Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Help solving 3 simultaneous equations including sin and cos I have been trying to solve a problem involving conservation of momentum, in the end I end up with 3 equations which I feel cannot be solved. I would really appreciate if someone could show me how to manipulate these as nothing I seem to do really helps.
*
*$v_x^2 + v_y^2 = 16$
*$v_x\sin{30°} = v_y \sin{\theta}$
*$v_x \cos{30°} + v_y \cos{\theta} = 4$
The base of the problem is that $v_x$ and $v_y$ are the speeds of two objects. One object moving at angle $\theta$ the other at ${30°}$.
I feel like it may not be solvable due to the fact the first equation does not have a reference to $\theta$, but everything up to this point is correct.
Working through using $v_x = {\sqrt 16-v_y^2}$
Gets to the equation $32-2v_y^2+8v_y({\sqrt 1-\frac{16-v_y^2}{4v_y^2}})=0$
| If the angles are expressed in degrees we have $$\sin 30^{\circ}=\frac{1}{2}\qquad\text{and}\qquad\cos 30^{\circ}=\frac{\sqrt 3}{2}$$
Then in the second equation we have
$$v_x=2v_y\sin \theta$$
So, plugging this into the third equation we get
$$(2v_y\sin \theta)\frac{\sqrt 3}{2}+v_y\cos\theta=4$$
Then$$v_y=\frac{4}{\sqrt{3}\sin \theta+\cos\theta}\qquad\text{and}\qquad v_x=\frac{8\sin \theta}{\sqrt{3}\sin\theta+\cos\theta}$$
So, first equation reduces to
$$\frac{1+4\sin^2\theta}{(\sqrt{3}\sin \theta+\cos \theta)^2}=1$$
Then, by dividing by $\cos^2 \theta$ both terms in the quotient
\begin{align}
\frac{\sec^2\theta+4\tan^2\theta}{(\sqrt{3}\tan \theta+1)^2}&=1\\[5pt]
\frac{5\tan^2 \theta+1}{(\sqrt{3}\tan \theta+1)^2}&=1\\[5pt]
5\tan^2\theta+1&=3\tan^2\theta+2\sqrt{3}\tan \theta+1\\[5pt]
2\tan^2\theta-2\sqrt{3}\tan\theta&=0
\end{align}
Then, $\tan \theta =0$ or $\tan\theta=\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof that doctors could relate to I am supposed to present a mathematical proof to a lecture hall full of doctors in order to show them how mathematicians think. I'm having trouble picking a proof that will be easily followed by people who haven't seen anything beyond basic calculus in years. Does anyone have any suggestions? I'm trying to think of something visual to keep their attention.
| Claim: $1+1/2+1/3+1/4+\ldots=\infty$. [sic!]
Proof:
\begin{align*}
&\,1+\frac{1}{2}+\underbrace{\frac{1}{3}+\frac{1}{4}}+\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}+\underbrace{\frac{1}{9}+\ldots\frac{1}{15}+\frac{1}{16}}+\underbrace{\frac{1}{17}+\ldots+\frac{1}{31}+\frac{1}{32}}+\ldots\\
\geq&\,1+\frac{1}{2}+\underbrace{\frac{1}{4}+\frac{1}{4}}_{\text{2 terms}}+\underbrace{\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}}_{\text{4 terms}}+\underbrace{\frac{1}{16}+\ldots\frac{1}{16}+\frac{1}{16}}_{\text{8 terms}}+\underbrace{\frac{1}{32}+\ldots+\frac{1}{32}+\frac{1}{32}}_{\text{16 terms}}+\ldots\\
=&\,1+\frac{1}{2}+\left(2\times\frac{1}{4}\right)+\left(4\times\frac{1}{8}\right)+\left(8\times\frac{1}{16}\right)+\left(16\times\frac{1}{32}\right)+\ldots\\
=&\,1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\ldots\\
=&\,\infty.
\end{align*}
Bam! Lecture hall full of doctors shocked! $\quad\blacksquare$
Then show numerical examples to further convince them on the intuitive level that the harmonic series actually does diverge, albeit very slowly. For example:
\begin{align*}
\sum_{k=1}^{10}\frac{1}{k}\approx&\,2\mathord.93,\\
\sum_{k=1}^{100}\frac{1}{k}\approx&\,5\mathord.19,\\
\sum_{k=1}^{1000}\frac{1}{k}\approx&\,7\mathord.49,\\
\vdots&\,\\
\sum_{k=1}^{10^6}\frac{1}{k}\approx&\,14\mathord.39,\\
\vdots&\,\\
\sum_{k=1}^{10^{100}}\frac{1}{k}\approx&\,230\mathord.84\\
\vdots&\,\\
\sum_{k=1}^{10^{(10^6)}}\frac{1}{k}\approx&\,2\mathord,302\mathord,586\\
\vdots&\,
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 9,
"answer_id": 1
} |
Find $\mathop {\lim }\limits_{x \to \infty } (\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} ) $ $$\mathop {\lim }\limits_{x \to \infty } (\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} )
$$
My try:
$${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})
$$
$$\mathop {\lim }\limits_{x \to \infty } \frac{{(\sqrt[3]{{{x^3} + 5{x^2}}} - \sqrt {{x^2} - 2x} )(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}}{{(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}} =
$$
$$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} + 5{x^2} - {x^2} - 2x}}{{(\sqrt[{\frac{3}{2}}]{{{x^3} + 5{x^2}}} + \sqrt[3]{{{x^3} + 5{x^2}}}\sqrt {{x^2} - 2x} + {x^2} - 2x)}}
$$
And what's next...?
This task in first and second remarkable limits. I think i can replace variable, but how i will calculate it...
| We can also solve Using $$\displaystyle \bullet\; (1+x)^{n} = 1+nx+\frac{n(n-1)x^2}{2}+..........+\infty$$
So Here Limit $$\displaystyle \lim_{x\rightarrow \infty}\left[\sqrt[3]{x^3+5x^2}-\sqrt{x^2-2x}\right] = \lim_{x\rightarrow \infty}x\cdot \left[\left(1+\frac{5}{x}\right)^{\frac{1}{3}}-\left(1-\frac{2}{x}\right)^{\frac{1}{2}}\right]$$
So limit $$\displaystyle \lim_{x\rightarrow \infty}x\cdot \left[\left(1+\frac{5}{3x}+\frac{1}{3}\cdot -\frac{2}{3}\cdot \frac{25}{x^2}+........\infty\right)-\left(1-\frac{2}{2x}+\frac{1}{2}\cdot -\frac{1}{2}\cdot \frac{4}{x^2}.......\infty\right)\right]$$
So we get $$\displaystyle \lim_{x\rightarrow \infty}x\cdot \left[\frac{5}{3x}+\frac{1}{x}\right] = \frac{8}{3}$$
Here we Negelect Higher power of $x$ in Denominator, bcz when $\lim_{x\rightarrow \infty}\;,$ Then $\displaystyle \frac{1}{x^n}\rightarrow 0\; \forall n\geq 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given that $\ln (2x^2+9x-5)=1+\ln (x^2+2x-15)$ find $x$ in terms of $e$ Given that $\ln (2x^2+9x-5)=1+\ln (x^2+2x-15)$ find $x$ in terms of $e$. Then the left side is $2x^2+9x-5=\ln (1)+x^2+2x-15$, is that true? How should I proceed?
| No, it is equal to
$2x^2+9x−5 = e^{1+ln(x^2+2x−15)} = e(x^2+2x-15)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Fundamental unit of real quadratic field Let $K=\mathbb{Q}(\sqrt d)$, where $d$ is squarefree and greater than $1$. Assume that $5|(d+1)$ or $5|(d-1)$. Let $\mathfrak{P}$ be a prime of $K$ over $5$. Let $\delta _o$ be a fundamental unit of $K$.
How to show that $\delta_o$ is a $5^{th}$ power in $K_{\mathfrak{P}}$ iff $\delta_o^2 \equiv \pm 1$ (mod $\mathfrak{P}^2)$?
Try: I have shown that $5$ splits in $K$ and $\delta_o^2\equiv \pm 1$ (mod $\mathfrak{P})$. Then I tried to use Hensel`s lemma but failed. Thank you for your help.
| The $5$th power map, $(1+5x) \mapsto (1+5x)^5 = 1 + 5^2x + O(5^3x^2)$ is a bijection between $1 + 5 \Bbb Z_5$ and $1 + 5^2\Bbb Z_5$ (to show this you can use Hensel's lemma on the polynomial with integer coefficients $((1+5X)^5-1)/5^2$)
In particular this means that anything congruent to $1$ modulo $5^2$ is a $5$th power in $\Bbb Z_5$, and so the invertible $5$th powers are an open subgroup of $\Bbb Z_5^*$ : an invertible element of $\Bbb Z_5$ is a $5$th power if and only if it is a $5$th power mod $5^2$.
The $5$th powers mod $5^2$ are $1^5=1,2^5=7,3^5=-7$ and $4^5=-1$.
Next, the squaring map on the invertibles of $\Bbb Z_5$ is a local isometry, and is $2$-to-$1$ because $1^2 = -1$, so $x = \pm 1, \pm 7 \pmod {5^2}$ if and only if $x^2 = 1^2$ or $x^2 = 7^2 = -1\pmod {5^2}$.
You can even go one step further and show that this is also equivalent to $x^4 = 1 \pmod {5^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $(x+y)^n = \sum\limits_{k=0}^n \binom{n}{k} x^k y^{n-k}$ I'm reading Serge Lang's 'Analysis I', and there's a problem I cannot figure out how to prove:
Problem: Prove by induction that $$(x+y)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k y^{n-k} . $$
Attempt at proof : I established the base case, which is easily verified. Now I want to prove the inductive step. So assume the assertion holds for $n \geq 1$, and we want to show it also holds for $n + 1$. So we have to proof that: $$(x+y)^{n+1} = \sum_{k=0}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} x^k y^{n+1-k} . $$
I started with the the LHS and applied the induction hypothesis:
\begin{align*} (x+y)^{n+1} &= (x+y)^n (x+y) \\ & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k y^{n-k} (x+y) \\ &= \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^{k+1} y^{n-k} + \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k y^{n-k+1} \\ &= \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \bigg[ x^{k+1} y^{n-k} + x^k y^{n-k+1} \bigg] \end{align*} Now I don't know how to proceed. Any help would be appreciated!
| Hint: $$\begin{align}
\sum_{k=0}^n &\binom n k \bigg[ x^{k+1} y^{n-k} + x^k y^{n-k+1} \bigg]\\
&=\color{red}{\sum_{k = 0}^n {n \choose k} x^{k+1}y^{n-k}} + \color{#00A000}{\sum_{k = 0}^n {n \choose k} x^{k} y^{n-k+1}}\\
&=\color{red}{{n \choose n} x^{n+1} + \sum_{k = 0}^{n-1} {n \choose k} x^{k+1} y^{n-k}} + \color{#00A000}{{n \choose 0} y^{n+1} + \sum_{k = 1}^{n} {n \choose k} x^{k} y^{n-k+1}}\\
&=x^{n+1} + y^{n+1} + \color{royalblue}{\sum_{k = 0}^{n-1} {n \choose k} x^{k+1} y^{n-k}} + \sum_{k = 1}^{n} {n \choose k} x^{k} y^{n-k+1}\\
&={n+1 \choose 0} x^{n+1} + {n+1 \choose n+1} y^{n+1} + \color{royalblue}{\sum_{k = 1}^n {n \choose k-1} x^{k} y^{n-k+1}} + \sum_{k = 1}^n {n \choose k} x^{k} y^{n-k+1}.\\
\end{align}$$
Can you continue?
| {
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"url": "https://math.stackexchange.com/questions/1436306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Show that $4\cos\frac\pi5\cdot\sin\frac\pi{10}=1$ We have to prove this identity : $$4\cos\frac\pi5\cdot\sin\frac\pi{10}=1$$
The book's hint is we somehow find out that $\displaystyle1+2\left(\cos\frac\pi5-\cos\frac{2\pi}5\right)$ equals something (from the geometry of a regular pentagon) and the result follows. But I can neither see why the book's hint is true nor how it could yield a result for the original problem. So could you help me proving it either by the book's hint or another way?
| HINT:
Notice $$LHS=2\cos \frac{\pi}{5}\sin\frac{\pi}{10}$$
$$=2\cos 36^\circ \sin18^\circ$$ Setting the values of $\cos 36^\circ$ & $\sin 18^\circ$
$$=2\left(\frac{\sqrt 5+1}{4}\right)\left(\frac{\sqrt 5-1}{4}\right)$$
$$=\frac{(\sqrt 5+1)(\sqrt 5-1)}{8}$$
$$=\frac{(\sqrt 5)^2-1^2}{8}$$ $$=\frac{5-1}{8}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solving a quadratic equation with parameters $$\frac { x+2 }{ a }- \frac { 2-a }{ x } =2$$
Steps I took:
$$\frac { x(x+2) }{ ax } -\frac { a(2-a) }{ ax } =\frac { 2ax }{ ax } $$
$$x^2+2x-2a+a^2=2ax$$
$$x^{ 2 }+2x-2a+a^{ 2 }-2ax=0$$
I get stuck at this point because I have no clue how to factor an equation in this form. Hints a more appreciated than the actual answer
| Notice, we have $$\frac{x+2}{a}-\frac{2-a}{x}=2$$
$$\frac{x(x+2)-a(2-a)}{ax}=2ax$$ $$x^2+2x-a(2-a)=2ax$$
$$x^2-2(a-1)x+a(a-2)=0$$ $$x=\frac{2(a-1)\pm\sqrt{4(a-1)^2-4a(1)(a-2)}}{2(1)}$$
$$=\frac{2(a-1)\pm\sqrt{4(a-1)^2-4a(1)(a-2)}}{2(1)}=\frac{2(a-1)\pm 2}{2(1)}$$
$$=\frac{2(a-1)\pm 2}{2}$$
$$\iff x=\frac{2(a-1)+2}{2}=a$$
$$\iff x=\frac{2(a-1)-2}{2}=a-2$$
Hence, we get
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x=a,\ (a-2)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solving Isosceles Triangle if we know a Median on Leg and a Perimeter I solved Isosceles Triangle if we know an Median on Leg and a Perimeter. But the solution seems excessively complicated to be computed other than by a computer.
I start with the following cubic equation. There are two valid results, both of which are for leg:
$4Px^3 − (5P^2 + m^2)x^2 + 2P^3x − (1/4)P^4 = 0$
where P is Perimeter and m is median on leg.
Is it possible either to find x in a easier way, or to calculate the problem in a completely different way?
| I don't see from where comes your equation.
Anyway: there are infinitely many triangle that satisfy the given condition, depending on the length of the side $AB$ where you put the midpoint.
I suggest an analytic approach.
Let $A=(-a,0)$ and $B=(a,0)$ so that the mid point is $M=(0,0)$. If $P=(x,y)$ is the other vertex of the triangle, we want: $PM=m$ so : $P=(x,\sqrt{m^2-x^2})$ ( here I ignore the symmetric solution) .
So we have:
$$
PA=\sqrt{(x+a)^2+(m^2-x^2)}=\sqrt{a^2+m^2+2ax}
$$
and, analogously:
$$
PB=\sqrt{a^2+m^2-2ax}
$$
and, if $p$ is the perimeter, we find the equation:
$$
p=PA+PB+AB \iff \sqrt{a^2+m^2+2ax}+\sqrt{a^2+m^2-2ax}+2a=p
$$
that is not so difficult to solve.
squaring the equation
$$
\sqrt{a^2+m^2+2ax}+\sqrt{a^2+m^2-2ax}=p-2a
$$
we have:
$$
a^2+m^2+2ax+a^2+m^2-2ax+2\sqrt{(a^2+m^2+2ax)(a^2+m^2-2ax)}=(p-2a)^2
$$
$$
2(a^2+m^2)+2\sqrt{(a^2+m^2)^2-4a^2x^2}=(p-2a)^2
$$
$$
2\sqrt{(a^2+m^2)^2-4a^2x^2}=(p-2a)^2-2(a^2+m^2)
$$
squaring again you find a second degree (pure) equation in $x$:
$$
4\left[(a^2+m^2)^2-4a^2x^2\right]=\left[(p-2a)^2-2(a^2+m^2)\right]^2
$$
can you solve this, and discuss the solutions?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2$ For positive integers Numbers $a, b, c $ prove that $$\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2$$
This inequality above take long time to prove it and still couldn't complete it. How I can prove this inequality? Any hint will help. Thanks
| The above inequality holds for any positive real numbers $a,b,c$.
Indeed, denote $x=1+b+c^2,y=1+c+a^2,z=1+a+b^2$, the inequality becomes
$$\frac{y-c}{x} + \frac{z-a}{y} + \frac{x-b}{z} \ge 2,$$
or equivalently
$$\frac{y}{x} + \frac{z}{y} + \frac{x}{z} \ge 2 + \frac{c}{x} + \frac{a}{y} + \frac{b}{z}.$$
By AM-GM inequality:
$$\frac{y}{x} + \frac{z}{y} + \frac{x}{z} \ge \sqrt[3]{\frac{y}{x} \frac{z}{y} \frac{x}{z}} =3.$$
It remains to prove $$\frac{c}{x} + \frac{a}{y} + \frac{b}{z} \le 1.$$
Since $x\ge 2c+b,y\ge 2a+c,z\ge 2b+a$ we only need to prove that
$$\frac{c}{2c+b} + \frac{a}{2a+c} + \frac{b}{2b+a} \le 1,$$ which is equivalent to
$$\left(\frac{c}{2c+b} - \frac{1}{2}\right) + \left(\frac{a}{2a+c} - \frac{1}{2}\right) + \left(\frac{b}{2b+a} - \frac{1}{2}\right) \le 1 - \frac{3}{2}$$
or
$$\frac{b}{2c+b} + \frac{c}{2a+c} + \frac{a}{2b+a} \ge 1.$$
The last inequality is true by Cauchy-Schwarz inequality:
$$\sum\frac{a}{2b+a} = \sum\frac{a^2}{2ab+a^2} \ge \frac{(a+b+c)^2}{\sum (2ab+a^2)}=1.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Linear system - number of solutions depending on the parameter k Determine for what value of $k$ the following system has
*
*unique solution,
*no solution and
*infinitely many solutions.
\begin{cases}
x+2y+z=3\\
2x-y-3x=5\\
4x+3y-z=k
\end{cases}
I did till
$$
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
1 & 1 & 1 & 1/5\\
0 & 0 & 1 & -k+42/5
\end{array}\right]
$$
I am not sure how to continue.
| 1) First step:
$$\begin{cases}
x+2y+z=3\\
2x-y-3x=5\\
4x+3y-z=k
\end{cases}\Longleftrightarrow$$
$x+2y+z=3 \Longleftrightarrow x=3-2y-z$
$$\begin{cases}
-3+y+z=5\\
12-5y-5z=k
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
-15+5y+5z=25\\
12-5y-5z=k
\end{cases}\Longleftrightarrow$$
$$-3=25+k \Longleftrightarrow$$
$$k=-28 $$
2) Second step:
$$\begin{cases}
x+2y+z=3\\
2x-y-3x=5\\
4x+3y-z=-28
\end{cases}\Longleftrightarrow$$
$2x-y-3x=5 \Longleftrightarrow y=-5-x$
$x+2y+z=3 \Longleftrightarrow z=28+4x+3y \Longleftrightarrow z=13+x$
$x+2y+z=3 \Longleftrightarrow x=3-2y-z \Longleftrightarrow x=z-13$
So we got:
$$k=-28,y=-x-5,x=z-13,z=13+x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find polynomial equation for a cardioid in $\mathbb{R}^2$ We have the cardioid with equations:
$$x(\theta)=\cos\theta+\frac{1}{2}\cos(2\theta)$$
$$y(\theta)=\sin\theta+\frac{1}{2}\sin(2\theta)$$
I have to show that you can define this cardioid with a polynomial in two variables.
My approach, I defined the auxiliary variables $u,v$ such that,
$$\begin{matrix}
u=\cos(2\theta)&\Rightarrow&\cos\theta =\frac{\sqrt{1+u}}{2}\\
v=\sin(2\theta)&\Rightarrow&\sin\theta =\frac{\sqrt{1+-u}}{2}
\end{matrix}$$
and since $u^2+v^2=1$, it is,
$$x^2+y^2-\frac{1}{4}=\frac{1}{2}+u\frac{\sqrt{1+u}}{2}+v\frac{\sqrt{1-u}}{2}=ux+vy$$
but I got stuck here, any tips on how can I continue?
| Given the two equations
$$x(\theta)=\cos\theta+\frac{1}{2}\cos(2\theta) \tag{1}$$
$$y(\theta)=\sin\theta+\frac{1}{2}\sin(2\theta) \tag{2}$$
By squaring both equations and adding (this gives the useful "expression" $\cos^2+\sin^2$ for both $\theta$ and $2\theta$):
$$\begin{align}
x^2+y^2&=1+\frac{1}{4}+\cos\theta \cos2\theta+\sin\theta\sin2\theta \\[1em]
&=\frac{5}{4}+\cos\theta \tag{3}
\end{align}$$
While from (1) and the identity $\cos2\theta=2\cos^2\theta-1$:
$$\begin{align}
\cos^2\theta+\cos\theta&=x+\frac{1}{2} \\[1em]
\cos^2\theta+\cos\theta+\frac{1}{4}&=x+\frac{3}{4} &(\text{completing the square})\\[1em]
\left(\cos\theta+\frac{1}{2}\right)^2&=x+\frac{3}{4} \tag{4}
\end{align}$$
From (3):
$$\begin{align}
x^2+y^2-\frac{3}{4}&=\cos\theta+\frac{1}{2} &\implies \\[1em]
\left(x^2+y^2-\frac{3}{4}\right)^2&=\left(\cos\theta+\frac{1}{2}\right)^2 &\stackrel{(4)}{\implies} \\[1em]
\left(x^2+y^2-\frac{3}{4}\right)^2&=x+\frac{3}{4}
\end{align}$$
which can be simplified further if you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Coordinate Geometry Parabola and Line Intersection Can someone please explain why $\alpha + \beta = -\frac{b}{a} = \frac{4}{3}$ for the image below?
Where $a$ and $b$ are constants that come from: $y^2-3y=5(\frac{1-y}{3})$
| If $\alpha$ and $\beta$ are roots of the quadratic polynomial $ay^2+by+c$, then
$$ay^2+by+c=a(y-\alpha)(y-\beta)$$
So
\begin{align}
ay^2+by+c&=ay^2-a(\alpha+\beta)y+a\alpha\beta
\end{align}
By equaling the coefficients of $y$ we get
\begin{align}
-a(\alpha+\beta)&=b\\
\alpha+\beta&=-\frac{b}{a}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I Approximate $\log{2}\approx 0.693$ without using the Maclaurin series? How do I approximate the value $\log{2}\approx 0.693$ without using the Maclaurin series?
The book gives the hint: consider $f(x)=e^x-e^{-x}-2x$.
| If $y=\ln(x)$, then $$y'=\frac{1}{x}\quad\text{and}\quad y'=e^{-y}$$ and you can numerically solve either differential equation to $x=2$, using initial condition $y(1)=0$.
With the first differential equation, using the Runge-Kutta method with only two steps:
$$
\begin{array}{rrr|rrrr}
n&x_n&y_n&k_{1,n}&k_{2,n}&k_{3,n}&k_{4,n}\\
0&1&0\\
&&&\frac{1}{1}=1&\frac{1}{1+\frac{1}{4}}=\frac{4}{5}&\frac{1}{1+\frac{1}{4}}=\frac{4}{5}&\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\\
1&\frac32&0+\frac{\frac{1}{2}}{6}\left(1+2\cdot\frac{4}{5}+2\cdot\frac{4}{5}+\frac{2}{3}\right)=\frac{73}{180}\\
&&&\frac{1}{\frac{3}{2}}=\frac{2}{3}&\frac{1}{\frac{3}{2}+\frac{1}{4}}=\frac{4}{7}&\frac{1}{\frac{3}{2}+\frac{1}{4}}=\frac{4}{7}&\frac{1}{\frac{3}{2}+\frac{1}{2}}=\frac{1}{2}\\
2&2&\frac{73}{180}+\frac{\frac{1}{2}}{6}\left(\frac{2}{3}+2\cdot\frac{4}{7}+2\cdot\frac{4}{7}+\frac{1}{2}\right)=\frac{1747}{2520}
\end{array}
$$
And $\frac{1747}{2520}=0.693\ldots$. For more accuracy, you have to start over with a smaller step size (and hence, more steps).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 3
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Fibonacci sequence developing For the sum $$\sum_i^n {n-i \choose i}$$
I evaluate it for $n=1,2,3,4,5$
For $n=1$ we have $$\sum_{i=0}^1 {1-i \choose i} = {1 \choose 0} + {0 \choose 1} = 1 + 0 = 1$$
For $n=2$ we have $$\sum_{i=0}^2 {2-i \choose i} = {2 \choose 0} + {1 \choose 1} + {0 \choose 2} = 1 + 1 + 0 = 2$$
For $n=3$ we have $$\sum_{i=0}^3 {3-i \choose i} = {3 \choose 0} + {2 \choose 1} + {1 \choose 2} + {0 \choose 3} = 1 + 2 + 0 + 0 = 3$$
For $n=4$ we have $$\sum_{i=0}^4 {4-i \choose i} = {4 \choose 0} + {3 \choose 1} + {2 \choose 2} + {1 \choose 3} + {0 \choose 4} = 1 + 3 + 1 + 0 + 0 = 5$$
For $n=5$ we have $$\sum_{i=0}^5 {5 - i \choose i} = {5 \choose 0} + {4 \choose 1} + {3 \choose 2} + {2 \choose 3} + {1 \choose 4} + {0 \choose 5} = 1 + 4 + 3 + 0 + 0 + 0 = 8$$
And we can see the pattern $$1,2,3,5,8,....$$ which makes me think about the Fibonacci sequence
One can also evaluate $n=0$ $$\sum_0^0 = {0 - i \choose i} = {0 \choose 0} = 1$$
And so we have $$1,1,2,3,5,8,....$$
Now to prove that it is really the Fibonacci sequence, Can I prove it by induction ?
By starting from $n =2$, we have $$\sum_{i=0}^n {n-i \choose i} = f(n) = f(n-1) + f(n-2)$$
Do we start with a base case $n=2$,
$$\sum_{i=0}^2 {2-i \choose i} = 2 = f(1) + f(0)$$
And so the base case works.
Now assume it works up to $n = k$ (Strong Mathematical Induction)
That is we assume it works up to $$\sum_{i=0}^k {k -i \choose i} = f(k-1) + f(k-2)$$
Now we consider $k + 1$, Now I am stuck here. Any suggestions ?
| Using Pascals identity we get
\begin{align*}
S_{n+2}=\sum_i^{n+2} {n+2-i\choose i}=\sum_i^{n+2}\Big({n+1-i\choose i}+{n-(i-1)\choose i-1}\Big)=S_{n+1}+S_n.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Does $\sum\limits_{n\geq 1}\frac{1}{(3n-1)(3n+1)}$ converge or diverge?
How would you prove convergence/divergence of the following series?
$$\sum\limits_{n\geq 1}\dfrac{1}{(3n-1)(3n+1)}$$
I'm interested in more ways of proving convergence/divergence for this series.
My thoughts:
$$\sum\limits_{n\geq 1}\dfrac{1}{(3n-1)(3n+1)}$$
I'm going to use Integral Test
let $f(n)=\dfrac{1}{(3n-1)(3n+1)}$
we note that $f$ is a continuous, positive and decreasing function on the interval $[1,+\infty ($ then
$$\sum_{n=1}^\infty f(n) \text{ Convergent } \iff \int_1^\infty f(x)\,dx \text{ Convergent }$$
so let's try to compute
$$\int_1^\infty f(n)\,dn=\int_1^\infty \dfrac{1}{(3n-1)(3n+1)}\,dn $$
Substitution for $n=\dfrac{1}{3}u$ and $dn=\dfrac{1}{3}du$ we got :
\begin{align}
\int_1^\infty f(n)\,dn &=\int_1^\infty \dfrac{1}{(3n-1)(3n+1)}\,dn \\
&=\int_1^\infty \dfrac{1}{(3(\dfrac{1}{3}u)-1)(3(\dfrac{1}{3}u)+1)}\,\dfrac{1}{3}du \\
&=\dfrac{1}{3}\int_1^\infty \dfrac{1}{(u-1)(u+1)}\,du \\
&=\dfrac{1}{3}\int_1^\infty \dfrac{1}{(u^2-1)}\,du \\
\end{align}
note that $\left(arctanh(u)\right)'=\left(\tanh^{-1}(h)\right)'=\dfrac{1}{1-u^{2}}$ so
\begin{align}
\int_1^\infty f(n)\,dn &=\dfrac{1}{3}\int_1^\infty \dfrac{1}{(u^2-1)}\,du \\
&=-\dfrac{1}{3}\int_1^\infty \dfrac{1}{(1-u^2)}\,du \\
&=-\dfrac{1}{3} \left(\tanh^{-1}(u)\right) \biggl|_{1}^{+\infty} \\
&=-\dfrac{1}{3} \left(\tanh^{-1}(3n)\right) \biggl|_{1}^{+\infty} \\
\end{align}
so can't go ahead beause i don't know the limit of $tanh^{-1}$ i can use woflrame to do that but suppose im in the contest what gonna do
so my questions:
*
*Is my proof correct
*since the calculation is not easy in this case I'm interested in more ways of proving convergence/divergence for this series.
| Checking Convergence
Applying a Telescoping Series, we get
$$
\begin{align}
\sum_{k=1}^N\frac1{(3n-1)(3n+1)}
&\le\sum_{k=1}^N\frac1{(3n-2)(3n+1)}\\
&=\frac13\sum_{k=1}^N\left(\frac1{3n-2}-\frac1{3n+1}\right)\\
&=\frac13\left(1-\frac1{3N+1}\right)
\end{align}
$$
Therefore,
$$
\sum_{k=1}^\infty\frac1{(3n-1)(3n+1)}\le\frac13
$$
Evaluating the Sum
Using $(9)$ from this answer,
$$
\begin{align}
\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n-1}\right)
&=\frac13\sum_{n=1}^\infty\left(\frac1n-\frac1{n-1/3}\right)\\
&=\frac13H_{-1/3}\\[3pt]
&=-\frac12\log(3)+\frac\pi{6\sqrt3}\tag{1}
\end{align}
$$
Subtracting $(8)$ from $(7)$ in that same answer gives
$$
\begin{align}
\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n+1}\right)
&=\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n-2}\right)+\overbrace{\sum_{n=1}^\infty\left(\frac1{3n-2}-\frac1{3n+1}\right)}^{\text{telescoping series}}\\
&=\frac13\sum_{n=1}^\infty\left(\frac1n-\frac1{n-2/3}\right)+1\\
&=\frac13H_{-2/3}+1\\[3pt]
&=-\frac12\log(3)-\frac\pi{6\sqrt3}+1\tag{2}
\end{align}
$$
Partial fractions and $(1)$ and $(2)$ yields
$$
\begin{align}
\sum_{n=1}^\infty\frac1{(3n-1)(3n+1)}
&=\frac12\sum_{n=1}^\infty\left(\frac1{3n-1}-\frac1{3n+1}\right)\\
&=\frac12\left[\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n+1}\right)
-\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n-1}\right)\right]\\[3pt]
&=\frac12-\frac\pi{6\sqrt3}\tag{3}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many nonnegative integer solutions does the following equation have: $x + y + z = 15$
How many nonnegative integer solutions does the following equation
have: $x + y + z = 15$
(1) if $1 \le x \le 6$
(2) if $x \ge 2$ and $y \le 3$
(3) if $x \ge 3$, $y \ge 2$, and $1 \le z \le 3$
I solved the first question (which is C(3+11-1,11)). Still very stuck on the next two
| (1) We wish to solve the equation $x + y + z = 15$ in the nonnegative integers subject to the restrictions that $1 \leq x \leq 6$. Let $w = x - 1$. Then $w$ is a nonnegative integer. Substituting $w + 1$ for $x$ in the equation $x + y + z = 15$ yields
\begin{align*}
w + 1 + y + z & = 15\\
w + y + z & = 14\tag{1}
\end{align*}
A particular solution corresponds to where two addition signs are placed in a row of $14$ ones. For instance,
$$1 1 1 1 1 1 1 + 1 1 1 + 1 1 1 1 1$$
corresponds to the solution $w = 6$, $y = 3$, and $z = 5$, while
$$+ 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1$$
corresponds to the solution $w = 0$, $y = 5$, and $z = 9$. The number of solutions of equation 1 in the nonnegative integers is the number of ways to select which two of the $16$ symbols (the $14$ ones and the two addition signs) will be addition signs, which is
$$\binom{14 + 2}{2} = \binom{16}{2}$$
However, the restriction that $1 \leq x \leq 6 \Rightarrow 0 \leq w \leq 5$. Thus, we must remove those solutions in which $w \geq 6$. Assume $w \geq 6$. Let $v = w - 6$. Then $v$ is a nonnegative integer. Substituting $v + 6$ for $w$ in equation 1 yields
\begin{align*}
v + 6 + y + z & = 14\\
v + y + z & = 8 \tag{2}
\end{align*}
Equation 2 has $$\binom{8 + 2}{2} = \binom{10}{2}$$ solutions in the nonnegative integers. Hence, the number of solutions of the equation $x + y + z = 15$ in the nonnegative integers subject to the restrictions that $1 \leq x \leq 5$ is $$\binom{16}{2} - \binom{10}{2}$$
(2) We wish to solve the equation $x + y + z = 15$ in the nonnegative integers subject to the restrictions that $x \geq 2$ and $y \leq 3$. Let $w = x - 2$. Then $w$ is a nonnegative integer. Substituting $w + 2$ for $x$ in the equation $x + y + z = 15$ yields
\begin{align*}
w + 2 + y + z & = 15\\
w + y + z & = 13 \tag{3}
\end{align*}
which is an equation with $$\binom{13 + 2}{2} = \binom{15}{2}$$ solutions in the nonnegative integers. From these, we must eliminate those solutions in which $y \geq 4$. Assume $y \geq 4$. Let $v = y - 4$. Substituting $v + 4$ for $y$ in equation 3 yields
\begin{align*}
w + v + 4 + z & = 13\\
w + v + z & = 9
\end{align*}
which is an equation with $$\binom{9 + 2}{2} = \binom{11}{2}$$ solutions in the nonnegative integers. Hence, the number of solutions of the equation $x + y + z = 15$ in the nonnegative integers subject to the restrictions that $x \geq 2$ and $y \leq 3$ is
$$\binom{15}{2} - \binom{11}{2}$$
Using the methods shown above should enable you to solve the third problem.
| {
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"url": "https://math.stackexchange.com/questions/1449051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Using substitution $p = x + \frac{1}{x}$ to reduce $2x^4+x^3-6x^2+x+2=0$ How do I use the substitution $p = x + \frac{1}{x}$ to show that the equation
$$2x^4+x^3-6x^2+x+2=0$$
reduces to $2p^2+p-10=0$.
Now, I can do the problem in reverse, but when I try solving $p = x + \frac{1}{x}$ for $x$ and substituting, the resulting equation is too complicated and nothing like the form required.
| HINT :
Dividing the both sides by $x^2$ gives
$$2x^2+x-6+\frac 1x+\frac{2}{x^2}=0,$$
i.e.
$$2\left(x^2+\frac{1}{x^2}\right)+\left(x+\frac 1x\right)-6=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the equation whose roots are Okay I found this one on a test and it must be something really silly I am missing $ \alpha$ is not equal to $\beta$ and $\alpha^2=5\alpha-3,5\beta-3=\beta^2$ then the equation whose roots are $\alpha/\beta$ and $ \beta/\alpha$ is $ (a)3x^2-25+3=0,(b)x^2-5x+3=0,(c)x^2+5x-3=0,(d)3x^2-19x+3=0$ Of course the answer is either a or d I tried solving for $\alpha$ and $\beta$ but the equation became too complicated
| Notice, the roots $\alpha$ & $\beta$ are distinct roots &
$$\alpha^2=5\alpha -3, \ \ 5\beta-3=\beta^2$$
Hence, the quadratic equation is $$x^2-5x+3=0$$
whose roots are determined by using quadratic formula $$\alpha, \beta=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-3)}}{2(1)}$$ $$=\frac{5\pm\sqrt{37}}{2}$$ $$\implies \alpha=\frac{5+\sqrt{37}}{2}, \ \ \beta=\frac{5-\sqrt{37}}{2}$$
Now, $$\frac{\alpha}{\beta}=\frac{\frac{5+\sqrt{37}}{2}}{\frac{5-\sqrt{37}}{2}}=-\frac{31+5\sqrt {37}}{6}$$
$$\frac{\beta}{\alpha}=\frac{\frac{5-\sqrt{37}}{2}}{\frac{5+\sqrt{37}}{2}}=-\frac{31+5\sqrt {37}}{6}=-\frac{31-5\sqrt {37}}{6}$$
Hence, the quadratic equation having roots $\frac{\alpha}{\beta}$ & $\frac{\beta}{\alpha}$ is given as $$\left(x-\frac{\alpha}{\beta}\right)\left(x-\frac{\beta}{\alpha}\right)=0$$
$$\left(x+\frac{31+5\sqrt {37}}{6}\right)\left(x+\frac{31-5\sqrt {37}}{6}\right)=0$$
$$\color{red}{3x^2+31x+3=0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Each of the two persons makes a single throw with a pair of unbiased dice.What is the probability that the throws are equal? Each of the two persons makes a single throw with a pair of unbiased dice.What is the probability that the throws are equal?
Since the same throws can result if either both of them get 1 or both of them get 2 or both of them get 3 or both of them get 4 or both of them 5 or both of them get 6.
So I calculated probability as $\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}=\frac{1}{6}$
But my answer is wrong and the correct answer is $\frac{73}{648}$.Please help me with the correct approach to solve it.Thanks.
| If I'm interpreting this correctly...
There are a total of $36$ combinations you can get when rolling a pair of $6$-sided dice. There is $1$ way to get a $2$, $2$ ways to get a $3$, $3$ ways to get a $4$, etc.
You have a total of $36\cdot 36= 1296$ events. Now, we do a bit of counting. There is $1$ way both people can get a $2$, $2 \cdot 2$ ways both people can get a $3$, $3 \cdot 3$ ways both people can get a $4$, etc.
Thus, the number of successes is (by symmetry)
$2(1+4+9+16+25)+36 = 146$.
The answer you are looking for is $\frac{146}{1296} = \frac{73}{648}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve $\int_{0}^{\frac{π}{2}} \frac{dx}{4\sin^2(x) +5\cos^2(x)} $ $?$ I apply the substitutions:
$$t=\tan(x), \sin(x)=\frac{t}{\sqrt{1+t^2}}, \cos(x)=\frac{1}{\sqrt{1+t^2}}\ \&\ dx=\frac{dt}{1+t^2}$$
(using $t=\tan(x)$ you can draw a right angled triangle to find the other substitutions)
So we get:
$$\int_{0}^{\frac{π}{2}} \frac{1}{\frac{4t^2}{1+t^2}+\frac{5}{1+t^2}}\frac{dt}{1+t^2}=\frac{1}{4}\int_{0}^{\frac{π}{2}} \frac{1}{t^2+\Big(\frac{\sqrt{5}}{2}\Big)^2} dt$$
This is in a standard integral form, thus:
$$\frac{1}{4}\cdot\frac{2}{\sqrt{5}}\tan^{-1}\Bigg(\frac{2t}{\sqrt{5}}\Bigg)=\frac{1}{2\sqrt{5}}\tan^{-1}\Bigg(\frac{2\tan(x)}{\sqrt{5}}\Bigg)$$
this is from $0$ to $π/2$.
But I can't substitute for $π/2$, because $\tan(x)$ is undefined for $π/2$. If I input this integral into Mathcad I get $\frac{π\sqrt{5}}{20}$. How can I get the right answer out of this? Thanks in advance!
| Hint: $4\sin^2x+5\cos^2x=4+\cos^2x$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 4
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Solving a Diaphontine Equation Find all integral solutions to the equation
$(x^2 + 1)(y^2 + 1) + 2(x − y)(1 − xy) = 4(1 + xy)$.
I've gotten this far:
$x^2y^2+x^2+y^2+1+2x-2y-2x^2y-2xy^2=4+4xy$
$(x-y)^2+x^2y^2-2x^2y-2xy^2+2x-2y-2xy=3$
What do I do from here?
| Let $A=(x^2+1)(y^2+1)$, $B=2(y-x)(1-xy)$, $C=4(1+xy)$. If $|x|,|y| \geq 4$, then we have $A > 4C$ and if $|x|,|y| \geq 6$, then we have $3A>4B$.
Thus for $|x|,|y| \geq 6$, there's no solution. Now the remaining cases can be solved by calculation.
Now I prove just one of the claims above, i.e. $A>4C$ for $|x|,|y| \geq 4$: Let $a=|x|,b=|y|$, then $4C \leq 16|C| \leq 16(1+ab) < (a^2+1)(b^2+1)$ because $a^2b^2 \geq 16ab$ and $a^2 \geq 16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462789",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ A question on submultiple angles states...
Prove that:$$\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$$
My efforts
I tried using the formula
$$a^3+b^3=(a+b)^3-3ab(a+b)$$
and
$$\cos^2{\frac{\theta}{2}=\frac{\cos\theta + 1}{2}}$$
Then I tried simplifying it:
$$\require{cancel} \begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\
& = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}\\
& = 1 - 3(1 - \cos^2{\frac{\theta}{2}})\cos^2{\frac{\theta}{2}}\\
& = 1 - \frac{3}{2}(\cos\theta+1) + \frac{3}{4}(\cos\theta+1)^2\\
& = \frac{4\cancel{-6\cos\theta}-2+3\cos^2+3+\cancel{6\cos\theta}}{4}\\
& = \frac{1}{4}(5+3\cos^2\theta)\end{align} $$
I suspect I must have messed up with some sign somewhere. The trouble is, I can't seem to find where. Please help me in this regard.
Update: I am not accepting an answer because all the answers are equally good. It would be an injustice to the other answers.
Note to the editors: I also suspect that my post is a little short on appropriate tags. If required, please do the needful.
| Hint: $$\sin^2\theta/2\cos^2\theta/2\\=\frac{1}{4}(1-\cos\theta)(1+\cos\theta)=\frac{1}{4}(1-\cos^2\theta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $2^{n-1}$ divides $\binom{n}{1} + \binom{n}{3}5 + \binom{n}{5}25 + \binom{n}{7}125 + \cdots$ for $n \geqslant 1$. Prove that $2^{n-1}$ divides $\binom{n}{1} + \binom{n}{3}5 + \binom{n}{5}25 + \binom{n}{7}125 + \cdots$ for $n \geqslant 1$.
Assume $\binom{n}{k} = 0$ if $k>n$.
Does anyone know an elementary proof without using the Binet formula?
| Note that $\binom{n-2}{j-1}+2\binom{n-2}{j}+\binom{n-2}{j+1}=\binom{n}{j+1}$. Define
$$
g(n)=\sum_j \binom{n}{2j+1}\cdot5^j.
$$
Then
$$
\begin{aligned}
4g(n-2)+2g(n-1)&=4\sum_j\binom{n-2}{2j+1}\cdot5^j+2\sum_j\binom{n-1}{2j+1}\cdot5^j\\
&=(5-1)\sum_j\binom{n-2}{2j+1}\cdot5^j+2\sum_j\left[\binom{n-2}{2j}+\binom{n-2}{2j+1}\right]\cdot5^j\\
&=\sum_j\binom{n-2}{2j+1}\cdot5^{j+1}+2\sum_j\binom{n-2}{2j}\cdot5^j+\sum_j\binom{n-2}{2j+1}\cdot5^j\\
&=\sum_j\left[\binom{n-2}{2j-1}+2\binom{n-2}{2j}+\binom{n-2}{2j+1}\right]\cdot5^j\\
&=\sum_j\binom{n}{2j+1}\cdot5^j\\
&=g(n).
\end{aligned}
$$
So $g(n)=2g(n-1)+4g(n-2)=2\left[g(n-2)+2g(n-2)\right]$ with the initial conditions $g(1)=1$, $g(2)=2$. The result follows by induction.
It would be satisfying to interpret this calculation combinatorially. I'll think about this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1469026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
A golden ratio series from a comic book The eighth installment of the Filipino comic series Kikomachine Komix features a peculiar series for the golden ratio in its cover:
That is,
$$\phi=\frac{13}{8}+\sum_{n=0}^\infty \frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}}$$
How might this be proven?
| First Approach: Catalan Numbers
Some straightforward manipulations of the sum brings it to the form
$$
S=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(2(n+1))!}{(n+2)!(n+1)! 4^{2n+3}}
$$
Using the definition of the Catalan numbers this nicely rewrites as
$$
S=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}C_{n+1}}{ 4^{2n+3}}=\frac{1}{8}\sum_{n=0}^{\infty}C_{n+1}\left(\frac{-1}{16}\right)^{n+1}
$$
Using the generating function of the Catalan numbers,
$$
\sum_{n=0}^{\infty}C_{n+1} x^{n+1}=\frac{1-\sqrt{1-4x}}{2x}-1
$$
setting $x=\frac{-1}{16}$ we may conclude that
$$
S=\frac{1}{8} \left(4 \sqrt{5}-9\right)
$$
Furthermore
$$
S+\frac{13}{8}=\frac{\sqrt{5}}{2}-\frac{9}{8}+\frac{13}{8}=\frac{1+\sqrt{5}}{2}=\phi
$$
------------------------------------------------------------------------------------------------------------------------------------------------
Second Approach: Contour Integration
OK, let's see what contour integration can do.
We can show by an straightforward application of Cauchy's integral formula, that
$$
\binom{n}{k}=\frac{1}{2 \pi i}\oint_C\frac{(1+z)^{n}}{z^{k+1}}dz \quad (1)
$$
where $C$ is a circle traversed counter-clockwise (with radius < $1/4$ in our case) .
Furhermore we may observe that ($C_n$ are again Catalan numbers)
$$
C_n=\frac{1}{n+1}\binom{2n}{n}=\binom{2n}{n}-\binom{2n}{n+1} \quad (2)
$$
Now let's apply (1) and (2) to the function $q(x)=\sum_{n=0}^{\infty}C_{n+1}x^{n+1}$
we get
$$
q(x)=\frac{1}{2 \pi i}\oint_C\sum_{n=0}^{\infty}x^{n+1}\left(\frac{(1+z)^{2n+2}}{z^{n+2}}-\frac{(1+z)^{2n+2}}{z^{n+3}}\right)dz
$$
where the exchange of summation and integration is justified as long as the poles lie not on the contour of integration (which will be the case)
the sums are now usual geometric series and yield
$$
q(x)=\frac{1}{2 \pi i}\oint_C\left(-\frac{1}{z}\frac{x (z+1)^2}{x (z+1)^2-z}+\frac{1}{z^2}\frac{x (z+1)^2}{x (z+1)^2-z}\right)dz
$$
It is easy to show that the zero's of the denominator are given by
$z_{\pm}=\frac{-2 x\pm\sqrt{1-4 x}+1}{2 x}$ and only $z_-$ lies in the unit circle if $x<1/4$ which is important because we want to set $x=\frac{-1}{16}$ in the end. Applying the residue theorem, we get
$$
q(x)=\mathrm{res}(z=0)+\mathrm{res}(z=z_{-})=-\frac{x-1}{x}+\frac{\sqrt{1-4 x}+1}{2 x}=\\
\frac{2 x+\sqrt{1-4 x}-1}{2 x}
$$
Recalling the sum you are looking for (see my first answer)is
$$
S=\frac{1}{8}q\left(\frac{-1}{16}\right)
$$
we obtain
$$S+\frac{13}{8}=\phi$$
as desired.
This also constitutes a proof of the generating functions for the Catalan numbers (it is $q(x)$)!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 5,
"answer_id": 1
} |
How to find the remainder of $\frac{2^{99}}{9}$? The question is solved in the book I read in a very odd way as,
$$\frac{2^{99}}{9} = \dfrac{{(2^{3})}^{33}}{2^3-(-1)}$$
Hence by remainder theorem the remainder is $-1$. In questions when remainder is negative than the number is subtracted from numbers like $2^3$ to get a positive number so the answer is $8-{-1}=9$.
I don't know which remainder theorem the book is talking of. I converted the question in mod notation as,
$$2^{99} \bmod 9 = (2^3)^{33} \bmod (2^3 - (-1))$$
But I don't think there is some formula for $a \bmod (c-d)$. Please tell me how to solve these type of questions.
| The theorem referred to is probably the polynomial remainder theorem. That is that the remainder of $p(x)/(x-a)$ is $p(a)$. So with $p(x) = x^{33}$ and $a=-1$ we have that the remainder of the division will be $p(-1) = (-1)^{33}$. That is:
$$p(x) = x^{33} = (x-(-1))q(x) + (-1)^{33} = (x-(-1))q(x) - 1$$
Inserting $x = 2^3$ into this results in
$$p(2^3) = (2^3-(-1))q(2^3) - 1 = 9q(2^3)-1$$
From this we can see that $2^{99} = 9q(2^3) - 1 = 9q(2^3)-9+8 = 9(q(2^3)-1) + 8$ and since $q$ is a polynomial with integer terms we have that $Q = q(2^3)-1$ is an integer so $2^{99} = 9Q + 8$
For proof of the polynomial remainder theorem let $p(x)$ be a polynomial and use euclid division with $x-a$ then we will get a result $q(x)$ and a remainder $r(x)$ of degree one less than $x-a$, that is r(x) is a constant expression $C$. Now the result of the division will be
$$p(x) = q(x)(x-a) + r(x) = q(x)(x-a) + C$$.
Now substitute $x$ with $a$ and we get
$$p(a) = q(a)(a-a) + C = 0q(a) + C = C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Question related to tan in a ratio in a triangle If in a triangle $\tan A:\tan B:\tan C = 1:2:3$ then, what are the ratio of the sides $a,b,c $?
| $$
sinA = \frac{a}{2R}
$$
where R is the radius of the circumcircle. It'll be cancelled out later.
$$
cosA = \frac{b^2+c^2-a^2}{2bc}
$$
So,
$$
tanA = \frac{abc}{R(b^2+c^2-a^2)}
$$
And similarly,
$$
tanB = \frac{abc}{R(a^2+c^2-b^2)},tanC = \frac{abc}{R(a^2+b^2-c^2}
$$
Canceling out $\frac{abc}{R}$ from the ratio, we get,
$$
\frac{1}{b^2+c^2-a^2}:\frac{1}{a^2+c^2-b^2}:\frac{1}{a^2+b^2-c^2} = 1:2:3
$$
Equating each of these terms to k, 2k and 3k respectively (so that the ratio 1:2:3 is maintained), we get the following equations
$$
b^2+c^2-a^2 = \frac{1}{k}
$$
$$
a^2+c^2-b^2 = \frac{1}{2k}
$$
$$
a^2+b^2-c^2 = \frac{1}{3k}
$$
On solving these, we get $a=\frac{\surd{5}}{2\surd{3}k}$, $ b=\frac{\surd{2}}{\surd{3}k} $ and $c=\frac{\surd{3}}{2k}$
Therefore $a:b:c = \surd{5}:2\surd{2}:3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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How to solve $x^2 + x = 0 (\mod 10^5)$? I've tried to solve it in the following way:
First let's solve $x^2+x\equiv 0\ (\mod10)$. Solutions are $-1, 0, 4, 5$.
So $-1$ and $0$ are obviously solutions of $x^2 + x \equiv 0\ (\mod 10^5)$.
Now let's solve this for $4$ and $5$:
Let $x = 10k + 4$, then put this $x$ into $x^2 + x \equiv 0\ (\mod 100)$. So after a few operation we get $k=-2\ (\mod 10)$. Now let $l = 10k + 2$ and $x = 10\cdot (10k + 2) + 4$. Now we can solve our equation by modulo $1000$. Then we keep doing similar steps until modulo $10^5$ solution is found.
The problem is we need to operate really big numbers. And we need to do it twice (for both roots $4$ and $5$). Is there any less complicated solution?
| $x^2 + x = x (x+1)$ At most one of $x$ or $x+1$ is divisible by 2, and likewise by 5. So either one of them is divisible by $10^5$ or one is divisible by $5^5$ and the other by $2^5$.
To find solutions, consider odd multiples of $5^5 = 3125$, add or subtract 1, and see how what the largest power of 2 is that divides the result.
A bit of checking shows that $3\times 5^5 = 9375$, and $9376 = 32\times293$, so $x = 9375$ is a third solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1476851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $w$ be nth root of unity , then $1+2w+3w^2+\dots+nw^{n-1}$ is equal to? I tried it by letting expression $1+2w+3w^2+\dots+nw^{n-1}= x$ and then multiplying $w$ both sides . I subtracted equation 1 from 2 but it does not seems to help me because i have just started learning this topic and i am weak in properties . Can anyone help me out here ?
| Recall that $$\frac{z^n-1}{z-1} = 1+z+\cdots+z^{n-1}, \quad z \ne 1$$ so upon multiplying both sides by $z$ and differentiating, we obtain $$\frac{d}{dz}\left[\frac{z(z^n-1)}{z-1}\right] = 1+2z+3z^2+\cdots+nz^{n-1}.$$ After computing the derivative, evaluate the LHS for a primitive root of unity $z = e^{2\pi i/n}$.
Alternatively, by Abel summation: let $$f(z) = 1+2z+3z^2+\cdots+nz^{n-1},$$ so that $$\begin{align*} zf(z) - f(z) &= (z+2z^2+3z^3+\cdots+nz^n) - (1+2z+3z^2+\cdots+nz^{n-1}) \\ &= nz^n - (1+z+z^2+\cdots+z^{n-1}) \\ &= nz^n - \frac{z^n-1}{z-1}. \end{align*}$$ It immediately follows that $$f(z) = \frac{nz^n(z-1) - (z^n-1)}{(z-1)^2},$$ and the same substitution as in the above method easily results in $$f(w) = \frac{n(w-1)}{(w-1)^2} = \frac{n}{w-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve the equation with radicals: $x^2- x \sqrt[4]{2} (1+ \sqrt[4]{2} ) + \sqrt[4]{8} = 0$ Solve this equation :
$ x^2- x \sqrt[4]{2} (1+ \sqrt[4]{2} ) + \sqrt[4]{8} = 0 $
I've calculated $ \Delta $ and it is something like this:
$ \sqrt[4]{4} + 4 \sqrt[4]{2} + 4 - 4 \sqrt[4]{8} $
I don't know what to do next.
| Let us denote $a=\sqrt[4]2$. (Maybe this will help to notice how coefficients are related.)
We can rewrite the original equation as
\begin{align*}
x^2-xa(1+a)+a^3&=0\\
x^2-xa-xa^2+a^3&=0\\
x(x-a)-a^2(x-a)&=0\\
(x-a^2)(x-a)&=0
\end{align*}
This only has solutions $x_1=a$ and $x_2=a^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of the infinite product $(3)^{\frac{1}{3}} (9)^{\frac{1}{9}} (27)^{\frac{1}{27}}$....
I'm not sure if this is meant to be a contradiction but if a product is an infinite product it does not mean that the value if infinity? Or is the word infinite product just misleading.
I let:
$ (3)^{\frac{1}{3}}(9)^{\frac{1}{9}}(27)^{\frac{1}{27}}....=x$
If i cube each side: I get:
$(3)(9)^{\frac{1}{3}}(27)^{\frac{1}{9}}...=x^3$
If you notice you can rewrite the inside as:
$(3)(3^2)^{\frac{1}{3}}(3^3)^{\frac{1}{9}}...=x^3$
and simplifying the last part, we get:
$(3)(3^2)^{\frac{1}{3}}(3)^{\frac{1}{3}}...=x^3$
However, by substitution we have:
$(3)(3^2)^{\frac{1}{3}}x...=x^3$
$(3)(3^2)^{\frac{1}{3}}...=x^2$
$(3)^{\frac{3}{3}} (3)^{\frac{2}{3}}=x^2$
$(3)^{\frac{5}{3}}=x^2$
I got up to here but for some reason it doesnt look right.
| Do all your work in the exponent:
\begin{align}
\prod_{k=1}^\infty (3^k)^\frac{1}{3^k}
& = \prod_{k=1}^\infty 3^\frac{k}{3^k} \\
& = 3^{\sum_{k=1}^\infty \frac{k}{3^k}} \\
& = 3^{\frac{3}{4}} \\
& = \sqrt[4]{27}
\end{align}
The sum can be obtained as follows:
\begin{align}
\sum_{k=1}^\infty k \sigma^k
& = \sigma \sum_{k=1}^\infty k \sigma^{k-1} \\
& = \sigma \sum_{k=1}^\infty \frac{d}{d\sigma} \sigma^k \\
& = \sigma \frac{d}{d\sigma} \sum_{k=1}^\infty \sigma^k \\
& = \sigma \frac{d}{d\sigma} \frac{\sigma}{1-\sigma} \\
& = \frac{\sigma}{(1-\sigma)^2}
\end{align}
When $\sigma = 1/3$, the sum is $3/4$.
ETA: As Arthur points out in the comments, this requires logarithms to hold in the infinite case. In this instance, this should be fine because the logarithms in this case are all positive (so the series is absolutely convergent).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Plane through a point and the line of intersection of other two planes
Find the equation of the plane through $(-1,4,2)$ and containing the line of intersection of the planes $$4x-y+z-2=0 \\ 2x+y-2z-3=0$$
My answer comes out to be:
$$-9x-67y+104=51$$
While the answer provided on the answer sheet is:
$$4x-13y+21z=-14$$
Could you please check if the answer calculated by me is correct or the one provided on the answer sheet.
| HINT:
$$\begin{cases}
4x-y+z-2=0\\
2x+y-2z-3=0
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=-4x+y+2\\
2x+y-2z-3=0
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=-4x+y+2\\
2x+y-2(-4x+y+2)-3=0
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=-4x+y+2\\
10x-y-7=0
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=-4x+y+2\\
y=10x-7
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=-4x+(10x-7)+2\\
y=10x-7
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=6x-5\\
y=10x-7
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
complex numbers equation i'm sturggling with the two following problems:
1) Find all complex numbers $z$ satisfying
$$|(1+i)z-2i|=2$$
2) Find all complex numbers $z$ satisfying
$$\left|\frac{z-3}{z-5}\right|=\frac{\sqrt{2}}{2}$$
Here is what i found :
1) : i write $z=a+ib$. Putting this is the equation i get that equation 1) implies that
$$a^2+b^2-2(a+b)=0$$
but i'm not able to simplify the solution more.
2) setting $z=a+ib$ and computing i get that the equation implies that $|z-3|=\frac{\sqrt{2}}{2}|z-5|$, hence $(a-3)^2+b^2=\frac{1}{2} (a-5)^2+b^2$. The problem is that it seems to lead to the fact that $a$ satisfies an equation of the second degree whch only admits non-real solutions... What did i do wrong?
| For the first one, note that
$$a^2+b^2-2(a+b)=0\iff (a-1)^2+(b-1)^2=2.$$
Try the second one in the similar way as above, but note that we have
$$(a-3)^2+b^2=\frac 12(a-5)^2+\color{red}{\frac 12}b^2.$$
You'll get
$(a-1)^2+b^2=8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The limit of $\frac{x^2 + y^2 - x^{3}y^{3}}{ x^2 + y^2}$ as $(x,y)\to(0,0)$ I have formula: $$ f(x,y) = \frac{x^2 + y^2 - x^{3}y^{3}}{ x^2 + y^2}$$
and I have to find whether $ \lim_{(x,y) \rightarrow (0,0) } f(x,y)$ exists .
My attempt was:
(1) set $x=0$; then $f(0,y) = 1$, set $y=0$, then $f(x,0) =1$.
I concluded that this function probably has $L = 1$.
By applying the definition of limit in two variables,
if $0 < \sqrt{x^2+y^2 }< \delta $, then there exists $|(-xy)^3 + x^2 + y^2) / 2 - 1 | < \beta$.
I simplified all the way up to
$(|(xy)^3|/(x^2+y^2) ) < \beta $, then I hit the brick wall.
It appears that $(xy)^3$ is unbounded, and I can't simplify this for the life of me. What am I doing wrong?
| Use polar coordinates: $x=r\cos(\theta), y= r\sin(\theta), 0\leq \theta < 2\pi$.
Then we get, for $(x,y)\neq(0,0)$, that $f(x,y)=\frac{x^2+y^2-(xy)^3}{x^2+y^2}=1-\frac{(xy)^3}{x^2+y^2}=1-\frac{(r^2\cos(\theta)\sin(\theta))^3}{r^2}=1-r^4(\cos(\theta)\sin(\theta))^3\rightarrow 1$ when $r\rightarrow 0, \theta$ varies freely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How large is the sum of the quadratic and the geometric mean? I conjecture the following inequality:
$$\frac{\sqrt{\frac{\sum_{i=1}^n x_i^2}{n}}+\sqrt[n]{\prod_{i=1}^n x_i}}{2} \leq \frac{\sum_{i=1}^n x_i}{n}$$ for all real $x_1, x_2, \cdots, x_n > 0$
I am able to prove it when $n=2$.
Apply the power mean inequality between the $\frac12$power-mean and the arithmetic mean to $\frac{k^2+1}{2}$ and $k$ to obtain the following.
Multiply by $x_1$ and let $x_2=kx_1$ to obtain the result.
$$\left(\frac{\sqrt{\frac{k^2+1}{2}}+\sqrt{k}}{2}\right)^2 \leq \frac{\frac{k^2+1}{2}+k}{2}$$
$$\left(\sqrt{\frac{k^2+1}{2}}+\sqrt{k}\right)^2 \leq k^2+1+2k$$
$$\sqrt{\frac{k^2+1}{2}}+\sqrt{k} \leq k+1$$
$$\sqrt{\frac{k^2x_1^2+x_1^2}{2}}+\sqrt{kx_1^2} \leq kx_1+x_1$$
$$\sqrt{\frac{x_2^2+x_1^2}{2}}+\sqrt{x_1x_2} \leq x_1+x_2$$
$$\frac{\sqrt{\frac{x_2^2+x_1^2}{2}}+\sqrt{x_1x_2}}{2} \leq \frac{x_1+x_2}{2}$$
I haven't made much progress with other $n$. I tried using the power mean inequality $\sqrt{\frac{\sum_{i=1}^n x_i^2}{n}} \leq \sqrt[n]{\frac{\sum_{i=1}^n x_i^n}{n}}$ for $n \geq 2$ to get the type of the radical the same on both sides, but that wouldn't work.
The conjecture is my own. Can someone prove or disprove it for general $n$?
| Your inequality is false for $n\ge 3$ : A counterexample is
$$(x_1,x_2,\cdots,x_{n-1},x_n)=(1,1,\cdots,1,2)$$
For $(x_1,x_2,\cdots,x_{n-1},x_n)=(1,1,\cdots,1,2)$, we have
$$\frac{\sqrt{\frac{n-1+2^2}{n}}+2^{1/n}}{2}\le \frac{n-1+2}{n}.$$
This is equivalent to
$$2+2x-\sqrt{1+3x}-2^x\ge 0$$
where $0\lt x=\frac 1n\le 1$.
Let $f(x)=2+2x-\sqrt{1+3x}-2^x.$ Then,
$$f'(x)=2-\frac{3}{2\sqrt{1+3x}}-2^x\ln 2.$$
For $0\lt x\le \frac{1}{10}$,
$$f'(x)=2-\frac{3}{2\sqrt{1+3x}}-2^x\ln 2\lt 2-\frac{3}{2\sqrt{\frac{3}{10}+1}}-\ln 2\lt 0.$$
With $f(0)=0$, we know that $f(x)\lt 0$ at least for $0\lt x\le \frac{1}{10}$.
Hence, we know that your inequality is false for $n\ge 10$. Checking each of $n=3,4,\cdots,9$ gives that your inequality is false for $n\ge 3$.
| {
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"url": "https://math.stackexchange.com/questions/1487893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric Substitution on $\frac{1}{x\sqrt{(x^2 +25)}}$ How can I find
$$\int\frac{1}{x\sqrt{(x^2 +25)}} \space dx$$
using trigonometric substitution?
| $$\int \frac{1}{x\sqrt{(x^2+25)}} \space \text{d}x =$$
Substitute $x=5\tan(u)$ and $\text{d}x=5\sec^2(u)\text{d}u$. Then $\sqrt{x^2+25}=\sqrt{25\tan^2(u)+25}=5\sec(u)$ and $u=\tan^{-1}\left(\frac{x}{5}\right)$:
$$5 \int \frac{\csc(u)}{25} \space \text{d}u =$$
$$\frac{5}{25} \int \csc(u) \space \text{d}u =$$
$$\frac{1}{5} \int \csc(u) \space \text{d}u =$$
$$\frac{1}{5} \int -\frac{-\cot(u)\csc(u)-\csc^2(u)}{\cot(u)+\csc(u)} \space \text{d}u =$$
Substitute $s=\cot(u)+\csc(u)$ and $\text{d}s=(-\csc^2(u)-\cot(u)\csc(u))\text{d}u$:
$$\frac{1}{5} \int -\frac{1}{s} \space \text{d}s =$$
$$-\frac{1}{5} \int \frac{1}{s} \space \text{d}s =$$
$$-\frac{1}{5} \cdot \ln\left(s\right) +C =$$
$$-\frac{\ln\left(\cot(u)+\csc(u)\right)}{5}+C =$$
$$-\frac{\ln\left(\cot\left(\tan^{-1}\left(\frac{x}{5}\right)\right)+\csc\left(\tan^{-1}\left(\frac{x}{5}\right)\right)\right)}{5}+C =$$
$$-\frac{\ln\left(\frac{5+\sqrt{x^2+25}}{x}\right)}{5}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1488324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $x$ and $y$ in a simple matrix I've just started learning matrices, but I'm not sure how to go about solving this:
$\begin{pmatrix} 1 & 2 & 0 \\ 2 & 0 & 1\\1&0&2\end{pmatrix}$
$\begin{pmatrix} 0\\ x\\ y\end{pmatrix}$ $=$ $\begin{pmatrix} 6\\2\\4\end{pmatrix}$
I'm told to use the gaussian method, but I'm getting confused with the $\begin{pmatrix} 0\\ x\\ y\end{pmatrix}$ column, every example video online that shows how to solve matrices, does not have such a column, I'm not sure what to do, any help would be appreciated.
| Just proceed with Row Reduction as normal. If it helps, replace the zero with $z$. If the system is consistent you will conclude that $z=0$.
Can you reduce:
$$\begin{pmatrix} 1 & 2 & 0 \\ 2 & 0 & 1\\1&0&2\end{pmatrix}\begin{pmatrix} z\\ x\\ y\end{pmatrix} = \begin{pmatrix} 6\\2\\4\end{pmatrix}$$
To
$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\0 & 0 & 1\end{pmatrix}\begin{pmatrix} z\\ x\\ y\end{pmatrix} = \begin{pmatrix} 0\\3\\2\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
General solution of differntial equation $(x^2)(y)dx = ( (x^3) + (y^3))dy$ I tried this problem by many ways , i took $dy$ to other side and tried to form bernaulli equation but it is not helping and not giving me proper answer .
| $$x^2ydx = ( x^3 + y^3)dy$$
$$\frac{dy}{dx}=\frac{x^2y}{x^3+y^3}$$
$$y'=\frac{y/x}{1+y^3/x^3}$$
now assume $y=ux$ or $u=\frac{y}{x}$
$$y'=\frac{u}{1+u^3}$$
this is first order homogeneous equation
$$y'=u+x\frac{du}{dx}$$
$$u+x\frac{du}{dx}=\frac{u}{1+u^3}$$
$$x\frac{du}{dx}=\frac{u}{1+u^3}-u$$
$$x\frac{du}{dx}=\frac{u-u-u^4}{1+u^3}$$
$$x\frac{du}{dx}=-\frac{u^4}{1+u^3}$$
$$-\frac{1+u^3}{u^4}du=\frac{dx}{x}$$
$$-(1/u^4+1/u)du=\frac{dx}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve ODEs by converting it to Clairaut's form through suitable substitutions. Sometimes it's really confusing to find out suitable substitutions.like this one:
$(xy'-y)(x+yy')=2y'$
Which substitution should I try to put this equation in Clairaut's form?
| Let $u=x^2+y^2$ ,
Then $\dfrac{du}{dx}=2x+2y\dfrac{dy}{dx}$
$\therefore\left(\dfrac{x}{2y}\left(\dfrac{du}{dx}-2x\right)-y\right)\dfrac{1}{2}\dfrac{du}{dx}=\dfrac{1}{y}\left(\dfrac{du}{dx}-2x\right)$
$\left(x\left(\dfrac{du}{dx}-2x\right)-2y^2\right)\dfrac{du}{dx}=4\left(\dfrac{du}{dx}-2x\right)$
$x\dfrac{du}{dx}-2x^2-2y^2=4\left(1-\dfrac{2x}{\dfrac{du}{dx}}\right)$
$x\dfrac{du}{dx}-2u=4-\dfrac{8x}{\dfrac{du}{dx}}$
$2u+4=x\dfrac{du}{dx}+\dfrac{8x}{\dfrac{du}{dx}}$
$u+2=\dfrac{x}{2}\dfrac{du}{dx}+\dfrac{4x}{\dfrac{du}{dx}}$
Let $v=x^2$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=2x\dfrac{du}{dv}$
$\therefore u+2=x^2\dfrac{du}{dv}+\dfrac{2}{\dfrac{du}{dv}}$
$u+2=v\dfrac{du}{dv}+\dfrac{2}{\dfrac{du}{dv}}$
Let $s=u+2$ ,
Then $\dfrac{ds}{dv}=\dfrac{du}{dv}$
$\therefore s=v\dfrac{ds}{dv}+\dfrac{2}{\dfrac{ds}{dv}}$
$s\dfrac{dv}{ds}=v+\dfrac{2}{\left(\dfrac{ds}{dv}\right)^2}$
$v=s\dfrac{dv}{ds}-2\left(\dfrac{dv}{ds}\right)^2$
Which reduces to Clairaut's ODE.
$\dfrac{dv}{ds}=s\dfrac{d^2v}{ds^2}+\dfrac{dv}{ds}-4\dfrac{dv}{ds}\dfrac{d^2v}{ds^2}$
$\dfrac{d^2v}{ds^2}\left(4\dfrac{dv}{ds}-s\right)=0$
$\therefore\begin{cases}\dfrac{d^2v}{ds^2}=0\\4\dfrac{dv}{ds}-s=0\end{cases}$
$\begin{cases}v=as+b\\v=\dfrac{s^2}{8}+c\end{cases}$
$\therefore\begin{cases}as+b=as-2a^2\\\dfrac{s^2}{8}+c=\dfrac{s^2}{4}-\dfrac{s^2}{8}\end{cases}$
$\begin{cases}b=-2a^2\\c=0\end{cases}$
$\therefore\begin{cases}v=as-2a^2\\v=\dfrac{s^2}{8}\end{cases}$
$\begin{cases}x^2=au+2a-2a^2\\x^2=\dfrac{(u+2)^2}{8}\end{cases}$
$\begin{cases}x^2=ax^2+ay^2+2a-2a^2\\x^2=\dfrac{(x^2+y^2+2)^2}{8}\end{cases}$
$\begin{cases}(a-1)x^2+ay^2=2a^2-2a\\x^2=\dfrac{(x^2+y^2+2)^2}{8}\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find slopes of tangent lines where $\frac{dy}{dx}$ has removable discontinuity. A colleague of mine wrote a worksheet where she asked students to find the derivative $\frac{dy}{dx}$, where the points $(x,y)$ are defined by the "Heart Curve,"
$$(x^2+y^2-1)^3-x^2y^3=0.$$
She asks students about points where the derivative is well-defined, but I wondered about the points where the derivative isn't well-defined.
A plot of the graph shows sharp points at $(0,\pm1)$, and seems to be a smooth curve at the points $(\pm1,0)$, yet the derivative is undefined at points where $x=0$ or $y=0$. At points where $x=0$, there should be VERTICAL tangent lines (I have since ran the zoom in on a point test to find this), and at points where $y=0$ there should be one distinct tangent line. It seems slopes here should be $\pm2$.
I have calculated $\frac{dy}{dx}$ as
$$\frac{dy}{dx}=\frac{2xy^3-6x(x^2+y^2-1)^2}{6y(x^2+y^2-1)^2-3x^2y^2},$$
and I feel like there should be a way to remove singularities for $x=0$ or $y=0$ in the expression but I do not know exactly how.
I thought about maybe using partial derivatives and L'Hopital's Rule, but not sure. My goal is to discuss this with students in a single-variable calculus course, so I'd like a solution that doesn't use partial differentiation.
| This solution works with implicit differentiation. The equation of the heart-curve is
$${({x^2} + {y^2} - 1)^3} - {x^2}{y^3} = 0\tag{1}$$
Using implicit differentiation, one can obtain the derivative to be
$${{dy} \over {dx}} = - {{6x{{\left( {{x^2} + {y^2} - 1} \right)}^2} - 2x{y^3}} \over {6y{{\left( {{x^2} + {y^2} - 1} \right)}^2} - 3\;{x^2}{y^2}}}\tag{2}$$
but from Eq.$(1)$ we can observe that
$${({x^2} + {y^2} - 1)^2} = {x^{{4 \over 3}}}{y^2}\tag{3}$$
Now combining $(2)$ and $(3)$ leads to
$$\eqalign{
& {{dy} \over {dx}} = - {{6x\left( {{x^{{4 \over 3}}}{y^2}} \right) - 2x{y^3}} \over {6y\left( {{x^{{4 \over 3}}}{y^2}} \right) - 3\;{x^2}{y^2}}} = - {{6{x^{{7 \over 3}}}{y^2} - 2x{y^3}} \over {6{x^{{4 \over 3}}}{y^3} - 3{x^2}{y^2}}} \cr
& \,\,\,\,\,\,\,\,\, = - {{6{x^{{7 \over 3}}} - 2xy} \over {6{x^{{4 \over 3}}}y - 3{x^2}}} = - {{6{x^{{4 \over 3}}} - 2y} \over {6{x^{{1 \over 3}}}y - 3x}} \cr}\tag{4}$$
and in summary
$${{dy} \over {dx}} = - {{6x\root 3 \of x - 2y} \over {6\root 3 \of x y - 3x}}\tag{5}$$
Finally, we can use Eq.$(5)$ to conclude that the slope at $\left( { \pm 1,0} \right)$ is $ \pm 2$ and also that the slope at $\left( {0, \pm 1} \right)$ goes to $ \pm \infty $ depending on whether we are approaching the point from the left or right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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} |
Show that sin(3x) = sin(x)(2cos(2x) + 1) I'm currently studying Trigonometric Functions and came to exercise where I am to find all possible values for x in the range of x[0, PI] where.
The equation is as follows:
sin(3x) = 0.39
I found two values possible for x but the solution in my exercise book shows 2 extra solutions.
Wolfram Alpha tells me that an alternate form of sin(3x) is sin(x)(2cos(2x)+1)
What I want to know is: how do I transform sin(3x) to be the alternate form, and how do I find all the possible solutions for x?
The measurements are in radians.
| $$\begin{align}\sin 3x &= \sin(x+2x)\\
&=\sin x\cos 2x + \sin 2x\cos x\\
&=\sin x\cos 2x +2\sin x\cos^2 x \\
&=\sin x(\cos 2x +2\cos^2x)\end{align}$$
But then remember $2\cos^2 x-1=\cos 2x$, so $2\cos^2 x=\cos 2x+1$
In general, when $n$ is an integer:
$$\frac{\sin nx}{\sin x}$$
can be written as a polynomial of degree $n-1$ in $\cos x$. This polynomial is called a Chebyshev polynomial of the second type. It turns out, you can write it instead in terms of a linear combination of $\cos(n-1)x,\cos(n-3)x,\dots$.
So $$\begin{align}\frac{\sin 2x}{\sin x} &= 2\cos x\\
\frac{\sin 3x}{\sin x} &= 2\cos 2x + 1\\
\frac{\sin 4x}{\sin x} &= 8\cos^3 x -4\cos x=2\cos 3x +2\cos x
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How is this is an identity? So we have started studying partial fractions.The book teaches two methods:
*
*By equating coefficients
*By utilizing the fact that when a rational fraction is decomposed to partial fractions it is an identity
I cannot quite understand how does the 2nd method works.If we make the denominator of one of the fractions zero , it would become undefined , so how will it be an identity then???
Thanks in Advance.
For example, take the following
$$
\frac{1}{(x+3)(x+4)}=\frac{a}{(x+3)} + \frac{b}{(x+4)}
$$
so if we put $x=-3$ then the left hand side of the equation would become undefined.
Also on the right hand side the term $\frac{a}{(x+3)}$ would also become undefined, so how would this equation be an identity?
| Partial Fractions : Steps
Step (1) Check the degrees of both numerator (n) and denominator (d) separately.
(i) If $n<d$ follow step (2)
(ii) If $n \geq d$ Add a polynomial of degree $(n-d)$, and follow (2)
Ex : $\frac{x^2}{x^3+1} \Rightarrow 2<3 \Rightarrow $ follow step (2)
Ex: $\frac{x^3+x+1}{x^3-2x-1} \Rightarrow 3=3 \Rightarrow \frac{x^3+x+1}{x^3-2x-1} =A+ $ .....follow step (2)
Ex: $\frac{3x^4+x^2+1}{3x^3-2x+1} \Rightarrow 4>3\Rightarrow \frac{3x^4+x^2+1}{3x^3-2x+1} =Ax+B+ $ ......follow step (2)
Ex: $\frac{x^5+x+1}{3x^3-2x+1} \Rightarrow 5>3\Rightarrow \frac{x^5+x+1}{3x^3-2x+1} =Ax^2+Bx+C +$ .......follow step (2)
Step (2) : Factorize the denominator and write each factor separately with a numerator of degree less than one with respect to the chosen factor.
Ex: $\frac{x^5+x+1}{3x^3-2x+1} \Rightarrow 5>3\Rightarrow \frac{x^5+x+1}{3x^3-2x+1} =\frac{x^5+x+1}{(x+1)(3x^2-3x+1)} =Ax^2+Bx+C + \frac {D}{(x+1)}+ \frac{Ex+F}{(3x^2-3x+1)}$
Step (3) : Now find $A,B,C,D,E,F$ by multiplying whole expression by $(x+1)(3x^2-3x+1)$ . You could use $x=-1$ to find constants and compare co-efficients.
Important Ex:
$1) \frac{x}{(x+1)^3} = \frac {A}{(x+1)}+\frac{B}{(x+1)^2}+\frac {C}{(x+1)^3}$
$2) \frac{x}{(x^2+1)^3} = \frac {Ax+B}{(x^2+1)}+\frac{Cx+D}{(x^2+1)^2}+\frac {Ex+F}{(x^2+1)^3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $\int (1-x^2)^{k-1} \ dx= x \cdot {_2F_1}(\frac 12, 1-k, \frac 32, x^2)$? I am reading a paper and there is written that
$$\int (1-x^2)^{k-1} \ dx = x \cdot {_2F_1}\left(\frac 12, 1-k, \frac 32, x^2\right)$$
where $ {_2F_1}$ is the hypergeometric function
(this should hold for $x \in [-1, 1]$)
How is this result proven?
I tried computing the taylor expansion of $(1-x^2)^{k-1}$ is neighborhood of $0$, then integrate term by term and find $a,b,c,z$ such that our geometric function has the same form. This may be doable but I find it hard to compute the $\displaystyle \frac{d^n}{x^n}\left(1-x^2\right)^{k-1}$
Are there easier ways? Thank you! :-)
| With $X=x^2$ :
$$\int (1-x^2)^{k-1}dx=\frac{1}{2}\int (1-X)^{k-1}X^{-1/2}dX=\frac{1}{2}B_{X}(\frac{1}{2},k)$$
$B$ is the Incomplete Beta function.
The Incomplete Beta function is a particular hypergeometric function, with relationship :
$$ _2F_1(a,b;b+1;X)=bX^{-b}B_X(b,1-a)$$
In this case $a=1-k$ and $b=\frac{1}{2}$:
$$ _2F_1(1-k,\frac{1}{2};\frac{1}{2}+1;X)=\frac{1}{2}X^{-\frac{1}{2}}B_{X}(\frac{1}{2},k)$$
$$ _2F_1(1-k,\frac{1}{2};\frac{3}{2};x^2)=\frac{1}{2}x^{-1}B_{X}(\frac{1}{2},k)$$
$$\frac{1}{2}B_{X}(\frac{1}{2},k)=x \: _2F_1(1-k,\frac{1}{2};\frac{3}{2};x^2)=x \: _2F_1(\frac{1}{2},1-k;\frac{3}{2};x^2)=\int (1-x^2)^{k-1}dx$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3
Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3
This means that:
$$\begin{align}
2\binom{n}{k} =\binom{n}{k+1}\\
3\binom{n}{k} =\binom{n}{k+2}\\
\end{align}$$
I simplefied these equations:
$$\begin{align}
2\binom{n}{k} =\binom{n}{k+1}\\
\end{align}$$
Is the same as:
$$\begin{align}
2k+2 = n-k\\
3k+2 = n
\end{align}$$
And:
$$\begin{align}
3\binom{n}{k} =\binom{n}{k+2}\\
\end{align}$$
Is the same as:
$$\begin{align}
3(k+1)(k+2) = (n-k-1)(n-k)
\end{align}$$
I do not know how I could go further so that I would end up with the value of n and k. This is because I can not simplify 3(k+1)(k+2) = (n-k-1)(n-k) enough.
How can I proceed so that I can get the value of n and k for:
$$\begin{align}
3(k+1)(k+2) = (n-k-1)(n-k) \\
3k+2 = n
\end{align}$$
| Alternatively, coming from this question, dividing the consecutive binomial coefficients yields
$$\frac21=\frac{\tbinom{n}{k+1}}{\tbinom nk}=\frac{n-k}{k+1}\qquad\text{ and }\qquad
\frac32=\frac{\tbinom{n}{k+2}}{\tbinom{n}{k+1}}=\frac{n-k-1}{k+2},$$
and solving both for $n$ yields $n=3k+2$ and $n=\tfrac52k+4$, respectively, so $k=4$ and $n=14$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding number of integers divisible by 2, 3 or 4 using inclusion-exclusion principle. I want to find number of integers from 1 to 19 (both included) which are divisible by 2 or 3 or 4. Lets denote it by N.
So counting and enumerating them gives N = 12. Integers are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18.
I thought of applying inclusion-exclusion principle for finding N. Here is how I proceeded:
Lets denote the followings:
N1 = (Number of integers which are divisible by 2) + (Number of integers which are divisible by 3) + (Number of integers which are divisible by 4)
N2 = (Number of integers which are divisible by 2*3 = 6) + (Number of integers which are divisible by 3*4 = 12) + (Number of integers which are divisible by 2*4 = 8)
N3 = (Number of integers which are divisible by 2*3*4 = 24)
Then using inclusion-exclusion principle we have: N = N1 - N2 + N3
Finding N1, N2 and N3 (denoting Greatest Integer Function using floor function):
N1 = $\lfloor \frac{19}{2} \rfloor + \lfloor \frac{19}{3} \rfloor + \lfloor \frac{19}{4} \rfloor = 9 + 6 + 4 = 19$
N2 = $\lfloor \frac{19}{6} \rfloor + \lfloor \frac{19}{12} \rfloor + \lfloor \frac{19}{8} \rfloor = 3 + 1 + 2 = 6$
N3 = $\lfloor \frac{19}{24} \rfloor$ = 0
Therefore, N = N1 - N2 + N3 = 19 - 6 + 0 = 13.
From here, I am getting N = 13 which is wrong as I counted them manually above (N = 12).
Can anyone point out the mistake I am making here and suggest the correct way of doing this using inclusion-exclusion principle?
| When performing IE with non-coprime elements, you need to use the LCM function.
So $N_1=19$, $N_2=\lfloor 19/6 \rfloor+\lfloor 19/4 \rfloor+\lfloor 19/12 \rfloor=3+4+1=8$ and $N_3=\lfloor 19/12 \rfloor=1$ which gives you your $12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Express a quadratic form in three variable in the format $x^tAx$ using a substitution $x=Py$ I was asked to determine if a quadratic form is positive definite. To do so I must convert in the format $x^tAx$ using a substitution x=Py. So that "it can be written in diagonal form".
$$Q(x,y,z) = 3x^2 + 8xz+2y^2+z^2$$
My idea is:
$$(x,y,z) \begin{pmatrix} 3x+4z \\ 2y \\ 4x+z^2 \end{pmatrix} $$
$$\begin{pmatrix} 3&0&4 \\0&2&0 \\ 4&0&1 \end{pmatrix} $$
Is this correct? And then it would be positive definite if all eigen values of this are equal to or larger than 0?
| Ummm, $$ (4x+z)^2 = 16 x^2 + 8 x z + z^2, $$ so
$$ -13 x^2 + (4x+z)^2 = 3 x^2 + 8 x z + z^2, $$
$$ -13 x^2 + (4x+z)^2 + 2 y^2 = 3 x^2 + 8 x z + z^2 + 2 y^2 = Q(x,y,z). $$
Note $Q(1,0,-1) = -4$
If
$$
D =
\left(
\begin{array}{ccc}
-13 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2
\end{array}
\right)
$$
and
$$
R =
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
4 & 0 & 1 \\
0 & 1 & 0
\end{array}
\right),
$$
what is
$$ R^T D R? $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$ Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$
For $n=1$ inequality holds.
$(*)$For $n=k$
$2!\cdot\cdot\cdot (2k)!\ge ((k+1)!)^k$
Multiplying LHS and RHS with $(2k+2)!$ gives
$$2!\cdot\cdot\cdot (2k)!(2k+2)!\ge ((k+1)!)^k(2k+2)!$$
Assume (by contradiction)$$2!\cdot\cdot\cdot (2k)!(2k+2)!< ((k+1)!)^k(2k+2)!$$
$$2!\cdot\cdot\cdot (2k)!(2k+2)!-((k+1)!)^k(2k+2)!<0$$
$$(2k+2)!(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)<0$$
$(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)\ge 0$ by $(*)$, thus inequality holds $\forall n\in\mathbb{N}$
Is this proof correct?
| Define $S$ the wanted number. Then
$$
S=\prod_{i=1}^n (2i)!=\prod_{i=1}^n (2(n+1-i))!.
$$
Therefore, based on $a!b!\ge \max(a,b)!\ge \frac{a+b}{2}!$ whenever $a,b$ are integers with same parity, we conclude that
$$
S=\sqrt{\prod_{i=1}^n(2(n+1-i))!(2i)!}\ge \sqrt{\prod_{i=1}^n(n+1)!^2}=(n+1)!^n.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $s^4-18s^2-8s$
Let $a,b,c$ be the roots of $x^3-9x^2+11x-1=0$, and let $s=\sqrt{a}+\sqrt{b}+\sqrt{c}$. Find $s^4-18s^2-8s$.
$s^4 - 18s^2 - 8s = (s)(s + 4)(s - 2 + \sqrt{6})(s - 2 - \sqrt{6})$
$P(x) = (x - a)(x - b)(x - c)$
From Vieta's formulas:
$a + b+ c = 9$
$abc = 1$
$ab + ac + bc = 11$
Actually, if $y = \sqrt{x}$ then if $y$ is a root, the function must map $y \to x$, which means, let $x \to x^2$ to get:
$P_2(x) = x^6 - 9x^4 + 11x^2 - 1 = 0$.
Then $s = \sum \text{roots } P_2(x) = 0$.
But then $s^4 - 18s^2 - 8s = 0$? What am I doing wrong?
| $$s=\sqrt{a}+\sqrt{b}+\sqrt{c}\\\Rightarrow s^2=a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\\\Rightarrow s^2-9=2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\\\Rightarrow s^4-18s^2+81=(s^2-9)^2=4(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2=4(ab+bc+ca)+8\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})\\\Rightarrow s^4-18s^2+81=4\cdot11+8\cdot 1\cdot s\\\Rightarrow s^4-18s^2-8s=44-81=-37$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $abc$ if one is given the values of $a + b + c,$ $a^2 +b^2+c^2, \ a^3+b^3+c^3$ and $ac+bc+ab,$ If I am given the values of $a + b + c,$ $a^2 +b^2+c^2, \ a^3+b^3+c^3$ and $ac+bc+ab,$ how do I find value of $abc \ ?$
I expanded $(a+b+c)^3$ to get $a^3+b^3+c^3 + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) +6abc $ but how do I manipulate $3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)$ to obtain a fixed value from it?
Appreciate all advice, thank you.
| HINT: $$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
| {
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Number of squarefree positive integers less than $100$
An integer is called squarefree if it is not divisible by the square
of a positive integer greater than $1$. Find the number of squarefree
positive integers less than $100$.
My attempt:
I apply the inclusion-exclusion principle directly.
*
*Total number of integers = 99
*Number of integers divisible by $2^2$ = $24$
*Number of integers divisible by $3^2$ = $11$
*Number of integers divisible by $4^2$ = $6$
*Number of integers divisible by $5^2$ = $3$
*Number of integers divisible by $6^2$ = $2$
*Number of integers divisible by $7^2$ = $2$
*Number of integers divisible by $8^2$ = $1$
*Number of integers divisible by $9^2$ = $1$
*Number of integers divisible by $2^2$ and $3^2$ = $2$
*Number of integers divisible by $2^2$ and $4^2$ = $1$
Then the required solution would be $99-(24+11+6+3+2+2+1+1)+2+1=52$. But the solution is $61$. Where is my mistake?
| If $4^2$ divides $n$ then so does $2^2$, so your method double-counts. A number is nonsquarefree iff it is divisible by a square of a prime, so we can proceed as follows:
Let $A_k$ denote the numbers (here, positive integers) less than $100$ divisible by $k^2$. Since $$ |A_k| = \left\lfloor \frac{100 - 1}{k^2} \right\rfloor ,$$ we have $|A_k| = 0$ for $k \geq 10$. Then, applying the inclusion-exclusion principle to the sets $A_p$, $p < 10$ prime, gives that the number of nonsquarefree numbers less than $100$ is
\begin{multline}|A_2| + |A_3| + |A_5| + |A_7| \\- (|A_2 \cap A_3| + |A_2 \cap A_5| + |A_2 \cap A_7| + |A_3 \cap A_5| + |A_3 \cap A_7| + |A_5 \cap A_7|) + \cdots ,\end{multline} where $\cdots$ denotes terms containing cardinalities of intersections of three and four sets.
Now, $n \in A_q \cap A_{q'}$ for $q, q'$ coprime iff $n$ is divisible by both $q^2$ and $(q')^2$, and hence by coprimality, by $q^2 (q')^2 = (qq')^2$, so $A_q \cap A_{q'} = A_{qq'}$. So, for example, $A_2 \cap A_5 = A_{10}$, which by the above is empty, and the same is true for all of the intersections of two sets except $A_2 \cap A_3 = A_6$. Similarly, by induction, the triple and quadruple intersections are all empty. This leaves that the number of nonsquarefree numbers less than $100$ is
$$|A_2| + |A_3| + |A_5| + |A_7| - |A_6| .$$
| {
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Prove for the inequality between arithmetic and quadratic means $\frac{x+y}2\leq \sqrt{\frac{x^2+y^2}2}$ Show that for $x,y\in\mathbb{R}$ with $x,y\geq 0$, the arithmetic mean-quadratic mean inequality
$$\frac{x+y}{2}\leq \sqrt{\frac{x^2+y^2}{2}}$$
holds.
After my calculations I'll get:
$$-x^2+2xy-y^2$$
which can't be $\leq 0$.
| We can square both sides of your inequality, to get
$$\frac{x^2+2xy+y^2}{4} \leq \frac{x^2+y^2}{2}$$
$$\frac{x^2+2xy+y^2}{4} - \frac{x^2+y^2}{4} \leq \frac{x^2+y^2}{2}-\frac{x^2+y^2}{4} $$
$$\frac{xy}{2} \leq \frac{x^2+y^2}{4} $$
$$2xy \leq x^2+y^2 $$
$$0 \leq (x-y)^2$$
This inequality is true, since $x-y$ is a real number and all squares of real numbers are non-negative. We have shown that your inequality is equivalent to one that we know is always true. Therefore yours is always true.
| {
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Evaluating the indefinite trigonometric integral $\int \frac{dx}{(3+2\sin x)^2}$
$$\int \frac{dx}{(3+2 \sin x)^2}$$
ATTEMPT:-
Re Writing the integral as:
$I=\int \frac{2\cos x \sec x \,dx}{2(3+2 \sin x)^2}$ and using by parts:-
$\int u\,dv= uv-\int v\,du$
Here $u=\sec x \implies du=\sec x \tan x \,dx$
$\quad$ $dv=\frac{2\cos x \,dx}{2(3+2sinx)^2} \implies v=\frac{-1}{2(3+2\sin x)}$
$I=\frac{-\sec x\,dx}{2(3+2\sin x)} +\int \frac{\sec x \tan x \,dx}{2(3+2\sin x)}$
Let $I'=\int \frac{\sec x \tan x \,dx}{2(3+2\sin x)}$
$\implies I'=\int \frac{\sin x \,dx}{2\cos^2x(3+2\sin x)}$
$\implies I'=\int \frac{\sin \,dx}{2(1-\sin^2x)(3+2\sin x)}$
$\implies I'=\int \frac{-dx}{4(1+\sin x)} +\frac{3dx}{5(2\sin x+3)} + \frac{-dx}{20(\sin x-1)}$ which can be easily done by weierstrass's substitution.
But I am not able to modify my answer as given in the text.
Text Ans:-$\frac{2\cos x\,dx}{5(3+2\sin x)^2} + \frac{2}{5\sqrt{5}} \arctan\frac{(3\tan(\frac{x}{2})+2)}{\sqrt{5}} +c.$
My Ans:-$\frac{-\sec x\,dx}{2(3+2\sin x)} + \frac{6}{5\sqrt{5}} \arctan\frac{(3\tan(\frac{x}{2})+2)}{\sqrt{5}}-10\tan x+15\sec x-15.$
| Note $\sin x=\cos (\frac\pi2-x)=\frac{1-\tan^2(\frac\pi4-\frac x2)}{1+\tan^2(\frac\pi4-\frac x2)}$
$$\int \frac{dx}{3+2\sin x}=-2\int \frac{d(\tan(\frac\pi4-\frac x2))}{\tan^2(\frac\pi4-\frac x2) +5}=-\frac2{\sqrt{5}}\tan^{-1}
\frac{\tan (\frac \pi4-\frac x2) }{ \sqrt{5}}
$$
$$\left( \frac{2\cos x}{3+2\sin x}\right)’
= -\frac{3}{3+2\sin x}+\frac{5}{(3+2\sin x)^2}
$$
Integrate both sides to obtain
\begin{align}
I=\int \frac{dx}{(3+2\sin x)^2}=\frac 2{5}\frac{\cos x}{3+2\sin x}- \frac{6}{5\sqrt5}\tan^{-1}
\frac{\tan (\frac\pi4-\frac x2 )}{ \sqrt{5}}+C
\end{align}
| {
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Equilateral triangle and another triangle with same perimeter. Which has larger area? There is an equilateral triangle with sides $a$ and another triangle with sides $p,q,r$, both having the same perimeter $S$.
How can we mathematically show which of them has a larger area?
| Have you heard of Heron's formula? Especially on the form
$$
A = \sqrt{\frac S2\left(\frac S2-a\right)\left(\frac S2-b\right)\left(\frac S2-c\right)},\quad a, b, c\text{ are the sides of the triangle}
$$we come very close to a full solution. What we need to get all the way is the AM-GM inequality, which states that for any three positive numbers $k,l,m$, we have
$$
\frac{k+l+m}{3}\geq \sqrt[3]{klm}
$$
Since you asked, here is a full work-out:
For the equilateral triangle, we have $k = l = m = (S/2 - a)$, and the AM-GM inequality is actually an equality:
$$
\frac{S/2 - a + S/2 - a + S/2 - a}{3} = \sqrt[3]{(S/2 - a)^3}\\
\frac{3S/2 - 3a}{3} = \sqrt[3]{(S/2 - a)^3}\\
\frac{3S/2 - S}{3} = \sqrt[3]{(S/2 - a)^3}\\
\frac S6 = \sqrt[3]{(S/2-a)^3}\\
\left(\frac S6\right)^3 = \left(\frac{S}2 - a\right)^3
$$
(Although, that could've been worked out without using AM-GM.)
Now, for a non-equilateral triangle, we set $k = (S/2 - p), l = (S/2 - q)$ and $m = (S/2 - r)$, and the AM-GM inequality is a strict inequality:
$$
\frac{S/2 - p + S/2 - q + S/2 - r}{3} > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\
\frac{3S/2 - (p+q+r)}{3} > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\
\frac{3S/2 - S}{3} > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\
\frac S6 > \sqrt[3]{(S/2 - p)(S/2 - q)(S/2 - r)}\\
\left(\frac S6\right)^3 > \left(\frac S2 - p\right)\left(\frac S2 - q\right)\left(\frac S2 - r\right)
$$
Now insert these two different products into Heron's formula, and you can see that the equilateral triangle gives the bigger area.
| {
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What would be the value of this joint probability?
We roll a die until we get 1 or 5. What is the probability that we will make odd number of rolls?
Let, A = 1 or 5, and, B = otherwise.
Therefore, $$\Omega = \{A, BA, BBA, BBBA, ...\}$$
So, $$P(A) = P(1) + P(5) = 2 \times \frac{1}{6} = \frac{1}{3}$$
$$P(BA) = P(B \cap A) = P(B) * P(A) = \frac{4}{6} \times \frac{1}{3} = \frac{2}{3} \times \frac{1}{3} $$
Likewise, $$P(B........A) = { (\frac{2}{3} ) }^{n-1} \times \frac{1}{3}$$
Therefore, $P($Odd number of rolls$) = \frac{1}{3} + \frac{1}{3} . (\frac{2}{3})^{2} + \frac{1}{3} . (\frac{2}{3})^{4}+... ... = \frac {\frac{1}{3}}{1 - (\frac{2}{3})^2} = \frac{3}{5}$
My question is, is there any alternative way to calculate this?
| Let $p$ be the required probability. We condition on the result of the first roll.
If we get a $1$ or a $5$, we are finished. If not, we need to roll something other than $1$ or $5$ the first and second time. Given that this happened, the conditional probability we then ultimately get a first $1$ or $5$ on an odd numbered roll is $p$. Thus
$$p=\frac{1}{3}+\frac{4}{9}p.$$
Solve for $p$.
| {
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Solve $\sin(x)-\sin(\frac{\pi}{3}-x)=\sqrt{\frac{3}{2}}$ As in the title, solve $\sin(x)-\sin(\frac{\pi}{3}-x)=\sqrt{\frac{3}{2}}$. I was trying to rewrite it into a simpler form, but without any luck.
| $$\sin(x)-\sin\left(\frac{\pi}{3}-x\right)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$\sin(x)-\cos\left(\frac{\pi}{6}+x\right)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$\frac{3\sin(x)}{2}-\frac{1}{2}\sqrt{3}\cos(x)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$-\sqrt{3}\left(\frac{\cos(x)}{2}-\frac{1}{2}\sqrt{3}\sin(x)\right)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$-\sqrt{3}\left(\cos\left(\frac{\pi}{3}\right)\cos(x)-\sin\left(\frac{\pi}{3}\right)\sin(x)\right)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$-\sqrt{3}\sin\left(\frac{\pi}{6}-x\right)=\sqrt{\frac{3}{2}}\Longleftrightarrow$$
$$\sin\left(\frac{\pi}{6}-x\right)=-\frac{1}{\sqrt{2}}\Longleftrightarrow$$
$$\frac{\pi}{6}-x=\frac{5\pi}{4}+2\pi n_1\Longleftrightarrow\space\space\vee\space\space\frac{\pi}{6}-x=\frac{7\pi}{4}+2\pi n_2\Longleftrightarrow$$
$$-x=\frac{13\pi}{12}+2\pi n_1\Longleftrightarrow\space\space\vee\space\space -x=\frac{19\pi}{12}+2\pi n_2\Longleftrightarrow$$
$$x=-\frac{13\pi}{12}-2\pi n_1\space\space\vee\space\space x=-\frac{19\pi}{12}+2\pi n_2$$
With $n_1,n_2\in\mathbb{Z}$
| {
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Show that $ax+by=1 $ implies $\frac{1}{x^2+y^2} \leq a^2+b^2$
Show that for real positive x,y,a and b: $ax+by=1 \implies \frac{1}{x^2+y^2} \leq a^2+b^2$
I used to use contradiction but no results ;
We show that
$$a^2+b^2 < \frac{1}{x^2+y^2} \implies ax+by \ne1$$
$$a^2+b^2 < \frac{1}{x^2+y^2} \iff (a^2+b^2)(x^2+y^2)<0 \iff (ax-by)^2-(ay-bx)^2 <0$$
I'm stuck here, I thought about proving it by putting something in a denominator which is not zero so it's not $1$.
| Hint:
For a nice geometric method, consider the circle $$x^2+y^2=\frac{1}{a^2+b^2}$$ and observe that any point on the line $$ax+by=1$$ is always outside of that circle, except for exactly one point.
Recall that the perpendicular distance of a point $(x_0,y_0)$ from a line $Ax+By+C=0$ is $$d=\frac{\left | Ax_0 + By_0 + C \right |}{\sqrt{A^2+B^2}}$$
You may like to consider the distance of the line $ax+by=1$ from the origin, and you'll almost certainly observe that it has something to do with the circle I mentioned earlier.
| {
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Primitive Pythagorean triple divisible by 3 Prove that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple
of 3, and c cannot be a multiple of 3.
My attempt:
Let a and b be relatively prime positive integers.
If $a\equiv \pm1 \pmod{3}$ and $b\equiv \pm1 \pmod{3}$,
$c^2=a^2+b^2\equiv 1+1\equiv 2 \pmod{3}$
This is impossible as the only quadratic residues modulo 3 are 0 and 1.
So far, so good.
If one of a, b is $\equiv 0 \pmod{3}$ and the other is $\equiv \pm1 \pmod{3}$,
$c^2=a^2+b^2\equiv 0+1\equiv 1 \pmod{3}$
This is the part I don't understand. Just because $c^2\equiv 1\pmod{3}$ doesn't mean that $c^2$ must be a perfect square. For example, $a=12$ and $b=13$ satisfy the above conditions but $c^2=a^2+b^2=313$, which isn't a perfect square.
| Your proof is complete. You are asked to show that one of $a,b,c$ is divisible by $3$. In the first part you show that $a$ and $b$ can't both be non-divisible by $3$. In the second part, you assume that one of $a,b$ is divisible by $3$ and show that $c^2\equiv 1$ (mod $3$) which implies that $c$ is not divisible by $3$ and hence exactly one of $a,b,c$ is divisible by $3$.
| {
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How to evaluate $\int \sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}\frac{\sec x}{\sqrt{1+2\sec x}}dx$ $$\int \sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}\frac{\sec x}{\sqrt{1+2\sec x}}dx$$
$$\int \frac{1}{\csc x+\cot x}\frac{\sec x}{\sqrt{1+2\sec x}}dx$$
This becomes $$-\int\frac{dt}{(t^2+t)\sqrt{t^2+2t}} $$ after putting $\cos x=t$. Solution to this is obtained by putting $t=\frac{1}{p}$
Hence finally I get $$\int \sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}\frac{\sec x}{\sqrt{1+2\sec x}}dx=ln(2\sec x+1)+ln\Big(\frac{\sqrt{2\sec x+1}+1}{\sqrt{2\sec x+1}-1}\Big)$$
But the answer given is $$\sin^{-1}{\Big( \frac{1}{2}\sec^2\frac{x}{2}\Big)}$$
| You can verify that the factor $\sqrt{\frac{\csc{\left(x\right)}-\cot{\left(x\right)}}{\csc{\left(x\right)}+\cot{\left(x\right)}}}$ is equal to $\left|\tan{\frac{x}{2}}\right|$. This strongly suggests that the tangent half-angle substitution would be a more natural choice here.
We find that for $\theta\in[0,\frac{\pi}{2}]$,
$$\begin{align}
I{\left(\theta\right)}
&=\int_{0}^{\theta}\frac{\sec{\left(x\right)}}{\sqrt{1+2\sec{\left(x\right)}}}\sqrt{\frac{\csc{\left(x\right)}-\cot{\left(x\right)}}{\csc{\left(x\right)}+\cot{\left(x\right)}}}\,\mathrm{d}x\\
&=\int_{0}^{\theta}\frac{\sec{\left(x\right)}}{\sqrt{1+2\sec{\left(x\right)}}}\sqrt{\tan^{2}{\left(\frac{x}{2}\right)}}\,\mathrm{d}x\\
&=\int_{0}^{\tan{\left(\frac{\theta}{2}\right)}}\frac{\frac{1+t^2}{1-t^2}\cdot t}{\sqrt{1+2\frac{1+t^2}{1-t^2}}}\cdot\frac{2}{1+t^2}\,\mathrm{d}t;~~~\small{\left[\tan{\left(\frac{x}{2}\right)}=t\right]}\\
&=\int_{0}^{\tan^{2}{\left(\frac{\theta}{2}\right)}}\frac{\mathrm{d}u}{\sqrt{\left(1-u\right)\left(3+u\right)}};~~~\small{\left[t=\sqrt{u}\right]}\\
&=\int_{\frac12}^{\frac12\sec^{2}{\left(\frac{\theta}{2}\right)}}\frac{\mathrm{d}w}{\sqrt{1-w^2}};~~~\small{\left[\frac{1+u}{2}=w\right]}\\
&=\arcsin{\left(\frac12\sec^{2}{\left(\frac{\theta}{2}\right)}\right)}-\arcsin{\left(\frac12\right)}.\blacksquare\\
\end{align}$$
| {
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Upper bound for $2(x+y+z)-3(xy+yz+zx)+4xyz$ Let $x,y,z\geq 0$ and $x+y+z\leq\frac12$. What is the maximum of $$S=2(x+y+z)-3(xy+yz+zx)+4xyz?$$
When $x=y=z\leq\frac{1}{6}$, we have $S=6x-9x^2+4x^3$, which is an increasing function in $[0,\frac16]$, so the maximum is attained when $x=y=z=\frac16$, which is $83/108$.
When $x=\frac12, y=z=0$, we have $S=1$. Is this the maximum?
It is not clear if we can assume without loss of generality that $x+y+z=\frac12$, since the effect of increasing one variable on $S$ is unclear.
| For $y=z=0$ and $x=\frac{1}{2}$ we get a value $1$.
We'll prove that it's a maximal value.
Indeed, let $x=\frac{a}{6}$, $y=\frac{b}{6}$ and $z=\frac{kc}{6}$, where $k>0$ and $a+b+c=3$.
Hence, the condition gives $$\frac{a}{6}+\frac{c}{6}+\frac{kc}{6}\leq\frac{1}{2}$$ or
$$3-c+kc\leq3,$$
which gives $k\leq1$ and we need to prove that
$$\frac{a+b+kc}{3}-\frac{ab+kac+kbc}{12}+\frac{kabc}{54}\leq1$$ or
$$36(a+b+kc)-9(ab+kac+kbc)+2kabc\leq108$$ or
$$k(36c-9ac-9bc+2abc)+36(a+b)-9ab\leq108.$$
But $36c-9ac-9bc+2abc=12c(a+b+c)-9ac-9bc+2abc\geq0$
and since $k\leq1$, it's enough to prove that
$$36(a+b+c)-9(ab+ac+bc)+2abc\leq108$$ or
$$4(a+b+c)^3-3(ab+ac+bc)(a+b+c)+2abc\leq4(a+b+c)^3,$$
which is obviously true.
Done!
| {
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If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$
If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$
This is a 1989 ARML problem. One, ugly way to solve this is:
Approximate this as $x^4=379^2$, so $x\approx \sqrt{379}\approx 19$ and guess and check around there to see that $18$ works.
What's a nicer way?
Hint
Difference of squares
| Note that
\begin{align*}
x(x+1)(x+2)(x+3) &=379^2-1\\
&=(380)(378) \\
&=(19)(20)(18)(21).
\end{align*}
Hence it follows that $x=18$.
| {
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Prove: for any $n$, $1+10^4+...+10^{4n}$ is a composite number. I saw the first three terms: $1$, $10000$, $100000000$. I thought about proving by induction, so I did:
Base case: At $n>0$, $10001$ is composite. Note that at $n=0$, we get $1$ which is not composite. And at $n<0$, we get decimals which are against composite number definition. So, $n$ cannot be $\le 0$.
Hypothesis: Let it be true for $k$, such that $m = 1+10^4+...+10^{4k}$ is a composite number.
Proof: Add $10^{4(k+1)}$ to $m$. So, m becomes $1+10^4+...+10^{4k}+10^{4k+4}$ which is a composite number + composite number. However, as I can trivially infer from $9+4=13$, a composite number + composite number is not always another composite.
Question: How to solve this question? Hints for getting started?
UPDATE:
So, as @Michael told me, this is a geometric series. So, using $\frac{a(1-r^m)}{1-r}$, so letting $a=1, r = 10^4$, I got: $$\frac{1*(1-(10^{4})^{n+1})}{1-10^4} \implies \frac{10^{4n+4}-1}{9999}$$ How to proceed further?
| We write the number as
$$\frac{10^{4m}-1}{10^4-1},$$
where $m \geqslant 2$.
Case 1: $m$ even. Write $m = 2r$, then our number is
$$\frac{10^{8r}-1}{10^4-1} = \frac{10^{4r}-1}{10^4-1}\cdot (10^{4r}+1).$$
If $r = 1$, then it is easy to check that $10^4+1$ is composite, and for $r > 1$ we note that $\frac{10^{4r}-1}{10^4-1} > 1$, which shows our number is composite.
Case 2: $m$ odd. Write
$$\frac{10^{4m}-1}{10^4-1} = \frac{10^{2m}+1}{101}\cdot \frac{10^m+1}{11}\cdot\frac{10^m-1}{9}.$$
Since $m$ is odd, we have $10^{2m} \equiv (-1)^m \equiv -1\pmod{101}$, and $10^m \equiv (-1)^m \equiv -1\pmod {11}$, so all factors are integers. Since $m > 1$, all three factors are $> 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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What is a combined function and what factors affect its domain. Can someone please explain to me what a combined function is and how it is different from a composite function.
From my understanding a combined function would be $(f+g)(x) = f(x)+g(x)$ and a composite function is $(f\circ g)(x)=f\big(g(x)\big)$
Also what factors affect a combined functions domain?
| Using the example functions $f(x)= 2x^2+5x+2$ and $g(x)= 2x+1$ whose domains are the real numbers.
A combined function is an algebraic combination of two function: sum, difference, product, quotient. $$\begin{align}(f+g)(x) & = f(x)+g(x) \\ & = (2x^2+5x+2)+(2x+1) \\ & = 2x^2+7x+3 \\[2ex] (f-g)(x) & = f(x)-g(x) \\ & = (2x^2+5x+2)-(2x+1) \\ &= 2x^2+3x+1 \\[2ex] (f\cdot g)(x) & = f(x)\cdot g(x) \\ & = (2x^2+5x+2)\cdot(2x+1)\\ & = 4x^3+12x^2+9x+2 \\[2ex] (f\div g)(x) & = \frac{f(x)}{g(x)} & \big[g(x)\neq 0\big] \\ & = \frac{ (2x^2+5x+2)}{(2x+1)} & \big[2x+1\neq 0\big]\\ &= x+2 & \big[x\neq -1/2\big]\end{align}$$
The domain of a combined function is the intersection of the domains of the functions which are combined. The domain of a quotient function is further restricted to exclude those points in the denominator function whose image is zero. (vis $g(x)\neq 0$).
Note: in particular, that although $x+2$ can be defined at $x=-\tfrac 1 2$, the domain of the quotient function actually excludes this point. Our $(f\div g)(x)$ is discontinuous at the limit point $x=-\tfrac 1 2$ because $g(-\tfrac 1 2)=0$. It's a subtle but important point when finding valid solutions to systems of equations.
A composite function is a composition of functions (the composition operator is $\circ$ ). $$\begin{align}(f\circ g)(x) & = f\big(g(x)\big) & \neq (\;g\circ f\;)(x) \\ & = f(2x+1) \\ & = 2(2x+1)^2+5(2x+1)+2 \\ & = 8x^2+18x+9\end{align}$$
The domain of a composite function is the domain of the outer function whose image intersects the domain of the inner function. (Order is important here, composition is not commutative).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The roots of $x^2-2x+3=0$ are $\alpha$ and $\beta$. Find the equation whose roots are: $\alpha+2$, $\beta+2$. Not sure of answer in book. The roots of $x^2-2x+3=0$ are $\alpha$ and $\beta$. Find the equation whose roots are: $\alpha+2$, $\beta+2$. Not sure of answer in book.
My working:
$\alpha+\beta=2, \alpha\beta=3$
$(\alpha+2)+(\beta+2)=\alpha+\beta+4=6$
$(\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)=7$
Therefore, equation whose roots are $\alpha+2, \beta+2$ is $x^2-6x+7=0$
Answer in book is $x^2-6x+11=0$
| Note that
$$(\alpha+2)(\beta +2)=\alpha\beta+2(\alpha+\beta)\color{red}{+4}$$
| {
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"url": "https://math.stackexchange.com/questions/1531826",
"timestamp": "2023-03-29T00:00:00",
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Calculate the density function of $Y=\frac{1}{X}-X$ where $X\sim U[0,1]$ I know that :
$$f_X(x)=\cases{1 & $x\in [0,1]$\\0 & $x\notin[0,1]$}$$
Then:
$$P(Y\leq y)=P(\frac{1}{X}-X\leq y)=P(X\leq\frac{1}{2}(\sqrt{y^2+4}-y))$$ as $$\frac{1}{x}-x=y\rightarrow x\frac{1}{x}-xx=yx\rightarrow 1-x^2=yx\rightarrow 1-x^2-xy=0\rightarrow \cases{\frac{1}{2}(\sqrt{y^2+4}-y)\\\frac{1}{2}(-\sqrt{y^2+4}-y)} $$ I assumed positive root only(dont know if it is right assumption). So I am stuck from now on. Any suggestions.
| As mentioned by @André and @Did, your inequality sign is the wrong way. We should have
$$F_Y(y)=
P(Y\leq y) = P\left(\dfrac{1}{X}-X \leq y \right) = P\left(X^2+yX-1 \geq 0 \right).
$$
You found the quadratic roots correctly. So the general solution to the quadratic inequality is $$X\leq \frac12\left(-y-\sqrt{y^2+4}\right)\quad\text{or}\quad X\geq\frac12 \left(-y+\sqrt{y^2+4}\right).$$ However, we must have $X\in (0,1)$ so we take the positive range only and we get, for any $y\gt 0$,
$$F_Y(y) = P\left(X\geq\frac12 \left(-y+\sqrt{y^2+4}\right)\right) = \dfrac{2+y-\sqrt{y^2+4}}{2}.$$
For $y\le0$, $F_Y(y)=0$. Differentiating $F_Y$ (and rearranging the result slightly) shows that the density $f_Y$ is $$f_Y(y) = \frac{\mathbf 1_{y>0}}{2}\left(1-\frac{y}{\sqrt{4+y^2}}\right) = \frac{\left(\sqrt{4+y^2}-y\right)\,\mathbf 1_{y>0}}{2\sqrt{4+y^2}} \cdot \frac{\sqrt{4+y^2}+y}{\sqrt{4+y^2}+y} = \frac{2\,\mathbf 1_{y>0}}{4+y^2+y\sqrt{4+y^2}}.$$
| {
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${{\tan x}\over x}\gt {{x}\over {\sin x}}$ Problem was asked as an application of MVT theorem. I got the derivative of $\sin x \tan x - x^2$ as $\sin x +\tan x \sec x - 2x$. How do I prove that the derivative is positive for $x \in (0,\pi/2)$ so as to prove the function is increasing and prove the question.
| Sorry I cannot find any relation about MVT and increasing/decreasing, but we may use monotonicity to deal with this problem.
Let $y>x$, for $x,y \in (0,\frac{\pi}{2})$
If $\sin{y}\tan{y}-y^2-\sin{x}\tan{x}+x^2>0$, function would be monotonically increasing.
$\because y-x>\sin({y-x})$, for $x,y \in (0,\frac{\pi}{2})$
$\therefore \sin{y}\tan{y}-y^2-\sin{x}\tan{x}+x^2$
$=\sin{y}\tan{y}-\sin{x}\tan{x}-(y^2-x^2)$
$=\sin{y}\tan{y}-\sin{x}\tan{x}-(y-x)(y+x)$
$>\sin{y}\tan{y}-\sin{x}\tan{x}-\sin({y-x})\sin({y+x})$
$=\frac{\sin^2{y}}{cos{y}}-\frac{\sin^2{x}}{cos{x}}+\frac{1}{2}(\cos{y}-\cos{x})$
$=\frac{(1-\cos^2{y})\cos{x}-(1-\cos^2{x})\cos{y}}{\cos{y}\cos{x}}+\frac{1}{2}(\cos{y}-\cos{x})$
$=\frac{(\cos{x}-\cos{y})(\cos{y}\cos{x}+1)}{\cos{x}\cos{y}}+\frac{1}{2}(\cos{y}-\cos{x})$
$=\frac{1}{2}\cos{x}\cos{y}+\frac{\cos{x}-\cos{y}}{\cos{y}\cos{x}}$
$\because \cos{x}\cos{y}>0$, for $x,y \in (0,\frac{\pi}{2})$
$\cos{x}-\cos{y}=-\frac{1}{2}\sin\frac{x+y}{2}\sin\frac{x-y}{2}=\frac{1}{2}\sin\frac{x+y}{2}\sin\frac{y-x}{2}>0$
$\therefore$ function is increasing on $(0,\frac{\pi}{2})$
| {
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"timestamp": "2023-03-29T00:00:00",
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integrate sin(x). OK. I have a doubt with this: I know $-\cos(x) + k =\int \sin x\,dx$ but doing
$\sin(x)=2\sin\frac{x}{2}\cos\frac{x}{2}$ I get $\int \sin x\,dx = \int 2\sin\frac{x}{2}\cos\frac{x}{2}\,dx $ if $ u = \sin \frac{x}{2}$ then $du = \cos \frac{x}{2} \frac{dx}{2}$ then $$\int \sin x\,dx = \int 2\sin \frac{x}{2}\cos \frac{x}{2}\, dx =4\int \sin \frac{x}{2} \cos\frac{x}{2}\frac{dx}{2}=4\int u\,du= 4\frac{u^2}{2}= 2u^2 = 2\sin^2\frac{x}{2}$$ then if $\theta = \frac{x}{2} \rightarrow \cos 2\theta =-2\sin^2(\theta)+k \rightarrow \frac{\cos(2\theta)}{2}=\cos^2 \theta -1+\frac{k}{2} $ if $k=0$, then $$\frac{\cos 2\theta}{2}=\cos^2 \theta-1.$$ Now, why isn't it a trigonometric identity? or Is it? Because I think that I found one.
| You mistakenly equate indefinite integrals, which is wrong. Keep in mind that indefinite integrals are defined up to a constant, and that "fixing" this constant the way you do makes no sense mathematically (an indefinite integral should be thought of a set of functions, not as a precise function).
| {
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computing the product $\prod_{n=1}^{2016} \frac{2^{2^{n-1}}+1}{2^{2^{n-1}}}$ how can i calculate the product: $\prod_{n=1}^{2016} \frac{2^{2^{n-1}}+1}{2^{2^{n-1}}}$?
I can see that in the denominator it's a geometric series, but in the numerator i can't see how to simplify.
Thanks.
| Note that
\begin{align*}
\prod_{n=1}^{2016} \frac{2^{2^{n-1}}+1}{2^{2^{n-1}}}&
=\frac{(2^{2^{0}}-1)(2^{2^{0}}+1)\cdots (2^{2^{2015}}+1)}{(2^{2^{0}}-1)2^{2^{0}}\cdots 2^{2^{2015}}}\\
&
=\frac{(2^{2^{1}}-1)(2^{2^{1}}+1)\cdots (2^{2^{2015}}+1)}{(2^{2^{0}}-1)2^{2^{0}+\cdots +2^{2015}}}\\
&
=\frac{(2^{2^{2}}-1)(2^{2^{2}}+1)\cdots (2^{2^{2015}}+1)}{2^{2^{0}+\cdots +2^{2015}}}\\
&
\cdots\\
&
=\frac{2^{2^{2016}}-1}{2^{2^{2016}-1}}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"Extending" the calculation of the golden ratio using square roots (not silver-ratio) I'm looking at the following formula:
$x =\frac{-n+\sqrt{n^{2}+4n}}{2}$
For $n=1$ this this gives $0.618...$ and then $\frac n x$ gives $1.618...$ which is $\phi$, the golden ratio.
What particularly interests me are the following values of $n$:
$1, 2, 6, 8, 12, 14, 18, 28, 32, 36$
The values (or ratios) for $\frac n x$ then give:
1.618033989
2.732050808
6.872983346
8.898979486
12.92820323
14.93725393
18.94987437
28.96662955
32.97056275
36.97366596
I can also get to these ratios using the following function:
$\dfrac n 2 + \sqrt a \times b$
Using these values:
n=1, a=5, b=0.5
n=2, a=3, b=1
n=6, a=15, b=5
n=8, a=6, b=2
n=12, a=12, b=2
n=14, a=7, b=3
n=18, a=11, b=3
n=28, a=14, b=4
n=32, a=8, b=6
n=36, a=10, b=6
Can someone explain if I'm just doing some algebra or if there is some other pattern which explains this?
And by 'this' I mean the correlation between: $\frac{n}{\frac{-n+\sqrt{n^{2}+4n}}{2}}$ and $\dfrac n 2 + \sqrt a \times b$)
| $$\frac{n}{x}=\frac{2n}{-n+\sqrt{n^2+4n}}=\frac{2n\left(-n-\sqrt{n^2+4n}\right)}{-4n}=\frac{n}{2}+\frac{\sqrt{n^2+4n}}{2}.$$
If $n$ is even, then this is
$$\frac{n}{2}+\frac{\sqrt{4\left[(\frac{n}{2})^2+n\right]}}{2}=\frac{n}{2}+\sqrt{\left(\frac{n}{2}\right)^2+n}.$$
Now, any surd $\sqrt{m}$ can be simplified to $a\sqrt{b}$ with $b$ squarefree, so apply with $m=(\frac{n}{2})^2+n$.
| {
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Height of a rotated ellipse If I have an ellipse, it is easy to find its height, twice the length of the major axis. But if the ellipse is rotated a certain number of degrees, how do you find the vertical height from top to bottom?
| The equation of a rotated ellipse is:
$$
\frac{(x\cos \theta+y\sin \theta)^2}{a^2}+\frac{(-x\sin \theta+y\cos \theta)^2}{b^2}=1
$$
Expand the equation:
$$
\left(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right)x^2+2\cos\theta\sin\theta \left(\frac{1}{a^2}-\frac{1}{b^2}\right)xy+\left(\frac{\sin^2\theta}{a^2}+\frac{\cos^2\theta}{b^2}\right)y^2=1
$$
This can be written as $Ax^2+Bxy+Cy^2=1$. Solve for x:
$$
x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}
$$
The maximum and minimum values of $y$ are where there is only one value of $x$, i.e. where $B^2-4AC=0$, so we solve that equation:
$$
\left(2\cos\theta\sin\theta \left(\frac{1}{a^2}-\frac{1}{b^2}\right)y-1\right)^2-4\left(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right)\left(\frac{\sin^2\theta}{a^2}+\frac{\cos^2\theta}{b^2}\right)y^2=0
$$
$$
a^2+b^2-2y^2-(a^2-b^2)\cos2\theta=0
$$
$$
y=\pm\sqrt{\frac{1}{2}\left(a^2+b^2-(a^2-b^2)\cos2\theta\right)}
$$
Finally, the height of the ellipse is twice this distance, $2y$:
$$
\sqrt{2\left(a^2+b^2-(a^2-b^2)\cos2\theta\right)}
$$
| {
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Series expansion for $x$, when $x$ is small Suppose that we are given the series expansion of $y$ in terms of $x$, where $|x|\ll 1$. For example, consider $$y=x+x^2+x^3+\cdots\qquad\qquad\qquad (1).$$ From this I would like to derive the series expansion of $x$ in terms $y$. Given that $|x|\ll 1$ we can take $y=x+O(x^2)$ such that $x\sim y$.
It then follows from $(1)$ that $$x=y-(x^{2}+x^{3}+\cdots)\qquad\qquad\qquad (2).$$ Given that $x\sim y$, $(2)$ becomes $$x=y-y^2-y^3+\cdots\qquad\qquad\qquad (3).$$
While the first two terms of $(3)$ are correct I suspect that the remaining terms are incorrect.
For instance given that $y=x+O(x^2)$ implies that $y^2=x^2+2xO(x^{2})+(O(x^{2}))^{2}$. Therefore, $2xO(x^{2})$ will yield a $y^{3}$ term in $(3)$ when computing $x^{2}$ in $(2)$.
How do I find the correct term containing $y^{3}$ in $(3)$, and so on.
| If we assume that the series continues as $y=\sum\limits_{k=1}^\infty x^k$, we have
$$
y=\frac{x}{1-x}\tag{1}
$$
then we can compute
$$
x=\frac{y}{1+y}=\sum_{k=1}^\infty(-1)^{k-1}y^k\tag{2}
$$
However, if we don't make this assumption, we can invert the series using Lagrange Inversion:
$$
\begin{align}
\left[\,y^k\,\right]x
&=\frac1k\left[\,x^{-1}\,\right]y^{-k}\\
&=\frac1k\left[\,x^{-1}\,\right]\left(x+x^2+x^3+\dots\right)^{-k}\tag{3}
\end{align}
$$
where $\left[x^k\right]f$ is the coefficient of $x^k$ in the power series expansion for $f$.
Since
$$
\begin{align}
\left(x+x^2+x^3+\dots\right)^{-1}&=\frac{\color{#C00000}{1}}x-1+0x+\dots\\
\left(x+x^2+x^3+\dots\right)^{-2}&=\frac1{x^2}\color{#C00000}{-}\frac{\color{#C00000}{2}}x+1+\dots\\
\left(x+x^2+x^3+\dots\right)^{-3}&=\frac1{x^3}-\frac3{x^2}\color{#C00000}{+}\frac{\color{#C00000}{3}}x+\dots
\end{align}\tag{4}
$$
$(3)$ gives
$$
x=y-y^2+y^3+\dots\tag{5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
} |
Evaluate the improper integral $ \int_0^1 \frac{\ln(1+x)}{x}\,dx $ I am trying to evaluate
$$
\int_0^1 \frac{\ln(1+x)}{x}\,dx
$$
I started by using the Taylor series for $\ln (1+x)$
$$\begin{align*}
\int_0^1 \frac{\ln(1+x)}{x}\,dx &= \int_0^1\frac{1}{x}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}dx \\&=\int_0^1 \frac{1}{x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+...\right)dx \\&=\int_0^1 \left(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\frac{x^4}{5}-\frac{x^5}{6}+...\right)dx\\&=1-\frac{1}{4}+\frac{1}{9} -\frac{1}{16}+\frac{1}{25}-\frac{1}{36}+... \\ &=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\end{align*}
$$
I'm aware of the fact that $$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$$ however the $(-1)^{n+1}$ is giving me trouble.
| Hint: If you take $\sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}$ what do you get?
| {
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Cubic identity involving the angles of a triangle If $A+B+C=180^\circ$, then prove that
$$\sin^3A+\sin^3B+\sin^3C=3\cos\frac{A}{2}\,\cos\frac{B}{2}\,\cos\frac{C}{2}+\cos\frac{3A}{2}\,\cos\frac{3B}{2}\,\cos\frac{3C}{2}$$
I could only get:
$A+B+C=\pi$ and $A+B=\pi-C$. After this, I don't get enough idea to solve.
| If $A+B+C=(2n+1)\pi,$
$\sin A+\sin B+\sin C=2\sin\dfrac{A+B}2\cos\dfrac{A-B}2+2\sin\dfrac C2\cos\dfrac C2$
Now $\sin\dfrac{A+B}2=\sin\dfrac{(2n+1)\pi-C}2=(-1)^n\cos\dfrac C2$
and $\cos\dfrac{A+B}2=\cos\dfrac{(2n+1)\pi-C}2=(-1)^n\cos\dfrac C2$
$\implies2\sin\dfrac{A+B}2\cos\dfrac{A-B}2+2\sin\dfrac C2\cos\dfrac C2$
$=2(-1)^n\cos\dfrac C2\cos\dfrac{A-B}2+2(-1)^n\cos\dfrac{A+B}2\cos\dfrac C2$
$=2(-1)^n\cos\dfrac C2\left(\cos\dfrac{A-B}2+\cos\dfrac{A+B}2\right)$
$$\implies\sin A+\sin B+\sin C=4(-1)^n\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2$$
if $A+B+C=(2n+1)\pi$
Here $n=0,1$
| {
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Proving $\sqrt{2}x-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}$ How can I prove that
$$
x\sqrt{2}-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}
$$
It's a derivation-based process if I remember correctly, however I was unable to prove it correctly.
| $$f(x):=x\sqrt{2}-\sqrt{x^2+1}-\frac{\sqrt{2}}{2}\ln x$$
Which is continuously differentiable on $(0,\infty)$
$$f'(x)=-\frac{x}{\sqrt{x^2+1}}-\frac{1}{\sqrt{2} x}+\sqrt{2}=\frac{-2 x^2+2 \sqrt{2} \sqrt{x^2+1} x-\sqrt{2} \sqrt{x^2+1}}{2 x \sqrt{x^2+1}}$$
We want to solve
$$
-2 x^2+2 \sqrt{2} \sqrt{x^2+1} x-\sqrt{2} \sqrt{x^2+1}=0$$
$$
\frac{2x^2}{2x-1}=\sqrt{2}\sqrt{x^2+1}\Rightarrow -2 + 8 x - 10 x^2 + 8 x^3 - 4 x^4=0
$$
Or:
$$
(x-1) \left(2 x^3-2 x^2+3 x-1\right)=0
$$
The second term has two complex roots, as its discriminant is negative, and a root between $0$ and $\frac{1}{2}$(by Bolzano-Weierstrass), but from
$$
\frac{2x^2}{2x-1}=\sqrt{2}\sqrt{x^2+1}
$$
we can see that it must be a false root. We have to check the second derivative:
$$f''(x)=\frac{x^2}{\left(x^2+1\right)^{3/2}}-\frac{1}{\sqrt{x^2+1}}+\frac{1}{\sqrt{2} x^2}$$
$$\left.f''(x)\right|_{x=0}=\frac{1}{2\sqrt{2}}>0$$
Thus it is a minimum, and the only possible one.
| {
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QR decompostion - how to finish decomposition? I try to find $QR$ decomposition, but I got stuck in some etap and I can't continue - how to finish it ?
$$ A= \left[\begin{matrix}
1 & -1 \\
1 & 0 \\
1 & 1 \\
1 & 2
\end{matrix}\right] $$
$v = \left[\begin{matrix}
1+2 \\
1 \\
1 \\
1
\end{matrix}\right] \cdot \frac{1}{\sqrt{9+1+1+1}} = \frac{1}{2\sqrt{3}}\cdot \left[\begin{matrix}
1+2 \\
1 \\
1 \\
1
\end{matrix}\right] $
$$Q_1 = I - 2vv^t = I - \frac16 A= \left[\begin{matrix}
9 & 3 & 3 &3 \\
3 & 1 & 1 &1 \\
3 & 1 & 1 &1 \\
3 & 1 & 1 &1 \\
\end{matrix}\right] = \left[\begin{matrix}
-\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} &-\frac{1}{2} \\
-\frac{1}{2} & \frac{5}{6} & -\frac{1}{6} &-\frac{1}{6} \\
-\frac{1}{2} &- \frac{1}{6} & \frac{5}{6} &-\frac{1}{6} \\
-\frac{1}{2} &- \frac{1}{6} & -\frac{1}{6} &\frac{5}{6} \\
\end{matrix}\right] $$
$$Q_1A =\left[\begin{matrix}
-2 & -1 \\
0 & 0 \\
0 &1 \\
0 &2 \\
\end{matrix}\right] $$
| Let me try again: using my $v_2$ from the answer compute the second Householder matrix. This matrix is $Q_2$.
You have already computed $Q_1$ in your question-post.
Then define a matrix $Q^T$ as the product of both in this very order:
$Q^T =Q_2Q_1$
Then you have that $Q^T$ is an orthogonal matrix and therefore $Q^TQ=I$ is the unit matrix.
You have by the computation the equation
$Q_2Q_1A=R$
This is nothing else (using the definition of $Q^T$ on the Left hand side and the orthogonality of Q on the Right hand side):
$Q^TA = Q^TQR$
And now you multiply both sides with Q to get
$\underbrace{QQ^T}_{=I}A = \underbrace{QQ^T}_{=I}QR$
So
$ A = QR$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the limit of a fraction: $\lim_{x \to 3} \frac{x^3-27}{x^2-9}$
Find $$\lim_{x \to 3} \frac{x^3-27}{x^2-9}$$
What I did is:
\begin{align}
\lim_{x \to 3} \frac{x^3-27}{x^2-9} &= \lim_{x \to 3} \frac{(x-3)^3+9x-27x}{(x-3)(x+3)} = \lim_{x \to 3} \frac{(x-3)^3+9(x-3)}{(x-3)(x+3)} \\
&= \lim_{x \to 3} \frac{(x-3)^3}{(x-3)(x+3)} + \lim_{x \to 3} \frac{9(x-3)}{(x-3)(x+3)} \\
&= \lim_{x \to 3} \frac{(x-3)^2}{(x+3)} + \lim_{x \to 3} \frac{9}{(x+3)} =0 + \frac{9}{6}
\end{align}
Wolfram factor the numerator to $(x-3)(x^2+3x+9)$ is there a quick way to find this?
| $$\lim_{x \to 3}\frac{x^3-27}{x^2-9}=\frac{0}{0}=\\\lim_{x \to 3}\frac{(x^3-27)'}{(x^2-9)'} =\\ \lim_{x \to 3}\frac{3x^2}{2x}=\\ \lim_{x \to 3}\frac{3x^2}{2x}\\=\lim_{x \to 3}\frac{3x}{2}=\frac{9}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
On the limit of $f(n)$, specifically having to do with integration of an iterated $\arctan$ Assume we are given that $A_n(x)$ denotes $n$ iterations of $\arctan(x)$, for example $A_2(x)=\arctan (\arctan(x))$
If $$f(n)=\int_{0}^n A_n(x)\space \text{d}x$$
I am looking for a rigorous proof that $\lim_{n\to\infty}f(n)$ diverges
I have worked on this problem for a while actually, and all I have gotten was a proof that $\forall x\in\Bbb{R},\lim_{x\to\infty}A_n(x)=0$
Beyond that information though, I am stuck. Thanks for any help.
| Step 1. Applying the mean value theorem to the function $x \mapsto x - \arctan x$, for some $\xi \in [0, A_{k-1}]$ we have
$$ A_{k-1} - A_k
= (A_{k-1} - \arctan A_{k-1})
= A_{k-1} \cdot \frac{\xi^2}{1 + \xi^2}
\leq \frac{A_{k-1}^3}{1 + A_{k-1}^2}.$$
This shows that
$$ A_{k-1} - A_k \leq A_{k-1}^3 \qquad \text{and} \qquad \frac{A_{k-1}}{1+A_{k-1}^2} \leq A_k \leq A_{k-1}. \tag{*} $$
Step 2. From $\text{(*)}$, for $n \geq 2$ we have
\begin{align*}
\frac{1}{A_n^2}
&= \frac{1}{A_1^2} + \sum_{k=2}^{n} \left( \frac{1}{A_k^2} - \frac{1}{A_{k-1}^2} \right) \\
&= \frac{1}{A_1^2} + \sum_{k=2}^{n} \frac{(A_{k-1} + A_k)(A_{k-1} - A_k)}{A_k^2 A_{k-1}^2} \\
&\leq \frac{1}{A_1^2} + \sum_{k=2}^{n} \frac{(1+A_{k-1}^2)^2}{A_{k-1}^2} \frac{(2A_{k-1})(A_{k-1}^3)}{A_{k-1}^2} \\
&= \frac{1}{A_1^2} + \sum_{k=2}^{n} 2(1+A_{k-1}^2)^2 \\
&\leq \frac{1}{A_1^2} + Cn
\end{align*}
for some absolute constant $C > 0$, where the last inequality follows from the fact that $A_i \leq \pi/2$ for $i \geq 1$. (Here, we may choose $C = 2(1+\frac{1}{4}\pi^2)^2$ though this particular value is not important.) This shows that
$$ \frac{\arctan x}{\sqrt{1 + Cn\arctan^2 x}} \leq A_n (x) $$
and in particular,
$$ A_n(x) \geq \frac{c}{\sqrt{1+C'n}} \quad \text{for } x \geq 1 $$
for some absolute constant $c, C > 0$. (Here, we may choose $c = \arctan 1$ and $C' = \frac{1}{2}\pi C$.) This gives
$$ \int_{0}^{n} A_n(x) \, dx \geq \int_{1}^{n} \frac{c}{\sqrt{1+C'n}} \, dx = \frac{c(n-1)}{\sqrt{1+C'n}}. $$
Therefore, as $n \to \infty$, the integral diverges to infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Double radical proof I'm trying to prove that
$$
\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
$$
With
$$
C=\sqrt{A^2 - B}
$$
How can I handle this?
Edit: obviously is easy that this holds when you know the r.h.s., but my question is: how to get the r.h.s. when you only know the l.h.s.
| I can derive the left side from the right, not the other way around.
The right side is the sum of the two positive roots (assuming $A^2>B$) of:
$$x^4-Ax^2+\frac{B}{4}=0$$
But this polynomial factors as:
$$x^4-Ax^2+\frac{B}{4}=\left(x^2+\frac{\sqrt{B}}2\right)^2-(A+\sqrt{B})x^2\\
=
\left(x^2-\sqrt{A+\sqrt{B}}x+\frac{\sqrt{B}}2\right)
\left(x^2+\sqrt{A+\sqrt{B}}x+\frac{\sqrt{B}}2\right)$$
Note that the positive values cannot be roots to the right factor, so they have to be roots of the left factor, and therefore their sum is $\sqrt{A+\sqrt{B}}$.
Reversing the direction is uglier:
$$A+\sqrt{B}$$ is the larger root of:
$$x^2-2Ax + C^2 = 0$$
Therefore $w=\sqrt{A+\sqrt{B}}$ is the largest root of $$x^4-2Ax^2+C^2 = 0$$
Now $$x^4-2Ax^2+ C^2= \left(x^2+C\right)^2-2(A+C)x^2\\
=\left(x^2-\sqrt{2(A+C)}x+C\right)\left(x^2+\sqrt{2(A+C)}x+C\right)$$
It can't be a root of the right side, which is positive at $w$, so we have:
$$w^2-\sqrt{2(A+C)}w+C=0$$
We can also factor the above by completing the square the other way:
$$x^4-2Ax^2+C^2 = (x^2-C)^2 - 2(A-C)x^2 =\\
=\left(x^2-\sqrt{2(A-C)}x-C\right)\left(x^2+\sqrt{2(A-C)}x-C\right)$$
The positive roots of this polynomial have to be split on the two sides, and we can see since that $\sqrt{A+\sqrt{B}}$ must be the largest root, it must be a root of the left factor.
A common root $u=\sqrt{A+\sqrt{B}}$ of the two equations:
$$w^2 - \sqrt{2(A+C)}w + C=0\\
w^2-\sqrt{2(A-C)}w - C=0$$
Thus, adding, then dividing by $w$ and re-arranging, we get: $$2u = \sqrt{2(A+C)} + \sqrt{2(A-C)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
what's the maximum value of the algebraic expression?
Suppose that $a,b,c>0$ and satisfy the equation
$$
a^2+b^2+4c^2=1,
$$
then what's the maximum of $F(a,b,c)=ab+2ac+3\sqrt{2}bc$ ?
**
My way: $F(a,b,c)=ab+\sqrt{2}bc+2(ac+\sqrt{2}bc)=\cdots$
| Use AM-GM inequality we have
$$\dfrac{\sqrt{2}}{2}a^2+\dfrac{\sqrt{2}}{4}b^2\ge ab\tag{1}$$
$$\sqrt{2}c^2+\dfrac{\sqrt{2}}{2}a^2\ge ac\tag{2}$$
$$\dfrac{3\sqrt{2}}{4}b^2+3\sqrt{2}c^2\ge 3\sqrt{2}bc\tag{3}$$
$(1)+(2)+(3)$ we have
$$ab+ac+3\sqrt{2}bc\le \sqrt{2}(a^2+b^2+4c^2)=\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How do I solve for $x$ in the equation : $\lvert \sin x \rvert = \lvert \cos 3x \rvert$? The exact question asked for the number of solutions to the equation in the interval $[-2\pi , 2\pi]$.
My understanding & approach :
$\lvert \sin x \rvert = \lvert \cos 3x \rvert$
$\Rightarrow \sin x = \cos 3x$ or $\sin x = -\cos 3x$
$\Rightarrow 4x = ( 2m \pm 1 ) \dfrac{\pi}{2}$ or $\Rightarrow 2x = (2n \pm 1 ) \dfrac{\pi}{2}$
But by this approach , I get only $8$ solutions after applying general rule.
The answer says : There are $24$ solutions.
| $$|a|=|b| \to \pm a=\pm b \to a^2=b^2$$hence
$$|\sin x|=|\cos 3x|\\ \sin^2x=\cos^23x $$wen know $$\sin^2a=\frac{1-\cos2a}{2},\cos^2a=\frac{1+\cos2a}{2}$$ so
$$\frac{1-\cos2x}{2}=\frac{1+\cos6x}{2}\\-\cos(2x)=\cos(6x)\\ \cos(2x+\pi)=\cos (6x)\\6x=\pm(2x+\pi)+2k\pi\\3x=\pm(x+\frac{\pi}{2})+k\pi $$
$$3x=(x+\frac{\pi}{2})+k\pi \to 2x=\frac{\pi}{2}+k\pi \\x=\frac{\pi}{4}+k\frac{\pi}{2} \to -2\pi \leq \-frac{\pi}{4}+k\frac{\pi}{2} \leq 2\pi \\-8 \leq 1+2k \leq 8 \to k=-4,-3,-2,-1,0,1,2,3$$there is 8 solution
$$3x=-x-\frac{\pi}{2}+k\pi\\x=\frac{-\pi}{8}+\frac{k\pi}{4} \\
-2\pi \leq \frac{-\pi}{8}+\frac{k\pi}{4} \leq 2\pi\\-7.5 \leq k \leq 8.5 \to k=-7,-6,...,0,1,...,7,8$$there is 16 solution
8+16 =24 solution overall
| {
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"url": "https://math.stackexchange.com/questions/1544193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find closed form of a sequence $2,5,11,23,...$ Find closed form of a sequence $2,5,11,23,\dots$
How to get generating function for this sequence (closed form)?
Explicit form is $f(x)=2+5x+11x^2+23x^3+\cdots$
Is it possible to get to geometric series representation?
I tried to derive the series multiple times, but that doesn't help.
Could someone give a hint?
| $2, 5, 11, 23, 47, \cdots$ Well, I think $a_n = 3 \cdot 2^{n - 1} - 1$ here.
OK, let me elaborate a bit. Observe that $d_n := a_{n + 1} - a_{n} = 3 \cdot 2^{n - 1}$. Since $$a_n - a_1 = \sum_{k = 1}^{n - 1} d_{k} = 3 \sum_{k = 1}^{n - 1} 2^{k - 1} = 3 \cdot (2^{n - 1} - 1) = 3 \cdot 2^{n - 1} - 3.$$ Since $a_1 = 2$, we have $a_n = 3 \cdot 2^{n - 1} - 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Taylor's Remainder $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) Prove that
$\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) <x-\frac{x^2}{2}+\frac{x^3}{3}$
My attempt: I can prove this by taking one side at a time and assuming $\displaystyle f(x) = \log (1+x)-x+\frac{x^2}{2}-\frac{x^3}{3}$
and then proving that it is a decreasing function and then the same for the other side.
But I am looking for a better solution using Lagrange's Mean Value Theorem or using Taylor's remainder.
| For the inequality on the right, note that the Lagrange form of the remainder gives
$$\log (1+x) = x- \frac{x^2}{2} +\frac{x^3}{3} -\frac{1}{4(1+c)^4},$$
where $0<c<x.$ Since the last term in negative, we get the strict inequality. In fact we get the strict inequality for $-1<x<0$ as well, by the same argument.
The inequality on the left seems easier for $-1<x<0$ to me. Here we get
$$\log (1+x) = x-x^2/2 + x^3/3 + \cdots = -(|x|+|x|^2/2 + |x|^3/3+ |x|^4/4\cdots )$$ $$> -(|x|+|x|^2/2 + (|x|^3/3)(1+|x| + |x|^2 + \cdots))$$ $$ = -(|x|+|x|^2/2 + (|x|^3/3)/(1-|x|) = x-x^2/2 + (x^3/3)/(1+x).$$
So there is good motivation for looking at $x-x^2/2 + (x^3/3)/(1+x)$ as below $\log (1+x)$ for all $x\in (-1,\infty), x\ne 0.$ At this point I would do exactly as you did in showing this is true for $x>0.$
| {
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"url": "https://math.stackexchange.com/questions/1546172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using Lagrange Multipliers to find the maximum of a asymmetric value This problem is from Korean Mathematical Olympiad 2015 P3.
The problem asks to find, with proof, the maximum value of $$(ax+by)^2+(bx+cy)^2$$ with the constraint of $$a^2+b^2+c^2+x^2+y^2=1$$
Now, I do know a solution with AM-GM and Cauchy-Schwarz - I posted on the link.
However, I want to know how to solve this problem using Lagrange Multipliers.
I managed to set up six equations - since there are $5$ variables $a,b,c,x,y$ and also $\lambda$.
Doing that, I got the following equations.
\begin{align*}a^2+b^2+c^2+x^2+y^2&=1\\
ax^2+bxy+a\lambda&=0\\
cy^2+bxy+c\lambda&=0\\
bx^2+axy+cxy+by^2+b\lambda&=0\\
a^2x+aby+bcy+b^2x+x\lambda&=0\\
c^2y+abx+bcx+b^2y+y\lambda&=0\end{align*}
I tried to prove $a=c$ and $x=y$ from there, but I couldn't do it.
How should I solve this monstrous system of equations? Thanks!
| Let's say that you guess $a=c$, e.g. by observing equations (2) and (3) as you have them above. Then from equations (5) and (6) you may see that $x=y$ works. For $a=c$ and $x=y$ equation (2) becomes equal to equation (3) and (5) equal to (6), and so you get the system of equations:
\begin{align*}2a^2+b^2+2x^2&=1\\
ax^2+bx^2+a\lambda&=0\\
2ax^2+2bx^2+b\lambda&=0\\
a^2x+2abx+b^2x+x\lambda&=0
\end{align*}
which may be rewritten as
\begin{align*}2a^2+b^2+2x^2&=1\\
(a+b)x^2+a\lambda&=0\\
2(a+b)x^2+b\lambda&=0\\
((a+b)^2+\lambda)x&=0
\end{align*}
Getting an expression for $x^2$ from the first equation and substituting in equations (2) and (3), and deriving that $λ=-(a+b)^2$ (or $x=0$) from equation (4), should give you a solution for $a$ and $b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing that a polynomial is a unit in a quotientring In this exercise I have a polynomial ring over a finite field $F_2$:$(F_2[X],+,*)$ There is then given an ideal : I $=<X^3+X+1>$.
I am then trying to show that $X^2+X+1 + I$ is a unit in $F_2[X]/I$ by using the extended Euclidean algorithm on $X^3+X+1$ and $X^2+X+1$.
I tried using this algorithm and then got that $(X^3+X+1)*r(X)+(X^2+X+1)*s(X)=gcd(X^3+X+1,X^2+X+1) => (X^3+X+1)*(X+1)+(X^2+X+1)*X^2=1$
I can't really figure out if this is right, and if it is what I am suppose to do from here to show that the given expression is an unit.
| You have already found the inverse, it is: $X^2 + I$.
You know that:
$(X^3 + X + 1)(X + 1) + (X^2 + X + 1)X^2 = [X^4 + X^3 + X^2 + (X+X) + 1] + [X^4 + X^3 + X^2]$
Now in $F_2[X]$, for any element, we have $f(x) + f(x) = (1 + 1)(f(x)) = 0(f(x)) = 0$.
So in the product above, we continue:
$[X^4 + X^3 + X^2 + (X+X) + 1] + [X^4 + X^3 + X^2] = [X^4 + X^3 + X^2 + 1] + [X^4 + X^3 + X^2]$
$= (X^4 + X^4) + (X^3 + X^3) + (X^2 + X^2) + 1 = 0 + 0 + 0 + 1 = 1$.
This shows your gcd calculation is correct.
Now take your calculation mod $I$:
$[(X^3 + X + 1) + I][(X + 1) + I] + [((X^2 + X + 1) + I][X^2 + I] = 1 + I$
$[0 + I][(X + 1) + I] + [((X^2 + X + 1) + I][X^2 + I] = 1 + I$
($(X^3 + X + 1) + I = 0 + I = I$, because $X^3 + X + 1$ is in $I$).
$[0 + I] + [((X^2 + X + 1) + I][X^2 + I] = 1 + I$
(because $0 + I = I$ is an ideal, and "absorbs" anything we multiply it by)
$[((X^2 + X + 1) + I][X^2 + I] = 1 + I$
(since $0 + I = I$ is the additive identity of the quotient ring, and we can omit it in any sum).
This last equation shows explicitly that $(X^2 + X + 1) + I$ is indeed a unit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On the decreasing sequences I do not know how to prove that these sequences are decreasing:
$$
a_n=\frac{1}{\sqrt{n}(n+1)}-\frac{1}{\sqrt{n+1}(n+2)},
$$
$$
b_n=\frac{1}{n(\sqrt{n}+1)}-\frac{1}{(n+1)(\sqrt{n+1}+1)}.
$$
Thank you for all kind help and comments.
My attemption. I considered the following function
$$
f(x)=\frac{1}{\sqrt{x}(x+1)}-\frac{1}{\sqrt{x+1}(x+2)},
$$
and calculated its derivative. But its derivative is rather complicated
$$
\nabla f(x)=\frac{3x+1}{2\sqrt{x}x(x+1)^2}-\frac{3x+4}{2\sqrt{x+1}(x+1)(x+2)^2}
$$
| Observe that for any $n$,
\begin{align}
\sum_{k=1}^n a_k &= \sum_{k=1}^n \left(\frac1{\sqrt k(k+1)} - \frac1{\sqrt{k+1}(k+2)}\right)\\
&= \frac12 - \frac1{\sqrt{n+1}(n+2)}\\
&\stackrel{n\to\infty}\longrightarrow\frac12
\end{align}
and similarly
\begin{align}
\sum_{k=1}^n b_k &= \sum_{k=1}^n \left( \frac1{k(\sqrt k + 1)}\right) - \left( \frac1{(k+1)(\sqrt{k+1} + 1)}\right)\\
&= \frac12 - \left( \frac1{(n+1)(\sqrt{n+1} + 1)}\right)\\
&\stackrel{n\to\infty}\longrightarrow\frac12.\end{align}
Since the sequences of partial sums are monotone increasing and convergent, the sequences $\{a_n\}$ and $\{b_n\}$ are monotone decreasing.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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integate $\int\sqrt{x^2+3x+3} dx$ Is there any universal method how to solve integrals like this?
$$\int\sqrt{x^2+3x+3} dx$$
Or this?
$$\int\sqrt{-x^2+3x+3} dx$$
I tried use first Euler subs, but it was not good idea.
$$\sqrt{ax^2+bx+c} = t\pm\sqrt{ax}$$
| By a linear transform of the argument, you can normalize the integrand to one of
$$\sqrt{1+x^2}\text{ or }\sqrt{1-x^2}.$$
Then by parts,
$$\int\sqrt{1\pm x^2}\,dx=x\sqrt{1\pm x^2}\mp\int\frac{x^2}{\sqrt{1\pm x^2}}dx=x\sqrt{1\pm x^2}-\int\frac{(1\pm x^2)-1}{\sqrt{1\pm x^2}}dx.$$
Move the first part of the integral to the LHS and remains to integrate
$$\int\frac{dx}{\sqrt{1\pm x^2}},$$ which is elementary.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})$ Like the question says
How to prove $$\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})$$
I have tried using many identity but in vain
For reference
$$\tanh ^{-1} x=\frac{1}{2} \log \frac{1+x}{1-x}$$
and $$\sinh^{-1} x=\log (x+\sqrt{x^2+1})$$
| We note
\begin{align*}
\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})&\iff\tan x=\sinh\left(\log \tan (\frac{\pi}{4}+\frac{x}{2})\right).
\end{align*}
The latter equality follows because
\begin{align*}
\sinh\left(\log \tan (\frac{\pi}{4}+\frac{x}{2})\right)&=\frac{1}{2}\left(\tan (\frac{\pi}{4}+\frac{x}{2})-\cot (\frac{\pi}{4}+\frac{x}{2})\right)\\
&=\frac{1}{2}\left(\frac{\tan(x/2)+1}{1-\tan(x/2)}+\frac{\tan(x/2)-1}{1+\tan(x/2)}\right)\\
&=\frac{2\tan(x/2)}{1-\tan^2(x/2)}\\
&=\tan(x).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding lim sup and lim inf Given the sequence
$$ (a_n)=\begin{cases} 3^{-n}, & \text{for even }n \\ 5^{-n}, & \text{for odd } n \end{cases} $$
How to find:
$$\limsup_{n\to\infty}\frac{a_{n+1}}{a_n}$$
$$\liminf_{n\to\infty}\frac{a_{n+1}}{a_n}$$
$\dfrac{a_{n+1}}{a_n}$ goes like this:
$$\frac{5^1}{3^2},\frac{3^2}{5^3},\frac{5^3}{3^4},\frac{3^4}{5^5},\frac{5^5}{3^6}$$
How can I find lim sup and lim inf?
| HINT: First write out a general expression for $\dfrac{a_{n+1}}{a_n}$::
$$\begin{align*}
\frac{a_{n+1}}{a_n}&=\begin{cases}
\dfrac{3^n}{5^{n+1}},&\text{if }n\text{ is even}\\
\dfrac{5^n}{3^{n+1}},&\text{if }n\text{ is odd}
\end{cases}\\\\
&=\begin{cases}
\dfrac15\left(\dfrac35\right)^n,&\text{if }n\text{ is even}\\
\dfrac13\left(\dfrac53\right)^n,&\text{if }n\text{ is odd}\;.
\end{cases}
\end{align*}$$
How do the sequences
$$\left\langle\frac{a_{2n+1}}{a_{2n}}:n\in\Bbb N\right\rangle$$
and
$$\left\langle\frac{a_{2n+2}}{a_{2n+1}}:n\in\Bbb N\right\rangle$$
behave?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Question on summation of the given series and applying limits. Question is
If the value of $\displaystyle \lim_{n\to\infty}\sum_{k=2}^{n}\cos^{-1}\left[\frac{1+\sqrt{(k-1)(k)(k+1)(k+2)}}{k(k+1)}\right]$ is equal to $\displaystyle\frac{120\pi}{m}$.
Then find the value of $m$.
What I did was to try putting the value of the variable $k$ form $1$ to infinity one by one and writing its correspondence value of $\arctan(x)$
And then tried to to connect all the terms.
I am stuck at that point.. Any further help or any other better method would be of immense help..
Thanks in advance.....
| Let $$\displaystyle S_{n} = \lim_{n\rightarrow \infty}\sum^{n}_{k=2}\cos^{-1}\left(\frac{1+\sqrt{(k-1)k(k+1)(k+2)}}
{k(k+1)}\right)\;$$
Now Let $$\displaystyle T_{k} = \cos^{-1}\left[\frac{1}{k}\cdot \frac{1}{k+1}+\frac{\sqrt{(k-1)k(k+1)(k+2)}}
{k(k+1)}\right]\;$$
Now Let $\displaystyle x=\frac{1}{k}$ and $\displaystyle y=\frac{1}{k+1}$
So $$\displaystyle \sqrt{1-x^2} = \sqrt{1-\frac{1}{k^2}} = \frac{\sqrt{(k-1)(k+1)}}{k}$$ and
$$\displaystyle \sqrt{1-y^2} = \sqrt{1-\frac{1}{(k+1)^2}} = \frac{\sqrt{k(k+2)}}{k+1}$$
So $\displaystyle T_{k}$ is of the form $$\cos^{-1}\left(xy+\sqrt{1-x^2}\sqrt{1-y^2}\right) = \cos^{-1}(y)-\cos^{-1}(x)\;,$$ because $(y<x)$
Now $$\displaystyle T_{k} = \cos^{-1}\left(\frac{1}{k+1}\right)-\cos^{-1}\left(\frac{1}{k}\right)$$
So $$\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=2}T_{k} = \lim_{n\rightarrow \infty}\left[\cos^{-1}\left(\frac{1}{n+1}\right)-\cos^{-1}\left(\frac{1}{2}\right)\right]$$
So we get $$\displaystyle \sum^{n}_{k=2}\cos^{-1}\left(\frac{1+\sqrt{(k-1)k(k+1)(k+2)}}
{k(k+1)}\right)= \cos^{-1}(0)-\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2}-\frac{\pi}{3} = \frac{\pi}{6}$$
So $$\displaystyle S_{n} = \frac{\pi}{6} = \frac{120\pi}{m}\;,$$ We get $$m=720$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
integer solution of $\frac{x^3+y^3+z^3-xy(x+y)-yz(y+z)-xz(x+z)-2xyz}{(x+y+z)(x+y-z)(x-y+z)(x-y-z)}=\frac{1}{2016}$ Let $x,y,z$ be positive integers such that $\frac{x^3+y^3+z^3-xy(x+y)-yz(y+z)-xz(x+z)-2xyz}{(x+y+z)(x+y-z)(x-y+z)(x-y-z)}=\frac{1}{2016}$.
How to find all solutions ?
I have no any idea.
Thanks in advance.
| This method uses Pythagorean triples $x^2+y^2=z^2$ to solve,
$$\frac{1}{-x+y+z}+\frac{1}{x-y+z}+\frac{1}{x+y-z}+\frac{1}{x+y+z}=\frac{1}{N}\tag1$$
It does not give all the solutions, but does give an easy way to tackle general $N$.
Let $x\leq y\leq z$. The method is based on the fortuitous observation that some $x,y,z$ (of a higher search range) were Pythagorean triples. Substitute,
$$x,\,y,\,z = a^2-b^2,\;2ab,\;a^2+b^2$$
into $(1)$, and one gets a much simpler equation,
$$(a-b)b = N\tag2$$
Thus, quick solutions can be found using the divisors of $N$. For $N = 1008$, this has thirty, namely,
$$b = \color{brown}{1}, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 28, 36, 42, 48, 56, 63, 72, 84, 112, 126, 144, 168, 252, 336, 504, \color{brown}{1008}.$$
For example, using $a,b = 1009,\color{brown}1$ yields,
$$x,\,y,\,z = 1018080,\; 2018,\; 1018082$$
while $a,b = 1009,\color{brown}{1008}$ gives,
$$x,\,y,\,z = 2017,\; 2034144,\; 2034145$$
and so on for thirty solutions to $(1)$ that also obey $x^2+y^2=z^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Derivative of $\sqrt{\frac{x-1}{x+1}}$ $$f(x)=\sqrt{\frac{x-1}{x+1}}$$
\begin{align}
f'(x) & ={1\over 2} \left(\frac{x-1}{x+1}\right)^{\frac{-1}{2}} \cdot {x+1-(x-1)\over (x+1)^2} \\[10pt]
& =\frac{1}{2}\left(\frac{x-1}{x+1}\right)^{\frac{-1}{2}}\cdot{2\over (x+1)^2} =\left(\frac{x-1}{x+1}\right)^{\frac{-1}{2}}\cdot{1\over (x+1)^2}
\end{align}
Is it valid to write $ (\frac{x-1}{x+1})^{\frac{-1}{2}}= \sqrt{\frac{x+1}{x-1}}$ ?
| Notice, you need to apply some constraints for the variable $x$,
for all $x>1$, it will be valid to write $$\left(\frac{x-1}{x+1}\right)^{-1/2}=\sqrt{\frac{x+1}{x-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1557071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
After solving for eigenvalues, how do you solve for eigenvectors if your matrix has free variables? Given the matrix
$$\begin{pmatrix} 3 & 0& 0 \\ -3& 4& 9 \\ 0 & 0& 3 \end{pmatrix}$$
you get eigenvalues $3$ (twice) and $4$. However, when solving for the eigenvector of $3$, you end up with
$$\begin{pmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$
How do you solve for the eigenvectors with the free variables?
| Your caracteristic equation is $(3-\lambda)^2(4-\lambda)$ so we need to find two vectors such that
$\begin{bmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{bmatrix}v = 0$
Fortunately, you have the freedom do to that.
I think the easiest way to do these is to plug one of your entries equal to 0, and then plug a different entry equal to 0
$\begin{bmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}v_1\\v_2\\0\end{bmatrix} = -3u_1 + u_2 = 0$
$u_2 = 3u_1$
$u = \begin{bmatrix}1\\3\\0\end{bmatrix}$
then
$\begin{bmatrix} 0 & 0 & 0 \\ -3 & 1 & 9 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}v_1\\0\\v_3\end{bmatrix} = -3v_1 + 9v_3 = 0$
$v_1 = 3v_3$
$v = \begin{bmatrix}3\\0\\1\end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1557775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Adding vector to span a space
Let $A=\left\{\left(\begin{array}{c} 18 \\ 6 \\ -4 \\ 12 \end{array}\right),\left(\begin{array}{c} 6 \\ 2 \\ 2 \\ -6 \end{array}\right)\right\}$
find the vectos to be added so A will span $\mathbb{R}^4$?
So what I did is :
\begin{pmatrix}
18 & 6 & -4 & 12 \\
6 & 2 & 2 & -6 \\
\end{pmatrix}
the row reduced echelon form is:
\begin{pmatrix}
1 & \frac{1}{3} & 0 & 0 \\
0 & 0 & 1 & -10 \\
\end{pmatrix}
What should I do next? I need to write the vectors that should be added with 1 and 0 only?
| Linear independence
First, we must check to see if our initial two vectors are linearly independent. In this instance, the signs provide essential clues. The sign patterns reveal they aren't proportional:
$$
\begin{array}{cccc}
+ & + & - & + \\
+ & + & + & - \\
\end{array}
$$
Gram-Schmidt orthogonalization
The Gram-Schmidt orthogonalization of the vector set
$$
\left\{ \,
% 1
\left[
\begin{array}{r}
18 \\ 6 \\ -4 \\ 12 \\
\end{array}
\right],
% 2
\left[
\begin{array}{r}
6 \\ 2 \\ 2 \\ -6 \\
\end{array}
\right],
% 3
\left[
\begin{array}{c}
1 \\ 0 \\ 0 \\ 0 \\
\end{array}
\right],
% 4
\left[
\begin{array}{r}
0 \\ 0 \\ 0 \\ 1 \\
\end{array}
\right]
%
\, \right\}
$$
provides one solution
$$
\mathbb{R}^{4} = \text{span }
\left\{ \,
% 1
\frac{1}{2\sqrt{130}}
\left[
\begin{array}{r}
9 \\ 3 \\ -2 \\ 6 \\
\end{array}
\right],
% 2
\frac{1}{\sqrt{130}}
\left[
\begin{array}{r}
6 \\ 2 \\ 3 \\ -9 \\
\end{array}
\right],
% 3
\frac{1}{\sqrt{10}}
\left[
\begin{array}{r}
1 \\ -3 \\ 0 \\ 0 \\
\end{array}
\right],
% 4
\frac{1}{\sqrt{10}}
\left[
\begin{array}{r}
0 \\ 0 \\ 1 \\ -3 \\
\end{array}
\right]
%
\, \right\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding $\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$ $$\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$$
I'm trying to understand trigonometric substitution better, because I never could get a good handle on it. All I know is that this integral is supposed to reduce to the integral of some power of cosine. I tried $x^2=\tan\theta$, but I ended up with $\sin\theta\cos^3\theta$ as my integrand. Can someone explain how to compute this?
| Notice $$\int_{-\infty}^\infty \frac{x^2}{x^4+1} dx
= 2 \int_0^\infty \frac{x^2}{x^4+1} dx
= 2 \int_0^\infty \frac{dx}{x^2+x^{-2}}
= 2 \int_0^\infty \frac{dx}{(x-x^{-1})^2+2}\\
= 2 \left(\color{red}{\int_0^1} + \color{blue}{\int_1^\infty}\right) \frac{dx}{(x-x^{-1})^2+2}
$$
Change variable from $x$ to $\frac1x$ for that part of integral on $(0,1)$,
this becomes
$$2\int_1^\infty \frac{1}{(x-x^{-1})^2+2}(\color{blue}{1}+\color{red}{x^{-2}})dx
= 2\int_1^\infty \frac{d(x-x^{-1})}{(x-x^{-1})^2+2} $$
Change variable once more to $y = x-x^{-1}$ and then to $y = \sqrt{2}z$, we get
$$\int_{-\infty}^\infty \frac{x^2}{x^4+1} dx
= 2 \int_0^\infty \frac{dy}{y^2+2}
= \int_{-\infty}^\infty \frac{dy}{y^2+2}
= \frac{1}{\sqrt{2}}\int_{-\infty}^{\infty} \frac{dz}{1+z^2}
= \frac{\pi}{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 13,
"answer_id": 2
} |
If $ \sum_{r=1}^{n} t_r $ = $ { n ( n + 1 ) ( n + 2 ) ( n + 3 ) } \over {8} $ , then what does $ \sum_{r=1}^{n} {1\over {t_r} } $ equal to? $ \sum_{r=1}^{n} t_r $ = $ { n ( n + 1 ) ( n + 2 ) ( n + 3 ) } \over {8} $
$ \sum_{r=1}^{n} {1\over {t_r} } $ = ?
My progress so far :
$ t_1 = {{1*2*3*4 } \over 8 } = 3 $
$ t_2 = 15 $
$ t_3 = 45 $
$ t_4 = 105 $
$ t_5 = 210 $
$ \sum_{r=1}^{n} {1\over {t_r} } $ = 1/3 + 1/12 + 1/30 + 1/60 + 1/105 .... So on , but I dont get a pattern .
How do I solve this ?
| You can verify the series simplification at the bottom via telescoping or induction.
$$\sum_{r=1}^{n}t_r=\frac{n(n+1)(n+2)(n+3)}{8}$$
$$\implies t_n=\sum_{r=1}^{n}t_r-\sum_{r=1}^{n-1}t_r=\frac{n(n+1)(n+2)(n+3)}{8}-\frac{(n-1)n(n+1)(n+2)}{8}=\frac{n(n+1)(n+2)}{8}\big((n+3)-(n-1)\big)=\frac{n(n+1)(n+2)}{2}$$
$$\implies \frac{1}{t_r}=\frac{2}{r(r+1)(r+2)}=\frac{1}{r}-\frac{2}{r+1}+\frac{1}{r+2}$$
$$\implies \sum_{r=1}^n\frac{1}{t_r}=\sum_{r=1}^n\frac{1}{r}-2\sum_{r=1}^n\frac{1}{r+1}+\sum_{r=1}^n\frac{1}{r+2}=\frac{n(n+3)}{2(n+1)(n+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the distribution law Throwing a cube twice,
let $X$ be the sum of the two throws
Find the distribution law
My attempt
$$
\begin{array}{c|lcr}
&2&3&4&5&6&7&8&9&10&11&12 \\
\hline
\text{P}_\text{X}(x) & 1/12 & 1/12 & 1/12&1/12&1/12&1/12&1/12&1/12&1/12&1/12&1/12 \\
\end{array}
$$
But shouldn't the sum be $1$? Currently the sum is $11/12$
| What you've tried is wrong:
You can get $2$ just getting a one in each die (1+1), so the probability is $\frac{1}{6} \frac{1}{6} = \frac{1}{36}$. Instead, you have more options to get $5$ (1+4,2+3,3+2,4+1), giving you a total probability of $\frac{2}{18}$
Doing this process for all the numbers you get the correct result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How can I solve complex number equation? $-\frac12·\frac{iz-2z}{z^2}+\frac{-1-i}{2i-2z}=
\frac{\frac{-3}4-\frac12i}{z}$
if $z=a+bi$,
How to find $a$ and $b$?
Thank you.
| The procedure is the same as solving a non-complex equation. First one has to set the domain to be $\mathbb{C} \setminus \{0,i\} $, else you can get a division by zero.
Then we multiply the equation with $2z^2(i-z)$ in order to get rid of the fractions. So we get
\begin{align*}
(2z-iz)(i-z) + (-1-i)z^2 &= -2z(i-z)(\dfrac{3}{4} + \dfrac{1}{2}i)\\
(-2z^2 + 2iz + z +iz^2) + (-z^2-iz^2) &= (z^2-iz)(\dfrac{3}{2} + i)\\
-3z^2+z + 2iz &= \frac{3}{2}z^2 + iz^2 -\frac{3}{2}iz + z \\
\dfrac{9}{2}z^2+iz^2-\dfrac{7}{2}iz &= 0 \\
z^2(\dfrac{9}{2}+i)-z(\dfrac{7}{2}i) &= 0
\end{align*}
We know that the $z$ can't be zero from the solution domain from $z$, so we can divide the equation by z:
\begin{align*}
z(\dfrac{9}{2}+i)-\dfrac{7}{2}i &= 0 \\
z(\dfrac{9}{2}+i)&= \dfrac{7}{2}i \\
z &= \frac{\frac{7}{2}i}{\frac{9}{2} + i}
\end{align*}
This complex division should not be a problem for you and you get your solution.
In the case of a quadratic equation at the end: use the quadratic formula as you would with a non-complex equation. Just pay attention to the complex square-root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $ \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}} = 27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$. Is my solution correct? Find the roots of the following equation, if any:
$$
\frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}
=
27\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}.
$$
My approach:
The following constraints should hold jointly for x:
*
*$1-x^2\geq0\iff x\in[-1,1]$
*$1+x\geq0 \iff x\geq-1$
*$1-x\geq0 \iff x\leq1$
*$\sqrt{1+x}\ne\sqrt{1-x}\iff x\ne0$
Consequently, $x\in[-1,0)\cup(0,1]$ and for such x's I solve and I reach at the following equation
$$
\left(\frac{x}{1+\sqrt{1-x^2}}\right)^3=27\implies
\frac{x}{1+\sqrt{1-x^2}}=3,
$$
which gives
$$
x = 3 + 3\sqrt{1-x^2}\implies
3\sqrt{1-x^2} = x-3\implies
9-9x^2=x^2-6x+9\implies
10x^2-6x=0\implies
x(10x-6)=0,
$$
and thus $x=0$ or $x=\frac{3}{5}$. The first one is rejected, but $x=\frac{3}{5}\in[-1,0)\cup(0,1]$.
However, if we plug $x=\frac{3}{5}$ into the original equation, we find out that this is impossible. So we should reject $x=\frac{3}{5}$. The original equation is impossible.
Is my approach correct? Thank you very much in advance.
| The left-hand side is less than 1; the right-hand side is more than 27.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compute$\sum_{k=2}^{n+5}\frac {1}{k(k-1)}\binom {n+1}{k-2}2^k$ Compute$\sum_{k=2}^{n+5}\frac {1}{k(k-1)}\binom {n+1}{k-2}2^k$ .
Well, here how I tried and got stuck:
| Note that
$$\begin{align*}
\frac1{k(k-1)}\binom{n+1}{k-2}&=\frac{(n+1)!}{k(k-1)(k-2)!\big((n+1)-(k-2)\big)!}\\
&=\frac{(n+1)!}{k!(n+3-k)!}\\
&=\frac1{(n+3)(n+2)}\cdot\frac{(n+3)!}{k!(n+3-k)!}\\
&=\frac1{(n+3)(n+2)}\binom{n+3}k\;,
\end{align*}$$
so
$$\sum_{k=2}^{n+5}\frac1{k(k-1)}\binom{n+1}{k-2}2^k=\frac1{(n+3)(n+2)}\sum_{k=2}^{n+5}\binom{n+3}k2^k\;.$$
Now use the binomial theorem and make a couple of adjustments, a bit like what you did in the previous problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A simple expansion Where I am doing wrong? if any one can tell!!!
I have done this equation as
$$y ={x\over 1+(x-x^4+x^7)^3} = x [1+(x-x^4+x^7)^3]^{-1}$$
$$y= x[ 1-(x-x^4+x^7)^3+...]$$
$$y= x[ 1-(x^3+x^{12}+x^{21}-3x^6+3x^{9}+3x^9+3x^{15}+3x^{15}-3x^{18})+...] $$
$$y= x[ 1-x^3-x^{12}-x^{21}+3x^6-3x^9-3x^9-3x^{15}-3x^{15}+3x^{18}+...]$$
$$y= x-x^4+3x^7+...$$
But in book it is written as,
$$y= x-x^4+4x^7+...$$
| Since $\frac1{1+x}=1-x+x^2-x^3+\dots$, we need to include at least one more term to get the series correct to $x^7$.
$$
\frac{x}{1+\left(x-x^4+x^7\right)^3}=x-x^4+4x^7+O\left(x^{10}\right)
$$
Using two terms, that is, $\frac1{1+x}=1-x$, we get
$$
x\left(1-\left(x-x^4+x^7\right)^3\right)=x-x^4+3x^7+O\left(x^{10}\right)
$$
However, if we include three terms, that is, $\frac1{1+x}=1-x+x^2$, we get
$$
x\left(1-\left(x-x^4+x^7\right)^3+\left(x-x^4+x^7\right)^6\right)=x-x^4+4x^7+O\left(x^{10}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
solution of an improper integral. I was solving following improper integral:
$$
\int\limits_0^\frac{\pi}{2}\frac{log~x}{x^a}dx
$$
where $a<1$.
My attempt:
$0$ is the only point of discontinuity. So,
$\frac{log~x}{x^a}\leq \frac{x}{x^a} = \frac{1}{x^{a-1}}$.
Now $f(x) =\frac{1}{x^{a-1}}$. And this is convergent if $a-1<1$.
So given integral converges if $a<2$.
Is my attempt correct? Are the steps to proceed to solution justified? Kindly rectify if I am wrong somewher. Thanks for the help.
| Integrate by parts using
\begin{align}
&&f &= \log(x) \qquad &dg &= x^{-a}\\
&&df&=\frac{1}{x} \qquad &g &= \frac{x^{1-a}}{1-a}
\end{align}
Then
$$\int_{0}^{\frac{\pi}{2}} \frac{\log(x)}{x^a} dx = \frac{x^{1-a} \log(x)}{1-a} - \frac{1}{1-a} \int_{0}^{\frac{\pi}{2}} x^{-a}dx.$$
It yelds
$$\int_{0}^{\frac{\pi}{2}} \frac{\log(x)}{x^a} dx = x^{1-a} \left[\frac{ \log (x)}{1-a}-\frac{1}{(1-a)^2}\right] \Bigg{|}_{0}^{\frac{\pi}{2}} = \left(\frac{\pi}{2}\right)^{1-a} \left[\frac{ \log \left(\frac{\pi}{2}\right)}{1-a}-\frac{1}{(1-a)^2}\right] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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