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... | 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*}
Th... | {
"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
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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 thi... | 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
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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{... | 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",
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"answer_id": 1
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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... | 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 contain... | {
"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 $y'$:
$$y'''=\frac{(y'')^2}{y'}-\fra... | Solving :
$$ \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)\... | {
"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
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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 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-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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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... | 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 $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$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} ... | $$
\frac{dz}{dx} = e^y \phi'\frac{x}{y}e^{\frac{x^2}{2y^2}}
$$
$$
\frac{dz}{dy} = e^y\phi + e^y\phi'\left(e^\frac{x^2}{2y^2} - \frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}\right)
$$
$$
(x^2-y^2)\frac{dz}{dx} + xy\frac{dz}{dy} = e^y \phi'\frac{x}{y}e^{\frac{x^2}{2y^2}}(x^2-y^2)+\left(e^y\phi + e^y\phi'\left(e^\frac{x^2}{2y^2} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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For which $n$ does $2^n+1$ divide $10^n+1$? This came up in a discussion of numbers that divide their own binary representation (when interpreted as a decimal). The question I'm asking zooms in on a special case:
For which $n$ does $2^n+1$ divide $10^n+1$?
I computed the remainder after division of $10^n+1$ by $2^n+1... | In comments on the question, Chris has proven all the cases other than $ $$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 $$
Assume $2^m||16k$ fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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How do I solve $\int\frac{dx}{\sin x+\cos x-1}$? Please help me find the following indefinite integral:
$$\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\fra... | {
"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... | $$S(n)=1-x+x^2-x^3+..(-1)^nx^n$$ 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
$$s(n)+xS(n)=1+(x-x)+(x^2-x^2)+....+((-1)^{n-1}x^n+(-1)^nx^n)+(-1)^nx^{n+1}\\S(n)(x+1)=1+(-1)^nx^{n+1}\\
s(n)=\frac{1+(-1)^nx^{n+1}}{1+x}=\\
\frac{1+(-1)(-1)^{n+1}x^{n+1}}{1+x}=\\
\to \\
S(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Simplify $5(3 \sin(x) + \sqrt3 \cos(x))$ So Wolfram tells me I can reach $10 \sqrt3 \sin(x + π/6)$ from this expression, but I cant grasp how to do it. Any help is appreciated.
| $a \sin x + b\cos x$
let $\phi = \arctan \frac{b}{a}$
$\sin {\phi} = \frac {b}{\sqrt {a^2+b^2}}\\
\cos {\phi} = \frac {a}{\sqrt {a^2+b^2}}$
$a \sin x + b\cos x = \sqrt {a^2 + b^2} (\cos\phi \sin x + \sin\phi\cos x) = \sqrt {a^2 + b^2} \sin (x+\phi)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Finding the eigenvalues of a $n\times n$ $A=(a_{ij})$ defined by $a_{ij}=1$ if $i+j=n+1$, $0$ elesewhere. If $A=(a_{ij})\in M_n(\mathbb{R})$ is a matrix defined by $a_{ij}=1$ if $i+j=n+1$, $0$ otherwise. Then what are the eigenvalues of $A$?
\begin{bmatrix}
0 & 0 & \ldots & \ldots & \ldots & ... | Note that, if $e_j=(0,0,\ldots,0,1,0,\ldots,0)$, where the $1$ is in the $j-$th place, then
$$
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}
$$
From $(1)$ we obtain $\lfloor\frac{n+1}{2}\rfloor$ linearly independent eigenvectors
corresponding to the eigenvalue $1$, while $(2)$ pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find $F(1) = -8$, of the antiderivative $\displaystyle f(x) = x^3 + 2 \sqrt{x}$. Consider the function $\displaystyle f(x) = x^3 + 2 \sqrt{x}$.
Let $F(x)$ be the antiderivative of $f(x)$ with $F(1) = -8.$
$\dfrac{x^4}{4} + \dfrac{4x^\frac{3}{2}}{3}+C$
$-8=\dfrac{(1)^4}{4} + \dfrac{4\cdot(1)^\frac{3}{2}}{3}+C$
$-8-.25... | Let
$$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}.$$
The constant can be found by applying $F(1) = -8$ for which
\begin{align}
F(1) = -8 &= \frac{1}{4} + \frac{4}{3} + c_{0} \\
c_{0} &= - \frac{115}{12}.
\end{align}
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Transform curve into canonical form and determine type I have the curve:
$$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$$
Which is purely the case.
$\textbf{Thoughts:}$
1.) I dete... | The equation $9x^2-6xy+y^2+6x-2y-3=0$ is equivalent with
$$(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$$
Now we apply the principal axis theorem: The eigenvalues of the matrix, say $A$, are $\lambda=0$ and $\mu=10$, the eigenvecto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Characteristic polynomial of a $7 \times 7$ matrix whose entries are $5$ Avoiding too many steps, what is the characteristic polynomial of the following $7 \times 7$ matrix? And why?
\begin{pmatrix}5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\end{pmatrix}
| As it was stated in the commentaries, the rank of this matrix is $1$; so it will have $6$ null eigenvalues, which means the characteristic polynomial will be in the form:
$p(\lambda)=\alpha\,\lambda^6(\lambda-\beta) = \gamma_6\,\lambda^6 +\gamma_7\,\lambda^7$
Using Cayley-Hamilton:
$p(A)=\gamma_6\,A^6+\gamma_7\,A^7 =0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Algebraic identities Given that $$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 $$
We can rearrange this to be more useful:
\begin{align}2 a^2 + 2 a b + 2 a c + 2 b^2 + 2 b c + 2 c^2 &= 2 a^2 + 2b^2+2c^2 + 2 a b + 2 a c + 2 b c\\
&=2(a^2+b^2+c^2)+2(ab+ac+bc)\end{align}
We know the value of $ab+ac+bc$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solving Pinter 7.B.4 with a program Here is exercise 7.B.4 from 'A Book of Abstract Algebra' by Charles C. Pinter.
A solution to this using a C# program is posted.
Is there another good approach using a computer program?
Any language is welcome.
The subgroup of $S_5$ generated by
$
f = \begin{pmatrix}1 & 2 & 3 & 4 & 5... | SageMath provides built-in methods for such computations
sage: S5 = SymmetricGroup(5)
sage: f = S5( (1,2)); g = S5( (3,4,5)) #cycle notation
# or f = S5([2,1,3,4,5]); g = S5( [1,2,4,5,3])
sage: S5fg = S5.subgroup([f,g])
sage: S5fg.multiplication_table(names='elements')
$$\scriptsize{
\begin{array}{c|*{6}{c}}
{\ast}&(... | {
"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
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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 $u = \sqrt{\frac{4}{3}}(x+\frac{1}{2})$
$$\frac{du}{dx} = ... |
This answer was posted prior to an edit made by the OP.
Note that we have
$$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$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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How to obtain the sum of the following series? $\sum_{n=1}^\infty{\frac{n^2}{2^n}}$ It seems that I'm missing something about this.
First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)
What I tried to do is to find a limit of a partial sum $\li... | $$
\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}
$$
Add... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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What is the probability of the sum of four dice being 22? Question
Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?
My Approach
I simplified it to the equation of the form:
$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\le... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "29",
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find a,b,c where $\sqrt{5^{1/3}-4^{1/3}}=\frac{1}{3}(a^{1/3}+b^{1/3}-c^{1/3})$ Is it ramanujan problems? $$\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}$$ or
... | {
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"url": "https://math.stackexchange.com/questions/2353420",
"timestamp": "2023-03-29T00:00:00",
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If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this ... | Use the inequality $x^2+y^2+z^2>zy+yz+zx$ for distinct $x,y,z$ repeatedly: $$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.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration by Partial Fractions, using $\sqrt2$ Integrate $$\int\frac{10x^2 + 13x + 9}{(x^2 - 2)(2x + 1)^2}\mathrm dx$$
The difficulty here is in $x^2 - 2$. In order to determine the coefficients A to D, $\sqrt2$ must be invoked, in factorizing. Firstly, put $x = \sqrt 2$, next $-\sqrt 2$, $-0.5$.
The coefficients com... | Note that the coefficient of $A$ of the term
$\frac{1}{x-\sqrt{2}}$ in the partial fraction expansion of $$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^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 serie... | 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
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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)(... | $$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$$
$$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc$$
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 $\f... | {
"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
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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
$$\begin{align}12\lambda y - 4y - \lambda^2y=0\end{align}$$
$\Leftrightarrow$
$$\begin{align}(\lamb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 6
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$\sqrt{9x^2-16}>3x+1$ I'm trying to solve the following inequality:
$$\sqrt{9x^2-16}>3x+1$$
Here's my attempt:
$\sqrt{9x^2-16}>3x+1$
$\longrightarrow 9x^2-16>9x^2+6x+1$
$\longrightarrow -16>6x+1$
$\longrightarrow x<-\frac{17}{6}$
Now, I need to check the constraints:
$9x^2-16 > 0$
$\longrightarrow (3x)^2 > 4^2$
$\lo... | *
*$3x+1\leq0$ and $9x^2-16\geq0$, which gives $x\leq-\frac{4}{3}$;
*$3x+1>0$ and $9x^2-16>(3x+1)^2$. The last gives $x<-\frac{17}{6}$, which is impossible.
Thus, the answer is $\left(-\infty,-\frac{4}{3}\right]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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determine for which values of $x$ the function $f(x)=\begin{cases} \frac{x^2-4}{x-2}+10, &\text{if }x < 2\\ 2x^3-x, &\text{if } x ≥ 2\\ \end{cases}$ is continuous...
$$f(x)=\begin{cases}
\frac{x^2-4}{x-2}+10, &\text{if }x < 2\\
2x^3-x, &\text{if } x ≥ 2\\
\end{cases}$$
These kind of problems are very unclear to me, I o... | As Zubin Mukerjee indicated in the comments, we must show that
*
*$f$ is continuous when $x < 2$;
*$f$ is continuous when $x > 2$;
*$\lim_{x \to 2+} f(x) = \lim_{x \to 2-} f(x) = f(2)$.
If $x < 2$, then
\begin{align*}
f(x) & = \frac{x^2 - 4}{x - 2} + 10\\
& = \frac{(x + 2)(x - 2)}{x - 2} + 10\\
& =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does $\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na}) $ converge? I want to check whether
$\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na}) $
converges or not. (a is a positive constant number.)
If it converges, how to find the value it converge... | This is the limit of a Riemann sum:
$$\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}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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If $\sin(\alpha)+\cos(\alpha)=a,$ denote $|\sin(\alpha)-\cos(\alpha)|$ in terms of $a$ I attempted to solve it using the following identities: $1.\ a^2-b^2.\ 2.\ (a\pm b)^2.\ 3. a^3\pm b^3.\ 4.\ (a\pm b)^3.$
Since none of my efforts led me to the correct answer (which is $\sqrt{2-a^2}$), I found it better not to write ... | $$\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}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1$ When I'm reading Computer Systems: A Programmer's Perspective, I met the sum of binary numbers and failed to prove it:
$$ 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1 $$
This might be preliminary knowledge, I'm not good at mathematics, any bod... | Notice $2^0=2^1-1$.
Then $2^0+2^1=2(2^1)-1=2^2-1$
Then $2^0+2^1+2^2=2(2^2)-1=2^3-1$.
We have $2^0+2^1+2^2...2^{n-1}=2(2^{n-1})-1=2^n-1$.
And to show the this holds for the $n+1$ case, notice that $(2^n-1)+2^n=2(2^n)-1=2^{n+1}-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$
Evaluate the following limit.
$$\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-... | 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 $x \to +\infty$:
$$ \Bigg( 1 - \frac{a}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of monomials in $a_n=a_{n-1}+(a_{n-2})^2$ with $a_1=a$, $a_2=b$ I was playing around with the sequence $a_1=a$, $a_2=b$ and the recurrence $a_n=a_{n-1}+(a_{n-2})^2$ and just listed out a few general terms.
For instance, $a_1=a$, $a_2=b$, $a_3=b+a^2$, $a_4=b+b^2+a^2$, $a_5=a^4+2a^2b+a^2+2b^2+b$, and $a_6=b^4+2b^3... | What an interesting question! The sequence of the number of monomial terms is $1,1,2,3,5,8,14,24,44,80,152,288,560,1088,2144,\dots$
-- not Fibonacci. For example $a_7$ =
$b + 4 b^2 + 6 b^3 + 5 b^4 + a^2 + 6 b a^2 + 10 b^2 a^2 + 8 b^3 a^2 + 3 a^4 + 6 b a^4 + 8 b^2 a^4 + 2 a^6 + 4 b a^6 + a^8$ has 14 terms.
The sequence ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Show that $\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
I want to prove that $\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
My ideas, I don't know if they lead anywhere:
Let's substitute $\cos(\theta)=\frac{e^{i\theta}+e^{-i\t... | As Dr. Sonnhard Graubner answered, expand $\cos(2x)$ and $\cos(3x)$ using the classical formulas and use the tangent half-angle substitution to get
$$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$$ Using partial fraction decomposit... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
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How to show $x^5-x+3$ is irreducible in $\Bbb Z_5[x]$? How to show $x^5-x+3$ is irreducible in $\Bbb Z_5[x]$?
I checked that there are no linear factors since there is no root. How to know that there is no quadratic factor? I checked from Mathematica that there are $32$ irreducible quadratic in $\Bbb Z_5[x]$.
| By long division, if an $x^2 + a x + b$ divided $x^5 - x + 3$, you would need $a^4 - 3 a^2 b + b^2 - 1 \equiv 0$ and $a^3 b - 2 a b^2 + 3 \equiv 0 \mod 5$.
You can "complete the square" in $a^3 b - 2 a b^2 + 3 \mod 5$ to obtain
$3 a (b + a^2)^2 + 2 a^5 + 3 \equiv 3 a \left((b+a^2)^2 +4 a^4 + 1/a\right) $. For this to... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How prove this inequality with $x+y+z=2$ Let $x,y,z\ge 0$,and such $x+y+z=2$,show that
$$\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{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 understa... | 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 $n^4 - n^2 + 64.$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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}... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the differential equation $x^2\frac{d^2y}{dx^2}+(x^3-x)\frac{dy}{dx}-3y=0$ $$x^2\frac{d^2y}{dx^2}+(x^3-x)\frac{dy}{dx}-3y=0$$
$$\sum_{n=0}^{\infty}(n+r)(n+r-2)c_nx^{n+r}+\sum_{n=2}^{\infty}(n+r-2)c_{n-2}x^{n+r}-3\sum_{n=0}^{\infty}c_nx^{n+r}=0$$
$$r(r-2)c_0x^r+(r^2-1)c_1x^{1+r}-3c_0x^r-3c_1x^{r+1}+\sum_{n=2}^{\in... | Wolfram Alpha gives the solutions as
$$
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$ a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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... | 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*}
*
... | {
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"url": "https://math.stackexchange.com/questions/2376267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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 re... | 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 + ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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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$$
Can anyone thro... | Wolog $a\neq 0.$ If we substitute $x=ay$, we get completely rid of the $a$:
$$
(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-... | {
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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 cont... | 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.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
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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
$(x^2 y + y^2 z + z^2 x) +(x^2 y + y^2 z + z^2 x) =2186+2188=4374$.
... | $$\sum_{cyc}(x^2y-x^2z)=(x-y)(x-z)(y-z)=-2$$
and the rest for you.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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$\,\,f(x)=a(x+\alpha)(x+\beta)$, ... | 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 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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,
$$a^2+b^2=2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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... | 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... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $$-2x^2y+y(1+x^2-y^2)=y(1-x^2-y^2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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",
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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 $x$ a... | 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:
$... | {
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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},\... | {
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"url": "https://math.stackexchange.com/questions/2396913",
"timestamp": "2023-03-29T00:00:00",
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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 ... | 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 $... | {
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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 in... | Let $u = \frac{1}{x}$
Then the problem becomes:
$$\lim_{u\to 0}\left(\sqrt[3]{\left(\frac{1}{u}\right)^3-5\left(\frac{1}{u}\right)^2+1}-\frac{1}{u}\right)$$
$$=\lim_{u\to 0}\left(\sqrt[3]{\frac{1-5u+u^3}{u^3}}-\frac{1}{u}\right)$$
$$=\lim_{u\to 0}\left(\frac{1}{u}(1-5u+u^3)^{\frac{1}{3}}-\frac{1}{u}\right)$$
$$=\lim_{u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the given limit: Evaluate the given limit:
$$\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 (\... | 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?
| {
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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.
$$\alpha + \beta = \dfrac {-b}... | Alternatively:
$$\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$$
Squaring and then dividing both sides by $4a$ will prove the claim.
| {
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Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able ... | $$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)$$
Now compute $(a^2+ab+b^2)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the expected number of tyres that are installed in their original positions? After summer, the winter tyres of a car (with four wheels) are to be put back. However, the owner has forgotten which tyre goes to which wheel, and the tyres are
installed `randomly', each of the $4! = 24$ permutations being equally li... | $0$ tires in their correct positions (derangements): The number of ways $k$ of the $4$ tires are in their correct positions is $\binom{4}{k}$. The remaining $4 - k$ tires can be arranged in $(4 - k)!$ orders. Hence, by the Inclusion-Exclusion Principle, the number of arrangements in which no tires are in their corre... | {
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"url": "https://math.stackexchange.com/questions/2401536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve the recurrence $T(n)=T(n-1)+\frac{n}{n-1}$?
Solve the recurrence $T(n)=T(n-1)+\frac{n}{n-1}$.
I thought to try solving it by iteration method:
$$
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}
$$
I know that $\sum^n_i i=0.5n(n+1)$ but I don't kn... | Well, at first let $T(1)=a\in\mathbb{R}$ be the first term of the sequence $T(n)$. Then, let us try to find some of the next terms of the sequence:
$$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\... | {
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"url": "https://math.stackexchange.com/questions/2402071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$, $\frac{P(x)}{x^3+x}$ 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^... | $P(x)=x^{100}+x^{50}-2x^4-x^3+x+1$
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 ... | {
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"timestamp": "2023-03-29T00:00:00",
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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... | 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 condit... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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... | 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:
$$a(x^2-x)+b= x(a(x-1)+b)\Longrigh... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{1}{x}<\frac{x}{2}<\frac{2}{x}$. First I consider the case $\frac{1}{x}<\frac{x}{2}$, manipulating to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x}<0\Leftrightarrow x < -\sqrt{2}.$ Did this by subtracting $\frac{x}{2}$ from both sides and writing everything with a common denominator.
Second case to consider is $\fra... | First, notice that $1/x < 2/x$ implies $1/x$ is positive, hence $x$ is positive. Now you know you can multiply the inequalities by $x$ without reversing the inequality signs. So you get $2<x^2$ and $x^2<4$. And so on...
| {
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"timestamp": "2023-03-29T00:00:00",
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find the probability of selecting exactly two women and at least two women when a six-person committee is selected from $7$ men and $4$ women? A committee of six members is formed from a group of $7$ men and $4$
women. What is the probability that the committee contains
a. exactly two women?
b. at least two women?
My a... |
A committee of six members is formed from a group of $7$ men and $4$ women. What is the probability that the committee contains exactly two women?
Since there are a total of $7 + 4 = 11$ people, the number of ways we can select a committee of six people is
$$\binom{11}{6}$$
A committee of six that contains exactly ... | {
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"url": "https://math.stackexchange.com/questions/2408038",
"timestamp": "2023-03-29T00:00:00",
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Repeated Linear Factors in denominator of fraction e.g. $\frac{2x^2 + 2x + 18}{x(x-3)^2}$ I have the following fraction below and I must find the partial fraction decomposition.
$$\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... | 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.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $a+b+c$, wh... | 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... | {
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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}\,\... | 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}\i... | {
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"answer_id": 1
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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}{... | 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,
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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:
$\dfrac{1}{x+1}+\dfrac{1}{y+1}$
$=\dfrac{x+y+2}{xy+x+y+1}$
$=\dfrac{2(k+1)}{(k+1)^2}$
$=\dfrac{2}{k+1}$... | 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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 the... | $$\frac{2x+1}{\sqrt{x^2+x+1}}=\frac{x\left(2+\dfrac{1}{x}\right)}{\sqrt{x^2\left(1+\dfrac{1}{x}+\dfrac{1}{x^2}\right)}}=\frac{x\left(2+\dfrac{1}{x}\right)}{|x|\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}}=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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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:
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(3)+(-2x-x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(2x+x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(1+2x+x^2) +1}} dx $$
$$\int \frac{x}{... | $$\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}}.$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What kind of series is the following what kind of series is the following and can I get a generalized formula for it?
$T_1 = (1-\frac{1}{2N^3})$
$T_2 = (1- T_1.\frac{1}{2N^3}) $
$T_3 = (1 - T_2.\frac{1}{2N^3})$
$T_4 = (1 - T_3.\frac{1}{2N^3})$
..
.
$T_x = (1 - T_{x-1}.\frac{1}{2N^3})$
$ P_{total} = T_1 + T_2 + T... | You have a recurrent formulation of a series. Let $a=\frac1{2N^3}$. You can see that the explicit formula
$T_n = \frac{1 + (-a)^n a}{1+a}$ solves your given recurrence $T_{n+1} = 1 - a T_{n}$, with $T_0=1$, because
$$ 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}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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let : $\forall n \in \mathbb{N}$ : $x_n=\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}$ then : $\displaystyle{\sum_{i=1}^{20}}\dfrac{1}{x_i}=?$ let : $\forall n \in \mathbb{N}$ : $x_n=\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}$
then :
$$\sum_{i=1}^{20}\dfrac{1}{x_i}=?$$
my try :
$$\dfrac{1}{\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{13}}+.... | 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",
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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}$$$$$$
Multiplying by $\frac{1}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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3, 5, 7 are the only three consecutive odd natural numbers that are prime. Can someone please help me proof this theorem. So far I assumed towards contradiction that there exist three consecutive odd natural numbers that are primes P, P+2, and P+4.
Theorem: 3, 5, 7 are the only three consecutive odd natural numbers th... | Let's get three numbers: $n, n+2, n+4$.
We have 2 cases:
*
*$n\not\equiv0\mod 3$:
Thus $n=3k+1$ or $n=3k+2$, so $n+2=3(k+1)$ or $n+4 = 3(k+2)$
*$n=3k$
In both cases at least one of our numbers is divisible by $3$.
The only prime that is divisible by $3$ is $3$.
The only sequention of prime numbers $n, n+2, n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $A$ be a $4\times2$ matrix and $B$ be a $2\times4$ matrix. Find $BA$ when $AB$ is given.
Let $A$ be a $4\times2$ matrix and $B$ be a $2\times4$ matrix so that
$$AB=
\begin{pmatrix}
1 & 0 & -1 & 0 \\
0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & -1 & 0 & 1
\end{pmatrix}.
$$
Find $BA$.
I have the ... | Write $A=\begin{pmatrix}A_1\\A_2 \end{pmatrix}$ and $B=\begin{pmatrix}B_1 &B_2 \end{pmatrix}$ where $A_1,A_2,B_1,B_2$ are $2\times2$ matrices.
Note that $AB=\begin{pmatrix}A_1B_1 & A_1B_2\\A_2B_1 & A_2B_2 \end{pmatrix}$, hence $A_1B_1= A_2B_2=I_2$.
Since a square matrix commutes with its inverse, we have as well $B_1A_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$ Question:
Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
My attempt:
Base case is trivial.
Suppose $ \ n \ge 2$ and $ \ 2^{n} + 3^{n} < 4^{n}$
Then,
$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n} $... | Hint: for $\,0 \lt a \lt 1\,$ the sequence $\,a^n\,$ is decreasing with $\,n\,$ since $\,a^{n+1} = a \cdot a^n \lt 1 \cdot a^n\,$, so:
$$\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$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to integrate following integral How to integrate following integral i am having physics background and i am trying calculating the induced charge in a specific problem.
The Integral need to be evaluted is
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{(x^2+y^2+b^2)^{3/2}}\,dxdy$$
| Let $b\not=0$, then, by letting $x=r\cos(t)$ and $y=r\sin(t)$ (polar coordinates) the integral becomes
\begin{align*}
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{(x^2+y^2+b^2)^{3/2}}\,dxdy&=
\int_{r=0}^{\infty}\int_{t=0}^{2\pi}\frac{1}{(r^2+b^2)^{3/2}}\,(rdrdt)\\
&=\left[-\frac{2\pi}{(r^2+b^2)^{1/2}}\right]_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
In the list of questions proposed in the "Meeting for Training for the Brazilian Olympiad", 2013. No answer provided. Could solve some problems in that list but got stuck in this one. My developments are going into v... | We can use the binomial theorem for the coefficients of $(a+b)^7$.
The coefficients are 1,7,21,35,35,21,7,1, but we minus the $a^7$ and $b^7$.
So we have $7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6$, and then, let $a=0$, and since the result is 0 when we insert it, we have $a$ as a root. $b$ and $a+b$ works too. W... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Two equivalent simple continued fractions $A\overset{\color{red}?}=B$ Fix a sequence $a_n$ of complex numbers for natural number $n$,such that $|a_n|\gt1$
and then define the following two simple continued fractions
$A= \cfrac{1}{a_{0}-\cfrac{1}{a_{1}+\cfrac{1}{a_{2}-\cfrac{1}{a_{3}+\cfrac{1}{a_{4}-\cfrac{1}{a_{5}+\cfr... | $$ 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}} $$
Applying this should prove that in the sequences defining your continued fractions, there is a common subsequence, so if b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Determine the derivative of $\arctan$ function Find $f'(x)$ for the function:
$f(x)= \arctan(\frac{a+x}{1-ax}))$ , $a\in R$
So this is what I've done:
$f(x) = \arctan x$
$f'(x) = \frac{1}{1+x^2}$
$x= \frac{a+x}{1-ax}$
$f'(x) = \frac{1}{1+\frac{(a+x)^2}{(1-ax)^2}}$
$f'(x) = \frac{(1-ax)^2}{(1-ax)^2+(a+x)^2}$
Is this co... | Hint:
$$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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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)\\
&=... | 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
$$
\log 2 + \frac{1}{n} \log \left( \frac{n+1}{2n} \times \frac{n+2}{2n} \ti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$f(x)=?$ if we have $f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$ $f(x)=?$
If we have $$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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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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... | 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 ... | {
"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
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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 $$\lim\limits_{x\rightarrow\infty}\frac{\ln(1+2x^2)}{\ln(x^4+3)}=\lim\limits_{x\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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If $\alpha,a,b$ are integers and $b\neq-1$, then prove that, if $\alpha$ satisfies the equation $x^2+ax+b+1=0$, $a^2+b^2$ must be composite.
Let $\alpha,a,b$ be integers such that $b\neq-1$. Assume that $\alpha$ satisfies the equation $x^2+ax+b+1=0$. Prove that the integer $a^2+b^2$ must be composite.
$\alpha=\frac{-... | Hint: $a^2-4(b+1)=a^2+b^2-(b^2+4b+4)=c^2$ for some integer $c$.
Do you know any facts about expressing primes as the sum of two squares?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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limit of trigonometric functions without L'Hôpital's rule $\lim_{x\to 0} [\csc^22x-\frac 1 {4x^2}]\ $
and
$\lim_{x\to y} \frac {\tan x- \tan y} {1-\frac{x}{y}(\tan x\tan y+1)}\ $
We have not learn L'Hôpital's rule in my class, we do the $\lim_{x\to 0}\frac{\sin x}{x}=1\ $
I have not came up with anything yet
| Let's prove some double-sided estimate for $1/\sin^2 x-1/x^2$ giving the first limit, using only $\lim_{h\to0}\sin h/h=1$ and $\lim_{h\to0}\cos h=1$:
$$\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}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 ... | $$\frac{dx}{xy}=\frac{dy}{-x^2}=\frac{dz}{-yz}$$
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",
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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:
$$x_{n} - 3x_{n-1} = -8 \\ x_n = 3(3x_{n-2} - 8) - 8 \\ = 3^2 x_{n-2} -8 ( 1+3) \\ = 3^3 x_{n-3} - 8(1+3+3^2) \\ = 3^n x_{0} - 8(1+3+3^2 + \ldots + 3^{n-1}) \\ = 2\times 3^n - 8\left(\frac{3^n - 1}... | To answer the question of why the method 2 answer is incorrect, the problem is that you try to figure out the coefficient on the homogenous portion of the solution before figuring out a particular solution. This gives an incorrect value because you haven't taken the inhomogeneous portion into account. Instead, once yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
modify sequence so it converges to pi Modify sequence by Archimedes
$s_{n+1}=\frac{2^{n+1}}{ \sqrt{2} } \sqrt{ 1 -\sqrt{1-(s_n/2^{n})^2}}; s_2=2\sqrt{2} $
so that it approaches $\pi$
the issue is that it is going to $0$
$$\lim_{n\to \infty} \sqrt{1-(s_n/2^n)^2}=1 $$
making $s_{n+1}$ go to $0$
Now
$$ \begin{aligned}... | 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 regu... | {
"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
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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... | 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\c... | {
"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
$$\begin{align*}
T(n)&=4\cdot (4T(n-4)+2)+2\\
&=4^{2... | Hint: This is a linear second order recurrence equation. You can solve it with methods similar to solution methods for differential equations. Note thate recurrence equations are sometimes called difference equations, because they are the discrete form of differential equations. Notice that the generating functions met... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \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... | $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^... | {
"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
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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}{... | 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 = ... | {
"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\... | $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).$$
Now, it's obvious that $(n-1)n(n+1)$ is divisible by $3$.
Let $n=3k-1$, where $k$ be a natural number.
Thus, $n^2-n+1$ is divisible by $3$.
Let $n=3k+1$, where $k$ be a natural number.
Thus, $n^2+n+1$ is divisible by $3$,
which says that $n^7-n$ is divisible by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability of getting correct answer in a multi-choice question.
In a multiple choice question there are 4 alternative answers of which
1, 2, 3, or all may be correct. A candidate decides to tick answers at random. If he is allowed upto 5 chances to answer the question, the probability that he will get the marks i... | There are four choices, each of which may be correct or incorrect. However, not all of them may be incorrect. Therefore, there are $2^4 - 1 = 15$ possible ways to answer the question. The candidate is allowed five guesses. That means the candidate chooses $5$ of the $15$ possible answers. The candidate receives cr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
if $\sin{(ax)}+\sin{(bx)}=\sin{(cx)}+\sin{(dx)}$,show $a=c$ or $a=d$
let $a,b,c,d>0$ and such for any real numbers $x\in R$ have
$$\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... | 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.