Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $a^ab^b+a^bb^a\le 1$.
Let $a$ and $b$ be positive real numbers such that $a+b=1$. Prove that $a^ab^b+a^bb^a\le 1$.
This is RMO 2012.
I got a solution, but I think it's flawed.
Consider $a+\dots +a (a$ $ \text{times})$ and $b+\dots +b(b$ $ \text{times})$. We get $a\cdot a$+ $b\cdot b \\ $.
Applying AM-GM , ... | Your idea is correct however its representation has to be fixed as you gussed the representation is only valid when $a,b$ are natural
Actually we have as a subcase the weighted AM-GM inequality as follows $$xa+yb\ge (x+y)\sqrt[x+y]{a^xb^y}$$ where $a,b,x,y$ are positive reals.
For more reference see here.
So instead o... | {
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"url": "https://math.stackexchange.com/questions/3974021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that: $ \sum_{i=0}^{\infty} \frac{\tan \frac{\theta}{2^i}}{2^i}= \frac{1}{\theta} - 2 \cot 2 \theta$ My attempt:
Consider the following series:
$$ S = \sum_{i=0}^n \ln( \sec \frac{x}{2^i} )$$
Notice that $ \lim_{n \to \infty} \frac{dS}{dx}$ is the required sum.
Simplfying S,
$$ S = - \ln \left( \cos x \cdot \cos... | For a calculus-free solution, prove by induction $\sum_{i=0}^n2^{-i}\tan\tfrac{\theta}{2^i}=2^{-n}\cot(2^{-n}\theta)-2\cot2\theta$ for $n\ge0$. Since $\tan x=\cot x-2\cot2x$, the left-hand side is a telescoping series.
| {
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"url": "https://math.stackexchange.com/questions/3978919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $ The question is $$\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $$ I know the answer is $\frac{1}{2}$ and I found it using this equality :
$$(\sqrt{x^2+1} - x)(x+1) = \frac{x+1}{\sqrt{x^2+1} + x}$$
But is there any other way to solve this? Any hints would be appreciated... | Use binomial series:
$\sqrt{1+\dfrac1{x^2}}=1+\dfrac1{2x^2}\cdots$
Addendum with further explanation:
$\lim\limits_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $
$=\lim\limits_{x\to\infty}\left(\sqrt{1+\dfrac1{x^2}}-1\right)x(x+1)$
$=\lim\limits_{x\to\infty}\left(1+\dfrac1{2x^2}\cdots-1\right)(x^2+x)$
$=\lim\limits_{x\to\infty}... | {
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"timestamp": "2023-03-29T00:00:00",
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Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$
I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.
Please... | Hint:
Let $x=a+b$ then $b=x-a$ so we get an quadratic equation on $a$: $$3a^2(x+2)-3a(x^2+2x)+x^3+11=0$$ Since $a$ is real it discriminant is non negative, so we have $$-3(x+2)(x^3-6x^2+44)\geq 0$$
Notice that $x\mapsto x^3$ and $x\mapsto -6x^2$ are increasing for $x \leq 0$.
Now if $x\leq -{7\over 3}$ then we get $... | {
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A sequence $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and find $x_{2020}$ The sequence is given by the formula $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and it is known that $x_{2017} + x_{2023} = 990$, then what is $x_{2020}$ = ?
My little approch:
It is given that $x_{2017} + x_{2023} = 990$ ----- (1) and $x_{n+1} = 3x_n + \... | We have
$$x_{n+1}-3x_n = \sqrt{8x_n^2+2}$$
$$(x_{n+1}-3x_n)^2 = 8x_n^2+2$$
$$x_{n+1}^2 -6x_{n+1} x_n + x_n^2 = 2 \tag{I}$$
And because
$$x_{n-1}^2 -6x_{n-1} x_n + x_n^2 = 2 \tag{II}$$
From $(I)$ and $(II)$, we have, for all $n$
$$(x_{n+1}-x_{n-1})(x_{n+1}+x_{n-1}-6x_{n})=0$$
Is it possible that $x_{n+1}=x_{n-1}$ ? ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Alternative approaches to prove the following inequality For $a,b,c \in \mathbb{R^+},$ prove that
$$\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3.$$
I managed to prove this problem using the technique of isolated fudging. In ... | Let $p=a+b+c,\,q=ab+bc+ca,\,r=abc.$ Using the Holder inequality, we have
$$\left[\sum \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}}\right]^3 \sum a^2(b+c)^2 \geqslant 4(a+b+c)^4.$$
It remain to show that
$$4(a+b+c)^4 \geqslant 27 \sum a^2(b+c)^2,$$
or
$$2p^4+27pr \geqslant 27q^2.$$
If $p^2 > 9q,$ then $$2p^4 > 2(9q)^2 > ... | {
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"timestamp": "2023-03-29T00:00:00",
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How do we prove the proposed expression is nonnegative? So the expression which I am interested in is given by
\begin{align*}
f(b,\theta) = \frac{2b-1}{b\sqrt{(1-\theta)(2b-1)^{2} + \theta}} + \frac{\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1}{2b^{2}(\theta - 1)}
\end{align*}
where $b\in(0,1]$, $\theta\in\mathbb{R}_{>0}$ ... | $$
f(b,\theta)\geq 0
$$
\begin{align*}
\implies\frac{2b-1}{b\sqrt{(1-\theta)(2b-1)^{2} + \theta}} + \frac{\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1}{2b^{2}(\theta - 1)}&\geq 0\\ \\
\implies 2b-1+\frac{\left(b\sqrt{(1-\theta)(2b-1)^{2}+\theta}\right)\cdot\left(\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1\right)}{2b^{2}(\th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Conditional probability - voting There is a population that consists of a mixture of "Unchangeables" and "Changeables". If you choose a person at random, then the probability that they're an Unchangeable is $p$ and the probability that they're a Changeable is $1-p$.
Every person is asked repeatedly to vote either "yes"... | I got a totally different result from yours.
Intro. 1: $$P(A \mid B) = P(A \mid BC) \cdot P(C \mid B) + P(A \mid B \overline C) \cdot P(\overline C \mid B)$$
Proof:
\begin{align}
P(A \mid B) P(B) &= P(AB) \\
&= P\left[AB(C \cup \overline C)\right] \\
&= P\left[(ABC) \cup (AB \overline C)\right] \\
&= P(ABC) + P(AB \ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluate $\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$
How to evaluate:
$$\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$$
Attempt:
$$\int_{0}^{1}\frac{\ln(1 + x + x^2 + \ldots + x^n)}{x} \mathrm dx
= \int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\mathrm d x$$
Any hints ... | Ok we have:
$$
I = \int_0^1 \frac{\ln(\sum_{k=0}^n x^n)}{x}dx
$$
Using:
$$
S_n = \frac{1-r^{n+1}}{1-r}
$$
$$
I = \int_0^1 \frac{\ln\left( \frac{1-x^{n+1}}{1-x} \right)}{x}dx
$$
Using the fact that $\ln(a/b) = \ln(a) - \ln(b)$
$$
I = \int_0^1 \frac{1}{x} \ln\left(1-x^{n+1} \right)dx - \int_0^1 \frac{1}{x} \ln(1-x) \, dx... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $\frac{a+b+c+ab+ac+bc}{1+abc}$ is a real number.
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $$x = \frac{a+b+c+ab+ac+bc}{1+abc}$$ is a real number.
I wanted to calculate $2 \cdot Im(x) = ... | $$x = \frac{(a+1)(b+1)(c+1)-(1+abc)}{1+abc} = \frac{(a+1)(b+1)(c+1)}{1+abc}-1$$
and using the property $a \bar{a} = |a|^2 = 1$ repeatedly:
$$\frac{(a+1)(b+1)(c+1)}{1+abc} = \frac{(1+1/a)(1+1/b)(1+1/c)}{1/abc+1} = \frac{(1+\bar{a})(1+\bar{b})(1+\bar{c})}{\overline{abc}+1}$$
so $x = \bar{x}$ (why?) and hence $x$ is real.... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $k^2+pk+q=0$ and $l^2-pl-q=0$ then between $k$ and $l$ there is a solution to $x^2-2px-2q=0$. Real number $k$ satisfies $x^2+px+q=0$ and $l$ satisfies $x^2-px-q=0$. Prove that between $k$ and $l$ there is a solution to the equation $x^2-2px-2q=0$.
I started the solution this way:
$$ k = \frac{-p\pm\sqrt{p... | The comparison argument you mentioned might run something like this. As Dhanvi Sreenivasan notes, all three parabolas for the quadratic equations have a mutual intersection at $ \ x = -\frac{q}{p} \ \ . $ If we inspect a graph, the zeroes
$$ k \ = \ \frac{-p}{2} \ + \ \frac{\sqrt{p^2 \ - \ 4q}}{2} \ \ , \ \ l \ =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008041",
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Minimizing trigonometric expression How can one get the minimum of a trigonometric expression if the first-order derivative is not in a nice form?
The expression is this:
$$
\left[\cos ^{2}\left(\dfrac{\theta }{2}\right) +
\,\sqrt{\,{1 - \gamma}\,}\,\sin^{2}\left(\dfrac{\theta }{2}\right)\right] ^{2} + \dfrac{\gamma }{... | Note $$\cos^2 \frac{\theta}{2} = \frac{1 + \cos \theta}{2}, \quad \sin^2 \frac{\theta}{2} = \frac{1 - \cos \theta}{2},$$ and $$\sin^2 \theta = 1 - \cos^2 \theta.$$ So if we let $x = \cos \theta$, we get the function $$\begin{align}f(x) &= \left( \frac{1 + x}{2} + \sqrt{1-\gamma} \frac{1 - x}{2} \right)^2 + \frac{\gamma... | {
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If $\left\{a_n\right\}$ is Fibonacci Number, prove that the sequence$\left\{\frac{\ln a_n}{\ln a_{n+1}}\right\}$ increase
If $a_0=1,a_1=1,a_{n+2}=a_{n+1}+a_{n}$, prove the sequence$\left\{\frac{\ln a_n}{\ln a_{n+1}}\right\},\,n\geq 1$ increase.
When $n$ is even this inequality is easy proved by Am-Gm.
If $n$ is even,... | There is a primary proof by a student in Fudan University(Not me).
First, it is easy to prove that $\left\{\frac{a_{n+1}}{a_{n}}\right\}$ strictly increase.
Second, we can also prove $$0\leq\ln(1+x)\leq x,\quad(\forall\,x\geq 0).$$
And now we begin the proof.
Consider $\forall\,n\geq 3$, we have $$\frac{a_{n+1}}{a_{n-1... | {
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"timestamp": "2023-03-29T00:00:00",
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Is it possible to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ without using trigonometric substitution? The normal approach to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ is using the substitution $x=2\tan\theta$. But I wonder is is possible to do it without using trigonometric substitution? I tried this approach:
$$\int\frac{dx... | $$\frac 1{\sqrt{x^2+4}}=\frac 14\frac{(x^2+4)-x^2}{\sqrt{x^2+4}}=\frac 14\left(\sqrt{x^2+4}-\frac{x^2}{\sqrt{x^2+4}}\right)$$
$\int\sqrt{x^2+a^2}$ can be computed using IBP without any trig sub.
For the second part, consider the sub $t:=\sqrt{x^2+4}$ so that $t^2=x^2+4$ and $t\,\mathrm dt=x\,\mathrm dx$
$$\int\frac{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Fibonacci sequence proof. Given the Fibonacci Sequence $F_{0}=0,F_{1}=1,F_{n+2}=F_{n+1}+F_{n+1};\;n\geq0$
Prove the following:
$\sum_{i=1}^{n}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{n+2}}{2^{n}}$
Base case:
$\frac{F_{0}}{2^{0}}= 1-\frac{F_{2}}{2^{0}}\\ \frac{0}{1} = 1-\frac{1}{1} \\ 0 =0$
Which is true.
Now, for the inducti... | Since $\sum_{i=1}^k\frac{F_{i-1}}{2^i}=1-\frac{F_{k+2}}{2^k}$,$$\sum_{i=1}^{k+1}\frac{F_{i-1}}{2^i}=1-\frac{F_{k+2}}{2^k}+\frac{F_k}{2^{k+1}}=1-\frac{2F_{k+2}-F_k}{2^{k+1}},$$so you just need to prove $2F_{k+2}-F_k=F_{k+3}$. With $a:=F_k,\,b:=F_{k+1}$ we get$$F_{k+2}=a+b,\,F_{k+3}=a+2b\implies2F_{k+2}-F_k=2(a+b)-a=a+2b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012485",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$
How to evaluate $$\int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$$
Attempt:
$$f(x) = \frac{x^8-1}{x^{10}+1}$$
$$I = \int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$$
Substituting $t = \frac{1}{x}$,
$$I = \int_{0}^{\infty} \dfrac{\frac{1}{x^8} -1}{\frac{1}{... | As @DinosaurEgg and @StefanLafon note, in fact $I=\int_0^\infty\frac{1-x^8}{x^{10}+1}dx=-I$ after the substitution, so $I$ is $0$ if it converges. Indeed, the integrand is $O(1)$ for small $x$ and $O(1/x^2)$ for large $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show two properties about the Cantor Set Define $C_0=[0,1]$ and for $n\in\mathbb{N}$, define $$C_n=C_{n-1}\setminus\bigg(\bigcup_{k=0}^{3^{n-1}-1}\bigg(\frac{1+3k}{3^n},\frac{2+3k}{3^n}\bigg)\bigg) $$ Then the Cantor set is defined as $$C=\bigcap_{n\in\mathbb{N}}C_n$$
Things I need to show:
(1) Each $C_n$ is the... | Here we look at the induction proof of the claim
Let
\begin{align*}
C_n&=C_{n-1}\setminus\bigcup_{k=0}^{3^{n-1}-1}\bigg(\frac{1+3k}{3^n},\frac{2+3k}{3^n}\bigg)\qquad\qquad n\geq 1\tag{1}\\
C_0&=[0,1]\\
\\
C&=\bigcap_{n\in\mathbb{N}}C_n
\end{align*}
then each $C_n$ is the disjoint union of $2^n$ closed sub-intervals of ... | {
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"timestamp": "2023-03-29T00:00:00",
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Stable and Unstable Manifolds Computation I want to compute (approximately) the stable and unstable manifolds around the origin of the following system.
$x' = 6x + 8y + (sinx)^2 - y^2$
$y' = 8x - 6y + x^2 + xy$
How can I solve it? Thank you very much.
| Making $y=h(x) = \sum_{k=1}^n a_k x^k$ over the manifolds we have
$$
\dot y = h_x(x)\dot x
$$
or
$$
8x-6h(x)+x^2 + x h(x) = h_x(x)\left(6x+8h(x)+\left(x^2-\frac{x^4}{3}\right)-h^2(x)\right)
$$
considering $n = 3$ we have after reducing to a null polynomial
$$
\left\{
\begin{array}{l}
4 \left(a_1+2\right) \left(2 a_1-1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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sum of floor function simplification Let $n \in \mathbb{N}$.
I'm trying to show $\sum_{i=1}^{\infty} \left(\left \lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}}\right \rfloor \right) = 2^k-1$.
I know that when $2^i > n-1$, the right floor function will become zero.
I'm not sure how I cancel ... | The point is that the LHS is a geometric series, because extraneous parts of the sums cancel. However, the result doesn't seem quite right, because you get some extra terms.
$$
\begin{align}
&\sum_{i=1}^{\infty}
\left( \left\lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find a matrix $A$ with no zero entries such that $A^3=A$ I took a standard $2 × 2$ matrix with entries $a, b, c, d$ and multiplied it out three times and tried to algebraically make it work, but that quickly turned into a algebraic mess. Is there an easier method to solve this?
| $$
\left(
\begin{array}{cc}
\cos t & \sin t \\
\sin t & - \cos t
\end{array}
\right)
$$
with eigenvectors as columns of
$$
\left(
\begin{array}{cc}
\cos \frac{t}{2} & -\sin \frac{t}{2} \\
\sin \frac{t}{2} & \cos \frac{t}{2}
\end{array}
\right) \; \; , \; \;
$$
this confirmed by the identity
$$
\left(
\begin{array}{... | {
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"timestamp": "2023-03-29T00:00:00",
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How to compute $1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$ How to compute $S = 1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$
Was thinking $\frac{S}{24} = {4\choose 4} + {6\choose 4} + {8\choose 4} + ... ... | I don't have reputation for comment. so trying to answer here.
${S_n= (2n-1)(2n)(2n+1)(2n+2)}$, simplifying it.
*
*${S_n = (2n) \cdot (4n^2-1) \cdot (2n+2)}$,
*${S_1+S_2+S_3+S_4= 24+360+1680+5040=7104}$
*Simple program gives sum till ${100 = 33211999680 }$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the total length of two line segments of two overlapping right triangles
Find the length of $AE+EB$ ?
(A) $\frac{128}{7}$
(B) $\frac{112}{7}$
(C) $\frac{100}{7}$
(D) $\frac{96}{7}$
(E) $\frac{56}{7}$
My solution:
For $\Delta AEB$ :
$\angle BAC = sin^{-1}(\frac{5}{13}) = 22.62^{\circ}$
$\angle ABD = sin^{-1}(\frac... | Note that $\triangle AED\sim\triangle CEB$. Then, letting $AE = x$ and $BE = y$, we have:
$$DE = \frac{9}{5}y,\ CE = \frac{5}{9}x$$
From the similar triangles. Now, we must have $AE + CE = AC$ and $BE + DE = BD$. Noting that $AC = 13$ and $BD = 15$:
$$x + \frac{5}{9}x = 13$$
$$y + \frac{9}{5}y = 15$$
Solving, we find ... | {
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Find the asymptotic analysis of $\frac{a_{n+ 1}^{2}}{\prod_{0}^{n+ 1}a_{i}}- \sqrt{5}$
Given a recursion $a_{n+ 1}= a_{n}^{2}- 2$ with $a_{0}= 3.$ Find the asymptotic analysis of
$$\frac{a_{n+ 1}^{2}}{\prod_{0}^{n+ 1}a_{i}}- \sqrt{5}$$
Maybe this helps - https://artofproblemsolving.com/community/c4h2308107p18323223 -
... | First of all I think it's strange your product goes up to $i=n+1$, as that factor cancels with the numerator.
Assuming that is what you mean, notice that
$$\prod_{i=0}^{n+1} a_i = F_{2^{n+3}},$$
where $F_k$ is the $k$th Fibonacci number. That alongside with the formula
$$ a_n = \left(\frac{2}{\sqrt{5}+3}\right)^{2^n}+\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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obtain parity check matrix in different ways We have the code words 111000, 100110, 110101 which result in the following generator matrix.
$G=\begin{bmatrix}1&1&1&0&0&0\\1&0&0&1&1&0 \\ 1&1&0&1&0&1\end{bmatrix}$
According to my book, the rows of the control matrix are solution vectors of the system of equations whose ro... | Having the generator matrix in systematic form, that is $G=[I_k|P]$, has of course the advantage as stated, which is the easy construction of the parity check matrix $H=[P^T|I_{n-k}]$.
Consider any word $x$ as a row vector, then the syndrome $s=Hx^T=0$ if and only if $x$ is a codeword of the given linear code. Since $H... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $-y' +y^2 = 1$ I must solve the ODE $-y' + y^2 = 1$. Rewriting this I can get: $$y' - y^2 + 1 = 0$$ However, because of the nonlinear term $y^2$ I do not know how to solve this. We cannot use the characteristic equation I think and I cannot see that this is seperable.
Can someone please lend a hand?
I get the... | Try $\displaystyle y' = \frac{dy} {dx} = y^2 - 1$ which is immediately separable and solvable by partial fraction decomposition. A hyperbolic trigonometric substitution will likely work as well, it's less elementary though.
Adding on, there is a very important subtlety here. The integral of $\displaystyle \frac 1x$ is ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A Numerical Sequences proof. Let $ 0<a<1 $ be a real number, and let $ a_n \in \left ( -1,0 \right )$ defined by the relation $ \sqrt[n]{a} = 1+a_n$ , $ n \in \mathbb{N}$.
Show the following inequality:
$ |a_n| \le \frac{1}{n} \bigl( \frac{1-a}{a}\bigr)$, $ n \ge 1$.
This problem was proposed by my professor of Ma... | By hypothesis it's $ \sqrt[n]{a} = 1 + a_n$ , in particular $ a_n=\sqrt[n]{a}-1<0$ $\to$ $|a_n|=1-\sqrt[n]{a}$.
Second member can be viewed as a division of polynomials:
$$ 1 - \sqrt[n]{a} = \frac{1 - a}{1 + \sqrt[n]{a} + \sqrt[n]{a^2} + \sqrt[n]{a^3} + ... +\sqrt[n]{a^{n-1}} }$$
Note that terms in denominator,
$$1 + \... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$ Evaluate
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$$
My attempt : I put $t= \sqrt{1-r^2}$ now $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$
$$= 2\int_{0}^{1} r^2... | With $r=\sin t$ it's$$\int_0^{\pi/2}\sin^3tdt=\int_0^{\pi/2}(\sin t-\sin t\cos^2t)dt=[\tfrac13\cos^3t-\cos t]_0^{\pi/2}=\tfrac23.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that for any integer $n>0$ which is a perfect square, $n+4$ is not a perfect square. I have been told that this proof is incorrect, however, I'm having a hard time seeing the issue.
Prove that for any integer $n > 0$ which is a perfect square, $n+4$ is not a perfect square:
Proof:
$n > 0$
Since $n$ is a perfect s... | What's wrong is no-one said it was the next perfect square.
So if $n = a^2$ and $n+4 = b^2$ we have no reason to assume $b = a+1$.
But we can assume $b \ge a+1$ and get a similar problem.
$n=a^2$ and $n+4 = b^2 \ge (a+1)^2 = a^2 + 2a + 1$
So $4 \ge 2a+1$ so $a \le 1.5$. Well nothing wrong with that. So $a = 1, 0$ and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Point within the interior of a given angle The point $M$ is within the interior of given angle $\alpha$. Find the distance between $M$ and the vertex of the angle ($OM=?$) if $a$ and $b$ are the distances from $M$ to the sides of the angle.
We can see that $$OM^2=OP^2+PM^2=OK^2+KM^2$$
Let $OP=x;PK=y$. Then $$2OM^2=OP^... | Denote the mid-point of $OM$ be $N$. Then $N$ is the center of the circle passing through the points $O,P,M,K$.
We have:
$$\angle PNK = 2\alpha,\quad \angle PMK = \pi-\alpha$$
By Cosine Theorem,
$$PK^2 = a^2+b^2-2ab\cos \angle PMK = PN^2 + KN^2 - 2PN\cdot KN \cos 2\alpha$$
Writing $PN = KN = \frac12 OM = r$, we have:
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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given $a+b+c=3$ prove that $abc(a^2+b^2+c^2) \leq 3$ I am preparing for inmo and I came accross this problem while solving a worksheet, but couldn't solve it, pl help me...
Problem-
Prove that if a,b,c are non negative real numbers such that a+b+c=3, then $abc(a^2+b^2+c^2)\leq3$ ... (0)
Developments-
Firstly applying... | WLOG $a\le b\le c$ so $\frac{b+c}{2}\ge 1$
let $$f(a,b,c)= abc(a^2+b^2+c^2)$$ now $$f(a,b,c)-f(a,\frac{b+c}{2},\frac{b+c}{2})=-\frac{1}{8} a (b - c)^2 (2 a^2 + b^2 - 2 b c + c^2)\le 0$$ let $t=\frac{b+c}{2}\ge 1$ So it suffices to show $$f(a,t,t)\le 3 \iff f(3-2t,t,t)\le 3$$ but $$f(3-2t,t,t)-3=-3 (t - 1)^2 (4 t^3 - ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Piecewise equivalence of trigonometric inverses In this website, it is a theorem that:
$$ 2 \tan^{-1}(x)= \begin{cases} \sin^{-1} \frac{2x}{1+x^2}, x \in [-1,1] \cr
\pi-\sin^{-1} \frac{2x}{1+x^2}, x >1 \cr -\pi - \sin^{-1} \frac{2x}{1+x^2}, x<1 \end{cases}$$
Why is there such a piecewise definition? I considered pro... | The arcsine function, $\arcsin : [-1,1] \to [-\frac \pi 2, \frac \pi 2]$ ($\color{red}{\mathit{note\ the \ codomain}}$ , and the function is often written as $\sin^{-1}$ but I prefer $\arcsin$) is defined as follows : if $a \in [-1,1]$ we can find a unique $x \in [-\frac \pi 2, \frac \pi 2]$ such that $\sin x =a$. We... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Collatz Conjecture: does an integer $4x+1$ resolve in $n$ steps if and only if an odd $x$ resolves in $n$ steps When analyzing the collatz conjecture in reverse. I came across this argument that attempts to show that an integer $4x+1$ resolves in $n$ steps if $x$ is odd and resolves in $n$ steps and likewise if $x$ is ... | If $x$ is odd then after one (×3+1) step we have $3x+1$. We then complete the op's step by applying $k$ (/2) steps. Where $k$ is the largest integer such that $2^k|3x+1$. $4x+1$ is also odd so after one (×3+1) step we have $12x+4$ then if we apply two (/2) steps we get $3x+1$ then we divide by $2^k$ and get to the same... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of continuous and differentiable of a given function with two variables I am trying to figure out why my function is differentiable and therefore continious.
Considering my function: $f(x, y) = x^3 - 3xy^2$
The partiel derivate of $x$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2 $$... | Let $(a,b)$ be any point in $\mathbb{R^2}$. Starting from where you left off, we need to show $f$ is differentiable at $(a,b)$. We calculate:
$f(a+h,b+k) - f(a,b)-((3a^2-3b^2)h+(-6ab)k)= (a+h)^3-3(a+h)(b+k)^2-a^3+3ab^2-3a^2h+3b^2h+6abk=3ah^2-3ak^2-6bhk+h^3-3hk^2$. Thus put $A = \left(3a^2-3b^2, -6ab\right)\implies \dfr... | {
"language": "en",
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Find the function $f(x)$. Let $f(x)$ be a polynomial function. If $f(x+2) - f(x) = 8x - 2$ and $f(0) = 5$, then what is $f(x)$?
I tried to replace $x$ with $0,2,4,\ldots $ for discovering some regular pattern but I have no idea after doing that.
| Let's try to determine the degree of the polynomial $f(x).$
If $\ f(x)\ $ has degree $0$, then $f(x) = a_0\ $ which implies $\ f(x+2)-f(x) = 0 \neq 8x-2.$
If $\ f(x)\ $ has degree $1$, then $f(x) = a_0 + a_1x\ $ which implies $\ f(x+2)-f(x) = 2a_1 \neq 8x-2.$
If $\ f(x)\ $ has degree $2$, then $f(x) = a_0 + a_1x + a_2x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Algebraic simplification of $(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)$ When trying to replicate a proof about Catalan numbers, I came across the following simplification $$(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)= a^2+a$$
This is confirmed by writing out all the terms (and by WolframAlpha). My question - is there a... | I found a solution from Lord Commander's working. As they note:
$$(a-b)(2b+1) = b(2a-2b-1) + a \tag{1}\label{eq1}$$
And:
$$(b+1)(2b-a) = b(2b-a+2) -a \tag{2}\label{eq2}$$
For simplicity, let $x = 2a-2b-1$ and $y = 2b-a+2$.
Thus $\eqref{eq1}$ becomes $(a-b)(2b+1) = bx + a$ and $\eqref{eq2}$ becomes $(b+1)(2b-a) = by - a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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2 numbers are chosen at random from set $\{1,2,3..5n\}$ (without replacement). Find the probability that $n_1^4-n_2^4$ is divisible by 5 All the numbers can be written in the form of $5k,5k+1,5k+2,5k+3,5k+4$
Also the given expression can be written as
$$(n_1-n_2)(n_1+n_2)(n_1^2 + n_2^2)$$
Case 1
Let $n_1-n_2$ be a mult... | Idea: separate into two sets $$S_1=\{5,10,15,..,5n\}\\S_2=\{1,2,3,4,6,7,8,9,11,...,5n-1\}$$now note that dividing by $5$ for each number (to the power of 4)has two remainders
$$a=5k+0\to r_1=\underbrace{a^4}_{mod 5}=0\\
a=5k+1\to r_2=\underbrace{a^4}_{mod 5}=1^4=1\\
a=5k+2\to r_3=\underbrace{a^4}_{mod 5}=2^4=16(mod \ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Compute $\sum\limits_{a=1}^{\infty} \sum\limits_{b=1}^{\infty} \sum\limits_{c=1}^{\infty}\frac{ab(3a+c)}{4^{a+b+c}((a+b)(b+c)(a+c)}$ Compute the following sum:
$$
\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty}\frac{ab(3a+c)}{4^{a+b+c}((a+b)(b+c)(a+c)}\
$$
I have no head or tail of how to even start this si... | The basic idea is that, the final value of the sum doesn't depend on the dummy indices, i.e. whether you choose $(a,b,c)$, $(x,y,z)$ or some other triplet - it simply does not matter. Take advantage of this, and swap the indices $(a,b,c)$ around. How many permutations of these $3$ exist? Exactly $3! = 6$ permutations. ... | {
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Finding a polynomial $f(x)$ that when divided by $x+3$ yields quotient $2x^2-x+7$ and remainder $10$ I'm struggling to grasp this particular question:
When a polynomial $f(x)$ is divided by $x+3$, the quotient is $2x^2-x+7$ and the remainder is $10$. What is $f(x)$?
This is what I did:
$$\begin{align}
f(x) &= (x+3)(2... | "Then, I realized that the remainder has to be 10 not 0...I have no idea how to do that"
Don't you?
If $f(x) = (x+3)(2x^2 -x + 7)$ then
$f(x) + r = (x+3)(2x^2-x +7) + r$
And $f(x) + 10 = (x+3)(2x^2-x +7) +10$.
And doesn't that mean the remainder of $f(x)\div (x+3) = 10$?
After all if we divide $(x+3)(2x^2 -x +7) + 10$ ... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $ How to determine the range of the function
$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$
It it is straightforward to verif... | Note: $f(1) \triangleq
\lim_{x\to 1} \tan ( \frac\pi2 \frac{(1+x)^2}{3+x^2})
\tan ( \frac\pi2 \frac{(1-x)^2}{3+x^2}) = 0$.
Since $f(x)$ is even, we only need to prove the case $x \in [0, \infty)$.
Clearly $f(x) < 0$ for all $x > 1$. Also, $f(1) = 0$.
Thus, we only need to prove the case $0\le x < 1$.
(Inspired by @piso... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the expression $\sqrt{8\cdot32\cdot(-3)^2}$ equal to? What is the expression $\sqrt{8\cdot32\cdot(-3)^2}$ equal to?
Sorry for the basic question. I am a little confused when solving such problems. They are very easy, I know, but still... Which is the easiest algorithm and how would you solve it? We are supposed... | It looks like you have the right idea... $8 = 2^3, 32 = 2^5, 8\cdot 32 = 2^8, (-3)^2 = 3^2...$
$\sqrt {2^8\cdot 3^2} = (2^8\cdot 3^2)^\frac 12 = 2^4\cdot 3 = 48$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let $m,n,b \in \mathbb{N} $ with $b > 1$ and $m \neq n$. Let $m,n,b \in \mathbb{N} $ with $b > 1$ and $m \neq n$.
If $b^{m}-1 $ and $b^{n}-1 $ have the same prime divisors, prove that $b+1$ is a power of 2.
I know that $b^{m}-1 = (b-1)(b^{m-1}+b^{m-2}+\cdots+ 1 )$ and same for $b^{n}-1$
Or maybe, I should consider $b^{... | Please note that this solution is due to Kamil Duszenko.
Consider first the case $n = 1$. So suppose first that $a - 1$ and $a^m - 1$ have the same prime divisors. If a prime p divides m, then $a - 1$ divides $a^p - 1$, and $a^p - 1$ divides $a^m - 1$, so a - 1 and $a^p - 1$ have the same prime divisors. But $a^p - 1 =... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{\sin{\frac{a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}$ (and another)
Compute
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{\sin{\frac{a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}\right)$$ and
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{\sin{\frac{a}{3^k}}\sin{\frac{2a}{3^k}}}{3^... | The second sum is telescoping, due to $$\frac{\sin x\sin 2x}{\sin 3x}=\frac14(\cot x-3\cot 3x).$$ This easily gives $$\sum_{k=1}^n\frac{\sin\frac{a}{3^k}\sin\frac{2a}{3^k}}{3^k\sin\frac{a}{3^{k-1}}}=\frac14\sum_{k=1}^n\left(\frac{1}{3^k}\cot\frac{a}{3^k}-\frac{1}{3^{k-1}}\cot\frac{a}{3^{k-1}}\right)=\frac14\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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probability distribution of a random variable that is uniformly distributed in [-1, 1] If $X$ is a random variable that is uniformly distributed between $-1$ and $1$, find the PDF of $\sqrt{\vert X\vert}$ and the PDF of $-\ln\vert X\vert$.
Solution or an explantion of approach wooulf work.
thank you in advance for your... | $X \sim U[-1, 1]$, $Y = \sqrt{|X|} \Rightarrow$
$$
F_Y(y) = \text{Pr}\left(Y \leq y\right), \quad y \geq 0,
$$
is a cumulative distribution function (c.d.f.) of $Y$ (by definition).
$$
\begin{aligned}
\text{Pr}\left(Y \leq y\right) &= \text{Pr}\left(\sqrt{|X|} \leq y\right) = \text{Pr}\left(|X| \leq y^2\right) = \text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof for The Limit $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4},\space n\in\Bbb{N}$ I am revising my knowledge on the topic of real analysis by attempting some simple proofs. The question requires for the proof of the Limit for $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4}$ using the definition of converg... | I think you need more words, and that as it stands the second line isn't quite right.
I would write
Given $\epsilon>0$ we want to find an $N$ such that for $n>N$ we have
$$|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\epsilon.\tag 1 $$
Let us see how big the left hand side can be:
$$|\frac{3n^2-n}{4n^2+1}-\frac{3n^2+\frac{3}{4}... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve one system of equations Solve the system of equations
$$ ax + by + cz = 0, $$
$$ bcx + cay + abz = 0, $$
$$ xyz + abc (a^3x + b^3y + c^3z) = 0 $$
I tried solving this using cross multiplication method but got stuck at one point :
$$x/ab^2-ac^2 = y/bc^2-ba^2 = z/ca^2-cb^2 = k (say) $$
I substituted the values in ... | Let us consider $b \neq c, a \neq b, a \neq c$. Then from the first two equation we get
\begin{align}
\left[\begin{array}{cc}b&c\\c&b\end{array}\right]\left[\begin{array}{c}y\\z\end{array}\right]=\left[\begin{array}{c}-a\\-\frac{bc}{a}\end{array}\right]x,
\end{align}
i.e.
\begin{align}
\left[\begin{array}{c}y\\z\end{ar... | {
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"timestamp": "2023-03-29T00:00:00",
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How to prove the Vandermonde's determinant for a $3\times 3$ matrix when rows and columns have been swapped? The problem is as follows:
The following determinant is named after french mathematician Alexandre-Théophile Vandermonde who lived in the late 18th century.
Prove this determinant
$\left|\begin{matrix}
1 & 1 & ... | Subtract the first column from the second column and factor out a common term from the second column.
$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=\left|\begin{matrix}
1 & 0 & 1\\
a & b-a & c\\
a^2 & b^2-a^2 & c^2
\end{matrix}\right|=(b-a)\left|\begin{matrix}
1 & 0 & 1\\
a & 1 ... | {
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Can $(x+y)^2$ be written in the form $f(x) g(y) + g(x)f(y)$? Are there functions $f$ and $g$ such that $f(x)g(y) + g(x)f(y) = (x+y)^2$. I'm asking this because in quantum mechanics, the symmetric wavefunction of two identical particles is written as $\psi(x_1, x_2) =\frac{1}{\sqrt{2}}[ f(x_1) g(x_2) + g(x_1) f(x_2) ]$.... | There are no functions $f,g$ defined on $\{0,1,2\}$ so that $f(x)g(y) + g(x)f(y) = (x+y)^2$ for all $x,y \in \{0,1,2\}$.
Proof. Assume such $f,g$ exist.
Take $x=y=0$,
$$
2f(0)g(0) = 0 .
$$
At least one of $f(0), g(0)$ is $0$. WLOG assume $f(0)=0$. Write
$a = g(0)$.
Take $x=0$,
$$
f(0)g(y)+g(0)f(y) = y^2\\
af(y) = y^... | {
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$\frac{(x-a)^2}{(x-2a)^2}=\left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert$ leading to contradictory solutions I have the following equation
$$\frac{(x-a)^2}{(x-2a)^2}=\left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert$$
And to find the critical values for the absolute value function I carried out polynomial long division and ... | If $$\frac{(x-a)^2}{(x-2a)^2}=\left|\frac{x-a}{x-2a} \right|$$
Let's $r = \left|\frac{x-a}{x-2a} \right|$.
This just says that $r^2=r$, hence $r=0$ or $r=1$.
If $r=0$, we have $x=a$.
Otherwise, $|x-a|=|x-2a|$, the distance from $a$ is equal to the distance from $2a$, hence $x=\frac{3a}{2}$.
Now, let's examine your sol... | {
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How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8,9$ which are divisible by $3$ and $5$? How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8$ and $9$ which are divisible by $3$ and $5$, without any of the digits repeating?
For the number to be divisible by $5$ it must end wit... | The two divisiblity rules you are assumed to use are: Divisible by $5$ if and only if last digit is $0$ or $5$ and; divisible by $3$ if and only if the sum of the digits add to a multiple of $3$.
Divisble by $5$ means last digit is $0$. That's the only option.
Divisble by $3$ means that first four digits, lets call th... | {
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Prove that $17$ divides $xy-12x+15y$
if $x,y$ are integers and $17$ divides both the expression $x^2-2xy+y^2-5x+7y$ and $x^2-3xy+2y^2+x-y$, then prove that $17$ divides $xy-12x+15y$
My attempt:
I tried to factories both expression and the result obtained is as followed
$x^2-2xy+y^2-5x+7y=(x-y)(x-y+5)+2y.....(1)$
$x... | Note: I'm going to introduce some integer variables like $k,n ,m $ etc. They are only for the purpose of showing the remainder of a number when divided by a particular number. For example if I say $x=17k+3 $ then what you need to focus on is the $3$ ( remainder) and not $k$ ( I'm stating this explicitly because at som... | {
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In which of the intervals is $\sqrt{12}$ In which of the intervals is $\sqrt{12}:$
a) $(2.5;3);$
b) $(3;3.5);$
c) $(3.5;4);$
d)
$(4;4.5)$?
We can use a calculator and find that $\sqrt{12}\approx3.46$ so the correct answer is actually b. How can we think about the problem without a calculator (if on exam for example)? I... | The smallest integer square greater than $12$ is $4^2 = 16$, and the largest integer square less than $12$ is $3^2 = 9$. Therefore, $3 < \sqrt{12} < 4$. This eliminates (a) and (d) as choices. The next thing to do is consider the value of $(3.5)^2$, which is $$(3 + 1/2)^2 = 9 + 2(3)(1/2) + (1/2)^2 = 12 + 1/4 > 12.$$... | {
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Expected value in rock paper scissor Two players play rock, paper, scissors for 3 wins ( one player wins 3 times). What is the expected value of the number of rounds?
I tried: for the n-th games probability : $$ p(x_n)=2\frac{\binom{n}{3}}{3^n} $$
Therefore my $E(x)\approx 1.8733$ which is too low...
| As suggested in comments, let's consider the states the game can be in after each round. $S(x,y)$ is the state where player A has won $x$ rounds so far and player B has won $y$ rounds so far. A tie round keeps the game in the same state. And let $T(x,y)$ be the distribution of the number of turns remaining in the game ... | {
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How can I prove this inequality using HM-GM-AM-QM inequalities? Given $a,b,c\in\mathbb{R}^{+}$, prove that $$(a+b+c)\left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c}\right) \geq \frac{9}{2}$$
I've been trying for a couple of hours in total, and I just can't seem to get it to work no matter what I do.
Edit: I also rea... | The AM-HM inequality states that for $x,y,z\in\Bbb{R}_{>0}$ $$\frac{x+y+z}{3}\geq\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\\\iff(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq9$$ Now take $x=(a+b),y=(b+c)$ and $z=(c+a)$
One More Proof Using the Famous Nesbitt's Inequality:
$$(a+b+c)\left(\frac{1}{a+b}+\... | {
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Evaluate $f(n)=\frac{(n-2)!!}{(n-3)!!}$, for $n \geq 3$ by Big O notation
Is there a better way than mine to evaluate this function
$$f(n)=\frac{(n-2)!!}{(n-3)!!},\quad \text{for}\quad n \geq 3$$
by Big O notation?
I got result below, but not so accurate. I know that:
$$n!!=\left\{ \begin{array}{ll}
2^{k}k! &,... | You should analyze the even and odd cases separately since they have different asymptotics.
Also, there is no need to approximate $2^{2n}$ since it cancels out.
If $n = 2k$ we have
$$\frac{(2k)!!}{(2k-1)!!}=\frac{(2^k k!)^2}{(2k)!} = \frac{4^k\left(\sqrt{2\pi k}\left(\frac{k}{e}\right)^k\left(1+\frac{1}{12k}+\frac{1}{2... | {
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Find the number of members in a sequence $X_n=\frac{2n-1}{4n+5}$, outside the interval $[\frac{1}{2}-\frac{1}{1000},\frac{1}{2}+\frac{1}{1000}]$ Find the number of members in a sequence $X_n=\frac{2n-1}{4n+5}$ outside the interval $ [\frac{1}{2}-\frac{1}{1000} , \frac{1}{2} + \frac{1}{1000}]$.
Step 1:
I remember these ... | You have found
$$|\frac{2n-1}{4n+5} -\frac{1}{2}|<\frac{1}{1000} \iff n > 873,75.$$
Since $ n \in \mathbb N$, this gives
$$|\frac{2n-1}{4n+5} -\frac{1}{2}|<\frac{1}{1000} \iff n \ge 874.$$
Hence
$$|\frac{2n-1}{4n+5} -\frac{1}{2}| \ge \frac{1}{1000} \iff n \le 873.$$
| {
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The maximum of $a+b+c$ satisfies $2^n=a!+b!+c!$ for $n,~a,~b,~c\in\Bbb N$ How to find the maximum of $a+b+c$ satisfies $2^n=a!+b!+c!$ for $n,~a,~b,~c\in\Bbb N$? I can only deduce that $a,~b,~c$ cannot be $\geq 3$ simultaneously.
| Wolog assume $a \le b \le c$. So $a!|b!, c!$ so $a!|a! + b! + c!=2^n$. But as $3\not \mid 2^n$ we must have $a < 3$ so so $a = 1, 2$.
If $a = 1$ then $2\le b! +c! = 2^n-1$ and that is odd . So $b!$ and $c!$ can't both be even. So $b! = 1$. And we have $1\le c! = 2^n - 2 = 2(2^n-1)$. $2^n - 1$ is odd so only a sing... | {
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Help understanding the solution of $m^2 = n^3 -4$. The material I'm studying has the solution for $m^2 = n^3 + 4$ but I can't quite understand it. It starts like this:
First suppose, for a contradiction, that $m$ is odd. Then $n^3 = m^2 − 4 = (m + 2)(m − 2)$. Any common factors of $m + 2$ and $m − 2$ divide $(m + 2) − ... |
a) How did they get to the conclusion gcd(m+2, m-2)=1?$.
Let $\gcd(m+2, m-2) = d$ then $d|m+2, m-2$ so $d|(m+2) - (m-2) = 4$. So $d = 1,2,4$. But $m$ is odd (so your example of $m=10$ won't work) so $m\pm 2$ is odd. So $d \ne 2,4$. So $d = 1$.
Note: the general idea that if $\gcd(a,b)$ divides $a,b$ it will divid... | {
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Solving the functional equation $ f \left( x ^ 2 + f ( y ) \right) = f \big( f ( x ) \big) + f \big( f ( y ) \big) + x ^ 2 f ( y ) + y ^ 2 f ( x ) $
Find all functions $ f : \mathbb R \to \mathbb R $ such that
$$ f \left( x ^ 2 + f ( y ) \right) = f \big( f ( x ) \big) + f \big( f ( y ) \big) + x ^ 2 f ( y ) + y ^ 2 f... | Suppose $f$ is not the constant zero function.
If $f(y_1)=f(y_2)$ for some $y_1, y_2$ then letting $y:=y_1$, $y:=y_2$ and $x$ such that $f(x)\neq 0$ we obtain $y_1=\pm y_2$.
Then it follows from $f(f(x))=f(x^2)$ that for all $x$ either $f(x)=x^2$ or $f(x)=-x^2$.
If $f(x)=-x^2$ for some $x$ then letting $y:=x$ we get $f... | {
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Determine all ordered pair (x,y) of positive integers such that $y^{2}−(x+ 2)2^x=1$. This is what I have come up so far,
*
*y has to be odd. Since $(x+2) 2^{x}$ is always even, hence $(x+2)2^x+1$ is always odd. Thus, y must be an odd number.
*$$y^2 \equiv1(mod 4)$$Since all square numbers either congruent to 0 or ... | Rearrange first to get
$$(x+2)2^x = y^2-1.$$
Then factoring the RHS gives
$$(x+2)2^x = (y-1)(y+1).$$
Now, this implies $y-1 = a2^{m}$ and $y+1 = b2^{n}$ for some $a,b \in \mathbb{N}$ such that $ab=(x+2)$, and some $m,n \in \mathbb{N} \cup \{0\}$ such that $m+n=x$. Now, note that $2^{\min\{m,n\}}$ is a factor of both $a... | {
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Calculate integral $\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$ I recently saw the integral problem
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$$
and tried to solve it. Below is what I did.
Interesting to look at other easier solutions.
$$\int\limits_{0}^{2\pi}\frac{dx}{\l... | Back to serious
At a point you wrote
$$\int \frac{1+n^2+y^2}{(1+y^2)^2}\, dy$$ Write $(1+y^2)=(y+i)(y-i)$ and use partial fraction decomposition to obtain
$$\frac{1+n^2+y^2}{(1+y^2)^2}=\frac{2+n^2}4 i\left(\frac 1{y+i} -\frac 1{y-i}\right)-\frac {n^2}4 \left(\frac 1{(y+i)^2} +\frac 1{(y-i)^2}\right)$$ and use the logar... | {
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How to find a vector component from another vector and the angle between the two vectors? I have two vectors, $(-2, 3, 1)$, and $(-1, 2, a)$. I know that the angle between these two vectors is $40^\circ$. How do I find a?
When I try to algebraically solve for a, I run into problems with the magnitudes. I understand tha... | We know that the scalar product is defined as:
$$\mathbf{v}\bullet \mathbf{u}=vu\cos \theta\equiv v_xu_x+v_yu_y+v_zu_z$$
i.e.
$$\sqrt{14}\cdot \sqrt{5+a^2}=\frac{8+a}{\cos(40^{\circ})} \iff \sqrt{14\cdot (5+a^2)}=\frac{8+a}{\cos(40^{\circ})}$$
Hence, squaring LHS and RHS,
$$70\cos ^2\left(40^{\circ}\right)+14\cos ^2\le... | {
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Find $S = a + b$ such that for $\forall m \in \left[a\sqrt{\frac{15}{7}} + b\sqrt{\frac{7}{15}}; 2\right)$ then $2x^2 + 2x - mf(x) + 5 = 0$ has root
Let $f(x)$ be continuos on $\mathbb R$ satisfy $f(0) = 2\sqrt{2}$ and $f(x) > 0, \forall x \in \mathbb R$ and $f(x) f'(x) = (2x+1)\sqrt{1+f^2(x)}$.
For all $m \in \left[a... | The question is equivalent to asking when $$f(x) = (4-m^2) x^4 +(8-2m^2)x^3 +(24-7m^2) x^2 +(20-6m^2) x+25 - 8m^2 $$ has a root.
Let’s consider
$$f’(x) = 4(4-m^2)x^3 +6(4-m^2)x^2 +2(24-7m^2) x+(20-6m^2) \\ = -2 (1 + 2 x) (-10 + 3 m^2 + (m^2 -4)x + (m^2 -4)x^2) $$
So, there is always a turning point at $x=-0.5$.
$$f(-0.... | {
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if p is prime then 1920 divides $p^4-10p^2+9$ if $p>6$ is prime then 1920 divides $p^4-10p^2+9$
Take out the factors of 1920 which are $$1920=2^7(3)(5)$$
and factor the expression that remains as
$$p^4-10p^2+9=(p-3)(p-1)(p+1)(p+3)$$
and I assumed as p is prime then p-1 and p + 1 is even then it is divisible by $2^7$
bu... | *
*Since $p$ is prime greater than $3$ then $\gcd(p,3)=1$ and by little Fermat theorem we have $p^3\equiv p\pmod 3$
This allows us to reduce $p^4-10p^2+9\equiv p^2-10p^2+0\equiv -9p^2\equiv 0\pmod 3$
*
*Similarly $p^4\equiv 1\pmod 5$
This allows us to reduce $p^4-10p^2+9\equiv 1-0p^2-1\equiv 0\pmod 5$
*
*For the d... | {
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Prove that the product of four consecutive natural numbers are not the square of an integer
Prove that the product of four consecutive natural numbers are not the square of an integer
Would appreciate any thoughts and feedback on my suggested proof, which is as follows:
Let $f(n) = n(n+1)(n+2)(n+3) $.
Multiplying out... | Hint: $n(n+1)(n+2)(n+3) +1$ is a square.
| {
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Fun Equilateral Triangle Problem $\Delta ABC$ is an equilateral triangle with a side length of $4$ units.$\: $$\: $
$\angle CAF = \angle EBC =\angle FAB$ .$\: $$\: $
$D \in \left | AF \right |\: ,\: E \in \left | CD \right |\: ,\: F \in \left | BE \right | $ $\: $
Find the length of $\left | AD \right |$
By given angl... | Let $AD = x$. Then, $CD = 2 + x$ and $AC = 4$. Because we know $\angle ADC = \frac{2\pi}{3}$, we can apply the Law of Cosines on $\triangle ACD$:
$$AC^{2} = AD^{2} + CD^{2} - 2(AD)(CD)\cos\bigg(\frac{2\pi}{3}\bigg)$$
$$4^{2} = x^{2} + (2 + x)^{2} + x(2 + x)$$
$$3x^{2}+6x - 12 = 0$$
$$x^{2} + 2x - 4 = 0$$
The solutions ... | {
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Prove that $\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A} \leq \frac{A}{B}+\frac{B}{A}$ for acute angles, $A$ and $B$.
Prove that $\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A} \leq \frac{A}{B}+\frac{B}{A}$ for acute angles, $A$ and $B$.
I'm confused about how to do this since we can't say $\frac{\sin A}{\sin B}\leq \fra... | $f(t) = t\sin t,\ g(t) = \frac{t}{\sin t}$ are both increasing functions on $(0,\pi/2)$.
Also note that: if $a \le b$ and $c\le d$, $$ad +bc \le ac+bd\tag{1}$$ which follows from $(a-b)(c-d) \ge 0$.
Suppose $x <y$. Then we have $x\sin x <y \sin y$ and $\frac{x}{\sin x} < \frac{y}{\sin y}$.
Then from $(1)$, it follows ... | {
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Prove $(1+\sqrt 2)^n - (1-\sqrt 2)^n$ is divisible by $2$ for all integers $n\ge0$ Prove $(1+\sqrt 2)^n - (1-\sqrt 2)^n$ is divisible by $2$ for all integers $n\ge0$
I am trying to prove this by induction and having a hard time doing so. What I have for the inductive step is $$(1+\sqrt 2)^{k+1} - (1-\sqrt 2)^{k+1}$$ t... | Statement: prove that for any $n \ge 1$, $(1+\sqrt{2})^n + (1-\sqrt{2})^n$ is divisible by $2$. ( Note we can start with $n = 1$ and that the sign of $+$ is used as the original statment by OP is wrong if the $-$ sign is used. )
Claim 2: For any $k \ge 1, \exists m \ge 1: (1+\sqrt{2})^k - (1-\sqrt{2})^k= m\sqrt{2}, k, ... | {
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How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involvin... | This is not an answer, but its too long to be a comment. Also, this isn't a closed-form solution, but I think its an interesting approach. Also, I won't be rigorous here, I'll just be pretending everything converges nicely
We start with $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$ with the goal of ending at
$$\sum_{n=3}^\... | {
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Value of $\sum_{r= 0}^{2019}\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{2021}{r-k} $ What is tried was : $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\... | In seeking to evaluate
$$\sum_{r=0}^n \sum_{k=0}^r
(-1)^k (k+1) (k+2) {n+2\choose r-k}$$
we write
$$2 \sum_{r=0}^n [z^r] (1+z)^{n+2}
\sum_{k=0}^r {k+2\choose 2} (-1)^k z^k.$$
The coefficient extractor enforces the range of the inner sum:
$$2 \sum_{r=0}^n [z^r] (1+z)^{n+2}
\sum_{k\ge 0} {k+2\choose 2} (-1)^k z^k
\\ = 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Divisibility of an expression by 11 How to prove that $5^{5n+1}+4^{5n+2}+3^{5n}$ where $n\in \mathbb{N}$ is divisible by $11$ using mathematical induction?
I have tried and got to this $$5 \cdot 25 \cdot 25 \cdot 5^{5k+1}+4\cdot 16 \cdot 16 \cdot 4^{5k+2}+3\cdot 9 \cdot 9 \cdot 3^{5k}$$
and I even skipped the first ste... | Hint:
Let $f(n)=5^{5n+1}+4^{5n+2}+3^{5n}$. Then
$$
f(n+1)-f(n)
=(5^5-1) \cdot5^{5 n + 1} + (4^5-1) \cdot 4^{5 n + 2} + (3^5-1) \cdot 3^{5n}
= 11(284\cdot5^{5 n + 1}+ 93 \cdot 4^{5 n + 2} +22\cdot 3^{5n})
$$
The crucial point is that $5^5-1,4^5-1,3^5-1$ are all divisible by $11$.
| {
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"source": "stackexchange",
"question_score": "2",
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Consecutive Number Divisibility While I was solving a practice problem, I became interested in coming to the conclusion about the following:
Is it possible for both $\frac{x+1}y$ AND $\frac x{y+1}$ to be integers, and if so, how would I find them. Looking at this, I was pretty sure there wasn't any, but I had no concre... | \begin{align}
\dfrac{x+1}y &= m \\
\dfrac x{y+1} &= n \\
\hline
x+1 &= my \\
x &= ny + n \\
\hline
ny + n + 1 &= my \\
my - ny &= n+1 \\
\hline
y &= \dfrac{n+1}{m-n} \\
x &= n(y+1) \\
\end{align}
So, for example, let $n=11$, then the possible values for $m-11$ are
$1,2,3,4,6,12$, the divisors of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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if $AB+BC+AC\le 2+\sqrt{3}$ find the maxmum of the value $P=AB\cdot BC\cdot AC$ For $\Delta ABC$ if the circumradius $R=1$,and such
$$AB+BC+AC\le 2+\sqrt{3}$$
find the maxmum of the value
$$P=AB\cdot BC\cdot AC$$
if let $a=BC,b=AC,c=AB$,then we have
$$a+b+c\le 2+\sqrt{3}$$
and if use sine theorem we have
$$a=2R\sin{A}... | We need to find the maximum of $P = 8\sin A \sin B \sin C$ under the conditions
$A, B, C \ge 0$, $A + B + C = \pi$, and $2\sin A + 2\sin B + 2\sin C \le 2 + \sqrt{3}$.
From the conditions, we have
$(2\sin A + 2\sin B)^2 \le (2 + \sqrt{3} - \sin C)^2$.
We have
\begin{align*}
(2 \sin A + 2\sin B)^2 &= 16 \sin^2 \frac{A ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Minimizing $a_1x_1^2 + a_2x_2^2$ for positive $a_i$, where $a_1x_1+a_2x_2=B$
Find $$\min\{a_1x_1^2 + a_2x_2^2\}$$
Where $ a_1x_1 + a_2x_2 = B$, and $a_1>0$ and $a_2>0 $. Find $x_1$ and $x_2$.
Can we do it usig AG mean inequality?
Let's say we have the problem to find the minimum value of $ x_1^2 + x_2^2 $.
From: $ (x... | Use standard algebra:
We have,
$$\begin{align}a_1x_1^2+a_2x_2^2&=a_1\left(\frac{B-a_2x_2}{a_1}\right)^2+a_2x_2^2\\
&=\frac{B^2-2Ba_2x_2+a_2^2x_2^2}{a_1}+a_2x_2^2\\
&=\left(a_2+\frac{a_2^2}{a_1}\right)x_2^2-\left(\frac{2Ba_2}{a_1}\right)x_2+\frac{B^2}{a_1}\end{align}$$
This is the quadratic polynomial. You can minimize ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is all possible value of a,b $\in\mathbb{R}$ so that the following integral converges:$\int_{0}^{+\infty}{dx\over x^a(4+9x)^{b+1}}$ What is all possible value of a,b $\in\mathbb{R}$ so that the following integral converges:
$\int_{0}^{+\infty}{dx\over x^a\space(4+9x)^{b+1}}$
I want to use the limit comparison test... | The two improprieties we must consider are $0$ and $\infty$.
For small $x$, we have $x^a (4 + 9x)^{b + 1} \sim x^a 4^{b + 1}$, and we know that $\int\limits_0^1 \frac{1}{x^a} dx$ converges iff $a < 1$. So we require that $a < 1$.
Edit: to be precise, $\lim\limits_{x \to 0} \frac{x^a (4 + 9x)^{b + 1}}{x^a 4^{b + 1}} = 1... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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How many ways are there to distribute $18$ distinguishable object into boxes?
How many ways are there to distribute $18$ distinguisable object into
a-) $5$ distinguishable boxes so that the boxes have $1,2,4,5,6$ objects in them, respectively.
b-) $5$ distinguishable boxes so that the boxes have $1,2,4,5,6$
objects in... | I only have comments on your working for (f) and (h).
(f) The idea is to create $5$ indistinguishable heaps of $1, 2, 4, 5$ and $6$ objects and that is simply $C(18,1) \times C(17,2) \times C(15,4) \times C(11,5) \times C(6,6)$, same as in your answer $(a)$.
(h) It is similar to (f) but we have heaps of $4, 2, 2, 5$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluating a Double Integral $\iint \sqrt{x(2a-x)+y(2b-y)}$ Please help me in evaluating the double integral
$$\iint \sqrt{x(2a-x)+y(2b-y)}$$
over the region bounded by the circle $x^2+y^2-2ax-2by=0$.
My thought was to change the variable by substituting $x=r(1+cos \theta)$ and $y=r(1+sin \theta)$. so that $0\leq r\leq... | Completing the square we have that the integrand is really
$$\sqrt{2ax-x^2+2by-y^2} = \sqrt{a^2+b^2-(x-a)^2-(y-b)^2}$$
which, when integrated over the circle, $$(x-a)^2+(y-b)^2=a^2+b^2$$ gives half the volume of the sphere $$(x-a)^2+(y-b)^2+z^2=a^2+b^2$$ Thus the answer is
$$I = \frac{2\pi}{3}(a^2+b^2)^{\frac{3}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the zero divisors and ideals of $ \left\{ \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \: : \: a,b \in \mathbb{Z}_{5} \right\}$ Let consider the following subset of $M_{2\times 2}(\mathbb{Z}_{5})$
$$A:= \left\{ \begin{pmatrix} a & b \\
-b & a \end{pmatrix} \: : \: a,b \in \mathbb{... | The $(a,b)$ pairs such that $a^2 + b^2 = 0$ in $\mathbb Z_5$ are $(0,0), (1,2), (1,3), (2,1), (2,4), (3,1),(3,4),(4,2), (4,3)$
If $a^2 + b^2 = 0$ then $\begin{bmatrix} a & b\\ -b & a\end{bmatrix}\begin{bmatrix} b & a\\ -a & b\end{bmatrix} = 0$
It is still open to prove the other direction.
The (non-trivial) ideals
If w... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving for integers $m>8, n > 0$ for $2^m - 3^n = 13$ Let $m,n$ be integers such that:
$$2^m - 3^n = 13$$
$m > 8$ since $2^8 - 3^5 = 13$.
I am trying to either find a solution or prove that no solution exists.
I tried to use an argument similar to this one for $2^m - 3^n = 5$ where $m > 5$.
$3^n \equiv -13 \pmod {512}... | $$2^{m} - 3^{n} = 13\tag{1}$$
An argument mod $3$ shows that $m$ is even, then let $m=2a.$
We can take the three cases $n=3b, n=3b+1,$ and $n=3b+2.$
The problem can be reduced to finding the integer points on elliptic curves as follows.
$\bullet n=3b$
Let $X=3^{b}, Y=2^{a}$, then we get
$Y^2 =X^3 + 13.$
According to LM... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Let a sequence ${[a_n]}$ be defined such that $a_1=1$ and when $n\ge1$, $a_{n+1}=a_n+\frac{1}{a_n}$. Then, show that:$12
Let a sequence ${[a_n]}$ be defined such that $a_1=1$ and when $n\ge1$, $$a_{n+1}=a_n+\frac{1}{a_n}$$
Then, show that:$$12<a_{75}<15$$
(the bracket in ${[a_n]}$ does not denote GIF, it is a general s... | Use
$$ a_n^2 + 3 \gt a_{n+1}^2 \gt a_n^2 + 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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System of two quadratics equation, $P(x)$ and $Q(x)$ If $P(x) = ax^2 + bx + c$ and $Q(x) = – ax^2 + dx + c$, $ac \ne 0$, then the equation $P(x) . Q(x) = 0$ has
(A) Exactly two real roots
(B) At least two real roots
(C) Exactly four real roots
(D) No real roots
My approach is as follow Let $T(x)=P(x).Q(x)$
$T\left( x \... | If the coefficient of $x^2$ and the constant term of a quadratic equation are of opposite sign $($both non-zero$)$ then discriminant of the quadratic equation is positive, which means roots are real.
So either $P(x)$ or $Q(x)$ has coefficient of $x^2$ and the constant term with opposite sign $($both are non-zero$)$. So... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Conjugate Harmonic $\displaystyle u=\frac{1}{2}\log(x^2+y^2)$ is harmonic and find its harmonic conjugate function
$\displaystyle\frac{\partial u}{\partial x}=\frac{x}{x^2+y^2}$
$\displaystyle\frac{\partial u}{\partial y}=\frac{y}{x^2+y^2}$
$\displaystyle dv=\int\frac{-\partial u}{\partial y}dx +\int\frac{\partial u}{... | $$\displaystyle\frac{\partial u}{\partial x}=\frac{x}{x^2+y^2}$$
One of the Cauchy-Riemann equations is:
$$\displaystyle\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$
so
$$\frac{\partial v}{\partial y}=\frac{x}{x^2+y^2}$$
$$v = \int \frac{x}{x^2+y^2} \, dy=\frac{1}{x} \int \frac{1}{1+(\frac{y}{x})^2} \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations a... | It is equivalent to show that $\log \mathcal{P}(N) < 0$ for $N \ge 1$. That is,
$$
\begin{align*}
\frac{2N+1}{2N+2}\log\left(\frac{2N}{2N + 1}\right) + \frac{\log(2)}{2N+2} &< 0 \\
\frac{2N+1}{2N+2}\log\left(1-\frac{1}{2N + 1}\right) + \frac{\log(2)}{2N+2} &< 0
\end{align*}
$$
Since for $0 < x < 1$ we have $\log(1-x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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Prove that $\sum_{n=1}^\infty \frac{2}{2n}-\frac{2}{2n+1}=2-2\ln(2)$ The question is simple, to prove that $\sum_{n=1}^\infty{\frac{2}{2n}-\frac{2}{2n+1}}=2-2\ln(2)$.
I did it by wirtting down some terms of the series, rearranging as follows:
$$\sum_{n=1}^\infty \frac{2}{2n}-\frac{2}{2n+1}$$
$$2\sum_{n=1}^\infty \frac... | Use $\frac{1}{m}=\int_{0}^{1} t^{m-} dt$ and sum of an infinite GP to get
$$S=2\sum_{n=1}^\infty \left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\sum_{n=1}^\infty \left(\int_{0}^{1}( t^{2n-1}-t^{2n})dt\right)=2\int_{0}^{1} \left( \frac{t^{-1} t^2}{1-t^2}-\frac{t^2 }{1-t^2}\right)dt$$ $$=2\int_{0}^{1} \left(\frac{t}{1+t}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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There exists an $\alpha \in R$, with $\alpha^2=2$. Reading Abott's Understanding Analysis. I'm trying to understand each part of this proof that is presented for a theorem. I'll only state the first part of it since it is a division of cases and I'm just confused about the first part.
Theorem 1.4.5. There exists a real... | First of all,
$$ \left( \alpha + \frac1n \right)^2 = \alpha^2 + \frac{2\alpha}{n}+\frac{1}{n^2}.$$
Second, you need to prove that this is $<2$ for some $n$. It doesn't matter how you prove it, but you do need to prove it one way or another.
One way to prove this is to simplify:
$$ \frac{2\alpha}{n}+\frac{1}{n^2} \le \f... | {
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"url": "https://math.stackexchange.com/questions/4152937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Solution Verification: Find Extrema points of the function $f(x,y)=2^{3x+8y}$ that are on $x^2+y^2=1$.
$f(x,y)=2^{3x+8y}$
$x^2+y^2=1$
Polar Coordinates approach:
Let $x=cos(t), y=sin(t), 0\le t \le 2\pi$ (We can see that this solved the circle equation).
Substituting them into my function:
$f = 2^{3\cos(t)+8\sin... | Since $\cos^2 t + \sin^2 t = 1$, can you maximise $3 \cos t + 8 \sin t$?
Without using calculus, you need $k (\sin(t+a)) = k(\sin t \cos \alpha + \cos t \sin \alpha)= 3 \cos t + 8 \sin t$, so $k \sin \alpha = 3, k \cos \alpha = 8$ which lets you find $\alpha$, then $k$.
This gives $\tan \alpha = \frac{8}{3}$ and since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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What is the length of side AB of the triangle? Point X is 3m from A, 4m from B and 5m from C. Point X is inside the triangle formed by ABC. If AB = BC and angle B is right angle. Find the length of side AB.
I have come up with the following figure and aware that this is an isosceles right triangle, unfortunately i'm no... |
Reflect point $X$ across the sides of $\triangle ABC$. Observe,
$$\begin{align*}
\angle DAF&=\angle DAX+\angle FAX=2(\angle BAX+\angle CAX)=90^{\circ}\;\implies DF=3\sqrt{2}\\
\angle DCE&=\angle DCX+\angle ECX=2(\angle ACX+\angle BCX)=90^{\circ}\;\implies DE=5\sqrt{2}\\
\angle FBE&=\angle FBX+\angle EBX=2(\angle ABX+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Convergence of the series $\sum_{n=1}^\infty \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} 2^n= \sum_{n=1}^\infty \dfrac{4^n n! n!}{(2n)!}$ I am trying to compute the interval of convergence of the power series $\sum_{n=1}^\infty \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} x^{2n+1}$. Letting $a_n= \dfrac{n!}{1\cdot... | One more, possibly easier, way to use estimation:
$$\sqrt{2n} \lt \frac{2}{1}\cdot\frac{4}{3}\cdot \frac{6}{5}\cdots \frac{2n}{2n-1} \leqslant 2\sqrt{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for real numbers $x$ and $y$, simultaneously the equations given by $xy^2 = 15x^2 + 17xy + 15y^2$ and $x^2y = 20x^2 + 3y^2$.
Solve for real numbers $x$ and $y$, simultaneously the equations given by
$$
\left\{
\begin{array}{l}
xy^2 &= 15x^2 + 17xy + 15y^2 \\
x^2y &= 20x^2 + 3y^2
\end{array}
\right.
$$
by taking... | Subbing $y = x t$ is a good idea. Since you've already found $x = y = 0$, let's assume $x \ne 0$. Doing the substitution and simplifying gives
$$
xt^2 = 15 t^2 + 17 t + 15 \\
x t = 20 + 3t^2.
$$
Multiply the bottom equation by $t$ and subtract to get
$$
0 = t(3t^2 + 20) - (15t^2 + 17 t + 15)= 3t^3 -15 t^2 + 3t - 15 = 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x^3-\frac1{x^3}=108+76\sqrt2$, find $x-\frac1x$
If $x^3-\frac1{x^3}=108+76\sqrt2$, find $x-\frac1x$
LHS = $(x-\frac1x)(x^2+\frac1{x^2}+1)=(x-\frac1x)((x-\frac1x)^2+3)$
Now, maybe RHS needs to be factorized so that some comparisons can be made, but not able to do so.
Or maybe LHS can be written as $(x-\frac1x)^3+3... | let $z = x - x^{-1}$; then we seek a "nice" solution of the form $z = a + b \sqrt{2}$ such that $z^3 + 3z = 108 + 76 \sqrt{2}$. This in turn implies $$a(3 + a^2 + 6b^2) = 108, \\ b(3 + 3a^2 + 2b^2) = 76.$$ If $a, b$ are integers, then we must have $a \in \{1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108\}$ and $b \in \{1, ... | {
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"url": "https://math.stackexchange.com/questions/4164309",
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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... | Always remember : "never simplify denominator unless you know the values."
Onto the answer :
just cross multiply to get $$\frac{1 + 4 \sin50 \sin10}{2 \sin50}$$
which would further simplify as $$\frac{1+2(\cos40 - \cos60)}{2\sin50}$$
which is easy to solve as we get $$\frac{\cos40}{\sin50} = 1$$
so the answer is 1
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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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 ... | See , $$\frac{1-\sin x}{1+\sin x}=4$$ $$1-\sin x=4+4\sin x$$ $$\sin x=-\frac{3}{5}$$ Hence , $x\approx-37^o$ and $\tan\frac{-37}{2}\approx-\frac{1}{3
}$
| {
"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": 4
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Maximizing the product of $k$ positive integers where the sum is equal to $n$ Question: Given an integer $n$, break it into the sum of $k$ positive integers, where $k \geq 2$, and maximize the product of those integers.
Return the maximum product you can get.
Comment:
The key observation is the fact that for any intege... | Take $n = 10$ and $k = 4$ for example. If we use your method, we have $3 \cdot 3 \cdot 3 \cdot 1 = 27$. If we try another way, we can have $2 \cdot 2 \cdot 3 \cdot 3 = 36$.
My hypothesis (without the $k$ constraint):
To maximize, break $n$ into $\lfloor\frac{n}{2}\rfloor$ and $\lceil\frac{n}{2}\rceil$. Then, for each ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the range of possible values of $\sqrt{a^2+a+1}-\sqrt{a^2-a+1}$
Let $x=\sqrt{a^2+a+1}-\sqrt{a^2-a+1},x\in \mathbb R$. Find range of possible values of $x$.
I tried drawing the graph and obtained this:
Through which the answer came out to be $(-1,1)$.
What should be the procedure through algebra?
| Since $f(-a)=-f(a)$, we only need to find the range of $f(a)$ for $a\ge0$. Now, we have $$g(a):=f(a)^2=2a^2+2-2\sqrt{(a^2+1)^2-a^2}=2a^2+2-2\sqrt{a^4+a^2+1},$$
so
$$g'(a)=4a-2\frac{4a^3+2a}{2\sqrt{a^4+a^2+1}}=4a\left(1-\frac{a^2+\frac12}{\sqrt{(a^2+\frac12)^2+\frac34}}\right)\ge0,$$ and hence $g(a)$ (and hence also $f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$ \lim_{n \to \infty} \sum_{k=1}^n \sqrt{2- \left( \frac{k}{n} \right)^2} \cdot \frac{k}{n^2}$ is my solution any good? First write
$$ \lim_{n \to \infty} \sum_{k=1}^n \sqrt{2- \left( \frac{k}{n} \right)^2} \cdot \frac{k}{n^2} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \sqrt{2- \left( \frac{k}{n} \right)^2} \cdot \f... | For
$I
=\int^1_0 \sqrt{2-x^2} \ x \ dx
$,
if
$u = 2-x^2$
then $du = -2xdx$
so
$x dx = -du/2$
so
$I
= \int_{\sqrt{2}}^1 \sqrt{u}(-du/2)
= \frac12\int_1^{\sqrt{2}} \sqrt{u}du
=\frac12\dfrac{x^{3/2}}{3/2}|_1^{\sqrt{2}}
=\frac13(2\sqrt{2}-1)
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solution Verification: Find the mass of $V$. (Triple integral)
We define $V=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2+z^2\le z$ }
Mass Density is given by $g(x,y,z)=\sqrt{x^2+y^2+z^2}$ ,
Find the Mass of $V$.
My Work:
So I need to find $I=\iiint_{V}\sqrt{x^2+y^2+z^2}dxdydz$, and in order to do that, I will move to spherica... | Your approach is correct but you have a mistake in your work.
$V=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2+z^2\le z\}$
As you defined $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$
$x^2+y^2+z^2 \leq z \implies \rho \leq \cos\phi, 0 \leq \phi \leq \dfrac{\pi}{2}$ and $0 \leq \theta \leq 2\pi$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$. For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$.
My attempt: We have
$$2(a^2+b^2)\geq (a+b)^2$$
so $$-2\leq a+b \leq 2$$
In other hand $$ab=\frac{(a+b)^2-2}{2}=(a+b)^2-1$$
| Let $x=a+1,y=b+1$. Then the statement is equivalent to $$(x-1)^2+(y-1)^2=2\Rightarrow 3(x-1)+3(y-1)+(x-1)(y-1)\geq -5,$$ or $$x^2+y^2=2x+2y\Rightarrow (2x+2y)+xy\geq 0,$$ which is clear, since $$(x^2+y^2)+xy\geq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find the value of $\cos105^\circ+\sin75^\circ$ Find the value of
$$\cos105^\circ+\sin75^\circ.$$
We can write the given trig expression as $$\cos(180^\circ-75^\circ)+\sin75^\circ=-\cos75^\circ+\sin75^\circ\\=\sin75^\circ-\cos75^\circ$$
I don't see what else I can do. Thank you!
| There's some transformation formula I'm listing below,
$\sin (a+b) + \sin (a-b)= 2 \sin a \cos b$
$\sin (a+b)-\sin (a-b) = 2\cos a \sin b$
$\cos (a+b)+ \cos (a-b) = 2\cos a \cos b$
$\cos (a-b) -\cos (a+b) = 2\sin a \sin b$
Now your problem reduces to $\sin 75° - \cos 75°$, to make use the above formula, we convert $\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $ without expansion It's easy to prove that if $a,b,c \neq 0 $: $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $$
as $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff \frac{ab+bc+ca}{ab... | We first start with a little rearrangement that is :
$$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} $$
implies :
$$ (ab+bc+ac)(a+b+c)-abc=0$$
Now note try setting $a=0$ we see then
$$bc(b+c)=0$$
So In order to make this vanish set ,
$$b=-c$$
But also observe that , this belongs to the solution set even if
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to check if an integer can be represented with all set bits for any given base? By all set bits, I mean all the set bits are placed consecutively together. e.g., for base 2, we have 1, 3, 7, 15, 31, 63, i.e., any integer $x$ such that $x = 2^m - 1$ for some integer $m$.
In binary, $1$ is $1$ and $3$ is $11$ and $7$... | I don't know if you'll find this intuitive, but it may, at least, make the result more memorable. First note that your number $x$ with $m$ bits "set" is $x = b^{m-1} + b^{m - 2} + \ldots + b + 1$. Then lay out the calculation of $(b- 1)x$ like this:
$$
\begin{align*}
(b - 1)(b^{m-1} + b^{m - 2} + \ldots + b + 1)
&= \... | {
"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": 1
} |
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... | $$f(x)=x^8+(x+2)^8\qquad f'(x)=8x^7+8(x+2)^7\qquad f''(x)=56x^6+56(x+2)^6$$
$f''(x)$is clearly positive so $f$ is convex and above its tangents, in particular since $f'(-1)=0$ it is above the horizontal line $y=f(-1)+(x-1)f'(-1)=2$.
Since this value is effectively reached for $x=-1$ then it is the global minimum (no ot... | {
"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": 3
} |
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.