Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What is the probablity that the sum of two dice is 4 or 6?
What is the probablity that the sum of two dice is 4 or 6?
The explanation I found is as follows:
Total number of outcomes $6 \times6 = 36$
Number of outcomes where the event occurs: $1+3, 2+2, 3+1, 1+5, 2+4, 3+3, 4+2$ and $5+1$ (Total $ 8$)
The probability t... | Here is a simple way to visualize the sample space:
\begin{array}{c|llllll}
+ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & 2 & 3 & \color{blue}{4} & 5 & \color{blue}{6} & 7\\
2 & 3 & \color{blue}{4} & 5 & \color{blue}{6} & 7 & 8\\
3 & \color{blue}{4} & 5 & \color{blue}{6} & 7 & 8 & 9\\
4 & 5 & \color{blue}{6} & 7 & 8 & 9 & 10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Proving $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ for positive $a$, $b$, $c$
For $a,b,c>0$ Prove that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$
My attempt: By AM-GM we obtain $$\frac{a}{b}+\frac{a}{b}+\fr... | Since $\prod\limits_{cyc}(a+b)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$, by AM-GM we obtain:
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3=\frac{\sum\limits_{cyc}(a^2b+a^2c+2abc)}{abc}=$$
$$=\frac{\sum\limits_{cyc}c(a+b)^2}{abc}\geq \frac{3\sqrt[3]{abc\prod\limits_{cyc}(a+b)^2}}{abc}\geq \frac{3\sqrt[3]{abc\cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
$f\colon\{1,2,3,4,5\}\longrightarrow\{1,2,3,4,5\}$ Find the total number of functions such that $f(f(x))=f(x)$ I tried to solve it by letting $f(x)=y$
then equation reduces to $f(y)=y$ but i didn't get it further.
Please give me hint to solve this from where I should start
| You have a very good starting point: for all $x$ in your set, $y=f(x)$ is a fixed point.
Conversely, suppose $f$ maps elements to fixed points, then it clearly satisfies $f(f(x))=f(x)$.
So the answers looks like: choose a subset of $\{1,2,3,4,5\}$ to be the fixed points of your function, and then map the remaining poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int\frac{dx}{(a+b\cos(x))^2},(a>b)$ Evaluate $$\int\frac{dx}{(a+b\cos(x))^2},(a>b)$$
I tried to write 1 in numerator as $p'(x)(a+b\cos(x))-p(x)(a+b \cos(x))'$,making something like quotient rule but did not get after that.
| $$I=\int \frac{dx}{(a+b\cos x)^2} \\=\int\frac{dx}{\left(a+b\cdot \frac{1-\tan^2 \frac x2}{1+\tan^2 \frac x2} \right)^2}\\=\int\frac{1+\tan^2 \frac x2}{\left(a(1+\tan^2 \frac x2) +b(1-\tan^2\frac x2)\right)^2} \sec^2\frac x2 \ dx \\ \overset{t=\tan \frac x2}=2\int\frac{1+t^2}{\left[(a-b)t^2 +a+b\right]^2} dt$$ Use part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine if the integral $ \intop_{0}^{\infty}\left(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\right)dx $ convergent So, I need to determine if the integral $ \intop_{0}^{\infty}\left(\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\right)dx $ convergent, for any $ C\in \mathbb{R} $.
I found that for C=1 the integral convergent. I ... | Note
\begin{eqnarray}
&&\frac{1}{\sqrt{x^{2}+4}}-\frac{C}{x+2}\\
&=&\frac{x+2-C\sqrt{x^{2}+4}}{(x+2)\sqrt{x^{2}+4}}\\
&=&\frac{(x+2)^2-C^2(x^{2}+4)}{(x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})}\\
&=&\frac{(1-C^2)x^2+4x+4(1-C^2)}{(x+2)\sqrt{x^{2}+4}(x+2+C\sqrt{x^{2}+4})}
\end{eqnarray}
If $1-C^2>0$ or $|C|<1$, then there i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=... | With $s=\sin \alpha$ and $c=\cos \alpha$ we have
$$\frac 1{s^3} -\frac 1{c^3}=\left(\frac 1 s - \frac 1c\right)\left(\frac 1{s^2} + \frac 1{sc} + \frac 1{c^2}\right)$$
$$ = \frac{c-s}{sc}\left(\frac 1{sc}+\frac 1{(sc)^2}\right)$$
Now, since $s-c=\frac 12$, you have $\frac 14 = 1-2sc \Leftrightarrow sc = \frac 38$. So, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{... | This is one case where you can solve the problem without any integration.
Because of the cube in denominator, assume that
$$\int\frac{u^3}{(u^2+1)^3}du=\frac{P_n(u)}{(u^2+1)^2}$$ Differentiate both sides and remove the common denominator to get
$$u^3=\left(u^2+1\right) P_n'(u)-4 u P_n(u)$$ Comparing the degrees $n=2$; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 4
} |
Find the $x$ intercepts of $x^3+x^2-4x-4$ I am to find the $x$ intercepts of $x^3+x^2-4x-4$.
The solution in my book shows these as: $(2,0)$, $(-2,0)$, $(-1,0)$ whereas I get just $(4, 0)$, $(-1, 0)$.
My working:
$x^3+x^2-4x-4$ =
$x^2(x+1)-4(x+1)$ =
$(x^2-4)(x+1)$
Thus, intercepts when $x$ is equal to $4$ and $-1$.
Whe... | Just as said in the comments, I will quickly put this into answer form.
The $x^2-4$ can be simplified to $(x+2)(x-2)$, so the roots are $2$, $-2$, and $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For $x≠y$ and $2005(x+y) = 1$; Show that $\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$ Problem:
Let $x$ and $y$ two real numbers such that $x≠0$ ; $y≠0$ ; $x≠y$ and $2005(x+y) = 1$
*
*Show that $$\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$$
*Calculate $l$:
$$l = \frac{y}{y-x} - \frac{y-... | For part 1, if $xy\not=0$, then
$$\begin{align}
1=2005(x+y)\implies{1\over xy}&=2005(x+y){1\over xy}\\
&=2005\left({x\over xy}+{y\over xy}\right)\\
&=2005\left({1\over y}+{1\over x}\right)\\
&=2005\left({1\over x}+{1\over y}\right)
\end{align}$$
For part 2, regroup the terms with denominators $y-x$, $y$, and $x$ and si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Extreme points of a function at domain ends Consider the following function: $f(x) = x\sqrt{9-x^2}$
$\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad $
$f'(x) = \frac{-2x^2+9}{\sqrt{9-x^2}}$ and $D(f) = [-3,3]$ therefore the critical points of the function are $x_{c_i} = \left\{ -3, -\frac{3\sqrt2}{2}... | The domain it's $[-3,-3].$
Let $0\leq x\leq 3$.
Thus, by AM-GM $$x\sqrt{9-x^2}=\sqrt{x^2(9-x^2)}\leq\frac{x^2+9-x^2}{2}=\frac{9}{2}.$$
The equality occurs for $$x^2=9-x^2$$ or
$$x=\frac{3}{\sqrt2},$$ which says that $\frac{9}{2}$ is a maximal value of $f$ on $[0,3]$.
The minimal value is $0$ and occurs for $x=0$ or $x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \t... | So we want to solve
$${\int \frac{x^3}{\left(4x^2 + 9\right)^{\frac{3}{2}}}dx}$$
The first thing I do when I see an integral like this is ask if there are any simple substitutions I can make. Well, notice if I make the substitution ${u=4x^2 + 9}$, then...
$${\Rightarrow \int \frac{x^3}{u^{\frac{3}{2}}}\times \frac{1}{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos ... | $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$
$=\lim_{x \to 0} \frac{(\sqrt{1 + x\sin x} - \sqrt{\cos x})(\sqrt{1 + x\sin x} + \sqrt{\cos x})}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
$=\lim_{x \to 0} \frac{1+x\sin x-\cos x}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
$=\lim_{x \to 0} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Why do some partial fractions have x or a variable in the numerator and others don't? Why do rational expressions like $\left(\frac{1}{(x-2)^3}\right)$ do not have x in the numerator of the partial fraction but a rational expression like $\left(\frac{1}{(x^2+2x+3)^2}\right)$ does have x in the numerator of its partial ... | You can integrate $\frac{1}{(x-2)^k}$ by itself, so it is not necessary to break it into the form $\frac{A}{x-2} + \frac{Bx+C}{(x-2)^2} + \dots.$ On the other hand, $\frac{1}{(x^2+2x+3)^k}$ does not have a simple antiderivative, so we must decompose it into the form $\frac{Ax+B}{x^2+2x+3} + \frac{Cx+D}{(x^2+2x+3)^2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the absolute extrema of $F(x) = 2x + 5\cos(x)$
Find the absolute extrema of $F(x) = 2x + 5\cos(x)$ on the interval $[0,2\pi]$ using the extreme value theorem.
Answer should be 2 ordered pairs.
I got $\arcsin(2/5)$ for the first value of $x$, but can’t figure out the second.
Thanks in advance.
| The question says absolute extreme.
$5\cos x \le 5$ and $2x \le 4\pi$ so $2x+ 5\cos x\le 5+4\pi$ so if $2x + 5\cos x$ ever equals $5+4\pi$, which it does at $x = 2\pi$, that will be an absolute (albeit not necessarily a local) maximum. So the absolute maximum is $(2\pi, 5+4\pi)$.
We can't do the same for minimum as $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Solving $\sin x\cos3x+\cos x\sin3x=\sqrt{3}/2$ in the interval $[0,2\pi]$
Solve the following trig equation:
$$\sin(x)\cos(3x)+\cos(x)\sin(3x)=\frac{\sqrt{3}}{2}$$
in the interval $[0,2\pi]$.
Using trig identities, it is now at
$$\sin(x+3x)=\sqrt{3}/2 \tag{1}$$
Next, it has become
$$\sin(x+3x)=\sin 60^\circ \tag{2}$$... | $\sin(\theta)=\frac{\sqrt{3}}{2}$ when $\theta=\frac{\pi}{3}$, as you pointed out, but also when $\theta=\frac{2\pi}{3}$. Not only that, because the sine function is periodic, we must also remember that adding any multiple of $2\pi$ to either of those angles gives another solution to $\sin(\theta)=\frac{\sqrt{3}}{2}$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Question invoving using binomial identities to determine $n$ and $k$ given $\binom{n}{k-1} = 2002$ and $\binom{n}{k} = 3003$ I have been trying to do a problem in a combinatorics textbook involving using binomial identities. The problem is :
"Determine $n,k \in \mathbb{N}$ from the equalities $\binom{n}{k-1} = 2002$ an... | Starting from $$\frac{n-k+1}{k} = \frac{3}{2},$$ we obtain the Diophantine equation $$5k - 2n = 2$$ where $k < n \in \mathbb Z^+$. Thus $k$ must be even, say $k = 2m$, and the above equation becomes $$5m - n = 1,$$ or $n = 5m-1$, hence $$3003 = \binom{n}{k} = \binom{5m-1}{2m} = \frac{(5m-1)!}{(2m)!(3m-1)!}.$$ Since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\... | Here is an alternative method. You can first use substitution to simplify the integral:
$$\int \dfrac{dx}{(x^2-4x+13)^2}$$
$$ =\int \dfrac{dx}{((x-2)^2+9)^2} = \int \dfrac{du}{(u^2+9)^2} \tag{$u=x-2, du = dx$}$$
$$=\frac{1}{81}\int \dfrac{du}{(\frac{u^2}{9}+1)^2} = \frac{1}{27} \int \dfrac{dv}{(v^2+1)^2} \tag{$u=3v, du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 4
} |
$a = \log_{40}100, b = \log_{10}20$.How can I express $b$ depending only on $a$? Let $a = \log_{40}{100}, b = \log_{10}{20}$. How can I express $b$ depending only on $a$? I tried using the formula to change the base from $40$ to $10$, but couldn't get it just depending on $a$.
I used the base change formula $\log_a b =... | $$a=\log_{40}{100}=2\log_{40}10=\frac{2}{\log_{10}40}=\frac{2}{1+2\log_{10}2},$$ which gives $$\log_{10}2=\frac{2-a}{2a}.$$
Id est, $$\log_{10}20=1+\log_{10}2=\frac{a+2}{2a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Newton's evaluation of $1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$ How might have Newton evaluated the following series?
$$\sqrt{2} \, \frac{\pi}{4} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$$
The method of the this thread applies by sett... | Looking at the text mentioned, we see something interesting on page $156$, in note $(48)$. Transcribing it (you can find a screencap of the text here):
On observing that $$1 + \frac 1 3 - \frac 1 5 - \frac 1 7 + \frac 1 9 + \text{etc.} = \int_0^1 \frac{1+x^2}{1+x^4}dx$$ by expanding the integrand as an ascending serie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Need help with even number problem Problem:
A random non-zero positive even integer, $E$ is picked.
We can call the bracketed section in the statement $E = \{A\cdot2^n\}$ the factored form of this number (because it is even, $n$ is at least $1$). $n$ is to be as large as possible ensuring $A$ is odd.
E.g.: $E= 80$ so $... | To find the expected value of $n$, we use the following method. Assume we want to find the probability that $n=n_0$ for a fixed $n_0$. We need $2^{n_0} \mid E$ and $2^{n_0+1} \nmid E$. This is equivalent to saying $E \equiv 2^{n_0} \pmod{2^{n_0+1}}$. Thus, we have one choice for $E$ in every $2^{n_0+1}$ numbers, giving... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Rolling Dice Game, Probability of Ending on an Even Roll The game is described as follows. $A$ and $B$ take turns rolling a fair six sided die. Say $A$ rolls first. Then if $A$ rolls {1,2} they win. If not, then $B$ rolls. If $B$ rolls {3,4,5,6} then they win. This process repeats until $A$ or $B$ wins, and the game st... | Thanks to the comment of @JMoravitz I realized my mistake. I was interpreting turns as the rolls $A$ AND $B$, as in $\{A_1,B_1\}, \{A_2,B_2\}, \dots$. In reality the question is merely asking what the probability of $B$ winning if $A$ rolls first.
The work is as follows:
We calculate the probability of $B$ winning. Den... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
A problem involving the roots of quartic polynomial $x^4+px^3+qx^2+rx+1$ Let $\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$ be the roots of the following polynomial
$$P(x)=x^4+px^3+qx^2+rx+1$$
Show that
$$(1+{\alpha_1}^4)(1+{\alpha_2}^4)(1+{\alpha_3}^4)(1+{\alpha_4}^4)=(p^2+r^2)^2+q^4-4pq^2r.$$
I came across this prob... | Rewrite $x^4+px^3+qx^2+rx+1=0$ as
$$x^2+\frac1{x^2}+q=-(px +\frac r x) $$
Square to get
$$x^4+\frac1{x^4} + (2+q^2-2pr)=(p^2-2q)x^2+ \frac{r^2-2q}{x^2}$$
Square again and rearrange to obtain the quartic equation in $x^4$
\begin{align}
f(x^4)=&x^{16}+[2(2+q^2-2pr)-(p^2-2q)^2]x^{12}\\
&+[2+(2+q^2-2pr)^2-2(p^2-2q)(r^2-2q)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Solving the system $\sqrt{x} + y = 7$, $x + \sqrt{y} = 11$ I want to solve the following nonlinear system of algebraic equations. Indeed, I am curious about a step by step solution for pedagogical purposes. I am wondering if you can come up with anything. I tried but to no avail.
\begin{align*}
\sqrt{x} + y &= 7 \\
x +... | By subtracting the first equation from the second equation, we have
$$\sqrt{y}-\sqrt{x} + x-y = 4$$
and then $\sqrt{y}-\sqrt{x} + (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) = 4$. So, we get $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-1) = 4$.
Put $t := \sqrt{x}-\sqrt{y}$. It follows from the above equation that $t(t+2\sqrt{y}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Show that $(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$ Let $a,b,c\ge 0$ be such that $a^2+b^2+c^2=3$. Show that
$$(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$$
I want to consider the function
$$f(x)=\ln{(x^{3/2}+x^{1/2}+1)}$$ Maybe it isn't the case $f''(x)\le 0$, so I can't use Jensen's inequality.
| Let $f,g$ be given by
$$
\left\lbrace
\begin{align*}
f(a,b,c)&=(a^3+a+1)(b^3+b+1)(c^3+c+1)\\[4pt]
g(a,b,c)&=a^2+b^2+c^2\\[4pt]
\end{align*}
\right.
$$
Using the method of Lagrange Multipliers, if a point $(a,b,c)$ satisfying the constraint $g(a,b,c)=3$ is such that $f$ has a local extremum at $(a,b,c)$, then for some $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x$ I was generalizing the following integral:
$$\int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x \hspace{40pt} n\geq 0$$
We can start by noting that $\sin x =\dfrac{e^{ix}-e^{-ix}}{2i}$ and thus $\displaystyle \sin^{2n+1}x=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n ... | One can use the recurrence relation of Gradshteyn and Ryzhik to give, for some $p, q, \in \mathbb{N}$
\begin{align}
\displaystyle \int_0^\infty \frac{\sin^p x}{x^q}\,\mathrm{d}x &= \frac{p}{q-1}\displaystyle \int_0^\infty \frac{\sin^{p-1} x}{x^{q-1}}\cos x\,\mathrm{d}x \qquad (p > q-1>0)\\
&= \frac{p(p-1)}{(q-1)(q-2)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Solving $x^8 - x^5 + x^2 - x + 1 > 0$ over $\mathbb{R}$ I could not find any decent approach to solve this inequality. I would appreciate any help, and input if this is even possible to solve(without a computer).
$$x^8-x^5+x^2-x+1>0$$
| Using the AM-GM inequality, we have
$$x^8 + \frac{x^2}{2} \geqslant 2\sqrt{x^8 \cdot \frac{x^2}{2}} = \sqrt{2} \cdot |x^5| \geqslant x^5,$$
and
$$1+\frac{x^2}{2} \geqslant 2\sqrt{\frac{x^2}{2}} = \sqrt 2 \cdot |x| \geqslant x$$
Therefore
$$x^8+x^2+1> x^5 + x.$$
Done.
Note. The SOS form
$$x^{8}+x^{2}+1-x^{5}-x=\left(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is this a valid proof for $I(n^2) \geq \frac{5}{3}$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$, and let $I(x)=\sigma(x)/x$ be the abundancy index of $x$.
Note that both $\sigma$ and $I$ are multiplicative functions.
A number $m$ ... | Not a complete answer, just some thoughts that recently occurred to me, which would be too long to fit in the Comments section.
Since the biconditionals
$$I(n^2) > \frac{5}{3} \iff q > 5$$
and
$$I(n^2) = \frac{5}{3} \iff \bigg(q = 5 \land k = 1\bigg)$$
hold, it remains to consider what happens to the bounds for $I(n^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$ Prove that $$(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$$
First of all, I don't really know if by proving it means finding the function Sum ... | Let $f(x)=(x-1)\log(1-2x)-2x$, then $f$ is differentiable and
$$ f'(x)=(x-1)\frac{-2}{1-2x}+\log(1-2x)-2=\log(1-2x)+\frac{1}{1-2x}-1$$
For $|x|<\frac{1}{2}$, we have
$$ \log(1-2x)=-\sum_{n=1}^{+\infty}\frac{(2x)^n}{n} \text{ and } \frac{1}{1-2x}=\sum_{n=0}^{+\infty}(2x)^n$$
thus
$$ f'(x)=\sum_{n=1}^{+\infty}(2x)^n\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Polar form of -$2$ - $2\sqrt{2i}$ -$2$ - $2\sqrt{2i}$
So I am confused on the $\sqrt{2i}$ part, so what I did was using the rule $\sqrt{ab}$ = $\sqrt{a} \sqrt{b}$ and that gives $\sqrt{2} \sqrt{i}$
-$2$ - $2\sqrt{2i}$ = -2 -2$\sqrt{2} \sqrt{i}$ multiply by $i^\frac{9}{2}$ to get an $i$ therefore it becomes -2 -2$\sqrt{... | Note
\begin{align}
-2-2\sqrt{2i}&= -2-2\sqrt2 (e^{i\frac\pi2})^{1/2}\\
&= -2-2\sqrt2e^{i\frac \pi4}\\
&= -4-2i \\
&= 2\sqrt5 e^{i\tan^{-1}\frac12}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}\tan x&\tan(x+h)&\tan(x+2h)\\\tan(x+2h)&\tan x&\tan(x+h)\\\tan(x+h)&\tan(x+2h)&\tan x\end{vmatrix}$ Evaluate
$$
\lim_{h\to 0}\frac{\Delta}{h^2}=\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}
\tan x&\tan(x+h)&\tan(x+2h)\\
\tan(x+2h)&\tan x&\tan(x+h)\\
\tan(x+h)&\tan(x+2h)&\t... | $$ \Delta = \begin{vmatrix} \tan x&\tan(x+h)&\tan(x+2h)\\
\tan(x+2h)&\tan x&\tan(x+h)\\
\tan(x+h)&\tan(x+2h)&\tan x \end{vmatrix} $$
$$=\begin{vmatrix} \tan x&\tan(x+h)-\tan x&\tan(x+2h)-\tan x\\
\tan(x+2h)&\tan x-\tan(x+2h)&\tan(x+h)-\tan(x+2h)\\
\tan(x+h)&\tan(x+2h)-\tan(x+h)&\tan x-\tan(x+h) \end{vmatrix} $$
$$=\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $x^2+y^2+xy=1$ then find minimum of $x^3y+xy^3+4$
If $x,y \in \mathbb{R}$ and $x^2+y^2+xy=1$ then find the minimum value of $x^3y+xy^3+4$
My Attempt:
$x^3y+xy^3+4$
$\Rightarrow xy(x^2+y^2)+4$
$\Rightarrow xy(1-xy)+4$ (from first equation)
$\Rightarrow xy-(xy)^2+4 =f(x)$
For minimum value, $\frac{df(x)}{dx}=0$.
$\... | By AM-GM, $x^2+y^2=|x|^2+|y|^2\ge2|xy|$ so $r^2=x^2+y^2$ has extrema $\tfrac23,\,2$ respectively achieved by$$2xy=x^2+y^2\implies1=(1+1/2)(x^2+y^2)$$and$$-2xy=x^2+y^2\implies1=(1-1/2)(x^2+y^2).$$Note the minimum of $r^2(1-r^2)$ on $[\tfrac23,\,2]$ occurs at $r^2=2$. In particular, $x=-y=1$ minimizes $r^2(1-r^2)+4$ as $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the minimum value of $\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$.
Given positives $a, b, c$ such that $abc = 1$, if possible, calculate the minimum value of $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \lef... | By AM-GM $$\sum_{cyc}\left|\frac{a^2-bc}{b-c}\right|=\sqrt{\left(\left|\sum\limits_{cyc}\frac{a^2-bc}{b-c}\right|\right)^2}=$$
$$=\sqrt{\left(\sum\limits_{cyc}\frac{a^2-bc}{b-c}\right)^2-2\sum_{cyc}\frac{(a^2-bc)(b^2-ac)}{(b-c)(c-a)}+2\sum_{cyc}\left|\frac{(a^2-bc)(b^2-ac)}{(b-c)(c-a)}\right|}\geq$$
$$\geq\sqrt{-2\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How can I integrate this function? $\frac{\sqrt{e^{2x+2y+z}}}{(1+e^x+e^y+e^{x+y+z})^2}$ I want to evaluate the following integration
$$
\int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy \int_{-\infty}^{\infty} dz \frac{\sqrt{e^{2x+2y+z}}}{(1+e^x+e^y+e^{x+y+z})^2}.
$$
According to Mathematica12, the answer is $\pi^2... | Let $x= \ln X$, $y=\ln Y$ and $z=2\ln Z$. Then
$$ I = \int_0^\infty\int_0^\infty\int_0^\infty\frac{XYZ}{(1+X+Y+XYZ^2)^2} \frac{dX}{X}\frac{dY}{Y}\frac{2\cdot dZ}{Z}.$$
This is a rational function, and thus can be easily integrated. For instance integrate first against X:
$$ I = 2\int_0^\infty\int_0^\infty \frac{1}{1+YZ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $\frac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$ Let $\dfrac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$
My attempt :
\begin{align*}
\dfrac{\tan2A}{2}=\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ\\
\tan2A=2... | Let me post a different route to the simplication, as OP has begun.
So we have
$$
\sin^2 20^\circ-\sin160^\circ\sin220^\circ+\sin^2 320^\circ=
\sin^2 20^\circ(1+2\cos 20^\circ + 4\cos^2 20^\circ)
$$
I claimed for all $\theta$,
$$
\sin^2\theta(1+2\cos\theta+4\cos^2\theta)=\frac12(2+\cos\theta-\cos2\theta-\cos3\theta-\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a^2+a b+b^2=40$ and $a^2-\sqrt{a b}+b=5$, then find $a^2+\sqrt{a b}+b$ I was given this problem to solve with elementary methods (High School level).
Knowing that
$$\begin{align}
a^2+a b+b^2 &=40 \\
a^2-\sqrt{a b}+b &=\phantom{0}5
\end{align}$$
find
$$a^2+\sqrt{a b}+b$$
I tried to look for $\sqrt{a b}$, since the... | One way to do this is to graph out these equations of lines.
Find the values of $a,b$ and plug into the desired output equation:
*
*Equation of a line:
see link here.
*Look at the possible values and
*plug answers into the final equation:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
Find all values of $a$ for which the maximum value of $f(x)=\frac{ax-1}{x^4-x^2+1}$equals $1$. Find all values of $a$ for which the maximum value of $$f(x)=\frac{ax-1}{x^4-x^2+1}$$equals $1$.
I equated $f(x)$ equal to $1$ to obtain a polynomial $x^4-x^2-ax+2=0$. Now this must have at least one repeated root.But I could... | $$f(x)=\frac{ax-1}{x^4-x^2+1}\implies f'(x)=\frac{a \left(-3 x^4+x^2+1\right)+4 x^3-2 x}{\left(x^4-x^2+1\right)^2}$$
So, we need
$$a \left(-3 x^4+x^2+1\right)+4 x^3-2 x=0\tag 1$$ for an extremum and $$a x-x^4+x^2-2=0\tag 2$$ to have the maximum value of $1$.
From $(2)$,we can get $a=\frac{x^4-x^2+2}{x}$. Plug it in $(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What is the minimum value of $x+y$?
Suppose $x,y$ are positive real numbers that satisfy $$xy(x+2y)=2$$ What is the minimum value of $x+y$?
My Thoughts
I’ve attempted using arithmetic-geometric mean inequality and got:
$\frac{x+y+x+2y}{3} \geq \sqrt[3]{2}$
Therefore $2(x+y)+y \geq 3\sqrt[3]{2}$, then I got trapped.
... | $xy(x+2y)=2$.
Let z=x+y,
$(z-y)y(z+y)=2$
$(z^2-y^2)y=2 $
$z^2-y^2=2/y $
$z^2=2/y+y^2$
$z=(2/y+y^2)^{0.5}$
$\frac {d}{dx} (2/y-y^2)^{0.5}=\frac{y^3-1}{y^2((y^3+2)/y)^{0.5}} $which equals $0$ at $y=1$
$z^2=2/1+1=3$, $z=3^{0.5}$, $x=3^{0.5}-1$,
$x+y=3^{0.5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Compute the Surface area of $S=\left\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=z^2,1\le z\le2\right\}$ $\def\hl#1#2{\bbox[#1,1px]{#2}}
\def\box#1#2#3#4#5{\color{#2}{\bbox[0px, border: 2px solid #2]{\hl{#3}{\color{white}{\color{#3}{\boxed{\underline{\large\color{#1}{\text{#4}}}\\\color{#1}{#5}\\}}}}}}}
\def\verts#1{\left\vert#1\r... | I checked it directly in cylindrical coordinates as parametric surface and obtain
$$\begin{cases}
E=\cos^2 \phi + \sin^2 \phi +1 = 2 \\
G=r^2 \sin^2 \phi + r^2 \cos^2 \phi +0 = r^2 \\
F = r\cos \phi (-\sin \phi) + r\sin \phi \cos \phi +0 =0
\end{cases}
$$
So $\sqrt{EG-F} = r\sqrt{2}$ so your integral is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sum\frac{(-1)^{n+1}} {{n}^r} \sum\frac{(-1)^{n+1}} {{n}^s} $ by Abel's rule forms a series that doesn't converge when r+s=1. It is a similar problem to that in Show that the series $\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots$ converges, and its square (formed by Abel's rule) doesn'... | So here we have, for example,
if r<s,
$
|-\frac{1} {{1}^r} \frac{1} {{(2k)}^s}
-\frac{1} {{2}^r} \frac{1} {{(2k-1)}^s}+\dots
-\frac{1} {{k}^r}\frac{1} {{(k+1)}^s}
-\frac{1} {{(k+1)}^r}\frac{1} {{k}^s}
\dots-\frac{1} {{(2k)}^r}\frac{1} {{1}^s}|>
|-\frac{1} {{1}^s} \frac{1} {{(2k)}^s}
-\frac{1} {{2}^s} \frac{1} {{(2k-1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove $\int_0^1 \frac{dx}{(x-2) \sqrt[5]{x^2{(1-x)}^3}} = -\frac{2^{11/10} \pi}{\sqrt{5+\sqrt{5}}}$ $$
\mbox{Prove}\quad
\int_{0}^{1}{\mathrm{d}x \over
\left(\,{x - 2}\,\right)\,
\sqrt[\Large 5]{\,x^{2}\,\left(\,{1 - x}\,\right)^{3}\,}\,}
=
-\,{2^{11/10}\,\pi \over \,\sqrt{\,{5 + \,\sqrt{\,{5}\,}}\,}\,}
$$
*
*Being h... | $$
\begin{aligned}
\text { Let } x&=\sin ^2 \theta, \quad \textrm{ then } d x=2 \sin \theta \cos \theta d \theta, \textrm{ and }\\
I&=\int_0^{\frac{\pi}{2}} \frac{2 \sin \theta \cos \theta d \theta}{\left(\sin ^2 \theta-2\right) \sqrt[5]{\sin ^4 \theta \cos ^6 \theta}}\\
&=2 \int_0^{\frac{\pi}{2}} \frac{\tan ^{\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
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Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using ... | Let $\dfrac{(k+1)^2}{3^k}=f(k)-f(k-1)$ where $f(n)=\dfrac{an^2+bn+c}{3^n}$
$\implies\dfrac{(k+1)^2}{3^k}=\dfrac{(ak^2+kb+c)-3(a(k-1)^2+b(k-1)+c)}{3^k}$
$\implies k^2+2k+1=-2ak^2+k(b+6a-3b)+c-3(a-b+c)$
Compare the coefficients of $k^2$ to find $$-2a=1$$
and by the coefficients of $k,$ $$6a-2b=2\iff b=?$$
and by the cons... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Check the convergence of the series : Which test do we apply? Check the convergence of the following series:
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^2+1}{n^4+n}}$
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^{n^2}}{(n+1)^{n^2}}}$
For the first series do we use the comparison test? But with which sequence do we... | For the second, it is better to use the root test.
$$\lim_{\infty}a_n^\frac 1n=\lim_{\infty}(\frac{n}{n+1})^n$$
think of
$$\lim_{\infty}(\frac{n+1}{n})^n=e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Double integration with Indicator function The integral of interest is:
$$Q = \int_{\mathbb{R}^2}\int_{\mathbb{R}^2} I\left(\frac{1}{2}\frac{x_2^2 - x_1^2 + y_2^2 - y_1^2}{x_2-x_1} \in [0,1]\right) \nonumber \\
\times I\left(2\arcsin\left(\frac{\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}{2\sqrt{y_1^2 + \left(x_1 - \frac{1}{2}\fr... | Start with the stretch-rotation transformations $$\begin{pmatrix} z_1 \\ z_2\end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2\end{pmatrix} \qquad \text{and} \qquad \begin{pmatrix} z_3 \\ z_4\end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
How to prove this equality of the determinant of matrix?
Prove that
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=(a^2-bc)(b^2-ca)(c^2-ab)\end{equation*}
My attempt:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end... | Try this. Multiply column $1$ by $c$, column $2$ by $a$ and column $3$ by $b$. Then
\begin{aligned}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
&=\frac{1}{abc}\det\begin{pmatrix}
a^2c & b^2a & c^2b\\
abc & abc & abc\\
b^2c& c^2a & a^2b
\end{pmatrix}\\
&=\det\begin{pmatrix}
a^2c ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Prove that this number is divisible by 7 Without using induction, how can it be proved that 7 divides $3^{2n+1}+2^{n+2}$ for each $n\in\mathbb{N}$?
I tried to expand it using $\frac{x^{n+1}-1}{x-1}=1+x+..+x^n$ but I had no success.
It would be great if more than one proof is provided.
| $3^{2n + 1} + 2^{n+2} = 3\cdot 3^{2n} + 2^2\cdot 2^n = 3\cdot(9)^n + 4\cdot s2^n\equiv 3\cdot(2)^n + 4\times 2^n = 7\cdot 2^n\equiv 0\pmod 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Prove/Disprove: $A - \lfloor A/B \rfloor - \lceil A/B \rceil \leq \lfloor A/B \rfloor \times (B+1)$ for $A \geq B$ For $A \geq B$, both are strictly positive integers, is the following true?
$$A - \lfloor A/B \rfloor - \lceil A/B \rceil \leq \lfloor A/B \rfloor \times (B+1)$$
I tried the technique used in proving a ver... | $\newcommand{f}[1]{\left\lfloor #1 \right\rfloor}$
Let $B = 100$ and $A = 199$. Then:
\begin{align*}
LHS &= 199 - 1 - 2 = 196 \\
RHS &= 1(100 + 1) = 101
\end{align*}
So the inequality is false.
EDIT: In response to OP's comment, suppose we restrict further that $\f{A/B} \geq N$ for some $N \in \Bbb{Z}^+$. Let $B = 3N ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Two inequalities with parameters $a,b,c>0$ such that $ca+ab+bc+abc\leq 4$
Let $a,b,c>0$ be such that $bc+ca+ab+abc\leq 4$. Prove the following inequalities:
(a) $8(a^2+b^2+c^2)\geq 3(b+c)(c+a)(a+b)$, and
(b) $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{a^2b}+\dfrac{2}{b^2c}+\dfrac{2}{c^2a}\geq 9$.
Prove al... | This is an approach using Lagrange multipliers.
For the first part, we can write the problem as \begin{align}\min&\quad8(a^2+b^2+c^2)-3(a+b)(a+c)(b+c)\\\text{s.t.}&\quad ab+ac+bc+abc=4-\epsilon\\&\quad a,b,c>0\quad\land\quad0\le\epsilon<4.\end{align} Then we have $\mathcal L=f-\lambda g$ where $f(a,b,c)=8(a^2+b^2+c^2)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$ Let $x=\tan y$, then
$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$
Given answer is $0$
What’s wrong here?
| Hint:
$$\sin^{-1}\dfrac{2(-10)}{1+(-10)^2}=\sin^{-1}\left(-\dfrac{20}{101}\right)$$
$$\tan^{-1}\dfrac{2(-10)}{1-(-10)^2}=\tan^{-1}\dfrac{20}{99}=u(\text{say})$$
$\implies\dfrac\pi2>u>0$
$\sec u=+\sqrt{1+\left(\dfrac{20}{99}\right)^2}=\dfrac{101}{99}$
$\implies\sin u=\dfrac{\tan u}{\sec u}=?$
$u=\sin^{-1}?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
solve for $x$, $(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a$ Find the value of $x$ when $$(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a.$$See, by hit and trial method it is clear that $x=2$ is a solution. But I failed to solve this explicitly to get the solutions.
My Attempt: \begin{align*} &(\s... | Hint : Let $y=a+\sqrt{a^2-1}$. The equation you have to solve is equivalent to
$$y^{x/2} + \frac{1}{y^{x/2}} = y +\frac{1}{y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the second derivative of the absolute function $\left|\frac{x+1}{x+2}\right|$? I calculated the derivative of $\left|\frac{x+1}{x+2}\right|$ in the same way that I would do with $ \frac{x+1}{x+2}$ in order to study the function.
But when I verified on wolfram, I noticed it is all wrong. Wolfram uses the chain r... | This is how I deal with absolute functions:
$$
\begin{align}
\left|\frac{x+1}{x+2}\right|&=\sqrt{\left(\frac{x+1}{x+2}\right)^{2}}\\
\\
\frac{d}{dx} \left|\frac{x+1}{x+2}\right|&=\frac{d}{dx} \sqrt{\left(\frac{x+1}{x+2}\right)^{2}}\\
&=\frac{1}{2 \sqrt{\left(\frac{x+1}{x+2}\right)^{2}}}\cdot 2 \left(\frac{x+1}{x+2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $ \lim_{ n \to \infty }\int_{0}^{ \infty } \frac{1}{(1+x^3)(1+x^n)} dx$ Evaluate $$\displaystyle \lim_{ n \to \infty }\displaystyle \int_{0}^{ \infty } \frac{1}{(1+x^3)(1+x^n)} dx$$
I tried some substitution like $t=x^3+1$ and $t=x^n+1$ but did not work, and also division $$\lim_{ n \to \infty }\displaystyle \... | $$I=\lim_{n \to \infty}\int_{0}^{\infty} \frac{1}{(1+x^3)(1+x^n)}dx=\int_{0}^{1} \frac{1}{1+x^3} dx+\int_1^{\infty} 0~ dx$$
As $\displaystyle \lim_{n \to \infty} x^n =0, 0<x<1$; $\displaystyle\lim_{n \to \infty} x^n= \infty ,x>1.$
$$I=\int_{0}^{1}\left( \frac{1}{3(1+x)}-\frac{2x-1-3}{6(1-x+x^2)}\right) dx$$ $$I=\left[\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find all $x\in \mathbb{R}$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$
Find all $x\in \mathbb{R}$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$
Letting $a=2^x$ and $b=3^x$ we get $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$
from the numerator we have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=7$$
since $7$ is a pr... | $$\frac{a^3 + b^3}{a^2b + ab^2} = \frac{7}{6} \implies 7 a^2b + 7ab^2 = 6 a^3 + 6b^3 \implies 7ab(a + b) = 6(a + b)(a^2 -a b + b^2)$$ $$ \implies (a + b)(6a^2 -13ab + 6b^2) =0 $$
Then either $(a + b) = 0$ or $(6a^2 -13ab + 6b^2) = (2a -3b)(3a - 2b) = 0 $
Hence, we have that
$$ a = - b \implies 2^x = - 3^x \implies \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove that $\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}$ given $n$ is a positive integer and $0 < x \lt 1$. Problem Statement: If $n$ is a positive integer and $0 < x \lt 1$, show that
$$ \frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n}.$$
My Solution:
$$
\frac{ 1- x^{n+1} }{n+1} \lt \frac{1-x^n}{n} ~~~~\text{is true} \\
... | We have that
$$\frac{1 - x^{n+1} }{n+1} \lt \frac{1-x^n}{n} \iff n-nx^{n+1} \lt n-nx^n+1-x^n$$
$$\iff nx^n(1-x) \lt 1-x^n \iff nx^n<\overbrace{1+x+x^2+\ldots+x^{n-1}}^{\color{red}{\text{n terms}\,> \,x^n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Solving $\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right)$ for $x\in(0,\pi/2)$
Solve this equation for $x\in (0 , \frac{\pi}{2})$
$$\cos(2x)\cos\left(x - \frac{\pi}{6}\right) = \sin(2x)\sin\left(\frac{\pi}{6} - x\right)$$
I gave a try using the $\cos(a-b)$ and $\sin(a-b)$ formul... | The equation is $\cos (2x) \cos (x-\frac {\pi} 6)+\sin (2x)\sin (x-\frac {\pi} 6)=0$. This is same as $\cos (2x-(x-\frac {\pi} 6))=0$ or $\cos (x+\frac {\pi} 6)=0$. So $x+\frac {\pi} 6=\frac {(2n+1)\pi} 2$ for some integer $n$. For $x \in (0,\frac {\pi} 2)$ we must have $n=0$ so $x =\frac {\pi} 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equ... | Your calculations look correct except for the last line which I don't understand.
One you get $x_1$ and $x_2$, you could simply write
$x=123^{456}=9+11k$ (from $x_1$)
so reducing mod $8$ yields
$x \equiv 1+3k \pmod{8} \equiv 1 \pmod{8}$ (from $x_2$)
therefore
$3k\equiv0 \pmod{8}$ and since $\gcd(3,8)=1$, $3$ is inverti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
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Evaluate integral $\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}} \, dx$ Evaluate the integral
$$\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}} \, dx$$
Edit:
Here is Quanto's brilliant method of doing this problem with the details filled out:
SOLUTION:
First, lets complete the square $-9x^2+6x+15$:
$-9x^2+6x+15$
$=-9(x^2-\fr... | Let $\sin t = \frac{1-3x}4$. Then,
\begin{align}
\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}}dx
&=-\frac1{432}\int \frac{(1-4\sin t)^2}{\cos^2t}dt\\
&=\frac1{432}\int\left( 16-17\sec^2 t +\frac{8\sin t}{\cos^2t}\right)dt
\\
&= \frac1{432}\left( 16t -17\tan t+\frac8{\cos t}\right) +C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Proving that inequality holds under condition.
Let $a$ and $b$ be positive numbers. Prove that inequality $$\frac{ax+by}{2} \leqslant \sqrt{\frac{ax^2+by^2}{2}}$$
holds for all real $x$ and $y$ only and only if $a+b \leqslant2$
Problem needs to be done using "basic" algebraic methods.
I tried expanding this into form... | Suppose $\frac{ax+by}2 \le \sqrt{\frac{ax^2+by^2}{2}}$ is true, let $x=y=1$,
$$\frac{a+b}2\le \sqrt{\frac{a+b}2}$$
$$\frac{(a+b)^2}{4}\le \frac{a+b}2$$
Hence we must have $a+b \le 2$.
Suppose we have $a+b \le 2$, we want to investigate when does
$$(2a-a^2)x^2+(2b-b^2)y^2-2abxy \ge 0, \forall x, y$$
View it as a quadra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Solving $\frac{dP}{dt} = P(a - b\sqrt P)$ using separation of variables I need to solve the following differential equation using separation of variables but I am unsure on how to do this
$$
\frac{dP}{dt} = P(a - b\sqrt P)
$$
Where $p(0) = 4$, $a = 0.302$ and $b =0.002$
I tried substituting $P = u^2$ as a change of var... | $$\begin{aligned}\frac{dP}{dt}& = P(a - b\sqrt P)\\
(\text{Replacing }P(t)=(u(t)^2))\implies 2u\frac{du}{dt} &=u^2(a-bu) \\ (u(t)\equiv_t 0,\text{ i.e.} u(t)=0 \ \forall t, \text{ or}\cdots ) \implies2\dfrac{du}{dt} & = u(a-bu) \\ \implies \int dt = \int \dfrac{2du}{u(a-bu)}&=\dfrac{2b}{a}\int \dfrac{a}{bu(a-bu)}du\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Simple String of 6-digits This is the problem:
A string of 6 digits, each taken from the set $[0, 1, 2]$ is to be formed. The string should not contain any of the substrings $012$, $120$, and $201$. How many such 6-digit strings can be formed?
I really have trouble in using PIE in here. I'm badly needing some clear ... | You want to avoid 12 properties:
\begin{matrix}
000xxx & x000xx & xx000x & xxx000 \\
111xxx & x111xx & xx111x & xxx111 \\
222xxx & x222xx & xx222x & xxx222 \\
\end{matrix}
The inclusion-exclusion formula is $$\sum_{k=0}^{12} (-1)^k N_k,$$
where $N_k$ is an overcount of the number of $6$-strings that satisfy $k$ propert... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812431",
"timestamp": "2023-03-29T00:00:00",
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Proving $2\left(b^2+c^2\right)-a^2\leqslant 12$ with some condition. Problem. Let $a,b,c\in\mathbb{R}$ such that $a+b+c=6,$ $a^2+b^2+c^2\in\left[12,\frac{68}3\right]$ and $a\geq b\geq c.$ Prove
$$2\left(b^2+c^2\right)-a^2\leqslant 12.$$
When do we have equality?
I can only prove it for $a,b,c \geqslant 0.$
From $a+b+c=... | Let $s = b - c \ge 0$. From $a = 6 - b - c \ge b$, we have $c \le 2 - \frac{2s}{3}$.
We have $a^2 + b^2 + c^2 = (6-b-c)^2 + b^2 + c^2 = 6(c - 2 + s/2)^2 + s^2/2 + 12$.
From $a^2 + b^2 + c^2 \in [12, 68/3]$, we have
$$6(c - 2 + s/2)^2 + s^2/2 + 12 \le \frac{68}{3}$$
and thus
$$0 \le s \le \frac{8}{\sqrt{3}}, \quad 2 - s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Binomial identity involving negative values I want to check this identity that appears in Feller's first book on probability:
$${2n\choose n}=(-1)^n4^n{-1/2\choose n}$$
So I apply this definition
$${x\choose r}=\frac{x(x-1)(x-2)\ldots(x-r+1)}{r!}$$
obtaining
$$\frac{1/2(1/2-1)(1/2-2)\ldots(1/2-n+1)}{n!}=\frac{(-1)(-1)^... | You got as far as $${-1/2\choose n}=\frac{(-1)^n\left(\frac{1}{2}\right)^n(1+0)(1+2)(1+4)(1+6)\ldots(1+2n-2)}{n!}$$ though you still need to multiply by $(-1)^n 4^n$. Writing this as
$$(-1)^n4^n{-1/2\choose n}=4^n\frac{\left(\frac{1}{2}\right)^n \times 1 \times 3 \times 5 \times 7 \ldots\times (2n-1)}{ n!} \\
= 2^n \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a$ and $b$ be positive integers such that $(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$ Find $a + b.$ Let $a$ and $b$ be positive integers such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$$ Find $a + b.$
We let $\sqrt[3]a, \sqrt[3]b$ be $x,y,$ respectively. We expand LHS to get $$x... | Okay. Hint: consider rational and irrational parts separately.
Solution:
Consider $(1+x\sqrt[3]{a}+y\sqrt[3]{b})^2$ for some rational $x,\,y$.
If $a$ or $b$ are other than $6$ and $6^2$ we would have $\sqrt[3]{a}$ and $\sqrt[3]{b}$ in the irrational part (as $(1+\color{blue}a+\color{red}b)^2=1+a^2+b^2+\color{blue}{2a}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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proof that $ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $
Proof that
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $$
by induction.
Proof
Base case: Statement clearly holds for $n = 1$. Now assume that st... | The base of induction cannot be $n=1$ because then $1/(n(n-1))$ is not defined. For this sum you do not need induction. The sum is equal to $$(1-1/2)+(1/2-1/3)...(1/(n-1)-1/n)=1-1/n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Comparing the expansions of $\sin ^3 (x)$ and $\cos ^3 (x)$ By using $(\cos x+i\sin x)^3=\cos 3x+ i \sin 3x$,
We have:
$\sin ^3x= \frac{3}{4} \sin x- \frac{1}{4} \sin 3x$
$\cos ^3 x=\frac{3}{4}\cos x + \frac{1}{4}\cos 3x$
If we compare the coefficients of the equations we see coefficient of $\sin x$ and $\cos x$ are bo... | Simply do the substitution $x \mapsto x + \frac \pi2$: Then $\sin x$ becomes $\sin (x + \frac\pi2)$, which is $\cos x$. And $\sin 3x$ becomes $\sin (3x + \frac{3\pi}2)$, which is $-\cos 3x$.
Note that similar substitutions do not exist for $\sin 2x$ and $\cos 2x$, so there's nothing similar for $\sin 2x$. That's where ... | {
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"url": "https://math.stackexchange.com/questions/3815376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to evaluate below expression of modular arithmetic? How to evaluate (10^18)%(10^9 + 7) using modular arithmetic? How to proceed and what will be the steps?
| (10^9 + 7) * (10^9 - k) = 10^18 + 7*10^9 - k*10^9 - 7k
k=7: (10^9 + 7) * (10^9 - 7) = 10^18 - 49
10^18 = (10^9 + 7) * (10^9 - 7) + 49
10^18 % (10^9 + 7) = 49
Thought process:
The form of this problem is a^2 % (a+k).
Recall (a+k)*(a-k) = a^2 - k^2
So a^2 % (a+k) = k^2 % (a+k)
Now because k is so small, k^2 % (a+k) = k^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to deduce this factorization of $x^5+x+1$ by looking at $\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$? The question is:
$$\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$$
I tried a lot but couldn't solve it so I looked at the solution which is:
$$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$$ and we can write $$3x^4+2x^3-2x+1=(x^3-x^2+1)+(3x^2-2... | Hmm. I'd look at it and say "There's no rational root" because $x = \pm 1$ doesn't work. So there's some irrational root, $\alpha$, and two complex-conjugate pairs.
So there's no nice obvious linear factor I can write down.
Then I'd say "Maybe there's a quadratic factor." I can assume it's monic, so I'm looking to writ... | {
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"url": "https://math.stackexchange.com/questions/3817273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding a volume using the method of cylindrical shells which is generated by a parabola This problem is from the 7th edition of the book "Calculus and Analytic Geometry" by George Thomas
and Ross Finney. It is problem number 3 of section 5.4
Problem:
Find the volume generated when the region bounded by the given curve... | You said that you would use shells but you have actually used washers. The volume should be $$V=\int_0^2 2\pi y f(x) dx$$ Here $f(x)=(3x-x^2)-x$ is the length of the rotated vertical element and $y=\frac12 ((3x-x^2)+x)$ is the distance of the centroid (= mid-point) of the rotated element from the $x$ axis.
If you were ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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contour integration with dogbone, branch cut Compute the following integral
$$\frac1{2\pi\mathrm{i}}\int_{|z|=2}\frac{\sqrt{z^2-1}}{z-3}\,\mathrm{d}z.$$
Taking a branch of $\sqrt{z^2-1}$, satisfying $\sqrt{z^2-1}>0$ for $z>0$, I tried this problem with a 'dogbone' contour and I get,
$$\int_C\frac{\sqrt{z^2-1}}{z-3}\,\m... | For $R>3$, Cauchy's Integral Theorem guarantees that
$$\begin{align}
\oint_{\text{Dogbone}}\frac{\sqrt{z^2-1}}{z-3}\,dz&=\oint_{|z|=2}\frac{\sqrt{z^2-1}}{z-3}\,dz\\\\
&=\oint_{|z|=R}\frac{\sqrt{z^2-1}}{z-3}\,dz-2\pi i \text{Res}\left(\frac{\sqrt{z^2-1}}{z-3},z=3\right)\\\\
&=-2\pi i \text{Res}\left(\frac{\sqrt{z^2-1}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the roots of $x^3 - 6x = 4$ This exercise is from the book Complex Analysis by Joseph Bak, and it says:
"Find the three roots of $x^{3}-6x=4$ by finding the three real-valued possibilities for $\sqrt[3]{2+2i}+\sqrt[3]{2-2i}$".
I know that these numbers were found by Cardan's method, but I don't understand why they... | I find exponential form useful for a case known to have 3 real roots. One can get to a real expression with $\cos\left(\frac{\theta}{3}+\frac{2\pi}{3}k\right)$ and keep everything real from that point forward:
$$\begin{align*}\sqrt[3]{2+2i}+\sqrt[3]{2-2i} &= \sqrt[3]{2\sqrt{2}\left(\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3821909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
for which values of $n$, $1+n+n^{2}+n^{3}+n^{4}$ is a perfect square? I saw a question my number theory notes:
for which values of $n$ , $1+n+n^{2}+n^{3}+n^{4}$ is a perfect square where $"n"$ is non negative integer?
It was solved my professor using inequalities,but i think that his method is too long.Because of that... | Following from the comments.
We want to find integers $(x,y)$ such that $x^4+x^3+x^2+x+1=y^2.$ Clearly if $x^4+x^3+x^2+x+1$ is a square then so is $f(x)=4x^4+4x^3+4x^2+4x+4.$
Note that $f(x)> (2x^2+x)^2$ $\forall x\in\mathbb R$ and also $f(x)=4x^4+4x^3+4x^2+4x+4=(2x^2+x+1)^2-(x+1)(x-3).$ Since the last term is non-posi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Predetermine the middle term of an quadratic equation when there are two possibilities
Factorise completely:
*
*$2x^2+13x+15$
*$3x^2-5x+2$
*
*As i figured, there could be two ways to break the middle term. Either, $15x-2x$ or, $10x+3x$.
If I continue with $15x-2x$:
$2x^2+13x+15$
$2x^2+15x-2x+15$
$2x(x-1)+15(x... | There is a way to factorize a general quadratic trinomial
$$Ax^2 + Bx + C = (Dx + E)(Fx + G),$$
that is, assuming that the trinomial can be factored in the first place. (This method will also let you determine whether the trinomial cannot be factored at all.)
Set
$$Ax^2 + Bx + C = 0$$
and solve for $x$ using the Quadr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3825217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying the following mathematical expression using a computer? I have this following beastly expression typed up very nicely in LaTeX formatting, as you can see. What is the easiest way that I can get a computer to simplify this expression for me? I have zero programming experience. I installed sagemath but it see... | Wolfram|Alpha supports Latex. Unfortunately, the engine didn't seem to understand your query, perhaps because of its length. If possible, I would suggest splitting the expression up into chunks, and entering them one by one into Wolfram|Alpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Real roots of $x^7+5x^5+x^3−3x^2+3x−7=0$? The number of real solutions of the equation,
$$x^7+5x^5+x^3−3x^2+3x−7=0$$
is
$$(A) 5 \quad (B) 7 \quad (C) 3 \quad (D) 1.$$
Using Descartes rule we may have maximum no. of positive real roots is $3$ and negative real root is $0.$ So there can be either $3$ real roots or $1$ re... | Descare's rukles say that there will be at most 3 positive roots and no negative root. Since $f(1)=0$, so $(x-1)$ is factor of $f(x)=x*7+5x^5+x^3-3x^2+3x-7$
Note that $f(x)=(x-1)(x^6+x^5+6x^3+7x^2+4x+7)=(x-1)g(x)$, It can be seen that
the number of sign changes in $g(x)$ are none, so $g(x)$ will have no positive roots.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y} = 0$.
If $a \neq 0$ , $b \neq 0$ , $c \neq 0$ and if :- $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}=0$ ; $\frac{1}{a} + \frac{1}{x... | From the first and the second equality we obtain:
$$x=-\frac{a(a+2b)}{a+b}$$ and $$y=-\frac{a(a+2c)}{a+c}.$$
Thus, $$\frac{1}{a}-\frac{a+b}{a(a+2b)}-\frac{a+c}{a(a+2c)}=0$$ or
$$(a+2b)(a+2c)=(a+b)(a+2c)+(a+c)(a+2b)$$ or
$$a(a+b+c)=0$$ or $$a+b+c=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Where did I go wrong in my proof that for all $n \in \mathbb{Z}^+$, $\sqrt{2} < a_n$ with $(a_n)$ being a particular recursive sequence? I am in Introduction to Abstract Math, and I have taken Calculus 1, Linear Algebra, and Discrete Math. I got stuck with an apparent contradiction in my proof and wanted to know where ... | You didn't really "mess up". You just got stuck because this approach leads nowhere. However, there is certainly something wrong in what you typed in Wolfram Alpha.
You could do this, assuming $a_n>0$:
$$a_{n+1}-\sqrt{2}=\frac{a_n}{2}+\frac{1}{a_n}-\sqrt{2}\\=\frac{1}{2}\left(a_n+\frac{2}{a_n}-2\sqrt{2}\right)=\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
An urn has $4$ red, $6$ white and $3$ blue marbles, what is the probability of drawing with replacement $2$ red, $2$ white and $2$ blues? An urn has $4$ red, $6$ white and $3$ blue marbles, what is the probability of drawing with replacement $2$ red, $2$ white and $2$ blues? The order for marbles been picked up does no... | The selections you are making are not equally likely to occur. For instance, there are $3^6 = 729$ ways for all six marbles to be blue since there are three ways to select a blue marble on each draw. However, there are
$$\binom{6}{2}4^2\binom{4}{3}6^3\binom{1}{1}3^1 = 1,036,800$$
ways to select two red, three white, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this absolute value inequality? I am new to absolute value inequalities.I was looking trough a book and I found this inequality,
I know a little bit about absolute value inequalities.
The inequality is given below:
$$
\left| \frac{n+1}{2n+3} - \frac{1}{2} \right| > \frac{1}{12}, \qquad n \in \mathbb{Z}
$$
| For case 1:
$ \frac{n+1}{2n+3} - \frac{1}{2} > \frac{1}{12} $
$ \frac{n+1}{2n+3} - \frac{2n+3}{2(2n+3)} > \frac{1}{12} $
$ \frac{-1}{2(2n+3)} > \frac{1}{12} $
$ \frac{-6}{2n+3} > 1 $
$ \Rightarrow 2n+3 < 0 $ and $ 2n+3 > -6 $
$ \frac{-9}{2} > n < \frac{-3}{2} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
$\frac{a}{a^{2}+b^{2}+2}+\frac{b}{b^{2}+c^{2}+2}+\frac{c}{c^{2}+d^{2}+2}+\frac{d}{d^{2}+a^{2}+2}\le 1$ Let be a,b,c,d non negative real numbers. Prove that :
$$\frac{a}{a^{2}+b^{2}+2}+\frac{b}{b^{2}+c^{2}+2}+\frac{c}{c^{2}+d^{2}+2}+\frac{d}{d^{2}+a^{2}+2}\le 1$$
I tried many attempts but still can't find the result. At... | By AM-GM twice we obtain:
$$\sum_{cyc}\frac{a}{a^2+b^2+2}=\sum_{cyc}\frac{a}{a^2+1+b^2+1}\leq\frac{1}{2}\sum_{cyc}\frac{a}{\sqrt{(a^2+1)(b^2+1)}}=$$
$$=\frac{1}{2}\sum_{cyc}\sqrt{\frac{a^2}{a^2+1}\cdot\frac{1}{b^2+1}}\leq\frac{1}{4}\sum_{cyc}\left(\frac{a^2}{a^2+1}+\frac{1}{b^2+1}\right)=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx=\frac{5\pi^3}{64}+\frac\pi{16}\ln^22$ Tried to evaluate the integral
$$I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx$$
and managed to show that
\begin{align}
I &= \int_0^1 \frac{\ln^2x}{(x+1)^2+1} \, dx + \int_0^1 \frac{\ln^2x}{(x+1)^2+x^2} \, dx\\
&= \int_0^1 \frac{\... | We can use the integral of
\begin{equation}
f(z)=\frac{\ln^3z}{(z+1)^2+2}
\end{equation}
along the classical keehole contour: above the positive real axis , along the large circle, back from below along the positive axis and avoiding the origin by a small circle. Except the horizontal parts, the other contributions ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Given $|b-c|\leq a\leq b+c$ show that $\frac{a}{1+a}\leq\frac{b}{1+b}+\frac{c}{1+c}$ I am trying to show the following:
Given that $|b-c| \leq a \leq b+c$ show that:
$$\frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c}$$
So far I have done the following:
$a/(1+a) \leq b/(1+b) + c/(1+c) \iff$
$1-1/(1+a) \leq 2 - 1/(1+b)... | I will assume that $a,b,c > 0$ because without such an assumption we cannot even have $1+b$ and $1+c$ in the denominator.
$f(x)=\frac x {1+x}$ is an increasing function of $x$ on $[0,\infty)$. Hence $\frac a {1+a} \leq \frac {b+c} {1+b+c} =\frac b {1+b+c}+\frac c {1+b+c} \leq \frac b {1+b} +\frac c {1+c} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area between $r=1$ and $r=3\cos\theta$
Find the area between $r=1$ and $r=3\cos\theta$.
I squared both sides to get $r^2 = 1$, then did $r^2(\cos^2 \theta + \sin^2 \theta) = (r \cos \theta)^2 + (r \sin \theta)^2$$ = x^2+y^2 = 1$ to get $x^2+y^2=1$.
For $r = 3 \cos \theta$, I multiplied by $r$ on both sides ... | Geometric solution:
Rewrite $x^2+y^2=3x$ as $x^2-3x+(\frac{3}{2})^2 + y^2 = (\frac{3}{2})^2 \Rightarrow (x-\frac{3}{2})^2 + y^2 = \frac{9}{4}$. Then solving for the intersections:
$$(x-\frac{3}{2})^2 + y^2 = \frac{9}{4} \tag{1}$$
$$x^2+y^2 = 1 \tag{2}$$
$(1) - (2)$ gives $-3x + \frac{9}{4} = \frac{5}{4} \Rightarrow x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x... | Following along your approach,
After expansion,
$$(\cos^2x + \sin^2x)^3 = \cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$$
$$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$$
$$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x = 1$$
Hence,
$$\cos^6x + \sin^6x = 1 - 3\cos^2x\sin^2x$$
$$ = 1 - \frac{3}{4}(4\sin^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 7
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Find $\sqrt a + \sqrt b + \sqrt c$ only in terms of $p$ .
If :- $$a^2x^3 + b^2y^3 + c^2z^3 = p^5$$
$$ax^2 = by^2 = cz^2$$
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{p}$$ Find $\sqrt a + \sqrt b + \sqrt c$ in terms of $p$ .
What I Tried :- How do you use the information here? I assumed that :-
$$ax^2 = by^2... | Now, for positive variables $a=\frac{k}{x^2},$ $b=\frac{k}{y^2}$, $c=\frac{k}{z^2}$ and $$a^2x^3+b^2z^3+c^2z^3=k^2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=p^5,$$ which gives $$k^2=p^6$$ or
$$k=p^3.$$
Id est, $$\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{k}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\sqrt{p}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is the representation of any prime of the form $6n+1$ as $a^2+3b^2$ essentially unique? Two well-known results in number theory are:
Fermat's $4n+1$ theorem: Every prime of the form $4n+1$ can be represented as $a^2+b^2 (a,b \in\mathbb{N})$.
Euler's $6n+1$ theorem: Every prime of the form $6n+1$ can be represented a... | If you require $a, b \in \mathbb{N}$ then the representation is literally unique, and we can argue as Daniel Fischer does in the comments: the ring of Eisenstein integers $\mathbb{Z}[\omega]$, where $\omega = \frac{-1 + \sqrt{-3}}{2}$ is a primitive third root of unity, is a unique factorization domain, and the primes ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
} |
find the largest integer $m$ such that $2^m$ divides $3^{2n+2}-8n-9$
find the largest integer $m$ such that $2^m$ divides $\space 3^{2n+2}-8n-9$ when $n$ is a natural number.
If the answer was known it will be easy induction.
I started out like this :
$\space 3^{2n+2}-8n-9=9(3^{2n}-1)-8n=9\underbrace{(3^n-1)(3^n+1)}-... | In such problems, it's common to check for some small values to see if there's a pattern early on. Let's to do that here:
$$\begin{align}
n=1&: 3^4 - 8- 9 = 64 = 2^6 \\
n=2&: 3^6 - 16 - 9 = 704 = 64\cdot 11 = 2^6 \cdot11 \\
n=3&: 3^8 - 24 - 9 = 6528 = 128\cdot 51 = 2^7 \cdot51 \\
n=4&: 3^{10} - 32 - 9 = 59008 = 128... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Advice on integrating $\int \frac{x^2 +n(n-1)}{(x \sin x +n \cos x)^2} dx$ Our integral is:
$$\int \frac{x^2 +n(n-1)}{(x \sin x + n\cos x)^2} dx$$
I considered that this can be turned into some sort of quotient differential:
$$ d \frac{u}{v} = \frac{ v du - u dv}{v^2}$$
Now, comparing this with our integral:
$$ v = x \... | Instead of looking for a function $u$ such that the integrand is the derivative of the quotient of two functions, I will present a different approach.
\begin{align*}
\int \frac{x^2 +n(n-1)}{(x \sin x + n\cos x)^2} \; \mathrm{d}x &= \int \frac{x^2 +n(n-1)}{\left(\sqrt{x^2+n^2} \cos \left(x-\arctan{\left(\frac{x}{n}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
prove that $\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$
prove that $$\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$$ for positives $a,b,c$
Attempt: By C-S; $$\left(\sum_{cyc}\frac{a}{b^2+c^2} \right) \left(\sum_{cyc} a(b^2+c^2) \right)\ge {(a+b+c)}^2$$ .
or as inequality is ... | Proof by full expanding.
After your work we need to prove that:
$$5\left(\left(\sum_{cyc}a\right)^3\sum_{cyc}ab-abc\left(\sum_{cyc}a\right)^2\right)\geq4\left(\left(\sum_{cyc}a\right)^2+\sum_{cyc}ab\right)\left(\sum_{cyc}a\sum_{cyc}ab-3abc\right)$$ or
$$5\sum_{cyc}(a^2+2ab)\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Prove $\int_0^\infty \frac{x^a-1}{x^2-1}\, dx=\frac{\pi}{2}\tan\frac{a\pi}{2}$ for $0\lt a\lt 1$ How could I prove
$$\int_0^\infty \frac{x^a-1}{x^2-1}\, dx=\frac{\pi}{2}\tan\frac{a\pi}{2}$$
where $0\lt a\lt 1$?
I only thought of splitting it like this:
$$\int_0^\infty \frac{x^a-1}{x^2-1}\, dx=\int_0^\infty \frac{x^a}{x... | Your integral is$$\int_0^1\frac{1-x^a}{1-x^2}dx+\int_0^1\frac{y^{-a}-1}{y^{-2}-1}\frac{dy}{y^2}=\int_0^1\frac{x^{-a}-x^a}{1-x^2}dx.$$For integer $n\ge0$,$$\int_0^1(x^{2n-a}-x^{2n+a})dx=\frac{1}{2n-a+1}-\frac{1}{2n+a+1}=\frac{2a}{(2n+1)^2-a^2}.$$Now finish with a result proven here, with $z=a/2$:$$\sum_{k\ge0}\frac{4a}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3855636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Definite integral evaluation of $\frac{\sin^2 x}{2^x + 1}$
Compute
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} \ dx.$$
So looking at the limits I first checked whether the function is even or odd by substituting $x = -x$ but this gave me:
$$\frac{\sin^2 x}{2^x + 1} \cdot 2^x$$
so the function is ... | Let $$\;I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{\sin^2 x}{2^x+1} dx\;.$$
By substituting $\;x\to -x\;$ we get that $$I=-\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \dfrac{\sin^2 x}{2^{-x}+1} dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{2^x\sin^2 x}{2^x+1} dx=\\=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin^2 x -\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Show that $\sum_{k=0}^{n} \binom{n}{k} ka^k = an(a+1)^{n-1}$ Problem:
Show that $\sum_{k=0}^{N} \binom{N}{k} ka^k = aN(a+1)^{N-1}$
Attempt:
I tried using induction but got stuck.
$N=0$ implies $\sum_{k=0}^{0} \binom{0}{k} ka^k = 0$
$N=1$ implies $\sum_{k=0}^{1} \binom{1}{k} ka^k = a = a(1)(a+1)^{1-1}$
$N=2$ implies $\s... | Just like Thomas Andrews suggested, we can use the fact that
$$
k\binom{n}{k}=n\binom{n-1}{k-1}.
$$
We have
\begin{align}
\sum_{k=0}^n\binom{n}{k}ka^k&=\sum_{k=0}^nn\binom{n-1}{k-1}a^k\\
&=n\sum_{k=0}^n\binom{n-1}{k-1}a^k \ \ \ \ \ \ (\text{setting}\ \ell=k-1),\\
&=n\sum_{\ell=0}^{n-1}\binom{n-1}{\ell}a^{\ell +1}\\
&=n... | {
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Find the limits between which $a$ must lie in order that $\frac{ax^2-7x+5}{5x^2-7x+a}$ can take all real values Find the limits between which $a$ must lie in order that $$\frac{ax^2-7x+5}{5x^2-7x+a}$$ may be capable of taking all real values for all possible real values that $x$ may take.
My Attempt:
Let $$y=\frac{ax^2... | If $$y=\frac{f(x)}{g(x)}$$ where $f(x)$ and $g(x)$ are both quadratics.
If $y$ is of the form $$y=\frac{(x-a)(x-c)}{(x-b)(x-d)}$$ where $a<b<c<d$ then from the plot of graph it is clearly seen that $-\infty<y<+\infty$ when $x\in (b,d)$
It appears reasonable enough to say that if one root of numerator(denominator) lies... | {
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"answer_count": 1,
"answer_id": 0
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Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$
Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$
Plugging in the Catalan Equation, we want to prove... | As somebody suggested the method of generating functions is good.
We have:
\begin{eqnarray}
g(x)&:=&\sum\limits_{i=0}^\infty \binom{2 i}{i} \frac{x^i}{i+1}\\
&=& \frac{1}{x} \int_0^x \sum\limits_{i=0}^\infty \binom{2 i}{i} \xi^i d\xi\\
&=& \frac{1}{x} \int_0^x \frac{d \xi}{\sqrt{1-4 \xi}} \\
&=& - \frac{-1+\sqrt{1-4 x... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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System of equations (Problem $50$ from $101$ algebra by Titu)
Let $a$ and $b$ be given real numbers.
Solve the system of equations
$$\begin{aligned} \frac{x-y \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}} &=a \\ \frac{y-x \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}} &=b \end{aligned}$$
for real $x$ and $y$.
Solution -
Let $... | It is $a^2-b^2=x^2-y^2$ hence $$\dfrac{\frac{u+v}{2}-\frac{u-v}{2}\sqrt{a^2-b^2}}{\sqrt{1-a^2+b^2}}=a$$ and $$\dfrac{\frac{u-v}{2}-\frac{u+v}{2}\sqrt{a^2-b^2}}{\sqrt{1-a^2+b^2}}=b$$
If you add them you get $u$ and if you subtract them you get $v$.
| {
"language": "en",
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"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving the equation $X^2 + (3 +i)X + 1 +i =0$ I have to find the roots of the equation
$$X^2 + (3 +i)X + 1 +i =0.$$
The first step is to find the discrimant which is $4 + 2i$. Then, I assume that the square root of the discriminant is in the form of $a+bi$, so we have to solve the below system
$$a^2-b^2=4\qquad\text{ ... | You're work so far is good, and indeed you want to find $a,b\in\Bbb{R}$ such that $(a+bi)^2=4+2i$, i.e.
$$a^2-b^2=4\quad\text{ and }\qquad 2ab=2.$$
The latter shows you that $a$ and $b$ are nonzero, and that $b=a^{-1}$. Plugging this into the former yields
$$a^2-a^{-2}=4,$$
and multiplying everything by $a^2$ and rearr... | {
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"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find an asymptotic equivalent of the sequence $(\int_{-\infty}^{+\infty} \frac{1}{\cosh^n(x)} dx)_n$.
Find an asymptotic equivalent of the sequence $(\int_{-\infty}^{+\infty} \frac{1}{\cosh^n(x)} dx)_n$.
I found the result using a trick (using $e^x = \tan(\theta/2)$ as mentioned here) but I wanted to know if there is... | There is a different approach: You can derive a recursive relation for $a_n:=\int_{\mathbb{R}} \cosh^{-n}(x) dx$ using the identity $\cosh^2-\sinh^2=1$ and integration by parts. In fact, for $n\geq 3$, we have
$a_n= \frac{n-2}{n-1}a_{n-2}$.
Indeed,
$a_n=\int_{\mathbb{R}} \frac{1}{\cosh^{n}(x)} dx= \int_{\mathbb{R}} \fr... | {
"language": "en",
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"answer_count": 4,
"answer_id": 1
} |
cyclic rational inequalities $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ when $a+b+c=1$ I've been practicing for high school olympiads and I see a lot of problems set up like this:
let $a,b,c>0$ and $a+b+c=1$. Show that
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$$
Any prob... | Another way.
Since $$\left(\frac{1}{x^2+3}\right)''=\frac{6(x^2-1)}{x^2+3)^2}<0$$ for $0<x<1$, by Jensen for the concave function $f(x)=\frac{1}{x^2+3}$ we obtain:
$$\sum_{cyc}(\frac{1}{a^2+3}\leq\frac{3}{\left(\frac{\sum\limits_{cyc}a}{3}\right)^2+3}=\frac{3}{\left(\frac{1}{3}\right)^2+3}=\frac{27}{28}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find $\sum_{k=1}^n P(z_k)$ where P is a polynomial and $z_k$ the $k$th root of $z^{13}=1$ Question: Given a complex number z such that $z^{13}=1$, find the sum of all possible values of $z+z^3+z^4+z^9+z^{10}+z^{12}$.
I know we have to use roots of unity and try to manipulate the polynomial we want to evaluate, but can'... | If $z^{13} = 1$ and $z \not= 1$, then $z = e^{2\pi i \frac{k}{13}}$ where $(k, 13) = 1$.
Since $z^{12} = z^{-1}$, $z^{10} = z^{-3}$ and $z^{9} = z^{-4}$ we have
$z+z^3 + z^4 + z^9 + z^{10}+z^{12} = 2(\cos(2 \pi \frac{k}{13}) + \cos(2 \pi \frac{3k}{13}) + \cos(2 \pi\frac{4k}{13}))$ where $z = e^{2\pi i \frac{k}{13}}$.
T... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
What I Tried: Here is a picture :-
I know the centroid divides each of the medians ... | From where you left,
$A = \displaystyle 3 \sqrt{{(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}}$
Take the first two terms, it is of the form $(a-b)(a+b)$ so we have,
$(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3}) = 2\sqrt2$
Next two terms can be taken as (a+... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Three vertices of a triangle lie on $y^2=x^3$
If ${P(a,b),Q(c,d),R(m,n)}$ are the centroid, orthocenter, circumcenter respectively of a scalene triangle with vertices on the curve $y^2=x^3$, then find the value of $$\frac{a}{b}+\frac{c}{d}+\frac{m}{n}$$
I know that the centroid, orthocenter and circumcenter lie on ... | $\begin{array}{} G=(a,\sqrt{a^3}) & H=(c,\sqrt{c^3}) & O=(m,-\sqrt{m^3}) \end{array}$
$\begin{array}{} \frac{x_{H}-x_{G}}{x_{G}-x_{O}}=2 & \frac{c-a}{a-m}=2 & a=\frac{2m+c}{3} \end{array}$
$\begin{array}{} \frac{y_{H}-y_{G}}{y_{G}-y_{O}}=2 & \frac{\sqrt{c^3}-\sqrt{a^3}}{\sqrt{a^3}+\sqrt{m^3}}=2 \end{array}$
$\frac{\sq... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$ I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$
Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = ... | Let $f\left( x \right) = \sum\limits_{n = 0}^\infty {a_n x^n } $ be the generating function of the solution of the recurrence.
$$
\sum\limits_{n = 0}^\infty {na_n x^n } = \sum\limits_{n = 0}^\infty {\left( {n + 1} \right)a_n x^{n + 1} - 4\sum\limits_{n = 0}^\infty {a_{n - 1} x^n } } +
12\sum\limits_{n = 5}^\inf... | {
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"answer_count": 5,
"answer_id": 4
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