Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Rewrite $\frac{125}{\left(\frac{1}{625}\right)^{-x-3}}=5^3$ in a common base then solve for $x$ I am to rewrite $\frac{125}{(\frac{1}{625})^{-x-3}}=5^3$ and then solve for x. My textbooks solutions section says the solution is -3. I gave it a shot and got 3.25. Here is my working:
$$\frac{125}{\left(\frac{1}{625}\right... | Your first mistake starts here:
$$5^3 \times 5^{4(x-3)}=5^3$$
This is not correct. Because,
$$\begin{align}\left(\frac{1}{625}\right)^{-x-3}&=\left(5^{-4}\right)^{-x-3}\\
&=5^{-4\times (-x-3)}\\
&=5^{4(x+3)}.\end{align}$$
Your second mistake is:
$$5^3 \times 5^{4(x-3)}=5^3$$
$$5^{12x-36}=5^3$$
The correct one is as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determine $\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$ Determine $$\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$$ | Hint:
For small enough $x,y$ such that the ordered pair $(x,y)\ne (0,0)$, we have $|\sin(xy^2)|\lt |xy^2|$
$|y|=\sqrt{y^2}\le\sqrt{y^2+x^2}$ and so by similar arguments, we have:
$\big|\frac{x^2 y\sin(xy^2)}{(x^4+y^4)\sqrt{x^2+y^2}}-0\big|\le \frac{|x^3|y^2}{x^4+y^4}\le \frac{|x|x^2}{\sqrt{x^4+y^4}}\le |x|\le \sqrt{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
If $P_n(x)=\sum_{k=0}^{[(n-1)/2]}(-1)^k\ _nC_{2k+1}\ x^k$, find $A(x)$ and $B(x)$ such that $P_{n+2}(x)=A(x)P_{n+1}(x)+B(x)P_n(x)$ A high school math problem:
For $n \in N$ (the natural number set), let us define the polynomial $P_n(x)$ as follows:
$$P_n(x) = \sum_{k=0}^{[(n-1)/2]}(-1)^k \ {}_nC_{2k+1} \ x^k$$
where $... | We obtain
\begin{align*}
\color{blue}{P_n(x)}&\color{blue}{=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^k\binom{n}{2k+1}x^k}\tag{1}\\
&=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}i^{2k}\binom{n}{2k+1}\left(\sqrt{x}\right)^{2k}\tag{2}\\
&=\frac{i\sqrt{x}}{i\sqrt{x}}\sum_{k=0}^{\left\lfloor\frac{n-1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4135453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ to prove $3^{n+2}$ does not divide $2^{3^n}+1$ Prove for all natural numbers $n$ such that $2^{3^n}+1$ is not divisible by $3^{n+2}$.
My working for not divisible: Induction proof
Base case: n = 0
$$3^{0+2}=9$$
and
$$2^{3^0}+1=3$$
9 cannot divide 3, so base case is true.
Assume n=k is ... | Hint: It is easy to see that $$3~|~(2^{3^k})^2-2^{3^k}+1$$ but $$9\nmid(2^{3^k})^2-2^{3^k}+1.$$
Note: To complete the proof, use the Unique Factorization Theorem of integers and count the multiplicity of the prime factor $3$. (This theorem requires only the Division Algorithm to prove.)
An alternative formal way to sho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Simplifying the sequence defined by $A_0=6$ and $A_n=\frac{8}{9}A_{n-1}+\frac{6}{5}(\frac{20}{9})^n$, using sigma notation
I have $A_0 = 6$ and
$$A_n = \left(\frac{8}{9}\right)A_{n-1} + \left(\frac{6}{5}\right) \left(\frac{20}{9}\right)^n$$
and I want to simplify it with sigma notation.
I have it up to
$$\begin{align... | (I have done this type of problem
so often,
I could almost do it in my sleep.
However,
each time I do it a little differently
and from scratch,
so here is the
latest version.)
If
$a_n
=ua_{n-1}+b_n
$
then,
dividing by $u^n$,
$\dfrac{a_n}{u^n}
=\dfrac{ua_{n-1}}{u^n}+\dfrac{b_n}{u^n}
=\dfrac{a_{n-1}}{u^{n-1}}+\dfrac{b_n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the limit $\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$ $$I=\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+n+1}}+\frac{1}{\sqrt{n^2+n+2}}+\cdots+\frac{1}{\sqrt{n^2+n+n}}\right)^n$$
I tried $$\frac{n}{\sqrt{n^2+n+n}}\leq\frac{1}{\sqrt{n^2+... | If you like generalized harmonic numbers
$$S_n=\sum_{k=1}^n \frac{1}{\sqrt{n^2+n+k}}=H_{n^2+2 n}^{\left(\frac{1}{2}\right)}-H_{n^2+n}^{\left(\frac{1}{2}\right)}$$
Using twice the asymptotics
$$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 p^{1/2}}-\frac{1}{24 p^{3/2}}+O\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
} |
Nonhomogeneous First order differential equation I'm trying to understand what is wrong with this solution, since I'm not getting the same answer in Matlab
$y'-xy=xy^{3/2}\, ,y(1)=4$
\begin{align*}
y'-xy=&xy^{3/2}&\\
\dfrac{dy}{dx}=&x(y+y^{3/2})&\\
\dfrac{dy}{(y+y^{3/2})}=&x\, dx&\\
-2\ln \dfrac{1+\sqrt{y}}{\sqrt{y}}=&... | I think $$y=\frac {1}{(1-e^{-\frac {x^2}{4}+c})^2}$$ instead of what you wrote. $y$ cannot be negative anyway, as $\sqrt y$ is present in the equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Perfect Square With Two Integer Variables I am trying to solve a number theory problem in general form. However, I got stuck in the following step:
$a,b,n \in \mathbb Z^{+}$ for which values of $n$, this equation is solvable $\frac{(n+1)(n+2a)}{2} = b^2$ ?
Can we make a general statement about $n$ ? By the way I have t... | Clearing denominators and expanding, the question is to find all $n\in\Bbb{Z}$ for which there exist $a,b\in\Bbb{Z}$ such that
$$2b^2=(n+1)(n+2a).$$
So let $n\in\Bbb{Z}$ be given and suppose $a,b\in\Bbb{Z}$ satisfy the above.
Clearly $2b^2$ is divisible by $n+1$, say
$$2b^2=m(n+1),$$
for some $m\in\Bbb{Z}$. It follows ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_{0}^{\infty} \frac{\left(\frac{\pi x}{2}-\log (x)\right)^3}{\left(x^2+1\right)^2 \left(\log^2(x)+\frac{\pi ^2}{4}\right)} = \pi$ proof Today @integralsbot at twitter posted an interesting integral relation for complicated integral that gives $\pi$ as a result. I don't know how this bot works or where doe... | Let
$$f(x)= \frac{\pi^3 x^3-6 \pi ^2 x^2 \log (x)-3 \pi ^3 x+2 \pi ^2 \log (x)}{2 \left(x^2+1\right)^2 \left(4\log ^2(x)+\pi ^2\right)} $$ that is to say
$$f(x)=\frac{\pi ^2 \left(\pi x \left(x^2-3\right)-2\left(3x^2-1\right) \log
(x)\right)}{2 \left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}$$
and write
$$I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
How do I find all values $x$ such that a vector is a linear combination of a nonempty set of vectors in vector space $\mathbb{R^3}$? In vector space $\mathbb{R^3}$.
Find all values $x$ such that $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}$ $\in$ span $\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\... | Here's another approach. Just for your own entertainment.
Using techniques from multivariable calculus you can write $$\text{span}\Bigg\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}\Bigg\}=\big\{(x,y,z)\in \mathbb{R}^3:2x+y-z=0\big\}$$ The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Summation to infinity with a general pattern $$
\sum_{n=1}^{\infty} \frac{(n-1)!}{(x+1)(x+2)..... (x+n)}.
$$
I tried solving it by inserting some values
$\frac{1}{x+1}$ +$\frac{1}{(x+1)(x+2)}$+$\frac{2}{(x+1)(x+2)(x+3)}$
$\frac{1}{x+1}$+$\frac{1}{x+1}$-$\frac{1}{x+2}$+$\frac{1}{x+1}$-$\frac{1}{x+2}$+$\frac{1}{x+3}$-$\f... | A more general result is true.
$\textbf{Lemma:}$ Let $(a_n)_{n\ge 1}$ be a positive sequence. Then for $x \ne 0$,
$$
\sum_{n=1}^{\infty} \frac{a_1a_2\cdots a_{n-1}}{(x+a_1)(x+a_2)\cdots(x+a_n)} = \frac{1}{x}
$$
Proof. Define $A_0 = \frac{1}{x}$ and $$A_N = \frac{a_1a_2\cdots a_{N}}{x(x+a_1)(x+a_2)\cdots(x+a_N)}$$
No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}$
My question:
Let $a,b,c$ be positive real numbers satisfy $a+b+c=3.$
Prove that $$\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}.$$
I have tried to change the LHS to $$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb... | Partial answer hint :
For $x,y,z\in[0.5,2]$ we have the inequality :
$$\frac{x}{y+z^{2}}-\frac{1}{x+y+z}\frac{x}{y+z}\ge0$$
Summing and using the constraint gives the inequality .
Now we have one variable less than $1/2$ come back further .
Extended comment :
I have almost if finish if $\max({a,b,c})=a\ge 2$ the ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations a... | We have
$$
(\mathcal{P}(N))^{2N + 2} = \left( {\frac{{2N}}{{2N + 1}}} \right)^{2N + 1} 2 = \frac{1}{{\left( {1 + \frac{1}{{2N}}} \right)^{2N} }}\frac{2}{{1 + \frac{1}{{2N}}}} < \frac{4}{9} \cdot 2 < 1,
$$
since $
{\left( {1 + \frac{1}{n}} \right)^n }
$ is increasing monotonically to $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Differential equation with really strange solution... I wish to solve the following differential equation $$\frac{dy}{dx}=\frac{\sqrt{y+7}}{(x+3)(x+8)}$$ with initial condition $y(0)=2$. I think my method is correct but my final solution seems like a really strange answer... This is what I got:
$$\Rightarrow y=2+\frac{... | You should have added $C$, the constant, before squaring. That made all the difference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the perimeter of triangle $ABC$. In triangle $ABC$, point $M$ is the midpoint of side $BC$. Suppose that the point
$D$ and $E$ on the $CA$ side so $CD = DE$ with $D$ between $C$ and $E$.
Suppose also points $F$ and $G$ on the side $AB$ so that $AM$, $BD$, and $CF$ meet at one point. Likewise with $AM$, $BE$, and $... | Let $AG/AB = AE/AC = r$. Then because $F$ is the midpoint of $GB$, we have $$\frac{AF}{AB} = \frac{AG + (AB - AG)/2}{AB} = \frac{1+r}{2}.$$ Similarly, $$\frac{CE}{AC} = \frac{AC - AE}{AC} = 1 - r, \quad \frac{DC}{AC} = 1 - \frac{1+r}{2} = \frac{1-r}{2}.$$ Therefore, $$17 = AF + BM + CD = \frac{1+r}{2} c + \frac{a}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Consider hyperboloid $z^2 = x^2 − 4y^2 + 5$. Find an equation of the line which belongs to this hyperboloid and the point $(0, 1, 1)$. Consider hyperboloid $z^2 = x^2 − 4y^2 + 5$. Find an equation of the line which belongs to this hyperboloid and the point $(0, 1, 1)$.
This is what I have so far:
$I(s) = OP + sv$
Letti... | We'll assume $s \neq 0$ to simplify things since we know $I(0) = (0,1,1)$. Notice then that
\begin{align*}
(cs+1)^2=(as)^2 -4(bs+1)^2+5 &\implies s \left(sa^2 - 4 b^2 s - 8 b - c^2 s - 2 c\right) = 0\\
& \implies s\left(a^2 -4b^2 -c^2\right) = 8b+2c \tag{1}
\end{align*}
From here, since it should hold that $I(s)$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4158405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Looking for other approaches to find the height $AH$ in triangle $ ABC $ where $A(1,5)$ , $B(7,3)$, $C(2,-2)$
We have a triangle with vertices $A(1,5)$ , $B(7,3)$, $C(2,-2)$. What
is the length of the height $AH$ in the triangle $ABC$ ?
$1)4\qquad\qquad2)3\sqrt2\qquad\qquad3)5\qquad\qquad4)4\sqrt2$
This is a problem ... | Area of triangle is \begin{align}\frac12|x_1y_2+x_2y_3+x_3y_1-x_2y_1-x_3y_2-x_1y_3|&=\frac12|3-14+10-35-6+2|\\
&=20 \end{align}
Hence
$$AH \times BC=40$$
$$AH \times \sqrt{5^2+5^2}=40$$
$$5\sqrt{2} AH = 40$$
$$AH=\frac{8}{\sqrt2}=4\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4160028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$\measuredangle C=120^\circ$ and two altitudes $AH$ and $BD$ are altitudes of $\triangle ABC$ and $\measuredangle ACB=120^\circ$. If $S_{\triangle HCD}=\dfrac{15\sqrt3}{4},$ find the area of $\triangle ABC$.
$$S_{\triangle HCD}=\dfrac12\cdot CH\cdot CD\cdot\sin\measuredangle HCD=\dfrac{\sqrt3}{4}CH\cdot CD=\dfrac{15\s... | As $\angle BCD = 60^0, BD = CD \tan 60^0 = CD \sqrt3$
$CH = AC \cos 60^0 \implies AC = 2 \cdot CH$ as $\angle ACH = 60^0$
Area $\triangle ABC = \dfrac{1}{2} \cdot AC \cdot BD = \dfrac{1}{2} \cdot 2 \cdot CH \cdot CD \sqrt3 = 15 \sqrt3$
(as you already found that $CH \cdot CD = 15$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Computing probabilities for each value of a random variable Assume you have a bag of 6 marbles, 2 of which are red and 4 of which are blue, and let the random variable Y represent the number of red marbles drawn after 3 are taken from the bag.
I'm trying to compute the probability for each value of Y.
What I've attempt... | I have a slightly different way of doing it, and my results are the same as yours.
$P(Y=0)=\frac{4\times 3\times 2}{6\times 5\times 4}=\frac{1}{5}$
$P(Y=1)=3\times \frac{2\times 4\times 3}{6\times 5\times 4}=\frac{3}{5}$
$P(Y=2)=3\times \frac{2\times 1\times 4}{6\times 5\times 4}=\frac{1}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac {x}2$?
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac
{x}2$?
$1)-3\qquad\qquad2)2\qquad\qquad3)-2\qquad\qquad4)3$
Here is my method:
$$1-\sin x=4+4\sin x\quad\Rightarrow\sin x=-\frac{3}5$$
We have $\quad\sin x=\dfrac{2\tan(\frac ... | Hint:
$ \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$
$1=\sin^2 x + \cos^2 x$
Complete the square... and divide by something... then you should be able to use tangent formula
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Proving upper bound on a term involving binomial coefficients I am trying to show:
$$\binom{n}{k}\left(1-\frac{k}{n}\right)^{2n-2k+2} \left(\frac{k-1}{n}\right)^{2k}\leq \frac{1}{n^2}$$ for $k\in \{2,3,\cdots,n-1\}$ for all $n$
I observe that numerically that is true but analytically, I am not able to look at the ter... | This is not a proof.
$$a_k=\binom{n}{k}\left(1-\frac{k}{n}\right)^{2n-2k+2} \left(\frac{k-1}{n}\right)^{2k}$$
Expanding as series for large $n$ and keeping the first term only, we have
$$a_2=\frac{1}{2 e^4 n^2}+O\left(\frac{1}{n^3}\right)\qquad \qquad
a_3=\frac{32}{3 e^6 n^3}+O\left(\frac{1}{n^5}\right)$$
$$a_4=\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Smallest $s$ such that $\sum_i^n (1-\frac{1}{i})^s<\epsilon$ Take $f(n)$ defined as follows
$$\begin{array}{lll}
g(s,n)&=&\frac{1}{n}\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s\\
f(n)&=&\text{smallest } s \text{ such that } g(s,n)<\epsilon
\end{array}$$
Empirically, $f(n)$ grows linearly in $n$, and in $-\log\epsilon$, i... | First, suppose $s \ge n\log\dfrac{1}{\epsilon}$. Then since $1-\dfrac{1}{k} \le 1-\dfrac{1}{n}$ for $1 \le k \le n$, and $\left(1-\dfrac{1}{n}\right)^n \le \dfrac{1}{e}$ for all $n \ge 1$, we have
\begin{align*}
g(s,n) &= \dfrac{1}{n}\sum_{k = 1}^{n}\left(1-\dfrac{1}{k}\right)^s
\\
&\le \dfrac{1}{n}\sum_{k = 1}^{n}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the area of the part of the sphere $x^2+y^2+z^2=2$ that lies inside the cylinder $y^2+z^2=1$ This is what I have tried so far. I am just not sure whether I got it right or not.
We have $y^2+z^2=1$, then $r=1,\quad x=\sqrt{2-y^2-z^2}\quad\text{and}\quad x=\sqrt{2-r^2}$
$$\frac{\partial{x}}{\partial{y}}=-\frac{y}{\s... | Yes it is correct.
Another method, that gives the answer faster id to use the spherical coordinates $(r,\theta,\phi)$ oriented not around $z$ axis as usual, but around $x$ axis (so that $y^2+z^2 = r^2 \sin^2\theta$).
The sphere is given by the condition $r=\sqrt{2}$. The element of the area of a sphere is $dA = r^2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Interesting Inequality using AM-GM and other identities. Let $a, b, c > 0$. Prove that $$\sqrt{a^2-ab+b^2} + \sqrt{b^2 - bc + c^2} + \sqrt{c^2 - ca + a^2} \le \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b}.$$ This should be solvable with AM-GM and a few other inequalities, but I am a little stuck on this problem.
My idea w... | The inequality
$$\frac{5}{2}\left(a+b+c\right)-3\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\right) \le \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b},$$
doesn't true, example $a = \frac 13,\, b = \frac 1 4,\,c = 1.$
Let $a = \frac{1}{x},\,b = \frac{1}{y},\,c = \frac{1}{z},$ the inequality become
$$x \sqrt{y^{2}-y z+... | {
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"url": "https://math.stackexchange.com/questions/4179434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Computing $\int_{-2}^{2}\frac{1+x^2}{1+2^x} dx$ I am trying to compute the following integral by different methods, but I have not been able to come up with the result analytically.
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx$$
First I tried something like: $2^{x}=e^{x\ln{2}}\Rightarrow u=x\ln{2} \iff x=\frac{u}{\ln{2}}$ $\Ri... | Note that $\frac{1}{1+2^x}= \frac12 - \frac12 \tanh (\frac x2\ln2)
$,
where the odd function $\tanh(\cdot)$ vanishes under integration. Thus
$$\int_{-2}^{2}\frac{1+x^2}{1+2^x}dx= \int_{-2}^{2}\frac12(1+x^2)dx=\frac{14}3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to graph and solve this equation? I'am trying to solve this equation.
\begin{equation}
x^{8}+(x+2)^{8}=2
\end{equation}
What I tried:
\begin{equation}g(x)=x^{8}+(x+2)^{8}\end{equation}
\begin{equation}
\begin{array}{l}
\text { }\\
y=x+1
\end{array}
\end{equation}
\begin{equation}
(y+1)^{8}+(y-1)^{8}=2
\end{equatio... | OK then, let's use calculus.
Consider the function $f(x) = x^8 + (2 + x)^8$. Note that $f(x) \to \infty$ as $x \to \pm \infty$, so the function must achieve a global minimum at a stationary point, i.e. where $f'(x) = 0$.
We have,
$$f'(x) = 8x^7 + 8(x + 2)^7.$$
This is $0$ if and only if
$$8x^7 + 8(x + 2)^7 = 0 \iff (x ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.
An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.
Here is a procedure how to do it analiticaly.
*
*If $T(x_0,y_0)$ is ... | The ellipse divides the plane into three regions.
*
*the exterior, where the sum of the distances to the foci is greater than $c$.
*the ellipse itself, where the sum of the distances to the foci is exactly $c$.
*the interior, where the sum of the distances to the foci is smaller than $c$.
Parametrize the tangent a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by induction that $3 \mid n^4-n^2 \forall n \in \mathbb{Z^+}, n \ge 2$. Proposition: $3 \mid n^4-n^2$ for all $n \in \mathbb{Z^+}, n \ge 2$
My attempt
Lemma: $3 \mid (m-1)m(m+1)$
Proof.
Suppose $m \in \mathbb{Z}$. By the QRT, we have $m=3q+r$ $\,\ni\, r \in \{0,1,2\}$, and $q=\lfloor{\frac{m}{3}}\rfloor$.
We want... | Your proof seems correct but messy.
There’s a much easier inductive style proof. You can show it for $0,1,2$ by calculation and then note that $3| (n+3)^4-(n+3)^2-n^4+n^2$ by basic modular arithmetic.
| {
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Find the largest number that $ n(n^2-1)(5n+2) $ is always divisible by? My Solution:
$$ n(n^2-1)(5n+2) = (n-1)n(n+1)(5n+2) $$
*
*This number is divisible by 6 (as at least one of 2 consecutive integers is divisible by 2 and one of 3 consecutive integers is divisible by 3.
*$ 5n+2 \equiv 5n \equiv n \mod 2 $ then $n... | Alternatively.
Let $p$ be an odd prime. If $p|n$ but $p^2 \not \mid n$ then we have $p\not \mid n-1, n-2, 5n + 2$ so we need never have any odd prime $p^2|n(n^2-1)(5n+2)$.
If $p=3$ we must have one of $n-1, n, n+1$ be divisible by $3$ so we must have $3|n(n^2-1)(5n+2)$ but we need not have $3^2|n(n^2-1)(5n+2)$.
If $p=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given an equation with trigonometric roots $\tan a$ and $\cot a$, determine $\sin a + \cos a$. Question: Given that $\tan a$ and $\cot a$ are two real roots of the equation $x^2 + k^2 - kx - 3 = 0$, and $3\pi < a < \frac{7\pi}{2}$, find the value of $\sin a + \cos a$.
My solution can be found below in the answers secti... | My solution:
First realize that the range for $a$ can be simplified to $\pi < a < \frac{3\pi}{2}$.
Since $\tan a$ and $\cot a$ have a product of $1$, using Vieta's formula, $k^2 - 3 = 1 \Longrightarrow k = \pm 2$.
So, either $x^2 + 2x + 1 = 0$ or $x^2 - 2x + 1 = 0$.
Hence, $x = \pm 1$.
So, using the fact that $\tan a$ ... | {
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Given a general region, find the double integral bounded between $y = x$ and $y=3x-x^2 $ $$ J = \iint_R (x^2-xy)\,dx \,dy, $$
Suppose region R is bounded between $y = x$ and $y=3x-x^2 $
My attempt using vertical integration:
$$ \int^{x=2}_{x=0} \int^{y=3x-x^2}_{y=x} \left({x^2-xy}\right)dy\ dx$$
$$\int^2_0 \left[x^2y-... | Note that in your second case, your region is divided into two (under and above the green line). When you find the inverse functions of $y(x)$'s you have $$x(y) = y$$
for $y = x$ and $$x(y) = \frac{3 \pm \sqrt{9-4y}}{2}$$ (separate them into two at $x = 3/2)$ for $y = 3x-x^2$
Now, move your pen from left to right to no... | {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$ As seen in the title I'm interested in a way to evaluate
$$\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k-1}}{n^2\left(k^2-2n^2\right)}$$
But I'm not sure what to do, I did attempt some... | Let $S_1(n)=\sum\limits_{k=1}^{\infty } \frac{\left ( -1 \right )^{k-1}}{ k^2-2n^2} $, then $S=\sum\limits_{n=1}^{\infty }\left (\sum\limits_{k=1}^{\infty } \frac{\left ( -1 \right )^{k-1}}{n^2\left ( k^2-2n^2 \right )} \right )=\sum\limits_{n=1}^{\infty }\frac{S_1(n)}{n^2}$
Let's consider the integral in the complex p... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to evaluate $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$?
Evaluate: $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$.
I've just started learning limits and calculus and this is an exercise problem from my textbook.
To solve the problem, I tried factorizing the numerator and denominator o... | When $x$ is very large the dominant term in Num is $2x^4$ and in den it is $6x^4$. So the required limit is $$\lim_{x \to \infty} \frac{2x^4}{6x^4}=\frac{1}{3}.$$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum a product of Bernoulli numbers and binomial coefficients Context: I am interested in developing the large-$x$ asymptotic series of the digamma function
$$\psi\Big(\frac{1}{2}+ix\Big)$$
for real positive $x$.
For this I am using the known asymptotic expansion of $\psi(x)$:
$$\psi(z) \sim \ln z - \frac{1}{2z} - \sum_... | We seek to evaluate
$$\sum_{q=1}^{\lfloor n/2\rfloor} \frac{B_{2q}}{2q}
{-2q\choose n-2q} 2^{2q}.$$
Taking advantage of the odd index Bernoulli numbers being zero except
$B_1$ this is
$$(-1)^{n+1} + \sum_{q=1}^n \frac{B_{q}}{q}
{-q\choose n-q} 2^q.$$
Now with the binomial coefficient we get
$$(-q)^{\underline{n-q}}/(n-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to make use of angle sum and difference identities to find the value of sine and cosine?
Calculate: $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$
What is the easiest way to find $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$ (without a calculator)? If I know that $\fra... | After practicing a lot of problems in trigonometry, I realized that I need to know the values of sine and cosine only at $\{ \pi/6, \pi/4, \pi/3\}$, and the double angle, half angle and sum and difference identities. @marwalix et al. answered the same.
By applying the double angle formula (i.e. $\sin \left( 2\theta \ri... | {
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"source": "stackexchange",
"question_score": "4",
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Finding the maximum of two variable Question :
If $\sqrt x + \sqrt y = \sqrt {135}$ for $x,y \in Z_{\ge 0}$
Find $max ~(xy)$
My first approach :
suppose $a = \sqrt{x}$ , $b = \sqrt{y}$
$ ab = \left(xy\right)^\frac{1}{2}, ~ a+b = \left(135\right)^\frac{1}{2}$
usage of AM-GM , equality hold if (and only if ) $a=b$:
$$ 0... | $$y = 135 -2\sqrt{135x}+x$$
So $x=15k^2$ ( where $k$ is a non negative integer) because $135x$ is a perfect square
$$\sqrt{y} = \sqrt{135}-\sqrt{x}=(3-k)\sqrt{15}$$
$$\implies y=15(3-k)^2$$
(Note: $k\le3$ otherwise $\sqrt{y}$ will be negative)
So the general solution is $(x,y)=(15k^2,15(3-k)^2)$
$$\implies xy=225(3k-k^... | {
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can $x^2+a=y^2$ where a is prime Let a sequence of the sum of prime numbers be $S_1=2$, $S_2=5$, $S_3=10$, $S_4=17...$
Can $S_n$ and $S_{n+1}$ both be square numbers $x^2$ and $y^2$ respectively?
I have already made the observation that $x^2+a=y^2$
$a=(y+x)(y-x)$
and therefore $y-x=1$ and $x+y=a$ for this to hold.
| If $S_n=x^2$, and $S_{n+1}=y^2,$ then, as you’ve noticed, $y+x=p_{n+1}, y-x=1.$ Solving these equations:
$$y=\frac{p_{n+1}+1}2,x=\frac{p_{n+1}-1}2$$
This also means that you need $$S_n=\left(\frac{p_{n+1}-1}2\right)^2.$$
Now for $n\geq 5:$
$$\begin{align}S_n&<1+3+5+7+9+11+\cdots +p_n\\&= \left(\frac{p_n+1}2\right)^2\\&... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving an inequality given $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\le 1$ Given that $a,b,c > 0$ are real numbers such that $$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\le 1,$$ prove that $$\frac{1}{b+c+1}+\frac{1}{c+a+1}+\frac{1}{a+b+1}\ge 1.$$
I first rewrote $$\frac{1}{a+b+1} = 1 - \frac{a+b}{a+b+1},$... | Since the two inequalities are completely symmetric in $a,b,c$. WLOG, we only need to study the case $a \ge b \ge c$.
Let $\Lambda = a + b + c + 1$. The two inequalities can be rewritten
as
$$\sum_{cyc}\frac{a}{\Lambda -a } \stackrel{def}{=}\frac{a}{b+c+1} + \frac{b}{c+a+1} + \frac{c}{a+b+1} \le 1\\
\sum_{cyc}\frac{1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Committee with 4 members
My school's Future Mathematicians of America club has 16 members, 7 boys and 9 girls. A president and a 3-person executive committee are chosen (where the president cannot serve on the committee). What is the probability that the president is the same gender as the majority of the committee?
... | Here's a different approach that yields the same result. First choose the four boys and girls, and then choose a president from the majority:
$$\frac{\binom{7}{0}\binom{9}{4}4+\binom{7}{1}\binom{9}{3}3+\binom{7}{3}\binom{9}{1}3+\binom{7}{4}\binom{9}{0}4}{\binom{16}{4}4}=\frac{479}{1040}$$
Note the absence of $\binom{7... | {
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"url": "https://math.stackexchange.com/questions/4196981",
"timestamp": "2023-03-29T00:00:00",
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Pardon my mistakes in asking t I have problems solving equations of LIMITS. To be specific the ones I have to rationalize. https://www.ck12.org/book/ck-12-precalculus-concepts/section/14.5/
Example 5.
The last example is what's giving me issues. The denominator at the $4$-th equation to be precise. I don't know how it ... | The answer is indeed wrong.
\begin{align}
\lim_{x \to 0} \left( \frac{(3-\sqrt{9-x}}{x\sqrt{9-x}} \cdot \frac{3+\sqrt{9-x}}{3+\sqrt{9-x}} \right) &= \lim_{x \to 0} \frac{1}{3+\sqrt{9-x}}\lim_{x \to 0}\frac{9-(9-x)}{x\sqrt{9-x}}\\
&= \frac16 \cdot \lim_{x \to 0}\frac{x}{x\sqrt{9-x}}\\
&=\frac16 \cdot \lim_{x \to 0}\frac... | {
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"url": "https://math.stackexchange.com/questions/4198756",
"timestamp": "2023-03-29T00:00:00",
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With $x^2+y^2=1$ find Minimum and Maximum of $x^5+y^5$ (do not use derivative) It's easy to see that the minimum is $-1$ and maximum is $1$.
My idea is put $x=\cos(a)$, $y=\sin(a)$ and $t=x+y$, so I have $-\sqrt{2}\le t \le \sqrt{2}$ then $0 \le t^2 \le 2$
When $t=x+y$ then $(x+y)^2=1+2xy$.
Then
\begin{aligned}
x^5+y... | $x^2+y^2=1$ implies $x \in [-1,1]$ and $y \in [-1,1]$
(You can verify this by the fact that all (x,y) lie on the unit circle of radius 1 so x and y must lie in the interval [-1,1])
Now, $x^2 \in [0,1]$ and $y^2 \in [0,1]$
Also $x^3 \in [-1,1]$ (since $x \in [-1,1]$)
and $y^3 \in [-1,1]$ (since $y \in [-1,1]$)
Now $$-1 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Project parabola onto another plane (edited) I am starting with a paraboloid in the form
$$A x^{2} + B x y + C y^{2} + D x + E y + F = z$$
and a plane
$$a x + b y + c z + d = 0$$
By equating the z terms in the two equations, I have the equation of the resulting 2nd degree curve in the world coordinate frame as
$$
f = \... | First, define $\mathbf{r} = [x, y, z]^T $ as the position vector. Then the equation
$ z= A x^2 + B x y + C y^2 + D x + E y + F $
can be written as
$ \mathbf{r}^T \mathbf{Q r} + \mathbf{b}^T \mathbf{r} + c = 0 \hspace{12pt}(1)$
where
$\mathbf{Q} = \begin{bmatrix} A && \frac{1}{2} B && 0 \\ \frac{1}{2} B && C && 0 \\ 0 ... | {
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How do i approach ahead in this question
Let $a,b \in\Bbb N$, $a$ is not equal to $b$ and the quadratic equations $(a-1)x^2 -(a^2 + 2)x +a^2 +2a=0$ and $(b-1)x^2 -(b^2 + 2)x +b^2 +2b=0$ have a common root, then the value of $ab/5$ is
So what I did was,
I subtracted the two equations and got
$x=(a+b+2)/(a+b)$
I tried ... | If you like using Vieta's formulas, here is the possible solution based on these formulas:
Assuming $x$ as a coefficient, you can observe that, $a$ and $b$ are root of the quadratic respect to the variable $u:$
$$u^2(1-x)+u(x^2+2)-(x^2+2x)=0, ~ x≠1$$
Then, using the Vieta's formulas, we have
$$\begin{align}&ab=\frac{x^... | {
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"timestamp": "2023-03-29T00:00:00",
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If $x^3 + \frac{1}{x^3} = 2\sqrt{5}$, then find $x^2 + \frac{1}{x^2}$. Question: If $x$ is a real number satisfying $x^3 + \frac{1}{x^3} = 2\sqrt{5}$, determine the exact value of $x^2 + \frac{1}{x^2}$.
My partial solution: $(x + \frac{1}{x})^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}.$
We know that $x^3 + \frac{1}{x^... | From $x^3+\frac{1}{x^3}=2\sqrt{5}$ we get $\left(x^3+\frac{1}{x^3}\right)^2=20$ or $x^6+\frac{1}{x^6}=18$
Let $y=x^2+\frac{1}{x^2}$. Then $y^3 = 3y + x^6+\frac{1}{x^6} = 3y +18.$
The only real solution of $y^3-3y-18=0$ is $y=3$. Therefore $x^2+\frac{1}{x^2}=3.$
By the way, $x = \frac{\sqrt{5}\pm 1}{2}$ can be used to s... | {
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Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of i... | Second approach : using only AM-GM
The second approach (which is correct thus far) does work out using only AM-GM. It works using CS quite clearly, but it comes out through AM-GM as well, although you have to sum a few instances of the two variable AM-GM.
We want to prove that $x^2+y^2+z^2 \geq 12$ for $x+y+z =6$ (Note... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution
Ok, so it's been a really long time since I've done a problem like this but after doing a little bit o studying, this is how far I've gotten.
$$\int\frac{x^... | This is a binomial differential integral of the type:
$I=\int x^{m}(A+B x^{n})^{q/r}dx$.
Chebiceff showed that:"such an integral can be traced back to an integral of rational function only when
$\frac{m+1}{n}$ or $\frac{m+1}{n}+\frac{q}{r}$,
are integers.
The second case is the one that interests us.
Let it be
$a^{2}-x... | {
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Solving the equation $3\cos x+\frac{1}{2}=\cos ^{2}x\left(1+2\cos x\left(1+4\sin ^2x\right)\right)$ Solve the equation:$$3\cos x+\frac{1}{2}=\cos ^{2}x\left(1+2\cos x\left(1+4\sin ^2x\right)\right)$$
To solve it, I tried writing the equation in term of $\cos x$. ( I denote $\cos x$ by $c$):
$$3c+\frac12=c^2(1+2c(5-4c^2... | Your calculations are right. We have $c=\pm\frac 12\sqrt 2$ as two of its roots. Hence, we get $x\in\{\pm\frac{\pi}{4}+k\pi\mid k\in\mathbb Z\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4211972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this proof regarding the nonexistence of odd perfect numbers correct? Preamble: This post is an offshoot of this earlier question, which was not so well-received in MathOverflow.
Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)... | Assuming your statements about the recursive estimates for $I(n^2)$ are correct, $I(n^2)$ may still not be greater than the limit of the recursive estimates. If we know $a>x_n$ for all $n$, and $x_n\to x$ as $n\to\infty$, we can only deduce that $a\ge x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expected value of $\overline{ABC} \times \overline{DEF}$ Here is a question from HMMT:
https://hmmt-archive.s3.amazonaws.com/tournaments/2013/nov/team/solutions.pdf
The digits $1$, $2$, $3$, $4$, $5$, $6$ are randomly chosen (without replacement) to form the three-digit numbers $M = \overline{ABC}$ and $N = \overline{... | The correct answer is 143745.
$$
\frac{1}{6}\cdot\frac{1\cdot(2+3+4+5+6)+2\cdot(1+3+4+5+6)+...+6\cdot(1+2+3+4+5)}{5}\cdot111\cdot111
$$
Your answer is incorrect because you average the numbers evenly
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find minimum value of $\frac{x^3}{\sqrt{2(y^4+1)}}+\frac{y^3}{\sqrt{2(z^4+1)}}+\frac{z^3}{\sqrt{2(x^4+1)}}$
Let $x,y,z>$ and $x+y+z=xy+yz+zx$
. Find the minimum value of $$P=\frac{x^3}{\sqrt{2(y^4+1)}}+\frac{y^3}{\sqrt{2(z^4+1)}}+\frac{z^3}{\sqrt{2(x^4+1)}}$$
My solution: I know the minimum value is $\frac{3}{2}$ whe... | The inequality, which you got is wrong. Try $x=y=2$ and $z\rightarrow0^+$.
For $x=y=z=1$ we obtain a value $\frac{3}{2}$.
We'll prove that it's a minimal value.
Indeed, by Holder $$\sum_{cyc}\frac{x^3}{\sqrt{y^4+1}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x^3}{\sqrt{y^4+1}}\right)^2\sum\limits_{cyc}x^3(y^4+1)}{\sum\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution Verification: Show that $\sum_{i=0}^{n}{\binom{n+a}{n-i}\binom{i+a+1}{i}}=2^{n}\binom{n+a}{a}+2^{n-1}\binom{n+a}{a+1}$ for $n\geq 1$ I solved it by counting the same object using different methods. Say that there are $n+a$ boys and $1$ girl. I want to form a team of $a+1$ kids with some reserve players, but I ... | OP asks for an alternate evaluation of
$$\sum_{q=0}^n {n+a\choose n-q} {a+1+q\choose q}.$$
We have by inspection that this is
$$[z^n] (1+z)^{n+a} \frac{1}{(1-z)^{a+2}}$$
which is in turn
$$\underset{z}{\mathrm{res}}\;
\frac{1}{z^{n+1}} (1+z)^{n+a} \frac{1}{(1-z)^{a+2}}.$$
Now we put $z/(1+z) = w$ so that $z=w/(1-w)$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of ways to sample $x$ beads from 3 red beads, 3 green beads, and 3 blue beads Suppose I wanted to sample $x$ beads from 3 green, 3 blue, and 3 red beads (apart from colour, the beads are not distinct). I've been trying to find one equation I can use to solve for something like this (likely involving combinatoric... | we will use generating functions Then , if the same color of beads are different from each other :
Generating function for different red beads = $$1 + C(3,1)x + C(3,2)x^2 + C(3,3)x^3 = 1+3x +3x^2 +x^3$$
Generating function for different blue beads = $$1 + C(3,1)x + C(3,2)x^2 + C(3,3)x^3 = 1+3x +3x^2 +x^3$$
Generating f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I find the value of $f(0)$? Let $f$ be a polynomial such that $f(0)>0$ and $f(f(x))=4x+1$ for all $x\in R$, then $f(0)$ is $?$
So this is what I've tried so far,
$f(f(0))=1$ so $f(f(0))=4f(0)+1$ (*)
Also, $f(f(1))=5$,this implies $f(5)=16f(0)+5$ Now how should I find the value of $f(5)$ in terms of $f(1)$ so tha... | $f(x) = ax + b$ $(*)$
$f(f(x)) = a^2x + ab + b = 4x + 1$
Comparing coefficients,
$b(a+1) = 1$
$a^2 = 4$
$\implies a = 2, b = \frac{1}{3}$
($a$ is not $-2$ by the fact that $f(0) > 0$, $a = -2$ would result in $b = -\frac{1}{3} \implies f(0) = -\frac{1}{3}$)
Therefore $f(x) = 2x + \frac{1}{3}$ and $f(0) = \frac{1}{3}$
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Interesting inequality with three positive reals which seems trivial Let $a,b,c$ be positive real numbers, such that $(ab)^2 + (bc)^2 + (ca)^2 = 3$. Prove that
$(a^2 - a + 1)(b^2 - b + 1)(c^2 - c + 1) \geq 1$.
First I multiplied both sides by $(a+1)(b+1)(c+1)$ and expand the inequality:
$$a^3b^3c^3+a^3b^3+b^3c^3+c^3a^3... | Now, since $$\prod_{cyc}(x+y)\geq\frac{8}{9}(x+y+z)(xy+xz+yz),$$ it's enough to prove that $$x+y+z\geq3,$$ which is for you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}$?
Calculate this limit
$$\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}$$
My attempt: using the limit development : we find
$$\exp(\sin(x)-x)=\exp(x-\frac{x^{3}}{3!}+o (x^{3})-x)=\exp(-\frac{x^3}{3!}+o(x^3))=1-\frac{x^3}{... | You don't really need to factor like that. You can just say $\sin(x)=x-x^3/6+o(x^3)$ so
$$\exp(\sin(x))=1+\sin(x)+\sin(x)^2/2+\sin(x)^3/6+o(\sin(x)^3) \\
= 1 + (x-x^3/6) + (x-x^3/6)^2/2 + (x-x^3/6)^3/6 + o(x^3) \\
= 1 + x + x^2/2 - x^3/6 + x^3/6 + o(x^3) \\
= 1 + x + x^2/2 + o(x^3)$$
so $\exp(x)-\exp(\sin(x))=x^3/6 + o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum value of a expression Problem: If $\alpha+\beta+\gamma=20$, then what is $\max(\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5})$?
My attempt: Assume $\alpha \geq \beta \geq \gamma$. Then $\alpha+\beta+\gamma \leq 3\alpha$ so $25 \leq 3\alpha+5$.
Also $\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}\leq 3\s... | Let $u=\sqrt{3\alpha+5}+\sqrt{3\beta+5}+\sqrt{3\gamma+5}$.
Denote:
$$\begin{cases}
3\alpha +5=a^2\\
3\beta+5=b^2\\
3\gamma +5=c^2
\end{cases}$$
Then, the problem becomes:
$$v=a+b+c\to max \ \ s.t.\ \ a^2+b^2+c^2=75$$and:
$$v^2=a^2+b^2+c^2+2(ab+bc+ca)=75+2(ab+bc+ca)\le \\
75+2(a^2+b^2+c^2)=225,$$
the equality occurs for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving an integral equivalence involving floor and ceiling functions During the course of looking at the Euler–Mascheroni constant I have run across the following result:
$$\gamma
= \int \limits_1^\infty \Bigg( \frac{1}{\lfloor x \rfloor} - \frac{1}{x} \Bigg) \ dx
= \int \limits_1^\infty \Bigg( \frac{\lceil x \rceil}... | Applying the suggestion (in comments) from Paramanand Singh we get:
$$\begin{align}
\int \limits_1^\infty \Bigg( \frac{\lceil x \rceil}{x^2} - \frac{1}{\lfloor x \rfloor} \Bigg) \ dx
&= \sum_{n=1}^\infty \int \limits_n^{n+1} \Bigg( \frac{\lceil x \rceil}{x^2} - \frac{1}{\lfloor x \rfloor} \Bigg) \ dx \\[6pt]
&= \sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Solution verification for limit problem $\lim\limits_{x\to 0}\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}.$ Find the value of
$$\lim\limits_{x\to 0}\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}.$$
I try as below.
\begin{align}
\lim\limits_{x\to 0}\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}... | More simply we can proceed as follows
$$\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}=\frac 4 3 x\cdot\dfrac{\frac{\sin(4x)}{4x}\frac{\tan^2(3x)}{(3x)^2}9x+\frac 3 2}{\frac 2 3x+\frac{\sin(3x)}{3x}\cos(2x)}\to 0\cdot \frac{1\cdot 1\cdot 0+\frac32}{0+1\cdot1}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is integral of $(x+y)^2 \,dx$ different than integral of $(x^2+ 2xy + y^2) \,dx$? I calculated the integral of $(x + y)^2 \,dx$ using the substitution method and got $\frac{1}{3}(x+y)^3$ as result.
Then I applied the distributive property to $(x + y)^2$, calculated the integral of $(x^2+2xy+y^2) \,dx$ and I got $\f... | Their difference is constant (it's $\frac{y^3}3$). So, you got the same answer by both methods.
To be more precise,$$\frac13(x+y)^3=\left(\frac13x^3+x^2y+xy^2\right)+\frac{y^3}3,$$and therefore\begin{align}\frac{\mathrm d}{\mathrm dx}\left(\frac13(x+y)^2\right)&=\frac{\mathrm d}{\mathrm dx}\left(\left(\frac13x^3+x^2y+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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evaluate $\int_0^{\pi/2} x^2\log(\sin x)\,dx$ I am a high school student , I know how to evaluate $\int_0^{\pi/2} x\log(\sin x)\,dx$.
It would be great if someone can help me evaluating $\int_0^{\pi/2} x^2\log(\sin x)\,dx$ and tell me if this integral is elementary or non elementary .
I tried using the "a-x" property b... | Another possibility is to write
$$\log[\sin(x)]=\sum_{n=1}^\infty (-1)^n \frac{2^{2 n-3}\,\, (E_{2 n-1}(1)-E_{2 n-1}(0))}{n \,(2 n-1)!}\left(x-\frac{\pi }{2}\right)^{2 n}$$ where appear the Euler polynomial and use
$$\int_0^{\frac \pi 2}x^2\left(x-\frac{\pi }{2}\right)^{2 n}\,dx=\frac{\left(\frac{\pi }2\right)^{2 n+3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4229391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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square root of $1 \pmod m$ for prime $m$ and non prime $m$ While doing square root $1$ moduli i.e. $x ^2 \equiv 1 \pmod m$ I noticed the following:
If the moduli $m$ is prime it seems that the square is always $1$ and $m - 1$
But if it is not prime, it is not the case and I am not sure if there is a general formula for... | Outline for an answer.
Let $g(m)$ be the number of distinct roots to $x^2\equiv1\pmod m.$
If $m,n$ are relatively prime, then show $g(mn)=g(m)g(n).$ (Hint: use Chinese remainder theorem.) (This property is called being “multiplicative.”)
This is true for any polynomial congruence in one variable. If $h(m)$ is the numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4229793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Complex number related problem Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
My solution is as follow
${z_1} = 2{e^{i{\theta _1}}};{z_2} = 2{e^{i{\theta _2}}};{z_3} = 2{e^{i{\theta _3}}}$ &... | Algebra + Geometry approach
Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
Let's try interpret the problem geometrically.
$|z_1| = |z_2 | = |z_3|$ means that the three points lie on a circ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate maximum and minimum when second partial derivative test fail
Calculate the maximum and minimum of function $z = f(x,y) = x^2 - y^2 +2$ subjected to the inequality constraint $D=\{(x,y)| x^2 + \frac{y^2}{4} \leq 1\}$.
My solution:
First form the function
$$
g(x,y) = x^2 + \frac{y^2}{4} - c, ~0 \leq c \leq 1.... | HINT: It is an absolute maximum (and minimum). It is easy to see that it is obtained on the boundary. Then you will have a finite number of possible points. Simply compare them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all analytic functions in the disk $|z-2|<2$ such that $f \left(2+\frac{1}{n} \right) = \frac{1-n}{5n+3} + \sin{\left(\frac{n \pi}{2} \right)}$ Find all analytic functions (or prove that no such exist) inside the disk $|z-2|<2$ that satisfy the following condition:
$$f \left(2+\frac{1}{n} \right) = \frac{1-n}{5n+3... | If $n$ is a multiple of $4$, then\begin{align}f\left(2+\frac1n\right)&=\frac{1-n}{5n+3}\\&=\frac{1/n-1}{5+3/n}\\&=\frac{2+1/n-3}{-1+3(2+1/n)}.\end{align}Now, for each $z\in D_2(2)$, let $g(z)=\frac{z-3}{-1+3z}$. Then, by the identity theorem, $g=f$. But that's impossible, since $g(3)\ne f(3)$: $g(3)=0$, whereas$$f(3)=f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove by contradiction that if $n^3$ is a multiple of $3$, then $n$ is a multiple of $3$ Problem statement:
Using proof by contradiction, prove that if $n^3$ is a multiple of $3$ , then $n$ is a multiple of $3.$
Attempt 1:
Assume that there is exist $n$ which is a multiple of $3$ such that $n^3$ is not a multiple of $... | If $P,Q$ are statements then $P \implies Q$ is same as $ \neg Q \implies \neg P $.
Here $P : n^3$ is a multiple of $3$ and $Q: n$ is a multiple of $3$.
So if you want to prove it by contradiction you have to assume that $n^3$ is multiple of $3$ but not $n$ and then get a contradiction.
You can do this by noting that e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral using the definition of integral is returning double the real answer? I'm trying to evaluate an integral using the definition of the definite integral but for some reason my answer is 12 when it should be 6. I assumed that I'd missed a 2 somewhere that would go in the denominator but I've tried this exercise m... | As Michael said, it's just a sign error:
$$\cdots= \frac{3}{n} \left( \frac{6}{n} \cdot \frac{n(n+1)}{2} \color{red}{-} n \right) = \frac{3}{n} \left( 3(n+1) \color{red}{-} n \right) = \frac{3}{n}(\color{red}{2}n + 3) = \frac{\color{red}{6}n + 9}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ivan has 3 red blocks, 4 blue blocks and 2 green blocks. He builds the tower with randomly selected blocks but only stops when the tower consists of all three colours.
*
*What is the probability that the tower is 4 blocks tall?
My approach is the following but I am not sure at all:
to make it 3 colours it builds a t... | We can easily solve using generating functions
For blocks of $4$ that have all three colors, the number of permutations is given by
coefficient of $x^4$ in $4!(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!})(x+\frac{x^2}{2!} + \frac{x^3}{3!})(x+\frac{x^2}{2!}) = 36$
while unrestricted permutations = $\dfrac{9!}{4!3!2!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244586",
"timestamp": "2023-03-29T00:00:00",
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How can I solve $\frac{(1-x)(x+3)}{(x+1)(2-x)}\leq0$ and express solution in interval notation? The inequality I need to solve is
$$\frac{(1-x)(x+3)}{(x+1)(2-x)}\leq0$$
My attempt:
Case I: $x<2$
Then we have $(1-x)(x+3)\leq0$
$x\leq-3, x\geq1$
We reject $x\leq-3$ (not sure why exactly, I just looked at the graph on des... | The key insight here is that for a fraction to be negative you need to have the numerator and denominator have opposite signs. If you have $$\frac{A}{B}\le 0$$ then either you have $A\le 0$ and $B> 0$ or you have $A\ge 0$ and $B< 0$. This gives us two cases, with several sub-cases:
Case $1$:
$$(1-x)(x+3)\le 0\quad\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$
With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
... | Denoting with $e_k(x,y,z), k\geq 0$ the elementary symmetric polynomials
\begin{align*}
e_1(x,y,z)&=x+y+z=0\\
e_2(x,y,z)&=xy+xz+yz=-1\tag{1}\\
e_3(x,y,z)&=xyz=-1
\end{align*}
and with $p_k(x,y,z), k\geq 0$ the $k$-th power sum
\begin{align*}
p_k(x,y,z)=x^k+y^k+z^k\tag{2}
\end{align*}
we recall Newtons identities
admit ... | {
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"url": "https://math.stackexchange.com/questions/4246285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
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Finding binomial Coefficient using expansion If ${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{2n - 1}}{x^{2n - 1}} + {a_{2n}}{x^{2n}}$, prove that $a_0=a_{2n},a_1=a_{2n-1},....a_n=a_{n+1}$
My approach is as follow ${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + ... + ... | $$(1+x+x^2)^n=\sum_{k=0}^{2n} A_k x^k$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:
$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \rig... | After playing around a bit more with the Weierstrass product method, I realized I could proceed to do the integration with partial fractions. Using that
$$
\int_{0}^{\infty} \frac{1}{(x^2+a^2)^n} \ dx = \frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2a^{2n-1}}, \quad a\ge0,\ n \in \mathbb{N}
$$
as shown in this answer, we get the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 1
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Find value of $x^{5} + y^{5} + z^{5}$ given the values of $x + y + z$, $x^{2} + y^{2} + z^{2}$ and $x^{3} + y^{3} + z^{3}$ If$$x+y+z=1$$
$$x^2+y^2+z^2=2$$
$$x^3+y^3+z^3=3$$
Then find the value of $$x^5+y^5+z^5$$
Is there any simple way to solve this problem ? I have tried all my tricks tried to multiply two equations ,... | Another way.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$,where $v^2$ can be negative, and $xyz=w^3$.
Thus, $u=\frac{1}{3}$ and since
$$2=x^2+y^2+z^2=9u^2-6v^2,$$
we obtain $$v^2=-\frac{1}{6}.$$
Also, since $$3=x^3+y^3+z^3=27u^3-27uv^2+3w^3=1+\frac{3}{2}+3w^3,$$ we get $$w^3=\frac{1}{6}.$$
Id est, $$x^5+y^5+z^5=243u^5-405u^3v^2+13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the minimum value of $a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ Let $a,b,c$ be the lengths of the three sides of the triangle, $a+b+c=3$. Find the minimum value of $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)$$
My attempts:
$\bullet$ The minimum value is $3$, equality holds iff $a=b=c=1$ so by AM-GM, we have: $$a^8+b^8+c^8\ge 8(a+b+c)-21=3... | The function $f(a,b,c)=a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ has only one extrema minimum point $f(1,1,1)=3$ on the domain $a+b+c=3$. To prove that this one is the global minimum we need to check the value of $f$ at the boundary of the following domain:
*
*$0<a<\frac 32$
*$\frac32-a<b<\frac 32$
*$c=3-a-b$
and we have
*
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Non-probabilistic/non-combinatoric proof of $\sum_{k=n}^{2n} \frac{{k \choose n}}{2^k}=1$ The summation $$S_n=\sum_{k=n}^{2n} \frac{{k \choose n}}{2^k}=1~~~~(1)$$
has been proved using probabilistic/combinatoric method in MSE earlier:
Combinatorial or probabilistic proof of a binomial identity
Here we give an analyti... | Using algebra.
$$\sum_{k=n}^{2n} \binom{k}{n} x^k=(1-x)^{-(n+1)} x^n\Bigg[1-\binom{2 n+1}{n} ((1-x) x)^{n+1} \, _2F_1(1,2 n+2;n+2;x) \Bigg]$$
Developed as a series around $x=\frac 12$
$$\sum_{k=n}^{2n} \binom{k}{n} x^k=1+ \left(4 n+2-\frac{4\, \Gamma
\left(n+\frac{3}{2}\right)}{\sqrt{\pi }\, \Gamma
(n+1)}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is it possible to find a closed-form expression for this summation? Is it possible to find a closed-form expression for this summation?
$$\sum\limits_{k = 0}^n {\left( \left( {\matrix{
2n \cr
k \cr
} } \right)+ \left( {\matrix{
2n \cr
n+k-1 \cr
} } \right) \left( \frac{1}{3}\right)^\left(2n-2k+1 \... | When you have this kind of summation, you cannot avoid at least gaussian hypergeometric functions and in most cases you will not obtain any closed form.
Relacing the $\frac 13$ by $x$ and splitting the sum, you have
$$\sum_{k=0}^n \binom{2 n}{k} x^k=(x+1)^{2 n}-\binom{2 n}{n+1} x^{n+1} \, _2F_1(1,1-n;n+2;-x)$$ The sec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4267115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Gambler's ruin recursive solution Suppose we have 2 players A and B, A starting with $a$ coins and B starting with $n-a$ coins. Assuming every game is independent, every time A wins, they win $2$ coins and every time they lose they lose 1 coin. What is the probability that A wins the game ?
I can solve this using a dif... | (Note that we need $W_{n + 1} = 1$ as an additional boundary condition.)
Your idea is good; I’ll write it more compactly as follows:
$$0 = \sum_{i=1}^a (W_i - pW_{i+2} - qW_{i-1}) = pW_1 + pW_2 + qW_a - pW_{a+1} - pW_{a+2}.$$
We can generalize it by throwing in a $c^i$ coefficient, where $c$ is any root of $p - c^2 + c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4267905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solving differential equation $af'(x)^2-f''(x)=0$ I want to solve the differential equation $$af'(x)^2-f''(x)=0$$ where $a $ is constant and we have the boundary conditions $f(0)=0, f (1)=c $ for some positive $c$.
If $a=0$ we get $f (x)=cx$ but what if $a\ne 0$. Could someone show how to solve this differential equati... | The differential equation can be written as
$$
\frac{f''(x)}{f'(x)^2} = a.
$$
Taking antiderivatives gives
$$
-\frac{1}{f'(x)} = ax + B,
$$
for some constant $B.$ Thus,
$$
f'(x) = -\frac{1}{ax+B}
$$
and taking another antiderivative gives
$$
f(x) = C - \frac{1}{a}\ln(ax+B)
$$
where $C$ is a constant.
Boundary condition... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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"solved" Confusion calculating $\mathop {\lim }\limits_{x\to 0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}}$ I can get the correct answer through one solution, but when I try the second method, it shows an obvious error, and I can't find where and why. Can someone know the reason,... | When calculating limits, you can not just replace $ x $ in a part of the expression and leave it in the other part, this is a very common mistake.
I'll provide a solution to a more generalized limit, and won't use series expansions or L'hopital's rule. We must know that $ \frac{1-\cos{x}}{x^{2}}\underset{x\to 0}{\longr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality between integral of $(1-x)^k P(x)^2$ and integral of $(1-x)^{k+1} P'(x)^2$ Given a polynomial $P$ with real coefficients and an integer $k \ge 0$, define
$$I_k(P) = \int_0^1 \frac{(1-x)^k}{k!} P(x)^2 dx.$$
I would like to show that if $P(0) = 0$ and $P$ is not identically zero, then
$$I_k(P) \le \frac{4}{k+2... | Using the inequality $I_k(P) \le 2 \sqrt{I_{k+1}(P) I_{k+1}(P')}$, it suffices to show $I_{k+1}(P) \le \left(\frac{2}{k+2}\right)^2 I_{k+1}(P')$.
Integrating by parts and applying Cauchy-Schwarz, we have
\begin{align}
I_{k+1}(P)
&= 2 \int_0^1 \frac{(1-x)^{k+2}}{(k+2)!} P(x) P'(x) dx \\
&\le 2 \left(\frac{k+3}{k+2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the function(s) $f$ satisfying the condition: $x^2f(x)+yf\left(y^2\right)=f(x+y)f\left(x^2-xy+y^2\right)$.
Find the function(s) $f$ satisfying the condition: $$x^2f(x)+yf\left(y^2\right)=f(x+y)f\left(x^2-xy+y^2\right)\text.$$
Of Course, I expect the function $f$ be $f(x)=x$ or $f\equiv0$.
My attempt:
\begin{alig... | You've correctly managed to show that if $ f ( x ) \ne 0 $ then $ f ( x ) = x $. As $ f \not \equiv 0 $, there is some $ x _ 0 \ne 0 $ with $ f ( x _ 0 ) = x _ 0 $. Consider an arbitrary $ y \ne 0 $. If $ f ( y ) = 0 $, by \eqref{B} you get $ f \left ( y ^ 2 \right ) = 0 $. Thus, $ P ( x _ 0 , y ) $ gives
$$ f ( x _ 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4280110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Induction inequality/ n in exponential I need to find out through induction for which $n \geq 0$ the following inequality holds
$$3n+2^{n} \leq 3^n$$
clearly it does not work for n=1 and n=2, so my induction hypothesis is that it works for $n \geq 3$,
however when performing the induction step, I am kind of stuck
becau... | So I think the necessary steps in the proof are all there, it is just better to do it slightly differently. You initially write $3(n+1) + 2^{n+1} \leq 3^{n+1}$ and then manipulate it. I know that's what you are trying to show, but you never want to manipulate what you are trying to show. Its nice form to start with one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4280652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Conditional expectation and expected value Given the joint distribution $f(x,y) = {x \choose y} (\frac{1}{2})^x \frac{x}{20}$ where $y=0,1,2 \cdots x$ and $x=2,3,4,5,6$.
Find $E(y\mid x)$ and $E(y)$
Wondering if I am missing something.
Using Binomial identity and direct definition of $f(x)$ (the limit was from the plo... | The easiest way to proceed is to observe that your joint pmf is
$$p(x,y)=p(y|x)p(x)=\binom{x}{y}\left(\frac{1}{2}\right)^x\times\frac{x}{20}$$
Thus $(Y|X=x)\sim Bin\left(x;\frac{1}{2}\right)$
And X is a discrete rv with support $x=\{2,3,4,5,6\}$ and pmf
$$p(X=x)=x/20$$
Thus
$$\mathbb{E}[Y|X=x]=x/2$$
And
$$\mathbb{E}[Y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Strange/Unexpected behavior of an Infinite product Some friends and I were playing around with this continued fraction:
We noticed when writing it out for each next step, the end behavior went either to 1 (when there was an even number of terms) or to a linear dependence on x (when there were odd). This was expected -... | Transform the product to a sum using logarithms
$$P_m=\prod_{n=0}^m \frac{x+2 n}{x+2 n+1} \implies \log(P_m)=\sum_{n=0}^m \log \left(\frac{x+2 n}{x+2 n+1}\right)$$ Ubsing Taylor for large values of $x$
$$\log \left(\frac{x+2 n}{x+2 n+1}\right)=-\frac 1 x+\frac{4 n+1}{2 x^2}-\frac{12 n^2+6 n+1}{3 x^3}+O\left(\frac{1}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Failure of L’Hospital’s rule to $\lim_{x\rightarrow\infty} x\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right)$? I rewrote the limit
$$\lim_{x\rightarrow\infty} x\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\right) =\lim_{x\rightarrow\infty} \frac{\left(\arctan\left(\frac{x+1}{x+2}\right)-\frac\pi 4\rig... | You forgot to use chain rule in the numerator:
$\frac{d}{dx}(arctan(\frac{x+1}{x+2})-\frac{\pi}{4}) = \dfrac{\frac{x+2-(x+1)}{(x+2)^2}}{1+(\frac{x+1}{x+2})^2} = \frac{1}{(x+2)^2+(x+1)^2}$
Now the limit exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4283482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Evaluating $\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$ in two ways gives different answers
Evaluate
$$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$$
I started by rationalising followed by dividing numerator and denominator by $x$.
$\lim_{x\rightarrow-\infty}{\sqrt{4x^2-x}+2x}$
=$\lim_{x\rightarrow-\infty}\frac{{(\... | The problem happens when you go from
$$\lim_{x\rightarrow-\infty}\frac{-x}{\sqrt{4x^2-x}-2x}$$
to
$$\lim_{x\rightarrow-\infty}\frac{-1}{\sqrt{4-\frac{1}{x}}-2}$$
The first line is actually not identical to the second line, and you can verify yourself by plugging in some value of $x$. For example, if $x=-20$, then
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4287655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $S_{N} = \sum_{i=0}^\infty \frac{i^N}{4^i}$ using recursion I am trying to obtain a formulae for a summation problem under section (d) given in a solutions manual for "Data Structures and Algorithm Analysis in C - Mark Allen Weiss", here's the screen shot
As the pdf is protected i could not download it.
He... | The technique given in the question gives the recurrence:
$$3S_N=\sum_{i}\frac{(i+1)^N-i^N}{4^i}=\sum_{j=0}^{N-1} \binom Nj S_j$$
This follows from the binomial theorem: $$(i+1)^N-i^n=\sum_{j=0}^{N}\binom Nj i^j$$
But “solving” this recurrence is very painful. I don’t know how to solve it. It’s not even clear the book ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4288188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying $\frac{p^2q^2(1-\epsilon^2\cos^2t)}{p^2\cos^2t+q^2\sin^2t}$, where $\epsilon=\sqrt{1-(q/p)^2}$ I am currently trying to show that the product of the distances from the focis of an ellipse to the tangent line at any point of the ellipse is a constant. While I thought that the computation is a straightforward... | Let us use that $\cos^2(t)+\sin^2(t)=1$ and that $\epsilon^2=1-(q/p)^2$:
$$\frac{p^2q^2(1 - \epsilon^2\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=\frac{p^2q^2(\cos^2(t)+ \sin^2(t) - (1-(q/p)^2)\cos^2(t))}{p^2\sin^2(t) + q^2\cos^2(t)}=$$
$$=\frac{p^2q^2(\cos^2(t)+\sin^2(t) - \cos^2(t)+(q/p)^2\cos^2(t))}{p^2\sin^2(t) + q^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Proving Big O and Big Omega of a Polynomial I have a polynomial:
$f(n)=\frac{1}{4}n^2-24n-16$
I am supposed to show that it is $\Omega(n^2)$ and $O(n^2)$. Once I prove those, it should be easy to show that it is also $\Theta(n^2)$ using the theorems given, but I'm having difficulties with $\Omega$ and $O$. Mainly the s... | For checking that $f(n)$ is $O(n^2)$ I think you have just made a simple mistake and it seems like you understand what to do: if $C = \frac{1}{4}$ then $f(n) \leq C g(n)$ becomes
$$
\frac{1}{4} n^2 - 24 n - 16 \leq \frac{1}{4} n^2
$$
which is automatically always true for $n \geq 0$.
They key point for showing that $f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4297177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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show this sequence always is rational number let $\{a_{n}\}$ such $a_{1}=-8$,and such
$$4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$$
show that
$$a_{n}\in Q,\forall n\in N^{+}$$
I try let $a_{2}=x$,and for $n=1$, then we have
$$-8+5\sqrt[3]{x}=3\sqrt[3]{-49(x+1)}\Longrightarrow x=-1/8$$
and fo... | By looking at the initial values, we hypothesize that
*
*$a_n = x_n^3, x_n \in \mathbb{Q}$
*$a_n + 1 = 7y_n^3, y_n \in \mathbb{Q}$
This suggests to study rational points on $7y^3 = x^3 + 1$.
We map $ a_i$ to the point $ ( \sqrt[3]{a_i}, \sqrt[3]{\frac{a_i + 1}{7} } )$.
We have the starting points $a_1: (-2, -1), a_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 0
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Prove that the sequence $(a_n)$ is Cauchy and find the limit. Let us define a sequence $(a_n)$ as follows:
$$a_1 = 1, a_2 = 2 \text{ and } a_{n} = \frac14 a_{n-2} + \frac34 a_{n-1}$$
Prove that the sequence $(a_n)$ is Cauchy and find the limit.
I have proved that the sequence $(a_n)$ is Cauchy. But unable to find the... | Given that $a_1=1$ and $a_2=2$ such that $\displaystyle a_n=\frac{1}{4}a_{n-2}+\frac{3}{4}a_{n-1}$
for $n\geq3$
Now $\displaystyle a_{n}-a_{n-1}=-\frac{1}{4}\left(a_{n-1}-a_{n-2}\right)=\left(-\frac{1}{4}\right)^2\left(a_{n-2}-a_{n-3}\right)$
$\displaystyle \dots=\left(-\frac{1}{4}\right)^{n-2}(a_2-a_1)=\left(-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4298951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving an absolute value inequality with fractions I'm having a hard time figuring out this inequality: $\bigg|\dfrac{x-4}{x+5}\bigg| \le 4$
I use one of the absolute value properties, which results in: $-4 \leq \dfrac{x-4}{x+5} \leq 4$.
From there, I get $x \geq -\frac{16}{5}$ and $x \geq -8$, yet when I look at Wolf... | From your steps, a couple of red flags jump out at me. First, as I predicted in my comment, it appears you have multiplied both sides by $x + 5$ without worrying whether $x + 5$ is positive or negative (it can't be zero, for obvious reasons!). Second, you are ignoring the other inequality $-4 \le \frac{x - 4}{x + 5}$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Expressing a solid in spherical coordinates I am trying to solve the following question:
The volume of the solid $E$ can be represented as
$$\int_{-3}^3 \int_0^{\sqrt{9-x^2}} 3 - \sqrt{x^2+y^2} dydx$$
Describe the solid in spherical coordinates.
I graphed the solid and it looks like this:
So clearly $0 \le \theta \l... | The solid is half of an inverted cone with vertex at $(0, 0, 3)$ and above $z = 0$.
Equation of the surface of the cone is $ ~\sqrt{x^2+y^2} = 3 - z$
That translates to $ \rho \sin\phi = 3 - \rho \cos\phi \implies \rho = \frac{3}{\cos\phi + \sin\phi}$
Please note that $0 \leq \phi \leq \pi/2$ as we are above $z = 0$.
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Do I have the right bounds and function for this integral? Find the integral of function $f(x,y,z)=(x^2+y^2+z^2)^{3/2}$ inside the sphere $(z-2)^2+x^2+y^2=4$
My approach: by changing it to spherical coordinates, we have $0\le\rho\le2\cos\varphi$ $(0\le\theta\le2\pi,\ 0\le\varphi\le\frac\pi2)$, and the function becomes ... | The region of integration is the sphere of radius $2$ centered at $(0,0,2)$
So in spherical coordinates, the change of variables is
$x = r \sin \theta \cos \phi$
$y = r \sin \theta \sin \phi$
$z = 2 + r \cos \theta$
The differential volume is $ r^2 \sin \theta dr d\theta d\phi $ and the limits of integration are $ r \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to separate $ \prod_{k=1}^{\infty} \frac{(16k-15)^{\frac{1}{16k-15}}}{(16k-1)^{\frac{1}{16k-1}}} $ from Log-gamma function. Using the Fourier series expansion for $0 < z < 1$, one has:
$$ \ln (\Gamma(z)) = (\tfrac{1}{2}-z)(\gamma+\ln(2)) + (1-z) \ln(\pi)-\frac{1}{2}\ln( \sin(\pi z)) + \frac{1}{\pi} \sum_{n=1}^{\inf... | Let $$f(z)=\pi\left(\ln\Gamma(z)+\left(z-\frac{1}{2}\right)(\gamma+\ln 2)+(z-1)\ln\pi+\frac{1}{2}\ln\sin(\pi z)\right)=\sum_{k=1}^{\infty} \frac{\sin(2 \pi kz)}{k} \ln k$$
Then assuming $\alpha\in\mathbb{N}$ and $\alpha>2$,
$$\ln\prod_{k=1}^{\infty} \frac{(\alpha k-\alpha+1)^{\frac{1}{\alpha k-(\alpha -1)}}}{(\alpha k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Find the minimum value of the constant term.
Let $f(x)$ be a polynomial function with non-negative coefficients such that $f(1)=f’(1)=f’’(1)=f’’’(1)=1$. Find the minimum value of $f(0)$.
By Taylor’s formula, we can obtain
$$f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3+\cdots+\frac{f^{(n)}(1)}{n!}(x-1)^n.$... | Motivation: First, consider $f(x) = a_0 + a_1 x + a_2x^2 + a_3 x^3 + a_4 x^4$.
Using $f(1) = f'(1) = f''(1) = f'''(1) = 1$,
we get $a_0 = \frac13 + a_4$.
Second, consider $f(x) = a_0 + a_1 x + a_2x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5$.
Using $f(1) = f'(1) = f''(1) = f'''(1) = 1$,
we get $a_0 = \frac13 + a_4 + 4a_5$.
Thus, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Ellipse in the triangle Find the equation of an ellipse if its center is S(2,1) and the edges of a triangle PQR are tangent lines to this ellipse. P(0,0), Q(5,0), R(0,4).
My attempt: Let take a point on the line PQ. For example (m,0). Then we have an equation of a tangent line for this point: $(a_{11}m+a_1)x+(a_{12}m+... | The triangle vertices are $P_1 (0,0), Q_1 (5, 0), R_1 (0, 4) $
The equation of the ellipse in matrix-vector form is
$ (r - C)^T Q (r - C) = 1 $
where $C = (2, 1) $ is the center , and $Q$ is a $2 \times 2$ symmetric matrix.
Drawing the triangle, we realize that the required ellipse is tangent to the $x$ axis, the $y$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4312797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find $a$ and $b$ such that $4(b^3-a^3)-3(b^4-a^4)=\frac{1}{2}$ and $b-a$ gets minimum. I was solving this problem that for $f(x)=12x^2-12x^3, 0<x<1$ that is probability density function we have $P(a<X<b)=\frac{1}{2}$. So we have
$$P(a<X<b)=\int_{a}^{b}f(x)dx=\frac{1}{2}$$
with calculation I reached to this: $4(b^3-a^3)... | Let's rewrite our function in terms of $s = a + b$ and $d = b-a$. Then we have
$$
4(b^3 - a^3) + 3(b^4 - a^4) = \frac{d}{2}\left[d^2(2-3s) + 3(2-s)s^2\right] = \frac{1}{2}
$$
Now we can proceed by implicit differentiation. We can define $d(s)$ as satisfying the equation $d[d^2(2-3s) + 3(2-s)s^2] =1$. Differentiating wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Write $x^k - y^k$ as the product of two factors. Write $x^k - y^k$ as the product of two factors.
$x^3 - y^3 = (x-y)(x^2+xy+y^2)$
$x^4 - y^4 = (x-y)(x^3+x^2y+xy^2+y^3)$
$x^5 - y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
$x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$
$x^k - y^k = (x-y)(x^{k-1}+\underbrace{x^{k-2}y + \c... | The sum you bracketed can be expressed as $$x^{k-2}y+\cdots +xy^{k-2} =\sum_{i=1}^{k-2}x^{k-2+1-i}y^i =\sum_{i=0}^{k-1}x^iy^{k-j-1} -x^{k-1}-y^{k-1}$$ For a formal proof: Notice that $$a^2-b^2 = a^2+ab-ab-b^2 = a(a+b)-b(a+b) = (a-b)(a+b)$$ so that our result holds for $n=2$.Now suppose that $$a^k -b^k = (a-b)(a^{k-1}+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that $\frac{a}{b^2 + c}+\frac{b}{c^2 + a}+\frac{c}{a^2 + b} \ge \frac{3}{2}$ when $a+b+c=3$ I've tried solving Pham Kim Hung's famous inequality problem which he posted on aops in April 2007 for 2 days now. This is the problem: Show that $$\frac{a}{b^2 + c}+\frac{b}{c^2 + a}+\frac{c}{a^2 + b} \ge \frac{3}{2} \tag... | $\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^4}{a^3b+a^3c^2}\geq\frac{(a^2+b^2+c^2)^2}{\frac{1}{3}(a^2+b^2+c^2)^2+\frac{1}{3}(a^2+b^2+c^2)^2}=\frac{3}{2}$
(For numerator we use Titu's Lemma and simplify denominator :) )
Thanks to arqady
For more information vist https://artofproblemsolving.com/community/c6h401290p2234... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the general value of $\theta$ for the inverse trigonometric function $\text { Find the general value of } \theta, \text { when } 9 \sec ^{4} \theta=16$
My work-
Given $\sec ^{4} \theta=\frac{16}{9}$
or, $\sec ^{4} \theta = \frac{(4)^{2}}{(3)^{2}}$
$\implies \sec ^{2} \theta=\frac{4}{3} \Longrightarrow \theta=\sec ... | $$
\begin{aligned}
9 \sec ^{4} \theta &=16 \\
\sec ^{4} \theta &=\frac{16}{9} \\
\sec \theta &=\pm \frac{2}{\sqrt{3}} \\
\cos \theta &=\pm \frac{\sqrt{3}}{2} \\
\theta &=n \pi \pm \frac{\pi}{6}=\frac{(6 n \pm 1) \pi}{6},
\end{aligned}
$$
where $n \in \mathbb{Z}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4318209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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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.