Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculating the length of a curve for $ y=\ln (1-x^2) $ In calculating the length of $ y=\ln (1-x^2) $ on the interval $ [0, {3\over4}]$, I found that for whatever reason, I end up with invalid domains in the indefinite integral using the given limits of integration.
Using the formula $ \int^b_a \sqrt{1+f'(x)^2} $, I was able to derive
$$ {dy\over dx}=f'(x)=-{2x\over1-x^2} $$
$$ \int^{3\over4}_0 \sqrt{1+(-{2x\over1-x^2})^2}\ dx = \int^{3\over4}_0 \sqrt{1+{4x^2\over1-2x^2+x^4}}\ dx = \int^{3\over4}_0 \sqrt{{x^4+2x^2+1}\over{x^4-2x^2+1}}\ dx $$
$$ = \int^{3\over4}_0 {{x^2+1}\over{x^2-1}}\ dx $$
as the integrand in question. Then by the rule of the sums of integrals:
$$ \int^{3\over4}_0 {{x^2}\ dx\over{x^2-1}} + \int^{3\over4}_0 {{dx}\over{x^2-1}}$$
The clearest line of attack was through the following trigonometric substitution:
$$ x=\sec \theta $$
$$ dx=\sec \theta \tan \theta\ d\theta$$
$$ \int {{\sec^3 \theta \tan \theta\ d\theta}\over{\tan^2 \theta}} + \int {{\sec \theta \tan \theta}\over{\tan^2 \theta}} = \int \sec^2 \theta \csc \theta\ d\theta + \int \csc \theta\ d\theta$$
The former I integrated by parts:
**I.**$$ u=\csc \theta\ \quad dv=\sec^2 \theta\ d\theta$$
$$ du=-\csc \theta \cot \theta\ d\theta\ \quad v=\tan \theta$$
$$ \int \sec^2 \theta \csc \theta\ d\theta = \boxed{\csc \theta \tan \theta + \int \csc \theta\ d\theta}$$
And the latter:
**II.**$${\int \csc \theta\ d\theta= \boxed{-\ln \lvert{\csc \theta + \cot \theta}\rvert}}$$
Hence the complete indefinite integral:
**I.+II.**$$ \int \sec^2 \theta \csc \theta\ d\theta + \int \csc \theta\ d\theta = \boxed{\csc \theta \tan \theta -2 \ln \lvert{\csc \theta + \cot \theta}\rvert}$$
Transforming back to the original variable:
$$ x=\sec \theta \quad\quad \sqrt{x^2-1} = \tan \theta$$
$$ {1\over\sqrt{x^2-1}} = \cot \theta \quad\quad {x\over\sqrt{x^2-1}} = \csc \theta$$
$$ {\csc \theta \tan \theta -2 \ln \lvert{\csc \theta + \cot \theta}\rvert} = \boxed{x - 2\ln \lvert{{x+1}\over{\sqrt{x^2-1}}}\rvert} $$
As one can easily see, for whatever reason the limits produce a negative term under the radical in the natural log. Any insight as to why this is the case would be much appreciated.
| ..........................................................................
The hitch is here :
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I prove remainder side of this inequality? Let $x$, $y$ be two positive numbers such that $x^4+y^4=x^2+y^2$. Prove that
$$1\leqslant x+y\leqslant 2.$$
With $x+y\leqslant 2$. I tried
We have $$x^4+y^4\geqslant \dfrac{(x^2+y^2)^2}{2}.$$
Therefore, $$x^2+y^2\geqslant \dfrac{(x^2+y^2)^2}{2}$$
or
$$(x^2+y^2)^2-2(x^2+y^2)\leqslant 0$$ Implies $x^2+y^2\leqslant 2.$
Another way
$$\dfrac{x+y}{2} \leqslant \sqrt{\dfrac{x^2+y^2}{2}}=1.$$
Thus $x+y\leqslant 2.$
How can I prove $x+y\geqslant 1$?
Is this true?
Let $x$, $y$ be two positive numbers such that $x^m+y^m=x^n+y^n$, where $m$, $n$, ($m \neq n$) be two positive integer numbers, we have
$$ x+y\leqslant 2.$$
| If $x+y\lt 1$ then $(x+y)^2\lt 1$, and therefore $x^2+y^2\lt 1$.
But from $0\lt x^2+y^2\lt 1$, we conclude that $(x^2+y^2)^2 \lt x^2+y^2$, and therefore $x^4+y^4\lt x^2+y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$n^{th}$ derivative of $y=x^2\cos x$ I am stuck with Leibniz formula
$$D^{n}y = \sum_{k=0}^{n} \binom{n}{k} \, x^{(2k)}\cos^{(n-k)}x$$
Could someone show how to do it?
| The Leibniz formula only has a few terms. This is seen by the following.
\begin{align}
D^{n} y &= D^{n}\{ x^{2} \, \cos(ax)\} = \sum_{k=0}^{n} \binom{n}{k} \, D^{k}\{x^{2}\} \, D^{n-k}\{\cos(ax)\}
\end{align}
Since $D(x^{2}) = 2 x$, $D^{2}(x^{2}) = 2$, and $D^{3+m}(x^{2}) = 0$ for $m \geq 0$, then
\begin{align}
D^{n} y &= \binom{n}{0} \, D^{0}(x^{2}) \, D^{n}(\cos(ax)) + \binom{n}{1} \, D^{1}(x^{2}) \, D^{n-1}(\cos(ax)) + \binom{n}{2} \, D^{2}(x^{2}) \, D^{n-2}(\cos(ax)) \\
&= x^{2} \, D^{n}(\cos(ax)) + 2 n x \, D^{n-1}(\cos(ax)) + n(n-1) \, D^{n-2}(\cos(ax))
\end{align}
For the case of $n$ being even then
\begin{align}
D^{2n} y &= (-1)^{n} \, a^{2n-2} \, \left[(a^{2} \, x^{2} - 2n(2n-1)) \, \cos(ax) + 4 a n x \, \sin(ax) \right]
\end{align}
and for $n$ being odd
\begin{align}
D^{2n+1} y &= (-1)^{n+1} \, a^{2n-1} \, \left[ (a^{2} x^{2} + 2n(2n+1)) \, \sin(ax) - 2(2n+1) a x \cos(ax) \right]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How to solve the trigonometric equation $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$ the question is:
Find the solutions of the equation: $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$.
Let $\sin x+\cos x=u \text{ and } \sin x \cos x=v \implies \sin^2x+\cos^2x+2\sin x\cos x=u^2 \implies v=\frac {u^2-1} 2$
similarly solving the above equation it comes out to be:
$$\sqrt2 u^2-u-\sqrt2=0=(u-\sqrt2)(\sqrt2u+1)=0 \implies \sin x+\cos x=\sqrt2 \quad(1)$$
and
$$\sin x+\cos x= \frac{-1} {\sqrt2}\quad (2)$$
so solving the results differently i got the answers:
$$x=2n\pi + \frac{\pi}4, 2n\pi +\frac{7\pi}{12}, 2n\pi-\frac{\pi}{12}$$
but the answers are:
$$x=2n\pi + \frac{\pi}4, 2n\pi -\frac{5\pi}{12}, 2n\pi+\frac{11\pi}{12}$$
I divided the eq(2) by $\sqrt2$
| my hints: $$\sin \theta=\sin \alpha\iff \theta=2k\pi+\alpha\ \text{or}\ \theta =(2k+1)\pi-\alpha$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve a complex equation Solve the following equation $$(4-3i)z^2-25z+31-17i= 0 $$
Dividing by 4-3i gives me $$z^2 \frac{-100z-75zi + 124 + 93i -68i -51i^2}{25}$$
which goes to $$z^2 -4z-3zi + 7+i$$
then i collect the terms so $$z - \left(\frac{(4-3i)}{2}\right)^2 = -7 -i + \left(\frac{4-3i}{2}\right)^2$$
and after that i can't get the expansion to work.
Can you help me out?
| Notice, $$(4-3i)z^2-25z+31-17i=0$$ Solving the above quadratic equation for $z$ as follows $$z=\frac{-(-25)\pm\sqrt{(-25)^2-4(4-3i)(31-17i)}}{2(4-3i)}$$
$$z=\frac{-(-25)\pm\sqrt{(-25)^2-4(4-3i)(31-17i)}}{2(4-3i)}$$ $$z=\frac{(4+3i)(25\pm\sqrt{333+644i})}{2(16+9)}$$
$$z=\frac{(4+3i)(25\pm\sqrt{333+644i})}{50}$$
$$z=\frac{(4+3i)\left(25\pm\sqrt{725}\left(\cos \frac{\alpha}{2}+i\sin\frac{\alpha}{2}\right)\right)}{50}$$ Where, $\alpha=\cos^{-1}\left(\frac{333}{725}\right)=\sin^{-1}\left(\frac{644}{725}\right)$
I hope you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Solve trigonometric equation $ 3 \cos x + 2\sin x=1 $ Solve trigonometric equation: $$ 3 \cos x + 2\sin x=1 $$
I tried to substitue $\cos x = \dfrac{1-t^2}{1+t^2}, \sin x = \dfrac{2t}{1+t^2}$. Yet with no results.
| If you have a sum $S=a\cos x+b \sin x$
define $r$ by $r^2=a^2+b^2$
and $\theta$ by $r\sin \theta =a$ and $r\cos \theta =b$ so that $\tan \theta =\cfrac ab$
Then $S=r\sin \theta \cos x+r\cos \theta \sin x=r\sin (x+\theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve the binomial equation Solve the binomial equation $$z^4 = -8$$
Below is the steps i have done
1: I have taken |-8| that is 8 and then done 8^(1/4) which is 2^(1/4).
2: Since $z=r(cos\alpha+isin\alpha)$ leads me to
$r^4(cos4\alpha+isin4\alpha)=-8(cos\pi/2+isin\pi/2)$
Divide by 4 since the z term is raised by four gives $2^{1/4}(cos\pi/8 + k * \pi/2 +sin\pi/8 + k * \pi/2) $
Is this the correct way to solve this problem ? I am asking since i just started with binomic equations and been stuck for some hours with the question.
| $$ z^4 = -8 \Longleftrightarrow $$
$$ z^4 = |-8|e^{\arg(-8)i} \Longleftrightarrow $$
$$ z^4 = 8e^{\pi i} \Longleftrightarrow $$
$$ z^4 = 8e^{\left( \pi + 2 \pi k \right) i} \Longleftrightarrow $$
$$ z = \left( 8e^{\left( \pi + 2 \pi k \right) i} \right)^{\frac{1}{4}} \Longleftrightarrow $$
$$ z = \sqrt[4]{8}e^{\frac{1}{4}\left( \pi + 2 \pi k \right) i} \Longleftrightarrow $$
$$ z = \sqrt[4]{8}e^{\left( \frac{\pi}{4} + \frac{\pi k}{2} \right) i} \Longleftrightarrow $$
$$ z = \sqrt[4]{8}e^{\frac{\pi(2k+1)}{4} i} $$
With $k \in \mathbb{Z}$ and $k$ goes from $0-3$
So the solutions are:
$$z_0=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 0+1)}{4} i}=\sqrt[4]{8}e^{\frac{\pi}{4}i}$$
$$z_1=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 1+1)}{4} i}=\sqrt[4]{8}e^{\frac{3\pi}{4}i}$$
$$z_2=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 2+1)}{4} i}=\sqrt[4]{8}e^{-\frac{3\pi}{4}i}$$
$$z_3=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 3+1)}{4} i}=\sqrt[4]{8}e^{-\frac{\pi}{4}i}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solve trigonometric inequality $\cos x \geq \sin^2 x - \cos^2 x $ Solve trigonometric inequality $$\cos x \geq \sin^2 x - \cos^2 x $$
My incorrect solution:
$$\cos^2 x-\sin^2 x \geq -\cos x $$
$$\cos 2x \geq \cos (\pi - x) $$
which means:
$$ 2x \geq -(\pi + x)$$
$$ x \geq -\pi $$
Which is wrong.
And
$$ 2x \leq 2\pi + (\pi - x)$$
$$ x \leq \pi$$
| \begin{align*}
\cos^2x − \sin^2x & \geq −\cos x\\
\cos^2x − \cos^2x-1 & \geq −\cos x\\
2\cos^2x + \cos x + 1 & \geq 0
\end{align*}
solving for cosx
$$\cos x = [-\infty,1] \cup \left[\frac{1}{2},1\right]$$
rejecting 1st part
$$\cos x \geq \frac{1}{2}$$
$$x \leq \frac{\pi}{3}$$
$$x \geq -\frac{\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Advanced Algebra Manipulation/Inequality Proof: $\frac{4x^3(x^2+y^2)-2x(x^4+y^4)}{(x^2+y^2)^2} \leq 6|x|$ I need to show that
$$\frac{4x^3(x^2+y^2)-2x(x^4+y^4)}{(x^2+y^2)^2} \leq 6|x|$$
by starting with the left side of the inequality and working from there.
Hints from the textbook said to work from these inequality "tricks,"
$$2|ab| \leq a^2 + b^2$$
$$|a| + |b| \leq \sqrt{2} \sqrt{a^2+b^2}$$
And the triangle inequality,
$$|a+b| \leq |a| + |b|$$
Expanding the numerator gives us
$$\frac{2x^5 + 4x^3y^2 - 2xy^4}{(x^2 + y^2)^2}$$
For starters, I use the triangle inequality theorem to make the numerator look like this:
$$|2x^5 + 4x^3y^2 - 2xy^4| \leq |2x^5|+|4x^3y^2|+|2xy^4|$$
Also, in regard to the denominator, since it is even, it is always positive and thus equal to its absolute value. Then, I multiply both sides by the denominator, $(x^2+y^2)^2$. This gives us
$$|2x^5|+|4x^3y^2|+|2xy^4| \leq 6|x||x^4 + 2x^2y^2+y^4|$$
$$|2x^5|+|4x^3y^2|+|2xy^4| \leq 6|x^5+2x^3y^2+xy^4|$$
My question is, since $|a+b| \leq |a|+|b|$, I feel as though I cannot violate that rule and am stuck. What trick(s) can I use from here? If there were no absolute value symbols, it would be as simple as simplifying the inequality since the orders match up. However, it would not be true for all (x,y) as the right side of the inequality is odd.
| We have
$$|4x^3(x^2+y^2)-2x(x^4+y^4)|\le4|x|x^2(x^2+y^2)+2|x|(x^2+y^2)^2\\\le6|x|(x^2+y^2)^2$$
and then the result is immediate by canceling the factor $(x^2+y^2)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
a very basic question on finding the discriminant for $x^2+2(a-3)x-3a-7=0$ Sorry for asking such a basic question.
In the following quadratic equation
$$x^2+2(a-3)x-3a-7=0$$
by my calculations,
$$D=\left(\frac{b}{2}\right)^2-ac=(a-3)^2-1(-3a-7)=a^2-6a+9+3a+7=a^2-3a+16$$
But in a PDF article that describes a solution to a problem concerning this equation the discriminant is calculated as
$$a^2-3a+2$$
Is there any error in my calculation?
| Notice, compare the given equation: $x^2+2(a-3)x-3a-7=0$ with $Ax^2+Bx+C=0$, we get $$A=1, \ B=2(a-3), \ C=-3a-7$$
Hence,
discriminant is calculated as follows $$D=B^2-4AC=\left(2(a-3)\right)^2-4(1)(-(3a+7))$$ $$=4a^2-24a+36+12a+28$$ $$=4a^2-12a+64$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the integration : $\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$ $$\int{\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx}$$
$$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx=\int \frac{(1-\sin x)(2-\sin x)}{\sqrt{(1-\sin x)(2-\sin x)(1+\sin x)(2+\sin x)}}dx$$
I am stuck. Please help me....
| Let $$\displaystyle I = \int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$$
We can write $$\displaystyle \sqrt{\frac{1-\sin x}{1+\sin x}} = \sqrt{\frac{1-\sin x}{1+\sin x}\times \frac{1+\sin x}{1+\sin x}} = \frac{\cos x}{1+\sin x}$$
So we get $$\displaystyle I = \int\frac{\cos x}{1+\sin x}\cdot \sqrt{\frac{2-\sin x}{2+\sin x}}dx$$
Now Let $1+\sin x= y\;,$ Then $\cos xdx = dy$
So Integral $$\displaystyle I = \int\frac{1}{y}\cdot \sqrt{\frac{3-y}{1+y}}dy$$
Now Put $$\displaystyle \frac{3-y}{1+y}=t^2\Rightarrow y=\frac{3-t^2}{1+t^2}$$
So we get $$\displaystyle y=-\left[1-\frac{4}{1+t^2}\right] = \left[\frac{4}{1+t^2}-1\right].$$ So $\displaystyle dy = -\frac{8t}{(1+t^2)^2}$
So Integral $$\displaystyle I = \int\frac{1+t^2}{3-t^2}\cdot t\cdot \frac{-8t}{(1+t^2)^2}dt = 8\int\frac{t^2}{(t^2-3)\cdot (1+t^2)}dt$$
So Integral $$\displaystyle I = 2\int \left[\frac{3(t^2+1)+(t^2-3)}{(t^2-3)\cdot (1+t^2)}\right]dt = 2\int \left[\frac{3}{t^2-(\sqrt{3})^2}+\frac{1}{1+t^2}\right]dt$$
So Integral $$\displaystyle I = 6\cdot \frac{1}{2\sqrt{3}}\cdot \ln\left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+2\tan^{-1}(t)+\mathcal{C}$$
So Integral $$\displaystyle I = \sqrt{3}\cdot \ln\left|\frac{\sqrt{2-\sin x}-\sqrt{3}\cdot \sqrt{2+\sin x}}{\sqrt{2-\sin x}-\sqrt{3}\cdot \sqrt{2+\sin x}}\right|+2\tan^{-1}\left(\sqrt{\frac{2-\sin x}{2+\sin x}}\right)+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$
(original image)
I think we need to simplify it writing it in summation sign as you can see here:
$$\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}$$
or in Wolfram Alpha input in comments.
I can compute it too! It's easy to write a script for this kind of question.
I need a way to solve it. How would you solve it on a piece of paper?
| For the Calculation of $$\displaystyle \frac{\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{\sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}} = $$
Let $$\displaystyle A_{n} = \sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}$$ and $$\displaystyle B_{n} = \sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}$$ , where $n>1$
Now $$\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right)^2 = 2n-2\sqrt{n^2-k}$$
So $$\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sqrt{2}\cdot \sqrt{n-\sqrt{n^2-k}}$$
So $$\displaystyle \sum_{k=1}^{n^2-1}\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sum_{k=1}^{n^2-1}\sqrt{2}\cdot \sqrt{n-\sqrt{n^2-k}}$$
So So $$\displaystyle \sum_{k=1}^{n^2-1}\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sum_{k=1}^{n^2-1}\sqrt{2}\cdot \sqrt{n-\sqrt{k}}$$
So $$A_{n}-B_{n} = B_{n}\sqrt{2}$$
So $$A_{n} = B_{n}\left(1+\sqrt{2}\right)$$
So $$\displaystyle \frac{A_{n}}{B_{n}} = 1+\sqrt{2}$$
Now Put $\displaystyle n^2-1 = 99\Rightarrow n= 10\;,$ So we get $$\displaystyle \frac{\sum_{k=1}^{99}\sqrt{n+\sqrt{k}}}{\sum_{k=1}^{99}\sqrt{n-\sqrt{k}}} =\frac{A_{10}}{B_{10}} = 1+\sqrt{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 2
} |
Simple coupled ODE Find two linearly independent solutions to the pair of coupled ODEs
$\frac{dx}{dt} = 2x + 3y$,
$\frac{dy}{ dt} =-3x+2y$.
I figured the eigenvalues/eigenvectors of the corresponding matrix to be $$e^{(2+3i)t} \quad\text{with}\quad \left[\begin{array}{r}
i \\
1
\end{array}\right] \quad\text{and} \quad e^{(2-3i)t} \quad\text{with}\quad\left[\begin{array}{r}
-i \\
1
\end{array}\right].$$
But apparently the solutions are $$\left[\begin{array}{r}
x \\
y
\end{array}\right] =e^{2t} \left[\begin{array}{r}
\cos3t \\
-\sin3t
\end{array}\right] ,\left[\begin{array}{r}
x \\
y
\end{array}\right] =e^{2t} \left[\begin{array}{r}
\sin3t \\
\cos3t
\end{array}\right].$$
Even after transforming the exponential eigenvalues to trigonometric functions I still don't get the desired results. I checked the eigenvectors/values with a computer and they seem to be correct. Please help.
| Notice, we have $$\frac{dx}{dt}=2x+3y\tag 1$$
$$\frac{dy}{dt}=-3x+2y\tag 2$$
*
*Diving (1) by (2), we get $$\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{2x+3y}{-3x+2y}$$ $$\frac{dx}{dy}=\frac{2\frac{x}{y}+3}{-3\frac{x}{y}+2}$$ Now, let $x=uy\implies \frac{dx}{dy}=u+y\frac{du}{dy}$, setting the values we get
$$u+y\frac{du}{dy}=\frac{2u+3}{-3u+2}$$ $$y\frac{du}{dy}=\frac{2u+3}{-3u+2}-u$$ $$y\frac{du}{dy}=\frac{3(1+u^2)}{2-3u}$$ $$\frac{2-3u}{3(1+u^2)}du=\frac{dy}{y}$$
$$\frac{2}{3}\int \frac{du}{1+u^2}-\frac{1}{2}\int \frac{2udu}{1+u^2}=\int \frac{dy}{y}$$ $$\frac{2}{3}\tan^{-1}(u)-\frac{1}{2}\ln(1+u^2)=\ln y+C_1$$
*Diving (2) by (1), we get $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3x+2y}{ 2x+3y}$$ $$\frac{dy}{dx}=\frac{-3+2\frac{y}{x}}{2+3\frac{y}{x}}$$
Now, let $y=ux\implies \frac{dy}{dx}=u+x\frac{du}{dx}$, setting the values we get
$$u+x\frac{du}{dx}=\frac{2+3u}{-3+2u}$$ $$x\frac{du}{dx}=\frac{2+3u}{-3+2u}-u$$ $$x\frac{du}{dx}=\frac{2+6u-2u^2}{-3+2u}$$ $$\frac{-3+2u}{2+6u-u^2}du=\frac{dx}{x}$$
I hope you can solve further.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergent/divergent series Is the following series divergent/convergent?
$$S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}-...$$
I think it is divergent since
$$
\begin{align}
S&>1-\frac{1}{2}-\frac{1}{2}+4\cdot\frac{1}{6}-\frac{4}{7}+\frac{5}{15}-\frac{6}{16}+...\\
&=1/3-4/7+1/3-6/16+...=\sum_{n=1}^\infty 1/3-\alpha_n
\end{align}
$$
where $\alpha_n=4/7, 6/16, 8/22$ which tends to 0, so the series on the right hand side is divergent. Is this the right answer? Thanks
| S is convergent ( Using Leibnitz's Test for convergence of Alternating series)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Average waiting time to guess a random number from $1$ to $100$ with equal probability and 1 second between each number is generated A random number between $1$ and $100$ is generated every second.
What would be the average waiting time for a specific number ($1$ for instance) to be generated?
Probability distribution is uniform.
Each number is generated independently of the others.
| Let $X$ denote the number of seconds until $1$ is generated:
*
*$P(X=0)=\frac{1}{100}$
*$P(X=1)=\frac{99}{100}\cdot\frac{1}{100}$
*$P(X=2)=\frac{99}{100}\cdot\frac{99}{100}\cdot\frac{1}{100}$
*$\dots$
*$P(X=n)=\left(\frac{99}{100}\right)^n\cdot\frac{1}{100}$
So the expected number of seconds until $1$ is generated is:
$$E(X)=\sum\limits_{n=0}^{\infty}n\cdot\left(\frac{99}{100}\right)^n\cdot\frac{1}{100}=99$$
Why is it that $\sum\limits_{n=0}^{\infty}n\cdot\left(\frac{99}{100}\right)^n\cdot\frac{1}{100}=99$:
*
*$|x|<1 \implies \sum\limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$
*Differentiate each side of the equation: $\sum\limits_{n=0}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}$
*Multiply by $x$ each side of the equation: $\sum\limits_{n=0}^{\infty}nx^n=\frac{x}{(1-x)^2}$
*Use $x=\frac{99}{100}$ on each side of the equation: $\sum\limits_{n=0}^{\infty}n\left(\frac{99}{100}\right)^n=9900$
*Multiply by $\frac{1}{100}$ each side of the equation: $\sum\limits_{n=0}^{\infty}n\cdot\left(\frac{99}{100}\right)^n\cdot\frac{1}{100}=99$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding all solutions of $x^2+2x-15\equiv0 \pmod{105}$- Proof strategy. Find all solutions of $x^2+2x-15\equiv0 \pmod{105}$. Now, I wanted to suggest a proof relying on the algorithm presented in class, and there are some parts where I could use some help or criticism.
First of all, let us write the prime factorization of $105$. Then let us find the solutions of this equation modulo every one of the coprime factors and then let us use the Chinese Remainder Theorem to find the set of solutions inclusive of them all(is that grammatically logical?).
$105=3\cdot 5 \cdot 7$. It is easy to see that $x=0$ is already a solution modulo $3$, $5$, and any integer divisible by 15.
Solutions modulo 3: if $f(x)=x^2+2x-15$ then $f(1)=-12\equiv 0 (3)$. Therefore we get $x\equiv 1(3)$. The algorithm now claims: $x-1|f$ and I get: $x+3$ (and a remainder I am led to believe is irrelevant. Why?). $x+3\equiv 0(3)$ means $x\equiv 3k$. That is, every integer divisible by $3$ is a solution modulo 3. (I could actually be noticed at the very beginning, couldn't it?).
Solutions modulo 5: $f(3)$ happens to be one. (Should I try other $f(x)$ or is dividing by $x-3$ sufficient?) Let us now divide:$f(x)=(x-3)(x-5)$ as I've just discerned. $x-5\equiv 0(5)\iff x\equiv 5k (5)$. (It already seems to me like $x=1$ is no longer an option(?).)
Solutions modulo 7:$f(2)$ is a solution. Dividing by $x-2$, ${f\over x-2}=x+4$. Now $x+4\equiv 0 (7)\iff x\equiv 3 (7)\iff x=7k+3$.
That is the problematic part: I was not guided well enough as for how to find an intersection of sets that fits all subequations(Is that even a term?). I am left with:
$x=3k$, $x=5k$, $x=7k+3$.
How can I find a way to find all integers fulfilling it? I would appreciate your help.
| To solve the equation modulo $3$:
$$
\begin{align}
x^2 + 2x - 15 &\equiv 0 \pmod 3\\
x^2 + 2x &\equiv 0 \pmod 3\\
x(x + 2) &\equiv 0 \pmod 3
\end{align}
$$
Therefore $x \equiv 0 \textrm{ or } 1 \pmod 3$. Similarly, working modulo 5, we'll again get $x(x+2) \equiv 0$, so $x \equiv 0 \textrm{ or } 3 \pmod 5$. Finally, in modulo 7, we have:
$$
\begin{align}
x^2 + 2x - 15 &\equiv 0 \pmod 7\\
x^2 + 2x - 1 &\equiv 0 \pmod 7\\
x^2 + 2x + 1 &\equiv 2 \pmod 7\\
(x + 1)^2 &\equiv 2 \pmod 7
\end{align}
$$
So $x+1 \equiv 3 \textrm{ or } 4 \pmod 7$, which means that $x \equiv 2 \textrm{ or } 3 \pmod 7$.
All of this leads to eight possibilities, for each of which you can use the Chinese Remainder Theorem.
There are eight possibilities because there are two choices for an equivalence modulo $3$, two choices modulo $5$, and two again modulo $7$, so in total we have $2^3=8$. Each set of choices will lead to a result modulo $105$. For instance, suppose we take the following set of equivalences:
$$x\equiv0\pmod3\\x\equiv0\pmod5\\x\equiv2\pmod7$$
Then we use the Chinese Remainder Theorem to find $x$:
$$x \equiv 0\cdot70 + 0\cdot21 + 2\cdot15 = 30 \pmod{105}$$
You can check that indeed $x=30$ satisfies the initial equation. Now do the same for the seven other combinations of equivalences.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Factorization of a polynomials in complex number. Factorize this expression:
$$a^2+b^2+c^2-ab-bc-ca.$$
The result is
$$(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)$$
How I can get $\Omega$ here?What's the approach?
| If we are given that
$a^2 + b^2 + c^2 - ab - ac - bc$
$ = (a + b\Omega + c\Omega^2)(a + b\Omega^2 + c\Omega), \tag{1}$
and we wish to find $\Omega$, we can proceed as follows: multiplying out the right-hand side, we find
that
$a^2 + b^2 + c^2 - ab - ac - bc$
$ = a^2 + \Omega^3 b^2 + \Omega^3 c^2
$
$+ (\Omega^2 + \Omega)ab + (\Omega^2 + \Omega)ac + (\Omega^4 + \Omega^2) bc; \tag{2}$
we see from (2) that the factorization succeeds provided that
$\Omega^3 = 1, \tag{3}$
$\Omega^2 + \Omega = -1, \tag{4}$
and
$\Omega^4 + \Omega^2 = -1; \tag{5}$
it follows from (3) that $\Omega$ must be one of the three cube roots of unity $1$, $e^{2\pi /3}$, $e^{4\pi i /3}$; it follows from (4), (5) that $\Omega \ne 1$; thus, since
$(\Omega - 1)(\Omega^2 + \Omega + 1) = \Omega^3 - 1 = 0, \tag{6}$
(4) must bind; the apparent discrepancy 'twixt (4) and (5) is resolved by observing that (3) yields
$\Omega^4 = \Omega \Omega^3 = \Omega(1) = \Omega; \tag{7}$
(4) and (5) are in reality the same equation. We thus see that the factorization (1) holds with $\Omega = e^{2 \pi i / 3}$, $\Omega = e^{4 \pi i / 3}$, since both satisfy (3)-(5).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Proving $\binom{n}{m}+2\binom{n-1}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}$
For $m,n\in\mathbb{N},\;n\geq m$, prove the following:
$$
\tag{i}\binom{n}{m}+\binom{n-1}{m}+\binom{n-2}{m}+......+\binom{m}{m} = \binom{n+1}{m+1}
$$
$$
\tag{ii}\binom{n}{m}+2\binom{n-1}{m}+3\binom{n-2}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}
$$
My Attempt:
For $(\mathrm{i})$, We can write $\binom{n}{m}$ as the coefficient of $x^m$ in $(1+x)^n$. Thus we can also write $\binom{n-1}{m}$ as the coefficient of $x^m$ in $(1+x)^{n-1}$ and $\binom{n-2}{m}$ as the coefficient of $x^m$ in $(1+x)^{n-2}$. So we have to find the coefficient of $x^m$ in
$$
(1+x)^n+(1+x)^{n-1}+(1+x)^{n-2}+........+(1+x)^{m}
$$
Using the formula for the sum of a geometric progression, this sum equals
$$\frac{(1+x)^{n+1}-(1+x)^m}{x}$$
So we now need to find the coefficient of $x^m$ in
$$
\frac{(1+x)^{n+1}-(1+x)^m}{x}
$$
or, equivalently, we need to find the coefficient of $x^{m+1}$ in
$$
(1+x)^{n+1}-(1+x)^m = \binom{n+1}{m+1}
$$
We can use a similar method to solve $(\mathrm{ii})$. Can these questions be solved using combinatorial methods instead?
| Suppose $n \geq m$ and there is a set $S$ of $n + 2$ objects, denoted as $o_1, o_2, \cdots, o_{n+2}$. Your task to sample $m + 2$ objects from $S$. Let $X_i$ denote the # of ways to choose $m + 2$ objects such that the object with second minimum id is $o_i$. Easy to see that
$$
X_i = (i-1) \cdot { n + 2 - i \choose m }
$$
So we have
$$
\sum_{i=2}^{n+2-m} X_i = {n \choose m} + 2\cdot{n-1 \choose m} + \cdots + (n + 1 -m) \cdot {m \choose m} = {n + 2 \choose m + 2}
$$
Similar derivation can be applied to (i) with some minor modification. Instead of enumerating the second minimum id, you need to enumerate the minimum id instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
reflecting a function about a line Say you have a function $f(x)$ and a line $g(x)=ax+b$. How do you reflect $f$ about $g$?
I am apparently supposed to write more text, but the line above is all I am after, hence I wrote this sentence as well.
| Represent the graph of f parametrically by $x=t, y=f(t)$.
If we reflect the point $(t, f(t))$ in the line $y=ax+b$ to get the point $(x,y)$, then
$\color{red}{y-f(t)=-\frac{1}{a}(x-t)}$
since the line and the line segment between $(t, f(t))$ and $(x,y)$ are perpendicular to each other.
We also have that $\frac{y+f(t)}{2}=a\left(\frac{x+t}{2}\right)+b$ and therefore $\color{red}{y+f(t)=a(x+t)+2b}$
since the midpoint of the line segment lies on the line.
Subtracting these equations gives $2f(t)=ax+at+\frac{1}{a}x-\frac{1}{a}t+2b=\left(\frac{a^2+1}{a}\right)x+\left(\frac{a^2-1}{a}\right)t+2b$,
and solving for x gives $\displaystyle \color{blue}{x=\frac{1-a^2}{1+a^2}t+\frac{2a}{a^2+1}\left(f(t)-b\right)}$.
Solving for x in the first equation gives $-ay+af(t)=x-t$ and so $\color{red}{x=-ay+af(t)+t}$.
Substituting into the second equation gives
$y+f(t)=a\big(-ay+af(t)+2t\big)+2b=-a^2y+a^2f(t)+2at+2b,$ and solving for y gives
$(a^2+1)y=(a^2-1)f(t)+2(at+b)$ so $\displaystyle\color{blue}{y=\frac{a^2-1}{a^2+1}f(t)+\frac{2}{a^2+1}(at+b)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Define g(x) as a function of $x$ Let $f(x)= \begin{cases}
-1 & ,-2\leq x\leq 0 \\
\\
|x-1| & ,0<x\leq2
\end{cases}$
and $g(x)=\int_{-2}^{x}f(t) dt$.Define g(x) as a function of $x$.
I tried to solve it.I redefined $f(x)= \begin{cases}
-1 & ,-2\leq x\leq 0 \\
1-x & ,0<x<1 \\
x-1 & ,1\leq x\leq2
\end{cases}$
$g(x)=\int_{-2}^{x}f(t) dt=\begin{cases}
\int_{-2}^{x}-1dx & ,-2\leq x\leq 0 \\
\int_{-2}^{x}(1-x)dx & ,0<x<1 \\
\int_{-2}^{x}(x-1)dx & ,1\leq x\leq2
\end{cases}$
$g(x)= \begin{cases}
-(x+2) & ,-2\leq x\leq 0 \\
x-\frac{x^2}{2}+4 & ,0<x<1 \\
\frac{x^2}{2}-x-4 & ,1\leq x\leq2
\end{cases}$
But the answer is given to be
$g(x)= \begin{cases}
-(x+2) & ,-2\leq x\leq 0 \\
x-\frac{x^2}{2}-2 & ,0<x<1 \\
\frac{x^2}{2}-x-1 & ,1\leq x\leq2
\end{cases}$
Have i done it wrong,what should be the correct method.Please guide me.
| For $0<x<1$ we have
\begin{align*}
g(x)&=\int_{-2}^{x}f(t)dt\\
&=\int_{-2}^0f(t)dt+\int_0^xf(t)dt\\
&=\int_{-2}^0(-1)dt+\int_0^x(1-t)dt\\
&=(-1)(0+2)+x-\frac{1}{2}x^2-0\\
g(x)&=-\frac{1}{2}x^2+x-2,\qquad \text{ for }\;\;0<x<1
\end{align*}
On the other hand, for $1\le x\le 2$ it follows
\begin{align*}
g(x)&=\int_{-2}^{x}f(t)dt\\
&=\int_{-2}^0f(t)dt+\int_0^1f(t)dt+\int_1^xf(t)dt\\
&=\int_{-2}^0(-1)dt+\int_0^1(1-t)dt+\int_1^x(t-1)dt\\
&=(-1)(0+2)+1-\frac{1}{2}(1)^2-0+\frac{1}{2}x^2-x-\left[\frac{1}{2}-1\right]\\
g(x)&=\frac{1}{2}x^2-x-1,\qquad \text{ for }\;\;1\le x\le 2
\end{align*}
Thus
\begin{equation*}
\color{blue}{g(x)=
\begin{cases}
-(x+2)&-2\le x\le 0\\
-\frac{1}{2}x^2+x-2& 0<x<1\\
\frac{1}{2}x^2-x-1& 1\le x\le 2
\end{cases}
}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $y^2 + 3xy - 10x^2 + y + 5x = 0$ for y in terms of x I'm given the following equation: $y^2 + 3xy - 10x^2 + y + 5x = 0$
and asked to solve $y$ in terms of $y$.
My attempt:
$y^2 + (3x+1)\times y - 10x^2 + 5x = 0$
$\Rightarrow (y+(3x+1)/2)^2 - ((3x+1)/2)^2 = (x - (\frac 12)\times x)^2 - (\frac 12)^2 $
But I cant seem to get any further and it's probably in the wrong direction.
| $$y^2 + 3xy - 10x^2 + y + 5x = 0\Rightarrow y^2+(3x+1)y-(10x^2-5x) = 0$$
So $$\displaystyle y = \frac{-(3x+1)\pm \sqrt{(3x+1)^2+4(10x^2-5x)}}{2}$$
$$\displaystyle = \frac{-(3x+1)\pm \sqrt{49x^2-14x+1}}{2}$$
so we get $$\displaystyle y = \frac{-(3x+1)\pm \sqrt{(7x-1)^2}}{3} = \frac{-(3x+1)\pm (7x-1)}{2}$$
So we get $\displaystyle y = 2x-1\;\;,-5x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
| HINT:
$$ac+bd=0\iff\dfrac ad=\dfrac b{-c}=\pm\sqrt{\dfrac{a^2+b^2}{d^2+(-c)^2}}$$
But if $a^2+b^2=d^2+c^2,$ not necessarily $=1$
$$\dfrac ad=\dfrac b{-c}=\pm1$$
So, either $a=d,b=-c$ or $a=-d,b=c$
The result should follow immediately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 2
} |
How to Derive this Matrix Equation - Converting Lagrangian in Proximal Operator Problem Form $$ \left< Z,X-L-S \right> \quad +\quad \frac { r }{ 2 } \left\| X-L-S \right\|_F^2 \quad =\quad \frac { r }{ 2 } { \left\| L-\left( X-S+\frac {Z}{r} \right) \right\| }_F^2 $$
I think $ \left< Z,X-L-S \right>$ should be $\operatorname{Tr}(Z^T(X-L-S))$. This equation is presented in a paper, but the author doesn't show the derivation. See Robust Principal Component Analysis on Graphs at page 11.
| Note that the left side of
$$ \left< Z,X-L-S \right> \quad +\quad \frac { r }{ 2 } \left\| X-L-S \right\|_F^2 \quad =\quad \frac { r }{ 2 } { \left\| L-\left( X-S+\frac {Z}{r} \right) \right\| }_F^2 $$
can be written as
$$
\operatorname{tr}(Z^T(X-L-S)) + \frac { r }{ 2 } \operatorname{tr}( (X-L-S)^T(X-L-S) ).$$
And this is the same as
$$
\operatorname{tr}(Z^T(X-L-S)) + \operatorname{tr}(\frac { r }{ 2 } (X-L-S)^T(X-L-S) ) =
\operatorname{tr}\left(Z^T(X-L-S)) + \frac { r }{ 2 } (X-L-S)^T(X-L-S) )\right)
= \operatorname{tr}\left(\left((Z^T) + \frac { r }{ 2 } (X-L-S)^T\right)(X-L-S) )\right)
= \frac { r }{ 2 }\operatorname{tr}\left(\left(\frac{2 Z^T}{r} + (X-L-S)^T\right)(X-L-S) )\right)
= \frac { r }{ 2 }\operatorname{tr}\left(\left(\frac{2 Z}{r} + X-L-S\right)^T(X-L-S) )\right).
$$
The right hand side on the other hand can be written as
$$\frac { r }{ 2 } { \left\| L-\left( X-S+\frac {Z}{r} \right) \right\| }_F^2 =
\frac { r }{ 2 } \operatorname{tr} \left( \left(L-X+S-\frac {Z}{r} \right)^T \left(L-X+S-\frac{Z}{r} \right) \right)
=
\frac { r }{ 2 } \operatorname{tr} \left( \left(L-X+S-\frac {2Z}{r}+\frac {Z}{r} \right)^T \left(L-X+S-\frac{2Z}{r} +\frac{Z}{r} \right) \right)
=
\frac { r }{ 2 } \operatorname{tr} \left( \left(-L+X-S+\frac {2Z}{r}-\frac {Z}{r} \right)^T \left(-L+X-S+\frac{2Z}{r} -\frac{Z}{r} \right) \right)
=
\frac { r }{ 2 } \operatorname{tr} \left( \left((-L+X-S+\frac {2Z}{r})-\frac {Z}{r} \right)^T \left((-L+X-S)+(\frac{2Z}{r} -\frac{Z}{r}) \right) \right)
=
\frac { r }{ 2 } \operatorname{tr} \left((-L+X-S+\frac {2Z}{r})^T(-L+X-S)-\frac {Z^T}{r}(-L+X-S) + (-L+X-S+\frac {2Z}{r})^T\frac{Z}{r}-\frac {Z^T}{r}\frac{Z}{r} \right)
=
\frac { r }{ 2 } \operatorname{tr} \left((-L+X-S+\frac {2Z}{r})^T(-L+X-S) + (\frac {2Z}{r})^T\frac{Z}{r}-\frac {Z^T}{r}\frac{Z}{r} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Poisson events distributed uniformly in a given time It is given that $4$ Poisson events occur between $12:00$ to $13:00 $ (interval denoted by T).
Intuitively, Why the probability of each event to occur at time $t \in T$ is uniform across $T$?
I'm well aware of the calculation for that:
Let's calculate the probability of $3$ (out of $ 4$) events to occur between $12:30$ to $13:00$ ($\frac{1}{2}$ an hour).
Let the following random variables:
*
*$X$ - the $4 $ events from $12:00$ to $13:00$.
*$X_1$ - the single event that has happened between $12:00 $ to $ 12:30$ .
*$X_2$ - the 3 events which happened between $12:30$ to $13:00$.
So their distributions are:
$$
\begin{align*}
& X \sim Poi(\lambda \cdot t) = Poi(4 \cdot 1) = Poi(4) \\
& X_1, X_2 \sim Poi(4 \cdot \frac{1}{2}) = Poi(2)
\end{align*}
$$
Given that a total of 4 events occurred at $T$, then:
$$
\begin{align*} P(X_2 = 3 \,|\, X = 4) &= \\
& = \frac{P(X_2 = 3 \cap X = 4)}{P(X = 4)} \\
& = \frac{P(X_2 = 3 \cap (X_1 = 1 \cup X_2 = 3))}{P(X = 4)} \\
& = \frac{P(X_2 = 3 \cap X_1 = 1)}{P(X = 4)} \\
& = \frac{P(X_2 = 3) \cdot P(X_1 = 1)}{P(X = 4)} \\
& = \frac{\frac{e^{-2} \cdot 2^3}{3!} \cdot \frac{e^{-2} \cdot 2^1}{1!}}{\frac{e^{-4} \cdot 4^4}{4!}} \\
& = \frac{4!}{3! 1!} \cdot \left(\frac{2}{4}\right)^1 \cdot \left(\frac{2}{4}\right)^3\\
& = \frac{2^4 \cdot 4!}{3! \cdot 4^4}\\
& = 4 \cdot \left(\frac{1}{2}\right)^4\\
& = 0.25
\end{align*}
$$
(Note that at the end we get a binomial distribution)
But I just can't understand this intuitively.
Edit:
The root of my confusion:
Indeed, it is given that in an hour there are precisely 4 events.
Therefore the total 4 events are NOT distributed like Poission.
But it is not given how much time each event takes to occur, so maybe the second, third and fourth events are distributed like Poission? Perhaps the time that each event takes to occurs is distributed exponentially (which indicates on a Poission distribution)?
| It has to do with the fact that it is given that there are exactly 4 events in the hour, whereas the Poisson distribution allows any number of events in a given time period. Both $X_1$ and $X_2$ are both greater than or equal to $1$ (there is at least one event in both half hours), so we need to divide the remaining two events over $X_1$ and $X_2$. This distribution is no longer Poisson. The probability that the final two events both fall in $X_2$ is simply $0.5^2 = 0.25.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Trying to solve $\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$ The equation is
$$\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$$
I solve it thus:
$$
\begin{cases}
2\cos^2(x)-\sqrt3=2\sin^2(x) \\
-\sqrt2 \sin(x)\ge 0 \iff \sin(x)\le 0
\end{cases}
$$
The first equation boils down to
$$2\cos^2(x)=2(1-\cos^2(x))+\sqrt3$$
$$4cos^2(x)-2=\sqrt3$$
$$2(2\cos^2(x)-1)=\sqrt3$$
$$2\cos(2x)=\sqrt3$$
$$\cos(2x)=\frac{\sqrt3}{2}$$
$$2x=\pm \arccos(\frac{\sqrt3}{2})+2n\pi$$
$$2x=\pm \frac{\pi}{6}+2n\pi$$
$$x=\pm \frac{\pi}{12}+n\pi$$
Considering the condition $\sin(x)\le 0$, we are left with
$$x=- \frac{\pi}{12}+n\pi$$
But the textbook's answer is
$$x=- \frac{\pi}{12}+2n\pi; x=- \frac{11\pi}{12}+2n\pi$$
What did I overlook?
P.S. The problem and the answer from the texbook:
| As Brent points out in his comment, your only mistake was applying the condition $\sin x<0$ at the end:
1) If $x=\frac{\pi}{12}+n\pi$, $\;\;\sin x>0$ for $n$ even and $\sin x<0$ for $n$ odd, so this gives
$\hspace{.4 in} x=\frac{\pi}{12}+(2n+1)\pi=\frac{13\pi}{12}+2n\pi=-\frac{11\pi}{12}+2n\pi$
2) If $x=-\frac{\pi}{12}+n\pi$, $\;\;\sin x<0$ for $n$ even and $\sin x>0$ for $n$ odd, so this gives
$\hspace{.4 in} x=-\frac{\pi}{12}+2n\pi=\frac{23\pi}{12}+2n\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Asymptotic behavior of elliptic integral (first kind) I came accross some obstacles in proving that the time $T(\delta)$ taken by a pendulum to travel from $\theta=\pi-\delta$ to a considerably distant angle $\theta=\theta_0\in(0,\pi/4)$ diverges logarithmic-ally with respect to $\delta$.
Here is my work so far.
The expression of $T(\delta)$ is
$$T(\delta)=-\int_{\pi-\delta}^{\theta_0}\frac{1}{\sqrt{\frac{2g}{L}[\cos{\theta}+\cos{\delta}]}}d\theta.$$
I was able to make a heuristic argument by taking $\cos{\delta}\approx 1$ (since a cosine function is 1 up to the second order) and directly doing the integration. It indeed is a logarithm blow-up.
Not being convinced by my own argument (okay for my physics-oriented self by skeptical for my math-oriented self because $\delta$ appears both in the integral limit and in the integrand, and thus taking the limit only in the integrand seems un-justified), I came to the solution of the integral, the elliptical integral of the first kind.
What I get so far is by a standard change-of-variable (learned from Zorich) $\sin{\theta}=\frac{\sin{\psi/2}}{\sin{\phi_0/2}}$ where $\phi_0=\pi-\delta$, I changed it to a complete elliptic integral (plus a finite constant)
$$T(\delta)=\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2\sin^2{\theta}}}d\theta=K(k)$$
where $k=\sin{\phi_0/2}=\sin{(\pi/2-\delta/2)}$.
It seems the remaining task is to show that
$$\lim_{\delta\rightarrow0}\frac{K(\sin{(\pi/2-\delta/2)})}{\ln{\delta}}=const\neq0.$$
I'm not so familiar with the elliptic integrals and did not get readable results about this asymptotic behavior of K in Google. I'd very appreciate it if there is any derivation on this.
Thanks in advance for any answer or remark!!
| I prefer to have the singular behaviour near $0$ rather than at $\frac{\pi}{2}$, so let's make the substitution $\varphi = \frac{\pi}{2} - \theta$. We obtain
$$K(k) = \int_0^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1 - k^2\cos^2\varphi}}.$$
Split the integral at $\frac{\pi}{4}$. The part
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1 - k^2\cos^2\varphi}}$$
remains harmless as $k \to 1$ and tends to
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{d\varphi}{\sin \varphi}.$$
For the other part, write $1 = \sin^2 \varphi + \cos^2 \varphi$ to obtain $1 - k^2\cos^2\varphi = \sin^2 \varphi + (1-k^2)\cos^2 \varphi$. Let $\varepsilon = \sqrt{1-k^2}$. Then
\begin{align}
\int_0^{\frac{\pi}{4}} \frac{d\varphi}{\sqrt{1 - k^2\cos^2 \varphi}} &= \int_0^{\frac{\pi}{4}} \frac{\cos^2 \varphi + \sin^2 \varphi}{\sqrt{\cos^2 \varphi + \sin^2 \varphi}\cdot \sqrt{\sin^2 \varphi + \varepsilon^2 \cos^2\varphi}}\,d\varphi\\
&= \int_0^{\frac{\pi}{4}} \frac{1 + \tan^2\varphi}{\sqrt{1 + \tan^2 \varphi}\cdot \sqrt{\varepsilon^2 + \tan^2 \varphi}}\,d\varphi \tag{$t = \tan \varphi$}\\
&= \int_0^1 \frac{dt}{\sqrt{1+t^2}\cdot \sqrt{\varepsilon^2 + t^2}}\\
&= \int_0^1 \frac{dt}{\sqrt{\varepsilon^2 + t^2}} - \int_0^1 \Biggl(1 - \frac{1}{\sqrt{1+t^2}}\Biggr) \frac{dt}{\sqrt{\varepsilon^2 + t^2}}.
\end{align}
Since
$$1 - \frac{1}{\sqrt{1+t^2}} = \frac{\sqrt{1+t^2}-1}{\sqrt{1+t^2}} = \frac{t^2}{\sqrt{1+t^2}\cdot (1 + \sqrt{1+t^2})},$$
the last integral remains bounded and tends to
$$\int_0^1 \frac{t \,dt}{1 + t^2 + \sqrt{1+t^2}}$$
as $\varepsilon \to 0$.
And the substitution $t = \varepsilon u$ gives us
\begin{align}
\int_0^1 \frac{dt}{\sqrt{\varepsilon^2 + t^2}} &= \int_0^{\frac{1}{\varepsilon}} \frac{du}{\sqrt{1+u^2}}\\
&= \operatorname{Ar sinh} \frac{1}{\varepsilon}\\
&= \log \biggl(\frac{1}{\varepsilon} + \sqrt{1 + \frac{1}{\varepsilon^2}}\biggr)\\
&= \log \frac{1}{\varepsilon} + \log 2 + \log \frac{1 + \sqrt{1+\varepsilon^2}}{2}.
\end{align}
Thus we have
$$K(k) = \log \frac{1}{\sqrt{1-k^2}} + O(1) = \frac{1}{2}\log \frac{1}{1-k} - \frac{1}{2}\log (1+k) + O(1) = \frac{1}{2}\log \frac{1}{1-k} + O(1).$$
In our specific case, with $k = \sin \bigl(\frac{\pi}{2} - \frac{\delta}{2}\bigr) = \cos \frac{\delta}{2}$, we have $\varepsilon = \sqrt{1-k^2} = \sin \frac{\delta}{2} = \frac{\delta}{2} + O(\delta^3)$, so
$$\log \frac{1}{\varepsilon} = \log \frac{2}{\delta} + O(\delta^2)$$
and overall
$$K\bigl(\cos \tfrac{\delta}{2}\bigr) = \log \frac{1}{\delta} + O(1),$$
where the $O(1)$ term is not only bounded, it in fact converges as $\delta \to 0$. We have the relevant terms:
$$K\bigl(\cos \tfrac{\delta}{2}\bigr) = \log \frac{1}{\delta} + 2\log 2 + \int_0^1 \frac{t\,dt}{1+t^2+\sqrt{1+t^2}} + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{d\varphi}{\sin \varphi} + o(1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Difficult sets of Equations, counting
Let $ m$ be the number of solutions in positive integers to the equation $ 4x+3y+2z=2009$, and let $ n$ be the number of solutions in positive integers to the equation $ 4x+3y+2z=2000$. Find the remainder when $ m-n$ is divided by $ 1000$.
Please no complete solutions.
$4x + 3y + 2z = 2009$ first.
$2z = 2009 - 4x - 3y$
$\implies 2z \equiv 2 - 4x \pmod{3} \implies z \equiv 1 - 2x \pmod{3}$.So $z = 1 - 2x + 3k$.
$4x + 3y + 2 - 4x + 6k = 2009$.
$3y = 2007 + 6k$.
$ y = (2007 + 6k)/3 = 2007/3 + 2k = 669 + 2k$.
But that doesnt work properly.
I literally cant do ANYTHING, I am very stuck!
EDIT:
$x, y, z = 1$ each.
$3y + 2z = 2005 \implies 3y \equiv 1 \equiv 3 \pmod{2} \implies y = 1 + 2k, z = \frac{2005 - 3(1 + 2k)}{2} = 1001 - 3k.$
$2x + z = 1003 \implies z \equiv 1 \pmod{2}, z = 1 + 2k, x = \frac{1002 - 2k}{2} = 501 - k$
$4x + 3y = 2007 \implies 3y \equiv 3 \pmod{4} \implies y = 1 + 4k, x = \frac{2007 - 3y}{4} = 1002 - 3k$
I just cannot find how many $k$ values will work.
| If $(x,y,z)=(a,b,c)$ is a solution of $4x+3y+2z=2000$, then $(x,y,z)=(a+1,b+1,c+1)$ is a solution of $4x+3y+2z=2009.$
Now find the number of the solutions $(x,y,z)$ of $4x+3y+2z=2009$ such that either $x,y$ or $z$ equals $1$. (note that this is $m-n$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that the function is continuous To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ with $f=\left\{\begin{matrix}
\frac{x^3-y^3}{x^2+y^2} & , (x,y) \neq (0,0)\\
0 & , (x,y)=(0,0)
\end{matrix}\right.$ is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$ so we have to show that $\left |\frac{x^3-y^3}{x^2+y^2}\right | \leq L \sqrt{x^2+y^2}$.
I have done the following:
$$\left |\frac{x^3-y^3}{x^2+y^2}\right |=\frac{|x-y||x^2+xy+y^2|}{x^2+y^2}\leq \dfrac{(|x-y|)(x^2+y^2+|xy|)}{x^2+y^2} \leq 2\dfrac{(|x-y|)(x^2+y^2)}{x^2+y^2} = 2(|x-y|)\leq 2(|x|+|y|)\leq 2\sqrt{(|x|+|y|)^2}=2\sqrt{|x|^2+|y|^2+2|xy|} \leq 2\sqrt{|x|^2+|y|^2+|x|^2+|y|^2}=2\sqrt{2(|x|^2+|y|^2)}=2\sqrt{2}\sqrt{|x|^2+|y|^2}=2\sqrt{2}\sqrt{x^2+y^2}=2\sqrt{2}||(x,y)||$$
Is this correct?
| Hint for a different, but elegant, approach: Try polar coordinates.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Number of polyhedron diagonals Suppose that I have a polyhedron with given number of faces, edges and vertices are given. Is there a formula that gives me the number of polyhedron diagonals, http://mathworld.wolfram.com/PolyhedronDiagonal.html ?
| In my comment, I failed to recognize that you were interested in just the "space diagonals". As you suspect, eliminating face diagonals requires knowing the shapes of the various faces.
Suppose, for $k = 3, 4, \dots$, there are $f_k$ faces with $k$ sides. Arguing as in my comment, we observe that the vertices of each such face determine $\frac12 k(k-1)$ segments, of which $k$ are edges; there are $\frac{1}{2} k(k-1)-k = \frac12 k(k-3)$ face diagonals per $k$-face.
Therefore, for a polyhedron with $v$ vertices, $e$ edges, and $f = f_3 + f_4 + \cdots$ faces, the number of space diagonals is given by
$$d = \frac12 v(v-1) - e - \frac12 \sum_{k=3}^{\infty}f_k\,k\,(k-3)$$
We can massage this expression a bit, using Euler's formula $v-e+f=2$ and writing $f$ in terms of the $f_k$s.
$$\begin{align}
d &= \frac12 v(v-1) - (v+f-2) - \frac12 \sum_{k=3}^{\infty} f_k k(k-3) \\[4pt]
&= 2 + \frac12 v(v-1) - v - \sum_{k=3}^\infty f_k - \frac12 \sum_{k=3}^{\infty} f_k k(k-3) \\[4pt]
&= 2 + \frac12 v(v-3) - \frac12 \sum_{k=3}^{\infty} f_k (k-1)(k-2)
\end{align}$$
This gives
$$d = \frac12\left(\;4 + v(v-3) - \sum_{k=3}^{\infty} f_k (k-1)(k-2)\;\right)$$
As a sanity check, here are some test cases:
Tetrahedron: $v = 4$; $f = f_3 = 4$; $f_{\neq 3} = 0$
$$d = \frac12\left(\; 4 + 4(4-3) - 4 (3-1)(3-2) \;\right) = 0\;\checkmark$$
Cube: $v = 8$; $f = f_4 = 6$; $f_{\neq 4} = 0$
$$d = \frac12\left(\; 4 + 8(8-3) - 6 (4-1)(4-2) \;\right) = 4\;\checkmark$$
Octahedron: $v = 6$, $f = f_3 = 8$, $f_{\neq 3} = 0$
$$d = \frac12\left(\; 4 + 6(6-3) - 8 (3-1)(3-2) \;\right) = 3\;\checkmark$$
Truncated Icosahedron (Soccer Ball): $v = 60$, $f_5 =12$, $f_6 = 20$, $f_{\text{other}} = 0$
$$d = \frac12\left(\; 4 + 60(60-3) - 12(5-1)(5-2) - 20( 6-1)(6-2) \;\right) = 1440 \;\checkmark ?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $(1+\frac{1}{n})^n+\frac{1}{n}$ is eventually increasing I would like to find a way to show that the sequence $a_n=\big(1+\frac{1}{n}\big)^n+\frac{1}{n}$ is eventually increasing.
$\hspace{.3 in}$(Numerical evidence suggests that $a_n<a_{n+1}$ for $n\ge6$.)
I was led to this problem by trying to prove by induction that $\big(1+\frac{1}{n}\big)^n\le3-\frac{1}{n}$, as in
$\hspace{.4 in}$ A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?
| Let
$$ \eqalign{f(n) = \dfrac{1}{n} + \left( 1 + \dfrac{1}{n}\right)^n &= \dfrac{1}{n} + \exp\left( n \ln\left(1+\dfrac{1}{n}\right)\right) \cr &=
\dfrac{1}{n} + \exp\left(1 - \dfrac{1}{2n} + \dfrac{1}{3n^2} + O\left(\dfrac{1}{n^3}\right)\right) \cr &= e - \dfrac{e-2}{2n} + \dfrac{11e}{24 n^2} + O\left(\dfrac{1}{n^3}\right) }$$
Then $$\eqalign{f(n+1) &= e - \dfrac{e-2}{2n+2} + \dfrac{11e}{24 (n+1)^2} + O\left(\dfrac{1}{n^3}\right)\cr
&= e - \dfrac{e-2}{2n} + \dfrac{23 e - 24}{24 n^2} + O\left(\dfrac{1}{n^3}\right) \cr
f(n+1) - f(n) &= \dfrac{12e-24}{24n^2} + O\left(\dfrac{1}{n^3}\right)}$$
and since $e > 2$, this is positive for sufficiently large $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
$x(x^2-2)=0$, The answers are $x = 0, \sqrt{2}$, how do I get there? $$x(x^2-2)=0$$
The answers are $x=\sqrt{2}, 0$ how do I get there?
| Let's start from the beginning.
You are given
$$x(x^2 - 2 ) = 0$$
What are they asking for here?
Well, they're asking for any value of $x$ which would cause the given statement to become true.
What does this mean?
It means that if you were to substitute a number in place of $x$, then solve out the equation, the resulting statement must make logical sense.
How do we find these values of $x$?
We can make a few observations that will aide us in the discovery of these values. For instance, we could separate the equation into two parts then solve each individually.
Observe,
$$x(x^2-2) = 0$$
$$x\qquad(x^2-2) = 0$$
$$x=0,\qquad(x^2-2) = 0$$
Why do we do this?
Well, we want to find a value for $x$ that makes the whole equation true, and since the equation is just a product, if a $0$ is the result anywhere in the equation the entire product will become $0$.
Less talk, more proof.
$$
\begin{align}
x(x^2-2) &= 0\qquad x(x^2-2) = 0\\
(0)(x^2-2) &= 0\qquad\qquad\; x(0) = 0\\
0 &= 0\qquad\qquad\quad\ \ \ 0 = 0
\end{align}
$$
$$\color{green}{True!}$$
So , we can break up this equation into two parts and just solve each individually for an $x$ that will be equal to $0$.
Now let's start actually doing this.
We have already divided up the equation above
$$x=0,\qquad(x^2-2) = 0$$
As you can see, we already have a value for $x$ given just by dividing up the equation, that value being $x=0$. Now, we have to see if there are any other possible solutions that can be given by solving the other part of the equation.
Let's look at
$$(x^2-2) = 0$$
All we have to do here is a bit of algebra to solve for $x$.
$$
\begin{align}
x^2-2 & = 0\\
x^2-2+2 & = 0+2\\
x^2 & = 2\\
\sqrt{x^2} & = \pm\sqrt{2}\\
x & = \pm\sqrt{2}\\
\end{align}
$$
Why is there a $\pm$ in front of $\sqrt{2}$?
That is a 'plus or minus' sign, it literally means that the following value can be either positive or negative.
Why do we need it?
Well, squaring a real number will always produce a positive number, regardless of whether the real number being squared is positive or negative.
$$(2)^2 = 4$$
$$(-2)^2 = 4$$
Therefore, when we are undoing a square, in other words taking a square root, we cannot assume that the result is just positive or negative, since it could be either and still be true. So, we have to split the result of a square root up into two components, its positive result and its negative result, and solve the resulting equations using each result in turn.
In summary,
$$
\begin{align}
x &= \pm\sqrt{2}\\
\implies x=\sqrt{2}&,\quad x=-\sqrt{2}\\
\end{align}
$$
Now that we have our values for $x$, we have to check them to make sure nothing was done incorrectly.
$$\color{red}{\text{Always check your work!}}$$
Given:
$$x(x^2 - 2 ) = 0$$
Tests:
$$x=0\qquad x=\pm\sqrt{2}$$
We see:
$$
\begin{align}
x(x^2 - 2 ) &= 0 \qquad\qquad\quad x(x^2 - 2 ) = 0 \qquad\qquad\qquad\ \ \ x(x^2 - 2 ) = 0\\
(0)((0)^2 - 2 ) &= 0 \qquad (\sqrt{2})((\sqrt{2})^2 - 2 ) = 0 \qquad (-\sqrt{2})((-\sqrt{2})^2 - 2 ) = 0\\
(0)(0-2) &= 0 \qquad\qquad (\sqrt{2})(2 - 2) = 0 \qquad\qquad\quad (-\sqrt{2})(2 - 2 ) = 0\\
(0)(-2) &= 0 \qquad\qquad\quad\ \ \ (\sqrt{2})(0) = 0 \qquad\qquad\qquad\ \ \ (-\sqrt{2})(0) = 0\\
0 &= 0 \qquad\qquad\qquad\qquad\ \ \ 0 = 0 \qquad\qquad\qquad\qquad\qquad\ \ 0 = 0\\
\end{align}
$$
$$
\color{green}{True!}
$$
Everything works out.
Tips hat, walks away.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
elementary problem in combinatorics Let $X=\{1,2,3,4,5,6,7,8,9,10\}$ and $R$ be a set defined by $$\{(x,y)\in X\times X: \text{$x$ and $y$ have the same remainder when divided by $3$}\}$$ Then what's the numbers of elements in $R$? My approach is
$\bar{0} = \{3,6,9\}$ so we can get $2!\binom{3}{2}$ numbers of elements
$\bar{0}= \{1,4,7,10\}$ we get $2!\binom{4}{2}$ elements
$\bar{2} = \{2,5,8\}$ we get $2!\binom{3}{2}$ elements
Now counting all the elements like $(1,1),(2,2)$ and so on..., I get the answer $34$. Is it correct?
| Your answer is correct.
$R\subset X\times X$ can be recognized as an equivalence relation. As comes forward in your question its equivalence classes are $\{3,6,9\}$, $\{1,4,7,10\}$ and $\{2,5,8\}$.
Then: $$R=\left[\{3,6,9\}\times\{3,6,9\}\right]\cup\left[\{1,4,7,10\}\times\{1,4,7,10\}\right]\cup\left[\{2,5,8\}\times\{2,5,8\}\right]$$
And its cardinality is $3^2+4^2+3^2=34$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<$...$<\frac{n-1}{n}$ In an attempt to find a pattern, I did this:
Let a,b,c,d be non-zero consecutive numbers. Then we have:
$a=a$
$b=a+1$
$c=a+2$
$d=a+3$
This implies:
$\frac{a}{b}=\frac{a}{a+1}$
$\frac{b}{c}=\frac{a+1}{a+2}$
$\frac{c}{d}=\frac{a+2}{a+3}$
I don't know how that helps. I'm greatly seeking your help. Thank you very much.
| If $a_n = \frac{n-1}{n}$,
$$a_{n+1} - a_n = \frac{n}{n+1} - \frac{n-1}{n} = \frac{n^2 - (n-1)(n+1)}{n(n+1)} = \frac{1}{n(n+1)} > 0$$
Also,
$$a_n = \frac{n-1}{n} = 1 - \frac1 n$$
It should be clear that $1/n$ is decreasing as $n$ increases, and so this sequence increases towards $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Primes of the form $x^2+n\cdot y^2$, given $n$? In an attempt to get to grips with algebra for a course I intend to follow, I was working through a bunch of exercise sheets. A series of questions got me wondering:
Given an integer $n$, is there a general method (or collection of methods) to tell which primes $p$ are of the form $p=x^2+n\cdot y^2$? Here $x$ and $y$ should also be integers.
For example, for $n=2$ these are precisely the primes $p\equiv1,2,3\pmod{8}$. For $n=5$ (for example) all such primes are congruent to $1$ or $9$ modulo $20$, but I'm unable to prove (or even guess) whether all primes $p\equiv1,9\pmod{20}$ are of this form. My proofs can't seem to avoid the rings $\Bbb{Z}[\sqrt{-n}]$, which makes me think this could related to class field theory, which I think is beyond my grasp (for now). Is there an 'easy' solution to this problem?
| well, no. if you pick a discriminant of binary quadratic forms, call it $\Delta,$ and you have an odd prime $p$ that does not divide $\Delta,$ finally $(\Delta|p)= 1,$ then there is some form of that discriminant that represents the prime. If the class number is larger than one, that may not be the principal form. To find the form that works, see How to find a quadratic form that represents a prime?
Usual example: discriminant $-108.$ Every prime $p \equiv 1 \pmod 3$ is represented by some form, out of $x^2 + 27 y^2,$ $4 x^2 + 2 xy + 7 y^2,$ $4 x^2 - 2 xy + 7 y^2.$ The last two represent the same numbers, of course, but this way you get the class group. How can you tell? Well, given $p \equiv 1 \pmod 3,$ if there exists a solution to $u^3 \equiv 2 \pmod p,$ then we can write $p = x^2 + 27 y^2.$ If not, we have $p =4 x^2 \pm 2 xy + 7 y^2.$ Indeed, $31 \equiv 1 \pmod 3,$ and
$$ 4^3 = 64 = 62 + 2 \equiv 2 \pmod {31}, $$ from which we know we will be able to find $$ 31 = 4 + 27 = 2^2 + 27 \cdot 1^2 $$
Let's see, the cubes $\pmod 7$ are $\pm 1,$ easy enough to confirm by hand. Two is not a cube, $7$ cannot be written as $x^2 + 27 y^2,$ but we can write it as $4x^2 + 2 xy + 7 y^2$ with $x=0,y=1.$ Cubes $\pmod {13}$ are $1,5,8,12,$ no $2,$ and we can write $13$ as $4x^2 + 2 xy + 7 y^2$ with $x=1,y=1.$ Or as $4x^2 - 2 xy + 7 y^2$ with $x=1,y=-1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Proof that expression is integer, $\frac{(3n)!}{6^nn!}$ Can you help me with this exercises?
Proof that expression is integer,
$$\frac{(3n)!}{6^nn!}$$
I've tried for induction!!
$p(1):\frac{(3)!}{6}=1 $
for $p(k)=\frac{(3k)!}{6^kk!}$
for $p(k+1)=\frac{(3k+3)!}{6^{k+1}(k+1)!}$
where,
$$\frac{(3k+3)(3k+2)(3k+1)(3k)!}{6^k.6(k+1)(k!)},$$For hypothesis:
$$\frac{(3k+2)(3k+1)}{2},$$
How can I follow??
help me??
| Let us begin with the basic case, $n=0$. The expression $\frac{(3n)!}{6^nn!}$ is an integer for $n=0$, so the basic case is valid.
Let us assume that the statement is true for $n$, and that the integer it results in is equal to $k$. Our goal is to prove it true for $n+1$.
We first plug in $n+1$ for the expression.
\begin{align*}
\frac{[3(n+1)]!}{6^{n+1}(n+1)!}
\end{align*}
Using factorial properties and properties of exponents, we can rewrite it like this:
\begin{align*}
\frac{(3n+3)(3n+2)(3n+1)(3n)!}{6\cdot6^n(n+1)n!}
\end{align*}
Using our inductive hypothesis, we can rewrite the expression as:
\begin{align*}
k\cdot\frac{(3n+3)(3n+2)(3n+1)}{6(n+1)}
\end{align*}
We can use algebraic simplification to completely remove the $3n+3$ out of the numerator along with $n+1$, leaving us with:
\begin{align*}
k\cdot\frac{(3n+2)(3n+1)}{2}
\end{align*}
Notice that $3n+2$ and $3n+1$ are consecutive integers, and the product of consecutive integers is always even, since one of the two integers must be even. This leads to the claim that
\begin{align*}
\frac{(3n+2)(3n+1)}{2}\in\mathbb{Z}
\end{align*}
This means that the entire expression is an integer, and the expression is an integer for $n+1$, finishing the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$ It is asked to calculate $e^A$, where
$$A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$$
I begin evaluating some powers of A:
$A^0= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\; ; A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \; ; A^2 = \begin{pmatrix} -4 & 0 \\ 0 & -4\end{pmatrix} \; ; A^3 = \begin{pmatrix} 0 & -4 \\ 16 & 0\end{pmatrix}\; ; $
$ A^4=\begin{pmatrix} 16 & 0 \\ 0 & 16\end{pmatrix},\; \ldots$
I've noted that, since
$$e^A = \sum_{k=0}^\infty \frac{A^k}{k!}$$
we will have the cosine series at the principal diagonal for $\cos(2)$. But couldnt get what we will have in $(e^A)_{12}$ and $(e^A)_{21}$.
Also, we know that if $B=\begin{pmatrix} 0 & \alpha \\ -\alpha & 0 \end{pmatrix}$, then $e^B = \begin{pmatrix} \cos(\alpha) & \sin(\alpha) \\ -\sin(\alpha) & \cos(\alpha) \end{pmatrix} $. Is there a general formula for
$$B=\begin{pmatrix} 0& \alpha \\ \beta & 0 \end{pmatrix}$$?
Thanks!
| By separating odd and even terms in the series, what you have found can be rewritten as:
$$
e^A=I \sum_{k=0}^\infty{(-1)^k4^k\over(2k)!}+A \sum_{k=0}^\infty{(-1)^k 4^k\over(2k+1)!}=I \cos2+A{1\over2}\sin2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Calculation of improper integral: $\int_{0}^{\infty}\frac{x^3\ln x}{(x^4+1)^3} \,dx$ One of an exam's task was to calculate the following integral: $$I=\int_{0}^{\infty}\frac{x^3\ln x}{(x^4+1)^3} \,dx$$
I tried integration by parts:
$$I=\frac{1}{4}\int_{0}^{\infty}\ln x \cdot (x^4+1)^{-3} \,d(x^4+1)$$
but then things got messy and I ultimately concluded that the improper integral diverges, which is wrong since the answer is:
$$I=-\frac{1}{32}$$
What should I do?
| Integrating by parts,
$$
\int\frac{x^3\log x}{(1+x^4)^3}\,dx=-\frac{\log x}{8(1+x^4)^2}+\int \frac{1}{8(1+x^4)^2x}\,dx
$$
Doing partial fraction decomposition,
$$
\frac{1}{8(1+x^4)^2x}=\frac{1}{8x}-\frac{x^3}{8(1+x^4)}-\frac{x^3}{8(1+x^4)^2}.
$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove by Induction: $8^n - 3^n$ is divisible by $5$ for all $n \geq 1$ Prove by Induction that for all $n \geq 1$ we have
$$8^n - 3^n \text{is divisible by 5} ...(*)$$
My proof so far
Step 1: For $n=1$ we have $8^1 - 3^1 = 8 - 3 = 5$ which is divisible by 5.
Step 2: Suppose (*) is true for some $n=k\geq 1$ that is $8^k-3^k$ is divisible by 5.
Step 3: Prove that (*) is true for $n=k+1$, that is $8^{k+1} - 3^{k+1}$ is divisible by 5. We have
$$8^{k+1}-3^{k+1} = 8*8^k - 3*3^k$$
Can anyone explain the next logical expansion?
Update:
$$8^{k+1}-3^{k+1} = 8*8^k - 3*3^k = 5*8^k + 3*8^k - 3*3^k = 5*8^k + 3(8^k - 3^k)$$
Now we can say that $5*8^k$ is divisible by 5 since it has the form $5*p$ where $p$ is an integer $\geq 1$
And it is assumed that $8^k-3^k$ is divisible by $5$, then $3(8^k-3^k)$ is divisible by $5$ which means it has the form $5p$
so we reduce the expression to
$$5p+5p = 5(p+p)$$
which is of the form $5p$, which is divisible by $5$.
Is my proof correct?
| Continue like this:
$8 \cdot 8^k - 3 \cdot 3^k = (5+3) \cdot 8^k - 3 \cdot 3^k = 5 \cdot 8^k + 3 \cdot (8^k - 3^k)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Prove that the group $G$ is abelian if $a^2 b^2 = b^2 a^2$ and $a^3 b^3 = b^3 a^3$
In a Group $G$, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds, $\forall a,b\in G$. Prove that the group $G$ is abelian.
My approach was the following:
Let $a,b\in G$
Then, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds.
Now, $$\begin{align}
a^3b^3=b^3a^3 \\
\implies aaabbb=&bbbaaa\\
\implies a\cdot a^2\cdot b^2&\cdot b=b\cdot b^2\cdot a^2\cdot a\\
\implies a\cdot b^2\cdot a^2&\cdot b =b\cdot a^2\cdot b^2\cdot a\\
\implies ab \cdot ba \cdot a&b=ba\cdot ab\cdot ba
\end{align}$$
I am unable to proceed further.
What I need is a simple proof using simple theorems on groups, better if could be done using elementary properties.
| Hint: If you can somehow show that $a^2b^3=b^3a^2$. First take $(a^2b^3)^2$ and then prove it equal to $(b^3a^2)^2$. After this show that $a^2b^3=b^3a^2$ implies the group to be abelian
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the number of possible solutions in $x+y+z=30$ under conditions.
Three variables $x,y,z$ have a sum of $30$. All three are Non-Negative integers.
If any $2$ variables don't have the same value and exactly one variable has
value less than or equal to $3$, find the number of possible solutions ?
$a.)\ 98 \\
b.)\ 285 \\
c.)\ 68 \\
\color{green}{d.)\ 294\\}
$
I did
$x=0,y+z=30\implies 31\ \text{ways}$
$x=1,y+z=29\implies 30\ \text{ways}$
$x=2,y+z=28\implies 29\ \text{ways}$
$x=3,y+z=27\implies 28\ \text{ways}$
Total ways=$118$
But the book is giving answer as $294$ .
I look for a short and simple way.
I have studied maths upto $12$th grade.
| X+Y+Z=30 ; given any one of the number ranges from 0-3 and all other numbers start from 4. Hence consider the following equations:
*
*X=0 ; Y+Z=30
The solution of the above equation is obtained from (n-1)C(r-1) formula.
Total solutions obtained = 29 of which (y,z): (2,27)(27,2)(2,28)(28,2)(1,29)(29,1) must not be counted since they contain 1,2,3 as a part of solution. So, total number of ways= 29-6= 23 ways.
*X=1; Y+Z=29
Total solutions obtained = 28 of which (y,z): (2,27)(27,2)(1,28)(28,1)(3,26)(26,3) must not be counted since they contain 1,2,3 as a part of solution. So, total number of ways= 28-6= 22ways.
*X=2 ; Y+Z=28
Total solutions obtained = 27 of which (y,z): (1,27)(27,1)(2,26)(26,2)(3,25)(25,3) must not be counted since they contain 1,2,3 as a part of solution. So, total number of ways= 27-6= 21ways.
*X=3 ; Y+Z= 27
Total solutions obtained = 26 of which (y,z): (1,26)(26,1)(2,25)(25,2)(3,24)(24,3) must not be counted since they contain 1,2,3 as a part of solution. So, total number of ways= 26-6= 20 ways.
Total number of ways = 23+22+21+20
= 84 ways
Total number of ways considering X,Y and Z = 84*3 = 252.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Solving $2\cot 2x\cos2x = 1-\sin 2x$ How would I solve the following trigonometric equation?
$$2\cot 2x\cos2x = 1-\sin 2x$$
I got to this stage: $2\cos^22x = \sin2x - \sin^22x$
How do I continue?
| $$2cot2xcos2x=1-sin2x\\ \frac { \cos ^{ 2 }{ 2x } }{ \sin { 2x } } =\frac { 1-\sin { 2x } }{ 2 } \\ 2\left( 1-\sin ^{ 2 }{ 2x } \right) +\sin ^{ 2 }{ 2x } -\sin { 2x } =0\\ \sin ^{ 2 }{ 2x } +\sin { 2x } -2=0\\ \sin { 2x } =\frac { -1\pm 3 }{ 2 } \\ \sin { 2x\neq -2 } ,\sin { 2x } =1\\ \sin { 2x } =1\Rightarrow 2x=\frac { \pi }{ 2 } +2n\pi \Rightarrow \\$$
so the final answer is:
$$ x=\frac { \pi }{ 4 } +n\pi $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ then $\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.$ Suppose that $\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ . Then , prove that $\displaystyle\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.$
Attempt :
From the given relation , $\displaystyle \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}=-2$.........(1)
Now I want to calculate , $\displaystyle \frac{a^5}{b^5}+\frac{b^5}{a^5}+\frac{b^5}{c^5}+\frac{c^5}{b^5}+\frac{a^5}{c^5}+\frac{c^5}{a^5}$. I tried by expanding pair of terms and putting the value of (1), but it can't help...
| Given $$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}\Rightarrow \frac{(ab+bc+ca)}{abc} = \frac{1}{a+b+c}$$
So $$\displaystyle (ab+bc+ca)(a+b+c) = abc\Rightarrow a^2b+a^2c+ab^2+b^2c^2+c^2a+3abc=abc$$
so we get $(a+b)(b+c)(c+a) = 0\Rightarrow (a+b) = 0$ or $(b+c) =0$ or $(c+a) =0$
So $a=-b$ or $b=-c$ or $c=-a$
So $a=c$ or $b=-c$
So $$\displaystyle \frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5} = \frac{1}{c^5}$$ and $$\displaystyle \frac{1}{a^5+b^5+c^5} = \frac{1}{c^5}$$
So $$\displaystyle \frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5} = \frac{1}{a^5+b^5+c^5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $g(x) = x^2 + 3x + 1$ and $f \circ g = g\circ f$, then $f$ and $g$ intersect on the line $y = x$.
Let $f:\mathbb R\to \mathbb R$, $g(x) = x^2 + 3x + 1$ and $f∘g=g∘f$.
Prove that $Cf, Cg$ and the line $y=x$ have at least one common point.
My solution:
$fog=gof$ then $f=g$, $Df = Dg = \mathbb R$ and $f(x) = g(x)$
I fear that the above assumption is wrong..
But if it is correct, here is the rest of the solution.
Then $f(x) = g(x) = x^2 + 3x + 1$
Let $f(x)=y$ then $x^2 + 3x + 1 = y$
for $x=y$ we get: $x^2 + 3x + 1 = x$
$x^2 + 2x + 1 = 0$
$(x+1)^2 = 0$
$x = -1$
Because $x=y$ the common point is $(-1,-1)$
| Notice that the graphs of $g(x) = x^2 + 3x + 1$ and $y = x$ intersect at precisely one point: $(-1, -1)$. In other words, we know that:
$$\boxed{
g(x) = x \iff x = -1
} \tag{$\star$}$$
To obtain our common point, it is enough to show that $f(-1) = -1$.
To this end, let $k = f(-1)$ and suppose that $f \circ g = g\circ f$. Then in particular, we know that both functions agree at $x = -1$ so that:
$$
g(k) = g(f(-1)) = f(g(-1)) = f(-1) = k
$$
Hence, by $(\star)$, we conclude that $f(-1) = k = -1$, as desired. $~~\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Express the given quantity as a single Logarithm $\frac{1}{5}\ln(x+2)^5+\frac{1}{2}[\ln x-\ln(x^2+3+2)^2)]$ Express the given quantity as a single Logarithm
$$\frac{1}{5}\ln(x+2)^5+\frac{1}{2}[\ln x-\ln(x^2+3x+2)^2)]$$
My answer is $\ln\dfrac{[(x+2)^5]^{\frac{1}{5}}\cdot x^2}{[(x^2+3x+2)^2]^{\frac{1}{2}}}$
What have I done wrong?
| Given $$\displaystyle \frac{1}{5}\ln(x+2)^5+\frac{1}{2}\left[\ln(x)-\ln(x^2+3x+2)^2\right]$$
$$\displaystyle \ln(x+2)^{\frac{5}{5}}+\ln(x)^{\frac{1}{2}}-\ln(x^2+3x+2)^{\frac{1}{2}} = \ln(x+2)+\ln (\sqrt{x})-\ln(x^2+3x+2)$$
So we get $$\displaystyle \ln\left[\frac{(x+2)\cdot \sqrt{x}}{x^2+3x+2}\right] = \ln\left[\frac{(x+2)\cdot \sqrt{x}}{(x+1)(x+2)}\right] = \ln\left[\frac{\sqrt{x}}{x+1}\right]$$
Above we have Used $$\bullet\; n\cdot \ln(m) = \ln(m)^{n}$$ and $$\displaystyle \bullet\; \ln(m)-\ln(n) = \ln\left(\frac{m}{n}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$ $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$
I tried to solve this question but no luck.
My try:
$$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx=\int x^4(x^8+x^4+1)(2x^8+3x^4+6)^{1/4}dx\\
\int x^4(x^8+x^4+1)x^2(2+3x^{-4}+6x^{-8})^{1/4}dx$$
Now i got stuck,please help me reach the answer.Answer is $$\frac{x^5}{30}(2x^8+3x^4+6)^{\frac54}+C$$
| We start by factoring out $x^4$:
$$
(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}=x^4(x^{8}+x^4+1)(2x^8+3x^4+6)^{1/4}.
$$
Next, we write
$$
x^{8}+x^4+1=\frac{1}{6}(6+3x^4+2x^8)+\frac{1}{2}x^4+\frac{2}{3}x^8,
$$
so the integrand can be written as
$$
\Bigl(\frac{1}{2}x^8+\frac{2}{3}x^{12}\Bigr)(2x^8+3x^4+6)^{1/4} +\frac{1}{6}x^4(2x^8+3x^4+6)^{5/4},
$$
or, factoring out $x^5/24$ in the first term,
$$
\frac{x^5}{24}\Bigl(12x^3+16x^{7}\Bigr)(2x^8+3x^4+6)^{1/4} +\frac{1}{6}x^4(2x^8+3x^4+6)^{5/4}.
$$
Hooray (this is really lucky!), this is a derivative of a product, since
$$
D(2x^8+3x^4+6)=16x^7+12x^3,
$$
we find that the expression above is
$$
\frac{x^5}{30}D\bigl[(2x^8+3x^4+6)^{5/4}\bigr]+\bigl[D(x^5/30)\bigr](2x^8+3x^4+6)^{5/4},
$$
i.e.
$$
D\Bigl[\frac{x^5}{30}(2x^8+3x^4+6)^{5/4}\bigr].
$$
Hence,
$$
\int (x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}\,dx=\frac{x^5}{30}(2x^8+3x^4+6)^{5/4}+c.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
How do I prove the above expression using mathematical induction? So far I have only been proving simpler stuff. The base case P(1) is easy enough but I am lost as to where I should even start with my inductive step. I really don't know what the steps for P(k + 1) should be, and so help would be greatly appreciated.
| We are to prove
$$\sum_{k=1}^n(-1)^{k-1}k^2=\frac{(-1)^{n-1}n(n+1)}{2} \tag 1$$
First, we establish a base case. So, for $n=2$, we note that the left-hand side of $(1)$ is $1^2-2^2=-3$ while the right-hand side of $(1)$ is $(-1)\frac{2\times 3}{2}=-3$. Thus, the case $n=2$ is verified.
Next, we assume that $(1)$ holds for some number $N$ and demonstrate that it holds for $N+1$.
To that end, we have
$$\begin{align}
\sum_{k=1}^{N+1}(-1)^{k-1}k^2&=\sum_{k=1}^{N}(-1)^{k-1}k^2+(-1)^N(N+1)^2\\\\
&=\frac{(-1)^{N-1}N(N+1)}{2} +(-1)^N(N+1)^2\\\\
&=\frac{(-1)^{N+1}}{2}\left(N(N+1)-2(N+1)^2\right)\\\\
&=\frac{(-1)^N(N+1)(N+2)}{2}
\end{align}$$
and this completes the proof by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Adding pieces on NxN chess board sequentially The problem is:
Given a $N\times N$ chess board, if you put $B$ pieces starting on square $1,1$ and going right, then down, then left, then up when possible (thus, forming an spiral), in which coordinate the last piece would be put?
Example: $8\times8$ board, $53$ pieces
Putting $8$ on first row, then $7$ down, then $7$ left, then $6$ up, then $6$ right and so on, would give me position (row $4$, col $6$)
Please note the iterative solution is too costly, as $N$ can go up to $2^{30}$ and $B$ up to $2^{60}$.
| Assuming we have completed $x$ spirals (and no more spirals can be completed)
Then we have
(Comparing number of pieces utilized by $x$ and $x+1$ spirals)
$$ n^2-(n-2x)^2\le B\le n^2-(n-2x-2)^2$$
Now,$$n^2-(n-2x)^2\le B\implies n^2-B\le (n-2x)^2\implies x\le\frac{n-\sqrt{n^2-B}}{2}$$
Or,
$$n^2-(n-2x-2)^2\ge B\implies n^2-B\ge (n-2x-2)^2\implies x\ge\frac{n-\sqrt{n^2-B}}{2}-1$$
We get
$$\frac{n-\sqrt{n^2-B}}{2}-1\le x\le\frac{n-\sqrt{n^2-B}}{2}$$
For our purpose we can safely say
$$\left\lceil{\frac{n-\sqrt{n^2-B}} {2}}-1\right\rceil=x=\left\lfloor{\frac{n-\sqrt{n^2-B}} {2}}\right\rfloor$$
Now we have number of completed spirals then it's not difficult to obtain position of last piece.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
Then, plugging the value $1$ for $x$, you get $\frac{1}{3}$.
I don't see how did they reach that conclusion. This is how I tried to tackle it:
$$\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} = \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} \cdot \frac{\sqrt[3]{x}+1}{\sqrt[3]{x}+1},$$
which then yields
$$\frac{x-1}{(2\sqrt{x}-2)(\sqrt[3]{x}+1)},$$
and that becomes
$$\frac{x-1}{2\cdot(\sqrt{x}-1)\cdot(\sqrt[3]x+1)}.$$
That's
$$\frac{x-1}{2\cdot(\sqrt[6]{x}+\sqrt{x}-\sqrt[3]{x}-1)},$$
and this will still evaluate to $\frac{0}{0}$.
How did they solve this, exactly?
| Why not to apply L-Hospital's rule for $\frac{0}{0}$ form
$$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{2\sqrt x-2}$$
$$=\lim_{x\to 1}\frac{\frac{d}{dx}\left(\sqrt[3]{x}-1\right)}{\frac{d}{dx}(2\sqrt x-2)}$$
$$=\lim_{x\to 1}\frac{\frac{1}{3}x^{-2/3}}{2\frac{1}{2}x^{-1/2}}$$
$$=\lim_{x\to 1}\frac{1}{3}x^{-1/6}=\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.
| In the equation $x^3+\frac{1}{x^3}=18$, multiply everything by $x^3$ to get $x^6-18x^3+1=0$. Then let $y=x^3$ and solve $y^2-18y+1=0$, then substitute back in to get $x$. Then you can input the value into $x^{11}+\frac{1}{x^{11}}$ directly.
It winds up being $x^{11}+\frac{1}{x^{11}}=39603$ (regardless of whether you use the plus or the minus in the quadratic formula).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Solutions to the Laplace Equation $\Delta u =0$, where $u= \log p$ Find all real solutions to the two dimensional Laplace equation $U_{xx} + U_{yy} =0$ of the form $u=\log p(x,y)$, where $p$ is a quadratic polynomial.
Solution:
Let $p(x,y) = Ax^2 + By^2 +Cxy + D$ be a quadratic polynomial such that $A, B \not= 0$. Then
$$U_{x} = \frac{2Ax + Cy}{\ln(10)(Ax^2 + By^2 +Cxy + D},$$
$$U_{xx} = \frac{2A ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2Ax + Cy)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2},$$
$$U_{y} = \frac{2By + Cx}{\ln(10)(Ax^2 + By^2 +Cxy + D)},$$
$$U_{yy} = \frac{2B \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2By + Cx)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2}.$$
This implies
$$U_{xx} + U_{yy} = \frac{2A \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2Ax + Cy)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2} + \frac{2B \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2By + Cx)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2} = 0. $$
I feel like I am not doing this right. Is there a simpler way? Thanks. And also how do I find such solutions.
| The solution that is proposed makes use of the wolfram language for symbolic computations.
In the following link you can download the notebook to explore all the details:
https://community.wolfram.com/groups/-/m/t/2108153
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving quartic equations Given the following quartic equation:
$$x^4-2x^3-7x^2+8x+12=0$$
Could anyone give some techniques required to solve any quartic equation (apart from this one) if they exist?
| As an A level student, I would first look at this with the Remainder Theorem, which states that if we sub in a value, say 1, and we get zero, then (x-1) is a factor.
Let $$p(x)=x^4-2x^3-7x^2+8x+12$$
Sub in -1:
$$p(-1)=(-1)^4-2(-1)^3-7(-1)^2+8(-1)+12 = 0$$
Therefore, (x + 1) must be a factor. We can then use algebraic division to gain a cubic equation. After doing the division, we are left with:
$$p(x)=(x+1)(x^3-3x^2-4x+12)$$
We can then use the remainder theorem to try and factorise the cubic. Subbing in -2, we get another 0. Therefore, (x+2) is a factor of the cubic. Long division gives that p(x) equals:
$$p(x)=(x+1)((x+2)(x^2-5x+6))$$
The quadratic can be factorised to give:
$$p(x)=(x+1)(x+2)(x-3)(x-2)$$
Since this equals 0, then:
$$x+1 = 0 \implies x = -1$$
$$x+2 = 0 \implies x = -2$$
$$x-3 = 0 \implies x = 3$$
$$x-2 = 0 \implies x = 2$$
To confirm, a graph gives:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Help factorising a sixth degree polynomial I have to factorise- $$x^6+5x^3+8$$Answer is $$(x^2−x+2)(x^4+x^3−x^2+2x+4)$$.I have also used factor theorem.Please help me.Thanks in advance.
| It can be factored with help of following identities (applied twice below, marked by a '*')
$$u^3 \pm v^3 = (u \pm v)(u^2 \pm uv + v^2)$$
Let $u = x^2 + 2$, we have
$$\begin{align}
x^6 + 5x^3 + 8
&= (x^2)^3 + 2^3 + 5x^3\tag{1}\\
&\stackrel{*}{=} (x^2+2)(x^4 - 2x^2 + 4) + 5x^3\\
&= u(u^2 - 6x^2) + 5x^3\tag{2}\\
&= (u^3 - x^3) - 6x^2(u-x)\\
&\stackrel{*}{=} (u-x)(u^2 + ux + x^2 - 6x^2)\\
&= (x^2 - x + 2)((x^2 + 2)^2 + x(x^2+2) - 5x^2)\\
&= (x^2 - x + 2)(x^4 + x^3 - x^2 + 2x + 4)
\end{align}
$$
Rationale behind the steps
*
*The motivation for step 1 is the non-zero coefficients of $x^k$ are symmetric
around $k = 3$ term. i.e
$$x^6 + 5x^3 + 8 = x^3 \left(x^3 + 5 + \left(\frac{2}{x}\right)^3\right)$$
I try to express everything in terms of $x + \frac{2}{x}$ and looks for simplification.
*Some where in the process, I notice the $1, -6, 5$ pattern in some
expression equivalent to $(2)$. This implies the existence of a factor $u - x$
in the original expression. The rest is more or less following the nose.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1437662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Diophantine Equation or Ellipse I guess this is something to do with circle
The question is:
"Given x and y are real numbers, such that $2x^2 + 3y^2 - 4x - 12y = -14$, find xy."
What is the trick ?
| Rewrite
$$2x^2-4x+3y^2-12y=-14$$
$$2(x^2-2x)+3(y^2-4y)=-14$$
$$2(x^2-2x+1-1)+3(y^2-4y+4-4)=-14$$
$$2(x^2-2x+1)-2+3(y^2-4y+4)-12=-14$$
$$2(x-1)^2+3(y-2)^2-14=-14$$
$$2(x-1)^2+3(y-2)^2=0$$
When is a sum of two squares equal to zero? Only if both squares are equal to zero. So one can conclude $x-1=0$ and $y-2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $a^2+ab+b^2\ge 0$ How to prove that $a^2+ab+b^2\ge 0$?
Obviously the squares are positive, but how can I be sure that $ab$ doesn't become too negative with a certain combination of $a$ and $b$?
| $a^2+ab+b^2= (a+ \frac{1}{2}b)^2 + \frac{3}{4}b^2$
One may prefer symmetry:
$2(a^2+ab+b^2)= a^2 +b^2+(a+b)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
How did this result come about? I was reading Chebyshev polynomials Wiki page and I could not understand one thing
$$ T_n(x) = x^n \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \binom{n}{2k} \left (1 - x^{-2} \right )^k \\$$
$$= \tfrac{n}{2} \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} (-1)^k \frac{(n-k-1)!}{k!(n-2k)!}~(2x)^{n-2k} \\ $$
How did this step came about? They probably are missing one step in between. As far as I know, on needs to expand $\left (1 - x^{-2} \right )^k$ and then change the order of summation but that may take quite a time and it is not this easy, I guess.
| Suppose we have the Chebychev polynomial
$$T_n(x) = x^n \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} \left(1-\frac{1}{x^2}\right)^k
\\ = x^n \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} \frac{1}{x^{2k}}
\sum_{q=0}^k {k\choose q} (-1)^{k-q} x^{2q}$$
and we seek to verify that
$$[x^{n-2p}] T_n(x)
= \frac{1}{2} n (-1)^p \frac{1}{n-p} {n-p\choose p} 2^{n-2p}.$$
For this coefficient we must have $2q-2k = -2p$
or $q = k - p$
getting
$$ \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} {k\choose k-p} (-1)^p
\\ = \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} {k\choose p} (-1)^p.$$
To evaluate this introduce for the sum term
$${n\choose 2k} = {n\choose n-2k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k+1}} (1+z)^n \; dz.$$
This is zero when $2k\gt n$ so we may extend $k$ to infinity,
getting for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^n
\sum_{k\ge 0} {k\choose p} (-1)^p z^{2k} \; dz
\\ = (-1)^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^n
\sum_{k\ge p} {k\choose p} z^{2k} \; dz
\\ = (-1)^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^n
\sum_{k\ge 0} {k+p\choose p} z^{2k+2p} \; dz
\\ = (-1)^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2p+1}} (1+z)^n
\frac{1}{(1-z^2)^{p+1}} \; dz
\\ = (-1)^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2p+1}} (1+z)^{n-p-1}
\frac{1}{(1-z)^{p+1}} \; dz.$$
Exract the coefficient to get
$$(-1)^p \sum_{q=0}^{n-2p}
{n-p-1\choose q} {n-2p-q+p\choose p}
\\ = (-1)^p \sum_{q=0}^{n-2p}
{n-p-1\choose n-p-q-1} {n-p-q\choose p}
\\ = (-1)^p \sum_{q=0}^{n-2p} \frac{n-p-q}{n-p}
{n-p\choose n-p-q} {n-p-q\choose p}
\\ = (-1)^p \sum_{q=0}^{n-2p} \frac{n-p-q}{n-p}
{n-p\choose p} {n-2p\choose q}
\\= (-1)^p {n-p\choose p} \frac{1}{n-p}
\sum_{q=0}^{n-2p} (n-p-q)
{n-2p\choose q}.$$
Hence it remains to show that
$$\frac{1}{2} n 2^{n-2p} =
\sum_{q=0}^{n-2p} (n-p-q) {n-2p\choose q}.$$
This is
$$(n-p) 2^{n-2p}
- \sum_{q=0}^{n-2p} q {n-2p\choose q}
= (n-p) 2^{n-2p}
- \sum_{q=1}^{n-2p} q {n-2p\choose q}
\\ = (n-p) 2^{n-2p}
- (n-2p) \sum_{q=1}^{n-2p} {n-2p-1\choose q-1}
\\ = (n-p) 2^{n-2p} - (n-2p) 2^{n-2p-1}
= n 2^{n-2p} - n 2^{n-2p-1} = \frac{1}{2} n 2^{n-2p}.$$
This concludes the argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove $2^{2n} = \sum_{k=0}^{n}\binom{2n+1}{k}$ I'm trying to prove the following equation above. So far I have:
\begin{align}
2^{2n} &= (1+1)^{2n}\\
&= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{n-k} = \sum_{k=0}^{2n}\binom{2n}{k} & \text{(By the Binomial Theorem)}
\end{align}
I know I have to use the following identity somehow:
$$\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$$
How do I split my summation to get what I'm looking for? Thanks!
EDIT: HERE IS MY SOLUTION
\begin{align*}
2^{2n} &= (1+1)^{2n}\\
&= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{2n-k} & \text{(By the Binomial Theorem)}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{k}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{2n-k} & \text{(Binomial Symmetry)}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=0}^{n-1}\binom{2n}{k}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\
&= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\
&= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n+1}{k} & \text{(By Identity listed above)}\\
&= \sum_{k=0}^{n}\binom{2n+1}{k}
\end{align*}
| $$ 2^{2n}=\frac12\cdot 2^{2n+1}=\frac12\sum_{k=0}^{2n+1}{2n+1\choose k}=\frac12\sum_{k=0}^n\left({2n+1\choose k}+{2n+1\choose 2n+1-k}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Proving that $\Gamma \left(n+ \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{2^{2n}n!}$. The proof I am dealing with is worded exactly as follows:
Prove $\Gamma\left(n+ \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{2^{2n}n!}$.
The proof itself can be done easily with induction, I assume. However, my issue is with the domain of the given $n$; granted, the factorial operator is only defined for positive integer values. However, the gamma function, as far as I know, is defined for all complex numbers bar $\mathbb{Z}^-$. I don't think induction will suffice for generalizing the proof for said domain. Would that be necessary, or should I be looking for other methods to tackle the proof?
| We can also exploit the fact that $\Gamma(x+1)=x\Gamma(x)$ to obtain
\begin{align*}
\Gamma\left(n + \frac{1}{2}\right)
& =\left(n-1+\frac{1}{2}\right)\Gamma\left(n-1+\frac{1}{2}\right) \\
& =\left(n-1+\frac{1}{2}\right)\left(n-2+\frac{1}{2}\right)\Gamma\left(n-2+\frac{1}{2}\right) \\
& = \ldots
= \left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\dots \ \frac{1}{2}\ \Gamma\left(\frac{1}{2}\right) \\
& = \frac{(2n-1)(2n-3)\cdots1}{2^n} \cdot \Gamma\left( \frac{1}{2}\right) \\
& =\frac{(2n-1)(2n-2)(2n-3)(2n-4)\cdots1}{2^n(2n-2)(2n-4)\dots2} \cdot\Gamma\left( \frac{1}{2}\right) \\
& =\frac{(2n-1)!}{2^{2n-1}(n-1)!}\Gamma\left( \frac{1}{2}\right)\\
& =\frac{2n!}{2^{2n}n!}\sqrt{\pi}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Limits of two variables functions I'm computing limits of two variables functions I do not manage to resolve them all.
4)$\lim_{(x,y) \to (0,0)} \dfrac{x^3-y^3}{x^2+2y^2} $
Here I use polar coordinates with $x=r\cos{\theta}$ and $y=r\sin{\theta}$ I finally get $\lim_{(r) \to (0)} \dfrac{rcos^3\theta - sin^3 \theta}{cos^2\theta +2sin^2 \theta} $ but I do not see how to continue.
6)$\lim_{(x,y) \to (0,0)} \dfrac{x^{1/3}y^2}{x^2+y^2+|x-y|} $
For 6 I began using polar coordinates but do not manage to determine it too, how would it be possible to compute them ?
Thank you
| Regarding 4) you have $x^2+2y^2 \ge x^2 + y^2$, $\vert x^3 \vert \le \vert x \vert (x^2+y^2)$ and $\vert y^3 \vert \le \vert y \vert (x^2+y^2)$. Consequently: $$\vert \frac{x^3-y^3}{x^2+2y^2} \vert \le \vert \frac{x^3-y^3}{x^2+y^2} \vert \le \frac{(\vert x \vert + \vert y \vert)(x^2+y^2)}{x^2+y^2} \le \vert x \vert + \vert y \vert$$ and $\lim\limits_{(x,y) \to (0,0)} \vert x \vert + \vert y \vert =0$ so $$\lim\limits_{(x,y) \to (0,0)} \frac{x^3-y^3}{x^2+2y^2} = 0.$$
5) is not well defined. How do you define the function when $x+y=1$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int\frac{1}{\sin^2 x+\sin x+1}dx$
Evaluation of $\displaystyle \int\frac{1}{\sin^2 x+\sin x+1}dx$
$\bf{My\; Try::}$ Using $$\; \bullet\; x^2+x+1 = (x-\omega)\cdot (x-\omega^2)\;,$$ where $\omega,\omega^2$ are cube root of unity
So we can write Integal $$\displaystyle I = \int\frac{1}{(\sin x-\omega)\cdot (\sin x-\omega^2)}dx$$
So we get $$\displaystyle I = \frac{1}{\omega-\omega^2}\int\frac{(\sin x-\omega^2)-(\sin x-\omega)}{(\sin x-\omega)\cdot (\sin x-\omega^2)}dx$$
So $$\displaystyle I = \frac{1}{\omega-\omega^2}\int \left[\frac{1}{\sin x-\omega}-\frac{1}{\sin x-\omega^2}\right]dx$$
Now Substitute $$\displaystyle \sin x= \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$$
Can we solve it above method our we directly put $$\displaystyle \sin x= \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$$
in $$\displaystyle \int\frac{1}{\sin^2 x+\sin x+1}dx$$ and then solve it.
Or is there is any other process by which we can solve it.
Help me , Thanks
| Substitute $y=\sec x - \tan x$, or $\sin x=\frac{1-y^2}{1+y^2}$
\begin{align}
&\int \frac{1}{\sin^2 x+\sin x+1}dx\\
=&-\int\frac{2+{2y^2}}{3+y^4}dy
= -\int\frac{\frac2{y^2}+{2}}{\frac3{y^2}+y^2}dy \\
=&\ \frac{\sqrt3+1}{\sqrt{6\sqrt3}}\tan^{-1}\frac{y-{\sqrt3}y^{-1}}{\sqrt{2\sqrt3}}
-\frac{\sqrt3-1}{\sqrt{6\sqrt3}}\coth^{-1}\frac{y+{\sqrt3}y^{-1}}{\sqrt{2\sqrt3}}\\
= &\ \frac{\sqrt3+1}{\sqrt{6\sqrt3}}\tan^{-1}\frac{(\sqrt3-1)\sec x+(\sqrt3+1)\tan x}{\sqrt{2\sqrt3}}\\
&+\frac{\sqrt3-1}{\sqrt{6\sqrt3}}\coth^{-1}\frac{(\sqrt3+1)\sec x+(\sqrt3-1)\tan x}{\sqrt{2\sqrt3}}+C\\
\end{align}
which leads to the definite integral below
\begin{align}
&\int_{-\frac\pi2}^{\frac\pi2} \frac{dx}{\sin^2 x+\sin x+1}
=\frac\pi3 \sqrt{3+2\sqrt3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{ c + 1} −\sqrt{c}$ for all real $c \geq 1$. Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{c + 1} −\sqrt{c}$ for all real $c \geq 1$.
Can anyone provide some form of guidance? So far all I have been able to think of is writing $c$ as $x^2$ for some $x$, or eliminating the radical on one side...
| To show
$\sqrt{ c} − \sqrt{c − 1}
\geq \sqrt{c + 1} −\sqrt{c}
$.
This is the same as
$\sqrt{c}
\ge \frac{\sqrt{c + 1}+\sqrt{c - 1}}{2}
$.
If
$f(x) = x^{1/2}
$,
then
$f'(x) = \frac12x^{-1/2}
$
and
$f''(x) = -\frac14x^{-3/2}
< 0
$
so
$f(x)$
is concave
which means that
$f(\frac{a+b}{2})
\ge \frac{f(a)+f(b)}{2}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
If $(a,b,M)$ is a Pythagorean triple, can $(b,b+a,N)$ be another triple? Does anyone know of a pair of Pythagorean triples of the form
$$(a, b, M) \quad\text{and}\quad(b, b+a, N)$$
Is such a pair possible?
| Using Wolfram Alpha to ensure accurate algegra, we start with Euclid's formula (modified) for the OP.
$$A=m^2-n^2\qquad B=2mn+(m^2-n^2)\qquad C=m^2+n^2$$
$$A^2=m^4 - 2 m^2 n^2 + n^4$$
$$B^2=m^4 + 4 m^3 n + 2 m^2 n^2 - 4 m n^3 + n^4$$
$$C^2=m^4 + 2 m^2 n^2 + n^4$$
$$A^2+B^2-C^2=m^4 + 4 m^3 n - 2 m^2 n^2 - 4 m n^3 + n^4=0$$
The only solutions to the last equation are:
$$n =\pm m\qquad n = 2 m \pm \sqrt{5} m\qquad n=m=0$$
$\therefore$ there are no integer solutions except $A,B,C=(0,0,0), (0,2,2), (0,8,8), (0,18,18), ...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How do I solve $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1})$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule as homework. So I tried rationalizing the denominator and numerator but it didn't work.
My best was: $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1}) = \lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x^2+x+1-2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{(x-1)(x+2)}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x+2}{(x^2+x+1)}) = 3 ???$
| First note that $x^3 - 1 = (x-1)(x^2 + x + 1)$. Therefore multiply $\frac{1}{x-1}$ by 1 in the form $\frac{x^2 + x + 1}{x^2 + x + 1}$ and combine the fractions as $\frac{(x^2 + x + 1) - 2}{(x-1)(x^2 + x + 1)} = \frac{x^2 + x -1}{(x-1)(x^2 + x + 1)}$. The numerator is now going to $1 + 1 - 1 = 1$, while the denominator is going to $(1-1)(1 + 1 + 1) = (0)(3) = 0$. Notice that $1/0$ is not an indeterminant form. Notice that if we look at the limit from the right,
$\lim_{x \to 1^+} \frac{x^2 + x -1}{(x-1)(x^2 + x + 1)}$, the top is going to $1$, while the bottom is going to $0$ through positive values, so the limit evaluates to $+\infty$. Checking the limit from the right in the same way,
the top is still going to $1$, but the bottom is going to $0$ through negative values, so the limit evaluates to $-\infty$. The limit does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Another beautiful arctan integral $\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx$ Do you think we can express the closed form of the integral below in a very nice and short way?
As you already know, your opinions weighs much to me, so I need them!
Calculate in closed-form
$$\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx.$$
I'm looking forward to your precious feedback!
Mathematica tells us the closed form is
$$\frac{1}{4} i \text{Li}_2\left(\frac{1}{5}+\frac{2 i}{5}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{2}{15}+\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{5}-\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{2}{15}-\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}+\frac{i}{8}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}-\frac{i}{8}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{10}+\frac{3 i}{10}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{10}-\frac{3 i}{10}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}+\frac{i}{12}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}-\frac{i}{12}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{4}{15}+\frac{8 i}{15}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{4}{15}-\frac{8 i}{15}\right)+\frac{1}{4} \log (4) \tan ^{-1}(6)-\frac{1}{4} \log (4) \tan ^{-1}(9)+\frac{1}{4} \log (4) \tan ^{-1}\left(\frac{1}{6}\right)-\frac{1}{4} \log (4) \tan ^{-1}\left(\frac{1}{9}\right)-\frac{1}{4} \log (9) \tan ^{-1}(2)+\frac{1}{4} \log (9) \tan ^{-1}(3)-\frac{1}{4} \log (9) \tan ^{-1}(6)+\frac{1}{4} \log (9) \tan ^{-1}(9)-\frac{1}{4} \log (9) \tan ^{-1}\left(\frac{3}{55}\right)+\frac{1}{4} \log (16) \tan ^{-1}(2)-\frac{1}{4} \log (16) \tan ^{-1}(3).$$
| Substituting $x\to\frac{1-x}{1+x}$ is usually a first reaction of mine. Here it works surprisingly well: the integral is equivalent to
$$I=\int_{1/2}^1 \frac{\tan^{-1}\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx=\frac12\int_0^{\frac13}\frac{\tan^{-1}\left(\frac{x}{x^2+6}\right)}{x} dx.$$
Integrating by parts,
$$I=-\frac12\ln3 \tan^{-1}\left(\frac{3}{55}\right)-\frac12\int_0^{\frac13} \frac{6-x^2}{x^4+13x^2+36}\,\ln x \, dx$$
This looks managable. We can decompose $\displaystyle \, \frac{6-x^2}{x^4+13x^2+36}=\frac{2}{x^2+4}-\frac{3}{x^2+9} $
so that
$$\int_0^{\frac13} \frac{6-x^2}{x^4+13x^2+36}\,\ln x \, dx=2\int_0^{\frac13} \frac{\ln x}{x^2+2^2}\,dx-3\int_0^{\frac13}\frac{\ln x}{x^2+3^2}\,dx$$
Now these integrals may also be done by partial fraction (whence emerge the complex stuff) and the antiderivative of the remaining parts is easily found in terms of logarithms and dilogarithms..
So the logarithm and $tan^{-1}$ cancel at the end, leaving:
$$ I=\frac12\Im\operatorname{Li_2}\left(\frac1{9i}\right)-\frac12\Im\operatorname{Li_2}\left(\frac1{6i}\right)$$
Edit
The final cancellation and Chris' comment below suggest a detour to the integration by parts and partial fractions decomposition:
We just notice that $\displaystyle \tan^{-1}\left(\frac{x}{x^2+6}\right)=\tan^{-1}\left(\frac{x}{2}\right)-\tan^{-1}\left(\frac{x}{3}\right)$, whence
$$I=\frac12\int_0^{\frac13}\frac{\tan^{-1}\left(\frac{x}{x^2+6}\right)}{x} dx=\frac12\int_0^{\frac13}\frac{\tan^{-1}\left(\frac{x}{2}\right)-\tan^{-1}\left(\frac{x}{3}\right)}{x}dx\\=\frac12\int_0^{\frac16}\frac{\tan^{-1}x}{x}dx-\frac12\int_0^{\frac19}\frac{\tan^{-1}x}{x}dx=\frac{1}{2}\operatorname{Ti}_2\left(\frac{1}{6}\right)-\frac12\operatorname{Ti}_2\left(\frac{1}{9}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Proving tan((x + y)/2) = (sin x + sin y)/(cos x + cos y) with the angle sum and difference identities I've recently come across this problem of proving
$$ \tan \frac{x + y}{2} = \frac{\sin x + \sin y}{\cos x + \cos y} $$
Not a difficult problem, I thought. I would have rewritten the RHS using the sum-to-product identities of sine and cosine.
But the solution given in the back of the book is
\begin{align}
\tan \frac{x + y}{2} &= \frac{\sin x \sin y}{\cos x + \cos y} \\
&= \frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})+2\sin(\frac{y}{2})\cos(\frac{y}{2})}{2\cos^2(\frac{x}{2})-1 + 1 - 2\sin^2(\frac{y}{2})} \\
&= \frac{\sin(\frac{x}{2})\cos(\frac{x}{2}) + \sin(\frac{y}{2})\cos(\frac{y}{2})}{\cos^2(\frac{x}{2}) - \sin^2(\frac{y}{2})} \tag{1} \\
&= \frac{\sin(\frac{x+y}{2})}{\cos(\frac{x+y}{2})} \tag{2} \\
&= \tan \frac{x + y}{2}
\end{align}
What I'm confused about is the step from (1) to (2). It appears they tried to use the angle sum identities, but Wikipedia says that the formulas are actually
$$ \sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\
\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta
$$
which don't actually match the numerator or denominator in (1). Can the angle sum identities be used like this?
I'm also interested in knowing how one would actually go from (1) to (2) without the angle sum identities. It seems like there would be a few intermediate steps.
| I would use the formulae that express the trigonometric functions in function of the tangent of the half-angle. So set $\;t=\tan\dfrac x2,\enspace u=\tan y2$. Then:
\begin{align*}
\frac{\sin x+\sin y}{\cos x+\cos y}&=\frac{\dfrac{2t}{1+t^2}+\dfrac{2u}{1+u^2}}{\dfrac{1-t^2}{1+t^2}+\dfrac{1-u^2}{1+u^2}}=\frac{2\bigl(t(1+u^2)+u(1+t^2)\bigr)}{(1-t^2)(1+u^2)+(1-u^2)(1+t^2)}\\
&=\dfrac{2(t+u)(1+tu)}{2(1-t^2u^2)}=\frac{t+u}{1-tu}= \tan\Bigl(\frac x2+\frac y2\Bigr).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
inverse of $\arcsin (\frac{x}{x-1})$ determine the inverse of
a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$
b) $y= \dfrac{1-2e^{-x}}{4}$
I learned you the steps for finding the inverse are
1) get it in a form of $x= \dots$
2) change $x$ and $y$
a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$
is $y=\arcsin \left(\dfrac{x}{x-1}\right)\quad \iff \quad \sin y = \dfrac{x}{x-1}$? and if so what next?
b) $y= \dfrac{1-2e^{-x}}{4} \quad \iff \quad 4y-1=-2e^{-x} \quad\iff\quad -2y + \frac{1}{2}= e^{-x}$ . but then I'm stuck.
I would really appreciate the help!
| a)
\begin{align}
\sin y &= \dfrac{x}{x-1}\\
x\sin y -\sin y &=x\\
(\sin y - 1)x&=\sin y\\
x&=\frac{\sin y}{\sin y -1}
\end{align}
So the inverse function is $x\mapsto \dfrac{\sin x}{\sin x -1}$
And for b) is similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is $\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta (2)$?
$$\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta(2)$$
By using numerical calculation, I found this relationship between the values of zeta function at even integers and $\zeta(2)$, but this needs proving, any help?
| Ok, let give me a slightly different answer the Jack,
Our sum is
$$
S=\frac{8}{3}\sum_{n=1}^{\infty}\frac{n}{2^{2n}}\sum_{k=1}^{\infty}\frac{1}{k^{2n}}=\frac{8}{3}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{n}{(2^2 k^2)^n}\underbrace{=}_{(1)}\frac{8}{3}\sum_{k=1}^{\infty}\frac{4 k^2}{\left(4 k^2-1\right)^2}\underbrace{=}_{(2)}\\\frac{2}{3}\left(\partial_x\cot(1/x)-1+\underbrace{2\sum_{k=1}^{\infty}\frac{1}{k^2\pi^2 x^2-1}}_{-Q(x)}\right)\big|_{x=\frac{2}{\pi}}\underbrace{=}_{(3)}\frac{2}{3}\left(\partial_x\cot(1/x)\right)\big|_{x=\frac{2}{\pi}}\underbrace{=}_{(4)}\frac{\pi^2}{6}
$$
in (1) we just differentiated a geometric series. In (2) we used the Mittag-Leffler expansion of $\cot(1/z)$ and differentiated it
$$
\partial_z\cot(1/z)=\partial_z\left(z+2z\sum_{k=1}\frac{1}{1-\pi^2 k^2 z^2}\right)=1+\underbrace{2\sum_{k=1}\frac{1}{1-\pi^2 k^2 z^2}}_{Q(z)}+\underbrace{4\sum_{k=1}\frac{\pi^2 k^2 z^2}{(1-\pi^2 k^2 z^2)^2}}_{\frac{3}{2}S \quad\text{if}\quad z=2/\pi}
$$
In (3) we again used the Mittag Leffler expansion to show that the sum on the left is $Q(2/\pi)=-1+\cot(1/x)/x\big|_{\frac{2}{\pi}}=-1$
In (4) we used $ \partial_z \cot(1/z)=\frac{z^2}{\sin^2(1/z)}$ and $\sin(\pi/2)=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function such that $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^{2}$ . Then $f(3)$ =? Options: (a)$4$, (b)$4f(0)$, (c)$4-f(0)$, (d)$4+f(0)$, (e)$16+f(0)$.
CORRECT ANSWER USING REDUCTION
Deep thanks to @martini and @A.S. , soo much respect .
Since we don't exactly know the nature of $f(x)$ we will start of by considering our function as $f(x)=g(x)+cx^2$ where $g(x)$ represents all the other functions (including constant values) inside $f(x)$ and $cx^{2}$ includes the no. of $x^{2}$ terms.
Now we have $cx^{2}-2\frac{cx^{2}}{4}+\frac{cx^{2}}{16}=x^{2}$
Comparing coefficients of $x^{2}$ in first equation we have $c=\frac{16}{9}$.
And
$g(x)-2g(\frac{x}{2})+g(\frac{x}{4})=0$
Now to solve this recurring equation we substitute $x=\frac{x}{2},\frac{x}{4},\frac{x}{8}......$ and add it up .
Yielding $g(\frac{x}{2^i})=ig(\frac{x}{2})-(i-1)g(x)$
Now as $i\to\infty$. Since $g$ is continuous we will have
$g(x)=g(x/2)$.
Now solving the recurring equation $g(x)-g(x/2)=0$ by substituting $x=\frac{x}{2},\frac{x}{4},\frac{x}{8}......$ and adding it up .
Now as $i\to\infty$. Since $g$ is continuous we will have
$g(x)=g(0)$
Hence $f(3)=g(3)+(16/9)3^2$.
Now as $g(3)=g(0)= f(0)-(16/9)0^2$.
We finally have $f(3)=f(0)+16.$
| Let $x = 3\cdot 2^{-n}$, giving
$$ f(3\cdot 2^{-n}) = 9 \cdot 2^{-2n} + 2 f(3 \cdot 2^{-(n+1)}) - f(3 \cdot 2^{-(n+2)}) $$
Write brevitatis causa $a_n := f(3 \cdot 2^{-n})$, we have
$$ a_{n+2} - 2a_{n+1} + a_n = 9 \cdot 2^{-2n} $$
The general solution of the homogenous recursion
$$ a_{n+2} - 2a_{n+1} + a_n = 0 $$
is $$ a_n^{\text{hom}} = \alpha n + \beta $$
to find a solution of the inhomogenous equation, we make the ansatz $\gamma \cdot 2^{-2n}$, giving
$$ \gamma 2^{-2n}(2^{-4} - 2\cdot 2^{-2} + 1) = 9 \cdot 2^{-2n} \iff \gamma = 16 $$
So we have
$$ f(3 \cdot 2^{-n}) = a_n = 16\cdot 2^{-2n} + \alpha n + \beta $$
for some $\alpha, \beta \in \mathbf R$. As $f$ is continuous, we must have $a_n \to f(0)$. As $16\cdot 2^{-2n}$ converges, this implies $\alpha = 0$, or
$$ f(3 \cdot 2^{-n}) = 16 \cdot 2^{-2n} + \beta $$
For $n \to \infty$ this gives $f(0) = \beta$, or
$$ f(3 \cdot 2^{-n}) = 16 \cdot 2^{-2n} + f(0) $$
for $n=0$ we have
$$ f(3) = 16 + f(0) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculus 1: limits using squeeze or sandwich theorem, verification. I have the next problem:
Suppose that a function $g$ satisfies that $-1\leq g(x) \leq 1$ for all $x\geq 0$.
Calculate $\lim_{x\to\infty}\dfrac{x+g(2x)}{g(3x)+4x}$
So: $$\lim_{x\to\infty}\dfrac{x+g(2x)}{g(3x)+4x}=\lim_{x\to\infty}\dfrac{\frac{x+g(2x)}{x}}{\frac{g(3x)+4x}{x}}=\lim_{x\to\infty}\dfrac{1+\frac{g(2x)}{x}}{\frac{g(3x)}{x}+4}$$
And because $-1\leq g(x) \leq 1$ then $\frac{g(2x)}{x}\to 0$ and $\frac{g(3x)}{x}\to 0$ as $x\to \infty$, therefore:
$$\lim_{x\to\infty}\dfrac{1+\frac{g(2x)}{x}}{\frac{g(3x)}{x}+4}=\frac{1}{4}$$
But I don't know in which part I am using the squeeze/sandwich theorem, this is the second bullet of an exercise and the first bullet is just to enunciate the theorem.
| You can get the limit with a single application of the squeeze theorem.
For the numerator at $x>1$,
$$0<1-\frac{1}{x}\le 1+\frac{g(2x)}{x} \le 1+\frac{1}{x}$$
and for the denominator at $x>1$,
$$0<4-\frac{1}{x} \le \frac{g(3x)}{x}+4 \le 4+\frac{1}{x}$$
So for all $x>1$
$$\frac{1-\frac{1}{x}}{4+\frac{1}{x}} \le \frac{1+\frac{g(2x)}{x}}{\frac{g(3x)}{x}+4} \le \frac{1+\frac{1}{x}}{4-\frac{1}{x}}$$
and $\lim_{x\to\infty}{\frac{1-\frac{1}{x}}{4+\frac{1}{x}}}=\frac{1}{4}$ while $\lim_{x\to\infty}{\frac{1+\frac{1}{x}}{4-\frac{1}{x}}}=\frac{1}{4}$, so by the squeeze theorem
$$\lim_{x\to\infty}{\frac{1+\frac{g(2x)}{x}}{\frac{g(3x)}{x}+4}}=\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$ Determine the value of $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$. I've tried forcing in a difference of squares, looked for various difference of $n$s or sum of odd powers that I could equate this to, but have yet to find a solution.
| Hint Dividing gives $$x - 3 + \frac{1}{x} = 0,$$ and then rearranging gives
$$x + \frac{1}{x} = 3.$$
Additional hint Squaring gives $$3^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2},$$ and rearranging again gives $$x^2 + \frac{1}{x^2} = 7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
How to evaluate the limit of $\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$ when $x\to1$ without using L'Hospital's rule?
How do I evaluate this without using L'Hospital's rule:$$\lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}\ ?$$
Note : I used L'Hospital's Rule I find $\frac{-4}{3}$ but in wolfram alpha is $0$
| We start by using the double-angle identity for the cosine function to write
$$2\cos (x-1)-2=-4\sin^2\left(\frac{x-1}{2}\right) \tag 1$$
Next, we factor the denominator of the term of interest to find
$$\begin{align}
x^2-2x^{1/2}+1&=(x^{1/2}-1)(x^{3/2}+x+x^{1/2}-1)\\\\
&=\frac{(x^{3/2}+x+x^{1/2}-1)(x-1)}{x^{1/2}+1} \tag 2
\end{align}$$
Then, using $(1)$ and $(2)$ we have
$$\frac{2\cos (x-1)-2}{x^2-2x^{1/2}+1}=\left(-4\frac{x^{1/2}+1}{x^{3/2}+x+x^{1/2}-1}\right) \times \left(\frac{\sin\left(\frac{x-1}{2}\right)}{x-1}\right)\times\left( \sin\left(\frac{x-1}{2}\right) \right)\tag 3$$
The first parenthetical term on the right-hand side of $(3)$ approaches $-4$ as $x\to 1$, the second term approaches $1/2$, and the third term approaches $0$.
Finally, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 1}\left(\frac{2\cos (x-1)-2}{x^2-2x^{1/2}+1}\right)=0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Showing $\pi/(2\sqrt3)=1-1/5+1/7-1/11+1/13-1/17+1/19-\cdots$ I am struggling to show that $$\dfrac \pi{2\sqrt3}=1-\dfrac 15+\dfrac 17-\dfrac 1{11}+\dfrac 1{13}-\dfrac 1{17}+\dfrac 1{19}-\cdots$$ by using the Fourier series $$\frac \pi2-\frac x2=\sum_1^\infty \dfrac {\sin(nx)}{n}.$$
Can somebody give me any hint?
| The approach we take is to find a discrete and finite Fourier-series of the periodic sequence $f(n)$ in the sum $\sum\frac{f(n)}{n} = \frac{\color{red}{1}}{1} + \frac{\color{red}{0}}{2} + \frac{\color{red}{0}}{3} + \frac{\color{red}{0}}{4} + \frac{\color{red}{-1}}{5} + \frac{\color{red}{0}}{6} \ldots$ we are trying to compute. This will allow us to use this (finite) Fourier series in combination with the Fourier series in the question to evaluate the sum.
Consider the function $f:\mathbb{N}\to\mathbb{N}$ given by (which is the Dirichlet character modulo $6$ hinted to by Paul in the comments)
$$f(n) = \left\{\matrix{0 & n\equiv 0\mod 6\\1 & n\equiv 1\mod 6\\0 & n\equiv 2\mod 6\\0 & n\equiv 3\mod 6\\0 & n\equiv 4\mod 6\\-1 & n\equiv 5\mod 6}\right.$$
This function is constructed such that your series can be written as
$$1 - \frac{1}{5} + \frac{1}{7} - \frac{1}{11} + \frac{1}{13} - \ldots = \sum_{n=1}^\infty \frac{f(n)}{n}$$
Since $f$ is periodic with period $6$ it has a Fourier series
$$f(n) = \sum_{k=0}^5c_k \sin\left(\frac{2\pi k}{6}n\right)$$
where the coefficients are given by
$$c_k = \frac{1}{6}\sum_{j=0}^5f(j)\sin\left(\frac{2\pi k}{6}j\right) = \frac{\sin\left(\frac{\pi k}{3}\right)-\sin\left(\frac{5\pi k}{3}\right)}{6}$$
The sum you are after can therefore be written as
$$\sum_{n=1}^\infty\frac{f(n)}{n} = \sum_{k=0}^5c_k \color{red}{\sum_{n=1}^\infty\frac{\sin\left(\frac{2\pi k}{6} n\right)}{n}} = \sum_{k=0}^5c_k\color{red}{\left(\frac{\pi}{2} - \frac{\left(\frac{2\pi k}{6}\right)}{2}\right)}$$
where we have used the Fourier series in the question to evaluate the infinite sums. This gives the desired result
$$\sum_{n=1}^\infty \frac{f(n)}{n} = \sum_{k=0}^5\frac{\sin\left(\frac{\pi k}{3}\right)-\sin\left(\frac{5\pi k}{3}\right)}{6} \frac{(3-k)\pi}{6} = \frac{\pi}{2\sqrt{3}}$$
The method we used above generalises to computing the sum of any series $\sum \frac{f(n)}{n}$, where $f(n)$ is a periodic function (with integer period).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
} |
Binomial Theorem Question (Expansion of Three Terms) I have the term: $(1 + 2x - x^2)^4.$
The question asks me to find the coefficient of $x^5$.
My solution:
$\sum\limits_{i=0}^4 {4 \choose r} (1)^{4-r}(2x-x^2)^r$
I then factored out x from $(2x-x^2)$, getting $x(2-x)$.
Then, since the terms with the x's are being raised to the $r$th power, I did:
$(x(2-x))^r$, or $x^r(2-x)^r$
I know that I'm dealing with x's, so since I want the exponent to be 5 as the question says, I focus on the x's and multiply them together to get $x^{2r}$, and then I equated 2r to 5. Solving, I got r = 5/2 which can't be because I'm dealing with a binomial coefficient - integers only.
What did I do wrong? My logic makes sense to me, but I don't see why I'm incorrect here.
Thanks.
| $$\begin{align}
&[x^5](1+2x-x^2)^4
\\=&[x^5]\sum_{r=0}^4\binom 4r (2x-x^2)^r
\\=&[x^5]\sum_{r=0}^4\binom 4r (2x)^r \left(1-\frac x2\right)^r
\\=&[x^5]\sum_{r=0}^4\binom 4r (2x)^r \sum_{j=0}^r \binom rj \left(-\frac x2\right)^j
\\=&\underbrace{\underbrace{\binom 43 2^3\cdot \binom 32 \left(-\frac 12\right)^2}_{r=3, j=2}+
\underbrace{\binom 44 2^4\cdot \binom 41 \left(-\frac 12\right)^1}_{r=4,j=1}}_{r+j=5; \quad 0\leq j\leq r\leq 4}
\\=&-8\qquad\blacksquare
\end{align}$$
Alternatively, using the multinomial theorem,
$$\begin{align}
&[x^5](1+2x-x^2)^4
\\
=&\text{coeff. of }1^1(2x)^1(-x^2)^2+\text{coeff. of }1^0(2x)^3(-x^2)^1
\\
=&\binom {4}{1,1,2}\cdot 1^1\cdot 2^1\cdot (-1)^2+\binom 4{0,3,1}\cdot 1^0\cdot 2^3\cdot (-1)^1
\\
=&\frac {4!}{1!\;1!\;2!}\cdot 2+\frac {4!}{0!\;3!\;1!}\cdot 8\cdot (-1)
\\
=&-8\qquad\blacksquare
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Cofactor expansion to check if matrices is invertible. I have a question regarding a co-factor expansion question. I understand that an easy way to check if a matrix is invertible is to do co-factor expansion and if $A \ne 0$ then it's invertible. I'm familiar with the basic co-factor expansion with $3 \times 3$ matrices although I'm unfamiliar with $4\times 4$ matrices.
Could someone please give me a hint, or tell me how to go about checking if this matrix is invertible?
Q: Use the inversion algorithm to find the inverse of the matrix (if the inverse exists)
$$\begin{bmatrix}0 & 0 & 2 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & -1 & 3 & 0\\ 2 & 1 & 5 & -3 \end{bmatrix}$$
| To check if matrices are invertible, you need to check the determinant is non-zero:
To find the determinant of this matrix we look for the row or column with the most zeros and do a Laplace development on that row or column.
The first row contains the most zeros so we Laplace develop that row:
$$\det=\begin{vmatrix}0 & 0 & 2 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & -1 & 3 & 0\\ 2 & 1 & 5 & -3 \end{vmatrix}$$
$$=0\times\begin{vmatrix} 0 & 0 & 1 \\ -1 & 3 & 0 \\ 1 & 5 & -3 \end{vmatrix}+0\times\begin{vmatrix} 1 & 0 & 1 \\ 0 & 3 & 0 \\ 2 & 5 & -3 \end{vmatrix}+2\times\begin{vmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 2 & 1 & -3 \end{vmatrix}+0\times\begin{vmatrix} 1 & 0 & 0 \\ 0 & -1 & 3 \\ 2 & 1 & 5 \end{vmatrix}$$$$=2\times\begin{vmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 2 & 1 & -3 \end{vmatrix}$$ $$=2\left( 1\times\begin{vmatrix}-1 & 0 \\ 1 & -3 \\ \end{vmatrix}+0\times\begin{vmatrix}0 & 0 \\ 2 & -3 \\ \end{vmatrix}+ 1\times\begin{vmatrix}0 & -1 \\ 2 & 1 \\ \end{vmatrix}\right)$$$$=2\times \left(1\times(3-0)+1\times(0--2)\right)= 2(3+2)=10 \ne 0$$ So your matrix is invertible (or has an inverse). By the way, when I say 'Laplace development', I think that's what you meant by co-factor expansion. They are not the same thing because the matrix of co-factors in something different. Does this help? Or is there something else I can do?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$x^{2000} + \frac{1}{x^{2000}}$ in terms of $x + \frac 1x$. If $x + \frac{1}{x} = 1$, then what is
$$ x^{2000} + \frac{1}{x^{2000}} = ?$$
| Assume that $x=e^{i\theta}$. Then you are trying to write $2\cos(2000\cdot\theta)$ in terms of $2\cos(\theta)$, so the answer is given by Chebyshev polynomials of the first kind:
$$ \left(x^{2000}+\frac{1}{x^{2000}}\right) = 2\cdot \widetilde{T}_{\!2000}\left(x+\frac{1}{x}\right) $$
where $\widetilde{T}_{n}(z) = T_n(z/2)$. In our very particular case, we may notice that $\frac{1}{x}+x=1$ implies $x=e^{\pm \pi i/3}$: since $2000\equiv 2\pmod{6}$, given $x+\frac{1}{x}=1$ we have:
$$ x^{2000}+\frac{1}{x^{2000}}=x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 = \color{red}{-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What surprises me is that the result implies
$$y=\frac{1}{2}\tan^{-1}x+C$$
Can anyone tell me how to see that directly?
| Let me rewrite the proof a bit differently.
We are required to prove (RTP) that $$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\frac{\pi}{2}$$
Proof:
Plan: We will use the formula F first, followed by the identity I.
Let us call the following identity I:
$\tan^{-1}(x) + \tan^{-1} (1/x) =$
\begin{cases}
\hfill \pi/2 & \text{if $x\geq 0$}\\
-\pi/2 & \text{if $x<0$}
\end{cases}
Next, call the following formula F:
$2 \tan^{-1} (x) =$
\begin{cases}
\hfill \tan^{-1} \frac{2x}{1-x^2} & \text{if $|x|\neq 1$}\\
\frac{\pi}{2} & \text{if $x = 1$}\\
-\frac{\pi}{2} & \text{if $x = -1$}
\end{cases}
Also note that $x - \sqrt{1+x^2} < 0$,
therefore $|x - \sqrt{1+x^2}| > 1$ is equivalent to $x - \sqrt{1+x^2} < -1 \implies x < 0$
Done with the ground work!
If $y < 0$,
LHS of RTP =
$$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\tan^{-1}\frac{1}{y} - \tan^{-1}(y) = -\frac{\pi}{2}$$
[we used formula F (first case) first, followed by identity I (second case)]
If $y > 0$,
LHS of RTP =
$$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\tan^{-1}\frac{1}{y} - \tan^{-1}(y) = -\frac{\pi}{2}$$
[we used formula F (first case) first, followed by identity I (first case)]
The case $y = 0$ is obvious!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 2
} |
separating equation $$
\begin{align}
\frac {dy}{dx} &= \frac{(y-1)(x+2)}{(y+1)(x-3)}\\
\frac{y+1}{y-1}dy &= \frac{x+2}{x-3}dx
\end{align}
$$
Integrate both sides :
$$
\begin{align}
\int\frac{y+1}{y-1}dy & =\int\frac{x+2}{x-3}dx\\
\int 1+\frac{2}{y-1}dy &=\int 1+\frac{5}{x-3}dx\\
y+2\ln|y-1|&=x+5\ln|x-3|+C\\
\end{align}
$$
trapped here because I cannot get rid of $2\ln|y-1|$, any techniques I should use to solve this?
| One first thing, you should put the absolute values there. The correct way to write down the solution is
$$y + 2\ln \left| {y - 1} \right| = x + 4\ln \left| {x - 3} \right| + C$$
At this situation, they say that the ordinary differential equation is solved. However, there is not always a way to explicitly find $y$ in terms of $x$ from a relation of the form $h(x,y)=0$. In your case, it seems to happen. If you want to just get rid of $\ln \left| {y - 1} \right|$, do the following
$$\eqalign{
& \ln {\left| {y - 1} \right|^2} - \ln {\left| {x - 3} \right|^4} = - y + x + C \cr
& \cr
& \ln \left( {\frac{{{{\left| {y - 1} \right|}^2}}}{{{{\left| {x - 3} \right|}^4}}}} \right) = - y + x + C \cr
& \cr
& \frac{{{{\left| {y - 1} \right|}^2}}}{{{{\left| {x - 3} \right|}^4}}} = {e^{ - y + x + C}} \cr
& \cr
& {\left| {y - 1} \right|^2}{e^y} = {e^C}{\left| {x - 3} \right|^4}{e^x} \cr} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Exponential conversion While solving Fourier series coefficients if found this problem. Anyone help me to tell that how they converted $\frac{1}{2}e^{j\pi/4}=\frac{\sqrt{2}}{4}(1+j)$
| Hint (assuming you have $j^2=-1$): Use $e^{jx}= \cos x + j \sin x$ and $\cos \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$
Edit- Here the completed answer after reading your comment:
$$\frac{1}{2}e^{j\pi/4}=\frac{1}{2}\left(\cos \frac{\pi}{4} + j \sin \frac{\pi}{4} \right)=
\frac{1}{2}\left(\frac{\sqrt{2}}{2} + j\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{4}\Big(1+j\Big)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
if $\sin{(aw)}+\sin{(bw)}+\sin{(cw)}=3$ find $w$ range let $w$ is postive integer,if there exist $a,b,c(\pi\le a<b<c\le 2\pi)$such
$$\sin{(aw)}+\sin{(bw)}+\sin{(cw)}=3$$
Find the $w$ range.
My attempt：
since
$$\sin{(wa)}\le 1,\sin{(wb)}\le 1,\sin{(wc)}\le 1$$
then $$\sin{(wa)}=\sin{(wb)}=\sin{(wc)}=1$$
$$aw=\dfrac{\pi}{2}+2k_{1}\pi,k_{1}\in Z$$
$$bw=\dfrac{\pi}{2}+2k_{2}\pi,k_{2}\in Z$$
$$cw=\dfrac{\pi}{2}+2k_{3}\pi,k_{3}\in Z$$
what approaches do you think, I could take to solving the next step?
| we also solve if $w$ be postive real numbers,I have done This result is $$w\in [\dfrac{17}{4},\dfrac{9}{2}]\bigcup [\dfrac{21}{4},+\infty) $$
This problem is equivl this:there $p,q,r\in N^{+}$,such
$$w\pi\le 2p\pi+\dfrac{\pi}{2}<2q\pi+\dfrac{\pi}{2}<2r\pi+\dfrac{\pi}{2}\le 2w\pi$$
case1 if $w\ge 6$ it is clear,because $(w\pi,2w\pi)\subset (6\pi,12\pi)$,and $\dfrac{12\pi-6\pi}{3}=2\pi$,
case 2: if $0<w<6$,then $(w\pi,2w\pi)\subset (0,12\pi)$,
then
2.1 if
$$w\pi\le\dfrac{\pi}{2}<\dfrac{5\pi}{2}<\dfrac{9\pi}{2}\le 2w\pi$$
then $$\Longrightarrow w\le\dfrac{1}{2},\rm{and} w\ge\dfrac{9}{4}$$
that's impossible.
2.2 if $$w\pi\le\dfrac{5\pi}{2}<\dfrac{9\pi}{2}\le\dfrac{13\pi}{2}\le 2w\pi$$
then
$$\Longrightarrow w\ge \dfrac{13}{4},\rm{and} w\le \dfrac{5}{2}$$
also impossible
$2.3$ if
$$w\pi\le\dfrac{9\pi}{2}<\dfrac{13\pi}{2}<\dfrac{17}{2}\pi\le 2w\pi$$
then
$$\Longrightarrow \dfrac{17}{4}\le w\le\dfrac{9}{2}$$
$2.4$ if
$$w\pi\le\dfrac{13\pi}{2}<\dfrac{17\pi}{2}<\dfrac{21\pi}{2}\le 2w\pi$$
then
$$\dfrac{21}{4}\le w\le\dfrac{13}{2}$$
so
$$w\in [\dfrac{17}{4},\dfrac{9}{2}]\bigcup [\dfrac{21}{4},+\infty) $$
Note: if $w\in Z^{+}$,then $w\ge 6,w\in Z^{+}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Help showing $\lim\limits _{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right) = \frac{5}{4}$ I am stuck solving the following limit. I know the answer is 5/4, I just can't get it. This is the steps I have done so far.
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right)$
Multiply by Conjugate
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right)\cdot \frac{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Multiply Out
$\lim\limits_{x\to -\infty \:}\cdot \frac{\left(4\cdot \:\:\:x^2-5\cdot \:\:\:x-4\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Combine Like Terms
$\lim\limits_{x\to -\infty \:}\cdot \frac{\left(4\cdot \:\:\:x^2-9\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Factor out x
$\lim\limits_{x\to -\infty \:}\cdot \frac{x\left(4\cdot \:x-9\right)}{\left(\sqrt{x^2\left(4-\frac{5}{x}\right)}-2\cdot \:\:x\right)}$
Pull out x of sqrt and factor again
$\lim \limits_{x\to -\infty \:}\cdot \frac{x\left(4\cdot \:x-9\right)}{x\left(\sqrt{\left(4-\frac{5}{x}\right)}-2\right)}$
Now cancel x terms
$\lim \limits_{x\to -\infty \:}\frac{4\cdot \:\:x-9}{\sqrt{\left(4-\frac{5}{x}\right)}-2}$
Now I don't know what to do next.
If I plug in I get
$\frac{4\cdot \:\:\:-\infty \:-9}{\sqrt{\left(4-0\right)}-2}=\:\frac{-\infty \:}{0}$ Which doesn't equal $\frac{5}{4}$?
| Expanding DirkGently's answer,
this is the point
where multiplying
by the conjugate
is useful:
$\begin{array}\\
\lim_{x\to-\infty}2x\left(1-\sqrt{1-\frac{5}{4x}}\right)
&=\lim_{x\to-\infty}2x\left(1-\sqrt{1-\frac{5}{4x}}\right)
\frac{1+\sqrt{1-\frac{5}{4x}}}{1+\sqrt{1-\frac{5}{4x}}}\\
&=\lim_{x\to-\infty}2x\frac{\left(1-(1-\frac{5}{4x})\right)
}{1+\sqrt{1-\frac{5}{4x}}}\\
&=\lim_{x\to-\infty}2x\frac{\left(\frac{5}{4x})\right)
}{1+\sqrt{1-\frac{5}{4x}}}\\
&=\lim_{x\to-\infty}\frac{\frac{5}{2}
}{2}\\
&= \frac54\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
My answer:
$$x+y=35$$
$x^2y^5$ is maximum
$$y=35-x$$
$$\frac{d}{dx} x^2(35-x)^5$$
Which rule to apply here after? I reached:
$$(35-x)^4(-5x^2+(35-x)2x)=0$$
Either $(35-4x)^4 =0$ or $x^2-7=0$
| Note that if $y \leq 0 \Rightarrow x^2y^5 \leq 0, \forall x \in \mathbb{R}$, thus assume $y > 0$, and since the max won't change if $x$ is replaced by $-x$, we can also assume $x > 0$. Then we have a much simpler problem to solve.
We have: $35 = x+y = \dfrac{x}{2}+\dfrac{x}{2}+ \dfrac{y}{5}+\dfrac{y}{5}+\dfrac{y}{5}+\dfrac{y}{5}+\dfrac{y}{5} \geq 7\sqrt[7]{\dfrac{x^2y^5}{2^25^5}}\Rightarrow x^2y^5 \leq 2^25^{12}=\text{max}$. Equality occurs when $\dfrac{x}{2}=\dfrac{y}{5}\Rightarrow x = 10,y=25$. Thus we conclude that $x = 10,y = 25$ will do the job.
Note: This answer is not complete and I want to post it as a hint rather than a full answer since you have one part to do: solve the case $x$ is negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
What are all functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$ self-reciprocal under Fourier transform? There are some functions that are self reciprocal under cosine Fourier transform:
\begin{equation}
\frac{1}{\cosh x}, \frac{\cosh x}{\cosh 2x},\frac{1}{1+2\cosh x}
\end{equation}
It seems that they have been discovered by Ramanujan. Detailed study of these three functions can be found in the book "Ramanujan's lost notebook, part IV" by Andrews and Berndt, and also in Titchmarsh's book "Introduction to the theory of Fourier integrals".
Recently I have found that
\begin{equation}
\frac{\cosh\frac{\sqrt{3}x}{2}}{\cosh x-\cos(\sqrt{3}\pi)} \qquad (1)
\end{equation}
is self-reciprocal under Fourier transform. Although I doubt that this hasn't been derived before, I can't find any information about this particular function in the literature. Can someone point me to the literature where this function is studied in more detail?
Is it possible to find all functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$ that are self-reciprocal? There are some functions which are similar to these three, but actually are sums of several functions, e.g. $\frac{\cosh x-(1+\sqrt{3})/2}{\cosh(3x/2)}$. The three functions found by Ramanujan are composed of a single term, and we are interested only in such functions.
Below I show how self-reciprocity of (1) can be established, which is pretty straightforward:
Let's substitute $\gamma=1,\beta=\frac{\sqrt{3}}{2}, b=(\sqrt{3}-1)\pi$ in the formula p. 539, formula 3.983, no. 6, from Gradshteyn and Ryzhik,
\begin{align}
\int_0^\infty &\frac{\cos ax\cosh \beta x}{\cosh \gamma x+\cos b}dx=\\
&\pi\frac{\cos\left[\frac{\beta}{\gamma}(\pi-b)\right]\cosh\left[\frac{a}{\gamma}(\pi+b)\right]-
\cos\left[\frac{\beta}{\gamma}(\pi+b)\right]\cosh\left[\frac{a}{\gamma}(\pi-b)\right]}
{\gamma\sin b\left(\cosh\frac{2\pi a}{\gamma}-\cos\frac{2\pi \beta}{\gamma}\right)}\\
&|\text{Re} \beta|<\text{Re} \gamma,\quad 0<b<\pi,\quad a>0
\end{align}
We see that since $\frac{\sqrt{3}}{2}<1, 0<\sqrt{3}-1<1$ all the conditions for the validity of this integral evaluation are satisfied, and we get
$$
\int_0^\infty \frac{\cos ax \cosh\frac{\sqrt{3}x}{2}}{\cosh x-\cos(\sqrt{3}\pi)}d x=\pi\frac{\cosh(a\sqrt{3}\pi)}{\cosh 2\pi a-\cos(\sqrt{3}\pi)}
$$
It can be written in the symmetrical form
$$
\sqrt{\frac{2}{\pi}}\int_0^\infty\frac{\cosh(\sqrt{\frac{3\pi}{2}}x)}{\cosh(\sqrt{2\pi}x) -\cos(\sqrt{3}\pi)}\cos ax dx=\frac{\cosh(\sqrt{\frac{3\pi}{2}}a)}{\cosh(\sqrt{2\pi}a) -\cos(\sqrt{3}\pi)}
$$
Also the following identity can be derived from the integral mentioned above
\begin{align}
\frac{\Gamma(\frac{1}{2}+\frac{s}{2})}{\pi^{\frac{s}{2}}}\sum_{n=-\infty}^\infty\text{sgn}(n)\frac{\sin(\sqrt{3}\pi p n)}{{\bigl\lvert n+\frac{\sqrt{3}}{2p}\bigr\rvert}^s}=
\frac{\Gamma(1-\frac{s}{2})}{\pi^{\frac{1-s}{2}}}\sum_{n=-\infty}^\infty\text{sgn}(n)\frac{\sin(\sqrt{3}\pi q n)}{{\bigl\lvert n+\frac{\sqrt{3}}{2q}\bigr\rvert}^{1-s}},\quad pq=1,\quad\frac{\sqrt{3}}{2}<p<\frac{2}{\sqrt{3}}.
\end{align}
This identity is an 8 term functional equation for Lerch zeta function and in fact it can be shown that it is a consequence of Lerch's functional equation. I have searched the literature about Lerch zeta function, but nobody mentions that the Lerch Zeta function has any symmetry at the point $\sqrt{3}/2$.
$\bf{Update:}$ The question can be reduced to finding all $0<b<\pi, \beta<1$ such that
$$
\frac{\cos\beta(\pi-b)\cdot \cosh a(\pi+b)-\cos\beta(\pi+b)\cdot \cosh a(\pi-b)}{\cosh 2\pi a-\cos 2\pi \beta}=\frac{\cosh\frac{\pi-b}{1-\beta}\beta a\cdot \cos\beta(\pi-b)}{\cosh\frac{\pi-b}{1-\beta}a+\cos b} \quad(2)
$$ for all $a\ge 0$.
The case $\beta=0$ doesn't lead to anything new. When $\beta\neq 0$ we can substitute $a=0$ and obtain:
$$
\sin\beta b\cdot(1+\cos b)=\cos\beta(\pi-b)\cdot\sin\pi\beta. \qquad (3)
$$
Then we simplify equation (2):
\begin{align}
&\cos\beta(\pi-b)\cosh\frac{2b-\pi\beta-b\beta}{1-\beta}a-\cos\beta(\pi+b)\cosh\frac{(\pi-b)(2-\beta)}{1-\beta}a+\\
&\cos(3\pi-b)\beta\cosh\frac{\pi-b}{1-\beta}\beta a+2\cos\beta(\pi-b)\cos b\cosh(\pi+b)a-\\
& 2\cos\beta(\pi+b)\cos b\cosh(\pi-b)a=\cos\beta(\pi-b)\cosh\frac{2\pi-3\pi\beta+b\beta}{1-\beta}a
\end{align}
Here we have 6 different exponents:
$$
\frac{|2b-\pi\beta-b\beta|}{1-\beta},\quad\frac{(\pi-b)(2-\beta)}{1-\beta},
\quad\frac{\pi-b}{1-\beta}\beta,\quad\pi+b,\quad\pi-b,\quad\frac{|2\pi-3\pi\beta+b\beta|}{1-\beta}.
$$
When $\beta<1/2$ the exponent $\frac{\pi-b}{1-\beta}\beta$ can be equal to only $\frac{|2b-\pi\beta-b\beta|}{1-\beta}$. When $\beta>1/2$ the exponent $\pi-b$ can equal either $\frac{|2b-\pi\beta-b\beta|}{1-\beta}$ or
$\frac{|2\pi-3\pi\beta+b\beta|}{1-\beta}$. Together with (3) we have two equations. I have solved this equations and besides recovering all of the Ramanujan's examples and the case $\beta=\sqrt{3}/2,b=(\sqrt{3}-1)\pi$ I found one more example: $\beta=3/4,b=2\pi/3$.
So the answer is there are 5 self-reciprocal functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$:
\begin{equation}
\frac{1}{\cosh x}, \frac{\cosh x}{\cosh 2x},\frac{1}{1+2\cosh x},\frac{\cosh\frac{3x}{4}}{2\cosh x-1},\frac{\cosh\frac{\sqrt{3}x}{2}}{\cosh x-\cos(\sqrt{3}\pi)}.
\end{equation}
| $$
\newcommand{\res}{\operatorname{Res}}
\newcommand{\re}{\operatorname{Re}}
$$
I cannot give you the references you are looking for but I will calculate the Fourier transform of your fifth function.
Put
$$
I = \dfrac{1}{2}\int_{-\infty}^{\infty}\dfrac{e^{iax}\cosh\frac{\sqrt{3}x}{2}}{\cosh x -\cos(\sqrt{3}\pi)}\, dx.
$$
Since $I$ is an even function of $a$ it is no essential restriction to assume that $a>0$.
We split $I$ into two terms $ I = \frac{1}{2}I_1+ \frac{1}{2}I_2$ where
$$
I_1 = \int_{-\infty}^{\infty}\dfrac{e^{iax}e^{\frac{\sqrt{3}x}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}\, dx
$$
and
$$
I_2 = \int_{-\infty}^{\infty}\dfrac{e^{iax}e^{-\frac{\sqrt{3}x}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}\, dx = \int_{-\infty}^{\infty}\dfrac{e^{-iax}e^{\frac{\sqrt{3}x}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}\, dx = \overline{I_1}.
$$
Consequently $\displaystyle I = \re(I_1).$
To be able to calculate $I_1$ we integrate in positive direction around a rectangle with corners in $(R,0), (R,i2\pi), (-R,i2\pi)$ and $(-R,0)$. Since $a>0$ and $\frac{\sqrt{3}}{2}<1$ there will be no contributions from the vertical edges as $R \to \infty$. There are two singularities $i\sqrt{3}\pi$ and $i(2-\sqrt{3})\pi$ inside the curve. The residue theorem then gives that
$$
\begin{gathered}
I_1-\int_{-\infty}^{\infty}\dfrac{e^{ia(x+i2\pi)}e^{\frac{\sqrt{3}(x+i2\pi)}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}\, dx = (1-e^{-2a\pi+i\sqrt{3}\pi})I_1 \\[2ex]
= 2{\pi}i\left(\underset{x=i\sqrt{3}\pi}{\res}\dfrac{e^{iax}e^{\frac{\sqrt{3}x}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}+\underset{x=i(2-\sqrt{3})\pi}{\res}\dfrac{e^{iax}e^{\frac{\sqrt{3}x}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}\right)\\[2ex]
= 2{\pi}i\left(\dfrac{e^{-a\sqrt{3}\pi}e^{\frac{i3\pi}{2}}}{2i\sin(\sqrt{3}\pi)}+ \dfrac{e^{-a(2-\sqrt{3})\pi}e^{\frac{i\sqrt{3}(2-\sqrt{3})\pi}{2}}}{2i\sin((2-\sqrt{3})\pi)}\right)
= \dfrac{-{\pi}i}{\sin(\sqrt{3}\pi)}(e^{-a\sqrt{3}\pi}+e^{-2a\pi +a\sqrt{3}\pi +i\sqrt{3}\pi}).
\end{gathered}
$$
We now make some preparations for the calculation of $\re(I_1).$
$$
\begin{gathered}
I_1 = \dfrac{-{\pi}i}{\sin(\sqrt{3}\pi)}(e^{-a\sqrt{3}\pi}+e^{-2a\pi +a\sqrt{3}\pi +i\sqrt{3}\pi})\dfrac{1-e^{-2a\pi-i\sqrt{3}\pi}}{(1-e^{-2a\pi+i\sqrt{3}\pi})(1-e^{-2a\pi-i\sqrt{3}\pi})}\\[2ex]
= \dfrac{-{\pi}i}{\sin(\sqrt{3}\pi)}\dfrac{e^{-a\sqrt{3}\pi}-e^{-2a\pi-a\sqrt{3}\pi -i\sqrt{3}\pi}+ e^{-2a\pi +a\sqrt{3}\pi +i\sqrt{3}\pi} - e^{-4a\pi+a\sqrt{3}\pi}}{1+e^{-4a\pi}-2e^{-2a\pi}\cos(\sqrt{3}\pi)}.
\end{gathered}
$$
Finally we have that
$$
\begin{gathered}
I =\re(I_1) = \dfrac{{\pi}}{\sin(\sqrt{3}\pi)}\dfrac{e^{-2a\pi +a\sqrt{3}\pi}\sin(\sqrt{3}\pi)+e^{-2a\pi-a\sqrt{3}\pi}\sin(\sqrt{3}\pi)}{1+e^{-4a\pi}-2e^{-2a\pi}\cos(\sqrt{3}\pi)}\\[2ex]
= \pi\dfrac{e^{a\sqrt{3}\pi}+e^{-a\sqrt{3}\pi}}{e^{2a\pi}+e^{-2a\pi}-2\cos(\sqrt{3}\pi)} = \pi\dfrac{\cosh(a\sqrt{3}\pi)}{\cosh(2a\pi)-\cos(\sqrt{3}\pi)}.
\end{gathered}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Calculate the integral of the function only with the immediate integrals I need to calculate this indefinite integral:
$$
\int{\frac{x^2-1}{x^2+1}}dx
$$
I realized that the technique to solve is to "transform" the numerator equal to the denominator of the fraction so I tried this:
$$
\int{\frac{x^2-1}{x^2+1}}dx = \int{\frac{(x^2-1)+2}{x^2+1}}dx = \int{\frac{x^2-1}{x^2+1}}dx + \int{\frac{2}{x^2+1}}dx
$$
My problem is that my equation is returning to the original problem!The book suggests the following resolution , but why it appears this integral dx?
$$
\int{\frac{x^2-1}{x^2+1}}dx = \int{\frac{(x^2-1)-2}{x^2+1}}dx = \int dx - \int{\frac{2}{x^2+1}}dx = x - 2\int{\frac{1}{x^2+1}}dx = x - 2arctanx + c
$$
| Try using long-division on your rational expression:
$$\frac{x^2-1}{x^2+1}=\frac{(x^2+1)-2}{x^2+1}=1-\frac{2}{x^2+1}$$
So now we can integrate the $1$ (that is, $\int 1\ dx$), and subtract the integral of $\frac{2}{x^2+1}$ (which is $2\arctan(x)+C$).
Whenever you have a rational function whose numerator's degree is not smaller than its denominator's, a good first step is to do polynomial long-division.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Algebraic values of sine at sevenths of the circle At the end of a calculation it turned out that I wanted to know the value of
$$\sin(2\pi/7) + \sin(4\pi/7) - \sin(6\pi/7).$$
Since I knew the answer I was supposed to get, I was able to work out that the the above equals $\sqrt{7}/2$, and I can confirm this numerically. How would I prove this? I suspect I want to use the seventh roots of unity in some way but I am not sure how to proceed.
| Solution without using trinometry, but only properties of roots of unity:
(This proof has been refined after reviewing this excellent post here. I knew there had to be a more direct approach!:))
Let $\omega=e^{i2\pi/7}$, i.e. the primitive $7$th root of unity. By definition,
$$1+\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6=0$$
Let
$$\begin{align}
S&=\omega^1+\omega^2+\omega^4\\
\Rightarrow\quad
S^2&=\omega^2+\omega^4+\omega^8+2(\omega^3+\omega^6+\omega^5)\\
&=2(\underbrace{\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=-1})-(\underbrace{\omega^1+\omega^2+\omega^4}_{=S})\\
&=-2-S\\
S^2+S+2&=0\\
S&=-\frac 12\pm\frac{\sqrt7}2i\\\\
\sin\frac{2\pi}7+\sin\frac{4\pi}7-\sin\frac{6\pi}7
&=\Im(\omega^1+\omega^2-\omega^3)\\
&=\Im(\omega^1+\omega^2+\omega^4)\\
&=\Im(S)\\
&=\frac{\sqrt7}2\quad\blacksquare
\end{align}$$
NB - The positive value is chosen as $\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7=\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {\pi}7>0$, since $\sin\frac \pi7<\sin\frac {2\pi}7$.
$\tiny\color{\lightgrey}{\text{Previous solution (now superceded) shown below.}}$
$$\tiny\color{lightgrey}{\begin{align}
\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7
&=\sin\theta+\sin 2\theta-\sin3\theta
\qquad\qquad (\theta=2\pi/7;\;\omega=e^{i\theta}=e^{i2\pi/7})\\
&=\frac 1{2i}\left[(\omega^1-\omega^{-1})+(\omega^2-\omega^{-2})-(\omega^3-\omega^{-3})\right]\\
&=\frac 1{2i}\left[(\omega^1-\omega^6)+(\omega^2-\omega^5)-(\omega^3-\omega^4)\right]\\
&=\frac 1{2i}\left[\underbrace{\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=-1}-2(\omega^3+\omega^5+\omega^6)\right]\\
&=-\frac 1{2i}\left[1+2(\omega^3+\omega^5+\omega^6)\right]\\
\left(\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7\right)^2
&=-\frac14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^3+\omega^5+\omega^6)^2\right]\\
&=-\frac 14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^6+\omega^{10}+\omega^{12}+2\omega^8+2\omega^{11}+2\omega^9)\right]\\
&=-\frac 14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^6+\omega^{3}+\omega^{5}+2\omega^1+2\omega^{3}+2\omega^2)\right]\\
&=-\frac 14\left[8(\underbrace{1+\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=0})-7\right]\\
&=\frac 74\\
\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7&=\frac{\sqrt7}{\;2}\quad\blacksquare
\end{align}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
How to solve a system of non linear equations of the form $2ax-2x(ax^2+by^2) = 0$ Solve this system for $x,y$:
$$2ax-2x(ax^2+by^2) = 0\tag{1}$$
$$2by-2y(ax^2+by^2) = 0\tag{2}$$
where $a$ and $b$ are constants such that $a>b>0$.
I re-arranged (1) to $x = \sqrt{\frac{a-by^2}{a}}$ and tried substituting it into (2) but I did not know where to go from there. I'm looking for real solutions.
| If $x = 0$, the first equation is ok, and:
$$
2by-2y(ax^2+by^2) = 0 \\
2by-2y(by^2) = 0 \\
by-by^3 = 0 \\
y-y^3 = 0 \\
y(1-y^2) = 0
$$
Thus, either $y=0$ or $y=1$ or $y=-1$. Here we have three solutions.
If $x \ne 0$:
$$
2ax-2x(ax^2+by^2) = 0 \\
ax = x(ax^2+by^2) \\
a = ax^2+by^2
$$
Replacing in the second equation, we have:
$$
2by-2y(ax^2+by^2) = 0 \\
by - y(ax^2+by^2) = 0\\
by - ya = 0 \\
y(b-a) = 0
$$
As $a > b$, the only solution is $y=0$. So:
$$
a = ax^2+by^2 \\
a = ax^2 \\
1 = x^2 \\
$$
Thus, either $x=1$ or $x=-1$.
Therefore, the solutions $(x, y)$ are: $\{ (0, 0), (0, 1), (0, -1), (1, 0), (-1, 0) \}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the coordinate of a point $P$ on the line $2x-y+5=0$ such that $|PA-PB|$ is maximum where $A=(4,-2)$ and $B=(2,-4)$ What is the coordinate of a point $P$ on the line $2x-y+5=0$ such that $|PA-PB|$ is maximum where $A=(4,-2)$ and $B=(2,-4)$.
Let the coordinates of the point $P$ be $(x_1,y_1).$
$P$ lies on the line $2x-y+5=0$,so $2x_1-y_1+5=0$
$|\sqrt{(x_1-4)^2+(y_1+2)^2}-\sqrt{(x_1-2)^2+(y_1+4)^2}|$ is maximum.
But i dont know,how to solve it further and find $(x_1,y_1).$Please help me.Thanks.
|
Now $PA=\sqrt{(x-2)^2+(y+4)^2}$ and $PB=\sqrt{(x-4)^2+(y+2)^2}$ Now using Cosine formula
We get $$\displaystyle \cos \theta = \frac{(PA)^2+(PB)^2-(AB)^2}{2\cdot (PA)\cdot (PB)}$$
Now we know that $$\displaystyle |\cos \theta | \leq 1\Rightarrow \left|\frac{(PA)^2+(PB)^2-(AB)^2}{2\cdot (PA)\cdot (PB)}\right|\leq 1$$
So we get $$|(PA)^2+(PB)^2-(AB)^2|=2|PA||PB|\Rightarrow \left |PA-PB\right|^2\leq |AB|^2$$
So we get $|PA-PB|\leq |AB|$ and equality hold when $\theta = \pi$
Means $A,P,B$ are Collinear.
So equation of line $PAB$ is $$\displaystyle y+4 = \frac{-2+4}{4-2}(x-2)\Rightarrow x-y=6$$
Now solving $2x-y=-5$ and $x-y=6\;,$ we get $(x,y) = (-11,-17)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$ $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
Base case: For $n=1$
$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$
Induction hypothesis: For $n=m$
$\sum\limits_{k=1}^{m}sin(kx)=\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}$
Induction step: $n=m+1$
$\sum\limits_{k=1}^{m+1}sin(kx)=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$
Prove: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}+sin(m+1)x=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$
Left side: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}+sin\frac{x}{2}sin(m+1)x}
{sin\frac{x}{2}}$
How to prove this equality? I used $sin(u)sin(v)$ identity but that didn't help.
| It's simply a change product into sum or difference of trigonometric function.Note that we have double angle formula and apply it to $$\sin(m+1)x$$,we'll get $$sin\frac{(m+1)x}2 *\sin \frac{mx}2+2 \sin\frac{ x}2 *\sin\frac{(m+1)x}2*\cos\frac{(m+1)x}2=\sin\frac{(m+1)x}2*(\sin\frac{mx}2+2\cos\frac{(m+1)x}2*\sin\frac{x}2)$$. Apply the "change product into sum of trigonometric function" twice and we'll get the answer:$$\sin\frac{mx}2+2\cos\frac{(m+1)x}2*\sin\frac{x}2=\sin\frac{(m+1)x-x}2+2\cos\frac{(m+1)x}2*\sin\frac{x}2=\sin\frac{(m+1)x+x}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to prove that the following system of equations has only one solution? $
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
I have to solve this system of equations. After substituting $y = -\frac{3}{4}x - \frac{13}{2}$ into $(x - 1)^2 + (y + 1)^2 = 25$, I got that $(x + 2)^2 = 0$, so $x = -2$, so $y = -\frac{3}{4} \cdot (-2) - \frac{13}{2} = -5$.
How to prove that this is the only solution?
| As the other answers said, your argument already proved there is only one solution.
You can make an independent check by solving the equations graphically.
$(x−1)^2 +(y+1)^2 =25$ is the equation of a circle, radius $5$, center $(1, -1)$. The second equation is another circle, and the circles have two intersection points. The third equation represents a straight line, which goes through just one of the intersection points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
How to solve $x^6-x^5+x^4-x^3+x^2-x+1=0$? Can anyone tell me how to solve this?
$x^6-x^5+x^4-x^3+x^2-x+1=0$
What I got to was $x^7+1=0$.
Thanks in advance.
| This equation has the same coefficients read backwards.
There is a technique of solving such equations:
If the degree is odd, then $-1$ is a root, and dividing by $x+1$ gives you an even degree equation with the same property.
For even degree: divide by $x$ to the power half degree, and make the substitution $t=x+\frac{1}{x}$.
In this case we get
$$x^3+\frac{1}{x^3}-(x^2+\frac{1}{x^2})+x+\frac{1}{x}-1=0$$
we have
$$t=x+\frac{1}{x} \\
t^2=x^2+\frac{1}{x^2}+2\\
t^3=x^3+\frac{1}{x^3}+3t$$
Therefore, your equation becomes
$$t^3-3t-t^2+2+t-1=0\\
t^3-t^2-2t+1=0$$
You can solve this by using the cubic formula, and then solve the corresponding quadratics.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
A serie about $\sum_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}}$ How to prove $$\sum\limits_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}} = \ln 3 - \frac{\pi }{4}.$$
Add: Maybe we can follow this!
| Since $\arctan(x)=\arg(1+ix)$ and we can factor
$$
1+\frac{10in}{\left(3n^2+2\right)\left( 9n^2-1\right)}
=\frac{\left(1-\frac in\right)\left(1+\frac i{3n-1}\right)\left(1+\frac i{3n+1}\right)\left(1+\frac i{3n}\right)}{1+\frac2{3n^2}}
$$
we have that
$$
\begin{align}
&\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\[6pt]
&=\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right)
\end{align}
$$
Now this becomes a telescoping series:
$$
\begin{align}
&\sum_{n=1}^\infty\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\
&=\lim_{m\to\infty}\sum_{n=1}^m\left[\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right)\right]\\
&=-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\arctan\left(\frac1n\right)\\
&=-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\left[\frac1n+O\left(\frac1{n^3}\right)\right]\\[6pt]
&=\log(3)-\frac\pi4
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What is $\angle AEB$? Let $E$ be a point inside the square $ABCD$. and $|EC|=3,|EA|=1,|EB|=2$
What is the angle $\widehat {AEB}$?
I can only find $|ED|=\sqrt{12}$
| Here is a less algebraicly intensive solution and only requires knowledge of the sine law and the cosine law, and solving quadratics of the form $x^2 = a$(super easy) all of which is material known by the average grade 10 math student in Canada.
From the sine law we know that
\begin{align}
\frac{\sin(ABE)}{1} = \frac{\sin(AEB)}{s} && \frac{\sin(CBE)}{3} =\frac{\sin(90 -ABE)}{3} = \frac{\cos(ABE)}{3} = \frac{\sin(BEC)}{s}
\end{align}
After re arranging and squaring both parts we get
\begin{align}
\sin^2(ABE) = \frac{\sin^2(AEB)}{s^2} && \cos^2(ABE) = \frac{9\sin^2(BEC)}{s^2}
\end{align}
Recall that $\cos^2(x) = 1 - \sin^2(x)$.
\begin{align}
\frac{9\sin^2(BEC)}{s^2} &= 1 - \frac{\sin^2(AEB)}{s^2}\\
9(1 - \cos^2(BEC)) &= s^2 - 1 + \cos^2(AEB)
\end{align}
Using the cosine law we know that $5 - 4\cos(AEB) = s^2$ and $13 - 12\cos(BEC) = s^2$. Setting the equations equal and isolating for $\cos(BEC)$ yields $\cos(BEC) = \frac{\cos(AEB)}{3} + \frac{2}{3}$. We can know solve our equation
\begin{align}
9(1 - (\frac{\cos(AEB)}{3} + \frac{2}{3})^2) &= 5 - 4\cos(AEB) - 1 + \cos^2(AEB)\\
9(1 - \frac{\cos^2(AEB)}{9} - \frac{4\cos(AEB)}{9} - \frac{4}{9}) &= 4 - 4\cos(AEB) + \cos^2(AEB)\\
5 - \cos^2(AEB) - 4\cos(AEB) &= 4 - 4\cos(AEB) + \cos^2(AEB)\\
\end{align}
And since the $4\cos(AEB)$ terms cancel out we get
$2\cos^2(AEB) - 1 = 0$ yielding
\begin{equation}
\cos(AEB) = \frac{\pm 1}{\sqrt{2}} \iff \angle AEB = 45, 135
\end{equation}
And since $AEB$ can't be 45 $\angle AEB = 135 \deg$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1489850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Of sum of cosines and the $7$th roots of unity In my solution here, it was shown that
$$\omega+\omega^2+\omega^4=-\frac 12\pm\frac{\sqrt7}2\qquad\qquad (\omega=e^{i2\pi/7})$$
from which we know that
$$\sin \frac{2\pi}7+\sin \frac{4\pi}7-\sin \frac{6\pi}7=\Im (\omega+\omega^2+\omega^4)=\frac{\sqrt7}2$$
This can also be verified easily by computation.
By the same token it would appear that
$$\cos \frac{2\pi}7+\cos \frac{4\pi}7-\cos \frac{6\pi}7=\Re (\omega+\omega^2+\omega^4)=\frac 12$$
However a quick computational check shows that the result is $1.3019...$ and not $\frac 12$.
Why is this so?
| In that answer there is the following equality
$$\begin{align}
\sin\frac{2\pi}7+\sin\frac{4\pi}7-\sin\frac{6\pi}7
&=\Im(\omega^1+\omega^2-\omega^3)\\
&=\Im(\omega^1+\omega^2+\omega^4)\\
&=\Im(S)\\
&=\dfrac{\sqrt{7}}{2}\\
\end{align}.$$
For the real part you have
$$\begin{align}
-\dfrac{1}{2}&=
\Re(S)\\
&=\Re(\omega^1+\omega^2+\omega^4)\\
&=\Re(\omega^1+\omega^2+\omega^3)\\
&=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{6\pi}7
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the indefinite integral: $\int { {\sqrt{x+1}} \over {\sqrt{x+2} - \sqrt{x-2}} }dx$ Find the indefinite integral:
$$\int { {\sqrt{x+1}} \over {\sqrt{x+2} - \sqrt{x-2}} }dx$$
I don't know how to start, multiplying by ${ {\sqrt{x+2} + \sqrt{x-2}} \over {\sqrt{x+2} + \sqrt{x-2}} }$ hasn't helped neither...
| Notice, $$\int\frac{\sqrt{x+1}}{\sqrt{x+2}-\sqrt{x-2}} \ dx$$
$$=\int\frac{\sqrt{x+1}(\sqrt{x+2}+\sqrt{x-2})}{(\sqrt{x+2}+\sqrt{x-2})(\sqrt{x+2}-\sqrt{x-2})} \ dx$$
$$=\int\frac{\sqrt{x^2+3x+2}+\sqrt{x^2-x-2}}{x+2-(x-2)} \ dx$$
$$=\int\frac{\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{1}{4}}+\sqrt{\left(x-\frac{1}{2}\right)^2-\frac{9}{4}}}{4} \ dx$$
$$=\frac{1}{4}\int\sqrt{\left(x+\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2} \ dx+\frac{1}{4}\int\sqrt{\left(x-\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2} \ dx$$
Now, you can use standard formula $\int\sqrt{x^2-a^2}\ dx=\frac{1}{2}\left(x\sqrt{x^2-a^2}-a^2\ln\left|x+\sqrt{x^2-a^2}\right|\right)$
Proof:
let $$I=\int \sqrt{x^2-a^2}\ dx=\int \underbrace{\sqrt{x^2-a^2}}_{I}\cdot \underbrace{1}_{II}\ dx$$ using integration by parts $$I=x\sqrt{x^2-a^2}-\int \frac{2x}{2\sqrt{x^2-a^2}}x\ dx$$
$$I=x\sqrt{x^2-a^2}-\int \frac{(x^2-a^2)+a^2}{\sqrt{x^2-a^2}}\ dx$$
$$I=x\sqrt{x^2-a^2}-\int \sqrt{x^2-a^2}\ dx-a^2\int \frac{1}{\sqrt{x^2-a^2}}\ dx$$
$$I=x\sqrt{x^2-a^2}-I-a^2\int \frac{1}{\sqrt{x^2-a^2}}\ dx$$$$\implies I=\frac{1}{2}\left(x\sqrt{x^2-a^2}-a^2\int \frac{1}{\sqrt{x^2-a^2}}\ dx\right)\tag 1$$
Now, let $x=a\sec\theta\implies dx=a\sec\theta\tan \theta\ d\theta$ hence,
$$\color{red}{\int \frac{1}{\sqrt{x^2-a^2}}\ dx}=\int\frac{a\sec\theta\tan\theta\ d\theta}{\sqrt{a^2\sec^2\theta-a^2}}=\int\frac{a\sec\theta\tan\theta\ d\theta}{a\tan\theta}=\int \sec\theta\ d\theta$$$$=\ln\left|\sec\theta+\tan\theta\right|+c=\ln\left|\sec\theta+\sqrt{\sec^2\theta-1}\right|+c=\ln\left|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2-1}\right|+c$$ $$=\ln\left|x+\sqrt{x^2-a^2}\right|+\ln|a|+c=\color{red}{\ln\left|x+\sqrt{x^2-a^2}\right|+c_1}$$
now, setting the value of integral: $\int\frac{1}{\sqrt{x^2-a^2}}\ dx$ in (1), we get $$I=\frac{1}{2}\left(x\sqrt{x^2-a^2}-a^2\ln\left|x+\sqrt{x^2-a^2}\right|\right)+C$$
hence, we get
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\int\sqrt{x^2-a^2}\ dx=\frac{1}{2}\left(x\sqrt{x^2-a^2}-a^2\ln\left|x+\sqrt{x^2-a^2}\right|\right)+C}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to Calculate $x^6+x^3y^3+y^6$ Given that $x,y$ real numbers such that :
$x^2+xy+y^2=4$
And
$x^4+x^2y^2+y^4=8$
How can one calculate :
$x^6+x^3y^3+y^6$
Can someone give me hint .
| why not.
$$ x^4 + x^2 y^2 + y^4 = (x^2 + xy + y^2)(x^2 - xy+y^2). $$
Worth memorizing. So
$$ x^2 - xy + y^2 = 2. $$ Also
$$ 2xy = 2, \; \; \; xy=1. $$
And
$$ x^2 + y^2 = 3. $$
$$ 27 = (x^2 + y^2)^3 = x^6 + 3 x^4 y^2 + 3 x^2 y^4 + y^6 = x^6 + y^6 + 3 x^2 y^2 (x^2 + y^2) = x^6 + y^6 + 3 \cdot 1 \cdot 3 = x^6 + y^6 + 9 $$
So,
$$x^3 y^3 = 1, \; \; \; x^6 + y^6 = 18, $$
$$ x^6 + x^3 y^3 + y^6 = 19. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the equation of the circle which cuts the circle $x^2+y^2+2x+4y-4=0\;$ and the lines $xy-2x-y+2=0\;$ orthogonally The equation of the circle which cuts the circle $x^2+y^2+2x+4y-4=0\quad$ and the lines $xy-2x-y+2=0\quad$ orthogonally, is
$a.\quad x^2+y^2-2x-4y-6=0\;$
$b.\quad x^2+y^2-2x-4y+6=0\;$
$c.\quad x^2+y^2-2x-4y+12=0\;$
$d.\quad$ (not possible to determine.)
Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0\quad$. As this is intersecting the circle $x^2+y^2+2x+4y-4=0\quad$ orthogonally, so $2g+4f=c-4\;$ by using the condition of orthogonality of Terri circles $2g_1g_2+2f_1f_2=c_1+c_2\;$
The second equation is an equation of pair of lines,$(x-1)(y-2)=0\quad$
But I dont know what is the condition of orthogonality of a circle and a pair of lines. I am unable to solve it further. Please help me.
| 1493485
For two figures to be orthogonal vis-à-vis each other means that at each of their points of intersection their slopes are perpendicular. In terms of analytic geometry, this would mean that the slope of one is the negative reciprocal of the slope of the other -- i.e., the product of the two slopes at the intersection equals $-1\;$.
$xy-2x-y+2=0\;$ is really the two lines $x=1$ and $y=2\;$. Any circle orthogonal to the two lines must therefore be centered at $A(1\mid 2)\;$ .
The two tangents from $A\;$ to the circle $B\equiv[x^2+y^2+2x+4y-4=0]\;$ can be constructed by drawing the circle whose diameter is the line joining the center of the given circle $C(-1\mid -2)\;$ to $A\;$. The equation of this new circle is $D\equiv[x^2+y^2=5]\;$. The two intersections $E_1\left(\frac{-1-6\sqrt{11}}{10}\mid\frac{-2+3\sqrt{11}}{10}\right)\;$ and $E_2\left(\frac{-1+6\sqrt{11}}{10}\mid\frac{-2-3\sqrt{11}}{10}\right)\;$ of the two circles $B\;$ and $D\;$ are the points of tangency.
Your required circle $[(x-1)^2+(y-2)^2=11]\;$, centered, as required, at $A\;$, passes through $E_1\;$ and $E_2\;$.
Oh, by the way, from the menu of choices you provided, the proper choice is $a.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ I had an example in the book given as follows:
Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .
Solution: $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$
$(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$
Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$
Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$.
Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$
So degree of $p(x)$ is atleast $4$.
But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .
So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$
But I can't get the step how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$
Kindly help with this.
| Assume that $m$ and $n$ are square-free coprime integers. Then, if we prove that $Q(\sqrt m +\sqrt n) = Q(\sqrt m, \sqrt n)$, we can conclude that that $f(x) = x^4 -2(m+n)x +(m-n)^2$ must be the minimal polynomial of $\sqrt m +\sqrt n$ since $Q(\sqrt m +\sqrt n)$ is a four-dimensional $Q$-vector space. But if $\xi = (\sqrt m + \sqrt n)$, then ${\frac {{\xi}^2-(m+n)} {2}}={\sqrt m}{\sqrt n}$. Thus, $({\frac {{\xi}^2-(m+n)} {2}})\xi = m{\sqrt n}+n{\sqrt m}$. Hence, $$({\frac {{\xi}^2-(m+n)} {2}})\xi -n{\xi} = (m-n){\sqrt n},$$ and $$({\frac {{\xi}^2-(m+n)} {2}})\xi -m{\xi} = (n-m){\sqrt n}.$$ Thus, $Q(\sqrt m +\sqrt n)$ contains $\{\sqrt m, \sqrt n \}$, and so $Q(\sqrt m +\sqrt n)\supset Q(\sqrt m ,\sqrt n)$. The reverse set inclusion is immediate. The case $m =2$, and $n = 3$ is of course what inspired this train of thought which is hopefully correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
What line is determined by the following complex equation? $$\left | z+1-i \right |= \frac{\Re z-\Im z}{\sqrt{2}}$$
This leads to the following equation, if we write $z=x+iy$ :
$$\sqrt{\left ( x+1 \right )^{2}+\left( y-1 \right)^2}=\frac{x-y}{\sqrt{2}}$$
which according to desmos, defines a set in $\mathbb{C}$ which is empty.
| This is indeed impossible: Using Cauchy-Schwarz, we have
$$ x-y = (1)(x+1) + (-1)(y-1) -2 \le \sqrt{2}\sqrt{(x+1)^2+(y-1)^2} - 2 $$
and hence $\frac{x-y}{\sqrt{2}} < \sqrt{(x+1)^2+(y-1)^2}$ for all $x$ and $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof that sum and product of geometric series is divisible by 10 I need to prove that for $x=7+7^2+\cdots+7^{2016}$, $y=8+8^2+\cdots+8^{2016}$, $10\mid x+y$ and $10\mid xy$. I think that if $10\mid(m_1+m_2)$, where $m_1$ is the remainder of $\frac{x}{10}$ etc., than $10\mid x+y$, but I can't determine the remainder. I also tried determining calculating the last digit of the geometric series, but I can't get past the division.
| $$
7 + 7^2 + 7^3 + 7^4 \equiv 0 \pmod {10}
$$
Therefore
\begin{align}
& \overbrace{7 + 7^2 + 7^3 + 7^4} + \overbrace{7^5 + 7^6 + 7^7 + 7^8} + \overbrace{7^9 + 7^{10} + 7^{11} + 7^{12}} + \cdots \\[10pt]
= {} & \Big(7 + 7^2 + 7^3 + 7^4\Big) + 7^4\Big(7 + 7^2 + 7^3 + 7^4\Big) + 7^8 \Big(7 + 7^2 + 7^3 + 7^4\Big) + \cdots \\[10pt]
= {} & (10\times\text{something}) + 7^4(10\times\text{something}) + 7^8 (10\times\text{something}) + \cdots
\end{align}
A similar thing works with $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding positive values
For how many ordered triples of unequal positive integers $(x,y,z)$
does the expression $$ \frac{x}{(x-y)(x-z)} + \frac{y}{(y-x)(y-z)} +
\frac{z}{(z-x)(z-y)} $$ take on positive values?
I started with $x=3, y=4$ and $z=5$ and got:
$\frac{3}{2}+\frac{4}{-1}+\frac{5}{2}=0$
Then I worked with $x=1, y=2$ and $z=3$
$\frac{1}{2}+\frac{2}{-1}+\frac{3}{2}=0$
Then I worked with: $x=4, y=5$ and $z=6$
$\frac{4}{2}+\frac{5}{-1}+\frac{6}{2}=0$
So I did a random triple: $x=8, y=9$ and $z=10$
$\frac{8}{2}+\frac{9}{-1}+\frac{10}{2}=0$
So the answer to this question would be none, but I could be wrong. Any ideas?
| $\dfrac{x}{(x-y)(x-z)} + \dfrac{y}{(y-x)(y-z)} + \dfrac{z}{(z-x)(z-y)}$
$\dfrac{x}{(x-y)(x-z)} + \dfrac{-y}{(x-y)(y-z)} + \dfrac{z}{(x-z)(y-z)}$
$\dfrac{x(y-z) -y(x-z) + z(x-y)}{(x-y)(x-z)(y-z)}= \dfrac{0}{(x-y)(x-z)(y-z)} = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving A Linear Recurrence Relation With Complex Roots Question:
For the given linear homogeneous difference equation, find the general solution:
$$y_{n+2} + y_{n+1} + y_n = 0$$
With the initial conditions of:
$$y(0)=\sqrt3, y(1) = 0$$
Attempted Answer:
I approached the problem normally as one would except by solving the auxiliary equation which yields:
$$m^2 + m + 1 = 0$$
$$\implies m = \frac{-1}{2} \pm \frac{i\sqrt3}{2}$$
Now the general solution is of form:
$$y_n = A \left(\frac{-1}{2} - \frac{i\sqrt3}{2}\right)^n
+ B \left(\frac{-1}{2} + \frac{i\sqrt3}{2}\right)^n $$
Here is the part where I get stuck when I substitute the initial conditions to form a system of equations:
$$A + B = \sqrt{3}$$
$$A \left(\frac{-1}{2} - \frac{i\sqrt3}{2}\right)
+ B \left(\frac{-1}{2} + \frac{i\sqrt3}{2}\right) = 0$$
Now to make my life easier (which it really didn't), I decided to take a 'shortcut' and rewrite the 2nd equation as the sum of an imaginary and real part:
$$\frac{-1}{2}(A + B) + i\frac{\sqrt3}{2}(-A + B) = 0 + 0i$$
Since the real part on the LHS must equal the real part on the RHS, and the same applies for the imaginary part, we obtain:
$$\frac{-1}{2}(A + B) = 0$$
$$\frac{\sqrt3}{2}(-A + B) = 0$$
So the above implies that $A = B = -B$ which can only be true if $A = B = 0$. However if that is the case, then the first equation in the system of equations ($A + B = \sqrt{3}$) implies that 0 = $\sqrt{3}$.
I know that there is a mistake somewhere in the logic of my reasoning (I think perhaps when I equated each real and imaginary part of the LHS to the RHS) but I don't know why and where exactly. Please point out my mistake.
| Use generating functions. Define $A(z) = \sum_{n \ge 0} y_n z^n$, multiṕly your recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$\begin{align}
\frac{A(z) - y_0 -y_1 z}{z^2}
+ \frac{A(z) - y_0}{z} + A(z)
&= 0 \\
\frac{A(z) - \sqrt{3}}{z^2}
+ \frac{A(z) - \sqrt{3}}{z} + A(z)
&= 0
\end{align}$
Solving for $A(z)$:
$\begin{align}
A(z)
&= \frac{(1 + z) \sqrt{3}}{1 + z + z^2} \\
&= \frac{1 - z^2}{1 - z^3} \sqrt{3}
\end{align}$
We need:
$\begin{align}
[z^n] A(z)
&= [z^n] \frac{1 - z^2}{1 - z^3} \sqrt{3} \\
&= \sqrt{3} [z^n] (1 - z^2) \sum_{k \ge 0} z^{3 k} \\
&= \sqrt{3} \left(
[z^n] \sum_{k \ge 0} z^{3 k}
- [z^{n - 2}] \sum_{k \ge 0} z^{3 k}
\right) \\
&= \begin{cases}
\sqrt{3} & \text{if \(n \equiv 0 \pmod{3}\)} \\
- \sqrt{3} & \text{if \(n \equiv 1 \pmod{3} \wedge n \ge 2\)} \\
0 & \text{otherwise}
\end{cases}
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove using induction that $2^n < \binom{2n}{n} < 4^n$ for $n \geq 2$
Trying to prove that, for $n\geq2$, $2^n < \binom{2n}{n} < 4^n$.
Inductive hypothesis: Assume $P(k)$ is true:
\begin{align}
2^k < \binom{2k}{k} < 4^n \\\\
\end{align}
Show $P(k+1)$
\begin{align}
2^{k+1} < \binom{2k+2}{k+1} < 4^{k+1} \\\\
\end{align}
Not sure what's the strategy here... since it's a compound inequality I think it make sense to break the two apart and prove them separately?
Start with the inductive hypothesis
\begin{align}
2^k < \binom{2k}{k} < 4^n \\\\
\end{align}
Trying to prove the first inequality...
\begin{align}
2^k < \binom{2k}{k} \\\\
(2)2^k < (2)\binom{2k}{k} \\\\
2^{k+1} < (2)\binom{2k}{k}
\end{align}
Stuck here now... not sure if I see what to do to the equation to get to
\begin{align}
2^{k+1} < \binom{2k+2}{k+1}
\end{align}
| For the induction step on the first inequality, you wish to show that if
$$2^k < \binom{2k}{k}$$
then
$$2^{k + 1} < \binom{2k + 2}{k + 1}$$
Using the definition of the binomial coefficient yields
\begin{align*}
\binom{2k + 2}{k + 1} & = \frac{(2k + 2)!}{(k + 1)!(k + 1)!}\\
& = \frac{(2k + 2)(2k + 1)(2k)!}{(k + 1)k!(k + 1)k!}\\
& = \frac{2(k + 1)(2k + 1)(2k)!}{(k + 1)k!(k + 1)k!}\\
& = 2 \cdot \frac{2k + 1}{k + 1} \cdot \frac{(2k)!}{k!k!}\\
& = 2 \cdot \frac{2k + 1}{k + 1}\binom{2k}{k}\\
& > 2\binom{2k}{k}\\
& > 2 \cdot 2^k & \text{by the induction hypothesis}\\
& = 2^{k + 1}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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.