Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $ Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $ in the plane $ x+y+z=1 $ with positive orientation . $$ $$ I have thought the parametriation:
\begin{align} x(t)=1+ 3 \cos (t) \hat j +3 \sin (t) \hat k \\ y(t)=1+3 \cos (t) \hat i+3 \sin (t) \hat k \\ z(t)=-1+3 \cos (t) \hat i +3 \sin (t) \hat j , \ \ 0 \leq t \leq 2 \pi \end{align} But I am not sure . Any help is there ?
| The plane can be parameterised as follows
\begin{eqnarray*}
x=1+t \\
y=1+s \\
z=-1-t-s.
\end{eqnarray*}
Now substitute this into the equation for the sphere $(x-1)^2+(y-1)^2+(z+1)^2=9$. We have
\begin{eqnarray*}
t^2+s^2+(t+s)^2=9 \\
\end{eqnarray*}
Rearrange this to
\begin{eqnarray*}
(2t+s)^2+3s^2=18
\end{eqnarray*}
This can be paramterised
\begin{eqnarray*}
s=\sqrt{6} \sin(\theta) \\
2t+s= \sqrt{18} \cos(\theta)
\end{eqnarray*}
Now substitute back into the parameterisation of the plane
\begin{eqnarray*}
x=1+ \frac{\sqrt{18} \cos(\theta) - \sqrt{6} \sin(\theta)}{2} \\
y=1+\sqrt{6} \sin(\theta) \\
z=-1-\frac{\sqrt{18} \cos(\theta)+\sqrt{6} \sin(\theta)}{2}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove by this inequality to with $x^2+y^2+z^2=1$ Let $x,y,z>0,x^2+y^2+z^2=1$ show that
$$\sum_{cyc}\sqrt{\dfrac{x^2+y}{2}}\ge\sqrt{2}$$
Here's what I have done. The expression $x^2+y\ge x^2+y^2$. it is is equivalent to
$$\sum_{cyc}\sqrt{1-x^2}\ge 2$$ shows that the function is neither concave or convex. So I don't think Jensen useful here.
| You have
$$
2 = \sum_{cyc} (x^2 + y^2) \le \sum_{cyc} \sqrt{x^2 + y^2}
\le \sum_{cyc} \sqrt{x^2 + y}
$$
because $a \le \sqrt a$ for $0 \le a \le 1$.
The inequality is sharp, equality holds exactly if $(x, y, z)$ is
any permutation of $(1, 0, 0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $\int \frac{1}{6+(x+4)^2} dx$. $\int \frac{1}{6+(x+4)^2} dx = 6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx$
Now $u=\frac{(x+4)}{\sqrt{6}}$ and $du=\frac{1}{\sqrt{6}}dx$.
$6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx=\frac{6}{\sqrt{6}}\int\ \frac{1}{1+u^2}=\frac{6}{\sqrt{6}}$ arctan$(u)$=$\frac{6}{\sqrt{6}}$ arctan$(\frac{(x+4)}{\sqrt{6}})+C$
However the result is $\frac{arctan(\frac{(x+4)}{\sqrt{6}})}{\sqrt{6}}+C$
Why is my result wrong? I can't see any mistake.
| Your first step is wrong $$ \frac{1}{6+(x+4)^2} \color{red}{\neq} \frac{6}{1+\frac{(x+4)^2}{6}}$$
Also, your substitution is wrong. You have substituted $$ \mathrm dx =\frac{1}{\sqrt 6} \mathrm du$$ Which actually is $$\mathrm du =\frac{1}{\sqrt 6} \mathrm dx$$
The given answer is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
number of one one function function from $P$ to $Q$ such that $g(1)\neq 0$
If $P = \{1,2,3,4,5\}$ and $Q = \{0,1,2,3,4,5\}$. The number of one one function function from
$P$ to $Q$ such that $g(1)\neq 0$ and $g(i)\neq i$ for $i=1,2,3,4,5$ is?
My attempt: Since $g(1) \neq 0,1$, $g(1)$ can take any $4$ values as $2,3,4,5$, which can be done in $4$ ways. Now $g(2),g(3),g(4),g(5)$ can take the remaining $5$ values which can be done in $\displaystyle \binom{5}{4}\cdot 4!$ ways. As such, the number of one-one functions such that $g(1)\neq0,1$ equals:
$$\displaystyle 4\cdot \binom{5}{4}\cdot 4! = 480$$
How can I find the number of functions for which $g(i) \neq i$ for $i = 2, 3, 4, 5$?
| You can use the inclusion-exclusion principle to solve this question. First consider all $6! = 720$ permutations of $0, 1, 2, 3, 4, 5$, and discard the last digit. We then remove disallowed permutations step by step, considering the $i^{th}$ digit in the permutation. In the first step we remove all permutations containing a 0 or 1 in $1^{st}$ position. Then, in the second step, we subtract all possible permutations with a 2 in $2^{nd}$ position, and add again all permutations containing a 0 or a 1 in $1^{st}$ position and a 2 in $2^{nd}$ position, since these were already eliminated in the first step. Removing invalid permutations one by one, we get:
*
*Permutations containing a 0 or 1 in $1^{st}$ position: $$720 - 2 \cdot 5! = 480$$
*Permutations containing a 2 in $2^{nd}$ position: $$480 - 5! + 2 \cdot 4! = 408$$
*Permutations containing a 3 in $3^{rd}$ position: $$408 - 5! + 2 \cdot 4! + 4! - 2 \cdot 3! = 348$$
*Permutations containing a 4 in $4^{th}$ position: $$348 - 5! + 2 \cdot 4! + 4! + 4! - 2 \cdot 3! - 2 \cdot 3! - 3 \cdot 2 + 2 \cdot 2 = 298$$
*Permutations containing a 5 in $5^{th}$ position: $$348 - 5! + 2 \cdot 4! + 4! + 4! + 4! - 2 \cdot 3! - 2 \cdot 3! - 2 \cdot 3! - 3 \cdot 2 - 3 \cdot 2 - 3 \cdot 2 + 2 \cdot 2 + 2 \cdot 2 + 2 \cdot 2 + 2 - 2 = 256$$
If you would like to verify these results, you can use the following Python script:
from itertools import permutations
d = [0] * 5
for p in permutations(range(6)):
for i in range(5):
if p[0] != 0 and p[i] != i + 1:
d[i] += 1
else:
break
for r in d:
print(r)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving non-linear ordinary differential equation: $(y'')^2-y'y'''=\left(\frac{y'}{x}\right)^2$ $$(y'')^2-y'y'''=\left(\frac{y'}{x}\right)^2$$
I have been struggling to solve this equation by doing following simplifications:
$$y'''y'=(y'')^2-\left(\frac{y'}{x}\right)^2$$
Dividing by $y'$:
$$y'''=\frac{(y'')^2}{y'}-\frac{y'}{x^2}$$
Dividing by $y''$:
$$\frac{y'''}{y''}=\frac{y''}{y'}-\frac{y'}{y''x^2}$$
I can't figure out what goes next since I have never encountered with the last fraction. I would really appreciate any help on the matter.
| Solving :
$$ \Bigg(\frac{d^2 y(x)}{dx^2}\Bigg) - \frac{dy(x)}{dx}\frac{d^3 y(x)}{ dx^3} = \frac{\frac{dy(x)}{dx}^2}{x^2}$$
Let :
$$\frac{dy(x)}{dx} = v(x)$$
which yields :
$$\frac{d^2 y(x)}{dx^2} = \frac{dv(x)}{dx}$$
$$\frac{d^3 y(x)}{ dx^3}= \frac{d^2 v(x)}{dx^2}$$
You get :
$$\Bigg( \frac{dv(x)}{dx}\Bigg)^2 -v(x)\frac{d^2 v(x)}{dx^2} = \frac{v(x)^2}{x^2}$$
$$\Rightarrow$$
$$\Bigg[-x^2 \Bigg( \frac{dv(x)}{dx}\Bigg)^2 + v(x)^2 + x^2\frac{d^2 v(x)}{dx^2}v(x)\Bigg]\frac{1}{x^2} = 0 $$
$$\Rightarrow$$
$$\frac{1}{x^2} + \frac{\frac{d^2 v(x)}{dx^2}}{v(x)} - \frac{\Bigg(\frac{dv(x)}{dx}\Bigg)^2}{v(x)^2} = 0 $$
$$\Rightarrow$$
$$\int \Bigg[\frac{1}{x^2} + \frac{\frac{d^2 v(x)}{dx^2}}{v(x)} - \frac{\Bigg(\frac{dv(x)}{dx}\Bigg)^2}{v(x)^2}\Bigg]dx = \int0dx$$
$$\Rightarrow$$
$$- \frac{1}{x} + \frac{\frac{dv(x)}{dx}}{v(x)} = c_1$$
$$\Rightarrow$$
$$\frac{dv(x)}{dx} = \frac{(c_1x+1)v(x)}{x}$$
$$\Rightarrow$$
$$\int \frac{dv(x)}{dx} dx = \int \frac{(c_1x+1)v(x)}{x} dx$$
$$\Rightarrow$$
$$\ln{v(x)} = \ln x + c_1x + c_2$$
$$\Rightarrow$$
$$v(x) = e^{c_1x + c_2}x$$
$$\Rightarrow$$
$$v(x) = c_2 c_1^x x$$
Substituting back for $\frac{dy(x)}{dx} = v(x)$, you get :
$$\frac{dy(x)}{dx} = c_2 c_1^x x $$
$$\Rightarrow$$
$$y(x) = \int c_2 c_1^x x dx = \frac{c_2 (\ln{(c_1)}x-1)c_1^x}{\ln^2c_1} + c_3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to find $\ker T$ from matrix representation over polynomial finite field?
Let $\mathbb F$ be a finite field of 5 elements (essentially $\mathbb Z_5$).
$T:\mathbb F_3[x] \to \mathbb F_3[x]$ is a linear map defined by the representation matrix:
$$
[T]_B=\begin{pmatrix}
1&2&3\\
1&0&4\\
0&1&2
\end{pmatrix}
$$
with the basis $B=(1, 1+x,1+x+x^2)$. Find $\text{ker}T$.
$\mathbb F_3[x]$ is a space of polynomials of form $ax^2+bx+c$ over $\mathbb F$.
We know that if a vector $v\in \text{ker}T$ then $[v]_B$ is in the nullity space of $[T]_B$.
If we row reduce $[T]_B$ and solve the system for $a_1x_1+a_2x_2+a_3x_3=0$:
$$
\begin{align}
\begin{pmatrix}
1&2&3\\
1&0&4\\
0&1&2
\end{pmatrix}\\Row_2\to Row_2-Row_1 \Rightarrow \begin{pmatrix}
1&2&3\\
0&3&1\\
0&1&2
\end{pmatrix}\\
Row_3\to Row_3-3^{-1}Row_2\Rightarrow \begin{pmatrix}
1&2&3\\
0&3&1\\
0&0&0
\end{pmatrix}\\
Row_2\to 3^{-1}Row_2\Rightarrow \begin{pmatrix}
1&2&3\\
0&1&2
\end{pmatrix}
\end{align}
$$
So:
$$
x_2=-2x_3=3x_3 \quad(??)\\
x_1=-2x_2-3x_3=3\cdot 3x_3+2x_3=4x_3+2x_3=x_3 \quad(??)
$$
and the general solution is: $(x_3,3x_3,x_3)$ from which follows that $\text{ker}T=span\{1,3,1\}$.
I'm not sure whether my calculations are correct especially when calculating the general solution.
Also I'm not sure how/whether to take into consideration that a vector belonging to $\text{ker}T$ is actually a polynomial.
| The row reduction is
\begin{align}
\begin{pmatrix}
1&2&3\\
1&0&4\\
0&1&2
\end{pmatrix}
&\to
\begin{pmatrix}
1&2&3\\
0&3&1\\
0&1&2
\end{pmatrix}
&& R_2\gets R_2-R_1
\\&\to
\begin{pmatrix}
1&2&3\\
0&1&2\\
0&1&2
\end{pmatrix}
&& R_2\gets 3^{-1}R_2
\\&\to
\begin{pmatrix}
1&2&3\\
0&1&2\\
0&0&0
\end{pmatrix}
&& R_3\gets R_3-R_2
\\&\to
\begin{pmatrix}
1&0&4\\
0&1&2\\
0&0&0
\end{pmatrix}
&& R_1\gets R_1-R_2
\end{align}
This means that the third variable is free, so a basis of the null space of the matrix is given by the single vector
$$
\begin{pmatrix} -4 \\ -2 \\ 1 \end{pmatrix}
=
\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}
$$
This vector gives the coordinates of the generator of the kernel with respect to $B$. Therefore a basis for the kernel is given by the single polynomial
$$
1\cdot 1+3\cdot(1+x)+1\cdot(1+x+x^2)=4x+x^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding Variance from a joint moment generating function The random vars X and Y have, for all real values of $T_1, T_2$, the joint mgf $M(T_1 , T_2) = \frac{1}{2} e^{T_1 +T_2} + \frac{1}{4} e^{2T_1 +T2} + \frac{1}{12}e^{T_2} + \frac{1}{6} e^{4T_1 +3T_2}$
I'm looking for V[X], but I believe that I am overlooking a way to simplify this problem. I know that variance is the second moment or the second derivative of the mgf,but I am not sure how to generate these derivatives and what to do after, given that it is a bivariate equation. Any pointers would be appreciated.
| Even more simply, you know from the form of the MGF that $X$ and $Y$ are joint discrete random variables: it is easy to see that
$$\begin{align*}\Pr[(X,Y) = (1,1)] &= \frac{1}{2}, \\
\Pr[(X,Y) = (2,1)] &= \frac{1}{4}, \\
\Pr[(X,Y) = (0,1)] &= \frac{1}{12}, \\
\Pr[(X,Y) = (4,3)] &= \frac{1}{6}, \\
\end{align*}$$ and $0$ for any other values. Thus, the variance of $X$ is simply $$\operatorname{Var}[X] = \left(\frac{1^2}{2} + \frac{2^2}{4} + \frac{4^2}{6} \right) - \left(\frac{1}{2} + \frac{2}{4} + \frac{4}{6}\right)^2 = \frac{25}{18}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$z = e^y \phi \left(y e^ \frac{x^2}{2y^2} \right)$ satisfies $(x^2 - y^2) \frac{\partial z}{\partial x} + xy \frac{\partial z}{\partial y} = xyz$ Show that the function $z = e^y \phi \left(y e^ \frac{x^2}{2y^2}\right)$ satisfies the equation $(x^2 - y^2) \frac{\partial z}{\partial x} + xy \frac{\partial z}{\partial y} = xyz$.
$\frac{\partial z}{\partial x} = \dfrac{x}{y} e^ {y + \frac{x^2}{2y^2}} \phi ' \left(y e^ \frac{x^2}{2y^2}\right) $. In the same way $\frac{\partial z}{\partial y}$ is calculated. And then I did the multiplications but the calculation becomes very long.
Is there another way, for example with a change of variable, so that it is not so long?
| $$
\frac{dz}{dx} = e^y \phi'\frac{x}{y}e^{\frac{x^2}{2y^2}}
$$
$$
\frac{dz}{dy} = e^y\phi + e^y\phi'\left(e^\frac{x^2}{2y^2} - \frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}\right)
$$
$$
(x^2-y^2)\frac{dz}{dx} + xy\frac{dz}{dy} = e^y \phi'\frac{x}{y}e^{\frac{x^2}{2y^2}}(x^2-y^2)+\left(e^y\phi + e^y\phi'\left(e^\frac{x^2}{2y^2} - \frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}\right)\right)xy
$$
$$
e^y\phi xy+e^y\phi'\left(\frac{x}{y}e^{\frac{x^2}{2y^2}}(x^2-y^2)+\left(e^\frac{x^2}{2y^2} - \frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}\right)xy\right)
$$
$$
e^y\phi xy+e^y\phi'\left(\frac{x^3}{y}e^{\frac{x^2}{2y^2}}-xye^{\frac{x^2}{2y^2}}+xye^\frac{x^2}{2y^2} - \frac{x^3}{y}e^{\frac{x^2}{2y^2}}\right) = xy e^y\phi = xyz
$$
$$
\therefore (x^2-y^2)\frac{dz}{dx} + xy\frac{dz}{dy} = xyz \quad z = e^y \phi
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For which $n$ does $2^n+1$ divide $10^n+1$? This came up in a discussion of numbers that divide their own binary representation (when interpreted as a decimal). The question I'm asking zooms in on a special case:
For which $n$ does $2^n+1$ divide $10^n+1$?
I computed the remainder after division of $10^n+1$ by $2^n+1$ for $1 \leq n \leq 300$, which led me to suppose that there are no such $n$.
| In comments on the question, Chris has proven all the cases other than $ $$8|n$.
If $ $ $ $$n=8k$,
Assume $2^n+1$ divides $5^n-1$ (that is equivalent to the main question as you can see easily).
Take any prime $p$ that divides
$2^{8k}+1$, then $p\neq 2,5$ obviously and
$$\Rightarrow p|2^{16k}-1 $$
Assume $2^m||16k$ for some $m\geq4$,
Because $p$ does not divide $2^{8k}-1$,
$$\Rightarrow 2^m|p-1$$
Because $2^{8k}+1$ divides $5^{8k}-1$,
$p$ divides $5^{8k}-1$,
$\Rightarrow$ 5 is square in modulo $p$.
$\Rightarrow \left (\dfrac {p}{5} \right ) \left (\dfrac {5}{p} \right )=(-1)^{\frac {p-1}{2} \frac {5-1}{2} }=1$
$\Rightarrow \left (\dfrac {p}{5} \right )=1 \Rightarrow$ $ p$ is square in modulo 5.
$\Rightarrow p\equiv 1,-1 \pmod {5}$
$\Rightarrow $ all prime divisors of $2^n+1$ are of the form $5k+1$ or $5k-1$.
$\Rightarrow 2^n+1 $ should be $\equiv 1,-1 \pmod {5}$,but $ 2^n+1=2^{8k}+1\equiv 2 \pmod {5} \Rightarrow \Leftarrow $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
How do I solve $\int\frac{dx}{\sin x+\cos x-1}$? Please help me find the following indefinite integral:
$$\int\dfrac{dx}{\sin x+\cos x-1}$$
I have tried many different substitutions to no avail. Any help is appreciated.
| $$\int\frac{1}{\sin{x}+\cos{x}-1}dx=\int\frac{1}{2\sin\frac{x}{2}\cos\frac{x}{2}-2\sin^2\frac{x}{2}}dx=$$
$$=\int\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}dx=\int\frac{\sin^2\frac{x}{2}+\sin\frac{x}{2}\cos\frac{x}{2}-\sin\frac{x}{2}\cos\frac{x}{2}+\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}dx=$$
$$=\int\frac{\sin\frac{x}{2}\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)+\cos\frac{x}{2}\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}{2\sin\frac{x}{2}\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}dx=$$
$$=\int\left(\frac{\sin\frac{x}{2}+\cos\frac{x}{2}}{2\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}+\frac{\cos\frac{x}{2}}{2\sin\frac{x}{2}}\right)dx=$$
$$=\int\left(-\frac{d\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}{\cos\frac{x}{2}-\sin\frac{x}{2}}+\frac{d\left(\sin\frac{x}{2}\right)}{\sin\frac{x}{2}}\right)=\ln\left|\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}\right|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
$1-x+x^2-x^3+..(-1)^nx^n$ I have the following sum:
$1-x+x^2-x^3+..(-1)^nx^n, x\neq -1$
So what I thought was separating it in two cases like this:
Case 1. n is even
$$
1+x^2+x^4+...+x^n-x(1+x^2+...+x^n)
$$
Which I can turn into $\frac{1-x^{n+2}}{1-x^2}-\frac{x(1-x^{n+2})}{1-x^2}=\frac{1-x^{n+2}}{1-x^2}(1-x)$
Case 2. n is odd
$$
1+x^2+x^4+...+x^{n-1}-x(1+x^2+...+x^{n-1})=\frac{1-x^{n+1}}{1-x^2}-x(\frac{1-x^{n+1}}{1-x^2})
$$
My question is: Assuming what I've written is correct, which I'm not entirely sure, how can I combine the two cases for n even and odd into one equation?
| $$S(n)=1-x+x^2-x^3+..(-1)^nx^n$$ multiply by $+x $
$$xS(n)=x-x^2+x^3...(-1)^{n-1}x^n+(-1)^nx^{n+1}$$ now find add $S(n) ,xS(n)$
so
$$s(n)+xS(n)=1+(x-x)+(x^2-x^2)+....+((-1)^{n-1}x^n+(-1)^nx^n)+(-1)^nx^{n+1}\\S(n)(x+1)=1+(-1)^nx^{n+1}\\
s(n)=\frac{1+(-1)^nx^{n+1}}{1+x}=\\
\frac{1+(-1)(-1)^{n+1}x^{n+1}}{1+x}=\\
\to \\
S(n)=\frac{1-(-1)^{n+1}x^{n+1}}{1+x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Simplify $5(3 \sin(x) + \sqrt3 \cos(x))$ So Wolfram tells me I can reach $10 \sqrt3 \sin(x + π/6)$ from this expression, but I cant grasp how to do it. Any help is appreciated.
| $a \sin x + b\cos x$
let $\phi = \arctan \frac{b}{a}$
$\sin {\phi} = \frac {b}{\sqrt {a^2+b^2}}\\
\cos {\phi} = \frac {a}{\sqrt {a^2+b^2}}$
$a \sin x + b\cos x = \sqrt {a^2 + b^2} (\cos\phi \sin x + \sin\phi\cos x) = \sqrt {a^2 + b^2} \sin (x+\phi)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Finding the eigenvalues of a $n\times n$ $A=(a_{ij})$ defined by $a_{ij}=1$ if $i+j=n+1$, $0$ elesewhere. If $A=(a_{ij})\in M_n(\mathbb{R})$ is a matrix defined by $a_{ij}=1$ if $i+j=n+1$, $0$ otherwise. Then what are the eigenvalues of $A$?
\begin{bmatrix}
0 & 0 & \ldots & \ldots & \ldots & 0 & 1 \\
0 & 0 & \ldots & \ldots & \ldots & 1 & 0 \\
\vdots & \vdots & \vdots & & & \vdots & \vdots \\
\vdots & \vdots & & \vdots & & \vdots & \vdots \\
\vdots & \vdots & & & \vdots & \vdots & \vdots \\
0 & 1 & \ldots & \ldots & \ldots & 0 & 0 \\
1 & 0 & \ldots & \ldots & \ldots & 0 & 0
\end{bmatrix}
I could find out that the determinant of $A$ is $(-1)^{n+1}$.
| Note that, if $e_j=(0,0,\ldots,0,1,0,\ldots,0)$, where the $1$ is in the $j-$th place, then
$$
A(e_j+e_{n-j})=e_j+e_{n-j} \tag{1}
$$
while
$$
A(e_j-e_{n-j})=-(e_j-e_{n-j}) \tag{2}
$$
From $(1)$ we obtain $\lfloor\frac{n+1}{2}\rfloor$ linearly independent eigenvectors
corresponding to the eigenvalue $1$, while $(2)$ provides $\lfloor\frac{n}{2}\rfloor$ linearly independent eigenvectors corresponding to the eigenvalue $-1.$
Note that
$$
\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n+1}{2}\right\rfloor=n,
$$
for all $n$.
So, the characteristic polynomial of $A$ is
$$
p_A(x)=-(-1)^n(x-1)^{\left\lfloor\frac{n}{2}\right\rfloor}(x+1)^{\left\lfloor\frac{n+1}{2}\right\rfloor}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $F(1) = -8$, of the antiderivative $\displaystyle f(x) = x^3 + 2 \sqrt{x}$. Consider the function $\displaystyle f(x) = x^3 + 2 \sqrt{x}$.
Let $F(x)$ be the antiderivative of $f(x)$ with $F(1) = -8.$
$\dfrac{x^4}{4} + \dfrac{4x^\frac{3}{2}}{3}+C$
$-8=\dfrac{(1)^4}{4} + \dfrac{4\cdot(1)^\frac{3}{2}}{3}+C$
$-8-.25-1.{\bar 3}=C$
$-9.58{\bar3} =C$
Why is this not correct?
| Let
$$F(x) = \int^{x} f(t) \, dt$$
where $f(x) = x^3 + 2 \sqrt{x}$. It is seen that
$$F(x) = \frac{x^{4}}{4} + \frac{4}{3} \, x^{\frac{3}{2}} + c_{0}.$$
The constant can be found by applying $F(1) = -8$ for which
\begin{align}
F(1) = -8 &= \frac{1}{4} + \frac{4}{3} + c_{0} \\
c_{0} &= - \frac{115}{12}.
\end{align}
The function $F(x)$ can now be stated as
$$F(x) = \frac{x^{4}}{4} + \frac{4}{3} \, x^{\frac{3}{2}} - \frac{115}{12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Transform curve into canonical form and determine type I have the curve:
$$9x^2-6xy+y^2+6x-2y-3=0$$
I have to transform it into the nnormal form and determine its type.
Well, we all know that such a curves are given by equation:
$$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F = 0$$
Which is purely the case.
$\textbf{Thoughts:}$
1.) I determined if given curve is central or not:
$$\Delta=\begin{vmatrix} A&B\\B&C \end{vmatrix} = \begin{vmatrix}9&-3\\-3&1 \end{vmatrix} = 9 -9=0$$
Therefore, the curve is not central.
2.) Now have to find an angle of transformation, by the formula:
$$\tan^2(\varphi)+\frac{a_{11}-a_{22}}{a_{12}}\tan(\varphi)-1=0$$
Where:
$$\begin{array}{|c|c|c|}\hline a_{11}&a_{12}&a_{22}\\ \hline 9 & -3 & 1 \\ \hline \end{array}$$
Therefore:
$$\tan^2(\varphi)-\frac{8}{3}\tan(\varphi)-1=0$$
or
$$3\tan^2(\varphi)-8\tan(\varphi)-3=0$$
which is square equation and substitute $t=\tan(\varphi)$ we get:
$$3t^2-8t-1=0$$
And roots are $t_1=3$ and $t_2=-\frac{1}{3}$ (which is strange for tangent, in my opinion).
3.) Find transformation, and appropriate $\sin x$ and $\cos x$
It is well known that there are transformations:
$$\begin{cases} x = x_1\cos(\varphi)-y_1\sin(\varphi) \\ y= x_1\sin(\varphi)+y_1\cos(\varphi) \end{cases} \tag{1}$$
So for $\cos (\varphi)$: $$\cos(\varphi)=\frac{1}{\sqrt{1+\tan^2(\varphi)}}$$
And for $\sin (\varphi)$: $$\sin(\varphi) = \frac{\tan(\varphi)}{\sqrt{1+\tan^2(\varphi)}}$$
As it does not matter which root to apply here I take $3$ and get:
$$\cos(\varphi) = \frac{1}{\sqrt{10}}$$
and
$$\sin(\varphi) = \frac{3}{\sqrt{10}}$$
or, if substitute results into $(1)$ we get:
$$\begin{cases}x = x_1\frac{1}{\sqrt{10}}-y_1\frac{3}{\sqrt{10}} \\ y = x_1\frac{3}{\sqrt{10}}+y_1\frac{1}{\sqrt{10}} \end{cases} \tag{2}$$
4.) System $(2)$ shows us new coordinates, which we should substitute into original equarion, and have:
$$9\left(x_1\frac{1}{\sqrt{10}}-y_1\frac{3}{\sqrt{10}}\right)^2-6\left(x_1\frac{1}{\sqrt{10}}-y_1\frac{3}{\sqrt{10}}\right)\cdot \left(x_1\frac{3}{\sqrt{10}}+y_1\frac{1}{\sqrt{10}} \right) + \left(x_1\frac{3}{\sqrt{10}}+y_1\frac{1}{\sqrt{10}}\right)^2 + 6\left( x_1\frac{1}{\sqrt{10}}-y_1\frac{3}{\sqrt{10}}\right) - 2\left(x_1\frac{3}{\sqrt{10}}+y_1\frac{1}{\sqrt{10}}\right) - 3 = 0 \tag{3}$$
And I am totally stuck here
| The equation $9x^2-6xy+y^2+6x-2y-3=0$ is equivalent with
$$(x,\; y)\begin{pmatrix} 9 & -3\\ -3 & 1\end{pmatrix}\begin{pmatrix} x\\ y\end{pmatrix}+(6,\;-2)\begin{pmatrix} x\\ y\end{pmatrix}-3=0$$
Now we apply the principal axis theorem: The eigenvalues of the matrix, say $A$, are $\lambda=0$ and $\mu=10$, the eigenvectors $v_{\lambda}=\binom{1}{3}$ and $v_{\mu}=\binom{3}{-1}$. For
$T=\frac{1}{\sqrt{10}}\begin{pmatrix}1&3 \\ 3& -1\end{pmatrix}$ we have $T=T^{-1}=T^{t}$. Consider the transformation $\binom{x}{y}=T\binom{x'}{y'}$, then the above equation is
$$(x',\; y')T^{t}AT\binom{x'}{y'}+(6,\;-2)T\binom{x'}{y'}-3=0$$
which is with $T^{t}AT=\mathrm{diag}(0,10)$ and $(6,\;-2)T=(0,\;20)$ the same as
$$10y'^2+20y'-3=0$$
After a small rearrangement, we get
$$10(y'+1)^2=13$$
For $y''=\frac{y'+1}{\sqrt{13}}$ we arrive at
$$10y''^2=1$$
This equation describes a pair of parallel lines.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Characteristic polynomial of a $7 \times 7$ matrix whose entries are $5$ Avoiding too many steps, what is the characteristic polynomial of the following $7 \times 7$ matrix? And why?
\begin{pmatrix}5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\end{pmatrix}
| As it was stated in the commentaries, the rank of this matrix is $1$; so it will have $6$ null eigenvalues, which means the characteristic polynomial will be in the form:
$p(\lambda)=\alpha\,\lambda^6(\lambda-\beta) = \gamma_6\,\lambda^6 +\gamma_7\,\lambda^7$
Using Cayley-Hamilton:
$p(A)=\gamma_6\,A^6+\gamma_7\,A^7 =0$
Any power of this matrix will have the same format, a positive value for all elements.
$B=\begin{bmatrix}1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\end{bmatrix}$
$A = 5\,B$
$A^2 = 5^2\,7\,B$
$...$
$A^6 = 5^6\,7^5\,B$
$A^7=5^7\,7^6\,B$
$p(A) = (\gamma_6+35\,\gamma_7)\,B=0\Rightarrow\gamma_6=-35\gamma_7$
So we have: $\alpha=\gamma_7$ and $\beta = 35$
$p(\lambda)=\alpha\,\lambda^6(\lambda-35)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
| We can expand $$(a+b)^2+(b+c)^2+(c+a)^2$$ to get
$$2 a^2 + 2 a b + 2 a c + 2 b^2 + 2 b c + 2 c^2 $$
We can rearrange this to be more useful:
\begin{align}2 a^2 + 2 a b + 2 a c + 2 b^2 + 2 b c + 2 c^2 &= 2 a^2 + 2b^2+2c^2 + 2 a b + 2 a c + 2 b c\\
&=2(a^2+b^2+c^2)+2(ab+ac+bc)\end{align}
We know the value of $ab+ac+bc$, so we can say that \begin{align}2(a^2+b^2+c^2)+2(ab+ac+bc)&=2(a^2+b^2+c^2)+2\end{align}
Now we need to find the value of $a^2+b^2+c^2$.
We can do this as follows:
\begin{align}(a+b+c)^2&=2^2\\
a^2 + 2 a b + 2 a c + b^2 + 2 b c + c^2&=4\\
a^2+b^2+c^2+2(ab+ac+bc)&=4\\
a^2+b^2+c^2+2\times 1&=4\\
a^2+b^2+c^2&=2\end{align}
So now we can say that \begin{align}2(a^2+b^2+c^2)+2&=2\times 2+2\\
&=6\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 0
} |
Solving Pinter 7.B.4 with a program Here is exercise 7.B.4 from 'A Book of Abstract Algebra' by Charles C. Pinter.
A solution to this using a C# program is posted.
Is there another good approach using a computer program?
Any language is welcome.
The subgroup of $S_5$ generated by
$
f = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\2 & 1 & 3 & 4 & 5\end{pmatrix}
\qquad
g = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\1 & 2 & 4 & 5 & 3\end{pmatrix}
$
has six elements. List them, then write the table of this group:
$\varepsilon = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\1 & 2 & 3 & 4 & 5\end{pmatrix}$
$f = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\2 & 1 & 3 & 4 & 5\end{pmatrix}$
$g = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\1 & 2 & 4 & 5 & 3\end{pmatrix}$
$h = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\ \, \end{pmatrix}$
$k = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\ \,\end{pmatrix}$
$l = \begin{pmatrix}1 & 2 & 3 & 4 & 5\\ \,\end{pmatrix}$
$
\begin{array}{c|ccc}
\circ & \varepsilon & f & g & h & k & l \\
\hline
\varepsilon \\ f \\ g \\ h \\ k \\ l
\end{array}
$
| SageMath provides built-in methods for such computations
sage: S5 = SymmetricGroup(5)
sage: f = S5( (1,2)); g = S5( (3,4,5)) #cycle notation
# or f = S5([2,1,3,4,5]); g = S5( [1,2,4,5,3])
sage: S5fg = S5.subgroup([f,g])
sage: S5fg.multiplication_table(names='elements')
$$\scriptsize{
\begin{array}{c|*{6}{c}}
{\ast}&()&(3,4,5)&(1,2)&(3,5,4)&(1,2)(3,4,5)&(1,2)(3,5,4)\\\hline
{}()&()&(3,4,5)&(1,2)&(3,5,4)&(1,2)(3,4,5)&(1,2)(3,5,4)\\
{}(3,4,5)&(3,4,5)&(3,5,4)&(1,2)(3,4,5)&()&(1,2)(3,5,4)&(1,2)\\
{}(1,2)&(1,2)&(1,2)(3,4,5)&()&(1,2)(3,5,4)&(3,4,5)&(3,5,4)\\
{}(3,5,4)&(3,5,4)&()&(1,2)(3,5,4)&(3,4,5)&(1,2)&(1,2)(3,4,5)\\
{}(1,2)(3,4,5)&(1,2)(3,4,5)&(1,2)(3,5,4)&(3,4,5)&(1,2)&(3,5,4)&()\\
{}(1,2)(3,5,4)&(1,2)(3,5,4)&(1,2)&(3,5,4)&(1,2)(3,4,5)&()&(3,4,5)\\
\end{array}}
$$
But as @anomaly says in their comment, you don't have to use a computer to do such problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Where was my mistake (integration by trig-substitution problem)? I am attempting to solve the problem
$$\int \frac{dx}{x^2+x+1}$$
First, I complete the square, then factor out a $\frac {3}{4}$:
$$\int \frac{dx}{\frac{3}{4}(\frac{4}{3}(x+\frac{1}{2})^2+1)}$$
Let $u = \sqrt{\frac{4}{3}}(x+\frac{1}{2})$
$$\frac{du}{dx} = \frac{2}{\sqrt{3}}$$
$$dx = \frac{\sqrt{3}}{2} du$$
Thus, we now have the integral:
$$\frac{4}{3} \frac{\sqrt{3}}{2} \int \frac{du}{u^2+1}$$
Let $u = \tan \theta$
$$du = \sec^2\theta \ d\theta$$
What follows is obvious now, and the solution should be:
$$\frac{4}{3} \frac{\sqrt{3}}{2} \theta + C$$
$$\theta = \tan^{-1}(u)$$
Thus, the final solution is:
$$\frac{4}{3} \frac{\sqrt{3}}{2} \tan^{-1} \left( \sqrt{\frac 4 3} \left(x+\frac 1 2 \right)\right) + C$$
However, according to online calculator integral-calculator, the answer is:
$$\frac 2 {\sqrt 3} \tan^{-1} \left( \frac{2x+1}{\sqrt 3} \right)+C$$
Any indication as to where my mistake falls would be very beneficial.
|
This answer was posted prior to an edit made by the OP.
Note that we have
$$x^2+x+1=(x+1/2)^2+3/4\ne \frac34 \left(\frac43 (x+1/2)^2+\frac34 \right)=x^2+2+5/2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How to obtain the sum of the following series? $\sum_{n=1}^\infty{\frac{n^2}{2^n}}$ It seems that I'm missing something about this.
First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)
What I tried to do is to find a limit of a partial sum $\lim_{n\rightarrow\infty}S_n$ as follows: $S_n=\frac{\frac{1}{6} n (n+1) (2 n+1)}{\frac{1-\left(\frac{1}{2}\right)^n}{2 \left(1-\frac{1}{2}\right)}}$. Still, the limit is $\infty$ and I'm clearly doing something wrong.
| $$
\begin{align}
\sum_{n=0}^\infty x^n=\frac1{1-x}&\implies\sum_{n=0}^\infty\frac1{2^n}=2\tag{1}\\
\sum_{n=0}^\infty nx^{n-1}=\frac1{(1-x)^2}&\implies\sum_{n=0}^\infty\frac{n}{2^{n-1}}=4\tag{2}\\
\sum_{n=0}^\infty n(n-1)x^{n-2}=\frac2{(1-x)^3}&\implies\sum_{n=0}^\infty\frac{n(n-1)}{2^{n-2}}=16\tag{3}
\end{align}
$$
Adding $\color{#CCC}{x^2=}\frac14$ times $(3)$ to $\color{#CCC}{x=}\frac12$ times $(2)$ gives
$$
\color{#CCC}{\sum_{n=0}^\infty n^2x^n=\frac{x(1+x)}{(1-x)^3}\implies}\sum_{n=0}^\infty\frac{n^2}{2^n}=6\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
What is the probability of the sum of four dice being 22? Question
Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?
My Approach
I simplified it to the equation of the form:
$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $
Solving this equation results in:
$x_{1}+x_{2}+x_{3}+x_{4}=22$
I removed restriction of $x_{i} \geq 1$ first as follows-:
$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
$\Rightarrow \binom{18+4-1}{18}=1330$
Now i removed restriction for $x_{i} \leq 6$ , by calculating the number of bad cases and then subtracting it from $1330$:
calculating bad combination i.e $x_{i} \geq 7$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
We can distribute $7$ to $2$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{2}$
We can distribute $7$ to $1$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{1}$ and then among all others .
i.e
$$\binom{4}{1} \binom{14}{11}$$
Therefore, the number of bad combinations equals $$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$
Therefore, the solution should be:
$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$
However, I am getting a negative value. What am I doing wrong?
EDIT
I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
The number of configurations that satisfies "the sum is $\ds{22}$" is given by:
\begin{align}
X & =
\sum_{d_{1} = 1}^{6}\sum_{d_{2} = 1}^{6}\sum_{d_{3} = 1}^{6}\sum_{d_{4} = 1}^{6}
\bracks{z^{22}}z^{d_{1} + d_{2} + d_{3} + d_{4}} =
\bracks{z^{22}}\pars{\sum_{d = 1}^{6}z^{d}}^{4} =
\bracks{z^{22}}\pars{z\,{z^{6} - 1 \over z - 1}}^{4}
\\[5mm] & =
\bracks{z^{18}}{1 - 4z^{6} + 6z^{12} - 4z^{18} + z^{24}\over \pars{1 - z}^{4}}
\\[5mm] & =
\bracks{z^{18}}\pars{1 - z}^{-4} - 4\bracks{z^{12}}\pars{1 - z}^{-4} + 6\bracks{z^{6}}\pars{1 - z}^{-4} - 4\bracks{z^{0}}\pars{1 - z}^{-4}
\\[5mm] & =
{-4 \choose 18}\pars{-1}^{18} - 4{-4 \choose 12}\pars{-1}^{12} +
6{-4 \choose 6}\pars{-1}^{6} - 4 =
{21 \choose 18} - 4{15 \choose 12} + 6{9 \choose 6} - 4
\\[5mm] & =
1330 -4 \times 455 + 6 \times 84 - 4 = \bbx{10}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 11,
"answer_id": 4
} |
find a,b,c where $\sqrt{5^{1/3}-4^{1/3}}=\frac{1}{3}(a^{1/3}+b^{1/3}-c^{1/3})$ Is it ramanujan problems? $$\sqrt{5^{1/3}-4^{1/3}}=\frac{1}{3}(a^{1/3}+b^{1/3}-c^{1/3})$$
find $a,b,c$
Any helps would be appreciated.
| By squaring of the both sides easy to see that
$$\sqrt{\sqrt[3]5-\sqrt[3]4}=\frac{1}{3}\left(\sqrt[3]2+\sqrt[3]{20}-\sqrt[3]{25}\right)$$
Since $\sqrt[3]2+\sqrt[3]{20}-\sqrt[3]{25}>0$, we need to prove that:
$$9(\sqrt[3]5-\sqrt[3]4)=\sqrt[3]4+\sqrt[3]{400}+\sqrt[3]{625}+2\sqrt[3]{40}-2\sqrt[3]{50}-2\sqrt[3]{500}$$ or
$$9(\sqrt[3]5-\sqrt[3]4)=\sqrt[3]4+2\sqrt[3]{50}+5\sqrt[3]{5}+4\sqrt[3]{5}-2\sqrt[3]{50}-10\sqrt[3]{4},$$
which is obvious.
Yes, it's the Ramanujan's problem.
For example, see here: Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this question, but not able to proceed. How do I solve this?
| Use the inequality $x^2+y^2+z^2>zy+yz+zx$ for distinct $x,y,z$ repeatedly: $$a^8+b^8+c^8>a^4b^4+b^4c^4+c^4a^4>a^4b^2c^2+a^2b^4c^2+a^2b^2c^4>a^3b^3c^2+a^2b^3c^3+a^3b^2c^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Integration by Partial Fractions, using $\sqrt2$ Integrate $$\int\frac{10x^2 + 13x + 9}{(x^2 - 2)(2x + 1)^2}\mathrm dx$$
The difficulty here is in $x^2 - 2$. In order to determine the coefficients A to D, $\sqrt2$ must be invoked, in factorizing. Firstly, put $x = \sqrt 2$, next $-\sqrt 2$, $-0.5$.
The coefficients come out as fractions, which is unusual.
| Note that the coefficient of $A$ of the term
$\frac{1}{x-\sqrt{2}}$ in the partial fraction expansion of $$f(x)=\frac{10x^2 + 13x + 9}{(x^2 - 2)(2x + 1)^2}$$ can be determined by evaluating the limit
$$A=\lim_{x\to \sqrt{2}} \left((x-\sqrt{2})\cdot\frac{(10x^2 + 13x + 9)}{ (x^2 - 2)(2x + 1)^2}\right)=
\left.\frac{10x^2 + 13x + 9}{ (x +\sqrt{2})(2x + 1)^2}\right|_{x=\sqrt{2}}=\frac{2+157\sqrt{2}}{196}.$$
Similarly for the coefficient of $B$ of the term $\frac{1}{x+\sqrt{2}}$ we get $\frac{2-157\sqrt{2}}{196}$.
Hence
$$\frac{A}{x-\sqrt{2}}+\frac{B}{x+\sqrt{2}}=\frac{(A+B)x+\sqrt{2}(A-B)}{x^2-2}=\frac{x+157}{49(x^2-2)}.$$
So far we have
$$f(x)=\frac{x+157}{49(x^2-2)}+\frac{C}{2x+1}+\frac{D}{(2x+1)^2}$$
and it remains to find the coefficients of $C$ and $D$. They should be
$C=-2/49$ and $D=-20/7$.
P.S. According to your given answer there is a little typo in the integrand (replace $(x^2-2)$ with $(x-2)$). The the integral is easier:
\begin{align*}
\int \frac{10x^2 + 13x + 9}{(x - 2)(2x + 1)^2}\,dx=
&\int\left(\frac{3}{x-2}-\frac{1}{2x+1} -\frac{2}{(2x+1)^2}\right)\,dx\\
\\&=\ln\left(\frac{(x - 2)^3}{\sqrt{2x + 1}}\right) - \frac{1}{2x + 1}+c.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A series in which some terms are in AP and other terms in GP In an increasing sequence of four positive integer, the first three terms are in AP, the last three terms are in GP and the fourth term exceed the first term by $30$, then the common difference of AP lying in the intervsl $[1,9]$ is:
MY ATTEMPT:
Let the series is $W,X,Y,Z$.
The first three terms are $a-d,a,a+d$ and the last three terms are $b/r$,$b$,$br$.
So we get $b=a+d$ and $$r=\frac{30 + a-d}{a+d}$$
Then I replaced $b$ and $r$ in terms of $a$ and $d$.
I obtained a quadratic equation. After this I could not do anything as I used all the data given.
| Let $a,b,c,d$ be the four positive integers.
The first three are in AP and the last three are in GP so we have
$$b-a=x,c-b=x,\dfrac{d}{c}=\dfrac{c}{b}\rightarrow bd=c^2$$
furthermore The 4th is the 1st plus 30, that is $d=a+30$
Let substitute the last information into the system
$$b-a=x;\;c-b=x;\;b(a+30)=c^2$$
we get
$$c=b+x;\;a=b+x\rightarrow b(b+x+30)=(b+x)^2\rightarrow b^2+bx+30=b^2+2bx+x^2\rightarrow $$
$$b=\frac{x^2}{30-3x}$$
as the numbers are positive integers the only value for $x$ is $x=9$ and the four terms are $a= 18,\;b= 27,\;c= 36,\;d= 48$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving equations $a+b+c=5,a^2+b^2+c^2=11,a^3+b^3+c^3=27$
Let $a, b, c $ be real numbers such that $a<b<c$ and satisfying
$$a+b+c=5;$$
$$a^2+b^2+c^2=11;$$
$$a^3+b^3+c^3=27.$$
Prove that $0<a<1<b<2<c<3$.
My understanding :
$(a+b+c)^2 - (a^2+b^2+c^2) = 14$
so, $ab+bc+ca = 7$
$a^2+b^2+c^2-ab-bc-ca = 4$
$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=20$
so, $abc = -\frac{7}{3}$
Thus $a, b, c$ are roots of the equation $P(x), x^3-5x^2+7x-\frac{7}{3}=0$.
We have,
$P(0) = -\frac{7}{3} <0$
$P(1) = 1-5+7-\frac{7}{3} >0$
$P(2) = 8-20+14-\frac{7}{3} <0$
$P(3) = 27-45+21-\frac{7}{3} >0$.
by Intermediate value theorem, $P(x)$ has a root in each of the interval $(0,1), (1, 2), (2,3)$.
Since $a<b<c$, we get $0<a<1<b<2<c<3$.
| $$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$$
$$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc$$
Get from here $ab+ac+bc$ and $abc$.
I got $ab+ac+bc=7$ and $abc=\frac{7}{3}$, which gives that $a$, $b$ and $c$ are roots of the equation
$$x^3-5x^2+7x-\frac{7}{3}=0$$ or
$$\frac{1}{(x-1)^3}-\frac{3}{x-1}=-\frac{3}{2}$$
Now, let $\frac{1}{x-1}=2\cos\phi$.
Hence, $\cos3\phi=-\frac{3}{4}$ and the rest is smooth for finding of values $a$, $b$ and $c$, but it's unnecessary.
Let $f(x)=x^3-5x^2+7x-\frac{7}{3}$ and check $f(0)$, $f(1)$, $f(2)$ and $f(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum of the value $3x^2-2xy$ if $\frac{x^2}{4}-y^2=1$ Let $x,y\in R$,such
$$\dfrac{x^2}{4}-y^2=1$$
find the minium of the
$$3x^2-2xy$$
I think $x=2\sec{t},y=\tan{t}$,then
$$3x^2-2xy=12(\sec{t})^2-4\sec{t}\tan{t}$$
| Rearranging B. Goddard's system of equations:
$$\begin{align}12x-4y-\lambda x=0\end{align}\tag1$$
$$\begin{align}x=\lambda y\end{align}\tag2$$
$$\begin{align}x^2-4y^2=4\end{align}\tag3$$
Inserting $(2)$ into $(1)$ yields
$$\begin{align}12\lambda y - 4y - \lambda^2y=0\end{align}$$
$\Leftrightarrow$
$$\begin{align}(\lambda^2-12\lambda + 4)y=0\end{align}.$$
The two solutions to the quadratic equation are
$$\begin{align}\lambda_1=-4\sqrt 2 + 6\end{align}$$
and
$$\begin{align}\lambda_2=+4\sqrt 2 + 6\end{align}.$$
Inserting $(2)$ into $(3)$ gives us
$$\begin{align}(\lambda^2-4)y^2=4\end{align}.$$
For $\lambda_1$ with
$$\begin{align}\lambda_1^2=-48\sqrt 2+68\end{align}$$
and
$$\begin{align}\lambda_1^2-4=16(-3\sqrt 2+4)\end{align}$$
so that
$$\begin{align}\dfrac{4}{\lambda_1^2-4}=\dfrac{1}{4(-3\sqrt 2+4)}=\dfrac{3\sqrt 2+4}{4\cdot-2}\end{align}$$
it follows that
$$\begin{align}-8y^2=+3\sqrt{2}+4\end{align}.$$
Unfortunately this is not solvable over the reals.
Likewise for $\lambda_2$ we have
$$\begin{align}-8y^2=-3\sqrt{2}+4\end{align}$$
which is solvable over the reals since $3\sqrt{2}\geq 4$.
The minimum is
$$\begin{align}3x^2-2xy=4\sqrt 2 + 6\end{align}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
$\sqrt{9x^2-16}>3x+1$ I'm trying to solve the following inequality:
$$\sqrt{9x^2-16}>3x+1$$
Here's my attempt:
$\sqrt{9x^2-16}>3x+1$
$\longrightarrow 9x^2-16>9x^2+6x+1$
$\longrightarrow -16>6x+1$
$\longrightarrow x<-\frac{17}{6}$
Now, I need to check the constraints:
$9x^2-16 > 0$
$\longrightarrow (3x)^2 > 4^2$
$\longrightarrow \pm3x > 4$
$\longrightarrow 3x > 4$; $-3x > 4$
$\longrightarrow x > \frac{4}{3}$; $x < -\frac{4}{3}$
Making sure the answer meets the constraints:
$\{(-\infty, -\frac{4}{3})\cup(\frac{4}{3}, \infty)\}\cap (-\infty, -\frac{17}{6}) = (-\infty, -\frac{17}{6})$
So, my answer is $x=(-\infty, -\frac{17}{6})$, however verifying on Wolfram|Alpha results in $x=(-\infty, -\frac{4}{3}]$.
Where, what, and why is wrong with my solution?
| *
*$3x+1\leq0$ and $9x^2-16\geq0$, which gives $x\leq-\frac{4}{3}$;
*$3x+1>0$ and $9x^2-16>(3x+1)^2$. The last gives $x<-\frac{17}{6}$, which is impossible.
Thus, the answer is $\left(-\infty,-\frac{4}{3}\right]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
determine for which values of $x$ the function $f(x)=\begin{cases} \frac{x^2-4}{x-2}+10, &\text{if }x < 2\\ 2x^3-x, &\text{if } x ≥ 2\\ \end{cases}$ is continuous...
$$f(x)=\begin{cases}
\frac{x^2-4}{x-2}+10, &\text{if }x < 2\\
2x^3-x, &\text{if } x ≥ 2\\
\end{cases}$$
These kind of problems are very unclear to me, I only know these kind of functions have several graphs, in this example There are 2, But I can't figure out what these intervals means and even worse I don't even know where to start from.
| As Zubin Mukerjee indicated in the comments, we must show that
*
*$f$ is continuous when $x < 2$;
*$f$ is continuous when $x > 2$;
*$\lim_{x \to 2+} f(x) = \lim_{x \to 2-} f(x) = f(2)$.
If $x < 2$, then
\begin{align*}
f(x) & = \frac{x^2 - 4}{x - 2} + 10\\
& = \frac{(x + 2)(x - 2)}{x - 2} + 10\\
& = x + 2 + 10 && \text{provided $x \neq 2$}\\
& = x + 12
\end{align*}
Since $x < 2$, we are not dividing by zero in the second step. Hence, the given expression is equivalent to $x + 12$ in the interval $(-\infty, 2)$. Since $x + 12$ is a polynomial, $f$ is continuous in the interval $(-\infty, 2)$.
If $x > 2$, then $f(x) = 2x^3 - x$. Since this is a polynomial, $f$ is continuous the interval $(2, \infty)$.
Since
$$\lim_{x \to 2+} f(x) = \lim_{x \to 2+} (2x^3 - x) = 2 \cdot 2^3 - 2 = 16 - 2 = 14$$
and
$$\lim_{x \to 2-} f(x) = \lim_{x \to 2-} (x + 12) = 2 + 12 = 14$$
and
$$f(2) = 2 \cdot 2^3 - 2 = 16 - 12 = 14$$
it is true that
$$\lim_{x \to 2+} f(x) = \lim_{x \to 2-} f(x) = f(2)$$
Since we have verified all three conditions, $f$ is continuous on $(-\infty, \infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Does $\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na}) $ converge? I want to check whether
$\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na}) $
converges or not. (a is a positive constant number.)
If it converges, how to find the value it converges?
And if not, why?
| This is the limit of a Riemann sum:
$$\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na})=\frac{1}{a}\lim_{n\rightarrow\infty}\frac{a}{n}\sum_{k=1}^n\frac{1}{1+\frac{ka}{n}}\\=\frac{1}{a}\int_0^a\frac{dx}{1+x}=\frac{\ln(a+1)}{a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
If $\sin(\alpha)+\cos(\alpha)=a,$ denote $|\sin(\alpha)-\cos(\alpha)|$ in terms of $a$ I attempted to solve it using the following identities: $1.\ a^2-b^2.\ 2.\ (a\pm b)^2.\ 3. a^3\pm b^3.\ 4.\ (a\pm b)^3.$
Since none of my efforts led me to the correct answer (which is $\sqrt{2-a^2}$), I found it better not to write my lengthy work towards solving it.
I mostly got the expression $2\sin(\alpha)\cos(\alpha).$ So, any helpful hints, comments or answers are welcome!
| $$\begin{align}
\sin b+\cos b&=a\\
\sin^2 b+2\sin b\cos b+\cos^2 b&=a^2\\
-2(\sin^2b+\cos^2b)&\quad -2\\
-\sin^2b+2\sin b\cos b-\cos^2 b&=a^2-2\\
-(\sin b-\cos b)^2&=a^2-2\\
(\sin b-\cos b)^2&=2-a^2\\
|\sin b-\cos b|&=\sqrt{2-a^2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1$ When I'm reading Computer Systems: A Programmer's Perspective, I met the sum of binary numbers and failed to prove it:
$$ 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1 $$
This might be preliminary knowledge, I'm not good at mathematics, any body could give me a hint?
| Notice $2^0=2^1-1$.
Then $2^0+2^1=2(2^1)-1=2^2-1$
Then $2^0+2^1+2^2=2(2^2)-1=2^3-1$.
We have $2^0+2^1+2^2...2^{n-1}=2(2^{n-1})-1=2^n-1$.
And to show the this holds for the $n+1$ case, notice that $(2^n-1)+2^n=2(2^n)-1=2^{n+1}-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 5
} |
Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$
Evaluate the following limit.
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
My Attempt:
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-a} + \sqrt {bx}}$$
$$=\lim_{x\to \infty} \dfrac {x-a-bx}{\sqrt {x-a} + \sqrt {bx}}.$$
How do I proceed?
| If $b \geq 0$ with $b \neq 1$:
$$ \lim \limits_{x \to +\infty} \sqrt{x-a} - \sqrt{bx} = \lim \limits_{x \to +\infty} \sqrt{x}\Bigg( \sqrt{1 - \frac{a}{x}} - \sqrt{b} \Bigg) = \begin{cases} +\infty & \text{if } 0 \leq b < 1 \\ -\infty & \text{if } b > 1 \end{cases} $$
because, as $x \to +\infty$:
$$ \Bigg( 1 - \frac{a}{x} \Bigg)^{1/2} - \sqrt{b} = 1 - \sqrt{b} - \frac{a}{2x} + o\Big( \frac{1}{x} \Big). $$
As a result:
$$ \lim \limits_{x \to +\infty} \Bigg( 1 - \frac{a}{x} \Bigg)^{1/2} - \sqrt{b} = 1 - \sqrt{b}. $$
If $b=1$:
$$ \lim \limits_{x \to +\infty} \sqrt{x-a} - \sqrt{bx} = 0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Number of monomials in $a_n=a_{n-1}+(a_{n-2})^2$ with $a_1=a$, $a_2=b$ I was playing around with the sequence $a_1=a$, $a_2=b$ and the recurrence $a_n=a_{n-1}+(a_{n-2})^2$ and just listed out a few general terms.
For instance, $a_1=a$, $a_2=b$, $a_3=b+a^2$, $a_4=b+b^2+a^2$, $a_5=a^4+2a^2b+a^2+2b^2+b$, and $a_6=b^4+2b^3+3b^2+2a^2b^2+4a^2b+b+2a^4+a^2.$
When I did that, I realized that the number of terms in $a_n$ appears to be $F_n$ where $F$ represents the Fibonacci numbers. I conjecture that this is true, but I am not sure how to even approach this.
| What an interesting question! The sequence of the number of monomial terms is $1,1,2,3,5,8,14,24,44,80,152,288,560,1088,2144,\dots$
-- not Fibonacci. For example $a_7$ =
$b + 4 b^2 + 6 b^3 + 5 b^4 + a^2 + 6 b a^2 + 10 b^2 a^2 + 8 b^3 a^2 + 3 a^4 + 6 b a^4 + 8 b^2 a^4 + 2 a^6 + 4 b a^6 + a^8$ has 14 terms.
The sequence is now in the OEIS as sequence A290075.
A recursion for the number of terms is $b_n = 2 b_{n-1} +2 b_{n-2} -4 b_{n-3}$ for $n\ge 6$ and there is more information in the OEIS link.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Show that $\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
I want to prove that $\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
My ideas, I don't know if they lead anywhere:
Let's substitute $\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and $z=e^{i\theta}$ right after:
$\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=-i\cdot\int_1^{-1}\frac{z^2+z^{-2}+\frac{1}{2}z^3+\frac{1}{2}z^{-3}}{5z+2z^2+2}dz$
This now gives me 4 new integrals, for example
$\displaystyle-i\int_1^{-1}\frac{z^2}{2z^2+5z+2}dz$, $\displaystyle-i\int_1^{-1}\frac{1}{2z^4+5z^3+2z^2}dz$ and so on.
But since I haven't been able to solve any of the new integrals, I'm a little lost.
Edit: Can't I do a partial fractions decomposition of all the 4 integrals and solve them seperately?
| As Dr. Sonnhard Graubner answered, expand $\cos(2x)$ and $\cos(3x)$ using the classical formulas and use the tangent half-angle substitution to get
$$I=\int\frac{\cos(2x)+\cos(3x)}{5+4\cos(x}\,dx=\int\frac{2 \left(t^6+5 t^4-25 t^2+3\right)}{\left(t^2+1\right)^3 \left(t^2+9\right)}dt$$ Using partial fraction decomposition $$\frac{2 \left(t^6+5 t^4-25 t^2+3\right)}{\left(t^2+1\right)^3 \left(t^2+9\right)}=\frac{3}{8 \left(t^2+9\right)}+\frac{13}{8
\left(t^2+1\right)}-\frac{9}{\left(t^2+1\right)^2}+\frac{8}{\left(t^2+1\right)^3
}$$ Integration now to get $$I=\frac{1}{8} \left(\frac{4 t-12 t^3}{\left(t^2+1\right)^2}+\tan
^{-1}\left(\frac{t}{3}\right)+\tan ^{-1}(t)\right)$$ and the bounds are $0$ and $\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
How to show $x^5-x+3$ is irreducible in $\Bbb Z_5[x]$? How to show $x^5-x+3$ is irreducible in $\Bbb Z_5[x]$?
I checked that there are no linear factors since there is no root. How to know that there is no quadratic factor? I checked from Mathematica that there are $32$ irreducible quadratic in $\Bbb Z_5[x]$.
| By long division, if an $x^2 + a x + b$ divided $x^5 - x + 3$, you would need $a^4 - 3 a^2 b + b^2 - 1 \equiv 0$ and $a^3 b - 2 a b^2 + 3 \equiv 0 \mod 5$.
You can "complete the square" in $a^3 b - 2 a b^2 + 3 \mod 5$ to obtain
$3 a (b + a^2)^2 + 2 a^5 + 3 \equiv 3 a \left((b+a^2)^2 +4 a^4 + 1/a\right) $. For this to be $0$ mod $5$, we need $ a^4 - 1/a$ to be a quadratic residue mod $5$ (thus $0$, $1$ or $4$); the possibilities are then
$(a,b) = (1,4), (3,3)$ and $(3,4)$. But none of those cases makes
$a^4 - 3 a^2 b + b^2 - 1 \equiv 0 \mod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How prove this inequality with $x+y+z=2$ Let $x,y,z\ge 0$,and such $x+y+z=2$,show that
$$\sum\sqrt{\dfrac{x}{y^2+z^2}}\ge \dfrac{2}{15}\sum\sqrt{\dfrac{2+47x}{2-x}}$$
I tried C-S,Holder but without success.$$\left(\sum_{cyc}\sqrt{\dfrac{x}{y^2+z^2}}\right)^2(\sum x(y^2+z^2))\ge (x+y+z)^3$$
| We need to prove that $$\sum_{cyc}\sqrt{\frac{x}{y^2+z^2}}\geq\frac{2\sqrt2}{15\sqrt{x+y+z}}\sum_{cyc}\sqrt{\frac{48x+y+z}{y+z}}.$$
Now, by Holder
$$\left(\sum_{cyc}\sqrt{\frac{x}{y^2+z^2}}\right)^2\sum_{cyc}x^2(y^2+z^2)\geq(x+y+z)^3$$ and by C-S
$$\left(\sum_{cyc}\sqrt{\frac{48x+y+z}{y+z}}\right)^2\leq\sum_{cyc}\frac{48x+y+z}{6x+y+z}\sum_{cyc}\frac{6x+y+z}{y+z}.$$
Thus, it remains to prove that
$$225(x+y+z)^4\geq16\sum_{cyc}x^2y^2\sum_{cyc}\frac{48x+y+z}{6x+y+z}\sum_{cyc}\frac{6x+y+z}{y+z},$$
which is
$$\sum_{sym}(30x^9y+237x^8y^2-42x^7y^3-237x^6y^4+12x^5y^5)+$$
$$+xyz\sum_{sym}(365x^7+3918x^6y+2142x^5y^2+3484x^4y^3)+$$
$$+x^2y^2z^2\sum_{sym}(11019x^4+39786x^3y+17141x^2y^2+34785x^2yz)\geq0,$$
for which it's enough to prove that
$$\sum_{sym}(30x^9y+237x^8y^2-42x^7y^3-237x^6y^4+12x^5y^5)\geq0$$ or
$$\sum_{cyc}xy(x^2-y^2)^2(10x^4+79x^3y+6x^2y^2+79xy^3+10y^4)\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understand what the author tries to say here. Can't this problem be done in another manner?
| If ( with integer $a \geq 1$) $$ (a-1)^2 < w < a^2, $$
then $w$ is not a square at all. I guess I can add that then
$$ a-1 < \sqrt w < a, $$
so that $\sqrt w$ is not an integer, it lies strictly between two consecutive integers.
As usual, there are a few cases to check for small $w$
You are given $n^4 - n^2 + 64.$ Now, $(n^2)^2 = n^4.$ Also $(n^2 - 1)^2 = n^4 - 2 n^2 + 1.$
For $n \geq 9$ we have $n^2 > 64$ so that $n^4 - n^2 + 64 < n^4.$
For $n \geq 8,$
$$ n^4 - n^2 + 64 - (n^2 - 1)^2 = n^2 - 63 > 0, $$ so
$$ n^4 - n^2 + 64 > (n^2 - 1)^2. $$
Alrighty, for $n \geq 9 $ we get
$$ (n^2 - 1)^2 < n^4 - n^2 + 64 < (n^2)^2 $$ so that $n^4 - n^2 + 64$ is NOT a square for $n \geq 9.$
You still need to check $n=0,1,2,3,4,5,6,7,8.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
} |
Arithmetic-geometric mean, prove that $c_n = 4me^{-\ell 2^n+\epsilon_n}$ Let $a$ and $b$ reals with $a > b > 0$. Let $(a_n)$ and $(b_n)$ with $a_0 = a$, $b_0 = b$ and
$$a_{n+1} = \tfrac{a_n+b_n}{2} \quad\text{;}\quad b_{n+1} = \sqrt{a_nb_n}$$
We know that $\lim_{n \rightarrow +\infty} a_n = \lim_{n \rightarrow +\infty} b_n = m$.
Let $c_n = \sqrt{a_n^2-b_n^2}$.
We know that $\lim_{n \rightarrow +\infty} c_n =0$ and that $c_{n+1} \leq \frac{c_n^2}{4m}$.
The question is to prove that there is $\ell > 0$ and $\epsilon_n > 0$ with $\lim_{n \rightarrow +\infty} \epsilon_n = 0$ such as
$$c_n = 4me^{-\ell 2^n+\epsilon_n}$$
An indication is to use $u_n = -2^{-n} \ln(c_n)$ and $\sum (u_{n+1}-u_n)$.
| As I assumed, the statement is not clear. Only the first two lines are the hypotheses. I'll do it tomorrow:
$a_{n+1} = \tfrac{a_n+b_n}{2}, b_{n+1} = \sqrt{a_nb_n}\rightarrow$ (arithmetic mean > geometric mean)
$0<b<b_{1}<a_{1}<a,\quad 0<b<b_{1}<b_{2}<a_{2}<a_{1}<a, \quad \text{etc.}\rightarrow \left\{ \begin{array}{lcc}
a_{n} \quad \text{is decreasing} \\
\\ b_{n} \quad \text{is increasing}
\end{array}
\right.$
$\{a_{n}\}$ and $\{b_{n}\}$ monotones and bounded $\rightarrow$ They are convergent $\rightarrow \left\{ \begin{array}{lcc}
\exists \lim_{n\to \infty}\{a_{n}\}=m \\
\\ \exists \lim_{n\to \infty}\{a_{n}\}=n
\end{array}
\right.$
$a_{n+1}=\dfrac{a_{n}+b_{n}}{2}\rightarrow \lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}\dfrac{a_{n}+b_{n}}{2}\rightarrow m=\dfrac{m+n}{2}$
$\rightarrow m=n\rightarrow \lim_{n\to \infty}\{a_{n}\}=\lim_{n\to \infty}\{b_{n}\}=m$
$a_{n}$ is increasing $\rightarrow a_{n}\geq m$
$c_{n}=\sqrt{a_{n}^{2}-b_{n}^{2}}\rightarrow c_{n}^{2}=(a_{n}-b_{n})(a_{n}+b_{n})$
$c_{n+1}=\sqrt{a_{n+1}^{2}-b_{n+1}^{2}}=\sqrt{\left( \frac{a_{n}+b_{n}}{2}\right)^{2}-(\sqrt{a_{n}b_{n}})^{2}}$
$\rightarrow c_{n+1}=\sqrt{\frac{a_{n}^{2}+2a_{n}b_{n}+b_{n}^{2}-4a_{n}b_{n}}{4}}=\dfrac{1}{2}\sqrt{(a_{n}-b_{n})^{2}}=\dfrac{1}{2}(a_{n}-b_{n})$
$\rightarrow c_{n+1}=\dfrac{1}{2}\dfrac{c_{n}^{2}}{(a_{n}+b_{n})}=\dfrac{1}{2}\dfrac{c_{n}^{2}}{2a_{n}}\leq \dfrac{c_{n}^{2}}{4m}\rightarrow c_{n+1}\leq \dfrac{c_{n}^{2}}{4m}$
You've understood?
EDIT:
It defines: $u_{n}=-2^{-n}\ln(c_{n})$
$\rightarrow u_{n+1}=-2^{-n-1}\ln(c_{n+1})$
$\rightarrow u_{n+1}=-2^{-n-1}\ln\left( \frac{c_{n}^{2}}{4a_{n+1}}\right)$
$\rightarrow u_{n+1}=-2^{-n-1}[2\ln(c_{n})-\ln(4a_{n+1})]=-2^{-n}\ln(c_{n})+2^{-n-1}\ln(4a_{n+1})$
$\rightarrow u_{n+1}=u_{n}+2^{-n-1}\ln(4a_{n+1})$
$\rightarrow u_{n+1}-u_{n}=2^{-n-1}\left[ \ln(4m)+\ln\left( \frac{4a_{n+1}}{4m}\right)\right]$
$\rightarrow u_{n+1}-u_{n}=2^{-n-1}\left[ \ln(4m)+\ln\left( \frac{a_{n+1}}{m}\right)\right]$;
we put: $a_{n+1}=\ln\left( \frac{a_{n+1}}{m}\right)\rightarrow \left\{ \begin{array}{lcc}
a_{n} \quad \text{decreases} \\
\\ a_{n}>0 \\
\\ \lim_{n\to \infty}a_{n}=0
\end{array}
\right.$
$u_{n}-u_{0}=\sum^{n-1}_{k=0}(u_{k+1}-u_{k})$
$\rightarrow -2^{-n}\ln(c_{n})-[-2^{-0}\ln(c_{0})]=\sum^{n}_{k=0}\left[2^{-k-1}[\ln(4m)+\alpha_{k+1}]\right]$
$\rightarrow -2^{-n}\ln(c_{n})+\ln \sqrt{a-b}=\ln(4m)\sum^{n-1}_{k=0}\dfrac{1}{2^{k+1}}+\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}$
$\rightarrow 2^{-n}\ln(c_{n})=-\ln(4m)\cdot \dfrac{\frac{1}{2}\left( 1-\frac{1}{2^{n}}\right)}{1-\frac{1}{2}}+\ln \sqrt{a-b}-\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}$
$\rightarrow 2^{-n}\ln(c_{n})=-\dfrac{2^{n}-1}{2^{n}}\ln(4m)-\ln \dfrac{1}{\sqrt{a-b}}-2^{n}\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}$
$\rightarrow \ln(c_{n})=\ln(4m)-2^{n}\left( \ln \dfrac{1}{\sqrt{a-b}}+\ln 4m\right)-2^{n}\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}$
$\rightarrow \ln(c_{n})=\ln(4m)-2^{n}\ln \dfrac{4m}{\sqrt{a-b}}-2^{n}\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}$
Where $\sum^{\infty}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}=K$, since
$\dfrac{\frac{\alpha_{k+1}}{2^{k+1}}}{\frac{\alpha_{k}}{2^{k}}}=\dfrac{\alpha_{k+1}}{2\alpha_{k}}<1$ (Quotient criterion)
$\rightarrow \beta_{n}=K-\sum^{n-1}_{k=0}\dfrac{\alpha_{k+1}}{2^{k+1}}, \quad \text{with}\quad \lim_{n\to \infty}\beta_{n}=0$
$\rightarrow \ln(c_{n})=\ln(4m)-2^{n}\ln \dfrac{4m}{\sqrt{a-b}}-2^{n}(K-\beta_{n})$
$\rightarrow \ln(c_{n})=\ln(4m)-2^{n}\left( \ln \frac{4m}{\sqrt{a-b}}+K\right)-2^{n}\beta_{n}$
Let $l=\ln \dfrac{4m}{\sqrt{a-b}}+K>0$:
$$c_{n}=4me^{-2^{n}l-2^{n}\beta_{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve the differential equation $x^2\frac{d^2y}{dx^2}+(x^3-x)\frac{dy}{dx}-3y=0$ $$x^2\frac{d^2y}{dx^2}+(x^3-x)\frac{dy}{dx}-3y=0$$
$$\sum_{n=0}^{\infty}(n+r)(n+r-2)c_nx^{n+r}+\sum_{n=2}^{\infty}(n+r-2)c_{n-2}x^{n+r}-3\sum_{n=0}^{\infty}c_nx^{n+r}=0$$
$$r(r-2)c_0x^r+(r^2-1)c_1x^{1+r}-3c_0x^r-3c_1x^{r+1}+\sum_{n=2}^{\infty}[(n+r)(n+r-2)c_n+(n+r-2)c_{n-2}-3c_n]x^{n+r}=0$$
The indicial equation is
$$(r-3)(r+1)=0$$
We have two possibility
Deal with the larger value $3$ first,
$$c_n=-\frac{(n+1)c_{n-2}}{n^2+4n}$$
Taking values for $n\geq2$
$$c_2=-\frac{c_0}{4}$$
$$c_3=0$$
$$c_4=-\frac{5c_2}{32}$$
$$c_5=0$$
$$y_1(x)=C_1x^3(1-\frac{x^2}{4}+\frac{5x^4}{128}+...)$$
The second possibility $r=-1$
$$c_n=-\frac{(n-3)c_{n-2}}{n^2-4n},n\neq4$$
we know that $c_4 is an arbitrary constant
We know that $c_4$ will make $c_2=0$ from the above equation
$$c_2=-\frac{c_0}{4}$$
We see that it contradicts our initial assumption that $c_0\neq0$
$$c_3=0$$
$$c_5=-\frac{2c_3}{5}=0$$
$$c_6=-\frac{c_4}{4}$$
$$c_7=0$$
$$c_8=-\frac{5c_6}{32}$$
$$y_2=(x^{-1})c_4(x^4-\frac{x^6}{4}+\frac{5x^8}{128}+...)$$
Obviously they are two sides of the same coin.
How to find another linearly independent solution $y_2$. Is there a closed form for the above series?
The thing seems very messy.
| Wolfram Alpha gives the solutions as
$$
y_1(x) = x e^{-x^2/4}~I_1(x^2/4) = \frac{x^3}{8} \left( 1 - \frac{x^2}{4} + \frac{5 x^4}{128} - \frac{7 x^6}{1536} + \dots\right)
$$
which is the one you found in the expansion. The other solution diverges at $x=0$ and is
$$
y_2(x) = x e^{-x^2/4}~K_1(x^2/4)
$$
$I_1$ and $K_1$ are modified Besselfunctions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Region of Convergence for Laurent Expansion Find the Laurent expansion of $\frac{z}{(z+1)(z+2)}$ about the singularity $z=-2$. Specify the region of convergence and the nature of singularity at $z = -2$.
The Laurent expansion I get is
$1+(z+2)+(z+2)^2+ \ldots + \frac{2}{z+2}$
The singularity is a simple pole.
But how to find the Region of Convergence.
| The function
\begin{align*}
f(z)&=\frac{z}{(z+1)(z+2)}\\
&=\frac{2}{z+2}-\frac{1}{z+1}\\
\end{align*}
is to expand around the center $z=-2$. Since there are simple poles at $z=-1$ and $z=-2$ we have to distinguish two regions of convergence
\begin{align*}
D_1:&\quad 0<|z+2|<1\\
D_2:&\quad |z+2|>1
\end{align*}
*
*The first region $D_1$ is a punctured disc with center $z=-2$, radius $1$ and the pole $-1$ at the boundary of the disc.
In the interior we have a representation of the fractions with pole at $z=-2$ as principal part of a Laurent series at $z=-2$, while the fraction with pole at $z=-1$ admits a representation as power series.
*The other region $D_2$ containing all points outside the closure of $D_1$ admits for all fractions a representation as principal part of a Laurent series at $z=-2$.
Expansion in $D_1$:
\begin{align*}
f(z)&=\frac{1}{1-(z+2)}+\frac{2}{z+2}\\
&=\frac{2}{z+2}+\sum_{n=0}^\infty(z+2)^n
\end{align*}
Expansion in $D_2$:
\begin{align*}
f(z)&=\frac{1}{1-(z+2)}+\frac{2}{z+2}\\
&=-\frac{1}{z+2}\cdot\frac{1}{1-\frac{1}{z+2}}+\frac{2}{z+2}\\
&=-\frac{1}{z+2}\sum_{n=0}^\infty\frac{1}{(z+2)^n}+\frac{2}{z+2}\\
&=-\sum_{n=2}^\infty\frac{1}{(z+2)^n}+\frac{1}{z+2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Factoring $x^{7} - 1$ into irreducibles over $\mathbb{F}_{2}[x]$ I know this breaks down into $(x - 1)(x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x +1)$, so the task is to show whether the second factor is irreducible over $\mathbb{F}_{2}[x]$. It's quick check that there are no roots, so one way might be to look at all remaining partitions of 6, then divide through by all polynomials of degree 4, 3 and 2. But there must be something faster.
| Factoring $x^{7} - 1$ is equivalent to factoring $x^{8} - x$.
Now, $x^{8} - x=0$ is the equation that defines the finite field with $8$ elements.
Therefore, $x^{8} - x$ has at least one irreducible factor of degree $3$ (here we use that $8=2^3$).
This answer gives a simple argument that $x^3 + x^2 + 1$ and $x^3 + x + 1$ are the only irreducible polynomials of degree $3$ over $\mathbb{F}_{2}$. By the uniqueness of finite fields, both polynomials must divide $x^{8} - x$. Indeed,
we can easily check that $x^{8} - x=x (x + 1) (x^3 + x + 1) (x^3 + x^2 + 1)$ and so
$x^{7} - 1=(x + 1) (x^3 + x + 1) (x^3 + x^2 + 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve for four values of x where x lies in complex plane. I came across this problem and have to find four roots of this equation. Direct multiplication was of no help and factorizing it is a mess.
I tried to factor it using $a^2+b^2=(a+ib)(a-ib) $
but was of no help.
$$x^2 + (\frac{ax}{x+a})^2= 3a^2$$
Can anyone through in some ideas?
| Wolog $a\neq 0.$ If we substitute $x=ay$, we get completely rid of the $a$:
$$
(ay)^2+\left(\frac{a^2y}{ay+a}\right)^2=3a^2
\;\;\Longleftrightarrow \;\;
y^2+\left(\frac{ay}{ay+a}\right)^2=3
\;\;\Longleftrightarrow \;\;
y^2+\left(\frac{y}{y+1}\right)^2=3
$$
Now we multiply this with $(y+1)^2$ and we get
$$
y^4+2y^3-y^2-6y-3 =0
$$
Now we can cheat a little bit and use e.g. an online solver for roots of polynomials, and we see that the golden ratio $1,618\ldots\ $ seems to be one of the roots, so we can try the factor $y^2-y-1$. And indeed we get
$$
y^4+2y^3-y^2-6y-3 =0
\;\;\Longleftrightarrow \;\;
(y^2-y-1)(y^2+3y+3)=0
$$
Now it is easy to get the $y$. From $x=ay$, we finally get the $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$. Question:
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$.
My attempt:
Proof by contradiction:
Assume $c$ is divisible by $3$ and $a$ or $b$ is not divisible by $3$.
Since $c$ is divisible by $3$ we can write $c$ as $ \ c = 3m \implies c^{2} = 9m^{2} \implies 9 | c^{2}$.
Since $a$ and $b$ are not divisible by $3$, $\ a = 3k+1$ and $ \ b = 3n+1$ for some integers $\ k,n.$
Then,
$ a^{2} + b^{2} = (3k+1)^{2} + (3n+1)^{2} = 9k^{2} + 6k +9n^{2} + 6n + 2$.
I am stuck here. I can't find a contradiction. How can I show that $ a^{2} + b^{2} $ is not divisible by $9$.
| A couple of points:
1) The negation of "$a$ and $b$ are both divisible by $3$" is "at least one of $a$ or $b$ is not divisible by $3$".
2) You don't need divisibility by $9$ to prove a contradiction. Divisibility by $3$ will do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
What is the value of $x^2 + y^2 + z^2$ if $x^2 y + y^2 z + z^2 x=2186$ and $xy^2 + y z^2 + z x^2=2188$
What is the value of $x^2 + y^2 + z^2$ if $x^2 y + y^2 z + z^2 x=2186$ and $xy^2 + y z^2 + z x^2=2188$, where $x,y,z$ are integers.
My Attempt
$(x^2 y + y^2 z + z^2 x) +(x^2 y + y^2 z + z^2 x) =2186+2188=4374$.
From this we can derive $(x+y+z)(xy+yz+zx)-3xyz=4374$ . We can subtract the given equations to get $xy(x-y)+yz(y-z)+zx(z-x)=-2$. But after this , I can't figure out how to proceed.
Any help is appreciated. Thanks in advance.
| $$\sum_{cyc}(x^2y-x^2z)=(x-y)(x-z)(y-z)=-2$$
and the rest for you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find triples (a,b,c) such that $f(n)f(n+1)=f(m)$ where $f(x)=ax^2+bx+c$
Find all triples of integers $(a,b,c)$ with $a\neq0$ such that the function $f(x)=ax^2+bx+c$ has the property that, for each positive integer $n$, there is an integer $m$ with $$f(n)f(n+1)=f(m).$$
Attempt: Writing$\,\,f(x)=a(x+\alpha)(x+\beta)$, we get $$f(n)f(n+1)=a^2\left(z+\alpha)(z+\beta\right)$$ where $z=n^2+n\left(1+\frac{b}{a}\right)+\frac{c}{a}$. Hence all triples $(1,b,c)$ work. Are these the only ones?
EDIT. I am able to show if $a=m^2, a\neq 1$, then the triples are exactly $(m^2,2mn,n^2)$ where $m$ divides $n(n-1)$. So now the question is whether $a$ must be a perfect square.
P.S. This is the problem 11964 from AMM problem section (March 2017) for which the deadline for submitting solutions has ended.
| Negative $a$ will certainly not work: It makes $f(m)$ bounded from above whereas $f(n)f(n+1)$ is not.
Your argument is correct in showing that all triples with $a=1$ work.
We vary the problem by demanding the desired property only for all sufficiently large $n$. Then $(a,b,c)$ has the property if and only if $(a,b\pm 2a,a\pm b+c)$ has it. Thus we may assume wlog that $|b|\le a$ (equivalently, $f$ has its minimum in $[-\frac12,\frac12]$).
Case 1: Assume $b=0$. Now the equation is
$$ a^2n^4+2a^2n^3+a(a+c)n^2+2acn+(a+c)c = am^2+c.$$
So $m^2 = an^4+O(n^3)$, hence $m=\pm n^2\sqrt a+O(n)$.
With $m=n^2\sqrt a+u n$,
$$ a^2n^4+2a^2n^3+a(a+c)n^2+2acn+(a+c)c= a^2n^4+2au\sqrt an^3+u^2an^2+c,$$
or,
$$ 2a\sqrt a(\sqrt a-u)= (u^2-a-c)an^{-1}-2acn^{-2}+(1-a-c)cn^{-3}$$
As $u$ is bounded, we conclude that $u=\sqrt a+O(n^{-1})$.
This makes $m=n^2\sqrt a+n\sqrt a+v$ with $v=O(1)$.
Now we arrive at
$$a(c-2v\sqrt a) = 2a(\sqrt a v-c) n^{-1} +v^2a+(1-a-c)cn^{-2}$$
and from this $v=\frac c{\sqrt a}+O(n^{-1})$,
$$m=\sqrt a\cdot(n^2+n+\tfrac ca)+O(n^{-1}).$$
Case 1.1: If $a$ is a perfect square, this implies that $c$ is a multiple of $\sqrt a$, and we can write $a=d^2$, $c=kd$; then for $n$ large enough, we have $m=dn^2+dn+k$ and then $f(n)f(n+1)=f(m)$ leads to $kd^3-kd$, i.e., $k=0$ or $a=1$, i.e., the solutions found by the OP and in the comments.
Case 1.2: If $a$ is not a perfect square ...
Case 2: If $b\ne 0$, ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $(a^2+b^2),$ where $(a+b)=\dfrac{a}{b}+\dfrac{b}{a}$ The question is the same .Find $a^2+b^2$. I think we have to find $a$ and $b$ firstly. Given that $a$ and $b$ are integers.
| $$a+b=\frac{a}{b} + \frac{b}{a}$$
therefore,$$a^2+b^2=ab(a+b)$$
or,$$a^2(1-b)+b^2(1-a)=0$$
from this relation we can say $1-b=1-a=0$
therefore,$$a=b=1$$
so finally,
$$a^2+b^2=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Check my solution of no.2 please 1.Show that the closest integer to $(2 + \sqrt{5})^{2017} -2^{2018}$ is divisible by $2017$. (Solved)
2.Suppose that $a,b,c$ are positive real numbers and that for some integer $2 \leq n$, $a^n + b^n = c^n$
Let $k$ be an integer with $1 \leq k <n$. Prove that there is triangle with side length $a^k,b^k,c^k$
My $Sol$ of 2) $f(n)=a^n +b^n ,g(n)=c^n,
f(n)-g(n) =h(n)=0, h'(n)=a^n log(a) + b^n log(b) - c^n log(c)$. $h'(n)$ is decreasing function, so by $k<n,h(k)=a^k+b^k-c^k>0 =h(n) $.
($a,b<c$ is trivial)
This proof is work?
| Hint for 1. $(2+\sqrt{5})^{2017}+(2-\sqrt{5})^{2017}$ is an integer and $(2-\sqrt{5})^{2017}$ is very small.
By setting $a_n=(2+\sqrt{5})^n+(2-\sqrt{5})^n$ we have $a_0=2, a_1=4$ and $a_{n+2}=4 a_{n+1}+a_n$.
$2017$ is a prime and $5$ is not a quadratic residue $\!\!\pmod{2017}$. By considering $\mathbb{F}_{2017}[x]/(x^2-4x-1)\simeq\mathbb{F}_{2017^2}$ and by taking $\alpha$ as a root of $x^2-4x-1$, by Vieta's Theorem and Frobenius map $x\to x^{2017}$ we get $\alpha+\alpha^p = \alpha^p+\alpha^{p^2}=4$.
It follows that
$$(2+\sqrt{5})^{2017}+(2-\sqrt{5})^{2017}-2^{2018}\equiv 4-4\equiv\color{red}{0}\pmod{2017}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to show that the number $w =\frac{ z}{1+z^2}$ is real only if $|z| = 1$ when $y\ne0$
If $z = x + iy$ where $y \ne 0$, show that the number $w = \frac{ z}{1+z^2}$ is real only if $|z| = 1 .$
Hi, so I've subbed $x+iy$ into the components into where $z$ is but where do I go from there?
Thanks.
| $$\dfrac z{1+z^2}=\dfrac{x+iy}{1+x^2-y^2+i(2xy)}=\dfrac{(x+iy)(1+x^2-y^2-i2xy)}{(1+x^2-y^2)^2+(2xy)^2}$$
The imaginary part of $(x+iy)(1+x^2-y^2-i2xy)$
is $$-2x^2y+y(1+x^2-y^2)=y(1-x^2-y^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove $1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$ for $A$, $B$, $C$ the angles of a triangle Given that $A$,$B$ and $C$ are angles of a triangle, show that
$$1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$$
| Note:
$$RHS=4\sin A\sin B\cos C=$$
$$-4\sin A\sin B\cos (A+B)=$$
$$-4\sin A\sin B(\cos A \cos B-\sin A \sin B)=$$
$$-\sin 2A\sin2B+(1-\cos2A)(1-\cos2B)=$$
$$1-\cos2A-\cos2B+\cos(2A+2B)=$$
$$1-\cos2A-\cos2B+\cos2C=LHS.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve the Diophantine equation $x^6 + 3x^3 + 1 = y^4$
Find all pairs $(x, y)$ of integers, such that $x^6 + 3x^3 + 1 = y^4$.
My solution:
Claim: The pair $(0, 1)$ are the only solutions.
Proof. Suppose there exists other solutions for $y \gt 1$ and $x \gt 0$, then I shall show that such pairs are impossible if $x$ and $y$ are integers. Let us Factorize the given equation as follows.
$x^6 + 3x^3 = y^4 - 1$
$x^6 + 3x^3 = (y^2)^2 - 1^2$
$x^6 + 3x^3 =(y^2 - 1)(y + 1)(y^2 + 1)$
$x^3(x^3+ 3) = (y - 1)(y + 1)(y^2 + 1) . . . (*)$
Because the numbers $(y - 1), (y + 1)$, and $(y^2 + 1)$ are all distinct, it follows that their products can never be a cube, hence, from $(*)$, we obtain the system.
$(y - 1)(y + 1)(y^2 + 1) = x^3 + 3$, and
$x^3 = 1$,
Which is equivalent to:
$(y - 1)(y + 1)(y^2 +) = 4 . . .(**)$
Since $(y - 1) \gt 0$, then $y$ is minimum if and only if $y = 2$ which clearly does not satisfy $(**)$. Hence we must have $x = 0$ and $y = 1$.
Please, Is there any mistake in my solution?.
| Case 1. If $x= 0$ we have $y^4=1$ thus $y=\pm 1$
Case 2. If $x>0$ then
$(x^3+1)^2= x^6+2x^3+1< x^6 + 3x^3 + 1 = y^4$
$ y^4 = x^6 + 3x^3 + 1< x^6+4x^3+4 = (x^3+2)^2$
So we have:$$(x^3+1)^2< y^4 <(x^3+2)^2$$
So $x^3+1< y^2 < x^3+2$ a contradiction.
Case 3. If $x<0$ then write $x=-t$ and $t>0$. Now we have to sove:
$$t^6 - 3t^3 + 1 = y^4$$ and this can be done in similar fashion:
$$(t^3-2)^2<y^4<(t^3-1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{\gamma}\frac{z^5}{z^6-1}$
Evaluate $$\int_{\gamma}\frac{z^5}{z^6-1}$$ where $\gamma=\{(x,y)\in\mathbb{R^2}\,|\,x^2+4y^2=16\}$.
How should I approach this? first to find the singularities by $z^6-1=0\iff z^6=1$
And then to plug each singularity in $\{(x,y)\in\mathbb{R^2}x^2+4y^2=16\}$?
| Since the curve $\gamma$ (ellipse centered at the origin with semiaxis $2$ and $4$) contains all the finite poles of the rational function $\frac{z^5}{z^6-1}$ ($6$ poles on the unit circle centered at the origin), it follows that
\begin{align*}\int_{\gamma}\frac{z^5}{z^6-1}dz&=-2\pi i\mbox{Res}\left(\frac{z^5}{z^6-1},\infty\right)=2\pi i\mbox{Res}\left(\frac{1/z^5}{z^2(1/z^6-1)},0\right)\\
&=2\pi i\mbox{Res}\left(\frac{1}{z(1-z^6)},0\right)=
2\pi i\mbox{Res}\left(\frac{1+z^6+o(z^6)}{z},0\right)=2\pi i
\end{align*}
where we used the notion of residue at infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof that $n!$ is divisible by $(n+1)^2$ for $n=xy+x+y$ I noticed the other day that for $n>8, n\in\mathbb{N}$, the factorial $n!$ seems to be divisible by $(n+1)^2$ when $n$ can be written in the form $xy+x+y$ (where $x,y\geq1$ and $\in\mathbb{N}$). Some examples:
*
*$n=14=2 \times 4+2+4$ (i.e. $x=2$, $y=4$), and we have $14!|15^2$
*$n=15=3 \times 3+3+3$ (i.e. $x=y=3$), and we have $15!|16^2$,
*$n=19=3 \times 4+3+4$ (i.e. $x=3$, $y=4$), and we have $19!|20^2$,
*...
This seems to hold for the first $15$ values I checked, so it seemed natural to try to prove it for all $n=xy+x+y$, however I could not find a proof. Here is my attempt:
Plugging $xy+x+y$ into the expressions above and noticing that $xy+x+y+1=(x+1)(y+1)$, we need to prove that
$$
(xy+x+y)!|(x+1)^2(y+1)^2,
$$
or
$$
\Gamma((x+1)(y+1))|(x+1)^2(y+1)^2.
$$
It is easy to prove that
$$
\Gamma((x+1)(y+1))|(x+1)(y+1)
$$
however I am struggling to prove the same for $(x+1)^2(y+1)^2$. Any hints?
| Wlog. $x\le y$.
If we are lucky, we find $x+1,2(x+1),y+1,2(y+1)$ as different elements among the factors $1,2,3,\ldots, n$ and we are done.
How can we fail to be lucky? The four candidates may be too large or not be different after all. As we assume $x\le y$, the following list is exhaustive:
*
*It may happen that $2(y+1)>n=xy+x+y$. This implies $x=1$. And indeed, $x=1$, $y=2$ gives $n=5$ and $36$ does not divide $5!$. More generally if $y+1$ is prime, then $(2y+1)!$ is not divisible by $(y+1)^2$.
*It may happen that $x+1=y+1$, which of course means that $x=y$. If $x\le 3$, we have $n\le 8$, which is excluded. If $x\ge 4$, we can use the factors $x+1,2(x+1),3(x+1),4(x+1)$ instead. These are guaranteed to be different and we have $4(x+1)\le xy+y<n$.
*It may happen that $2(x+1)=y+1$. Then we can use $x+1,2(x+1)=y+1, 3(x+1), 2(y+1)$ instead.
Thus
If $n=xy+x+y>8$ with positive integers $x,y$ then $(n+1)^2\mid n!$ except when $n+1$ is twice a prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving $\lim_{x\to \infty}(\sqrt[3]{x^3-5x^2+1}-x)$ with series? In "Problem-Solving Through Problems" from Loren C Larson , there is a problem (5-4-30) that says find the below limit with infinite series .
$$\lim_{x\to \infty}(\sqrt[3]{x^3-5x^2+1}-x)$$ this is easy limitation, but what is the idea to solving with infinite series?
I don't have a clue. I thankful for any hint in advance.
| Let $u = \frac{1}{x}$
Then the problem becomes:
$$\lim_{u\to 0}\left(\sqrt[3]{\left(\frac{1}{u}\right)^3-5\left(\frac{1}{u}\right)^2+1}-\frac{1}{u}\right)$$
$$=\lim_{u\to 0}\left(\sqrt[3]{\frac{1-5u+u^3}{u^3}}-\frac{1}{u}\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}(1-5u+u^3)^{\frac{1}{3}}-\frac{1}{u}\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[(1-5u+u^3)^{\frac{1}{3}}-1\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[(1-\frac{5u}{3}+O(u^2))-1\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[-\frac{5u}{3}+O(u^2)\right]\right)$$
$$=\lim_{u\to 0}\left(-\frac{5}{3}+O(u)\right)$$
$$=-\frac{5}{3}$$
EDIT: $$=\lim_{u\to 0}\left(\frac{1}{u}\left[(1-5u+u^3)^{\frac{1}{3}}-1\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[-1+\sum_{n=0}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}({u^3-5u})^n\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[\sum_{n=1}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}({u^3-5u})^n\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}\left[\frac{1}{3}(u^3-5u) + \sum_{n=2}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}u^n({u^2-5})^n\right]\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{3}(u^2-5) + \sum_{n=2}^{\infty}{\,\dbinom{\tfrac{1}{3}}{n}}u^{n-1}({u^2-5})^n\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{3}u^2-\frac{5}{3} + u\left[\sum_{m=0}^{\infty}{\,\dbinom{\tfrac{1}{3}}{m+2}}u^{m}({u^2-5})^{m+2}\right]\right)$$
$$=-\frac{5}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\dfrac {\pi x}{\sin \pi x} \times \dfrac {1}{\pi^2 x^2})$$
| Set $\pi(1-x)=2y$
and use $\cos(\pi-A)=-\cos A,\tan(\pi-B)=-\tan B$
and $\cos2y=1-2\sin^2y,\sin2y=2\sin y\cos y$
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}=\lim_{y\to0}\dfrac{1-\cos2y}{\tan^22y}$$
$$=\lim_{y\to0}\dfrac{2\sin^2y}{(2\sin y\cos y)^2}\cdot\lim_{y\to0}\cos^22y=?$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
If one root of the equation $ax^2+bx+c=0$ be the square of the other. If $a \neq 0$ and if one root of the equation $ax^2+bx+c=0$ is the square of the other, prove that: $$b^3+a^2c+ac^2=3abc.$$
My Attempt:
Given:
$$ax^2 + bx + c=0$$
Let $\alpha $ and $\beta $ be the roots of the equation.
$$\alpha + \beta = \dfrac {-b}{a}$$
$$\alpha . \beta = \dfrac {c}{a}$$
According to the question:
$$\alpha = \beta^2 $$
| Alternatively:
$$\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)^2=\frac{-b+\sqrt{b^2-4ac}}{2a} \Rightarrow$$
$$b^2-2ac+ab=(a-b)\sqrt{b^2-4ac} \Rightarrow$$
Squaring and then dividing both sides by $4a$ will prove the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful direction after that.
Note: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question.
| $$c^4=(-a-b)^4=(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
Therefore,
$$2(a^4+b^4+c^4)=4(a^4+2a^3b+3a^2b^2+2ab^3+b^4)$$
Now compute $(a^2+ab+b^2)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
} |
What is the expected number of tyres that are installed in their original positions? After summer, the winter tyres of a car (with four wheels) are to be put back. However, the owner has forgotten which tyre goes to which wheel, and the tyres are
installed `randomly', each of the $4! = 24$ permutations being equally likely.
What is the expected number of tyres that are installed in their
original positions?
Expected no. of tyres installed in original position = $P(1\ \text{tyre}) + 2P(2\ \text{tyre}) + 3P(3\ \text{tyre}) + 4P(4\ \text{tyre})$
I'm stuck after that. When counting $P(1\ \text{tyre})$ am I allowed to take the combination that all 4 tyres are in their right position ? Because that also includes 1 tyre in its original position...
If yes then $P(1\ \text{tyre}) = 13/24$
Or must I find the probability that exactly 1 tyre is in the original position ? If so, then what is the probability that 3 tyres are in their original position ? Because having 3 tyres in the original position would mean the 4th tyre must be in the original position ?
| $0$ tires in their correct positions (derangements): The number of ways $k$ of the $4$ tires are in their correct positions is $\binom{4}{k}$. The remaining $4 - k$ tires can be arranged in $(4 - k)!$ orders. Hence, by the Inclusion-Exclusion Principle, the number of arrangements in which no tires are in their correct positions is
$$4! - \binom{4}{1}3! + \binom{4}{2}2! - \binom{4}{3}1! + \binom{4}{4}0! = 9$$
exactly $1$ tire is in its correct position: By the generalized Inclusion-Exclusion Principle, the number of arrangements in which exactly one tire is in its correct position is
$$\binom{1}{1}\binom{4}{1}3! - \binom{2}{1}\binom{4}{2}2! + \binom{3}{1}\binom{4}{3}1! - \binom{4}{1}\binom{4}{4}0! = 8$$
where the factor $\binom{m}{1}$ counts the number of ways we can designate $1$ of the $m$ tires that are in their correct positions as being the one in its correct position.
exactly $2$ tires are in their correct positions: By the generalized Inclusion-Exclusion Principle, the number of arrangements in which exactly two tires are in their correct positions is
$$\binom{2}{2}\binom{4}{2}2! - \binom{3}{2}\binom{4}{3}1! + \binom{4}{2}\binom{4}{4}0! = 6$$
where the factor $\binom{m}{2}$ denotes the number of ways we can designate $2$ of the $m$ tires that are in their correct positions as being the ones that are in their correct positions.
exactly $3$ tires are in their correct positions: If three tires are in their correct positions, so is the fourth. Hence, there are no arrangements with exactly three tires in their correct positions.
exactly $4$ tires in their correct positions: There is only one arrangement in which all four tires are in their correct positions.
Hence, the expected number of tires in their correct positions is
$$\frac{9 \cdot 0 + 8 \cdot 1 + 6 \cdot 2 + 0 \cdot 3 + 1 \cdot 4}{4!} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to solve the recurrence $T(n)=T(n-1)+\frac{n}{n-1}$?
Solve the recurrence $T(n)=T(n-1)+\frac{n}{n-1}$.
I thought to try solving it by iteration method:
$$
T(n)=\frac{n}{n-1}+\frac{n-1}{n-2}+\frac{n-2}{n-3}+...+\frac{3}{2}+\frac{2}{1}=\sum_{i=0}^n \frac{i}{i-1}
$$
I know that $\sum^n_i i=0.5n(n+1)$ but I don't know how to develop the series from here.
| Well, at first let $T(1)=a\in\mathbb{R}$ be the first term of the sequence $T(n)$. Then, let us try to find some of the next terms of the sequence:
$$T(2)=T(1)+\frac{2}{2-1}=a+2$$
$$T(3)=T(2)+\frac{3}{3-1}=a+2+\frac{3}{2}=a+\frac{7}{2}(=a+3.5)$$
$$T(4)=T(3)+\frac{4}{4-1}=a+\frac{7}{2}+\frac{4}{3}=a+\frac{29}{6}(=a+4.8\overline{3})$$
And, in general:
$$T(n)=a+\sum_{k=2}^n\frac{k}{k-1}$$
Now, we have:
$$\sum_{k=2}^n\frac{k}{k-1}=\sum_{k=2}^n\frac{k-1+1}{k-1}=\sum_{k=2}^n\left(1+\frac{1}{k-1}\right)=n-1+\sum_{k=2}^n\frac{1}{k-1}$$
So, we have to calculate the sum:
$$\sum_{k=2}^n\frac{1}{k-1}=\sum_{k=1}^{n-1}\frac{1}{k}=H_{n-1}$$
Where $H_n$ is the $n$-th harmonic number.
So, we finally have:
$$T(n)=a+n-1+H_{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$, $\frac{P(x)}{x^3+x}$ remainder Given the polynomial:
$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$
What is the remainder of $\frac{P(X)}{x^3+x}$?
I don't think the long division is efficient the way to go, and the remainder theorem doesn't seem to be applicable here as $x^3+x$ is not linear. Could I have some hints on how to approach this? Thank you.
| $P(x)=x^{100}+x^{50}-2x^4-x^3+x+1$
Let's divide $x^{100}+x^{50}$ by $x^3+x$
-> $Q=x^{97},r=-x^{98}+x^{50}$
Divide $-x^{98}+x^{50}$ by $x^3+x$
-> $Q=-x^{95}, r=x^{96}+x^{50}$
Divide $x^{96}+x^{50}$ by $x^3+x$
-> $Q=x^{93}, r=-x^{94}+x^{50}$
...
Can you see a pattern?
...
The pattern ends here:
Divide $x^{52}+x^{50}$ by $x^3+x$
-> $Q=x^{49}, r=0$
Now return to the original polynomial (without the first two term):
$-2x^4-x^3+x+1$
And do polynomial division as you do normally!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a+b)^2 = 1^2 = 1 \implies (a+b)-(a+b)^2 = 1-1 =0$ (1)
$(a-b)^2 \ge 0$
So by (1) we have:
$(a-b)^2 \ge (a+b)-(a+b)^2$
$(a^2-2ab+b^2) \ge (a+b) - (a^2+2ab+b^2)$
$(a^2-2ab+b^2) + (a^2+2ab+b^2) \ge (a+b) $
$ a^2+a^2+b^2+b^2+2ab-2ab \ge (a+b)$
$2(a^2+b^2) \ge (a+b)$
$2(a^2+b^2) \ge 1$
$(a^2+b^2) \ge \frac{1}{2} $
$\blacksquare$
| For an alternative proof, let $a=\frac{1}{2}+u\,$, then $a+b=1 \implies b = 1-a = \frac{1}{2} - u$. It follows that:
$$\require{cancel}
a^2+b^2 = \left(\frac{1}{2}+u\right)^2+\left(\frac{1}{2}-u\right)^2 = \frac{1}{4}+\bcancel{u}+u^2+\frac{1}{4}-\bcancel{u}+u^2 = \frac{1}{2}+2u^2 \ge \frac{1}{2}
$$
Note that the condition that $a,b$ be positive is not used, or required for the conclusion to hold true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Find all polynomials : $ P(x^2-x)=xP(x-1)$ Find all polynomials $P(x) \in\mathbb{R}[x]$ satisfying $$ P(x^2-x)=xP(x-1)$$
Please check my answer :
$P(0) =0$, so $0$ is root of $P(x)$, there exists $Q(x)$ such that $P(x)=xQ(x)$
then $(x^2-x)Q(x^2-x)=x(x-1)Q(x-1)$, so $Q(x^2-x)=Q(x-1)$ where $x \not= 0, 1$
Substitute $x=2$, we have $Q(2) = Q(1)$
Substitute $x=3$, we have $Q(6) = Q(2)$
Substitute $x=7$, we have $Q(42) = Q(6)$
Substitute $x=43$, we have $Q(43^2-43) = Q(42)$
Since there are infinitely many $x$ such that $Q(x) = Q(1)$ so $Q(x)$ is constant polynomial.
Therefore, $P(x)=cx$ where $c$ is constant.
| You can go also with canonical form of polynomial: $p(x) =a_nx^n+...$, where $a_n\ne 0$. So we have $$a_n(x^2-x)^n+...= xa_n(x-1)^n+...$$ so $a_nx^{2n}=a_nx^{n+1}$. Since $a_n\ne 0$ we have $n=1$ or $n=- \infty $. So $p(x)=ax+b$ for some $a,b$. Pluging in to starting equation we have:
$$a(x^2-x)+b= x(a(x-1)+b)\Longrightarrow ax^2-ax+b = ax^2+(-a+b)x$$
So $p(x)=ax$ is solution for any $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Solve $\frac{1}{x}<\frac{x}{2}<\frac{2}{x}$. First I consider the case $\frac{1}{x}<\frac{x}{2}$, manipulating to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x}<0\Leftrightarrow x < -\sqrt{2}.$ Did this by subtracting $\frac{x}{2}$ from both sides and writing everything with a common denominator.
Second case to consider is $\frac{x}{2}<\frac{2}{x}\Leftrightarrow\frac{(x-2)(x+2)}{2x}<0\Leftrightarrow x<-2.$
Since both of these inequalities have to be satisfied simultaneously, one can combine them to get $x<-2.$ Correct answer is $x\in(\sqrt{2},2)$
| First, notice that $1/x < 2/x$ implies $1/x$ is positive, hence $x$ is positive. Now you know you can multiply the inequalities by $x$ without reversing the inequality signs. So you get $2<x^2$ and $x^2<4$. And so on...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
find the probability of selecting exactly two women and at least two women when a six-person committee is selected from $7$ men and $4$ women? A committee of six members is formed from a group of $7$ men and $4$
women. What is the probability that the committee contains
a. exactly two women?
b. at least two women?
My attempt : given $P(A) = 7/11$ and $P(B) = 4/11$
option a) probability that the commitee contains
exactly two women
$$= \frac{P(AB)}{P(B)} = \frac{P(A)P(B)}{P(B)} = \frac{\frac{7}{11} \cdot \frac{4}{11}}{\frac{4}{11}} = \frac{7}{11}$$
option b)the probability that the committee contains at least two women =
$$\frac{P(AB)}{P(A \cup B)} = \frac{\frac{7}{11} \cdot \frac{4}{11}}{1- \frac{7}{11} \cdot \frac{4}{11}} = \frac{28}{93}$$
If my answer is correct or not, I would be more thankful to those rectifying my mistakes......
|
A committee of six members is formed from a group of $7$ men and $4$ women. What is the probability that the committee contains exactly two women?
Since there are a total of $7 + 4 = 11$ people, the number of ways we can select a committee of six people is
$$\binom{11}{6}$$
A committee of six that contains exactly two women must contain four of the seven men and two of the four women, so it can be selected in
$$\binom{7}{4}\binom{4}{2}$$
ways. Therefore, the probability that the committee contains exactly two women is
$$\frac{\dbinom{7}{4}\dbinom{4}{2}}{\dbinom{11}{6}}$$
Under the same conditions as above, what is the probability that the committee contains at least two women?
If the committee contains at least two women, it must contain two, three, or four women. Since there are six people on the committee, a committee that contains exactly $k$ women contains $6 - k$ of the seven men and $k$ of the four women. Thus, there are
$$\binom{7}{5}\binom{4}{2} + \binom{7}{4}\binom{4}{3} + \binom{7}{3}\binom{4}{4}$$
such committees, so the probability of selecting a committee with at least two women is
$$\frac{\dbinom{7}{5}\dbinom{4}{2} + \dbinom{7}{4}\dbinom{4}{3} + \dbinom{7}{3}\dbinom{4}{4}}{\dbinom{11}{6}}$$
Alternatively, the number of committees that contain fewer than two women is
$$\binom{7}{6}\binom{4}{0} + \binom{7}{5}\binom{4}{1}$$
so the probability that a committee contains fewer than two women is
$$\frac{\dbinom{7}{6}\dbinom{4}{0} + \dbinom{7}{5}\dbinom{4}{1}}{\dbinom{11}{6}}$$
Thus, the probability that the committee contains at least two women is
$$1 - \frac{\dbinom{7}{6}\dbinom{4}{0} + \dbinom{7}{5}\dbinom{4}{1}}{\dbinom{11}{6}}$$
An indication that you made an error is that the probability of selecting at least two women should be at least the probability that a committee contains exactly two women.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Repeated Linear Factors in denominator of fraction e.g. $\frac{2x^2 + 2x + 18}{x(x-3)^2}$ I have the following fraction below and I must find the partial fraction decomposition.
$$\frac{2x^2 + 2x + 18}{x(x-3)^2}$$
Now, I thought I could simplify this into the following...
$$\frac{2x^2 + 2x + 18}{x(x-3)^2} = \frac{A}{x} + \frac{B}{(x-3)^2}$$
However, my professor said that this is incorrect, and instead it should be written as:
$$\frac{2x^2 + 2x + 18}{x(x-3)^2} = \frac{A}{x} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$
I'm confused.
I can't seem to find an explanation for why my method is wrong and why my professor's solution is correct, using algebraic reasoning.
Is there a reason why this is the case?
| Consider -
$$\frac{2x^2 + 2x + 18}{x(x-3)^2} = \frac{A}{x} + \frac{Bx + C}{(x-3)^2} = \frac{A}{x} + \frac{Bx}{(x-3)^2} + \frac{C}{(x-3)^2} =
\frac{A}{x} + \frac{B'}{x-3} + \frac{C}{(x-3)^2}$$
Where B is a constant that can be modified.
Hence the professor's solution is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the minimum value of $a^2+b^2+c^2$
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.
I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, which is unknown. How can I solve this?
| You can solve this problem by your starting step.
Indeed, by C-S
$$3(a^2+b^2+c^2)\geq(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc).$$
Thus,
$$3(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+ac+bc)$$ or
$$a^2+b^2+c^2\geq ab+ac+bc$$ and since $ab+ac+bc\geq2,$ we obtain:
$$a^2+b^2+c^2\geq2.$$
The equality occurs for $(1,1,1)||(a,b,c)$ and $ab+ac+bc=2$,
which says that $2$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Triple Integral With Spherical Coordinates - Napkin Ring? I wish to calculate the integral bellow, where $T$ is the region bounded by $x^2 + y^2 = 1$ and $x^2 + y^2 + z^2 = 4.$ It looks to me that it represents a Napkin ring.
$$\iiint_T\bigl(x^2 + y^2\bigr)\,\text{d}V.$$
The answer is $\dfrac{\bigl(256 - 132\sqrt{3}\,\bigr)\pi}{15}.$
| Looks to me like the region is the interior, and not the napkin ring.
converting to cylindrical
$\displaystyle\int_0^{2\pi}\int_0^1 \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r^3 \ dz\ dr\ d\theta$
$\displaystyle\int_0^{2\pi}\int_0^1 2r^3\sqrt{4-r^2} \ dr\ d\theta\\
r^2=4-u^2 ,\ 2r\ dr = -2u\ du\\
\displaystyle\int_0^{2\pi}\int_\sqrt{3}^{2} 2(4-u^2)u^2 \ du\ d\theta\\
(4\pi)(\frac 43 u^3 - \frac 15 u^5 )|_\sqrt{3}^2\\
(4\pi)(\frac {64}{15} - \frac {33\sqrt {3}}{15})\\
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Induction: Bounds on sum of inverses of first $n$ square roots I am trying to show by induction that $2(\sqrt{n}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$.
For the upper bound, I have that if $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$, then $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}<2\sqrt{n}+\dfrac{1}{\sqrt{n+1}}=$
$2\sqrt{n}+\dfrac{2}{2\sqrt{n+1}}<2\sqrt{n}+\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\sqrt{n}+2(\sqrt{n+1}-\sqrt{n})=2\sqrt{n+1}$.
However, despite hours of trying, I cannot figure out how to show that $2(\sqrt{n+1}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}$.
| Observe that: $\dfrac{1}{2\sqrt{k}}> \dfrac{1}{\sqrt{k+1}+\sqrt{k}}= \sqrt{k+1}-\sqrt{k}$. Taking summation for $k$ from $1$ to $n$ and get:
$\dfrac{S_n}{2} > \displaystyle \sum_{k=1}^n\left(\sqrt{k+1} - \sqrt{k}\right)= \sqrt{n+1}-1\implies S_n > 2\left(\sqrt{n+1}-1\right)> 2\left(\sqrt{n}-1\right)$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $x(2+\sqrt3)=y(2-\sqrt3)$ then find the value of $\frac{1}{x+1}+\frac{1}{y+1}$ My attempt:
Let $\frac{x}{2-\sqrt3}=\frac{y}{2+\sqrt3}=k$,
so $x=(2-\sqrt3)k$ and $y=(2+\sqrt3)k$
Substituting these values we get:
$\dfrac{1}{x+1}+\dfrac{1}{y+1}$
$=\dfrac{x+y+2}{xy+x+y+1}$
$=\dfrac{2(k+1)}{(k+1)^2}$
$=\dfrac{2}{k+1}$
How do I find the value of $k$?
Options for the answers are $1, \sqrt3, 2\sqrt3, 2$
If I perform backcalculation, that gives me options for k as $1, \dfrac{2-\sqrt3}{\sqrt3}$ and $\dfrac{2-2\sqrt3}{\sqrt3}$.
If we can claim that k is an integer, then we can say $k=1$ and we are done. But can we say k is an integer, and not any rational?
| With a missing data, it could actually attain any value
From there you've stopped $\frac{1}{x+1}+\frac{1}{y+1} = \frac{4k+2}{k^2+4k+1}$
Now, suppose $\frac{4k+2}{k^2+4k+1} = t$ for some $t \in \mathbb{R}$, let show there is a solution for this specific t. From here $tk^2 + 4(t-1)k + (t-2) = 0$ we have $k_{1,2} = \frac{-4(t-1) \pm \sqrt{16(t-1)^2-4t(t-2)}}{2t}$.
The expression under square root is $12t^2-24t+16$ which is $12(t-1)^2 + 4$, i.e. always positive, hence there are even two values of $k$, which solves for any given $t$. (except for $t = 0$, then you get $k=-\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can one find $\lim_{x \to \pm\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$ without graphing? I'm trying to find both of these:
$$\lim_{x \to +\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$$
$$\lim_{x \to -\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$$
I know that they end up being $2$ and $-2$ by graphing, but what process can I use to find them without graphing (in the simplest way possible)?
| $$\frac{2x+1}{\sqrt{x^2+x+1}}=\frac{x\left(2+\dfrac{1}{x}\right)}{\sqrt{x^2\left(1+\dfrac{1}{x}+\dfrac{1}{x^2}\right)}}=\frac{x\left(2+\dfrac{1}{x}\right)}{|x|\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}}=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Finding $\int\frac{x}{\sqrt{3-2x-x^2}} dx$. I was looking for the integral of
$$\int\frac{x}{\sqrt{3-2x-x^2}} dx$$
My work:
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(3)+(-2x-x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(2x+x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(1+2x+x^2) +1}} dx $$
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(4)-(x+1)^2}} dx $$
Then I often remember this integral $\frac{u}{\sqrt{a^2 - x^2}} du$. So I modified the above integral to
look like the integral $\frac{u}{\sqrt{a^2 - x^2}}$.
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x+1-1}{\sqrt{(4)-(x+1)^2}} dx $$
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x+1}{\sqrt{(4)-(x+1)^2}} dx + \int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$$
I recognized the the last integral $\int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$ has the form $\int \frac{1}{\sqrt{a^2-u^2}} du$, where $a =2$ and $u = x+1$.
It's corresponding integral would be $\arcsin \left( \frac{u}{a}\right) + c$.
Evaluating $\int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$, it would be $-\arcsin \left( \frac{x+1}{2}\right)$
Here's the problem: I couldn't find the integral of $\int \frac{x+1}{\sqrt{(4)-(x+1)^2}} dx, $ because my Table of Integral doesn't show what is the
integral of $\frac{u}{\sqrt{a^2 - x^2}} du$.
How to evaluate the integral of $\frac{x}{\sqrt{3-2x-x^2}} dx$ properly?
| $$\frac{x}{\sqrt{3-2x-x^2}} = -\frac{1}{2}\left(\frac{-2x}{\sqrt{3-2x-x^2}}\right)\\= -\frac{1}{2}\left(\frac{-2-2x+2}{\sqrt{3-2x-x^2}}\right)\\ =-\frac{1}{2}\left(\frac{-2-2x}{\sqrt{3-2x-x^2}}\right) - \frac{1}{\sqrt{3-2x-x^2}} \\= -\frac{1}{2}\left(\frac{-2-2x}{\sqrt{3-2x-x^2}}\right) - \frac{1}{\sqrt{4-(x+1)^2}}.$$
Taking the integral, the first integrand will be an integral in the form
$$\frac{u'}{\sqrt{u}}$$
and the second is a standard integral that evaluates to a sine inverse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
What kind of series is the following what kind of series is the following and can I get a generalized formula for it?
$T_1 = (1-\frac{1}{2N^3})$
$T_2 = (1- T_1.\frac{1}{2N^3}) $
$T_3 = (1 - T_2.\frac{1}{2N^3})$
$T_4 = (1 - T_3.\frac{1}{2N^3})$
..
.
$T_x = (1 - T_{x-1}.\frac{1}{2N^3})$
$ P_{total} = T_1 + T_2 + T_3 + ... T_x $
| You have a recurrent formulation of a series. Let $a=\frac1{2N^3}$. You can see that the explicit formula
$T_n = \frac{1 + (-a)^n a}{1+a}$ solves your given recurrence $T_{n+1} = 1 - a T_{n}$, with $T_0=1$, because
$$ 1 - a T_n = 1 - \frac{1 + a - a - a\cdot(-a)^n\cdot a}{1+a} = \frac{1 + (-a)^{n+1} a}{1+a} = T_{n+1}$$
and of course $T_0 = \frac{1 + (-a)^0 a}{1+a} = 1$.
For your $P_\mathrm{total}$ you have
$$P_\mathrm{total} = \sum_{n=1}^x T_n = \frac{\sum_{n=1}^x1 + a\sum_{n=1}^x(-a)^n}{1+a} = \frac{x + a\left(\frac{1-(-a)^{x+1}}{1-(-a)} - 1\right)}{1+a} = \frac{x + a \biggl(\bigl((-a)^x - 1\bigr) a + x\biggr)}{(1 + a)^2}.$$
by using the formula for a geometric sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
let : $\forall n \in \mathbb{N}$ : $x_n=\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}$ then : $\displaystyle{\sum_{i=1}^{20}}\dfrac{1}{x_i}=?$ let : $\forall n \in \mathbb{N}$ : $x_n=\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}$
then :
$$\sum_{i=1}^{20}\dfrac{1}{x_i}=?$$
my try :
$$\dfrac{1}{\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{13}}+...$$
But boring .
| Let me try.
$$\frac{1}{x_n} = \frac{1}{\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}}$$
$$= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{\frac{4}{n}}$$
$$= \frac{1}{4}\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2}\right)$$
Then, $$\sum_{i=1}^{20}\frac{1}{x_i} = \frac{1}{4}\left(\sqrt{20^2+21^2}-\sqrt{1^2+0^2}\right) = 7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I evalute this product? How can I evalute this product??
$$\prod_{i=1}^{\infty} {(n^{-i})}^{n^{-i}}$$
Unfortunately, I have no idea.
| Taking log
$$\ln\Big( \prod_{i=1}^{\infty} (n^{-i})^{n^{-i}}\Big)$$$$$$
$$\sum_{i=1}^{\infty} \frac{i}{n^i} \ln n$$$$$$
$$\ln n \sum_{i=1}^{\infty} \frac{i}{n^i} $$$$$$
Let$$$$$$S= \sum_{i=1}^{k} \frac{i}{n^i}$$ $$$$
$$S=\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}+\cdots+\frac{k}{n^k}$$$$$$
Multiplying by $\frac{1}{n}$$$$$
$$\frac{S}{n}=\frac{1}{n^2}+\frac{2}{n^3}+\frac{3}{n^4}+\cdots+\frac{k}{n^{k+1}}$$$$$$
Now $S-\frac{S}{n}$$$$$
$$\frac{n-1}{n} S =\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\cdots+\frac{1}{n^k}-\frac{k}{n^{k+1}}$$$$$$
Using GP formula,
$$\frac{n-1}{n} S =\frac{1}{n}\frac{\Big(\frac{1}{n}\Big)^k-1}{\frac{1}{n}-1} -\frac{k}{n^{k+1}}$$$$$$
$$S=\frac{n}{n-1} \Biggr(\frac{1}{n}\frac{\Big(\frac{1}{n}\Big)^k-1}{\frac{1}{n}-1} -\frac{k}{n^{k+1}}\Biggr)$$$$$$
For $k=\infty$ if $n\gt 1$$$$$
$$S=\frac{n}{n-1}\frac{1}{n} \Big(\frac{n}{n-1}\Big)$$$$$$
$$S=n\Big(\frac{1}{n-1}\Big)^2$$$$$$
Then substitute it back
$$n \ln n \Big(\frac{1}{n-1}\Big)^2$$Taking antilog,$$n^{n\Big(\frac{1}{n-1}\Big)^2}$$
This is only for $n\gt 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
3, 5, 7 are the only three consecutive odd natural numbers that are prime. Can someone please help me proof this theorem. So far I assumed towards contradiction that there exist three consecutive odd natural numbers that are primes P, P+2, and P+4.
Theorem: 3, 5, 7 are the only three consecutive odd natural numbers that are prime.
| Let's get three numbers: $n, n+2, n+4$.
We have 2 cases:
*
*$n\not\equiv0\mod 3$:
Thus $n=3k+1$ or $n=3k+2$, so $n+2=3(k+1)$ or $n+4 = 3(k+2)$
*$n=3k$
In both cases at least one of our numbers is divisible by $3$.
The only prime that is divisible by $3$ is $3$.
The only sequention of prime numbers $n, n+2, n+4$ that contains $3$ is $3,5,7$.
So the only sequention of prime numbers $n, n+2, n+4$ is $3,5,7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Let $A$ be a $4\times2$ matrix and $B$ be a $2\times4$ matrix. Find $BA$ when $AB$ is given.
Let $A$ be a $4\times2$ matrix and $B$ be a $2\times4$ matrix so that
$$AB=
\begin{pmatrix}
1 & 0 & -1 & 0 \\
0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & -1 & 0 & 1
\end{pmatrix}.
$$
Find $BA$.
I have the answer, which should be
$BA=
\begin{pmatrix}
2 & 0 \\
0 & 2
\end{pmatrix}
$, but how do I show that this is the only possible solution or is it sufficient (according to some property which I'm not aware of) to have one match for $A$ and $B$ and therefore no other outcome for $BA$ is possible.
| Write $A=\begin{pmatrix}A_1\\A_2 \end{pmatrix}$ and $B=\begin{pmatrix}B_1 &B_2 \end{pmatrix}$ where $A_1,A_2,B_1,B_2$ are $2\times2$ matrices.
Note that $AB=\begin{pmatrix}A_1B_1 & A_1B_2\\A_2B_1 & A_2B_2 \end{pmatrix}$, hence $A_1B_1= A_2B_2=I_2$.
Since a square matrix commutes with its inverse, we have as well $B_1A_1=B_2A_2=I_2$.
Finally, note that $BA=\begin{pmatrix}B_1A_1+B_2A_2 \end{pmatrix}=\begin{pmatrix}2I_2 \end{pmatrix}=\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$ Question:
Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
My attempt:
Base case is trivial.
Suppose $ \ n \ge 2$ and $ \ 2^{n} + 3^{n} < 4^{n}$
Then,
$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n} $, by I.H.
I am stuck here. how do I show that this expression is $ < 4^{n+1}$?
| Hint: for $\,0 \lt a \lt 1\,$ the sequence $\,a^n\,$ is decreasing with $\,n\,$ since $\,a^{n+1} = a \cdot a^n \lt 1 \cdot a^n\,$, so:
$$\left(\frac{2}{4}\right)^n+\left(\frac{3}{4}\right)^n \le \left(\frac{2}{4}\right)^2+\left(\frac{3}{4}\right)^2 = \frac{13}{16} \lt 1 \quad\quad\text{for}\;\; \forall n \ge 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to integrate following integral How to integrate following integral i am having physics background and i am trying calculating the induced charge in a specific problem.
The Integral need to be evaluted is
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{(x^2+y^2+b^2)^{3/2}}\,dxdy$$
| Let $b\not=0$, then, by letting $x=r\cos(t)$ and $y=r\sin(t)$ (polar coordinates) the integral becomes
\begin{align*}
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{(x^2+y^2+b^2)^{3/2}}\,dxdy&=
\int_{r=0}^{\infty}\int_{t=0}^{2\pi}\frac{1}{(r^2+b^2)^{3/2}}\,(rdrdt)\\
&=\left[-\frac{2\pi}{(r^2+b^2)^{1/2}}\right]_0^{\infty}=\frac{2\pi}{|b|}.
\end{align*}
P.S. The integral can be done also by using cartesian coordinates
\begin{align*}4\int_{0}^{\infty}\left(\int_{0}^{\infty}\frac{dx}{(x^2+y^2+b^2)^{3/2}}\right) dy
&=4\int_{0}^{\infty}\left[\frac{x}{(y^2+b^2)(x^2+y^2+b^2)^{1/2}}\right]_{x=0}^{\infty}dy\\
&=4\int_{0}^{\infty}\frac{dy}{y^2+b^2}
=\frac{4}{|b|}\left[\arctan\left(\frac{y}{|b|}\right)\right]_{y=0}^{\infty}\\&=\frac{2\pi}{|b|}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
In the list of questions proposed in the "Meeting for Training for the Brazilian Olympiad", 2013. No answer provided. Could solve some problems in that list but got stuck in this one. My developments are going into very complicated expressions, and are most likely wrong. Hints or solutions are welcomed. Sorry if this is a duplicate.
| We can use the binomial theorem for the coefficients of $(a+b)^7$.
The coefficients are 1,7,21,35,35,21,7,1, but we minus the $a^7$ and $b^7$.
So we have $7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6$, and then, let $a=0$, and since the result is 0 when we insert it, we have $a$ as a root. $b$ and $a+b$ works too. We now have $7ab(a+b)(a^4+2a^3b+3a^2b^2+2ab^3+b^4)$, which equals $7ab(a+b)(a^2+ab+b^2)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Two equivalent simple continued fractions $A\overset{\color{red}?}=B$ Fix a sequence $a_n$ of complex numbers for natural number $n$,such that $|a_n|\gt1$
and then define the following two simple continued fractions
$A= \cfrac{1}{a_{0}-\cfrac{1}{a_{1}+\cfrac{1}{a_{2}-\cfrac{1}{a_{3}+\cfrac{1}{a_{4}-\cfrac{1}{a_{5}+\cfrac{1}{a_{6}-\cfrac{1}{a_{7}+\dots}}}}}}}}\tag1$
which converges by Śleszyński-Pringsheim Theorem
and
$B= \cfrac{1}{a_{0}-1+\cfrac{1}{1+\cfrac{1}{a_{1}-1+\cfrac{1}{a_{2}-1+\cfrac{1}{1+\cfrac{1}{a_{3}-1+\cfrac{1}{a_{4}-1+\cfrac{1}{1+\cfrac{1}{a_5-1+\cfrac{1}{a_{6}-1+\cfrac{1}{1+\dots}}}}}}}}}}}\tag2$
How do we prove that $A=B$ ?
Special case
(i) Define $R(q)=q^{1/5}\frac{(q;q^5)_{\infty}(q^4;q^5)_{\infty}}{(q^2;q^5)_{\infty}(q^3;q^5)_{\infty}}$ the Rogers-Ramanujan continued fraction where $|q|\lt1$ ,then
$R(q)=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\cfrac{q^4}{1+\cfrac{q^{5}}{1+\dots}}}}}}$
and
$R(q)=\cfrac{q^{1/5}}{1+\cfrac{q}{1-q+\cfrac{q}{1-\cfrac{q}{1+q+\cfrac{q^3}{1+q^2-\cfrac{q^2}{1+\cfrac{q^2}{1-q^2+\cfrac{q^5}{1-q^3+\cfrac{q^3}{1-\cfrac{q^3}{1+q^3+\cfrac{q^7}{1+q^4-\cfrac{q^4}{1+\dots}}}}}}}}}}}}$
(ii) Let $a_n=2$,then
$A= \cfrac{1}{2-\cfrac{1}{2+\cfrac{1}{2-\cfrac{1}{2+\cfrac{1}{2-\cfrac{1}{2+\dots}}}}}}=\frac{1-\sqrt{5}}{2}$
and
$B=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\dots}}}}}}=\frac{1-\sqrt{5}}{2}$
Remark: this general identity leads to another continued fraction for the Generalized Rogers-Ramanujan continued fraction
Edited 17 Oct 2017:I've just noticed that this oeis article also discusses the same continued fraction for the special case of the Generalized Rogers-Ramanujan continued fraction $R(-1,q)$
| $$ 1-\frac1{1+\frac 1 {x-1}} = 1-\frac {x-1}{x-1+1} = 1-\frac{x-1}x = \frac {x-x+1}x = \frac 1x$$
And in particular,
$$\frac 1 {a-1 + \frac 1 {1 + \frac 1 {b-1 + x}}} =
\frac 1 {a-\frac 1{b+x}} $$
Applying this should prove that in the sequences defining your continued fractions, there is a common subsequence, so if both have a limit, then their limit are equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determine the derivative of $\arctan$ function Find $f'(x)$ for the function:
$f(x)= \arctan(\frac{a+x}{1-ax}))$ , $a\in R$
So this is what I've done:
$f(x) = \arctan x$
$f'(x) = \frac{1}{1+x^2}$
$x= \frac{a+x}{1-ax}$
$f'(x) = \frac{1}{1+\frac{(a+x)^2}{(1-ax)^2}}$
$f'(x) = \frac{(1-ax)^2}{(1-ax)^2+(a+x)^2}$
Is this correct?
| Hint:
$$\dfrac{d}{dx}f=\dfrac{d}{d \frac{a+x}{1-ax}}\left[\arctan\frac{a+x}{1-ax}\right]\dfrac{d}{dx}\left[\frac{a+x}{1-ax}\right]=\dfrac{d}{d u}\left[\arctan u\right]\bigg|_{u=\frac{a+x}{1-ax}}\dfrac{d}{dx}\left[\frac{a+x}{1-ax}\right]$$
$$=\frac{1}{1+\left(\frac{a+x}{1-ax} \right)^2}\dfrac{d}{dx}\left[\frac{a+x}{1-ax}\right]$$
EDIT1: Explaining the derivative
$$=\frac{1}{1+\left(\frac{a+x}{1-ax} \right)^2}\frac{1\cdot(1-ax)-(a+x)(-a)}{(1-ax)^2}=\frac{1+a^2}{(1-ax)^2+(a+x)^2}$$
$$=\frac{1+a^2}{(1+a^2)(1+x^2)}=\frac{1}{1+x^2}$$
EDIT2: Explaining $(1-ax)^2+(a+x)^2=(1+x^2)(1+a^2)$
$$(1-ax)^2+(a+x)^2=1-2ax+a^2x^2+a^2+2ax+x^2$$
$$=1+a^2x^2+a^2+x^2=(1+x^2)+a^2(1+x^2)=(1+x^2)(1+a^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving an equation $z + |z|^2 = 10 - 2i$ How do you solve the equation $z+ |z|^2 = 10 - 2i$ and express the solution in the form $a+bi$?
I feel like this should be easy but I'm having a blank.
| Write $z=a+ib$ and extract two equations for the real part and imaginary part. Solve for a and b.
Edit:
$|z|^2+z=10-2i$, we get:
$a^2+b^2+a+ib=10-2i$, hence
$a^2+b^2+a+ib-10+2i=0$
$a^2+b^2+a-10+i(b+2)=0$, therefore the two equations
I. $a^2+b^2+a-10=0$ which is the real part and
II. $b+2=0$ which is the imaginary part.
From II. we get $b=-2$ immediately and can use it in I.
This gives us:
$a^2+a+4-10=0$
$a^2+a-6=0$ use a formula for quadratic equations, or see the solutions immediately, since
$a^2+a-6=(a+3)(a-2)=0$ therefore we get two solutions for $a$ namley $a_1=-3$ and $a_2=2$
And we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
calculate sum of limit using riemann sum calculate this sum of limit using riemann sum, $\lim_\limits {n\to \infty} \frac{1}{n}{\left({\dfrac{(2n)!}{(n!)}}\right)}^{1/n} $
\begin{align*}
&=\ln \frac{1}{n}+\ln\left ( \frac{1}{n} + \frac{1}{n}{\log 2n+\log (2n-1)+....+\log 1-(\log n+\log(n-1)+...+\log 1)} \right)\\
&=**\log \frac{1}{n}+ \frac{1}{n}\sum_{k=1}^n \ln(n+k)\\&= \frac{1}{n}\left[\sum_{k=1}^n\ln \frac{1}{n}+log(n+k)=\frac{1}{n}\sum_{k=1}^n\ln\left( 1+\frac{k}{n} \right)\right]**\\
.
\end{align*}
$\frac{1}{n}*{{\log 2n+\log (2n-1)+....+\log 1-(\log n+ \log(n-1)+....+\log 1)}}$
i was confused how can i get the starred part, how to manipulate it to $\sum \ln(n+k)$? is this using property of power series? can someone explain this so clearly?
| Let's compute the log and do some algebra:
$$
\log \frac{1}{n}{\left({\dfrac{(2n)!}{(n!)}}\right)}^{1/n}
=
\log \left[ 2 \times \left( \frac{n!}{n!} \times \frac{(n+1)(n+2)\cdots(2n-1)(2n)}{2^n \times n^{n}} \right)^{1/n} \right]
,
$$
which is
$$
\log 2 + \frac{1}{n} \log \left( \frac{n+1}{2n} \times \frac{n+2}{2n} \times \cdots \times \frac{2n}{2n} \right)
,
$$
and rewrinting as a sum
$$
\log 2 + 2 \times \left[ \frac{1}{2n} \left( \log \frac{n+1}{2n} + \log\frac{n+2}{2n} + \cdots + \log\frac{2n}{2n} \right) \right]
.
$$
Inside the bracket we have a Riemann sum for
$$
\int_{1/2}^1 \log x\, \mathrm{d} x = \frac{-1+\log 2}2
.
$$
Therefore the log of the limit is
$$
\log 2 + \log 2 - 1
$$
so the limit is
$$
4/e
.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$f(x)=?$ if we have $f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$ $f(x)=?$
If we have $$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$$ to fractions are very similar. I don't have an idea to find $f(x)$. Can someone show me a clue ?
| HInt: you can take $$x+\frac 1x =u $$ and turn all the expression into $u$ ,or do like below
$$\quad{f(\frac{x}{x^2+x+1})=\frac{x}{x^2-x+1}\\
\frac{x}{x^2+x+1}=u\\\frac{x^2+x+1}{x}=\frac 1u\\x+\frac{1}x +1=\frac 1u \\\to x+\frac{1}x =\frac 1u -1\\
\frac{x}{x^2-x+1}=\dfrac{1}{\dfrac{x^2-x+1}{x}}=\\\dfrac{1}{x+\dfrac{1}{x}-1}=\dfrac{1}{(\frac 1u -1)-1}}$$can you go on ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
If six different balls are placed in three different boxes, what is the probability that exactly two balls are placed in the first box? If six different balls are placed in three different boxes, what is the probability that exactly two balls are placed in the first box?
My attempt: $k_1+k_2+k_3=6$, $k_i$ - quantity of balls in $i$ box. $k_1=2$ hence there are 4 sets for $k_2, k_3:$ $\{0,4\},\{1,3\},\{3,1\},\{4,0\}$.
Adding set $\{2,2\}$ we get all combinations when there are 2 balls in the first box $\Rightarrow$ possibility $= \frac{4}{5}$?
| There are $3^6$ ways to distribute six balls to three boxes since there are three choices for each of the six balls.
There are $\binom{6}{2}$ ways to select exactly two of the six balls to be placed in the first box and $2$ ways to select the box in which each of the four remaining balls is placed. Hence, there are
$$\binom{6}{2}2^4$$
favorable cases.
Therefore, the probability that exactly two balls are placed in the first box when six balls are distributed to three different boxes is
$$\frac{\binom{6}{2}2^4}{3^6}$$
Note: You attempted to count cases by considering how many balls are placed in each box. However, the
$$\binom{6 + 3 - 1}{3 - 1} = \binom{8}{2}$$
solutions to the equation
$$k_1 + k_2 + k_3 = 6$$
in the nonnegative integers are not equally likely to occur. There is only one way to place all six balls in the first box. However, there are
$$\binom{6}{2}\binom{4}{2}\binom{2}{2} = 90$$
ways to place two balls in each box.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $\ln(1+2x^2 )/\ln(x^4+3)$ I've tried finding the limit
$$\lim_{x \to \infty} \frac{\ln(1+2x^2)}{\ln(x^4+3)}$$
using L'Hôpital's rule and got $0$. My CAS says that the limit should be $\frac 12$.
Can you please give me a hint on what to do?
| $$\frac{\text{d}}{\text{d}x}[\ln(1+2x^2)]=\frac{\text{d}}{\text{d}x}[1+2x^2]\cdot\frac{1}{1+2x^2}=\frac{4x}{1+2x^2}$$
$$\frac{\text{d}}{\text{d}x}[\ln(x^4+3)]=\frac{\text{d}}{\text{d}x}[x^4+3]\cdot\frac{1}{x^4+3}=\frac{4x^3}{x^4+3}$$
Then $$\lim\limits_{x\rightarrow\infty}\frac{\ln(1+2x^2)}{\ln(x^4+3)}=\lim\limits_{x\rightarrow\infty}\frac{\frac{4x}{1+2x^2}}{\frac{4x^3}{x^4+3}}=\lim\limits_{x\rightarrow\infty}\frac{4x}{1+2x^2}\frac{x^4+3}{4x^3}=\lim\limits_{x\rightarrow\infty}\frac{4x}{4x^3}\frac{x^4+3}{1+2x^2}=\lim\limits_{x\rightarrow\infty}\frac{1}{x^2}\frac{x^4+3}{1+2x^2}=\lim\limits_{x\rightarrow\infty}\frac{x^4+3}{2x^4+x^2}=\lim\limits_{x\rightarrow\infty}\frac{1+\frac{3}{x^4}}{2+\frac{1}{x^2}}=\frac{\lim\limits_{x\rightarrow\infty}(1+\frac{3}{x^4})}{\lim\limits_{x\rightarrow\infty}(2+\frac{1}{x^2})}=\frac{1+\lim\limits_{x\rightarrow\infty}\frac{3}{x^4}}{2+\lim\limits_{x\rightarrow\infty}\frac{1}{x^2}}=\frac{1}{2}$$
(There are better ways to tackle this problem as shown by the other answers, but this is how you correctly do it with L'Hopital)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $\alpha,a,b$ are integers and $b\neq-1$, then prove that, if $\alpha$ satisfies the equation $x^2+ax+b+1=0$, $a^2+b^2$ must be composite.
Let $\alpha,a,b$ be integers such that $b\neq-1$. Assume that $\alpha$ satisfies the equation $x^2+ax+b+1=0$. Prove that the integer $a^2+b^2$ must be composite.
$\alpha=\frac{-a\pm\sqrt{a^2-4(b+1)}}{2}$is an integer. But manipulating this expression is leading me to nowhere. Please help.
| Hint: $a^2-4(b+1)=a^2+b^2-(b^2+4b+4)=c^2$ for some integer $c$.
Do you know any facts about expressing primes as the sum of two squares?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
limit of trigonometric functions without L'Hôpital's rule $\lim_{x\to 0} [\csc^22x-\frac 1 {4x^2}]\ $
and
$\lim_{x\to y} \frac {\tan x- \tan y} {1-\frac{x}{y}(\tan x\tan y+1)}\ $
We have not learn L'Hôpital's rule in my class, we do the $\lim_{x\to 0}\frac{\sin x}{x}=1\ $
I have not came up with anything yet
| Let's prove some double-sided estimate for $1/\sin^2 x-1/x^2$ giving the first limit, using only $\lim_{h\to0}\sin h/h=1$ and $\lim_{h\to0}\cos h=1$:
$$\frac13\le\frac1{\sin^2 x}-\frac1{x^2}\le\frac1{3\cos^2 x/2}\tag1$$ for all $x\in(0,\pi)$.
Proof: We have $$\sin x=2\sin x/2\,\cos x/2,$$ implying
$$\frac1{\sin^2 x}=\frac14\left(\frac1{\sin^2 x/2}+\frac1{\cos^2 x/2}\right)$$
Replacing $x$ by $x/2^k$ and multiplying by $4^{-k}$ gives
$$\frac{4^{-k}}{\sin^2 x/2^k}-\frac{4^{-k-1}}{\sin^2 x/2^{k+1}}=\frac{4^{-k-1}}{\cos^2 x/2^{k+1}}$$ and thus (by monotony of $\cos x$)
$$4^{-k-1}\le\frac{4^{-k}}{\sin^2 x/2^k}-\frac{4^{-k-1}}{\sin^2 x/2^{k+1}}\le\frac{4^{-k-1}}{\cos^2 x/2}$$
Summing from $k=0$ to $k=n-1$ gives
$$\frac13(1-4^{-n})\le\frac1{\sin^2 x}-\frac{4^{-n}}{\sin^2 x/2^n}\le\frac{\frac13(1-4^{-n})}{\cos^2 x/2}$$
Now since $$\frac{4^{-n}}{\sin^2 x/2^n}=\frac1{x^2}\cdot\left(\frac{\sin x/2^n}{x/2^n}\right)^{-2},$$ letting $n\to\infty$ proves our claim (1). qed
It's clear that (1) together with the squeeze lemma means
$$\lim_{x\to0}\left(\frac1{\sin^2 x}-\frac1{x^2}\right)=\frac13,$$ and that won't change if we replace $x$ by $2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Characteristic method for solving a linear PDE $xyz_x-x^2z_y=-yz$.
Find the solution of the linear PDE
$$xyz_x-x^2z_y=-yz$$
Chacteristic equations:
$$\frac{dx}{xy}=\frac{dy}{-x^2}=\frac{dz}{-y}$$
from here, I found the characteristic $x^2+y^2=c_1$ and $lnx+z=c_2$.
since $c_2=f(c_1)$, we have $lnx+z=f(x^2+y^2)$ and solution is $z=f(x^2+y^2)-lnx$.
It is right?
if it is wrong, where am I wrong?
| $$\frac{dx}{xy}=\frac{dy}{-x^2}=\frac{dz}{-yz}$$
then with $\dfrac{dx}{xy}=\dfrac{dy}{-x^2}$ we find $x^2+y^2=C_1$ and with $\dfrac{dx}{xy}=\dfrac{dz}{-yz}$ we have $xz=C_2$ thus
$$z=\frac{f(x^2+y^2)}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $x_{n} - 3x_{n-1} = -8$ with $n\geq 1$ and $x_0 = 2$ I tried two methods which gave different answers:
Method 1:
$$x_{n} - 3x_{n-1} = -8 \\ x_n = 3(3x_{n-2} - 8) - 8 \\ = 3^2 x_{n-2} -8 ( 1+3) \\ = 3^3 x_{n-3} - 8(1+3+3^2) \\ = 3^n x_{0} - 8(1+3+3^2 + \ldots + 3^{n-1}) \\ = 2\times 3^n - 8\left(\frac{3^n - 1}{3-1}\right)\\ = 2\times 3^n - 4(3^n - 1) \\ = -2\times 3^n +4.$$
Method 2:
Solving the homogenous equation using $x_n = r^n$,
$$r^n - 3r^{n-1} = 0 \\ \implies r = 3$$
So the homogenous solution is $h_n = a\times 3^n$ for some $a\in\mathbb{R}$.
With the initial condition, $h_n = 2\times 3^n$.
Also, by guessing the particular solution with $x_n = C$,
$$C - 3C = -8 \implies C = 4$$.
So, the final solution is
$$x_n = 2\times 3^n + 4.$$
I'm unsure why the two methods differ. It looks obvious that the first one gives the correct result.
| To answer the question of why the method 2 answer is incorrect, the problem is that you try to figure out the coefficient on the homogenous portion of the solution before figuring out a particular solution. This gives an incorrect value because you haven't taken the inhomogeneous portion into account. Instead, once you've found $C=4$ in the particular solution, you plug in to $x_n=a\times 3^n+4$ to match $x_0=2$: $a\times 3^0+4=2$; $a\times 1=2-4=-2$; $a=-2$, so $x_n=(-2)\times 3^n+4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
modify sequence so it converges to pi Modify sequence by Archimedes
$s_{n+1}=\frac{2^{n+1}}{ \sqrt{2} } \sqrt{ 1 -\sqrt{1-(s_n/2^{n})^2}}; s_2=2\sqrt{2} $
so that it approaches $\pi$
the issue is that it is going to $0$
$$\lim_{n\to \infty} \sqrt{1-(s_n/2^n)^2}=1 $$
making $s_{n+1}$ go to $0$
Now
$$ \begin{aligned}
&\sqrt{1-\sqrt{1-(s_n/2^n)^2}} * \sqrt{1-\sqrt{1-(s_n/2^n)^2}}
\\&=\sqrt{(1-\sqrt{1-(s_n/2^n)^2}) * (1-\sqrt{1-(s_n/2^n)^2)}}
\\&=\sqrt{1 +\sqrt{1-(s_n/2^n)^2} - \sqrt{1-(s_n/2^n)^2}+1 - (\frac{s_n}{2^n})^2}
\\&=\sqrt{2-(\frac{s_n}{2^n})^2}
\end{aligned}$$
multiplying it $s_{n+1}$ by $\frac{ \sqrt{1+(s_n/2^n)^2} }{\sqrt{1+(s_n/2^n)^2}} $
$$s_{n+1}=\frac{2^{n+1}}{ \sqrt{2} } \sqrt{ 1 -\sqrt{1-(s_n/2^{n})^2}}* \frac{ \sqrt{1+(s_n/2^n)^2} }{\sqrt{1+(s_n/2^n)^2}}$$
is
$$S_{n+1} = \frac{\frac{2^{n+1}}{\sqrt{2}} * \sqrt{2-(\frac{s_n}{2^n})^2}}{\sqrt{1+\sqrt{1-(\frac{s_n}{2^n})^2}}}$$
need to see if that goes to $\pi$ better. need to throw it to a for loop and see that it dosent go to zero
Appreciate feedback
| In the end what you should get is $$s_n=2^n\sin\frac\pi{2^n}$$ so that indeed
$$
\lim s_n=\lim_{x\to0}\frac{\sin(\pi x)}{x}=\frac{d\sin(\pi x)}{dx}\Big|_{x=0}=\pi\cos(0)=\pi.
$$
As said in the comments, the formula arises by inscribing first a square into the circle and then refine it by halving the angles in the regular polygon. $s_n$ is half the perimeter of the $2^n$-gon.
Let's see how the formula emerges. For $n=2$ we have $s_2/4=(\sqrt{2})/2=\sin(\pi/4)$. Then if one sets $s_n/2^n=\sin\theta_n$, then the given formula leads to
$$
(1/\sqrt{2})\sqrt{1-\sqrt{1-\sin^2θ_n}}=\sqrt{(1-\cosθ_n)/2}=\sqrt{\sin^2(θ_n/2)}=\sin(θ_n/2)
$$
which matches the required formula for $(s_{n+1})/(2^{n+1})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to proof $\forall n>0: 5|(6^n-1)$ is true using mathematical induction This is what i did so far, i got stuck on the later steps of using mathematical induction
I started with a base of $1$ which gave me $5|5$ which is $1$. Then I used $n = k$, and i got $5|(6^k-1)$ and i assumed this was true, but when i got to $n = k+1$ i didnt knew how to implement $5|(6^k-1)$ to $5|(6^{k+1}-1)$ to show that the theorem is true. For the real assignment, check this link: https://imgur.com/e6Z4tZW
| With induction. If $5|6^n-1$, then there is $q_n$ such that $6^n-1=5\cdot q_n$. Note that
$$
6^{n+1}-1=6\cdot 6^n-1=5\cdot 6^n+6^n-1=5(6^n+q_n)
$$
Without induction.
\begin{align}
6^n-1
=&
(5+1)^n-1
\\
=&
(5^{n}+n\cdot 5^{n-1}+\frac{n(n-1)}{2}\cdot 5^{n-2}+\cdots+\frac{n(n-1)}{2}\cdot 5^2+ n\cdot5+1)-1
\\
=&
5^{n}+n\cdot 5^{n-1}+\frac{n(n-1)}{2}\cdot 5^{n-2}+\cdots+\frac{n(n-1)}{2}\cdot 5^2+ n\cdot5
\\
=&
5\cdot (5^{n-1}+n\cdot 5^{n-2}+\frac{n(n-1)}{2}\cdot 5^{n-3}+\cdots+\frac{n(n-1)}{2}\cdot 5+ n)
\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Solving Recurrence Relation of $T(n)=4T(n-2)+2$ Question
Solve Recurrence Relation of $T(n)=4T(n-2)+2$
Base case-: $T(1)=1,T(2)=2$
My Approach/solution
$$T(n)=4T(n-2)+2$$
$$T(n-2)=4T(n-4)+2 \tag{1}$$
$$T(n-4)=4T(n-6)+2 \tag{2}$$
Using $(1)$ and $(2)$ in my equation
$$\begin{align*}
T(n)&=4\cdot (4T(n-4)+2)+2\\
&=4^{2}\cdot T(n-2\cdot 2)+2\cdot 4^{1}+2\cdot 4^{0}\\
&=4^{2}\cdot(4T(n-6)+2)+2\cdot 4^{1}+2\cdot 4^{0}\\
&=4^{3}\cdot T(n-2\cdot 3)+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}\\
\vdots \\
&=4^{k}\cdot T(n-2\cdot k)+2\cdot 4^{k-1}+...+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}
\end{align*}$$
Substituing $T(n-2\cdot k)$ by $2$, i.e $T(2)=2$
$$n-2\cdot k=2 \Rightarrow k=\frac{n-2}{2}$$
So our equation will look like
$$\begin{align*}
T(n)&=2\cdot 4^{k}+2\cdot 4^{k-1}+...+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}\\
T(n)&=2\cdot \left(4^{0}+4^{1}+4^{2}+...+4^{k-1}+4^{k}\right)\\
T(n)&=2\cdot \left(4^{0}\cdot \frac{(4^{k+1}-1)}{4-1}\right)
\end{align*}$$
$k=\frac{n-2}{2}$
$$\begin{align*}
T(n)&=2\cdot \left(\frac{(4^{k+1}-1)}{4-1}\right)\\
T(n)&=2\cdot \frac{2^{n}-1}{3}
\end{align*}$$
Is it correct? Also if it is correct, can anyone hint me another approach as it is bit lengthy.
Thanks!
| Hint: This is a linear second order recurrence equation. You can solve it with methods similar to solution methods for differential equations. Note thate recurrence equations are sometimes called difference equations, because they are the discrete form of differential equations. Notice that the generating functions method (as provided in the other answers) is also nothing else than the power series method used for differential equations.
The general solution is given as a superposition of the solution of
$$T_h(n+2)=4T_h(n)$$
which is the homogeneous equation and the particular solution. The homogeneous solution can be obtained by an exponential ansatz $T_h(n)= A^n$. Note you will get two solutions for $A$. They should be $A_{1,2}=\pm 2$.
Then find the particular solution by using the method of undetermined coefficients. Ansatz is a constant $T_p(n)=c=\text{const.}$. You should obtain $T_p(n)=-2/3$.
The general solution is given by:
$$T(n)=c_1A_1^n+ c_2A_2^n+(-2/3).$$
Determine the “constants of integration” $c_{1,2}$ by using your “initial values” for $T(1)$ and $T(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \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} $$
I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in.
Is there a more elegant way to prove this, or a way to find the right "trick"?
| $M= (a^2 + b^2 + c^2)(a^3 + b^3+c^3)=$
$a^5 + b^5 + c^5 + a^3b^2 +a^3c^2 +a^2b^3 + b^3c^2 + a^2c^3 +b^2c^3$
$= (a^5 + b^5 + c^5) + L$ where
$L = a^3b^2 +a^3c^2 +a^2b^3 + b^3c^2 + a^2c^3 +b^2c^3$
$= a(a^2b^2 + a^2c^2) + b(a^2b^2 + b^2c^2) + c(a^2c^2 + b^2b^2)$
$= (a+b+c)(a^2b^2 + a^2c^2 + b^2c^2) - ab^2c^2 -a^2bc^2 - a^2b^2c$
$= - ab^2c^2 -a^2bc^2 - a^2b^2c= -abc(ab + ac + bc)$
On the other hand:
$a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)=-2(ab+ac + bc)$
$= \frac {2L}{abc}$
while $0=(a+b+c)^3 = (a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3$
$= a^3 + 3a^2b + 3ab^2 + b^3 +3a^2c + 6abc + 3b^2c + 3ac^2 + 3bc^2 + c^3$
$= a^3 + b^3 + c^3 + 3a(ab + ac) +3b(ab + bc) + 3c(ac+ bc) + 6abc$
$=a^3+b^3+c^3 +3(a+b+c)(ab + ac + bc) - 9abc+6abc$
$=a^3 + b^3 +c^3 -3abc$
So $a^3 + b^3+c^3 = 3abc$.
So $M =(a^2+b^2 + c^2)(a^3 + b^3 + c^3) = \frac {2L}{abc}*3abc = 6L$
So $M = a^5 + b^5 + c^5 + L$ and $M = 6L$ so $5L=a^5 + b^5 +c^5$.
And we have:
$\frac {a^2+b^2+c^2}2\times\frac {a^3+b^3 + c^3}3 = \frac M6 = L = \frac{a^5+b^5 + c^5}5$
Yes. Long and tedious and little insight. But methodical.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Showing the cusps in $\Gamma_1(4)$: [$\infty$], [0], [1/2] equal $\mathbb{P_1(Q)}$ I am trying to show that $\Gamma_1(4)$ only has three cusps: $0, 1/2$ and $\infty$. To do so, I need to show $\Gamma_{0},\Gamma_{\infty}, \Gamma_{1/2}$ equal $\mathbb{P_1(Q)}$.
Here is what I have so far:
$\Gamma_{\infty}= \{\frac{4a+1}{4c} | a,c \in \mathbb{Z}, (4a+1,4c)=1 \}$
$\Gamma_{0}= \{\frac{b}{4d+1} | b,d \in \mathbb{Z}, (4d+1,b)=1 \}$
$\Gamma_{1/2}= \{\frac{4a+1+2b}{4c+8d+2} | a,b,c,d \in \mathbb{Z}, (4a+1)(4d+1)-4bc=1 \}= \{\frac{4a+1}{4c} | a,b,c,d \in \mathbb{Z}, (4a+1)(4d+1)-4bc=1 \}$
Now how do I show their union equals $\mathbb{P_1(Q)}$?
| Let $\frac{r}{s}$ be an element of $\mathbb{Q}$ such that $\gcd(r,s) = 1$.
We shall distinguish the following cases:
*
*$r$ odd, $4\mid s$
*$r$ any number, $s$ odd
*$r$ odd, $s\equiv 2 \mod 4$
Note that if $A \equiv 3 \mod 4$, then $A^2 \equiv 1\mod 4$.
For the first case: If $r\equiv 1 \mod 4$, choose $a = r, b = $anything, $ c = s, d = 1$. Then
$$\begin{pmatrix} a&b\\c&d \end{pmatrix}.\infty = \frac{a}{c} = \frac{r}{s}.$$
Similarly, if $r\equiv 3\mod 4$, then $a = r^2, b = $anything, $c = rs, d = 1$ does the job:
$$\begin{pmatrix} a&b\\c&d \end{pmatrix}.\infty = \frac{a}{c} = \frac{r^2}{sr} = \frac{r}{s}.$$
Now consider the second case: Let $r$ be any integer, $s\equiv 1\mod 4$. Then
$$\begin{pmatrix} a&b\\c&d \end{pmatrix}.0 = \frac{b}{d} = \frac{r}{s}$$
if $a=1,b=r, c=0,d=s$. If $s\equiv 3\mod 4$, choose $a=1,b=rs,c=0,d=s^2$ instead and proceed as above:
$$\begin{pmatrix} a&b\\c&d \end{pmatrix}.0 = \frac{b}{c} = \frac{rs}{s^2} = \frac{r}{s}.$$
Finally, if $s\equiv 2\mod 4$, and $r\equiv 1 \mod 4$, then we can write $s = 4l+2, r = 4k+1$. Hence,
$$\begin{pmatrix} a&b\\c&d \end{pmatrix}.\frac{1}{2} = \frac{a/2+b}{c/2+d} = \frac{1/2+2k}{4l/2+1} = \frac{1+4k}{4l+2} = \frac{r}{s}$$
for $a = 1, b=2k, c=4l, d=1$. If $s\equiv 2\mod 4$ and $r\equiv 3\mod 4$, then we can write $s=4l+2, r = 4k+3$. Therefore,
$$\begin{pmatrix} a&b\\c&d \end{pmatrix}.\frac{1}{2} = \frac{a/2+b}{c/2+d} = \frac{1/2+2k+1}{4l/2+1} = \frac{1+4k+2}{4l+2} = \frac{r}{s}$$
for $a=1,b=2k+1,c=4l,d=1$.
We have accounted for all possible cases for $r$ and $s$ and thus we are done showing that the cusps are $[0],[\frac{1}{2}],$ and $[\infty]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $n\in\mathbb N$ so that: $\sqrt{1+5^n+6^n+11^n}\in\mathbb N$ Find $n\in\mathbb N$ so that: $\sqrt{1+5^n+6^n+11^n}\in\mathbb N$
My attempt led me to have $\quad n=2k+1:\quad k\in\mathbb N$
The expression under square root is odd, so the square root's value is also odd.
I assumed $\sqrt{1+5^n+6^n+11^n}=2a+1:\quad a\in\mathbb N \Rightarrow$
$5^n+6^n+11^n=(2a+1)^2-1 \Rightarrow 5^n+6^n+11^n=4a(a+1)$
Then the expression on the left must be divisible by 8
$5^n\equiv 5\pmod 8$ if $n$ is odd, $5^n\equiv 1\pmod 8$ if $n$ is even.
$6^n\equiv 0\pmod 8$ if $n\ge3$.
$11^n\equiv 3\pmod 8$ if $n$ is odd, $11^n\equiv 1\pmod 8$ if $n$ is even.
Then $n$ must be odd and $\ge3$ for the expression to be divisible by 8.
I'm stuck here, would anyone give a hand please?
| $1+5^n+6^n+11^n$ has the unit digit = $3$ for any $n$ and therefore cannot be a square
Indeed $11^n$ ends always with $1$,
$6^n$ ends with $6$,
$5^n$ ends with $5$ adding $1$ makes $13$ so we get the unit digit to be $3$ in any case
Hope this can be useful
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
A question on Fermat's Little Theorem Let $n$ be an integer not divisible by $3$. Show that $n^7 ≡ n (\mod 63)$.
I know that we can split $63$ into $3^2 \cdot 7$
So we have $n^7=n (\mod 7\cdot3^2)$
$n^7=n (\mod 3^2)$ and $n^7 = n (\mod 7)$
And I am stuck how to go about solving this question after this
| $$n^7-n=n(n^3-1)(n^3+1)=(n-1)n(n+1)(n^2-n+1)(n^2+n+1).$$
Now, it's obvious that $(n-1)n(n+1)$ is divisible by $3$.
Let $n=3k-1$, where $k$ be a natural number.
Thus, $n^2-n+1$ is divisible by $3$.
Let $n=3k+1$, where $k$ be a natural number.
Thus, $n^2+n+1$ is divisible by $3$,
which says that $n^7-n$ is divisible by $9$.
Also, $n^7-n$ is divisible by $7$ for all integer $n$ by the Fermat's little theorem, which says that it's divisible by $7$ for all integers $n$, which not divisible by $3$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability of getting correct answer in a multi-choice question.
In a multiple choice question there are 4 alternative answers of which
1, 2, 3, or all may be correct. A candidate decides to tick answers at random. If he is allowed upto 5 chances to answer the question, the probability that he will get the marks in the question is?
It is equally likely for 1, 2, 3, or all to be correct so probability is 1/4.
Case1 when only 1 option is correct.
Since the candidate is allowed 5 chances, probability of getting correct answer is 1.
Case2 when 2 options are correct.
Toltal ways in which 2 options can be correct is $\left(^4_2\right)$, which is 6. Out of these only 1 is correct. So probability of selecting correct answer is 1/6. Since he has 5 chances the probability of getting marks is $1 - \left(\frac{5}{6}\right)^5$
Case3 when 3 options are correct.
Total ways in which 3 options can be correct is $\left(^4_3\right)$, which is 4. Since he has 5 chances, probability of getting correct answer is 1.
case4 when all options are correct.
Only one way in which all can be correct. Probality of getting marks is 1.
So answer should be $\left(\frac{1}{4}\times1\right) + \frac{1}{4}\times\left(1 - \left(\frac{5}{6}\right)^5\right) + \left(\frac{1}{4}\times1\right) + \left(\frac{1}{4}\times1\right)$.
Which is indeed wrong.
A friend of mine did this question as follows.
Total options: $\left(^4_1\right) + \left(^4_2\right) + \left(^4_3\right) + \left(^4_4\right)$, i.e 15. Since he has 5 chances to answer, probability would be 5/15. I know this is wrong (or not?) but beacause my textbook says answer is 1/3 I couldn’t argue.
Please help. Thanks in advance.
| There are four choices, each of which may be correct or incorrect. However, not all of them may be incorrect. Therefore, there are $2^4 - 1 = 15$ possible ways to answer the question. The candidate is allowed five guesses. That means the candidate chooses $5$ of the $15$ possible answers. The candidate receives credit if one of those guesses is correct, so the probability that the candidate guesses the correct answer is
$$\frac{5}{2^4 - 1} = \frac{5}{15} = \frac{1}{3}$$
Your friend counted the number of possible answers in a different way. We have the option of choosing one, two, three, or four of the answers. There are $\binom{4}{k}$ ways to choose exactly $k$ of the four answers. Hence, the number of possible ways to answer the question is
$$\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15$$
Edit: To understand why the probability is the number of guesses divided by the number of answers, suppose that you are given $k$ guesses and $n$ answers, of which only one is correct. There are $\binom{n}{k}$ ways to make $k$ guesses. If you select the correct answer with one of those $k$ guesses, then you also select $k - 1$ of the $n - 1$ incorrect answers. Therefore, your probability of selecting the right answer among your $k$ guesses is
\begin{align*}
\frac{\dbinom{1}{1}\dbinom{n - 1}{k - 1}}{\dbinom{n}{k}} & = \frac{\dfrac{(n - 1)!}{(k - 1)![(n - 1) - (k - 1)]!}}{\dfrac{n!}{k!(n - k)!}}\\
& = \frac{(n - 1)!}{(k - 1)!(n - k)!} \cdot \frac{k!(n - k)!}{n!}\\
& = \frac{(n - 1)!}{n!} \cdot \frac{k!}{(k - 1)!}\\
& = \frac{k}{n}
\end{align*}
In our problem, $k = 5$ and $n = 15$, so the probability of obtaining the correct answer is
$$\frac{5}{15} = \frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
if $\sin{(ax)}+\sin{(bx)}=\sin{(cx)}+\sin{(dx)}$,show $a=c$ or $a=d$
let $a,b,c,d>0$ and such for any real numbers $x\in R$ have
$$\sin{(ax)}+\sin{(bx)}=\sin{(cx)}+\sin{(dx)}$$
show that
$a=c$ or $a=d$
I try it,becasue
$$\sin{\left(\dfrac{(a+b)x}{2}\right)}\cos{\left(\dfrac{(a-b)x}{2}\right)}=\sin{\left(\dfrac{(c+d)x}{2}\right)}\cos{\left(\dfrac{(c-d)x}{2}\right)}$$
then I don't known how to continue
| If you differentiate and put $x=0$, you get $a+b=c+d$. Derivating two more time and putting again $x=0$, you get $a^3+b^3=c^3+d^3$. Simplifying by $a+b=c+d>0$ gives $a^2-ab+b^2=c^2-cd+d^2$, hence $(a+b)^2-3ab=(c+d)^2-3cd$ and $ab=cd$. It is easy to finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $a,b,c,a^2+b^2+c^2$ are primes, then $a$ or $b$ or $c$ is equal to $3$ Given primes $a,b,c$ such that $a^2+b^2+c^2$ is prime, then $3\in\{a,b,c\}$.
Tested for $a,b,c<500$.
| If $p\neq3$, $p$ a prime, then $p^2\equiv1\pmod3$. So if $3\notin\{a,b,c\}$ then $3\mid a^2+b^2+c^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.