Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. So here is the Question :-
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find $f(-3)$ .
What I tried:- Since it's given that $f(x)$ is a polynomial of degree $3$ , I can assume $f(x) = ax^3 + bx^2 + cx + d$ for some integers $a,b,c,d$ and $a\neq 0$. Then we have :-
$$ax^3 + bx^2 + cx + d = (x^2 - 1)y + (2x - 5)$$
$$ax^3 + bx^2 + cx + d = (x^2 - 4)z + (-3x + 4)$$
This gives that $(x^2 - 1)y + (2x - 5) = (x^2 - 4)z + (-3x + 4)$ . But I am not sure how to proceed further since we have $3$ variables to deal with , and I am stuck here.
Any hints or explanations for this problem will be greatly appreciated !!
| Hint:
Let $$\dfrac{f(x)}{(x-1)(x+1)(x-2)(x+2)}=\dfrac A{x-1}+\dfrac B{x+1}+\dfrac C{x-2}+\dfrac D{x+2}$$
$$\implies f(x)=?$$
where $A,B,C,D$ which are arbitrary constants which can be found by putting $x=1,-1,2,-2$ one by one.
For example, put $x=1,$ $$2\cdot1-5=A(1+1)(1-2)(1+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Correct way to integrate $\int x(x^2-16)dx$
Evaluate:
$$\int x(x^2-16)dx$$
I have noticed that this integral can be solved using two different methods, but I am not sure which one is the correct one.
Way 1: Using $u$-subtitution
Let $u=x^2-16, du = 2xdx$
Then, we have $$\int x(x^2-16)dx$$ $$= \frac{1}{2}\int udu$$ $$= \frac{1}{2}(\frac{1}{2}u^2)+C$$
$$= \frac{1}{4}(x^2-16)^2+C$$
Way 2:
$$\int x(x^2-16)dx$$ $$= \int(x^3-16x)dx$$ $$= \frac{1}{4}x^4-\frac{16}{2}x^2+C$$ $$= \frac{1}{4}x^4-8x^2+C$$
| Alternatively, use hyperbolic trigonometric substitution.
Let $x = 4\cosh t\Rightarrow x^2 - 16 = 16\sinh^2 t$ and $\frac{dx}{dt} = 4\sinh t$.
The integral becomes
$$\begin{array} {r c l }
\displaystyle \int 4\cosh t \cdot (16\sinh^2 t)(4\sinh t) \, dt
&=& 4^4 \displaystyle \int (\cosh t)(\sinh t)^3 \, dt = 4^3 \sinh^4 t + C \\
&=& 64(\cosh^2 + 1)^2 + C \\
&=& 64\left(\left ( \frac x4 \right)^2 + 1\right)^2 + C
\end{array} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $ED$ in the triangle $ABC$
$AB = 6$, $AB \| EG$, $BC = 8$, $AC = 10$
I only have this:
From here, we know $BF =FG, AF=DA, GC=DC, OG=OF=OD$, also $$\frac{6}{GE}=\frac{8}{GC}=\frac{10}{EC}$$
And, we can see $$\frac{OG}{OC} = \frac{GC}{EC} \implies OG = \frac{4}{5}OE$$
From the same $$\frac{4}{5}EC=CG=DC=EC - ED \implies DE =\frac{1}{5}EC$$
|
Let
$BC=8=a$,
$AC=10=b$,
$|AB|=6=c$,
$\angle BAC=\angle GEC=\alpha$.
Since $b^2=a^2+c^2=100$,
$\angle CBA=\angle CGE=90^\circ$.
Inradius of the right triangle
$r=\tfrac12(a+c-b)=2$.
\begin{align}
|ED|&=
r\,\cot\alpha
=r\cdot\frac ca
=\frac32
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0)$ and $(-2,0)$ and $y$ intercept of $(0,18)$. Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0) $ and $ (-2,0)$ and $y$ intercept of $(0,18)$.
The root ($\frac{1}{2},0)$ has multiplicity $2$.
I am to write the factored form of the polynomial with the above information. I get:
$f(x)=-6\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
Whereas the provided solution is:
$f(x)=-\frac{3}{2}(2x-1)^2(x+2)(x-6)$
Here's my working:
Write out in factored form:
$f(x) = a\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
I know that $f(0)=18$ so:
$$18 = a\big(-\frac{1}{2}\big)^2(2)(-6)$$
$$18 = a\big(\frac{1}{4}\big)(2)(-6)$$
$$18 = -3a$$
$$a = -6$$
Thus my answer:
$f(x)=-6\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
Where did I go wrong and how can I arrive at:
$f(x)=-\frac{3}{2}(2x-1)^2(x+2)(x-6)$ ?
| $$\Big(x-\dfrac{1}{2}\Big)^2 = \Big(\color{red}{\frac{1}{2}}\Big(2x-1\Big)\Big)^2=\color{red}{\frac{1}{4}}\Big(2x-1\Big)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$? I'm confused about a question I posted this morning.
I am trying to understand if $-6(x-\frac{1}{2})^2$ can be rewritten as $-\frac{3}{2}(2x-1)^2$?
I tried multiplying out the expression $-6(x-\frac{1}{2})^2$ to a polynomial form $36x^2-36x+9$ but that didn't take me closer to understanding my goal.
I noticed that I can remove the fraction inside $-6(x-\frac{1}{2})^2$ by doubling the contents:
$-6(x-\frac{1}{2})^2$ <> $-6(2x-1)^2$ # used <> for does not equal
I don't think I can simply half the factor -6 to get $-3(2x-1)^2$
As you can no doubt see, I am confused.
Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$ ?
If it does could someone show me how to transform from $-6(x-\frac{1}{2})^2$ to $-\frac{3}{2}(2x-1)^2$ in granular baby steps?
| $$-6\left(x-\frac{1}{2} \right)^2 = -\frac{3}{2}\left(2x-1 \right)^2$$
$$-6\left( x^2 -x +\frac{1}{4}\right) = -\frac{3}{2} \left(4x^2-4x+1 \right)$$
$$-6x^2+6x - \frac{6}{4} = -6x^2+6x -\frac{3}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
For which values of $\eta\in\mathbf{N}$, $\frac{10^2\cdot\eta}{\varphi\cdot\eta-1}$ is an integer ($\varphi\in\mathbf{N}_0$)? Recently, I have found this problem:
Determine for which values of $\eta\in\mathbf{N}$ and $\varphi\in\mathbf{N}_0$, the quantity:
$$\gamma=\frac{10^2\cdot\eta}{\varphi\cdot\eta-1}$$
is a positive integr ($\gamma\in\mathbf{N}$).
To solve this problem, I began studying the case where $\gamma=0$. Here $\varphi$ can be every number and $\eta=0$.
After that, I passed to the case where $\varphi=1$. So, I have:
$$\gamma=\frac{10^2\cdot\eta}{\eta-1}$$
The tecnique I used, is trying to transform the numerator into the denominator, so:
$$\gamma=\frac{10^2\cdot\eta-10^2+10^2}{\eta-1}=\frac{10^2\cdot(\eta-1)+10^2}{\eta-1}=10^2+\frac{10^2}{\eta-1}$$
Now, $10^2$ is an integer, so also $\frac{10^2}{\eta-1}$ must be an integer. In conclusion if $\varphi=1$ then the solutions are: $\eta=\{2,3,5,6,11,21,26,51,101\}$.
Then, I studied the case when $\varphi\mid 10^2$. Here, I can use the tecnique shown above. In fact:
$$\gamma=\frac{10^2\cdot\eta-\frac{10^2}{\varphi}+\frac{10^2}{\varphi}}{\varphi\cdot\eta-1}=\frac{\frac{10^2\cdot\varphi\cdot\eta-10^2\cdot\varphi}{\varphi}+\frac{10^2}{\varphi}}{\varphi\cdot\eta-1}=\frac{10^2}{\varphi}+\frac{10^2}{\varphi\cdot(\varphi\cdot\eta-1)}$$
In the hypotesis written abvove $\frac{10^2}{\varphi}$ is an integer, so $\frac{10^2}{\varphi\cdot(\varphi\cdot\eta-1)}$ must be an integer for a fixed $\varphi$.
The case where $\varphi\nmid10^2$, I think, is more complicated.
So, can we build a general method to find $\eta$ when $\varphi$ varies over the positive integers?
| $\eta$ must be a factor of $\gamma$ (because $\phi \eta - 1$ and $\eta$ are relatively prime). Then $\gamma$ is a positive integer iff $\eta \phi - 1$ is a positive factor of 100. Clearly, there are only finitely many such combinations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Set of prime integers Let $S$ be a set of primes such that $a,b\in S$ ($a$ and $b$ need not be distinct) implies $ab+4\in S$. Show that $S$ must be empty. (Hint use modulo 7)
I don't have an idea how to use the hint, any further hint will be appreacited
| Suppose $p \in S$.
Suppose $p\equiv 1 \pmod 7$ and $p\in S$. Then $p_2=p^2 +4 \equiv 5 \pmod 7$ is prime and in $S$. so $p_3 =p_1p_2 + 4\equiv 9\equiv 2\pmod 7$. So $p_4 = p_3p_2 +4\equiv 2*5+4 \equiv 0 \pmod 7$ is in $S$. But $7|p_4$ and $p_4$ is prime so $p_4 = p_3p_2 + 4 =7$ so $p_3p_2=3$ but $3$ is primes so that's impossible as $p_3p_2$ are prime so neither are equal to $1$.
So $S$ contains no primes $\pmod 1 \pmod 7$.
suppose $p\equiv 2\pmod 7$. Then $p_2 = p^2 +4 \equiv 8 \equiv 1 \pmod 7$ which we proved is impossible..
Suppose $p \equiv 3\pmod 7$ then $p_2 = p^2 + 4\equiv 6\pmod 7$. Then $p_3=pp_2 + 4\equiv 18\equiv 4\pmod 7$. And $pp_3 + 4\equiv 16 \equiv 2\pmod 7$ which was impossible.
Suppose $p\equiv 4 \pmod 7$. Then $p_2= p^2+4 \equiv 6$ And $p_3=pp_2 + 4 \equiv 28\equiv 0$ so $7|p_3$ so $p_3$ (which is prime) is equal to $7$ and $pp_2 \equiv 3 \pmod 7$ which was impossible two paragraphs up and is still impossible.
Suppose $p\equiv 5 \pmod 7$ then $p_2=p^2 + 4\equiv 29 \equiv 1$. That's out.
Suppose $p \equiv 6 \pmod 7$ then $p_2 = p^2 + 4 \equiv 5\pmod 7$. Nope.
So that only leaves $p\equiv 0 \pmod 7$ and as $p$ prime then $7$ is the only element in $S$. But $7^2 + 4 =53\in S$ which is a contradiction.
so $S$ is empty.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Determine the number of ways to color a 1-by-n chessboard with the colors red, blue, green and orange. Determine the number of ways to color a 1-by-n chessboard, using the colors red, blue, green and orange if an even number of squares is to be colored red and an even number is to be colored green.
Current result:
I will use exponential generating functions.
\begin{align*}
g^{(e)}(x)&=
\left(1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\dots\right)^2\left(1+\frac{x^{1}}{1!}+\frac{x^{2}}{2!}+\dots\right)^2\\
&=\left(\frac{e^x+e^{-x}}{2}\right)^2e^{2x}
\\&=\frac{1}{4}\left(e^{2x}+2e^xe^{-x}+e^{-2x}\right)e^{2x}\\
&=\frac{1}{4}(e^{4x}+2e^{2x}+1)\\
&=\frac{1}{4}\left(\sum_{n=0}^{\infty}4^n\frac{x^n}{n!}+2\sum_{n=0}^{\infty}2^n\frac{x^n}{n!}+1\right)\\
&=\sum_{n=0}^{\infty}\left(\frac{4^n+2\times 2^n}{4}\right)\frac{x^n}{n!}+\frac{1}{4}
\end{align*}
I'm wondering how I can move $1$ into the sum? I've been suggested to use $x^0$ but I'm not certain how to incorporate that as a sum since I want to find $h_n$ which will be the coefficient of $\frac{x^n}{n!}$.
Thanks in advance!
| You can simply read off the coefficients $h_n$ from the sum
$$\sum_{n=0}^\infty\left(\frac{4^n+2\cdot 2^n}4\right)\frac{x^n}{n!}+\frac14\;:$$
$h_0=\frac{1+2}4+\frac14=1$, and
$$h_n=\frac{4^n+2\cdot 2^n}4=4^{n-1}+2^{n-1}=2^{n-1}\left(2^{n-1}-1\right)$$
for $n\ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Coordinate Geometry Question using matrices
Let on the x-y plane, the distance between the points $A(x_1,y_1)$ and $B(x_2,y_2)$ be $d$. Another point $P(a,b)$ satisfies the equations $x_1a+y_1b=1$ and $x_2a+y_2b=1$ and the distance between point $P$ from the origin,$O(0,0)$ is $p$. Find the area of triangle $\Delta OAB$.
My solution: Define a non-singular square matrix $$A=\begin{bmatrix} x_{1} & y_{1} \\ x_{2} & y_{2} \end{bmatrix}$$.
The required area, $\Delta = \frac{1}{2} \begin{vmatrix} x_{1} & y_{1} \\ x_{2} & y_{2} \end{vmatrix} = \frac{1}{2} \det(A)$
The given condition of point $P$ can be written as
$$\begin{bmatrix} x_{1} & y_{1} \\ x_{2} & y_{2} \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
$$\implies \begin{bmatrix} a \\ b \end{bmatrix} = A^{-1} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ (on pre multiplication of both sides with $A^{-1}$)
The inverse of matrix $A$,
$A^{-1} = \frac{1}{2 \Delta} \begin{bmatrix} y_{2} & -y_{1} \\ -x_{2} & x_{1} \end{bmatrix}$
$$\implies \begin{bmatrix} a \\ b \end{bmatrix} = \frac{\begin{bmatrix} y_{2}-y_{1} \\ -x_{2}+x_{1} \end{bmatrix}}{2 \Delta}$$.
Comparing element wise and taking sum of their squares,
$p^2 = \frac{d^2}{4 \Delta^2} \implies \Delta = \frac{d}{2p}$
Is there any other way to solve this question?
| Points $A$ and $B$ lie on the line $\ell \ldots ax+by-1=0$. The distance of $\ell$ from the origin is
$$d(\ell,0) = \frac{|a\cdot 0+b\cdot 0-1|}{\sqrt{a^2+b^2}}=\frac1p$$
since $\sqrt{a^2+b^2} = d(P,O) = p$. The area of your triangle is
$$\text{area} = \frac12 \,\text{base}\cdot \text{height} = \frac12 d(A,B)\cdot d(\ell,O) = \frac12 \cdot d \cdot \frac1p = \frac{d}{2p}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof that $2^{ab} - 1$ is divisible by $2^a - 1$ Could you please comment on the correctness and quality of the following proof?
I'm a bit suspicious of the line $n = ab, a = 1, r = 2^a$. Should it be $r = 2$? And are there any "off by one errors? E.g. $n = ab$ vs $n = ab -1$?
To prove:
For $a, b > 1$
$2^{ab} - 1$ is divisible by $2^a - 1$
Explanation:
$2^{ab} - 1$ can be expressed as $\sum_{i=0}^{ab-1}2^i$
Considering the formula for the sum of a geometric series
$S_n = a \frac{r^n-1}{r-1}$
with
$n = ab, a = 1, r = 2^{a}$
we have
$S_n = \frac{(2^a)^b-1}{2^a-1}$
From which it can be seen that $2^a - 1$ is indeed a factor of $2^{ab} - 1$.
$\square$
| If we pick the common ratio to be $2^a$,
\begin{align}
\sum_{i=0}^{ab-1}2^i = \sum_{t=0}^{b-1}\sum_{i=0}^{a-1}2^{at+i} =\sum_{t=0}^{b-1}(2^a)^t \left(\sum_{i=0}^{a-1}2^i\right)
\end{align}
the first term is $\left(\sum_{i=0}^{a-1}2^i\right)$ and we have $b$ terms in total.
That is $$2^{ab}-1=\sum_{t=0}^{b-1}(2^a)^t \left(2^a-1\right)$$
$$\frac{2^{ab}-1}{2^a-1}=\sum_{t=0}^{b-1}(2^a)^t$$
Hence $2^a-1 | 2^{ab}-1$
I usually prove it in the following way:
*
*$x^b-1$ is clearly divisible by $x-1$ since $1^b-1=0$.
*Now, we just have to let $x=2^a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
List all the pairs $(x,y)$ s.t. $x^2 - y^2 = 2020$ List all of the pairs $(x,y) \in \mathbb{Z}^2$ s.t. $^2 - ^2 = 2020$.
The prime factorization of $2020$ is $2^2 \cdot 5 \cdot 101$. I used the fact that there exists a solution $(x,y) \in \mathbb{Z}^2$ to the Diophantine equation $x^2 - y^2 = n$ if and only if $n$ is odd or $n$ is a multiple of $4$. Since $n$ is a multiple of $4$, there exist a solution.
I know there is a solution but I am neither able to obtain the number of solution nor able to get the list all the pairs $(x,y)$.
| Hint
$x+y,x-y$ have the same parity as $x+y+x-y$ or $x+y-(x-y)$ is even
Now as $2020$ is even, both must be even
$$\implies\dfrac{x+y}2\cdot\dfrac{x-y}2=\dfrac{2020}4$$
Now if $x,y>0$ $$x+y>x-y$$
As $505=1\cdot505=5\cdot101,$
$\dfrac{x-y}2=1$ or $5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Estimating Error due to replacing the sum $\sum\limits_{n=1}^{\infty} \frac{1}{n!} (\frac{1}{2})^n$ by the first $n$ terms Question: Estimating Error due to replacing the sum $\sum\limits_{n=1}^{\infty} \frac{1}{n!} (\frac{1}{2})^n$ by the first $n$ terms
All I can really say at this point is that the remainder $R_n$ when summing the first $n$ terms will be:
$$R_n = a_n [\frac{1}{2}\frac{1}{n+1} + (\frac{1}{2})^2\frac{1}{(n+1)^2} + (\frac{1}{2})^3\frac{1}{(n+1)^3} + ...]$$
Where $a_n = \frac{1}{n!} (\frac{1}{2})^n$. I can technically restrict $R_n$ by the following:
$$R_n < a_n[\frac{1}{2}\frac{1}{n+1} + (\frac{1}{2})^2 \frac{1}{(n+1)^2} + ...] = a_n\sum(\frac{1}{2(n+1)})^p$$
I am suppose to conclude that the error bounds $R_n < \frac{a_n}{2n+1}$, can anyone help me make that conclusion?
| Note that we have
$$\begin{align}
\sum_{m=n+1}^\infty \frac{1}{m!2^m}&=\frac{1}{n!2^n}\sum_{m=n+1}^\infty \frac{n!2^n}{m!2^m}\\\\
&=\frac{1}{n!2^n}\sum_{m=n+1}^\infty \frac{n!}{m!2^{m-n}}\\\\
&=\frac{1}{n!2^n}\sum_{m=1}^\infty \frac{n!}{(m+n)!2^{m}}\\\\
&\le \frac1{n!2^n}\sum_{m=1}^\infty \frac1{(n+1)^m2^m}\\\\
&=\frac1{n!2^n}\frac{\frac1{2(n+1)}}{1-\frac1{2(n+1)}}\\\\
&=\frac1{(2n+1)n!2^n}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\over2}+{c^2\over2}\right)$$
$$ \geq\left(\sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4}\right)^3\geq8 $$
$$ \Rightarrow \sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4} \geq2 \Rightarrow \sqrt[3]{2(ab)^2}+\sqrt[3]{2(bc)^2}+\sqrt[3]{2(ca)^2}\geq4 $$
I reached till here but can't take it forward.
Approach 2:
$$ \prod_{cyc} {(1+a^2)}=\prod_{cyc} \sqrt{(1+a^2)(1+b^2)} \geq \prod_{cyc} {(1+ab)}\ge8 $$
but it failed as
$$ \sum_{cyc}{ab}=3 \Rightarrow \sum_{cyc}{(1+ab)}=6 \Rightarrow 8\ge \prod_{cyc} {(1+ab)} $$
This approach is surely weak, but I think that the first approach is unfinished.
Probably brute-force would help but other solutions are always welcome.
Thanks!
| Another way.
Since $$(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2,$$ we obtain:
$$(1+a^2)(1+b^2)(1+c^2)=((1-ab)^2+(a+b)^2)(1+c^2)=$$
$$=(1-ab-(a+b)c)^2+((1-ab)c+a+b)^2=$$
$$=(1-ab-ac-bc)^2+(a+b+c-abc)^2=4+(a+b+c-abc)^2.$$
Id est, it's enough to prove that
$$a+b+c-abc\geq2,$$ which is true by Maclaurin.
Indeed, let $a+b+c=3u$, $ab+ac+bc=v^2$, where $v>0$ and $abc=w^3$.
Thus, we need to prove that $$3u-w^3\geq2$$ or
$$3u^2v-2v^3\geq w^3,$$ which is true because $$u\geq v\geq w.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$ Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
My attempt:
Square both sides three times
$$\begin{align*}
36(x^2+x)&=4(\sqrt{x^2+x})(2x+1+\sqrt{x^2+x})\\
(\sqrt{x^2+x})(35\sqrt{x^2+x}-4(2x+1))&=0
\end{align*}$$
This means $0,-1$ are solutions but I can't make sure that these are the only solutions. Also I'm not sure that squaring three times is a good approach or not.
| It is trivial that for any $a>0$ and $b>0$ (just raise both side to the $n^{th}$ power)
$$a^\frac1n +b^\frac1n >(a+b)^\frac1n$$
therefore $\forall x>0$
$$\sqrt[4]{x}+\sqrt[4]{x+1}> \sqrt[4]{2x+1}$$
and for $x=0$
$$\sqrt[4]{x}+\sqrt[4]{x+1}= \sqrt[4]{2x+1}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Simplifying $\frac{(\ln a)^3+(\ln b)^3+(\ln c)^3+(\ln d)^3}{\ln ab\;(\ln c\ln d-\ln a\ln b)}$ for positive reals with $abcd=1$ The problem is as follows:
Given the condition for $a$, $b$, $c$, and $d$ to be real positive numbers whose product equals to $1$, find the value of $B$.
$$B=\frac{(\ln a)^3+(\ln b)^3+(\ln c)^3+(\ln d)^3}{\ln ab\cdot\left(\ln c\cdot\ln d-\ln a\cdot\ln b\right)}$$
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{4}\\
2.&\textrm{2}\\
3.&\textrm{3}\\
4.&\textrm{6}\\
\end{array}$
I'm not very sure exactly how to simplify this expression. So far what I've could spot was this:
$abcd=1$
$\ln ab=\ln a + \ln b$
$(\ln a + \ln b)(\ln^2a -\ln a \ln b+\ln^2b)$
$(\ln c + \ln d)(\ln^2c -\ln c \ln d+\ln^2d)$
Taking the natural logarithm to the mentioned expression would become into:
$\ln abcd = \ln 1 = 0$
$\ln a + \ln b + \ln c + \ln d = 0$
But from here is where it becomes quite confusing on how to organize these expressions so that can achieve simplication of the before mentioned equation. Can someone help me here?.
From what I could also spot is:
$(\ln c + \ln d)((\ln c + \ln d)^2 - 3 \ln c \ln d)$
$-(\ln a+\ln b)((\ln a + \ln b)^2 - 3 \ln c \ln d)$
But that's how far I went. Does it exist other manipulation which can be done?
| Following the idea given in the comments, let $a=e^A$, $b=e^B$, $c=e^C$, $d=e^D$ then
$$\frac{A^3+B^3+C^3+D^3}{(A+B)\cdot\left(CD-AB\right)}=\frac{(A+B)(A^2-BC+B^2)+(C+D)(C^2-CD+D^2)}{(A+B)\cdot\left(CD-AB\right)}=$$
$$=\frac{(A+B)(A^2-AB+B^2)-(A+B)(C^2-CD+D^2)}{(A+B)\cdot\left(CD-AB\right)}=$$
$$=\frac{(A^2-AB+B^2)-(C^2-CD+D^2)}{\left(CD-AB\right)}=$$
$$=\frac{(CD-AB)+A^2+B^2-C^2-D^2}{\left(CD-AB\right)}=$$
$$=\frac{(CD-AB)+2CD-2AB}{\left(CD-AB\right)}=3$$
using that
$$A+B=-(C+D) \implies \\(A+B)^2=(C+D)^2 \implies A^2+B^2-C^2-D^2=2CD-2AB$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve $(\cos2x+1)^2=1/2$? $x$ belongs to $$[0,\pi/2[$$
$$(\cos(2x+1))^2 = \frac{1}{2}$$
I tried to find $x$ using
$$2x+1=\frac{\pi}{4} +\frac{n\pi}{2}
\qquad\text{or}\qquad
2x+1=\frac{3\pi}{4} + n\pi$$
but I didn't find my answer in MCQ which is
$$\frac{3\pi-4}{8}
\qquad\text{and}\qquad
\frac{5\pi-4}{8}.$$
I just need a hint on how to solve such equation.
| $$\implies\cos2(2x+1)=2\cos^2(2x+1)-1=0$$
$2(2x+1)= \dfrac{(2n+1)\pi}2 $ where $n$ is any integer
Now $1\le2x+1\le1+\pi$
If $1\le\dfrac{(2n+1)\pi}4\le1+\pi\iff\dfrac4\pi-2\le n\le\dfrac4\pi\implies n=0,1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Diophantine Equation with Square Root I want to resolve the diophantine equation:
$\sqrt{x^2+5x+12} ≡ x-2\pmod 5$
I have thought 2 ways:
1.
$(\sqrt{x^2+5x+12})^2 ≡ (x-2)^2\pmod 5$
$ x^2+5x+12 ≡ x^2 -4x+4\pmod 5$
$ 9x+8 ≡ 0\pmod 5$
$ 4x+3 ≡ 0\pmod 5$
$ 4x+3 = 5y$
$ 4x-5y = -3$
I resolve:
$x=5n+3$,
n
∈
Z
$y=4n+3$,
n
∈
Z
2.
$\sqrt{x^2+5x+12} ≡ x-2\pmod 5$
$\sqrt{x^2+5x+12} = x-2 + 5y$
$(\sqrt{x^2+5x+12})^2 = (x-2 + 5y)^2$
$x^2+5x+12 = 10xy+x^2-4x-20y+25y^2+4$
$25y^2+10xy-9x-20y-8 = 0$ and resolve that diophantine equation.
My questions are: Are both procedures valid? If so, does the development of each procedure preserve the value of the initial x and y unknowns?
Thank you
| Writing the congruence as an equation in the field $\Bbb F_5$ we obtain
$$
x^2+5x+12=(x-2)^2=x^2-4x+4,
$$
which gives $4x=2$. In $\Bbb F_5$ this has the unique solution $2^{-1}=3$, because $2\cdot 3=6=1$, so that $3$ is the inverse of $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $f(x)=\begin{cases} \cos (x^3 )& x<0 \\ \sin (x^3 )- |x^3-1| & x\ge 0 \end {cases}$, find the number of points where $f(|x|)$ is non differentiable Since we are dealing with $f(|x|)$, the $\cos (x^3)$ part can be ignored. (And $\cos$ is differentiable everywhere anyway)
For $\sin |x|^3$, the graph would break its smooth flow at $x=0$, so that is a potential point
$||x^3|-1|$ would be non differentiable at $x=\pm 1$, and that gives two more potential points
So there are a total of three points.
The given answer is 2 though (points are unknown)
Have I picked the right points ?
| $$f(x)=\begin{cases}\cos(x^3 )& x<0 \\ \sin (x^3 )- |x^3-1| & x\geq 0 \end {cases}=\begin{cases}\cos(x^3 )& x<0 \\ \sin (x^3 )- 1+x^3 & 0\leq x<1\\ \sin(x^3)-x^3+1&x\geq1\end {cases}$$
$$f(|x|)=\begin{cases}\sin(x^3)-x^3+1&x\geq1\\\sin(x^3)+x^3-1&0\leq x<1\\-\sin(x^3)-x^3-1&-1\leq x<0\\-\sin(x^3)+x^3+1&x<-1\end {cases}$$
There are three points that lie on the seam between two pieces of a piecewise smooth function, and as such, we need to check if the function is continuous and differentiable only at the seams.
For the seam $x=1$, the function is continuous at $f(1)=\sin(1)$. However, it is not differentiable, as the left handed derivative is $3\cos(1)+3$ and the right handed derivative is $3\cos(1)-3$.
For the seam $x=0$, the function is continuous at $f(0)=1$. The left handed derivative is $0$ as is the right handed derivative, so the function is differentiable (at least for the first derivative) there as well.
For the seam $x=1$, the function is continuous at $f(1)=\sin(1)$. However, it is not differentiable, as the left handed derivative is $-3\cos(1)-3$ and the right handed derivative is $-3\cos(1)+3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$. Question: Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$.
My attempt: It can be easily seen that $x_1+x_3=\frac{a+b}{2}$ and $x_2+x_4=\frac{a-b}{2}$. Further, the expression $[x_1^2+x_2^2+x_3^2+x_4^2]$ can be written as $[(x_1+x_3)^2+(x_2+x_4)^2-2(x_1x_3+x_2x_4)].$ I'm having trouble eliminating $(x_1x_3+x_2x_4)$ from this expression. Failing to make any sense out of this, I manipulated the existing expressions to deduce $$x_1x_2+x_1x_4+x_2x_3+x_3x_4=\frac{a^2-b^2}{4}$$and $$(x_1^2+x_3^2)-(x_2^2+x_4^2)+2(x_1x_3-x_2x_4)=a\cdot b$$Beyond this, I cannot make sense of the expressions anymore. I have no idea how to proceed with simplifying the expressions further, and would appreciate hints in the same direction.
| Using algebra.
Use the two equality constraints to get $x_3$ and $x_4$ as linear functions of $x_1$ and $x_2$.
This makes
$$x_1^2+x_2^2+x_3^2+x_4^2=x_1^2+x_2^2+\frac{1}{4} (a+b-2 x_1)^2+\frac{1}{4} (-a+b+2 x_2)^2$$
Compute the partial derivatives wrt $x_1$ and $x_2$ and set them equal to $0$. This would give $x_1=\frac {a+b}4$ and $x_2=\frac {a+b}4$. So, for the minimum
$$x_1^2+x_2^2+x_3^2+x_4^2=\frac {a^2+b^2}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
One matrix identity implies the other I have matrices $A\in\mathbb R^{n\times n}$, $C\in\mathbb R^{m\times m}$, and $B\in\mathbb R^{n\times m}$. I know that $A$ and $C$ are symmetric and positive semidefinite and that $K_A = A^2+BB^T$ and $K_C = C^2+B^TB$ are positive definite. Now assume that $AB = BC$. I can prove indirectly that
$$
AK_A^{-1}B = BK_C^{-1}C,
$$
but I can't prove it. Can I maybe express $AK_A^{-1}B = BK_C^{-1}C$ by using $AB-BC$?
Here is the indirect proof: We have
\begin{align}
Q
&:=\begin{pmatrix}AK_A^{-1}A+BK_C^{-1}B^T & AK_A^{-1}B-BK_C^{-1}C\\B^TK_A^{-1}A-CK_C^{-1}B^T & CK_C^{-1}C+B^TK_A^{-1}B\end{pmatrix}\\
&= \begin{pmatrix}A & B\\B^T & -C\end{pmatrix}\begin{pmatrix}K_A^{-1}A & K_A^{-1}B\\K_C^{-1}B^T & -K_C^{-1}C\end{pmatrix}\\
&= \begin{pmatrix}A & B\\B^T & -C\end{pmatrix}\begin{pmatrix}K_A^{-1} & 0\\0 & K_C^{-1}\end{pmatrix}\begin{pmatrix}A & B\\B^T & -C\end{pmatrix}.
\end{align}
Hence, $Q = MR^{-1}M$ with the matrices $M$ and $R^{-1}$ in the last row. Now,
$$
M^2 = \begin{pmatrix}A & B\\B^T & -C\end{pmatrix}\begin{pmatrix}A & B\\B^T & -C\end{pmatrix} = \begin{pmatrix}K_A & AB-BC\\(AB-BC)^T & K_C\end{pmatrix}.
$$
Hence, if $AB=BC$, then $M^2 = R$. Therefore, we have
$$
M(Q-I)M = MQM-M^2 = M^2R^{-1}M^2-M^2 = 0.
$$
Now, the fact that $K_A$ and $K_C$ are positive definite implies that $M$ is invertible (easy proof). Therefore, we get $Q = I$. But then the upper right corner in the matrix $Q$ must be zero.
| I have it! If $AB=BC$, then also $B^TA = CB^T$ and thus $CB^TB = B^TAB = B^TBC$. Also,
$$
BK_C = BC^2 + BB^TB = A^2B + BB^TB = K_AB.
$$
Hence, $BK_C^{-1} = K_A^{-1}B$. Therefore,
$$
AK_A^{-1}B = ABK_C^{-1} = BCK_C^{-1} = BK_C^{-1}C,
$$
where the last identity follows from $CB^TB = B^TBC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many four digits numbers are there not containing zero and multiplication of its digits divisible by 7? I saw a question in my math book, it seems very trivial, it says that:
How many four digits numbers are there not containing zero and
multiplication of its digits divisible by 7?
I thought of:
(all four digits numbers not containing zero) minus (all four digits numbers not containing 7 and 0)
in order to find all four digits number not containing zero and multiplication of its four digits divisible by 7.
Then $(9^4)-(8^4)=2465$. However the answer is $4904$. What am I missing?
| Your first answer is right in regard of the statement you gave, indeed:
Let $(a,b,c,d) \in \{1,2,3,4,5,6,7,8,9\}^4$. So $A = 1000a+100b+10c+d $ is a four-digit number.
Morevover, $a\cdot b \cdot c \cdot d $ is divisible by $7$, if and only if, its prime factorization contains at least one time $7$ so, if and only if, at least one of $A$'s digit is equal to $7$. Hence the answer is $9^4-8^4 = 2465$ as you said.
However if you are looking for the number of four-digit numbers such that the product of their digits is divisible by $7$ the answer is $4904 = 8(10^3-8^3) + 10^3$. You can check that: in order for a four-digit number $A$ to have the product of its digits divisible by $7$, it must contain $0$ or $7$.
Let $A = 1000a+100b+10c+d$ where $0\leq a,b,c,d \leq 9$ are integers and $a \neq0$.
If $a=7$ then you can have all the combinations possible for $b,c$ and $d$. Thus, it gives you $10^3$ choices.
If $a \neq 7$, then you are looking for the number $n$ of possibilities to have at least $b,c$ or $d$ equals to $0$ or $7$. Morevover, you have exactly $8^3$ possibilities for $b$, $c$ and $d$ not to be equal to $0$ nor $7$. Hence $n = 10^3-8^3$. Finally there are only $8$ possibilities for $a$ to be different from $7$.
Therefore the number you are looking for is $8(10^3-8^3)+10^3 = 4904$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $ \int\frac{2x-\sqrt{4x^{2}-x+1}}{x-1}dx $ Im trying to integrate $ \int\frac{2x-\sqrt{4x^{2}-x+1}}{x-1}dx $.
I've tried integration by parts, and any resonable substitution that came to my mind. Neither seem to work. I tried to use integral online calculator, the result was too complicated, and I'm pretty sure there is a resonable way to solve it, as we had this question in an exam.
Thanks in advance.
| Take,$$t=\sqrt{4x^2-x+1}-2x$$
Then,$$\sqrt{4x^2-x+1}=2x+t \;\;\Longrightarrow\;\;4x^2-x+1=(2x+t)^2 \;\;\Longrightarrow\;\; x=\frac{1-t^2}{4t+1}\,.$$
And $$dx = -\frac{4t^2+2t+4}{(4t+1)^2}dt\,.$$
So,
$$I=\int\frac{2x-\sqrt{4x^2-x+1}}{x-1}dx=\int\frac{t}{(1-t^2)/(4t+1)-1}\cdot\frac{4t^2+2t+4}{(4t+1)^2}dt\,,$$
$$\begin{align} & =-\int\frac{4t^2+2t+4}{(4t+1)(t+4)}dt\\
&=-\int\frac{4t^2+2t+4}{4t^2+17t+4}dt\\
&=-\int\left(1-\frac{15t}{4t^2+17t+4}\right)dt\\
&=-t+\int\frac{15t}{(4t+1)(t+4)}dt\,.
\end{align}$$
Splitting into partial fractions,
$$\frac{15t}{(4t+1)(t+4)} = \frac{4}{t+4}-\frac{1}{4t+1}\,,$$
We get,
$$I=-t+\int\frac{4}{t+4}dt-\int\frac{1}{4t+1}dt=-t+4\ln(t+4)-\frac{1}{4}\ln(4t+1)+C\,,$$
so, therefore,
$$\Rightarrow \bbox[5px,border:2px solid red]{I=-t+4\ln(t+4)-\frac{1}{4}\ln(4t+1)+C\,,}$$
On rewriting we get,
$$\Rightarrow \bbox[5px,border:2px solid red]{\begin{align}I\;=\;&2x-\sqrt{4x^2-x+1}+4\ln(\sqrt{4x^2-x+1}-2x+4)\\
&-\frac{1}{4}\ln(4\sqrt{4x^2-x+1}-8x+1)+C\,.
\end{align}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does $-8^{\frac{1}{3}}$ have $2$, $e^{\frac{\pi}{3}}$ and $e^{\frac{5\pi}{3}}$? Use DeMoivre’s theorem to find $-8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. –2
b. – 2, 2 cis ($\pi$/3)
c. – 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
d. 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
e. None of these
The correct answer is : $2, 2 e^{\pi/3}, 2 e^{5\pi/3}$
My calculation is here:
$r=\sqrt{-8^{2}}=8$
Then,
$= 2\ cis\ \frac{2\pi k}{n}$
If k is $0$,
$= 2\ cis\ 0=2$
If k is $1$,
$= 2\ cis\ \frac{\pi }{3}$
If k is $2$,
$= 2\ cis\ \frac{4\pi }{3}$
Therefore, the results are $2$, $= 2\ cis\ \frac{\pi }{3}$, and $= 2\ cis\ \frac{4\pi }{3}$.
So, why the correct answer is $2$, $2\ cis (\frac{\pi}{3})$, $2\ cis (\frac{5\pi}{3})$?
| $-8 = 8(-1) = 8\operatorname{cis} (\pi) = 8\operatorname{cis}( (2k+1)\pi)$
So $(-8)^{\frac 13} = \sqrt[3]{8}\operatorname{cis}(\frac {2k+1}3\pi)=$.
$2\operatorname{cis}(\frac \pi 3 + \frac {2k\pi}3); k=0,1,2$
$2\operatorname{cis}(\frac \pi 3), 2\operatorname{cis}(\pi)= 2\cdot (-1)=-2, 2\operatorname{cis}(\frac {5\pi} 3)$
You assumed that we needed $3*\theta = 2k \pi$ and you didn't realize we actually needed $3*\theta = (2k+1)\pi$.
.....
And I don't quite understand we you did $r= \sqrt{(-8)^2} 8$. Is that the $r$ that "belongs to" $-8$ or to $(-8)^{\frac 13}$? Either way, I'm not sure what you were doing.
To solve $z^{\frac 1m}$ you need to:
*
*Express $z = r\operatorname{cis}({\theta})$. where $r = |z|$ and $\theta =\operatorname{Arg}(z)= \operatorname{Arg}(\frac z{|z|})$.
*$z^{\frac 1m} = \sqrt[m]{r}\operatorname{cis}(\frac \theta m + \frac {2k\pi}m)$.
so to do $1$ we have $r = |-8| = 8$ and so $\theta = \operatorname{Arg}(-8)= \operatorname{Arg}\frac{-8}{8} = \operatorname{Arg}(-1) = \pi$.
I think somehow you assumed our Argument was $0$ and not $\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Let $L=\lim_{x\to 0} \frac{ a-\sqrt {a^2-x^2} -\frac{x^2}{2}}{x^4}$, $a>0$. If $L$ is finite, find $a$ and $L$ I need a hint to start this question, because I have no idea how to do it. It’s a $\frac 00$ form, so L’Hospital can be applied, but that would be extremely tedious. Expansion can’t be used because there is no function to use it for. How do I start it?
| You have
$$\frac{a - \sqrt{a^2-x^2}- \frac{x^2}{2}}{x^4} = \frac{a-a\sqrt{1-\frac{x^2}{a^2}} - \frac{x^2}{2}}{x^4} = \frac{a-a\left( 1- \frac{x^2}{2a^2}-\frac{x^4}{8a^4} + o(x^5)\right) - \frac{x^2}{2}}{x^4}$$
$$= \frac{\frac{x^2}{2} \left( \frac{1}{a}-1\right)+\frac{x^4}{8a^3} + o(x^5) }{x^4}$$
You see that if $a \neq 1$, the limit is infinite.
For the limite to be finite, you must have $a=1$, and then you have $$\frac{1 - \sqrt{1^2-x^2}- \frac{x^2}{2}}{x^4} = \frac{1}{8} + o(x)$$
which tends to $\frac{1}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Show for any odd prime $p\geq 5,$ $(-3/p)=1$ or $ -1$ Show for any odd prime $$p\geq 5,$$
$$\left ( \frac{-3}{p} \right ) =\begin{cases}
1 & \text{ if } p\equiv 1,-5\pmod{12} \\
-1& \text{ if } p\equiv -1,5\pmod{12}
\end{cases}$$
So far I have that
(1) Let $$p\equiv 1\pmod{4}$$ then $$p\equiv 1\pmod{3}$$ to get
$$\left ( \frac{-3}{p} \right )=-\left ( \frac{p}{3} \right )=-\left ( \frac{1}{3} \right )=-1$$
(2)Let $$p\equiv 1\pmod{4}$$ then $$p\equiv 2\pmod{3}$$ to get
$$\left ( \frac{-3}{p} \right )=-\left ( \frac{p}{3} \right )=-\left ( \frac{2}{3} \right )=1$$
(3)Let $$p\equiv 3\pmod{4}$$ then $$p\equiv 1\pmod{3}$$ to get
$$\left ( \frac{-3}{p} \right )=\left ( \frac{p}{3} \right )=\left ( \frac{1}{3} \right )=1$$
(4)Let $$p\equiv 3\pmod{4}$$ then $$p\equiv 2\pmod{3}$$ to get
$$\left ( \frac{-3}{p} \right )=\left ( \frac{p}{3} \right )=\left ( \frac{2}{3} \right )=-1$$
After solving CRT systems I get,
$$\left ( \frac{-3}{p} \right ) =\begin{cases}
1 & \text{ if } p\equiv 5,-5\pmod{12} \\
-1& \text{ if } p\equiv 1,-1\pmod{12}
\end{cases}.$$
So I'm not sure where I'm messing up. Any help would be appreciated.
| You messed up some calculations. When $p\equiv1\pmod4$, $\left(\dfrac{-3}p\right)=\left(\dfrac{-1}p\right)\left(\dfrac{3}p\right)=1\left(\dfrac p3\right)$,
so in those cases $\left(\dfrac{-3}p\right)=1$ when $p\equiv1\pmod3$ and $\left(\dfrac{-3}p\right)=-1$ when $p\equiv2\pmod3$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
limit of $\lim\limits_{x \to \infty} (1+ \frac{\pi}{2} - \arctan(x))^x$ Question: Limit of $\lim\limits_{n \to \infty} (1+ \frac{\pi}{2} - \arctan(x))^x$
The first things I notice are : $\lim\limits_{x \to \infty} \arctan(x) = \frac{\pi}{2}$ and the limit looks something of the form $(1 + \frac{1}{x})^x$. Unfortunetly, I can not seem to apply these ideas to solve the limits. I am not sure if it is right:
$$\ln(y) = \lim\limits_{x \to \infty} x \ln(1 + \frac{\pi}{2} - \arctan(x)) = \lim\limits_{x \to \infty}\frac{\ln(1+\frac{\pi}{2}- \arctan(x))}{\frac{1}{x}}$$
Apply l'hopital's rule ... ?
Could someone confirm this approach is right, or if wrong provide a correct approach?
| $$A=\left(1+ \frac{\pi}{2} - \tan^{-1}(x)\right)^x=\left(1+\tan ^{-1}\left(\frac{1}{x}\right)\right)^x$$
$$\log(A)=x \log\left(1+\tan ^{-1}\left(\frac{1}{x}\right)\right)$$
By Taylor
$$\tan ^{-1}\left(\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{3 x^3}+O\left(\frac{1}{x^5}\right)$$
$$\log\left(1+\tan ^{-1}\left(\frac{1}{x}\right)\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{12
x^4}+O\left(\frac{1}{x^5}\right)$$
$$\log(A)=1-\frac{1}{2 x}+\frac{1}{12 x^3}+O\left(\frac{1}{x^4}\right)$$
$$A=e^{\log(A)}=e \left(1-\frac{1}{2 x}+\frac{1}{8 x^2}+\frac{1}{16
x^3} \right)+O\left(\frac{1}{x^4}\right)$$
Edit
Consider $x=\frac {11}{24}\pi$ (this is quite far away from $\infty$) for which the arctangent is $\left(1+\sqrt{2}\right) \left(\sqrt{2}+\sqrt{3}\right)$.
the exact value is $1.97993$ while this truncated expression gives $1.99516$.
In fact, the relative error is smaller than $0.01$% if $x\geq3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $\lim_{x\rightarrow +\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) = 0$ without L'Hôpital's rule How would you solve the limit
$$\lim_{x\rightarrow +\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) = 0$$
without using L'Hôpital's rule?
| \begin{align}
\lim_{x\to+\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right)
&= \lim_{x\to+\infty}\left(\frac{1}{x}\cdot\frac{1+x^2}{x+1}\right)\left[\frac{x+1}{1+x^2}\log\left(\frac{x+1}{1+x^2}\right)\right]=\\
&= \lim_{x\to+\infty}\frac{1+x^2}{x^2+x}\cdot\lim_{y\to0}y\log y=1\cdot0=0\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$ $A$ is a matrix with elements
$$\begin{pmatrix}
\frac{2}{3} & -\frac{2}{3}& -\frac{1}{3} \\
-\frac{2}{3}& 1 & \frac{2}{3} \\
\frac{4}{3} & 0 & -\frac{1}{3}
\end{pmatrix}$$
Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $$\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$$
A's eigenvalues are $-\frac{1}{3}$,1 and $\frac{2}{3}$ and eigenvectors are $\left(
\begin{array}{c}
0 \\
-\frac{1}{2} \\
1
\end{array}
\right),\left(
\begin{array}{c}
1 \\
-1 \\
1
\end{array}
\right),\left(
\begin{array}{c}
\frac{3}{4} \\
-\frac{1}{2} \\
1
\end{array}
\right)$ for each.
I tried to get $A^n$ at first then got $$\begin{pmatrix}
(\frac{2}{3})^n & -2+\frac{2^{n+1}}{3}& 0 \\
-\frac{2+2^{n+1}}{3^{n+1}}& \frac{1-2^{n+2}}{3^{n+1}}-2 & \frac{2}{3^n} \\
\frac{4+2^{n+2}}{3^{n+1}} & \frac{2-2^{n+3}}{3^{n+1}} & -\frac{1}{3^n}
\end{pmatrix}$$
Then I tried find $$\displaystyle \lim_{ n \to \infty }(\frac{2}{3})^nx_1+(-2+\frac{2^{n+1}}{3})x_2=x_1 \\\displaystyle \lim_{ n \to \infty }-\frac{2+2^{n+1}}{3^{n+1}}x_1+(\frac{1-2^{n+2}}{3^{n+1}}-2)x_2+\frac{2}{3^n}x_3=x_2\\\ \displaystyle \lim_{ n \to \infty }\frac{4+2^{n+2}}{3^{n+1}}x_1+\frac{2-2^{n+3}}{3^{n+1}}x_2-\frac{1}{3^n}x_3=x_3$$
But I could not lead meaningful conclusion from these equations but $(0,0,0)$.
Where did I get wrong?
| Let $e_1, e_2, e_3$ be three linearly independent eigenvectors which are of unit norm. (You have shown that these exist.)
Suppose that they are corresponding to the eigenvalues $1, -1/3, 2/3$, respectively.
Let $\mathbf{x} \in \Bbb R^3$ be any such with the property you stated. Since $\{e_1, e_2, e_3\}$ form a basis, we can write $$\mathbf{x} = a_1e_1 + a_2e_2 + a_3e_3$$ for some reals $a_1, a_2, a_3$.
Then, applying $A^n$ to both sides, we get $$A^n\mathbf{x} = 1^na_1e_1 + \left(-\dfrac{1}{3}\right)^n a_2e_2 + \left(\dfrac{2}{3}\right)^na_3e_3.$$
As $n \to \infty$, the last two terms on the right tend to $0$ and thus, we get $$\lim_{n \to \infty}A^n\mathbf{x} = a_1e_1.$$
Thus, $$\mathbf{x} = a_1e_1.$$ Thus, we get that $a_2 = a_3 = 0.$
Conversely, it is easy to check that any $\mathbf{x} \in \operatorname{span}\{e_1\}$ does have the property you want.
Thus, we conclude that $\operatorname{span}\{e_1\}$ is precisely the set of all such vectors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $a = \frac{1+\sqrt5}2$, then what is $a^{18} + \frac{323}{a^6}$? My question is this:
If $a = \frac{1+\sqrt5}2,\frac{1-\sqrt5}2$, then what is $a^{18} + \frac{323}{a^6}$?
It is an AMC style question and is timed, so I will not be able to use solutions with a lot of case work.
I defined $a$ from part a of a 2 part question and would like to know how to do this without the use of a calculator.
|
Lemma. Define $$L_n = a^n + (-a)^{-n}.$$ Then $L_n$ is a Lucas number satisfying the recurrence $$L_{n+1} = L_n + L_{n-1}, \\ L_0 = 2, \quad L_1 = 1.$$
Proof. Since $a^2 = a + 1$, or equivalently, $$a - a^{-1} = 1,$$ we have $$(a^n + (-a)^{-n})(a - a^{-1}) = a^{n+1} + (-a)^{-(n+1)} - \left(a^{n-1} + (-a)^{-(n+1)}\right),$$ hence $$L_n L_1 = L_{n+1} - L_{n-1}.$$
Along with $L_0 = a^0 + (-a)^0 = 2$, this proves the claim.
We next use this recursion to compute the table
$$\begin{array}{c|ccccccccccccc}
n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
L_n & 2 & 1 & 3 & 4 & 7 & 11 & 18 & 29 & 47 & 76 & 123 & 199 & 322
\end{array}$$
Therefore
$$\begin{align}
a^{18} + 323 a^{-6}
&= a^6 (a^{12} + 323a^{-12}) \\
&= a^6(a^{12} + a^{-12} + 322a^{-12}) \\
&= a^6 (L_{12} + 322 a^{-12}) \\
&= a^6 (322 + 322 a^{-12}) \\
&= 322 (a^6 + a^{-6}) \\
&= 322 L_6 \\
&= 322 (18) \\
&= 5796.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to prove $\frac{a}{b}=\frac{c}{d} \implies \frac{a+c}{b+d}=\frac{a-c}{b-d}$? My question:
How do I prove the following property of ratio?
$$\frac{a}{b}=\frac{c}{d} \implies \frac{a+c}{b+d}=\frac{a-c}{b-d} \tag{1}$$
$a,b,c,d \in \mathbb{R} \backslash \{ 0 \}$
I want to use the result in argument below.
Use the sine rule to establish the following identities for triangles:
$$\frac{a+b}{c}=\frac{\sin(A) +\sin(B)}{\sin(C)}$$ and \begin{equation}\frac{a-b}{c}=\frac{\sin(A) -\sin(B)}{\sin(C)}\end{equation}
To prove both these identities the equations rearrange to the following
$$\frac{\sin(C)}{c}=\frac{\sin(A)+\sin(B)}{a+b}=\frac{\sin(A)-\sin(B)}{a-b} $$
From here, $(1)$ would complete the argument.
Thanks in advance
| But why not use this simple solution:
$$
\frac{a}{b}=\frac{c}{d}=x \implies a=bx , c = dx
$$
$$
\begin{align}
\frac{a+c}{b+d}&=\frac{bx+dx}{b+d}\\
\\
&=x\\
\\
\frac{a-c}{b-d}&=\frac{bx-dx}{b-d}\\
\\
&=x
\end{align}
$$
For the desired result, use sine rule
$$
\begin{align}
a&=2R\sin{(A)}\\
b&=2R\sin{(B)}\\
c&=2R\sin{(C)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $r$ such that the equation $x^4+x^2(1-2r)-2x+1=0$ has only one real solution I'm looking for $r$ such that the equation $$x^4+x^2(1-2r)-2x+1=0$$
has only one real solution. I've made some attempts to this problem, but it seems that I even didn't get close to the solution. The approximation for $r$ is 0.3347498
Is it possible to find analytic solution for $r$ and if yes then how?
Thanks for all the help.
| We can write the function in following form where $c>0$.
$(x-a)^2((x-b)^2+c)=0$
$(x^2-2ax+a^2)(x^2-2bx+b^2+c)=0$
$x^4-(2a+2b)x^3+(a^2+b^2+4ab+c)x^2-(2ab^2+2ac+2a^2b)x+a^2b^2+a^2c=0$
Therefore,
$b=-a, c={1-a^4\over a^2} > 0$
$2ab^2+2ac+2a^2b=2a^3+{2-2a^4\over a}-2a^3=-2,a^4-a-1=0$
$1-2r=a^2+b^2+4ab+c=2a^2-4a^2+{1-a^4\over a^2}={1-3a^4\over a^2}=-{3a+2\over a^2}$
$r={a^2+3a+2\over 2a^2}$
We take the real root where $a^4<1$ in $a^4-a-1=0$ which is around $-0.72449$ and gives $r$ around $0.33476$. And yes, there is a closed formula for quartic equation using radicals so $r$ also has an exact formula as well, but it should be pretty messy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum value of $f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$ Find the minimum value of
$$f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$$ for $x > 0.$
I'm not sure how to start this problem. I think AM-GM is the best way, but I'm not sure.
| Here is your solution by A.M-G.M
$$f(x)=x+\frac{x}{x^2+1}+\frac{x(x+4)}{x^2+2}+\frac{2(x+2)}{x(x^2+2)}$$
$$f(x)=x+\frac{x}{x^2+1}+\frac{x^2+4x}{x^2+2}+\frac{2}{x^2+2}+\frac{4}{x(x^2+2)}$$
$$f(x)=\frac{x(x^2+2)}{x^2+1}+\frac{x^2}{x^2+2}+\frac{4x}{x^2+2}+\frac{2}{x^2+2}+\frac{4}{x(x^2+2)}$$
$$f(x)=\frac{x(x^2+2)}{x^2+1}+1+\frac{4x}{x^2+2}+\frac{4}{x(x^2+2)}$$
$$f(x)=1+\frac{x(x^2+2)}{x^2+1}+\frac{4(x^2+1)}{x(x^2+2)}\geq 1+2\sqrt{\frac{x(x^2+2)}{x^2+1}.\frac{4(x^2+1)}{x(x^2+2)}}$$
$$f(x)\geq 5$$
Equality occurs when$$\frac{x(x^2+2)}{x^2+1}=\frac{4(x^2+1)}{x(x^2+2)}$$
or$$x^3-2x^2+2x-2=0$$ which has a real root lying between $1$ and $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Does the series $\frac{2}{4-1}+\frac{4}{16-1}+\dots+\frac{2k}{4k^2-1}$ have a sum up to $\infty$? If there's one, then how do I find the sum?
I tried to rewrite the common term into partial fractions to see if it's a telescoping series but got a dead end there. How can I proceed now?
| To expand on @TheSilverDoe's answer...
The sum is approximately $$\sum_{k=1}^\infty \frac{1}{2k}.$$
Note that this is $$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\dots$$
Group the sum as follows:
$$\left(\frac{1}{2}\right)+\left(\frac{1}{4}\right)+\left(\frac{1}{6}+\frac{1}{8}\right)+\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}\right)+\dots$$
$$\frac{1}{2}\ge\frac{1}{4}$$
$$\frac{1}{4}\ge\frac{1}{4}$$
$$\frac{1}{6}+\frac{1}{8}\ge\frac{1}{4}$$
$$\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}\ge\frac{1}{4}$$
And so on. The reason for the last two is because $\frac{1}{6}>\frac{1}{8}$, and $2\cdot \frac{1}{8} = \frac{1}{4}$. Same logic for the next one, and on and on, for infinity.
Therefore, the series is divergent and leads to $\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that$(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ Prove that $(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ where $\omega$ is the cube root of unity
My attempt:
$(1+\omega^n+\omega^{2n})=0$
$\Rightarrow (1-\omega^n+\omega^{2n})=-2\omega^n$
$\Rightarrow \prod_{n=1 \to 2n}(-2\omega^n) =2^{2n}\omega^{1+2+3…2n}$
$\Rightarrow \prod_{n=1 \to 2n}(-2\omega^n) =2^{2n}\omega^{n(2n+1)}$
If $n$ is $3k$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{3m}=2^{2n}$
If $n$ is $3k+1$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{3m}=2^{2n}$
If $n$ is $3k+2$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{(3k+2)(6k+5)} =2^{2n}\omega^{(3m+1)}=2^{2n}\omega^{3m}\omega=2^{2n}\omega$
What have I done wrong here?
| Hint : Evaluate first few factors and see.
$$
1st : (1-\omega+\omega^2)=-2\omega \\
2nd : (1-\omega^2+\omega^4)=-2\omega^2 \\
1st \times 2nd = \ldots \\
3rd : (1-\omega^4+\omega^8)=-2\omega \\
4th : (1-\omega^8+\omega^{16})= \ldots
$$
Can you complete?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is this alternative proof of the inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ correct? Prove that for all positive real numbers:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}$$
This is same as this question but a different approach is used there whereas I want to verify my approach to this problem.
My Approach:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\Big(\dfrac{a}{b+c}+1\Big)+\Big(\dfrac{b}{c+a}+1\Big)+\Big(\dfrac{c}{a+b}+1\Big)-3$$
$$=(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3$$
By AM-HM inequality:
$$\dfrac{3}{\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}}\leq\dfrac{2(a+b+c)}{3}\Rightarrow (a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]\geq \dfrac{9}{2}$$
$$(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3\geq \dfrac{3}{2}$$
$\therefore \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}\space \forall\ a,b,c\in \mathbb R$ and $a,b,c>0$
Please check this approach and provide suggestions. Also please provide alternative solutions if available.
THANKS
| Your solution is right.
Also, SOS helps:
$$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum_{cyc}\frac{2a-b-c}{2(b+c)}=$$
$$=\sum_{cyc}\frac{a-b-(c-a)}{2(b+c)}=\sum_{cyc}\left(\frac{a-b}{2(b+c)}-\frac{c-a}{2(b+c)}\right)=$$
$$=\sum_{cyc}\left(\frac{a-b}{2(b+c)}-\frac{a-b}{2(c+a)}\right)=\sum_{cyc}(a-b)\left(\frac{1}{2(b+c)}-\frac{1}{2(c+a)}\right)=$$
$$=\sum_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0.$$
Now we see that the starting inequality is true for any reals $a$, $b$ and $c$ such that $ab+ac+bc>0.$
Also, there is a solution by AM-GM, by C-S, by TL, by $uvw$ and by more and more and more.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
SOS proof for $\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$ I need an SOS(sum of squares) proof for $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
if $a,b,c>0$
I already have a am-gm proof but is there a way to use SOS.
Am-gm proof :
$\frac{a^3}{bc}+b+c\ge 3a$ .....by(AM-GM ineq.)
thus $$\sum \frac{a^3}{bc}+2\sum a \ge 3\sum a$$
or $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
| The idea of the SOS's proof it's the following.
Let $P$ be a symmetric function of three variables $a$, $b$ and $c$ and let we can get:
$$P(a,b,c)=\sum_{cyc}((a-b)Q(a,b,c)-(c-a)Q(a,c,b)).$$
Thus, $$P(a,b,c)=\sum_{cyc}((a-b)Q(a,b,c)-(c-a)Q(a,c,b))=$$
$$=\sum_{cyc}((a-b)Q(a,b,c)-(a-b)Q(b,a,c))=\sum_{cyc}(a-b)(Q(a,b,c)-Q(b,a,c))$$ and if $Q$ is a rational function we obtain a factor $a-b$ again.
There are some expressions, which we need to learn:
$$2a-b-c=a-b-(c-a),$$
$$a^2-bc=\frac{1}{2}((a-b)(a+c)-(c-a)(a+b))$$ and more similar.
This idea helps to prove inequalities by SOS without computer.
I hope now it's clear, how it works:
$$\sum_{cyc}\frac{a^3}{bc}-\sum_{cyc}a=\sum_{cyc}\frac{a^3-abc}{bc}=\frac{1}{2}\sum_{cyc}\tfrac{a((a-b)(a+c)-(c-a)(a+b))}{bc}=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{a(a+c)}{bc}-\frac{b(b+c)}{ca}\right)=\frac{1}{2}\sum_{cyc}\tfrac{(a-b)^2(a^2+b^2+ab+ac+bc)}{abc}\geq0.$$
We saw before that we can get the expession $a^2-bc$ and after this we ended the proof.
Another example.
Let we need to prove the Nessbitt:
$$\sum_{cyc}\frac{a}{b+c}\geq\frac{3}{2}.$$
We see that easy to get the expession $2a-b-c$ and it ends the proof by SOS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
A hint on finding all solutions of $S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$? I've been working at this for awhile but I haven't been able to figure out the right approach. The question is to find all values of $S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$ for $a,b,c,d > 0$. Anyone have a hint (no solutions, please) at a more principled way to approach the problem?
I've been just throwing stuff at the wall and trying to find something that sticks.
What I've tried so far:
Plugging in some quick values. Putting $a=b=c=d=1$ yields $S = 4/3$. Taking $a=c=1, b=d=2$ yields $7/5$ so $4/3$ is not the only answer. Note: if $(a,b,c,d)$ yields $k$ then $(sa,sb,sc,sd)$ yields $k$ for any $s > 0$.
If I fix $a=b=c=1$ then $S = 4/3$ regardless of the value of $d$.
If I fix $a=c, b=d$, I can write $b = ka$, which gives $\frac{2}{2k+1} + \frac{2k}{k+2} = S$. This is a quadratic in $k$, with positive discriminant for any value of $S$. But $k$ is not necessarily positive, so I can't claim that all values of $S$ are valid.
Getting some quick bounds: $S > \frac{a}{a+b+c+d} + \frac{b}{a+b+c+d} + \frac{c}{a+b+c+d} + \frac{d}{a+b+c+d} = 1$, and $S < a/a + b/b + c/c + d/d = 4$, so I know $1 < S < 4$. The thought is to either try and tighten these bounds somehow (not sure how to approach this) or figure out how to represent an arbitrary $x$ in this range with some choice of $a,b,c,d$.
Some substitutions: Imposing the constraint $a+b+c+d = 1$ or $u = a+b+d, v = a+b+c, w = b+c+d, x = a+c+d$ come to mind, both attempts at removing the sums from the denominator. Respectively those give $S = \frac{a}{1-c} + \frac{b}{1-d} + \frac{c}{1-a} + \frac{d}{1-b}$ and $S = \frac{u+v+w-2x}{3u} + \frac{u+v+w-2x}{3v} + \frac{v+x+w-2u}{3w} + \frac{u+w+x-2v}{3x}$.
Another way of looking at things: $\frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$ is continuous in $(a,b,c,d)$. And I think we can get arbitrarily close to 1 by setting $a$ very large, and then $c$ very small relative to $b + d$. So 1 is probably the $\inf$ of $S$. If we do the opposite and set $a$ large and $c$ large, we can get close to 2. My suspicion is that $2$ is the $\sup$ but I'm not sure how to prove it.
| For $c=d\rightarrow0^+$ we see that $$S\rightarrow\frac{a}{a+b}+\frac{b}{a+b}=1$$ and
$$\sum_{cyc}\frac{a}{a+b+d}>\sum_{cyc}\frac{a}{a+b+c+d}=1.$$
For $b=d\rightarrow0^+$ we see that $S$ close to $2$.
We'll prove that $$\sum_{cyc}\frac{a}{a+b+d}<2.$$
Indeed, we need to prove that
$$\sum_{cyc}\left(\frac{a}{a+b+d}-1\right)<-2$$ or
$$\sum_{cyc}\frac{b+d}{a+b+d}>2,$$ which is true by C-S:
$$\sum_{cyc}\frac{b+d}{a+b+d}=\sum_{cyc}\frac{(b+d)^2}{(b+d)(a+b+d)}\geq\frac{4(a+b+c+d)^2}{\sum\limits_{cyc}(b+d)(a+b+d)}=$$
$$=\frac{4(a+b+c+d)^2}{\sum\limits_{cyc}(2ab+2a^2+2ac)}=\frac{2(a+b+c+d)^2}{\sum\limits_{cyc}(a^2+ab+ac)}>2.$$
Since $S$ is a continuous function, we got the best estimations:
$$1<S<2.$$
Another way to get $$\sum_{cyc}\frac{b+d}{a+b+d}>2.$$
$$\sum_{cyc}\frac{b+d}{a+b+d}=(b+d)\left(\tfrac{1}{a+b+d}+\tfrac{1}{c+d+b}\right)+(a+c)\left(\tfrac{1}{a+b+c}+\tfrac{1}{a+c+d}\right)>$$
$$>(b+d)\left(\tfrac{1}{a+b+c+d}+\tfrac{1}{a+b+c+d}\right)+(a+c)\left(\tfrac{1}{a+b+c+d}+\tfrac{1}{a+b+c+d}\right)=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the difference of square root of conjugate complex number Find the imaginary part of $\left( {{{\left( {3 + 2\sqrt { - 54} } \right)}^{\frac{1}{2}}} - {{\left( {3 - 2\sqrt { - 54} } \right)}^{\frac{1}{2}}}} \right)$
(1) $-\sqrt 6$
(2) $-2\sqrt 6$
(3) $\sqrt 6$
(4) $6$
My Approach is as follow and none of the answer is matching, I cross checked it
$T = 3 + 2\sqrt { - 54} = 3 + i6\sqrt { 6} \to {I^{st}} - Quadrant - Angle = \theta $
$U = 3 - 2\sqrt { - 54} = 3 - i6\sqrt { 6} \to I{V^{th}} - Quadrant - Angle = - \theta $
$\Rightarrow \left( {{{\left( {3 + i6\sqrt 6 } \right)}^{\frac{1}{2}}} - {{\left( {3 - i6\sqrt 6 } \right)}^{\frac{1}{2}}}} \right)$
$r\cos \theta = 3$ & $r\sin \theta = 6\sqrt 6 \Rightarrow {r^2} = 225 \Rightarrow r = 15 \Rightarrow \tan \theta = 2\sqrt 6 $
$ \Rightarrow \left( {\sqrt {15} {e^{\frac{{i\theta }}{2}}} - \sqrt {15} {e^{ - \frac{{i\theta }}{2}}}} \right) \Rightarrow \sqrt {15} \left( {{e^{\frac{{i\theta }}{2}}} - {e^{ - \frac{{i\theta }}{2}}}} \right) = i\sqrt {15} \left( {2\sin \frac{\theta }{2}} \right)$
$ \Rightarrow \frac{{2\tan \frac{\theta }{2}}}{{1 - {{\tan }^2}\frac{\theta }{2}}} = 2\sqrt 6 \Rightarrow {\tan ^2}\frac{\theta }{2} + \frac{2}{{\sqrt {24} }}\tan \frac{\theta }{2} + \frac{1}{{24}} = \frac{{25}}{{24}} \Rightarrow \left( {\tan \frac{\theta }{2} + \frac{1}{{\sqrt {24} }}} \right) = \frac{5}{{\sqrt {24} }} \Rightarrow \tan \frac{\theta }{2} = \frac{4}{{\sqrt {24} }} = \frac{{\sqrt 2 }}{{\sqrt 3 }}$
$\sin \frac{\theta }{2} = \frac{{\sqrt 3 }}{{\sqrt 5 }} \Rightarrow i\sqrt {15} \left( {2\sin \frac{\theta }{2}} \right) = i\sqrt {15} \left( {2 \times \frac{{\sqrt 2 }}{{\sqrt 5 }}} \right) = 2\sqrt 6 i$
| Let $$z=\sqrt{3+6\sqrt{6}i)}=x+iy ~~~(1)$$
Squaring we get $3+6\sqrt{6}i=x^2-y^2+2ixy$
$$\implies x^2-y^2=3, xy=3\sqrt{6}~~~~(2)$$
$$\bar z=\sqrt{3-6\sqrt{6}i}=x-iy~~~~(3),$$
multiplying the two (1) and (3)we get $$x^2+y^2=\sqrt{9+36.6}=15~~~(4)$$
Using (4) in (2) we get $x=\pm 3, y=\pm \sqrt{6}.$
So $z-\bar z=2y=\pm 2\sqrt{6}.$
Hence, option (B) is one correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$
prove $$a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b),$$$a,b,c>0$
Obviously this is a direct consequence of the third degree schur's inequality.
I was wondering if this could be proved without this theorem ,or uvw but through basic methods like AM-GM,C-S etc.
| Suppose $a \geqslant b \geqslant c,$ we get
$$a^3+b^3+c^3+3abc - \sum ab(a+b)$$
$$=(c^3+b^2c+ca^2-abc-ac^2-bc^2)+(a^3+b^3+4abc-a^2b-ab^2-2ca^2-2b^2c)$$
$$=(a^2+b^2+c^2-ab-bc-ca)c+(a+b-2c)(a-b)^2 \geqslant 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Solving 2-degree equations in 3 variables. We are given 3 equations:
$x^2+\sqrt3 xy + y^2 = 25$
$y^2 + z^2 = 9$
$x^2 +xz+ z^2 = 16$.
$x,y,z$ are positive real numbers.
Then we have to find value of $xy + 2yz + \sqrt3 xz$.
| Based on system of equations being sides of a right angled triangle and a point P
inside the triangle such that -
$\angle BPC = 90^0, \angle APC = 150^0, \angle APB = 120^0 $
and, $AP = x, CP = y, BP = z$
We know area of a triangle is $\frac{1}{2} \times$ length of side 1 $\times$ length of side 2 $\times \sin \theta$
where $\theta$ is the angle between side 1 and side 2.
Now sum of area, $\triangle APC + \triangle BPC + \triangle APB = \triangle XYZ$
$\frac{1}{2}(xy\sin150^0 + yz\sin90^0 + xz\sin 120^0) = \frac{1}{2} \times 3 \times 4$
$xy \times \frac{1}{2} + yz + xz \times \frac{\sqrt3}{2} = 12$
$xy + 2yz + \sqrt3 xz = 24$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding $k$ such that $\frac{3x^2+kx-44}{x^2-121}$ simplifies to $\frac{3x-4}{x-11}$
If the rational expression $\frac{3x^2+kx-44}{x^2-121}$, where $k$ is an element of $\mathbb{W}$, simplifies to $\frac{3x-4}{x-11}$, then the value of $k$ must be?
my work:
$$\frac{3x^2+kx-44}{x^2-121}= \frac{3x-4}{x-11}$$
so the answer = value of $k$ is $11$.
| Notice $x^2-121=(x+11)(x-11)$. Then we have
$$
3x^2+kx-44=(3x-4)(x+11)=3x^2+29x-44.
$$
So $k=29$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Inequality manipulation: $\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}}$ I'm reading Analytic Inequalities by Nicholas D. Kazarinoff. On page 5, we are trying to use induction to prove the inequality
$$ \frac{1}{\sqrt{4n + 1}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n} < \frac{1}{\sqrt{3n + 1}} . $$
For the inductive step, we want to show that it holds for $n + 1$, i.e.
$$ \frac{1}{\sqrt{4n + 5}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 1}{2n} \cdot \frac{2n + 1}{2n + 2} < \frac{1}{\sqrt{3n + 4}} . $$
Kazarinoff says that this is true if
$$ \frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}} $$
is also true. I'm trying to figure out why this is the case, because it's not obvious to me.
Edit: this is completely nonsensical because it's circular.
What I've tried:
\begin{align}
\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} &> \frac{1}{\sqrt{4n + 5}} \\
\implies \frac{1}{\sqrt{4n + 1}} &> \frac{1}{\sqrt{4n + 5}} \cdot \frac{2n + 2}{2n + 1} \\
&< \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n} \\
&> \frac{1}{\sqrt{4n + 1}}
\end{align}
That's a pretty useless result that hasn't gotten me anywhere. Any ideas?
| although an elementary solution is aesthetically preferable (and a good algebraic solution was posted and then for some reason deleted by Soumyadwip Chanda), it may be of interest to see the number-theoretic result posted by OP as a corollary of a more general theorem of real analysis:
suppose that for $a, b \in (0,1)$ we had:
$$
\bigg(\frac{1-a}{1+a}\bigg)^b \lt \frac{1-ba}{1+ba} \tag{1}
$$
then, as a special case with $b=\frac12$ and $a = \frac2{4n+3}$, we would have:
$$
\bigg(\frac{1-\frac2{4n+3}}{1+\frac2{4n+3}}\bigg)^{\frac12} \lt \frac{1-\frac1{4n+3}}{1+\frac1{4n+3}}
$$
which simplifies to:
$$
\sqrt{\frac{4n+1}{4n+5}} < \frac{2n+1}{2n+2}
$$
the formula (1) can be verified by taking the Taylor series for the logarithms of the two sides, which gives:
$$
-2\bigg(ab + \frac{a^3b}3 + \frac{a^5b}5+...\bigg) \lt -2\bigg(ab + \frac{a^3b^3}3 + \frac{a^5b^5}5+...\bigg)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3823022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How does this division work? $\frac{\;\frac{6^6}{1}\;}{2^{-3}}\cdot2^{-10}$ I came across this division and can't wrap my head around how it is solved:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}$$
They subtract the exponent of $2^{-10}$ from the denominator's exponent $2^{-3}$:
$$2^{-3-(-10)}$$
Which gives us:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^7}$$
If anyone knows what is actually being done here I would appreciate it!
| First off, the way in which the problem is typeset is almost intended to cause confusion. Adding some parentheses and not using \dfrac everywhere should make it easier to understand the expression:
$$\frac{\left(\frac{6^6}{1}\right)}{2^{-3}} \cdot 2^{-10}. $$
Division is the inverse operation of multiplication and so, rather than thinking of the problem as a division problem, it might be easier to think of it as the multiplication of several fractions. Specifically,
$$\frac{\left(\frac{6^6}{1}\right)}{2^{-3}} \cdot 2^{-10}
= 6^6 \cdot \frac{1}{2^{-3}} \cdot 2^{-10}. $$
One of the fundamental properties of exponents is that (under appropriate hypotheses) $a^{-n} = \frac{1}{a^n}$, and so we can further simplify this expression to
$$ 6^6 \cdot \frac{1}{2^{-3}} \cdot 2^{-10}
= 6^6 \cdot 2^{3} \cdot \frac{1}{2^{10}}
= 6^6 \cdot \frac{2^{3}}{2^{10}}. $$
Finally, another property of exponents is that
$$ \frac{a^m}{a^n} = a^{m-n}. $$
Applying this to the last result gives
$$ 6^6 \cdot \frac{2^{3}}{2^{10}}
= 6^6 \cdot 2^{3-10}
= 6^6 \cdot 2^{-7}
\overset{\text{or}}{=} \frac{6^6}{2^7}. $$
Personally, I would also probably use the fact that $(ab)^n = a^n b^n$ to rewrite $6^6 = 2^6\cdot 3^6$, and conclude by writing
$$ 6^6 \cdot 2^{-7} = 2^6\cdot 3^6 \cdot 2^{-7}
= 2^{6-7} \cdot 3^6
= 2^{-1} \cdot 3^6
\overset{\text{or}}{=} \frac{3^6}{2}, $$
but perhaps this is not part of the intended exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
Epsilon Delta proof for rational function containing radicals I am having trouble constructing the delta-epsilon proof for this limit. I am currently trying to find the delta in terms of epsilon:
$$\lim_{x \to 1} \frac{x^3-1}{\sqrt{x}-1}$$
So far I have gotten to
$$6-\epsilon < (x-1)*\frac{x^2+x+1}{\sqrt{x}-1} < 6+\epsilon$$
I am having trouble bounding the fraction $\frac{x^2+x+1}{\sqrt{x}-1}$ so that I can isolate $x-1$ and find the delta. It has a vertical asymptote at x=1, which makes it difficult to find an upper/lower bound.
Am I on the right track to this problem? If not, how would I go about finding the delta?
| We have that for $x\neq 1$
$$\frac{x^3-1}{\sqrt{x}-1}=\frac{(x-1)(x^2+x+1)}{\sqrt{x}-1}=$$
$$=\frac{(\sqrt{x}-1)(\sqrt{x}+1)(x^2+x+1)}{\sqrt{x}-1}=(\sqrt{x}+1)(x^2+x+1)$$
and since for $|x-1|<1 \iff 0<x<2$, using triangle inequality and that $\forall a,b \:a\ge b\ge 0 \implies |x^a-1|\le |x^b-1|$, we obtain
$$\left|\frac{x^3-1}{\sqrt{x}-1}-6\right|=\left|(\sqrt{x}+1)(x^2+x+1)-6\right|\le\left|x^2\sqrt{x}+x^2+x\sqrt x+x+\sqrt{x}-5\right|\le$$
$$\leq |x^2\sqrt{x} -1|+\ldots +|\sqrt{x}-1| \leq 5|x^3 - 1| \le 5|x-1||x^2+x+1|\le 35|x-1|<\varepsilon$$
and it suffices to assume $\delta =\min\left(1,\frac{\varepsilon}{35}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Convergence of the series $ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$ The series $ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$
(A) is uniformly convergent for all x
(B) is convergent for all x but the convergence is not uniform
(C) is convergent only for $|x|\le \frac{1}{2}$ but the convergence is not uniform
(D) is uniformly convergent on $[-\frac{1}{2},\frac{1}{2}]$
My approach is if we take $ f(x)=\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$
then $$ \int f(x)dx=\log((1+x^2)(1+x^4)(1+x^8)...)=\log\left(\frac{1-x^{2^n}}{1-x^2}\right)$$
Hereafter, I am stuck.
| We have
$$f(x)=\sum_{k=1}^\infty \frac{2^kx^{2^k-1}}{1+x^{2^k}}$$
and by root test
$$\sqrt[k]{\frac{2^kx^{2^k-1}}{1+x^{2^k}}}=2\sqrt[k]{\frac{x^{2^k-1}}{1+x^{2^k}}}$$
and for $|x|<1$
$$2\sqrt[k]{\frac{x^{2^k-1}}{1+x^{2^k}}} \to 0$$
$|x|\ge1$
$$2\sqrt[k]{\frac{x^{2^k-1}}{1+x^{2^k}}} \to 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Alternative approach for proving that for any $x\in\mathbb R^+$, $x^2+3x+\frac{1}{x} \ge \frac{15}{4}$. Let $x \in \mathbb{R^+}$. Prove that:
$$x^2+3x+\frac{1}{x} \ge \frac{15}{4}.$$
While this is indeed easily proven using derivatives, where the minimum is obtained when $x=\frac{1}{2}$, is it possible to prove it through other means, say by AM-GM? Any hints would be much appreciated.
| If we know equality when $x=\frac{1}{2}$ then find AM-GM solution very easy.
For $x=\frac 12,$ then
$$x^2 = \color{blue}{\frac{1}{4}}, \quad 3x = \frac{3}{2} = \color{red}{6} \cdot \color{blue}{\frac{1}{4}}, \quad \frac 1x = 2 = \color{red}{8} \cdot \color{blue}{\frac{1}{4}}.$$
Now, using the AM-GM inequality, we have
$$x^2+3x+\frac1x = x^2 + 6 \cdot \frac x 2+8 \cdot \frac{1}{8x} \geqslant 15\sqrt[15]{x^2\left(\frac x 2\right)^6\left(\frac{1}{8x}\right)^8}=\frac{15}{4}.$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for even $n$? How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for all positive even integer $n$?
Here are my efforts:
Let $f(n)=5^{2n+1}-3^{2n+1}-2$.
For $n=2$,
$$f(1)=5^{2\cdot 2+1}-3^{2\cdot2+1}-2=2880=30\times 96$$
So, the statement is true for $n=2$.
Assume the statement is true for some integer $k\geq 2$.
$$f(k)=5^{2k+1}-3^{2k+1}-2=30t$$
For $n=k+2$,
\begin{align}
f(k+2)&=(5^{2(k+2)+1})-(3^{2(k+2)+1})-2\\
&=(5^4)\cdot (5^{(2k+1)})-(3^4)\cdot (3^{(2k+1)})-2
\\&=30(80t)+160+544\cdot (5^{2k+1})
\end{align}
I'm stuck here. Any help would be appreciated!
| Since $$5^{2n+1}-3^{2n+1}-2=5^{2n+1}-5-3\cdot3^{2n}+3=5^{2n+1}-5-3\left(81^{\frac{n}{2}}-1\right)=$$$$=5^{2n+1}-5-3\cdot80\left(81^{\frac{n}{2}-1}+...+1\right),$$
we see that $5^{2n+1}-3^{2n+1}-2$ is divisible by $5$.
Also, $$5^{2n+1}-3^{2n+1}-2=5^{2n+1}+1-\left(3^{2n+1}+3\right)=(5+1)(5^{2n}-...+1)-\left(3^{2n+1}+3\right),$$
which says that $5^{2n+1}-3^{2n+1}-2$ is divisible by $6$.
Id est, $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that $\lim_{(x,y)\to(1,2)}\frac{x}{x+y}=\frac{1}{3}$ How to prove that $$\lim_{(x,y)\to(1,2)}\frac{x}{x+y}=\frac{1}{3}$$ with $\varepsilon-\delta$ limit definition?
I know so far that $\forall \varepsilon > 0 \ \exists \delta > 0: 0 < d((x, y), (1, 2)) < \delta \Rightarrow \Big|\frac{x}{x+y} - \frac{1}{3}\Big| < \varepsilon$, so we get
$d = \sqrt{|x-1|^2 + |y-2|^2} \Rightarrow d^2 = |x-1|^2 + |y-2|^2 \Rightarrow \begin{cases}
|x - 1| < d < \delta\\
|y - 2| < d < \delta\\
\end{cases}$
But I don't know how to express $\Big|\frac{x}{x+y} - \frac{1}{3}\Big| = \Big|\frac{3x-(x+y)}{3(x+y)}\Big| = \Big|\frac{3x-x-y}{3x+3y}\Big| = \Big|\frac{2x-y}{3x+3y}\Big|$ using $|x-1|$ and $|y-2|$.
Can anyone help me with that?
| By $|x-1|<\delta$ and $|y-2|<\delta$ assuming wlog $\delta\le\frac12$ we have
$$\left|\frac{x}{x+y} - \frac{1}{3}\right|=\left|\frac{2x-y}{x+y}\right|=\left|\frac{2(x-1)-(y-2)}{x+y}\right| \le 2\left|\frac{x-1}{x+y}\right|+\left|\frac{y-2}{x+y}\right|\le$$
$$\le 2\left|\frac{x-1}{x+y}\right|+\left|\frac{y-2}{x+y}\right| \le|x-1|+\frac12|y-2|=\frac32 \delta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
how to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
How to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
My attempt:
\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\frac{\pi}{3} - x\right) \\
&= \frac{1}{\sin x} - \frac{1}{\sin\left(\frac{\pi}{3} + x\right)} + \frac{1}{\sin\left(\frac{\pi}{3} -x\right)} \\
&= \frac{\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x }{\sin x \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right)} \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(\sin x - \sin\left(\frac{\pi}{3} + x\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(2\sin\frac{-\pi}{6}\cos\left(x + \frac{\pi}{6}\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\right) \\
\end{align}
How should I proceed? Or did I make some mistakes somewhere? Thanks in advance.
| If $\csc3x=\csc3y\iff \sin3y=\sin3x$
$$3y=3x+(-1)^nn\pi$$ where $n$ is any integer
$y=x+\dfrac{2n\pi}3; n=-1,0,1$
Now as $\sin3y=3\sin y-4\sin^3y,$
the roots of $$4\sin^3y-3\sin y+\sin3x=0$$ are
$\sin\left(x-\dfrac{2\pi}3\right)=\sin\left(-\pi+\left(x+\dfrac\pi3\right)\right)=-\sin\left(x+\dfrac\pi3\right)=x_1,$
$\sin x=x_2,$
$\sin\left(x+\dfrac{2\pi}3\right)=\sin\left(\pi-\left(\dfrac\pi3-x\right)\right)=\left(\dfrac\pi3-x\right)=x_3$
By Vieta's formula,
$$\dfrac1{x_1}+\dfrac1{x_2}++\dfrac1{x_3}=\dfrac{x_1x_2+x_2x_3+x_3x_1}{x_1x_2x_3}=\dfrac{\dfrac{-3}4}{-\dfrac{\sin3x}4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right) $ I am having trouble finding the infinite sum
$$
\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right).
$$
I know that
$$
\lim_{n\to\infty}\frac1{(3n+1)(3n+4)} =0,
$$
but I have no ideas for a further solution.
| You want to compute $$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac{1}{(3i+1)(3i+4)}.$$
Using partial fractions we want to find constants $A$ and $B$ with $$\frac{1}{(3n+1)(3n+4)}=\frac{A}{3n+1}+\frac{B}{3n+4}=\frac{A(3n+4)+B(3n+1)}{(3n+1)(3n+4)}$$
So comparing numerators we want $1=A(3n+4)+B(3n+1)$. Then choosing $n=-\frac{1}{3}$ we obtain $1=3A$ so $A=\frac{1}{3}$ and similarly $B=-\frac{1}{3}$.
We then have $$\frac{1}{(3n+1)(3n+4)}=\frac{1}{3}\big(\frac{1}{3n+1}-\frac{1}{3n+4}\big)$$
So lets see what happens to the partial sum $$\sum_{i=1}^{n}\frac{1}{(3i+1)(3i+4)}=\frac{1}{3}\sum_{i=0}^{n}\big[\frac{1}{3i+1}-\frac{1}{3i+4}\big]$$
$$=\frac{1}{3}\big[(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})+\frac{1}{10}+...+(\frac{1}{3n-2}-\frac{1}{3n+1})+(\frac{1}{3n+1}-\frac{1}{3n+4})\big]$$
$$=\frac{1}{3}\big[\frac{1}{4}-\frac{1}{3n+4}\big]$$
Thus taking the limit we obtain
$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac{1}{(3i+1)(3i+4)}$$
$$=\lim_{n\rightarrow\infty}\frac{1}{3}\big[\frac{1}{4}-\frac{1}{3n+4}\big]=\frac{1}{3}\cdot\frac{1}{4}=\frac{1}{12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\lim_{x\to \infty} \frac{e^{\frac{1}{x^2}}-1}{2\tan^{-1} (x^2)-\pi}$ $$\lim_{x\to \infty} \frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \frac{1}{(2\tan^{-1}(x^2)-\pi)x^2}$$
$$=\lim_{x\to \infty} \frac{1}{x^2(2\tan^{-1} x^2 -\pi)}$$
How do I proceed?
| Apply l'Hôpital's rule:
$$L=\lim\limits_ {x \to \infty} \dfrac {x^{-2}} {2\arctan x^2 -\pi}=\lim\limits_ {x \to \infty} \dfrac {-2x^{-3}} {2\dfrac {2x}{x^4+1}}$$
$$L==\lim\limits_ {x \to \infty} -{\dfrac {x^4}{2x^4}}=-\dfrac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the minimum of $P = (a - b)(b - c)(c - a)$
Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
$$P = (a - b)(b - c)(c - a)$$
My solution:
*
*We have:
$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$
$$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$
$$\implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 12$$
*
*Using AM-GM Inequality, we have:
$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$
$$\implies 3 \sqrt[3]{P^2} \leq 12$$
$$\implies -8 \leq P \leq 8$$
*
*Therefore, $\min P = -8$
Is this solution correct? If not, then why?
| Using $x=a-b,y=b-c$, then $-(x+y)=c-a$, so $P=-xy(x+y)=-(x^2y+xy^2)$.
The condition $(a−b)^2+(b−c)^2+(c−a)^2=12$ becomes $g=x^2+y^2+xy=6$.
Now using Lagrange multipliers:
\begin{align}
\nabla P&=-\langle2xy+y^2,x^2+2xy\rangle\\
&=-\langle y(2x+y),x(x+2y)\rangle\\
\\
\nabla g&=\langle2x+y,x+2y\rangle
\end{align}
Now letting
$$\nabla P=\lambda\cdot\nabla g$$
$$\lambda=-y,\,\lambda=-x$$
So $x=y$, plugging this back in $g$:
$$3x^2=6$$
$$x=\pm\sqrt2$$
Then the extreme values of $P$ are:
$$P=-2\cdot\pm2\sqrt2=\pm4\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
For any real positive numbers $a, b, c$, prove that $3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq abc(a+b+c)^3$ My progress is that I applied Hölder’s for this,
$3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} \geq (a+b+c)^3$
whereas $3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} = (1+1+1)(a^2b+b^2c+c^2a)(\frac{b}{c} + \frac{c}{a} + \frac{a}{b})$
And I was lost here. Can anyone check or continue this proof?
| Now, you can use Holder so:
$$(1+1+1)\left(a^2b+b^2c+c^2a\right)\left(\frac{b}{c} + \frac{c}{a} + \frac{a}{b}\right)=\sum_{cyc}1\sum_{cyc}a^2b\sum_{cyc}\frac{a}{b}\geq$$
$$\geq\left(\sum_{cyc}\sqrt[3]{1\cdot a^2b\cdot\frac{a}{b}}\right)^3=(a+b+c)^3.$$
Another way.
We need to prove that:
$$\sum_{cyc}(3a^3b^3+3a^4bc+3a^2b^2c^2)\geq\sum_{cyc}(a^4bc+3a^3b^2c+3a^3c^2b+2a^2b^2c^2)$$ or
$$\sum_{cyc}(3a^3b^3+2a^4bc-3a^3b^2c-3a^3c^2b+a^2b^2c^2)\geq0,$$ which is true by Schur:
$$3\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0$$ and Muirhead:
$$\sum_{cyc}(2a^4bc-2a^2b^2c^2)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating a definite integral as the limit of a Riemann sum I'm having trouble evaluating the following problem using the limit of a Riemann sum:
$\int_1^4x^2-4x+2dx$
Using $\lim_{n->\infty}\sum_{i=1}^n f(x_i)\Delta x$ where $a=1$ and $b=4$, $\Delta x = \frac{b-a}{n}$, and $x_i = a + i\Delta x$,
I come up with:
$\lim_{n->\infty}\sum_{i=1}^n[\frac{(3i+n)^2}{n^2}-\frac{12i+4n}{n}+2][\frac{3}{n}]$
which I simplify:
$\lim_{n->\infty}\sum_{i=1}^n[\frac{27i^2+18ni+3n^2}{n^3}-\frac{36i+12n}{n^2}+\frac{6}{n}]$
$\lim_{n->\infty}\sum_{i=1}^n[\frac{27i^2+18ni+3n^2}{n^3}-\frac{36ni+12n^2}{n^3}+\frac{6n^2}{n^3}]$
$\lim_{n->\infty}\sum_{i=1}^n[\frac{27i^2-18ni-3n^2}{n^3}]$
$\lim_{n->\infty}\sum_{i=1}^n\frac{27i^2}{n^3}-\sum_{i=1}^n\frac{18i}{n^2}-\sum_{i=1}^n\frac{3}{n}$
$\lim_{n->\infty}\frac{27}{n^3}\sum_{i=1}^ni^2-\frac{18}{n^2}\sum_{i=1}^ni-\frac{3}{n}\sum_{i=1}^n1$
$\lim_{n->\infty}[\frac{27}{n^3}][\frac{n(n+1)(2n+1)}{6}]-[\frac{18}{n^2}][\frac{n(n+1)}{2}]-[\frac{3}{n}]$
which renders $\lim_{n->\infty}\frac{15n+9}{2n^2}$ which is $0$.
The integral evaluates to $-3$, so I know I have made a mistake or more, however I've checked my arithmetic several times each step and can't find the error, so unless my brain is fried, I must have made a mistake in step.
| Wow this confused me for a bit also until I realized we were both making the same mistake. In your case, you wrote
$$\sum_{i=1}^n\frac3n=\frac3n\sum_{i=1}^n1=\frac3n,$$
but $\sum_{i=1}^n1=n$ and this should have been
$$\sum_{i=1}^n\frac3n=\frac3n\sum_{i=1}^n1=3.$$
Since $x_i=1+\frac{3i}{n}=\frac{n+3i}{n}$, the sum is
\begin{align}
\sum_{i=1}^nf\left(\frac{n+3i}{n}\right)\left(\frac3n\right)&=\sum_{i=1}^n\left(\frac{(n+3i)^2}{n^2}-4\frac{n+3i}{n}+2\right)\left(\frac3n\right)\\ \\
&=\sum_{i=1}^n \left(\frac{27i^2}{n^3}-\frac{18i}{n^2}-\frac{3}{n}\right) \\ \\
&=\frac{27}{n^3}\sum_{i=1}^n(i^2)-\frac{18}{n^2}\sum_{i=1}^n(i)-\sum_{i=1}^n\frac3n\\ \\
&=\frac{27}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)-\frac{18}{n^2}\left(\frac{n(n+1)}{2}\right)-3\\ \\
&=\frac{9}{2n^2}+\frac{3}{2n}-3.
\end{align}
As $n\to\infty$, the sum approaches $-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$(Sweden 1971) If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$
I was trying to do the question above using homogenization. I was trying to do it as follows:
$\sqrt{xy+yz+xz}=\sqrt{26}$
Hence $x+y+z>\frac{35}{4*\sqrt{26}}*\sqrt{xy+yz+xz}$
So it is enough to prove that $16*26(x^2+y^2+z^2+2xy+2yz+2xz)>35*35*(xy+yz+xz)$. Which is true since $x^2+y^2+z^2+2xy+2yz+2xz\ge 3(xy+yz+xz)$ Could you please share other simple ways to solve the question, as I am intrigued by its simplicity.
| Maybe the following is better.
Let $x=\sqrt{\frac{26}{3}}a$, $y=\sqrt{\frac{26}{3}}b$ and $z=\sqrt{\frac{26}{3}}c$.
Thus, $$ab+ac+bc=3$$ and $$x+y+z=\sqrt{\frac{26}{3}}(a+b+c)\geq\sqrt{\frac{26}{3}}\sqrt{3(ab+ac+bc)}=\sqrt{78}.$$
Thus, it's enough to prove that $$\sqrt{78}>\frac{35}{4}$$ or
$$1248>1225$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3847360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Venn Diagram Conditional Probability Questions Solution I was having a go at this question, I got an answer of $\frac{2}{29}$, not sure if this is correct, I would appreciate if someone could explain the solution to it.
| You are correct, please include your working next time.
From the definition of conditional probability we have $$P(B\cap A'|C)=\frac{P(B\cap A'\cap C)}{P(C)}$$
$$=\frac{\frac{x}{5}}{x+\frac{x}{5}+\frac{x}{2}+2x^2+x}=\frac{\frac{x}{5}}{\frac{27}{10}x+2x^2}=\frac{\frac{1}{5}}{\frac{27}{10}+2x}.$$
But the sum of the probabilities must be $1$ so we have $$x+x+x^2+2x+\frac{x}{2}+x+\frac{x}{5}+2x^2+x+13x-1=1 $$
$$3x^2+\frac{197}{10}x-2=0$$
and solving gives $x=\frac{1}{10}$ and $x=-\frac{20}{3}$, but $x$ cannot be negative.
Thus we have $$P(B\cap A'|C)=\frac{\frac{1}{5}}{\frac{27}{10}+\frac{1}{5}}=\frac{\frac{1}{5}}{\frac{29}{10}}=\frac{2}{29}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3851301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use the $\varepsilon - \delta$ definition of the limit to verify that $\lim_{(x,y)\to(2,5)} xy = 10$.
Question:
Use the $\varepsilon - \delta$ definition of the limit to verify that
$\lim\limits_{(x,y)\to(2,5)}xy = 10$.
Hint given:
$xy − 10 = (x − 2)(y − 5) + 5(x − 2) + 2(y − 5).$
My solution
$\forall \space\varepsilon\gt0\space\space\exists\space\delta(\varepsilon)\gt0:|xy-10|\lt\varepsilon$ where $(x,y)\in D$, whenever $0\lt\sqrt{(x-2)^2+(y-5)^2}\lt\delta$
Note:
$(x-2)^2 \le (x-2)^2+(y-5)^2$ $\implies |x-2|\le\sqrt{(x-2)^2+(y-5)^2}$, similarly, $|y-5|\le\sqrt{(x-2)^2+(y-5)^2}$
So if $(x,y)\in D\space$ and $\space0\lt\sqrt{(x-2)^2+(y-5)^2}\lt\delta$. We choose $ \delta :=\frac{-7+\sqrt{49+4\varepsilon}}2$.
We obtain...
$\begin{aligned}|xy-10|&=|(x − 2)(y − 5) + 5(x − 2) + 2(y − 5)|\\&\le|x − 2||y − 5|+ 5|x − 2| + 2|y − 5|\\&\le \left((x-2)^2 +(y-5)^2\right) +5\sqrt{(x-2)^2+(y-5)^2}+2\sqrt{(x-2)^2+(y-5)^2}\\&=\left(\sqrt{(x-2)^2+(y-5)^2}\right)^2+7\sqrt{(x-2)^2+(y-5)^2}\\&\lt \delta^2+7\delta\\&=\frac{49\pm\sqrt{49-4\varepsilon}+49+4\varepsilon}4+\frac{-98\pm14\sqrt{49+4\varepsilon}}4\\&=\frac{4\varepsilon}4\\&=\varepsilon\end{aligned}$
Comments
This is a question on my Analysis 2 module. Would be great if someone could check my solution. Also, I realise you have to use a minimum for $\delta$, but unsure on how to unpack it when writing the solution, would be great if anyone can make my understanding of this clearer.
| First observe that:
$$
|xy-10|=|xy-2y+2y+10|\le |y||x-2|+2|y-5|
$$
Then set the first restriction: $\delta_1=1$
$$
|y-5|\le \big((x-2)^2+(y-5)^2)\big)^{½}<1=\delta_1,
$$
which implies, that
$$
|y|\le |y-5|+5<6,
$$
and hence, if $\big((x-2)^2+(y-5)^2)\big)^{½}<1$, then
$$
|xy-10|=|xy-2y+2y+10|\le |y||x-2|+2|y-5|\le 6|x-2|+2|y-5|\\ \le (6^2+2^2)^{1/2}\big((x-2)^2+(y-5)^2)\big)^{½} \le 7\big((x-2)^2+(y-5)^2)\big)^{½}
$$
Hence, if
$$
\big((x-2)^2+(y-5)^2)\big)^{½}<\delta=\min\{\delta_1,\varepsilon/7\}=\min\{1,\varepsilon/7\},
$$
then
$$
|xy-25|\le 8\big((x-2)^2+(y-5)^2)\big)^{½}<\varepsilon.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Please give an algebraic proof of the following inequality: $\frac{a}{a+b} \gt \frac{a-1}{a+b-1}$ for $a,b \gt 0$. For $ a,b > 0 $ and $\left(\frac{a}{a+b}\right)\cdot\left(\frac{a-1}{a+b-1}\right)=\frac{1}{3}$, show that $$\frac{a}{a+b} > \frac{a-1}{a+b-1}$$
I am not getting how to do this. Please enlighten me. What i need is an algebraic proof.
My thought, let $f(x) = \frac{x}{x+b}$, and $f'(x) = \frac{b}{(x+b)^2}$, this implies $f(x)$ is increasing, i.e, $f(x) > f(x-1)$. But can we solve it simply, I tried solving just using inequalities I can't find how.
| You have
$$
\frac{a}{a+b}-\frac{a-1}{a+b-1}=\frac{b}{(a+b)(a+b-1)}
$$
Given that $a,b>0$, your statement is equivalent to $a+b-1>0$.
If $a$ and $b$ are integer, as you seem to hint in the comments, then the statement is completely obvious.
For general real $a,b$, we'd like to see whether this follows from
$$
\frac{a}{a+b}\frac{a-1}{a+b-1}=\frac{1}{3}
$$
It's simpler if we set $a+b=c$, so the equality becomes
$$
3a^2-3a=c^2-c
$$
We'd like to see whether the equation $c^2-c-3a^2+3a=0$ has or not solutions satisfying $a<c<1$, which would prove the conjecture false.
The roots are
$$
\frac{1\pm\sqrt{12a^2-12a+1}}{2}
$$
We have
$$
a<\frac{1-\sqrt{12a^2-12a+1}}{2}
$$
if and only if $1-2a>\sqrt{12a^2-12a+1}$, which reduces to $0<a<(3-\sqrt{6})/6$. In this case, the root is obviously less than $1$.
Suppose $a=0.01$. Then the chosen root is $\approx0.03$ and your conjecture is false.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that if x,y,z are positive integers such that $x^3+y^3+z^3=3(x+y+z+xyz)$ then they must be consecutive numbers I was thikning going for contradiction, assuming they're not consecutive but I found it very hard to find a statement that contradicts that condition. Maybe stating that there exist some $m, n \in \Bbb{Z}$ such that $|mn|\ne 1$ and $y=x+m$ and $z=x-n$? Another way would be assuming at least 2 of them are non-consecutive?
I would like to see general ideas on how to proceed.
| First, subtract $3xyz$ from both sides of the equation:
$$x^3+y^3+z^3=3(x+y+z+xyz) \\
x^3+y^3+z^3-3xyz = 3(x+y+z)$$
Now, using this factorisation, observe:
$$
(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=3(x+y+z)$$
Since $x,y$, and $z$ are positive integers which implies $x+y+z\neq 0$, you can divide the equation by $x+y+z$ to get
$$ x^2+y^2+z^2 - xy-yz-zx = 3$$
Now, from the above equation, it's easy to observe that
$$(x-y)^2+(y-z)^2+(z-x)^2=6$$
Now, if two of the differences in the brackets are greater than $1$ then the sum will be greater than $2^2 + 2^2 = 8 > 6$ (contradiction!).
Also, if all are equal to $1$, then the sum will be $3 \neq 6$ (contradiction!).
Hence, one of the differences is $2$ and the other two are equal to $1$ (so that $2^2+1^2+1^2=6$ as desired) which implies $x,y$, and $z$ are consecutive integers.
Why is that? Now goes the explanation:
Since you have already shown that one of the differences is $2$ and the other two are $1$, without loss of generality, you may take
$$
\begin{cases}
x - y = 1 \\
y - z = 1 \\
x - z = 2
\end{cases} \implies
\begin{cases}
x = y + 1 \\
y = z + 1 \\
x = z + 2
\end{cases}
$$
Hence, $$(x,y,z) = (z+2, ~z+1, z)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3856174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $a+b+c=3$, then show $\sqrt{a^2 + ab +b^2}+ \sqrt{b^2 + bc +c^2}+\sqrt{c^2 + ac +a^2} \geq \sqrt{3}$ If $a+b+c=3$, and $a,b,c$ are positive real numbers, then show $\sqrt{a^2 + ab +b^2} + \sqrt{b^2 + bc +c^2} +\sqrt{c^2 + ac +a^2}\geq \sqrt{3}$
Normally when I do inequalities I try to first find where equality would be achieved, but in this case I have no idea where to start to find it.
From $a^2 - 2ab +b^2 \geq 0$ we can obviously get $a^2 -ab +b^2\geq 3ab$, and after substituting that, we can change the inequality to showing that $\sqrt{ab} + \sqrt{bc} +\sqrt{ac} \geq 1$, but clearly if a=b=c, then equality is not achieved. I wanted to try to use Cauchy-Schwarz or something of the sort, but those inequalities generally are used to find the upper bound of sums of roots.
| $$\left(\frac{a+b}{2}\right)^2+\frac{(a-b)^2}{12}=\frac{a^2+ab+b^2}{3}$$
$$\therefore\ \sqrt{\frac{a^2+ab+b^2}{3}}\ge\frac{a+b}{2}$$
$$\therefore\ \sum_{cyc}\sqrt{\frac{a^2+ab+b^2}{3}}\ge\sum_{cyc}\frac{a+b}{2}=a+b+c=3$$
So in fact a stronger identity is possible $\sum_{cyc}\sqrt{a^2+ab+b^2}\ge 3\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Closed form $\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin(xyz)\,dx\,dy\,dz$ This is part of an assignment from a multivariable calculus course. We've only just defined triple integrals and we're not using special functions or anything like that, so I'm pretty sure this was a mistake. Still, I'm interested to see if there exists some nice closed form.
The only progress I've managed to get is
$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin(xyz)\,dx\,dy\,dz=\int_{0}^{\frac{\pi}{2}}\dfrac{1}{z}\left(\ln\left(\dfrac{\pi^2}{4}z\right)-\text{Ci}\left(\dfrac{\pi^2}{4}z\right)+\gamma\right)dz\,,$$
where $\text{Ci}(x)$ is the cosine integral, but nothing else. Any ideas?
| Incomplete (?) solution.
\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\sin(xyz)dxdydz\\
&\overset{u=\frac{2x}{\pi},v=\frac{2y}{\pi},w=\frac{2z}{\pi}}=\frac{\pi^3}{8}\int_0^1\int_0^1\int_0^1\sin\left(\frac{\pi^3uvw}{8}\right)dudvdw\\
&=-\frac{\pi^3}{8}\int_0^1\int_0^1 \ln x\sin\left(\frac{\pi^3xy}{8}\right)dxdy\\
&=-\frac{\pi^3}{16}\int_0^1\int_0^1 \ln (xy)\sin\left(\frac{\pi^3xy}{8}\right)dxdy\\
&=\frac{\pi^3}{16}\int_0^1 \ln^2 (z)\sin\left(\frac{\pi^3z}{8}\right)dz\\
&=\frac{\pi^3}{16}\int_0^1 \ln^2 z\left(\sum_{n=0}^\infty \frac{(-1)^n\pi^{6n+3}z^{2n+1}}{2^{6n+3}(2n+1)!}\right)dz\\
&=\frac{\pi^3}{16}\sum_{n=0}^\infty\left(\int_0^1 \frac{(-1)^n\pi^{6n+3}z^{2n+1}\ln^2 z}{2^{6n+3}(2n+1)!}\right)\\
&=\frac{\pi^3}{8}\sum_{n=0}^\infty \frac{(-1)^n\pi^{6n+3}}{2^{6n+3}(2n+2)^3(2n+1)!}\\
&=\frac{\pi^3}{8}\sum_{n=0}^\infty \frac{(-1)^n\pi^{6n+3}}{2^{6(n+1)}(n+1)^3(2n+1)!}\\
\end{align}
NB: I use the following result: If $f$ is a continuous function on $[0;1]$ then $\displaystyle \int_0^1 \int_0^1 f(xy)dxdy=-\int_0^1 f(x)\ln xdx$
(proof: change of variable $u(y)=xy$ then IBP)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3863508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Have I done my homogenization correctly for this question and if so how do I finish it off? We have $a,b,c>0$ with $a^2+b^2+c^2=1$ prove that:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}>1$
As soon as I saw this question, I immediately thought of using homogenization in the following manner:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}>\frac{1}{\sqrt{a^2+b^2+c^2}}$
Then I tried using Andreescu:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{(1+bc)(1+ac)(1+ab)}$
Which didn't work out.
Could you please explain to me if I have done my homogenization correctly? If I have done it correctly, could you please show me how to finish it off and if I haven't how to think of doing the correct homogenization?
| we will prove $$\sum_{cyc}\frac{a}{1+bc}=3-\sum_{cyc}\frac{abc}{1+bc}>1$$
or $$\sum\frac{abc}{1+bc}<2$$
we know will make denominator homogenous in degree 2
$1+bc=a^2+b^2+c^2+bc\ge 4a^{1/2}b^{3/4}c^{3/4}$
thus $$\sum\frac{abc}{1+bc}\le \frac{1}{4}a^{1/2}b^{1/4}c^{1/4}\le \frac{3}{4}abc\le \frac{1}{4\sqrt{3}}<2$$\
here we used $$abc\le \frac{1}{3\sqrt{3}}$$ by am-gm using $a^2+b^2+c^2=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Triple integration with $x^2+y^2-z^2$ I'm having this problem.
$$\iiint_{K}(x^2+y^2-z^2)dxdydz,\quad K: x^2+y^2+z^2\leq 1$$
My solution.
$$x=\rho \sin\phi \cos\Theta,\ y=\rho \sin\phi \sin\Theta, z=\rho \cos \phi\\
dxdydz=\rho^2 \sin\phi d\rho d\phi d\theta\\
0\leq \rho \leq 1\\
0 \leq \phi \leq \pi\\
0 \leq \theta \leq 2\pi\\
$$
$$\int _0^{2\pi }\int _0^{\pi }\int _0^1\left(\left(\rho \sin\left(\phi\right)\cos\left(\theta\right)\right)^2+\left(\rho \sin\left(\phi\right)\sin\left(\theta\right)\right)^2+\left(\rho \cos\left(\phi\right)\right)^2\right)d\rho d\phi d\theta$$
$$\int _0^1\left(\rho\sin \left(\phi\right)\cos \left(\theta\right)\right)^2+\left(\rho\sin \left(\phi\right)\sin \left(\theta\right)\right)^2-\left(\rho \cos \left(\phi\right)\right)^2d\rho\\ =-\cos ^2\left(\phi\right)\frac{1}{3}+\cos ^2\left(\theta\right)\sin ^2\left(\phi\right)\frac{1}{3}+\sin ^2\left(\phi\right)\sin ^2\left(\theta\right)\frac{1}{3}\\
\int _0^{\pi }\left(-\cos ^2\left(\phi\right)\frac{1}{3}+\cos ^2\left(\theta\right)\sin ^2\left(\phi\right)\frac{1}{3}+\sin ^2\left(\phi\right)\sin ^2\left(\theta\right)\frac{1}{3}\right)d\phi\\
=-\frac{\pi }{6}+\frac{\pi }{6}\cos ^2\left(\theta\right)+\frac{\pi }{6}\sin ^2\left(\theta\right)=\frac{\pi }{6}\left(\cos ^2\left(\theta\right)+\sin ^2\left(\theta\right)-1\right)=\frac{\pi }{6}\left(-1+1\right)=0\\
\int _0^{2\pi }0d\theta=0
$$
It's the minus sign in $-z^2$ that creates multiplication with $0$, and therefore the answer is $0$. Is this correct?
When I plot this in Geogebra I get a sphere inside a hyperboloid.
Thanks for all the help!
| $\int \int \int _{K}(x^2+y^2-z^2)dxdydz $
$= \displaystyle \int _0^{2\pi }\int _0^{\pi} \int_0^1 ((\rho \sin \phi cos \theta)^2 + (\rho \sin \phi \sin\theta)^2 - (\rho \cos \phi)^2) \, \rho^2 \, sin\phi \, d\rho \, d\phi \, d\theta$
$= \displaystyle \int _0^{2\pi} \int_0^{\pi} \int_0^1 (\rho^2 \sin^2 \phi - \rho^2 \cos^2 \phi) \, \rho^2 \, sin\phi \, d\rho \, d\phi \, d\theta$
$= \displaystyle -\int _0^{2\pi} \int_0^{\pi} \int_0^1 \rho^4 \cos 2\phi \, sin\phi \, d\rho \, d\phi \, d\theta$
$= \displaystyle - \frac{1}{5}\int _0^{2\pi} \int_0^{\pi} \cos 2\phi \, sin\phi \, d\phi \, d\theta$
Substituting $t = \cos \phi$ for integral, can you do rest of the steps?
It comes to $\frac{4 \pi}{15}$. Checked WolframAlpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3866733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
System of equations from roots of polynomial I'm given the equation $3072x^4-2880x^3+840x^2-90x+3=0$ and told that its roots are $\alpha, \alpha r, \alpha r^2, \alpha r^3,$ for some $r\in \mathbb{R}$.
By considering the sum of the roots, the product, etc. I've found that \begin{gather}\alpha(1+r+r^2+r^3)=\frac{15}{16} \\ \alpha^2r(1+r+2r^2+r^3+r^4)=\frac{35}{128} \\ \alpha^3 r^3(1+r+r^2+r^3)=\frac{15}{512} \\ \alpha^4 r^6=\frac{1}{1024}\end{gather}
But this looks like a rather complex system and I can't see any obvious way solve this for $\alpha$ and $r$.
How can this system be solved?
EDIT
I can see that all of the denominators are powers of $2$, but I can't see how that will help me here.
| Going with your equations... Dividing third equation by first, we get $\alpha^2 r^3=\frac{1}{32}$. This implies $r > 0$ because $\alpha$ must be real as well (follows from the first equation). Plugging this into the second equation we get
$$
\frac{35}{128}=\frac{1}{32}\left(\frac{1}{r^2}+\frac{1}{r}+2+r+r^2\right)=\frac{1}{32}\left(\left(r+\frac{1}{r}\right)^2+\left(r+\frac{1}{r}\right)\right).
$$
Letting $u=r+1/r$ gives quadratic equation
$$
u^2+u-\frac{35}{4}=0.
$$
This yields $u=\frac{5}{2}$ as we must have $u>0$. Then solving corresponding quadratic equation given by $\frac{5}{2}=r+1/r$ we see $r \in \{\frac{1}{2},2\}$. From the first equation we get $\alpha$ and so the two solutions are $r=\frac{1}{2}, \alpha=\frac{1}{2}$ and $r=2, \alpha=\frac{1}{16}$.
Clearly both solutions generate the same set of roots $\{\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16}\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Generating primitive Pythagorean triples Given this equation $a^2 + b^2 = c^2$ , in order to generate all primitive Pythagorean triples all we nedd to do is : write (a,b,c) as :
$a = 2mn $
$b = m^2-n^2$
$c= m^2+n^2$
with conditions : $\gcd(m,n)=1$ and $(m+n)$%$2= 1 $
But what about if we add a parameter $K$ to the equation such that : $a^2 + b^2 = c^2 - K$ , with $K$ being a positive integer , how can we generate $(a,b,c)$ in this case satisfying the equation above such that $\gcd(a,b,c) =1$ ?
| We need only show how side $C$ of one triple is the same as side $A$ of another triple. For example, if we have $(3,4,5)$ and $(5,12,13)$, then
$$A_1^2+B_1^2=C_2^2-K_2^2\quad\text{ is simply }\quad 3^2+4^2=13^2-12^2$$
In this "easiest" case, $GCD(A_1,B_1,C_2)=1$.
There is a formula to identify these matches, if they exist. Let us begin with $(33,56,65)\text{ and }(63,16,65)\quad $ and find matches where $A_2=C_1=65$.
\begin{equation}
A=m^2-n^2\implies n=\sqrt{m^2-A}\qquad\text{for}\qquad \sqrt{A+1} \le m \le \frac{A+1}{2}
\end{equation}
$$A=65\implies \lfloor\sqrt{65+1}\rfloor=8\le m \le \frac{65+1}{2} =33\quad\land\quad m\in\{9,33\}\implies n \in\{4,32\} $$
$$F(9,4)=(65,72,97)\qquad \qquad F(33,32)=(65,2112,2113)$$
What this means is $$(33^2+56^2)=(63^2+16^2)=(97^2-72^2)=(2113^1-2112^2)$$
There are an infinite number of these because side $A$ includes any odd number greater than $1$. Also, I haven't proven it yet but all of the examples I have seen have shown that the $GCD(A,B,C)=1$ requirement is also met.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all $n,x \in \Bbb N$ such that $3\cdot2^x+4=n^2$
Find all $n,x \in \Bbb N$ such that $3\cdot2^x+4=n^2$
Arranging the equation a bit one has that $$(n-2)(n+2)=3\cdot2^x.$$
Now considering the cases $n-2=3, n+2=2^x$ from $n-2=3 \Rightarrow n=5$. Substituting this to $n+2=2^x \Rightarrow 7=2^x$ which isn't an integer solution.
Considering the opposite $n+2=3$ and $n-2=2^x$ one gets that $n+2=3 \Rightarrow n=1$ and substituting this to $n-2=2^x$ we have that $-1=2^x$ which also doesn't have any solutions.
What am I missing here since this seemed to have $3$ pairs of solutions?
| You're off to a good start; writing
$$(n-2)(n+2)=3\cdot2^x,$$
shows, by unique factorization of integers, that
$$n-2=2^a3^b,\qquad\text{ and }\qquad n+2=2^c3^d,$$
for nonnegative integers $a$, $b$, $c$ and $d$ with $a+c=x$ and $b+d=1$. Of course $b+d=1$ leaves very few options for $b$ and $d$. Can you continue from here?
Solution:
It follows that $$2^{x-a}3^{1-b}-2^a3^b=(n+2)-(n-2)=4,$$ so if $x\geq3$ we see that either $a\geq2$ or $x-a\geq2$. Then reducing mod $4$ shows that in fact $a\geq2$ and $x-a\geq2$, and so $$2^{x-a-2}3^{1-b}-2^{a-2}3^b=(n+2)-(n-2)=1.$$ Then precisely one of the two factors on the left hand side is odd, meaning that either $x-a-2=0$ or $a-2=0$, and so correspondingly either $$3^{1-b}-2^{a-2}3^b=1\qquad\text{ or }\qquad2^{x-4}3^{1-b}-3^b=1.$$ For the former, note that $3^{1-b}\in\{1,3\}$. Clearly $3^{1-b}=1$ does not yield a solution, so $3^{1-b}=3$ and hence $b=0$. Then $2^{a-2}=2$ and so $a=3$, yielding the solution $x=5$ with $n=10$. For the latter, note again that $3^b\in\{1,3\}$. A quick check shows that $3^b=1$ does not yield a solution, so $3^b=3$ and hence $b=1$. Then $2^{x-4}=4$, yielding the solution $x=6$ with $n=14$. Finally we are left with the case $x\leq3$. Another quick check yields the last solution $x=2$ with $n=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove by epsilon delta $\lim\limits_{(x,y) \to (0,0)} \frac{(x-1)^2\sin(x^2)y}{x^2+y^4} = 0$ I need to prove this using $\epsilon$ $\delta$
$$\lim\limits_{(x,y) \to (0,0)} \frac{(x-1)^2\sin(x^2)y}{x^2+y^4} = 0$$
I know that $|x| < \delta $ and $|y| < \delta $, so then
$|\frac{(x-1)^2sin(x^2)y}{x^2+y^4}-0|$ $\lt$ $\frac{(x-1)^2|x||y|}{x^2+y^4}$ $\lt$
$\frac{(x-1)^2\delta^2}{\delta^2+\delta^4}$ $\lt$ $\frac{(x-1)^2}{1+\delta^2}$
And I have no idea how to continue from this. Any help would be appreciated!
| $|\frac {(x-1)^{2} \sin (x^{2}) y} {x^{2}+y^{4}}| \leq (x-1)^{2} \frac {x^{2}|y|} {x^{2}}$ since $|\sin (x^{2})| \leq x^{2}$. Hence $|\frac {(x-1)^{2} \sin (x^{2}) y} {x^{2}+y^{4}}| <\epsilon$ if $|x|<1$ and $|y|<\frac {\epsilon } 4 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of digits of square number raised to itself From testing a few different square numbers, it seems to be the case that when raising a square number to the power of itself, the sum of the digits of the result satisfy the property that the sum of their digits is the square number itself.
I realise the above sentence is quite wordy, so as an example, consider $4^4$. We know that $4^4=256$ and that $2+5+6=13$. It is also the case that $1+3=4$, i.e., the square number itself.
Is this true for any square number? And if so, how can one prove it?
| Your original claim is poorly worded in that it does not adequately distinguish between sums of digits and sums of sums of digits and what numbers should be compared against one another. As written the claim is false, as pointed out by other users, since you have the sum of digits or the sum of the sum of the digits not equaling the original number itself in some cases as is the case for example for $25^{25}$ not having sum of digits or sum of sum of digits equal to $25$.
If you were to instead talk about the repeated sum of digits for both the original number and the number to the power of itself, then we do actually have a true statement.
Claim: for perfect square $x=n^2$ one has that $x^x\equiv x\pmod{9}$
Proof by cases:
As $x=n^2$ it follows that $x$ is equivalent to one of $0,1,4,$ or $7$ modulo $9$
In the first case, we have $x^x\equiv 0^x\equiv 0\pmod{9}$ trivially.
Similarly in the second case we have $x^x\equiv 1^x\equiv 1\pmod{9}$
In the third case, $x^x\equiv 4^x\equiv 4^{9k+4}\equiv (4^{3})^{3k}\cdot 4^3\cdot 4\equiv 1^{3k}\cdot 1\cdot 4\equiv 1\pmod{9}$ noting that $4^3=64=9\cdot 7 + 1$
Finally, for the fourth case we have $x^x\equiv 7^{9k+7}\equiv (7^3)^{3k}\cdot 7^3\cdot 7\equiv 1^{3k}\cdot 1\cdot 7\equiv 7\pmod{9}$ just like in the previous case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the Limit.? What is the limit of
$Y_{n} := \frac{1}{3}Y_{n-1} + \frac{2}{3}Y_{n-2} $
for $n > 2$ and $Y_{1}<Y_{2}$.
I have tried solving, but the limit tends to exist between $y_{1}$ and $y_{2}$ and hence I am not able to evaluate what would be the ratio between which they would lie?
| Let me show you the general method for dealing with these linear recurrences. The idea is to convert the linear recurrence into a matrix recurrence: $v_{n} = Av_{n-1}$ so that $v_n = A^nv_0$.
Here, you have
$$ \begin{pmatrix} Y_n \\ Y_{n - 1} \end{pmatrix} = \begin{pmatrix} \frac13 & \frac23 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} Y_{n - 1} \\ Y_{n - 2} \end{pmatrix} $$
So by induction,
$$ \begin{pmatrix} Y_n \\ Y_{n - 1} \end{pmatrix} = \begin{pmatrix} \frac13 & \frac23 \\ 1 & 0 \end{pmatrix}^{n - 2} \begin{pmatrix} Y_{2} \\ Y_{1} \end{pmatrix}. $$
The matrix
$$ P = \begin{pmatrix} \frac13 & \frac23 \\ 1 & 0 \end{pmatrix}$$
is a regular, row-stochastic matrix. Thus by standard Markov chain theory,
$$ P^n \to \begin{pmatrix} \frac35 & \frac25 \\ \frac35 & \frac25 \end{pmatrix} $$
where $(3/5, 2/5)$ is the probability vector in the 1-eigenspace of $P^T$. So $Y_n \to \frac25Y_1 + \frac 35Y_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3878008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
minimum value of $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$ if $x^2+y^2+z^2=1$ If $x^2+y^2+z^2=1$ what is the minimum value of $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$ for $x,y,z \gt 0$ ?
I would like to know if the minimum could be found using simpler ways.(like $AM \ge GM \ge HM$).
knowing $xy+yz+xz \geq \frac{-1}{2}$ for real $x,y,z$ might help.
| First, notice that $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y} = \frac{xyz}{z^2}+\frac{xyz}{x^2}+\frac{xyz}{y^2}=xyz({\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}})$
Using the inequality of arithmetic and geometric means, we have:
$$ \frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2}\frac{1}{y^2}}\iff \frac{1}{x^2}+\frac{1}{y^2}\ge \frac{2}{xy}$$
Do that for every pair of variables and add everything together:
$$\frac{1}{x^2}+\frac{1}{y^2}\ge \frac{2}{xy}$$
$$\frac{1}{x^2}+\frac{1}{z^2}\ge \frac{2}{xz}$$
$$\frac{1}{y^2}+\frac{1}{z^2}\ge \frac{2}{yz}$$
We have: $$2({\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}})\ge2(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}) \iff {\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}\ge\frac{x+y+z}{xyz}$$
Which simplifies to: $$xyz({\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}})\ge x+y+z$$
Now, we just need to find the minimum value of $x+y+z$ when $x^2+y^2+z^2=1$, which I am not sure how to do. But since knowing $xy+xz+yz>-0.5$ might help, I suggest you to try using that $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3880366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do I solve $(3x^2 \tan y- 2y^3/x^3 )dx + (4y^3 + x^3 \cos^2 y+ 3y^2/x^2 ) dy=0$ Somebody please help me to solve this Differential equation.
$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0 $
| It is an exact DE, solution is $x^3\tan{y}+y^4+\frac{y^3}{x^2}+C=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove a limit in two variables is $0$ I have the limit
\begin{align}
\lim_{(x,y) \rightarrow(0,0)} \frac{1-\cos(xy)}{\sqrt{x^2+y^2}(x^2+y^2)}
\end{align}
First thing I did is to use Taylor, so $$1-\cos(xy)=\frac{(xy)^2}{2} -\frac{(xy)^4}{4} + o((xy)^4)$$
Therefore the limit is $$\lim_{(x,y) \rightarrow (0,0)} \frac{1}{2} \frac{x^2y^2}{\sqrt{x^2+y^2}(x^2+y^2)}$$
Now, $\frac{y^2}{x^2+y^2} \leq 1$, therefore I have
$$\left|\frac{1}{2} \frac{x^2y^2}{\sqrt{x^2+y^2}(x^2+y^2)}\right| \leq \left|\frac{x^2}{\sqrt{x^2+y^2}}\right| \leq \frac{x^2}{|x|} = |x|$$
Since $|x| \rightarrow 0 $ as $(x,y) \rightarrow (0,0)$, the limit is $0$, correct?
| It seems correct, to avoid Taylor we can use that
$$\frac{1-\cos(xy)}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{1-\cos(xy)}{(xy)^2}\frac{(xy)^2}{\sqrt{x^2+y^2}(x^2+y^2)}$$
with $\frac{1-\cos(xy)}{(xy)^2} \to \frac12$ by standard limits.
For the conclusion by squeeze theorem, we should also consider a part the case $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Factorizing the determinant Here, in this question, we have to factorize this determinant. I understood the (a-b), (b-c), (c-a) part but I am not able to understand
*
*how to find if determinant is symmetrical in a, b, c.
*I know that rest factor will be of degree 2 as we have got (a-b), (b-c), (c-a), but as here (a^2 + b^2 + c^2) and (ab + bc + ca) are possible factors of degree 2, can't be (a + b + c)^2 will be the 2nd degree factor?
Book HK Das Engineering Mathematics page 388 example no. 36
Book HK Das Engineering Mathematics page 388 example no. 36
| (1) If you put $a=b,b=c$ or $c=a$ then, $\Delta=0$.
Being symmetric about $a,b,c$ means that if $(a_0,b_0,c_0)$ is a solution to $\Delta=0$, then $(b_0,a_0,c_0)$ is also a solution and so are all other permutations of $(a_0,b_0,c_0)$. This can be seen easily. For now, we verify it for the $3$ permutations $(a_0,b_0,c_0),(b_0,a_0,c_0),(c_0,b_0,a_0)$:
$$\begin{vmatrix}1 & 1 & 1\\a_0^2&b_0^2&c_0^2\\a_0^3&b_0^3&c_0^3\end{vmatrix}=-\begin{vmatrix}1 & 1 & 1\\b_0^2&a_0^2&c_0^2\\b_0^3&a_0^3&c_0^3\end{vmatrix}=\begin{vmatrix}1 & 1 & 1\\c_0^2&b_0^2&a_0^2\\c_0^3&b_0^3&a_0^3\end{vmatrix}\mathrm{~~and~~}\begin{vmatrix}1 & 1 & 1\\a_0^2&b_0^2&c_0^2\\a_0^3&b_0^3&c_0^3\end{vmatrix}=0$$
$$\Rightarrow\begin{vmatrix}1 & 1 & 1\\a_0^2&b_0^2&c_0^2\\a_0^3&b_0^3&c_0^3\end{vmatrix}=\begin{vmatrix}1 & 1 & 1\\b_0^2&a_0^2&c_0^2\\b_0^3&a_0^3&c_0^3\end{vmatrix}=\begin{vmatrix}1 & 1 & 1\\c_0^2&b_0^2&a_0^2\\c_0^3&b_0^3&a_0^3\end{vmatrix}=0$$
$\therefore a=a_0,b=b_0,c=c_0$ and $a=b_0,b=a_0,c=c_0$ and $a=c_0,b=b_0,c=a_0$ are all solutions to $\Delta=0$.
(2) $(a+b+c)^2$ may or may not be a possible factor. Notice that if it is then
$$(a+b+c)^2=1\cdot(a^2+b^2+c^2)+2\cdot(ab+bc+ca).$$
The point of taking $(a^2+b^2+c^2)$ and $(ab+bc+ca)$ is that any $2$ degree factor in terms of $a,b,c$ can be written in the form
$$k\cdot(a^2+b^2+c^2)+l\cdot(ab+bc+ca)$$
whereas, there are possible $2$ degree factors of $\Delta=0$ that may not be multiples of $(a+b+c)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Combinatorics problem - $5$-digit code with $3,4,5$ present I have a combinatorics problem that I'm struggling with. Here it is: "How many 5-digit codes have among their digits each of $3,4,5$?"
I do understand the general strategy, where we first see how we can permute $3,4,5$ with the other empty slots ($\frac{5!}{2} = 60$ ways) and then we multiply by $10^2$ (both of the other two slots can have any of the $10$ digits). However now we end up double counting, as some of those other two digits will be $3,4,5$ and will result in duplicating combinations between each other. I am struggling to come up with a formula for excluding these.
| There are a couple of strategies we could employ here.
Method 1: We consider cases depending on how often each digit appears.
Case 1: The digits $3$, $4$, $5$ each appear exactly once.
Choose one of the positions for the $3$, one of the remaining four positions for the $4$, and one of the remaining three positions for the $5$. Then each of the remaining two positions may be filled with one of the remaining $10 - 3 = 7$ digits. Hence, there are
$$5 \cdot 4 \cdot 3 \cdot 7^2$$
such codes.
Case 2: Exactly one of the digits $3$, $4$, $5$ appears twice and each of the others appears exactly once.
Choose which of the digits $3$, $4$, $5$ appears twice. Choose two of the positions for that digit. Choose one of the remaining three positions for the smaller of the two remaining digits from the set $\{3, 4, 5\}$ and one of the remaining two positions for the remaining digit from the set $\{3, 4, 5\}$. Choose which of the remaining seven digits fills the remaining position. There are
$$\binom{3}{1}\binom{5}{2}\cdot 3 \cdot 2 \cdot 7$$
such codes.
Case 3: Exactly two of the digits $3, 4, 5$ appear twice and the other one appears once.
Choose which of the three digits $3, 4, 5$ appears exactly once. Choose which of the five positions that digit fills. Choose two of the remaining four remaining positions for the smaller of the two remaining digits from the set $\{3, 4, 5\}$, then fill the remaining two positions with the remaining digit from the set $\{3, 4, 5\}$. There are
$$\binom{3}{1}\binom{5}{1}\binom{4}{2}$$
such codes.
Case 4: Exactly one of the three digits $3, 4, 5$ appears three times and each of the others appears once. Choose which of the three digits appears three times, then choose three of the five positions for that digit. Choose one of the remaining two positions for the smaller of the remaining digits from the set $\{3, 4, 5\}$, then fill the final position with the other digit. There are
$$\binom{3}{1}\binom{5}{3}\binom{2}{1}$$
such codes.
Total: Since these positions are mutually exclusive and exhaustive, add the above cases.
Method 2: We use the Inclusion-Exclusion Principle.
There are $10^5$ codes. We wish to exclude from these those in which at least one of the digits $3, 4, 5$ is missing.
We choose which of the three digits $3, 4, 5$ to exclude, which leaves nine ways to fill each of the five positions. Thus, we subtract
$$\binom{3}{1}9^5$$
from the total.
However, if we do so, we will have subtracted each case in which two of the digits $3, 4, 5$ are missing twice, once for each way we could have designated one of those digits as the missing digit. We only want to subtract such cases once, so we must add them to the total.
We choose which two of the three digits $3, 4, 5$ to exclude, which leaves us eight ways to fill each of the five positions. Thus, we add
$$\binom{3}{2}8^5$$
to our running total.
However, if we first subtract those cases in which one of the digits $3, 4, 5$ is excluded and then add those cases in which two of the digits $3, 4, 5$ are excluded, we will not have excluded those cases in which all three digits $3, 4, 5$ are excluded at all. This is because we first subtracted them three times, once for each way we could have designated one of those three digits as the excluded digit. We then added these cases three times, once for each of the $\binom{3}{2}$ ways we could have designated two of those three digits as the excluded digits. Thus, we must them from the total.
If all three digits $3, 4, 5$ are excluded, then we have seven choices for each of the five positions. Thus, there are
$$\binom{3}{3}7^5$$
cases in which all the digits $3, 4, 5$ are excluded.
By the Inclusion-Exclusion Principle, there are
$$10^5 - \binom{3}{1}9^5 + \binom{3}{2}8^5 - \binom{3}{3}7^5$$
admissible codes.
As you can verify, the two methods yield the same answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3894230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$ Here is my attempt:
Let $(x, y)$ be any such point.
Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we have the equalities
$$
\lvert x \rvert = \lvert y \rvert = \sqrt{ (x-4)^2+(y-2)^2}.
$$
From $\lvert x \rvert = \lvert y \rvert$, we obtain $y = \pm x$.
Thus we have the equations
$$
\sqrt{ (x-4)^2 + (\pm x -2)^2} = \lvert x \rvert,
$$
which implies
$$
(x-4)^2 + ( \pm x - 2)^2 = x^2,
$$
which simplifies to
$$
x^2 - 2(4 \pm 2)x + 20 = 0.
$$
Thus we have the following two quadratic equations
$$
x^2 - 12x + 20 = 0 \qquad \mbox{ and } \qquad x^2 -4x + 20 = 0,
$$
and the solutions of these quadratic equations are
$$
x = \frac{12 \pm 8 }{ 2 } \qquad \mbox{ and } \qquad x = \frac{ 4 \pm 8 \iota }{ 2 },
$$
that is,
$$
x = 10, 2 \qquad \mbox{ and } \qquad x = 2 \pm 4 \iota.
$$
We will of course only need the real values for our $x$.
Thus there are eight possible points satisfying the condition given in the problem, namely
$$
(10, 10), (10, -10), (-10, 10), (-10, -10), (2, 2), (2, -2), (-2, 2), (-2, -2).
$$
Is my solution correct in terms of the technique employed as well as the answers obtained? Or, are there any mistakes?
| If you had actually checked each candidate point for equality of distance from the coordinate axes and $(4,2)$, you would have seen that only $(2,2)$ and $(10,10)$ satisfy the conditions of the original question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3896578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Determine convergence and limit I want to determine whether each of the sequences converges, and if so, to what limit.
$$\begin{align} 1. \quad \quad & a_n=\frac{n^{2}}{n+1}-\frac{n^{2}+1}{n}\\
2. \quad \quad & a_n=\sqrt{n+1}-\sqrt{n}
\end{align}$$
For the first one, I tried to use ratio test, but it's not working. For the second, I tried to multiply it by $\sqrt{n+1}+\sqrt{n}$ and then devided it. But I'm stuck again and don't know which test should I use to determine its convergence.
| For the first sequence, note that
$$\begin{align}
a_n & = \frac{n^2}{n+1} - \frac{n^2+1}{n}\\
& = \frac{n^3-n^3-n-n^2-1}{n^2+n}\\
& = \frac{-n^2-n-1}{n^2+n}.
\end{align}$$
And taking the limit you obtain
$$\lim_{n \to \infty} a_n = -1 \quad (\in \mathbb{R})$$
so it converges.
For the second sequence, as you said, multiply the by $\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}$ and you will obtain
$$\begin{align}
a_n & = (\sqrt{n+1}-\sqrt{n}) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\
& = \frac{(\sqrt{n+1})^2-(\sqrt{n})^2}{\sqrt{n+1} + \sqrt{n}}\\
& = \frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\\
& = \frac{1}{\sqrt{n+1}+\sqrt{n}}.
\end{align}$$
Taking the limit you get
$$\lim_{n \to \infty} a_n = 0 \quad (\in \mathbb{R})$$
so it converges too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Showing that a sequence is decreasing I need to show that the sequence defined by $a_1 = \frac{5}{2} $ and $a_{n+1} = \frac{1}{5}(a_n^2 + 6) $
is decreasing. The fact that $2 < a_n < 3 $ has been proven already.
I now just need to show that $a_n$ is decreasing.
I have tried the usual ways of showing $a_{n+1} - a_n \le 0 $ and $\frac{a_{n+1}}{a_n} \le 1 $
but could not get anywhere with these.
| Using the definition of a decreasing sequence, we starter by computing $a_{n+1}-a_n.$
$$\begin{align}
a_{n+1}-a_n & \implies \frac{1}{5}(a_n^2+6)-a_n\\
& \implies \frac{a_n^2-5a_n+6}{5}\\
& \implies \frac{(a_n - \frac{5}{2})^2-\frac{1}{4}}{5}\\
& \implies \frac{(\frac{2a_n-5}{2})^2-\frac{1}{4}}{5}\\
& \implies \frac{(2a_n^2-5)-1}{20}
\end{align}$$
Next, since you have proved that $2 < a_n < 3,$ observe that
$$\begin{align}
2 < a_n < 3 & \implies 4 < 2a_n < 6\\
& \implies -1 < 2a_n -5 < 1\\
& \implies (2a_n-5)^2<1\\
& \implies (2a_n-5)^2-1<0\\
& \implies \frac{(2a_n-5)^2-1}{20}<0
\end{align}$$
Therefore you have that
$$a_{n+1}-a_n < 0$$
and by definition you conclude that $a_n$ is a decreasing sequence. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3899466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the residue of $1/(z^2+1)^{n+1}$ at $z=i$. I'm trying to find the residue of $\frac{1}{(z^2+1)^{n+1}}$ at $z=i$. I believe this should be approached combinatorially but I am getting stuck in the combinatorics. Here's what I have so far.
$$ \frac{1}{(z^2+1)^{n+1}} = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(z-i+2i)^{n+1}} $$
$$ = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(2i)^{n+1}(1-[-z+i]/(2i))^{n+1}} $$
$$ = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(2i)^{n+1}}\cdot \left[\sum_{k=1}^\infty \left((-z+i)/(2i)\right)^k\right]^{n+1} $$
$$ = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(2i)^{n+1}}\cdot \left[\sum_{k=1}^\infty \left(\frac{-1}{2i}\right)^k(z-i)^k\right]^{n+1} $$
To get the residues we want the $(z-i)^{-1}$ term, which we can determine once we know the $(z-i)^n$ term of the factor which is expressed above as a series. This will come from making $n+1$ choices of a term from each of the $n+1$ copies of the series, such that the exponents sum to $n$. However, this now gets into a pretty famous combinatorial problem of finding the ways to partition an integer. That sounds like a lot of tricky combinatorial stuff when you're learning about residues. Am I doing this the hard way--is there an easier one?
| There is agreement between the residue and the integral of $f(x)$ along the real axis. We show that $$\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n+1}} = \frac{\pi}{2^{2n}a^{2n+1}} \frac{(2n)!}{n!n!}.$$
This is true for $n=0$:
$$\int_{-\infty}^\infty \frac{dx}{x^2+a^2} = \frac{1}{a}\int_{-\infty}^\infty \frac{\frac{dx}{a} }{\left(\frac{x^2}{a^2}+1\right)}=\frac{1}{a} \int_{-\infty}^\infty \frac{dq}{q^2+1}=\frac{1}{a} \left.\tan^{-1} q\right|_{-\infty}^\infty=\frac{\pi}{a}.$$
Assume true for $n-1$ and differentiate with respect to $a$.
$$\frac{d}{da}\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n}} = \frac{d}{da}\left[\frac{\pi}{2^{2n-2}a^{2n-1}} \frac{(2n-2)!}{(n-1)!(n-1)!}\right].$$
$$-2 n a\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n+1}} =-(2n-1)\left[\frac{\pi}{2^{2n-2}a^{2n}} \frac{(2n-2)!}{(n-1)!(n-1)!}\right].$$
$$\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n+1}} =\frac{2n}{2n}\left[\frac{\pi}{2^{2n-1}a^{2n+1}} \frac{(2n-1)!}{n!(n-1)!}\right].$$
$$\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n+1}} =\frac{\pi}{2^{2n}a^{2n+1}} \frac{(2n)!}{n!n!}.$$
With $a=1$,
$$\int_{-\infty}^\infty \frac{dx}{(x^2+1)^{n+1}} =\frac{\pi}{2^{2n}} \frac{(2n)!}{n!n!}=2\pi i \cdot \text{Res}_{z=i}\, f(z).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A bag contains 7 blue balls and 3 green balls. What is the probability of the following events? A bag contains 7 blue balls and 3 green balls.
I am not sure about my answer but please allow me to share.
*
*If 5 balls are drawn in succession at random without replacement, what is the probability of drawing 1 green, 1 blue, 1 green, and 1 blue in order?
There are 10 balls in the bag initially, 7 of which are blue and 3 balls are green. All balls are equally likely to be drawn from the bag.
Since the order is important, we need to compute the probability of getting 1 green, 1 blue, 1 green, and 1 blue in order.
Hence,
$GBGB = (\frac{3}{10})(\frac{7}{9})(\frac{2}{8})(\frac{6}{7}) = \frac{1}{20}$
The probability of getting 1 green, 1 blue, 1 green, and 1 blue in order is $\frac{1}{20}$.
*If 3 balls are drawn at the same time, what is the probability that all the 3 balls drawn are blue?
$BBB = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8}) = (\frac{7}{24})$
*If 5 balls are drawn at the same time, what is the probability that 3 balls are blue and 2 balls are green?
Work it out first where the order they are drawn is important.
$BBBGG = (\frac{7}{10})(\frac{6}{9})(\frac{5}{8})(\frac{3}{7})(\frac{2}{6}) = (\frac{1}{24})$
Then work out possible different orders they could be drawn, there are 20 permutations $P(5,2)$. so the answer is
$20 × (\frac{1}{24}) = (\frac{5}{6})$
Any comments or suggestions will be much appreciated. Thank you in advance.
| Problem 1.
Your answer to problem 1. is wrong, because you have computed the probability
of getting G-B-G-B in the first four drawings. The problem specifically
asked for the probability of the satisfying sequence occuring anywhere
within the first five drawings.
The actual problem is much more complicated, and is best attacked by the
following approach: the answer will be
$$\frac{N\text{(umerator)}}{D\text{(enoninator)}}$$
where
$$D = 10 \times 9 \times 8 \times 7 \times 6.$$
Because of the constraints of the problem, the easiest approach is to identify
each of the 4 possible satisfying sequences, and enumerate each possibility.
$N_1~: G-B-G-B-G ~:~N_1 = 3 \times 7 \times 2 \times 6 \times 1.$
$N_2~: G-B-G-B-B ~:~N_2 = 3 \times 7 \times 2 \times 6 \times 5.$
$N_3~: G-G-B-G-B ~:~N_3 = 3 \times 2 \times 7 \times 1 \times 6.$
$N_4~: B-G-B-G-B ~:~N_4 = 7 \times 3 \times 6 \times 2 \times 5.$
Final answer:
$$ \frac{N_1 + N_2 + N_3 + N_4}{D}.$$
Problem 2.
Your answer is correct.
Problem 3.
Your answer is almost right. You overlooked overcounting.
There are only $\binom{5}{2} = 10$ ways of selecting which slots the 2 blue
balls will go in. So you need to apply a scaling factor of (1/2) to your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
On the nearest-square function - Part 2 and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number (Note: This question has been cross-posted to MO.)
This question is an offshoot of this earlier one and this other question.
Let $n = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
It was conjectured in Dris (2008) and Dris (2012) that the inequality $p^k < m$ holds.
Brown (2016) showed that the Dris Conjecture (that $p^k < m$) holds in many cases.
It is trivial to show that $m^2 - p^k \equiv 0 \pmod 4$. This means that $m^2 - p^k = 4z$, where it is known that $4z \geq {10}^{375}$. (See this MSE question and answer, where the case $m < p^k$ is considered.) Note that if $p^k < m$, then $$m^2 - p^k > m^2 - m = m(m - 1),$$
and that
$${10}^{1500} < n = p^k m^2 < m^3$$
where the lower bound for the magnitude of the odd perfect number $n$ is due to Ochem and Rao (2012). This results in a larger lower bound for $m^2 - p^k$. Therefore, unconditionally, we have
$$m^2 - p^k \geq {10}^{375}.$$
We now endeavor to disprove the Dris Conjecture.
Consider the following sample proof argument:
Theorem If $n = p^k m^2$ is an odd perfect number satisfying $m^2 - p^k = 8$, then $m < p^k$.
Proof
Let $p^k m^2$ be an odd perfect number satisfying $m^2 - p^k = 8$.
Then $$(m + 3)(m - 3) = m^2 - 9 = p^k - 1.$$
This implies that $(m + 3) \mid (p^k - 1)$, from which it follows that
$$m < m + 3 \leq p^k - 1 < p^k.$$
We therefore conclude that $m < p^k$.
QED
So now consider the equation $m^2 - p^k = 4z$. Following our proof strategy, and the formula in the accepted answer to the first hyperlinked question, we have:
$$m^2 - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2 = p^k + \Bigg(4z - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg).$$
So the only remaining question now is whether it could be proved that
$$\Bigg(4z - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg) = -y < 0$$
for some positive integer $y$?
In other words, is it possible to prove that it is always the case that
$$\Bigg((m^2 - p^k) - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg) < 0,$$
if $n = p^k m^2$ is an odd perfect number with special prime $p$?
(Additionally, note that it is known that $m^2 - p^k$ is not a square, if $p^k m^2$ is an OPN with special prime $p$. See this MSE question and the answer contained therein.)
If so, it would follow that
$$\Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg)\Bigg(m - \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) = p^k - y$$
which would imply that
$$\Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) \mid (p^k - y)$$
from which it follows that
$$m < \Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) \leq p^k - y < p^k.$$
Update (November 11, 2020 - 10:21 PM Manila time) Please check out the recently posted answer for a minor adjustment to the logic that should make the general proof argument work.
| If you don't have a proof that the smallest square larger than $m^2-p^k$ is not $m^2$, then your method does not work.
Otherwise, your method works.
Using your idea, one can prove that if $\lfloor\sqrt{4z}+1\rfloor\lt m$, then $m\lt p^k$.
Proof :
Subtracting $\lfloor\sqrt{4z}+1\rfloor^2$ which is the smallest square larger than $4z$ from the both sides of
$$m^2=p^k+4z$$
gives
$$m^2-\lfloor\sqrt{4z}+1\rfloor^2=p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z$$
which can be written as
$$(m-\lfloor\sqrt{4z}+1\rfloor)(m+\lfloor\sqrt{4z}+1\rfloor)=p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z\tag1$$
So, we can say that
$$m+\lfloor\sqrt{4z}+1\rfloor\mid p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z\tag2$$
If $\lfloor\sqrt{4z}+1\rfloor\lt m$, then LHS of $(1)$ is positive, so RHS of $(1)$ is positive. So, we can say that$$(2)\implies m+\lfloor\sqrt{4z}+1\rfloor\le p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z$$from which we can have$$m\lt m+\lfloor\sqrt{4z}+1\rfloor\le p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z\lt p^k.\quad\blacksquare$$
If $m=\lfloor\sqrt{4z}+1\rfloor$, then letting $\sqrt{4z}=N+a$ where $N\in\mathbb Z$ and $0\le a\lt 1$, we have
$$p^k-m=(N+1)^2-(N+a)^2-N-1=(1-2a)N-a^2$$
whose sign depends on $a$ and $N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$ This is the $y=2^k$ case of this question.
Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?
Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?
| Following from the OP's self-answer, the quadratic Diophantine equation $$nx^2-4(n-1)y^2=1\tag1$$ means that $nx^2\equiv1\pmod4$ which forces $n=4N+1$ as $x^2\equiv0,1\pmod4$. Thus $$(4N+1)x^2-NY^2=1$$ where $Y=4y$ so $Y^2\equiv4\pmod{4N+1}$ and $x^2\equiv1\pmod N$.
Suppose that $Y\equiv\pm2\pmod{4N+1}$. Then $y=(rn\pm1)/2$ and substituting into $(1)$ gives $$nx^2-(n-1)(rn\pm1)^2=1\implies x^2=r^2n^2-(r^2\mp2r)n\mp2r+1.$$ Let $x=rn-a$ so $$n=\frac{a^2\pm2r-1}{2ra-r^2\pm2r}=\frac1{4r^2}\left(2ra+r^2\mp2r+\frac{r^2(r\pm4)}{2a-r\pm2}\right)$$ which reduces to $$2^{k+3}=m+2r+\frac{r(r\pm4)}m$$ since $rn\pm1=2y$ and $y=2^k$.
Notice that this formulation is quite similar to your equivalency statement. The latter is derived from the system $sx=ty=4^k-1$ and $x+y=2^{k+1}$, which in turn is equivalent to solving $st=c(4^k-1)$ and $s+t=c\cdot2^{k+1}$. In the formulation above, we are looking for integers $s,t$ such that $st=r(r\pm4)$ and $s+t=2^K-2r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 2
} |
Find the probability that a 4 will appear on the second dice. Two dice are rolled. If the sum of the two dice is 7 or less than 7. Find the probability that a 4 will appear on the second dice.
I am trying to do this exercise. But my result is different from the one provided by the book. In the book it appears that the answer is $1/4$. This is what I have done. It's okay?
Let $P(A)$ be The probability that the sum is equal to 7 or less than $7$ and let $P(B)$ be the probability that a 4 appears on the second die, then
$P(A)=\frac{21}{36}=\frac{7}{12}$
$P(B)=\frac{6}{36}=\frac{1}{6}$
So $P(A\cap B)=\frac{3}{36}=\frac{1}{12}$
Thus $\displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{12}}{\frac{7}{12}}=\frac{1}{7}$
If I try $P(A|B)$ and I get $1/2$
| Your working is correct.
With one of the dice being $1$, you can have max sum of $7$ with total of $6 \times 2 - 1 = 11$ possibilities. (Numbers can interchange places on dice and subtract one as we did count $(1, 1)$ twice.)
Now with $2$, there are $10$ possibilities (but subtract 2 possibilities for $(1,2)$ which is covered and one for $(2, 2)$). Now you have $7$ possibilities.
With $3$, $3$ possibilities.
No point in checking with $4$, as all the possible cases summing to $7$ or less have been covered already.
So total of $21$ possibilities to get sum of $7$ or less on two dice.
$4$ on the second dice with sum being less than equal to $7$ - $(1, 4), (2, 4), (3, 4) \,$ - 3 possibilities.
So $\frac{1}{7}$ is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3905414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Changing bounds of integration question; spot the mistake. Suppose we wanted to compute $$\int \frac{1}{x^2 + x + 1} ~ dx.$$ One such way is to fix $c$ and let $x$ vary as bounds, thus we have that $$\int \frac{1}{x^2 + x + 1} ~ dx = \int_c^x \frac{1}{t^2 + t + 1} ~ dt = \int_c^x \frac{1}{(t + 1/2)^2 + 3/4} ~ dt.$$ We will now shift the integral by $1/2,$ so we have $$\int_c^x \frac{1}{(t + 1/2)^2 + 3/4} ~ dt = \int_{c+1/2}^{x + 1/2} \frac{1}{t^2 + 3/4} ~ dt.$$ The expansion theorem for integrals tells us that $$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{ac + k}^{bc + k} f(x/c) ~ dx.$$ Using the aforementioned theorem, we have that $$\int_{c+1/2}^{x + 1/2} \frac{1}{t^2 + 3/4} ~ dt = \frac{\sqrt{3}}{2} \int_{\frac{2}{\sqrt{3}}c+1/2}^{\frac{2}{\sqrt{3}}x + 1/2} \frac{1}{(3/4)t^2 + 3/4} ~ dt.$$ Factoring, we have $$\frac{\sqrt{3}}{2} \int_{\frac{2}{\sqrt{3}}c+1/2}^{\frac{2}{\sqrt{3}}x + 1/2} \frac{1}{(3/4)t^2 + 3/4} ~ dt = \frac{\sqrt{3}}{2}\cdot\frac{4}{3} \int_{\frac{2}{\sqrt{3}}c+1/2}^{\frac{2}{\sqrt{3}}x + 1/2} \frac{1}{t^2 + 1} ~ dt.$$ This of course evaluates to $$\frac{2\sqrt{3}}{3} \arctan\left( \frac{2}{\sqrt{3}}x + \frac{1}{2} \right) + \text{constant}.$$ But then notice, taking the derivative of the expression before will not give us $1/(x^2 + x + 1).$ But, taking the derivative of $$\frac{2\sqrt{3}}{3} \arctan\left( \frac{2}{\sqrt{3}}x + \frac{1}{\sqrt{3}} \right) + \text{constant}$$ does give us $1/(x^2 + x + 1).$ Where is the mistake?
| $$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{ac + k}^{bc + k} f(x/c) ~ dx.$$ is wrong. It should be:
$$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{a(c + k)}^{b(c + k)} f(x/c) ~ dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3906621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is $h(n)$ independent of $n$? Let $n$ be a positive integer.
Let $f(n) = \lfloor\frac{8n+13}{25}\rfloor$ and $g(n) = \lfloor\frac{n-17}{25}\rfloor$.
Now consider $$h(n) = f(n) - \lfloor\frac{n - 12 - g(n)}{3}\rfloor$$
Then I calculated values of $h(n)$ for various values of $n$ and I'm getting $h(n)=4 $ everytime. Infact I ran a C code also and verified it till $n = 10^6$. But I am not able to prove this equality. How can I prove this?
| Let $n=25k+r$ for some non-negative integers $k,r$ with $r<25$. We then have
$$\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor = \left\lfloor\frac{24k+r-12-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\right\rfloor = 8k-4+\left\lfloor\frac{r-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\right\rfloor.$$
We also have
$$\left\lfloor\frac{8n+13}{25}\right\rfloor=8k+\left\lfloor\frac{8r+13}{25}\right\rfloor.$$
So it suffices to show that for any non-negative integer $r<25$,
$$\left\lfloor\frac{8r+13}{25}\right\rfloor - \left\lfloor\frac{r-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\right\rfloor=0.$$
Try showing that this is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the smallest possible number of different numbers between $a + b, b + c, c + a, ab + 1, bc + 1, ca + 1, abc$
Find the smallest possible number of different real positive number
between $a + b, b + c, c + a, ab + 1, bc + 1, ca + 1, abc$
$a$, $b$, $c$ are real positive numbers and they're all
different. From that, I can tell that $a+b \ne b+c \ne c+a$, and $ab + 1 \ne bc + 1 \ne ca + 1$, but I couldn't get any further with the exercise. How can I find the answer and prove it?
P. S. I am not very good at these kinds of exercises but couldn't find any good resources. If you know about any, I'd appreciate it if you told me about some.
| Consider the inequality chain:
$$ a+b =ab+1 < a+c = ac+1 = abc < b+c < bc+1,$$
which has general solutions $ a = 1, 1 < b < \frac{1+\sqrt{5}}{2} , c = \frac{1}{b-1}$.
As an explicit example, with $a = 1, b = 1.5, c = 2$, we get 4 distinct values: $2.5, 3, 3.5, 4.$
We will show that 4 distinct values is the best we can do.
Proof by contradiction. Suppose we can get 3 distinct values.
WLOG $ a < b < c$.
Notice that $ a+ b < a+c < c + b $, so this must be the 3 distinct values in order.
and $ ab + 1 < ac + 1 < bc + 1$, so this must be the 3 distinct values in order.
Thus, we have $ a + b = ab+ 1, a+c = ac+1, c+b = bc+1$.
This yields $(a-1)(b-1) = 0$, $(a-1)(c-1) = 0$, $(b-1)(c-1) = 0$.
This can be satisfied if and only if (at least) 2 of the variables are equal to 1, which contradicts the assumption that they are distinct.
To come up with the example of 4 distinct values, study the possible chains of intertwining the 2 inequalities, along with $abc$. Several of these lead to solutions. (This is not an exhaustive list)
*
*The initial inequality chain of
$$ a+b =ab+1 < a+c = ac+1 = abc < b+c < bc+1,$$
with solutions $ a = 1, 1 < b < \frac{1+\sqrt{5}}{2} , c = \frac{1}{b-1}$.
*We could have
$$ a+b =ab+1 < a+c = ac+1 < abc = b+c < bc+1,$$
with solutions $ a = 1, 1 < b <2, c = 1 + \frac{1}{ b - 1 } $.
*We could have
$$ a+b = ab+ 1 = abc , a+c, ac+1, b+c = bc+1, $$
which has solutions $0 < a < 1, b=1, c = \frac{a+1}{a}$.
As an explicit example, $ a = \frac{1}{2}, b = 1, c = 3$.
*We could have
$$ a+b < ab+1 = a+c < ac+1 = b+c < bc+1 = abc,$$
with solution $a\approx1.33826, b\approx1.61803, c\approx1.82709$ taken from Wolfram.
*We could have
$$ a+b < ab+1 = a+c =abc < ac+1 = b+c < bc+1,$$
with solution $a\approx 1.27716, b \approx 1.42902, c \approx 1.54792$ taken from Wolfram
*We could have
$$ a+b = abc < ab+1 = a+c < ac+1 = b+c < bc+1,$$
with solution $a\approx 1.26523, b\approx 1.39919, c\approx 1.50507$ taken from Wolfram.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Bessel function, characteristic function, semicircle distribution I am trying to prove that $\dfrac{1}{x} J_1(2x)$, where $J_n$ is the Bessel function of order n, is the characteristic function of the semicircle distribution, i.e. $σ(x)=\dfrac{1}{2π}\sqrt{4-x^2}
\textbf{1}_{|x|<2} $. Basically, I would like to calculate the integral
$$\dfrac{1}{2π} \displaystyle\int_{\mathbb{R}}{\dfrac{1}{t}J(2t)}e^{-itx} dt $$
So, I do not want to calculate the characteristic equation of $σ$, I want to find $σ$ when I am given the characteristic function. I have tried some integral forms of $J_1$ but can't seem to find what I am looking for. Any hints?
| $$J_1(z) = \dfrac{1}{iπ} \int_{0}^π e^{iz\cos θ}\cos θ dθ $$
Now, we want to calculate the integral shown above to find the density function corresponding to the characteristic function $ \dfrac{1}{x} J_1(2x) $.
\begin{align*}
f(x) &= \dfrac{1}{2π} \int_{\mathbb{R}} φ(t) e^{-itx} dt = \dfrac{1}{2π} \int_{\mathbb{R}} \dfrac{1}{t} J_1\left(2t\right) e^{-itx} dt \\[7pt]
&= \dfrac{1}{2iπ^2} \int_{\mathbb{R}}\int_{0}^π \dfrac{1}{t} e^{2it\cos θ }\cos θ e^{-itx} dθ dt \\[7pt]
&= \dfrac{1}{2iπ^2} \int_{0}^π \cos θ \underbrace{ \int_{\mathbb{R}} \dfrac{1}{t} e^{it(2\cos θ-x)} dt }_{-iπ \text{sgn}(2cosθ-x) } dθ \\[7pt]
&= - \dfrac{1}{2π} \int_{0}^π \cos θ \text{sgn}(2\cos θ-x) dθ
\end{align*}
If $ x > 2 $ then $ \text{sgn} (2 \cos θ-x) = -1$, and if $ x < -2 $ then $ \text{sgn} (2 \cos θ-x) = 1$ and so, since $ \int_{0}^π \cos θ dθ = 0$, $ f(x) = 0$ as well. Now, let $ -2\le x\le 2 $. If we set $ u=2\cos θ-x $, then
\begin{align*}
&-\dfrac{1}{2π}\int_{-2-x}^{2-x} \dfrac{u+x}{2} \dfrac{1}{\sqrt{1-\left(\frac{u+x}{2}\right)^2}} \text{sgn}(u) du \\[7pt]
&= \dfrac{1}{2π}\int_{-2-x}^{0} \dfrac{u+x}{2} \dfrac{1}{\sqrt{1-\left(\frac{u+x}{2}\right)^2}} du-\dfrac{1}{2π}\int_{0}^{2-x} \dfrac{u+x}{2} \dfrac{1}{\sqrt{1-\left(\frac{u+x}{2}\right)^2}} du\\[7pt]
&= \dfrac{1}{2π} \left( 2 \int_{-1}^{\frac{x}{2}} \dfrac{y}{\sqrt{1-y^2}} dy - 2\int_{\frac{x}{2}}^{1} \dfrac{y}{\sqrt{1-y^2}} dy \right)\\[7pt]
&=\dfrac{1}{2π} \left( 2 \sqrt{1-y^2} \bigg\rvert_{-1}^{x/2} - 2 \sqrt{1-y^2} \bigg\rvert_{x/2}^{1} \right) = \dfrac{1}{2π} \sqrt{4-x^2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$a_n = u^n+v^n+(-1)^n$ Anyone has solutions for this problem?
Let $u, v$ are distinct roots of equation:
$$x^2-tx+1=0 (t \in \mathbb{N}, t>2)$$
and sequence $$(a_n): a_n = u^n+v^n+(-1)^n$$
Prove that: $$a_{3^m}\mid a_{3^mn} \forall m,n \in \mathbb{N} , gcd(n,3)=1$$
Notes
$S_n=\frac{a_{3^mn}}{a_{3^m}}\\$
$a=u^{3^m}$ then $\frac{1}{a}=v^{3^m}$
+If $n$ is odd
$$S_n = \frac{a^{2n}-a^n+1}{a^{n-1}(a^2-a+1)}$$
Because $gcd(n,6)=1$, polynomial $a^{2n}-a^n+1$ is divided by $a^2-a+1$, then $\frac{a^{2n}-a^n+1}{a^2-a+1}$ is a symmetric polynomial with degree $2(n-1)$
So $S_n=\sum\limits_{i=0}^{n-1}s_i(a^i+\frac{1}{a^i})$
+If $n$ is even, then $n=2^tn'$
$S_n = S_{n'}.a_{\frac{n}{2}}a_{\frac{n}{4}}...a_{\frac{n}{2^t}}$
| By either using Vieta's formulas or just expanding $(x - u)(x - v) = x^2 - tx + 1$ and comparing coefficients, we get
$$\begin{equation}\begin{aligned}
u + v & = t \\
uv & = 1
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
For non-negative integers $i$, define $f_i(t)$ to be $u^i + v^i$ expressed as a function of $t$. Thus, $f_0(t) = u^0 + v^0 = 2$ and $f_1(t) = u + v = t$. For any $i \ge 2$, using \eqref{eq1A}, we get
$$\begin{equation}\begin{aligned}
(u^{i-1} + v^{i-1})(u + v) & = u^{i} + u^{i-1}(v) + v^{i-1}(u) + v^{i} \\
t(f_{i-1}(t)) & = u^{i} + v^{i} + uv(u^{i-2} + v^{i-2}) \\
t(f_{i-1}(t)) & = f_{i}(t) + f_{i-2}(t) \\
f_{i}(t) & = t(f_{i-1}(t)) - f_{i-2}(t)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
This confirms $u^{i} + v^{i}$, for all $i \ge 0$, can be expressed as a polynomial in $t$ with integral coefficients and, thus, is an integer. Next, for any odd positive integer $j$, your sequence gives
$$a_j = u^j + v^j - 1 \tag{3}\label{eq3A}$$
For conciseness, with any integer $k \ge 0$, let
$$w_k = u^{kj} + v^{kj} \tag{4}\label{eq4A}$$
With $k = 0$ and $k = 1$, we get
$$\begin{equation}\begin{aligned}
& w_0 = u^{0} + v^{0} = 1 + 1 \equiv 2 \pmod{a_j} \\
& w_1 = u^{j} + v^{j} \equiv 1 \pmod{a_j}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
For $k \ge 2$,
$$\begin{equation}\begin{aligned}
(u^{(k-1)j} + v^{(k-1)j})((u^j + v^j) - 1) & \equiv 0 \pmod{a_j} \\
u^{kj} + u^{(k-1)j}(v^{j}) + v^{(k-1)j}(u^j) + v^{kj} & \equiv u^{(k-1)j} + v^{(k-1)j} \pmod{a_j} \\
u^{kj} + v^{kj} + (uv)^{j}(u^{(k-2)j} + v^{(k-2)j}) & \equiv w_{k-1} \pmod{a_j} \\
w_{k} + (u^{(k-2)j} + v^{(k-2)j}) & \equiv w_{k-1} \pmod{a_j} \\
w_k & \equiv w_{k-1} - w_{k-2} \pmod{a_j}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
This therefore gives
$$\begin{equation}\begin{aligned}
& w_2 \equiv w_1 - w_0 \equiv 1 - 2 \equiv -1 \pmod{a_j} \\
& w_3 \equiv w_2 - w_1 \equiv -1 - 1 \equiv -2 \pmod{a_j} \\
& w_4 \equiv w_3 - w_2 \equiv -2 - (-1) \equiv -1 \pmod{a_j} \\
& w_5 \equiv w_4 - w_3 \equiv -1 - (-2) \equiv 1 \pmod{a_j} \\
& w_6 \equiv w_5 - w_4 \equiv 1 - (-1) \equiv 2 \pmod{a_j} \\
& w_7 \equiv w_6 - w_5 \equiv 2 - 1 \equiv 1 \pmod{a_j}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Due to \eqref{eq6A} showing $w_k$ depends only on the previous $2$ smaller indice values, then since $w_6 \equiv w_0 \pmod{a_j}$ and $w_7 \equiv w_1 \pmod{a_j}$, the congruence values will repeat cyclically with a period of $6$. Thus, using \eqref{eq5A} and \eqref{eq7A}, for all non-negative integers $r$,
$$\begin{equation}\begin{aligned}
& a_{(6r)j} \equiv w_0 + 1 \equiv 3 \pmod{a_j} \\
& a_{(6r + 1)j} \equiv w_1 - 1 \equiv 0 \pmod{a_j} \\
& a_{(6r + 2)j} \equiv w_2 + 1 \equiv 0 \pmod{a_j} \\
& a_{(6r + 3)j} \equiv w_3 - 1 \equiv -3 \pmod{a_j} \\
& a_{(6r + 4)j} \equiv w_4 + 1 \equiv 0 \pmod{a_j} \\
& a_{(6r + 5)j} \equiv w_5 - 1 \equiv 0 \pmod{a_j}
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
For all $n \in \mathbb{N}$ with $\gcd(n, 3) = 1$, \eqref{eq8A} shows $a_{nj} \equiv 0 \pmod{a_j} \implies a_j \mid a_{nj}$. Since $j$ was any odd positive integer, this includes your question's specific case of $j = 3^m$. In addition, apart from the one case of $t = 4$ and $j = 1$, since in all other cases $a_j \gt 1$ and $a_j \neq 3$, we also have for $\gcd(n, 3) = 3$ that $a_{j} \not\mid a_{nj}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3927015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Moment generating function of Laplace distribution step by step There are two similar posts but none of them helped me to get through the full derivation of the very simple MGF that should be according to Wikipedia:
$$
\frac {\exp(\mu t)}{1-b^{2}t^{2}}
$$
Here's my attempt:
Definition of MGF
$$
M_X(t) = \mathbb{E}\left[\exp(t X)\right]
$$
by LOTUS
$$
= \int \exp(t x) \cdot p(x) dx
$$
Plugging-in PDF
$$
= \int \exp(t x) \cdot \frac{1}{2b}\exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$
$$
= \frac{1}{2b} \int \exp(t x) \cdot \exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$
$$
= \frac{1}{2b} \int \exp \left(t x - \frac{\mid x - \mu \mid }{b} \right) dx
$$
getting rid of abs. value
$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(t x - \frac{ x - \mu }{b} \right) dx
+ \frac{1}{2b} \int_{\mu}^{\infty} \exp \left(t x - \frac{- x + \mu }{b} \right) dx
$$
the left integral first
$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x - \mu }{b} \right) dx
$$
$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x }{b} \right) \exp \left(\frac{- \mu }{b} \right) dx
$$
$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu }{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x }{b} \right) dx
$$
$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu }{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{ (bt - 1) }{b} x \right) dx
$$
Now since $\int e^{ax}\,dx={\frac {1}{a}}e^{ax}+C$
$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu }{b} \right) \cdot \left| \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} x \right) \right|_{-\infty}^{\mu}
$$
Let's evaluate the integral first
$$
\left| \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} x \right) \right|_{-\infty}^{\mu}
$$
$$
= \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right)
-
\frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} (-\infty)\right)
$$
Since $\lim_{x \rightarrow 0} e^{-x} = 0$, the second term is 0
$$
= \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right)
- 0
$$
Double check from WorlframAlpha:
$$
= \frac{b}{bt-1} \exp \left( \mu \left( t - \frac{1}{b} \right) \right)
$$
for $\frac{1}{b} < t$
Now the right-hand side integral:
$$
\frac{1}{2b} \int_{\mu}^{\infty} \exp \left(t x - \frac{- x + \mu }{b} \right) dx
$$
$$
\frac{1}{2b} \int_{\mu}^{\infty} \exp \left(\frac{b t x + x - \mu }{b} \right) dx
$$
$$
= \frac{1}{2b} \int_{\mu}^{\infty} \exp \left(\frac{b t x + x }{b} \right) \exp \left(\frac{-\mu }{b} \right)dx
$$
$$
= \frac{1}{2b} \cdot \exp \left(\frac{ -\mu }{b} \right) \int_{\mu}^{\infty} \exp \left(\frac{b t x + x }{b} \right) dx
$$
$$
= \frac{1}{2b} \cdot \exp \left(\frac{ -\mu }{b} \right) \int_{\mu}^{\infty} \exp \left( \frac{b t + 1 }{b} x\right) dx
$$
Now evaluate the integral
$$
\int_{\mu}^{\infty} \exp \left( \frac{b t + 1 }{b} x\right) dx
$$
According to wolfram alpha
$$
= - \frac{b}{bt + 1}\exp\left(\mu \frac{1}{b + t}\right)
$$
for $\frac{1}{b} + t < 0$
Now sum-up these two integrals:
$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu }{b} \right) \cdot \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right) + \\
\frac{1}{2b} \cdot \exp \left(\frac{ -\mu }{b} \right) \cdot \left( - \frac{b}{bt + 1}\exp\left(\mu \frac{1}{b + t}\right) \right)
$$
And simplify
$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu }{b} \right) \cdot \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right) - \\
\frac{1}{2b} \cdot \exp \left(\frac{ -\mu }{b} \right) \cdot \frac{b}{bt + 1} \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$
$$
= \frac{1}{2b} \cdot \frac{b}{ (bt - 1) } \cdot \exp \left(\frac{- \mu }{b} \right) \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right) - \\
\frac{1}{2b} \cdot \frac{b}{bt + 1} \cdot \exp \left(\frac{ -\mu }{b} \right) \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$
$$
= \frac{1}{b} \cdot \frac{1}{ (bt - 1) } \cdot \exp \left(\frac{- \mu }{b} \right) \cdot \exp \left(\frac{ (bt - 1) }{b} (\mu) \right) - \\
\frac{1}{b} \cdot \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu }{b} \right) \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$
$$
= \frac{1}{b} \cdot \frac{1}{ (bt - 1) } \cdot \exp \left(\frac{- \mu }{b} +\frac{ (bt - 1) }{b} (\mu) \right) - \\
\frac{1}{b} \cdot \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu }{b} + \mu \frac{1}{b + t}\right)
$$
$$
= \frac{1}{b} \cdot \frac{1}{ (bt - 1) } \cdot \exp \left( \mu \left(t - \frac{2}{b} \right) \right) - \\
\frac{1}{b} \cdot \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu }{b} + \mu \frac{1}{b + t}\right)
$$
But I cannot move away from this ugly formula to the simple beauty. Is there just a stupid mistake above or am I doing something substantially wrong?
| Alright, here's the full correct solution if anyone is interested.
Definition of MGF
$$
M_X(t) = \mathbb{E}\left[\exp(t X)\right]
$$
by LOTUS
$$
= \int \exp(t x) \cdot p(x) dx
$$
Plugging-in PDF
$$
= \int \exp(t x) \cdot \frac{1}{2b}\exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$
$$
= \frac{1}{2b} \int \exp(t x) \cdot \exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$
$$
= \frac{1}{2b} \int \exp \left(t x - \frac{\mid x - \mu \mid }{b} \right) dx
$$
For integrating absolute value, we split the integral at $\mu$, such that it change the absolute values to
*
*for $x < \mu$: $\mid x - \mu \mid = -(x - \mu)$
*for $x \geq \mu$: $\mid x - \mu \mid = x - \mu$
$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(t x - \frac{-(x - \mu) }{b} \right) dx +
\frac{1}{2b} \int_{\mu}^{\infty} \exp \left(t x - \frac{x - \mu}{b} \right) dx
$$
$$
= \frac{1}{2b} \left( \int_{-\infty}^{\mu} \exp \left(\frac{b t x + x - \mu }{b} \right) dx +
\int_{\mu}^{\infty} \exp \left(\frac{btx - x + \mu}{b} \right) dx \right)
$$
$$
= \frac{1}{2b} \left( \int_{-\infty}^{\mu} \exp \left(\frac{b t x + x - \mu }{b} \right) dx +
\int_{\mu}^{\infty} \exp \left(\frac{btx - x + \mu}{b} \right) dx \right)
$$
Let's solve the left-hand side integral first
$$
\int_{-\infty}^{\mu} \exp \left(\frac{b t x + x - \mu }{b} \right) dx
$$
$$
= \exp \left( \frac{-\mu}{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{b t x + x}{b} \right) dx
$$
$$
= \exp \left( \frac{-\mu}{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{x (b t + 1)}{b} \right) dx
$$
Now since $\int e^{ax}\,dx={\frac {1}{a}}e^{ax}+C$
$$
= \exp \left( \frac{-\mu}{b} \right) \frac{b}{bt + 1} \left| \exp \left(\frac{x (b t + 1)}{b} \right) \right|_{-\infty}^{\mu}
$$
When evaluating for $-\infty$, we need to have this limit converge:
$$
\lim_{x \rightarrow - \infty} \exp \left(x \frac{(b t + 1)}{b} \right)
$$
The constants have to be positive, so
$$
\frac{(b t + 1)}{b} > 0 \\
t > -\frac{1}{b}
$$
So let's evaluate and move on
$$
= \exp \left( \frac{-\mu}{b} \right) \frac{b}{bt + 1} \left( \exp \left(\frac{\mu (b t + 1)}{b} \right) - 0 \right)
$$
$$
= \exp \left( \frac{-\mu + \mu (b t + 1)}{b} \right) \frac{b}{bt + 1}
$$
$$
= \exp \left( \frac{-\mu + \mu b t + \mu}{b} \right) \frac{b}{bt + 1}
$$
$$
= \exp \left( \frac{\mu b t}{b} \right) \frac{b}{bt + 1}
$$
$$
= \exp \left( \mu t \right) \frac{b}{bt + 1}
$$
Now the right-hand side integral:
$$
\int_{\mu}^{\infty} \exp \left(\frac{btx - x + \mu}{b} \right) dx
$$
$$
= \exp \left( \frac{\mu}{b} \right) \int_{\mu}^{\infty} \exp \left(\frac{btx - x}{b} \right) dx
$$
$$
= \exp \left( \frac{\mu}{b} \right) \int_{\mu}^{\infty} \exp \left(\frac{x(bt - 1)}{b} \right) dx
$$
And again since $\int e^{ax}\,dx={\frac {1}{a}}e^{ax}+C$
$$
= \exp \left( \frac{\mu}{b} \right) \frac{b}{bt - 1} \left| \exp \left(\frac{x(bt - 1)}{b} \right) \right|_{\mu}^{\infty}
$$
Again, for the integral to converge, we need to make sure that this goes to zero
$$
\lim_{x \rightarrow \infty} \exp \left(x \frac{(b t - 1)}{b} \right)
$$
So the constant have to be negative such that
$$
\frac{(b t - 1)}{b} < 0 \\
t < \frac{1}{b}
$$
Let's evalate and move on
$$
= \exp \left( \frac{\mu}{b} \right) \frac{b}{bt - 1} \left( 0 -\exp \left(\frac{\mu(bt - 1)}{b} \right) \right)
$$
$$
= - \exp \left( \frac{\mu}{b} \right) \frac{b}{bt - 1} \exp \left(\frac{\mu(bt - 1)}{b} \right)
$$
$$
= - \exp \left( \frac{\mu}{b} \right) \frac{b}{bt - 1} \exp \left(\frac{\mu bt - \mu)}{b} \right)
$$
$$
= - \exp \left( \frac{\mu + \mu bt - \mu}{b} \right) \frac{b}{bt - 1}
$$
$$
= - \exp \left( \mu t \right) \frac{b}{bt - 1}
$$
Now plug these two results in the original equation
$$
\frac{1}{2b} \left( \int_{-\infty}^{\mu} \exp \left(\frac{b t x + x - \mu }{b} \right) dx +
\int_{\mu}^{\infty} \exp \left(\frac{btx - x + \mu}{b} \right) dx \right)
$$
$$
= \frac{1}{2b} \left( \exp \left( \mu t \right) \frac{b}{bt + 1}
- \exp \left( \mu t \right) \frac{b}{bt - 1} \right)
$$
$$
= \frac{1}{2b} \exp \left( \mu t \right) \left( \frac{b}{bt + 1}
- \frac{b}{bt - 1} \right)
$$
$$
= \exp \left( \mu t \right) \left( \frac{b}{2b(bt + 1)}
- \frac{b}{2b(bt - 1)} \right)
$$
$$
= \exp \left( \mu t \right) \left( \frac{1}{2(bt + 1)}
- \frac{1}{2(bt - 1)} \right)
$$
$$
= \frac{1}{2 }\exp \left( \mu t \right) \left( \frac{1}{bt + 1}
- \frac{1}{bt - 1} \right)
$$
$$
= \frac{1}{2 }\exp \left( \mu t \right) \left( \frac{(bt - 1) - (bt + 1)}{b^2 t^2 - 1} \right)
$$
$$
= \frac{1}{2 }\exp \left( \mu t \right) \left( \frac{bt - 1 - bt - 1)}{b^2 t^2 - 1} \right)
$$
$$
= \frac{1}{2 }\exp \left( \mu t \right) \left( \frac{- 2}{b^2 t^2 - 1} \right)
$$
$$
= \exp \left( \mu t \right) \left( \frac{- 1}{b^2 t^2 - 1} \right)
$$
$$
= \exp \left( \mu t \right) \left( \frac{1}{- b^2 t^2 + 1} \right)
$$
$$
= \frac{\exp ( \mu t )}{1 - b^2 t^2} \qquad \square
$$
for $t < \frac{1}{b}$ and $t > -\frac{1}{b}$, which we can rewrite as $t < \frac{1}{b}$ and $- t < \frac{1}{b}$ and thus simplify to $|t| < \frac{1}{b}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3929288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Eigenvalues of an almost diagonal matrix I know that the eigenvalue of a diagonal matrix is simply the values in the diagonal. However, if I have a matrix of the following form:
$$
\begin{bmatrix}
a & b & 0 & 0 \\
b & c & 0 & 0 \\
0 & 0 & d & e \\
0 & 0 & e & f
\end{bmatrix}.
$$
Is there a closed form way to express the eigenvalues of this matrix? I can derive the eigenvalue of the smaller blocks along the diagonal, but how does it relate to the overall matrix?
| The eigenvalues of a block diagonal matrix are the eigenvalues of each block. The corresponding eigenvectors are the eigenvectors of each block padded with zeros. For example:
The eigenvalues of the matrix $$A = \begin{bmatrix}4 & 3 \\ 3 & 4 \end{bmatrix}$$ are $7$ and $1$, and the corresponding eigenvectors are respectively $$\begin{bmatrix}1/\sqrt{2}\\ 1/\sqrt{2} \end{bmatrix} \quad \text{and} \quad \begin{bmatrix}1/\sqrt{2}\\ -1/\sqrt{2} \end{bmatrix}.$$
The eigenvalues of the matrix $$B = \begin{bmatrix}2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2\end{bmatrix}$$ are $2+\sqrt{2}$, $2$, and $2-\sqrt{2}$ and the corresponding eigenvectors are respectively $$\begin{bmatrix}1/2 \\ -1/\sqrt{2} \\ 1/2\end{bmatrix}, \quad \begin{bmatrix}-1/\sqrt{2} \\ 0 \\ 1/\sqrt{2}\end{bmatrix}, \quad \text{and} \quad \begin{bmatrix}1/2 \\ 1/\sqrt{2} \\ 1/2\end{bmatrix}.$$
The eigenvalues of the matrix $$\begin{bmatrix}A & 0 \\ 0 & B \end{bmatrix} = \begin{bmatrix}4 & 3 & 0 & 0 & 0 \\ 3 & 4 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 1 & 2 \end{bmatrix}$$ are $7$, $1$, $2+\sqrt{2}$, $2$, and $2-\sqrt{2}$, and the corresponding eigenvectors are respectively $$\begin{bmatrix}1/\sqrt{2}\\ 1/\sqrt{2} \\ 0 \\ 0 \\ 0\end{bmatrix}, \quad \begin{bmatrix}1/\sqrt{2}\\ -1/\sqrt{2} \\ 0 \\ 0 \\ 0\end{bmatrix}, \quad \begin{bmatrix}0 \\ 0 \\ 1/2 \\ -1/\sqrt{2} \\ 1/2\end{bmatrix}, \quad \begin{bmatrix}0 \\ 0 \\ -1/\sqrt{2} \\ 0 \\ 1/\sqrt{2}\end{bmatrix}, \quad \text{and} \quad \begin{bmatrix}0 \\ 0 \\ 1/2 \\ 1/\sqrt{2} \\ 1/2\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3932481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\left.\int_0^{\pi/2}\sqrt{1+\frac1{\sqrt{1+\tan^nx}}}\text dx \middle/\int_0^{\pi/2}\sqrt{1-\frac1{\sqrt{1+\tan^nx}}}\text dx\right.$
Evaluate the integral ratio$$\dfrac{I_1}{I_2}=\dfrac{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}$$
Using Wolframalpha, I found out that the ratio of the above two integrals is $\sqrt{2}+1$, that is, independent of $n$. I couldn't think of a way to solve it.
I don't think any simple substitutions would work here. Any useful hints would be appreciated.
| Note
\begin{align}
\frac{I_1}{I_2}&=1+ \frac{I_1-I_2}{I_2} \\
&=1+\frac1{I_2}\int_0^{\frac\pi2} \left( \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}}- \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \right)dx \\
&=1+\frac1{I_2}\int_0^{\frac\pi2}
\frac{\sqrt{\sqrt{\sin^nx+\cos^nx}+\sqrt{\cos^nx}}
- \sqrt{\sqrt{\sin^nx+\cos^nx}-\sqrt{\cos^nx}}}{\sqrt[4]{\sin^nx+\cos^nx}}dx \\
&=1+\frac1{I_2} \int_0^{\frac\pi2}
\frac{\sqrt{2\sqrt{\sin^nx+\cos^nx}-2\sqrt{\sin^nx}}
}{\sqrt[4]{\sin^nx+\cos^nx}}dx \\
&= 1+\frac{\sqrt2}{I_2} \int_0^{\frac\pi2} \sqrt{1-\frac{1}{\sqrt{1+\cot^nx}}}dx\>\>\>\>\>\>\>\>(x\to \frac\pi2-x)\\
&=1+\frac{\sqrt2}{I_2}\int_0^{\frac\pi2} \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}}dx \\ &=1+\sqrt2 \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3939493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 1,
"answer_id": 0
} |
Range of quadratic function using discriminant
Let $x^2-2xy-3y^2=4$. Then find the range of $2x^2-2xy+y^2$.
Let $2x^2-2xy+y^2=a$.
Then $ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2\implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$.
We divide both side by $y^2$ and let $t=\frac{x}{y}$.
Then it implies $(a-8)t^2-(2a-8)t-(3a+4)=0$.
Since its discriminant is not negative, $\frac{\Delta}{4}\ge 0\implies a^2-7a-4\ge 0$. It gives us $a$ can have negative values like $-1$. But if clearly contracts $a-4=x^2+4y^2\ge 0\implies a\ge 4$. Where did I mistake?
| Consider the curves $x^2-2xy-3y^2=4 \ (C)$ and $2x^2-2xy+y^2=k^2+4 \ (E)$ At the points where they intersect, we can substitute the value of $-2xy$ from the first into the second to get $$x^2+4y^2=k^2 \ (E)$$ This represents an ellipse centered at the origin. Now, either with a bit of graph sketching, or by considering $(C)$ as a quadratic in $x^2$, deduce that $y$ can range anywhere in the real numbers, and that $(C)$ is a hyperbola. Increasing $k$ in $(E)$ just enlarges the ellipse, and it’s not hard to see that there is no upper bound on $k$ for $(E)$ and $(C)$ to intersect. For obtaining a lower bound, we need to find $k$ such that the two curves touch. Note that if $(x,y)$ lies on the two curves, then so does $(-x,-y)$. So we’re looking for exactly two intersections. Then substituting $x=\pm\sqrt{k^2-4y^2}$ in $(C)$ gives a quadratic in $y^2$: $$65y^4-y^2(18k^2-56)+(k^2-4)^2 = 0 $$ Setting the discriminant equal to zero will give $k^2=\frac{\sqrt{65}-1}{2}$ and hence we have the desired range: $$\frac{\sqrt{65}+7}{2} \le 2x^2-2xy+y^2 =k^2+4\lt \infty $$
What was wrong in your approach? What you stated was a necessary condition on $a$, but that doesn’t mean that $a$ could really equal anything in that range.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
A polynomial has the same remainder when divided by $x+k$ or $x-k$; what is $k$? Question
Given that $y = 3x^3 + 7x^2 - 48x + 49$ and that $y$ has the same remainder when it is divided by $x + k$ or $x - k$, find the possible values of $k$.
My attempt
Let $f(x) = 3x^3 + 7x^2 - 48x + 49$
$\text{Using Remainder Theorem,}$
\begin{align}
f(-k) &= f(k) \\
3(-k)^3 + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\
-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\
-3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\
-6k^3 + 14k^2 + 96k &= 0 \\
\frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\
6k^2 + 14k - 96 &= 0
\end{align}
$\text{Comparing } 6k^2 + 14k - 96 = 0 \text{ with } ak^2 + bk + c = 0, a = 6, b = 14, c = -96$
\begin{align}
k = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } &= \frac{ -(14) \pm \sqrt{(14)^2 - 4(6)(-96)} }{ 2(6) } \\
k &= 3 \text{ and } -5\frac{1}{3}
\end{align}
$\therefore k = 3, -5\frac{1}{3} \text{ or } 0 $
My answer is incorrect. The correct answer is $k = 0, 4 \text{ or } -4$
| \begin{align}
f(-k) &= f(k) \\
3(-k)^3 + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\
\bf-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\
-3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\
-6k^3 + 14k^2 + 96k &= 0 \\
\frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\
6k^2 + 14k - 96 &= 0
\end{align}
Right there. $7(-k)^2 = 7k^2$. Looking through it, there are more sign errors throughout. $-7k^2 - 7k^2$ would be $-14k^2$, not $14k^2$.
Another one: $-3k^3 - -3k^3 = 0$.
Dividing by $k$ isn't exactly a kosher operation. What if $k = 0$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$ I've tried to solve this limit:
$$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
Here,
$$\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$ $$\sim \frac{(x \arctan x-x^2)(\sqrt{1+2x^4}+1)}{2x^4}$$ $$\sim \frac{(x \arctan x-x^2)}{x^4}= \frac{ \arctan x-x}{x^3} \sim \frac{x-x}{x^3} \sim 0$$
But the final result should be $-\frac{4}{3}$.
Any help would be appreciated.
| The handling of denominator is easy. Multiplying the expression under limit with $$\frac{\sqrt{1+2x^4}+1}{\sqrt{1+2x^4}+1}$$ we can see that the limit in question is equal to the limit of the expression $$\frac{\log(1+x\arctan x) - e^{x^2}+1}{x^4}$$ Split this expression into three parts as $$\frac{\log(1+x\arctan x) - x\arctan x} {(x\arctan x) ^2}\cdot\frac{(\arctan x) ^2}{x^2}+\frac{\arctan x - x}{x^3}-\frac{e^{x^2}-1-x^2}{x^4}$$ For the first term substitute $u=x\arctan x$ and conclude that it tends to $-1/2$. You can also find the limit of second term easily as $-1/3$. For third term just put $t=x^2$ and observe that it tends to $1/2$. The desired answer is thus $$-\frac{1}{2}-\frac{1}{3}-\frac{1}{2}=-\frac{4}{3}$$
The issue with your approach is the wrong use of equivalents ($\sim$ notation). Unless you are familiar with the rules of this notation avoid using it and instead stick to the usual limit laws/theorems/rules.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$4q=a^2+27b^2$ elementary proof of uniqueness I’m trying to prove that if $q$ is prime then the representation $$4q=a^2+27b^2$$ is unique (up to sign). Any help would be appreciated.
Edit: By elementary I mean only using divisibility rules, modular arithmetic, etc.
| Suppose that $4q=a^2+27b^2=c^2+27d^2$ and $a,b,c,d> 0$ (equals 0 is not possible).
We have $a^2d^2-b^2c^2\equiv 0 $ mod $q$.
Then $(ad-bc)(ad+bc)\equiv 0$ mod $q$.
If $ad+bc\equiv 0$ mod $q$, then by $16q^2=(a^2+27b^2)(c^2+27d^2)=(ac-27bd)^2+27(ad+bc)^2$, we must have $ad+bc=0$. This is impossible, so we have $ad-bc \equiv 0 $ mod $q$.
By $16q^2=(a^2+27b^2)(c^2+27d^2)=(ac+27bd)^2+27(ad-bc)^2$,
we have $ad-bc=0$. Thus, $a/c=b/d=t$ gives
$a^2+27b^2=t^2(c^2+27d^2)=c^2+27d^2$. Hence, $t^2=1$ and $t=\pm 1$. Since $a,b,c,d> 0$, we must have $a=c$ and $b=d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Generalize a sum given first elements Given $\rho\in(0,1)$ and $f\colon\mathbb N\to\mathbb R$ defined as
$$\begin{aligned}
f(0)&\mapsto1\\
f(1)&\mapsto2+\rho\\
f(2)&\mapsto3+2\rho+\rho^2\\
f(3)&\mapsto4+3\rho+2\rho^2+\rho^3
\end{aligned}$$
generalize $f$ for an arbitrary $n$ and simplify the summator (if possible)
$$\sum_{n=0}^{\infty_+}(1+r)^{-n}f(n)$$
I think I could simplify the summator but I can't get my head around generalizing $f(n)$. What should be the approach to this? Thanks.
| I would like to add one more approach.
Consider
$$
f(n) = (n+1) + n \rho + (n-1) \rho^2 + \cdots + 1 \rho^n = \sum_{k=1}^{n+1} k \rho^{n+1-k}.
$$
Obviously $f(0)= 1$, $f(1) = 2 + \rho$ and so on.
Now make this function a bit more complicated
$$
f(n,x) = (n+1) x^n + n x^{n-1} \rho + (n-1) x^{n-2} \rho^2 + \cdots + 1 \rho^n.
$$
You may see that $f(n,1) = f(n)$.
Let's represent our new function as a derivative and do few transformations
$$
\begin{align}
f(n,x)
&= \frac{\partial}{\partial x} \left(x^{n+1} + x^{n} \rho + x^{n-1} \rho^2 + \cdots + x \rho^n\right) =\\
&= \frac{\partial}{\partial x} \left( x \left[x^{n} + x^{n-1} \rho + x^{n-2} \rho^2 + \cdots + \rho^n \right] \right) =\\
&= \frac{\partial}{\partial x} \frac{x (x - \rho) (x^{n} + x^{n-1} \rho + x^{n-2} \rho^2 + \cdots + \rho^n)}{x - \rho} =\\
&= \frac{\partial}{\partial x} \frac{x (x^{n+1} - \rho^{n+1})}{x - \rho}
\end{align}
$$
Doing all the boring stuff with derivatives and setting $x=1$, at the end of the day we get
$$
f(n) = \frac{(n+1) - \rho(n+2) + \rho^{n+2}}{(\rho - 1)^2}
$$
One may check that
$$
f(0) = \frac{1 - 2\rho + \rho^2}{(\rho - 1)^2} = 1,
$$
$$
\begin{align}
f(1) &= \frac{2 - 3\rho + \rho^3}{(\rho - 1)^2}
= \frac{2 - 4\rho + 2\rho^2 + \rho^3 - 2\rho^2 + \rho}{(\rho - 1)^2} =\\
&= \frac{2(1-\rho)^2+ \rho(\rho - 1)^2}{(\rho - 1)^2}
= 2 + \rho
\end{align}
$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3946005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Find a polynomial $P$ of the smallest degree such that $\int_{-1}^{1}\left|x^{2 / 3}-P(x)\right|^{2} d x<0.01$ Find a polynomial P of the smallest degree such that $$\int_{-1}^{1}\left|x^{2 / 3}-P(x)\right|^{2} d x<0.01$$
I am ususally solve this kind of "best approximation" problems by taking using $GS$ but I think this time it should be solved by Chebyshev polynomials since that was the topic of the lecture, I am not sure this is the idea here but I would like for some help here (Im not even sure if I can tag it under "chebyshev polynomials"
| First, let us try $\mathrm{deg} P(x) \le 2$.
Let $P(x) = ax^2 + bx + c$.
We have $$\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x =
\frac{2}{5}a^2 + \frac{4}{3}ac + \frac{2}{3}b^2 + 2c^2 - \frac{12}{11}a - \frac{12}{5} c + \frac{6}{7}.$$
It is easy to minimize a quadratic form. The best $P(x)$
is given by $P(x) = \frac{9}{11}x^2 + \frac{18}{55}$
with $\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x = 384/21175 \approx 0.01813 > \frac{1}{100}$.
Similarly, we try $\mathrm{deg} P(x) \le 3$. The best $P(x)$ is the same as that of $\mathrm{deg} P(x) \le 2$.
Then we try $\mathrm{deg} P(x) \le 4$.
The best $P(x)$ is given by
$P(x) = -\frac{189}{187}x^4 + \frac{315}{187}x^2 + \frac{45}{187}$
with $\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x = 1536/244783 \approx 0.006275 < \frac{1}{100}$.
Thus, the smallest degree is $4$.
Remark: How to find the best $P(x)$ if $\mathrm{deg} P(x) \le 2n$?
Let $P(x) = a_{2n}x^{2n} + \cdots + a_0$. We set
$$\frac{\partial }{\partial a_k}
\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x = 0, \ k=0, 1, \cdots, 2n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Best way to integrate this? $$ \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{\sin^2(x)}{\sqrt{1+\cos(x)}}dx $$
What would be the best way to go about integrating and evaluating this integral?
| Substitute $\cos x =t \implies -\sin x dx =dt $: $$\int_{0}^{0.5} \frac{\sqrt{1-t^2}}{\sqrt{1+t}} dt =\int_0^{0.5} \sqrt{1-t}dt = \left[ -\frac 23 (1-t)^{3/2} \right]_0^{0.5} = \frac 23-\frac 23 \cdot \frac{1}{2\sqrt 2} = \frac 23 -\frac{1}{3\sqrt 2}$$
Second way: $$\frac{\sin^2x}{\sqrt{1+\cos x}}= \frac{4\sin^2(x/2)\cos^2 (x/2) }{\sqrt 2 \cos(x/2)} = 2\sqrt 2 \sin^2(x/2) \cos(x/2) $$ Then you have something of the form $u^2 du$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int \sqrt{16-x^2} \,dx$ Are the steps below correct? Can you tell me another way to solve it, just hint?
$ \int \sqrt{16-x^2} \,dx $
$ x = 4 \sin u \rightarrow u=\arcsin(\frac{x}{4}) \rightarrow dx = 4\cos u \,dx$
$ \int \sqrt{16-x^2} \,dx = 16 \int \cos^2 u \,du = 16 \int \frac{\cos 2u +1}{2} \,du $
$ v = 2u \Rightarrow dv = 2 du \Rightarrow du = \frac{1}{2}dv $
$ 16 \int \frac{\cos 2u +1}{2} \,du = 16 \int \frac{\cos v}{2} +\frac{1}{2} \,dv = 16(\frac{\sin v}{2} +\frac{v}{2}) = 8\sin v +8v = 8\sin 2u +16u $
$ \Downarrow $
$8\sin(2\arcsin(\frac{x}{4})) + 4\arcsin\frac{x}{4} $
| Let $I = \int\sqrt{16-x^2} \ dx$. Then applying integration by parts with $u = \sqrt{16-x^2}, du = -\frac{x}{\sqrt{16-x^2}}$ and $dv = 1, v = x$:
$$I = x \sqrt{16-x^2} + \int \frac{x}{\sqrt{16-x^2}} \ dx$$
and now let $w = 16-x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove by definition that $\frac{1-2n^3}{4n^3-2n^2-1} \to -\frac12$ $a_n = \dfrac{1-2n^3}{4n^3-2n^2-1}$
My attempt (of the scratch):
$\left| \dfrac{1-2n^3}{4n^3-2n^2-1} + \dfrac{1}{2}\right|=\left| \dfrac{2-4n^3+4n^3-2n^2-1}{8n^3-4n^2-2}\right|=\left|\dfrac{-2n^2+1}{8n^3-4n^2-2} \right|=\dfrac{2n^2-1}{8n^3-4n^2-2}<\dfrac{2n^2}{8n^3-4n^2-2}=\dfrac{n^2}{4n^3-2n^2-1}<\epsilon$
But I don't know what to do next to choose my $N \in \Bbb{N}$
| Let $N \in \mathbb{N}$ s.t. $4N-2>n'+\delta$ for $n' \in \mathbb{N}$ and $\delta >0$, then $n \geq N$ implies $4n-2 \geq 4N-2 > n'+\delta$.
Let $N' \in \mathbb{N}$ s.t. $\frac{1}{N'^2}< \delta$ for $\delta >0$, then $n\geq N'$ implies $\frac{1}{n^2} \leq \frac{1}{N'^2} < \delta$.
If $n \geq \text{max}\{N, N'\}$ then $4n-2-\frac{1}{n^2} > n'+ \delta - \delta= n'$ and $\frac{n^2}{4n^3 -2n^2 -1} =\frac{1}{4n-2-\frac{1}{n^2}}<\frac{1}{n'}.$
Let $n'$ be such that $\frac{1}{n'}<\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3955581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$ Solve for $x$ in
$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$
I'm really not good with 4th degree equations but since the 1st term in the RHS looked a simple (a+b)(a-b) application, I tried solving that but I'm really not able to reach to the final answer as it only gets more complicated...
can anyone please help me out?
The equation I got was:
$$x^4-3x^3-22x^2-48x-32=0$$
| Note that the given equation
$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$
is purposefully structured to allow convenient factorization as follows
$$3x^3=x^4-(\sqrt{18}x+\sqrt{32})^2 -4x^2$$
$$x^4-3x^3-4x^2-(\sqrt{18}x+\sqrt{32})^2 =0$$
$$x^4-(3x+4)x^2-2(3x+4)^2 =0$$
$$(x^2+(3x+4))(x^2-2(3x+4))=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.