Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
$\int_{0}^{\pi/2}\ln\left(1+4\sin^4 x\right)\mathrm{d}x$ and the golden ratio We already know that, for any real number $t$ such that $t\geq-1$,
$$
\int_{0}^{\pi/2} \ln \left(1+t \sin^2 x\right) \mathrm{d}x = \pi \ln \left( \frac{1+\sqrt{1+t}}{2} \right).
$$
Prove that
$$
\int_{0}^{\pi/2} \ln \left(1+4\sin^4 x\right) \mathrm{d}x = \pi \ln \left( \frac{\varphi+\sqrt{\varphi}}{2} \right)
$$
where $\displaystyle \varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio.
|
Let's put it with real numbers.
We have
$$
\int_{0}^{\pi/2} \ln \left(1+t\sin^4 x\right) \mathrm{d}x
= \pi \ln \left( \frac{1}{4} \sqrt{ 1 + \sqrt{1+t} }
\left( \sqrt{1 + \sqrt{1+t} } + \sqrt{2} \right) \right), \quad t \geq -1.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/873846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
}
|
Prove that: $\sum\limits_{cyc} \frac{a^2+2bc}{(b+c)^2}\geq \sum\limits_{cyc} \frac{3}{2}\frac{a}{b+c}$ Let $a, b, c > 0$.Prove that: $\sum\limits_{cyc} \frac{a^2+2bc}{(b+c)^2}\geq \sum \frac{3}{2}\frac{a}{b+c}$
p/s:
I tried to solve the problem by $S.O.S$. But I cannot solve it !!
I have:
The inequatily $\Leftrightarrow \sum\limits_{cyc} S_{c}(a-b)^2\geq 0$ with $S_c=\frac{(a+b)(a+b+2c)}{2(b+c)^2(a+c)^2}-\frac{1}{4(a+b)^2}-\frac{3}{4(a+b)(b+c)}$
However I cannot find a way....
|
There is a good reason why you couldn’t solve it : your inequality is false
as claimed. Let $\varepsilon\in[-\frac{1}{2},\frac{1}{2}]$ be a variable tending to zero. Putting $a=1-2\varepsilon,b=1+\varepsilon,c=1+\varepsilon$, we have (using Landau’s $O$-
notation)
$$
\begin{array}{lcl}
\sum\limits_{cyc}\frac{a^2+2bc}{(b+c)^2} &=&
\frac{9}{4}+\frac{27}{8}\varepsilon^2
+\frac{243}{16}\varepsilon^3+\frac{3}{32}\varepsilon^4+O(\varepsilon^5) \\
\sum\limits_{cyc}\frac{a}{b+c} &=&
\frac{3}{2}+\frac{9}{4}\varepsilon^2
-\frac{9}{8}\varepsilon^3+\frac{27}{16}\varepsilon^4+O(\varepsilon^5) \\
\end{array}
$$
Thus, when $\varepsilon$ is positive and very small, we have
$$
\sum_{cyc}\frac{a^2+2bc}{(b+c)^2}-\frac{3}{2}
\sum_{cyc}\frac{a}{b+c}=-\frac{81}{16}\varepsilon^3+\frac{81}{16}\varepsilon^4+O(\varepsilon^5) < 0.
$$
Note that when $\varepsilon$ is negative and very small, the LHS above becomes positive, so that the question in the
OP cannot be fixed by reversing the inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/876745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Polynomial Division - "Define the largest natural number..." Would someone mind helping me with this question? The more detailed possible so I can have 100% of understanding. Thanks.
Question: Define the largest natural number m such that the polynomial
$$P(x) = x^5-3x^4+5x^3-7x^2+6x-2$$
be divisible by $(x-1)^m$.
|
Using the Euclidean division of $x^5-3x^4+5x^3-7x^2+6x-2$ and $x-1$ we get:
$$x^5-3x^4+5x^3-7x^2+6x-2=(x^4-2x^3+3x^2-4x+2)(x-1)$$
Then apply the Euclidean division of $x^4-2x^3+3x^2-4x+2$ and $x-1$.
Then we get $x^4-2x^3+3x^2-4x+2=q(x-1)$.
Then apply the Euclidean division of $q$ and $x-1$ and so on.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/877267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Is there also an other way to show the equality: $\left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$? I want to show that:
$$ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$$
That's what I have tried:
*
*$ \left\lfloor \frac{n}{2}\right\rfloor=\max \{ m \in \mathbb{Z}: m \leq \frac{n}{2}\}$
*$\left\lceil \frac{n}{2} \right\rceil= \min \{ m \in \mathbb{Z}: m \geq \frac{n}{2}\}$
If $n=2k,k \in \mathbb{Z}$,then: $\frac{n}{2} \mathbb{Z}$,so
$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n}{2}, \frac{n-2}{2}, \dots \right\}=\frac{n}{2} \\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n}{2}, \frac{n+2}{2}, \dots \right\}=\frac{n}{2}$$
Therefore, $ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$.
If $n=2k+1, k \in \mathbb{Z}$,then $\frac{n}{2} \notin \mathbb{Z}$.So:
$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n-1}{2}, \frac{n-3}{2}, \dots \right\}=\frac{n-1}{2}\\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n+1}{2}, \frac{n+3}{2}, \dots \right\}=\frac{n+1}{2}$$
Therefore, $ \lfloor \frac{n}{2}\rfloor + \lceil \frac{n}{2} \rceil=\frac{n-1}{2}+\frac{n+1}{2}=n$
Is there also an other way to show the equality or is it the only one?
|
Well, here's another one. In fact, it is possible to prove that for any integer $n$ and any real number $\alpha$, we have
$$n=\lfloor (1-\alpha) n\rfloor +\lceil \alpha n\rceil.$$
First, we have $$\alpha n\leq\lceil\alpha n\rceil,$$ therefore, $$n-\lceil\alpha n\rceil\leq (1-\alpha)n,$$ but then, since $n-\lceil\alpha n\rceil$ is an integer, this means $$n-\lceil\alpha n\rceil\leq \lfloor (1-\alpha)n\rfloor.$$
Similarly, we have $$\lfloor (1-\alpha)n\rfloor\leq (1-\alpha)n,$$ therefore, $$\alpha n\leq n-\lfloor (1-\alpha)n\rfloor,$$ but then again, $$\lceil\alpha n\rceil\leq n-\lfloor (1-\alpha)n\rfloor,$$
from the above, we get $$\lfloor (1-\alpha) n\rfloor +\lceil \alpha n\rceil\leq n\leq \lfloor (1-\alpha) n\rfloor +\lceil \alpha n\rceil.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/878854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Can you factor before finding derivative? Say the function is $y=\frac{x^2-1}{x-1}$
Can you factor functions before finding the derivative or does that not work?
|
The functions $f(x) = \dfrac{x^2-1}{x-1} = \dfrac{(x-1)(x+1)}{x-1}$ and $g(x) = x+1$ are equal at every real number $x$ except $x = 1$, where $f(x)$ is undefined, but $g(x)$ is defined.
Therefore, $f'(x) = g'(x) = 1$ at every real number $x$ except $x = 1$.
If you used the quotient rule, you get
$f'(x) = \dfrac{(x-1)(2x)-(x^2-1)(1)}{(x-1)^2} = \dfrac{x^2-2x+1}{(x-1)^2} = \dfrac{(x-1)^2}{(x-1)^2} = 1$ (provided $x \neq 1$).
So, yes, you can factor and simplify the function before computing the derivative, but you must be careful to not enlarge the domain of the function or its derivative.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/878913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Finding a mistake in the computation of a double integral in polar coordinates I have to find $P\left(4\left(x-45\right)^2+100\left(y-20\right)^2\leq 2 \right) $
$f(x)$ and $f(y)$ are given, which I will use in my solution below .
$$P\left(4\left(x-45\right)^2+100\left(y-20\right)^2\leq 2 \right) = \int\int_D f(x)\cdot f(y)\hspace{1mm}dydx$$
Where D is the region inside $4\left(x-45\right)^2+100\left(y-20\right)^2\leq 2$
Substitute $x = 45+\dfrac{r}{2}\cos\theta$ and $y = 20+\dfrac{r}{10}\sin\theta$
The Jacobian will be $r\cdot \dfrac{1}{2}\cdot \dfrac{1}{10} = \dfrac{r}{20}$
Then we can say $D \in \left( r, \theta\right)\hspace{1mm}|\hspace{2mm} 0\leq r\leq \sqrt{2},\hspace{2mm}0\leq \theta \leq 2\pi$
$$\int\int_D f(x)\cdot f(y)\hspace{1mm}dydx = \int_0^{2\pi} \int_{0}^{\sqrt{2}}\left[ \dfrac{2}{\sqrt{2\pi}}\hspace{1.4mm} e^{-0.5r^2\cos^2\theta}\right]\cdot \left[\dfrac{10}{\sqrt{2\pi}} e^{-0.5r^2\sin^2\theta} \right] \cdot \left[ \dfrac{r}{20}\right]drd\theta$$
$$= \int_0^{2\pi} \int_{0}^{\sqrt{2}}\dfrac{r}{2\pi}\hspace{1.4mm} e^{-0.5r^2\cos^2\theta-0.5r^2\sin^2\theta}\hspace{1mm}drd\theta$$
$$= \dfrac{1}{2\pi}\left[\int_0^{2\pi} d\theta \right]\left[\int_{0}^{\sqrt{2}}r\hspace{1.4mm} e^{-0.5r^2}\hspace{1mm}dr \right]$$
Substitute $-0.5r^2 = u$ and $-r dr = du$
$$\text{Limits of Integration will change from $\int_0^{\sqrt{2}}$ to $\int_{-0^2}^{-(\sqrt{2})^2} = \int_0^{-2}$}$$
$$= \dfrac{1}{2\pi}\left[{2\pi} \right]\left[-\int_{0}^{-2}e^{u}\hspace{1mm}du \right]$$
$$=\left[-e^{u}\right]_{0}^{-2} = 1-e^{-2} = 1-\dfrac{1}{e^2}\approx 0.8647$$
Answer at the back of the book is $0.632$!
|
Upper limit $-\color{red}{0.5}(\sqrt2)^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/879899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Trignometry-Prove that $(\csc\theta - \sec\theta )(\cot \theta -\tan\theta )=(\csc\theta +\sec\theta )(\sec\theta ·\csc\theta -2)$ Prove that $$(\csc\theta - \sec\theta )(\cot \theta -\tan\theta )=(\csc\theta +\sec\theta )(\sec\theta ·\csc\theta -2)$$
I tried solving the LHS and RHS seperately but they were not coming out to be equal.
Please help me answer this question.
And also how does one go about proving such questions?
Thanks in advance
|
$(\csc\theta-\sec\theta)(\cot\theta-\tan\theta)=\displaystyle\big(\frac{1}{\sin\theta}-\frac{1}{\cos\theta}\big)\big(\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}\big)$
$\displaystyle=\big(\frac{\cos\theta-\sin\theta}{\sin\theta\cos\theta}\big)\big(\frac{\cos^{2}\theta-\sin^2\theta}{\cos\theta\sin\theta}\big)$
$\displaystyle=\frac{(\cos\theta-\sin\theta)^2(\cos\theta+\sin\theta)}{\sin^2\theta\cos^2\theta}$
$\displaystyle=\frac{(\cos^2\theta-2\sin\theta\cos\theta+\sin^2\theta)(\cos\theta+\sin\theta)}{\sin^2\theta\cos^2\theta}$
$\displaystyle=\frac{(1-2\sin\theta\cos\theta)(\cos\theta+\sin\theta\big)}{\sin^2\theta\cos^2\theta}$
$\displaystyle=\big(\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}\big)\big(\frac{1-2\sin\theta\cos\theta}{\sin\theta\cos\theta}\big)$
$\displaystyle=\big(\frac{1}{\sin\theta}+\frac{1}{\cos\theta}\big)\big(\frac{1}{\sin\theta\cos\theta}-2\big)$
$\displaystyle=(\csc\theta+\sec\theta)(\csc\theta\sec\theta-2)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/880899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Find the derivative of $y=x\sqrt{9-x}$ "Find the derivative of $y=x\sqrt{9-x}$."
So this is what I have and now I'm stuck.
\begin{align}
y' &= x \frac{d}{dx}\left[(9-x)^{1/2}\right] + (9-x)^{1/2} \frac{d}{dx}(x)\\
&= x \left[\frac{1}{2}(9-x)^{-1/2}\right] + (9-x)^{1/2} (1)
\end{align}
So I now that I need to multiply and simplify but I don't know where to start. Help!
This problem is actually part of a homework question where I have to analyze a graph and find critical points and min and max.
|
$$y=x\sqrt{9-x}$$
$$y'=x'\sqrt{9-x}+x(\sqrt{9-x})'=\sqrt{9-x}+x\frac{1}{2\sqrt{9-x}}(9-x)'=$$
$$=\sqrt{9-x}+x\frac{1}{2\sqrt{9-x}}(-1)=\sqrt{9-x}+\frac{-x}{2\sqrt{9-x}}=$$
$$=\frac{2(9-x)-x}{2\sqrt{9-x}}=\frac{18-3x}{2\sqrt{9-x}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/882867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
}
|
Show that the inequality holds $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$ We have to show that:
$\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$
To be honest I don't have idea how to deal with it. I only suspect there will be need to consider two cases for $n=2k$ and $n=2k+1$
|
For $n = 2k$
$$\frac{1}{n} + \ldots + \frac{1}{2n} = \frac{1}{2k} + \frac{1}{2k+1} + \ldots + \frac{1}{3k} \stackrel{\downarrow}{+} \frac{1}{3k+1} + \ldots + \frac{1}{4k} \geq\\
\geq \overbrace{\frac{1}{3k} + \frac{1}{3k} + \ldots + \frac{1}{3k}}^{k+1 \text{times}} + \overbrace{\frac{1}{4k} + \frac{1}{4k} + \ldots + \frac{1}{4k}}^{k \text{times}} = \frac{k+1}{3k} + \frac{k}{4k} \geq \frac{1}{3} + \frac{1}{4} = \frac{7}{12}
$$
For $n = 2k+1$
$$\frac{1}{n} + \ldots + \frac{1}{2n} = \frac{1}{2k+1} + \frac{1}{2k+2} + \ldots + \frac{1}{3k} \stackrel{\downarrow}{+} \frac{1}{3k+1} + \ldots + \frac{1}{4k+2} \geq\\
\geq \overbrace{\frac{1}{3k} + \frac{1}{3k} + \ldots + \frac{1}{3k}}^{k \text{times}} + \overbrace{\frac{1}{4k} + \frac{1}{4k} + \ldots + \frac{1}{4k}}^{k \text{times}} + \frac{1}{4k+1} + \frac{1}{4k+2} \geq \\ \geq \frac{k}{3k} + \frac{k}{4k} + \geq \frac{1}{3} + \frac{1}{4} = \frac{7}{12}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/883670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
}
|
show that there are $1\leq j_1< j_2< j_3\leq 13$ such that $\left | A_{j1}\cap A_{j2}\cap A_{j3} \right |\geq 3$. Given $A_i,A_2,...,A_{13}$ $\subset [10]$ when $\left | A_i \right |=5$ for all $i$.
Need to show that there are $1\leq j_1< j_2< j_3\leq 13$ such that $\left | A_{j_1}\cap A_{j_2}\cap A_{j_3} \right |\geq$ $3$.
I tried to put the numbers but didn't get how to prove.
Thanks.
|
Your claim is false as currently stated (though I think
it becomes true if $13$ is replaced by $14$). Here is a
counterexample :
$$
\begin{array}{lcllcllcl}
A_1 &=& \lbrace 1,2,3,4,5 \rbrace, &
A_2 &=& \lbrace 1,2,3,4,6 \rbrace, &
A_3 &=& \lbrace 1,2,5,6,7 \rbrace, \\
A_4 &=& \lbrace 1,2,7,8,9 \rbrace, &
A_5 &=& \lbrace 1,3,5,6,7 \rbrace, &
A_6 &=& \lbrace 1,3,7,8,10 \rbrace, \\
A_7 &=& \lbrace 1,4,5,9,10 \rbrace, &
A_8 &=& \lbrace 2,3,5,6,8 \rbrace, &
A_9 &=& \lbrace 2,3,7,8,10 \rbrace, \\
A_{10} &=& \lbrace 2,4,5,7,9 \rbrace, &
A_{11} &=& \lbrace 2,4,6,7,10 \rbrace, &
A_{12} &=& \lbrace 3,4,6,7,9 \rbrace, \\
A_{13} &=& \lbrace 4,5,6,8,10 \rbrace. & & & & & \\
\end{array}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/885482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find side BC of a triangle given AB, AC, and a relation between $\angle A$ and $\angle B$ A question from my class:
In triangle $ABC$, $3\angle A+2\angle B=180$ and $AB=10, AC=4$. So question is, what all can we comment on side $BC$. Can we find its exact length?
I have a crude solution involving trigonometry and a equation, but it's too large. So, can anybody help me?
|
Let $\frac{A}{2} = \beta$
$10(4\cos^3\beta - 3\cos \beta)= 4\cos \beta$
Since $\cos\beta$ is not equal to zero (if it were equal to zero, then A would be $180^\circ$ and the triangle would not exist),
$40\cos^2\beta=34$
$\cos^2\beta=\frac{17}{20}$, $\sin^2\beta=\frac{3}{20}$
$\cos A= \cos 2 \beta = \cos^2\beta - \sin^2\beta = 0.7$
$\angle A = 45.6^\circ$
$\angle B = 21.6^\circ$
It is not necessary to know the value of $\angle B$ to determine the value of $BC$, but it helps to verify that $3\angle A + 2\angle B = 180^\circ$
$$\frac{10}{\cos \left(\frac{45.6^\circ}{2}\right)} = \frac{BC}{\sin 45.6^\circ}$$
$BC =7.75$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/885687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How do I go from this $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$? So I am doing $\int\frac{x^2-3}{x^2+1}dx$ and on wolfram alpha it says the first step is to do "long division" and goes from $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$. That made the integral much easier, so how would I go about doing that in a clear manner? Thanks in advance for the help!
|
$\left( \frac{x^2 -3}{x^2 +1}\right) \Rightarrow \left(\frac {x^2}{x^2+1} - \frac {3}{x^2+1} \right) \Rightarrow \left(\frac{x^2+1}{x^2+1} - \frac {1}{x^2+1} -\frac{3}{x^2+1} \right)\Rightarrow \left(1 -\frac {4}{x^2+1}\right) $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/885999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
}
|
$\sin x+\sqrt{5} \cos x$ in a form $c\cdot \sin (x+d)$ How can I rewrite $\sin x + \sqrt{5}\cdot \cos x$ in a form $c \cdot \sin (x+d)$??? How can I find the values for $c$ and $d$? I have no idea how to solve that algebraically.
Is there also a possibility to rewrite it in termes of $\cos$ instead of $\sin$? Or maybe even with $\tan$?
|
In general, if we have: $a\sin x + b\cos x$, you can write that as: $$\sqrt{a^2 + b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin x + \frac{b}{\sqrt{a^2 + b^2}}\cos x \right)$$
and call $y = \arccos\left(\frac{a}{\sqrt{a^2 + b^2}}\right)$, to finally use the formula for $\sin(x+y)$ on the other direction. Applying that to your specific problem, we have: $$\sin x + \sqrt{5}\cos x = \sqrt{6}\left(\frac{1}{\sqrt{6}}\sin x + \frac{\sqrt{5}}{\sqrt{6}}\cos x\right) = \sqrt{6}\sin(x + \theta)$$
where $\theta = \arccos(1/\sqrt{6})$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/886494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $ \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $ Can someone please help me with this question?
$ \large \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $
My steps so far:
$ \large \frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{x-2}.4^{x}}{2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{2x-2}}{2^{x-2}} = 2^{3x-2}$
I get stuck here but I am assuming I need to get that 4 to a 2 somehow so I can combine them... so:
$ \large \frac {(2^2)^{2x-2}}{2^{x-2}} = 2^{3x-2}$
I feel like I am not on the right track here as I have no idea where to go now. Could anyone help please? Thank you!
|
$\frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2}$
$\frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2}$
$\frac {4^{2x-2}}{2^{x-2}} = 2^{3x-2}$
$\frac{2^{2x-2}}{2^{x-2}} 2^{2x-2} = 2^{3x-2}$
Then use the fact that : $\frac{a^{b}}{a^{c}} = a^{b-c}$ with $a\ne{0}$
So : $2^{2x-2-(x-2)} 2^{2x-2} = 2^{3x-2}$
$\Rightarrow$ $2^{x}2^{2x-2} = 2^{3x-2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/887631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form EDIT: Please see EDIT(2) below, thanks very much.
I want to prove by infinite descent that the positive divisors of integers of the form $a^2+3b^2$ have the same form. For example, $1^2+3\cdot 4^2=49=7^2$, and indeed $7,49$ can both be written in the same form: $49$ is already shown, and $7$ is $2^2+3\cdot 1^2$.
I'm looking for a general proof of this using infinite descent. I tried to let $a^2+3b^2=xy$, where $x,y$ are positive integer factors of the LHS, but couldn't continue from here I'm afraid. However, I have managed to prove that integers of the form $a^2+3b^2$ are closed under multiplication, i.e. the product of two of them is of the same form. I was hoping this would help.
I also had another question if you have time: Is the representation in the form $a^2+3b^2$ unique? For example, is it possible for an integer, say $N$, to be $N=a^2+3b^2=x^2+3y^2$ with $a\ne x$ and $b\ne y$?
EDIT: Sorry, here is the "closed under multiplication" proof I found:
Using Diophantus' Identity we have $(a^2+3b^2)(x^2+3y^2)=(a^2+(\sqrt{3}b)^2)(x^2+(\sqrt{3}y)^2=(ax+\sqrt{3}\cdot \sqrt{3} y)^2)+(a\cdot \sqrt{3} y - \sqrt{3} b\cdot x)^2=(ax+3by)^2+3(ay-bx)^2$ as required.
EDIT (2): I am very sorry. The question should be:
If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form.
|
EDIT: added second answer proving the main question.
Since we've proven that the family of numbers in the form of $a^2+3b^2$ is closed unfer product, we just jave to prove that any prime divisor $d$ of $a^2 + 3b^2$ can be represented as $r^2 + 3s^2$ for some integers $r,s$.
Let $d\ \vert \ (a^2 + 3b^2)$ then, $dk = a^2 +3b^2$ for a certain $k$.
If $d=1$ or $k = 1$ then the representation is obvious.
($d = 1^2+3·0^2$ for the first case and $d = a^2 + 3b^2$ for the second)
By integer division:
$$a = md + r \hspace{2cm} b=nd+s$$
If $r >d/2$ then $a = (m+1)d + (r-d) = m'd + r'$ where $m' = m+1$ and $r' = r-d$ (yes, $r'$ is negative, that's fine, we just want that either $r < d/2$ or $-r' < d/2$
(we use strict inequalities since $d$ is odd and therefore $d/2 \notin \mathbb{Z}$)
Similarly, if $s > d/2$ we construct $n'$ and $s'$ acordingly.
(use $m', r'$ and/or $n', s'$ from here on if you constructed them)
So $$a^2 +3b^2 = (md+r)^2 + (nd+s)^2 = $$
$$m^2d^2+2mdr+r^2+3n^2d^2+6nds+3s^2 =
d(m^2d+2mr+3n^2d+6ns)+r^2+3s^2$$
But $d\ \vert\ (a^2 + 3b^2)$, therefore $d\ \vert\ (r^2 + 3s^2)$
But $r < d/2$ and $s < d/2$ therefore $r^2 + 3s^2 < \displaystyle\frac{d^2}{4} + \frac{3d^2}{4} = d^2$, so $r^2 + 3s^2 = d$
(remember that $d$ is prime, so the only multiples of $d$ lower than $d^2$ are $d$ and $0$ and $r^2 + 3s^2\neq 0$ because otherwise $r = s = 0$ what would mean that $d$ divides both $a$ and $b$)
Thus any odd divisors, $d$ of $a^2 + 3b^2$ can be written as $r^2 + 3s^2$ for some integers $r$ and $s$.
NOTE: If you don't asume that $d$ is prime, then $d$ is a divisor of $r^2 + 3s^2 < a^2 + 3b^2$ and you can create the descent argument you wanted.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/888272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
Introductory Induction proof that $n(n^2 +5)$ is divisible by $6$ I am in currently in a discrete mathematics class, and I've done well on every problem I've encountered. Unfortunately, I find myself weak at some of the seemingly straight forward induction problems. As is the case of the following corollary. I will attempt to show as far as I've gotten in the problem and where I have hit a road block.
Theorem: $n(n^2 +5)$ is divisible by 6 for every integer $n\ge0$
Proof: Let the property $P(n)$ be the sentence "$n(n^2 +5)$ is divisible by 6."
For our base case we will show that $P(0)$. That is, $0(0^2 +5)$ is divisible by 6. But, $0(0^2 +5)=0$ so $P(0)$ says that $0$ is divisible by 6 which is clearly true because $0$ is divisible by all integers as was previously shown in our text.
Now, for our inductive step we must show that $P(k) \Rightarrow P(k+1)$. Now, by definition of divisibility, $P(k)$ says that $k(k^2 +5) =6r$ for some $r\in \mathbb Z$ (EDIT AFTER HINT). Also, $P(k+1)$ says that $(k+1)[(k+1)^2 +5]=6p$ for some $p\in \mathbb Z$. After expansion and substitution $(k+1)[(k+1)^2 +5 = k^3 + 3k^2 +8k +6 = (k^3 +5k)+3k^2 +3k +6 = 6r +3k^2 +3k + 6= 3(2r +k^2 + k +2)=3(2r+k(k+1)+2)$
However, note that $k(k+1)$ represents the pruduct of two consecutive integers and is thus even. That is, $k(k+1)=2q$ for some $q\in \mathbb Z$. Now, $2r+k(k+1)+2 = 2r+2q+2$, but this is simply the sum of three even numbers and is thus an even number itself. That is, $2r+2q+2 = 2p$ for some $p\in\mathbb Z$. Therefore, $(k+1)[(k+1)^2 +5] =6q$.
EDIT:PROBLEM SOLVED - This is where I get stuck I've tried to expand $(k+1)[(k+1)^2 +5]$ which gives me $k^3 + 3k^2 +8k +6$. I've tried to manipulate it so that $k^3 + 3k^2 +8k +6=(k^3 +5k)+3k^2 +3k +6 = 6r +3k^2 +3k + 6= 3(2r +k^2 + k +2)$. However, this only shows divisibility by 3.
Any help would be much appreciated. Thank you!
|
Method 1:
$$0\left(0^2+5\right)=0$$
$$1\left(1^2+5\right)=6$$
$$2\left(2^2+5\right)=18$$
$$3\left(3^2+5\right)=42$$
$$4\left(4^2+5\right)=84$$
$$5\left(5^2+5\right)=150$$
Since all of these are $0 \pmod 6$, every other number must be $0 \pmod 6$ as well.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/889232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Interesting combinatorial identities Let $n$ be a strictly positive integer and let $j=0,\dots,n-1$.
By using Mathematica I managed to guess the following identities:
\begin{eqnarray}
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m}{j} &=& \frac{1}{2} \binom{2 n}{2j + 1} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+1}{j+1} &=& \frac{1}{2} \binom{2 n+1}{2j + 2} + \binom{n-1}{j} \binom{n+1}{j+1} \frac{n(n-j)}{2 (n+1)(j+1)} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+2}{j+2} &=& \frac{1}{2} \binom{2 n+2}{2j + 3} + \binom{n-1}{j} \binom{n+2}{j+2} \frac{n(n-j)}{(n+2)(j+1)} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+3}{j+3} &=& \frac{1}{2} \binom{2 n+3}{2j + 4} + \binom{n-1}{j} \binom{n+3}{j+3} \frac{n(n-j)(11+5 j+(7+3 j) n)}{(n+2)(n+3)(4+ 2j)(j+1)} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+4}{j+4} &=& \frac{1}{2} \binom{2 n+4}{2j + 5} + \binom{n-1}{j} \binom{n+4}{j+4} \frac{2 n(n-j)(5+2 j+(3+ j) n)}{(n+3)(n+4)(2+ j)(j+1)} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+5}{j+5} &=& \frac{1}{2} \binom{2 n+5}{2j + 6} + \binom{n-1}{j} \binom{n+5}{j+5} \cdot \\
&&\frac{n(n-j)((137+93 j+16 j^2)+(264+161j+25 j^2)n/2+(62+35 j+ 5 j^2)n^2/2)}{(n+3)(n+4)(n+5)(j+1)(j+2)(j+3)}
\end{eqnarray}
The temptation is strong to write a general conjecture:
\begin{equation}
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+a}{j+a} = \frac{1}{2} \binom{2 n+a}{2j + 1+a} + \binom{n-1}{j} \binom{n+a}{j+a} \cdot\left(\cdots\right)
\end{equation}
Can anyone advise me how to tackle such a problem?. I think that in this particular case the method of generating functions does not lead to a closed form result.
|
We provide a closed form expression for our sum.
\begin{eqnarray}
&&\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+a}{j+a}=\\
&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\binom{a+2 n}{a+2 j+1}-\binom{a+n}{a+j+1} \binom{a+2 n}{j} \, \cdot _3F_2(-j,a+j+1,a+n+1;a+j+2,a-j+2 n+1;1) =\\
&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\binom{a+2 n}{a+2 j+1}-\binom{a+n}{a+j+1} \binom{-j+2 n-1}{j} \, \cdot _3F_2(-j,a+j+1,j-n+1;a+j+2,j-2 n+1;1)
\end{eqnarray}
We obtained this result by expanding the first binomial factor on the left hand side into a linear combination of binomial factors that are "adjacent" to the second binomial factor , then absorbing the resulting binomial factors into the second binomial factor and finally doing the sum over $m$ using the telescoping property of the binomial factor. Note that the hypergeometric function on the right hand side is a rational function in $n$.
Note: Interestingly enough Gosper's algorithm fails on this sum. If I have time I will explain why this is the case.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/890261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Number of irrational roots of the equation $3^x8^{\frac{x}{x+1}}=36$ Find the number of irrational solutions of the equation
$$3^x8^{\frac{x}{x+1}}=36.$$
|
Write exponents in LHS of ${3}^{x} {8}^{\frac{x}{x+1}} = 36$ in terms of a common base:
$${3}^{x} {8}^{\frac{x}{x+1}} = {e}^{\ln{3^x}} e^{\ln{{8}^{\frac{x}{x+1}}}} = {e}^{x \ln{3}} {e}^{\frac{{x} \ln{8}}{x+1}} = {e}^{x \ln{3}+\frac{x \ln{8}}{x+1}} = 36$$
Taking the natural log of both sides gives $x\ln{3} + \frac{x\ln{8}}{x+1} = \ln{36}$, and writing the LHS as a single fraction gives $$\frac{x\ln{8}+{x}^{2}\ln{3}+x\ln{3}}{x+1} = \ln{36}$$
By clearing fractions and bringing together like terms we get a quadratic in $x$:$$(\ln{3}){x}^{2}+(\ln{3}+\ln{8}-\ln{36})x-\ln{36} = 0$$
And after applying the quadratic formula we get the roots
$$x = \frac{-\ln{3}-\ln{8}+\ln{36}\pm\sqrt{{(\ln{3}+\ln{8}-\ln{36})}^{2}+4\ln{3}\ln{36}}}{2\ln{3}}$$
Which, after drastic simplification, reduce to the trivial root $2$ and the irrational root $-\frac{\ln{2}+\ln{3}}{\ln{3}}$.
Since there is one irrational root, your answer is $\boxed{1}$.
I realize there is probably a more efficient way to solve this problem, but I don't feel like thinking of one as of the moment.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/890737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
condition for tuple of non-zero integers $(a,b,c,d)$ such that $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$ As the title says, what would be the condition for tuple of non-zero integers $(a,b,c,d)$ such that $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$? Would there be infinitely many tuples that satisfy the condition above?
|
Let $i = \sqrt{-1}$. Then, $(a+bi)(c+di) = (ac-bd)+(ad+bc)i = 0+0i = 0$, so either $a+bi = 0$ or $c+di = 0$, i.e. either $a = b = 0$ or $c = d = 0$.
If $a = b = 0$, then we are left with $cd = c^2-d^2 = 0$, from which we get $c = d = 0$.
If $c = d = 0$, then we are left with $ab = a^2-b^2 = 0$, from which we get $a = b = 0$.
Therefore, the only solution to $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$ is the trivial solution $(a,b,c,d) = (0,0,0,0)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/890819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Generalization of a formula for 2x2-matrices It is well known that $$|det(v_1,...,v_n)|\le ||v_1||_2...||v_n||_2$$
with equality if and only if the vectors are pairwise orthogonal. For n = 2,
the following formula holds : $$det(\pmatrix {a&b\\c&d})^2=(a^2+b^2)(c^2+d^2)-(ac+bd)^2$$
How can this be generalized to matrices with higher size ?
I checked if $$(a^2+b^2+c^2)(d^2+e^2+f^2)(g^2+h^2+i^2)-det(\pmatrix {a&b&c\\d&e&f\\g&h&i})^2$$ has the factor $$(ad+be+cf)^2+(ag+bh+ci)^2+(dg+eh+fi)^2$$ because the
above expression is $0$ if and only if all the scalar products are $0$. But
according to wolfram this is not the case, but perhaps there are too many
variables to check it with the online-version of wolfram.
|
Let me explain this formula in geometric language. Assume $a=(a_1,a_2),b=(b_1,b_2)$ be two vectors, then the determinant of $\det (a,b)$ is the area of the parallelogram $S$ with two sides $a,b$. Note that
$$S^2:=\left(\det (a,b)\right)^2=(ab_\perp)^2=a^2\left(b^2-\frac{(a \cdot b)^2}{a^2}\right)=a^2b^2-(a\cdot b)^2$$
where $b_\perp$ is the perpendicular component of $b$ with respect to $a$. We uses the Pythagoras theorem in the third identity.
Similarly in 3 dimension, assume $a,b,c\in\mathbb R^3$, and $S$ the parallelogram (instead of its area) with two sides $a,b$, we have
$$V^2:=(\det(a,b,c))^2=(Sc_\perp)^2=S^2\left(c^2-\frac{(c\cdot S)^2}{S^2}\right)=S^2c^2-(c\cdot S)^2$$
where $c_\perp$ is similarly defined, $c\cdot S=(c\cdot a)b-(c\cdot b)a$ is the projection of $c$ in $S$ up to an area |S|. Replace $S$ by the above, we obtain
$$(\det(a,b,c))^2=a^2b^2c^2-(a\cdot b)^2c^2-(b\cdot c)^2a^2-(c\cdot a)^2b^2+2(a\cdot b)(b\cdot c)(c\cdot a)$$
which is a 3 dimensional generalization. So the essence of the formula is to compute a volume by computing perpendicular components literally.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/890884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Dealing with absolute values after trigonometric substitution in $\int \frac{\sqrt{1+x^2}}{x} \text{ d}x$. I was doing this integral and wondered if the signum function would be a viable method for approaching such an integral. I can't seem to find any other way to help integrate the $|\sec \theta|$ term in the numerator of the integrand.
$$ \begin{aligned} \int \dfrac{\sqrt{x^2 + 1}}{x} \text{ d}x \ \ & \overset{x=\tan \theta}= \int \dfrac{\sqrt{\sec^2 \theta} \sec^2 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \int \dfrac{|\sec \theta| \sec^2 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \int \dfrac{\text{sgn} (\sec \theta) \sec^3 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \text{sgn} (\sqrt{1+x^2}) \left( - \log \left| \dfrac{\sqrt{1+x^2} + 1}{x} \right| + \sqrt{1+x^2} \right) + \mathcal{C} \end{aligned} $$
It's clear that $\text{sgn} (\sqrt{1+x^2}) = 1$ since the sign of the argument of that function is always positive and the signum function extracts the sign. So I'd leave the integral as: $$ \begin{aligned} \int \dfrac{\sqrt{x^2 + 1}}{x} \text{ d}x = \log \left| x \right| - \log \left( \sqrt{1+x^2} + 1 \right) + \sqrt{1+x^2} + \mathcal{C} \end{aligned} $$
Would that be ok?
Also, apparently Wolfram has suggested that my final result should have $x$ as opposed to $|x|$ in the argument of the first logarithm. Why is that?
Any help would be greatly appreciated!
|
To avoid absolute values
\begin{align}
\int \frac{\sqrt{1+x^2}}{x} dx
= &\int \frac x{\sqrt{1+x^2}} + \frac 1{x\sqrt{1+x^2}} \ dx\\
=&\ \sqrt{1+x^2} - \text{coth}^{-1}\sqrt{1+x^2}
\end{align}
which is valid for all domain $x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/892496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
How to prove: $\left(\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt[4]{25}-\sqrt[4]{125}}}-1\right)^{4}=5$? Question:
show that: the beautiful ${\tt sqrt}$-identity:
$$
\left({2 \over \sqrt{\vphantom{\Large A}\, 4\ -\ 3\,\sqrt[4]{\,5\,}\
+\ 2\,\sqrt[4]{\,25\,}\ - \,\sqrt[4]{\,125\,}\,}\,}\ -\ 1\right)^{4}
=5
$$
Can you someone have methods to prove this by hand? (Maybe this problem have many methods?because this result is integer. It's a surprise to me.)
Thank you
Because I found this
$$4\ -\ 3\sqrt[4]{\,5\,}\ +\ 2\sqrt[4]{\,25\,}\ -\ \sqrt[4]{\,125\,}$$ is not square numbers.
|
Let $x = 4-3\sqrt[4]{5}+2\sqrt[4]{25}-\sqrt[4]{125}$. Then, we have:
(1) $x = 4 - 3 \cdot 5^{1/4} + 2 \cdot 5^{2/4} - 5^{3/4}$
(2) $5^{1/4}x = 4\cdot 5^{1/4} - 3 \cdot 5^{2/4} + 2 \cdot 5^{3/4} - 5$ (Multiply (1) by $5^{1/4}$)
(3) $(5^{1/4}+1)x = -1 + 5^{1/4} - 5^{2/4} +5^{3/4}$ (Add (1) and (2))
(4) $5^{1/4}(5^{1/4}+1)x = -5^{1/4} + 5^{2/4} - 5^{3/4} +5$ (Multiply (3) by $5^{1/4}$)
(5) $(5^{1/4}+1)^2x = 4$ (Add (3) and (4))
Therefore, $x = \dfrac{4}{(5^{1/4}+1)^2}$. Hence, $\left(\dfrac{2}{\sqrt{x}}-1\right)^4 = \left(\dfrac{2}{\tfrac{2}{5^{1/4}+1}}-1\right)^4 = (5^{1/4}+1-1)^4 = 5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/892739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 0
}
|
How to compute $\int \frac{x}{(x^2-4x+8)^2} \mathrm dx$? Can someone help me to compute:
$$\int \frac{x}{(x^2-4x+8)^2}\mathrm dx$$
And, in general, the type:
$$\int \frac{N(x)}{(x^2+px+q)^n}\mathrm dx$$
with the order of polynomial $N(x)<n$ and $n$ natural greater than 1?
|
$$\int \frac{x}{(x^2-4x+8)^2} dx = \int\frac{x - 2 + 2}{(x^2 - 4x + 8)^2}\,dx $$
$$= \frac 12\int \frac{2x - 4}{(x^2 - 4x + 8)^2}\,dx + \int \frac 2{((x-2)^2 + 2^2)^2}\,dx$$
For the first integral, use $u = x^2 - 4x + 8 \implies du = (2x-4)\,dx$.
For the second integral, put $2\tan \theta = (x-2)\implies 2\sec^2 \theta\,d\theta = dx$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/892986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Asymptotic behaviour of $\prod_{p \leq x} (1 + 4/(3p) + C p^{-3/2})$ I'm reading Montgomery and Vaughan and in it they state quite simply
\begin{equation}
\prod_{p \leq x} \left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}} \right) \ll (\log x)^{4/3}
\end{equation}
as $x \rightarrow \infty$ and where $C$ is some constant. It states that it's using Mertens' formula so there should probably be some relation of the form
\begin{equation}
\prod_{p \leq x} \frac{1 + \frac{4}{3p} + \frac{C}{p^{3/2}}}{\left(1 - \frac{1}{p}\right)^{4/3}} \ll 1
\end{equation}
but I can't quite get it.
|
Basically you should just mimic the proof of Mertens' formula. We have that
\[\sum_{p \leq x} \log\left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}}\right) = \frac{4}{3} \sum_{p \leq x} \frac{1}{p} + \sum_{p \leq x}\left(\log\left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}}\right) - \frac{4}{3p}\right),\]
and the first term is asymptotic to $\frac{4}{3} \log \log x$, while one can write the logarithm in terms of its power series in order to show that the second term is $o(1)$, from which the result follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/896373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
What is the non-trivial, general solution of these equal ratios? Provide non-trivial solution of the following:
$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}$$
$a=?, b=?, c=?$
The solution should be general.
|
Besides to the good answers posted for this question (this answer and this one), I think finding eigenvalues and eigenvectors of such system will help you to find the solution.
$$\begin{array}{l}\frac{a}{{b + c}} = \frac{b}{{c + a}} = \frac{c}{{a + b}} = t\\\left[ {\begin{array}{*{20}{c}}1&{ - t}&{ - t}\\{ - t}&1&{ - t}\\{ - t}&{ - t}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\\\left[ C \right]\left\{ X \right\} = \left\{ 0 \right\}\\\det \left[ {C - \lambda I} \right] = 0\\\det \left[ {\begin{array}{*{20}{c}}{1 - \lambda }&{ - t}&{ - t}\\{ - t}&{1 - \lambda }&{ - t}\\{ - t}&{ - t}&{1 - \lambda }\end{array}} \right] = 0\\\left( {1 - \lambda } \right)\left\{ {{{\left( {1 - \lambda } \right)}^2} - {{\left( { - t} \right)}^2}} \right\} = 0\\\lambda = 1,1 - t,1 + t\\\begin{array}{*{20}{c}}{\lambda = 1}\\{\lambda = 1 - t}\\{\lambda = 1 + t}\end{array}\left| {\begin{array}{*{20}{c}}{ + \left( {1 - \lambda } \right)a - \left( t \right)b - \left( t \right)c = 0}\\{ - \left( t \right)a + \left( {1 - \lambda } \right)b - \left( t \right)c = 0}\\{ - \left( t \right)a - \left( t \right)b + \left( {1 - \lambda } \right)c = 0}\end{array}} \right\rangle \left\{ {\left. {\begin{array}{*{20}{c}}{a = b = c = 0}\\{a = b = c = 0}\\{a + b + c = 0}\end{array}} \right\rangle } \right.\left. {\underline {\,
{a + b + c = 0} \,}}\! \right| \end{array}$$
As it seems the answer is compatible with the other answers to this question.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/897118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
}
|
Determinant of a matrix with $t$ in all off-diagonal entries. It seems from playing around with small values of $n$ that
$$
\det \left( \begin{array}{ccccc}
-1 & t & t & \dots & t\\
t & -1 & t & \dots & t\\
t & t & -1 & \dots & t\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t & t & t & \dots& -1
\end{array}\right) = (-1)^{n-1}(t+1)^{n-1}((n-1)t-1)
$$
where $n$ is the size of the matrix.
How would one approach deriving (or at least proving) this formally?
Motivation
This came up when someone asked what is the general solution to:
$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b},$$
and for non-trivial solutions, the matrix above (with $n=3$) must be singular. In this case either $t=-1\implies a+b+c=1$ or $t=\frac{1}{2}\implies a=b=c$.
So I wanted to ensure that these are also the only solutions for the case with more variables.
|
Using elementary operations instead of induction is key.
$$\begin{align}
&\begin{vmatrix}
-1 & t & t & \dots & t\\
t & -1 & t & \dots & t\\
t & t & -1 & \dots & t\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t & t & t & \dots& -1
\end{vmatrix}\\
&=
\begin{vmatrix}
-t-1 & 0 & 0 & \dots & t+1\\
0 & -t-1 & 0 & \dots & t+1\\
0 & 0 & -t-1 & \dots & t+1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t & t & t & \dots& -1
\end{vmatrix}\\
&=
\begin{vmatrix}
-t-1 & 0 & 0 & \dots & 0\\
0 & -t-1 & 0 & \dots & 0\\
0 & 0 & -t-1 & \dots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t & t & t & \dots& (n - 1)t -1
\end{vmatrix}\\
&=
(-1)^{n - 1}(t + 1)^{n - 1}((n - 1)t - 1)
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/897469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
}
|
Verifying an antiderivative found in any integral table If $a > 0$, and $0 < b < c$.
\begin{equation*}
\int \frac{1}{b + c\sin(ax)} \, {\mathit dx}
= \frac{-1}{a\sqrt{c^{2} - b^{2}}} \, \ln\left\vert\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)}\right\vert .
\end{equation*}
(This is the antiderivative given in any calculus book.) With calculations that I made, I showed that
\begin{align*}
\int \frac{1}{b + c\sin(ax)} \, {\mathit dx}
&= \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert
,
\end{align*}
and since the sum of an antiderivative of $1/[b + c\sin(ax)]$ and a constant is another antiderivate of $1/[b + c\sin(ax)]$, and since
\begin{equation*}
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b}
\qquad \text{and} \qquad
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b}
\end{equation*}
are reciprocals of each other,
\begin{align*}
&\int \frac{1}{b + c\sin(ax)} \, {\mathit dx} \\
&\qquad \qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\dfrac{c - \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert \\
&\qquad \qquad \qquad\qquad - \frac{1}{a\sqrt{c^{2} - b^{2}}} \ln\left( \frac{c + \sqrt{c^{2} - b^{2}}}{b} \right) \\
&\qquad \qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
\ \
\sin(ax) + \dfrac{c + \sqrt{c^{2} - b^{2}}}{b} + \dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \cos(ax)
\ \
}
{
\dfrac{c + \sqrt{c^{2} - b^{2}}}{b} \, \sin(ax) + 1 + \cos(ax)
}
\right\vert \\
&\qquad\qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
\right\vert .
\end{align*}
Furthermore, we have the following trigonometric identity:
\begin{align*}
&\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)} \\
&\qquad\qquad =\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
} .
\end{align*}
So, the antiderivative given in any calculus book is the same function as the second antiderivative that I obtained:
\begin{align*}
&\frac{-1}{a\sqrt{c^{2} - b^{2}}} \, \ln\left\vert\frac{c + b\sin(ax) + \sqrt{c^{2} - b^{2}}\cos(ax)}{b + c\sin(ax)}\right\vert \\
&\qquad = \frac{-1}{a\sqrt{c^{2} - b^{2}}}
\ln \left\vert
\frac{
c + b\sin(ax) + \sqrt{c^{2} - b^{2}} \, \cos(ax) + \sqrt{c^{2} - b^{2}} + c \cos(ax)
}
{
b + c\sin(ax) + \sqrt{c^{2} - b^{2}} \, \sin(ax) + b\cos(ax)
}
\right\vert .
\end{align*}
Here are my questions. Is it evident to anyone that the function on the right side of the trigonometric identity simplifies to the function on the left side? (It is surprising that the two sides are equal. The numerators and denominators on each side are "almost" the same: the numerator and denominator on the right side have two more terms than those on the left side.) If it is not evident, can someone give me calculations, starting from integration using the technique of trigonometric substitution, that show that the antiderivative of $1/[b + c\sin(ax)]$ is the function that is found in the integral tables of any calculus book - calculations that avoid all the algebraic manipulations?
|
For the simplicity let your calculation be $f(x)$ and let the antiderivative given in calculus book be $g(x)$. Assuming that your calculation is correct, then the proper way to express the indefinite integral using your calculation is
$$
\int \frac{1}{b + c\sin(ax)}\ dx=f(x)+C_1,
$$
where $C_1$ is a constant of integration. Similarly, the proper way to express the indefinite integral using table in calculus book is
$$
\int \frac{1}{b + c\sin(ax)}\ dx=g(x)+C_2,
$$
where $C_2$ is also a constant of integration. But you cannot deduce $C_1=C_2$, therefore you cannot equate $f(x)$ with $g(x)$ since $f(x)\neq g(x)$. The problem is you tried to set the constant equal to zero but it doesn't always make sense. Here is the classic example (you may also refer to here), $2\sin x\cos x$ can be integrated in at least three different ways:
\begin{align}
\int 2\sin x\cos x\,dx &= \sin^2(x) + C = -\cos^2x + 1 + C = -\frac12\cos2x + C\\[12pt]
\int 2\sin x\cos x\,dx &= -\cos^2(x) + C = \sin^2(x) - 1 + C = -\frac12\cos2x + C\\[12pt]
\int 2\sin x\cos x\,dx &= -\frac12\cos2x + C = \sin^2x + C = -\cos^2x + C
\end{align}
So setting $C$ to zero can still leave a constant. This means that, for a given function, there is no "simplest antiderivative" and of course $-\frac12\cos2x\neq-\cos^2x$. Note that, the constant of integration is sometimes omitted in lists of integrals for simplicity.
Here is an approach to obtain the given antiderivative in the calculus book. Start with setting $u=ax$, then
$$
\int\frac{1}{b+c\sin ax}\ dx=\frac1a\int\frac{1}{b+c\sin u}\ du.\tag1
$$
Using the Weierstrass substitution by setting $t=\tan\frac u2$, we have
$$
\sin u=\frac{2t}{1+t^2}\quad\text{and}\quad du=\frac{2}{1+t^2}\ dt.
$$
Therefore, $(1)$ becomes
\begin{align}
\int\frac{1}{b+c\sin ax}\ dx&=\frac2a\int\frac{1}{bt^2+2ct+b}\ dt\\
&=\frac2a{\Large\int}\frac{1}{b\left(t+\frac{c+\sqrt{c^2-b^2}}{b}\right)\left(t+\frac{c-\sqrt{c^2-b^2}}{b}\right)}\ dt\\
&=\frac2{ab}\cdot\frac{1}{2\frac{\sqrt{c^2-b^2}}{b}}{\Large\int}\left[\frac{1}{\left(t+\frac{c-\sqrt{c^2-b^2}}{b}\right)}-\frac{1}{\left(t+\frac{c+\sqrt{c^2-b^2}}{b}\right)}\right]\ dt\\
&=\frac1{a\sqrt{c^2-b^2}}\left[\ln{\left|t+\frac{c-\sqrt{c^2-b^2}}{b}\right|}-\ln{\left|t+\frac{c+\sqrt{c^2-b^2}}{b}\right|}\right]\\
&=\frac1{a\sqrt{c^2-b^2}}\ln{\left|\frac{bt+c-\sqrt{c^2-b^2}}{bt+c+\sqrt{c^2-b^2}}\right|}\\
&=\frac1{a\sqrt{c^2-b^2}}\ln{\left|\frac{b\tan\frac{ax}{2}+c-\sqrt{c^2-b^2}}{b\tan\frac{ax}{2}+c+\sqrt{c^2-b^2}}\right|}.
\end{align}
Notice that
$$
\tan\frac{ax}{2}=\frac{\sin ax}{1+\cos ax}.
$$
It can easily be proven by using double angle formula.
$$\color{red}{\text{to be continued...}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/897548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Im trying to find what I got wrong in simplifying The question states " Simplify: $$\frac{6\left ( 27^{2n+3} \right )}{9^{3n+6}}$$
What I did was factoring $9$ and and $27$ to make it $3$. In the end I got a $\frac{6}{27}$ answer, simplified: $\frac{2}{9}$. but my problem is that in the book, the answer is stated as $\frac{27}{2}$. What could be the solution?
|
$$\frac{6 \cdot27^{2n+3}}{9^{3n+6}}=\frac{2\cdot3\cdot[(3^3)]^{2n+3}}{(3^2)^{3n+6}}$$
$$=\frac{2\cdot3\cdot3^{6n+9}}{3^{6n+12}}=\frac{2\cdot3}{3^3}=\frac2{3^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/898915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Factoring a quadratic with number in front of $x^2$ I have not yet understood how to factor a quadratic that contains a number in front of $x^2$, without using the quadratic equation. I am used to just brute forcing numbers such that AB and A + B are solutions to it.
Example:
$7x^2 = 25x + 12$
....
$7x^2 - 25x - 12 = 0$
How steps do I need to take to proceed from here ?
Cheers.
|
Method I: Completing the square:
$$7x^2 - 25x - 12
=7\left(x^2-\frac{25}7x-\frac{12}7\right)\\
=7\left(\left(x-\frac{25}{14}\right)^2-\left(\frac{12}7+\frac{25^2}{14^2}\right)\right)\\
=7\left(\left(x-\frac{25}{14}\right)^2-\left(\frac{961}{196}\right)\right)\\
=7\left(\left(x-\frac{25}{14}\right)^2-\left(\frac{31^2}{14^2}\right)\right)\\
=7\left(x-\frac{25}{14}-\frac{31}{14}\right)\left(x-\frac{25}{14}+\frac{31}{14}\right)\\
=7(x-4)\left(x+\frac{3}{7}\right)\\
=(x-4)(7x+3)\\
$$
Method II: Quadratic formula:
Using quadratic formula, roots are:
$$\frac{25\pm\sqrt{625+336}}{14}=\frac{25\pm31}{14}=4,\frac{-3}7$$
$$7x^2 - 25x - 12 = 0\equiv(x-4)\left(x+\frac37\right)=0$$
Now just multiply coefficient of $x^2$ as:
$$7(x-4)\left(x+\frac37\right)=0\equiv(x-4)(x+3)=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/899139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Find out the no of digits in product between some prime. How many digits are there in?
$2^{17}*3^{2}*5^{14}*7$. help me.
|
As André Nicolas has pointed out in the comments,
$$2^{17}\times 5^{14} = 2^3\times 2^{14}\times 5^{14} = 2^3\times 10^{14} = 8\times 10^{14}.$$
Therefore,
$$2^{17}\times 3^2\times 5^{14}\times 7 = 3^2\times 7\times 8\times 10^{14} = 504\times 10^{14} = 5.04\times 10^{16}.$$
So $2^{17}\times 3^2\times 5^{14}\times 7$ has $17$ digits.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/900303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Sum of the series $\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$ How do I find the sum of the following infinite series:
$$\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$$
I think the sum can be converted to definite integral and calculated but I don't know how to proceed from there.
|
The $n$-th term in the series is $\dfrac{2 \cdot 6 \cdots (2n-2)}{5 \cdot 10 \cdots 5n} = \dfrac{1}{n+1}\dbinom{2n}{n}\dfrac{1}{5^{n+1}} = \dfrac{C_n}{5^{n+1}}$, where $C_n = \dfrac{1}{n+1}\dbinom{2n}{n}$ is the $n$-th Catalan number. Thus, the sum is $\displaystyle\sum_{n = 1}^{\infty}\dfrac{C_n}{5^{n+1}}$.
The generating function of the Catalan numbers is $c(x) = \displaystyle\sum_{n = 0}^{\infty}C_nx^n = \dfrac{1-\sqrt{1-4x}}{2x}$.
Multiply by $x$ and subtract $x$ to get $xc(x)-x = \displaystyle\sum_{n = 1}^{\infty}C_nx^{n+1} = \dfrac{1-\sqrt{1-4x}}{2} - x$
Then, plug in $x = \dfrac{1}{5}$ to get $\displaystyle\sum_{n = 1}^{\infty}\dfrac{C_n}{5^{n+1}} = \dfrac{3-\sqrt{5}}{10}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/900687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Prove $ x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ So what I am trying to prove is for any real number x and natural number n, prove $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$
I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis. My base case is when $n=2$ we have on the left side of the equation $x^2-1$ and on the right side: $(x-1)(x+1)$ which when distributed is $x^2-1$. So my base case holds.
Now I assume that $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ for some $n$. However, this is where I am stuck. Am I trying to show $x^{n+1}-1=(x-1)(x^n + x^{n-1}+x^{n-2}+...+x+1)$? I am still a novice when it comes to these induction proofs. Thanks
|
To conclude your induction proof, just multiply x both sides :
$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1) $
multiply $x$ both sides :
$\begin{align} \\ x^{n+1}-x &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\
x^{n+1}-1 -(x-1) &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\ x^{n+1}-1 &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x)+(x-1) \\ \end{align}$
factor $(x-1)$ and you're done !
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/900869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
}
|
Problems with trigonometry getting the power of this complex expression I'm here because I can't finish this problem, that comes from a Russian book:
Calculate $z^{40}$ where $z = \dfrac{1+i\sqrt{3}}{1-i}$
Here $i=\sqrt{-1}$. All I know right now is I need to use the Moivre's formula
$$\rho^n \left( \cos \varphi + i \sin \varphi \right)^n = \rho^n \left[ \cos (n\varphi) + i \sin (n\varphi) \right]$$
to get the answer of this.
First of all, I simplified $z$ using Algebra, and I got this:
$$z = \dfrac{1-\sqrt{3}}{2} + i \left[ \dfrac{1+\sqrt{3}}{2} \right]$$
Then, with that expression I got the module $|z| = \sqrt{x^2 + y^2}$, and its main argument $\text{arg}(z) = \tan^{-1} \left( \dfrac{y}{x} \right)$.
I didn't have problems with $|z| = \sqrt{2}$, but the trouble begins when I try to get $\text{arg}(z)$. Here is what I've done so far:
$$\alpha = \text{arg}(z) = \tan^{-1} \left[ \dfrac{1+\sqrt{3}}{1-\sqrt{3}} \right]$$
I thought there's little to do with that inverse tangent. So, I tried to use it as is, to get the power using the Moivre's formula.
$$z^{40} = 2^{20} \left[ \cos{40 \alpha} + i \sin{40 \alpha} \right]$$
As you can see, the problem is to reduce a expression like: $\cos{ \left[ 40 \tan^{-1} \left( \dfrac{1+\sqrt{3}}{1-\sqrt{3}} \right) \right] }$.
And the book says the answer is just $-2^{19} \left( 1+i\sqrt{3} \right)$.
I don't know if I'm wrong with the steps I followed, or if I can reduce those kind of expressions. I'll appreciate any help from you people :)
Thanks in advance!
|
Oh my. We have:
$$1+i\sqrt{3} = 2\exp\left(i\arctan\sqrt{3}\right)=2\exp\frac{\pi i}{3}$$
$$\frac{1}{1-i} = \frac{1}{2}(1+i) = \frac{1}{\sqrt{2}}\exp\frac{\pi i}{4},$$
hence:
$$z=\frac{1+i\sqrt 3}{1-i} = \sqrt{2}\exp\frac{7\pi i}{12},$$
so:
$$ z^{40} = 2^{20}\exp\frac{70\pi i}{3}=2^{20}\exp\frac{4\pi i}{3}=-2^{20}\exp\frac{\pi i}{3}=-2^{19}(1+i\sqrt{3}).$$
As an alternative way, if you set $a=1+i\sqrt{3}$ and $b=\frac{1}{1-i}$ you have:
$$ a^3 = -2^3,\qquad b^4 = -2^{-2}, $$
hence:
$$ z^{40} = a^{40} b^{40} = a(a^3)^{13} (b^4)^{10} = a\cdot(-2^{39})\cdot(2^{-20}) = -2^{19}(1+i\sqrt{3}).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/901475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Families of Idempotent $3\times 3$ Matrices I did the following analysis for $2\times2$ real idempotent (i.e. $A^2=A$) matrices:
$$
\begin{bmatrix}a&b\\c&d\end{bmatrix}^2=\begin{bmatrix}a^2+bc&(a+d)b\\(a+d)c&bc+d^2\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}
$$
So in particular we have $(a+d)c=c$ and $(a+d)b=b$ so if either $b$ or $c$ is nonzero we have $a+d=1$. We also see that $a$ and $d$ both satisfy the equation $x^2+bc=x\iff x^2-x+bc=0$ which is a quadratic equation having solutions
$$
x=\frac{1\pm\sqrt{1-4bc}}{2}=0.5\pm\sqrt{0.25-bc}
$$
But this is only possible if $bc\leq 0.25$ for otherwise the above expression is not real. This gives us the following cases:
CASE 1: If $b,c=0$ we have $x\in\{0,1\}$ and since $a+d=1$ is unnecessary we have four possibilities: $(a,d)\in\{(0,0),(1,0),(0,1),(1,1)\}$.
CASE 2: If $bc=0.25$ we have $x=0.5$ so $a=d=0.5$.
CASE 3: If $bc<0.25$ yet $(b,c)\neq(0,0)$ we have $x\in L=\{0.5-\sqrt{0.25-bc},0.5+\sqrt{0.25-bc}\}$ and to have $a+d=1$ we must have $\{a,d\}=L$ so that if $a$ is one solution, then $d$ is forced to be the other solution. Or the other way around.
The cases can be illustrated via the following diagram graphing the hyperbola $xy=0.25$ corresponding to CASE 2, the area $xy<0.25$ corresponding to CASE 3, and the point $(0.0)$ corresponding to CASE 1:
The blue bands show the graphs of $xy=k$ for $k=0.05$ to $0.20$ and the cyan bands show $xy=k$ for $k=-0.05,-0.10,...$ For instance one could choose $(b,c)=(3.75,-1)$ so that $\sqrt{0.25-bc}=2$ thus rendering $x=0.5\pm 2=-1.5$ and $2.5$ and form the matrix
$$
A=\begin{bmatrix}-1.5&3.75\\-1&2.5\end{bmatrix}
$$
which will then be idempotent, as an example of CASE 3.
QUESTIONs:
*
*Can similar descriptions be derived for $3\times 3$ matrices?
*Is this a well known description of idempotent $2\times 2$ matrices?
|
This is a partial answer, but something that you may find useful.
Let $A$ be a $3\times3$ matrix and denote $t=\text{tr}(A)$, $d=\det(A)$ and $a=a_1+a_2+a_3$, where $a_k=\det(A_{\hat k\hat k})$ is the subdeterminant corresponding to the $k$th diagonal element of $A$.
The characteristic polynomial of $A$ is
$$
p(x)
=
\det(A-xI)
=
-x^3+tx^2-ax+d.
$$
By the Cayley-Hamilton theorem $p(A)=0$, when the polynomial is naturally interpreted for matrices.
That is,
$$
-A^3+tA^2-aA+dI=0.
$$
We can factor this polynomial:
$$
-(A^2-A)(A-(t-1))+(t-a-1)A+dI=0.
$$
Suppose that $A$ is idempotent: $A^2=A$.
The above equation gives now $(t-a-1)A+dI=0$.
If $t-a-1\neq0$, this means that $A$ is a multiple of the identity.
It is easy to see that the only such solutions are $A=I$ and $A=0$ (and these are of course idempotent matrices).
All other idempotent matrices must therefore satisfy $t-a-1=0$ and $d=0$.
That is, $\det(A)=0$ and
$$
\text{tr}(A)
=
1+\det(A_{\hat1\hat1})+\det(A_{\hat2\hat2})+\det(A_{\hat3\hat3}).
$$
The second condition can be alternatively expressed as $\text{tr}(A)=1+\text{tr}(\text{cof}(A))$, where $\text{cof}(A)$ is the cofactor matrix of $A$.
Trying to find explicit solutions to these equations seems messy.
Let me stress that these conditions are necessary for all idempotent matrices other than $0$ and $I$, but not sufficient.
The characteristic polynomial does not retain all information about the original matrix.
The following matrix satisfies both of my conditions but is not idempotent:
$$
\begin{pmatrix}
1&0&0\\
0&0&1\\
0&0&0
\end{pmatrix}.
$$
As a remark, the same approach works better for a $2\times2$ matrix $A$.
Now $p(x)=x^2-tx+d$, so $A^2-tA+dI=0$.
Using $A^2=A$ gives $(1-t)A+dI=0$.
Again, either $A$ is a multiple of the identity (these two cases we know), or $1-t=0$ and $d=0$.
The equations are much easier so solve now.
Suppose
$
A
=
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}.
$
We have $ad-bc=0$ and $a+d=1$.
If $c=0$, we get $a\in\{0,1\}$ and $d=1-a$, and $b$ can be anything.
If $c\neq0$, we get $a=1-d$ and $b=(d-d^2)/c$, and $d$ can be anything.
It is easy to check which of these solutions are actually idempotent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/902401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
}
|
Improper integral containing $\sqrt{\cos x-\frac1{\sqrt 2}}$ in the denominator How do I find the value of this integral--
$$I=\displaystyle\int_{0}^{\pi/4} \frac{\sec^2 x \ dx}{\sqrt {\cos x-\dfrac{1}{\sqrt 2}}}$$
I came across this integral too in physics.
|
With $t=\cos(x)$, we see it is an elliptic integral,
$$
I = \int_{1/\sqrt{2}}^1 \frac{dt}{t^2\sqrt{(1-t^2)(t-1/\sqrt{2})}}
$$
added
Maple gets: if $0<a<1$, then
$$
\int _{a}^{1}\!{\frac {dt}{{t}^{2}\sqrt { \left( 1-t^2 \right)
\left( t-a \right) }}}{}
=-{\frac {\sqrt {2}{\bf K} \left( 1
/2\,\sqrt {-2\,a+2} \right) }{a}}
+{\frac {\sqrt {2} \left( a+1
\right) {\bf E} \left( 1/2\,\sqrt {-2\,a+2} \right) }{a
\left( 1+a \right) }}
+\frac{1}{\sqrt {2}} \left( a+1 \right) {\bf \Pi} \left({\frac {a-1}{2a}},
\frac{\sqrt {-2\,a+2}}{2} \right)
{a}^{-2}
$$
and in particular with $a=1/\sqrt{2}$,
$$
I = 2\,{\bf E} \left( 1/2\,\sqrt {2-\sqrt {2}} \right) -2\,{\bf K} \left( 1/2\,\sqrt {2-\sqrt {2}} \right) + \left( 1+\sqrt {2
} \right) {\bf \Pi} \left( 1/4\, \left( -2+\sqrt {2} \right)
\sqrt {2},1/2\,\sqrt {2-\sqrt {2}} \right) \approx 3.338954
$$
where Maple's notation uses the modulus $k$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/904320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
}
|
Why is $ A_1 x + ... + A_n x^n $ a solution of $ \sum_0^{n} (-1)^n \frac{x^n}{n!} \frac{d^n y}{d x^n} = 0 $? I was playing(/fiddling) around with some maths and I saw this pattern(
where $ A_n $ is a constant.):
$ A_1 x $ is a soultion of:
$$ \frac{y}{x} - \frac{dy}{dx} = 0 $$
$ A_1 x + A_2 x^2 $ is a solution of:
$$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} =0 $$
$ A_1 x + A_2 x^2 + A_3 x^3 $ is a solution of:
$$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} - \frac{x^2}{3!} \frac{d^3y}{dx^3} =0 $$
It continues so on. Can someone prove the solution of
$ A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n $ is:
$$ \sum_{k=0}^{n} (-1)^k \frac{x^{k-1}}{k!} \frac{d^k y}{d x^k} = 0 $$
|
Let $z=\frac yx$. Then we have
$$\dfrac{d^nz}{dx^n}=\dfrac{d^n(yx^{-1})}{dx^n}=\sum_{k=0}^n\binom nk\dfrac{d^ky}{dx^k}\dfrac{d^{n-k}(x^{-1})}{dx^{n-k}}=$$
$$\sum_{k=0}^n\binom nk\dfrac{d^ky}{dx^k}(-1)^{n-k}(n-k)!x^{-1-n+k}=$$
$$\sum_{k=0}^n(-1)^{n-k}\dfrac{n!}{k!}x^{-1-n+k}\dfrac{d^ky}{dx^k}$$
So we have
$$(\pm1)n!x^{n+1}\dfrac{d^nz}{dx^n}=\sum_{k=0}^n(-1)^k\dfrac{x^k}{k!}\dfrac{d^ky}{dx^k}$$
Setting this to $0$, since $x=0$ won't work, we have $\dfrac{d^nz}{dx^n}=0$. Integrate $n$ times, then $y=zx$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/904424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
Circles - point of intersection of tangents Question:
Let $A$ be the center of the cricle $x^2 + y^2 - 2x-4y-20=0$. Suppose that the tangents at the points $B(1,7)$ and $D(4,-2)$ on the cricle meet at point $C$. Find the area of the quadrilateral $ABCD$.
What I have done:
Well I have found the center of the circle and its radius. Upon drawing the diagram, it is obvious that the quadrilateral formed can be split into two right angled triangles. The only thing I need is the distance between the point of contact and the point of intersection of the tangents. How would I obtain this?
|
The center of the circle is quickly found by derivatives of the equation
$$ \left. \begin{aligned}
\frac{{\rm d}}{{\rm d}x} (x^2+y^2-2x -4 y-20) &= 0 \\
\frac{{\rm d}}{{\rm d}y} (x^2+y^2-2x -4 y-20) &= 0
\end{aligned} \right\}
\begin{aligned} x &= 1 \\ y &= 2 \end{aligned} $$
I like to use homogeneous coordinates, and so $A=(1,2,1)$ is the center of the circle. We also know $B=(1,7,1)$ and $D=(4,-2,1)$.
The tangent lines through B and D are found by
$$ \begin{aligned}
L_B &= U\, B = (0,5,-35) \} 0x+5y-35 =0 \\
L_D & = U\, D =(3,-4,-20) \} 3x-4y-20=0 \\
U & = \begin{bmatrix} 1 & 0 & -x_A \\ 0 & 1 & -y_A \\ -x_A & -y_A & -x_A^2-y_A^2-r^2 \end{bmatrix}
=\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ -1 & -2 & -20 \end{bmatrix}
\end{aligned} $$
where $U$ is a 3×3 matrix representing the circle such that the equation of the circle is found by the quadratic form
$$ \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ -1 & -2 & -20 \end{bmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} =0 \} x^2+y^2 -2 x - 4 y -20 =0$$
Point $C$ is found where $L_A$ and $L_B$ meet
$$ \left. \begin{aligned}
C & = L_A \times L_B \\
& = (0,5,-35) \times (3,-4,-20) = (-240,-105,-15)
\end{aligned} \right\} \begin{aligned} x_C & = 16 \\ y_C & = 7 \end{aligned} $$
Now the area of the triangle $$\triangle_{BAD} = \frac{1}{2} \frac{ A \cdot (B \times D) }{|A| |B| |D|} = \frac{15}{2}$$
where $\cdot$ is the inner product, $\times$ is the cross product and $|(a,b,c)|=c$ correspond to the scalar component of the homogeneous coordinates.
Also the area of the triangle $$\triangle_{DCB} = \frac{1}{2} \frac{ C \cdot (D \times B) }{|C| |D| |B|} = \frac{135}{2}$$
Combined you have $\boxed{\triangle_{BAD}+\triangle_{DCB} = 75}$
The beauty of homogeneous coordinates is that a) no trigonometry is needed and b) if the coordinates are rational then the result is a rational number. So integer algebra is sufficient to produce results.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/905086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.
Thank you every one.
But consider this problem.
Find the polynomial with degree 3 such that $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots
Note that $\dfrac{\pi}{12}$, $\dfrac{9\pi}{12}$, $\dfrac{17\pi}{12}$ are solution of equation $\cos3\theta=\dfrac{1}{\sqrt{2}}$ and $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are distinct number.
We have $\cos3\theta=4\cos^3\theta-3\cos\theta$. Let $x=\cos\theta$, therefore
$\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots of $4x^3-3x=\dfrac{1}{\sqrt{2}}.$
I want method similar to this to find $P(x)$.
Thank you.
|
May be not the answer you wanted given that it is of degree 8. But it has integer coefficients, so may be of interest.
If $R_n(x)=T_n(\sqrt{1-x^2})$, where $T_n$ is the Chebyshev polynomial of degree $n$, then
$$
T_n(\sin t)=\cos nt
$$
for all $t$. Because $\cos \alpha=0$, iff $\alpha$ is an odd multiple of $\pi/2$, the 12 zeros of
$$
R_{12}(x)=1 - 72 x^2 + 840 x^4 - 3584 x^6 + 6912 x^8 - 6144 x^{10} + 2048 x^{12}
$$
are the numbers $\sin((2j+1)\pi/24), j=0,1,2,\ldots,23$. Each zero occurs here twice, because $\sin x=\sin (\pi-x)$. We can throw away the zeros that
correspond to $3\mid 2j+1$, for those are also zeros of
$$
R_4(x)=1-8x^2+8x^4.
$$
This leaves us with
$$
P(x)=\frac{R_{12}(x)}{R_4(x)}=1-64x^2+320x^4-512x^6+256x^8.
$$
In addition to the prescribed zeros $P$ vanishes at the negatives of those sines. Observe that $\sin(5\pi/24)=\sin(19\pi/24)$ et cetera.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/905303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
}
|
If $\sin^2 \theta + 2\cos \theta – 2 = 0$, then find the value of $\cos^3 \theta + \sec^3 \theta$ If $\sin^2 \theta + 2\cos \theta – 2 = 0$, then find the value of $\cos^3 \theta + \sec^3 \theta$.
|
HINT:
We have $$1-\cos^2\theta+2\cos\theta-2=0\iff(\cos\theta-1)^2=0$$
Alternatively , $$1-\cos^2\theta+2\cos\theta-2=0\iff\cos^2\theta+1=2\cos\theta$$
Dividing by $\displaystyle\cos\theta$ (which is clearly $\ne0$) $$\cos\theta+\sec\theta=2$$
Can you calculate $$\cos^3\theta+\sec^3\theta=\cos^3\theta+\frac1{\cos^3\theta}$$ from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/905980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
The number $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is an integer Prove by induction that this number is an integer:
$$u_n=(3+\sqrt{5})^n+(3-\sqrt{5})^n$$
Progress
I assumed that it holds for $n$ and I tried to do it for $n+1$ but the algebra gets quite messy and I'm unable to prove that the following term is an integer: $\sqrt{5}((3+\sqrt{5})^n-(3-\sqrt{5})^n)$
|
Outline: For the (strong) induction step, we can use the fact that
$$(3+\sqrt{5})^{n+1}+(3-\sqrt{5})^{n+1}=\left[(3+\sqrt{5})^{n}+(3-\sqrt{5})^{n}\right]\left[(3+\sqrt{5})+(3-\sqrt{5})\right]-(3+\sqrt{5})(3-\sqrt{5})\left[(3+\sqrt{5})^{n-1}+(3-\sqrt{5})^{n-1}\right].$$
Note that $(3+\sqrt{5})+(3-\sqrt{5})$ and $(3+\sqrt{5})(3-\sqrt{5})$ are integers.
Remark: There are better "non-induction" ways. For example, imagine expanding each of $(3+\sqrt{5})^n$ and $(3-\sqrt{5})^n$, using the Binomial Theorem. Now add. The terms in odd powers of $\sqrt{5}$ cancel.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/906584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
}
|
Evaluation of $ \int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx$
$(1)$ Evaluation of $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{a\sin x+b\cos x}{\sin \left(x+\frac{\pi}{4}\right)}dx$
$(2)$ Evaluation of $\displaystyle \int_{-1}^{1}\ln\left(\frac{1+x}{1-x}\right)\cdot \frac{x^3}{\sqrt{1-x^2}}dx$
$(3)$ Evaluation of $\displaystyle \int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx$
$\bf{My\; Try::}$ For $(1)$ one
Let $\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{a\sin x}{\sin (x+\frac{\pi}{4})}dx+\int_{0}^{\frac{\pi}{2}}\frac{b\cos x}{\sin (x+\frac{\pi}{4})}dx$
Now Let $\displaystyle J = \int_{0}^{\frac{\pi}{2}}\frac{a\sin x}{\sin (x+\frac{\pi}{4})}dx$ and $\displaystyle K = \int_{0}^{\frac{\pi}{2}}\frac{b\cos x}{\sin (x+\frac{\pi}{4})}dx$
Now We will Calculate $\displaystyle J = \int_{0}^{\frac{\pi}{2}}\frac{a\sin x}{\sin (x+\frac{\pi}{4})}dx$
Using $\displaystyle \left(x+\frac{\pi}{4}\right)=t\;,$ Then $dx = dt$.
So $\displaystyle J = a\int_{0}^{\frac{\pi}{4}}\frac{\sin \left(t-\frac{\pi}{4}\right)}{\sin t}dt = a\cdot \frac{1}{\sqrt{2}}\cdot \frac{\pi}{2}$
Similarly we will Calculate $\displaystyle K = \int_{0}^{\frac{\pi}{2}}\frac{b\cos x}{\sin (x+\frac{\pi}{4})}dx$
Using $\displaystyle \left(x+\frac{\pi}{4}\right)=t\;,$ Then $dx = dt$.
So $\displaystyle K = b\int_{0}^{\frac{\pi}{4}}\frac{\cos \left(t-\frac{\pi}{4}\right)}{\sin t}dt = b\cdot \frac{1}{\sqrt{2}}\cdot \frac{\pi}{2}$
So $\displaystyle I = J+K = \frac{\pi}{2\sqrt{2}}\cdot (a+b)$
Is there is any Shorter Solution for $(1)$ one and How can I calcultae $(2)$ and $(3)$ one
Help me
Thanks
|
We will work on the third one.
Set $x=2a t$ then the third one becomes
$$I_3=\int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx=(2a)^2\int_{0}^{1}\sin^{-1}\left(\sqrt{\frac{1-t}{2}}\right)tdt \tag{1}$$
Set $\sin s=\sqrt{(1-t)/2}$, then $t=\cos(2s)$, $dt=-2\sin(2s)ds$. So (1) becomes
$$I_3=(2a)^2\int_{0}^{\pi/4}\sin^{-1}\left(\sin s\right)\cos(2s)2\sin(2s)ds \implies $$
$$ I_3=(2a)^2\int_{0}^{\pi/4}s\sin(4s)ds\tag{2}$$
Set $u=4s$ in (2) we obtain:
$$ I_3=(a/2)^2\int_{0}^{\pi}u\sin udu=(a/2)^2 \pi \tag{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/907086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Solve a limit with radicals I don't know how to solve this limit. What should I do?
$$
\lim_{x\to 0} {\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1} \over \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}}
$$
Thank you!!
|
Multiply the given expression by a special kind of $1$ :
$1 = \frac{\sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}} . \frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1} }$
$\lim \limits_{x\to 0} \frac{\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}}
\\= \lim \limits_{x\to 0} {\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1} \over \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}} .\frac{\sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}} .\frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1} }
\\= \lim \limits_{x\to 0} \frac{x^3-x^2+5x}{-2x^3-2x^2-8x} .\frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1} }
\\= \lim \limits_{x\to 0} \frac{x^2-x+5}{-2x^2-2x-8} .\frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1} }
$
plugin x = 0
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/907412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
}
|
To find maximum value If $A>0,B>0$ and $C>0$ and further it is known that $A+B+C=\frac{5\pi}{4}$,then find the maximum value of $\sin A+\sin B+\sin C$
|
We can use the method of Lagrange Multipliers. Note that the constraint function is $$ g(A,B,C) = A+B+C $$ and the maximization function is $$ f(A,B,C) = \sin A + \sin B + \sin C. $$We can find $ \nabla f $ and $ \nabla g $ and we get $$ \begin {eqnarray*} \nabla g &=& \left< 1, 1, 1 \right>, \\ \nabla f &=& \left< \cos A, \cos B, \cos C \right>. \end {eqnarray*} $$
Then, setting $ \nabla f \propto \nabla g $, we get $ \cos A = \cos B = \cos C $. We can go back to the restrictions that $A,B,C>0$ and $A+B+C=\frac{5\pi}{4}$ and we see that the only solution here is $$ (A,B,C) = \left( \frac{5\pi}{12}, \frac{5\pi}{12}, \frac{5\pi}{12} \right), $$ at which $ f(A,B,C) = \boxed{\frac {3 \sqrt{2+\sqrt{3}}}{2}} $. $\blacksquare$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/907695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
The inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/( x^3+2x^2+2)$ What is the independent coefficient in the inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/(x^3+2x^2+2)$ ?
I have been calculating some combinations, but I don't know how I can calculate the inverse.
|
$$
(ax^2 + bx + c)(2x^2 + 2) + (dx + e)(x^3 + 2x + 2) = 1\\
(2a + d)x^4 + (2b + e)x^3 + (2a + 2c + 2d)x^2 + (2b + 2d + 2e)x + (2c + 2e) = 1\\
$$
$$\left[ \begin{array}{ccccc}
2 & 0 & 0 & 1 & 0 \\
0 & 2 & 0 & 0 & 1 \\
2 & 0 & 2 & 2 & 0 \\
0 & 2 & 0 & 2 & 2 \\
0 & 0 & 2 & 0 & 2 \\
\end{array} \right]
\left[ \begin{array}{c}
a \\ b \\ c \\ d \\ e \\ \end{array} \right]
=
\left[ \begin{array}{c}
0 \\ 0 \\ 0 \\ 0 \\ 1 \\ \end{array} \right]
$$
Solve for $a$, $b$, and $c$ and you find the inverse.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/907789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
|
How do I simplify $\frac{\sqrt{21}-5}{2} + \frac{2}{\sqrt{21} - 5}$?
How do I simplify the following equation?
$$\frac{\sqrt{21}-5}{2} + \frac{2}{\sqrt{21} - 5}$$
I have no idea where to start. If I multiply either fraction by its denominator I will still end up with a square root. I know the end result should be $-5$.
|
$$\begin{align} \frac{\sqrt{21} - 5}{2} + \frac{2}{\sqrt{21} - 5} &= \frac{\sqrt{21} - 5}{2} + \frac{1}{\frac{\sqrt{21} - 5}{2}} \\ &= \frac{\big(\frac{\sqrt{21} - 5}{2}\big)^2 + 1}{\frac{\sqrt{21} - 5}{2}} \\ &= \frac{2\big(\frac{(\sqrt{21} - 5)^2}{4} + 1\big)}{\sqrt{21} - 5} \\ &= \frac{\frac{21 + 25 - 10\sqrt{21}}{2} + 2}{\sqrt{21} - 5} \\ &= \frac{\frac{46 + 2(2 - 5\sqrt{21})}{2}}{\sqrt{21} - 5} \\ &= \frac{23 + 2 - 5\sqrt{21}}{\sqrt{21} - 5} \\ &= \frac{5(5 - \sqrt{21})}{\sqrt{21} - 5}\tag1\end{align}$$.
$$\because \forall (x, y), \ \frac{x - y}{y - x} = \frac{x - y}{-(x - y)} = -1,$$
$$(1) \iff \frac{5(5 - \sqrt{21})}{\sqrt{21} - 5} = 5\times \frac{5 - \sqrt{21}}{\sqrt{21} - 5} = 5\times (-1) = -5$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/908027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
}
|
If $ab+bc+ca=0$ then the value of $1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is If $ab+bc+ca=0$
then the value of
$1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is...
|
$ab+bc+ca=0$
$a(b+c)= -bc$
$b+c= -bc/a$
Adding both side $a$ then $b+c+a=a-bc/a$
$a(a+b+c)= a^2-bc$
$1/(a^2-bc)= 1/a(a+b+c)$
Similarly $1/(b^2-ca) = 1/b(a+b+c)$
and $1/(c^2-ab) = 1/c(a+b+c)$.
By adding all eqution we get
$1/a(a+b+c)+ 1/b(a+b+c)+ 1/c(a+b+c)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/910709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Why a particular ring of integers is not generated by a single element It says here in the Sage documentation that the ring of integers in the number field obtained from
$$f(x) = x^3 + x^2 - 2x + 8$$
is not generated by a single element. How would one go about showing that this is the case?
|
Let $\alpha$ be a root of $f(x) = x^{3} + x^{2} - 2x + 8$. We will show that $\mathcal{O}_{K} = \mathbb{Z}[\alpha, \beta]$, where $\beta = (\alpha + \alpha^{2})/2$. Granting this, assume that $\mathbb{Z}[\vartheta]$ for some $\vartheta = a + b\alpha + c\beta$. Since $\{1, \vartheta, \vartheta^{2}\}$ is an integral basis iff $\{1, (\vartheta + 1), (\vartheta + 1)^{2}\}$ is an integral basis iff $\{1, (\vartheta - a), (\vartheta - a)^{2}\}$ is an integral basis, we may assume that $a = 0$. We find that (writing $\alpha^{2}, \alpha\beta, \beta^{2}$ in terms of $\alpha, \beta$)
$$(b\alpha + c\beta)^{2} = -8bc - 6c^{2} + \left(2bc - b^{2} - \frac{c^{2}}{2}\right)\alpha - 2b^{2}\beta,$$
so that the change of basis matrix is given by
$$A = \begin{bmatrix}
1 & 0 & -8bc - 6c^{2}\\
0 & b & 2bc - b^{2} - \frac{c^{2}}{2}\\
0 & c & -2b^{2}
\end{bmatrix}.$$
If $\mathbb{Z}[\vartheta]$ is indeed an integral basis, then this matrix must have determinant $\pm 1$. As
$$\pm 1 = \det A = -2b^{3} - 2bc^{2} + b^{2}c + \frac{c^{3}}{2},$$
we must have $c$ even. But then the determinant is even, a contradiction, so that no such $\vartheta$ exists.
We now show that $\mathcal{O}_{K} = \mathbb{Z}[\alpha, \beta]$. Note that the linear transformation of multiplication by $a + b\alpha + c\alpha^{2}$ is given by
$$L = \begin{bmatrix}
a & -8c & -8(b - c)\\
b & 2c + a & 2b - 10c\\
c & b - c & a - b + 3c
\end{bmatrix}.$$
Consider $\mathcal{O} = \mathbb{Z}[\alpha]$. First, $f$ is indeed the minimal polynomial since it is irreducible mod 3 (no roots). Then the discriminant of $\{1, \alpha, \alpha^{2}\}$ is
$$d(1, \alpha, \alpha^{2}) = -N\Big(f'(\alpha)\Big) = -N(3\alpha^{2} + 2\alpha - 2) = -\rm{det}\begin{bmatrix}
-2 & -24 & 8\\
2 & 4 & -26\\
3 & -1 & 7
\end{bmatrix} = -4 \cdot 523.$$
As a result, since $2^{2} \mid d$, if $\mathcal{O}_{K} \not= \mathcal{O}$, (by a theorem) there exists an algebraic integer of the form
$$\frac{1}{2}(\lambda_{1} + \lambda_{2}\alpha + \lambda_{3}\alpha^{2}),$$
where $\lambda_{i} \in \{0, 1\}$.
If such an algebraic integer $\gamma$ of this form exists, we have
$$\rm{Tr}\, \gamma = \rm{tr}\, L = \frac{3\lambda_{1} - \lambda_{2} + 5\lambda_{3}}{2} \in \mathbb{Z}.$$
Examining the parity of the numberator, we see that for $\gamma \not= 0$, we must have exactly two $\lambda_{i} = 1$. Thus, the possibilities for $\gamma$ are
$$\frac{1 + \alpha}{2}, \frac{1 + \alpha^{2}}{2}, \frac{\alpha + \alpha^{2}}{2}.$$
However, for the first two possibilities,
$$N(\gamma) = \rm{det} \begin{bmatrix}
\frac{1}{2} & 0 & -4\\
\frac{1}{2} & \frac{1}{2} & 1\\
0 & \frac{1}{2} & 0
\end{bmatrix} = -\frac{5}{4} \notin \mathbb{Z}, \qquad N(\gamma) = \rm{det} \begin{bmatrix}
\frac{1}{2} & -2 & 4\\
0 & \frac{3}{2} & -5\\
\frac{1}{2} & -\frac{1}{2} & 2
\end{bmatrix} = \frac{9}{4} \notin \mathbb{Z}.$$
For the third possibility, which we denoted $\beta$ earlier, the norm is indeed an integer and we further find (upon examining $\beta, \beta^{2}, \beta^{3}$) that
$$\beta^{3} - \beta^{2} + 6\beta - 8 = 0,$$
so that $\beta \in \mathbb{O}_{K}$ and $\mathbb{Z}[\alpha, \beta] \in \mathcal{O}_{K}$.
Now, $\mathbb{Q} \oplus \mathbb{Q}\alpha \oplus \mathbb{Q}\alpha^{2} \approx \mathbb{Q} \oplus \mathbb{Q}\alpha \oplus \mathbb{Q}\beta$ and we have a change of basis matrix ($\{1, \alpha, \alpha^{2}\} \rightarrow \{1, \alpha, \beta\}$)
$$B = \begin{bmatrix}
1 & 0 & 0\\
0 & 1 & -1\\
0 & 0 & 2
\end{bmatrix}, \qquad \rm{det} B = 2.$$
Hence, $d(1, \alpha, \beta) = d(1, \alpha, \alpha^{2})/2^{2} = 523$. Since $523$ is squarefree, $\{1, \alpha, \beta\}$ is a basis for $\mathcal{O}_{K}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/912148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Induction Proof: $\sum_{k=1}^{n}\frac{a-1}{a^k}=1-\frac{1}{a^n}, a \ne 0$ I'm having trouble showing equality for the $A(n+1)$ portion of the proof.
Prove by Induction: $$\sum_{k=1}^{n}\frac{a-1}{a^k}=1-\frac{1}{a^n}, a \ne 0$$
Base Case $(n=1)$: $$\frac{a-1}{a^1}=1-\frac{1}{a^1}$$
$$1-\frac{1}{a}=1-\frac{1}{a}$$
Therefore, $A(1)$ is true.
Assume $A(n+1)$ is true for some $n\ge1$. Then:
$$\sum_{k=1}^{n+1}\frac{a-1}{a^k}=1-\frac{1}{a^{n+1}}$$
Because of the lack of $n$ on both sides I am confused as to show we show $A(n+1)$ to be true. Any hints would be great.
|
Hint:
The idea behind induction is that once the case $A(1)$ is proved you assume $A(k)$ is true for all $1\leq k\leq n$ to conclude that it musy be true for $A(n+1)$.
$$A(n+1) = \sum_{k=1}^{n+1} \frac{a-1}{a^k} = \sum_{k=1}^n \frac{a-1}{a^k} + \frac{a-1}{a^{n+1}} = 1-\frac{1}{a^n} + \frac{a}{a^{n+1}} - \frac{1}{a^{n+1}} = 1 - \frac{1}{a^{n+1}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/913059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
How do I solve $x^5 +x^3+x = y$ for $x$? I understand how to solve quadratics, but I do not know how to approach this question. Could anyone show me a step by step solution expression $x$ in terms of $y$?
The explicit question out of the book is to find $f^{-1}(3)$ for $f(x) = x^5 +x^3+x$
So far I have reduced $x^5 +x^3+x = y$ to $y/x - 3/4 = (x^2 + 1/2)^2$ or $y = x((x^2+1/2)^2 + 3/4)$ but Im still just as lost.
|
$$x^5+x^3+x-3=0\\(x^5-1)+(x^3-1)+(x-1)=0\\(x-1)(x^4+x^3+2 x^2+2x+3)=0$$
So the solution you wanted is $x=1$.
Sorry but it is just a solution ad hoc.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/913487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 7
}
|
How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$
let $x,y,z>0$, find the minimum of the value
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4]{xy^3}$$
$$3x+z=x+x+x+z\ge 4\sqrt[4]{x^3z}$$
but this is not true,because not all four equalities can hold at once.
This problem is from china middle school test,so I think have without Lagrange methods,so I think this inequality have AM-GM inequality
|
Let
$$
f(x,y,z)=\log(5y+2)+\log(2z+5)+\log(x+3y)+\log(3x+z)-\log(xyz).
$$
Then
$$
f_x=\frac{-1}{x}+\frac{1}{x+3y}+\frac{3}{3x+z};\\
f_y=\frac{-1}{y}+\frac{3}{x+3y}+\frac{5}{2+5y};\\
f_z=\frac{-1}{z}+\frac{1}{3x+z}+\frac{2}{5+2z}.
$$
So the FOC's yield
$$
x=1,\quad y=\sqrt{2/15}, \quad z = \sqrt{15/2}.
$$
With these values, the original function evaluates to $241+44\sqrt{30}$. To be rigorous, you can verify the SOC's but I think it will probably work. According to Mathematica's FindMinimum, the values of $x$, $y$, and $z$ above solve the problem.
P.s. A commment on why naive AM-GM may not work: Suppose you split the 4 expressions according to
\begin{gather}
\frac{x}{m}+\cdots+\frac{x}{m}+\frac{3y}{n}+\cdots+\frac{3y}{n}\tag{i}\\
\frac{3x}{p}+\cdots+\frac{3x}{p}+\frac{z}{q}+\cdots+\frac{z}{q},\tag{ii}\\
\frac{5y}{r}+\cdots+\frac{5y}{r}+\frac{2}{s}+\cdots+\frac{2}{s},\tag{iii}\\
\frac{2z}{c}+\cdots+\frac{2z}{c}+\frac{5}{d}+\cdots+\frac{5}{d}.\tag{iv}
\end{gather}
The first 2 equations say
$$
x=\frac{3m}{n}y,\quad z=\frac{3q}{p}x.
$$
You want the exponent of $x$ after applying AM-GM to match the exponent of $x$ in the denominator, which is $1$, so $\frac{m}{m+n}+\frac{p}{p+q}=1$, which simplifies to $\frac{m}{n}=\frac{q}{p}$. So (i) and (ii) imply
$$
z=\frac{9m^2}{n^2}y.\tag{A}
$$
Next, (iii) implies $y=\frac{2r}{5s}$. And since we want $\frac{n}{n+m}+\frac{r}{r+s}=1$, it must be that $\frac{r}{s}=\frac{m}{n}$. So (iii) implies
$$
y=\frac{2}{5}\frac{m}{n}.\tag{B}
$$
Similarly, (iv) implies $z=\frac{5c}{2d}$ with $\frac{c}{d}=\frac{p}{q}$ due to $\frac{c}{c+d}+\frac{q}{p+q}=1$. But $\frac{p}{q}=\frac{n}{m}$, so (iv) implies
$$
z=\frac{5}{2}\frac{n}{m}.\tag{C}
$$
Putting together (A), (B), and (C), we see that
$$
9\frac{m^2}{n^2}=\frac{z}{y}=\frac{5n}{2m}/\frac{2m}{5n}\implies\frac{9m^2}{n^2}=\frac{25n^2}{4m^2}.
$$
Taking square roots and simplifying yield
$$
6m^2=5n^2.
$$
But now we have a problem because $\sqrt{\frac{5}{6}}$ cannot be rational.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/913664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
}
|
Simplify rational expression How do I simplfy this expression?
$$\dfrac{\frac{x}{2}+\frac{y}{3}}{6x+4y}$$
I tried to use the following rule $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$
But I did not get the right result.
Thanks!!
|
$$\frac{\frac{x}{2}+\frac{y}{3}}{6x+4y}=\frac{6 \cdot \left ( \frac{x}{2}+\frac{y}{3}\right ) }{6 \cdot (6x+4y)}=\frac{3x+2y}{6 \cdot 2 \cdot (3x+2y)}=\frac{1}{12}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/915154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equation*} \dfrac{5}{3} = 1 + \dfrac{2}{3}\text{.}
\end{equation*} This is because when 5 is divided by 3, 3 goes into 5
once (hence the $1$ term) and there is a remainder of $2$ (hence the
$\dfrac{2}{3}$ term). Note the following: every division problem can
be decomposed into an integer (the $1$ in this case) plus a fraction,
with the denominator being what you divide by (the $3$ in this
case).
So, when $n$ is divided by 14, the remainder is 10. This can be
written as \begin{equation*} \dfrac{n}{14} = a + \dfrac{10}{14}
\end{equation*} where $a$ is an integer.
We want to find the remainder when $n$ is divided by 7, which I'll
call $r$. So \begin{equation*} \dfrac{n}{7} = b + \dfrac{r}{7}\text{,}
\end{equation*} where $b$ is an integer.
Here's the key point to notice: notice that \begin{equation*}
\dfrac{n}{7} = \dfrac{2n}{14} = 2\left(\dfrac{n}{14}\right)\text{.}
\end{equation*} This is because $\dfrac{1}{7} = \dfrac{2}{14}$.
Thus, \begin{equation*} \dfrac{n}{7} = 2\left(\dfrac{n}{14}\right) =
2\left(a + \dfrac{10}{14}\right) = 2a + 2\left(\dfrac{10}{14}\right) =
2a + \dfrac{10}{7} = 2a + \dfrac{7}{7} + \dfrac{3}{7} = (2a+1) +
\dfrac{3}{7}\text{.} \end{equation*} So, since $a$ is an integer, $2a
+ 1$ is an integer, which is our $b$ from the original equation. Thus, $r = 3$.
To her, this method was not very intuitive. She did understand the explanation. Are there any suggestions for how I can explain this in another way?
|
We have $n=14a+10$ for some integer $a$. If we take this equation modulo $7$ we have that $n \equiv 10 \mod 7 \equiv 3 \mod 7$ (This because $14 \equiv 0 \mod 7$ and $10 \equiv 3 \mod 7$). You should look up modular arithmetic
where you will see how to deal with this sort of questions. Modular arithmetic is very important in much branches of mathematics mostly in structures that have a finite number of elements in them (finite groups, rings, fields,...).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/915238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
}
|
If $\frac{1}{x^2+x+1}$, then find $y_n$ ($n^{th} $differention of the equation). The answer:
$$\frac{2(-1)^n \cdot n!}{\sqrt{3}r^{n+1}}\sin(n+1)\theta,$$
where $r=\sqrt{x^2+x+1}$ and $\theta=\cot^{-1}\frac{2x+1}{\sqrt{3}}$.
How I have tried:
$$y=\frac{1}{x^2+x+1+\frac{1}{4}-\frac{1}{4}}$$
$$y=\frac{1}{(x^2+x+\frac{1}{4}-\frac{3}{4})}$$
$$y=\frac{1}{(x+\frac{1}{2})^2-\frac{3}{4}}$$
$$y=\frac{4}{(2x+1)^2-3}$$
I don't know how to proceed further, someone please help.
|
HINT:
Let $x^2+x+1=(x-a)(x-b)$
So, we can write $\displaystyle a,b=\frac{-1\pm\sqrt3i}2$
$$\implies\frac1{x^2+x+1}=\frac1{a-b}\left(\frac1{x-b}-\frac1{x-a}\right)$$
Now from this and many others, $$\frac{d^n(1/(x-a))}{dx}=(-1)^n\frac{n!}{(x-a)^{n+1}}$$
For $\displaystyle x-a=x-\left(\dfrac12+i\dfrac{\sqrt3}2\right)$ set $x-\dfrac12=r\cos\phi,$ and $-\dfrac{\sqrt3}2=r\sin\phi$
and apply de Movire Theorem
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Show that two expressions are equivalent I am trying to prove a hyperbolic trigonometric identity and I ran into the following expression:
$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} \quad.$$
This expression is supposed to be equivalent to
$$\sqrt{x^2+1} \quad.$$
I tried to algebraically manipulate the original expression to get the required expression and got to:
$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} = x+\frac{1}{\sqrt{x^2+1} + x} \quad.$$
but I don't know how I can show that
$$x+\frac{1}{\sqrt{x^2+1} + x} = \sqrt{x^2+1}\quad .$$
|
$$\\an-inovative-solution\\ \\x=tan(a) \Rightarrow \sqrt{x^2+1}=sec(a) \\\frac{(\sqrt{x^2+1}+x)^2+1}{2(\sqrt{x^2+1}+x)}=\\\frac{(sec(a)+tan(a))^2+1}{2(sec(a)+tan(a))}=\\\frac{(sec^2(a)+tan^2(a)+2sec(a)tan(a))+1}{2(sec(a)+tan(a)}=\\\frac{(sec^2(a)+tan^2(a)+2sec(a)tan(a))+1}{2(sec(a)+tan(a))}=\\as-we-know-(sec^2(a)=1+tan^2(a))\\so\\\frac{(sec^2(a)+sec^2(a)+2sec(a)tan(a))}{2(sec(a)+tan(a)}=\\\frac{2sec^2(a)+2sec(a)tan(a)}{2(sec(a)+tan(a)}=\\\frac{2sec^2(a)+2sec(a)tan(a))}{2(sec(a)+tan(a))}=\\\frac{2sec(a)(sec(a)+tan(a)}{2(sec(a)+tan(a))}=\\sec(a)=\sqrt{x^2+1}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Find the limit of $\frac{1}{n^2}\sum_{k=1}^n k^2\sin\left(\frac{\theta}{k}\right)$ I am trying to find the limit of
$$\frac{1}{n^2}\sum_{k=1}^n k^2\sin\left(\frac{\theta}{k}\right)\qquad \theta>0$$
as $n\to \infty$. I know that since $x-\frac{x^3}{6}\leq \sin x \leq x$ then
$$\frac{1}{n^2}\sum_{k=1}^n k^2\sin \left( \frac{\theta}{k}\right)\leq \frac{\theta}{n^2}\sum_{k=1}^n k=\theta\frac{(n+1)}{2n}$$
Since the $\lim_{n\to \infty}\theta\frac{(n+1)}{2n}=\frac\theta2$, then we know that
$$\frac{1}{n^2}\sum_{k=1}^n k^2\sin \left( \frac{\theta}{k}\right)\leq \frac \theta2 \qquad n \to \infty$$
Now I concentrate on the other side and, with the same logic, I obtain
$$\frac{\theta}{n^2}\sum_{k=1}^nk -\frac{\theta^3}{6n^2}\sum_{k=1}^n\frac1k\leq\frac{1}{n^2}\sum_{k=1}^n k^2\sin \left( \frac{\theta}{k}\right)$$
How to prove that also the left side tends to $\frac\theta2$? Can you give me a little hint please?
|
We have:
$$\frac{\theta}{n^2}\sum_{k=1}^{n} k = \theta\frac{n+1}{2n}= \frac{\theta}{2}+O\left(\frac{1}{n}\right) $$
while:
$$\frac{\theta^3}{6n^2}\sum_{k=1}^{n}\frac{1}{k}=\frac{\theta^3}{6}\cdot\frac{H_n}{n^2}=O\left(\frac{1}{n}\right),$$
since, obviously:
$$H_n = \sum_{k=1}^{n}\frac{1}{k}\leq\sum_{k=1}^{n} 1 = n.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
$\lim_{x \to 0} \left ({e^x+e^{-x}-2\over x^2} \right )^{1\over x^2}$ Could anyone give me hints on how to solve this limit without L'Hospital rule?
$$\lim_{x \to 0} \left ({e^x+e^{-x}-2\over x^2} \right )^{1\over x^2}$$
I tried using standard formualaes but got nowhere.
|
If $L$ is the desired limit then $$\begin{aligned}\log L &= \log\left\{\lim_{x \to 0}\left(\frac{e^{x} + e^{-x} - 2}{x^{2}}\right)^{1/x^{2}}\right\}\\
&= \lim_{x \to 0}\log\left(\frac{e^{x} + e^{-x} - 2}{x^{2}}\right)^{1/x^{2}}\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(\frac{2\cosh x - 2}{x^{2}}\right)\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\log\left(1 + \frac{2\cosh x - 2 - x^{2}}{x^{2}}\right)\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\dfrac{\log\left(1 + \dfrac{2\cosh x - 2 - x^{2}}{x^{2}}\right)}{\dfrac{2\cosh x - 2 - x^{2}}{x^{2}}}\cdot\dfrac{2\cosh x - 2 - x^{2}}{x^{2}}\\
&= \lim_{t \to 0}\frac{\log(1 + t)}{t}\cdot\lim_{x \to 0}\frac{2\cosh x - 2 - x^{2}}{x^{4}}\\
&= \lim_{x \to 0}\frac{2\cosh x - 2 - x^{2}}{x^{4}}\\
&= \lim_{x \to 0}\frac{1}{x^{4}}\left\{2\left(1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \cdots\right) - 2 - x^{2}\right\}\\
&= \frac{1}{12}\end{aligned}$$ hence $L = e^{1/12}$. In the above derivation we have used the substitution $$\begin{aligned}t &= \frac{2\cosh x - 2 - x^{2}}{x^{2}}\\
&= \frac{1}{x^{2}}\left\{2\left(1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \cdots\right) - 2 - x^{2}\right\}\\
&= \frac{x^{2}}{12} + \cdots\end{aligned}$$ and clearly $t \to 0$ as $x \to 0$. I am not sure if we can do this limit problem without the use of Taylor's series or L'Hospital's rule.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
The remainder of $1^1+2^2+3^3+\dots+98^{98}$ mod $4$ How can I solve this problem:
If the sum $S=(1^1+2^2+3^3+4^4+5^5+6^6...+98^{98})$ is divided by $4$ then what is the remainder?
I know that all the even terms I can ignore since $(2n)^{2n}=4^nn^{2n})$ which is divisible by $4$,but i dont know what to do next it. Also I know we can write each of them as $(99-98)^1+(99-97)^2+(99-96)^3...+(99-1)^{98}$.
|
$$S=1^1+2^2+3^3+4^4+5^5+6^6+7^7+...97^{97}+98^{98}\equiv \\S\equiv1^1+0+3^3+0+5^5+0+7^7+...+97^{97}+0\\\equiv1^1+3^3+5^5+7^7+...+97^{97}\\\equiv1^1+0+3^3+0+5^5+0+7^7+...+97^{97}+0\\\equiv1^1+3^3+5^5+7^7+...+97^{97}\equiv\\S\equiv1+(4k-1)^3+(4k+1)^5+(4k-1)^7+(4k+1)^9+(4k-1)^{11}...+(4k+1)^{97}\\\equiv1+(-1)^3+1^5+(-1)^7+...+(-1)^{97}\\(1,3,5,..97)=49 ,term\\so\\\equiv1+(-1+1)+(-1+1)+....=1
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Solve $x+3y=4y^3,y+3z=4z^3 ,z+3x=4x^3$ in reals
Find answers of this system of equations in reals$$
\left\{
\begin{array}{c}
x+3y=4y^3 \\
y+3z=4z^3 \\
z+3x=4x^3
\end{array}
\right.
$$
Things O have done: summing these 3 equations give $$4(x+y+z)=4(x^3+y^3+z^3)$$
$$x+y+z=x^3+y^3+z^3$$
I've also tried to show that $x,y,z$ should be between $1$ and $-1$(As $(-1,-1,-1)$ and $(1,1,1)$ are answers of this system of equations) but I was not successful.
|
Suppose that we had $x>1$. Then, since $4x^3-3x > x$, we have $z>x$, and similarly $y>z$, $x>y$, contradiction. By symmetry, we conclude that $x,y,z\in [-1,1]$.
So there exist $\alpha,\beta,\gamma \in [0,\pi]$ with $x=\cos \alpha$, $y=\cos \beta$, $z=\cos \gamma$. By the formula for $\cos 3\alpha$, we can rewrite the system of equations as:
$$
\left\{
\begin{array}{c}
\alpha \equiv \pm 3 \beta\ (\operatorname{mod}\ 2\pi) \\
\beta = \pm 3 \gamma\ (\operatorname{mod}\ 2\pi) \\
\gamma = \pm 3\alpha\ (\operatorname{mod}\ 2\pi)
\end{array}
\right.
$$
So we have $\pm 27\alpha \equiv \alpha$, so either $26\alpha \equiv 0$ or $28\alpha \equiv 0$. We find that $\alpha = \pi k / 13$ or $\alpha = \pi k / 14$ for some nonnegative integer $k$.
This gives $27$ solutions,* $x=\cos{\frac{\pi k}{13}}$ for $0\leq k\leq 13$, and $x=\cos{\frac{\pi k}{14}}$ for $1\leq k\leq 13$.
For example, one solution is $(\cos\frac{\pi}{14},\cos\frac{9\pi}{14},\cos\frac{3\pi}{14})$.
* This is exactly the number we expect, up to multiplicity, when intersecting three cubic surfaces; in particular there are no more solutions over $\mathbb{C}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/924596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
}
|
Solve equation with two unknowns I have these equations.
$2\pi r_1+2\pi r_2=24$
and $\pi r_1^2+\pi r_2^2=20$
and to solve them I did the following steps
Step 1 : $\frac {2\pi r_1+2\pi r_2}{2}= \frac {24}{2}$
Step 2 : $\pi r_1+\pi r_2= 12$
Step 3 : $\pi r_1= 12 - \pi r_2$
Step 4 : $\pi r_1^2+(12 - \pi r_2)^2=20$
Step 5 : $\pi r_1^2+(12 - \pi r_2)^2=20$
Step 6 : $\pi r_1^2+144 - 24\pi r_2 + \pi r_1^2=20$
Step 7 : $2\pi r_1^2- 24\pi r_2+124=0$
Step 8 : $\pi r_1 = - \frac {-24}{2} \pm \sqrt{(\frac {-24}{2})^2 - 124}$
I get the result $6 \pm \sqrt{10}$ but is this correct? Something feels wrong about this answer but don't know what.
Thanks!
|
$\begin{cases}2\pi r_1+2\pi r_2=24 & (1) \\ \pi r_1^2+\pi r_2^2=20 & (2)\end{cases}$
Equation $(1)$ gives $r_1=\frac{12-\pi r_2}{\pi}$
Plug that into $(2)$, we have $\pi\big(\frac{12-\pi r_2}{\pi}\big)^2+\pi r_2^2=20 \Rightarrow 2\pi r_2^2-24r_2+\frac{144}{\pi}-20=0$.
This quadratic equation has no solutions, and therefore there is no solution to the system of equations.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Determinant of a $2\times 2$ block matrix I would like to know the proof for:
The determinant of the block matrix
$$\begin{pmatrix} A & B\\ C& D\end{pmatrix}$$
equals
$$(D-1) \det(A) + \det(A-BC) = (D+1) \det(A) - \det(A+BC)$$
when $A$ is a square matrix, $D$ is a scalar, $C$ is a row vector and $B$ is a column vector.
I appreciate your help.
|
We add an extra column and row to this matrix:
$$\left(\begin{array}{cc} A & B & 0 \\ C & D & 0 \\ 0 & 0& 1\end{array}\right).$$
This new matrix has the same determinant as the original.
Now we perform some row and column operations that don't change the determinant. First add $D-1$ times the last row to the second-to-last, and then subtract the last column from the second-to-last column. The result is
$$ \left(\begin{array}{ccc} A & B & 0 \\ C & 1 & D-1\\ 0 & -1 & 1\end{array}\right).$$
By linearity of the determinant in each row and column, we can write the determinant of the above as
$$ \begin{align}
&\det\left(\begin{array}{ccc} A & B & 0 \\ C & 1 & 0\\ 0 & -1 & 1\end{array}\right) + \det\left(\begin{array}{ccc} A & B & 0 \\ C & 1 & D-1\\ 0 & -1 & 0\end{array}\right) \\
&=\det \left(\begin{array}{cc} A & B \\ C & 1\end{array}\right) + (-1)(D-1)\det(\left(\begin{array}{cc} A & B \\ 0 & -1\end{array}\right) \\
&= \det(A-BC) + (D-1) \det(A).\end{align}$$
The other formula can be derived by replacing $D-1$ with $D+1$ in the second step and proceeding in a similar way.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
}
|
Trignometric problem (using De Movier's Theorem)
Ok so this question, I started out writing tan as sin and cos in the right side of the equation, simplified as much as possible and ended up with a very (sort of) fascinating equation which is
Where s = sin thetha and c= sin thetha
As you can see the denominator and numerator looks very simple where on top sin x is with the power 5 and in the bottom cos x is with the power 5.
Then I went onto applying the De Movire theorem
And then ended up with-
As you can see the required terms sin 5 thetha and cos 5 thetha is there, but now all I need is to cancel out those two other terms in both denominator and numerator. I simply cant find a way in which those cancel off, My question is-
How do I proceed from here?? And if I had done something wrong (if those dont cancel) then where did I go wrong? Please help
|
Using DeMoivre's Theorem and the Binomial Theorem, you get:
$\cos 5\theta + i\sin 5\theta = (\cos \theta + i\sin \theta)^5$
$= \cos^5\theta + 5i\cos^4\theta\sin\theta + 10i^2\cos^3 \theta\sin^2\theta + 10i^3\cos^2 \theta\sin^3\theta + 5i^4\cos \theta \sin^4 \theta + i^5\sin^5\theta$
$= \cos^5\theta + 5i\cos^4\theta\sin \theta - 10\cos^3 \theta\sin^2\theta - 10i\cos^2 \theta\sin^3\theta + 5\cos\theta\sin^4 \theta + i\sin^5\theta$
$= (\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4 \theta)+i(5\cos^4\theta\sin \theta - 10\cos^2 \theta\sin^3\theta + \sin^5\theta)$
Equate real and imaginary parts to get expressions for $\cos 5\theta$ and $\sin 5\theta$. Can you finish from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
What is the sum of $\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$? Consider the power sequence
$$\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$$
What is the function to which it sums to?
My reasoning is to differntiate the sum with respect to $x$, then to integrate with respect to x from $0$ to $x$ after variation of the sum into a Taylor series form.
I mean:
$$\int _0 \frac{d}{dx}\sum_{n=1}^\infty (n^2+n^3)x^{n-1}dx=\int_0 \sum_{n=1}^\infty \frac{n^2+n^3}{n-1}x^{n-2}dx$$
But it doesn't seem to work in this exercise.
|
Another way to solve it....
First:
Note that
$$\frac{d^2}{dx^2}\Bigl(nx^{n+1}\Bigr)=n^2(n+1)x^{n-1}=(n^2+n^3)x^{n-1}\qquad(1)$$
we will use this later.
Second:
We know that
$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\qquad|x|<1$$
Taking the derivative respect to $x$, we have:
$$\begin{array}{rcl}
\frac{d}{dx}\Bigg(\sum_{n=0}^{\infty}x^n\Bigg)&=&\frac{d}{dx}\Bigg(\frac{1}{1-x}\Bigg)\\
\sum_{n=0}^{\infty}\frac{d}{dx}x^n&=&\frac{1}{(1-x)^2}\\
\sum_{n=1}^{\infty}nx^{n-1}&=&\frac{1}{(1-x)^2}\\
\sum_{n=0}^{\infty}nx^{n-1}&=&\frac{1}{(1-x)^2}\qquad(2)
\end{array}$$
where in the LHS of $(2)$ I summed a $0$, that is equivalent to the term for $n=0$.
Third:
Take $(2)$ and multiply by $x^2$:
$$\sum_{n=0}^{\infty}nx^{n+1}=\frac{x^2}{(1-x)^2}\qquad|x|<1\qquad(3)$$
Now we have the general term $nx^{n+1}$ in the sum, so if we derive respect to $x$ two times, we will have our function because of the relation of $(1)$. Then:
$$\begin{array}{rcl}
\frac{d^2}{dx^2}\Bigg(\sum_{n=0}^{\infty}nx^{n+1}\Bigg)&=&\frac{d^2}{dx^2}\Bigg(\frac{x^2}{(1-x)^2}\Bigg)\\
\sum_{n=1}^{\infty}\frac{d^2}{dx^2}\Big(nx^{n+1}\Big)&=&\frac{d}{dx}\Bigg(\frac{-2x}{(1-x)^3}\Bigg)\\
\sum_{n=1}^{\infty}(n^2+n^3)x^{n-1}&=&\frac{2(2x+1)}{(1-x)^4}\\
\end{array}$$
Note we had to drop the constant term that happens at $n=0$ for the 1st derivative of the series (second line).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
Calculus Limits Problem L'Hopital's Rule is not allowed.
Question 1:
$$\lim_{x\to -2} \frac{\sqrt{6+x}-2}{\sqrt{3+x}-1} = \ ?$$
I tried to cross multiply $\frac{\sqrt{6+x}-2}{\sqrt{3+x}-1}$with $\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ and I got $x+2$ on both LHS and RHS thus I conclude
$\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ is equal to $\frac {\sqrt{6+x}-2)}{\sqrt{3+x}-1}$.
Thus when I sub $x = -2$ into $\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ I got $0.5$.
Question 2:
$$\lim_{x\to \pi} \sin\frac{x+\pi}{x-\pi}\sin\frac{x-\pi}{x+\pi} = \ ?$$
I said that $\sin\frac{x-\pi}{x+\pi}$ tends to $0$.
For $\sin\frac{x+\pi}{x-\pi}$, I use squeeze theorem and said that it can lies between $1$ and $-1$.
Thus, $\sin\frac{x+\pi}{x-\pi}\sin\frac{x-\pi}{x+\pi} = 0$.
|
For real $x, \left|\sin\dfrac{x+\pi}{x-\pi}\right|\le1$
$$\lim_{x\to\pi}\sin\frac{x-\pi}{x+\pi}\sin\dfrac{x+\pi}{x-\pi}$$
$$=0\cdot\text{ a finite real number}=0$$
$$\lim_{x\to-2}\frac{\sqrt{6+x}-2}{\sqrt{3+x}-1}=\lim_{x\to-2}\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}\cdot\lim_{x\to-2}\frac{6+x-4}{3+x-1}$$
$$=\frac{\sqrt{3+(-2)}+1}{\sqrt{6+(-2)}+2}\cdot1$$ as $x+2\ne0$ as $x\ne-2$ as $x\to-2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/930498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How to prove the $\Theta$ notation? I know that to prove that f(n) = $\Theta$(g(n)) we have to find c1, c2 > 0 and n0 such that
$$0 \le c_1 g(n) \le f(n) \le c_2 g(n)$$
I'm quite new with the proofs in general.
Let assume that we want to prove that $$an^2+bn+c=\Theta(n^2)$$
where a,b,c are constants and a > 0
I'll start with
$$0 \le c_1 \le a + \frac{b}{n} + \frac{c}{n^2} \le c_2$$
In the book I'm reading, they found $c_1 = a/4, c_2 = 7a/4, n_0 = 2*max(|b|/a, \sqrt{|c|/a})$
How ? I'm really lost. What is the approach for this kind of problems ?
|
The intuition is to make $\frac{\left|b\right|}{n} \le \frac{a}{2}$ and to make $\frac{|c|}{n^2} \le \frac{a}{4}$. That way, you can show the following inequality:
$$ \frac{a}{4} = a - \frac{a}{2} - \frac{a}{4} \le a+\frac{b}{n}+\frac{c}{n^2} \le a + \frac{a}{2} + \frac{a}{4} = \frac{7a}{4}$$
Therefore, let $c_1 = \frac{a}{4}$ and $c_2 = \frac{7a}{4}$. It suffices to find an appropriate $n_0$.
In order to satisfy $\frac{\left|b\right|}{n} \le \frac{a}{2}$, it follows that $n \ge \frac{2|b|}{a}$. Similarly, in order to satisfy $\frac{|c|}{n^2} \le \frac{a}{4}$, it follows that $n^2 \ge \frac{4|c|}{a}$ (or equivalently $n \ge 2\sqrt{\frac{|c|}{a}}$). Therefore, we must have:
$$n \ge \max\left(\frac{2|b|}{a}, 2\sqrt{\frac{|c|}{a}}\right) = 2 \cdot \max\left(\frac{|b|}{a}, \sqrt{\frac{|c|}{a}}\right)$$
The General Case
This approach works very well for arbitrary polynomials. Let
$$p(n) = \sum_{i=0}^k a_i n^i$$
where $a_k > 0$. Then, to show that $p(n) \in \Theta\left(n^k\right)$, one can show that $c_1 = \frac{1}{2^k}a_k$ and $c_2 = \left(2-\frac{1}{2^k}\right)a_k$ are both valid constants for $n \ge n_0$, where
$$n_0 = 2 \cdot \max\left\{\sqrt[i]{\frac{\left|a_i\right|}{a_k}} \mathrel{}\mathclose{}\middle|\mathopen{}\mathrel{} 0 \le i \le k-1 \right\}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/932619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solve the inequality: $|x^2 − 4| < 2$ This is a question on a calculus assignment our class received, I am a little confused on a few parts to the solution, can someone clear a few things up with it?
Since $x^2-4 = 0$ that means $x = 2$ and $x = -2$ are turning points.
When $x < -2$ :
$x^2-4<2$
$x^2<6$
$x < \sqrt{6}$ and $x < - \sqrt{6}$
But the solution says that $x > -\sqrt{6}$, what am I not getting?
|
$$|x^2 - 4| < 2 \implies -2 < x^2 - 4 < 2 \implies 2 < x^2 < 6 \implies \sqrt{2} < |x| < \sqrt{6}$$
So, the solutions are $x$ such that $-\sqrt{6} < x < -\sqrt{2}$ or $\sqrt{2} < x < \sqrt{6}$. Drawing the number line is very helpful for these exercises.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/932930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
More Generating Functions problems (a) For this problem, define a nonstandard die as a 6-sided die that is equally likely to come up on each side, but has a different set of numbers than the usual 1,2,3,4,5,6 on its sides. A standard die will be the usual fair die bearing the numbers 1,2,3,4,5,6.
Is it possible to design a pair of nonstandard dice, with positive integers on the faces (not necessarily all distinct), so that the probability of rolling any given total on those two dice is the same as the probability of rolling that total on two standard dice? The two dice do not need to be identical. If it is possible, do it. If it is not possible, prove that it is not possible.
Some hints I got were to try to construct a pair of nonstandard dice such that the generating function for their sum matches the corresponding g.f. for a pair of standard dice and to note that the g.f. for a pair of standard dice can be factored quite a bit (you can reduce it to factors of degree 2 or less).
I still can't figure it out after rereading the textbook and looking at a few examples. Can anyone help me?
|
Use a generating function for the dice. Normal dice have faces with 1 to 6 dots, so they are represented by the polynomial:
$\begin{align}
D(z)
&= z + z^2 + z^3 + z^4 + z^5 + z^6 \\
&= z \frac{1 - z^6}{1 - z}
\end{align}$
Throwing two dice gives the distribution:
$\begin{align}
D^2(z)
&= z^2 + 2 z^3 + 3 z^4 + 4 z^5 + 5 z^6 + 6 z^7
+ 5 z^8 + 4 z^9 + 3 z^{10} + 2 z^{11} + z^{12}
\end{align}$
We want polynomials $D_1(z)$ and $D_2(z)$ that give:
$\begin{align}
D_1(z) D_2(z) = D^2(z)
\end{align}$
It is clear that $D_i(1)$ is the number of faces of the die, so we want:
$\begin{align}
D_1(1) = D_2(1) = 6
\end{align}$
If we want no blank faces, it means $D_i(0) = 0$ (no constant term).
Let's factor:
$\begin{align}
D(z)
&= z (z + 1) (z^2 - z + 1) (z^2 + z + 1)
\end{align}$
If we evaluate the factors at $z = 1$:
$\begin{align}
D(1)
&= 1 \cdot 2 \cdot 1 \cdot 3
\end{align}$
Thus, in order to have $D_i(1) = 6$, each has to get a factor $z + 1$ and a factor $z^2 + z + 1$. As we want $D_i(0) = 0$, each one has to get a factor $z$ too. This leaves two factors $z^2 - z + 1$ (one from each normal die) to distribute, giving the dice:
$\begin{align}
D_1(z)
&= z (z + 1) (z^2 - z + 1)^2 (z^2 + z + 1) \\
&= z + 2 z^2 + 2 z^3 + z^4 \\
D_2(z)
&= z (z + 1) (z^2 + z + 1) \\
&= z + z^3 + z^4 + z^5 + z^6 + z^8
\end{align}$
These is dice with faces $\{1, 2, 2, 3, 3, 4\}$ and $\{1, 3, 4, 5, 6, 8\}$
The only other solution is normal dice (and exchanging them, obviously).
They are called Sicherman dice.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/934624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Prove by induction that $n(n+1)(n+5)$ is multiple of $3.$ $n(n+1)(n+5) = 3d$
I cannot figure out how to solve this homework question. A friend gave me a solution I couldn't make sense of, and I hope there's something easier out there. Also, what would be the general approach towards questions of this form?
|
The proof can be done without induction, just see that $n$ is of one of the three forms
$n=3k$, $n=3k+1$ or $n=3k+2$. and check each of these cases
Suppose $n=3k$ and $n(n+1)(n+5)=3k(3k+1)(3k+5)=3(k(3k+1)(3k+5))=3d$ where $d=k(3k+1)(3k+5)$,
Next suppose that $n=3k+1$, hence
$n(n+1)(n+5)=(3k+1)(3k+2)(3k+6)=3((3k+1)(3k+2)(k+2))=3d$ where $d=(3k+1)(3k+2)(k+2)$,
Finally suppose that $n=3k+2$, hence
$n(n+1)(n+5)=(3k+2)(3k+3)(3k+7)=3((3k+2)(k+1)(3k+7))=3d$ where $d=(3k+2)(k+1)(3k+7)$.
Since it works in all three cases it must be true for any $n$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/934977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
}
|
How to represent Fermat number $F_n$ as a sum of three squares? Let $F_n=2^{2^n}+1$ be the Fermat number. How to represent the Fermat number $F_n$ for $n \geq 3$ as a sum of three squares of different natural numbers? For example for $n=3$ we have
$$
F_3=257=5^2+6^2+14^2.
$$
Is there any simple procedure to write out such representations for another $n$?
|
Let $X=2^{2^{n-1}}$. Then $X$ is congruent to $1$ modulo $3$, so
$$
F_n=X^2+1=\bigg(\frac{2X+1}{3}\bigg)^2+
\bigg(\frac{2X-2}{3}\bigg)^2+
\bigg(\frac{X+2}{3}\bigg)^2
$$
The values are all distinct, except when $X=1$ or $4$ (which are excluded since
$n\geq 3$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/935172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
}
|
Two methods to integrate? Are both methods to solve this equation correct?
$$\int \frac{x}{\sqrt{1 + 2x^2}} dx$$
Method One:
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}} du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{1}{4}\int\frac{du}{\sqrt{u}}$$
$$\frac{1}{2}\sqrt{u}+C$$
$$\frac{1}{2}\sqrt{1+2x^2}+C$$
I am confused why I get two different answers.
|
The second answer is correct. The mistake in the first method is that your computation of the integral (after substitution) is wrong.
It holds:
$\int\frac{x}{\sqrt{1+2x^2}}dx=\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}\int(1+u)^{-\frac{1}{2}}u=\frac{2}{4} \sqrt{1+u} +C_2=\frac{1}{2}\sqrt{1+2x^2}+C_2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/936324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
}
|
How to show this equals 1 without "calculations" We have $$ \sqrt[3]{2 +\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} = 1 $$
Is there any way we can get this results through algebraic manipulations rather than just plugging it into a calculator?
Of course, $(2 +\sqrt{5}) + (2-\sqrt{5}) = 4 $, maybe this can in some way help?
|
Let $a=\sqrt[3]{2+\sqrt{5}}$ $b=\sqrt[3]{2-\sqrt{5}}$. Obviously $a b =-1$.
Now let's forget what values of $a$ and $b$ and solve the system:
$$
ab=-1\\
a+b=1
$$
They imply
$$
a^2-a-1=0
$$
namely $a=\frac {1\pm \sqrt{5}}{2}$. So $a=\frac {1+\sqrt{5}}{2}, b=\frac {1-\sqrt{5}}{2}$. Now taking $a^3,b^3$ we can easily get $2\pm\sqrt{5}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/937073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Does the series converge or diverge? I want to check, whether $$\sum\limits_{n=0}^{\infty }{\frac{n!}{(a+1)(a+2)...(a+n)}}$$
converges or diverges.
$a$ is a constant number
Ratio test
$$\begin{align}
& \frac{a_{n}}{a_{n-1}}=\frac{n!}{(a+1)(a+2)...(a+(n-1))(a+n)}\cdot \frac{(a+1)(a+2)...(a+(n-1))}{(n-1)!}=\frac{n}{a+n} \\
& \underset{n\to \infty }{\mathop{\lim }}\,\frac{a_{n}}{a_{n-1}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{a+n}=1 \\
\end{align}$$
We know nothing
Root test
$$\sqrt[n]{a_{n}}=\sqrt[n]{\frac{n!}{(a+1)(a+2)...(a+(n-1))(a+n)}}=$$ I can't :(
I can only resolve the serie if "a" is a integer number
$$\begin{align}
& \text{if }a\in \mathbb{Z} \\
& (a+1)(a+2)...(a+(n-1))(a+n)=(n+a)(a+(n-1))...(a+2)(a+1)= \\
& (n+a)(n+a-1)(n+a-2)...(a+2)(a+1)=\frac{(n+a)!}{a!} \\
& \Rightarrow a_{n}=\frac{a!n!}{(n+a)!} \\
& \text{if }a<0\Rightarrow \text{(}n+a)<n,a\in \mathbb{Z} \\
& \underset{n\to \infty }{\mathop{\lim }}\,a_{n}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{a!n!}{(n+a)!}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{a!n(n-1)...(n+a)!}{(n+a)!}= \\
& \underset{n\to \infty }{\mathop{\lim }}\,a!n(n-1)...(n+a-1)=\infty \\
& \Rightarrow \sum\limits_{n=0}^{\infty }{a_{n}=\infty } \\
& \text{if }a>1,a\in \mathbb{Z} \\
& a_{n}=\frac{a!n!}{(n+a)!}=\frac{a!n!}{(n+a)(n-1+a)...\underbrace{(n-a+a)!}_{n!}}=\frac{a!}{(n+a)(n-1+a)...(n+1)} \\
& \Rightarrow \\
& \frac{a!}{\underbrace{(n+a)(n-1+a)...(n+1)}_{a\text{ times}}}<\frac{a!}{(n+1)^{a}}\Rightarrow \\
& \sum\limits_{n=0}^{\infty }{\frac{a!}{(n+a)(n-1+a)...(n+1)}<\sum\limits_{n=0}^{\infty }{\underbrace{\frac{a!}{(n+1)^{a}}<\infty }_{a>1}}} \\
& \text{if }a=1 \\
& \sum\limits_{n=0}^{\infty }{\frac{a!n!}{(n+a)!}}=\sum\limits_{n=0}^{\infty }{\frac{a!n!}{(n+1)!}}=\sum\limits_{n=0}^{\infty }{\frac{a!}{n+1}}=\infty \\
\end{align}$$
But how i can resolve if "a" is not a integer number?
|
Since:
$$(a+1)(a+2)\cdot\ldots\cdot(a+n)=\frac{\Gamma(a+n+1)}{\Gamma(a+1)}$$
assuming $\Re(a)>1$ we can write the original series as:
$$ S = \sum_{n\geq 1}\frac{\Gamma(n+1)\Gamma(a+1)}{\Gamma(a+n+1)}=\sum_{n\geq 1}n\cdot B(n,a+1)=\sum_{n\geq 1}n\int_{0}^{1}x^{n-1}(1-x)^a\,dx
\tag{1}$$
but $\sum_{n\geq 1}n x^{n-1}=\frac{1}{(1-x)^2}$ gives:
$$ S = \int_{0}^{1}(1-x)^{a-2}\,dx = \int_{0}^{1}x^{a-2}\,dx = \frac{1}{a-1}.$$
In order to prove that the condition $\Re(a)>1$ is necessary for the convergence of the series, just notice that the Euler product for the $\Gamma$ function gives:
$$\frac{\Gamma(n+1)}{\Gamma(n+a+1)}=\Theta\left(\frac{1}{n^a}\right)$$
hence the criterion for the convergence of the generalized harmonic series applies.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/938791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal. Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal.
Part A:
$$T(x,y,z)=\begin{pmatrix} 1 & 1 & 0 \\ 1 & -1 & 0 \\ 1 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z\\ \end{pmatrix}$$
\begin{equation*}
\begin{split}
det(T(x,y,z)-I_n \lambda ) & =\begin{vmatrix} 1-\lambda & 1 & 0 \\ 1 & -1-\lambda & 0 \\ 1 & 0 & 1-\lambda \\ \end{vmatrix} \\
& =(1-\lambda)\begin{vmatrix} -1-\lambda & 0 \\ 0 & 1-\lambda \\ \end{vmatrix}-\begin{vmatrix} 1 & 0 \\ 1 & 1-\lambda \\ \end{vmatrix}=(1-\lambda)^2(-1-\lambda)-(1-\lambda) \\
& =\begin{vmatrix} 1-\lambda & 1 & 0 \\ 1 & -1-\lambda & 0 \\ 1 & 0 & 1-\lambda \\ \end{vmatrix} \\
& =-(1-\lambda)[(1-\lambda)(1+\lambda)+1] \\
& =-(1-\lambda)[1-\lambda^2+1] \\
& =(\lambda-1)[2-\lambda^2]=0 \\
\end{split}
\end{equation*}
Hence $\lambda=1, \pm \sqrt{2}$.
When $\lambda=1$:
$\begin{pmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,1,0)^T, (0,0,1)^T \}$.
When $\lambda=\sqrt{2}$:
$\begin{pmatrix} 1-\sqrt{2} & 1 & 0 \\ 1 & -1-\sqrt{2} & 0 \\ 1 & 0 & 1-\sqrt{2} \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 1-\sqrt{2} \\ 0 & 1-\sqrt{2} & 1-\sqrt{2} \\ 0 & 0 & 0 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,\sqrt{2}-1,0)^T, (1-\sqrt{2},1+\sqrt{2},0)^T \}$.
When $\lambda=-\sqrt{2}$:
$\begin{pmatrix} 1+\sqrt{2} & 1 & 0 \\ 1 & -1+\sqrt{2} & 0 \\ 1 & 0 & 1+\sqrt{2} \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 1+\sqrt{2} \\ 0 & -1+\sqrt{2} & 1+\sqrt{2} \\ 0 & 0 & 0 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,1-\sqrt{2},0)^T, (1+\sqrt{2},1-\sqrt{2},0)^T \}$.
Are my eigenvectors correct? Is there a better way to find see if the two vectors are orthogonal other than using the cross-product method?
|
Eigenvectors solve the equation $(A-\lambda I) v = 0$; eigenvectors are vectors, not matrices.
For $\lambda = 1$, you should have
$$\begin{align*}
(A-I )v &= 0 \\
\begin{pmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{align*}$$
This means that the vector $v$ is in the null space of the matrix $\begin{pmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 1 & 0 & 0 \end{pmatrix}.$ It should be apparent that any vector of the form $\begin{pmatrix} 0 \\ 0 \\ v_3\end{pmatrix}$ satisfies this relation. We shall choose $v_3 = 1$, giving us the eigevector $v = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ corresponding to $\lambda = 1$.
Repeat this process for your other matrices, and be sure to perform the matrix subtraction properly. You should find that the eigenvectors are simply the unit vectors.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/939393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to find this integral $I=\int_{-\pi}^{\pi}\frac{x\cdot \sin(x) \cot^{-1}{(2014^x)}}{1+\cos^4(x)}dx$ Question:
Find this integral
$$I=\int_{-\pi}^{\pi}\frac{x\cdot \sin(x) \cot^{-1}{(2014^x)}}{1+\cos^4(x)}dx$$
let $x\to -x$,so
$$I=\int_{-\pi}^{\pi}\dfrac{x\sin(x) \cot^{-1}{(2014^{-x})}}{1+\cos^4(x)}dx$$
and note
$$\cot\cot^{-1}{2014^x}+\cot\cot^{-1}{2014^{-x}}=\dfrac{\pi}{2}$$
so
$$2I=\dfrac{\pi}{2}\int_{-\pi}^{\pi}\dfrac{x\sin(x)}{1+(\cos{x})^4}dx=\pi\int_{0}^{\pi}\dfrac{x\sin{x}}{1+(\cos{x})^4}dx=\pi^2\int_{0}^{\pi/2}\dfrac{\sin{x}}{1+(\cos{x})^4}dx$$
so
$$I=\dfrac{\pi^2}{2}\int_{0}^{1}\dfrac{1}{1+x^4}dx$$
I feel it's very ugly. Could you please help me using easy methods? It is said that we can use the Gamma function?
|
We have
\begin{equation}
I=\frac{\pi^2}{2}\int_0^1\frac{1}{1+x^4}dx\tag{1}
\end{equation}
Substituting $t=x^4$ yields
\begin{equation}
I=\frac{\pi^2}{8}\int_0^1\frac{t^{-\frac{3}{4}}}{1+t}dt
\end{equation}
According to Part 11. Scientia 18, 2009, 61-75 by Khristo N. Boyadzhiev, Luis A. Medina, and Victor H. Moll, the above integral is defined as the incomplete beta function
\begin{equation}
\beta(a)=\int_0^1\frac{t^{a-1}}{1+t}dt=\frac{1}{2}\left[\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a}{2}\right) \right]
\end{equation}
It can easily be proved by multiplying the integrand by $\dfrac{1-t}{1-t}$ and expanding each of the integrands as a geometric series. The details proof can be seen in the cited paper. So
\begin{align}
I&=\frac{\pi^2}{8}\beta\left(\frac{1}{4}\right)=\frac{\pi^2}{16}\left[\psi\left(\frac{5}{8}\right)-\psi\left(\frac{1}{8}\right) \right]\tag{2}
\end{align}
Both results in $(1)$ and $(2)$ are numerically agreed to 50 digits precision
\begin{equation}
I\approx4.2783402057377873840521430671212757030095999944703
\end{equation}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/942379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Evaluate $\lim_{x\ \to\ \infty}\left(\,\sqrt{\,x^{4} + ax^{2} + 1\,}\, - \,\sqrt{\,x^{4} + bx^{2} +1\,}\,\right)$
I rewrote the function to the form
$$
x^{2}\left(\,
\sqrt{\,1 + \dfrac{a}{x^{2}} + \dfrac{1}{x^{4}}\,}\,-\,
\sqrt{\,1 + \dfrac{b}{x^{2}} + \dfrac{1}{x^{4}}\,}\,\right)
$$
and figured that the answer would be $0$, but apparently this is wrong.
The correct answer is $\displaystyle{{1 \over 2}\left(\,a - b\,\right)}$.
|
Hint:
$$\lim_{x \to \infty} \frac{(\sqrt{x^4 + ax^2 +1} - \sqrt{x^4 + bx^2 +1})(\sqrt{x^4 + ax^2 +1} + \sqrt{x^4 + bx^2 +1})}{\sqrt{x^4 + ax^2 +1} + \sqrt{x^4 + bx^2 +1}}$$
The correct answer is $\frac{1}{2}(a-b)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/942514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
Verify Demorgan's Law Algebraically If $\overline X \equiv \text { not }X$,
De Morgan's Laws are stated as:
*
*$ \overline{(A + B)}= \overline A\cdot \overline B$
*$ \overline{(A\cdot B)} = \overline A + \overline B$
Verify the above laws algebraically.
I can prove this using truth tables and logic gates but algebraically, I don't know any intuitive way to prove it.
$$
\begin{array}{ |c|c| }
\hline
{\text{Axioms}}\\
\hline
\text{Property of 0} & X + 0 = X \space ;\space X\cdot0 = 0 \\
\text{Property of 1} & X + 1 = 1 \space;\space X\cdot1 = X\\
\text{Idempotence Law} & X + X = X \space;\space X\cdot X = X\\
\text{Involution Law} & \overline{\overline X} = X\\
\text{Complementarity Law} & X + \overline X = 1 \space;\space X\cdot\overline X = 0\\
\text{Commutative Law} & X+Y = Y+X \space;\space X\cdot Y = Y\cdot X\\
\text{Associative Law} & (X+Y)+Z = X+(Y+Z) \space;\space (X\cdot Y)\cdot Z = X\cdot (Y\cdot Z)\\
\text{Distributive Law} & X(Y+Z) = XY + XZ \space;\space X + YZ = (X+Y)(X+Z)\\
\text{Absorption Law} & X + XY = X \space;\space X(X+Y) = X\\
\text{Other (3rd Distributive)} & X + \overline XY = X+Y\\
\hline
\end{array}
$$
|
Lemma: $ A + B = 1 \land AB = 0 \implies A = \overline B $
Proof:
Given $ A + B = 1 $, then:
$ (A + B) \overline B = 1 \cdot \overline B = \overline B $
$\implies A \overline B + B \overline B = A \overline B + 0 = A \overline B = \overline B $ (1)
Given $ AB = 0 $, we combine it with (1) to obtain:
$ A \overline B + AB = \overline B + 0 $
$\implies A (\overline B + B) = A \cdot 1 = A = \overline B$ QED
First law:
$ (X + Y) + \overline X \overline Y = Y + (X + \overline X \overline Y) = $
$ = Y + (X + \overline Y) = (Y + \overline Y) + X = $
$ = 1 + X = 1 $ (2)
$ (X + Y) \cdot \overline X \overline Y = X \cdot \overline X \overline Y + Y \cdot \overline X \overline Y = $
$ = (X \overline X) \cdot \overline Y + (Y \overline Y) \cdot \overline X = 0 \cdot \overline Y + 0 \cdot \overline X = $
$ = 0 + 0 = 0 $ (3)
By (2), (3), and the lemma, we have $ \overline { X + Y } = \overline X \overline Y $ -- QED
Second law:
$ \overline { X Y } = \overline { { \overline { \overline X } } \cdot { \overline { \overline Y } } } = $ (by First law)
$ = \overline { \overline { \overline X + \overline Y } } = \overline X + \overline Y $ QED
Notes
*
*Contrary to the approach in the first answer, it is necessary to prove both that $ A + B = 1 $ and $ AB = 0 $ in order to claim that $ A = \overline B $.
*Absorption Law and 3rd Distributive are redundant as axioms, following from Property of 1, Complementarity Law, Idempotence Law, and Associative Law.
*Idempotence Law is redundant as an axiom, following from Property of 1 and Distributive Law
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/943164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Minimum of $x^2+y^2+z^2$ subject to $ax+by+cz=1$
If $ax+by+cz=1$, what is the minimum of $x^2+y^2+z^2$?
It is obvious that we can do Lagrangian multiplier
$$W=x^2+y^2+z^2-\lambda (ax+by+cz-1)$$
|
Using the Cauchy-Schwartz Inequality:
$(x^2+y^2+z^2)\cdot (a^2+b^2+c^2)\geq (ax+by+cz)^2$ , Now Given $(ax+by+cz) = 1$
So $\displaystyle (x^2+y^2+z^2)\geq \frac{1}{(a^2+b^2+c^2)}$ and equality hold when $\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/943838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Proving that $x-\left\lfloor \frac{x}{2} \right\rfloor\; =\; \left\lfloor \frac{x+1}{2} \right\rfloor$ How would you prove that if $x$ is an integer, then
$$x-\left\lfloor \frac{x}{2} \right\rfloor\; =\; \left\lfloor \frac{x+1}{2} \right\rfloor$$
I tried to start by saying that if $x$ is an even integer, then:
$$\left\lfloor \frac{x}{2} \right\rfloor = \frac{x}{2}.$$
However, I am stuck on showing that $$\left\lfloor \frac{x+1}{2} \right\rfloor$$ is also $\frac{x}{2}$. Intuitively It makes intuitive sense just by playing around with some sample numbers, but I don't know how to make it mathematically rigorous. Further, what do you do in the odd case? Is this even the right way to go about it?
|
If $x$ is even, then $x=2k$ where $k \in \mathbb{Z}$.
We then have: $$x-\left\lfloor \frac{x}{2} \right\rfloor=2k-\lfloor k \rfloor=k=\left\lfloor k+\frac{1}{2} \right\rfloor=\left\lfloor \frac{2k+1}{2} \right\rfloor=\left\lfloor \frac{x+1}{2} \right\rfloor$$
If $x$ is odd, then $x=2k+1$ where $k \in \mathbb{Z}$.
We then have $$x-\left\lfloor \frac{x}{2} \right\rfloor=2k+1-\left\lfloor k+{1\over 2} \right\rfloor=k+1=\left\lfloor k+1 \right\rfloor=\left\lfloor \frac{2k+2}{2} \right\rfloor=\left\lfloor \frac{x+1}{2} \right\rfloor$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/945138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Estimating $\int_0^{\sqrt 2 / 2} \sin (x^2) dx$ with Taylor Series I seem to be having trouble with part of this question (Reference: Apostol Volume 1, Section 7.8, Question 8). The full question states:
(a) If $0 \leq x\leq \frac{1}{2}$, show that $\sin x = x - x^3/3! + r(x)$, where $|r(x)| \leq (\frac{1}{2})^5/5!$
(b) Use the estimate in part (a) to find an approximate value for the integral $\int_0^{\sqrt{2}/2} \sin(x^2)\, dx$. Make sure you give an estimate for the error.
I have successfully solved part (a) and the approximate value for the integral in part (b), but it's the error estimate in part (b) that's got me stumped. My working for part (b) has been as follows:
Since $\sin x = x - x^3/3! + r(x)$, we can say $\sin x^2 = x^2 - x^6/3! + r(x^2)$. We already know that
$$
r(x^2) = E_4(x^2) \leq \frac{(x^2)^5}{5!}
$$
so it then follows:
$$
\int_0^{\sqrt{2}/2} \sin(x^2)\, dx = \left(\int_0^{\sqrt{2}/2} x^2 - \frac{x^6}{3!}\, dx\right) + E_4(x^2)
$$
$$
= \left[\frac{x^3}{3} - \frac{x^7}{42}\right]_0^{\sqrt{2}/2}+ E_4(x^2)
$$
$$
= \sqrt{2}(\frac{55}{672}) + E_4(x^2)
$$
Since
$$
E_4(x^2) \leq \frac{(x^2)^5}{5!}
$$
then (AFAIK) the error estimate would be $((\sqrt{2}/2)^2)^5/120$ which is $\approx 2.6\times 10^{-4}$. This doesn't match the answer in Apostol, which gives the error estimate as $\frac{\sqrt{2}}{7680} < 2\times 10^{-4}$.
I'm clearly missing the point somewhere so any pointers in the right direction would be appreciated.
|
Write $\sin x = x - x^3/3! + E(x)$, where $E(x)$ is the error and you have shown that $\lvert E(x) \rvert \leq 1/2^5 5!$ for $0 \leq x \leq \frac{1}{2}$. Equivalently, $\sin x^2 = x^2 - x^6/3! + E(x^2)$, where $\lvert E(x^2) \rvert \leq 1/2^5 5!$ for $0 \leq x \leq 1/\sqrt 2$.
Then
$$\int_0^{1/\sqrt{2}} \sin x^2 dx = \int_0^{1/\sqrt 2} [x^2 - x^6/3!] dx + \int_0^{1/\sqrt 2} E(x^2) dx.$$
You know how to evaluate the first integral. The error comes from the second interval. A good start is to simply try to bound the error, which leads us to try to understand
$$ \left \lvert \int_0^{1/\sqrt 2} E(x^2) dx \right \rvert \leq \int_0^{1/\sqrt 2} \lvert E(x^2) \rvert dx \leq \int_0^{1/\sqrt 2} \frac{1}{2^5 5!} dx = \frac{1}{2^{5.5} 5!} = \frac{\sqrt 2}{2^6 5!}.$$
So we can conclude that the error is bounded by $\sqrt 2 / 2^6 5! = \sqrt 2 / 7880$. $\diamondsuit$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/952139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $ In an acute triangle with angles $ A, B $ and $ C $, show that
$ \cos {A} \cdot \cos {B} \cdot \cos {C} \leq \dfrac{1}{8} $
I could start a semi-proof by using limits: as $ A \to 0 , \; \cos {A} \cos {B} \cos {C} $ becomes big (as we want), but $ A+B+C $ becomes too small. Also, as $ A \to \frac{\pi}{2}, A+B+C \to \pi $ (as we want), but $ \cos {A} \cos {B} \cos {C} \to 0 $.
How can I proceed?
|
NTS $\cos A\cos B\cos C \leq 1/8$
Now, since $ABC$ is a triangle, only one of the angles can possibly be equal or greater than $\pi/2$. So, without loss of generality, let $B, C<\pi/2$. Then $\cos B>0,\cos C>0.$
Case 1: $A\ge\pi/2$
$\Rightarrow \cos A<0 \Rightarrow \cos A\cos B\cos C<0<1/8$
Case 2: $A<\pi/2$ $\Rightarrow \cos A>0$
Then, by the AM-GM inequality:
$$\sqrt[3]{\cos A\cos B\cos C}\leq {\frac 1 3}(\cos A+\cos B+\cos C)$$
Now, consider the function $f(x)=\cos x$ where $x \in(0,\pi/2)$. $f$ is a concave function because $f''(x)=-\cos x>0$ for $x\in(0,\pi/2)$. Then, by the Jenson's inequality: $$f(\frac1 3A +\frac 1 3B+\frac 13 C)\ge\frac 13f(A) +\frac13f(B)+\frac13f(C) \\$$ Then:$$\cos(\frac \pi3)\ge \frac13(\cos A+\cos B+\cos C)$$ Which is: $$\frac12\ge\frac13(\cos A+\cos B+\cos C)$$ Combining that result with the result from the AM-GM inequality, we get: $$\sqrt[3]{\cos A\cos B\cos C}\leq {\frac 1 3}(\cos A+\cos B+\cos C)\le\frac12$$ Cube each portion:$$\cos A\cos B\cos C\leq {\frac 1 {27}}(\cos A+\cos B+\cos C)^3\le\frac18$$
$$\therefore \cos A\cos B\cos C\leq \frac18$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/952893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Integrating $\cos^3(x)$ My attempt at integrating $\cos^3(x)$:
$$\begin{align}\;\int \cos^3x\mathrm{d}x &= \int \cos^2x \cos x \mathrm{d}x
\\&= \int(1 - \sin^2 x) \cos x \mathrm{d}x
\\&= \int \cos x dx - \int \sin^2x \cos x \mathrm{d}x
\\&= \sin x - \frac {1}{3}\sin^3x + C\end{align}$$
My question is how does integrating $\sin^2(x) \cos x $ become $\frac {1}{3}\sin^3(x) + C$? What mathematical process is being done? Why does the $\cos x$ disappear?
|
If by $\sin(x)^2$ you mean $(\sin(x))^2$, we could follow
$$\cos(2x)=\cos(x)^2-\sin(x)^2=1-2\sin(x)^2$$
to have
$$\int \sin(x)^2 = \int \frac{1}{2} - \frac{1}{2}\cos(2x) = \frac{1}{2}x-\frac{1}{4}\sin(2x)$$
Then go ahead and use integration by parts we done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/958586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
}
|
What am I doing wrong? Finding a limit as $x$ approaches $0$ $$\lim_{x\to 0} {a-\sqrt{a^2-x^2}\over x^2} =$$
$${a-\sqrt{a^2-x^2}\over x^2}\cdot{a+\sqrt{a^2-x^2}\over a+\sqrt{a^2-x^2}} = $$
$$a^{2} - a^{2} - x^{2}\over ax^{2} + x^{2}\sqrt{a^{2}-x^{2}} $$
$$-x^{2}\over ax^{2}+x^{2}\sqrt{a^{2}-x^{2}}$$
$$-1\over a+\sqrt{a^{2}-x^{2}}$$
Can anyone tell me where I'm wrong?
|
Should be
$${a^{2} - (a^{2} - x^{2})\over ax^{2} + x^{2}\sqrt{a^{2}-x^{2}}}=
{a^{2} - a^{2} +x^{2}\over ax^{2} + x^{2}\sqrt{a^{2}-x^{2}}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/960015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
the sum of $1-\frac{1}{5}+\frac{1}{9}-\frac{1}{13}+.....$ I thought this was the real part of the series: $\sum_{n=0}^\infty \frac{i^n}{1+2n}$, with $i=\sqrt{-1}$. When taking the real part I am left with: $\sum_{n=0}^\infty \frac{\cos(n\pi/2)}{1+2n}$. I know this sum is around 0.866973, but I have no idea how to come to this answer. Can someone please help me?
|
We can write $$\displaystyle I = 1-\frac{1}{5}+\frac{1}{9}-\frac{1}{13}+......\infty = \int_{0}^{1}\left(1-x^{4}+x^{8}-x^{12}+.......\infty\right)$$
So we get $$\displaystyle I =\int_{0}^{1}\frac{1}{1+x^4}dx=\frac{1}{2}\int_{0}^{1}\frac{(x^2+1)-(x^2-1)}{x^4+1}dx =\frac{1}{2}\int_{0}^{1}\frac{x^2+1}{x^4+1}dx-\frac{1}{2}\int_{0}^{1}\frac{x^2-1}{x^4+1}dx$$
Now Let $$\displaystyle J = \int_{0}^{1}\frac{x^2+1}{x^4+1}dx = \int_{0}^{1}\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+\left(\sqrt{2}\right)^2}dx = \frac{1}{\sqrt{2}}\left[\tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)\right]_{0}^{1}$$
So we get $$\displaystyle J = \frac{1}{\sqrt{2}}\left(0+\frac{\pi}{2}\right) = \frac{\pi}{2\sqrt{2}}$$
Similarly Let $$\displaystyle K = \int_{0}^{1}\frac{x^2-1}{x^4+1}dx = \int_{0}^{1}\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-\left(\sqrt{2}\right)^2}dx =\frac{1}{2\sqrt{2}}\left[\ln\left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|\right]_{0}^{1}$$
So we get $$\displaystyle K = \frac{1}{2\sqrt{2}}\left[\ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)\right] = -\frac{1}{2\sqrt{2}}\left[\ln \left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)\right]=-\frac{1}{\sqrt{2}}\ln(\sqrt{2}+1)$$
So we get $$\displaystyle I = \frac{1}{2}J-\frac{1}{2}K = \frac{\pi}{4\sqrt{2}}+\frac{1}{2\sqrt{2}}\ln(\sqrt{2}+1) = {\frac{\pi+2\ln(\sqrt{2}+1)}{4\sqrt{2}}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/960944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
How many (decimal) digits does $2^{3021 377}$ have? I was wondering, how many (decimal) digits does $2^{3021377}$ have?
We have $2^4=16,\, 2^5=32,\, 2^6=64$ and $2^7=128,\, 2^8=256, \, 2^9=512$ but $2^{10}=1024,\, 2^{11}=2048, \, 2^{12}=4096, \, 2^{13}=8192$, so there is no periodicity in the power.
On the other hand,
\begin{align}
2^{3021377}=& 2^{3000000 + 20000 + 1000 + 300 + 70 +7} \\
=& (2^{10^6})\times{2^3} \times (2^{10^4})\times{2^2} \times 2^{10^3} \times ( 2^{10^2})\times{2^3} \times (2^{10})\times{2^7} \times2^7 \\
\end{align}
using the fact $2^{10}=1024$. Can we deduce the number of digits from this? Mathematica gives:
IntegerLength[2^{3021377}]
{909526}
|
$$\left\lfloor 3021377\log_{10}2\right\rfloor +1 $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/962177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How do I solve an equation with three terms, with the unknown inside a square root, inside a third root, in two of them? The equation is $$\\ \sqrt[3]{\sqrt{a}+b}+\sqrt[3]{-\sqrt{a}+b}=k.$$
How do I find$\ a$?
|
$$k=(b+c)^{1/3}+(b-c)^{1/3}$$ where $c=\sqrt a$. We use $(x+y)^3=x^3+y^3+3xy(x+y)$.
$$k^3=b+c+b-c+3(b^2-c^2)^{1/3}k=2b+3k(b^2-c^2)^{1/3}$$
Now subtract $2b$, divide by $3k$, cube, subtract from $b^2$, and you have $c^2$, which is $a$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/962924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Stereographic projection proof that is geometrical. Given tangents $VP$ and $ZP$ on circle intersecting at $P$.
Prove: $YX=XW$
Heres what I do know. Obviously segments $ON, OV, OS$ are all congruent since they are just radii. I also know that $VP$ and $ZP$ are congruent since they are tangent lines intersecting. There is also an isosceles triangle $ONV$ since two of the sides are just radii which means angles $ONV$ and $NVO$ are congruent. I also know $OVZ$ is an isosceles triangle. Based off of all the information I have figured out I am not sure how to use it to show that $YX = XW$. Thoughts? Ideas? Solutions?
|
This is a coordinate-based proof. Choose (w.l.o.g.) the following coordinates for your points:
\begin{align*}
N&=\begin{pmatrix}0\\1\end{pmatrix} &
V&=\frac{1}{a^2+1}\begin{pmatrix}2a\\a^2-1\end{pmatrix}\\
S&=\begin{pmatrix}0\\-1\end{pmatrix}&
Z&=\frac{1}{b^2+1}\begin{pmatrix}2b\\b^2-1\end{pmatrix}
\end{align*}
This is using a rational parametrization of the unit circle, i.e. using the tangens of half the angle as the parameter. Then you have the tangents
$$2ax + (a^2-1)y = a^2+1\qquad 2bx + (b^2-1)y = b^2+1$$
which intersect in
$$P=\frac1{ab+1}\begin{pmatrix}a+b\\ab-1\end{pmatrix}$$
The stereographic projections of these points are
\begin{align*}
W&=\begin{pmatrix}2a\\-1\end{pmatrix} &
Y&=\begin{pmatrix}2b\\-1\end{pmatrix} &
X&=\begin{pmatrix}a+b\\-1\end{pmatrix} &
\end{align*}
So you can see that they are equidistant, as claimed.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/964460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
The integral of $x^3/(x^2+4x+3)$ I'm stumped in solving this problem. Every time I integrate by first dividing the $x^3$ by $x^2+4x+3$ and then integrating $x- \frac{4x^2-3}{x+3)(x+1)}$ using partial fractions, I keep getting the wrong answer.
|
Hint
If you start doing the long division, you should arrive to $$\frac{x^3}{x^2+4 x+3}=x-4+\frac{13 x+12}{x^2+4 x+3}$$ For the last term, decompose in simple fractions and get $$\frac{x^3}{x^2+4 x+3}=x-4-\frac{1}{2 (x+1)}+\frac{27}{2 (x+3)}$$
I am sure that you can take from here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/964731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to solve this exponential equation? $2^{2x}3^x=4^{3x+1}$. I haven't been able to find the correct answer to this exponential equation:
$$\eqalign{
2^{2x}3^x&=4^{3x+1}\\
2^{2x} 3^x &= 2^2 \times 2^x \times 3^x\\
4^{3x+1} &= 4^3 \times 4^x \times 4\\
6^x \times 4 &= 4^x \times 256\\
x\log_6 6 + \log_6 4 &= x\log_64 + \log_6 256\\
x + \log_6 4 &= x\log_64 + \log_6 256\\
x-x\log_6 4 &= \log_6 256 - \log_6 4\\
x(1-\log_6 4) &= \log_6 256 - \log_6 4\\
x &= \dfrac{\log_6 256 - \log_6 4}{1-\log_6 4}\\
x &= 10.257}$$
so when I checked the answer I wasn't able to make them equal, I have tried variants of this method but I feel I'm missing something..
|
Your second step goes wrong
$2^{2x}=2^x+2^x$ but not $2^2*2^x$
simpler way
$3^x=(2^2)^{3x+1}/2^{2x}$
$3^x=2^{6x+2-2x}$
$3^x=2^{4x+2}$
Then apply log to the base e on both sides and use your calculator.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/964962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
}
|
Simplification of geometric series. Can someone please help simplify this series?
$$\sum_{k=1}^\infty k\left(\frac 12\right)^k$$
In general, $$\sum_{k=1}^\infty\left(\frac 12\right)^k = \frac{1}{1-\frac{1}{2}} =2.$$
However, I am confused with the $k$ in front of the term $k\big(\frac 12\big)^k$.
I understand if the problem is $\sum_{k=1}^\infty\big(\frac 12\big)^k$. However, it is the $k$ term that I don't understand. The answer is suppose to be $2$.
Can someone help me? Thank you.
|
We can try to solve like a geometric series:
$$
S_n = \frac{1}{2} + 2\frac{1}{2^2} + 3\frac{1}{2^3} + \dots + (n - 1)\frac{1}{2^{n - 1}} + n\frac{1}{2^n} \\
\frac{1}{2}S_n = \frac{1}{2^2} + 2\frac{1}{2^3} + 3\frac{1}{2^4} + \dots + (n - 1)\frac{1}{2^{n}} + n\frac{1}{2^{n+1}} \\
S_n - \frac{1}{2}S_n = \frac{1}{2} + (2 - 1)\frac{1}{2^2} + (3 - 2)\frac{1}{2^3} + \dots + (n - (n - 1))\frac{1}{2^n} - n\frac{1}{2^{n+1}}
$$
This gives:
$$
S_n - \frac{1}{2}S_n = \left(\sum_{i = 1}^{i = n} \frac{1}{2^i}\right) - \frac{n}{2^{n+1}} = \frac{1}{2}\left(\sum_{i = 0}^{i = n - 1} \frac{1}{2^i}\right) - \frac{n}{2^{n+1}}
$$
We know the geometric sum: $\lim_{n\rightarrow\infty}\sum_0^n \frac{1}{2^i} = \frac{1}{1 - \frac{1}{2}} = 2$, which gives:
$$
\lim_{n\rightarrow\infty} S_n = \frac{\frac{1}{2}\cdot 2}{1 - \frac{1}{2}} = 2
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/967410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Use row reduction to show that the determinant is equal to this variable. Show
determinant of:
\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}
is equal to $(b - a)(c - a)(c - b)$
I'm not sure if you can use squares or square roots hmmm.. please help me. I'm sure it's a simple question. Much appreciated.
|
We see immmediatley that the determinant is a polynomial in $a,b,c$, homgenous of degree $3$. And if $a=b$ or $b=c$ or $c=a$, two columns are identical and make the determiniant zero.
On the other hand, if $a,b,c$ are pairwise distinct, the determinant must be nonzero as any vanishing linear combination of rows would result in a quadratic polynomial having three roots $a,b,c$.
If these arguments are not convincing enough, just employ Gauss's algorithm:
$$\begin{align}\det\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}
&=\det\begin{pmatrix}1&1&1\\0&b-a&c-a\\0&b^2-a^2&c^2-a^2\end{pmatrix}\\
&=\det\begin{pmatrix}1&1&1\\0&b-a&c-a\\0&0&c^2-a^2-(b+a)(c-a)\end{pmatrix}\\
&=(b-a)\cdot(c^2-a^2-(b+a)(c-a))\\
&=(b-a)(c-a)(c+a-(b+a))\\&=(b-a)(c-a)(c-b) \end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/969304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Solve $\sqrt{3x}+\sqrt{2x}=17$ This is what I did:
$$\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3}\sqrt{x}+\sqrt{2}\sqrt{x}=17$$
$$\implies\sqrt{x}(\sqrt{3}+\sqrt{2})=17$$
$$\implies x(5+2\sqrt{6})=289$$
I don't know how to continue. And when I went to wolfram alpha, I got:
$$x=-289(2\sqrt{6}-5)$$
Could you show me the steps to get the final result?
Thank you.
|
$$x(5+2\sqrt{6})=289\\
\Rightarrow x=\frac{289}{(5+2\sqrt{6})}\\
\Rightarrow x=\frac{289(5-2\sqrt{6})}{(5+2\sqrt{6})(5-2\sqrt{6})}\\
\Rightarrow x=\frac{289(5-2\sqrt{6})}{25-24}\\
\Rightarrow x=-289(2\sqrt{6}-5)\\$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/977429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\begin{align*}
(a + ib)^3 &= 8\\
a + \sqrt{4-a^2} &= 2\\
\sqrt{4-a^2} &= 2 - a\\
2 - a &= 2 - a
\end{align*}
$$
Therefore if $(a + ib)^3 = 8$, then $a^2 + b^2 = 4$.
|
Hint: The question asks: If $z^3=8$, show that $|z|^2=4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/979252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 6
}
|
Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and similarly for $b,c$. Hence it suffices to prove that $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$$. From $a+b+c=1$ and $a,b,c>0$ we have $0<a,b,c<1$, so we have $$\frac{a}{1-a}=a+a^2+a^3+...$$ and similarly for $b,c$, so it suffices to prove that $$\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}$$, or equivalently (by $a+b+c=1$) $$\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}$$, where $\sum_{cyc} a=a+b+c$ similarly for $\sum_{cyc}a^n=a^n+b^n+c^n$. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the $a^2+b^2+c^2\ge (a+b+c)^2/3$ inequality.
"stuff below": Now, from $0<a<1$ we have $a^3>a^4$, $a^5>a^8$, $a^7>a^8$, $a^9>a^{16}$, and so on, so it suffices to prove that $$\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}$$, or, multiplying by 2, $$2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1$$, which by a simple inequality (i.e. recursively using $a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}$) is equivalent to $$2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1$$. But then http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B5859879%7D+2%5Ei+*%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed.
|
Let S = $\frac{a}{b+c}$ + $\frac{b}{c+a}$ + $\frac{c}{a+b}$
M = $\frac{b}{b+c}$ + $\frac{c}{c+a}$ + $\frac{a}{a+b}$
N = $\frac{c}{b+c}$ + $\frac{a}{c+a}$ + $\frac{b}{a+b}$
Now M + N = 3.
M + S >= 3 (AM-GM Property)
N + S >= 3 (AM-GM Property)
M + N + 2S >= 6 (Adding above inequalities)
2S >= 3
S >= $\frac{3}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/980751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
}
|
Solve differential equation, $x'=x^2-2t^{-2}$ Solve differential equation:
$x'=x^2- \frac{2}{t^2}$
Maybe is it sth connected with homogeneous equation?
I have no idea how to solve it.
|
Let $x=\dfrac{1}{u}$ ,
Then $x'=-\dfrac{u'}{u^2}$
$\therefore-\dfrac{u'}{u^2}=\dfrac{1}{u^2}-\dfrac{2}{t^2}$
$u'=\dfrac{2u^2}{t^2}-1$
Let $v=\dfrac{u}{t}$ ,
Then $u=tv$
$u'=tv'+v$
$\therefore tv'+v=2v^2-1$
$tv'=2v^2-v-1$
$t\dfrac{dv}{dt}=(2v+1)(v-1)$
$\dfrac{dv}{(2v+1)(v-1)}=\dfrac{dt}{t}$
$\int\dfrac{dv}{(2v+1)(v-1)}=\int\dfrac{dt}{t}$
$\int\left(\dfrac{1}{3(v-1)}-\dfrac{2}{3(2v+1)}\right)dv=\int\dfrac{dt}{t}$
$\dfrac{\ln(v-1)-\ln(2v+1)}{3}=\ln t+c_1$
$\ln\dfrac{v-1}{2v+1}=3\ln t+c_2$
$\dfrac{v-1}{2v+1}=Ct^3$
$v=\dfrac{1+Ct^3}{1-2Ct^3}$
$\dfrac{1}{xt}=\dfrac{1+Ct^3}{1-2Ct^3}$
$x=\dfrac{1-2Ct^3}{t(1+Ct^3)}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/982149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
|
Integral of rational function with higher degree in numerator How do I integrate this fraction:
$$\int\frac{x^3+2x^2+x-7}{x^2+x-2} dx$$
I did try the partial fraction decomposition:
$$\frac{x^3+2x^2+x-7}{x^2+x-2} = \frac{x^3+2x^2+x-7}{(x-1)(x+2)}$$
And:
$$\frac{A}{(x-1)}+\frac{B}{(x+2)}=\frac{A(x+2)}{(x-1)(x+2)}+\frac{B(x-1)}{(x-1)(x+2)}$$
Then:
$$\ A(x+2) + B(x-1)= x^3+2x^2+x-7$$
When I do this I gets a wrong answer, so I figured out that this may only work when the degree of the denominator is greater than the degree of the numerator...
So how do I integrate such a fraction?
|
You need to use polynomial long division, first, so the degree in the numerator is less than that of the denominator to get
$$I = \int \left(x+1 + \dfrac{2x-5}{x^2 + x-2}\right)\,dx$$
THEN you can use partial fraction decomposition given the factors you found for the denominator.
$$I = \frac {x^2}{2} + x + \left( I_2 = \int \frac{(2x-5)\,dx}{(x+2)(x-1)}\right)$$
$$I_2 = \int \left(\dfrac A{x+2} + \frac B{x-1}\right) \,dx$$
So we want to solve for the coefficients $A, B$, given the fact that
$$A(x-1) + B(x+2) = 2x - 5$$
At $x = 1,\quad A(0) + B(3) = 2-5=-3 \implies B = -1$
At $x = -2, \quad -3A = -9 \implies A = 3$.
Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/982391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
}
|
Base conversion between base r^3 to r^2 The problem asks the following:
If $(\alpha2)_{r^3} = (r3\beta)_{r^2},$ find $\alpha, \beta,$ and r.
First off, I assume $r^2$ > 3 and $r^3 > 2$.
I know that $(r3\beta)_{r^2}$ = (10 _ _ _ _ )$_{r}$
I've attempted to convert to base 10 and set them equal to each other:
$(\alpha2)_{r^3} = \alpha ({r^3})^1 + 2({r^3})^0 $
= $ (\alpha {r^3} + 2)_{10}$
$(r3\beta)_{r^2} = r({r^2})^2 + 3({r^2})^1 + \beta ({r^2})^0 = (r({r^4}) + 3{r^2} + \beta)_{10}$
and converting to base r:
$\alpha = Mr^2 + Nr + O$
$ 2 = Xr^2 + Y$
$ r = 1r + 0$
$ 3 = Sr + T$
$ \beta = Vr + U$
But I'm pretty much stuck at this point. Don't see any logical progression from here (if I'm even headed in the right direction).
Any help is greatly appreciated.
|
Let:
\begin{align*}
\alpha &= Ar^2 + Br + C \\
2 &= Dr^2 + Er + F \\
3 &= Gr + H \\
\beta &= Ir + J
\end{align*}
where each uppercase coefficient is some integer in $\{0, 1, \ldots, r - 1\}$. Then by substituting into our expressions in base $10$, we have that:
$$
(Ar^2 + Br + C)r^3 + (Dr^2 + Er + F) = r^5 + (Gr + H)r^2 + (Ir + J)
$$
Comparing coefficients, we have:
\begin{align*}
\boxed{r^5}:\quad A &= 1 \\
\boxed{r^4}:\quad B &= 0 \\
\boxed{r^3}:\quad C &= G \\
\boxed{r^2}:\quad D &= H \\
\boxed{r^1}:\quad E &= I \\
\boxed{r^0}:\quad F &= J \\
\end{align*}
Now let's assume that $r \geq 2$. Then $2 \geq Dr^2 \geq 4D$, which implies that $D = 0$. Hence, $H = 0$ so that $3 = Gr$. But then since $3$ is prime and $r \neq 1$, we have that $G = 1$ and $r = 3$.
Now since $G = 1$, we know that $C = 1$ so that $\alpha = 1(3)^2 + 0(3) + 1 = 10$.
Likewise, since $r = 3$, we know that $D = E = H = I = 0$ and $F = J = 2$ so that $\beta = 0(3) + 2 = 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/983114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $a$ cannot be a prime The sides of a triangle are of length $a,b,c$ where $a,b,c$ are integers and $a>b$, angle opposite to $c$ is $60$ degrees. Prove that $a$ Cannot be a prime
|
By the Law of Cosines, we have
$$
c^2=a^2+b^2-2ab\cos(\angle ACB)=a^2+b^2-ab\implies(c+b)(c-b)=a(a-b).
$$
Now, suppose $a$ is prime. Then, either $a|c+b$ or $a|c-b$. But by the triangle inequality, $a>c-b$ so we must have
$$
a|c+b.
$$
Next, because the angle opposite $c$ is $60$ degrees, either of the remaining angles is at least as large as $60$. Thus, by the Law of Sines, we can infer that $c\leq\max\{a,b\}=a$. This, together with $b<a$, implies that $c+b<2a$. But $a|c+b$ so we must have $a=c+b$, violating the triangle inequality and giving us the desired contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/983781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $A^{10}$ is equal to linear combination of $A^k, k = 1,...,9$ and identity matrix. Let $A=\begin{bmatrix}2&0&1\\0&1&0\\1&0&1\end{bmatrix}$.
I did this with brute force and it was messy, is there a more theoretical way?
|
The characteristic polynomial of $A$ is
$$P_A(t)= \det (t I_3 - A) = t^3- 4 t^2 + 4 t -1$$
The matrix $A$ satisfies the polynomial equality
$$P_A(A) = A^3 - 4A^2 + 4A - 1 =0$$
and therefore $A^3 = 4A^2 - 4A + 1$ and so $A^{10} = A^7(2A-1)^2$, a polynomial of degree $9$.
We can find a unique polynomial $R(t)$ of degree $2$ that satisfies $A^{10}= R(A)$; $R(t)$ is the remainder of the division of $t^{10}$ by $P_A(t)$. We have
$$t^{10} = ( t^3- 4 t^2 + 4 t -1) ( t^7+4 t^6+12 t^5+33 t^4+88 t^3+232 t^2+609 t+1596)+\\ +\,4180 t^2-5775 t+1596$$
and so
$$A^{10} = 4180\, A^2-5775\, A+1596$$
Obs: $4180 t^2-5775 t+1596$ is the unique polynomial of degree $\le 2$ that coincides with $t^{10}$ on the set of zeroes $\{1, \frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}\}$ of the polynomial $P_A(t)$ -- the eigevalues of $A$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/985341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Why is $x^5 \sin x$ an odd function? Why is $x^5 \sin x$ an odd function?
Is the result just wrong? Because $f(-x)= (-x)(-x)(-x) \sin(-x) = (-x)(-x)(-x)(-x)(-x) (-\sin x) = (-x^5)(-\sin x) = x^5 \sin x$
|
If $f(x) = x^5 \sin x$, then
$$f(-x) = (-x)^5 \sin (-x) = -x^5 \left( - \sin x\right) = x^5 \sin x = f(x),$$
and so $f$ is actually even.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/988264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
$\lim_{n\rightarrow \infty}n^{-\left(1+1/n\right)/2}\times \left(1^1\times 2^2\times 3^3\times\cdots\times n^n\right)^{1/{n^2}}$
Evaluate the limit
$$
y=\lim_{n\rightarrow \infty}n^{-\left(1+1/n\right)/2}\times \left(1^1\times 2^2\times 3^3\times\cdots\times n^n\right)^{1/{n^2}}
$$
My Attempt:
When $n\rightarrow \infty$, then $n^{-\left(1+1/n\right)/2}\rightarrow n^{-1/2}$. So the limit becomes
$$
y=\lim_{n\rightarrow \infty}n^{-1/2}\times \left(1^1\times2^2\times3^3\times\cdots\times n^n\right)^{1/{n^2}}
$$
Taking $\ln()$ on both side, we get
$$
\begin{align}
y&=\lim_{n\rightarrow \infty}\ln\left\{n^{-1/2}\times\left(1^1\times2^2\times3^3\times\cdots\times n^n\right)^{1/{n^2}}\right\}\\
&=\lim_{n\rightarrow \infty}\left\{-\frac{1}{2}\ln(n)+\frac{1}{n^2}\ln\left(1^1\times2^2\times3^3\times\cdots\times n^n\right)\right\}
\end{align}
$$
How can I complete the solution from this point?
|
As already commented by Lucian, $$\prod_{k=0}^n k^k=H(n)$$ which is the hyperfactorial function which also has a Stirling-like series $$H(n)=A e^{-\frac{n^2}{4}} n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}} \left(1+\frac{1}{720
n^2}-\frac{1433 }{7257600 n^4}+O\left(\left(\frac{1}{n}\right)^6\right)\right)$$ where $A$ is the Glaisher-Kinkelin constant.
Using it, the expression you are dealing with has then an asymptotic expansion $$\frac{1}{\sqrt[4]{e}}+\frac{12 \log (A)+\log (n)}{12
\sqrt[4]{e} n^2}+O\left(\left(\frac{1}{n}\right)^{7/2}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/989873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Finding $\sum \frac{1}{n^2+7n+9}$ How do we prove that $$\sum_{n=0}^{\infty} \dfrac{1}{n^2+7n+9}=1+\dfrac{\pi}{\sqrt {13}}\tan\left(\dfrac{\sqrt{13}\pi}{2}\right)$$
I tried partial fraction decomposition, but it didn't work out after that. Please help me out. Hints and answers appreciated. Thank you.
|
Start with the infinite product expansion of $\cos x$
$$\cos x = \prod_{k=0}^\infty \left(1 - \frac{x^2}{(k+\frac12)^2\pi^2}\right)$$
Taking logarithm of $\cos(\pi x)$ and differentiate, we have
$$-\pi\tan(\pi x) = \sum_{k=0}^\infty \frac{2x}{x^2-(k+\frac12)^2}
\quad\implies\quad
\sum_{k=0}^\infty\frac{1}{(k+\frac12)^2-x^2} = \frac{\pi}{2x}\tan(\pi x)
$$
This lead to
$$
\sum_{k=0}^\infty \frac{1}{k^2+7k+9}
= \sum_{k=0}^\infty \frac{1}{(k+\frac72)^2 - \frac{13}{4}}
= \sum_{k=3}^\infty \frac{1}{(k+\frac12)^2 - \frac{13}{4}}\\
= \frac{\pi}{2\cdot\frac{\sqrt{13}}{2}}\tan\left(\frac{\pi\sqrt{13}}{2}\right) - \sum_{k=0}^2 \frac{1}{(k+\frac12)^2 - \frac{13}{4}}
= 1 + \frac{\pi}{\sqrt{13}}\tan\left(\frac{\pi\sqrt{13}}{2}\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/990241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.