Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solving limit without L'Hôpital's rule: $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$ How can I solve this limit without L'Hôpital's rule?
$$\begin{align}\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\lim\limits_{x \to 0} \frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x... | Hint:
$$\tan x- \sin x=\frac{x^3}{3}-\frac{x^3}{3!}+......$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\fr... | Note that$$4+\frac1x-\frac1{x^2}=\frac1{x^2}\left(4x^2+x-1\right).$$So, solve the equation $4x^2+x-1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Solving second-order differential equations. I have a few problems that I'm trying to work through. Want to see if these few are correct.
*
*$$3y'' + 4y' - 3y = 0$$
auxiliary equation is: $$3r^2 + 4r -3 = 0$$ where $a = 3$, $b = 4$, $c = -3$
can't really find roots by factoring so gonna use quadratic:
$$r = \frac{-4... | For this type of problems you can also follow the table given below.
I think, this table will help you to deal with all linear differential equations with constant coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Largest circle in basin of attraction of the origin. We're given dynamical system:
$$
\dot x = -x + y + x (x^2 + y^2)\\
\dot y = -y -2x + y (x^2 + y^2)
$$
Question is what's the largest constant $r_0$ s.t. circle $x^2+y^2 < r_0^2$ lies in the origins basin of attraction.
So far with relatively easy algebra I've got:
$$... | After solving the system
$$
\dot r = \frac{r}{2}(-2-\sin(2 \phi)+2r^2) \\
\dot \phi = -(1+\cos^2(\phi))
$$
we have
$$
\left\{
\begin{array}{rcl}
r & = & \frac{\sqrt{3} \sqrt{3-\cos \left(2 \sqrt{2} \left(t-2 c_1\right)\right)}}{\sqrt{3 e^{2 t} c_2-\cos \left(2 \sqrt{2} \left(t-2
c_1\right)\right)+\sqrt{2} \sin \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Solving a set of multivariate polynomials; four equations, four unknowns, maximum degree 2 I am currently trying to solve a set of four polynomial expressions for a geometric purpose. in short they take on this form:
$A_0=\frac{w_0^2+w_1 w_0+w_2 w_0+w_3 w_0+27 w_0+6 w_1+6 w_2+6 w_3+117}{\left(w_0+w_1+w_2+w_3+27\right){... | No matter how you attack this, generally, you will have one variable the other three depend on, another variable that the remaining two depend on, and of these last two, one will depend on the other. (I say "generally" because very, very rarely, the variables decouple. Consider $x^2 - x - y + 1 = 0$. This (surprise!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Different methods give different answers Solve $ \sec x- 1 = (\sqrt 2 - 1) \tan x $ Solve $ \sec x- 1 = (\sqrt 2 - 1) \tan x $
Case 1)
Square both the sides and using $ \sec ^2 x = 1+ \tan^2 x. $ And solving the quadratic we get answer $\tan x = 1$ or $\tan x = 0$.
Putting them back in also solves the equation.
Thus $... | Squaring both sides of an equation can introduce extraneous solutions. Substitution demonstrates that if $n$ is odd, then $x = n\pi$ is not a solution as $-2 \neq 0$ and that $x = n\pi + \frac{\pi}{4}$ is not a solution as the LHS is negative while the RHS is positive. On the other hand, direct substitution shows tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3218129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof that $ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $ for $p>5$ Proof that $$ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $$
for $p>5$ and $p$ is prime.
$\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}$
My try
Let show that
$$p^2 - 1 \equiv 0 \Mod{30} \vee p^2 - 1\equiv 18 \Mod{30}$$
Let check $$p^2 -1 = (p... | Any prime $> 5$ is congruent to $1$ mod $2$, to $1$ or $2$ mod $3$, and to $1, 2, 3$ or $4$ mod $5$.
Its square is congruent to $1$ mod $2$, to $1$ mod $3$, and to $1$ or $4$ mod $5$. The numbers that are congruent to $1$ mod $2$, to $1$ mod $3$ and to $1$ mod $5$ are congruent to $1$ mod $30$, while those that are co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left... | Looks good, and thumbs up for taking the time to provide your own work!
$$\lim\limits_{x \to 0}\frac{e^{(1+x)^{1/x}}-(1+x)^{e/x}}{x^2}$$
Not sure if this is any better, but taking $t=\left(1+x\right)^{1/x}$ turns the numerator into $e^t-t^e$ with $t \to e$ when $x \to 0$. Its first non-zero term in the series expansi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Why is it that exponents that are divisors of φ(N) capable of generating the identity element as a power when N is prime? Let $p=x$ where $a^x \equiv 1\pmod N$. When $N$ is prime I can check whether $p=N-1$ and if it is, there are a full set of remainders. However, if $p<N-1$, I need to keep checking. For example, if $... | Let $N$ be a prime (as you specify) and let $(a,N) = 1$.
Let $\omega$ be the smallest positive integer such that
$$ a^{\omega} \equiv 1 \pmod{N}.$$
Then $\omega$ is called the order of $a$ mod $N$.
Since $\omega | \phi(N)$, it follows that $\phi(N) = k\omega$, for some positive integer $k$. Hence,
$$a^{\phi(N)} \equiv ... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\sum_{n=1}^{\infty}{\frac{1}{n 2^{n-1}}}$ I need to find $$S = \sum_{n=1}^{\infty}{\frac{1}{n 2^{n-1}}}$$
Attempt:
$$f'(x) = \sum_{n=1}^{\infty}\frac{x^n}{2^n} = \frac{x}{2-x}$$
Which is just evaluating geometric series
$$f(x) = \sum_{n=1}^{\infty}\frac{x^{n+1}}{(n+1)2^{n}}$$
Now, by finding antiderivative of... | $$S=2\sum_{n=1}^{\infty} \frac{2^{-n}}{n}=2\sum_{k=1}^{\infty}~ 2^{-n} \int_{0}^{1} x^{n-1} ~dx=2 \int_{0}^{1}\sum_{n=1}^{\infty} \frac{dx}{x}\left(\frac{x}{2}\right)^n =\int_{0}^{1}\frac{2}{x} \frac{x/2}{1-x/2} ~dx=2 \ln 2. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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How to find the sum: $\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) $ $$S_n=\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) = \frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{r}\right)$$
I fail to see how this:
$$S_n = \frac{1}{r-s}\left( \frac... | Consider the partial sum
$$S_p=\sum_{n=0}^p \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right)=\frac{1}{r-s}\left(H_{p+s+1}-H_{p+r+1}+H_r-H_s \right)$$ and use the asymptotics of harmonic numbers to get
$$S_p=\frac{H_r-H_s}{r-s}-\frac{1}{p}+\frac{r+s+3}{2
p^2}+O\left(\frac{1}{p^3}\right)$$ Then, the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A Problem Concerning the Series Problem
Let $\{a_n\},\{b_n\}$ be two sequences such that $a_n>0(\forall n \geq 1)$, $\sum\limits_{n=1}^{\infty}b_n$ is absolutely convergent, and $$\frac{a_n}{a_{n+1}}\leq 1+\frac{1}{n}+\frac{1}{n\ln n}+b_n(\forall n \geq 2).\tag{1}$$
Prove
*
*$\dfrac{a_n}{a_{n+1}}< \dfrac{n+1}{n}\cdot... | Step 1 is correct though the simple estimate $\log{(1+\frac{1}{n})} > \frac{1}{n} - \frac{1}{2n^2} \ge \frac{1}{n+1}$ immediately shows (4)
Now let $x>A>0, |y|<\frac{x}{2}$, then $\frac{1}{x+y}=\frac{1}{x}+z, |z| < \frac{2|y|}{A^2}$ since obviosuly $z=-\frac{y}{x(x+y)}$ and $x(x+y)>\frac{x^2}{2}> \frac{A^2}{2}$ by our ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inserting random numbers from 1 to $n^2$ in a matrix of size $n \times n$ I have two matrices of size nxn with random numbers that are in range of $1$ to $n^2$.
I'm trying to calculate the probability of :
*
*the numbers 1 and 9 are present in the same indices in the two matrices,for example($n=4$):
\begin{bmatrix}
... | I'll solve the first part of the problem and leave the rest for you to solve
First place 1 and 9 at any two locations out of $n^2$ available locations. Then fill the rest of $2n^2-4$ with any number between 1 and $n^2$. So the total number of desired possibilities $=^{n^2}\!\!\!P_{2}(n^2)^{2n^2-4}$.
The number of possi... | {
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"url": "https://math.stackexchange.com/questions/3230059",
"timestamp": "2023-03-29T00:00:00",
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How can I order these numbers without a calculator?
*
*Classify the following numbers as rational or irrational. Then place them in order on a number line:
$$\pi^2, -\pi^3, 10, 31/13, \sqrt{13}, 2018/2019, -17, 41000$$
I know $\pi$ is irrational so $\pi^2$ and $-\pi^3$ are irrational. 10, 31/30, 2018/2019, -17, 410... | $\sqrt{13}$ is irrational, by user.
The order of your first question should be $-\pi^3,-17,2018/2019,31/13,\pi^2,10,41000$.
$-\pi^3<-3^3 = -27<-17$, so we can compare two negative numbers.
We now compare $\pi^2$ and $10$, others are easy to compare. Note that $\pi = 3.1415926\dots$ and so $\pi<3.15$. Now $3.15^2 = 9.92... | {
"language": "en",
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How to prove De Moivre's theorem inductively It is given that $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ where $n\in Z^+$.
I can show it works for $n=1$ but I am stuck in showing it inductively. I have got as far as below but are stuck in the rearranging:
$$(\cos\theta + i\sin\theta)^{k+1}= (\cos k\t... | By induction,
$(\cos \theta + i\sin \theta )^{k+1} = (\cos \theta + i\sin \theta )^k.(\cos \theta + i\sin \theta ) = (\cos k\theta + i\sin k\theta )(\cos \theta + i\sin \theta )$
$(\cos \theta + i\sin \theta )^{k+1} = \cos k\theta\cos \theta + i\sin k\theta\cos \theta + i\sin \theta \cos k\theta - \sin k\theta\sin \th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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convergence of a fibonacci-like sequence I posted a question earlier on finding a formula for the sequence
$$t_1, t_2, t_1+t_2, t_1+2t_2,....$$
This is the question I posted earlier
I want to show that as $n\rightarrow \infty$, $\frac{t_{n+1}}{t_n} \rightarrow \phi$
The relationship is $T_n = A_nF_{n-2} +B_n F_{n_1}$
... | From linear recurrence theory, as characteristic polynomial of the sequence is $x^2 - x - 1$ and it's roots are $\frac{1 \pm \sqrt{5}}{2}$, we have $t_n = A \left(\frac{1 + \sqrt{5}}{2}\right)^n + B \left(\frac{1 - \sqrt{5}}{2}\right)^n$ where $A$ and $B$ can be found from $t_1$ and $t_2$. If $A \neq 0$ then $\lim\lim... | {
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"timestamp": "2023-03-29T00:00:00",
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How many solutions are there to $x_{1} + x_{2} + x_{3} + x_{4} = 15$ How many solutions are there to (I think they mean non-negative integer solutions)
$x_{1} + x_{2} + x_{3} + x_{4} = 15$
where
$1\leq x_{1} \leq 4$
$2 \leq x_{2} \leq 5$
$7\leq x_{3}$
$2\leq x_{4}$
My "Solution"
I use the method shown in this video: ht... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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The coefficient of $x^n$ in the expansion of $\frac{2-3x}{1-3x+2x^2}$ is? The coefficient of $x^n$ in the expansion of $\frac{2-3x}{1-3x+2x^2}$ is?
Working: $1 – 3x + 2x^2 = (1 – x)(1 – 2x)
=> (1 – x)^{-1} = 1 + (-1)(-x) + {(-1)(-1 -1)/2!}(-x)^2 + {(-1)(-1-1)(-12)/3!}(-x)^3 + … = 1 + x + x^2 + x^3 + …. $
$(1 – 2x)^{-1}... | $\dfrac{2-3x}{1-3x+2x^2}=\dfrac{1}{1-x}+\dfrac1{1-2x}$
The coefficient of $x^n$ is $1^n+2^n=2^n+1$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to compute gcd of two polynomials efficiently I have two polynomials $A=x^4+x^2+1$
And $B=x^4-x^2-2x-1$
I need to compute the gcd of $A$ and $B$ but when I do the regular Euclidean way I get fractions and it gets confusing, are you somehow able to use a SylvesterMatrix to find the gcd or am I probably doing somethi... | $$ \left( x^{4} + x^{2} + 1 \right) $$
$$ \left( x^{4} - x^{2} - 2 x - 1 \right) $$
$$ \left( x^{4} + x^{2} + 1 \right) = \left( x^{4} - x^{2} - 2 x - 1 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 2 x^{2} + 2 x + 2 \right) $$
$$ \left( x^{4} - x^{2} - 2 x - 1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Limit of powers of $3\times3$ matrix Consider the matrix
$$A = \begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$$
What is $\lim_{n→\infty}$$A^n$ ?
A)$\begin{bmatrix} 0 & 0 & 0\\ 0& 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
B)$\begin{bmatrix} \frac{1}{4} &\f... | By this question, we know that
\begin{equation}
A^n=
\begin{pmatrix}
2^{-n} & n\cdot 2^{-n-1} - 2^{-n-1} + \frac12 & {1-\frac{n+1}{2^n}\over2}\\
0 & {2^{-n}+1\over2} & {1-2^{-n}\over2} \\
0 & {1-2^{-n}\over2} & {2^{-n}+1\over2}
\end{pmatrix}.
\end{equation}
It is thus clear that $\lim_{n\to\infty} A^n = \begin{pmatrix... | {
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Calculate $\lim _{n\rightarrow +\infty} \sum _{k=0}^{n} \frac{\sqrt{2n^2+kn-k^2}}{n^2}$
Calculate $$\lim _{n\rightarrow +\infty} \sum _{k=0}^{n} \frac{\sqrt{2n^2+kn-k^2}}{n^2}$$
My try:
$$\lim _{n\rightarrow +\infty} \sum _{k=0}^{n} \frac{\sqrt{2n^2+kn-k^2}}{n^2}=\lim _{n\rightarrow +\infty} \frac{1}{n} \sum _{k=0}^{... | Hint. Just make the change of variable
$$
u=\frac 32 \sin x,\qquad du = \frac 32 \cos x \,dx,
$$ giving
$$
\int^{\frac{1}{2}}_{-\frac{1}{2}} \sqrt{\frac{9}{4}-u^2} _,du=\frac 94\int^{\arcsin\frac{1}{3}}_{-\arcsin\frac{1}{3}} \cos^2 x \,dx=\frac 98\int^{\arcsin\frac{1}{3}}_{-\arcsin\frac{1}{3}}\left(1+ \cos 2 x\right) d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Find all roots of the equation :$(1+\frac{ix}n)^n = (1-\frac{ix}n)^n$ This question is taken from book: Advanced Calculus: An Introduction to Classical Analysis, by Louis Brand. The book is concerned with introductory real analysis.
I request to help find the solution.
If $n$ is a positive integer, find all roots of ... | Hint:
Put $$z=\frac{1+i\frac{x}{n}}{1-i\frac{x}{n}}$$ then $z$ will be a $n$-root of unity and solve for $x:$ $$z= \frac{1+i\frac{x}{n}}{1-i\frac{x}{n}}=\exp{\left(i\frac{2k\pi}{n}\right)},\quad k\in\{0,1,...,n-1\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculate limit in use of integrals. Calculate limit in use of integrals:
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k+n}{3k^2+n^2+1} $$
Solution:
$$\sum_{k=1}^{n} \frac{k+n}{3k^2+n^2+1} = \frac{1}{n} \sum_{k=1}^{n} \frac{\frac{k}{n}+1}{3(k/n)^2+1 + n^{-2}} = \\
\frac{1}{n}\cdot \sum_{k=1}^{n} \frac{\frac{k}{n... | For any $\varepsilon>0$ and all $n>1/\sqrt\varepsilon$, we have
$$
\frac{k+n}{3k^2+n^2}\geq
\frac{k+n}{3k^2+n^2+1}\geq
\frac{k+n}{3k^2+(1+\varepsilon)n^2}.
$$
So we have
$$
\int_0^1\frac{x+1}{3x^2+1}\,\mathrm{d}x\geq\lim_{n\to\infty}\frac{k+n}{3k^2+n^2+1}\geq\int_0^1\frac{x+1}{3x^2+(1+\varepsilon)}\,\mathrm{d}x
$$
and ... | {
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How do I see if $a^3+b^3=c^3+d^3$ has any solutions where $ 1 \le a,b,c,d \in \mathbb{Z} \le 1000$ and $a \ne b \ne c \ne d$? How do I see if $a^3+b^3=c^3+d^3$ has any solutions where $ 1 \le a,b,c,d \in \mathbb{Z} \le 1000$ and $a \ne b \ne c \ne d$ ?
I know I can write a program to brute force this and find out, bu... | You are asking about Taxicab Numbers, of which the most famous is
$$1729 = 12^3 + 1^3 = 10^3 + 9^3.$$
There is no (known) algebraic method to solve this equation. All of our current understand of these types of numbers come from computer calculation. There is also an OEIS entry dedicated to them.
If you're curious, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Theorem of Pythagoras - Incorrect Derivation I am trying to solve below question from Coursera Intro to Calculus (link)
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and
$2ab$ units respectively, where $a$ and $b$ are positive real numbers
such that $a$ is greater than $b$. Find an exact expre... | You are done. just multiply out $(a^2 + b^2)^2$ and check that you get $x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove for all positive a,b,c that $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$ Prove for all positive a,b,c $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$$
My Try
I tried taking common denominator of the expression,
$\frac{a^2b+ab^2+b^2c+c^2b+ac^2+a^2c}{abc}$
How to proceed? Is there a way to write them as per... | Using Am-Gm for 6 terms we get: $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\geq 6\sqrt[6]{\frac{a}{b} \frac{b}{a} \frac{b}{c} \frac{c}{b} \frac{c}{a}\frac{a}{c}}= 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$.
$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}... | I have an alternative solution, by CS:
$$A=\sum\limits_{cyc} \frac{a^2}{b+c}=\sum\limits_{cyc} \frac{a^4}{a^2b+a^2c}\geq \frac{\left(\sum\limits_{cyc}a^2\right)^2}{\sum\limits_{sym}a^2b} $$
By Jensen:
$$1=\sum\limits_{cyc}\sqrt{a^2+b^2}\leq 3\sqrt{\frac{2}{3}(a^2+b^2+c^2)}$$
So
$$A\geq \frac{\left(\sum\limits_{cyc}a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Let $A, B$ and $C$ be the angles of an acute triangle. Show that: $\sin A+\sin B +\sin C > 2$. Let $A, B$ and $C$ be the angles of an acute triangle. Show that: $\sin A+\sin B +\sin C > 2$.
I started from considering
$$\begin{align}\sin A+\sin B+\sin (180^o-A-B) &= \sin A+\sin B+\sin(A+B)
\\&=\sin A+\sin B+\sin A\cos ... | Note that since $\sin(x)$ is concave and $\left ( \frac{\pi}{2} , \frac{\pi}{2}, 0 \right ) \succ (A, B, C)$, then by Karamata's Inequality
$$\sin(A) + \sin(B) + \sin(C) > \sin \left ( \frac{\pi}{2} \right ) + \sin \left ( \frac{\pi}{2} \right ) + \sin(0) = 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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The value of $I=\int_{|z|=2}z^2\sin\frac{1}{z}dz$? I want to get the value of $I=\int_{|z|=2}z^2\sin\frac{1}{z}dz$.
My idea is to let $t=\frac{1}{z}$, then
$$I=\int_{|z|=2}z^2\sin\frac{1}{z}dz=\int_{|t|=\frac{1}{2}}\frac{\sin t}{t^4}dt=2\pi i \times-\frac{1}{6}=-\frac{\pi i}{3}$$ Is this right?
| Your result has the wrong sign because you did not take into account that the substitution $t=-\frac{1}{z}$ reverts the orientation of the circle.
You could also apply the residue theorem directly to
$$
z^2 \sin \frac 1z = z^2 \left( \frac 1z - \frac{1}{3!z^3} + \ldots \right) = z -\frac{1}{6z} + \ldots
$$
so that
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum of $\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$ $$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
The question is divided into three parts:
1. Determine its radius of convergence
2. By using the power series of $\frac{1}{1-x}$, show that for all x $\in$ ]-1,1[ , we get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
3.... | Note that $$\sum \limits_{n=1}^{\infty} \frac{-1}{2(n+2)}x^n = \frac{1}{x^2}\sum \limits_{n=1}^{\infty} \frac{-1}{2(n+2)}x^{n+2} = -\frac{1}{x^2}(\frac{1}{2}x + \frac{1}{4}x^2 +\sum \limits_{n=1}^{\infty} \frac{1}{2n}x^n).$$
But you have already found $\sum \limits_{n=1}^{\infty} \frac{1}{2n}x^n$, so the rest is straig... | {
"language": "en",
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"source": "stackexchange",
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Simplifying $\sum_{cyc}\tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)$. I get $0$, but the answer is $\pi$. So the question is
$$ \tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)+\tan^{-1}\left(\sqrt{\frac{y(x+y+z)}{xz}}\right)+\tan^{-1}\left(\sqrt{\frac{z(x+y+z)}{yx}}\right) =\ ? $$
So my take on the question is to... | Like Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$\tan^{-1}\sqrt{\dfrac{x(x+y+z)}{yz}}+\tan^{-1}\sqrt{\dfrac{y(x+y+z)}{zx}}$$ $$=\begin{cases} \tan^{-1}\left(\dfrac{\sqrt{\dfrac{x(x+y+z)}{yz}}+\sqrt{\dfrac{y(x+y+z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $-\frac{1}{a}<\int_a^b \sin(x^2) dx<\frac{1}{a}$ I have encountered a question:
Prove
$$-\frac{1}{a}<\int_a^b \sin(x^2) dx<\frac{1}{a}$$
There are plenty of solutions to $\int_0^{\infty} \sin(x^2) dx$ online, but there seems to be no solution to the boundary of $\int_a^b \sin(x^2) dx$.
Could anyone help ... | By change of variables $u=x^2$, we have $\displaystyle{\int_{a}^{b}\sin(x^2)\,{\rm d}x=\frac{1}{2}\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}\,{\rm d}u}$.
Now, by integration by parts we have
$$\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}\,{\rm d}u=-\int_{a^2}^{b^2}\frac{1}{\sqrt{u}}{\rm d}(\cos u)=\frac{\cos(a^2)}{a}-\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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How to evaluate $\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$
How to evaluate $$\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$$
I tried to integrate by parts, but no way so far, help me, thanks.
| different approach to evaluate $\displaystyle\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx$ :
from here , we have $\displaystyle\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1-x)}{x}\ dx=\frac{\pi^3}{16}\tag{1}$
and from here , we have $\displaystyle \ 3\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Ordinary Differential Equation getting two different answers Solve $y(y^2-2x^2)dx+x(2y^2-x^2)dy=0$ and find a particular curve passing through $(1,2)$
My attempt:
1st Solution: Rewrite as
$y(y^2dx+x2ydy)-x(x^2dy+y2xdx)=0$
$\implies xy^2d(xy^2)-x^2yd(x^2y)=0$ (multiplying $xy$)
$\implies (xy^2)^2-(x^2y)^2 = c\,\,\,$ ... | I got $$\frac{y}{x}\left(\frac{y^2}{x^2}-2\right)-\left(2\frac{y^2}{x^2}-1\right)y'(x)=0$$ now let $$\frac{y}{x}=u$$ and then we get
$$-u^3-u=u'x(2u^2-1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluate $(\sqrt{3}+i)^{14}+(\sqrt{3}-i)^{14}$ Evaluate $(\sqrt{3}+i)^{14}+(\sqrt{3}-i)^{14}$
I tried by using the De'moivers theorem but I didn't get proper value I get a mess value 4..but I am not sure about the answer can anyone please tell me
| Observe that
$$
(\sqrt{3} + i)^{14} = ((\sqrt{3} + i)^2)^7
$$
Also, $ (\sqrt{3} + i)^2 = 2 + i 2\sqrt{3} = 2 (2 e^{j\theta}) = 4 e^{j\theta}$, where $ \theta = \pi / 3 $.
Similarly, for the second term we have that $ (\sqrt{3} - i)^2 = 4 e^{-j\theta} $.
You problem reduces to finding
$$
M = 4^7 \cdot e^{j 7\theta} + ... | {
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"timestamp": "2023-03-29T00:00:00",
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Find maximum and minimum of $\sin x + \sin y$ I am working on my scholarship exam practice but I am stuck on finding the minimum. Pre-university maths background is assumed.
When $x + y = \frac{2\pi}{3}, x\geq0, y\geq0$, the maximum of
$\sin x+\sin y$ is ....., and the minimum of that is .....
Let me walk you thro... | For the minimum, note that since $x,y\ge0\implies y\le\dfrac{2\pi}3$, we have $$\dfrac{x-y}2=\dfrac{x+y-2y}2=\dfrac{\dfrac{2\pi}3-2y}2=\frac\pi3-y$$ so $$\sin x+\sin y = \sqrt{3}\cos\frac{x-y}{2}=\sqrt3\cos\left(\frac\pi3-y\right)\ge\begin{cases}\sqrt3\cos\left(\frac\pi3-0\right)\\\sqrt3\cos\left(\frac\pi3-\frac{2\pi}3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Does $\int_{0}^{\infty} \sin(x) \cdot\sin(x^2)\,dx$ converge? My try: Let $x = \sqrt{t}$, then $dx = \frac{1}{2\sqrt{t}}dt$. We get the following integral: $\int_{0}^{\infty}\frac{\sin(t)\sin(\sqrt{t})}{2\sqrt{t}}dt$. Now I tried to use Dirichlet's test: The function $g(x) = 1/2\sqrt{t}$ has limit $0$ at infinity and i... | As usual, trig identities are the answer:
\begin{align}
\int_0^\infty \sin(x^2)\sin(x)dx =& \frac{1}{2}\int_0^\infty\left[ \cos(x^2-x)-\cos(x^2+x)\right]dx \\ =& \frac{1}{2}\int_0^\infty\left( \cos\left[\left(x-\frac{1}{2}\right)^2 -\frac{1}{4}\right]-\cos\left[\left(x+\frac{1}{2}\right)^2 -\frac{1}{4}\right]\right)dx
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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$\sqrt{2} x^2 - \sqrt{3} x +k=0$ with solutions $\sin\theta$ and $\cos\theta$, find k
If the equation $\sqrt{2} x^2 - \sqrt{3} x +k=0$ with $k$ a constant has two solutions $\sin\theta$ and $\cos\theta$ $(0\leq\theta\leq\frac{\pi}{2})$, then $k=$……
My approach is suggested below but I am not sure how to continue.
Sin... | By the Viete we obtain: $$1=\left(\sqrt{\frac{3}{2}}\right)^2-\sqrt2k.$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given remainders from other polynomial divisions. Find the remainder in a polynomial division Let $F(x)$ be a polynomial.
If $F(x)$ is divided by $(x-1)^2$ the remainder will be $x+1$
and if $F(x)$ is divided by $x^2$ the remainder will be $2x+3$.
What is the remainder if $F(x)$ is divided by $x^2(x-1)$?
My solution ... | Your answer is correct.
The second answer cannot be also true, as $(-3x^2 + 2x + 3) - (-3x^2 + 2x + 1) =2 $ is not divisible by $(x-1)x^2$.
Here is another elementary solution:
*
*$(1):F(x) = (x-1)^2p(x) + x+1$
*$(2):F(x) = x^2q(x) + 2x+3$
*Setting $p(x) =x^2f(x)+ax+b$ you get
$$(x-1)^2p(x) + x+1 = x^2q(x) -2bx+... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$ Find the solution set of $\frac{3\sqrt{2-x}}{x-1}<2$
Start by squaring both sides
$$\frac{-4x^2-x+14}{(x-1)^2}<0$$
Factoring and multiplied both sides with -1
$$\frac{(4x-7)(x+2)}{(x-1)^2}>0$$
I got
$$(-\infty,-2)\cup \left(\frac{7}{4},\infty\right)$$
Since $x\leq2$ ... | Define $f(x)=\frac{3\sqrt{2-x}}{x-1}-2$. Being continuous on its domain $(-\infty,2]\setminus\{1\}$,the function may change its sign only at its zero $7/4$ or at its singularity, namely at $1$. Now check the sign of $f$ in the corresponding intervals $(-\infty,1)$, $(1,7/4)$ and $(7/4,2]$; you want $f(x)<0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find remainder of division of $x^3$ by $x^2-x+1$ I am stuck at my exam practice here.
The remainder of the division of $x^3$ by $x^2-x+1$ is ..... and that of $x^{2007}$ by $x^2-x+1$ is .....
I tried the polynomial remainder theorem but I am not sure if I did it correctly.
By factor theorem definition, provided by W... | Another way if you're familiar with modular arithmetic is to work modulo $x^2-x+1$, in which case we have $x^2\equiv x-1$ and thus
$$x^3\equiv xx^2\equiv x(x-1)\equiv x^2-x\equiv (x-1)-x\equiv -1.$$ This can be extended to your other question by noting that $x^{2007}=\left(x^3\right)^{669}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+... | Hint: Writing your equation in the form
$$\sqrt{3x+7}=1+\sqrt{x+2}$$
squaring gives
$$3x+7=1+x+2+2\sqrt{x+2}$$ so
$$x+2=\sqrt{x+2}$$ squaring again:
$$(x+2)^2=x+2$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 0
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Determine fractional transformation which maps $\mathbb{R}$ to $\mathbb{R}$ and $S^1$ to $S^1$ I want to determine all fractional transformation $F:\mathbb{C}\to\mathbb{C}$ which maps $\mathbb{R}$ on $\mathbb{R}$ and the unit circle on the unit circle. Let $\begin{pmatrix}a&b\\c&d
\end{pmatrix}$ be such a linear transf... | As you have noted in your comment, $F$ either fixes $1$ and $-1$ or swaps them.
Replacing $F$ by $-F$ if necessary, we can assume $F$ fixes $1$ and $-1$. Then we have
$$
\frac{a\cdot 1 + b}{c \cdot 1 + d} = 1 \quad\quad \frac{a\cdot -1 + b}{c \cdot -1 + d} = -1
$$
So:
$
a + b = c + d
$
and
$
b - a = c - d
$
Adding thes... | {
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"question_score": "2",
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Prime factor inversion Define a function $f(n)$ for $n \in \mathbb{N}$ to "invert" the prime
factorization of $n$ in the following sense. Let me start with an example.
If $n= 3564 = 2^2 \cdot 3^4 \cdot 11^1$,
then $f(n) = 2^2 \cdot 4^3 \cdot 1^{11} = 256$,
inverting the base primes and their exponents.
In general, if t... | Q1: Yes, for example $p^q q^p$, where $p$ and $q$ are primes, is a fixed point (or products that consist of these). More generally you should be able to use permutations of order $2$, i.e products of disjoint transpositions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integrating quadratics in denominator I'm following a book on Calculus that introduces partial fraction expansion. They discuss common outcomes of the partial fraction expansion, for example that we are left with an integral of the form:
$$
\int \frac{dx}{x^2+bx+c}
$$
And then we can use complete the square and $u$-sub... | For integrating the Quadratics in the denominator, there are two cases,
$b^2-4c>0:$
$x^2+bx+c= (x+\frac{b}{2})^2-\frac{b^2-4c}{4} = u^2 - \beta^2 \Rightarrow $ This will give you: $\frac{1}{2 \beta} \ln \frac{u- \beta} {u+ \beta}; (\beta> 0)$
$4c-b^2>0:$
$x^2+bx+c= (x+\frac{b}{2})^2+\frac{4c-b^2}{4} = u^2 + \alpha^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Form groups of people using combinatorics I have this problem:
How many groups of 5 people can be formed between 4 boys and 7 girls
if there should be at least 2 girls included?
My attempt was:
Ways of select two girls of a total of $7$ is: $\binom{7}{2} = 21$
Ways to select the $3$ remaining people, out of a tota... | There are four cases, with 2, 3, 4, or 5 girls in the 5-group.
The solution is:
$$ {4 \choose 3}{7 \choose 2} +
{4 \choose 2}{7 \choose 3} +
{4 \choose 1}{7 \choose 4} +
{4 \choose 0}{7 \choose 5} = 455
$$
What you have counted is :
$$ {4 \choose 3}{7 \choose 2} {\color{red} {2 \choose 2}} +
{4 \choose ... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^{3} = 6+ 3xy - 3 ( \sqrt{2}+2 )^{{1}/{3}} , y^{3} = 9 + 3xy(\sqrt{2}+2)^{{1}/{3}} - 3(\sqrt{2}+2)^{{2}/{3}}$ Solve the system of equations for $x,y \in \mathbb{R}$
$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}} $
$ y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}$
I j... | $$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}$$
$$y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}} $$
Let $A=\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}$, then re-write the equations as
$$x^{3} = 6+ 3xy - 3A\tag{1}$$
$$y^{3} = 9 + 3xyA - 3A^2\tag{2} $$
Solving (1) for $3xy$
$$ 3xy=... | {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $H'(x)$ given $H(x) = \int_{x^3 + 1}^{x^2 + 2x} e^{-t^2} dt$ Given $\displaystyle H(x) = \int_{x^3 + 1}^{x^2 + 2x} e^{-t^2} dt$, we want to find $H'(x)$.
First, we rewrite $H(x)$ as follows:
$$\begin{align}
&= \int_0^{x^2 + 2x} e^{-t^2} dt + \int_{x^3 + 1}^0 e^{-t^2} dt \qquad &\text{Properties of integral... | Use Lebnitz rule:
$$\frac{d}{dx} \int_{L(x)}^{U(x)} f(t) dt= U'(x) f(U(x)) -L'(x) f(L(x))$$
So in your case you get $$H'(x)=(2x+2) e^{-(x^2+2x)^2}-3x^2 e^{-(x^3+1)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Let $x, y, z$ be real numbers. If $x + y + z = 1$ and $x^2 + y^2 + z^2 = 1$, then what is the minimum value of $x^3 + y^3 + z^3$? Through manipulation, I got as far as $x^3 +y^3 + z^3 = 1 + 3xyz$
I tried solving it using AM - GM inequality but it only gave me the maximum value.
GM - HM does not to help either.
| The intersection of sphere and plane is a smooth curve, one can apply standard optimization methods.
With the Lagrange functional
$$
L(x,y,z,\lambda,\mu)=x^3+x^3+z^3+λ(x^2+y^2+z^2-1)+μ(x+y+z-1)
$$
one gets the equilibrium conditions
\begin{align}
3x^2+2λx+μ&=0\\
3y^2+2λy+μ&=0\\
3z^2+2λz+μ&=0
\end{align}
As now $x,y,z$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Radius of convergence of $\sum_{n=0}^\infty a_n x^n$, with $a_{n+2} = \frac{n(n+1) a_{n+1} - a_n}{(n+2)(n+1)}, a_2 = -a_0/2$ Problem
Find the radius of convergence of the power series
$$
\sum_{n=0}^\infty a_n x^n
$$
where $a_n$'s are defined by the following recurrence relation
$$
\begin{aligned}
a_{n+2} &= \frac{n(... | A possible way to approach the problem is to consider
$$F(x):=\sum_{n\ge0}a_nx^n$$
as a purely formal series (so without thinking about convergence radii etc) and try to understand what function does it actually represent. Playing around with the recursive formula for the coefficients, you can derive (up to mistakes on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if so... | Hint:
You can rewrite this fraction as
$$\frac{\sin(x^2+2)-\sin(x+2)}{x}=\frac{\sin(x^2+2)-\sin(2)}{x}-\frac{\sin(x+2)-\sin(2)}{x},$$
which is the difference of two rates of variation. Can take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Find modulus and argument of $\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i\cos^2(\theta)}}$ A guideline on which identity to use would be greatly appreciated, as $1-\cos2A=2\sin^2A$ identity isn't giving me the correct answer I think.
Given that:
$$\zeta = {\frac {1-\cos4(\theta) + i \sin4(\th... | Method 1: Using the $1-\cos2A=2\sin^2A$ identity
For $|\zeta| = 2\sin(\theta)$:
$$\zeta = {\frac {|1-\cos4(\theta) + i \sin4(\theta)|} {|\sin2(\theta)+ 2i \cos^2(\theta)|}} $$
$$\zeta = {\frac {\sqrt{(1-\cos4\theta)^2 + \sin^24(\theta)}} {\sqrt{\sin^22(\theta)+ 4\cos^4(\theta})}} $$
$$\zeta = {\frac {\sqrt{(1-2\cos4\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) ... | The number $n$ first appears in the sequence at position $n$ until position $2n-1$. So, the number $1024$ appears in the sequence at position $1024$ until $2047$. Therefore the number will be $1024$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 0
} |
Solve inequality Given positive numbers $a,b,c$ satisfying $a^2+b^2+c^2=1$, prove the following inequality
$$\frac{a}{\sqrt{1-bc}} + \frac{b}{\sqrt{1-ac}}+\frac{c}{\sqrt{1-ab}}\le\frac{3}{\sqrt{2}}$$
Thanks
I have tried using CS, try to make use of $a+b+c\leq\sqrt3$, $abc\leq\frac{1}{3\sqrt3}$, but got nowhere –
| hint:
$\dfrac{a}{\sqrt{2-2bc}} \le \dfrac{a}{\sqrt{1+a^2}}$
if you can prove $f(x)=\sqrt{\dfrac{x}{1+x}}$ is concave function, then the problem is solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Given $(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$, find$(a,b,c,d)$
Given $(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$,
find$(a,b,c,d)$
my attempt:
$$(x+1)=(x-1)\frac{(x+1)}{(x-1)}$$
but this seems useless?
I want to use synthetic division but I don't know how
| In case you don't know how to solve this most elegantly, there is still the straightforward possibility by expanding both sides. The LHS is given by
$$
x^3-5x,
$$
whereas the RSH is
$$
ax^3 + (3a + b)x^2 + (3a + 2b + c)x + a + b + c + d.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\omega = \frac{1 + \sqrt3 i}{2}$ , $ \omega^5 = ? $
$\omega = \frac{1 + \sqrt3 i}{2} $, $ \omega^5 = ? $
$\omega^3 = 1$ by definition?
So, $\omega^5 = \omega^2$
But why do i get wrong answer?
| Because $\omega=\cos\left(\frac\pi3\right)+i\sin\left(\frac\pi3\right)$ and therefore it is not true that $\omega^3=1$. Actually, $\omega^3=-1$. And$$\omega^5=-\omega^2=-\cos\left(\frac{2\pi}3\right)-i\sin\left(\frac{2\pi}3\right)=\frac{1-i\sqrt3}2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to show the matrix has Rank $\le 5$
I want to show that the following matrix has Rank $\le 5$.
The matrix is
\begin{bmatrix}
2&1&1&1&0&1&1&1\\
1&2&1&1&1&0&1&1\\
1&1&2&1&1&1&0&1\\
1&1&1&2&1&1&1&0\\
0&1&1&1&2&1&1&1\\
1&0&1&1&1&2&1&1\\
1&1&0&1&1&1&2&1\\
1&1&1&0&1&1&1&2
\end{bmatrix}
I found that there is a submatri... | By inspection, the vectors $(1,-1,1,-1,1,-1,1,-1)$, $(1,0,-1,0,1,0,-1,0)$, and $(0,1,0,-1,0,1,0,-1)$ are all in the nullspace, and are linearly independent, so the rank is at most $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find $n$ if $\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$ Find $n$ if $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$$
In this video they show a shortcut and say $n=-1/2$ without any explanation.
Key observation here is that the geometric mean of $9$ and $4$ is $6$.
It seems numerator and denominator are partial sums of geometric ser... | $$\dfrac{9(9^n) + 4(4^n)}{9^n + 4^n} = 6 $$ divide numerator and denominator by $9^n$:
$$ \dfrac{9 + 4\left(\frac{4}{9}\right)^n}{1 + \left(\frac{4}{9}\right)^n} = 6$$
Let $\left(\frac{4}{9}\right)^n = a$:
$$ \dfrac{9+ 4a}{1 + a} = 6 $$
$$a = \frac{3}{2} $$
$$ \left(\frac{4}{9}\right)^n =\left(\frac{2}{3}\right)^{2n} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 6
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Maximizing area of rectangle inscribed in circle sector of radius 2 Question:
A rectangle is inscribed in a circle sector. The top two corners of the rectangle lies on the radius of the circle sector and the bottom two corners lie on the arc of the circle sector.
The radius is 2 and angle is $\frac{2\pi}{3}$.
Find the... | $A'(\alpha) = 8(-\sin^{2}{\alpha} + \cos^{2}{\alpha} - \frac{2}{\sqrt{3}} \sin{\alpha}\cos{\alpha})$. Setting this to zero gives the family of solutions $\alpha = \frac{1}{6}(3 \pi n + \pi)$, $n \in \mathbb{Z}$. Taking $n = 0$ gives $A(\pi/6) = \frac{4}{\sqrt{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Minimize $\left\lVert \mathbf{x} \right\rVert_1 - \left\lVert \mathbf{x} - \overline x \,\mathbf{1}_n\right\rVert_1$ over the unit $n$-cube I want to solve the following minimization problem
$$\min_{\mathbf{x} \in [0, 1]^n} \left\lVert \mathbf{x} \right\rVert_1 - \left\lVert \mathbf{x} - \overline x \,\mathbf{1}_n\rig... | We need to minimize the following expression
$$
S = \sum_{i=1}^n (x_i - |x_i - \overline{x}|)
$$
where $x_i \in [0,1]$.
We can rewrite it as
$$
S = \sum_{i\in I_{>}} (x_i - (x_i - \overline{x})) + \sum_{i \in I_{<}} (x_i - (\overline{x} - x_i)).
$$
Here
$$
I_{>} = \{i\in \{1,2,\ldots n\}: x_i \ge \overline{x} \},
$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$a+b+c=-3\sqrt{ac}$
If $a,b,c$ be in Geometic Progression, and $b-c,c-a,a-b$ in Harmonic Progression then prove that $a+b+c=-3\sqrt{ac}$.
My attempt: $$c-a=\frac{2(b-c)(a-b)}{b-c+a-b}$$$$(c-a)^2+2(b-c)(a-b)=0$$$$c^2+a^2-2ac+2(ab-b^2-ac+bc)=0$$$$a^2+c^2-2b^2-4ac+2ab+2bc=0$$$$a^2+c^2-6ac+2b(a+c)=0(b^2=ac)$$$$a^2+c^2+2a... | Let $$a,b,c \equiv k,kr,kr^2 \text{ be in GP } $$
Then $$c-a = \frac{2(b-c)(a-b)}{b-c+a-b} \implies- (kr^2-k)^2 = 2(kr-kr^2)(k-kr) $$
$$\implies-(r^2-1)^2 = 2(r)(1-r)(1-r) \implies-(r-1)^2(r+1)^2 = 2r(r-1)^2$$
$\color{red}{r\neq1} $ as we have a GP
$$ -(r+1)^2 = 2r \text{ or } r^2+1 =- 4r$$
Now $$a+b+c = k(1+r+r^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trying to find Pythagorean triples for a given ratio of $\frac{C}{A}$ I want to solve a ratio equation for $n$ in terms of $m$ and $R$ hoping, given a Ratio and a range of $m$ values, that finding an integer $n$ will give me the $m,n$ needed to generate a Pythagrean triple where $\frac{C}{A}$ has that ratio.
Given $A=m... | In
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{\pm\sqrt{4*(R+1)*m^2(R-1)}}{2(R+1)}=\frac{2m\sqrt{R^2-1}}{2(R^2-1)}=\frac{m\sqrt{R^2-1}}{R^2-1}$$
Going from the second to the third part, you introduced an extra factor of $R-1$ into the denominator, i.e., it went from $2(R+1)$ to $2(R^2-1)$. Thus, with handling for both both ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Evaluating $\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx$ My first instinct was to evaluate the indefinite form of the integral, which I did by substituting $x=\tan t$, therefore yielding
\begin{align}
\int \frac{x\ln x}{(1+x^2)^2} \,dx
&= \int \frac{\tan t \sec^2 t \ln\tan t}{(1+\tan^2 t)^2} \,dt
&& \text{by substi... | Different approach:
\begin{align}
I&=\int_0^\infty\frac{x\ln x}{(1+x^2)^2}\ dx\overset{x^2=y}{=}\frac14\int_0^\infty\frac{\ln y}{(1+y)^2}\ dy\overset{1+y=z}{=}\frac14\int_1^\infty\frac{\ln(z-1)}{z^2}\ dz\\
&\overset{\large z\ \mapsto\ \frac{1}{z}}{=}\frac14\int_0^1\ln\left(\frac{1-z}{z}\right)\ dz=\frac14\underbrace{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Calculate $\int \frac{dx}{1-\sin^4x}$
Calculate $$\int \frac{dx}{1-\sin^4x}$$
My try:
\begin{align}
\int \frac{dx}{1-\sin^4x}&=\int \frac{(1-\sin^2x)+(1+\sin^2x)dx}{1-\sin^4x}
\\&=\int \frac{dx}{1-\sin^2x}+\int \frac{dx}{1+\sin^2x}
\\&=\tan x+\int \frac{dx}{1+\sin^2x}
\end{align}
How to deal with the second one?
| $$\dfrac{dx}{a\sin x\cos x+b\sin^2x+c\cos^2x}$$
Divide numerator and denominator
by $\cos^2x$ and set $\tan x=u$
Or by $\sin^2x$ and set $\cot x=v$
Here $(1)+\sin^2x=(\cos^2x+\sin^2x)+\sin^2x=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3313468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Determine the maximum possible volume, excluding the volume of the legs. A table is to be constructed by gluing together 68 cubes of dimension $1\times 1\times 1$. All four legs and the rectangular top will be formed by the cubes. The four legs must be the same length and must be one cube thick, and the top is one cube... | Noting that subtracting $68-4=64$, we can see that, if the legs are only $1$ cube high, that leaves $8\times 8\times 1=64$ for the table top. The space under the table (minus the legs) is $64-(4\times 1\times 1)=60$ so the volume from the table top to the floor is $2*64-4=124$.
If we let the legs be $4\times 2=8$, the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the number of polynomials $P(x)$ with coefficients $0,1,2,3$ such that $P(2) = n$.
Find the number of polynomials $P(x)$ with coefficients $0,1,2,3$ such that $P(2) = n$.
Only what I currently know is that I should consider polynomials with the smallest degree $k$, where $k$ is first integer such that
$$ 3 \cdot... | Let your polynomial be $P(x) = \sum_{k=0}^n c_i x^i$.
Let $A_k$ be the sets of nonnegative integers $i$ such that $c_i = k$, $k=0,1,2,3$. Thus $$
\eqalign{n &= \sum_{i \in A_1} 2^i + 2 \sum_{i \in A_2} 2^i + 3 \sum_{i \in A_3} 2^i\cr
&= \sum_{i \in A_1 \cup A_3} 2^i + 2\sum_{i \in A_2 \cup A_3} 2^i}$$
Given any nonneg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\epsilon - N$ proof of $\sqrt{4n^2+n} - 2n \rightarrow \frac{1}{4}$ I have the following proof for $\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$ and was wondering if it was correct. Note that $\sqrt{4n^2+n} - 2n = \frac{n}{\sqrt{4n^2+n} + 2n}$.
$$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| ... | You did the scratch work correctly. But I wouldn't call this a proof (of course it contains all ingredients of a good proof!).
You didn't introduce $\epsilon$. Of course, everyone knows what you mean but one should write it out if one wants to be fully rigorous.
So, your proof should read:
Let $\epsilon >0$. Let $N$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Convert a double integral into polar coordinates If $a$ is a positive number, compute
$$
\int_0^{2a}\int_{-\sqrt{2ay-y^2}}^0 \sqrt{x^2+y^2} \, dx \, dy.
$$
I wanted to use polar coordinates to compute this double integral. However, the lower $x$-limit
$$
x=-\sqrt{2ay-y^2}
$$
is equivalent to saying that $x$ satisfies
$... | Let $x=r\cos\theta,y=r\sin\theta$. Since $x\leq 0$, $\frac{\pi}2\leq \theta\leq\pi$. Since $x^2+y^2\leq 2ay$, $r^2\leq 2ar\sin\theta$ thus $r\leq 2a\sin\theta$. Hence
\begin{align*}
\int_0^{2a}\int_{-\sqrt{2ay-y^2}}^0 \sqrt{x^2+y^2} \, dx \, dy&=\int\limits_{x^2+y^2\leq 2ay\\x\leq 0}\sqrt{x^2+y^2} \, dx \, dy\\&=\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Value of a and P(x) when P(x) is a rational number when satisfies a certain equation This is the question as I still don't have permission to post picture, but it is basically when P(x) = x^3+x^2+ax+1, when a is a rational number, P(X) is also rational number for every x that satisfy x^2+2x-2=0
Consider the integral e... | Hint: By a direct computation, $x^2+2x-2=0$ has two solutions, namely
$$
x=\pm \sqrt{3}-1.
$$
For $x=\sqrt{3}-1$ we have
$$
P=x^3+x^2+ax+1=(a+4)\sqrt{3}-a-5.
$$
Hence for $a=-4$, this is rational. For $x=-\sqrt{3}-1$ we have
$$
P=-(a+4)\sqrt{3}-a-5.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the area of a region bounded by a simple closed curve? I have the following equation:
$$
\frac{p}{(a-x)^2+y^2}+\frac{1-p}{(b-x)^2+y^2}=1 \text{ where } 0\leq p\leq 1
$$
Which represent a simple close curve. Obviously, when $p=0,p=1$ or $a=b$ we recover a unit circle. However the shape of the curve is more i... | By clearing denominators, we see that our curve is a quartic $p(x, y) = 0$. For generic values of $p, a, b$, it is elliptic and so does not admit a rational parameterization. Probably the areas can be computed in terms of elliptic functions.
In the special case $p = \frac{1}{2}$, the curve is symmetric not only about t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Express a vector as a linear combination of two other vectors. Let $A = \begin{bmatrix}
1 & -1 & 1 \\
1 & 0 & 2 \\
-1 & 2 & 0
\end{bmatrix}$ and $B = \begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}$. $A$ row reduces to $B$ but you are asked not to verify this.
1. Find a basis for the column space $\te... | Just subtract $\begin{bmatrix} 0 \\ -1 \\ -1 \end{bmatrix}$ from $\begin{bmatrix} 2 \\ -2 \\ 0 \end{bmatrix}$. That vector must be in $\text{col}(A)^{\perp}$, by properties of projections. Indeed, you can check its perpendicularity to your basis for col$(A)$ by computing the dot product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help differentating $f(x) = \sqrt\frac{x^2-1}{x^2+1}$ The equation I'm trying to differentiate is, $ f(x) = \sqrt\frac{x^2-1}{x^2+1}$ and I know the answer is meant to be
$$=\frac{\frac{x\sqrt {x^2+1}}{\sqrt {x^2-1}}-\frac{x\sqrt {x^2-1}}{\sqrt {x^2+1}}}{x^2+1}$$
But when I do the working out I get this
$$=\frac{(x^2-... | Just a small trick.
When you face expressions which contains products, quotients, powers, logarithmic differentiation makes life easier.
$$y= \sqrt\frac{x^2-1}{x^2+1}\implies \log(y)=\frac 12 \log(x^2-1)-\frac 12 \log(x^2+1)$$ Differentiate both sides
$$\frac {y'} y=\frac 12 \frac {2x}{x^2-1}-\frac 12 \frac {2x}{x^2+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$ Problem:
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$$
$$\lim_{x \to -\infty} \sqrt{\frac{1}{x^6}}=0$$ so...
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}=\frac{1}{0}$$
The answer is $-1$ and I kn... | Yes, $\lim_{x\to-\infty}\sqrt{\frac{1}{x^6}}=0$. But $\lim_{x\to-\infty}\sqrt{x^6+4}=\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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If $a+b+c = 4, a^2+b^2+c^2=7, a^3+b^3+c^3=28$ find $a^4+b^4+c^4$ and $a^5+b^5+c^5$ I have tried to solve it but cannot find any approach which would lead me to the answer
| Use $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
and $$a^3+b^3+c^3-3bc=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}$$ to find $ab+bc+ca=u, abc=v$(say)
So, $a,b,c$ are the roots of $$t^3-4t^2+ut-v=0$$
Use Newton's Sums
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Computing a circle from a given set with a tangent line condition I was asked to compute the circles' equations from the set
$$x^2 + y^2 -3x + (k-6)y + (9-3k)=0$$
that fulfill the following request: the circles must be tangent to the line
$$x+y-3=0$$
I started to compute
$$
\begin{cases}
x^2 + y^2 -3x + (k-6)y + (9-3k)... | You have occoured in an error when you say: "obtaining: " $$\begin{cases}
2y^2 +(k-3)y +(9-3k)=0\\
x=3-y
\end{cases}$$
You should have obtained: $$\begin{cases}
2y^2 +(k-9)y +(9-3k)=0\\
x=3-y
\end{cases}$$
From here: $(k-9)^2-8(9-3k)=0$ and $k^2+6k+9=0$, so $k=-3$. The equation of the circle is: $x^2+y^2-3x-9y+18=0$.
H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How can we not use Muirhead's Inequality for proving the following inequality? There was a question in the problem set in my math team training homework:
Show that $∀a, b, c ∈ \mathbb{R}_{≥0}$ s.t. $a + b + c = 1, 7(ab + bc + ca) ≤ 2 + 9abc.$
I used Muirhead's inequality to do the question (you can try out yourself):... | A proof by SOS:
$$2+9abc-7(ab+ac+bc)=2(a+b+c)^3+9abc-7(a+b+c)(ab+ac+bc)=$$
$$=\sum_{cyc}(2a^3+6a^2b+6a^2c+4abc+3abc-7a^2b-7a^2c-7abc)=$$
$$=\sum_{cyc}(a^3-a^2b-ab^2+b^3)=\sum_{cyc}(a-b)^2(a+b)\geq0.$$
Also, $uvw$ kills it immediately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
tangent inequality in triangle Let $a$, $b$ and $c$ be the measures of angles of a triangle (in radians).
It is asked to prove that
$$\tan^2\left(\dfrac{\pi-a}{4}\right)+\tan^2\left(\dfrac{\pi-b}{4}\right)+\tan^2\left(\dfrac{\pi-c}{4}\right) \ge 1$$
When does equality occur ?
My try :
Letting $u:= \tan\left(\dfrac{\pi... | $$\left(\tan^2\frac{\pi-x}{4}\right)''=\frac{2-\sin\frac{x}{2}}{8\cos^4\frac{\pi-x}{4}}>0.$$
Thus, by Jensen
$$\sum_{cyc}\tan^2\frac{\pi-\alpha}{4}\geq3\tan^2\frac{\pi-\frac{\alpha+\beta+\gamma}{3}}{4}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Inverse of rational function $y= \frac{3-x}{1+x^2}$ I have the function $$y= \frac{3-x}{1+x^2}$$ and I want to find the inverse of this function.
I know that
$$x= \frac{1 \pm \sqrt{1-4y(3-y)}}{2y}$$
My question is how do I find the domain where the function is
$$x= \frac{1- \sqrt{1-4y(3-y)}}{2y}$$
and
$$x= \frac{1... | Domain of $x$ is the range of $y$, and vice versa.
You have a square root for $x=f(y)$, so you should solve the condition for the expression within that square root is bigger than or equal $0$. Or:
$1-12y+4y^2\ge 0$
Solving $1-12y+4y^2=0$ yields $(1/2)(3+2\sqrt{2})$ and $(1/2)(3-2\sqrt{2})$. These two are real values, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Asymptotes of hyperbola by considering x tends to infinity I saw this derivation of equations of the asymptotes of hyperbola and it goes like this...
For a standard hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, rearranging the terms we get
$y=\pm\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}$
So as $x\rightarrow\infty$, $\frac{... | I think you are right to be suspicious of this method.
Let's apply it to the hypberbola given by the equation
$$ \frac{x^2}{2^2} + x -\frac{y^2}{2^2}=1 .$$
Solving for $y$ as in the method in the question, we get
$$ y = \pm x \sqrt{1 + \frac 4x - \frac 4{x^2}},$$
and as $x\to\infty,$ we find that $\sqrt{1 + \frac 4x - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Probability Question 3.17 in Wackerly, Mendenhall, and Scheaffer, 5th Ed.
Problem 3.17. Two construction contracts are to be randomly assigned to one or more of three firms:
I, II, and III. Any firm may receive both contracts. If each contract will yield a profit of
$\$90,000$ for the firm, find the expected profi... | There's a problem with your reasoning: each combination is not just as likely to occur. If each contract has a probability of $\frac{1}{3}$ to go to one of the firms, than the probabilities are as follows:
$$P(0) = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}$$
$$P(1) = \frac{2}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
All possible antiderivatives for Integration by Parts Just had a quick inquiry with regards to the formula for Integration by Parts. If I'm not mistaken, the formula states that
$$\int f'(x)g(x) = f(x)g(x)- \int f(x)g'(x)$$
However, in the case that I try to substitute an antiderivative with a valid constant, the formu... | Say you pick $f(x) = \frac{7}{2}x^2 + 5$, $g(x) = x^2$, and are doing
$$\int f'(x)g(x)\,dx = \int (7x)x^2\,dx = \int 7x^3\,dx = \frac{7}{4}x^4 + C.$$
Now, if you try using integration by parts with the anti-derivative you pick, you have
$$\begin{align*}
\int f'(x)g(x)\,dx &= f(x)g(x) - \int g'(x)f(x)\,dx\\
&= \left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\int_{1-\sqrt{1-t^2}}^{1+\sqrt{1-t^2}}(y^2-2y+t^2)dy$, where t is constant I need to evaluate the following one. Can't understand the method in my textbook.
$$\int_{1-\sqrt{1-t^2}}^{1+\sqrt{1-t^2}}(y^2-2y+t^2)dy$$
My textbook is to let $\alpha=1-\sqrt{1-t^2}$, $\beta=1+\sqrt{1-t^2}$, so it becomes
$$\... | The author set up the limits so that
\begin{align*}
\alpha + \beta & = 1 - \sqrt{1 - t^2} + 1 + \sqrt{1 - t^2}\\
& = 2\\
\alpha\beta & = (1 - \sqrt{1 - t^2})(1 + \sqrt{1 - t^2})\\
& = 1 - (1 - t^2)\\
& = t^2
\end{align*}
Therefore,
\begin{align*}
(y - \alpha)(y - \beta) & = y^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to compare $\pi, e\cdot 2^{1/3}, \frac{1+\sqrt{2}}{\sqrt{3}-1}$ This is in the GRE exam where we are supposed to answer fast so I think there might be some trick behind this to allow us to do that. But so far the best I can do is to write $\frac{1+\sqrt{2}}{\sqrt{3}-1}=\frac{1+\sqrt{6}+\sqrt{2}+\sqrt{3}}{2}$ and co... | Here's a dirty decimal arithmetic method that presumes knowledge only of the bounds $1.41 < \sqrt{2} < 1.42$, $1.73 < \sqrt{3} < 1.74$---which you probably know if you're taking the GRE subject test---and the not-too-obscure fact $e^3 > 20$: Since $\sqrt 6 = \sqrt 2 \sqrt 3$ multiplying gives $2.43 < \sqrt{6} < 2.47$ T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Question about the particular part in a non homogeneous recurrence Question about the particular part in the following non homogeneous recurrence :
$$a_n - 6a_{n-1} + 9a_{n-2} = n * 3^n $$
I have the following particual part : $$ a_n = n * 3^n$$
Now the solution of the homogenous part is $$x_1 = 3, x_2 = 3$$ and is of... | OK, now with the full recurrence. For the homogeneous part $a_{n + 2} - 6 a_{n + 1} + 9 a_n = 0$ we have a characteristic equation $r^2 - 6 r + 9 = 0$, which is just $(r - 3)^2 = 0$. You have two equal roots, so the solution to the homogeneous part is $a_n^h = (c_1 n + c_2) \cdot 3^n$.
The forcing function is $n \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the method to factor $x^3 + 1$? In the solution to a problem, it's stated that
We see that $x^3+1=(x+1)(x^2-x+1)$.
Why is this, and what method can I use for similar problems with different coefficients?
The full problem is
Find the remainder when $x^{81}+x^{48}+2x^{27}+x^6+3$ is divided by $x^3+1$.
| $x^3+1=(x+1)(x^2-x+1)$ is something one just knows, similar to how one knows that
$
x^2+2x+1=(x+1)^2
$, or $
x^2-1=(x+1)(x-1)
$
If you don't already know it, note that $(-1)^3+1=0$, which implies that $x-(-1)$ is a factor of $x^3+1$. Finding the second factor can be done through long division.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Finding generating series from equation How can I solve $f(x)=(1+x+x^2)f(x^2)$, where $f$ is a generating series of a set? I just need the technique, not necessarily the solution.
| $f(x) = a_0 + a_1 x + a_2 x^2 + \cdots$
$f(x^2) = a_0 + a_1 x^2 + a_2 x^4 + \cdots$
so $(1 + x + x^2) f(x^2) = a_0 + a_1 x + a_2 x^2 + \cdots + a_0 x + a_1 x^2 + a_2 x^3 + \cdots + a_0 x^2 + a_1 x^3 + a_2 x^4 + \cdots$
Given $f(x) = (1 + x + x^2) f(x^2)$ we equate the following powers of $x$:
$x^0$ therefore: $a_0 = a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Note: $x$ and $y$ are obtuse angles.
My attempt that is not simple is as follows.
Expand both known constraints, so we have
\be... | Try to solve the first equation for $x$ and plug this in the second equation .
I got this for $y$:
$$\cos ^{-1}\left(\frac{1}{3}\right)=2
y+\cos
^{-1}\left(\frac{1}{5}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Calculate $\sum_{k=1}^n k2^{n-k}$ I want to Calculate $\sum_{k=1}^n k2^{n-k}$. Here's my attempt:
$$\begin{align}
&\sum_{k=1}^n k2^{n-k} = \sum_{k=1}^n k2^{n+1}2^{-k-1} =2^{n+1}\sum_{k=1}^n k2^{-k-1} & &\text{(1)} \\
&k2^{-k-1} = -\frac{d}{dk}2^{-k} & &\text{(2)}
\end{align}$$
Plugging $(2)$ to $(1)$,
$$\begin{align... | Let $S=\sum_{k=1}^n k 2^{n-k}$
$ S = 2^{n-1} + 2(2^{n-2}) + 3(2^{n-3}) + \cdots + n(2^0)$
$ {S \over 2} = \quad\quad\quad1(2^{n-2}) + 2(2^{n-3}) + \cdots + (n-1)(2^0) + n(2^{-1})$
$\begin{align} (1-{1\over2})S
&= (2^{n-1} + 2^{n-2} + 2^{n-3} + \cdots + 1) - n(2^{-1})\cr
{S\over2}&= (2^n-1) - n(2^{-1})\cr\cr
S &= 2(2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Integration simple question I want to integrate $$\int\left(\frac{y}{2}\right)^\frac{2}{3}dy$$
Doing this I get $$\cfrac{3}{5}\left(\cfrac{y}{2}\right)^\cfrac{5}{3}$$
But the answer is $$\cfrac{6}{5}\left(\cfrac{y}{2}\right)^\cfrac{5}{3}$$
What step am I missing
| $$\int (\frac{y}{2})^\frac{2}{3} dy = \frac{1}{2^\frac{2}{3}} \int y^\frac{2}{3} dy = \frac{1}{2^\frac{2}{3}} \frac{3}{5} (y)^\frac{5}{3} = \frac{2 \cdot 3}{2^\frac{5}{3} \cdot 5} (y)^\frac{5}{3} = \frac{6}{5}(\frac{y}{2})^\frac{5}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Eigen Decomposition Check I am following the wiki entry on eigen dicomposition with the following matrix:
$$A = \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix}$$
I wish to find a diagonalizing matrix T.S.
$$T^{-1}AT=\Lambda$$
$$AT=T\Lambda$$
where,
$$\Lambda = \begin{pmatrix}
x & 0 \\
0 & y \\
\end{pmatri... | As a good check you can compute $TT^{-1}$, and you'll find you don't get the identity matrix. So the mistake must be in your computation of $T^{-1}$.
It looks to me like you've set it up wrong. Seems you're trying to use the formula $$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ always square Let $n\in \mathbb{N}$, if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ is a perfect square.
For example $n=1$, it is clear.
and $n=13$ because $$ 4\cdot 13^2-1=3\cdot 15^2$$ and $$2n-1=25=5^2$$
| An Experimental approach:
There are infinitely many integers like n such that:
$2n-1=k^2$; $k∈N$
They make following series:
$n=13, 41, 61, 181, 221 . . .$
therefore we may write:
$4\times 13^2-1=3\times 15^2$
$4\times 41^2-1=3\times 47.34^2$
$4\times 61^2-1=3\times 70.4^2$
$4\times 181^2-1=3\times 209^2$
$4\times 221... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are there integer solutions such that the sum of the roots of x and y equal the root of a prime p? How could one prove whether or not integer pairs (x, y) exist such that $\sqrt{x} + \sqrt{y} = \sqrt{p}$, where p is prime? `
| A slightly less elementary approach. Consider the polynomial whose roots are the pairwise sums of the roots of the minimal polynomials of $\sqrt x$ and $\sqrt y$: $$(z - \sqrt x - \sqrt y)(z - \sqrt x + \sqrt y)(z + \sqrt x - \sqrt y)(z + \sqrt x + \sqrt y) \\
= ((z - \sqrt x)^2 - y) ((z + \sqrt x)^2 - y) \\
= (z^2 + x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Formulas for Sequences Removing Multiples of 2, 3, and 5 First off, I am a programmer so please excuse if some of the terms I use are not the correct mathematical terms. I was working on devising a function to improve one of my prime number generation algorithms. With this in mind, I first set out to find the formulas ... | The first one is:
$$y=5+6\bigg\lfloor \frac{x}{2} \bigg\rfloor + 2(x\mod2)$$
Although arguably $x(0)=1$, so:
$$y=1+6\bigg\lfloor \frac{x+1}{2} \bigg\rfloor - 2(x\mod2)$$
The second, with $x(0)=1$, is:
$$y=15+30\bigg\lfloor \frac{x}{8} \bigg\rfloor +4(x\mod8-3.5)-\frac{(x\mod8-3.5)}{|x\mod8-3.5|}\cdot2(\bigg\lfloor\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now ... | $$
\frac{n^2(n^2-1)(n^2+1)}{60} = 2n\binom{n+2}5 + 2\binom{n+1}4 + \binom{n+1}3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Solving Diophantine Equations Using Inequalities Prove that if $x$ and $y$ are positive integers such that $xy$ divides $x^2+y^2+1$ then prove that the quotient is $3$.
Now as $x^2+y^2 +1 > 2xy$ we have $$\frac{x^2+y^2+1}{xy} \geq 3$$ So now we have to prove that $$\frac{x^2+y^2+1}{xy} \leq 3$$
Now if we prove that for... | Suppose $x y \mid (x^2 + y^2 + 1)$ and without loss of generality assume that $x < y$.
Then $(x, y)$ is a pair of positive integers with $x < y$ such that $x \mid (y^2 + 1)$ and $y \mid (x^2 + 1)$. Notice that this implies that $\gcd (x, y) = 1$.
Therefore there exist positive integers $w, z$ such that $x^2 + 1 = w y$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all natural numbers $n$ such that $n+1$ divides $3n+11$ Following the example of my teacher:
Find all natural numbers $n$ such that $n-2$ divides $n+5$.
$$n+5 = n-2+7$$
As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
*
*$n-... | You can use polynomial long division, which is useful if you have a more complex expression:
$$
\require{enclose}
\begin{array}{rll}
3 && \\[-3pt]
n+1 \enclose{longdiv}{\ 3n+11}\kern-.2ex \\[-3pt]
\ \ - \underline{\ (3n+3)} && \\[-3pt]
8 && \\[-3pt]
\end{array}
$$
Therefore $\frac{3n+11}{n+1} = 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How is this Taylor expansion computed? I am reading a paper, where we consider the following function,
$$F(r) = \frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)$$
where $b>0$ is a constant such that $r>2/b.$ After stating this definition the author writes that the taylor expression of $F$ is,
$$F(r) = ... | $$F(r) = \frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)\tag{1}$$
Expand the square-root term with $1/(b^2r^2)$ as the small variable,
$$\sqrt{b^2r^2-4}=br\sqrt{1-\frac{4}{b^2r^2}}=br \left( 1-\frac{2}{b^2r^2} \right) \tag{2}=br - \frac{2}{br} $$
Similarly, expand the log term as
$$\ln(\sqrt{b^2r^2-4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Real Analysis Inf and Sup question I am hung up on this question for real analysis ( intro to anaylsis ).
Find $\inf D$ and $\sup D$
$$\mathrm{D}=\left\{\frac{m+n\sqrt{2}}{m+n\sqrt{3}} :m,n\in\Bbb{N}\right\}$$
I have spent enough time staring at this thing that I know the $\sup D=1$ and $\inf D=\frac{\sqrt{2}}{\sqrt{... | In general if $\alpha$ is a lower bound of $X$ and $Y$ is a nonempty subset of $X$ with $\alpha = \text{inf}(Y)$, then $\alpha = \text{inf}(X)$.
By fleablood's opening argument, we know that $\alpha = \frac{\sqrt 2}{\sqrt 3}$ is a lower bound for $D$.
Let $\mathrm{E}=\left\{\frac{1+n\sqrt{2}}{1+n\sqrt{3}} :n\in\Bbb{N}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Factorizing $x^5+1$ as a product of linear and quadratic polynomials. I am encountering some trouble with this question:
Factorize $$x^5+1$$ as a product of real linear and quadratic polynomials.
I know that if we subtract 1 from $x^5+1$, we get that $x^5 = -1$, but I am unsure where to go from here.
Can anyone help wi... | If you remember the formula for the sum of a geometric progression, you will have
$$\begin{align} & 1 + x + x^2 + \cdots + x^{n-1} = \frac{1-x^n}{1-x}\\
\iff & x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1)
\end{align}
$$
When $n = 2k+1 $ is odd, substitute $x$ by $-x$ and move the minus sign around, you will get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Given three positive numbers $a,b,c$. Prove that $\sum\limits_{sym}\frac{a+b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$ .
(A problem due to Mr. Le Khanh Sy). Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{sym}\frac{a+ b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{s... | By the Vornicu Schur, we have
$$\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} \right)\left( {\frac{a}{c} + \frac{c}{b} + \frac{b}{a}} \right) - \left( {a + b + c} \right)\left( {\frac{a}{{bc}} + \frac{b}{{ca}} + \frac{c}{{ab}}} \right) = \sum \frac{(a-b)(a-c)}{a^2} \geqslant 0.$$
Theforere we will show that
$$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
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Fractions in Questions and Answers
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