Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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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 ... | Other than applying law of cosine or using distance formula from a point to a line that I mentioned in comment, another approach that I would consider is to recognize that $\triangle ABC$ is an isosceles triangle with,
$AC = BC = 5\sqrt2, \ AB = 2 \sqrt{10}$. If $G$ is midpoint of $AB$ then we know that $CG^2 = AC^2 - ... | {
"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": 2
} |
Finding $n>2$ such that $a_1$, $a_2$, $a_3$ are in arithmetico-geometric progression and $\left(1-x^3\right)^n=\sum_{r=0}^na_rx^r(1-x)^{3n-2r}$ This is the strangest, mind-boggling question I came across while doing binomial theorem.
Let
$$\displaystyle\left(1-x^3\right)^n=\sum_{r=0}^na_rx^r(1-x)^{3n-2r},\quad n\gt2$$... | Dividing both sides by $(1-x)^{3n}$ shows that the equation is equivalent to
$$
\left(1+\frac{3x}{(1-x)^2}\right)^n=\sum_{r=0}^na_r\left(\frac{x}{(1-x)^2}\right)^r\tag1
$$
Set $u=\frac{x}{(1-x)^2}$, then we have
$$
\left(1+3u\right)^n=\sum_{r=0}^na_ru^r\tag2
$$
So $a_r=3^r\binom{n}{r}$, as ZAhmed computed.
Suppose tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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How to solve $\frac{1}{2\sin50^\circ}+2\sin10^\circ$ I want to solve the expression $\frac{1}{2\sin50^\circ}+2\sin10^\circ$ to get a much simpler and neater result. I have tried to manipulate this expression such as using sum/difference formulas, but it didn't help (and made the expression even more messy).
Here is wha... | $$\begin{align}\frac1{2\sin50^\circ}+2\sin10^\circ&=\frac{1+4\sin50^\circ\sin10^\circ}{2\sin50^\circ}\\
&=\frac{1+2(\cos40^\circ-\cos60^\circ)}{2\cos40^\circ}\\&=1
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"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 ... | $$\tan \frac{x}{2} = \frac{\sin x/2}{\cos x/2} = \frac{2 \sin x/2 \cos x/2}{2 \cos^2 x/2} = \frac{\sin x}{\cos x + 1}$$
hence with $\sin x = -\frac{3}{5}, \cos x = ±\sqrt{1 - \sin^2 x} = ±\frac{4}{5}$, $\tan \frac{x}{2} = -\frac{1}{3}, -3$.
| {
"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": 3
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Use the definition of a limit to prove the following: $\lim_{x\to4}\frac{x-2}{\sqrt{x}+2}=\frac{1}{2}$ Use the definition of a limit to prove the following:
$$\lim_{x\to4}\frac{x-2}{\sqrt{x}+2}=\frac{1}{2}$$
I'm trying to prove that:
$$\forall\varepsilon>0,\exists\delta>0; \\~\\
0<|x-4|<\delta\Longrightarrow\left|\fra... | Note that\begin{align}\frac{x-2}{\sqrt x+2}-\frac12&=\frac{2x-\sqrt x-6}{2\sqrt x+4}\\&=\frac{2(x-4)-\left(\sqrt x-2\right)}{2\sqrt x+4}\\&=\left(\sqrt x-2\right)\frac{2\sqrt x+3}{2\sqrt x+4}\\&=(x-4)\frac{2\sqrt x+3}{\left(\sqrt x+2\right)\left(2\sqrt x+4\right)}\\&\leqslant(x-4)\frac{2\sqrt x+3}8,\end{align}and there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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For $a,b,c>0$ and $a+b+c=6$. Prove that $\sum^{}_{cyc} \frac{ab}{\sqrt{a^2+b^2+2c^2}}\leq 3$ For $a,b,c>0$ and $a+b+c=6$. Prove that $$\sum^{}_{cyc} \frac{ab}{\sqrt{a^2+b^2+2c^2}}\leq 3$$
And this attempt $$\sum \frac{ab}{\sqrt{a^2+b^2+2c^2}}$$
$$=\sum \sqrt{\frac{a^2b^2}{\sqrt{a^2+b^2+2c^2}}}$$
$$\leq \sqrt{3}.\sqrt{\... | From $(a+b+c)^2 \geqslant 3(ab+bc+ca),$ we get $ab+bc+ca \leqslant 12.$
Now, using the Cauchy-Schwarz and AM-GM inequality, we have
$$\left(\sum \frac{ab}{\sqrt{a^2+b^2+2c^2}}\right)^2 \leqslant (ab+bc+ca)\sum \frac{ab}{a^2+b^2+2c^2}$$
$$ \leqslant \frac{1}{4}(ab+bc+ca) \sum \frac{(a+b)^2}{a^2+b^2+2c^2}$$
$$ \leqslant... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Diagonalizing the matrix $A$ then finding $A^{10}$ Diagonalize the matrix
$$A= \begin{pmatrix} -3 & -14 & -10\\ 2 & 13 & 10\\ -2 & -7 & -4 \end{pmatrix}$$
Then find $A^{10}.$
We have the characteristic polynomial of $A:$
$$\left | A- \lambda I \right |= \left | \begin{pmatrix} -3 & -14 & -10\\ 2 & 13 & 10\\ -2 & -7 & -... | Observe that if $A=SJS^{-1}$ we also have $A^n = SJ^n S^{-1}$ for all $n \in \mathbb N$. We can see this by induction with $A^2= SJS^{-1}SJS^{-1} = SJ^2S^{-1}$ and $A^n = SJS^{-1} SJ^{n-1}S^{-1} = SJ^nS^{-1}$
Now $J^{10} = \text{diag}(1,1, 6^{10})$
and $A^{10} = SJ^{10}S^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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prove that $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $ $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $
the only idea I have is using the series expansion of $x^{-x} \approx 1 - x\log x + \dfrac{(x\log... | Uniform convergence of the power series allows to interchange summation and integration:
\begin{aligned}
&\int_{0}^{1} x^{-x} \, \mathrm d x=\int_{0}^{1} e^{-x \ln x}\, \mathrm d x \Rightarrow \text { Applying Power Series of } e^{k x}=\sum_{n=0}^{\infty} \frac{(k x)^{n}}{n !} \\
&\Rightarrow \int_{0}^{1} \sum_{n=0}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $ without expansion It's easy to prove that if $a,b,c \neq 0 $: $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $$
as $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff \frac{ab+bc+ca}{ab... | $p(x)=(x-a)(x-b)(x-c)=x^3-ux^2+vx-w$ has roots $a,b,c$ with sum $a+b+c=u$.
$(a+b)(b+c)(c+a)=0 \iff (u-c)(u-a)(u-b)=0 \iff p(u) = 0\,$ thus:
$$
\require{cancel}
\cancel{u^3}-\cancel{u\,u^2} + v\,u-w=0 \;\;\iff\;\; \frac{v}{w}=\frac{1}{u} \;\;\iff\;\; \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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How to check if an integer can be represented with all set bits for any given base? By all set bits, I mean all the set bits are placed consecutively together. e.g., for base 2, we have 1, 3, 7, 15, 31, 63, i.e., any integer $x$ such that $x = 2^m - 1$ for some integer $m$.
In binary, $1$ is $1$ and $3$ is $11$ and $7$... | Your formula for $x$ is a geometric sum of $m$ digits of all $1$s,
$$
{\overbrace{111\cdots1}^m}_b = b^0+b^1+\cdots + b^{m-1} = \frac{b^m-1}{b-1} = \frac{10_b^m-1}{10_b-1}$$
To apply the usual proof of geometric sum specifically to $x$ in base $b$,
$$\begin{array}{crcrl}
&10_b x &= &{\overbrace{111\ldots 1}^m0}_b\\
-& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Throwing a coin - expected value of tails There are $10$ coins: $8$ of them are fair (equal probability for heads and tails) and for $2$ probability of heads is two times higher than for tails (so $\frac{2}{3}$ for heads). We draw one coin and throw it three times. Let $X$ denote a number of tails - I need to find a $\... | You know that the expectation of a binomial is $\mu=np$ thus you have
$$E(X)=\frac{8}{10}\times 3 \times \frac{1}{2}+\frac{2}{10}\times 3 \times \frac{1}{3}=\frac{14}{10}$$
this easy method works as you pmf is a mixture of two pmf's
$$p_X(x)=0.8p_{X_1}(x)+0.2p_{X_2}(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to graph and solve this equation? I'am trying to solve this equation.
\begin{equation}
x^{8}+(x+2)^{8}=2
\end{equation}
What I tried:
\begin{equation}g(x)=x^{8}+(x+2)^{8}\end{equation}
\begin{equation}
\begin{array}{l}
\text { }\\
y=x+1
\end{array}
\end{equation}
\begin{equation}
(y+1)^{8}+(y-1)^{8}=2
\end{equatio... | Let's try
$$ 1 + 1 = 2.$$
$$ \left\{
\begin{aligned}
x^8=1\\
(x+2)^8=1
\end{aligned}
\right.
$$
A solution is $x=-1.$
Actually, $x=-1$ is a double root:
Let $q(x) = x^8 + (x+2)^8 -2$. So we have $$q(x)=0.$$ We can factor q(x):
$$q(x) = 2 (x+1)^2 (127 + 258 x + 253 x^2 + 132 x^3 + 43 x^4 + 6 x^5 + x^6).$$
The equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the value of $α^3+β^3+γ^3+δ^3$ given that $α,β,γ,δ$ are roots of $x^4+3x+1=0$ Let $α,β,γ,δ$ be the roots(real or non real) of the equation $x^4-3x+1=0$. Then find the value of $α^3+β^3+γ^3+δ^3$.
I tried this question as $S_1=0, S_2=0, S_3=3, S_4=1$ and then I used $S_1^3$ to find the the value of asked question, b... | Using Vieta's theorem for $x^4+3x+1=0$, we get:
$$a+b+c+d=0 \Rightarrow b+c+d=-a\\
ab+ac+ad+bc+bd+cd=0 \Rightarrow bc+bd+cd=a^2\\
abc+abd+acd+bcd=-3\Rightarrow a^3=-bcd-3\\
abcd=1$$
Hence:
$$a^3+b^3+c^3+d^3=(-bcd-3)+(-acd-3)+(-abd-3)+(-abc-3)=\\
-(abc+abd+acd+bcd)-12=3-12=-9.$$
Note: $x^4+3x+1=0$ is given in the title,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Alternative approaches to maximize $y=x\sqrt{100-x^2}$ I could find three good approaches to find maximum of the function $y=x\sqrt{100-x^2}$. I will explain them briefly :
First: Finding $x$ satisfies $y'=0$ then plugging it in the function.
Second: Using the substitution $x=10\sin\theta$ (or $x=10\cos\theta)$ for
$\t... | Let $f(x) = x\sqrt{100-x^{2}}$. Then $x\in[-10,10]$.
In order to find the maximum value, we can consider that $x\geq 0$ (otherwise $f(x) \leq 0$).
For such values of $x$, one has that
\begin{align*}
f(x) & = x\sqrt{100 - x^{2}}\\\\
& = \sqrt{100x^{2} - x^{4}}\\\\
& = \sqrt{2500 - (2500 - 100x^{2} + x^{4})}\\\\
& = \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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How to convert $\alpha = \cos(\frac{8\pi}{11})+ i \sin(\frac{8\pi}{11})$ into $\text{n}^{th}$ roots of unity format? Context: Preparing for JEE and it is one of the practice problems.
Question:
If $\alpha = cos(\frac{8\pi}{11})+ i sin(\frac{8\pi}{11})$ then $\Re({\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5})$ is equa... | Roots of unity is indeed the correct approach here, rather than geometric series.
$a^{11} = 1$ and $a \bar{a} = 1$. Thus the conjugate of $a$, $\bar{a} = \frac{1}{a} = \frac{1}{a} \cdot a^{11} = a^{10}$, which has the same real part as $a$. Now since:
$$1 + (a + a^2 + \cdots + a^5) + (a^6 + a^7 + \cdots + a^{10})$$
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A Problem dealing with a Deck of Cards and the Hypergeometric Distribution Problem:
If $13$ cards are to be chosen at random(without replacement), determine
the probability that (a) $6$ will be picture cards, (b) none will be
picture cards.
Answer: (a)
The probability that a card drawn from a deck of cards will be a
pi... | Your answer is correct, and the book is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding all polynomials $P(x)$ satisfying $(P(x)+P(\frac{1}{x}))^2 =P(x^2)P(\frac{1}{x^2})$
Find all polynomials $P(x)$ satisfying
$$\left(P(x)+P\left(\frac{1}{x}\right)\right)^2 =P(x^2)P\left(\frac{1}{x^2}\right)$$
If $P'(x)\neq 0$, then $P(x) = a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$
$\Rightarrow$ ($a_nx^n+a_{n-1}x^{n... | We claim that the only solutions of the equation is $P(x) = 0$.
Let $P(x) = \sum_{k} a_k x^k$ satisfy the given equation. Write $d = \deg P $.
Lemma. $P(x)$ is an even polynomial.
Proof. The conclusion is obvious if $P(x)$ is constant, so we only consider the case where $P(x)$ is non-constant.
First, we show that $P(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Different answers of same differentiation Question with two different methods
Question:- Find $\frac{dy}{dx}$ if $$\arccos\bigg({\frac{x^2-y^2}{x^2+y^2}}\bigg)=\arctan( a)$$
$$\arccos\bigg({\frac{x^2-y^2}{x^2+y^2}}\bigg)=\arctan( a)$$
$$\implies \frac{x^2-y^2}{x^2+y^2}=\cos(\arctan(a))$$
Taking derivative on both s... | In more detail, suppose that for some constant $c$
$$ c =\frac{x^2-y^2}{x^2+y^2}. \tag{1} $$
Note that this is undefined iff $\,x^2+y^2 = 0.\,$
Solve this for $\,y\,$ explicitly to get
$$ y = \pm \sqrt{\frac{1-c}{1+c}}x = \pm C\, x. \tag{2}$$
Take the implicit differential of equation $(1)$ to get
$$ 0 = \frac{4 x y (x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{y+1}$ correct even if $y=-1$? I was trying to solving this question:
If roots of the equation $a x^{2}+b x+c=0$ are $\alpha$ and $\beta$, find the equation whose roots are $\frac{1-\alpha}{1+\alpha}, \frac{1-\beta}{1+\beta}$
I was not able to solve it so I... | Is $\frac{1-\alpha}{1+\alpha}=y \Rightarrow \alpha=\frac{1-y}{y+1}$ correct even if $y=-1$?
The tag algebra-precalculus is given. The answer to the question is: No.
But with tag riemann-sphere, the answer to the question would be: Yes.
Calculating in the Riemann sphere, we have
$$
\frac{1-\infty}{1+\infty} = -1\qquad\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How to evaluate the following limit $\lim_{n\to \infty} \sum_{k=2}^n\log_{\frac{1}{3}} \left(1-\frac{2}{k(k+1)}\right)$? In order to evaluate this limit :
$$\lim_{n\to \infty} \sum_{k=2}^n\log_{\frac{1}{3}} \left(1-\frac{2}{k(k+1)}\right)$$
I have to compute the following sum :
$$ \sum_{k=2}^n\log_{\frac{1}{3}} \left(... | In this answer I'm considering $b=\frac{1}{3}$.
\begin{align}
\sum_{k=2}^n \log_b\left(1-\frac{2}{k(k+1)}\right)&= \sum_{k=2}^n \log_b\left(\frac{k^2+k-2}{k(k+1)}\right)\\
&=\sum_{k=2}^n \log_b \left(\frac{k+2}{k+1}\right) +\log_b \left(\frac{k-1}{k}\right)\\
&=\sum_{k=2}^n \log_b (k+2)-\log_b(k+1) +\sum_{k=2}^n \log(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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With $x^2+y^2=1$ find Minimum and Maximum of $x^5+y^5$ (do not use derivative) It's easy to see that the minimum is $-1$ and maximum is $1$.
My idea is put $x=\cos(a)$, $y=\sin(a)$ and $t=x+y$, so I have $-\sqrt{2}\le t \le \sqrt{2}$ then $0 \le t^2 \le 2$
When $t=x+y$ then $(x+y)^2=1+2xy$.
Then
\begin{aligned}
x^5+y... | My solution using AM -GM.
It's more complex than @Aman Kushwaha. His solution is very simple and naturally. Sorry for not using MathJax
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the minimum of $P(X=0)$ when $E[X]=1,E[X^2]=2,E[X^3]=5$ using the probability generating function I was given the following exercise.
Let $X$ be a random variable that takes non-negative natural number
values such that $E[X]=1,E[X^2]=2,E[X^3]=5$. Find the minimum value of
$P(X=0)$ using the taylor expansion of th... | Conditional on a given value $X=N\geq 4$ we get the tail sum
$$\sum_{n=4}^N {N\choose n} (-1)^n=- ({N\choose 0}-{N\choose 1} +{N\choose 2}-{N\choose 3})=$$
$$-(1-N+N(N-1)/2+N(N-1)(N-2)/6)$$
$$=\frac{1}{6} (N-1)(N-2)(N-3)={N-1\choose 3}>0.$$
(We re-derived a special case of a known identity for binomial coefficients.)
O... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Converting $\intop_{0}^{a}\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}dx$ to elliptic integral Tried using $x=a\sin\left(\theta\right)$ $$\rightarrow \intop_{0}^{\pi/2}\sqrt{\frac{a^{2}-\left(a\sin\left(\theta\right)\right)^{2}}{1-\left(a\sin\left(\theta\right)\right)^{2}}}a\cos{\left(\theta\right)d\theta}$$
$$\iff \intop_{0}^{\... | You can convert this integral in terms of elliptic functions. You need to make an assumption about the parameter $a$:
If $0<a<1$ we can continue with your approach:
First, note
$$ J=\intop_{0}^{\pi/2} \frac{\sin^{2}\left(\theta\right)}{\sqrt{1-k^2\sin^2\left(\theta\right)}}{d\theta} = \frac{K(k)-E(k)}{k^2}$$
To show th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
I don't know how to exactly compute this determinant I've tried to compute this determinant by row transformations and column transformations, but it gives me a formula that doesn't work. The determinant is:
\begin{vmatrix}
x & a & b & c & d\\
a & x & b & c & d\\
a & b & x & c & d\\
a & b & c & x & d\\
a & b & c & d & ... | Another approach: Consider it as a polynomial in $x$. Note that it will vanish when $x$ is any of $a,b,c,d$, since then we see a repeated row. It will also vanish when $x = -a-b-c-d$, since then all rows sum to $0$, and so we have the eigenvalue $0$ with eigenvector $(1,1,1,1,1)$. Furthermore, we know that the leading... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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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... | A classic fact is that if $a+b+c=3$ then $a^{p}+b^{p}+c^{p}\ge 3$ for every $p\ge 1$ this can be proved by Hôlder, C-S when $p=2$ or Power Mean inequality.
Here suppose the converse $a^2+b^2+c^2+ab+bc+ac<6$, $(p=2)$ necessarily $ab+bc+ac<3$ adding this you get $a^2+b^2+c^2+2(ab+bc+ac)=(a+b+c)^2<9$ a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
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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.
$$\int\frac{x^4}{\... | You need to substitute $\theta=\arcsin(\frac{x}{a})$, or as you said just find $\sin(4\theta) $ and $\sin (2\theta)$ using the double angle properties and put the values in the final answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Conditional inequality $2a^3+b^3≥3$
Non-negative $a$ and $b$ such that $a^5+a^5b^5=2$. How then do I prove
the following inequality $2a^3+b^3≥3$?
So, we can try using the Lagrange multiplier method:
Let $f(a, b)=2 a^{3}+b^{3}+\lambda(a^{5}+a^{5} b^{5}-2), \quad a, b \geq 0$
$$\tag1 \frac{\partial f}{\partial a}=6 a^{... | $b^5=\frac{2-a^5}{a^5},$ where $0<a<\sqrt[5]{2}$ and we need to prove that
$$2a^3+\left(\frac{2-a^5}{a^5}\right)^{\frac{3}{5}}\geq3$$ and since for $3-2a^3\leq0$ the inequality is obvious, we need to prove that $f(a)\geq0$ for any $0<a<\sqrt[3]{\frac{3}{2}},$ where $$f(a)=\frac{3}{5}\ln(2-a^5)-3\ln{a}-\ln(3-2a^3).$$
We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$(\frac{s-a_1}{n-1})^{a_1} \cdot (\frac{s-a_2}{n-1})^{a_2}\cdots (\frac{s-a_n}{n-1})^{a_n} \le (\frac{s}{n})^s$ If $a_1,a_2,\ldots,a_n$ be $n$ positive rational numbers and $s=a_1+a_2+\cdots+a_n$, prove that
$$\left(\frac{s-a_1}{n-1}\right)^{a_1} \cdot \left(\frac{s-a_2}{n-1}\right)^{a_2}\cdots \left(\frac{s-a_n}{n-1}\... | AM-GM gives:
$\frac{\left(\frac{s-a_1}{n-1}\right)a_1 + \left(\frac{s-a_2}{n-1}\right)a_2+ \cdots +\left(\frac{s- a_n}{n-1}\right)a_n}{a_1 + a_2 +\cdots+a_n} \ge \left(\left(\frac{s-a_1}{n-1}\right)^{a_1} \cdot \left(\frac{s-a_2}{n-1}\right)^{a_2}\cdots \left(\frac{s-a_n}{n-1}\right)^{a_n}\right)^{\frac{1}{s}}$
LHS $= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$. I want to find limit of $a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}$ as $n\to \infty$.
$\displaystyle a_{n} =\frac{\left( 2n^{\frac{1}{n}} -1\right)^{n}}{n^{2}} =\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right)^{n}$
$\displaystyle \begin{array}{{>{\displaystyl... | Power Series Approach
Using the power series $e^x=1+x+O\!\left(x^2\right)$, we get
$$
\begin{align}
2n^{1/n}-1
&=2e^{\frac1n\log(n)}-1\tag{1a}\\[6pt]
&=2\left(1+\frac1n\log(n)+O\!\left(\frac{\log(n)^2}{n^2}\right)\right)-1\tag{1b}\\
&=1+\frac2n\log(n)+O\!\left(\frac{\log(n)^2}{n^2}\right)\tag{1c}
\end{align}
$$
and
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Solve the integral of heat equation with only boundary condition $u\left(x,0\right)=\dfrac{1}{x^2+1}$ with $x \in R$ The most thing I want is solve this integral
$$\frac{1}{{16t\sqrt \pi }}\int\limits_{ - \infty }^{ + \infty } {\frac{{\exp \left( { - {a^2}} \right)}}{{{{\left( {a + \frac{x}{{4\sqrt t }}} \right)}^2} +... | We convolve the initial condition with the fundamental solution:
$$u(x,t)=\int_{\mathbb{R}}\frac{1}{1+y^2}\frac{1}{\sqrt{\pi 16t}}\exp\bigg\{-\frac{(x-y)^2}{16t}\bigg\}dy=\mathbb{E}_{Y \sim\mathcal{N}(x,8t)}\bigg[\frac{1}{1+Y^2}\bigg]$$
This has been solved on stats.stackexchange. The solution to the PDE is
$$u(x,t)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two sides of quadrilateral $ABCD$ intersect at $E$ Quadrilateral $ABCD$ is inscribed in a circle. The extensions of $AD$ and $BC$ intersect at $E$, such that $D$ is between $A$ and $E$. If $BC=2DE,AD:EC=7:2$ and $\cos\measuredangle AEB=\dfrac78$, find $\cos\measuredangle ABC$.
Since $\measuredangle ABC+\measuredangle ... | Hint:
From the similarity of $\triangle ABE$ and $\triangle CDE$ we have
$$\frac{x}{2y} = \frac{2x+2y}{x+7y}$$
$$x^2+7xy = 4xy + 4y^2 $$
$$(x+4y)(x-y) = 0$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Divisibility rules with prime number Let $n \in \mathbb{Z}$ with the property that
$$
7 \mid\left(n^{3}+1\right)
$$
but $7$ does not divide $\left(n^{2}-2 n-3\right) .$ Prove that $7 \mid(4 n+1)$.
So far I got:
$n^3+1 = (n+1)\cdot(n^2-n+1)$, since $7$ is a prime, so
$7\mid(n+1)$ or $7\mid(n^2-n+1)$.
And since $7$ does ... | Two methods:
Method 1
If $n^3\equiv-1\bmod 7$ then $n\in\{3,5,6\}\bmod 7$. But $n\in\{3,6\}$ renders $n^2-2n-3=(n+1)(n-3)\equiv0$, denied by hypothesis, and the last possibility $n\equiv5$ renders $4n+1\equiv0$.
Method 2
Multiply $n^2-2n-3$ by $4n+1$:
$(n^2-2n-3)(4n+1)=4n^3-7n^2-14n-3\equiv4(n^3+1)\bmod 7$
Thus if the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Is there a technique in how to write certain expression in certain times of integration as in case of $\int \frac {(3x + 2)} {(5x + 1)^2}dx$
I can solve this integration $$\frac{3} {5}\int\frac{1}{5x + 1}dx + \frac{7} {5}\int\frac{1}{(5x + 1)^2}dx$$ but thing is I dont under how that online calculator wrote $$3x + 2$$... | It is just partial fraction decomposition. The key step is to write $3x + 2 = A(5x + 1) + B$, as you want the $5x+1$ terms to cancel so you can integrate $\frac{A}{5x+1} + \frac{B}{(5x+1)^2}$, which can be done by $u$-substitution.
Comparing coefficients will be easier than substituting $x = -\frac{2}{3}, -\frac{1}{5}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $m^3-n^3=2mn+8$ We have to find integer solutions to the given equation, this is what i tried :-
For ease, let us denote $x=-m, \enspace n=-y$, and then we are basically considering $x,m,y,n$ as nonnegative wherever they occur below :-
We have four cases :-
CASE $1$ : $mn<0$ and $x^3+n^3=-(8+2xn)$
This case is... | Observe that $x^3 - (x-1)^3 > 2x(x-1) + 8$ for all $x$ except $-2 ≤ x ≤ 3$ and that $x^3 - x^3 < 2x(x)+ 8$ for all $x$. From this, the equation lies between $n = m-1$ and $n = m$, hence the only integer solutions are when $-2 ≤ m ≤ 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
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... | If $a$ and $b$ are two coprime positive integers and
$$x_1^2\equiv1\pmod{a}\qquad\text{ and }\qquad x_2^2\equiv1\pmod{b},\tag{1}$$
then by the Chinese remainder theorem there exists a unique integer $x$ with $0\leq x<ab$ such that
$$x\equiv x_1\pmod{ab}\qquad\text{ and }\qquad x\equiv x_2\pmod{ab},$$
which consequently... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4229793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I show that $\frac {|s|} {\sigma} \leq \sqrt {1 + \tan \left (\frac {\pi} {2} - \delta \right )}\ $? Show that in the region $\left \{s\ \big |\ \left |\arg \left (s \right ) \right | \leq \frac {\pi} {2} - \delta \right \}$ $$\frac {|s|} {\sigma} \leq \sqrt {1 + \tan \left (\frac {\pi} {2} - \delta \right )}$$ ... | Your derivation of
$$
\frac {|s|} {\sigma} \leq \sqrt {1 + \tan^2 \left (\frac {\pi} {2} - \delta \right )}
$$
is correct, and equality holds if $|\arg(s)| = \pi/2 - \delta$.
If $|\arg(s)| = \pi/2 - \delta$ and $0 <\delta < \pi/4$ then
$$
\frac {|s|} {\sigma} = \sqrt {1 + \tan^2 \left (\frac {\pi} {2} - \delta \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4229936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating the Image of $A.$ Where did I err? $$A=\begin{pmatrix}
4 &-1& 1\\
8&-2&2\\
-6&1&-2\\
\end{pmatrix}$$
I have to show $p=\begin{pmatrix}
1 \\
2\\
-2 \\
\end{pmatrix} \in \mathrm{Im}A=\left\{Ax \mid x\in \mathbb{R^3} \right\}$
If I let $x=\begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix},$ then $Ax=\begin{pmatrix}
... | Your mistake is that $\mathrm{C\left(A\right) ≠ span}\left(\begin{bmatrix}1\\0\\0\end{bmatrix},\ \begin{bmatrix}0\\1\\0\end{bmatrix}\right)$.
Elimination actually changes the column space of $\mathrm A$, so $\mathrm{C\left(A\right) ≠ C\left(rref\left(A\right)\right)}$.
By reducing $\mathrm A$ you just revealed a number... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cardinal of the set $A=\Big\{x=\frac pq\in\mathbb{Q} :|\sqrt 2-x|<\frac1{q^3}\Big\}$ Let $A=\Big\{x=\frac pq\in\mathbb{Q} :|\sqrt 2-x|<\frac1{q^3}\Big\}$
We want to prove that $A$ is finite and find its cardinal? we can prove that it's finite by using directly Roth’s theorem which is a generalized theorem of Lionville'... | Note that $|2 - \frac{p^2}{q^2}| \ge \frac{1}{q^2}$, and $\sqrt{2}+\frac{p}{q}< \sqrt{2} + (\sqrt{2} + \frac{1}{q^3})$ therefore
$$\frac{1}{q^3} >|\sqrt{2} - \frac{p}{q}| = \frac{|2 - \frac{p^2}{q^2}|}{\sqrt{2}+\frac{p}{q}}> \frac{\frac{1}{q^2}}{2 \sqrt{2}+\frac{1}{q^3}} $$
and so
$$2\sqrt{2}+\frac{1}{q^3}> q$$
We co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simplifiying $A \cos(x-\phi)$ I have to proof that we can transform this expression : $$a \cos x + b \sin x$$
to this one : $$A \cos (x-\phi)$$
Indeed, I can do the reverse path using the formula of $\cos(x-y)= \cos x \cos y + \sin x \sin y$
but I get stuck on the initial path.
Which trig formula should I use?
Thanks
| You have
\begin{align*}
A\cos(x-\phi)&=A\cos x\cos \phi+A\sin x\sin \phi\\
&=(A\cos \phi)\cos x+(A\sin \phi)\sin x\\
&=a\cdot \cos x+ b\cdot \sin x
\end{align*}
where $a=A\cos \phi$ and $b=A\sin \phi$
Now, let us try to solve the equations
$$a=A\cos \phi\\ b=A\sin \phi$$
Squaring the equations and adding them, we get
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Number of ternary sequences of length $n$ with same number of $1$'s and $0$'s My attempt is considering the number of $1$'s to be $k$ then for each $k$ we choose choose the inner order of the $1$'s and $0$'s which has ${2k \choose k }$ options then choosing the indices of the $2$'s which has ${ n \choose n -2k } = { n ... | We find using OPs approach
\begin{align*}
\color{blue}{\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{2k}{k}\binom{n}{2k}}
&=\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{k!k!(n-2k)!}=\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{k!(n-k)!}\,\frac{(n-k)!}{k!(n-2k)!}\\
&\color{blue}{=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluate $\int_0^\infty \frac{e^{-x}}{x}\ln\big(\frac{1}{x}\big) \sin(x)dx$ I'm having trouble with this integral
Evaluate $$\int_0^\infty \frac{e^{-x}}{x}\ln\left(\frac{1}{x}\right) \sin(x)dx$$
$$I=\int_0^\infty \frac{e^{-x}}{x}\ln\left(\frac{1}{x}\right) \sin(x)dx=\Im\left[\int_0^\infty \frac{e^{-x+ix}}{x}\ln\left(... | Using the identities
\begin{align*}
\frac{1}{x} &= \int_0^\infty \mathrm{e}^{-x t}\, \mathrm{d} t, \\
\ln x &= \int_0^\infty \frac{\mathrm{e}^{-s} - \mathrm{e}^{-xs}}{s}\,\mathrm{d} s,
\end{align*}
we have
\begin{align*}
I &= \int_0^\infty \mathrm{e}^{-x}\sin x
\left(\int_0^\infty \mathrm{e}^{-x t}\, \mathrm{d} t\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
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
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
... | The following is an elementary, entirely self-contained solution which does not assume knowledge of Vieta's relations, Newton's identities, or anything other than simple algebra. (This is not the best or recommended way to solve it - see the other answers and comments for that - but it answers OP's edit asking for a mo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 4
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How to solve $\sin(2\theta)$ questions Given that: $\sin\theta=\displaystyle{}\frac{12}{13}$ and $0<\theta<\displaystyle{}\frac{\pi}{2}$ the value of $\sin(2\theta)$ is:
I figured out a way to solve it, though I'm not sure if it is the best solution.
Here we will combine two different trigonemtric identities. First:
$\... | The line
$$\cos^2\theta = \frac{25}{169}$$
simplifies to
$$|\cos\theta| = \frac{5}{13}$$
since $\sqrt{x^2} = |x|$. Since $\cos\theta > 0$ if $0 < \theta < \dfrac{\pi}{2}$, $|\cos\theta| = \cos\theta$ in this interval, which allows you to conclude that
$$\cos\theta = \frac{5}{13}$$
The rest of your work is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Integrate $\int^{\pi}_{0}\left\{\frac{\tan^2\left(\frac{x}{2}\right)}{\sin^2(x)\cdot\,\cos^2(x)}\right\}^{\frac{1}{9}}\,dx$ $\displaystyle\int^{\pi}_{0}\left\{\dfrac{\tan^2\left(\dfrac{x}{2}\right)}{\sin^2(x)\cdot\,\cos^2(x)}\right\}^{\frac{1}{9}}\,dx$
$\sf{\color{blue}{My\,\,approach\,}:}$
$=\displaystyle\int^{\pi}_{0... | $I = \dfrac{{0,028 \cdot {{2,496}^3}}}{{12}} + 2 \cdot \left( {\dfrac{{0,62 \cdot {{0,052}3}}}{{12}} + 0,62 \cdot 0,052 \cdot {{1,274}2}} \right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Maximum value of $(ab+1)(bc+1)(cd+1)(da+1)$: mistake in solution? Find maximum value of $(ab+1)(bc+1)(cd+1)(da+1)$ if $abcd=1$ and $\frac{1}{2}≤a,b,c,d≤2$
$$ab+cd≥2(abcd)^{1/2}=2$$
$$da+bc≥2(abcd)^{1/2}=2$$
$$(ab+1)(bc+1)(cd+1)(da+1)≤(\frac{ab+1+bc+1+cd+1+da+1}{4})^4$$
$$\implies ≤(\frac{2+2+4}{4})^4=16$$
However, $(2,... | The direction of your second inequlity sign in the followilng should be reversed
$$(ab+1)(bc+1)(cd+1)(da+1)≤(\frac{ab+1+bc+1+cd+1+da+1}{4})^4$$
$$\implies ≤(\frac{2+2+4}{4})^4=16$$
Thus your implication is not valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4249430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to prove $\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}$ Prove that
$$
\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}
$$
My attempt :I tried to use the beta function, but I couldn't.
| In search of a simple solution and after many attempts, I found this solution. Let $\cos \theta = \tan \alpha$ and $d\theta = -\dfrac{\sec^2 \alpha \ d\alpha}{\sqrt{1 - \tan \alpha}}$. Thus,
$$
I = \int_{0}^{\pi/2}\dfrac{\sqrt{\cos \theta}}{1 + \cos^2 \theta}d\theta = \int_{\pi/4}^{0}\dfrac{\sqrt{\tan \alpha}(-\sec^2 \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the minimum value of: $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$ Let $a,b,c>0$, find the minimum value of: $$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$$
I have tried:
$\bullet$ The minimum value is $\dfrac{3}{4}$ occur when $a=b=c$
$\bullet \dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}= \dfrac{a^2}... | The value $\frac 3 4$ is a maximum, indeed we have that by $x=\frac b a$, $y=\frac cb$, $z=\frac a c$ with $xyz=1$
$$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c} = \dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3} \le \frac 3 4$$
and
$$\frac34-\left(\dfrac{1}{x+3}+\dfrac{1}{y+3}+\dfrac{1}{z+3}\right) =\frac{3 x y z+ 5 y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Number of different value of given expression for $x,y,z \in \{1,2,3,4,5\}$ Let $x,y,z \in \{1,2,3,4,5\}$, then find number of different values of
$$\dfrac{x^3}{(x-z)(x-y)}+\dfrac{y^3}{(y-x)(y-z)}+\dfrac{z^3}{(z-x)(z-y)}$$
Could someone please share the approach to start this question. I tried to factorize the expressi... | First, by observing the form of the denominator, we can immediately pose the condition that $x \neq y \neq z$ for the fraction to be well-defined.
Next, because all $3$ fractions are identical, permutation of a set of $(x,y,z)$ will all result in the same value (i.e. $(x,y,z) = (1,2,3)$ will give the same value as $(x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4252656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:
$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \rig... | A somewhat simpler approach using contour integration is to first notice that $$ \begin{align} \int_{-\infty}^{\infty} \frac{\operatorname{sech}^{2}\left(\frac{\pi x}{2}\right)}{\left(x+i \right)^{2}} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{x^{2}-1}{(1+x^2)^{2}} \operatorname{sech}^{2} \left(\frac{\pi x}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 3
} |
problem with power series $\sum \frac{x^n}{n+3}$ I need to evaluate the following power series, and I don’t really know how to do it, this is the series $$\sum \frac{x^n}{n+3}$$
This is how I tackled this problem, but the final solution looks very dumb.
So we know that $$\frac{1}{1-x}=\sum x^n$$
But we also know that... | Note that
$$-\int\frac{x^2}{x-1}dx=-\int\frac{x^2-1+1}{x-1}dx=-\int\frac{(x+1)(x-1)+1}{x-1}dx$$
$$=-\int(x+1)dx-\int\frac{1}{x-1}dx=-\frac{x^2}{2}-\color{red}{x}-\ln|x-1|+C$$
So
$$\sum_{n=0}^{\infty}\frac{x^{n+3}}{n+3}=-\frac{x^2}{2}-\color{red}{x}-\ln|x-1|$$
and we have
$$\sum_{n=0}^{\infty}\frac{x^n}{n+3}=-\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are relations of $S_4$, if generators are $a=(12)$, $b=(1234)$? I think, relations could be $a^2=e$, $b^4=e$, $ababab=e$, $b^2 a b^2 a b = b a b^2 a$.
By two first relations and $b^3 = ababa$ (it is result of our relations), we can present any element that is generated by $a$ and $b$, as a sequence, where there a... | As Derek Holt mentions in the comments you have $S_4 \cong \langle x,y | x^2 = y^4 = (xy)^3 = 1 \rangle$, so for the generators $a = (1\ 2)$ and $b = (1\ 2\ 3\ 4)$ three relations $a^2 = b^2 = (ab)^3 = 1$ suffice.
Let $G = \langle x,y | x^2 = y^4 = (xy)^3 = 1 \rangle$. In $S_4$ you have a normal subgroup $V \cong C_2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find all functions $f: \mathbb{R} \to\mathbb{R} $ such that if $a+f(b)+f^2(c) = 0$ then $f(a)^3 + bf(b)^2 + c^2f(c) = 3abc$
Find all functions $ f : \mathbb R \to \mathbb R $ such that for all $ a , b , c \in \mathbb R $, if
$$ a + f ( b ) + f ^ 2 ( c ) = 0 \tag {Eq1} \label {eqn1} $$
then
$$ f ( a ) ^ 3 + b f ( b ) ^... | First of all, note that the problem can be stated in the following equivalent form.
Find all functions $ f : \mathbb R \to \mathbb R $ such that
$$ f \left( - f ( x ) - f ^ 2 ( y ) \right) ^ 3 + 3 x y \left( f ( x ) + f ^ 2 ( y ) \right) + x f ( x ) ^ 2 + y ^ 2 f ( y ) = 0 \tag 0 \label 0 $$
for all $ x , y \in \mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$ I'm trying to evaluate the following expression$$(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})$$
I'm not really used to these types of problems, so I first tried using logarithms but I'm no... | Sayan Dutta shows a clever way. Another way is to note that when you multiply the parentheses, each term will be of the form $2^n 3^{127-n}$:
$$
(2^1 + 3^1)(2^2 + 3^2)(2^4 + 3^4)(2^8 + 3^8)(2^{16} + 3^{16})...(2^{64} + 3^{64})
= \prod_{k=0}^{6} (2^{2^k}+3^{2^k})
\\
= \sum_{n=0}^{127} 2^n 3^{127-n}
= 3^{127} \sum_{n=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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: $$a^8+b^8+c^8\ge 8(a+b+c)-21=3... | Let $a=b=c=1$.
Thus, we obtain a value $3$.
We'll prove that it's a minimal value, for which it's enough to prove that $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)\geq3.$$
We'll prove that this inequality is true even for any reals $a$, $b$ and $c$ such that $a+b+c=3$.
Indeed, by AM-GM $$\sum_{cyc}a^8\geq\sum_{cyc}(2a^4-1)$$ and it'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Range of $\frac{x-y}{1+x^2+y^2}=f(x,y)$ I have a function $\frac{x-y}{1+x^2+y^2}=f(x,y)$. And, I want to find the range of it. I analyzed this function by plotting it on a graph and found interesting things. Like if the level curve is $0=f(x,y)$, then I get $y=x$ which is a linear function. But if the level curve is so... | The denominator of the expression for the function $ \ z \ = \ f(x,y) \ = \ \frac{x-y}{1+x^2+y^2} \ $ has "four-fold" symmetry about the origin, which is "broken" by the "diagonal" (anti-)symmetry in the numerator. So this function has the symmetry property $ \ f(x,-y) \ = \ -f(-x,y) \ \ $ and it is reasonable to e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4262138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
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How to find numbers like 174 We know Ramanujan number:
$1729 = 1^3+12^3 = 9^3+10^3$
The smallest number expressible as the sum of cubes of two positive integers in two different ways.
We also know how to find other Ramanujan numbers:
$n^3 +(12n)^3 = (9n)^3 + (10n)^3$
In the same way, I want to find $174$
$$\begin{align... | There's a lot you could throw at it actually. Some of these include :
$$y=mx+b\land p\mid m, p\in \mathbb{P} \implies y^p\equiv b^p\pmod {pm}$$
$$y=a^2+b^2+c^2+d^2\implies a\leq b\leq c\leq d$$
The fact that distinct $(a,b,c,d)$ for each sum is only possible above $576=24^2$ because there will be $24$ squares involved.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4262462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find $a$ and $b$ given this limit if $f(x)=ax+b$? Question
Suppose $f(x)=ax+b$ and $\lim\limits_{x \to 25} \dfrac{\sqrt{x}-5}{f(x)}=\dfrac{1}{20}$.
Find $a$ and $b$.
Attempt
I figured out that
$\lim\limits_{x \to 25} \sqrt{x}-5=0\space$ but $0$ divided by any denominator will give $0.\quad$ So I do not know how ... | Following the hints in the comments, $ax+b$ must go to $0$ when $x$ goes to 25. So $$\lim_{x\to 25}ax+b=25a+b=0$$
That way the limt has the form of $0/0$:$$\lim_{x\to 25}\frac{\sqrt x-5}{f(x)}\to\frac 00$$
Since $25$ is a root of $ax+b$, we can write $$f(x)=a(x-25)=a(\sqrt x -5)\sqrt x+5)$$
It is equivalent of saying $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Binomial Expansion related problem If ${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - {}^n{C_3} + .. + {\left( { - 1} \right)^r}{}^n{C_r} = 28$. Then find the value of $n$.
My approach is as follow ${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - {}^n{C_3} + .. + {\left( { - 1} \right)^r}{}^n{C_r} = 28 = {}^8{C_2} = {}^8{C_6}$
${\left( {1 ... | Pascal's identity is sufficient :
$$ {N-1 \choose R-1} + {N-1 \choose R} = {N \choose R}$$
Since $${n \choose 0} = 1 = {n-1 \choose 0}$$
given expression is
$${n-1 \choose 0} - {n \choose 1} + {n \choose 2} - {n \choose 3} + \ldots + (-1)^r {n \choose r}$$
Operating on first two terms every step,
$$= \color{red}{- {n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4264207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Does the (Stirling number of the second kind) equality ${2n\brace 2} = 2^{2n-1}-1$ hold? I filled in from the definition of a Stirling number of the second kind that the following holds.
$${2n\brace 2} = \frac{1}{2} \sum_{i=0}^{2} (-1)^i \binom{2}{i} (2-i)^{2n}$$
And I've visually confirmed in Desmos that the following... | I guess the most straight-forward way is to expand the summation:
\begin{align}
{2n\brace 2} = &\frac{1}{2} \sum_{i=0}^{2} (-1)^i \binom{2}{i} (2-i)^{2n} = \\
& \frac{1}{2} \Bigg( \Big[(-1)^0 \binom{2}{0} (2-0)^{2n}\Big] + \Big[(-1)^1 \binom{2}{1} (2-1)^{2n}\Big] + \Big[(-1)^2 \binom{2}{2} (2-2)^{2n}\Big] \Bigg) = \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4264856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to solve for the zeros of $3x^4 - 7x^3 - 12x^2 + 12x + 11$? I’ve tried to see if a rational factor would work from $q/p$, but that hasn’t worked.
I tried grouping and then factoring but that does not work either.
I do not know any further methods that I could use since it doesn’t seem like I could use the quadratic... | Hints: You can convert the equation such that factorization becomes easier:
$3x^4-7x^3-12x^2+12x+11=0$
rewrite as:
$x^4-\frac 73 x^3-\frac 4x^2+4x+\frac {11}3=0$
Let y=kx :
$y^4-\frac 73 \cdot \frac{y^3}{k^3}-4\cdot \frac{y^2}{k^2}+4\cdot\frac yk +\frac{11}3=0$
let $k=3$:
$y^4-7y^3-36y^2+108y+297=0$
Now you can factor ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving central limit theorem in a specific case( fourier analysis course) I am stuck with the following problem: Show that for any given $R > 0$
$$\lim_{N \to \infty}\sup_{\left|\xi \right| \le R} \left |\left(\frac{1}{6}\sum_{k=1}^{6}e^{2\pi i\xi \frac{k-\frac{7}{2} }{\sqrt{N}} }\right)^N -e^{\frac{-35\pi^2\... | I will show what a possible start could be and let you fill in the technical details.
Using $z + z^* = 2 \mathrm{Re}(z)$ we have
$$
\left(\frac{1}{6} \sum_{k = 1}^{6}{e^{2 \pi i \xi \frac{k - \frac{7}{2}}{\sqrt{N}}}} \right)^N
= \left(\frac{1}{3} \sum_{k \in \{1, 3, 5\}} \frac{e^{\pi i \xi \frac{k}{\sqrt{N}}} + e^{-\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is the proof for variance of triangular distribution? In Wikipedia, the formula for the variance of the triangular distribution is given here. However, I don't know how to find it. I have tried a brute force method but the formula is quite complicated (polynomial of degree 5 in a, b, c) and I can't simplify it (I ... | It's a little easier to use the identity $\ \sigma^2=\mathbb{E}\big(Z^2\big)-\mathbb{E}(Z\,)^2\ $:
\begin{align}
\mathbb{E}\big(Z^2\big)&=\frac{2}{(b-a)(c-a)}\int_a^cx^2(x-a)\,dx\\
&\hspace{2em}+\frac{2}{(b-a)(b-c)}\int_c^bx^2(b-x)\,dx\\
&=\Bigg(\frac{2}{(b-a)(c-a)}\Bigg)\Bigg(\frac{c^4-a^4}{4}-\frac{a\big(c^3-a^3\big)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Let $x$,$y$ be rationals such that $\frac{x^2+x+\sqrt{2}}{y^2+y+\sqrt{2}}$ is also rational, prove that either =, or +=−1. I followed one convincing proof in this related post but I don't understand the assumption and conclusion.
The proof goes as follows:
Let $r\in\mathbb{Q}$ such that
$\frac{x²+x+\sqrt{2}}{y²+y+\sqr... | No, it's not allowed to even consider $x$ and $y$ integers.
When you arrive at
$$
x(x+1)=y(y+1)
$$
you're almost done, because this can be rewritten as
$$
x^2-y^2+x-y=0
$$
hence
$$
(x-y)(x+y+1)=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Factoring a third degree polynomial with a given root My professor gave us the following polynomial:
$f(x) = 3x^3-4x^2-x+2$
Given is that $x = 1$ is a root of this function. We are asked to find the other ones.
He then told us that, given $x=1$ is a root, we now know that we can factorize this polynomial into $(x-1)$, ... | Yes, that is correct. If $\alpha$ is the root of the polynomial $P(x)$, then $x-\alpha \mid P(x)$.
Also, we have several methods to factorise your polynomial:
*
*Polynomial long division
*Synthetic division
Finally, I can suggest the following general method:
$$\begin{align}&3x^3-4x^2-x+2=(x-1)(3x^2+ax+b)\\
\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4284582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you prove by induction that $\frac{1}{2} + \frac{2}{2^2} + \ldots + \frac{n}{2^n} = 2 - \frac{n+2}{2^n}$? For $n=1$ this is true because $\frac{1}{2^{1}}=2-\frac{1+2}{2^{1}}=\frac{1}{2}$. Further, it is a little more complicated, can we now assume that this is true up to the number $n-1$? Then do the induction s... | \begin{align*}
\frac{1}{2} + \frac{1}{2^2} + ... + \frac{n-1}{2^{n-1}}+\frac{n}{2^n}
&=2-\frac{n+1}{2^{n-1}} + \frac{n}{2^n}\\
&= 2+\frac{-2n-2 + n}{2^{n}}\\
&= 2- \frac{n+2}{2^n}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now... | When $2x=3$, then the expression becomes $$4x^2-2xy-4x+3y-3=9 - 3y -6 + 3y - 3 = 0$$
Hence the expression can be factored by $(2x-3)$, and you can see easily that
$$4x^2-2xy-4x+3y-3 = (2x-3)(2x-y+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 1
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Prove the inequality $\displaystyle\sum_{s y m} x^{4} y^{2} z \geqslant 2 \sum_{s y m} x^{3} y^{2} z^{2}$ Prove for positive $x,y,z$ the inequality $x^{4} y^{2} z+x^{4} z^{2} y+y^{4} x^{2} z+y^{4} z^{2} x+z^{4} y^{2} x+z^{4} x^{2} y \geqslant 2(x^{3} y^{2} z^{2}+x^{2} y^{3} z^{2}+x^{2} y^{2} z^{3})$.
I found the follow... | The last term of the left side of the second line should be $z^3y$ and not $z^3x$. Its a typo. Also, they used the AM-GM inequality:
$2x^3(y+z) + y^3z + z^3y = x^3y+x^3y+x^3z+x^3z+y^3z+z^3y \ge 6\sqrt[6]{x^3y\cdot x^3y\cdot x^3z\cdot x^3z\cdot y^3z\cdot z^3y}=6x^2yz$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What's the measure of the side of the rombus in the figure below? For rereence:In a rhombus the sum of the measures of its diagonals is $70$ cm
and the radius of the inscribed circle is $12$ cm. Calculate the measure of the rhombus side,
My progress:
$CO +BO = \frac{70}{2} = 35$
$\frac{1}{h^2}=\frac{1}{a^2}+\frac{1}{c^... | $\mathsf{
AO(x)+OB(y) = \frac{70}{2}=35\\
(x+y)^2 = 35^2\implies x^2+y^2+2xy = 1225\\
Por~ métrica: x^2+y^2 = AB(a)^2\\
a.h = x.y\implies 12a = xy\\
\therefore : a^2+24a-1225 = 0\\
\therefore \boxed{\color{red}a =AB = 25} }$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Mistake computing $\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}} \right)\frac{dx}{x}$ I am looking to evaluate the integral
$$\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}} \right)\frac{dx}{x}=\ln\left(\frac{\pi}{\Gamma^2\left(\frac{1+a}{2b}\right)\cos\left(\frac{a\pi}{2}\right)}\righ... | With a simple renaming of variables, I solved the OP's question using Ramanujan's generalization of Frullani's integral. The OP's method should work as well.
Evaluate $\int_0^{\infty } \Bigl( 2qe^{-x}-\frac{\sinh (q x)}{\sinh \left(\frac{x}{2}\right)} \Bigr) \frac{dx}x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
integral with disks We are looking for the volume of the space between $y=\frac{-1}{2}x+\frac{3}{2}$ and $y=x$ as rotated about the x-axis. I used the disk/washer method ($\pi r^2$) and broke it up into 3 pieces. I notice that there is a bit of overlap under the x-axis, so I have excluded it in my computation. Please s... | You split up the integral into three parts and these parts should be
$$ \pi \int_{-2}^{0}\left(\frac{-x}{2}+\frac 3 2\right)^2dx$$
$$\pi \int_{0}^1\left(\frac{-x}{2}+\frac 3 2\right)^2 - x^2dx$$
$$\pi \int_{1}^3 x^2 - \left(\frac{-x}{2}+\frac 3 2\right)^2dx$$
Notice that your third term has an extra $-x^2$ inside the i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4297180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Generating Function of Riordan numbers I would like to find generating function of $f(n)$, where $f(n)$ is defined as following:
$$f(n)=\sum_k^n \binom{n}{k}(-1)^{n-k}C_k\text{.}$$
With $C_k=\frac{1}{k+1}\binom{2k}{k}$($C_k$ is the $k^{th}$ Catalan's number).
Thanks in advantage for Your help
Good Evening
| We have for the sum
$$\sum_{k=0}^n {n\choose k} (-1)^{n-k} C_k
= \sum_{k=0}^n {n\choose k} (-1)^k C_{n-k}
\\ = [z^n] \frac{1-\sqrt{1-4z}}{2z}
\sum_{k=0}^n {n\choose k} (-1)^k z^k
= [z^n] \frac{1-\sqrt{1-4z}}{2z}
(1-z)^n
\\ = \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{n+1}} (1-z)^n
\frac{1-\sqrt{1-4z}}{2z}.$$
Now put ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that if $0 \leq a \leq b$ then $f_m(a) \leq f_n(b)$ for $m \leq n.$
Let $n \in \mathbb N$ and let $f_n$ be the function $t \mapsto t \left (t + \frac {1} {n} \right )^{-1}$ on $\mathbb R_{+}.$ Let $a$ and $b$ be two positive elements in a $C^{\ast}$-algebra $A$ such that $a \leq b$ and let $m \leq n.$ Then $f_m (... | As you showed, what you want is to prove that $f_n(a)\leq f_n(b)$ if $a\leq b$. The terminology for this is that $f_n$ is operator monotone.
For notational simplicity we may take $f(t)=t(t+c)^{-1}$, with $c>0$. As a motivation, how do we show that $f$ is (number) monotone? We want to show that
$$
s<t\implies\frac s{s+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4306048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$f(n) = f(2n)$, and $f(2n + 1) = f(n) + 1$, find expression of such $f$ The question is
a. $f(n) = f(2n)$
b. $f(2n + 1) = f(n) + 1$,
with $f(1)=1$, find expression of such $f$ that defined on positive integers
Got an initial idea about the pattern but find the difficulty to find the exact form. It's related to exponent... | Expanding on my comment, let the binary representation of $a$ be $\,\overline{a_na_{n-1}\dots a_1a_0}\,$ with $\,a_k \in \{0,1\}\,$, then $\,a = \sum_{k \ge 0} 2^k a_k\,$ and $\,a_k = \left\lfloor \frac{a}{2^k}\right\rfloor \bmod 2\,$, so the sum of binary digits can be written as $\,\sum_{k \ge 0} \left(\left\lfloor \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x... | Rewrite the equation as $$\left(\sqrt[4]{3x + 2} - \frac x{\sqrt[4]{3x+2}}\right)^2 = 0.$$ It follows that $\sqrt{3x+2} = x$ and then $x^2 - 3x - 2 = 0$. Note that we require $x \geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that...
For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that $$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$$
Here's what I've done so far:... | Hint:
$$(\frac{a}{b}+\frac{c}{b})+(\frac{a}{c}+\frac{b}{c})+(\frac{b}{a}+\frac{c}{a})+6$$
$$ = \frac{1-b}{b}+\frac{1-c}{c}+\frac{1-a}{a}+2+2+2$$
Now check if $$\frac{1-x}{x}+2\geq 2\sqrt2 \sqrt\frac{1-x}{x}$$ is true?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Lagrange inversion theorem and Legendre polynomials generating function I have been trying to solve the following problem (taken from Spiegel's Complex variables, problem 6.105).
By considering the equation $z=a+w\frac{(z^2-1)}{2}$, show that
$$\frac{1}{\sqrt{1-2aw+w^2}}=1+\sum_{n=1}^{\infty} \frac{w^n}{2^n n!}\frac{d^... | Starting with $$z = a + \frac{1}{2} w (z^2 -1) \tag{1}$$
and applying Lagrange's expansion, we have
$$z = a + \sum_{n=1}^{\infty} \frac{w^n}{2^n n!} \frac{d^{n-1}}{da^{n-1}} (a^2-1)^n \tag{2}$$
On the other hand, $(1)$ is also a quadratic equation which we can solve for $z$, with result
$$z = \frac{1 \pm \sqrt{1-2aw+w^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Probability a random spherical triangle has area $> \pi$ From Michigan State University's Herzog contest:
Problem 6, 1981
Three points are taken at random on a unit sphere. What is the probability that the area of the spherical triangle exceeds the area of a great circle?
I assume we always take the unique proper sp... | The substitution $u=\cot(a/2)\cot(b/2)$ yields $$I(a)=\int_{\pi-a}^\pi\arccos\left(\cot\frac a2\cot\frac b2\right)\sin b\,db=\int_0^1\frac{4u\tan^2(a/2)\arccos u}{(1+u^2\tan^2(a/2))^2}\,du$$ so that $$P(\sigma>\pi)=\frac1{4\pi}\int_0^\pi I(a)\sin a\,da=\frac4\pi\int_0^{\pi/2}J(a)\sin^2a\tan a\,da$$ where $\displaystyle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Find the quadratic equation with real coefficients and solutions $x_1,x_2$ if you know that $\Delta =b^2-4ac=-36$ and $x_1+x_2=6m$ What I've done so far is:
We know that $\Delta \lt 0$, which means that $x_1$ and $x_2$ are two conjugate complex numbers with the form: $x_1=r+n\cdot i \space\text{ and }\space x_2=r-n\cdo... | \begin{align*}
b^2-4ac=-36\implies 4ac\ge36\implies ac&\ge9\tag{1}\\
\\
x_1+x_2=
\bigg(\dfrac{-b+\sqrt{-36}}{2a}\bigg)+
\bigg(\dfrac{-b-\sqrt{-36}}{2a}\bigg)=
\dfrac{-b}{a}&=6m\tag{2}\\
\\
x_1\cdot x_2=
\bigg(\dfrac{-b+\sqrt{-36}}{2a}\bigg)
\bigg(\dfrac{-b-\sqrt{-36}}{2a}\bigg)
= \dfrac{b^2+36}{4 a^2}& \tag{3}
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4318399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Number of solution of ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ Number of solution of the equation ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ in the interv... | Having $a = \sin x - 1$ and $c = \sin x$ is redundant, it hides potential cancellation and simplification, and leaves you trying to factor a polynomial in three variables instead of just two. Instead, everywhere you have a $c$, replace it with $a+1$.
\begin{align*}
a^3 + b^3 + (a+1)^3 &= (a+b+a+1)^3 \\
a^3 + b^3 + (a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4322038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What is the result of $\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+\sqrt{-\frac{1}{4}+...}}}}$ Given the golden ratio is equal to $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}$ you get $f(x) = \sqrt{1+f(x)}$.
Solving for $f(x)$:
$f(x)^2 = x + f(x)$
$f(x)^2 - f(x) = x$
$f(x)^2 - f(x) + \frac{1}{4} = x + \frac{1}... | The trick is to see if there exists a sequence that would work. The trouble is that many times, recursive sequences can be sensitive to their starting seeds (exhibit chaotic behavior), thus these sequences can not be said to have well defined limits, only regularized values useful for specific applications. In this cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the sum $ \sum\limits_{a,b=0}^ \infty \begin{vmatrix} x^{a+b} & y^{a+b} & z^{a+b}\\ x^b & y^b & z^b \\ 1 & 1 & 1 \end{vmatrix} u^a v^b. $ Find a close determinant expression for the sum $$ \sum_{a,b=0}^ \infty \begin{vmatrix}
x^{a+b} & y^{a+b} & z^{a+b}\\
x^b & y^b & z^b \\
1 & 1 & 1
\end{vmatrix} u^a v^b.
$$
Is ... | $$
\sum_{a,b=0}^ \infty \begin{vmatrix}
x^{a+b} & y^{a+b} & z^{a+b}\\
x^b & y^b & z^b \\
1 & 1 & 1
\end{vmatrix} u^a v^b = \\
\sum_{a,b=0}^ \infty \begin{vmatrix}
u^ax^{a+b} & u^ay^{a+b} & u^az^{a+b}\\
x^b & y^b & z^b \\
1 & 1 & 1
\end{vmatrix} v^b =\\
\sum_{b=0}^ \infty \begin{vmatrix}
\frac{1}{1-xu}x^{b} & \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4328308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Strange inequality in an NT problem If $a\neq b$ are positive integers and $a^2+ab+b^2 | ab(a+b)$ then show that $|a-b|>\sqrt[3]{ab}$.
WLOG $a>b$. $a^2+ab+b^2 | ab(a+b)-a(a^2+ab+b^2)=-a^3$, so we have $a^2+ab+b^2$ divides $a^3$ as well as $b^3$. From here I tried lots of things, but I don't think they are worth of ment... | Notice that $$a^2+ab+b^2\mid ab(a+b)\iff a^2+ab+b^2\mid a^3\iff a^2+ab+b^2\mid b^3$$ $$\iff a^2+ab+b^2\mid \gcd(a,b)^3$$
Thus, let $d:=\gcd(a,b)$, and let $x:=\frac{a}d, y:=\frac{b}d$. It follows that $$a^2+ab+b^2\mid \gcd(a,b)^3\iff x^2+xy+y^2\mid d\implies d>xy$$
You are left to show that $\lvert a-b\rvert>\sqrt[3]{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Power of Stochastic Matrix Given the stochastic matrix
$$Q = \begin{pmatrix}0&2/3&1/3\\1/3&0&2/3\\2/3&1/3&0\\\end{pmatrix}\in \mathbb{R}^{3\times3}$$
I wish to compute $Q_{1,1}^n$ (the entry in the first row and first column) for $n\in \mathbb{N}$.
I did so by diagonalizing $Q=PDP^{-1}\implies Q^n=PD^nP^{-1}$. The comp... | Following the comment by @kimchi lover, since the matrix is stochastic and circulant, the eigenvalues are $1, \lambda = \frac{2}{3} e^{2 \pi i/3} + \frac{1}{3} e^{4 \pi i/3} = \frac{1}{2}\left(-1 + \frac{i}{\sqrt{3}} \right)$, and $\overline{\lambda} = \frac{1}{2}\left(-1 - \frac{i}{\sqrt{3}} \right)$. Thus $D^n = \te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $1/(a+b) + 1/(b+c) + 1/(c+a) > 3/(a+b+c)$ for positive $a, b, c\,$? I have to prove that:
$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} > \frac{3}{a+b+c},$$
where $a, b , c$ are positive real numbers.
I am thinking about using arithmetical and geometrical averages:
$$A_{3} = \frac{a_{1}+a_{2}+a_{3}}{3},$$
... | By AM-HM inequality
$$\frac13 \left(\frac1{a+b}+\frac1{b+c}+\frac1{c+a}\right)\ge
\frac{3}{(a+b)+(b+c)+(c+a)}.$$
Hence $\displaystyle \frac1{a+b}+\frac1{b+c}+\frac1{c+a}\ge \frac{9}2\frac1{a+b+c}$. Because $\dfrac92>3$ the problem is solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating an improper integral using the Residue Theorem I have:
\begin{equation}
\int_{-\infty}^{\infty}\frac{x^2}{(x^2+1)(x^2+9)}dx
\end{equation}
and I want to solve it using a complex closed contour on C. I do the following:
\begin{equation}
\int_{-\infty}^{\infty}\frac{z^2}{(z^2+1)(z^2+9)}dz
\end{equation}
which ... | Also, using Calculus
$$
\frac{x^2}{(x^2+1)(x^2+9)}=\frac{1}{8}\left(\frac{9}{x^2+9}-\frac{1}{x^2+1}\right)
$$
and
$$
\int_{-\infty}^\infty\frac{dx}{x^2+1}=\tan^{-1}(x)\Big|_{x=-\infty}^{x=\infty}=\pi, \qquad
\int_{-\infty}^\infty\frac{dx}{x^2+9}=\frac{1}{3}\tan^{-1}(x/3)\Big|_{x=-\infty}^{x=\infty}=\frac{\pi}{3},
$$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Given two distinct intersecting circles, length of that chord of larger circle which is bisected by the smaller circle is equal to? Two circles whose centres lie on the x axis, whose radii are $\sqrt2cm$ and $1cm$ and whose centres are 2 cm apart intersect at a point A.The chord AC of the larger circle cuts the smalle... | Yes there is a different path using trigonometry:
Let, with your coordinate system:
$$B=(\cos \alpha, \sin \alpha) \ \text{and} \ C=(2+\sqrt{2}\cos \beta, \sqrt{2}\sin \beta)$$
We just have to express that B is the midpoint of [AC] by writing that $$2B=A+C \ \iff \ \begin{cases}2\cos \alpha&=& \dfrac34+2+\sqrt{2}\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
relation between roots and coefficient in a cubic polynomial If $\alpha,\beta,\gamma$ are roots of the cubic equation
$$2x^{3}+3x^2-x-1=0$$ then I want to find the equation whose roots are $\frac{\alpha}{\beta+\gamma}, \frac{\beta}{\gamma+\alpha}, \frac{\gamma}{\alpha+\beta}$.
I have $\alpha+\beta+\gamma$= - $\frac{3}{... | While this is not a general method for zeroes of a cubic polynomial, this particular polynomial is amenable to familiar methods. We can apply the Rational Zeroes Theorem to find that one zero is $ \ \alpha \ = \ -\frac12 \ \ , $ so we may factor $ \ 2x^3 + 3x^2 - x - 1 \ = \ \left(x + \frac12 \right)·2·(x^2 + x - 1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac{1}{2^{2n+1}}\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j} \right]= \frac{1}{2}$ I have a hard time showing that that
$$ \frac{1}{2^{2n+1}}\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j} \right]= \frac{1}{2}$$
Namely, I try to show hat
$$\sum_{i=0}^n \le... | We have
$$
\begin{split}
S=\sum_{0\le i<j\le n+1}\binom{n}{i}\binom{n+1}{j}&=\sum_{0\le i\le n}\sum_{0\le j\le n+1}\binom{n}{i}\binom{n+1}{j}-\sum_{0\le j\le i\le n+1}\binom{n}{i}\binom{n+1}{j}\\
&=2^n\cdot 2^{n+1}-\sum_{0\le j\le i\le n+1}\binom{n}{n-i}\binom{n+1}{n-j+1}\\
&=2^{2n+1}-\sum_{0\le i'<j'\le n+1}\binom{n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Number of zeros of $f(x)= \frac{1}{2} E\left[ \tanh \left( \frac{x+Z}{2} \right) \right]-\tanh(x)+\frac{x}{2}$ where $Z$ is standard normal Consider the following function:
\begin{align}
f(x)= \frac{1}{2} E\left[ \tanh \left( \frac{x+Z}{2} \right) \right]-\tanh(x)+\frac{x}{2},
\end{align}
where $Z$ is standard normal.
... | Not a complete proof, but rather a sketch. First, observe that as $x\to +\infty$, than $\tanh (x) \to 1$, so $f(x)$ asymptotically asymptotically equivalent to $\frac{1}{2} E [1] - 1 + \frac{x}{2} = \frac{x-1}{2}$, so for some large enough positive value of $x$ one has $f(x) > 0$. After that, let's look at the graph of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Solve system of 2 equations with 3 unknowns We are given a triangle $ABC$ with sides $a, b, c$ respectively and for which the following relationships hold:
$a^2+bc\sqrt 3 = b^2+c^2$,
$c^2+ba = a^2+b^2$
We want to prove that angle $B$ is right.
I am trying to express sides $b$ and $c$ in relation to $a$ and then prove ... | Comparing the first equation with cosine rule:
$$a^2=b^2+c^2-2bc\cos A,$$
$$a^2=b^2+c^2-\sqrt3bc,$$
$$\implies\cos A=\frac{\sqrt3}2\implies \measuredangle A=30^\circ.$$
Now using the two given equations we get, $2b=\sqrt3c+a$.
Applying sin rule, $$2\sin B=\sqrt3\sin C+\sin A$$ $$2\sin B=\sqrt3\sin(150^\circ-B)+\sin30^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Struggling to compute a power series for a complex value function I am struggling to compute the power series expansion of $$f(z) = \frac{1}{2z+5}$$ about $z=0$, where $f$ is a complex function. I tried comparing it to the geometric series as follows,$$ f(z) = \frac{1}{2z+5} = \frac{1}{1-\omega} = \sum_{n=0}^\infty\ome... | $$\begin{array}{lcl}
\dfrac{1}{2 z + 5} & = & \dfrac{1}{5} \dfrac{1}{1 + \dfrac{2 z}{5}} \\[3mm]
& = & \displaystyle \dfrac{1}{5} \sum_{n = 0}^{+\infty} (-1)^n \left(\dfrac{2 z}{5}\right)^n \\[3mm]
& = & \displaystyle \sum_{n = 0}^{+\infty} \dfrac{2^n}{5^{n + 1}} (-1)^n z^n \\[3mm]
\end{array}$$
with the condition :
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find $a$ such that ${x_1}^2+{x_2}^2$ takes the minimal value where $x_1, x_2$ are solutions to $x^2-ax+(a-1)=0$ DO NOT USE CALCULUS My thinking:
Let $x_1 = \frac{a+\sqrt{a^2-4a+4}}{2}$ and $x_2 = \frac{a-\sqrt{a^2-4a+4}}{2}$
By the AGM (Arithmetic-Geometric Mean Inequality):
We have
$x_1\cdot x_2\le \left(\frac{x_1\cdo... | Another way to go... From $x^2 - a x + (a-1)$, we know that any root satisfies
$$ x^2 = ax - (a-1) \text{,} $$
so \begin{align*}
x_1^2 + x_2^2 &= (ax_1 - (a-1)) + (ax_2 - (a-1)) \\
&= a (x_1 + x_2) - 2a + 2 \text{.}
\end{align*}
From \begin{align*}
x^2 -ax +(a-1) &= (x-x_1)(x-x_2) \\
&= x^2 -(x_1 + x_2)x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4357090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
The solution set of equation $\sin^{-1}{\sqrt{1-x^2}}\;+\cos^{-1}x=\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}-\sin^{-1}x$
Find the solution set of equation $\sin^{-1}{\sqrt{1-x^2}}\;+\cos^{-1}x=\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}-\sin^{-1}x$
My Approach:
$\sin^{-1}{\sqrt{1-x^2}}\;+\cos^{-1}x \; +\sin^{-1}x=\cot^{-1}{\dfrac{\s... | The part where you went wrong is:
$\implies$$\sin^{-1}{\sqrt{1-x^2}}\;+\tan^{-1}{\dfrac{\sqrt{1-x^2}}{x}}=0$
$\implies$$2\sin^{-1}{\sqrt{1-x^2}}=0$
Because, $\sin^{-1}{\sqrt{1-x^2}}=$$\tan^{-1}{\dfrac{\sqrt{1-x^2}}{|x|}}$
.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to solve the Diophantine equation $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$? How can one solve the following Diophantine equation in $x, y \in \mathbb{Z}$?
$$x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$$
| $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0\iff(x^2+x)^2-4(x^2+x)=(y+1)^2+5$
Putting $X=x^2+x$ we get $(X-2)^2=(y+1)^2+9$ whose only solutions are clearly
$$(X-2,y+1)=(\pm3,0),(\pm5,\pm4)$$ in both cases, since $X=x^2+x$, we have the equations for the unknown $x$
$$x^2+x-5=0\text{ and } x^2+x+1=0\\x^2+x-7=0\text{ and } x^2+x+3=0$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Determine the greatest of the numbers $\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$ Determine the greatest of the numbers $$\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$$ The least common multiple of $2,3,4,5$ and $6$ is $LCM(2,3,4,5,6)=60$, so $$\sqrt2=\sqrt[60]{2^{30}}\\\sqrt[3]3=\sqrt[60]{3^{20}}\\\sqrt[4]4=\sq... | It is also possible to work with logarithms. We may write for each of these numbers
$$ \log_2 \sqrt2 \ \ = \ \ \log_2 2^{1/2} \ \ = \ \ \frac12 \ \ , \ \ \log_3 \sqrt[3]3 \ \ = \ \ \frac13 \ \ , \ldots , \ \ \log_6 \sqrt[6]6 \ \ = \ \ \frac{1}{6} \ \ . $$
It will be more helpful to our purpose to express these i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Solving the system $\tan x + \tan y = 1$ and $\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$ How can I solve this system of trigonometric equations:
$$\tan x + \tan y = 1$$
$$\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$$
I tried to write tangent as $\sin/\cos$ and then multiply the first equation with the second one but it is ... | Substitute $u=\cos x, \sqrt{1-u^2}=\sin x, v=\cos y, \sqrt{1-v^2}=\sin y$.
The second equation:
$$ u\sqrt{1-v^2} = \frac{1}{\sqrt2} $$
$$\Rightarrow u=\frac{1}{\sqrt{2-2v^2}}$$
The first equation:
$$ \frac{\sqrt{1-u^2}}{u} + \frac{\sqrt{1-v^2}}{v} = 1 $$
$$ \Rightarrow \sqrt{1-2v^2} + \frac{\sqrt{1-v^2}}{v} = 1 $$
$$\R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve ODE $y'(y'+y)=x(x+1)$ Solve ODE $$y'(y'+y)=x(x+1)$$
I tried to remove $y'^2$ term by differentiate it wrt x and then replace value in hope that it will turn out some exact form but got stuck after
$$2y'y''-yy'+y''=x-x^2+1$$
How i proceed further or my method is wrong ?
Edit:
Exact problem
If $y'-x\neq0$ is a solu... | The equation can be rewritten as $(y')^2+yy'-x^2-x=0.$ One can solve $y'$ in terms of $y$ and $x.$ Notice that the above equation is equivalent to $(2y')^2+2y(2y')-4x^2-4x=0=(2y')^2+2y(2y')+y^2-(y^2+4x^2+4x)=(2y'+y)^2-(y^2+4x^2+4x),$ hence $(2y'+y)^2=y^2+4x^2+4x,$ implying $y^2\geq-(4x^2+4x).$ Notice that $4x^2+4x=4x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4368299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
In a triangle ABC, if certain areas are equal then P is its centroid Let $P$ be a point in the interior of $\triangle ABC$. Extend $AP$, $BP$, and $CP$ to meet $BC$, $AC$, and $AB$ at
$D$, $E$, and $F$, respectively. If $\triangle APF$, $\triangle BPD$, and $\triangle CPE$, have equal areas, prove that $P$ is the centr... | I will use barycentric coordinates, an excellent reference is
Barycentric Coordinates for the Impatient, Max Schindler, Evan Chen
and it is by chance compact, and easily accesible/readable. We will use the formula in Theorem 10 in it.
Let $P$ be $P=(x,y,z)$ in barycentric coordinates w.r.t. $\Delta ABC$, $x,y,z >0$, $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Inequality $2a^2 + 8b^2 + \frac{1}{2ab} \ge 4$ I saw the following question:
Prove that the least value of
$$2a^2+8b^2+\frac{1}{2ab}$$
is 4, where $a,b$ are positive numbers.
I believe that this can be shown using the criterion of the saddle and critical points that is taught usually in Calculus 3.
Can this be solved u... | Differently from the proposed solutions, you can also proceed as follows:
\begin{align*}
2a^{2} + 8b^{2} + \frac{1}{2ab} & = 2(a^{2} + 4b^{2}) + \frac{1}{2ab}\\\\
& = 2(a^{2} - 4ab + 4b^{2}) + 8ab + \frac{1}{2ab}\\\\
& = 2(a - 2b)^{2} + \frac{1}{2ab} + 8ab\\\\
& \geq \frac{1}{2ab} + 8ab
\end{align*}
Finally, according ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4378260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Triple integration $\iiint_{x^2+y^2+z^2 \leq 2z} x^2y^2 dxdydz$ $$\iiint_{x^2+y^2+z^2 \leq 2z} x^2y^2 dxdydz$$
I want to solve it with cylindrical coordinates:
So we get:
$$\iiint_{r^2+z^2 \leq 2z} r^5\sin^2(\phi)\cos^2(\phi) dxdydz$$
I see that the bounds are actually a circle with radius of $\sqrt{2z}$.
How can I use... | It's better to use spherical coordinates
The region is $ x^2 + y^2 + z^2 \le 2 z $, i.e., $x^2 + y^2 + (z - 1)^2 \le 1 $
So we can define the spherical coordinates $r, \theta, \phi $ such that
$ x = r \sin \theta \cos \phi $
$ y = r \sin \theta \sin \phi $
$ z = 1 + r \cos \theta $
And the integral becomes
$ I = \displ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4378984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.