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Sides $\frac{|b - c|}{\sqrt{(b^2 + 1)(c^2 + 1)}}, \frac{|c - a|}{\sqrt{(c^2 + 1)(a^2 + 1)}}, \frac{|a - b|}{\sqrt{(a^2 + 1)(b^2 + 1)}}$ of a triangle.
Prove that for all pairwise distinct $a, b, c \in \mathbb R$, $$\frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}}, \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}, \frac{|a - b|}{\sqrt{a^2 + 1}\sqrt{b^2 + 1}}$$ are always sides of a triangle.
For all $\triangle MNP$ where $m = MP, n = PM, p = MN$, we have that $$n + p > m, p + m > n, m + n > p$$
We need to obtain that $$\frac{|a - b|}{\sqrt{a^2 + 1}\sqrt{b^2 + 1}} + \frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}} > \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}$$
First attempt, we have that $$\frac{(a - b)^2}{|a - b|\sqrt{a^2 + 1}\sqrt{b^2 + 1}} + \frac{(b - c)^2}{|b - c|\sqrt{b^2 + 1}\sqrt{c^2 + 1}}$$
$$ \ge \frac{(c - a)^2}{\sqrt{b^2 + 1} \cdot \left(|b - c|\sqrt{c^2 + 1} + |a - b|\sqrt{a^2 + 1}\right)}$$
and $$\left(|b - c|\sqrt{c^2 + 1} + |a - b|\sqrt{a^2 + 1}\right)^2 \le \left[(b - c)^2 + (a - b)^2\right] \cdot (c^2 + a^2 + 2)$$
It is needed to prove that $$\sqrt{\left[(b - c)^2 + (a - b)^2\right] \cdot (b^2 + 1)(c^2 + a^2 + 2)} < |c - a|\sqrt{c^2 + 1}\sqrt{a^2 + 1}$$
Second attempt, it is to prove that $$|a - b|\sqrt{c^2 + 1} + |b - c|\sqrt{a^2 + 1} > |c - a|\sqrt{b^2 + 1}$$
According to the Cauchy - Schwarz inequality, we have that $$\left(|a - b|\sqrt{c^2 + 1} + |b - c|\sqrt{a^2 + 1}\right)^2 \ge 2|(a - b)(b - c)|\sqrt{(c^2 + 1)(a^2 + 1)}$$
What needs to be established is $$2|(a - b)(b - c)|\sqrt{(c^2 + 1)(a^2 + 1)} > (c - a)^2(b^2 + 1)$$
Third attempt, let $a = \tan\alpha, b = \tan\beta, c = \tan\gamma$ $\left(\alpha, \beta, \gamma \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\right)$, it could be easily deducted that $$\frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}} = \frac{|\tan\gamma - \tan\alpha|}{\sqrt{\tan\gamma^2 + 1}\sqrt{\tan\alpha^2 + 1}} = \frac{\left|\dfrac{\sin(\gamma - \alpha)}{\cos\gamma\cos\alpha}\right|}{\dfrac{1}{\cos\gamma\cos\alpha}} = \pm\sin(\gamma - \alpha)$$
For all of the above attempts, there need to be considered multiple cases of $a, b, c$, whether they're positive and negative, and their arrangements from littlest to greatest.
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Let $a>b>c\geq0$.
Thus, easy to see that $$\frac{a-c}{\sqrt{(a^2+1)(c^2+1)}}>\frac{a-b}{\sqrt{(a^2+1)(b^2+1)}}$$ and
$$\frac{a-c}{\sqrt{(a^2+1)(c^2+1)}}>\frac{b-c}{\sqrt{(b^2+1)(c^2+1)}}$$ because $$a-c>\frac{a}{b}(b-c)$$ and it's enough to prove that:
$$\frac{a-b}{\sqrt{(a^2+1)(b^2+1)}}+\frac{b-c}{\sqrt{(b^2+1)(c^2+1)}}>\frac{a-c}{\sqrt{(a^2+1)(c^2+1)}}$$ or
$$(a-b)\sqrt{c^2+1}+(b-c)\sqrt{a^2+1}>(a-c)\sqrt{b^2+1}$$ or
$$(a-b)(b-c)\sqrt{(a^2+1)(c^2+1)}>(a-b)(b-c)(ac+1),$$ which is true by C-S.
The equality in C-S here does not occur because our variables are different.
Since the given expressions are not changed after substitution $a$ at $-a$, $b$ at $-b$ and $c$ at $-c$,
it remains to assume $a>b\geq0>c,$ which we can end by the similar way.
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Proving $\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin\frac{ab\pi}{2}$ for $a,b\in\mathbb{N}$? I was playing with the graphs of $f:\,\mathbb{R}\to\mathbb{R},\, f(x)=\sin axt\pi$ and $g:\,\mathbb{R}\to\mathbb{R},\, g(x)=\sin xt\pi\sin at\pi$.
What I conjectured is the following:
If $t=\frac{m}{2}$ where $m$ is $1$ above a multiple of $4$ (or if $m$ is even), then $f=g$ for all $a,x\in\mathbb{N}$.
I'm the most interested in the simplest non-trivial case, which is $m=1$:
$$\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin \frac{ab\pi}{2}\quad a,b\in\mathbb{N}.$$
If this is true, how could it be rigorously proved?
I tried rewriting $\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}$ as
$$\frac{1}{4}\left(e^{\frac{\pi i}{2}(a-b)}+e^{-\frac{\pi i}{2}(a-b)}-e^{\frac{\pi i}{2}(a+b)}-e^{-\frac{\pi i}{2}(a+b)}\right)$$
and rewriting $\sin\frac{ab\pi}{2}$ as
$$\frac{1}{2i}\left(e^{\frac{ab\pi i}{2}}-e^{-\frac{ab\pi i}{2}}\right),$$
but that doesn't seem to help when I want to couple $a$ and $b$ so that there is a product of them in both expressions.
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Suppose $a$ or $b$ is even. Without loss of generality, $a=2k$. Then
$$\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{b\pi}{2}\right)=\sin\left(\frac{2k\pi}{2}\right)\sin\left(\frac{b\pi}{2}\right)=\sin\left(k\pi\right)\sin\left(\frac{b\pi}{2}\right)=0$$
and
$$\sin\left(\frac{ab\pi}{2}\right)= \sin\left(\frac{2kb\pi}{2}\right)=\sin\left(kb\pi\right)=0$$
Suppose $a$ and $b$ are both odd. Splitting into further cases, suppose
$$a\equiv b\equiv 1\ (\text{mod }4)$$
Then
$$a=4k+1\text{ and }b=4j+1$$
This implies
$$\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{b\pi}{2}\right)=\sin\left(\frac{(4k+1)\pi}{2}\right)\sin\left(\frac{(4j+1)\pi}{2}\right)=\sin\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right)=1$$
and
$$\sin\left(\frac{ab\pi}{2}\right)= \sin\left(\frac{(4k+1)(4j+1)\pi}{2}\right)=\sin\left(\frac{\pi}{2}\right)=1$$
The cases where
$$a\equiv b\equiv 3\ (\text{mod }4)$$
and
$$a\equiv 1\ (\text{mod }4)$$
$$b\equiv 3\ (\text{mod }4)$$
can be worked out in a similar manner.
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|
finding bounds for $\int_0^X\lfloor x^2\rfloor \, dx$ I am trying to find bounds for:
$$I=\int_0^X\lfloor x^2\rfloor \, dx$$ and through integrating two different ways I ended up with the sums:
$$I=(X^2+1)^{3/2}-\sum_{i=1}^{X^2+1}\sqrt{i}$$
$$I=X^3-\sum_{i=1}^{X^2}\sqrt{i}$$
I can't verify if either of these are correct firstly, and also is it possible for me to use these to create bounds? Thanks
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$\newcommand{\d}{\text{d}}$Assuming $X$ is an integer, we have :
$\begin{align}
I(X)&=\int_0^X\lfloor x^2\rfloor\d x\\
&=\sum_{n=0}^{X^2-1}\int_{\sqrt{n}}^{\sqrt{n+1}}\lfloor x^2\rfloor\d x\\
&=\sum_{n=0}^{X^2-1}(\sqrt{n+1}-\sqrt{n})n\\
&=\sum_{n=0}^{X^2-1}n\sqrt{n+1}-\sum_{n=0}^{X^2-1}n\sqrt{n}\\
&=\sum_{n=1}^{X^2}(n-1)\sqrt{n}-\sum_{n=1}^{X^2-1}n\sqrt{n}\\
&=X(X^2-1)-\sum_{n=1}^{X^2-1}\sqrt{n}.
\end{align}$
Now, according to Is there any way to approximate a sum of square roots
, we have : $$\frac{2}{3}N^{3/2}\leqslant\sum_{n=1}^N\sqrt{n}\leqslant\frac{2}{3}\left[(N+1)^{3/2}-1\right],$$
which can be used with $N=X^2-1$ to provide : $$X(X^2-1)-\frac{2}{3}(X^3-1)\leqslant I(X)\leqslant X(X^2-1)-\frac{2}{3}(X^2-1)^{3/2}.$$
Now, in the case where $X$ is not necessarily an integer, by assuming that $X\geqslant0$, we have $\lfloor X\rfloor\leqslant X<\lfloor X\rfloor+1$, from which we obtain : $$I(\lfloor X\rfloor)\leqslant I(X)<I(\lfloor X\rfloor+1).$$
In the end, the previous bounds can be used to provide : $$\lfloor X\rfloor(\lfloor X\rfloor^2-1)-\frac{2}{3}(\lfloor X\rfloor^3-1)\leqslant I(X)\leqslant (\lfloor X\rfloor+1)\left[(\lfloor X\rfloor+1)^2-1\right]-\frac{2}{3}\left[(\lfloor X\rfloor+1)^2-1\right]^{3/2}.$$
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Advanced Inequality with Floor Functions For positive real numbers $a,$ $b,$ $c,$ and $d,$ find the minimum value of
$$\left\lfloor \frac{b + c + d}{a} \right\rfloor + \left\lfloor \frac{a + c + d}{b} \right\rfloor + \left\lfloor \frac{a + b + d}{c} \right\rfloor + \left\lfloor \frac{a + b + c}{d} \right\rfloor.$$
I'm not seeing a technique to easily solve this problem; at this moment, I'm thinking a property of some basic inequalities such as Cauchy or AM-GM will help. Other than that, I really have no idea. Solutions?
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By C-S $$\sum_{cyc}\left[\frac{a+b+c}{d}\right]>\sum_{cyc}\frac{a+b+c}{d}-4=\sum_{cyc}a\sum_{cyc}\frac{1}{a}-8\geq16-8=8.$$
Now, make an example for $$\sum_{cyc}\left[\frac{a+b+c}{d}\right]=9.$$
I got that $(a,b,c,d)=(5,5,5,4)$ is valid.
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Prove that $e^x > 1 + (1 + x)\log(1 + x), x > 0$ using power series expansion. Prove that $e^x > 1 + (1 + x)\log(1 + x), x > 0$ using power series expansion.
I am a bit puzzled by this statement because the power series for $\log(1+x)$ only converges iff $|x|<1$. Is the problem sound? Should it be $1>x>0$?
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For $0<x<1$, does the following proof work?
$e^x>1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$ and $\log(1+x)<x-\dfrac{x^2}{2}+\dfrac{x^3}{3}$ and so,
$$1+(1+x)\log(1+x)<1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{3}$$
We observe that $$1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{3}<1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$$
iff
$x^3>x^4$ which is true for $x\in (0,1)$.
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|
Finding the exponential form of $e^{17\pi i/60} + e^{27\pi i/60} + e^{37\pi i /60} + e^{47 \pi i /60} + e^{57 \pi i/60}$
Find the exponential form of the complex number
$$ e^{17\pi i/60} + e^{27\pi i/60} + e^{37\pi i /60} + e^{47 \pi i /60} + e^{57 \pi i/60}$$ with proof.
I know that I have to combine some of the exponents and simplify them, but I am having some trouble with it. For example, $e^{17 \pi i/60} + e^{57 \pi i/60} = e^{17 \pi i/60} (1+e^{40})$ but I'm having trouble connecting it with the other terms.
Can anybody help? Thanks!
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My sol.
We first look at $e^{17 \pi i/60} + e^{57 \pi i/60}$ and represent it in polar form.
We note that $$e^{17 \pi i/60} = \cos \frac{17\pi}{60} + i\sin\frac{17\pi}{60}$$ and $$e^{57 \pi i/60} = \cos \frac{57\pi}{60} + i\sin\frac{57\pi}{60}.$$
Therefore, by adding the cosines and sines we get $$e^{17 \pi i/60} + e^{57 \pi i/60} = 2\cos\frac{37\pi}{60}\cos\frac{\pi}{3} + i(2\sin\frac{37\pi}{60}\cos\frac{\pi}{3})$$ which simplifies to $\text{cis} \frac{37\pi}{60}$ after evaluating the cosines and sines of common angles.
We can simplify $e^{27 \pi i/60} + e^{47 \pi i/60}$ in the same way.
We can first note that $$e^{27 \pi i/60} = \cos \frac{27\pi}{60} + i\sin\frac{27\pi}{60}$$ and $$e^{47 \pi i/60} = \cos \frac{47\pi}{60} + i\sin\frac{47\pi}{60}.$$ Therefore after collecting like terms and simplifying cosines and sines that we know on the top of our heads, we get that $$e^{27 \pi i/60} + e^{47 \pi i/60} = \sqrt{3}(\text{cis}\frac{37\pi}{60}).$$
Finally $e^{37\pi i /60}$ can be simply simplified to $$\text{cis} \frac{37 \pi}{60}.$$
Adding the individual terms, we finalize to $$(\sqrt{3}+2) \text{cis} \frac{37 \pi}{60}.$$
Therefore, $$e^{17\pi i/60} + e^{27\pi i/60} + e^{37\pi i /60} + e^{47 \pi i /60} + e^{57 \pi i/60} = \boxed{(\sqrt{3}+2) e^{\frac{37\pi i}{60}}}.$$
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Special case ($3\times 3$ and $4\times 4$) of USAMO 1998 problem #$4$ This is a special case ($3\times 3$ and $4\times 4$) of USAMO 1998 problem #4.
A $98\times 98$ chess board has the squares colored alternately black and white in the usual way. A move consists of selecting a rectangular subset of the squares (with boundary parallel to the sides of the board) and switching their colors. What is the smallest number of moves required to make all the squares black?
I am trying to obtain optimal moves for $3\times 3$ and $4\times 4$ cases. (I couldn't understand the answers of linked post) I read somewhere that the optimal moves for first one is 2 and for second case is 4. But I tried almost all ways without reaching these numbers. For example for $3\times 3$ one possible solution is as follows:
$$\begin{bmatrix}
\color{red}{1} & 0&1\\
\color{red}{0}&1&0\\
\color{red}{1} & 0&1
\end{bmatrix}\implies \begin{bmatrix}
0 & 0&\color{red}{1}\\
1&1&\color{red}{0}\\
0 & 0&\color{red}{1}
\end{bmatrix}\implies \begin{bmatrix}
0 & 0&0\\
\color{red}{1}&\color{red}{1}&\color{red}{1}\\
0 & 0&0
\end{bmatrix}\implies \begin{bmatrix}
0 & 0&0\\
0&0&0\\
0 & 0&0
\end{bmatrix}$$
which is 3 moves and not 2. What is the best way with 2 moves? and the $4\times 4$ case one possible solution is as follows which is 5 moves:
$$\begin{bmatrix}
\color{red}{1} & 0&1&0\\
\color{red}{0}&1&0&1\\
\color{red}{1} & 0&1&0\\
\color{red}{0}&1&0&1
\end{bmatrix}\implies \begin{bmatrix}
0 & 0&\color{red}{1}&0\\
1&1&\color{red}{0}&1\\
0 & 0&\color{red}{1}&0\\
1&1&\color{red}{0}&1
\end{bmatrix}\implies \begin{bmatrix}
0 & 0&0&\color{red}{1}\\
1&1&1&\color{red}{0}\\
0 & 0&0&\color{red}{1}\\
1&1&1&\color{red}{0}
\end{bmatrix}\implies \begin{bmatrix}
0 & 0&0&0\\
\color{blue}{1}&\color{blue}{1}&\color{blue}{1}&\color{blue}{1}\\
0 & 0&0&0\\
\color{red}{1}&\color{red}{1}&\color{red}{1}&\color{red}{1}
\end{bmatrix}\stackrel{2 rows=2times}{\implies}\begin{bmatrix}
0 & 0&0&0\\
0&0&0&0\\
0 & 0&0&0\\
0 & 0&0&0
\end{bmatrix}$$
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For the $3\times3$ case you could change two middle rows and get all $1$'s (check that). For the $4\times4$ case the possible sequence of moves is "$3$rd row", "$3$rd column", "$1$st row", "$1$st column".
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|
Evaluating double sum $\sum_{k = 1}^\infty \left( \frac{(-1)^{k - 1}}{k} \sum_{n = 0}^\infty \frac{1}{k \cdot 2^n + 5}\right)$
Find $$\sum_{k = 1}^\infty \left( \frac{(-1)^{k - 1}}{k} \sum_{n = 0}^\infty \frac{1}{k \cdot 2^n + 5}\right)$$
So far, I've gotten that the sum of the left is equal to $\log(2),$ meaning we have to evaluate $\displaystyle \sum_{n=0}^{\infty} \frac{\log(2)}{k\cdot2^n+5},$ but I don't know how to proceed. I don't think it's geometric or we can use Partial Fraction Decomposition on it.
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The inner sum is bounded by $\frac{2}{k},$ so the full sum is absolutely convergent, which means we can rearrange and combine terms.
If $m=2^i p$ for $p$ odd, the coefficient of $\frac1{m+5}$ is $$\frac{1}{p}-\left(\frac{1}{2p}+\frac{1}{4p}+\cdots +\frac{1}{2^ip} \right)=\frac{1}{m}.$$
So your sum is: $$\sum_{m=1}^{\infty}\frac{1}{m(m+5)}\tag1$$
Since $$\frac1{m(m+5)}=\frac15\left(\frac1m-\frac1{m+5}\right)$$
$(1)$ is equal to:
$$\frac{1}5\left(\frac{1}1+ \frac{1}2+ \frac{1}3+ \frac{1}4+ \frac{1}5\right)$$
This gives us more generally, for any positive integer $p:$ $$\sum_{k = 1}^\infty \left( \frac{(-1)^{k - 1}}{k} \sum_{n = 0}^\infty \frac{1}{k \cdot 2^n + p}\right)=\frac{1}{p}H_p,$$ where $H_p$ is the harmonic number, $\frac11+\dots+\frac1p.$
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Easier way for probability problem?
Question: Bob takes a math test with $10$ questions. He has a $70\%$ chance of getting each question right. What is the probability that they get exactly $60\%$ on the test? Express your answer as a common fraction.
My bashy way: We know that the successful outcomes (numerator of probability) is $\binom{10}{6} \left(\frac{7}{10}\right)^6 \left(\frac{3}{10}\right)^4$ since we choose $6$ to be correct and the rest $4$ to be wrong. Following in vein, the denominator of our probability is just
$$\sum_{n=1}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}.$$
Is there a slicker way to do this probability problem?
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We know that the successful outcomes (numerator of probability)
look here: Binomial distribution.
The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function: $$\Pr(X=k)={\binom {n}{k}}p^{k}(1-p)^{n-k}$$
Therefore your score is the exact probability of 6 successes in 10 independent trials. Following:
$$\sum_{n=1}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}=\Pr(X=1)+\Pr(X=2)+...+\Pr(X=3)=\Pr(X\ge1)$$
the result of which the probability that Bob has a result greater than or equal to 10%. So that Bob will answer at least one question correctly. Look: 0.999994 which is the opposite of the fact that Bob will answer each question wrong: $$\Pr(\text{all wrong})=\Pr(X=0)=\left(\frac{3}{10}\right)^{10}$$ $$\sum_{n=1}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}=1-\Pr(\text{all wrong})$$ $$\sum_{n=1}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}=1-\left(\frac{3}{10}\right)^{10}$$ $$\left(\frac{3}{10}\right)^{10}+\sum_{n=1}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}=1$$ $$\sum_{n=0}^{10} \binom{10}{n} \left(\frac{7}{10}\right)^n \left(\frac{3}{10}\right)^{10-n}=1$$
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solve : $xy''+2y'+xy=1$ given that $y=\frac{\sin x}{x}$ is a solution for $xy''+2y'+xy=0$, solve $xy''+2y'+xy=1$.
My try:
$\begin{aligned}xy''+2y'+xy&=0\\y''+\frac2xy'+y&=0\\\begin{pmatrix}
\frac{\sin x}x& u\\
\frac{x\cos(x)-\sin(x)}{x^2} & u'
\end{pmatrix}&=\frac2x\\\frac{\sin x }{x}u'-\frac{x\cos(x)-\sin(x)}{x^2}u&=\frac2x\\\left(u\frac{x}{\sin x}\right)'&=\frac{2x}{\sin^2x}\\u\frac{x}{\sin x}&=2\left(-x\cot x+\log\left(\sin x\right)\right)+c\\u&=2\left(-\cos x+\frac{\sin x\log\sin x}x\right)+c\frac{\sin x}x\\\end{aligned}$.
So it we found another solution.
Now let's try to find a specific solution for :
$y''+\frac2xy'+y=\frac1x$
The undetermined coefficient method seems to be inefficient here, does anyone have an efficient solution for this.
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$$y''-\frac{2}{x}y'+y=0~~~(1)$$, if $y_1=\frac{\sin x}{x}$ is a solution of (1), then the other solution $y_2(x)$ is given by
$$y_2(x)=y_1(x) \int \frac{e^{\int p(x) dx}}{y^2_1(x)}dx,~ p(x)=-2/x$$
$$\implies y_2(x)= \frac{\sin x}{x} \int \frac{1}{x^2}\frac{x^2}{\sin ^2 x} dx$$
One may ignor the multiplicative $-$ sign for $y_2(x)$.
$$\implies y_2(x)=\frac{\sin x}{x} \int \csc^2 x dx=-\frac{\cos x}{x}.$$
For solving the inhomogeneous equation $$Y''-\frac{2}{x}Y'+Y=\frac{1}{x}~~~(2)$$
the method of variation of parameter can be used which requires $y_1(x), y_2(x)$, the solution of (2) is given by $$Y(x)=C_1(x) y_1(x)+ C_2(x) y_2(x)~~~(3)$$
where $$C_1(x)=-\int \frac{y_2(x)/x}{W(x)} dx+D_1= \int \cos x dx+D_1=\sin x +D_1.$$
The Wronskian $W(x)$ of $y_1,y_1$ is $-1/x^2$.
Similarly, we have $$C_2=\int \frac{y_1(x)/x}{W(x)} dx+D_2=-\int \sin x dx+D_2=\cos x +D_2.$$
Inserting , $y_1(x), y_2(x). C_1(x), C_2(x)$ in (3), we get
$$Y(x)=\frac{1}{x}[D_1 \sin x+ D_2 \cos x]+\frac{1}{x}[\sin^2 x+ \cos^2 x].$$
$$\implies Y(x)=\frac{1}{x}[D_1 \sin x+ D_2 \cos x]+\frac{1}{x},$$
which is the solution of the reqyured ODE (2).
For the method of Variation of Parameters see:
https://en.wikipedia.org/wiki/Variation_of_parameters
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"timestamp": "2023-03-29T00:00:00",
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How to find $\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$? I tried to solve it, but the answer I got was different from the answer given.
Answer given:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)} = \frac{n(2n+1)}{4(2n-1)(2n+3)}$$
My working:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$= \sum_{r=1}^{n} \Biggl[\frac{1}{16(2r-1)}+\frac{1}{8(2r+1)}-\frac{3}{16(2r+3)}\Biggl]$$
$$= \frac{1}{16} + \frac{1}{24} - \require{cancel} \cancel{\frac{3}{80}}$$
$$+ \frac{1}{48} + \require{cancel} \cancel{\frac{1}{40}} - \require{cancel} \cancel{\frac{3}{112}}$$
$$+ \require{cancel} \cancel{\frac{1}{80}} + \require{cancel} \cancel{\frac{1}{56}} - \require{cancel} \cancel{\frac{3}{144}}$$
. . .
. . .
. . .
$$+ \require{cancel} \cancel{\frac{1}{16(2n-3)}} + \require{cancel} \cancel{\frac{1}{8(2n-1)}} - \frac{3}{16(2n+1)}$$
$$+ \require{cancel} \cancel{\frac{1}{16(2n-1)}} + \frac{1}{8(2n+1)} - \frac{3}{16(2n+3)}$$
$$= \frac{1}{16} + \frac{1}{24} + \frac{1}{48} - \frac{3}{16(2n+1)} + \frac{1}{8(2n+1)} - \frac{3}{16(2n+3)}$$
$$= \frac{2n^{2}+6n+3}{4(2n+1)(2n+3)}$$
|
Hint:
Another way:
Let $$\dfrac r{(2r-1)(2r+1)(2r+3)}=f(r)-f(r+1)$$ where $f(m)=\dfrac{am+b}{(2m-1)(2m+1)}$
so that $$\sum_{r=1}^n\dfrac r{(2r-1)(2r+1)(2r+3)}=\cdots=f(n+1)-f(1)$$
Now comparing the numerator of $f(r)-f(r+1)$ with $r,$
$$2a=1, a+4b=0\iff b=-\dfrac a4$$
|
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|
What is the average distance between the first occurence of an element and the first occurence of another after that one? We have a set of Na elements 'a' and Nb elements 'b'.
For example for Na=2 and Nb=2 it would be (a, a, b, b).
We can generate all possible permutations with them. In our example we will end up having 24, some could be the same.
Now, for each of this "vectors" we can measure the distance between the first occurence of 'a' and the first occurence of 'b' happenning after that 'a'.
(We have capitalized that occurrences)
(A,B,a,b) = 2-1 = 1
(A,B,a,b) = 2-1 = 1
(A,B,a,b) = 2-1 = 1
(A,B,a,b) = 2-1 = 1
(A,B,b,a) = 2-1 = 1
(A,B,b,a) = 2-1 = 1
(A,B,b,a) = 2-1 = 1
(A,B,b,a) = 2-1 = 1
(A,a,B,b) = 3-1 = 2
(A,a,B,b) = 3-1 = 2
(A,a,B,b) = 3-1 = 2
(A,a,B,b) = 3-1 = 2
(b,A,B,a) = 2-1 = 2
(b,A,B,a) = 2-1 = 2
(b,A,B,a) = 2-1 = 1
(b,A,B,a) = 2-1 = 1
(b,A,a,B) = 3-1 = 2
(b,A,a,B) = 3-1 = 2
(b,A,a,B) = 3-1 = 2
(b,A,a,B) = 3-1 = 2
(b,b,A,a)
(b,b,A,a)
(b,b,A,a)
(b,b,A,a)
Now we can count how many times we have at least a 'b' after any previous 'a', here 20, and we can calculate the average of these distances.
averagedistance = 30/20 = 1.5
What is the general formula for any Na and Nb?
I have only been able to calculate it (simulating numbers) for
Na=1 -> averagedistance= 1
or
Nb=1 -> averagedistance=((Na+1)/2)
but not a general formula.
and I think the complete solution has a factor Na!Nb!/(NaNb)! and something else.
|
Let $\mathbf X$ be the distance we want.
Then there are $\binom{N_a+N_b}{N_b}$ ways to choose a permutation of the as and bs to consider. Of them, $\binom{N_a+N_b}{N_b}-1$ have an a followed by a b at some point.
Among them, there are $\binom{N_a + N_b-k+1}{N_b}-1$ ways to choose a permutation in which $\mathbf X \ge k$, because to get such a permutation, we can take any permutation of $N_a-k+1$ as and $N_b$ bs (except for the one we're excluding) then replace the first a we see by a block (a,a,...,a) of length $k$. This is a bijection.
Therefore we have
$$
\mathbb E[\mathbf X] = \sum_{k=1}^{N_a} \Pr[\mathbf X \ge k] = \sum_{k=1}^{N_a} \frac{\binom{N_a + N_b - k +1}{N_b}-1}{\binom{N_a + N_b}{N_b}-1}.
$$
To simplify the sum, set $j = N_a - k + 1$. Then we have
$$
\mathbb E[\mathbf X] = \frac1{\binom{N_a+N_b}{N_b}-1} \sum_{j=1}^{N_a} \binom{N_b+j}{N_b} - \frac{N_a}{\binom{N_a+N_b}{N_b}-1}= \frac{\binom{N_a+N_b+1}{N_b+1} - N_a - 1}{\binom{N_a+N_b}{N_b}-1}
$$
by the hockey-stick identity.
|
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|
Floor function bounding From CMC:
What is the sum of the square of the real numbers $x$ for which $x^2 - 20\lfloor x\rfloor + 19 = 0$?
We use $\lfloor x\rfloor\le x<\lfloor x\rfloor+1$ and eventually get the bounds $1\le x\le19$ and $x\ge 18,x\le 2.$ Of course, it's possible for $x$ not to be an integer, so how do we find the other solutions, other than $19$ and $1$?
Someone wrote this solution:
$x^2 - 20\lfloor x \rfloor + 19 = 0$
Cleary $x\geq \lfloor x \rfloor$ for all real $x$. Thus,
$$x^2-20x+19 \leq x^2 - 20\lfloor x \rfloor + 19=0.$$
Which leads to
$$1 \leq x \leq19.$$Also $x^2=20\lfloor x\rfloor - 19$ which implies $\lfloor x \rfloor=1,17,18,19$.
I'm not sure how we get $\lfloor x\rfloor=17,18$ from this.
|
$$x^2 - 20 \lfloor x \rfloor + 19 = 0$$
The intuition is that the solutions does not get too far away from the solutions of $x^2-20x+19=0$, namely $x=1, 19$. So go ahead and express that intuition!
I am not used to fiddling with $x-1 < \lfloor x \rfloor \le x$. So let's go for a more granular method.
Let $n = \lfloor x \rfloor , u = x-n$. So $0\le u < 1$. $$(n+u)^2 - 20n + 19=0.$$ Expanding we get $$u^2 + 2nu + (n^2-20n + 19)=0.$$ We know that $u \in [0,1)$. So this equation got to have a solution in that interval. Will it have two? it would mean that the sum of these two solutions is greater than 0. But from Vieta's theorem it is not possible. Therefore, there is exactly 1 solution in the interval $[0,1)$. If that solution is exactly 0, then $0^2 + 2n\cdot 0 + (n^2-20n+19)=0.$ So $n = 1,19$. If not, we see that the function $f_n(u) = u^2 + 2nu + (n^2-20n + 19)$ must change sign exactly once on the interval $(0,1)$. So $f(0)f(1) < 0.$ That means $$(n^2-20n+19)(n^2-18n + 20)<0,$$ which we can factor: $$(n-19)(n-1)(n-1.1897..)(n-16.8102..) < 0.$$ (In practice you don't need that much precision, you just need to calculate the integral part.) Since $n$ is an integer, $n=17, 18.$ (Can you see why? I shall explain that further on request.)
Substituting $n=17$ and $n=18$, we get equations for $u$ respectively. And this becomes regular quadratic equations.
|
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|
Necklace problem with Burnside Lemma How many necklaces can be made with two red beads, two green beads, and four violet beads?(8 total)
Using Burnside lemma is complicated for me due to my lack of understanding of the lemma. I want to know the method step-by-step.
|
We may as well deploy PET here since we need the cycle index $Z(D_8)$ of
the dihedral group $D_8$ of order $16$ to apply Burnside. We compute
and average the number of assignments of colors to the eight slots fixed
by permutations from each conjugacy class in $D_8$, taking into account
the order of the class. This means the assignment is constant on the
cycles, so we can place exactly one color in the slots on a given cycle,
substituing $a_d$ from the index with $R^d + G^d + V^d,$ which is PET.
Consulting the following fact sheet on necklaces and
bracelets
we get for the cycle index of the dihedral group $D_8$
$$Z(D_8) = \frac{1}{16} a_1^8
+ \frac{1}{4} a_1^2 a_2^3
+ \frac{5}{16} a_2^4
+ \frac{1}{8} a_4^2
+ \frac{1}{4} a_8.$$
We seek $$[R^2 G^2 V^4] Z(D_8; R+G+V).$$
Here we assume that the OP asks for the full symmetry i.e. dihedral,
which means that the label to use is bracelet. Working through the five
terns in the cycle index we obtain
$\frac{1}{16} a_1^8$
$$[R^2 G^2 V^4] \frac{1}{16} (R+G+V)^8
= \frac{1}{16} {8\choose 2,2,4}$$
$\frac{1}{4} a_1^2 a_2^3$
$$[R^2 G^2 V^4] \frac{1}{4} (R+G+V)^2 (R^2 + G^2 + V^2)^3
\\ = [R^2 G^2 V^4] \frac{1}{4} (R^2 + G^2 + V^2) (R^2 + G^2 + V^2)^3
\\ = [R^2 G^2 V^4] \frac{1}{4} (R^2 + G^2 + V^2)^4
\\ = [R G V^2] \frac{1}{4} (R + G + V)^4
= \frac{1}{4} {4\choose 2,1,1}.$$
$\frac{5}{16} a_2^4$
$$[R^2 G^2 V^4] \frac{5}{16} (R^2 + G^2 + V^2)^4
= \frac{5}{16} {4\choose 2,1,1}.$$
$\frac{1}{8} a_4^2$
$$[R^2 G^2 V^4] \frac{1}{8} (R^4 + G^4 + V^4)^2
= 0.$$
$\frac{1}{4} a_8$
$$[R^2 G^2 V^4] \frac{1}{4} (R^8 + G^8 + V^8)
= 0.$$
We thus get for our answer
$$\frac{1}{16} {8\choose 2,2,4} + \frac{9}{16} {4\choose 2,1,1}
= \bbox[5px,border:2px solid #00A000]{33.}$$
|
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|
How does this division work? $\frac{\;\frac{6^6}{1}\;}{2^{-3}}\cdot2^{-10}$ I came across this division and can't wrap my head around how it is solved:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}$$
They subtract the exponent of $2^{-10}$ from the denominator's exponent $2^{-3}$:
$$2^{-3-(-10)}$$
Which gives us:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^7}$$
If anyone knows what is actually being done here I would appreciate it!
|
\begin{align}
\frac{\;\;\dfrac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}
&=\frac{6^6}{2^{-3}}\cdot 2^{-10}\qquad&\text{(simplify the denominator)}\\
&=6^6\cdot 2^3\cdot 2^{-10}\qquad &(\frac{1}{2^{-3}}=2^3)\\
&=6^6\cdot 2^{3+(-10)}\qquad &(2^a2^b=2^{a+b})\\
&=6^6\cdot 2^{-7}\qquad &\text{(simplify)}\\
&=\frac{6^6}{2^7}\qquad &(2^{-7}=\frac{1}{2^7})
\end{align}
|
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|
Solving for $x$ and $y$ when $(3x + y)(x + 3y)\sqrt{xy} = 14$ Solve for $x$ and $y$ when $$(3x + y)(x + 3y)\sqrt{xy} = 14$$ $$(x+y)(x^2 + 14xy + y^2) = 36.$$
I was thinking of squaring the first equation and moving on from there, but I think it'll be a bit too messy. Is there a better way to start this problem?
|
Setting $p=x+y$ and $q=\sqrt{xy}$ (almost as suggested by Alexey in the comments, but $\sqrt{xy}$ looks like it will be more symmetric) and expanding gives
\begin{align}
3p^2q + 4q^3 &= 14 \\
p^3 + 12pq^2 &= 36
\end{align}
From here, a lucky coincidence is that
$$
(p + 2q)^3 = p^3 + 6p^2q + 12pq^2 + 8q^3 = 36 + 2 \cdot 14 = 64
$$
and
$$
(p - 2q)^3 = p^3 - 6p^2q + 12pq^2 - 8q^3 = 36 - 2 \cdot 14 = 8
$$
which gives us $p+2q = 4$ and $p-2q = 2$. Therefore $p=3$ and $q = \frac12$, giving us $x,y = \frac{3 \pm 2\sqrt2}{2}$.
|
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|
Solving for $x$ when $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x$ Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$
I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can someone give me a hint?
|
Let $a = \sqrt{3}$, so $x = \sqrt{a-\sqrt{a+x}}$. Let's call this equation $(*)$.
Let $y = \sqrt{a+x}$ so $x = \sqrt{a-y}$. Therefore
*
*$x^2 = a-y$
*$y^2 = a+x$
Subtracting 2. from 1. we have $x^2 - y^2 = -(x+y)$. Let's consider the following cases:
*
*If $x+y = 0$, then $x = -\sqrt{a+x}\le 0$, but from $(*)$ we have $x\ge 0$, so we must have $x = 0$. It's easy to see that this doesn't work.
*If $x+y \neq 0$, we have $\begin{aligned}\frac{x^2-y^2}{x+y} = -1 \therefore x-y = -1\end{aligned}$, i.e., $x+1 = \sqrt{a+x}$, so $(x+1)^2 = x^2+2x+1 = a+x$, therefore $x^2+x+1-a = 0$, which can be solved as a quadratic equation to get $\begin{aligned}x = \frac{-1+ \sqrt{1-4(1-a)}}{2} = \frac{-1+\sqrt{4a-3}}{2}\end{aligned}$ (notice that this is the only positive solution).
|
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|
Does there exist a natural number pair $(a,b)$ such that $a^2b+a+5$ divide $ab^2+a+b$? I tried that if I can show $\gcd(a^2b+a+5,ab^2+a+b) = 1$, then there is no solution. Am I true?
Any hints /idea?
I can't factorize them out such that each is multiple of each other.
|
Since $a^2b+a+5$ divides $ab^2+a+b$, we can conclude that it also divides:
$$a(ab^2+a+b)-b(a^2b+a+5)=a^2-5b$$
This means that we must have $a^2-5b=0$ or we must have $|a^2-5b| \geqslant a^2b+a+5$. In the first case, substituting $a=5k$ and $b=5k^2$ gives:
$$(125k^4+5k+5) \mid (125k^5+5k^2+5k)$$
which is obviously true. Next, in the second case, we either have:
$$|a^2| \geqslant |a^2-5b| \implies a^2 \geqslant a^2b+a+5$$
which is impossible, or we have:
$$5b \geqslant |a^2-5b| \geqslant a^2b+a+5 \implies a \leqslant 2$$
If $a=1$, then $b+6$ divides $b^2+b+1$. Performing the same deleting technique gives $b+6 \mid b+37 \implies b+6 \mid 31$ which gives $b=25$.
If $a=2$, then $4b+7$ divides $2b^2+b+2$, and the trick gives $(2,11)$. Can you see that?
Thus, the only solutions are:
$$(a,b)=(5k,5k^2),(1,25),(2,11)$$
for positive integers $k$.
Motivation : We must try to delete the first term in the number divisible by
$ab^2+a+b$ so that we can perform bounding.
|
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|
How to integrate hyperbolic function $\frac{1-\sin(x)}{1+\sin(x)}$ I'm trying to find a primitive of $ \frac{1-\sin(x)}{1+\sin(x)} $. By changing the variable to $t=\tan(\frac{x}{2})$ and letting $\sin(x) = \frac{2t}{1+t^2}$ I get the following integral:
\begin{align}
& \int \frac{1-\sin(x)}{1+\sin(x)} \, dx \\[8pt] = {} & \int \frac{ 1-\frac{2t}{1+t^2} } { 1+ \frac{2t}{1+t^2} } \frac{2}{1+t^2} \, dt \\[8pt]
= {} & 2\int \frac{ t^2 -2t +1 }{ (t^2 +2t+1)(t^2 + 1) } \, dt \\[8pt]
= {} & 2\int \frac{t^2 -2t +1 }{(t+1)^2(t^2+1)} \, dt \end{align}
Now I know that I could do a partial fraction expansion. I would get 3 simpler fraction but I also know that the result contains only two fraction by calculating it in xcas: $$ \int \frac{1-\sin(x)}{1+\sin(x)} \, dx = 2\left(-\frac{2}{\tan(\frac{x}{2}) +1} - \frac{x}{2}\right) $$
Is there an easier way to calculate this primitive ?
|
$\displaystyle \frac{1-sinx}{1+sinx}=\frac{2}{1+sinx}-1=\frac{2}{(sin\frac{x}{2}+cos\frac{x}{2})^{2}}-1$
$\displaystyle =\frac{1}{cos^{2}(\frac{x}{2}-\frac{\pi}{4})}-1=tan^{2}(\frac{x}{2}-\frac{\pi}{4})$
$\displaystyle \theta=\frac{x}{2}-\frac{\pi}{4}\Rightarrow 2d\theta=dx$
$\displaystyle z=tan\theta\Rightarrow dz=(1+tan^{2}\theta)d\theta $
$\displaystyle dz-d\theta=tan^{2}\theta d\theta=d(z-\theta)$
$\displaystyle \int\frac{1-sinx}{1+sinx}dx=2\int tan^{2}\theta d\theta=2\int d(z-\theta)=2z-2\theta+c$
$\displaystyle \int\frac{1-sinx}{1+sinx}=2tan(\frac{x}{2}-\frac{\pi}{4})-2(\frac{x}{2}-\frac{\pi}{4})+c=2tan(\frac{x}{2}-\frac{\pi}{4})-x+C$
|
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|
Is $(x^2+3)^3-4(x-1)^3$ an irreducible polynomial in $Q[x]$? I was trying to solve the problem below
Let $K \subset \mathbb{C}$ be a splitting field of $f(x) = x^3 − 2$ over $\mathbb{Q}$. Find a complex number $z$, such that $K = \mathbb{Q}(z)$.
All the roots of the polynomial $f(x)$ are $2^{1/3}$, $2^{1/3}\omega$ and $2^{1/3}\omega^2$, where $\omega = (-1+i\sqrt{3})/2$ is a cube root of unity. Thus $\mathbb{Q}(2^{1/3},\omega)$ or $\mathbb{Q}(2^{1/3},i\sqrt{3})$ is a splitting field of $f(x)$. $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$ as irreducble polynomial of $2^{1/3}$ is $x^3-2$(Eisenstien's criteria for $p=2$) and $[\mathbb{Q}(\omega):\mathbb{Q}]=2$, as irreducible polynomial of $\omega$ is $g(x) = x^2+x+1$( $g(x+1)$ is irreducble by eisenstien's criteria for $p=3$). Therefore, if we have a splitting field $\mathbb{Q}(z)$, then degree of irreducible polynomial of $z$ is $6$ by compositum of fields. Now we need to find $z$.
Me and my friend got the idea of taking $z$ as $2^{1/3} + i\sqrt{3}$.
$$\begin{eqnarray*} x & = & 2^{1/3} + i\sqrt{3} \\ \left( x - 2^{1/3} \right)^2 & = & -3
\\ -2^{2/3}x+2^{2/3} & = & -x^2 -3 \\ 4(x-1)^3 & = & (x^3+3)^3 \end{eqnarray*}$$
Thus $h(x) = (x^2+3)^3-4(x-1)^3$ is a degree six polynomial having root $2^{1/3} + i\sqrt{3}$. How do i show from here that it is irreducible in $Q[x]$? It's expanded form is $x^6+9x^4-4x^3+39x^2-12x+31$.
|
Another way is to notice $h(x)$ factors as
$$(x^4 + 4x^3 + 3x^2 + 3)(x^2 + x + 2)\mbox{ in }\mathbb{Z_5}[x]\tag{1}$$
and
$$(x^3 + 4x^2 + 6x + 5)(x^3 + 3x^2 + 5x + 2)\mbox{ in }\mathbb{Z_7}[x]\tag{2}.$$
(the $2$-nd and $3$-th degrees factors are irreducible by testing finitely many roots in $\mathbb{Z_5}[x]$ and $\mathbb{Z_7}[x]$, similarly the $4$-th degree polynomial has no linear factor, and we don't need to care about its quadratic factors as the argument below would still go through).
Now any factorization in $\mathbb{Z}[x]$ would reduce to a factorization in $\mathbb{Z_5}[x]$ and $\mathbb{Z_7}[x]$, yet the degrees of $(1)$ and $(2)$ are compatible only with factors of degree $1$ and $6$ in $\mathbb{Z}[x]$. Hence $h(x)$ is irreducible in $\mathbb{Z}[x]$.
|
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|
Find the supremum of $|z|^2+\text{Re}(\overline{z}w)$ subject to $|z|^2+|w|^2=1$. I am trying to find the operator norm of $$A=\begin{pmatrix}3 & 1 \\
1 & 1\end{pmatrix}\in M_n(\mathbb{C}).$$This is what I have so far: If $|z|^2+|w|^2=1$, then $$||A\begin{pmatrix}z\\w\end{pmatrix}||_2=\sqrt2\sqrt{1+4(|z|^2+\text{Re}(\overline{z}w))}$$ Since $|z|^2+\text{Re}(\overline{z}w)\in\mathbb{R}$ and $\sqrt{2}\sqrt{1+4x}$ is an increasing function of $x$, we need to maximise $|z|^2+\text{Re}(\overline{z}w)$. In other words, $$||A||=\sqrt2\sqrt{1+4m},$$ where $m=\text{sup}\lbrace|z|^2+\text{Re}(\overline{z}w): |z|^2+|w|^2=1\rbrace$.
But how do I go about finding $m$? I'm a bit rusty on this stuff, do I need to use Lagrange multipliers?
|
Let $z=x+iy$ and $w=u+iv$ so that $|z|^2+\Re(\overline zw)=x^2+y^2+xu+yv$ is subject to $x^2+y^2+u^2+v^2=1$. Then we have $\mathcal L=x^2+y^2+xu+yv-\lambda(x^2+y^2+u^2+v^2-1)$ so that \begin{align}\mathcal L_u&=x-2\lambda u=0\implies x=2\lambda u\\\mathcal L_v&=y-2\lambda v=0\implies y=2\lambda v\\\mathcal L_x&=2x+u-2\lambda x=0\implies(2-2\lambda)2\lambda u+u=0\\\mathcal L_y&=2y+v-2\lambda y=0\implies(2-2\lambda)2\lambda v+v=0.\end{align} Either $u=v=0$ or $(2-2\lambda)2\lambda+1=0$. If the former then $w=0$ so $|z|^2+\Re(\overline zw)=|z|^2=1$.
If $(2-2\lambda)2\lambda+1=0$ then $\lambda=(1\pm\sqrt2)/2$ so that $|z|^2+\Re(\overline zw)=4\lambda^2(u^2+v^2)+2\lambda(u^2+v^2)$ subject to $4\lambda^2(u^2+v^2)+u^2+v^2=1$. This means that $u^2+v^2=1/(1+4\lambda^2)$ so $|z|^2+\Re(\overline zw)=(2\lambda+4\lambda^2)/(1+4\lambda^2)=\lambda$. Hence $\max\{|z|^2+\Re(\overline zw)\}=(1+\sqrt2)/2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all positive integer values $(x, y, n)$ such that $x^n+1=y^{n+1}$ and $gcd(x, n+1)=1$ My approach to this problem is as follows:
First, I attempt to prove that $x^n+1=y^{n+1}$ has solutions on the integers only for $n=1$. Since $y>x$, it follows that $y>1$. Thus If $n\ge2$, we have $$y^{n+1}-x^n>y^n-x^n=(y-x)(y^{n-1}+y^{n-2}x+...+x^{n-1})> 1$$
My reasoning is based on the fact that the factor $(y^{n-1}+y^{n-2}x+...+x^{n-1})$ is at least the sum of $n-1$ powers of $y\geq2$. Therefore, there are no integer solutions for the equation $x^n+1=y^{n+1}$.
Now, since $gcd(x,n+1)=1$, it can be concluded that $x$ has to be an odd number, and therefore, $y$ is an even number.
Let $y=2k$ for some integer $k$, then $x=(2k)^2-1=4k^2-1$. That is, the answer to the problem is the triplet $(4k^2-1, 2k, 1)$.
EDIT: I just realized the sentence
Since $y>x$, it follows that $y>1$
Is not correct, since it is possible for $y$ to be lesser than $x$. Therefore, the whole argument is incorrect. However, I'd like to find a way to prove $y^{n+1}-x^n>1$.
Thanks in advance
|
One guy trying to prove Fermat’s Last Theorem send a telegram: “We should move $y^{n}$ to the right-hand side. The details will be sent a letter”. So let us follow this advice.
We have $x^n=y^{n+1}-1=(y-1)z$, where $z=y^n+y^{n-1}+\dots+1$. We have $\gcd(y-1,z)=(y-1,n+1)$. Since $(y-1)z=x^n$ is coprime with $n+1$, we have that $\gcd(y-1,z)=1$. It follows $y-1=u^n$ and $z=v^n$ for some positive integers $u$ and $v$. If $n>1$ then $y^n<z<(y+1)^n$, a contradiction. So $n=1$, and this case you already solved.
|
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|
Find a mistake in a computation of a determinant. I recently have to solve one exercise in which I need to compute the determinant of the matrix
$$ A= \begin{pmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{pmatrix} ,$$
where $a\in\mathbb{R}$. This is how I proceed using famous properties of the determinant
$$\det A = \begin{vmatrix}
a & 1 & 1 & 1 \\
a-1 & 1-a & 0 & 0 \\
0 & a-1 & 1-a & 0 \\
\textbf{0} & \textbf{0} &\textbf{a-1} & \textbf{1-a} \\
\end{vmatrix} = (1-a) \begin{vmatrix} a & 1 & 1 \\ a-1 & 1-a & 0 \\ 0 & a-1 & 0 \end{vmatrix} + (1-a)\begin{vmatrix} a & 1 & 1 \\ a-1 & 1-a & 0 \\ 0 & a-1 & 1-a \end{vmatrix} ; $$
$$ \det A = (1-a) \begin{vmatrix} a & 1 & \textbf{1} \\ a-1 & 1-a & \textbf{0} \\ 0 & a-1 & \textbf{0} \end{vmatrix} + (1-a)\begin{vmatrix} a & 1 & 2 \\ a-1 & 1-a & 1-a \\ \textbf{0} & \textbf{a-1} & \textbf{0} \end{vmatrix} ; $$
$$ \det A = (1-a) \begin{vmatrix} a-1 & 1-a \\ 0 & a-1 \end{vmatrix} + (1-a)^2\begin{vmatrix} a & 2 \\ a-1 & 1-a \end{vmatrix} ; $$
$$ \det A = (1-a)(a-1)^2 + (1-a)^2[-a^2-a+2] = (1-a)(a-1)^2 - (1-a)^2[a^2 + a -2] ; $$ $$\det A = (1-a)(a-1)^2 - (1-a)^2(a+2)(a-1);$$
$$\det A = -(a-1)^3 - (a-1)^2 (a+2)(a-1) = -(a-1)^3 - (a-1)^3(a+2); $$
$$\det A = -(a-1)^3 [1 + (a+2)] = \boxed{-(a-1)^3(a+3)} $$
I can't see any mistake in my procedure. However, calculating $\det A$ with a computer, I obtained $$\det A= (a-1)^3(a+3)$$
It is the same result except for a minus sign...
EDIT
This is what I am doing in the first steps, basically substracting rows as follows:
$$\det A = \begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{vmatrix} =\begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
0 & 0 & a-1 & 1-a \\
\end{vmatrix} =\begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
0 & a-1 & 1-a & 0 \\
0 & 0 & a-1 & 1-a \\
\end{vmatrix} = \begin{vmatrix}
a & 1 & 1 & 1 \\
a-1 & 1-a & 0 & 0 \\
0 & a-1 & 1-a & 0 \\
0 & 0 & a-1 & 1-a \\
\end{vmatrix} $$
So where is exactly my mistake?
|
We consider a square matrix $A$ and recall two rules regarding calculation with $\det A$:
*
*Multiplication of a row of $A$ with a scalar $c$ results in $c\cdot\det A$
*Addition of a scalar multiple of a row with another row leaves the value of the determinant unchanged.
Looking at OPs example we obtain
\begin{align*}
\det A = \begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{vmatrix} =
\color{blue}{(-1)} \begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
\color{blue}{-1} & \color{blue}{-1} & \color{blue}{-1} & \color{blue}{-a} \\
\end{vmatrix} =
(-1)\begin{vmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
0 & 0 & a-1 & 1-a \\
\end{vmatrix}
\end{align*}
In the first step we multiply the last row with $-1$ and have to compensate it by a factor $(-1)$ in front of the determinant. In a second step we add the third row to the fourth row.
|
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|
Maximizing $\frac{y+1}{x+2}$ when $(x-3)^2 + (y-3)^2 = 6$ Suppose $x$ and $y$ are real numbers such that $(x-3)^2 + (y-3)^2 = 6.$ Than, maximize $\frac{y+1}{x+2}.$
I do in fact realize that this is a double post, but it's a 5 year old question and I don't feel as if it is appropriate to bump it. I did as the original post hinted towards, setting $k = \frac{y+1}{x+2}$ and than writing the given equation as $$(x-3)^2 + (k(x+2) - 4)^2 = 6.$$ I than proceeded to expand and simplify, which gave me $$x^2(k^2 + 1) + x(4k^2 - 8k - 6) + (4k^2 - 16k + 19).$$ However, I am unsure where to go from here. Should I use the discriminant now?
|
A bit of geometry.
Introduce new coordinates X,Y:
$Y := y+1;$ $X := x+2;$
1)$Y=CX;$
2)$(X-5)^2 + (Y-4)^2=6;$
Need to find the slope $C$ of the line $Y=CX$ s.t. it is tangent to the circle 2).
3)Distance of tangent line $Y=CX$ to the centre of circle $(5,4):$
$\dfrac{|-5C+4|}{\sqrt{1+C^2}}=√6;$
$(-5C+4)^2=6(1+C^2);$
A quadratic equation for $C$. Choose the larger solution.
|
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|
How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for even $n$? How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for all positive even integer $n$?
Here are my efforts:
Let $f(n)=5^{2n+1}-3^{2n+1}-2$.
For $n=2$,
$$f(1)=5^{2\cdot 2+1}-3^{2\cdot2+1}-2=2880=30\times 96$$
So, the statement is true for $n=2$.
Assume the statement is true for some integer $k\geq 2$.
$$f(k)=5^{2k+1}-3^{2k+1}-2=30t$$
For $n=k+2$,
\begin{align}
f(k+2)&=(5^{2(k+2)+1})-(3^{2(k+2)+1})-2\\
&=(5^4)\cdot (5^{(2k+1)})-(3^4)\cdot (3^{(2k+1)})-2
\\&=30(80t)+160+544\cdot (5^{2k+1})
\end{align}
I'm stuck here. Any help would be appreciated!
|
This is not an induction proof, but I'll nevertheless leave it here.
To show that $5^{2n+1}-3^{2n+1}-2$ is divisble by $30=2\cdot 3\cdot 5$, you can show that $2,3,5$ divides it.
*
*$2\mid 5^{2n+1}-3^{2n+1}-2\iff 2\mid 5^{2n+1}-3^{2n+1}$, but this is clear since the sum of two odd numbers is even.
*$3\mid 5^{2n+1}-3^{2n+1}-2\iff 3\mid 5^{2n+1}-2$, write $5^{2n+1}-2$ as $$(6-1)^{2n+1}-2$$ Upon expanding, every term is a multiple of $6$, and the last term is $(-1)^{2n+1}-2=-3$. So everything is divisible by $3$.
*$5\mid 5^{2n+1}-3^{2n+1}-2\iff 5\mid -3^{2n+1}-2$. Same trick, rewrite $-3^{2n+1}-2$ as $$-3(10-1)^n-2$$ Upon expanding every term except the last term is divisible by $5$, the last term is $-3(-1)^n-2$, so the result is only true for even $n$.
The claim is only true for even $n$.
|
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|
Where did I go wrong in solving this equation?
The value of the expression $ax^2 + bx + 1$ are $1$ and $4$ when $x$ takes the values of $2$ and $3$ respectively. Find the value of the expression when $x$ takes the value of $4.$
Here is my first attempt at solving this question:
\begin{align}
2a^2 + 2b + 1 &= 1 \leftarrow\text{(1)} \\
3a^2 + 3b + 1 &= 4 \leftarrow\text{(2)}
\end{align}
$$ \text{From (1):} $$
\begin{align}
2a^2 + 2b + 1 &= 1 \\
2a^2 &= -2b + 1 - 1 \\
2a^2 &= -2b \\
a^2 &= -b \leftarrow\text{(3)}
\end{align}
$$ \text{Substitute (3) into (2):} $$
\begin{align}
3(-b) + 3b + 1 &= 4 \\
-3b + 3b + 1 &= 4 \\
1 &= 4
\end{align}
Here is my second attempt at solving it:
\begin{align}
ax^2 + bx + 1 &= 1 \\
2a^2 + 2b + 1 &= 1 \\
a^2 + b + \frac{1}{2} &= \frac{1}{2} \\
a^2 + b &= \frac{1}{2} - \frac{1}{2} \\
a^2 + b &= 0 \\
4a^2 + 4b &= 0 \\
4a^2 + 4b + 1 &= 1 \leftarrow\text{(1)}
\end{align}
\begin{align}
3a^2 + 3b + 1 &= 4 \\
3a^2 + 3b &= 4 - 1 \\
3a^2 + 3b &= 3 \\
a^2 + b &= 1 \\
4a^2 + 4b &= 4 \\
4a^2 + 4b + 1 &= 5 \leftarrow\text{(2)}
\end{align}
$$ \text{(1) = (2): } 1 = 5 $$
Where did I go wrong?
|
We have
*
*$p(2)=1 \implies 4a + 2b + 1=1$
*$p(3)=4 \implies 9a + 3b + 1=4$
then solve for $a$ and $b$.
|
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|
Determining whether a subset is a subspace The problems are as follows:
Determine whether the following subsets of $\mathbb{R}^3$ are subspaces of $\mathbb{R}^3$.
*
*$A = \{(u^2, v^2, w^2) \,|\, u, v, w \in \mathbb{R} \}$,
*$B = \left\{(a, b, c) \,|\,
\begin{pmatrix}
a & b & c\\
1 & 2 & 0\\
0 & 1 & 2
\end{pmatrix} \text{is not invertible}\right\}$,
*$C = \left\{(x, y, z) \,|\, \begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\right\}$.
Here's what I've tried so far:
*
*$A$ is a subspace of $\mathbb{R}^3$ as it contains the $0$ vector (?).
*The matrix is not invertible, meaning that the determinant is equal to $0$. With this in mind, computing the determinant of the matrix yields $4a - 2b + c = 0$. The original subset can thus be represented as $B = \left\{\left(\frac{2s - t}{4}, s, t\right) \,|\, s, t \in \mathbb{R}\right\}$; i.e. $B = \text{span}\left\{(\frac{1}{2}, 1, 0), (-\frac{1}{4}, 0, 1)\right\}$, a plane in $\mathbb{R}^3$.
*Solving for the linear system,
$$
\begin{aligned}
\begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} &=
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\\
x
\begin{pmatrix}
1\\
4\\
7
\end{pmatrix}
+ y
\begin{pmatrix}
2\\
5\\
8
\end{pmatrix}
+ z
\begin{pmatrix}
3\\
6\\
9
\end{pmatrix}
&=
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\\
x
\begin{pmatrix}
-1\\
4\\
7
\end{pmatrix}
+ y
\begin{pmatrix}
2\\
3\\
8
\end{pmatrix}
+ z
\begin{pmatrix}
3\\
6\\
7
\end{pmatrix}
&=
\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}
\end{aligned}$$
Converting to row echelon form gives the trivial solution $x = 0, y = 0, \text{ and } z = 0$; $C$ only contains the $0$ vector.
Is my reasoning correct? Also, quite a bit of the computations have been omitted in this post for brevity. Regardless of this, are my answers well justified?
Thank you.
|
Most of it is great.
In 1 the vectors in $A$ can not be multiplied by negative scalar, so it is not a subspace.
In 2 and 3 your reasoning is perfect, but maybe you can state it in a more concise manner, depending on what your learned already. Once you get that the subset is defined by linear equations you are done - it immediately implies that it is in fact a subspace. You don't have to actually solve the equations.
|
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|
how to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
How to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
My attempt:
\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\frac{\pi}{3} - x\right) \\
&= \frac{1}{\sin x} - \frac{1}{\sin\left(\frac{\pi}{3} + x\right)} + \frac{1}{\sin\left(\frac{\pi}{3} -x\right)} \\
&= \frac{\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x }{\sin x \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right)} \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(\sin x - \sin\left(\frac{\pi}{3} + x\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(2\sin\frac{-\pi}{6}\cos\left(x + \frac{\pi}{6}\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\right) \\
\end{align}
How should I proceed? Or did I make some mistakes somewhere? Thanks in advance.
|
We have $$\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right) \\$$
$$=\sin(x)[\sin(\frac{\pi}{3})\cos(x)+\cos(\frac{\pi}{3})\sin(x)]$$
$$+(\sin(\frac{\pi}{3})\cos(x)-\cos(\frac{\pi}{3})\sin(x))(\cos(\frac{\pi}{6})\cos(x)-\sin(\frac{\pi}{6})\sin(x))$$
$$=\sin(x)[\frac{\sqrt{3}}{2}\cos(x)+\frac{1}{2}\sin(x)]$$
$$+(\frac{\sqrt{3}}{2}\cos(x)-\frac{1}{2}\sin(x))(\frac{\sqrt{3}}{2}\cos(x)-\frac{1}{2}\sin(x))$$
$$=\frac{\sqrt{3}}{2}\sin(x)\cos(x)+\frac{1}{2}\sin^2(x)$$
$$+\frac{3}{4}\cos^2(x)-\frac{\sqrt{3}}{2}\cos(x)\sin(x)+\frac{1}{4}\sin^2(x)$$
or $$\frac{3}{4}(\sin^2(x)+\cos^2(x))=\frac{3}{4}.$$
|
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|
ABC is a triangle and the line YCX is parallel to AB such that AX and BY are the angular bisectors of angle A and I'm interested in the following problem:
$ABC$ is a triangle and the line $YCX$ is parallel to $AB$ such that $AX$ and $BY$ are the angular bisectors of angle $A$ and angle $B$ respectively. If $AX$ meets $BC$ at $D$ and $BY$ meets $AC$ at $E$ and if $YE =XD$, then prove that $AC=BC$.
My attempt. I tried using some angle chasing then I used cosine formula for finding the two equal segment but that goes too lengthy.
Any help will be highly appreciated!
|
Since $XY||AB$, in the standard notation we obtain:
$\measuredangle CXD=\frac{\alpha}{2},$ $\measuredangle CYE=\frac{\beta}{2},$ $\measuredangle DCX=\beta$, $\measuredangle ECY=\alpha$, $CX=b$ and $CY=a.$
Thus, by the law of sines for $\Delta DCX$ we obtain:
$$\frac{XD}{\sin\beta}=\frac{b}{\sin\left(\beta+\frac{\alpha}{2}\right)},$$ which gives
$$XD=\frac{b\sin\beta}{\sin\left(\beta+\frac{\alpha}{2}\right)}=\frac{bl_a}{c}.$$
By the same way we can get:
$$YE=\frac{al_b}{c},$$ which gives $$al_b=bl_a.$$
Now, $$l_a=\frac{2bc\cos\frac{\alpha}{2}}{b+c}=\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2}}{2}}}{b+c}=\frac{2\sqrt{(a+b+c)(b+c-a)bc}}{b+c}.$$
By the same way:
$$l_b=\frac{2\sqrt{(a+b+c)(a+c-b)ac}}{a+c}.$$
Thus, $$a^3(a+c-b)(b+c)^2=b^3(b+c-a)(a+c)^2.$$
Now, let $a>b$.
Thus, $$a+c-b>b+c-a,$$
$$a^2(b+c)^2>b^2(a+c)^2,$$ which gives $$a^3(a+c-b)(b+c)^2>b^3(b+c-a)(a+c)^2,$$
which is a contradiction.
By the same way if $a<b$ so
$$a^3(a+c-b)(b+c)^2<b^3(b+c-a)(a+c)^2,$$ which is a contradiction again.
Id est, $a=b$ and we are done!
|
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|
Limit of multiple absolute values Let $f(x) = \frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}$. Find $\lim_{x\rightarrow -1}f(x)$.
I broke up each absolute value into parts:
$|1-x^2| = \begin{cases}
1-x^2 & -1 \leq x\leq 1 \\
-(1-x^2) & x>1,x<-1
\end{cases}
$ ,
$|x+1| = \begin{cases}
x+1 & x\geq -1 \\
-(x+1) & x<-1
\end{cases}
$,
$|x^2+x| = \begin{cases}
x^2+x & x\geq0,x\leq -1 \\
-(x+1) & -1<x<0
\end{cases}
$
Thus, when $x\rightarrow -1$, the function will approach the positive value, because by my definitions of the absolute values,each has the positive value at $x=-1$. So then you can take
$\lim_{x\rightarrow -1}\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\lim_{x\rightarrow -1}\frac{x^2+2x-3(1-x^2)+1}{2(x+1)-(x^2+x)}=\lim_{x\rightarrow -1}\frac{(4x-2)(x+1)}{(x+1)(-x+2)}=-2$.
But looking at the graph, the limit is -6. So I must have messed up in my absolute value declarations, most likely in the third one.
So, how would I correctly declare and solve this limit?
|
Note that\begin{align}\require{cancel}f(x)&=\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}\\&=\frac{|x+1|^{\cancel2}-3\cancel{|x+1|}|x-1|}{\cancel{|x+1|}(2-|x|)}\\&=\frac{|x+1|-3|x-1|}{2-|x|}\\&\to_{x\to-1}\frac{-6}1\\&=-6.\end{align}
|
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|
Asymptotic expansion, solving roots to an equation with dominant balance, what went wrong in my approach? So I wanted to compute the asymptotic expansion of the roots to, as $\epsilon \to 0$,
$$\epsilon x^3-x^2+2x-1=0$$
Now when I tried to find $x\sim x_0+\epsilon x_1+\epsilon^2x_2+...$ I ran into trouble as at $O(\epsilon)$, I had $0=2x_1-2x_1+1.$ Now I just wanted to know why did I reach such contradiction? Is it because I implicitly balanced $x$ with $1$? In other words, is it because if I assume $x=O(1)$ then $x^2$ would also have order $1$ and thus the two terms I balanced are not dominant enough?
Moreover, I applied dominant balance to $\epsilon x^3$ and $x^2$ to reach one of the root, how could I reach the other two roots via dominant balance? This is because if I balance any other two terms in the above expression, there is always another term that is larger or have the same order as $\epsilon \to 0.$
Many thank in advance!
|
Another way to obtain the expansion is by using series inversion. We have
\begin{align*}
\varepsilon = \frac{{(x - 1)^2 }}{{x^3 }} & = \frac{{(x - 1)^2 }}{{1 + 3(x - 1) + 3(x - 1)^2 + (x - 1)^3 }} \\ & = (x - 1)^2 (1 - 3(x - 1) + 6(x - 1)^2 - \cdots ),
\end{align*}
and thus
\begin{align*}
\pm \sqrt \varepsilon & = (x - 1)\left( {1 - \frac{3}{2}(x - 1) + \frac{{15}}{8}(x - 1)^2 - \cdots } \right) \\ & = (x - 1) - \frac{3}{2}(x - 1)^2 + \frac{{15}}{8}(x - 1)^3 - \cdots \,.
\end{align*}
By series inversion
$$
x - 1 = \pm \sqrt \varepsilon + \frac{3}{2}\varepsilon \pm \frac{{57}}{8}\varepsilon ^{3/2} + \cdots ,
$$
i.e.,
$$
x = 1 \pm \sqrt \varepsilon + \frac{3}{2}\varepsilon \pm \frac{{57}}{8}\varepsilon ^{3/2} + \cdots \,.
$$
Addendum. Following David's comment, I add the expansion for the third root. By Viète's formula and simple series manipulation, we find
\begin{align*}
x & = \frac{1}{\varepsilon }\frac{1}{{1 + \sqrt \varepsilon + \frac{3}{2}\varepsilon + \frac{{57}}{8}\varepsilon ^{3/2} + \cdots }}\frac{1}{{1 - \sqrt \varepsilon + \frac{3}{2}\varepsilon - \frac{{57}}{8}\varepsilon ^{3/2} + \cdots }}
\\ & = \frac{1}{\varepsilon }\frac{1}{{1 + 2\varepsilon + \cdots }} = \frac{1}{\varepsilon }(1 - 2\varepsilon + \cdots ) = \frac{1}{\varepsilon } - 2 + \cdots \,.
\end{align*}
|
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|
The limit and Asymptotic Behavior Wallis-like Integral It is known that $$I_n=\int^{\frac{\pi}{2}}_{0} x \cos^n xdx \quad (n=0,1,2,3,\dots),$$ then
$$\lim_{n \to \infty} nI_{n}=1.$$
My question is: does there exist $k$ such that
$$\lim_{n \to \infty} n^k(nI_{n}-1)$$
converges to a non-zero real number?
|
You can use Laplace's method to establish the asymptotics
$$
\int_0^{\pi /2} {x\cos ^n xdx} = \int_0^{\pi /2} {xe^{n\log \cos x} dx} = \frac{1}{n} - \frac{2}{3} \frac{1}{n^2} + \mathcal{O}\!\left( {\frac{1}{{n^{5/2} }}} \right).
$$
Thus
$$
n\cdot (nI_n - 1) \to - \frac{2}{3} ,
$$
i.e., $k=1$.
Addendum 1. A more straightforward way is to use Watson's lemma. By the change of integration variable $x = \arccos e^{ - t}$,
$$
I_n = \int_0^{ + \infty } {e^{ - nt} \frac{{\arccos (e^{ - t} )}}{{\sqrt {e^{2t} - 1} }} dt} .
$$
Near the origin,
$$
\frac{{\arccos (e^{ - t} )}}{{\sqrt {e^{2t} - 1} }} = 1 - \frac{2}{3}t + \frac{2}{{15}}t^2 + \frac{4}{{315}}t^3 + \cdots .
$$
Therefore, by Watson's lemma,
$$
I_n \sim \frac{1}{n} - \frac{2}{3}\frac{1}{{n^2 }} + \frac{4}{{15}}\frac{1}{{n^3 }} + \frac{8}{{105}}\frac{1}{{n^4 }} + \cdots
$$
as $n\to +\infty$.
Addendum 2. Just for fun. You can rewrite $I_n$ in the form
$$
I_n = \int_0^{ + \infty } {e^{ - (n + 1)t} \frac{{\arcsin (\sqrt {1 - e^{ - 2t} } )}}{{\sqrt {1 - e^{ - 2t} } }}dt} .
$$
Substituting the Taylor series of the inverse sine around the origin yields the series
\begin{align*}
I_n & = \sum\limits_{k = 0}^\infty {\frac{{(2k)!}}{{(2^k k!)^2 }}\frac{1}{{2k + 1}}\int_0^{ + \infty } {e^{ - (n + 1)t} (1 - e^{ - 2t} )^k dt} } \\ & = \sum\limits_{k = 0}^\infty {\frac{{(2k)!}}{{2^k k!}}\frac{1}{{2k + 1}}\frac{1}{{(n + 1)(n + 3) \cdots (n + 2k + 1)}}} ,
\end{align*}
which coverges for all positive integer $n$.
Addendum 3. An elementary argument. You make the substitution as in Addendum 1. Because the function tends to $0$ exponentially fast at infinity, taking into account its Taylor series, it is readily seen that there is a $C>0$ such that $$
\left| \frac{\arccos (e^{ - t} )}{\sqrt {e^{2t} - 1} } - 1 + \frac{2}{3}t\right| \le Ct^2
$$ for all $t>0$. Then
\begin{align*}
\left| {I_n - \frac{1}{n} + \frac{2}{3}\frac{1}{{n^2 }}} \right| & = \left| {\int_0^{ + \infty } {e^{ - nt} \left( {\frac{{\arccos (e^{ - t} )}}{{\sqrt {e^{2t} - 1} }} - 1 + \frac{2}{3}t} \right)dt} } \right| \\ & \le \int_0^{ + \infty } {e^{ - nt} \left| {\frac{{\arccos (e^{ - t} )}}{{\sqrt {e^{2t} - 1} }} - 1 + \frac{2}{3}t} \right|dt} \le C\int_0^{ + \infty } {e^{ - nt} t^2 dt} = \frac{{2C}}{{n^3 }}
\end{align*}
for all $n\geq 1$.
|
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|
How do you calculate phase of $\text{DTFT}$ when exponential is in denominator? $$Xe^{jw}=\frac{1}{1-ae^{jw}}$$
How is the phase of this derived as $\tan ^{-1}\left(a*\sqrt{\left(1-a^2\right)}\right)$?
|
$$\begin{align*} X\left(e^{j\omega}\right) &= \dfrac{1}{1-ae^{-j\omega}}\\
\\
&= \dfrac{1}{1-ae^{-j\omega}}\cdot\dfrac{1-ae^{j\omega}}{1-ae^{j\omega}} \\
\\
&= \dfrac{1-ae^{j\omega}}{1 -ae^{-j\omega} -ae^{j\omega} + a^2}\\
\\
&= \dfrac{1-ae^{j\omega}}{1 -2a\cos\omega + a^2}\\
\\
&= \dfrac{1-a\cos\omega}{1 -2a\cos\omega + a^2}+ j\dfrac{-a\sin\omega}{1 -2a\cos\omega + a^2}\\
\end{align*}$$
So the phase is given by:
$$\begin{align*}\angle X\left(e^{j\omega}\right) &= \mathrm{arctan2}\left(\dfrac{-a\sin\omega}{1 -2a\cos\omega + a^2}, \dfrac{1-a\cos\omega}{1 -2a\cos\omega + a^2}\right)\\
\end{align*}$$
And the magnitude is given by:
$$\begin{align*}\left|X\left(e^{j\omega}\right)\right| &= \dfrac{\sqrt{(1-a\cos\omega)^2+(-a\sin\omega)^2}}{1 -2a\cos\omega + a^2}\\
\\
&= \dfrac{\sqrt{1-2a\cos\omega+a^2\cos^2\omega+a^2\sin^2\omega}}{1 -2a\cos\omega + a^2}\\
\\
&= \dfrac{1}{\sqrt{1 -2a\cos\omega + a^2}}\\
\end{align*}$$
|
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|
The result of $\int{\sin^3x}\,\mathrm{d}x$ $$\int{\sin^3x}\,\mathrm{d}x$$
I find that this integration is ambiguous since I could get the answer with different approaches. Are these answers are valid and true? Could someone tell me why and how? And also, is there any proof stating that these two method I use results the same value/answer?
Here how I work, please correct me if I'm wrong
First method :
\begin{align}
\int{\sin^3x}\,\mathrm{d}x & = \int{\sin x \cdot \sin^2x}\,\mathrm{d}x
\\ &= \int{\sin x (1 - \cos^2x)}\,\mathrm{d}x
\\& = \displaystyle\int{(\sin x - \sin x\cos^2x)}\,\mathrm{d}x
\\& = \dfrac{1}{3}\cos^3x - \cos x + C
\end{align}
Second method :
First, we know that $$\sin 3x = 3\sin x - 4\sin^3x$$
Therefore, $$\sin^3x = \dfrac{3}{4}\sin x - \dfrac{1}{4}\sin 3x$$
\begin{align}
\int{\sin^3x}\,\mathrm{d}x & = \int{\left(\frac{3}{4}\sin x - \frac{1}{4}\sin 3x\right)}\,\mathrm{d}x\\
& = \frac{1}{12}\cos 3x - \frac{3}{4}\cos x + C
\end{align}
|
Yes, they are both valid and true. Actually,$$(\forall x\in\Bbb R):\frac13\cos^3(x)-\cos(x)=\frac1{12}\cos(3x)-\frac34\cos(x)$$since$$(\forall x\in\Bbb R):\cos(3x)=4\cos^3(x)-3\cos(x).$$
|
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|
Find a and c in the equation Find a and c in the equation
$$y=ca^x$$
when it is given that $x = \frac{1}{2}$ gives $y = 6$ and $x = 2$ gives $y = 162$
I already tried to solve when $6=ca^{\frac{1}{2}}$ and managed to get $a=\frac{36}{c^2}$ and when I insert $a$ in the equation $6=c\cdot \left(\frac{36}{c^2}\right)^{\frac{1}{2}}$, then I end up getting $0$ for $c$. Would appreciate help :)
|
Suppose we have $y=ca^{x}$. Then substituting the points $(x,y)=(\frac{1}{2},6)$ and $(x,y)=(2,162)$ gives $$6=ca^{\frac{1}{2}}$$
$$162=ca^{2}$$
Dividing the two we obtain $a^{\frac{3}{2}}=\frac{162}{6}=27$ so $a=9$. Now substituting into the first we obtain $$6=c(9)^{\frac{1}{2}}=3c$$
so $c=2$. Thus we have $y=2\cdot 9^{x}$.
|
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|
How to show that $\frac{n^3}{\sqrt{n^6 + 1}}$ converges to 1. I want to show that $\frac{n^3}{\sqrt{n^6 + 1}}$ converges to 1. I've tried using $\epsilon-N$ to show that it converges but I can't isolate $n$. Is there a way to bound this so I can apply the squeeze theorem? Or am I not on the right track?
|
If you want to use the sqeeze theorem you can use
$\frac {n^3}{(n+1)^3} = \frac {n^3}{\sqrt{(n+1)^6}}=\frac {n^3}{n^6+ 6n^5 + .... + 1} < \frac {n^3}{\sqrt{n^6 + 1}} < \frac {n^3}{\sqrt{n^6}}=\frac {n^3}{n^3} =1$ (but you will have to prove $\frac {n^3}{(n+1)^3} \to 1$).
But to do an $\epsilon-N$ proof:
$|\frac {n^3}{\sqrt{n^6 + 1}} - 1| < \epsilon\Leftarrow$
$1-\epsilon < \frac {n^3}{\sqrt{n^6+1}} < 1 + \epsilon$
$1 -2\epsilon + \epsilon^2 < \frac {n^6}{n^6 +1} < 1 + 2\epsilon + \epsilon^2$ (assuming $\epsilon < 1$)
$1-3\epsilon < 1-\epsilon=1-2\epsilon +\epsilon < 1-2\epsilon + \epsilon^2$ and $1+2\epsilon + \epsilon^2 < 1+3\epsilon$. So if we assumme $\epsilon < \frac 13$ then
$|\frac {n^6}{n^6 + 1}-1| < 3\epsilon$
$|\frac {n^6}{n^6 + 1} -\frac {n^6+1}{n^6+1}| < 3\epsilon$
$|-\frac {1}{n^6+1}|< 3\epsilon$
$1 < 3(n^6 +1) \epsilon$
$\frac 1{3\epsilon} -1 < n^6$ so if $N \ge \sqrt[6]{\frac 1{3\epsilon'} -1}$ (where $\epsilon' = \min(\epsilon, \frac 13)$ we are done.
$n > N\implies n^6 > \frac 1{3\epsilon} -1\implies |\frac {n^3}{\sqrt{n^6+1} -1}|<\epsilon$.
|
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|
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$
Also I got $2(xy+yz+xz)+(x^2+y^2+z^2)=4$.
But I think it is a wrong method to proceed...
|
$xy+yz+xz-xyz=(1-x)(1-y)(1-z)+x+y+z-1=(1-x)(1-y)(1-z)+1$.
And $x+y+z=2$ gives $(1-x)+(1-y)+(1-z)=1$.
$1-x,1-y,$ and $1-z$ are all strictly positive (otherwise we couldn't make a triangle out of $x,y,$ and $z$). So let $a=1-x,b=1-y,$ and $c=1-z$. We have
$$a,b,c\in(0,1)$$
$$a+b+c=1$$
From $AM\ge GM$, their product satisfies $abc\in(0,\frac{1}{27}]$. You can take it from there.
|
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|
$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists.
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
My proof is here:
Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}}$.
So, $\lim_{h\to 0+} \frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = -\infty$.
So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
I want an easier proof.
Is there any general theorem or general proposition to prove the above fact?
By the way, the following proposition is not true. $f(x) = x^{\frac{1}{3}}$ is not differentiable at $x = 0$ and $g(x) = x^3$ is differentiable at $x = 0$ but $f(g(x)) = x$ is differentiable at $x=0$.
If $f(x)$ is not differentiable at $x = a$ and $g(x)$ is differentiable at $x = b$ and $g(b) = a$, then $f(g(x))$ is not differentiable at $x = b$.
So, we cannot prove as follows:
$\sqrt{x}$ is not differentiable at $x = 0$. And $1-x^2$ is differentiable at $x = 1$. So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
|
Let $g(x) = 1 - x^2$. Then
\begin{align}
g(f(x))
&= 1 - \left(\sqrt{1-x^2} \right)^2\\
&= 1 - \left(1-x^2 \right)\\
&= x^2\\
\end{align}
so
$$
(g \circ f)'(x) = 2x \tag{1}
$$ for every $x$.
$f$ is evidently differentiable for $-1 < x < 1$.
Suppose that $f$ were differentiable at $x = 1$ as well.
Then (by a version of the chain rule extended to deal with one-sided limits), we'd have
$$
(g\circ f)'(1) = g'(f(1)) \cdot f'(1)
$$
Using equation $1$, this becomes
$$
2 = g'(0) \cdot f'(1)
$$
Now $g'(x) = -2x$, so $g'(0) = 0$, so we would have (assuming $f$ is differentiable!) that
$$
2 = 0 \cdot f'(1)
$$
which is impossible. Hence our assumption that $f$ is differentiable (at $1$) must be incorrect.
Is this "simpler"? Probably not. After all, you have to review your proof of the chain rule to confirm that it works when some/all limits are one-sided. That's straightforward, but involves some writing.
|
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|
How to solve $\cos x-\sin 3x=\cos 2x$? My attempt:
$$\begin{align} (\cos x- \cos 2x) - \sin 3x &= 0 \\
2\sin \frac{3x}{2}\sin \frac{x}{2} - 2 \sin \frac{3x}{2}\cos \frac{3x}{2}&=0\\
\sin\frac{3x} {2} \left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)&=0 \end{align}$$
Now either $\sin\frac{3x} {2} = 0$ or $\left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)=0$. Solving for the former,
$$\frac{3x}{2}=n\pi\rightarrow x = \frac{2n\pi}{3}$$
How can I solve $\left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)=0$? I know the elementary identities involving trigonometric ratios, but not complex numbers or calculus.
|
I think you can use the fact that $\sin(x) = \cos(\frac{\pi}{2}-x)$ to translate the equation into one involving only sines (or cosines) on both sides, and then compare directly.
|
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|
Hard limit $\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$ Prove that :
$$\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$$
I can prove that :
$$\lim_{x\to\infty}\frac{(x(x+1)(x+2))^{\frac{1}{3}}-2(x+1)}{-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}}=1$$
Using the Hospital rule but it doesn't help here .Moreover I have tried power series without success .
Any helps is welcome .
Thanks in advance
|
We have that
$$\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=$$
$$=x\left(\left(1+\frac1x\right)^{\frac{1}{3}}\left(1+\frac2x\right)^{\frac{1}{3}}-\left(2\left(1+\frac1x\right)-\left(\left(1+\frac1x\right)^{x+1}\left(1+\frac2x\right)^{x+2}\right)^{\frac{1}{3x+3}}\right)\right)=$$
and since by binomial expansion we have
*
*$\left(1+\frac1x\right)^{\frac{1}{3}}\left(1+\frac2x\right)^{\frac{1}{3}}=\left(1+\frac1{3x}+O\left(\frac1{x^2}\right)\right)\left(1+\frac2{3x}+O\left(\frac1{x^2}\right)\right)=1+\frac1x+O\left(\frac1{x^2}\right)$
*$\left(\left(1+\frac1x\right)^{x+1}\left(1+\frac2x\right)^{x+2}\right)^{\frac{1}{3x+3}}=\left(1+\frac1x\right)^{\frac13}\left(1+\frac2x\right)^{\frac13}\left(1+\frac2x\right)^{\frac1{3x+3}}=1+\frac1x+O\left(\frac1{x^2}\right)$
then
$$=x \left(1+\frac1x+O\left(\frac1{x^2}\right)-2\left(1+\frac1x\right)+1+\frac1x+O\left(\frac1{x^2}\right)\right)=O\left(\frac1{x}\right) \to 0$$
|
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|
Volume with spherical polar coordinates Determine the volume between the surface $z=\sqrt{4-x^2-y^2}$ and the area of the xy plane determined by $x^2+y^2\le 1,\ x+y>0,\ y\ge 0$.
I convert to spherical polar coordinates.
$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$
$$\iiint_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi drd\phi d\theta$$
I can't figure out how to take $\int_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi dr$, which makes me think I made a mistake somewhere.
EDIT: Thanks for all the answers.
Now I understand how the limits of $\theta ,r,z$ works.
I don't fully understand where the function "disappear".
$\sqrt {4-x^2-y^2} =\sqrt {4-r^2}$
Why isn't it then:
$\int \int \int _{K} {\sqrt {4-r^2}rdzdrd\theta }$
|
This is much easier to solve in cylindrical coordinates. $$x=r\cos\theta\\y=r\sin\theta\\z=h$$
Then the limits for $r$ are $0$ and $1$, the limits for $\theta$ are from $-\frac\pi4$ to $\frac{3\pi}4$, and the limits for $h$ are $0$ and $4-r^2$.
With these, $$V=\int_{-\frac\pi4}^{\frac{3\pi}4}d\theta\int_0^1dr\cdot r\int_0^{\sqrt{4-r^2}}dh$$
Note see comment below. Since $y>0$, the lower limit for $\theta$ is $0$, not $-\pi/4$
|
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|
prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$
Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$
My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$
As inequality is cyclic:
let $a\ge b\ge c$ then $$\frac{1}{a^2+2a}\le \frac{1}{b^2+2b}\le \frac{1}{c^2+2c}$$.The rest follows by rearrangement inequality.
The case $a\ge c\ge b$ is analogous.
Thus Proved!
Is it correct?...And any other alternative ways possible?
|
Your proof is nice and right.
Another way.
Let $c=\min\{a,b,c\}$.
Thus, we need to prove that:
$$\frac{a}{b}+\frac{b}{a}-2+\frac{c}{a}-\frac{b}{a}+\frac{b}{c}-1\geq\frac{a+2}{b+2}+\frac{b+2}{a+2}-2+\frac{c+2}{a+2}-\frac{b+2}{a+2}+\frac{b+2}{c+2}-1$$ or
$$\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\geq\frac{(a-b)^2}{(a+2)(b+2)}+\frac{(c-a)(c-b)}{(a+2)(c+2)},$$ which is obvious.
|
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|
Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$ So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$
So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that the solution is $n^3$, do I have to prove $(n+1)^3$ is equal to that or any other method or I prove that $(n+1)^3-n = 2m^2+2m+2+n_m$ (the differience between $(n+1)$ and $n$
|
Approaching by pattern,
*
*We know that sum of first $k$ odd numbers is $k^2$.
*We have $1,2,3,4\ldots$ terms which may be expressed as difference of triangular number of terms
$$ 1 = \underset{1}{\underbrace{1}} - \underset{0}{\underbrace{0}} $$ $$ 3+5 = \underset{\text{first 3 odd}}{\underbrace{(1+3+5)}} - \underset{\text{first odd}}{\underbrace{1}}$$ $$ 7+9+11 = \underset{\text{first 6 odd}}{\underbrace{(1+3+5+7+9+11)}}-\underset{\text{first 3 odd}}{\underbrace{(1+3+5)}}$$
In particular, $n^{th}$ sum is $$\text{sum of first $T_{n}$ odd numbers - sum of first $T_{n-1}$ odd numbers}$$
*
*Thus $n^{th}$ sum is $$ 1 = 1^2 - 0^2 $$ $$ 3+5 = 3^2 - 1^2$$ $$ 7+9+11 = 6^2-3^2$$ $$ 13+15+17+19 = 10^2-6^2$$
In particular, $n^{th}$ sum is $$T_{n}^2-T_{n-1}^2$$
$$=\dfrac{n^2(n+1)^2}{4}-\dfrac{n^2(n-1)^2}{4}$$
$$=\dfrac{n^2}{4}(4n)$$
$$=n^3$$
Note : Once one makes these observations by naked eye, one can simply finish the problem by writing down step $3$ directly.
|
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|
Which positive integers $a$ and $b$ make $(ab)^2-4(a+b) $ a square of an integer? Which positive integers
$a$ and $b$ make
$(ab)^2-4(a+b)
$
a square of an integer?
I saw this in quora,
and found that
the only solutions with
$a \ge b > 0$
are
$(a, b, (ab)^2–4(a+b)) = (5, 1, 1)$
and $(3, 2, 16)$.
Another “solution” is
$a = b = 2, (ab)^2–4(a+b) = 0$.
My solution is
messy and computational,
and I wonder if
there is a more elegant solution.
Here is my solution.
Assume $a \ge b$
and write
$n^2 = (ab)^2–4(a+b)$ so $n < ab$.
Let $n = ab-k$
where $ab > k>0$ so
$(ab)^2–4(a+b) = (ab-k)^2 = (ab)^2–2kab+k^2$
or
$k^2–2kab+4(a+b) = 0$.
Then
$\begin{array}\\
k
&= \dfrac{2ab-\sqrt{4a^2b^2–16(a+b)}}{2}
\qquad \text{(use "-" since } k < ab)\\
&= ab-\sqrt{a^2b^2–4(a+b)}\\
&=(ab-\sqrt{a^2b^2–4(a+b)})\dfrac{ab+\sqrt{a^2b^2–4(a+b)}}{ab+\sqrt{a^2b^2–4(a+b)}}\\
&=\dfrac{4(a+b)}{ab+\sqrt{a^2b^2–4(a+b)}}\\
\end{array}
$
Therefore,
since $k \ge 1, 4(a+b) \ge ab$
so
$0 \ge ab-4(a+b) = ab-4(a+b)+16–16
=(a-4)(b-4)-16$
or $16 \ge (a-4)(b-4)$.
This gives a finite number of possible $a, b$,
all at least $4$.
Computation shows that
none of these are solutions.
To get the possible values of
$a$ and $n$ in terms of $b$
for any fixed $b$,
do this:
Since
$n^2
= a^2b^2-4(a+b)$,
$\begin{array}\\
b^2n^2
&= a^2b^4-4b^2a-4b^3\\
&= a^2b^4-4b^2a+4-4b^3–4\\
&=(b^2a-2)^2–4(b^3+1)\\
\end{array}
$
so
$4(b^3+1)
= (b^2a-2)^2-b^2n^2
= (b^2a-2-bn)(b^2a+bn)
$.
For each factorization
$r*s = 4(b^3+1)$, try
$r=b^2a-2-bn, s=b^2a-2+bn$.
This gives
$s-r=2bn$,
so if $2b$ divides $s-r$,
then
$n=\dfrac{s-r}{2b}$.
Adding $s$ and $r$,
$2b^2a-4=s+r$ so if
$2b^2$ divides $s+r+4$,
then $a = \dfrac{s+r+4}{2b^2}$.
This allows us to compute all solutions for any fixed value of b.
Running this for $1 \le b \le 16$
gives the solutions above.
For $a \ge b \ge 5$,
the restriction
$16 \ge= (a-4)(b-4)$
gives a finite set of possibilities which computation shows yields no additional solutions.
I sure would like to see
a more elegant solution.
Also,
this messy algebra
provides lots of opportunities
for errors.
|
We can find an upper bound for their "product" in the following way:
\begin{align}
{\left( {ab} \right)^2} - 4\left( {a + b} \right) &< {\left( {ab} \right)^2} \ \ (\because a,b>0) \\
{\left( {ab} \right)^2} - 4\left( {a + b} \right) &\le {\left( {ab - 1} \right)^2} \\
(2ab-1)-4(a+b) &\le 0\\
ab - 2(a+b) - \frac 12 \color{blue}{+4} &\le 0 \color{blue}{+4}\\
\left( {a - 2} \right)\left( {b - 2} \right) &\le \frac{9}{2} \\
\left( {a - 2} \right)\left( {b - 2} \right) &\le 4 \\
\end{align}
Now, seeing some cases should finish the work.
|
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|
Prime elements in $\mathbb{Z}[\sqrt{5}]$ I need to determine which of the elements $3+2\sqrt{5}$, $9+4\sqrt{5}$ and $4-\sqrt{5}$ are prime elements in $\mathbb{Z}[\sqrt{5}]$, respectively which are associated.
My ansatz is as follows:
So let $x=3+2\sqrt{5}$ divide $ab$ for $a,b \in R$. Thus, there are $u,v \in \mathbb{Z}$, such that
$$ab=(a_1+b_1\sqrt{5})(a_2+b_2\sqrt{5})=(a_1a_2+5b_1b_2)+(a_1b_2+a_2b_1\sqrt{5})=(u+v\sqrt{5})(3+2\sqrt{5})=(3u+10v)+(3v+2u)\sqrt{5}.$$
How do I go from here in order to check whether $x|a$ or $x|b$?
|
$3+2\sqrt{5}$ has norm $-11$ and $4-\sqrt{5}$ has norm $11$. Perhaps they are associates. Indeed
$$
\frac{3+2\sqrt{5}}{4-\sqrt{5}} = 2+\sqrt{5}
$$
and $2+\sqrt{5}$ is a unit because it has norm $-1$. In fact, $(2+\sqrt{5})(-2+\sqrt{5})=1$.
$9+4\sqrt{5}$ is a unit because it has norm $1$ and so $(9+4\sqrt{5})(9-4\sqrt{5})=1$.
It remains to decide whether $3+2\sqrt{5}$ is prime. Consider
$$
\frac{\mathbb Z[\sqrt{5}]}{\langle 3+2\sqrt{5} \rangle}
=
\frac{\mathbb Z[\sqrt{5}]}{\langle 3+2\sqrt{5},11 \rangle}
=
\frac{\mathbb Z[X]}{\langle X^2-5,3+2X,11 \rangle}
\cong
\frac{\mathbb Z_{11}[X]}{\langle X^2-5,3+2X \rangle}
=
\frac{\mathbb Z_{11}[X]}{\langle X^2-5,2(7+X) \rangle}
=
\frac{\mathbb Z_{11}[X]}{\langle X^2-5,X+7 \rangle}
=
\frac{\mathbb Z_{11}[X]}{\langle X+7 \rangle}
\cong
\mathbb Z_{11}
$$
which is a domain. Therefore, $3+2\sqrt{5}$ is prime.
|
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|
A unique question in combinatorics and non-commutative variables. Let $x,y$ be two variables that satisfy:
$xy=yx+1$ (they are not commutative).
Find $(xy)^2 ,(xy)^3, (yx)^2, (yx)^3$ as a linear combination in terms of $y^jx^j$.
Then find a formula for $(xy)^n$ and $(yx)^n$.
After some calculations we get:
$(xy)^2=y^2x^2+3yx+1$.
$(xy)^3=y^3x^3+6y^2x^2+7yx+1$.
$(xy)^4=y^4x^4+10y^3x^3+25y^2x^2+15yx+1$.
$(yx)^2=y^2x^2+yx$.
$(yx)^3=y^3x^3+3y^2x^2+y$
Now, given $xy=yx+1$ we can use the binomial formula:
$(xy)^n=\sum_{k=0}^{n} {n \choose k} (yx)^k$.
Denote:
$(yx)^{n}=\sum a_{n,k} y^kx^k$.
Then,
$(yx+1)^n=(yx)(yx)^n=\sum {n\choose k} yxy^kx^k$.
After computing we get,
$(yx)^{n+1}= \sum a_{n,k} (ky^kx^k+y^{k+1}x^{k+1})$.
By comparing the coefficient of $[y^kx^k]$ in $(yx)^{n+1}$:
$a_{n,k}=ka_{n-1,k}+a_{n-1,k-1}$
|
I found different formulas, namely
\begin{align*}
(xy)^1 & = yx+1 \\
(xy)^2 & = y^2x^2+3yx+1\\
(yx)^3 & = y^3x^3+6y^2x^2+7yx+1\\
(yx)^4 & = y^4x^4+10y^3x^3+25y^2x^2+15yx+1\\
(yx)^5 & = y^5x^5+15y^4x^4+65y^3x^3+90y^2x^2+31yx+1
\end{align*}
if correct, can you guess a general formula?
|
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|
Show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges How would you show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges?
My approach
At first glance I would go for the alternate series criterion and show that $\frac{\left(1+\frac{1}{n}\right)^n}{n}$ is a monotone sequence with $\lim\limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}=0$. So after a few algebraic manipulations I get:
$$
\frac{\frac{\left(1+\frac{1}{n}\right)^n}{n}}{\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{n+1}}=
\frac{\left(1+\frac{1}{n}\right)^n(n+1)}{\left(1+\frac{1}{n+1}\right)^{n}\left(1+\frac{1}{n+1}\right)n}=
\left(\frac{1+\frac{1}{n}}{1+\frac{1}{n+1}}\right)^{n}\cdot \frac{n+1}{\frac{n^2+2n}{n+1}}=
\cdots=\frac{n^2+2n+1}{n^2+2n}\left(\frac{n^2+2n+1}{n^2+2n}\right)^n\geq1.
$$
So the expression is monotonically decreasing. Further, we know that $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e^1$. Hence, $\left(1+\frac{1}{n}\right)^n$ is bounded, say $S\geq\Big| \left(1+\frac{1}{n}\right)^n\Big|$. This allows us to conclude:
$$
\Big| \frac{\left(1+\frac{1}{n}\right)^n}{n}-0\Big|\leq \frac{S}{n}<\epsilon,\text{ where n is sufficiently large enough}.
$$
Hence, $\lim\limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}=0$ and by the alternate series crterion the series $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges.
Is this correct? Is there a faster way?
|
HINT
We have that
$$\left(1+\frac{1}{n}\right)^n= e^{n\log\left(1+\frac{1}{n}\right)}=e^{1-\frac1{2n}+O\left(\frac1{n^2}\right)}=e-\frac e{2n}+O\left(\frac1{n^2}\right)$$
therefore
$$\frac{\left(1+\frac{1}{n}\right)^n}{n}=\frac e n-\frac e{2n^2}+O\left(\frac1{n^3}\right)$$
|
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|
Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$ Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$.
I tried using substitution as follows:
$$ a = x^2 $$
$$ b = (m + 1)^2$$
Using the substitution:
$$ a^2 - (1 + b)a + b \geq 0$$
Substituting in the second degree equation:
$$ \frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
Solving the binomial coefficients $(m + 1)^4$ and $(m + 1)^2$, I got:
$$ \frac{-(1 + (m + 1)^2) + \sqrt{m^4 + 4m^3 + 8m^2 + 8m + 8}}{2} $$
From here I don't know how to proceed...
|
Here's how it's done in high school, using Vieta's relations and results on the sign of a quadratic polynomial:
First set $y=x^2$ and solve for $y$: you have the quadratic inequation
$$y^2-\bigl(1+(m+1)^2\bigr)y+(m+1)^2 \ge0,\qquad y\ge 0.$$
The discriminant is
$$\Delta=\bigl(1+(m+1)^2\bigr)^2-4(m+1)^2=\bigl(1-(m+1)^2\bigr)^2,$$
therefore the quadratic polynomial in $y$ has two roots $y_0$ and $y_1\;(y_0\le y_1)$.
Furthermore these roots are nonnegative, since $y_0+y_1=\bigl(1+(m+1)^2\bigr)^2>0\:$ and $\:y_0y_1=(+1)^2\ge 0$.
Conclusion; the original inequation is equivalent to
$$0\le y\le y_0\;\text{ or }\;y\ge y_1\iff \begin{cases}-\sqrt{y_0\mathstrut}\le x\le \sqrt{y_0\mathstrut},\\
\quad\text{ or }\\ x\le -\sqrt{y_1\mathstrut},\enspace x\ge\sqrt{y_1\mathstrut}. \end{cases}$$
|
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|
Proving the duplication formula I'm trying to prove the famous Legendre duplication formula:
$$ \Gamma(2z) = \frac{2^{2z-1} \Gamma(z) \Gamma(z+\frac{1}{2})}{\sqrt{\pi}} $$
I have to prove this using these two identities/theorems:
$$ \Gamma(z)= \lim_{n \rightarrow \infty} \frac{n! n^{z}}{z(z+1)...(z+n)}$$
and
$$ (2n)! \sim \frac{2^{2n} (n!)^2}{\sqrt{\pi n}} $$ as $n \rightarrow \infty $
I put in $ z = 2z $ in the first one and obtained:
$$ \Gamma(2z) = \lim_{n \rightarrow \infty} \frac{n! n^{2z}}{2z(2z+1)...(2z+n)} $$.
I then tried multiplying both the numerator and denominator by various factors to try to incorporate the other identity, but it was of no use. Can anyone tell me what direction I'm supposed to go from here?
|
We have (with all limits for $n\to\infty$):
$$
\begin{aligned}
\Gamma(z)
&=
\lim \frac{n!\; n^{z}}{z(z+1)(z+2)\dots(z+n)}\ ,
\\
\Gamma\left(z+\frac 12\right)
&=
\lim
\frac{n!\; n^{z+\frac 12}}
{\left(z+\frac 12\right)\left(z+\frac 32\right)\left(z+\frac 52\right)\dots\left(z+\frac {2n+1}2\right)}\ ,
\\[3mm]
\Gamma(z)\Gamma\left(z+\frac 12\right)
&=
\lim \frac{n!^2\; n^{2z+\frac 12}}{z\left(z+\frac 12\right)(z+1)\left(z+\frac 32\right)(z+2)\left(z+\frac 52\right)\dots(z+n)\left(z+\frac {2n+1}2\right)}\ ,\\
&=
\lim \frac{n!^2\; n^{2z+\frac 12}\; 2^{2n+2}}{(2z)(2z+1)(2z+2)\dots(2z+2n)(2z+2n+1)}\ ,
\\[3mm]
\Gamma(2z)
&=
\lim \frac{n!\; n^{2z}}{(2z)(2z+1)(2z+2)\dots(2z+n)}\ ,
\\
&=
\lim \frac{(2n)!\; (2n)^{2z}}{(2z)(2z+1)(2z+2)\dots(2z+2n)}\ ,\qquad\text{(subsequence)}
\\[3mm]
\frac
{\Gamma(z)\Gamma\left(z+\frac 12\right)}
{\Gamma(2z)}
&=
\lim\frac
{n!^2\; n^{2z+\frac 12}\; 2^{2n+2}}
{(2n)!\; (2n)^{2z}}
\cdot
\frac{1}{2z+2n+1}\qquad\text{ (now use Stirling)}
\\
&=
\lim\frac
{\displaystyle
{\color{blue}{\left(\frac ne\right)^n}} \sqrt {2\pi n}\;
{\color{blue}{\left(\frac ne\right)^n}} \sqrt {2\pi n}\;
{\color{green}{n^{2z}}}\cdot n^{\frac 12}\; \color{red}{2^{2n+\color{navy}2}} }
{\displaystyle
{\color{blue}{\left(\frac {{\color{red}{2}}n}e\right)^{2n}}}\sqrt {2\pi \;2n}\;
\; 2^{2z}\;
{\color{green}{n^{2z}}}}
\cdot
\frac{1}{2z+2n+1}
\\
&=
\lim\frac{\sqrt {2\pi}\cdot \sqrt {2\pi}}{\sqrt {2\pi}\cdot\sqrt 2}
\cdot
\frac{\sqrt n\cdot\sqrt n\cdot n^{\frac 12}}{\sqrt n(2z+2n+1)}
\cdot
\frac{\color{red}2^{\color{navy}2}}{2^{2z}}
\\
&=\frac{\sqrt \pi}{2^{2z-1}}\ ,
\end{aligned}
$$
hence the formula.
|
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|
Implicit differentiation $(\sin x +y)/(\cos y +x)=1$ I am currently stuck on this problem
Find $\frac{dy}{dx}$ when
$$\frac{\sin x+y}{\cos y+x}=1$$
I've been using the quotient rule to try and differentiate this, but my online homework rejected my answer of:
$$\frac{-x\cos x - \cos x\cos y +y+\sin x}{\cos y +x-\sin y \sin x+y\sin y}$$
Can anyone tell me where I went wrong, or if there is another way to do this besides the quotient rule?
|
The use of the Quotient Rule doesn't need to be complicated, as long as one resists differentiating any expressions that don't really need that (or writing out components until we finally need them). If we call the ratio $ \ \large{\frac{\sin x \ + \ y}{\cos y \ + \ x}} \ \normalsize{= \ \frac{f(x)}{g(x)}} \ = \ 1 \ \ , $ then we can write
$$ \frac{g(x) · (\cos x \ + \ y') \ - \ f(x) · (-\sin y · y' \ + \ 1)}{[ \ g(x) \ ]^2} \ \ = \ \ 0 $$
$$ \Rightarrow \ \ \frac{g(x) · y' \ + \ f(x) · \sin y · y' }{[ \ g(x) \ ]^2} \ \ = \ \ \frac{-g(x) · \cos x \ + \ f(x) }{[ \ g(x) \ ]^2} $$
$$ \Rightarrow \ \ y' \ \ = \ \ \frac{f(x) \ - \ g(x) · \cos x }{g(x) \ + \ f(x) · \sin y} \ \ = \ \ \frac{ (\sin x \ + \ y) \ - \ (\cos y \ + \ x) · \cos x }{ (\cos y \ + \ x) \ + \ (\sin x \ + \ y) · \sin y} $$
$$ = \ \ \frac{ \sin x \ + \ y \ - \ \cos x \cos y \ - \ x \cos x }{ \cos y \ + \ x \ + \ \sin x \sin y \ + \ y \sin y} \ \ , $$
so you were nearly correct. Raffaele's comment about the sign error in the denominator is confirmed and this may well be all that the on-line system was "rejecting."
To "bridge" this to ultralegend5385's much simpler expression for the derivative, we will want to use the original equation, which tells us that $ \ \sin x \ + \ y \ = \ \cos y \ + \ x \ \ . $ Upon checking the above result for the appearance of such sums, we observe that
$$ y' \ \ = \ \ \frac{ ( \ \sin x \ + \ y \ ) \ - \ [ \ \cos x \ · \ ( \ \cos y \ + \ x \ ) \ ] }{ ( \ \cos y \ + \ x \ ) \ + \ [ \ ( \ \sin x \ + \ y \ ) · \ \sin y \ ]} \ \ = \ \ \frac{ ( \ \sin x \ + \ y \ ) \ - \ [ \ \cos x \ · \ ( \ \sin x \ + \ y \ ) \ ] }{ ( \ \sin x \ + \ y \ ) \ + \ [ \ ( \ \sin x \ + \ y \ ) · \ \sin y \ ]} $$
$$ = \ \ \frac{ ( \ \sin x \ + \ y \ ) \ · \ ( \ 1 \ - \ \cos x \ ) }{ ( \ \sin x \ + \ y \ ) \ · \ ( \ 1 \ + \ \ \sin y \ ) } \ \ . $$
|
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|
System of equations involving 4 variables
If $$a + b = 6$$
$$ax + by = 10$$
$$ax^2 + by^2 = 24$$
$$ax^3 + by^3 = 62$$
then $$ax^4 + by^4 = ?$$
I got $$a(x-1) + b(y-1) = 4$$ $$ax(x-1) + by(y-1) = 14$$ $$ax^2(x-1) + by^2(y-1) = 38$$ by subtracting the given equations and also got $$a(x-1)^2 + b(y-1)^2 = 10$$ $$ax^2(x-1)^2 + by^2(y-1)^2 = 24$$ by further subtracting the three equations.
I don't think I'm going anywhere with this process and so I'm not sure how to approach this. Any help is appreciated.
|
Use succssive eliminations.
From $(1)$, $b=6-a$. PLug in $(2)$ to get $y=\frac{a x-10}{a-6}$. Plug in $(3)$ to get $a=\frac{22}{3 x^2-10 x+12}$. Plug in $(4)$ and obtain
$$22\frac{x^2-3x+1}{3 x-5}=0 \implies x= ???$$
Now, go backward.
|
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Maxima and minima of $\frac{x^2-3x+4}{x^2+3x+4}$ without calculus How to find the minimum and max values of $y=\frac{x^2-3x+4}{x^2+3x+4}$ for all real values of $x$ without using calculus?
Perhaps it could be done graphically by noting the fact that the numerator and denominator are a pair of parabolas symmetric about the $x$ axis, but I do not know how to continue.
Thanks!
|
Let the maximum of $f(x) = \frac{x^2-3x+4}{x^2+3x+4}$ be $m$. Then:
$$x^2-3x+4 = mx^2 + 3mx + 4m$$
$$(m-1)x^2 + (3m+3)x + (4m - 4) = 0$$
We want this equation to have only one real root (a double root), so:
$$\Delta = 0 \Rightarrow (3m+3)^2-4(m-1)(4m-4) = 0.$$
A similar process for the minimum ($n$) yields the same equation, as multiplying it by $-1$ does not change the values of $m$. Therefore, both the maximum and minimum values are given by this equation.
|
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|
Integration of $\frac{1}{x(x+1)(x+2)...(x+m)}$ I came across this question,
$$\int \frac{1}{x(x+1)(x+2)...(x+m)} dx$$
I tried to split the rational function into partial fractions
$$ \frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}+...+\frac{Z}{x+m}$$
Not really sure how to proceed from here.
Can someone enlighten me with step by step explanations? I kinda get confused easily when certain steps are skipped.
|
Note
$$\frac{1}{x(x+1)(x+2)...(x+m)}=\sum_{k =0}^{m} \frac{a_k}{x+k}
$$
where $a_k$ are obtained as follows
\begin{align}
a_k &=\lim_{x\to -k} \frac{x+k}{x(x+1)(x+2)...(x+k)...(x+m)}\\
&= \frac{1}{[(-k)(1-k)(2-k)(-2)(-1)]\cdot[(1)(2)...(m-k-1)(m-k)]}\\
&=\frac1{(-1)^k k!(m-k)!}
\end{align}
Thus
$$\int \frac{dx}{x(x+1)(x+2)...(x+m)}
=\int \sum_{k =0}^{m} \frac{a_k}{x+k}dx
= \sum_{k =0}^{m} \frac{(-1)^k\ln|x+k|}{k!(m-k)!}+C
$$
|
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|
How to prove elegantly this inequality? Let $a$, $b$ and $c$ three reals greater than or equal to 1. Put $a+b+c=3u$, $ab+bc+ca=3v^2$ and $abc=w^3$. Show that
$$w-\frac{1}{w}\ge\frac{w^3}{v^2}-\frac{1}{u}$$
I am able to prove it by assuming without loss of generality that $c\ge b\ge a$ and posing $b=a+u$ and $c=a+u+v$ for $u,v\ge0$. The problem here is that when plug this two expression and develop, I get a huge polynomial of $a$ whose coefficients are all positive. I am wondering if there a possibility to obtain a more elegant solution.
|
Write the question as
$$
\sqrt[3]{abc} - \frac{1}{\sqrt[3]{abc}} \ge\!\!? \; \sqrt[3]{abc} \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} - \frac{1}{\sqrt[3]{abc}} \frac{3 \sqrt[3]{abc}}{a+b+c}
$$
By AM-GM, both following coefficients are at most $1$:
$$
A = \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} \le 1\qquad
B = \frac{3 \sqrt[3]{abc}}{a+b+c}\le 1$$
Again, rewrite the question as
$$
\sqrt[3]{abc} - \frac{1}{\sqrt[3]{abc}} \ge\!\!? \; \frac{B+A}{2}\Big(\sqrt[3]{abc} - \frac{1}{\sqrt[3]{abc}} \Big) - \frac{B-A}{2}\Big(\sqrt[3]{abc} + \frac{1}{\sqrt[3]{abc}} \Big)
$$
So the inequality is immediately proved for the case that
$$
B = \frac{3 \sqrt[3]{abc}}{a+b+c} \ge \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} = A\\
\leftrightarrow ab + bc + ca \ge (a+b+c)\sqrt[3]{abc}
$$
Now this is not generally true. By symmetry, let w.l.o.g. $c \ge b \ge a$ and realize this by writing $b=a x$ and $c = a x^2 q^3$. Since, by the original question, $a,b,c \ge 1$, we have $a \ge 1$ and $ x\ge 1$ and $ q^3 x\ge 1$ . Then we get the equivalent
$$
1 + x q^3 + x^2 q^3- (1 + x + x^2 q^3)q \ge 0 \\
\leftrightarrow q^3(1-q)(x-1/q)(x-1/q^2) \ge 0
$$
which is true (since $ x\ge 1$) for
$q \le 1$ and either $x<1/q$ or $x >1/q^2$.
So we need to consider the remaining cases 1.) $q > 1$, and 2.) $q < 1$ and $1/q<x <1/q^2$.
Taking a closer look at case 2.) we can write the condition as $q^2< q^3 x < q$ but, since also $q <1$ in this case, this never meets the required $ q^3 x\ge 1$. So case 2.) needs not be considered.
We have case 1.) $q > 1$: Using our substitutions, the original inequality gets
$$
a^2x^2q^2\Big( 1 - \frac{3xq^2}{1 + x q^3 + x^2 q^3} \Big) \ge\!\!? \; 1 - \frac{3xq}{1 + x + x^2 q^3}
$$
Again with $a \ge 1$ and $ x\ge 1$, we can prove the stronger inequality
$$
q^2 - \frac{3xq^4}{1 + x q^3 + x^2 q^3} \ge\!\!? \; 1 - \frac{3xq}{1 + x + x^2 q^3}
$$
Consider first $q$ fixed and $x$ variable. If $x$ increases, the inequality gets less tight. Inspecting the derivatives w.r.t. $x$, the difference between the LHS and the RHS increases for $x > q^{-3/2}$ and as we have $q>1$ and $x\ge1$, this is always the case. So we only need to consider the tightest case for $x=1$. This leaves us to show
$$
q^2\Big( 1 - \frac{3q^2}{1 + 2 q^3 } \Big) - \Big( 1 - \frac{3q}{2 + q^3} \Big)\ge\!\!? \; 0
$$
or
$$
\frac{(q-1)^3(2q^5 + 3q^4 + q^3 + q^2 + 3q + 2)}{(1 + 2 q^3) (2 + q^3)} \ge\!\!? \; 0
$$
which is true, since we consider $q>1$.
$\qquad \Box$
|
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|
Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$ represent the sides of a triangle whose centroid is G, then the equation of the circle which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is
|
Hint:
$k=\dfrac{11-2h}4$
For any circle, the distance of a tangent from the center $(h,\dfrac{11-2h}4)=$radius
$$r^2=\dfrac{(h-k+1)^2}{1^2+1^2}=\dfrac{\left(h+1-\dfrac{11-2h}4\right)^2}2$$
Again $r^2=(h-1)^2+(k-1)^2=(h-1)^2+\left(\dfrac{11-2h}4-1\right)^2$
Equate the two values of $r^2$ to find $h$
|
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|
Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$ Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
I couldn't do this question and hence I looked at the solution which goes as follows:
if $x=0$ then $x^2+2xy=0\ne5$ hence $x\ne0$
I state that $y=lx$. Hence the system becomes:
$x^2(1+2l)=5$, $x^2(l^2-3l)=-2$ which becomes $5l^2-11l+2=0$, hence $l=2$ or $l=\frac{1}{5}$.
Hence the possible solutions are:
$(1, 2), (-1,-2), (\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}), (-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}})$.
My question is why was I supposed to think of substituting $y=lx$? Why was this supposed to be intuitive and is there a more intuitive approach?
|
$x^2+2xy=5$, $y^2-3xy=-2$
Multiply the first by $2$, the second by $5$ and add
$$2 x^2-11 x y+5 y^2=0$$
Divide all terms by $y^2$ and set $\frac{x}{y}=z$
$$2z^2-11z+5=0\to z=\frac{1}{2};\;z=5$$
So we have $\frac{x}{y}=\frac{1}{2}\to y=2x$ and $\frac{x}{y}=5\to x = 5y$
Plug $y=2x$ into the first equation
$$x^2+2x(2x)=5\to x=\pm 1;\; (1,2);\;(-1,-2)$$
and then $x=5y$
$$25y^2+10y^2=5\to y=\pm\frac{1}{\sqrt 7};\; \left(\frac{5}{\sqrt 7},\frac{1}{\sqrt 7}\right);\;\left(-\frac{5}{\sqrt 7},-\frac{1}{\sqrt 7}\right)$$
|
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|
Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$? Consider the integral domain $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$?
I know the following elementary facts. We have
\begin{equation}
\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right] = \left\{ \frac{m + n \sqrt{5}}{2} : m, n \in \mathbb{Z} \text{ are both even or both odd} \right\}.
\end{equation}
For every $\frac{m + n \sqrt{5}}{2} \in \mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$, define its norm as usual:
\begin{equation}
N\left(\frac{m + n \sqrt{5}}{2}\right)=\frac{m^2-5n^2}{4}.
\end{equation}
Since $m, n$ are both even or both odd, it is easy to see that the norm is an integer. From this fact it is easily seen that $\frac{m + n \sqrt{5}}{2}$ is a unit of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ if and only if $m^2 - 5n^2=4$ or $m^2 - 5n^2=-4$. Now since $N(4+\sqrt{5})=11$ we easily get that $4+\sqrt{5}$ is an irreducible element of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. If $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ were a unique factorization domain, we could conclude that $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. But I do not know if $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ is a unique factorization domain. Does someone know if it is?
Thank you very much in advance for your attention.
|
Call $A = \mathbb Z \left[ \frac{1 + \sqrt 5}2\right]$. We can show that $A / (4+\sqrt 5) \cong \mathbb Z/11 \mathbb Z$, so that the ideal $(4 + \sqrt 5)$ is maximal.
*
*As $N(4 + \sqrt 5) = 11$, it is clear that the elements $0, 1, \ldots, 10$ are pairwise incongruent modulo $4 + \sqrt 5$.
*Every element of $A$ is congruent to an integer modulo $4 + \sqrt 5$: indeed, if it is of the form $a + b \sqrt 5$ with $a, b \in \mathbb Z$ we can subtract a suitable integer multiple of $4 + \sqrt5$ to land in $\mathbb Z$. If it is of the form $(a+b\sqrt5)/2$ with $a, b$ odd, we can subtract
$$\frac{1 + \sqrt5}2 \cdot (4 + \sqrt5) = \frac{9 + 5\sqrt5}2$$
to land in $\mathbb Z + \mathbb Z\sqrt5$.
Consider the ring homomorphism
$$\mathbb Z / (11) \to A / (4+\sqrt5) \,.$$
By the first observation, it is injective. By the second, it is surjective.
|
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|
How to prove $\lim_{n \to \infty} \frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1})= -\infty$? I tried to prove by using definition for divergence:
We say that a sequence $a_n$ diverges to $-\infty$ if for every $M>0$, there is an integer $n_0$ such that $a_n < -M$ whenever $n>n_0$
So I start:
$$\frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1}) < -M$$
Because $|cos x|\leq 1:$
$$-\frac{n}{2} \leq -\frac{n}{1+\sqrt{n}} \leq \frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1}) < -M$$
$$-\frac{n}{2} < -M$$
$$n>2M$$
Therefore $n_0$ exists and we can say (for example): $n_0=2M+1$.
Is this correct? Is there some other way we could prove this?
|
Hi I have like no idea about this being proof enough for you but we usually do this:
$
\lim_{n \to \infty} \frac{n}{1+\sqrt{n}}\cos\left(\frac{\pi n}{n+1}\right)= -\infty
$
$
1. \frac{\pi n}{n+1} = \frac{\pi }{1+\frac{1}{n} }
$ // divide by n
for $n\to
\infty , \frac{1}{n} = 0
$
$
\frac{\pi }{1+\frac{1}{n} } = \frac{\pi }{1+0}
$
so $\cos(\pi) = -1$ // so to get to $-\infty$ we only need to show now that:
$
2. \frac{n}{1+\sqrt{n}} \to \infty
$ // divide by n again
$
\frac{n}{1+\sqrt{n}} = \frac{n}{n\cdot\left( \frac{1}{n} + \frac{\sqrt{n}}{n}\right) }
$
for
$n-> \infty, \frac{1}{n} = 0
$
and
$ \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}=0
$
so we have
$
\frac{1}{1\cdot( 0 + 0) } = \frac{1}{0}
$ that goes to infinity
so we end with
$
\infty \cdot -1 = -\infty$
Take care! I did not write the lim in every row because of fatal lazyness
|
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|
Solving Differential Equation $y'=\cos(xy)$ My question is about solving differential equation $y'=\cos(xy)$.
I tried to changing variable $u=xy$
\begin{align*}
u &= xy \\
\frac{du}{dx} &= y+ x\frac{dy}{dx}\\
\frac{dy}{dx} &= \frac{1}{x}\frac{du}{dx}-\frac{y}{x}\\
\end{align*}
then since $u=xy\to \frac{y}{x}=\frac{u}{x^2}$
\begin{align*}
\frac{1}{x}\frac{du}{dx}-\frac{u}{x^2}-\cos(u)=\;0\\
\end{align*}
finally
\begin{align*}
x\,du-(u+x^2\cos(u))\,dx=0\\
\end{align*}
since this is not an Exact Differential Equation$($Because of the existence of $\cos(u))$, what is multiplicative "integrating" factor? Is this equation solvable or not?
|
Hint:
Let $u=xy$ ,
Then $y=\dfrac{u}{x}$
$\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}$
$\therefore\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=\cos u$
$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{x^2\cos u+u}{x^2}$
$(x^2\cos u+u)\dfrac{dx}{du}=x$
Let $v=x^2$ ,
$\dfrac{dv}{du}=2x\dfrac{dx}{du}$
$\therefore\dfrac{(x^2\cos u+u)}{2x}\dfrac{dv}{du}=x$
$(x^2\cos u+u)\dfrac{dv}{du}=2x^2$
$(v\cos u+u)\dfrac{dv}{du}=2v$
This belongs to an Abel equation of the second kind.
Let $w=v+u\sec u$ ,
Then $v=w-u\sec u$
$\dfrac{dv}{du}=\dfrac{dw}{du}-(u\tan u+1)\sec u$
$\therefore(\cos u)w\left(\dfrac{dw}{du}-(u\tan u+1)\sec u\right)=2(w-u\sec u)$
$(\cos u)w\dfrac{dw}{du}-(u\tan u+1)w=2w-2u\sec u$
$(\cos u)w\dfrac{dw}{du}=(u\tan u+3)w-2u\sec u$
$w\dfrac{dw}{du}=(u\tan u\sec u+3\sec u)w-2u\sec^2u$
|
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|
How can I rewrite $\frac{1}{(1-z)^4}$ as a power series? I'm supposed to rewrite $\frac{1}{(1-z)^4}$ as a power series, using the Cauchy-Product. It is given that $|z| < 1$.
So far I have
\begin{aligned}
\frac{1}{(1-z)^4} &= \left( \frac{1}{(1-z)^2} \right)^2\\
&= \left( \sum_{n=0}^{\infty} (n+1)z^n \right)^2\\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{n} (n-k+1)z^{n-k}kz^k \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{n} k(k+1)z^{n} \\
\end{aligned}
First of all, I'm not sure whether every step was correctly so far. Additionally, I don't know how I can reduce that to a power series.
|
The third line should be
\begin{eqnarray*}
&= \sum_{n=0}^{\infty} \sum_{k=0}^{n} (n-k+1)z^{n-k} \color{red}{(k+1)}z^k \\
\end{eqnarray*}
So we need to perform the sum
\begin{eqnarray*}
\sum_{k=0}^{n} (n-k+1)(k+1) = -\sum_{k=0}^{n} k^2+ n \sum_{k=0}^{n} k +(n+1) \sum_{k=0}^{n} 1.
\end{eqnarray*}
Recall that
\begin{eqnarray*}
\sum_{k=0}^{n} k^2= \frac{n(n+1)(2n+1)}{6}
\end{eqnarray*}
and we have
\begin{eqnarray*}
\sum_{k=0}^{n} (n-k+1)(k+1) = \frac{(n+1)(n+2)(n+3)}{6}.
\end{eqnarray*}
|
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|
critical points of $f(x,y) = \frac{\sin(x)}{1+y^2}$ so I determined: $f'(x,y) = \left(\begin{array}{c} \frac{\cos\left(x\right)}{y^2+1}\\ -\frac{2\,y\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$
and: hess-f = $\left(\begin{array}{cc} -\frac{\sin\left(x\right)}{y^2+1} & -\frac{2\,y\,\cos\left(x\right)}{{\left(y^2+1\right)}^2}\\ -\frac{2\,y\,\cos\left(x\right)}{{\left(y^2+1\right)}^2} & \frac{8\,y^2\,\sin\left(x\right)}{{\left(y^2+1\right)}^3}-\frac{2\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$
My first problem: What $x$ and $y$ do I plug in the hessian-matrix? If I solve $f'(x,y)$ for $(0,0)$ I get:
$x = \pi \cdot (k+1/2)$ or $x = \pi \cdot k$
Now, do I have to substitute these values for $x$ and solve for $y$? Seems like a huge case-differentiation to me.
...There must be an easier way, right?
|
Trying plugging in those values of $x$, you'll notice that the Hessian of $f$ doesn't change for any value of $k$.
You could consider studying the critical points of $f$ only in a $\{(x,y)\mid -\frac{\pi}2<x<\frac{\pi}2\}$ strip first, and deduce later that all the other critical points are just translated copies of these one.
|
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|
Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$ I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.
The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$
So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so that'd be a go-to. Also, since the degree of th enumerator is one less than the denominator we can maybe treat this as $du/u$ which would imply a natural log, though we don't quite know that yet.
So we decompose: I know that $x^2+3x-4$ can be factored as $(x-1)(x+4)$.
That gets me
$$\int{\frac{x}{(x-1)(x+4)}dx}$$
so I can do this:
$$\frac{Ax}{x-1}+\frac{B}{x+4} = \frac{x}{(x-1)(x+4)}$$
Which implies
$$A(x+4)+B(x-1) = x$$
and since the roots are at $x=1$ and $x=-4$, I can set it up like this:
$$5A = 1;-5B=1 \text{ and } A = \frac{1}{5}, B=-\frac{1}{5}$$
Leading to:
$$\frac{x}{5(x-1)}-\frac{1}{5(x+4)}$$
Which I can set up the integral like so:
$$\int{\frac{x}{5(x-1)}-\frac{1}{5(x+4)}dx}=\int{\frac{x}{5(x-1)}dx-\int{\frac{1}{5(x+4)}dx}}$$
I can now integrate by addition here $$\int{\frac{x}{x-1}}dx=\int{\frac{x-1+1}{x-1}}dx=x+\int{\frac{1}{x-1}}dx=x-\ln(x-1)$$
And doing the same thing for the second term and bringing back my $\frac{1}{5}$
$$\frac{1}{5}(x-\ln(x-1)+\ln(x+4))$$
I suspect there is a further simplification I could do. On a problem like this I also saw it integrated as an arctangent, but that seemed needlessly complex? In any case I was curious if I did this correctly.
|
$$\int{\frac{xdx}{x^2 + 3x -4}}=\frac{1}{2}(\int{\frac{(2x+3)dx}{x^2 + 3x -4}}-3\int{\frac{dx}{x^2 + 3x -4}})=\frac{1}{2}(\int{\frac{(2x+3)dx}{x^2 + 3x -4}}+\frac{3}{5}(\int{\frac{dx}{x +4}}-\int{\frac{dx}{x -1}}))=\frac{1}{2}\ln{|x^2+3x-4|}+\frac{3}{5}\ln{|x+4|} -\frac{3}{5}\ln{|x-1|}+C=\frac{1}{2}\ln{|x+4|}+\frac{1}{2}\ln{|x-1|}+\frac{3}{10}\ln{|x+4|} -\frac{3}{10}\ln{|x-1|}+C=\frac{4}{5}\ln|x+4|+\frac{1}{5}\ln|x-1|+C$$
|
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|
Question on limit of Riemann Sum I'm trying to follow the following calculation of the definite integral using a limit of Riemann Sums. I have a question on the reasoning for a step in derivation of the limit which I will detail.
The original post is here.
$$\int_0^3 x^3 dx = \lim \limits_{n \to \infty} \sum_{i=1}^n f(x^*_i) \Delta x$$
$$= \lim \limits_{n \to \infty} \sum_{i=1}^n f\left(\frac{3i}{n}\right)\left(\frac 3n\right)$$
$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right) \sum_{i=1}^n \left(\frac{3i}{n}\right)^3$$
$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right)^4 \sum_{i=1}^n i^3$$
My first question is how is $\left(\frac{n(n+1)}{2}\right)^2$ multiplied to both sides when the RHS is unknown. Is the RHS assumed to be 1?
$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right)^4 \left(\frac{n(n+1)}{2}\right)^2 \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$
My second question is how are the next two equations derived from the prior?
$$= \left(\frac {3^4} 4\right) \lim \limits_{n \to \infty} \left(\frac 1 {n^4}\right) \left(\frac{n(n+1)}{2}\right)^2$$
$$= \left(\frac {3^4} 4\right) \lim \limits_{n \to \infty} \left(1 + \frac 1 n\right)^2$$
$$= \frac {3^4} 4$$
|
$= \lim \limits_{n \to \infty} (\frac 3 n)^4 (\frac{n(n+1)}{2})^2 \sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$ should be
$= \lim \limits_{n \to \infty} (\frac 3 n)^4 (\frac{n(n+1)}{2})^2$ makes use of $\sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$.
(Also, I saw Riemann spelled incorrectly there.)
Addendum to answer additional question posted as comment on this question:
It should be
$$= \lim \limits_{n \to \infty} \left(\frac 3 n\right)^4 \left(\frac{n(n+1)}{2}\right)^2 = \left(\frac {3^4} 4\right) \lim \limits_{n \to \infty} \left(\frac 1 {n^4}\right) \left(\frac{n(n+1)}{\color{red}1}\right)^2,$$
because $$\left(\dfrac3n\right)^4 \left(\dfrac{n(n+1)}{2}\right)^2=3^4\left(\dfrac1n\right)^4\dfrac1{2^2}(n(n+1))^2 .$$
Finally,
$$\left(\dfrac1n\right)^4(n(n+1))^2=\dfrac{n^2}{n^4}(n+1)^2=\dfrac1{n^2}(n+1)^2=\left(\dfrac{n+1}n\right)^2=\left(1+\dfrac1n\right)^2.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Combinatorial proof of $ \sum\limits_{k=0}^n(-1)^k\frac{2^{n-k}\binom{n}{k}}{(m+k+1)\binom{m+k}{m}}=\sum\limits_{k=0}^n\frac{\binom{n}{k}}{m+k+1}$ While answering the following question : Solution to a definite integral as $\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$, I came accross a strange identity involving binomials.
Using the binomial theorem, we can find our first expression to that integral : $ \int_{0}^{1}{x^{m}\left(1+x\right)^{n}\,\mathrm{d}x}=\sum\limits_{k=0}^{n}{\frac{\binom{n}{k}}{m+k+1}} $, while solving it the way I did leads to another expression : $ \int_{0}^{1}{x^{m}\left(1+x\right)^{n}\,\mathrm{d}x}=\sum\limits_{k=0}^{n}{\left(-1\right)^{k}\frac{2^{n-k}\binom{n}{k}}{\left(m+k+1\right)\binom{m+k}{m}}} $.
I was wondering if we could prove, using combinatorials, that for any $ n,m\in\mathbb{N} $ : $$ \sum\limits_{k=0}^{n}{\left(-1\right)^{k}\frac{2^{n-k}\binom{n}{k}}{\left(m+k+1\right)\binom{m+k}{m}}}=\sum\limits_{k=0}^{n}{\frac{\binom{n}{k}}{m+k+1}}$$
|
Here is an algebraic proof by way of enrichment. We seek to verify that
$$\sum_{k=0}^n (-1)^k
\frac{2^{n-k} {n\choose k}}{(m+k+1) {m+k\choose k}}
= \sum_{k=0}^n \frac{{n\choose k}}{m+k+1}.$$
We can re-write this as
$$\frac{m! n!}{(n+m+1)!} 2^n
\sum_{k=0}^n (-1)^k 2^{-k} {n+m+1\choose n-k}
= \sum_{k=0}^n \frac{{n\choose k}}{m+k+1}$$
or
$$2^n \sum_{k=0}^n (-1)^k 2^{-k} {n+m+1\choose n-k}
= (m+1) {n+m+1\choose n}
\sum_{k=0}^n \frac{{n\choose k}}{m+k+1}.$$
We get for the LHS
$$2^n \sum_{k=0}^n (-1)^k 2^{-k} {n+m+1\choose m+k+1}
= 2^n \sum_{k=0}^n (-1)^k 2^{-k} [z^{n-k}]
\frac{1}{(1-z)^{m+k+2}}
\\ = 2^n [z^n] \frac{1}{(1-z)^{m+2}}
\sum_{k=0}^n (-1)^k 2^{-k} z^k \frac{1}{(1-z)^k}.$$
Here the coefficient extractor enforces the range
and we find
$$2^n [z^n] \frac{1}{(1-z)^{m+2}}
\sum_{k\ge 0} (-1)^k 2^{-k} z^k \frac{1}{(1-z)^k}
= 2^n [z^n] \frac{1}{(1-z)^{m+2}}
\frac{1}{1+z/(1-z)/2}
\\ = [z^n] \frac{1}{(1-2z)^{m+2}}
\frac{1}{1+z/(1-2z)}
= [z^n] \frac{1}{(1-2z)^{m+1}}\frac{1}{1-z}.$$
On the other hand we have
$${n+m+1\choose n} {n\choose k}
= \frac{(n+m+1)!}{(m+1)! \times k! \times (n-k)!}
= {n+m+1\choose n-k} {m+k+1\choose m+1}$$
which gives for the RHS
$$\sum_{k=0}^n {n+m+1\choose n-k}
\frac{m+1}{m+k+1} {m+k+1\choose m+1}
= \sum_{k=0}^n {n+m+1\choose m+k+1} {m+k\choose m}
\\ = \sum_{k=0}^n {m+k\choose m}
[z^{n-k}] \frac{1}{(1-z)^{m+k+2}}
= [z^n] \sum_{k=0}^n {m+k\choose m}
\frac{1}{(1-z)^{m+k+2}} z^k.$$
We once more have the coefficient extractor enforcing the range
and we get
$$[z^n] \frac{1}{(1-z)^{m+2}} \sum_{k\ge 0} {m+k\choose m}
\frac{1}{(1-z)^k} z^k
\\ = [z^n] \frac{1}{(1-z)^{m+2}} \frac{1}{(1-z/(1-z))^{m+1}}
= [z^n] \frac{1}{1-z} \frac{1}{(1-2z)^{m+1}}.$$
The LHS is the same as the RHS which concludes the argument. The
coeffcient extractor evaluates to
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n {k+m\choose m} 2^k.}$$
|
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|
Solve for $a, b$ if $|x^2 + ax + b| \le \frac 12, \forall 0 \le x \le 2$. This is from a recently closed question. I do not think the accepted answer is rigorous because it did not address the cases when the discriminant of the two quadratics is negative, even though the final answer was correct.
My solution relies heavily on plotting several inequalities. I wonder if there is an alternative yet clean way of doing it without the plots. Another idea is whether it is helpful by replacing $x$ with $y+1$ so that $-1 \le y \le 1$ may simplify things a little bit.
My solution:
Denote $f(x)=x^2+ax+b$. It achieves its minimum at $x=-\frac a2$ and $f(-\frac a2)=b-\frac{a^2}{4}$.
First we must have $|f(0)| \le \frac 12 \text{ and } |f(2)| \le \frac 12$, or equivalently
$$|b| \le \frac 12 \tag1$$
$$|4+2a+b|\le \frac 12 \tag2$$
From the plots on the $a,b$-plane, we see that $(a,b)$ must reside in the parallelogram $BCDE$ and that $-\frac a2 \in [\frac{1.5}{2}, \frac{2.5}{2}] \subset [0,2]$, therefore we must also have
$$|f(-\frac a2)|=|b-\frac{a^2}{4}|\le \frac 12 \tag 3$$
After we add $(3)$ to the plots we found the only intersection of $(1), (2), (3)$ is at the point $C$ where $a=-2, b=\frac 12$. And indeed, when $0 \le x \le 2$, we have $|f(x)|=\left|(x-1)^2-\frac 12\right| \le \frac 12$.
|
The polynomial $x^2 - 2 x + \frac{1}{2}$ takes values $\frac{1}{2}$, $-\frac{1}{2}$, $\frac{1}{2}$ at $0$, $1$, $2$. We conclude that the linear function $(x^2-2x+ \frac{1}{2}) - (x^2+ a x + b)$ takes values $\ge 0$, $\le 0$, $\ge 0$ at $0$, $1$, $2$, so it must be the $0$ function. We get $a=-2$, $b=\frac{1}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Domain of $\sqrt{x}$ function when solving $x^2 - 2=0$ I've been going through this solution and got stuck on explaining myself why $x^2 = 2 \Rightarrow \sqrt{x^2} = \sqrt{2}\Rightarrow x = \pm \sqrt2$. Domain of $f(x)=\sqrt{x}$ is $[0,\infty)$. How taking square root both sides give $\pm \sqrt{2}$ ?
|
There are two distinct concepts at play here. $\sqrt{x}$ is a function denoting the nonnegative square root of $x$. However, the general solution to $a^2 = b^2$ is $a = \pm b$. So $\sqrt{25}=5$, but if $x^2=25$ then $x=\pm5$. As Gae. S. has already mentioned, the easiest way to see that $a^2=b^2 \iff a= \pm b$ is through factorisation:
\begin{align}
&a^2=b^2 \\
\iff&a^2-b^2=0 \\
\iff&(a+b)(a-b)=0 \\
\iff&a=-b \text{ or } a=b \, .
\end{align}
So, in your example, we have $x^2=(\sqrt{2})^2 \iff x=\pm\sqrt{2}$.
|
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|
Determinant and inverse matrix calculation of a special matrix Is there any smart way of calculating the determinant of this kind matrix?
\begin{pmatrix}
1 & 2 & 3 & \cdots & n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmatrix}
I encountered this in a problem for the case $n=4$. I need to find the inverse matrix.
I doubt the idea of this problem is to calculate all the cofactors and then the inverse matrix in the usual way.
I see the pattern here and some recursive relations... but I am not sure if this helps for calculating the determinant of the existing matrix.
|
$$\begin{vmatrix}
1 & 2 & 3 & \cdots & n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{vmatrix}$$
Notice the sum of an element from the first column and an element from the last column (with the same row) will always be equal to $n+1$,thus we can do $(C_1 = C_1+C_n)$
$$ = \begin{vmatrix}
n+1 & 2 & 3 & \cdots & n \\
n+1 & 1 & 2 & \cdots & n-1 \\
n+1 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n+1 & n-1 & n-2 & \cdots & 1 \end{vmatrix} = (n+1)
\begin{vmatrix}
1 & 2 & 3 & \cdots & n \\
1 & 1 & 2 & \cdots & n-1 \\
1 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & n-1 & n-2 & \cdots & 1 \end{vmatrix}$$
From here, we can do as follows:
go from the last row towards the first and decrease each row's value with the one above it (for any row but the first one).
($\forall i \neq 1, R_i=R_i-R_{i-1}$, Starting with $i=n$ then $i=n-1 ... i=2$).
$$ = (n+1)\begin{vmatrix}
1 & 2 & 3 & \cdots & n \\
0 & -1 & -1 & \cdots & -1 \\
0 & 1 & -1 & \cdots & -1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 1 & 1 & \cdots & -1 \end{vmatrix}$$
Expand $C_1$:
$$ = (n+1)\begin{vmatrix}
-1 & -1 & -1 & \cdots & -1 \\
1 & -1 & -1 & \cdots & -1 \\
1 & 1 & -1 & \cdots & -1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & -1 \end{vmatrix}$$
Now Add the first row to all of the other rows ($\forall i \neq 1, R_i = R_i + R_1$).
$$ = (n+1)\begin{vmatrix}
-1 & -1 & -1 & \cdots & -1 \\
0 & -2 & -2 & \cdots & -2 \\
0 & 0 & -2 & \cdots & -2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & -2 \end{vmatrix} = (n+1)[-(-2)^{n-2}]$$
|
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|
How to evaluate $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}$ without using antiderivative? Someone gives a solution here:
\begin{align*} \int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}&=\int_0^{\frac{\pi}{2}} \sum_{n=0}^{\infty} (-\cos\alpha\cos\beta)^n{\rm d}\alpha\\ &=\sum_{n=0}^{\infty}(-\cos\beta)^n\int_0^{\frac{\pi}{2}} \cos^n\alpha{\rm d}\alpha\\ &=\sum_{n=0}^{\infty}(-\cos \beta) ^n\frac{\sqrt{\pi}\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{2\Gamma\left(\frac{n}{2}+1\right)}\\ &=\frac{\pi-2\arcsin\cos \beta}{2\sqrt{1-\cos^2\beta}}\\ &=\frac{\beta}{\sin \beta}. \end{align*}
Is it correct? How to obtain the fourth equlity?
|
Using tangent half-angle substitution $t=\tan \frac{x}{2} $, we transform the integral into
$$
I=2 \int_0^1 \frac{d t}{(1-\cos \beta) t^2+1 +\cos \beta},\quad \textrm{ where }\beta \textrm{ is not a multiple of }\pi.
$$
By the half-angle formula of sine and cosine, we have
$$
\begin{aligned}
I&=\int_0^1 \frac{d t}{t^2 \sin ^2 \frac{\beta}{2}+\cos ^2 \frac{\beta}{2}} \\
&=\frac{1}{\sin \frac{\beta}{2} \cos \frac{\beta}{2}}\left[\tan ^{-1}\left(\frac{t \sin \frac{\beta}{2}}{\cos \frac{\beta}{2}}\right)\right]_0^1 \\
&=\frac{2}{\sin \beta} \cdot \tan ^{-1}\left(\tan \frac{\beta}{2}\right) \\
&=\frac{\beta}{\sin \beta}
\end{aligned}
$$
|
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|
What is inverse of a matrix whose diagonal elements are all zero. I am a science researcher and I got a problem to find the generic inverse of the following matrix:
$$
A_n = \left(\begin{array}{ccc}
0 & a_2 & a_3 & ... & a_{n-1} & a_n \\
a_1 & 0 & a_3 & ... & a_{n-1} & a_n \\
a_1 & a_2 & 0 & ... & a_{n-1} & a_n \\
... & ... & ... & ... & ... & ... \\
... & ... & ... & ... & ... & ... \\
a_1 & a_2 & a_3 & ... & 0 & a_n \\
a_1 & a_2 & a_3 & ... & a_{n-1} & 0 \\
\end{array}\right)
$$
I figured out that for n=2,3
$$
A_2^{-1} = \left(\begin{array}{ccc}
0 & a_2^{-1} \\
a_1^{-1} & 0 \\
\end{array}\right)
$$
$$
A_3^{-1} = \frac{1}{2} \left(\begin{array}{ccc}
-a_1^{-1} & a_1^{-1} & a_1^{-1} \\
a_2^{-1} & -a_2^{-1} & a_2^{-1} \\
a_3^{-1} & a_3^{-1} & -a_3^{-1} \\
\end{array}\right)
$$
but can we extend it to a general case?
Can anyone help?? Thanks!
[later]
It looks like it is
$$
m_{ij} = -\frac{n-2}{n-1} a_{i}^{-1}\ (i=j)
$$
$$
= \frac{1}{n-1} a_{i}^{-1}(else)
$$
|
Your matrix can be written as $D_n + u_n \, v_n^T$:
$$u_n = \left[ \begin{matrix} 1 & ... & 1 \end{matrix} \right]^T$$
$$v_n = \left[ \begin{matrix} a_1 & ... & a_n \end{matrix} \right]^T$$
$$D_n = - \text{diag}(v_n) = - \left[ \begin{matrix}
a_1 & ... & 0 \\
\vdots & ... & \vdots \\
0 & ... & a_n \\
\end{matrix} \right]$$
Which reminds me of the Sherman–Morrison formula:
$$\left(A + u \, v^T\right)^{-1} = A^{-1} - {A^{-1} \, u \, v^T \, A^{-1} \over 1 + v^T \, A^{-1} \, u}$$
[EDIT]
Your pattern seems to be right. Defining $w_n = \left[ \begin{matrix} a_1^{-1} & ... & a_n^{-1} \end{matrix} \right]^T$:
$$ $$
$$1 + v^T \, A^{-1} \, u = 1-n$$
$$A^{-1} \, u = - w_n$$
$$ v^T \, A^{-1} = - u_n^T$$
$$\left(A + u \, v^T\right)^{-1} = - \text{diag}(w_n) - { w_n \, u_n^T \over 1-n}$$
$$ \left(A + u \, v^T\right)^{-1} =
- \left[ \begin{matrix}
a_1^{-1} & ... & 0 \\
\vdots & ... & \vdots \\
0 & ... & a_n^{-1} \\
\end{matrix} \right]
-
\frac{1}{1-n}
\left[ \begin{matrix}
a_1^{-1} & ... & a_1^{-1} \\
\vdots & ... & \vdots \\
a_n^{-1} & ... & a_n^{-1} \\
\end{matrix} \right]
$$
$$ $$
$$
m_{ij} = -a_{i}^{-1} -\frac{a_{i}^{-1}}{1-n} = \frac{n-2}{1-n} \, a_{i}^{-1} = - \frac{n-2}{n-1} \, a_{i}^{-1} \, (i=j)
$$
$$
= -\frac{a_{i}^{-1}}{1-n} = \frac{1}{n-1} \, a_{i}^{-1} \, (else) \hphantom{aaaaaaaaaaaaaaa}
$$
|
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|
Show factorization $ x^{2n}-1=(x^2-1) \prod_{k=1}^{n-1}(x^2-2x \cos \frac{\pi k}{n} + 1) $ I'm interested in how to show that
$$
x^{2n}-1=(x^2-1) \prod_{k=1}^{n-1}(x^2-2x \cos \frac{\pi k}{n} + 1)
$$
I've seen this equality too often, but have no idea how to derive it. I've tried the following:
$$
x^{2n}-1=(x^n-1)(x^n+1)
$$
We all know that $x^n-1=(x-\xi^0)(x-\xi^1) \dots (x-\xi^{n-1})$, where $\xi=\exp(i \cdot \frac{2 \pi}{n})$.
Here $x-\xi^0$ gives us the desired $x-1$(for $x^2-1$ in RHS).
But the problem is, we can not do the same with $x^n+1$, because as I understand, there is no such general factorization of $x^n+1$. Firstly, it depends whether $n$ is even or not. Secondly, if for the sake for simplicity $n$ is odd, then we can decompose the polynomial such a way:
$$x^n+1=(x+1)(1-x+x^2-x^3 \dots -x^{n-2}+x^{n-1})$$
I don't see here anything to continue with. Any suggestions?
|
Thanks to @CalvinLin for this post. The idea was to apply direct substitution
$$
x^{2n}-1=\prod_{k=0}^{2n-1} (x-\xi^k)
$$
where $\xi$ is a base root of the equation $x^{2n}-1=0$, e.g. $\xi=\exp({i \cdot \frac{\pi}{n}})$.
It gives us the following:
$$
\prod_{k=0}^{2n-1} (x-\xi^k)=(x-\xi^0)(x-\xi^n) \prod_{k=1}^{n-1} (x-\xi^k)(x-\xi^{2n-k})
$$
$(x-\xi^0)(x-\xi^n)=(x-1)(x+1)=(x^2-1)$ since $\xi^n=\exp(i \cdot \pi)=-1$
Now let's take a deeper look at multiplier inside the product notation.
$$
(x-\xi^k)(x-\xi^{2n-k})=x^2-(\xi^k+\xi^{2n-k})x+\xi^k \cdot \xi^{2n-k}
$$
Obviously, $\xi^k \cdot \xi^{2n-k}=\xi^{2n}=1$. Consider the coefficient at $x$:
$$
\xi^k+\xi^{2n-k} \\
=\cos(k\cdot \frac{\pi}{n})+i\sin(k\cdot\frac{\pi}{n})+\cos((2n-k)\cdot \frac{\pi}{n})+i\sin((2n-k)\cdot \frac{\pi}{n})\\
=\cos(k\cdot \frac{\pi}{n})+i\sin(k\cdot\frac{\pi}{n})+\cos(2\pi - k\cdot \frac{\pi}{n})+i\sin(2\pi - k\cdot \frac{\pi}{n}) \\
=2\cos(k\cdot \frac{\pi}{n})
$$
The last equality comes from the periodicity and evenness/oddness of $\cos$/$\sin$.
So, for now we have:
$$
=(x^2-1)\prod_{k=1}^{n-1}(x^2-2x\cos(k \cdot \frac{\pi}{n})+1)
$$
And we are done!
|
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|
How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$ I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}$$
and
$$I_n=\int_0^1\frac{x-1}{x^{n+1}-1}dx$$
It makes sense to me that, for all values of $n$, $I_n$ is convergent since the bottom of the function is always above zero and the integral exists for $n\to\infty$ however I cannot seem to find a nice closed form for this.
One thought I did have was using:
$$\sum\ln(x_i)=\ln\left(\prod x_i\right)$$
but I cannot seem to make it work. Does anyone have any hints for this type of problem as I would like to try and complete it myself. Thanks :)
|
Decompose the integrand as
$$\frac{1}{1+x+x^2+...+x^n}=\sum_{k=1}^{n}\frac{a_k}{x-x_k},\>\>\>\>\> a_k=\frac{x_k(x_k-1)}{n+1}
$$
where $x_k= e^{i \frac{2\pi k}{n+1}},\> k=1,2...n$. Then
\begin{align}
I_n&=\int_0^1\frac{dx}{1+x+x^2+...+x^n}\\
&=\int_0^1 \sum_{k=1}^{n}\frac{a_k}{x-x_k}dx
=\frac1{n+1}\sum_{k=1}^{n} x_k(x_k-1)\ln\left(1-\frac1{x_k}\right)
\end{align}
Substitute $x_k= e^{i \frac{2\pi k}{n+1}}$ into above summation to derive the close-form
$$\color{blue}{I_n = \frac\pi{2(n+1)} \csc\frac{2\pi}{n+1}-\frac4{n+1}\sum_{k=1}^{[\frac n2]}
\sin\frac{\pi k}{n+1}\sin\frac{3\pi k}{n+1}\ln\left(\sin\frac{\pi k}{n+1}\right)}
$$
Listed below are some sample results
\begin{align}
I_2 &= \frac{\pi}{3\sqrt3}\\
I_3 &= \frac\pi8 +\frac14\ln2\\
I_4 &= \frac\pi{10}\csc\frac{2\pi}{5}+\frac1{\sqrt5}\ln\left(2\cos\frac\pi5\right)\\
I_5 &= \frac{\pi}{6\sqrt3}+\frac13\ln2\\
I_7 &= \frac{\pi}{8\sqrt2}+\frac18\ln2 + \frac{1}{8\sqrt2}\ln \frac{\sqrt2+1}{\sqrt2-1}\\
I_9 &= \frac\pi{20}\csc\frac{\pi}{5}+\frac15\ln2
+\frac1{2\sqrt5}\ln\left(2\cos\frac\pi5\right)\\
\end{align}
\begin{align}
I_6
&=\frac\pi{14}\csc\frac{2\pi}{7}-\frac1{2\sqrt7}\bigg( \frac{\ln\sin\frac\pi7}{\sin\frac{2\pi}7}
+\frac{\ln\sin\frac{2\pi}7}{\sin\frac{3\pi}7}-\frac{\ln\sin\frac{3\pi}7}{\sin\frac{\pi}7}\bigg)
\\
\end{align}
\begin{align}
I_8 &= \frac\pi{18}\csc\frac{2\pi}{9}+\frac{2}{3\sqrt3} \bigg(\frac{\ln\csc\frac\pi9}{\csc\frac{\pi}9}
+\frac{\ln\csc\frac{2\pi}9}{\csc\frac{2\pi}9}-\frac{\ln\csc\frac{4\pi}9}{\csc\frac{4\pi}9}\bigg)
\\
\end{align}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3959724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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|
Cannot find limit using epsilon delta definition Prove
$$\lim\limits_{x\to 2} x^3 = 8$$
using epsilon delta definition.
I try as below.
Let $\varepsilon>0$. We choose $\delta>0$.
Consider that
\begin{align}
\vert x^3-8\vert &= \vert (x-2) (x^2+2x+4)\vert\\
&=\vert (x-2) \vert\vert(x^2+2x+4) \vert \\
&=\vert (x-2) \vert\vert(x-2)^2+6x \vert .
\end{align}
Now I don't know how to continue this answer. I confused with $6x$.
Anyone can help me?
EDIT: I have tried as below as JC12's answer.
Let $\vert x-2\vert <1$, then $\vert x\vert -2< \vert x-2\vert <1$
then we have $$\vert x\vert -2<1 \iff \vert x\vert<3.$$
Now, $$ \vert(x-2)^2+6x \vert < \vert(3-2)^2+6\cdot 3 \vert =19. $$
Thus, \begin{align} \vert x^3-8\vert&= \vert (x-2)
\vert\vert(x-2)^2+6x \vert < 19 \vert (x-2) \vert. \end{align}
Now choose $\delta=\min(1,\frac{\varepsilon}{19})$. We have
\begin{align} \vert x^3-8\vert < 19 \vert (x-2) \vert< 19
\frac{\varepsilon}{19} = \varepsilon. \end{align}
So, we can conclude $\lim\limits_{x\to 2} x^3 =8$.
|
More generally,
if you want to show that
$\lim_{x \to c} x^n=c^n$,
note that
$x^n-c^n
=(x-c)\sum_{k=0}^{n-1} x^kc^{n-1-k}
$.
If
$c-r < x < c+r$
where
$0 < r< \min(1, x)$
then
$\sum_{k=0}^{n-1} x^kc^{n-1-k}
\lt \sum_{k=0}^{n-1} (c+r)^kc^{n-1-k}
\lt \sum_{k=0}^{n-1} (c+r)^k(c+r)^{n-1-k}
=n(c+r)^n
$
so
$\begin{array}\\
|x^n-c^n|
&=|(x-c)\sum_{k=0}^{n-1} x^kc^{n-1-k}|\\
&=|x-c|\,|\sum_{k=0}^{n-1} x^kc^{n-1-k}|\\
&<|x-c|n(c+r)^{n-1}\\
\end{array}
$
so if
$|x-c|
\le \dfrac{\delta}{n(c+r)^{n-1}}
$
then
$|x^n-c^n|
\lt \delta$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3962012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$ Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how to deal with the equation. Thank you in advance!
|
From $45x^2-30x+1=(7+6x-9x^2)^2$, you have that $$81x^{4}-108x^{3}-135x^{2}+114x+48 = 0$$
This factors as $$3(x-1)(3x+1)(9x^2-6x-16) = 0$$
So the roots of that polynomial are $x = 1, -\frac{1}{3}, \frac{1\pm\sqrt{17}}{3}$. Out of these, the only ones in the range $\left[\frac{1-2\sqrt{2}}{3} ,\frac{1+2\sqrt{2}}{3} \right]$ are $x = 1$ and $x = -\frac{1}{3}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3962818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $a ( n ) = b( n + 2)$
let $a(n)$ denotes the number of ways of expressing the positive integer $n$ as an ordered sum of 1's and 2's. Let $b(n)$ denote the number of ways of expressing n as an ordered sum of integers greater than 1. prove that $a(n) = b(n+2)$. for $n=1,2,3...$
My approach:
$a(1) = 1$ (only) , $b(3) = 3$
$a(2) = 2$, because $2=1+1=2$ ,and $b(4)=2$, because $4=2+2=4$
$a(3) = 3$, because $3=1+1+1=1+2=2+1$ and, $b(5) = 3$ , because $5=2+3=3+2=5$
$a(4)= 5$, because $4=1+1+1+1=2+2=1+1+2=1+2+1=2+1+1$ and, $b(6)=5$, because $6=3+3=2+4=4+2=2+2+2=6$
$a(5)=8$, because $5=1+1+1+1+1=2+1+1+1=1+2+1+1=1+1+2+1=1+1+1+2=2+2+1=2+1+2=1+2+2$ and, $b(7)=8$, because
$7=3+2+2=2+3+2=2+2+3=3+4=4+3=2+5=5+2=7$
By this way i am able to show that $a(n)=b(n+2)$. but is there any general method for this problem. I mean any recursion relation which i can understand.
Background:-This problem is from pathfinder for Olympiad mathematics.
|
It is also possible to exhibit a bijection between compositions of $n$ using only $1$ and $2$ and compositions of $n+2$ that do not use $1$.
Suppose that $n=c_1+c_2+\ldots+c_k$, where $c_i\in\{1,2\}$ for $i=1,\ldots,k$. Let $c_0=2$; then $n+2=c_0+c_1+\ldots+c_k$. Let $i_1<\ldots<i_\ell$ be the set of indices $i$ such that $c_i=2$, and let $i_{\ell+1}=k+1$. For $j=1,\ldots,\ell$ let
$$d_j=\sum_{i=i_j}^{i_{j+1}-1}c_i\;;$$
then $n+2=d_1+\ldots+d_\ell$ is a composition of $n+2$ that does not use $1$.
Example: The composition $1+2+2+1+1+2$ of $9$ is first extended to the sum $2+1+\color{blue}2+\color{red}{2+1+1}+\color{green}2$, which is then reduced to the composition $3+\color{blue}2+\color{red}4+\color{green}2$ of $11$.
This mapping is clearly injective, and it’s not hard to see that it is surjective as well. Let $n+2=d_1+\ldots+d_\ell$ be a composition of $n+2$. Replace each $d_i$ by its composition
$$2+\underbrace{1+\ldots+1}_{d_i-2}$$
and drop the leading $2$ to get the composition
$$n=\underbrace{1+\ldots+1}_{d_1-2}+2+\underbrace{1+\ldots+1}_{d_2-2}+\ldots+2+\underbrace{1+\ldots+1}_{d_\ell-2}\;;$$
applying the original procedure to this composition of $n$ recovers the composition $n+2=d_1+\ldots+d_\ell$ of $n+2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{(\frac{x+1}{x-2})^{\frac{2}{3}}-(\frac{x-1}{x-2})^{\frac{2}{3}}}{(\frac{x+2}{x-2})^{\frac{2}{3}}-1}$$
L'Hopital
$$L=\lim_{x \to \infty}\frac{\frac{2}{3}(\frac{x+1}{x_2})^{-\frac{1}{3}}(\frac{x-2-(x+1)}{(x-2)^2})-\frac{2}{3}(\frac{x-1}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x-1)}{(x-2)^2})}{\frac{2}{3}(\frac{x+2}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x+2)}{(x-2)^2})}=\lim_{x\ \to \infty}{\frac{3(x+1)^{-\frac{1}{3}}-(-1)(x-1)^{-\frac{1}{3}}}{(x+2)^{-\frac{1}{3}}(-4)}}$$
$$=\lim_{x \to \infty}{\frac{3(1+\frac{1}{x})^{-\frac{1}{3}}-(1-\frac{1}{x})^{-\frac{1}{3}}}{4(1+\frac{2}{x})^{-\frac{1}{3}}}}=\frac{3-1}{(4)(1)}=\frac{1}{2}.$$
Is this correct and is there a more elegant way of doing it?
EDIT: strictly speaking L'Hopital is not applicable with $x \to \infty$ so just got lucky here...
|
Hint:
$$\lim_{x\to\infty}\dfrac{(x+1)^{2/3}-(x-1)^{2/3}}{(x+2)^{2/3}-(x-2)^{2/3}}=\dfrac12\cdot\lim_{t\to0}\dfrac{\dfrac{(1+t)^{2/3}-(1-t)^{2/3}}t}{\dfrac{(1+2t)^{2/3}-(1-2t)^{2/3}}{2t}}=\dfrac12\cdot\dfrac{\lim_{t\to0}f(t)}{\lim_{t\to0}f(2t)}$$
Now for $\lim_{t\to0}f(t)=\lim_{t\to0}\dfrac{(1+t)^{2/3}-(1-t)^{2/3}}t,$ set $(1+t)^{1/3}=p,(1-t)^{1/3}=q$
$\implies p^3-q^3=2t$ and $\implies p\to q, q\to1$
$$\lim_{t\to0}f(t)=\lim_{p\to q, q\to1}\dfrac{2(p-q)}{p^3-q^3}=\lim_{q\to1}\dfrac2{3q^2}=?$$
|
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"url": "https://math.stackexchange.com/questions/3963376",
"timestamp": "2023-03-29T00:00:00",
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|
Are there infinitely many $p$ such that $\frac{p_k+p_{k+1}}{2} = m^2 - n^2$ has no integer solutions? $\frac{p_k+p_{k+1}}{2} = m^2 - n^2$ with $p_k$ as the $k$-th prime has no integer solutions for $9$ values of $k \leq 25$, $17$ values of $k \leq 50$, and $33$ values of $k \leq 100$. This means that roughly $33\%$ of each sample consists of such primes. Though this is a small sample, the ratio of primes with this property to those which do not seems to be consistent. Are there infinitely many primes with this property, and if so, what proportion of primes have it?
|
Write $m^2-n^2=(m+n)(m-n)$ The two factors on the right have the same parity, so any number that is equivalent to $2 \bmod 4$ cannot be expressed this way. Because you divide the sum by $2$, this is any pair of primes that sum to a number that is equivalent to $4 \bmod 8$. For example, the sequential primes $5,7$ give $\frac 12(5+7)=6$, which cannot be expressed as $m^2-n^2$ because it only factors into $1,6$ or $2,3$, and neither pair has the same parity. $\frac 12(7+11)=9$, which we can factor as $1,9$ or $3,3$ and $m=5,n=4$ works for the first and $m=3,n=0$ for the second. $\frac 12(11+13)=12$, which we can factor as $2,6$ getting $m=4,n=2$. If the left side is odd, both factors will be odd, while if the left side is a multiple of $4$ there will be at least one factorization with both factors even.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Large N limit for $P(\mathbf{x} \cdot\mathbf{y}\cdot \sum\mathbf{x}\cdot \sum \mathbf{y}>0)$ for two binary vector $\mathbf{x}$ and $\mathbf{y}$ I have two binary vectors $\mathbf{x}$, and $\mathbf{y}$, each with $N$ elements; Each element of $\mathbf{x}$ and $\mathbf{y}$ belongs to {-1, 1}, and is drawn from uniform random distribution.
Now I would like to derive the large $N$ limit for $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})>0)$ and $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})<0)$, where $\mathbf{x} \cdot \mathbf{y}$ is the inner product between $\mathbf{x}$ and $\mathbf{y}$, and $\sum\mathbf{x}$, $\sum\mathbf{y}$ are the sum of all elements in $\mathbf{x}$ and $\mathbf{y}$, respectively.
I have derived the explicit form for $P(\mathbf{x} \cdot \mathbf{y} =k, \sum\mathbf{x}=x, \sum \mathbf{y}=y)$, which is
\begin{align}
P(\mathbf{x} \cdot \mathbf{y}=k, \sum\mathbf{x} = x, \sum \mathbf{y} = y) =\frac{{N \choose (N + x)/2} \cdot{ (N+x)/2 \choose (N+x+y+k)/4 } \cdot {(N-x)/2 \choose (N+k-x-y)/4}}{2^{2N}}
\end{align}
From the above formula, I computed $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})>0)$ and $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})<0)$ from nested sum over all possible $x$, $y$ and $k$ values from $n=2$ to $n=400$, and the results suggest that $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})>0)$ and $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})<0)$ are both converging to $0.5$, see the plot below:
However, I have hard time trying to derive the large $N$ limit in a formal way.
Another side question is that, is above $P(\mathbf{x} \cdot \mathbf{y}=k, \sum\mathbf{x} = x, \sum \mathbf{y} = y)$ an known probability distribution?
|
Edited
Let $a^T=[1, 1, ..., 1]$, that is an all-one vector of length $N$. Then you can rewrite the random variable as:
$Z=(Y^T X)(X^T a)(a^T Y)$,
in which $T$ represent transpose, so $Y^TX$ is the inner product of $X$ and $Y$, $X^T a$ is the sum of the elements of $X$, and $a^T Y$ is the sum of the elements of $Y$. The reason I wrote it this way will be clear in a few moments!
You are interested in large $N$, so I suggest investigating the mean of $Z$ first, as follows.
$\mu = E\{Z\} = E\{Y^T X X^T a a^T Y\}$
Since $Y^T X X^T a a^T Y$ is a scalar, it equals to its trace. I will use the fact that $tr(AB)=tr(BA)$, also the fact that trace and expectation are both linear operators and interchangeable:
$\mu = E\{Y^T X X^T a a^T Y\} = tr(E\{Y^T X X^T a a^T Y\}) = E\{tr(Y^T X X^T a a^T Y)\} = E\{tr(Y Y^T X X^T a a^T)\} $
where I took $A=Y^T X X^T a a^T$ and $B=Y$ in $tr(AB)=tr(BA)$. So,
$\mu = E\{tr(Y Y^T X X^T a a^T)\} = tr(E\{Y Y^T X X^T a a^T\})
= tr(E\{Y Y^T\} E\{X X^T\} a a^T\})$.
Since $E\{Y Y^T\}=E\{X X^T\} = I_N$, where $I_N$ is the identity matrix of size $N$, and also $a a^T=1_N$ is an all-one matrix of size $N$:
$\mu = tr(I_N I_N 1_N) = N$.
The expected value of your random variable diverges with $N$. I suspect that this might change your question.
|
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|
What is the remainder when $1^{2016} + 2^{2016} + ⋯ + 2016^{2016}$ is divided by $2016$? What is the remainder when $1^{2016} + 2^{2016} + ⋯ + 2016^{2016}$ is divided by $2016$?
My Approach:
started breaking 2016 into product of pairwise coprime positive integers to get $2016 = 32 * 7 * 9$
This suggests that we want to find the sum modulo 32,7, 9 separately so that we can apply the chinese remainder theorem.
but I couldn't solve it any further than this so please help me out.
|
Define $$S = \sum_{i=1}^{2016} i^{2016}$$
Notice that $2016 = 32\cdot 9\cdot7$
and $\phi(32) | 2016,\; \phi(9) |2016,\; \phi(7) | 2016$, so we have
$$i^{2016} \equiv \begin{cases}
1 \pmod n \text{ if } \gcd(i,n)=1\\ 0 \pmod n \text{ if } \gcd(i,n)>1 \end{cases}$$
for $n=32,9,7$.
Then each group of $n$ has $\phi(n)$ values coprime to $n$ and there are $2016/n$ such groups in each case so:
$$S \equiv \frac{2016}{32}\cdot \phi(32) \equiv 7\cdot 9 \cdot \phi(32) \pmod{32} \tag 1$$
$$S \equiv \frac{2016}{9}\cdot \phi(9) \equiv 32\cdot 7 \cdot \phi(9) \pmod{9} \tag 2$$
$$S \equiv \frac{2016}{7}\cdot \phi(7) \equiv 9 \cdot 32 \cdot \phi(7) \pmod{7} \tag 3$$
Can you end it now?
Update:
There's a shortcut to CRT. Simply add RHS of $(1), (2), (3)$,
$$x = 7 \cdot 9 \cdot \phi(32) + 32 \cdot 7 \cdot \phi(9) + 9 \cdot 32 \cdot \phi(7) = 4080$$
Then $S \equiv x \pmod{32,9,7} \implies S\equiv x = 4080 \equiv 48 \pmod{2016}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Problem proving $\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx$ I am trying to show that the value of the following integral is:
$$\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx=2^{n-1}\pi^{\frac{1}{2}}(2n+1)n!\delta_{mn}+2^n\pi^{\frac{1}{2}}(n+2)!\delta_{(n+2)m}+2^{n-2}\pi^{\frac{1}{2}}n!\delta_{(n-2)m},$$
where $H_n$ is the $n$-th Hermite polynomial.
My attempt is the following:
Suppose that the value of the integral is already known and that:
\begin{equation*}
\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx=\lambda
\end{equation*}
From the recurrence relation we have to:
\begin{equation*}
H_{n+1}(x)=2xH_{n}(x)-2nH_{n-1}(x)
\end{equation*}
Therefore, since it is valid for any $ n $ then it will also be valid for any $ m $, that is:
\begin{equation*}
H_{m+1}(x)=2xH_m(x)-2mH_{m-1}(x)
\end{equation*}
Solving for the terms $2xH_ {m} (x)$ and $2xH_ {n} (x)$ we have:
\begin{align}
2xH_ {n} (x) & = H_ {n + 1} (x) + 2nH_ {n-1} (x) \\
2xH_ {m} (x) & = H_ {m + 1} (x) + 2mH_ {m-1} (x)
\end{align}
And multiplying the \ textcolor {red} {\ textbf {equation 1}} by $ 2xH_m (x) $ and substituting the cleared value we have:
\begin{align*}
4x^2H_{n}(x)H_{m}(x)&=2xH_m(x)\left[H_{n+1}(x)+2nH_{n-1}(x)\right]\\
4x^2H_{n}(x)H_{m}(x)&=\left[H_{m+1}(x)+2mH_{m-1}(x)\right]\left[H_{n+1}(x)+2nH_{n-1}(x)\right]\\
4x^2H_{n}(x)H_{m}(x)&= H_{m+1}(x)H_{n+1}(x)+2nH_{m+1}(x)H_{n-1}(x)+2mH_{m-1}(x)H_{n+1}(x)+4mnH_{m-1}(x)H_{n-1}(x)\\
\end{align*}
And from what we have just found we can conclude that:
{\begin{align*}
4\lambda&=\int_{-\infty}^{\infty}e^{-x^2}\left[H_{m+1}(x)H_{n+1}(x)\right]dx+\int_{-\infty}^{\infty}e^{-x^2}2nH_{m+1}(x)H_{n-1}(x)dx+\int_{-\infty}^{\infty}2me^{-x^2}H_{m-1}(x)H_{n+1}(x)+\int_{-\infty}^{\infty}4mne^{-x^2}H_{m-1}(x)H_{n-1}(x)dx
\end{align*}
So going back to the integral that we already had:
\begin{align*}
\cfrac{1}{4}\times4\lambda&=\cfrac{1}{4}\times \sqrt{\pi}\left\{2^{n+1}(n+1)!\left[\delta_{(m+1)(n+1)}+2m\delta_{(m-1)(n+1)}\right]+2^{n-1}(n-1)!\left[\delta_{(m+1)(n-1)}2n+4mn\delta_{(m-1)(n-1)}\right]\right\}\\
\lambda&=\cfrac{\sqrt{\pi}}{4}\left\{2^{n+1}(n+1)!\left[\delta_{(m+1)(n+1)}+2m\delta_{(m-1)(n+1)}\right]+2^{n-1}(n-1)!\left[\delta_{(m+1)(n-1)}2n+4mn\delta_{(m-1)(n-1)}\right]\right\}\\
\end{align*}
And as it was supposed from the beginning:
\begin{equation*}
\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx=\lambda
\end{equation*}
Then finally it will come to:
$\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx=\cfrac{\sqrt{\pi}}{4}\left\{2^{n+1}(n+1)!\left[\delta_{(m+1)(n+1)}+2m\delta_{(m-1)(n+1)}\right]+2^{n-1}(n-1)!\left[\delta_{(m+1)(n-1)}2n+4mn\delta_{(m-1)(n-1)}\right]\right\}$
But I can't find the way to give the answer $2^{n-1}\pi^{\frac{1}{2}}(2n+1)n!\delta_{mn}+2^n\pi^{\frac{1}{2}}(n+2)!\delta_{(n+2)m}+2^{n-2}\pi^{\frac{1}{2}}n!\delta_{(n-2)m}$
|
I am going to work with the statistician's Hermite polynomials, which can be related to the physicists' (which appear in this question) via
$$
He_{\alpha}(x) = 2^{-\frac{\alpha}{2}}H_\alpha\left(\frac{x}{\sqrt{2}}\right)
$$
and the desired integral becomes
$$
I_{nm} = \int_{-\infty}^\infty x^2e^{-x^2}H_n(x)H_m(x)dx \\
= 2^{\frac{n+m}{2}-1}\sqrt{\pi}\int_{-\infty}^\infty x^2He_n(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}.
$$
We can use the identity $x^2 = He_2(x) + He_0(x)$ to split the integral into two parts
$$
I_{nm}=2^{\frac{n+m}{2}-1}\sqrt{\pi}\left(\int_{-\infty}^\infty He_n(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}+\int_{-\infty}^\infty He_2(x)He_n(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}\right)
$$
the first integral in the parentheses is equal to $n!\delta_{n,m}$ by orthogonality (of the statistician's Hermite polynomials)
$$
\int_{-\infty}^\infty He_{\alpha}(x)He_m(\beta)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}=n!\delta_{\alpha,\beta}
$$
and we will focus on the second
$$
J_{nm}=\int_{-\infty}^\infty He_2(x)He_n(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}.
$$
To calculate $J_{nm}$ we will make use of the Hermite linearization formula
$$
He_\alpha(x)He_\beta(x)=\sum_{k=0}^{\min(\alpha,\beta)}{\alpha \choose k}{\beta \choose k}k!He_{\alpha+\beta-2k}(x)
$$
which is proven in this paper (and elsewhere).
Using the linearization formula to write the product $He_2(x)He_n(x)$ in terms of a sum of individual Hermite polynomials gives
$$
J_{nm}=\sum_{k=0}^{\min(n,2)}{n \choose k}{2 \choose k}k!\int_{-\infty}^\infty He_{2+n-2k}(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}.
$$
There are at maximum three terms in this summation, therefore we can write out each term
$$
J_{nm}={n \choose 0}{2 \choose 0}0!\int_{-\infty}^\infty He_{n+2}(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}\\+{n \choose 1}{2 \choose 1}1!\int_{-\infty}^\infty He_{n}(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}\\+{n \choose 2}{2 \choose 2}2!\int_{-\infty}^\infty He_{n-2}(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}.
$$
Only the first term occurs when $n=0$, the first two when $n=1$ and all three terms otherwise. These integrals can all be done by orthogonality with result
$$
J_{nm}=(n+2)!\delta_{n+2,m} + 2nn!\delta_{n,m} + n(n-1)(n-2)!\delta_{n-2,m}.
$$
Putting this together with the first term above leads to your sought after answer
$$
I_{nm}=\sqrt{\pi}2^{n}(n+2)!\delta_{n+2,m} + \sqrt{\pi}2^{n-1}(2n+1)n!\delta_{n,m} + \sqrt{\pi}2^{n-2}n!\delta_{n-2,m}.
$$
Note that the third term is not present if $n<2$
|
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"url": "https://math.stackexchange.com/questions/3972541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Evaluate integral $\int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx $ I tried to solve it by the following method:
\begin{align*}
\int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx&= \int \frac{(1-x^2)e^x+\frac{e^x}{2}+\frac{e^x}{2}}{(1-x)(\sqrt{1-x^2}) }dx \\ &= \int \frac{(1-x^2)e^x+\frac{(1-x)e^x}{2}+\frac{(1+x)e^x}{2}}{(1-x)(\sqrt{1-x^2}) }dx \\ &=\int \frac{\sqrt{(1-x^2)}e^x+\frac{\sqrt{(1-x)}e^x}{2\sqrt{x+1}}+\frac{\sqrt{(1+x)}e^x}{2\sqrt{1-x}}}{(1-x) }dx \\&=
\int \frac{\sqrt{1-x}(\sqrt{x+1}\cdot e^x+\frac{e^x}{2\sqrt{x+1}})+\sqrt{x+1}\cdot e^x \cdot \frac{1}{2\sqrt{1-x}}}{1-x}dx \\&= \int\left(\frac{\sqrt{x+1}\cdot e^x}{\sqrt{1-x}}\right)'dx \\ &=\left(\frac{\sqrt{x+1}\cdot e^x}{\sqrt{1-x}}\right)+C
\end{align*}
Is this solution right?
if it is right are there other methods to solve it.
|
Recognize
$$\frac{2-x^2}{(1-x)(\sqrt{1-x^2})}= \frac{(1-x^2)+1}{(1-x)(\sqrt{1-x^2})}\\=\sqrt{\frac{1+x}{1-x} }+ \frac{1}{(1-x)^{3/2}\sqrt{1+x})}
= \sqrt{\frac{1+x}{1-x} }+ \left( \sqrt{\frac{1+x}{1-x} }\right)’
$$
and then apply to $f(x)= \sqrt{\frac{1+x}{1-x} }$ the general integral result
$$\int e^x(f(x)+f’(x)) dx= e^x f(x)$$
to obtain
$$\int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx
=\sqrt{\frac{1+x}{1-x} }\>e^x +C
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all natural numbers $n$ where $n<1000$ such that cube of sum of the digits of $n$ be equal to $n^2$
Find all natural numbers $n$ where $n<1000$ such that cube of sum of
the digits of $n$ be equal to $n^2$
For one digit numbers $n=1$ is the only answer. For two digits numbers $\overline{ab}$:
$$(10a+b)^2=(a+b)^3$$
$$100a^2+20ab+b^2=a^3+3a^2b+3ab^2+b^3$$
But it seems this method doesn't work because I can't simplify above equation.
|
Use @lulu's notation for $S(n)$ to denote the sum of digits of $n$. Using @NeatMath's observation that $(S(n))^3 = n^2$, we see that $n$ is a perfect cube, so it suffices to check the set $\{1^3,2^3,\dots,9^3\}$, due to the constraint $n < 1000 = 10^3$. You might include $0$ if you have an affirmative answer to Is $0$ a natural number?. $0$ and $1$ are trivial solutions.
@lulu's criterion $n^3 \equiv n^2 \pmod9$ restricts the searching to $1^3$, $3^3$, $6^3$ and $9^3$ because if $\gcd(n,9) = 1$, then the cancellation law gives $n\equiv1\pmod9$.
$n$
$S(n)$
$(S(n))^3$
$n^2$
$1^3 = 1$
$1$
$1$
$1$
$3^3 = 27$
$2+7 = 9$
$9^3 = 729$
$729$
$6^3 = 216$
$2+1+6 = 9$
$9^3 = 729$
$46656$
$9^3 = 729$
$7+2+9 = 18$
$18^3 = 5832$
$531441$
Hence, there's only two solutions $1$ and $27$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
For what $a$ and $b$ are there explicit expressions for $I(a, b) =\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b} $? For what $a$ and $b$
are there explicit expressions for
$I(a, b)
=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}
$?
This is inspired by
the answer to
https://www.quora.com/What-is-displaystyle-int_-0-1-int_-0-1-frac-1-1-xy-3-mathrm-d-x-mathrm-d-y
where it is shown that
$I(1, 3)
=\dfrac34\left(\ln(3)+\dfrac{\pi\sqrt{3}}{9}\right)
$.
Here's what I've done
so far.
$\begin{array}\\
I(a, b)
&=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}\\
&=\int_0^1 \int_0^1 dx\,dy\sum_{n=0}^{\infty} (x^ay^b)^n\\
&=\sum_{n=0}^{\infty} \int_0^1 \int_0^1 dx\,dy(x^ay^b)^n\\
&=\sum_{n=0}^{\infty} \int_0^1 x^{an}dx \int_0^1 y^{bn}dy\\
&=\sum_{n=0}^{\infty} \dfrac{x^{an+1}}{an+1}\big|_0^1 \dfrac{y^{bn+1}}{bn+1}\big|_0^1\\
&=\sum_{n=0}^{\infty} \dfrac{1}{an+1} \dfrac{1}{bn+1}\\
&=ab\sum_{n=0}^{\infty} \dfrac{1}{abn+b} \dfrac{1}{abn+a}\\
\\
I(a.a)
&=\sum_{n=0}^{\infty} \dfrac{1}{(an+1)^2}\\
&=\dfrac1{a^2}\sum_{n=0}^{\infty} \dfrac{1}{(n+1/a)^2}\\
&=\dfrac1{a^2}\psi^{(1)}(1/a)\\
\text{If } a \ne b\\
I(a, b)
&=\dfrac{ab}{a-b}\sum_{n=0}^{\infty} \left(\dfrac{1}{abn+b} -\dfrac{1}{abn+a}\right)\\
&=\dfrac{ab}{a-b}\sum_{n=0}^{\infty} \int_0^1 \left(x^{abn+b-1} -x^{abn+a-1}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(\sum_{n=0}^{\infty}x^{abn+b-1} -\sum_{n=0}^{\infty}x^{abn+a-1}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(x^{b-1}\sum_{n=0}^{\infty}x^{abn} -x^{a-1}\sum_{n=0}^{\infty}x^{abn}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(x^{b-1}\dfrac1{1-x^{ab}} -x^{a-1}\dfrac1{1-x^{ab}}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(\dfrac{x^{b-1} -x^{a-1}}{1-x^{ab}}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 x^{b-1}\left(\dfrac{1 -x^{a-b}}{1-x^{ab}}\right)dx\\
\end{array}
$
But,
in general,
I can't go further.
Special cases can be handled.
For example,
if $b=1$
and $a$ is a positive integer,
we can show that
$I(a, 1)
=\dfrac{a}{(a-1)^2} \left(\ln(a)-\int_0^1 \left(\dfrac{\sum_{k=0}^{a-3}(a-2-k)x^k}{\sum_{k=0}^{a-1}x^k}\right)dx\right)
$.
This gives,
as above,
$I(3, 1)
=\dfrac34\left(\ln(3)-\int_0^1 \left(\dfrac{1}{1+x+x^2}\right)dx\right)
$
and completing the square
gives the answer above.
Also
$I(2, 1)
=2\ln(2)
$.
Similarly,
$\begin{array}\\
I(4, 1)
&=\dfrac49\left(\ln(4)-\int_0^1 \left(\dfrac{2+x}{1+x+x^2+x^3}\right)dx\right)\\
&=\dfrac49\left(\ln(4)- \dfrac{3 π + \ln(4)}{8}\right)
\quad\text{(According to Wolfy)}\\
&=\dfrac49\left(\dfrac{7\ln(4)-3 π}{8}\right)\\
\end{array}
$
We could get
$I(5, 1)$
since Wolfy gives
$$\dfrac1{1+x+x^2+x^3+x^4}
=\dfrac{-2 x + \sqrt{5} - 1}{\sqrt{5} (2 x^2 - (\sqrt{5}-1) x + 2)} + \dfrac{2 x + \sqrt{5} + 1}{\sqrt{5} (2 x^2 + (\sqrt{5}+1) x + 2)}
$$
but I'm not going to bother
to work it out.
|
May be, we could start with
$$I(a, b)=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}=\int_0^1 \, _2F_1\left(1,\frac{1}{b};\frac{1}{b}+1;x^a\right)\,dx$$ and consider
$$J_a=\int_0^1 \, _2F_1\left(1,\frac{1}{b};\frac{1}{b}+1;x^a\right)\,dx$$ and discarding the ones which contain logarithms of complex arguments, we can obtain a few of them
$$J_1=\frac{\psi \left(\frac{1}{b}\right)+\gamma }{1-b}$$
$$J_2=\frac{\psi \left(\frac{1}{b}\right)+\gamma +\log (4)}{2-b}$$
$$J_3=\frac{6 \psi \left(\frac{1}{b}\right)+\sqrt{3} \pi +6 \gamma +9 \log (3)}{18-6 b}$$
$$J_4=\frac{2 \left(\psi \left(\frac{1}{b}\right)+\gamma \right)+\pi +\log (64)}{8-2 b}$$
$$J_6=\frac{2 \psi \left(\frac{1}{b}\right)+\sqrt{3} \pi +2 \gamma +\log
(432)}{12-2 b}$$
$$J_8=\frac{2 \psi \left(\frac{1}{b}\right)+\sqrt{2} \pi +\pi +2 \gamma +\log
(256)+2 \sqrt{2} \coth ^{-1}\left(\sqrt{2}\right)}{16-2 b}$$
$$J_{12}=\frac{2 \psi \left(\frac{1}{b}\right)+\sqrt{3} \pi +2 \pi +2 \gamma +2
\sqrt{3} \log \left(2+\sqrt{3}\right)+\log (1728)}{24-2 b}$$
For the particular case where $b=a$, the result seems to be quite simpler
$$K_a=\int_0^1 \, _2F_1\left(1,\frac{1}{a};\frac{1}{a}+1;x^a\right)\,dx=\frac 1 {a^2}\psi ^{(1)}\left(\frac{1}{a}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Find all positive integer solutions for the following equation $5^a + 4^b = 3^c$:
Find all positive integer solutions for the following equation:
$$5^a + 4^b = 3^c$$
My first guess would be to study the equation in mod, but I tried modulo 3, 4, 5, and 9 and I can't find anything.
|
Hint: $$1^a + 0 \equiv _4 (-1)^c \implies c \equiv_2 0$$
so $c=2d$, $d\in \mathbb{N}$.
So $$ 5^a = (3^d-2^b)(3^d+2^b)$$
so $3^d-2^b = 5^x$ and $3^d+2^b = 5^y$. Now add/substract both equations...
...we see that $x=0$. Now if $b>1$ then we have $$(-1)^d -0 \equiv _41 $$ so $d$ is even. So we have $$(3^m-1)(3^m+1)= 2^b$$ Now $(3^m-1)$ and $(3^m+1)$ are two consecutive even numbers so exactly one is divisible by $4$ so the other number is exactly $2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3976046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$ I tried:$$I=\int\frac{x^4}{(x^2-1)^3}dx=\int\frac{(x^2-1)(x^2+1)+1}{(x^2-1)^3}dx=\int\frac{x^2+1}{(x^2-1)^2}+\frac{1}{(x^2-1)^3}dx$$
For first fraction we can write it as $\frac{1}{x^2-1}+\frac{2}{(x^2-1)^2}$. therefor we have:
$$I=\frac12\ln\left|\frac{x-1}{x+1}\right|+\int\frac{2}{(x^2-1)^2}dx+\int\frac{1}{(x^2-1)^3}dx$$
I don't know how to evaluate remaining integrals.
|
Another approach: with $x=-\csc t$ the original integral becomes $\int\sec^5tdt$, then use this together with $\sec t\tan t=\frac{x}{1-x^2},\,(\sec t+\tan t)^2=\frac{2x-1-x^2}{1-x^2}$.
|
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"url": "https://math.stackexchange.com/questions/3981505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Given $\frac{dS}{dt}=-\frac{SI}{S + I + R}$ and $R = -(SI + S + I)/ \frac{dS}{dt}$ Determine $\frac{d^2S}{dt^2}$ I am given a system of equations:
\begin{align}
\frac{dS}{dt} &= -\frac{SI}{S + I + R}\\
\frac{dI}{dt} &= \frac{SI}{S + I + R} -I \\
\frac{dR}{dt} &= I - R\\
\end{align}
I am to subsitute $R = -\left(SI + S + I\right) / \frac{dS}{dt} $ to determine $\frac{d^2S}{dt^2}$
I chose to work only with the first equation.
When I substitute what is given for $R$ into the denominator I get:
\begin{align}
S + I + R &= S + I + R = -\left(SI + S + I\right) / \frac{dS}{dt}\\
&=\frac{\frac{dS}{dt}(S+I)-(SI+S+I)}{\frac{ds}{dt}}
\end{align}
I substitute this into the equation for $\frac{dS}{dt}$ and I get:
\begin{align}
\frac{dS}{dt} &= \frac{-SI \left(\frac{dS}{dt}\right)}{\frac{ds}{dt}\left(S+I\right)-(SI+S+I)}\\
\left(\frac{dS}{dt}\right)^2(S+I)-\frac{dS}{dt}\left(SI+S+I\right) &= -SI\left(\frac{dS}{dt}\right)\\
\left(\frac{dS}{dt}\right)^2(S+I) &= \left(\frac{dS}{dt}\right)\left(S+I\right)\\
\left(\frac{dS}{dt}\right)^2 &= \left(\frac{dS}{dt}\right)\\
\end{align}
How do I get the desired result of
$$
\frac{d^2S}{dt^2}= \frac{2}{S}\left(\frac{dS}{dt}\right)^2
$$
|
I acknowledge this answer is very similar to that of our colleague Ak, with perhaps some differences in formatting and added words of explanation.
We start with the given equations
$\dot S = -\dfrac{SI}{S + I + R}, \tag 1$
$\dot I = \dfrac{SI}{S + I + R} - I, \tag 2$
$\dot R = I - R; \tag 3$
we differentiate (1) (here I have for the moment used "${}^\prime$" in lieu of "$\dot {}$" on the right-hand side):
$\ddot S = -\dfrac{(SI)'(S + I + R) - (SI)(S + I + R)'}{(S + I + R)^2}, \tag 4$
whence
$\ddot S = -\dfrac{(\dot S I + S\dot I)(S + I + R) - (SI)(\dot S + \dot I + \dot R)}{(S + I + R)^2}; \tag 5$
we add (1) and (2):
$\dot S + \dot I = -I, \tag 6$
and to this add (3):
$\dot S + \dot I + \dot R = -R, \tag 7$
and substitute this into (5):
$\ddot S = -\dfrac{(\dot S I + S\dot I)(S + I + R) + RSI}{(S + I + R)^2}; \tag 8$
again from (1) and (2):
$\dot SI + S\dot I = -\dfrac{SI^2}{S + I + R} + \dfrac{S^2I}{S + I + R} - IS$
$= \dfrac{S^2I - I^2S}{S + I + R} - IS, \tag 9$
whence
$(\dot SI + S\dot I)(S + I + R) = S^2I - I^2S - IS(S + I + R)$
$= S^2I - I^2S - IS^2 - I^2S - ISR = -2I^2S - ISR; \tag{10}$
thus,
$(\dot SI + S\dot I)(S + I + R) + RSI = -2I^2S; \tag{11}$
finally we substitute this into (8) and execute some very simple algebraic maneuvers:
$\ddot S = \dfrac{2I^2S}{(S + I + R)^2} = \dfrac{2I^2S^2}{S(S + I + R)^2}, \tag{12}$
$\ddot S = \dfrac{2}{S} \left (\dfrac{IS}{(S + I + R)} \right )^2 = \dfrac{2}{S} \dot S^2, \tag{13}$
$OE\Delta$.
|
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|
What is a more rigorous method of solving modulus function inequalities? I have the inequality
$$\vert x-3\vert < 3x-4$$
To solve this I first break the modulus containing function into the two cases using the critical value
$$\vert x-3 \vert \left\{\begin{array}{cc}
x-3 & x > 3 \\
3-x & x<3
\end{array}\right. $$
For the case where $x>3$:
$$x-3<3x-4$$
$$x>\frac{1}{2}$$
For the case where $x<3$:
$$3-x<3x-4$$
$$x>\frac{7}{4}$$
However it is apparent that these two statements are contradictory and by simply looking at the graph of the two functions it is clear that the value $x>\frac{1}{2}$ does not hold.
The value does however seem to hold for the function $f(x)=x-3$
My question is this, why has my method given a false answer and what improvements can be made to this method to make it such that these false answers no longer appear?
|
Note that the inequality is true whenever $3x-4\le 0.$ Therefore assume $3x-4>0$ and square both sides to obtain $$(x-3)^2>(3x-4)^2,$$ or $$(x+3)^2-(3x-4)^2>0,$$ which gives $$(x+3-3x+4)(x+3+3x-4)>0,$$ simplifying to $(-2x+7)(4x-1)>0,$ or $$(x-3.5)(x-0.25)<0.$$ This implies $0.25<x<3.5$ for $x>4/3,$ or in short $$4/3<x<3.5,$$ so that taking the union of $x\le 4/3$ and $4/3<x<3.5$ gives the result $$(-\infty, 3.5).$$
|
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|
Finding the posterior PMF using Bayes Theorem The problem is as follows
Nefeli. a student in a probability class, takes a multiple-choice test with 10 questions and 3 choices per question. For each question. there are two equally likely possibilities, independent of other questions: either she knows the answer, in which case she answers the question correctly. or else she guesses the answer with probability of success $1/3$.
Given that Nefeli answered correctly 6 out of the 10 questions, what is the posterior PMF of the number of questions of which she knew the answer?
The textbook solution is as follows:
Let $K$ be the r.v. she knows the answer and $C$ be the r.v. she answers the question correctly
$$P(K|C) = \frac{P(C|K)P(K)}{P(C|K)P(K) + P(C|K^C)P(K^C)} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{3}\frac{1}{2}} = \frac{3}{4}$$
The probability that Nefeli knows the answer to a question that she answered correctly is 3/4, so the posterior PMF is binomial with $n = 6$ and $p = 3/4$.
I decided I would try to approach this problem using Bayes Theorem, that is let $X$ be the r.v. for the number of questions correctly answered and $Y$ be the r.v. for the number of questions she knows the answers to. Thus we have:
$$p_{Y|X}(y|6) = \frac{p_{X|Y}(6|y)p_Y(y)}{p_X(6)}$$
$p_X(6)$ is the probability of answering 6 questions correctly which is just a binomial with $n=10$ and $p=P(C)=\frac{1}{2} + \frac{1}{3}\frac{1}{2} = \frac{2}{3}$:
$$\binom{10}{6}\left(\frac{2}{3}\right)^6\left(1-\frac{2}{3}\right)^4 $$
$p_Y(y)$ is the probability that she knows the answer for $y$ questions which is a binomial with $n=10$ and $p=\frac{1}{2}$
$$p_Y(y) = \binom{10}{y}\left(\frac{1}{2}\right)^y\left(1-\frac{1}{2}\right)^{10-y} $$
$p_{X|Y}(6|y)$ is the probability of answering 6 questions correctly given that she knows the answer for $y$ questions. This means that there will be $10-y$ questions to choose from and she will only need to answer $6-y$ more correctly with probability $P(C)$:
$$p_{X|Y}(6|y) = \binom{10-y}{6-y}\left(\frac{2}{3}\right)^{6-y}\left(1-\frac{2}{3}\right)^{10-6} $$
However, when I plug in say $y=3$ just to get a number for this result, it is not the same as the result I get from the book solution. Does anyone know what I may be doing wrong here? I seem to be off by a factor of 2, as in my answer is half the probability of the answer from the binomial with $n = 6$, $p = 3/4$, $k = 3$.
|
Correcting a small error in your last expression, you should have $$p_{X\mid Y}(6\mid y) = \binom{10-y}{6-y}\left(\frac{1}{3}\right)^{6-y}\left(1-\frac{1}{3}\right)^{10-6}$$
You then get $$p_{Y\mid X}(y\mid 6) = \frac{\binom{10-y}{6-y}\left(\frac{1}{3}\right)^{6-y}\left(1-\frac{1}{3}\right)^{10-6} \binom{10}{y}\left(\frac{1}{2}\right)^y\left(1-\frac{1}{2}\right)^{10-y}}{\binom{10}{6}\left(\frac{2}{3}\right)^6\left(1-\frac{2}{3}\right)^4}$$ and with cancellation this is $$=\frac{6!}{y!(6-y)!}\frac{3^y}{2^{12}} =\binom{6}{y}\left(\frac{3}{4}\right)^{y}\left(\frac{1}{4}\right)^{6-y} $$
|
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|
Factoring of a particular polynomial It is given that $n$ is an odd integer greater than $3$ but not a multiple of $3$. Prove that $x^3 +x^2 +x$ is a factor of $(x+1)^n -x^n -1$.
I tried to write $x^3+x^2+x=x(x^2+x+1) =x(x-\omega^2)(x-\omega)$ where $\omega$ and $\omega^2$ are cube roots of unity $≠1$. Then I tried to do it by putting values but could not get the answer
|
Modulo $x$, $(x+1)^n-x^n-1\equiv 1^n-0-1=0$, so $(x+1)^n-x^n-1$ is divisible by $x$.
It remains to be shown that $(x+1)^n-x^n-1$ is divisible by $x^2+x+1$. Well,
$(\omega+1)^n-\omega^n-1=(-\omega^2)^n-\omega^n-1=-(\omega^{2n}+\omega^n+1)$ for $n$ odd.
Furthermore, for $n$ not a multiple of $3$, $(\omega^{2n}+\omega^n+1)(\omega^{n}-1)=\omega^{3n}-1=0$ and $\omega^n-1\ne0$,
so $\omega^{2n}+\omega^n+1=0$. Therefore, $\omega$ is a zero of $(x+1)^n-x^n-1$.
|
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|
Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that $|ba^{-1}| = |b^3a^3|$. Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that
$|ba^{-1}| = |b^3a^3|$.
attempt:
Write $a*b=ab$. Let $a,b \in G$.
Then, $ab=b^5a^3$. Note that
$ba^{-1} = a^{-1}b^5a^2$,
$ab^{-1} = b^2(b^3a^3)b^{-2}$, and
$b^3a^3 = b^{-2}(ab^{-1})b^2$.
Denote the identity element of $G$ by $e_G$.
Fact: for all $a \in G$,
$|a| = |a^{-1}|$.
Let $|ba^{-1}| = m$ and $|b^3a^3| = p$.
To show $m \mid p$:
\begin{align*}
(b^3a^3)^p &= e_G \\
(b^{-2}(ab^{-1})b^2)^p &= e_G \\
b^{-2}(ab^{-1})^pb^2 &= e_G \\
b^{-2}((ba^{-1})^{-1})^pb^2 &= e_G \\
((ba^{-1})^{-1})^p &= e_G
\end{align*}
By fact, $|(ba^{-1})^{-1}| = m$.
Hence, $m \mid p$ as desired.
To show $p \mid m$:
\begin{align*}
(ba^{-1})^m &= e_G \\
((ab^{-1})^{-1})^m &= e_G \\
((b^2(b^3a^3)b^{-2})^{-1})^m &= e_G \\
((b^2(b^3a^3)b^{-2})^m)^{-1} &= e_G
\end{align*}
Here I got a little bit confuse.
Does it true that in:
\begin{equation*}
((b^2(b^3a^3)b^{-2})^m)^{-1} = e_G
\end{equation*}
implies
$((b^2(b^3a^3)b^{-2})^m) = e_G$? (This is come from fact.) If yes, then
$|b^3a^3| = m$. Hence, $p \mid m$.
And thus, $m=p$.
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You don't need to go through divisibility arguments. Since $ab^{-1}=(b^2)(b^3a^3)(b^{-2})$, $ab^{-1}$ and $b^3a^3$ are conjugate and have the same order. Using the fact about orders of inverses, the original claim follows.
|
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|
Matrices equation with one unknown matrix Let us consider following equation:
$$
2X^2+2X=
\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 & -3
\end{array} \right).
$$
I have to show that there is no real matrix which satisfy my equation.
I consider three diffrent way to do this.
*
*Direct method. Let $$X=\left( \begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i
\end{array} \right).
$$ Then we can write euqation in equivalent form $X(X+I)=\frac{1}{2}\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 & -3
\end{array} \right)$, and we obtain system of nonlinear equations
$$
\left\{ \begin{array}{llll}
a^2+a+bd+cg=-\frac{1}{2} \\
ab+be+b+ch=\frac{5}{2}\\
ac+bf+ci+c=\frac{3}{2}\\
da+d+ed+fg=-1\\
db+e^2+e+fh=\frac{1}{2}\\
dc+ef+fi+f=1\\
ga+g+hd+ig=0\\
gb+he+h+ih=-2\\
gc+hf+i^2+i=-\frac{3}{2}
\end{array} \right.
$$
I was trying to deal with it, but it was impossible to solve for me (I can not use Mathematica, Matlab,etc)
*Determinant method. From the fact, that $X(X+I)=\frac{1}{2}\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 & -3
\end{array} \right)$, we have that $\det X \det (X+I)=-\frac{11}{8}$, and then using direct method i was traing to write formula for $\det (X+I)$ in terms of $\det X$, but here also appear formulas,to difficult for me to interpret it.
*My last idea to solve is to use formula for inverse matrix $3\times3$ ($X$ is invertible, which is consequence of the fact, that $\det X\neq 0$) and from relation $X+I=X^{-1}\frac{1}{2}\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 & -3
\end{array} \right)$ we also obtain system of equations, which is much more complicated, that this previous one.
I would be grateful if you give me some hints.
|
If there were a real matrix $X$ satisfying ...
Scalar-multiply the given equation with $2$ and add the identity matrix to get
$$(2X+\mathbb 1)^2 \;=\; \begin{pmatrix} -1& 10& 6\\
-4& 3& 4\\
0& -8& -5 \end{pmatrix}$$
Apply the determinant which produces $\,-25\,$ on the RHS, hence a contradiction because the resulting square on the LHS cannot get negative if $X$ is a real matrix.
|
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|
find angle x in isosceles triangle Considering the attached image, is that possible to compute for $\angle x$ in the isosceles triangle below, with no further known values?
$\\$I have found $\angle B$ and $\angle C$ are $50^o$. So $\angle B$ is divided by $\overline{BP}$ to $10^o$ and $40^o$. But I can't find how the $\angle C$ is divided by $\overline{CP}$. Also its clear that $\overline{BP}$ is perpendicular to leg $\overline{AC}$.
|
Construct equilateral triangle $ABX$ as in the picture.
Note that the concave angle $BXA$ equals $300^\circ = 2\cdot 150^\circ = 2\angle BPA$, hence $P$ lies on the circle with center $X$ and radius $XA=XB$. It follows that $\angle PXB = 2\angle PAB = 40^\circ$.
On the other hand, $AB=AX=AC$, so $A$ is the circumcenter of triangle $BXC$, hence $\angle CXB = \frac 12 \cdot \angle CAB = \frac 12 \cdot 80^\circ = 40^\circ$.
Since $\angle PXB = 40^\circ = \angle CXB$, points $X, P, C$ are collinear.
Since $AC=AX$, we have $\angle PCA = \angle AXP = 2\cdot \angle ABP = 20^\circ$, hence $$\angle APC = 180^\circ - \angle CAP - \angle PCA = 180^\circ - 60^\circ - 20^\circ = 100^\circ.$$
|
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|
Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$.
Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$
Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:
\begin{align}
n = 2 \Longrightarrow f(x) &= x^x \\
f'(x) &= \frac{d}{dx}\left(e^{x\ln \left(x\right)}\right) \\
&= e^{x\ln \left(x\right)}\frac{d}{dx}\left(x\ln \left(x\right)\right) \\
&= e^{x\ln \left(x\right)}\left(\ln \left(x\right)+1\right) \\
&= x^x\left(\ln \left(x\right)+1\right)
\end{align}
\begin{align}
n = 3 \Longrightarrow f(x) &= x^{x^{x}} \\
f'(x) &= \frac{d}{dx}\left(e^{x^x\ln \left(x\right)}\right) \\
&= e^{x^x\ln \left(x\right)}\frac{d}{dx}\left(x^x\ln \left(x\right)\right) \\
&= e^{x^x\ln \left(x\right)}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right) \\
&= x^{x^x}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)
\end{align}
\begin{align}
n = 5 \Longrightarrow f(x) &= x^{x^{x^{x^{x}}}} \\
f'(x) &= ... \\
&= x^{x^{x^{x^x}}}\left(x^{x^{x^x}}\ln \left(x\right)\left(x^{x^x}\ln \left(x\right)\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)+x^{x^x-1}\right)+x^{x^{x^x}-1}\right)
\end{align}
However, I am not sure if I see a pattern here that can help solve the question.
|
Let $f_1(x)=x$ and $f_n(x)=x^{f_{n-1}(x)}=x^{x^{\ldots^{x}}}$ ($n$ $x's$ on the exponent).
Then $\ln(f_n(x))=f_{n-1}(x)\cdot\ln(x)$ so: $$\frac{f'_n(x)}{f_{n}(x)}=f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\implies f'_{n}(x)=f_{n}(x)\cdot\left(f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\right).$$
|
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|
Divisibility Rule for 21 similar to rule for 12 I already understand the divisibility rule for 12. $$10\equiv -2 \pmod{12} \implies 10^n\equiv (-2)^n \pmod{12}$$
Then for some number $n = abcd = 1000a + 100b + 10c + d, abcd\equiv1000a + 100b + 10c + d \pmod{12}$
Then $abcd\equiv (-2)^3a + (-2)^2b + (-2)c + d$
Example: $7368\equiv -8*7 + 4*3 -2*6 + 8\equiv -48\equiv 0 \pmod{12}$
But why does it work for 21 as well?
If I switch the order that the power rises in, so that now it is $a + (-2)b + (-2)^2c + (-2)^3d$ now, all of a sudden, it works for mod 21
Example: $2331\equiv 2 + (-2)*3 + (-2)^2*3 + (-2)^3*1 \equiv 0 \pmod{21}$
It's obvious why this works for numbers like $48$ and $84$ since with both you are multiplying $4 * -2$ and $8 * 1$ and their sum is then $0$ for both.
But why does it still work for numbers like $105$.
$501$ is obviously not divisible by $12$ and yet: $105\equiv 1 + (-2)*0 + (-2)^2*5\equiv 21 \equiv 0 \pmod{21}$
I'm also fairly sure this is part of the reason that $f$ works in Proof of general divisibility rule
|
For $21$, the reason this works is that $10\times(-2)=1\pmod{21}$, so $10\equiv(-2)^{-1}\pmod{21}$. Hence
$$1000a+100b+10c+d\equiv(-2)^{-3}a+(-2)^{-2}b+(-2)^{-1}c+(-2)^{0}d\pmod{21},$$
so multiplying the RHS by $(-2)^3$ implies the result.
You will observe a similar result for all teen numbers $\overline{1m}=10+m$, because
$$10\times(-m)\equiv-\overline{m0}\equiv1\pmod{\overline{m1}}.$$
|
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|
Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
If I would start from the right expression I would be able to reduce it. But I'm not used to do the reverse.
I can see that Wolframalpha expands the left expression in many ways, one of which is the one I want, but are there any trick to do this expansion by hand?
This question is linked to this one.
|
As we know the RHS, let's try to express $\cos^2(2\theta)$ in terms of $\cos(4\theta)$
$\begin{align}(\sin\theta-\cos2\theta)^2 & = \sin^2\theta - 2\sin\theta\cos2\theta + \cos^22\theta\\& = \sin^2\theta-2\sin\theta\cos2\theta + \frac{1 + \cos4\theta}2 \end{align}$
Now $\cos2\theta = 1 - 2\sin^2\theta$ we use this to get $4\sin^3\theta$
$\begin{align}(\sin\theta-\cos2\theta)^2 & = \sin^2\theta - 2\sin\theta + 4\sin^3\theta +\frac12+\frac{\cos4\theta}2 \\
&=\sin^2\theta +\sin\theta -( 3\sin\theta - 4\sin^3\theta) +\frac12+\frac{\cos4\theta}2 \\&=\sin\theta - \sin3\theta+\frac{\cos4\theta}2 + 1 - \frac12+\sin^2\theta \\&=\sin\theta-\sin3\theta+\frac{\cos4\theta}2 + 1 - \frac{1-2\sin^2\theta}{2} \\&=\boxed{1+\sin\theta-\sin3\theta+\frac{\cos4\theta}2 - \frac{\cos2\theta}{2} }\end{align}$
|
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|
Finding the limit of $a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2})$ for arbitrary $a_{0}$ and $a_{1}$. Here is the problem statement:
Let $a$, $b$ $\in$ $\Bbb{R}$. A sequence $(a_{n})_{n \in \Bbb{N}}$ is defined recursively by
$$a_{0}:=a, \qquad a_{1}:=b, \qquad a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2}) \quad \text{for} \quad n \geq 2.$$
Prove that $\lim_{n\to\infty}a_{n}$ exists and compute its value.
Now I have proved that the limit exists. My idea was to use the Monotone Convergence Theorem as follows:
Proof: Without loss of generality, assume $a<b$. By inspection of the first two terms of $a_{n}$, we see that
$$a_{3} = \frac{1}{2}(a_{2}+a_{1})=\frac{1}{2}\left[\frac{1}{2}(a_{1}+a_{0}) + a_{1}\right] = \frac{1}{2}(a_{1}+a_{0})+\frac{1}{4}(a_{1}-a_{0})=a_{2}+\frac{1}{4}(a_{1}-a_{0})>a_{2} \quad (\because a_{1}>a_{0}).$$
Therefore we need to show that the sequence is increasing, so our inductive hypothesis is $a_{n}<a_{n+1}$.
$$a_{n}<a_{n+1}\implies a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{n+1}+a_{n})=a_{n+2}.$$
Thus $a_{n}$ is increasing for $n\geq 2$. Now we need to show that it is bounded. By inspection, $a_{3}=\frac{1}{2}(a_{2}+a_{1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}$, and $a_{4}<a_{1}$ and so on. So again assume by induction $a_{n}<a_{1}$ and we have
$$a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}.$$
Thus $a_{n}$ is bounded above. Hence by the Monotone Convergence Theorem, sequence is convergent and $\lim a_{n}$ exists.$\quad\square$
However, I'm stuck in finding the limit. Any hints would be appreciated.
|
The characteristic equation is
$$\lambda^2=\frac{1}{2}(\lambda+1)$$
Its solutions are
$$\lambda=1;\;\lambda=-1/2$$
The general solution of the recurrence is
$$a_n=m+n\left(-\frac{1}{2}\right)^n$$
for $n=0$ we get $a_0=m+n=a$
for $n=1$ we have $a_1=m-\frac12n=b$
so we get
$$m= \frac{1}{3} (a+2 b),n= \frac{2 (a-b)}{3}$$
and the recurrence can be written as
$$a_n=\frac{1}{3} (a+2 b)+ \frac{2 (a-b)}{3}\left(-\frac{1}{2}\right)^n$$
as $n\to\infty$ we have $a_n\to \frac{1}{3} (a+2 b)$.
Hope this is useful
|
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|
Show that there is only one real root of the equation $1 + a^2 + ax - x^3$ I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?
Thanks.
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Write $f(x)=x^3-ax-a^2-1.$ Then we have that $f'(x)=3x^2-a.$ Hence the function has its local extrema at $$x=\pm\sqrt {\frac a3}.$$ It follows that there is exactly one real root if the relative extrema have equal sign. Hence we compute $$f\left(\sqrt {\frac a3}\right)f\left(-\sqrt {\frac a3}\right)$$ and confirm that it is positive. This gives $$\left[a^2+1 +a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right]\left[a^2+1-\left(a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right)\right]=(a^2+1)^2-\left[a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right]^2=(a^2+1)^2-\frac a3\left(a-\frac a3\right)^2=a^4-\frac{4}{27}a^3+2a^2+1=\left(a^2+\frac{2a}{27}\right)^2+\left(2-\frac{4}{27^2}\right)a^2+1,$$ which is clearly positive for all $a.$
|
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|
Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$ $a$ and $b$ are both positive Real Numbers .
I saw this in a math olympiad, and within two days i couldn't solve it.
There is a simple case where:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
We can solve it like this:
$$(a-1)^{2}\geq 0$$
$$\Leftrightarrow a^{2}-2a+1 \geq0$$
$$\Leftrightarrow a^{2}+1 \geq2a$$
$$\Leftrightarrow \frac{a^{2}+1}{a} \geq2$$
With the same method:
$$\frac{b^{2}+1}{b} \geq2 $$
And when we add the two inequalities:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
But in case where:
$$\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$$
I can’t solve it,
Thank you for your help.
|
You can use your method, making $(a-1)^2$ and $(b-1)^2$ appearing was a nice move.
$$\frac{a^2+1}{b}+\frac{b^2+1}{a}-4=\frac 1{\underbrace{ab}_{>0}}(\underbrace{a^3+a+b^3+b-4ab}_E)$$
Now use $2ab\le a^2+b^2$
$E\ge a^3+a+b^3+b-2a^2-2b^2=a(a-1)^2+b(b-1)^2\ge 0$
|
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|
Is it possible to solve $\frac{x(x-1)}{(x+y)(x+y-1)} =\frac12$, with $x>y$ and $100 \leq x+y\leq500$, without writing some program? I am reading a puzzle that ends up based on the given info to require the solution of the following equation:
$$\frac{x(x-1)}{(x + y)(x + y -1)} =\frac12$$
also knowing that:
$$x \gt y \qquad\text{and}\qquad 100 \leq x + y \leq 500$$ and $x >= 0 \land y >= 0$
The solution states that it requires a computer to figure out the values of $x$ and $y$ (which are $x=85$ and $y = 35$).
I was wondering is this only possible to be solved with a computer program or is there a way to progress the equation even further?
Trying out I could reach nowhere to be honest:
$$\begin{align}
2x(x-1) &= (x + y) (x + y -1) \\[4pt]
2x^2 - 2x &= x^2 + xy - x + yx + y^2 - y \\[4pt]
x^2 - x &= y^2 +2xy -y
\end{align}$$
but I don't see how this helps at all.
Is there any way to do some progression?
|
This extends @Empy2's answer.
Starting from $2x(x-1) = (x+y)(x+y - 1) := z(z-1)$, we complete the squares:
$$2(x^2-x+\frac14)-\frac12 = z^2-z + \frac14 - \frac14$$
$$2(x-\frac12)^2 = (z-\frac12)^2 + \frac14$$
$$2(2x-1)^2 = (2z-1)^2+1$$
Equations of the form $x^2 - Dy^2 = \pm1$, where $D$ is not a square, are the simplest Pell equations (as opposed to the generalized version). Their solutions are related to the convergents of $\sqrt D$ (here, $\sqrt 2$.)
http://oeis.org/A000129 gives some details of these convergents, whose numerators and denominators give the solutions:
$$1/1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741\dots$$
These fractions have a pattern:
$$1^2-2\times 1^2 = -1$$
$$3^2-2\times 2^2 = +1$$
$$7^2 - 2\times 5^2 = -1$$
$$17^2-2\times 12^2 = +1$$
In particular:
$$239^2 - 2\times169^2 = -1$$
which gives the solution $x=85, z = 120$, and is the only one in range ($199 < 2z - 1 < 999$).
|
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I have to show that the sum of this double series is $\frac{1}{2}$ i have to solve this double series. i tried it, but i am not sure, that it is enough.
$$\sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \left(\left(\frac{1}{k+1} \cdot \left(\frac{k}{k+1}\right)^{i}\right) - \left(\frac{1}{k+2} \cdot \left(\frac{k+1}{k+2}\right)^{i}\right)\right)$$
$$= \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \frac{(1)^{i}}{k+1} - \frac{(1)^{i}}{k+2} $$
$$= \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{k+1} - \frac{1}{k+2} $$
this is a telescoping series. Because of that, the Series is convergence to $\lim_{n\to\infty} \frac{1}{2} - \frac{1}{n+2} = \frac{1}{2}$
Is this solution right and enough?
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Given that
$$\color{red}{\sum_{i=1}^{\infty}} \sum_{k=1}^{\infty} \bigg(\bigg(\frac{1}{k+1} \cdot \bigg(\frac{k}{k+1}\bigg)^{i}\bigg) - \bigg(\frac{1}{k+2} \cdot \bigg(\frac{k+1}{k+2}\bigg)^{i}\bigg)\bigg)$$
let's deal with the black sum first.
$$\begin{align}
&\sum_{k=1}^{\infty} \bigg(\underbrace{\bigg(\frac{1}{k+1} \cdot \bigg(\frac{k}{k+1}\bigg)^{i}}_{a_k}\bigg) - \bigg(\underbrace{\frac{1}{k+2} \cdot \bigg(\frac{k+1}{k+2}\bigg)^{i}\bigg)}_{a_{k+1}}\bigg) \\
&= \lim_{n\to\infty}\sum_{k=1}^n(a_k - a_{k+1}) \\
&= \lim_{n\to\infty}(a_1 - a_{n+1}) \\
&= \lim_{n\to\infty}\left(\frac{1}{2}\left(\frac{1}{2}\right)^i - \left(\frac{1}{n+2}\left(\frac{n+1}{n+2}\right)^i\right)\right) \\
&= \left(\frac{1}{2}\right)^{i+1}
\end{align}$$
Now, we will move to the red sum:
$$\color{red}{\sum_{i=1}^\infty} \left(\frac{1}{2}\right)^{i+1} = \frac{\left(\frac{1}{2}\right)^2}{1 - \frac{1}{2}} = \frac{1}{4}\cdot 2 = \frac{1}{2}$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/4000278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Ideas on how to integrate $\int \frac{dx}{\sqrt{(a-x)(x-b)(x^2+q+px)}}$ where $a>b$ I am trying to integrate $$\int \frac{dx}{\sqrt{(a-x)(x-b)(x^2+q+px)}},$$ where $a>b$, and $p$ and $q$ are real positive constants.
I would be very grateful if I can get some hints on how to proceed. Wolfram Mathematica tells me that this will be giving me an ellpitic integral.
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Yes, it has the structure of an Elliptic integral. You cannot solve it in terms of elementary functions, so you have to accept W. Mathematica's result, which is the only "true" result you can get.
By the way, for the curious people, the result is:
$$\frac{2 (a-x) \left(a \left(x-\sqrt{-\text{px}-q}\right)+x \sqrt{-\text{px}-q}+\text{px}+q\right) \sqrt{\frac{a \left(-x \sqrt{-\text{px}-q}+\text{px}+q\right)-(\text{px}+q) \left(\sqrt{-\text{px}-q}+x\right)}{(a-x) (\text{px}+q)}} \sqrt{\frac{(b-x) \left(a^2+\text{px}+q\right)}{(a-x) \left(a \left(b-\sqrt{-\text{px}-q}\right)+b \sqrt{-\text{px}-q}+\text{px}+q\right)}} F\left(\sin ^{-1}\left(\frac{\sqrt{\frac{(\text{px}+q) \left(\sqrt{-\text{px}-q}-x\right)+a \left(\text{px}+q+x \sqrt{-\text{px}-q}\right)}{(\text{px}+q) (a-x)}}}{\sqrt{2}}\right)|-\frac{2 (a-b) \sqrt{-\text{px}-q}}{\text{px}+q+a \left(b-\sqrt{-\text{px}-q}\right)+b \sqrt{-\text{px}-q}}\right)}{\left(a^2+\text{px}+q\right) \sqrt{\frac{a \left(x \sqrt{-\text{px}-q}+\text{px}+q\right)+(\text{px}+q) \left(\sqrt{-\text{px}-q}-x\right)}{(a-x) (\text{px}+q)}} \sqrt{(a-x) (x-b) \left(\text{px}+q+x^2\right)}}$$
EDIT
As suggested by @Claude Leibovici, one can write $x^2+px+q$ as $(x-c)(x-d)$ (in a way such that $q = cd$ etc...), and get the more suitable solution:
$$\frac{2 (x-b) (x-d) \sqrt{\frac{(a-b) (c-x)}{(a-x) (c-b)}} F\left(\sin ^{-1}\left(\sqrt{\frac{(a-d) (b-x)}{(b-d) (a-x)}}\right)|\frac{(a-c) (b-d)}{(b-c) (a-d)}\right)}{(b-d) \sqrt{\frac{(a-b) (a-d) (b-x) (x-d)}{(a-x)^2 (b-d)^2}} \sqrt{(a-x) (x-b) (x-c) (x-d)}}$$
EDIT 2
Due to some considerations, we can simplify the non Elliptic terms to get
$$-\frac{2 (b-x) (x-d) \sqrt{\frac{x-c}{(x-a) (c-b)}}}{\sqrt{\frac{(a-d) (x-b) (d-x)}{(a-x)^2}} \sqrt{(x-a) (b-x) (x-c) (x-d)}} \times \text{Elliptic Term}$$
Due to the unknown nature of $x$ and of $a$ wrt $b$ and $c$ wrt $d$, we cannot simplify it more, I believe.
EDIT 3 - A Very Special Case
If we assume that $x>c$, $x>a$, $c>b$, $d>a$, we get
$$\frac{-2\text{sgn}(d-x) F\left(\sin ^{-1}\left(\sqrt{\frac{(a-d) (b-x)}{(b-d) (a-x)}}\right)|\frac{(a-c) (b-d)}{(b-c) (a-d)}\right)}{\sqrt{(a-d) (-(b-c))}}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4006685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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