Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Integration using different substitutions resulting in different answers? I'm going through the MIT online calc course, and I'm trying to find the following from an Integration Problem set:
$$\int \frac{x^2+1}{x^2+2x+2}dx$$
I could use the substitution u = x+1, which results in the correct answer:
$$\int \frac{(u-1)^2+... | The mistake is in thinking $\cos\theta=x^2+2x+2$.
Remark: I would probably want to divide first, getting $1-\dfrac{2x+1}{(x+1)^2+1}$.
Then write $\dfrac{2x+1}{(x+1)^2+1}$ as $\dfrac{2x+2}{(x+1)^2+1}-\dfrac{1}{(x+1)^2+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Implicit differentiation question Given that $x^n + y^n = 1$, show that $$\frac{d^2y}{dx^2} = -\frac{(n-1)x^{n-2}}{y^{2n-1}}.$$
I found that $\displaystyle nx^{n-1}+ny^{n-1}\frac{dy}{dx} = 0$ so that $\displaystyle y'=\frac{-x^{n-1}}{y^{n-1}}$.
Then $$n(n-1)x^{n-2}+n(n-1)y^{n-2}\left(\frac{dy}{dx}\right)^2 + \frac{d^2y... | Differentiating $y^{n-1}y_1+x^{n-1}=0$ wrt $x,$
$\displaystyle y^{n-1}y_2+(n-1)y^{n-2}y_1^2+(n-1)x^{n-2}=0$
$\implies \displaystyle -\frac{y^{n-1}y_2}{n-1}=x^{n-2}+y^{n-2}\left(-\frac{x^{n-1}}{y^{n-1}}\right)^2$ as $y_1=-\frac{x^{n-1}}{y^{n-1}}$
$$\implies -\frac{y^{n-1}y_2}{n-1}=x^{n-2}+\frac{x^{2n-2}}{y^n}=\frac{x^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Give $3$ coterminal angles
Give three angle measures in radians which are coterminal with each of the following. Include at least one positive and one negative angle measure.
*
*$$\pi/4\quad \text{rad}$$
*$$5\pi/3\quad \text{rad}$$
| Firstly let us have a look at the definition of coterminal:
Two angles are coterminal if they are drawn in the standard position and both have their terminal sides in the same location.
The angles can be anything as long as the lines are aligned. Here are some examples:
$\angle ABC$ and $\angle DBC$ are coterminal ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Effective method to solve $ \frac{x}{3x-5}\leq \frac{2}{x-1}$ I need to solve this inequality. How can I do so effectively?
$$ \frac{x}{3x-5}\leq \frac{2}{x-1}$$
| Make $0$ one side of the inequality:
$\displaystyle\frac{x}{3x-5}-\frac{2}{x-1}\le 0$
Write the expression in $x$ as a single fraction:
$\displaystyle\frac{x(x-1)-2(3x-5)}{(3x-5)(x-1)}\le 0$
Expand the numerator:
$\displaystyle\frac{x^2-7x+10}{(3x-5)(x-1)}\le 0$
Factorise the numerator:
$\displaystyle\frac{(x-2)(x-5)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/482518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solutions of $x^2 + 7y^2 = 2^n$ where $x$ and $y$ are odd numbers Is it true that for any $n\geq 2$ the equation $x^2 + 7y^2 = 2^n$ has a solution with $x$ and $y$ odd ??
| The result is true for $n=3$, since $2^3=x^2+7y^2$ with $x=y=1$. Moreover, $x\equiv y\pmod{4}$.
We show that if there exist $x$ and $y$, odd, with $x\equiv y\pmod{4}$, such that $x^2+7y^2=1$, then there exist $x_1,y_1$, odd, with $x_1\equiv y_1\pmod{4}$, such that $x_1^2+7y_1^2=2^{n+1}$. Let
$$x_1=\frac{x+7y}{2}\quad\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
How find the value $\lim_{x\to 0}\dfrac{1}{x^2}\left(1-\cos{x}\cdot\sqrt{\cos{(2x)}}\cdots\sqrt[n]{\cos{(nx)}}\right)$ find the value
$$I_{n}=\lim_{x\to 0}\dfrac{1}{x^2}\left(1-\cos{x}\cdot\sqrt{\cos{(2x)}}\cdots\sqrt[n]{\cos{(nx)}}\right)$$
This is my methods:
\begin{align*}I_{n+1}-I_{n}&=\lim_{x\to 0}\dfrac{1}{x^2}\l... | $$
\cos(nx) = 1 - \frac{n^2 x^2}{2} + O(x^4)
$$
as $x \to 0$ so
$$
\sqrt[n]{\cos(nx)} = 1 - \frac{n x^2}{2} + O(x^4)
$$
by the binomial theorem, thus
$$
\begin{align}
\cos(x)\sqrt{\cos(2x)}\cdots\sqrt[n]{\cos(nx)} &= 1 - \frac{x^2}{2} \sum_{k=1}^{n} k + O(x^4) \\
&= 1 - \frac{n(n+1)}{4} x^2 + O(x^4).
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Evaluation Of Maximum Value Without Calculus $x$,$y$,$z$ are non-negative numbers.
$x+y+z=3$
Find the maximum value of $~$ $x^{2}y+y^{2}z+z^{2}x$ $~$ without calculus.
| Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Hence, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$
$$=b(a^2+ac+c^2)\leq b(a^2+2ac+c^2)=b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq$$
$$\leq4\left(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/489292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In a numerical system of base $r$, the polynomial $x^2 − 11x + 22 = 0$ has the solutions $3$ and $6$. What is the base r of the system? From Algebra, the statement is equivalent to say that $(x^2− 11x +
22)_{r}$ = $(x − 3)_{r} \cdot (x − 6)_{r}$. Doing operations we arrive at $3 + 6 = 11_{r} = r + 1$, and $(3)(6) = 22_... | If the roots are $3$ and $6$, the equation is $(x-3)(x-6)=x^2-(3+6)x+3\cdot 6=0$ So $11_r=3+6$ and $11$ in base $r$ is $r+1$ because the leading digit is $r$. Similarly we have $3\cdot 6 = 22_r=2r+2$ Each of these gives us $r=8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/489969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Summing random numbers with different domain This should be simple, but my math skills are rusty.
Given the random integers $X_0,\ldots, X_i,\ldots, X_n$, each of which is uniformly distributed on a domain $0,\ldots,J_i,\ldots,J_n$, what is the probability that the sum of the numbers will be greater than or equal to so... | I will show you how to do the two case example, and hopefully you can then do it for a three case scenario. So, suppose that $X$ models the first dice throw, and therefore has a discrete uniform distribution on $[1,6]$, while $Y$ models the second dice throw, thus being discrete uniform on $[1,10]$. We are interested i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/491802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$ Knowing that $1 - \frac 12 + \frac 13 - \cdots = \ln 2$ and $1 - \frac 13 + \frac 15 - \cdots = \frac{\pi}{4}$, compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}... | $$
{\cal F}\left(x\right)
\equiv
\sum_{n = 0}^{\infty}{x^{4n + 2} \over \left(4n +1\right)\left(4n + 2\right)}
$$
\begin{align}
{\cal F}''\left(x\right)
&\equiv
\sum_{n = 0}^{\infty}x^{4n}
=
{1 \over 1 - x^{4}}
=
{1/2 \over 1 - x^{2}} + {1/2 \over 1 + x^{2}}
=
{1 \over 4}\sum_{\sigma = \pm}{1 \over 1 + \sigma\,x} + {1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
what is the largest number here? what is the largest number here? how to find these with out inspection but a proper mathematical rule?
$$\Large 2^{3^4},2^{4^3},3^{2^4},3^{4^2},4^{2^3},4^{3^2}$$
Thank you
| As $\displaystyle a^{m^n}=a^{(m^n)}$ and $\displaystyle (a^m)^n=a^{m\cdot n}$
$2^{3^4}=2^{81}$
$2^{4^3}=2^{64}$
$3^{2^4}=3^{16}$
$3^{4^2}=3^{16}$
$4^{2^3}=(2^2)^{(2^3)}=(2^2)^8=2^{16}$
$4^{3^2}=(2^2)^{(3^2)}=(2^2)^9=2^{18}$
As we are interested only in finding the largest, we can safely consider $2^{81}$ and $3^{16}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493554",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Minimum and maximum value of function $f(x,y,z) = x^3 + y^3 + z^3$ Find minimum and maximum value of function $$f(x,y,z) = x^3 + y^3 + z^3$$ on set $$ \left\{ (x,y,z): x^2 + y^2 + z^2 = 1 \wedge x+y+z = \sqrt{3} \right\} $$
I don't know what is this set. We have sphere and plane so I suppose that it may be circle or... | note
$$(x^2+y^2+z^2)(1+1+1)\ge (x+y+z)^2$$
so if
$$x^2+y^2+z^2=1,x+y+z=\sqrt{3}$$
then we have
$$3(x^2+y^2+z^2)=(x+y+z)^2$$
$$\Longrightarrow x^2+y^2+z^2-xy-yz-xz=0$$
so
$x=y=z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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What is the sum of $4\sqrt{28}$ and $3\sqrt{7}$ ? As far as I can simplify it -
$$4\sqrt{7*4} + 3\sqrt{7}
= 8\sqrt{7} + 3 \sqrt{7}
= \sqrt{7} * 11$$
However , The options for the correct answer are -
A) $ 8/3$
B) $ 16/3$
C) $ 18/3$
D) $24/3$
I am a ninth grader so please try to explain in simple terms .
| I think there was a misprint , most probably they meant - $$\dfrac{4\sqrt{28}} {3\sqrt{7}}
= \dfrac{8\sqrt{7}} {3\sqrt{7}}
= \dfrac83 $$
In this case , the answer would be A) 8/3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.
I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
| $x^4+x^3+x^2+x+1=0$. Divide by $x^2$, and substitute $z=x+x^{-1}$, resulting in $z^2+z-1=0$. This has two solutions: $z_1=2cos72^\circ=\dfrac{-1+\sqrt5}{2}$, and $z_2=2cos144^\circ=\dfrac{-1-\sqrt5}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Solving a system of equation modul0 5 Consider the system of linear equations $$\begin{pmatrix} 6 & -3\\ 2 & 6 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}=\begin{pmatrix} 3\\ 1 \end{pmatrix} $$
a) Solve the system in $\mathbb{F}_5$
I just want to make sure my solution is correct:
We have: $$A=\begin{pmatrix} ... | This method is legitimate since, if a solution exists, the determinant is non-zero mod $5$, and thus is invertible mod $5$.
To continue, $21 \equiv 1 \pmod 5$ and $\frac{1}{2}=2^{-1} \equiv 3 \pmod 5$.
So the solution becomes $$\begin{bmatrix} 3 \\ 0 \\ \end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trying to calculate the probability that one RV exceeds another RV I am running into a silly mistake when trying to calculate the probability that a random variable, $U$ is less than another random variable, $V$. I am hoping that someone can help me spot my mistake.
My setup is as follows:
*
*$U$ ~ $U(0,1)$
*$W = \... | It looks like the integral does evaluate to a meaningful result. I just had a silly algebra mistake.
We can proceed by substitution. Define $x = 1-2w$, so that $w = \frac{1-x}{2}$ and $dx = \frac{dw}{-2}$. Then:
$$\begin{align} P(U<W) &= \int_{0}^{1/2} 2nw(1-2w)^{n-1} dw \\
&= \int 2n \frac{1-x}{2} x^{n-1} \frac{dx}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How to solve the following equation $\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$ I am trying to solve this equation:
$$\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$$
I would like to get some advice, how to solve it.
Thanks.
| W'll use $$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2.$$
Since $x+3=x=-3-8x$ is impossible, our equation is equivalent to
$$x+3+x-3-8x+3\sqrt[3]{x(x+3)(8x+3)}=0$$ or
$$\sqrt[3]{x(x+3)(8x+3)}=2x$$ or
$$x((x+3)(8x+3)-8x^2)=0,$$
which gives $x=0$ or $x=-\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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Show that the odd prime divisors of the integer $n^2+n+1$ which are different from $3$ are of the form $6k+1$. 1)The odd prime divisors of the integer $n^4+1$ are of the form $8k+1$.
My attempt: Let $p$ be odd divisor of $n^4+1$.Then $n^4+1 \equiv 0$ (mod p) $\Rightarrow n^4 \equiv -1$ (mod p) $\Rightarrow n^8 \equiv 1... | Hint: $$4(n^2+n+1)=(2n+1)^2+3$$
Alternatively:
$$(n-1)(n^2+n+1)=n^3-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrating $\sec^2 x$ from first principles Is it possible to solve $\int \sec^2 x ~dx$ without knowing that $\frac{d}{dx}\tan x = \sec^2 x$?
| Yes, this can be done, but the method is far longer than if you know the derivative of $\tan x$ already. The point is to convert this trigonometric integral into the integral of a rational function, integrate that with partial fractions, and then convert back. We use the substitution $u = \tan(x/2)$, for which
$$
\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
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Number theory proof from AoPS http://www.artofproblemsolving.com/Resources/articles.php?page=htw.readers
In the above link, he gives a problem, namely
Let $S(n)$ be the sum of the digits of $n$. Find
$S(S(S(4444^{4444}))).$
Then in the lemma he presents is where I get confused. He states
Every integer $n$, writte... | Let's define a $k$-digit number $n$ to have the decimal (base-10) digits $d_0, d_1, \ldots d_{k-1}$. We can then write $n$ as:
$$n = 10^0d_0 + 10^1d_1 + \cdots + 10^{k-1}d_{k-1}$$
Note that we can re-arrange this as:
$$\begin{align}
n &= d_0 + 10^1d_1 + \cdots + 10^{k-1}d_{k-1}\\
&= d_0 + (9d_1+d_1) + \cdots + (10^{k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Trigonometric Equation $\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$ How can I solve this trigonometric equation?
$$\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$$
| Ok, there might be nicer ways to do this but here it goes. Let $\theta=\pi/15$ so you have
$$
\tan(\theta)\tan(4\theta)\tan(\frac{9}{2}\theta)\tan(6\theta)
$$
then you use $\tan()=\frac{\sin()}{\cos()}$, and then you use the equivalences
$2\sin(\alpha)\sin(\beta)=\cos(\alpha-\beta)-\cos(\alpha+\beta),$
and
$2\cos(\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Combinatorial interpretation of an alternating binomial sum Let $n$ be a fixed natural number. I have reason to believe that $$\sum_{i=k}^n (-1)^{i-k} \binom{i}{k} \binom{n+1}{i+1}=1$$ for all $0\leq k \leq n.$ However I can not prove this. Any method to prove this will be appreciated but a combinatorial solution is gr... | Suppose we seek to verify that
$$\sum_{q=k}^n (-1)^{q-k} {q\choose k} {n+1\choose q+1} = 1$$
where $n\ge k.$
We first treat the case when $k\gt 0$ and introduce
$${q\choose k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} (1+z)^q \; dz.$$
Observe that this is zero when $0\le q\lt k$ so that we may extend th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
} |
Simplest or nicest proof that $1+x \le e^x$ The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is acc... | The shortest proof I could think of:
$$1 + x \leq 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = e^x.$$
However, it is not completely obvious for negative $x$.
Using derivatives:
Take $f(x) = e^x - 1 - x$. Then $f'(x) = e^x - 1$ with $f'(x) = 0$ if and only if $x = 0$. But this is a minimum (global in this case) si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/504663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "113",
"answer_count": 27,
"answer_id": 2
} |
Integrate $\int x \sqrt{2 - \sqrt{1-x^2}}dx $ it seems that integration by parts with some relation to substitution...
$$
\int x \sqrt{2-\sqrt{1-x^2}}
= \frac{2}{5} \sqrt{2-\sqrt{1-x^2}} \cdot \sqrt{1-x^2}+\frac{8}{15}\sqrt{2-\sqrt{1-x^2}}+c
$$
How can I get that?
| Let $x=\sqrt{1-y^2}$, then then integral becomes
$$-\int dy \, y \, \sqrt{2-y}$$
Integrate by parts:
$$\frac{2}{3}y (2-y)^{3/2} + \frac{2}{3} \int dy \, (2-y)^{3/2}$$
which is
$$\frac{2}{3}y (2-y)^{3/2} - \frac{4}{15} (2-y)^{5/2} + C$$
Therfore
$$\int dx \, x \, \sqrt{2-\sqrt{1-x^2}} = \frac{2}{3} \sqrt{1-x^2} \left ( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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A definite integral $\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx$ I need to find a value of this definite integral:
$$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx.$$
Its numeric value is approximately $0.7875720991394284$, and lookups in Inverse Symbolic Calc... | There is an alternate way to compute this integral w/o a summation over $n$ in the middle steps.
Notice
$$\frac{2-\cos z}{5 - 4\cos z} = \frac12 \left[\frac{(2-e^{iz})+(2-e^{-iz})}{(2-e^{iz})(2-e^{-iz})}\right]
= \frac12\left[\frac{1}{2-e^{iz}} + \frac{1}{2-e^{-iz}}\right]
$$
and $\displaystyle\;\frac{1}{1+z^4}$ is an ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
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How prove this $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1 $ Show that
$$\sum_{k=0}^{n}\dfrac{\binom{2n-k}{n}}{2^{2n-k}}=1$$
I think this problem can be solved with nice methods, such as algebraic ones. Or can I use probability methods? Thank you
| Suppose we seek to show that
$$\sum_{k=0}^{n} {2n-k\choose n} 2^k = 2^{2n}.$$
Introduce
$${2n-k\choose n}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-k}}{z^{n+1}} \; dz.$$
This gives for the sum that
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}}
\sum_{k=0}^n \frac{2^k}{(1+z)^k}\; dz
\\ =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
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Form of a harmonic function Given that $\phi(x^2+y^2)$ is harmonic, where $\phi: (0, \infty)\to \mathbb{R}$, find the form of $\phi$.
I do not know what they mean by form nor could I find anything online... My book says that for the complex function $f(z)= (y^3-3x^2y)+i(-3xy^2+x^3+C)$ the form $f(z) = i(z^3+c)$ is easi... | Hint: Note that $\phi$ is a function of one variable, hence $\psi_{xx} + \psi_{yy} = 0$ implies
$$4\phi'(x^2+y^2)+4(x^2+y^2)\phi''(x^2+y^2) = 0,$$
i.e.
$$\phi'(t)+t\phi''(t) = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Approximate $\int_0^{\pi /2} \frac{ds}{\sqrt{1-x\sin^2s}}$ I am trying to approximate the following integral
$$K(x)=\int\limits_0^{\pi /2} \frac{ds}{\sqrt{1-x\sin^2s}}$$
with $0<x<1$.
I need to show that for x close to one that $K(x)\sim -\frac{1}{2}\ln(1-x)$.
My first attempt was to Taylor expand the function $f(x)= ... | The integral as written is the
complete elliptic integral of the first kind and it can be expressed in term of hypergeometric function:
$$K(x) = \frac{\pi}{2}\,_2F_1(\frac12,\frac12;1;x)\tag{*1}$$
We know when $\gamma = \alpha + \beta$, $\alpha$ and $\beta \ne 0, -1, -2, \ldots$, $|\arg(1-x)| < \pi$, $|1-x| < 1$, the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/511202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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How do I decompose this partial fraction case? Decompose $$\dfrac{2x}{1+x} $$
Looking at this case, it looks like any simple partial fraction. But it is trickly. This is how I attempted:
$$\dfrac{2x}{1+x} = \dfrac{A}{1+x}$$
multiply by LCD $(1+x)$ to get $2x = A$
How to I reduce this to give me the value of $A$?
What ... | $$\frac{x}{1+x} = \frac{1+x}{1+x} - \frac{1}{1+x} = 1-\frac{1}{1+x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Extended euclidean algorithm So I am trying to figure this out.
And for one of the problem the question is x*41= 1 (mod 99)
And the answer lists
x | 41x mod 99
0 99
1 41
-2 17
5 7
-12 3
29 1
And conclude x=29
How did they get this value(can someone explain)? To better put, how do you calculate thi... | A better table would probably be the following
$$\begin{matrix}
99 & 41\\
\hline
1 & 0 & | & 99\\
0 & 1 & | & 41\\
1 & -2 & | & 17\\
-2 & 5 & | & 7\\
5 & -12 & | & 3\\
-12 & 29 & | & 1
\end{matrix}$$
where each line
$$\begin{matrix}
a & b & | & r\\
\end{matrix}$$
means
$$
99 \cdot a + 41 \cdot b = r,
$$
and you go from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/513217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by induction that $1^3 + \dots + n^3 = (1 + \dots + n)^2$ I'm suppose to prove by induction:
$1^3 + \dots + n^3 = (1 + \dots + n)^2$
This is my attempt; I'm stuck on the problem of factoring dots.
| Hint:
$$
(n+1)^3 =(n+1)(n+1+n(n+1))=(n+1)(n+1+2(1+2+\dots+n))\\
$$
Then
$$
(1 + \dots + n)^2+(n+1)^3=(n+1)^2+2(n+1)(1+2+\dots+n)+(1+2+\dots+n)^2\\
=(1+2+\dots+n+(n+1))^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/516044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solve for $x: 2/\sqrt{2-x^2} = 4- 2x^2?$ How do you expand and solve for $x$?
It is $2 = 4\sqrt{2-x^2} - 2x^2\sqrt{2-x^2}$
Thank you!
How would I solve for x?
| \begin{align*}
\frac{2}{\sqrt{2-x^2}} &= 4 - 2x^2\\
2 &= 2(2-x^2)^{3/2}\\
1 &= (2-x^2)^{3/2}\\
1 &= 2-x^2\\
1 &= x^2\\
\pm1 &= x
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How is it, that $\sqrt{x^2}$ is not $ x$, but $|x|$? As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.
But then I remember that $\sqrt{x}$ is the same as $x^{1... | It is true that $\sqrt{x} = x^{\frac{1}{2}}$ and $(\sqrt{x})^2 = (x^{\frac{1}{2}})^2 = x$. This is true for all $x$ in the domain of $\sqrt{x}$, namely $x \in [0, \infty)$.
The absolute value comes from composing the square root and square in the opposite order; that is $\sqrt{x^2} = (x^2)^{\frac{1}{2}}$. For any $x \i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Given that $gcd(a,b)=1$, prove that $gcd(a+b,a^2-ab+b^2)=1$ or $3$, also when will it equal $1$? It is an exercise on the lecture that i am unable to prove.
Given that $gcd(a,b)=1$, prove that $gcd(a+b,a^2-ab+b^2)=1$ or $3$, also when will it equal $1$?
| Let $d|a+b \quad(1)$ and $d|a^2-ab+b^2 (2)$
$ d|a^2-ab+b^2 \quad (2)\implies d|(a+b)^2 - 3ab \qquad \qquad(3)$
And as $d|a+b$ then, $(3) \land (1) \implies d|-3ab \implies d|3ab \implies d|3 \vee d|a \vee d|b$
On the other hand, as $d|a+b$, if $d|a \implies d|b$, and vice verse. And as $gcd(a,b)=1$, then $d=1$.
The ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to have this definite integral?
Suppose that a function $f$ of $x$ and $y$ be defined as follows:$$f(x,y) =
\begin{cases}
\frac{21}{4}x^2y & \text{for $x^2 \leq y\leq 1$,} \\
0 & \text{otherwise. } \\
\end{cases}$$
I have to determine the value of integral for which $y\leq x$ also holds.
The answer is $\frac{... | The given condition $y\le x$and $x^2\le y\le1$ implies $x^2\le y\le x$.
Since $x^2\le x$, $0\le x\le 1$.
$$\int_{0}^1\int_{x^2}^x\frac{21}{4}x^2ydydx=\int_{0}^1\left[\frac{21}{8}x^2y^2\right]^x_{x^2}dx=\int_{0}^1\frac{21}{8}x^2(x^2-x^4)=\frac{21}{8}\left[\frac{x^5}{5}-\frac{x^7}{7}\right]^1_{0}=\frac{3}{20}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluating $\int_{-2}^{2} 4-x^2 dx$ with a Riemann sum I'm having problems with a Riemann sum ... I need to find the integral:$$\int_{-2}^2 (4-x^2)\;dx$$Clearly we have $$\int_{-2}^{2}(4-x^2)\;dx=4x-\frac{x^3}{3}\mid_{-2}^{2}=(4\cdot2-\frac{2^3}{3})-(4\cdot(-2)-\frac{(-2)^3}{3})=\frac{32}{3}$$OK.
On the other hand, we... | Check the step where you expand $n \left( n+1 \right) \left( 2\,n+1 \right)$. I believe the mistake is in that step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/525189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $x^3 \equiv x \bmod 6$ for all integers $x$
Prove that $x^3 \equiv x \bmod 6$ for all integers $x$
I think I got it, but is this proof correct?
We can write any integer x in the form: $x = 6k, x = 6k + 1, x = 6k + 2, x = 6k + 3, x = 6k + 4$, and $x = 6k + 5$.
If $x = 6k$, then $x^3 = 216k^3$. Then $x^3 - ... | One has
$$x^3-x=x(x+1)(x-1)\qquad\forall x\ .$$
When $x$ is an integer then at least one factor on the right is even, and exactly one factor on the right is divisible by $3$. It follows that for any $x\in{\mathbb Z}$ the right hand side is divisible by $6$, and so is the left hand side. That is to say: $x^3=x$ mod $6$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the inequality $xyz^3 \leq 27(\frac{x+y+z}{5})^5$. For $x,y,z>0$, prove the inequality $xyz^3 \leq 27(\frac{x+y+z}{5})^5$.
Any ideas? I am stuck.
| Write $f(x,y,z) = 27 \left( \frac{x+y+z}{5} \right)^5 - x y z^3$. Then
$$\frac{\partial f}{\partial x}(x,y,z) = 27 \left(\frac{x+y+z}{5}\right)^4 - y z^3$$
$$\frac{\partial f}{\partial y}(x,y,z) = 27 \left(\frac{x+y+z}{5}\right)^4 - x z^3$$
$$\frac{\partial f}{\partial z}(x,y,z) = 27 \left(\frac{x+y+z}{5}\right)^4 - 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to solve the inequality $\frac {5x+1}{4x-1}\geq1$ Please help me solve the following inequality.
\begin{eqnarray}
\\\frac {5x+1}{4x-1}\geq1\\
\end{eqnarray}
I have tried the following method but it is wrong. Why?
\begin{eqnarray}
\\\frac {5x+1}{4x-1}&\geq&1\\
\\5x+1&\geq& 4x-1\\
\\x &\geq& -2
\end{eqnarray}
Th... | $$
\frac{5x+1}{4x-1}\geq 1\Leftrightarrow \frac{5x+1}{4x-1}-1\geq 0\Leftrightarrow\frac{5x+1-(4x-1)}{4x-1}\geq 0 \Leftrightarrow \frac{5x+1-4x+1}{4x-1}\geq 0
$$
$$
\Leftrightarrow \frac{x+2}{4x-1}\geq 0
$$
Now $x+2=0 \Rightarrow x=-2$, and $4x-1=0\Rightarrow x=\frac{1}{4}$
For $x\in (-\infty, -2]$, and $x\in(\frac{1}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Finding $\int \frac{x^2}{(a^2-x^2)^{\frac{3}{2}}}dx$ using trigonometric substitution. Where did I go wrong? Evaluate the following integral using trigonometric substitution
$$\int \frac{x^2}{(a^2-x^2)^{\frac{3}{2}}}dx$$
I used the substitution $x=a \sin(u)$, then $dx = a \cos(u) du$. The integral then becomes:
$$\int ... | The bottom should be $a^3\cos^3 u$, so you are essentially integrating $\tan^2 u$, that is, $\sec^2 u-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/529088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral of quartic function in denominator I'm sorry, I've really tried to use MathJaX but I can't get integrals to work properly.
indefinite integral
$$\int {x\over x^4 +x^2 +1}$$
I set it up to equal
$$x\int {x\over x^4 +x^2 +1} - \int {x\over x^4 +x^2 +1}$$
$$\text{so } (x-1)\int {1\over x^4 +x^2 +1}$$
OKAY, now I ... | HINT:
As $\displaystyle x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$ and $\displaystyle (x^2+x+1)-(x^2-x+1)=2x$
$$\frac x{x^4+x^2+1}=\frac12\frac{(x^2+x+1)-(x^2-x+1)}{x^4+x+1}=\frac12\left(\frac1{x^2-x+1}-\frac1{x^2+x+1}\right)$$
Now as, $\displaystyle x^2+x+1=\frac{(2x+1)^2+(\sqrt3)^2}4$ put $2x+1=\sqrt3\tan\theta$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/531322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Showing that a system of Diophantine equations will have irrational solutions as well as integers Solve $\begin{cases} 3xy-2y^2=-2\\ 9x^2+4y^2=10 \end{cases}$
Rearranging the 2nd equation to $x^2=\dfrac{10-4y^2}{9} \Longrightarrow 0\leq x^2 \leq 1$ if $x^2=1$ than $y=\pm\dfrac{1}{2}$ and $x=\pm1$ but how do I show the... | If you add five times the first equation plus the second, you get $9 x^2 + 15 x y - 6 y^2 = 0,$ or
$$ 3 (3x - y)(x+ 2 y) = 0. $$
So the choices lie along either the line $$ y = 3x $$ or the line $$ y = \frac{-x}{2}. $$
From the ellipse equation $9 x^2 + 4 y^2 = 10,$ we get either $45 x^2 = 10$ and $9 x^2 = 2$ and $(3x... | {
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"url": "https://math.stackexchange.com/questions/532786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How prove this $x+y=0$ if $\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$ Question:
let $x,y$ are real numbers,and such
$$\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$$
show that
$$x+y=0\tag{1}$$
before I have solve following problem:
if
$$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1,\L... | Rather than use $x$ directly, we will use $z=-x$.
Suppose that $y>0$. Let
$$
f_1(z)=z+\sqrt{y^2+z^3} \ (\text{for} \ z\geq{-y^{\frac{2}{3}}}), \ \
f_2(z)=\frac{y^3}{-y+\sqrt{y^3+z^2}}
\ (\text{for} \ z\in{\mathbb R}), \
$$
We must then show that $f_1(z) \neq f_2(z)$ when $z\neq y$.
It is easy to see that $f_1$ is in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the natural numbers $a$ and $b$ so that $a\cdot b$ has the largest possible value but $a + b = x$ must hold. Is there a way to find the natural numbers $a$ and $b$ so that $a\cdot b$ has the largest possible value but $a + b = x$ must hold. It's easy small numbers but is there any way, through calculus or otherwis... | Take $(a-b)^2$. Any square is positive (except $0^2=0$) so you can conclude that
\begin{align}
0&\le(a-b)^2 &&\text{expand square binom:}\\
0&\le a^2-2ab+b^2 &&\text{add $4ab$ at both sides:}\\
4ab&\le a^2+2ab+b^2 &&\text{factorize square binom:}\\
4ab&\le(a+b)^2 &&\text{replace $x$:}\\
4ab&\le x^2 &&\text{divid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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How to prove this inequality $\sin{\left(\frac{\pi}{2}ab\right)}\le\sin{\left(\frac{\pi}{2}a\right )}\sin{\left(\frac{\pi}{2}b\right)}$? Let $$0\le a\le 1,0\le b\le 1$$
Prove or disprove
$$\sin{\left(\dfrac{\pi}{2}ab\right)}\le\sin{\left(\dfrac{\pi}{2}a\right )}\sin{\left(\dfrac{\pi}{2}b\right)}$$
My try:
Since
$$\... | The statement is true.
Proof: Given $a\in[0,1]$, define
$$f(x)=\frac{\sin( ax)}{\sin x},\quad x\in (0,\frac{\pi}{2}],\quad\text{and}\quad g(x)=a\tan x-\tan(ax),\quad x\in[0,\frac{\pi}{2}).$$
From $g(0)=0$ and
$$g'(x)=a \big((\sec x)^2-(\sec (ax))^2\big)\ge 0,\quad\forall x\in (0,\frac{\pi}{2}),$$
we know that $g(x)\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 0
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Factorize the polynomial $x^3+y^3+z^3-3xyz$ I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using Mathematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?
| $$\underbrace{x^3+y^3}+z^3-3xyz = \underbrace{(x+y)^3-3xy(x+y)}+z^3-3xyz$$
$$=\underbrace{(x+y)^3+z^3}-\underbrace{3xy(x+y)-3xyz} $$
$$=\underbrace{\{(x+y)+z\}}\{(x+y)^2-(x+y)z+z^2\}-3xy\underbrace{\{(x+y)+z\}} \left(\text{ using } a^3+b^3=(a+b)(a^2-ab+b^2)\text{ for the first two terms }\right)
=(x+y+z)\{(x+y)^2-(x+y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 0
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Proper method for solving quadratic equations with exponents $(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}$ + $(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}$ = $2^{(x+4)/4}$
I have found out, by trial and error method, that $x=0$ and $x=4$ satisfy this equation. But is there a proper way to solve this equation and get the solut... | Let's start by putting $$y_1=\left(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}\right)^{x/4}$$ and $$y_2=\left(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4}\right)^{x/4}.$$ Now, observe that $$\begin{align}2 &= (x^2-5x+6)-(x^2-5x+4)\\ &= \left(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4}\right)\left(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4}\right),\end{align}$$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/546947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Product of two complementary error functions (erfc) I believe that (i.e., it would be convenient if, and visually appears that) the product of the two complementary error functions:
$$\operatorname{erfc}\left[\frac{a-x}{b}\right]\operatorname{erfc}\left[\frac{a+x}{b}\right]$$
will have a solution, or can be approximate... | In this paper, the authors provide the following simple approximation of the error function:
$$\operatorname{erf}(z) = 1- \exp\{-c_1z-c_2z^2\},\; z\ge 0 $$
with
$$ c_1 = 1.09500814703333,\;\; c_2 = 0.75651138383854$$
Set $\frac{a-x}{b} \equiv z_1$, and $\frac{a+x}{b} =\equiv z_2$. Given also that $\operatorname{erfc} =... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving trigonometry equation Please help me understand how to solve this for $0\leq x\leq360 $
I seem to have a problem with equations with powers.
$$3\sin^2 x-3\cos^2x+\cos x-1=0 $$
thinking that I would start by simplifying:
$$3 (\sin^2 x- \cos^2x)+\cos x - 1=0 $$
How I wish the equation in the bracket was in fo... | $3\sin^2 x-3\cos^2 x+ \cos x - 1= 0$
$6\cos^2 x-\cos x - 2 = 0$
By the quadratic formula
$cos x = \dfrac{1 \pm \sqrt{1-4(6)(-2)}}{12} =\dfrac{-1\pm 7}{12}$
So $cos x = \frac{-1}{2}$ and $cos x = \frac{2}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/547554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $... | Let the minimum value of $a+b+\frac{1}{ab} = k$. Both curves must be tangent to each other, which means they must have the same gradient. Thus $a^2 + b^2 = 1 \implies \frac{db}{da} = -\frac{a}{b}$, and:
$$a+b+\frac{1}{ab} = k \implies 1 + \frac{db}{da} - \frac{1 + db/da}{(ab)^2} = 0$$
$$\implies (ab)^2 \left(1 - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/547928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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How to prove this inequality $\frac{x^y}{y^x}+\frac{y^z}{z^y}+\frac{z^x}{x^z}\ge 3$
let $x,y,z$ be positive numbers, and such $x+y+z=1$
show that
$$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}\ge 3$$
My try:
let
$$a=\ln{\dfrac{x^y}{y^x}},b=\ln{\dfrac{y^z}{z^y}},c=\ln{\dfrac{z^x}{x^z}}$$
so
$$a=y\ln{x}-x\... | This is to prove $a^{b-1}b^{1-a}\ge 1$ as needed in Ron Ford's answer above. Let $f(b)=(1-b)(-\log a)-(1-a)(-\log b)$, $b\in [a,1]$. $f(a)=f(1)=0$. $f$ is concave, as $f''(b)=-\frac{1-a}{b^2}<0$. So $f(b)>0,\,\forall b\in(a,1)$ and the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/549971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
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Denesting Phi, Denesting Cube Roots I have been looking into denesting square roots but I have found that $\sqrt[3]{2+\sqrt{5}}$ equals $(1+\sqrt{5})/2$. The same is true for $\sqrt[3]{2-\sqrt{5}}$ and $(1-\sqrt{5})/2$. I cannot figure how this is true. I proved this by setting both equal to x and forming polynomial e... | Hint : As a general rule, when dealing with nested radicals of the form $\sqrt[n]{A+B\sqrt[m]C}$ , you write $A+B\sqrt[m]C=(a+b\sqrt[m]C)^n$, and then employ Newton's binomial theorem. In our case, we have
$$\left(a+b\sqrt5\ \right)^3=a^3+3a^2(b\sqrt5)+3a(b\sqrt5)^2+(b\sqrt5)^3=\underbrace{(a^3+15ab^2)}_2+\underbrace{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/550121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
write formula to predict nth term of sequence $1, 1\cdot3, 1\cdot3\cdot5, 1\cdot3\cdot5\cdot(2n-1)$ How can I write a formula for a sequence with the following behavior:
{$1, 1\cdot3, 1\cdot3\cdot5, 1\cdot3\cdot5\cdot7, 1\cdot3\cdot5\cdot7\cdot9$}
1st term is $1$
2nd term is $1 \cdot 3 = 3$
3rd term is $1 \cdot 3 \cdot... | The classic trick is to multiply and divide by the even terms:
$$
a_n=1\cdot3\cdot\dots\cdot(2n-1)=\frac{1\cdot2\cdot3\cdot \dots\cdot 2n}{2\cdot 4\cdot (2n)}
$$
The numerator is $(2n)!$. As for the denominator, we can factor each of the $n$ terms by $2$, so the denominator is $2^nn!$.
Finally, $a_n=\frac{(2n)!}{2^nn!}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Non-Homogeneous System [Problem] "Find a general solution of the system and use that solution to find a general solution of the associated homogeneous system and a particular solution of the given system."
$\begin{bmatrix}3 & 4 & 1 & 2 \\ 6 & 8 & 2 & 5\\9 & 12 & 3 & 10 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ ... | For the augmented RREF matrix, I get:
$$\begin{bmatrix}1 & \dfrac{4}{3} & \dfrac{1}{3} & 0 & \dfrac{1}{3} \\ 0 & 0 & 0 & 1 & 1\\0 & 0 & 0 & 0 & 0\end{bmatrix}$$
This means we can write:
$$x_4 = 1$$
$$x_1 = \dfrac{1}{3}-\dfrac{4}{3}x_2 - \dfrac{1}{3}x_3$$
So, we have two free variables, $x_2$ and $x_3$.
Can you now wri... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you find the smallest possible value of aan equation with two unknowns? I'm solving a list of problems where I'm given an equation and I find the smallest possible value by comparing the equation to a quadratic equation and completing the square, however the next one involves another unknown $y$:
$$ x^2 - 3x + ... | copper.hat is right that the easiest way to solve this is to view it as composed from two functions and take calculate the derivate of each. Then set $f'(x)=0$ and solve for $x$ for both functions.
But this problem is from Spivak's Calculus (1.717(b), p. 17), before Spivak has introduced differentiation. I think Spivak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/554499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$$
| Define the function $I(s)$ for $s > 0$ by
$$ I(s) = \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - e^{s}\sqrt{x+1} )}{x+1} \, dx. $$
By observing that $I(\infty) = 0$, we have
$$ I(s) = -\int_{s}^{\infty} I'(t) \, dt. $$
Applying Leibniz's integral rule,
$$ I'(s) = \int_{0}^{\infty} \frac{e^{s}\sqrt{x+1}}{1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
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which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$ Which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$?
Squaring both sides will give me something but I could not go any further.
| $$\frac{f(5)+f(7)}{2} \lt f(\frac{5+7}{2}) $$
for a concave function $f$, and so for $f(x)=\sqrt x$.
But even if you proceed on the way you have started you will get the result.
$$\begin{eqnarray}
\sqrt{7}-\sqrt{6} &\lt& \sqrt{6}-\sqrt{5} &\mid& ^2 \\
13-2\,\sqrt{6}\,\sqrt{7} &\lt& 11-2\,\sqrt{5}\,\sqrt{6} &\mid& -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/562002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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"answer_id": 11
} |
how is a factorial fraction equal to the product notation How is the $\prod_{k=2}^n(2k-3)={(2n-3)!\over 2^{n-2}(n-2)!}$, where $n \geq 2$
Note:
I know that the $(2n-3)!$ is equal to the product of $2k-3$ from $k=2$ to $n$, but I can't figure out the bottom half of how diving by $2^{n-2}(n-2)!$ equals the product notati... | Your equation $\prod_{k=2}^n(2k-3)$ can be expanded to $(2n-3)(2n-5)\cdots(3)(1)$. Clearly, we have every other term for (2n-3)! since we only have the odd values.
Hence,$(2n-3)! = (2n-3)(2n-4)\cdots(2)(1) = (2n-4)(2n-6)\cdots(4)(2) \prod_{k=2}^n(2k-3)$.
Taking the addition terms on the RHS, $(2n-4)(2n-6)\cdots(4)(2) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/563541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there a closed form for this sum? While generalizing the previous result, I conjectured that the series expansion of
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \arctan \left( \frac{2x \sin\theta}{1-x^{2}} \right) \arctan \left( \frac{2y \sin\theta}{1-y^{2}} \right) \arctan \left( \frac{2z \sin\theta}{1-z^{2}} \right) \... | The OP already gave the series expansion for $\arctan \left( \frac{2x \sin\theta}{1-x^{2}} \right) $ in here. Using that expansion, we have
\begin{align*}
&\int_{0}^{\frac{\pi}{2}} \arctan \left( \frac{2x \sin\theta}{1-x^{2}} \right) \arctan \left( \frac{2y \sin\theta}{1-y^{2}} \right) \arctan \left( \frac{2z \sin\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/564087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 1,
"answer_id": 0
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Use partial fractions to find the integral. Find the integral using partial factions.
$$\int\frac{(2x^2+5x+3)}{(x-1)(x^2+4)}\,dx$$
So do I do...$$\frac{2x^2+5x+3}{(x-1)(x^2+4)}=\frac{A}{x-1} + \frac{Bx+C}{x^2+4},$$
then get
\begin{align*}
2x^2+5x+3 &= A(x^2+4)+(Bx+C)(x-1) \\
2x^2+5x+3 &= Ax^2+4A+Bx^2-Bx+Cx-C?
\end{ali... | It seems that you are stuck at the actual partial fraction decomposition, rather than at the integration. So let's pick up at $$2x^2+5x+3= A(x^2+4) + (Bx+C)(x-1)$$
While this must hold true for all values of $x$, certain values of $x$ will lead us to the values of $A,B,C$ more quickly...
$x = 1$ is a natural choice, as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/564845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How find this inequality minimum $\sum\limits_{cyc}\sqrt{a^2+b^2+ab-2a-b+1}$
Let $0<a,b,c<1$, find this follow minimum
$$\sqrt{(a+b)^2-(a+1)(b+2)+3}+\sqrt{(b+c)^2-(b+1)(c+2)+3}+\sqrt{(c+a)^2-(c+1)(a+2)+3}$$
My try:
since
$$(a+b)^2-(a+1)(b+2)+3=a^2+b^2+2ab-ab-2a-b-2+3=a^2+b^2+ab-2a-b+1$$
so we only find this follow ... | $a^2+b^2+ab-2a-b+1=(b+\dfrac{a-1}{2})^2+\dfrac{3}{4}(a-1)^2$
$\sum\limits_{cyc}\sqrt{a^2+b^2+ab-2a-b+1} \ge \sqrt{(\sum\limits_{cyc} (b+\dfrac{a-1}{2})^2+\dfrac{3}{4}(\sum\limits_{cyc}(a-1))^2}=\sqrt{(\dfrac{3}{2}(\sum\limits_{cyc} a-1)^2+\dfrac{3}{4}(\sum\limits_{cyc} a-3)^2}=\sqrt{3((\sum\limits_{cyc} a)^2-3\sum\limi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/564929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding derivative using product and chain rule I need to find first derivative of $x\sqrt{2-x^2}$. My approach
*
*Using product rule: $(2-x^2)^{1/2} + x\frac{\operatorname{d}(2-x^2)^{1/2}}{\operatorname{d}x}$
*Using chain rule: $(2-x^2)^{1/2} + x\left[\frac{1}{2}(2-x^2)^{-1/2} (-2x)\right]$
*Result: $(2-x^2)^{1/... | It is, indeed, correct. A way to see that the answers are the same is to note that $$\begin{align}(2-x^2)^\frac12-x^2(2-x^2)^{-\frac12} &= (2-x^2)^{1+-\frac12}-x^2(2-x^2)^{-\frac12}\\ &= (2-x^2)(2-x^2)^{-\frac12}-x^2(2-x^2)^{-\frac12}\\ &= \bigl((2-x^2)-x^2\bigr)(2-x^2)^{-\frac12}\\ &= \frac{2-2x^2}{(2-x^2)^\frac12}\\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing $m+n\sqrt 2$ is a unit in set $R$ Let $R=\{m+n\sqrt 2|m,n\in\mathbb{Z}\}$. Show that $m+n\sqrt2$ is a unit in $R$ iff $m^2-2n^2=\pm 1$. Hint: Show that if $(m+n\sqrt2)(x+y\sqrt 2)=1$ then $(m-n\sqrt 2)(x-y\sqrt2)=1$ I needed help with the foward direction
This is what I have so far: ($\rightarrow$). Suppose tha... | Let $(x,y)$ denote $x+y\sqrt 2$. Then, we have multiplication defined as $(x,y)(x',y')=(xx'+2yy',xy'+x'y)$. If we define $N(x,y)=x^2-2y^2$, we have that $$\begin{align}N(x,y)N(x',y')&=(x^2-2y^2)(x'^2-2y'^2)\\&=x^2x'^2-2(x'^2y^2+x^2y'^2)+4y'^2y^2\\&=(xx')^2+(2y'y)^2-2((x'y)^2+(xy')^2)\\&=(xx')^2+\color{red}{2(xx')(2yy')... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Double Integral $\int\limits_0^b\int\limits_0^x\sqrt{a^2-x^2-y^2}\,dy\,dx$ What is the best method for evaluating the following double integral?
$$
\int_{0}^{b}\int_{0}^{x}\,\sqrt{\,a^{2} - x^{2} - y^{2}\,}\,\,{\rm d}y\,{\rm d}x\,,
\qquad a > \sqrt{\,2\,}\,\,b
$$
Is there exist an easy method?
My try:
$$\int_0^b\int_... | Suppose
$$I=\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}dy\,dx$$
Let $y=\sqrt{a^2-x^2}\sin \theta$ then $dy=\sqrt{a^2-x^2}\cos \theta\,d\theta$, so
$$I=\int_0^b\int_0^{\arcsin\frac{x}{\sqrt{a^2-x^2}}}(a^2-x^2)\cos^2 \theta\,d\theta\,dx$$
$$=\frac{1}{2}\int_0^b(a^2-x^2)\left[\theta+\frac{1}{2}\sin 2\theta\right]_0^{\arcsin\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
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Expressing $\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$ in closed form I want to express $$\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$$ in closed form.
I know that $$\pi z\cot(\pi z)=1+2z^2\sum_{n=1}^\infty\dfrac{1}{z^2-n^2}$$ which looks close, but I don’t know how to use it.
| Here is a closed form by Maple
$$ \frac{2}{3}\,{\frac {\pi \, \left( -\cos \left( 2\,\pi \,z \right) +\sqrt {3}
\sin \left( \pi \,z \right) \sinh \left( \pi \,z\sqrt {3} \right) +
\cos \left( \pi \,z \right) \cosh \left( \pi \,z\sqrt {3} \right)
\right) }{{z}^{2} \left( 2\,\sin \left( \pi \,z \right) \cosh \left(
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/576343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Boolean Algebra simplify minterms I have this equation
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
and need to simplify it. I have got as far as I can and spent a good 2 hours at it. I've realized I now need to use De Morgan's law to continue however I am baffl... | The $\mathrm{ExOR}$ function can be denoted by: $\oplus$ ; $X \oplus Y=\overline {X}\cdot Y +X \cdot \overline{Y} $.
Also $\overline{X \oplus Y}=\overline {X}\cdot \overline{Y} +X \cdot {Y} $
Hence $$\begin{align}
\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C &=\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial? Stirling's approximation of the factorial for even numbers is given by
$$ (2n)! \sim \left(\frac{2n}{e}\right)^{2n}\sqrt{4 \pi n}. \tag{1} $$
Further, the Euler numbers grow quite rapidly for large indices as they have the follo... | There's more to it, we have
$$\frac{1}{\cos z} = \sum_{n=0}^\infty (-1)^n \frac{E_{2n}}{(2n)!}z^{2n}.\tag{1}$$
Since $\dfrac{1}{\cos z}$ has poles in $(k+\frac12)\pi$ for $k\in\mathbb{Z}$ and is holomorphic everywhere else, the series $(1)$ has a radius of convergence of $\dfrac{\pi}{2}$. The Cauchy-Hadamard formula no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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How many ways are there for $2$ teams to win a best of $7$ series? Case $1$: $4$ games: Team A wins first $4$ games, team B wins none = $\binom{4}{4}\binom{4}{0}$
Case $2$: $5$ games: Team A wins $4$ games, team B wins one = $\binom{5}{4}\binom{5}{1}-1$...minus $1$ for the possibility of team A winning the first four.
... | For another approach, if the series ends before the seventh game, extend it to seven games by having the losing team win the rest. The series will now be four games to three. There are $2$ ways to choose the losing team, and ${7 \choose 3}$ ways to choose which game the losing team wins. The total is then $2{7 \choo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Why is this binomial coefficient bounded thus? Source: Miklos Bona, A Walk Through Combinatorics.
$$ \forall k\geq 2,\binom{2k-2}{k-1}\leq4^{k-1}.$$
The RHS is the upper bound of the Ramsey number $R(k,k)$.
How can I prove the inequality without using mathematical induction? I've merely expanded the LHS to obtain $\fra... | Let $n=k-1$.
We have $\frac {2n!}{n!n!}$.
Then $2n\cdot (2n-1)\cdot(2n-2)\cdot(2n-3)\dotsm 3\cdot 2\cdot 1$.
Taking even terms $2n\cdot(2n-2)\cdot(2n-4)\dotsm 8\cdot 6\cdot 4\cdot 2$
we have 2 common in all term, so it is $\underbrace {2\cdot 2\dotsm2}_{\text{$n$ times}} \cdot n\cdot (n-1)\cdot(n-2)\dotsm 3\cdot2\cdot1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate $\frac{4x+4}{x^4+x^3+2x^2}$? Please could anyone help me to integrate
$\quad\displaystyle{4x + 4 \over x^4 + x^3 + 2x^2}.\quad$
I know how to use partial fraction and I did this:
$$
x^{4} + x^{3} + 2x^{2} = x^{2}\left(x^{2} + x + 2\right)
$$
And then ?.$\quad$ Thanks all.
| Hint:
$$
\begin{align}
\frac{4x+4}{x^4+x^3+2x}
&=\frac{4x+4}{x^2(x^2+x+2)}\\
&=\frac1x+\frac2{x^2}-\frac{x+1/2}{x^2+x+2}-\frac{5/2}{x^2+x+2}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/582954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$ I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I al... | Use the series for cotangent
$$\cot x = \frac{1}{x}-\frac{1}{3}x-\frac{1}{45}x^3\pm\dots$$
$$\cot^2 x = \frac{1}{x^2}-\frac{2}{3}+\frac{1}{15}x^2\pm\dots $$
$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)=
\lim_{x\to 0}\left(\frac{1}{x^2}-\cot^2 x\right)=
\lim_{x\to 0}\left(\frac{2}{3}-\frac{1}{15}x^2 \pm \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/584563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Fibonacci sequence proof Prove the following:
$$f_3+f_6+...f_{3n}= \frac 12(f_{3n+2}-1) \\ $$
For $n \ge 2$
Well I got the basis out of the way, so now I need to use induction: So that $P(k) \rightarrow P(k+1)$ for some integer $k \ge 2$
So, here are my first steps:
$$ \begin{align} & \frac 12(f_{3k+2}-1) + f_{3k+3} = ... | A combinatorial argument:
The fibonacci number $f_n$ represents the number of paths from 0 to $n - 1$ by taking steps of 1 or 2.
Now for any path for $f_{3n + 2}$,
*
*it's all ones
*for $0 \leq k < 3n$, it starts with k ones, followed by a 2, followed by one of the paths for $f_{3n - k}$
so
\begin{align*}
f_{3n ... | {
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Calculate the determinant of the matrix. Calculate the determinant.
\begin{bmatrix} C_{n}^{p+n} & C_{n}^{p+n+1} & \dots & C_{n}^{p+2n} \\
C_{n}^{p+n+1} & C_{n}^{p+n+2} & \dots & C_{n}^{p+2n+1} \\
\vdots & \vdots & \dots & \vdots \\
C_{n}^{p+2n} & C_{n}^{p+2n+1} & \dots & C_{n}^{p+3n} \end{bmatrix}
| For $n=0$, we have
$$\det(C(p,0)) = 1$$
For $n=1$, we have
$$\det\left(\begin{bmatrix} p+1 & p+2\\ p+2 & p+3\end{bmatrix} \right) = -1$$
For $n=2$, we have
$$\det\left(\begin{bmatrix} \dfrac{(p+1)(p+2)}2 & \dfrac{(p+2)(p+3)}2 & \dfrac{(p+3)(p+4)}2\\ \dfrac{(p+2)(p+3)}2 & \dfrac{(p+3)(p+4)}2 & \dfrac{(p+4)(p+5)}2\\ \dfr... | {
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"source": "stackexchange",
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"answer_count": 2,
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How prove this congruence equation has four zeros solution Question:
let congruence equation
$$\begin{cases}
\left(\overline{a_{1}a_{2}\cdots a_{m}}\right)^2\equiv \overline{a_{m}}(\mod 10)\\
\left(\overline{a_{1}a_{2}\cdots a_{m}}\right)^2\equiv \overline{a_{m-1}a_{m}}(\mod 10^2)\\
\cdots\cdots\cdots\cdots\\
\left(\ov... | As @Hurkyl pointed out, if we let $x=\overline{a_1a_2\ldots a_m}$, it suffices to prove that $x^2 \equiv x \pmod{10^m}$ has exactly four solutions $\pmod{10^m}$.
This is straightforward; we get $2^m(5^m)=10^m \mid x(x-1)$. Note that $\gcd(x, x-1)=1$, so we have four cases:
\begin{align}
x \equiv 0 \pmod{2^m}, x \equiv ... | {
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Find all triples of positive integers (x,y,z) such that Find all triples of positive integers (x,y,z) such that
$x^{z+1} \ - \ y^{z+1}=2^{100}$
The RHS is even, then x and y must be odd and $x^{z+1}>y^{z+1}$, but how to find out them all ?
| $x \ge 0 \land y \ge 0 \rightarrow x \ge y$.
$$x^{z+1} - y^{z+1} = \left(x - y\right)\left(x^z + x^{z-1}y + x^{z-2}y^2 + \dots +y^z\right) = 2^{k+1} \tag{T1}$$
So we can conclude $x - y = 2^{r_1}$, and that $x$ and $y$ have the same parity.
If $z$ is even, rewrite $(T1)$:
$$x^z + y^z + xy\underbrace{\left(x^{z-2} + x... | {
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Find a limit $\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$ I am to find the limit of
$$\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$
so I used:
$$\lim_{x \to -\infty} = \lim_{x \to \infty}f(-x)$$
but I just can't solve it to the end.... | $$\left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$
$$\left(\frac{4^{x}*16- 2\cdot3^{-x}}{4^{-x}+2\cdot3\cdot3^{x}}\right)$$
$$\left(\frac{4^{x}*16- 2\cdot\frac{1}{3^x}}{\frac{1}{4^x}+2\cdot3\cdot3^{x}}\right)$$
$$\left(\frac{16\cdot4^{x}\cdot3^x- 2 }{3^x}\right)\cdot\left(\frac{4^x}{1+6\cdot3^x\cdot4^x... | {
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"source": "stackexchange",
"question_score": "1",
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Integer $a$ , If $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots If the equation $x ^ 2 + (a-6) x + a = 0 (a ≠ 0)$ has two integer roots. Then the integer value of $a$ is
$\bf{My\; Try}::$ Let $\alpha,\beta\in \mathbb{Z}$ be the roots of the equation . Then $\alpha+\beta = (6-a)$ and $\alpha\cdot \beta = a$
Now... | From the first line of your solution, $\alpha.\beta+\alpha+\beta=6-a+a=6$, so $(\alpha+1)(\beta+1)=7$
What are the factors of 7?
| {
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"question_score": "1",
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How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy... | Lagrange multipliers seems like a good approach. Solve the system of equations $$\displaystyle \frac{\partial }{\partial a}\left(a+b+\sqrt{a^2+b^2}-\lambda \left(\frac{2}{a}+\frac{1}{b}-1\right)\right)=0\textrm,$$,
$$\frac{\partial }{\partial b}\left(a+b+\sqrt{a^2+b^2}-\lambda \left(\frac{2}{a}+\frac{1}{b}-1\right)\r... | {
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"answer_id": 4
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Prove that if $p>2$ is prime then $\left(\prod_{k=1}^{p-1}k^k\right)^2\equiv (-1)^{\frac{p+1}2}\pmod p.$ Prove that if $p>2$ is prime then
$$\left(\prod_{k=1}^{p-1}k^k\right)^2\equiv (-1)^{\frac{p+1}2}\pmod p.$$
I find this by computer but cannot prove it, thank you!
| Let's deal with the stuff inside the square first. Since $p-1$ is even we have that:
$\prod_{k=1}^{p-1} k^k \equiv \left(\prod_{k=1}^{\frac{p-1}{2}} k^k\right) \left(\prod_{k=1}^{\frac{p-1}{2}} (-k)^{p-k}\right) = (-1)^{\frac{p-1}{2}} \prod_{k=1}^{\frac{p-1}{2}}k^p \equiv (-1)^{\frac{p-1}{2}}\prod_{k=1}^{\frac{p-1}{2}}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this $1\times 3\times 5\times 7\cdots\times 2009+2\times4\times6\cdots 2010\equiv 0 (\mod2011)$ Question:
show that
$$1\times 3\times 5\times 7\cdots\times 2009+2\times4\times6\cdots 2010\equiv 0 (\mod2011)$$
my try: since
\begin{align*}&1\times 3\times 5\times 7\cdots\times 2009+2\times4\times6\cdots \times2... | Ok so for the first term:
$1\times 3 \times 5 \times ... \times 2009$
$\equiv 1 \times 3 \times 5 \times ... \times 1005 \times (-1004) \times... \times (-4) \times (-2)$
$\equiv (1 \times 2 \times ... \times 1005)$
$\equiv 1005! \bmod 2011.$
The second term is similar:
$2\times 4\times 6 \times ... \times 2010$
$\e... | {
"language": "en",
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sums of squares of integers We have to prove that there exists infinitely many integers $a,b,c$ such that $a^2 + b^2 = c^2 + 3$ .
This looked like a very straight-forward question . I did some algebraic manipulations but couldn't reach the conclusion . Help me here
| You need $c^2 - a^2 = b^2 - 3 \iff (c-a)(c+a) = b^2- 3$
Let $b=3k$, then we need $(c-a)(c+a)= 3(3k^2-1)$, so let us try $c-a = 3, c+a = 3k^2-1$
Adding and subtracting, we get $c = \frac{3k^2+2}{2}, a = \frac{3k^2-4}{2}$, both would be integer if $k$ is even.
Hence one set of solutions is $a = 6k^2-2, b = 6k, c = ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Expanding brackets of the form $(a+b)^n$ If we have the equation
$
(a+b)^n = (a+b)\times(a+b)\times(a+b)\times \ldots \times (a+b)
$
expanding the right hand side of the above, is the sum of terms in the form $a^n$, $a^{n-1}b$,$a^{n-2}b^2$, $\ldots$, $b^n$. That is $a^{n-1}b^k$ for $k=0,1,\ldots,n$
Now, my textbook sta... | To simplify talking about the statement a bit, let $C_i = (a + b)$ for $1\le i \le b$, so
$$(a + b)^n = C_1 \times \dots \times C_n.$$
For the case where $n = 3$ and the term $a^2b$:
If we choose $b$ from $C_1$, then we choose $a$ from $C_2$ and $C_3$.
If we choose $b$ from $C_2$, then we choose $a$ from $C_1$ and $C_3... | {
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Induction: $n^{n+1} > (n+1)^n$ and $(n!)^2 \leq \left(\frac{(n + 1)(2n + 1)}{6}\right)^n$ How do I prove this by induction:
$$\displaystyle n^{n+1} > (n+1)^n,\; \mbox{ for } n\geq 3$$
Thanks.
What I'm doing is bunch of these induction problems for my first year math studies.
I tried using Bernoulli's inequality at som... | Your inequality is the same as
$$
n\left(\frac{n}{n+1}\right)^n\gt1
$$
Notice that
$$
\begin{align}
(n+1)\left(\frac{n+1}{n+2}\right)^{n+1}
&=\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+1}n\left(\frac{n}{n+1}\right)^n\\
&=\left(1+\frac1{n(n+2)}\right)^{n+1}n\left(\frac{n}{n+1}\right)^n\\[9pt]
&\gt1\cdot1
\end{align}
$$
| {
"language": "en",
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"question_score": "4",
"answer_count": 5,
"answer_id": 2
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finding the value of $f(\frac{1}{7})$ $f$ is a function mapping positive reals between $0$ and $1$ to reals. Let $f$ be given by, $f( \frac{x+y}{2} ) = (1-a)f(x)+af(y)$ where $y > x$ and $a$ being a constant. Also,$f(0) = 0$ and $f(1) = 1$. Find $f( \frac{1}{7} )$.
I have tried some things but it is not working out. Ho... | For each $n$ with $1\leq n\leq 6$ we have the equation $f(\frac{n}{7}) = (1-a)f(\frac{n-1}{7}) + af(\frac{n+1}{7})$. Since we know $f(0)$ and $f(1)$, this gives us a system of $6$ linear equations in $6$ unknowns.
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Prove the following trigonometric identity. $$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\sin x +1}{\cos x}$$
I tried substituting $\sin^2x+\cos^2x = 1$ but I cannot solve it.
| The above method is really verifying and always quick. Another method to arrive at the answer is by rationalising denominator (mainly when the answer [or RHS] is not known or one is asked to work out only from LHS to RHS):
$$\frac{\sin x - \cos x + 1 }{\sin x + \cos x - 1 }\cdot \frac{\sin x + \cos x + 1}{\sin x + \cos... | {
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"source": "stackexchange",
"question_score": "2",
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Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^22^n}$ How can I prove that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2).$$
Can anyone help me please?
| \begin{align*}
\operatorname{Li}_3\left(\frac{1}{2}\right)&=\int _0^{\frac{1}{2}}\frac{\operatorname{Li}_2\left(x\right)}{x}\:dx=-\operatorname{Li}_2\left(\frac{1}{2}\right)\ln \left(2\right)+\int _{\frac{1}{2}}^{1}\frac{\ln \left(x\right)\ln \left(1-x\right)}{1-x}\:dx\\[2mm]
&=-\frac{1}{2}\ln \left(2\right)\zeta \left... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
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Fibonacci proof question: $f_{n+1}f_{n-1}-f_n^2=(-1)^n$ Show that
$$f_{n+1}f_{n-1}-f_n^2=(-1)^n$$
when $n$ is a positive integer and $f_n$ is the $n$th Fibonacci number.
| If this is a homework exercise then this is likely not how you're meant to prove it, since you'll first need to prove that the $n$-th Fibonacci number can be written as $$f_n=\frac{\varphi^n-(-\frac{1}{\varphi})^{n}}{\sqrt{5}}$$ where $\varphi=\frac{1+\sqrt{5}}{2}$. But if you have that then
$$\begin{align}f_{n-1}f_{n... | {
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"source": "stackexchange",
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Ellipse problem : Find the slope of a common tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and a concentric circle of radius r. Problem :
Find the slope of a common tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and a concentric circle of radius r.
Few concepts about Ellipse :
Equation of Ta... | If we use parametric forms of the ellipse $(a\cos\phi,b\sin\phi)$ and of the circle $(r\cos\psi,r\sin\psi)$
we get the tangents to be $\displaystyle x\frac{\cos\phi}a+y\frac{\sin\phi}b=1\iff x b\cos\phi+y a\sin\phi=ab$
and for the circle $x\cos\psi+y\sin\psi=r$ with slope $\displaystyle-\frac{\cos\psi}{\sin\psi}=-\cot... | {
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How to calculate the size of any rectangle to fit into an ellipse? It's been a long time I do not review my math knowledge, please help me out.
I have an ellipse image with fix size, let's say it has the bounding rect with width=w1, height=h1, and I have any random image with random size let's say width=w2, height=h2, ... | For your ellipse centred on (0,0) the equation is:
$$
\frac{x^2}{w1^2}+\frac{y^2}{h1^2} = 1 \Rightarrow y^2 = h1^2 \cdot \left( 1 - \frac{x^2}{w1^2} \right)
$$
Now imagine a box centred on (0,0) of width $w2'$ and height $h2'$ where $\frac{h2'}{w2'} = \frac{h2}{w2}$.
Finally imagine a straight line passing through (0,0... | {
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"source": "stackexchange",
"question_score": "4",
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What is $\lim_{x\to0} \frac{(\cos x + \cos 2x + \dots+ \cos nx - n)}{\sin x^2}$? What is the limit of $$\lim_{x\to0} \frac{\cos x + \cos 2x + \dots+ \cos nx - n}{\sin x^2}$$
| We have by taylor series
$$\cos(kx)\sim_0 1-\frac{k^2}{2}x^2$$
and
$$\sin x^2\sim_0 x^2$$
hence
$$\lim_{x\to0} \frac{(\cos x + \cos 2x + ...+ \cos nx - n)}{\sin x^2}= -\frac{1}{2}\sum_{k=1}^n k^2=-\frac{n(n+1)(2n+1)}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/610729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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How to prove continued fraction convergents of a number Let $x=1+\sqrt{3}$. Prove that in pairs the continued fraction convergents of $x$ are $a_n$/$b_n$ < x < $c_n$/$d_n$ where $a_1$ = 2, $b_1$ = 1, $c_1$ = 3, $d_1$ = 1, $a_{n+1}$ = 2$c_n$ + $a_n$, $b_{n+1}$ = 2$d_n$ + $b_n$, $c_{n+1}$ = 3$c_n$ + $a_n$, $d_{n+1}$ = 3$... | Let $x = [a_0; a_1 ,a_2 , \ldots]$. Find the continued fraction using the common algorithm. You will get $x = [2;\overline{1,2}]$.
Sidenote: If you are confused about the inequalities $a_n/b_n < x < c_n/d_n$, remember that $C_{2k} < C_{2k+1}$, and $C_{2k} < C_{2(k+1)}$ so that $C_0 < C_2 < \ldots < C_n < C_{n-1} < \ldo... | {
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"source": "stackexchange",
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All solutions of $a+b+c=abc$ in natural numbers I was observing some nice examples of equalities containing the numbers $1,2,3$ like $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ and $\log 1+\log 2+ \log 3=\log (1+2+3)$. I found out this only happens because $1+2+3=1*2*3=6$. I wanted to find other examples in small numbers, b... | leaving aside the solution in which $a=b=c=0$ order the numbers so $a \le b \le c$
as OP shows, $c|(a+b)$ so we have:
$$
\frac{a+b}{c} \le 2
$$
thus only the values $1$ and $2$ are possible for $\frac{a+b}c$. these give $a+b=c$ and $a+b=2c$ respectively. however the latter is only possible if all three numbers are equa... | {
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"source": "stackexchange",
"question_score": "12",
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How to simplify $\tan{\arcsin{\frac{y}{R}}/2}$? I have verified with Mathematica that, for $R>0, y \in \mathbb{R}$:
$$ \tan{\frac{\arcsin{\frac{y}{R}}}{2}} = \frac{R - \sqrt{R^2 -y ^2}}{y}$$
using
Assuming[Element[y, Reals] && R > 0,
FullSimplify[TrigToExp[Tan[ArcSin[y/R]/2]]]]
How can I prove this withouse messing... | Let $\displaystyle\arcsin\frac yr=\phi\implies \sin\phi=\frac yR\ \ \ \ (1)$
and based on the definition of principal value, $\displaystyle-\frac\pi2\le\phi\le\frac\pi2\ \ \ \ (2)$
So, we need to find $\displaystyle\tan\frac\phi2$
Using Weierstrass substitution, $\displaystyle\sin\phi=\frac{2\tan\frac\phi2}{1+\tan^2\... | {
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"source": "stackexchange",
"question_score": "5",
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Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
My approach :
If I divide numerator and denominator by $n^2$ I get :
$$\lim_{ n \to \infty} \dfrac{\frac{1}... | $$ S_n=\sum_{k=1}^n \frac{k}{k+n^2} = \sum_{k=1}^n \left( 1 -\frac{n^2}{k+n^2} \right) \\
= n - \sum_{k=1}^n \left(\frac{n^2}{k+n^2} \right)
$$
but
$$\sum_{k=1}^n \left(\frac{n^2}{k+n^2} \right) = \sum_{k=1}^n \frac1{1+\frac{k}{n^2}}$$
$$ = \sum_{k=1}^n \sum_{j=0}^{\infty} \left(\frac{-k}{n^2} \right)^j \\
= n -\fra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
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Need help simplifiying a rational expression There's a math question on an online test which asks the following
Multiply the following expression, and simplify:
$\frac{x^2-16y^2}{x} * \frac{x^2+4xy}{x-4y}$
But no matter how I try I keep getting the answer incorrect with a message telling me to simplify my answer. I ca... | $$\frac{x^2+16y^2}{x} \cdot \frac{x^2+4xy}{x-4y}=\frac{x^2+16y^2}{x} \cdot \frac{x(x+4y)}{x-4y}=(x^2+16y^2)\cdot \frac{(x+4y)}{x-4y}$$
$$=(x+i4y)(x-i4y) \frac{x+4y}{x-4y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/617594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to get the simplest form of this radical expression: $3\sqrt[3]{2a} - 6\sqrt[3]{2a}$. How to get the simplest form of this radical expression:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$
Here is my work:
$$3\sqrt[3]{2a} - 6\sqrt[3]{2a}$$
Since the radicands are the same, we just add the coefficients.
$$-3\sqrt[3]{2a} \sqrt... | Your middle step is incorrect, it should be $-3\sqrt[3]{2a}$ not $-3\sqrt[3]{2a}\sqrt[3]{2a}$. It should be $$3\sqrt[3]{2a} - 6\sqrt[3]{2a} = 3\times\sqrt[3]{2a} - 6\times\sqrt[3]{2a} = (3 - 6)\times\sqrt[3]{2a} = -3\times\sqrt[3]{2a} = -3\sqrt[3]{2a}.$$ I don't think $-3\sqrt[3]{2}\sqrt[3]{a}$ is any simpler than $-3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$ I was solving an exercise, so I realized that the one easiest way to do it is using a "weird", but nice identity below. I've tried to found out it on internet but I've founded nothingness, and I ... | Observe:
This equation is a cubic(degree 3) in "a".So at maximum it can have 3 roots if it is not an identity. But if you show that it has 4 roots , then it becomes an identity.
Roots to try :
b,c(easy ones),0(also easy),-a(think it over).
Now you have shown that it has 4 roots. Thus it becomes an identity.
If b,c ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 0
} |
What is the shortest way to compute the last 3 digits of $17^{256}$? What is the shortest way to compute the last 3 digits of $17^{256}$ ?
My solution:
\begin{align}
17^{256} &=289^{128} \\
&=(290 - 1)^{128}\\
&=\binom{128}{0}290^{128} - ... +\binom{128}{126}290^2 - \binom{128}{127}290 + \binom{128}{128}
\end{align}
C... | Use Euler's Totient Theorem and the Chinese Remainder Theorem.
We have that $ 17 \equiv 1 \mod 8 $ and hence $ 17^{256} \equiv 1 \mod 8 $ as well.
Because $ 17 $ is coprime to $ 125 $, we know that $ 17^{100} \equiv 1 \mod 125 $.
We are left to calculate $ 17^{56} \mod 125 $, which can be quickly be done by hand via th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Indefinite integral $\int(\sin x)^2/(x\cdot\cos x-\sin x)^2\,dx$ How to find the antiderivative of $\dfrac{(\sin x)^2}{(x\cdot\cos x-\sin x)^2}$? I know that series theory can solve the problem. But I wish to solve the problem by the antiderivative way. The numerator and denominator divide the $x^2$ at the same time, d... | As $\displaystyle \frac{d(x\cos x-\sin x)}{dx}=-x\sin x,\int\frac{-x\sin x}{(x\cos x-\sin x)^2}dx=-\frac1{x\cos x-\sin x}$
Integrating by parts,
$$\int\frac{\sin^2x}{(x\cos x-\sin x)^2}dx=\int \frac{(-x\sin x)}{(x\cos x-\sin x)^2}\cdot\frac{-\sin x}x dx$$
$$=\frac{-\sin x}x\int\frac{(-x\sin x)}{(x\cos x-\sin x)^2}dx-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Inequality in three variables $ \frac{3-(xy+yz+zx)}{2} \geq \sum\limits_{\text{cyc}}\frac{1-x^2y^2}{2+x^2+y^2}$ If $x, y, z$ are positive real numbers with the property $ xy, yz, zx \leq 1 $, then prove that $$ \frac{3-(xy+yz+zx)}{2} \geq \sum_{\text{cyc}}\frac{1-x^2y^2}{2+x^2+y^2}.$$
| Use AM-GM to get:
$$RHS = \sum_{\text{cyc}}\frac{1-x^2y^2}{2+x^2+y^2} \le \sum_{cyc}\frac{1-x^2y^2}{2+2xy}= \frac12\sum_{cyc}(1-xy) = LHS $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.