Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving an exponential with three different bases $$2^x+4^x=8^x$$ Solve for $x$. I reduced the bases to $2$ and try to use logarithms but could not get passed the logarithm of an expression. When typing into online calculators they say there is no solution but graphing shows an answer.
| Set $y = 2^x$, then you get
\begin{align}
y+y^2= y^3 \ \ \implies \ \ \ y(1+y-y^2) = 0 \ \ \implies \ \ y=0 \ \ \text{ or } \ \ y = \frac{1\pm\sqrt{5}}{2}.
\end{align}
Hence it follows that
\begin{align}
y = 2^x=\frac{1+\sqrt{5}}{2} \ \ \implies \ \ x = \log_2\left(\frac{1+\sqrt{5}}{2}\right).
\end{align}
Note that $y \neq 0$ nor does it equal $(1-\sqrt{5})/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find the natural solutions of $a^3-b^3=999$ I want to find the natural solutions of $a^3-b^3=999$.
I got $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$, so if we consider the equation in $\mathbb{Z}/3\mathbb{Z}$ we get
$$(a-b)\cdot(a^2+ab+b^2) \equiv0 \text{ mod }3$$
and because $\mathbb{Z}/3\mathbb{Z}$ is a domain, we get
$$a\equiv b \text{ mod } 3 \text{ or } a^2+ab+b^2\equiv0 \text{ mod } 3.$$
Besides, the prime factorization of $999=3^3\cdot37$, but I don't know how to go on.
I would appreciate any hints.
| By Fermat Little Theorem
$$x^3 \equiv x \pmod{3}$$
Therefore
$$0 \equiv a^3-b^3 \equiv a-b \pmod{3}$$
This shows that $a-b=3k$.
Then
$$3^3 \cdot 37 =a^3-b^3=(a-b)(a^2+ab+b^2)=(a-b)((a-b)^3+3ab)=3k(9k^2+3ab) \Rightarrow \\
3 \cdot 37=k(3k^2+ab)$$
Since $k <3k^2+ab$ the only posibilities are
$$k=1 \\
3k^2+ab=3 \cdot 37$$
or
$$k=3 \\
3k^2+ab= 37$$
This leads to
$$a-b=3k=3 \\
ab=108$$
or
$$a-b=9 \\
ab=10$$
which are easy to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
What is $\lim_{x\to 2} \frac{\sqrt{x+2}-2}{x-2}$? I tried multiplying by the conjugate which gave:
$$\frac{x-2}{(x-2)\sqrt{x+2}+2x-4}$$
But i'm still gettting $\frac{0}{0}$. According to my textbook the answer should be $\frac{1}{4}$, but how do I get there?
| Let $x=2+h$ and let $h \rightarrow 0$ hence
$$
\frac{\sqrt{4+h}-2}{h}=\frac{2\sqrt{1+\frac{h}{4}}-2}{h}=\frac{2\left(1+\frac{h}{8}-1+o\left(h\right)\right)}{h} \underset{h \rightarrow 0}{\rightarrow}\frac{1}{4}
$$
Hence
$$
\frac{\sqrt{x+2}-2}{x-2} \underset{x \rightarrow 2}{\rightarrow}\frac{1}{4}]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Number of solution is twice $(x,y)$ Problem: Count the number of $2 \times 2$ matrices $A$ with $A^TA=-I$ in $Z_p$ for $p>2$.
Answer: if $p$ is an odd prime, the number of such matrices $A$ is twice the number of solutions $(x,y)$ to the congruence $x^2+y^2 \equiv -1 \pmod p$.
What's the reason behind "twice"?
| Let
\begin{eqnarray*}
A=
\begin{bmatrix}
a &b \\c &d \\
\end{bmatrix} .
\end{eqnarray*}
Then we require
\begin{eqnarray*}
\begin{bmatrix}
a &c \\b &d \\
\end{bmatrix}
\begin{bmatrix}
a &b \\c &d \\
\end{bmatrix} =
\begin{bmatrix}
-1 &0 \\0 &-1 \\
\end{bmatrix} .
\end{eqnarray*}
So
\begin{eqnarray*}
a^2+c^2 \equiv -1 \pmod{p} \\
ab+cd \equiv 0 \pmod{p} \\
b^2+d^2 \equiv -1 \pmod{p} \\
\end{eqnarray*}
After a little algebra with these
\begin{eqnarray*}
a^2b^2=b^2(-1-c^2)=c^2d^2 \\
c^2(\underbrace{d^2+b^2}_{-1})=-b^2
\end{eqnarray*}
So $b=c$ or $b=p-c$. Giving two solutions for each pair $(a,c)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Stretching an ellipse along major or minor axis Consider the ellipse given by:
$$
Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0.
$$
What is the equation of an ellipse which has major and minor axis equal to $p$ times the major and minor axis length of the above ellipse.
My attempt is as follows:
We can remove rotation, increase axis length and then rotate back. An example of rotation is given below:
Rotating a conic section to eliminate the $xy$ term.
I am wondering if there is less complicated intuition into this problem or less complicated way.
| Referring to the standard results here, the centre is given by
$$(h,k)=
\left(
\frac{2CD-BE}{B^2-4AC}, \frac{2AE-BD}{B^2-4AC}
\right)$$
and the transformed conics is
$$\frac{A+C \color{red}{\pm} \sqrt{(A-C)^{2}+B^{2}}}{2} X^2+
\frac{A+C \color{red}{\mp} \sqrt{(A-C)^{2}+B^{2}}}{2} Y^2+
\frac
{\det
\begin{pmatrix}
A & \frac{B}{2} & \frac{D}{2} \\
\frac{B}{2} & C & \frac{E}{2} \\
\frac{D}{2} & \frac{E}{2} & F
\end{pmatrix}}
{\det
\begin{pmatrix}
A & \frac{B}{2} \\
\frac{B}{2} & C \\
\end{pmatrix}}=0$$
It's just simply re-scaling the constant term $F$, that is
$$Ax^2+Bxy+Cy^2+Dx+Ey+\color{blue}{F'}=0$$ where
$$p^2
\det
\begin{pmatrix}
A & \frac{B}{2} & \frac{D}{2} \\
\frac{B}{2} & C & \frac{E}{2} \\
\frac{D}{2} & \frac{E}{2} & F
\end{pmatrix}
=
\det
\begin{pmatrix}
A & \frac{B}{2} & \frac{D}{2} \\
\frac{B}{2} & C & \frac{E}{2} \\
\frac{D}{2} & \frac{E}{2} & \color{blue}{F'}
\end{pmatrix}$$
On solving,
$$\color{blue}{F'}=
\frac{1}
{\det
\begin{pmatrix}
A & \frac{B}{2} \\
\frac{B}{2} & C \\
\end{pmatrix}}
\left[
\frac{AE^2+CD^2-BDE}{4}+p^2
\det
\begin{pmatrix}
A & \frac{B}{2} & \frac{D}{2} \\
\frac{B}{2} & C & \frac{E}{2} \\
\frac{D}{2} & \frac{E}{2} & F
\end{pmatrix}
\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Irrational Computation of a Rational Cubic Root. \begin{align}
& x^3+5x-18=0\\
\text{Let } & x=a+b\\
& (a+b)^3+5(a+b)-18 =0\\
& a^3+b^3+(3ab+5)(a+b)-18=0
\end{align}
Taking $a$ and $b$ such that $3ab+5=0$.
We have, $a^3+b^3-18=0$
\begin{align}
& a^3-\frac{5^3}{(3a)^3}-18=0\\
& a^6-18a^3-\frac{125}{27}=0\\
& a^3=9+\sqrt{81+\frac{125}{27}}\\
& a=\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}
\end{align}
Therefore, $x=\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}+\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}$.
I can't solve the last equation without computing the value of the square-root within the cube root. But, by doing that I can only get an approximate value of $x$, not the real value, which is $2$. Is there a way to solve the above equation so that one arrives at the value $2$ and not an approximation?
| $$x^3+5x-18=x^3-2x^2+2x^2-4x+9x-18=(x-2)(x^2+2x+9),$$
which gives $x=2$ because $$x^2+2x+9=(x+1)^2+8>0.$$
We need to prove that $$\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}+\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}=2$$ or
$$9+\sqrt{81+\frac{125}{27}}+9-\sqrt{81+\frac{125}{27}}-2^3+3\cdot2\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}\cdot\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}=0$$ or
$$10+6\sqrt[3]{9^2-9^2-\frac{125}{27}}=0,$$
which is obvious.
I used $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and the following fact:
$$a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}\sum_{cyc}(a-b)^2=0$$ for $a=b=c$ only.
In our case $a=\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}$, $b=\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}$ and $c=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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check if $\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k$ converges How to check if $\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k$ converges?.
$\begin{align}
\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k &=
\lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k \\
&= \lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7\frac{k}{k}-2\frac{1}{k}}{8\frac{k}{k}-3 \frac{\sqrt{k}}{k}}\right)}^k \\
&= \lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7-2\frac{1}{k}}{8-3 \frac{\sqrt{k}}{k}}\right)}^k \\
\end{align}$
Am I on the right track ? If so, how should I go on ? I need some help to do it without the root test.
| HINT
Note that eventually
$$0<\frac{7k-2}{8k-3 \sqrt{k}}\le c<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Addition of $2$ Events
Let $X$ and $Y$ be independent, each uniformly distributed on $\{1, 2, ..., n\}$. Find $P(X + Y = k)$ for $2 \le k \le 2n$.
\begin{align}P(X + Y = k) &= \sum_{(x,y)\,:\,x+y=k} P(x, y) \\
&= \sum_{(x,y)\,:\,x+y=k} \frac{1}{n^2} \\
&= (k - 1)\frac{1}{n^2} \\
&= \frac{k-1}{n^2} \end{align}
When $k = 2: (1, 1)$
When $k = 3: (1, 2), (2, 1)$
When $k = 4: (1, 3), (3, 1), (2, 2)$
When $k = 5: (1, 4), (4, 1), (2, 3), (3, 2)$
$$\#(x, y) = k - 1$$
Textbook Answer:
$\frac{k-1}{n^2}\,\,\,$ for $\,\,\,2 \le k \le n+1$
$\frac{2n-k+1}{n^2}\,\,\,$ for $\,\,\,n+2 \le k \le 2n$
Why are there $2$ intervals being considered?
| We have $x+y = k$ where $x,y \in \{ 1, \ldots, n\}$.
We need not always have $k-1$ pairs of $(x,y)$. In particular, if we let $x=1$, we will need to let $y=k-1$, however, this need not always be possible, as it is possible that $n<k-1$.
If $n\geq k-1$ and $k\geq $, there are $k-1$ pairs, this case has been discussed.
Now, if $k \geq n+2$, the smallest value of $x$ that can be taken would be $x=k-y=k-n$ and the largest value it can take would be $n$. Hence there are a total of $n-(k-n)+1=2n-k+1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$ I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$:
$$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$
which gives (Wolfram Alpha)
$$D= \left[
\frac{\sin \theta(\cos \theta - a)}{2(a^2-1)(a \cos\theta-1)^2}
+\frac{\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{(a^2-1)^{3/2}}\right]_0^{2\pi}
.$$
which can be expressed as
$$D= \left[
\frac{(a^2-1)^{1/2}\sin \theta(\cos \theta - a)+2 (a \cos\theta-1)^2\tanh^{-1}\left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right)}{2(a^2-1)^{3/2}(a \cos\theta-1)^2}
\right]_0^{2\pi}
.$$
This expression involves discontinuities and complex numbers which is beyond my present abilities to handle.
| Let $z=e^{i\theta}$ and then
$$ \cos\theta=\frac12(z+z^{-1}),\sin\theta=\frac1{2i}(z-z^{-1}),d\theta=\frac{1}{iz}dz. $$
So
\begin{eqnarray}
&&\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta\\
&=&\int_{|z|=1}\frac{[\frac1{2i}(z-z^{-1})]^2}{(1-\frac a2(z+z^{-1}))^3}\frac{1}{iz}dz\\
&=&-\int_{|z|=1}\frac{2 i \left(z^2-1\right)^2}{\left(a z^2+a-2 z\right)^3}dz\\
&=&-\frac{2i}{a^3}\int_{|z|=1}\frac{\left(z^2-1\right)^2}{\left(z^2-\frac{2}{a} z+1\right)^3}dz\\
&=&-\frac{2i}{a^3}\int_{|z|=1}\frac{\left(z^2-1\right)^2}{\left(z-\frac{1+\sqrt{1-a^2}}{a}\right)^3\left(z-\frac{1-\sqrt{1-a^2}}{a}\right)^3}dz\\
&=&-\frac{2i}{a^3}\cdot2\pi i\cdot\frac12\frac{d^2}{dz^2}\frac{\left(z^2-1\right)^2}{\left(z-\frac{1+\sqrt{1-a^2}}{a}\right)^3}\bigg|_{z=\frac{1-\sqrt{1-a^2}}{a}}\\
&=&\frac{4\pi}{a^3}\cdot\frac{a^3}{4(1-a^2)^{3/2}}\\
&=&\frac{\pi}{(1-a^2)^{3/2}}.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
U-substitution of 2x in trigonometric substitution Find
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}} \,\mathrm{d}x.$$
The text says to use substitution of $u = 2x$. How did they get $u = 2x$ and not $u = x^3$?
| The hint:
$$\frac{x^3}{\sqrt{(4x^2+9)^3}}=\frac{x^3+\frac{9}{4}x-\frac{9}{4}x}{\sqrt{(4x^2+9)^3}}=\frac{x}{4\sqrt{4x^2+9}}-\frac{9x}{4\sqrt{(4x^2+9)^3}}.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$ This is a question from Math Olympiad.
If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.
I basically do not know how to approach this question. Please let me know how to approach this question, and if you attach full explanation, I will appreciate it. Thanks.
| The given equations $(1),(2),(3)$ can be expressed as:
$$\begin{cases} x^3-y^3=3(x-y) \\ y^3-z^3=y-z \\ z^3-x^3=4(z-x)\end{cases} \stackrel{+}{\Rightarrow} x=3z-2y \ \ (4)$$
Plug $(4)$ in $(1)$ and consider it with $(2)$:
$$\begin{cases} 3z^2-3zy+y^2=1 \\ y^2+yz+z^2=1\end{cases} \Rightarrow 2z^2-4zy=0 \Rightarrow z=2y \ \ (5)$$
Put $(5)$ to $(4)$ to find $x=4y$.
Now you can put $z=2y$ and $x=4y$ to $(1)$ to find $y=\frac{1}{\sqrt{7}}$. Hence:
$$xy+yz+zx=\frac{4}{7}+\frac{2}{7}+\frac{8}{7}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 6,
"answer_id": 5
} |
Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can verify this through expansion, but if I were a mathematician without access to CAS, how might I notice that this is the case?
I suppose what I'm asking is how one should "see" that $P$ can be simplified to $(1 + x + x^2 + x^3 + x^4 + x^5)^2$? Is it a multinomial thing (which seems a bit too complicated for someone to "notice"), or is there something simpler about the polynomial that one could use to factor it?
| When you multiply two polynomials, the result is the sum of each pair of terms, one from the "left" and one from the "right", multiplied together.
With $(1+x+x^2+x^3+x^4+x^5)^2$, you'll get $1*1$ once (there's only one way to pick "1" from each side). But you'll get $x$ twice, once from taking the left's $1$ and the right's $x$, and once the other way around. Hence you get a $2x$ in the product. Then there's 3 ways to get $x^2$ -- $1*x^2$, $x*x$, and $x^2*1$. It's the same from the other end, with a maximum in the middle.
Recognizing the factoring is, like many complex polynomial factorings, just a matter of being familiar with the pattern.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 1
} |
Kernel/Image expression
Express the kernel of the 1 × 4 matrix A = \begin{bmatrix}1&2&3&4 \end{bmatrix}
as the
image of a 4 × 3 matrix B.
I understand that the kernel of a matrix is solving the system for A$\vec{x}$ = 0, but I have no idea what this question is asking nor how to do it. Are there any kind souls who can walk me through it?
| $Ax=0\Rightarrow [1 \ 2 \ 3 \ 4][x \ y \ z \ v ]^t=0\Rightarrow x+2y+3z+4v=0\Rightarrow x=-2y-3z-4v$
$\Rightarrow \begin{pmatrix}
x \\
y \\
z \\
v
\end{pmatrix}=\begin{pmatrix}
-2y \\
y \\
0 \\
0
\end{pmatrix}+\begin{pmatrix}
-3z \\
0 \\
z \\
0
\end{pmatrix}+\begin{pmatrix}
-4v \\
0 \\
0 \\
v
\end{pmatrix}\Rightarrow ker(A)=<\begin{pmatrix}
-2 \\
1 \\
0 \\
0
\end{pmatrix},\begin{pmatrix}
-3 \\
0 \\
1 \\
0
\end{pmatrix},\begin{pmatrix}
-4 \\
0 \\
0 \\
1
\end{pmatrix}>
$
Can you continue now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$ It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$
Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}{4})$
I was thinking of making polar transformations $X=r \cos \theta, Y=r \sin \theta$
Then I am getting stuck at ranges of $\theta$
| If you have already proved $\frac{X Y}{\sqrt{X^2 + Y^2}} $ and $\frac{X^2 - Y^2}{\sqrt{X^2 + Y^2}}$ are gaussian, as long as that pair of them is jointly gaussian, then you may use their property: $u,\, v \text{ independent} \iff \operatorname{cov}(u,v) = 0$.
$$ \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\frac{X^2 - Y^2}{\sqrt{X^2 + Y^2}} \right)$$
$$ = \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\frac{X^2}{\sqrt{X^2 + Y^2}} \right) - \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\frac{Y^2}{\sqrt{X^2 + Y^2}} \right)$$
Now make use of the symmetry of the expressions we have:
$$ \operatorname{cov}(\cdots) = 0 $$
To make this more rigorous one may rewrite the covariance as follows:
$$ \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\frac{X^2}{\sqrt{X^2 + Y^2}} \right) - \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\frac{Y^2}{\sqrt{X^2 + Y^2}} \right) $$
$$ = E \left( \frac{X^3 Y}{X^2 + Y^2} \right) - E \left( \frac{X Y^3}{X^2 + Y^2} \right) $$
Now renaming $X \to Y, \, Y \to X$ under the first expectation sign (note that this is the same as renaming variables under integral) we get the result.
Covariance method can be carried even further, following the last line and the trick $ X = -X $ ($X$ is symmetric distribution) which we apply to the second $E$:
$$ = E \left( \frac{X^3 Y}{X^2 + Y^2} \right) - E \left( \frac{X Y^3}{X^2 + Y^2} \right) = E \left( \frac{X^3 Y}{X^2 + Y^2} \right) + E \left( \frac{X Y^3}{X^2 + Y^2} \right) $$
$$ = E \left( \frac{X Y(X^2 + Y^2)}{X^2 + Y^2} \right) = E(XY) = 0$$
The last equation is due to independence of $X, Y$ $\implies 0=\operatorname{cov}(X,Y)=E(XY) - E(X)E(Y)=E(XY).$
| {
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"url": "https://math.stackexchange.com/questions/2647442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculus sine proof Suppose that $a, b, c$ are non-zero acute angles such that
$$\frac{\sin(a − b)}{\sin(a + b)} + \frac{\sin(b − c)}{\sin(b + c)} + \frac{\sin(c − a)}{\sin(c + a)}= 0$$
Prove that at least two of $a, b, c$ are equal.
I have no idea how to begin.
| It's just trigonometric manipulation.
$$\frac{\sin(a-b)}{\sin(a+b)}+\frac{\sin(b-c)}{\sin(b+c)}=\frac{\sin(a-b)\sin(b+c)+\sin(b-c)\sin(a+b)}{\sin(a+b)\sin(b+c)}$$
We can now use $\sin x\sin y=\frac{1}{2}(\cos(x-y)-\cos(x+y))$ to gwtite the above expression as
$$\frac{\sin(a-b)\sin(b+c)+\sin(b-c)\sin(a+b)}{\sin(a+b)\sin(b+c)}=\frac{\cos(a-2b-c)-\cos(a+c)+\cos(a+c)-\cos(a+2b-c)}{2\sin(a+b)\sin(b+c)}$$
The use $\cos x-\cos y=2\sin\frac{x-y}{2}\sin\frac{x+y}{2}$ to get
$$\frac{\sin(2b)\sin(a-c)}{\sin(a+b)\sin(b+c)}$$
If we go back to the given equation, we see that we have $\sin(a-c)$ as a common term, so $a-c=0$ is one of the solutions. So the only thing left to do is to show that $$\sin(2b)\sin(c+a)+\sin(a+b)\sin(b+c)=0$$ contains the other two solutions.
Using the same identity as above we get $$\sin(2b)\sin(c+a)-\sin(a+b)\sin(b+c)=\frac{1}{2}(\cos(a-2b+c)-\cos(a+2b+c)-\cos(a-c)+\cos(a+2b+c))=-\sin(a-b)\sin(b-c)=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How is $x(2x+7)+3$ equal to $(2x+1)(x+3)$? For some reason, $x(2x+7)+3$ seems like it should be equal to $(2x+7)(x+3)$ instead of $(2x+1)(x+3)$. How does $(2x+1)$ factor out of here?
The original equation was $2x^2+7x+3$.
Proof of this: https://www.desmos.com/calculator/2xpcqznkio
Notice how $x(2x+7)+3$ and $(2x+1)(x+3)$ overlap, but $(2x+7)(x+3)$ does not!
| It is impossible for $2x^2+7x+3$ to factor into $(2x+7)(x+3)$, because when you expand the brackets, the constant term is $21$ — not $3$.
A way to factor quadratics without the quadratic formula, where the leading coefficient $a \ne 1$ is like this:
$$
\begin{array} {c|c|c}
2x^2 & ? \\ \hline
? & 3 \\
\end{array}
$$
Here we have to 'split' the $x$ term, or $7x$ into the remaining two boxes. Try splitting $x$ in different ways - one might be $x$ and $6x$, like this:
$$
\begin{array} {c|c|c}
2x^2 & x \\ \hline
6x & 3 \\
\end{array}
$$
However, $2x^2$ factors into $x(2x+1)$, and $6x+3$ factors into $3(x+2)$, which doesn't match. Can you find a way to split the $7x$, so that the two rows have the same factor?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there any way to prove $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $ by induction since $ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $ we have that for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $
my problem is can we prove the statement
"for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $"
only using induction ?(without using the fact that convergence of the above series)
any ideas, thanks!
| Take a look at this proof of Daniel J. Velleman.
For $n\geq 1$, we define the positive numbers,
$$I_n:=\int_0^{\pi/2}(\cos(x))^{2n}\,dx\quad\text{and}\quad
J_n:=\int_0^{\pi/2}x^2(\cos(x))^{2n}\,dx.$$
By integration by parts we show that
$$I_n=(2n-1)(I_{n-1}-I_n)\implies I_n=\frac{2n-1}{2n}I_{n-1}$$
and
$$I_n=n(2n-1)J_{n-1}-2n^2J_n.$$
Hence, by dividing the last one by $n^2I_n$, we get
$$\frac{1}{n^2}=\frac{(2n-1)J_{n-1}}{nI_n}-\frac{2J_n}{I_n}=
2\left(\frac{J_{n-1}}{I_{n-1}}-\frac{J_n}{I_n}\right).$$
It follows by induction (see the telescopic sum) that for $K\geq 1$,
$$\frac{\pi^2}{6}-\sum_{n=1}^K\frac{1}{n^2}=\frac{\pi^2}{6}-2\sum_{n=1}^K\left(\frac{J_{n-1}}{I_{n-1}}-\frac{J_n}{I_n}\right)=\frac{\pi^2}{6}-2
\left(\frac{J_0}{I_0}-\frac{J_K}{I_K}\right)=\frac{2J_K}{I_K}>0$$
where we used that $I_0=\pi/2$ and $J_0=\pi^3/24$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x \cos\theta+y\sin\theta=a$ and $x\sin\theta-y\cos\theta=b$, then $\tan\theta=\frac{bx+ay}{ax-by}$. (Math Olympiad) I tried to solve it but I can’t get the answer. Please help me in proving this trig identity:
If
$$x \cos\theta+y\sin\theta=a$$
$$x\sin\theta-y\cos\theta=b$$
then $$\tan\theta=\frac{bx+ay}{ax-by}$$
I've spent many hours trying.
Thanks in advance.
| We may write the given equations as:
$$\begin{pmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix}x \\ - y \end{pmatrix} = \begin{pmatrix}a \\ b \end{pmatrix} $$
and recognise that the first matrix represents a counterclockwise rotation by $\theta$ about the origin.
We can then use the formula for the tangent of the difference of two angles to see that
$$\tan \theta = \frac{\frac{b}{a}- \frac{-y}{x}}{1+\frac{b}{a} \frac{-y}{x}}=\frac{bx+ay}{ax-by}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Radical with pattern Let $a =111 \ldots 1$, where the digit $1$ appears $2018$ consecutive times.
Let $b = 222 \ldots 2$, where the digit $2$ appears $1009$ consecutive times.
Without using a calculator, evaluate $\sqrt{a − b}$.
| \begin{align}
a &= \frac{10^{2018}-1}{9} \\
b &= \frac{2(10^{1009}-1)}{9} \\
a-b &= \frac{10^{2018}-2\times10^{1009}+1}{9} \\
&= \frac{(10^{1009}-1)^{2}}{9} \\
\end{align}
Can you proceed?
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
Secant and Tangent identity i've been stuck on this question too long
$x = \sec A + \tan A$
show $x + \frac{1}{x} = 2\cdot \sec A$
I've been using $\tan^2 \theta + 1 = \sec^2 \theta$
and $\tan\theta = \frac{\sin \theta}{\cos\theta}$
help would be much appreciated
$x=\sec A +\tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}$
| Enough to show:
$$\cos{A}\left( x + \frac{1}{x}\right) = 2$$
$$\cos{A} \cdot x + \frac{\cos{A}}{x} = 1 +\sin{A} + \frac{\cos^2{A}}{1 +\sin{A}} = 1 +\sin{A} + \frac{1-\sin^2{A}}{1 +\sin{A}} = 2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654388",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$. Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.
Attempt at a solution:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \cos x - 3 \sin x \tan y) \sin x$$
$$\tan y + 3 \sin^2 x + \tan y = 3 \sin x \cos x$$
$$\tan y = \frac{3 \sin x \cos x}{1 + 3 \sin^2 x}$$
I have also tried substituting $0$, $30$, $45$, $60$, $90$ to the values of $x$.
| Starting from marty cohen's answer
$$\tan (y)=\dfrac{3\sin(2 x)}{5-3\cos(2x)}$$ let $t=\tan(x)$ to get
$$\tan (y)=\frac{3 t}{4 t^2+1}=f(t)$$ So
$$f'(t)=\frac{3-12 t^2}{\left(4 t^2+1\right)^2}\qquad \text{and} \qquad f''(t)=\frac{24 t \left(4 t^2-3\right)}{\left(4 t^2+1\right)^3}$$ The first derivative cancels for $t_{\pm}=\pm \frac 12$. For $t_+$, $f''(t+)=-3$ corresponding to a maximum and for $t_-$, $f''(t_-)=3$ corresponding to a minimum.
So, $t=\frac 12$ and $\tan(y)=\frac34$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $3\sin^2 x - \cos^2 x - 2 =0$
Find all the angles between $0$ and $360^\circ$ that satisfy $$3\sin^2 x - \cos^2 x - 2 =0$$
My attempt -
$3\sin^2 x - (1-\sin^2x) - 2 =0$
$ 3 \sin^2 x + \sin^2 x = 3 $
$4\sin^2 x = 3 $
$ \sin x= \frac{\sqrt{3}}{2} $
I found that $x= 60,120 $
Why is the answer for this $60,120,240,300$ ? How do I find 240 and 300?
| The original definition of $\sin \theta$ of a right-angled triangle is just $\sin \theta =\dfrac {\text {opposite side}}{\text {hypotenuse}}$.
Later, because of its in-adequacy in handling angles larger than $90^0$, the definition is expanded to $\sin \theta = \dfrac {y-ordinate}{\text {radius of the unit circle}} = \dfrac {y-ordinate}{1}$ for the point P = (x, y).
As seen in the figure, P' = (-x, y) also satisfies that definition. Therefore, for a single valued ($\dfrac {\sqrt 3}{2}$), there corresponds two solutions of $\theta$. Another two for ($- \dfrac {\sqrt 3}{2})$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $ Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $
Find the minimum value of $h(\theta)$
$h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} = 5 \sin (\theta + 53.13) + \sqrt{2} $
Minimum value -
$5\sin (\theta + 53.13) + \sqrt{2} = -5 $
Therefore min value is = $ -5/5 - \sqrt{2} $
Why am I wrong ? And how should I do this question..
| Apply Buniakowski inequality: $(3\sin \theta + (-4)\cos \theta)^2 \le (3^2+(-4)^2)(\sin^2 \theta+\cos^2 \theta) = 25\implies 3\sin \theta - 4\cos \theta \ge -5\implies $ min value $ = -5+\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Fourier series for $f(x+4) = f(x)$, $f(x)=1$ for $x\in (0,2), f(x)=-1$ for $x \in (-2,0)$ Given that $f(x+4) = f(x)$ and
$f(x) = -1$ if $-2 < x < 0$, and
$f(x)=1$ if $0 < x < 2$
Find the Fourier series.
I tried it out but I get all $0$ for $a_0$, $a_n$ and $b_n$. Can anyone help me out? I can attach the working if you need it.
| Since $f(x)$ is odd, you should get $a_n =0$ for all $n \ge 0$. The $a_n$ coefficients correspond to the $\cos$ terms which represent the even part of $f(x)$. However, you should not get $b_n = 0$. We see that $$\int_{-2}^2 f(x) \sin\left(\frac{n\pi x}{2}\right)dx = 2 \int^2_0 \sin\left(\frac{n\pi x}{2}\right)dx = 2 \left[-\frac{2}{n\pi} \cos\left(\frac{n\pi x}{2}\right)\right]_{x=0}^{x=2} = -\frac{4}{n\pi}((-1)^n - 1),$$ which is non-zero when $n$ is odd. The coefficients $b_n$ are $\frac 1 2$ of this integral, so $$b_n = -\frac{4}{n\pi} \frac{((-1)^n- 1)}{2} = \left\{\begin{matrix}\tfrac{4}{n\pi}, & n \text{ odd}, \\ 0, & n \text{ even}. \end{matrix} \right.$$
In the integral, I used $$\int_{-2}^2 f(x) \sin\left(\frac{n\pi x}{2}\right)dx = 2 \int^2_0 f(x) \sin\left(\frac{n\pi x}{2}\right)dx$$ wihch holds since $f(x)$ is odd and thus $f(x) \sin(\alpha x)$ is even for any $\alpha \in \mathbb R$ and then we can plug in $f(x) = 1$ on the interval $0 \le x \le 2$.
| {
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Find the length of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ which is intercepted between the axes.
Find the length of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ which is intercepted between the axes.
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\implies \frac{dy}{dx}={(\frac{-y}{x})}^{\frac{1}{3}}$.
Slope at $(p,q)$ will be ${(\frac{-q}{p})}^{\frac{1}{3}}$.
So equation of tangent at $(p,q);$ $y-q={(\frac{-q}{p})}^{\frac{1}{3}}(x-p)$.
How to find the length of tangent intercepted in between the axes from here?
| Any point on $$x^{2/n}+y^{2/n}=a^{2/n}\ \ \ \ (1)$$ can be chosen as $(a\cos^nt,a\sin^nt)$
Differentiating $(1)$ wrt $x,$ $$\dfrac{dy}{dx}=-\dfrac{x^{(2-n)/n}}{y^{(2-n)/n}}$$
$$\dfrac{dy}{dx}_{\text{ at }(a\cos^nt,a\sin^nt)}=-\dfrac{x^{(2-n)/n}}{y^{(2-n)/n}}=-\dfrac{\cos^{2-n}t}{\sin^{2-n}t}$$
So, the equation of the tangent at $(a\cos^nt,a\sin^nt)$ will be $$\dfrac{y-a\sin^nt}{x-a\cos^nt}=-\dfrac{\cos^2t\sin^nt}{\sin^2t\cos^nt}$$
$$\iff x\cos^2t\sin^nt+y\sin^2t\cos^nt=a\cos^nt\sin^nt$$
$$\iff\dfrac x{\cos^{n-2}t}+\dfrac y{\sin^{n-2}t}=1$$
Here $n=3$
Can you take it from here?
See also: intercept form
| {
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Prove that if $3\mid(a^2+b^2)$,then $3\mid a$ and $ 3\mid b$ I am trying to prove this by contradiction. So if $3$ doesn't divide $a$ or $3$ doesn't divide $b$, then the remainder is either $1$ or $2$. I am struggling on what to do next. How do I get the remainder of $a^2$ and $b^2$ for these cases?
Any help is greatly appreciated. Thank you!
| Assume $a,b$ are not divisible by $3$. Then:
$$a=3m+1 \ \ \text{or} \ \ a=3m+2; \\
b=3n+1 \ \ \text{or} \ \ b=3n+2.$$
Note:
$$(3k+1)^2=3(3k^2+2k)+1; \ \ (3k+2)^2=3(3k^2+4k+1)+1.$$
Then:
$$a^2+b^2 \equiv 2 \ (\mod 3).$$
Now assume only one of them is divisible by $3$. Then:
$$a^2+b^2 \equiv 1 \ (\mod 3).$$
Contradiction, hence both $a$ and $b$ must be divisible by $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why doesn't $A^2=I$ imply $A=\pm I$? Im having trouble believing this T/F Question:
if $\mathrm A^2=I$ then $\mathrm A = \pm \mathrm I$
The answer is False but why?
If the matrix is $\mathrm A = \mathrm I,$ say
\begin{bmatrix}1& 0\\
0 & 1 \end{bmatrix}
then $\mathrm A^2$ is also that. And if $\mathrm A = -\mathrm I,$ then it is
\begin{bmatrix}-1 & 0\\
0 &-1
\end{bmatrix}
and that squared is also the same? Where am i going wrong?
| for matrices in $\mathbb{M}_2(\mathbb{R})$
$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix}^2 =
\begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{pmatrix} =
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
$$
requires either $b=c=0, a,d= \pm1$ or $a=-d, bc=1-a^2$
the first type of solution gives $4$ solutions which form a multiplicative group isomorphic to $V_4$ (Klein's Viergruppe).
the second type of solution gives a two-parameter family of matrices:
$$
\begin{pmatrix} k & r \\ \frac{1-k^2}r & -k \end{pmatrix}
$$
in case $|k| \le 1$ we may also (after re-scaling $r$) write these latter matrices as:
$$
\begin{pmatrix} \cos \theta & r \sin \theta \\ \frac{\sin \theta}r & -\cos \theta \end{pmatrix}
$$
looking at the subfamily with $r=1$ we obtain the solutions
$$
\begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{pmatrix}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:
If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
$$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}.$$
Here's how I tried solving the problem:
$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}$
$R_2 \to R_2 - R_1$
$R_3 \to R_3 -R_1$
$= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\sin^2A & \cot C-\cot A & 0\end{vmatrix}$
Expanding the determinant along $C_3$
\begin{align}
&= (\sin^2B-\sin^2A)(\cot C-\cot A)-(\cot B-\cot A)(\sin^2C-\sin^2A) \\
&= \sin(B+A) \sin(B-A) \left[\frac {\cos C}{\sin C} - \frac {\cos A}{\sin A}\right] - \left[\frac {\cos B}{\sin B} - \frac {\cos A}{\sin A}\right]\sin(C+A) \sin(C-A) \\
&= \frac {\sin(B+A) \sin(B-A) \sin(A-C)} {\cos A \cos C} - \frac {\sin(A-B) \sin(C+A) \sin(C-A)} {\cos A \cos C} \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin(A+B)} {\cos C} - \frac {\sin(A+C)} {\cos B}\right] \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin C} {\cos C} - \frac {\sin B} {\cos B}\right] \\
&= \frac {\sin(B-A) \sin (A-C) \sin (C-B)} {\cos A \cos B \cos C}
\end{align}
I tried solving further but the expression just got complicated. I don't even know if the work I've done above is helpful. My textbook gives the answer as $0$. I don't have any clue about getting the answer. Any help would be appreciated.
| $$F=\begin{vmatrix} \sin^2B-\sin^2A & \cot B-\cot A \\ \sin^2C-\sin^2A & \cot C-\cot A \end{vmatrix}$$
$$=\begin{vmatrix} \sin^2B-\sin^2A & -\dfrac{\sin(B-A)}{\sin A\sin B} \\ \sin^2C-\sin^2A & -\dfrac{\sin(C-A)}{\sin C\sin A} \end{vmatrix}$$
$$=\dfrac1{\sin B\sin^2A\sin C}\begin{vmatrix}\sin(B-A)\sin(B+A)\sin B\sin A&-\sin(B-A)\\ \sin(C-A)\sin(C+A)\sin C\sin A&-\sin(C-A)\end{vmatrix}$$
Using $A+B+C=\pi,\sin(B+A)=\sin C$ etc.,
$$F=\dfrac{\sin A\sin B\sin C}{\sin^2A\sin B\sin C}\begin{vmatrix}\sin(B-A)&-\sin(B-A)\\\sin(C-A)&-\sin(C-A)\end{vmatrix}=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $f(x) = \frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$ Prove $f(x) =\frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$.
My attempt...
Proof
$$\left| f(x) - f(y) \right| = \left| \frac{1}{1+x^2} - \frac{1}{1+y^2}\right| = \frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\right| $$
By the triangle inequality
$$\frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\right| \leq \left(\frac{\left|x\right|}{\left(1+x^2\right)\left(1+y^2\right)} +\frac{\left|y\right|}{\left(1+x^2\right)\left(1+y^2\right)} \right) \left|x-y\right| \tag{$\star$}$$
Note that for all $x\in \mathbb{R}$
$$\left|x\right| < 1 +x^2 \implies \left|x\right| < \left(1 +x^2\right)\left(1+y^2\right)$$
Therefore
$$\frac{\left|x\right|}{\left(1 +x^2\right)\left(1+y^2\right)} \leq 1$$
Applying this fact to $(\star)$ we see that
$$\left(\frac{\left|x\right|}{\left(1+x^2\right)\left(1+y^2\right)} +\frac{\left|y\right|}{\left(1+x^2\right)\left(1+y^2\right)} \right) \left|x-y\right| \leq \left(1 + 1\right)\left|x-y\right| \leq 2\left|x - y \right|$$
Therefore $f$ is a Lipschitz function, which implies $f$ is uniformly continuous on $\mathbb{R}$.
Please comment on validity and/or readability, thank you.
| Apply mean value theorem in an interval $[x,y], x,y\in \mathbb R$. So, $\exists \xi \in [x,y]: f(x)-f(y)=f'(\xi)(x-y)$. Also we know that $|f'(\xi)|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve $x^4 - 8x^3 + 21x^2 - 20x + 5 = 0$ given that the sum of two of its roots is $4$ Here's what I tried:
Let the roots be $a$, $b$, $c$ and $d$, $a+b=4$. Then,
$$a + b + c + d = 8 \Longrightarrow 4 + c+ d = 8 \Longrightarrow a+b = c+d = 4$$
$$(a + b)(c + d) + ab + cd = 21$$
$$ab (c + d) + cd (a + b) = 20 \Longrightarrow 4ab + 4cd = 20 \Longrightarrow ab + cd = 5$$
$$abcd = 5$$
I can't figure out how to proceed.
| make the Ansatz $$x^4-8x^3+21x^2-20x+5=(x^2+Ax+B)(x^2+Cx+D)$$
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Using summation by parts to evaluate an alternating sum I want to evaluate
$$
\sum_{k=0}^n (-1)^k \binom{n}{k} k.
$$
I tried summation by parts, i.e. the formula
$$
\sum_{k=0}^n (f(k+1) - f(k))g(k)
= f(n+1)g(n+1) - f(0)g(0) - \sum_{k=0}^n f(k+1) (g(k+1) - g(k))
$$
with $f(k+1) - f(k) = (-1)^k \binom{n}{k}$ and $g(k) = k$. As
$$
(-1)^{k+1}\binom{n-1}{k+1} - (-1)^k \binom{n-1}{k}
= (-1)^{k+1} \left( \binom{n-1}{k+1} + \binom{n-1}{k} \right)
= (-1)^{k+1} \binom{n}{k+1}
$$
we have $f(k) = (-1)^k \binom{n-1}{k}$. Plugging into the formula
\begin{align*}
\sum_{k=0}^n (-1)^k \binom{n}{k} k
& = (-1)^{n+1} \binom{n-1}{n+1} (n+1) - (-1)^0 \binom{n-1}{0}\cdot 0
- \sum_{k=0}^n (-1)^{k+1} \binom{n-1}{k+1} \\
& = \sum_{k=0}^n (-1)^{k} \binom{n-1}{k+1}.
\end{align*}
But for example if $n = 3$ then
$$
\sum_{k=0}^n (-1)^k \binom{n}{k} k = -3 + 6 -3 = 0
$$
but
$$
\sum_{k=0}^n (-1)^{k} \binom{n-1}{k+1}
= \binom{2}{1} - \binom{2}{2}
= 1
$$
which is not equal, but I cannot see whats wrong with the above derivation??
| You need $f(k+1) - f(k) = (-1)^k \binom{n}{k}$. But
$$
(-1)^{k+1}\binom{n-1}{k+1} - (-1)^k \binom{n-1}{k}
= (-1)^{k+1} \left( \binom{n-1}{k+1} + \binom{n-1}{k} \right)
= (-1)^{k+1} \binom{n}{k+1}
$$
So
$$
(-1)^{k}\binom{n-1}{k} - (-1)^{k-1} \binom{n-1}{k-1}
= (-1)^{k} \left( \binom{n-1}{k} + \binom{n-1}{k-1} \right)
= (-1)^{k} \binom{n}{k}
$$
$f(k)=(-1)^{k-1} \binom{n-1}{k-1}$.
Of course, When $k=0$ and $k=n+1$, we have special cases and should consider seperately.
$f(1)=\binom{n-1}{0}=1$
$f(1)-f(0)=(-1)^0\binom{n}{0}=1$ and so $f(0)=0$
$f(n)=(-1)^{n-1}\binom{n-1}{n-1}=(-1)^{n-1}$
$f(n+1)-f(n)=(-1)^n\binom{n}{n}=(-1)^n$ and so $f(n+1)=(-1)^n+(-1)^{n-1}=0$
\begin{align*}
\sum_{k=0}^n (-1)^k \binom{n}{k} k
& =(0) (n+1) -(0)(0)
- \sum_{k=1}^{n-1} (-1)^{k} \binom{n-1}{k}-f(1)(g(1)-g(0))-f(n+1)(g(n+1)-g(n)) \\
& = - \sum_{k=0}^{n-1} (-1)^{k} \binom{n-1}{k}+(-1)^0\binom{n-1}{0}-1\\
&=0
\end{align*}
My work:
$\displaystyle (1+x)^n=\sum_{k=1}^n\binom{n}{k}x^k+1$
Differentiating,
$\displaystyle n(1+x)^{n-1}=\sum_{k=1}^nk\binom{n}{k}x^{k-1}$
Put $x=-1$.
\begin{align*}
\sum_{k=1}^nk\binom{n}{k}(-1)^{k-1}&=0\\
\sum_{k=1}^nk\binom{n}{k}(-1)^{k}&=(-1)(0)\\
\sum_{k=0}^nk\binom{n}{k}(-1)^{k}&=0+0=0\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $\sum\limits_{n=0}^\infty\frac{5^n(3^{5^{n+1}}-5\cdot3^{5^n}+4)}{(729)^{5^n}-(243)^{5^n}-5\cdot3^{5^n}+1}=\frac12$ This problem is taken from Algerian Olympiad and asks to prove that
$$\sum_{n=0}^{\infty} \dfrac{5^n(3^{5^{n+1}} -5\cdot3^{5^n} + 4)}{(729)^{5^n} - (243)^{5^n}-5\cdot3^{5^n}+1} = \frac 12.$$
Noticing that $729=3^6$ and $243 = 3^5$, I tried simplifying the general terms by setting $x=3^{5^n}$ but it seems to give no simplifications.
Thanks in advance for any advice / ideas.
| I think the "$5 \cdot 3^{5^n}$" in the denominator is a typo and it can be proved that$$
\sum_{n = 0}^\infty \frac{5^n (3^{5^{n + 1}} - 5 \cdot 3^{5^n} + 4)}{3^{6 \cdot 5^n} - 3^{5^{n + 1}} - 3^{5^n} + 1} = \frac{1}{2}.
$$
In fact,$$
\sum_{n = 0}^\infty \frac{5^n (3^{5^{n + 1}} - 5 \cdot 3^{5^n} + 4)}{3^{6 \cdot 5^n} - 3^{5^{n + 1}} - 3^{5^n} + 1} = \sum_{n = 0}^\infty \left( \frac{5^n}{3^{5^n} - 1} - \frac{5^{n + 1}}{3^{5^{n + 1}} - 1} \right) = \frac{5^0}{3^{5^0} - 1} = \frac{1}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Find the eigenvalues of block matrix $C$
Find the eigenvalues of $$C = \begin{bmatrix}\begin{array}{c|c} 0 & A\\ \hline A^T & 0\end{array}\end{bmatrix}$$ where $$A = \begin{bmatrix}
0&0&0&1&1&1&1\\0& 1& 1& 0& 1& 0& 1\\0 &1 &1 &1 &0 &1 &0\\1& 0& 1& 0& 0& 1& 1\\1 &0 &1& 1& 1& 0& 0\\1 &1 &0& 0& 1& 1& 0\\1 &1& 0& 1& 0& 0 &1
\end{bmatrix}$$
The characteristic polynomial of $C$ is $$\det(xI-C)=x^2-A^2$$
Will the eigenvalues of $A$ be related to those of $C$ in some way?
I am unable to proceed here.
| Something you have written is wrong but can be corrected. We have$$xI-C=\begin{bmatrix}
\begin{array}{c|c}
xI & -A \\
\hline
-A^T & xI
\end{array}
\end{bmatrix}$$therefore$$|xI-C|=|x^2I-A^TA|$$therefore $$x=\pm|\lambda|$$where $\lambda$ is an eigenvalue of $A$ therefore the eigenvalues of $C$ are the $\pm$ singular values of $A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sequence and two limits to calculate Let $(a_n)_{n \ge 0}$ with $a_0>0$ and $a_{n+1}=\frac{a_n}{1+\sqrt{1+a_n^2}}$, for every $n \ge 0$. Find $\lim_{n\to \infty} a_n$ and $\lim_{n\to \infty} (1+a_0^2)(1+a_1^2)...(1+a_n^2)$.
I proved $(a_n)_n$ is convergent and I found $\lim_{n\to \infty} a_n=0$ and for the second limit I write $1+a_n^2=(\frac{a_n-a_{n+1}}{a_{n+1}})^2$ and I obtained $(1+a_0^2)(1+a_1^2)...(1+a_n^2)=(\frac{(a_0-a_1)(a_1-a_2)...(a_n-a_{n+1})}{a_1a_2...a_{n+1}})^2$, but I don't know how to continue. Please help me!
| If $a_n=\tan\theta$ for some $\theta\in\left(0,\frac{\pi}{2}\right)$, then $a_{n+1}=\tan\frac{\theta}{2}$ (familiarity with the bisection formulas).
By induction $a_n = \tan\frac{\arctan a_0}{2^n}=\tan\frac{\varphi}{2^n}$ and $1+a_n^2=\sec^2\frac{\varphi}{2^n}$.
By the duplication formula for the sine function
$$ \sin(\theta)=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=4\sin\frac{\theta}{4}\cos\frac{\theta}{4}\cos\frac{\theta}{2}=\ldots =2^n\sin\frac{\theta}{2^n}\prod_{k=1}^{n}\cos\frac{\theta}{2^k} $$
hence
$$ \prod_{k=1}^{n}\sec\frac{\theta}{2^k} = \frac{2^n\sin\frac{\theta}{2^n}}{\sin\theta}\to\frac{\theta}{\sin\theta}\text{ as }n\to+\infty $$
and
$$ \prod_{n\geq 0}(1+a_n^2) = \prod_{k\geq 0}\sec^2\frac{\varphi}{2^n} = \frac{\varphi^2}{\sin^2\varphi\cos^2\varphi}=\left(a_0+\frac{1}{a_0}\right)^2\arctan^2 a_0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Checking if a functional $F(x)$ is a norm in $\mathbb{R}^2$ There is a functional given:
$$F(x) = \sqrt{2x_1^2 + 3x_2^2}$$
Of course $x\in\mathbb{R}^2 \rightarrow x = (x_1, x_2)$
It is easy to check that:
1) $\forall x \in \mathbb{R}^2$ $F(x) \ge 0$
2) $F(x) = 0 \iff x = 0$
3) $ F(\lambda x) = | \lambda | F(x)$
I failed however to show that $F(x+y) \le F(x) + F(y)$. How should it be proven?
| You can just use the Cauchy-Schwartz inequality:
\begin{align}
F(x+y)^2 &= 2(x_1+y_1)^2+3(x_2+y_2)^2 \\
&= \underbrace{2x_1^2 + 3x_2^2}_{F(x)^2} + \underbrace{2y_1^2+3y_2^2}_{F(x)^2}+2(2x_1y_1 + 3x_2y_2)\\
&\stackrel{CSB}{\le} F(x)^2 + F(y)^2 + 2\sqrt{2x_1^2+3x_2^2}\sqrt{2y_2^2+3y_2^2}\\
&= F(x)^2 + F(y)^2 + 2F(x)F(y)\\
&= (F(x) + F(y))^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Factorize $\det\left[\begin{smallmatrix}yz-x^2&zx-y^2&xy-z^2\\zx-y^2&xy-z^2&yz-x^2\\xy-z^2&yz-x^2&zx-y^2\end{smallmatrix}\right]$ using factor theorem Factorize and prove that
$$
\Delta=\begin{vmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{vmatrix}\\=\frac{1}{4}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2
$$
using factor theorem.
My Attempt:
$\Delta$ is a homogeneous symmetric polynomial of degree $6$.
When $(x-y)^2+(y-z)^2+(z-x)^2=0$, i.e. $x=y=z$
$$
\Delta=\begin{vmatrix}
0&0&0\\
0&0&0\\
0&0&0\\
\end{vmatrix}=0
$$
Thus, $(x-y)^2+(y-z)^2+(z-x)^2$ is a factor.
How do I extract the other $(x-y)^2+(y-z)^2+(z-x)^2$ from $\Delta$ $\color{red}{?}$
Does this have anything to do with all rows (or columns) being zero when $(x-y)^2+(y-z)^2+(z-x)^2=0$ $\color{red}{?}$
If I can extract that then i think I know how to proceed. The remaining factor must be a homogeneous quadratic symmetric polynomial, i.e. $p(x,y,z)=a(x^2+y^2+z^2)+b(xy+yz+zx)$
$$
\Delta(x,y,z)=\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.a(x^2+y^2+z^2)+b(xy+yz+zx)
$$
$$
\Delta(1,0,0)=\begin{vmatrix}
-1&0&0\\
0&0&-1\\
0&-1&0\\
\end{vmatrix}=1=4.a\implies a=\frac{1}{4}
$$
$$
\Delta(1,1,0)=\begin{vmatrix}
-1&-1&1\\
-1&1&-1\\
1&-1&-1\\
\end{vmatrix}=\begin{vmatrix}
0&0&1\\
-2&0&-1\\
0&-2&-1\\
\end{vmatrix}\\
=\begin{vmatrix}
-2&0\\
0&-2\\
\end{vmatrix}=4=4.(2a+b)=4(1/2+b)=2+4b\\
\implies b=\frac{1}{2}
$$
$$
\Delta(x,y,z)=\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.\frac{1}{4}(x^2+y^2+z^2)+\frac{1}{2}(xy+yz+zx)\\
=\frac{1}{4}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.(x^2+y^2+z^2+2xy+2yz+2zx)\\
=\frac{1}{4}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2(x+y+z)^2
$$
Note:
I am trying to factorize the determinant using factor theorem given the fact that the determinant is a homogeneous symmetric polynomial of degree 6.
| By taking elementary operation:$$\Delta=\begin{vmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{vmatrix}=\begin{vmatrix}
xy+yz+zx-x^2-y^2-z^2&zx-y^2&xy-z^2\\
xy+yz+zx-x^2-y^2-z^2&xy-z^2&yz-x^2\\
xy+yz+zx-x^2-y^2-z^2&yz-x^2&zx-y^2
\end{vmatrix}\\=-(x^2+y^2+z^2-xy-yz-zx)\begin{vmatrix}
1&zx-y^2&xy-z^2\\
1&xy-z^2&yz-x^2\\
1&yz-x^2&zx-y^2
\end{vmatrix}\\=-\frac{1}{2}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]\begin{vmatrix}
0&(x+y+z)(z-y)&(x+y+z)(x-z)\\
0&(x+y+z)(x-z)&(x+y+z)(x-z)\\
1&yz-x^2&zx-y^2
\end{vmatrix}\\=-\frac{1}{2}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]\begin{vmatrix}
(z-y)&(x-z)\\
(x-z)&(x-z)\\
\end{vmatrix}=\frac{1}{2}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2,
$$ as we know $x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.
Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.
I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides which would give me a simple quadratic to solve for $x$, but I don't know how to simplify $\frac{(2^{x^2}-1)}{(2^x-1)}$. Am I missing a simple trick here, or am I on a completely wrong path to solving this?
| Using Mathematica
eq = 2 (2^x - 1) x^2 + (2^(x^2) - 2) x == 2^(x + 1) - 2;
sol = Solve[eq, x, Reals]
(* {{x -> -1}, {x -> 0}, {x -> 1}} *)
Verifying the solutions
And @@ (eq /. sol)
(* True *)
Showing the solutions on a plot
Plot[{2 (2^x - 1) x^2 + (2^(x^2) - 2) x, 2^(x + 1) - 2},
{x, -2, 2}, PlotLegends -> Placed["Expressions", {.22, .75}],
Epilog -> {Red, AbsolutePointSize[5],
Point[{x, 2^(x + 1) - 2} /. sol]}]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find an equation of the tangent plane to the given function at the given point. $z = \sin(xy)$ at the point $(x,y) = (2,3\pi /4)$
Work I have done so far:
Partial derivatives:
$$f_x(x,y) = \cos(xy)y \implies f_x\left(2,\frac{3\pi}{4}\right) = \frac{3\pi}{4}\cos\frac{6\pi}{4}\\
f_y(x,y) = \cos(xy)x \implies f_y\left(2,\frac{3\pi}{4}\right) = 2\cos\frac{6\pi}{4}\\
z-0 = \frac{3\pi}{4}\cos\frac{6\pi}{4}\left(x-2\right)+2\cos\frac{6\pi}{4}\left(x-\frac{3\pi}{4}\right)$$
answer: link to incorrect answer
| On the left-hand side, you should not have $z-0,$ but $z-f(2,3\pi /4).$ Also, $\cos\frac{6\pi}{4} = \cos\frac{3\pi}{2} = 0,$ so you can make your life a lot easier.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\frac{\partial^6 f}{\partial x^4 \partial y^2}$
Find $$\frac{\partial^6 f}{\partial x^4 \partial y^2}(0,0)$$ Of the $f(x,y)=\frac{1}{1-x^2y}$
$$\frac{1}{1-x^2y}=\sum_{n=1}^{\infty}(x^2y)^n$$ as $|x^2y|<1$
So we have $\frac{1}{1-x^2y}\approx1+x^2y+x^4y^2$
But how should we continue from here?
| Your method is correct:
$$z=\sum_{n=\color{red}{0}}^{\infty} (x^2y)^n=1+x^2y+x^4y^2+x^6y^3+\cdots$$
$$z_{x^4}=4!y^2+4!x^2y^3+\cdots$$
$$z_{x^4y^2}=4!\cdot 2!+4!x^2\cdot 2!y+\cdots$$
$$z_{x^4y^2}(0,0)=4!2!=48.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2686731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $\cos^2x+\cos^22x+\cos^23x=1$
Solve the equation: $$\cos^2x+\cos^22x+\cos^23x=1$$
IMO 1962/4
My first attempt in solving the problem is to simplify the equation and express all terms in terms of $\cos x$. Even without an extensive knowledge about trigonometric identities, the problem is solvable.
$$
\begin{align}
\cos^22x&=(\cos^2x-\sin^2x)^2\\
&=\cos^4x+\sin^4x-2\sin^2\cos^2x\\
&=\cos^4+(1-\cos^2x)^2-2(1-\cos^2)\cos^2x\\
&=\cos^4+1-2\cos^2x+\cos^4x-2\cos^2x+2\cos^4x\\
&=4\cos^4x-4\cos^2x+1
\end{align}
$$
Without knowledge of other trigonometric identities, $\cos3x$ can be derived using only Ptolemy's identities. However for the sake of brevity, let $\cos 3x=4\cos^3x-3\cos x$:
$$
\begin{align}
\cos^23x&=(4\cos^3x-3\cos x)^2\\
&=16\cos^6x+4\cos^2x-24\cos^4x
\end{align}
$$
Therefore, the original equation can be written as:
$$\cos^2x+4\cos^4x-4\cos^2x+1+16\cos^6x+4\cos^2x-24\cos^4x-1=0$$
$$-20\cos^4x+6\cos^2x+16\cos^6x=0$$
Letting $y=\cos x$, we now have a polynomial equation:
$$-20y^4+6y^2+16y^6=0$$
$$y^2(-20y^2+6y+16y^4)=0\Rightarrow y^2=0 \Rightarrow x=\cos^{-1}0=\bbox[yellow,10px]{90^\circ}$$
From one of the factors above, we let $z=y^2$, and we have the quadratic equation:
$$16z^2-20z+6=0\Rightarrow 8z^2-10z+3=0$$
$$(8z-6)(z-\frac12)=0\Rightarrow z=\frac34 \& \ z=\frac12$$
Since $z=y^2$ and $y=\cos x$ we have:
$$\biggl( y\rightarrow\pm\frac{\sqrt{3}}{2}, y\rightarrow\pm\frac{\sqrt{2}}2 \biggr)\Rightarrow \biggl(x\rightarrow\cos^{-1}\pm\frac{\sqrt{3}}{2},x\rightarrow\cos^{-1}\pm\frac{\sqrt{2}}2\biggr)$$
And thus the complete set of solution is:
$$\bbox[yellow, 5px]{90^\circ, 30^\circ, 150^\circ, 45^\circ, 135^\circ}$$
As I do not have the copy of the answers, I still hope you can verify the accuracy of my solution.
But more importantly...
Seeing the values of $x$, is there a more intuitive and simpler way of finding $x$ that does away with the lengthy computation?
| Hint:
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$0=\cos^2x+\cos^22x+\cos^23x-1$$
$$=\cos(3x+x)\cos(3x-x)+\cos^22x$$
$$=\cos2x(\cos4x+\cos2x)$$
Use
http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
| {
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"url": "https://math.stackexchange.com/questions/2687769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Verfication of deduction made using the Cauchy-Schwarz inequality Is the following proof correct?
Show that $$16\leq(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$$ for all positive numbers $a,b,c,d$.
Proof. Let $\mathbf{R}^4$ be the inner product space with the inner product defined as in the euclidean product in $\textbf{6.4}$. Now let $a,b,c,d$ be arbitrary positive numbers and let $u = (|\sqrt{a}|,|\sqrt{b}|,|\sqrt{c}|,|\sqrt{d}|)$ and $v = (\frac{1}{|\sqrt{a}|},\frac{1}{|\sqrt{b}|},\frac{1}{|\sqrt{c}|},\frac{1}{|\sqrt{d}|})$ then by appealing to the Cauchy Schwartz inequality we have\begin{align*}
|\langle u,v\rangle| &= |4| = |\langle (|\sqrt{a}|,|\sqrt{b}|,|\sqrt{c}|,|\sqrt{d}|),(\frac{1}{|\sqrt{a}|},\frac{1}{|\sqrt{b}|},\frac{1}{|\sqrt{c}|},\frac{1}{|\sqrt{d}|})\rangle|\\
&\leq \sqrt{|\sqrt{a}|^2+|\sqrt{b}|^2+|\sqrt{c}|^2+|\sqrt{d}|^2}\cdot\sqrt{\frac{1}{|\sqrt{a}|^2}+\frac{1}{|\sqrt{b}|^2}+\frac{1}{|\sqrt{c}|^2}+\frac{1}{|\sqrt{d}|^2}}\\
&= \sqrt{(a+b+c+d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})} = \|u\|\cdot\|v\|.
\end{align*}
Squaring both sides yields the required result.
NOTE The reference to $6.4$ above is simply to what is commonly understood to be the dot product.
$\blacksquare$
| it is $$\frac{a}{b}+\frac{b}{a}+\frac{c}{a}+\frac{a}{c}+\frac{a}{d}+\frac{d}{a}+\frac{b}{c}+\frac{c}{b}+\frac{d}{c}+\frac{c}{d}+\frac{b}{d}+\frac{d}{b}+4\geq 12+4=16$$
since $$x+\frac{1}{x}\geq 2$$ for $x>0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Simplifying calculation for searching for a matrix The exercise it the following: I should find all $$A \in M^{}_{22}(\mathbb{R}) $$ with
$$
A
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
A
$$
I solved it like the following but I wonder whether there is a more elegant way:
$$
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}
$$
$$
=> \begin{pmatrix}
a & a + b \\
c & c + d \\
\end{pmatrix}
=
\begin{pmatrix}
a + c & b + d \\
c & d \\
\end{pmatrix}
$$
=>
I)
$$
a = a + c => c = 0
$$
II)
$$
a + b = b + d => a = d
$$
III)
$$
c = c
$$
IV)
$$
d = c + d => d = d = a
$$
$$
A
=
\begin{pmatrix}
a & b \\
0 & a \\
\end{pmatrix}
$$
with
$$
a, b \in (\mathbb{R})
$$
| Perhaps not more elegant, but you can simplify the problem a little further. Since $$\begin{pmatrix}1&1\\0&1\end{pmatrix}=\mathbf{I}_2+\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ you have, using that $\mathbf{A}\mathbf{I}_2=\mathbf{I}_2\mathbf{A}$,
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ which simplifies to $$\begin{pmatrix}0&a\\0&c\end{pmatrix}=\begin{pmatrix}c&d\\0&0\end{pmatrix}$$
This means that $a=d$ and $c=0$.
| {
"language": "en",
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"source": "stackexchange",
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Finding the the derivative of $y=\sqrt{1-\sin x}; 0A question I'm attempting is:
Find the derivative of $ y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$.
I did this:
$y = \sqrt {1 - \sin x} = \sqrt {\cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} = \sqrt { (\sin \frac{x}{2}-\cos \frac{x}{2})^2} = \sin \frac{x}{2} - \cos \frac{x}{2}$
So, $\frac{dy}{dx} = \frac{1}{2} \cdot (\cos\frac{x}{2} + \sin\frac{x}{2})$.
But apparently this is wrong. The correct solution is:
$\frac{dy}{dx} = -\frac{1}{2}\cdot(\cos\frac{x}{2} + \sin\frac{x}{2})$.
So I want to know what I have done wrongly here. Why is my answer not right?
| Let apply chain rule
$$\left( \sqrt{f(x)} \right)'=\frac{f'(x)}{2\sqrt{f(x)}}=\frac{-\cos x}{2\sqrt {1 - \sin x}}=\frac12\frac{\sin^2 x/2-\cos^2 x/2}{(\cos x/2-\sin x/2)}=\\=\frac12\frac{ (\sin x/2+\cos x/2)(\sin x/2-\cos x/2) }{(\cos x/2 - \sin x/2)}=-\frac12(\sin x/2+\cos x/2)$$
Note indeed that for $0<x<\pi/2$ since
*
*$\cos x/2 >0$
*$0<\tan x/2<1$
$$\cos x/2-\sin x/2=\cos x/2\cdot(1-\tan x/2)>0$$
| {
"language": "en",
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An Example of the Chi-Square Test Here is another problem dealing with the Chi-Square Test. I am hoping that somebody can confirm that I did it correctly or tell me where I went wrong.
Thanks,
Bob
Problem:
An urn contains $6$ red marbles and $3$ white ones. Two marbles are
selected at random from the urn, their colors are noted and then the
marbles are replaced in the urn. This process is performed a total
of $120$ times, and the result are given below. (a) Determine the
expected frequencies. (b) Determinate at a level of significance of $0.05$ whether
the results obtained are consistent with those expected.
There were $6$ times when $0$ red balls and $2$ white balls were drawn.
There were $53$ times when $1$ red ball and $1$ white ball were drawn.
There were $61$ times when $2$ red balls and $0$ white balls were drawn.
Answer: (a)
Let $p_{r2}$ be the probability that we draw two red balls.
\begin{eqnarray*}
p_{r2} &=& \frac{6}{9} \Big( \frac{5}{8} \Big) = \frac{30}{72} \\
p_{r2} &=& \frac{5}{12} \\
\end{eqnarray*}
Hence the expected number of times we draw two red balls
is $120(\frac{5}{12})$ or $50$. Let $p_{w2}$ be the probability that we draw two white balls.
\begin{eqnarray*}
p_{w2} &=& \frac{3}{9} \Big( \frac{2}{8} \Big) = \frac{6}{72} \\
p_{w2} &=& \frac{1}{12} \\
\end{eqnarray*}
Hence the expected number of times we draw two white balls
is $120(\frac{1}{12})$ or $10$. Let $p_{rw}$ be the probability that we draw one white ball and one red ball
in either order.
\begin{eqnarray*}
p_{rw} &=&
\frac{6}{9} \Big( \frac{3}{8} \Big) + \frac{3}{9} \Big( \frac{6}{8} \Big) \\
p_{rw} &=& \frac{18 + 18}{72} = \frac{1}{2} \\
\end{eqnarray*}
Hence the expected number of times we draw one white ball and one red ball is $120(\frac{1}{2})$ or $60$.
(b)
I am now going to apply the Chi-Square Test to determine how good of a fit we have.
\begin{eqnarray*}
\chi^2 &=& \frac{(61-50)^2}{50} + \frac{(53-60)^2}{60} + \frac{(6-10)^2}{10} \\
\chi^2 &=& \frac{121}{50} + \frac{49}{60} + \frac{16}{10} \\
\chi^2 &=& 4.8366667 \\
\end{eqnarray*}
From a table we find that with two degrees of freedom at the 0.95 confidence level
$\chi^2_{0.95} = 5.99$. Therefore we cannot accept the hypothesis that the above
procedure was not used. We also cannot accept the hypothesis that it was.
| For $(a)$, yes, your reasoning is correct.
For $(b)$, you set up your hypotheses the wrong way around. With $\chi^2$ test, we set our null hypothesis to be that our results are consistent with the expected results. Furthermore, in statistics, we never talk about accepting the null hypothesis; only whether we reject or fail to reject it.
Let $$H_0 : p_{r2}=\frac{5}{12}, p_{w2}=\frac{1}{12}, p_{rw}=\frac{1}{2}$$
$$H_a :\text{ at least one of these probabilities does not hold}$$
> observed <- c(61,6,53)
> expected <- c(5/12,1/12,1/2)
> chisq.test(x=observed,p=expected)
Chi-squared test for given probabilities
data: observed
X-squared = 4.8367, df = 2, p-value = 0.08907
Since $0.05\lt0.08907$ we fail to reject the null hypothesis. We do not have significant evidence to conclude that the results are inconsistent with the expectations.
| {
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"source": "stackexchange",
"question_score": "1",
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How does this transformation happen? $$
ax^2+2hxy+by^2 = a\left(x+ \frac{h}{a}y \right)^2 + \frac{ab-h^2}{a}y^2
$$
How do I go from the equation on the LHS to the equation on the right hand side? My study material mentions "completing the square", which I know but I can't understand how it applies to this particular equation.
| For completing the square the coefficient of $\,x^2$ should be one ,so factor out $\,a$ then complete the square as normal with the coefficient of $\,y$ being $\,\frac{2hx}{a}$
ie.
$\,ax^2+2hxy+by^2 $
$= a(x^2+\frac{2hxy}{a}+\frac{by^2}{a}) $
$= a(x^2+\frac{2hxy}{a}+\frac{by^2}{a}+(\frac{hy}{a})^2-(\frac{hy}{a})^2)$
$ = a\big((x +\frac{h}{a}y)^2 -\frac{h^2y^2}{a^2} + \frac{by^2}{a})$
$ = a\big((x +\frac{h}{a}y)^2 + \frac{bay^2-h^2y^2}{a^2})$
$ = a(x +\frac{h}{a}y)^2 + y^2\frac{ba-h^2}{a}$
| {
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Factoring $x^2-y^2-z^2+2yz+x+y-z$ This is a factorization problem on polynomials. I can't find a way to solve it, neither can a math program called MathWay. Help me, please.
$$x^2-y^2-z^2+2yz+x+y-z$$
My book says that the answer is $(x+y-z)(x-y+z+1)$.
| $$\begin{align}x^2-y^2-z^2+2yz+x+y-z&=x^2-(y-z)^2+x+y-z\\&=(x+(y-z))(x-(y-z))+x+y-z\\&=(x+y-z)(x-y+z+1)\end{align}$$
| {
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$2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$ Related to solving $~~2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$
I've tried some arithmetic to find something like
$a^x=b~\Longrightarrow~x=\log_{a}b$
But what I've found is that
$2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$
$2^8\cdot2^{x-5}+3^{x-5}=2^8\cdot2^{3(x-5)}+3^{2(x-5)}$
$256\cdot2^{x-5}+3^{x-5}=256\cdot(2^{x-5})^3+(3^{x-5})^2$
$256\cdot a +b=256\cdot a^3+b^2$
$256 \cdot a(a^2-1)+b(b-1)=0$
I don't know how to solve this equation above. Maybe it doesn't the better way to solve my question (tittle) or maybe it is wrong.
Can someone help me to solve that?
| As @JohnHughes hinted at in the comments, the trick to solving $2^{x+3}+3^{x−5}=2^{3x−7}+3^{2x−10}$ is in realizing that powers of two are always even, and powers of three are always odd, so a power of two won't ever affect the value of a power of three. (Edit: This is only true when you are multiplying powers of two by powers of three, not adding them, so this is one solution, there might be others.)
So, basically,
$$\begin{align*}
2^{x+3}+3^{x−5}=2^{3x−7}+3^{2x−10}
&\Rightarrow \begin{cases}
2^{x+3} = 2^{3x-7} \\
3^{x-5} = 3^{2x-10}
\end{cases} \\
&\Rightarrow
\begin{cases}
x+3 = 3x-7 \\
x-5 = 2x-10
\end{cases} \\
&\Rightarrow
\begin{cases}
2x = 10 \\
x = 5
\end{cases}\\
&\Rightarrow x=5
\end{align*}$$
| {
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"source": "stackexchange",
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Formula for discriminant of a polynomial of degree 2 in 3 variables Which is the correct way/ method/Formula to find the discriminant of a quadratic equation $f$ in 3 variables?
i.e., a quadratic form in 3 variables.
Also, how to conclude that, whether this $f$ is reducible or irreducible?
Some one knows this please help. Thank you very much
| Let us write the general 2nd degree equation under the form:
$$\tag{1}ax^2+2bxy+cy^2+2dxz+2eyz+fz^2.$$
In order for (1) to be expressible under the form of a product of two first degree polynomials:
$$\tag{2}(px+qy+rz)(sx+ty+uz),$$
coefficients $a,b,c,d,e,f$ must satisfy the pair of constraints :
$\tag{3}\begin{cases}b^2 + d^2 + e^2 - ac - af - cf & = & 0\\
acf + 2bde - cd^2 - fb^2 - ae^2 & = & 0\end{cases} \ \ \ \text{with} \ \ \ a + c + f \neq 0.$
Here is a linear algebra explanation.
Let us write equation (1) under its classical matrix form :
$$\begin{pmatrix}x&y&z\end{pmatrix}\underbrace{\begin{pmatrix}a&b&d\\b&c&e\\d&e&f\end{pmatrix}}_A\begin{pmatrix}x\\y\\z\end{pmatrix}$$
Relationship (2) is known to be true (see Appendix) if and only if matrix $A$ has rank one, i.e., if and only if it has exactly two zero eigenvalues.
Expressing that $A$ has exactly 2 zero eigenvalues is the same as saying that its characteristic polynomial is equal to $-\lambda^3+k\lambda^2$ for some nonzero $k$.
Identifying $-\lambda^3+k\lambda^2$ to the characteristic polynomial of $A$:
$$\det(A-\lambda I_3)= - \lambda^3 + (a + c + f) \lambda^2 +(b^2 + d^2 + e^2 - ac - af -cf)\lambda + (acf + 2bde - cd^2 - fb^2 - ae^2),$$
we obtain conditions (3).
Appendix : Rank 1 property is due to the fact that (2) can expressed under the following form:
$$\begin{pmatrix}x&y&z\end{pmatrix}\begin{pmatrix}p&&\\&q&\\&&r\end{pmatrix}\begin{pmatrix}0&&\\&0&\\&&1\end{pmatrix}\begin{pmatrix}s&&\\&t&\\&&u\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}.$$
Remark : A quadratic form in three variables set equal to 0 is the equation of a quadric surface passing through the origin. For example :
$$x^2+y^2-z^2=0 \ \ \text{is the equation of a circular cone with axis Oz.}$$
Expression (2), set equal to $0$, instead of a surface, corresponds to the degenerated case of 2 planes passing through the origin.
Remark: Let
$$\underbrace{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}}_U, \ \ \underbrace{\begin{pmatrix}1&0&1\\-1&1&1\\0&-1&1\end{pmatrix}}_P, \ \
\underbrace{\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}}_D \ \ $$
Rank-one matrix $U$ could have taken instead of rank-one matrix $D$ because
$$\tfrac13U=PDP^{-1}$$
(Columns of matrix $P$ are eigenvectors of $U$ resp. associated with eigenvalues $0,0,1$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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4 digit numbers divisible by 11 Four digit numbers are formed using the
digits 1,2,3,4 (repetition is allowed). The
number of such four digit numbers divisible
by 11 is-
(1) 22 (2) 36 (3) 44 (4) 52
I know for a number to be divisible by 11 the sum of digits at even places must be equal to the sum of those at odd places. But how do I use this to get the answer?
| Let $[abcd]$ be the number. Then $[abcd] \equiv [ab]+[cd] \pmod{99}$ which implies $[abcd] \equiv [ab]+[cd] \pmod{11}$. Because $a,b,c,d \in \{1,2,3,4\}$,
$[abcd] \equiv [(a+c)(b+d)] \pmod{11}$. Hence $[abcd]$ is a multiple of $11$ if and only if $a+c=b+d$.
\begin{array}{|r|r|r|}
\text{sum} &\text{ac and bd events} & \text{counts} \\
\hline
2 & 11 & 1\\
3 & 12, \ 21 & 4\\
4 & 13, \ 22, \ 31 & 9 \\
5 & 14, \ 23, \ 32, \ 41 & 16\\
6 & 24, \ 33, \ 42 & 9\\
7 & 34, \ 43 & 4\\
8 & 44 & 1 \\
\hline
\text{total} && 44
\end{array}
| {
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Visualizing $3^3+4^3+5^3=6^3$ I was wondering how would the visualization of $3^3+4^3+5^3=6^3$ look? I tried some 3D sketches but when calculating none work. Any ideas would be highly appreciated.
| No complete answer but some thoughts:
Subtracting a 5-cube from a 6-cube leaves
$$6^3-5^3=216-5^3=91.$$
Imagine you place the 5-cube into the 6-cube and check what is left over. Lets assume we rotate it so that the 5-square "on ground" is inscribed into the 6-square.
Now we extend the 5-cube by one unit to the top and subtract that as well, thats
$$91-5^2=91-(3^2+4^2)=66.$$
We take these additional 25 volume units we have subtracted from one layer of the 3-cube and one layer of the 4-cube, thats Pythagoras and that can be visualised convincingly. Now we are left with showing
$$3\cdot3\cdot2 + 4\cdot4\cdot3=66.$$
$3\cdot3\cdot2$ and $4\cdot4\cdot3$ is what is left over from the 3- and 4-cube and 66 is apprently the volume of the remaining 4 trigonal prisms (="wedges" and when they are inscribed as suggested wil have sidelengths $3+\sqrt{\frac 72}$, $3-\sqrt{\frac 72}$ and 6 each).
So what remains to show is $$3\cdot3\cdot2 + 4\cdot4\cdot3=66,$$
Since the wedges have hight $6$ and both remainders of the smaller cubes contain 6 as a factor we could divide each figure by $6$ getting four bodies of height 1, and then we can collapse the bodies to 2D polygons arriving at four area $\frac{11}{4}$ triangles each and rectangles with sides $3\times4$, and $4\times2$, representing the equation
$$3 \cdot 1 + 4\cdot 2 = 4 \frac12 \bigg[3-\sqrt{\frac 72}\bigg]\cdot\bigg[3+\sqrt{\frac 72}\bigg]=11$$
I did not come up with a nice diagram to show that but maybe it can be somehow derived from a construction of $\sqrt{\frac{7}{2}}$, somehow in the style of the linked Pythagoras proofs.
| {
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"question_score": "3",
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How is this nested radical expression for $\cos(2\pi /7)$ proven? On the Wolfram Mathworld page for the silver constant, the following identity is present, how is it proven?
$$2\cos\Big(\frac{2\pi}{7}\Big) = \frac{2+\sqrt[3]{\,7+7\sqrt[3]{\,7+...}}}{1+\sqrt[3]{\,7+7\sqrt[3]{\,7+...}}}$$
Link: http://mathworld.wolfram.com/SilverConstant.html
| This is the same as
$$ 2\cos\left(\frac{2\pi}{7}\right)-1=\frac{1}{1+\sqrt[3]{7+7\sqrt[3]{7+\cdots}}}. $$
Write $T=2\cos\left(\frac{2\pi}{7}\right)=\zeta+\zeta^{-1}$ where $\zeta=e^{2\pi i/7}$ and $x=\sqrt[3]{7+7\sqrt[3]{7+\cdots}}$. Then observe
$$ \begin{array}{ll} T & = \zeta+\zeta^{-1} \\ T^2 & = \zeta^2+2+\zeta^{-2} \\ T^3 & = \zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}. \end{array} $$
We know $0=\zeta^7-1=(\zeta-1)(\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1)$. Since $\zeta-1$ isn't $0$, that means the other factor is, and dividing by $\zeta^3$ yields the symmetric form
$$ \zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=0. $$
Therefore, $T^3+T^2-2T-1=0$. Next, observe $x^3=7+7x$ so $x^3-7x-7=0$. Define
$$ A=T-1, \quad B=1+x. $$
We want to show these are inverse of each other. Observe
$$ (A+1)^3+(A+1)^2-2(A+1)-1=0 \\ \implies f(A)=A^3+4A^2+3A-1=0 $$
$$ (B-1)^3-7(B-1)-7=0 \\ \implies g(B)=B^3-3B^2-4B-1=0. $$
In the case of $f(t)$, notice $f'(t)=3t^2+8t+3$ is always positive for $t>0$, and $f(0)=-1$, therefore $f$ has exactly one positive root (namely $t=A$). Now, $B$ is a root of $g(t)$, but $g(t)=-t^3f(1/t)$ is the opposite reciprocal polynomial, so $1/B$ must also be a (necessarily positve) root of $f(t)$. Conclude $A=1/B$, as desired.
| {
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Prove $2 \cdot \sum\limits_{k = 0}^{n} \binom{2n}{2k} = 2^{2n}$ without induction I am aware of
$$
2^{2n} = (1 + 1)^{2n} =
\sum_{k = 0}^{2n} \binom{2n}{k}
$$
I then tried grouping the terms and using that $\binom{n}{k} = \binom{n}{n-k}$, to obtain
$$
\begin{align}
\sum_{k = 0}^{2n} \binom{2n}{k} &=
\binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{n - 1} + \binom{2n}{n} + \binom{2n}{n + 1} + \ldots + \binom{2n}{2n} \\
&=
2 \left[ \binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{n - 1}\right] + \binom{2n}{n} =
\left( 2 \sum_{k = 0}^{n-1} \binom{2n}{k} \right) + \binom{2n}{n}
\end{align}
$$
But I don't know if I'm on the right track or how to progress from here.
| I would also use
$$0 = (1 - 1)^{2n} = \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}.$$
Adding this to the previous sum,
$$2^{2n} = \sum_{k=0}^{2n}\left[(-1)^k\binom{2n}{k} + \binom{2n}{k}\right].$$
When $k$ is odd, the term is $0$, leaving only even $k$ terms (but doubled!). Therefore,
$$2 \sum_{k=0}^{n} \binom{2n}{2k} = 2^{2n},$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the number of ways in which 6 persons out of 5 men and 5 women can be seated at a round table such that 2 men are never together. Find the number of ways in which 6 persons out of 5 men and 5 women can be seated at a round table such that 2 men are never together.
My attempt:
6 people may be 3 men and 3 women, 2 men and 4 women or 1 man and 5 women.
Then by the gap method, 3 men and 3 women can be seated in $\binom{5}{3}2!3!=120$ ways.
2 men and 4 women can be seated in $\binom{5}{4}3!\frac{4!}{2!}=360$ ways
1 men and 5 women can be seated in $\binom{5}{5}4!\frac{5!}{4!}=120$ ways
So the total number of ways is 600, but the answer given in my book is 5400. Where did I go wrong?
| Label the chairs from $1$ to $6$ in increasing order, label the men and the women from $1$ to $5$.
Consider three cases:
1) There is 1 man: choose a man in $5$ ways, place the man at chair $1$, and permute the $5$ women in the remaining chairs in $5!$ ways. Therefore the number of ways is $5\cdot 5!=600$.
2) There are 2 men: choose two men in $\binom{5}{2}$ ways, place the man with the smallest label at chair 1 and the other at chair $3$, $4$ or $5$ ($3$ ways), choose four women in $\binom{5}{4}$ ways and arrange them in the remaining chairs in $4!$ ways. Therefore the number of ways is $\binom{5}{2}\cdot 3\cdot \binom{5}{4}\cdot 4!=3600$.
3) There are 3 men: choose three men in $\binom{5}{3}$ ways, place the man with the smallest label at chair 1, and the other at chair $3$ and $5$ ($2$ ways), choose three women in $\binom{5}{3}$ ways and arrange them in the remaining chairs in $3!$ ways.
Therefore the number of ways is $\binom{5}{3}\cdot2 \cdot \binom{5}{3}\cdot 3!=1200$.
Hence the total number of ways is $600+3600+1200=5400$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)d\left(\frac{\sin x}{1+\cos x}\right)$$
$$=\int_{0}^{\frac{\pi}{2}}x^2d\log\left(\frac{\sin x}{1+\cos x}\right)=x^2\log\left(\frac{\sin x}{1+\cos x}\right)|_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{\sin x}{1+\cos x}\right)dx$$
$$=0+2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{1+\cos x}{\sin x}\right)dx=2\int_{0}^{\frac{\pi}{2}}x\log\left(1+\cos x\right)dx-2\int_{0}^{\frac{\pi}{2}}x\log\left(\sin x\right)dx$$
$$=2\int_{0}^{\frac{\pi}{2}}x\log\cot \left(\frac{x}{2}\right)dx=8\int_{0}^{\frac{\pi}{4}}x\log\cot xdx$$
but I can't proceed next step,help me,thanks.
| $$since\ I=\int_{0}^{\frac{\pi }{2}}\frac{x^2}{sinx}dx=-2\int_{0}^{\frac{\pi }{2}}xln(tan(\frac{x}{2}))dx=-8\int_{0}^{\frac{\pi }{2}}t \ln(tan(t))dt\\
\\
=-8\int_{0}^{1}\frac{lnt\arctan(t)}{1+t^2}dt=-4\pi \int_{0}^{1}\frac{ln(t)}{1+t^2}dt+8\int_{0}^{1}\frac{ln(t)arctan(t)}{1+t^2}dt\\
\\
=4\pi G-8\int_{1}^{\infty }\frac{ln(t)arctan(t)}{1+t^2}dt\\\
\therefore 2I=4\pi G-8\int_{0}^{\infty }\frac{ln(t)arctan(t)}{1+t^2}dt\ , let\ Q=\int_{0}^{\infty }\frac{ln(t)arctan(t)}{1+t^2}dt\\
\\
\therefore Q(a)=\int_{0}^{\infty }\frac{ln(t)arctan(at)}{1+t^2}dt\ \therefore Q'(a)=\frac{1}{4}\int_{0}^{\infty }\frac{ln(t)dt}{(1+t)(1+a^2t)}=\frac{1}{4(1-a^2)}\int_{0}^{\infty }[\frac{1}{1+t}-\frac{a^2}{1+a^2t}]lntdt\\
\\
\therefore Q'(a)=-\frac{\pi }{4(1-a^2)}\frac{d}{da}(\frac{1-a^-2a}{sin(\pi a)})_{a}^{=0}=\frac{\pi }{8(1-a^2)}lim[\frac{-2a.a^{-2a}ln^{2}a}{a^2}]=\frac{lna^{2}}{2(1-a^2)}\ \therefore Q=\frac{1}{4}\int_{0}^{1}\frac{(lna)^{2}}{1-a}da+\int_{0}^{1}\frac{(lna)^{2}}{1+a}da=\frac{7}{8}\zeta 3\\
\\
\therefore I=2\pi G-\frac{7}{2}\zeta(3)\\
\\
ahmed hegazi$$
| {
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"url": "https://math.stackexchange.com/questions/2714146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of five digit numbers that can be formed using the digits 1,2,3,4,5,6,7,8,9 in which one digit appears once and two digits appear twice Find the number of five digit numbers that can be formed using the digits 1,2,3,4,5,6,7,8,9 in which one digit appears once and two digits appear twice (e.g.41174 is one such number but 75355 is not)
The three digits can be chosen from 9 digits in $\binom{9}{1}\binom{8}{1}\binom{7}{1}$ ways.These three can be arranged in $\frac{5!}{2!2!}$ways.So total ways are
$\binom{9}{1}\binom{8}{1}\binom{7}{1}\frac{5!}{2!2!}=15120$ but the correct answer is 7560.
Please help
| Choose two numbers to appear twice and then one number to appear once. Now we have five numbers. Choose two places for the smaller number that appears twice and then the two places for the larger number that appears twice.
$$\binom{9}{2}\binom{7}{1}\binom{5}{2}\binom{3}{2}=7560$$
| {
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How to prove a sequence is strictly increasing and bounded above? For this proof, I'm not sure if I am doing it right. Here is what I have so far. Can anyone please help me out?
Let $\{a_n\}$ be a sequence defined recursively by
$a_1 = \sqrt{6}$
$a_{n+1} = \sqrt{6+5a_n}, n = 1,2,3,...$
Prove that $\{a_n\}$ is strictly increasing and strictly bounded above
$a_2 = \sqrt{6+5\sqrt{6}}< \sqrt{6+5*15} = 9$
$a_3 = \sqrt{6+5a_2}$
$a_3 = \sqrt{6+5\sqrt{6+5 \sqrt{6}}}$
wts there exists an M st $a_n \le M$
Choose $M = 9$
Base Case: $a_1 = \sqrt{6}<\sqrt{81} = 9$
Induction step:
Let k in the natural numbers be arbitrary
Assume $a_k \le 9$
$a_{k+1} = \sqrt{6+5a_k} < \sqrt{6+75} = \sqrt{81} = 9$
Therefore by induction, $a_n \le 9$
$a_{n+1} = \sqrt{6+5a_n}$
$a_{n+2} = \sqrt{6+5a_{n+1}}$
$a_{n+1} \le a_{n+2}$
Therefore, the sequence is strictly increasing and strictly bounded above
| It is correct globally, but I think you are not expressing yourself clearly about why the sequence is strictly increasing. The base case is $a_1<a_2$, which is clearly true. Andn, if $a_n<a_{n+1}$, then $6+5a_n<6+5a_{n+1}$ and therefore $\sqrt{6+5a_n}<\sqrt{6+5a_{n+1}}$. But this means that $a_{n+1}<a_{n+2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)$
Evaluate $$\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)$$
I know that $\arccos\frac12$ is $60^\circ$. I don't know how to continue.
| $$\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)=$$
$$\sin(x+y) = \sin x \cos y + \cos x \sin y$$
Where $$x=\arccos\frac12$$ and $$y= \arccos\frac{7}{25}$$
Note that $$ \cos x =\frac {1}{2}$$ and $$ \cos y =\frac {7}{25}$$
We can find $$ \sin x= {\sqrt 3}/2 $$ and $$\sin y = 24/{25}$$
upon substitution we get $$\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)=$$
$$ {7\sqrt 3}/50 +24/{50} = \frac { 7\sqrt 3 + 24}{50} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717882",
"timestamp": "2023-03-29T00:00:00",
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Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$ Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$
i am trying to solving integrating combination
since given that
$\cot (x^2+y^2)dy+xdx+ydy=0$
then $\cot (x^2+y^2)dy+d(xy)=0$ is it correct way ? and we can apply integration from here? can any one help me this problem
| Dividing both sides of $\cot (x^2+y^2)dy+xdx+ydy=0$ by $dy$ we get $\cot (x^2+y^2)+x\frac {dx}{dy}+y=0$
Then,
$$x\frac {dx}{dy}=-y-\cot (x^2+y^2) \iff \frac {dx}{dy}=\frac{-y-\cot (x^2+y^2)}{x} \iff \frac {dy}{dx}=-\frac{x}{y+\cot (x^2+y^2)}$$
Now you can integrate this: $\frac {dy}{dx}=-\frac{x}{y+\cot (x^2+y^2)}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum value of $\frac{a+b}{2} + \frac{2}{ab-b^{2}}$, with $a,b \in \mathbb{R}$, $a>b>0$ Find the minimum value of $$ \frac{a+b}{2} + \frac{2}{ab-b^{2}},$$
where $a,b \in \mathbb{R}$, $a>b>0$.
Attempt : The only method I knew was using partial derivatives.
Let
$$f(a,b) = \frac{a+b}{2} + \frac{2}{ab-b^{2}},$$
then the partial derivatives are
$$ f_{a} = \frac{1}{2} - \frac{2b}{(ab-b^{2})^{2}},\\
f_{b} = \frac{1}{2} - \frac{2(a-2b)}{(ab-b^{2})^{2}}.$$
Setting them to $0$ implies
$$ \frac{2b}{(ab-b^{2})^{2}} = \frac{2(a-2b)}{(ab-b^{2})^{2}},$$
so
$$ 3b = a $$
with $ ab - b^{2} \ne 0$.
Subtitute this to $f(a,b)$:
$$ f(b) = 2b + \frac{1}{b^{2}} \implies f'(b) = \frac{2b^{3} - 2}{b^{3}}.$$
so we must have $b^{3} = 1 \implies b = 1$, for $f'(b)=0$. So this means $a=3$. Then
$$f(a,b) = 3.$$
How to check whether this is valid minimum global of the expression, preferably without testing $D(a,b) = f_{aa} f_{bb} - (f_{ab})^{2}$?
How to solve another way without multivariable calculus? One way perhaps is by letting $a = kb, k > 1$.
| Using A.M. >= G.M.
$$\cfrac{a-b}{2} + {b} + \cfrac{2}{b(a-b)} >=3\sqrt[3]{\cfrac{a-b}{2}*{b}* \cfrac{2}{b(a-b)}} = 3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{9/2}{2}\right)^2+\frac{9}{2}\right)$$
What am I missing and what is the best way to factorize when we have coefficients?(Is this the right term, ($a x^2$)?)
| Make the polynomial equal to $0$ to find the roots so that you can factorise.
$$2x^2-9x+9=0$$
$$x^2-\frac{9x}{2}+\frac{9}{2}=0$$
$$x^2-\frac{9x}{2}=\frac{-9}{2}$$
To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of b.
$$(\frac{b}{2})^2=(\frac{-9}{4})^2$$
Add the term to each side of the equation.
You get:
$$(x-\frac{9}{4})^2=\frac{9}{16}$$
$$x=\frac{9}{4} \pm \frac{3}{4}$$
$$\implies x=3,\frac{3}{2}$$
Hence, ($x$-$\frac{3}{2}$)($x$-$3$) is your answer.
| {
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If the integers 1-9 are randomly distributed into three sets of 3 integers, what is the probability that at least one set contains only odd integers? The problem https://www.cut-the-knot.org/Probability/ThreeGroupsOfThree.shtml has been stated here,
If the integers 1-9 are randomly distributed into three sets of 3 integers, what is the probability that at least one set contains only odd integers?
My approach is that as there are 1, 3 , 5, 7 , 9 = 5 odd numbers, selecting 3 numbers from here would be 5C3 and total number of groups would be 9C3*6C3 so the probability is 5C3/
| The number of ways of forming three labeled sets of three numbers from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is
$$\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
since we must select three of the nine numbers to be in the first set, three of the remaining six numbers to be in the second set, and all three of the remaining three numbers to be in the third set.
However, if the sets are not labeled, the order in which the sets are selected does not matter. Hence, the number of ways of forming three sets of three numbers from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is
$$\frac{1}{3!}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
Since there are five odd numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, there can be at most one set with three odd numbers since $2 \cdot 3 = 6 > 5$. If three odd numbers are selected to be in one of the sets, then the remaining two odd numbers are placed in one set or two different sets.
Three odd numbers in one set and the other two odd numbers in another set: There are $\binom{5}{3}$ ways to select the three odd numbers that appear in the same set and $\binom{4}{1}$ to select which even number will be placed in the set with the remaining two odd numbers. The remaining three even numbers form the third set. Hence, there are
$$\binom{5}{3}\binom{4}{1}$$
such cases.
Three odd numbers in one set and the other two odd numbers are placed in two different sets: There are $\binom{5}{3}$ ways to select the three odd numbers that appear in the same set. There are $\binom{4}{2}$ ways to select which two of the four even numbers will be placed in the same set as the smaller of the two remaining odd numbers. The remaining three numbers form the third set. Hence, there are
$$\binom{5}{3}\binom{4}{2}$$
such cases.
Thus, the probability that one of the three sets of three numbers will contain only odd integers is
$$\frac{\dbinom{5}{3}\dbinom{4}{1} + \dbinom{5}{3}\dbinom{4}{2}}{\dfrac{1}{3!}\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the $\lim\limits_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$ The task is to find $$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
What I've tried is dividing both the numerator and the denominator by $x$, but I just can't calculate it completely.
I know it should be something easy I just can't see.
Thanks in advance.
| Let $y=-x\to \infty$ then
$$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}=\lim_{y\rightarrow \infty}\frac{\sqrt{y^2+a^2}-y}{\sqrt{y^2+b^2}-y}$$
and
$$\frac{\sqrt{y^2+a^2}-y}{\sqrt{y^2+b^2}-y}\frac{\sqrt{y^2+a^2}+y}{\sqrt{y^2+a^2}+y}\frac{\sqrt{y^2+b^2}+y}{\sqrt{y^2+b^2}+y}=\frac{a^2}{b^2}\frac{\sqrt{y^2+b^2}+y}{\sqrt{y^2+a^2}+y}\\=\frac{a^2}{b^2}\frac{\sqrt{1+b^2/y^2}+1}{\sqrt{1+a^2/y^2}+1}\to \frac{a^2}{b^2}$$
| {
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Is it true that $b^n-a^n < (b-a)nb^{n-1}$ when $0 < a< b$? A Real Analysis textbook says the identity
$$b^n-a^n = (b-a)(b^{n-1}+\cdots+a^{n-1})$$ yields the inequality
$$b^n-a^n < (b-a)nb^{n-1} \text{ when } 0 < a< b.$$
(Note that $n$ is a positive integer)
No matter how I look at it, the inequality seems to be wrong. Take for instance, the inequality does not hold for $n=1$ when one tries mathematical induction. It does not hold for other values of $n$ too. I guess there is something I am missing here and I will appreciate help.
| \begin{align}
b^n-a^n & = (b-a)(b^{n-1}+ b^{n-2}a + b^{n-3}a^2 + b^{n-4}a^3 + b^{n-5} a^4 +\cdots+a^{n-1}) \\[10pt]
& < (b-a)(b^{n-1} + b^{n-2} b + b^{n-3}b^2 + b^{n-4}b^3+ b^{n-5}b^4 + \cdots + b^{n-1}) \\[10pt]
& = (b-a)(b^{n-1} + b^{n-1} + b^{n-1} + b^{n-1} + b^{n-1} + \cdots + b^{n-1}) \\[10pt]
& = (b-a) n b^{n-1}.
\end{align}
The only positive integer $n$ for which this does not work is $n=1,$ where the second factor has only one term, which is $1.$ And in that case it works if you say $\text{“}\le\text{''}$ instead of $\text{“}<\text{''}.$
\begin{align}
b^2-a^2 & = (b-a)(b+a) < (b-a)(b+b) & & = (b-a)2b. \\[10pt]
b^3-a^3 & = (b-a)(b^2 + ba + a^2) < (b-a)(b^2+b^2+b^2) & & = (b-a)3b^2. \\[10pt]
b^4 - a^4 & = (b-a)(b^3+b^2a+ba^2+a^3) \\
& < (b-a)(b^3+b^3+b^3+b^3) & & = (b-a)4b^3. \\[10pt]
b^5-a^5 & = (b-a)(b^4 + b^3a + b^2 a^2 + ba^3 + a^4) \\
& < (b-a)(b^4+b^4+b^4+b^4+b^4) & & = (b-a)5b^4. \\[10pt]
& \qquad\qquad\text{and so on.}
\end{align}
| {
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Calculate the limit: $\lim_{n\to\infty}n^2\left(\left(1+\frac{1}{n}\right)^8-\left(1+\frac{2}{n}\right)^4\right)$ Calculate the limit:
$\lim\limits_{n\to\infty}n^2\left(\left(1+\dfrac{1}{n}\right)^8-\left(1+\dfrac{2}{n}\right)^4\right)$
My first suggestion was that $\lim\limits_{n\to\infty} = 0$. As in both brackets as ${n\to\infty}$: $\dfrac{1}{n}$ and $\dfrac{2}{n}$ will ${\to0}$, so it was going to be $\lim\limits_{n\to\infty}n^2\left(\left(1\right)^8-\left(1\right)^4\right) => \lim\limits_{n\to\infty} = 0$.
My second suggestion was using the properties of $\lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n = e$ and $\lim\limits_{n\to\infty}\left(1+\dfrac{k}{n}\right)^n = e^k$ to find limits of enclosing brackets expressions:
1. $\lim\limits_{n\to\infty}(1+\dfrac{1}{n})^8 => \lim\limits_{n\to\infty}((1+\dfrac{1}{n})^n)^{\frac{8}{n}} => e^{\frac{8}{n}}$
2. $\lim\limits_{n\to\infty}(1+\dfrac{2}{n})^4 => \lim\limits_{n\to\infty}((1+\dfrac{2}{n})^n)^{\frac{4}{n}} => (e^2)^{\frac{4}{n}} => e^{\frac{8}{n}}$
It brought me again to $\lim\limits_{n\to\infty}n^2 *(e^{\frac{8}{n}} - e^{\frac{8}{n}}) => \lim\limits_{n\to\infty} = 0$
However, $0$ is a wrong answer. How to find the limit?
P.S. I am self-study calculus newbie, so please answer as easy as possible (don't know L'Hôpital's rule yet).
| Let $x=1+\frac{1}{n}$ and $y=1+\frac{2}{n}$ then
$$x^8-y^4=(x^2-y)(x^2+y)(x^4+y^2)$$ and
$$x^2-y=\frac{1}{n^2}$$
Thus the limit is
$$(x^2+y)(x^4+y^2) \to 4$$
| {
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Recurrence relation from Generating function - how to find recurrence relation?? Given a generating function equation of the form -
$$ f(x) = A.x.f(x)^2 + B.x.f(x) + C $$
for the sequence $ f(x) = \sum_{n=0}^{\infty} a_n.x^n $. How to find the corresponding recursive relation?
This type of generating function is quite common, for ex- Schröder number's generating function equation is of the form -
$$ f(x) = x.f(x)^2 + x.f(x) + 1 $$, i.e. the former equation with $A=B=C=1$.
It has the recursive relation as follows -
$$ S(n)=\frac{1}{n+1}\left((6n-3)S(n-1)-(n-2)S(n-2)\right) $$ with $S(0)=1$ and $S(1)=2$.
I was reading a paper where the author told of differentiating the equation w.r.t to $x$ and then expand to power series in $x$. But, the complete procedure was not clear. Any help is appreciated.
| $$f(x) = x.f(x)^2 + x.f(x) + 1 \implies x.f(x)^2 + (x-1)f(x) + 1 = 0$$
$$f(x) = \frac{(1-x) \pm \sqrt{(x-1)^2 - 4x}}{2x} = \frac{(1-x) \pm \sqrt{1-6x+x^2}}{2x}$$
To get recurrence from this (and taking the negative root):
$$(1-x) - \sum_{n=0}^{\infty} 2xf_nx^n = (1-3x) - \sum_{n=2}^{\infty} 2f_{n-1}x^n = \sqrt{1-6x+x^2}$$
Squaring we get:
$$(1-3x)^2 + (\sum_{n=2}^{\infty} 2f_{n-1}x^n)^2 - 2(1-3x)\sum_{n=2}^{\infty} 2f_{n-1}x^n = 1 - 6x + x^2$$
$$2x^2 + (\sum_{n=2}^{\infty} f_{n-1}x^n)^2 - \sum_{n=2}^{\infty} f_{n-1}x^n +3x\sum_{n=2}^{\infty} f_{n-1}x^n = 0$$
Rearranging indices, we get:
$$(\sum_{n=2}^{\infty} f_{n-1}x^n)^2 - \sum_{n=3}^{\infty} f_{n-1}x^n +3\sum_{n=3}^{\infty} f_{n-2}x^n = 0$$
Consider first term:
$$(\sum_{n=2}^{\infty} f_{n-1}x^n)^2 = (f_1x^2 + f_2x^3 + \ldots f_{n-3}x^{n-2} + ...)(f_1x^2 + f_2x^3 + \ldots f_{n-3}x^{n-2} + ...)$$
The coefficient of $x^n$ will be of the form:
$$\sum_{i=1}^{n-3} f_if_{n-i-2}$$
So we will get:
$$f_{n-1} - 3f_{n-2} = \sum_{i=1}^{n-3} f_if_{n-i-2} = \sum_{i=0}^{n} f_{i+1}f_{n-i+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$? I'm trying to find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$ in $\mathbb{Z}[\sqrt[3]{2}]$. I know it is a unit, so there is an inverse, but I feel like I may be doing too much work in the wrong direction. Here's what I have so far:
Let $\alpha = 5+4\sqrt[3]{2}+3\sqrt[3]{4}$ and $\alpha^{-1} = a+b\sqrt[3]{2}+c\sqrt[3]{4}$ for some $a,b,c \in \mathbb{Z}$.
$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(5+4\sqrt[3]{2}+3\sqrt[3]{4})=1$
$=5a+6b+8c+4a\sqrt[3]{2}+5b\sqrt[3]{2}+6c\sqrt[3]{2}+3a\sqrt[3]{4}+4b\sqrt[3]{4}+5c\sqrt[3]{4}=1$
$=a(5+4\sqrt[3]{2}+3\sqrt[3]{4})+b(6+5\sqrt[3]{2}+4\sqrt[3]{4})+c(8+6\sqrt[3]{2}+5\sqrt[3]{4})=1$
and trying to solve for a,b, and c, but I don't know how?
Edit: Regrouping to $(5a+6b+8c)+(4a+5b+6c)\sqrt[3]{2}+(3a+4b+5c)\sqrt[3]{4}=1$
| Alt. hint: let $\,x=5+4\sqrt[3]{2}+3\sqrt[3]{4}\,$, then successively multiplying with $\,\sqrt[3]{2}\,$:
$$
\begin{cases}
\begin{align}
5-x+4\sqrt[3]{2}+3\sqrt[3]{4} &= 0 \\
12+(5-x)\sqrt[3]{2}+4 \sqrt[3]{4} &= 0 \\
8 + 12 \sqrt[3]{2}+(5-x)\sqrt[3]{4} &= 0
\end{align}
\end{cases}
$$
Eliminating $\,\sqrt[3]{2}\,$ and $\,\sqrt[3]{4}\,$ between the equations gives in the end:
$$
x^3 - 15 x^2 + 3 x - 1 = 0 \tag{1}
$$
Dividing the polynomial by the known factor $\,x-5-4\sqrt[3]{2}-3\sqrt[3]{4}\,$ gives the factorization:
$$
x^3 - 15 x^2 + 3 x - 1 \\
= \big(x-5-4\sqrt[3]{2}-3\sqrt[3]{4}\big)\cdot\big(x^2 + (-10 + 4 \sqrt[3]{2} + 3 \sqrt[3]{4}) x + 1 - 2\sqrt[3]{2}+\sqrt[3]{4}\big)
$$
Finally, identifying the free terms between the two sides gives:
$$
-1 = \big(-5-4\sqrt[3]{2}-3\sqrt[3]{4}\big)\cdot\big(1 - 2\sqrt[3]{2}+\sqrt[3]{4}\big) \;\;\iff\;\; \frac{1}{5+4\sqrt[3]{2}+3\sqrt[3]{4}} = 1 - 2\sqrt[3]{2}+\sqrt[3]{4}
$$
[ EDIT ] Another way to conclude, while avoiding the polynomial division, is to write $(1)$ as:
$$
\frac{1}{x}=x^2 - 15 x + 3 = \big(5+4\sqrt[3]{2}+3\sqrt[3]{4}\big)^2 - 15 (5+4\sqrt[3]{2}+3\sqrt[3]{4}\big) + 3 = \ldots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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A problem from Makarov Selected Problems in Real Analysis This is a problem from Makarov’s Selected Problems in Real Analysis. Put $S(n) = \sum_{k=0}^{2n}{\frac{k}{k+n^2}}$ Find the limit as n tends to infinity. The answer is $\frac {1}{2} \ln 5$. My solution goes as follows.
Since $y=\frac{x}{x+n^2}$ is increasing on $[0, \infty)$ , we have
$$\int_{0}^{2n}\frac{x dx}{x+n^2}\leq S(n) \leq \int_{0}^{2n+1}\frac{x dx}{x+n^2}$$
Integrating both sides and letting n tend to infinity we get the limit 2. (LHS becomes $2n+n^2\log \frac{n}{n+2}$ and RHS $2n+1+2n^2\log{\frac{n}{2n+1}}$) Could you please point out what is wrong with my answer?
| It is true that
$\sum\limits_{k=0}^{2n}{\frac{k}{k^2+n^2}}\lt S(n)\lt \sum\limits_{k=0}^{2n}{\frac{k}{n^2}}$
Both sides can be formed to Riemann integrals:
$2\frac{1}{2n}\sum\limits_{k=0}^{2n}{\frac{2\frac{k}{2n}}{4\big(\frac{k}{2n}\big)^2+1}}\lt S(n)\lt 4\frac{1}{2n} \sum\limits_{k=0}^{2n}{\frac{k}{2n}}$
$\frac{1}{2}\int\limits_0^1 \frac{8x}{4x^2+1}dx\lt S(n)\lt 4 \int\limits_0^1xdx$
$\frac{1}{2}ln({4x^2+1})\lt S(n)\lt 4 \frac{x^2}{2}$ between $0$ and $1$.
So I got the following bounds: $\frac{1}{2}ln{5}\lt S(n)\lt 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving $45x-3795x^3 +95634x^5 - \cdots + 945x^{41}-45x^{43}+x^{45} = N$? We have the following excerpt from John Stillwell's Mathematics and its History:
My question is, what are the hints which give away the fact that this comes from the expansion of $\sin 45 \theta$?
| Perhaps, it is this:
${\sin n \theta = \dbinom{n}{1}\cos^{n-1}\theta\sin \theta- \dbinom{n}{3}\cos^{n-3}\theta \sin^3 \theta + \dbinom n 5\cos^{n-5}\theta\sin ^{5}\theta...}\\= \color{blue}{\displaystyle\sum_{r=0, 2r+1\le n}(-1)^r\dbinom{n}{2r+1}\cos^{n-2r-1}\theta \sin^{2r+1}\theta} $
Proof:
$(\cos \theta+ i\sin \theta)^n = \cos n\theta + i\sin n\theta$
Writing the binomial expansion of LHS we get:
$\dbinom{n}{0}\cos^n \theta + i\dbinom{n}{1}\cos^{n-1}\theta\sin \theta -\dbinom{n}{2}\cos^{n-2}\theta \sin^2\theta - i\dbinom {n}{3}\cos^{n-3}\theta \sin \theta+\dbinom{n}{4}\cos^{n-4}\theta\sin ^4 \theta+ i\dbinom{n}{5}\cos^{n-5}\theta\sin \theta$
Comparing real and imaginary parts,
${\sin n \theta = \dbinom{n}{1}\cos^{n-1}\theta\sin \theta- \dbinom{n}{3}\cos^{n-3}\theta \sin^3 \theta + \dbinom n 5\cos^{n-5}\theta\sin ^{5}\theta...}$... which is the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Adding $k^2$ to $1^2 + 2^2 + \cdots + (k - 1)^2$. In his book, Calculus Vol. 1, Tom Apostol mentions that adding $k^2$ to the predicate
$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 < \frac{k^3}{3}$$
gives the inequality
$$ 1^2 + 2^2 + \cdots + k^2 < \frac{k^3}{3} + k^2.$$
Why does the RHS $$1^2 + 2^2 + \cdots + (k - 1)^2 $$ become $$ 1^2 + 2^2 + \cdots + k^2 $$ and not $$ 1^2 + 2^2 + \cdots + (k - 1)^2 + k^2$$
when adding $k^2$?
Thank you.
| The notation does not matter. $1^2 + 2^2 + \cdots + (k - 1)^2 + k^2$ and $1^2 + 2^2 + \cdots + k^2$ both stand for the same value (under the appropriate interpretation of $\cdots$). Both stand for $\displaystyle \sum_{j=1}^{k}j^2$.
Example:
$1^2+2^2+3^2 + \cdots + 7^2 = 1^2+2^2+3^2 + \cdots + 6^2 + 7^2 = 1^2+2^2+3^2 + 4^2+5^2 + 6^2 + 7^2 = \displaystyle \sum_{j=1}^{7}j^2=140$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Mini-Challenge on a condition Hello I would like to purpose to you an enigma this is the following :
Let $a,b,c$ be positive real number find the condition on $abc$ and $ab+bc+ca$ and $a+b+c$ to have $a^9+b^9+c^9=3$
I have a solution using the following identity :
$$a^9+b^9+c^9= 3a^3b^3c^3−45abc(ab+bc+ca)(a+b+c)^4+ 54abc(ab+bc+ca)^2(a+b+c)^2−27a^2b^2c^2(ab+bc+ca)(a+b+c) + (a+b+c)^9−9(ab+bc+ca)(a+b+c)^7+ 9(ab+bc+ca)^4(a+b+c)−30(ab+bc+ca)^3(a+b+c)^3+ 18a^2b^2c^2(a+b+c)^3+ 27(ab+bc+ca)^2(a+b+c)^5+ 9abc(a+b+c)^6−9abc(ab+bc+ca)^3$$
But I would like a solution without this because it's too ugly .
So could you help me ?
| We can use the following identity to solve the problem in a more efficient manner than remembering a gigantic identity:
$$a^3+b^3+c^3= \sigma_1^3-3\sigma_1\sigma_2+3\sigma_3$$
where $\sigma_1=a+b+c, \ \sigma_2=ab+bc+ca$, and $\sigma_3=abc$ are the Elementary Symmetric Polynomials. Next we find the cubed versions of the symmetric polynomials as follows
$$a^3b^3+b^3c^3+c^3a^3=(ab+bc+ca)^3-3(ab+bc+ca)(a^2bc+b^2ac+c^2ab)+3a^2b^2c^2$$
$$a^3b^3+b^3c^3+c^3a^3=\sigma_2^3-3\sigma_1\sigma_3+3\sigma_3^2$$
and $$a^3b^3c^3=\sigma_3^3$$
Now we can plug these cubed versions back into our original identity to get
$$a^9+b^9+c^9=(\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3)^3-3(\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3)(\sigma_2^3 - 3\sigma_1\sigma_3+3\sigma_3^2)+\sigma_3^9=3$$
You can then expand if you'd like to obtain your version, but I think I'll leave it here.
| {
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"url": "https://math.stackexchange.com/questions/2736757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Definite integral question with rotation Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
$y=x^2, x = y^2;$ rotate about $y=1$.
So the first thing that jumps out to me is that we're rotating this around $y=1$. So it seems like using vertical washers might do the trick. So I'm going to represent all the equations in terms of x. So $y = x^2$ and $y = \sqrt{x}$
radius outer: $1 - x^2$
radius innter: $1 - \sqrt{x}$
$$ \pi \int_0^1 (1 - x^2)^2\,dx - \pi \int_0^1 (1 - \sqrt{x})^2 \,dx$$
$$ = \pi \int_0^1 (1 - 2x^2 + x^4) \,dx - \pi \int_0^1 (1 - 2\sqrt{x} + x)\,dx$$
$$ = \pi \int_0^1 (-2x^2 + 2\sqrt{x} + x^4 - x)\,dx$$
Is this setup right so far? Is the answer $\frac{\pi * 11}{30}$
| You've done very well. Your work is correct, and indeed, the answer is $\frac{11\pi}{30}$.
Now it's simply a matter of using the power rule for integration, evaluating at $x=1$, and $x=0$, finding a common denominator. I edited your post to include the missing $dx$ in each integral.
$$\pi \int_0^1 (-2x^2 + 2\sqrt{x} + x^4 - x)\,dx = \pi\int_0^1(-2x^2 + 2x^{1/2} + x^4 - x^1)\,dx $$
$$= \left(-\frac 23 x^3 + \frac 43x^{3/2} + \frac 15 x^5- \frac 12x^2\right) \Big|_0^1= \frac {11\pi}{30}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Alternate solutions to similar integrals? So, when I started this problem I couldn't see any reasonable trig substitutions that would make it simpler, and I don't know of a better method, so I attempted to solve it using repeated use of integration by parts. Is there a much easier, better, alternate solution to integrals of this nature in general? I'd also like to know where I went wrong with this solution.
$$
\begin{align*}
\int\sin^4x\,\mathrm{d}x&\rightarrow\\
u_1&=\sin^4x\\
u_1'&=4\cos x\sin^3x\\
v_1&=x\\
v_1'&=1\\
\leftarrow\int\sin^4x\,\mathrm{d}x&=x\sin^4x-4\int x\cos x\sin^3x\,\mathrm{d}x\\
\rightarrow\int x\cos x\sin^3x\,\mathrm{d}x&\rightarrow\\
u_2&=\sin^3x\\
u_2'&=3\cos x\sin^2x\\
v_2&=\int x\cos x\,\mathrm{d}x\\
v_2'&=x\cos x\\
\rightarrow\int x\cos x\,\mathrm{d}x\rightarrow\\
u_3&=x\\
u_3'&=1\\
v_3&=\sin x\\
v_3'&=\cos x\\
\leftarrow\int x\cos x\,\mathrm{d}x&=x\sin x-\int\sin x\,\mathrm{d}x\\
&=x\sin x+\cos x\\
\leftarrow\int x\cos x\sin^3x\,\mathrm{d}x&=\left(\sin^3x\right)\left(x\sin x+\cos x\right)-\int\left(3\cos x\sin^2x\right)\left(x\sin x+\cos x\right)\,\mathrm{d}x\\
&=x\sin^4x+\cos x\sin^3x-3\int x\cos x\sin^3 x\,\mathrm{d}x+3\int\cos^2
x\sin^2x\,\mathrm{d}x\\
4\int x\cos x\sin^3x\,\mathrm{d}x&=x\sin^4
x+\cos x\sin^3 x+3\int\cos^2x\sin^2x\,\mathrm{d}x\\
&=x\sin^4
x+\cos x\sin^3 x+3\int\left(1-\sin^2x\right)\sin^2x\,\mathrm{d}x\\
&=x\sin^4
x+\cos x\sin^3 x+3\int\sin^2x\,\mathrm{d}x-\int\sin^4x\,\mathrm{d}x\\
\rightarrow\int\sin^2x\,\mathrm{d}x\rightarrow\\
\cos2x&=1-2\sin^2x\\
\sin^2x&=\frac{1}{2}(1-\cos2x)\\
\leftarrow\frac{1}{2}\int1-\cos2x\,\mathrm{d}x&=\frac{1}{2}x-\frac{1}{4}\sin{2x}\\
\leftarrow4\int x\cos x\sin^3x\,\mathrm{d}x&=x\sin^4
x+\cos x\sin^3 x+3\left[\frac{1}{2}x-\frac{1}{4}\sin{2x}\right]-\int\sin^4x\,\mathrm{d}x\\
&=x\sin^4 x+\cos x\sin^3 x+\frac{3}{2}x-\frac{3}{4}\sin{2x}-\int\sin^4x\,\mathrm{d}x\\
\leftarrow\int\sin^4x\,\mathrm{d}x&=x\sin^4x-\left[x\sin^4 x+\cos x\sin^3 x+\frac{3}{2}x-\frac{3}{4}\sin{2x}-\int\sin^4x\,\mathrm{d}x\right]\\
&=x\sin^4x-x\sin^4x-\cos x\sin^3x-\frac{3}{2}x+\frac{3}{4}\sin2x+\int\sin^4x\,\mathrm{d}x\\
\end{align*}
$$
As you can see, I can't continue because the integral cancels itself.
| Consider:$$\int\sin^4(x)\,\mathrm{d}x$$Apply the reduction formula:$$=\frac{3}{4}\cdot\int\sin^2(x)\,\mathrm{d}x\,-\frac{\cos(x)\sin^3(x)}{x}$$$$=\frac{3}{4}\cdot\left(\frac{1}{2}x-\frac{\cos(x)\sin(x)}{x}\right)-\frac{\cos(x)\sin^3(x)}{x}$$$$\boxed{=\frac{\sin(4x)-8\sin(2x)+12x}{32}+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to find Jordan blocks from minimal polynomial If the characteristic polynomial is $p=(x-λ)^6$ and the minimal polynomial is $p=(x-λ)^4,$ how do we find all the Jordan blocks?
| From your characteristic polynomial, we see that the matrix has size six-by-six (so the sum of the lengths of the Jordan blocks is six) and only one eigenvalue. From the minimal polynomial, we see that one Jordan block must have length four. That leaves two possibilities for the Jordan blocks:
$$\begin{pmatrix}
\lambda & 1 & 0 & 0\\
0 & \lambda & 1 & 0\\
0 & 0 & \lambda & 1\\
0 & 0 & 0 & \lambda\\
\end{pmatrix},
\begin{pmatrix}
\lambda & 1\\
0 & \lambda\\
\end{pmatrix} \mbox{or}$$
$$\begin{pmatrix}
\lambda & 1 & 0 & 0\\
0 & \lambda & 1 & 0\\
0 & 0 & \lambda & 1\\
0 & 0 & 0 & \lambda\\
\end{pmatrix},
\begin{pmatrix}
\lambda
\end{pmatrix},
\begin{pmatrix}
\lambda
\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Limit of improper integral $ \lim_{x\to \infty} \ \int_1^x x\,e^{t^2-x^2} \,dt$ I need to calculate the limit of the following improper integral:
$$ \lim_{x\to \infty} \ \int_1^x x\,e^{t^2-x^2} \,dt$$
My solution is $\infty$, but when I enter it in WolframAlpha the solution is $\frac{1}{2}$, so I guess mine is wrong, but I can't figure out where. This is my current solution:
$$
First\ calculate\ the\ integral\ (leaving\ away\ the\ limit\ for\ now)\\[10pt]
Let \
u = t^2-x^2 \\
\frac{du}{dt} = 2t \leftrightarrow dt = \frac{du}{2t} \\[20pt]
\
Substitution:\\[5pt]
\begin{align*}
x \ \int_{1-x^2}^0 x\,e^{u} \frac{du}{2t}
&= \frac{x}{2t} \ \int_{1-x^2}^0 e^{u} \,du \\
&= \frac{x}{2t} \ {\bigl (}{e^u{\bigl )}{\bigl \vert }\,}_{1-x^2}^{0} \\
&= \frac{x}{2t} \ {\bigl (}{e^0-e^{1-x^2}{\bigl )}\,} \\
&= \frac{x}{2t} - \frac{e^{1-x^2} \ x}{2t} \\
&= \frac{x - e^{1-x^2} \ x}{2t} \\
&= \frac{x \ (1-e^{1-x^2})}{2t}
\end{align*}\\
\
\\[20pt]
Now\ calculate\ the\ limit:\\[10pt]
\lim_{x\to \infty} \ \frac{x \ (1-e^{1-x^2})}{2t}
= \infty
$$
What have I done wrong, that I don't get the solution $\frac{1}{2}$
| It is
$$
\int \limits_1^x x e^{t^2-x^2} dt = \frac{x}{e^{x^2}} \int \limits_1^x e^{t^2} dt = \frac{x}{e^{x^2}} \cdot (G(x) - G(1))
$$
for some differentiable function $G: [1, \infty) \rightarrow \mathbb{R}$ by the fundamental theorem of calculus. Furthermore, $G'(x) = e^{x^2}$, hence $G$ is strictly increasing and $\lim \limits_{x \to \infty} G(x) = + \infty$. Therefore, we are to calculate the limit
$$
\lim \limits_{x \to \infty} \frac{x \cdot G(x)}{e^{x^2}}
$$
This limit is of the form $\frac{\infty}{\infty}$, hence by L'Hospital's rule:
$$
\lim \limits_{x \to \infty} \frac{x \cdot G(x)}{e^{x^2}} \overset{L'H}{=} \lim \limits_{x \to \infty} \frac{x \cdot G'(x) + G(x)}{2x \cdot e^{x^2}} = \frac{1}{2} + \lim \limits_{x \to \infty} \frac{G(x)}{2x \cdot e^{x^2}}
$$
The second limit is again evaluated using L'Hospital:
$$
\lim \limits_{x \to \infty} \frac{G(x)}{2x \cdot e^{x^2}} = \frac{1}{2} \lim \limits_{x \to \infty} \frac{G'(x)}{e^{x^2} + 2 x^2 \cdot e^{x^2}} = \frac{1}{2} \lim \limits_{x \to \infty} \frac{e^{x^2}}{e^{x^2} + 2 x^2 \cdot e^{x^2}} = \frac{1}{2} \lim \limits_{x \to \infty} \frac{1}{1 + 2 x^2} = 0
$$
All in all, we arrive at
$$
\lim \limits_{x \to \infty} \int \limits_1^x x e^{t^2-x^2} dt = \lim \limits_{x \to \infty} \frac{x \cdot G(x)}{e^{x^2}} = \frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $a,b\in\mathbb N_+$ satisfy $\sqrt{ab},\sqrt{\frac{a^2+b^2}2}\in\mathbb N_+$, is it necessarily true that $a=b$? This question is related to another question where it asks to find pairs of positive integers $(a, b)$ such that
$$
\frac{a + b}{2},\ \sqrt{ab},\ \frac{2ab}{a + b} \in \mathbb{N}_+.
$$
In that question I gave a set of solutions, which is actually all the possible solutions. When I was trying to further incorporate the requirement that $\sqrt{\dfrac{a^2 + b^2}{2}} \in \mathbb{N}_+$, it seems, but I failed to prove, that all pairs such that$$
\sqrt{ab},\ \sqrt{\frac{a^2 + b^2}{2}} \in \mathbb{N}_+ \tag{1}
$$
are trivial, i.e. $a = b$, whereas the requirements$$
\frac{2ab}{a + b},\ \sqrt{\frac{a^2 + b^2}{2}} \in \mathbb{N}_+ \tag{2}
$$
still yields non-trivial solutions (See below).
Here is what I have done so far: I first derived that all positive integer solutions to $ab = c^2$ are$$
(a, b, c) = (km^2, kn^2, kmn), \tag{3}
$$
all positive integer solutions to $2ab = c(a + b)$ are$$
(a, b, c) = (km(m + n), kn(m + n), 2kmn) \tag{4}
$$
or$$
(a, b, c) = (k(2m - 1)(m + n - 1), k(2n - 1)(m + n - 1), k(2m - 1)(2n - 1)), \tag{4'}
$$
and all positive integer solutions to $a^2 + b^2 = 2c^2$ are$$
(a, b, c) = (k\,|m^2 + 4mn + 2n^2|, k\,|m^2 - 2n^2|, k\,|m^2 + 2mn + 2n^2|), \tag{5}
$$
or the positions of $a$ and $b$ swapped.
Thus\begin{align*}
a &= k\,|m^2 + 4mn + 2n^2| (|m^2 + 4mn + 2n^2| + |m^2 - 2n^2|)\\
b &= k\,|m^2 - 2n^2| (|m^2 + 4mn + 2n^2| + |m^2 - 2n^2|)
\end{align*}
are non-trivial solutions to (2). To get solutions to (1), I combined (3) and (5) to get\begin{align*}
k_1 m_1^2 &= k\,|m^2 + 4mn + 2n^2|\\
k_1 n_1^2 &= k\,|m^2 - 2n^2|
\end{align*}
Without loss of generality, assume that $(m, n) = (m_1, n_1) = 1$, then it can be proved that$$
(|m^2 + 4mn + 2n^2|, |m^2 - 2n^2|) = 1 \text{ or } 2.
$$
Now it reduces to two cases:$$
\begin{cases}
|m^2 + 4mn + 2n^2| = m_1^2\\
|m^2 - 2n^2| = n_1^2
\end{cases} \text{ or } \begin{cases}
|m^2 + 4mn + 2n^2| = 2m_1^2\\
|m^2 - 2n^2| = 2n_1^2
\end{cases}.
$$
Now I am not sure if it is the right way to proceed. If there were to be only one equation, it could be transformed to a Pell's equation and explicit solutions are known. However, here it is a system of quadratic equations and I have no idea how to deal with it.
Incidentally, for $|m|, |n| \leqslant 30000$, all pairs of $(m, n)$ satisfying either system of equations are of the form $m + n = 0$, $m + 2n = 0$, $m = 0$, or $n = 0$, all of which lead to $a = b$.
Thus, is it necessarily true that all positive integer solutions to (1) satisfy $a = b$?
| First, it is easy to argue that, since $\sqrt{ab} \in \mathbb N$, there exists a square free integer $k$ and integers $m,n$ such that
$$a=m^2k\\
b=n^2k$$
($k$ is simply the product of the primes which in the prime decomposition of $a$ appear at an odd power).
Then,
$$\sqrt{\frac{a^2+b^2}{2}}=\sqrt{k^2\frac{m^4+n^4}{2}} \in \mathbb N$$
It follows from here that $\sqrt{2( m^4+n^4) }$ is rational and hence integer. Thus there must exists some integer $l_1$ such that
$$2(m^4+n^4)=l_1^2$$
Now, $l_1=2l$ for some integer $l$, and hence
$$m^4+n^4=2l^2$$
Your claim follows now from this post:
$x^4+y^4=2z^2$ has only solution, $x=y=z=1$ .
Indeed, since $m=n=1$, we have
$$a=k=b$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is this set of solutions complete? Let $p,q,r$ be distinct primes and $a,b,c\ge 2$ integers.
The equation $$p^a+q^b=r^c$$ has the following solutions :
$$2^4+3^2=5^2$$
$$2^5+7^2=3^4$$
$$2^2+11^2=5^3$$
$$2^7+17^3=71^2$$
$$7^3+13^2=2^9$$
Can we prove without using any unproven conjecture that this list is complete ?
If not, does the explicit abc-conjecture imply that the list is complete ?
With the explicit abc-conjecture, I mean the following :
Let $a,b$ be coprime positive integers not both $1$, let $c=a+b$ , let $n=rad(abc)$ be the product of the distinct prime factors of $abc$ and $\omega=\omega(n)$ the number of distinct prime factors of $n$. Then, the inequality $$c<\frac{6}{5}\cdot n\cdot \frac{(\ln(n))^{\omega}}{\omega!}$$ holds.
We can assume $p<q$ and one of the primes must be $2$ because otherwise the left side would be even and the right side odd.
| In this answer, I'll assume Explicit abc-Conjecture is true and will show that the set of solution is finite (the upper bound is very big). Since exactly one of $p$, $q$ and $r$ is $2$, we can split the equation into two cases.
i) $r=2$.
The equation is now $p^a+q^b=2^c$. When $a=b=2$, $p^a+q^b=p^2+q^2\equiv 2\pmod 4$, therefore $c=1$ and no solution. Therefore, $\frac{1}{a}+\frac{1}{b}\le\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$. Let $p^a+q^b=2^c=x$, then $p<x^{1/a}$ and $q<x^{1/b}$. It is obvious that $pq<x^{5/6}$. Therefore by Explicit abc,$$x=2^c<\frac{6}{5}\cdot2pq\cdot\frac{(\log{2pq})^3}{6}<\frac{2x^{5/6}\times\log^3(2x^{5/6})}{5}$$and solving it for $x$ gives $x<1.489\times10^{29}$.
ii) $q=2$.
The equation is now $p^a+2^b=r^c$. Similarly when $a=c=2$, the equation becomes $2^b=(r+p)(r-p)$ and one can show that $3^2+2^4=5^2$ is the only solution. Now we can assume $\frac{1}{a}+\frac{1}{c}\le\frac{5}{6}$ and if we let $p^a+2^b=r^c=x$, then $pr<x^{5/6}$. By Explicit abc,$$x=r^c<\frac{6}{5}\cdot2pr\cdot\frac{(\log{2pr})^3}{6}<\frac{2x^{5/6}\times\log^3(2x^{5/6})}{5}$$and this is exactly the same inequality as the first case.
I verified by computer that there is no other solutions assuming explicit abc conjecture. It took ~15 hours with my machine, and the code is available at https://github.com/didgogns/number_theory/blob/master/2748541.py .
| {
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"question_score": "3",
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Generalizing $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}$ If $a+b+c=0$ as discussed in this, this, and this post, then,
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}\tag1 $$
$$ \frac{a^3+b^3+c^3}{3} \times \frac{a^4+b^4+c^4}{2} = \frac{a^7+b^7+c^7}{7}\tag2 $$
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^5+b^5+c^5}{5} = \frac{a^7+b^7+c^7}{7}\tag3 $$
Using these basic identities, we can prove the nice squared identities here,
$$ \frac{a^3+b^3+c^3}{3}\times \frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2 $$
and here for $(a^7+b^7+c^7)^2$. Some investigation shows that,
$$\frac{(a^5+b^5+c^5)}{5\times18}\times\big(9(a^6+b^6+c^6) -(a^3+b^3+c^3)^2\big)= \frac{a^{11}+b^{11}+c^{11}}{11}\tag4$$
$$\frac{(a^7+b^7+c^7)}{7\times18}\times\big(9(a^6+b^6+c^6) +(a^3+b^3+c^3)^2\big)= \frac{a^{13}+b^{13}+c^{13}}{13}\tag5$$
Q: What would be the corresponding identities, as concise as possible, for $p=17$ and $p=19$?
| Define $\;s_k := (a^k+b^k+c^k)/k.\;$ Then
$\;s_{17}=2s_{13}s_4+3s_9s_5s_3,\; s_{19}=8s_8s_7s_4+3s_{11}s_5s_3.$
Not as nice as the ones for $\;s_{11}\;$ and $\;s_{13}\;$ but they are two terms with positive integer coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $x$, $y$, and $z$ are real numbers such that $x+y+z=8$ and $x^2+y^2+z^2=32$, what is the largest possible value of $z$? I tried swapping $z$ from the first equation to the second, and got $$x^2 + x y - 8 x + y^2 - 8 y + 16=0$$ Not sure where to go from there, and if I'm on the right track at all.
| Writing the root-mean square inequality in the form $\;\displaystyle\frac{a^2+b^2}{2} \ge \left(\frac{a+b}{2}\right)^2\;$ which holds for all $\,a,b \in \mathbb{R}\,$ regardless of signs, and using that the largest of $\,x,y,z\,$ must be $\,z \gt 0\,$:
$$
\begin{alignat*}{3}
\frac{x^2+y^2}{2} \ge \left(\frac{x+y}{2}\right)^2 \;\;&\iff\;\; \frac{32-z^2}{2} \ge \left(\frac{8-z}{2}\right)^2 \;\;&&\iff\;\; 2(32-z^2) \ge (8-z)^2 \\
&\iff\;\; z(16-3z) \ge 0 \;\;&&\iff\;\; z \le \frac{16}{3}
\end{alignat*}
$$
The equality case $\,z = \dfrac{16}{3}\,$ is attained for $\,x = y\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I derive the formula for the reciprocal of a hypotenuse? Given $a^2 + b^2 = c^2$,
Why is it that the equation below yields the reciprocal of the hypotenuse c, ($\frac{1}c$)?
$\sqrt{(\frac{a}{a^2+b^2})^2 + (\frac{b}{a^2+b^2})^2}$
Worked example:
$3^2 + 4^2 = c^2$
$c = 5$
$\sqrt{(\frac{3}{3^2+4^2})^2 + (\frac{4}{3^2+4^2})^2}$
= $\sqrt{(\frac{3}{25})^2 + (\frac{4}{25})^2}$
= $\sqrt{(.0144) + (.0256)}$
= $\sqrt{.04}$
= .2 or $1/5$
| Expand the equation:
$$
\begin{align}
\sqrt{\left(\frac{a}{a^2+b^2}\right)^2+\left(\frac{b}{a^2+b^2}\right)^2}&=\sqrt{\left(\frac{a}{c^2}\right)^2+\left(\frac{b}{c^2}\right)^2}\\
&=\sqrt{\frac{a^2}{c^4}+\frac{b^2}{c^4}}\\
&=\sqrt{\frac{c^2}{c^4}}\\
&=\sqrt{\frac{1}{c^2}}\\
&=\frac{1}{c}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $3$ equations with parameters $a, b, c$, evaluate $a^2+b^2+c^2$
Given the equations
$$\begin{align}
x&=cy+bz \\
y&=az+cx \\
z&=bx+ay
\end {align}$$
where $x,y,z$ are not all zero, evaluate $a^2+b^2+c^2$.
I don't understand the way to relate $a,b,c$ and $x,y,z$.
Can someone explain this?
| We can rearrange the equations as $-x+cy+bz=0$, $cx-y+az=0$, $bx+ay-z=0$, and so
$$\begin{pmatrix}-1&c&b\\c&-1&a\\b&a&-1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.$$
This has a solution with $x,y,z$ not all zero if and only if the determinant
$$\begin{vmatrix}-1&c&b\\c&-1&a\\b&a&-1\end{vmatrix}=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Notation/simplification: $(n-1)(n-3)(n-5)...(3)(1)$ How would one write this in simplified form? I am aware that the answer is $\frac{n!}{2^{n/2}(n/2)!}$. How to arrive at this answer?
| The number $n$ is requiered to be even.
$$1 \cdot 3 \cdot 5 \cdot \ldots \cdot (n-5)(n-3)(n-1) \\[1em]
=\frac{1 \cdot \color{blue}{2} \cdot 3 \cdot \color{blue}{4} \cdot 5 \cdot \color{blue}{6} \cdot \ldots \cdot (n-5) \color{blue}{(n-4)} (n-3) \color{blue}{(n-2)} (n-1) \color{blue}{n} }{\color{blue}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (n-4) (n-2) n} }$$
Then factor a $2$ out of each one of the $\tfrac{n}2$ even numbers in the denominator:
$$=\frac{n!}{ 2^{n/2} \big(1 \cdot 2 \cdot 3 \cdot \ldots \cdot (\tfrac{n}2-2) (\tfrac{n}2-1) \tfrac{n}2\big)} \\[1em]
=\frac{n!}{2^{n/2} (\tfrac{n}2)!}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve a trigonometric equation: $|1-2\sin^2 x|=|\cos x|$ I have difficulty in solving this equation:
$$|1-2\sin^2 x|=\lvert\cos x\rvert.$$
What are the indications to solve the exercise correctly?
| Since both sides are absolute values, you can square both sides:
$$\begin{align}
1-4\sin^2x+4\sin^4x&=\cos^2x \\
0&=-3\sin^2x+4\sin^4x \\
&=\sin^2x\left(4\sin^2x-3\right) \\
\sin x&=0,\pm\frac{\sqrt3}2
\end{align}$$
Following a comment by @labbhattacharjee, we note that $\sin(3x)=3\sin(x)-4\sin^3x$, so
$$\begin{align}
0&=-\sin(3x)\sin x \\
x&=\frac{\pi}3k , \quad k\in\mathbb Z.
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve the inequality $x^{12}−x^9+x^4−x+1>0$ using intervals? My book has the question $x^{12}−x^{9}+x^{4}−x+1>0$. The solution given gives three cases, when $x \le 0$, when $0 < x \le 1$ and when $x > 1$. How did they get these intervals? What is the method used?
| Hint: When $x=0$ we get $1>0$ and when $x<0$ then $-x^9,-x>0$ so $$x^{12}-x^9+x^4-x+1>0$$
For $$0<x\le 1$$ we get $$x=1$$ then we get $$1>0$$ and we substitute $$x=\frac{1}{y}$$ so we obtain
$$1+y^3(y^5-1)+y^{11}(y-1)>0$$
and for $x>1$ we have $$x^9(x^3-1)+x(x^3-1)+1>0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Without using a calculator, is $\sqrt[8]{8!}$ or $\sqrt[9]{9!}$ greater? Which is greater between $$\sqrt[8]{8!}$$ and $$\sqrt[9]{9!}$$?
I want to know if my proof is correct...
\begin{align}
\sqrt[8]{8!} &< \sqrt[9]{9!} \\
(8!)^{(1/8)} &< (9!)^{(1/9)} \\
(8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\
(8!)^{(9/72)} - (9!)^{8/72} &< 0 \\
(9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) &< 0 \\
\left(\frac{8!}{9!}\right)^{(1/72)} - 1 &< 0 \\
\left(\frac{8!}{9!}\right)^{(1/72)} &< 1 \\
\left(\left(\frac{8!}{9!}\right)^{(1/72)}\right)^{72} &< 1^{72} \\
\frac{8!}{9!} < 1 \\
\frac{1}{9} < 1 \\
\end{align}
if it is not correct how it would be?
| Another way to see this is to convert both sides into two "averages".
*
*"Uniform distribution on $\{\log1,\dots,\log8\}$": $$\frac{\log1 + \dots + \log8}{8}$$
*"Uniform distribution on $\{\log1,\dots,\log9\}$": $$\frac{\log1 + \dots + \log9}{9}$$
Intuitively, the later has a greater "expectation", so the result follows. If you're not satisfied with this probabilistic interpretation, break the later into a sum of two terms and regroup them as
\begin{align}
& \frac{\log1 + \dots + \log8}{8} < \frac{\log1 + \dots + \log8}{9} + \frac{\log9}{9} \\
&\iff (\log1 + \dots + \log8) \left(\frac18-\frac19\right) < \frac{\log9}{9} \\
&\iff \frac{\log1 + \dots + \log8}{72} < \frac{\log9}{9} \\
&\iff \log1 + \dots + \log8 < 8 \log9
\end{align}
The last inequality is true since $\log$ is strictly increasing.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
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Definite Integral involving Max function. Here is an integral I was attempting to solve $\int\limits_{0}^{\ln t}{\max{\left(1,x\right)dx}}$ but my answer is not coming to be correct. What is wrong in my attempt?
Attempt 1. Let's call the integral $I$
Case 1. Let $0<t<e$ $\implies$ $\ln t < 1$
This means the given integral becomes $\int\limits_{0}^{\ln t}{1 dx}$ because from $(0,1)$ the maximum between $1$ and $x$, is $1$ $$\implies I=\ln t \tag{1}$$
Case 2. Let $e<t< \infty$ $\implies$ $\ln t >1$
This means the given integral becomes $I= \int\limits_{1}^{\ln t}{\max{\left(1,x\right)dx}} = \int\limits_{1}^{\ln t}x dx$ because from $(0,\infty)$ the maximum between $1$ and $x$, is $x$ $$\implies I=\left[\frac{x^2}{2}\right]_{1}^{\ln t}= \frac{(\ln t)^2-1}{2} \tag{2}$$
Adding $(1)$ and $(2)$ we get $$I = \ln t + \frac{(\ln t)^2-1}{2}$$
But this is not the correct answer.
Attempt 2. We know that $\max{\left(f(x),g(x)\right)} = \frac{f(x)+g(x)}{2}+\frac{|f(x)-g(x)|}{2}$
Hence $I=\int\limits_{0}^{\ln t}{\max{\left(1,x\right)dx}}=\int\limits_{0}^{\ln t}{(\frac{1+x}{2}+\frac{|1-x|}{2})dx}$
$$\implies \int\limits_{0}^{\ln t}{\frac{1}{2}dx}+\int\limits_{0}^{\ln t}{\frac{x}{2}dx}+\int\limits_{0}^{\ln t}{\frac{|1-x|}{2}dx}$$
The first two integrals will give us $\frac{\ln t}{2} + \frac{(\ln t)^2}{4} \tag{1}$
Let's calculate the third integral $I_3 = \int\limits_{0}^{\ln t}{\frac{|1-x|}{2}dx}$
Case 1. Let $0<t<e$ $\implies$ $\ln t < 1$
$I_3 = \int\limits_{0}^{1}{\frac{x-1}{2}dx}=\left[\frac{x^2}{4}-\frac{x}{2}\right]_{0}^{1}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4} \tag{2}$
Case 2. Let $e<t< \infty$ $\implies$ $\ln t >1$
$I_3 = \int\limits_{1}^{\ln t}{\frac{1-x}{2}dx} = \left[\frac{x}{2}-\frac{x^2}{4}\right]_{1}^{\ln t} = \frac{\ln t}{2} - \frac{(\ln t)^2}{4} - \frac{1}{4} \tag{3}$
Adding $(1)$,$(2)$ and $(3)$, we get $$I=\ln t -\frac{1}{2}$$
But this also is not the correct answer.
Please tell me where am I doing wrong?
| Correct Attempt 1. Let's call the integral $I$
Case 1. Let $0<t\le e$ $\implies$ $\ln t \le 1$
This means the given integral becomes $\int\limits_{0}^{\ln t}{1 dx}$ because from $(0,1)$ the maximum between $1$ and $x$, is $1$ $$\implies I=\ln t \tag{1}$$
Case 2. Let $e\le t< \infty$ $\implies$ $\ln t \ge 1$
This means the given integral becomes $\color{red}{I= \int\limits_{0}^{\ln t}{\max{\left(1,x\right)dx}} = \int\limits_{0}^{1}{1 dx} + \int\limits_{1}^{\ln t}x dx}$ because from $(1,\infty)$ the maximum between $1$ and $x$, is $x$ $$\implies I=1 + \left[\frac{x^2}{2}\right]_{1}^{\ln t}= 1 + \frac{(\ln t)^2-1}{2}=\frac{(\ln t)^2+1}{2} \tag{2}$$
We get combining $(1)$ and $(2)$ $$I = \begin{cases} \ln t & \text{if } 0<t \le e, \\ \frac{(\ln t)^2+1}{2} & \text{if } e\le t<\infty. \end{cases}$$
Correct Attempt 2. We know that $\max{\left(f(x),g(x)\right)} = \frac{f(x)+g(x)}{2}+\frac{|f(x)-g(x)|}{2}$
Hence $I=\int\limits_{0}^{\ln t}{\max{\left(1,x\right)dx}}=\int\limits_{0}^{\ln t}{(\frac{1+x}{2}+\frac{|1-x|}{2})dx}$
$$\implies \int\limits_{0}^{\ln t}{\frac{1}{2}dx}+\int\limits_{0}^{\ln t}{\frac{x}{2}dx}+\int\limits_{0}^{\ln t}{\frac{|1-x|}{2}dx}$$
The first two integrals will give us $$\frac{\ln t}{2} + \frac{(\ln t)^2}{4} \tag{1}$$
Let's calculate the third integral $I_3 = \int\limits_{0}^{\ln t}{\frac{|1-x|}{2}dx}$
Case 1. Let $0<t\le e$ $\implies$ $\ln t \le 1$
$\color{red}{I_3 = \int\limits_{0}^{\ln t}{\frac{1-x}{2}dx}=\left[\frac{x}{2}-\frac{x^2}{4}\right]_{0}^{\ln t}=\frac{\ln t}{2} - \frac{(\ln t)^2}{4}} \tag{2}$
Case 2. Let $e\le t< \infty$ $\implies$ $\ln t \ge 1$
$$\color{red}{I_3 = \int\limits_{0}^{1}{\frac{1-x}{2}dx} + \int\limits_{1}^{\ln t}{\frac{x-1}{2}dx} = \left[\frac{x}{2}-\frac{x^2}{4}\right]_{0}^{1} + \left[\frac{x^2}{4}-\frac{x}{2}\right]_{1}^{\ln t}}$$ $$\implies I_3 = \left[\frac{1}{4}\right] + \left[\frac{(\ln t)^2}{4} - \frac{\ln t}{2} + \frac{1}{4}\right] = \frac{(\ln t)^2}{4} - \frac{\ln t}{2} + \frac{1}{2} \tag{3}$$
Combining $(1)$and$(2)$, and $(1)$and$(3)$ separately, we get $$I = \begin{cases} \ln t & \text{if } 0<t\le e, \\ \frac{(\ln t)^2+1}{2} & \text{if } e\le t<\infty. \end{cases}$$
Both attempts give us the same solution.
$\color{red}{\text{The corrections have been pointed out in red.}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplifying the derivative of $x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$ $$x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$$
So I get:
$$-x^{\frac{2}{3}} \cdot \frac{1}{3} (6-x) ^{\frac{-2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {\frac{-1}{3}}$$
How does one go about simplifying this?
I guess I can pull out common terms like this:
$$\frac{1}{3} x ^{-\frac{1}{3}} (6-x)^{\frac{-2}{3}} ( -x + (6-x) \cdot 2)$$
Is that right?
| $$f^3 (x)=x^2 (6-x) $$
by differentiation
$$3f^2 (x)f'(x)=2x (6-x)-x^2=3x (4-x) $$
thus
$$f'(x)=\frac {x (4-x)}{f^2 (x)} $$
$$=\frac {x (4-x)}{x^\frac43 (6-x)^\frac23} $$
$$=\frac {4-x}{(6-x)^\frac23}x^{\frac {-1}{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum power of $2$ which divides $3^{1024}-1$ What is the maximum power of $2$ which completely divides $3^{1024}-1$?
I proceeded thus:
$\phi(2^n)=2^{n-1}$ for all $n\ge1$
$$3^{1024}=3^{2^{10}}\equiv1\pmod {2^{11}}$$
$$3^{1024}-1\equiv0\pmod {2^{11}}$$
Since $\phi(2^{11})=2^{10}$. So, maximum power of $2$ must be $11$.
But the answer says it is $12$.
Where am I wrong and how to solve it correctly?
| Just to give a different approach, since $3^4=81=1+80=1+5\cdot2^4$, we have
$$3^{1024}-1=(1+80)^{256}-1=256\cdot80+{256\choose2}80^2+{256\choose3}80^3+\cdots\\
=2^8(5\cdot2^4)+(2^7\cdot255)(5^2\cdot2^8)+(2^7\cdot85\cdot254)(5^3\cdot2^{12})+\cdots\\
\equiv5\cdot2^{12}\mod2^{13}$$
all terms after the first having higher and higher powers of $2$. Thus $3^{1024}-1$ is divisible by $2^{12}$ but not by $2^{13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2770741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find polynomial over Z2, that has the least degree, for which the field P=GF(64) is the minimum decomposition field. How much such polynomial are? Find polynomial over Z2, that has the least degree, for which the field P=GF(64) is the minimum decomposition field. How much such polynomials are?
Please, answer in detail! I think that these polynomials are all sixth irreducible polynomials, but I don't know how to validate this fact.
| This late answer uses computer power to find explicitly the irreductible polynomials of degree six in $\Bbb F_2[X]$.
There exists up to isomorphy only one field with $2^6$ elements, $F=\Bbb F_{64}$, its elements are fixed by the corresponding Frobenius isomorphism, so $F$ contains the roots of
$$x^{2^6}-x\ .$$
sage gives the following factorization for it:
sage: R.<x> = PolynomialRing( GF(2) )
sage: for f, multiplicity in ( x^(2^6) - x ).factor():
....: print f
....:
x
x + 1
x^2 + x + 1
x^3 + x + 1
x^3 + x^2 + 1
x^6 + x + 1
x^6 + x^3 + 1
x^6 + x^4 + x^2 + x + 1
x^6 + x^4 + x^3 + x + 1
x^6 + x^5 + 1
x^6 + x^5 + x^2 + x + 1
x^6 + x^5 + x^3 + x^2 + 1
x^6 + x^5 + x^4 + x + 1
x^6 + x^5 + x^4 + x^2 + 1
How many elements in $F$ are roots of a polynomial of degree exactly $6$? There are, by inclusion/exclusion $|F_{2^6}|-|F_{2^3}|-|F_{2^2}|+|F_{2^1}|=2^6-2^3-2^2+2=64-8-4+2=54$ such elements, so we expect $54/6=9$ (edited division...) irreducible polynomials of degree $6$ as factors above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove this stronger inequality
Let $a,b,c>0$, and $a+b+c=1$,show that
$$\sum_{cyc}\dfrac{a^3+b^2}{b+c}\ge\dfrac{2}{3}+\dfrac{5+\sqrt{2}}{12}\sum_{cyc}(a-b)^2\tag{1}$$
I have prove $$\sum_{cyc}\dfrac{a^3+b^2}{b+c}\ge\dfrac{2}{3}$$
because Use Holder inequality
$$\sum_{cyc}\dfrac{a^3}{b+c}\cdot\sum(b+c)\sum_{cyc}(1)\ge (a+b+c)^3$$
so $$\sum_{cyc}\dfrac{a^3}{b+c}\ge\dfrac{1}{6}$$
and
$$\sum_{cyc}\dfrac{b^2}{b+c}\ge\dfrac{(a+b+c)^2}{2(a+b+c)}=\dfrac{1}{2}$$
For $(1)$ I can't,
| We'll prove a stronger inequality:
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=1$. Prove that
$$\sum_{cyc}\frac{a^3+b^2}{b+c}\geq\frac{2}{3}+\frac{2}{3}\sum_{cyc}(a-b)^2.$$
Indeed, by C-S
$$\sum_{cyc}\frac{b^2}{b+c}\geq\frac{(a+b+c)^2}{2(a+b+c}=\frac{1}{2}.$$
Thus, it's enough to prove that
$$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{1}{6}+\frac{4}{3}\sum_{cyc}(a^2-bc)$$ or
$$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c)^2}{6}+\frac{4}{3}\sum_{cyc}(a^2-bc)$$ or
$$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{1}{2}\sum_{cyc}(3a^2-2ab)$$ or
$$\sum_{cyc}\left(\frac{2a^3}{b+c}-3a^2+ab+ac\right)\geq0$$ or
$$\sum_{cyc}\frac{a(2a^2+b^2+c^2-3ab-3ac+2bc)}{b+c}\geq0$$ or
$$\sum_{cyc}\frac{a((c-a)(c+b-a)-(a-b)(b+c-a))}{b+c}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{b(a+c-b)}{a+c}-\frac{a(b+c-a)}{b+c}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2(a^2+b^2-c^2)}{(a+c)(b+c)}\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)(a^2+b^2-c^2)\geq0.$$
Now, let $a\geq b\geq c$.
Thus,
$$\sum_{cyc}(a-b)^2(a+b)(a^2+b^2-c^2)\geq$$
$$\geq(a-c)^2(a+c)(a^2+c^2-b^2)+(b-c)^2(b+c)(b^2+c^2-a^2)\geq$$
$$\geq(b-c)^2(a+c)(a^2-b^2)+(b-c)^2(b+c)(b^2-a^2)=(b-c)^2(a-b)^2(a+b)\geq0.$$
| {
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"url": "https://math.stackexchange.com/questions/2774801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Piecewise function as initial condition in Heat Equation exercise - Confirm solution I have $u_t = u_{xx}$ for $0<x<1$,
$u(0,t) = 0, \; 0 \leq t < \infty$
$u(1,t) = 0, \; 0 \leq t < \infty$
$u(x,0) = \varphi(x), \; 0 \leq x \leq 1$ where $\varphi(x) = \begin{cases}
\frac{5x}{2}, 0 < x < \frac{2}{3} \\
3-2x, \frac{2}{3} < x < 1
\end{cases}$
I know the general solution to the heat equation written as $u_t = \alpha^2 u_{xx}$ for $0 < x< l$, $t > 0$
$u(0,t) = 0, \; u(l,t) = 0$ for all $t > 0$ and $u(x,0) = f(x)$ for all $0<x<l$ is $u(x,t) = \sum_{k=1}^{\infty} \beta_k \sin(\frac{k\pi}{l}x)e^{\frac{-\alpha^2 \pi^2 k^2 t}{l^2}}$ where $\beta_k = \frac{2}{l}\int_0^l sin(\frac{k \pi x}{l})dx$. This is derived directly from applying the method of separation of variables.
I've never had a piece-wise function as the initial condition for this sort of exercise so I want to confirm my thoughts for the solution. I think $u(x,t) = \sum_{k=1}^{\infty} \beta_k \sin(\frac{k\pi}{1}x)e^{\frac{-\alpha^2 \pi^2 k^2 t}{1^2}}$ where $\beta_k = \begin{cases} \frac{2}{\frac{2}{3}}\int_0^\frac{2}{3} \frac{5x}{2}\sin(\frac{k \pi}{1}x)e^{\frac{-1^2 \pi^2 k^2 t}{1^2}} ,\; 0 < x<\frac{2}{3} \\ \frac{2}{1}\int_\frac{2}{3}^1 (3-2x)\sin(\frac{k \pi}{1}x)e^{\frac{-1^2 \pi^2 k^2 t}{1^2}} ,\; \frac{2}{3} < x<1\end{cases}$.
Is this correct?
| Yes, this is correct, it is just an integration by part.
| {
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"timestamp": "2023-03-29T00:00:00",
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Binomial Inequality in which Binomial coefficient is in square root.
$$1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\leq \frac{1}{2}\cdot (2^{100}-1)+\frac{20301}{12}$$
where $\displaystyle C_{r}=\binom{n}{r}$
Try: Using Cauchy Schwarz Inequaity
$$\bigg(1^2+2^2+\cdots \cdots +100^2\bigg)\bigg(C_{1}+C_{2}+\cdots \cdots +C_{100}\bigg)\geq \bigg(1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\bigg)^2$$
$$\bigg(1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\bigg)\leq \bigg[\frac{100\cdot 101\cdot 201}{6}\cdot (2^{100}-1)\bigg]^{\frac{1}{2}}$$
i am not understand how can i prove my original inequality,
could some help me , Thanks
| Hint.
$$
2\bigg[\frac{100\cdot 101\cdot 201}{6}\cdot (2^{100}-1)\bigg]^{\frac{1}{2}}\le \frac{100\cdot 101\cdot 201}{6} + (2^{100}-1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrix satisfying $A-I = A^{-1}$ Recall the (infinitely) continued fraction definition of the golden ratio
\begin{align}
\phi = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}}}
\end{align}
This is equivalent to the expression
\begin{align}
\phi = 1+\frac{1}{\phi}
\end{align}
Saying that the inverse of $\phi$ is equal to $\phi-1$. Inspired by this fact, consider a square matrix $A$ satisfying:
\begin{align}
A-I = A^{-1}
\end{align}
Does this relationship have any significance?
| $$\begin{align}A-I = A^{-1}\end{align}$$
$$\begin{align}A = I + A^{-1}\end{align}$$
$$\begin{align}A^2 = A + I\end{align}$$
$$\begin{align}A^3 = A^2 + A\end{align}=2A+I$$
$$\begin{align}A^4 = A^3 + A^2\end{align}=3A+2I$$
The pattern is now suggesting,
$$A^n = F_n A+F_{n-1}I$$ where $F_n$ is the Fibonacci's sequence.
| {
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"url": "https://math.stackexchange.com/questions/2778724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Complex Number Series If
$$
z + 1/z = -1
$$
where $z$ is a complex number then value of the sum from r = 1 to r = 99 of $$
( z^r + 1/z^r)^2
$$
is equal to
A) 198
B) 3
C) 99
D) 0
I tried by putting z = x + iy but that gives, |z| = 1
| Define $$f_n(z) = z^n + z^{-n}.$$ Then note $$f_1(z) f_n(z) = (z+z^{-1})(z^n + z^{-n}) = z^{n+1} + z^{-(n+1)} + z^{n-1} + z^{-(n-1)} = f_{n+1}(z) + f_{n-1}(z).$$ In the case where $f_1(z) = -1$, we find $$f_{n+1}(z) = -f_n(z) - f_{n-1}(z).$$ Also note that $$f_0(z) = z^0 + z^{-0} = 1 + 1 = 2.$$ Unfolding the first few terms of this recurrence, we obtain $$f_2(z) = -(-1 + 2) = -1 = f_1(z), \\
f_3(z) = -(-1 + (-1)) = 2 = f_0(z).$$
This shows that the sequence is periodic: we will have $f_4(z) = f_5(z) = -1$, $f_6(z) = 2$, etc., and in general, $$f_{3k+1}(z) = f_{3k+2}(z) = -1, \quad f_{3k}(z) = 2.$$ It follows that $$\sum_{r=1}^{99} f_r^2(z) = \sum_{k=1}^{33} f_{3k-2}^2(z) + f_{3k-1}^2(z) + f_{3k}^2(z) = \sum_{k=1}^{33} (-1)^2 + (-1)^2 + 2^2 = 33(1 + 1 + 4) = 198.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find recurrence for $f(x)$ coefficient in the expression $f(x)\cdot(1+2x+2x^2+x^3)=\frac{1}{1-x^3}$ using generating functions?
Let $f(x)=\sum_{i=0}^{\infty} a_ix^i$. Also $f(x)\cdot(1+2x+2x^2+x^3)=\frac{1}{1-x^3}$. Find numbers $b,c,d$ such that:
$$
a_n={3+n-1 \choose n}-ba_{n-1}-ca_{n-2}-da_{n-3}
$$
for $n\ge 3$.
The expression can be modified to the following using generating functions:
$$
f(x)=\frac{1}{1-x^3} : (1+2x+2x^2+x^3)=\frac{1}{1-x^3}:(1+x)(1+x+x^2)=\frac{(1-x)^2}{(1-x^3)^2(1-x^2)}
$$
Then:
$$
f(n)=\sum_{n=0}^2 {2\choose n}(-1)^nx^n\cdot \sum_{n=0}^{\infty}{2+n-1\choose n}x^{3n}\cdot \sum_{n=0}^{\infty}x^{2n}
$$
For example:
$$
a_0={2\choose 0}{1\choose 0}\cdot 1=1\\
a_1=-{2\choose 1}{2\choose 1}\cdot 1=-4\\
a_2={2\choose 2}{3\choose 2}\cdot 1=1
$$
but what happens with $\sum_{n=0}^2 {2\choose n}(-1)^nx^n$ for $a_n$ where $n\ge 3$? Its range for $n$ is only between $0$ and $2$ so does it become $1$ for all $n\ge 3$? If it does then:
$$
a_3 = {4\choose 3}\cdot 1=4
$$
Then:
$$
a_3={5\choose 3}-3b+4c-d=10-3b+4c+d=4\implies 3b-4c-d=6 \qquad \ast
$$
But there's an infinite amount of solutions for the above equation so I don't think I'm in the right direction.
Note: as noticed by @ancientmathematician there was a mistake in the conditions to the problem it should be $f(x)\cdot(1+2x+2x^2+x^3)=\frac{1}{(1-x)^3}$
| I believe Claude is correct about the statement of the problem, but I will attempt to provide some useful commentary as is.
So if we substitute $\sum_{i=0}^{\infty} a_{i}x^i$ in for $f(x)$, we obtain
$$\sum_{i=0}^{\infty} a_{i}x^i + 2x\sum_{i=0}^{\infty} a_{i}x^i + 2x^2\sum_{i=0}^{\infty} a_{i}x^i + x^3\sum_{i=0}^{\infty} a_{i}x^i = \sum_{i=0}^{\infty}x^{3i}$$
If you insert the $x$'s into the sums, and peel off all terms with degree $2$ or less, we obtain
$$a_0 + (2a_0+a_1)x + (2a_0 + 2a_1 + a_2)x^2 + \sum_{i=3}^{\infty} (a_{i}+ 2a_{i-1} + 2a_{i-2} + a_{i-3})x^i = \sum_{i=0}^{\infty} x^{3i}$$
Equating coefficients, we can say
$$a_0 = 1$$
$$2a_0+a_1 = 0$$
$$2a_0 + 2a_1 + a_2 = 0$$
$$a_{i}+ 2a_{i-1} + 2a_{i-2} + a_{i-3} = 1, i\geq3$$
Where the last equation holds if $i$ divides $3$, and is $0$ otherwise.
The first three equations give initial values $a_0, a_1, a_2$, and the third statement leads me to believe Claude's comment, since that would yield the desired binomial coefficient in place of the $1$ in the recurrence relation. Hope this helps!
Edit: Notice
$$\frac{1}{(1-x)^3} = \sum_{n\geq 0} \dbinom{-3}{n}(-1)^nx^n = \sum_{n\geq 0} \frac{(-3)(-3-1)...(-3-n+1)}{n!}(-1)^nx^n$$
$$=\sum_{n\geq 0} \frac{(3)(4)...(3+n-1)}{n!}x^n =\sum_{n\geq 0}\frac{(2)(3)(4)...(3+n-1)}{(3-1)n!}x^n = \sum_{n\geq 0}\dbinom{3+n-1}{n}x^n$$
So if you were to have equated coefficients as above, you would get the desired recurrence relation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Maximum Likelihood estimator for $r$ I have a set of data
$$X \sim N_2(0,\Sigma) \quad \text{with} \quad \Sigma = \pmatrix{ \sigma^2 & r\sigma^2 \\ r\sigma^2 & \sigma^2}$$
and I am asked to find the Maximum Likelihood estimates of $r$.
Firstly I find the log likelihood of my data which is
$$-n\log(2\pi) - 2n\log(\sigma) - \frac{n}{2} \log(1-r^2) - \frac{1}{2\sigma^2(1-r^2)}\sum_i (x_{i1}^2 - 2rx_{i1}x_{i2} + x_{i2}^2)$$
Then I want to differentiate this w.r.t. to $r$ which gives
$$-\frac{n}{2}\bigg(\frac{-2r}{1-r^2}\bigg) - \frac{1}{2\sigma^2}\bigg(\frac{2r}{(1-r^2)^2}\bigg)\sum_i (x_{i1}^2 - 2rx_{i1}x_{i2} + x_{i2}^2) - \frac{1}{2\sigma^2(1-r^2)}\sum_i -2x_{i1}x_{i2}$$
Now the question says show that the MLE for $r$ is given as
$$\hat r = 2\frac{\sum_i x_{i1}x_{i2}}{\sum_i (x_{i1}^2 + x_{i2}^2)}$$
but I can't seem to get to this. Can someone please put me out of my misery. I've spent far too long on this.
Edit: I had some mistypes. It's $x_{i2}^2$ not $2x_{i2}^2$
| Write the log-likelihood as
$$
\ell(r,\sigma^2)=-\frac{n}{2}\log(1-r^2)-\frac{1}{2\sigma^2}\frac{1}{1-r^2}\left(A-2rB\right) - n\log\sigma^2,
$$
where
$$
A = \sum_{i}x_i^2+y_i^2, \qquad B\sum_{i}x_iy_i.
$$
Then $\frac{\partial \ell}{\partial r}=0$ implies that
$$
nr\sigma^2+\frac{1}{1-r^2}\left((r^2+1)B - Ar \right) =0 \tag{1}.
$$
Also $\frac{\partial \ell}{\partial \sigma^2} = 0$ further implies
$$
n\sigma^2 = \frac{1}{2(1-r^2)}\left(A - 2rB\right) \tag{2}.
$$
Put $(2)$ into $(1)$ to get
\begin{align}
\frac{r}{2(1-r^2)}\left(A-2rB \right) + \frac{1}{1-r^2}\left((r^2+1)B - Ar \right) = 0,
\end{align}
or
$$
rA - 2r^2B + 2r^2B + 2B - 2Ar = 0 \iff r=\frac{2B}{A}.
$$
Couple more details on the derivation of $(1)$
\begin{align}
\frac{\partial \ell}{\partial r}&=n\frac{r}{1-r^2} - \frac{1}{\sigma^2}\frac{r}{(1-r^2)^2}\left[ A-2rB \right] + \frac{1}{\sigma^2(1-r^2)}B \\
&=\frac{1}{\sigma^2(1-r^2)}\left(nr\sigma^2 +\frac{1}{1-r^2}\left[2r^2B - rA\right] + \frac{1-r^2}{1-r^2}B\right) \\
&=\frac{1}{\sigma^2(1-r^2)}\left(nr\sigma^2 + \frac{1}{1-r^2}\left[ (r^2 + 1)B + rA\right]\right).
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to rewrite a Sum? How to find the following sum?
$$
\frac{2}{n} \sum_{i=1}^{n} \left(\frac{4i}{n} +1\right)
$$
For example
$$
\sum_{i = 1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.
$$
I need to find the value of which this would converge to and I know that it must be six because the integral from $1$ to $3$ of the function $(2x-1)$ is $6$.
Any help is appreciated!!!
| $$\frac{2}{n} \sum_{i=1}^{n} \left(\frac{4i}{n} +1\right) = \frac{2}{n}\left( \sum_{i=1}^{n} \frac{4i}{n} + \sum_{i=1}^{n} 1\right) = \frac{8}{n^2}\sum_{i=1}^{n} i + \frac{2}{n}\cdot n = \frac{8}{n^2}\cdot \frac{n(n+1)}{2} + 2 = 4 \left(1+\frac{1}{n} \right) + 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
My attempt:
We have: $f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}\implies x = 1$ and $x = -1$ points of extrema.
then for $a\geq 1$:
so then $$2-a=0\implies a=2$$ and $$\frac {a+2}3=2\implies a=4.$$
and for $a\leq1:$
so then $$2-a=3\implies a=-1$$ and $$\frac {a+2}3=0\implies a=-2.$$
Now my answers are in the type of interval. How do I know which interval to choose?
| You're doing well. The derivative is
$$
f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}=(1-a)\frac{(x^2-1)}{(x^2+x+1)^2}
$$
Leaving aside $1-a$, the fraction is negative for $|x|<1$.
If $1-a>0$, the function has a maximum at $-1$ and a minimum at $1$. The situation is reversed for $1-a<0$. For $a=1$ the function is constant and satisfies the requirement.
Since $\lim_{x\to\pm\infty}f(x)=1$, we just need to compute
$$
f(-1)=2-a
\qquad
f(1)=\frac{2+a}{3}
$$
Thus we must have
$$
\begin{cases}
1-a>0 \\[4px]
2-a\le2 \\[4px]
2+a\ge0
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
1-a<0 \\[4px]
2-a\ge0 \\[4px]
2+a\le6
\end{cases}
\qquad\text{or}\qquad a=1
$$
Solving this gives $0\le a\le 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that $a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2$. Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that
$$a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2.$$
In my opinion, I do this problem by th following:
$$\sum_{cyc}a\sqrt{a^2+1}=\sum_{cyc}a\sqrt{(a+b)(c+a)},$$
but I do not know how to continue following this way.
It seems easy to enlarge this quantity, but how to reduce this quantity.
Any help and hint will welcome!!
| I think your idea works, let's proceed:
Introduce the following substitution: $$a + b = u^2$$
$$b + c = v^2$$
$$c + a = w^2$$
Then we have:
$$2a = u^2 + w^2 - v^2$$
$$2b = v^2 + u^2 - w^2$$
$$2c = w^2 + v^2 - u^2$$
$ab + bc + ca = 1$ becomes:
$$4 = 2u^2v^2 + 2v^2w^2 + 2w^2u^2 - u^4 - v^4 - w^4$$
So the inequality becomes:
$$\sum_{cyc}{(u^2 + w^2 - v^2)uw} \geq 2\sum_{cyc}{u^2v^2} - \sum_{cyc}u^4$$
In Muirhead-like notation:
$$T[3, 1, 0] + \frac{1}{2}T[4, 0, 0] \geq T[2, 2, 0] + \frac{1}{2}T[2, 1, 1]$$
This is just pure Muirhead with $T[3, 1, 0] \geq T[2, 2, 0]$ and $T[4, 0, 0] \geq T[2,1,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$. Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$.
First, observe that $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!}=\frac{(2n)!}{(n!)^2}=\frac{1\cdot 2\cdot 3\cdot...\cdot 2n-1\cdot 2n}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 4\cdot 6\cdot ... \cdot 2n)}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 1\cdot 2\cdot 2\cdot 2\cdot 3\cdot ... \cdot 2n)}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot 2^n\cdot n!}{(n!)^2}=2^n\cdot\frac{1\cdot 3\cdot 5\cdot ... \cdot 2n-1}{(n!)}.$
For $n=0$ we have:
$\frac{(2*0)!}{(0!)^2}=1=2^0.$
Now to our inductive step $n+1$:
\begin{align*}
& \frac{(2(n+1))!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(2n+2)!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{1\cdot 2\cdot 3\cdot...\cdot 2n+1 \cdot 2n+2}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)\cdot(2\cdot 4\cdot 6\cdot ... \cdot 2n \cdot 2n+2)}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)\cdot 2^{n+1} \cdot (n+1)!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & 2^{n+1}\cdot \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)}{((n+1)!)}\geq2^{n+1}\\
\end{align*}
Is this proof correct/sufficient? Was induction even needed here or should I have argued instead that $\frac{1\cdot 3\cdot 5\cdot ... \cdot 2n-1}{(n!)}>0$? Is this step necessary even for the induction?
| From this identity and inequality, the proof becomes obvious
$$
\binom{2n}n = \sum_{k=0}^n {\binom nk}^2\ge \sum_{k=0}^n \binom nk.
$$
This identity is by comparison of coefficients of $x^n$ in
$$
(1+x)^{2n}=(1+x)^n (1+x)^n.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
This function has two different integrals? $f(x)=∫\frac{1}{x^2}dx$
Integrating by u-substitution:
$u=x^2$
$du=2dx$
$\frac{1}{2}du = dx$
$∫\frac{1}{x^2}dx=$ $∫\frac{1}{u}\times\frac{1}{2}du$
$\frac{1}{2}$∫ $\frac{1}{u}du$
$=\frac{1}{2}ln u+c$
$=\frac{1}{2}ln x^2+c$
$=lnx+c$
Another way:
$∫\frac{1}{x^2}dx=∫x^{-2}dx $
$∫x^{-2}dx$
$=\frac{x^{-1}}{-1} + c$
$=-\frac{1}{x} + c$
Where have I gone wrong?
| Your mistake is in changing the $dx$: it should be $du = 2x \, dx$, or $dx = \frac{1}{2}u^{-1/2} \, du$, which gives
$$ \int \frac{dx}{x^2} = \int \frac{1}{u} \frac{1}{2u^{1/2}} \, du = \frac{1}{2} \int u^{-3/2} \, du = -u^{-1/2}+C = -\frac{1}{x}+C. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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