Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
| Let $a = c + \delta$. Then as $a+b = c+\delta + b = c+d$ we have $b= d-\delta$.
So we ahve $c^2 + d^2 = (c+\delta)^2 + (d-\delta)^2$ so
$2\delta(c-d) + 2\delta^2= 0$.
Case 1: $\delta = 0$.
The we go back to near the beginning. Then $a =c+\delta =c$ and $b=d-\delta = d$.
Case 2: $\delta \ne 0$ then
$(c-d) + \delta = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
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Solving $3z^2 + (2+3i)z + (5i-5) = 0$ In a Precalculus book, Complex numbers chapter, I am given the following exercise:
Solve the equation $3z^2 + (2+3i)z + (5i-5) = 0$, giving your answers in the form $z = a + bi$
From the way that the exercise is shown, I thought about using the quadratic formula to solve it. Star... | Suppose $a+bi\ne 0$. Suppose we want to solve $(x+yi)^2 = a+bi$.
Then $(x+yi)^2 =x^2 + 2xyi -y^2 = (x^2 -y^2) + 2xyi = a+bi$. As $a,b,x^2-y^2, 2xy\in \mathbb R$ we must have
$x^2 - y^2 =a$ and $2xy =b$.
So solve that.
$x = \frac b{2y}$ so $\frac{b^2}{4y^2} - y^2 = a$ so $y^4 +y^2a -\frac 14b^2=0$ so $y^2 = \frac {-1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Various ways to calculate $\int \sin(x) \cos(x) \, \mathrm{d}x$ Consider the integral
$$\mathcal{I} := \int \sin(x) \cos(x) \, \mathrm{d} x$$
$
\newcommand{\II}{\mathcal{I}}
\newcommand{\d}{\mathrm{d}}
$
This is one of my favorite basic integrals to think about as an instructor, because on the face of it, there are a l... | $$\begin{align*}
\int\sin{x}\cos{x}dx
&= \frac{1}{4}\int\frac{4\tan{x}\sec^2{x}}{\sec^2{x}\sec^2{x}}dx\\
&= \frac{1}{4}\int\frac{4\tan{x}\sec^2{x}}{(1+\tan^2{x})^2}dx\\
&=\frac{1}{4}\int\frac{2\tan{x}\sec^2{x}((1+\tan^2{x})-(\tan^2{x}-1))}{(\tan^2{x}+1)^2}dx\\
&=\frac{1}{4}\int\frac{2\tan{x}\sec^2{x}(1+\tan^2{x})-(\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4262190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
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"answer_id": 0
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Are there infinitely many primes $q$, such that $\sigma(p^k) = 2q$, where $p$ is prime and $p \equiv k \equiv 1 \pmod 4$? Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Here is my question:
Are there infinitely many primes $q$, such that $... | It seems that it is not known if there are infinitely many primes $q$ such that $2q−1$ is also prime according to OEIS/A005382 and Solve $2p=q+1$ where $p$ and $q$ are prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4262319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the probability of $P(Y=6)$ Anna writes down a random sequence created by the following process. She repeatedly rolls a fair
6-sided die. If the number she rolls is larger than all of the numbers she has previously rolled (if
any), then she writes the new number down and then continues rolling. Otherwise, s... | $Y = 6$ means we have maximum of six rolls. If none of the six rolls is $6$, we would have at least one of the numbers between $1$-$5$ appear twice breaking the sequence and the process would stop.
Probability of getting $6$ in the first roll is $\frac{1}{6}$
Probability of getting $6$ in the second roll is $ \frac{5}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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What are the sine and cosine of dyadic angles? What are the values of sine and cosine of dyadic angles? We know
$$\cos\pi = -1 \qquad \cos\frac{\pi}{2} = 0 \qquad \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}\,,$$
and we can calculate sine by appealing to symmetry. But I don't think I've seen the sine and cosine values, prese... | This answer has some approaches that were not the easiest ways to solve the problem. I'm recording them here in a separate answer so as not to lose the information (even if what this tells us is not to do it this way.)
For a complete table of sines and cosines of $\pi/16,$ we only need to compute values for $0,$ $\pi/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4267002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\frac{1}{a_3+a_4}+\frac{1}{a_1+a_2}\ge \frac{4}{a_1+a_2+a_3+a_4} $ Prove that $$\frac{1}{a_3+a_4}+\frac{1}{a_1+a_2}\ge \frac{4}{a_1+a_2+a_3+a_4} $$
I am supposed to use AM-HM but I don't how I should. By AM-HM
We have $$\frac{1}{a_3+a_4}+\frac{1}{a_1+a_2}\ge \frac{2}{a_1+a_2+a_3+a_4} $$
Did I do something wrong?
| Hint : Let $a_1+a_2=x$ and $a_3+a_4=y$. It remains to prove,
$$\frac{1}{x}+\frac{1}{y}\ge \frac{4}{x+y}\;\; \Leftrightarrow\;\; (x+y)\left(\frac{1}{x}+\frac{1}{y}\right)\ge 4$$
The A.M-H.M inequality for two terms,
$$\frac{x+y}{2}\ge \dfrac{1}{\dfrac{\frac{1}{x}+\frac{1}{y}}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can trigonometric substitution be used to solve this integral? The integral in question is $$\int \frac{x+1}{9x^2+6x+5}dx.$$ I first completed the square in the denominator giving $(3x+1)^2+4$ and proceeded to perform a $u$-substitution with $u = 3x+1$, $du=3~dx$, and $x=(u-1)/3$. After simplifying, I was left with $$\... | Since $u=2\tan\theta$, you have$$u^2=4\tan^2\theta=4(\sec^2\theta-1).$$So,$$\sec\theta=\sqrt{\frac{u^2+4}4}$$and therefore$$\frac19\ln(\sec\theta)=\frac1{18}\ln\left(\frac{u^2+4}4\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the maximum value of the following product.
Let $x,y,z$ be three reals which satisfies $x^2+y^2+z^2=1$. Find the maximum value of $$P = (x^2-yz)(y^2-zx)(z^2-xy).$$
I think expanding would be a bad idea. I tried to apply the Weirstrass product inequality but it may give the minimum not maximum value.
| We claim that $P \leq \dfrac{1}{8}.$
Case $1:$ All three terms are greater than or equal to zero.
By AM-GM,
\begin{align}
\sqrt[3]{P} & \leq \dfrac{x^2+y^2+z^2-xy-yz-xz}{3} \\
&= \dfrac{1-(xy+yz+xz)}{3}.
\end{align}
It thus suffices to prove $\dfrac{1-(xy+yz+xz)}{3} \leq \dfrac{1}{2} \iff xy+yz+xz \geq \dfrac{-1}{2}.$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the partial sum $\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$ Evaluate the partial sum $$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$$
What I have tried:
Calculate the partial fractions which are (for sake of brevity) :
$$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}$$
So we get:
$$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)} = \sum_{n=1}... | Calling $S_n = \sum_k^n \frac{1}{k}$ we have
$$
T_n = \frac 12 S_n - \left(S_n-1+\frac{1}{n+1}\right)+\frac 12\left(S_n-1-\frac 12+\frac{1}{n+1}+\frac{1}{n+2}\right)= \frac 14+\frac 12\left(\frac{1}{n+2}-\frac{1}{n+1}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4272062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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For all $a, b$ in $\mathbb{Z}$, show $a +b$ is a factor of $a^{2n} - b^{2n}$. Given solution approach for first problem below, unable to approach the second one.
#1. For all $a, b$ in $\mathbb{Z}$, show $a - b$ is a factor of $a^n - b^n$.
By weak form of induction.
Base step: $n=1$.
Hypothesis step: Let, $n=k$; then:
$... | Not an answer, but too long for a comment:
You can get Heropup's result by applying the induction step you did in 1) twice. Note that $$\begin{aligned}
a^{2(n + 1)} - b^{2(n + 1)} &= a^{2n + 1}(a - b) + (a^{2n + 1} - b^{2n + 1})b \\
&= a^{2n + 1}(a - b) + (a^{2n}(a - b) + (a^{2n} - b^{2n})b)b
\end{aligned}
$$ Which aft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4272344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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In triangle. Prove that: $2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$ Problem: Given a,b,c are length of triangle. Prove that: $$2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$$
Ha... | I found a stronger one:
Problem: Given a,b,c are length of triangle. Prove that: $$2\sum_{cyc}{\frac{a+b-c}{c}}\sum_{cyc}{\frac{a+b-c}{2(a+b)-c}}+3\ge\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$$
Note that: $\frac{c}{2(a+b)-c}+\frac{a}{2(b+c)-a}+\frac{b}{2(c+a)-b}\ge1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that: $2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$ Problem: For $a,b,c\ge0: ab+bc+ca>0.$ Prove that: $$2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$$
Recently, i have seen a post on AoPS link
My approach: After squaring both side,... | Similarly: If $a,b,c\ge0$ then: $$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge \sqrt{4bc+ca+ab}+\sqrt{4ca+ab+bc}+\sqrt{4ab+bc+ca}$$
$$2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge \sqrt{bc+4ca+4ab}+\sqrt{ca+4ab+4bc}+\sqrt{ab+4bc+4ca}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to get roots in the form of a quadratic from the quartic $x^4 - 4x^2 + 16$? So I need to factor this function into quadratics:
$$x^4 - 4x^2 + 16$$
I know that there are only complex solutions to this question, however, it is still possible to obtain quadratic factors without requiring the imaginary unit to be prese... | We see that, the roots of the equation are not real numbers. So, if we want to factor polynomials whose coefficients are real numbers, and if that's possible, then we have
$$\begin{align}x^4-4x^2+16=(x^2+ax+b)(x^2-ax+b) \end{align} \tag 1$$
This factorization uses the following fact:
If $x_1,x_2$ are the root of the b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve this infinite system of equations? I'm trying to solve an infinite set of coupled equations. j is a real integer index where $-\infty<j<\infty$. {a,b,c,d,f} are variables, and K's are constants. I need to find closed form solutions for $\{a_{j},b_{j},c_{j}\}$ in terms of only the K's. I believe the respe... | Given
$$
\left\{ \begin{array}{l}
a_j + b_j + c_j = K_j \\
3a_j + 2b_j + c_j = c_{j + 1} \\
6a_j + 3b_j + d_j + 3f_j = 0 \\
3a_j + 2b_j + d_j + 2f_j = 0 \\
d_j + f_j - f_{j - 1} = 0 \\
\end{array} \right.
$$
we can use 3rd and 4th rows to enucleate $d$ anf $f$ as a combination of $a$ and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Finding $\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx $ Is there a way to show $$\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx = 2C$$ where $C=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} $ is Catalan’s constant, preferably without using complex analysis?
The following is an attempt to expand it as as a series:
\begin{align*... | 1st Solution. Applying the Weierstrass substitution $t=\tan(x/2)$, it follows that
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \, \mathrm{d}x
&= \int_{0}^{1} \frac{2\arctan(t)}{2t/(1+t^2)} \, \frac{2\mathrm{d}t}{1+t^2}
= 2 \int_{0}^{1} \frac{\arctan t}{t} \, \mathrm{d}t.
\end{align*}
Now note that we have
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Why does $|\sin(\sqrt{4\pi^2n^2+x})| =|\sin(\sqrt{4\pi^2n^2+x})-\sin(2\pi n)|$ I need in a demonstration that $$|\sin(\sqrt{4\pi^2n^2+x})|
=|\sin(\sqrt{4\pi^2n^2+x})-\sin(2\pi n)|$$
but I don't understand why this equality is true. Can someone explain it to me?
| From the angle-sum formulas $$\sin (A\pm B)=\sin A \cos B \pm \cos A \sin B$$ $$ \cos (A\pm B)=\cos A \cos B\mp\sin A\sin B$$ we obtain $$\sin C-\sin D=2\sin A\cos B$$ where $A=(C-D)/2$ and $B=(C+D)/2.$
If $y>0$ and $4\pi^2y^2+x>0,$ let $C=\sqrt {4\pi^2y^2+x}$ and $D=2\pi y.$ We have $$|A|=\frac {1}{2}|C-D|=\frac {1}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to determine regions of positive and negative concavity of $h(g)= \frac{(g-a)(g-b)}{(g+a)(g+b)}$? We may assume that $a<0,b<0, |b|>|a|$
How to check where the concavity of the function $h(g)= \frac{(g-a)(g-b)}{(g+a)(g+b)}$ is positive or negative?
$h''(g)=\frac{4 \, {\left(a^{3} b + 2 \, a^{2} b^{2} + a b^{3} - {\l... | I don't think there is anything simpler (at least not to come up with) than just checking the second derivative. Here is the second derivative fully factored (over $\mathbb{Q}$):
$$\frac{4 (a+b) \left(a^2 b+a b^2+3 a b g-g^3\right)}{(a+g)^3(b+g)^3}$$
The main issue is that $a^2 b+a b^2+3 a b g-g^3$ is irreducible (exce... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can the vertex of an elliptic paraboloid be recovered from its quadric matrix form? A circular paraboloid with unit axis $\mathbf{a}$, vertex $\mathbf{c}\in\mathbb{R}^3$, and quadratic coefficient $b\in\mathbb{R}$ is defined implicitly as the set of points $\mathbf{r}\in\mathbb{R}^3$ such that the following equality ho... | Thanks intelligenti-pauca for pointing me in the right direction! I'll add some details on my approach here.
Let $\mathbf{P} = \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{a} \end{bmatrix}$ be the matrix of eigenvectors of $\mathbf{Q}_r$. I want to compose $\mathbf{c}$ in the eigenbasis of $\mathbf{Q}_r$ as $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Fourth Degree Pell Equation I got stuck at the following of my research problem:
Prove that only solution to equation $4b^2-3a^4=1$ for odd positive integers $a$, $b$ is $(1,1)$.
I made factorization -->$3a^4 = (2b-1)(2b+1)$
from here since 2b-1 and 2b+1 are coprime,
$$2b-1 = 3x^4, 2b+1 = y^4$$
or
$$2b-1 = x^4, 2b+1 ... | Let $a$ and $b$ be positive integers such that
$$4b^2-3a^4=1.\tag{1}$$
Then indeed, as you note
$$3a^4=4b^2-1=(2b+1)(2b-1),$$
where the greatest common divisor of the two factors divides their sum, which is $2$. Of course it is clear from $(1)$ that $3a^4$ is odd, and hence the two factors are coprime. This means that... | {
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"timestamp": "2023-03-29T00:00:00",
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How to get the value of t For each number t, consider the triangle T with corners at, A = (1, 2, 3), B = (1, 0, −2),
C = (t, t, 1).
For which value of t does T have a right angle at A?
I know how it has to do with AB*CD = something
Rest i don't know how to solve this??
| As I commented earlier, you can use the Pythagoras theorem: for a right angle we know that $a^2+b^2=c^2$. We can find the lengths first:
$AB = \sqrt{0^2+2^2+5^2} = \sqrt{29}$
$BC = \sqrt{(1-t)^2 + t^2 +3^2} = \sqrt{10-2t+2t^2}$
$AC = \sqrt{(1-t)^2 + (2-t)^2 + 2^2} = \sqrt{9-6t+2t^2}$
Now we can plug in these values for... | {
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"timestamp": "2023-03-29T00:00:00",
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Non-linear 1st order differential equation of $~x\frac{dy}{dx}+y=x\sqrt{y}~~~$where$~\sqrt{y}~$exists I think this is the first time when I handle of non-linear differential equation.
$$ \underbrace{x \frac{dy}{dx} + y = x\sqrt{ y } }_{x > 0} \tag{1} $$
$$ \frac{dy}{dx} + \frac{1}{ x } y = \sqrt{ y } $$
$$ \u... | $$y'+\frac{y}{x} = \sqrt{y}$$
$$\frac{y'}{\sqrt{y}} +\frac{\sqrt{y}}{x} = 1$$
Using
$$z=\sqrt{y}$$
$$\frac{dz}{dx} = \frac{1}{2\sqrt{y}}\cdot y'$$
$$2\cdot\frac{dz}{dx} = \frac{y'}{\sqrt{y}}$$
Replace
$$2\cdot\frac{dz}{dx} +\frac{z}{x} = 1$$
And now you have built a linear system
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to find the probability that the number of boxes which must be selected to obtain all three blue boxes is greater than or equal to $5$. I am trying to solve this problem.
There are four identical red boxes and three identical blue boxes. If the boxes are randomly selected one at a time without replacement, find th... | New Interpretation
There is disagreement about what the OP is asking. It has been suggested that my original interpretation is wrong. In hindsight, it does seem very plausible that I was mistaken. Therefore, I left my original interpretation at the very end of my answer, just in case it turns out to be what the OP i... | {
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Finding all primes $p$ satisfying $p=a^2+b^2$ and $p$ divides $a^3+b^3-4$ .
Determine all primes $p$ such that there exist integers $a, b$ satisfying $p=a^2+b^2$ and $a^3+b^3-4$ is divisible by $p$.
So, $p | ab(a+b)+4$ after writing $a^3+b^3-4=(a+b)(p-ab)-4$ and $p=2$ is a solution if $p|4$. If $p$ doesn't divide $4$... | Observe, $$\begin{align*} 2ab(a+b)+8&\equiv 0 &&\pmod p \\ 2ab&\equiv (a+b)^2&&\pmod p\\ \therefore\;\; (a+b)^3+8&\equiv 0 &&\pmod p \end{align*}$$
So, $$p|(a+b+2)\; \;\text{or}\; \;p|(a+b)^2-2(a+b)+4$$
Case 1: $$p|a+b+2$$
$$a^2+b^2|a+b+2\implies a^2+b^2\le a+b+2 \le \sqrt{2(a^2+b^2)}+2$$
$$\therefore\;\; p\le \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Solving $ a \cos x + b \sin x = u $, $ -a \sin x + b \cos x = v $ in a different way gives me a solution that does not work for $ x = n\pi $ I am solving the following equations to find $ a $ and $ b $:
\begin{align*}
a \cos x + b \sin x &= u \\
-a \sin x + b \cos x &= v
\end{align*}
Solution 1
Multiplying the first eq... | All you have to do to complete the second solution is work in cases:
Case 1: $\sin x\ne 0$
.
.
.
$$b = \frac{u - u \cos^2 x + v \sin x \cos x}{\sin x} = \frac{u \sin^2 x + v \sin x \cos x}{\sin x} = u \sin x + v \cos x$$
Case 2: $\sin x = 0$. Then $\cos^2 x =1$ and the original equations become
\begin{align*}
a \cos x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that: $\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$ Given 3 positive real numbers $x, y, z$ satisfies $xy+yz+xz=1$. Prove that: $$\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\fra... | Let $x=\tan\frac{\alpha}{2},$ $y=\tan\frac{\beta}{2}$ and $z=\tan\frac{\gamma}{2},$ where $\{\alpha,\beta,\gamma\}\subset\left(0,\frac{\pi}{2}\right).$
Thus, $\alpha+\beta+\gamma=\pi$ and since $$\left(\sin\frac{\alpha}{2}+\cot^2\frac{\alpha}{2}\right)''=\frac{2\cos{\alpha}+4-\sin^5\frac{\alpha}{2}}{4\sin^4\frac{\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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In the figure, determine $x$ as a function of $R$. For reference: In the figure, determine $x$ as a function of $R$.
My progress: I don't know and I can affirm that $\angle FGD$ is $90^o$ ...I think you would need to demonstrate ...but I can't
$\triangle FDB: (R+r)^2 = R^2+(2R-2r)^2\\R^2+2Rr+r^2=R^2+4R^2-4Rr+4r^2\\6Rr... | If you are not sure about the perpendicularity, let's follow the lengthy method.
Call the centres of small circle, small half circle and big half circle $C$, $E$ and $F$ respectively. Also draw the perpendicular lines to $AB$ and $BC$ to meet each side at $D$ and $G$. Thus we see, $CE=r+x$ and $CF=R+x$.
Let $CD=a$, $C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to turn into a product of $2$ factors the expression : $ x^4 -3x^2+1$? Source : Lebossé & Hémery, Algèbre et Analyse ( Classe de seconde , 1965).
The exercice requires : turn into a product of $2$ factors the expression $ x^4 -3 x^2 +1$.
The only way I see is to use the property : $P(x)$ being a polynomial , if $... | You can stop earlier:
$$
x^4 - 3x^2 + 1
= X^2 - 3X + 1
= (X - \frac{3+\sqrt5}{2})(X - \frac{3-\sqrt5}{2})
= (x^2 - \frac{3+\sqrt5}{2})(x^2 - \frac{3-\sqrt5}{2}).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a problem in Coordinate Geometry Suppose, $2x \cos \alpha - 3y \sin \alpha = 6$ is equation of a variable straight line. From two point $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$ foot of the altitude on that straight line is $P$ and $Q$ respectively. Show that the product of the length of two line segment $AP$ and ... | One missing factor is the locus of points $P$ and $Q$.
For point P we have the system of equations
$2\cos\alpha x - 3\sin\alpha y=6$
$3\sin\alpha x + 2\cos\alpha y=3\sqrt5\sin\alpha$
Square both equations and add. Cross-product terms cancel and we end with
$(4\cos^2\alpha+9\sin^2\alpha)(x^2+y^2)=36+45\sin^2\alpha$
Elim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculate $\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x$ I'm supposed to compute
$$\lim_{x\rightarrow \infty}\left( \frac{1}{\ln(x+1)-\ln(x)}-x\right).$$
However, I keep getting the wrong answer, so I'll present my solution for you, and I hope you can give me any tips on how to solve it.
Rewriting using logar... | I shall try to evaluate the limit by using L’Hospital Rule twice.
$$
\begin{array}{l}
\displaystyle \quad \lim _{x \rightarrow \infty}\left(\frac{1}{\ln (x+1)-\ln x}-x\right)\\
\begin{aligned}=& \lim _{x \rightarrow \infty} \frac{1-x \ln \left(\frac{x+1}{x}\right)}{\ln \left(\frac{x+1}{x}\right)} \quad\left(\frac{0}{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
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Showing the formal identity $(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots =\frac{1}{1+x+x^2}$
How to show this formal identity (or you can assume $|x|<1$)?
$$(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots =\frac{1}{1+x+x^2}$$
I can show that the latter is $$=1-x+x^3-x^4+x^6-x^7+\cdots$$ but how to show this is equal to the infinite ... | Notice that $$(1+x^n+x^{2n})(1-x^n+x^{2n}) = 1+x^{2n}+x^{4n}$$
Now if $p_n$ is product of first $n$ brackets on LHS then $$p_n ={1+x^{2^n}+x^{2^{n+1}}\over 1+x+x^2}$$
Taking $\displaystyle \lim _{n\to \infty}p_n$ for $|x|<1$ we get the desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a exponential floor flunction equation I am trying to solve this question but can't simplify this further.
Question:
$$ 2^{\lfloor log_2{(x)} + \frac{1}{2} \rfloor} = 2^{\lfloor log_2(x-2^{\lfloor{ log_2{(\frac{x}{2})} + \frac{1}{2}}\rfloor}) + \frac{1}{2}\rfloor} + 2^{\lfloor log_2{(\frac{x}{2})+\frac{1}{2}}\... | ${first}... \\ $
$\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} {x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} =\mathrm{2}.\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} \\ $
$. \\ $
${so}... \\ $
$. \\ $
$\mathrm{2}^{\left[\left(\mathrm{log}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the set can describe all rational numbers in (0, 1). Let A be a subset of (0, 1) ∩ Q, where Q is the set of all rational numbers.
Given that $\frac{1}{2}\in A$ and that $\frac{x}{x+1},\frac{1}{x+1}\in A$ if $x\in A$, show that A = (0, 1) ∩ Q.
What I've tried to do:
For simplicity, let (1) be the operation $x\... | You can do it by induction on the denominator. That is, we can show that, for all $0 < a < n$, we have $\frac{a}{n} \in A$, and we show this by induction on $n \ge 2$. Clearly the result holds for $n = 2$, by our assumption that $\frac{1}{2} \in A$.
Suppose that, for some $n \ge 3$, we have $\frac{a}{k} \in A$, wheneve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the distance $ID$ in the triangle below For reference: In the triangle $ABC. \angle A = 2\angle C, IG|| AC$, where $I$ is the incenter and $G$ the centroid. If $IG =x$, calculate the distance $ID$ if $BD$ is an interior angle bisector. (Answer: $3x\sqrt2$)
My progress:
I was able to see these relationships but I ... | Question says $IG = a$. To avoid confusion, I will instead use $IG = x$ and standard notation for the side lengths of the triangle.
$IG \parallel AC \implies r = \dfrac h 3$ where $h$ is altitude from $B$ to $AC$
So using $A = r \cdot s, $ where $r$ is inradius, $s$ is sub-perimeter and $A$ is area,
$\displaystyle \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating $\lim_{x \to 0}\frac{x \tan2x-2x\tan x}{(1-\cos2x)^2}$ I'm a bit confused regarding this question. I've been trying to solve it and have gotten to the same answer ($3/8$) thrice now. I have no idea where I'm going wrong and would really appreciate some help figuring it out. Here's my solution:
$$\lim_{x \to... | Your problem arises here:
\begin{align*}
\lim_{x \to 0}&\frac{(\frac{\tan2x}{2x})+(\frac{2x(\sec^22x)}{2x})-(\frac{2\tan x}{2x})-(\frac{2x\sec^2x}{2x})}{4(1-\cos2x)\frac{(\sin2x)}{2x}}\\
&=\lim_{x \to 0}\frac{1+\sec^22x-1-\sec^2x}{4(1-\cos2x)}\\
&=\lim_{x \to 0}\frac{\sec^22x-\sec^2x}{4(1-\cos2x)}
\end{align*}
particul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
unbiased estimator for the negative binomial distribution Let be $P(X=x)= \binom{x-1}{r-1} \theta^r (1-\theta)^{x-r}$
Show that $\frac{r-1}{x-1}$ is an unbiased estimator for
My attempt;
E$[\frac{r-1}{x-1}]$ $=$ $\sum_{x=0}^{\infty}\frac{r-1}{x-1}\binom{x-1}{r-1} \theta^r (1-\theta)^{x-r} = \sum_{x=0}^{\infty} \binom{... | There are some errors in your calculation. Note that the lower index of summation should be $r$, not $0$. Then we have
$$\begin{align}
\frac{r-1}{x-1} \binom{x-1}{r-1} \theta^r (1-\theta)^{x-r}
&= \frac{r-1}{x-1} \frac{(x-1)!}{(r-1)! (x-r)!} \theta^r (1-\theta)^{x-r} \\
&= \theta \frac{(x-2)!}{(r-2)! ((x-2)-(r-2))!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4321610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left... | As an alternative by $\cos x= e^t$ with $t\to 0$ and then $\sin^2 x=1-e^{2t}$ we have
$$\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}=\\=\frac1t+\frac2{1-e^{2t}}=\frac{1-e^{2t}+2t}{t(1-e^{2t})}=\frac{1-1-2t-2t^2+2t+o(t^2)}{t(1-1-2t+o(t))}=\frac{-2t^2+o(t^2)}{-2t^2+o(t^2)}=\frac{1+o(1)}{1+o(1)}\to 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
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Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not? Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?
$$a^2+ab+b^2=(a+b)^2-ab=0$$ iff $$(a+b)^2=ab \tag{1}$$
but $(a+b)^2 = a^2+2ab+b^2 $ so equation 1 couldn't possibly be true.
Also, w... | Fix $b\neq0$ and try to solve in $a$. This a second order equations so the possible values of $a$ in terms of $b$ are
$$ a_{1,2}=\frac{-b\mp \sqrt{-3b^2}}{2}$$
So, for any non zero real value $b$ the possible values of $a$ are complex conjugated because $-3b^2<0$. Therefore you cannot find two non-zero reals such that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Solving $z^3 =-1$ I need to find the solution of $z^{3} =-1$.
Let$$z=[ r,\theta ]$$
then using de Moivre's theorm$$ z^{3} =\left[ r^{3} ,3\theta \right] $$
write the number $-1$ in modulus argument from,
$$[ 1,( 2n-1) \pi ]$$
the we can write$$\left[ r^{3} ,3\theta \right] =[ 1,( 2n-1) \pi ]$$
Therefore,$$ r^{3} =1 \sp... | You can find the complex roots of unity as below. The equation you have solved is not the one that will give you complex roots of unity.
$$z^3=1 \implies z^3-1=0$$
Factor the LHS (using any of a variety of methods):
$$(z-1)(z^2+z+1)=0$$
Use the zero-product property:
$$z-1=0\implies z=1$$
$$z^2+z+1=0\implies z=\frac{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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A limit with geometrical progression I got stuck trying to solve this limit.
$$ \lim_{n\to\infty} \frac{1 + \frac{1}{2} + \frac{1}{4} + ....+ \frac{1}{2^{n}}}{1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}} $$
The first thing I did was calculating the sums of the geometrical progressions found in both numerator and... | We see, that $\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+ \frac{1}{2^{n}}\right)$ and $\left(1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}\right)$ are the geometric series.
Therefore,
$$
\lim_{n\to\infty}\left(1+\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}\right)=\sum_{n=0}^{\infty}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\sqrt3\;\cos(x)= \sin\;(\frac {x}{2})$ I'm currently stuck on this trigonometric equation:
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
Here's what I've tried so far (2 methods):
Method n.1
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
$$\sqrt3 \;(1-\sin^2x)= \sin\;(\frac {x}{2})$$
$$\sqrt3-\sqrt3\;\sin^2x= \sin\;(... | hint
If we make the substitution$$x=\pi+2y$$
the equation becomes
$$-\sqrt{3}\cos(2y)=\cos(y)=\sqrt{3}(1-2\cos^2(y))$$
or
$$2\sqrt{3}\cos^2(y)+\cos(y)-\sqrt{3}=0$$
with
$$\Delta=25$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4337047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding x which maximizes the volume of any cuboid missing one side So my grandpa and me where doing some maths for fun and we stumbled upon this problem:
Given a rectangle with unknown sidelenghts a and b, we need to cut off a square in each corner (call the side length of the square x) so that we get a net of a cuboi... | Why $x_{1,2}=\frac{-(-(4a+4b) \pm \sqrt{(-(4a+4b))^2-4\cdot12\cdot ab}}{2ab}$? You made a serious mistake since using the root fomula of quadratic equation.
Since the equation is $12x^2-4(a+b)x+ab=0$, so...
$$
x_{1,2}=\frac{-(-4(a+b))\pm\sqrt{(-4(a+b))^2-4\cdot12\cdot ab}}{2\cdot12}
\\
=\frac{a+b\pm\sqrt{a^2-ab+b^2}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Power series of $1/(1-x-x^2)$ I have a difficulty in finding the power series of: $$\frac{1}{1-x-x^2}$$ At first I did a polynomial division and I found out that: $$\frac{1}{1-x-x^2}=1+x+2x+3x^2+5x^3+8x^4+13x^5+...$$ where the coefficients of this Taylor Series are the numbers of the Fibonacci Sequence.
However, this i... | It is possible to continue from your approach. Let's fix a natural number $N$ and compute its coefficient. Given $n$, the inner sum can only contribute terms with powers $n, n+1, \ldots, 2n$. So to compute the coefficient of $x^{N}$, we should consider those terms with $n = N/2, N/2 + 1, \ldots N$. It now makes a diffe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Geometry problem to prove Euler line of a triangle $I$ is the incenter of $\triangle{ABC}$ and $D$ is the tangent point of the incircle with side $BC$. Points $E$ and $F$ are on side $BC$ so that $\angle{BAE}=\angle{ACB}, \angle{CAF}=\angle{ABC}$. $G$ and $H$ are incenters of $\triangle{ABE}$ and $\triangle{ACF}$ respe... | $\begin{array}{} ∠(ACB)=∠(BAE)=γ & ∠(ABC)=∠(CAF)=β & ∠(AEK)=β+γ \end{array}$
$\begin{array}{} A=\left( c\,cos(β),c\,sin(β) \right) & B=(0,0) & C=(a,0) \end{array}$
$\begin{array}{} ΔABC & cos(β)=\frac{a^2-b^2+c^2}{2ac} & cos(γ)=\frac{a^2+b^2-c^2}{2ab} \\ sin(β)=\frac{2S}{ac} & sin(β)=\frac{2S}{ab} & S=Δ(ABC)=\sqrt{s(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$? I am stuck with this equation. All I could do is this:
$\lfloor x \rfloor$ = $\lfloor n + m \rfloor$ such that $n \in N$ and $m<1$.
We get:
$\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$
$\lfloor n + m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1... | In general, to solve $\lfloor a \rfloor + \lfloor b \rfloor =c$ you can consider the pairs of values that $a$ and $b$ can take.
In this case your possibilities are very small: $a=0, b=1$ or $a=1, b=0$.
Can you proceed from there?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How many ways to deal with the integral $\int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x$ and $\int_{0}^{\frac{\pi}{4}} \ln (\sin x) d x$? After finding that $\displaystyle \int_{0}^{\frac{\pi}{2}} \ln (\tan x) d x =0$ in my post, I was curious about the value of the integral with different upper limit $\dfrac{\pi}{4} $.
The... | We can also find the integrals using the Fourier series of $\ln (\sin x)$ on $(0, \pi)$,
$$
\ln (\sin x)=-\ln 2-\sum_{n=1}^{\infty} \frac{1}{n} \cos (2 n x) \quad \forall x \in(0, \pi).
$$
Integrating both sides from $0$ to $\dfrac{\pi}{4} $ yields
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} \ln (\sin x) d x &=-\int_{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Power series solution of $y'-y=x^2$ $y'-y=x^2$
I try to get a power series solution.
$y(x)=\sum_{n=0}^{\infty}a_{n}X^{n},y'(x)=\sum_{n=1}^{\infty}na_{n}X^{n-1}=\sum_{n=0}^{\infty}(n+1)a_{n+1}X^{n}$.
$\sum_{n=0}^{\infty}(n+1)a_{n+1}X^{n}-\sum_{n=0}^{\infty}a_{n}X^{n}=X^{2}$
Then $ \space a_{n+1}=\frac{a_n}{n+1} \space ... | Maxima (Linux version of Macsyma) gives this solution to your
differential equation: $y(x) = C\exp(x)-(x^2+2x+2)$. You need a start value to get the integration constant $'C'$. Then you can proceed with the standard power series for $\exp(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4354247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{4b^2+b+1}{4\left|b\right|}\ge \sqrt{b+1}$, $b\ge -1,\:b\ne 0$ I've literally tried everything for this question. I think I'm supposed to use the Arithmetic-Geometric Mean Inequality (AGM) but I don't know how to apply AGM to this question so I tried algebraic manipulation and it feels like I'm getting nowh... | Hint: Remember that $x^2 =|x|^2$ and write LHS as $${4b^2\over 4|b|} +{b+1\over 4|b|}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area inside the curve $r^2=2\cos(5\theta)$ and outside the unit circle. I found the area of one full rose-petal($A_1$) and the area enclosed by the petal and the unit circle($A_2$), subtracted these from one another to get the area enclosed by the curve outside of the circle, and multiplied the answer by the a... | The intersection points of one of the petals of the curve $r^2 = \sqrt{2 \cos(5\theta)}$ and unit circle $r = 1$ are indeed $r = 1, \theta = \pm \frac{\pi}{15}$.
But as you need area outside the unit circle, the integral should be
$A = 6 \cdot \displaystyle \frac 12 \int_{-\pi/15}^{\pi/15} \left(2 \cos(5\theta) - 1\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A good explanation about this limit (Hierarchy of Infinities) I am searching for a good explanation (whether it be analytical or graphical) for why if I have this limit
$$\lim_{x\to+\infty} \sqrt{x^2 - 2x + 1} - x$$
I cannot use the hierarchy of infinities like
$$\sqrt{x^2\left(1 - \frac{2}{x} + \frac{1}{x^2}\right)} -... | The other answers already explain why the initial method does not work, so I will only give an easiest way to derive this limit : just notice that for $x > 1$, one has
$$\sqrt{x^2 - 2x + 1} - x = \sqrt{(x-1)^2} - x = -1$$
So $$\boxed{\lim_{x\to+\infty} \sqrt{x^2 - 2x + 1} - x = -1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving an infinite alternating product Idea is to alternately multiply infinite many values from $1$ to $x$, with const distance $h$.
Reweriting it to math:
$$
h = \frac{x-1}{n}
$$
$$
f(x) = \lim_{n \to \infty} \left( \prod_{m=0}^{n} (x - mh)^{(-1)^{m}} \right)
$$
I have played with the idea in Python and it seems for... | $$P_n=\prod_{m=0}^n\left(x-\frac{m }{n}(x-1)\right)^{(-1)^m}$$
$$P_n=\frac{n x \left(\frac{2(1- x)}{n}+x\right) \left(\frac{2(1-x)}{n}\right)^{\left\lceil
\frac{1-n}{2}\right\rceil +\left\lfloor \frac{n}{2}\right\rfloor -1}
\left(\frac{(n-4) x+4}{2(1- x)}\right)_{\left\lfloor \frac{n}{2}\right\rfloor
-1}}{((n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finding a linear combination of a linear combination of vectors Problem
Assume $\vec x, \ \vec y$ are linearly independent.
Let $\vec a = 3\vec x + 6\vec y$ and $\vec b = 9 \vec x + 21\vec y$.
Find an expression for $\vec x + \vec y$ as a linear combination of $\vec a, \vec b$.
I.e., find $c_1, \ c_2 \ \in \ \mathbb R$... | Eliminate the $x$ term in row two by replacing row two with (row two minus three times row one).
$$
\boxed{\begin{align*}
3x + 6y &= a \\
(9x + 21y) - 3(3x + 6y) &= (b) - 3(a)
\end{align*}}
\implies
\boxed{\begin{align*}
3x + 6y &= a \\
3y &= -3a + b
\end{align*}}
$$
Conveniently, the coefficients of $x$ and $y$ will m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ equals $(a+b)/2$ If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ is equal $(a+b)/2$
Multiplying by the conjugate, simplifying and clearing I arrive at the following expression
$$\frac{n}{\sqrt{(n+a)(n+b)}+n} < \frac{\epsilon+(a+b)-ab}{(a+b)}$$
| Another fun solution involves the HM-GM-AM-Inequality:
By the mentioned inequality, we obtain
$$
\frac{2}{\frac1{n+a} + \frac1{n+b}} \leq \sqrt{(n+a)(n+b)} \leq \frac{n+a+n+b}2
$$
Now note that
$$
\frac{2}{\frac1{n+a} + \frac1{n+b}} = \frac{2(n+a)(n+b)}{n+a+n+b} = \frac{n (2n + a + b) + n(a+b)+ab}{2n+a+b} \geq n + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$? As mentioned in my post, I started to investigate the integral $$
I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}},
$$
where $m$ and $n$ are natural number and... | Glad to see that there are several nice solutions to the problem, now I want to add one more by discovering a reduction formula using integration by parts.
In order to use integration by parts, I applied the inversion substitution to the original integral.
For any fixed natural number $n$, we now define
$$
J_{m}:=\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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In $\frac{1}{a^2-x^2}$, how can $a\sin(\theta)$ be substituted for $x$ when finding the anti-derivative? $$\int{\frac{1}{a^2-x^2}}dx\tag{1}$$
When finding the above anti-derivative, $x$ is substituted with $a\sin(\theta)$. However, the range of $x$ is $\mathbb{R}-\{-a,a\}$ while the range of $a\sin(\theta)$ is $[-a,a]$... | When $x\in (a, +\infty)$, you can substitute $x = a \cosh \theta$, thus
\begin{equation}
\int \frac{d x}{a^2 - x^2} = \int\frac{1}{a^2}\frac{a\sinh \theta d\theta}{1 -\cosh^2 \theta}
= -\frac{1}{a}\int\frac{d\theta}{\sinh\theta}
\end{equation}
You can then use $u = e^\theta$ or $t = \tanh\frac{\theta}{2}$.
Of course, y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sqrt{3}a \cos x + 2b \sin x=c$ has 2 distinct real roots, then what is the range of $2c/(3a+2b)$? Let $a$,$b$,$c$ be three non-zero real numbers such that the equation $\sqrt{3}a\cos x + 2b \sin x = c$ ,($x$ belongs to $[-\pi/2,\pi/2]$) has two distinct real roots m and n with $m+n = \pi/3$. If range of values of ... | Well... think it reversely. If we write the two roots in advance, can we determine $a,b,c$? Of course, we can.
Let the roots to be $m,n$ so we have
$$\sqrt{3}a\cos m + 2b \sin m = c, \sqrt{3}a\cos n + 2b \sin n = c.$$
So we have $$\sqrt{3}a(\cos m-\cos n) + 2b (\sin m-\sin n)=0.$$
And using sum-to-difference we have $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If you are given the values of m, $2^a$ mod m, and $2^b$ mod m, is there a way to find the value of $2^{ab}$ mod m? If you are given the values of m, $2^a$ mod m, and $2^b$ mod m, is there a way to find the value of $2^{ab}$ mod m?
| Yes.
Let us first assume that $n$ is odd. If you know $2^a \pmod n$, $2^b \pmod n$, and $n$ [but not the integers $a$ and $b$], then let $a'$ and $b'$ be any two integers that satisfy the equations $2^{a'} \pmod n =2^{a} \pmod n$ and $2^{b'} \pmod n = 2^b \pmod n$. [Now, finding such integers $a'$ and $b'$, can be done... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why should I only pick the positive number for $C$ in this IVP (separable differential equation)? I'm practicing some problems for an upcoming DEs test. I tried the following initial value problem:
$\displaystyle\frac{1}{2}\displaystyle\frac{dy}{dx}=\sqrt{y+1}\cos x,y(\pi)=0$
Here's my work:
$\begin{align}
\displayst... | First, note that since $\dfrac{dy}{dx}=2\sqrt{y+1}\cos(x)$ is a function which is continuous on an open subset of the domain of $f(x)=2\sqrt{y+1}\cos(x)$ containing the point $(\pi,0)$ and so is $\partial f/\partial y$, we are guaranteed that there exists a unique solution containing the point $(\pi,0)$. So it cannot b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $f(x)=x^2-2$ with $x\in [-2,2]$. Show that the equation $f^{n}(x)=x$ has $2^{n}$ real roots. [Where $f^{n}(x)=f(f^{n-1}(x))$] Let $f(x)=x^2-2$ with $x\in [-2,2]$. Show that the equation $f^{n}(x)=x$ has $2^{n}$ real roots. [Where $f^{n}(x)=f(f^{n-1}(x))$]
My solution:
Let $x=2\cos(\theta)$ for $\theta\in [0,\pi]$
$... | I think that your idea is very nice.
I would add proofs for the following claims :
Claim 1 : $f^{n}(2\cos\theta)=2\cos(2^n\theta)$ (your idea is very nice, but I would prove this claim by induction rigorously)
Claim 2 : The number of integers $m$ such that $0\leqslant\dfrac{2m\pi}{2^{n}-1}\leqslant\pi$ is $2^{n-1}$, an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Calculate the limit: $ \lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}$
Calculate the limit:
$$\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}
$$
I tried to change to polar coordinates like that:
\begin{align*}
\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}... | Hint. For now, suppose that the limit exists and is equal to $L$. Then,
$$L = \lim\limits_{r\to0}{\frac{e^{-\frac{1}{r^2}}}{r^4(1-2\cos^2\theta \sin^2\theta)}}$$
Take $\theta = \frac\pi 2$ first. Then, $$L = \lim\limits_{r\to0}{\frac{e^{-\frac{1}{r^2}}}{r^4}}$$
Now, take $\theta = \frac\pi 4$. Then,
$$L = \lim\limits_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$.
Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$. Then $z+\frac{1}{y}=\frac{m}{n}$, where $m$ and $n$ are coprime. Find... | Another solution: we have $\frac{1}{x}=yz$, $\frac{1}{y}=xz$, and $\frac{1}{z}=xy$. So
\begin{align*}
x+xy&=5\\
y+yz&=29\\
z+xz&=Q
\end{align*}
where Q is the quantity we're trying to determine.
Multiplying these three equations together gives $145Q=xyz(x+1)(y+1)(z+1)$ and hence $145Q=(x+1)(y+1)(z+1)$.
Expanding, we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int {\frac{{dx}}{{{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^2}}}} $ My first solution was:
$$\begin{array}{l}
\int {\dfrac{{dx}}{{{{(\sin \dfrac{x}{2} + \cos \dfrac{x}{2})}^2}}}} \\
= \int {\dfrac{{dx}}{{{{\sin }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2}}}} \\
= \int {\d... | $$\tan x-\frac{1}{\cos x}=\frac{2t}{1-t^2}-\frac{1+t^2}{1-t^2}-\frac{(1-t)^2}{(1+t)(1-t)}--\frac{1-t}{1+t}=1-\frac{2}{1+t}$$
where $t=\tan\frac12x$
so the answers are the same, modulo a constant
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve derivative/limit of $f(x)=x\sqrt{4-x^2}$ I'm trying to differentiate $x\sqrt{4-x^2}$ using the definition of derivative.
So it would be something like
$$\underset{h\to 0}{\text{lim}}\frac{(h+x) \sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}$$
I was trying to solve and I just can end up with somethi... | I think your algebra could look more like:
$$\begin{align}
&\frac{(x+h)\sqrt{4-(x+h)^2}-x\sqrt{4-x^2}}{h}\cdot\frac{(x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}}{(x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}}\\
&=\frac{(x+h)^2(4-(x+h)^2)-x^2(4-x^2)}{h\left((x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}\right)}\\
\end{align}$$
This leaves no radical... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $a+b=8$, $ax+by=9$, $ax^2+by^2=33$, $ax^3+by^3=60$, find $ax^4+by^4$
Given $$\begin{align}a+b&=8\\ax+by&=9\\ax^2+by^2&=33\\ax^3+by^3&=60\end{align}$$
Find $ax^4+by^4$
I really have no idea here.
Without coefficients $a$ and $b$, we can simply figure out $xy$ and find what we want by linear recurrences, but that... | Following the hint in comments, letting $x+y=s$ and $xy=p$
we can write the following linear system:
$$60=33s-9p\\33=9s-8p$$
Solving gives $s=1,p=-3$ and
$$ax^4+by^4=60s-33p=159$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Decomposing $\sum_{i = 0}^{2n} x^i$ as a simple sum of squares As we have $\sum_{i = 0}^{2n} x^i = (x^{2n + 1} - 1) / (x - 1)$, the polynomial is positive.
So we know that there is a decomposition as a sum of squares. Is there a closed simple form for such a decomposition? For small value of $n$ we have
$$ x^2 + x + 1 ... | Note that
$$\begin{split}
\sum_{k=0}^{2n}x^k
&= \frac{x^{2n+1}-1}{x-1} \\
&= (2n+1)\int_0^1 (1-t + tx)^{2n}\mathrm{d}t \\
&= \frac{2n +1}{2^{2n+1}}
\int_{-1}^1(x+1 + (x-1)t)^{2n}\mathrm{d}t \\
&= \sum_{k=0}^{2n}\frac{_{2n}C_k(2n +1)}{2^{2n+1}}
\int_{-1}^1(x+1)^{2n-k}(x-1)^{k}t^k\mathrm{d}t \\
&= \sum_{k=0}^{n}\frac{_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Area of shaded region in a square
My approach
The area of shaded region = 8×the area of
Given that the points of the shaded region are closer to the center than the boundary of the square.
Let's talk about the boundary of the shaded region
The boundary of the shaded region therefore must be the locus of all the point... | $$\int_0^{\frac a2(\sqrt 2 -1)}(\frac a4-\frac{x^2}a-x)dx=\frac{ax}4-\frac{x^3}{3a}-\frac {x^2}2\bigg|_0^{\frac a2(\sqrt 2 -1)}\\=a^2\left(\frac{\sqrt 2-1}{8}-\frac{(\sqrt 2-1)^3}{24}-\frac{(\sqrt 2-1)^2}8\right)\\=\frac{a^2}{24}(3\sqrt2-3-2\sqrt2+6-3\sqrt 2+1-3(2+1-2\sqrt 2))\\=\frac{a^2}{24}(4\sqrt 2-5)$$
Your error ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Reduce precision of fraction Say I have a reduced fraction where the numerator and denominator can only be integers:
$$ \frac{1071283}{28187739} $$
and I want to reduce it more, accepting the lose of precision.
I could just remove an equal numbers of integers from the right:
$$ \frac{107}{2818} $$
However playing aroun... | As suggested in comments, use the continued fraction decomposition.
$$\frac{1071283}{28187739} = \frac{1}{26+\frac{1}{3+\frac{1}{4+\frac{1}{1+\frac{1}{9+
\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{1+\frac{1}{1+
\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{14+\frac{1}{2+
\frac{1}{3}}}}}}}}}}}}}}}}$$
Halting the decomposi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Weird Problem on Polynomial Roots The polynomial $f(x)=x^3-3x^2-4x+4$ has three real roots $r_1$, $r_2$, and $r_3$. Let $g(x)=x^3+ax^2+bx+c$ be the polynomial which has roots $s_1$, $s_2$, and $s_3$, where
\begin{align*}
s_1 &= r_1+r_2z+r_3z^2, \\
s_2 &= r_1z+r_2z^2+r_3, \\
s_3 &= r_1z^2+r_2+r_3z,
\end{align*}and $z=\d... | You know the symmetric polynomials in the roots:
$\sigma_1 = \displaystyle\sum_{i=1}^{3} r_i$,
$\sigma_2 = \displaystyle\sum_{1\leq i < j\leq 3} r_i r_j$,
and $\sigma_3 = \displaystyle\prod_{i=1}^{3} r_i$.
Think of combining these into products of degree 3, namely:
$\sigma_1^3, \sigma_1 \sigma_2, \sigma_3$.
Some linear... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to calculate optimal strategy for bimatrix game Consider the bimatrix game given by
$$A = \begin{bmatrix} 3 &2 &-2 \\
-1 &3 &0 \\
1 &-2 &0 \\ \end{bmatrix} $$
$$ B = \begin{bmatrix} -2 &3 &-1 \\
-1 &-2 &2 \\
-1 &-1 &-3 \\ \end{bmatrix} $$
Taking $y^*$ to be the optimal strategy for $P_2$, I get $Ay^{*T} = u_1$ an... | What you appear to be doing is trying to find a strategy for each player that makes his or her opponent's payoff independent of the pure strategy chosen by that opponent. While a pair of such strategies will constitute a Nash equilibrium if they exist, it's unfortunately very possible (and very likely, in fact, if the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the minimum value of $\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}$ given $ a+b+c=2$? What is the minimum value of $\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}$ given $ a+b+c=2$?
I know the answer is going to be $\frac{6}{5}$ because it will occur when there is equality between th... | This is the most basic solution I can think of. This method is useful even if you only know AM-GM. Also, when I starts to learn the inequalities, I do them like this.
To prove
$$\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}\ge 6/5$$
Cancelling out the denominators
$$5(2a(b+2c+2a)(c+2a+2b)+2b(a+2b+2c)(c+2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find $\sum_{n=0}^\infty \big( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\big)$ without power series or integration? In a calculus, I have to find a direct way to compute the sum
$$\sum_{n=0}^\infty \left( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\right).$$
I tried to reform a partial sum:
$$\sum_{n=0}^N a_n = \fra... | First note that
$$\sum_{n=0}^\infty z^n = \frac{1}{1-z}$$
implies
$$\sum_{n=1}^\infty \frac{z^n}{n} = \log\frac{1}{1-z} = -\log(1-z),$$
and so
\begin{align}
\sum_{n=0}^\infty \frac{z^n}{n+1} &= \frac{-\log(1-z)}{z} \tag1\\
\sum_{n=0}^\infty \frac{z^n}{n+2} &= \frac{-\log(1-z)-z}{z^2} \tag2\\
\sum_{n=0}^\infty \frac{z^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate this integral with square trinomial and root? I have to evaluate this integral:
$$\int{\frac{1-x+x^2}{x\sqrt{1+x-x^2}}dx}$$
I have experience with methods such as:
*
*Using substitute for x;
for example, $${\frac{1}{x}=t;dx = -\frac{dt}{t^2}}$$
*Method of adding a new variable;
*Substitution metho... | $\int{\frac{1-x+x^2}{x\sqrt{1+x-x^2}}dx}$
The rooted is a trinomial ($a x^{2}+2b x+c$), which has two real roots $\alpha$ and $\beta$, decomposing it we have:
$\sqrt{a x^{2}+2b x+c}=\sqrt{a(x-\alpha)(x-\beta)}=(x-\beta)\sqrt{\frac{a(x-\alpha)}{x-\beta}}$.
Let's say
$\frac{a(x-\alpha)}{x-\beta}=t^{2}$, and we get :
$x=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4402337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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why substitution works in general Consider the equations $x^2 +y^2=1$ and $y=x^2-1$. If $y=x^2-1$ is substituted into $x^2 +y^2=1$ to obtain $x^2 +(x^2-1)^2=1$, then the $x$ values that result from solving this equation will be the $x$ values that produce equal $y$ values in the original equations $y=x^2-1$ and $x^2 ... | Let $(a,b)$ be a solution of the system which consists of the equations$$x^2+y^2=1\tag1$$and$$y=x^2-1.\tag2$$Asserting that $(a,b)$ is a solution of both $(1)$ and $(2)$ means that $a^2+b^2=1$ and that $b=a^2-1$. But then $a^2+(a^2-1)^2=1$, and therefore $a$ is indeed a solution of the equation $x^2+(x^2-1)^2=1$. And $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4403804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Can we prove that the equation $2^n+1=5^m$ has no integral solution other than $(2, 1)$? $\textrm{Assume, on the contrary, that }n\geq 4,\text{then }$
\begin{aligned} 2^{n} &=5^{m}-1 \\ &=(5-1)\left(5^{m-1}+5^{m-2}+\cdots+1\right) \\ &=4\left(5^{m-1}+5^{m-2}+\cdots+1\right) \\ 2^{n-2} &=5^{m-1}+5^{m-2}+\cdots+1 \\ 2^{n... | We suspect the largest solution is $4+1 = 5.$ Therefore write $2^n - 4 = 5^m - 5.$ Make new letters $x +2 = n$ and $y+1 = m. $ This becomes
$$ 4 (2^x - 1) = 5 (5^y-1) $$
ASSUME that $x \geq 1, y \geq 1.$
Since $2^x \equiv 1 \pmod 5,$ we see that $4 | x.$ Furthermore $2^4 - 1 | 2^x - 1.$ And $15 = 3 \cdot 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How do I find the centralizer for this matrix?
Let $B=\begin{pmatrix}
2 & 2 \\
0& 1 \\
\end{pmatrix} \in GL_2(\mathbb{R})$. Determine the centralizer for $B$.
Here's what I've done so far:
Let $D=\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} \in GL_2(\mathbb{R})$.
$\begin{align}
BD&=DB\\
\begin{pmatrix}
2 & 2 \\
... | This means that an invertible $2 \times 2$ matrix $D$
commutes with $B$ if and only if it is of the form
$$D=\begin{pmatrix}
a & 2a-2d \\
0 & d \\
\end{pmatrix}$$
for $a,d \in \mathbb{R}\setminus\{0\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving by Induction that $x_{2n+1} = 1+1/2+1/2^3+\cdots+1/(2^{2n-1})$ Let $x_1 = 1$, $x_2 = 2$ and $x_n = \frac{x_{n-1}+x_{n-2}}{2}$. The sequence $(x_n)$ is Cauchy, which I can easily prove. To find its limit I must first show (by Induction) the assertation
$$
P(n): x_{2n+1} = 1+\frac{1}{2}+\frac{1}{2^3}+\cdots+\frac... | As you already noted, $$x_{2n+1} = 1 + \sum_{i=1}^{n-1}\frac1{2^{2i-1}} = \frac13\left(5-\frac1{2^{2n-1}}\right).$$
You might also note that $$x_{2n+2} = 3 - \sum_{i=0}^n\frac1{2^{2i}} = \frac13\left(5+\frac1{2^{2n}}\right).$$
Combine these two, you have
$$x_n = \frac13\left(5 + 4\left(-\frac12\right)^n\right) \qquad(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4406976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to evaluate $\int_{0}^{1} \sqrt{\frac{x^{m}}{1-x^{n}}} d x$? Stuck by the integral
$$\int_{0}^{1} \sqrt{\frac{x}{1-x^{3}}} d x,$$
I finally solve the the problem using Beta Function.
Then I generalize the result to
$$
I(m,n)=\int_{0}^{1} \sqrt{\frac{x^{m}}{1-x^{n}}} d x
$$
can be tackled by the Beta function by let... | Your first integral has an elementary antiderivative.
$$\int_0^1\sqrt{\frac{x}{1-x^3}}=\int_0^1\frac{x^\frac{1}{2}}{\sqrt{1-(x^\frac{3}{2})^2}}dx=\frac{2}{3}\int_0^1\frac{d(x^\frac{3}{2})}{\sqrt{1-(x^\frac{3}{2})^2}}=\frac{2}{3}\arcsin(x^\frac{3}{2})\Bigg|_0^1=\frac{2}{3}\frac{\pi}{2}=\frac{\pi}{3}$$
I doubt there is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding roots of the equation $8x^3+4x^2-4x-1=0$ which are in form of cosine of angles. So, I start this by reducing it to a cubic with 0 coefficient of the 2nd degree term which gives me :
$$\left(x + \frac{1}{6}\right)^3 - \frac{7}{12}\left(x + \frac{1}{6}\right)-\frac{7}{216}=0$$
Replacing $\left(x + \frac{1}{6}\rig... | From the book in the chat, we can spot that the wrong thing is $r$, it should be $\sqrt{7}/3$ not $\sqrt{7/3}$.
So, your $\theta$ should be (I think you have a wrong calculation there!)
$$\theta = \cos^{-1}\left(\frac{\sqrt{7}}{{\color{red} 3}}{\color{red}\cos}\left( {\color{red}{\frac13}} \cos^{-1}\left(\frac{1}{2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving that $-\ln(x-\sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})$ Show $-\ln(x-\sqrt{x^2-1})=ln(x+\sqrt{x^2-1})$
Can somebody verify that my solution is correct (or incorrect!) and also show me alternative methods to do this? Thanks!
Let $r=-\ln(x-\sqrt{x^2-1})$ and $s=ln(x+\sqrt{x^2-1})$
$\rightarrow e^r=(x-\sqrt{x^2-1})^{-1}$ ... | Looks good.
But note that it's really just a matter of observing that $x \pm \sqrt{x^2-1}$ are reciprocals, since
$$\frac{1}{x + \sqrt{x^2-1}} = \frac{1}{x + \sqrt{x^2-1}}\cdot\frac{x - \sqrt{x^2-1}}{x - \sqrt{x^2-1}}$$
$$=\frac{x - \sqrt{x^2-1}}{x^2-(x^2-1)}$$
$$= \frac{x - \sqrt{x^2-1}}{1}$$
$$=x - \sqrt{x^2-1}$$
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
... | The first term
$$
\eqalign{
& \left( {a_1 + a_2 + \cdots + a_n } \right)^2 = R^2 \quad \Rightarrow \cr
& \Rightarrow \quad \left( {{{a_1 } \over R} + {{a_2 } \over R} + \cdots
+ {{a_n } \over R}} \right)^2 - 1 = 0\quad \Rightarrow \cr
& \left( {\left( {b_1 + b_2 + \cdots + b_n } \right) + 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 6,
"answer_id": 3
} |
$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$
For $a,b,c\ge0: ab+bc+ca+abc=4$ then: $$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$$
I used the condition and get: $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2$
So we need to prove that:... | Now I find a proof by AM-GM: $$4=2\sqrt{\frac{a(b+c)+bc+abc}{a+\sqrt{bc+abc}}(a+\sqrt{bc+abc})}\le a+\sqrt{bc+abc}+\frac{a(b+c)+bc+abc}{a+\sqrt{bc+abc}}=\frac{a(a+b+c)}{a+\sqrt{bc+abc}}+2\sqrt{bc+abc}$$
$$\implies \frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}\ge\frac{2(b+c)}{a+b+c}$$
Or: $$\frac{2-a}{a+\sqrt{bc+abc}}\ge\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}\text{ for }n\ge 2$ I'm completely lost on this question:
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\g... | For $k\geq2,$ \begin{align} & \frac{2}{k^2}\left(1-\frac{1}{\sqrt{k+1}}\right)\leq \frac{2}{(k+1)^2}\\
\iff & 1-\frac{k^2}{(k+1)^2}\leq \frac{1}{\sqrt{k+1}}\\
\iff & \frac{2k+1}{(k+1)^2}\leq \frac{1}{\sqrt{k+1}}\\
\iff & (2k+1)^2\leq (k+1)^3\\
\iff & k(k^2-k-1)\geq0\\
\iff & k^2-k-1\geq0\\
\iff & k\in\left(-\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Writing a polynomial as polynomial powers I want to write the polynomial
$$p(x)=x^4+3x^3+4x^2-7x+6$$ as a polynomial in powers of $x-1$.
What I am trying to do is to determine the taylor series around $x=1$.To do so, consider that $p(1)=7, p'(1)=14, p''(1)=38, p'''(1)=42, p''''(1)=24$
From this it follows that
$p(x)=7+... | Your expansion is correct.
Here is an unconventional way to do this.
$P(x) = x^4 + 3x^3 + 4x^2 - 7x + 6\\
P(x+1) = (x^4 + 4x^3 + 6x^2 + 4x + 1) + 3(x^3+3x^2+3x+1) + 4(x^2 + 2x+1) - 7(x+1) + 6\\
P(x+1) = x^4 + 7x^3 + 19x^2 + 14x + 7\\
P((x-1)+1) = P(x) = (x-1)^4+7(x-1)^3 + 19(x-1)^2 + 14(x-1)+7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How would you find the inverse of a complex function? Let's say you had a function $f \colon \mathbb{C} \longrightarrow \mathbb{C}$ defined by
$$ f(x+ \iota y) = (3+6x)+ \iota (2−3y). $$
How would you go about finding its inverse? This provoked my interest and curiosity, as it seems you can't simply manipulate the equa... | Note that, for any complex numbers $x+\iota y$ and $u + \iota v$, we have $f ( x+ \iota y) = u + \iota v$ if and only if $x+ \iota y = f^{-1}(u+\iota v)$, or $f^{-1}(u+\iota v) = x+\iota y$.
Suppose that $f(x +\iota y) = u + \iota v$, then we have
$$
(3+6x)+\iota (2−3y) = u + \iota v,
$$
which implies
$$
\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
When should I use partial fractions in generating functions I am currently studying generating functions and I don't understand why should I use partial fraction decomposition when solving $x^n$ coefficient of a question.
For example, in this function
$$\frac{1}{(1-x)(1-2x)^2} $$
Why should I use partial fractions to d... | First example:
The partial fraction decomposition in the first case is a convenient representation since we can directly derive a closed formula (without sums) from it. Denoting with $[x^n]$ the coefficient of $x^n$ of a series we obtain at a glance
\begin{align*}
[x^n]\left(\frac{1}{1-x} - \frac{2}{1-2x} + \frac{2}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I prove algebraically that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$? I found out that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$, by reverse engineering:
$$\sqrt{6}-\sqrt{2} =
\sqrt{ (\sqrt{6}-\sqrt{2} )^{2} } = \sqrt{ (\sqrt{6}-\sqrt{2} ) (\sqrt{6}-\sqrt{2} )}= \sqrt{ 6-\sqrt{6×2}-\sqrt{6×2}+2}=\sqrt{ 8-2\sqrt{12}... | If you didn't know the answer already for $\sqrt{8-4\sqrt{3}}$, you could start with $$8-4\sqrt{3}=(\sqrt{a}-\sqrt{b})^2$$
$$\implies8-4\sqrt{3}=a-2\sqrt{ab}+b$$
$$\implies a+b=8$$ and $$\sqrt{ab}=2\sqrt{3}\implies ab=12$$
and solve for $a$ and $b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A $17$-digit number and the number formed by reversing its digits are added together. Show that the sum has a even digit. A $17$ digit number is chosen, and it's digits are reversed, forming a new number. These two numbers are added together. Show that there sum has at least one even digit.
The solution given in the bo... | $E$ indicates even; $O$ indicates odd; $d_{n}$ indicates digit $n$; column $k$ indicates the column number (with $1$ being the rightmost column and $17$ being the leftmost column); $\checkmark$ indicates that the condition of at least $1$ even digit in the sum has been satisfied.
$$\begin{array}{|c|c|}
\hline
&d_1& d_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is $d(\mathbb{Q}(\sqrt{2},\sqrt{3}))$? $K=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ and $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4$. We also know the conjugates of $\sqrt{2}+\sqrt{3}$ are,
$$x_1=\sqrt{2}+\sqrt{3}$$
$$x_2=\sqrt{2}-\sqrt{3}$$
$$x_3=-\s... | Well, my favorite method of defining (and calculating!) the discriminant of a free algebra is as the determinant of the trace pairing. Before I launch into this, however, in case your ring is of the form $\Bbb Q[a]$, so that a basis is the powers $\{a^m\}_{0\le m<n}$, where the rank is $n$, there’s a quick and dirty wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4420178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Given that $ a,b,c$ are positive numbers and $(a+b)(b+c)(c+a)=1$, find the maximum value of $P=ab+bc+ca$ I am stuck with this problem:
Given that $ a,b,c$ are positive numbers and $(a+b)(b+c)(c+a)=1$, find the maximum value of $P=ab+bc+ca$
I tried to use: $(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ca)-abc \implies P=\frac{1+abc... | We have $(a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) - abc$ which results in
$$P = \frac{1 + abc}{a + b + c}.$$
Using AM-GM, we have
$$(a + b)(b + c)(c + a)
\ge 2\sqrt{ab}\cdot 2\sqrt{bc}\cdot 2\sqrt{ca} = 8abc$$
and
$$(a + b)(b + c)(c + a)\le \left(\frac{a + b + b + c + c + a}{3}\right)^3 = \frac{8}{27}(a + b + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Solving an exponential diophantine equation I came across the following problem:
Find all positive integers $a,b$ such that
The following equation is satisfied by the two integers:
$5^b+2=7^c$
My Attempt: I tried equating both sides modulo 6. We know $7^c\cong 1\mod 6$. Hence $5^b\cong 5\mod 6$ which means $b=2k+1$. Th... | Better use a modulo where you can really look at the specifics of the case $b \ge 2$ or $c \ge 2$, because without either of those, you won't manage to prove much looking only at possible remainders (since the equality can stand with $b=c=1$).
For instance, modulo $25$, the LHS is always $2 \pmod{25}$, but you can chec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Express $2\cos^2x - 3\sin^2x$ in terms of $\cos(2x)$ I need to express $2\cos^2x - 3\sin^2x$ in terms of $\cos(2x)$.
I have substituted the $\sin^2x$ using trig identities and have reached:
$5\cos^2x - 3$.
But now I'm really stuck.
| So $2\cos^{2}x-3\sin^{2}x=5\cos^{2}x -3$ but also $\cos 2x=2\cos^{2}x-1$ substitution give \begin{align*} 2\cos^{2}x-3\sin^{2}x&=5\cos^{2}x -3,\\&=5\left(\frac{1}{2}+\frac{1}{2}\cos2x\right)-3,\\&=-\frac{1}{2}+\frac{5}{2}\cos 2x.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4431956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Relation between roots and coefficients of equation If the roots of the equation $x^4 - x^3 +2x^2+x+1 = 0 $ are given by $a,b,c,d$ then find the value of $(1+a^3)(1+b^3)(1+c^3)(1+d^3)$
I found out that:
$$ (1+a^3)(1+b^3)(1+c^3)(1+d^3) = (abcd)^3+\sum(abc)^3 +\sum(ab)^3+\sum(a)^3 +1$$
But how do I calculate $\sum(abc)^3... | Something seemed a bit suspicious about that polynomial: poking around a little, I found that
$$ ( x^4 \ - \ x^3 \ + \ 2x^2 \ + \ x \ + \ 1)·(x^2 \ + \ x \ - \ 1 ) \ \ = \ \ x^6 \ + \ 4x^3 \ - \ 1 \ \ . $$
Treating this as a "quadratic in $ \ x^3 \ \ , $ " we find that there are two "triplets" of zeroes given by $ \ x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Given $x^4+y^4+z^4=1$ Find the minimum value of $\sum_{cyc}\frac{x^3}{1-x^8}$. Let $x,y,z$ be positive reals such that $x^4+y^4+z^4=1$. Find the minimum value of
$$\sum_{cyc}\frac{x^3}{1-x^8}$$
First I tried Jensen. Let $$f(x)=\frac{x^3}{1-x^8}$$
Then $f$ is convex on $\mathbb R^+$, using Jensen $$\frac{1}{3}\sum_{cyc... | We have
\begin{align*}
\frac{1}{1 - x^8}
&= \frac{1}{8/9 + (1/9 - x^8)}\\
&= \frac{9}{8}\cdot \frac{1}{1+ (9/8)(1/9 - x^8)}\\
&\ge \frac98 \left(1 - \frac{9}{8}(1/9 - x^8)\right) \tag{1}\\
&= \frac{9}{64}(7 + 9x^8)\\
&\ge \frac{9}{64}\cdot 8\sqrt[8]{1^7 \cdot 9x^8} \tag{2}\\
&= \frac98 \sqrt[8]{9} \, x
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Generating Function Approach giving wrong combination count while normal brute force casework approach giving correct answer I am solving a problem: How many words are less than four letters long and contain only the letters A, B, C, D, and E? Here, 'word' refers to any string of letters.
My Solution 1 uses Generating ... | Ordinary Generating functions does not calculate the permutations. It gives the combinations and thus order is not taken into account. So, second appraoch gives right answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$A^2 - B^2$ is an invertible matrix and $A^5 = B^5$ and $A^3B^2 = A^2B^3$ . Then what is the determinant of the matrix $A^3 + B^3$? $A,B$ are two $3 \times 3$ real-matrices with following three properties.
*
*$A^2 - B^2$ is an invertible matrix
*$A^5 = B^5$
*$A^3B^2 = A^2B^3$
Then what is the determinant of the mat... | Let $V=\mathbb R^3$. The condition $A^5=B^5$ implies that $A^5V=B^5V$. Since $A$ is $3\times 3$, the descending chain $V\supseteq AV\supseteq A^2V\supseteq\cdots$ must have been stabilised at $A^3V$ and the similar is true for $B^3V$. Therefore from $A^5V=B^5V$ we get $W:=A^3V=B^3V$.
Furthermore, as the descending chai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4438063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to find $x$ such that $\frac{1}{2^{p-1}} -\frac{1}{a^p}>x$? Let $a, p\in\mathbb{R}$ be such that $a\ge 2$ and $p>1$. Consider the quantity
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}.$$
Since $a\ge 2$, it seems clear to me that
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}>0.$$
My question is: it is possible to find a positive real nu... | $$
f(a) = \frac{1}{2^{p-1}} -\frac{1}{a^p}
$$
is increasing in $a$, therefore is
$$
f(a) \ge f(2) = \frac{1}{2^{p-1}} -\frac{1}{2^p}= \frac{1}{2^p}
$$
for $a \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculate the minimum value of $T = 3x + 2y$ with positives $x$ and $y$ satisfying $\log_3\frac{xy + 2}{x^2 + 1} = \frac{3^x + 3x^2 + 1}{3^x + xy}$.
Consider two positives $x$ and $y$ such that $\log_3\dfrac{xy + 2}{x^2 + 1} = \dfrac{3^x + 3x^2 + 1}{3^x + xy}$. Calculate the minimum value of $T = 3x + 2y$.
We have th... | Build the auxiliary Lagrangian function
$$\mathscr L=3x+2y-\lambda\left(\log_3\frac{xy+2}{x^2+1}-\frac{3^x+3x^2+1}{3^x+xy}\right)+\mu_1x+\mu_2y.$$
The first order conditions for a minimum (see KKT multipliers) are
\begin{align}
\frac{\partial\mathscr L}{\partial x}&=0\\
\frac{\partial\mathscr L}{\partial y}&=0\\
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
provided $f( x+1) =\ \lim _{n\rightarrow \infty }\left(\frac{n+x}{n-2}\right)^{n}$, what is f(x)? $$
f( x+1) =\ \lim _{n\rightarrow \infty }\left(\frac{n+x}{n-2}\right)^{n}
$$
Here is one of the solution from my workbook:
$$
\begin{array}{l}
f( x+1) \ =\ \lim _{n\rightarrow \infty }\left[\left( 1+\frac{2+x}{n-2}\right... | First, $$\frac{n+x}{n-2} = 1 + \frac{1}{\frac{n-2}{2+x}}$$
from rearranging.
Then they use the following fact $$\lim_{k \to \infty} \left(1 + \frac{1}{k}\right)^k = e$$
which implies
$$\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x}} = e.$$
By continuity, we can raise both sides to the $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find a any matrix $C$ satisfying the below conditions. Let the $A,B$ and $C$ are $3\times 3$ matrices.
For the vectors $v_1= \begin{bmatrix} 1 \\ 1 \\ -1 \\ \end{bmatrix}$,$v_2= \begin{bmatrix} 0 \\ 1 \\ -1 \\ \end{bmatrix}$ and $v_3= \begin{bmatrix} 1 \\ 1 \\ -2 \\ \end{bmatrix}$, $Av_1=v_1
Av_2=v_2, Av_3=-2v_3, Bv_1... | $$C = P^{-1}DP=\left(\begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & -1 \\\end{bmatrix}\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0& -1 \\\end{bmatrix}\right)\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ -1 & -1& -2 \\\end{bmatrix}$$
$$=\left(\begin{bmatrix} 0 & 1 & -1 \\ 0 & 1 & 0\\ 0& -1& 1 \\\end{bmatrix}\right)\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$
Find $x$ in the equation $\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$.
My attempt:
$$\log\left(x\right)\ -\ \frac{1}{2}\ \log\left(x-\frac{1}{2}\right)\ =\ \log\... | Simplifying $$\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$$
into
$$\log\left(\frac{x}{(x-\frac{1}{2})^\frac{1}{2}}\right) = \log\left(\frac{x+\frac{1}{2}}{(x+\frac{1}{8})^\frac{1}{2}}\right)$$
And taking the exponential,
$$\frac{x}{(x-\frac{1}{2})^\frac{1}{2}} = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.