Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solving the definite integral: $\int_0^1 x\sqrt{px + 1}\,dx$ with $p>0$?
How to integrate $$I = \int_0^1 x\sqrt{px + 1}\,dx$$ , $p>0, p\in\mathbb{R}$ ?
I tried this:
$$ t = \sqrt{px + 1} \implies x = \frac{t^2 - 1}{p}$$
$$ x = 0 \implies t = 1$$
$$ x = 1 \implies t = \sqrt{p+1}$$
so I have:
$$ I = \int_1^\sqrt{p+1} \... | You may write
$$
\begin{align}
I &= \int_0^1 x\sqrt{px + 1}\,dx\\\\
&= \frac1p\int_0^1 ((px+1)-1)\sqrt{px + 1}\,dx\\\\
&= \frac1p\int_0^1 (px+1)^{3/2}\,dx-\frac1p\int_0^1 (px+1)^{1/2}\,dx\,dx\\\\
&= \frac1{p^2}\int_0^1 (px+1)^{3/2}\,d(px+1)-\frac1{p^2}\int_0^1 (px+1)^{1/2}\,d(px+1)\\\\
&= \frac1{p^2}\left[\frac{(px+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Parameter Matrix Determinant $A=\begin{bmatrix} 7x+42&x-21&x-21&x-21&x-21\\x-21&7x+42&x-21&x-21&x-21\\x-21&x-21&7x+42&x-21&x-21\\x-21&x-21&x-21&7x+42&x-21\\x-21&x-21&x-21&x-21&7x+42\end{bmatrix}$
I am trying to find the determinant of $A$, so far without success. I tried reducing it to a triangular matrix but wasn't su... | Let me write $a=7x+42$ and $b=x-21$. Then by subtracting the first columns from all other columns we obtain
$$
\det(A) = \det\pmatrix{ a & b & b & b & b \\ b & a & b & b & b \\ b & b & a & b & b \\ b & b & b & a & b \\ b & b & b & b & a }
= \det\pmatrix{ a & b-a & b-a & b-a & b-a \\ b & a-b & 0 & 0 & 0 \\ b & 0 & a-b &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Sum of real solutions on equation $\sqrt{\sin^2{x} + {1 \over 2}} + \sqrt{\cos^2{x} + {1 \over 2}} = 2$ in interval $[0,2\pi]$ is? I know that solution is $4\pi$ but I do not know how do they get to this solution. I always get that $x \in R$ and that $-1 < \cos 2x < 1$ when converting it to double angle.
EDIT : So ok,... | Using the Cauchy-Schwarz inequality. which states that for any 2 vectors $( a, b )$ and $( c, d )$ with $a,b,c,d \ge 0$:
$$(ac + bd)^2 \le (a^2 + b^2)(c^2 + d^2) $$
Let $a = \sqrt{\sin^2 x + \frac{1}{2}}$, $b = \sqrt{\cos^2 x + \frac{1}{2}}$ and $c = d = 1$ we have
$$ \left(\sqrt{\sin^2 x + \frac{1}{2}} + \sqrt{\cos^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$ then quotient $\frac xy$ is equal to?
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$, then quotient $\frac xy$ is equal to?
Other conditions ... | $$2x+\sin3a+i(1-2x\sin3a)=y-\sin a+i(1+y\sin a)$$
Eauting the imaginary parts, $1-2x\sin3a=1+y\sin a\iff \dfrac xy=\dfrac{\sin a}{-2\sin3a}$
Now $\sin3a=\sin a(3-4\sin^2a)$ and $\cos2a=1-2\sin^2a\iff2\sin^2a=1-\cos2a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to integrate $\frac{x^2}{x^2+9}$? I know you can use division to get $1-\frac{p}{x^2+9}$ and then use the result for arctan, but I was trying to do this using substitution and nothing seems to work? Is there a substitution that could be used for this?
| If one chooses not to write $\frac{x^2}{x^2+9}=1-\frac{9}{x^2+9}$, then one can make still directly substitute $x=3 \tan y$ with $dx=3 \sec^2y \,dy$. Thus, we have
$$\begin{align}
\int \frac{x^2}{x^2+9}dx&=\int\frac{9\tan^2y}{9\tan^2y+9}3\sec^2y\,dy\\\\
&=3\int \tan^2y\,dy\\\\
&=3(\tan y-y)+C\\\\
&=x-3\arctan (x/3)+C
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve the follwing system of equations for $x, y$ and $z$ $$\frac{y+z}{5}=\frac{z+x}{8}=\frac{x+y}{9}$$ and $$6(x+y+z)=11$$
My teacher told me that I would have to get $3$ different equations to get $x, y$ and $z$. I've tried many methods and I'm confused as to how to do this problem.
| Hint
$$\frac{y+z}{5} =\frac{z+x}{8} = \frac{x+y}{9}$$ is the same as
$\color{red}{\large{\frac{y+z}{5} =\frac{z+x}{8}}}$ and $\color{red}{\large{\frac{z+x}{8} = \frac{x+y}{9}}}$
Which is also the same as $\color{blue}{\large{\frac{y+z}{5} = \frac{x+y}{9}}}$ and $\color{blue}{\large{\frac{z+x}{8} = \frac{y+z}{5}}}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find coefficient of $x^{12}$ in the expansion of $(1+x+x^2+x^3+...+x^n)^4$ How to find coefficient of $x^{12}$ in the expansion of $(1+x+x^2+x^3+...+x^n)^4$
I tried this : Since $(1+x+x^2+x^3+...+x^n)$ is in GP its sum will be $(x^{n+1}+1)(x-1)^{-1}$ now ACQ we have to expand $((x^{n+1}+1)(x-1)^{-1})^4$ next i a... | Continuing from where you left off but with a corrected formula for the geometric series, use the negative binomial theorem:
$$
(1-x)^{-n}=\sum_{k=0}^\infty \binom{n+k-1}{k} x^k
$$
In your case, we have
$$
(1-x)^{-4}=\sum_{k=0}^\infty \binom{k+3}{3} x^k = 1+\frac{2\cdot 3 \cdot 4}{1 \cdot 2\cdot 3}x + \frac{3\cdot 4 \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
find coordinate on line at given distance from given coordinate I got two coordinates of a straight line $(-2,-4)$ and $(3,4)$. How can i find a coordinate that lies on this line and is $5$ units away from the $(-2,-4)$ coordinate?
| The equation of the line passing through the points $(-2, -4)$ & $(3, 4)$ is given as
$$y-(-4)=\frac{4-(-4)}{3-(-2)}(x-(-2))$$ $$\implies y+4=\frac{8}{5}(x
+2)$$ $$\implies 5y+20=8x+16$$ $$\implies 8x-5y-4=0$$
Let the coordinates be $(p, q)$ on the line: $8x-5y-4=0$ Then this point will satisfy the equation of the lin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Alternative Quadratic Formula Well the formula for solving a Quadratic equation is :
$$\text{If }\space ax^2+bx+c=0$$
then
$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$
But looking at this : [Wolfram Mathworld] (And also in other places)
They give An Alternate Formula:
$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$
How does one... | Consider the quadratic $ax^2 + bx + c$. If $a = 0$, then finding the root is easy and boring, so we only look at the case when $a \neq 0$.
One way to get at the original quadratic formula is to complete the square. That is, note that
$$\begin{align}
ax^2 + bx + c &= a\left(x^2 + \frac{b}{a} x + \frac{c}{a}\right) \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Finding an isomorphism between polyomial quotient rings Let $F_1 = \mathbb{Z}_5[x]/(x^2+x+1)$ and $F_2 = \mathbb{Z}_5[x]/(x^2+3)$. Note neither $x^2+x+1$ nor $x^2+3$ has a root in $\mathbb{Z}_5$, so that each of the above are fields of order 25, and hence they are isomorphic from elementary vector space theory, howeve... | As per Jyrki Lahtonen's hint, I will instead try $\phi(x^2+x+1) = c(x^2+3)$. So then since $x \mapsto ax+b$, we have $a^2x^2 + (2ab+a)x + b^2+b+1 = cx^2 + 3c$, so $a^2 = c$, $2ab+a = 0$, and $b^2+b+1 = 3c$. Since $a^2 = c$ and the only squares mod 5 are 1 and 4, we have that $c = 1,4$. Having already gleaned that 1 wil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the last two digits of $33^{100}$
Find the last two digits of $33^{100}$
By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of @Lucian:
$33^2\equiv-11 \pmod{100}$ t... | Hint: $33^2\equiv-11\bmod100$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
Find the value of this series what is the value of this series $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots$$
I really tried, but I couldn't, help guys?
| Since $n^2 = 2\binom{n}{2}+\binom{n}{1}$ we have:
$$ S=\sum_{n\geq 1}\frac{n^2}{2^n}=\left.\left(2\frac{x^2}{(1-x)^3}+\frac{x}{(1-x)^2}\right)\right|_{x=\frac{1}{2}}=\color{red}{6}.$$
As an alternative to the negative binomial series, we may also use:
$$ S = 2S-S = \sum_{n\geq 1}\frac{n^2}{2^{n-1}}-\sum_{n\geq 1}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
What is the maximum value of the LCM of three numbers $\leq n$, as a function of $n$? Given $n \geq 3$, what maximum LCM of any three numbers $\leq n$ can we obtain?
Now, if $n$ is odd, the answer would be
$$n(n - 1)(n - 2)$$
because $\newcommand{\lcm}{\operatorname{lcm}}$
$$\begin{align*}
\lcm(a, b, c) &= \lcm(a, \lcm... | If $n$ is even and $\ge 4$, and not divisible by $3$, use $n(n-1)(n-3)$.
If $n$ is even and divisible by $3$, use $(n-1)(n-2)(n-3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Simultaneous equation with fractional solutions. How do you get to find $x$ when $y$ is a fraction ? Anyone mind to explain it step by step for the clearest explanation.=)
$$-7x +2y = 2$$
$$14x + 3y = -5$$
Answer: $x=?, y=-1/7$
| Inserting $y=-\frac{1}{7}$ in the first equation
$-7x-2\cdot \frac{1}{7}=2$
Transforming 2 into $\frac{14}{7}$ to have a common denominator with $2\cdot \frac{1}{7}$
$-7x-\frac{2}{7}=\frac{14}{7} \quad \quad |+\frac{2}{7}$
Adding $\frac{2}{7}$ on both sides.
$-7x\underbrace{-\frac{2}{7}+\color{blue}{\frac{2}{7}}}_{=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding $\sum\limits_{k=0}^n k^2$ using summation by parts Sorry to bother you guys again, but I still have some doubts. I do think I'm making some progress, though.
So, again, the formula that I'm using for summation by parts is
$\sum\limits_{k=o}^n f(k)g(k) = g(n)(\sum\limits_{k=0}^n f(k)) - \sum\limits_{k=0}^{n-1} ... | I would solve it like this: $p_3(n+1)^3 + p_2(n+1)^2 + p_1(n+1) + p_0 - (p_3n^3 + p_2n^2 + p_1n + p_0) = 1n^2$
Then treat the polynomial as a vector space, applying binomial theorem and writing on matrix form solving a linear equation system, we get:
$${\bf p} = \left[\begin{array}{r}p_3\\p_2\\p_1\\p_0\end{array}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Determine whether $\sum \frac{2^n + n^2 3^n}{6^n}$ converges For the series $$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n},$$ I was thinking of using the root test? so then I would get $(2+n^2/n+3)/6$ but how do I find the limit of this?
| $$\sum\limits_{n=1}^{\infty}\frac{2^n+n^2 3^n}{6^n}$$
I wouldn't recommend the root test for this series. However, here are the steps
$$r=\limsup\limits_{n\to\infty}\sqrt[n]{\left|\frac{2^n+n^2 3^n}{6^n}\right|}$$
$$=\limsup\limits_{n\to\infty}\frac{\sqrt[n]{2^n+n^2 3^n}}{\sqrt[n]{6^n}}$$
$$=\frac16\limsup\limits_{n\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Let $a$, $b$, and $c$ be positive real numbers. Let $a$, $b$, and $c$ be positive real numbers. Prove that
$$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$
Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?
| The expressions $a^2 - ab + b^2$, $a^2 - ac + c^2$, and $b^2 + bc + c^2$ should remind you of the law of cosines. Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be vectors such that $\vec{a}$ and $\vec{b}$ are separated by 60 degrees, and $\vec{a}$ and $\vec{c}$ are separated by 60 degrees. Consider the triangle formed by the thr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Parabolic range conditions proof This problem is getting the better of me, since I have no idea where to start:
The equation of a curve is $y=ax^2-2bx+c$, where a, b and c are constants with $a>0$.
Given that the vertex of the curve lies on the line $y=x$, find an expression for $c$ in terms of $a$ and $b$. Show that i... | $$c=\frac{b^2+b}{a}=\frac{4b^2+4b}{4a}=\frac{(4b^2+4b+1)-1}{4a}=\frac{(2b+1)^2-1}{4a}\ge -\frac{1}{4a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How can I find the minimum value for $F(x,y,z,w)=x^2+y^2+z^2+w^2+xy+zw-xz-yw-yz$ Let $x,y,z,w$ be integer numbers,and $xw=yz+1$
Find this minimum of the value
$$x^2+y^2+z^2+w^2+xy+zw-xz-yw-yz$$
This is how did it and I would like to know if I made a mistake
Let $$F(x,y,z,w)=x^2+y^2+z^2+w^2+xy+zw-xz-yw-yz$$
we have tha... | let $z=-v ,xw=1-yv,f=x^2+y^2+v^2+w^2+xy-vw+xv-yw+yv=(x-w)^2+2xw+(x-w)(y+v)+(y+v)^2-yv=(x-w)^2+(x-w)(y+v)+(y+v)^2+2-3yv, \\ yv \le \dfrac{(y+v)^2}{4} \implies -3yv \ge -\dfrac{3(y+v)^2}{4} \implies f \ge (x-w)^2+(x-w)(y+v)+\dfrac{(y+v)^2}{4}+2=(x-w+\dfrac{y+v}{2})^2+2 \ge 2 $
when $y=v ,x-w+y=0 \cap xw=1-y^2 ,f $ get mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Probability that no two consecutive heads occur?
A fair coin is tossed $10$ times. What is the probability that no two
consecutive tosses are heads?
Possibilities are (dont mind the number of terms):
$H TTTTTTH$, $HTHTHTHTHTHTHT$.
But except for those,
let $y(n)$ be the number of sequences that start with $T$
$T _$... | $F (n)$ is in fact a fibonacci sequence:
\begin{align}
F (n+1) &= 2 y (n) + y (n-1)\\
&= 2 [y (n-1) +y (n-2)] + y (n-2) + y (n-1)\\
&= 2 y (n-1) + y (n-2) + 2 y(n-2) +y (n-3)\\
&= F (n) +F (n-1)\\
\end{align}
Now $F (1) = |\{H,T\}|=2$
And $F (2) = |\{HT,TH,TT\}|=3$
So $ F (10) = 144$
There are $ 2^{10}=1024$ possible ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Differentiate the Function: $y=e^{k\ tan\sqrt{x}}$ $y=e^{k\tan\sqrt{x}}$
$=e^{k\tan\sqrt{x}}\cdot [{k\tan\sqrt{x}}]'$
$=e^{k\tan\sqrt{x}}\cdot\ (k)\cdot[\tan x^{\frac{1}{2}}]'+(\tan x^{\frac{1}{2}})\cdot[k]'$
$=e^{k\tan\sqrt{x}}\cdot\ (k)\cdot (\frac{1}{2}tanx)\cdot \sec^2x + (\tan x^{\frac{1}{2}})\cdot (1)$
Is my proc... | Your work is not quite correct.
$$\begin{align}
(e^{k\tan\sqrt x})' &= e^{k\tan\sqrt x}\cdot(k\tan\sqrt x)' \\
&= e^{k\tan\sqrt x}\cdot k\cdot (\tan\sqrt x)' \\
&= e^{k\tan\sqrt x}\cdot k\cdot\sec^2\sqrt x\cdot (\sqrt x)' \\
&= e^{k\tan\sqrt x}\cdot k\cdot\sec^2\sqrt x\cdot \frac 1{2\sqrt x} \\
&= \frac{k\sec^2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the roots of the summed polynomial
Find the roots of: $$x^7 + x^5 + x^4 + x^3 + x^2 + 1 = 0$$
I got that:
$$\frac{1 - x^8}{1-x} - x^6 - x = 0$$
But that doesnt make it any easier.
| Clever factoring: $x^7+x^5+x^4+x^3+x^2+1 =(x+1)(x^2 - x + 1)(x^4+x^2+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Help with this Trigonometry integral I've to find this integral:
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx$$
So far as I know to go:
$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\p... | $$\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx$$
$$\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$
$$\int_{\frac{\pi}{365}}^{\frac{365}{\pi}}\left(4^\pi\cdot \frac{1}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
Let $0\le a\le b\le c,abc=1$, then show that
$$a+b^2+c^3\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}$$
Things I have tried so far:
$$\dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}=\dfrac{b^2c^3+ac^3+ab^2}{bc^2}$$
Since $abc=1$, it suffices to ... | hint: $c\ge 1, a \le 1$
when $b\le 1 $. replace $a$,get $f(c)=$LHS-RHS,prove $f'(c)\ge0$ with
$ bc=\dfrac{1}{a} \ge 1,$ then use $ c\ge \dfrac{1}{b^2} $ to prove $f(c=\dfrac{1}{b^2} )\ge 0$
Edit: here is full solution:
case 1: $b \le 1, c \ge \dfrac{1}{b^2},bc=\dfrac{1}{a} \ge 1$
LHS-RHS$=b^2c^6-b^3c^4+b^4c^3-c^3+bc^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove that if $a^2+b^2$ is a multiple of three, then a and b are multiples of three I have attempted to prove the above. I am uncertain about the correctness of my proof:
Both numbers have to be multiples of three, i.e. $3a+3b=3n$, $\ 3(a+b)=3n$
It is not possible to arrive at an integer that is a multiple of three wit... | Working in $\pmod{3}$ we can just break this into cases.
If $a\equiv b \equiv 0\pmod 3$, then $a^2+b^2\equiv 0 \pmod 3$
If $a=0$, $b=1$,then $a^2+b^2 \equiv 1 \pmod 3$, and same if $a$ and $b$ are switched.
If $a=0$, $b=2$, then $a^2+b^2 \equiv 1 \pmod 3$
Just fill in the rest and note that the only time $a^2+b^2 \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
The maximum and minimum values of the expression Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$
My try:I have just normally squared the expression and got
$u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2... | For $\min$
Using $\triangle$ Inequality::
Let $z_{1} = a\cos x+i b\sin x$ and $z_{2} = b\cos x+i a\sin x$
So $$|z_{1}|+|z_{2}|\geq |z_{1}+z_{2}|$$
So $$\sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\geq \sqrt{(a+b)\cos^2 x+(a+b)^2\sin^2 x}=|a+b|$$
For $\max$ Same as Deepsea
$$\left[\left(\sqrt{a^2\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
How can I prove the following equality I have the following equality :
$$I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}$$
$$I_2=-\frac{ab}{2\pi}\int_0^\pi \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=0$$
where $0 <b \leq a$.
I used the residues but I could not prov... | Let $$I=I_1+iI_2.$$
Then
\begin{eqnarray}
I&=&-\frac{ab}{2\pi}\int_0^\pi \frac{e^{2it}}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\
&=&-\frac{ab}{4\pi}\int_0^{2\pi} \frac{e^{it}}{a^2\frac{1-\cos t}{2}+b^2\frac{1+\cos t}{2}}dt\\
&=&-\frac{ab}{2\pi}\int_0^{2\pi} \frac{e^{it}}{(a^2+b^2)+(b^2-a^2)\cos t}dt\\
&=&-\frac{ab}{2\pi i}\int_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$
| \begin{align}
x&=\sqrt[3]{17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}
\end{align}
Note that
\begin{align}
17\sqrt{5}-38&=\frac{1}{17\sqrt{5}+38}.
\end{align}
Let $a=\sqrt[3]{17\sqrt{5}+38}$.
Then we have
\begin{align}
x^3&=\left(a-\frac1a\right)^3
\\
x^3&= a^3-3a+\frac3a-\frac{1}{a^3}
\\
x^3+3\left(a-\frac1a\right)&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Nature of the roots of quadratic equation Here is the problem that I need to prove:
If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$
Here is what I did:
\begin{align*}
p(2x-1)=3(x^2+1) \\
3x^2 - 2px + (p+3)=0 \\
b^2 - 4ac = 4(p^2-3(p+3))
\end{align*}
By inspection I c... | It must be $b^2-4ac\geq 0$ because it is given $x$ is real so equation must have real roots which may be either equal or unequal
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Vector equation with no solution There exists a real number $k$ such that the equation
$\begin{pmatrix} -1 \\ -2 \end{pmatrix} + t\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} + s\begin{pmatrix} -4 \\ k \end{pmatrix}$
does not have any solutions in $t$ and $s$. Find $k$.
I am confused as ... | Given $$\begin{bmatrix}-1 \\ -2\end{bmatrix} + t\begin{bmatrix}3\\-2\end{bmatrix} = \begin{bmatrix}5 \\ 0\end{bmatrix} + s\begin{bmatrix}-4\\k\end{bmatrix}$$ implies
$$-s\begin{bmatrix}-4\\k\end{bmatrix} + t\begin{bmatrix}3 \\- 2\end{bmatrix} = \begin{bmatrix}5\\0\end{bmatrix} - \begin{bmatrix}-1\\-2\end{bmatrix}$$ whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prime factorization number theory Let $n$ be a positive integer, and let
$ 1=d_1<d_2<\dots <d_6=n $
be all of its divisors. Find all $n$ that satisfy
$ \frac 1{d_1} +\frac 1{d_2} + \dots + \frac 1{d_6 } = 2. $
I started by noting $n=a^2b$ where $a,b$ are primes, and tried to manipulate the second equation. Not getting ... | As you wrote, $n$ can be expressed as $n=a^2b$ where $a,b$ are distinct primes. Now we have
$$\frac{1}{1}+\frac{1}{a}+\frac{1}{b}+\frac{1}{a^2}+\frac{1}{ab}+\frac{1}{a^2b}=2,$$
i.e.
$$a^2b+ab+a^2+b+a+1=2a^2b,$$
i.e.
$$b=1+\frac{2a+2}{a^2-a-1}$$
Here, since $(2a+2)/(a^2-a-1)\lt 1$ for $a\ge 4$, we have $a=2,3$.
Thus, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Fixed points of $\phi_a(z)$
Prove that $\phi_a(z)=\frac{a-z}{1-\bar az}$ , $0<|a|<1$ has exactly two fixed points ; one inside the unit disc and the other outside the unit disc.
Putting $\phi_a(z)=z$ I find that there are exactly two fixed points which are $$\frac{1\pm \sqrt{1-|a|^2}}{\bar a}.$$
I am unable to find o... | $z_1\overline{a}=1-\sqrt{1-x^2}, x = |a|\to (|z_1||\overline{a}|)^2 =1-2\sqrt{1-x^2}+1-x^2\to bx^2= 2-x^2-2\sqrt{1-x^2}$. Thus we can prove: $2-x^2-2\sqrt{1-x^2} < x^2 \iff 1-x^2 < \sqrt{1-x^2}$ which is true since $0 < 1-x^2 < 1$. Thus $b = |z_1| < 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is this a correct solution to the linear congruence? I want to solve this linear congruence:
$$2x \equiv 5 \pmod{9}$$
Backward substitution:
$$9 = 4 \cdot 2 + 1$$
$$4(-2) + 9 = 1$$
Therefore, the inverse is: $-2$
Now multiply the linear congruence with $-2$
$$(2)(-2)x \equiv (-2)5 \pmod{9}$$
$$x \equiv -10 \pmod{9}$$
S... | $$
2x\equiv 5\pmod 9 \Longrightarrow 2x=9n + 5, n\in\mathbb Z
$$
We can solve linear diophantine equation now, but LHS divides by $2$; hence,
$$
0\equiv 9n + 5 \equiv 1 \cdot n + 1\pmod 2\Longrightarrow n = 2m + 1, m\in\mathbb Z
$$
So,
$$
2x=9(2m+1)+5\Longrightarrow x = 9m + 7
$$
If you know that $2^{-1}\equiv 5 \pmod ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve triangle point given base, point height and difference of sides I have the intuition that one should be able to calculate the position of the circle in the image below (or the equivalent, solve a and b).
We have the following information:
h and d is known as well as the difference between the lengths a and b.
Loo... | Let $a=x,b-a=t$. Here, note that $b=x+t$ and that $b\lt a+d\Rightarrow t\lt d$.
Then, by the Pythagorean theorem for the biggest triangle, we have
$$\begin{align}&(x+t)^2=h^2+\left(\sqrt{x^2-h^2}+d\right)^2\\&\Rightarrow x^2+2tx+t^2=h^2+x^2-h^2+2d\sqrt{x^2-h^2}+d^2\\&\Rightarrow 2tx+t^2-d^2=2d\sqrt{x^2-h^2}\\&\Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a system of five polynomials I am trying to solve the following system of equations for tuple $\left(a,b,c,d,t\right) \in \mathbb{R}^{4} \times [0,1]$, with parameter $\ell\in\mathbb{R}$.
$$
\begin{eqnarray}
a\frac{t^{2}}{2} - bt + 1 = 0 \qquad (1)\\
a \frac{t^{3}}{6} - b \frac{t^{2}}{2} + t - \ell = 0 \qquad (... | Seems like a perfect opportunity to use computer algebra (such as Mathematica), where one imposes the range of values of $l$ explicitly:
Assuming[1/6 <= l <= 1/4,
Solve[{a t^2/2 - b t + 1 == 0, a t^3/6 - b t^2/2 + t - l == 0,
c t^2/2 - d t + (d - c/2 - 1) == 0,
c t^3/6 - d t^2/2 + (d - c/2 - 1) t + (c/3 - d/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find th... | Required,
$= a/b + b/c + c/a$
By Cross multiplication,
$= (a^2bc+b^2ac+c^2ab)/(abc)$
$= abc (a+b+c) /(abc)$
$ = (a+b+c) ..........1$
Well known
Property Relation:-
{If α1, α2,α3 ... αn are the roots of the equation
$f(x)= a_0x_n +a_1x_{n-1} +a_2x_{n-2} +...+a_{n-1}x + a_n =0$, then
$f(x)= a_0 (x-α_1)(x-α_2)(x-α... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
The expression $(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ where $q\ne 1$, equals The expression
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$
where $q\ne 1$, equals
(A) $\frac{1-q^{128}}{1-q}$
(B) $\frac{1-q^{64}}{1-q}$
(C) $\frac{1-q^{2^{1+2+\dots +6}}}{1-q}$
(D) none of the fore... | Let $$P = (1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64}).$$
One has $$(1-q)P = (1-q)(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})
= 1 - q^{128}.
$$
So, one gets $$P = \frac{1-q^{128}}{1-q}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Solution of a quadratic diophantine equation I try to solve the Diophantine quadratic equation:
$$X^2+Y^2+Z^2=3W^2.$$ Obviously, there is a non-trivial solution: $(1,1,1,1)$. So I tried to apply Jagy's method: Solutions to $ax^2 + by^2 = cz^2$ .
I consider integers $t,p,q,r$ and the point $P=(1,1,1)$ of the sphere $X^2... | From above, the below mentioned equation has parametric solution,
$x^2+y^2+z^2=3w^2$
$x=k^2-6k+1$
$y=k^2-2k+5$
$z=5k^2-2k+1$
$w=3k^2-2k+3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_0^1\frac{\log(x)\log(1+x)}{\sqrt{1-x}}\,dx$ I'm trying to evaluate this definite integral:
$$\int_0^1\frac{\log(x) \log(1+x)}{\sqrt{1-x}} dx$$
It's clear that the result can be expressed in terms of derivatives of a hypergeometric function with respect to its parameters. I obtained the following form:
$$... | The integral may readily be decomposed into a sum of integrals of products of log-linear terms:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\frac{\ln{\left(x\right)}\ln{\left(1+x\right)}}{\sqrt{1-x}}\,\mathrm{d}x\\
&=2\int_{0}^{1}\ln{\left(1-y^2\right)}\ln{\left(2-y^2\right)}\,\mathrm{d}y;~~~\small{\left[\sqrt{1-x}=y\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem :
Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.
My approach:
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But ... | For a method which is intermediate in cleverness between the one given by the other answers and just solving the system of simultaneous equations, you could use successive differences. We write the following table:
$$
\begin{array}{ccccccccccccccc}
h & & 1 & & 2 & & 3 & & 4 & & 5 & & 6 & & x_6 \\
& 1-h & & 1 & & 1 & & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 0
} |
Complex number identity by trigonometry Show that $\lvert e^{i\theta} - 1\rvert = 2\lvert\sin(\theta/2)\rvert$ by using the geometry of the triangle with vertices 0, 1, and the midpoint of the line joining 0 and $e^{i\theta}$.
I have been able to show this identity through other means, however I am stuck on how to util... | $|e^{i\theta}-1|=|\cos\theta+i\sin\theta-(1+0i)|=|(\cos\theta-1)+i\sin\theta|=\sqrt{(\cos\theta-1)^2+(\sin\theta)^2}=\sqrt{\cos^2\theta+1-2\cos\theta+\sin^2\theta}=\sqrt{2-2\cos\theta}=\sqrt{2(1-\cos\theta)}=\sqrt{2(\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}-\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2})}=\sqrt{4sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
| We have, $$\underbrace{\frac {1}{1+\sqrt2+\sqrt3} + \frac {1}{1-\sqrt2+\sqrt3}} + \underbrace{\frac {1}{1+\sqrt2-\sqrt3} + \frac {1}{1-\sqrt2-\sqrt3}}$$ $$=\left(\frac {1}{1+\sqrt2+\sqrt3} + \frac {1}{1-\sqrt2+\sqrt3}\right) +\left( \frac {1}{1+\sqrt2-\sqrt3} + \frac {1}{1-\sqrt2-\sqrt3}\right)$$ $$=\left(\frac {1-\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Modular Quadratic Equation I'm trying to solve that equation:
$x^2-3x-5\equiv0\pmod{343}$
I've completed the square as follows:
$x^2-3x-5 \equiv x^2+340x-5\equiv(x+170)^2-170^2-5\pmod{343}\\
(x+170)^2 \equiv 93\pmod{343}\\
y^2 \equiv 93 \pmod{343}$
But I have no idea how to move on. How can I use the fact that $343=7^3... | $$y^2\equiv93\pmod{343}\implies y^2\equiv93\pmod{49}\implies y^2\equiv93\pmod{7}$$
From the last congruence we get $y\equiv\pm3\pmod 7$ or $y=7z\pm 3$. Now, plug this into the second congruence to get
$$\pm21z+9\equiv-5\pmod{49}$$
or
$$\pm3z\equiv-2\pmod 7\iff z\equiv\mp3\pmod 7$$
so $z=7u\mp 3$ and $y=7(7u\mp 3)\pm 3=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sum of cosines of complementary/suplementary angles Why are $(\cos(2^{\circ})+\cos(178^{\circ})), (\cos(4^{\circ})+\cos(176^{\circ})),.., (\cos(44^{\circ})+\cos(46^{\circ}))$ all equal zero?
Could you prove it by some identity?
| Notice, the following trigonometric equation $$\color{blue}{\cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}$$ Hence, if angles $A$ & $B$ are supplementary angles i.e. $A+B=180^\circ$ $$\implies \cos\left(\frac{A+B}{2}\right)=\cos\left(\frac{180^\circ}{2}\right)=\cos\left(90^\circ\right)=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Find the area of a triangle given the radius of its incircle and a tangential point A friend gave me recently the following interesting problem and I would like to share a couple of solutions. Any additional contributions are welcome.
A triangle $\vartriangle ABC$ is given and we know the radius $r$ of its incircle $(O... | Solution 1: Draw $AO$, $AO$ and $CO$. We know that these are the bisectors of $\angle BAC$, $\angle ABC$ and $\angle BCA$ respectively.
Let $\angle OAC = \frac{1}{2} \angle BAC =: \phi_A$ and $\angle OBE = \frac{1}{2} \angle ABC =: \phi_B$ and $\angle OCA = \frac{1}{2}\angle BCA =: \phi_C$.
We have that $\angle ODA$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Does $\frac{x+y}{2}>\frac{a+b}{2}$ hold? $a$ and $b$ are two real positive numbers. Given that $x=\sqrt{ab}$ and $y=\sqrt{\frac{a^2+b^2}{2}}$, which one has a higher value, $\frac{x+y}{2}$ or $\frac{a+b}{2}$?
We know that $y=\sqrt{\frac{a^2+b^2}{2}}>\frac{a+b}{2}>x=\sqrt{ab}$ by inequality, and at this point I'm stuck.... | we know $$\sqrt{\frac{x^2+y^2}{2}} \geq \frac{x+y}{2} $$ so plug x,y $$ \sqrt{\frac{x^2+y^2}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{(\sqrt{ab})^2+(\sqrt{\frac{a^2+b^2}{2}})^2}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{ab+\frac{a^2+b^2}{2}}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{\frac{a^2+b^2+2ab}{2}}{2}} \geq \frac{x+y}{2}\\\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to integrate $\int_{l1}^{l2}\frac{e^{\pm i a x}}{\sqrt{bx^2+cx+d}}dx$ I have the above mentioned integral
$$
\int_{l_1}^{l_2}\frac{e^{\pm i a x}}{\sqrt{bx^2+cx+d}}dx
$$
which I want to solve. I expect some special functions in its solution, but so far I am out of ideas now.
I have tried the following:
I substitute ... | Too long for a comment (a work in progress)
$$
\int \frac{\mathrm{e}^{\pm iax}}{\sqrt{bx^2+cx+d}}dx
$$
Focusing on the denominator
$$
b\left(x^2+\frac{c}{b}x+\frac{d}{b}\right) =b\left[\left(x+\frac{c}{2b}\right)^2+\frac{d}{b}-\left(\frac{c}{2b}\right)^2\right]
$$
lets change variables
$$
t = \frac{x+\frac{c}{2b}}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $P$ such that $P^TAP$ is a diagonal matrix
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_2(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
So here's the solution:
$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \sim \left(\begin{ar... | Hermite Reduction.
SEE ALSO Orthogonal basis for this indefinite symmetric bilinear form
Transforming quadratic forms, how is this theorem called?
What is the difference between using $PAP^{-1}$ and $PAP^{T}$ to diagonalize a matrix?
When you have a symmetric matrix of integers, you may use Hermite's method for diagona... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
simplifying complex fractions I have this expression
${2 \over x^2 - 4}$ + ${1 \over x + 2}$
So I would take the LCM to ${x^2 - 4}$ or using the difference of 2 squares to be ${(x + 2)(x - 2)}$
So I would simplify this to:
${2 + (x - 2) \over (x - 2)(x + 2)}$ which would simplify further to
${2\over x + 2}$
But the ans... | Notice, we have $$\frac{2}{x^2-4}+\frac{1}{x+2}$$ $\color{red}{\text{apply} \ a^2-b^2=(a-b)(a+b)}$ $$=\frac{2}{(x-2)(x+2)}+\frac{1}{x+2}$$ $\color{red}{\text{take L.C.M of denominators }}$ $$=\frac{2+x-2}{(x-2)(x+2)}$$$$ =\frac{x}{(x-2)(x+2)}$$ Thus we have following simplified form $$\bbox[5px, border: 2px solid #C0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How does this algebraic trick regarding partial fraction works? Suppose I have to evaluate the integral $$\int \frac{x}{(x-1)(2x+1)(x+3)} \, dx $$
I write it as $$\frac{a_1}{x-1} +\frac{a_2}{2x+1} +\frac{a_3}{x+3}$$
where $a_1$, $a_2$, $a_3$ are constants. once I have foud relationaships between them I solve for the $... | As you wrote, $$\frac x {(x-1)(2x+1)(x+3)}=\frac {a_1}{x-1}+\frac {a_2}{2x+1}+\frac {a_3}{x+3}$$ As Peter commented, multiply everything by $(x-1)(2x+1)(x+3)$; this gives $$x=a_1(2x+1)(x+3)+a_2(x-1)(x+3)+a_3(x-1)(2x+1)$$ First make $x=1$ so $$1=a_1\times3\times 4=12a_1$$ Do the same with $x=-3$ and $x=-\frac 12$. You i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$
and derivative of denominator is
$$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(... | For $a>0,$
$$\lim_{x\to0}\dfrac{a-\sqrt{a^2-x^2}}{x^2}=\lim_{x\to0}\dfrac{a^2-(a^2-x^2)}{x^2(a+\sqrt{a^2-x^2})}=\dfrac1{2a}$$ as $x\ne0$ as $x\to0$
Or set $x=a\sin2y\implies y\to0\implies\sin y\to0$
$$\lim_{x\to0}\dfrac{a-\sqrt{a^2-x^2}}{x^2}=a\lim_{y\to0}\dfrac{1-\cos2y}{(a\sin2y)^2}=\dfrac1a\lim_{y\to0}\dfrac{1-\cos2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Find all solutions of the equation $\cos 2θ + 7 \cos θ = 8$ Find all solutions of the equation in the interval [0, 2π) (express your answer in terms of $k$, where $k$ is any integer).
$$\cos 2\theta + 7 \cos\theta = 8$$
Did I make a mistake somewhere? My final answer was $0+2\pi k$
Here is my work. I went from step 1 ... | Notice, we have $$\cos 2\theta+7\cos \theta=8$$ $$2\cos^2\theta-1+7\cos \theta=8$$ $$2\cos^2\theta+7\cos \theta-9=0$$ $$\underbrace{2\cos^2\theta+9\cos \theta}\underbrace{-2\cos \theta-9}=0$$
$$\cos\theta(2\cos \theta+9)-(2\cos \theta+9)=0$$ $$(\cos \theta-1)(2\cos \theta+9)=0$$ $$\implies \cos\theta-1=0\iff \cos\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Distance formula for points in the Poincare half plane model on a "vertical geodesic". In comment at https://math.stackexchange.com/a/1381829/88985 at
Distance of two hyperbolic lines
is says (as i interpreted it) that the distance between two points $(a,r)$ and $(a, R)$ in the Poincaré half-plane model ( https://en.... | Compute first the inverse of
$$y=\operatorname {arcosh}(x)=\frac{e^x+e^{-x}}2$$
for positive $x$'s.
Whit the substitution $u=e^x$ we have $2y=u+\frac1u$. Multiplying both sides by $u$ we get the following quadratic equation:
$$u^2-2yu+1=0,$$
the solutions of which are
$$u^+=y+\sqrt{y^2-1}\ \text{ and } \ u^-=y-\sqrt{y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Given functional equation $f(x,y)=f(x,y+1)+f(x+1,y)+f(x-1,y)+f(x,y-1)$, show that $f(0,0)=0$ Let $f:\mathbb{Z}\times \mathbb{Z}\to \mathbb{R}$, such that
for all $x,y$
$$f(x,y)=f(x,y+1)+f(x+1,y)+f(x-1,y)+f(x,y-1),$$
and if $m,n\in \mathbb{Z},mn\neq 0$,we have $f(2m,2n)=0$.
Show that
$$f(0,0)=0.$$
| Sketching a solution since there is a lot of tedium required. This method finds all solutions $f$; perhaps there is a shortcut to show only $f(0, 0) = 0$?
*
*Show that if $f$, $g$ are solutions, then $f + g$ is a solution as well. Easy from the condition.
*Show that if$$f(3, 4) = f(4, 3) = f(3, 6) = f(6, 3) = 0,$$t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Solve this trigonometric equation $\cos3x\cos x=\sin 3x$ Solve this equation $$ \cos 3x\cos x=\sin 3x$$
I tried converting product into sum but with no results. I think they forgot to add $\sin x$ after $\sin 3x$.
| Hint:
Notice, $$\cos 3x\cos x=\sin 3x$$ $$(4\cos^3x-3\cos x)\cos x=3\sin x-4\sin^3x$$
$$4\cos^4x-3\cos^2 x=3\sin x-4\sin^3x$$ $$3\sin x-4\sin^3x=4(1-\sin^2 x)^2-3(1-\sin^2x)$$ $$3\sin x-4\sin^3x=4+4\sin^4 x-8\sin^2x-3+3\sin^2x$$
$$4\sin^4 x+4\sin^3 x-5\sin^2x-3\sin x+1=0$$ Let, $\sin x=t$, we get $$4t^4+4t^3-5t^2-3t+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390616",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluate the limit $\lim_{x \to 0} \left(\frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$
Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$
My attempt
So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$
$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\... | I think this limit would be considerably easier using Taylor Series instead of LHR.
$ \frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} \approx \frac{(x-\frac{x^3}{6})^2-x^2(1-\frac{x^2}{2})^2}{x^2(x)^2}=\frac{\frac{2x^4}{3}+O(x^5)}{x^4} \rightarrow \frac{2}{3} $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Finding the equations of the tangents where a quadratic equation cuts the $x$-axis and the angle between the tangents (differentiation involved) Calculate the equations of the tangent where $y=x^2-5x-24$ cuts the $x$-axis.
$(x-8)(x+3)$ factorising
$x=8, x=-3 $
$y'(x)=2x-5$
$y'(8)=11$
$y'(-3)=-11$
$y=11x+c$
$0=11(8)+c... | Notice, the slope of tangent at general point of the curve $y=x^2-5x-24$ $$\frac{dy}{dx}=\frac{d}{dx}(x^2-5x-24)=2x-5$$ Now, the point where curve $y=x^2-5x-24$ cuts the x-axis has $y=0$ thus we have $$0=x^2-5x-24$$ Solving the quadratic equation for the values of $x$ as follows $$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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maximum value of $a+b+c$ given $a^2+b^2+c^2=48$? How can i get maximum value of this $a+b+c$ given $a^2+b^2+c^2=48$ by not using AM,GM and lagrange multipliers .
| We prove if $a+b+c=r$ then $a^2+b^2+c^2>\frac{r^2}{3}$ and equality is reached only when $a=b=c$.
Suppose the minimum is achieved with values $a,b,c$ and we do not have $a=b=c$. then without loss of generality we have $a\neq b$ let $a+b=m$ and write $a$ as $\frac{m+n}{2}$ and $b$ as $\frac{m-n}{2}$. Then the sum of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $ I have been trying to algebraically solve this limit problem without using L'Hospital's rule but was whatsoever unsuccessful:
$$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $$
Thanks in advance!
| A scheme sometimes useful is to make the limit at zero...
Let $x=1+t$ so that $x \to 1$ means $t \to 0$. Then
$$
\frac{\sqrt{x}+\sqrt[3]{x}-2}{x-1} = \frac{\sqrt{1+t}+\sqrt[3]{1+t}-2}{t}
$$
so compute some asymptotics:
$$
\sqrt{1+t} = 1+\frac{1}{2}t+o(t)
\\
\sqrt[3]{1+t} = 1+\frac{1}{3}t+o(t)
\\
\sqrt{1+t}+\sqrt[3]{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Prove $\sum\limits^m_{k=0} \frac{2n-k\choose k}{2n-k\choose n}\frac{2n-4k+1}{2n-2k+1}2^{n-2k}=\frac{n\choose m}{2n-2m\choose n-m}2^{n-2m}$ for- Let $n$ be a positve integer. Prove that$$\sum\limits^m_{k=0} \frac{2n-k\choose k}{2n-k\choose n}\frac{2n-4k+1}{2n-2k+1}2^{n-2k}=\frac{n\choose m}{2n-2m\choose n-m}2^{n-2m}$$ f... | This reduces to a telescoping sum
$$
\begin{align}
&\sum_{k=0}^m\frac{\binom{2n-k}{k}}{\binom{2n-k}{n}}\frac{2n-4k+1}{2n-2k+1}2^{n-2k}\\
&=\sum_{k=0}^m\frac{(2n-k)!}{k!\,(2n-2k)!}\frac{n!\,(n-k)!}{(2n-k)!}\left(1-\frac{2k}{2n-2k+1}\right)2^{n-2k}\\
&=\sum_{k=0}^m\left[\frac{n!\,(n-k)!}{k!\,(2n-2k)!}2^{n-2k}
-\frac{n!\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 0
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Convergence of improper integral $\int_{0}^{\frac{\pi}{6}}\dfrac{x}{\sqrt{1-2\sin x}}dx$ I'm trying to determine whether the following improper integral is convergent or divergent.
$$
\int_{0}^{\pi/6}\frac{x}{\sqrt{1-2\sin x}}dx
$$
At first, I substituted $t=\dfrac{\pi}{2} - x $
and then I used $1-\dfrac{1}{2}x^2 \le ... | Using the right hand side of the well known Jordan inequality that states $\sin x < x$ for all $x \in \left[{0,\frac{\pi}{2}}\right]$ we have $1-2\sin x > 1-2x$ so that $\frac{1}{1-2\sin x} < \frac{1}{1-2x}$ so that for all $0\le x\le \frac{\pi}{6}$ we have $\frac{x}{\sqrt{1-2\sin x}} < \frac{x}{\sqrt{1-2x}}$.
Thus,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Evaluation of Gaussian integrals. In Bender and Orszag's section on applying Laplace's method they state the equality
$$\int_{0}^{\infty}e^{-xt^2}\frac{1}{3}xt^4 dt = \frac{1}{2}\sqrt{\frac{\pi}{x}}\left(\frac{1}{4x}\right)$$
How does one calculate this integral ?
| Notice, the Laplace transform $$L[t^n]=\int_{0}^{\infty}e^{-st}t^ndt=\frac{\Gamma(n+1)}{s^{n+1}}$$ & $\Gamma\left(\frac{1}{2}\right)=\sqrt {\pi}$
Now, we have $$\int_{0}^{\infty}e^{-xt^2}\frac{1}{3}xt^4dt$$
$$=\frac{1}{3}x\int_{0}^{\infty}e^{-xt^2}t^4dt$$
Let, $t^2=u\implies 2tdt=du\iff dt=\frac{du}{2\sqrt u}$, we get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Coeff. of $x^{97}$ in $f(x) = (x-1)\cdot (x-2)\cdot (x-3)\cdot (x-4)\cdot ........(x-100)$
If $f(x) = (x-1)\cdot (x-2)\cdot (x-3)\cdot (x-4)\cdot ........(x-100)\;,$ Then Coefficient of $x^{99}$ and
Coefficient of $x^{98}$ and Coefficient of $x^{97}$ in $f(x).$
$\bf{My\; try::}$ We can write $f(x)$ as
$$\displaystyl... | For the coefficient of $x^{97}$ you can apply a method similar to the one you used to find the coefficient of $x^{98}$. You already noted that the coefficient $C$ of $x^{97}$ is
$$C:=-\mathop{\sum^{100}\sum^{100}\sum^{100}}_{i=1\ j=1\ k=1\ i<j<k}i\cdot j\cdot k=-\sum_{i=1}^{98}\sum_{j=i+1}^{99}\sum_{k=j+1}^{100}i\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
| Writing the polynomial $x^4+3x^3+3x^2-3x+1$ as a product of two polynomials of degree two, i.e,. $$x^4+3x^3+3x^2-3x+1= (x^2+Ax+B)(x^2+Cx+D)$$, then we will get 4 equations $A+C=3,B+D+AC=3,AD+BC=-3,BD=1$, using wolframalpha.com to solve this eqs. we get
Thus, $$\int{\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx}=\int{\frac{1+x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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$\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$ Show that $\displaystyle \int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$
What... | Let $$\displaystyle I = \int_{0}^{\infty} f\left(\frac{a}{x}+\frac{x}{a}\right)\cdot \frac{\ln x}{x}dx\;,$$ Now Put $\displaystyle \frac{a^2}{x} = t\;,$ Then $\displaystyle x= \frac{a^2}{t}$ and $\displaystyle dx = -\frac{a^2}{t^2}dt$
and Changing Limit, We get
$$\displaystyle I = \int_{\infty}^{0}f\left(\frac{t}{a}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Upper and lower bounds for $S(n) = \sum_{i=1}^{2^{n}-1} \frac{1}{i} = 1+\frac{1}{2}+ \cdots +\frac{1}{2^n-1}.$
For a positive integer $n$ let $S(n) = \sum_{i=1}^{2^{n}-1} \frac 1i = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{2^n-1}.$
Then which of the following are true.
*
*(a) $S(100)\leq 100$.
... | This is not an answer but it is too long for a comment.
Since $$S(n)=\sum_{i=1}^{2^n-1}\frac 1i=H_{2^n-1}$$ you could be interested by this paper in which the author, Mark B. Villarino, developed extremely sharp bounds for the harmonic numbers.
The first and simplest set of bounds (Theorem $1$) is given by $$\frac{1}{2... | {
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"url": "https://math.stackexchange.com/questions/1404602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of powers of a matrix with primitive polynomial modulo $2^{r}$ I need to prove an statement in the matrix form, which leads to the following equality modulo $2^{r}$. Which I couldn't prove but with computer simulation for lots of primitive polynomial, it seems to be true.
Question.Prove
$$ (I+A+A^2+\ldots+A^{2^r... | $A$ is the companion matrix of $P(x)$. As $P(x)$ was assumed to be primitive
$$
B:=A^{2^m-1}\equiv I\pmod 2.
$$
Furthermore, $2^m-1$ is the lowest positive exponent for which this congruence holds. So $B=I+2C$ for some integer matrix $C$. We prove by induction that
$$
B^{2^\ell}\equiv I\pmod{2^{\ell+1}}.
$$
The base ca... | {
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"url": "https://math.stackexchange.com/questions/1405668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the following If tangent at $(a,b)$ passes through $(a_1,b_1)$ for the curve $x^3+y^3=c^3$ and $\frac{a_1}{a} + \frac{b_1}{b} = -k$ then find the value of k.(a_1,b_1) lies on curve as well?
solution :$ \frac{b_1-b}{a_1-a} = -\frac{a^2}{b^2} , a^3+b^3=c^3$ and $a_1^3+b_1^3 = c^3$
I got stuck here how to solve it?... | We have $$ 0= (a^3+b^3)-(a_1^3+b_1^3)$$ $$=(a^3-a_1^3)+(b^3-b_1^3)$$ $$=(a-a_1)(a^2+aa_1+a_1^2)+(b-b_1)(b^2 +bb_1+b_1^2).$$Now ,as $ (a-a_1)/(b-b_1)=b^2/(-a^2)$, this gives$$ 0=b^2(a^2+aa_1+a_1^2)+(-a^2)(b^2+bb_1-b_1^2)$$ $$=(ba_1-ab_1)(ba+ba_1+ab_1)$$ $$=(ba_1-ab_1)(ba)(1-k).$$ I believe you can do the rest.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Maximize $xy^2$ on the ellipse $x^2+4y^2=4$ I was using Lagrange multiplier, any steps gone wrong?
$$f(x,y)=xy^2$$
$$c(x,y)=x^2+4y^2$$
Partial Derivatives
$$\frac {\partial f}{\partial x} = y^2 $$
$$\frac {\partial f}{\partial y} = 2xy $$
$$\frac {\partial c}{\partial x} = 2x $$
$$\frac {\partial c}{\partial y} = 8y $$... | $$f = xy^2$$
with
$$4=x^2+4y^2$$
becomes
$$f = xy^2 = x\left(1-\frac{1}{4}x^2\right) = x - 0.25x^3$$
maximum via derivative
$$\frac{d}{dx}f = 0 = 1 - 0.75x^2$$
$$x_{1,2}=\pm\sqrt{\frac{4}{3}}$$
$$\frac{d^2}{dx^2}f(x_1) < 0 $$
$$\frac{d^2}{dx^2}f(x_2) > 0 $$
and with $4=x^2+4y^2$ again:
$$y_{1,2} = \pm\sqrt{\frac{2}{3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $\lim \frac{3^n + 2^n}{5\cdot3^n + 7\cdot2^n} = \frac{1}{5}$ Prove using limit definition.
$$\lim \frac{3^n + 2^n}{5\cdot 3^n + 7\cdot 2^n} = \frac{1}{5}
$$
My try:
$$\left| {\frac{3^n + 2^n}{5\cdot 3^n + 7\cdot 2^n} - \frac{1}{5}} \right| < \varepsilon
$$
$$ \Leftrightarrow \left| \frac{3^n + 2^n - 5\cdot 3^n - ... | If you know equivalents in Asymptotic analysis, that is easy:
$\;3^n+2^n\sim3^n$, $\sim5\times3^n$, hence
$$\frac{3^n+2^n}{5\times3^n+7\times2^n}\sim\frac{3^n}{5\times3^n}=\frac15.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Thue equation $ x^4 - 6 x^3 y - x^2 y^2 + 6 x y^3 - y^4=-1 $ I need help solving the Thue equation
$ x^4 - 6 x^3 y - x^2 y^2 + 6 x y^3 - y^4=-1 $.
It can be written as
$ x(x-y)(x+y)(x-6y) = (y-1)(y+1)( y^2 +1) $.
From this I found 8 solutions (0,1),(0,-1),(1,1),(-1,-1),(-1,1),(1,-1),(6,1) and (-6,-1). But there are tw... | OP seems to be looking for a simple method that would ensure an easy discovery of all solutions listed. One such simple method is to let $y$ run through $\pm 1, \pm 2, \pm 3, ...$ and calculate $(y−1)(y+1)(y^2+1) = y^4-1$. From the left hand side, we can see that $x$ must be a divisor of $y^4-1$ and we simply check all... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$ I'm having trouble on starting this induction problem.
The question simply reads : prove the following using induction:
$$1^{2} + 2^{2} + ...... + (n-1)^{2} < \frac{n^3}{3} < 1^{2} + 2^{2} + ...... + n^{2}$$
| Perhaps simpler, the induction step is done if you show
$$3n^2<(n+1)^3-n^3<3(n+1)^2$$
But $(n+1)^3-n^3=3n^2+3n+1$ which is clearly in between $3n^2$ and $3(n+1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Why doesn't quadratic formula lead to a the correct factored form of the original equation? Applying the quadratic formula to $2x^2-3x+1$ we have
\begin{eqnarray*}
a&=&2 \\
b&=&-3 \\
c&=&1
\end{eqnarray*}
which gives me two roots:
\begin{eqnarray*}
x_1&=&1 \\
x_2&=&\tfrac{1}{2}
\end{eqnarray*}
Therefore you can re-wr... | The quadratic formula gives the solution to the equation
$$ax^2+bx+c=0.$$
But the formula is derived by completing the square, and to do so the coefficient of $x^2$ must be 1. Therefore, the first step is to divide by $a$ (it is assumed without loss of generality that $a>0$). This gives
$$x^2+\tfrac{b}{a}x+\tfrac{c}{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What is $\lim_{n\to\infty}2^n\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt{p}}}}$ for $negative$ and other $p$? This was inspired by similar posts like this one. Define the function,
$$F(p) = \lim_{n\to\infty}2^n\sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\dots+\sqrt{p}}}}_{n \textrm{ square roots}}}$$
We know that,
$$F(2) = \frac{\pi}{... | The (hyperbolic) cosine bisection formula gives:
$$2\cos\frac{x}{2}=\sqrt{2+2\cos x},\qquad 2\cosh\frac{x}{2}=\sqrt{2+2\cos x}$$
hence assuming $a_0=2\cosh(u_0)=\sqrt{p}$ and $a_{n+1}=\sqrt{2+a_n}$ we have:
$$ a_n = 2 \cosh\left(\frac{u_0}{2^n}\right),\quad \sqrt{2-a_n}= 2\sinh\left(\frac{u_0}{2^{n+1}}\right)$$
so:
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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$a,b,c\in \Bbb Z$ and $a\cdot b\cdot c$ is a root of $ax^2+bx+c$. I was curious if there are quadratic equations where $a,b,c\in \Bbb Z$ and $a\cdot b\cdot c$ is a root of $ax^2+bx+c$.
So trivially if $c=0$, $a$ and $b$ can be arbitrary, and if either $a$ or $b$ is zero, this implies that $c=0$, and the other arbitrary... | The proof looks correct to me.
However, it might be good to say "However, we assumed $c \neq 0$" after point 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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When is y increasing? Arctan is the inverse of tan.
I tried to normally differentiate and solve for y' but the LCM is complicated.
Is there any easy substitution simplifies the problem?
$$y=(\frac{1}{3})\textrm{log}(\frac{x+1}{\sqrt{x^2-x+1}}) + \frac{1}{\sqrt{3}}\textrm{arctan}(\frac{2x+1}{\sqrt3})$$
| $$y=\frac{1}{3}\ln{\frac{x+1}{\sqrt{x^2-x+1}}}+\frac{1}{\sqrt{3}}\tan^{-1}{\frac{2x+1}{\sqrt{3}}}$$
Before differentiating, we will manipulate the function to make it easier to differentiate.
$$y=\frac{\ln(x+1)}{3}-\frac{\ln(x^2-x+1)}{6}+\frac{1}{a}\tan^{-1}\frac{u}{a}$$where $a=\sqrt3$
$$\frac{\delta y}{\delta x}=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to find a formula for the expression $2^0+2^1+\cdots+2^n$? Is there a way to simply this equation and express it in terms of $n$?
$$2^0+2^1+2^2+2^3+\cdots+2^n$$
| $$\sum\limits_{i=0}^{n}2^i = 1+2+2^2+2^3+\cdots + 2^n = 1 + 2\left(1+2+2^2+\cdots + 2^{n-1}\right)=1+2\sum\limits_{i=0}^{n-1}2^i\text{.}$$
Now subtract $\sum\limits_{i=0}^{n-1}2^i$ from both sides:
$$\sum\limits_{i=0}^{n}2^i-\sum\limits_{i=0}^{n-1}2^i=2^n$$
and
$$1+2\sum\limits_{i=0}^{n-1}2^i-\sum\limits_{i=0}^{n-1}2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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A simple double inequality with roots Prove or disprove for integer $n$ greater than $1$:
$$\sqrt{1} + \sqrt{2} + ...+ \sqrt{n-1} < \frac{2n\sqrt{n}}{3} < \sqrt{1} + \sqrt{2} + ...+ \sqrt{n} $$
I think I know a solution, but I am looking for different approaches.
EDIT: Inspired by Prove that $1^2 + 2^2 + ..... + (n-1)... | by induction :
for LHS: if $n=k ,\sum _{i=1}^k \sqrt{i-1} \le \dfrac{2k\sqrt{k}}{3}$
when $n=k+1$, we have to prove $\dfrac{2k\sqrt{k}}{3}+\sqrt{k} \le \dfrac{2(k+1)\sqrt{k+1}}{3} \iff 3k\ge 4$ which is true when $k\ge2$
similar method to RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Number Theory - Remainders A number is of the form $13k_1+12$ and of the form $11k_2+7$
That is $N = 13k_1 + 12 = 11k_2 + 7$
Now why must N also equal $(13 \times 11)k_3 + 51$ ?
Thanks
| Finding a Solution
To find an $N$ so that
$$
\begin{align}
N&\equiv12&\pmod{13}\\
N&\equiv7&\pmod{11}
\end{align}\tag{1}
$$
we can start by solving
$$
13x+11y=1\tag{2}
$$
using the Extended Euclidean Algorithm. The implementation described in this answer gives
$$
\begin{array}{r}
&&1&5&2\\\hline
1&0&1&-5&11\\
0&1&-1&6&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a perpendicular plane Question
Plane $B$ contains the points $(4, 2, 1)$ and $(4, 1, -6)$, and it is perpendicular to the plane $7x+9y+4z=18$ . What is the equation of the plane $B$?
What I tried:
I know that to find the equation of plane $B$ I need a point on the plane and a perpendicular vector. The point par... | Your approach is both correct and simple.
Your plane has to be parallel to both the vectors
$$\begin{pmatrix}4\\2\\1\end{pmatrix}-\begin{pmatrix}4\\1\\-6\end{pmatrix}=\begin{pmatrix}0\\1\\7\end{pmatrix}
\qquad\text{and}\qquad
\begin{pmatrix}7\\9\\4\end{pmatrix}$$
so its normal vector can be computed as
$$\begin{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A sequence related to squares of Fibonacci nubers Let $f(n)$ be defined by
$f(n)=f(n-1)+f(n-3)+f(n-4)$, for $n \ge 5$,
$f(1)=1, f(2)=1, f(3)=2, f(4)=4$.
First few terms of the sequence $(f(1), f(2), f(3), \ldots$) look like
$(1, 1, 2, 4, 6, 9, 15, 25, 40, 64, 104, 169, \ldots)$
Here we recognize that the subsequence c... | This is simple to solve, surprisingly. Define $g(z) = \sum_{n \ge 0} f(n) z^n$,
shift indices to get:
$\begin{align}
f(n + 4)
= f(n + 3) + f(n + 1) + f(n)
\end{align}$
Multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:
$\begin{align}
\frac{g(z) - f(0) - f(1) z - f(2) z^2 - f(3) z^3}{z^4}
= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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In how many ways can a selection be done of $5$ letters?
In how many ways can a selection be done of $5$ letters out of $5 A's, 4B's, 3C's, 2D's $ and $1 E$.
$ a) 60 \\
b) 75 \\
\color{green}{c) 71} \\
d.) \text{none of these} $
Number of ways of selecting $5$ different letters = $\dbinom{5}{5} = 1$ way
Number of... | Let $x_i$ be the number of letter $i$ chosen for $1\le i\le 5$, where we number the letters in alphabetical order.
We want to find the number of solutions in nonnegative integers to the equation $x_1+\cdots+x_5=5$
$\hspace{.5 in}$with the restrictions $x_1\le5,\; x_2\le4, \;x_3\le3,\; x_4\le2, \;x_5\le1$.
Let $S$ be th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Consider the following system and find the values of b for which the system has a solution So I have this system:
$$\left\{\begin{array}{c}
x_1 &− x_2 &+ 2x_3 &= 2 \\
x_1 &+ 2x_2 &− x_3 &= 2 \\
x_1 &+ x_2 & &= 2 \\
x_1 & & +x_3 &= α
\end{array}\right.$$
And we are asked to find the values of $\alpha$ for ... | Gaussian elimination.
Downwards:
$$\tag 1 \begin{array} {l}
x_1 & -x_2 & + 2 x_3 & 0 & = 2 \\
x_1 & +2x_2 & - x_3 & 0 & = 2 \\
x_1 & +1x_2 & 0 & 0 & = 2 \\
x_1 & 0 &+1 x_3 & -\alpha & = 0 \\
\end{array} $$
Subtract row 1 from rows 2 to 4:
$$\tag 2 \begin{array} {l}
x_1 & -x_2 & + 2 x_3 & 0 & =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Assumption about the form of solutions to a recurrence relation Basically, when solving such recurrence relations, we try to find solutions of the form $a_n = r_n$, where $r$ is a constant.
$a_n = r^n$ is a solution of the recurrence relation
$a_n = c_1a_{n-1} + c_2a_{n-2} + … + c_ka_{n-k}$ if and only if
$r^n = c_1r^... | One clean explanation (and a uniform way to solve such recurrences) is to use generating functions. Say you have:
$\begin{align}
a_{n + k} = c_{k - 1} a_{n + k - 1}
+ \dotsb + c_0 a_n
\end{align}$
Define the generating function $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$ and sum over ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Simplifying $\tan100^{\circ}+4\sin100^{\circ}$ The answer is $-\sqrt3$.
I was wondering if this is just a coincidence?
Also, is there a relation between $$\tan(100^{\circ}+20^{\circ})=\frac{\tan100^{\circ}+\tan20^{\circ}}{1-\tan100^{\circ}.\tan20^{\circ}}=-\sqrt3$$ and the given expression? Or is there a more elegant... | One has $$\tan 100^\circ + 4\sin 100^\circ = \frac{\sin 100^\circ + 2\sin 200^\circ}{\cos 100^\circ} = \frac{\sin 100^\circ - 2\sin 20^\circ}{\cos 100^\circ} = \frac{2\cos 60^\circ\sin 40^\circ - \sin 20^\circ}{\cos 100^\circ} = \frac{\sin 40^\circ - \sin 20^\circ}{\cos 100^\circ} = \frac{2\cos 30^\circ\sin 10^\circ }{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can I show $\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$ by induction? It is known that $a_{n+1} > a_n$. I tried to prove
$$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$$
via induction.
Base case: $a_0 = \frac{1}{0+1} = 1 \leq 1$
Inductive step: assume $a_n \leq 1$, prove $a_{n+1} \leq ... | According to Wolfram Alpha, this series is equivalent to $ \psi^{(0)}(2n+1)-\psi^{(0)}(n+1)$ where $ \psi^{(0)}(x)$ is the $n^{th}$ derivative of the digamma function. Using this, you could show the continuity for the function for values of $n > 0$, show that the limit as n approaches infinity is $log(2)$ (easily prove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the domain of the function $f(x) = \frac{x}{\sqrt{x^2+9}}$ For some reason I can't find domain to this function, but it is pretty clear that it's (ALL X)
$$f(x) = \frac{x}{\sqrt{x^2+9}}$$
I can say that: $$\space x^2+9 > 0\space$$
$$x^2 > -9$$
$$x = \pm \sqrt{-9}$$
I don't get it..
| $$f(x) = \frac{x}{\sqrt{x^2+9}}$$
Only $\sqrt{x^2+9}=0$ could be a problem for your domain. But as $x^2+9>0$ (sum of two squares), this cannot happen. Hence, the domain of $f$ is $R$.
If you also need to find the map of your function:
First of all we evaluate the derivative of $f(x)$
$$f'(x)=\frac{1\cdot\sqrt{x^2+9}-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Legendre functions $Q_n(x)$ of the second kind Legendre functions $Q_n(x)$ of the second kind
\begin{equation*}
Q_n(x)=P_n(x) \int \frac{1}{(1-x^2)\cdot P_n^2(x)}\, \mathrm{d}x
\end{equation*}
what to do after this step?
how can I complete ?
I need to reach this formula
\begin{equation*}
Q_n(x)=\frac{1}{2} P_n(x)\ln\le... | So you want to prove that
$$Q_n(x)= P_n(x) \int^x \frac{1}{(1-x^2) \, \left[P_n(x)\right]^2} \, dx = \frac{1}{2} P_n(x) \ln\left(\frac{1+x}{1-x}\right)$$.
Let's first compute for the case $n=0$, knowing that $P_0(x) = 1$ we would have
$$Q_0(x) = \int^x \frac{1}{(1-x^2)} \, dx = \frac{1}{2} \int \left[\frac{1}{1+x} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problem with multivariable calculus: $\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$ Anyone can help me with this limit?
$$\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$$
I'm having trouble with proving that this limit really goes to $0$
thank you
| Let $y=(x^3-x^6)^2-x^2$. $\,$Then
\begin{align*}
\frac{x^3+y^3}{x^2+y} &= \frac{x^3+(x^3-x^6)^6 - x^6 + 3x^4(x^3-x^6)^2 - 3 x^2 (x^3-x^6)^4}{(x^3-x^6)^2}\\
&= \frac{1}{x^3-x^6} + (x^3-x^6)^4+3x^4-3x^2(x^3-x^6)^2\\
&\rightarrow \infty.
\end{align*}
That is,
\begin{align*}
\lim_{(x,y)\rightarrow (0, 0)} \frac{x^3+y^3}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Implicit differentiation of $3x^2 + 4y^2 = 12$
The equation $3x^2 + 4y^2 = 12$ defines $y$ implicitly as two functions of $x$ if $|x| < 2$ and $|x|=2$. Assuming
the second derivative $y''$ exists, show that it satisfies the equation $4y^3 y'' = -9$.
Progress. I derived $6x+8yy'=0$ then $y'= \frac{-3x}{4y}$ then de... | It turns out that
if
$ax^2+by^2 = c$,
then
$b^2y^3y''
=-ac
$.
In this case,
$16y^2y''
= -36$
or
$4y^2y''
= -9$.
I use implicit differentiation
twice.
From
$ax^2 + by^2 = c
$,
as you did
$0
=2ax+2byy'
=ax+byy'
$,
so
$y'
=-\frac{ax}{by}
$.
Doing it again,
$0
= a+b((y')^2+yy'')
$
or
$-\frac{a}{b}
=(y')^2+yy''
=(\frac{ax}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How many pairs of positive integers $(x,y)$ satisfy the equation $x^2 - 10! = y^2$? I have a question about number theory:
How many pairs of positive integers $(x,y)$ satisfy the following equation? $$x^2 - 10! = y^2$$
My attempt:
Move the $y^2$ from right to the left and 10! From left to the right such that:
$$x^2-y... | $105$ pairs.
As marty cohen said: "$(x+y)(x-y)=10!=ab$
with
$x+y=a$ and $x-y=b$.
Then
$x = (a+b)/2$ and
$y = (a-b)/2$.
So $a$ and $b$
have to have the same parity."
$10!=2^{\color{red}{8 }}\cdot 3^{\color{red}{4}}\cdot 5^{\color{red}{2}} \cdot 7^{\color{red}{1}} \implies (\color{red}{8}+1)(\color{red}{4}+1)(\color{red... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
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Evaluation of $ \int\frac{x^2-1}{\sqrt{x^4+4x^3+7x^2+4x+1}}dx$
Evaluation of $\displaystyle \int\frac{x^2-1}{\sqrt{x^4+4x^3+7x^2+4x+1}}dx$
$\bf{My\; Try::}$ Let $\displaystyle I = \int\frac{x^2-1}{\sqrt{x^4+4x^3+7x^2+4x+1}}dx = \int\frac{x^2-1}{\sqrt{(x+1)^4+x^2}}dx$
So $\displaystyle I = \int\frac{x^2-1}{x\cdot \sqr... | It looks to me that an elliptic integral is unavoidable.
By replacing $\sqrt{x}+\frac{1}{\sqrt{x}}$ with $y$ the problem boils down to computing:
$$ \int \left(y^3-2y+y^2\sqrt{y^2-4}\right)\,\frac{dy}{\sqrt{1+y^4}}=\frac{\sqrt{1+y^4}}{2}+\text{arcsinh}(y^2)+\frac{1}{2}\int\sqrt{\frac{z^2-4z}{z^2+1}}\,dz $$
where in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sqrt{\frac{n+15}{n+1}}\in\mathbb Q$, then $n=17$ How can one show that :
If $\sqrt{\frac{n+15}{n+1}}\in\mathbb Q$ so $n=17$
I tried using the fact that any number $a\in\mathbb Q$ so $a=\frac{x}{y}$ such that $\gcd(x,y)=1$
So $\frac{n+15}{n+1}=\frac{x^2}{y^2}$
But here I'm stuck.
| Essentially, you know $x^2(n+1)=y^2(n+15)$, let $m=n+1$ for convenience, then you know that since $x$ and $y$ are relatively prime, $y^2 \mid m$
and $x^2\mid m+14$. Since $\frac{m+14}{m}=\frac{x^2}{y^2}$, we then have
that $m=y^2a$ for some $a$ and $m+14=x^2a$ for that same $a$. Then $y^2a+14=x^2a$. Then $14=(x^2-y^2)a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove by induction $\sum \frac {1}{2^n} < 1$ Prove by induction $\sum \frac {1}{2^n} < 1$
Well supposing the base case has been shown to be true, I start with the induction step:
Suppose true for n = k:
$$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} < 1$$
Want to show this is true for:
$$ \frac{1}{2} + \frac{1}{4} +... | Use induction to prove that $$\frac 12 + \frac 14 + \cdots + \frac 1{2^n} = 1 - \frac{1}{2^n}$$ for all $n$. This will give you the result immediately since $1 - \frac{1}{2^n} < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate limit as x approaches infinity of $\lim_{x\to\infty}\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}$ I am having trouble figuring out how to answer this question by determining the degree of the numerator and/or denominator:
$$\lim_{x\to\infty}\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}$$
I have tried deriving the first coeffic... | You factor out the dominant term from the numerator such that it cancels with the dominant term in the denominator. In this case we take out a factor of $x^3$ since $x^3$ is the highest degree or dominant term in both the numerator and denominator.
Hence, showing all intermediate steps:
$\color{blue}{\lim_{x\rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I've been trying to solve this over and over without L'Hopital but keep on failing:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$$
My first attempt involved rationalizing:
$$\frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} = \frac{1-\cos x}{x\cdot... | After your first rationalization, you showed that the original expression is equivalent to
$$\lim_{x\to 0} \frac{1-\cos x}{x^2(1+\sqrt{\cos x})}=\lim_{x \to 0}\color{green}{\frac{1-\cos x}{x^2}} \cdot \color{blue}{\frac {1}{1+\sqrt{\cos x}}}$$
Since the $\color{green}{\text{green}}$ part of the limit converges to $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Binomial transform of Catalan numbers formula How to prove that OEIS A007317 Binomial transform of Catalan numbers $a_{n}: 1, 2, 5, 15, 51, 188, 731, 2950, 12235, 51822, .. (n = 1, 2, ..)$ has a recurrence formula: $(n+2)a_{n+2} = (6n+4)a_{n+1} - (5n)a_{n}$ ?
The sequence is defined as: $a_{n+1} = \sum_{k=0}^{n} \bino... | We prove the recurrence formula with the help of generating functions. Let $A(z)$ be the generating functions of the $a_n$
\begin{align*}
A(z)=\sum_{n=1}^\infty a_nz^n=\sum_{n=0}^\infty a_{n+1}z^{n+1}
=\sum_{n=0}^\infty \left(\sum_{k=0}^n\binom{n}{k}C_k \right)z^{n+1}
\end{align*}
and let $C(z)$ be the (well-known) gen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplify $\sin^4 x+\cos^4 x$ using trigonometrical identities? $\sin^{4}x+\cos^{4}x$
I should rewrite this expression into a new form to plot the function.
\begin{align}
& = (\sin^2x)(\sin^2x) - (\cos^2x)(\cos^2x) \\
& = (\sin^2x)^2 - (\cos^2x)^2 \\
& = (\sin^2x - \cos^2x)(\sin^2x + \cos^2x) \\
& = (\sin^2x - \c... | Let $$\displaystyle y=\sin^4 x+\cos^4 x = \left(\sin^2 x+\cos^2 x\right)^2-2\sin^2 x\cdot \cos^2 x = 1-\frac{1}{2}\left(2\sin x\cdot \cos x\right)^2$$
Now using $$ \sin 2A = 2\sin A\cos A$$
So, we get $$\displaystyle y=1-\frac{1}{2}\sin^2 2x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Convert $r^2= 9 \cos 2 \theta$ into a Cartesian equation This is how I tried so far...
$r^2= 9 \cos 2 ( \theta)$
$\cos (2 \theta) = \cos ^2 (\theta) - \sin^2 (\theta)$ and
$r^2= x^2 + y^2$
so, it will become
$x^2 + y^2 = 9 [\cos^2 (\theta) - \sin^2 (\theta) ]$
| multiply both side by $r^2$ and replace $r.cos\theta=x$ and $r.sin\theta=y$
and $r^2=x^2+y^2$ as $cos^2\theta-sin^2\theta=(cos\theta+sin\theta)(cos\theta-sin\theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Mathematical Induction Question About A Basis Step Use mathematical induction to show that:
$1^3 + 3^3 + 5^3 + ... + (2n + 1)^3 = (n+1)^2(2n^2+4n+1)$ whenever $n$ is a positive integer.
I am able to solve this problem, but the thing that confuses me on this is the Basis Step.
P(1): $(2(1) + 1)^3$ $\neq$ $4(7)$, does th... | Method -1: Mathematical Induction
Notice, the following steps by Mathematical Induction
*
*Setting $n=1$, we get
$$1^3+(2\cdot 1+1)^3=(1+1)^2(2(1)^3+4(1)+1)$$
$$28=28$$
The identity holds good for $n=1$.
*Assume that it holds for $n=k$ then we have
$$1^3+3^3+5^3+\ldots +(2k+1)^3=(k+1)^2(2k^2+4k+1)$$
*Setting $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.