Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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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 , we get $$\sqrt[a+b]{a^a\cdot b^b}=a^a\cdot b^b\le a\cdot a+b\cdot b.$$
Considering $a+\dots +a (b$ $\text{times})$ and $b+\dots+b(a$ $\text{times})$, we get
$$\sqrt[a+b] {a^bb^a}=a^bb^a \le 2\cdot ab \Rightarrow a^a\cdot b^b +a^b\cdot b^a\le (a-b)^2=1$$
I have a lot of flaws. Like $a$ is real so we can't actually consider this $a+\dots +a (b$ $\text{times})$ . Can this be corrected somehow?
| 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 of writing $$a+a+a...(a \space \text{times})+b+b+.....(b \space \text{times})=a^2+b^2\ge \sqrt[a+b]{a^ab^b}$$ i would just write it as by weighted AM-GM $$a\cdot a+b\cdot b\ge \sqrt[a+b]{a^ab^b}$$
Similarly for the other...:)
| {
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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,
$$ S = - \ln \left( \cos x \cdot \cos \frac{x}{2} ... \cos \frac{x}{2^n} \right)$$
or,
$$ S = - \ln \left( \frac{ \sin(2x)}{2^{n+1} \sin (\frac{x}{2^n})} \right)= \ln(2^{n+1}) + \ln( \sin \frac{x}{2^n}) - \ln( \sin(2x) )$$
Now,
$$ \frac{dS}{dx} = \frac{1}{2^n} \cot \frac{x}{2^n} - 2 \cot(2x)$$
The problem I'm having is proving that
$$ \lim_{ n \to \infty} \frac{1}{2^n} \cot \frac{x}{2^n} = \frac{1}{x}$$
| 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",
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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) $
$=\lim\limits_{x\to\infty}\left(\sqrt{1+\dfrac1{x^2}}-1\right)x(x+1)$
$=\lim\limits_{x\to\infty}\left(1+\dfrac1{2x^2}\cdots-1\right)(x^2+x)$
$=\lim\limits_{x\to\infty}\left(\dfrac1{2x^2}\cdots\right)(x^2+x).$
Can you take it from here?
| {
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"url": "https://math.stackexchange.com/questions/3988183",
"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 give some idea/hint for the other part.
| 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 $$\Big(-{7\over 3}\Big)^3-6 \Big(-{7\over 3}\Big)^2 +44\geq 0$$ A contradicition.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$
So,
$x_{2017} = x_{2016+1} = 3 x_{2016} + \sqrt{8x^2_{2016} + 2}$ -------(2)
$x_{2023} = x_{2022+1} = 3 x_{2022} + \sqrt{8x^2_{2022} + 2}$ -------(3)
Then from (1),(2),(3) =>
$3 x_{2016} + \sqrt{8x^2_{2016} + 2} + 3 x_{2022} + \sqrt{8x^2_{2022} + 2} = 990$
$3 (x_{2016}+x_{2022} ) + \sqrt{8x^2_{2016} + 2} + \sqrt{8x^2_{2022} + 2} = 990$
$3 (x_{2016}+x_{2020} + x_{2021}) + \sqrt{8x^2_{2016} + 2} + \sqrt{8x^2_{2020} + 2}+ \sqrt{8x^2_{2021} + 2}+ \sqrt{8x^2_{2022} + 2} = 990$
I stuck here and can't go further. Please help me with it.
| 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 $x_{n+1}=x_{n-1}$ ? No, because $x_n$ is an increasing series. Effectively, we have $x_{n+1}-x_n = \sqrt{8x_n^2+2}+2x_n > \sqrt{4x_n^2}+2x_n \ge 0$.
So, $x_{n+1}+x_{n-1}-6x_{n}=0$ for all $n$ or $x_n = \frac{x_{n+1}+x_{n-1}}{6}$
Now, we
$$
\begin{align}
x_n &= \frac{x_{n-1}+x_{n+1}}{6} \\
&= \frac{ \frac{x_{n-2}+x_{n}}{6} +\frac{x_{n}+x_{n+2}}{6}}{6} \\
&= \frac{x_{n-2}+x_{n+2}}{6^2} + \frac{2x_n}{6^2} \\
&= \frac{\frac{x_{n-3}+x_{n-1}}{6} +\frac{x_{n+1}+x_{n+1}}{6}}{6^2} + \frac{2x_n}{6^2} \\
&= \frac{x_{n-3}+x_{n+3}}{6^3}+\frac{x_{n-1}+x_{n+1}}{6^3} + \frac{2x_n}{6^2} \\
&= \frac{x_{n-3}+x_{n+3}}{6^3}+\frac{x_n}{6^2} + \frac{2x_n}{6^2} \\
\end{align}
$$
So,
$$(1-\frac{3}{6^2})x_n= \frac{x_{n-3}+x_{n+3}}{6^3}$$
Or
$$x_n = \frac{x_{n-3}+x_{n+3}}{6(6^2-3)}$$
Hence, if $x_{2017}+x_{2023} = 990$, then $x_n = \frac{990}{6(6^2-3)} = 5$.
| {
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"url": "https://math.stackexchange.com/questions/3996456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 particular, one can prove that $\dfrac{1}{3}\left(\frac{2a}{b+c} \right)^{\frac{2}{3}} \geq \dfrac{a}{a+b+c} $ (motivation is explained below) using AM-GM, as follows:
\begin{align*}
\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} \geq \dfrac{a}{a+b+c} & \iff \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} \geq \dfrac{3a}{a+b+c} \\
& \iff \left(\dfrac{2a}{b+c} \right)^{2} \geq \left(\dfrac{3a}{a+b+c}\right)^3 \\
& \iff 4a^2(a+b+c)^3 \geq 27a^3(b+c)^2.
\end{align*}
The preceding inequality is homogeneous, hence W.L.O.G. set $a+b+c=3$. It thus suffices for us to prove that $4 \geq a(3-a)^2 \iff 8 \geq 2a(3-a)^2$. But this is obvious from AM-GM: $2a(3-a)^2 \leq \left(\dfrac{2a+(3-a)+(3-a)}{3}\right)^3=8.$ Similarly, we have $\dfrac{1}{3}\left(\dfrac{2b}{a+c} \right)^{\frac{2}{3}} \geq \dfrac{b}{a+b+c} $ and $\dfrac{1}{3}\left(\dfrac{2c}{a+b} \right)^{\frac{2}{3}} \geq \dfrac{c}{a+b+c} $. Thus, summing cyclically, we obtain:
$$\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \dfrac{1}{3}\left(\dfrac{2b}{a+c} \right)^{\frac{2}{3}} + \dfrac{1}{3}\left(\dfrac{2c}{a+b} \right)^{\frac{2}{3}} \geq 1 \Rightarrow \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3 .$$
And we are done.
Some motivation:
Let $f(a,b,c)=\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} - \dfrac{a^r}{a^r+b^r+c^r}$, for some $r \in \mathbb{R}$. Ideally, we want $f(a,b,c) \geq 0$ for all values of $a,b,c$. Note that $f(1,1,1)=0$, which suggests that we should set $(1,1,1)$ as a local minimum of $f$. Hence, take the partial derivative of $f$ with respect to $a$, and set it to zero at $(1,1,1)$. By solving the resulting equation, we find a corresponding value of $r$:
$$\dfrac{\partial f}{\partial a}= \dfrac{1}{3} \cdot \sqrt[^3]{4} \cdot \left(\dfrac{1}{b+c} \right)^{\frac{2}{3}} \cdot \dfrac{2}{3} \cdot a^{\frac{-1}{3}} - \dfrac{ra^{r-1}(a^r+b^r+c^r)-a^r(ra^{r-1})}{(a^r+b^r+c^r)^2}$$
Hence,
$$\frac{\partial f}{\partial a }\Bigr|_{(1,1,1)} =0 \Rightarrow \dfrac{2}{9}- \dfrac{2r}{9} =0 \Rightarrow r=1.$$
My question is, is there any other way to solve this inequality besides isolated fudging? I.e. would methods such as Cauchy / Holder / Jensen also work here? I would love to see any other alternative approach.
| 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 $$2p^4 > 2(9q)^2 > 27q^2.$$
If $p^2 \leqslant 9q,$ by the Schur inequality, we have
$$r \ge \frac{4pq-p^3}{9}.$$
We will show that
$$2p^4+3p^2(4q-p^2) \geqslant 27q^2,$$
equivalent to
$$(9q - p^2)(p^2-3q) \geqslant 0.$$
This is true. The proof is completed.
| {
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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 $b\in(0,1]$, $\theta\in\mathbb{R}_{>0}$ and $\theta\neq 1$.
It is part of my research project and I would like to know a good way to approach it.
Any hint is appreciated.
EDIT
According to WA, the solution is given by $0 < b \leq 1$ when $\theta > 0$.
| $$
f(b,\theta)\geq 0
$$
\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}(\theta - 1)}&\geq 0\\\\
\implies 2b-1+\frac{(1-\theta)(2b-1)^{2}+\theta-\sqrt{(1-\theta)(2b-1)^{2}+\theta}}{2b(\theta-1)}&\geq 0\\
\implies 2b(\theta-1)(2b-1)+(1-\theta)(2b-1)^{2}+\theta-\sqrt{(1-\theta)(2b-1)^{2}+\theta}&\geq 0\\\\
\implies 2b\theta-2b+1-\sqrt{(1-\theta)(2b-1)^{2}+\theta}&\geq 0
\end{align*}
This implies the following :
*
*First Case : $b=0$ which is not the case since $b\in(0,1]$.
*Second Case : $0<b \leq 0.5$ and $\theta >\displaystyle \frac{4b^{2}-4 b+1 }{4 b^{2}-4 b}$
*Third Case : $0.5<b\leq 1$ and $\theta> 0$
Hence it follows that $\forall b\in(0,1]$ and $\forall\theta\in\mathbb{R}_{>0}$ with $\theta\neq 1$ we have that $f$ is non-negative.
| {
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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" or "no" for a proposal (which doesn't change). Unchangeables will always vote the same way on the same proposal. However, each time a Changeable votes, the probability that they will change their mind from their previous vote is $r$ and the probability that it will stay the same is $1-r$. (That is for example if a Changeable votes "yes" one time, then the probability that they vote "yes" next time is $1-r$ and "no" is $r$).
A randomly chosen person is noticed to have voted the same way on the proposal twice in succession. What is the probability that they will vote in the same way next time?
So far I've got that there are 4 possible categories of people, each combination of Unchangeable and Changeable that votes either twice the same or twice different in succession (at a particular instance), so this table should give the probabilities for getting each type of person if you pick someone randomly from the population.
twice same
twice different
Unchangeable
$p$
$0$
Changeable
$(1-r)(1-p)$
$r(1-p)$
That way, you'd expect to get $$P(\text{unchangeable } |\text{ voted same way twice}) = \frac{p}{(1-r)(1-p)+p}$$ and $$P(\text{changeable } |\text{ voted same way twice}) = \frac{(1-r)(1-p)}{(1-r)(1-p)+p}$$
and then combine these linearly to get $$P(\text{will vote the same next time}) = \frac{p}{(1-r)(1-p)+p} \times 1 + \frac{(1-r)(1-p)}{(1-r)(1-p)+p} \times (1-r). $$
Is this a correct approach? This seems to be correct to me, but I'm not completely sure that I haven't overlooked anything.
| 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 \overline C) \\
&= P(A \mid BC) \cdot P(BC) + P(A \mid B \overline C) \cdot P(B \overline C) \\
&= P(A \mid BC) \cdot P(C \mid B) \cdot P(B) + P(A \mid B \overline C) \cdot P(\overline C \mid B) \cdot P(B) \\
&= P(B) \cdot \left[ P(A \mid BC) \cdot P(C \mid B) + P(A \mid B \overline C) \cdot P(\overline C \mid B) \right]
\end{align}
Intro. 2: $$P(A B \mid C) = P(A \mid C) \cdot P(B \mid AC)$$
Proof:
\begin{align}
P(AB \mid C) &= \frac{P(ABC)}{P(C)} \\
&= \frac{P(B \mid AC) \cdot P(AC)}{{P(AC)}/{P(A \mid C)}} \\
&= P(A \mid C) \cdot P(B \mid AC)
\end{align}
Say person A is the randomly selected one, whose proposals could be noted as $a_1, a_2, \ldots{}$ et cetera. $A_i$ stands for "yes" and $\overline A_i$ stands for "no".
If A is unchangeable, we say $X$; otherwise, $\overline X$.
If unchangeable A proposes "yes" every time, we say $Y$; otherwise, $\overline Y$. Obviously, $Y, \overline Y \subseteq X$, $Y \cup \overline Y = X$.
We can say $P(X) = p$, $P(\overline X) = 1 - p$.
Furthermore, we have
$P(a_{k+1} = a_k \mid X) = 1$,
$P(a_{k+1} \neq a_k \mid X) = 0$,
$P(a_{k+1} \neq a_k \mid \overline X) = r$,
$P(a_{k+1} = a_k \mid \overline X) = 1 - r$.
Suppose $P(A_1 \mid \overline X) = q$, $P(\overline A_1 \mid \overline X) = 1 - q$.
Suppose $P(A_1 \mid X) = s$, $P(\overline A_1 \mid X) = 1 - s$.
Now we want to calculate:
\begin{align}
P &= P(A_3 \mid A_1 A_2) + P(\overline A_3 \mid \overline A_1 \overline A_2) \\
&= P(A_3 \mid A_1 A_2 X) \cdot P(X \mid A_1 A_2) + P(A_3 \mid A_1 A_2 \overline X) \cdot P(\overline X \mid A_1 A_2) \\
&+ P(\overline A_3 \mid \overline A_1 \overline A_2 X) \cdot P(X \mid \overline A_1 \overline A_2) + P(\overline A_3 \mid \overline A_1 \overline A_2 \overline X) \cdot P(\overline X \mid \overline A_1 \overline A_2)
\end{align}
\begin{align}
P(A_1 A_2 \mid X) &= P(A_1 \mid X) \cdot P(A_2 \mid A_1 X) = P(A_1 \mid X) \cdot P(a_{k+1} = a_k \mid X) = s \\
P(A_1 A_2 \mid \overline X)
&= P(A_1 \mid \overline X) \cdot P(A_2 \mid A_1 \overline X)
= P(A_1 \mid \overline X) \cdot P(a_{k+1} = a_k \mid \overline X)
= q (1-r) \\
P(A_1 A_2) &= P(X) \cdot P(A_1 A_2 \mid X) + P(\overline X) \cdot P(A_1 A_2 \mid \overline X) = ps + (1-p)q(1-r) \\
P(X \mid A_1 A_2) &= \frac{P(X) \cdot P(A_1 A_2 \mid X)}{P(A_1 A_2)} = \frac{ps}{ps + (1-p)q(1-r)} \\
P(\overline X \mid A_1 A_2) &= \frac{P(\overline X) \cdot P(A_1 A_2 \mid \overline X)}{P(A_1 A_2)} = \frac{(1-p)q(1-r)}{ps + (1-p)q(1-r)} \\
P(\overline A_1 \overline A_2 \mid X) &= P(\overline A_1 \mid X) \cdot P(\overline A_2 \mid \overline A_1 X) = 1 - s \\
P(\overline A_1 \overline A_2 \mid \overline X) &= P(\overline A_1 \mid \overline X) \cdot P(\overline A_2 \mid \overline A_1 \overline X) = (1-q)(1-r) \\
P(\overline A_1 \overline A_2) &= P(X) \cdot P(\overline A_1 \overline A_2 \mid X) + P(\overline X) \cdot P(\overline A_1 \overline A_2 \mid \overline X) = p(1-s) + (1-p)(1-q)(1-r) \\
P(X \mid \overline A_1 \overline A_2) &= \frac{P(X) \cdot P(\overline A_1 \overline A_2 \mid X)}{P(\overline A_1 \overline A_2)} = \frac{p(1-s)}{p(1-s)+(1-p)(1-q)(1-r)} \\
P(\overline X \mid \overline A_1 \overline A_2) &= \frac{P(\overline X) \cdot P(\overline A_1 \overline A_2 \mid \overline X)}{P(\overline A_1 \overline A_2)} = \frac{(1-p)(1-q)(1-r)}{p(1-s)+(1-p)(1-q)(1-r)}
\end{align}
Therefore,
\begin{align}
P &= \frac{ps}{ps+(1-p)q(1-r)} + (1-r)\frac{(1-p)q(1-r)}{ps+(1-p)q(1-r)} + \frac{p(1-s)}{p(1-s)+(1-p)(1-q)(1-r)} + (1-r)\frac{(1-p)(1-q)(1-r)}{p(1-s)+(1-p)(1-q)(1-r)} \\
&= \frac{ps + (1-p)q(1-r)^2}{ps + (1-p)q(1-r)}+\frac{p(1-s)+(1-p)(1-q)(1-r)^2}{p(1-s)+(1-p)(1-q)(1-r)}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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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 would be appreciated.
Edit: Testing it with different values of $n$, it seems like the integral evaluates to be $\frac{n \pi^2}{6(n+1)}$
| 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 $\ln(a/b) = \ln(a) - \ln(b)$
$$
I = \int_0^1 \frac{1}{x} \ln\left(1-x^{n+1} \right)dx - \int_0^1 \frac{1}{x} \ln(1-x) \, dx
$$
We now use $\ln$'s taylor series:
$$
\ln(1-x) = \sum_{k=1}^\infty \frac{x^k}{k}
$$
We obtain
$$
I = \int_0^1\sum_{k=1}^\infty \frac{x^{(n+1)k-1}}{k} dx - \int_0^1 \sum_{k=1}^\infty \frac{x^{k-1}}{k} dx
$$
Switching the bounds with the summation because of monotone convergence and integrating:
$$
I = \left[ \sum_{k=1}^\infty \frac{x^{k(n+1)}}{(n+1)k^2} \right]_0^1 -\left[ \sum_{k=1}^\infty \frac{x^k}{k^2} \right]_0^1
$$
$$
I= \frac{\zeta(2)}{n+1} - \zeta(2)
$$
$$
I = \left(\frac{-n}{n+1}\right) \frac{\pi^2}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
} |
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 $2 \cdot Im(x) = x- \overline x$ and show that it's equal to zero:
$$\frac{a+b+c+ab+ac+bc}{1+abc} - \frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{ac}+\overline{bc}}{1+\overline{abc}} =$$ $$=a+b+c+ab+ac+bc-\overline{a}-\overline{b}-\overline{c}-\overline{ab}-\overline{ac}-\overline{bc} = $$ $$=2(Im(a) + Im(b)+Im(c)+Im(ab)+Im(ac)+Im(bc))=$$$$=2(Im(a+b+c+ab+bc+ac))$$But I have no idea what to do next or what am I missing.
| $$x = \frac{(a+1)(b+1)(c+1)-(1+abc)}{1+abc} = \frac{(a+1)(b+1)(c+1)}{1+abc}-1$$
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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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:
$$ k = \frac{-p\pm\sqrt{p^2-4q}}{2} $$
$$ l = \frac{p\pm\sqrt{p^2+4q}}{2} $$
$$ x = p\pm\sqrt{p^2+2q} $$
But I failed to compare these solutions. What should be done next?
| 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
$$ k \ = \ \frac{-p}{2} \ + \ \frac{\sqrt{p^2 \ - \ 4q}}{2} \ \ , \ \ l \ = \ \frac{p}{2} \ - \ \frac{\sqrt{p^2 \ + \ 4q}}{2} \ \ , \ \ \text{and} \ \ r \ = \ p \ - \ \sqrt{p^2 \ + \ 2q} $$
appear to lie in the order $ \ k < r < l \ $ , so we will attempt to prove that $ \ r \ $ always lies in the interval $ \ [k \ , \ l] \ \ . $ The condition for the proposition to be meaningful is that for $ \ q > 0 \ \ , $ we must have $ \ p^2 > 4q \ \ , $ and for $ \ q < 0 \ \ , \ \ p^2 > -4q \ \ . $ This can be summarized as $ \ p^2 > 4·|q| \ \ ; $ however, since the situation of the parabolas is symmetrical about the $ \ y-$axis for these two cases, we only need to consider $ \ q > 0 \ \ . $ It is also clear that it will suffice to investigate $ \ p > 0 \ \ , $ as taking $ \ p < 0 \ $ produces a mirror-image arrangement of the geometrical situation.
The $ \ q = 0 \ $ case is not mentioned above, as it is "trivial": the zeroes cited become $ \ k = l = r = -\frac{q}{p} = 0 \ \ . $
For $ \ p^2 = 4q \ \ , $ we have $ \ k \ = \ \frac{-p}{2} \ , \ l \ = \ \frac{p}{2} \ - \ \frac{\sqrt{8q}}{2} \ , \ $ and $ r \ = \ p \ - \ \sqrt{6q} \ \ . $ A test of the requisite inequalities yields
$$ \frac{-p}{2} \ < \ p \ - \ \sqrt{6q} \ \ [?] \ \ \Rightarrow \ \ \sqrt{6q} \ < \ \frac{3p}{2} \ \ \Rightarrow \ \ 6q \ < \ \frac{9p^2}{4} \ \ \Rightarrow \ \ \frac{24}{9}·q \ < \ p^2 \ = \ 4q \ \ [!] $$
and
$$ p \ - \ \sqrt{6q} \ < \ \frac{p}{2} \ - \ \frac{\sqrt{8q}}{2} \ \ [?] \ \ \Rightarrow \ \ \frac{p}{2} \ < \ \sqrt{6q} \ - \ \sqrt{2q} $$ $$ \Rightarrow \ \ p^2 \ = \ 4q \ = \ 16 · \frac14 · q $$ $$ < \ \ 4 · (\sqrt{6} \ - \ \sqrt{2})^2 · q \ = \ 4 · 2 · (\sqrt{3} \ - \ 1)^2 · q \ = \ 16 · (2 - \sqrt3) · q \ \ [!] \ \ . $$
Thus, $ \ r \ $ lies in $ \ [k \ , \ l] \ $ for this case.
In the general case, we will write $ \ p^2 = \alpha·q \ \Rightarrow \ q = \frac{p^2}{\alpha} \ \ , \ \ \alpha > 4 \ \ , $ making the zeroes under discussion
$$ k \ = \ \frac{-p}{2} \ + \ \frac{p·\sqrt{1 \ - \ \frac{4}{\alpha}}}{2} \ , \ l \ = \ \frac{p}{2} \ - \ \frac{p·\sqrt{1 \ + \ \frac{4}{\alpha}}}{2} \ , \ r \ = \ p \ - \ p·\sqrt{1 \ + \ \frac{2}{\alpha}} \ \ . $$
Again, checking the inequalities, we find
$$ \frac{p}{2} · \left( -1 \ + \ \sqrt{1 \ - \ \frac{4}{\alpha}} \right) \ < \ p · \left( 1 \ - \ \sqrt{1 \ + \ \frac{2}{\alpha}} \right) \ \ [?] $$
$$ \Rightarrow \ \ -1 \ + \ \sqrt{1 \ - \ \frac{4}{\alpha}} \ < \ 2 \ - \ 2 · \sqrt{1 \ + \ \frac{2}{\alpha}} \ \ \Rightarrow \ \ 2 · \sqrt{1 \ + \ \frac{2}{\alpha}} \ + \ \sqrt{1 \ - \ \frac{4}{\alpha}} \ < \ 3 $$
$$ \Rightarrow \ \ 4 \ + \ \frac{8}{\alpha} \ + \ 1 \ - \ \frac{4}{\alpha} \ + \ 4 · \sqrt{1 \ + \ \frac{2}{\alpha}} · \sqrt{1 \ - \ \frac{4}{\alpha}} \ < \ 9 $$ $$ \Rightarrow \ \ \frac{1}{\alpha} \ + \ \sqrt{1 \ + \ \frac{2}{\alpha}} · \sqrt{1 \ - \ \frac{4}{\alpha}} \ < \ 1 \ \ \Rightarrow \ \ \sqrt{1 \ + \ \frac{2}{\alpha}} · \sqrt{1 \ - \ \frac{4}{\alpha}} \ < \ 1 \ - \ \frac{1}{\alpha} $$
$$ \Rightarrow \ \ 1 \ - \ \frac{2}{\alpha} \ - \ \frac{8}{\alpha^2} \ < \ 1 \ - \ \frac{2}{\alpha} \ + \ \frac{1}{\alpha^2} \ \ \Rightarrow \ \ - \frac{8}{\alpha^2} \ < \ \frac{1}{\alpha^2} $$
and
$$ p · \left( 1 \ - \ \sqrt{1 \ + \ \frac{2}{\alpha}} \right) \ < \ \frac{p}{2} · \left( 1 \ - \ \sqrt{1 \ + \ \frac{4}{\alpha}} \right) \ \ [?] $$
$$ \Rightarrow \ \ 2 \ - \ 2 · \sqrt{1 \ + \ \frac{2}{\alpha}} \ < \ 1 \ - \ \sqrt{1 \ + \ \frac{4}{\alpha}} \ \ \Rightarrow \ \ 1 \ < \ 2 · \sqrt{1 \ + \ \frac{2}{\alpha}} \ - \ \sqrt{1 \ + \ \frac{4}{\alpha}} $$
$$ \Rightarrow \ \ 1 \ < \ 4 \ + \ \frac{8}{\alpha} \ + \ 1 \ + \ \frac{4}{\alpha} \ - \ 4 · \sqrt{1 \ + \ \frac{2}{\alpha}} · \sqrt{1 \ + \ \frac{4}{\alpha}} $$
$$ \Rightarrow \ \ \sqrt{1 \ + \ \frac{2}{\alpha}} · \sqrt{1 \ + \ \frac{4}{\alpha}} \ < \ 1 \ + \ \frac{3}{\alpha} $$
$$ \Rightarrow \ \ 1 \ + \ \frac{6}{\alpha} \ + \ \frac{8}{\alpha^2} \ < \ 1 \ + \ \frac{6}{\alpha} \ + \ \frac{9}{\alpha^2} \ \ \Rightarrow \ \ \frac{8}{\alpha^2} \ < \ \frac{9}{\alpha^2} $$
So these inequalities are certainly true for $ \ \alpha > 4 \ \ . $
We may therefore conclude that $ \ k \ < \ r \ < \ l \ \ . $ [It can also be shown by corresponding arguments that this also applies for the other (more widely-separated) roots
$$ k' \ = \ \frac{-p}{2} \ - \ \frac{\sqrt{p^2 \ - \ 4q}}{2} \ < \ r \ < \ l' \ = \ \frac{p}{2} \ + \ \frac{\sqrt{p^2 \ + \ 4q}}{2} \ \ . ] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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:
$$
\left[\cos ^{2}\left(\dfrac{\theta }{2}\right) +
\,\sqrt{\,{1 - \gamma}\,}\,\sin^{2}\left(\dfrac{\theta }{2}\right)\right] ^{2} + \dfrac{\gamma }{4}\,\sin^{2}\left(\theta\right)\quad
\mbox{where}\quad 0 \le \gamma \le 1.
$$
I have tried to solve this but I reach a term which is not easily solvable.
| 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 $$\begin{align}f(x) &= \left( \frac{1 + x}{2} + \sqrt{1-\gamma} \frac{1 - x}{2} \right)^2 + \frac{\gamma}{4} (1 - x^2) \\
&= ax^2 + bx + c,
\end{align}$$ where $$\begin{align} a &= \frac{1 - \gamma - \sqrt{1-\gamma}}{2}, \\ b &= \frac{\gamma}{2}, \\ c &= \frac{1 + \sqrt{1-\gamma}}{2}. \end{align}$$ Now compute the critical point(s) of $f$ with respect to $x$ and consider where these have absolute value less than or equal to $1$, since we require $|x| = |\cos \theta| \le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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,we have $$\frac{\ln a_n}{\ln a_{n+1}}<\frac{\ln a_{n+1}}{\ln a_{n+2}}\Leftrightarrow \ln a_n\cdot\ln a_{n+2}<(\ln a_{n+1})^2$$
By Am-Gm, and $a_{n-1}a_{n+1}-a_n^2=(-1)^n$, we have$$LHS<\left(\frac{\ln a_n+\ln a_{n+2}}{2}\right)^2=\left(\frac{\ln(a_n\cdot a_{n+2})}{2}\right)^2=\left(\frac{\ln(a_{n+1}^2-1)}{2}\right)^2<RHS.$$
But when n is odd, the inequality is a little difficult, or I didn't use a right method.
Supplement:
Here we prove that $$a_{n-1}a_{n+1}-a_n^2=(-1)^n.$$
It is not hard to prove that if $$A=\begin{pmatrix}
0&1\\
1&1
\end{pmatrix}$$then$$A^n=\begin{pmatrix}
a_{n-1}&a_n\\
a_n&a_{n+1}
\end{pmatrix}$$by induction. So $$a_{n-1}a_{n+_1}-a_n^2=\det\begin{vmatrix}
0&1\\
1&1
\end{vmatrix}^n=\left(\det\begin{vmatrix}
0&1\\
1&1
\end{vmatrix}\right)^n=(-1)^n$$
| 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 $$\frac{a_{n+1}}{a_{n-1}}=\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n+1}}{a_n}\geq\frac{3}{2}\cdot\frac{3}{2}=\frac{9}{4}.$$
So
$$
\ln a_{n+1}-\ln a_{n-1}\geq\ln\frac{9}{4}=2\ln\frac{3}{2}>\frac{4}{5}\Rightarrow(\ln a_{n+1}-\ln a_{n-1})^2>\frac{16}{25}
$$
By this inequality when $n\geq 5$ we can get
$$
\begin{align*}
\ln a_{n+1}\cdot\ln a_{n-1}&=\frac{1}{4}(\ln a_{n+1}+\ln a_{n-1})^2-\frac{1}{4}
(\ln a_{n+1}-\ln a_{n-1})^2\\
&< \frac{1}{4}(\ln a_{n+1}+\ln a_{n-1})^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_{n+1}\cdot a_{n-1}))^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_{n}^2+(-1)^{n-1}))^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_{n}^2+(-1)^{n-1}))^2-\frac{1}{4}(\ln a_n^2)^2+(\ln a_n)^2-\frac{4}{25}\\
&\leq\frac{1}{4}(\ln(a_n^2+1))^2-\frac{1}{4}(\ln a_n^2)^2+(\ln a_n)^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_n^2+1)-\ln a_n^2)(\ln(a_n^2+1)+\ln a_n^2)+(\ln a_n)^2-\frac{4}{25}
\\
&\leq\frac{1}{4}\left(\ln(a_n^2+1)-\ln a_n^2\right)\cdot2\ln((a_n+1)^2)+(\ln a_n)^2-\frac{4}{25}\\
&=\ln(1+\frac{1}{a_n^2})\ln(1+a_n)+(\ln a_n)^2-\frac{4}{25}\\
&\leq \frac{1}{a_n^2}\cdot a_n-\frac{4}{25}+(\ln a_n)^2\\
&\leq\frac{1}{8}-\frac{4}{25}+(\ln a_n)^2\\
&<(\ln a_n)^2
\end{align*}
$$
In a word, it is $\ln a_{n-1}\cdot\ln a_{n+1}<(\ln a_n)^2$, which is equal to$$\frac{\ln a_{n-1}}{\ln a_n}<\frac{\ln a_n}{\ln a_{n+1}},\quad n\geq 5.$$
For the rest situation ,we just need to prove it by caculating it.
| {
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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:
$$\int\frac{dx}{\sqrt{x^2+4}}=\int\frac{xdx}{\sqrt{x^2(x^2+4)}}=\frac12\int\frac{du}{\sqrt{u(u+4)}}\quad\text{where}\quad u=x^2$$
But I can't see a way to evaluate final integral without completing square and using the substitution $u+2=2\sec t$ (which is a trigonometric substitution!)
| $$\frac 1{\sqrt{x^2+4}}=\frac 14\frac{(x^2+4)-x^2}{\sqrt{x^2+4}}=\frac 14\left(\sqrt{x^2+4}-\frac{x^2}{\sqrt{x^2+4}}\right)$$
$\int\sqrt{x^2+a^2}$ 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 $t\,\mathrm dt=x\,\mathrm dx$
$$\int\frac{x^2}{\sqrt{x^2+4}}\,\mathrm dx=\int\frac{t^2-4}{t}\frac t{\sqrt{t^2-4}}\,\mathrm dt=\int\sqrt{t^2-4}\,\mathrm dt$$
which is again computable using IBP without any trig sub.
| {
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"url": "https://math.stackexchange.com/questions/4010133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 inductive step with $n=k$ we have
$\sum_{i=1}^{k}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{k+2}}{2^{k}}$
We want to prove then that for $n=k+1$
$\sum_{i=1}^{k+1}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{k+3}}{2^{k+1}}$
We rewrite the LHS such that
$\frac{F_{k}}{2^{k+1}}+\sum_{i=1}^{k}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{k+3}}{2^{k+1}}$
From the inductive hypothesis it follows that
$\frac{F_{k}}{2^{k+1}}+ 1-\frac{F_{k+2}}{2^{k}}= 1-\frac{F_{k+3}}{2^{k+1}}$
I get stuck here. Any hint will be appreciated.
| 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$$F_{k+2}=a+b,\,F_{k+3}=a+2b\implies2F_{k+2}-F_k=2(a+b)-a=a+2b=F_{k+3}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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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 $t = \frac{1}{x}$,
$$I = \int_{0}^{\infty} \dfrac{\frac{1}{x^8} -1}{\frac{1}{x^{10}} + 1} x^2dx$$
$$I = \int_{0}^{\infty} \frac{x^{4} - x^{12}}{x^{10} + 1} dx$$
Adding the two versions of $I$,
$$2I = \int_{0}^{\infty} \frac{x^8 - x^{12} -1 + x^4}{x^{10} + 1} dx = \int_{0}^{\infty} \frac{(x^8-1)(1 - x^{4}) }{x^{10} + 1} dx$$
However, that does not help much.
Putting it in WolframAlpha gives $I = 0$
I am pretty sure there is an elegant solution which I am not able to find it.
| 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 $x$.
| {
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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 disjoint union of $2^n$ closed sub-intervals of length $3^{-n}$ and that the endpoints of each $C_n$ are in $C$.
(2) For any distinct $x,y\in C$, there exists non-empty, disjoint, open sets (open in in $C$) $A,B\subset C$ such that $A\cup B=C$ with $x\in A$ and $y\in B$.
For (1), using induction seemed to me to be the best way to go about proving (1), since I am unsure how to go about proving it for an arbitrary $n\in\mathbb{N}$. However, using the inductive hypothesis with this definition turned out to be unwieldy and awkward, and did not yield any results with the methods I tried.
Proof outline for (2):
Let $x,y\in C$ and WLOG, let $x<y$. Choose $N\in\mathbb{N}$ such that $$3^{N-1}(y-x)>2$$ Now $$3^{N-1}x<3^{N-1}x+2<3^{N-1}y $$ so there is an integer $k\in\mathbb{N}$ such that $$3^{N-1}x<k<k+1<y3^{N-1}$$ Then $$x<\frac{3k}{3^{N}}<\frac{3(k+1)}{3^{N}}<y $$ and finally, we have $$x<\frac{3k}{3^{N}}<\frac{1+3k}{3^N}<\frac{2+3k}{3^N}<\frac{3(k+1)}{3^N}<y $$
The idea here was to find one of the "deleted" intervals which separated $x$ and $y$. If we set $A=(-1,z)$ and $B=(z,2)$ for some $z$ between $\frac{1+3k}{3^N}$ and $\frac{2+3k}{3^N}$, then $A\cap C$ and $B\cap C$ will be open in $C$, they will be disjoint, and their union will equal $C$.
Questions:
How should I proceed in proving (1)? Is this proof for (2) sufficient?
| 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 length $3^{-n}$ and that the endpoints of each $C_n$ are in $C$.
Base step: $n=0$
$C_0=[0,1]$ is the disjoint union of $2^0=1$ closed sub-intervals of length $3^{0}=1$. We also see the endpoints $0$ and $1$ are in $C$, since the open intervals which are subtracted in the definition of $C_n$ do not contain these points.
Induction hypothesis: $n=N$
We assume the claim is valid for $n=N$. In order to apply the hypothesis we look somewhat closer at $C_n$ and derive a further representation. We see (more or less clearly) that in each step the middle third of intervals is subtracted. This indicates to use a representation of the reals at base $b=3$. We start with small $n=0,1,2$ and get
\begin{align*}
\color{blue}{C_0}&=[0,1] \color{blue}{= [0_3,0.\dot{2}_3]}\tag{2.0}\\
\color{blue}{C_1}&=[0,1]\setminus\left(\frac{1}{3},\frac{2}{3}\right)
=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3},1\right]\color{blue}{=[0_3,0.0\dot{2}_3]\cup[0.2_3,0.\dot{2}_3]}\tag{2.1}\\
\color{blue}{C_2}&=C_1\setminus
\left(\left(\frac{1}{9},\frac{2}{9}\right)\cup\left(\frac{4}{9},\frac{5}{9}\right)
\cup\left(\frac{7}{9},\frac{8}{9}\right)\right)\\
\\
&\,\,\color{blue}{=\left[0_3,0.00\dot{2}_3\right]\cup\left[0.02_3,0.0\dot{2}_3\right]
\cup\left[0.2_3,0.20\dot{2}_3\right]\cup\left[0.22_3,0.\dot{2}_3\right]}\tag{2.2}
\end{align*}
We observe in (2.0) - (2.2) the numbers in $C_n$ can be written in ternary representation (indicated with an index $3$) without the digit $1$ at the first $n$ positions after the comma. We can generally write
\begin{align*}
C_n=\left\{z\in[0,1]\bigg|z=\sum_{j=1}^\infty a_j\cdot 3^{-j}, a_k\in\{0,2\}, 1\leq k\leq n\right\}\tag{3}
\end{align*}
We are now well prepared for the induction step.
Induction step: $n=N+1$
We obtain
\begin{align*}
C_{N+1}&=C_N\setminus\bigcup_{k=0}^{3^{N}-1}\left(\frac{1+3k}{3^{N+1}},\frac{2+3k}{3^{N+1}}\right)\\
&=C_N\setminus\bigcup_{k=0}^{3^{N}-1}\left(\frac{k}{3^N}+\frac{1}{3^{N+1}},\frac{k}{3^N}+\frac{2}{3^{N+1}}\right)\tag{4.1}\\
&=\left\{z\in[0,1]\bigg|z=\sum_{j=1}^\infty a_j\cdot 3^{-j}, a_k\in\{0,2\}, 1\leq k\leq N+1\right\}\tag{4.2}
\end{align*}
and the claim follows.
Comment:
*
*In (4.1) we observe the intervals $\left(\frac{k}{3^N}+\frac{1}{3^{N+1}},\frac{k}{3^N}+\frac{2}{3^{N+1}}\right)$ which are subtracted from $C_N$ have length $\frac{1}{3^{N+1}}$. Furthermore these are intervals which are precisely the numbers which can be written with a $1$ at position $N+1$ in ternary representation.
According to the induction hypothesis $C_N$ is the union of $2^N$ disjoint intervals. Since we subtract from each of these intervals the middle part, the number of disjoint intervals is doubled, giving $2^{N+1}$ intervals.
*In (4.2) we use the result from (4.1) together with the induction hypothesis to obtain this representation. We observe the endpoints of the intervals in $C_{N+1}$ can be written without using the digit $1$ and are this way justified as points in $C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
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
$$
\left\{
\begin{array}{l}
4 \left(a_1+2\right) \left(2 a_1-1\right) =0\\
-a_1^3+6 \left(4 a_1+3\right) a_2-1 =0\\
16 a_2^2+\left(1-4 a_1^2\right) a_2+8 \left(4 a_1+3\right) a_3 =0\\
\end{array}
\right.
$$
and after solving we have
$$
\cases{
h_1(x) = -\frac{1183 x^3}{18000}+\frac{7 x^2}{30}-2 x\\
h_2(x) = -\frac{9 x^3}{16000}+\frac{3 x^2}{80}+\frac{x}{2}
}
$$
Those are two representations for unstable manifolds (in red) depicted in the following figure
NOTE
$\left(x^2-\frac{x^4}{3}\right)$ represents the first two term series representation for $(\sin x)^2$
| {
"language": "en",
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"answer_count": 1,
"answer_id": 0
} |
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 out the $n$'s.
I also tried solving the sum of each floor function separately, but didn't get anywhere.
Any suggestions will be appreciated!
Thanks.
| 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.
$$
\begin{align}
&\sum_{i=1}^{\infty}
\left( \left\lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)\\
&= \sum_{i=1}^{k}
\left( \left\lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)
+\sum_{i=k+1}^{\infty}
\left( \left\lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)\\
&= \sum_{i=1}^{k}
\left( \frac{2^k}{2^i} +\left\lfloor{\frac{n-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}} \right\rfloor \right)
+\sum_{i=k+1}^{\infty}
\mathbb{1}(n-1<2^i, n+2^k-1\ge 2^i)\\
&= \sum_{i=1}^{k}2^{k-i}+ \text{something}\\
&= \sum_{j=0}^{k-1}2^j + \text{something}\\
&= 2^k-1 + \text{something}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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
$$
\left(
\begin{array}{cc}
\cos t & \sin t \\
\sin t & - \cos t
\end{array}
\right)
\left(
\begin{array}{cc}
\cos \frac{t}{2} & -\sin \frac{t}{2} \\
\sin \frac{t}{2} & \cos \frac{t}{2}
\end{array}
\right) =
\left(
\begin{array}{cc}
\cos \frac{t}{2} & \sin \frac{t}{2} \\
\sin \frac{t}{2} & - \cos \frac{t}{2}
\end{array}
\right)
$$
It does not need to be symmetric, although the reflection is what first came to mind for $M^2 = I,$ which happens when $M^3 = M$ and $M$ is nonsingular. For real matrices, we need just trace zero and determinant $-1$
As long as $$ a^2 + bc = 1 $$ we can use
$$
\left(
\begin{array}{cc}
a & b \\
c & - a
\end{array}
\right)
$$
for example
$$
\left(
\begin{array}{cc}
7 & 8 \\
-6 & - 7
\end{array}
\right)
$$
We see the eigenvectors as columns
$$
\left(
\begin{array}{cc}
a & b \\
c & - a
\end{array}
\right)
\left(
\begin{array}{cc}
1+a & -b \\
c & 1+ a
\end{array}
\right) =
\left(
\begin{array}{cc}
1+a & -b \\
c & -1- a
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 8
} |
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 $\frac{S}{24} = {4\choose 4} + {6\choose 4} + {8\choose 4} + ... + {100\choose 4}$ but how do I sum this up?
| 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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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 $\Delta AEB$ :
$\angle BAC = sin^{-1}(\frac{5}{13}) = 22.62^{\circ}$
$\angle ABD = sin^{-1}(\frac{9}{15}) = 36.87^{\circ}$
$\angle AEB = 180 - \angle BAC - \angle ABD = 120.52^{\circ}$
Use Law of Sines:
$\frac{AE}{sin(\angle ABD)} = \frac{AB}{sin(\angle AEB)}$
$AE = \frac{AB}{sin(\angle AEB)} \times sin(\angle ABD)$
$AE = \frac{12 \times \frac{9}{15}}{sin 120.51} = 8,372$
$\frac{EB}{sin(\angle BAC)} = \frac{AB}{sin(\angle AEB)}$
$EB = \frac{AB}{sin(\angle AEB)} \times sin(\angle BAC)$
$EB = \frac{12 \times \frac{5}{13}}{sin 120.51} = 5.367$
$AE+EB = 8.372 + 5.367 = 13,7387 \approx \frac{96}{7}$ Answer: (D)
My question:
Is there a solution without having to compute both arcsin $\angle BAC$ & $\angle ABD$? The reason I'm asking is because the choices all have 7 as denominators. So I'm guessing there may be a solution that contains only integers.
| 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 $\displaystyle x = \frac{117}{14}$ and $\displaystyle y = \frac{75}{14}$. Thus, $\boxed{AE + BE= x + y = \frac{96}{7}.}$
| {
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"source": "stackexchange",
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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 -
And this is my way of thinking, I use two inequalities
$$\sqrt{5}\leq \frac{a_{n+ 1}^{2}}{\prod_{0}^{n+ 1}a_{i}}\leq\sqrt{5}+ \frac{2}{3}\cdot\left ( \varphi- 1 \right)^{-2^{n+ 1}}$$
with $\varphi= \frac{3+ \sqrt{5}}{2}.$ From the recurrence sequence $a_{n+ 1}= a_{n}^{2}- 2\Rightarrow a_{n+ 1}= \varphi^{2^{n}}+ \varphi^{-2^{n}},$ we have
$$\frac{a_{n+ 1}^{2}}{\prod_{i= 0}^{n+ 1}a_{i}}= \frac{\varphi^{2^{n+ 1}}+ \varphi^{-2^{n+ 1}}}{\prod_{i= 1}^{n}\left ( \varphi^{2^{i}}+ \frac{1}{\varphi^{2^{i}}} \right )}= (\varphi- \frac{1}{\varphi})\left ( 1+ \frac{2\varphi^{-2^{n+ 2}}}{1- \varphi^{-2^{n+ 2}}} \right )> \varphi- \frac{1}{\varphi}= \sqrt{5}$$
Because I'm busy now.. OK, I'll get it done by tonight, I haven't posted the solution for the rightside inequality, and how can we use it to find the $o\left ( 1 \right ).$ I need to the help, thanks a real lot !
Edit. The rightside inequality is just single variable of $f\left ( n \right )\leq g\left ( n \right )$ with
$$f\left ( n \right )= \sqrt{5}+ \frac{2}{3}\cdot\varphi^{-2^{n+ 1}}, g\left ( n \right )= \sqrt{5}\left ( 1+ \frac{2\varphi^{-2^{n+ 2}}}{1- \varphi^{-2^{n+ 2}}} \right ), 0< \varphi^{-2^{n+ 1}}\leq\frac{7-3\sqrt{5}}{2}$$
| 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
$$ a_n = \left(\frac{2}{\sqrt{5}+3}\right)^{2^n}+\left(\frac{1}{2} \left(\sqrt{5}+3\right)\right)^{2^n} $$
should do the trick.
| {
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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 rows are formed by the base of the given code, i.e.
\begin{eqnarray}
y_1 + y_2 + y_3 = 0\\
y_1 + y_4 + y_5 = 0 \\
y_1 + y_2 + y_4 + y_6 = 0 \\
\end{eqnarray}
If we choose $y_1$, $y_2$ und $y_4$ as free variables, we get the rows of the following matrix as the basis of the solution space:
$H=\begin{bmatrix}1&0&1&0&1&1\\0&1&1&0&0&1 \\ 0&0&0&1&1&1\end{bmatrix}$
where columns 1, 2 and 4 are the identity matrix.
*
*Why are the colums 1, 2 and 4 the identity matrix?
*How did he use the generator matrix to obtain the parity check matrix?
*Why did he choose $y_1$, $y_2$ und $y_4$ and what does this mean?
I thought the way to get the parity check matrix is the following:
If the generator matrix is in the form $G = [I_k | P]$, then its parity check matrix is $H = [P^T | I_{n-k}]$
So we swap columns 1&3, 2&5 and 3&6
$G'=\begin{bmatrix}1&0&0&0&1&1\\0&1&0&1&0&1 \\ 0&0&1&1&1&1\end{bmatrix}$
$H'=\begin{bmatrix}0&1&1&1&0&0\\1&0&1&0&1&0 \\ 1&1&1&0&0&1\end{bmatrix}$
Then we permutate it back (3&6, 2&5 and 1&3) and obtain:
$H= \begin{bmatrix}0&0&0&1&1&1\\0&1&1&0&0&1 \\ 1&0&1&0&1&1\end{bmatrix}$
Does it matter in which order the identity matrix appears?
| 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 $H$ is a $(n-k)\times n$ matrix, $x^T$ is a $n \times 1$ vector. It is easy to see, that swapping rows does not have any effect, but swapping columns changes the code, since bit positions are permuted.
Consider the valid codeword $100011$ (first row of $G'$). Using $H$ to check: $H\cdot \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\1\end{pmatrix}=\begin{pmatrix} 0 \\ 1 \\ 1\end{pmatrix}$, hence the syndrome is not equal to the zero vector. However, performing the same for $H'$ yields to $H'\cdot \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1\end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$ as intended.
| {
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"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 inverse of what WolframAlpha says the answer is. I get $y = \frac{e^{2x+2c} + 1 }{1 - e^{2x + 2c}}$.
From $\frac{dy}{dx} = y^2 - 1$ I write $\frac{dy}{y^2 - 1} = dx$ and then integrate: $$\int \frac{1}{y^2 - 1} dy = \int 1 dx = x + c_1$$
The integral on the left I decompose into partial fractions: $$\frac{1}{y^2 -1} = \frac{1}{(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y+1}$$ and I get $A = \frac{1}{2}$ and $B = -\frac{1}{2}$
So then I have $\frac{1}{2}\int \frac{1}{y-1} - \frac{1}{2} \int \frac{1}{y+1} = x + c_1$ which is $$\frac{1}{2}ln(y-1)-\frac{1}{2}ln(y+1) = x + c$$ This simplifies to $$\frac{1}{2}ln(\frac{y-1}{y+1}) = x+c$$ Raising both sides by $e$: $$(\frac{y-1}{y+1})^{\frac{1}{2}} = e^{x+c}$$
Now squaring... $$\frac{y-1}{y+1} = e^{2x+2c}$$
Then: $$y - 1 = e^{2x+2c}y + e^{2x+2c} \implies y(1-e^{2x+2c}) = e^{2x+2c}+1$$
and finally $$y = \frac{e^{2x+2c} + 1}{1 - e^{2x+2c}}$$
| 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 not simply $\displaystyle \ln x + c$, it is actually $\displaystyle \ln |x| + c$. That absolute value sign is actually critical. It means that for the domain $\displaystyle x<0$, you should take $\displaystyle \ln (-x) + c_1$, while for $\displaystyle x>0$, you should take $\displaystyle \ln x + c_2$.
After integration, you wrote: $\displaystyle \frac 12\ln(y-1) - \frac 12\ln(y+1) = x + c$
This is imprecise (and not correct without domain restriction). For the general solution, consider three cases (note that $\displaystyle y$ cannot be equal to either $\displaystyle -1$ or $\displaystyle 1$):
Case 1: $\displaystyle y<-1$:
In this case, integration yields $\displaystyle \frac 12\ln(-(y-1)) - \frac 12\ln(-(y+1)) = x + c$ which, after you do the remaining working, will get you the exact solution you got, namely $\displaystyle y = \frac{1+e^{2(x+c)}}{1- e^{2(x+c)}}$
Case 2: $\displaystyle -1<y<1$:
In this case, integration yields $\displaystyle \frac 12\ln(-(y-1)) - \frac 12\ln(+(y+1)) = x + c$
Working through this would get you the same form as the Wolfram Alpha solution:
$\displaystyle y = \frac{1-e^{2(x+c)}}{1+e^{2(x+c)}}$
And finally,
Case 3: $\displaystyle y>1$ would get you, after integration:
$\displaystyle \frac 12\ln(+(y-1)) - \frac 12\ln(+(y+1)) = x + c$
Working through which you would get the exact same form as for case 1:
$\displaystyle y = \frac{1+e^{2(x+c)}}{1- e^{2(x+c)}}$
Splitting it up into cases gives you the full picture. You can visualise exactly what is going on by graphing.
| {
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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 Mathematical Analysis $2$. The following is my personal solution, different from the one expected.
| 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,
$$1 + \sqrt[n]{a} + \sqrt[n]{a^2} + \sqrt[n]{a^3} + ... +\sqrt[n]{a^{n-1}}$$
are all $\ge$ $a$: being $ 0<a<1$, indeed, elevating $a$ to an exponent $ \in \left [ 0,1 \right]$ (note that exponents of $ a$ in denominator are $\frac{1}{n}$, $\frac{2}{n}$, ... , $\frac{n-1}{n} \le 1$) will get a value $\ge a$, i.e. closer to $1$.
There are $n$ terms in denominator (note that all powers of $ a $ from $0$ to $ n-1 $ are present, and $ n-1+1=n$). Then it's
$$ |a_n| = 1 - \sqrt[n]{a} = \frac{1 - a}{1 + \sqrt[n]{a} + \sqrt[n]{a^2} + \sqrt[n]{a^3} + ... +\sqrt[n]{a^{n-1}} } \le \frac{1-a}{na}$$
from which follows: $ |a_n| \le \frac{1}{n} \bigl( \frac{1-a}{a} \bigr)$
$\square$
| {
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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 $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$
$$= 2\int_{0}^{1} r^2 dt$$
$t= \sqrt{1-r^2}\implies r^2= t^2-1$
$$2\int_{0}^{1} t^2-1 dt= 2[\frac{t^3}{3} -t]_{0}^{1}==-4/3$$
| With $r=\sin t$ it's$$\int_0^{\pi/2}\sin^3tdt=\int_0^{\pi/2}(\sin t-\sin t\cos^2t)dt=[\tfrac13\cos^3t-\cos t]_0^{\pi/2}=\tfrac23.$$
| {
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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 square, $a^2=n$
The subsequent perfect square can be written as ${b^2=\left(a+1\right)}^2$
The difference between consecutive squares is then ${|b}^2-a^2|=|2a+1|$
$\forall a\in\mathbb{Z}(|2a+1|\neq 4)$
$\square$
Edit: Thanks everyone, appreciate it :)
| 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 $a=0\implies n =a^2 = 0$ that's out (but it is a case where $n =0^2$ and $n+4 = 2^2$) and $a=1\implies n=1^2 = 1$ and $n+4 = 5$ not a perfect square.
So that can fix the proof.
Or we can do:
You have $n=a^2$ and let's suppose $n+4 = b^2$ then
$b^2 - a^2 = 4$ and
$(b-a)(b+a) =4$ and... now we are talking. Assume $a,b$ are positive and $b > a$ we get $(b-a)(b+a) = (1,4), (2,2)$.
The first means $2b = 5; 2a = 3$ and $a = 1.5$ and $b = 2.5$ (and indeed $1.5^2 = 2.25$ and $2.5^2 = 6.25 = 2.25 + 4$. But $a, b$ are not integers.
The second means $a=0$ and $b =2$. So $n = 0^2 = 0 \not > 0$. So that is out.
| {
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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 $$2OM^2=OP^2+PM^2+OK^2+KM^2\\=x^2+b^2+y^2+a^2.$$ I can't approach the problem further. How can we use the given angle $\alpha$? Thank you in advance!
| 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:
$$a^2 + b^2 +2ab\cos \alpha = 2r^2(1-\cos 2\alpha)$$
or
$$r = \sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(1-\cos2\alpha)}$$
To find $OM$ we just double this value:
$$OM = 2r = 2\sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(1-\cos2\alpha)} = 2\sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(2 \sin^2 \alpha)} = \frac {\sqrt{a^2+b^2+2ab\cos \alpha}}{\sin \alpha}$$
| {
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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 am-gm on $a+b+c$ to get $abc\leq1$
Then applying quadratic mean arthematic mean to get
$a^2+b^2+c^2 \ge 3$
Then writing the expansion of $(a+b+c)^2$ we get
$a^2+b^2+c^2 = 9-2(ab+bc+ca)$ ...(I)
Thn applying am-hm on $ab+bc+ca$ to get
$ab+bc+ca \ge 3abc$
Then substituting in (I) we get
$a^2+b^2+c^2 \leq 9-6(abc)$
Then we substituting in (0)
$LHS \leq abc(9-6abc)$
Whose max value is $27/8$ but is not less than three, so I am stuck for this point onwards
| 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 $$f(3-2t,t,t)-3=-3 (t - 1)^2 (4 t^3 - 6 t^2 + 2 t + 1)\le 0$$ which is true because $4t^3-6t^2+2t=2t(t-1)(2t-1)\ge 0$
Done!
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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 proving equivalence of tan inverse and sine inverse and I was only able to achieve the first definition.
My proof:
We begin with $ \sin^{-1} \frac{2x}{1+x^2}$ , substitute $ x = \tan \phi$ , the previous result simplfies as $ \sin^{-1} ( 2 \tan \phi \cos^2 \phi) = \sin^{-1} ( 2 \sin \phi \cos \phi) = 2 \phi = 2\tan^{-1} (x)$. Where is my mistake?
| 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 let $x = \arcsin a$.
Since there may be many $x$ such that $\sin x = a$, it is important to specify the codomain of the arcsine, so that we need not worry about multiple values.
Similarly, $\arctan : \mathbb R \to (-\frac \pi 2, \frac \pi 2)$ is defined as : for $a \in \mathbb R$ there is a unique $-\frac \pi 2 < x <\frac \pi 2$ with $\tan x =a$. We let $x = \arctan a$.
Now, we must prove the given statement. First of all, note that $-1\leq \frac{2x}{1+x^2} \leq 1$ for all values of $x$, so certainly $\arcsin \frac{2x}{1+x^2}$ is defined.
Now let $x \in \mathbb R$ and $\phi = \arctan x$ so that $x = \tan \phi$. Note that $-\frac \pi 2 < \phi < \frac \pi 2$. The argument you make shows that $$
\frac{2x}{1+x^2} = \sin 2 \phi
$$
and therefore, $$
\arcsin\left(\frac{2x}{1+x^2}\right) = \arcsin(\sin 2 \phi)
$$
Now, we would like to simplify the RHS. For this, we cannot simply cancel out the two functions, because of codomain issues.
For example, let $x = 2$. Then the LHS is $\arcsin \frac 45 = 53.13^\circ$. On the RHS : we have $\phi = \arctan 2 = 63.43^\circ$ so $2 \phi = 126.86^\circ$. So clearly, $LHS \neq 2 \phi$ always, which is what one would have expected if one could cancel the $\sin$ and $\arcsin$.
In fact, we have $\arcsin(\sin 2 \phi) = \theta$ where $\theta$ is the unique angle in $[-\frac \pi 2, \frac \pi 2]$ such that $\sin 2 \phi = \sin \theta$. Then we can write $\arcsin(\frac{2x}{1+x^2}) = \theta$. Now, we need the dependence between $2 \phi$ and $\theta$.
For this, note that the points of problem are those where the codomain breaks. For example at $\phi = \frac \pi 4$, we have $2 \phi = \frac \pi 2$ which is still in the $\arcsin$ codomain. Take $\phi$ a little higher and that is lost. So the relation between $\phi$ and $\theta$ is expected to change at the point $\frac \pi 4$, and similarly at $\phi = \frac {-\pi}4$ as well, for the lower end of the codomain breaks then.
So we can break into three cases depending upon these break points :
*
*If $2 \phi \in [-\frac \pi 2 , \frac \pi 2]$ then of course $\theta = 2 \phi$.
*If $2\phi \in (\frac \pi 2 , \pi]$ then $\sin(\pi - x) = \sin x$ gives $\sin(\pi - 2\phi) = \sin 2 \phi$ so $\theta = \pi - 2\phi$ (which lies in the codomain).
*If $2 \phi \in [-\pi , -\frac \pi 2)$ then $\sin(-\pi-x) = -\sin(\pi + x) = \sin x$ so $\sin(-\pi - 2\phi) = \sin(2 \phi)$, and hence $\theta = -\pi - 2 \phi$.
Now, we need to express $\theta$ in terms of $x$, not $\phi$, and the break needs to be in terms of $x$. From our usual trigonometry relations, we know that $x>1, x<1,x \in [-1,1]$ if and only if $2 \phi > \frac \pi 2 , 2\phi < - \frac \pi 2, 2\phi \in [-\frac \pi 2, \frac \pi 2]$ respectively.
Therefore, we get :
*
*If $-1 \leq x \leq 1$ then $\arcsin\left(\frac{2x}{1+x^2}\right) = 2 \phi$.
*If $x>1$ then $\arcsin\left(\frac{2x}{1+x^2}\right) =\pi- 2 \phi$.
*If $x<1$ then $\arcsin\left(\frac{2x}{1+x^2}\right) =-\pi- 2 \phi$
Bring $2\phi$ to one side in each of these equations makes it the subject of the equation, and you have the definition given by the website. So the piecewise nature of the definition reflects the breaks in the arcsin codomain which can only be compensated on a case-by-case basis.
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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 odd an resolves in $n$ steps, then $4x+1$ likewise resolves in $n$ steps.
Is my reasoning correct?
Let:
*
*$\nu_2(x)$ be the 2-adic valuation of $x$
*$x_1, x_2, \dots, x_n$ be $n$ distinct odd integers such that:
*
*$x_{i+1} = \dfrac{3x_i + 1}{2^{\nu_2(3x_i+1)}}$
*$x_1$ resolves in $n$ steps if $\dfrac{3x_n+1}{2^{\nu_2(3x_n + 1)}} = 1$
Observation 1: if an odd integer $x$ resolves to $1$ in $n$ steps, then there exist $n$ positive integers $a_1, a_2, \dots, a_n$ where $x = f(a_1, a_2, \dots, a_n)$
(1) If $n=1$, then $\dfrac{3x+1}{2^{a_1}} = 1$ and $x = f(a_1) = \dfrac{2^{a_1}-1}{3}$
(2) Assume that for any $x$ that resolves in $n-1$ steps, there exist $n-1$ integers $a_1, \dots, a_{n-1}$ with $f_{n-1}(a_1, a_2, \dots, a_{n-1})= x$
(3) If $x$ resolves in $n$ steps, there exists a positive integer $a_n$ such that $\dfrac{3x+1}{2^{a_n}}$ is an integer that resolves in $n-1$ steps and $x = f(a_1, a_2, \dots, a_n) = \dfrac{3f_{n-1}(a_1, a_2, \dots, a_{n-1}) + 1}{2^{a_n}}$
Observation 2: if an odd integer $x$ resolves in $n$ steps, then it follows that $4x+1$ also resolves in $n$ steps
(1) From Observation 1, there exist $a_1, a_2, \dots, a_n$ such that $\dfrac{3x+1}{2^{a_n}} = f_{n-1}(a_1, \dots, a_{n-1})$
(2) $\dfrac{3(4x+1)+1}{2^{a_n+2}} = \dfrac{12x+4}{2^{a_n+2}} = \dfrac{4(3x+1)}{2^{a_n+2}} = f_{n-1}(a_1, \dots, a_{n-1})$
Corollary: If $x$ is a positive odd integer and $4x+1$ is a solution in $n$ steps, then $x$ is also a solution in $n$ steps:
By assumption, there exist $a_1, a_2, \dots, a_n$ such that:
$$\frac{3(4x+1)+1}{2^{a_n}} = \frac{4(3x+1)}{2^{a_n}} = \frac{3x+1}{2^{a_n-2}}= f_{n-1}(a_1, a_2, \dots, a_{n-1})$$
Update 1: corrected an ambiguity -- my definitions only apply to odd numbers.
Update 2: I wrote a simple java app to validate. I have been able to validate this result up to $90\times10^6$.
My code iterates through odd numbers $x$ and validates that the count of steps for $x$ matches the count of steps for $4x+1$.
| 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 number. So after one of the op's steps $x$ and $4x+1$ arrive at the same number $\frac{3x+1}{2^k}$.
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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:
$$ \frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2 $$
The partiel derivate of $y$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial y}(x,y) = -6xy $$
Therefore I get: $$\nabla f(x,y) = \begin{pmatrix} \frac{\partial f }{\partial x}\\ \frac{\partial f }{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2-3y^2 \\ -6xy \end{pmatrix} $$
The limit value investigation at a certain point
from the left and from the right only makes sense for a function with only one variable.
How can I proof if my function is differentiable?
| 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 $A = \left(3a^2-3b^2, -6ab\right)\implies \dfrac{\left|f(a+h,b+k) - f(a,b) - A\cdot \begin{pmatrix} h\\k \end{pmatrix}\right|}{|(h,k)|}=\dfrac{|3a(h^2-k^2)+ h(h^2-6bk-3k^2)|}{\sqrt{h^2+k^2}}\le \dfrac{|3a||(h^2+k^2)}{\sqrt{h^2+k^2}}+ \dfrac{|h|}{\sqrt{h^2+k^2}}\cdot|h^2-6bk-3k^2|\le |3a|\sqrt{h^2+k^2} + |h^2-6bk-3k^2|\to 0$ when $(h,k) \to (0,0) \implies f'(a,b) = A$ as claimed.
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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^2\ $ which, after some cancelling, implies $\ f(x+2)-f(x) = 4a_2x + (2a_1+4a_2).$ For this to equal $8x-2,\ $ we must have $a_2 = 2\ $ and $a_1 = -5.$ So $f(x) = a_0 - 5x + 2x^2.\ $ The last condition, $f(0) = 5\ $ implies that $a_0 = 5.$ So $f(x) = 5 - 5x + 2x^2\ $ is certainly a solution - the only solution of degree 2, since we got there by deduction.
Are there any other solutions (of degree $\geq 3$)?
Well if the degree of $f(x) \geq 3,\ $ then $$f(x+2)-f(x) = a_0 + a_1(x+2) + a_2(x+2)^2 + ... + a_n(x+2)^n - (a_0 + a_1x+a_2x^2+...+a_nx^n),\ $$ where $n\geq3.\ $ From the binomial expansion, the $x^{n-1}\ $ coefficient of $f(x+2)-f(x)\ $ is equal to $a_{n-1} + 2na_n - a_{n-1} = 2na_n.\ a_n \neq 0\ $ and $n \neq 0,\ $ therefore the $x^{n-1}\ $ term in $f(x+2)-f(x)\ $ has at degree of at least $2\ $, and so $f(x+2)-f(x)\neq 8x-2.$ So no, there are no other solutions.
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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 clever substitution or factorisation that makes this obvious / less tedious? (Or can anyone provide a more succinct explanation of the answer linked above?)
Thanks
| 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 $(b+1)(2b-a) = by - a$
Applying the equations:
\begin{align} (a-b)(2b+1)(2b-a+2) - (b+1)(2b-a)(2a-2b-1) &= y(bx + a) - x(by-a) \\
&= bxy + ay -bxy + ax \\
&= a(x+y) \\
&= a(2a-2b-1+2b-a+2) \\
&= a(a+1) \\
&= a^2 + a
\end{align}
Thanks Lord Commander!
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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 multiple of 5
Which implies that both numbers have to be of the from $5k$, ie. $n$ possible numbers
Case 2
Let $n_1+n_2$ be a multiple of 5
So $(n_1,n_2)$ can be $(5k,5k), (5k+1, 5k+4), (5k+4, 5k+1), (5k+2,5k+3), (5k+3, 5k+2)$
With $n$ cases for each ordered pair
Case 3
Let $n_1^2 + n_2^2$ be a multiple of 5
So here cases can be 5k and 5k only (IMO)
The cases with $(5k,5k)$ are repeated 3 times, so count that as one case only
So
$$P=\frac {\binom n2 + 2\times \binom n2 +2\times \binom n2}{\binom {5n}{2}}$$
But that doesn’t match with the given answer $\frac{17n-5}{5(5n-1)}$
Where am I going wrong?
| 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
$$a=5k+0\to r_1=\underbrace{a^4}_{mod 5}=0\\
a=5k+1\to r_2=\underbrace{a^4}_{mod 5}=1^4=1\\
a=5k+2\to r_3=\underbrace{a^4}_{mod 5}=2^4=16(mod \ 5)=1\\
a=5k+3\to r_4=\underbrace{a^4}_{mod 5}=3^4=81(mod \ 5)=1\\
a=5k+4\to r_5=\underbrace{a^4}_{mod 5}=2^4=256(mod \ 5)=1\\$$ so $n_1^4,n_2^4$ can choose from $S_1$ and $S_2$ such that $5|n_1^4-n_2^4$ now
$$P=\frac { \binom n2 + \binom {4n}2}{\binom {5n}{2}}$$
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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 since the summation is interlinked.
| 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. I leave it to you to write them down. Now that you know that these $6$ different ways of writing the sum mean exactly the same thing - we're good to go. Denote the sum by $S$, and add all these (six) different representations of the same sum together. Get a nice expression for $6S$, and thus find $S$. See the magic unfold!
$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{ab(3a+c) + ac(3a+b) + bc(3b+a) + ba(3b+c) + ca(3c+b) + cb(3c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}$$
$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{3a^2b+3a^2c+3b^2c+3b^2a+3c^2a+3c^2b+6abc}{4^{a+b+c} (a+b)(b+c)(c+a)}$$
$$6S = \sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac{3(a+b)(b+c)(c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}$$
$$S = \frac12\sum_{a = 1}^{\infty} \sum_{b = 1}^{\infty} \sum_{c = 1}^{\infty} \frac1{4^{a+b+c}} = \frac12\left(\sum_{a=1}^{\infty}\frac1{4^a}\right)^3 = \frac12\left(\frac13\right)^3 = \boxed{\frac1{54}}$$
If you want to practice this trick that I just mentioned, try to compute the following sum in a similar fashion (it has definitely been asked on this site before):
$$\sum_{m=1}^\infty\sum_{n=1}^\infty\dfrac{m^2n}{3^m(n3^m+m3^n)} = \frac{9}{32}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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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:
$$\begin{align}
f(x) &= (x+3)(2x^2-x+7) \\
&= 2x^3-x^2+7x+6x^2-3x+21 \\
&= 2x^3+5x^2+4x+21
\end{align}$$
Making sure it is correct:
Then, I realized that the remainder has to be 10 not 0...I have no idea how to do that
| "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 $(x+3)(2x^2 -x +7) + 10$ we'd get $\frac {(x+3)(2x^2 - x+ 7)}{x+3} = 2x^2 -x + 7$ and then .... we'd be left with the remainder of $10$.
So if you are told $f(x)$ divided by $d(x)$ has a quotient of $q(x)$ and a remainder of $r$ then you can just do $f(x) = d(x)q(x) + r$.
And so $g(x) = (x+3)(2x^2 - x+7) + 10=$
$(2x^3+5x^2+4x+21) + 10=$
$2x^3 + 5x^2 +4 + 31$
has to do it and if we did that we'd have:
| {
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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 verify that $f(x)$ is even and
$$f(0)= \frac13, \>\>\>\>\> \lim_{x\to\pm \infty} f(x) \to -\infty$$
which implies $f(x) \in (-\infty,\frac13]$, i.e.
$$\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le \frac13
$$
and is visually confirmed below
However, it is not obvious algebraically that $f(x)$ monotonically decreases away from $x=0$. The standard derivative tests are not viable due to their rather complicated functional forms.
So, the question is how to prove the inequality $f(x) \le \frac13$ with rigor.
Note that it is equivalent to proving
$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right)
\cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13
$$
| 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 @pisoir's 1st equation) Let $A = \frac\pi2 \frac{(1+x)^2}{3+x^2}, B = \frac\pi2 \frac{(1-x)^2}{3+x^2}$.
Clearly $A, B \in (0, \frac{\pi}{2})$.
It suffices to prove that $\sin A \sin B \le \frac{1}{3}\cos A \cos B$ or
$$\frac{1}{2}[\cos (A - B) - \cos ( A + B)] \le \frac{1}{3}\cdot \frac{1}{2}[\cos (A - B) + \cos (A + B)]$$
or
$$\cos \frac{2\pi x}{x^2 + 3} \le 2 \cos \frac{\pi (x^2 + 1)}{x^2 + 3}. $$
Let $C = \frac{2\pi x}{x^2 + 3}$ and $D = \frac{\pi (x^2 + 1)}{x^2 + 3}$.
We split into two cases:
*
*$\frac{1}{3} \le x < 1$:
Clearly, $\frac{\pi}{3} \le D \le \frac{\pi}{2}$ and $0 \le \frac{\pi}{2} - C \le 2(\frac{\pi}{2} - D) \le \frac{\pi}{2}$. We have
\begin{align}
\cos C &= \sin (\tfrac{\pi}{2} - C) \\
&\le \sin [2(\tfrac{\pi}{2} - D)]\\
&= 2\sin (\tfrac{\pi}{2} - D) \cos (\tfrac{\pi}{2} - D) \\
&\le 2\sin (\tfrac{\pi}{2} - D) \\
&= 2\cos D.
\end{align}
*$0 \le x < \frac{1}{3}$:
We give the following auxiliary results (Facts 1-2). The proofs are easy and thus omitted.
Fact 1: $\cos u \le 1 - \frac{1}{3}u^2$ for all $u$ in $[0, \pi/4]$.
Fact 2: $\cos v \ge \frac{1}{2} - \frac{5}{9}\sqrt{3}\, (v - \tfrac{\pi}{3})$ for all $v \in [\pi/3, \pi/2]$.
Let us proceed. Clearly $C \in [0, \pi/4]$ and $D \in [\pi/3, \pi/2]$.
By Facts 1-2, it suffices to prove that
$$1 - \frac{1}{3}C^2 \le
2\left[\frac{1}{2} - \frac{5}{9}\sqrt{3}\, \left(D - \frac{\pi}{3}\right)
\right]$$
that is
$$\frac{4\pi x^2 [9\pi - 5\sqrt{3}(x^2 + 3)]}{27(x^2 + 3)^2}\ge 0$$
which is true.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051812",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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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 to use the rule that $$\sqrt{a^2b}=a\sqrt{b},a\ge0, b\ge0,$$ right? What would you do to solve the problem? I was thinking about $$8=4\times2\\32=8\times4\\(-3)^2=3^2$$ but are we supposed to continue this to $$8=4\times2=2\times2\times2=2^3\\32=8\times4=2^3\times2^2=2^5?$$ How can we still use $\sqrt{a^2b}=a\sqrt{b}$ then? Thank you in advance!
| 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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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 $b^{m}-1 -b^{n}+1 = b^{m} - b^{n}$.
I really have no idea on how to tackle this problem.
| 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 $a^p - 1 = (a - 1) L$, where $L = a^{p-1} + a^{p-2 }+ ... + a + 1$. So if a prime q divides L, then it must divide $a^p - 1$ and hence $a - 1$. But $L = (a - 1) (a^{p-2} + 2 a^{p-3} + ... + p-1) + p$, so any prime q dividing L and $a - 1$ must also divide p. So L must be a power of p, m must be a power of p, and p must divide $a - 1$. In other words, $a = 1 \, mod \, p$. Hence $(a^{p-2}+ 2 a^{p-3 }+ ... + p-1) = 1 + 2 + ... + p-1 = p(p-1)/2 \, mod \, p$. But if $p > 2 \, , p(p-1)/2 = 0 \, mod \, p$, so $L = p^2 + p \, mod \, p^2$. But L is a power of p, so $L = p$. Contradiction, since L is obviously > p (or p > 2). Hence p must be 2. Hence $L = a + 1$, and $ a + 1$ is a power of 2.
Now take the general case. Suppose d is the greatest common divisor of m and n. Put $m = dM \, , n = dN$ and $ a = b^d$, so that $a^M - 1$ and $a^N - 1$ have the same prime divisors, and M, N are relatively prime. Take positive integers h, k such that $hM - kN = 1$, then $(a^{hM}- 1) - (a^{kN} - 1) = a^{hM} - a^{kN} = a^{kN}(a - 1)$. So if s > 1 divides $a^M - 1$ and $a^N - $1, then it also divides $a^{hM} - 1$ and $a^{kN} - 1$ and hence also $a^{kN}(a - 1)$. It cannot divide $a^{kN}$, so it must divide $a - 1$. So the greatest common divisor of $a^M - 1$ and $a^N - 1$ must divide $a - 1$. But $a - 1$ divides both, so the greatest common divisor is $a - 1$. So $a- 1$ must also have the same prime divisors as $a^M - 1$. It follows from the special case that $a + 1 = b^d + 1$ is a power of 2. Since $b > 1, b^d + 1 > 2$, so $ b^d + 1$ is certainly divisible by 4. If d is even, then $odd^d = 1 \, mod \, 4$ and $even^d = 0 \, mod \, 4$, so $b^d + 1 = 2 $ or $1 \, mod \, 4$. Hence d must be odd. Hence $b + 1$ divides $b^d + 1$, so $b + 1$ is also a power of 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060658",
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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
$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{\sin{\frac{a}{3^k}}\sin{\frac{2a}{3^k}}}{3^k\sin{\frac{a}{3^{k-1}}}}\right)$$
where $a$ is a positive real number.
I think the second limit may be solved by using the comparison criteria(limit comparison test), putting $-1 \le \sin{\frac{2a}{3^k}} \le 1$ and using the first limit. I suspect the first limit is $0$, so the second one should be $0$ too, but I might be wrong. I'm not sure how would you do the first one - I tried using the comparison test on the first one, but I didn't succeed. Can you help?
| The second sum is telescoping, due to $$\frac{\sin x\sin 2x}{\sin 3x}=\frac14(\cot x-3\cot 3x).$$ This easily gives $$\sum_{k=1}^n\frac{\sin\frac{a}{3^k}\sin\frac{2a}{3^k}}{3^k\sin\frac{a}{3^{k-1}}}=\frac14\sum_{k=1}^n\left(\frac{1}{3^k}\cot\frac{a}{3^k}-\frac{1}{3^{k-1}}\cot\frac{a}{3^{k-1}}\right)=\frac14\left(\frac{1}{3^n}\cot\frac{a}{3^n}-\cot a\right).$$
The first sum doesn't seem to. Trying to replace $\cot$ by $-\csc$ yields$$\frac14(3\csc 3x-\csc x)=\frac{\sin^2 x}{\sin 3x},$$ so it would telescope with the sine in the numerator squared.
| {
"language": "en",
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"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 help
| $X \sim U[-1, 1]$, $Y = \sqrt{|X|} \Rightarrow$
$$
F_Y(y) = \text{Pr}\left(Y \leq y\right), \quad y \geq 0,
$$
is a cumulative distribution function (c.d.f.) of $Y$ (by definition).
$$
\begin{aligned}
\text{Pr}\left(Y \leq y\right) &= \text{Pr}\left(\sqrt{|X|} \leq y\right) = \text{Pr}\left(|X| \leq y^2\right) = \text{Pr}\left(-y^2 \leq X \leq y^2\right) = \\
&= \left|\begin{aligned}
&\text{since }X\sim U[-1, 1] \Leftrightarrow \text{c.d.f. of } X\text{ is }\\
&F_X(x) =
\left\{
\begin{array}{rl}
0, & x \leq -1 \\
\frac{x-(-1)}{1-(-1)} = \frac{x+1}{2}, & -1 < x \leq 1 \\
1, & x > 1
\end{array}
\right.
\end{aligned}\right| = \\
&= F_X(y^2) - F_X(-y^2) =
\left\{
\begin{array}{rl}
\frac{y^2+1}{2}-\frac{-y^2+1}{2} = y^2, & 0 \leq y^2 \leq 1 \Leftrightarrow 0 \leq y \leq 1\\
1, & y^2 > 1 \Leftrightarrow y > 1
\end{array}
\right.
\end{aligned}
$$
So, the c.d.f. of $Y$, $F_Y(y) = \text{Pr}(Y \leq y)$, $\quad y \geq 0$, is equal to
$$
F_Y(y) =
\left\{
\begin{array}{rl}
y^2, & 0 \leq y \leq 1\\
1, & y > 1
\end{array}
\right.
$$
Then, the probability density function (p.d.f.) of $Y$ equals to
$$
f_Y(y) = F'_Y(y) =
\left\{
\begin{array}{rl}
2y, & 0 \leq y \leq 1\\
0, & y > 1
\end{array}
\right.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 convergence: $$\forall\epsilon>0\space\space\exists N\in\Bbb{R}\space s.t\space\space \forall n>N,|s_n-L|<\epsilon$$ Below is my working: $$|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\epsilon$$ $$|\frac{3n^2-n}{4n^2+1}-\frac{3n^2+\frac{3}{4}}{4n^2+1}|=|\frac{-n-\frac{3}{4}}{4n^2+1}|=\frac{|-n-\frac{3}{4}|}{4n^2+1}<\frac{4n}{4n^2}=\frac{1}{n}<\epsilon $$ $$\frac{1}{n}<\epsilon\rightarrow\frac{1}{\epsilon}<n$$ $$\forall n>N=\frac{1}{\epsilon},\space|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\frac{1}{n}<\epsilon$$ By the definition of convergence,$\space\frac{3n^2-n}{4n^2+1}\rightarrow\frac{3}{4}$ as $n\rightarrow\infty.$
Is this correct? Any feedback would be greatly appeciated.
| 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:
$$|\frac{3n^2-n}{4n^2+1}-\frac{3n^2+\frac{3}{4}}{4n^2+1}|=|\frac{-n-\frac{3}{4}}{4n^2+1}|=\frac{|-n-\frac{3}{4}|}{4n^2+1}<\frac{4n}{4n^2}=\frac{1}{n},$$
so that (1) will be true for all $n>N:=\frac{1}{\epsilon}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 the third equation, and after simplifying ended up being here :
$$ k(abc)[a^2c^2(b^2 + a^2c^2 - b^2c^2 - a^2)] $$
| 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{array}\right]=\frac{1}{a(b^2-c^2)}\left[\begin{array}{c}b(c^2-a^2)\\c(a^2-b^2)\end{array}\right]x.
\end{align}
By substitution to the third Eq. we get either $x=0$ or
\begin{align}
x^2=\frac{a^2(b^2-c^2)\left[a^4(c^2-b^2)+b^4(a^2-c^2)+c^4(b^2-a^2)\right]}{(c^2-a^2)(a^2-b^2)}\equiv a^2(b^2-c^2)^2.
\end{align}
Let us check the solution. For $a=1,b=2,c=3$ we get solution $x=5,y=-16,z=9$ (and also $x=-5,y=16,z=-9$), which are the correct solutions of the system of equations given above. $\spadesuit$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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
$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=(b-a)(c-b)(c-a)$
What I've attempted to do was to use this well known identity on matrices:
$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=1\left|\begin{matrix}b&c\\b^2&c^2\end{matrix}\right|-1\left|\begin{matrix}a&c\\a^2&c^2\end{matrix}\right|+1\left|\begin{matrix}a&b\\a^2&b^2\end{matrix}\right|$
Then working in the right side of this equation I'm getting:
(Here I'm using the notation of capital $V$ for Vandermonde's determinant)
$V=(bc^2-cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$
Factoring the common terms I'm getting
$V=bc(c-b)-ac(c-a)+ab(b-a)$
But now what? I'm stuck, what else should it be done here? Can someone help me here on how to prove this determinant? Could it be that my method wasn't that right? Please help.
| Subtract the first column from the second column and factor out a common term from the second column.
$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=\left|\begin{matrix}
1 & 0 & 1\\
a & b-a & c\\
a^2 & b^2-a^2 & c^2
\end{matrix}\right|=(b-a)\left|\begin{matrix}
1 & 0 & 1\\
a & 1 & c\\
a^2 & b+a & c^2
\end{matrix}\right|$
Now do that again.
$(b-a)\left|\begin{matrix}
1 & 0 & 1\\
a & 1 & c\\
a^2 & b+a & c^2
\end{matrix}\right|=
(b-a)\left|\begin{matrix}
1 & 0 & 0\\
a & 1 & c-a\\
a^2 & b+a & c^2-a^2
\end{matrix}\right|=
(b-a)(c-a)\left|\begin{matrix}
1 & 0 & 0\\
a & 1 & 1\\
a^2 & b+a & a+c
\end{matrix}\right|$
Last, subtract the second column from the third column
$=
(b-a)(c-a)\left|\begin{matrix}
1 & 0 & 0\\
a & 1 & 0\\
a^2 & b+a & c-b
\end{matrix}\right|$
That final discriminant is obviously $c-b$, and you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4075925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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 $x=0$,
$$
f(0)g(y)+g(0)f(y) = y^2\\
af(y) = y^2
$$
Now $y \ne 0$ sometimes, so $a \ne 0$ and $f(y) = y^2/a$ for all $y$.
Take $x=1,y=1$,
$$
2f(1)g(1) = 4\\
\frac{2g(1)}{a} = 4\\
g(1)=2a
$$
Take $x=1,y=2$,
$$
f(1)g(2)+g(1)f(2)=9
\\
\frac{g(2)+8a}{a} = 9
\\
g(2) = a
$$
Take $x=2, y=2$,
$$
2f(2)g(2) = 16\\
\frac{8a}{a}=16\\
8=16
$$
Contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 ended up with
$$1+ \frac{a}{x-2a}, x>2a$$
$$-(1+ \frac{a}{x-2a}), x<2a$$
I first attempted to solve this using the original form of both equations
For the case where $x>2a$:
$$\frac{(x-a)^2}{(x-2a)^2} =\frac{x-a}{x-2a} $$
$$\frac{x-a}{x-2a} =1$$
$$x-a=x-2a$$
However this lead me to the answer
$$-a=-2a$$
Which is self contradictory, I am extremely confused on where I messed up, have I made a problem cross multiplying or is it something else?
| 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 solution.
If $x> 2a$, we know that $x -a > a$, it is unclear if $|x-a| = x-a$ isn't it?
In fact, if $x=a$, then you divided by $x-a$, which is $0$.
If $x=\frac32a$ and $\frac32 a > 2a$, then we have $a<0$. $x-a=\frac12a$ which is negative.
| {
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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 with $0$ (or $5$, but we don't have it in the given problem). So we have $V_6^4$ numbers which are divisible by $5$ but how can we exclude those which aren't divisible by $3$?
$120$
| 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 them $a,b,c,d$ add up to a multiple of $3$
We have already used $0$ so there we must pick from the remaining $6$ for digits that add to a multiple of $3$.
Insight is... we must remove $2$ and which $2$ can we remove? Suppose we keep $a,b,c,d$ and remove $e,f$. Then $a+b+c+d$ must be a multiple of $3$ and $(a+b+c+d) +(e+f) = 1+2+6 + 7 + 8 +9 = 33$. So we must have $e+f = 33 -$ a multiple of $3=$ a multiple of $3$.
So we can have $e$ and $f$ be $6$ and $9$ the only two multiples of $3$. Or we can have $e$ or $f$ be a pair that aren't multiples of three but add to multiples of $3$. That would mean one has remainder of $1$ and the other has a remainder of $2$. So one is $\{1,7\}$ then other is $\{2,8\}$. So that is $4$ possible pairs. Or it could be $\{e,f\}=\{6,9\}$ make $5$ possible pairs.
(If that was too confusing. To have $e+f$ be a multiple of $3$ we can have $\{e,f\} = (1,2), (1,8), (7,2), (7,8), (6,9)$. Five options. And $a,b,c,d$ can be...everything left. Five options: $(6,7,8,9), (2,6,7,9) ,(1,6,8,9),(1,2,6,9), (1,2,7,8)$.)
So there are $5$ choices for $(a,b,c,d)$ and there are $4! = 24$ ways to arrange them and there is only one option for the last digit. There are $5*24 =120$ such numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
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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
$x^2-2xy+y^2-5x+7y=(x-y)(x-y+5)+2y.....(1)$
$x^2-3xy+2y^2+x-y=(x-y)(x-2y+1) ...... (2)$
From second expression I concluded that $17$ either divides $(x-y)$ or $(x-2y-1)$
Case 1: Let $17$ divides $(x-y)$ then from equation $(1)$ i concluded that $17$ divides $2y$ and hence $17$ divides $x$ so the $17$ divides $(xy-12x+15y)$
Is my argument correct that $17$ will divide $2y$?
Also what if $17$ does not divides $(x-y)$ instead it divides $(x-2y-1)$?
I want solution without using modulo arithmetic for, what if $17$ does not divides $(x-y)$ instead it divides $(x-2y-1)$.
Any other proof will also help which does not involve modulo arithmetic.
| 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 some places I've written $3(17k+3)=17k+9$ , which actually isn't correct but the $k$ is only introduced to show divisibility). Actually this is the idea behind using modular arithmetic as well.
*
*Yes, your argument that $17$ divides $ 2y$ if it divides $x-y$ is correct. From there you deduced that $17$ divides $y $ and hence also $x$ and therefore your case $1$ is correct.
case 2: $17$ divides $x-2y+1$
Write the first equation as
$x^2-2xy+y^2-5x+7y=(x-2y+1)(x-6)+y^2-5y+6$.
Since $17$ divides $x-2y+1$, it divides $y^2-5y+6$.
$y^2-5y+6=(y-2)(y-3) \implies 17$ divides atleast one of $y-2$ and $y-3$. That means $y=17k+2$ or $y= 17k+3$.
Since $17$ divides $ x-2y+1$, it is equal to $17n$.
Sub case 1: $y=17k+2$
$\implies x-17k-4+1=17n \implies x= 17k+3$
Now we can see that $xy-12x+15y= 17k+6-36+30=17k$, therefore $17$ divides it.
Sub case 2: $y=17k+3$
$\implies x-17k-6+1=17n\implies x=17k+5$
And again $xy-12x+15y=17k+15-60+45=17k$, hence $17$ divides it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086665",
"timestamp": "2023-03-29T00:00:00",
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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 was able to conclude that $$\sqrt{9}=3<\sqrt{12}<\sqrt{16}=4,\\3<\sqrt{12}<4,\\\sqrt{12}\in\left(3;4\right).$$ How can I further constrict the interval? Thank you in advance!
| 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 $$(3 + 1/2)^2 = 9 + 2(3)(1/2) + (1/2)^2 = 12 + 1/4 > 12.$$ Hence $$3 < \sqrt{12} < 3.5.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 at state $S(x,y)$.
Since the rules and stop condition are symmetric for both players, $T(y,x) = T(x,y)$. When either win count reaches $3$, our overall count is done: $T(x,3)=T(3,y)=0$. When $x<3$ and $y<3$, we get the recursion
$$ E\{T(x,y)\} = 1 + \frac{1}{3} E\{T(x,y)\} + \frac{1}{3} E\{T(x+1,y)\} + \frac{1}{3} E\{T(x,y+1)\} \\
E\{T(x,y)\} = \frac{3}{2} + \frac{1}{2} E\{T(x+1,y)\} + \frac{1}{2} E\{T(x,y+1)\} $$
From here the expected values are easy to compute, going from largest $x$ and $y$ to smallest:
$$E\{T(2,2)\} = \frac 32 \\
E\{T(2,1)\} = \frac 32 + \frac 34 = \frac 94 \\
E\{T(2,0)\} = \frac 32 + \frac 98 = \frac{21}{8} \\
E\{T(1,1)\} = \frac 32 + \frac 98 + \frac 98 = \frac{15}{4} \\
E\{T(1,0)\} = \frac 32 + \frac{21}{16} + \frac{15}{8} = \frac{75}{16} \\
E\{T(0,0)\} = \frac 32 + \frac{75}{32} + \frac{75}{32} = \frac{99}{16}$$
The last line is the expected total number of rounds for the whole game, $99/16$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 realize this is a general problem-solving issue I have, as it requires me to pick the right option out of thousands of possible paths I could take with the proof. Is there a generalized way to approach these kinds of questions?
| 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:
$$(a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\\=\left(\frac{a}{a+b}+\frac{b}{a+b}\right)+\left(\frac{b}{b+c}+\frac{c}{b+c}\right)+\left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\\=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge3+\frac{3}{2}=\frac{9}{2}$$
The last step follows from the very well-known Nesbitt's inequality
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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:
$$n!!=\left\{ \begin{array}{ll}
2^{k}k! &,n=2k\\
\frac{(2k-1)!}{2^{k-1}(k-1)!} & , n=2k-1\\
\end{array} \right.$$
So, I put this in $f(x)$ and got (for $x=\frac{n-2}{2}$, because for $n=2k$ function is always bigger in this case):
$$f(x)=\frac{2^{2x-1}x!(x-1)!}{(2x-1)!}$$
When I use asymptotic equation for $x$!:
$$x! = \sqrt{2\pi x}\bigg(\frac{x}{e}\bigg)^x\bigg(1+\frac{1}{12x}+\frac{1}{288n}+O(x^{-3})\bigg)$$
And notice that $e\gt2$ (I must somehow evalute $2^{2x-1}$ by Big O):
$$e^{2x-1} = 1 + 2x-1 + \frac{(2x-1)^2}{2!}+ O(x^{-3})$$
So now I connect everything, but for $n = 6$ i got error $0,3$ and it's too much
(because of my "fancy" evaluation $e\gt 2$). When I paste this into Wolfram it looks like a $O(\log)$ or $O(\sqrt{n})$ but I can't figure this out how to get it.
| 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
$$\frac{(2k)!!}{(2k-1)!!}=\frac{(2^k k!)^2}{(2k)!} = \frac{4^k\left(\sqrt{2\pi k}\left(\frac{k}{e}\right)^k\left(1+\frac{1}{12k}+\frac{1}{288k^2}+O\left(\frac{1}{k^3}\right)\right)\right)^2}{\sqrt{2\pi 2k}\left(\frac{2k}{e}\right)^{2k}\left(1+\frac{1}{12\cdot 2k}+\frac{1}{288\cdot 4k^2}+O\left(\frac{1}{k^3}\right)\right)}=\sqrt{\pi k}\frac{\left(1+\frac{1}{12k}+\frac{1}{288k^2}+O\left(\frac{1}{k^3}\right)\right)^2}{1+\frac{1}{12\cdot 2k}+\frac{1}{288\cdot 4k^2}+O\left(\frac{1}{k^3}\right)}=\sqrt{\pi k}\left(1+\frac{1}{8k}+\frac{1}{128k^2}+O\left(\frac{1}{k^3}\right)\right)$$
A similar calculation can be done in the case of $n=2k+1$.
| {
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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 three definitions.
$ \lim X_{n\to \infty} = \ell $
$ |X_n -\ell|\lt\varepsilon$
$(\ell+\varepsilon), (\ell - \varepsilon) $
Step 2:
I substitute the desired values.
$ \lim X_{n\to \infty} = \frac{1}{2} $
$ |\frac{2n-1}{4n+5} -\frac{1}{2}|\lt\frac{1}{1000} $
Step 3:
I calculate
$|\frac{2n-1}{4n+5} -\frac{1}{2}|\lt\frac{1}{1000} \Rightarrow$
$ |\frac{2(2n-1)}{2(4n+5)} -\frac{1(4n+5)}{2(4n+5)}|\lt\frac{1}{1000} \Rightarrow$
$ |\frac{4n-2-4n-5}{8n+10}|\lt\frac{1}{1000} \Rightarrow$
$ |\frac{-7}{8n+10}|\lt\frac{1}{1000} \Rightarrow$
$ \frac{7}{8n+10}\lt \frac{1}{1000} \Rightarrow$
$ 6990\lt 8n $
Step 4:
I have no idea what to do next.
$n$ by definition, the number of the element with which the elements that lie inside the interval begin. I am asked to find how many elements are outside the interval.
That is, I need something like $n-1$. Since it is out of range.
That is, if I know that $8$ elements will lie inside the interval, then $7$ will lie outside the interval.
But before that, I need to find the extreme $n$, which will determine the members of the sequence that lie within the interval and also satisfy the inequality. After that, find the desired $n - 1$, from which the terms will lie outside the interval.
Step 5:
$ 6990\lt 8n \Rightarrow 873.75\lt n $
$ 873.75 \approx 874 $
$ 873.75 \lt n $
$ 873.75 \lt 874 $
$ n-1 \Rightarrow 874 - 1 = 873 $
Answer: $873$
The main question: this is that number $873$ that is outside the range? Did I do everything right or was I wrong? Please, say Me.
| 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 single power of of $2|c!$ so $c \le 3$. If $c = \color{red}1,2,3$ we can have $2 + c! = \color{red}3, 4, 8= \text{no power of two}, 2^2, 2^3$.
If $a= 2$ then $4 \le b! + c! = 2(2^n-1)$ with $2^n-1$ being odd. $b!|b! +c!$ so $b \le 3$ so $b = 2, 3$.
If $a = b=2$ then $6\le c! = 2^n -4 = 4(2^{n-1}-1)$ where $2^{n-1}-1$ odd so only $2^2$ can divide $c!$ and $2|3!$ and $8|4!$ so $c< 4$ and $c = 2,3$. $2!+2!+2! = 6 \ne 2^n; 2! + 2! +3! = 10\ne 2^n$.
$a =2; b=3$ then $14 \le 2! + 3! + c! = 2^n$ and $c! = 2^n - 8= 8(2^{n-1}-1)$ and argument should be familiar now. $2^3|c!$ but $2^4 \not \mid c!$ so $4 \le c \le 5$. So $a=2; b=3; c=4,5$ so $a! + b! + c! = 2!+3!+4! = 2+6 + 24=32^5$ or $a!+b!+c! = 2!+3!+5! = 2+6 + 120=128= 2^7$.
So that's that. $n = 7$ is the highest possible result.
| {
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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 $(m + 2) − (m − 2) = 4$ and so, $\gcd(m + 2, m − 2) = $1$.
It follows that $m + 2$ and $m − 2$ are both cubes.
a) How did they get to the conclusion $\gcd(m+2, m-2)=1$? I mean, at this stage of the proof, I don't know.
$m\neq 10$ so I could think if $m=10$ then gcd(12,8) $=4\neq 1$.
b) Why does it follow that $m+2$ and $m-2$ are both cubes, and not that their multiplication is a cube?
Let's carry on the proof.
[...] No two odd cubes differ by 4, giving a contradiction.
Since $m$ is even there exists $m'$ such that $m = 2m'$. We thus have that $4m'^2 = n^3 + 4$ and $n$ is divisible by $2$ so we write $n = 2n'$. Now, $m'^2 = 2n'^3 + 1$ and so $m'$ is odd so let $m' = 2r + 1$. Substituting this into the previous equation tells us that $2r^2 + 2r = n'^3$.
No problems with the above, let's carry on
Finally,$r^2 + r = r (r + 1)$ and since $r$ and $r + 1$ are coprime we must have that one of $r$ and $r + 1$ is a cube and the other is four times a cube.
c) Why are we considering $r^2+r$ when we have $2r^2+2r=n'^3$? Yes, they have a common factor of $2$, but we don't know this about $n'$ do we? we can't cancel them.
d) ``since $r$ and $r + 1$ are coprime we must have that one of $r$ and $r + 1$ is a cube and the other is four times a cube'' ????
Any help with the above is really appreciated.
|
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 divide any sum or difference of $a,b$ leads to the Lemma
Lemma: $\gcd(a,b) = \gcd(a\pm b, b)$
and so $\gcd(m+2, m-2) = \gcd((m+2)-(m-2), m-2) = \gcd(4, m-2)$ which must be a divisor of $4$.
b) Why does it follow that m+2 and m−2 are both cubes, and not that their multiplication is a cube?
Because $m+2$ and $m-2$ have no factors in common.
Any prime factor $p$ of $m+2$ or of $m-2$ will divide into $n^3$ a power of a multiple of $3$. But as that prime factor belongs to $m+2$ or to $m-2$ alone and not to the other it must divide into $m+2$ or $m-2$ a power of a multiple of $3$ times.
So $m+2$ and $m-2$ are cubes.
c) Why are we considering r2+r when we have 2r2+2r=n′3?
$2r^2 + 2r = n'^3$ so
$r^2 + r = \frac {n'^3}2$
$r(r+1) = \frac {n^3}2$.
That's why we are considering them.
d)``since r and r+1 are coprime we must have that one of r and r+1 is a cube and the other is four times a cube'' ????
$2|n'^3$ so $2|n'$. And so $8|n'^3$.
Let $n' = 2\overline n$. so $n'^3 = 8\overline n^3$
So $r(r+1) = \frac {8\overline n^3}2 = 4\overline n^3$.
Now one of $r, r+1$ is odd and the other is even. And the even one is divisible by $4$. Let $r_1 = $ the odd one, and let $4r_2 =$ the even one we have $4r_1r_2 =4\overline{n}^3$ or $r_1r_2 = \overline n^3$.
Now we just redo what we did in b). $r, r+1$ are relatively prime so $r_1, 4r_2$ are relatively prime and $r_1, r_2$ are relatively prime. And if we redo the argument of b) we know that $r_1, r_2$ are both perfect cubes.
So $r_1$ and $4r_2$, which are $r, r+1$ (although we do not know which is which) are so that $r_1$ is a perfect cube and $4r_2$ is $4$ times a perfect cube.
| {
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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
$$ f \left( x ^ 2 + f ( y ) \right) = f \big( f ( x ) \big) + f \big( f ( y ) \big) + x ^ 2 f ( y ) + y ^ 2 f ( x ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $.
It's straightforward to check that the constant zero function and the squaring function are both solutions. I suspect that those are the only ones.
Plugging $ x = y = 0 $ in \eqref{0} gives $ f \big( f ( 0 ) \big) = 0 $, and the putting $ x = 0 $ and $ y = 1 $ in \eqref{0} shows that $ f ( 0 ) = 0 $. Then, setting $ y = 0 $ in \eqref{0} gives
$$ f \left( x ^ 2 \right) = f \big( f ( x ) \big) \tag 1 \label 1 $$
for all $ x \in \mathbb R $. We can use \eqref{1} to rewrite \eqref{0} as
$$ f \left( x ^ 2 + f ( y ) \right) = f \left( x ^ 2 \right) + f \left( y ^ 2 \right) + x ^ 2 f ( y ) + y ^ 2 f ( x ) \tag 2 \label 2 $$
for all $ x , y \in \mathbb R $. Substituting $ - x $ for $ x $ in \eqref{2} and comparing the result with \eqref{2} itself (for example when $ y = 1 $), we see that $ f $ is an even function; i.e.
$$ f ( - x ) = f ( x ) \tag 3 \label 3 $$
for all $ x \in \mathbb R $. Finally, exchanging $ x $ and $ y $ in \eqref{0}, we get
$$ f \left( x ^ 2 + f ( y ) \right) = f \left( y ^ 2 + f ( x ) \right) \tag 4 \label 4 $$
for all $ x , y \in \mathbb R $.
This is as far as I could go in this direction. Some messy calculations show that the only real analytic solutions are in fact those of the form $ f ( x ) = 0 $ and $ f ( x ) = x ^ 2 $, but this stil leaves open the possibility of another (non-analytic) solution.
Source:
Number eight on the list at the end of this page.
| 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 $f(x^2)=x^4$. In particular, $f(1)=1$.
Letting $y:=1$ we obtain $f(x^2+1)=f(x^2)+1+x^2+f(x)$ (*). If $x\neq 0$ satisfies $f(x)=-x^2$ then we have two possibilities:
(i) $f(x^2+1)=(x^2+1)^2$. Then (*) reduces to $(x^2+1)^2=x^4+1$, a contradiction.
(ii) $f(x^2+1)=-(x^2+1)^2$. Then (*) reduces to $-(x^2+1)^2=x^4+1$, again a contradiction.
Hence $f(x)=x^2$ for all $x\neq 0$. Therefore $f(x)=x^2$ for all $x$ which clearly works.
| {
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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 1 (mod 4), but $y^2$ is odd, it has to be congruent to 1 (mod 4).
After re-writing $y=2k+1$, I am able to use quadratic formula to get k in term of x $$k=\frac{-4\pm\sqrt{16+(4)(4)(x+2)(2^x)}}{8}$$ So far, I only know that $x=5,y=15$ is a solution. I would greatly appreciate if anyone can help.
| 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 $a2^m$ and $b2^n$. And so $a2^m$ and $b2^n$ must differ by a multiple of $2^{\min\{m,n\}}$. However, as $y-1 = a2^m$ and $y+1 =b2^n$ differ by precisely 2 this implies that $2^{\min\{m,n\}}$ is either 1 or 2 and so $\min\{m,n\}$ is either 0 or 1 and so at least one of $m,n$ is 0 or 1.
Case 1: If $m \in \{0,1\}$ then on the one hand $y-1 \le a 2^m \le 2(x+2)$, and also $y+1 \ge 2^n = 2^{x-m} \ge 2^{x-1}$. But on the other hand, $y+1 = (y-1)+2$ so putting this together gives $$2^{x-1} \le y+1 = (y-1) +2 \le 2(x+2) +2,$$ which yields
$$2^{x-1} \le 2(x+2)+2.$$
There are no solutions to this for $x \ge 6$, as the LHS is larger than the RHS for all such $x$.
Case 2: If $n \in \{0,1\}$ then on the one hand $y+1 \le 2(x+2)$ and also $y-1 \ge 2^{x-1}$. But on the other hand $(y+1)=(y-1)+2$ so putting this together gives
$$2^{x-1} \le y-1 = (y+1) - 2 \le 2(x+2) -2,$$
which yields
$$2^{x-1} \le 2(x+2)-2.$$
There are no solutions to this for $x \ge 5$.
So now all that is left to do is to check whether or not there is a solution to this for $x \in \{1,2,3,4,5\}$. This can be done easily by just checking for each such $x$. [I got $x=5,y=15$ as the one solution.]
| {
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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.
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}=4\int_{0}^{\pi /2}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}\\\overset{t=\operatorname{tg} x}{=}\int\limits_{0}^{\infty }\frac{1+t^2}{\left ( 1+\left ( 1+n^2 \right )t^2 \right )^2}dt\\
\overset{t=\frac{y}{\sqrt{1+n^2}}}{=}\frac{4}{\left ( 1+n^2 \right )\sqrt{1+n^2}}\int\limits_{0}^{\infty }\frac{1+n^2+y^2}{\left ( 1+y^2 \right )^2}dy\\ \overset{y=\operatorname{tg} \theta }{=}\frac{4}{\left ( 1+n^2 \right )\sqrt{1+n^2}}\int_{0}^{\pi /2}\left ( 1+n^2\cos^2 \theta \right )d\theta \\
=\frac{\pi \left ( 2+n^2 \right )}{\left ( 1+n^2 \right )\sqrt{1+n^2}}$$
| 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 logarithmic representations to get
$$\int \frac{1+n^2+y^2}{(1+y^2)^2}\, dy=\frac{2+n^2}2 \tan^{-1}(y)+\frac{ n^2}{2 }\frac{y}{1+y^2}$$
| {
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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 that $X \cdot Y = |X| |Y| cos(Z)$. The problem is the magnitude of $Y$ in this case turns out to be $\sqrt{5 + a^2}$, and the dot product of $X$ and $Y$ is a + 8, so I can't figure out how to resolve the two to find a.
$a + 8 = \sqrt{14} \cdot \sqrt{5 + a^2} \cdot \cos(40^\circ)$
| 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,
$$70\cos ^2\left(40^{\circ}\right)+14\cos ^2\left(40^{\circ }\right)a^2=64+16a+a^2$$
and the solution are (equation of second degree or quadratic formula), being the $70+14a^2>0, \, \forall a\in \Bbb R$,
$$a_1=\frac{-16+\sqrt{-3920\cos ^4\left(40^{\circ}\right)+3864\cos ^2\left(40^{\circ}\right)}}{2\left(1-14\cos ^2\left(40^{\circ}\right)\right)}, $$ $$a_2=-\frac{\sqrt{-3920\cos ^4\left(40^{\circ}\right)+3864\cos ^2\left(40^{\circ}\right)}+16}{2\left(1-14\cos ^2\left(40^{\circ }\right)\right)}$$
| {
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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 $m \in \left[a\sqrt{\dfrac{15}{7}} + b\sqrt{\dfrac{7}{15}}; 2\right)$, then $2x^2 + 2x - mf(x) + 5 = 0$ has at least a root. Find $S = a + b$
*
*Here is what I did so far
$$f(x) f'(x) = (2x+1)\sqrt{1+f^2(x)}$$
$$\implies \dfrac{f(x) f'(x)}{\sqrt{1+f^2(x)}} = 2x+1$$
$$\implies \int \dfrac{f(x) f'(x)}{\sqrt{1+f^2(x)}} \,dx = \int2x+1 \,dx = x^2 + x + C$$
Let $t = f^2(x) + 1 \implies dt = 2f'(x)f(x) dx$. Therefore:
$$\int \dfrac{1}{2\sqrt{t}} \,dx = x^2 + x + C$$
$$\implies \sqrt{t} = x^2 + x + C$$
$$\implies \sqrt{1 + f^2(x)} = x^2 + x + C \, (2)$$
Since $f(0) = 2\sqrt{2}$, plug in $(2)$ we get $C = 3 \implies f(x) = \sqrt{(x^2+x+3)^2 - 1}$
I can't proceed further from here
| 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 $x=-0.5$.
$$f(-0.5) =\frac{81}{4}-\frac{105m^2}{16} \le 0 \iff m\ge \sqrt{\frac{108}{35}} $$
Now, the leftover quadratic has discriminant $$-11m^4+80m^2-144 \ge 0 \iff \frac{6}{\sqrt{11}} \le m \le 2 $$
You can check that $\sqrt{\frac{108}{35}} \lt \frac{6}{\sqrt{11}} $, and so there’s no need to consider the case where $f$ has three turning points, as $f$ will already have atleast one root for $m\ge \sqrt{\frac{108}{35}} $. Now it only remains to write this square root in the form in the question. For that, manipulate $$a\sqrt {\frac{15}{7}} +b\sqrt{\frac{7}{15}}= \frac{15a+7b}{\sqrt{105}} \overset{\text{must be}}= \sqrt{\frac{108}{35}} \\ \implies \color{magenta}{15a+7b = 18 } $$
Unless you specify more conditions, $a+b$ can take infinitely many values.
| {
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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 $2^7$
but I think it is not correct, can you help me?
| *
*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 divisibility by $2^7$ your work is a good starting point
Let explicit $p=2k+1$ then $p^4-10p^2+9=16(k-1)k(k+1)(k+2)$
This is a multiple of $4$ consecutive numbers, so they belong to the set $\{4n,\ 4n+1,\ 4n+2,4n+3\}$ not necessarily in that order.
The product of the two $4n(4n+2)=16n^2+8n$ is divisible by $8$.
So overall the expression is divisible by $16\times 8=2^7$
| {
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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 the expression and refactoring it in a slightly different way gives
$$f(n) = n^4 + 6n^3+11n^2+6n \\= n^4 + 6n^3 + 9n^2 + 2n^2 + 6n = (n^2 + 3n)^2 + 2n(n+3). \tag{1}\label{1} $$
We want to show that the only possible way for $ f(n) $ to be the square of an integer is if $ f(n) = (n^2 + 3n +1 ) ^2. $ We show this by proving that $ (n^2+3n)^2 < f(n) < (n^2+3n+2)^2 $. The left-hand side follows immediately from $(1)$, since $ 2n(n+3) > 0 $ for all $ n \geq 1 $, and the right-hand side can be verified by multiplying out both sides:
$$
\begin{align}
(n^2+3n)^2 + 2n(n+3) &< (n^2+3n+2)^2 \\ \iff
n^4 + 6n^3 + 11n^2 + 6n &< n^4 + 9n^2 + 4 + 6n^3+4n^2+12n \\ \iff
0 &< 2n^2 + 6n + 4
\end{align}
$$
which is true for all $n \geq 1 $. Now we note that $n^2+3n = n(n+3)$ is even since one of the factors $n$ or $n+3$ is even for all $n$. It follows that $ n^2+3n+1$ must be odd, and so $ (n^2+3n+1)^2 $ must be odd. But $ f(n) $ must be even, since either $n$ and $(n+2)$ is even, or $(n+1)$ and $(n+3)$ is even and an even number multiplied by an odd number is an even number. So $f(n) \neq (n^2 + 3n +1)$ and therefore $f(n)$ cannot be the square of an integer for all $n \geq 1 $.
| 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 angles, its easy to see that $\Delta DEF$ is an equilateral triangle and by similarity, $\left | EF \right |$ is $2$. Area of $\Delta ABC = 4S = 4\sqrt3 \Rightarrow S = \sqrt{3} $.
I couldn't get any further from this point
| 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 are $x = -1\pm\sqrt{5}$. Taking the positive solution, we find $\boxed{AD = \sqrt{5}-1}$
| {
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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 $\frac{\sin A}{\sin B}\leq \frac{A}{B}$. So I simplified and got $$\frac{\sin^2 A+ \sin^2 B}{\sin A \sin B} \leq \frac{A^2+B^2}{AB}$$
Using $\sin x \leq x$ we can say $\sin^2 A+ \sin^2 B \le A^2+B^2$ but since we cannot divide, this doesn't work either.
| $f(t) = t\sin t,\ g(t) = \frac{t}{\sin t}$ 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 that
$$(x \sin x)\frac{y}{\sin y} +(y \sin y)\frac{x}{\sin x} \le (x \sin x)\frac{x}{\sin x} + (y \sin y)\frac{y}{\sin y} = x^2+y^2$$
Your inequality follows after dividing both sides by $xy$.
| {
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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 $$(1+\sqrt 2)^{k+1} - (1-\sqrt 2)^{k+1}$$ then $$(1+\sqrt 2)(1+\sqrt 2)^k - (1-\sqrt 2)(1-\sqrt 2)^k$$
I do not know where to go from here.
| 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 $k \ge 1, \exists m \ge 1: (1+\sqrt{2})^k - (1-\sqrt{2})^k= m\sqrt{2}, k, m \in \mathbb{N}$ .
Proof: $k = 1 \implies m = 2$. Assume assertion is true for $k$, we show it's trur for $k+1$. Indeed, $(1+\sqrt{2})^{k+1} - (1-\sqrt{2})^{k+1}= (1+\sqrt{2})(1+\sqrt{2})^k-(1-\sqrt{2})(1-\sqrt{2})^k= (1+\sqrt{2})^k -(1-\sqrt{2})^k +\sqrt{2}((1+\sqrt{2})^k + (1-\sqrt{2})^k)= c\sqrt{2}+\sqrt{2}\cdot p= (c+p)\sqrt{2}, c, p\in \mathbb{N}$.
Claim 1: for all $n \ge 1, (1+\sqrt{2})^n + (1-\sqrt{2})^n$ is an integer.
Proof: $n = 1$ is clear as $2$ is an integer. Assume it's true for up to $n$. We show it's true for $n+1$. Indeed, $(1+\sqrt{2})^{n+1} + (1-\sqrt{2})^{n+1}= 2((1+\sqrt{2})^n + (1-\sqrt{2})^n)+ ((1+\sqrt{2})^{n-1}+(1-\sqrt{2})^{n-1}) = 2e+f$ which is an integer since $e,f$ are integers by inductive step. Thus the claim is true for all $n$.
Return to your inductive step above: $(1+\sqrt{2})^{k+1}+(1-\sqrt{2})^{k+1}= (1+\sqrt{2})^k+(1-\sqrt{2})^k+\sqrt{2}((1+\sqrt{2})^k - (1-\sqrt{2})^k)= 2d+\sqrt{2}(s\sqrt{2})= 2d+2s= 2(d+s)$ which is divisible by $2$. Thus by induction we're done.
Note: in doing this proof,we use the following identity: $a^{n+1}+b^{n+1} = (a+b)(a^n+b^n) - ab(a^{n-1}+b^{n-1})$. Apply this for $a = 1+\sqrt{2}, b = 1 - \sqrt{2}$ to get the identity in the proof of claim 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4122715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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, involving the natural logarithm, that is similar to the identity that:
$$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{\psi(a) - \psi(b)}{a-b}$$
Also potentially related, the Lüroth analogue of Khintchine’s constant can be defined as the following:
$$\sum_{n=1}^{\infty} \frac{\ln (n)}{n(n+1)}$$
as mentioned here.
After some work, the following can be shown:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{5\ln(2) + 4\ln(3)}{16} + \frac{1}{2} \sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right)$$
and furthermore:
$$\sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right) = \int_{0}^{2} \left( \frac{\left(1-\pi x \cot(\pi x) \right)}{2x^2} + \frac{1}{x^2 - 1} + \frac{1}{x^2 -4} \right) \, dx$$
EDIT
I have derived yet another form for my sum of interest, however, I found this one interesting as it seems like it could potentially be solvable?
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \int_{0}^{\infty} \left( \frac{\psi^{(0)} (s+3) + \gamma}{(s+2)(s-2)} - \frac{25}{16 (s-2)(s+1)} \right) \, ds$$
From this, it is possible to obtain the following:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4} i + \frac{25}{48} (\ln (2) - i \pi) - \frac{1}{8} + \frac{1}{16} i \pi + \frac{1}{4} \int_{0}^{i \pi} \psi^{(0)} \left( \frac{4}{1+ e^{u}} \right) \, du$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4}i+\frac{25}{48} (\ln (2)-i \pi )+\frac{7 i \pi }{48}-\frac{1}{8}-\frac{\ln (2)}{3} -2 \int_0^{\infty } \frac{t \ln (\Gamma (1-i t))}{\left(t^2+4\right)^2} \, dt$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = -\frac{1}{8}-\frac{i \pi }{4}+\frac{i \gamma \pi }{4}-\frac{\ln (2)}{16} - 2 \int_{0}^{\infty} \frac{t \ln (\Gamma (-i t)) }{(4+t^2)^2} \, dt$$
$$\implies \sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} =\frac{25}{48} \ln (2) -\frac{1}{8} + \int_{1}^{\infty} \frac{\ln (v-1) \text{li} (v^2)}{v^5} \, dv$$
Where $\text{li}$ is the logarithmic integral function.
$$\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} = \frac{3 \ln (2)}{16} - \frac{\pi^2+1}{8} - \frac{\pi}{2} \int_{0}^{\infty} \sin(4\pi x) (\psi (x) - \ln (x)) \, dx$$
| 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
$$\sum_{n=3}^\infty \frac{\ln(k)}{k^2-4}$$
We will need k, k-2, and k+2 on the bottom, so I'll integrate to get those terms
$$\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right) = \sum_{n=3}^\infty x^{n-1}$$
$$\int\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right)dx = \sum_{n=3}^\infty \frac{x^{n}}{n}$$
$$\int x\int\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right) = \sum_{n=}^\infty \frac{x^{n+2}}{n(n+2)}$$
$$\int \frac{1}{x^5}\int x\int\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right) = \sum_{n=3}^\infty \frac{x^{n-2}}{n(n+2)(n-2)}$$
$$x^2\int \frac{1}{x^5}\int x\int\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right)$$
To clean things up, lets just allow that $$f(x)=x^2\int \frac{1}{x^5}\int x\int\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right)$$ and solve for it later.
Now notice that if we take the derivative and multiply by x we get a new n term. In particular
$$x\frac{d}{dx}f(x)=\sum_{n=3}^\infty \frac{nx^{n}}{n(n+2)(n-2)}$$
If we let that process be its own operation, with $D = x\frac{d}{dx}$, then applying that operation k times will give
$$D^k(f(x))=\sum_{n=3}^\infty \frac{n^{k}x^{n}}{n(n+2)(n-2)}$$
Then taking the derivative will respect to k gives.
$$\frac{d}{dk}D^k(f(x))=\sum_{n=3}^\infty \frac{\ln(n)n^{k}x^{n}}{n(n+2)(n-2)}$$
Then letting k=1, and x=1 will give
$$\frac{d}{dk}D^k(f(x))=\sum_{n=3}^\infty \frac{\ln(n)}{(n+2)(n-2)}$$
Simplification (I will fill this section out more if I decide to make this more than simply a sketch of how to get of the sum). For the initial sums, I have mainly used wolfram, because it is very tedious to compute otherwise. (Edit, left out a factor somewhere, so I'll need to edit the below calculations)
$$x\int_{0}^{x}\left(\frac{1}{1-t}-1-t-t^{2}\right)dt = \frac{x}{6}\left(-2x^{3}-3x^{2}-6x-6\ln\left(1-x\right)\right)$$
$$\frac{1}{x^5}\int_{0}^{x}\frac{t}{6}\left(-2t^{3}-3t^{2}-6t-6\ln\left(1-t\right)\right)dt = \frac{1}{120x^5}\left(x\left(-8x^{4}-15x^{3}-40x^{2}+30x+60\right)-60\left(x^{2}-1\right)\ln\left(1-x\right)\right)$$
$$x^2\int_0^x\left(\frac{1}{120t^5}\left(t\left(-8t^{4}-15t^{3}-40t^{2}+30t+60\right)-60\left(t^{2}-1\right)\ln\left(1-t\right)\right)\right)dt = -\frac{x^{2}\left(30\left(x^{2}-1\right)^{2}\ln\left(1-x\right)+x\left(16x^{4}-50x^{2}+15x+30\right)\right)}{240x^{4}} =f(x)$$
With that out of the way, I'll now look at how to define $D$ with fractional values.
Since we know we will only be taking one derivative, and then looking at the neighborhood around that one derivative, we can just take the first derivative and then add on the fractional part. So:
$$ x^k\frac{d^k}{dx^k} xf'(x)= \frac{x^k}{\Gamma(1-k)}\frac{d}{dx}\int_0^x\frac{f(t)}{(x-t)^k}dt $$
Edit: I have defined D^k slightly wrong, I will edit this more when I have fixed the definition
Edit: I haven't yet checked this, but I believe this is the correct way to define D^k
We start with the fact that $D^k(x^n) = n^kx^n$ and extend this to any function which has a power series.
We have an easy way to get the natural powers of $n$ from using natural number applications of $D^k$ (i.e. $D^1$, $D^2$, etc. are all well-defined). We can use this craft $n^\alpha$ from these.
Notice that $$n^\alpha = \sum_{k=0}^{\infty} a_k n^k$$
Now, we just need to get the $n^k$ out of a function with a power series.
If $f(x) = \sum_{k=0}^{\infty} b_k x^k$, then applying $D^k$ gives
$$D^k(f(x)) = \sum_{n=0}^{\infty} b_n n^kx^n$$
Since we would like to just get the $n^k$ term, where $n=m$, then, we can do
$$\frac{D^k(f(x))}{b_mx^{m+1}} = \sum_{n=0}^{\infty} \frac{b_n}{b_m} n^kx^{(n-(m+1))}$$
Then take the residue, so
$$Res\left(\frac{D^k(f(x))}{b_mx^{m+1}}\right)=n^k$$
And so $$D^\alpha(f(x)) = \sum_{k=0}^{\infty} a_k Res\left(\frac{D^k(f(x))}{b_nx^{n+1}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 6,
"answer_id": 2
} |
Value of $\sum_{r= 0}^{2019}\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{2021}{r-k} $ What is tried was : $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\mbox{Note that inside summation is }\quad\left.
\sum_{k = 0}^{r}\pars{-1}^{k}\pars{k + 1}\pars{k + 2}{2021 \choose r - k}
\right\vert_{\ 2021\ >\ 3} =
\left.\partiald[2]{}{x}\sum_{k = 0}^{r}{2021 \choose r - k}x^{k + 2}
\right\vert_{\ x\ =\ -1}
\end{align} , now how to simplfiy the summation part which needs to be double differentiated or there is some other method to solve for that summation value too ?
| 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:
$$2 \sum_{r=0}^n [z^r] (1+z)^{n+2}
\sum_{k\ge 0} {k+2\choose 2} (-1)^k z^k
\\ = 2 \sum_{r=0}^n [z^r] (1+z)^{n+2}
\frac{1}{(1+z)^3}
= 2 \sum_{r=0}^n [z^r] (1+z)^{n-1}
\\ = 2 \sum_{r=0}^n {n-1\choose r}
= 2 \times 2^{n-1} = 2^n$$
where we have used $n\ge 1,$ for $n=0$ we find $2[z^0] (1+z)^{-1} = 2.$
Addendum. We may also change the order of summation as asked in the
comments to get
$$2 \sum_{k=0}^n {k+2\choose 2} (-1)^k
\sum_{r=k}^n {n+2\choose r-k}
= 2 \sum_{k=0}^n {k+2\choose 2} (-1)^k
\sum_{r=0}^{n-k} {n+2\choose r}
\\ = 2 \sum_{k=0}^n {k+2\choose 2} (-1)^k
[z^{n-k}] \frac{1}{1-z} (1+z)^{n+2}
\\ = 2 [z^n] \frac{1}{1-z} (1+z)^{n+2}
\sum_{k=0}^n {k+2\choose 2} (-1)^k z^k.$$
We once more have a coefficient extractor enforcing the upper limit
of the sum and we find
$$2 [z^n] \frac{1}{1-z} (1+z)^{n+2}
\sum_{k\ge 0} {k+2\choose 2} (-1)^k z^k
= 2 [z^n] \frac{1}{1-z} (1+z)^{n+2} \frac{1}{(1+z)^3}
\\ = 2 [z^n] \frac{1}{1-z} (1+z)^{n-1}
= 2 \sum_{r=0}^{n-1} {n-1\choose r} = 2 \times 2^{n-1} = 2^n.$$
Here we see that keeping the original order of the two summations was
the better method moreover no new features appeared on changing the
order.
| {
"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 step.
Thanks in advance.
(I am sorry if the tags are not relevant to this question)
| 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 $11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4125397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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 concrete mathematical proof. I still don't have a conclusion, which is why I was wondering if any of you all did.
| \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 $n+1=12$.
\begin{array}{rrr| rr | rr}
m-11 & m & n & x & y & \frac{x+1}y & \frac{x}{y+1}\\
\hline
1 & 12 & 11 & 143 & 12 & 12 & 11 \\
2 & 13 & 11 & 77 & 6 & 13 & 11 \\
3 & 14 & 11 & 55 & 4 & 14 & 11 \\
4 & 15 & 11 & 44 & 3 & 15 & 11 \\
6 & 17 & 11 & 33 & 2 & 17 & 11 \\
12 & 23 & 11 & 22 & 1 & 23 & 11
\end{array}
| {
"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
} |
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
$$a=2R\sin{A}=2\sin{A},b=2\sin{B},c=2\sin{C}$$
so
$$\sin{A}+\sin{B}+\sin{C}\le\dfrac{2+\sqrt{3}}{2}$$
so we want to find the maximum of the value
$$P=AB\cdot BC\cdot AC=abc=8\sin{A}\sin{B}\sin{C}$$
if we use AM-GM
$$\sin{A}\sin{B}\sin{C}\le\left(\dfrac{\sin{A}+\sin{B}+\sin{C}}{3}\right)^3\le\left(\dfrac{2+\sqrt{3}}{2}\right)^3$$
the $=$ iff $A=B=C$ but this is clear wrong.so How to solve this problem? Thanks
| 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 + B}{2} \cos^2 \frac{B - A}{2}\\
&= 4(\cos C + 1)(\cos(B - A) + 1).
\end{align*}
Then, we have
$$4(\cos C + 1)(\cos(B - A) + 1) \le (2 + \sqrt{3} - 2\sin C)^2$$
which results in
$$\cos(B - A) \le \frac{(2 + \sqrt{3} - 2\sin C)^2}{4(\cos C + 1)} - 1.$$
On the other hand, we have
\begin{align*}
P &= 8\sin A \sin B \sin C\\
&= 4[\cos (B - A) - \cos(B + A)]\sin C\\
&= 4[\cos (B - A) + \cos C]\sin C.
\end{align*}
We split into two cases:
(1) $C \ge 2\pi/3$:
We have
\begin{align*}
P &\le 4(1 + \cos C)\sin C \\
&= 4\left(1 + \frac{1 - u^2}{1 + u^2}\right)\frac{2u}{1 + u^2}\\
&= \frac{16u}{(1 + u^2)^2}\\
&\le \sqrt{3}
\end{align*}
where $u = \tan \frac{C}{2} \ge \sqrt{3}$
(with equality if $A = B = \pi/6, \ C = 2\pi/3$).
(2) $C < 2\pi/3$:
We have
\begin{align*}
P &\le 4\left[\frac{(2 + \sqrt{3} - 2\sin C)^2}{4(\cos C + 1)} - 1 + \cos C \right]\sin C \\
&= -(8\sqrt{3} + 16)\frac{u^2}{1 + u^2} + (4\sqrt{3} + 7) u \\
&= \sqrt {3}-{\frac { \left( 4\,\sqrt {3}+7 \right) \left( \sqrt {3} - u
\right) \left( u-2+\sqrt {3} \right) ^{2}}{{u}^{2}+1}}\\
&\le \sqrt{3}
\end{align*}
where $u = \tan \frac{C}{2} \in (0, \sqrt{3})$.
So, the maximum of $P$ is $\sqrt{3}$ at $A = B = \pi/6, \ C = 2\pi/3$.
| {
"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_1 - x_2)^2 ≥0 $,
$ x_1^2 + x_2^2 ≥ 2x_1x_2 $
So the minimum value is: $2x_1x_2 $ for $x_1=x_2$.
Can be this done in a simillar manner for the starting problem.
why cannot we put $ x^2 =a_1 x_1^2, y^2 =a_2 x_2^2$ and solve it like root mean square inequality (without generalization?) We get:
$x^2 = y^2 $
.
$ a_1 x_1^2 = a_2 x_2^2 $
but not $x_1 = x_2$?
| 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 the quadratic polynomial with the following method:
$$ax^2+bx+c=a(x-m)^2+n$$
where, $$m=-\frac{b}{2a}, ~ n=-\frac{b^2-4ac}{4a}$$
In this case, we have
$$\begin{align}a:=a_2+\frac{a_2^2}{a_1}, ~b:=-\frac{2Ba_2}{a_1} , ~c:=\frac{B^2}{a_1}\end{align}$$
Hence we get,
$$m:=\frac{B}{a_1+a_2},~ n:=\frac{B^2}{a_1+a_2}$$
Thus,
$$\begin{align}a_1x_1^2+a_2x_2^2=\left(a_2+\frac{a_2^2}{a_1}\right)\left(x_2-\frac{B}{a_1+a_2}\right)^2+\frac{B^2}{a_1+a_2}\end{align}$$
Finally, we conclude that
$$\min\left\{a_1x_1^2+a_2x_2^2 \mid a_1x_1+a_2x_2=B\right\}=\frac{B^2}{a_1+a_2}$$ which is attained at the point $x_1=x_2=B/(a_1+a_2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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. Since $\lim_{x\to \infty}{{x^a\space (4+9x)^{b+1}}\over{9^{b+1}
\space\space x^{a+b+1}}} = constant$, then the improper integral $\int_{0}^{\infty}{dx\over{9^{b+1}
\space\space x^{a+b+1}}}$ and $\int_{0}^{\infty}{dx\over x^a\space(4+9x)^{b+1}}$ should both converge or diverge.
Then we consider ${1 \over {x^{a+b+1}}}\space $
if $a+b+1 = 1$ then the improper integral $\int_{0}^{\infty}{1 \over {x^{a+b+1}}}\space $diverges;
if $a+b+1<1$ then the improper integral diverges on ($1$,$\infty$);
if $a+b+1>1$ then the improper integral diverges on ($0$,$1$)
So it seems that the applicable set of (a,b) is empty. Is that right?
| 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, $\lim\limits_{x \to 0} \frac{x^a (4 + 9x)^{b + 1}}{x^a 4^{b + 1}} = 1$. Thus, $\int\limits_0^1 \frac{1}{x^a (4 + 9x)^{b + 1}} dx$ converges iff $\int\limits_0^1 \frac{1}{x^a} dx$ converges, which occurs iff $a < 1$.
For large $x$, we have $x^a (4 + 9x)^{b + 1} \sim x^{a + b + 1} 9^{b + 1}$, and we know that $\int\limits_1^\infty \frac{1}{x^n} dx$ converges iff $n > 1$. So we require that $a + b + 1 > 1$: that is, $a + b > 0$.
So the requirements are $a < 1$ and $a + b > 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 them.
c-) $5$ distinguishable boxes so that the boxes have $4,2,2,5,5$
objects in them, respectively.
d-) $5$ distinguishable boxes so that the boxes have $4,2,2,5,5$
objects in them.
e-) $5$ indistinguishable boxes so that the boxes have $1,2,4,5,6$
objects in them, respectively.
f-) $5$ indistinguishable boxes so that the boxes have $1,2,4,5,6$
objects in them.
g-) $5$ indistinguishable boxes so that the boxes have $4,2,2,5,5$
objects in them, respectively.
h-) $5$ indistinguishable boxes so that the boxes have $4,2,2,5,5$
objects in them.
My attempt:
a-) $ C(18,1) \times C(17,2) \times C(15,4) \times C(11,5) \times C(6,6) $
b-) $ 5! \times (C(18,1) \times C(17,2) \times C(15,4) \times C(11,5) \times C(6,6)) $
c-) $C(18,4) \times C(14,2) \times C(12,2) \times C(10,5) \times C(5,5) $
d-) $(\frac{5!}{2!\times 2! \times 1!})\times(C(18,4) \times C(14,2) \times C(12,2) \times C(10,5) \times C(5,5)) $
e-) I think this question is invalid , because if the boxes are indistinguishable then we cannot talk about respectivity.
f-) In this question , i firstly thought them like distinguishable boxes, after that i divide them by
$5!$ to obtain indistinguishable form such that $\frac{(C(18,1) \times C(17,2) \times C(15,4) \times C(11,5) \times C(6,6))}{5!} $
g-)I think this question is invalid, because if the boxes are indistinguishable then we cannot talk about respectivity.
h-) In this question, i firstly thought them like distinguishable boxes, after that i divide them by
$\frac{5!}{2!\times 2! \times 1!}$ such that $\frac{C(18,4) \times C(14,2) \times C(12,2) \times C(10,5) \times C(5,5)}{\frac{5!}{2!\times 2! \times 1!}} $
Is my solutions correct? If not, can you correct me, please ?
| 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 $4, 2, 2, 5$ and $5$ objects. Now note that $C(18,4) \times C(14,2) \times C(12,2) \times C(10,5) \times C(5,5)$ will order two heaps of $2$ objects and will order two heaps of $5$ objects but we do not want that.
So the answer to this is $ \ \displaystyle \frac{1}{2 \cdot 2} \cdot C(18,4) \times C(14,2) \times C(12,2) \times C(10,5) \times C(5,5)$
| {
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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 $0\leq r\leq \sqrt{a^2+b^2}$ and $0\leq \theta \leq 2\pi$. But i could not evaluate the outcome of the substitution. Any help in this regard is highly appreciated. Thank you in advance.
| 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
$$I = \frac{2\pi}{3}(a^2+b^2)^{\frac{3}{2}}$$
| {
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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 $M_{2\times 2}(\mathbb{Z}_{5})$
$$A:= \left\{ \begin{pmatrix} a & b \\
-b & a \end{pmatrix} \: : \: a,b \in \mathbb{Z}_{5} \right\}.$$
(1) Prove that $A$ is subring of $M_{2\times 2}(\mathbb{Z}_{5})$.
(2) Prove that $\begin{pmatrix} a & b \\
-b & a \end{pmatrix}$ is a zero divisor if and only if $a^{2}+b^{2}=0$, and compute all the zero divisors of $A$.
(3) Compute all the ideals of $A$.
I proved (1) straightforward, but the problem arives trying to prove (2). For this I take a matrix in $A$: $\begin{pmatrix} a & b \\
-b & a \end{pmatrix}$, such that $a^{2}+b^{2}=0$. The I want to show there is another matrix lets say $B=\begin{pmatrix} c & d \\
-d & c \end{pmatrix} \neq \begin{pmatrix} 0 & 0 \\
0 & 0 \end{pmatrix}$ so that
$$\begin{pmatrix} a & b \\
-b & a \end{pmatrix} \begin{pmatrix} c & d \\
-d & c \end{pmatrix}=\begin{pmatrix} 0 & 0 \\
0 & 0 \end{pmatrix}.$$
I tried to compute all the entries of this matrix product which are all equal to $0$ in $\mathbb{Z}_{5}$. But I dont know how to take the values of this matrix $B$. And not sure how to use the hypothesis $a^{2}+b^{2}=0$ for either two implications. And to compute all the zero divisors of $A$ I guess I should test all the six elements of $\mathbb{Z}_{5}$ satisfying $a^{2}+b^{2}=0$. And for (3) Im run out of ideas.
EDIT: For (2) I´ve just noticed the following pairs $(a,b)$ where $a,b \in \mathbb{Z}_{5}$ satisfy $a^{2}+b^{2}=0$: $(1,2),(2,1),(1,3), (3,1), (0,0), (2,4), (4,2)$. Let me know if there if this is correct? If correct, are there any other choices for $\mathbb{Z}_{5}$?
| 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 we pick one of the pairs... e.g. (1,2) and start multiplying by all the elements in $A.$ It might be worth noting that this ring is commutative.
(1,2)(1,0) = (1,2)
(1,2)(2,0) = (2,4)
(1,2)(3,0) = (3,1)
(1,2)(4,0) = (4,3)
(1,2)(2,1) = (0,0)
(1,2)(3,1) = (1,2)(2,1) + (1,2)(1,0) = (1,2)
(1,2)(4,1) = (2,4)
(1,2)(0,1) = (3,1)
(1,2)(1,1) = (4,3)
(1,2)(4,2) = (0,0), etc.
It looks likes $\{(0,0),(1,2),(2,4),(3,1),(4,3)\}$ is a non-trivial ideal.
And it should be the case that $\{(0,0),(2,1),(4,2),(1,3),(3,4)\}$ will also form and ideal. And their union will form an ideal.
And the trivial ideals.
| {
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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 $m > 5$.
$3^n \equiv -13 \pmod {512}$ if and only if $n \equiv{69} \pmod {128}$
Is there a straight forward way to complete the argument? Is there a better way to answer the question for $2^m - 3^n = 13$ with $m > 8$?
Edit:
Adding context for question:
I have been trying to understand why it is so difficult to establish a lower bound to $2^m - 3^n$ when $2^m > 3^n$. This came out of thinking about the Collatz Conjecture.
It occurs to me that for $m \ge 3$, $2^m - 3^n$ is congruent to either $5$ or $7$ modulo $8$ (since $-3^{2i} \equiv 7 \pmod 8$ and $-3^{2i+1} \equiv 5 \pmod 8$)
For $2^m - 3^n \equiv 7 \pmod {12}$, $m$ and $n$ are even, so the lower bound is at least $3^{n/2}$ since:
$$2^m - 3^n = (2^{m/2} - 3^{n/2})(2^{m/2} + 3^{m/2}) > 0$$
and
$$2^{m/2} - 3^{m/2} \ge 1$$
So, to reach a lower bound, I need to better understand the implications of $2^m - 3^n \equiv 5 \pmod 8$ and $2^m - 3^n \equiv 7 \pmod 8$ when $2^m - 3^n \not\equiv 7 \pmod {12}$.
I am also working to better understand this blog post by Terence Tao.
Edit 2:
I think that I can complete the argument using the congruence classes of $257$.
Here's my thinking:
Since $n \equiv 69 \pmod {128}$, $3^n \equiv 224$ or $33 \pmod {257}$
Then $2^m \equiv 3^n + 13 \pmod {257}$ which means $2^m \equiv 237$ or $46 \pmod {257}$
But there is no such solution of $2^m$ since for each $2^m$, there exists an integer $i$ such that $2^m \equiv \pm 2^{i} \pmod {257}$ and there is no such $i$ where $\pm 2^{i} \equiv {237} \pmod {257}$ or $\pm 2^{i} \equiv {46} \pmod {257}$
| $$2^{m} - 3^{n} = 13\tag{1}$$
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 LMFDB, this elliptic curve has no integral solution.
$\bullet n=3b+1$
Let $X=3\cdot3^{b}, Y=3\cdot2^{a}$, then we get
$Y^2 =X^3 + 117.$
This elliptic curve has integral solution $(X,Y)=(3,\pm 12).$
From this solution, we get $(m,n)=(4,1).$
$\bullet n=3b+2$
Let $X=9\cdot3^{b}, Y=9\cdot2^{a}$, then we get
$Y^2 =X^3 + 1053.$
This elliptic curve has integral solutions $(X,Y)=(-9,\pm 18)$, $(27,\pm 144).$
From $(27,\pm 144)$ we get $(m,n)=(8,5).$
Hence there are only two integral solutions $(m,n)=(4,1),(8,5).$
| {
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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 sequence)
All I got to know was that using the recursive relation we get:$$a_{75} = a_1+(\frac{1}{a_1} + \frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+...+\frac{1}{a_{74}})$$
I do not know how to proceed further. When I tried to express everything in form of $a_1$, it is becoming too messy and complicated.
Like:
$$a_2 = a_1 + \frac{1}{a_1}$$
$$a_3 = a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}} $$
$$a_4 = a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}} + \frac{1}{a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}}}$$
$$a_5 = a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}} + \frac{1}{a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}}}+\frac{1}{a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}} + \frac{1}{a_1+\frac{1}{a_1} +\frac{1}{a_1 + \frac{1}{a_1}}}}$$
And so on...
Any hints or help would be appreciated!
This question has to be done without using induction.
| Use
$$ a_n^2 + 3 \gt a_{n+1}^2 \gt a_n^2 + 2$$
| {
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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 $T(x)=P(x).Q(x)$
$T\left( x \right) = - a^2{x^4} + a\left( {d - b} \right){x^3} + \left( {bd} \right){x^2} + c\left( {d + b} \right)x + {c^2}$
Not able to approach from here
| 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 one of them has $2$ real roots.
Therefore $P(x)Q(x)=0$ has at least $2$ real roots.
We can also solve using your approach
$T\left( x \right) = - a^2{x^4} + a\left( {d - b} \right){x^3} + \left( {bd} \right){x^2} + c\left( {d + b} \right)x + {c^2}$
$T(0)>0$ and for sufficiently large $k$( $k$ is positive) we have $T(k)<0$ and $T(-k)<0$.
So there should be a real root in the interval $(-k,0)$ and $(0,k)$ so at least $2$ real roots exist.
| {
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Conjugate Harmonic $\displaystyle u=\frac{1}{2}\log(x^2+y^2)$ is harmonic and find its harmonic conjugate function
$\displaystyle\frac{\partial u}{\partial x}=\frac{x}{x^2+y^2}$
$\displaystyle\frac{\partial u}{\partial y}=\frac{y}{x^2+y^2}$
$\displaystyle dv=\int\frac{-\partial u}{\partial y}dx +\int\frac{\partial u}{\partial x}dy$
then
$\displaystyle v= \tan^{-1}\frac{y}{x} - \tan^{-1}\frac{x}{y} +c$
What did I do wrong here ?
| $$\displaystyle\frac{\partial u}{\partial x}=\frac{x}{x^2+y^2}$$
One of the Cauchy-Riemann equations is:
$$\displaystyle\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$
so
$$\frac{\partial v}{\partial y}=\frac{x}{x^2+y^2}$$
$$v = \int \frac{x}{x^2+y^2} \, dy=\frac{1}{x} \int \frac{1}{1+(\frac{y}{x})^2} \,dy= \tan^{-1} \frac{y}{x} + \text{const.} $$
Similarly,
$$v =- \int \frac{y}{x^2+y^2} \, dx=- \tan^{-1} \frac{x}{y} + \text{const.} $$
Since $\tan^{-1} \frac{y}{x} $ and $ - \tan^{-1} \frac{x}{y}$ differ by a constant $(|\tan^{-1} \frac{y}{x} +\tan^{-1} \frac{x}{y}| = \frac{\pi}{2})$, either answer is acceptable.
| {
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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 and worked out examples convince me that this inequality indeed holds true. I have tried to solve this problem by induction. Clearly, for $N=1$ we have,
$$\mathcal{P}(1) = \left(\frac{2}{3}\right)^{\tfrac{3}{4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{4}} \approx 0.8774 < 1.$$
Now assume the inequality holds for $N$, then for $N+1$ we have,
\begin{align}
\mathcal{P}(N+1) &=\left(\frac{2N+2}{2N+3}\right)^{\tfrac{2N+3}{2N+4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{2N+4}} \\
&= \left(\left(\frac{2N}{2N+1}\right)\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)\right)^{\left(\frac{2N+1}{2N+2}\right)\left(\frac{2N+2}{2N+1}\cdot\frac{2N+3}{2N+4}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\frac{2N+2}{2N+4}}\\[1em]
&= \left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)\left(1 + \frac{1}{2(N+1/2)(N+2)}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\left(1 - \frac{1}{N+2}\right)}\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \small\left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}}\cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \mathcal{P}(N) \cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \cdot\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}
\end{align}
Now from here we know that the first three terms are all smaller than 1 ($\mathcal{P}(N) < 1$ by induction hypothesis). However the last term is larger than one. For the proof by induction to work out, we need that this last term cancels against,
$$\left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}}.$$ But I do not see how it does. Any help is greatly appreciated.
| 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 $\log(1-x) < -x - \frac{x^2}{2}$,
$$
\log\left(1-\frac{1}{2N + 1}\right) < -\frac{1}{2N+1} - \frac{1}{2(2N+1)^2} < -\frac{1}{2N+1} - \frac{1}{(2N+1)^2}
$$
Thus
$$
\begin{align*}
\frac{2N+1}{2N+2}\log\left(1-\frac{1}{2N + 1}\right) + \frac{\log(2)}{2N+2} &< \frac{2N+1}{2N+2}\left(-\frac{1}{2N+1} - \frac{1}{(2N+1)^2}\right) + \frac{\log(2)}{2N+2}\\
&= \frac{(2N+1)(\log(2)-1) - 1}{(2N+1)(2N+2)} < 0
\end{align*}
$$
since $\log(2) - 1 < 0$.
| {
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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:
$$\sum_{n=1}^\infty \frac{2}{2n}-\frac{2}{2n+1}$$
$$2\sum_{n=1}^\infty \frac{1}{2n}-\frac{1}{2n+1}$$
$$2\left[\frac 12 - \frac 13 + \frac 14 - \frac 15+ \frac 16- \frac 17+ \frac 18- \frac 19+ \ldots\right]$$
$$2\left[1-\left(1-\frac 12 + \frac 13 -\frac 14 +\frac 15-\frac 16+\frac 17-\frac 18+\frac 19-\ldots\right)\right]$$
and noticing that the series inside the round parenthesis corresponds to $$1-\frac 12 + \frac 13 -\frac 14 +\frac 15-\frac 16+\frac 17-\frac 18+\frac 19-\ldots=\ln(2)$$ follows that:
$$2[1-\ln(2)]=2-2\ln(2)$$
So what I'm wanting to clarify is if is there any other way to improve this answer, a more "elegant" way to come up to the infinite sum of the series.
| Use $\frac{1}{m}=\int_{0}^{1} t^{m-} dt$ and sum of an infinite GP to get
$$S=2\sum_{n=1}^\infty \left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\sum_{n=1}^\infty \left(\int_{0}^{1}( t^{2n-1}-t^{2n})dt\right)=2\int_{0}^{1} \left( \frac{t^{-1} t^2}{1-t^2}-\frac{t^2 }{1-t^2}\right)dt$$ $$=2\int_{0}^{1} \left(\frac{t}{1+t}\right)dt=2[1-\ln 2].$$
| {
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"url": "https://math.stackexchange.com/questions/4152561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 number $\alpha \in R$ satisfying $\alpha^2=2$.
$\textbf{Proof}$. Consider the set
$T=\{t\in R : t^2 \lt 2 \}$
and set $\alpha = \sup A$. We are going to prove $\alpha^2 = 2$ by ruling out the possibilities $\alpha^2 \lt 2$ and $\alpha^2 \gt 2$. Lets see what happens if we assume $\alpha^2 \lt 2$. In search of an element of $T$ that is larger than $\alpha$, write
$(\alpha + \frac{1}{n})^2 = \alpha^2 + \frac{2}{\alpha} + \frac{1}{n^2}$
So far, I think I get what is going on. We are using the Archimedean Property to find a number that is greater than $\alpha$, but less than 2 (If we have a real number $\alpha$, there exists a natural number such that $\frac{1}{n} \lt \alpha$. We know $\frac{1}{n}$ is smaller than $\alpha$ so we add them in search of $t\in T$, such that t $\gt \alpha$.).
$\textbf{Proof cont.}$
\begin{align*}
(\alpha + \frac{1}{n})^2 &= \alpha^2 + \frac{2}{\alpha} + \frac{1}{n^2} \\
&\lt \alpha^2 + \frac{2}{\alpha} + \frac{1}{n} \\
&= \alpha^2 + \frac{2\alpha + 1}{n}
\end{align*}
This is where I get a little confused. I understand that $(\alpha + \frac{1}{n})^2 \lt \alpha^2 + \frac{2}{\alpha} + \frac{1}{n}$, but why is this necessary to say? Wouldn't it suffice to say $(\alpha + \frac{1}{n})^2 = \alpha^2 + \frac{2}{\alpha} + \frac{1}{n^2} \lt 2$? This is my question.
$\textbf{Proof cont.}$
But now assuming $\alpha^2 \lt 2$ gives us a little space in which to fit the $\frac{(2\alpha+1)}{n}$ and keep the total less than 2. Specifically, chose $n_0 \in N$ large enough so that
$\frac{1}{n_0} \lt \frac{2- \alpha^2}{2\alpha +1}$.
This implies $\frac{(2 \alpha +1)}{n_0} \lt 2 - \alpha^2$, and consequently that
$(\alpha + \frac{1}{n_0})^2 \lt \alpha^2 + (2- \alpha) = 2$ $\square$.
| 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:
$$ \frac{2\alpha}{n}+\frac{1}{n^2} \le \frac{2\alpha}{n}+\frac{1}{n} = \frac{2\alpha + 1}{n}$$
Because we know already that $c/n \to 0$ for any constant $c$.
| {
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"url": "https://math.stackexchange.com/questions/4152937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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:
$f = 2^{3\cos(t)+8\sin(t)} \Longrightarrow f=e^{(3\cos(t)+8\sin(t))ln(2)}$
Taking derivative to find critical points:
$f' = \ln(2)(-3\sin(t)+8\cos(t))e^{(3\cos(t)+8\sin(t))ln(2)}=0$
Now $e^x>0$, so I need to find when
$-3\sin(t)+8\cos(t)=0 \Longrightarrow 3 \sin(t)=8\cos(t) \Longrightarrow \tan(t) = \frac{8}{3} \Longrightarrow t=\arctan(\frac{8}{3}), \arctan(\frac{8}{3}) + \pi $
So I get two Points:
$(\cos(\arctan(\frac{8}{3})) ,\sin(\arctan(\frac{8}{3})))$
$(\cos(\arctan(\frac{8}{3}) + \pi), \sin(\arctan(\frac{8}{3}))$
and now I need to find $f_{xx}, f_{yy}, f{xy}$ in order to know which point is minimum and which is maximum:
$f_{x}=8^x3ln(2)$
$f_y = 256^y 8ln(2)$
$f_{xx} = 9\ln^2(2) 8^x$
$f_{yy} = 64ln^2(2) 256^y$
$f_{xy} = 0$
$f_{yx} = 0$.
for first point: $f_{xx} = 8.97$, $f_{yy} = 5530.131$. So $f_{xx}*f_{yy} - 0 > 0$ , and $f_{xx} > 0 $, which means it is a local minimum point.
for second point: $f_{xx} = 2.03$ , $f_{yy} = 0.17$. and same as above, it is a minimum point.
Would appreciate any help in approving this solution or finding mistakes in it, and if there's any more efficient approaches, thanks in advance.
EDIT: Approach from Parcly Taxel answer:
Checking maximum and minimum for $3x+8y$:
$\phi (x,y,\lambda) = 3x+8y + \lambda(x^2+y^2-1)$
$\phi_x = 3 + 2\lambda x = 0 \Longrightarrow x= \frac{-3}{2\lambda}$
$\phi_y = 8 + 2 \lambda y = 0 \Longrightarrow y= \frac{-4}{\lambda}$
$\phi_{\lambda} = x^2 +y^2 -1 = 0 \Longrightarrow \frac{9}{4\lambda^2} + \frac{16}{\lambda^2}-1=0 \Longrightarrow 9+64=4\lambda^2 \Longrightarrow \lambda^2 = \frac{73}{4} \Longrightarrow \lambda = \pm \frac{\sqrt{73}}{2}$
So: $x= \pm 0.3511 $
$y = \pm 0.9363$
$(0.3511, 0.9363) , (-0.3511, -0.9364)$
| 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 $\cos \arctan \alpha = \frac{1}{\sqrt{1+\alpha^2}}$, $k$ is $\frac{8}{1/\sqrt{1+(8/3)^2}} = \sqrt{73}$, hence the maximum value is $2^{\sqrt{73}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 not sure what to do next.
|
Reflect point $X$ across the sides of $\triangle ABC$. Observe,
$$\begin{align*}
\angle DAF&=\angle DAX+\angle FAX=2(\angle BAX+\angle CAX)=90^{\circ}\;\implies DF=3\sqrt{2}\\
\angle DCE&=\angle DCX+\angle ECX=2(\angle ACX+\angle BCX)=90^{\circ}\;\implies DE=5\sqrt{2}\\
\angle FBE&=\angle FBX+\angle EBX=2(\angle ABX+\angle CBX)=180^{\circ}\implies F,\;B,\;E \;\;\text{are collinear.}
\end{align*}
$$
Using the cosine rule, in $\triangle EDF,$
$$\begin{align*}
EF^2&=DE^2+DF^2-2\cdot DE\cdot DF\cdot\cos\angle EFD \\
64&=18+50-60\cdot\cos\angle EFD \\
\cos \angle EFD&=\frac{1}{15} \implies \sin\angle EFD=\frac{4\sqrt{14}}{15}
\end{align*} $$
Using the cosine rule, in $\triangle ADC,$
$$\begin{align*}
AC^2 &=AD^2+CD^2-2\cdot AD\cdot CD\cdot \cos\angle ADC\\
&=9+25-30\cos(90^{\circ}+\angle EFD)\\
&=34+30\cdot\sin\angle EFD\\
&=34+8\sqrt{14} \\ \\
\therefore\; AB&=\sqrt{\frac{AC^2}{2}}=\boxed{\sqrt{17+4\sqrt{14}}\approx 5.6539}
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $a_n= \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} x^{2n+1}$, we have $\lim_{n\to \infty} |a_{n+1}/a_n|= x^2/2$, so the radius of convergence is $\sqrt{2}$. What I only need to check now is that checking the convergence of the series above when $x=\pm \sqrt{2}$. Since $\sum_{n=1}^\infty \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} \sqrt{2}^{2n+1}=-\sum_{n=1}^\infty \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} (-\sqrt{2})^{2n+1}$, I only need to consider the case $x=\sqrt{2}$. In the case $x=\sqrt{2}$, the series is given by $\sqrt{2} \sum_{n=1}^\infty \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} 2^n= \sqrt{2} \sum_{n=1}^\infty \dfrac{4^n n! n!}{(2n)!}$, but I can't see whether or not this series converges. Any hints?
| 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 $y = xt$, I get $x=0$ and $y=0$. Is there some other way or some other solution? Thanks in advance.
| 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
$$
0 = t(3t^2 + 20) - (15t^2 + 17 t + 15)= 3t^3 -15 t^2 + 3t - 15 = 3(t-5)(t^2+1)
$$
which has the real solution $t = 5$. From this, we find the solution $(x, y) = (19, 95)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $(x-\frac1x)^3+3(x-\frac1x)$. Now, RHS can be broken down into two terms. One could be the cube of one third of the other term, but not able to do this either.
Any ideas how to approach such questions?
| 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 $b \in \{1, 2, 4, 19, 38, 76\}$ but clearly we can eliminate most of these possibilities, for if $a > 108^{1/3} > 4$ or $b > 38^{1/3} > 3$ neither equation can be met. This leaves easy casework to check, and we find that $a = 3, b = 2$ works, hence $z = 3 + 2\sqrt{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 what I did:
\begin{align}
\frac{1}{2\sin50^\circ}+2\sin10^\circ&=\frac{1}{2\sin(60-10)^\circ}+2\sin10^\circ \\
&= \frac{1}{2\left(\frac{\sqrt{3}}{2}\cdot\cos10^\circ-\frac{1}{2}\cdot\sin10^\circ\right)}+2\sin10^\circ \\
&= \frac{1}{\sqrt{3}\cdot\cos10^\circ-\sin10^\circ}+2\sin10^\circ \\
&= \frac{\sqrt{3}\cdot\cos10^\circ+\sin10^\circ}{\left(\sqrt{3}\cdot\cos10^\circ-\sin10^\circ\right)\left(\sqrt{3}\cdot\cos10^\circ+\sin10^\circ\right)}+2\sin10^\circ \\
&= \frac{\sqrt{3}\cdot\cos10^\circ+\sin10^\circ}{3\cos^210^\circ-\sin^210^\circ}+2\sin10^\circ
\end{align}
But I don't know how to continue at this point. Multiplying in $2\sin10^\circ$ into the fraction is clearly unrealistic as it would result in trignometry of third power. Any help or hint would be appreciated. According to a calculator, the result of this expression should come to a nicely $1$, but I just want to know how to algebraically manipulate this expression to show that it is equal to $1$.
| 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
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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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 $\quad\sin x=\dfrac{2\tan(\frac x2)}{1+\tan^2(\frac{x}2)}=-\frac35\quad$. by testing the options we can find out $\tan(\frac x2)=-3$ works (although by solving the quadratic I get $\tan(\frac x2)=-\frac13$ too. $-3$ isn't the only possible value.)
I wonder is it possible to solve the question with other (quick) approaches?
| 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 $\tan\frac{-37}{2}\approx-\frac{1}{3
}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 integer $n \geq 4$, it has the property such that $3\cdot(n - 3) > n$ which means breaking it into two integers $3$ and $n - 3$ makes the product larger while keeping the sum unchanged.
If $n - 3$ is still greater than $4$, we can break it into $3$ and $n - 6$, resulting in $3 \cdot 3 \cdot (n - 6)$ and so on, until we cannot break it (less than or equal to $4$) anymore.
I was just wondering, why is the above observation true/valid, from an academic/mathematical perspective
| 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 number, repeat until the number is either $2$, $3$, or $4$.
The reason why $2$ should not be broken is because it breaks into two $1$'s, and the product is $1$ which is less than $2$. For $3$, the maximum product if broken is $2$ which is less than $3$. For $4$, the maximum product is $4$ since $2 + 2 = 4$ and $2 \cdot 2 = 4$.
Let's have $n = 25$. Breaking it, we have $12$ and $13$. Breaking $12$, we have two $6$'s, where breaking $6$ gives two $3$'s. Hence, $12$ breaks into $[3,3,3,3]$. Breaking $13$, we have $6$ and $7$. Breaking $6$ gives two $3$'s and breaking $7$ gives $3$ and $4$. Hence, $13$ breaks into $[3, 3, 3, 4]$.
Multiplying all, we have $3^{7}\cdot 4 = 8748$.
I am still trying to prove that this is true. Hopefully, someone might do it.
With the $k$ constraint, the maximum product is $(\lfloor\frac{n}{k}\rfloor)^{k - (n \bmod k)}(\lfloor\frac{n}{k}\rfloor + 1)^{(n \bmod k)}$ where $0 \leq (n \bmod k) < n$. In my example where $n = 10$ and $k = 4$, $10 \bmod 4 = 2$, hence the maximum product is $2^{4 - 2}(2 + 1)^{2} = 2^{2}3^{2}$. In @user64494 's example where $n = 40$ and $k = 6$, $40 \bmod 6 = 4$, hence $6^{6 - 4}(6 + 1)^{4} = 6^{2}7^{4}$. I don't know how to prove this if this gives the maximum product. This was already mentioned by @MatthewLeingang .
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $f(a)$) increases on $[0,\infty)$. Note also that $g'(a)>0$ whenever $a>0$ so $f(a)$ is strictly increasing.
Since $f$ is continuous, the range of $f$ on $[0,\infty)$ is $[0,A)$ where $$A=\lim_{a\to\infty}f(a)=\lim_{a\to\infty}\frac{2a}{\sqrt{a^2+a+1}+\sqrt{a^2-a+1}}=1.$$
P.S. an alternative method, which is probably way better
Let $h(a)=a+1-\sqrt{a^2+a+1}$ so that $g(a)=2h(a^2)$. Now, we wish to prove $h(a)<\frac12$ for $a>0$, or equivalently, $\sqrt{a^2+a+1}>a+\frac12$. This follows from $a^2+a+1=\sqrt{(a+\frac12)^2+\frac34}$.
Now, again by computing $\lim_{a\to\infty}f(a)$ you arrive at the same conclusion.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \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
$$ \lim_{n \to \infty} \sum_{k=1}^n \sqrt{2- \left( \frac{k}{n} \right)^2} \cdot \frac{k}{n^2} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \sqrt{2- \left( \frac{k}{n} \right)^2} \cdot \frac{k}{n}.$$
From here I use the fact that
$$
\lim_{n \to \infty} \frac{b-a}{n} \sum^n_{k=1}f(a + k \frac{b-a}{n})= \int^b_a f(x) \ dx,
$$
to get
$$
\lim_{n \to \infty} \sum_{k=1}^n \sqrt{2- \left( \frac{k}{n} \right)^2} \cdot \frac{k}{n^2} = \int^1_0 \sqrt{2-x^2} \ x \ dx.
$$
Now write $u = 2 - x^2$, $du = -2x$, and $dx = -\frac{du}{2x}$
$u = 2- x^2$, so for $x = 0 \to u = 2 $ and for $ x = 1 \to u = 1$
$$
\int^1_0 \sqrt{2-x^2} \ x \ dx \to \int^2_1 \sqrt{u} \ x - \frac{du}{2x} = -\frac{1}{2}\int^2_1 \sqrt{u} \ du = -\frac{1}{2} \cdot \ \left[\frac{2}{3}u^{\frac{3}{2}} \right]^2_1 = \\ -\frac{1}{2} \cdot \ \left[\frac{2}{3}(2-x^2)^{\frac{3}{2}} \right]^1_0 = -\frac{1}{2} \cdot \ \left[\frac{2}{3}(2-1^2)^{\frac{3}{2}} \right] -\left(-\frac{1}{2} \cdot \ \left[\frac{2}{3}(2-0^2)^{\frac{3}{2}} \right] \right)= \frac{2\sqrt{2}-1}{3}.
$$
| 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)
$.
| {
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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 spherical coordinations:
$x=rsin\phi cos\theta$
$y=rsin\phi sin\theta$
$z=rcos\phi$.
And so $x^2+y^2+z^2=r^2\le z = rcos\theta \Longrightarrow r\le cos\theta \Longrightarrow -\frac{\pi}{2} \le\theta \le \frac{\pi}{2}$ ($cos\theta \ge 0)$.
and so there's no restrictions on $0 \le \phi \le \pi$. and $0 \le r\le cos\theta$.
So:
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{\pi}\int^{cos\theta}_{0}r.r^2sin\phi drd\phi d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\int_0^{\pi}\frac{cos^4\theta}{4}sin\phi ]d\phi d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\int^{\pi}_0\frac{1+cos^2(2\theta)}{2}sin\phi]d\phi d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\int^{\pi}_0\frac{3+cos(4\theta)}{4}sin\phi]d\phi d\theta = \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\int^{\pi}_0(3sin\phi+cos(4\theta)sin\phi)]d\phi d\theta = \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[(\pi3sin\phi+sin(4\pi)sin\phi)]d\phi d\theta = \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[(\pi3sin\phi)]d\phi=\frac{3}{4}\pi (-cos(-\pi) - cos(\pi))=\frac{6\pi}{4}$$I would really appreciate any feedback about my solution, even the smallest mistakes, and would love to get an approval of my answer, Thanks in advance!
| 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 the integral should be,
$\displaystyle \int_0^{2\pi} \int_0^{\pi/2} \int_0^{\cos\phi} \rho^3 \sin\phi \ d\rho \ d\phi \ d\theta = \frac{\pi}{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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",
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"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 $\cos 75°= \sin 15°$ and we write 75 = 45+30 and 15 = 45-30.
So we have $\sin (45°+30°) - \sin (45°-30°) = 2\cos 45° \cdot \sin 30° = 2\cdot \dfrac{1}{√2}\cdot \dfrac{1}{2}= \dfrac{1}{\sqrt{2}}$
Alter: You need not break the given form even, as follows
$\cos 105° + \sin 75°= \cos 105° +\cos 15°= \cos(60+45)+\cos (60-45)$
Check from the above given formulas, and you should get the same answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
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}{abc}=\frac{1}{a+b+c}\iff (ab+bc+ca)(a+b+c)=abc \iff (a+b)(b+c)(c+a)=0 $
I'm curious to prove this some other ways. I tried to use function and inequality but still no progress.
| 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
$$a\neq 0$$
that is :
$$(a(b)+b(-b)+a(-b))(a+b-b)-ab(-b)=0$$
Hence by symmetry the other roots must be
$$a=-b , c= -a$$
Therefore we have that :
$$(a+b)(b+c)(c+a)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $7$ is $111$ and $15$ is $1111$ and so on.
I want to be able to do this with any base. Based on wolfram, I am inducing that the formula for checking an integer $x$ and some base $b$ and some non-negative exponent $m$ is
$$
x = \frac{1}{b - 1}\left(b^m - 1 \right)
$$
How can you intuitively derive this formula?
After I wrote out $x = b^0 + b^1 + b^2 + \ldots + b^m$, it became obvious to me how to derive this expression.
Let $f(m) = x = b^0 + b^1 + b^2 + \ldots + b^m$, then we have
$$
f(m) / b = 1/b + b^0 + b^1 + \ldots + b^{m - 1} \\
= 1/b + f(m) - b^m \\
\implies f(m) = 1 + bf(m) - b^{m + 1} \\
\therefore f(m) = \frac{b^{m + 1} - 1}{b - 1}
$$
| 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:
$$
\begin{align*}
(b - 1)(b^{m-1} + b^{m - 2} + \ldots + b + 1)
&= \begin{array}[t]{ccccccc}
b^m &+b^{m-1} &+ &\ldots &+ b^2 &+ b\\
&-b^{m-1} &- &\ldots &- b^2 &- b &- 1
\end{array}\\
&= b^m - 1
\end{align*}
$$
Now divide both sides by $b - 1$. This is how I remember the formula for the sum of a geometric progression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to graph and solve this equation? I'am trying to solve this equation.
\begin{equation}
x^{8}+(x+2)^{8}=2
\end{equation}
What I tried:
\begin{equation}g(x)=x^{8}+(x+2)^{8}\end{equation}
\begin{equation}
\begin{array}{l}
\text { }\\
y=x+1
\end{array}
\end{equation}
\begin{equation}
(y+1)^{8}+(y-1)^{8}=2
\end{equation}
\begin{equation}
(y+1)^{8}=y^{8}+a_{7} y^{7}+a_{6} y^{6}+\cdots+a_{1} y+1
\end{equation}
\begin{equation}
(y-1)^{8}=y^{8}-a_{7} y^{7}+a_{6} y^{6}+\cdots-a_{1} y+1
\end{equation}
\begin{equation}
\begin{array}{l}
2\left(y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}+1\right)=2 \Leftrightarrow \\
y^{8}+a_{6} y^{6}+a_{4} y^{4}+a_{2} y^{2}=0 \Leftrightarrow y^{2}=0 \Leftrightarrow y=0
\end{array}
\end{equation}
y=0, x= -1
Maybe there is another way of solving this...
| $$f(x)=x^8+(x+2)^8\qquad f'(x)=8x^7+8(x+2)^7\qquad f''(x)=56x^6+56(x+2)^6$$
$f''(x)$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 other because of convexity), and you can generalize to any $2n$ exponent instead of $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
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