Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to prove this inequality with the following condition? Let $x$, $y$, $z$ be three positive real numbers satisfying
\begin{equation}
x + y +z + 1 =4xyz.\tag{1}
\end{equation}
Prove that
\begin{equation}
xy + yz + zx \geqslant x + y + z.\tag{2}
\end{equation}
I don't know how to start?
| Here is a boring, uninspiring but elementary proof ^_^. By $(1)$ and A.M.$\ge$G.M., we get $xyz = (x+y+z+1)/4 \ge (xyz)^{1/4}$. Hence $xyz\ge1$. WLOG, suppose $0<x\le y\le z$. There are three cases. Case 1: $x\ge1$. Clearly $(2)$ holds. Case 2: $0<x<1\le y< z$. Then $(2)$ also holds because
$$1-xyz\le0\le(1-x)(1-y)(1-z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/272830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
What is the solution number of the equation $x^2-x+1\equiv 0 \pmod{p^e}$ What is the solution number of the equation
$$x^2-x+1\equiv 0\pmod{p^e}$$
I know when $e=1$, it is $1+\left(\frac{-3}{p}\right)$, and I guess it is the same for $e>1$, but can anyone provide a proof?
updated:
I know when $e=1$, the number is
$$
1... | If $p>2,$$$p^e\mid (x^2-x+1)\iff p^e\mid (2x-1)^2+3$$
So, $$(2x-1)^2\equiv-3\pmod{p^e}$$
Applying Discrete Logarithm w.r.t some primitive root $g\pmod {p^e}$,
$2ind_g(2x-1)\equiv ind_g(-3)\pmod{p^{e-1}(p-1)}$ as $\phi(p^e)=p^{e-1}(p-1)$
Using Linear congruence theorem, the last equation is solvable iff $(2,p-1)\mid in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Decomposition of polynomial of n-degree into irreducible polynomials. I'm getting ready for the exam in algebra. And I have problem that I can't solve again.
I'm understand what I need to do if I have a polynomial with a given degree. But I very confused by arbitrary degree in this problem.
Given polynomial
$$x^{2n... | $$x^{2n}-2x^n+2=(x^n-1)^2+1=(x^n-1+i)(x^n-1-i)=(x^n-2^{0.5}e^{\frac{-i\pi}{4}})(x^n-2^{0.5}e^{\frac{i\pi}{4}})$$
Now note that:
$$x^n-2^{0.5}e^{\frac{-i\pi}{4}}=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{-i\pi}{4}+\frac{-2\pi ik}{n}})$$
$$x^n-2^{0.5}e^{\frac{i\pi}{4}}=\prod_{k=1}^n(x-2^{\frac{1}{2n}}e^{\frac{i\pi}{4}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$ Some time ago I came across to the following integral:
$$I=\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$$ What are the hints on how to compute this integral?
| You may like this method. Note
$$ \int_0^1x^ada=-\frac{1-x}{\ln x}, \text{ for }x>0, H_a=\sum_{k=1}^\infty\frac{a}{k(a+k)}, $$
and hence
\begin{eqnarray}
\int_0^1\frac{1-x}{1+x}\frac{dx}{\ln x}&=&-\int_0^1\int_0^1\frac{x^a}{1+x}dadx\\
&=&-\int_0^1\int_0^1\frac{x^a}{1+x}dxda\\
&=&-\frac{1}{2}\int_0^1(H_{\frac{a}{2}}-H_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 5,
"answer_id": 3
} |
Finding the integer solutions Find all integer solutions of
$$(a + b^2)(a^2 + b) = (a − b)^3.$$
Obviously $b = 0$ is one. But how to get other solutions?
| Here is one approach.
Expand both sides and equate terms, so we have:
$$(a + b^2)(a^2 + b) = (a - b)^3$$
$$a^{3} + a^{2}b^{2} + ab + b^{3} = a^{3} - 3a^{2}b + 3a b^2 - b^{3}$$
We can eliminate the $a^{3}$ terms, so we are left with:
$$2b^{3} + a^{2}b^{2} - 3ab^{2} + 3a^{2}b +ab = 0$$
Now, we have a Cubic in the form of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The last digit of $2^{2006}$ My $13$ year old son was asked this question in a maths challenge. He correctly guessed $4$ on the assumption that the answer was likely to be the last digit of $2^6$. However is there a better explanation I can give him?
| The last digit of the number is the remainder of division by 10, since any number is represented as:
$$ a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0 = \overline{a_na_{n-1}\ldots a_1a_0} $$ $$\quad \Leftrightarrow \quad $$ $$a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/284902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 8,
"answer_id": 7
} |
How to expand $\tan x$ in Taylor order to $o(x^6)$ I try to expand $\tan x$ in Taylor order to $o(x^6)$, but searching of all 6 derivative in zero (ex. $\tan'(0), \tan''(0)$ and e.t.c.) is very difficult and slow method.
Is there another way to solve the problem?
Any help would be greatly appreciated :)
| This answer is inspired by the answer of coffeemath.
We know the following (from e.g. Wikipedia):
\begin{align}
&\sin(x) = x -\frac{x^3}{6} +\frac{x^5}{120}+ O(x^7) \\
&\cos(x) = 1 -\frac{x^2}{2} +\frac{x^4}{24} +\frac{x^6}{720} +O(x^8) \\
&\tan(x) = a +bx+ cx^2+dx^3+ex^4+fx^5+gx^6+ O(x^7)
\end{align}
Furthermore it h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/286529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Taylor expansion for $\sqrt{x+2}$ I'm enrolled in Coursera's calculus with a single variable and am trying to solve one of the homework problems.
In lecture, it was stated that to expand $\sqrt x$ about $x=a$, you would have:
$$\sqrt{x} = \sqrt{a} + {1 \over 2 \sqrt{a}}(x-a)- {1\over 8 \sqrt{a^3}}(x-a)^2 + H.O.T$$
The ... | This is what I have now:
Evaluating derivatives for $\sqrt{x+2}$ at x=2, I get:
$ f(0) = \sqrt{2+ (2)} = 2 $
$ f'(0) = {1\over2} {1 \over \sqrt{2+(2)} }= 1/4$
$ f''(0) = -{1\over 4} { 1 \over (\sqrt{ 2+(2)})^{3} }= -1/32$
The taylor series expansion for $ \sqrt{x+2} \space \space at \space \space x=2 :$
$2 + {1\over 4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/286718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Working with exponent on series Hi have this sequence:
$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}$$
I understand that this is a Geometric series so this is what I've made to get the sum.
$$\sum\limits_{n=1}^\infty (-1)^n\frac{3^{n}\cdot 3^{-2}}{4^n}$$
$$\sum\limits_{n=1}^\infty (-1)^n\cdot 3^{-2}{(\frac{3}{4}... | $$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}=\sum\limits_{n=1}^\infty (-1)^n\frac{1}{9}\left(\frac{3}{4}\right)^n=\frac{1}{9}\sum\limits_{n=1}^\infty \left(-\frac{3}{4}\right)^n=
\frac{1}{9}\left(\frac{1}{1+3/4}-1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Determinants of $3\times 3$ Matrix Suppose that $a,b,c,d,e,f$ are numbers such that
$$\det\left(\begin{matrix}
a&1&d\\b&1&e\\ c&1&f
\end{matrix}\right)=7$$ and
$$\det\left(\begin{matrix}
a&1&d\\b&2&e\\ c&3&f
\end{matrix}\right)=11.$$
How do you find the determinant of the Matrix
$$\begin{pmatrix}
a&3&d\\b&5&e\\ c&7... | Using multilinearity of the determinant (by columns), we get:
$$\begin{vmatrix}
a&3&d\\b&5&e\\ c&7&f
\end{vmatrix}=\begin{vmatrix}
a&1+2&d\\b&1+4&e\\ c&1+6&f
\end{vmatrix}=\begin{vmatrix}
a&1&d\\b&1&e\\ c&1&f
\end{vmatrix}+2\begin{vmatrix}
a&1&d\\b&2&e\\ c&3&f
\end{vmatrix}=7+2\cdot 11=29$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$ i cannot figure out a way to find this limit. $$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$$
Its undeterminate form $0\cdot\infty$ so i tried using $$\displaystyle\lim_{n\to \infty}\frac{a^{X_n}-1}{X_n}=\ln{a}$$ this leads to $0\cdot\i... | We have:
$
\begin{align*}
\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2}) & = \lim_{n\to \infty} n^2\left( 2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right) = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n}-\frac{1}{n+1}} - 1 \right) \\
& = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/288002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Weird question pertaining to HCF I encountered this question which seems weird/incomplete to me :
Q: H.C.F. of 3240, 3600 and a third number is 36, and their L.C.M. is
$2^4 \cdot 3^5 \cdot 5^2 \cdot 7^2$ . The third number is?
Can anyone please teach me concept wise how to solve it?
| \begin{align}
3240 & = 2\cdot2\cdot2\cdot3\cdot3\cdot3\cdot3\cdot5 & & = 2^3\cdot3^4\cdot5\\
3600 & = 2\cdot2\cdot2\cdot2\cdot3\cdot3\cdot5\cdot5 & & = 2^4\cdot3^2\cdot5^2 \\
36 & = 2\cdot2\cdot3\cdot3 & & =2^2\cdot3^2
\end{align}
The third number cannot have more than two $2$s in its prime factorization, since then th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/288142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
closest point to on $y=1/x$ to a given point I feel like I'm missing something basic - given a point $(a,b)$ how do I find the closest point to it on the curve $y=1/x$? I tried the direct approach of pluggin in $y=1/x$ into the distance formula but it leads to an order-4 polynomial...
| The slope of $y = 1/x$ at $x$ is $-1/x^2$,
so the tangent at $(p, 1/p)$ is
$(y-1/p)/(x-p) = -1/p^2$
or $y = 1/p - (x-p)/p^2$.
At $y = 0$, $(x-p)/p^2 = 1/p$
or $x = 2p$;
at $x = 0$, $y = 1/p + p/p^2 = 2/p$.
The intercept points are thus
$(0, 2/p)$ and $(2p, 0)$.
I will minimize the distance by making
the line from $(a, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Does an integer $9
Does an integer $9<n<100$ exist such that the last 2 digits of $n^2$ is $n$? If yes, how to find them? If no, prove it.
This problem puzzled me for a day, but I'm not making much progress. Please help. Thanks.
| We are solving $n(n-1)=n^2-n\equiv0\pmod{100}$. Since $\gcd(n,n-1)=1$, one of $n$ or $n-1$ must be a multiple of $4$ while the other must be a multiple of $25$.This leads to the equations
$$
\begin{align}
4x-25y=+1\tag{1}\\
4x-25y=-1\tag{2}
\end{align}
$$
For $(1)$, $n=4x$ and $n-1=25y$. For $(2)$, $n=25y$ and $n-1=4x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/292651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Implicit Differentiation of $x^2y = 1$ I have a slight problem with this: Given the equation
$$ x^2y = 1$$
and asked to find $y''$, I attempted to apply implicit differentiation by differentiation w.r.t. $y$.
$$2xy'y + x^2y' = 0$$
However, it does not seem to be right
UPDATE Solved:
Differentiate w.r.t. $x$
$$2xy... | We are given that $x^2y = 1 \,\,\,\,\, (\spadesuit)$.
Hence, we have $$\dfrac{d(x^2y)}{dx} = 0 \implies \dfrac{d(x^2)}{dx}y + x^2 \dfrac{dy}{dx} = 0\implies 2xy + x^2 \dfrac{dy}{dx} = 0 \implies 2y + x\dfrac{dy}{dx}=0 \,\,\, (\star)$$
Now differentiate again to get
$$\dfrac{d}{dx} \left(2y + x \dfrac{dy}{dx}\right) = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/293406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Application for mean value theorem $f(x)$ is three-times differentiable on $[a,b]$, how to show that there is $\varepsilon\in(a,b)$ such that
$$f(b)=f(a)+\cfrac{1}{2}(b-a)[f'(a)+f'(b)]-\cfrac{1}{12}(b-a)^3f'''(\varepsilon)$$
| Let $$F(x) := f(b)-f(x)- \frac{1}{2} (b-x) \cdot (f'(b)+f'(x)) - K \cdot (b-x)^3$$ where $$K := \frac{f(b)-f(a)-\frac{1}{2} (b-a) \cdot (f'(b)+f'(a))}{(b-a)^3} \tag{1}$$
Then $F$ is differentiable on $[a,b]$ and $F(a)=F(b)=0$. ($K$ is chosen such that $F(a)=0$.) By Rolle's theorem there exists $\varrho \in (a,b)$ such ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/295246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solving complicated fraction for X I realize this rather rudimentary but it has been over a decade since my algebra classes and now I have problem that I can't figure out. I would like someone to walk me through the steps in solving the "X" in this problem:
$$
1.20 \cdot 10^6 = \frac{1}{\left(\dfrac{1}{6 \cdot 10^6}\ri... | $$1.20 \cdot 10^6 = \dfrac{1}{\left(\frac{1}{6\cdot10^6}\right)X} \iff 1.2\cdot 10^6 = \frac{6\cdot 10^6}{X}$$
$$\iff X =\frac{6\cdot 10^6}{1.2\cdot 10^6} = \frac {6}{1.2} = 5$$
The nice thing about the problem is that there is no need to evaluate numerically until the very end, where we need only compute $X =\dfrac{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/297240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The maximum of $|x^3 + ax^2 + bx + c|$ on $[-1, 1]$ is at least $1/4$ Let $f(x)=x^{3}+ax^{2}+bx+c$ with a, b, c real.
Show that
$$\frac{1}4 \le \max_{-1 \le x \le 1\hspace{2mm}} |f(x)|=M$$
and find all cases where equality occurs.
| Note that
$$
\max_{[-1,1]}\,\left|\,x^3-\tfrac34x\,\right|=\tfrac14\tag{1}
$$
Considering the symmetry about $0$ of the domain, we have for any $t\in[0,1]$
$$
\max_{\{t,-t\}}\,\left|\,x^3+ax^2+bx+c\,\right|=\left|\,t^3+bt\,\right|+\left|\,at^2+c\,\right|\tag{2}
$$
Using $(2)$, it is obvious that
$$
M_b=\max_{[-1,1]}\,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/297539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$\det(\exp X)=e^{\mathrm{Tr}\, X}$ for 2 dimensional matrices I want to prove that for $X\in M_2(\mathbb{R})$ the formula $\det(\exp X)=e^{\mathrm{Tr}\, X}$ holds, writing $X$ in normal form gives $X=PJP^{-1}$, where $J$ is the Jordan matrix, now $\exp (PJP^{-1})=P(\exp J)P^{-1}$ and $\det P(\exp J)P^{-1}=\det \exp J$.... | $$
\exp\begin{pmatrix}0&1\\0&0\end{pmatrix}=I+\begin{pmatrix}0&1\\0&0\end{pmatrix}+\frac1{2!}\begin{pmatrix}0&1\\0&0\end{pmatrix}^2+\ldots
=I+\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}1&1\\0&1\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rolling three dice...am I doing this correctly? Tree dice are thrown. What is the probability the same number appears on exactly two of the three dice?
Since you need exactly two to be the same, there are three possibilities:
1. First and second, not third
2. First and third, not second
3. Second and third, not first
F... | Yes your answer is correct. $${3\choose 2}\cdot\frac{1}{6}\cdot\frac{5}{6} = \frac{5}{12}$$
Good job!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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Another integral with Catalan Show that:
$$\int_0^1\frac{\arcsin^3 x}{x^2}\text{d}x=6\pi G-\frac{\pi^3}{8}-\frac{21}{2}\zeta(3)$$
I evaluated this by some Fourier series. Is there any other method?
Start with substitution of $$u=\arcsin x$$
Then we have to integrate $$\int_0^{\frac{\pi}{2}}\frac{u^3\cos u}{\sin^2 u}\t... | Integrating by parts twice, we get
$$
\int_0^{\pi/2}x^2\,e^{ikx}\,\mathrm{d}x
=i^{k-1}\frac{\pi^2}{4k}+i^k\frac\pi{k^2}+\frac2{k^3}\left(i^{k+1}-i\right)
$$
Therefore, using $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$,
$$
\begin{align}
&\int_0^1\frac{\arcsin^3(x)}{x^2}\mathrm{d}x\\
&=\int_0^{\pi/2}\frac{x^3}{\sin^2(x)}\mathrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/302005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Irreducible elements of $\mathbb{Z}[i\sqrt{5}]$ Is there any neat way to show that $9$ and $3-3i\sqrt{5}$ are irreducible elements of $\mathbb{Z}[i\sqrt{5}]$, while $1+4i\sqrt{5}$ and $5-2i\sqrt{5}$ are reducible?
| $9 = 3 \cdot 3$ is definitely reducible, and so is $3-3i\sqrt{5} = 3 \cdot (1-i\sqrt{5})$.
The norm of $a = 1+4i\sqrt{5}$ is $81$. There are no elements of norm $3$, so if it reducible, $a$ could only be the product of two elements of norm $9$, and these are $\pm 3, \pm 2 \pm i \sqrt{5}$. And in fact $a = (-2 + i \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/302506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Can $n(n+1)2^{n-2} = \sum_{i=1}^{n} i^2 \binom{n}{i}$ be derived from the binomial theorem? Can this identity be derived from the binomial theorem?
$$n(n+1)2^{n-2} = \sum_{i=1}^{n} i^2 \binom{n}{i}$$
I tried starting from $2^n = \displaystyle\sum_{i=0}^{n} \binom{n}{i}$ and dividing it by $4$
in order to get $2^{n-2}... | Start from the binomial theorem in the form
$$(x+1)^n=\sum_{k=0}^n\binom{n}kx^k$$
and differentiate with respect to $x$:
$$\begin{align*}
n(x+1)^{n-1}&=\sum_{k=0}^n\binom{n}kkx^{k-1}\\\\
&=\sum_{k=1}^n\binom{n}kkx^{k-1}\;.
\end{align*}$$
Differentiate again:
$$\begin{align*}
n(n-1)(x+1)^{n-2}&=\sum_{k=1}^n\binom{n}kk(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/302651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Identify a formula for each entry of the matrix $\small \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}^n$ Identify a formula for each entry of the matrix $\begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}^n$.
It's easy to find a solution by just looking at the first few results:
\begin{pmatrix} 7^{n-1} & 3\cdot 7^{n-1} \\ 2\c... | Fleshing out a little what the comments hint you:
$$\det(tI-A)=\begin{vmatrix}t-1&-3\\-2&t-6\end{vmatrix}=t(t-7)\Longrightarrow$$
the matrix's eigenvalues are $\,0\,,\,7\,$ , and eigenvectors for these values can be found as follows:
$$(1)\,\,\lambda=0:\;\;\;\;\begin{cases}\;\,-x-3y=0\\{}\\-2x-6y=0\end{cases}\;\;\Longr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/303906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Finding coefficient of generating function
Find the coefficient of $x^{52}$ in $$(x^{10} + x^{11} + \ldots + x^{25})(x + x^2 + \ldots + x^{15})(x^{20} + x^{21}+ \ldots + x^{45})$$
One thing I tried doing was factoring out $x^{10}, x, x^{20}$ from each of the products, respectively, then using the identity of a produ... | The original problem, after factoring out the terms as stated in the post, we have
$$
x^{10}(1+x+\ldots+x^{15})\cdot x(1 + x + \ldots+x^{14})\cdot x^{20}(1+x+\ldots+x^{25}) \tag{1}
$$
Applying some identities, $(1)$ gives
$$ x^{31}\cdot\frac{1-x^{16}}{1-x}\cdot\frac{1-x^{15}}{1-x}\cdot\frac{1-x^{26}}{1-x} \tag{2}$$
Rea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/305328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
linear algebra, show B is a basis for $\mathbb{R}^3$ Let $B = \{(1,0,1), (1,1,2), (1,2,4)\}$
a) Show that $B$ is a basis for $\mathbb{R}^3$
b) Compute the coordinate vector $[v]_B$ with respect to $B$ of the vector $v = (3, -1, 3).$
| To show it is a basis you have to show it generates $\mathbb{R}^3$, so every $$v=\begin{pmatrix} x \\y \\z \end{pmatrix}$$ can be written as
$$\lambda_1 \begin{pmatrix} 1\\ 0\\1 \end{pmatrix} + \lambda_2 \begin{pmatrix} 1\\1\\2\\ \end{pmatrix} + \lambda_3 \begin{pmatrix} 1\\2\\ 4\\ \end{pmatrix}$$
A basis is minimal,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/306544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof of triangle inequality I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :)
| A simple proof of the triangle inequality that is complete and easy to understand (there are more cases than strictly necessary; however, my goal is clarity, not conciseness).
Prove the triangle inequality $| x | + | y| ≥ | x + y|$.
Without loss of generality, we need only consider the following cases:
*
*$x = 0$
*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/307348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "75",
"answer_count": 11,
"answer_id": 1
} |
Computing limits which involve square roots, such as $\sqrt{n^2+n}-n$ Is there a general strategy for this? For example I'm working on the limit
$$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n $$
I have a simple argument to show that this limit is less than or equal to 1/2, but I can't get much further because it's dif... | $\begin{align}
\sqrt{n^2 + n} - n & = (\sqrt{n^2 + n} - n) \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} +n} \\
& = \frac{n^2+n-n^2}{\sqrt{n^2 + n} + n} \\
& = \frac{1}{\frac{\sqrt{n^2 + n} + n}{n}} \\
& = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} \rightarrow \frac{1}{2}\\
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to integrate $\int\frac{x}{(x^2-2x+5)^2} \, dx$? $$\int\frac{x}{(x^2-2x+5)^2} \, dx$$
I tried to complete the square of the bottom like this $\int\frac{x}{((x-1)^2+4)^2} \, dx$ but I'm still not sure what to do.
Any help would be appreciated.
Thanks
| A good strategy, now that you've completed the square is to also "complete the differential":
$$\int\frac{x}{((x-1)^2+4)^2} \, dx=$$
$$\frac 1 2\int\frac{2(x-1)}{((x-1)^2+4)^2} \, dx+\int\frac{1}{((x-1)^2+4)^2} \, dx=$$
$$\frac 1 2\int\frac{du}{u^2} +\int\frac{1}{(u^2+4)^2} \, du=$$
$$-\frac 1 2\frac{1}{u} +\int\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/307713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find $\sum\limits_{k\, \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}$ How to find $$\sum_{k \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}$$ Here we find $\displaystyle\sum_{k=1}^{\infty} \frac{2(k^2-1)}{k^4+k^2+1}=1$ and we know that $\displaystyle\sum_{k \,\text{odd}} + \sum_{k \,\text{even}}=1.$ Can we use this information to fi... | I write here so everyone can see. I finally solve it.
First, note that $$\sum_{k \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1},$$ because $$\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1}=\sum_{k\, \text{odd}}\left(\frac{2k-1}{k^2-k+1}-\frac{2(k+1)-1}{(k+1)^2-(k+1)+1}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Using Taylor's Theorem to show that $\ln(1 + x^2) \leq x^2$ Can we show that if $\operatorname{abs}(x) \lt 1$, then $$\ln(1+x^2) \leq x^2\;,$$
using Taylor's Theorem?
I am thinking of expanding it about $x=0$ but I got something like
$$f(x) = -x^2 + \frac{x^4}{2} - \dots$$
Is my approach correct? Could you give me som... | $$
\begin{array}{rcl}
\ln(1 + x^2) & \le & x^2 \\
e^{\ln(1 + x^2)} & \le & e^{x^2} \\
1 + x^2 & \le & e^{x^2} \\
1 + x^2 & \le & 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots \\ {}
0 & \le & \frac{x^4}{2!} + \frac{x^6}{3!} + \frac{x^8}{4!} + \dots \\ {}
\end{array}
$$
Which is true for all real $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Complete induction: $\sum^n_{i=1}\frac{1}{(2i-1)(2i+1)}=\frac{n}{2n+1}$ I am very confused with complete induction. Because in every task there is something different to do, and I never know what to insert (thats my biggest problem).
Here's the example:
Proof with complete induction. Please please help me, because I ha... | For a full solution, proceed like this:
$n=1$: $$\sum_{i=1}^1 \frac{1}{(2i-1)(2i+1)} = \frac{1}{(2-1)(2+1)} = \frac{1}{3} = \frac{1}{2 \cdot 1 +1},$$
so it holds for $n=1$.
Assume next that it holds for some generic $n$. You need to show that then it also holds for $n+1$. As it holds for $n$, you can assume that
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/309300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Trigonometry tangent line question How would I figure this out.
Find all x values between $0$ and $2\pi$ where the line tangent to the graph of
$y=\frac{\cos x}{2+ \sin(x)}$ is horizontal.
I did the deriavative
$\frac{(2+\sin(x)-\sin(x)+\cos(x)\cos(x)}{(2+\sin x)^2}$ but I think I need to find $x$
| For a horizontal tangent line you need the derivative to be zero. After all, $\operatorname{d}\!y/\!\operatorname{d}\!x$ gives you the gradient of the tangent line to the graph $y=f(x)$. We are told that
$$y = \frac{\cos x}{2+\sin x}$$
Using the quotient rules, and the standard trig identity $\sin^2x+\cos^2x \equiv 1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/310642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integrating :$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$ How to integrate :
$$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$$
| $$\begin{aligned}
\int \sqrt{\sin x} \cos^{3/2}x\,\mathrm{d}x &=\int\sqrt{\sin x\cos x}\cos x\,\mathrm{d}x\\
&=\frac{1}{\sqrt{2}}\int\sqrt{\sin 2x}\cos x\,\mathrm{d}x\\
&\overset{(1)}{=}\frac{1}{\sqrt{2}}\left(\sin x\sqrt{\sin 2x} - \int\sin x\frac{\cos 2x}{\sqrt{\sin 2x}}\,\mathrm{d}x\right)\\
&=\frac{\sin x\sqrt{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Solving the nonlinear ODE: $ \frac{1}{y^{\prime}}+\left(x-\frac{y}{y^{\prime}}\right)^{2}+1=0$ $$\frac{1}{y^{\prime}}+\left(x-\frac{y}{y^{\prime}}\right)^{2}+1=0$$
I am trying to solve this non-linear ODE and have tried all sorts of substitutions; any hints on how I should progress?
Thanks.
| Let $y = a x + b$, then $y' = a$, and you have the equation:
$$\begin{align} \dfrac{1}{a} + \left(x - x - \dfrac{b}{a}\right)^2 + 1 & = 0\\ \dfrac{1}{a} + \frac{b^2}{a^2} + 1 & = 0 \\ a + b^2 + a^2 & = 0 \\ \left(a + \frac{1}{2}\right)^2 + b^2 & = \left(\frac{1}{2}\right)^2 \end{align} $$
So the solutions are li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
what is the no. of possible parellelograms? A parallelogram having an acute angle of 30 degrees whose area is equal to the perimeter and the sides are whole numbers then number of such parallelograms possible are?
| We interpret the question to mean that we want to find the number of (unordered) pairs $a$, $b$ of integers that can be the sides. The perimeter is $2a+2b$. The area is twice the area of the triangle we get by joining the two vertices at which there is a $150^\circ$ angle.
The area of each of these triangles is $(1/2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/313790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Abstract Algebra: Equivalence Classes Under ~(conjugate) Relation Sorry. Don't know the latex.
Let $g=(1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8)$ in $S_8$
Find a specific permutation $f$ in $S_8$ so that $fgf^{-1} =(1 \ 5 \ 6 \ 3 \ 8 \ 7 \ 2 \ 4)$ (note: $fgf^{-1}$ is short for composition)
| Use the conjugation rule $fgf^{-1} = (f(1)\; f(2)\; f(3)\; f(4)\; f(5)\; f(6)\; f(7)\; f(8))$.
EDIT: Upon request, I give more details (which in principle are already given by Berci in the comments):
If f and g are any permutations, you can compute $fgf^{-1}$ by simply writing down $g$ as a product of disjoint cycles, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/315208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Prove $\frac{1}{2\sqrt{2}+1}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+\cdots+\frac{1}{100\sqrt{100}+99\sqrt{99}}<\frac{9}{10}$ What would you suggest for the following inequality?
$$\frac{1}{2\sqrt{2}+1}+\frac{1}{3\sqrt{3}+2\sqrt{2}}+\cdots+\frac{1}{100\sqrt{100}+99\sqrt{99}}<\frac{9}{10}$$
Thanks in advance!
Sis.
EDIT: Based upo... | $$
\sum_{m=1}^{99} \frac{1}{(m+1)^{3/2} + m^{3/2}} < \sum_{m=1}^{99} \left(\frac{1}{\sqrt{m}} - \frac{1}{\sqrt{m+1}} \right) = \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{99+1}} = \frac{9}{10}
$$
The above inequality is true since:
$$ \begin{eqnarray}
\frac{1}{(m+1)^{3/2} + m^{3/2}} &=& \frac{1}{\sqrt{m}\sqrt{m+1}} \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/316187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
} |
$xy=1 \implies $minimum $x+y=$? If $x,y$ are real positive numbers such that $xy=1$, how can I find the minimum for $x+y$?
| $x+y=x+y-2\sqrt{xy}+2\sqrt{xy}=(\sqrt{x}-\sqrt{y})^2+2\ge2$
The value will be attained when $(\sqrt{x}-\sqrt{y})^2=0\Rightarrow \sqrt{x}=\sqrt{y}\Rightarrow x=y=1$
Please note that this is the basis for the inequality of A.M.$\ge$ G.M.
If $x,y\ge 0$ then we know that $(\sqrt{x}-\sqrt{y})^2\ge0$
$\Rightarrow x+y-2\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/317831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 13,
"answer_id": 4
} |
Finding f intervals How would I find the f intervals for the following two functions.
$f(x)=(x-2)^2(x+1)^2$
using the chain rule I got $(x-2)^2(2)(x+1)(1)+(2)(x-2)(1)(x+1)^2$
then I got f decrease $(-\infty,-1]$
and f increase $[2,\infty)$
but the area between -1 and 2 in confusing me.
My second function is
$f(x)=x+\f... | As you did $f'(x)$ correctly, we have $$f'(x)=2(x-2)(x+1)(2x-1)$$ When:
*
*$x\le -1~~$, $(x-2)\leq0$ and then two terms $(x+1)$ and $(2x-1)$ are negative. So $f'(x)$ is negative.
*$x\le\frac{1}2~~$, $(x-2)\leq0$ and $(x+1)>0$ and $(2x-1)<0$ is negative, so $f'(x)$ is positive.
*$x\le2~~$, $(x-2)\leq0$ and then t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to determine equation for $\sum_{k=1}^n k^3$ How do you find an algebraic formula for $\sum_{k=1}^n k^3$? I am able to find one for $\sum_{k=1}^n k^2$, but not $k^3$. Any hints would be appreciated.
| One last Proof :
$$\sum_{k=1}^{n}k^{3}=1+8+27+64+...+n^{3}$$
$$\sum_{k=1}^{n}k^{3}=\underbrace{1}_{1^{3}}+\underbrace{3+5}_{2^{3}}+\underbrace{7+9+11}_{3^{3}}+\cdots+\underbrace{\left(n(n-1)+1\right)+\cdots+\left(n(n+1)-1\right)}_{n^{3}}$$
$$\sum_{k=1}^{n}k^{3}=1+3+5+\cdots+\left(n(n+1)-1\right)$$
Since $n(n+1) -1 $ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/320985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 10,
"answer_id": 3
} |
Construct a matrix transform consider
$\frac{dx}{dt} = Ax$ where $A$ is the matrix
$$
\begin{bmatrix}
1 & 0 & 1 \\
0 & 0 & -2 \\
0 & 1 & 0 \\
\end{bmatrix}
$$
construct a real matrix $P$ such that the change of coordinates $x=Py$ transforms our real equation to
$\frac{dy}{dt}=By$... | First of all, your eigenvalues (for both $A$ and $B$) should be $1$ and $\pm i\sqrt{2}$. Since you have three distinct eigenvalues, the eigenvectors of either matrix are guaranteed to be linearly independent, and thus constitute a basis.
(By necessity, the eigenvectors corresponding to the complex eigenvalues must be c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/321790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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$n^2$ for bases to prove that the last digit of base b ends with 0 Consider numbers written in base $b$, where $x\leq q\leq z$. For which bases $q$ $n^2$ ends with if $n$ ends with $0$.
| Hint: For any base $b$, things will go bad if there is a positive integers $x$ such that $x^2$ is divisible by $b$ but $x$ isn't. For example, $30^2$ is divisible by $50$ but $30$ isn't, so things will go bad for base $50$.
Added: Let us think about the specific bases $5$ to $9$. Let's see what we need to make sure th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/327885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Gambling puzzle A math friend of mine showed me this strange gambling puzzle. There is a button in a casino and every time you press it you can win either $1$ or $0$ dollars. The probability of winning $1$ dollar depends on how many times you have pressed it so far. If you press it for the $x$-th time you have probab... | Caution: This answer is WRONG. But I leave it here since my mistake may help to understand the question correctly. See edit in the last.
Suppose you play $L$ times. If $L > M$ then you will get
$$ \begin{align*}
E(L) &= \frac{1}{M} + \frac{2}{M} + \cdots + \frac{M}{M} + 1 + \cdots + 1 \\
&= \frac{1}{2}(M + 1) + (L - M... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/329854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
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Generating Functions: Solving a Second-Order Recurrence I'm self-studying generating functions (using GeneratingFunctionology as a text). I came across this programming problem, which I immediately recognized as a modification of the Fibonacci sequence. I wanted to place my newly found generating function techniques ... | $$1-x-kx^2=-kx^2-x+1\Rightarrow x_{1,2}=\frac{1\pm\sqrt{1+4k}}{2}$$so
$$1-x-kx^2=-kx^2-x+1=(-k)(x-x_1)(x-x_2)=(-k)(x-\frac{1-\sqrt{1+4k}}{2})(x-\frac{1+\sqrt{1+4k}}{2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/330292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$ let $a,b,c,d\ge 0$,and $a^2+b^2+c^2+d^2=3$,prove that
$ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$
I find this inequality are same as Crux 3059 Problem.
| Let $a=\sqrt 3\cos x\cos y,b=\sqrt 3\cos x\sin y,c=\sqrt 3\sin x\cos y,d=\sqrt 3\sin x\sin y, \pi/2>x,y>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/331004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int\frac{6^x}{9^x-4^x}dx $ How to solve this integral:
$$\int\frac{6^x}{9^x-4^x}dx $$
(I notice that $\frac{6^x}{9^x-4^x}=\frac{2^x3^x}{(3^x-2^x)(3^x+2^x)}$)
Thank you!
| Divide top and bottom by $ 4 ^ x $.
We have:
$$\int\dfrac{\left(\frac{3}{2}\right)^x}{\left(\frac{9}{4}\right)^x-1}dx=\int\dfrac{\left(\frac{3}{2}\right)^x}{\left(\left(\frac{3}{2}\right)^x\right)^2-1}dx$$
is replaced:
$$u=\left(\frac{3}{2}\right)^x$$
$$du=\left(\frac{3}{2}\right)^x\cdot\ln\left(\frac{3}{2}\right)dx$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Wolframalpha step-by-step of $\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$ I wonder, where the minus sign goes after the first $u$-substitution of integral $\displaystyle\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$?
| Rewrite $2+\cos 2x$ as $3-2\sin^2 x$ and let $u=\sin x$. We end up at
$$\int \frac{du}{\sqrt{3-2u^2}}.$$
Let $\sqrt{2}u=\sqrt{3}{v}$. Then $du=\frac{\sqrt{3}}{\sqrt{2}}\,dv$, and we end up with
$$\int\frac{1}{\sqrt{2}}\frac{dv}{\sqrt{1-v^2}}.$$
Thus our integral is $\frac{1}{\sqrt{2}}\arcsin v+C$. Replace $v$ by $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Computing $\int \frac{6x}{\sqrt{x^2+4x+8}} dx$ I am trying to compute an indefinite integral. Thanks!
$$\int \frac{6x}{\sqrt{x^2+4x+8}} dx.$$
| $$
\begin{align}
\int \frac{6x +12 -12}{\sqrt{x^2+4x+8}} dx
&= \int \frac{6x +12}{\sqrt{x^2+4x+8}} dx - \int \frac{12}{\sqrt{x^2+4x+8}} dx \\
&= \int \frac{6(x+2)}{\sqrt{x^2+4x+8}} dx - \int \frac{12}{\sqrt{x^2+4x+8}} dx \\
&= \int \frac{3(2x+4)}{\sqrt{x^2+4x+8}} dx - \int \frac{12}{\sqrt{x^2+4x+8}} dx.
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/332473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
$ \sqrt{2-\sqrt{2}} $ simplified So I have a (nested?) square root $ \sqrt{2-\sqrt{2}} $. I know that $ \sqrt{2-\sqrt{3}} = \frac{\sqrt{6}-\sqrt{2}}{2} $. I know how to turn the simplified version into the complex one, but not vice versa. What is $ \sqrt{2-\sqrt{2}} $ simplified in this fashion and what are the steps?
| A general way to solve such problems:
$$\sqrt{2-\sqrt{2}}=a$$
then
$a^2=2-\sqrt{2}\rightarrow(2-a^2)^2=2\rightarrow 4-4a^2+a^4=2\rightarrow a^4-4a^2+2=0$
Let $b=a^2$
then
Solve $$b^2-4b+2=0$$
As @Martin suggested, this is a way but the conslusion is not right. The right way should be
$$\sqrt{a+b\sqrt{c}}=\sqrt{d}+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 3
} |
Express trigonometric expressions in terms of one trigonometric function What is the general process to solve problems such as this:
I'm preparing for this type of exam problem.
For Reference, similar/more advanced problem (with solution): http://prntscr.com/ws6xl
| (Answered before the edit.) For the first question: If $x=\sin \theta $, then $\cos ^{2}\theta =1-\sin ^{2}\theta =1-x^{2}$ and $\cos \theta =\pm \sqrt{1-x^{2}}$. So $$\sin 2\theta =2\sin \theta \cos \theta =\pm 2x\sqrt{1-x^{2}}.\tag{1}$$
As for the second question divide $2\sin \theta \cos \theta$ by $1=\sin ^{2}\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Quadratic Congruence (with Chinese Remainder Thm) How do we solve quadratic congruences such as:
$x^2 \equiv11 \pmod{39}$
I know I must use the chinese remainder theorem with $p = 13, 3$ but I've only done linear examples and am unsure about how to do quadratic ones.
| $$x^2 \equiv 11 \pmod{39} \implies x^2 \equiv 2 \pmod 3 \implies \text{No solution}$$
EDIT
In general, when you want to solve for $$x^2 \equiv a \pmod n \,\,\, (\spadesuit)$$ and if $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$, the idea is to first solve for $$x^2 \equiv a \pmod {p_l^{a_l}} \,\,\, (\clubsuit)$$ You have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/335179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a ... | I'm not sure if "proof with Mathematica" counts, but here it is:
In order to find the local extrema of function $f(a,b,c) = \sqrt{3a+b^3}+\sqrt{3b+c^3}+\sqrt{3c+a^3}$ under the constraint $a+b+c=3$, we differentiate the corresponding lagrangian and arrive at a system of equations
$$\left\{\begin{aligned}& (a^2c^2-a^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 6
} |
Write in polynomial in factored form in complex number Write the following polynomial in factored form(in complex number):
$$1+z+z^2+z^3+z^4+z^5+z^6$$
Also, is there general solution of factoring for $1+z+z^2...z^n$ types of polynomial?
| $$1+z+z^2+z^3+z^4+z^5+z^6=\frac{1-z^7}{1-z}$$
$$1+z+z^2+z^3+z^4+z^5+...+z^n=\frac{1-z^{n+1}}{1-z}$$
Denote $$1+z+z^2+...+z^n=S$$ then multiply by z we get
$$z+z^2+...+z^{n+1}=Sz$$
from firs equation subtract the second then solve by $S$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim\limits_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
*
*Evaluate $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$.
*Examine whether $x^{1/x}$ possesses a maximum ... | This begs to be viewed as a Riemann sum:
$$
\frac 1n\left(\frac{1^3}{n^3}+\cdots+\frac{n^3}{n^3}\right) \to \int_0^1 x^3\,dx.
$$
Your second question seems quite different from your first and should be posted separately.
| {
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"url": "https://math.stackexchange.com/questions/338121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding determinant (using row reduction) I'm trying to calculate the following
$\det \begin{pmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ -1 & -1 & x-1 & -1 & -1 \\ -1 & -1 &-1 & x-1 & -1 \\ -1 & -1 & -1 & -1 & x-1 \end{pmatrix}$
and I'm sure there must be a way to get this as an upper/lower triangular... | Hints: make zeros below the entry 2-1 on the first column and develop wrt it and etc.:
$$\begin{vmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ -1 & -1 & x-1 & -1 & -1 \\ -1 & -1 &-1 & x-1 & -1 \\ -1 & -1 & -1 & -1 & x-1 \end{vmatrix}=\begin{vmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/340179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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What is the first non zero digit in 50 factorial (50!)? What is the first non zero digit in 50 factorial (50!)?
Any Help or hint will be appreciated.
| Since $50_{\text{ten}}=200_{\text{five}}$, $\sigma_5(50)=2$. Thus, the number of factors of $5$ in $50!$ is $\frac{50-2}{5-1}=12$.
Since $50_{\text{ten}}=110010_{\text{two}}$, $\sigma_2(50)=3$. Thus, the number of factors of $2$ in $50!$ is $\frac{50-3}{2-1}=47$.
Thus, $\frac{50!}{10^{12}}$ has $35$ factors of $2$.
Lit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/341578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stu... | An idea:
$$\frac{4}{x+4}\le 2\stackrel{\text{div. by 2}}\iff \frac{2}{x+4}\le 1\iff \frac{2}{x+4}-1\le 0\stackrel{\text{common denom.}}\iff $$
$$\iff \frac{2-x-4}{x+4}\le 0\stackrel{\text{mult. by (-1)}}\iff \frac{x+2}{x+4}\ge0\stackrel{\text{mult. by}\; (x+4)^2}\iff(x+2)(x+4)\ge 0\iff$$
$$\iff x<-4\,\,\vee\,\,x\ge-2\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/344268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 3
} |
Show that $\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$ Show that
$$\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$$
| Hint::
$$\frac{\sin x}{\cos 3x}=$$
$$= \frac{1}{2} \cdot \frac{2 \cdot \sin x \cdot \cos x}{\cos x \cdot \cos 3x}$$
$$= \frac{1}{2} \cdot \frac{\sin 2x}{\cos x \cdot \cos 3x}$$
$$= \frac{1}{2} \cdot \frac{ \sin(3x - x) }{ \cos x \cdot \cos3x }$$
$$= \frac{1}{2} \cdot \frac{ \sin 3x \cdot \cos x - \cos 3x \cdot \sin x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/344505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
$f(x - 1) + f(x − 2) $ and the sum of coeficients If $f(x-1)+f(x-2) = 5x^2 - 2x + 9$
and
$f(x)= ax^2 + bx + c$
what would be the value of $a+b+c$?
I was doing
$f(x-1)+f(x-2)= f(x-3)$
then
$f(x)$
a = 5
b = -2
c = 9
$(5-3)+(-2-3)+(9-3)$
But do not think is is correct
What would be correct approach?
| $$f(x-1)+f(x-2) = 5x^2 - 2x + 9$$and $$f(x)=ax^2+bx+c$$
for $x=1,2,3$ we get the system
$$f(0)+f(-1)=a-b+2c =12$$
$$f(1)+f(0)=a+b+2c=25$$
$$f(2)+f(1)=5a+3b+2c = 48$$
with solutions
$$a=5/2,b=13/2,c=8$$
so $$f(x)=\frac{5}{2}x^2+\frac{13}{2}x+8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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What goes wrong in this derivative? $$ f(x) = \frac{2}{3} x (x^2-1)^{-2/3} $$
and f'(x) is searched.
So, by applying the product rule $ (uv)' = u'v + uv' $ with $ u=(x^2-1)^{-2/3} $ and $ v=\frac{2}{3} x $, so $ u'=-\frac{4}{3} x (x^2-1)^{-5/3} $ and $ v' = \frac{2}{3} $, I obtain
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3... | Doing what you said, you should find
$$f'(x)=-\frac{8}{9}(x^2-1)^{-5/3}x^2+\frac{2}{3}(x^2-1)^{-2/3}.$$
Now
$$
(x^2-1)^{-2/3}=(x^2-1)(x^2-1)^{-5/3}.
$$
So
$$
f'(x)=-\frac{8}{9}(x^2-1)^{-5/3}x^2+\frac{2}{3}(x^2-1)(x^2-1)^{-5/3}
$$
$$
= \left(\frac{2}{3}-\frac{8}{9}\right)(x^2-1)^{-5/3}x^2-\frac{2}{3}(x^2-1)^{-5/3}
$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/346797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Weird calculus limit How to find the following limit?
$$
\lim_{n \to \infty} \dfrac{ 5^\frac{1}{n!} - 4^\frac{1}{n!} }{ 3^\frac{1}{n!} - 2^\frac{1}{n!} }
$$
Edit done to the question.
Thank you!
| $$\frac{5^x-4^x}{3^x-2^x}$$
$$=\frac{4^x\left(\left(\frac54\right)^x-1\right)}{2^x\left(\left(\frac32\right)^x-1\right)}$$
$$=2^x\frac{\frac{ \left(\frac54\right)^x-1 }x}{\frac{\left( \frac32\right)^x-1 }x}$$
So, $$\lim_{x\to0}\frac{5^x-4^x}{3^x-2^x}$$
$$=\lim_{x\to0}2^x\frac{\frac{ \left(\frac54\right)^x-1 }x}{\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$ \sin^{2000}{x}+\cos^{2000}{x} =1$ equation explanation Solve the equation:
$$ \sin^{2000}{x}+\cos^{2000}{x} =1.$$
What I did:
$\sin^2{x} =1 \land \cos^2{x}=0$ when $x=\frac \pi2 + \pi k $
$\cos^2 {x} =1 \land \sin^2{x}=0$ when $x= \pi k$
I think that these solutions apply for this equation as well but I don't reall... | Since we are concerned with the case when $\theta \neq k\pi/2$, we get
$$0 < \sin^2\theta < 1$$ and $$0 < \cos^2\theta < 1$$
Now observe that for $0 < \theta < \pi/2$
$$1 = (\sin^2\theta + \cos^2\theta)^2 = \sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta > \sin^4\theta + \cos^4\theta \quad (1)$$
Now observe th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/348085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Prove $\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$ If $a,b$ and $c \ge 0$ and $ab + bc + ca = 1$, prove that the following inequality holds:
$$\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$$
I've tried two aproaches, but it seems like both doesn't work. Here they are:
Cauch... | If none of the terms $a+bc$, $b+ca$, and $c+ab$ is greater than one, then each summand of the LHS is at least $\frac{1}{3}$, and thus the inequality holds. It remains to show the inequality also holds when at least one of $a+bc$, $b+ca$, or $c+ab$ is strictly greater than one. WOLOG, suppose that $c+ab>1$. Since $c=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Generating function for the solutions of equation $ 2x_1 + 4x_2 = n$ Let's denote $h_n$ as the number of soulutions of the following equation:
$$ 2x_1 + 4x_2 = n$$
where $x_i \in \mathbb N$.
Find generating function of the sequence $h_n$ and calculate $h_{2000}$.
I've found the generating function:
$$\frac{1}{1-x^2}\cd... | As stated, the generating function is:
\begin{align}
f(z) &= \frac{1}{1 - z^2} \cdot \frac{1}{1 - z^4} \\
&= \frac{1}{4 (1 - z^2)} + \frac{1}{4 (1 + z^2)} + \frac{1}{2 (1 - z^2)^2}
\end{align}
(this results from recognizing the generating function as a fraction in $z^2$, and splitting as such into partial fraction... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$ Base case: n=1. $1/1\le 2-1/1$. So the base case holds.
Let $n=k\ge1$ and assume
$$1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k$$
We want to prove this for $k+1$, i.e.
$$(1/1^2+1/2^2+1/3^2+\cdots+1/k^2)+1/(k+1)^2\le 2-\frac{1}{k+1}$$
This i... | For your induction step, all you need is
$$
2-\frac{1}{k}+\frac{1}{(k+1)^2}\leq 2-\frac{1}{k+1}.
$$
That's equivalent to
$$
\frac{1}{k}-\frac{1}{(k+1)^2}-\frac{1}{k+1}\geq 0
$$
i.e.
$$
\frac{(k+1)^2-k-k(k+1)}{k(k+1)^2}=\frac{1}{k(k+1)^2}\geq 0.
$$
So it holds.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Sum of all numbers x such that $(3x^2+9x-2012)^{(x^3-2012x^2-10x+1)} = 1$? What is the sum of all $x$ such that $(3x^2 + 9x - 2012)^{(x^3-2012x^2-10x+1)} = 1$?
| Note that $x^3-2012x^2-10x+1$ has no rational root (it would have to be $\pm1$, which can be checked explicitly).
Also note that yb polynomial division
$$\begin{align}x^3-2012x^2-10x+1 &= (3x^2+9x-2012)\cdot\frac{x-2015}3 + \frac{20117 x -4054177}3, \\
x^3-2012x^2-10x+k &= (3x^2+9x-2013)\cdot\frac{x-2015}3 + 6706 x -13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/353362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Residue Integral
Verify the integral with the aid of residues:
$$\int^{\infty}_0\frac{x^2+1}{x^4+1}dx=\frac{\pi}{\sqrt 2}$$
I got:
$f(z)=\frac{z^2+1}{z^4+1}$ and now I must find the residues for $f(z)$ and I got that the poles are: $e^{i\frac{\pi}{4}}$ and $e^{i\frac{3\pi}{4}}$. But I do not know how to finish.
| An easy way to get the solution is just to use partial fractions in Calculus. Here is the solution:
\begin{eqnarray*}
\int_0^\infty\frac{x^2+1}{x^4+1}dx&=&\int_0^\infty\frac{x^2+1}{(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)}dx\\
&=&\frac{1}{2}\int_0^\infty\frac{1}{x^2+\sqrt{2}x+1}dx+\frac{1}{2}\int_0^\infty\frac{1}{x^2-\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/353429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Show that $\frac{1}{1+x}H(\frac{x}{1+x})=\sum^\infty_{k=0}[\Delta^kh_0]x^k$ For a sequence $\{h_n\}_{\geq 0}$, let $H(x)=\sum_{n\geq0}h_nx^n$. Show that:
$$\frac{1}{1+x}H(\frac{x}{1+x})=\sum^\infty_{k=0}[\Delta^kh_0]x^k$$
What I did was that by proving the $$\Delta^k h_o=\sum^k_{j=0}(-1)^{k-j}{k \choose j}h_j$$
But no... | What we seek to show here is that
$$\sum_{k=0}^n {n\choose k} (-1)^{n-k} h_k
= [z^n] \frac{1}{1+z} H\left(\frac{z}{1+z}\right)$$
where $$H(z) = \sum_{q\ge 0} h_q z^q.$$
The RHS is given by the integral
$$\frac{1}{2\pi i}\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{1+z} H\left(\frac{z}{1+z}\right) \; dz
\\ = \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/354076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Find the Laurent expansion in powers of $\,z\,$ and $\,1/z\,$ $f(z)=\large\frac{z+2}{z^2-z-2}$ in $1<|z|<2$ and then in $2<|z|<\infty$
I am unsure how the two different regions of $z$ affect the series expansion.
Any help would be appreciated.
| Hints:
$$1<|z|<2\implies \begin{cases}\frac{1}{2}<\frac{1}{|z|}<1\\{}\\\frac{|z|}{2}<1\end{cases}\;\;\;,\;\;\;\;\text{so}$$
$$\frac{z+2}{(z-2)(z+1)}=\frac{1}{3}\left(\frac{4}{z-2}-\frac{1}{z+1}\right)=$$
$$=-\frac{1}{3}\left(\frac{2}{1-\frac{z}{2}}+\frac{1}{z}\frac{1}{1+\frac{1}{z}}\right)=-\frac{1}{3}\left(4\left(1+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/354345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ a+b+c = \frac{9}{2}$ and $a,b,c>0$, then what is the minimum value of $\frac{a}{b^3+54}+\frac{b}{c^3+54}+\frac{c}{a^3+54}$
If $a+b+c = \dfrac{9}{2}$ and $a,b,c>0$, then what is the minimum value of $$\dfrac{a}{b^3+54}+\dfrac{b}{c^3+54}+\dfrac{c}{a^3+54} \qquad ?$$
My try:
$$\begin{align*}
\frac{a}{b^3+54}+\frac... | Let $\sum$ denote the cyclic sum in the answer below. Also, let $f$ denote the expression to be minimised. Then,
$$f = \sum \frac{a}{b^3+54} = \frac{1}{54}\left(\sum a - \frac{ab^3}{b^3+54}\right)$$
By AM-GM, $\qquad b^3+54 = b^3 + 27 + 27 \ge 27 b$
So $$54f \ge \frac{9}{2} - \frac{1}{27} \sum ab^2$$
We prove below ... | {
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"url": "https://math.stackexchange.com/questions/357988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find this limit: $\lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2$ $$
\lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2
$$
The answer is
$$
\frac{-3}{2}
$$
according to Wolfram alpha.
| first, substitute : $t=x^2$, you get :
$$\lim_{t\to +\infty} \sqrt{t^2-3t-1}-t=\lim_{t\to \infty} \frac{-3t-1}{\sqrt{t^2-3t-1}+t}=\lim_{t\to \infty} \frac{-3-\frac{1}{t}}{\sqrt{1-\frac{3}{t} -\frac{1}{t^3}}+1} =\frac{-3}{2}. $$
Explanation : first we use this identity for $a\neq -b$ : $a-b =\frac{a^2-b^2}{a+b}$, then w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/359441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Angle between two 3D vectors is not what I expected. Using the definition of the dot product, you can find the angle between two vectors. I am experiencing an unexpected result, so my question is where did I go wrong.
I have two unit vectors in 3 dimensions. So the angle between these vectors would be just the inverse ... | Hint: Calculate
$\displaystyle \cos \theta = \frac{a . b}{|a||b|}.$
What do you get?
$\displaystyle a . b = \left(\frac{\sqrt{3} \sqrt{2}}{3 \times 2}\right) + \left(\frac{\sqrt{3} \sqrt{2}}{3 \times 2}\right) + (0) = \frac{\sqrt{6}}{3} = \sqrt{\frac{2}{3}}$
$\displaystyle |a| .|b| = \left|~\sqrt{\left(\frac{\sqrt{3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/360998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding this solution to a recurrence relation So, I know that the recurrence relation $a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3$ with $a_0 = 1$ and $a_1 = 4$ has the solution of $a_n = -4(2^n) - n^2 / 4 - 5n / 2 + 1/8 + (39/8)(3^n)$. I just wanted to know how we arrived at this solution. Thank you!
| Write:
$$
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 4 \cdot 2^n + n + 5 \quad a_0 = 1, a_1 = 4
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$. If you multiply the recurrence by $z^n$ and sum over $n \ge 0$ you get:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 4 \frac{A(z) - a_0}{z} - 3 A(z)
+ \frac{4}{1 - 2 z} + \frac{z}{(1 - z)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Can someone check the solution to this recurrence relation? Here's the recurrence relation: $a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3$ with $a_0 = 1$ and $a_1 = 4$
Here's the solution:Write:
$$
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 2^n + n + 3 \quad a_0 = 1, a_1 = 4
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$. If you multiply... | But
$a_{n+2}=4a_{n+1}−3a_n+2^{n+2}+(n+2)+3\,$!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/364671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Iterative recurrence.. Iteration method Here is the Equation and how far I got into solving this problem using the iteration method:
$$T(1) = 8 \\
T(n) = 3T(n-1) - 15$$
Iterations:
$i=1, $
$$T(n) = 3(3T(n-2) - 15) -15$$
$i=2, $
$$ 3(3(3T(n-3) - 15) -15) - 15$$
$i=3,$
$$3(3(3(3T(n-4) - 15) -15) - 15) - 15$$
$i=4,$
$... | It is easier to use generating functions. Define $g(z) = \sum_{n \ge 0} T(n + 1) z^n$. Write:
$$
T(n + 2) = 3 T(n + 1) - 15 \quad T(1) = 8
$$
Multiply by $z^n$ and add over $n \ge 0$:
$$
\frac{g(z) - T(1)}{z} = 3 g(z) - 15 \frac{1}{1 - z}
$$
From here:
$$
g(z) = \frac{8 - 23 z}{1 - 4 z + 3 z^2}
= \frac{1}{2} \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the closest point to the origin On the line,Find the closest point to the origin
$y=\frac{2}{\sqrt{x}}$
What I did so far is :
First point is : $(x,\frac{2}{\sqrt{x}})$ and point two is : (0,0)
$d=\sqrt{x^2+(\frac{2}{\sqrt{x}})^2}$
then - > $d' = 2x-\frac{4}{{x^2}}$
I found the X but what then? I do not think I ... | You are almost there. First note that $y = \dfrac2{\sqrt{x}}$ is not a line. If $r$ is the distance from the origin, we have
$$r^2 = x^2 + \left(\dfrac2{\sqrt{x}} \right)^2 = x^2 + \dfrac4x$$
Now setting $\dfrac{dr^2}{dx} = 0$, we get that
$$2x - \dfrac4{x^2} = 0 \implies x^* =\sqrt[3]{2}$$
Hence, the value of $r^2$ at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/369022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find expansion around $x_0=0$ into power series and find a radius of convergence My task is as in the topic, I've given function $$f(x)=\frac{1}{1+x+x^2+x^3}$$
My solution is following (when $|x|<1$):$$\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)+(x^2+1)}=\frac{1}{1-(-x)}\cdot\frac{1}{1-(-x^2)}=$$$$=\sum_{k=0}^{\infty}(-x)^k\c... | Use the Cauchy product:
$$\sum_{k=0}^\infty a_kx^k\cdot \sum_{k=1}^\infty b_kx^k=\sum_{k=0}^\infty c_kx^k$$
where $$c_k=\sum_{n=0}^k a_n\cdot b_{k-n}$$
In your case: $a_k=(-1)^k$ and $$b_k=\begin{cases}0 & ,k =2l+1 \\(-1)^k&,k=2l\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove that $x^{12}-x^9+x^4-x+1>0$ for all $x\in \mathbb{R}$
Prove that the expression $x^{12}-x^9+x^4-x+1>0\; \forall x\in \mathbb{R}$
My try:: Using Interval method::
$\bullet \; $If $x\leq 0$, Then $x^{12}-x^9+x^4-x+1>0$
$\bullet \; $If $0<x\leq 1$, Then $x^{12}+x^4.(1-x^5)+(1-x)>0$
$\bullet \; $If $x>1$, Then $x^9... | $$\begin{aligned}
\frac{1}{4}x^{12}+x^6&\ge |x^9|\ge x^9\\
\frac{3}{4}x^{12}+\frac13&\ge x^6\\
x^4+\frac14&\ge x^2\\
x^2+\frac14&\ge |x|\ge x
\end{aligned}$$
Add them all up, you get that
$$x^{12}-x^9+x^4-x+1\ge \frac16.$$
Another proof:
$$\begin{aligned}
\frac{x^{12}+x^{12}+x^{12}+1}4&\ge \sqrt[4]{x^{36}}\ge x^9\\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this resu... | Yikes. I am hopind I didn't do any mistake, but that would be a miracle :)
Let $x= \left( p+q\sqrt{3}\right) ^{1/3}\,;\, y= \left(
p-q\sqrt{3}\right) ^{1/3}$.
Then
$$xy= (p^2-3q^2)^\frac{1}{3} $$
Hence
$$2p=x^3+y^3=(x+y)^3-3xy(x+y)=n^3-3n\sqrt[3]{p^2-3q^2}$$
This shows that $\sqrt[3]{p^2-3q^2}$ must be rational, hence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/374619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 6,
"answer_id": 2
} |
integrating fraction/completing the square Does anyone know how to integrate the following?
$\frac{dx}{9x^2 + 6x + 17}$
I have been trying for ages and cannot get an answer anywhere close to the answer I get on maple?
| Without many words and since you've already been given some ideas. We're looking for an arctangent (general set up for this particular case of integrals), so try to justify each step:
$$9x^2+6x+17=9\left(x^2+\frac{2}{3}x\right)+17=9\left(x+\frac{1}{3}\right)^2+16=$$
$$16\left[1+\left(\frac{3}{4}x+\frac{1}{4}\right)^2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/375140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to compute the integral $\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$ I want to compute this integral $$\displaystyle\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$$
What I did was the following. I substituted $x=t^{6}$, so that my $dx= 6t^{5} \ dt$ and so the integral changes to $$\int_{0}^{\infty} \frac{t^2}{1... | Substitute $x^{1/3} = t$, i.e., $x = t^3$, i.e., $dx = 3t^2 dt$. Hence,
$$I = \int_0^{\infty} \dfrac{t}{1+t^6} 3t^2 dt = 3 \int_0^{\infty} \dfrac{t^3}{1+t^6}dt$$
$$\int_0^{\infty} \dfrac{t^3}{1+t^6}dt = \int_0^{1} \dfrac{t^3}{1+t^6}dt + \int_1^{\infty} \dfrac{t^3}{1+t^6}dt$$
$$\int_1^{\infty} \dfrac{t^3}{1+t^6}dt = \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A continued fraction involving Bernoulli numbers Let $B_n$ be the Bernoulli numbers. Then, we can write a function $F(x)$, expressed as a continued fraction involving those $B_n$, such that it gives the form,
$$ \displaystyle
\displaystyle F(x)= B_0+ \cfrac{(1/x)}{B_1+ \cfrac{(2/x)}{B_2+ \cfrac{(3/x)}{B_3+ \cfra... | Let's start with your definition:
$$F(x)= B_0+ \cfrac{(1/x)}{B_1+ \cfrac{(2/x)}{B_2+ \cfrac{(3/x)}{B_3+ \cfrac{(4/x)}{B_4+ \cfrac{(5/x)}{B_5+ \cfrac{(6/x)}{B_6+ \cfrac{(7/x)}{B_7+ \cfrac{(8/x)}{\dots}}}}}}}}$$
The only non-zero odd Bernoulli number is $B_1$, so we have
$$F(x)= B_0+ \cfrac{(1/x)}{
B_1+ \cfrac{(2/x)}{
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simultaneous solution(s) to $a^2+4b^2+4ab=0$ and $a^2+4b^2+32+16a-8b=0$? Could you tell me just how should I solve this system:
$$
a^2+4b^2+4ab=0\\
a^2+4b^2+32+16a-8b=0
$$
I can't remember the proceeding and it's driving me crazy.
Thanks a lot
| We have:
$$
a^2+4b^2+4ab=0 \tag{1}$$
$$
a^2+4b^2+32+16a-8b=0\tag{2}
$$
$\bf (I)$ Subtract equation $(2)$ from $(1)$:
$$\begin{align}
& a^2+4b^2+4ab & =0\\
- & a^2+4b^2+32+16a-8b &=0 \\
& \hline \\
= & 4ab -16a + 8b - 32 & = 0 \\
4 & (ab - 4a + 2b - 8) & = 0 \\ \\
= & ab - 4a + 2b -8 = 0 \tag{3}\\
\end{align}
$$
$\bf (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2... | $$
\left(2 \over 5\right)^{2/5}
=
0.69314\color{#ff0000}{\Large 4}843155146\ldots\,,
\qquad\qquad
\ln\left(2\right)
=
0.69314\color{#ff0000}{\Large 7}180559945\ldots
$$
So$\ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/380302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "95",
"answer_count": 8,
"answer_id": 4
} |
Trig and algebra problem: Finding sides of a triangle Let $ABC$ be a triangle such that $\angle ACB = \pi/6$ and let $a,b,c$ denote the lengths of the sides opposite to $A,B,C$, respectively. What are the value(s) of x for which $a = x^2 + x + 1, b = x^2-1$ and $c = 2x+1$?
I used law of cosines, then saw that I should ... | HINT:
Applying Law of Cosines, $$\cos \frac\pi6=\frac{a^2+b^2-c^2}{2ab}$$
Now, $$b^2+a^2-c^2=(x^2-1)^2+(x^2+x+1)^2-(2x+1)^2$$
$$=(x^2-1)^2+(x^2-x)(x^2+3x+2)\text{ applying } (a^2-b^2)=(a+b)(a-b)\text{ for the last two terms}$$
$$=(x^2-1)^2+x(x-1)(x+1)(x+2)$$
$$=(x^2-1)\{x^2-1+x(x+2)\}=(x^2-1)(2x^2+x-1)$$
As $x^2-1\ne0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Homework: Maclaurin Power Series Help I'm trying to find the Maclaurin Power Series for
$$f(x)=\frac{3x-8}{3x^2+5x-2}$$
but each degree of differentiation gets more complex with no discernible pattern. Any help is appreciated, thanks.
| $$\begin{align*} f(x)&=\frac{3x-8}{3x^2+5x-2}=\frac{3x-8}{(x+2)(3x-1)}=\{\text{partial fraction decomposition}\}=\\
&=\frac{2}{2+x}-\frac{3}{3x-1}=\frac{1}{1-\left(-\frac{x}{2}\right)}+\frac{3}{1-3x}=\{\text{geometric series for }|x|<\frac{1}{3}\}=\\
&=\sum_{n=0}^\infty\left(-\frac{x}{2}\right)^n+3\sum_{n=0}^\infty(3x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\sin x \cdot \cos x$ I've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different?
$ \int \sin x \cos x dx $
1) via subsitution $ u = \sin x $
$ u = \sin x; du = \cos x dx \Rightarrow \int udu = \frac12 u^2 \Righta... | Note: You are calculating indefinite integral and constants can be anything(they may differ). In fact the general solution to that would be just $C+\dfrac{\sin^2 x}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/381243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Permutation & Combination - how many numbers smaller than $2.10^8$ and are divisible by $3$ can be written by means of the digits $0$,$1$ and $2$ How many numbers smaller than $2\times10^8$ and divisible by $3$ can be written by means of the digits $0$,$1$ and $2$?
Left zero padding not allowed.
I am getting this as -
... | We solve the simpler problem of counting the numbers from $0$ to $10^8-1$ of the right shape that are divisible by $3$. We do allow zero padding to make all the numbers into $8$-digit numbers. This makes no difference to the count.
The same method works for the other half of our interval.
We use a general approach, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/382897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A concise distance problem A falsely simple Euclidian geometry problem:
Points $A$, $B$, $C$ are collinear; $\|AB\|=\|BD\|=\|CD\|=1$; $\|AC\|=\|AD\|$.
What is the set of possible $\|AC\|$ ?
I'm after a concise answer, with reasoning, that would get maximum points to an 11th-grader.
A related question asks an approp... | W.l.o.g. choose coordinates as follows:
\begin{align*}
A &= \begin{pmatrix}0\\0\end{pmatrix} &
B &= \begin{pmatrix}1\\0\end{pmatrix} &
C &= \begin{pmatrix}x\\0\end{pmatrix} &
D &= \begin{pmatrix}\frac{x+1}2\\
\sqrt{x^2-\left(\frac{x+1}{2}\right)^2}\end{pmatrix}
\end{align*}
Point $D$ is choosen on the perpendicular bis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/382978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
If $x^2-xy+2y^2=\frac{a^2}{7}$, find $y'''$.
If $x^2-xy+2y^2=\frac{a^2}{7}$, find $y'''$.
For our 1st derivative we got $$y'=\frac{2x-y}{x-4y}.$$
For the second derivative we got $$y''=\frac{14x^2-14xy+28y^2}{(x-4y)^3}.$$
And for the final answer we got $$y'''=\frac{4(-84x^3+119x^2y-154xy^2-84y^3)}{(x-4y)^5}.$$
Took ... | As Will Jagy noted, $$y''=\frac{2a^2}{(x-4y)^3}.$$
Then $$y'''=2a^2(-3)(x-4y)^{-4}(1-4y')$$ Now $$1-4y'=\frac{4y-8x-4y+x}{x-4y}=\frac{7x}{x-4y},$$ so your $y'''$ is $$y'''=\frac{42a^2x}{(x-4y)^5}.$$
I used your answer for second derivative to find the third derivative and I got $$y'''=\frac{(x-4y)(28x-14y-14xy'+56yy')-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/383896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Another trigonometric proof...? ...sigh..another problem how shall I prove the following?
$$ {\cot A\over1- \tan A} + {\tan A \over 1- \cot A} = 1 + \tan A + \cot A$$
so what now? the following's what I've done:
$$\cot A - \cot^2 A + \tan A- \tan^2 A \over 2 - \tan A - \cot A$$
| $$
\begin{align}
\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}&=\frac{\cos A\cot A - \sin A \tan A}{\cos A-\sin A}
\\\\&=\frac{\frac{\cos^2 A}{\sin A} - \frac{\sin^2 A}{\cos A}}{\cos A-\sin A}
\\\\&=\frac{\cos^3 A - \sin^3 A}{(\cos A \sin A)(\cos A-\sin A)}
\\\\&=\frac{\cos^2 A + \sin^2 A + \sin A\cos A}{\cos A \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/385582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
number of solutions to an equation? Given $x$ and $y$ are multiples of $2$ satisfying
$$x^2 - y^2 = 27234702932$$
Find the number of solutions to $x$ and $y$.
| As $27234702932=2^2\cdot181\cdot37616993$ where the last two factors are primes
If $x=2X,y=2Y$
$$X^2-Y^2=181\cdot37616993$$
If $X^2-Y^2=p\cdot q$ where $p,q$ are primes,
the possible cases for $X+Y,X-Y$ are $\pm pq, \pm 1$ and $\pm p,\pm q $
For example, if $X+Y=1,X-Y=pq$
The number of positive factors of $p\cdot q$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Improper integral $\sin(x)/x $ converges absolutely, conditionally or diverges? Improper integral of $\sin(x)/x $ converges absolutely, conditionally or diverges?
We have
$$\int_1^{\infty}\frac{\sin x}{x}\text{d}x$$
Integrating by parts
$$u=\frac{1}{x}$$
$$\text{d}u=-\frac{1}{x^2}\text{d}x$$
$$\text{d}v=\sin x\;\text{d... | Let $N \in \Bbb N, N > 1$, we have:
\begin{align}
\int_0^{2\pi N} \left|\frac{\sin x}{x}\right|\,dx &= \sum_{n=0}^{N-1} \int_{2\pi n}^{2\pi(n+1)} \left|\frac{\sin x}{x}\right|\,dx \\
&\ge \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{2\pi n}^{2\pi(n+1)} \left|\sin x\right|\,dx \\
&= \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "48",
"answer_count": 4,
"answer_id": 0
} |
Using Spherical coordinates find the volume: Inside the surfaces $z=x^2+y^2$ and $z=\sqrt{2-x^2-y^2}$
I integrated over the ranges:
$0 \leq \theta \leq 2\pi$
$ 0 \leq \phi \leq \frac{\pi}{2}$
$0 \leq r \leq \sqrt{2}$
I get $\frac{\pi}{2}(4\sqrt{2} -4).$
There answer is the same except a $-\frac{7}{2}$ instead of the 4 ... | This problem is actually better suited for cylindrical coordinates:
$$\begin{align}\int_{0}^{2\pi}\int_0^1\int_{r^2}^{\sqrt{2-r^2}}rdzdrd\theta&=2\pi\int_0^1(r\sqrt{2-r^2}-r^3)dr\\&=2\pi(-\frac{1}{3}(2-r^2)^{3/2}-\frac{r^4}{4})\mid_{0}^1\\&=2\pi(-\frac{1}{3}-\frac{1}{4}+\frac{2\sqrt{2}}{3})\\&=\frac{\pi}{3}(4\sqrt2-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluating Complex Line Integrals Calculate $\int_{\gamma}\frac{\Re(z)}{z-\frac{1}{2}}dz$ and $\int_{\gamma}\frac{\Im(z)}{z-\frac{1}{2}}dz$ when $\gamma$: $|z|=1$ is positively oriented.
This is what I have tried to do, starting with the first line integral.
Since $\frac{\Re(z)}{z-\frac{1}{2}}$ is not analytical/holo... | You have some problem on partial fraction $\displaystyle \frac{z^2 + 1}{z(z-1/2)} \neq 1 - \frac{\frac z 2 + 1}{z(z-1/2)}$.
This should be $\displaystyle \frac{z^2 + 1}{z(z-1/2)} = 1 + \frac{\frac z 2 + 1}{z(z-1/2)} = 1+\frac{5/2}{z - 1/2} - \frac 2 z$.
And this should give you $\displaystyle \frac{2 \pi i}{2} ( \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of a specific determinant.
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.
$$A_2
\begin{pmatrix} 1& 1\\
1& 2
\end{pmatrix};
A_3 = \begin{pmatrix} 1& 1& 1\\
1& 2& 2\\
1& 2& 3
\end{pmatrix};
A_4 = \begin{pmatrix} ... | If you consider the matrix
$$
B_n=\begin{bmatrix}
1&0&\cdots&0\\
1&1&0&\cdots\\
&&\cdots\\
1&1&\cdots &1
\end{bmatrix}
$$
(i.e. the $i,j$ entry of $B_n$ is $1$ is $i\geq j$ and $0$ otherwise), then the $k^{\rm t h}$ column of $A_n$ is obtained by adding the first $k$ columns of $B_n$. So
$$
\det A_n=\det B_n=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Integrate by parts: $\int \ln (2x + 1) \, dx$ $$\eqalign{
& \int \ln (2x + 1) \, dx \cr
& u = \ln (2x + 1) \cr
& v = x \cr
& {du \over dx} = {2 \over 2x + 1} \cr
& {dv \over dx} = 1 \cr
& \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr
& = x\ln (2x + 1) - \int 1 - {1 \over... | Here is a cute variant. Let $u=\ln(2x+1)$ and let $dv=dx$. Then $du=\frac{2}{2x+1}$ and (this is the cute part) we can take $v=x+\frac{1}{2}$. It follows that
$$\int \ln(2x+1)\,dx=\left(x+\frac{1}{2}\right)\ln(2x+1)-\int \left(x+\frac{1}{2}\right)\frac{2}{2x+1}\,dx.$$
But the remaining integrand is just $1$! It follows... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
derive using the chain rule Given the polinomyal $f(x)=\frac{x^3}{(4-x^2)^3}$ find $f'(x)$
So, If I try to derive this, first I must to apply the chain rule in the denominator and then derive of the division (...)
$$f'(x)=\frac{x^3}{3(4-x^2)^2(-2)} = \frac{x^3}{-6(4-x^2)^2}$$
(...)?
Or there is another way to do this?
| You're doing wrong. A good way to do this when still not expert in derivatives, is to write
$$
f(x)=\frac{g(x)}{h(x)}
$$
where
$$
g(x)=x^3,\qquad h(x)=(4-x^2)^3.
$$
Then, first of all, you have to differentiate the quotient:
$$
f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{(h(x))^2}
$$
So you need to compute $g'(x)$ and $h'(x)$. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
solution to a root inequality I have the inequality
$$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc} \geq \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab}.$$I tried to do $u=a^2+b^2+c^2$ and $v=ab+ac+bc$ and $x=a^2+2bc$, $y=b^2+2ac$, $z=c^2+2ab$ ...but I did not find any solution. Any help is appreciated.
| We need to prove that
$$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc}\geq\sum\limits_{cyc}\sqrt{a^2+2bc}$$ or
$$\sum\limits_{cyc}\left(\sqrt{a^2+b^2+c^2}-\sqrt{c^2+2ab}\right)\geq2\left(\sqrt{a^2+b^2+c^2}-\sqrt{ab+ac+bc}\right)$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2}{\sqrt{a^2+b^2+c^2}+\sqrt{c^2+2ab}}\geq\sum\limits_{cyc}\frac{(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
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