+O(z^{6})$
Overall you have $\frac{z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7})}{\frac{z^2}{2}-\frac{z^4}{24}+O(z^{6})}$.
Now look at the denominator: $(\frac{z^2}{2}-\frac{z^4}{24}+\frac{z^6}{720}+O(z^8))^{-1}=(\frac{z^2}{2})^{-1}(1-\frac{z^2}{12}+\frac{z^4}{360}+O(z^{6}))^{-1}=\frac{2}{z^2}(1-(-\frac{z^2}{12}+\frac{z^4}{360})+\frac{z^4}{144}+O(z^6))
=\frac{2}{z^2}(1+\frac{z^2}{12}+\frac{z^4}{240})=\frac{2}{z^2}+\frac{1}{6}+\frac{z^2}{120}$
Multiply everything together to get: $(z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7}))$ $(\frac{2}{z^2}+\frac{1}{6}+\frac{z^2}{120} )$ $ =\frac{2}{z}-\frac{z}{6}-\frac{z^3}{360} +O(z^5).$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3182920",
"timestamp": "2023-03-29T00:00:00",
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|
Proving the independence of an equation, from certain variables
Assume that $x, y, z$ are three distinct nonzero real numbers satisfying the equation
$$x^3+y^3+m(x+y)=y^3+z^3+m(y+z)=z^3+x^3+m(z+x)$$
for some real $m$. Prove that
$$K=\left(\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}\right)\left(\frac{z}{x-y}+\frac{x}{y-z}+\frac{y}{z-x}\right)$$
is not dependent upon $x, y, z, m$.
This problem comes from a leaflet given for the preparation for the international exam of JBMO. I don't know how to solve it. I've been trying it for a while, but I can't.
I know that the equation
$$x^3+y^3+m(x+y)=y^3+z^3+m(y+z)=z^3+x^3+m(z+x)$$
can be rewritten as
$$m(x+y)-z^3=m(y+z)-x^3=m(z+x)-y^3 ,$$
which is shorter but not necessarily more useful.
|
$$
x^3 + y^3 + m(x+y) = y^3 + z^3 + m(y+z) = z^3 + x^3 + m(z+x)
$$
holds for some $m \in \mathbb R$, if and only if
$$
\begin{cases}
x^3 - y^3 = -m(x-y) \\
y^3 - z^3 = -m(y-z)
\end{cases}
$$
holds for some $m \in \mathbb R$. If we presume that $x,y,z$ are pairwise distinct, this is equivalent to saying that, considering the curve $Y = X^3$ on the plane, the three points $(x, x^3), (y, y^3)$ and $(z, z^3)$ are co-linear.
Now we look a bit geometrically. If the line $Y = aX + b$ intersects $Y = X^3$ at three points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$, then $x_1, x_2, x_3$ are the roots of the equation $X^3 = aX + b$, so $-x_1-x_2-x_3$ are the $x^2$ coefficient, which is zero. So if we move back to our problem, this is equivalent to saying that
$$
x+y+z=0.
$$
The second half of the problem is to show that
$$
K = \left(\frac{x-y}z + \frac{y-z}x + \frac{z-x}y\right)\left(\frac z{x-y} + \frac x{y-z} + \frac y{z-x}\right)
$$
is constant whenever $x+y+z=0$. You need to know that the constant is, if such constant exists. Taking $(x,y,z) = (1,-3,2)$ for instance, you see that $K$ is $c = 9$. So we need to show that $K$ is $c=9$ all the time.
Observe that as $K$ is homogeneous of degree 0, multiplying $x,y,z$ simultaneously by a constant $c$ does not change $K$. Therefore we can scale $x$ to $1$, assume $y = t$ and $z = -t-1$. This gives us an expression
$$
K = \frac{p(t)}{q(t)},
$$
where $p(t)$ and $q(t)$ has degree at most6 and 5 in $t$, respectively.
To show that $K$ is constantly 9, you need to show that $F(t) = p(t) - 9q(t)$ is constantly zero. You can expand it, which will be brutal; an easier way is to observe that $F$ is of degree at most 6, so if you can find seven different $t$ such that $F(t) = 0$, then we are done. The pair $(1,-3,2)$ we started with, we see that $F(-3) = 0$, with a bit more insight we also see that $F(-2/3) = F(1/2) = 0$ (why?), and we can also tell that $F(2) = F(-1/3) = F(-3/2) = 0$. Now you get six zeros, and try a new set of test value to get the seventh.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the partial sum formula of $\sum_{i=1}^n \frac{x^{2^{i-1}}}{1-x^{2^i}}$ Given next series:
$$\frac{x}{1 - x^2} + \frac{x^2}{1 - x^4} + \frac{x^4}{1 - x^8} + \frac{x^8}{1 - x^{16}} + \frac{x^{16}}{1 - x^{32}} + ... $$
and $|x| < 1$. Need to derive $S_n$ formula from series partial sums.
I could only find that $S_{k+1}=\frac{S_k}{1-x^{2^k}} + \frac{x^{2^k}}{1-x^{2^{k+1}}}$. But this is incorrect answer, of course, but I don't know what to do next...
Thank you in advance!
|
By induction you can proof easily $$\sum\limits_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^k}} = \frac{1}{1-x^{2^n}}\sum\limits_{k=1}^{2^n-1}x^k$$ and with $\enspace\displaystyle \sum\limits_{k=1}^{2^n-1}x^k = \frac{x-x^{2^n}}{1-x}\enspace$ the formula is complete.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$
Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$
In the task which I do I need record $\sin (\sin x-x)$ in a way to have $ax^3$. So: $$\sin (\sin x -x)=\sin x -x +r(\sin x -x)=x-\frac{x^5}{6}+\frac{x^5}{120}+r(x)-x +r(\sin x -x)$$ I know that $r(x)=o(x^5)$ and it is easy. However how to prove that $r(\sin x -x)=o(x^5)$? I tried to do it but then I have: $$\frac{r(\sin x -x)}{\sin x -x}\cdot \frac{\sin x -x}{x^5}=\frac{r(\sin x -x)}{\sin x -x}\cdot(\frac{\sin x}{x}-1)\cdot \frac{1}{x^4} \rightarrow 0\cdot(1-1)\cdot(-\infty)$$Can you help me how can I prove it?
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I would be stupid about it and work formally, in the ring $\Bbb Q[[x]]/(x^6)$.
Now, in that ring, $\sin(x)=x-\frac{x^3}6+\frac{x^5}{120}$. Let me write $g(x)=\sin(x)-x$, which is divisible by $x^3$. Then $\sin(g(x))=g(x)+g(g(x))$, but since $g(g(x))$ is divisible by $x^9$, we can ignore it. Thus in $\Bbb Q[[x]]/(x^6)$, we get $\sin(\sin(x)-x)=-\frac{x^3}6+\frac{x^5}{120}$, just the same as $\sin(x)-x$.
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Hard combinatorics question The origin of the coordinates is a pixel. After 1 second, it splits into two particles, one
shifts to the left and the other to the right. This process is repeated every second, and the two particles at the same point are mutually destroyed (so that, for example,
two seconds later, two particles remain.) How many particles will there be in $n$ seconds?
I calculated first values using programm ($1,$
$2,$ $2,$ $4,$ $2,$ $4,$ $4,$ $8,$ $2,$ $4,$ $4,$ $8,$ $4,$ $8,$ $8,$ $16,$ $2,$ $4,$ $4,$ $8,$ $4,$ $8,$ $8,$ $16,$ $4,
$ $8,$ $8,$ $16,$ $8,$ $16,$ $16,$ $32,$ $2,$ $4,$ $4,$ $8,$ $4,$ $8,$ $8,$ $16,$ $4,$ $8,$ $8,$ $16,$ $8,$ $16,$ $16,$
$32,$ $4,$ $8,$ $8,$ $16,$ $8,$ $16,$ $16,$ $32,$ $8,$ $16,$ $16,$ $32,$ $16,$ $32,$ $32,$ $64,$ $2,$ $4,$ $4,$ $8,$ $4,
$ $8,$ $8,$ $16,$ $4,$ $8,$ $8,$ $16,$ $8,$ $16,$ $16,$ $32,$) but I still cant find any formula.
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This is called Gould's sequence, or Dress' sequence, or Glaisher's sequence; it appears in the OEIS. The $n$-th term is the number of odd entries entries in row $n$ of Pascal's triangle. The specific operation in OP is Rule 90 for cellular automata, and we are counting how many $1$'s there are in every stage. Much more is available in the OEIS link above.
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Finding rank-$1$ matrix
Let $$S = \frac{1}{12} \begin{pmatrix} 1 & 10 & 1 \\ 5 & 2 & 5\\ 1 & 2 & 9\end{pmatrix}$$ Find a rank-$1$ matrix $R$ so that $$ M = S + R $$ will have the same eigenvalues as $S$ and all the diagonal elements equal.
I have found the eigenvalues of $S$ to be: $$\{ 1, \frac{\sqrt{2}}{3},\frac{-\sqrt{2}}{3} \}$$
So I now have
$$M = \frac{1}{12} \begin{pmatrix}
1 & 10 & 1 \\
5 & 2 & 5\\
1&2&9\\
\end{pmatrix} + R $$
and $M$ has eigenvalues $$\{ 1, \frac{\sqrt{2}}{3},\frac{-\sqrt{2}}{3}\}$$
How do I find the rank-$1$ matrix $R$ with diagonal elements equal?
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I'm assuming you meant the three diagonal elements of $S+R$ are equal.
Let $R = u v^\top$. You have $5$ equations to solve for $6$ variables $u_1,u_2,u_3,v_1,v_2,v_3$ (of course there is redundancy here): the coefficients of $\lambda^0$ to $\lambda^2$ in $\det(S+R-\lambda I) - \det(S-\lambda I)$ are $0$, $S_{11} + R_{11}-S_{22} - R_{22}=0$ and $S_{11}+R_{11}-S_{33} - R_{33} = 0$.
One of the solutions is
$$ u_1 = \frac{1}{4},\; u_2 = \frac{5}{48},\; u_3 = \frac{-5}{36}, \;v_1 = 1, \;v_2 = \frac{8}{5}, \;v_3 = 3 $$
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|
Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$, at $a = 0$ Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$ , at $a = 0$
So I start of with:
$f(0) = \frac{1}{(1-0)}=1$
$f'(0) = \frac{1}{(1-0)^2 }= 1$
$f''(0) = \frac{2}{(1-0)^3 }= 2$
$f'''(0) = \frac{6}{(1-0)^4 }= 6$
$f^{(4)}(0) = \frac{24}{(1-0)^5 }= 24$
$f^{(5)}(0) = \frac{120}{(1-0)^6 }= 120$
and I get
$1 + x + \frac{2x^2}{2} + \frac{6x^3}{3} + \frac{24x^4}{4} + \frac{120x^5}{5} $
Answer = $1 + x + x^2 + 2x^3 + 5x^4 + 24x^5 $
The problem is that a calculator i used to check it up said the answer was
answer = $1 + x + x^2 + x^3 + x^4 + x^5 $
So im confused, im I right or is the calculator wrong? Thanks!
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i see a problem. The taylor series in this case is given by :
$$f(x)=f(a)(x-a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}+f'''(a)\frac{(x-a)^3}{3!}+...$$
I think you forgot the $!$...
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What is the correct way to solve the equation: $x^4-x^3+x^2-x+1=0$ Given the equation: $x^4-x^3+x^2-x+1=0$ we need to find both its real and complex roots. What is the easiest and correct method for solving the equation?
Here is my approach, but it gives wrong result on the end. Since the equation is symmetric we can group the terms.
$$x^4+1 - (x^3+x)+x^2=0 \text{ Divide everything by } x^2 \\(x^2+\frac{1}{x^2})-(x+\frac{1}{x})+1=0\\ \text{ Let } t = x + \frac{1}{x}, \text{ we can see that } x^2 + \frac{1}{x^2} = t^2 - 2 \\ \text{back in our equation: } t^2 - 2 - t + 1 = 0 \\ t_{12} = \frac{1 \pm \sqrt{5}}{2} \\ \text{however if we go back in } x+\frac{1}{x} = t_{12} \text{ we don't get the correct result }$$.
As given in the textbook the solutions are: $x_{12}=\frac{1+\sqrt5 \pm \sqrt{10-2\sqrt5}}{4}, x_{34}=\frac{1-\sqrt5 \pm \sqrt{10-2\sqrt5}}{4}$ Can someone say if those are the correct solutions or not?
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The solutions in your textbook are wrong; you can plug them in to verify this yourself.
The easiest way to solve is to note that if $x\neq-1$ then
$$\frac{x^5+1}{x+1}=x^4-x^3+x^2-x+1,$$
and so writing $x=re^{\theta i}$ quickly yields $r=1$ and $\theta=\tfrac k5\pi$ with $k$ odd.
Your approach is also fine; you get the same solutions by solving the two quadratics
$$x+\frac1x=\frac12\pm\frac{\sqrt{5}}{2}.$$
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Solving $3x^2 - 4x -2 = 0$ by completing the square I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$
\begin{align}
3x^2 - 4x -2 = 0 \\
3x^2 - 4x = 2
\end{align}
$$
Now, divide both sides by three:
$$x^2 - \frac{4}{3}x = \frac{2}{3}$$
Next, the authors add to both sides the square of the coefficient of $x$, completing the square on the LHS:
$$x^2 - \frac{4}{3}x + \left(\frac{2}{3}\right)^2 = \frac{2}{3} + \left(\frac{2}{3}\right)^2$$
Now, the next two steps (especially the second step) baffle me. I understand the right-hand side of the first quation (how they get the value of $\frac{10}{9}$), but the last step is a complete mystery to me:
$$
\begin{align}
x^2 - \frac{4}{3}x + \frac{4}{9} = \frac{10}{9} \\
\left(x - \frac{2}{3}\right)^2 = \frac{10}{9}
\end{align}
$$
Can anyone please explain how they went from the first step to the second step?
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Note that it is better to multiply by $3$, thus transforming the given equation as follows:
$3x^2 - 4x - 2 = 0$
$9x^2 - 12x - 6 = 0$
$(3x-2)^2 = 10$
To get the last line, you want $(3x+?)^2$ to match the $9x^2$ so you twiddle the $?$ until you match the $9x^2-12x$.
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|
Sets of Polynomial roots For a real number $k,$ let $A$ be the set of roots of $$x^2 + (k - 1) x - 2(k + 1) = 0,$$and let $B$ be the set of roots of $$(k - 1) x^2 + kx + 1 = 0.$$(These roots may be complex.) Find the number of values of $k$ so that $|A \cup B| = 3.$
I don't know how to start this. Can someone help?
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The roots of the two equations are
$$\frac{1}{2}\left(-(k-1) \pm \sqrt{(k-1)^2 + 8 (k+1)}\right) = \frac{1}{2}(-(k-1) \pm (k+3)) = \{2, -(k+1)\}$$
$$\frac{1}{2(k-1)} \left(-k \pm \sqrt{k^2 - 4(k-1)}\right) = \frac{1}{2(k-1)}(-k \pm (k-2)) = \{-\frac{1}{k-1}, -1\}$$
so you want
$$\{2, -(k+1), -\frac{1}{k-1}, -1\}$$
to have cardinality $3$.
|
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|
Solving $\int \frac{6x^{3}+7x^2-12x+1}{\sqrt{x^2+4x+6}}dx$ I have previously asked here the following integral:$$\int \frac{6x^{3}+7x^2-12x+1}{\sqrt{x^2+4x+6}}dx.$$
In the meantime I have received an answer from Achille Hui here, but
there are still some points that I don't understand from his answer.
At some point there is:
*
*Set $B(y)$ to $ay^2 + by + c$, RHS($*1$) becomes $(6-3a)y^3 + \cdots$. We should fix $a$ to $2$.
*Repeat this procedure once more, we find $c$ should be fixed to $24$.
I don't understand how does he fix these numbers? Does he let them equal $0$ to obtain the final result?
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I have used a slightly different approach using a substitution. This makes the integral into a routine computational effort.
$$I = \int \frac{6x^{3}+7x^2-12x+1}{\sqrt{x^2+4x+6}}dx$$
$$ = \int \frac{6x^{3}+7x^2-12x+1}{\sqrt{(x+2)^2+2}}dx$$
Let $x + 2 = \sqrt2\tan\theta \implies dx = \sqrt2\sec^2\theta\ d\theta\text{ and } x=\sqrt2\tan\theta-2$
$$\implies I = \int\frac{12\sqrt2\tan^3\theta -58\tan^2\theta+32\sqrt2\tan\theta+5}{\sqrt2\sec\theta}\sqrt2\sec^2\theta\ d\theta$$
$$= \int (12\sqrt2\tan^3\theta\sec\theta -58\tan^2\theta\sec\theta+32\sqrt2\tan\theta\sec\theta+5\sec\theta)\ d\theta$$
$$ = \color{red}{12\sqrt2\int (\sec^2\theta-1)\tan\theta\sec\theta\ d\theta} - \color{green}{58\int(\sec^2\theta-1)\sec\theta\ d\theta} + 32\sqrt2\int\tan\theta\sec\theta\ d\theta + 5\int\sec\theta\ d\theta$$
$$ = \color{red}{12\sqrt2\int (\sec^2\theta)\tan\theta\sec\theta\ d\theta -12\sqrt2\int\tan\theta\sec\theta\ d\theta} - \color{green}{58\int(\sec^2\theta)\sec\theta\ d\theta +\underline{58\int\sec\theta\ d\theta}} + 32\sqrt2\int\tan\theta\sec\theta\ d\theta + \underline{5\int\sec\theta\ d\theta}$$
$$ = 12\sqrt2\left(\frac{\sec^3\theta}{3} -\sec\theta\right) - 58\int\sec^3\theta\ d\theta + 32\sqrt2\sec\theta + 63\int\sec\theta\ d\theta$$
$$ = 4\sqrt2\sec^3\theta + 20\sqrt2\sec\theta - 58\int\sec^3\theta\ d\theta+ 63\int\sec\theta\ d\theta$$
$$ = 4\sqrt2\sec^3\theta + 20\sqrt2\sec\theta \color{green}{- 29}\left(\sec\theta\tan\theta + \color{green}{\ln(\sec\theta+\tan\theta)}\right) + \color{green}{63\ln(\sec\theta+\tan\theta)}$$
Using $\ln(z+\sqrt{z^2+1}) = \sinh^{-1}z$, we get
$$I = 4\sqrt2\sec\theta\left(\sec^2\theta + 5 -\frac{29}{4\sqrt2}\tan\theta\right) + \color{green}{34\sinh^{-1}(\tan\theta)}$$
Using $\tan\theta = \dfrac{x+2}{\sqrt2}\text{ and } \sec\theta = \dfrac{\sqrt{x^2+4x+6}}{\sqrt2}$, we get
$$I = 4\sqrt{x^2+4x+6}\left(\frac{x^2+4x+6}{2} + 5 - \frac{29}{4\sqrt2}\frac{x+2}{\sqrt2}\right) + 34\sinh^{-1}\left(\frac{x+2}{\sqrt2}\right)$$
$$ = \sqrt{x^2+4x+6}\left(2x^2+8x+12+20-\frac{29}{2}x-29\right) + 34\sinh^{-1}\left(\frac{x+2}{\sqrt2}\right)$$
$$ = \boxed{\sqrt{x^2+4x+6}\left(2x^2-\frac{13}{2}x+3\right) + 34\sinh^{-1}\left(\frac{x+2}{\sqrt2}\right)}$$
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|
Minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12$ If $x,y\in\mathbb R^+$, $x+y=12$, what is the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$?
I got the question from a mathematical olympiad competition (of China) for secondary 2 student, so I don't expect an "analysis" answer.
The answer should be $15$, when $x=\frac{20}{3}$ and $ y=\frac{16}{3}$ (just answer, no solution :< ).
I try to use AM-GM inequality, but I couldn't manage to get the answer.
$$\sqrt{x^2+25}+\sqrt{y^2+16}=\sqrt{x^2+25}+\sqrt{(x-12)^2+16}=\sqrt{x^2+25}+\sqrt{x^2-24x+160}$$
Can this help?
I also tried to plot the graph, see here.
Any help would be appreciated. Thx!
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We are asked for the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12;$
that is, the minimum value of $\sqrt{x^2+25}+\sqrt{(12-x)^2+16}.$
Let $\overrightarrow a=(x,5)$ and $\overrightarrow b=(12-x,4)$ in $\mathbb R^2,$ so $\overrightarrow a + \overrightarrow b=(12,9).$
By the triangle inequality, $|\overrightarrow a+\overrightarrow b|\le|\overrightarrow a|+|\overrightarrow b|.$
Therefore, $\mathbf{15}=\sqrt{12^2+9^2}\le\sqrt{x^2+25}+\sqrt{(12-x)^2+16}.$
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|
$a, b$ are positive whole numbers such that $a + b = a/b + b/a$. How many possible values can $a^4 + b^4$ be? $a, b$ are positive whole numbers such that $a + b = a/b + b/a$. How many possible values can $a^4 + b^4$ be?
I tried using $(a^2 + b^2)^2$ but I don't know what to do after.
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$a+b=\frac ab+\frac ba\iff a+b=\frac{a^2+b^2}{ab}\implies a^2b+ab^2=a^2+b^2\implies a^2\underbrace{(b-1)}_{\ge 0}+b^2\underbrace{(a-1)}_{\ge 0}=0$
Since the equation has $a,b$ both on denominator, we can reject zero as an admissible value for $a$ or $b$.
The only possibility for the sum above to be $0$ is that each term is zero, leading to $a=1$ and $b=1$.
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|
Simplifying $\sec^2 \frac{2\pi}{7} + \sec^2 \frac{4\pi}{7} + \sec^2 \frac{8\pi}{7}$
Simplify the following expression:
$$y =\sec^2 \frac{2\pi}{7} + \sec^2 \frac{4\pi}{7} + \sec^2 \frac{8\pi}{7}$$
Note. The source asks the value of $y/3$, which, according to the instructions, has to be an integer from $0$ to $9$.
My Attempt:
Man! I tried everything I could. Converted it into sin's and cosine's, tan's and sec's. Applied every identity I could find in my book. But to no avail.
All I was aiming is to somehow make the squares disappear and converting everything in sin's and cosine's since we only know to simplify such expressions
like ($\cos x + \cos 2x + \cos 3x+\cdots$ and $\cos x\cos 2x\cos 4x\cdots$ etc) in tems of sin's and cosine's. (Sorry for all the sin's and cosine's being repeated too many times :) )
Any help would be appreciated.
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Like If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$,
the roots of
$$t^3-21t^2+35t-7=0$$ are $\tan^2\dfrac{2n\pi}7, n=1,2,4$
Using Vieta's Formula,
$$\tan^2\dfrac{2\pi}7+\tan^2\dfrac{4\pi}7+\tan^2\dfrac{8\pi}7=\dfrac{21}1$$
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Problem with $\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}$ How to simplify $$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}?$$
Rationalise the denominator
$$\frac{\sqrt{6+4\sqrt{2}}}{4}(2-\sqrt{2})$$
This is still not simplify.
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$\sqrt{6+4\sqrt{2}} = \sqrt{(2 + \sqrt{2})^{2}} = {2+\sqrt{2}}$ ,
as $6+4\sqrt{2}= 4 +2 +2.2\sqrt{2} = (\sqrt{2})^{2} + 2 .2. \sqrt{2} + 2^ 2 = (2 + \sqrt{2})^{2}$
Now $\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}} = \frac{2+\sqrt{2}}{2(2+\sqrt{2})}= \frac{1}{2}$
|
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How do I evaluate this trigonometric limit involving infinite product? Tried a lot but couldn't solve it. How do I begin approaching this limit?
$ \displaystyle\lim _{n\to\infty} \displaystyle\prod_{k=3}^{n} 1 - \tan^4\left( \frac{\pi}{2^k}\right)$
|
There isn't that many trick to create a horrible looking and yet doable limit.
In school, if you are asked to evaluate a series or product which involve trigonometry functions whose arguments contain terms like $\frac{(\cdots)}{2^k}$. One thing you should do is lookup the double-angle formula for various trigonometry functions and see whether you can use them to turn your series/product into a telescoping one.
In this case, it do work.
Let $t = \tan\theta$, we have
$$1 - t^4 = (1-t^2)^2\frac{1+t^2}{1-t^2}
= \left(\frac{2t}{\frac{2t}{1-t^2}}\right)^2\frac{1+t^2}{1-t^2}$$
Recall
$$\frac{2t}{1-t^2} = \tan(2\theta)\quad\text{ and }\quad
\frac{1-t^2}{1+t^2} = \cos(2\theta) = \frac12\frac{\sin(4\theta)}{\sin(2\theta)}
$$
We obtain
$$1 - t^4 = 8\left(\frac{\tan(\theta)}{\tan(2\theta)}\right)^2\frac{\sin(2\theta)}{\sin(4\theta)}
= 8\frac{\tan^2\theta\sin(2\theta)}{\tan^2(2\theta)\sin(4\theta)}
$$
The product is telescoping and
$$\begin{align}\prod_{k=3}^n\left(1 - \tan^4\frac{\pi}{2^k}\right)
&= 8^{n-2} \frac{\tan^2\frac{\pi}{2^n}\sin \frac{\pi}{2^{n-1}}}{\tan^2\frac{\pi}{4}\sin\frac{\pi}{2}}\\
&= \frac{1}{32} \left(2^n\tan \frac{\pi}{2^n}\right)^2\left(2^{n-1}\sin\frac{\pi}{2^{n-1}}\right)\end{align}
$$
Using
$$\lim_{N\to\infty} N\tan\frac{\pi}{N} = \lim_{N\to\infty} N\sin\frac{\pi}{N} = \pi$$
We obtain
$$\lim_{n\to\infty} \prod_{k=3}^n\left(1 - \tan^4\frac{\pi}{2^k}\right) = \frac{\pi^3}{32}$$
|
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|
Calculating $\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$ Calculate $$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
Here is my attempt:
$$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}= \left(\frac{4\infty^2+5\infty-6}{4\infty^2+3\infty-10}\right)^{3-4\infty}$$
$$=\left(\frac{\infty(4\infty+5)}{\infty(4\infty+3)}\right)^{-4\infty}=\left(\frac{4\infty}{4\infty}\right)^{-4\infty} = 1^{-4\infty} = \boxed{1}$$
However, when I try to graph the function, I can't reliably get my answer due to precision limitations, and I feel that this method of calculating limits is less than ideal. How can I confirm that this is indeed the limit?
|
Let us suppose that you want to compute accurately the value of the expression for large values of $n$ and not only the limit.
$$a_n=\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}=\left(1+\frac{2}{4 n-5}\right)^{3-4 n}$$ Take logarithms
$$\log(a_n)=(3-4n) \log\left(1+\frac{2}{4 n-5}\right)$$ and use the Taylor expansion
$$\log(1+\epsilon)=\epsilon -\frac{\epsilon ^2}{2}+\frac{\epsilon ^3}{3}-\frac{\epsilon
^4}{4}+O\left(\epsilon ^5\right)$$
Make $\epsilon=\frac{2}{4 n-5}$ and continue with Taylor expansion (or long division) to get
$$\log(a_n)=(3-4n)\left(\frac{1}{2 n}+\frac{1}{2 n^2}+\frac{49}{96 n^3}+\frac{17}{32
n^4}+O\left(\frac{1}{n^5}\right)\right)$$ that is to say
$$\log(a_n)=-2-\frac{1}{2 n}-\frac{13}{24 n^2}-\frac{19}{32
n^3}+O\left(\frac{1}{n^4}\right)$$ COntinue with Taylor using
$$a_n=e^{\log(a_n)}=\frac 1{e^2}\left(1-\frac{1}{2 n}-\frac{5}{12 n^2}-\frac{11}{32
n^3} \right)+O\left(\frac{1}{n^4}\right)$$
Try with $n=5$ (which is quite far away from $\infty$. The exact value is
$$a_5=\frac{98526125335693359375}{827240261886336764177}\approx 0.119102$$ while the above truncated expression gives
$$\frac{10567}{12000 e^2}\approx 0.119174$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3222029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Find antiderivative $\int (2x^3+x)(\arctan x)^2dx $
Find antiderivative $$\int (2x^3+x)(\arctan x)^2dx $$
My try:
$$\int (2x^3+x)(\arctan x)^2dx =(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int \frac{2\arctan x}{1+x^2}(\frac{1}{2}x^4+\frac{1}{2}x^2)dx=(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int (\arctan x) (x^2)dx= (\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\arctan x\cdot \frac{1}{3}x^3-\int (\frac{1}{1+x^2}) (\frac{1}{3}x^3)dx$$And then I don't know how I can find $\int (\frac{1}{1+x^2}) (\frac{1}{3}x^3)dx$. Can you help me with it?
|
Note that$$\frac{x^3}{1+x^2}=\frac{x^3+x}{1+x^2}-\frac x{1+x^2}=x-\frac x{1+x^2}.$$Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3224715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Partial fraction expansion of $\frac{1}{(s+1)^{2}(s-1)(s+5)}$ I'm seeking a partial-fraction expansion of $A=\frac{1}{(s+1)^{2}(s-1)(s+5)}.$
I was solve equation differential using Laplace transform, but I need use partial fraction of $A$.
|
The partial fraction expansion is:
$$\frac{1}{(s+1)^2(s-1)(s+5)} = \frac{1}{24(s-1)} -\frac{1}{32(s+1)} - \frac{1}{8(s+1)^2} - \frac{1}{96(s+5)} $$
The fraction expansion can be expressed as:
$$\frac{1}{(s+1)^2(s-1)(s+5)} = \frac{A}{s-1} + \frac{B}{s+1} + \frac{C}{(s+1)^2} + \frac{D}{s+5}$$
$$ \implies A(s+1)^2(s+5) + B(s+1)(s-1)(s+5) + C(s-1)(s+5) + D(s-1)(s+1)^2 = 1$$
Using $s=1$, we get:
$$24A = 1 \implies A = \frac{1}{24}$$
Using $s=-1$, we get:
$$-8C = 1 \implies C = -\frac{1}{8}$$
Using $s = -5$, we get:
$$-96D = 1 \implies D = -\frac{1}{96}$$
For $B$, none of these will work. So we can choose something that makes life easier. Using $s=0$, we get
$$5A -5B -5C -D = 1 $$
$$\implies \frac{5}{24} - 5B + \frac{5}{8} +\frac{1}{96} = 1$$
$$\implies B = -\frac{1}{5}\left(1 - \frac{20}{96} - \frac{60}{96} - \frac{1}{96}\right) = -\frac{1}{32}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3226373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Expectation of dependent Bernoulli sum I want to estimate the expected value of the following sum of random variables,
$$ Y = \sum_{i=1}^N X_i $$
where each $X_i$ is a Bernoulli random variable. In particular,
$$ X_1 = \begin{cases} 1, & \text{with prob. } 1/N \\
0, & \text{with prob. } 1 - 1/N \end{cases} $$
and for all $2 \leq i \leq N$
$$ X_i = \begin{cases} 1, & \text{with prob. } p_i \\
0, & \text{with prob. } 1 - p_i \end{cases} $$
where
$$ p_i = \begin{cases} p_{i-1} + 1/N, & \text{if } X_{i-1} = 0 \\
1/N, & \text{if } X_{i-1} = 1 \end{cases} $$
I have computed the value for $ N=2 $ and $ N=3 $. For $ N = 2 $,
$$
\mathbb{E}[Y] = 1 \left[ \left( 1 - \frac{1}{N} \right) \frac{2}{N} + \frac{1}{N} \left( 1 - \frac{1}{N} \right) \right] + 2 \frac{1}{N^2} = \frac{5}{4}
$$
where the first term corresponds to the sequences $(01)$ and $(10)$, while the second to $(11)$.
For $ N=3 $, here are the calculations
$$
\begin{gather}
\mathbb{E}[Y] = \left(1-\frac{1}{N}\right)\left(1-\frac{2}{N}\right)\frac{3}{N} + 2 \times \left(1-\frac{1}{N}\right)\frac{2}{N} \frac{1}{N} + 2 \times \frac{1}{N}\left(1-\frac{1}{N}\right)\frac{2}{N} + 3 \times \frac{1}{N} \frac{1}{N} \frac{1}{N} + \left(1 - \frac{1}{N}\right)\frac{2}{N}\left(1-\frac{1}{N}\right) + \frac{1}{N}\left(1-\frac{1}{N}\right)\left(1-\frac{2}{N}\right) + 2 \times \frac{1}{N} \frac{1}{N} \left(1-\frac{1}{N}\right) \\
= \frac{3}{N^3} - \frac{6}{N^2} + \frac{6}{N} \\
N=3 \rightarrow = \frac{39}{27}
\end{gather}
$$
This results from the sequences (001), (011), (101), (111), (010), (100), (110).
Is there any way to proceed farther?
|
MY TRY
We first look at $ P\{ Y=i\}$, i.e., $Y$ has exactly $i$ ones and $N-i$ zeros. The probability corresponding to ones is $N^{-i}$. The probability corresponding to zeros depends on number and size of blocks of zeros. There can be at most $i+1$ blocks of size between 0 and $N-i$, as we already have $i$ ones. Further, the probability corresponding to a block of size $a_k$ is $(a_k+1)!N^{-a_k} $. Thus, we have
\begin{align}
P\{ Y=i\} &= N^{-i}\prod_{\substack{a_1,a_2,\ldots,a_{i+1}\geq 0\\ \sum_{k=1}^{i+1} a_k=N-i} }1-\frac{(a_k+1)!}{N^{a_k}}\\
&= N^{-N}\prod_{\substack{a_1,a_2,\ldots,a_{i+1i}\geq 0\\ \sum_{k=1}^{i+1} a_k=N-i} } N^N-(a_k+1)!
\end{align}
Also, we have the following:
\begin{align}
E\{Y\} = \sum_{i=0}^N i P\{ Y=i\} = N^{-N} \sum_{i=0}^N i \prod_{\substack{a_1,a_2,\ldots,a_{i+1}\geq 0\\ \sum_{k=1}^{i+1} a_k=N-i} } N^N-(a_k+1)!
\end{align}
I don't know how to simplify it further!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3228206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Why is $\mathbf{x}$ yielding a product of $\mathbf{y}$ and the right inverse for $\mathbf{A}$? When you solve the equation $\mathbf{Ax}=\mathbf{y}$ where $\mathbf{x} \in \mathbb{R}^{5}$, $\mathbf{y} \in \mathbb{R}^{4}$ and $\mathbf{A}$ is a $4 \times 5$ matrix with a right inverse, I understand that the $\mathbf{x}$ is $\mathbf{A}^{-1}\mathbf{y}$ but how come it can happen that $\mathbf{x}$ is not $\mathbf{A}^{-1}\mathbf{y}$ but some right inverse times $\mathbf{y}$? So if $\mathbf{X}$ were a right inverse to $\mathbf{A}$, $\mathbf{x}$ is $\mathbf{X}\mathbf{y}$?
I would like some algebraic proof such as for the fact that $\mathbf{x}$ is $\mathbf{H}^{-1}\mathbf{y}$ for some square matrix $\mathbf{H}$
$$
\mathbf{H}^{-1}\mathbf{Hx} = (\mathbf{H}^{-1}\mathbf{H})\mathbf{x} = \mathbf{I}_4 \mathbf{x} = \mathbf{x} = \mathbf{H}^{-1}\mathbf{y}
$$
I do not think this is valid statement
$$
\mathbf{AXx} = (\mathbf{AX})\mathbf{x} = \mathbf{I}_4 \mathbf{x} = \mathbf{x} = \mathbf{X}\mathbf{y}
$$
so that does probably not that why $\mathbf{x}$ in my case is a the result of post-multiplying a right inverse of $\mathbf{A}$ by $\mathbf{y}$
To give details about my case where $\mathbf{x}$ is $\mathbf{X}\mathbf{y}$?, I have
$$
\mathbf{A} =
\left[\begin{array}{rrrrr}
2&-4&-1&-3&2\\
-1&2&1&0&1\\
1&-2&-1&-3&-1\\
-1&4&-1&0&5
\end{array}\right]
$$ and $\mathbf{y}$ is a column vector of four unknowns. Then I have the solution for $\mathbf{x}$ as
$$
\mathbf{x} =
\left[\begin{array}{r}
x_1\\
x_2\\
x_3\\
x_4\\
x_5
\end{array}\right]
=
\left[\begin{array}{r}
3y_1+y_4-3y_3+y_2-15t\\
\frac{1}{2}y_4+y_1-y_3+\frac{1}{2}y_2-6t\\
-y_3+y_1+y_2-4t\\
-\frac{1}{3}y_2-\frac{1}{3}y_3\\
t
\end{array}\right]
\quad
\text{where }
t \in \mathbb{R}
$$
For exapmle when $t = 0$, we have that
$$
\mathbf{x} =
\left[\begin{array}{r}
x_1\\
x_2\\
x_3\\
x_4\\
x_5
\end{array}\right]
=
\left[\begin{array}{r}
3y_1+y_4-3y_3+y_2-15 \cdot 0\\
\frac{1}{2}y_4+y_1-y_3+\frac{1}{2}y_2-6 \cdot 0\\
-y_3+y_1+y_2-4 \cdot 0\\
-\frac{1}{3}y_2-\frac{1}{3}y_3\\
0
\end{array}\right]
=
\left[\begin{array}{rrrr}
3&1&-3&1\\
1&\frac{1}{2}&-1&\frac{1}{2}\\
1&1&-1&0\\
0&-\frac{1}{3}&-\frac{1}{3}&0\\
0&0&0&0
\end{array}\right]
\left[\begin{array}{r}
y_1\\
y_2\\
y_3\\
y_4
\end{array}\right]
$$
We now have $\mathbf{x} = \mathbf{Xy}$ where $\mathbf{X}$ is a right inverse to $\mathbf{A}$ - so why this relationship ($\mathbf{x} = \mathbf{Xy}$ with $\mathbf{X}$ being a right inverse)?
|
Claim. Let $V,W$ be vector spaces, $A\colon V\to W$ and $B\colon W\to V$ be linear maps with $AB=1_W$. Also suppose that $Ax=y$ for certain $x\in V$, $y\in W$ with $y\ne 0$.
Then there exists a linear map $B'\colon W\to V$ such that $AB'=1_W$ and $x=B'y$.
Proof: It suffices to find $C\colon W\to V$ such that $AC=0$ and $Cy=x-By$ as we then can take $B'=B+C$.
Let $v:=x- By$. Note that $v\in\ker A$. By extending $y$ to a to a basis of $W$, we find a linear map $W\to \ker A$ that maps $y\mapsto v$. Composed with the inclusion $\ker A\to V$, we obtain a linear map $C\colon W\to V$ such that $AC=0$ and $Cy=x-By$. $\square$
Remark. If we allow $y=0$, the claim is no longer true because there may exist non-zero $x$ with $Ax=0$.
|
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"url": "https://math.stackexchange.com/questions/3229798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Finding value of $K$ when applying the error bound The problem is asking to use the error bound to find a value of $n$ for which the given inequality is satisfied.
$$\left|\sqrt{1.3}-T_{n}(1.3) \right| \leq 10^{-6}, \quad a=1 $$
Now this is how I started my attempt to solve this, we use the error bound, where,
$$\left|\sqrt{1.3}-T_{n}(1.3) \right| \leq K \frac{\left|1.3-1\right|^{n+1}}{(n+1)!}= K \frac{\left|0.3\right|^{n+1}}{(n+1)!}$$
What I am struggling with is how to find the value of $K$, this is how the solution manual found $K$:
For $f(x)=\sqrt{x}$, $|f^{(n+1)}(x)|$ is decreasing for $x>1$, hence the maximum value of $|f^{(n+1)}(x)| $ occurs at $x=1$:
\begin{align*}
K=|f^{(n+1)}(1)|&=\frac{1\cdot 3 \cdot 5 \cdots (2n+1)}{2^{n+1}}\\&
=\frac{1\cdot 3 \cdot 5 \cdots (2n+1)}{2^{n+1}} \times \frac{2\cdot 4 \cdot 6 \cdots (2n+2)}{2\cdot 4 \cdot 6 \cdots (2n+2)}\\
&=\frac{(2n+2)!}{(n+1)!2^{(2n+2)}}
\end{align*}
I have no idea why it choose $K$ to be this, I am only stuck on this I know how to continue on with the problem but I don't understand how the value of $K$ became this
|
It seems like you're missing where the $K$ comes from in general. One form of Taylor's theorem is that, for all $x$ there is some $z$ between $a$ and $x$ such that
$$
f(x) - T_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{n+1}
$$
Therefore, if you can find $K$ such that
$$
|f^{(n+1)}(z)| \leq K
$$
for all $z$ in the interval $[a,x]$, then you can conclude that
$$
|f(x) - T_n(x)| \leq \frac{K}{(n+1)!}|x-a|^{(n+1)}
$$
Let's take derivatives and look for a pattern.
\begin{align*}
f(x) &= x^{1/2} &
f(1) &= 1 \\
f'(x) &= \frac{1}{2}x^{-1/2} &
f'(1) &= \frac{1}{2} \\
f''(x) &= \left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) x^{-3/2} &
f''(1) &= \frac{(-1)}{2^2}\\
f'''(x) &= \left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) x^{-5/2} &
f'''(1) &= \frac{(-1)^2 1 \cdot 3}{2^3}\\
f^{(4)}(x) &= \left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) x^{-7/2} &
f^{(4)}(1) &= \frac{(-1)^3 1 \cdot 3 \cdot 5}{2^4} \\
\end{align*}
So each derivative $f^{(k)}(1)$ has:
*
*a power of $(-1)$ in the numerator,
*the product of odd numbers up to a certain point, and
*a power of $2$ in the denominator.
Furthermore, $f^{(k)}(x)$ is $f^{(k)}(1)$ times a fractional power of $x$
To determine the dependence of each of these terms on $k$, just check the first few terms we've already counted. The exponent on $(-1)$ is $k-1$; the exponent on $2$ is $k$, and all odd numbers up through $2k-3$ are multiplied together.
The power on $x$ is $-(2k-1)/2$.
$$
f^{(k)}(1) = \frac{(-1)^{k-1} 1 \cdot 3 \cdot 5 \cdots (2k-3)}{2^k}
$$
as long as $k \geq 2$. Therefore
$$
f^{(n+1)}(1) = \frac{(-1)^n 1\cdot 3 \cdot 5 \cdots (2(n+1)-3)}{2^{n+1}}
= \frac{(-1)^n 1\cdot 3 \cdot 5 \cdots (2n-1)}{2^{n+1}}
$$
This is what the solution author did to get to the first line there. Actually, we disagree there on the form of the final odd factor in the numerator. Is is $2n-1$ or $2n+1$? I showed my work so I think I'm right.
Now what? It would be good to represent that product of odd numbers in terms of factorials. We can do this by multiplying and dividing by all those missing even factors:
\begin{align*}
\frac{(-1)^n 1\cdot 3 \cdot 5 \cdots (2n-1)}{2^{n+1}}
&= \frac{(-1)^n 1\cdot 3 \cdot 5 \cdots (2n-1)}{2^{n+1}}
\times \frac{2 \cdot 4 \cdot 6 \cdots 2n}{2 \cdot 4 \cdot 6 \cdots 2n}\\
&= \frac{(-1)^n (2n)!}{(2 \cdot 4 \cdot 6 \cdots 2n)2^{n+1}}
\end{align*}
As for the denominator, each of those even factors is a multiple of two, and when you factor out the twos, you get another factorial:
\begin{align*}
\frac{(-1)^n (2n)!}{(2 \cdot 4 \cdot 6 \cdots 2n)2^{n+1}}
&= \frac{(-1)^n (2n)!}{(2 \cdot 1 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdots 2 \cdot n)2^{n+1}} \\
&= \frac{(-1)^n (2n)!}{2^n (\cdot 1 \cdot 2 \cdot 3 \cdot n)2^{n+1}} \\
&= \frac{(-1)^n (2n)!}{2^n \cdot n!\cdot 2^{n+1}} \\
&= \frac{(-1)^n (2n)!}{n!\cdot 2^{2n+1}} \\
\end{align*}
Therefore,
\begin{align*}
|f^{(n+1)}(x)| &= \left|\frac{(-1)^n (2n)!}{n!\cdot 2^{2n+1}} x^{-(2n-1)/2}\right| \\
&= \frac{(2n)!}{n!\cdot 2^{2n+1}} x^{-(2n-1)/2} \\
&\leq \frac{(2n)!}{n!\cdot 2^{2n+1}}
\end{align*}
As long as $x \geq 1$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
$$ $$
Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x = a + \sqrt{b}, z = c$. We can find $y$ as
$$ y = \sqrt{c^{2}- (a + \sqrt{b})^{2}} $$
so we have
$$ \cos(18) = \frac{y}{z} = \frac{\sqrt{c^{2}- (a + \sqrt{b})^{2}}}{c}$$
I also found out that
$$b = (c \sin(18) - a)^{2} = c^{2} \sin^{2}(18) - 2ac \sin(18) + a^{2}$$
I got no clue after this.
The solution says that $$ \sin(18) = \frac{-1 + \sqrt{5}}{4} $$
I gotta intuition that we must find $A,B,C$ such that
$$ A \sin(18)^{2} + B \sin(18) + C = 0 $$
then $\sin(18)$ is a root iof $Ax^{2} + Bx + C$, and $a = -B, b = B^{2} - 4AC, c = 2A$.
Totally different. This question is not asking to prove that $sin(18)=(-1+\sqrt{5})/4$, that is just part of the solution.
|
Let $ABCDE$ be a regular pentagon.
In right triangle $GHC$ we see $\sin 18^{\circ} = {y\over 2x} $. Let $k=y/x$.
From triangle similarity of $GFC$ and $ACE$ we have $${x\over y} = {2x+y\over a}$$
and from triangle similarity of $BFC$ and $EDC$ we have $${a\over x} = {2x+y\over a}$$
Eliminate $a = {x^2\over y}$ and we get $$x^3 = y^2(2x+y)\implies k^3+2k^2-1 =(k-1)(k^2+k-1)=0 $$
Clearly $k\ne -1$ so $$k_{2,3} = {-1\pm \sqrt{5}\over 2} $$
and so $$\sin 18^{\circ} = {-1+ \sqrt{5}\over 4} $$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3230730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Kronecker Product Interpretation The algebraic expression for a Kronecker product is simple enough. Is there some way to understand what this product is?
The expression for matrix-vector multiplication is easy enough to understand. But realizing that the multiplication yields a linear combination of the columns of the matrix is a useful insight.
Is there some analogous insight for the Kronecker product?
|
A nice motivating example is provided in Kronecker Products & Matrix Calculus with Applications by A. Graham.
We consider two linear transformations
\begin{align*}
\boldsymbol{x}=\boldsymbol{A}\boldsymbol{z}\qquad\text{and}\qquad \boldsymbol{y}=\boldsymbol{B}\boldsymbol{w}
\end{align*}
and the simple case
\begin{align*}
\begin{pmatrix}
x_1\\x_2
\end{pmatrix}
=
\begin{pmatrix}
a_{11}&a_{12}\\
a_{21}&a_{22}\\
\end{pmatrix}
\begin{pmatrix}
z_1\\z_2
\end{pmatrix}
\qquad\text{and}\qquad
\begin{pmatrix}
y_1\\y_2
\end{pmatrix}
=
\begin{pmatrix}
b_{11}&b_{12}\\
b_{21}&b_{22}\\
\end{pmatrix}
\begin{pmatrix}
w_1\\w_2
\end{pmatrix}
\end{align*}
The Kronecker product:
Now we consider the two transformations simultaneously by defining two vectors
\begin{align*}
\color{blue}{\boldsymbol{\mu}=\boldsymbol{x}\otimes\boldsymbol{y}}
=
\begin{pmatrix}
x_1y_1\\x_1y_2\\x_2y_1\\x_2y_2
\end{pmatrix}
\qquad\text{and}\qquad
\color{blue}{\boldsymbol{\nu}=\boldsymbol{z}\otimes\boldsymbol{w}}
=
\begin{pmatrix}
z_1w_1\\z_1w_2\\z_2w_1\\z_2w_2
\end{pmatrix}
\end{align*}
We want to find the transformation between $\boldsymbol{\mu}$ and $\boldsymbol{\nu}$. We look at the relation between the components of the two vectors and show it exemplarily for $x_1y_1$.
We obtain
\begin{align*}
x_1y_1&=\left(a_{11}z_1+a_{12}z_2\right)\left(b_{11}w_1+b_{12}w_2\right)\\
&=a_{11}b_{11}\left(z_1w_1\right)+a_{11}b_{12}\left(z_1w_2\right)+a_{12}b_{11}\left(z_2w_1\right)\\
&\qquad+a_{12}b_{12}\left(z_2w_2\right)
\end{align*}
It follows
\begin{align*}
\color{blue}{\boldsymbol{\mu}}&=
\begin{pmatrix}
a_{11}b_{11}&a_{11}b_{12}&a_{12}b_{11}&a_{12}b_{12}\\
a_{11}b_{21}&a_{11}b_{22}&a_{12}b_{21}&a_{12}b_{22}\\
a_{21}b_{11}&a_{21}b_{12}&a_{22}b_{11}&a_{22}b_{12}\\
a_{21}b_{12}&a_{21}b_{22}&a_{22}b_{21}&a_{22}b_{22}\\
\end{pmatrix}\boldsymbol{\nu}\\
&\color{blue}{=\left(\boldsymbol{A}\otimes\boldsymbol{B}\right)\boldsymbol{\nu}}
\end{align*}
We find
\begin{align*}
\color{blue}{\boldsymbol{Az}\otimes\boldsymbol{Bw}
=\left(\boldsymbol{A}\otimes \boldsymbol{B}\right)
\left(\boldsymbol{z}\otimes \boldsymbol{w}\right)}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3231073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Why is this integration method not valid? Let $$I=\int \frac{\sin x}{\cos x + \sin x}\ dx \tag{1}$$
Now let $$u=\frac{\pi}{2} - x \tag{2}$$ so $$I=\int \frac{\sin (\frac{\pi}{2} - u)}{\cos (\frac{\pi}{2} - u)+\sin (\frac{\pi}{2} - u)}\ du \tag{3}$$
$$=\int\frac{-\cos u}{\sin u + \cos u} \ du \tag{4}$$
$$= \int\frac{-\cos x}{\sin x + \cos x} \ dx \tag{5}$$
and hence $$2I=\int\frac{\sin x - \cos x}{\sin x + \cos x} \ dx \tag{6}$$
$$=-\ln\ |\sin x + \cos x| + c \tag{7}$$
$\implies I=-\frac{1}{2}\ln|\sin x + \cos x| + c \tag{8}$
But the actual answer is $$I= \frac{1}{2}x -\frac{1}{2}\ln|\sin x + \cos x| + c \tag{9}$$
according to Wolfram Alpha and supported by a different method.
Why does my method not yield the correct result?
|
Let's ignore the missing minus sign in line (3) (since $\mathrm{d}u = - \mathrm{d}x$). Instead let's inspect the method.
The short version: Once you establish a relation between $x$ and $u$, you have established it. You do not get to change that relation without implicitly reducing the domain of validity to the simultaneous solution set of the multiple relations.
Let's start with something we all know.
$$ \int x \,\mathrm{d}x = \frac{1}{2}x^2 + C \text{.} $$
And let's perform your change of variable (and because I noted it above, I'll include the minus sign), $x = \frac{\pi}{2} - u$, so $\mathrm{d}x = - \mathrm{d}u$:
\begin{align*}
-\int \left(\frac{\pi}{2} - u \right) \,\mathrm{d}u = \frac{-\pi u + u^2}{2} + C \text{.}
\end{align*}
This is very different from the previous answer (even accounting for the "elastic" nature of "${}+C$" to produce and absorb specific constant expressions). Why? Because you have already established that $x = \frac{\pi}{2} - u$. We are not free to switch to $u = x$ (unless we really only want an expression that works at only one point, where $u = x = \frac{\pi}{2} - u$, that is, at $u = \pi/4$).
So a customer walks into our math shop asking for the antiderivative of something in terms of $x$. We give them an answer involving $u$s. They presume we have had a stroke since we are speaking gibberish. To respond to the customer, we must switch back to the variables they asked about and we have already established the relation between $x$ and $u$. Therefore, we have \begin{align*}
\int x \,\mathrm{d}x
&= -\int \left(\frac{\pi}{2} - u \right) \,\mathrm{d}u \\
&= \frac{-\pi u + u^2}{2} + C \\
&= \frac{-\pi \left( \frac{\pi}{2} - x \right) + \left( \frac{\pi}{2} - x \right)^2}{2} + C \\
&= \frac{-\pi^2 + 4x^2}{8} + C \\
&= \frac{x^2}{2} + C \text{,}
\end{align*}
where here we have let $C$ absorb the constant $-\pi^2/8$.
|
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|
real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me please
|
You are right that $f(f(x))=x$
This gives $f(x)=f^{-1}(x)$
Do you know anything about the relationship between a function and its inverse?
Try sketching both $y=f(x)$ and $y=f^{-1}(x)$ on the same axes...
|
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|
Evaluate $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
Evaluate the following integral:
$ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
I've tried to apply Chebyshev theorem on the integration of binomial differentials.
We have $ m=4,a=2,b=-1,n=2,p=-3/2$.
$\frac{m+1}{n}+p$ is integer then we do the substitution
$t^2=2x^{-2}-1$, $x^2=\frac{2}{t^2+1}$
It go me there: $\int\frac{2}{t^2+1}^2(2-\frac{2}{t^2+1})^{-3/2}-\sqrt{2}t(\frac{1}{t^2+1})^{3/2}dt$ which is just more complicated expression. Where I went wrong?
|
Hint:
Another way:
Use Trigonometric substitution
As $2-x^2\ge0,$ WLOG $x=\sqrt2\sin t$
$\sqrt{2-x^2}=\sqrt2\cos t, dx=?$
|
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|
Why does $ \lim_{x \to 2} \frac{x^2-4}{x-2} =4 $ if x cannot be 2? I know that $ \lim_{x \to 2} \frac{x^2-4}{x-2} $ is evaluated as follows :-
$$ \lim_{x \to 2} \frac{x^2-4}{x-2} \\ = \lim_{x \to 2} \frac{(x+2)(x-2)}{x-2} \\ = \lim_{x \to 2} x+2 \\ = 2+2 \\ = 4 $$
By looking at the function $ \frac{x^2 - 4}{x - 2} $, I can see that 2 is not in its domain. Therefore, I am not able to understand how $ \lim_{x \to 2} x+2 = 2+2 $.
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It simply means that $f(x)$ approaches $4$ from both sides as $x$ approaches $2$. The function does not need to be defined at that point, although it could. The function $f(x) = \frac{x^2-4}{x-2}$ is exactly the "same" as the function $f(x) = x+2$ at all points except $x = 2$, where there is a removable discontinuity, or a gap, in the former function. Therefore, whatever value $x+2$ takes at $x=2$ will be equivalent to the value $\frac{x^2-4}{x-2}$ approaches as $x \to 2$. The fact that the function is undefined at that point doesn't change this.
|
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|
Calculate $\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}$ The question:\,
Calculate $$\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}.$$
Book's final solution: $\dfrac\pi 2$.
My mistaken solution: I don't see where is my mistake because my final solution is $-\dfrac{\pi i}2$:
$$\begin{align}
\text{A}:\int_{-\infty}^\infty {x^2\over (1+x^2)^2}dx &= 2\int_{0}^\infty {x^2\over (1+x^2)^2}dx \quad\text{as the function is even}\\&=
-2\left(\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,i\right)+\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,-i\right)\right)\end{align}
$$
Calculate the residues: Since $\pm i$ are poles of order $2$, then
$$\begin{align}\text{B}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,i\right)&={(2z\ln z+z^2\cdot z^{-1})(z+i)^2-z^2\ln z\cdot 2(z+i)\over (z+i)^4} \\&= \frac{(2i\ln i+i)(-4)+\ln i\cdot 4i }{16} \\ &= {1\over 16} (-8i\ln i -4i+4i\ln i)={-1\over 4}i(\ln i+1)\end{align}
$$
whereas
$$\begin{align}\text{C}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,-i\right)&={(2z\ln z+z^2\cdot z^{-1})(z-i)^2-z^2\ln z\cdot 2(z-i)\over (z-i)^4} \\ &=\frac{(-2i\ln(- i)-i)(-4)+\ln (-i)\cdot (-4i)}{16} \\ &= {1\over 16} (8i\ln (-i) +4i-4i\ln(- i))={1\over 4}i(\ln (-i)+1)\end{align}
$$
Combining $\text A$, $\text B$ and $\text C$, we get
$$
\int_{-\infty}^\infty {x^2\over (1+x^2)^2}\,dx=-2\left({i\over 4}(\ln(-i)+1)-{i\over 4}(\ln i+1)\right)={-i\over 2}\left({3\pi\over 2}-{\pi\over 2}\right)={-\pi i \over 2}
$$
Where is my mistake? Thanks in advance!
|
That $\ln z$ makes no sense. That integral is equal to $2\pi i$ times the sum of the residues at the singularities of $\frac{z^2}{(1+z^2)^2}$ in the upper half-plane. There is actually only one such singularity: at $i$. So\begin{align}\int_{-\infty}^\infty\frac{x^2}{(1+x^2)^2}\,\mathrm dx&=2\pi i\operatorname{res}_{z=i}\left(\frac{z^2}{(1+z^2)^2}\right)\\&=2\pi i\times\left(-\frac i4\right)\\&=\frac\pi2.\end{align}
|
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positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root
but the actual answer is simply $x=\sqrt3$
I am unable to perform the simplification any help would we helpful
|
Rather than pulling $\sqrt{13+4\sqrt3}=1+2\sqrt3$ magically out of a hat, write $\sqrt{13+4\sqrt3}=a+b\sqrt3,$
as suggested by Parcly Taxel in a comment. Squaring both sides, $13+4\sqrt3=(a^2+3b^2)+2ab\sqrt3.$
$13=a^2+3b^2$ and $4=2ab$ means $b=2/a$ and $13=a^2+\dfrac{12}{a^2},$ i.e., $a^4-13a^2+12=0$, i.e., $(a^2-1)(a^2-12)=0$, i.e., $a=\pm1$ or $a=\pm2\sqrt3$ and respectively $b=\pm2$ or $\pm\dfrac 1{\sqrt3}. $ Note that $\sqrt{13+4\sqrt3}>0.$ Can you take it from here?
|
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|
Find the minimum value of $\frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$ where $a, b > 0$ and $a + b \le 1$.
$a$ and $b$ are two positives such that $a + b \le 1$. Find the minimum value of $$\large \frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$$
We have that $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} = \dfrac{a^2 + b^2 + a + b}{ab + a + b + 1} \ge \dfrac{\dfrac{(a + b)^2}{2} + (a + b)}{\dfrac{(a + b)^2}{4} + 2(a + b)} = \dfrac{2(a + b) + 4}{(a + b) + 8}$
$\implies \dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b} \ge \dfrac{2(a + b) + 4}{(a + b) + 8} + \dfrac{1}{a + b}$
Let $x = 1 - (a + b) \implies x \ge 0$.
Thus $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b} \ge \dfrac{2(1 - x) + 4}{(1 - x) + 8} + \dfrac{1}{1 - x} = \dfrac{2x^2 - 9x + 15}{x^2 - 10x + 9}$
$= \dfrac{x(x + 23)}{3(9 - x)(1 - x)} + \dfrac{5}{3} \ge \dfrac{5}{3}$
The equality sign happens when $a = b$ and $x = 0$ or $a + b = 1 \implies a = b = \dfrac{1}{2}$.
I want to ask if the above solution is correct and if there are any other solutions that are more rational and reasonable.
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Let $y=a+b$. Your expression is the same as $$A=\frac{y+1}{b+1}-1+\frac{y+1}{a+1}-1+\frac{1}{y}=\frac{(y+1)(y+2)}{ab+y+1}+\frac{1}{y}-2$$
Now, by $GM\leq AM$ we have $ab\leq (y/2)^2$. And therefore
$$A\geq \frac{(y+1)(y+2)}{(y/2)^2+y+1}+\frac{1}{y}-2=\frac{(y+1)(y+2)}{(y/2+1)^2}+\frac{1}{y}-2=\frac{4(y+1)}{y+2}+\frac{1}{y}-2$$
Using that $\frac{4(y+1)}{y+2}=4-\frac{4}{y+2}$. We conclude
$$A\geq f(y)=2+\frac{1}{y}-\frac{4}{y+2}$$
Since $f'(y)=\ln(\frac{y}{(y+2)^4})\leq \ln(1)=0$ when $0<y\leq 1$ we conclude that $f(y)$ is decreasing in this interval, so
$$A\geq f(y)\geq f(1)=3/5.$$
The equality happens when $a=b$ and $a+b=y=1$, i.e. when $a=b=1/2$.
|
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|
Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1}$$
upon expanding them we get:
$$f_1(x)=1 + x + x^2$$
$$f_2(x)=1 + 2 x + 3 x^2 + 4 x^3 + 3 x^4 + 2 x^5 + x^6$$
$$f_3(x)=1 + 3 x + 6 x^2 + 10 x^3 + 15 x^4 + 18 x^5 + 19 x^6 + 18 x^7 +
15 x^8 + 10 x^9 + 6 x^{10} + 3 x^{11} + x^{12}$$
$$f_4(x)=1 + 4 x + 10 x^2 + 20 x^3 + 35 x^4 + 56 x^5 + 80 x^6 + 104 x^7 +
125 x^8 + 140 x^9 + 146 x^{10} + 140 x^{11} + 125 x^{12} + 104 x^{13} +
80 x^{14} + 56 x^{15} + 35 x^{16} + 20 x^{17} + 10 x^{18} + 4 x^{19} + x^{20}$$
$$\vdots$$
$$f_{n-1}(x)=1 + ?x + ?x^2 + ?x^3 +?x^4+ ?x^5+\cdots+?x^{n(n-1)}$$
I'm wondering how to determine the coefficients for the n-th order? I can observe that the coefficients are symmetric.
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Putting (probably far too many) details around Gerry Myerson's comment and avoiding the awkward upper bounds in IV_'s answer's sums...
\begin{align*}
f_n(x) &= \left( \sum_{k=0}^{n+1} x^{k} \right)^n \\
&= \left( \frac{1}{1-x} - \frac{x^{n+2}}{1-x} \right)^n \\
&= \sum_{k=0}^n (-1)^k \binom{n}{k} \left( \frac{1}{1-x} \right)^{n-k} \left( \frac{x^{n+2}}{1-x} \right)^k \\
&= \sum_{k=0}^n (-1)^k \binom{n}{k} x^{(n+2)k} \left( 1-x \right)^{-n} \\
&= \sum_{k=0}^{n+1} (-1)^k \binom{n}{k} x^{(n+2)k} \sum_{d = 0}^{n(n+1) - (n+2)k} \binom{n+d-1}{d} x^d \\
&= \sum_{i=0}^{n(n+1)} \left( \sum_{k=0}^{n+1} (-1)^k \binom{n}{k} \binom{n+i-(n+2)k-1}{i-(n+2)k} \right.\\
&\qquad \left. \phantom{\sum_{i=0}^{n(n+1)}} \left[ 0 \leq i-(n+2)k \leq n(n+1)-(n+2)k \right] \right) x^i \\
&= \sum_{i=0}^{n(n+1)} \left( \sum_{k=0}^{n+1} (-1)^k \binom{n}{k} \binom{n+i-(n+2)k-1}{i-(n+2)k} \right.\\
&\qquad \left. \phantom{\sum_{i=0}^{n(n+1)}} \left[ (n+2)k \leq i\right] \right) x^i \text{.}
\end{align*}
Above, we have used (roughly in top-to-bottom order) geometric sums and series, the binomial theorem (twice), the fact that we need not retain powers of $x^d$ that end up producing powers above the largest power of $x$ in $f_n(x)$ (since such contributions will conspire to cancel to zero), forcing $d$ to satisfy $(n+2)k+d = i$, the Iverson bracket, and the facts that $(n+2)k \geq 0$ and $i$ is upper bounded.
There's really an exclusion-inclusion hiding here, fairly explicit in the fourth line. As an example, per that line,
$$ f_3(x) = \frac{1}{(1-x)^3} - \frac{3x^5}{(1-x)^3} + \frac{3x^{10}}{(1-x)^3} - \frac{x^{15}}{(1-x)^3} \text{,} $$
where we can see that what we want is the sum of four degree-shifted and scaled copies of polynomials constructed directly from the third diagonal of Pascal's triangle.
We can also frame this example as counting lattice points in a finite cube. We include all the points in triangular numbers until we reach the first corner (at $x^4$), where we need to subtract off contributions from the $3$ missing cubes starting at those corners, until they overlap -- causing double-deletion of points past the second "layer" of vertices -- and we add in $3$ cubes starting at those vertices, until we reach the last vertex and deal with the final batch of overlapping and overcounting.
These coefficients count lattice points in layers (of constant coordinate sum) of:
*
*$n=1$: lattice points in layers of a line segment of length $3$,
*$n=2$: lattice points in layers of a square of side $4$,
*$n=3$: lattice points in layers of a cube of side $5$,
*$n=4$: lattice points in layers of a tesseract of side $6$,
*$n=5$: lattice points in layers of a pentaract of side $7$,
*and so on ...
|
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Calculate the maximum value of $\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$ where $3a + 4b + 5c = 12$
$a$, $b$ and $c$ are positives such that $3a + 4b + 5c = 12$. Calculate the maximum value of $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$$
I want to know if there are any other solutions that are more practical. This came from my homework my teacher gave today. I have given my solution down below if you want to check out.
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We have that $$3a + 4b + 5c = 12$$
$$\implies (a + b) + 2(c + a) + 3(b + c) = 12 \implies \sqrt{ab} + 2\sqrt{ca} + 3\sqrt{bc} \le 6$$
In addition to that, $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c} \le \frac{ab}{ab + 2\sqrt{ab}} + \frac{2ca}{ca + 2\sqrt{ca}} + \frac{3bc}{bc + 2\sqrt{bc}}$$
$$ = \frac{\sqrt{ab}}{\sqrt{ab} + 2} + \frac{2\sqrt{ca}}{\sqrt{ca} + 2} + \dfrac{3\sqrt{bc}}{\sqrt{bc} + 2} = 6 - 2\left(\frac{1}{\sqrt{ab} + 2} + \frac{2}{\sqrt{ca} + 2} + \frac{3}{\sqrt{bc} + 2}\right)$$
$$\le 6 - \frac{72}{\sqrt{ab} + 2\sqrt{ca} + 3\sqrt{bc} + 12} \le 6 - \frac{72}{6 + 12} = 2$$
The equality sign occurs when $a = b = c = 1$.
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Find the value of $a_4-a_2$
Let $a_1<a_2<a_3<a_4$ be positive integers such that $\sum_{i=1}^{4}\frac{1}{a_i}=\frac{11}{6}$. Find the value of $a_4-a_2.$
I do not know how to proceed. I have tried to simplify the summation but did not obtain anything useful. Please help.
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$\dfrac{11}{6}=\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}
\nleq\dfrac12+\dfrac13+\dfrac14+\dfrac15=\dfrac{77}{60} \quad \Longrightarrow \quad a_1=1$
$\dfrac{5}{6}=\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}
\nleq\dfrac13+\dfrac14+\dfrac15=\dfrac{47}{60} \quad \Longrightarrow \quad a_2=2$
$\dfrac{1}{3}=\dfrac{1}{a_3}+\dfrac{1}{a_4} \quad \Longrightarrow \quad a_3 a_4=3(a_3+a_4)$
In positive integers solutions of last equation is only $(4,12),(12,4),(6,6)$.
$a_3<a_4 \quad \Longrightarrow \quad (a_3,a_4)=(4,12)$.
$a_4-a_2=10$
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Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$ and $k \le 2m+1$.
My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}.$$ I need prove that for all real $x \ge 1$ and for all positive integers $m$ the inequality $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$$ holds.
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Incomplete answer to date
Let $$f(x,k)=\frac{x^{m+1}+1}{x^m+1}-\left(\frac{x^k+1}2\right)^{\frac1k}\tag1$$ so that $$f_x(x,k)=\frac{\partial f(x,k)}{\partial x}=\frac{x^{m-1}(x^{m+1}+(m+1)x-m)}{(x^m+1)^2}-\frac{x^{k-1}}{x^k+1}\left(\frac{x^k+1}2\right)^{\frac1k}\tag2.$$ Clearly, $$f_x(1,k)=\frac2{2^2}-\frac12=0\tag3$$ and since \begin{align}\small\!\!\!\!\!\!\!\!\!\!\!\!f_{xx}(x,k)=\frac{\partial^2f(x,k)}{\partial x^2}&=\small\frac{mx^{m-2}(x^m+1)^2(2x^{m+1}+m(m+1)x-m(m-1))}{(x^m+1)^4}\\&\small \,\,\,\,\,\,\,\,-\frac{2mx^{2m-2}(x^{m+1}+(m+1)x-m)}{(x^m+1)^4}-\frac{(k-1)x^{k-2}}{(x^k+1)^2}\left(\frac{x^k+1}2\right)^{\frac1k}\tag4,\end{align} we have that $$f_{xx}(1,k)=\frac{m\cdot2^2(2+2m)-2m\cdot2}{2^4}-\frac{k-1}{2^2}=\frac{m(1+2m)-k+1}{2^2}>0\tag5$$ for $m\ge1$ due to the fact that $k<2m+2$. Therefore, $x=1$ is a minimum with $$f(1,k)=\frac22-1=0\tag6.$$ It remains to show that $f(x,k)>0$ for $x>1$. However, it suffices to prove that $$f_m(x)=\frac{x^{m+1}+1}{x^m+1}-\left(\frac{x^{2m+1}+1}2\right)^{\frac1{2m+1}}>0\tag7$$ since $$\left(\frac{x^k+1}2\right)^{\frac1k}\le\left(\frac{x^{2m+1}+1}2\right)^{\frac1{2m+1}}\tag8$$ from your attempt. Now $(7)$ is implied by $$2(x^{m+1}+1)^{2m+1}>(x^m+1)^{2m+1}(x^{2m+1}+1)\tag9$$ which is equivalent to $$2\sum_{n=0}^{2m+1}\binom{2m+1}nx^{nm+n}>(x^{2m+1}+1)\sum_{n=0}^{2m+1}\binom{2m+1}nx^{nm}\tag{10}.$$ Collecting the summation terms yields $$\sum_{n=0}^{2m+1}\left[\binom{2m+1}nx^{nm}(2x^n-x^{2m+1}-1)\right]>0\tag{11}$$ which is almost certainly the case.
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"url": "https://math.stackexchange.com/questions/3254553",
"timestamp": "2023-03-29T00:00:00",
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|
Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
This has become the norm now... This problem is adapted from a recent competition.
We have that $6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) \ge 4(x + y + z)(xy + yz + zx)$
Furthermore, $$(x + y + z)^3 - 6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 9xyz$$
$$ = x^3 + y^3 + z^3 - 3(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 15xyz$$
In addition, $x^3 + y^3 + z^3 + 3xyz \ge x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2$.
So now we need to prove that $6xyz \ge x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2$, which I don't even know if it is true or not.
|
We need to prove that $$\sum_{cyc}(x^3+3x^2y+3x^2z+2xyz)+\sum_{cyc}3xyz\geq 4\sum_{cyc}(x^2y+x^2z+xyz)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.
Your way is wrong because you took too strong estimation, that got a wrong inequality.
Here happens like the following.
Let we need to prove that $2>1$.
We know that $1<3$, but it does not say that $2>3$ can help because it's just wrong.
The Schur's inequality we can proof so:
Let $x\geq y\geq z.$
Thus, $z(z-x)(z-y)\geq0$ and $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)=\sum_{cyc}x(x-y)(x-z)\geq$$
$$\geq x(x-y)(x-z)+y(y-x)(y-z)=(x-y)^2(x+y-z)\geq0.$$
|
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|
How can I prove that $ (3/p) = -1$ if $ p \equiv \pm 5 \pmod {12}$ I know how to prove that $ (3/p) = 1$ if $ p \equiv \pm 1 \pmod {12}$ but I need to prove that $ (3/p) = -1$ if $ p \equiv \pm 5 \pmod {12}$, which the book write it as it is without explaining why after explaining the first case thoroughly. Could anyone explain for me why this is true please?
|
A possible way is as follows. Both rely on the following fact: Let $p>3$ be prime. $-3$ is a quadratic residue modulo $p$ if and only if $p\equiv 1\pmod{6}$. Now, let me first give a self-contained proof of this fact. Let $x\equiv 2a+1\pmod{p}$ such that $x^2\equiv -3\pmod{p}$ (note that such an $a$ must indeed exist as $p\neq 2$). Now, $4a^2+4a+4\equiv 0\pmod{p}$, that is, $p\mid a^2+a+1$. This implies, $a^3\equiv 1\pmod{p}$. Now, if $p\equiv 5\pmod{6}$, then letting $p=6\ell+5$, we have $a^{p-1}=a^{6\ell+4}\equiv 1\pmod{p}$. Moreover, $a^3\equiv 1\pmod{p}$ also implies $a^{6\ell+3}\equiv 1\pmod{p}$, and thus, $a\equiv 1\pmod{p}$, which yields $a^2+a+1\equiv 3\pmod{p}$, which is clear contradiction. Thus. $p\not\equiv 5\pmod{6}$, and since $p>3$ we deduce $p\equiv 1\pmod{6}$. Now,
Case 1 If $p\equiv 5\pmod{12}$, we also enjoy $p\equiv 1\pmod{4}$, and thus, $(\frac{3}{p})=(\frac{-3}{p})=-1$ (where the second equality follows from the fact above) this clearly yields $3$ is not a quadratic residue modulo $p$.
Case 2 Now, let $p\equiv 7\pmod{12}$. Then, $-3$ is a quadratic residue modulo $p$, whereas $-1$ is not, and thus, $3$ is not a q. residue either.
Edit: (to address a comment by @rtybase below). I believe this argument should also extend to composite $p$. If $p=p_1^{\beta_1}\cdots p_k^{\beta_k}$, then $-3$ is a quadratic residue modulo $p$ if and only if $-3$ is a quadratic residue modulo $p_i$ for every $i $ (if $-3$ is a q. residue modulo $p$, then it must also be so for any $p_i$. conversely, we can establish by Chinese remainder theorem that, if $x^2\equiv -3\pmod{p_i}$ holds, then we can pass to $p$. From $p_i\to p_i^{\beta_i}$ is also not hard, and is well-known.
|
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|
$a+OA\lt b+OB\lt c+OC$ when $a\lt b\lt c$ in a triangle In the triangle $\triangle ABC$ of sides $a,b,c$ let $O$ be the incenter. If $a\lt b\lt c$ then (it is easy to prove that) $OC\lt OB\lt OA$.
Prove that
$$\max \{a+OA, b+OB, c+OC\}=c+OC$$
|
We need to prove that
$$c+\sqrt{\left(\frac{a+b-c}{2}\right)^2+r^2}>b+\sqrt{\left(\frac{a+c-b}{2}\right)^2+r^2}$$ and$$c+\sqrt{\left(\frac{a+b-c}{2}\right)^2+r^2}>a+\sqrt{\left(\frac{b+c-a}{2}\right)^2+r^2}.$$
We'll prove a first inequality. The second inequality can be proved by the same way.
We need to prove that
$$c-b>\frac{1}{2}\left(\sqrt{(a+c-b)^2+4r^2}-\sqrt{(a+b-c)^2+4r^2}\right)$$ or
$$2(c-b)>\frac{(a+c-b)^2-(a+b-c)^2}{\sqrt{(a+c-b)^2+4r^2}+\sqrt{(a+b-c)^2+4r^2}}$$ or
$$\sqrt{(a+c-b)^2+4r^2}+\sqrt{(a+b-c)^2+4r^2}>2a,$$ which is true because
$$\sqrt{(a+c-b)^2+4r^2}+\sqrt{(a+b-c)^2+4r^2}>a+c-b+a+b-c=2a.$$
|
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|
Limit of $\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}$
Find limit of $ \underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}$
How can I do that? It is interesting due to mathematica says that
$$\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}} = 0 $$
but wolfram that limit doesn't exists. What is more I am not sure too about existance of limit due to
$$\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}} $$ doesn't exists too...
|
Assuming $y\ne 0$,
\begin{align*}
&\lim_{(x,y)\to(0,0)}
\frac
{-\frac{x y}{2}+\sqrt{x y+1}-1}
{y \sqrt{x^2+y^2}}
\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}
\cdot
\frac
{\frac{x y}{2}+\sqrt{x y+1}+1}
{\frac{x y}{2}+\sqrt{x y+1}+1}
\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
\frac
{\left(\sqrt{x y+1}\right)^2-\left(\frac{x y}{2}+1\right)^2}
{2y \sqrt{x^2+y^2}}
\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
\frac
{(xy+1)-\left(\frac{x^2y^2}{4}+xy+1\right)}
{2y \sqrt{x^2+y^2}}
\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
-\frac{x^2y^2}{8y\sqrt{x^2+y^2}}\\[4pt]
=\;&\lim_{(x,y)\to(0,0)}
-\frac{x^2y}{8\sqrt{x^2+y^2}}\\[4pt]
=\;&\;\;0\\[4pt]
\end{align*}
|
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|
Matrix complicated equation Let $$A = \begin{bmatrix}
1 & 3 & 4\\
3 & 6 & 9\\
1 & 6 & 4
\end{bmatrix},$$
$B$ be a $3\times 3$ matrix and $$A \cdot A^{T} \cdot A +3B^{-1} =0$$
What would be the value of
$ \det( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}}))$ ?
|
No need to find the value of matrix $B$ since
\begin{equation}
B^{-1}=-\frac{1}{3} A A^T A
\end{equation}
then
\begin{equation}
B=-\frac{1}{3}(A A^T A)^{-1} = 3 A^{-1} A^{-T} A^{-1}
\end{equation}
put into the desired system
\begin{align}
A^{-1}(B^{-1})2B^T &=A^{-1} (-\frac{1}{3} A A^T A) 2 (3 A^{-1} A^{-T} A^{-1})
\notag \\
&= 2(A^{-1}A)A^T (A A^{-1}) A^{-T} A^{-1}
\notag \\
&= 2A^T A^{-T} A^{-1}
\notag \\
&= 2A^{-1}
\end{align}
where
\begin{equation}
A^{-1}=-\frac{1}{3}
\begin{pmatrix}-10 & 4 & 1\\ -1 & 0 & 1 \\ 4 & -1 & -1
\end{pmatrix}
\end{equation}
then find the result of $det(adj(A^{-1}))$.
|
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|
How to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$. Please tell if the problem can be solved using telescoping technique or not.
If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using that. It is given that $a,b \in \mathbb{R}{+},\, a\gt b,\, n \in \mathbb{N}.$
I tried as follows, but was unsuccessful to pursue:
$a^n − b^n = a^n+\sum_{i=1}^{n-1}(a^ib^{n-i}-a^ib^{n-i})-b^n=a^n+\sum_{i=1}^{n-1}a^ib^{n-i}-\sum_{i=1}^{n-1}a^ib^{n-i}-b^n$
Edit : based on the selected answer's comment.
Writing a few terms of the series, $\sum_{i=1}^n (a^{n+1-i}b^{i-1}-a^{n-i}b^i)$ get:
For $n =5$, get the terms as:
$i=1, \,\, a^{5+1-1}b^{1-1}-a^{5-1}b^1 = a^5-a^4b.$
$i=2, \,\, a^{5-1}b^{2-1}-a^{5-2}b^2 = a^4b-a^3b^2.$
$i=3, \,\, a^{5-2}b^{3-1}-a^{5-3}b^3 = a^3b^2-a^2b^3.$
$i=4, \,\, a^{5-3}b^{4-1}-a^{5-4}b^4 = a^2b^3-a^1b^4.$
$i=5, \,\, a^{5-4}b^{3-1}-a^{5-3}b^5 = a^1b^4-b^5.$
Adding all the terms, get:
$a^5-a^4b+ a^4b-a^3b^2+a^3b^2-a^2b^3+a^2b^3-a^1b^4+a^1b^4-b^5 = a^5 - b^5$
|
\begin{align}
(a-b)\sum_{i=1}^n (a^{n-i}b^{i-1}) &=\sum_{i=1}^n (a^{n+1-i}b^{i-1}-a^{n-i}b^i )\\
&=a^n+\sum_{i=2}^n a^{n+1-i}b^{i-1}-\sum_{i=1}^{n-1}a^{n-i}b^i - b^n \\
&= a^n+\sum_{i=1}^{n-1}a^{n-i}b^i-\sum_{i=1}^{n-1}a^{n-i}b^i-b^n\\
&=a^n-b^n
\end{align}
You might like to read the working backward to be similar to what you attempted.
$b<a$, then
$$b^{i-1}\le a^{i-1}$$
$$a^{n-i}b^{i-1}\le a^{n-1}$$
$$\sum_{i=1}^na^{n-i}b^{i-1}\le \sum_{i=1}^na^{n-1}=na^{n-1}$$
|
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|
Is this series for Pi correct?
The idea was to use an infinite series of triangles. The red then green then the... to get the area of this sector then the area of the circle is 16 times this. If it is a unit circle than area should equal Pi.
Here is the series I got using Pythagorean’s theorem , is it correct?
$$\begin{align}
A&=3r^{2} + 12\sum_{ n=0}^{\infty}2^{n-1}x_{n}\left(1-\sqrt{r^{2}-\frac{x{_{n}}^{2}}{4}}\right),\\ x_{0}&=r\sqrt{2-\sqrt{3}}
,\\x_{n+1}&=\sqrt{2r^{2}-2r\sqrt{r^{2}-\frac{x{_{n}}^{2}}{4}}}
\end{align}$$
So-for-a-unit-circle
$$\begin{align}
\pi&=3 + 12\sum_{ n=0}^{\infty}2^{n-1}x_{n}\left(1-\sqrt{1-\frac{x{_{n}}^{2}}{4}}\right),\\ x_{0}&=\sqrt{2-\sqrt{3}}
,\\x_{n+1}&=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}}
\end{align}$$
|
Note that, in
$x_{n+1}
=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}}
$
if
$x_n = 2\sin(t)
$
then
$\begin{array}\\
x_{n+1}
&=\sqrt{2-2\sqrt{1-\frac{x{_{n}}^{2}}{4}}}\\
&=\sqrt{2-2\sqrt{1-\sin^2(t)}}\\
&=\sqrt{2-2\cos(t)}\\
&=2\sqrt{\dfrac{1-\cos(t)}{2}}\\
&=2\sin(\dfrac{t}{2})\\
\end{array}
$
|
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|
Finding the sum $\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\cdots$ The sum of series is
$${1\over(3\times5)}+{1\over(5\times7)}+{1\over(7\times9)}+\cdots$$
I used to solve this problem as its $n$th term
$${1\over (2n+1)(2n+3)}$$
Now how can I proceed??
|
Let there are functions $$f(x) = \sum_{n=1}^\infty \frac{x^n}{\alpha n + \beta}$$
and $$\psi(x) = \frac{f(x,\alpha,\beta) - f(x,\alpha,\alpha + \beta)}{\alpha} = \sum_{n=1}^\infty \frac{x^n}{(\alpha n + \beta)(\alpha (n+1) + \beta)}$$
where $\alpha,\beta > 0$ and $x \in [-1,1]$.
Thereby $$\psi(x) = \frac{1}{\alpha}\left(\frac{x}{\alpha + \beta} + \lim_{n\to \infty} \frac{x^n}{\alpha(n+1) + \beta}\right) = \frac{x}{\alpha(\alpha + \beta)} .$$
For instance that $\alpha = 2,\beta = 1$ and $x = 1$, $$\psi(1) = \frac{1}{6}.$$
|
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|
Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^2} + \frac{1}{n+1} &= \frac{1+(n+1)}{(n+1)^2} \\
&=\frac{n+2}{(n+1)^2} \\
&=\frac{n+2}{n^2 + 2n + 1} \\
&<\frac{n+2}{n^2 + 2n} (\text{since } n \geq 1)\\
&=\frac{n+2}{n(n+2)}\\
&=\frac{1}{n} \end{align} $$
I am just wondering if there is a simpler way of doing this.
|
Your way is right, but you need to open the sumation from $k=2$:
$$\sum_{k=1}^n\frac{1}{k^2}=1+\sum_{k=2}^n\frac{1}{k^2}<1+\sum_{k=2}^n\frac{1}{k(k-1)}=1+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=2-\frac{1}{n}.$$
|
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|
How to correctly solve $\sqrt{1+x}+\sqrt{1-x}>1$? $$\sqrt{1+x}+\sqrt{1-x}>1, x\geq-1 \wedge x\leq1$$
$$\sqrt{1+x}>1-\sqrt{1-x}$$
$$1+x>1-2\sqrt{1-x}+1-x$$
$$0>-2\sqrt{1-x}+1-2x$$
$$2\sqrt{1-x}>1-2x$$
*
*Lets say both sides of inequality are positive, then we can easily $2\sqrt{1-x}>1-2x$ square and arrange. What we get$\implies x^2<\frac{3}{4} \implies \frac{-\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$Now, when we make set intersection with $x\geq-1 \wedge x\leq1$ we get the solution $\frac{-\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$.
*We also have to look at the possibility that $1-2x$ can be negative: $0>1-2x\implies x>\frac{1}{2}$We determine the solution with $x\geq-1 \wedge x\leq1$ and we get the complete solution $\frac{1}{2}<x<1$.
Now I want to get the solution for the original inequality and therefore I unite the final solutions from 1. and 2.case. I get the interval $$(-\frac{\sqrt{3}}{2}, 1]$$
This solution is wrong. The correct one is $[-1,1]$. I would really appreciate if someone could check the complete process and point out where I made a mistake. Also, is this solving process good and effective or I complicate things too much?
|
The domain of this inequation is $[-1,1]$. On its domain, both sides are non-negative, so comparing them amounts to comparing their squares:
$$\bigl(\sqrt{1+x}+\sqrt{1-x}\bigr)^2=1+\not x+1-\not x+2\sqrt{1-x^2}>1\iff 1+2\sqrt{1-x^2}>0,$$
which is satisfied by any $x$ in the domain since $1+2\sqrt{1-x^2}\ge 1$.
|
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|
$\sin(\frac{\pi}{n})\sin(\frac{2\pi}{n})...\sin(\frac{(n-1)\pi}{n})=\frac{n}{2^{n-1}}$ Prove that $$\sin\left(\frac{\pi}{n}\right)\sin\left(\frac{2\pi}{n}\right)\sin\left(\frac{3\pi}{n}\right).....\sin\left(\frac{(n-1)\pi}{n}\right)=\frac{n}{2^{n-1}}$$
Is there a proof without using complex numbers and $n-th$ roots of unity.
|
The following is the simplest proof I know. We have the identity
\begin{equation}
x^{2n} - 2x^n y^n \cos n\theta + y^{2n} = \bigg\{x^2 -2xy \cos \theta + y^2\bigg\}\bigg\{x^2-2xy \cos \bigg(\theta+\frac{2\pi}{n}\bigg)+y^2\bigg\}\cdots
\end{equation}
to $n$ factors adding $2\pi/n$ to each angle successively. This can be seen by noting the LHS and RHS share the same roots in $x$ using complex numbers, but given complex numbers are a trigonometric convenience I imagine there's a non-complex way to arrive at the identity. Let $x=y=1$, $\theta = 2\phi$, and apply $1-\cos\theta = 2\sin(\theta/2)$
\begin{equation}
\sin n\phi = 2^{n-1} \sin \phi \sin\bigg(\phi + \frac{\pi}{n}\bigg)\sin\bigg(\phi + \frac{2\pi}{n}\bigg) \cdots \sin\bigg(\phi + \frac{(n-1)\pi}{n}\bigg)
\end{equation}
Divide by $\sin \phi$ and let $\phi \rightarrow 0$ to get the equation
|
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|
Expanding the Polynomial Using Taylor Series? Expand the polynomial
$$f(x) = x^3-2x^2-3x+5$$
In power of $(x-2)$
This might be a simple question but all we have to do to solve this question is to expand using the Taylor series with $x =2 $ , as opposed to the Maclaurin at $x=0$.
$f(x) = f(a) + \frac{f'(a)(x-a)^1}{1!}+\frac{f''(a)(x-a)^2}{2!}+ \cdot \cdot \cdot$
$f(2)= -1+1(x-2)+\frac{8(x-2)^2}{2!}+\frac{6(x-2)^3}{3!}$
$f(2) = -1 +(x-2) +4(x-2)^2+(x-2)^3$
So in the end this would be the solution correct?
|
Yes that's correct.
Moreover for a polynomial of degree $n$, the Taylor series of order $n$ is exact (because all the remaining derivatives are nulls), thus you have:
$$
f(x)= -1+(x-2)+4 (x-2)^2+(x-2)^3
$$
(no need for the $O\left((x-2)^4\right)$ term here, this is a true equality)
|
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|
Does any (right) triangle exist such that $a^3+b^3=c^3$?
Does any right triangle exist such that $a^3+b^3=c^3$? Does any triangle exist such that $a^3+b^3=c^3$?
I'm stuck on this problem; I tried applying the Pythagorean theorem in three dimensions, but in vain. Any tips?
|
If $a^2+b^2=c^2,$ with $a,b,c$ positive, you'd have: $c>a,b>0$ and thus $ca^2>a^3$ and $cb^2>b^3.$ So you get:
$$c^3=c\cdot c^2=c(a^2+b^2)=ca^2+cb^2>a^3+b^3.$$
So $c^3>a^3+b^3.$
You can prove more generally for any triple $(a,b,c)$ of positive reals, there is at most one positive $n$ such that $a^n+b^n=c^n.$ If $a^n+b^n=c^n$ then $c>a,b>0$ and for $m>n$ you have:
$$\begin{align}c^m&=c^{m-n}\cdot c^n
\\&=c^{m-n}(a^n+b^n)\\&=c^{m-n}a^n+c^{m-n}b^n\\
&>a^{m-n}a^n+b^{m-n}b^n\\
&=a^m+b^m.\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{n\to \infty} \ \frac{1}{n} \Bigl[(a+\frac{1}{n})^2+(a+\frac{2}{n})^2+\cdots+(a+\frac{n-1}{n})^2\Bigr]$ without L'Hopital Find limit:
$$a_n = \lim_{n\to \infty} \ \frac{1}{n} \Bigl[\Bigl(a+\frac{1}{n}\Bigr)^2+\Bigl(a+\frac{2}{n}\Bigr)^2+\cdots+\Bigl(a+\frac{n-1}{n}\Bigr)^2\Bigr]$$
I tried limiting it with $$n\cdot\frac{1}{n}\Bigl(a+\frac{1}{n}\Bigr)^2\leqslant a_n \leqslant n \cdot \frac{1}{n}\Bigl(a+\frac{n-1}{n}\Bigr)^2$$
but that got me to the answer that left side limits to $a^2$ and right side to $(a+1)^2$
and so I didn't squeeze it the right way. Help, and keep it simple.
|
$\lim_{n\to \infty} \ \frac{1}{n} \Bigl[\Bigl(a+\frac{1}{n}\Bigr)^2+\Bigl(a+\frac{2}{n}\Bigr)^2+\cdots+\Bigl(a+\frac{n-1}{n}\Bigr)^2\Bigr] =\displaystyle\int_{a}^{a+1} x^2\, dx=\dfrac{x^3}{3}|_a^{a+1}=\dfrac{(a+1)^3-a^3}{3}=a^2+a+\dfrac{1}{3}$
|
{
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|
Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$ Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$
I tried to induct on n:
For $n = 0$ we have $3+5 = 8$ and $8 \equiv 1 \pmod{7^{n+1}}$.
Suppose it is true for $n = k$:
$$3^{7^k}+5^{7^k}\equiv 1 \pmod{7^{k+1}}$$
so $3^{7^k}+5^{7^k}=7^{k+1}*q_1+1$
For $n = k+1$:
$$3^{7^{k+1}}+5^{7^{k+1}}\equiv r \pmod{7^{k+2}}$$
so $3^{7^{k+1}}+5^{7^{k+1}}=7^{k+2}*q_2+r$
Now if we subtract them we get that:
$$3^{7^{k+1}}+5^{7^{k+1}}-3^{7^k}-5^{7^k}=7^{k+1}(7q_2-q1)+r-1$$
From Euler's theorem we know that $a^{\phi(7^{k+1})}\equiv 1 \pmod{7^{k+1}}$
and $\phi(7^{k+1}) = 6\cdot7^{k}$ so $3^{6\cdot7^{k}}\equiv 1 \pmod{7^{k+1}}$ and $5^{6 \cdot 7^{k}}\equiv 1 \pmod{7^{k+1}}$.
So $3^{7^{k+1}}+5^{7^{k+1}}-3^{7^k}-5^{7^k}\equiv3^{7^k}\cdot3^{6\cdot{7^{k}}}+5^{7^k}\cdot5^{6\cdot{7^{k}}}-3^{7^k}-5^{7^k}\equiv 0 \pmod{7^{k+1}}$
Finally I get that $r\equiv1\pmod{7^{k+1}}$.
Here I got stuck. I don't know how to show that $r=1$.
|
It is known that
$$(a+b)^7=a^7+b^7+7(a^6b+ab^6)+21(a^5b^2+a^2b^5)+35(a^4b^3+a^3b^4)$$ and $(A+B)^7= {A^7+B^7}$ in fields of characteristic $7$. We apply induction.
►$3^{7^1}+5^{7^1}=80312=1639\times7^2+1\iff3^{7^1}+5^{7^1}\equiv1\pmod{7^2}$
►$3^{7^n}+5^{7^n}\equiv1\pmod{7^{n+1}}\iff3^{7^n}+5^{7^n}=7^{n+1}m+1$ assumed to be true.
►$(7^{n+1}m+1)^7=7^{7(n+1)}m^7+7(7^{6(n+1)}m^6+7^{n+1}m)+21(7^{5(n+1)}m^5+7^{2(n+1)}m^2)+35(7^{4(n+1)}m^4+7^{3(n+1)}m^3)+1=1+7^{n+2}\color{red}{[}7^{6n+5}b^7+(7^{5n+6}m^6+\color{red} m)+21(7^{4n+3}m^5+7^nm^2)+35(7^{3n+2}m^4+7^{2n+1}m^3)\color{red}]$
Then $(3^{7^n}+5^{7^n})^7=(7^{n+1}m+1)^7\equiv1\pmod{7^{n+2}}$
But $(3^{7^n}+5^{7^n})^7\equiv{3^{7^{n+1}}+5^{7^{n+1}}}\equiv1\pmod{7^{n+2}}$
We are done.
|
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|
What is the correct solution of $\sqrt[7]{(-\sqrt{3}-i)^5}$? $\sqrt[7]{(-\sqrt{3}-i)^5}=(-\sqrt{3}-i)^\frac{5}{7}=
2^\frac{5}{7}(\cos(\frac{5}{7}\alpha)+i\sin(\frac{5}{7}\alpha)=$
$\tan\alpha=\frac{-1}{-\sqrt{3}} \implies \alpha=\frac{\pi}{6}+2k\pi$
$=2^\frac{5}{7}(\cos(\frac{5\pi}{42}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{42}+\frac{2k\pi}{7})$
The problem I have currently is that I have no idea where I made an error. The correct solution is $2^\frac{5}{7}(\cos(\frac{5\pi}{6}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{6}+\frac{2k\pi}{7})$.
|
$(-\sqrt3-i)^5$ is a specific complex number, but it can be written in multiple ways, making use of the fact that $e^{i2\pi k}=1$ for any $k\in\mathbb{Z}$: Since
$$-\sqrt3-i=-2\left(\sqrt3+i\over2\right)=-2\left(\cos\left(\pi\over6\right)+i\sin\left(\pi\over6\right)\right)=-2e^{i\pi/6}$$
we have
$$(-\sqrt3-i)^5=-2^5e^{i5\pi/6}=2^5e^{i11\pi/6}=2^5e^{i23\pi/6}=2^5e^{i35\pi/6}=2^5e^{i47\pi/6}=\cdots$$
Now among the numbers $11$, $23$, $35$, $47$, etc., we see one that's nicely divisible by $7$, so writing
$$(-\sqrt3-i)^5=2^5e^{i(35\pi/6+2\pi k)}$$
we get
$$\sqrt[7]{(-\sqrt3-i)^5}=2^{5/7}e^{i({5\pi/6}+2\pi k/7)}=2^{5/7}\left(\cos\left({5\pi\over6}+{2\pi k\over7}\right)+i\sin\left({5\pi\over6}+{2\pi k\over7}\right)\right)$$
|
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|
Solve for $x,y,z \in \mathbb{R^+}$ ,$x^2+y^2+z^2=xyz+4$ and $xy+yz+zx=2(x+y+z)$
Find all positive real numbers such that
$$x^2+y^2+z^2=xyz+4$$
And $$xy+yz+zx=2(x+y+z)$$.
I substitute $x,y,z$ by $(a+\frac{1}{a}),(b+\frac{1}{b}),(c+\frac{1}{c})$ respectively where $abc=1$ which comes from the first relation. Putting the values in the second relation it becomes complex.
I tried another way. Multiplying second Equation by 2 and adding with the first equation we get ,$(x+y+z-2)^2=xyz+8$.I failed to find any method. Any ideas guys?
|
After homogenization we obtain:
$$\frac{(x^2+y^2+z^2)(xy+xz+yz)}{2(x+y+z)}=xyz+\frac{(xy+xz+yz)^3}{2(x+y+z)^3}.$$
We'll prove that
$$\frac{(x^2+y^2+z^2)(xy+xz+yz)}{2(x+y+z)}\geq xyz+\frac{(xy+xz+yz)^3}{2(x+y+z)^3}.$$
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$, and $xyz=w^3$.
We see that it's enough to prove this inequality for a maximal value of $w^3$, which happens for equality case of two variables.
Since the inequality is symmetric and homogeneous, we can assume $y=z=1$, which gives
$$(2x+7)(x^2-1)^2\geq0.$$
The equality occurs for $x=y=z,$ which gives
$$3x^2=6x$$ or $$x=y=z=2.$$
Another way.
We need to prove that: $$\frac{(9u^2-6v^2)\cdot3v^2}{6u}\geq w^3+\frac{27v^6}{54u^3}.$$
Now, by AM-GM $$v^4\geq uw^3,$$ which says that it's enough to prove that
$$\frac{(9u^2-6v^2)v^2}{2u}\geq\frac{v^4}{u}+\frac{v^6}{2u^3}$$ or
$$9u^4-8u^2v^2-v^4\geq0$$ or
$$(u^2-v^2)(9u^2+v^2)\geq0$$ and since $$u^2-v^2\geq0$$ it's $$\sum_{cyc}(x-y)^2\geq0,$$ our inequality is true.
|
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|
Special Factorization Consider the natural numbers that are sum of a perfect square plus the product of consectutive natural numbers. For example, $97 = 5^{2} + 8\cdot 9$. What is the smallest multiple of 2019 that is not as described above?
Someone can help me? Thank you in advance.
|
consider the following theorem:
theorem: An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no prime congruent to $3(mod4)$ raised to an odd power.
Now, observe that
$$N = m^{2} + n\cdot (n + 1) \Leftrightarrow$$
$$4N = 4m^{2} + 4n^{2} + 4n \Leftrightarrow$$
$$4N + 1 = 4m^{2} + 4n^{2} + 4n + 1 \Leftrightarrow$$
$$4N + 1 = (2m)^{2} + (2n + 1)^{2}$$
Then, observe that
$4 \cdot 2019 + 1 = 8077$, and the prime decomposition of $8077$ is
$$8077 = 41 \cdot 197$$
where $41 \equiv 1 (mod4)$ and $197 \equiv 1 (mod 4)$. Therefore, $8077$ can be written as a sum of two squares.
Consider
$4 \cdot (2 \cdot 2019) + 1 = 16153$, and the prime decomposition of $16153$ is
$$16153 = 29 \cdot 557$$
where $29 \equiv 1 (mod4)$ and $557 \equiv 1 (mod4)$. Therefore, $16153$ can be written as a sum of two squares.
Now consider
$4 \cdot (3 \cdot 2019) + 1 = 24229 (prime)$, and $24229 \equiv 1 (mod4)$
Lastly,
$4 \cdot (4 \cdot 2019) + 1 = 32305$ and the prime decomposition of $32305$ is
$$32305 = 5 \cdot 7 \cdot 13 \cdot 71$$
Thus, we have $7 \equiv 3(mod 4)$ and $7$ is raised to an odd power. Hence $32305$ is not a sum of two squares, thus $4 \cdot 2019$ is the smallest multiple of 2019 that is not $m^{2} + n \cdot (n + 1)$
|
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|
Show that $\sum^{\infty}_{n=1}{\frac{2(n+1)\cdot3^n}{\sqrt[3]{n!}}}$ is Convergent
Show that $\sum^\infty_{n=1}\frac{n+1}{\sqrt[3]{n!}}\cdot(x^n+x^{-n})$ is uniformly convergent, when $-\frac{1}{3}\leq x\leq3$.
So I`m trying to show with Weierstrass M-test:
$|\frac{n+1}{\sqrt[3]{n!}}\cdot(x^n+x^{-n})|\leq|\frac{n+1}{\sqrt[3]{n!}}\cdot(3^n+3^{n})|={\frac{2(n+1)\cdot3^n}{\sqrt[3]{n!}}}=b_n$
In order to done, we need to show that the following
$\sum^{\infty}_{n=1}{\frac{2(n+1)\cdot3^n}{\sqrt[3]{n!}}}$
is convergent, but it didn`t work for me with root\ratio tests.
Edit: It is working with ratio test and the solution for this is posted.
|
Ratio test:
$\lim_{n\to\infty} \frac{\frac{2(n+2)3^{n+1}}{\sqrt[3]{(n+1)!}}}{\frac{2(n+1)3^n}{\sqrt[3]{n!}}}$= $\lim_{n\to\infty}\frac{3\cdot(n+2)\cdot\sqrt[3]{n!}}{(n+1)\sqrt[3]{(n+1)!}}$=$\lim_{n\to\infty}\frac{3\cdot(n+2)}{\sqrt[3]{n+1}(n+1)}$=$3\cdot\lim_{n\to\infty}\sqrt[3]{\frac{(n+2)^3}{(n+1)^4}}$=$0 <1$
|
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|
How many matrices satisfy this equality? How many matrices $A\in\mathcal{M}_{3\times 3} (\mathbb{N})$ satisfy this equality?
$$\begin{pmatrix}
1 \ \ 2 \ \ 4
\end{pmatrix}\cdot A=\begin{pmatrix}
3 \ \ 2 \ \ 1
\end{pmatrix}$$
I tried with examples and I found just one but I want to know how to approach this exercise.The right answer is $3$
|
Let $A=[x_{i,j}]$ Then we get
$$x_{1,1} +2 x_{2,1} +4 x_{3,1} = 3$$
$$ x_{1,2} + 2x_{2,2} + 4 x_{3,2} = 2 $$
$$x_{1,3} + 2 x_{2,3} + 4 x_{3,3} = 1$$
We shall count the number of solutions for each equation.
First equation has 2 solutions:
Since $4>3$ we have that $x_{3,1}=0$. Similarly $x_{2,1}<2$. If $x_{2,1}=1$ then $x_{1,1}=1$ and if $x_{2,1}=0$ then $x_{1,1}=3$. Therefore there are two solutions for the first equation.
Second equation has 2 solutions:
As before, $x_{3,2}=0$. Also $x_{2,2}$ is at most $1$, hence if $x_{2,2}=1$ then $x_{1,2}=0$ and if not then $x_{2,2}=0$ and $x_{1,2}=1$.
Third equation has 1 solution:
$4,2>1$ so $x_{2,3},x_{3,3}=0$ and so the only solution is $x_{1,3}=1$.
Therefore there is a total of $4$ solutions for all of the equations.
$$A_1 =\begin{bmatrix} 3 & 0 &1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}, A_2=\begin{bmatrix} 1 & 0 &1 \\ 1 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}$$
$$A_3 =\begin{bmatrix} 3 & 2 &1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}, A_4=\begin{bmatrix} 1 & 2 &1 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$
|
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|
Let $a,b,c$ be the sides of a triangle. The find maximum value $ \frac{A}{S^{2}}$ Let $a,b,c$ be the sides of a triangle, $A$ is the area and $S$ is the semi-perimeter $(a+b+c)/2$.
Find the maximum value $\frac{A}{S^{2}}$.
My Approach:
Method 1:
Applying AM-GM inequality on $S,S-a,S-b,S-c$
$$\frac{4S-2S}{4} \ge \sqrt[4]{S(S-a)(S-b)(S-c)}$$
$$\frac{1}{4}\ge \frac{A}{S^2}$$
Method 2:
Applying AM-GM inequality on $S-a,S-b,S-c$
$$\frac{3S-2S}{3} \ge \sqrt[3]{(S-a)(S-b)(S-c)}$$
$$\frac{1}{3\sqrt{3}}\ge \frac{A}{S^2}$$
For equality in method 1 $a=b=c=0$ which is not true hence we get maximum value from method 2. But not sure if $\frac{1}{3\sqrt{3}}$ is the maximum because some other method may give me some other maximum value. So is there a method which gives me the exact maximum value.
|
Your first way does not give a solution because the equality
$$s=s-a=s-b=s-c$$ is impossible.
By the way, by AM-GM
$$\frac{A}{s^2}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s^2}\leq\frac{\sqrt{s\left(\frac{s-a+s-b+s-c}{3}\right)^3}}{s^2}=\frac{1}{3\sqrt3}.$$
The equality occurs for $$s-a=s-b=s-c$$ or $$a=b=c,$$ which says that we got a maximal value.
|
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How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$. I am getting x = 4 but the book gets x = 81. What am I doing wrong?
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-\log_{3}(3)} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-1} + 8\\
\log_{3}(x) & = -\log_{3}(x) + 8\\
2\log_{3}(x) & = 8\\
\log_{3}(x) & = 4\\
x & = 4
\end{align*}
What am I doing wrong?
|
First, $\log_{1/3}x = \log_{3}\frac{1}{x}$; hence,
$$ \log_{3}x - \log_{3} \left( \frac{1}{x} \right) = \log_{3} \left(\frac{x}{\frac{1}{x}} \right) = \log_{3}x^{2} = 8$$
which is equivalent to
$$ 3^{8} = \left( 3^{4} \right)^{2} = x^{2}.$$
And so,
$$x = 3^{4} = 81.$$
|
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|
Is every prime contained in the set $\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$?
Is every prime contained in the set
$\,\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}\,$ for $k=1,2,3,...$?
Let $p\notin\{2,5\}$ be a prime number. Then $\frac{1}{p}$ has a decimal period of length at most $p-1$. Denote $\frac{1}{p}$ by $$\frac{1}{p}=0.\overline{d_1d_2...d_r}$$ where $1\leq r \lt p$ and $d_i\in\{0,1,2,...,9\}$.
Let $d=0.d_1d_2...d_r\cdot10^r\;$ (e.g., if $p=7$, then $\frac{1}{7}=0.\overline{142857}$, $r=6$, and $d=142857$).
Let's consider $\frac{1}{10^m-1}$ for some $m\in\mathbb{N}$:
$$
\frac{1}{10^m-1}=\frac{1}{\underbrace{99...9}_{\text{$m$ times}}}=0.\overline{\underbrace{00...0}_{m-1 \\ \text{times}}1}=\sum_{i=1}^\infty \frac{1}{10^{i\cdot m}}
$$
So
\begin{align}
\frac{1}{p}&=0.\overline{d_1d_2...d_r}=d\cdot \left(\frac{1}{10^r}\right)+d\cdot \left(\frac{1}{10^{2r}}\right)+d\cdot \left(\frac{1}{10^{3r}}\right)+\cdots \\ &=d\cdot\left(\frac{1}{10^r}+\frac{1}{10^{2r}}+\frac{1}{10^{3r}}+\cdots\right) \\ &=d\cdot\sum_{i=1}^\infty \frac{1}{10^{i\cdot r}} \\ &=\frac{d}{10^r-1}
\end{align}
Rearranging terms, we have $\,10^r-1=p\cdot d \, \Longrightarrow \, p|10^r-1$.
Does this prove that the set of primes is contained in the set $\{n\in\mathbb{N}:n|10^k-1\}\cup\{2,5\}$?
Clearly, $2$ and $5$ are missing from $\{n\in\mathbb{N}:n|10^k-1\}$. Are there any other primes not contained in this set?
|
Yes: it's true that every prime $p \neq 2$ or $5$ divides some number of the form $10^k-1$.
There is a simple way to prove this fact.
Look at the remainders of $10^k-1$ when divided by $p$. Clearly there are only finitely many remainders, namely $0,1,2, \dots, p-1$. Since the numbers of the form $10^k-1$ are infinitely many, there are two of them which give the same remainder when divided by $p$.
Let's say there are $k<l$ such that $10^k-1$ and $10^l-1$ have the same remainder when divided by $p$. This is equivalent on saying that
$$(10^l-1)-(10^k-1)$$
is a multiple of $p$. But
$$(10^l-1)-(10^k-1) = 10^l-10^k = 10^k (10^{l-k}-1)$$
Since $p$ divides $10^k (10^{l-k}-1)$ and does not divide $10^k$, necessarily $p$ divides $(10^{l-k}-1)$. This concludes the proof.
|
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|
If $a,b,c$ are positives such that $a+b+c=\pi/2$ and $\cot(a),\cot(b),\cot(c)$ is in arithmetic progression, find $\cot(a)\cot(c)$
Suppose $a,b,c$ are positive reals such that $a+b+c=\pi/2$ and $2\cot(b)=\cot(a)+\cot(c)$.
Find $\cot(a) \cdot \cot(c)$.
I have tried writing $b=\pi/2-a-c$ into the given equation and play with trigonometric identities, but it just got too messy and I don't know what else to try.
|
As
$$\cot(x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}$$
$$\cot(a+c) = \cot\left(\frac{\pi}{2}-b\right) = \tan(b) = \frac{1}{\cot b}= \frac{\cot a\cot c -1}{\cot a+ \cot c}$$
$$\frac{1}{\cot b} = \frac{\cot a\cot c - 1}{2\cot b}$$
So,
$$\cot a \cot c = 1+2 = 3$$
|
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|
Minimizing distance between an ellipse and a point Problem
An ellipse has the formula:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
What is the shortest distance from the ellipse to the point $P = (a,0)?$
Attempted solution:
I have previously solved the same problem with P = (1,0) and P = (2,0). For P = (1,0), the minimum where x = 1.8 and y = 1.6 and a distance of $\frac{4\sqrt{5}}{5}$.
For P = (2,0), the local minimum was larger than the distance between P and the ellipse at y = 0., which was equal to 1. So there is some threshold for $a$ for which the local minimum is not truly the smallest distance.
Basic strategy:
1) Solve the ellipse formula for y and put that into the distance formula for the distance between the ellipse and the point (Pythagorean theorem).
2) Take the derivative.
3) Set the derivative to 0 and solve for x.
4) Use the ellipse formula to solve for y.
5) Put in x and y in the distance formula and get the distance.
However, at some threshold the distance between the point and the ellipse where y = 0 is even shorter. So the shortest distance will have one of two cases, one between 0 and just below the threshold and one for the threshold value and above.
I figured that I could apply the same method I did for the previous cases of P = (1,0) and P = (2,0) and see if I can find some way to find this threshold.
Use the ellipse formula and solve for y:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1 \Rightarrow y = \sqrt{4\Big(1-\frac{x^2}{9}\Big)}$$
Write down distance formula
$$D = \sqrt{(x-a)^2 + y^2}$$
Put in y from the ellipse formula
$$D = \sqrt{(x-a)^2 + \Big(\sqrt{4\Big(1-\frac{x^2}{9}\Big)}\Big)^2}$$
Simplify:
$$D = \sqrt{\frac{5}{9} \cdot x^2 - 2ax +4}$$
Take the derivative:
$$D' = \frac{\frac{10}{9}x-2a}{2\sqrt{\frac{5}{9}x^2 - 2ax +4}}$$
Set the derivative to zero and solve for x:
$$\frac{10}{18}x -a = 0 \Rightarrow x = 1.8a$$
Use ellipse function and solve for y:
$$y = \sqrt{4\Big(1-\frac{(1.8a)^2}{9}\Big)} = a\sqrt{0.76}$$
Putting in x and y to the distance formula
$$D = \sqrt{(1.8a-a)^2 + (a\sqrt{0.76})^2} = \sqrt{0.64a^2 + 0.76a^2} = \sqrt{1.4a^2} = a\sqrt{1.4}$$
...but I feel like I am not getting anywhere. This is not the correct distance and the threshold is nowhere to be seen.
Expected answer is:
$$\sqrt{4-\frac{4}{5}a^2}$$
for $0 < a < \frac{5}{3}$ and
$$|a-3|$$
if $a \geq \frac{5}{3}$
|
The problem may be solved with the method of Lagrange multiplier.
$$L = (x-a)^2 + y^2 - \lambda \left(\frac{x^2}{9} + \frac{y^2}{4} -1\right)$$
$$ 0 = \frac{\partial L}{\partial x} = 2(x-a) - \frac{2}{9} \lambda x$$
$$ 0 = \frac{\partial L}{\partial y} = 2y - \frac{1}{2} \lambda y \phantom{00000}$$
From the second equation, $\lambda=4$ or $y=0$.
When $|a|\ge \frac{5}{3}$, we have $y=0$ and the distance is $||a|-3|$. [Please see the discussion in the comments below.]
Otherwise, $\lambda=4$, which implies $x=\frac{9}{5}a$, so the distance squared is
$$\delta^2 = (x-a)^2+y^2 = \frac{16 a^2}{25} + 4\left( 1-\frac{9a^2}{25}\right)=4-\frac{4}{5} a^2$$
$$\delta = 2\sqrt{1-\frac{a^2}{5}}.$$
|
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|
Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
My solution:
Set $f(z)=3$.
For every $0<\delta<1$ and $|z|=\delta$, we have $|P(z)-f(z)|\le \delta^5+2\delta^3<3\delta<3=|f(z)|$. Hence $P(z)$ and $f(z)=3$ has the same number of zeros, which is $0$, inside the open unit disk.
Now consider the behavior of $P(z)$ when $|z|=1$. If we have $P(z)=0$, then $z^5=-2z^3-3$, and $1=|z^5|=|-2z^3-3|$ when $|z|=1$ which forces $-2z^3=2$ and $z^3=-1$.
Clearly, $z=-1$ is a zero of $P(z)$.
For the other two roots of $z^3=-1$, note that: $$ P(z)=z^2\cdot z^3+2z^3+3=z^2-2+3=z^2+1=0\implies z=\pm 1 $$
Therefore we conclude that there is only one root, $z=-1$, of $P(z)$ in the closed unit disk.
However, I am asking for other solutions, such as applying the symmetric version of Rouché''s theorem like in this post. Thank you.
|
Let $z^5+2z^3+3=0$ with $|z|\le 1. $ Then $3=|-3|=|z^5+2z^3|=|z|^3\cdot |z^2+2|\le |z^2+2|\le |z|^2+|2|=|z|^2+2\le 3.$
So $|z|=1$. So let $z=e^{it}$ with $t\in \Bbb R.$ Then $-3=Re (e^{5it}+2e^{3it})=\cos 5t+2\cos 3t.$
Since $\cos 5t\ge -1$ and $2\cos 3t\ge -2,$ therefore $\cos 5t=\cos 3t=-1$. This implies $z^5=z^3=-1,$ hence $z^2=1,$ hence $z=\pm 1.$
So with $P$ standing for $(\,|z|\le 1\land z^5+2z^3+3=0\,)$ we have $$P\implies (z\in \{-1,1\}\land z^5+2z^3+3=0)\implies (z=-1)\implies P.$$
|
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|
Writing a proper inductive argument The algebra is simple for some reason I am having a hard time setting up the argument. Here is my attempt
Suppose we have:
$\sum_{k=1}^n (2k-1)^2 = \frac{n(4n^2-1)}{3}$ for all $n \in \mathbb{N}$
Let see if this holds true for $n = 1$
LHS: $= 1$
RHS $= 1$
Thus this formula holds for $n=1$
Lets see if this formula holds true for $n+1$
$\sum_{k=1}^{n+1} (2k-1)^2 = ((2(n+1)-1)^2) + \sum_{k=1}^n (2k-1)^2$
$ = (2n+2-1)^2 + \frac{n(4n^2-1)}{3}$
Then after all the calculations I show that:
$(2n+2-1)^2 + \frac{n(4n^2-1)}{3}$ is equal to $\frac{n(4n^2-1)}{3}$ with $n+1$ in place of $n$ meaning:
$(2n+2-1)^2 + \frac{n(4n^2-1)}{3} = \frac{(n+1)(4(n+1)^2-1)}{3}$
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Given the statement in this question, I will assume that by the natural numbers, you mean the set of positive integers. If you instead meant the set of nonnegative integers, you need to prove that the statement holds when $n = 0$ as well, as shown by Antoine Mathys in the comments.
In writing a proof by mathematical induction about a property of the positive integers, you should:
*
*State the property $P(n)$ you wish to prove.
*Prove the base case $P(1)$ to establish that the property holds for $n = 1$.
*State the induction hypothesis $P(m)$.
*Prove that $P(m) \implies P(m + 1)$ for each positive integer $m$.
*Conclude that since $P(1)$ holds and $P(m) \implies P(m + 1)$ for each positive integer $m$, then $P(n)$ holds for each positive integer $n$.
If you are instead proving a statement $P(n)$ for all nonnegative integers $n$, replace $1$ by $0$ and positive integer by nonnegative integer. More generally, if you wish to prove a statement $P(n)$ holds for each integer $n \geq n_0$, replace $1$ by $n_0$ and show that $P(m) \implies P(m + 1)$ for each integer $m \geq n_0$.
Let $P(n)$ be the statement
$$\sum_{k = 1}^{n} (2k - 1)^2 = \frac{n(4n^2 - 1)}{3}$$
We wish to prove that this statement for each positive integer $n$.
Proof. Let $n = 1$. Then
$$\sum_{k = 1}^{1} (2k - 1)^2 = (2 \cdot 1 - 1)^2 = (2 - 1)^2 = 1^2 = 1 = \frac{1 \cdot 3}{3} = \frac{1(4 \cdot 1 - 1)}{3} =\frac{1(4 \cdot 1^2 - 1)}{3}$$
Thus, $P(1)$ holds.
Since $P(1)$ holds, we may assume there is some positive integer $m$ such that
$P(m)$ holds. Then
$$\sum_{k = 1}^{m} (2k - 1)^2 = \frac{m(4m^2 - 1)}{3}$$
This is our induction hypothesis.
Let $n = m + 1$. Then
\begin{align*}
\sum_{k = 1}^{m + 1} (2k - 1)^2 & = \sum_{k = 1}^{m} (2k - 1)^2 + \sum_{k = m + 1}^{m + 1} (2k - 1)^2\\
& = \sum_{k = 1}^{m} (2k - 1)^2 + [2(m + 1) - 1]^2\\
& = \sum_{k = 1}^{m} (2k - 1)^2 + [2m + 2 - 1]^2\\
& = \sum_{k = 1}^{m} (2k - 1)^2 + (2m + 1)^2\\
& = \sum_{k = 1}^{m} (2k - 1)^2 + 4m^2 + 4m + 1\\
& = \frac{m(4m^2 - 1)}{3} + 4m^2 + 4m + 1 && \text{by the induction hypothesis}\\
& = \frac{m(4m^2 - 1)}{3} + \frac{12m^2 + 12m + 3}{3}\\
& = \frac{4m^3 - m + 12m^2 + 12m + 3}{3}\\
& = \frac{4m^3 + 12m^2 + 11m + 3}{3}\\
& = \frac{(m + 1)(4m^2 + 8m + 3)}{3}\\
& = \frac{(m + 1)(4m^2 + 8m + 4 - 1)}{3}\\
& = \frac{(m + 1)[4(m^2 + 2m + 1) - 1]}{3}\\
& = \frac{(m + 1)[4(m + 1)^2 - 1]}{3}
\end{align*}
Thus, $P(m) \implies P(m + 1)$ for each positive integer $m$.
Since $P(1)$ holds and $P(m) \implies P(m + 1)$ for each positive integer $m$, then $P(n)$ holds for each positive integer $n$.$\blacksquare$
|
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|
Prove That $3^n + 8^n$ is Not Divisible by $5$ (Using Induction) Prove that $3^n+8^n$ is not divisible by 5.
I know that this can be proved by using congruence and I am providing the proof by congruence below. But is there any way to Prove It By Induction.
The proof by congruence goes like this:
$3\equiv 3\pmod 5 \\ 3^2 \equiv 4\pmod 5 \\ 3^3\equiv 7\pmod 5 \\ 3^4\equiv 1\pmod 5 \\ 3^5\equiv 3\pmod 5$
Also,
$8\equiv 3\pmod 5 \\ 8^2 \equiv 4\pmod 5 \\ 8^3\equiv 7\pmod 5 \\ 8^4\equiv 1\pmod 5 \\ 8^5\equiv 3\pmod 5$
Adding the congruence up (since the same cycle repeats after the 4th power) none of them are divisible by 5 or equal to 0.
But I need a proof by Induction.
Any help will be appreciated.
|
Yes, you can do it by induction as well. Note that:
$3^{n+1}+8^{n+1}=3^n\times 3+8^n\times (3+5)=(3^n+8^n)\times 3+5\times 8^n$
By induction hypothesis the first term is not divisible by $5$ while the second term is obviously divisible by $5$. The required result follows.
|
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|
Jacobi symbol:$(\frac{7m^2-1}{18m^2+1})=(\frac{25m^2}{18m^2+1})=1$? By definition of Jacobi symbol, $(\frac{q}{n})=1$ if there exists $x$ such that $x^2 \equiv q \pmod n$ where $n$ is odd. So, $(\frac{7m^2-1}{18m^2+1})$ is equivalent to-
$x^2 \equiv 7m^2-1 \pmod {18m^2+1} \implies x^2+18m^2+1 \equiv 7m^2-1+18m^2+1 \pmod {18m^2+1}$
[adding $18m^2+1$ in both sides],
$\implies x^2+18m^2+1 \equiv 25m^2 \pmod {18m^2+1}$, clearly, $(\frac{25m^2}{18m^2+1})=1$,
we got, $(\frac{25m^2}{18m^2+1})$ from $(\frac{7m^2-1}{18m^2+1})$,
Does this allow us to write$(\frac{7m^2-1}{18m^2+1})=(\frac{25m^2}{18m^2+1})=1$? Can any one explain a bit more elaborately?
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If $a\equiv b\pmod n$ then $\left(\dfrac a n\right)=\left(\dfrac b n\right)$.
Now $7m^2-1\equiv 7m^2-1 +(18m^2+1)\equiv 25m^2\pmod{18m^2+1}.$
Therefore $\left(\dfrac{7m^2-1}{18m^2+1}\right)=\left(\dfrac{25m^2}{18m^2+1}\right).$
Furthermore, $25m^2=(5m)^2$, so $\left(\dfrac{25m^2}{18m^2+1}\right)=1$.
|
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|
Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$. $$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\} \tag{$\star$}$$
My proof of $(\star)$ is as follows:
\begin{align}
10^{2n+1}+1
&= 10\cdot10^{2n}+1 \\
&= (11-1)\cdot10^{2n}+1 \\
&= 11\cdot10^{2n}-10^{2n}+1 \\
&= 11\cdot10^{2n}-\left(10^{2n}-1\right) \\
&= 11\cdot10^{2n}-\left(100^{n}-1\right) \\
&= 11\cdot10^{2n}-\left((99+1)^{n}-1\right) \\
&= 11\cdot10^{2n}-\left(1+\binom{n}{1}99+\binom{n}{2}99^2+\cdots+\binom{n}{n-1}99^{n-1}+99^n-1\right) \\
&= 11\cdot10^{2n}-\underbrace{99}_{11\cdot9} \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right) \\
&= 11\left(10^{2n}-9 \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right)\right)
\end{align}
Is there an easier way to prove $(\star)$? The expansion of $(99+1)^n$ seems unnecessarily complicated, but I wasn't sure how else to go from there. Easier proofs are welcome!
|
Use modular arithmetic:
$$10\equiv-1\mod11$$
$$10^2\equiv1\mod11$$
$$10^{2n}\equiv1\mod11$$
$$10^{2n+1}\equiv-1\mod11$$
$$11|10^{2n+1}+1$$
|
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|
shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{25}- \frac{8}{5})}^2} = \frac{\sqrt{52}}{25}$
but my answer is wrong. Where am i wrong?
|
Let $$x_c^2+y_c^2=4$$ and $$3x_s+4y_s=12$$ then $$d=\sqrt{(x_c-x_s)^2+(y_c-y_s)^2}$$ you can eliminate two variables.
|
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|
Integral $ \int_0^\infty \frac{\ln x}{(x+c)(x-1)} dx$ I've been trying to solve the following integral for days now.
$$P = \int_0^\infty \frac{\ln(x)}{(x+c)(x-1)} dx$$
with $c > 0$. I figured out (numerically, by accident) that if $c = 1$, then $P = \pi^2/4$. But why? And more importantly: what's the general solution of $P$, for given $c$? I tried partial fraction expansions, Taylor polynomials for $ln(x)$ and more, but nothing seems to work. I can't even figure out where the $\pi^2/4$ comes from.
(Background: for a hobby project I'm building a machine learning algorithm that predicts sports match scores. Somehow the breaking point is this integral, so solving it would get things moving again.)
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A different approach using polylogarithms
For this solution the following identities are used
$\displaystyle\text{Li}_2\left(z\right)=-\int _0^z\frac{\ln \left(1-t\right)}{t}\:dt$, $\displaystyle \int _0^1\frac{c\ln ^n\left(x\right)}{1-cx}\:dx=\left(-1\right)^nn!\text{Li}_{n+1}\left(c\right)$, $\displaystyle \text{Li}_2\left(-z\right)+\text{Li}_2\left(-\frac{1}{z}\right)=-\zeta \left(2\right)-\frac{1}{2}\ln ^2\left(z\right)$
$$\int_0^{\infty}\frac{\ln\left(x\right)}{\left(c+x\right)\left(x-1\right)}\:dx$$
$$=\int _0^1\frac{\ln \left(x\right)}{\left(c+x\right)\left(x-1\right)}\:dx+\underbrace{\int _1^{\infty }\frac{\ln \left(x\right)}{\left(c+x\right)\left(x-1\right)}\:dx}_{x=\frac{1}{x}}$$
$$\hspace{-5mm}=\frac{2}{1+c}\int _0^1\frac{\ln \left(x\right)}{x-1}\:dx-\frac{1}{1+c}\underbrace{\int _0^1\frac{\ln \left(x\right)}{c+x}\:dx}_{x=ct}-\frac{c}{1+c}\int _0^1\frac{\ln \left(x\right)}{1+cx}\:dx$$
$$\hspace{-8mm}=\frac{2}{1+c}\zeta \left(2\right)-\frac{\ln \left(c\right)}{1+c}\int _0^{\frac{1}{c}}\frac{1}{1+t}\:dt-\frac{1}{1+c}\underbrace{\int _0^{\frac{1}{c}}\frac{\ln \left(t\right)}{1+t}\:dt}_{\text{IBP}}-\frac{1}{1+c}\text{Li}_2\left(-c\right)$$
$$\hspace{-2mm}=\frac{2}{1+c}\zeta \left(2\right)-\frac{\ln \left(c\right)}{1+c}\ln \left(1+\frac{1}{c}\right)+\frac{1}{1+c}\ln \left(c\right)\ln \left(1+\frac{1}{c}\right)$$
$$+\frac{1}{1+c}\underbrace{\int _0^{\frac{1}{c}}\frac{\ln \left(1+t\right)}{t}\:dt}_{t=-t}-\frac{1}{1+c}\text{Li}_2\left(-c\right)$$
$$=\frac{2}{1+c}\zeta \left(2\right)-\frac{1}{1+c}\text{Li}_2\left(-\frac{1}{c}\right)-\frac{1}{1+c}\text{Li}_2\left(-c\right)$$
$$=\frac{2}{1+c}\zeta \left(2\right)-\frac{1}{1+c}\left(\text{Li}_2\left(-\frac{1}{c}\right)+\text{Li}_2\left(-c\right)\right)$$
$$=\frac{2}{1+c}\zeta \left(2\right)-\frac{1}{1+c}\left(-\zeta \left(2\right)-\frac{1}{2}\ln ^2\left(c\right)\right)$$
Thus
$$\int_0^{\infty}\frac{\ln\left(x\right)}{\left(c+x\right)\left(x-1\right)}\:dx=\frac{3}{1+c}\zeta \left(2\right)+\frac{1}{2\left(1+c\right)}\ln ^2\left(c\right)$$
|
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|
$\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx$ Trying to compute Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$
I was facing:
\begin{align}J=\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx\end{align}
I want to prove that $\displaystyle J=0$, or equivalently, that,
\begin{align}\int_0^1 \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx=-\dfrac{1}{2}\text{G}\end{align}
$\text{G}$ being the Catalan constant.
Read carefully please.
I know, using so-called Feynman's trick, how to prove this.
I would like to obtain a proof, using only integration by parts and change of variable in simple integrals (that is, no multiple integrals)
I don't know, if, under these restrictions, such computation is possible.
NB:
You're probably wondering what is the link between :
$\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$ and $J$.
The link is, for $x\in\left[0;\frac{\pi}{2}\right]$, $(\sin{x}+\cos{x}+\sqrt{\sin{2x}})(\sin{x}+\cos{x}-\sqrt{\sin{2x}})=1$ and $\sin(2x)=2\sin x\cos x$
|
Let $
K=\int_0^\infty \frac{\ln(1+x+\sqrt{2x})}{x^2+1}dx$
and note that $J+K=\int_0^\infty \frac{\ln(x^2+1)}{x^2+1}\ {dx}$
\begin{align}
J-K=&\int_{0}^\infty \frac{\ln({1+x-\sqrt{2x}})-\ln({1+x+\sqrt{2x}})}{x^2+1}\ \overset{x=t^2}{dx}\\
=&\int_{-\infty}^\infty \frac{t\ln({1+t^2-\sqrt{2}t})-t\ln({1+t^2+\sqrt{2}t})}{t^4+1}\ dt\\
=& \ \frac1{\sqrt2}\int_{-\infty}^\infty \overset{x=\sqrt2t-1}{\frac{\ln(t^2-\sqrt2t+1)}{t^2-\sqrt2t+1}}+ \overset{x=\sqrt2t+1}{ \frac{\ln(t^2+\sqrt2t+1)}{t^2+\sqrt2t+1}}-\overset{x={(t-t^{-1})}/{\sqrt2}}{\frac{(1+t^2)\ln(t^4+1)}{t^4+1}}\ dt\\
=& \ 2\int_{-\infty}^\infty \frac{\ln \frac{x^2+1}2}{x^2+1}dx
- \int_{-\infty}^\infty \frac{\ln (2(x^2+1))}{x^2+1}dx
=2 \int_{0}^\infty \frac{\ln \frac{x^2+1}8}{x^2+1}dx
\end{align}
As a result
\begin{align}
J =\frac12 (J+K) + \frac12 (J-K)
=\ \frac32 \int_0^\infty \frac{\ln \frac{x^2+1}4}{x^2+1}dx=0
\end{align}
|
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|
Finding maximum of a given function
Show that $f(x)=\sin x(1+\cos x)$ attains its maximum at $x= \pi/3$.
I differentiated the function $f$ and got $f'(x)=\cos x(1+2\cos x)$. After equating with $0$, I got $x=\pi/2$ and $x =\pi/3 + n\pi$ with $n\neq 0$. So I did not get $x= \pi/3$ even as an extreme value.
|
$$\cos x(1+2\cos x)=0$$
$$\cos x=0\ \ \ \text{or} \ \ 1+2\cos x=0$$
$$x=\frac{(2n+1)\pi}{2}\ \ \ \text{or} \ \ \cos x=-\frac12$$
$$x=\frac{(2n+1)\pi}{2}\ \ \ \text{or} \ \ \cos x=\cos\frac{2\pi}{3}$$
$$x=\frac{(2n+1)\pi}{2}\ \ \ \text{or} \ \ \ x=2n\pi\pm\frac{2\pi}{3}$$
$$x=\ldots, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}\ldots \ \ \ \text{or} \ \ \ x=\ldots, -\frac{4\pi}{3}, -\frac{2\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \ldots $$
|
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|
Express $x=0.\overline{31}_5$ as a fraction in lowest terms
Given $x=0.\overline{31}_5$, find the value of $x$, expressed as a fraction in lowest terms.
I tried to change it into base $10$, but I don't think it's possible with fractions. So please help I'll appreciate it. Also I'm in 7th grade (easy solutions please) and no copying other people's answer (I've seen it in other problems).
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Updated solution:
$0.31_5 = \frac{3}{5} + \frac{1}{25} = \frac{16}{25}$.
$0.\overline{31}_5 = 0.31_5 + 0.0031_5 + 0.000031_5 + \cdots$. Now using the formula for an infinite geometric series:
$$0.\overline{31}_5 = \frac{0.31_5}{1 - 0.01_5} = \frac{16/25}{1-1/25} = \frac{16}{24} = \frac{2}{3}.$$
Old solution:
$x = 0.3_5 + 0.0\overline1_5$.
Now $0.\overline 4_5 = 1$. $0.0\overline1 \cdot 10_5 \cdot 4_5 = 1$, so $0.0\overline1 = \frac{1}{5 \cdot 4} = \frac{1}{20}$
Using the fact that $0.3_5 = \frac{3}{5}$, $x = \frac{3}{5} + \frac{1}{20} = \frac{13}{20}$.
Note:
$0.\overline 4_5 = 1$ is the equivalent of $0.\overline{99} =1$. To prove this you can break it into $0.4_5 + 0.04_5 + 0.004_5 + \cdots$, which is an infinite geometric series and evaluates to $\frac{0.4_5}{1 - 0.1_5} = \frac{4/5}{1 - 1/5} = 1$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
When going from $(x+2)^2=5$ to $x+2=\pm \sqrt{5}$, why isn't there also a $\pm(x+2)$? Say I am solving the following equation:
$$(x+2)^2 = 5$$
$$x + 2 = \pm \sqrt{5}$$
$$x = -2 \pm \sqrt{5}$$
However, when I took the positive and negative square root of $5$ in the second line, I did not take the positive and negative square root of $(x+2)^2$, which would be $\pm (x+2)$. Why is this?
|
Hint: Better is to write
$$(x+2)^2-\sqrt{5}^2=0$$ and this is, using that $$a^2-b^2=(a+b)(a-b)$$
$$(x+2-\sqrt{5})(x+2+\sqrt{5})=0$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the best solution to an inconsistent system $A\mathbf{u} = \mathbf{b}$. Let $A = \begin{bmatrix}
-1 & 1 \\
2 & -1 \\
1 & 1
\end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}$.
1. Find a "best solution" to the inconsistent system $A\mathbf{u} = \mathbf{b}$.
2. Find the orthogonal projection of $\mathbf{b}$ onto the column space of $A$.
For the second question the column space of $A$ has vectors that are all linearly independent. We first find the projection matrix given by $P = A(A^TA)^{-1}A^T$. Lets first calculate $A^TA$
\begin{equation*}
A^TA = \begin{bmatrix}
-1 & 2 & 1 \\
1 & -1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 1 \\
2 & -1 \\
1 & 1
\end{bmatrix} = \begin{bmatrix}
6 & -2 \\
-2 & 3
\end{bmatrix}.
\end{equation*}
This means that
\begin{equation*}
(A^TA)^{-1} = \frac{1}{14}\begin{bmatrix}
3 & 2 \\
2 & 6
\end{bmatrix}.
\end{equation*}
Hence,
\begin{equation*}
\begin{split}
P = A(A^TA)^{-1}A^T &= \frac{1}{14}\begin{bmatrix}
-1 & 1 \\
2 & -1 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
3 & 2 \\
2 & 6
\end{bmatrix}
\begin{bmatrix}
-1 & 2 & 1 \\
1 & -1 & 1
\end{bmatrix} \\
&= \frac{1}{14}\begin{bmatrix}
-1 & 1 \\
2 & -1 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 4 & 5 \\
4 & -2 & 8
\end{bmatrix} \\
&= \frac{1}{14}\begin{bmatrix}
5 & -6 & 3 \\
-6 & 10 & 2 \\
3 & 2 & 13
\end{bmatrix}.
\end{split}
\end{equation*}
So the orthogonal projection of $\mathbf{b} = (1,2,0)$ onto the column space of $A$ is
\begin{equation*}
\frac{1}{14}\begin{bmatrix}
5 & -6 & 3 \\
-6 & 10 & 2 \\
3 & 2 & 13
\end{bmatrix}
\begin{bmatrix}
1 \\
2 \\
0 \\
\end{bmatrix} = \frac{1}{14}\begin{bmatrix}
-7 \\
14 \\
7
\end{bmatrix} = \begin{bmatrix}
-1/2 \\
1 \\
1/2
\end{bmatrix}.
\end{equation*}
Not sure what is meant by the "best solution" to the system. Any would help would be great!!!
|
As you said $A\mathbf{u} = \mathbf{b}$ is an inconsistent system that has no solution. So the best you can hope for is to find a $\mathbf{u}$ that makes $A\mathbf{u}$ "close to" $\mathbf{b}$. Or to say the same thing you can try to make $A\mathbf{u} - \mathbf{b}$ "as close to $\mathbf{0}$ as possible".
So the problem boils down to minimizing the size of $A\mathbf{u} - \mathbf{b}$ i.e. $||A\mathbf{u} - \mathbf{b}||$. Though in practice it is more convenient to minimize $F(\mathbf{u}) = ||A\mathbf{u} - \mathbf{b}||^2$ instead.
Now if you set $\mathbf{u} = \begin{bmatrix}x \\ y \end{bmatrix}$ you can multiply things out to get:
$$
A\mathbf{u} - \mathbf{b} = \begin{bmatrix} -x + y - 1 \\ 2x - y - 2 \\ x + y \end{bmatrix}
$$ So that
$$
F(x, y) = ||A\mathbf{u} - \mathbf{b}||^2 = (-x + y -1)^2 + (2x - y - 2)^2 + (x + y)^2
$$
A standard way to try to minimize a multivariable function like $F(x, y)$ is to take the partial derivatives w.r.t. to $x$ and $y$ and set the resulting equations to $0$:
$$
\partial_x F(x, y) = 12x - 4y - 6 = 0
$$
$$
\partial_y F(x, y) = -4x + 6y + 2 = 0
$$
And solving these gives you $x = \frac{1}{2}$ and $y = 0$. If you want to convince yourself that these values of $x$ and $y$ do in fact minimize $F(x, y) = ||A\mathbf{u} - \mathbf{b}||^2$, you can use the Second Partial Derivative Test:
$$
\partial_{xx} F(x, y) = 12
$$
$$
\partial_{xy} F(x, y) = -4 = \partial_{yx} F(x, y)
$$
$$
\partial_{yy} F(x, y) = 6
$$ And note that the matrix of these second partials i.e. the Hessian of $F(x, y)$
$$
\begin{bmatrix} \partial_{xx} F(x, y) &\partial_{xy} F(x, y) \\
\partial_{yx} F(x, y) &\partial_{yy} F(x, y) \end{bmatrix} = \begin{bmatrix} 12 &-4 \\
-4 &6 \end{bmatrix}
$$ has positive determinant $56 > 0$ and furthermore, at $(\frac{1}{2}, 0)$ the value of $\partial_{xx} F(x, y)$ is always positive: $12 > 0$. So the Second Partial Derivative Test says that $(\frac{1}{2}, 0)$ does indeed minimize $F(x, y) = ||A\mathbf{u} - \mathbf{b}||^2$ and hence
$$
\mathbf{u} = \begin{bmatrix}\frac{1}{2} \\ 0 \end{bmatrix}
$$ is the best solution to our linear system.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Decompose $33+11\sqrt{-7}$ into irreducible integral elements of $\mathbb{Q}(\sqrt{-7})$ This is exercise I.3.1 in Neukrich's Algebraic Number Theory. Writing
$$ 33+11\sqrt{-7} = 11(3+\sqrt{-7}), $$
I have managed (by trying small values) to find
$$ 11 = (2+\sqrt{-7})(2-\sqrt{-7}) $$
and
$$ 3+\sqrt{-7} = 2 \left(\frac{3+\sqrt{-7}}{2}\right) = \left(\frac{1+\sqrt{-7}}{2}\right)\left(\frac{1-\sqrt{-7}}{2}\right)\left(\frac{3+\sqrt{-7}}{2}\right). $$
This shows that $(2\pm\sqrt{-7})$ and $\frac12(1\pm\sqrt{-7})$ are irreducible (because they are prime elements?) but I have a hard time continuing.
How does one determine whether an element (e.g. $\frac12(3+\sqrt{-7})$ here) is irreducible?
It seems that showing that an element is prime is easy, but what about irreducible elements that are not prime?
Also, how does one find the factorization of a reducible element in general (other than trial and error)?
|
I find that simplifying the notation helps with the calculations. This is what I did:
Since $-7 \equiv 1 \hspace{2pt}\mathrm{mod}\hspace{2pt} 4$ the ring of integers is $\mathbf{Z}[\alpha]$, where $\alpha :=\tfrac{1}{2}(1 + \sqrt{-7})$, and we know that $\{1, \alpha\}$ is an integral basis and that $\alpha^2 -\alpha +2 =0$. The norm form is $$\mathrm{Nm}(x+y\alpha) = x^2 +xy + 2y^2.$$ Starting with $33 + 11\sqrt{-7} = 11\cdot(3 + \sqrt{-7})$ we find that $1 + 2\alpha$ has norm equal to $11$, so in particular it is irreducible. Thus $11 = (1+2\alpha)\cdot(3-2\alpha)$. On the other hand, $3 + \sqrt{-7} = 2 + 1 + \sqrt{-7} = 2 + 2\alpha = 2\cdot(1+\alpha)$. We observe that $\alpha$ has norm $2$, so that $2 = \alpha \cdot (1-\alpha)$. The norm of $1 + \alpha$ is $4$, so we check if one of the elements with norm $2$ is a factor. Luckily, $1 + \alpha = -(1-\alpha)^2$. Summarizing:
\begin{equation*}
33 + 11\sqrt{-7} = -(1 + 2\alpha)\cdot(3-2\alpha)\cdot (\alpha)\cdot (1-\alpha)^3 .
\end{equation*}
|
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|
Euclidean distance of two uniform random variables Two random variables $X$ and $Y$ are uniformly distributed, the pdfs of which are given by $f_{X}\left(x\right) = f_{Y}\left(y\right) = 1/r$. I am trying to obtain $Z = \sqrt{X^2 + Y^2}$.
I tried the approach shown below, but I want to avoid having the $\tan^{-1}$ term in $f_{Z}\left(z\right)$.
Can anybody help me in finding an easier (algebraic) form of $f_{Z}\left(z\right)$? Further, it will be most desired that the resulting $f_{Z}\left(z\right)$ ends up with a "known" distribution (e.g., Rayleigh).
My approach:
The cdf of $Z$ is found as
\begin{align}\label{eq_F_L}
F_{Z}\left( z \right) &= \mathbb{P} \left( Z \le z \right)\nonumber\\
&= \mathbb{P} \left(\sqrt{X^2 + Y^2} \le z\right)\nonumber\\
&= \displaystyle \int_{0}^{r} \mathbb{P} \left(Y \le \sqrt{Z^2 - X^2} \right) f_{X}\left(x\right) \text{d}x\nonumber\\
&\stackrel{(a)}{=} \displaystyle \frac{1}{r^2} \int_{0}^{r} \sqrt{z^2 - x^2} \text{d}x\nonumber\\
&= \frac{1}{r^2} \left[ \frac{1}{2} \left( x \sqrt{z^2 - x^2} + z^2 \tan^{-1}\left( \frac{x}{\sqrt{z^2 - x^2}} \right) \right) \right]_{0}^{r}\nonumber\\
&= \frac{1}{2r^2} \left[ r \sqrt{z^2 - r^2} + z^2 \tan^{-1}\left( \frac{r}{\sqrt{z^2 - r^2}} \right) \right].
\end{align}
(a) follows from $F_{Y}\left(y\right) = \int f_{Y}\left(y\right) \text{d}y = y/r$.
Then the pdf of of $Z$ can be identified as
\begin{align}\label{eq_f_L_proof}
f_{Z}\left( z \right) &= \frac{\text{d}}{\text{d}z} F_{Z}\left( z \right)\nonumber\\
&= \frac{z}{r^2} \tan^{-1} \left( \frac{r}{\sqrt{z^2 - r^2}} \right).
\end{align}
|
Looking to the problem geometrically, it is clear that the pdf /cdf cannot but be defined piecewise
and it is easy to derive the pdf from the area of the segment of the circular annulus between $z$ and $z+dz$ intercepted by the square,
divided by the area of the whole square ($r^2$), i.e.
$$
p(z\,;\,r) = {1 \over {r^{\;2} }}\left\{ {\matrix{
{{\pi \over 2}\,z} & {0 \le z < r} \cr
{\left( {{\pi \over 2}\, - 2\arccos \left( {{r \over z}} \right)} \right)z} & {r \le z \le \sqrt 2 \,r} \cr
} } \right.
$$
giving the cdf as
$$
\eqalign{
& P(z\,;\,r)
= \left[ {0 \le z \le \sqrt 2 \,r} \right]\left( {{\pi \over 4}\left( {{z \over r}} \right)^{\;2} - 2\left[ {r \le z} \right]{1 \over {r^{\;2} }}\int_{t = r}^z {\arccos \left( {{r \over t}} \right)tdt} } \right) = \cr
& = \left[ {0 \le {z \over r} \le \sqrt 2 } \right]\left( {{\pi \over 4}\left( {{z \over r}} \right)^{\;2} - 2\left[ {1 \le {z \over r}} \right]\int_{t/r = 1}^{z/r}
{\arccos \left( {{1 \over {t/r}}} \right)\left( {{t \over r}} \right)d\left( {{t \over r}} \right)} } \right) = \cr
& = \left[ {0 \le {z \over r} \le \sqrt 2 } \right]\left( {{z \over r}} \right)^{\;2} \left( {{\pi \over 4} - \left[ {1 \le {z \over r}} \right]\left( {\arccos \left( {{1 \over {z/r}}} \right)
- {1 \over {z/r}}\sqrt {1 - \left( {{1 \over {z/r}}} \right)^{\;2} } } \right)} \right) \cr}
$$
where the square brackets indicate the Iverson bracket.
|
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|
Help on residue: $3^x + 22^y \equiv 15^z \equiv 15 \pmod{40} $ I am reading this note (click here and go to page 1994 for detail, in the proof of lemma 8), and found-
$$3^x + 22^y \equiv 15^z \equiv 15 \pmod{40} $$
now, I can derive $3^x + 22^y \equiv 15^z \ \pmod{40} $ but I cann't figure out how $3^x + 22^y \equiv 15^z\pmod{40}$ is $ \equiv 15 \pmod{40}$.
How to prove that the congruence has residue 15?
|
If the assumptions of your prior question hold here, then we know that $z$ is odd.
But for odd $z$, $15^z\equiv z \pmod {40}$.
To see this, We note that $15^2=225\equiv 25\pmod {40}$ and $$15^3\equiv 15\times 25\equiv 15\pmod {40}$$ Then $$15^4=15^3\times 15\equiv 15^2\equiv 25 \pmod {40}$$ and $$15^5\equiv 15^4\times 15\equiv 25\times 15\equiv 15 \pmod {40}$$ and so on.
By induction, we deduce that $15^{2n}\equiv 25\pmod {40}$ and $15^{2n+1}\equiv 15 {\pmod {40}}$.
|
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|
Evaluating a Trigonometric Integral without Substitutions I have been tasked with evaluating the integral $$I=\int\frac{\sin(2x)+\sin(4x)-\sin(6x)}{\cos(2x)+\cos(4x)+\cos(6x)+1}dx$$ After substituting first $u=2x$ and then $v=\cos(u)$ and a messy partial fraction decomposition, I get the answer $$\frac{4 \log(\cos(x)) - 2 \log(1 - 2 \cos(2 x)) + 3 \log(\cos(2 x))}{6} + C$$
But given how similar the numerator is to the derivative of the denominator, I suspect there is a much shorter way of going about this involving the identity $$I=\log(\cos(2x)+\cos(4x)+\cos(6x)+1) + \int\frac{3\sin(2x)+5\sin(4x)+5\sin(6x)}{\cos(2x)+\cos(4x)+\cos(6x)+1}dx$$ or something similar and some carefully chosen trigonometric identities. How do I proceed?
|
First express the integrand in the form
$$\displaystyle\frac{4\sin 3x \sin 2x \sin x}{4\cos 3x \cos 2x \cos x}.$$
For example, the denominator requires $\cos 2x+\cos 4x\equiv 2\cos 3x \cos x$, $\cos 6x+1\equiv 2\cos^2 3x$, and $\cos 3x+\cos x\equiv 2\cos 2x\cos x$.
Thus the integrand is $\tan 3x\tan 2x\tan x$.
What helps now is the little-known identity
$$\tan 3x\tan 2x\tan x\equiv \tan 3x-\tan 2x-\tan x.$$
This starts with
$$\sin 3x\sin 2x\sin x
\equiv \sin 3x(\cos 2x\cos x-\cos 3x)$$
$$\equiv \sin3x\cos 2x\cos x-\cos3x(\sin 2x\cos x+\cos 2x\sin x),$$
and so on.
Thus the result is
$$I=\ln(\cos x)+\frac12\ln(\cos 2x)-\tfrac13\ln(\cos 3x)+c.$$
The identity $\cos x(1-2\cos 2x)\equiv\cos 3x$ implies the equivalence of this answer and the one given above.
The identity $\tan(a+b)\equiv\displaystyle\frac{\tan a+\tan b}{1-\tan a\tan b}$ yields
$$\tan a\tan b\tan(a+b)=\tan(a+b)-\tan a-\tan b,$$
so this provides an easier route to the main identity.
|
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|
Inverse of $y = Ax^3 + Bx + C$ I know there already is a question and a bunch of excellent answers in How to find the inverse of $y=x^3-5x^2+3x+c$.
But here I am having a more general (and thus complicated) form to invert. I am stuck in obtaining the inverse, $g^{-1}$, of a polynomial
\begin{align}
y = g\left(x\right) = -\frac{x^3}{8r} - r\left(1 + \frac{1}{\sqrt{2}}\right)x + 2\sqrt{2}r^2, {\rm{~~~~}} 0 \le x \le \sqrt{2}r
\end{align}
where $r > 0$ is a constant and a real number.
The function $g$ is certainly invertible as being strictly decreasing when drawn:
Question:
(1) I would like to ask you guys if there would be a way to find a closed-form expression for $x = g^{-1}\left(y\right)$.
(2) Also, any approximation of $y$ for making it easier to be inverted will be most welcome as well.
Thanks a lot guys. I love you all!
|
Consider that you need to solve for $x$ the cubic equation
$$ -\frac{x^3}{8r} - r\left(1 + \frac{1}{\sqrt{2}}\right)x + \left(2\sqrt{2}r^2-y\right)=0$$
Applying the method for cubic equations, you would find
$$\Delta=-\frac{27 \left(2 \sqrt{2} r^2-y\right)^2}{64 r^2}-\frac{1}{2}
\left(1+\frac{1}{\sqrt{2}}\right)^3 r^2$$ and, since $\Delta <0$, then only one real root. On the other hand $$ p=8 \left(1+\frac{1}{\sqrt{2}}\right) r^2\qquad \text{and}\qquad q=-8 r \left(2 \sqrt{2} r^2-y\right)$$
Using the hyperbolic solution for one real root
$$x=4 \sqrt{\frac{2+\sqrt{2}}{3} } \sinh \left(\frac{1}{3} \sinh
^{-1}\left(\frac{3\sqrt{3} \left(2 \sqrt{2} r^2- y\right)}{2
\left(2+\sqrt{2}\right)^{3/2} r^2}\right)\right)\,r$$
Edit
It would have been simpler to let $x= X r \sqrt 2$ and $y= 2Y {r^2}\sqrt 2$ which make the equation to be dimensionless and rescaled
$$\sqrt{2} X^3+4 \left(1+\sqrt{2}\right) X-8 \sqrt{2} (1-Y)=0$$
the solution of which being
$$X=2 \sqrt{\frac{2\left(2+\sqrt{2}\right)}{3} } \sinh \left(\frac{1}{3} \sinh
^{-1}\left(\frac{3 \sqrt{6} }{\left(2+\sqrt{2}\right)^{3/2}}(1-Y)\right)\right)$$
|
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|
joint and conditional probability This is table of conditional probability with $X$ is for red coin, and $Y$ for blue coin.
$$\begin{array}{rc}
& X_{\text{red}} \\
Y_{\text{blue}} & \begin{aligned}
\ P (X \mid Y)&& X=\text{Tail} && X=\text{Head}
\\Y=\text{Head} &&\frac{6}{10} && \frac{4}{10}
\\ Y=\text{Tail} &&\frac{2}{10}&&\frac{8}{10}
\end{aligned}
\end{array}$$
It is known that $P(Y_{\text{blue}} = \text{Head} )=0.45$ and $P(X_{\text{red}} = \text{Tail})=0.55$
I'm trying to find the $p(x,y)$ joint probability of $x$ and $y$
such as
$$P(X=\text{Head},Y=\text{Tail})=P(Y=\text{Tail})\cdot P(X = H \mid Y= T)=0.55 \cdot 0.2 = 0.11$$
and
$$P(X= H,\, Y=H)=P(Y=H)\cdot P(X=H \mid Y=H)=0.6 \cdot 0.45=0.27$$
$$\begin{array}{rc}
& X_{\text{red}} \\
Y_{\text{blue}} & \begin{aligned}
\ P (X ,\, Y)&& X=\text{Tail} && X=\text{Head}
\\Y=\text{Head} &&\frac{27}{100} && \frac{18}{100}
\\ Y=\text{Tail} &&\frac{11}{100}&&\frac{44}{100}
\end{aligned}
\end{array}$$
is this right?
Also I want to know why for $p(x \mid y)$ , probability of $ Y(\text{Head}) =\frac{6}{10} + \frac{4}{10} = 1 $ and $Y(\text{Tail}) = 1$ but for $X( \text{Head})= \frac{4}{10} + \frac{8}{10}$ not equal to $1$?
|
\begin{align}
P(X=H, Y=H) &= P(Y=H) P(X=H|Y=H)\\
&= 0.45 \cdot \color{red}{0.4}
\end{align}
For the table of conditional probability.
Summing up the rows give you $1$, because given that $Y$ has occured, and the probability of possible outcome of $X$ must sum to $1$.
$$\sum_xP(X=x|Y=y)=1$$
However, for the columns,
$$\sum_y P(X=x|Y=y),$$
there is no reason for it to sum to $1$.
|
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|
Finding the length of a Curve The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney. It is relatively
early in the book, so I would expect the integration to be easy.
Find the length of the curve:
$$ 9x^2 = 4y^3$$
from $(0,0)$ to $\left(2\sqrt{3},3\right)$.
Answer:
The formal for the length of a curve is:
$$ L = \int_a^b \sqrt{ 1 + {f'(x)}^2 } \, dx $$
In this case, we have:
\begin{align*}
a &= 0 \\
b &= 2\sqrt{3} \\
y^3 &= \frac{9x^2}{4} \\
f(x) &= \left( \frac{9x^2}{4} \right) ^ {\frac{1}{3}} \\
f'(x) &= \frac{1}{3} \left( \frac{18x}{4} \right) \left( \frac{9x^2}{4} \right) ^ {-\frac{2}{3}} \\
f'(x) &= \left( \frac{3x}{2} \right) \left( \frac{9x^2}{4} \right) ^ {-\frac{2}{3}} \\
\end{align*}
\begin{align*}
L &= \int_0^{2\sqrt{3}} \sqrt{ 1 + \left( \frac{9x^2}{4} \right) \left( \frac{9x^2}{4} \right) ^ {-\frac{4}{3}} } \, dx \\
L &= \int_0^{2\sqrt{3}} \sqrt{ 1 + \left( \frac{9x^2}{4} \right) ^ {-\frac{1}{3}} } \, dx \\
L &= \int_0^{2\sqrt{3}} \sqrt{ 1 + \left( \frac{4}{9x^2} \right) ^ {\frac{1}{3}} } \, dx \\
\end{align*}
The book's answer is $$ \frac{14}{3} $$
Using an online integral calculator, my integral did not match. What did I do wrong?
I used the following website to do the integration:
https://www.integral-calculator.com/
Their answer is:
$$ \left( \frac{4^\frac{1}{3}}{9^\frac{1}{3}x^{\frac{2}{3}}} + 1 \right) ^ \frac{3}{2} x + C $$
|
It's possible to solve the integral but it's quite unwieldy if you don't use the right substitution.
$$\begin{aligned}
I&=\int_0^{2\sqrt{3}}\sqrt{1+\left(\frac{4}{9x^2}\right)^{\frac{1}{3}}}\ \mathrm{d}x\\
&=\int_1^4\sqrt{u}\ \mathrm{d}u\\
&=14/3
\end{aligned}$$
With the substitution $u=\left(\frac32x\right)^{2/3}+1$.
|
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|
Determine an explicit form for the generating function and a closed form given a recurrence Let $f_n = 4f_{n-1} - 4f_{n-2}$ be a recurrence relation for $f$ with initial conditions $f_0 = 1$ and $f_1 = 4$. I want to determine an explicit form for the generating function of $f$, $F(x)$ as well as a closed form for $f_n$. I'd appreciate someone verifying if my attempt at a solution is correct.
Rearranging the recurrence relation yields $f_n - 4f_{n-1} + 4f_{n-2} = 0$ and then I can take the generating function of both sides to get:
$$\sum_{n \geq 2}(f_n - 4f_{n-1} + 4f_{n-2})x^n=0.$$
Splitting this up yields:
$\begin{align*} 0 &= \sum_{n \geq 2}f_nx^n - 4\sum_{n \geq 2}f_{n-1}x^n + 4\sum_{n \geq 2}f_{n-2}x^n\\
&= \sum_{n \geq 0}f_nx^n - 4x - 1 - 4\sum_{n \geq 2}f_{n-1}x^n + 4\sum_{n \geq 2}f_{n-2}x^n\\
&= F(x) - 4x - 1 -4x\sum_{n \geq 2}f_{n-1}x^{n-1} + 4x^2\sum_{n \geq 2}f_{n-2}x^{n-2}\\
&= F(x) - 4x - 1 -4x(F(x) - 1) + 4x^2F(x).\\
\end{align*}$
Now we can solve for $F(x)$ which yields $F(x) = \frac{1}{(1-2x)^2}$ so this is the generating function. But using the binomial theorem for negative powers we have:
$$F(x) = (1- 2x)^{-2} = \sum_{k=0}^{\infty}\binom{2 + k - 1}{k}(-1)^k(-2x)^k.$$
Extracting the $n$th coefficient yields $f_n = \binom{n + 1}{n} \cdot 2^n$.
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Define the generating function $F(z) = \sum_{n \ge 0} f_n z^n$, multiply the recurrence shifted by $2$ by $z^n$ and sum over $n \ge 0$, recognize some sums:
$\begin{align*}
\sum_{n \ge 0} f_{n + 2} z^n
&= 4 \sum_{n \ge 0} f_{n + 1} z^n - 4 \sum_{n \ge 0} f_n z^n \\
\frac{F(z) - f_0 - f_1 z}{z^2}
&= 4 \frac{F(z) - f_0}{z} - 4 F(z)
\end{align*}$
Solve for $F(z)$:
$\begin{align*}
F(z)
&= \frac{1}{(1 - 2 z)^2} \\
&= \sum_{n \ge 0} (-1)^n \binom{-2}{n} (2 z)^n \\
&= \sum_{n \ge 0} \binom{n + 2 - 1}{2 - 1} 2^n z^n
\end{align*}$
Extracting the coefficient of $z^n$ you get:
$\begin{align*}
f_n
&= [z^n] F(z) \\
&= \binom{n + 1}{1} \cdot 2^n \\
&= (n + 1) \cdot 2^n
\end{align*}$
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"timestamp": "2023-03-29T00:00:00",
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|
Find the min value of $\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$, if $x\ne y \ne z$ and $x,y,z\in {1,2,3,4,5}$ My answer is $\frac{5}{29}$, I just use logic to substitute numbers in the expression, but I can't prove my answer If this expression be minimum then the denominator should be the greatest, so I just let x=5 and I need to let $\frac{1}{y+\frac{1}{z}}$ be the greatest so just need this denominator be smallest so y=1 and z=4, this is what I thought.
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$$
f(x,y,z) = \frac{1}{x+\frac{1}{y+\frac{1}{z}}}
$$
is decreasing in $x$ and $z$, and increasing in $y$. Since $x,z$ are distinct integers we have either
$$
x < z \implies x \le 4, y \ge 1, z \le 5 \implies f(x, y, z) \ge f(4, 1, 5) = \frac{6}{29}
$$
or
$$
z < x \implies x \le 5, y \ge 1, z \le 4 \implies f(x, y, z) \ge f(5, 1, 4) = \frac{5}{29}
$$
so that the minimal value is the smaller of these two, which is $f(5, 1, 4) = \frac{5}{29}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Generating function of ordered partitions What is exactly generating function of ordered partitions and how can I get number of ordered partitions from that?
Example:
$$ 4 = 1+1+1+1 \\ = 2+2 \\ = 1+1+2 \\ = 1+2+1 \\ = 2+1+1 \\ = 1+3 \\ = 3+1 \\ = 4 $$
so we have $8$ partitions. I was thinking about exponential generating function:
$$(1+t+\frac{t^2}{2!} + ...)(1+\frac{t^2}{2!} +\frac{t^4}{4!} +...)...(1+\frac{t^n}{n!} + ... ) = e^x e^{2x} e^{3x} \cdots e^{nx} = e^{n(n+1)/2} = \sum_{k \ge 0}\frac{\left(\frac{n(n+1)}{2}x\right)^k}{k!} $$
so the number of ordered partitions of $n$ is
$$\left(\frac{n(n+1)}{2}\right)^n $$
bot for $n=4$ it is
$$10^4$$ It seems to be completely wrong.
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We can do this by generating functions which you want to.
For partition $n$,
If we separate $n$ into $n$ numbers, we can generate the function $x+x^2+\cdots+x^n$ and the coefficient of $x^n$ is the number of combination.
If we separate $n$ into $\left(n-1\right)$ numbers, we can generate the function $\left(x+x^2+\cdots+x^n\right)^2$ and the coefficient of $x^n$ is the number of combination.
...
If we separate $n$ into $2$ numbers, we can generate the function $\left(x+x^2+\cdots+x^n\right)^{n-1}$ and the coefficient of $x^n$ is the number of combination.
If we separate $n$ into $1$ number, we can generate the function $\left(x+x^2+\cdots+x^n\right)^n$ and the coefficient of $x^n$ is the number of combination.
Therefore the total number of combination of partition of $n$ is the coefficient of $x^n$ of this function $$\small f:\left(x+x^2+\cdots+x^n\right)+\left(x+x^2+\cdots+x^n\right)^2+\cdots+\left(x+x^2+\cdots+x^n\right)^{n-1}+\left(x+x^2+\cdots+x^n\right)^n$$
However, this function $$\small g:\left(x+x^2+\cdots+x^n+\cdots\right)+\left(x+x^2+\cdots+x^n+\cdots\right)^2+\cdots+\left(x+x^2+\cdots+x^n+\cdots\right)^n+\cdots$$ has the same coefficient of $x^n$ as $f$. Then, we can simplify $g$ like that:
\begin{align}\small\dfrac{x}{1-x}+\left(\dfrac{x}{1-x}\right)^2+\cdots+\left(\dfrac{x}{1-x}\right)^n+\cdots&\small=\dfrac{\dfrac{x}{1-x}}{1-\dfrac{x}{1-x}}\\&\small=\dfrac{x}{1-2x}\\&\small=x\left(1+\left(2x\right)+\left(2x\right)^2+\cdots+\left(2x\right)^{n-1}+\cdots\right)\\&\small=x+2x^2+4x^3+8x^4+\cdots+2^{n-1}x^n+\cdots\end{align}
Therefore, the number of partition of $n=2^{n-1}$
The cases you have shown is $n=4$, which the number of partition is $8$, same as counting.
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Find $y$ in $\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$ I would like to solve this equation for $y$. Any tips? It seems like you cant really do it analytically?
$$\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$$
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You can factor out $2$ at the second summand.
$$\sqrt{\underbrace{4+(y-6)^2}_{=5}}+2\cdot \sqrt{\underbrace{4+\left(\frac{y-3}2\right)^2}_{=5}}=\sqrt{5}+2\cdot\sqrt{5}$$
Now we see that the following equations has to be true at the same time.
*
*$(y-6)^2=1 \Rightarrow y_1=7,y_2=5$
*$\left(\frac{y-3}2\right)^2=1\Rightarrow y_1=5,y_2=1$
Thus the solution is $y=...$
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