Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $\tan(75^\circ) = 2 + \sqrt{3}$ My (very simple) question to a friend was how do I prove the following using basic trig principles:
$\tan75^\circ = 2 + \sqrt{3}$
He gave this proof (via a text message!)
$1. \tan75^\circ$
$2. = \tan(60^\circ + (30/2)^\circ)$
$3. = (\tan60^\circ + \tan(30/2)^\circ) / (1 - \ta... | $$
\begin{array}{l}
\tan (75^{\circ})=\frac{\sin (75^{\circ})}{\cos (75^{\circ})}=\frac{\cos (15^{\circ})}{\sin (15^{\circ})}=\frac{\cos (30^{\circ} / 2)}{\sin (30^{\circ} / 2)}=\sqrt{\frac{1+\cos (30^{\circ})}{1-\cos (30^{\circ})}} \\
=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}}=\sqrt{\frac{2+\sqrt{3}}{2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/360747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Power Series Solution for $e^xy''+xy=0$ $$e^xy''+xy=0$$
How do I find the power series solution to this equation, or rather, how should I go about dealing with the $e^x$? Thanks!
| $e^xy''+xy=0$
$y''+xe^{-x}y=0$
Let $y=\sum\limits_{n=0}^\infty\dfrac{a_nx^n}{n!}$ ,
Then $y'=\sum\limits_{n=0}^\infty\dfrac{na_nx^{n-1}}{n!}=\sum\limits_{n=1}^\infty\dfrac{na_nx^{n-1}}{n!}=\sum\limits_{n=1}^\infty\dfrac{a_nx^{n-1}}{(n-1)!}$
$y''=\sum\limits_{n=1}^\infty\dfrac{(n-1)a_nx^{n-2}}{(n-1)!}=\sum\limits_{n=2}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/362089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$f: \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2)=5$ The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2) = 5$.
| Hint 1: $3=(x+1)-(x-2)$
$(x-2)f(x)-(x+1)f(x-1)=3\Leftrightarrow (x-2)f(x)-(x+1)f(x-1)=(x+1)-(x-2)\Leftrightarrow (x-2)(f(x)+1)-(x+1)(f(x-1)+1)=0$
Hint 2: Is there a function closely related to $f$ that would verify a simpler equation?
$g(x)=f(x)+1$
$ $
$(x-2)g(x) - (x+1)g(x-1)=0 \Leftrightarrow (x-2)g(x)=(x+1)g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/367473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Difficulties performing Laurent Series expansions to determine Residues The following problems are from Brown and Churchill's Complex Variables, 8ed.
From §71 concerning Residues and Poles, problem #1d:
Determine the residue at $z = 0$ of the function $$\frac{\cot(z)}{z^4} $$
I really don't know where to start with th... | $$\cos z=1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\;,\;\;\;\sin z=z-\frac{z^3}{6}+\frac{z^5}{120}-\ldots\implies$$
$$\cot z=\frac{\left(1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\right)}{z\left(1-\left(\frac{z^2}{6}-\frac{z^4}{120}\right)+\mathcal O(z^6)\right)}=$$
$$=\frac{1}{z}\left(1-\frac{z^2}{2}+\frac{z^4}{24}-\ldots\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/367940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finding the limit of function - exponential one Find the value of $\displaystyle \lim_{x \rightarrow 0}\left(\frac{1+5x^2}{1+3x^2}\right)^{\frac{1}{\large {x^2}}}$
We can write this limit function as :
$$\lim_{x \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\frac{1}{\large{x^2}}}$$
Please guide further how to pro... | $$\lim_{x \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\frac{1}{x^2}}$$
$$=\lim_{x \rightarrow 0}\left(\left(1+ \frac{2x^2}{1+3x^2}\right)^{\huge{\frac{1+3x^2}{2x^2}}}\right)^{\huge{\frac2{1+3x^2}}}$$
$$=\left(\lim_{\frac{2x^2}{1+3x^2} \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\huge{\frac{1+3x^2}{2x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Hyperbolic Functions
Hey everyone, I need help with questions on hyperbolic functions.
I was able to do part (a).
I proved for $\sinh(3y)$ by doing this:
\begin{align*}
\sinh(3y) &= \sinh(2y +y)\\
&= \sinh(2y)\cosh(y) + \cosh(2y)\sinh(y)\\
&= 2\sinh(y)\cosh(y)\cosh(y) + (\cosh^2(y)+\sinh^2(y))\sinh(y)\\
&= 2\sinh(y)(... | In the identity that you proved, put $\sinh 3y=2$. Then if $x=\sinh y$, the identity says that $4x^3+3x-2=0$. So we are almost finished, we have shown this $x$ is a solution of the equation.
Note that $\sinh t$ is a strictly increasing function, and that $\sinh t$ is large negative when $t$ is large negative, and large... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/369339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding coefficient of generating functions I have the equation
$$(1+x+x^2+\ldots+x^k+\ldots)(1+x^2+x^4+\ldots+x^{2k}+\ldots)(x^2+x^3)$$
how of I find the coefficent of $x^{24}$. I know to condense this down to
$$\frac1{1-x}\cdot\frac1{1-x^2}\cdot x(1+x)$$
but I don't know what to do after that
| Hint: There is cancellation, $(1-x^2)=(1-x)(1+x)$. And $x^2+x^3=x^2(1+x)$.
So we end up with $\dfrac{x^2}{(1-x)^2}$.
Now everything is straightforward. The coefficients for $\dfrac{1}{(1-x)^2}$ can be found by a direct computation of $(1+x+x^2+\cdots)(1+x+x^2+\cdots)$, or by noticing that $\dfrac{1}{(1-x)^2}$ is the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What will be the units digit of $7777^{8888}$? What will be the units digit of $7777$ raised to the power of $8888$ ?
Can someone do the math with explaining the fact "units digit of $7777$ raised to the power of $8888$"?
| $$7777^{1} \equiv 7 \pmod{10}$$
$$7777^{2} \equiv 9 \pmod{10}$$
$$7777^{3} \equiv 3 \pmod{10}$$
$$7777^{4} \equiv 1 \pmod{10}$$
$$7777^{5} \equiv 7 \pmod{10}$$
And the relation continues, in general you see that:
$$7777^{4n + 1} \equiv 7 \pmod{10}$$
And
$$7777^{4n} \equiv 1 \pmod{10}$$
$$\frac{8888}{4} = 2222$$
Hence,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove this inequality $a \sqrt{1-b^2}+b\sqrt{1-a^2}\le1 $ Prove that for $a,b\in [-1,1]$:
$$a\sqrt{1-b^2}+b\sqrt{1-a^2}\leq 1$$
| Method 1:
HINT:
As $1-b^2\ge 0\implies -1\le b\le 1,$ let $ b=\sin B$
Similarly, $a=\sin A$
$\implies a\sqrt{1-b^2}+b\sqrt{1-a^2}=\sin A\cos B+\cos A\sin B=\sin(A+B)$
Method 2:
Let $a\sqrt{1-b^2}+b\sqrt{1-a^2}=y$
Squaring we get, $y^2=a^2(1-b^2)+b^2(1-a^2)+2ab\sqrt{(1-a^2)(1-b^2)}$
So, $1-y^2$
$=(1-a^2)(1-b^2)+(ab)^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/375260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Fractional Trigonometric Integrands
*
*$$∫\frac{a\sin x+b\cos x+c}{d\sin x+e\cos x+f}dx$$
*$$∫\frac{a\sin x+b\cos x}{c\sin x+d\cos x}dx$$
*$$∫\frac{dx}{a\sin x+\cos x}$$
What are the relations between the numerator in the denominator, and what is the general pattern to solve these type of questions?
| I did not treat this question like a homework question, since it is not tagged homework.
What are the relations between the numerator in [and] the denominator
Both are linear functions of $\sin x,\cos x$
what is the general pattern to solve these type of questions?
The universal standard substitution to evaluate ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$
Find the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$.
Please brief about the concept behind this to solve such problems. Thanks.
| Fermat's little theorem says that $a^6\equiv 1 \pmod 7$ whenever $7$ does not divide $a$. So $2^{100}+3^{100}+4^{100}+5^{100}\equiv 2^4+3^4+4^4+5^4 \pmod 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/377378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Find all matrices $A$ of order $2 \times 2$ that satisfy the equation $A^2-5A+6I = O$
Find all matrices $A$ of order $2 \times 2$ that satisfy the equation
$$
A^2-5A+6I = O
$$
My Attempt:
We can separate the $A$ term of the given equality:
$$
\begin{align}
A^2-5A+6I &= O\\
A^2-3A-2A+6I^2 &= O
\end{align}
$$
This impl... | $A^2 - 5A + 6 = 0$ is equivalent to $(A-2)(A-3) = 0$, which is equivalent to $Sp(A) \subset \{2, 3\}$.
Three cases are possible :
*
*$Sp(A) = \{2\}$, i.e. $A = 2I$
*$Sp(A) = \{3\}$, i.e. $A = 3I$
*$Sp(A) = \{2, 3\}$, i.e. $A$ is similar to $\begin{pmatrix}
2 & 0\\
0 & 3
\end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
How do evaluate an inequality that involves a fractional part? I am stuck on how to evaluate whether the following condition is true:
Let $\{k\}$ be the fractional part of a real number such that $\{k\} = k - \lfloor{k}\rfloor$.
if $\{\frac{x}{2}\} < \frac{1}{2} + \frac{\{x\}}{2}$, does it then follow that:
$$\{\frac{x... | Let $x = n + r,$ where $n = \lfloor x \rfloor$ and $r = \{x\}.$ If $n$ is odd, then $\left\{\frac{n+r}{2}\right\} = \frac{1}{2} + \frac{r}{2},$ so the inequality fails. Hence $n$ is even. Then $n + 1$ is odd, whence $\left\{\frac{n+1+r}{2}\right\} = \frac{1}{2} + \frac{r}{2} = \frac{1}{2} + \frac{\{n+1+r\}}{2},$ so in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit $\frac{\tan^{-1}x - \tan^{-1}\sqrt{3}}{x-\sqrt{3}}$ without L'Hopital's rule. Please solve this without L'Hopital's rule? $$\lim_{x\rightarrow\sqrt{3}} \frac{\tan^{-1} x - \frac{\pi}{3}}{x-\sqrt{3}}$$
All I figured out how to do is to rewrite this as $$\frac{\tan^{-1} x - \tan^{-1}\sqrt{3}}{x-\sqrt{3}}$$
Any help... | Putting $$\frac\pi3-\tan^{-1}x=\theta\implies \tan^{-1}x=\frac\pi3-\theta$$
So, $$x=\tan\left(\frac\pi3-\theta\right)\text{ and }x-\sqrt3=\tan\left(\frac\pi3-\theta\right)-\tan\frac\pi3=\frac{\sin(\frac\pi3-\theta-\frac\pi3)}{\cos\frac\pi3\cos \left(\frac\pi3-\theta\right)}=-\frac{2\sin\theta}{\cos\left(\frac\pi3-\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/382684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove by mathematical induction for any prime number$ p > 3, p^2 - 1$ is divisible by $3$? Prove by mathematical induction for any prime number $p > 3, p^2 - 1$ is divisible by $3$?
Actually the above expression is divisible by $3,4,6,8,12$ and $24$.
I have proved the divisibility by $4$ like:
$$
\begin{align}
p^2 -1 ... | If $p>3$ is a prime then $3$ does not divide $p$ i.e. $3$ and $p$ are relatively prime , now $p-1$ , $p , p+1$ are three consecutive integers , so one of them must be divisible by $3$ , hence $3$ must divide their product i.e. , $p(p-1)(p+1)=p(p^2-1)$ , but $3$ and $p$ are relatively prime , so $3$ must divide $p^2-1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/383018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating Complex Integral. I am trying to evaluate the following integrals:
$$\int\limits_{-\infty}^\infty \frac{x^2}{1+x^2+x^4}dx $$
$$\int\limits_{0}^\pi \frac{d\theta}{a\cos\theta+ b} \text{ where }0<a<b$$
My very limited text has the following substitution:
$$\int\limits_0^\infty \frac{\sin x}{x}dx = \frac{1}{2i}... | $f(z) := \frac{1}{a\cos z+b}$
Let $I$ be the integral in question. Let's double the integral, to get
$$2I =\int_0^{2\pi} f(z)\, dz$$
Let $C$ be the contour of the unit circle $|z|=1$ in the positive direction. Then using this, we have
$$2I = \oint_C \frac{1}{a\left(\frac{z+z^{-1}}{2}\right)+b}\frac{dz}{iz} = -2 i \oi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/383874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Maclaurin expansion of Sin(Sin(x)) I want to calculate the limit,
$$ \lim_{x\to 0} \frac{sin(sin x) - sin x}{x^3}$$
and doing so using Maclaurin expansion.
Now $sin x$ expands to $x -\frac{x^3}{3!}x^3 + O(x^5)$
Which would give $sin(sinx)= (x -\frac{1}{3!}x^3 + O(x^5)) -\frac{1}{3!}(x -\frac{x^3}{3!}x^3 + O(x^5))^3 + O... | You are on the right track.
Since $\sin x=x-\frac{x^3}{6}+O(x^5)$, we have
$$
\begin{align}
\sin(\sin x)&=x-\frac{x^3}{6}+O(x^5)-\frac{1}{6}(x-\frac{x^3}{6}+O(x^5))^3+O((x-\frac{x^3}
{6}+O(x^5))^5)\\
&\text{Expanding $(x-\frac{x^3}{6}+O(x^5))^3=x^3+O(x^5)$, you get}\\
&=x-\frac{x^3}{6}-\frac{1}{6}x^3+O(x^5).
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/384289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrating a school homework question. Show that $$\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \frac{a\sqrt{3}+b-\pi}{6},$$ where $a$ and $b$ are constants to be found.
Answer is: $$\frac{24\sqrt3-48-\pi}{6}$$
Thank you in advance!
| Note first that the derivative of the quadratic in the denominator is $-2x+2$. It would be great if the numerator were $4x-4$, because then we could set $u=3+2x-x^2$, $du=(-2x+2)dx$, and write the indefinite integral as
$$\int\frac{-2}{\sqrt u}du=-2\int u^{-1/2}du\;.$$
Unfortunately, the numerator is actually $4x-5=-2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/384530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Closed form for $\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}$ Please help me to find a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$
| A related problem. Here is another approach. Recalling the generating function of the harmonic numbers
$$ \sum_{n=1}^{\infty} H_n x^n = \frac{\ln(1-x)}{x-1} \implies (xD)^4\sum_{n=1}^{\infty} H_n x^n=(xD)^4 \frac{\ln(1-x)}{x-1}, $$
$$ \implies \sum_{n=1}^{\infty} H_n n^4 x^n = {\frac {x \left( 1+11\,x+11\,{x}^{2}+{x}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/385067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 1
} |
Please correct my work, finding Eigenvector Determine whether matrix A is a Diagonalizable. if it is , determine matrix P that Diagnolizes it and compute $P^{-1}AP$.
$$A=
\begin{bmatrix}
+3 & +2\\
-2 & -3\\
\end{bmatrix}
$$
$$A-\ell I =
\begin{bmatrix}
+3-\ell & +2\\
... | The eigenvalues are correct. However you go wrong in the computation of the eigenvectors:
$$
\begin{bmatrix}
3-\sqrt{5} & 2\\
-2 & -3-\sqrt{5}
\end{bmatrix}
$$
Divide the first row by $3-\sqrt{5}$, which is the same as multiplying it by $(3+\sqrt{5})/2$, getting
$$
\begin{bmatrix}
1 & \frac{3+\sqrt{5}}{2}\\
-2 & -3-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/385972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to simplify $\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta} $ How to simplify the following expression :
$$\frac{(\sec\theta -\tan\theta)^2+1}{\sec\theta \csc\theta -\tan\theta \csc \theta} $$
| Here is a detailed solution.Ready, set, go!
$$\require{cancel}\begin{align}\frac{\left(\sec\theta-\tan\theta\right)^2+1}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{\sec^2\theta-2\sec\theta\tan\theta+\color{blue}{\tan^2\theta+1}}{\sec\theta\csc\theta-\tan\theta\csc\theta}\\&=\frac{\sec^2\theta-2\sec\theta\tan\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove $\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx$ Prove that:
$(1)$$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \ dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \ dx$$
$(2)$$$\int_0^{\infty } \frac{1}{\sqrt{8 x^3+x+7}} \ dx>1$$
What I do for $(1)$ is (s... | I modified the lower bound for the integral (2) though it is still not exactly 1. Still I'm giving my bounds.
\begin{equation*}
\int_{0}^\infty \frac{1}{\sqrt{(8x^3+x+7)}}dx= \int_{0}^1 \frac{1}{\sqrt{(8x^3+x+7)}}dx+\int_{1}^\infty \frac{1}{\sqrt{(8x^3+x+7)}}dx=I_1+I_2
\end{equation*}
Now, for $x \in [0,1],\ x^2\geq x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 0
} |
Calculate the distance between the points $(1, 2, \dots, n)$ and $(2, 3, \dots, n, 1)$ I know that the operation to find the distance between two vectors is:
$$\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+...+(b_n-a_n)^2}.$$
So the distance between $(7, 5, 3, 1)$ and $(1, 3, 5, 7)$ is:
$$\sqrt{(1-7)^2+(3-5)^2+(5-3)^2+(7-1)^2} = \sqrt... | Note that
$$\underbrace{1 + 1 + \dots + 1}_{n-1} + (1 - n)^2 = (n - 1) + (n - 1)^2 = (n-1)[1 + (n - 1)] = (n-1)n$$
where the first equality used the fact that
$$(1 - n)^2 = [(-1)(n-1)]^2 = (-1)^2(n-1)^2 = (n-1)^2.$$
Therefore your calculation shows that
$$\|(2, 3, \dots, n, 1) - (1, 2, \dots, n-1, n)\| = \sqrt{n(n-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/388836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}$? How would you compute$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}\, \, ?$$
| $$\int_0^{\frac \pi 2} \frac{1}{1 + \sin^2x}dx = \int_0^{\frac \pi 2} \frac{2}{3 - \cos 2x} dx = \int_0^\pi \frac{d\theta}{3 - \cos \theta } = \frac 12\int_0^{2\pi}\frac{d\theta}{3 - \cos \theta}$$
To evaluate $\displaystyle \int_0^{2\pi}\frac{1}{3 - \cos \theta}d\theta$ let $\displaystyle z = e^{i\theta} \implies \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391537",
"timestamp": "2023-03-29T00:00:00",
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$x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$ I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.
Given the function
$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$
Prove that the taylor polynomial of degree $4$ of $f$ is equal to
$5+x^4+y^4$.
First,... | Hint: The difference left minus right is $\frac12(x^4-2x^2y^2+y^4)$
| {
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How to integrate $\int \sqrt{x^2+a^2}dx$ $a$ is a parameter. I have no idea where to start
| Integrating by parts
$$I=\int \sqrt{x^2+a^2}\cdot1dx$$
$$=\sqrt{x^2+a^2}\int dx-\int\left( \frac{d \sqrt{x^2+a^2}}{dx}\int dx\right)dx$$
$$=x\sqrt{x^2+a^2}-\int\frac{x^2 dx}{\sqrt{x^2+a^2}}$$
Now, $$\int\frac{x^2 dx}{\sqrt{x^2+a^2}}=\frac{x^2+a^2-a^2}{\sqrt{x^2+a^2}}dx=I-a^2\int\frac{dx}{\sqrt{x^2+a^2}}$$
Put $x=a\tan... | {
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If $2n+1$ and $4n+3$ are prime, then $2n-1$ and $4n+1$ are not when $n>2$ How do you prove that, for $n>2$, if $2n+1$ and $4n+3$ are prime numbers, then $2n-1$ and $4n+1$ are composite numbers?
| HINT: $3$ divides $2n(2n+1)(2n-1)$ and $3$ divides $(4n+1)(4n+2)(4n+3)$
As $n>2, 4n+1>2n+1>3$
Hence, $4n+1, 2n+1$ are not divisible by $3$
If $4n+3$ is prime, $3$ does not divide $n\implies 3$ divide $2n-1$
If $2n+1$ is prime, $3$ does not divide $2(2n+1)\implies 3$ divide $4n+1$
which can also be derived from $4n+1... | {
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"timestamp": "2023-03-29T00:00:00",
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General Solution of Diophantine equation Having the equation:
$$35x+91y = 21$$
I need to find its general solution.
I know gcf $(35,91) = 7$, so I can solve $35x+917 = 7$ to find $x = -5, y = 2$. Hence a solution to $35x+91y = 21$ is $x = -15, y = 2$.
From here, however, how do I move on to finding the set of general... | On division by $7$ we get $5x+13y=3$
Now, by observation $13-5\cdot2=3$
So, $5x+13y=3=13-5\cdot2$
$\implies 5(x+2)=13(1-y)\implies \frac{5(x+2)}{13}=1-y$ which is an integer
$\implies 13$ divides $5(x+2)$
$\implies 13$ divides $(x+2)$ as $(5,13)=1$
So, $x+2=13a$ where $a$ is any integer, $x=13a-2$
So, $13y=3-5x=3-5(13... | {
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Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$ Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?
$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$
Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell with... | This is connected to some of the other answers, but I thought it might be worth mentioning to show the utility of approximation formulas. The first two derivatives of $ \ f(x) = x^{1/2} \ $ at $ \ a = 1000 \ $ are $ \ f'(1000) = \frac{1}{2} \cdot 1000^{-1/2} \ $ and $ \ f''(1000) = -\frac{1}{4} \cdot 1000^{-3/2} \ $ ... | {
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"source": "stackexchange",
"question_score": "25",
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Help me prove this inequality : How would I go about proving this?
$$ \displaystyle\sum_{r=1}^{n} \left( 1 + \dfrac{1}{2r} \right)^{2r} \leq n \displaystyle\sum_{r=0}^{n+1} \displaystyle\binom{n+1}{r} \left( \dfrac{1}{n+1} \right)^{r}$$
Thank you! I've tried so many things. I've tried finding a series I could compare... | And here's the proof why $(1+1/x)^x$ is concave for $x\le 1$ (the case $x\ge 1$ is already proven by @TCL).
We study the second derivative with respect to $x$, it is equal to (thanks to TCL) to
$\displaystyle \left(1+ \frac 1x\right)^x\left(\left(\ln (1+1/x)-\frac{1}{1+x}\right)^2-\frac{1}{x(1+x)^2}\right)$;
First of ... | {
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Calculating $\sqrt{28\cdot 29 \cdot 30\cdot 31+1}$ Is it possible to calculate $\sqrt{28 \cdot 29 \cdot 30 \cdot 31 +1}$ without any kind of electronic aid?
I tried to factor it using equations like $(x+y)^2=x^2+2xy+y^2$ but it didn't work.
| \begin{align}
&\text{Let }x=30
\\ \\
\therefore&\ \ \ \ \ \sqrt{(x-2)(x-1)x(x+1)+1}
\\ \\
&=\sqrt{[(x-2)(x+1)[(x-1)x]+1}
\\ \\
&=\sqrt{(x^2-x-2)((x^2-x)+1}
\\ \\
&=\sqrt{(x^2-x)^2-2(x^2-x)+1}
\\ \\
&=\sqrt{(x^2-x-1)^2}
\\
&=x^2-x-1
\\
&=30^2-30-1
\\
&=\boxed{869}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this integral easily: $\int \frac{x\cdot \sqrt[3]{x+2}}{x+\sqrt[3]{x+2}} dx$ I am trying to solve this integral $$\int\frac{x\cdot \sqrt[3]{x+2}}{x+\sqrt[3]{x+2}} dx$$
I can do it by brute force (means using a substitution then long division and then substitutions again) but it's too long (suspiciously lon... | Substitute $\sqrt[3]{x+2}=u$ therefore $\frac{du}{dx} =\dfrac{1}{3\sqrt[3]{(x+2)^2}}$
Your integral will become : 3$\int\dfrac{u^3(u^3-2)}{u^3+u-2}du $
= $3\int [u^3+ \dfrac{5u-2}{4(u^2+u+2)}-u-\dfrac{1}{4(u-1)}]du$
=$3\int u^3du + \dfrac{3}{4}\int(\dfrac{5(2u+1)}{2(u^2+u+2)}-\dfrac{9}{2(u^2+u+2)})du - \dfrac{3}{4}\int... | {
"language": "en",
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"question_score": "1",
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Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$
It has two poles at $\pm i$ and branch point of $-1$ while the integral is to b... | Letting $x=\tan \theta$ yields
$$\begin{aligned}\int_{0}^{\infty} \frac{\ln (1+x)}{1+x^{2}}dx&=\int_{0}^{\frac{\pi}{2}}[\ln (\cos \theta+\sin \theta) d \theta-\ln (\cos \theta)] d \theta \\&= \underbrace{\int_{0}^{\frac{\pi}{4}}2\ln (\cos \theta+\sin \theta)}_{L} d \theta-\underbrace{\int_{0}^{\frac{\pi}{2}} \ln (\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 8
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How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ... | Consider $a^3 =35+ 18i\sqrt{3}$ and $b^3=35- 18 i\sqrt{3}$
You are supposed to find $a+b$
$(a+b)^3=a^3+b^3+3ab(a+b) \implies(a+b)^3-3ab(a+b)=a^3+b^3$
Let $a+b=x \implies x^3-3ab(x)=70$ and $ab= [(35-18i\sqrt3)(35+18i\sqrt3)]^{1/3}=13$
You get $x^3-39x-70 =0$, product of the roots is $70$ and sum of the roots is $0$. Y... | {
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"source": "stackexchange",
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How prove this $\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$ Prove that
$$\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$$
I have some question. Using this, find this integral is not converge, I'm wrong?
Thank you... | Numerical calculation shows that the graph of
$$ y = \int_{0}^{x} \sin t \sin \sqrt{t} \, dt $$
is given by
Though a graph cannot constitute a proof, it strongly suggests that $y$ cannot converge as $x \to \infty$. Indeed, we can show that
$$ y(x) = -\cos x \sin \sqrt{x} + \frac{\sqrt{\pi}}{2} \sin \left( \frac{3\pi -... | {
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Solve the equation $\sqrt{3x-2} +2-x=0$
Solve the equation: $$\sqrt{3x-2} +2-x=0$$
I squared both equations $$(\sqrt{3x-2})^2 (+2-x)^2= 0$$
I got $$3x-2 + 4 -4x + x^2$$
I then combined like terms $x^2 -1x +2$
However, that can not be right since I get a negative radicand when I use the quadratic equation.
$x = 1/2... | If you meant
$$\sqrt{3x-2}+2-x=0\implies \sqrt{3x-2}=x-2\implies3x-2=(x-2)^2=x^2-4x+4\implies$$
$$\implies x^2-7x+6=0$$
Now just check that $x^2-7x+6=(x-6)(x-1)$ ...and remember to check at the end whether both solutions of this quadratic are actually solutions of your original equation, since when squaring some mess c... | {
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"source": "stackexchange",
"question_score": "2",
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How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$ How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?
It is part of a question about finding limits at mooculus.
| Here's a harder and similar way, but it still makes sense if you don't have what Julien said memorized.
$$x^4 - 1296 = x^4 - 6^4 \Rightarrow x^4 = 6^4$$Now two roots that seem obvious are $6$ and $-6$. But $x^4 = x^4 \cdot 1 = x^4 \cdot i^4 = (xi)^4$. So two more roots are $6i$ and $-6i$.
Hence, the factorization is ... | {
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"source": "stackexchange",
"question_score": "5",
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How can I prove this closed form for $\sum_{n=1}^\infty\frac{(4n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}$ How can I prove the following conjectured identity?
$$\mathcal{S}=\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}\s... | This is a very interesting question. Since it is different from similar question of this kind in that a new technique is involved I will provide the answer. We have:
\begin{eqnarray}
&&\sum\limits_{n=1}^\infty \frac{(4n)!}{\Gamma(n+\frac{2}{3}) \Gamma(n+\frac{4}{3})(n!)^2 (-256)^n}=\\
&&
\frac{1}{(-\frac{1}{3})!}\frac... | {
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"source": "stackexchange",
"question_score": "51",
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Solving $\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2 $ How can I solve the equation $$\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2$$
I know that it has two roots: $x=1$ and $x=3$.
| Consider the functions: $f_{1},f_{2}:D\rightarrow\textbf{R}$, where $D = (-\infty,-\sqrt{\frac{2}{3}}]\cup[\sqrt{\frac{2}{3}}, \infty) $,
$f_{1}(x)={\sqrt{3X^{2}-2}}+\sqrt[3]{x^{2}-1}$ and $f_{2}(x)=3x-2$.
For $x\leq-\sqrt{\frac{2}{3}}$ $f_{1}$ is decreasing, $f_{2}$ is increasing and $f_{1}(-\sqrt{\frac{2}{3}})>f_{2}(... | {
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Show that $n \ge \sqrt{n+1}+\sqrt{n}$ (how) Can I show that:
$n \ge \sqrt{n+1}+\sqrt{n}$ ?
It should be true for all $n \ge 5$.
Tried it via induction:
*
*$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.
*$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$
Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1... | Here is another way:
Define $f(x)=x-\sqrt{x}-\sqrt{x+1}$. We need to show that $f(x)\ge 0$ for all $x\ge 5$.
Since
$$f'(x)=1-\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x+1}} \ge 0, \quad \forall x\ge 1$$
the function is increasing on $[1,\infty)$. As $f(5)\ge 0$, the result follows.
| {
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If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$. If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.
I think it is trivial because they are distinct.
So I wonder just saying "Since they are distinct" is enough to prove it?
Of course there could be several more detailed versions but I just wan... | HINT:
$$a^3+b^3+c^3-3abc$$
$$=(a+b)^3-3ab(a+b)+c^3-3abc=(a+b)^3+c^3-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2\}-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2-3ab\}$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
$$=(a+b+c)\frac{\{(a-b)^2+(b-c)^2+(c-a)^2\}}2$$
which will be $>=<0 $ according as $a+b+c>=<0$ as for real distinct $a,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Does $\int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx $ converge? Does $ \displaystyle \int_{0}^{\infty} \ \frac{\sin (\tan x)}{x} dx $ converge?
$ \displaystyle \int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin (\tan x)}{x} \ dx + \sum_{n=1}^{\infty} \int_{\pi(n-\frac{1}{2})}^{\pi(n+\... | Yes.
Note that
$$\begin{align} I_n:=\int_{\pi(n-\frac12)}^{\pi(n+\frac12)}\frac{\sin(\tan(x))}{x}\,\mathrm dx&=\int_0^{\frac\pi2}\left(\frac1{n\pi+x}-\frac1{n\pi-x}\right)\sin(\tan(x))\,\mathrm dx\\&=\int_0^{\frac\pi2}\frac{-2x}{n^2\pi^2-x^2}\sin(\tan(x))\,\mathrm dx\\
\end{align}$$
With $A:=\int_0^{\frac\pi2}\max\{-2... | {
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Is it possible to simplify $\frac{\Gamma\left(\frac{1}{10}\right)}{\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$? Is it possible to simplify this expression?
$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$$
Is the... | Amazingly, this can be greatly simplified. I'll state the result first:
$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\Gamma\left(\frac{7}{15}\right)} = \frac{\sqrt{5}+1}{3^{1/10} 2^{6/5} \sqrt{\pi}}$$
The result follows first from a version of Gauss's multiplication... | {
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Last digits of a power of 2
Prove that there exists a power of 2 such that the last 1000 digits in its decimal representation are all 1 and 2.
One fact that I think can be used in this problem: if $2^{n}=\cdots dn$ where $d$ is the digit to the left of $n$, then $2^{dn}=\cdots dn$ (A concept that was used in MMO 1978... | Let $n = 1000$. Since $\Bbb Z/10^n \Bbb Z = (\Bbb Z/2^n \Bbb Z)\times(\Bbb Z/5^n \Bbb Z)$, a first step is to inquire about the possible values of powers of $2$ modulo $2^n$ and $5^n$.
The first one is easy : If $k\ge n$, then $2^k \equiv 0 \pmod {2^n}$. And with enough luck, we may find a solution with $k \ge 1000$ so... | {
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"url": "https://math.stackexchange.com/questions/406496",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Prove inequality: $74 - 37\sqrt 2 \le a+b+6(c+d) \le 74 +37\sqrt 2$ without calculus
Let $a,b,c,d \in \mathbb R$ such that $a^2 + b^2 + 1 = 2(a+b),
c^2 + d^2 + 6^2 = 12(c+d)$, prove inequality without calculus (or
langrange multiplier): $$74 - 37\sqrt 2 \le a+b+6(c+d) \le 74
+37\sqrt 2$$
The original problem is ... | HINT:
So, $(a-1)^2+(b-1)^2=1^2$ and we can set $a=1+\cos A,b=1+\sin A$
So, $a+b=2+(\cos A+\sin A)=2+\sqrt2\cos\left(A-\frac\pi4\right)$
As $-1\le \cos\left(A-\frac\pi4\right)\le 1, 2-\sqrt2\le a+b\le 2+\sqrt2 $
Similarly, $(c-6)^2+(d-6)^2=6^2$ and we can set $c=6+6\cos B,d=6+6\sin B$
| {
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Evaluate : $\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$ Evaluate: $$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$$
First approach :
$$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4(1-\cos^2x)}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{\cos^2xdx}{4 - 3\cos^2x}$$
$$=\int^{\frac{\pi}{2}}_0 \f... | Use the fact that
$$3 \int_0^{\pi/2} dx \frac{\cos^2{x}}{\cos^2{x}+4 \sin^2{x}} = 3 \int_0^{\pi/2} dx \frac{\cos^2{x}}{4-3 \cos^2{x}} = 4 \int_0^{\pi/2} \frac{dx}{4-3 \cos^2{x}} - \int_0^{\pi/2} dx$$
So consider the first integral on the RHS:
$$4 \int_0^{\pi/2} \frac{dx}{4-3 \cos^2{x}} = 2 \int_0^{2 \pi} \frac{dy}{5-3 ... | {
"language": "en",
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Prove $x^2+y^2+z^2 \ge 14$ with constraints Let $0<x\le y \le z,\ z\ge 3,\ y+z \ge 5,\ x+y+z = 6.$ Prove the inequalities:
$I)\ x^2 + y^2 + z^2 \ge 14$
$II)\ \sqrt x + \sqrt y + \sqrt z \le 1 + \sqrt 2 + \sqrt 3$
My teacher said the method that can solve problem I can be use to solve problem II. But I don't know what m... | I) We have $2x+y=12-(y+z)-z \leq 12-5-3=4$. Thus by AM-GM inequality $2xy \leq (\frac{2x+y}{2})^2 \leq 4$. Finally by QM-AM inequality $$x^2+y^2+z^2=(x+y)^2+z^2-2xy \geq 2(\frac{(x+y)+z}{2})^2-2xy=18-2xy \geq 14$$
II) Let $s=\sqrt{x}+\sqrt{y}+\sqrt{z}$. Then $$s^2=x+y+z+2(\sqrt{xy}+\sqrt{xz}+\sqrt{yz})=6+2(\sqrt{xy}+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/410845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding the root of a degree $5$ polynomial $\textbf{Question}$: which of the following $\textbf{cannot}$ be a root of a polynomial in $x$ of the form $9x^5+ax^3+b$, where $a$ and $b$ are integers?
A) $-9$
B) $-5$
C) $\dfrac{1}{4}$
D) $\dfrac{1}{3}$
E) $9$
I thought about this question for a bit now and can anyone p... | If you want to try process of elimination, an easy place to start is by assuming that either $a = 0$ or $b = 0$. (Hey, zero is an integer!)
If you set $a = 0$, then you get $b = -9x^5$. Because $\mathbb{Z}$ is closed under multiplication, if $x$ is an integer, then so is $b$, and so you have a valid $(a, b)$ pair wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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If $a, b$ are positive integers, does $\;b\mid(a^2 + 1)\implies b\mid (a^4 + 1)\quad?$ If $a, b$ are positive integers, does $\;b\mid(a^2 + 1)\implies b\mid (a^4 + 1)\;$?
Explain if this is true or not. If no, give a counterexample.
| Suppose $b = 5, a = 3$:
$$5\mid (3^2 + 1)\;\;\text{but}\;\;5 \not\mid (3^4 + 1)$$
Or, simply choose $a = 2, b= 5$ and again, $$5\mid (2^2 + 1)\;\;\text{but}\;\;5 \not\mid (2^4 + 1)$$
What is true is that $b\mid (a^2 + 1) \implies b\mid(a^2 + 1)^2$,
but note that $$\begin{align}\;(a^2 + 1)^2 &= (a^4 + 2a^2 + 1) \\ \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac... | Another approach for evaluating the second integral using contour integration that avoids having to deform the contour around branch cuts is to consider $$ \displaystyle f(z) = \frac{\log(z+ e^{i \pi /4})}{(1+z^{2})^{2}}$$ and integrate around a contour that consists of the line segment $[-R,R]$ and the upper half of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 6,
"answer_id": 1
} |
Show that $\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z} \ge 8$.
If $x,y,z$ are positive proper fractions satisfying $x+y+z=2$, prove that $$\dfrac{x}{1-x}\cdot\dfrac{y}{1-y}\cdot\dfrac{z}{1-z}\ge 8$$
Applying $GM \ge HM$, I get $$\left[\dfrac{x}{1-x}\cdot\dfrac{y}{1-y}\cdot\dfrac{z}{1-z}\right]^{1/3}\ge \dfrac{... | Two proofs of the inequality have been posted. The following is simply a comment. In the attempted solution of the OP, the inequality
$$\left(\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot \frac{z}{1-z}\right)^{1/3}\ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3}\tag{1}$$
has been proved.
One can complete things by showing, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/424529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Show that $(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z)$.
If $x>0,y>0,z>0$ and $x+y+z=1$, prove that $$(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z).$$
Trial: Here $$(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z) \\ \implies (1+x)(1+y)(1+z)\ge 8(y+z)(x+z)(x+y)$$ I am unable to solve the problem. Please help.
| since
$$t=(x+y)(y+z)(x+z)\le\left(\dfrac{2(x+y+z)}{3}\right)^3=\dfrac{8}{27}$$
and we
have
$$(1-x)(1-y)\ge 0$$
then
$$(1+x)(1+y)\ge2(x+y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. Can anyone give me a hint to evaluate this integral?
$$\int_0^\infty \frac{dx}{1+x^4}$$
I know it will involve the gamma function, but how?
| HINT:
Putting $x=\frac1y,dx=-\frac{dy}{y^2}$
$$I=\int_0^\infty\frac{dx}{1+x^4}=\int_\infty^0\frac{-dy}{y^2\left(1+\frac1{y^4}\right)}$$
$$=-\int_\infty^0\frac{y^2dy}{1+y^4}=\int_0^\infty\frac{y^2dy}{1+y^4} \text{ as } \int_a^bf(x)dx=-\int_b^af(x)dx$$
$$I=\int_0^\infty\frac{y^2dy}{1+y^4}=\int_0^\infty\frac{x^2dx}{1+x^4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 6
} |
Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$ We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$
I have tried and it gets confusing.
| multiply numerator and denominator by $(1-\sin\theta)$.
$$\dfrac {\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}\times \dfrac{1-\sin\theta}{1-\sin\theta}$$
$$\dfrac {\sin\theta-\sin^2\theta-\cos\theta+\sin\theta\cdot\cos\theta+1-\sin\theta}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$
$$\dfrac {1-\sin^2\theta-\cos\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Lagrange multiplier method, find maximum of $e^{-x}\cdot (x^2-3)\cdot (y^2-3)$ on a circle I attempted to design an exercise for my engineer students and couldn't solve it myself. Maybe here are some experts in calculus who have some better tricks than I do:
The exercise would be to find the maxima of $e^{-x}(x^2-3)(y... | Let us parametrize the circumfererence:$x=2\cos(\phi),y=1+2\sin(\phi)$. Then the function
$$f(\phi):=e^{-2\cos(\phi)}(4\cos^2(\phi)-3)(4\sin^2(\phi)+4\sin(\phi)-2)$$ should be maximized on $[0,2\pi]$.
The equation $f'(\phi)=0$ is equivalent to
$$
32 \sin^3 ( \phi ) \cos ^2(
\phi ) -32\cos ( \phi ) \sin^3
\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/429939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Is there an inverse to Stirling's approximation? The factorial function cannot have an inverse, $0!$ and $1!$ having the same value. However, Stirling's approximation of the factorial $x! \sim x^xe^{-x}\sqrt{2\pi x}$ does not have this problem, and could provide a ballpark inverse to the factorial function. But can thi... | It is better to start with
$$
x! \approx \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right)^{x + \frac{1}{2}} \sqrt {2\pi } .
$$
Setting
$$
y = \left( {\frac{{x + \frac{1}{2}}}{\mathrm{e}}} \right)^{x + \frac{1}{2}} \sqrt {2\pi } ,
$$
we find
$$
\frac{1}{\mathrm{e}}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/430167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Calculating 7^7^7^7^7^7^7 mod 100 What is
$$\large 7^{7^{7^{7^{7^{7^7}}}}} \pmod{100}$$
I'm not much of a number theorist and I saw this mentioned on the internet somewhere. Should be doable by hand.
| $7^4 = 2401 \equiv 1 \pmod{100}$, so you only need to calculate $7^{7^{7^{7^{7^7}}}} \pmod{4}$. We know that $7 \equiv -1 \pmod 4$ and $7^{7^{7^{7^7}}}$ is odd, so $7^{7^{7^{7^{7^7}}}} \equiv -1 \equiv 3 \pmod 4$, and then $$7^{7^{7^{7^{7^{7^7}}}}} \equiv 7^3 \equiv 43 \pmod {100}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ Problem statement:
Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $.
, $n\in \mathbb{N}$
My progress
LHS is e... | About the original attempt:
Your first "rewrite" (immediately above (1))is incorrect:
$$\frac{1}{\sqrt{3n+1}} \times \frac{2(n+1)-1}{2(n+1)} \leq \frac{1}{\sqrt{3(n+1)+1}}$$
is NOT equivalent to, nor does it entail the rewrite I referenced:
$$\underbrace{\frac{\sqrt{3n+1}}{3n+1}}_{\large =\,\frac 1{\sqrt{3n+1}}} \times... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability Question about Tennis Games! $2^{n}$ players enter a single elimination tennis tournament. You can assume that the players are of equal ability.
Find the probability that two particular players meet each other in the tournament.
I could't make a serious attempt on the question, hope you can excuse me this ... | First off, an even number of players is a necessary but not sufficient condition for a full balanced tournament. Fir this you need $P=2^R$ where $R$ is the number of rounds (including finals, semis etc). I will assume this is what you are after.
Without loss of generality, let Player 1 occupy the first slot in the draw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/433430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the determinant by using elementary row operations I'm having a problem finding the determinant of the following matrix using elementary row operations. I know the determinant is -15 but confused on how to do it using the elementary row operations.
Here is the matrix
$$\begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 ... | Note that the determinant of a lower (or upper) triangular matrix is the product of its diagonal elements. Using this fact, we want to create a triangular matrix out of your matrix
\begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix}
So, I will start with the last row and subtract it from the sec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/433870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Finding Potential with d dimensions terms Lagrangian for a spherically-symmetric, real scalar field in
d spatial dimensions,
$$L=c_d \int r^{d-1}dr\left[ \frac{1}{2} \dot\phi^2 - \frac{1}{2} \left(\frac{\partial \phi}{\partial r} \right)^2 -V(\phi)\right] \tag{1}$$
where $$v= m^2\phi^2$$, $$c_d = 2π^{d/2} /Γ(d/2)$$ i... | I think I have understood what you want to achieve. You want to obtain a Lagrangian for $A(t)$ with a suitable potential. Using $\int\limits_{0}^{\infty}r^{d-1}e^{-ar^{2}} dr=\frac{1}{2} a^{-\frac{d}{2}}\Gamma\left( \frac{d}{2} \right)$ we find by inserting $\phi(r,t)=A(t)e^{-\frac{r^{2}}{R^{2}}}$ that
$$ \begin{aligne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
simple limit but I forget how to prove it I have to calculate the following limit
$$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - x$$
it is in un undeterminated form.
I tried to rewrite it as follows:
$$\lim_{x\rightarrow -\infty} \sqrt{x^2+2x+2} - \sqrt{|x|^2}$$
but seems a dead road.
Can anyone suggest a solution?
... | Assuming you meant $\sqrt{x^2+2x+2} + x$ (as $\sqrt{x^2+2x+2} - x \to +\infty$ when $x\to-\infty$):
Another option would be to use asymptotics and known Taylor expansions (at $0$): for $x\to-\infty$,
$$
\begin{align*}
\sqrt{x^2+2x+2} + x &= |x|\sqrt{1+\frac{2}{x}+\frac{2}{x^2}} - |x| \\
&= |x|\left( 1+\frac{1}{2}\cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
The golden ratio and a right triangle Assume the square of the hypotenuse of a right triangle is equal to its perimeter and one of its legs is $1$ plus its inradius(the radius inside the circle inscribed inside the triangle.) Find an expression for the hypotenuse $c$ in terms of the golden ratio.
| Draw a picture. Let $r$ be the radius of the incircle, and let the legs be $r+x$ and $r+y$. We are told that one of the legs, say $r+x$, is equal to $r+1$, so $x=1$. Thus the hypotenuse is $1+y$.
The condition that the square of the hypotenuse is equal to the perimeter says that
$(1+y)^2=2r+2y+2$, which simplifes to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Partial fraction integration $\int \frac{dx}{(x-1)^2 (x-2)^2}$ $$\int \frac{dx}{(x-1)^2 (x-2)^2} = \int \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}\,dx$$
I use the cover up method to find that B = 1 and so is C. From here I know that the cover up method won't really work and I have to plug in values... | We can use the following method
$$\frac1{(x-1)^2(x-2)^2}=\frac{\{(x-1)-(x-2)\}^2}{(x-1)^2(x-2)^2}=\frac1{(x-2)^2}+\frac1{(x-1)^2}-2\frac1{(x-1)(x-2)}$$
$$\frac1{(x-1)(x-2)}=\frac{(x-1)-(x-2)}{(x-1)(x-2)}=\frac1{x-2}-\frac1{x-1}$$
Alternatively,
Put $x-2=y$ to ease of calculation
$$1=A(y+1)y^2+By^2+C(y+1)^2y+D(y+1)^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/436855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Show that $4x^2+3$ has at least a prime divisor of the form $12n+7$ If $x$ is not divisible by $3$, how to prove that $4x^2+3$ has at least a prime divisor of the form $12n+7$?
Thanks.
| *
*Fact $0$: A number of the form $4x^2 + 3$ is odd, hence all its prime divisors must necessarily be odd.
*Fact $1$: If $x$ is not a multiple of $3$, then $x^2 \equiv 1 \pmod{3}$, and therefore $4x^2 + 3$ is not a multiple of $3$.
*Fact $2$: If $p$ is a prime dividing a number of the form $4x^2 + 3$ where $x$ is no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Trigonometry Equations. Solve for $0 \leq X \leq 360$, giving solutions correct to the nearest minute where necessary,
a) $\cos^2 A -8\sin A \cos A +3=0$
Can someone please explain how to solve this, ive tried myself and no luck. Thanks!
| $$\frac{1+\cos2A}2-4(\sin2A)+3=0$$
$$\implies \cos2A-8\sin2A+7=0$$
Putting $1=r\cos B,8=r\sin B $ where $r>0$
Squaring we get $r^2=8^2+1^2=65\implies r=\sqrt{65}$ and $\cos B=\frac1{\sqrt{65}}$
So, $\cos2A-8\sin2A=r(\cos 2A\cos B-\sin2A\sin B)=\sqrt{65}\cos(2A+\arccos \frac1{\sqrt{65}})$
$\implies \cos(2A+\arccos \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Find volume of the body $V = \{ z = \sqrt{6-x^2-y^2}, z = x^2 + y^2 \}$
Find volume of the body $V = \{ z = \sqrt{6-x^2-y^2}, z = x^2 + y^2 \}$
Now what I said is:
$$V = \iint_{D} {\sqrt{6-x^2-y^2} - x^2 - y^2 dxdy}$$.
But when I wanted to get what $D$ is, I intersected the two $z$ functions to get $\sqrt{6-x^2-y^2}... | An idea: the intersection of
$$z=x^2+y^2\;,\;\;z=\sqrt{6-x^2-y^2}\implies z=\sqrt{6-z}\implies 0=z^2+z-6=(z+3)(z-2)$$
Since $\,z=-3\,$ is absurd (why?), we then have $\;2=z=x^2+y^2\;$ and you can now get easily your limits in the $\,xy-plane\;$ by means of this canonical circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Why does this polynomial related to Hamming Codes have integer coefficients? Question: Why does
$$
\frac{(1 + x)^{2^k - 1} - (1 - x^2)^{2^{k-1} - 1}(1-x)}{2^k}
$$
have integer coefficients?
Details: For a question I'm thinking about, I needed to know all the real numbers $c$ such that the generating function
$$
p(x) = ... | Note that
$$
\frac{(1+x)^{2^{k+1}-1}-(1-x^2)^{2^k-1}(1-x)}{2^{k+1}}
=(1+x)^{2^k-1}\frac{(1+x)^{2^k}-(1-x)^{2^k}}{2^{k+1}}
$$
$\frac{(1+x)^{2^k}-(1-x)^{2^k}}{2}$ is the odd part of $(1+x)^{2^k}$. Therefore, we just need to show that $\left.2^k\,\middle|\,\binom{2^k}{j}\right.$ when $j$ is odd. The number of factors of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/441881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$? $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$
This is a competition problem which I got from the book "Art of Problem Solving Volume 2". I'm not sure how to solve it because there... | Big hint of the day:
*
*Always try to simplify the expression first, before doing anything else
*What's the value of $\sin 30^\circ$?
*You can simplify $\sin 100^\circ$ by noticing that $\sin(90^\circ + \alpha) = \cos \alpha$.
Everything should be easy from here. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/442209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Proof that $ \lim_{x \to \infty} x \cdot \log(\frac{x+1}{x+10})$ is $-9$ Given this limit:
$$ \lim_{x \to \infty} x \cdot \log\left(\frac{x+1}{x+10}\right) $$
I may use this trick:
$$ \frac{x+1}{x+1} = \frac{x+1}{x} \cdot \frac{x}{x+10} $$
So I will have:
$$ \lim_{x \to \infty} x \cdot \left(\log\left(\frac{x+1}{... | By the Lagrange mean value theorem, for $0<a\le b$, we have
$$
\frac{b-a}{b}\le\log(b)-\log(a)\le\frac{b-a}{a}
$$
So,
$$
L(x)=x\log\left(\frac{x+1}{x+10}\right)=-x\log\left(\frac{x+10}{x+1}\right)=
-x(\log(x+10)-\log(x+1))\\
\implies -9\frac{x}{x+10}\le L(x)\le -9\frac{x}{x+1}
$$
which shows $L(x)\stackrel{x \to \infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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$x \sin x=2$ why is my proof that there no solutions wrong? $\frac 12 x \sin x=1$ . Let's look at a right triangle with base $x$ and altitude $\sin x$ . Then our equation is for the area of this triangle. Let the sides of the triangle be $a=x$ , $b=\sqrt {x^2+sin^2 x}$ , and $c= \sin x$ . According to wikipedia, Heron'... | This particular wikipedia formula is wrong.
It should be correctly either
$$A=\large \frac { \sqrt {4a^2b^2-(a^2+b^2-c^2)^2}}{4}$$
or
$$A=\large \frac { \sqrt {4a^2c^2-(a^2-b^2+c^2)^2}}{4}\,.$$
Mind the symmetry..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/442404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the coefficient of $x^{20}$ in $(x^{1}+⋯+x^{6} )^{10}$ I'm trying to find the coefficient of $x^{20}$ in
$$(x^{1}+⋯+x^{6} )^{10}$$
So I did this :
$$\frac {1-x^{m+1}} {1-x} = 1+x+x^2+⋯+x^{m}$$
$$(x^1+⋯+x^6 )=x(1+x+⋯+x^5 ) = \frac {x(1-x^6 )} {1-x} = \frac {x-x^7} {1-x}$$
$$(x^1+⋯+x^6 )^{10} =\left(\dfrac {x-x^7}... | The coefficient of $x^{10}$ in $(1+ x + \ldots + x^5)^{10}$ is equal to the number of integers $0 \leq x_i \leq 5 $ such that $\sum_{i=1}^{10} x_i= 10$.
We apply the Principle of Inclusion and exclusion, to deal with the restriction of $x_i \leq 5$.
If the only restriction is $0 \leq x_i$ then there are ${10 + 9 \choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Rationalizing a numerator I'm having trouble rationalizing a numerator with radicals. After multiplying the conjugate I get 0. Does anyone know where I went wrong?
\begin{align}
\frac{\sqrt{2+y} + \sqrt{2 - y}}{y} & = \left(\frac{\sqrt{2+y} + \sqrt{2 - y}}{y}\right) \left(\frac{\sqrt{2+y} - \sqrt{2 - y}}{\sqrt{2+y} - \... | Careful:
$$(a-b)(a+b)=a^2-b^2\implies $$
$$\left(\sqrt{2+y}+\sqrt{2-y}\right)\left(\sqrt{2+y}-\sqrt{2-y}\right)=(2+y)-(2-y)=2+y-2+y=2y$$
you forgot to change the second$\,-y\,$ into $\,+y\,$ ....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/444039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$
Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$
I have got as far as showing that:
$\cos(x+y)\cos(x-y) = \cos^2x\cos^2y -\sin^2x\sin^2y$
and
$\sin(x+y)\sin(x-y) = \sin^2x\cos^2y - \cos^2x\sin^2y$
I get stuck at showing:
$\cos^2x\co... | Looking at
$\cos^2x\cos^2y -\sin^2x\sin^2y - \sin^2x\cos^2y + \cos^2x\sin^2y$ (The last is a negative times a negative that forms a positive I believe.
You could pull $\sin^2x$ from the middle two terms to get this expression:
$ -\sin^2x\sin^2y - \sin^2x\cos^2y
= -\sin^2x(\sin^2y+\cos^2y) = -\sin^2x $
There is a sim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/444407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int\cos\theta~e^{−ia\cos\theta}~\mathrm{d}\theta$ Is anybody able to solve this indefinite integral :
$$
\int\cos\theta~e^{\large −ia\cos\theta}~\mathrm{d}\theta
$$
The letter $i$ denotes the Imaginary unit;
$a$ is a constant;
Mathematica doesn't give any result.
Thanks for any help you would like to pro... | $\int\cos\theta~e^{−ia\cos\theta}~d\theta$
$=\int\cos\theta\cos(a\cos\theta)~d\theta-i\int\cos\theta\sin(a\cos\theta)~d\theta$
$=\int\cos\theta\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n}\cos^{2n}\theta}{(2n)!}d\theta-i\int\cos\theta\sum\limits_{n=0}^\infty\dfrac{(-1)^na^{2n+1}\cos^{2n+1}\theta}{(2n+1)!}~d\theta$
$=\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/445011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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convert ceil to floor Mathematically, why is this true?
$$\left\lceil\frac{a}{b}\right\rceil= \left\lfloor\frac{a+b-1}{b}\right\rfloor$$
Assume $a$ and $b$ are positive integers.
Is this also true if $a$ and $b$ are real numbers?
| $$\left\lceil \frac { a }{ b } \right\rceil =\frac { a }{ b } +1-\left( \frac { a }{ b } \mod\ 1 \right) ,\quad \quad (1)$$
$$\left\lfloor \frac { a+b-1 }{ b } \right\rfloor =\frac { a+b-1 }{ b } -\left( \frac { a+b-1 }{ b } \mod\ 1 \right) =\frac { a }{ b } +1-\frac { 1 }{ b } -\left( \frac { a }{ b } -\frac { 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Proving : $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4(a+b+c)$ For $a,b,c > 0$ and $ab+bc+ca+2abc=1$, how to prove that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4(a+b+c) \, ?$$
| The condition is equivalent to $$\frac{a}{a + 1} + \frac{b}{b + 1} + \frac{c}{c + 1} = 1$$ then take $x = \frac{a}{a + 1}$, $y = \frac{b}{b + 1}$, $z = \frac{c}{c + 1}$ and the condition becomes $$x + y + z = 1$$ So we have $a = \frac{x}{1 - x} = \frac{x}{y + z}$ and similarly (by cyclic permutations) for other variabl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$5^m = 2 + 3^n$ help what to do how to solve this for natural numbers $5^m = 2 + 3^n$ i did this
$5^m = 2 + 3^n \Rightarrow 5^m \equiv 2 \pmod 3 \Rightarrow m \equiv 1 \pmod 2$
now if i put it like this
$ 5^{2k+1} = 2 + 3^n $
what to do ??
now is this right another try :
$ m = n \Rightarrow 5^n - 3^n = 2 = 5 -... | Assume there is a solution with $n>1$. Then $ 5^m \equiv 2 \pmod 9 \Rightarrow m \equiv 5 \pmod 6$.
Then $ 5^m \equiv 3 \pmod 7 \Rightarrow 3^n \equiv 1 \pmod 7 \Rightarrow n \equiv 0 \pmod 3.$
Also, $ 5^m \equiv 5 \text { or } 8 \pmod {13} \Rightarrow 3^n \equiv 3\text { or } 6 \pmod {13} \Rightarrow n \equiv 1 \pmod ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/450365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Power series for $(1+x^3)^{-4}$ I am trying to find the power series for the sum $(1+x^3)^{-4}$ but I am not sure how to find it. Here is some work:
$$(1+x^3)^{-4} = \frac{1}{(1+x^3)^{4}} = \left(\frac{1}{1+x^3}\right)^4 = \left(\left(\frac{1}{1+x}\right)\left(\frac{1}{x^2-x+1}\right)\right)^4$$
I can now use
$$\frac... | \begin{gather}(1+x^3)^{-4}=(1+t)^\alpha=1+\alpha t+\frac{\alpha(\alpha-1)}{2!}t^2+\dots=\\ =1+\frac{(-4)}{1!}x^3+\frac{(-4)(-4-1)}{2!}x^6+\dots
=\sum_{k=0}^\infty(-1)^{k}\frac{(3+k)!}{3!k!}x^{3k}\end{gather}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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for every integer $n \ge 1$ one has the equality Could any one help me?
For every integer $n \ge 1$ one has the equality:
$$ 1-{1\over 2}+ {1\over 3}-{1\over 4}+\dots+{1\over 2n-1}-{1\over 2n}={1\over n+1}+{1\over n+2}+\dots +{1\over 2n}.$$
Should I proceed by Induction?
| Here's another way to do the problem,$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{2n}\\=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right)\\=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Double Integral Over Region Common to Two Circles Evaluate $\iint\frac{(x^2+y^2)^2}{x^2y^2} dx dy$ over the region common to the circles $x^2+y^2=7x$ and $x^2+y^2=11y$.
| You may use the Cylindrical coordinates to find the value of the integrals. This, as @Ron's Caesarian's approach, will get a bit messy but it works as well. We have two circles on $z=0$ intersected each other, so $$7r\cos\theta=7x=x^2+y^2=11y=11r\sin\theta\longrightarrow\theta_{0}=\tan^{-1}\left(\frac{7}{11}\right)\con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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show that $\int_{-\infty}^{\infty} \frac {(\sin x) (x^2+a^2)}{x(x^2+b^2)}dx=\frac{\pi(a^2+e^{-b}(b^2-a^2))}{b^2}$ show that $$\int_{-\infty}^{\infty} \frac {(\sin x) (x^2+a^2)}{x(x^2+b^2)}dx=\frac{\pi(a^2+e^{-b}(b^2-a^2))}{b^2}$$
for every $a,b>0$
thanks for all
| Consider the contour integral
$$\oint_C dz \frac{(z^2+a^2) e^{i z}}{z (z^2+b^2)}$$
where $C$ is the semicircular contour of radius $R$ in the upper half-plane, with an additional semicircular contour of radius $\epsilon$ centered at the origin, jutting into the upper half-plane.
The contour integral is equal to
$$\i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$24\mid n(n^{2}-1)(3n+2)$ for all $n$ natural problems in the statement. "Prove that for every $ n $ natural, $24\mid n(n^2-1)(3n+2)$"
Resolution:
$$24\mid n(n^2-1)(3n+2)$$if$$3\cdot8\mid n(n^2-1)(3n+2)$$since$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)\Rightarrow3\mid n(n^{2}-1)(3n+2)$$and$$8\mid n(n^{2}-1)(3n+2)?$$$$$$
Would n... | To prove $8|n(n^2-1)(3n+2)$ you can simply break the problem in three cases:
Case 1 $n$ odd.
Then $n^2-1=(n-1)(n+1)$ is the product of two consecutive even numbers. Thus one must be divisible by $4$ and the other by $2$.
Case 2 $n$ multiple of 4.
Then $4|n$ and $2|3n+2$.
Case 3 $n$ even but not divisible by 4.
Then $2|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$
Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$
I have tried $\sin^3\alpha(3\sin\alpha - 4 \sin^3\alpha) = 3\sin^4\alpha - 4\sin^6\alpha$ and $\cos^3\alpha(4\cos^3\alpha - 3\cos\alpha) = 4\cos^6\alpha - 3\c... | Let $\cos^2\alpha=a,\sin^2\alpha=b\implies a+b=1, a-b=\cos2\alpha$
$ \sin3\alpha\sin^3\alpha + \cos3\alpha\cos^3\alpha = $
$ = 3\sin^4\alpha - 4\sin^6\alpha + 4\cos^6\alpha - 3\cos^4\alpha = $
$ = 3b^2-4b^3+4a^3-3a^2=4(a^3-b^3)-3(a^2-b^2)=(a-b)\{4(a^2+ab+b^2)-3(a+b)\} $
Now, $4(a^2+ab+b^2)-3(a+b)=4\{(a+b)^2-ab\}-3=4(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Recurrence relation for $n$ numbers in which no 3 consecutive digits are the same. I am stuck on trying to find (and solve) a recurrence relation to find all n-digit numbers in which no 3 consecutive digits are the same. These numbers are in decimal expansion.
Now I first imagine trying this for finding the number o... | We can capture the essence of this problem by introducing two sequences, namely the sequence $a_n$ that counts the number of $n$-digit numbers that do not end in a repeated digit, starting with $a_2 = 9\times 10 -9 = 81$ and the number $b_n$ of $n$-digit numbers that end in two repeated digits, starting with $b_2 = 9.$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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find the inverse of $x^2 + x + 1$ In $\mathbb{F}_2[x]$ modulo $x^4 + x + 1$
find the inverse of $x^2 + x + 1$
not 100% sure but here what i have:
user euclid algorithm:
$x^4 + x + 1 = (x^3 + 1)(x + 1) + x$
$(x^3 + 1) = x * x * x + 1$
$1 = (x^3 + 1) - x * x * x $
| Note that:
$(x^{2}+x+1)(x^{2}+x+1)=x^{2}+x$
$(x^{2}+x)(x^{2}+x+1)=x^{4}+x=1$. To get this I used that $x^{4}+x+1=0$ and that the coefficients are in $\mathbb{F}_{2}$.
So the inverse is $(x^{2}+x+1)^{2}=x^{2}+x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Find $\int \sin(2x)\sqrt{1+3\cos^{2}(x)}\ dx$ Is it humanly possible to integrate the following equation?
$$\int \sin(2x)\sqrt{1+3\cos^{2}(x)}\ dx$$
| Let $u=\cos^2(x)$, therefore $du=-2\cos(x)\sin(x)\ dx$:
$$\int\sin(2x)\sqrt{1+3\cos^2(x)}\ dx=-\int\sqrt{1+3u}\ du$$
Now, let $v=1+3u$ and $dv=3\ du$:
\begin{align*}
-\int\sqrt{1+3u}\ du&=-\frac13\int\sqrt v\ dv\\
&=-\frac{2v^{3/2}}9+C,\text{ where $C$ is a constant}\\
&=-\frac29(1+3u)^{\frac32}+C\\
&=\boxed{-\frac29(1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Approximation to the Lambert W function If:
$$x = y + \log(y) -a$$
Then the solution for $y$ using the Lambert W function is:
$$y(x) = W(e^{a+x})$$
In a paper I'm reading, I saw an approximation to this solution, due to "Borsch and Supan"(?):
$$y(x) = W(e^{a+x}) \approx x\left(1 - \frac{\log x - a}{1+x}\right)$$
Any id... | We can use a procedure known as "bootstrapping" to determine an approximation for the Lambert $W$ function. Let's go back to its definition.
For $x > 0$ the equation
$$
we^w = x
$$
has exactly one positive solution $w = W(x)$ which increases with $x$. Note that $(w,x) = (1,e)$ is one such solution, so if $x > e$ then... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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"answer_id": 0
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Find the sum : $\frac{1}{\cos0^\circ\cos1^\circ}+\frac{1}{\cos1^\circ \cos2^\circ} +\frac{1}{\cos2^\circ \cos3^\circ}+......+$ Find the sum of the following :
(i) $$\frac{1}{\cos0^\circ \cos1^\circ}+\frac{1}{\cos1^\circ\cos2^\circ} +\frac{1}{\cos2^\circ \cos3^\circ}+......+\frac{1}{\cos88^\circ \cos89^\circ}$$
I tried... | HINT:
$$\frac{\sin(A-B)}{\cos A\cos B}=\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}=\tan A-\tan B$$
If $A= (n+1)^\circ,B=n^\circ$
$$\frac{\sin 1^\circ}{\cos (n+1)^\circ\cos n^\circ}=\tan(n+1)^\circ-\tan n^\circ $$
Put $n=0,1,2,\cdots,87,88$ and add to find the series to be Telescopic
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Please help on this Probability problem
A bag contains 5 red marbles and 7 green marbles. Two marbles are drawn randomly one at a time, and without replacement. Find the probability of picking a red and a green, without order.
This is how I attempted the question: I first go $P(\text{Red})= 5/12$ and $P(\text{Green})... | Your method is correct. For more complicated problems of the same general kind, one might take a slightly different approach.
Imagine that the $12$ marbles are distinct, they have different driver license numbers. There are $\binom{12}{2}$ equally likely ways to choose $2$ marbles from the $12$.
There are $\binom{5}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Expected number of people sitting in the right seats. There was a popular interview question from a while back: there are $n$ people getting seated an airplane, and the first person comes in and sits at a random seat. Everyone else who comes in either sits in his seat, or if his seat has been taken, sits in a random un... | Correct answer to incorrect question: please see second answer
The answer is $\frac{1}{2}$ as was said.
The general pattern is that for $n$ people, there is a $\frac{1}{n}$ probability of success, $\frac{1}{n}$ probability of failure, and an $\frac{n-2}{n}$ probability that the problem repeats itself on the $n-1$ scale... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $ The question is to simplify $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
My friend has given me ... | Given that your original numbers are square roots and your expanded equation involves only even powers of your original numbers, you could say:
$$A=a^2=5, B=b^2=6, C=c^2=7$$
and simplify your equation to:
$$2(AB+AC+BC) - (A^2+B^2+C^2) = 2(30+35+42) - (25+36+49) = 214-110 = 104$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/465103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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How prove this $|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\frac{1}{20n^3}$ Prove that
$$|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\dfrac{1}{20n^3}$$
let $t=\{n\sqrt{2}\}-\{n\sqrt{3}\}$ and $k=[n\sqrt{3}]-[n\sqrt{2}]$
then we have $$t=k-(\sqrt{3}-\sqrt{2})n=k-\sqrt{5-2\sqrt{6}}n\neq 0$$
so
\begin{align*}
t&=\dfrac{(k-(\sqrt{3}-\sqrt{2})n)(... | If the statement is false, then there exists a $n \in \mathbb{Z}_{+}$ such that
$$\left|\{n\sqrt{3}\} - \{n\sqrt{2}\}\right| \le \frac{1}{20n^3}\tag{*1}$$
This in turn implies existences of $m \in \mathbb{Z}$ and
$\delta \in [ -\frac{1}{20n^3}, \frac{1}{20n^3} ]$ such that:
$$n (\sqrt{3} - \sqrt{2}) = m + \delta$$
Con... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find this $3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}+\sqrt{5}(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2})$ find this follow minimum
$$3\sqrt{x^2+y^2}+5\sqrt{(x-1)^2+(y-1)^2}+\sqrt{5}\left(\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}\right)$$
I guess This minimum is $6\sqrt{2}$
But I can't prove,Thank you
| Too long for a comment, I wonder if the problem can be solved geometrically.
Let $A,B, C, D$ be the points of coordinates $(0,0), (1,1), (1,0)$ and $(0,1)$ and let $P$ be the point in plane of coordinates $(x,y)$.
The problem asks you to find the minimum of
$$3PA+5PB+\sqrt{5}(PC+PD) (*)$$
The issue now is that we cann... | {
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"url": "https://math.stackexchange.com/questions/466244",
"timestamp": "2023-03-29T00:00:00",
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Problem 2-7 in Spivak One is asked to show that $ \sum\limits_{i=1}^{n} k^{p}$ (typo on $i$?) can always be written in the form
$$\frac{n^{p+1}}{p+1}+An^{p}+Bn^{p-1}+Cn^{p-2}+\cdots.$$
The solution states:
The proof is by complete induction on $p$. The statement is true for $p=1$, since
$$\sum\limits_{k=1}^{n} k= \f... | If I understand correctly you are concerned that the last part of the solution gives:
$\sum_{k=1}^{n}k^{p}=\frac{(n+1)^{p+1}}{p+1}+$ terms involving powers of $n\le p$
but the original question asked for:
$\sum_{k=1}^{n}k^{p}=\frac{n^{p+1}}{p+1}+$ terms involving powers of $n\le p$.
But this is not a problem since $(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
System of quadratic equations $x^2 + y = 4$ and $x + y^2 = 10$ How would you solve the following system of equations:
$$
x^2 + y = 4 \\
x + y^2 = 10
$$
Thanks very much!
I tried defining y in terms of x and then inserting in to the second equation:
$$
y = 4 - x^2 \\
x + (4 - x^2)^2 = 10
$$
Expand the second equation:... | $$y=4-x^2$$
put it in 2nd equation
$$x+(4-x^2)^2=10$$
$$x+16+x^4-8x^2=10$$
$$x^4-8x^2+x+6=0$$
$$x^4-x^3+x^3-7x^2-x^2+7x-6x+6=0$$
$$x^4-x^3+x^3-x^2-7x^2+7x-6x+6=0$$
$$x^3(x-1)+x^2(x-1)-7x(x-1)-6(x-1)=0$$
$$(x-1)(x^3+x^2-7x-6)=0\implies x-1=0\implies x=1$$
solve the above equation
one of the value of x=1 and y=3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Between Mertens' theorems It is well-known that
$$
\sum_{p\le x}\frac{\log p}{p}=\log x+O(1)
$$
and
$$
\sum_{p\le x}\frac1p=\log\log x+M+o(1).
$$
What is the order of
$$
\sum_{p\le x}\frac{\sqrt{\log p}}{p}
$$
?
| Let's calculate:
$$\begin{align}
\sum_{p\leqslant x} \frac{\sqrt{\log p}}{p} &= \sum_{n \leqslant x} \frac{\sqrt{\log n}}{n}\left(\pi(n) - \pi(n-1)\right)\\
&= \sum_{n \leqslant x}\frac{\sqrt{\log n}}{n}\pi(n) - \sum_{n \leqslant x-1}\frac{\sqrt{\log (n+1)}}{n+1}\pi(n)\\
&= \sum_{n \leqslant x} \left(\frac{\sqrt{\log n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derivative of $\frac {x\cdot\left(1 - 3x\right)}{\sqrt{x-1}}$ Problem. Find the first derivative of $$ \dfrac {x \left( 1 - 3x \right)}{\sqrt{x-1}} $$
Work. Let $u = x-1$ and $y = \dfrac {(u+1)(-3u-2)}{\sqrt{u}} $
Using the chain rule, I got$$\dfrac{(-9x^2-5x+2)}{(2(x-1)^\frac{3}{2})}$$
But the answer is $$\dfrac{(-9x... | You don't have to use the Chain Rule. You can, instead, use the Product Rule.
$ f(x) = \dfrac {x \cdot (1 - 3x)}{\sqrt{x-1}} = \dfrac {x-3x^2}{\sqrt{x-1}} $
$ f'(x) = \dfrac {\sqrt {x-1} \cdot (1-6x) - (x-3x^2) \cdot \left( \sqrt{x-1} \right)'}{x-1} $
$ f'(x) = \dfrac {\sqrt{x-1} \cdot (1-6x) - (x-3x^2) \cdot \dfrac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can someone explain this trigonometric limit without L'Hopital? I can not solve this limit:
$$\lim \limits_{x\to 0}\frac{x^2}{1-\sec(x)}$$
$$\lim \limits_{x\to 0} \frac{x^2}{1-\sec(x)}=\lim \limits_{x\to 0}\frac {x^2}{1-\sec(x)}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{1-\sec^2(x)}=... | $\lim \limits_{x\to 0} \frac{x^2}{1-\sec x}= \lim \limits_{x\to 0} \frac {x^2}{1-\sec x}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0} \frac{x^2(1+\sec(x))\cos^2 x}{\cos^2 x-1}=\lim \limits_{x\to 0} \frac{x^2(1+\sec(x))\cos^2 x}{-\sin^2(x)}$
Now if you know the limit of $\frac {\sin x}x$ the other terms are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/468674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Calculate $\int \frac{dx}{x\sqrt{x^2-1}}$ I am trying to solve the following integral
$$\int \frac{dx}{x\sqrt{x^2-1}}$$
I did the following steps by letting $u = \sqrt{x^2-1}$ so $\text{d}u = \dfrac{x}{\sqrt{{x}^{2}-1}}$ then
\begin{align}
&\int \frac{\sqrt{x^2-1} \, \text{d}u}{x \sqrt{x^2-1}} \\
&\int \frac{1}{x} \tex... | $$u^{2}=x^{2}-1\Rightarrow xdx=udu\\
\int \frac{dx}{x\sqrt{x^2-1}}=\int \frac{xdx}{(x^{2}-1+1)\sqrt{x^2-1}}\\
=\int\frac{udu}{(u^{2}+1)u} =\int\frac{du}{u^{2}+1}=\arctan(u)+c=\arctan\sqrt{x^{2}-1}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/468727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.