Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$ How do I show that:
$$\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$$
This is actually problem B $4371$ given at this link. Looks like a... | We can prove (below), the roots of $$z^3-z^2-2z+1=0 \ \ \ \ (1)$$
are $2\cos\frac{(2r+1)\pi}7$ where $r=0,1,2$
So,if we set $\displaystyle t=\frac1{\sin^2{\frac{(2r+1)\pi}{14}}}$ where $r=0,1,2$
$\implies 2\cos\frac{(2r+1)\pi}7=2\left(1-2\sin^2{\frac{(2r+1)\pi}{14}}\right)=2\left(1-\frac2t\right)=\frac{2(t-2)}t$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/45144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 5,
"answer_id": 3
} |
Integral with $\sqrt{2x^4 - 2x^2 + 1}$ in the denominator $$\int\frac{x^{2}-1}{x^{3}\sqrt{2x^{4}-2x^{2}+1}} \: \text{d}x$$
I tried to substitute $x^2=t$ but I am unable to solve it and I also tried to divide numerator and denominator by $x^2$ and do something but could not get anything.
| Well, using mathematica I can see that this function the given answer is the derivative of the integral.
Consider
\begin{align*}
f(x) &= \frac{\sqrt{2x^{4}-2x^{2}+1}}{x^{2}} \\ &= \frac{\frac{x^{2} \cdot (8x^{3}-4x)}{2 \sqrt{2x^{4}-2x^{2}+1}} - 2x \cdot \sqrt{2x^{4}-2x^{2}+1}}{x^{4}} \quad \ \Bigl[ \text{Note this is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/45474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Is $\frac{a_1+\cdots+a_n}{\sqrt{n(b_1+\cdots+b_n)}} \le \frac{1}{n}\left(\frac{a_1}{\sqrt{b_1}} +\cdots+\frac{a_n}{\sqrt{b_n}}\right)$? For $a_1>0$, $a_2>0,\dots,a_n>0$, and $b_1>0$, $b_2>0,\dots,b_n>0$
I want to prove:
$$\frac{a_1+a_2+\dots+a_n}{\sqrt{n(b_1+b_2+...+b_n)}} \le \frac{1}{n}\left(\frac{a_1}{\sqrt{b_1}}+\f... | For $n=2$ your inequality reduces to
$$\frac{a_1+a_2}{\sqrt{2(b_1+b_2)}} \leq \frac{1}{2} (\frac{a_1}{\sqrt{b_1}}+\frac{a_2}{\sqrt{b_2}})$$
Lets the $a_1=1,a_2=\frac{1}{2},b_1=1,b_2=\frac{11}{16}$ then you have
$$\sqrt{\frac{2}{3}} \leq \frac{1}{2}+\frac{1}{\sqrt{11}}$$
This is false, so the inequality doesn't hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/47616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain ... | Yet another take. Start with the definition of falling factorial powers:
$\begin{align}
x^{\underline{r}}
= x (x - 1) \dotsm (x - r + 1)
\end{align}$
so that:
$\begin{align}
\Delta n^{\underline{r}}
&= (n + 1)^{\underline{r}} - n^{\underline{r}} \\
&= r n^{\underline{r - 1}} \\
\sum_{0 \le k < n} k^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/48080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "145",
"answer_count": 32,
"answer_id": 9
} |
Inverse Laplace Transform help Is the information below correct?
Find the inverse Laplace transform of $$ F(s) = \frac{s}{s^2 + 4s + 13}$$
Soln:
a) Complete the squares to simplify our denominator
$$ s^2 + 4s + 13 = (s+2)^2 + 9 = (s+2)^2 + 3^2$$
$$\mathscr{L}^{-1}\left\{F(s)\right\} = \frac{s}{(s+2)^2 + 3^2}. $$ From ... | HINT:
$$
\frac{s}{{(s + 2)^2 + 3^2 }} = \frac{{s + 2 - 2}}{{(s + 2)^2 + 3^2 }} = \frac{{s + 2}}{{(s + 2)^2 + 3^2 }} - \frac{2}{3}\frac{3}{{(s + 2)^2 + 3^2 }}.
$$
Now note that the inverse transforms of $\frac{{s + \alpha }}{{(s + \alpha )^2 + \omega ^2 }}$ and $\frac{\omega}{{(s + \alpha)^2 + \omega^2 }}$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/52555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Proving that $ 30 \mid ab(a^2+b^2)(a^2-b^2)$ How can I prove that $30 \mid ab(a^2+b^2)(a^2-b^2)$ without using $a,b$ congruent modulo $5$ and then
$a,b$ congruent modulo $6$ (for example) to show respectively that $5 \mid ab(a^2+b^2)(a^2-b^2)$ and
$6 \mid ab(a^2+b^2)(a^2-b^2)$?
Indeed this method implies studying nu... | Hint for the divisors 2 and 3.
Note the factorisation $ab(a+b)(a-b)(a^2+b^2)$ which is divisible by $ab(a-b)(a+b)$
Either one of $a,b$ is divisible by 2 (or 3) or not. If not, look at the other factors to complete the proof.
Studying cases does not look too onerous to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/53135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{... | As Willie Wong observed, including an expression of the form $\displaystyle \frac{|\alpha|}{\alpha}$ is a way of ensuring that $\alpha > 0$. (As $\sqrt{|\alpha|/\alpha}$ is $1$ if $\alpha > 0$ and non-real if $\alpha < 0$.)
The ellipse $\displaystyle \left( \frac{x}{7} \right)^{2} + \left( \frac{y}{3} \right)^{2} - 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/54506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "466",
"answer_count": 10,
"answer_id": 9
} |
Olympiad Inequality Problem Consider three positive reals $x,y,z$ such that $xyz=1$.
How would one go about proving:
$$\frac{x^5y^5}{x^2+y^2}+\frac{y^5z^5}{y^2+z^2}+\frac{x^5z^5}{x^2+z^2}\ge \frac{3}{2}$$
I really dont know even where to begin! It looks a BIT like Nesbitts? Maybe?
| Edit. I posted a new "proof" (now deleted), before I realized I was adddressing the wrong question. I think the original proof is ok, I messed up something in trying to simplify the approach.
Hint.
Since this is an olympiad problem, it is likely that there is a proof without using calculus. I am sketching one here, bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/61289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 1
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Limit of difference of two square roots I need to find the limit, not sure what to do.
$\lim_{x \to \infty} \sqrt{x^2 +ax} - \sqrt{x^2 +bx}$
I am pretty sure I have to divide by the largest degree which is x^2 but that gets me some weird numbers that don't seem to help.
| This answer may be a bit longer, but I've tried to include every step in the process.
\begin{equation} \label{eq1}
\begin{split}
\lim_{x \to \infty} \sqrt{x^2 + ax} - \sqrt{x^2 + bx} & = \lim_{x \to \infty} \left(\sqrt{x^2 + ax} - \sqrt{x^2 + bx}\right)\left(\frac{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}}{\sqrt{x^2 + ax} + \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How can the following be calculated? How can the following series be calculated?
$$S=1+(1+2)+(1+2+3)+(1+2+3+4)+\cdots+(1+2+3+4+\cdots+2011)$$
| Note that $1$ occurs $2011$ times; $2$ occurs $2010$ times; $3$ occurs $2009$ times, and so on, until $2011$ occurs only once. Hence we can rewrite the sum as
$$(2012-1)(1)+(2012-2)(2)+(2012-3)(3)+\cdots+(2012-2011)(2011).$$
Split and regroup terms:
$$2012(1+2+3+\cdots+2011)-(1^2+2^2+3^2+\cdots+2011^2).$$
Now using the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/65465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Fourier transform integral I'm trying to calculate the 3D fourier transform of this function:
$$\frac{1}{(x^2+y^2+z^2)^{1/2}}$$
Any help would be appreciated, thanks.
| Inserting the Jacobian $r^2\sin\theta$
and $\sqrt{x^2+y^2+z^2}=r$ in polar coordinates gives
\begin{equation}
\int_0^\infty r^2 dr \int_0^{2\pi} d\phi \int_0^\pi \sin\theta d\theta \frac{1}{r} e^{i\mathbf{k}\cdot \mathbf{r}}
\end{equation}
\begin{equation}
=
\int_0^\infty r^2 dr \int_0^{2\pi} d\phi \int_0^\pi \sin\thet... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find extremes of function $f(x,y,z) = x^2y + y^2z + x - z$ I am preparing for an exam tuesday morning and I would like to ask you, if someone could please review my solution for the following excercise. I don't have the correct answer so I am unable to check whether it is OK.
Find the extremes of polynomial function $f... | You wrote $f''_{xz} = 2y$, which is incorrect; it should be 0. However in your later calculations you correctly treat the term as zero.
For your second Hessian, the first subdeterminant is -2, not 2. Even if it were 2, H would not be negative-definite: Sylvester's Criterion states that a matrix is positive-definite (an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/74911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$ I want to find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$.
My thoughts so far: I want to find $p$ such that $ \left( \frac{15}{p} \right) = 1$. By multiplicativity of the Legendre symbol, this is equivalent to $ \left( \fr... | Note that there is no such thing as conditions mod 3 and mod 5 being incompatible; see The Chinese Remainder Theorem. In particular, 7 is 2 mod 5 and 1 mod 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Relations between coefficient and exponent of Proth prime form $k\cdot 2^n+1$? Definition: Proth number is a number of the form :
$$k\cdot 2^n+1$$
where $k$ is an odd positive integer and $n$ is a positive integer such that : $2^n>k$
My question : If Proth number is prime number are there some other known relations in ... | There are many simple relationships that involve congruences. They have a flavour much like results you have mentioned in earlier posts.
For example, if $n>1$ is odd, then $k$ must be of the form $6a-1$ or $6a+3$. If $n$ is even, then $k$ must be of the form $6a+1$ or $6a+3$. The arguments are the familiar ones. For ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/83305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with convergence in distribution $Y$ is a random variable with $$M(t) = \frac{1}{(2-\exp(t))^s}.$$
Does $$\frac{Y-E(Y)}{\sqrt{\operatorname{Var}(Y)}}$$ converge in distribution as $s$ tends to infinity?
I let $Z = \frac{Y-E(Y)}{\sqrt{\operatorname{Var}(Y)}}$.
Differentiating the MGF of $Y$ and let $t = 0$ we have ... | As you have determined:
$$
M_Z(t) = \mathbb{E}(\exp(t Z ) ) = \frac{\exp\left(-t \frac{s}{\sqrt{2s}}\right)}{\left( 2 - \exp\left(\frac{t}{\sqrt{2s}} \right) \right)^s} = \left( \frac{\exp\left(-t \frac{1}{\sqrt{2s}}\right)}{ 2 - \exp\left(\frac{t}{\sqrt{2s}} \right) } \right)^s
$$
In order to determine large $s$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/87061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to find the area of the region, bounded by various curves?
Find the area of the region bounded by the curves $y=x^2$ and $y=x$.
Find the area of the region bounded by the curves $y=x^2+1$ and $y=2$
I have a ton of questions like this and I have been graphing them and then splitting them into intervals and adding ... | First we will find the area of the region bounded by the curves:
$y = x^2$ ... (i)
and $y = x $ ... (ii)
To determine the shaded area between these two curves, we need to sketch these curves on a graph.
Now, we will find the area of the shaded region from O to A.
Area of Shaded Region Between Two Curves :
$A = \displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/87149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Help with integrating $\int \frac{t^3}{1+t^2} ~dt$ What am I doing wrong on this integration problem?
$$
\begin{align*}
\int\frac{t^3}{1+t^2} &=
\frac14 t^4 (\ln(1+t^2) (t+\frac13 t^3))
\\ &= \frac14 t^4(t \ln(1+t^2)+\frac13 t^3 \ln(1+t^2)
\\ &= \frac14 t^5 \ln(1+t^2)+\frac{1}{3}t^7 \ln(1+t^2)
\end{align*}$$
Ans... | Your integral is:
$$\int \frac{t^3}{1+t^2} dt$$
Substitute: $x = 1+t^2$ and thus $dx = 2t dt$. Then the above transforms to:
$$\int \frac{t^3}{1+t^2} dt = \frac{1}{2} \int \frac{t^2 \ 2t dt}{(1+t^2)}$$
Using the transformation suggested earlier, we can re-write the right hand side as:
$$\frac{1}{2} \int \frac{(x-1) \ d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/87425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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How to prove that $2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}$ I want to prove that$$2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}, x\geq 0$$
I have started from the arcsin part and I tried to end to the arctan one but I failed.
Can anyone help me solve it?
| Let $\arctan{\sqrt{x}} = \theta$. Then we have $x = \tan^2 (\theta)$. Hence, $$\frac{x-1}{x+1} = \frac{\tan^2 (\theta)-1}{\tan^2 (\theta)+1} = \sin^2(\theta) - \cos^2(\theta) = - \cos(2 \theta)= \sin \left(2 \theta - \frac{\pi}{2} \right)$$
Hence, $$2 \arctan{\sqrt{x}} = 2 \theta = \arcsin \left(\frac{x-1}{x+1} \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/89759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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How to find the equation of the tangent line to $y=x^2+2x-4$ at $x=2$?
I'm given a curve $$y=x^2+2x-4$$
How do I find the tangent line to this curve at $x = 2$?
| General formulas
$y=f(x)=x^2+2x-4$
$m=f'(x)=2x+2$ Slope equation
Point-Slope equation of a line
$(y-y_0)=m_0\cdot(x-x_0)$
Compute actual values
$x_0 = 2$
$y_0=f(x_0)=f(2)=(2)^2+2(2)-4=4+4-4=4$
$m_0=f'(x_0)=f'(2)=2(2)+2=4+2=6$
Substitute values into the line equation
$y-4=6\cdot(x-2)$
$y=6\cdot(x-2)+4$
$y=6x-8$
Gener... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/92165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
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Proving that $ \frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+...+\frac{1}{\sin(133°)\sin(134°)}=\frac{1}{\sin(1°)}$ I would like to show that the following trigonometric sum
$$ \frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+\cdots+\frac{1}{\sin(133°)\sin(134°)}$$
telescopes to $$\frac{1}{\sin(1°)... | $$\frac{\sin(1^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}=\frac{\sin((x+1)^\circ-x^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}=$$
$$\frac{\sin((x+1)^\circ) \cos (x^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}-\frac{\sin(x^\circ) \cos(x+1)^\circ}{\sin(x^\circ) \sin(x+1)^\circ}= \cot(x^\circ)-\cot(x+1)^\circ$$
Add them and you get your... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Show $|x|<|x^2+b|$ for real $x$ and $b>\frac{1}{4}$ How do I show $|x|<|x^2+b|$ for real $x$ and $b>\frac{1}{4}$ ?
I mean it's clear for $x=0$ and $|x|>1$ but what is if $0<|x|<1$?
Please help!
| You only have to prove it for $x\geq 0$, since if true for $x$ it is true for $-x$.
If $x\geq 0$, then $|x|=x$ and $|x^2+b|=x^2+b$ and you only need to prove that $x^2+b>x$.
But $$x^2-x+b = x^2 - x + \frac{1}{4} + (b-\frac{1}{4}) = (x-\frac{1}{2})^2 + (b-\frac{1}{4})$$
Since $b>\frac{1}{4}$, this means that $x^2-x+b > ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/96424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit $\lim_{n\to+\infty} (1-\frac1{2^2})(1-\frac1{3^2})\cdot \cdots \cdot(1-\frac{1}{n^2})$ and series $\sum_{n=2}^{\infty} \ln(1-\frac1{n^2})$
Possible Duplicate:
Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$
Compute:
\begin{align*}
\lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2}... | If you write it like
$$
\lim_{n\to \infty}\prod_{j=2}^n \frac{(j+1)(j-1)}{j^2}
$$
you get
$$
\lim_{n\to \infty}\frac{1}{2}\frac{(n+1)!(n-1)!}{(n!)^2},
$$
where factorials cancel in the limit and give $\frac{1}{2}$, because a $2$ is missing in the numerator to get $(n+1)!$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/99537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Compute: $\int_{0}^{1}\frac{x^4+1}{x^6+1} dx$ I'm trying to compute: $$\int_{0}^{1}\frac{x^4+1}{x^6+1}dx.$$
I tried to change $x^4$ into $t^2$ or $t$, but it didn't work for me.
Any suggestions?
Thanks!
| Note that
$$I=\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx\quad\stackrel{\large x\,\mapsto\,\frac{1}{x}}\Longrightarrow\quad I=\int_{1}^{\infty}\frac{x^4+1}{x^6+1}\,dx\quad\Longrightarrow\quad I=\frac{1}{2}\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,dx$$
Using (proof can be seen here)
$$\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 5
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prove for all $n\geq 0$ that $3 \mid n^3+6n^2+11n+6$ I'm having some trouble with this question and can't really get how to prove this..
I have to prove $n^3+6n^2+11n+6$ is divisible by $3$ for all $n \geq 0$.
I have tried doing $\dfrac{m}{3}=n$ and then did $m=3n$
then I said $3n=n^3+6n^2+11n+6$ but now I am stuck.
| We have
$$
\begin{align}
n^3+6n^2+11n+6
&=6\binom{n}{3}+18\binom{n}{2}+18\binom{n}{1}+6\binom{n}{0}\\
&=6\left(\binom{n}{3}+3\binom{n}{2}+3\binom{n}{1}+\binom{n}{0}\right)
\end{align}
$$
so $6\mid(n^3+6n^2+11n+6)$ for all $n\in\mathbb{Z}$.
Of course, since $3\mid 6$, we have $3\mid(n^3+6n^2+11n+6)$, as requested.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 4
} |
about Pythagorean quadruples Respected Mathematicians,
I would like to prepare a function, which will generate Pythagorean quadruples (a, b, c, d) = $d^2$ = $a^2$ + $b^2$ + $c^2$...-> (1). How far I am correct I don't know. For some set of integers a and b, consider $a^2$ + $b^2$ = m and d = c + ε. Now by (1), c = (m ... | Let us only deal with primitive triples where the forms are
$A=2n+1\quad B=4n\quad CC=4n=1$. We begin with Euclid's formula shown here as $ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2.\quad$
Now we choose any odd number greater that one $(15?)$ for the $A$-value and solve the $A$-function for $k$, testing a defined ra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/105558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Factorizing polynomial $x^5+x+1$ I'm given a problem to factorize
$$ P(x)=x^5+x+1 $$
I've done the following:
$$ P(x)=(x^5+x^4+x^3)-(x^4+x^3+x^2)+(x^2+x+1)= (x^2+x+1)(x^3-x^2+1)$$
Is it possible to prove that this cannot be factorized any further?
| It depends what you are factoring over. Notice that if we use the quadratic formula, $x^2 + x + 1$ has roots $-\frac{1}{2} \pm \frac{i\sqrt{3}}{2}$, so it can be factorized as
$$(x^2 + x + 1) = (x - (-\frac{1}{2} + \frac{i\sqrt{3}}{2}))(x - (-\frac{1}{2} - \frac{i\sqrt{3}}{2}))$$
If you are factoring over just the real... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/106837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$.
I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = ... | The extended Euclidean algorithm gives
$$
9 = (2 n + 1)(n^3+1)+(-2 n^2 - n + 4)(n^2+2)
$$
and so $\gcd(n^3 + 1, n^2 + 2)$ divides $9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 5
} |
Square on the curve A square is drawn on the curve $y = x^3 + 27\cdot x^2 + 8\cdot x + 91$. What would be the area of the square. All the points of the square should lie on the curve.
All four points lie on the curve (not the sides)
| Time for some reverse engineering. Suppose we start with a square; translate it so that it is centred at the origin. Then for some $s$ and $\theta$, the vertices are $(s \cos(\theta), s \sin(\theta))$, $(s \sin(\theta), -s \cos(\theta))$, $(-s \cos(\theta), -s \sin(\theta))$ and $(-s \sin(\theta), s \cos(\theta))$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/111005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
solving some complex equations I want to solve the following three equations for $x,y,z$:
$$\begin{eqnarray}k_1&=&\frac{x+y+z}{yz}\\
k_2&=&\frac{x^2+y^2+z^2}{y^2z^2}\\
k_3&=&\frac{x^3+y^3+z^3}{y^3z^3}\end{eqnarray}$$
where $k_1,k_2,k_3$ are constants.
Is there any kind of standard methods?
| If you set $\frac{x}{yz} = p$, $\frac{1}{y} = q$, $\frac{1}{z} = r$, then your equations are
$p + q + r = k_1$
$p^2 + q^2 + r^2 = k_2$
$p^3 + q^3 + r^3 = k_3$
Using $(p+q+r)^2 - (p^2 + q^2 + r^2) = 2(pq + qr + rp)$ we get
$pq + qr + rp = (k_1^2 - k_2)/2$
Similarly we can find the value of $pqr$.
Thus we can find the po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using double angle formulas in integration, trouble following an example. I have just started looking at integration and I am having trouble understanding what has been done in one of the examples in the book I am working through.
It involves using the double angle formula for $\sin(2\theta)$ to provide a rearrangeme... | Consider $\theta = \frac{1}{2}x$. Then
$$ \cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right)
= \cos \theta \sin \theta$$
Notice that the double angle formula could be written:
$$ \sin \theta \cos \theta = \frac{1}{2}\sin(2\theta)$$ So the integrand is now:
$$ \cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How can one prove that $\sum_{k=1}^{\infty}\prod_{m=1}^{2k}\text{ctg}\frac{m\pi}{2k+1}=\frac{\pi}{4}-1$? A question from some Russian book, about different summations an integrals, by Prudnikhov, Brichkhov and Marichev.
Page 746, 20, they write:
$$ \sum_{k=1}^{\infty}\prod_{m=1}^{2k}\text{ctg}\frac{m\pi}{2k+1}=\frac{... | I believe we can show that
$$\prod_{m=1}^{2k} \cot \frac{m\pi}{2k+1} = (-1)^k \frac{1}{2k+1}$$
Using Chebyshev Polynomials $\displaystyle T_n(x)$ (of degree $\displaystyle n$, and leading coefficient $\displaystyle 2^{n-1}$), which satisfy
$$T_{n}(\cos x) = \cos (n x)$$
and
$$T_{2n+1}(\sin x) = (-1)^k\sin ((2n+1) x)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/113706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Integrate using Partial Fraction decomposition, completing the square The given problem is $\int{x\over x^3-1}dx$.
I know this equals
$${1\over3}\int {1\over x-1}-{x-1\over x^2+x+1}dx,$$
which can be separated into
$${1\over3}\int {1\over x-1}dx - {1\over3}\int{x+(1/2)-(3/2)\over x^2+x+1}dx.$$ This can further be s... | The purpose is to get an integral of the form $\displaystyle\int \frac{1}{v^{2}+1}\,dv=\arctan v$. Using the substitution indicated by you $u=x+1/2$, $du=dx$,
we obtain
$$\begin{eqnarray*}
\frac{1}{2}\int \frac{1}{\left( x+1/2\right) ^{2}+3/4}dx &=&\frac{1}{2}\int
\frac{1}{u^{2}+3/4}\,du \\
&=&\frac{1}{2\cdot 3/4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/114918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$ The following expression is given:
$$\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$$
Simplify it, knowing that $x+y+z=0$.
| Note: Using Newton's identities, we can calculate the below expressions more easily, following the easy recursive definition.
But, your idea of writing as roots of third degree polynomial works I believe, but requires some work and we show that here:
Let $\displaystyle x,y,z$ be roots of $\displaystyle t^3 + at - b = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/115520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Zeroes of a Particular Function I'm looking for the zeroes of $f(k) = e^{\sqrt{k}}[\frac{s}{k} - \frac{d}{\sqrt{k}}] - 1$ on the set $k > 0$. Is there a nice way to describe the set of solutions for given $s$ and $d$? Thanks!
| When the ratio $s/d$ is small, we can use the Lagrange inversion formula (see, e.g., my answer here) to solve $e^x(s-dx) = x^2$ (as per Arturo's comment) for $x$ as a power series in either $s$ or $d$:
$$\begin{align}x &= \frac{s}{d}-\frac{s^2}{d^3}+ \left(\frac{1}{d^4}+\frac{2}{d^5}\right) s^3-\left(\frac{1}{2 d^5}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/118705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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How to simplify this square $(3 \times 4 + 2)^2$? If I have this $(3 \times 4 + 2)^2$,
How can I simplify it with out the final result.
Do I distribute the $^2$ over each number like this:
$(3^2 \times 4^2 + 2^2)$?
What is the rule?
| $(3 \times 4 + 2)^2 = (12 + 2)^2 = 14^2 =196$ while $(3^2 \times 4^2 + 2^2) = 9 \times 16 +4 = 144+4 = 148$, so that does not work.
If you want a rule for squares of sums, try: $$(x+y)^2 = x^2 + 2 x y +y^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
For complex $z$, find the roots $z^2 - 3z + (3 - i) = 0$
Find the roots of:
$z^2 - 3z + (3 - i) = 0$
$(x + iy)^2 - 3(x + iy) + (3 - i) = 0$
$(x^2 - y^2 - 3x + 3) + i(2xy -3y - 1) = 0$
So, both the real and imaginary parts should = 0. This is where I got stuck since there are two unknowns for each equation. How do I p... | $$\begin{cases}
x^2-y^2-3x+3=0 \\
2xy-3y-1=0
\end{cases}$$
$$x = \frac {3y+1}{2y} \Rightarrow \left(\frac {3y+1}{2y}\right)^2-y^2-3 \cdot \frac {3y+1}{2y}+3=0 \Rightarrow$$
$$\Rightarrow (3y+1)^2-4y^4-3(3y+1)\cdot 2y+12y^2=0 \Rightarrow$$
$$\Rightarrow 4y^4-3y^2-1=0$$
Substitute $~y^2=t~$ and solve for $t$ over ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How do we get the result of the summation $\sum\limits_{k=1}^n k \cdot 2^k$?
Possible Duplicate:
Formula for calculating $\sum_{n=0}^{m}nr^n$
Can someone explain step by step how to derive the following identity? $$\sum_{k=1}^{n} k \cdot 2^k = 2(n \cdot 2^n - 2^n + 1) $$
| Although you got great answers, I will add another way to look at the expression you are looking for
Let us denote
$$ S = \sum_{k=1}^{n} k \cdot 2^k $$
This can be expanded as
$$
\begin{align*}
S &= 2 + 2 . 2^2 + 3 . 2^3 + \cdots n 2^n\\
2S &= \hspace{8pt}+1.2^2+2.2^3 + \cdots (n-1) . 2^n+n . 2^{n+1}
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Using antiderivative to calculate complex integral $$\int_{1}^{3}(z-2)^3 dz $$
I get the following -
$$\frac{1}{4}[(3-2)^3 - (1-2)^3] = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$
However the answer sheet I have just show it reduced to $\frac{1}{4} - \frac{1}{4} = 0$
Cant see how they are getting a - instead of a +...wha... | The integral of $(z-2)^3$ is $\frac14 (z-2)^4+c$ not $\frac14 (z-2)^3+c$, so you should get $$\tfrac{1}{4}(3-2)^4 - \tfrac{1}{4}(1-2)^4 =0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove for any positive real numbers $a,b,c$ $\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} \geq \frac{a+b+c}{3}$ Since the problem sheets says I should use Cauchy-Schwarz inequality, I used
$\frac{{a_1}^2}{x_1}+\frac{{a_2}^2}{x_2}+\frac{{a_3}^2}{x_3}$
$\geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}$
I... | Let $S$ be the sum on the left hand side and $T$ be the sum:
$\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}$
Then $S-T = (a-b)+(b-c)+(c-a)=0$
So it suffices to show $S+T \geq \frac{2(a+b+c)}{3}$. We can achieve this by showing $\frac{a^3+b^3}{a^2+ab+b^2} \geq \frac{a+b}{3}$
The inequality is equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/122741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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The net signed area between $t=0, y=0, t=x$, and $y=f(t)$ f(t) is continuous function.So
I know that $\int _0^x {f(t) dt}=$ "The net signed area between $t=0, y=0, t=x$, and $y=f(t)$"
And I can find the same result with endless small rectangulars areas method.
"The net signed area between $t=0, y=0, t=x$, and $y=f(t)$... | I suppose you're assuming $f$ is continuous. The statement is easily seen to be true for $f(t)=t$, and by subtracting a multiple of this we can assume $f(0) = f(x)$. By the Stone-Weierstrass theorem, trigonometric polynomials with period $x$ are uniformly dense in the continuous functions with period $x$, and so it s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$
*
*$\displaystyle
0\cdot \binom{n}{0} + 1\cdot \binom{n}{1} + 2\binom{n}{2}+\cdots+(n-1)\cdot \binom{n}{n-1}+n\cdot \binom{n}{n}$
*$\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}$
Ho... | The answer from lhf has already dealt with the second question posed here. Here is a Wikipedia article about it.
Here's a probabilistic approach to the first question. When you toss a coin $n$ times, the probability that a "head" appears exactly $k$ times is $\dbinom n k (1/2)^n$. The average number of times a "head... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
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Show $ I = \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = \frac{\pi}{\sqrt 2}$ Show $$ I = \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = \frac{\pi}{\sqrt 2}$$
| Based on KV Raman,$\displaystyle{\frac{1}{1+\cos^2 x}}$ is even function.
$$\int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = 2 \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+\cos^2 x}$$
$$
\begin{align*}
\int_0^{\frac{\pi}{2}} \frac{1}{1+\cos^2 x} \,\mathrm{d}x &=\int_0^{\frac{\pi}{2}}\frac{\mathrm{d}x}{\sin^2 x+2\cos^2 x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but ... | You are almost done: If $n \equiv 0, \pm 1 \pmod 5$ we are done (as we have $5 \mid n$, $5 \mid n-1$ or $5 \mid n+1$ then). So suppose $n \equiv \pm 2 \pmod 5$, but then $n^2 + 1 \equiv (\pm 2)^2 + 1 \equiv 0 \pmod 5$, hence $5 \mid n^2 + 1$, and so $5 \mid n^5 -n$ also in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 21,
"answer_id": 0
} |
How do I solve $(x-1)(x-2)(x-3)(x-4)=3$ How to solve $$(x-1) \cdot (x-2) \cdot (x-3) \cdot (x-4) = 3$$
Any hints?
| Look at $(x-1)(x-4)$ and $(x-2)(x-3)$ they multiply as $(x^{2}-5x+4)$ and $x^{2}-5x+6$. Now put $t= x^{2}-5x$ and reduce it to a quadratic equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Taking the derivative of $y = \dfrac{x}{2} + \dfrac {1}{4} \sin(2x)$ Again a simple problem that I can't seem to get the derivative of
I have $\frac{x}{2} + \frac{1}{4}\sin(2x)$
I am getting $\frac{x^2}{4} + \frac{4\sin(2x)}{16}$
This is all very wrong, and I do not know why.
| If $f(x) = \frac{x}{2} + \frac{1}{4}\sin(2x)$, then $f'(x) =$ $\frac{1}{2} + \frac{cos(2x)}{2}$. If you want $\int{f(x)}$, then we have $\frac{x^2}{4} - \frac{cos(2x)}{8}+C$. Here, we differentiate/integrate $\frac{x}{2}$ and $\frac{1}{4}\sin(2x)$ separately. Do you have an initial condition?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Tiling pythagorean triples with minimal polyominoes Given a Pythagorean triple $(a,b,c)$ satisfying $a^2+b^2=c^2$, how to calculate the least number of polyominoes of total squares $c^2$, needed, such that both the square $c^2$ can be build by piecing them together, as well as the two separate squares of side length $a... | WLOG let $(a, b, c)=1$. Then there is an upper bound of $2+a+b-c$. This bound is sharp for the pair $(3, 4, 5)$, and all the other pairs I've tested. It is attainable as follows:
Let one piece be a $a\times a$ square, and another be a $b\times b$ square with a $(a+b-c)\times (a+b-c)$ square removed from a corner. Now, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/136019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Let $a_{n}$ be a sequence such that $(a_{n})^{2}=ca_{n-1}$ where ($c>0,a_{1}>0$).Prove that $a_n$ converges to $c$. Let $a_{n}$ be a sequence such that $(a_{n})^{2}=ca_{n-1}$ where ($c>0,a_{1}>0$).Prove that the sequence converges to $c$.
My first problem was to find some terms of the sequence to verify that point and ... | Note that calculating some values gives:
$$\Large \eqalign{
& {a_1} = a \cr
& {a_2} = \sqrt c \sqrt a \cr
& {a_3} = \sqrt c \root 4 \of c \root 4 \of a \cr
& {a_4} = \sqrt c \root 4 \of c \root 8 \of c \root 8 \of a \cr} $$
In general you can prove that
$$\Large {a_n} = {c^{\frac{1}{2} + \frac{1}{4} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/137033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Curve In a Closed Interval with an Infinite Length How can I show that the length of the curve
$$f(x)=\begin{cases}
\sqrt{x}\cos(\pi/x) & \text{ if } x\neq0\\
0 & \text{ if } x=0
\end{cases}$$
is infinite on $[0,1]$?
I tried using the arc length formula, but ended up with the very nasty integral:
$$L=\int_0^1\sqrt{1... | Choosing $x$ so that
$$\pi/x \;\; = \;\; \frac{3 \pi}{2}, \; \frac{4 \pi}{2},\; \frac{5 \pi}{2},\; \frac{6 \pi}{2},\; \frac{7 \pi}{2},\; \frac{8 \pi}{2},\; \frac{9 \pi}{2},\; \frac{10 \pi}{2}, \; \frac{11 \pi}{2},\; \frac{12 \pi}{2},\; \frac{13 \pi}{2},\; ...$$
so that
$$x \;\; =\;\; \frac{2}{3}, \; \frac{2}{4}, \; \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/137725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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The roots of $t^5+1$ Just a quick question, how do we go about finding the roots of $t^5 +1$?
I can see that since $t^5=-1$ that an obvious root is $\sqrt[5]{-1}$.
I am assuming that since there is a $-1$ involved, some of the factors will be complex?
Any help would be welcome.
| Since no one has yet pointed this out, it may be of interest to show how J.M.'s comment leads to a purely algebraic solution (i.e. no use made of trigonometric or exponential function ideas).
Using
$$t^5 + 1 \;\; = \;\; \left(t+1\right)\left(t^4 - t^3 + t^2 - t + 1\right),$$
it follows that the solutions to $t^5 + 1 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/138382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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More highschool math $\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3} = \frac{n^3+3n^2+3n+1}{3n^3} \to \frac{1}{3}$ So the question I am trying to work through is:
Test the series
$$\frac{1}{3}+\frac{2^3}{3^2}+\frac{3^3}{3^3}+\frac{4^3}{3^4}+\frac{5^3}{3^5}+\cdot\cdot\cdot$$
for convergence.
The solution (using D'Alember... | *
*You have a factor of $3^n$ in the numerator, and a factor of $3^{n+1}$ in the denominator. So
$$\frac{3^n(\text{stuff})}{3^{n+1}(\text{other stuff})} = \frac{3^n(\text{stuff})}{3\times 3^{n}\text{(other stuff)}} = \frac{\text{stuff}}{3(\text{other stuff})}.$$
Since $3^{n+1}=3\times 3^n$.
*Dividing numerator and de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/138600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Equation with trigonometry To solve $\sin^3x+\cos^3x=1$
So I just thought of a solution like:
Let $\sin x=t$. Then we have:
$$t^3+(1-t^2) \sqrt{1-t^2} =1 \\ (1-t^2) \sqrt{1-t^2} =1-t^3 \\ (1-t^2)^3=(1-t^3)^2 \\ (1-t)^3(1+t)^3=(1-t)^2(1+t+t^2)^2 \\ (1-t)^2\left[ (1-t)(1+t)^3-(1+t+t^2)^2 \right]=0$$
Which is followed by ... | Note that $\sin x$ and $\cos x$ are always at most $1$, so $\sin^3 x+\cos^3 x\leq \sin^2 x+\cos^2 x=1$ with equality only $\sin x,\cos x$ are $0$ or $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/138992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determining the Smith Normal Form
Consider the integral matrix
$$R = \left(\begin{matrix} 2 & 4 & 6 & -8 \\ 1 & 3 & 2 & -1 \\ 1 & 1 & 4 & -1 \\ 1 & 1 & 2 & 5 \end{matrix}\right).$$
Determine the structure of the abelian group given by generators and relations.
$$A_r = \{a_1, a_2, a_3, a_4 | R \circ \vec{a} = 0\}$$
I ... | We do want to use a kind of Gaussian elimination, but you have to be careful since you should not multiply a row by anything other than $1$ and $-1$, and you should not add non-integer multiples of one row to another row. So we can get started simply enough:
$$\begin{align*}
\left(\begin{array}{rrrr}
2 & 4 & 6 & -8 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/144589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Interesting Determinant Let $x_1,x_2,\ldots,x_n$ be $n$ real numbers that satisfy $x_1<x_2<\cdots<x_n$.
Define \begin{equation*}
A=%
\begin{bmatrix}
0 & x_{2}-x_{1} & \cdots & x_{n-1}-x_{1} & x_{n}-x_{1} \\
x_{2}-x_{1} & 0 & \cdots & x_{n-1}-x_{2} & x_{n}-x_{2} \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
x_{n-... | Expanding Robert solution.
Let $det(A) = P(x)$. Let the polynomial on the right is a multi-variable polynomial $P(x)$.
If $x_1 = x_2$ then $det(A) = 0$ i.e $P(x) = 0$ i.e. $(x_1 - x_2)$ is a factor of $P(x)$.
If $x_2 = x_3$ then $det(A) = 0$ i.e $P(x) = 0$ i.e. $(x_2 - x_3)$ is a factor of $P(x)$.
etc. We calculate pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/144818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\cos(x)$ and $\arccos(x)$ couple limit
Find the value of the following limit:
$$\lim_{n\to\infty} \frac {\cos 1 \cdot \arccos \frac{1}{n}+\cos\frac
{1}{2} \cdot \arccos \frac{1}{(n-1)}+ \cdots +\cos \frac{1}{n} \cdot
\arccos{1}}{n}$$
| What follows is a little hand-wavy, and I wish I had more rigorous demonstration, but the post is too big for a comment.
$$ \begin{eqnarray}
\frac{1}{n} \sum_{k=1}^n \cos\left(\frac{1}{k}\right) \arccos\left( \frac{1}{n+1-k} \right) &=& \frac{1}{n} \sum_{k=1}^n \left( 1 - 2 \sin^2\left(\frac{1}{2 k}\right) \right) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/149734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Finding $\int \frac {dx}{\sqrt {x^2 + 16}}$ I can not get the correct answer.
$$\int \frac {dx}{\sqrt {x^2 + 16}}$$
$x = 4 \tan \theta$, $dx = 4\sec^2 \theta$
$$\int \frac {dx}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{4 \sec \theta}$$
$$\in... | Cosmetically nicer taking $$ x = 4 \sinh t \; , $$
so that $$ dx = 4 \cosh t \; dt $$
The quadratic formula is enought ot give us
$$ t = \log \left( \frac{x + \sqrt{x^2 + 16}}{4} \right) = \log \left( x + \sqrt{x^2 + 16} \right) - \log 4 $$
Then we get
$$ \int \frac{dx}{x + \sqrt{x^2 + 16} } = \int 1 dt = t + C = \lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Integral of $\int \frac{5x+1}{(2x+1)(x-1)}$ I am suppose to use partial fractions
$$\int \frac{5x+1}{(2x+1)(x-1)}$$
So I think I am suppose to split the top and the bottom. (x-1)
$$\int \frac{A}{(2x+1)}+ \frac{B}{x-1}$$
Now I am not sure what to do.
| Suppose $\frac{5x+1}{(2x+1)(x-1)} = \frac{A}{2x+1} + \frac{B}{x-1}$. Then $5x+1 = A(x-1)+B(2x+1) = (A+2B)x + (B-A)$. So by comparing coefficients, we get $A+2B = 5$ and $B-A = 1$. Solving this gives $A = 1$ and $B = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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how many 5-digit numbers satisfy the following conditions How many five-digit numbers divisible by 11 have the sum of their digits equal to 30?
I am able to get the 5-digit numbers divisible by 11
and
I am also able to get the five-digit numbers whose sum of their digits equal to 30.
But i am not able to get how i can ... | An integer written in ordinary base ten notation is divisible by $11$ if and only if the alternating sum of its digits is divisible by $11$. That is, if the digits of a five-digit number are $d_1,d_2,d_3,d_4,d_5$ from left to right, the number is a multiple of $11$ if and only if $d_1-d_2+d_3-d_4+d_5$ is a multiple of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Equivalent of $ u_{n}=\sum_{k=1}^n (-1)^k\sqrt{k}$ I'm trying to show that $$ u_{n}=\sum_{k=1}^n (-1)^k\sqrt{k}\sim_{n\rightarrow \infty} (-1)^n\frac{\sqrt{n}}{2}$$ when $n\rightarrow\infty$
How can I first show that $$u_{2n}\sim_{n\rightarrow \infty} \frac{\sqrt{2n}}{2}$$ and then deduce the equivalent of $u_{n}$?
| Pair consecutive terms together:
$$u_{2n} = \sum_{k=1}^{2n} (-1)^k \sqrt{k} = \sum_{k=1}^n \left( \sqrt{2k}-\sqrt{2k-1}\right) .$$
Since $\displaystyle \sqrt{1 - \frac{1}{2k}} = 1 - \frac{1}{4k} + \mathcal{O}(k^{-2})$ and $\displaystyle \sum_{k=1}^n k^p = \frac{n^{p+1}}{p+1} + \frac{n^p}{2} + \mathcal{O}(n^{p-1})$ we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/155252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 ... | Your integral is OK.
$${\ln |\sec x| + \frac{{\cos 2x}}{4} + C}$$
$${\ln \left|\frac 1 {\cos x}\right| + \frac{{1+\cos 2x}}{4} + C}-\frac 1 4$$
$${-\ln \left| {\cos x}\right| + \frac 1 2\frac{{1+\cos 2x}}{2} + K}$$
$${-\ln \left| {\cos x}\right| + \frac 1 2 \cos ^2 x + K}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 0
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Formula to estimate sum to nearly correct : $\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$ Estimate the sum correct to three decimal places :
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$$
This problem is in my homework. I find that n = 22 when use Maple to solve this. (with some programming) But, in my homework, teacher said find th... | With a little bit of approximation, we can achieve the desired $5\times 10^{-4}$ accuracy with fewer terms.
Notice
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^3} = -\left[\sum_{n=1}^{\infty}\frac{1}{n^3}- \sum_{n=1}^\infty\frac{2}{(2n)^3}\right] = -\frac34\sum_{n=1}^\infty \frac{1}{n^3} = -\frac34\zeta(3)$$
If we want to estim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Evaluating $\int \frac{dx}{x^2 - 2x} dx$ $$\int \frac{dx}{x^2 - 2x}$$
I know that I have to complete the square so the problem becomes.
$$\int \frac{dx}{(x - 1)^2 -1}dx$$
Then I set up my A B and C stuff
$$\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{-1}$$
With that I find $A = -1, B = -1$ and $C = 0$ which I know is w... | I suppose that you could have also used trigonometric substitutions. Completing the square as you have done, we get that
$$\displaystyle \int \frac{dx}{x^2-2x} = \int \frac{dx}{(x-1)^2 - 1}.$$
We now let $x-1 = \sec \theta.$ Notice then that $dx = \tan \theta \sec \theta \ d\theta.$ We now have
$$ \int \frac{dx}{(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Another Congruence Proof I've been asked to attempt a proof of the following congruence. It is found in a section of my textbook with Wilson's theorem and Fermat's Little theorem. I've pondered the problem for a while and nothing interesting has occurred to me.
$1^23^2\cdot\cdot\cdot(p-4)^2(p-2)^2\equiv (-1)^{(p+1)/2... | Here $p$ must be an odd prime.
There are two cases to consider, $p\equiv 1 \pmod 4$ and $p\equiv 3 \pmod{4}$. We deal with the first case.
We know that $(p-1)!\equiv -1\pmod{p}$. Rearrange the numbers from $1$ to $p-1$, so that we get the odd numbers from $1$ on going up, interleaved with the even numbers from $p-1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/156823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Given $3$ dice, which number is the most likely to appear Given $3$ dice, what is the value of the sum of the number of the $3$ dice most likely to appear? I know that by symmetry, there would always be $2$ different values with the same probability to appear, however I don't know which pairs appear with the highest ch... | Note that there are a total of $6 \times 6 \times 6 = 216$ options. The possible sums are from $3$ to $18$.
Also note that the distribution has to be symmetric since if a roll gives $x,y,z$ adding to $n$, then $7-x,7-y,7-z$ adds to $21-n$.
Hence, the number of ways of getting $n$ is same as the number of ways of gettin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A linear differential equation Find the general solution to:
$$x'' + 2c \; x' + \left( \frac{2}{\cosh^2 t} - 1\right)x =0$$
where $c$ is a constant.
| When $c=0$ we can guess that $x_1 = \frac{1}{\cosh t}$ is a solution. Then by inserting $x = x_1 \cdot u$ into equation it simplifies to:
$$u'' - 2 \tanh t \;u' = 0$$
From which we conclude that: $x(t) = \frac{C_1}{\cosh t} + C_2 \left(\sinh t + \frac{t}{\cosh t} \right)$.
For more general case when $c \neq 0$ we use s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the limit at x=0 Find the limit of $f(x)=\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x}-\sqrt{a+x}}$ (at x=0) so that $f(x)$ becomes continuous for all $x$. My answer is $2\sqrt{a}$. Am I right?
Sorry to state the question incorrectly. We have to define $f(x)$ at $x=0$ such that $f(x)$ is continuous fo... | The answer should be $\sqrt{a}$ assuming $a \geq 0$. Compute the limit of $f(x)$ as $x \to 0$. So for the function to be continuous at $x=0$, the functional value must be set equal to $\lim_{x \to 0} f(x)$.
Move your mouse over the gray area for the complete answer.
Note that $f(x)$ is not defined at $x=0$. Assuming $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/162531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Simplest method to find $5^{20}$ modulo $61$ What is the simplest method to go about finding the remainder of $5^{20}$ divided by $61$?
| $5^3 = 125 \equiv 3 \mod 61$, so $5^5 \equiv 25 \times 3 = 75 \equiv 14 \mod 61$,
$5^{10} \equiv 14^2 = 196 \equiv 13 \mod 61$, $5^{20} \equiv 13^2 = 169 \equiv 47 \mod 61$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/163186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
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Some doubt on Linear Diophantine equation We know $ax+by=c$ is solvable iff $(a,b)|c$ where $a,b,c,x,y$ are integers.
If $x=2$, $a=\dfrac{k(k+5)}{2}$, $y=k$, $b=k+3$ and $c=2k$, where $k$ is any integer, then
$$2 \frac{k(k+5)}{2} - k(k+3) = 2k.$$
So, $\left(\dfrac{k(k+5)}{2}, (k+3)\right) | 2$ for all integral value o... | $2k(k+5)/2 - k(k+3)=2k$ with $a=k(k+5)/2$, $x=k$, $y=k$, would mean $b=k+3$, not $k(k+3)$.
The conclusion is that $(k(k+5)/2,k+3) | 2k$, not $2$. For example, with $k=3$, $(24,6)=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/165577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$. Let $a,b,c>0$ how to prove that :
$$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$$
I find that
$$\ \frac{ab}{a^{2}+3b^{2}}=\frac{1}{\frac{a^{2}+3b^{2}}{ab}}=\frac{1}{\frac{a}{b}+\frac{3b}{a}} $$
By AM-GM
$... | From the question it is obvious that equality is achieved when $a=b=c$, this hints at the use of AM-GM. Firstly, use the fact that $ab \le \frac{1}{2} (a^2+b^2)$, we just need to show
$$ \frac{a^2+b^2}{a^2+3b^2} + \frac{b^2+c^2}{b^2+3c^2} + \frac{c^2+a^2}{c^2+3a^2} \le \frac{3}{2},$$ which is equivalent to (after remo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
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How many natural numbers $x, y$ are possible if $(x - y)^2 = \frac{4xy}{(x + y - 1)}$. How many natural numbers $x$, $y$ are possible if $(x - y)^2 = \frac{4xy}{x + y - 1}$.
Does this system has infinite solutions which can be generalized for some integer $k \geq 2?$
$(x - y)^2(x + y) = (x + y)^2 ;$
since $(x + y)$ can... | It is easy to verify that $(x, y) = (\frac{k(k + 1)}{2},\frac{k(k - 1)}{2})$ is solution of $(x - y)^2 = \frac{4xy}{x + y - 1}$.
we substitute $x$ and $y$ in $(x - y)^2 = \frac{4xy}{x + y - 1}$.
$$\left(\frac{k(k+1)}{2}-\frac{k(k-1)}{2}\right)^2 = \frac{4.[k(k-1)/2][k(k+1)/2]}{[k(k-1)/2]+[k(k+1)/2]-1}$$
after simplif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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How to prove that $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$ Help me prove $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
| $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
Since
$(a-bc)^2\geq 0$,
$(b-ac)^2\geq 0$,
$(c-ab)^2\geq 0$,
then
$a^2+b^2c^2\geq 2abc$,
$b^2+a^2c^2\geq 2abc$,
$c^2+a^2b^2\geq 2abc$.
By of collectted through for through three inequalities last will be obtained
$a^2+b^2c^2+b^2+a^2c^2+c^2+a^2b^2\geq6abc$,
or
$a^2+a^2b^2+b^2+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question about arithmetic progressions, a.k.a. arithmetic sequences What is the relation between the general formula of the sum of $n$ terms of an arithmetic progression, $a n^2+b n+c$, and the first term $a+b+c$ and the common difference?
| Suppose that the terms we are adding up are $a_1,a_2,a_3$, and so on. Let $s_1=a_1$, $s_2=a_1+a_2$, $s_3=a_1+a_2+a_3$, and so on.
If $a_1,a_2,a_3,\dots$ is an arithmetic sequence, let $d$ be the common difference. So $a_1=a_1$, $a_2=a_1+d$, $a_3=a_1+2d$, and so on. We therefore have $a_k=a_1+(k-1)d$. It follows that
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/172001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Evaluating :$\int \frac{1}{x^{10} + x}dx$ $$\int \frac{1}{x^{10} + x}dx$$
My solution :
$$\begin{align*}
\int\frac{1}{x^{10}+x}\,dx&=\int\left(\frac{x^9+1}{x^{10}+x}-\frac{x^9}{x^{10}+x}\right)\,dx\\
&=\int\left(\frac{1}{x}-\frac{x^8}{x^9+1}\right)\,dx\\
&=\ln|x|-\frac{1}{9}\ln|x^9+1|+C
\end{align*}$$
Is there complete... | Not really different, but even simpler:
$$\begin{align}
\int\frac{1}{x^{10}+x} dx=&\int\frac{x^{-10}}{1+x^{-9}} dx
=-\frac 1 9 \log |1+x^{-9}| + C
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
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Proving $\left\lfloor \frac{x}{ab} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor}{b} \right\rfloor$ for $a,b>1$ I'm trying to prove rigorously the following:
$$\left\lfloor \frac{x}{ab} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor}{b} \right\rfloor$$ for integ... | Let $\lfloor x/a \rfloor$ = c(say)=>x=ca+d 0≤d
Let $\lfloor c/b \rfloor$=e(say)=>c=be+f 0≤f
=>x=ca+d=(be+f)a+d=abe+af+d.
$\frac{x}{a}$=be+f+ $\frac{d}{a}$
$\frac{\frac{x}{a}}{b}$=e+$\frac{f}{b}$+$\frac{d}{ab}$
=>$\lfloor x/a/b \rfloor$ = e = $\lfloor c/b \rfloor$ = $\lfloor \lfloor x/a \rfloor /b \rfloor$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
What can be the possible value of $a+b+c$ in the following case? What can be the possible value of $a+b+c$ in the following case?
$$a^{2}-bc=3$$
$$b^{2}-ca=4$$
$$c^{2}-ab=5$$
$0, 1, -1$ or $1/2$?
After doing $II-I$, $III-I$ and $III-II$, I got,
$$(a+b+c)(b-a)=1$$
$$(a+b+c)(c-a)=2$$
$$(a+b+c)(c-b)=1$$
I'm unable to s... | You have
$$\left\{
\begin{align*}
&(a+b+c)(b-a)=1\\
&(a+b+c)(c-a)=2\\
&(a+b+c)(c-b)=1\;.
\end{align*}\right.\tag{1}$$
These clearly imply that $a+b+c\ne 0$, so the first and third of these imply that $b-a=c-b$. In other words, $\langle a,b,c\rangle$ is an arithmetic progression. (The second equation of $(1)$ confirms t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A circle is tangent to the $y$-axis at $y=3$ and has one $x$-intercept at $x=1$. Find the other $x$-intercept
A circle is tangent to the $y$-axis at $y=3$ and has one $x$-intercept at $x=1$. Find the other $x$-intercept
Like previously mentioned, I'm not all too familiar with circles. So, I plotted the two points an... | If (a,b) be the centre of the circle and radius=r and
clearly the circle passes through (1,0)
then $r^2=(1-a)^2+b^2$
The equation of the circle $(x-a)^2+(y-b)^2=(1-a)^2+b^2$
The gradient of the circle at (x,y) = $\frac{dy}{dx} = \frac{(a-x)}{(y-b)}$
The gradient of the circle at (0,3) =$ \frac{(a-0)}{(3-b)}$
As y-ax... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Compute integral $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ I want to solve $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ but I get the wrong results:
$$
\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x =
\int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \math... | You had these steps ok:
$$
\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x =
\int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x
$$
After that, there are a number of choices. It looks like you forgot to integrate the solution.
You could do this:
$$\int_{-6}^6 {\frac{16e^{4x} + 16e^{2x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluating complicated sum Evaluate for a fixed $m\neq 1$ ( $m\in \mathbb{N}$ )
$$\sum _{k=1}^{n}\left[\left( \sum _{i=1}^{k}i^{2}\right) \left(\sum _{k_{1}+k_{2}+...+k_{m}=k}\dfrac {\left( k_{1}+k_{2}+\ldots +k_{m}\right) !} {k_{1}!k_{2}!...k_{m}!}\right)\right]$$
| As Shaktal showed earlier it is remains to find closed form for
$$
\sum\limits_{k=1}^n\frac{m^k k^3}{3}
$$
Consider the following equality
$$
\sum\limits_{k=1}^n x^k=\frac{x(x^n-1)}{x-1}
$$
After triple differentiation we get
$$
\sum\limits_{k=1}^n kx^{k-1}=\frac{d}{dx}\frac{x(x^n-1)}{x-1}\\
\sum\limits_{k=1}^n k(k-1)x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$ This is the sequel of my previous question
$$I(a)=\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$$ I want to use differentiation under the integral sign with respect to parameter "a" but so far without success.
Any hint?
| Thanks for the nice question.
The answer is
$$
I(a) = \frac{\pi}{\sqrt{2}} \cdot \frac{a}{ \sqrt{1 + \sqrt{1+a^2}}}
$$
The sketch of the proof: expand $\arctan$ in series, and integrate term-wise (can do this for small enough $a$, since the sine is bounded):
$$
\arctan\left(a \sin^2(x)\right) = \sum_{n=0}^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
What values can $n$ assume, when $z^2+n= x^2+y^2$? $x^2+y^2 =n +z^2$ where $x,y,z$ are different natural numbers. What values can $n$ assume?
What if x,y,z>0 considering the confusion of 0 as natural number?
| Any odd number : $0^2+(n+1)^2=(2n+1)+n^2$
Any positive even number : $1^2+(n+1)^2=(2n+2)+n^2$
Any negative odd number : $0^2+n^2=-(2n+1)+(n+1)^2$
Any negative even number : $1^2+n^2=-2n+(n+1)^2$
If you really need to have different numbers x y z just remark that
$5^2+11^2=2+12^2$
and $3^2+5^2= -2 + 6^2$
If you need x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine the sum $T=a_0+a_1+a_2+...+a_{2012}$ Let ${a_n}$, $n \ge 0$ be a sequence of positive real numbers, given by
$a_0=1$ and
$a_m<a_n$ for all $m,n \in \mathbb{N}, m<n$ with
$a_n=\sqrt{a_{n+1}a_{n-1}}+1$ and $4\sqrt{a_n}=a_{n+1}-a_{n-1}$ for all $n \in \mathbb{N}, n\neq 0$.
Help me, determining the sum $T=a_0+a_... | $(a_{n+1}+a_{n-1})^2=(a_{n+1}-a_{n-1})^2+4a_{n+1}.a_{n-1}$
$=16a_n+4(a_n-1)^2=4(a_n+1)^2$
=>$a_{n+1}+a_{n-1}=2(a_n+1)$ as $a_n$ is increases with n and $a_0=1$
Or, $a_{n+1}-2 a_n +a_{n-1} -2 =0$
Putting n=m+1 and m,
$a_{m+2}-2 a_{m+1} +a_{m} -2 =0$ and
$a_{m+1}-2 a_m +a_{m-1} -2 =0$
Subtracting $a_{m+2}-3a_{m+1}+3a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do you solve this radical equation $\sqrt{2x+5} + 2\sqrt{x+6} = 5$? I have a radical equation
$$
\sqrt{2x+5} + 2\sqrt{x+6} = 5
$$
and I am having trouble calculating an answer. I keep on getting weird numbers that are not correct as my answer. How do you solve this? A step-by-step procedure would be highly apprec... | First we require $x \geq -\frac{5}{2}$ (for both radicals to be well defined). Now we have that $\sqrt{2x+5}=5-2\sqrt{x+6}$.
If $5-2\sqrt{x+6}\ge 0\Rightarrow -\frac{5}{2}\le x\le \frac{19}{4}$ then
\begin{equation}\sqrt{2x+5}^2=(5-2\sqrt{x+6})^2\Leftrightarrow 2x+5=25+4(x+6)-20\sqrt{x+6}\end{equation}
Can you contin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/175524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
For which angles we know the $\sin$ value algebraically (exact)? For example:
*
*$\sin(15^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$
*$\sin(18^\circ) = \frac{\sqrt{5}}{4} - \frac{1}{4}$
*$\sin(30^\circ) = \frac{1}{2}$
*$\sin(45^\circ) = \frac{1}{\sqrt{2}}$
*$\sin(67 \frac{1}{2}^\circ) = \sqrt{ \frac{\sqrt... | $\sin 3^\circ=\frac{(\sqrt{3}+1) (\sqrt{5}-1)}{8 \sqrt{2}}-\frac{(\sqrt{3}-1) \sqrt{5+\sqrt{5}}}{8}$.
Solving a cubic equation you can get a huge expression for $\sin 1^\circ$ in radicals, and therefore, for any $\sin n^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/176889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 2
} |
recurrence solution to gambler's ruin From DeGroot 2.4.2, let $a_i$ be the conditional probability that the gambler wins all $k$ given gambler is at $i$.
$a_i = pa_{i+1} + (1 - p)a_{i-1} $
It's not clear from the text what steps are taken to solve for the general form $a_i$ (maybe Gaussian elimination). How do you so... | This is best solved using generating functions. Let $f(x) = \sum_{k=0}^\infty a_k x^k$. Now multiply the recurrence equation with $x^k$ and perform the summation:
$$
f(x) - a_0 = \sum_{k=1}^\infty a_k x^k = \sum_{k=1}^\infty (p a_{k+1} + (1-p) a_{k-1}) x^k =
\frac{p}{x} \sum_{k=1}^\infty a_{k+1} x^{k+1} + (1-p) x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/178500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplify the difference quotient $\frac{f(x+h)-f(x)}{h}$.
Simplify the difference quotient $\frac{f(x+h)-f(x)}{h}$ where
a) $f(x)=2x+3,$
b) $f(x)=\frac{1}{x+1},$
c) $f(x)=x^2.$
I believe that if anyone can help me out with the first one, the other two might come clearer to me. But I started out this problem b... | $f(x+h)$ means replace $x$ with $x+h$ in your function definition. If $f(x) = 2x+3$, then $f(x+h) = 2(x+h)+3$, not $2x+3+f(h)$.
Therefore, $$\frac{f(x+h)-f(x)}{h} = \frac{2(x+h)+3-\left(2x+3\right)}{h} = \frac{2x+2h+3-2x-3}{h} = \cdots$$
Edit: To answer your second question, how do you handle just $\frac{1}{x+h+1}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to construct a $2\times 2$ real matrix $A$ not equal to Identity such that $A^3=I$? How to construct a $2\times 2$ real matrix $A$ not equal to Identity such
that $A^3=I$?
There is a correspondence between the ring of complex numbers and the
ring of $2\times2$ matrices (0 matrix
is included!) i.e.,$$a+ib\leftrigh... | Find the three cube roots of $1$. Let $a+bi$ be one of those. They are solutions of $x^3=1$. Hence $x^3-1=0$.
Since $1$ is one of the solutions, $x-1$ must be one of the factors, thus:
$$
x^3-1 = (x-1)(\cdots\cdots\cdots).
$$
Fill in the blanks by doing long division. You should get
$$
(x-1)(x^2+x+1).
$$
So the equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/182632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Hypergeometric functions & integral I'm having difficulty re-deriving a result a calculation from a paper. The integral is
$$\int_0^{2\pi} \int_0^{2\pi} \frac{\sinh\eta}{(\cosh\eta-\cos\theta)^2}\left(1-c\sinh^2\eta\sin\phi\right)^\frac12d\theta d\phi,$$
where $\eta$ and $c$ are parameters such that $\sinh^2\eta = 2$ a... | Going off Peter's comment, note that your integral is separable, and can thus be factored into a product of two one-dimensional integrals:
$$\begin{split}&\int_0^{2\pi} \int_0^{2\pi} \frac{\sinh\eta}{(\cosh\eta-\cos\theta)^2}\left(1-c\sinh^2\eta\sin\phi\right)^\frac12\mathrm d\theta\mathrm d\phi=\\&\quad\left(\color{gr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Why is the last digit of $n^5$ equal to the last digit of $n$? I was wondering why the last digit of $n^5$ is that of $n$? What's the proof and logic behind the statement? I have no idea where to start. Can someone please provide a simple proof or some general ideas about how I can figure out the proof myself? Thanks.
| $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$
$$=n(n^2-1)(n^2-4)+5n(n^2-1)$$
$$=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{ product of }5\text{ consecutive integers }}+5\cdot \underbrace{(n-1)n(n+1)}_{\text{ product of }3\text{ consecutive integers }}$$
Now, we know the product $r$ consecutive integers is divis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 7,
"answer_id": 4
} |
Evaluating a simple definite integral I'm currently teaching myself calculus and am onto the Mean Value Theorem for Integration.
I am finding the value of $f(c)$ on the function $f(x)=x^3-4x^2+3x+4$ on the interval $[1,4]$.
So, with the equation $(b-a)\cdot f(c)=\int_1^4f(x)dx $, you get
$(4-1)\cdot f(c)=\int_1^4(x^3-4... | We have the following string of equalities:
$$\begin{align*}\int_1^4(x^3-4x^2+3x+4)dx &= \left.\frac{x^4}{4}-\frac{4x^3}{3}+\frac{3x^2}{2}+4x\right|_{x=1}^4\\
&= \left(\frac{4^4}{4}-\frac{4\cdot4^3}{3}+\frac{3\cdot4^2}{2}+4\cdot4\right)-\left(\frac{1^4}{4}-\frac{4\cdot1^3}{3}+\frac{3\cdot1^2}{2}+4\cdot1\right)\\
&= \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$t=\frac{30^{65}-29^{65}}{30^{64}-29^{64}}$, find the closest pair of integers, a and b, such that, $a \lt t \lt b$. $t=\frac{30^{65}-29^{65}}{30^{64}-29^{64}}$
find the closest pair of integers, a and b, such that, $a \lt t \lt b$.
$30=1+29$
$(1+29)^{65}=(1+29)(1+29)^{64}$
| Let $A,B,a,b,n$ be positive numbers.
$\frac{A^{n+1}-B^{n+1}}{A^n-B^n}-A=\frac{B^n(A-B)}{A^n-B^n}$ will be $>0$ if $A>B$ and $n ≥ 1$
Putting $A=a+1,B=a$, $\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}>(a+1)$ for $n ≥ 1$.
Alternatively,
$\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}-(a+1)$
$=\frac{(a+1)^{n+1}-a^{n+1}-(a+1)^{n+1}+(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$ Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$
prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$
| Using Hölder's inequality we have:
$$\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2/3} (a(a+b)+b(b+c)+c(c+a))^{1/3}\geq a+b+c.$$
i.e.
$$\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2} \geq \frac{(a+b+c)^{3}}{a^2+b^2+c^2+ab+bc+ca}.$$
We have to prove that:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/185825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 0
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Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility
*
*$9^n$ $-$ $2^n$ is divisible by 7.
*$4^n$ $-$ $1$ is divisible by 3.
*$9^n$ $-$ $4^n$ is divisible by 5.
Can these be generalized as
$a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer?
But why is $a^n$ $-$ $b^n$$ ... | Let $d=a-b$. By the binomial theorem, $a^n = (d+b)^n = dt+b^n$. So $a^n-b^n=dt=(a-b)t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 8,
"answer_id": 5
} |
Inequality $(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\ge 16$ For every real positive number $a,b,c$ such that $ab+bc+ca=1$, how to prove that:
$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2\ge 16$$
| This is problem J57 of mathematical reflections, vol.2007, issue 4;
$a^2+b^2+c^2 \ge ab+bc+ca =1$
$2\left(\frac ab+\frac bc+\frac ca\right)\ge6$ (as BenjaLim does)
It remains
$\frac 1{a^2} + \frac 1{b^2} + \frac 1{c^2} \ge 9$ which follows by (use AGM) $\frac 3{(abc)^{2/3}}\ge9$ or $(abc)^{2/3}\le \frac13.$ This is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Inequality. $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2} \geq 3+n$ I can't find any solution for this inequality which can be found here
Exercise 1.3.5 Let $a,b,c$ be positive real numbers such that $a+b+c=ab+bc+ca$ and $n \leq 3$. Prove that
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{... | Making use of given hint,
$$ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} = {(a + b +c)(a^2 + b^2 + c^2)\over (a+b+c)} = a^2 + b^2 +c^2$$
The minimum value of $a^2+b^2+c^2 \geq a+b+c \geq 3 $ for given criteria.
Let ${a^2 + b^2 + c^2 = z}$ then $f(z, n) = z + {3n \over... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
solving for a coefficent term of factored polynomial.
Given: the coefficent of $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$ is $48,$ find the value of the constant $a.$
I expanded it and got
$64-64\,x-144\,{x}^{2}+320\,{x}^{3}-260\,{x}^{4}+108\,{x}^{5}-23\,{x}^{
6}+2\,{x}^{7}+64\,a{x}^{2}-192\,a{x}^{3}+240\,a{x}^{4... | You need not expand completely as only low powers play a role:
$(2-x)^6 = 2^6-6\cdot 2^5\cdot x + \frac{6\cdot5}2\cdot 2^4\cdot x^2+\ldots = 64-192x+240x^2+\ldots$
where the dots represent anything involving $x^3$ or even higher powers.
After this
$(1+2x+ax^2)(2-x)^6 = (64-192x+240x^2+\ldots) + (128x-384x^2+\ldots)+ (6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Can I get better approximation of $\sum_{k=1}^{n} k^k$ Is it possible to get approximation$f(n)$ of $\sum_{k=1}^{n} k^k$ with
\begin{align}
\lim_{n\to +\infty }\left(f(n)-\sum_{k=1}^{n} k^k\right)=0
\end{align}
Thanks for your attention!
| I have no idea how $f(n)$ should look like, and actually I suspect that there is no such simple formula for $f(n)$. But at least we can improve Sasha's asymptotics. Indeed, a moment of thought gives that for any fixed $m$, we have
\begin{align*}
\frac{1}{n^n} \sum_{k=1}^{n} k^k
&= \sum_{k=0}^{n-1} \left(1 - \frac{k}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
An intriguing definite integral: $\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$ I need some hints, suggestions for the following integral
$$\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$$
Since it's a high school problem, I thought of some variable change, integration ... | Let $u = 2\tan^{-1}\left(\frac{x}{4}\right) - x$. Then
$$ du = -\frac{x^2 + 8}{x^2 + 16}\,dx. $$
Now since
$$ \begin{align*}
\frac{x^2 + 8}{(x^2 - 16)\sin x + 8x \cos x}
&= \frac{\frac{x^2 + 8}{x^2 + 16}}{\frac{8x}{16+x^2} \cos x - \frac{16-x^2}{16+x^2}\sin x} \\
&= \frac{\frac{x^2 + 8}{x^2 + 16}}{\sin\left(2\tan^{-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to solve $x^3=-1$? How to solve $x^3=-1$? I got following:
$x^3=-1$
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
| The real solution:
$$x^3+1=0<=>$$
$$x^3=-1<=>$$
Take cube roots of both sides:
$$x=-1$$
Complex solution:
$$x^3+1=0<=>$$
$$x^3=-1<=>$$
$$x^3=|-1|e^{arg(-1)i}<=>$$
$$x^3=e^{(\pi +2\pi k)i}<=>$$
(with k is the element of Z)
$$x=\left(e^{(\pi +2\pi k)i}\right)^{\frac{1}{3}}<=>$$
$$x=e^{\left(\frac{1}{3}\pi +\frac{2}{3}\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 8,
"answer_id": 5
} |
More and more limits for sequences So here goes a bit of homework:
$$\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}$$
Well, this would trivially lead to:
$$\lim_{n\to\infty}{\left(\frac{3+\frac{2}{n}+\frac{1}{n^2}}{3+\frac{5}{n^2}}\right)^{\frac{n\left(1+\frac{2}{n^2}\right)}{2+\frac{1}{n... | $$\frac{3n^2+2n+1}{3n^2-5}=1+\frac{2n+6}{3n^2-5}=1+\frac{2}{\frac{3n^2-5}{n+3}}\Longrightarrow$$
$$\Longrightarrow\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}=\left[\left(1+\frac{2}{\frac{3n^2-5}{n+3}}\right)^{\frac{3n^2-5}{n+3}}\right]^\frac{n^3+3n^2+2n+6}{6n^3+3n^2-10n-5}\Longrightarrow$$
Well, now the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
how to calculate the exact value of $\tan \frac{\pi}{10}$ I have an extra homework: to calculate the exact value of $ \tan \frac{\pi}{10}$.
From WolframAlpha calculator I know that it's $\sqrt{1-\frac{2}{\sqrt{5}}} $, but i have no idea how to calculate that.
Thank you in advance,
Greg
| If $10x=\pi$
$\sin 2x=\cos 3x$ as $2x+3x=5x=\frac{\pi}{2}$
$\implies2\sin x \cos x=4\cos^3x-3\cos x$
$\implies 2\sin x=4\cos^2x-3$ as $\cos x≠0$
If $\sin x=t, 2t=4(1-t^2)-3\implies 4t^2+2t-1=0$
$$\implies t=\frac{-1±\sqrt{5}}{4}$$, but $\sin x>0$ as $0<x<\pi$
$$\sin \frac{\pi}{10}=\frac{\sqrt{5}-1}{4}$$
(1)So, $$\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/196067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
$\int\frac{x^3}{\sqrt{4+x^2}}$ I was trying to calculate
$$\int\frac{x^3}{\sqrt{4+x^2}}$$
Doing $x = 2\tan(\theta)$, $dx = 2\sec^2(\theta)~d\theta$, $-\pi/2 < 0 < \pi/2$ I have:
$$\int\frac{\left(2\tan(\theta)\right)^3\cdot2\cdot\sec^2(\theta)~d\theta}{2\sec(\theta)}$$
which is
$$8\int\tan(\theta)\cdot\tan^2(\theta)\... | $$
\begin{aligned}
\int \frac{x^{3}}{\sqrt{4+x^{2}}} d x &=\int x^{2} d \sqrt{4+x^{2}} \\
&\stackrel{IBP}{=} x^{2} \sqrt{4+x^{2}}-\int \sqrt{4+x^{2}} d\left(x^{2}\right) \\
&=x^{2} \sqrt{4+x^{2}}-\frac{2}{3}\left(4+x^{2}\right)^{\frac{3}{2}}+C \\
&=\frac{\sqrt{4+x^{2}}}{3}\left[3 x^{2}-2\left(4+x^{2}\right)\right]+C \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/197744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
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