Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Applications of the inequality $e^x\geqslant 1+x$ I am interested in proofs of famous theorems or inequalities which can be proved by the inequality
$$
e^x\geqslant 1+x.
$$
For example, the divergence of harmonic series can be proved by assigning $x=\frac{1}{k}$.
$e^{\pi}>\pi^e$ can be proved by assigning $x=\frac{\pi}{e}-1$.
The inequality of the arithmetic average of n-th degree can be proved by assigning $x=\frac{a_k}{(a_1*a_2*\cdots*a_n)^{\frac{1}{n}}} -1$.
The divergence of $\frac{e^t}{t^n}$ can be proved by assigning $x=\frac{t}{n+1}$.
I want to know other possible proofs.
| Define $p_i$ as the i-th prime number. Taking $x=\frac{1}{p_i -1}$ and we have
$$e^{\frac{1}{p_i-1}}≧1+\frac{1}{p_i -1}=1+\frac{1}{p_i} +\frac{1}{p_i ^{2}}+\cdots.$$
The product from $i=1$ to $n$ is
$$e^{1+\sum_{i=2}^n \frac{1}{p_i-1}}≧\prod_{i=1}^n \left(1+ \frac{1}{p_i} +\frac{1}{p_i ^{2}}+\cdots \right).$$
And here,
$$\prod_{i=1}^n \left(1+ \frac{1}{p_i} +\frac{1}{p_i ^2}+\cdots \right)≧\sum_{k=1}^{x(n)} \frac{1}{k},$$
$$\sum_{i=2}^n \frac{1}{p_i-1}≦ \sum_{i=1}^{n-1} \frac{1}{p_i}.$$
So, we can say
$$e^{\sum_{i=1}^{n-1} \frac{1}{p_i}}≧ \sum_{k=1}^{x(n)} \frac{1}{k},$$
$$\Leftrightarrow\sum_{i=1}^{n-1} \frac{1}{p_i}≧\log \sum_{k=1}^{x(n)} \frac{1}{k}.$$
When $n\to\infty$, $x(n)\to\infty$ and $\lim_{n\to\infty} \log \sum_{k=1}^{x(n)} \frac{1}{k}=\infty$. So,
$$\sum_p \frac{1}{p}=\infty.$$
That is how we can prove the sum of prime numbers diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Induction proof: $2^n + 3^n ≡ 5^n (mod 6)$ I'm trying to prove that $2^n + 3^n ≡ 5^n\ (mod\ 6)$ using induction.
$n=1$:
$2+3≡5\ (mod\ 6)$
$n=k$:
$2^k + 3^k ≡ 5^k\ (mod\ 6)$
$n=k+1$:
$2^{k+1} + 3^{k+1} ≡ 5^{k+1}\ (mod\ 6)$
$2*2^k + 3*3^k ≡ 5*5^k\ (mod\ 6)$
$6\ |\ 5*5^k - 2*2^k - 3*3^k$
$6\ |\ (2+3)*5^k - 2*2^k - 3*3^k$
$6\ |\ 2*5^k + 3*5^k - 2*2^k - 3*3^k$
$6\ |\ 2*(5^k - 2^k) + 3*(5^k - 3^k)$
Not quite sure if I'm going in the right direction or where to go from here...
| Multiply both sides of $$a^k+b^k\equiv(a+b)^k\pmod{ab},$$ by $a+b$ to find $$(a+b)^{k+1}\equiv(a+b)(a^k+b^k)\equiv a^{k+1}+b^{k+1}+a\cdot b^k+b\cdot a^k\pmod{ab}$$
Use the face $a\cdot b^k,b\cdot a^k,$ are both divisible by $ab$ for $k\ge1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the equation of the circle circumscribing the triangle formed by the three points $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$. Assuming the equation of the circle circumscribing the triangle formed by the three given points is given by the sphere through the three points and the plane through the three points.
Plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
and assuming the equation of the sphere is $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.
I've not been able to find the value for $d$, putting the three points in the equation of sphere I got $3$ equations with $4$ variables.
| Given the points
$$
\cases{
A=(a,0,0)\\
B=(0,b,0)\\
C=(0,0,c)\\
p = (x,y,z)
}
$$
the sphere center $p_0$ is determined by the intersection of plane $\Pi$ with the line $L$ such that
$$
\cases{
\Pi\to (p-A)\cdot\vec n=0\\
\vec n = (A-B)\times(B-C)\\
L\to \left((p-\frac{A+B}{2})\cdot(A-B)=0\right)\cap \left((p-\frac{A+C}{2})\cdot(A-C)=0\right)\\
L\to p=(0,\frac{b^2-a^2}{2b},\frac{c^2-a^2}{2c})+\lambda(1,\frac ab,\frac ac)
}
$$
then
$$
p_0 = \frac{1}{2(a^2b^2+a^2c^2+b^2 c^2)}\left(a^3 \left(b^2+c^2\right),b^3 \left(a^2+c^2\right),c^3\left(a^2+b^2\right)\right)
$$
and
$$
r = \|p_0-A\| = \frac{1}{2} \sqrt{\frac{\left(a^2+b^2\right) \left(a^2+c^2\right) \left(b^2+c^2\right)}{a^2b^2+a^2c^2+b^2 c^2}}
$$
Attached a plot for $a=1,b=2,c=3$. In pink the plane $(p-A)\cdot \vec n=0$. In green the plane $(p-\frac{A+B}{2})\cdot(A-B)=0$ and in yellow the plane $(p-\frac{A+C}{2})\cdot(A-C)=0$. In red the line $p=(0,\frac{b^2-a^2}{2b},\frac{c^2-a^2}{2c})+\lambda(1,\frac ab,\frac ac)$
| {
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How to solve $a^3 + 39 ab^2 - 18 = 0$, $3a^2 b + 13 b^3 - 5 = 0$ In this answer, the user @123 has claimed that by solving the system
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
we give $ a = \dfrac 32$ and $ b = \dfrac12$. Could anyone explain for me that how one can solve such a system, please?
| Disclaimer: This answer is probably not appropriate considering (algebra-precalculus) tag, but I'll write it anyway since it might be useful to others stumbling upon this question.
The system is easier to solve when considering original equation $$(a+b\sqrt{13})^3=18+5\sqrt{13}.$$
If we look at norm $N(x+y\sqrt{13}) = x^2-13y^2$, it turns out that $N(18+5\sqrt{13}) = -1$. By multiplicativity of $N$ it follows that $N(a+b\sqrt{13}) = -1$, which gives us $a^2-13b^2 = -1$. Substitute $13b^2 = a^2 + 1$ in $a^3+39ab^2 - 18 = 0$ to get equation $$4a^3+3a-18 = 0.$$
By rational root theorem, the only rational root of the last equation is $a=\frac 32$. It follows that $13b^2 = (3/2)^2+1$, or $b^2 = \frac 14$. From the equation $3a^2b + 13b^3 = 5$, we know that $b$ must be positive, so $b = \frac 12$.
Note that there are complex solutions as well.
| {
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Computing the determinant of a large matrix? How would I go about computing the determinant of large matrices, such as $6 \times 6$. I believe that I need to use multilinear maps, but I am not sure how I can go about computing the determinant in a nice and efficient way.
Can anyone show me how I can determine the determinant of the matrix below in a simple and efficient way?
\begin{pmatrix}
0 & 0& 1& 1& 1 & 1\\
1 & 0 & 0 & 0& 0& 1\\
1 & 0 & 1& 1 & 1 &1 \\
0 & 1 & 1 & 1 & 0 &1 \\
0 & 1 & 0& 1 & 0& 0\\
0 & 0& 1& 0 & 0& 0
\end{pmatrix}
| First, note that row 6 is mostly '$0$'s. Next note the red values in row 2. The non-zero part of the determinant has to go through that $1$, so swap rows 2 and 5,
\begin{align}
\begin{vmatrix}
0&0&1&1&1&1 \\
1&\color{red}0&0&\color{red}0&0&1 \\
1&0&1&1&1&1 \\
0&1&1&1&0&1 \\
0&1&0&1&0&0 \\
0&0&\color{red}1&0&0&0
\end{vmatrix}& = -
\begin{vmatrix}
0&0&1&1&1&1 \\
0&1&0&1&0&0 \\
1&0&1&1&1&1 \\
0&1&1&1&0&1 \\
1&\color{red}0&0&\color{red}0&0&1 \\
\color{red}0&\color{red}0&1&\color{red}0&\color{red}0&\color{red}0
\end{vmatrix}= 0
\end{align}
which shows that the determinant is zero.
The column solution is even easier:
\begin{align}
\begin{vmatrix}
0&\color{red}0&1&1&1&1 \\
1&\color{red}0&0&0&0&1 \\
1&\color{red}0&1&1&1&1 \\
\color{red}0&1&1&1&0&1 \\
\color{red}0&1&\color{red}0&1&0&0 \\
\color{red}0&0&1&0&0&0
\end{vmatrix}& = 0
\end{align}
This is a medium sized matrix at most - to find the determinant for a actual large matrix ($n>100$), look up RRQR.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$? How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$?
I've spent quite some time on this problem. So far, I've simplified the right-hand side to $\frac{1}{30}(n+1)[(2n+3)(3n^3)+n(n-1)]$. But then, the algebra becomes very complicated when I add $(n+1)^4$ to both sides of the inductive hypothesis.
| $$\underbrace{1^4+2^4+3^4\cdots\ + n^4}_{\sum_{k = 1}^n k^4} = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$$
See the Faulhaber's formula and the Bernoulli number: $\text{B}_{j}(x)$
\begin{align}
\sum_{k=0}^{n} k^p = \frac{1}{p+1} \sum_{j=0}^p (-1)^j {p+1 \choose j} B_j n^{p+1-j}=\frac{\text{B}_{p+1}(n+1)-\text{B}_{p+1}(0)}{p+1}
\end{align}
In your case
\begin{align}
\sum_{k = 1}^n k^4 &= \frac{\text{B}_{4+1}(n+1)-\text{B}_{4+1}(0)}{4+1}\\
&= \frac{1}{5}\text{B}_{5}(n+1)\\
&=\frac{1}{5}\left(\frac16 (-1 - n) + \frac53 (1 + n)^3 - \frac52 (1 + n)^4 + (1 + n)^5\right)\\
&=\frac{1}{5}\left(\frac{\left(-1-n\right)}{6}+\frac{5\left(1+n\right)^3\cdot 2}{6}-\frac{5\left(1+n\right)^4\cdot 3}{6}+\frac{\left(1+n\right)^5\cdot 6}{6}\right)\\
&=\frac{1}{5}\left(\frac{6n^5+15n^4+10n^3-n}{6}\right)\\
&=\frac{6n^5+15n^4+10n^3-n}{30}\\
&= \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n \tag*{$\Box$}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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calculating $\int_0 ^3 x\sqrt{|1-x^2|}dx$ I tried to calculate $\int_0 ^3 x\sqrt{|1-x^2|}dx$ using $x=\sin t$, but couldn't solve it and got stuck. Is $x=\sin t$ correct or should I have used another value for $x$? don't know how to solve it.
hoping you can help me with that.
thank you very much for helping.
| From x = 0 to x = 1,
$$1 - {x^2} \ge 0$$
so
$$\left| {1 - {x^2}} \right| = 1 - {x^2}$$
From x = 1 to x = 3
$$\left| {1 - {x^2}} \right| \le 0$$
so
$$\left| {1 - {x^2}} \right| = {x^2} - 1$$
$$\int_0^3 {x\sqrt {\left| {1 - {x^2}} \right|} dx} = \int_0^1 {x\sqrt {1 - {x^2}} dx} + \int_1^3 {x\sqrt {{x^2} - 1} } dx$$
Focusing on the first integral..
$$\int_0^1 {x\sqrt {1 - {x^2}} dx} $$
Let $$\boxed{u = 1-x^2}$$. It follows that:
$$\frac{{du}}{{dx}} = \frac{d}{{dx}}\left( {1 - {x^2}} \right) = - 2x$$
$$du = - 2xdx$$
$$xdx = - \frac{{du}}{2}$$
$$x = 0 \to u = 1 - {0^2} = 1$$
$$x = 1 \to u = 1 - {1^2} = 0$$
\begin{equation}
\int_{x = 0}^{x = 1} {x\sqrt {1 - {x^2}} dx = } - \frac{1}{2}\int_{u = 1}^{u = 0} {\sqrt u du} = - \frac{1}{2}\int_{u = 1}^{u = 0} {{u^{1/2}}du}
\end{equation}
If we switch the limits on the far right integral, that cancels out the $-$ sign...
$$\boxed{\int_{x = 0}^{x = 1} {x\sqrt {1 - {x^2}} dx = } \frac{1}{2}\int_{u = 0}^{u = 1} {\sqrt u du}}$$
The second integral... Let
$$\boxed{v = x^2 - 1}$$
It follows that:
$$\frac{{dv}}{{dx}} = \frac{d}{{dx}}\left( {{x^2} - 1} \right) = 2x$$
$$dv = 2xdx$$
$$xdx = \frac{{dv}}{2}$$
$$x = 1 \to v = {1^2} - 1 = 0$$
$$x = 3 \to v = {3^2} - 1 = 8$$
$$\boxed{\int_{x = 1}^{x = 3} {x\sqrt {{x^2} - 1} dx = } \frac{1}{2}\int_{v = 0}^{v = 8} {\sqrt v dv} = \frac{1}{2}\int_{v = 0}^{v = 8} {{v^{1/2}}dv}}$$
So from the boxed equations above... we have
$$\boxed{\int_{x = 0}^{x = 3} {x\sqrt {\left| {1 - {x^2}} \right|} dx = } \frac{1}{2}\int_{u = 0}^{u = 1} {\sqrt u du} + \frac{1}{2}\int_{v = 0}^{v = 8} {\sqrt v dv}}$$
Do you know how to finish from there ^^ ?
| {
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"timestamp": "2023-03-29T00:00:00",
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Residue of $\frac{\cos (\pi z)}{z^2-1}$ I want to find the residue $\displaystyle\frac{\cos (\pi z)}{z^2-1}$ at $z=1$.
The function has two poles, at $-1,1$, both of order $1$. The series expansion for the function gives $\displaystyle\frac{1}{z^2-1}\left(1- \frac{z^2\pi^2}{2!}+O(z^4z\pi^4) \right)$.
(1) Is then Laurent series the following?
$$\frac{1}{z^2-1}- \frac{z^2\pi^2}{2!(z^2-1)}+\frac{z^4}{4!(z^2-1)}+\dots$$
(2) How do I find the pole of order $1$ from the Laurent series?
(3) I know that the residue should be given by $\displaystyle\lim_{z \to 1}(z-1)\frac{\cos (\pi z)}{z^2-1}=\frac{1}{z+1}- \frac{z^2\pi^2}{2!(z+1)}+\frac{z^4\pi^4}{4!(z+1)}+\dots=\frac{1}{2}-\frac{\pi^2}{2\cdot2!}+\frac{\pi^4}{2\cdot4!}+\dots$
Is the residue $1/2$? I am not sure of what term in the Laurent series I should pick up to be the residue.
| You have the function $f(z) = \frac{g(z)}{z^2-1}$, where $g(z)= \cos(\pi z)$ and therefore holomorf. The residue at $z=1$:
Using limits.
\begin{align}
\text{Res}_{z=1} f(z) =\lim_{z\to 1} (z-1)\frac{g(z)}{(z-1)(z+1)} = \frac{g(1)}{2} = \frac{-1}{2}
\end{align}
Using Laurent series. You must have powers of $z-1$ in your laurent series:
\begin{align}
\frac{\cos(\pi z)}{z^2-1} &= \frac{\cos(\pi(z-1) + \pi)}{(z-1)(z+1)} \\&
= -\frac{\cos(\pi( z-1))}{(z-1)((z-1)+2)}\\
& = -\frac{1}{2}\frac{\cos(\pi( z-1))}{(z-1)((z-1)/2+1)} \\
&= -\frac{1}{2}\cdot \frac{\cos(\pi(z-1))}{(z-1)} \cdot \sum_{k=0}^\infty (-(z-1))^k \\
&= -\frac{1}{2} (1 + O((z-1)^2) \sum_{k=0}^\infty (-1)^k(z-1)^{k-1}\\
&=-\frac{1}{2} (1 + O((z-1)^2) \cdot ((z-1)^{-1}+O(1))\\
&= -\frac{1}{2} (z-1)^{-1} + O(1)
\end{align}
We used the geometric series. Now you can see the coefficient of $(z-1)^{-1}$ is $-\frac 1 2 $ and therefore the residue at $z=1$ is $-\frac 1 2 $.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to simplify solutions for $y'' + 4y = 2 \tan x$? The original equation is $$y'' + 4y = 2 \tan x$$
What I did so far:
$$\lambda^2+4 = 0$$
$$\lambda_1 = -2i \quad \lambda_2 = 2i$$
$$y(x) = C_1\cos2x+C_2\sin2x$$
Write down the system:
$$\begin{cases} C_1'\cos2x + C_2'\sin2x = 0 \\
2C'_2\cos2 x - 2C_1'\sin2x = 2\tan x \end{cases}$$
Applying Krammer's formulas
$$\Delta = \begin{vmatrix}\cos2x & \sin2x \\ 2\cos2x & -2\sin2x \end{vmatrix} = -2\cos4x$$
$$\Delta_2 = \begin{vmatrix} 0 & \sin2x \\ 2\tan x & -2\sin2x\end{vmatrix} = -2\tan x\sin2x = -4\sin^2x$$
$$\Delta_3 = \begin{vmatrix}\cos2x & 0 \\ 2\cos2x & 2\tan x \end{vmatrix} = -2\cos2x\tan x$$
but when I tried to find solutions I got different expressions which I do not know how to integrate, what is the best way to simplify solutions here to get better integrals?
| Mutiply the first equation by $\cos(2x)$ and the second by $\sin(2x)$ and subtract them from each other to get:
$$C_1'\cos^2(2x) + C_2'\cos(2x)\sin(2x) - C_2'\cos(2x)\sin(2x) + C_1'\sin^2(2x) = - \tan(x)\sin(2x)$$
$$C_1' = - 2\frac{\sin x}{\cos x} \cos(x)\sin(x) = - 2\sin^2(x) = \cos(2x) - 1$$
This is easily integrable.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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A deck of 40 cards, 4 suits from 1 to 10, 2 cards extracted, probability the sum is 9? You have a deck of $40$ cards, $4$ suits from $1$ to $10$, pick randomly $2$ cards, no reimmission, what is the probability you get $9$ as a sum?
How can I solve using the basic principle of counting?
The favorable outcome are $(8+1)*4$, $(7+2)*4$, $(6+3)*4$ and $(5+4)*4$, so $64$ possible combination?
But what about the all the possible outcome? Are $40*39$?
From a MC simulation the results seems $~0.08$, but I can't do analytically.
EDIT:
The numerator is wrong, a combination can be also 1 and then 8, 2+7 and so on, the combinations are, actually, doubled, 128.
$$ \frac{128}{40\cdot39} = .0821 $$
Thank you all for the answer.
| What is the probability that we select $1$ and $8$? There are $4$ $1$'s and $4$ $8$'s and we select $2$ cards from the $40$ so we have
$$\frac{{4\choose{1}} \cdot {4\choose{1}}}{40\choose{2}}$$
This is the same as the probability of selecting a $2$ and a $7$, a $3$ and a $6$, and a $4$ and $5$.
So the probability that the $2$ cards sum to $9$ is
$$\frac{4 \cdot {4\choose{1}} \cdot {4\choose{1}}}{40\choose{2}}\approx.082$$
Another way to approach this is to consider the probability of selecting a $8$ and a $1$. We can select $8$ and then $1$ or $1$ then $8$ giving a probability of
$$2\cdot \frac{4}{40} \cdot \frac{4}{39}$$
Then accounting for $1$ and $8$, $2$ and $7$, $3$ and $6$, and $4$ and $5$, we get
$$4 \cdot 2\cdot \frac{4}{40} \cdot \frac{4}{39} \approx .082$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520623",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Evaluating limits using Taylor's expansion. Evaluate the following limit : $\lim_{x \to 0}(\frac{1}{\log(\cos(x))}+\frac{2}{\sin^2(x)})$
Generally speaking I have no idea at what order should I stop the Taylor's expansion.
I've tried this using two different orders and i've gotten both the results 1 and 0.
If anyone could point out how I should pick the order of the expansion i would be grateful!
| $$\lim \limits_{x \to 0}\left(\frac{1}{\log(\cos(x))}+\frac{2}{\sin^2(x)}\right)=\lim \limits_{x \to 0}\frac{\sin^2x+2\ln\cos x}{\sin^2x\ln\cos x}$$
We can stop Taylor's expansion at the denominator to the first (different from zero) term, because it's a product and nothing is going to cancel. So if $x$ tends to $0$ we have
$$\sin^2x\ln\sin x=\left(x^2+o(x^2)\right)\ln\left(1-\frac{x^2}{2}+o(x^2)\right)=\left(x^2+o(x^2)\right)\left(-\frac{x^2}{2}+o(x^2)\right)=$$
$$=-\frac{x^4}{2}+o(x^4)$$
So we need $x^4$ "precision" on the numerator too, to avoid approximation errors. Hence
$$\sin^2x+2\ln\cos x=\left(x-\frac{x^3}{6}+o(x^3)\right)^2+2\ln\left(1-\frac{x^2}{2}+\frac{x^4}{24}+o(x^4)\right)=$$
$$=x^2-\frac{x^3}{6}-\frac{x^4}{3}+o(x^4)+2\left(-\frac{x^2}{2}+\frac{x^4}{24}-\frac{1}{2}\left(-\frac{x^2}{2}\right)^2+o(x^4)\right)$$
So finally we have
$$\lim \limits_{x \to 0}\left(\frac{1}{\log(\cos(x))}+\frac{2}{\sin^2(x)}\right)=\lim \limits_{x \to 0}\frac{-\frac{x^4}{2}+o(x^4)}{-\frac{x^4}{2}+o(x^4)}=1$$
| {
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} |
Prove that, if $a,b,c$ are positive real numbers $\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$
Prove that, if $a,b,c$ are positive real numbers:
$$\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$$
Can the proof be written without the need for high mathematics?
| We'll prove a stronger inequality:
$$\sum_{cyc}\frac{a}{2a+b+c}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{1}{4}-\frac{a}{2a+b+c}\right)\geq0$$ or
$$\sum_{cyc}\frac{b+c-2a}{2a+b+c}\geq0$$ or
$$\sum_{cyc}\frac{c-a-(a-b)}{2a+b+c}\geq0$$ or
$$\sum_{cyc}\left(\frac{c-a}{2a+b+c}-\frac{a-b}{2a+b+c}\right)\geq0$$ or
$$\sum_{cyc}\left(\frac{a-b}{2b+c+a}-\frac{a-b}{2a+b+c}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2}{(2a+b+c)(2b+a+c)}\geq0.$$
Done!
Also, we can use C-S:
$$\sum_{cyc}\frac{a}{2a+b+c}\leq\frac{1}{(1+3)^2}
\sum_{cyc}a\left(\frac{1^2}{a}+\frac{3^3}{a+b+c}\right)=$$
$$=\frac{1}{16}\sum_{cyc}\left(1+\frac{9a}{a+b+c}\right)=\frac{3}{16}+\frac{9}{16}=\frac{3}{4}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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For each $a \in \mathbb{R}$ evaluate $ \lim\limits_{n \to \infty}\left(\begin{smallmatrix}1&\frac{a}{n}\\\frac{-a}{n}&1\end{smallmatrix}\right)^n$
If $a \in \mathbb{R}$, evaluate $$ \lim_{n \to \infty}\left(\begin{matrix} 1&\frac{a}{n}\\\frac{-a}{n}&1\end{matrix}\right)^{n}$$
My attempt: Let $$A = \left(\begin{matrix} 0&a\\-a&0\end{matrix}\right) = -a\left(\begin{matrix} \cos(\frac{\pi}{2})&-\sin(\frac{\pi}{2})\\\sin(\frac{\pi}{2})&\cos(\frac{\pi}{2})\end{matrix}\right)$$ so that $$A^k = (-a)^k \left(\begin{matrix} \cos(\frac{k\pi}{2})&-\sin(\frac{k\pi}{2})\\\sin(\frac{k\pi}{2})&\cos(\frac{k\pi}{2})\end{matrix}\right)$$
Thus,
\begin{align}\displaystyle \lim_{n \to \infty}\left(\begin{matrix} 1&\dfrac{a}{n}\\\dfrac{-a}{n}&1\end{matrix}\right)^{n} &=\displaystyle \lim_{n \to \infty} \left(I+\dfrac{A}{n}\right)^n =e^A=\displaystyle \sum_{k=0}^{\infty}\dfrac{A^k}{k!}\\&= \sum_{k=0}^{\infty} \left(\begin{matrix} \dfrac{(-a)^k\cos(\frac{k\pi}{2})}{k!}&-\dfrac{(-a)^k\sin(\frac{k\pi}{2})}{k!}\\\dfrac{(-a)^k\sin(\frac{k\pi}{2})}{k!}&\dfrac{(-a)^k\cos(\frac{k\pi}{2})}{k!}\end{matrix}\right) \end{align}
and since $\displaystyle \sum_{k=0}^{\infty}\dfrac{(-a)^k\cos(\frac{k\pi}{2})}{k!}=1+0-\dfrac{a^2}{2!}+0+\dfrac{a^4}{4!}+\cdots= \cos a$ and
$\displaystyle \sum_{k=0}^{\infty}\dfrac{(-a)^k\sin(\frac{k\pi}{2})}{k!}=0-a+0+\dfrac{a^3}{3!}+0-\dfrac{a^5}{5!}+\cdots= -\sin a$ therefore the required answer is
$\left(\begin{matrix} \cos a&\sin a\\-\sin a&\cos a\end{matrix}\right).$
However the above answer does not match the choices provided which are $I, 0$ and none of the above. So my question is: Is my answer correct?
| The best approach is to diagonalize your matrix. Both $(1,i)$ and $(1,-i)$ are eigenvectors. So, let$$T=\begin{pmatrix}1&1\\i&-i\end{pmatrix}.$$Then$$T^{-1}.\begin{pmatrix}1&\frac an\\-\frac an&1\end{pmatrix}.T=\begin{pmatrix}1+\frac ani&0\\0&1-\frac 1ni\end{pmatrix}.$$Therefore$$T^{-1}.\begin{pmatrix}1&\frac an\\-\frac an&1\end{pmatrix}^n.T=\begin{pmatrix}1+\frac ani&0\\0&1-\frac 1ni\end{pmatrix}^n=\begin{pmatrix}\left(1+\frac{ai}n\right)^n&0\\0&\left(1-\frac{ai}n\right)^n\end{pmatrix}$$ and so$$\lim_{n\to\infty}T^{-1}.\begin{pmatrix}1&\frac an\\-\frac an&1\end{pmatrix}^n.T=\begin{pmatrix}e^{ai}&0\\0&e^{-ai}\end{pmatrix}.$$So$$\lim_{n\to\infty}\begin{pmatrix}1&\frac an\\-\frac an&1\end{pmatrix}^n=T.\begin{pmatrix}e^{ai}&0\\0&e^{-ai}\end{pmatrix}.T^{-1}=\begin{pmatrix}\cos a&\sin a\\-\sin a&\cos a\end{pmatrix}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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Simplify $\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$
Which one is equivalent to: $$\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$$
1)$\dfrac{\cos(x)}{1+\sin(x)}$
2)$\dfrac{\cos(x)}{1-\sin(x)}$
3)$\dfrac{1-\sin(x)}{\cos(x)}$
4)$\dfrac{1+\sin(x)}{\cos(x)}$
Of course we can find the correct choice by testing each of them,but I'm looking for an analytical solution assuming we don't know the final expression.
I tried the following techniques,all failed: multiplying by $\frac{\sin(x)}{\sin(x)}$ , multiplying by $\frac{\cos(x)}{\cos(x)}$ , using $\sin^2(x)+\cos^2(x)=1$
| Multiply numerator and denominator by $1-\sin x+\cos x$: you get
\begin{align}
\frac{1-\sin x+\cos x}{1+\sin x+\cos x}
&=\frac{(1-\sin x+\cos x)^2}{(1+\cos x)^2-\sin^2x} \\[6px]
&=\frac{1+\sin^2x+\cos^2x-2\sin x+2\cos x-2\sin x\cos x}{1+2\cos x+\cos^2x-(1-\cos^2x)}\\[6px]
&=\frac{2(1-\sin x+\cos x-\sin x\cos x)}{2\cos x(1+\cos x)}\\[6px]
&=\frac{(1-\sin x)(1+\cos x)}{\cos x(1+\cos x)}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $6^{273} + 8^{273}\pmod{49}$
The number $6^{273} + 8^{273}$ divided by $49$ has a remainder, what is its value?
I used the totient function to compute for modulo 49.
$6^{42}$ and $8^{42}$ are $-1$ and $1$ mod $49$ respectively, $273/49$ is equal to $5$ with a remainder of $21$.
We would then look for the remainder of $-6^{21}$ + $8^{21}$ which I do not know how to solve.
I am aware of other solutions such as factoring odd exponents, but I wanted to know if we can use this kind of approach.
| Note that $6\cdot 8\equiv 48 \equiv -1 \bmod 49$, so $8^{-1}\equiv -6 \bmod 49$.
Then $6^{273}+8^{273} \equiv 6^{273}+(-6)^{-273} \equiv 6^{273} -6^{-273}\bmod 49$ (since $-1^{-1} \equiv -1$)
Since $\phi(49)=42$ Euler gives us $6^{42}\equiv 1\bmod 49$. Then note that $6^{273}\not\equiv 0 \bmod 49$ and $6^{273} (6^{273} -6^{-273}) \equiv 6^{546} -1 \equiv (6^{42})^{13} - 1 \equiv 0 \bmod 49 $ and thus $6^{273} -6^{-273} \equiv 6^{273} + 8^{273} \equiv 0 \bmod 49$
| {
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"timestamp": "2023-03-29T00:00:00",
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Derive an $O(h^4)$ five point formula to approximate $f'(x_0)$ that uses $f(x_0 - h), f(x_0), f(x_0 + h), f(x_0 + 2h), f(x_0 + 3h)$ This is a Numerical Analysis textbook question on Numerical Differentiation (like https://en.wikipedia.org/wiki/Numerical_differentiation):
Official Textbook Question
Q: Derive an $O(h^4)$ five point formula to approximate $f'(x_0)$ that uses $f(x_0 - h), f(x_0), f(x_0 + h), f(x_0 + 2h), f(x_0 + 3h)$
Hint: Consider the expression $A f(x_0 - h) + B f(x_0 + h) + C f(x_0 + 2h) + D f(x_0 + 3h)$. Expand in fourth Taylor polynomials, and choose $A, B, C, D$ appropriately.
My Work
I can follow the full textbook derivation of the "Three Point" Formulas, the Three Point Endpoint Formula:
\begin{align*}
f'(x_0) &= \frac{1}{2h} \left[ -3 f(x_0) + 4 f(x_0 + h) - f(x_0 + 2h) \right] + \frac{h^2}{3} f^{(3)}(\xi_0) \\
\end{align*}
where $\xi_0$ is between $x_0$ and $x_0 + 2h$
and the Three-Point Midpoint Formula:
\begin{align*}
f'(x_0) &= \frac{1}{2h} \left[ f(x_0 + h) - f(x_0 - h) \right] - \frac{h^2}{6} f^{(3)}(\xi_1) \\
\end{align*}
where $\xi_1$ is between $x_0 - h$ and $x_0 + h$
These derivations involve calculating the Lagrange interpolating polynomial of three evenly spaced points, calculating the derivative, and solving for the derivative at each point. Using either the first or last point gives you the three point endpoint formula and using the midpoint gives you the three point midpoint formula.
Obviously, the same technique would work for five points, but the equations get very tedious and hard to work with, and the problem hint obviously doesn't want us to solve the problem that way.
I don't quite understand how the hint wants me to proceed. Is a fourth Taylor polynomial something like this:
\begin{align*}
f(x) &= f(x_0) + \frac{f'(x_0)}{1!} (x - x_0) + \frac{f^{(2)}(x_0)}{2!} (x - x_0)^2 + \frac{f^{(3)}(x_0)}{3!} (x - x_0)^3
\end{align*}
I don't see how that would help solve this problem. Thanks!
| Consider that your formula must give the correct values for $f (x) = 1$, $f (x) = x$, $f (x) = x^2$, $f (x) = x^3$ and $f (x) = x^4$. The hint is weird, since it assumes that you are not using f (x).
So you get the equations:
$A + B + C + D + E = 0$, $A(x-h) + B(x) + C(x+h) + D(x + 2h) + E(x + 3h) = 1$, $A(x-h)^2 + B(x)^2 + C(x+h)^2 + D(x + 2h)^2 + E(x + 3h)^2 = 2x$, $A(x-h)^3 + B(x)^3 + C(x+h)^3 + D(x + 2h)^3 + E(x + 3h)^3 = 3x^2$, $A(x-h)^4 + B(x)^4 + C(x+h)^4 + D(x + 2h)^4 + E(x + 3h)^4 = 4x^3$.
Consider that your formula must give the correct values for $f (x) = 1$, $f (x) = x$, $f (x) = x^2$, $f (x) = x^3$ and $f (x) = x^4$. The hint is weird, since it assumes that you are not using f (x).
These equations look a bit difficult. So we use different polynomials:
$f (x) = 1$, $f (x) = (x-x_0)/h$, $f (x) = ((x-x_0)/h)^2$, $f (x) = ((x-x_0)/h)^3$ and $f (x) = ((x-x_0)/h)^4$.
(1) $A + B + C + D + E = 0$
(2) $-A + C + 2D + 3E = 1/h$
(3) $A + C + 4D + 9E = 2/h^2$
(4) $-A + C + 8D + 27E = 3/h^3$
(5) $A + C + 16D + 81E = 4/h^4$.
This is easy enough to solve for A, B, C, D and E by subtracting (4) - (2) and (5) - (3), calculating E and D, subtracting (3) - (2) and substituting D, E to get C, adding (2) + (3) and substituting D, E to get A, and substituting all in (1) to get B. To follow the original "hint" set B = 0 instead.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of integer reciprocals The following problem results in a cubic, but is there a way to simplify it easier?:
The sum of the reciprocals of three consecutive integers is 47/60. What is the sum of these integers?
I ended up with $47{ x }^{ 3 }-180{ x }^{ 2 }-47x+60 =0$. However, unlike other methods where I've learned to "Let ${ x }^{ 2 }$ = k, $\therefore { x }^{ 4 }={ k }^{ 2 }$", ${x}^{3}$ is not a multiple of 2, so I cannot apply this same method. How should I proceed? (Other than just substitute, which the answers from this & this website did).
| This answer is based upon symmetry and a guess.
We try to exploit symmetry and use the approach
\begin{align*}
\color{blue}{\frac{47}{60}}&=\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}\\
&=\frac{n(n+1)+n^2-1+(n-1)n}{n(n^2-1)}\\
&\color{blue}{=\frac{3n^2-1}{n(n^2-1)}}
\end{align*}
Let's take a closer look at
\begin{align*}
\frac{3n^2-1}{n(n^2-1)}=\frac{47}{60}
\end{align*}
Hoping for a simple proportion we equate the numerators and get
\begin{align*}
3n^2-1=47\quad\Rightarrow\quad 3n^2=48\quad\Rightarrow\quad n\in\{4,-4\}
\end{align*}
The denominator $n(n^2-1)$ evaluated at $n=4$ gives $4(16-1)=60$ and we get the solution $$\color{blue}{n=4}$$
Note: Observe, it was the symmetry which enabled us to make the lucky guess.
| {
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"url": "https://math.stackexchange.com/questions/2537557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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if $\frac{m}{n},\frac{m+1}{n+2},\frac{m+2}{n+4},\frac{m+3}{n+6}$be positive integers, find the minimum of the m let $m$ be a positive integer,and $n$ be an positive integers,such $$\dfrac{m}{n},\dfrac{m+1}{n+2},\dfrac{m+2}{n+4},\dfrac{m+3}{n+6}$$be positive integers,find the minimum of the $m$.
It seem use CRT to solve it? But I can't find it so far
| If $\frac{m}{n}, \frac{m+1}{n+2}, \frac{m+2}{n+4}, \frac{m+3}{n+6}$ are all integers, so are $\frac{2m}{n}, \frac{2m+2}{n+2}, \frac{2m+4}{n+4}, \frac{2m+6}{n+6}$, and so are $\frac{2m-n}{n}, \frac{2m-n}{n+2}, \frac{2m-n}{n+4}, \frac{2m-n}{n+6}$. Hence $2m-n$ is a multiple of $\mbox{lcm}(n, n+2, n+4, n+6)$.
By Sridhar's comment, $n$ has to be odd, and by looking at the least common multiples of small $n$'s, the minimal l.c.m. is at $n=1$, hence $2m-1 \geq 105$, and $m\geq 53$. From Oscar's post we see that actually $m=53$ is the minimal solution!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find this maximum of the $\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$ Let $x\in \mathbb{R}$, find the function maximum of the value
$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
my attemp
$$x^2=5+4\sin{t},t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$
then
$$f=\dfrac{5\sqrt{3}}{4}+2\sin{\left(t+\frac{\pi}{6}\right)}\le 2+\dfrac{5}{4}\sqrt{3}$$
My Question:this function have other methods to find this maximum? such as AM-GM,Cauchy-Schwarz inequality and so on?
| Consider the function and its first derivative$$f(x)=\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
$$f'(x)=\frac{1}{2} x \left(\frac{5-x^2}{\sqrt{-x^4+10 x^2-9}}+\sqrt{3}\right)$$ The first derivative cancels for $x=0$ which has to be excluded.
Now, let $x^2=y$ and solve
$$\frac{5-y}{\sqrt{-y^2+10 y-9}}+\sqrt{3}=0\implies 5-y=-\sqrt{3}\sqrt{-y^2+10 y-9}$$ and square to get $$4 y^2-40 y+52=0 \implies y=5\pm2 \sqrt{3}$$ Only $y=5+2 \sqrt{3}$ must be kept because of the real domain.
$$f(\sqrt{5+2 \sqrt{3}})=2+\frac{5 \sqrt{3}}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $h_n = 3h_{n-1} -4n$ using generating functions Solve $h_n = 3h_{n-1} -4n$, where $h_0 = 2$ using generating functions.
I am struggling to figure out how to solve this using generating functions. I know the answer should be $h_n = -3^n +2n + 3$.
Here is the method that both my textbook and professor used.
Let $g(x) =\sum_{n=0}^{\infty} h_nx^n $. Then let $h_n = 0$ if $n>0$.
Then $3x g(x)= \sum_{n=0}^{\infty} 3h_nx^{n+1} = \sum_{n=0}^{\infty} 3h_{n-1} x^n $
Subtracting these two equations we get
$(1-3x)g(x) = \sum_{n=0}^{\infty}(h_n - 3h_{n-1})x^n $
$= h_0 + (h_1-3h_0)x + ... + (h_n -3h_{n-1})x^n + ... $
$= 2-4x-8x^2 -...-4nx^n -...$
From here we are suppose to find some sort of pattern to simplify, but all the example have coefficients that can be written as a number to the power n, but here we have -4n instead. How do I compensate for this?
| From where you left off, the first problem is to find the generating function for sequence $a_n = -4n$. This is not so hard if know the relation between operations on generating functions and operations on their corresponding sequences (see here): $\langle a_n \rangle = -4( \langle 1, 1, 1, \cdots \rangle * \langle 1, 1, 1, \cdots \rangle - \langle 1, 1, 1, 1 \cdots \rangle)$, where $*$ is the discrete convolution, which corresponds to multiplication of generating functions. So
$$\sum_{n = 1}^{\infty} -4nx^n =-4\left( \frac{1}{1-x}\frac{1}{1-x} - \frac{1}{1-x}\right) = \frac{-4x}{(1-x)^2}$$
Another way to get this is to notice that $nx^n = (x^{n+1})' - x^n$, so $$\sum_{n = 0}^{\infty} nx^n = \sum_{n = 0}^{\infty} (x^{n+1})' - \sum_{n = 0}^{\infty} x^n = (\sum_{n = 0}^{\infty} x^{n+1})' - \sum_{n = 0}^{\infty} x^n = (\sum_{n = 1}^{\infty} x^{n})' - \sum_{n = 0}^{\infty} x^n$$
So this evaluates to $(\frac{1}{1-x} - 1)' - \frac{1}{1-x} = \frac{1}{(1-x)^2}$.
So plug this into your result, we get
$$g(x) = \frac{1}{1-3x}\left(2 - \frac{-4x}{(1-x)^2}\right) = \frac{2x^2 -8x + 2}{(1-x)^2(1-3x)}$$
The only way I know to proceed here is to decompose this function into the following form:
$$\frac{2x^2 -8x + 2}{(1-x)^2(1-3x)} = \frac{Ax + B}{(1-x)^2} + \frac{C}{1-3x}$$
where after computing the right-hand-side out and matching the coefficient, I get $A = -1, B = 3, C = -1$. So, in fact,
$$g(x) = \frac{3}{(1-x)^2} - \frac{x}{(1-x)^2} - \frac{1}{1-3x}$$
The first term represents $3(n+1)$, the second $n$, and the third $n^3$.
Thus $$h_n = 3(n+1) - n - n^3 = -n^3 + 2n + 3$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find value of $x^2+y^2$ using given expression
Consider two real numbers $x , y$ such that $$\left(x^2+1\right)\left(y^2+1\right)+9=6\left(x+y\right)$$ Hence find the value of $x^2+y^2$.
At first I tried to factorise the condition but the $(xy)^2$ created much problems. I also tried to create a complete square to get a simpler expression but to no avail. And just out of curiosity, can this question be interpreted geometrically so as to find the required value?
| Continuing from Dylan's answer, $xy = 1$ and $x+y = 3$. Then:
$$x^2 + y^2 = (x+y)^2 - 2xy = (3)^2 - 2(1) = \boxed{7}.$$
Moreover, note that we can swap the values of $x$ and $y$ without changing the expression. Therefore from user16394's answer, $x = \frac{1}{2}(3 + \sqrt{5}), y = \frac{1}{2}(3 - \sqrt{5})$ or vice versa. Thus the value of $x^2+y^2$ is again:
$$\frac{1}{4} \left((3 + \sqrt5)^2 + (3 - \sqrt{5})^2 \right) = \frac{1}{4} \cdot 2 \left(3^2 + (\sqrt{5})^2 \right) = \boxed{7}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Number of seven digit numbers without repetition of digits divisible by $3$ Find number of seven digit numbers divisible by $3$ with
$1.$ Repetition
$2.$ Without Repetition
For Part $1.$ The least seven digit number divisible by $3$ is $1000002$ and highest seven digit number is $9999999$
So total is $3000000$
For Part $2$ The least such number is $1023456$ and the next is $1023459$
Now sum of the digits of $1023456$ is $21$.
So if we include $7$ by removing one of the digits in $1023456$ then the digits that can be removed is $1$ or $4$.
If we include $8$ then digits that can be removed is $2$ or $5$.
but we get more cases here, any better way to approach?
| Let $N_7 = \text{what you are looking for}$
However, this consists of one set that doesn't use any zeros - $(NZ)_7$ - and another that does. The number in the set that uses zeros can be defined recursively as $6(NZ)_6$ because the zero can be placed in 6 different places.
Notice that $(NZ)_7 = \frac{\text{pair of two numbers that are divisible by 3}}{9c2} \cdot(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3)$
This is because the total sum of $1,2, ..8, 9$ is $45$ and we need to remove two numbers such that the sum of the others is still a multiple fo $3$ ie the two numbers we remove must be a multiple of $3$
Therefore, $(NZ)_7 = \frac{(NZ)_2}{9P2}(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3)$
Similarly, $(NZ)_6 = \frac{(NZ)_3}{9P3}(9\cdot8\cdot7\cdot6\cdot5\cdot4)$
$(NZ)_2$ can be counted, but also it can be expressed as $9\cdot3 - \frac{1}{9}\cdot 9 \cdot 3$. This is found because if the first is $0 \mod 3$, then there is an overcounting (1/3 of such cases are false $1/3 *1/3 = 1/9$)
Similarly, $(NZ)_2=9\cdot 8 \cdot 3 - \frac{2}{8} \cdot \frac{2}{3} \cdot 9 \cdot 8\cdot 3$. Here, the second part is found by considering the probability that the second number chosen is the same modulus as the first ( $2/8$), which results in only $1/3$ of the possible last digits (so we subtract out $2/3$)
Therefore,
$$N_7 = \frac{9 \cdot 3 - 3}{72}(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3) + 6 \cdot \frac{9\cdot8\cdot3-4\cdot9}{504}(9\cdot8\cdot7\cdot6\cdot5\cdot4) = 190080$$
This is definitely not my most eloquent answer, so please ask me questions if something is confusing, and I'll try to explain my thought process.
| {
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} |
Division by $0$ and its restrictions Consider the following expression:
$$\frac{1}{2} \div \frac{4}{x}$$
Over here, one would state the restriction as $x \neq 0 $, as that would result in division by $0$.
But if we rearrange the expression, then:
$$\begin{align}
\frac12\div\frac4x &= \frac{1}{2} \times \frac{x}{4} \\
&= \frac{x}{8}
\end{align}$$
In this expression, there are no restrictions. If we substitute $x = 0$, then the answer is $\frac{0}{8} = 0$.
So, how come in the first unsimplified expression, when we substitute $ x =0$, we get undefined, whereas in the simplified expression we get $0$?
| When we make a simplification like that, we are implicitly assuming that $x\neq 0$. Really, one should write that
$$
\frac{1}{2} \div \frac{4}{x}=\begin{cases}
x/8&\text{if $x\neq 0$}\\
\text{undefined}&\text{if $x=0$}.
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "100",
"answer_count": 9,
"answer_id": 4
} |
Simplifying expression of algebra I am having trouble simplifying the following,
$$\frac{-1}{2(3+n\pi i)} \left( e^{-(3+n\pi i )}- e^{3+n\pi i }\right)$$
I am wondering how it reduces to the following, looking for someone to show me the steps,
$$\frac{1}{3+n\pi i} \left( \cos(n\pi) + i\sin (n\pi) \right )\frac{e^3-e^{-3}}{2}$$
Here is my attempt and I'm not quite sure where I am going wrong,
$$\frac{1}{2(3+n\pi i)} \left( -e^{-(3+n\pi i )}+ e^{3+n\pi i }\right)$$
$$\frac{1}{(3+n\pi i)} \left( -e^{-n\pi i }+ e^{n\pi i }\right) \frac{e^3 - e^{-3}}{2}$$
$$\frac{1}{(3+n\pi i)} \left( -e^{-n\pi i }+ e^{n\pi i }\right) \sinh3$$
$$\frac{1}{(3+n\pi i)} \left( { -[\cos(n\pi) - i\sin(n\pi)] }+ [cos(n\pi) + i\sin(n \pi)]\right) \sinh3$$
| The mistake is in the first line of your attempt:
$$\frac{1}{2(3+n\pi i)} \left( -e^{-(3+n\pi i )}+ e^{3+n\pi i }\right)$$
$$\frac{1}{(3+n\pi i)} \left( -e^{-n\pi i }+ e^{n\pi i }\right) \frac{e^3 - e^{-3}}{2}$$
The two expressions are not the same (multiply the second one out and check for yourself). In particular, you cannot separate the powers of $e$ in this manner.
Instead, observe that
$$
\frac{e^{3+n\pi i} - e^{-(3 + n\pi i)}}{2} = \sinh(3 + n \pi i).
$$
Now, use the formula for $\sinh(z)$, for $z = x+iy$,
$$
\sinh(z) = \sinh(x)\cos(y) + i\cosh(x)\sin(y)
$$
to get
$$
\begin{align*}
\frac{-1}{2(3+n\pi i)}(e^{-(3+n\pi i)}-e^{3+n\pi i}) &= \frac{1}{3+n\pi i}(\sinh(3)\cos(n\pi) + i \cosh(3)\sin(n\pi))\\
&=\frac{1}{3+n\pi i}\left( \frac{e^{3}-e^{-3}}{2} \right) \cos(n \pi) \\
&=\frac{1}{3+n\pi i}\left( \frac{e^{3}-e^{-3}}{2} \right) (\cos(n \pi) + i \sin (n\pi)).
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove for any integer $a$ show that $a$ and $a^{4n+1}$ have the same last digit
For any integer $a$, show that $a$ and $a^{4n+1}$ have the same last digit
I know that if $a^{4n+1} \equiv a\pmod{10}$ then $10|a^{4n+1}-a$, so $2|a^{4n+1}-a$ and $5|a^{4n+1}-a$, but I'm not sure where to go from here.
| The most elementary method is simply to observe patterns of the last digits ( there are only 10 possible digits to check so it is quite fast) by multiplying last digit by $a,a^2,a^3,a^4 $, for greater powers - the pattern is repeating ..
$0 \rightarrow 0,0,0,0,.. $
$1 \rightarrow 1,1,1,1,.. $
$2 \rightarrow 4,8,6,2,.. $
$3 \rightarrow 9,7,1,3,.. $
$4 \rightarrow 6,4,6,4,.. $
$5 \rightarrow 5,5,5,5,.. $
$6 \rightarrow 6,6,6,6,.. $
$7 \rightarrow 9,3,1,7,.. $
$8 \rightarrow 4,2,6,8,.. $
$9 \rightarrow 1,9,1,9,.. $
From this you can devise even the more general proposition - if $a$ and $b$ have the same last digit then $ab^{4m}$ or $ba^{4m}$ have the same last digit like $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Operating with Exponential Generating Function I am having troubles understanding how to operate with generating functions to obtain a final formula. I have been looking to the answer to this question: Exponential generating function and number of balls .
$$\color{red}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}\color{green}{\frac12\left(e^x+e^{-x}\right)}\color{blue}{e^x}.$$
This expands to
$$\frac12\cdot\frac{x^2}{2!} + \frac12\cdot\frac{x^4}{4!} + \sum_{n=2}^\infty \frac1{16}2^nn(n-1)\frac{x^n}{n!} + \sum_{n=4}^\infty\frac1{768}2^nn(n-1)(n-2)(n-3)\frac{x^n}{n!}. $$
Simplification yields
$$\frac{x^2}{2!}+3\frac{x^3}{3!}+13\frac{x^4}{4!} + \sum_{n=5}^\infty \frac1{768} 2^nn(n-1)(n^2-5n+54)\frac{x^n}{n!}. $$
What are the steps to obtain this and the simplification? I can't see how to get from one to another
I have tried a different approach, since
$$ e^x + e^{-x} =
\sum_{n=0}^\infty {2x^{2n}\over (2n)!}; \; e^x =
\sum_{n=0}^\infty {x^{n}\over n!}, $$
Multiplying everything, I obtain:
$$
\frac{1}{2}\frac{x^2}{2!}\left(\sum_{n=0}^\infty\frac{2x^{2n}}{(2n)!}\right)\cdot \left(\sum_{n=0}^\infty \frac{x^n}{n!}\right) + \frac{1}{2}\frac{x^4}{4!}\left(\sum_{n=0}^\infty\frac{2x^{2n}}{(2n)!}\right)\cdot \left(\sum_{n=0}^\infty \frac{x^n}{n!}\right)
$$
Removing the 2 of the numerator of the fractions and simplifying $x^{2n}/2n!$ for $x^{n}/n!$; and if I understand multiplication correctly, I should have:
$$
\sum_{k=0}^{n}\binom{n}{k}\frac{x^n}{n!}
$$
as a result of both the multiplications, but doesn't look like any form of the right answer.
PS: Decided to make another question instead of commenting the answer because it is 2 years old already.
| The term with $\frac{1}{16}$ arises as follows:
\begin{align*}
\frac{x^2}{2}\cdot\frac{1}{2}e^{2x}
&= \frac{1}{4}x^2\sum_{n=0}^\infty \frac{2^nx^n}{n!} \\
&= \frac{1}{4}\sum_{n=0}^\infty \frac{2^nx^{n+2}}{n!} \\
&= \frac{1}{16}\sum_{n=0}^\infty \frac{2^{n+2}x^{n+2}}{n!} \\
&= \frac{1}{16}\sum_{n=2}^\infty \frac{2^nx^n}{(n-2)!} \\
&= \frac{1}{16}\sum_{n=2}^\infty n(n-1)\frac{2^nx^n}{n!}.
\end{align*}
The other term arises similarly, starting from $\frac{x^4}{24}\cdot\frac{1}{2}e^{2x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solution of: Let $k,n \in \mathbb{N}$ and $\frac{n(n-1)}{4} = \frac{k(k-1)}{2} $ Let $k,n \in \mathbb{N}$, $k,n < 101 $ and $\frac{n(n-1)}{4} = \frac{k(k-1)}{2} $.
How can I solve this and find $n$ and $k$ in an analytic way without trying some solutions?
| solving the given equation $$n^2-n=2k^2-2k$$ for $n$ we get
$$n=\frac{1}{2}+\sqrt{\frac{1}{4}+2k^2-2k}$$
or
$$n=\frac{1}{2}-\sqrt{\frac{1}{4}+2k^2-2k}$$
now you Can solve the inequality
$$n=\frac{1}{2}+\sqrt{\frac{1}{4}+2k^2-2k}<101$$
writing the inequality in the form
$$\sqrt{\frac{1}{4}+2k^2-2k}<101-\frac{1}{2}$$
squaring this we get
$$2k^2-2k<101^2-101$$ and from here we get the interval
$$\frac{1}{2}\left(1-\sqrt{20201}\right)<k<\frac{1}{2}\left(1+\sqrt{20201}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit $\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$ Can someone please check my conclusion to the evaluation of the following limit?
$$\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$$
I got that the limit is equal to infinity. If limit is equal to infinity does this mean that limit does not exist?
| $\text{HINT}$
$$x(16x^4 + x^2+1)^{1/4}-2x^2 $$ $$=x^2(16+ \frac{1}{x^2}+\frac{1}{x^4})^{1/4}-2x^2$$ $$=x^2(\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}-2)$$ $$=\frac{\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}-2}{\frac{1}{x^2}}$$
You can multiply the last fraction with $$\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}+2$$ both numerator and denominator and then you will multiply the resulting fraction with $$\sqrt{16 + \frac{1}{x^2}+\frac{1}{x^4}}+4$$ both numerator and denominator.
For more convinience with your calculations put $u=\frac{1}{x^2} \to 0$ as $x \to +\infty$
So calculate the limit of the function $f(u)$ as $u \to 0$ where $u=\frac{1}{x^2}$
Thus you will get the result $\frac{1}{32}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Taylor series expansion of sin(x) I understand that Taylor series expansion for $\sin(x)$ is derived as follow:
$$
\sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-...
$$
Now, what exactly is the first, second, and third term?
Is the first term just $\sin(x) = x$?
Is the second term $\sin(x) = x-\frac{x^3}{3!}$?
| In the Taylor expansion at $0$ of the function $\sin(x)$, the even powers of $x$, i.e. the "missing" terms, are zero because $\sin(x)$ is an odd function:
$$
\begin{align}\sin(x)&=\sum_{k=0}^{\infty}\frac{D^k(\sin(x))_{x=0}}{k!}\cdot x^k
\\&=\sin(0)+\cos(0)x+\frac{-\sin(0)}{2!}\cdot x^2+\frac{-\cos(0)}{3!}\cdot x^3+
\frac{\sin(0)}{4!}\cdot x^4+\frac{\cos(0)}{5!}\cdot x^5+o(x^5)
\\&=\underbrace{0+x+0\cdot x^2-\frac{x^3}{3!}+0\cdot x^4+\frac{x^5}{5!}}_{\text{polynomial expansion up to the $5$-th term }}+o(x^5).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$ I saw this problem somewhere recently and I was having some difficulty getting started on it.
The problem is twofold. The first is to evaluate:
$$\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$$
and once this is done, to explain what this has to do with the construction of a pentagon (maybe some other polygon?) using a compass and straight edge.
In terms of evaluating the series, I tried writing each $n$ as $m \cdot 2^k$ and evaluating the summation there since $2^k$ will alternate between + and - mod 5. However, this leads to a divergent series and I think this is not a valid thing to do since the original series is not absolutely convergent so we can't rearrange terms like that.
| Note
\begin{eqnarray}
&&\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)\\
&=&\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2}\right) -\sum_{k=0}^\infty\left( \frac{1}{5k+3}-\frac{1}{5k+4} \right)\\
&=&\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+2)}-\sum_{k=0}^\infty\frac{1}{(5k+3)(5k+4)}.
\end{eqnarray}
Let
$$ f(x)=\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+2)}x^{5k+2}, g(x)=\sum_{k=0}^\infty\frac{1}{(5k+3)(5k+4)}x^{5k+4}. $$
So
\begin{eqnarray}
f'(x)&=&\sum_{k=0}^\infty\frac{1}{5k+1}x^{5k+1}, f''(x)&=&\sum_{k=0}^\infty x^{5k}=\frac{1}{1-x^5}, \\
g'(x)&=&\sum_{k=0}^\infty\frac{1}{5k+3}x^{5k+3},
g''(x)&=&\sum_{k=0}^\infty x^{5k+2}=\frac{x^2}{1-x^5}
\end{eqnarray}
and hence
\begin{eqnarray}
&&\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)\\
&=&\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+2)}-\sum_{k=0}^\infty\frac{1}{(5k+3)(5k+4)}\\
&=&f(1)-g(1)\\
&=&\int_0^1\int_0^t\frac{1}{1-x^5}dxdt-\int_0^1\int_0^t\frac{x^2}{1-x^5}dxdt\\
&=&\int_0^1\int_x^1\frac{1}{1-x^5}dtdx-\int_0^1\int_x^1\frac{x^2}{1-x^5}dtdx\\
&=&\int_0^1\frac{1-x}{1-x^5}dx-\int_0^1\int_x^1\frac{(1-x)x^2}{1-x^5}dx\\
&=&\int_0^1\frac{1-x^2}{1+x+x^2+x^3+x^4}dx\\
&=&\int_0^1\frac{\frac{1}{x^2}-1}{\frac{1}{x^2}+\frac{1}{x}+1+x+x^2}dx\\
&=&\int_0^1\frac{\frac{1}{x^2}-1}{(x+\frac{1}{x})^2+(x+\frac{1}{x})-1}dx\\
&=&-\int_0^1\frac{1}{(x+\frac{1}{x})^2+(x+\frac{1}{x})-1}d(x+\frac{1}{x})\\
&=&\int_2^\infty\frac{1}{u^2+u-1}du\\
&=&\frac{\log \left(\frac{1}{2} \left(7+3 \sqrt{5}\right)\right)}{2\sqrt{5}}.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 3
} |
Taylor series for $\sqrt{1+x^2}$ I want to expand
$$f(x)=\sqrt{1+x^2}$$ in powers of $x-2$
I started by getting the maclaurin series
$$\sqrt{1+x}=1+\frac{1}{2}x+\frac{1}{2} \left( \frac{-1}{2} \right) \frac{x^2}{2!} + \frac{1}{2} \left( \frac{-1}{2}\right) \left(\frac{-3}{2}\right)\frac{x^3}{3!}$$
$$\sqrt{1+x^2}=1+\frac{1}{2}x^2+\frac{1}{2}\left(\frac{-1}{2}\right)\frac{x^4}{2!}+\frac{1}{2} \left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)\frac{x^6}{3!}$$
Then
$$\sqrt{1+x^2}=\sqrt{1+(x-2)^2-4+4x}=\sqrt{(x-2)^2-3+4x}=\sqrt{(x-2)^2-3+4(x-2)+8}$$
I could not complete , what can we do then ?
(I know that we can differentiate the function and substitute in the Taylor formula , but I want a shorter way)
for example :
if we want to expand $$g(x)=\frac{1}{1-x}$$ around $x=2$ we can start by
\begin{align}
g(x) & =\frac{1}{1-x}=\frac{1}{1-(x-2)-2}=\frac{-1}{1+(x+2)} \\[10pt]
& =-[1-(x-2)+(x-2)^2-(x-2)^3+\cdots]
\end{align}
So I want to get convert $f(x)$ to a form that we can write its expansion without getting derivatives, like I did with $g(x)$ above
| To give the general term expansion let $y=x-2$, then
\begin{align}
\sqrt{1+x^2}
&= \sqrt{1+ (y+2)^2}
= \sqrt{5 + 4y + y^2}
= \sqrt{5} \sqrt{1 + \frac{y}{5}(4+y)}\\
&= \sqrt{5}\sum_{n=0}^{\infty} \binom{1/2}{n} \left[\frac{y}{5}(4+y)\right]^n
= \sqrt{5}\sum_{n=0}^{\infty} \binom{1/2}{n} \frac{y^n}{5^n}\sum_{k=0}^n\binom{n}{k}4^k y^{n-k}
\end{align}
Extracting the $m$th coefficient, $[y^m]\sqrt{1+(y+2)^2}$, which may be found by considering which multiples of $y^n$ and $y^{n-k}$ equal $y^m$, i.e., when $n+(n-k) = 2n-k = m$, so we only take the $k=2n-m$ term from the inner summation above. Hence the $m$th coefficient is
\begin{align}
[y^m] \sqrt{5}\sum_{n=0}^{\infty} \binom{1/2}{n} \frac{y^n}{5^n}\sum_{k=0}^n\binom{n}{k}4^k y^{n-k}
&= \sqrt{5} \sum_{n=0}^{\infty} \binom{1/2}{n} \frac{1}{5^n} \binom{n}{2n-m}4^{2n-m}\\
&= \frac{\sqrt{5}}{4^m} \sum_{n=0}^{\infty} \binom{1/2}{n} \binom{n}{2n-m} \left(\frac{16}{5}\right)^n
\end{align}
Moreover, the binomial coefficient $\binom{n}{2n-m}$ is $0$ for $2n-m<0$ and for $n < 2n-m$, so we need only consider $n \geq m/2$ and $n\leq m$. Removing zero terms and a bit of algebra gives the Taylor series
$$\sqrt{1+x^2} = \sum_{m=0}^{\infty} c_m y^m = \sum_{m=0}^{\infty}c_m (x-2)^m,\quad\text{where}\;
c_m \equiv\frac{\sqrt{5}}{4^m}\sum_{n=\lceil m/2\rceil }^m\binom{1/2}{n} \binom{n}{2n-m} \left(\frac{16}{5}\right)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int ydx + zdy + xdz$ using Stokes' Theorem?
Evaluate $\int ydx + zdy + xdz$ where $C $ is intersection of $x+y=2$
and $x^2+y^2+z^2=2(x+y) $ traversed counterclockwise as viewed from
origin
I am using Stokes' theorem to solve this question so
We want $\int \int curl F.N \; dS$ where $N$ is the normal unit vector to surface S, where S is a surface bounded by $C$
$F = \langle y,z,x\rangle$
$curl F = \langle -1,-1,-1 \rangle $
I take $S $ on the plane $x+y = 2$
$\nabla (x+y) = \langle 1,1,0 \rangle = A(say)$
Then unit normal vector $N = \langle -1/\sqrt2,-1/\sqrt2, 0\rangle $ {Multiplied by $-1$ because we are viewing it from origin }
$curl F.N= \sqrt(2)$
Now intersection of $x+y=2 $ and $x^2+y^2+z^2 = 2(x+y) $ gives
$x^2 + y^2 + z^2 = 4$
To get projection onto $xy$ plane $z=0$ we get
$x^2 +y^2 =4$
Now I am stuck in finding $dS$
How do I get dS = $\sqrt{z_x^2 +z_y^2 +1}dA$ where $A:x^2 +y^2 =4$
This is because my $S:x+y=2$ has no $z$ term
| Since $\nabla\times F\cdot \vec{N}=\sqrt{2}$, you only need to evaluate
$$\sqrt{2}\iint_S dS$$
which is $\sqrt{2}$ multiplied by the area of the disk. (Note that the intersection is a disk, and it passes through the center of the original ball.) The original ball can be written as
$$(x-1)^2+(y-1)^2+z^2=2$$
whose center is $(1,1,0)$. The plane $x+y=2$ also passes through the center $(1,1,0)$. Hence the area of the disk is $\pi r^2=2\pi$, and the answer for the integral is $2\sqrt{2}\pi$.
Alternatively, if we project this surface to $xz$-plane, (we cannot project it to $xy$-plane since it is perpendicular to the $xy$-plane) we substituting $y=2-x$ into the other equation to obtain
$$x^2+(2-x)^2+z^2=4\implies (x-1)^2+\frac{z^2}{2}=1.$$
Now setting $y=2-x$, we have
$$\sqrt{1+y_x^2+y_z^2}=\sqrt{2}.$$
Computing
$$\sqrt{2}\cdot\sqrt{2}\iint_A dA$$
where $A$ is region enclosed by the ellipse, we obtain $2\sqrt{2}\pi$, same as above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving the Differential equation: $y'=\frac{2}{x}y+x^3$ We have the differential equation $$y'=\frac{2}{x}y+x^3$$ and we know $x \in (0, \infty)$.
My attempt with variation of constants
\begin{align}
\phi(x) &= \exp \left(\int \frac{2}{x} dx \right) \\
&= \exp(2\ln|x|) \\
&= x^2c
\end{align}
and
\begin{align}
\psi(x) &= (x^2c) \cdot \int \frac{x^3}{x^2} dx \\
&= (x^2c) \cdot \frac{x^2}{2}
\end{align}
but this solution is wrong. Where is the mistake?
| Observe that
$$\phi\left(x\right)=\exp\left(\int\frac{2}{x}{\rm d}x\right)=\exp\left(2\ln\left|x\right|+C\right)=Ax^{2}$$
with $A\equiv\exp C$. Thus the particular solution is
$$y_{\rm p}=Ax^{2}\int\frac{x^{3}}{Ax^{2}}{\rm d}x=\frac{x^{4}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
| Alright, you want difference of squares have some difference of squares
$$(\frac{1+i \sqrt7}{2})^4 + (\frac{1-i \sqrt7}{2})^4=$$
Lets add $0$ cleverly
$$=(\frac{1+i \sqrt7}{2})^4 -2(\frac{1+i \sqrt7}{2})^2(\frac{1-i \sqrt7}{2})^2+ (\frac{1-i \sqrt7}{2})^4 + 2(\frac{1+i \sqrt7}{2})^2(\frac{1-i \sqrt7}{2})^2$$
$$=((\frac{1+i \sqrt7}{2})^2 - (\frac{1-i \sqrt7}{2})^2)^2 + 2(\frac{1+7}{4})^2 $$
The $2(\frac{1+7}{4})^2$ comes from the final term of the second line, it can be viewed as a difference of squares, squared. Next deal with the difference of tsquares in the squared parenthesis
$$ = (\frac{2\cdot2 i \sqrt{7}}{4})^2 + 8 $$
$$ =-7+8=1$$
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the limits of find the limits :
$$\lim_{x\to 0}\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}$$
i know that :
$$\sin 5x=5\sin x-20\sin^3x+16\sin^5x$$
$$5\sin 3x=15\sin x-20\sin^3x$$
so we have :
$$\sin 5x-5\sin 3x+10\sin x=16\sin^5x$$
$$\lim_{x\to 0}\frac{16\sin^5x}{\sin (\sin x)+\tan x-2x}=\frac{16\sin^5 x}{x^5}\cdot\frac{x^5}{\sin(\sin x)+\tan x-2x}$$
$$\lim_{x\to0}\frac{(\sin (\sin x)-\sin x)+(\sin x-x)+(\tan x-x)}{x^5}=0$$
thus :
$$\lim_{x\to 0}\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}=\infty$$
it is right?
| Use Taylor's formula at order $5$:
\begin{align}
&\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}=\\[1ex]
&\frac{5x-\frac{125x^3}6+\frac{625x^5}{24}+o(x^5)-5\Bigl(3x-\frac{27x^3}6+\frac{81x^5}{40}+o(x^5)\Bigr)+10\Bigl(x-\frac{x^3}6+\frac{x^5}{120}+o(x^5)\Bigr)}{\sin\Bigl(x-\frac{x^3}6+\frac{x^5}{120}+o(x^5)\Bigr)+x+\frac{x^3}3+\frac{2x^5}{15}+o(x^5)-2x}\\[1ex]
=&\frac{16x^5+o(x^5)}{x-\frac{x^3}3+\frac{x^5}{10}+o(x^5)+x+\frac{x^3}3+\frac{2x^5}{15}+o(x^5)-2x}=\frac{16x^5+o(x^5)}{\frac{7x^5}{30}+o(x^5)}\\[1ex]
&=\frac{16+o(1)}{\frac7{30}+o(1)}\to\frac{16}{\frac7{30}}=\color{red}{\frac{480}{7}}.
\end{align}
Added (following @ParamanandSingh's strategy):
Set $s=\sin x$. As the O.P. pointed, Chebyshev' polynomials yield the formula
$$\sin 5x-5\sin 3x+10\sin x=16s^5.$$
On the other hand, in the denominator, we have $\tan^2x=\dfrac{s^2}{1-s^2}$, so in $\bigl(-\frac\pi2,\frac\pi2\bigr)$,
$$\tan x=\frac s{\sqrt{1-s^2}}=s\Bigl(1+\frac{s^2}2+\frac{3s^4}8+o(s^4)\Bigr).$$
Also, we have
$$x=\arcsin s=s+\frac12\, \frac{s^3}3 +\frac{1\cdot 3}{2\cdot 4}\,\frac{x^5}5$$
whence the following expansion at order $5$ for the denominator:
\begin{alignat}{2}
\sin(\sin x)+\tan x-2x&= \sin s+\frac s{\sqrt{1-s^2}}-2\arcsin s&
&=s -\frac{s^3}6+\frac{s^5}{120} \\[1ex]
&&&+s+\frac{s^3}2+\frac{3s^5}8 \\[1ex]
&&&-2s-\frac{s^3}3-\frac{3s^5}{20}+o(s^5)\\
&=\frac {7s^5}{30} +o(s^5),
\end{alignat}
and finally
$$\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}\sim_0\frac{16s^5}{\dfrac {7s^5}{30}}=\frac{480}7.$$
| {
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Calculate the limit: $\lim_{x\to+\infty}(\frac{x^2 -x +1}{x^2})^{\frac{-3x^3}{2x^2-1}}$ without de l'Hôpital rule I was wondering how can I calculate the limit:
$$\lim_{x\to+\infty}\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}$$
without de l'Hôpital rule.
I tried to reconduct the limit at the well known one:
$$\lim_{x\to+\infty}\left(1+\frac1x\right)^x = e$$
Now I'm concentrating only on the part of the limit with still the Indeterminate Form, I reached this form elevating $e$ to the neperian logarithm of the function, trying to get rid of the $1^\infty$ I.F.
but, at the end of the day, I could only obtain:
$$\begin{align}\lim_{x\to+\infty}\ln\left(\frac{1}{x^2}(1+x^2-x)\right) &= \lim_{x\to+\infty} \ln\left(\frac{1}{x^2}\right) + \lim_{x\to+\infty} \ln\left(1 + \frac{1}{\frac{1}{x^2-x}}\right) \\&= \lim_{x\to+\infty} \ln\left(\frac{1}{x^2}\right) + \lim_{x\to+\infty} (x^2-x) \ln\left(\left(1 + \frac{1}{\frac{1}{x^2-x}}\right)^{\frac{1}{x^2-x}}\right)\end{align}$$
But then defining
$$t= \frac{1}{x^2-x}$$
the limit
$$\lim_{t\to 0} \ln\left(\left(1 + \frac{1}{t}\right)^{t}\right)$$
goes no more to
$$ \ln(e)$$
because now $$t \to 0$$
Can you please give me some help?
| Since $e^x$ and $\ln$ are continuous functions, we obtain:
$$\lim_{x\to+\infty}\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}=\lim_{x\to+\infty}\left(1+\frac{x^2 -x +1}{x^2}-1\right)^{\frac{1}{\frac{x^2 -x +1}{x^2}-1}\cdot\left(\frac{x^2 -x +1}{x^2}-1\right)\frac{-3x^3}{2x^2-1}}=$$
$$=
\lim_{x\to+\infty}\left(1+\frac{x^2 -x +1}{x^2}-1\right)^{\frac{1}{\frac{x^2 -x +1}{x^2}-1}\cdot\frac{3(x^2-x)}{2x^2-1}}=\lim_{x\rightarrow+\infty}e^{\frac{3(x^2-x)}{2x^2-1}\cdot\ln\left(1+\frac{x^2 -x +1}{x^2}-1\right)^{\frac{1}{\frac{x^2 -x +1}{x^2}-1}}}=$$
$$=e^{\lim\limits_{x\rightarrow+\infty}\frac{3(x^2-x)}{2x^2-1}\cdot\ln\lim\limits_{x\rightarrow+\infty}\left(1+\frac{x^2 -x +1}{x^2}-1\right)^{\frac{1}{\frac{x^2 -x +1}{x^2}-1}}}=e^{\frac{3}{2}}.$$
| {
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Argument principle in the evaluation of integral
Evaluate $$I=\int_{-\infty}^{\infty}\frac{x}{x^{2}+x+1}dx$$
Attempt: Note that $$\int_{-\infty}^{\infty}\frac{x}{x^{2}+x+1}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2x+1}{x^{2}+x+1}dx-\int_{-\infty}^{\infty}\frac{1}{x^{2}+x+1}dx$$
Denote $\Gamma $ as the circle with radius $R\rightarrow \infty$ and let $h=\frac{f{}'(z)}{f(z)}=\frac{2z+1}{z^{2}+z+1}$ be a holomorphic function in and on $\Gamma$. The conditions for the Argument principle are valid, and thus
$$\Rightarrow \int_{\Gamma }\frac{f{{}'}}{f}dz=2\pi i\left ( Z-P \right )=4\pi i$$. The second integral is easily evaluated to $\frac{2\pi}{\sqrt{3}}$, and finally our integral computes to $$I=2\pi i-\frac{2\pi}{\sqrt{3}}.$$ I am certain the wrongdoing is in using the argument principle, but how?
*Is it that $f$ is not holomorphic as $R\rightarrow \infty$?
| Using, as you did,$$I=\int\frac{x}{x^{2}+x+1}\,dx=\frac{1}{2}\int\frac{2x+1}{x^{2}+x+1}\,dx-\int\frac{1}{x^{2}+x+1}\,dx$$
$$I=\frac{1}{2} \log \left(x^2+x+1\right)-\frac{1}{\sqrt{3}}\tan ^{-1}\left(\frac{2
x+1}{\sqrt{3}}\right)$$ making, after simplifications,
$$J=\int_a^a\frac{x}{x^{2}+x+1}\,dx=\frac{1}{\sqrt{3}}\tan ^{-1}\left(\frac{a\sqrt{3} }{a^2-1}\right)+\frac{1}{2} \log \left(\frac{a^2+a+1}{a^2-a+1}\right)$$ For infinitely large values of $a$
$$J=-\frac{\pi }{\sqrt{3}}+\frac{2}{a}-\frac{2}{3
a^3}+O\left(\frac{1}{a^5}\right)$$
| {
"language": "en",
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I don't know how to solve this integral
Could you please help me compute the following integral: $$\int \frac{(x^2+x-4)}{x(x^2+4)}dx$$
What I've done so far is:
$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{A}{x}dx+\int \frac{B}{x^2+4}dx$$
So
$$x^2+x-4 = A(x^2-4)+Bx$$
And I did this so I could figure out the value of A and B but I am having a rough time trying to calculate A because of the $x^2$.
I've also seen that the solution solves this exercise by separating the fraction's numerator like this:
$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{x^2}{x(x^2+4)}dx+\int \frac{x}{x(x^2+4)}dx + \int \frac{-4}{x(x^2+4)}dx $$
When do I know that I have to use this? Is what I though correct? If yes, how do I continue it?
Thank you very much. Please, let me know if something is not very clear in my question.
Agapita.
| The partial fractions decomposition is not what you think: it should be
$$\frac{x^2+x-4}{x(x^2+4)}=\frac Ax+\frac{Bx+C}{x^2+4},$$
because the condition on the numerators is not they're constants, but they're polynomials with degree less than the degree of the irreducible factor in the denominator. Here the second irreducible factor has degree $2$, so the numerator has degree $<1$.
To find the coefficients, multiply both sides of this equality by $x(x^2+4)$:
$$x^2+x-4=A(x^2+4)+(Bx+C)x.$$
Then set successively
*
*$x=0$, which yields $\;-4=4A+0$, whence $A=-1$,
*$x=2i$, which results in $\;2i-8= 0-4B+2iC$, whence $B=2$, $C=1.$
Finally the integral becomes
\begin{align}
\int\frac{x^2+x-4}{x(x^2+4)}\,\mathrm dx&=-\int\frac{\mathrm dx}x+\int\frac{2x}{x^2+4}\,\mathrm dx + \int\frac{\mathrm dx}{x^2+4}\\&=-\ln|x|+\ln(x^2+4)+\frac12\,\arctan \frac x2.
\end{align}
| {
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"source": "stackexchange",
"question_score": "2",
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Factorising $x^2-x+1$ I attempted to factorise $x^2-x+1$, which seemed fairly straightforward:
$$
\begin{align}
x^2-x+1 &= (x+1)^2-3x \\
&= (x+1-\sqrt{3x})(x+1+\sqrt{3x}) \\
&= (x-\sqrt{3x}+1)(x+\sqrt{3x}+1)
\end{align}$$
Technically neither $(x-\sqrt{3x}+1)$ nor $(x+\sqrt{3x}+1)$ are linear functions, which is why I didn't get the "correct" answer:
$$(x-\frac{1}{2}-i\frac{\sqrt{3}}{2})(x-\frac{1}{2}+i\frac{\sqrt{3}}{2})$$
But I am curious as to why Desmos only considers the positive cases of $x$? If $(x-\sqrt{3x}+1)(x+\sqrt{3x}+1)$ does expand perfectly into $x^2-x+1$, why isn't it graphed the same way?
| What you propose is not a polynomial factorisation. The polynomial function $x^2-x+1$ is defined for all $x$, whereas your ‘factorisation’ requires $x\ge 0$.
Anyway, since the roots of $x^2+x+1$ are the non-real cubic roots of unity $j$ and $j^2$, the roots of $x^2-x+1$ are $-j$ and $-j^2$n, whence the factorisation
$$x^2-x+1=(x+j)(x+j^2).$$
| {
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Show that $(x + y\sqrt{-5})$ must be a prime in $\mathbb{Z}[\sqrt{-5}]$ Got these problems as separate sections of a question in a book's chapter on 'Divisibility & primes'.
*
*Show that if $x^2 + 5y^2 =1$, then $x = \pm 1$.
Can state it in terms of two factors as : $(x + y\sqrt{-5})(x-y\sqrt{-5}) =x^2 + 5y^2$, with
(i) $(x + y\sqrt{-5}) = 1$, (ii) $(x - y\sqrt{-5}) = 1$
Adding both (i) & (ii), get: $x = 1$, Subtracting (ii) from (i), get: $(y\sqrt{-5}) = 0$.
Unable to pursue after that, as $x=-1$ is not possible.
*Show that $(x + y\sqrt{-5})$ must be a prime in $\mathbb{Z}[\sqrt{-5}]$
The hint given is to use the unique factorization theorem for the integers, by supposing $(x + y\sqrt{-5}) = (a + b\sqrt{-5})(c + d\sqrt{-5})$. The hint asks to show that: $(x^2 + 5y^2) = (a^2 + 5b^2)(c^2 + 5d^2)$. I need some more hint or help to pursue, as squaring $(x + y\sqrt{-5}) = (a + b\sqrt{-5})(c + d\sqrt{-5})$ does not lead to $(x^2 + 5y^2) = (a^2 + 5b^2)(c^2 + 5d^2)$. My attempt is stated for squaring both sides below:L.H.S.: $(x + y\sqrt{-5})(x + y\sqrt{-5}) => x^2 + 2xy(-5) -5y^2$ R.H.S.: $(a + b\sqrt{-5})^2(c + d\sqrt{-5})^2 => (a + b\sqrt{-5})(a + b\sqrt{-5})(c + d\sqrt{-5})(c + d\sqrt{-5}) => (a^2 -5b^2 +2ab\sqrt{-5})(c^2 -5d^2 +2cd\sqrt{-5})$
*Find all primes less than 50 in integers that can be written in the form $x^2 + 5y^2$.
No clue except to first find the primes: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$.
Then trying to see if the factorization works, so starting with $2 = x^2 + 5y^2$, but cannot think further. Do I need to have $y$ as a imaginary number only, as $\sqrt{-5}$, or anything will work.
| For 3, the squares modulo $5$ are $-1,0,1$, so the primes are: $5,11,19,29,31,41$.
Question 2 doesn't make sense. Need provide context.
| {
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"timestamp": "2023-03-29T00:00:00",
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Extremal problem with positive integer numbers Let $a,b$ be two positive integer numbers such that $a\sqrt{3}-b\sqrt{7}>0$. Find the minimum value of
$$
S=(a\sqrt{3}-b\sqrt{7})(a+b).
$$
Attempt I have tried and guess that the minimum value of $S$ is $(55+36)(55\sqrt{3}-36\sqrt{7})$, where $(55,36)$ is the integer solution of the Pell equation $3a^2-7b^2=3$.
| We need to minimise
$$S = \sqrt3 \left(1+\frac{a}b\right)\cdot \color{blue}{b^2\left(\frac{a}b-\sqrt{\frac73} \right)}$$
There are infinitely many convergents from the regular continued fraction expansion of $\alpha = \sqrt{\frac73}$ which keep the terms in blue bounded, i.e. satisfy $b^2\left(\frac{a}b-\alpha \right) < \frac12$, and only convergents satisfy this. Further as we take higher convergents the terms not in blue gets lower. Hence it is enough to seek the minimum (or greatest lower bound) from these convergents.
We have the bounds for convergents $a_n/b_n$
$$\frac1{2b_n^2\alpha-1}> \left|\frac{a_n}{b_n} - \alpha\right| > \frac1{2b_n^2\alpha+1} $$
$$\implies \frac{1}{2\alpha -1/b_n^2} > b_n^2\left|\frac{a_n}{b_n} - \alpha\right| > \frac1{2\alpha+1/b_n^2}$$
$$\implies \frac{\sqrt3 (1+a_n/b_n)}{2\alpha-1/b_n^2} > S_n > \frac{\sqrt3 (1+a_n/b_n)}{2\alpha+1/b_n^2}$$
As we take better convergents, we get $a_n/b_n \to \alpha$ and $b_n \to \infty$, so the greatest lower bound is
$$S_n \to S_{glb} = \frac{\sqrt3(1+\alpha)}{2\alpha} = \frac{\sqrt3}2 + \frac3{2\sqrt7}\approx 1.432972113$$
| {
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"source": "stackexchange",
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Can I use the method of seperation variables to solve the following ODE Can I use the method of seperation variables to solve the following ODE
$$F'(x)=\frac{a}{(F(x))^2}+c$$
where $a,c>0$
If without $c>0$, I can solve it , but now how can I solve this equation?
| Rearrange to get $$\frac{F^2}{a+cF^2} dF = dx$$ which is a separable equation. Integrate both sides to get $$x = \int \frac{F^2}{a+cF^2} dF = \frac 1a \int \frac{F^2}{1+\frac caF^2} dF.$$ Let $\tan u = \sqrt{\frac ca} F$, then $\sec^2u \,du = \sqrt{\frac{c}{a}} \, dF$ so we get $$x = \frac 1a \int \frac{\frac ac \tan^2u}{1+\tan^2u} \sqrt{\frac ac}\sec^2 u \, du = \frac 1c \sqrt{\frac ac}\int \frac{\tan^2u}{\sec^2 u} \sec^2 u \, du = \frac 1c \sqrt{\frac ac}\int \tan^2u \, du$$ and thus $$x = \frac 1c \sqrt{\frac ac} \left ( \tan u - u\right)+K = \frac 1c \sqrt{\frac ac} \left ( \sqrt{\frac ca}F - \tan^{-1} \left( \sqrt{\frac ca}F\right) \right) +K$$ Therefore the solution is given implicitly by
$$x = \frac 1c F - \sqrt{\frac{a}{c^3}}\tan^{-1} \left( \sqrt{\frac ca}F\right)+K$$ where $K$ is our integration constant.
| {
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Show that $4^{3x+1} + 2^{3x+1}+ 1$ is divisible by 7 I want to show that $4^{3x+1} + 2^{3x+1} + 1$ is divisible by 7, I am trying to show this with modular arithmetic.
If I break up each part of the equation, I can see that
$4^{3x+1} = 4$ x $2^{6x}$
which implies that $4^{3x+1}mod(7) = 4$
I can't quite find a nice factorization of $2^{3x+1}$
Any help specifically on how to treat $2^{3x+1}$ would be appreciated.
| $$4^3\equiv 2^3\equiv 1\pmod{7} \tag{A}$$
$$ 4^{3x}\equiv 2^{3x}\equiv 1\pmod{7}\tag{B}$$
$$ 4^{3x+1}+2^{3x+1}+1\equiv 4+2+1 \equiv 0\pmod{7}.\tag{C}$$
| {
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How to prove that 2 is not a Gaussian prime? To prove an algebraic number is not a Gaussian prime, need find factors of it.
Let us assume, $(a + bi)(x +yi) = 2, \exists a,b,x,y \in \mathbb{Z}$; so $(ax -by) + i(ay + bx) =2$.
As imaginary part is null in $2$, so $ax - by =2, ay + bx =0$.
Now, how to reach from here, the fact that $a=x=1, b=i, y=-i$.
I read in here in a comment by @StevenStadnicki, that an easier way is to multiply both sides of $(a + bi)(x +yi) = 2$ by the complex conjugates, but is not clear, as my below attempt shows.
$$
\begin{align}
& ax - by =2, ay + bx =0 & \ \\
\implies & axy - by^2 =2y, axy + bx^2 =0 \text{ }{ -(i)} & \ \\
\implies & b = 0
\end{align}
$$
By the answer given below, my attempt is stated below:
Two set of values are possible, one in which $(a^2 + b^2) = 4, (x^2 + y^2) = 1$, second in which $(a^2 + b^2) = 2, (x^2 + y^2) = 2$.
Possible combinations for each of the two cases are stated below:
Case (i): $(a^2 + b^2) = 4, (x^2 + y^2) = 1$ $\implies a= \pm 2, b = 0, x = \pm 1, y = 0$, or $a= \pm 2, b = 0, x = 0, y = \pm 1$. But, as $ax -by =2, ay + bx = 0$. But being algebraic number, need have non-zero imaginary part, hence no algebraic factors of $2$.
Case (ii) $(a^2 + b^2) = 2, (x^2 + y^2) = 2$ $\implies$ possible values are : $ a= \pm 1, b = \pm 1, x = \pm 1, y = \pm 1$.
But, as : $ax -by =2, ay + bx = 0$, so the valid combinations are: 1. $a=1, b=-1, x=1, y=1$; 2. $a=1, b=1, x=1, y=-1$, 3. $a=-1, b=1, x=-1, y=-1$.
Here, combination 3 is same as 1, so can be ignored also valid.
So, the solution has values of $a =1,b=1, x = 1, y = -1$.
If there is a redundancy or unnecessary steps, please tell.
| Let $(a+ib)(c+id)=2$.
There is obviously no solution with real integers, so $b$ and $d$ are both nonzero (if only one is zero, the product is complex or 0, so it also fails). Also, since the product is real, their arguments are opposite: they are multiples of a conjugate pair of gaussian integers.
So it's really $k(a+ib)(a-ib)=2$
Hence
$$k(a^2+b^2)=2$$
So $k$ must be positive. And $k,a,b$ are integers. If $k=2$, it fails because then $b=\pm1$ and $a=0$ (remember $b\ne0$), and the factorization if $2i\times -i=2$ or $-2i\times i=2$, and this proves nothing as you have just multiplied $2$ by a unit.
Hence let $k=1$ and $a^2+b^2=2$, but then $a^2=b^2=1$. But the sign of $a$ is useless, because $(-a+ib)(-a-ib)=(a+ib)(a-ib)$. Hence, you can pick $a=1$, and the solution has to be
$$(1+i)(1-i)=2$$
Which you can check.
Of course, another solution is
$$(-1+i)(-1-i)=2$$
| {
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Solve the recurrence relation $y_{n+2}+2y_{n+1}+2y_n=n^3.$ First I solved the characteristic equation $r^2+2r+2=0$ with $r_1=-1+i$ and $r_2=-1-i.$ In polar form this becomes $r_{1,2}=\sqrt{2}e^{\pm i3\pi/4},$ thus the general solution to the homogenous equation is $$y_h=\sqrt{2}^n\left(A\cos{\frac{3\pi}{4}n}+B\sin{\frac{3\pi}{4}n}\right).$$
Since $r_1\neq 1$ and $r_2\neq 1$, for the particular solution I can use the ansatz $y_n^{p}=an^3+bn^2+cn+d.$ I have that
$$y_{n+2}+2y_{n+1}+2y_n=a(n+2)^3+b(n+2)^2+c(n+2)+d+2(a(n+1)^3+b(n+1)^2+c(n+1)+d)+2(an^3+bn^2+cn+d) \\ = 5an^3+(12a+5b)n^2+(18a+8b+5c)n+10a+6b+4c+5d.$$
I simplified this with a computer, it feels tedious and because of that, and that my constants become quite small (for example $d=86/625$), I think my ansatz is wrong.
Can someone confirm that my ansatz is wrong and tell me what the proper one should be?
| You might do better to choose $n+1, n, n-1$ which simplifies the arithmetic and promises more cancellation. Also you might organise the computation so that you get $$(n-1)^3=a\left((n+1)^3+2n^3+2(n-1)^3\right)+b\left((n+1)^2+2n^2+2(n-1)^2\right)+c\left((n+1)+2n+2(n-1)\right)+5d=$$$$=5an^3+(-3a+5b)n^2+(8a-4b+5c)n+(-a+3b-c+5d)$$
From which $5a=1$
$5b-3a=-3$ so that (multiply by $5$) we have $25b=-12$
Also $8a-4b+5c=2$ (multiply by $25$) $125c=50-40-4=6$
And $-a+2b-c+5d=-1$ (multiply by $125$) $625d=-125+25+180+6=86$
Note for future reference that the fractions are predictable from $1+2+2=5$ so that the new term (as you run down the coefficients) is always multiplied by $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
To find the Value of $\tan A + \cot A$, if the value of $\sin A + \cos A$ is given To Find -
$$\tan A + \cot A$$
Given,
$$\sin A + \cos A = \sqrt2$$
My progress as far -
1st way-
$$\Rightarrow \sin A = \sqrt2 - \cos A$$
$$\Rightarrow \tan A = \frac{\sqrt2 - \cos A}{\cos A}$$
$$\Rightarrow \tan A = \frac{ \sqrt 2 }{\cos A } - 1$$
$$\Rightarrow \tan A + \cot A =\frac{ \sqrt 2 }{\cos A} -1 + \cot A $$
and the 2nd way as -
$$(\sin A + \cos A)^2 = 2$$
$$\sin ^2 A + \cos ^2 A + 2\sin A\cos A = 2 $$
$$\Rightarrow 2\sin A\cos A=1$$
$$\Rightarrow \sin A\cos A=\frac12$$
As we can see the first way is unable to give an answer in absolute Real Number, and the second way doesn't go even near to what is required to proof.
I know few trigonometry identities as per my textbook, those are
*
*$\sin^2 A + \cos^2 A = 1$
*$1 + \cot^2 A = \csc^2 A$
*$\tan^2A + 1 = \sec^2 A$
| $$\begin{align}\tan A+\cot A&=\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}\\&=\dfrac{\sin^2A+\cos^2A}{\sin A\cos A}\\&=\dfrac1{\sin A\cos A}\\&=\dfrac2{2\sin A\cos A}\\&=\dfrac2{(\sin A+\cos A)^2-1}\\&=\dfrac2{(\sqrt2)^2-1}\\&=2\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find: $\lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
Find: $\displaystyle \lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
After developing the expression, by multiplying the fraction by the conjugates and rearranging, I found:
$$f(x)=x\times \frac{1+\sqrt{1+1/x^6}}{1+\sqrt{1+1/x^4}}$$
I can solve it for the limit when $x\to \infty$, with answer $\infty$ which is correct (by the book and Wolfram Alpha). But when the limit is $x\to -\infty$ the answer is zero, but from this expression, I get $-\infty$ as the answer.
I'm certainly missing something.
Hints and answers appreciated. Sorry if this is a duplicate.
| Let $y=-x:$
$F(y) = \dfrac{y^2 -\sqrt{y^4 +1}}{-y^3 - \sqrt{y^6+1}}=$
$(-1)\dfrac{y^2 - \sqrt{y^4+1}}{y^3 +\sqrt{y^6+1}}.$
Conjugates and rearrangement , and then considering $\lim_{y \rightarrow \infty}$ should get you there .
Or:
$0\le |F(y)| \le \dfrac{2\sqrt{y^4+1}}{2y^3} \lt \dfrac{4y^2}{2y^3}$.
And the limit is?.
Used:
$ |y^2 - \sqrt{y^4+1}| \lt 2\sqrt{y^4+1} \lt 4y^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Differences between using partial fractions or completing the square to solve an integral? I have this question
$$\int \frac{1}{4x^2-4x-3}\, dx$$
I tried to solve it by using completing square method, and I got
$$\frac{1}{4} \arctan\left(\frac{2x-1}{2}\right),$$
but when I saw the answers, I found that it should be solved by partial fraction.
So what are the differences between these two methods?
| $$\int { \frac { dx }{ 4x^{ 2 }-4x-3 } } =\int { \frac { dx }{ \left( 2x+1 \right) \left( 2x-3 \right) } } =-\frac { 1 }{ 4 } \int { \left[ \frac { 1 }{ 2x+1 } -\frac { 1 }{ 2x-3 } \right] dx } \\ =-\frac { 1 }{ 4 } \left[ \int { \frac { dx }{ 2x+1 } } -\int { \frac { dx }{ 2x-1 } } \right] =-\frac { 1 }{ 8 } \left[ \int { \frac { d\left( 2x+1 \right) }{ 2x+1 } } -\int { \frac { d\left( 2x-1 \right) }{ 2x-1 } } \right] =\\ =-\frac { 1 }{ 8 } \left[ \ln { \left| 2x+1 \right| -\ln { \left| 2x-1 \right| } } \right] =-\frac { 1 }{ 8 } \ln { \left| \frac { 2x+1 }{ 2x-1 } \right| } +C=\\ \\ or\\ -\frac { 1 }{ 8 } \ln { \left| \frac { 2x+1 }{ 2x-1 } \right| } +C=-\frac { 1 }{ 4 } \tanh ^{ -1 }{ \left( \frac { 2x+1 }{ 2x-1 } \right) } =\frac { 1 }{ 4 } \tanh ^{ -1 }{ \left( \frac { 2x+1 }{ 1-2x } \right) } \\ \\ $$
Obviously,here should be $\tanh ^{ -1 }{ x } $ not $\tan ^{ -1 }{ x } $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integral by change of variable $\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $ What is $$\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $$
Putting $\sin(x) = t$, then $\cos(x) = \sqrt{1-t^2}$ and $\cos(x)\,dx = dt$
At $x = 0$, $\sin(x) = 0$, $\therefore\ t = 0$
At $x = \pi$, $\sin(x) = 0$, $\therefore\ t = 0$
Integral becomes
$$\int_0^0 t^4\cdot\sqrt[5]{1-t^2}\,dt = 0$$
Integral should not evaluate to zero
What is going on here, i cannot find any mistake....
| $$\sin^4x\cos^6x=\frac{1}{16}\cos^2x\sin^42x=\frac{1}{128}(1+\cos2x)(1-\cos4x)^2=$$
$$=\frac{1}{128}(1+\cos2x)\left(1-2\cos4x+\frac{1+\cos8x}{2}\right)=$$
$$=\frac{1}{256}(1+\cos2x)(3-4\cos4x+\cos8x),$$
which says that our integral it's
$$\int\limits_0^{\pi}\frac{3}{256}dx=\frac{3\pi}{256}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Geometry problem involving circles and triangles This is the question:
$E$ is a point on the side $AB$ of a rectangle $ABCD$ such that $DE=6$ cm, $DA= 8cm$ and $DC=6cm$. If $CE$ extended meets the circumcircle of the rectangle at $F$, then what is the value of $BF$ ? (use $\sqrt{2}$=$1.414$).
Here is how I tried to solve it
In $\Delta BCD$, $\frac{CD}{BC} = \frac{DG}{BG}$ (Since CG is the Angle bisector of $\langle C$ )
From this, DG = $\frac{30}{7} cm$; BG= $\frac{40}{7} cm $
$\langle FCD $ = $\langle FAD$ = $\langle FBD$ = $45°$
Therefore, $\Delta CED$ ~ $\Delta AEF$ ; i.e. $\Delta AEF$ is a right isosceles triangle like $\Delta CED $
Hence, EF = AF = $\sqrt 2 cm $
In $\Delta CED $, CE = 6$\sqrt2$ cm
Now consider $\Delta CGD$ & $\Delta BGF $
$\Delta CGD$ ~ $\Delta BGF $
Assume the value of $GE = x$ $cm$
$\Delta GCD$ ~ $\Delta GBF$
$\frac {GC}{GB}$ = $\frac {CD}{BF}$ = $\frac {GD}{GF}$
$\frac {(6 \sqrt 2 -x)}{\frac{40}{7}}$ = $\frac {6}{BF}$ = $\frac {\frac{30}{7}}{(x+\sqrt 2)}$
$(x+\sqrt 2)(6 \sqrt 2 - x)$ = $\frac{30}{7}$ .$\frac{40}{7}$
My plan was to find the value of $x$ so that I could get the ratio and then evaluate $BF$. However, it looks like I am going wrong as I am getting complicated values for $x$. Could anyone please guide?
|
If $DA=8$ and $DE=6$, then $AE=2$.
Since $AC$ and $BD$ are diagonals of a $6 \times 8$ rectangle, $AC = BD = 10.$
$\triangle CDE$ is an isosceles right triangle. So $CE = 6 \sqrt 2.$
$\triangle CDE \sim \triangle AFE$. So $\triangle AFE$ is an isosceles right triangle. Since $AE=2$, then $EF = AF = \sqrt 2.$
$\triangle CAE \sim \triangle DFE \implies
\dfrac{CA=10}{\color{red}{DF=5\sqrt 2}}
= \dfrac{CE=6\sqrt 2}{DE=6}
= \dfrac{AE=2}{FE=\sqrt 2}$
Since $BD$ is a diameter of the circle, $\angle BFD$ is a right angle.
Hence $\triangle BFD$ is a right triangle. So
$BF = \sqrt{BD^2 - DF^2} = \sqrt{10^2 - (5\sqrt 2)^2} = 5\sqrt 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Verify that for $k=3/2 ,$ $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$ This is a part of a proof I am studying on:
Let $f_k(x) = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \qquad (k\notin\{0,-1,-2,\ldots\}) $
For $k=3/2 ,$ it's shown that $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$
How is it concluded this way?
| Look at the Taylor expansion $$\sin t = t - \frac{t^3}{3!} +\frac{t^5}{5!} - \frac{t^7}{7!} + \cdots$$Now put $t = 2x$:$$\sin 2x = 2x - \frac{8x^3}{3!} +\frac{32x^5}{5!} - \frac{128x^7}{7!} + \cdots$$Divide by $2x$: $$\frac{\sin 2x}{2x} = 1 - \frac{4x^2}{3!} +\frac{16x^4}{5!} - \frac{64x^4}{7!} + \cdots$$Plugging $k=3/2$ for your $f_k(x)$ series gives precisely the above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Remainder when $27^{40}$ is divided by $12$ What is the remainder when ${27}^{40}$ is divided by $12$ ?
The answer is supposed by $9$ but i’m getting it’s as $3$ . Please correct me in my approach to the problem.
This is how I did it :-
${27}^{40} = {(3^3)}^{40} = {3}^{120}$
$\frac{{3}^{120}}{12} = \frac{3^{119}}{4}$
Thus
$\frac{{3}^{119}}{4} = \frac{(3^{4})^{29}3^3}{4}$
We can now expand ${81}^{29}$ by binomial theorem
$27(\binom{29}{0}(80)^{29} + \binom{29}{1}(80)^{28} + \binom{29}{2}(80)^{27}......+\binom{29}{29}1)$
As all the numbers is the parentheses except $\binom{29}{29}$ are multiples of $4$ we can remove it out of the parentheses
Thus we will have ,
$27(4K) + 27$
$4m + 4(6) +3$
$4n +3$
Where $k,m,n$ will be multiple of $4$ this gives remainder as 3
Please help me find the flaw . Thank you
| $27(4K) + 27$ is the numerator in your fraction $\frac{(3^{4})^{29}3^3}{4}$. Did you forget about the $4$ in the denominator? Here is another approach to solve the problem. The remainder of 27 in dividing by $12$ is $3.$ Thus the remainder of $27^{40}$ is the same as the remainder of $3^{40}$ which is $3(27^{13}).$ Replacing 27 by its remainder 3 we get to $3(3^{13})$ which is $3^{14}= 9(27^4)$. Replacing 27 by 3 we get to $9(3^4)=3^6=27^2$ Replacing $27$ by $3$ and we get the correct answer which is 9.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2}\ge 1$ for real numbers $x$, $y$, and $z$ with $xyz=1$. I've been thinking to prove each one individually then sum them up which seems too complex to because I haven't made any notable success proving $\frac{x^2}{(x-1)^2}> 1$.
I know that $(x-1)^2\ge 0$ , but I'm not able to transfer it to what I'm looking to prove.
Is there is any techniques to look for when solving these kind of inequalities because I'm totally stuck here.
| If
$x=1+a, y=1+b, z=1+c$
then
$a, b, c \ge 0$
since all are positive integers
and
$\begin{array}\\
\dfrac{x^2}{(x-1)^2} + \dfrac{y^2}{(y-1)^2} + \dfrac{z^2}{(z-1)^2}
&=\dfrac{(1+a)^2}{a^2} + \dfrac{(1+b)^2}{b^2} + \dfrac{(1+c)^2}{c^2}\\
&=1+\dfrac{1+2a}{a^2} +1+ \dfrac{1+2b}{b^2} +1+ \dfrac{1+2c}{c^2}\\
&=3+\dfrac{1}{a^2}+ \dfrac{1}{b^2}+ \dfrac{1}{c^2}+2(\dfrac{1}{a}+ \dfrac{1}{b} + \dfrac{1}{c})\\
&\ge 3\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dividing polynomial fractions with varying term quantities I'm working through an old algebra book as a refresher and I've come across what should be a simple polynomial division. The exercise prompts the reader to perform the following operation:
$$
\frac{a^2-9}{a^2+3a} \div \frac{a-3}{4}
$$
I started off by inverting the $\div$ sign by instead multiplying by the reciprocal which results in:
$$
\frac{a^2-9}{a^2+3a} \cdot \frac{4}{a-3}
$$
I spent nearly 2 hours at this point trying everything my mind could conjure in terms of factoring, simplifying, and multiplying, but none of my attempts ever arrived at the listed answer:
$$
\frac{4}{a}
$$
If someone could help me through the steps required to solve this, you'll have taught a man to fish.
Solution from below:
$$
\frac{a^2-9}{a^2+3a} \cdot \frac{4}{a-3}=\frac{4(a-3)(a+3)}{a(a+3)(a-3)}=\frac{4}{a}
$$
I had not factored $a^2 + 3a$ to $a(a+3)$ correctly in any of my attempts.
| Because $$\frac{a^2-9}{a^2+3a} \cdot \frac{4}{a-3}=\frac{4(a-3)(a+3)}{a(a+3)(a-3)}=\frac{4}{a}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all ordered pairs such that $x + y^2 = 2$ and $y + x^2 = 2$ How can I find all ordered pairs $(x,y)$ such that $x + y^2 = 2$ and $y + x^2 = 2$?
A solution/explanation would be greatly appreciated.
If it helps, the solutions given in the textbook are
x=-2, y=-2; x=1, y=1; x= (1+root 5)/2, y=(1-root 5)/2; x= (1-root 5)/2, y=(1+root 5)/2
| HINT
$$y+x^2= x+y^2\iff y-x-y^2+x^2=0\implies y-x-(y-x)(y+x)=0\implies (y-x)(1-x-y)=0$$
thus
$$y-x=0 \implies y=x \implies x^2+x-2=0$$
or
$$x+y=1\implies x^2-x-1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $\lim_{x\rightarrow\infty}x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right).$ This is what my prof did:
Rewrite as $$x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)=\frac{x}{e}\left(1-e\cdot e^{-x\ln{\left(1+\frac{1}{x}\right)}}\right).$$
Here he uses that
$$\ln{(1+t)}=t-\frac{t^2}{2}+O(t^3),$$
so
$$\frac{x}{e}\left(1-e\cdot e^{-x\left(\frac{1}{x}-\frac{1}{2x^2}+O(\frac{1}{x^3})\right)}\right)=\frac{x}{e}\left(1-e^{\frac{1}{2x}+O\left(\frac{1}{x^3}\right)}\right).$$
So far so good. Now, he uses that $e^t=1+t+O(t^2)$ so that the expression aove becomes
$$\frac{x}{e}\left(1-\left(1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)\right)\right)=-\frac{1}{2e}+O\left(\frac{1}{t}\right)$$
and now he lets $x\rightarrow\infty$ to get the answer. However, when he introduces the expansion of $e^t$, how does he get $O(1/x^2)$?
It is my understanding that I should plug in $t=\frac{1}{2x}+O\left(\frac{1}{x^3}\right)$ everywhere in $1+t+O(t^2),$ doing that I get
$$1+\frac{1}{2x}+O\left(\frac{1}{x^3}\right)+O\left(\left(\frac{1}{2x}+O\left(\frac{1}{x^3}\right)\right)^2\right),$$
Squaring that last ordo-term, the largest power of $x$ is $1/x^2$ and the smallest is $1/x^6.$ Why choose $1/x^2$?
| $$\lim_{x\rightarrow\infty}x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)=\lim_{x\rightarrow0}\frac{\frac{1}{e}-\left(\frac{\frac{1}{x}}{\frac{1}{x}+1}\right)^{\frac{1}{x}}}{x}=\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{x}}-e}{ex(1+x)^{\frac{1}{x}}}=$$
$$=\frac{1}{e^2}\lim_{x\rightarrow0}\left((1+x)^{\frac{1}{x}}\right)'=\frac{1}{e^2}\lim_{x\rightarrow0}\left(e^{\frac{\ln(1+x)}{x}}\cdot\left(\frac{1}{x(1+x)}-\frac{\ln(1+x)}{x^2}\right)\right)=$$
$$=\frac{1}{e}\lim_{x\rightarrow0}\frac{x-(1+x)\ln(1+x)}{x^3+x^2}=\frac{1}{e}\lim_{x\rightarrow0}\frac{1-\ln(1+x)-1}{3x^2+2x}=$$
$$=-\frac{1}{e}\lim_{x\rightarrow0}\left(\frac{\ln(1+x)}{x}\cdot\frac{1}{3x+2}\right)=-\frac{1}{2e}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find the equation of the line through the point $(1,-1)$ Find the equation of the line through the point $(1,-1)$ which cuts off a chord of length $4\sqrt {3}$ from the circle $x^2+y^2-6x+4y-3=0$
My Attempt:
Let the equation of line be $y=mx+c$
It passes through $(1,-1)$ so
$$-1=m+c$$
$$c=-1-m$$
| $$(x-3)^2+(y+2)^2=3+9+4=4^2$$
Let the center be $O(3,-2)$
Let one of the intersections of the line with the circle be $P$
So, $|OP|=4$
Let the perpendicular from the center to the line intersect at $Q$
Using The perpendicular from the center of a circle to a chord bisects the chord (Proof)
$|PQ|=2\sqrt3$
$$OP^2=PQ^2+OQ^2\iff OQ^2=4^2-(2\sqrt3)^2\implies|OQ|=?$$
Now the equation of the line $$mx-y-1-m=0$$
The perpendicular distance from $O$ to the line $$\dfrac{|3m+2-1-m|}{\sqrt{1+m^2}}=|OQ|$$
Square both sides to find $m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to evaluate the integral $\int_0^{\pi/2}x^2(\sin x+\cos x)^3\sqrt{\sin x\cos x} \, dx$? How to evaluate the integral $$\int_0^{\pi/2}x^2(\sin x+\cos x)^3(\sin x\cos x)^{1/2} \, dx \text{ ?}$$
I tried to subsititution $x=\frac{\pi}{2}-t$, but it doesn't work. can someone
help me, any hint or trick are appreciated.
| $$\color{blue}{\int_0^{\pi/2} {{x^2}{{(\sin x + \cos x)}^3}\sqrt {\sin x\cos x} dx} = \frac{\pi}{384\sqrt{2}}\left(35 \pi ^2-84 \ln ^22+132 \ln 2-150 \right)}$$
Note that $$\int_0^{\pi/2} {{x^2}{{(\sin x + \cos x)}^3}\sqrt {\sin x\cos x} dx} = \int_0^{\pi/2} {{x^2}{{(\sin x + \cos x)}^3}\sin x\sqrt {\cot x} dx}$$
and the identity
$$(\sin x + \cos x)^3 \sin x = \frac{1}{4} (4 \sin (2 x)-\sin (4 x)-2 \cos (2 x)-\cos (4 x)+3)$$
Thus it suffices to find
$$I_1 = \int_0^{\pi/2} {{x^2}\sqrt {\cot x} dx}
\qquad I_2 = \int_0^{\pi/2} {{x^2}\sqrt {\cot x} \cos (2x)dx} \qquad
I_3 = \int_0^{\pi/2} {{x^2}\sqrt {\cot x} \cos (4x)dx}$$
$$I_4 = \int_0^{\pi/2} {{x^2}\sqrt {\cot x} \sin (2x)dx}\qquad
I_5 = \int_0^{\pi/2} {{x^2}\sqrt {\cot x} \sin (4x)dx}$$
Whenever the integral converges,
$$\tag{1}\int_0^\pi \frac{e^{i ax}}{\sin^b x} dx = \frac{{{2^b}\Gamma (1 - b)\pi }}{{\Gamma (1 + \frac{{a - b}}{2})\Gamma (1 - \frac{{a + b}}{2})}}{e^{ia\pi /2}}$$ this is a direct consequence of contour integration
Now, note that
$$\begin{aligned}\int_0^{\frac{\pi }{2}} {{x^2}\sqrt {\cot x} \cos rxdx} &= \sqrt 2 \int_0^{\frac{\pi }{2}} {{x^2}\frac{{\cos x}}{{\sqrt {\sin 2x} }}\cos rxdx} \\&= \frac{{\sqrt 2 }}{8}\int_0^\pi {{x^2}\frac{{\cos \frac{x}{2}}}{{\sqrt {\sin x} }}\cos \frac{{rx}}{2}dx} \\&= \frac{{\sqrt 2 }}{{16}}\int_0^\pi {\frac{{{x^2}}}{{\sqrt {\sin x} }}\left[ {\cos \frac{{(1 + r)x}}{2} + \cos \frac{{(1 - r)x}}{2}} \right]dx}\end{aligned}$$
the RHS can be evaluated by taking $b=1/2$ then differentiates $(1)$ with respect to $a$ twice.
Substituting suitable values of $r$ gives
$$I_1 = \frac{{\sqrt 2 \pi }}{{96}}(5{\pi ^2} - 12\pi \ln 2 - 12{\ln ^2}2)$$
$$I_2 = \frac{\pi}{96\sqrt{2}} \left(5 \pi ^2-12 \pi +24-12 \ln ^22-12 \pi \ln 2-24 \ln 2 \right)$$
$$I_3 = \frac{\pi}{192\sqrt{2}} \left(5 \pi ^2-18 \pi +54-12 \ln ^22-12 \pi \ln 2-36 \ln 2 \right)$$
Similar trick works for $I_4, I_5$, with respective values:
$$I_4 = \frac{\pi}{96\sqrt{2}} \left(5 \pi ^2-12 \pi -24-12 \ln ^22+12 \pi \ln 2+24 \ln 2 \right)$$
$$I_5 = \frac{\pi}{196\sqrt{2}} \left(5 \pi ^2-30 \pi -42-12 \ln ^22+12 \pi \ln 2+60 \ln 2 \right)$$
| {
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"url": "https://math.stackexchange.com/questions/2597905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$= $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$ QuestionProve that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$=
$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$
My Approach R.H.S
$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$=$\frac{x^{2^{k-1}\left(1-x^{2}\right)}}{\left(1-x^{2^{k}}\right)\left(1-x^{2^{k-1}}\right)}$
i just don't know what else i can do?
| $x^{2^k}=x^{2\times2^{k-1}}=(x^{2^{k-1}})^2$
⇒$1-x^{2^k}=(1-x^{2^{k-1}})(1+x^{2^{k-1}})$
⇒ $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^k}}=\frac{1+x^{2^{k-1}}-1}{1-x^{2^k}}=\frac{x^{2^{k-1}}}{1-x^{2^k}}$
| {
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"source": "stackexchange",
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Solve $(5+2\sqrt{6})^{\frac{x}{2}} + ( 5-2\sqrt{6})^{\frac{x}{2}} = 10$ I wish to solve the equation
$$(5+2\sqrt{6})^{\frac{x}{2}} + ( 5-2\sqrt{6})^{\frac{x}{2}} = 10$$
I tried factorizing until I reached
$(\sqrt{2}+\sqrt{3})^x + (\sqrt{2}-\sqrt{3})^x = 10$
But from there I don't know what to do any help would be welcome
Thanks in advance
| Since $\sqrt{3} - \sqrt{2} > 0$, your equation should simplify to
$$ (\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10 $$
Also note that
$$ \sqrt{3} - \sqrt{2} = \frac{1}{\sqrt{3}+\sqrt{2}} $$
Let $t = (\sqrt{3}+\sqrt{2})^x$, then
$$ t + \frac{1}{t} = 10 $$
$$ t^2 - 10t + 1 = 0 $$
which gives $t = 5 \pm 2\sqrt{6}$
Therefore $x = \pm 2$
| {
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Alternating summation and subtraction of square roots I encountered a problem, to find the integer part of: $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{3} + \sqrt{4}} +...+\frac{1}{\sqrt{99} + \sqrt{100}}$$.
I multiplied the conjugate of each denominator. Meaning, for $\frac{1}{\sqrt{a} + \sqrt{b}}$, I multiply $\sqrt{a} - \sqrt{b}$. I get $\sqrt{100} - \sqrt{99} + \sqrt{98} - \sqrt{97} + ... + \sqrt{2} - \sqrt{1}$. I am stuck here. How do I simplify this? Or is it that my method is wrong?
| For an approximation, we might start (using a not generally adopted, but not unfamiliar notation) with$$H^{(-1/2)}_n=1 + \sqrt{2} + \ldots + \sqrt{n}=\frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + \zeta\left(-\frac12\right)+o(1),$$ mentioned in this question ($\zeta$ is the famous Riemann zeta function).
Then, $$\sqrt{2n}-\sqrt{2n-1}\pm\ldots=-H^{(-1/2)}_{2n}+2\sqrt{2}\,H^{(-1/2)}_n=\frac12\sqrt{2n}+(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)+o(1).$$
For $n=50$, the main term would be $\displaystyle5+(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)=4.619895\ldots$, while the original sum is $4.6323951\ldots$, as @Henry pointed out, already, $\displaystyle(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)=-0.380105\ldots$ being slightly more accurate. The error (of the order $O(1/n)$) is no surprise.
| {
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Negafibonacci Identity Consider the sequence of the negafibonacci numbers: $$..., \overbrace{-8}^{F_{-6}}, \overbrace5^{F_{-5}}, \overbrace{-3}^{F_{-4}}, \overbrace2^{F_{-3}}, \overbrace{-1}^{F_{-2}}, \overbrace1^{F_{-1}}$$
I have discovered that the following identity holds for integers I've tried:
$$\sum_{n=1}^kF_{-n}=1-F_{1-k}$$
How could this be shown for any $k\in\mathbb{N}$?
| Here is an answer based upon generating functions. But at first note that the identity
\begin{align*}
\sum_{k=1}^nF_{-k}=1-F_{1-n}\qquad\qquad n\geq 1\tag{1}
\end{align*}
gives for $n=1$: $F_{-1}=1-F_{0}$ and so we have to add the initial condition
\begin{align*}
F_{0}=0
\end{align*}
in order to fully specify the identity.
We use the generating function of the Fibonacci numbers:
\begin{align*}
\frac{x}{1-x-x^2}=x+x^2+2x^3+5x^4+8x^5+13x^6+\cdots
\end{align*}
and the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. This way we can write e.g.
\begin{align*}
F_n=[x^n]\frac{x}{1-x-x^2}\qquad\qquad n\geq 1
\end{align*}
Since $F_{-n}=(-1)^{n+1}F_n$ with $F_n$ the Fibonacci numbers, the claim (1) can be written as
\begin{align*}
\sum_{k=1}^n(-1)^{k+1}F_k=1-(-1)^nF_{n-1}\qquad\qquad n\geq 1\tag{2}
\end{align*}
We start with the left-hand side of (2) and obtain
\begin{align*}
\color{blue}{\sum_{k=1}^n(-1)^{k+1}F_k}&=\sum_{k=1}^n (-1)^{k+1}[x^k]\frac{x}{1-x-x^2}\tag{3}\\
&=\sum_{k=1}^n[x^k]\frac{x}{1+x-x^2}\tag{4}\\
&=[x^0]\frac{x}{1+x-x^2}\sum_{k=1}^n\frac{1}{x^k}\tag{5}\\
&=[x^0]\frac{x}{1+x-x^2}\cdot\frac{1-x^{n}}{x^n(1-x)}\tag{6}\\
&=[x^n]\frac{x}{(1+x-x^2)(1-x)}\tag{7}\\
&=[x^n]\left(\frac{x}{1-x}-\frac{x^2}{1+x-x^2}\right)\tag{8}\\
&=1-[x^n]\frac{x^2}{1+x-x^2}\\
&=1-(-1)^n[x^n]\frac{x^2}{1-x-x^2}\\
&\color{blue}{=1-(-1)^nF_{n-1}}
\end{align*}
and the claim follows.
Comment:
*
*In (3) we write $F_k$ using the coefficient of operator.
*In (4) we note that multiplication of the coefficient of $x^n$ with $(-1)^n$ means replacing $x$ with $-x$.
*In (5) we use the linearity of the coefficient of operator and apply the rule $[x^{p+q}]A(x)=[x^p]x^{-q}A(x)$.
*In (6) we use the *finite geometric series formula and do some simplifications.
*In (7) we again apply the rule $[x^{p+q}]A(x)=[x^p]x^{-q}A(x)$ as we did in (5) and observe that we can skip $x^{n+1}$ in the numerator, since it does not contribute to $[x^n]$.
*In (8) we do a partial fraction decomposition as preparation for the final steps.
| {
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Derive the conditions $xy<1$ for $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$ and $xy>-1$ for $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$
$$
\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy} \text{, }xy<1\\
\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy} \text{, }xy>-1
$$
But, How do I reach the conditions $xy<1$ for the first expression and $xy>-1$ for the second from the domain and range of the functions, provided we are only considering the principal value branch ?
My Attempt
$$
\tan^{-1}:\mathbb{R}\to \Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)
$$
$$
\text{Taking, }\alpha=\tan^{-1}x, \quad\beta=\tan^{-1}y\implies x=\tan\alpha,\quad y=\tan\beta\\
\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{x+y}{1-xy}\\
\text{We have, }-\pi<\tan^{-1}x+\tan^{-1}y=\alpha+\beta<\pi
$$
If $\tfrac{-\pi}{2}<\alpha+\beta<\tfrac{\pi}{2}$ we have,
$$
\alpha+\beta=\tan^{-1}\bigg(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\bigg)\implies \tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}
$$
For the first expression, $xy\neq{1}$ as the denominator can not be equal to zero.
$$
\frac{-\pi}{2}<\tan^{-1}\frac{x+y}{1-xy}<\frac{\pi}{2}\text{ and }-\pi<\tan^{-1}x+\tan^{-1}y<\pi\\\implies\frac{-\pi}{2}<\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}<\frac{\pi}{2}
$$
I really dont see any clue which leads to the condition $xy<1$. I checked a similar question asked Inverse trigonometric function identity doubt, but it does not seem to clear how to get to the given conditions from the above proof.
Note: I am not looking for proving the statement is correct. I'd like to see how to reach the given conditions from the domain and range of the functions involved.
| We prove $xy<1$ for $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$. The other inequality is likewise. Suppose not. Therefore $xy>1$. The case $xy=1$ would be discussed later. Also the case where at least one of them is zero is trivial, so we assume both to be nonzero. Then we have four cases to check:
Case 1: $x,y>0$ so $\frac{x+y}{1-xy}<0$
$$0<\tan^{-1}x+\tan^{-1}y<\pi$$$$-\frac{\pi}{2}<\tan^{-1}\frac{x+y}{1-xy}<0$$
$$\to\tan^{-1}x+\tan^{-1}y\ne\tan^{-1}\frac{x+y}{1-xy}$$
Case 2: $x,y<0$ so $\frac{x+y}{1-xy}>0$
$$-\pi<\tan^{-1}x+\tan^{-1}y<0$$$$0<\tan^{-1}\frac{x+y}{1-xy}<\frac{\pi}{2}$$
$$\to\tan^{-1}x+\tan^{-1}y\ne\tan^{-1}\frac{x+y}{1-xy}$$
Case 3: $x>0,y<0$ and $x+y<0$
$$-\frac{\pi}{2}<\tan^{-1}x+\tan^{-1}y<0$$$$0<\tan^{-1}\frac{x+y}{1-xy}<\frac{\pi}{2}$$
$$\to\tan^{-1}x+\tan^{-1}y\ne\tan^{-1}\frac{x+y}{1-xy}$$
Because of symmetry that's exactly the case where $x<0,y>0$ and $x+y<0$.
Case 4: $x>0,y<0$ and $x+y>0$
$$0<\tan^{-1}x+\tan^{-1}y<\frac{\pi}{2}$$$$-\frac{\pi}{2}<\tan^{-1}\frac{x+y}{1-xy}<0$$
$$\to\tan^{-1}x+\tan^{-1}y\ne\tan^{-1}\frac{x+y}{1-xy}$$
Because of symmetry that's exactly the case where $x<0,y>0$ and $x+y>0$.
Here we deduce that $\tan^{-1}x+\tan^{-1}y\ne\tan^{-1}\frac{x+y}{1-xy}$ where $xy>1$. The case $xy=1$ is either $xy=1^-$ or $xy=1^+$. Here since $x={1\over y}$ we investigate 2 cases:
Case 1: $x>0$
Then $\tan^{-1} x+ \tan^{-1} \frac{1}{x}=\frac{\pi}{2}$ which leads to $xy=1^-$ since $$\tan^{-1} \frac{x+y}{1-xy}=\tan^{-1} \frac{x+y}{1-1^-}=\tan^{-1} +\infty=\frac{\pi}{2}$$
Case 2: $x<0$
This case similarly gives us $xy=1^+$.
Also you can prove $xy>1$ for $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$ by substituting $y\to -y$.
| {
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"timestamp": "2023-03-29T00:00:00",
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existence of solutions of $a^n+b^n+c^n=6^n$ $$2^2+4^2+4^2=6^2$$
$$3^3+4^3+5^3=6^3$$
There is no such thing for exponent 4. It's not hard to prove that the only integer solutions of $a^4+b^4+c^4=6^4$ are trivial ones.
For which $n$ there exist $a,b,c\in\mathbb{N}$ such that $a^n+b^n+c^n=6^n$?
(own question, a very probable answer is: $2$ and $3$ only).
| There are only solutions for $n=2,3$ as you suspect.
I assume you're looking for non-trivial solutions, so $0\leq a,b,c \leq 5$. Assuming that $a,b,c$ satisfy your equation, then
$$6^{n}=a^{n}+b^{n}+c^{n}
\leq 5^{n}+5^{n}+5^{n} = 3\times 5^{n} \\
\Rightarrow n \leq \frac{\log(3)}{\log(6)-\log(5)}\approx 6.0257$$
So that leaves us with the two cases $n=5$ and $n=6$. For $n=6$, Fermat's little theorem implies that $a^{6}\equiv 1 (\textrm{mod}\,7)$ if $1\leq a \leq 5$ and $a^{6}\equiv 0 (\textrm{mod}\,7)$ if $a=0$. As $6^{6}\equiv 1 (\textrm{mod}\,7)$, we must have that two of $a,b,c$ are zero, which leaves us with no solution.
Finally for $n=5$, first observe that at least one of $a,b,c$ must be $5$. Otherwise
$$6^{5}=a^{5}+b^{5}+c^{5}
\leq 3\times 4^{5}$$
which can be checked to be false. So
$$6^{5}-5^{5}=a^{5}+b^{5}$$
Now we just have to plug in the numbers $0$ to $5$ for $a$ and check if the result can be satisfied for $b$. It is easily seen that there are no such solutions.
| {
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Prove $a^2+b^2+c^2\gt \frac {1}{2018}$ given $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$ Let $a, b, c$ be real numbers such that $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$. Prove that $a^2+b^2+c^2\gt \frac {1}{2018}$.
It looks like an easy question, but I thought for a while and could not figure it out. Can anyone give me some hints? Thank you.
| By C-S $$a^2+b^2+c^2=$$
$$=\frac{\sqrt{(3^2+28^2+35^2)(a^2+b^2+c^2)\cdot(20^2+23^2+33^2)(a^2+b^2+c^2)}}{2018}\geq$$
$$\geq\frac{(3a+28b+35c)(20a+23b+33c)}{2018}=\frac{1}{2018}.$$
| {
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Is this : $\sqrt{3+\sqrt{2+\sqrt{3+\sqrt{2+\sqrt{\cdots}}}}}$ irrational number? I'm confused how do i show if the bellow number could be irrational or no
$$\sqrt{3+\sqrt{2+\sqrt{3+\sqrt{2+\sqrt{\cdots}}}}}$$ ? can I use irrationality of $\sqrt{2}+\sqrt{3}$ ?
| As Masacroso has already pointed out in the comments, you have to find rational solutions to $c=\sqrt {3+ \sqrt {2+c}}$.
$c^2=3+\sqrt {2+c}$
$c^2-3=\sqrt {2+c}$
$(c^2-3)^2=2+c$
$c^4-6c^2+9=2+c$
$c^4-6c^2-c+7=0$
So the answer to your question is whether there are any rational solutions to $c^4-6c^2-c+7=0$ or not.
Actually not. Answer courtesy WA.
| {
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$\tan{A} \cdot \tan{B} \cdot \tan{C}=9$, find $\tan^2{A}+\tan^2{B}+ \tan^2{C}$ In $\triangle{ABC}$,
$$\tan{A}\cdot \tan{B}\cdot \tan{C}=9$$
$$\tan^2{A}+\tan^2{B}+ \tan^2{C}=\lambda$$
then,
$\lambda$ lies in the interval?
| As lab bhattacharjee noted in comments, for triangle angles we have
$$\tan A\cdot \tan B \cdot \tan C = \tan A + \tan B + \tan C.\tag{1}$$
Denote $\;\;x=\tan A,\quad y=\tan B, \quad z=\tan C$.
We have
$$x+y+z=9;\\
xyz=9;\tag{2}$$
and we try to estimate expression $$x^2+y^2+z^2.$$
Let use $z$ as parameter ($0<z<9$). Then
$$x+y=9-z;\\
xy=9/z;$$
and we try to estimate expression
$$(x^2+y^2)+z^2=(x+y)^2-2xy+z^2=(9-z)^2-\dfrac{18}{z}+z^2.\tag{3}$$
Now, $x$ and $y$ (according to Vieta's theorem) are solutions of quadratic equation
$$a^2-(9-z)a+\dfrac{9}{z}=0.\tag{4}$$
This equation has (real) positive solutions if $(9-z)^2z\ge36$.
So, bounds for $z$ are (approximately):
$$z\in I, \mbox{ where } I=[0.498040,6.678221].\tag{5}$$
Considering expression $(3)$ on segment $(5)$, we reach minimal value when $z\approx 4.25098$: $$\min_{z\in I}\{x^2+y^2+z^2\}\approx 36.3897;\tag{min}$$
and we reach maximal value when $z\approx 6.678221$:
$$\max_{z\in I}\{x^2+y^2+z^2\}\approx 47.29396.\tag{max}$$
Minimizing parameters: $x \approx 0.4980402$, $y=z\approx 4.250979.$
Maximizing parameters: $x =y\approx 1.160889$, $z\approx 6.678221.$
This result can be obtained using Lagrange multipliers as well.
| {
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How and where can I calculate $\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\cdots\right)?$ We have
$$A=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)=\ln(2)$$
$$B=\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\cdots\right)$$
Here in $A$ $+-+-$, in $B$ $++--$, so
$$B-A=\frac{1}{2}\left(\psi^{(0)}\left(\frac{3}{4}\right)-\psi^{(0)}\left(\frac{1}{2}\right)\right)$$
and
$$AB=\ln(2)\left(\frac{1}{2}\left(\psi^{(0)}\left(\frac{3}{4}\right)-\psi^{(0)}\left(\frac{1}{2}\right)\right)+\ln(2)\right)\approx\frac{\pi}{4}$$
Here $\psi^{(0)}(z)$ - digamma function.
How and where can I calculate this constant?
| $$B=\sum_{k\geq 0}\left[\frac{1}{4k+1}+\frac{1}{4k+2}-\frac{1}{4k+3}-\frac{1}{4k+4}\right]=\int_{0}^{1}\sum_{k\geq 0}x^{4k}(1+x-x^2-x^3)\,dx $$
$$ B = \int_{0}^{1}\frac{1+x-x^2-x^3}{1-x^4}\,dx = \int_{0}^{1}\frac{1+x}{1+x^2}\,dx = \color{red}{\frac{\pi}{4}+\frac{\ln 2}{2}}.$$
| {
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Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{1}{6} - 1) (x^\frac{1}{6} + 1) } $$
$$\lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{2}{6} - 1) } $$
| Just to avoid L' Hospital's rule, consider the following:
$$\frac{x^{1/5}-1}{x^{1/6}-1}=\frac{x^{6/30}-1}{x^{5/30}-1}=\frac{(x^{1/30})^6-1}{(x^{1/30})^5-1}=\frac{(x^{1/30}-1)(x^{5/30}+x^{4/30}+\dots+1)}{(x^{1/30}-1)(x^{4/30}+x^{3/30}+\dots+1)}$$
So:
$$\lim_{x\to1}\frac{x^{1/5}-1}{x^{1/6}-1}=\lim_{x\to1}\frac{(x^{1/30}-1)(x^{5/30}+x^{4/30}+\dots+1)}{(x^{1/30}-1)(x^{4/30}+x^{3/30}+\dots+1)}=\lim_{x\to1}\frac{x^{5/30}+x^{4/30}+\dots+1}{x^{4/30}+x^{3/30}+\dots+1}=\frac{6}{5}$$
Hope this provided an alternative! :)
| {
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Finding minimum of $\max(\arg(a),\arg(b),\arg(c))$ in complex number
If $a,b,c$ be $3$ complex number (not all real ) such that $|a|=|b|=|c|=1$ and $2(a+b+c)-3abc$ is real , Then minimum of $\max(\arg(a),\arg(b),\arg(c))$ Given argument of $a,b,c$ all are positive.
Try: Assuming $z =2(a+b+c)-3abc$. Then $z=\bar{z}$
So $$2(a+b+c)-3abc=2(\overline{a}+\overline{b}+\overline{c})-3\overline{abc}$$
Could some help me to solve it , thanks
| We claim that the answer is $\dfrac{\pi}{6}$ and proceed to argue by contradiction.
Let $a = e^{ix}, b = e^{iy}, c = e^{iz}$ and assume to the contrary that there exists $x,y,z \in [0, \dfrac{\pi}{6})$, such that $$2(e^{ix}+e^{iy}+e^{iz})-3e^{i(x+y+z)} = 2(e^{-ix}+e^{-iy}+e^{-iz})-3e^{-i(x+y+z)}$$ or $$2(\sin x+\sin y +\sin z) = 3\sin(x+y+z)\,\ \dagger. $$
However, we see that $t = x+y+z<\dfrac{\pi}{2}$ and on $(0, \dfrac{\pi}{2})$, we will prove the following inequality:
$$\sin t>2\sin\dfrac{t}{3}\,\,\ddagger.$$
Consider $f(x) = \sin 3x - 2\sin x = \sin x - 4\sin^3x = \sin x(1-2\sin x)(1+2\sin x)>0$ when $0<x<\pi/6$, so $\ddagger$ is proven.
Finally, since $\sin t$ is concave on our interval, we apply the Jensen's inequality and combine with $\ddagger $ to obtain:
$$3\sin(x+y+z)> 6\sin\left(\dfrac{x+y+z}{3}\right)\geq 2(\sin x+\sin y+\sin z) = 3\sin(x+y+z),$$
a contradiction.
Therefore, $\max\{x,y,z\}\geq\dfrac{\pi}{6}$ and the minimum of that max is actually $\pi/6$, when $x,y,z = \dfrac{\pi}{6}.$
| {
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Jordan's Canonical Form of a Matrix I'm a little tied up about Jordan's canonical form of a square matrix. How would Jordan's Canonical form of the following square matrix be obtained?
$$A = \begin{pmatrix}
2 & 1 & 1 \\
-2 & -1 & -2\\
1 & 1 & -2
\end{pmatrix}$$
| You have to find a Jordan basis for $A$. As $-1$ is a double eigenvalue, to determine whether $A$ is diagonalisable, we have toknow whether the eigenspace for this value has dimension $2$ or not, and in the latter case determine a basis for the generalised eigenspace.
Now $\ker(A+I)$ has dimension $1$ since
$$A+I= \begin{pmatrix}
3 & 1 & 1 \\
-2 & 0 & -2\\
1 & 1 & -1
\end{pmatrix}$$
has rank $2$. The last two rows are linearly independent, and an eigenvector for the eigenvalue $-1$ satisfy the equations $x+z=0$, $\;x+y-z=0$.
$\ker (A+I)^2$ has dimension $2$ since
$$(A+I)^2= \begin{pmatrix}
8 & 4 & 0 \\
-8 & -4 & 0\\
0 & 0 & 0
\end{pmatrix}$$has rank $1$, and is defined by the equation $2x+y=0$.
To have a Jordan basis , we begin with taking $v_3\in\ker (A+I)^2 \setminus\ker (A+I)$. The simplest is to choose $v_3=(0,0,1)$.
Next set $v_2=(A+I)v_3=(1,-2,-1)$. This is an eigenvector for the eigenvalue $-1$, and by construction we have:
$$Av_3=-v_3+v_2.$$
Last step: find an eigenvector for the simple eigenvalue $1$, i.e. a vector in the kernel of
$$ A-I= A = \begin{pmatrix}
1 & 1 & 1 \\
-2 & -2 & -2\\
1 & 1 & -3
\end{pmatrix}.$$
It has to satisfy the independent linear equations:
$$x+y+z=0,\enspace x+y_3z=0 \iff z=0,\enspace x+y =0, $$
so we may choose $v_1=(1,-1,0)$.
By construction, in the basis $(v_11,v_2,v_3)$ the matrix of the endomorphism associated to $A$ is
$$J(A)= \begin{pmatrix}1&0&0\\0&-1&1\\0&0&-1\end{pmatrix}.$$
| {
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Trig substitution for $\int \frac{x^2dx}{\sqrt{4 - x^2}}$ According to my textbook the answer is: $2\arcsin(\frac{1}{2}x) - \sin(2\arcsin(\frac{1}{2}x))$, but I'm getting something slightly different.
First I tried setting $x = 2 \sin \theta, dx = 2 \cos \theta, \theta = \arcsin(\frac{x}{2})$
$$\int \frac{4\sin^2 \theta}{\sqrt{4(1-\sin^2 \theta})}*2\cos \theta d\theta $$
$$ \int \frac{4\sin^2 \theta}{2\cos \theta}2cos\theta = \int4\sin^2\theta{d\theta}$$
Then I let $u = 2\theta$
$$4 \int \frac{\sin 2\theta + 1}{2} = 4 [\frac{1}{2}\theta + \frac{1}{4} \int \sin udu]$$
$$4[\frac{1}{2}\theta - \frac{\cos u}{4}] = 4 [\frac{\arcsin(\frac{x}{2})}{2} - \cos(2\arcsin(\frac{x}{2}))]$$
Yielding:
$$2\arcsin(\frac{x}{2}) - \frac{\cos(2\arcsin(\frac{x}{2}))}{4}$$
What's wrong with this?
| You've apparently used an incorrect trigonometric identity,
$$\sin^2 t = \frac{1 + \sin 2t}{2}.$$
This is correct only if you erase all the sines and write cosines instead; it is false if $t = 0$, for example.
The correct identity for $\sin^2 t$ is (among several others, of course)
$$\sin^2 t = \frac{1 - \cos 2t}{2}$$
leading to
$$\int \cos 2\theta \, d\theta = \frac 1 2 \sin 2\theta = \frac 1 2 \sin(2 \arcsin x/2)$$
which has the correct form.
| {
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What is wrong with my solution to this equation : $346 x+ 1250 \equiv 49 \pmod{105}$? I'm not sure how to get $x$ but with my way $x = 39$. The solution for $x$ is 29 though. Could anyone possibly help me with the calculating method?
Mine:
$\gcd(346,105) = 1 \Rightarrow x = a^{-1} \cdot b \pmod m \Rightarrow 346^{-1} \pmod{105} = 61 \Rightarrow (61\cdot49)+1250 \pmod{105} = 39$
| Note that
$$346x+ 1250 \equiv 49 \pmod {105} \iff 31x \equiv 59 \pmod {105}$$
thus since $gdc(31,105)=1$ we can find the inverse of $31 \pmod {105}$ by Euclidean's algorithm.
As an alternative by CRT we get
$$\begin{cases}31x \equiv 59 \pmod {3}\iff x \equiv 2 \pmod {3} \\31x \equiv 59 \pmod {5} \iff x \equiv 4 \pmod {5} \\31x \equiv 59 \pmod {7} \iff 3x \equiv 3 \pmod {7} \iff x \equiv 1 \pmod {7} \end{cases}$$
thus
*
*$ x \equiv 1 \pmod {7} \implies x=1+7k$
*$x=1+7k \equiv 4 \pmod {5}\implies 1+2k \equiv 4\pmod 5 \implies 2k
\equiv 3\pmod 5 \\\implies k \equiv 4\pmod 5 \implies x=1+7(4+5h)=29+35h$
*$x=29+35h \equiv 2 \pmod {3} \implies h\equiv 0 \pmod 3$
thus $$x=29 \pmod {105}$$
| {
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Evaluating $\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}$
Evaluate:
$$\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}$$
I have been trying to solve this for $15$ minutes but sin(sin(x)) part has me stuck.
My attempt:
I tried multiplying with $x$ inside the $\sin$ as $\sin{(\frac{x\sin{x}}{x})}$. No leads.
| $$\sin(u)=u-\frac{u^3}{6}+\frac{u^5}{120}+o(u^6)$$
Then, $$\sin^2(x)=(x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6))^2 =x^2-\frac{x^4}{3} +o(x^6)$$
and
$$\sin(\sin(x))=\sin\left(x-\frac{x^3}{6}+\frac{x^5}{125}+o(x^6)\right) \\=x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6) -\frac{1}{6}\left(x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6)\right)^3 + \frac{1}{120} \left(x + o(x^3)\right)^5 \\=x-\frac{x^3}{3}+\frac{x^5}{10}+o(x^6) $$
Hence
$$\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}\\=\frac{x^2-\frac13x^4+\frac1{10}x^6-x^2+\frac13x^4-\frac2{45}x^6+o(x^6)}{x^6}=\frac1{18}+o(1)$$
Then the limit is $\frac1{18}$
| {
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Given is markov chain - Determine the probability $f_1(n)$
Given is markov chain $\left\{X_n\right\}_{n \in \mathbb{N}}$ with
transition probabilities
$$M= \begin{pmatrix}
1/2 & 1/2 & 0 & 0 & 0 & 0 \\
1/4 & 3/4 & 0 & 0 & 0 & 0 \\
1/4 & 1/4 & 1/4 & 1/4 & 0 & 0 \\
1/4 & 0 & 1/4 & 1/4 & 0 & 1/4 \\
0 & 0 & 0 & 0 & 1/2 & 1/2 \\
0 & 0 & 0 & 0 & 1/2 & 1/2
\end{pmatrix}$$
Determine the probability $f_1(n)$ where you return to state $1$ after
$n$ steps (for the first time).
I'm not sure how you can solve this because if I understood it correctly, we have $n$ steps and we are looking for a probability, so we have two unknowns...
Anyway, I think the correct way of calculating it is (don't miss the little exponent $n$ of that huge matrix!)
$$f_1(n) = \begin{pmatrix}
1/2 & 1/2 & 0 & 0 & 0 & 0 \\
1/4 & 3/4 & 0 & 0 & 0 & 0 \\
1/4 & 1/4 & 1/4 & 1/4 & 0 & 0 \\
1/4 & 0 & 1/4 & 1/4 & 0 & 1/4 \\
0 & 0 & 0 & 0 & 1/2 & 1/2 \\
0 & 0 & 0 & 0 & 1/2 & 1/2
\end{pmatrix}^n \cdot
\begin{pmatrix}
1\\
0\\
0\\
0\\
0\\
0
\end{pmatrix}$$
Here I'm stuck again for the same reason, I see no way of calculating the probability...? : /
| You need this:
$$
f_1(n) = \begin{pmatrix}
1, &
0, &
0, &
0, &
0, &
0
\end{pmatrix} \begin{pmatrix}
1/2 & 1/2 & 0 & 0 & 0 & 0 \\
1/4 & 3/4 & 0 & 0 & 0 & 0 \\
1/4 & 1/4 & 1/4 & 1/4 & 0 & 0 \\
1/4 & 0 & 1/4 & 1/4 & 0 & 1/4 \\
0 & 0 & 0 & 0 & 1/2 & 1/2 \\
0 & 0 & 0 & 0 & 1/2 & 1/2
\end{pmatrix}^n \cdot
$$
What tells you it's done that way is that the rows add up to $1$ and the columns don't.
Then you can see that you have a Markov chain in which, once you're in state $1$ or state $2,$ you can never get to other states than $1$ and $2.$ And that means you only need to pay attention to the first two rows and the first two columns.
| {
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Solve cubic equation $x^3 - 15x - 4 = 0$ with Cardano's method Suppose that we wanted to solve
$$x^3 - 15x - 4 = 0$$
by using Cardano's method. In doing so, I end up with
$$\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i}$$
However, by checking my answer, it seems there are two other answers I haven't accounted for: $\sqrt{3} - 2$ and $-2 - \sqrt{3}$. Is there a way to generate these solutions from this method, or would I need to find them in other ways, e.g., by factoring?
| The Cardano’s formula is normally used to calculate the real root of the a cubic equation with one real root and a pair of complex roots. In the case of three real roots, e.g. $x^3 - 15x - 4 = 0$, the formula results in the expression
$$\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i}$$
which is of multiple complex values. However, it can still be used to obtain the three real roots if their real parts are taken, i.e.
$$x=Re(\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i})= 2Re (\sqrt[3]{2 + 11i})\tag1 $$
Write the cubic radicant as
$$2+11i=\sqrt{125}e^{i \cos^{-1}\frac2{5\sqrt5}}
= \sqrt{125}e^{i \theta}$$
where
$$\cos\theta = \frac2{5\sqrt5}=4\cos^3\frac{\theta}3-3\cos\frac{\theta}3
$$
which factorizes as
$$\left(\cos\frac{\theta}3- \frac2{\sqrt5}\right)\left(\cos\frac{\theta}3+\frac{\sqrt3+2}{\sqrt5}\right)\left(\cos\frac{\theta}3- \frac{\sqrt3-2}{\sqrt5}\right)=0
$$
Then
$$\cos\frac{\theta}3=\frac2{\sqrt5},\>\frac{\sqrt3-2}{2\sqrt5},\>
\frac{-\sqrt3-2}{2\sqrt5}
$$
and the corresponding roots (1) are
\begin{align}
x&
=2Re \left( \sqrt{125}e^{i \theta}\right)^{\frac13} = 2\sqrt5 \cos\frac{\theta}3=4,\>\sqrt3-2,\>-\sqrt3-2
\end{align}
| {
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Using induction to prove sum of certain number-theoretic function. For any positive integer $ n > 1$, let $P(n)$ denote the largest prime not exceeding $n$.
Let $N(n)$ denote the next prime larger than $P(n)$. (For example $P(10) = 7$ and
$N(10) = 11$, while $P(11) = 11$ and $N(11) = 13$.) If $n + 1$ is a prime number, prove
that the value of the sum
$$ \frac{1}{P(2)N(2)} + \frac{1}{P(3)N(3)} + \cdot\cdot\cdot + \frac{1}{P(n)N(n)} = \frac{n-1}{2n+2}$$
Can I use induction on two consecutive primes $k+1$ and $k+q+1$?
If so, then here is my proof:
Since the result holds true for $n=2$ and $n=4$, suppose that it holds true for $n=k$ with $k+1$ being prime. Then,
$$\sum_{2}^{k}{\frac{1}{P(n)N(n)}} = \frac{k-1}{2(k+1)}$$
Then suppose that $k+q+1$ is the next prime greater than $k+1$, then $\forall m$ such that $ k+1 \leq m < k+q+1$,
$P(m) = k+1$ and $N(m) = k+q+1$.
Therefore,
$$\sum_{2}^{k+q}{\frac{1}{P(n)N(n)}} = \sum_{2}^{k}{\frac{1}{P(n)N(n)}} + \sum_{k+1}^{k+q}{\frac{1}{P(n)N(n)}} = \frac{k-1}{2(k+1)} +\frac{q}{(k+1)(k+q+1)}$$
Which, on manupulation yields
$$\sum_{2}^{k+q}{\frac{1}{P(n)N(n)}}=\frac{k+q-1}{2(k+q+1)}$$ Hence completes the proof.
Is this correct or am I missing something?
| I checked it and didn't find any mistakes. Nice identity!
I don't think you used any facts about primes anywhere, so if it's true it would be true for any sub-sequence of natural numbers.
| {
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How to approach this question on choosing balls from a box?
A box contains $2n$ balls of $n$ different colors, with 2 of each color. Balls are picked at random from the box with replacement until two balls of the same color have appeared. Let $X$ be the number of draws made.
a) Find a formula for $P(X > k)$, $k = 2, 3, ...$
$P(X>k) = 1 - P(X\le k) = 1- (P(X=2)+P(X=3)+P(X=4)+...P(X=k))$
So I manaully find $P(X=2)$, $P(X=3)$, $P(X=4)$, ... and try to find a pattern to make an equation?
Edit 1:
For each $n$ color, trials = 2, success rate = $\frac{2}{2n}$, # of needed success = 2, ordering doesn't matter, with replacement
$P(X = 2) = n \cdot Binomial(2, \frac{2}{2n}) = \binom{2}{2} (\frac{2}{2n})^2(1-\frac{2}{2n})^0$
For each $n$ color, trials = 3, success rate = $\frac{2}{2n}$, # of needed success = 2, ordering doesn't matter, with replacement
$P(X = 3) = n \cdot Binomial(3, \frac{2}{2n}) = \binom{3}{2} (\frac{2}{2n})^2(1-\frac{2}{2n})^1$
For each $n$ color, trials = 4, success rate = $\frac{2}{2n}$, # of needed success = 2, ordering doesn't matter, with replacement
$P(X = 4) = n \cdot Binomial(4, \frac{2}{2n}) = \binom{4}{2} (\frac{2}{2n})^2(1-\frac{2}{2n})^2$
For each $n$ color, trials = 5, success rate = $\frac{2}{2n}$, # of needed success = 2, ordering doesn't matter, with replacement
$P(X = 5) = n \cdot Binomial(5, \frac{2}{2n}) = \binom{5}{2} (\frac{2}{2n})^2(1-\frac{2}{2n})^3$
$$P(X>k)=1-(n \cdot \sum_{i=2}^k \binom{i}{2} (\frac{2}{2n})^2(1-\frac{2}{2n})^{i-2})$$
Edit 2:
Last ball has to be a color already chosen.
For X = 2, no restrictions on the 1st ball, 2nd ball must be a color already chosen.
$P(X = 2) = \frac{2n}{2n} \cdot \frac{2}{2n}$
For X = 3, no restrictions on the 1st ball, no restrictions on the 2nd ball, 3rd ball must be a color already chosen.
$P(X = 3) = \frac{2n}{2n} \cdot \frac{2n}{2n} \cdot \frac{4}{2n}$
For X = 4, no restrictions on the 1st ball, no restrictions on the 2nd ball, no restrictions on the 3rd ball, 4th ball must be a color already chosen.
$P(X = 4) = \frac{2n}{2n} \cdot \frac{2n}{2n} \cdot \frac{2n}{2n} \cdot \frac{6}{2n}$
$$P(X > k) = 1 - (\sum_{i=2}^k \frac{2^{i-1}}{2n})$$
Edit 3:
Last ball has to be a color already chosen AND other balls must be different!
For X = 2, no restrictions on the 1st ball, 2nd ball must be a color already chosen.
$P(X = 2) = \frac{2n}{2n} \cdot \frac{2}{2n}$
For X = 3, no restrictions on the 1st ball, 2nd ball must be different from the 1st ball, 3rd ball must be a color already chosen.
$P(X = 3) = \frac{2n}{2n} \cdot \frac{2n-2}{2n} \cdot \frac{4}{2n}$
For X = 4, no restrictions on the 1st ball, 2nd ball must be different from the 1st ball, 3rd ball must be different from the 1st and 2nd ball, 4th ball must be a color already chosen.
$P(X = 4) = \frac{2n}{2n} \cdot \frac{2n-2}{2n} \cdot \frac{2n-4}{2n} \cdot \frac{6}{2n}$
$$P(X>k)=1-(\frac{1}{n} + \frac{2(n-1)}{n^2} + \frac{3\left(n-1\right)\left(n-2\right)}{n^3} + ... + \frac{(k-1)(n-1)(n-2)\cdot\cdot\cdot(n-(k-2))}{n^{k-1}})$$
Answer from Textbook:
$\frac{2n-2}{2n} \cdot \frac{2n-4}{2n} \cdot ... \cdot \frac{2n-2(k-1)}{2n}$
| You interpreted the problem as if you pre-selected one of the colors before drawing any balls and require to draw two balls of that color, for example if there are red balls you might choose red and then you will stop only when you have drawn red twice. I would interpret it differently--if any color is drawn that has been drawn before, you stop. By the pigeonhole principle the maximum possible value of $X$ is $n +1.$
The textbook apparently interpreted the question the way I did.
For example, you draw one ball and observe its color.
Put it back and draw again.
In order for $X > 2,$ the second ball you draw must be different from the first. There are $2n - 2$ such balls, so
$P(X> 2) = \frac{2n - 2}{2n}.$
If the second ball is a different color from the first, then on the third draw there are now only $2n - 4$ balls that are different from any color already drawn.
(Either of the two previously seen colors is a match.)
To have $X > 3$ you must avoid a match on draw number $2$ and again on draw number $3,$ so
$P(X > 3) = \frac{2n - 2}{2n} \cdot \frac{2n - 4}{2n}.$
And so forth.
| {
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How to simplify the equation $ a\sin x\cos y + b\cos x\sin y = c\sin(2x)$? I am trying to derive the relation between x and y which are two physical parameters.I am trying to express x in terms of y, a and b. Is there a way to do it?
| You can rewrite your equation as
$$b \cos x \sin y = \sin x \;( 2 \cos x - a \cos y ) $$
Squaring, we have
$$b^2 \cos^2 x \sin^2 y = \sin^2 x (2 \cos x - a \cos y )^2$$
So, writing $p:= \cos x$ and $q := \cos y$,
$$b^2 p^2 (1-q^2) = (1-p^2) (2p - a q )^2$$
This gives a quartic polynomial in $p$:
$$4 c^2 p^4 - 4 a c p^3 q + ( b^2 - 4 c^2 + a^2 q^2 - b^2 q^2 ) p^2 + 4 a c p q - a^2 q^2 = 0 $$
which does not have nice roots.
| {
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Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$ I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$:
$$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$
which gives (Wolfram Alpha)
$$D= \left[
\frac{\sin \theta(\cos \theta - a)}{2(a^2-1)(a \cos\theta-1)^2}
+\frac{\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{(a^2-1)^{3/2}}\right]_0^{2\pi}
.$$
which can be expressed as
$$D= \left[
\frac{(a^2-1)^{1/2}\sin \theta(\cos \theta - a)+2 (a \cos\theta-1)^2\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{2(a^2-1)^{3/2}(a \cos\theta-1)^2}
\right]_0^{2\pi}
.$$
This expression involves discontinuities and complex numbers which is beyond my present abilities to handle.
| This calls for Kepler's angle! Let
$$\sin\theta=\frac{\sqrt{1-a^2}\sin\psi}{1+a\cos\psi}$$
Then
$$\begin{align}\cos\theta&=\frac{\cos\psi+a}{1+a\cos\psi}\\
d\theta&=\frac{\sqrt{1-a^2}}{1+a\cos\psi}d\psi\\
1-a\cos\theta&=\frac{1-a^2}{1+a\cos\psi}\end{align}$$
Also when $\theta$ makes a full cycle, so does $\psi$, so
$$\begin{align}\int_0^{2\pi}\frac{\sin^2\theta}{\left(1-a\cos\theta\right)^3}d\theta&=\int_0^{2\pi}\frac{\left(1-a^2\right)\sin^2\psi}{\left(1+a\cos\psi\right)^2}\frac{\left(1+a\cos\psi\right)^3}{\left(1-a^2\right)^3}\frac{\sqrt{1-a^2}}{1+a\cos\psi}d\psi\\
&=\frac1{\left(1-a^2\right)^{3/2}}\frac12\int_0^{2\pi}\left(1-\cos2\psi\right)d\psi\\
&=\frac1{2\left(1-a^2\right)^{3/2}}\left.\left[\psi-\frac12\sin2\psi\right]\right|_0^{2\pi}=\frac{\pi}{\left(1-a^2\right)^{3/2}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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} |
A simple problem with a simple and a nonsimple solution I have this seemingly simple problem to solve. Its statement is straightforward:
"If $x+y+z=3,x^2+y^2+z^2=5, x^3+y^3+z^3=7$ show that $x^4+y^4+z^4=9,x^5+y^5+z^5\neq 11$".
There is a highschool approach, that is I expanded $(x+y+z)^2, (x+y+z)^3$ and managed to get $$xyz=-2/3, xy+yz+zx=2$$ etc, which I guess I could employ to expand $(x+y+z)^4, (x+y+z)^5$ and get some results (I guess).
The thing is I don't think this is the proper approach. Is there a better way than this tedious, long calculation (if it is a solution that is)
Say we could divide the polynomials
$$f=x^4+y^4+z^4,g=x^3+y^3+z^3, h=x^2+y^2+z^2,k=x+y+z$$
Wouldn't this give us what we want by using the remainder polynomial?
Thanks in advance for your time.
| $$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+x^2z^2+y^2z^2)=$$
$$=25-2((xy+xz+yz)^2-2xyz(x+y+z))=$$
$$=25-2\left(4-2\left(-\frac{2}{3}\right)3\right)=9.$$
$$x^5+y^5+z^5=(x^2+y^2+z^2)(x^3+y^3+z^3)-\sum_{sym}x^3y^2=$$
$$=35-(x^2y^2+x^2z^2+y^2z^2)(x+y+z)+xyz(xy+xz+yz)=$$
$$=35-8\cdot3-\frac{2}{3}\cdot2=\frac{29}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Tangent substitution in trigonometric substitution Find:
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}}\;dx$$
Let $u=2x$ and with tangent substitution we have $x=\frac{3}{2}\,tan\,\theta$ and now we have $dx=\frac{3}{2}\,sec^2\,d\theta$.
Also, $\sqrt{4x^2+9}=3\,sec\,\theta$.
When $x=0$, $tan\,\theta=0$, so $\theta=0$; when $x=3\sqrt{3}/2$, $tan\,\theta=\sqrt{3}$ so $\theta=\pi/3$
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}}\,dx=\int^{\pi/3}_0\frac{\frac{27}{8}\,tan^3\,\theta}{27\,sec^3\,\theta}\,\frac{3}{2}\,sec^2\,\theta\,d\theta$$
$$=\frac{3}{16}\int^{\pi/3}_0\frac{1-cos^2\,\theta}{cos^2\,\theta}sin\,\theta\,d\theta$$
With u substitution $u=cos\,\theta$ and $du=-sin\,\theta\,d\theta$. When $\theta=0$, $u=1$; when $\theta=\pi/3$, $u=\frac{1}{2}$
Therefore:
$$\frac{3}{16}\int^{\pi/3}_0\frac{1-cos^2\,\theta}{cos^2\,\theta}sin\,\theta\,d\theta=\frac{-3}{16}\int^{\frac{1}{2}}_1\frac{1-u^2}{u^2}du$$
My question lies in the next part:
$$=\frac{3}{16}\int^{\frac{1}{2}}_1(1-u^{-2})du$$
Why does $\frac{-3}{16}$ become positive? Also what did they do to get $1-u^{-2}$?
| $$\frac{1-u^2}{u^2}=\frac{1}{u^2}-\frac{u^2}{u^2}=u^{-2}-1=-(1-u^{-2})$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve in $\mathbb{Z}$ the equation $x^4 + 1 = 2y^2$. Find all pairs of intergers $(x,y)$ such that $x^4 + 1 = 2y^2$.
I'm thinking of Gaussian integers, since the LHS can be factored in $\mathbb{C}$. But I don't know how to continue here.
| We consider $x, y$ positive (if $x$ is solution then $-x$ is solution, same with $y$). From $x^4+1=2y^2$ we deduce that $2y^2=t^2+1$, where $t=x^2$ must be odd. Let $t=2k+1$, so $y^2=k^2+(k+1)^2$. We can now use the fact that $y=(m+n)^2$ and $k=(m-n)^2$ and $(k+1)=2mn$ or $k=2mn$ and $(k+1)=(m-n)^2$, where $m, n\in \mathbb Z$.
Case 1. $k=(m-n)^2$ and $(k+1)=2mn$.
Then $k+2(k+1)=(m+n)^2=y$, so $y=3k+2$ so $$9k^2+12k+4=2k^2+2k+1$$ and we find k.
Case 2. $k=2mn$ and $(k+1)=(m-n)^2$
Then $2k+k+1=(m+n)^2=y$, so $y=3k+1$ so $$9k^2+6k+1=2k^2+2k+1$$ and we find k.
EDIT: It is NOT right, sorry for the confusion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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} |
Derivative of natural log with chain rule. Is there a better way? I'm a bit stuck on taking these two derivatives:
$$h(x) = \ln(x + \sqrt{x^2-1})$$
\begin{align}
h'(x) &= \frac{1}{x + \sqrt{x^2 - 1}} \cdot \frac{d }{dx} (x + \sqrt{x^2 -1} )\\
&= \frac{1}{x + \sqrt{x^2 - 1}} \biggl(1 + \frac{2x}{2\sqrt{x^2 - 1}}\biggr)
\end{align}
I'm a bit stuck on how to simplify from here? Was there a simpler way somewhere?
Also this one is giving me problems. I am thinking of changing the log forms first?:
$$\ln \frac{(2y+1)^5}{\sqrt{y^2 + 1}}= \ln(2y+1)^5 - \ln \sqrt{y^2 +1}=5 \ln(2y+1) - \frac{1}{2}\,\ln (y^2 +1)$$
so:
$$G'y = \frac{5}{2y+1} \cdot 2 - \frac{2y}{2(y^2 + 1)} = \frac{10}{2y+1} - \frac{y}{2 (y^2 +1)}$$
Is that right?
| Here is a sneaky way to deal with one of those derivatives. Notice that $h(x)=\cosh^{-1}(x)$, therefore
$$
h’(x)=\frac{1}{\sinh\left(\cosh^{-1}(x)\right)}
= \frac{1}{\sqrt{\cosh^2\left(\cosh^{-1}(x)\right)-1}}=\frac{1}{\sqrt{x^2-1}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Dividing balls into boxes with capacity limits Problem:
In how many ways can you divide $13$ identical balls into $3$ different boxes
$k_1$, $k_2$, $k_3$, such that $k_1$ contains no more than $5$ balls, $k_2$ contains no more than $6$ balls and $k_3$ contains no more than $4$ balls?
My idea:
So my idea is to use the following theorem:
"There are $C(n+r-1,r)$ r-combinations from a set with $n$ elements when repetition of elements is allowed." But I'm not sure.
Any great hints would be appreciated.
| The answer can simply be given using generating functions. We simply need to calculate the coefficient of $x^{13}$in the expansion of $$(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^6)$$
$$=\frac {(1-x^5)}{(1-x)}\cdot \frac {(1-x^6)}{(1-x)}\cdot \frac {(1-x^7)}{(1-x)}$$
$$=\frac {1-x^5-x^6-x^7+x^{11}+x^{13}+x^{12}-x^{18}}{(1-x)^3}$$
Hence the answer simply goes as $$\binom {15}{13} -\binom {10}{8}- \binom {9}{7}-\binom {8}{6}+\binom {4}{2}+1+\binom {3}{1}=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
On the convergence of the series $ \sum_n \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} $ I've got some problem with this series
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)}
$$
I know that the series diverges by a comparison test, but I've got a lot of trouble to prove it. I've tried some algebraic tricks like
$$
a - b = \frac{a^{3} - b^{3}}{a^{2} + ab + b^{2}},
$$
where $ a = \sqrt[3]{(n^{3}+n)} $ and $ b = \sqrt[3]{(n^3-n)} $.
It gave me
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} = \sum_{n=1}^{\infty} \frac{2n}{\sqrt[3]{(n^{3}+n)^{2}} + \sqrt[3]{n^{6} - n^{2}} + \sqrt[3]{(n^3-n)^{2}}}
$$
but for me it looks like a blind valley. I would appreciate every help, thank you.
| Your idea was good:
$$\sqrt[3]{(n^3 + n)^2} = n^2\sqrt[3]{(1 + 1/n^2)^2},$$
$$\sqrt[3]{n^6 - n^2} = n^2\sqrt[3]{1 - 1/n^4},$$
$$\sqrt[3]{(n^3 - n)^2} = n^2\sqrt[3]{(1 - 1/n^2)^2}.$$
So
$$
\sqrt[3]{(n^3 + n)^2} + \sqrt[3]{n^6 - n^2} + \sqrt[3]{(n^3 - n)^2}\sim 3n^2
$$
and
$$
\frac{2n}{\sqrt[3]{(n^3 + n)^2} + \sqrt[3]{n^6 - n^2} + \sqrt[3]{(n^3 - n)^2}}\sim\frac{2}{3n}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Calculate the limitation:$I=\lim_{n\rightarrow \infty}\int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x $ $$I=\lim_{n\rightarrow \infty}\int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x
$$
the answer is 0. This is what I've done so far.
$$ \begin{split}
\int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x&=\frac{1}{n}\int_0^{2\pi}{\frac{1}{x+1}}\text{d}\left( \sin nx \right)\\
&=\left. \frac{\sin nx}{n}\cdot \frac{1}{1+x} \right|_{0}^{2\pi}+\int_0^{2\pi}{\frac{\sin nx}{n}\cdot \frac{1}{\left( 1+x \right) ^2}}\text{d}x\\
&=\int_0^{2\pi}{\frac{\sin nx}{n}\cdot \frac{1}{\left( 1+x \right) ^2}}\text{d}x
\end{split}$$
| Following your intuition, an inequality is useful at the point you got to, namely $|\sin(x)| \leq 1$ Using that, we have
$$\left|\frac{1}{n}\int_0^{2\pi} \frac{\sin(nx)}{(1+x)^2}dx\right| \leq\frac{1}{n}\int_0^{2\pi}\frac{1}{(x+1)^2}dx=\frac{1}{n}\frac{-1}{x+1}\bigg|_0^{2\pi}=\frac{1}{n}\left(1-\frac{1}{2\pi+1}\right)<\frac{1}{n}$$
Now letting $n$ go to infinity we get
$$\lim_{n \to \infty} |I| <\lim_{n \to \infty} \frac{1}{n} = 0$$
So $I=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $x \cos\theta+y\sin\theta=a$ and $x\sin\theta-y\cos\theta=b$, then $\tan\theta=\frac{bx+ay}{ax-by}$. (Math Olympiad) I tried to solve it but I can’t get the answer. Please help me in proving this trig identity:
If
$$x \cos\theta+y\sin\theta=a$$
$$x\sin\theta-y\cos\theta=b$$
then $$\tan\theta=\frac{bx+ay}{ax-by}$$
I've spent many hours trying.
Thanks in advance.
| $x \cos\theta + y\sin\theta = a, \,\, x \sin\theta - y\cos\theta = b$. Therefore:
$$\dfrac{x \cos\theta + y\sin\theta}{x \sin\theta - y\cos\theta} = \dfrac{a}{b}$$
$$\implies \dfrac{x + y\tan\theta}{x \tan\theta - y} = \dfrac{a}{b}$$
$$\implies bx + by\tan\theta = ax \tan\theta - ay$$
$$\implies (ax - by) \tan\theta = bx + ay$$
$$\implies \tan\theta = \dfrac{bx + ay}{ax - by}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
On near-Pythagorean triples $(n^5-2n^3+2n)^2 + (2n^4-2n^2+1)^2 = n^{10} + 1$ We have,
$$\begin{aligned}
(n^3-2n)^2 + (2n^2-1)^2 &= n^6 + 1\\
(n^5-2n^3+2n)^2 + (2n^4-2n^2+1)^2 &= n^{10} + 1\\
(n^7-2n^5+2n^3-2n)^2 + (2n^6-2n^4+2n^2-1)^2 &= n^{14} + 1\end{aligned}$$
and so on.
Q: What is a clever way to show that this pattern does in fact go on?
| here is one way
$n^{4k+2}+1 = (n^{2k+1}-i^{2k+1})(n^{2k+1}-(-i)^{2k+1}) \\
= (n-i)(n^{2k}+ \cdots + i^{2k})(n+i)(n^{2k}+\cdots + (-i)^{2k+1}))\\
= (n+i)(n^{2k}+ \cdots + i^{2k})(n-i)(n^{2k}+\cdots + (-i)^{2k+1}))\\
= (n^{2k+1}+2in^{2k}+\cdots +2i^{2k}n+i^{2k+1})(n^{2k+1}+2(-i)n^{2k}+\cdots +2(-i)^{2k}n+(-i)^{2k+1}) \\
= (n^{2k+1}-2n^{2k-1}+\cdots+2(-1)^kn)^2 + (2n^{2k}- \dots + (-1)^{k-1}2n^2+(-1)^k)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Generator of $SL(2,\mathbb{R})$ Find all 2x2 matrices $ A \in SL(2,\mathbb {R}) $such that $det(A) =1$ and $A\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix} $. Do these matrices form a group?
My thoughts:
I have considered the matrices \begin{pmatrix} 1 & 0\\ x & 1 \end{pmatrix} and \begin{pmatrix} 1 & x\\ 0 & 1 \end{pmatrix} I have realised that the matrices will not satisfy the condition if $x \neq 0$. These two have the identity element, and are invertible. Another matrix of interest was \begin{pmatrix} 0 & -1\\ 1 & 2 \end{pmatrix}. My problem is, how does one compactly answer this question, and I believe that the group is finite.
| A nonempty subset $S$ of a group $G$ forms a subgroup if for all $A,B\in S$, $AB^{-1}\in S$. In this case, $S$ is nonempty because the identity is in $S$. You then assume that $A$ and $B$ are two matrices with the given property and consider (the $\frac{1}{\sqrt{2}}$ doesn't matter in this problem since it's a scaling factor).
$$
AB^{-1}\begin{pmatrix}1\\-1\end{pmatrix}.
$$
Since
$$
B\begin{pmatrix}1\\-1\end{pmatrix}=\begin{pmatrix}1\\-1\end{pmatrix},
$$
$$
\begin{pmatrix}1\\-1\end{pmatrix}=B^{-1}\begin{pmatrix}1\\-1\end{pmatrix}.
$$
Moreover, since
$$
A\begin{pmatrix}1\\-1\end{pmatrix}=\begin{pmatrix}1\\-1\end{pmatrix},
$$
the product has the correct property and the set $S$ forms a group. I'm leaving the details at this point for you to fill in.
If you want to describe the group, write a general group element as
$$
\begin{pmatrix}a&b\\c&d\end{pmatrix}
$$
and observe that the following three equalities must hold:
\begin{align*}
a-b&=1\\
c-d&=-1\\
ad-bc&=1
\end{align*}
You can write $a=1+b$ and $c=-1+d$. By substituting these into the determinant equation, you can solve for $c$ in terms of $a$. This gives a one-parameter family of solutions for different $a$'s.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Secant and Tangent identity i've been stuck on this question too long
$x = \sec A + \tan A$
show $x + \frac{1}{x} = 2\cdot \sec A$
I've been using $\tan^2 \theta + 1 = \sec^2 \theta$
and $\tan\theta = \frac{\sin \theta}{\cos\theta}$
help would be much appreciated
$x=\sec A +\tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}$
| $$
x=\sec A + \tan A= \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}
$$
so we have:
$$
x+\frac{1}{x}= \frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=\frac{1+\sin^2 A+2\sin A+\cos^2 A}{(\cos A)(1+ \sin A)}=
$$
$$
=\frac{2(1+\sin A)}{(\cos A)(1+ \sin A)}=\frac{2}{\cos A}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Divisors and Decimals
For positive integers $n$, let $f(n)$ return the smallest positive integer $k$ such that $\frac1k$ has exactly $n$ digits after the decimal point. How many positive integer divisors does $f(2010)$ have?
I understand what the question is asking for, but I don't understand how to solve it. Here's what I have so far. The question is asking for a number $n$ that when taking the reciprocal of $n$ has 2010 digits after the decimal point. This means that $\frac1n$ is a terminating decimal. Given this piece of information, I know that $n$ can be expressed in the form $2^a\cdot5^b$. Help is greatly appreciated.
| Suppose $n$ is such that the decimal expansion of ${\large{\frac{1}{n}}}$ has exactly $2010$ digits after the decimal point.
Since the decimal expansion terminates, it follows that $n=2^a5^b$ where $a,b$ are nonnegative integers.
By hypothesis,
$$\frac{1}{n}=\frac{x}{10^{2010}}$$
where $x$ is a positive integer.
Note that $x$ can't be a multiple of $10$, else by removing a common factor of $10$ from the numerator and denominator of
$${\large{\frac{x}{10^{2010}}}}$$
it would follow that the decimal expansion of ${\large{\frac{1}{n}}}$ has less than $2010$ digits.
\begin{align*}
\text{Then}\;\;&\frac{1}{n}=\frac{x}{10^{2010}}\\[4pt]
\iff\;&n=\frac{10^{2010}}{x}\\[4pt]
\end{align*}
hence $x$ divides $10^{2010}$.
Thus, the only possible prime factors of $x$ are $2$ and $5$.
Since $x$ is not a multiple of $10$, it follows that $x$ can't have both $2$ and $5$ as prime factors, hence $x$ must be a power of $2$ or a power of $5$.
To minimize $n$, we want to maximize $x$.
The largest power of $2$ which divides $10^{2010}$ is $2^{2010}$, and the largest power of $5$ which divides $10^{2010}$ is $5^{2010}$.
It follows that the largest possible value of $x$ is $5^{2010}$, hence the smallest possible value of $n$ is $2^{2010}$.
Finally, note that $2^{2010}$ has exactly $2011$ positive integer divisors, namely$2^0,2^1,2^2,...,2^{2010}$.
| {
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"url": "https://math.stackexchange.com/questions/2654911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve $3\sin^2 x - \cos^2 x - 2 =0$
Find all the angles between $0$ and $360^\circ$ that satisfy $$3\sin^2 x - \cos^2 x - 2 =0$$
My attempt -
$3\sin^2 x - (1-\sin^2x) - 2 =0$
$ 3 \sin^2 x + \sin^2 x = 3 $
$4\sin^2 x = 3 $
$ \sin x= \frac{\sqrt{3}}{2} $
I found that $x= 60,120 $
Why is the answer for this $60,120,240,300$ ? How do I find 240 and 300?
| Hint: $y^2 = a \implies y = \color{red}{\pm}\sqrt{a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the determinant of the following $5\times 5$ real matrix:
Let $A\in\mathbb{R^{5\times5}}$ be the matrix: $\left(\begin{array}{l}a&a&a&a&b\\a&a&a&b&a\\a&a&b&a&a\\a&b&a&a&a\\b&a&a&a&a\end{array}\right)$
Find the determinant of $A$.
Hey everyone. What I've done so far: $det\left(\begin{array}{l}a&a&a&a&b\\a&a&a&b&a\\a&a&b&a&a\\a&b&a&a&a\\b&a&a&a&a\end{array}\right)=det\left(\begin{array}{l}b&a&a&a&a\\a&b&a&a&a\\a&a&b&a&a\\a&a&a&b&a\\a&a&a&a&b\end{array}\right)$ (since switching two pairs of rows does not change the determinant)
$= det\left(\begin{array}{l}b-a&0&0&0&a-b\\0&b-a&0&0&a-b\\0&0&b-a&0&a-b\\0&0&0&b-a&a-b\\a&a&a&a&b\end{array}\right)$ (since adding a multiple of one row to another does not change the determinant) for all $1\le i\le 4 \rightarrow R_i-R_5$
Now I am quite stuck. I wanted to obtain a triangular matrix so I can calculate its determinant by the diagonal entries, but I don't know what to do with the last row. I've tried some column operations as well, but have had no success.
Would be happy to get your help, thank you :)
| ...now take out common factor from rows $1$ to $4$ :
$$(b-a)^4\begin{vmatrix}1&0&0&0&-1\\0&1&0&0&-1\\0&0&1&0&-1\\0&0&0&1&-1\\a&a&a&a&b\end{vmatrix}\;\;(*)$$
and now take $\;aR_1\;$ from row $\;R_5\;$ , then $\;aR_2\;$ for $\;R_5\;$, etc.
$$(*)=(b-a)^4\begin{vmatrix}1&0&0&0&-1\\
0&1&0&0&-1\\
0&0&1&0&-1\\
0&0&0&1&-1\\
0&0&0&0&b+4a\end{vmatrix}=(b-a)^4(b+4a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int \sqrt{1+x^2}dx$ I was trying to do this integral
$$\int \sqrt{1+x^2}dx$$
I saw this question and its' use of hyperbolic functions. I did it with binomial differential method since the given integral is in a form of $\int x^m(a+bx^n)^p\,dx$ and I spent a lot of time on it so I would like to see if it can be done this way and where did I go wrong.
$$\int(1+x^2)^\frac{1}{2}dx$$
$m=0$, $n=2$, $p=\frac{1}{2}$
Because $\frac{m+1}{n}+p \in \mathbb Z$ I used substitution $x^{-2}+1=t^2$.
From there I got:
$$-\frac{dx}{x^3}=t\,dt$$
$$x=\frac{1}{\sqrt {t^2-1}}$$
$$t=\frac{\sqrt{1+x^2}}{x}$$
I expanded the original with $x^4$:
$$\int \frac{x^4\sqrt{1+x^2}dx}{x^4}=\int \left(\frac{1}{\sqrt{t^2-1}}\right)^4t(-tdt)=\int\frac{-t^2dt}{(t^2-1)^2}$$
Now I used partial integration:
$u=t$,
$du=dt$,
$dv=\frac{-tdt}{(t^2-1)^2}$,
$v=\frac{1}{2(t^2-1)}$
Then
$$\begin{align}
\int\frac{-t^2dt}{(t^2-1)^2}
&=\frac{t}{2(t^2-1)}-\frac{1}{2}\int \frac{dt}{t^2-1}= \\
&=\frac{t}{2(t^2-1)}-\frac{1}{2}\frac{1}{2}\ln\frac{t-1}{t+1}= \\
&=\frac{\frac{\sqrt{1+x^2}}{x}}{2\left(\left(\frac{\sqrt{1+x^2}}{x}\right)^2-1\right)}-\frac{1}{4}\ln\frac{\frac{\sqrt{1+x^2}}{x}-1}{\frac{\sqrt{1+x^2}}{x}+1}= \\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{4}\ln\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}+x}=\\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{4}\ln\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}+x}\cdot\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}-x}=\\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{4}\ln(\sqrt{1+x^2}-x)^2=\\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{2}\ln(\sqrt{1+x^2}-x)+C
\end{align}$$
The solution is
$$\frac{x\sqrt{1+x^2}}{2}+\frac{1}{2}\ln(x+\sqrt{1+x^2})+C$$
My solution looks very similar, so where did I go wrong?
| What don't you do the substitution $x=\sinh(x)$ and use $$\cosh^2(x)=\frac{1+\cosh(2x)}{2}\ \ ?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2660140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
$\sum_{k=1}^n\frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2n}\right)}$ How to calculate this sum
$$S_n = \sum_{k=1}^n\frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2n}\right)} ?$$
Note :
1)
$$S_n = n+\sum_{k=1}^n{\rm cotan}^2\left(\frac{(2k+1)\pi}{2n}\right)$$
2) $S_n$ seems to be equal to $\frac{n^2}{4}$.
3) I kwow how to calculate
$$\sum_{k=1}^n\frac{1}{\sin^2\left(\frac{k\pi}{2n}\right)}$$
with the roots of $(X+1)^{2n} - (X-1)^{2n}$.
| Take a polynomial $P(t)=t^n$ and consider the function $$F_t(z)=\frac{P(tz)-P(t)}{z-1}=t^n\frac{z^n-1}{z-1}.$$
Lagrange interpolation of $F_t$ with basis points on the roots $z_0,z_1,...,z_{n-1}$ of $z^n+1$ (so that $z_k=e^{\frac{\pi(1+2k)}{n}}$) gives:
$$
F_t(z)=\sum_{k=0}^{n-1} F_t(z_k)\frac{z^n+1}{nz_k^{n-1}(z-z_k)}=-\frac{1}{n}\sum_{k=0}^{n-1}t^n\frac{z_k^n-1}{z_k-1}\frac{z^n+1}{z-z_k}z_k=\\
=-\frac{1}{n}\sum_{k=0}^{n-1}t^n\frac{2}{z_k-1}\frac{z^n+1}{z_k-z}z_k.
$$
On one hand $F_t(1)=tP'(t)=nt^n$ and on the other $F_t(1)=-\frac{1}{n}\sum_{k=0}^{n-1}t^n\frac{4z_k}{(z_k-1)^2}$.
Thus
$$
n^2=-4\sum_{k=0}^{n-1}\frac{z_k}{(z_k-1)^2}=\sum_{k=0}^{n-1}\frac{1}{\sin^2\left(\frac{\pi(1+2k)}{2n}\right)}
$$
since $\frac{e^{i\phi}}{(e^{i\phi}-1)^2}=-\frac{1}{4\sin^2 \phi/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2660233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $3\mid(a^2+b^2)$,then $3\mid a$ and $ 3\mid b$ I am trying to prove this by contradiction. So if $3$ doesn't divide $a$ or $3$ doesn't divide $b$, then the remainder is either $1$ or $2$. I am struggling on what to do next. How do I get the remainder of $a^2$ and $b^2$ for these cases?
Any help is greatly appreciated. Thank you!
| We have the fact that
If $a \equiv k \mod m$ and $b \equiv l \mod m$ then $ab \equiv kl \mod m$.
So, if $a \equiv 1 \mod 3$ then $a^2 \equiv 1^2 \mod 3$. And if $a \equiv 2 \mod3$ then $a^2 \equiv 2^2\mod3$. And same goes for $b$ as well. Using these, we don't have many cases to consider. Assuming $3|(a^2+b^2)$, we have
*
*For $a^2 \equiv 1 \mod 3$ and $b^2 \equiv 1 \mod 3$, we have $a^2+b^2 \equiv 2 \mod 3$ which is a contradiction as required.
*For $a^2 \equiv 0 \mod 3$ and $b^2 \equiv 1 \mod 3$ (Checking the "and" condition), we have $a^2+b^2 \equiv 1 \mod 3$ which is a contradiction as required.
Therefore $3 | a$ and $3 | b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Leading asymptotic behaviour about irregular singular point I am trying to find the leading asymptotic behaviours, using the method of dominant balance, about the irregular singular point $x=0$ for the equation $$x^4\frac{d^2y}{dx^2}+\frac{1}{4}y=0.$$
First let $y=e^{S(x)}, y'= S'e^{S(x)}, y''=[(S')^2+S'']e^{S(x)}$. Our equation then looks like $$x^4[(S')^2+S'']e^{S(x)}+\frac{1}{4}e^{S(x)}=0$$ which implies that $$x^4[(S')^2+S'']=-\frac{1}{4}.$$
We then make the assumption that $S'' << (S')^2$, allowing us to then say $$x^4(S')^2 \sim -\frac{1}{4}.$$
This then implies that $$S' \sim \pm \frac{i}{2x^2} \implies S \sim \mp \frac{i}{2x}.$$
We can see that $S'' \sim \mp \frac{i}{x^3}$ and $(S')^2=-\frac{1}{4x^4}$ which show our assumption to be true as $x \to 0$.
Now we assume that $$S(x) = \frac{i}{2x} + C(x), $$ where $C(x) = o(x^{-1})$.
Sub this into $$ S'' + (S')^2 + \frac{1}{4x^4} = 0 $$ to find $$ \frac{i}{x^3} + C'' -\frac{1}{4x^4}+\frac{i}{x^2}C'+(C')^2
+\frac{1}{4x^4} = 0. $$
After cancelling out and removing small terms we are left with $$ \frac{i}{x^3} + \frac{i}{x^2}C' \sim 0. $$
This then implies that $C \sim -\ln(x)$.
We then have that $$ S(x) = \frac{i}{2x} - \ln(x) + D(x),$$ where $D(x) = o(\ln(x)).$
Sub this into $$ S'' + (S')^2 + \frac{1}{4x^4} = 0 $$ to find $$ \frac{2}{x^2}+\frac{2i}{x^3}-\frac{2D'}{x}-\frac{1}{4x^4}+(D')^2-\frac{iD'}{x^2}+\frac{1}{4x^4}=0.$$
Again we cancel and remove very small terms to find $$\frac{2}{x^2}+\frac{2i}{x^3}-\frac{2D'}{x} \sim 0.$$
This then implies that $D' \sim \frac{1}{x} + \frac{i}{2x}.$
Hence, $D \sim \ln(x) + \frac{i}{2}\ln(x).$
Since this does not tend to $0$ as $x \to 0$ I figured I must continue with the process but the next stage gave me some very questionable results and I realised I didn't really understand what I was looking for or trying to achieve.
Any feedback on current work or clarifications on what I am supposed to be doing would be greatly appreciated.
| First, I think $C \sim \log(x)$ not $-\log(x)$, because in the equation above, we have
$$ \frac{i}{x^3} + C'' -\frac{1}{4x^4} \color{red}{\text{ minus }} \frac{i}{x^2}C'+(C')^2
+\frac{1}{4x^4} = 0 $$
instead of plus.
Next, let us get the equation for $D$ and as you said. The equation is
$$\big(\color{blue}{S''} + C'' + \color{orange}{D''}\big) + \big(\color{red}{S'^2} + C'^2 +\color{orange}{D'^2} + \color{blue}{2S'C'} + 2 S' D' + \color{orange}{2 C' D'} \big)+ \color{red}{\frac{1}{4x^4}} = 0$$
where $S = \frac{i}{2x}$, $C = \log(x)$.
*
*The $\color{red}{\text{ color red terms }}$ will cancel out because this is how we solved $S$, i.e.
$$S'^2 + \frac{1}{4x^2} \sim 0 \Longrightarrow S \sim \pm\frac{i}{2x}$$
and we can easily check
$$S'^2 + \frac{1}{4x^2} = 0 \Longleftarrow S = \pm\frac{i}{2x}.$$
*The $\color{blue}{\text{ color blue terms }}$ will cancel out because this is how we solved for $C$.
*The $\color{orange}{\text{ color orange terms }}$ are small compared to the rest of the terms.
Finally, we get
$$C'' + C'^2 + 2S'D' \sim 0$$
We get
$$D' \sim 0.$$
Hope my calculations are correct...
Furthermore, if as you calcuated $C = -\log(x)$, then you should get
$$D \sim -2ix.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The 4 digit base $6$ number $abcd$ with $a>0$ and $d$ odd is a perfect square.
The 4 digit base $6$ number $abcd$ with $a>0$ and $d$ odd is a perfect square. List all possible values of $c$. (The letters are the digits of the base 6 number.)
I've rewritten $abcd_6$ into $216a + 36b + 6c + d$, and I know that $d$ can be $1,3,5$, but now I'm stuck. I guess I could start plugging in some numbers since $a,b,c,d \le 5$, but I feel like there would be a more efficient way.
| To look at the last two digits in base $6$, we want to look at $4k^2+4k+1$ mod $36$. Since $k^2+k$ mod $9$ repeats with period $9$, $4k^2+4k+1$ mod $36$ also repeats with period $9$. Therefore, the the last two digits of odd squares in base $6$ also repeat with period $9$.
Perhaps more concisely,
$$
\begin{align}
(2(k+9)+1)^2
&=(2k+19)^2\\
&=4k^2+76k+361\\
&=\left(4k^2+4k+1\right)+(72k+360)\\
&=(2k+1)^2+36(2k+10)
\end{align}
$$
Thus, $(2k+1)^2$ repeats mod $36$ with period $9$. That is, the last two digits of $(2k+1)^2$ in base $6$ repeat with period $9$.
Thus, all the possible values of the $6^1$ digit in base $6$ are given in red:
$$
\begin{array}{c|c|r}
k&(2k+1)^2&\text{base $6$}\\
\hline
0&1&\color{#C00}{0}1\\
1&9&\color{#C00}{1}3\\
2&25&\color{#C00}{4}1\\
3&49&1\color{#C00}{2}1\\
4&81&2\color{#C00}{1}3\\
5&121&3\color{#C00}{2}1\\
6&169&4\color{#C00}{4}1\\
7&225&10\color{#C00}{1}3\\
8&289&12\color{#C00}{0}1\\
\end{array}
$$
These are all the possible values for any odd square in base $6$, in particular from $15^2$ to $35^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ How do I solve this equation I found in my textbook:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
This is what I tried:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
$\mapsto \sqrt{(x+4)(x-4)} - (x-4) = \sqrt{(x-1)(x-4)}$
Dividing both sides by $\sqrt{x-4}$
$\mapsto \sqrt{x+4} - \sqrt{x-4} = \sqrt{x-1} $
Squaring both Sides
$\mapsto x - 2 \sqrt{x^2 - 16}= -1$
$\mapsto \frac{x+1}{2} = \sqrt{x^2 - 16}$
Squaring both sides
$ \mapsto \frac{x^2 + 2x + 1}{4} = x^2 - 16$
$\mapsto 3x^2 -2x - 65 = 0$
Solving the quadratic equation
$$x = 5 OR \frac{-13}{3}$$
Checking in the initial equation we can see that $5$ is a valid root. But the second value that is given in the key to the book is $4$. How do I obtain $4$?
| HINT.-Your equation is the same as
$$\sqrt{(x-4)(x+4)}-\sqrt{(x-4)(x-4)}=\sqrt{(x-4)(x-1)}$$ which gives besides of $x-4=0$ the equation
$$\sqrt{x+4}-\sqrt{x-4}=\sqrt{x-1}\iff4(x^2-16)=(x+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
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.