Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ Finding the inverse of $2+\sqrt{5}+2\sqrt{7}$ in the field $\mathbb{Q}(\sqrt{5},\sqrt{7})$.
I know that all the elements of $\mathbb{Q}(\sqrt{5},\sqrt{7})$ are of the form:
$a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35}$, where $a,b,c,d \in \mathbb{Q} $
So I could simply solve: $(2+\sqrt{5}+2\sqrt{7})(a+b\sqrt{5}+c\sqrt{7}+d\sqrt{35})=1$
Which I believe works. Is there a better way of solving this problem?
|
Hint: Consider the product $(2+\sqrt{5}+2\sqrt{7})(2-\sqrt{5}+2\sqrt{7})(2+\sqrt{5}-2\sqrt{7})(2-\sqrt{5}-2\sqrt{7})$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1236171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Find multivariable limit $\frac{x^2y}{x^2+y^3}$ Find multivariable limit of: $$\lim_{ \left( x,y\right) \rightarrow \left(0,0 \right)}\frac{x^2y}{x^2+y^3}$$
How to find that limit? I was trying to do the following, but i am not able to find a proper inequality:
$$| \frac{x^2y}{x^2+y^3} | = |y-\frac{y^4}{x^2+y^3}| \le$$
|
Note that for $x \neq 0$ and $y \neq 0$ we have
$$ \left | \frac{x^2y}{x^2 + y^3} \right| = \left| \frac{y}{1 + \frac{y^3}{x^2}} \right| = \left| \frac{1}{\frac{1}{y} + \frac{y^2}{x^2}} \right| \leq \left| \frac{1}{\frac{1}{y}} \right| = \left| y \right|. $$
If $x = 0$ or $y = 0$ (but not $x = y = 0$), then we also have
$$ \left | \frac{x^2y}{x^2 + y^3} \right| = 0 \leq |y|. $$
Thus, by definition or by squeeze theorem, we have
$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^2y}{x^2 + y^3} = 0. $$
ERRATA: Let $f(x,y) = \frac{x^2y}{x^2 + y^3}$. Note that in the "solution" above, in the first line, we have used $y > 0$ for the inequality to be true. Indeed, if we restrict our function to the upper half plane, then the limit of the function is zero. However, if we consider $f$ as a function that is defined on its natural domain of definition $\mathbb{R}^2 \setminus \{ (x,y) \, | \, x^2 + y^3 \neq 0 \}$, then the function doesn't have a limit as $(x,y) \to (0,0)$. You can see this if you plot the graph of the function. Analytically,
$$ \lim_{n \to \infty} f \left( \frac{1}{n^3}, -\frac{1}{n^2} + \frac{1}{n^4} \right) = \frac{\frac{1}{n^6} \left( -\frac{1}{n^2} + \frac{1}{n^4} \right)}{\frac{1}{n^6} + \left(-\frac{1}{n^2} + \frac{1}{n^4} \right)^3} = \frac{\frac{1}{n^{10}} - \frac{1}{n^6}}{\frac{3}{n^8} - \frac{3}{n^{10}} + \frac{1}{n^{12}}} \\
= \lim_{n \to \infty} \frac{\frac{1}{n^4} - 1}{\frac{3}{n^2} - \frac{3}{n^4} + \frac{1}{n^6}} = -\infty.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1237309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
}
|
Proving $ \frac{1}{c} = \frac{1}{a} + \frac{1}{b}$ in a geometric context
Prove or disprove
$$
\frac{1}{c} = \frac{1}{a} + \frac{1}{b}.
$$
I have no idea where to start, but it must be a simple proof.
Trivia. This fact was used for determination of resistance of two parallel resistors in some circumstances long time ago.
|
I'm not sure what the protocol is for adding a separate answer when it uses a different approach. If there is an issue with doing this, I can move this to my existing entry.
Here is a method using analytic geometry. I have changed the labeling of two points, since I wish to place the origin of the coordinate system at a different point from $ \ O \ $ in the original diagram. For convenience, I will place the apex of $ \ \Delta ACB \ $ at $ \ (0, c) \ $ , with line $ \ OC \ $ on the $ \ y-$ axis, and lines $ \ AC \ $ and $ \ BC \ $ making equal angles to the $ \ y-$ axis. The equations for the lines along which the sides of the triangle lie are $ \ AC : \ y \ = \ c \ + \ mx \ , \ BC : \ y \ = \ c \ - \ mx \ , \ AB : \ y \ = \ kx \ , $ with $ \ k \ < \ m \ $ . (This condition proves to be important.)
We find the locations of the intersection points $ \ A \ $ and $ \ B \ $ from
$$ k \ x_A \ = \ c \ + \ m \ x_A \ \ \Rightarrow \ \ x_A \ = \ \frac{c}{k - m} \ \ , \ \ y_A \ = \ c \ + \ \frac{mc}{k - m} \ \ , $$
$$ k \ x_B \ = \ c \ - \ m \ x_B \ \ \Rightarrow \ \ x_B \ = \ \frac{c}{k + m} \ \ , \ \ y_A \ = \ c \ - \ \frac{mc}{k + m} \ \ . $$
[We will see shortly that it is preferable to express the $ \ y-$ coordinates in this manner.]
The squares of the lengths of sides $ \ AC \ $ and $ \ BC \ $ are then obtained
$$ a^2 \ = \ \left( \ \frac{c}{k - m} \ - \ 0 \ \right)^2 \ + \ \left( \ c \ - \ \left[ \ c \ + \frac{mc}{k - m} \ \right] \ \right)^2 \ = \ \ c^2 \ \left[ \ \frac{1 + m^2}{(k - m)^2} \ \right] \ \ , $$
$$ b^2 \ = \ \left( \ \frac{c}{k + m} \ - \ 0 \ \right)^2 \ + \ \left( \ c \ - \ \left[ \ c \ - \frac{mc}{k + m} \ \right] \ \right)^2 \ = \ \ c^2 \ \left[ \ \frac{1 + m^2}{(k + m)^2} \ \right] \ \ . $$
We can write directly that $ \ \frac{c}{b} \ = \ \frac{k + m}{\sqrt{1 + m^2}} \ $ . However, since we are constructing ratios of positive lengths, we must write $ \ \frac{c}{a} \ = \ \frac{\sqrt{(k - m)^2}}{\sqrt{1 + m^2}} \ = \ \frac{|k - m|}{\sqrt{1 + m^2}} \ = \ \frac{m - k}{\sqrt{1 + m^2}} \ $ . (I found this to be a little trap for the unwary... I'll also note at this point that forming these ratios now links this approach to those using trigonometry in some fashion.)
Summing the two ratios produces
$$ \frac{c}{a} \ + \ \frac{c}{b} \ = \ \frac{m - k}{\sqrt{1 + m^2}} \ + \ \frac{k + m}{\sqrt{1 + m^2}} \ = \ \frac{2m }{\sqrt{1 + m^2}} \ \ . $$
For the special case of $ \ m(\angle ACO) \ = \ m(\angle BCO) \ = \ \frac{\pi}{3} \ $ , we have the slope $ \ m \ = \ \frac{1}{\sqrt{3}} \ $ . So our general relation reduces to
$$ \frac{c}{a} \ + \ \frac{c}{b} \ = \ \frac{2 \ \cdot \ \frac{1}{\sqrt{3}} }{\sqrt{1 + \frac{1}{3}}} \ = \ \frac{\frac{2}{\sqrt{3}}}{\sqrt{\frac{4}{3}}} \ = \ 1 \ \ . $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1237390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
}
|
Compute $\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$ Given
$$\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$$
I couldn't evaluate this integral. My only idea here was evaluating this as integration by parts.
\begin{align}
\int\frac{x \log(x)}{(1+x^2)^2}dx & = \frac{1}{2} \int\frac{ \log(x)}{(x^2+1)^2}d(x^2+1)\\
& = \frac{1}{2} \frac{ \log(x)}{x^2+1} - \frac{1}{2}\int \left[(x^2+1)\frac{\frac{(x^2+1)^2}{x} - 4x(x^2+1) \log(x)}{(x^2+1)^4} \right ]dx\\
& = \frac{ \log(x)}{2(x^2+1)}-\frac{1}{2}\int \left [ \frac{x^2+1-4x^2 \log(x)}{x(x^2+1)^2} \right ] dx\\
& = \frac{ \log(x)}{2(x^2+1)} - \frac{1}{2}\int \frac{dx}{x(x^2+1)} + 2\int\frac{x \log(x)}{(x^2+1)^2}dx
\end{align}
Or this method doesn't work here, or I have done a mistake somewhere. However, I have also tried doing $u = x^2+1$ substitution, but this also didnt gave me any good results.Thank you.
|
Here is a general technique. Let's consider the integral
$$ I = \int_{0}^{\infty} \frac{x^{s-1}}{(1+x^2)^2} dx = -\frac{1}{4}\,{\frac { \left( s-2 \right) \pi }{\sin \left( \pi \,s/2
\right) }},$$
which is the Mellin transform of $\frac{1}{(1+x^2)^2}$. Your integral can be evaluate using $I$ as
$$ \int_{0}^{\infty} \frac{x\ln x }{(1+x^2)^2} dx = \lim_{s\to 2}\frac{dI}{ds} = 0. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1239241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
computing an integration with a floor function I am trying to compute
$$\int_0^1 \left(\frac{1}{x} - \biggl\lfloor \frac{1}{x}\biggr\rfloor\right) dx$$
with no success.
Any hints?
|
$$\int_0^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right) dx=\lim_{k\to +\infty}\int_{\frac{1}{k}}^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right) dx$$
Now to compute the other integral use the decomposition:
$$\int_{\frac{1}{k}}^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right) dx=\sum_{i=1}^{k-1} \int_{\frac{1}{i+1}}^{\frac{1}{i}}\left(\frac{1}{x} -i\right)dx=\sum_{i=1}^{k-1}\left[\ln(x)-ix\right]_{\frac{1}{i+1}}^{\frac{1}{i}}$$
and you can compute this sum and determine the limit, After computation you obtain:
$$\sum_{i=1}^{k-1}\left[\ln(x)-ix\right]_{\frac{1}{i+1}}^{\frac{1}{i}}=\ln(k)-\sum_{i=2}^{k}\frac{1}{i}$$
and this gives you:
$$\int_0^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right) dx=1-\gamma$$
with $\gamma$ denotes Euler–Mascheroni constant
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1239735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Primitive of the function $(\sin x)/x$ I know that for some functions, for instance $f(x) = e^{-x^2}$, there does not exist a primitive.
Does there is a primitive for the function $f(x) = \frac{\operatorname{sin}(x)}{x}$?
|
No It is not In terms of elementary functions, But we can write in Infinite Series form.
Using $$\displaystyle \sin x= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+..........$$
So $$\displaystyle \frac{\sin x}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{4!}-\frac{x^6}{6!}.........$$
So $$\displaystyle \int \frac{\sin x}{x}dx = \int \left\{1-\frac{x^2}{3!}+\frac{x^4}{4!}-\frac{x^6}{6!}.........\right\}dx$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1241021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding roots of $z^3 = 8$ I am having trouble finding the cubed roots of $8$ as a complex number.
$$z^3 = 8+0i$$
$$z^3 = r^3 e^{3\theta i}=8e^{2i\pi k},\quad k\in \Bbb Z$$
$$\implies r=2,3\theta = 2\pi k\implies \theta = \frac{2\pi k}{3}$$
$$z=2e^{\frac{2i\pi k}{3}}$$
I would think I have then $$z=2,2e^{\frac{2i\pi}{3}},2e^{\frac{4i\pi}{3}}$$
But the answer says: $z=2,-1+\sqrt{3}i, -1-\sqrt{3}i$
What's going on?
|
$z^3-2^3=(z-2)(z^2+2z+4)=0$
So first term has solution $z=2$.
Now-
$z^2+2z+4=0$
$\frac{-2 \pm \sqrt{4-16}}{2}=z$
So, $z=-\sqrt{3}i-1 \ and\ z=\ \sqrt{3}i-1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1241720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Algebraic Manipulation What is the best method to get the LHS equal to RHS?
$\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3}$
Thank you.
|
Firstly, take $(n+1)(n+2)$ out as a factor:
$((n+1)(n+2))[\frac n3 +1] = \frac{n+3}{3} (n+1)(n+2)$
Then divide through by $(n+1)(n+2)$:
$(\frac n3 +1) = \frac{n+3}{3}$
Multiply both sides by $3$:
$n+3 = n+3$
Therefore LHS = RHS
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1243683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Prove equality of two numbers written in complex polar form. Show that these two numbers are equal:
$$
z_1=\frac{e^{\tfrac{2\pi i}{9}}-e^{\tfrac{5\pi i}{9}}}{1-e^{\tfrac{7\pi i}{9}}}
$$
and
$$z_2=\frac{e^{\tfrac{\pi i}{9}}-e^{\tfrac{3\pi i}{9}}}{1-e^{\tfrac{4\pi i}{9}}}=\frac{1}{e^{\tfrac{\pi i}{9}}+e^{\tfrac{-\pi i}{9}}}.
$$
In case it helps, I do know that they are both real.
Thanks in advance for any suggestions!
|
Let $w = \exp(i\pi/9)$. Then
$$
z_1 = \frac{w^2-w^5}{1-w^7} = \frac{w^2(1-w^3)}{1-w^7}
$$
and
$$
z_2 = \frac{w-w^3}{1-w^4} = \frac{w(1-w^2)}{(1-w^2)(1+w^2)} = \frac{w}{1+w^2}.
$$
Hence
$$
z_1 - z_2 = \frac{w^2(1-w^3)(1+w^2)-w(1-w^7)}{(1-w^7)(1+w^2)} =
\frac{w(1-w)(w^6-w^3+1)}{(1-w^7)(1+w^2)}.
$$
It remains to show that $w^6 - w^3 + 1 = 0$, but
$$
w^6 - w^3 + 1 = w^{-3}(w^9 - w^6 + w^3) = w^{-3}(1-w^6+w^3)
$$
which establishes that. There are probably shorter ways to see this.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1245471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Evaluate $\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$ The task is to evaluate $$\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$$
My best approach has been substitution of $u=x+\frac{1}{2}$, and from there onto (some terrible) trig sub - finally arriving at a messy answer.
Is there a slicker way of doing this?
|
We have
\begin{align}
\dfrac{x^2}{\sqrt{1+x+x^2}} & = \dfrac{x^2+x+1}{\sqrt{x^2+x+1}} - \dfrac{x+1}{\sqrt{x^2+x+1}}\\
& = \sqrt{x^2+x+1} - \dfrac12 \dfrac{2x+1}{\sqrt{x^2+x+1}} - \dfrac12 \dfrac1{\sqrt{x^2+x+1}}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1245761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Use de Moivre's theorem to obtain an expression for $\sin^6x$ as a sum of terms in the form $\cos ax$ I'm not exactly sure if I'm on the right lines but I've started with a binomial expansion:
$(\cos x+i\sin x)^6=\cos 6x +i \sin 6x= \cos^6 x + i(6\cos^5x \sin x)-15\cos^4x \sin^2x-i(20\cos^3x \sin^3 x)+15 \cos^2x \sin^4x+i(6\cos x \sin^5 x)-\sin^6x$
If I rearranged this for $\sin^6x$ I'd get $\cos^kx$ terms, when the question asks for $\cos ax$
What should I do?
|
Let $z = e^{ix}$. $\sin x = \frac1{2i} (z-z^{-1})$ and
$$\sin^6 x = -\frac1{64} (z - z^{-1})^6 = -\frac1{64} (z^6 - 6z^4 + 15z^2 - 20 + 15 z^{-2} - 6 z^{-4} + z^{-6})$$
Now remember that $\cos ax = \frac12 (z^a + z^{-a})$ to get
$$\sin^6 x = -\frac1{32} \cos 6x + \frac3{16} \cos 4x - \frac{15}{32} \cos 2x + \frac5{16}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1247104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$\bigg(\frac{-2}{p}\bigg)= \begin{cases} 1 & \text{ if $p\equiv 1$ or $3 \mod 8$} \\ -1 & \text{ if $p\equiv 5$ or $7 \mod 8$} \\\end{cases}$ Show that
$$\bigg(\frac{-2}{p}\bigg)=
\begin{cases}
1 & \text{ if $p\equiv 1$ or $3 \mod 8$} \\
-1 & \text{ if $p\equiv 5$ or $7 \mod 8$} \\\end{cases}$$
$\textbf{Proof:}$
\begin{equation*}
\begin{aligned}
\bigg(\frac{-2}{p}\bigg) &=\bigg(\frac{-1}{p}\bigg)\bigg(\frac{2}{p}\bigg)
=(-1)^\frac{p-1}{2}(-1)^\frac{p^2-1}{8}
=(-1)^\frac{4(p-1)+p^2-1}{8} \\
& =(-1)^\frac{p^2+4p-5}{8}
=(-1)^\frac{(p-1)(p+5)}{8}
\end{aligned}
\end{equation*}
$\textbf{Case $1$:}$ \begin{equation*}
\begin{aligned}
(-1)^\frac{(p-1)(p+5)}{8}=1 & \iff \frac{(p-1)(p+5)}{8} \text{ is even} \\
& \iff \frac{(p-1)(p+5)}{8}=2k \text{ for $k\in \mathbb{Z}$}
\iff (p-1)(p+5)=16k \\
& \iff p-1=16k \text{ or } p+5=16k
\iff p=16k+1 \text{ or } p=16k-5 \\
& \iff p\equiv 1 \mod 8 \text{ or } p\equiv -5\equiv 3 mod 8
\end{aligned}
\end{equation*}
$\textbf{Case $2$:}$ \begin{equation*}
\begin{aligned}
(-1)^\frac{(p-1)(p+5)}{8}=-1 & \iff \frac{(p-1)(p+5)}{8} \text{ is odd} \\
& \iff \frac{(p-1)(p+5)}{8}=2k+1 \text{ for $k\in \mathbb{Z}$}
\iff (p-1)(p+5)=16k+8 \\
& \iff p-1=16k+8 \text{ or } p+5=16k+8
\iff p=16k+9 \text{ or } p=16k+3 \\
& \iff p\equiv 1 \mod 8 \text{ or } p\equiv 3 \mod 8
\end{aligned}
\end{equation*}
Am I right so far? If so how can I finish case 1? Note I am using Jacobi symbols.
|
1) Recall that if $p$ is an odd prime, then $2$ is a QR of $p$ if $p\equiv 1$ or $7 \pmod{8}$, and $2$ is an NR of $p$ if $p\equiv 3$ or $5\pmod{8}$.
2) Also, $-1$ is a QR of $p$ if $p\equiv 1\pmod{4}$ and is an NR of $p$ if $p\equiv 3\pmod{4}$. This can be restated as $-1$ is a QR of $p$ if $p\equiv 1$ or $5 \pmod{8}$, and is an NR of $p$ if $p\equiv 3$ or $7\pmod{8}$.
Looking at 1) and 2), we see that $2$ and $-1$ are both QR of $p$ if $p\equiv 1\pmod{8}$, and they are both NR of $p$ if $q\equiv 3\pmod{8}$.
Thus $-2$ is a QR of $p$ if and only if $p\equiv 1$ or $3\pmod{8}$.
Remark: In the OP as revised, you have reduced the problem to finding out when $(p-1)(p+5)$ is divisible by $16$.
If $p=8k+1$, then $p-1$ is divisible by $8$, and $p+5$ is even, so $(p-1)(p+5)$ is divisible by $16$.
If $p=8k+3$, then $p+5$ is divisible by $8$, and $p-1$ is even, so $(p-1)(p+5)$ is divisible by $16$.
If $p=8k+5$, then $(p-1)(p+5)=(8k+4)(8k+10)$. The highest power of $2$ that divides $8k+4$ is $4$, and the highest power of $2$ that divides $8k+10$ is $2$, so the highest power of $2$ that divides the product is $(4)(2)=8$. It follows that the exponent $\frac{(p-1)(p+5)}{8}$ is odd.
The case $p=8k+7$ goes along lines similar to the case $8k+5$.
Note that your current arguments are incorrect. It is not true that $(p-1)(p+5)$ is divisible by $16$ if and only if $p-1$ is divisible by $16$ or $p+5$ is divisible by $16$. For each of $p-1$ and $p+5$ is even, so each makes a contribution to divisibility by powers of $2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1247595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x)$
If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x).$
My Solution:: Let $$\displaystyle y = \sin^4 x+\cos^2 x \leq \sin^2 x+\cos^2 x=1$$
And for Minimum, We take $$\displaystyle y = \sin^4 x+\cos^2 x=(1-\cos^2 x)^2+\cos^2 x$$
So $$\displaystyle y=\cos^4 x-\cos^2 x+1 = \left(\cos^2 x-\frac{1}{2}\right)^2+\frac{3}{4}\geq \frac{3}{4}$$
So We get $\displaystyle y=\sin^4 x+\cos^2 x\in \left[\frac{3}{4}\;,1\right]$
My question is How can we find Min. of $f(x)$ other then that method,
Something Like Using Inequality.,Thanks
|
We can also solve this problem as follows.
Let $f(x)=\sin^4x+\cos^2x$. This simplifies $f(x)=\frac{1}{8}(7+\cos 4x).$ Then $f_{min}=\frac{3}{4}$ which occurs when $\cos 4x=-1$ and $f_{max}=1$ which occurs when $\cos 4x=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1252666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$ Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$
I am stuck at understanding why the constraint $xy\gt 1$. Here is my work so far
let $\arctan x =a\implies x=\tan a$
let $\arctan y =b\implies y=\tan b$
therefore $\frac{x+y}{1-xy}=\frac{\tan a+\tan b}{1-\tan a\tan b}=\tan(a+b)$
$\implies a+b=\arctan \frac{\tan a+\tan b}{1-\tan a\tan b} $
$\implies \arctan x+\arctan y = \arctan\frac{x+y}{1-xy}$
I know $\pi$ is the period of $\tan x$ so $xy\gt 1$ constraint must have something to do with this. But I am not able to figure out how exactly these period and $xy\gt 1$ are related. Any help is appreciated. Thanks!
|
Since $\arctan x >- \frac{\pi}{2}$ and the function is increasing, $xy<1$ would yield $$\arctan x + \arctan y < \arctan x + \arctan \frac{1}{x}=\frac{\pi}{2}< \pi + \arctan \frac{x+y}{1-xy}$$ if none of $x$ and $y$ is negative, or WLOG only $y$. If they are both negative we would have a negative LHS, while the RHS remains larger than $\frac{\pi}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1255968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
}
|
Why is $n^2+4$ never divisible by $3$? Can somebody please explain why $n^2+4$ is never divisible by $3$? I know there is an example with $n^2+1$, however a $4$ can be broken down to $3+1$, and factor out a three, which would be divisible by $4$.
|
Well, You have two cases
1st case $n$ is odd then $n=2k+1$ for some integer $k$
Then $n^2 + 4 = (2k+1)^2 + 4 = 4k^2 + 4k + 5 = 4k(k+1) +5$
Here we have either $k$ or $k+1$ is even and the other is odd or vice versa and we know that even $\times$ odd = even and so $4k(k+1) + 5 = 4(2m)+5$ for some integer $m$ and so have $8m +5$ Now it's obvious that $3 \nmid 8m+5$ for any integer $m$
2nd case $n$ is even then $n=2k$ for some integer $k$ and so $n^2 +4 = (2k)^2 + 4 = 4k^2 +4 = 4(k^2 +1)$
You should also deduce that 3 doesn't divide $4(k^2 + 1)$ and you are basically done !
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1256915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
}
|
Summation rule index In the simplification shown below, The index are changed but when i follow the index change rule, i cant find the same final answer.
$$\sum^{n-1}_{i=0} \frac{n}{n-i}=n\sum^{n}_{i=1} \frac{1}{i}$$
Thanks
|
Write it as
$$\sum^{n-1}_{i=0} \frac{n}{n-i}= \frac{n}{n} + \frac{n}{n-1} + \dots + \frac{n}{1}$$
Read from the end to the beginning.
$$\frac{n}{1} + \frac{n}{2} + \dots + \frac{n}{n} = n \Big(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n}\Big) = n\sum^{n}_{i=1} \frac{1}{i}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1258301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Indefinite Integral of a function $$\int \left(\frac15 x^3 - 2x + \frac3x + e^x \right ) \mathrm dx$$
I came up with
$$F=x^4-x^2+\frac{3x}{\frac12 x^2}+e^x$$
but that was wrong.
|
\begin{align}
\int \frac{1}{5}(x^3)-2x+\frac{3}{x}+e^x dx & = \int \frac{1}{5}x^3 dx- \int 2x dx+ \int \frac{3}{x} dx+ \int e^x dx \\ &= \frac{1}{20}x^4 - x^2 + 3 \ln(x) + e^x + C
\end{align}
Important rules:
•Derivative of $x^n$ is $\frac{1}{n+1}x^{n+1}$ for $n \neq -1$
•Derivative of $x^{-1}$ is $\ln(x)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1261067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2, $or $6$
Prove that if $n$ is a product of two consecutive integers, its units digits must be $0,2,$ or $6$.
I'm having a hard time with the $0,2,6$ and part but here is what I have so far.
Since $n$ is the product of two consecutive integers then one must be odd and the other must be even.
Let $m=2k+1$ represent the odd integer where $k \in \mathbb{Z}$, and let $q=2k$ represent the even integer where $k \in \mathbb{Z}$. Then
$$n=m \times q=(2k+1)(2k)=(2k)^2+2k$$
This tells me that $n$ is even. So how can I show that units digit must be $0,2,$ or $6$?
|
When you multiply two numbers, the units place of the product is determined only by the units' places of the two numbers. So it is sufficient to check the cases one by one.
$$
0 \times 1 \text{ gives } 0 \\
1 \times 2 \text{ gives } 2\\
2 \times 3 \text{ gives } 6\\
3 \times 4 \text{ gives } 2\\
4 \times 5 \text{ gives } 0\\
5 \times 6 \text{ gives } 0\\
6 \times 7 \text{ gives } 2\\
7 \times 8 \text{ gives } 6\\
8 \times 9 \text{ gives } 2\\
9 \times 0 \text{ gives } 0\\
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1264164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
}
|
How to prove that $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$? Let $a,b,c>0: (a+b)(b+c)(c+a)=ab+bc+ca$. How to prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$$
|
By Cauchy-Schwarz $\left(\sum\limits_{cyc}\frac{a}{\sqrt{a+b}}\right)^2\leq\sum\limits_{cyc}\frac{a}{(a+b)(a+b+2c)}\sum\limits_{cyc}a(a+b+2c)$.
Hence, it remains to prove that $\frac{27(a+b)(a+c)(b+c)}{16(ab+ac+bc)\sum\limits_{cyc}(a^2+3ab)}\geq\sum\limits_{cyc}\frac{a}{(a+b)(a+b+2c)}$, which after full expanding gives
$\sum\limits_{cyc}(6a^7b^2+22a^7c^2+9a^6b^3+89a^6c^3+29a^5b^4+109a^5c^4+28a^7bc+59a^6b^2c+155a^6c^2b-42a^5b^3c+246a^5c^3b+46a^4b^4c+66a^5b^2c^2-359a^4b^3c^2-231a^4c^3b^2-232a^3b^3c^3)\geq0$,
which is true because by AM-GM easy to show that
$288\sum\limits_{cyc}a^5c^3b\geq288\sum\limits_{cyc}a^4b^3c^2$,
$16\sum\limits_{cyc}a^7c^2\geq16\sum\limits_{cyc}a^4c^3b^2$,
$80\sum\limits_{cyc}a^6c^3\geq80\sum\limits_{cyc}a^4c^3b^2$,
$96\sum\limits_{cyc}a^6c^2b\geq96\sum\limits_{cyc}a^4c^3b^2$,
$80\sum\limits_{cyc}a^5c^4\geq80\sum\limits_{cyc}a^3b^3c^3$ and
$32\sum\limits_{cyc}a^4c^3b^2\geq32\sum\limits_{cyc}a^3b^3c^3$.
Thus, it remains to prove that
$\sum\limits_{sym}(6a^7b^2+9a^6b^3+29a^5b^4+14a^7bc+59a^6b^2c-42a^5b^3c+23a^4b^4c+33a^5b^2c^2-71a^4b^3c^2-60a^3b^3c^3)\geq0$,
which is true by Muirhead.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1264369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
}
|
Computing $\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3}dx$ I wish to compute
$$\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3}dx, \quad a>0$$
but have no contour to work with. Does anyone have ideas on how to compute this integral?
|
This integral can be evaluated with calculus only:
\begin{align*}
\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3} \, dx
&= \frac{1}{a^{3/2}} \int_{0}^{\infty} \frac{x^2 + 1}{x^6 + 1} \, dx \tag{1} \\
&= \frac{1}{a^{3/2}} \int_{0}^{\infty} \frac{1}{(x - \frac{1}{x})^2 + 1} \, \frac{dx}{x^2} \\
&= \frac{1}{2a^{3/2}} \int_{0}^{\infty} \frac{1+\frac{1}{x^2}}{(x - \frac{1}{x})^2 + 1} \, dx \tag{2} \\
&= \frac{1}{2a^{3/2}} \int_{-\infty}^{\infty} \frac{1}{u^2 + 1} \, du \quad (u = x-1/x) \\
&= \frac{\pi}{2a^{3/2}}.
\end{align*}
Explanation.
*
*For (1), we utilized the substitution $x \mapsto \sqrt{a} x$.
*For (2). the substitution $x \mapsto 1/x$ gives
$$ I := \int_{0}^{\infty} \frac{1}{(x - \frac{1}{x})^2 + 1} \, \frac{dx}{x^2} = \int_{0}^{\infty} \frac{1}{(x - \frac{1}{x})^2 + 1} \, dx. $$
Thus it follows that
$$ I = \frac{1}{2}(I + I) = \frac{1}{2} \int_{0}^{\infty} \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 1} \, dx. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1265905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Prove the sum of any $n$ consecutive numbers is divisible by $n$ (when $n$ is odd).
Let $n \in \mathbb N$ be odd. Prove that the sum of any $n$ consecutive numbers is divisible by $n$.
I started out with $s = x + (x + 1) + (x + 2) + … + (x + n) = kx + n.$ What I am interested in is if that's a right way to sum $x + (x + 1) + (x + 2) + … + (x + n)$ meaning does it equal $kx + n?$
|
$(x+1)+(x+2)+\cdots+(x+n)=nx+(1+2+\cdots+n)$
$=nx+\frac{n(n+1)}{2}=n\left(x+\frac{n+1}{2}\right)$. Since $n$ is odd, $x+\frac{n+1}{2}$ is an integer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1268499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Finding the number of real roots of an unusual(!) equation How many real roots does the below equation have?
\begin{equation*}
\frac{x^{2000}}{2001}+2\sqrt{3}x^2-2\sqrt{5}x+\sqrt{3}=0
\end{equation*}
A) 0 B) 11 C) 12 D) 1 E) None of these
I could not come up with anything.
(Turkish Math Olympiads 2001)
|
Consider the discriminant of $f(X) = 2\sqrt{3}x^2-2\sqrt{5}x+\sqrt{3}=0$:
$$(-2\sqrt{5})^2 - 4(2\sqrt{3})\sqrt{3} = 20 - 24 < 0.$$
Therefore $f(x)$ has no real roots.
But $f(0) = \sqrt{3} > 0$, so $f(x) > 0$ everywhere.
Now combine this with $$\frac{x^{2000}}{2001} \geq 0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1270107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Dice outcomes probability
Two dice are rolled. What is the probability that the sum of the
numbers on the dice is at least 10
Let $Z$ denote the set of successful outcomes:
$Z=\{(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)\}\\
\text{Sample space: }
S=6^2$
So the answer gives the probability as, $P=\frac{6}{36}$.
My question is why are the pairs $(5,5),(6,6)$ only included once? The way I see it for two dice $D_1,D_2$ is it may appear as $D_1: 5,D_2: 5$ or $D_2: 5,D_1: 5$, so why don't we count for this as two outcomes?
|
Your sample space $S=\{1,2,3,4,5,6\}^2$ has exactly 36 unique outcomes, with each having probability $\frac{1}{36}$. $(4,6)$ and $(6,4)$ are separate outcomes, whereas $(5,5)$ and $(5,5)$ are the same.
Intuitively, to get an outcome involving one $4$ and one $6$, I could roll either a $4$ or a $6$ for the first die, which has a probability of $\frac{2}{6}$ of success. If I was successful with the first die, I have a $\frac{1}{6}$ chance of rolling the number I'm missing with the second die. The total probability of success here is $\frac{2}{6}\cdot\frac{1}{6}=\frac{2}{36}$.
On the other hand, if I want an outcome with two $5$'s, I need to roll a $5$ on the first outcome, with probability $\frac{1}{6}$ of success. Given I succeed, I need to roll a $5$ again with the second die, with probability $\frac{1}{6}$. The total probability of success here is $\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1270218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Determining all the positive integers $n$ such that $n^4+n^3+n^2+n+1$ is a perfect square. I successfully thought of bounding our expression examining consecutive squares that attain values close to it, and this led to the solution I'll post as an answer, which was the one reported. However, before that, I had briefly tried manipulating $$n^4+n^3+n^2+n+1=m^2.$$Since $n=1$ is not a solution, I rewrote this as $$\frac{n^5-1}{n-1}=m^2 \\ n^5-1=nm^2-m^2 \\ m^2-1=n(m^2-n^4) \\ (m-1)(m+1)=n(m-n^2)(m+n^2),$$ but in vain. Does manipulation lead somewhere? Is there a different approach from both of mine?
|
There are only two additional non-positives that also give perfect squares.
Case 1: $\boldsymbol{n}$ odd
$$
\begin{align}
\left(n^2+\frac{n-1}2\right)^2
&=n^4+n^2(n-1)+\frac{n^2-2n+1}4\\
&=n^4+n^3-\frac34n^2-\frac12n+\frac14
\end{align}
$$
smaller than $n^4+n^3+n^2+n+1$ if $\frac74n^2+\frac32n+\frac34\gt0$
$$
\begin{align}
\left(n^2+\frac{n+1}2\right)^2
&=n^4+n^2(n+1)+\frac{n^2+2n+1}4\\
&=n^4+n^3+\frac54n^2+\frac12n+\frac14
\end{align}
$$
bigger than $n^4+n^3+n^2+n+1$ if $\frac14n^2-\frac12n-\frac34\gt0$.
Only leaves $n=-1,1,3$. Only $-1,3$ give squares.
Case 2: $\boldsymbol{n}$ even
$$
\begin{align}
\left(n^2+\frac{n}2\right)^2
&=n^4+n^2n+\frac{n^2}4\\
&=n^4+n^3+\frac14n^2
\end{align}
$$
smaller than $n^4+n^3+n^2+n+1$ if $\frac34n^2+n+1\gt0$
$$
\begin{align}
\left(n^2+\frac{n+2}2\right)^2
&=n^4+n^2(n+2)+\frac{(n+2)^2}4\\
&=n^4+n^3+\frac94n^2+n+1
\end{align}
$$
bigger than $n^4+n^3+n^2+n+1$ if $\frac54n^2\gt0$.
Only leaves $n=0$, which gives a square.
Thus, only $n\in\{-1,0,3\}$ give squares.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1270599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
}
|
For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a).$$
Clearly, $$\left\{\det(A)\neq0\left|\begin{matrix}c\neq b\\a\neq c\\b\neq a\\a,b,c\neq 0\end{matrix}\right.\right\}\\$$
Is it sufficient to say that the matrix is invertible provided that all 4 constraints are met? Would Cramer's rule yield more explicit results for $a,b,c$ such that $\det(A)\neq0$?
|
Yes, this is enough. Basically, if you are trying to solve:
$$\begin{pmatrix}d_1&d_2&d_3\end{pmatrix}A = \begin{pmatrix}y_1&y_2&y_3\end{pmatrix}$$ you are looking for a polynomial $f(x)=d_1+d_2x+d_3x^2$ such that $f(a)=y_1$, $f(d_2)=y_2$, and $f(d_3)=y_3$.
This we can find by simple interpolation of polynomials - given any $n$ distinct $a_i$ and $n$ distinct $b_i$ we can find a polynomial $p(x)$ of degree at most $n$ such that $p(a_i)=b_i$ for all $i$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1270699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
}
|
How do you add two series together How do you add the series
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{2^{n}}{(z-3)^{n+1}} + \sum_{n=0}^{\infty}\frac{(z-3)^{n}}{4^{n+1}}\right)$$
?
is this right?
$$\begin{aligned}
&\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2^{n}}{(z-3)^{n+1}} + \frac{(z-3)^{n}}{4^{n+1}}\right)\\
=\> & \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{n} (z-3)^{n+1}}{(z-3)^{n+1} 4^{n+1}} \right)\\
=\>& \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{n+1+n}}{(z-3)^{n+1} 4^{n+1}}\right)\\
=\>& \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{2n+1}}{(z-3)^{n+1} 4^{n+1}}\right)\end{aligned}$$
|
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{2^{n}}{(z-3)^{n+1}} + \sum_{n=0}^{\infty}\frac{(z-3)^{n}}{4^{n+1}}\right)$$
$$=\frac{1}{2(z-3)}\sum_{n=0}^{\infty} \left( \frac{2}{z-3}\right) ^n +
\frac18 \sum_{n=0}^{\infty} \left( \frac{z-3}{4}\right) ^n $$
evaluate the two sums using the formula for geometric series and add the results.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1270808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Points of intersection of $\sin x$ and $\cos x$ I'm trying to find the points of intersection of $\sin x$ and $\cos x$ between $0$ and $2\pi$. I've tried but I keep getting 4 solutions... Would someone please be able to take me through the process?
I squared $\sin x$ and $\cos x$ and then got $\sin x = \frac{\pm 1}{\sqrt{2}}$
which gave the solutions $\pi\over 4$, $3\pi \over 4$, $5\pi \over 4$ and $7\pi\over 4$.
|
$\sin x = \cos x \implies \sin^2x = \cos^2 x $ together with the proviso that $\sin x$ and $\cos x$ must share the same sign.
So the only two solutions are
$\sin x=\frac{1}{\sqrt 2}$, $\cos x=\frac{1}{\sqrt 2}$ giving $x=\frac{\pi}{4}$
and
$\sin x=\frac{-1}{\sqrt 2}$, $\cos x=\frac{-1}{\sqrt 2}$ giving $x=\frac{5\pi}{4}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1272404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
The value of ${\sum_{k=0}^{20}}(-1)^k\binom{30}{k}\binom{30}{k+10}$ $\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\c}[1]{{}^{30}{\mathbb C}_{#1}}
\newcommand{\r}[1]{\frac1{x^{#1}}}$
The value of $$\sum_{k=0}^{20}(-1)^k\binom{30}{k}\binom{30}{k+10}$$
It is also the coefficient of $x^{10}$ in:
$$\b{\c0\r0-\c1\r1+\c2\r2-....\c{20}\r{20}}\b{\c{10}x^{10}+\c{11}x^{11}+...\c{30}x^{30}}$$
Adding some terms maybe won't harm:
$$\b{\c0\r0-\c1\r1+\c2\r2-....\c{30}\r{30}}\b{\c{0}x^{0}+\c{1}x^{1}+...\c{30}x^{30}}$$
Contracting:
$$\b{1-\r1}^{30}\b{1+x}^{30}=\b{x-\r1}^{30}=\sum_{n=0}^{30}(-1)^{n}\c nx^{2n-30}$$
The $20$th term interests me, giving:
$$\c{20}=\c{10}$$
Well are there any other interesting ways?
|
Suppose we are interested in the value of
$$S(n,m) =
\sum_{k=0}^n (-1)^k {n+m\choose k} {n+m\choose k+m}.$$
Introduce
$${n+m\choose k+m}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+m}}{z^{k+m+1}} \; dz.$$
This integral controls the range being zero when $k>n$ so we can
extend the summation to infinity to obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+m}}{z^{m+1}}
\sum_{k\ge 0} {n+m\choose k}
(-1)^k \frac{1}{z^k} \; dz.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+m}}{z^{m+1}}
\left(1-\frac{1}{z}\right)^{n+m} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(z^2-1)^{n+m}}{z^{n+2m+1}} \; dz.$$
Now if $n+2m$ is odd i.e. $n$ is odd we are extracting a coefficient
from a polynomial that is even in $z$, so the sum is zero. If $n$ is
even we get
$$[z^{n+2m}] (z^2-1)^{n+m}
= (-1)^{n/2} {n+m\choose n/2 + m}.$$
With $n=20$ and $m=10$ this yields
$${30\choose 20}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1272945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Determinant properties Prove without expanding:
\begin{equation}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3 & c^3\end{vmatrix} = (ab + ac + bc)(b - a)(c - a)(c - b)\end{equation}
*I tried to zero some elements and expand until I reach the Right hand side.
*Also tried C1-C3, C2-C3 then decompose the determinant into two determinants and taking common factors. But I couldn't get (ab + ac + bc) part.
*I can only use the properties shown here http://www.vitutor.com/alg/determinants/properties_determinants.html
|
First note that the determinant is cyclic. Hence, it is of the form $f(a,b,c)$, where $f$ is a polynomial of degree $5$. Further, we have
$f(a,a,c) = f(a,b,b) = f(c,b,c) = 0$, which means $(a-b)$, $(b-c)$ and $(c-a)$ are factors, i.e., the determinant is $g(a,b,c)(a-b)(b-c)(c-a)$, where $g(a,b,c)$ is a cyclic polynomial of degree $2$. Any cyclic polynomial of degree $2$ is of the form $x(a^2+b^2+c^2)+y(ab+bc+ca)$. Setting $a=0$, we see that the determinant is $b^2c^3-b^3c^2 = b^2c^2(c-b)$. We also have
$$f(0,b,c) = g(0,b,c)(-b)(b-c)c = \left(x(b^2+c^2)+ybc \right)bc(c-b)$$
This means we need
$$\left(x(b^2+c^2)+ybc \right)bc(c-b) = b^2c^2(c-b) \implies x(b^2+c^2)+ybc = bc \implies x=0,y=1$$
Hence, we obtain that the determinant is
$$(ab+bc+ca)(a-b)(b-c)(c-a)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1273025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Can anyone explain why this series converges? Can anyone explain why this series converges?
$1+\frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6}...$
The answer is given, but i do not understand it:
$$**|S_{6n}-S_{3n}|= \frac{1}{3n+1}+\frac{1}{3n+2}-\frac{1}{3n+3}+...+\frac{1}{6n-2}+\frac{1}{6n-1}-\frac{1}{6n}**$$ where $S_{6n}$ and $S_{3n}$ are subseries of $S_n$ partial sums in the respective order, that is
$$**S_{6n}-S_{3n}> \frac{1}{3n+1}+ \frac{1}{3n+4}+...+ \frac{1}{6n-2}>\frac{n}{6n-2}>\frac{1}{6}**$$
I highlighted what I dont understand, and would like if someone could simplify, explain..
|
This inequality holds because each of the elements $\frac{1}{3n+1}.....\frac{1}{6n-2}$ is greater or equal to the last term. ie $\frac{1}{3n+1} \gt \frac{1}{6n-2}$ etc for all the terms. So the sum $\frac{1}{3n+1}+\frac{1}{3n+4}...$ (n terms) is definitrly less than $\frac{1}{6n-2}+\frac{1}{6n-2}$...(n terms)..$\frac{1}{6n-2}=\frac{n}{6n-2}$. Is that what you did not understand?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1276245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
number of subsets of even and odd Let $A$ be a finite set. Prove or disprove: the number of subsets of $A$ whose size is even is equal to the number of subsets of $A$ whose size is odd.
Example: $A = {1,2}$. The subsets of $A$ are {},{1},{2}, and {1,2}. Since there are two subsets of odd size ({1} and {2}) and two subsets of even size ({} and {1,2}) the claim holds for this particular example.
|
The number of subsets of size $k$ in a set with $n$ elements is $\binom{n}{k}$.
The Binomial Theorem states that
$$(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k}a^{n - k}b^k$$
In particular, if we let $a = b = 1$, we obtain
$$2^n = (1 + 1)^n = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}1^k = \sum_{k = 0}^{n} \binom{n}{k}$$
Therefore, the number of subsets of a set with $n$ elements is $2^n$. Furthermore, if we set $a = 1$ and $b = -1$ in the Binomial Theorem, we obtain
\begin{align*}
0 & = 0^n\\
& = [1 + (-1)]^n\\
& = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}(-1)^k\\
& = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \cdots + (-1)^{n}\binom{n}{n}\\
& = \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} - \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1}
\end{align*}
where $\lfloor x \rfloor$ represents the largest integer less than or equal to $x$. Thus,
$$\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} = \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1}$$
Hence, the number of subsets of a set with $n$ elements that contain an even number of elements is equal to the number of subsets that contain an odd number of elements.
Moreover, since a set with $n$ elements has $2^n$ subsets and the number of subsets with an even number of elements is equal to the number of subsets with an odd numbers of elements, the set has $\frac{1}{2} \cdot 2^n = 2^{n - 1}$ subsets with an even number of elements and $2^{n - 1}$ subsets with an odd number of elements.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1278169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Find numbers $a, b, c$ given that $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$
Let $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$. Find $a,b,c$
Suppose $a, b, c$ are roots of $P(x)$.
$$P(x) = k(x - a)(x - b)(x - c)$$
But then I get $(k = 1)$
$$P(x) = x^3 - 12x^2 + x(ab + ac + bc) - abc$$
Cant go further...
|
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$$
So $ab+ac+bc=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=47$
Similarly
$$a^3+b^3+c^3=(a+b+c)^3+3abc-3(a+b+c)(ab+ac+bc)$$
So $abc=\frac{a^3+b^3+c^3+3(a+b+c)(ab+ac+bc)-(a+b+c)^3}{3}=44$
And $a,b,c$ are solutions of the cubic
$$X^3-12X^2+47X-44=0$$
No simple solution except Cardan formulaes
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1278394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Integrate $\int \frac{x^5 dx}{\sqrt{1+x^3}}$ I took $1+x^3$ as $t^2$ . I also split $x^5$ as $x^2 .x^3$ . Then I subsituted the differentiated value in in $x^2$ . I put $x^3$ as $1- t^2$ . I am getting the last step as $2/9[\sqrt{1+x^3}x^3 ]$ but this is the wrong answer , i should get $2/9[\sqrt{1+x^3}(x^3 +2)]$.
Please help me. Thanks
|
This is an integration by parts problem, but first, set $u=x^3$. Then $ du = 3x^2 dx $, so the integral becomes
$$ \frac{1}{3}\int \frac{u}{\sqrt{1+u}} \, du, $$
which you can see how to do parts on:
$$ \int \frac{u}{\sqrt{1+u}} \, du = 2u\sqrt{1+u} - 2\int \sqrt{1+u} \, du = 2u\sqrt{1+u} - \frac{4}{3}(1+u)^{3/2} +C, $$
which you can also write as $\frac{2}{3}(u-2)\sqrt{1+u} + C$. Then the final answer is
$$ \frac{2}{9}(x^3-2)\sqrt{1+x^3} + C. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1279090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
Quadratic solutions puzzle The equation $x^2+ax+b=0$, where $a\neq b$, has solutions $x=a$ and $x=b$. How many such equations are there?
I'm getting $1$ equation as I can only find $a=b=0$ as an equation, which is not allowed.
$$x=\frac{±\sqrt{a^2-4 b}-a}2$$
$x=a$ or $b$ so these are the equations
$$a=\frac{\sqrt{a^2-4 b}-a}2$$
$$b=\frac{-\sqrt{a^2-4 b}-a}2$$
$$a=\frac{-\sqrt{a^2-4 b}-a}2$$
$$b=\frac{\sqrt{a^2-4 b}-a}2$$
The only solution for all of these is $a=b=0$, but is this right?
|
if $x=a$ is a solution to $x^2+ax+b=0$ then $b=-2a^2$
if $x=b$ is a solution $b^2+ab+b=0$ so either $b=0$ or $a+b+1=0$
the second solution gives $=-2a^2+a+1=0$
this equation has solutions $a=1$ and $a=-\frac12$
so it looks like we can have $x^2+x-2$ with solutions of 1 and -2
and $x^2 -\frac12 x -\frac 12$ which has $x=-\frac12$ as a solution.
so $(a,b) = (1,-2)$ and $(a,b) = (-\frac 12,-\frac 12)$ are both valid solutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1279477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
How do we find eigenvalues from given eigenvectors of a given matrix? For instance let
$$A=\begin{pmatrix}
3 & -1 & -1 \\
2 & 1 &-2 \\
0 & -1 & 2 \\
\end{pmatrix}$$
be a matrix and
$$u_1=\begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix},$$
$$u_2=\begin{pmatrix}
1 \\
0 \\
1\\
\end{pmatrix},$$
$$u_3=\begin{pmatrix}
0 \\
-1 \\
1 \\
\end{pmatrix}.$$
its eigenvectors. What are its eigenvalues?
Is there anything more simple than doing $A-λI$?
|
If $u$ is an eigenvector of $A$ and $\lambda$ is the corresponding eigenvalue, you know the following:
$$Au = \lambda u$$
So in your example, you can do the following (I’ll take the second one):
$$Au_2 = \begin{pmatrix}3 & -1 & -1 \\ 2 & 1 & -2 \\ 0 & -1 & 2\end{pmatrix}\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix} = \begin{pmatrix}2 \\ 0 \\ 2\end{pmatrix} = 2 u_2$$
So $\lambda_2 = 2$. Do the same for $u_1$ and $u_3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1280050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
}
|
Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$
Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $$a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$$
The inequality can be written in the condensed form
$$\sum\limits_{Sym}a\ge\sum\limits_{Sym}a^3+\sum\limits_{Sym}ab$$
I was told that this is a pretty inequality to prove, but I have been unable to do so.
I've tried naive things like multiplying both sides by $a+b+c+d$, and rewriting $(a^2+b^2+c^2+d^2)^2$, but nothing panned out (and the computations were relatively time-consuming). I also tried looking for clever applications of Cauchy-Schwarz (which seems like the way to go) and AM-GM, but nothing sprung out at me.
|
I got some hints from Crux Problem 3059, click here for more details=)
Based on the hints (which tries to relate the inequality with a constrained optimization problem), I worked out a proof as follows:
We have
\begin{align}
\frac{1}{2}(a+b+c+d)^2=&\frac{1}{2}(a^2+b^2+c^2+d^2)+ab+ac+ad+bc+bd+cd\\
=&\frac{1}{2}+ab+ac+ad+bc+bd+cd
\end{align}
by assumption. Thus, in order to prove the inequality, it is equivalent to prove
$$
a+b+c+d\geq a^3+b^3+c^3+d^3+\frac{1}{2}(a+b+c+d)^2-\frac{1}{2}.
$$
This can further be simplified as proving
$$
a^3+b^3+c^3+d^3+\frac{1}{2}(a+b+c+d-1)^2\leq1.~~~~(*)
$$
Now we try to maximize the LHS under the constraint, i.e.
\begin{align}
\max&~~f\triangleq a^3+b^3+c^3+d^3+\frac{1}{2}(a+b+c+d-1)^2\\
s.t.&~~a^2+b^2+c^2+d^2=1.
\end{align}
Now we try to use the Lagrangian multiplier method. Let
$$
L=f+\lambda(a^2+b^2+c^2+d^2-1).
$$
Take the derivative of $L$ regarding $a,b,c,d$ respectively and let the derivative equal to $0$ gives the following set of equations:
\begin{align}
L_a=&a+b+c+d+3a^2+2a\lambda=1\\
L_b=&a+b+c+d+3b^2+2b\lambda=1\\
L_c=&a+b+c+d+3c^2+2c\lambda=1\\
L_d=&a+b+c+d+3d^2+2d\lambda=1,
\end{align}
Notice that these four equations share exactly the same form. Thus, either
$$
3x^2+2\lambda x=0,~~x=a,b,c,d,~~~~(case1)
$$
or
$$
a=b=c=d.~~~~(case2)
$$
We analyse these two cases separately:
Case 1:
Under this case, we have
$$
a+b+c+d=1.
$$
Since
$$
a^2+b^2+c^2+d^2=1,
$$
the only possibility is one of four elements $a,b,c,d$ equal to $1$ and others are all $0$, which gives $\lambda=-3/2$ and $f=1$.
Case 2:
By the assumption
$$
a^2+b^2+c^2+d^2=1,
$$
it follows
$$
a=b=c=d=\frac{1}{2},
$$
which gives $\lambda=-7/4$ and $f=1$.
Above all we have the maximum of $f$ is $1$ and this proves $(*)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1282810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Mathematically Expressing the Sum of... I noticed that $$\large{2x-1 = \frac{x}{2} + \frac{\frac{x}{2}}{2} + \frac{\frac{\frac x2}2}2} + \cdots$$ until the output of one of the steps in the pattern equals $1$. Or in other words $2x-1$ is equal to the sum of $\large{\frac{x}{2}}$ plus that result divided by two plus that result divided by two and so forth.
How do I notate this with a Sigma? Or would I not use Sigma and if so what is the best way to notate this?
I cannot figure it out; any help will be appreciated!
|
The partial sum of this geometric series is $\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}$, if you start at k=0. Thus in your case it is $\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}-1$, because $2^0=1$.
$=\frac{1-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}-\frac{1-\frac{1}{2}}{1-\frac{1}{2}}$=$\frac{\frac{1}{2}-\left( \frac{1}{2} \right) ^n}{1-\frac{1}{2}}$
If $n \to \infty$ then $\lim_{n \to \infty} \frac{\frac{1}{2}-\left( \frac{1}{2} \right) ^n}{\frac{1}{2}}=\frac{\frac{1}{2}-0}{\frac{1}{2}}=1$.
Consider that $\lim_{n \to \infty} a^n=0$, if $|a|<1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1284088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
find the solutions to the equation $4\sin^2\theta + 1 = 6\sin\theta$ in the interval $0^\circ \leq \theta < 360^\circ$ Find the solutions to the following equation for $0^\circ \leq θ < 360^\circ$:
$$4\sin^2 θ + 1 = 6\sin θ$$
My work:
$$4\sin^2\theta - 6\sin\theta + 1 = 0$$
Factor
$$\sin\theta= \frac{1}{4}(3+ \sqrt{5}) \qquad \text{and} \qquad \sin\theta = \frac{1}{4}(3- \sqrt{5})$$
$$\theta = \arcsin\left[\frac{1}{4}(3+ \sqrt{5})\right] \qquad \text{and} \qquad \theta = \arcsin\left[\frac{1}{4}(3- \sqrt{5})\right]$$
When I take the inverse sin they both come out >1 (no solution), what did I do wrong?
\begin{align*}
2\cos^2\theta - 2\cos\theta - 1 & = 0\\
\cos\theta & = \frac{2 \pm \sqrt{4+8}}{4} && \text{quadratic formula}\\
\cos\theta & = \frac{1 \pm \sqrt{3}}{2}\\
\theta & = \arccos\left[\frac{1-\sqrt{3}}{2}\right]\\
\theta & = 111.5^\circ
\end{align*}
Is this correct?
|
what do you mean? $\frac{3 -\sqrt 5}4 = 0.19098$ and $\theta = 11.01^\circ, 168.989^\circ.
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1284163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find the 2x2 matrix given 2 equations. Find the $2 \times 2$ matrix ${A}$ such that ${A}^2 = {A}$ and
${A} \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$
I have tried to express A as a matrix with variables a,b,c, and d, but it gets too messy with the equations. Any help is appreciated.
|
First, $A^2-A=0$ means $A$ must have eigenvalues 1 or 0. $A$ is not a zero matrix, so the two eigenvalues of $A$ cannot be all 0. The two eigenvalues cannot be all 1 either for otherwise $A$ would either be (1) the identity matrix which is certainly the case (2) a nondiagonalizable matrix but judging from the polynomial $A$ satisfies this cannot be the case. This leaves the only possibility that $A$ has eigenvalues 0 and 1, and so it's of rank 1.
Since $A\begin{pmatrix}7\\-1\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$ its range is $\text{Span\{(3, 1)\}}$ so $A=\begin{pmatrix}3a&3b\\a&b\end{pmatrix}$ for some $a$ and $b$. Its null space is spanned by $\begin{pmatrix}7\\-1\end{pmatrix}-\begin{pmatrix}6\\2\end{pmatrix}=\begin{pmatrix}1\\-3\end{pmatrix}$. So
\begin{eqnarray}\begin{pmatrix}3a&3b\\a&b\end{pmatrix}\begin{pmatrix}1\\-3\end{pmatrix}=0\end{eqnarray}
implies $a=3b$. Finally
\begin{eqnarray}\begin{pmatrix}9b&3b\\3b&b\end{pmatrix}\begin{pmatrix}7\\-1\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}\end{eqnarray}
implies that $b=\frac{1}{10}$. So $A=\frac{1}{10}\begin{pmatrix}9&3\\3&1\end{pmatrix}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1285881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$
I broke the top into prime mods:
$$x \equiv 8^{38} \pmod 3$$
$$x \equiv 8^{38} \pmod {70}$$
But $x \equiv 8^{38} \pmod {70}$ can be broken up more:
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod {10}$$
But $x \equiv 8^{38} \pmod {10}$ can be broken up more:
$$x \equiv 8^{38} \pmod 5$$
$$x \equiv 8^{38} \pmod 2$$
In the end,I am left with:
$$x \equiv 8^{38} \pmod 5$$
$$x \equiv 8^{38} \pmod 2$$
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod 3$$
Solving each using fermat's theorem:
*
*$x \equiv 8^{38}\equiv8^{4(9)}8^2\equiv64 \equiv 4 \pmod 5$
*$x \equiv 8^{38} \equiv 8^{1(38)}\equiv 1 \pmod 2$
*$x \equiv 8^{38} \equiv 8^{6(6)}8^2\equiv 64 \equiv 1 \pmod 7$
*$x \equiv 8^{38} \equiv 8^{2(19)}\equiv 1 \pmod 3$
So now, I have four congruences. How can i solve them?
|
just apply Chinese remainder theorem (See Burton)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1286716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 5
}
|
Can $\sin(\pi/25)$ be expressed in radicals I suspect that sin(pi/25) is not expressible in elementary forms in radicals because it is the root of some quintic (or rather cos(pi/25) is). Can anyone prove that that particular quintic has no solution in radicals?
I know that (from chebyshev polynomials) if $$u=\cos{\frac{\pi}{25}}$$
then
$$u\left(16u^4-20u^2+5\right)=\frac{\phi}{2}$$ where $\phi$ is the golden ratio.
|
As requested, R. Israel's Maple answer can be simplified. That complicated expression,
$$u = -\frac{-3125\sqrt{-10}+12500\sqrt{5}\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}+15625\sqrt{-2}-25000\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}}{2048(\sqrt{5}-1)^2\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}}$$
is just equal to,
$$u=\big(\tfrac{5}{4}\big)^5\,\Big(\tfrac{1-\sqrt{5}}{4}-\sqrt{\tfrac{5+\sqrt{5}}{8}}i\Big)$$
Hence, after some simplification, his answer reduces to the ratio of arithmetic mean (AM) and geometric mean(GM),
$$\sin\big(\tfrac{\pi}{25}\big) = \frac{AM(a,b)}{GM(a,b)} = \frac{(a+b)/2}{\sqrt{ab}} = 0.1253332\dots$$
where,
$$a = e^{4\pi i/5} = \tfrac{-1-\sqrt{5}}{4}+\sqrt{\tfrac{5-\sqrt{5}}{8}}\,i$$
$$b = (e^{7\pi i/5})^{1/5} = e^{47\pi i/25} = e^{-3\pi i/25} = \left(\tfrac{1-\sqrt{5}}{4}-\sqrt{\tfrac{5+\sqrt{5}}{8}}\,i\right)^{1/5}$$
which should be prettier enough.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1288769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
}
|
$1989 \mid n^{n^{n^{n}}} - n^{n^{n}}$ for integer $n \ge 3$ Before anyone comments, yes this is kind of a duplicate of Prove that $1989\mid n^{n^{n^{n}}} - n^{n^{n}}$ . The problem that I'm having I don't see the $n=5$ as a counterexample. Also if anyone wants to know where I got this problem from here.
I'm looking at the problem $\color{red} {\text{A10}}$. This is not a homework. This is a question I chose to do for fun and I'm totally not sure how to do this problem after playing with for hours. I have made a conjecture that I cannot prove. I believe $n^n \equiv k \mod 1989$ while $n^{n^n} \equiv k \mod 1989$ while $n^{n^{n^n}}\equiv k \mod 1989$ for integer $n \ge 4$. Anyways right now I'm just looking for a hint. I still want to try. You can put spoilers in your answers if you want to. Also we can use whatever we want to prove this. Though I do warn you my number theory skills are still a work in progress. And what I'm looking for is to prove this: $1989 \mid n^{n^{n^{n}}} - n^{n^{n}}$ for integer $n \ge 3$
|
Lemma 1
$$ \forall n \in \mathbb{N}_{>0} \hspace{1.5em} n^{n^n} \equiv n^n \pmod3 $$
This is trivial.
Lemma 2
$$\forall n \in \mathbb{N} \hspace{1.5em} n\ge 3 \implies n^{n^n} \equiv n^n \pmod{16} $$
Proof. If $n$ is even, then $n\ge4$ and thus $16 \mid n^n, n^{n^n}$. If $n$ is congruent to $\pm 1$ modulo $16$, the proposition above holds. Then, we only need to check $\pm 3, \pm 5, \pm 7$. By Euler's theorem,
\begin{align}
n\equiv 3 \pmod{16} \implies 3^{n^n} - 3^n &\equiv 3^{n^n-n \pmod8}-1 \\
&\equiv 3^{3^{n\pmod4}-3}-1 \\
&\equiv 3^{3\cdot 8} -1 \equiv 0 \pmod{16}
\end{align}
\begin{align}
n\equiv 5 \pmod{16} \implies 5^{n^n} - 5^n &\equiv 5^{n^n-n \pmod8}-1 \\
&\equiv 5^{5^{n\pmod4}-5}-1 \equiv 0 \pmod{16}
\end{align}
\begin{align}
n\equiv 7 \pmod{16} \implies 7^{n^n} - 7^n &\equiv 7^{n^n-n \pmod8}-1 \\
&\equiv 7^{7^{n\pmod4}+1}-1 \\
&\equiv 7^{7^3+1} -1 \equiv 7^{43\cdot 8} -1 \equiv 0 \pmod{16}
\end{align}
The cases $-3,-5,-7$ are now trivial.
Theorem
$$\forall n \in \mathbb{N} \hspace{1.5em} n\ge 3 \implies n^{n^{n^n}} \equiv n^{n^n} \pmod{1989} $$
Proof. We first split the congruence as follows.
$$\begin{align} n^{n^{n^n}} \equiv n^{n^n} \pmod{1989} &\iff \begin{cases}
n^{n^{n^n}} \equiv n^{n^n} \pmod{9} \\
n^{n^{n^n}} \equiv n^{n^n} \pmod{13} \\
n^{n^{n^n}} \equiv n^{n^n} \pmod{17}
\end{cases}
\end{align}$$
Since $2,2,3$ are primitive roots of $9,13,17$ respectively, then
$$\begin{align}
n^{n^{n^n}} \equiv n^{n^n} \pmod{9} &\iff \begin{cases} n^{n^n}\cdot\operatorname{ind}_2n \equiv n^n\cdot\operatorname{ind}_2n \pmod{6} & 3 \nmid n \\ 0 \equiv 0 & 3 \mid n\end{cases} \\[1ex]
n^{n^{n^n}} \equiv n^{n^n} \pmod{13} &\iff \begin{cases} n^{n^n}\cdot \operatorname{ind}_2n \equiv n^n \cdot\operatorname{ind}_2n \pmod{12} & 13 \nmid n \\ 0 \equiv 0 & 13 \mid n\end{cases} \\[1ex]
n^{n^{n^n}} \equiv n^{n^n} \pmod{17} &\iff \begin{cases} n^{n^n}\cdot \operatorname{ind}_3n \equiv n^n \cdot\operatorname{ind}_3n \pmod{16} & 17 \nmid n \\ 0 \equiv 0 & 17 \mid n\end{cases}
\end{align}$$
By the lemmas, these congruences hold for all $n\ge 3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1288942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
}
|
Find the limit of: $\lim_{x\to 1}\left(\frac{\sqrt{2x-1}-1}{x^2-1}\right)$
Find the limit of: $\lim\limits_{x\to 1}\left(\frac{\sqrt{2x-1}-1}{x^2-1}\right)$
I tried multiplying by the conjugate of $\sqrt{2x-1}-1$ ,but I obtain
$\frac{2x-2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{2x-1}+1\right)}$,
which is again zero over zero indeterminate form.
How to approach this type of problems?
|
\begin{align}
L & = \lim_{x \to 1} \dfrac{\sqrt{2x-1}-1}{x^2-1} = \lim_{x \to 1} \dfrac{\sqrt{2x-1}-1}{x^2-1} \cdot \dfrac{\sqrt{2x-1}+1}{\sqrt{2x-1}+1}\\
& = \lim_{x \to 1} \dfrac{2x-1-1}{(x^2-1)(\sqrt{2x-1}+1)} = \lim_{x \to 1} \dfrac{2(x-1)}{(x^2-1)(\sqrt{2x-1}+1)}\\
& = \lim_{x \to 1} \dfrac2{(x+1)(\sqrt{2x-1}+1)} \left(\because \text{Canceling off }x-1\text{ from numerator and denominator}\right)\\
& = \dfrac2{2\cdot2}\\
& = \dfrac12
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1289783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Describe all odd primes p for which 7 is a quadratic residue I need to describe all odd primes $p$ for which $7$ is a quadratic residue.
Now let $\left(\frac{a}{b}\right)$ be the Legendre Symbol.
Then if $7$ is a quadratic residue $p$ we must have:
$$1=\left(\frac{7}{p}\right)=(-1)^{\frac{3(p-1)}{2}} \left(\frac{p}{7}\right)$$
Here I have used Gauss theorem on Quadratic Reciprocity.
This implies that $\left(\frac{p}{7}\right)=1$ and $(-1)^{\frac{3(p-1)}{2}} = 1$, or $\left(\frac{p}{7}\right)=-1$ and $(-1)^{\frac{3(p-1)}{2}} = -1$.
HOWEVER at this step in the solutions, we are given that:
$\left(\frac{p}{7}\right)=1$ and $p\equiv 1\bmod 4$
OR $\left(\frac{p}{7}\right)=-1$ and $p\equiv -1\bmod 4$
Why is this equivalent? So what I am basically asking is the following:
Why is $(-1)^{\frac{3(p-1)}{2}} = 1$ equivalent to $p\equiv 1\bmod 4$?
And $(-1)^{\frac{3(p-1)}{2}} = -1$ equivalent to $p\equiv -1\bmod 4$?
|
$$(-1)^n\equiv\begin{cases}
\hphantom{-}1 & \text{if }n\equiv 0\bmod 2\\
-1 & \text{if }n\equiv 1\bmod 2
\end{cases}$$
Since $p$ is odd, we know that $p-1$ is even, so $\frac{p-1}{2}$ is an integer.
By definition, $\frac{p-1}{2}\equiv 0\bmod 2$ if and only if $\frac{p-1}{2}=2x$ for some integer $x$, which is equivalent to saying $p=4x+1$ for some integer $x$, which by definition is equivalent to $p\equiv 1\bmod 4$.
Similarly, $\frac{p-1}{2}\equiv 1\bmod 2\iff p\equiv 3\bmod 4$.
Now observe that
$$(-1)^{\tfrac{3(p-1)}{2}}=\left((-1)^3\right)^{\tfrac{p-1}{2}}=(-1)^{\tfrac{p-1}{2}}=\begin{cases}
\hphantom{-}1 & \text{if }\frac{p-1}{2}\equiv 0\bmod 2\\
-1 & \text{if }\frac{p-1}{2}\equiv 1\bmod 2
\end{cases}=\begin{cases}
\hphantom{-}1 & \text{if }p\equiv 1\bmod 4\\
-1 & \text{if }p\equiv 3\bmod 4
\end{cases}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1290029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Given that $1 = 27 \times 11 - 74 \times 4$, solve the following equations in modulo $74$: $ 3x - y = 1$; $2x + 3y = 0$ Given that $1 = 27 \times 11 - 74 \times 4$, solve the following equations in modulo $74$:
$3x - y = 1$;
$2x + 3y = 0$.
Thank you.
|
In mod $74$, we have
$$27\times 11\equiv 1\iff 3\times 27\times 11\equiv 3.$$
Then,
$$\begin{align}y\equiv 3x-1&\Rightarrow 3y\equiv 9x-3\\&\Rightarrow -2x\equiv 9x-3\\&\Rightarrow 11x\equiv 3\\&\Rightarrow 11x\equiv 3\times 27\times 11\\&\Rightarrow x\equiv 3\times 27\equiv 81\equiv \color{red}{7}\end{align}$$
Hence, $$y\equiv 3x-1\equiv 3\times 7-1\equiv \color{red}{20}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1290876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Evaluate $7^{8^9}\mod 100$ I'm preparing myself for discrete math exam and here's one of the preparation problems:
Evaluate $$7^{8^9}\mod 100$$
Here's my solution:
$7^2\equiv49 \mod 100\implies (7^2)^2\equiv49^2=2401\equiv 1\mod 100$
So, as it turns out $7^4\equiv 1\mod 100$. It will be useful later. Now let's examine $8^9\mod 100$.
We have $512=8^3\equiv 12 \mod 100$, so $(8^3)^3\equiv 12^3 = 1728 \equiv 28 \mod 100$. So, as $8^9\equiv 28 \mod 100$, then $8^9=100x+28$ for some natural $x$. So:
$7^{8^9}=7^{100x+28}=7^{100x}7^{28}=(7^4)^{25x}(7^4)^7\equiv 1^{25x}1^7 \equiv 1 \mod 100$
Which means the final answer is that $7^{8^9}\mod 100=1$.
That's my solution, but I'm curious about other ones. Has anybody have any ideas?
|
Carmichael's function $\lambda(100) = \text{lcm}(\lambda(25),\lambda(4)) = \text{lcm}(20,2) = 20$.
$7$ is coprime to $100$, so $7^{8^9} \equiv 7^{8^9 \bmod 20} \bmod 100$
$\lambda(20) = \text{lcm}(\lambda(5),\lambda(4)) = \text{lcm}(4,2) = 4$
and $9 \equiv 1 \bmod 4$
$8$ has a higher exponent than $20$ in $2$, their only shared prime factor, so $8^9 \equiv 8 \bmod 20$
This gives $7^{8^9} \equiv 7^{8^9 \bmod 20} \equiv 7^{8} \bmod 100$
So we just need to square $7$ three times to get our answer... but when we get to $49^2$ we happily discover that $49^2 = (50-1)^2 \equiv 1 \bmod 100$.
So $7^{8^9} \equiv 7^{8} \equiv ((7^2)^2)^2 \equiv (49^2)^2 \equiv 1^2 \equiv 1 \bmod 100$
It's perfectly legitimate, of course, to find out $7^8 \bmod 100$ first in the hope that it will be something convenient, but this is a method that will work in the absence of good fortune.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1294684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
How to solve $12-\sin(\theta)=\cos(2\theta)$? $$12-\sin(\theta)=\cos(2\theta)$$
What's the correct answer on the $[0,2\pi]$?
I started with $12-\sin(\theta)=1-2\sin^2(\theta)$ and then i cant get anything sensible as i end up with $12=\sin(\theta)+2\sin^2(\theta)$
|
Note that
\begin{eqnarray*}
12\sin \theta &=&\cos (2\theta ) \\
&=&1-2\sin ^{2}\theta
\end{eqnarray*}
then
\begin{equation*}
2\sin ^{2}\theta +12\sin \theta -1=0.
\end{equation*}
Let $X=\sin \theta ,$ then
\begin{equation*}
2X^{2}+12X-1=0
\end{equation*}
This quadratic equation have two roots: $X_{1}=\frac{1}{2}\sqrt{38}-3$ and $%
X_{2}=-\frac{1}{2}\sqrt{38}-3.$ Note that $X_{2}<-3$ then $X_{2}\notin \left[
-1,1\right] ,$ then it cannot provide a $\theta -solution$ to the original
equation. Next, $\frac{1}{2}\sqrt{38}-3\simeq 0.082\,2\in \left[ -1,1\right]
$ then one has
\begin{equation*}
\sin \theta =\frac{1}{2}\sqrt{38}-3
\end{equation*}
and then
\begin{equation*}
\theta =\arcsin \left( \frac{1}{2}\sqrt{38}-3\right) \simeq 0.0823\ rad
\end{equation*}
Remark. It is known that when $\theta $ (in rad) is so small then $\sin \theta
\simeq \theta .$ Since our $\sin \theta =X_{1}\simeq 0.082\,2$ is so small
then it is not surprising to find $\theta \simeq \sin \theta .$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1294901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
}
|
Let $X$ be a set of primes $p$ so that $5^{p^2}+1 \equiv 0 \pmod {p^2}$ Which of these sets is $X$ equal to? $5^{p^2}+1\equiv 0\pmod {p^2}$
$1.$ $\emptyset $
$2.$ {$3$}
$3.$ All primes of the form $4k+3$
$4.$ All primes except $2$ and $5$
$5.$ All primes
This one is pretty easy to get right through the process of elimination, which is how I solved it (it's $3$). What would be a more rigorous way of solving this?
|
In the congruance holds you also have : $5^{p^2}+1\equiv 0 [\rm mod p]$.
On the other hand $5^p\equiv 5 [\rm mod p]$ ($equation : x^p \equiv x [\rm mod p]$) so we should have $5+1\equiv 0 [\rm mod p]$ (apply equation twice).
So the only possible primes are 2 or 3. 2 is not a solution and 3 is a solution indeed:
$$5^4+1 \equiv (4+1)^4+1 \equiv 2 [\rm mod 4]$$
$$5^9+1 \equiv (9-4)^{9} +1 \equiv -(4)^{9} +1 \equiv -(8)^6+1 \equiv -1(-1)^6+1 \equiv 0 [ \rm mod 9] $$.
Hence 3 is the only solution
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1297392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
If $a_1,a_2,a_3$ are roots $x^3+7x^2-8x+3,$ find the polynomial with roots $a_1^2,a_2^2,a_3^2$ If $a_1,a_2,a_3$ are the roots of the cubic $x^3+7x^2-8x +3,$ find the cubic polynomial whose roots are:
$a_1^2,a_2^2,a_3^2$ and the polynomial whose roots are $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}.$ Not really sure where to go. Hints appreciated.
|
HINT:
For the first let $y=x^2$
$\implies x^2(x+7)=8x-3\implies y=x^2=\dfrac{8x-3}{x+7}$
Express $x$ in terms of $y$ and put the value of $x$ in the given cubic equation and simplify.
For the second, let $z=\dfrac1x\implies x=\dfrac1z$
and put the value of $x$ in the given cubic equation and simplify.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1299134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$, partial fraction braindead decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$
the way my teacher wants us to solve is by substitution values for x,
I set it up like this:
(after setting the variables to the common denominator and getting rid of the denominator in the original equation)
$x^2-2x+3= \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+4}$
1)Let $x=1$, after plugging in for $x$ I got $2=5B$, or $B=\frac{2}{5}$
2)let the $x=0$, I get $3=-4A+4B+D$, and if I substitute B and simplify I get $D=4A-\frac{7}{5}$. Furthermore, no matter what value I plug in I still end up with two variables and cant seem to find a way to eliminate one. It seems like I have to set it up as the triple system of equations but I just dont know how to apply it here. Would really appreciate if you guys could give me a hint on how to go about it.
|
If you substitute $x=2i$, this gives
$\;\;\;-1-4i=((2Ci+D)(2i-1)^2=(-3-4i)(2Ci+D)=(-3D+8C)-(6C+4D)i$
Therefore $8C-3D=-1\;\;$ and $\;\;6C+4D=4,\;$ so
$\hspace{.6 in}16C-6D=-2\;\;$ and $\;\;9C+6D=6\implies 25C=4\implies C=\frac{4}{25}$.
Then $D=1-\frac{3}{2}C=1-\frac{6}{26}=\frac{19}{25},\;\;$ and $\;\;A=-\frac{4}{25}\;$ since $0=A+C$ from the coefficient of $x^3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1301773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Trigonometry Identity proving If $\sin(x-y) =\cos y$ prove that $\tan y = \frac{1+ \sin y}{\cos y} $.
Is there an error with the question? I don't seem to be able to get the answer. Should it be $\tan x$ instead of $\tan y = \frac{1+\sin y}{\cos y}$ ?
$$\tan x\times \cos y = 1 + \sin y $$
$$\frac{\sin x}{\cos x} \times \cos y = \frac{\sin x\times\cos y}{\cos x}$$
$$\frac{\sin x\times\cos y}{\cos x}= 1 + \sin y $$
$$\frac{\sin x\times\cos y}{\sin(x-y)}= 1 + \sin y $$
I am stuck at this step.
|
Since $\sin(x-y)= \cos(y)$, expanding it gives
\begin{equation*}
\sin (x) \cos(y)-\cos(x) \sin(y)=\cos(y).
\end{equation*}
Dividing both sides by $\cos(y)$ we have
\begin{equation*}
\sin(x)-\cos(x)\tan(y)=1.
\end{equation*}
So from here, we get
\begin{equation*}
\tan(y)=\dfrac{\sin(x)-1}{\cos(x)}.
\end{equation*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1302239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
First order differential equation: did i solve this equation right So i'm trying to solve:
$$x^2\frac{dy}{dx} + 2xy = y^3$$
I'm given this differential equation, that Bernoulli equation:
$$\frac{dy}{dx} + p(x)y = q(x)y^{n} $$
I think i've solved it and got
$$ u = \frac{2}{5x} +Cx^4$$
I'm just not sure i am right i will show you how i get there but firstly...
This was part of another question which i've already solved
Show that if $y$ is the solution of the above Bernoulli differential
equation and $u = y^{1−n}$, then $u$ satisfies the linear differential
equation:
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
Applying the chain rule to $u = y^{1-n}$ we obtain that
\begin{align}
\frac{d u}{dx}(x)&= \frac{du}{dy}\cdot\frac{dy}{dx}\\
&= (1-n)y^{-n}\cdot\frac{dy}{dx}
\end{align}
Futhermore using the Bernoulli equation we have
$$
\frac{dy}{dx}=q(x)y^n-p(x)y
$$
and
\begin{align}
\frac{d u}{dx}&= (1-n)y^{-n}\cdot\frac{dy}{dx}\\
&=(1-n)y^{-n}\cdot q(x)y^n - (1-n)y^{-n}\cdot p(x) y\\
&=(1-n)q(x) -(1-n)p(x)y^{1-n}\\
&=(1-n)q(x) -(1-n)p(x)u
\end{align}
Hence U satisfies the equation
$$
\frac{du}{dx}+(1-n)p(x)u = (1-n)q(x)
$$
$$x^2\frac{dy}{dx} + 2xy = y^3$$
Divide both sides by $x^2$
$$\frac{dy}{dx} + \frac{2}{x}y = x^{-2} y^3$$
Consider
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
We know that
*
*n = 3
*1- n = 1-3 = -2
*p(x) = $ \frac{2}{x}$
*q(x) = $x^{-2}$
*u = $y^{1-3} = y^{-2}$
Subbing these in...
$$
\frac{du}{dx} + (-2)\frac{2}{x}u = (-2)x^{-2}
$$
$$
\frac{du}{dx} + \left(-\frac{4}{x}\right)u = (-2)x^{-2}
$$
So...
$$ \text{integrating factor} = e^{\int p(x) \, dx} $$
- p(x) dx = $-\frac{4}{x}$
$$ -4 \int \frac{1}{x} = -4log(x) = log (x^{-4}) $$
$$ \text{integrating factor} = e^{log (x^{-4})}= x^{-4} = \frac{1}{x^4}$$
So multiply this to the equation
$$\frac{1}{x^4}\frac{du}{dx} + \left(\frac{-4}{x^5} \right)u = \frac{-2}{x^6}$$
So we want to solve
$$ \frac{d}{dx}\frac{1}{x^4}u = \frac{-2}{x^6} $$
$$ \int \frac{d}{dx}\frac{1}{x^4}u = \int \frac{-2}{x^6} $$
$$ \frac{1}{x^4}u = -2\int \frac{1}{x^6} $$
$$ \frac{1}{x^4}u = -2\frac{1}{-5x^5} + c $$
$$ \frac{1}{x^4}u = \frac{2}{5x^5} + c $$
$$ \therefore u= \frac{2}{5x} + cx^{4} $$
is this fine? Or do i need to somehow equate this y or sub $u=y^{1-n}$
As $$u=y^{-2}$$
$$\frac{1}{y^2}= \frac{2}{5x} + cx^{4} $$
$$y^2= \frac{5x}{2} + \frac{1}{cx^{4}} $$
$$y= \sqrt{\frac{5x}{2} + \frac{1}{cx^{4}}} $$
|
$$x^2\frac{dy}{dx} + 2xy = y^3$$
Divide both sides by $x^2$
$$\frac{dy}{dx} + \frac{2}{x}y = x^{-2} y^3$$
Consider
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
We know that
*
*n = 3
*1- n = 1-3 = -2
*p(x) = $ \frac{2}{x}$
*q(x) = $x^{-2}$
*u = $y^{1-3} = y^{-2}$
Subbing these in...
$$
\frac{du}{dx} + (-2)\frac{2}{x}u = (-2)x^{-2}
$$
$$
\frac{du}{dx} + \left(-\frac{4}{x}\right)u = (-2)x^{-2}
$$
So...
$$ \text{integrating factor} = e^{\int p(x) \, dx} $$
- p(x) dx = $-\frac{4}{x}$
$$ -4 \int \frac{1}{x} = -4log(x) = log (x^{-4}) $$
$$ \text{integrating factor} = e^{log (x^{-4})}= x^{-4} = \frac{1}{x^4}$$
So multiply this to the equation
$$\frac{1}{x^4}\frac{du}{dx} + \left(\frac{-4}{x^5} \right)u = \frac{-2}{x^6}$$
So we want to solve
$$ \frac{d}{dx}\frac{1}{x^4}u = \frac{-2}{x^6} $$
$$ \int \frac{d}{dx}\frac{1}{x^4}u = \int \frac{-2}{x^6} $$
$$ \frac{1}{x^4}u = -2\int \frac{1}{x^6} $$
$$ \frac{1}{x^4}u = -2\frac{1}{-5x^5} + c $$
$$ \frac{1}{x^4}u = \frac{2}{5x^5} + c $$
$$ \therefore u= \frac{2}{5x} + cx^{4} $$
is this fine? Or do i need to somehow equate this y or sub $u=y^{1-n}$
As $$u=y^{-2}$$
$$\frac{1}{y^2}= \frac{2}{5x} + cx^{4} $$
$$y^2= \frac{5x}{2} + \frac{1}{cx^{4}} $$
$$y= \sqrt{\frac{5x}{2} + \frac{1}{cx^{4}}} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1302602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Solution to $y'=y^2-4$ I recognize this as a separable differential equation and receive the expression:
$\frac{dy}{y^2-4}=dx$
The issue comes about when evaluating the left hand side integral:
$\frac{dy}{y^2-4}$
I attempt to do this integral through partial fraction decomposition using the following logic:
$\frac{1}{(y+2)(y-2)} = \frac{A}{y+2}+\frac{B}{y-2}$
Therefore,
$1=Ay-2A+By+2B$.
Since the coefficients must be the same on both sides of the equation it follows that:
$0=A+B$ and $1=-2A+2B$.
Hence, $A=-B$, $B=\frac14$, $A=-\frac14$.
Thus the differential equation should be transformed into:
$-\frac{1}{4} \frac{dy}{y+2} + \frac14 \frac{dy}{y-2} = x+C$
Solving this should yield:
$-\frac14 \ln|y+2| + \frac14 \ln|y-2| = x+C$
which simplifies as:
$\ln(y-2)-\ln(y+2)=4(x+c)$
$\ln[(y-2)/(y+2)]=4(x+c)$
$(y-2)/(y+2)=\exp(4(x+c))$
$y-2=y*\exp(4(x+c)+2\exp(4(x+c))$
$y-y\exp(4(x+c))=2+2\exp(4(x+c))$
$y(1-\exp(4(x+c)))=2(1+\exp(4(x+c)))$
$y= 2(1+\exp(4(x+c)))/(1-\exp(4(x+c)))$
However, when done in Mathematica/Wolfram Alpha the result is given as (proof in the attached image)
$\frac14 \ln(2-y) -\frac14 \ln(2+y) = x + C$
and returns an answer of:
$y= 2(1-\exp(4(x+c)))/(1+\exp(4(x+c)))$.
Can anyone figure out where I have made an error? The only thing I can think of is something with evaluating the absolute values of the natural logarithms.
|
Saturday morning: I do like robjohn's point about $\log |x|$ giving the student some incorrect expectations. What I wrote here is correct and careful, but the same conclusions come from dropping the absolute value signs and doing the calculations three times, $y < -2,$ $-2 < y < 2,$ $y > 2.$ At some point the vertical asymptotes will be revealed, how solutions with $y > 2$ reach a vertical asymptote and then jump below to $y < -2.$ That is how I usually do things, split into cases early.
Lost my train of thought. However, if I want $\int \frac{1}{x} dx,$ without absolute values, the answer would be $\log x$ for $x > 0,$ but $\log (-x)$ for $x < 0.$
$-\frac14 \log|y+2| + \frac14 \log|y-2| = x+C$
which simplifies as:
$\log|y-2|-\log|y+2|=4(x+c),$ or
$$ \left| \frac{y-2}{y+2} \right| = k e^{4x} $$
with $k > 0$ for the nonconstant solutions.
The item inside absolute value signs is a linear fractional or Mobius transformation, once we decide about the $\pm$ signs the inverse is given by the matrix
$$
\left(
\begin{array}{rr}
2 & 2 \\
-1 & 1
\end{array}
\right)
$$
If $y > 2$ or $y < -2,$ those inequalities hold true forever, we have
$$ \frac{y-2}{y+2} = k e^{4x} $$
and we get robjohn's
$$ y = \frac{2k e^{4x} + 2}{-k e^{4x} + 1}. $$
There is a jump discontinuity: for some value of $x,$ we find that
$-k e^{4x} + 1= 0,$ indeed $e^{4x} = 1/k,$ $4x = - \log k,$ x = $-(1/4) \log k.$ For $x < -(1/4) \log k,$ we have $y > 2.$ There is a vertical asymptote, then for $x > -(1/4) \log k,$ we have $y < -2.$
In all cases, the curves are asymptotic to the lines $y=2$ and $y=-2.$ Also, note that the ODE is autonomous. Given one of the solutions described, we get another solution by shifting $x$ by any constant we like. If preferred, we can drop the multiplier $k$ by emphasizing the horizontal shifts.
If, however, we begin with $-2 < y < 2,$ then those inequalites hold forever, we have
$$ \frac{y-2}{y+2} = - k e^{4x} $$ and we get the continuous
$$ y = \frac{-2k e^{4x} + 2}{k e^{4x} + 1}. $$
In the picture below, I took $k=1.$ To find all other solutions, replace $x$ by some $x - x_0.$ Put another way, we can take $e^{-4 x_0}= k;$ thus, we can account for all solutions either by varying $k,$ or by deleting $k$ and adjusting $x_0.$ Note also that the part with $y>2$ and the part with $y < -2$ are tied together, the same vertical asymptote.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1304633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
}
|
decimal to fractions When being asked how to solve the Arithmetic Means of $8, 7, 7, 5, 3, 2,$ and $2$, I understand that adding these numbers then dividing by $7$ (the amount of numbers) gives me the decimal $4.85714...$ But when being asked to change this number into a fraction, I do not understand where the $\frac{6}{7}$ comes from in the answer of $4\frac{6}{7}$. Can someone please help me understand?
|
Arithmetic mean $= \frac{8+7+7+5+3+2+2}{7} = \frac{34}{7}$
Now, we calculate $\frac{34}{7}$ by using long division and soon we will get a remainder of $6$, the quotient as $4$, and the divisor as $7$ ($x\frac{y}{z}$ ---- $y$ as the remainder, $x$ as the quotient, and $z$ as the divisor). As the result, we will get: $4\frac{6}{7}$
We can also cross-check this division by converting mixed number to improper fraction: $$4\frac{6}{7} =\frac{(4\times 7)+6}{7} = \frac{28+6}{7}= \frac{34}{7}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1304824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Finding the Standard Matrix for Linear Transformation We are asked to find the standard matrix $A$ for $T$:
Consider the transformation $T : \mathbb{R^3} \rightarrow \mathbb{R^4}$ given by
$$T(x_1, x_2, x_3) = (x_1 + x_2 + x_3, x_2 + x_3, 3x_1 + x_2, 2x_2 + x_3)$$
for every $$(x_1,x_2,x_3) \in \mathbb{R^3}$$
I am confused. I do not know how to start the problem. I have researched the description but I cannot find anything definitive for this type of problem.
Update:
$$ A = \left[\begin{array}{rrr}
1 & 1 & 1\\
0 & 1 & 1\\
3 & 1 & 0\\
0 & 2 & 1\\
\end{array}\right]
$$
$$ ref(A) = \left[\begin{array}{rrr}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1\\
0 & 0 & 0\\
\end{array}\right]
$$
From $T(e_1)=(1,0,3,0)^T, T(e_2)=(1,1,1,2)^T, T(e_3)=(1,1,0,1)^T$
Is this correct? Or am I completely lost?
How would I find the dimension of the kernel of $T$ and the range of $T$?
|
We are asked to find the standard matrix $A$ for $T$:
Consider the transformation $T : \mathbb{R^3} \rightarrow \mathbb{R^4}$ given by
$$T(x_1, x_2, x_3) = (x_1 + x_2 + x_3, x_2 + x_3, 3x_1 + x_2, 2x_2 + x_3)$$
for every $$(x_1,x_2,x_3) \in \mathbb{R^3}$$
Your update is correct $A=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 3 & 1 & 0 \\ 0 & 2 & 1 \end{bmatrix}$.
We are basically solving for $T(x)=Ax$ where $x$ is the vector containing entries $(x_{1},x_{2},x_{3})$ and $A$ is the standard matrix for $T$.
$$\begin{align}T(x) &= T(x_1, x_2, x_3) \\
&= (x_1 + x_2 + x_3, x_2 + x_3, 3x_1 + x_2, 2x_2 + x_3)\\
&=\begin{bmatrix}x_{1}+x_{2}+x_{3} \\ 0x_{1}+x_{2}+x_{3} \\3x_{1} +x_{2} +0x_{3} \\ 0x_{1}+2x_{2}+x_{3}\end{bmatrix}\\
&=x_{1}\begin{bmatrix} 1\\0\\3\\0\end{bmatrix}+x_{2}\begin{bmatrix}1\\1\\1\\2\end{bmatrix} +x_{3}\begin{bmatrix}1\\1\\0\\1\end{bmatrix}\\
&=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 3 & 1 & 0 \\ 0 & 2 & 1 \end{bmatrix}\begin{bmatrix} x_{1}\\x_{2}\\x_{3}\end{bmatrix} \\
&=Ax \\
\end{align}$$
Thus, $A=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 3 & 1 & 0 \\ 0 & 2 & 1 \end{bmatrix}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1305884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Basis of eigenspace For eigenvalue $1$ I did $(A-I)x=0$ which is
$$\left(\begin{array}{ccc}
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0\\
2 & 2 & 1 & 0\\
-1 & 1 & -1 & 1
\end{array}\right).$$
I got row reduced echelon form
$$\left(\begin{array}{ccc}
1 & 0 & 0 & 1/3\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & -2/3\\
0 & 0 & 0 & 0
\end{array}\right).$$
I'm a little confused with the free variables that I have and how to find a basis for this eigenspace.
If I set first column $x$, second $y$, third $z$ and fourth $w$:
$x = -1/3t$
$y = s$
$z = 2/3 t$
$w = t$
Is this correct?
|
Your reduced row echelon matrix is not correct. It should be
$$RREF(A-I) = \left[\begin{array}{cccc}1 & 0 & 0 & 1\\0 & 1 & 0 & 0\\0 & 0 & 1 & -2\\0 & 0 & 0 & 0\end{array}\right]$$
Now to see how to find the eigenspace from this let's rewrite this matrix as a set of linear equations:
$$\left[\begin{array}{cccc}
1 & 0 & 0 & 1\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & -2\\
0 & 0 & 0 & 0
\end{array}\right] \implies \begin{cases} x+w=0 \\ y=0 \\ z-2w=0 \\ 0=0\end{cases}$$
The only column without a pivot position is the $4^{th}$ column, so there's only $1$ free variable. If we call the $4^{th}$ variable $w$, then we first start by setting $w=t$. Then we see that the solutions are of the form $$\begin{bmatrix} x \\ y \\ z \\ w\end{bmatrix} = \begin{bmatrix}-t \\0 \\2t \\ t\end{bmatrix} = t\begin{bmatrix}-1 \\0 \\2 \\ 1\end{bmatrix}$$
for any $t\in \Bbb R$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1306182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Bisector of two lines in the euclidean space $\mathbb{E}_3$ Let $$r: \begin{cases} x + z = 0 \\ y + z + 1 = 0\end{cases}$$ and $$s: \begin{cases} x - y - 1 = 0 \\ 2x - z -1 = 0\end{cases}$$
be two lines in the euclidean space $\mathbb{E}_3$. It is easily seen that their intersection is one point.
How do I find the cartesian equation of the bisector of the angle that these lines form?
|
The unit direction vectors of the first line and the second line are
$$\overline a=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$$
and
$$\overline b=\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}\right),$$
respectively. (The scalar product of these two vectors is $0$. That is, our lines are perpendicular.)
(Proof: Take $x=0$ and $x=1$. In the case of the first line we have $(0,-1,0)$ for $x=0$ and $(1,0,-1)$ for $x=1$. Subtracting the first vector from the second one: a direction vector is $(1,1,-1)$; the length of this vector is $\sqrt{3}$. Then, in the case of the second line if $x=0$ then $(0,-1,-1)$ , if $x=1$ then we have $(1,0,1)$ . Subtracting the first vector from the second one we get $(1,1,2)$, the length is then $\sqrt{6}$.)
Summing $\overline a$ and $\overline b $ we get the following vector
$$\overline c=\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}},\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{6}}\right).$$
I claim that $(\overline c, \overline a, \overline b)$ are coplanar and $\overline c$ is their angle bisector. This follows from the paralelogramma rule but it can be checked by taking the following scalar products $$(\overline a,\overline c)=(\overline b,\overline c)=1.$$ So, these angles equal. (Don't be mistaken, the length of $\overline c$ is not $1$. So, the value $(1)$ of the scalar products does not tell exactly the value of the angle, it tells only that the angles equal. But we know that the angle is $\frac{\pi}{4}.$
The two lines meet at the point
$$\overline m=\left(\frac{1}{3},-\frac{2}{3},-\frac{1}{3}\right).$$
As a result the angle bisector line in parametric form is
$$\overline l= \overline m+ t\overline c, \ -\infty \le t \le \infty.$$
Or
$$x(t)=\frac{1}{3}+t\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}}\right) \tag 1,$$
$$y(t)=-\frac{2}{3}+t\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}}\right) \tag 2,$$
$$z(t)=-\frac{1}{3}+t\left(-\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{6}}\right) \tag 3.$$
In order to get the Cartesian equation of the angle bisector line, first, we express $t$ from $(1)$:
$$t=\frac{x-\frac{1}{3}}{\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}}} \tag 4$$
Subtracting $(2)$ from $(1)$ we get
$$x-y=1. \tag 5$$
Comparing $(3)$ and $(4)$ gives that
$$z=-\frac{1}{3}+\left(x-\frac{1}{3}\right)\frac{2-\sqrt 2}{1+\sqrt 2} \tag 6.$$
Finally, $(5)$ and $(6)$ are the Cartesian equations of the angle bisector.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1308388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prove that $\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c$ How to prove that
\begin{equation*}\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c,\ where \ a,b,c>0\end{equation*}
I tried the following:
\begin{equation*}abc(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge a+b+c\end{equation*}
Using Chebyshev's inequality
\begin{equation*}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\le3(\frac{1}{a}\frac{1}{a}+\frac{1}{b}\frac{1}{b}+\frac{1}{c}\frac{1}{c})\end{equation*}
from first inequality follows
\begin{equation*}\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2abc\ge a+b+c\end{equation*}
equivalent to
\begin{equation*}\frac{abc}{3(a+b+c)}\ge (\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^2\end{equation*}
and by amplifying both members by 9
\begin{equation*}\frac{3abc}{a+b+c}\ge (\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^2\end{equation*}
now using mean inequality
\begin{equation*}\sqrt[3]{abc}\ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\end{equation*}
the inequality in question becomes
\begin{equation*}3abc\ge (a+b+c)(\sqrt[3]{abc})^2\end{equation*}
which yields
\begin{equation*}3\sqrt[3]{abc}\ge a+b+c\end{equation*}
not what I wanted.
|
The inequality which you want to prove is symmetric, so we can assume without loss of generality that $a\geq b\geq c$.
Then $ab\geq ac\geq bc$ and $\frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a}$.
Thus, from the rearrangement inequality, we get that
$$\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq\frac{ab}{b}+\frac{ac}{a}+\frac{bc}{c}=a+c+b$$
which is what we wanted to show.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1309677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$? So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then:
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 173$$
Can anyone explain why this is the case?
|
$\left(2x^2+x\right)^2=\left(2x^2+x\right)\left(2x^2+x\right)=4x^4+4x^3+x^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1309780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
}
|
simple 2 sides inequality
$$2<\frac{x}{x-1}\leq 3$$
Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
|
The simplest, as we have a homographic function, is to write it in canonical form:
\begin{align*}
2<\frac{x}{x-1}\leq 3&\iff 2 < 1+\frac1{x-1}\leq 3 \iff 1< \frac1{x-1}\leq 2\\
&\iff \frac12\le x-1 <1 \iff \frac32 \le x <2
\end{align*}
The third equivalence is valid because all numbers at the end of the first line have the same sign.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1310939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
}
|
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
|
Below is a simple way to implement the (Euclidean) denominator descent implicit in Ivo Terek's answer (which was John Conway's favorite way to present this proof).
Theorem $\quad \rm r = \sqrt{n}\:$ is an $\rm\color{#c00}{integer}$ if rational,$\:$ for $\:\rm n\in\mathbb{N}$
Proof $\ \ \ $ Put $\ \ \displaystyle\rm r = \frac{A}B ,\;$ least $\rm\; B>0.\,$ $\ \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A.\ $ Taking fractional parts: $\rm\displaystyle\ \frac{b}B = \frac{a}A\ $ for $\rm\ 0 \le b < B.\ $ If $\,\rm\displaystyle\ \color{#c00}{B\nmid A}\,$ then $\rm\ b\ne 0,\ $ so $\rm\,\ \displaystyle \frac{A}B = \frac{a}b,\ $ contra leastness of $\,\rm B$.
Remark $\ $ See here for a conceptual view of the proof (principality of denominator ideals). Though the proof may seem unusual to those who have not yet studied advanced number theory, it is quite natural once one learns about conductor and denominator ideals.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 18
}
|
Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$. Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$.
What are all integral solutions of this equation?
I can only find $x=y=z=1$.
|
From the identity
$$(x + y + z)^3 = x^3 + y^3 + z^3 + 3(x + y)(y + z)(z + x)$$
we obtain $8 = (x+y)(y+z)(z+x)$. It follows that $(3−x)(3−y)(3−
z) = 8$. On the other hand,
$$(3−x)+(3−y)+(3−z)−3(x+y+z) = 6,$$
implying that either $|3−x|, |3−y|, |3−z|$ are all even, or exactly one of
them is even. In the first case, we get $|3 −x| = |3−y| = |3−z| = 2$,
yielding $x, y, z \in \{1, 5\}$. Because $x+y+z = 3$, the only possibility is
$x = y = z = 1$.
In the second case, one of $|3−x|, |3−y|, |3−z|$ must be $8$, say $|3−x| = 8$, yielding $x \in \{−5, 11\}$ and $|3−y| = |3−z| = 1$, from which $y, z \in \{2, z\}$. Taking into account that $x+y +z = 3$, the only possibility is $x = −5$ and $y = z = 4$. In conclusion, the desired
triples are
$$(1, 1, 1), (−5, 4, 4), (4,−5, 4), \text{ and }(4, 4,−5).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1312183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Power Series Coefficients Find the sum of the coefficients of $x^{20}$ and $x^{21}$ in the power series expansion of $\frac 1{(1-x^3)^4}$.
I don't know a lot on power series at the moment, and I was wondering how do I find the coefficients?
Thanks
|
Recall that
$$\frac1{1-s} = \sum_{n=0}^\infty s^n. $$
Differentiating we have
$$\frac{\mathsf d^3}{\mathsf ds^3}\left[\frac1{1-s}\right] = \frac6{(1-s)^4},
$$
and
$$\begin{align*}
\frac{\mathsf d^3}{\mathsf ds^3}\left[\sum_{n=0}^\infty s^n\right] &= \sum_{n=0}^\infty \frac{\mathsf d^3}{\mathsf ds^3}[s^n]\\
&= \sum_{n=0}^\infty n(n-1)(n-2)s^{n-3}\\
&= \sum_{n=0}^\infty (n+1)(n+2)(n+3)s^n.
\end{align*}$$
Hence,
$$\frac1{(1-s)^4} = \sum_{n=0}^\infty \frac16(n+1)(n+2)(n+3)s^n. $$
Put $s=x^3$, then
$$\frac1{(1-x^3)^4} = \sum_{n=0}^\infty \frac16(n+1)(n+2)(n+3)x^{3n}. $$
It is clear that $x^{20}=0$ and
$$x^{21} = \frac16(7+1)(7+2)(7+3) = 120. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1313649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Evaluate $\lim\limits_{n\to\infty}\prod\limits_{k=2}^{n}\frac{k^2+k-2}{k^2+k}$ I can't find the product of a sequence.
We have
$$\frac{(2+2)(2-1)}{2(2+1)}\frac{(3+2)(3-1)}{3(3+1)}...\frac{(k+2)(k-1)}{k(k+1)}$$
I am stuck with $$P=\frac{2(n+2)}{n^2(n-1)}$$ but that isn't correct. Can the squeeze theorem be used?
|
$$\prod_{k=2}^{K}\frac{k^2+k-2}{k^2+k}=\prod_{k=2}^{K}\frac{k-1}{k}\prod_{k=2}^{K}\frac{k+2}{k+1}=\frac{1}{K}\cdot\frac{K+2}{3}$$
hence the limit equals $\color{red}{\displaystyle\frac{1}{3}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1313718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Find the asymptotic behavior of solutions of the equation Find the asymptotic behavior of solutions $y$ of the equation
$$x^5 + x^2y^2=y^6,$$
which tends to $0$ when $x$ tends to $0$.
My solution: if $y=Ax^n$, then
$$x^5 + A^2x^{2+2n}=A^6x^{6n}.$$
If $2+2n\ge 5$, we should have $5=6n$, but $2+2\cdot5/6 \not> 5$.
If $2+2n< 5$, we have $2+2n=6n$, $n=1/2$; $A^2=A^6\Longrightarrow A=1$, and $y\sim \sqrt x$ when $x\to 0$.
Am I right? How to find the next terms of the expansion of $y$?
|
Now lets use a correction term
$$y=\sqrt x + f$$
Insert and use only the first two terms in the binomial expansion
$$x^5+x^2(x+2x^{0.5} f)=x^3+6x^{2.5}f$$
$$x^5=4x^{2.5}f$$
$$f=\frac{x^{2.5}}{4}$$
Now, you may want to insert $y=\sqrt x +\frac{x^{2.5}}{4}+f$
Adding (or multiplying) a correction term and inserting into the equation usually works well.
The order of the colors are blue orange green red
Beware, I simply guessed the 0.0275
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1314681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
$\frac{1}{{9\choose r}} -\frac{1}{{10\choose r}} = \frac{11}{6{11\choose r}}$. Is there a way to find $r$ without using algebra?
$$\frac{1}{\dbinom 9r} -\frac{1}{{\dbinom{10}r}} = \frac{11}{6\times \dbinom{11}r}$$
I guess directly applying algebra for this problem would be enough. But are there any simpler and prettier approaches to finding $r$ ?
|
This method is very simple :
$\frac{1}{{9\choose r}} -\frac{1}{{10\choose r}} =\frac{{10\choose r}-{9\choose r}}{{10\choose r}{9\choose r}}=\frac{\frac{{10\choose r}-{9\choose r}}{{9\choose r}}}{{10\choose r}}=\frac{(\frac{{10\choose r}-{9\choose r}}{{9\choose r}})\frac{11*6}{11-r}}{{10\choose r}\frac{11*6}{11-r}}=\frac{11}{6{11\choose r}}\Rightarrow \\
11=(\frac{{10\choose r}-{9\choose r}}{{9\choose r}})\frac{11*6}{11-r}\Rightarrow \\
1=(\frac{{10\choose r}-{9\choose r}}{{9\choose r}})\frac{6}{11-r}\Rightarrow \\
1=(\frac{10}{10-r}-1)\frac{6}{11-r}\Rightarrow \\
1=(\frac{r}{10-r})\frac{6}{11-r}\Rightarrow \\
r^2-21r+110=6r\Rightarrow...$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1316148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Solving for n in the equation $\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$
Solving for $n$ in the equation $$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$$
Can anyone show me a numerical method step-by-step to solve this?
Thanks
|
Classic problem for Newton's Method.
Note that $F(n) = \dfrac{1}{2^n} + \dfrac{1}{4^n} + \dfrac{3^n}{4^n}-1$ is continuously differentiable. Furthermore, $F(0) = 2$ and $\lim_{n \rightarrow \infty} F(n) =-1$ so by the Intermediate Value Theorem, there is definitely a zero.
I used Maple to calculate the root to a tolerance of $10^{-15}$ because I can.
with(Student[NumericalAnalysis]):
F(z):= n-> 1/2^n + 1/4^n + 3^n/4^n:
Newton(F(z),n=0.2,tolerance=10^-15)
>>>1.73050735785764
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1316957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 6
}
|
Integrate $f(y)=\int_{0}^{\frac{\pi}{2}} \ln(y^2 \cos^2x+ \sin^2x) .dx$ This Integral came up while attempting another question:
$$f(y)=\int_{0}^{\frac{\pi}{2}} \ln(y^2 \cos^2x+ \sin^2x) .dx$$
The suggested solution was as follows:
$$f'(y) = 2y \int_{0}^{\pi/2}\frac{cos^{2}x}{sin^{2}x + y^{2}cos^{2}x}dx$$
$$= 2y \int_{0}^{\pi/2}\frac{dx}{tan^{2}x + y^{2}}$$
$$= 2y \int_{0}^{\pi/2}\frac{sec^{2}x - tan^{2}x }{tan^{2}x + y^{2}}dx$$
$$= 2y . \frac{1}{y} tan^{-1}( \frac{1}{y}) |_{0}^{\infty} -2y\frac{\pi}{2} + y^{2}f'(y)$$
$$f'(y) = \frac{\pi}{1 + y}$$
Unfortunately, I was unable to understand the how to get the last 3 steps. As per my understanding,
$$f'(y)= 2y \int_{0}^{\pi/2}\frac{sec^{2}x - tan^{2}x }{tan^{2}x + y^{2}}dx$$
Splitting into 2 integrals,
$$= 2y\int_0^{\pi/2}\dfrac{\sec^2 x}{\tan^2 x + y^2}dx-2y \int_{0}^{\pi/2}\frac{tan^{2}x }{tan^{2}x + y^{2}}dx$$
Substituting $\tan(x)=u$ in the first integral,
$$ 2y\int_0^{\pi/2}\dfrac{\sec^2 x}{\tan^2 x + y^2}dx$$
$$= 2y\int_0^{\infty}\dfrac{du}{u^2 + y^2}$$
$$=2y\times \dfrac{1}{y}\tan^{-1}(\dfrac{u}{y})|_0^\infty$$
However, this does not seem to agree with the given solution. Also, I have no idea as to how to integrate $-2y \int_{0}^{\pi/2}\frac{tan^{2}x }{tan^{2}x + y^{2}}dx$
ould somebody please be so kind as to point out my error in computing the first integral and also help me integrate the second integral? Many, many thanks in advance!
|
Your calculation of $2y\int_0^{\pi/2}\sec^2 x/(\tan^2 x + y^2)\, dx$ is correct so far, and the value is $\pi$. Now
\begin{align}2y\int_0^{\pi/2} \frac{\tan^2 x}{\tan^2 x + y^2}\, dx &= 2y\int_0^{\pi/2} \left(1 - \frac{y^2}{\tan^2 x + y^2}\right)\, dx \\
&= 2y\cdot\frac{\pi}{2} - y^2\cdot 2y\int_0^{\pi/2} \frac{1}{\tan^2 x + y^2}\, dx\\
&= \pi y - y^2 f'(y),
\end{align}
and therefore
$$2y\int_0^{\pi/2} \frac{\sec^2 x - \tan^2 x}{\tan^2x + y^2}\, dx = \pi - \pi y + y^2f'(y),$$
that is,
$$f'(y) = \pi - \pi y + y^2 f'(y).$$
Solving for $f'(y)$,
$$f'(y) = \frac{\pi - \pi y}{1 - y^2} = \frac{\pi(1 - y)}{(1 - y)(1 + y)} = \frac{\pi}{1 + y}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1317167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
a limit problem? $$\lim_{x\to 0}{e^x+ \ln{1-x\over e }\over \tan x-x} $$
i tried doing l'hospital. but couldn't do ! could anyone help ?
|
This solution uses the standard limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{e^{x}-1-x-\frac{1}{2}x^{2}}{x^{3}} &=&\frac{1}{6}
\\
\lim_{x\rightarrow 0}\frac{\ln (1-x)+x+\frac{1}{2}x^{2}}{x^{3}} &=&-\frac{1}{%
3} \\
\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3}.
\end{eqnarray*}
Re-write the original expression as follows
\begin{eqnarray*}
\frac{e^{x}+\ln \left( \frac{1-x}{e}\right) }{\tan x-x} &=&\frac{e^{x}+\ln
(1-x)-\ln e}{\tan x-x} \\
&=&\frac{\left( e^{x} {\color{red}{-1-x-\frac{1}{2}x^{2}}}\right) +
{\color{red}{ \left( 1+x+\frac{1}{2}x^{2}\right)}}+\left( \ln (1-x)+
\color{blue}{ x+\frac{1}{2}x^{2}}\right) -\color{blue}{
(x+\frac{1}{2}x^{2}) }
-1}{\tan x-x} \\
&=&\frac{\left( e^{x}-1-x-\frac{1}{2}x^{2}\right) +\left( \ln (1-x)+x+\frac{1%
}{2}x^{2}\right) }{\tan x-x} \\
&=&\left( \frac{e^{x}-1-x-\frac{1}{2}x^{2}}{x^{3}}+\frac{\ln (1-x)+x+\frac{1%
}{2}x^{2}}{x^{3}}\right) \left( \frac{x^{3}}{\tan x-x}\right)
\end{eqnarray*}
passing to the limit one gets
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{e^{x}+\ln \left( \frac{1-x}{e}\right) }{\left(
\tan x-x\right) } &=&\lim_{x\rightarrow 0}\left( \frac{e^{x}-1-x-\frac{1}{2}%
x^{2}}{x^{3}}+\frac{\ln (1-x)+x+\frac{1}{2}x^{2}}{x^{3}}\right) \left( \frac{%
x^{3}}{\tan x-x}\right) \\
&=&\left( \lim_{x\rightarrow 0}\frac{e^{x}-1-x-\frac{1}{2}x^{2}}{x^{3}}%
+\lim_{x\rightarrow 0}\frac{\ln (1-x)+x+\frac{1}{2}x^{2}}{x^{3}}\right)
\left( \lim_{x\rightarrow 0}\frac{x^{3}}{\tan x-x}\right) \\
&=&\left( \frac{1}{6}-\frac{1}{3}\right) \left( \frac{3}{1}\right) \\
&=&-\frac{1}{2}.
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1324019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$
without using L'Hospital's rule.
Attempted solution:
First we multiply with the conjugate expression:
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) = \lim_{x \to \infty} \frac{x^2 + 3x - (x^2 + 1)}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Simplifying gives:
$$\lim_{x \to \infty} \frac{3x - 1}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Breaking out $\sqrt{x^2}$ from the denominator and $x$ from the numerator gives:
$$\lim_{x \to \infty} \frac{x(3 - \frac{1}{x})}{x(\sqrt{1 + 3x} + \sqrt{1 + 1})} = \lim_{x \to \infty} \frac{3 - \frac{1}{x}}{\sqrt{1 + 3x} + \sqrt{2}}$$
The result turns out to be $\frac{3}{2}$, but unsure how to proceed from here.
|
Your extraction of $x$ is incorrect: what you need is
$$ \sqrt{x^2+3x} = \sqrt{x^2(1+3/x)} = x\sqrt{1+3/x}. $$
Then you have
$$ \lim_{x \to \infty} \frac{3x-1}{x(\sqrt{1+3/x}+\sqrt{1+1/x^2})} = \lim_{x \to \infty} \frac{3}{\sqrt{1+3/x}+\sqrt{1+1/x^2}} - \frac{1}{x(\sqrt{1+3/x}+\sqrt{1+1/x^2})} $$
The square roots in this expression all tend to $1$, so the first term tends to $3/2$, the second to $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1324388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
How does one calculate: $\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$ How does one calculate:
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$$
Is the best way to just take the first term times the following two, and the second two times the next two to see the pattern?
First term will contribute:
$$\left(\frac{z^2}{2!2!}-\frac{2z^4}{2!4!}+\frac{2z^6}{2!6!}-\cdots\right)$$
$$\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
Second term will contribute:
$$\left(-\frac{z^4}{4!}+\cdots\right)$$
This already seems wrong, so expanding via the first term seems good, and I hazard a guess that
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2=\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
|
Instead of working forward, work backward:
Calculate the $z^2$ term of the result, then the $z^3$ term, and so on.
Clearly the coefficients of the $z^0$ and $z^1$ terms are zero. The $z^2$ term must be the result of squaring the $\frac z{2!}$ term, so the result is $\frac{z^2}{2!^2}$.
Now let's consider the $z^3$ term. To obtain one we'd need to multiply a $z^0$ term by a $z^3$ term, or a $z^1$ term by a $z^2$ term. But there is no $z^0$ or $z^2$ term, so there is no $z^3$ term in the result.
The $z^4$ term is more interesting. To obtain a $z^4$ term, we'd need to multiply a $z^0$ term by a $z^4$ term, a $z^1$ term by a $z^3$ term, or a $z^2$ term by a $z^2$ term. Only the second of these works, and we can take the $z^1$ term either from the left factor (getting $\frac z {2!}\cdot -\frac{z^3}{4!}$) or the right factor (getting $ -\frac{z^3}{4!}\cdot \frac z {2!}$). The $z^4$ term of the result is the sum of these two products, $$r_4 = -2\frac{z^4}{2!4!}.$$
Similarly, the $z^6$ term is obtained as $$\frac z{2!}\frac{z^5}{6!} + \left(-\frac{z^3}{4!}\right) \left(-\frac{z^3}{4!}\right) +\frac{z^5}{6!} \frac z{2!} \\
= \left(\frac{2}{2!6!}+\frac1{4!4!}\right)z^6 = \ldots
$$
In general, for any product of infinite (or finite!) power series, one has $$\left(\sum_{i=0}^\infty a_iz^i\right) \left(\sum_{i=0}^\infty b_iz^i\right) = \sum_{i=0}^\infty \left(\sum_{j=0}^i a_jb_{i-j}\right)z^i$$
where the important thing to notice is that the inner-sum on the right-hand side is finite and is a constant; it is the coefficient of the $z^i$ term in the product.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1324472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
How to retrieve the expression of $a^3+b^3+c^3$ in terms of symmetric polynomials? I recently had to find without any resources the expression of $a^3+b^3+c^3$ in terms of $a+b+c$, $ab+ac+bc$ and $abc$.
Although it's easy to see that $a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$, I couldn't come up myself with $a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc$.
Can someone explain to me how to retrieve this formula with only pen and paper, or how to memorize it for good ?
|
In terms of the products of the three polynomials, the only possible way of getting an $a^3$ term is from $$(a+b+c)^3=a^3+b^3+c^3+3\sum a^2b+6abc$$
Where the sum is of all the expressions of the same kind. To obtain $a^2b$ from the symmetric polynomials, you need $a\cdot ab$ which comes from $$(a+b+c)\cdot(ab+bc+ca)=\sum a^2b+3abc$$ and substituting the expression for $\sum a^2b$ into the first equation gives $$(a+b+c)^3=a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc$$ Which is what you want. Note that we started by eliminating the term in which $a$ appeared to the highest power - $a^3$ - and then went to $a^2$ and then we were done.
For higher degrees and more variables there is an entirely systematic process which works as a generalisation of this method.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1325263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Prove that $a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3$
Prove that $a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3$
Values $a,b,c$ are all positive reals. I tried Muirhead and a few AM $\geq$ GM.
This problem is equivalent to proving $a^4b^3 + b^4c^3 + c^4a^3 \geq c^2a^5 + b^2c^5+a^2b^5$.
|
The inequality does not change if $(a, b, c)$ are permuted
cyclically, therefore we can assume
that $a \le b \le c$ or $a \ge b \ge c$.
In the first case $a \le b \le c$ we can
apply the Rearrangement inequality to
$$
x_1 = a^4, x_2 = b^4, x_3 = c^4 \\ y_1 = a^3, y_2 = b^3, y_3 = c^3 \, .
$$
The rearrangement inequality states that
$$
x_1y_2 + x_2y_3 + x_3y_1 \le
x_1y_1 + x_2y_2 + x_3y_3
$$
which is exactly the desired inequality.
In the second case $a \ge b \ge c$ we can do the same with
$$
x_1 = c^3, x_2 = b^3, x_3 = a^3 \\ y_1 = c^4, y_2 = b^4, y_3 = a^4 \, .
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1325652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Is it true that two $3 \times 3$ matrices in $R$ are similar $\iff$ they have same determinant and same trace? Is it easy to cite some example of $3 \times 3 $ non-similar matrices of real entries with equal determinant and trace?
|
Even simpler examples than Chappers' one would be
$$ \begin{pmatrix} 0&0&0\\0&0&0\\0&0&0 \end{pmatrix} \quad\text{and}\quad
\begin{pmatrix} 0&0&1\\0&0&0\\0&0&0 \end{pmatrix} $$
(determinant and trace both $0$, but the zero matrix is similar only to itself) or
$$ \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1 \end{pmatrix} \quad\text{and}\quad
\begin{pmatrix} 1&1&0\\0&1&1\\0&0&1 \end{pmatrix} $$
(determinant $1$, trace $3$, but the identity matrix is similar only to itself).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1326370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Points of intersection of two functions Through the following steps I found the x-coordinates of the intersection points of two functions:
$(x)= -x^{2}+3x+1\: and\: g(x)=3/x $
The numbers I found are x=(1, 2, 3)
But on the graph, one of the points has a negative x value, could you guys point me to anything I have missed in my calculations.
$-x^{2}+3x+1=3/x
\\x(-x^{2}+3x+1)=3
\\x=3
\\and
\\-x^{2}+3x+1=3
\\x^{2}-3x-1=-3
\\x^{2}-3x-1+3=-3+3
\\x^{2}-3x+2=0
\\(x-1)(x-2)=0
\\So
\\x=1 \:and\: x=2
$
|
$x(-x^{2}+3x+1)=3\quad$ doesn't imply $x=3$ or $-x^{2}+3x+1=3$. Instead of that we have
$$x(-x^{2}+3x+1)=3\iff -x^3+3x^2+x-3=0\iff-(x+1)(x-1)(x-3)=0$$
Therefore the graphs of the functions mets at $x=-1$, $x=1$ and $x=3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1326758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site
states:
Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
“The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”
In other words, according to Example $1$: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.$$
Should:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}$$
not be:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}$$
as everything in the left-hand side is cubed?
|
$$1^3 +2^3 +\ldots +n^3 =\left(\frac{n(n+1)}{2}\right)^2.$$
We will prove by induction on $n$, that
$$\sum_{k=1}^n k^3 = 1^3 +2^3 +\ldots +n^3 =\frac{n^2 (n+1)^2}{4}.$$
For $n=1$, we have $1^3 = \dfrac{1^2 2^2}{4} = 1$.
We shall prove that
$$\sum_{k=1}^{n+1} k^3 = 1^3 +2^3 +\ldots +n^3 +(n+1)^3 =\frac{(n+1)^2 (n+2)^2}{4},$$
by assuming that
$$\sum_{k=1}^n k^3 = 1^3 +2^3 +\ldots +n^3 =\frac{n^2 (n+1)^2}{4}.$$
From the induction supposition, we have to prove that
$$\frac{n^2 (n+1)^2}{4} +(n+1)^3 = \frac{(n+1)^2 (n+2)^2}{4},$$
or
$$(n+1)^3 = \frac{(n+1)^2 (n+2)^2}{4} -\frac{n^2 (n+1)^2}{4} = \frac{(n+1)^2}{4} ((n+2)^2 -n^2) = (n+1)^3.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1328798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
}
|
List the elements of the field $K = \mathbb{Z}_2[x]/f(x)$ where $f(x)=x^5+x^4+1$ and is irreducible Since $\dim_{\mathbb{Z}_2} K = \deg f(x)=5$, $K$ has $2^5=32$ elements. So constructing the field $K$, I get:
\begin{array}{|c|c|c|}
\hline \text{polynomial} & \text{power of $x$} & \text{logarithm} \\\hline
0 & 0 & -\infty \\\hline
1 & 1 & 0 \\\hline
x & x & 1 \\\hline
x^2 & x^2 & 2 \\\hline
x^3 & x^3 & 3 \\\hline
x^4 & x^4 & 4 \\\hline
x^4+1 & x^5 & 5 \\\hline
x^4+x+1 & x^6 & 6 \\\hline
x^4+x^2+x+1 & x^7 & 7 \\\hline
x^4+x^3+x^2+x+1 & x^8 & 8 \\\hline
x^3+x^2+x+1 & x^9 & 9 \\\hline
x^4+x^3+x^2+x & x^{10} & 10 \\\hline
x^3+x^2+1 & x^{11} & 11 \\\hline
x^4+x^3+x & x^{12} & 12 \\\hline
x^2+1 & x^{13} & 13 \\\hline
x^3+x & x^{14} & 14 \\\hline
x^4+x^2 & x^{15} & 15 \\\hline
x^4+x^3+1 & x^{16} & 16 \\\hline
x+1 & x^{17} & 17 \\\hline
x^2+x & x^{18} & 18 \\\hline
x^3+x^2 & x^{19} & 19 \\\hline
x^4+x^3 & x^{20} & 20 \\\hline
\end{array}
Have I constructed the field correctly? I'm sure the claim that $K$ has $32$ elements is true, yet when I actually construct the field, I only get $22$ elements as $x^{21} = x^0 = 1$.
|
Elaborating more on the answer of Burde and generalizing:
Let $F$ be any finite field with $q$ elements, and $f(x)$ an irreducible polynomial from $F[X]$, say of degree $m$. Then the quotient ring by the principal ideal generated by $f(x)$ is in fact a field, has $q^m$ elements and the elements of this field is the set of all polynomials of degree${}<m$ with coefficients in $F$ (rather the cosets represented by them).
They are $q^m$ in number: this follows from the fact that two polynomials of degree less than $m$have their difference also a polynomial of degree less than $m$ and hence cannot be a multipleof a polynomial of degree $m$.
They are all the elements of the field because: the coset represented by a general polynomial $g(x)$ of degree$ \geq m$ is the same as the coset of $r(x)$ where $r(x)$ is obtained by Euclidean division with remainder: $g(x) = h(x) f(x) + r(x)$. (That is try to divide $g(x)$ by $f(x)$, getting quotient $h(x)$ and a remainder polynomial of degree less than the degree of $f(x)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1329867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Solving the definite integral: $\int_0^1 x\sqrt{px + 1}\,dx$ with $p>0$?
How to integrate $$I = \int_0^1 x\sqrt{px + 1}\,dx$$ , $p>0, p\in\mathbb{R}$ ?
I tried this:
$$ t = \sqrt{px + 1} \implies x = \frac{t^2 - 1}{p}$$
$$ x = 0 \implies t = 1$$
$$ x = 1 \implies t = \sqrt{p+1}$$
so I have:
$$ I = \int_1^\sqrt{p+1} \frac{t^2 - 1}{p}t \,dt = \frac{1}{p}\int_1^\sqrt{p+1} t^3 - t \,dt = \frac{1}{p}\bigg[\frac{t^4}{4}\bigg\vert_1^\sqrt{p+1} - \frac{t^2}{2}\bigg\vert_1^\sqrt{p+1}\bigg]$$
$$ I = \frac{1}{p} \bigg[\frac{1}{4}(p^2 + 2p + 1 - 1) - \frac{1}{2}(p + 1 - 1) \bigg] = \frac{1}{p}\frac{1}{4}p^2 = \frac{1}{4}p $$
So for $p=4$, the integral should have the value $1$, but WA doesn't agree: http://wolfr.am/5p-i9uOA
I looked at it for quite a while but I can't locate the mistake. Since $p>0$, everything seems fine to me but apparently it isn't.
|
$$I(p)=\int_{0}^{1}x\sqrt{px+1}\,dx = \frac{1}{p^2}\int_{0}^{p}x\sqrt{x+1}\,dx $$
but integration by parts gives:
$$ \int x\sqrt{x+1}\,dx = \frac{2}{15}(3x-2)(1+x)^{3/2} $$
hence:
$$ I(p) = \frac{4}{15p^2}+\frac{2}{15}\sqrt{1+p}\left(3+\frac{1}{p}-\frac{2}{p^2}\right).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1330334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Compute $\prod\limits _{r=3 }^{\infty }{ \frac { { r }^{ 3 }-{ 8 } }{ { r }^{ 3 }+{ 8 } } } $ How should I go about evaluating this product? I have not been able to figure out.
$$\lim_{n\to\infty}\prod _{r=3 }^{n }{ \frac { { r }^{ 3 }-{ 8 } }{ { r }^{ 3 }+{ 8 } } } $$
|
We have
$$\begin{align}
\prod_{n=3}^{N}\frac{n^3-8}{n^3+8}&=\prod_{n=3}^{N}\frac{(n-2)(n^2+2n+4)}{(n+2)(n^2-2n+4)}\\\\
&=\prod_{n=3}^{N}\frac{(n-2)}{(n+2)}\prod_{n=3}^{N}\frac{(n+1)^2+3}{(n-1)^2+3}\tag 1
\end{align}$$
Now, analyzing the individual partial products we see that the first partial product reduces to
$$\begin{align}
\prod_{n=3}^{N}\frac{(n-2)}{(n+2)}&=\frac{1}{5}\frac{2}{6}\frac{3}{7}\frac{4}{8}\frac{5}{9}\cdots\frac{N-6}{N-2}\frac{N-5}{N-1}\frac{N-4}{N}\frac{N-3}{N+1}\frac{N-2}{N+2}\\\\
&=\frac{(1)(2)(3)(4)}{(N-1)(N)(N+1)(N+2)} \tag 2
\end{align}$$
while the second partial product reduces to
$$\begin{align}
\prod_{n=3}^{N}\frac{(n+1)^2+3}{(n-1)^2+3}&=\frac{4^2+3}{2^2+3}\frac{5^2+3}{3^2+3}\frac{6^2+3}{4^2+3}\cdots \frac{(N-1)^2+3}{(N-3)^2+3}\frac{(N)^2+3}{(N-2)^2+3}\frac{(N+1)^2+3}{(N-1)^2+3}\\\\
&=\frac{(N)^2+3}{2^2+3}\frac{(N+1)^2+3}{3^2+3}\tag 3
\end{align}$$
Putting $(2)$ and $(3)$ into $(1)$ yields
$$\begin{align}
\prod_{n=3}^{N}\frac{n^3-8}{n^3+8}&=\frac27 \frac{(N^2+3)((N+1)^2+3))}{(N+2)(N+1)(N)(N-1)}
\end{align}$$
from which we can easily see that
$$\begin{align}
\bbox[5px,border:2px solid #C0A000]{\prod_{n=3}^{\infty}\frac{n^3-8}{n^3+8}=\lim_{N\to \infty}\prod_{n=3}^{N}\frac{n^3-8}{n^3+8}=\frac27}
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1333130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$ then quotient $\frac xy$ is equal to?
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$, then quotient $\frac xy$ is equal to?
Other conditions are ($\alpha \neq k\pi,\ \alpha \neq \frac \pi2+ k\pi,\ k\in\mathbb Z,\ i^2 = -1$). I tried this by making nominator the difference of squares but it does not lead me anywhere. Solution for this task is $-2-4\cos{2\alpha}$
|
$$(2x+i)(1-i\sin 3\alpha)=(y+i)(1+i\sin \alpha) \Rightarrow (2x+\sin 3\alpha)+i(1-2x\sin 3\alpha) = (y-\sin\alpha)+i(1+y\sin \alpha)$$
Equate the imaginary parts,
$$\Rightarrow \frac{x}{y}=-\frac{\sin \alpha}{2\sin 3\alpha}=-\frac{\sin \alpha}{2(\sin \alpha \cos 2\alpha+\cos \alpha \sin 2\alpha)}$$
Use $\sin2\alpha =2\sin\alpha\cos\alpha$ and $\cos 2\alpha=2\cos^2\alpha-1$
$$\Rightarrow \frac{x}{y}=-\frac{1}{2(\cos 2\alpha +2\cos^2\alpha)}=-\frac{1}{2(2\cos 2\alpha+1)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1333611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Find $\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t $ without using L'hopital's rule Find $$\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t $$ without using L'hopital's rule. Using the rule, I got it as 1/12. What's a way to do it without L'hopital?
|
Integration by parts is a reasonable choice. Since:
$$ \int \frac{t}{t^4+4}\,dt = \frac{1}{4}\arctan\left(\frac{t^2}{2}\right)\tag{1}$$
we have:
$$ \int_{0}^{x}\frac{t\log(1+t)}{t^4+4}=\frac{\log(1+x)}{4}\arctan\left(\frac{x^2}{2}\right)-\frac{1}{4}\int_{0}^{x}\frac{\arctan\frac{t^2}{2}}{1+t}\,dt \tag{2}$$
and by considering the Taylor series of $\log(1+x)$ and $\arctan\left(\frac{x^2}{2}\right)$ in a neighbourhood of the origin we easily get:
$$\lim_{x\to 0}\frac{1}{x^3} \int_{0}^{x}\frac{t\log(1+t)}{t^4+4}=\frac{1}{4}\cdot\frac{1}{2}-\frac{1}{4}\cdot\frac{1}{6}=\color{red}{\frac{1}{12}}\tag{3}$$
as wanted.
As pointed by Patrick Da Silva in the comments, we can also skip the integration-by-parts step, since we are just integrating $\frac{t^2}{4}+o(t^2)$ between $0$ and $x$, hence we trivially have $\frac{x^3}{12}+o(x^3)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1334221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
}
|
How to proceed with evaluating $\int\frac{dx}{\sqrt{9+4x^2}}$ and $\int\tan^2(3x)dx$
*
*$\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}$
*$\displaystyle\int\tan^2(3x)dx$
For the first one i'm not sure if I did it correctly, here is what I did:
Let $2x=3\tan(t)$, so $x=\frac{3}{2}\tan(t)$ and $dx=\frac{3}{2}\sec^2(t)dt$. So by substitution,
$$\begin{align}\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}&=\int\frac{\frac{3}{2}\sec^2(t)}{3\sec(t)}dt\\
&=\frac{1}{2}\int\sec(t)dt\\
&=\frac{1}{2}\left(\ln|\sec(t)+\tan(t)| + C\right)\\
&=\frac{1}{2}\left(\ln\bigg|\frac{\sqrt{9+4x^2}}{3}+\frac{2x}{3}\bigg|+C\right)\\
\end{align}$$
For the second one, i'm unable to proceed, what I did was
$\displaystyle\int\tan^2(3x)dx=\int\frac{\sin^2(3x)}{\cos^2(3x)}dx=\int\frac{\frac{1}{2}(1-\cos(6x)}{\frac{1}{2}(1+\cos(6x)}dx=\int\frac{(1-\cos(6x)}{(1+\cos(6x)}dx$
is this the right way to proceed?
Thanks
|
Put $3x=u
$ to get $$\frac{1}{3}\int\tan^{2}\left(u\right)du=\frac{1}{3}\int\left(\sec^{2}\left(u\right)-1\right)du=\frac{1}{3}\tan\left(u\right)-\frac{u}{3}+C=\frac{1}{3}\tan\left(3x\right)-x+C.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1336662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Quadrature on segment Is there a quadrature formula on the segment [0,1], such that on [0,1/2] the points and weights are symetric with respect to 1/4, on [1/2,1] they are symetric with respect to 3/4, and such that the formula exactly integrates polynomials of degree 5. If not, what maximal order can be expected ?
|
A simple example of such quadrature is
$$
x = \frac{1}{2} + \frac{1}{4}\begin{pmatrix}
-1 - \sqrt\frac{3}{5}&
-1&
-1 + \sqrt\frac{3}{5}&
1 - \sqrt\frac{3}{5}&
1&
1 + \sqrt\frac{3}{5}
\end{pmatrix}\\
w = \frac{1}{36}\begin{pmatrix}5 & 8 & 5 & 5 & 8 & 5\end{pmatrix}
$$
which is just a combination of two three-point Gaussian rules (which are exact for polynomials of degree 5).
It is easy to check that five-point rule of the form you've requested
$$
x = \begin{pmatrix}\frac{1}{4}-a & \frac{1}{4} + a & \frac{3}{4} - b & \frac{3}{4} & \frac{3}{4} + b\end{pmatrix}\\
w = \begin{pmatrix}
w_1 & w_1 & w_2 & w_3 & w_2
\end{pmatrix}
$$
gives an inconsistent system of moment equations, solvable only if requested degree is 4.
$$
w_1 = \frac{1}{4}\quad w_2 = \frac{5}{52}\quad w_3 = \frac{4}{13}\\
a = \frac{1}{4\sqrt{3}}\qquad b = \frac{\sqrt{13}}{4\sqrt{15}}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1339971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find $f\circ f$ if $f(t)=\dfrac{t}{(1+t^2)^{1/2}},\ \ t\in \mathbb{R}$
Find $f\circ f$ if $f(t)=\dfrac{t}{(1+t^2)^{1/2}},\ \ t\in \mathbb{R}$
$ a.)\ \dfrac{1}{(1+2t^2)^{1/2}} \\
\color{green}{ b.)\ \dfrac{t}{(1+2t^2)^{1/2}}}\\~\\
c.)\ (1+2t^2)\\~\\
d.)\ \text{none of these} \\$
I tried to put the value $t=1$ and after checking concluded that it's $d.)$ but the book is giving option $b.)$.
I look for a short and simple way.
I have studied maths up to 12th grade.
|
Our formula for $f$ is
$$f(\Box)=\frac{\Box}{\sqrt{1+\Box^2}}$$
With this in mind we have
\begin{align*}
f\bigl(f(t)\bigr)
&=f\left(\fbox{$f(t)$}\right) \\
&=\frac{\fbox{$f(t)$}}{\sqrt{1+\fbox{$f(t)$}^2}} \\
&=\frac{\frac{t}{\sqrt{1+t^2}}}{\sqrt{1+\left(\frac{t}{\sqrt{1+t^2}}\right)^2}} \\
&=\frac{t}{\sqrt{1+\frac{t^2}{1+t^2}}\cdot\sqrt{1+t^2}} \\
&=\frac{t}{\sqrt{\left(1+t^2\right)+\frac{t^2}{1+t^2}\cdot\left(1+t^2\right)}} \\
&=\frac{t}{\sqrt{1+t^2+t^2}} \\
&= \frac{t}{\sqrt{1+2\,t^2}}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1340714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Given primitive solution to $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$, show $a+b$ is a perfect square If $a,b,c$ are positive integers and
$\gcd(a,b,c)$ is $1$. Given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ then prove that $a+b$ is a perfect square.
I was trying to get something useful from the information given in the question but was unable to get that.
|
Let $d=(a,b)$ be the g.c.d. of $a$ and $b$. We will show that $a+b=d^2$. Write $a=a_1d$ and $b=b_1d$. You also have $(a_1,b_1)=1$. We thus have $$\frac{1}{a}+\frac{1}{b}=\frac{1}{d}\left(\frac{1}{a_1}+\frac{1}{b_1}\right)=\frac{1}{c}$$$$\frac{a_1+b_1}{a_1b_1}=\frac{d}{c}$$ Observing that $(d,c)=1$ and $(a_1+b_1,a_1b_1)=1$, we get that $a_1+b_1=d$ and $a_1b_1=c$. But then $$a+b=d(a_1+b_1)=d^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1341162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Finding an isomorphism between polyomial quotient rings Let $F_1 = \mathbb{Z}_5[x]/(x^2+x+1)$ and $F_2 = \mathbb{Z}_5[x]/(x^2+3)$. Note neither $x^2+x+1$ nor $x^2+3$ has a root in $\mathbb{Z}_5$, so that each of the above are fields of order 25, and hence they are isomorphic from elementary vector space theory, however I'm tasked with constructing an explicit isomorphism between the two fields, so I applied the technique I've learned here and elsewhere as follows.
First, let $\phi: F_1 \rightarrow F_2$ by $x \mapsto ax+b$ (as the base field is fixed it suffices to find the image of $x$). Then we should have that $\phi(x^2+x+1) = x^2 + 3$, and the left hand side reduces to $\phi(x)^2 + \phi(x) + 1$ as $\phi$ is a homomorphism. Simplifying, we have $(ax+b)^2 + (ax+b) + 1 = a^2x^2 + 2abx + b^2 + ax+b+1 = a^2x^2 + (2ab+a)x + (b^2+b+1) = x^2+3$.
However, this appears to be a problem. Notably, since $a^2 = 1$ mod 5, we have $a = 1,4$. Picking the former, we have $2b+1 = 0$ and thus $b = 2$, a problem as $2^2+2+1 = 7 \neq 1$ mod 5. Trying $a=4$, we have $3b+4 = 0$ so $b = 2$, and again the same problem.
Are there flaws in my technique or arithmetic that I don't see? If not, what other options do I have to attempt to construct the isomorphism?
|
Hint $\ $ Complete the square $\ 4(x^2\!+\!x\!+\!1) = (2x\!+\!1)^2\!+\!3 = X^2\!+\!3,\ $ for $\,X = 2x\!+\!1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1341458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Find the last two digits of $33^{100}$
Find the last two digits of $33^{100}$
By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of @Lucian:
$33^2\equiv-11 \pmod{100}$ then $33^{100}\equiv(-11)^{50}\pmod{100}\equiv (10+1)^{50}\pmod{100}$
By using the binomial expansion, we have:
$33^{100}\equiv (10^{50}+50\cdot 10^{49}+ \cdots + 50\cdot 10+1)\pmod{100}$
$\implies 33^{100}\equiv (50\cdot 10+1)\pmod{100}\equiv 01 \pmod{100}$
|
Another approach:
$$\text{Euler's formula: }\quad a^{\phi(n)}\equiv 1 \pmod{n} \text{ when} \gcd(a,n)=1$$
$$\phi(100)=\phi(2^2)\phi(5^2)=2\cdot 20=40$$
$$\gcd(100,33)=1$$
$$33^{40}\equiv 1 \pmod {100}$$
$$33^{100}\equiv (33^{40})^{2}3^{20}\equiv 3^{20}\pmod {100}$$
$$\equiv (3^5)^4\equiv43^4\equiv 49^2\equiv 01\pmod {100}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1342127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
}
|
Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ where $\omega^3 = 1$ If $\omega^3 = 1$ and $\omega \neq 1$, then compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$
I'm pretty lost, I don't really know where to start.
Thanks
|
$$(1 - \omega + \omega^2)(1 + \omega - \omega^2)$$
$$=((1 + \omega^2) - \omega)(1 + \omega - \omega^2)$$
$$=(-\omega - \omega)(-\omega^2 - \omega^2)$$
$$=(-2\omega)(-2\omega^2)$$
$$=4\omega^3=4$$
Where $1+\omega+\omega^2=0$ and $\omega^3=1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1342333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Trigonometry - log/ln and absolute sign in equations Will this equation still hold if the absolute sign is being used at different places
For example,
This trigonometry identity;
$\frac{-1}{3}log|\frac{cos3x+1}{cos3x-1}|=\frac{2}{3}ln|(\frac{sin\frac{3x}{2}}{\cos\frac{3x}{2}})|$.
My question is
Does $\frac{-1}{3}log\frac{|cos3x|+1}{|cos3x-1|}=\frac{2}{3}ln(\frac{sin\frac{3x}{2}}{\cos\frac{3x}{2}})$?
Or even $\frac{-1}{3}log\frac{cos3x+1}{|cos3x-1|}=\frac{2}{3}ln(\frac{sin\frac{3x}{2}}{\cos\frac{3x}{2}})$?
Quite confused. Thank you in advance :)
|
$$1+cos(x)=2cos^(\frac{x}{2})\\1+cos(x)=2sin^2(\frac{x}{2})$$ so $$\frac{|cos(3x)+1|}{|cos(3x)-1|} =|\frac{cos(3x)+1}{cos(3x)-1}|=\frac{|2cos^2(\frac{3x}{2})|}{|-2sin^2(\frac{3x}{2})|}$$ as we know $|ab|=|a||b| \rightarrow |-a|=|-1||a|=|a|$
$$\frac{|2cos^2(\frac{3x}{2})|}{|-2sin^2(\frac{3x}{2})|}=\frac{|2cos^2(\frac{3x}{2})|}{|+2sin^2(\frac{3x}{2})|}=\\\frac{|cos^2(\frac{3x}{2})|}{|sin^2(\frac{3x}{2})|}=\frac{cos^2(\frac{3x}{2})}{sin^2(\frac{3x}{2})}=\\(\frac{cos(\frac{3x}{2})}{sin(\frac{3x}{2})})^2$$and now $$\log (\frac{cos(\frac{3x}{2})}{sin(\frac{3x}{2})})^2=\log(|\frac{cos(\frac{3x}{2})}{sin(\frac{3x}{2})}|)^2=2log(|\frac{cos(\frac{3x}{2})}{sin(\frac{3x}{2})}|)=-2\log (|\frac{sin(\frac{3x}{2})}{cos(\frac{3x}{2})}|)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1345587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
How to solve $(z^1+z^2+z^3+z^4)^3$ using Pascals Triangle? In an exercise it seems I must use Pascal's triangle to solve this $(z^1+z^2+z^3+z^4)^3$. The result would be $z^3 + 3z^4 + 6z^5 + 10z^ 6 + 12z^ 7 + 12z^ 8 + 10z^ 9 + 6z^ {10} + 3z^ {11} + z^{12}$. But how do I use the triangle to get to that result? Personally I can only solve things like $(x+y)^2$ and $(x+y)^3$.
Thanks for any tips that may be given.
|
The expression factors as
$$z^3(1+z)^3(1+z^2)^3=z^3(1+3z+3z^2+z^3)(1+3z^2+3z^4+z^6).$$
Then I see no better way than to perform the multiply (though there is a symmetry)
$$\begin{align}
&1+3z+&3z^2+&z^3\\
&&3z^2+&9z^3+&9z^4+&3z^5\\
&&&&3z^4+&9z^5+&9z^6+&3z^7\\
&&&&&&z^6+&3z^7+&3z^8+z^9\\
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1345643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Quadratic polynomials describe the diagonal lines in the Ulam-Spiral I'm trying to understand why is it possible to describe every diagonal line in the Ulam-Spiral with an quadratic polynomial $$2n\cdot(2n+b)+a = 4n^2 + 2nb +a$$ for $a, b \in \mathbb{N}$ and $n \in 0,1,\ldots$.
It seems to be true but why?
Wikipedia says: "The pattern also seems to appear even if the number at the center is not 1 (and can, in fact, be much larger than 1). This implies [WHY?] that there are many integer constants b and c such that the function: $4n^2+bn+c$ as $n$ counts up $\{1, 2, 3, ...\}$, a number of primes that is large by comparison with the proportion of primes among numbers of similar magnitude."
I can't find a source with a detailed explanation.
I found these equations:
So here is the solution:
\begin{align*}
y_t - y_{t+1} - (y_{t+1} - y_{t+2}) &= 8\\
y_t - 2y_{t+1} + y_{t+2} &= 8\\
y_{t+2} - 2y_{t+1} + y_t &= 8
\end{align*}
1) We solve $y_{t+2} - 2y_{t+1} + y_t = 0.$
Let $y_t = A\beta^t$ \begin{align*}
A\beta^{t+2} - 2A\beta^{t+1} + A\beta^t &= 0\\
A\beta^{t}\cdot (\beta^2 - 2\beta + 1) &= 0
\end{align*}
$\beta^2 - 2\beta + 1 = 0$ has two identical solutions $\beta_{1,2} = 1$.
So with $A_1$ and $A_2t$ we get $$y_t = A_1 + A_2t.$$
2) $1 + a_1 + a_2 = 0$ and $a_1 = -2$ so let $y_t = ct^2$
\begin{align*}
c\cdot(t+2)^2 - 2c\cdot(t+1)^2 + ct^2 &= 8\\
c\cdot\big(t^2+4t+4 - 2\cdot(t^2+2t+1) + t^2\big) &= 8\\
c\cdot(t^2+4t+4 - 2t^2-4t-2 + t^2) &= 8\\
2c &= 8\\
c &= 4
\end{align*}
So $y_t = 4t^2$
3) The complete solution is $$y_t = 4t^2 + A_2t + A_1.$$
The "exclusion lines" seem to be interesting too:
$$4n^2+n$$ $$4n^2+3n$$ $$4n^2+3n-1$$ $$4n^2-n$$ seem not to have any primes at all.
Useful website I found a bit late: http://ulamspiral.com
|
Reading georgmierau comment, I've been trying to reach this "arbitrary" 8, so here what I've done:
Let $A(n)$ be the cardinal number of the $n^{th}$ ring in Ulam's spiral.
$$A(1)=1\\A(2)=8=(2\cdot2-1)^2-1^2\\A(3)=16=(2\cdot3-1)^2-3^2\\ \vdots \\ A(n)=(2n-1)^2-(2n-3)^2\Rightarrow A(n)=8n-8, n\neq1$$
So, $A(n+1)-A(n)=8$
That means that each time we pass to the next ring, its cardinal number will be the same of the last one added by 8.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1347560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
What the difference between the smallest two numbers from these numbers? There are infinitely many integers $n$ bigger than $1$, such that if we divide $n$ by any integer $k$ where $2\leq k\leq 11$, the remainder is equal to $1$.
What the difference between the smallest two such integers?
Help guys please. this is my try, the smallest number will be $lcm+1$ $$= 2^3 * 3^2 * 5*7*11+1$$ and the next number will be $$2*lcm+1$$
the difference between them $=icm=27720$
|
What you wan can be set up as:
$x\equiv1\bmod 2$
$x\equiv 1\bmod 3$
$x\equiv1\bmod 4$
$x\equiv 1\bmod 5$
$x\equiv1\bmod 6$
$x\equiv 1\bmod 7$
$x\equiv1\bmod 8$
$x\equiv 1\bmod 9$
$x\equiv1\bmod 10$
$x\equiv 1\bmod 11$
which is equivalent to
$x\equiv 1 \bmod 2^3$
$x\equiv 1 \bmod 3^2$
$x\equiv 1 \bmod 5$
$x\equiv 1 \bmod 7$
$x\equiv 1 \bmod 11$
which by CRT has unique solution $\bmod 2^3\cdot3^2\cdot 5\cdot 11$. The answer thus is $2^3\cdot3^2\cdot 5\cdot7 \cdot11$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1349779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Simultaneous equation with fractional solutions. How do you get to find $x$ when $y$ is a fraction ? Anyone mind to explain it step by step for the clearest explanation.=)
$$-7x +2y = 2$$
$$14x + 3y = -5$$
Answer: $x=?, y=-1/7$
|
The correct solution to your system of equations is $y = -\frac{1}{7}$ and you want to find $x$. To do so, substitute this value of $y$ into the first equation to get $$-7x - 2\cdot \frac{1}{7} = 2 \iff -7x = \frac{16}{7} \iff x = -\frac{16}{49}$$ by re-arranging and solving for $x$.
Alternatively, you could substitute your $y$ value into the second equation to get $$14x - 3\cdot \frac{1}{7} = -5 \iff 14x = -\frac{32}{7} \iff x = -\frac{16}{49}$$ by re-arranging and solving for $x$.
So the solution to your system of equations is $x = -\frac{16}{49}, y = -\frac{1}{7}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1350126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Finding $\sum\limits_{k=0}^n k^2$ using summation by parts Sorry to bother you guys again, but I still have some doubts. I do think I'm making some progress, though.
So, again, the formula that I'm using for summation by parts is
$\sum\limits_{k=o}^n f(k)g(k) = g(n)(\sum\limits_{k=0}^n f(k)) - \sum\limits_{k=0}^{n-1} [\Delta g(k) \sum\limits_{i=0}^k f(i)]$
Using this, I'm trying to find $\sum\limits_{k=0}^n k^2$. I think I almost got it, but there are some problems that I still can't get around.
Plugging in $g(k) = k$ and $f(k)=k$, we end up with the following equations:
\begin{align}
\sum\limits_{k=0}^n k^2 &= \sum\limits_{k=0}^n kk\\
\sum\limits_{k=0}^n k^2 &= n (\sum\limits_{k=0}^n k) - \sum\limits_{k=0}^{n-1}[\sum\limits_{i=0}^k i]\\
\sum\limits_{k=0}^n k^2 &= n(\frac{n(n+1)}{2}) - \sum\limits_{k=0}^{n-1}[\frac{k(k+1)}{2}]\\
\sum\limits_{k=0}^n k^2 &= \frac{n^3 + n^2}{2} - \frac{\sum\limits_{k=0}^{n-1} k^2+k}{2}\\
2\sum\limits_{k=0}^n k^2 &= n^3 + n^2 - \sum\limits_{k=0}^{n-1} k^2+k\\
2\sum\limits_{k=0}^n k^2 &= n^3 + n^2 - \sum\limits_{k=0}^{n-1} k^2+ \sum\limits_{k=0}^{n-1}k
\end{align}
Anyway, my problem is, I need the upper indexes of the sums in the right hand side to be $n$, not $n-1$ (I've already checked that, if the indexes are $n$, then I can find the desired result). Is there any trick I'm missing to make those indexes go up by $1$? Or am I missing an obvious step?
|
As you probably know:
$$\sum_{k=0}^{n-1} k = \frac{(n-1)^2+(n-1)}{2} = \frac{n^2-2n+1+n-1}{2} = \frac{n^2-n}{2}$$
You also know:
$$\sum_{k=0}^{n}k^2=n^2+\sum_{k=0}^{n-1}k^2 \rightarrow \sum_{k=0}^{n-1} k^2=\sum_{k=0}^{n} k^2 - n^2$$
So now you can substitute:
$$2\sum_{k=0}^n k^2=n^3+n^2-\left(\sum_{k=0}^n k^2-n^2\right)-\frac{n^2-n}{2}$$
Pass this over and you are good to go!
$$3\sum_{k=0}^n k^2=n^3+n^2+n^2-\frac{n^2-n}{2}$$
$$\sum_{k=0}^n k^2=\frac{n^3+\frac{3}{2}n^2-\frac{n}{2}}{3}=\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1351843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
}
|
Let $a$, $b$, and $c$ be positive real numbers. Let $a$, $b$, and $c$ be positive real numbers. Prove that
$$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$
Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?
|
An Algebraic Solution:
Note that $$\sqrt{a^2-ab+b^2}=\sqrt{\left(a-\frac{b}{2}\right)^2+\left(\frac{\sqrt{3}b}{2}\right)^2}$$ and $$\sqrt{a^2-ac+c^2}=\sqrt{\left(\frac{c}{2}-a\right)^2+\left(\frac{\sqrt{3}c}{2}\right)^2}\,.$$ Using the Triangle Inequality in the form $\sqrt{\sum_{i=1}^nx_i^2}+\sqrt{\sum_{i=1}^ny_i^2}\geq \sqrt{\sum_{i=1}^n\left(x_i+y_i\right)^2}$ for real numbers $x_i$'s and $y_i$'s, we have $$\sqrt{a^2-ab+b^2}+\sqrt{a^2-ac+c^2}\geq \sqrt{\Bigg(\left(a-\frac{b}{2}\right)+\left(\frac{c}{2}-a\right)\Bigg)^2+\left(\frac{\sqrt{3}b}{2}+\frac{\sqrt{3}c}{2}\right)^2}\,.$$ Expanding the right-hand side, we get $\sqrt{a^2-ab+b^2}+\sqrt{a^2-ac+c^2}\geq\sqrt{b^2+bc+c^2}$. The equality holds if and only if there exists $\lambda \geq 0$ such that $\lambda\left(a-\frac{b}{2},\frac{\sqrt{3}b}{2}\right)=\left(\frac{c}{2}-a,\frac{\sqrt{3}c}{2}\right)$ or $\left(a-\frac{b}{2},\frac{\sqrt{3}b}{2}\right)=(0,0)$ (equivalently, $(a,b,c)=(0,0,c)$ or $(a,b,c)=\left(\frac{\lambda b}{\lambda+1},b,\lambda b\right)$ for some $\lambda \geq 0$).
Although the geometric solution by thkim1011 is a more elegant solution, it is more complicated if $a$, $b$, or $c$ can be negative (although you can prove geometrically with essentially the same method, you have to deal with many cases). My algebraic solution is better in this sense.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1353184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
What are the ways to solve trig equations of the form $\sin(f(x)) = \cos(g(x))$? if I have the following trig equation:
$$\sin(10x) = \cos(2x)$$
I take the following steps to solve it:
*
*I rewrite $\cos(2x)$ as $\sin\left(\frac{\pi}{2} + 2x\right)$ or as $\sin\left(\frac{\pi}{2} - 2x\right)$ cause $\sin\left(\frac{\pi}{2} - a\right) = \sin\left(\frac{\pi}{2} + a\right) = \cos(a)$;
*Let's say that I have chosen $\sin\left(\frac{\pi}{2} + 2x\right)$, the equation becomes:
$$\sin(10x) = \sin\left(\frac{\pi}{2} + 2x\right).$$
*
*Then, I know that:
$$\sin(f(x)) = \sin(g(x)) \Leftrightarrow f(x) = g(x) + 2\pi n, n \in \mathbb{Z} \lor f(x) = (\pi - g(x)) + 2\pi k, k \in \mathbb{Z}$$
This means that (for $f(x) = 10x$ and $g(x) = \frac{\pi}{2} + 2x$):
$$\sin(10x) = \cos(2x) \Leftrightarrow 10x = \frac{\pi}{2} + 2x + 2\pi n, n \in \mathbb{Z} \lor 10x = (\pi - (\frac{\pi}{2} + 2x)) + 2\pi k, k \in \mathbb{Z}$$
Solving, I get the following results:
$$x_{1} = \frac{\pi}{16} + \frac{\pi}{4}n,\,\,x_{2} = \frac{\pi}{24} + \frac{\pi}{6}k,\,\,\,\,n,k \in \mathbb{Z}$$
Now, are there any other methods for solving such equations or could this one be just fine?
|
You may prefer to transform the sine into cosine:
$$
\cos\left(\frac{\pi}{2}-10x\right)=\cos(2x)
$$
This splits into two:
$$
\frac{\pi}{2}-10x=2x+2k\pi
$$
or
$$
\frac{\pi}{2}-10x=-2x+2k\pi
$$
The trick is that $\cos\alpha=\cos\beta$ if and only if $\alpha=\beta+2k\pi$ or $\alpha=-\beta+2k\pi$ (with integer $k$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1354442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
How do I evaluate this integral :$\int \frac{1}{ \sqrt{1-x^2} } \arctan( \frac{\sqrt{1-x^2}}{2})dx$? Is there somone who can show me how do I evaluate this integral :$$\int \frac{1}{ \sqrt{1-x^2} } \arctan\left(\frac{\sqrt{1-x^2}}{2}\right)dx$$
Note: I took : $x=\cos t$ , but it didn't work
Thank you for any help .
|
For the integral
\begin{align}
I = \int \frac{1}{ \sqrt{1-x^2} } \, \arctan\left(\frac{\sqrt{1-x^2}}{2}\right) \, dx
\end{align}
let $u^{2} = 1 - x^{2}$ to obtain
\begin{align}
I = - \, \int \tan^{-1}\left(\frac{u}{2}\right) \, \frac{du}{\sqrt{1-u^{2}}}.
\end{align}
Using $2 \tan^{-1}(x) = \ln(1 - ix) - \ln(1+ix)$ then after the transformation $2 t = i u$ the integral becomes
\begin{align}
I = i \, \int \ln\left( \frac{1-t}{1+t} \right) \, \frac{dt}{\sqrt{1 + 4 t^{2}}}.
\end{align}
Making use of integration by parts where
\begin{align}
u = \ln\left(\frac{1-x}{1+x}\right) \hspace{10mm} dv = \frac{dt}{\sqrt{1 + 4 t^{2}}}
\end{align}
to obtain
\begin{align}
-i \, I = \frac{1}{2} \, \ln\left(\frac{1-t}{1+t}\right) \, \sinh^{-1}(2t) - \int \frac{\sinh^{-1}(2t) \, dt}{t^{2} -1}.
\end{align}
From here use Wolfram Alpha to reduce the necessary terms.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1359461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
Let $0\le a\le b\le c,abc=1$, then show that
$$a+b^2+c^3\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}$$
Things I have tried so far:
$$\dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}=\dfrac{b^2c^3+ac^3+ab^2}{bc^2}$$
Since $abc=1$, it suffices to prove that
$$c+b^3c^2+bc^5\ge b^2c^3+ac^3+ab^2$$
then the problem is solved. I stuck in here.
|
we can write the inequality as
$$\left(a-\dfrac{1}{c^3}\right)+\left(c^3-\dfrac{1}{a}\right)\ge\dfrac{1}{b^2}-b^2$$
or
$$(c^3a-1)\left(\dfrac{1}{a}+\dfrac{1}{c^3}\right)\ge\dfrac{1-b^4}{b^2}\tag{1}$$
since $c\ge 1,bc\ge 1$,then
$$c^3a\ge c\cdot abc=c\ge 1$$
so it only prove $b\le 1$case.
$1$.case if $b^2\ge a$，then we have
$$c^3a+b^4-2\ge a(c^3+b^3)-2\ge abc(b+c)-2=b+c-2\ge 0$$
then we have
$$c^3a-1\ge 1-b^4$$
and
$$\dfrac{1}{a}\ge\dfrac{1}{b^2}$$
so $(1)$ prove done.
2. if $a\ge b^2$,we have
$$c^3a-1-c(1-b^4)\ge c^3b^2+cb^4-c-1\ge c^3b^2+cb^4-c(b^2c)^{4/3}-(b^2c)^{5/3}\ge 0$$
so we have
$$c^3a-1\ge c(1-b^4)$$
and
$$\dfrac{c}{a}-\dfrac{1}{b^2}\ge\dfrac{b-a}{ab^2}\ge 0$$
so
$$(c^3a-1)\left(\dfrac{c}{a}+\dfrac{1}{c^2}\right)\ge\dfrac{c(1-b^4)}{b^2}$$
so $(1)$ prove by done
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1360349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
If $A$, $B$ are $3\times 3$ matrices, and all elements are different from each other and greater in their absolute value than 3, then is $AB \ne 0$?
Let $A$ and $B$ be two $3\times 3$ matrices. All entries of $A$ are distinct and all entries of $B$ are distinct. All entries in both matrices are greater in their absolute value than 3. Then $AB \ne 0$.
Is this true or false, and why?
|
$$\begin{pmatrix} 4 & 8 & 12 \\ 5 & 10 & 15 \\ 7 & 14 & 21 \end{pmatrix}
\begin{pmatrix} -28 & -35 & -49 \\ 8 & 10 & 14 \\ 4 & 5 & 7 \end{pmatrix}
=
\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
The idea was to start with
$$\begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}
\begin{pmatrix} -7 & -7 & -7 \\ 2 & 2 & 2 \\ 1 & 1 & 1 \end{pmatrix} $$
and then experiment multiplying rows and columns respectively by constants until the elements in each respective matrix is distinct. You can read of those constants from the first column and last row: $4, 5, 7$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1360753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Way to make this homogoneous ODE seperable? Is my algebra correct, turning this:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4y-3x}{2x-y}$$
Into this:
I split into the difference of the two fractions,
then factored x out of the left fraction, and factored y
out of the right fraction, getting:
$$\frac{4\frac{y}{x}}{x(1-\frac{y}{x})} - \frac{3\frac{x}{y}}{2\frac{x}{y}-1}$$
Now I can substitute $v = \frac{y}{x}, v^{-1}= \frac{x}{y}$, right?
Then separate? Then integrate?
|
The usual way of doing such a question is to divide the top and bottom of the fraction by $x$ yielding $$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4\frac{y}{x} - 3}{2 - \frac{y}{x}}$$ and then use the substitition $v = \frac{y}{x} \implies y = vx$ so that $\frac{\mathrm{d}y}{\mathrm{d}x} = v + x\frac{\mathrm{d}v}{\mathrm{d}x}$ to get $$v + x\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{4v}{2-v} - \frac{3}{2-v}$$ can you take it from there? If not, you can continue by subtracting $v$ from both sides, giving you $$x \frac{\mathrm{d}v}{\mathrm{d}x} = - \frac{(v+3)(v-1)}{v-2} \implies \int \frac{2-v}{(v-1)(v+3)} \, \mathrm{d}v = \ln |x| + c$$ where partial fraction decomposition on the rational function of $v$ yields $$\frac{2-v}{(v-1)(v+3)} = \frac{1}{4(v-1)} - \frac{5}{4(v+3)}$$ Hence we get $$\frac{1}{4} \ln |v-1| - \frac{5}{4}\ln |v+3| = \ln |x| + c$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1362653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.