Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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A circle tangent to two circles touching internally and line Find the radius of a circle touching two circle $x^2+y^2+3\sqrt{2}(x+y)=0$ and $x^2+y^2+5\sqrt{2}(x+y)=0$ and also touching the common diameter of the two given circles.
The two circles touch internally and the common diameter is $x-y=0$.
Let centre of requir... | I am posting a diagram based on my understanding of the question.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove (or check) the expression is positive given constraints on variables? The following proof problem have taken me a few days. Perhaps it is too hard for me to overcome it. Can you help me?
The expression is by the following:
\begin{equation}
\begin{split}
&2\,x{c}^{x-1}\ln \left( c \right) -{2}^{x}\ln \left( 2 \r... | The following is a false proof. However, the method may be right. So it still has certain reference significance.
Combine term $9$, $10$ and $11$, we get
$$
2\,{c}^{x-1}+{2}^{x}-2\,{c}^{x}=\left( \frac{2}{c}-1 \right) {c}^{x}+\left({2}^{x}-{c}^{x}\right)>0
$$;
Combine term $4$ and $8$, we get
$$
{c}^{x}{2}^{x}\ln \lef... | {
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How does one parameterize $x^2 + xy + y^2 = \frac{1}{2}$?
Parameterize the curve $C$ that intersects the surface
$x^2+y^2+z^2=1$ and the plane $x+y+z=0$.
I have this replacing equations:
$$ x^2+y^2+(-x-y)^2=1$$
and clearing have the following:
$$ x^2+xy+y^2=1/2$$
which it is the equation of an ellipse but I find it... | Since $2 x^2 + 2 x y + 2 y^2$ is a quadratic form, we write
$$\begin{bmatrix} x\\ y\end{bmatrix}^T \begin{bmatrix} 2 & 1\\ 1 & 2\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix} = 1$$
We now compute the eigendecomposition of the matrix above
$$\begin{array}{rl} \begin{bmatrix} 2 & 1\\ 1 & 2\end{bmatrix} &= \begin{bmatri... | {
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Integrate $ \int \frac{1}{1 + x^3}dx $ $$ \int \frac{1}{1 + x^3}dx $$
Attempt:
I added and subtracted $x^3$ in the numerator but after a little solving I can't get through.
| You can factor the denominator using the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$, and in our case $a=x$ and $b=1$
$$\int \dfrac{1}{(x+1)(x^2-x+1)} dx$$
Assume $\dfrac{1}{(x+1)(x^2-x+1)} = \dfrac {A}{x+1} + \dfrac {Bx+C}{x^2-x+1} $
Can you take it from here?
| {
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Find the sum of the infinite series $\sum n(n+1)/n!$ How do find the sum of the series till infinity?
$$ \frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+\cdots$$
I know that it gets reduced to $$\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}$$
But I don't know how to proceed further.
| As you observed $$\sum_{n=1}^{\infty} \frac{n(n+1)}{n!}$$ further reduces to,
$$\begin{align}
&\sum_{n=1}^{\infty} \frac{n(n+1)}{n(n-1)!} \\
=&\sum_{n=1}^{\infty} \frac{n+1}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{n}{(n-1)!}+\frac{1}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{(n-1)}{(n-1)!} + \frac{2}{(n-1)!}\\
=&\sum_{n=1}^{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluation of $\lim\limits_{x\rightarrow 0}\frac1x\left((1+2x+3x^2)^{1/x}-(1+2x-3x^2)^{1/x}\right) $
Evaluation of $$\lim_{x\rightarrow 0}\frac{(1+2x+3x^2)^{\frac{1}{x}}-(1+2x-3x^2)^{\frac{1}{x}}}{x} $$
$\bf{My\; Try::}$ Let $$l=\lim_{x\rightarrow 0}\frac{e^{\frac{\ln(1+2x+3x^2)}{x}}-e^{\frac{\ln(1+2x-3x^2)}{x}}}{x}$... | Consider $$A=\frac{(1+2x+3x^2)^{\frac{1}{x}}-(1+2x-3x^2)^{\frac{1}{x}}}{x} =\frac{B^{\frac{1}{x}}-C^{\frac{1}{x}}}x$$ using $$B=(1+2x+3x^2) \qquad , \qquad C=(1+2x-3x^2)$$ So $$\log(B^{\frac{1}{x}})=\frac{1}{x}\log(B)$$ and now, using Taylor series $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ replace $... | {
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"url": "https://math.stackexchange.com/questions/1792724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show $\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$? Prove that:
$$\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$$
The LHS is irrational number and RHS is rational number. May ... | Here's the proof I would use... note that I skip a few steps, specifically on simplifying the far right term on the LHS. Likewise, I apologize for any typos... my computer crashed thrice while writing this
$$\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Best way to expand $(2+x-x^2)^6$
I've completed part $(a)$ and gotten:
$64+192y+240y^2+160y^3+...$
Using intuition I substituted $x-x^2$ for $y$ and started listing the values for :
$y, y^2 $ and $y^3,$ in terms of $x$.
$y=(x-x^2)\\y^2=(x-x^2)^2 = x^2-2x^3+x^4;\\y^3 = (x-x^2)^3 = (x-x^2)(x^2-2x^3+x^4) = \;...$
Everyth... | Here is another proposal about how to calculate example 4b). We start with
Some considerations
*
*At first it was a good idea from OP to start with 4a.). Due to the similarity of the somewhat simpler expression
\begin{align*}
(2+y)^6=64+192y+240y^2+160y^3+\cdots\tag{1}
\end{align*}
compared with $(2+x-x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$ Is there an easy to solve the problem? The way I did it is to find the value of $a$ from the second expression and then use it to find the value of the first expression. I believe there must be an simple and elegant approach to tackle the prob... | Hint. From the equation, one easily gets
$$
a^2=5a-1, \quad a^4=(5a-1)^2=25a^2-10a+1=115a-24
$$ giving in the first expression
$$
\frac{a^2}{a^4+a^2+1}=\frac{5a-1}{120a-24}=\frac{1 \times\color{red}{(5a-1)}}{24\times\color{red}{(5a-1)}}=\frac1{24}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solution of $4 \cos x(\cos 2x+\cos 3x)+1=0$ Find the solution of the equation:
$$4 \cos x(\cos 2x+\cos 3x)+1=0$$
Applying trigonometric identity leads to
$$\cos (x) \cos \bigg(\frac{x}{2} \bigg) \cos \bigg(\frac{5x}{2} \bigg)=-\frac{1}{8}$$
But I can't understand what to do from here. Could some suggest how to proceed... | Using the identities
$$
\cos(2x)=2\cos^2x-1\qquad \cos(3x)=4\cos^3x-3\cos x
$$
yields
$$
4\cos x(2\cos^2x-1+4\cos^3x-3\cos x)+1=0
$$
that is
$$
16\cos^4x+8\cos^3x-12\cos^2x-4\cos x+1=0
$$
Set $t=\cos x$. Our equation is
$$
\begin{array}{c}
16t^4+8t^3-12t^2-4t+1=0\\
(2t+1)(8t^3-6t+1)=0\\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
My solution: Consider
$$x^2 + y^2 + z^2 < 2(xy + yz + xz)$$
Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$
Hence... | Hint: In my personal experience, every inequality about sides of a triangle can be solved using the fact that there are positive real numbers $a,b,c$ such that $x=a+b,y=b+c,z=c+a$, i.e. the tangent points of the inner-circle.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with Change of Variable for the function $f(x,y)=e^{\frac{x}{2x+3y}}$ Let $D$ be the open triangle with the vertices $(0,0), (3,0), (0,2)$. For $f(x,y)=e^{ \frac{x}{2x+3y}}$ show that $f$ is integrable on $D$ and prove that $\iint_Df(x,y)dxdy=6\sqrt{e}-6$.
I was able to prove that $f$ is integrable on $D$, since $... | I think you had the upper limit for $y$ the wrong way around, unless you switched the vertices on the axes. If the vertices are on $(3,0)$ and $(0,2)$, the equation of the line joining them is $2x+3y=6$ so $y$ runs from $0$ to $-\tfrac{2}{3}x+2$.
If you let $u=x$ and $v=2x+3y$, then $u$ keeps the limits of $x$ and $v$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $(x+y)(y^2+z^2)(z^3+x^3) < \frac92$ for $x+y+z=2$ $x,y,z \geqslant 0$ and $x+y+z=2$, Prove
$$(x+y)(y^2+z^2)(z^3+x^3) < \frac92$$
While numerical method can solve this problem, I am more interested in classical solutions. I tried this problem for the past few months, using all kinds of AM-GM and CS, but still cann... | Let $f(x,y,z)=(x+y)(y^2+z^2)(z^3+x^3)$.
Assume that $y$ is largest of $x,y,z$. Then $$f(x,y,z)-f(x,z,y)=(y^2+z^2)(x + y) (x + z) (y - z) (x - y - z)\le0$$and therefore we may assume $x$ or $z$ is largest of $x, y, z$.
If $x$ is largest, then
$$f(x,y,z)-f(z,y,x)=(z^3+x^3)(z-x)(xy+xz-y^2+yz)\le0$$and therefore we may ass... | {
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"timestamp": "2023-03-29T00:00:00",
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Let $n \in \mathbb{N}$. Proving that $13$ divides $(4^{2n+1} + 3^{n+2})$ Let $n \in \mathbb{N}$. Prove that $13 \mid (4^{2n+1} + 3^{n+2} ). $
Attempt: I wanted to show that $(4^{2n+1} + 3^{n+2} ) \mod 13 = 0. $ For the first term, I have $4^{2n+1} \mod 13 = (4^{2n} \cdot 4) \mod 13 = \bigg( ( 4^{2n} \mod 13) \cdot ( 4... | Try induction. For $n=0$ the term $4^{2n+1} + 3^{n+2}$ becomes $4 + 9 = 13$, so that's OK.
Then suppose 13 divides $4^{2n+1} + 3^{n+2}$, then consider $4^{2(n+1)+1} + 3^{(n+1)+2} = 4^{2n+3} + 3^{n+3}$.
This equals $16 \cdot 4^{2n+1} + 3 \cdot 3^{n+2} = 13 \cdot 4^{2n+1} + 3\cdot( 4^{2n+1} + 3^{n+2})$, which is divisib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\fr... | $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
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Prove that $\phi \left(\frac{a^b-1}{c}\right)$ is divisible by $b$ What is the difference between the two questions below?
Prove that if $a,b,$ and $c$ are positive integers and $c \leq a$ where $\dfrac{a^b-1}{c}$ is an integer, then $\phi \left(\frac{a^b-1}{c}\right)$ is divisible by $b$.
Consider the integer $d = \... | Let $d=\frac{a^b-1}{c}$. Notice $\varphi(d)$ is the order of the multiplicative group of $\mathbb Z_d$.
What is the order of $a$ in $\mathbb Z_d^*$?
We have $a^b-1|a^b-1\implies \frac{a^b-1}{c}|a^b-1 \implies a^b \equiv 1 \bmod \frac{a^b-1}{c}$
Now notice $a^{b-1}-1\leq \frac{a^b}{c}-1< \frac{a^b-1}{c}$, so the order c... | {
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"timestamp": "2023-03-29T00:00:00",
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How is $\lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right) = n\times a^{n-1}$? In my book this is termed as a theorem and the proof given is as follows :-
$$\begin{align}
\lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right)
&=\lim_{x \to a}\left(\frac{(x - a)*(x^{x-1} + x^{n-2}*a + x^{n-3}*a^2 + x^{n-4}*a^3 + \cdots + x^... | To adress your question. In the expression
$$
(x - a)\cdot(x^{n-1} + x^{n-2}\cdot a + x^{n-3}\cdot a^2 + x^{n-4}\cdot a^3 + \cdots + x^1\cdot a^{n-2} + a^{n-1})
$$ you may just expand it as
$$
x \cdot \left(x^{n-1} + x^{n-2}\cdot a + x^{n-3}\cdot a^2 + x^{n-4}\cdot a^3 + \cdots + x^1\cdot a^{n-2} + a^{n-1}\right)
$$ gi... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that: $97|2^{48}-1$ Show that: $97|2^{48}-1$
My work:
$$\begin{align}
2^{96}&\equiv{1}\pmod{97}\\
\implies (2^{48}-1)(2^{48}+1)&=97k\\
\implies (2^{24}-1)(2^{24}+1)(2^{48}+1) &=97k\\
\implies (2^{12}-1)(2^{12}+1)(2^{24}+1)(2^{48}+1)&=97k\\
\implies (2^6-1)(2^6+1)(2^{12}+1)(2^{24}+1)(2^{48}+1) &=97k
\end{align}$... | Hint:
$2^{24}+1 = 16777217 = 172961 \times 97$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I tho... | If $n$ is odd, then as in the answer by @Semiclassical, $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n\equiv5\pmod{10}.$
If $n$ is even, say $n=2m,$ then first observe $4^m\equiv 5+(-1)^m\pmod{10}$ and $9^m\equiv(-1)^m\pmod{10}.$
So we have
$\begin{align}1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n&\equiv2(1+2^n+3^n+4^n)+5^n\\&\equiv2(1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that there is a unique positive integer $n$ such that $2^8 + 2^{11} + 2^n$ is a perfect square. I think I have almost got to the solution:
Let us start by considering two cases: $n\geq8$ and $n<8$.
1: $n\geq8$
$2^8+2^{11}+2^n$=$2^8(1+2^3+2^{n-8})=2^8(9+2^{n-8})$.......Factoring $2^8$ and ... | For n < 8.
$2^8 + 2^{11}$ is $48^2$. That means $48^2 + 2^n$ is a perfect square. Since $2^n$ is even and $48^2$ is even, and even plus even is even, the square also has to be even. The next greatest even square after $48^2$ is $50^2$ which is $196$ greater than $48^2$. $2^7$ is $128$, which is less than $196$ which p... | {
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Find $\sqrt{8+6i}$ in the form of $a+bi$ I need help with changing $\sqrt{8+6i}$ into complex number standard form.
I know the basics of complex number such as the value of $i$ and $i^2$, equality of complex number, conjugate and rationalizing method. This is my first encounter with such term and its only been a week s... | $$
8+6i = \sqrt{8^2+6^2} \cdot (\cos\theta + i\sin\theta) = 10(\cos\theta+i\sin\theta)
$$
where $\cos\theta = \dfrac 8 {\sqrt{8^2+6^2}} = \dfrac 8 {10} = \dfrac 4 5$ and $\sin\theta = \dfrac 6 {\sqrt{8^2+6^2}} = \dfrac 6 {10} = \dfrac 3 5$.
The square roots are
$$
\pm\sqrt{10} \left( \cos\frac\theta 2 + i \sin \frac\t... | {
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"timestamp": "2023-03-29T00:00:00",
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Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Given that $\tan 2x+\tan x=0$, show that $\tan x=0$
Using the Trigonometric Addition Formulae,
\begin{align}
\tan 2x & = \frac{2\tan x}{1-\tan ^2 x} \\
\Rightarrow \frac{2\tan x}{1-\tan ^2 x}+\tan x & = 0 \\
\ 2\tan x+\tan x(1-\tan ^2 x) & = 0 \\
2+1-\tan ^2 x & = 0 ... | In line 4 you divided both side with $\tan x$ assuming that $\tan x\neq0$
($1$) So, if $\tan x\neq0$, you are on right track. $x=60^\circ$
($2$) But, if $\tan x=0$, you didn't consider this one. So, $\tan x=0\implies x=0^\circ$
| {
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"url": "https://math.stackexchange.com/questions/1817022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Find the equations of the tangent planes to the sphere $x^2+y^2+z^2+2x-4y+6z-7=0,$ which intersect in the line $6x-3y-23=0=3z+2.$ Find the equations of the tangent planes to the sphere $x^2+y^2+z^2+2x-4y+6z-7=0,$ which intersect in the line $6x-3y-23=0=3z+2.$
Let the tangent planes be $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+... | Given sphere:$$
x^2+y^2+z^2+2x-4y+6z-7=0.
$$
Required planes which intersect in the given line are given by$$
6x-3y-23+λ(3z+2)= 0. \tag{1}
$$
The plane (1) will be tangent to the given sphere when radius of the sphere is equal to the perpendicular distance of the plane from the center $(-1, 2, -3)$ of the sphere. This ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determine $\frac{f''(\frac{1}{2})}{f'(\frac{1}{2})}$ if $f(x) = \sum_{k=0}^{1000} \ {2015 \choose k}\ x^k(1-x)^{2015-k}$
Problem : Determine $\frac{f''(\frac{1}{2})}{f'(\frac{1}{2})}$ if $f(x) = \sum_{k=0}^{1000} \ {2015 \choose k}\ x^k(1-x)^{2015-k}$
Trying to simply brute force the problem, yields the following der... | Define $$f_{m,n}(x) = \sum_{k=0}^m \binom{n}{k} x^k (1-x)^{n-k}, \quad 0 < m \le n, \quad 0 < x < 1.$$ Then
$$f_{m,n}'(x) = \sum_{k=0}^m \binom{n}{k} k x^{k-1} (1-x)^{n-k} - (n-k) \binom{n}{k} x^k (1-x)^{n-k-1}.$$
Using the identities
$$k \binom{n}{k} = n \binom{n-1}{k-1}, \quad (n-k)\binom{n}{k} = n \binom{n-1}{k},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
$(\sin^{-1} x)+ (\cos^{-1} x)^3$ How do I find the least and maximum value of $(\sin^{-1} x)+ (\cos^{-1} x)^3$ ?
I have tried the formula $(a+b)^3=a^3 + b^3 +3ab(a+b)$ , but seem to reach nowhere near ?
| $$\frac{d}{dx}\left(\arcsin x +\arccos^{3}x\right)$$
$$\frac{d}{dx}\left(\arcsin x\right) +\frac{d}{dx}\left(\arccos^{3}x\right)$$
$$\frac{1}{\cos(\arcsin x)}+3\arccos^2x\cdot\left(-\frac{1}{\sin(\arccos x)}\right)$$
$$\frac{1}{\sqrt{1-x^2}}+3\arccos^2x\cdot\left(-\frac{1}{\sqrt{1-x^2}}\right)$$
$$\frac{1-3\arccos^2 x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\f... | A faster approach:
$$ J(a,b)=\int_{0}^{+\infty}\frac{x^b}{a^2+x^2}\,dx = a^{b-1} \frac{\pi}{2\cos\left(\frac{\pi b}{2}\right)}\qquad (a>0,b\in(-1,1))\tag{1}$$
is a consequence of the reflection formula for the $\Gamma$ function and Euler's beta function properties.
By considering:
$$ \lim_{b\to 1^-}\lim_{a\to 1^-}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Why my calculations aren't right? (Maclaurin series) Good evening to everyone!
I tried to calculate $ \cos\left( x- \frac{x^3}{3} + o(x^4)\right) $ using the MacLaurin series but instead of getting the final result equal to $1 - \frac{x^2}{2}+\frac{3x^4}{8} + o(x^4)$ I got this:
$$
\cos\left( x- \frac{x^3}{3} + o(x^4)\... | You are making two errors: (1) your series for $\cos x$ is wrong; (2) you are not treating the $x^4$ term correctly.
We have $\cos x=1-\frac{1}{2}x^2+\frac{1}{24}x^4+O(x^5)$.
When we replace $x$ by $x-\frac{1}{3}x^3+O(x^5)$ the $\frac{1}{2}x^2$ gives us $\frac{1}{2}(x^2-\frac{2}{3}x^4)+O(x^5)=\frac{1}{2}x^2-\frac{1}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{n}{2^n} = 2 - \frac{n+2}{2^n} $ I need help with this exercise from the book What is mathematics? An Elementary Approach to Ideas and Methods. Basically I need to proove:
$$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{n}{2^n} = 2 - \frac{n+2}{2^n} $... | You don't need any induction for this. This Series is simple AGP. Procedure for solving these series is assume
$$s=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}.....\frac{n-1}{2^{n-1}}+\frac{n}{2^n}$$ Now multiply whole series by $\frac{1}{2}$ and write the whole series as one term shifted $$\frac{s}{2}= \ \ \ \ \ \ \ \ \ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Derivative of $x^y=y^x$ defines: $y=y(x)$ I need to find the derivative. given that:
$$x^y=y^x$$
defines:
$$y=y(x)$$
Thank you!
| Method 1
\begin{align*}
y^{x} &= x^{y} \\
x\ln y &= y\ln x \\
\ln y+\frac{xy'}{y} &= y'\ln x+\frac{y}{x} \\
\left( \frac{x}{y}-\ln x \right)y' &= \frac{y}{x}-\ln y \\
y' &= \frac{y(y-x\ln y)}{x(x-y\ln x)}
\end{align*}
Method 2
Let $y=(t+1)x$, then
\begin{align*}
[(t+1)x]^{x} &= x^{(t+1)x} \\
(t+1)^{x}x^{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I find the domain/range of functions algebraically? I've been having trouble when trying to find the domain/range of functions algebraically. Here is an example:
$P(x)=\frac{1}{3+\sqrt{x+1}}$
Finding the domain:
$x+1\ge0$
$x\ge-1$
Therefore, $x \in [-1,+\infty)$
Finding the range: Let $y=P(x)=\frac{1}{3+\sqrt{x+... | I don't fully follow what you are doing to determine the range. In any case, when you have:
$(1/y -3)^2\ge0$
The LHS is a square and thus always positive, this inequality is satisfied for all $y$...
In the formula:
$$y = \frac{1}{3+\sqrt{x+1}}$$
the range of the monotonically increasing part $\sqrt{x+1}$ is (clearly)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$ Can someone point me in the right direction how to solve this?
$3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$
I guess I have to get to logarithms of the same base. But how? What principle should I use here?
Thx
| A hint to get you started:
$$3^{x+2}\cdot 4^{-(x+3)}+3^{x+4}\cdot 4^{-(x+3)}=4^{-(x+3)}\cdot\left(3^{x+2}+3^{x+4}\right)=\frac{1}{4^{x+3}}\cdot\left(3^{x+3-1}+3^{x+3+1} \right)$$
Now try to factor out a term with $3^{(\ldots)}$ and see if you can apply some exponent laws.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find number of solutions of an equation? Given $n$, how to count the number of solutions to the equation $$x + 2y + 2z = n$$ where $x, y, z, n$ are non-negative integers?
| We can restate this as:
$$x=n-2y-2z=n-2(y+z)$$
$y$ and $z$ are non-negative integers, so $y+z \geq 0$. We also have $x$ is a non-negative integer, so $n-2(y+z) \geq 0$, or $y+z \leq \lfloor \frac{n}{2} \rfloor$. Thus, we need to figure out the number of positive integers $y, z$ that satisfy:
$$0 \leq y+z \leq \left\lfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\... | $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 5
} |
How to prove this Taylor expansion of $\frac{1}{(1+x)^2}=-1\times\displaystyle\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$? I came across this series of the Taylor Expansion-
$$\frac{1}{(1+x)^2}=1 - 2x + 3x^2 -4x^3 + \dots.=-1\times\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$$
But I have no idea how to prove this...
Thanks for any help!... | Here isn't a VERFICATION of hte fact, but a full derivation, as if you had to come up with this question all by yourself to ask someone else.
Recall that taylor's theorem states:
Given an smooth function $f(x)$ around a point $a$ (that is the function is infinitely differentiable around a, so $f(a)$, $f'(a)$, $f''(a)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Determine $f(x)$ which satisfies given condition Suppose $f(x)$ is real valued function of degree $6$ satisfying the following conditions:
$1.$ $f(x)$ has minimum at $x=0$ and $x=2$
$2.$ $f(x)$ has maximum at x=1
$3.$ $lim(x \to 0)$ $\frac{ln(\Delta)}{x}=2$ where $\Delta$ is
\begin{vmatrix}
\frac{f(x)}{x} & 1 & 0 &\\
... | For the limit to exist we need
\begin{align*}
&\lim_{x\rightarrow 0} \ln \Delta(0) = 0\\
\implies &\lim_{x\rightarrow 0} \Delta(0) = 1\\
\implies &\lim_{x\rightarrow 0} f(x)/x^3+1 = 1\\
\implies &\lim_{x\rightarrow 0} f(x)/x^3 = 0.
\end{align*}
Since $f(x)$ is continuous this is only possible if $f(0) = 0$. By L'hopi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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A very curious rational fraction that converges. What is the value? Is there any closed form for the following limit?
Define the sequence
$$ \begin{cases}
a_{n+1} = b_n+2a_n + 14\\
b_{n+1} = 9b_n+ 2a_n+70
\end{cases}$$
with initial values $a_0 = b_0 = 1$. Then $\lim_{n\to\infty} \frac{a_n}{b_n} = ? $
The lim... | The requested limit is:
$$ \frac{4 \sqrt{57} - 20}{4 \sqrt{57} + 44} \approx 0.1374586 $$
This is
$$ \frac{ \sqrt{57} - 5}{ \sqrt{57} + 11} $$
and rationalizing the denominator gives
$$ \frac{ \sqrt{57} - 7}{ 4} $$
$$ a_{n+2} = 11 a_{n+1} - 16 a_n - 42 $$
$$ b_{n+2} = 11 b_{n+1} - 16 b_n - 42 $$
The separate lin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 5,
"answer_id": 0
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Partial fraction integration problem I'm trying to solve this integral by partial fraction:
$$
\int \frac{2x-6} {(x-2)^2(x^2+4)} dx \
$$
i think i have to write the expression like
$$
2\int \frac{x-3} {(x-2)^3(x+2)} dx \
$$
Then i don't know how i should resolve the partial fraction!
| Careful!
$$
\int \frac{2x-6} {(x-2)^2(x^2+4)} dx \
$$
i think i have to write the expression like
$$
2\int \frac{x-3} {(x-2)^3(x+2)} dx \
$$
You seem to have replaced $x^2+4$ by $(x-2)(x+2)$, but these are not the same! You might be confusing with $x^2\color{red}{-}4=(x-2)(x+2)$...
You can take out the (constant) f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$ I want to show that
$$\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$$
Expand $(x^4-x+\pi)^2=x^4-2x^3+2x^2-2x\pi+\pi{x^2}+\pi^2$
Let see (substitution of $y=x^2$)
$$\int_{-\infty}... | Trick: the integral over $\mathbb{R}$ of a non-vanishing meromorphic function, $O\left(\frac{1}{\|z\|^2}\right)$ at infinity, is just $2\pi i$ times the sum of the residues for the poles in the upper half-plane. In our case such poles are located at $z=e^{\pi i/6}$ and $z=e^{5\pi i/6}$ (roots of $z^2-iz-1$) and the sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + 1/β$ and $β + 1/α$. Quadratic equation question, as specified in the title.
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + \frac{1}{β}$ and $β + \frac{1}{α}$.
I gather that ... | $$(x-(\alpha + 1/\beta))(x - (\beta + 1/\alpha)) = x^2 - (\alpha + \beta + 1/\alpha + 1/\beta) x + (\alpha + 1/\beta)(\beta + 1/\alpha $$
You know $\alpha + \beta = 6$ and $\alpha \beta = 7$.
$$ \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha \beta} = \dfrac{6}{7}$$
$$\left( \alpha + \frac{1}{\beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer.
I'm stuck at the basis step.
If I started with $1$. I get the right hand side is $1... | Proof by induction on $n$;
Set your base induction for $n = 1$; which gives you $18 = 18.$
Suppose your equation is true for $n$. We are interested to prove its correctness for $n+1.$ So
$3 + 3 \times 5 + \cdots + 3 \times 5^n + 3 \times 5^{n+1} = \frac{3 \times 5^{n+1} - 3}{4} + 3 \times 5^{n+1} = \frac{3 \times 5^{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Feasible point of a system of linear inequalities Let $P$ denote $(x,y,z)\in \mathbb R^3$, which satisfies the inequalities:
$$-2x+y+z\leq 4$$ $$x \geq 1$$ $$y\geq2$$ $$ z \geq 3 $$ $$x-2y+z \leq 1$$ $$ 2x+2y-z \leq 5$$
How do I find an interior point in $P$?
Is there a specific method, or should I just try some rando... | Label the planes and their intersection points as follows
$\begin{array}{llllllll}
&\Pi_1:&x=1 & A & B & C & & & W & X\\
&\Pi_2:&y=2 & A & B & & D & & W & & Y\\
&\Pi_3:&z=3 & A & & C & D & & & X & Y & Z\\
&\Pi_4:&-2x+y+z=4 & & B & C & & E & & & Y & Z\\
&\Pi_5:&x-2y+z=1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
How to solve the limit of this sequence $\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$ $$\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$$
I have tried to split the subset into telescopic series but got no result.
I also have tried to use the sque... | Alternatively one may recall the series representation for the digamma function
$$\begin{equation}
\psi(x+1) = -\gamma - \sum_{n=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n}
\right), \quad \Re x >-1, \tag1
\end{equation}
$$ where $\gamma$ is the Euler-Mascheroni constant.
Then by partial fraction decomposition we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1839102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. So I have to prove
$$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$
I rearranged it
$$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$
My idea from there is somehow using the AM-GM inequality. Not sure how t... | Hint:
Set $1/a=x$ etc.
and use $(x-y)^2\ge0$ for real $x,y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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For how many 3-digit prime numbers $\overline{abc}$ do we have: $b^2-4ac=9$?
For how many 3-digit prime numbers $\overline{abc}$ do we have: $b^2-4ac=9$?
The only analysis I did is:
$(b-3)(b+3)=4ac \implies\ b\geq3 $
$b=3\implies\ c=0\implies impossible!!$
So I deduced that $b\gt3$
Is there any better way for quickly... | $(b-3)(b+3) = 4ac$
Now, $4ac$ is an even number. Hence, $(b-3)(b+3)$ is also and even number. That means, b cannot be even number else, $(b-3)(b+3)$ will if odd.
Hence, $b > 3$ and b is odd. Thus, b can be 5, 7, 9.
Putting, the values for b:
When $b = 5$, we get $4ac = (5-3)(5+3) = 2 * 8 = 16$ or $ac = 4 = 2*2 = 1*4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$
\sin 2x = \sin x; \\ 0 \le x < 2 \pi $$
My method:
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$ \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\le... | $$\sin(2x)=\sin(x)\implies 2\cos(x)\sin(x)=\sin(x).$$
The elements of the form $k\pi$ are solutions (notice that if you divide the equation by $\sin(x)$ with not paying attention to this condition, you will loose solutions). Suppose $x\neq k\pi$. Then,
$$2\cos(x)\sin(x)=\sin(x)\implies 2\cos(x)=1\implies \cos(x)=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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If $x\in \left(0,\frac{\pi}{4}\right)$ then $\frac{\cos x}{(\sin^2 x)(\cos x-\sin x)}>8$
If $\displaystyle x\in \left(0,\frac{\pi}{4}\right)\;,$ Then prove that $\displaystyle \frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$
$\bf{My\; Try::}$ Let $$f(x) = \frac{\cos x}{\sin^2 x(\cos x-\sin x)}=\frac{\sec^2 x}{\tan^2 x(1-\ta... | $\bf{My\; Solution::}$ Using $\bf{A.M\geq G.M}$
$$2\sqrt{\sin x(\cos x-\sin x)}\leq \cos x$$
so $$4\sin x(\cos x-\sin x)\leq \cos^2 x\Rightarrow \frac{\sin^2 x(\cos x-\sin x)}{\cos x}\leq \frac{\cos x\sin x}{4}$$
So $$\frac{\sin^2 x(\cos x-\sin x)}{\cos x}\leq \frac{\sin 2x}{8}< \frac{1}{8}$$
So $$\frac{\cos x}{\sin^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Prove $\frac{a}{b} + \frac{b}{c}+\frac{c}{a} \geq \frac{c+a}{c+b} + \frac{a+b}{a+c} + \frac{b+c}{b+a}$ Prove that $\frac{a}{b} + \frac{b}{c}+\frac{c}{a} \geq \frac{c+a}{c+b} + \frac{a+b}{a+c} + \frac{b+c}{b+a}$ with a,b,c > 0
| Hint:
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-3=\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)+\left(\dfrac{b}{c}+\dfrac{c}{a}-\dfrac{b}{a}-1\right)=\dfrac{(a-b)^2}{ab}+\dfrac{(a-c)(b-c)}{ac}\tag{1}$$
Similarly, $$\dfrac{a+c}{b+c}+\dfrac{b+c}{a+b}+\dfrac{a+b}{a+c}-3=\dfrac{(a-b)^2}{(b+c)(a+b)}+\dfrac{(a-c)(b-c)}{(b+c)(a+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Decreasing sequence numbers with first digit $9$
Find the sum of all positive integers whose digits (in base ten) form a strictly decreasing sequence with first digit $9$.
The method I thought of for solving this was very computational and it depended on a lot of casework. Is there a nicer way to solve this question?... | For each digit $a$ from $0$ to $9$, let us count how many numbers there are in our sum with a digit of $a$ in the $10^n$ place. First, suppose $a<9$. A number with a digit of $a$ in the $10^n$ place has a subset of $\{0,1,\dots,a-1\}$ for its last $n$ digits, so there are $\binom{a}{n}$ choices for the last $n$ digit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Find $\sin \theta $ in the equation $8\sin\theta = 4 + \cos\theta$ Find $\sin\theta$ in the following trigonometric equation
$8\sin\theta = 4 + \cos\theta$
My try ->
$8\sin\theta = 4 + \cos\theta$
[Squaring Both the Sides]
=> $64\sin^{2}\theta = 16 + 8\cos\theta + \cos^{2}\theta$
=> $64\sin^{2}\theta - \cos^{2}\theta= ... | $$8\sin\theta=4+\cos\theta$$
$$8\sin\theta-\cos\theta=4$$
$$\sqrt{65}\sin\alpha\sin\theta-\sqrt{65}\cos\alpha\cos\theta=4$$
where $\alpha=\tan^{-1}8$
$$-\sqrt{65}\cos(\theta+\alpha)=4$$
Can you continue from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
implicit derivative for equation $y^5 = x^8$ I'm having a bit of trouble finding the second implicit derivative for the equation $y^5 = x^8$.
I have the first derivative, which is $8x^7 / 5y^4$, but I'm having trouble after that. I did the quotient rule and got $280x^6y^4 - 160x^7y^3 / (20y^3)^2$. I'm not quite sure wh... | $${ y }^{ 5 }-x^{ 8 }=0\\ \\ 5{ y }^{ 4 }\frac { dy }{ dx } -8{ x }^{ 7 }=0\\ \frac { dy }{ dx } =\frac { 8{ x }^{ 7 } }{ 5{ y }^{ 4 } } \\ \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\frac { d }{ dx } \left( \frac { 8{ x }^{ 7 } }{ 5{ y }^{ 4 } } \right) =\frac { 56{ x }^{ 6 }\cdot 5{ y }^{ 4 }-8{ x }^{ 7 }\cdot 20{ y }^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Possible assistance with a monstrous pair of integrals In my study of ODEs I have recently encountered this monster of an integral
the sum of two integrals:
$ \int_{0}^{2\pi} \frac{\sin(x)\sin\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}... | It's simpler than it appears at first :
Let $\quad f(x)=\frac{\sin(x)\cos\left(-\frac{1+A}{\sqrt{A}}\omega \tanh^{-1}\left(\frac{\cos\left(\frac{x}{2}\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}} \right)\right)}{\sqrt{A\sin^2\left(\frac{x}{2}\right)+1}}$
$$\int_0^{2\pi} f(x)dx = \int_0^{\pi} f(x)dx +\int_{\pi}^{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the integral $\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3}dx$ with substitution - how to think?
Find
$$\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3}dx$$
In the above question, I was literally stumped, and wasn't able to solve it for a long time. Turns out that you had to divide the numerator and the denomina... | Generally these type of questions we simply put $\displaystyle x = \frac{1}{t}$ and then use normal substution method
Now let $$I = \int \frac{2x^{12}+5x^9}{(x^5+x^3+1)^3}dx$$
Put $\displaystyle x= \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So $$I = -\int\frac{2+5t^3}{\left(t^5+t^2+1\right)^3}\cdot \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Finding the generating function of a recurrence relation in dependence of a variable Given this inhomogeneous linear recurrence relation of 2nd order :
$F_n = F_{n-2} + a$ for $n \geq 2$
with $F_1 = 1$ and $F_0 = 0$
How do I find the generating function of this recurrence relation in dependence of the variable a? I t... | It’s easy enough to derive a closed form directly from the recurrences:
$$\begin{align*}
F_{2n}&=an\\
F_{2n+1}&=an+1
\end{align*}$$
Thus,
$$\begin{align*}
\sum_{n\ge 0}F_{2n+1}x^{2n+1}&=\sum_{n\ge 0}(an+1)x^{2n+1}\\
&=x\sum_{n\ge 0}anx^{2n}+\sum_{n\ge 0}x^{2n+1}\\
&=x\sum_{n\ge 0}F_{2n}x^{2n}+\frac{x}{1-x^2}\;,
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the modulus of $|z-5|/|1-3z|$ when z is given If $z = 3-2i$ then find $$\frac { \left| z-5 \right| }{ \left| 1-3z \right| } $$
I've substituted z by $|z|^2/z$ conjugate but still cant figure out what to do,
Thanks in advance
| $$\frac { \left| z-5 \right| }{ \left| 1-3z \right| } =\frac { \left| 3-2i-5 \right| }{ \left| 1-9+6i \right| } =\frac { \left| -2-2i \right| }{ \left| -8+6i \right| } =\frac { \sqrt { { 2 }^{ 2 }+{ 2 }^{ 2 } } }{ \sqrt { { \left( -8 \right) }^{ 2 }+{ 6 }^{ 2 } } } =\frac { 2\sqrt { 2 } }{ 10 } =\frac { \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Another formula for the angle bisector in a triangle I have seen in an old geometry textbook that the formula for the length of the angle bisector at $A$ in $\triangle\mathit{ABC}$ is
\begin{equation*}
m_{a} = \sqrt{bc \left[1 - \left(\frac{a}{b + c}\right)^{2}\right]} ,
\end{equation*}
and I have seen in a much older ... | We may prove it by avoiding the usual path, and proving other interesting things along the way.
Let $I$ be the incenter and $I_A$ the $A$-excenter. $\widehat{IBI_A}=\widehat{ICI_A}=\frac{\pi}{2}$, hence $IBI_A C$ is a cyclic quadrilateral. By Van Obel's theorem we have
$$ \frac{AI}{IL_A}=\frac{b+c}{a}, $$
hence it is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Solving a system of linear congruences
Find all positive integer solutions to \begin{align*}x &\equiv -1 \pmod{n} \\ x&\equiv 1 \pmod{n-1}. \end{align*}
I rewrote the system as $x = nk_1-1$ and $x = (n-1)k_2+1$. Thus, we have $nk_1-1 = (n-1)k_2+1$ and so $n(k_1-k_2) = 2-k_2 \implies n = \frac{2-k_2}{k_1-k_2}$. How do... | The two congruences are equivalent to $$\begin{cases}x=nu-1\\x=(n-1)v+1\end{cases}$$
It follows the diophantine equation $$nu-(n-1)v=2$$ which admits the particular solution $(u,v)=(2,2)$ hence the general solution is $$\begin{cases}u=(n-1)t+2\\v=nt+2\end{cases}$$ Finally one has
$$x=n((n-1)t+2)-1$$ where $t$ is an ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find the probability of getting two sixes in $5$ throws of a die.
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial?
My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for ... | So both the fourth and the fifth rolls need to be sixes:
$$P(4^{th},5^{th}\mbox{ rolls are sixes}) = \frac{1}{6^2}$$
There are following possible combinations of rolls for the game to not end until the fifth roll:
$$(XXXOO), (XOXOO),(OXXOO)$$
where $X,O$ represent non-six and six, respectively.
Thus, the probability is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $\lim_{x\to 0}\frac1{x^3}\left(\frac1{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)=l$, then what is the value of $\frac{1}{a}-\frac{2}{l}+\frac{3}{b}$? If the function
$$\lim_{x\to 0}\frac1{x^3}\left(\frac1{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)$$ exists and has a value equal to $l$ then what will be the value of $\frac{1}{a... | Let $$f(x)=\frac{1}{\sqrt{1+x}}-\frac{1+ax}{1+bx}$$ and let $$g(x)=x^3.$$ Apply L'Hospital's rule.
First derivatives
We have $$f'(x)=-\frac{1}{2(1+x)^{3/2}}-\frac{a}{1+bx}-\frac{b(1+ax)}{(1+bx)^2}$$ so that $f'(0)=b-a-1/2$. Also $g'(x)=3x^2$ so that $g'(0)=0$. Thus since there is a finite limit $l$ it must be that $f'(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The $25$th digit of $100!$ I want to find The $25$th digit of $100!$.
My attempt:It is easy to know it has $24$ zeroes.Because:
$\lfloor {\frac{100}{5}} \rfloor+\lfloor {\frac{100}{25}} \rfloor =24$
By getting the fist digits(after deleting all $5$ factors and $24$,$2$ factors)and multiplying them to each other we get ... | If we exclude the numbers that are divisible by 5, we see a cycle that repeats.
$4! = 24\\
9!/6!\equiv 24 \mod 100$
$100! = \frac {100!}{5^{20} 20!} (5^{20} 20!) = \frac {100!}{5^{20} 20!}\frac {20!}{5^4 4!} (4!)(5^{24})$
$(24^{25})(5^{24}) = (12^{25})(10^{24})(2)$
$12^{25} \equiv 12^5 \mod 100\equiv 32 \mod 100$
The l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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solutions of the equation $x^3-y^3=z!-18$ What are the solutions of the equation $x^3-y^3=z!-18$? Here $x,y,z$ are non-negative integers. I have tried brute force but is there a better method?
| You can even solve this if $x,y,z\in\mathbb Z$ can also be negative. Note that $z!$ is never defined when $z<0$.
By Fermat's Little Theorem for all $a\in\mathbb Z$ we have $a^6\equiv 0,1\pmod{7}$ and so $a^3\equiv -1,0,1\pmod{7}$.
If $z\ge 7$, then $$x^3-y^3\equiv 0-18\equiv 3\pmod{7},$$ impossible, because $$x^3-y^3\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Show that $a_n = [n \sqrt{2}]$ contains an infinite number of integer powers of $2$
Show that the sequence $\{a_n\}_{n \geq 1}$ defined by $a_n = [n \sqrt{2}]$ contains an infinite number of integer powers of $2$. ($[x]$ is the integer part of $x$.)
I tried listing out the first few values, but I didn't see a pattern... | Let $1/\sqrt{2} = \sum_{j=1}^\infty d_j 2^{-j}$ be the base-2 expansion of $1/\sqrt{2}$, where each $d_j$ is $0$ or $1$. Since $1/\sqrt{2}$ is irrational, there are infinitely many $0$'s and infinitely many $1$'s.
Let $x_N = \sum_{j=1}^N d_j 2^{-j}$ and $n_N = 1 + 2^N x_N = 1 + \sum_{j=1}^N d_j 2^{N-j}$ which is a pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Analytical expressions for extreme values of $f(x):=\log(2)\left(\sum_\limits{k=-\infty}^\infty 2^{k+x}e^{-2^{k+x}}\right)-1$ The function $f(x):=\log(2)\left(\sum_\limits{k=-\infty}^\infty 2^{k+x}e^{-2^{k+x}}\right)-1$ is a periodic function. Numerical optimization shows that the minimum and maximum of $f$ are approxi... | The Fourier series of the function $f(x)$, which has period equal to 1, is
\begin{equation}
f(x) = \Re\!\left(\sum_{l=1}^{\infty}
a_l \exp(2\pi lx \rm{i})\right)
\end{equation}
with coefficients
\begin{multline}
a_l = 2\int_0^1 f(x) e^{-2\pi lx \rm{i}} dx
=
2\log(2)\int_0^1 \sum_{k=-\infty}^\infty 2^{k+x}e^{-2^{k+x}}e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Circles in complex plane.
Find the real value of a for which there is at least one complex number satisfying $|z+4i|=\sqrt{a^2-12a+28}$ and $|z-4\sqrt{3}|\lt a$.
My solutions:-
*
*Graphical solution:-
$|z+4i|=\sqrt{a^2-12a+28}$ represents a circle with center at $A\equiv(0,-4)$ and radius $r_1=\sqrt{a^2-12a+28}$... | The answer should be $a\ge 6+2\sqrt 2$.
($a\gt 9$ is not correct since, for $a=9$, $z=-3i$ satisfies the conditions.)
In your Algebraic Solution, it seems that you have some errors.
Case 1:-
$$8 \lt \sqrt{a^2-12a+28} + a \implies a\gt 9$$
Case 2:-
$$8 \gt \sqrt{a^2-12a+28} -a \implies a\gt -\frac{9}{7}$$
Ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Tangent and Circle in Complex Plane
Question:-
Three points represented by the complex numbers $a,b$ and $c$ lie on a circle with center $O$ and radius $r$. The tangent at $c$ cuts the chord joining the points $a$ and $b$ at $z$. Show that $$z=\dfrac{a^{-1}+b^{-1}-2c^{-1}}{a^{-1}b^{-1}-c^{-2}}$$
Attempt at a solutio... | Consider $O$ to be the origin, then $|a|=|b|=|c|=r$
We know that the equation of a line passing through points $z_1$ and $z_2$ is represented by
$$\begin{vmatrix}
z & \overline{z} & 1 \\
z_1 & \overline{z_1} & 1 \\
z_2 & \overline{z_2} & 1 \\
\end{vmatrix}= 0$$
Now as per the question $a, b$ and $c$, lie on a circ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\sum_{r=0}^{\left\lfloor\frac{n-1}3\right\rfloor}\binom{n}{3r+1}$
Show that $$\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\ldots=\dfrac{1}{3}\left[ 2^{n-2} + 2\cos{\dfrac{(n-2)\pi}{3}}\right]$$
My solution:-
$$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\ldots=\sum_{r=0}^{n}{\bi... | $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Sum to infinity of trignometry inverse: $\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)$ If we have to find the value of the following (1)
$$
\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)
$$
I know that
$$
\arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2... | We already have plenty of slick answers through creative telescoping, so I will go for the overkill.
By crude estimations we have $\sum_{r\geq 3}\frac{4}{r^2+3}\leq\frac{\pi}{2}$, hence:
$$ \sum_{r\geq 3}\arctan\left(\frac{4}{r^2+3}\right) = \text{Arg}\prod_{r\geq 3}\frac{r^2+3+4i}{r^2}\tag{1}=\text{Arg}\prod_{r\geq 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Integer solutions to $x^3+y^3+z^3 = x+y+z = 8$
Find all integers $x,y,z$ that satisfy $$x^3+y^3+z^3 = x+y+z = 8$$
Let $a = y+z, b = x+z, c = x+y$. Then $8 = x^3+y^3+z^3 = (x+y+z)^3-3abc$ and therefore $abc = 168$ and $a+b+c = 16$. Then do I just use the prime factorization of $168$?
| From $\sqrt[3]{abc}\approx 5.5 > 5\tfrac13=\frac{a+b+c}{3}$, we know that $a,b,c$ cannot all be positive (for in that case the AM-GM inequality states a "$\le $").
As $abc>0$, exactly two of $a,b,c$ must be negative.
Also, either all three numbers are even or exactly two are odd. In the first case, $\frac a2\frac b2\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Solving an equation involving an integral: $\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$ Determine a pair of number $a$ and $b$ for which
$$\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$$
I tried putting $x$ as $1-x$ as the integral wouldn't change but could not move forward from there so can you please suggest ... | Let $$I = \int_{0}^{1}\frac{ax+b}{(x^2+3x+2)^2}dx = \frac{a}{2}\int_{0}^{1}\underbrace{\frac{2x+3}{(x^2+3x+2)^2}dx}_{J}+\left(b-\frac{3a}{2}\right)\underbrace{\int_{0}^{1}\frac{1}{(x^2+3x+2)^2}dx}_{K}$$
So $$J = \int_{0}^{1}\frac{2x+3}{(x^2+3x+2)^2}dx = -\left[\frac{1}{x^2+3x+2}\right]_{0}^{1} = \left(\frac{1}{2}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Algebraic Expressions with Fractions can someone review this and see if i've done it correctly please.
$$\frac{\frac {3x} {y}}{\frac {2x}{7}}
$$
$$= \frac{3x}{y} . \frac{7}{2x}$$
$$= \frac{21x}{2xy}
$$
$$= \frac {21}{2y}
$$
Thank you for your time.
| Your answer and method look correct (but as pointed out in the comments, you need to add the restriction $x\ne 0$). I would have liked to have seen more steps though to demonstrate understanding.
$$\frac{\frac{3x}{y}}{\frac{2x}{7}}$$
$$= \color{red}{\frac{\frac{3x}{y}}{\frac{2x}{7}}\cdot\frac{\frac{7}{2x}}{\frac{7}{2x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Integration of the following trignometry We have to find the integration of the following ,
I tried but got stuck , can anyone help me
| Use Weierstrass substitution
By setting $t= \tan \left( \frac x2 \right) $
The integral is equivalent to:
$$2\int \frac{(1+t^2)^2}{(3-t^2)^3} dt$$
Decompose the fraction into:
$$ 2 \int \left( \frac 1{3-t^2} - \frac 8{(3-t^2)^2} + \frac{16}{(3-t^2)^3} \right)dt$$
Since $\frac 1{3-t^2} = \frac 1{2\sqrt 3} \left( \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to factorize the polynomial $a^6+8a^3+27$?
I would like to factorize $a^6+8a^3+27$.
I got different answers but one of the answers is
$$(a^2-a+3)(a^4+a^3-2a^2+3a+9)$$
Can someone tell me how to get this answer? Thanks.
| One may write
$$
\begin{align}
a^6+8a^3+27&=a^3\left(a^3+\frac{27}{a^3}+8 \right)
\\&=a^3\left(\left(a+\frac3a\right)^3-9\left(a+\frac3a\right)+9-1 \right)
\\&=a^3\left(\left[\left(a+\frac3a\right)^3-1\right]-9\left[\left(a+\frac3a\right)-1\right] \right)
\\&=a^3\left(a+\frac3a-1\right)\left(\left(a+\frac3a\right)^2+\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Finding the value of an expression using substitution How to find the value of $x^{4000} + \frac{1}{x^{4000}}$ if $x+\frac{1}{x}=1$.
I think binomial theorem will be useful in this.
Here's my proceedings:
$$x+\frac{1}{x}=1$$
Raising both sides to the power of 4000
$$(x+\frac{1}{x})^{4000} = 1^{4000}$$
$${4000 \choose... | HINT:
$$x^2-x+1=0\implies x^3+1=(x+1)(x^2-x+1)=0\implies x^3=-1$$
Now $4000=3(2n+1)+1$
Now $x^{3(2n+1))}=(x^3)^{2n+1}\cdot x=(-1)^{2n+1}\cdot x=-x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Spivak's Calculus - Chapter 1 Question 1.5 - Proof by Induction In Spivak's Calculus Fourth Edition, Chapter 1 Question 1.5 is as follows:
Prove $x^n - y^n = (x - y)(x^{n-1}+x^{n-2}y+ \cdots + xy^{n-2}+y^{n-1})$ using only the following properties:
It's easy enough for me to use P9 to expand the right-hand side:
$$... | If you're planning to do a proof by induction, the first step is to write down your inductive hypothesis, typically expressed as a proposition like "$P(n): (x^n - y^n) = (x-y) \sum_{i=0}^{n-1} x^i y ^ {n-1-i}$.
You then establish the truth of $P(0)$ or $P(1)$ or some other useful starting point. Then you say "Suppose ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $ Find $$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $$
Any hints please?
Could'nt think of any approach till now...
| Let $$I = \int\frac{x^2}{(x\sin x+\cos x)\cdot (x\cos x-\sin x)}dx$$
$\displaystyle \bullet\;\; x\sin x+1\cdot \cos x = \sqrt{x^2+1}\left[\sin x\cdot \frac{x}{\sqrt{1+x^2}}+\cos x\cdot \frac{1}{\sqrt{1+x^2}}\right]$
$$=\sqrt{x^2+1}\sin \left(x+\alpha\right)$$
Where $\displaystyle \cot \alpha = x\Rightarrow \alpha = \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\{f(n) = \frac{1}{4n \tan \frac{\pi}{n}} \}$ is a monotonically increasing sequence Could anyone is able to give me a hint why $\{f(n) = \frac{1}{4n \tan \frac{\pi}{n}} \}$ is a monotonically increasing sequence?
| Function
$$f(x)=\frac{1}{4x\tan{\frac{\pi}{x}}}$$ is increasing, since $f'(x)>0$, which is easily proved:
$$f'(x)=\frac{-4\tan{\frac{\pi}{x}}+\frac{4\pi}{x\cos^2{\frac{\pi}{x}}}}{4x^2\tan^2{\frac{\pi}{x}}}>0$$
is equivalent with $$-4\tan{\frac{\pi}{x}}+\frac{4\pi}{x\cos^2{\frac{\pi}{x}}}>0,$$
which is again equivalent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show $1+z=2\cos \frac{1}{2} x(\cos \frac{1}{2}x +i \sin \frac{1}{2}x)$, where $z=\cos x+i \sin x$ Let $z=\cos x+i \sin x$. Show that
$$1+z=2\cos \frac{1}{2} x(\cos \frac{1}{2}x +i \sin \frac{1}{2}x)$$
| Since
\begin{eqnarray}
\cos x&=&2\cos^2\frac{x}{2}-1\\
\sin x&=&2\sin\frac{x}{2}\cos\frac{x}{2},
\end{eqnarray}
we have
\begin{eqnarray}
1+z&=&1+\cos x+i\sin x\\
&=&1+\Big(2\cos^2\frac{x}{2}-1\Big)+i\Big(2\sin\frac{x}{2}\cos\frac{x}{2}\Big)\\
&=&2\cos^2\frac{x}{2}+2i\sin\frac{x}{2}\cos\frac{x}{2}\\
&=&2\cos\frac{x}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A tangent to the ellipse meets the $x$ and $y$ axes . If $O$ is the origin, find the minimum area of triangle $AOB$.
The question is that : A tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the $x$ and $y$ axes respectively at $A$ and $B$. If $O$ is the origin, find the minimum area of triangle $AOB$... | $A = \frac {a^2 b^2}{2xy}$
constrained by
$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$
$\frac{dA}{dx} = \frac {-2a^2b^2 (y + x y')}{(2xy)^2} = 0\\
y + x y' = 0\\
y' = -\frac yx$
Differentiating the constraint.
$\frac x{a^2} + \frac {y y'}{b^2} = 0\\
y' = -\frac {b^2x}{a^2y}\\
\frac {b^2x}{a^2y} = \frac yx\\
b^2 x^2 = a^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1881826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$
$$3x^2dx = \sec^2(y)dy$$
$$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$
How should I proceed after this?
EDITED: C... | Hint:
note that
$$
x^6+1=(x^2+1)(x^2-\sqrt{3}+1)(x^2+\sqrt{3}+1)
$$
than use partial fraction decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Evaluation of this series $\sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{4{{n}^{2}}-1}}=??$ I start first to use $u=n+1$ then
$$\sum_{u=2}^{\infty}
{\frac{{{\left( -1 \right)}^{u}}}{4{{\left( u-1 \right)}^{2}}-1}}.$$
| Another way to do it is converting the series into a double integral:
Let $$I=\int_{0}^{1}\int_{0}^{1} \frac{x^2}{1+x^2y^2}dydx.$$ Because the region of integration implies $$0<x,y<1$$ we can convert the integrand into a geometric series as such:
$$\frac{x^2}{1+x^2y^2}=\sum_{n=0}^{\infty}x^2(-x^2y^2)^n=\sum_{n=0}^{\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Do two nonsingular matrices exist such that this linear system can be solved? (I've posted this question on MathOverflow and was sent here. This is a copy of that submission)
I am trying to solve a system of linear equations. In this system I have two separate systems involving an unknown tension vector $\overrightarro... | So, if I understood correctly your symbolism, we are
dealing with matrices of the type
$$
\begin{gathered}
\mathbf{C}\left( \mathbf{x} \right) = \left[ {\begin{array}{*{20}c}
{x_1 } & { - x_2 } & 0 & \cdots & 0 \\
0 & {x_2 } & { - x_3 } & \cdots & 0 \\
0 & 0 & {x_3 } & \cdots & 0 \\
\vdots & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Exponential generating functions balls problem I had a question about an exercise that I couldn't solve on my exam.
It was the following:
You have 3 boxes. 1 with an unlimited amount of identical red balls, 1 with an unlimited amount of identical green balls and 1 with an unlimited amount of identical blue balls. We n... | To start off, we can find the exponential generating function where the coefficient of $\frac{x^r y^g z^b}{n!}$ is the number of lists containing $r$ red balls, $g$ green balls, and $b$ blue balls, where $r + g + b = n$. This is given by
$$
\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right)
\left(1 + y + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral trigonometry $\int\sin3x\sin^2x\,dx$
$=\int \sin3x \frac{(1-cos2x)}{2}dx$
$=\frac{1}{2}(\int \sin 3x dx - \int \sin 3x \cos 2x dx)$
$I=\frac{-1}{2}\frac{1}{3}cos 3x-1/4(\int \sin 5x dx+ \int \sin x dx)$
So my question is how do i get from:
$\int \sin (3x)\cos (2x) dx$ to $\frac{1}{2}(\int \sin (5x) dx+ \int ... | by the formula $\sin { \left( x \right) \cos { \left( y \right) =\frac { 1 }{ 2 } \left[ \sin { \left( x+y \right) +\sin { \left( x-y \right) } } \right] } } $
in your case we have $$\\ \sin { \left( 3x \right) \cos { \left( 2x \right) =\frac { 1 }{ 2 } \left[ \sin { \left( 3x+2x \right) +\sin { \left( 3x-2x \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How many different ways can a number be written as a sum of 1, 3, and 5? I had a programming question where we had to write code to output, given some number x, all the different ways x can be written as a sum of 1, 3 and 5. So for instance if x=6, then the answer is 4, as x=5+1=3+3=3+1+1+1=1+1+1+1+1+1.
This made me wo... | The generating function approach is to note that if you write:
$$\begin{align}
f(x)&=\frac{1}{1-x}\frac{1}{1-x^3}\frac{1}{1-x^5}\\
&=(1+x+x^2+x^3+\cdots)(1+x^3+x^6+x^9+\cdots)(1+x^5+x^{10}+\cdots)\\
&=a_0+a_1x+a_2x^2+\cdots
\end{align}$$
Then $a_n$ counts the number of ways to partition $n$ into values in $1,3,5.$
Then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the value of the constants If $\displaystyle
\frac{\left( \frac{2x^2}{3a} \right)^{n-1}}
{\left( \frac{3x}{a} \right)^{n+1}} =
\left( \frac{x}{4} \right)^3$, determine the values of the constants $a$ and $n$
I could find the value of $a$, i.e, $\displaystyle \frac{\sqrt{x^6 \times 3^{2n}}}{2^n x^n 2^5 }$ ... | Expand the LHS as
$$
\frac{{\left( {\frac{{2x^{\,2} }}
{{3a}}} \right)^{n - 1} }}
{{\left( {\frac{{3x}}
{a}} \right)^{n + 1} }} = \frac{{2^{n - 1} x^{2\left( {n - 1} \right)} }}
{{3^{n - 1} a^{n - 1} }}\;\frac{{a^{n + 1} }}
{{3^{n + 1} x^{n + 1} }} = \frac{{2^{n - 1} a^2 x^{n - 3} }}
{{3^{2n} }}\;
$$
The equating it to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Two limits involving integrals: $\lim_{\varepsilon\to 0^+}\left(\int_0^{1-\varepsilon}\frac{\ln (1-x)}{x\ln^p x}dx-f_p(\varepsilon)\right)$, $p=1,2$. By applying the Taylor series expansion to $\ln x$, as $x \to 1$, one has the Laurent series expansion,
$$
\frac1{\ln x}=-\frac1{1-x}+\frac{1}{2}+O\left(1-x\right)
$$
t... | We can try to use your idea for the other integral. Note that $$\frac{\log\left(1-x\right)}{x\log^{2}\left(x\right)}=\frac{\log\left(1-x\right)}{\left(1-x\right)^{2}}+\frac{\log\left(1-x\right)}{x}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)
$$ so we have $$\int_{0}^{1-\ep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Volume of ellipsoid outside sphere I have the ellipsoid $\frac{x^2}{49} + y^2 + z^2 = 1$ and I want to calculate the sum of the volume of the parts of my ellipsoid that is outside of the sphere $x^2+y^2+z^2=1$
How to do this? I know the volume of my sphere, $\frac{4\pi}{3}$, and that I probably should set up some doubl... | Hints.
1) If $(x,y,z)$ satisfies $x^2+y^2+z^2\leq 1$ then
$$\frac{x^2}{49} + y^2 + z^2 \leq x^2+y^2+z^2\leq 1.$$
What does this inequality mean?
2) $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}=1$ is the equation of an ellipsoid centered at the origin of semi-principal axes of length $a$, $b$, $c$, and its volu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Minimum and Maximum value of |z| This is a question that I came across today:
If $|z-(2/z)|=1$...(1) find the maximum and minimum value of |z|, where z represents a complex number.
This is my attempt at a solution:
Using the triangle inequality, we can write:
$||z|-|2/z||≤|z+2/z|≤|z|+|2/z|$
Let $|z|=r$ which implies... | Since $|z-2/z| = 1$, we have
$$\begin{aligned}
1 = |z- 2/z|^2 &= (z - 2/z)(\overline{z} - 2/\overline{z}) \\
&= |z|^2 - 2z/\overline{z} - 2\overline{z}/z + 4/|z|^2
\end{aligned}$$
Write $z = re^{i\theta}$. Then $\overline{z} = re^{-i\theta}$. Substituting into the above, we obtain
$$\begin{aligned}
1 &= r^2 - 2e^{i2\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
When does this equation have positive roots? I am reading a paper where the author states (without proofing) that the equation
$$
{x}^2 \left(1-2 \sqrt{2}\, {x} \right) \sin ^2(\psi )-{y}^2
\left(1-2 \sqrt{2} \,{y} \right)=0
$$
has no positive roots for $y<3\sqrt{2}$ unless $\sin(\psi)>3\sqrt{2}\,y\sqrt{3(1-2\sqrt{2... | Suppose $y<\frac{1}{2\sqrt{2}}$ is fixed. Then define
\begin{align*}
f(x)&={x}^2 \left(1-2 \sqrt{2}\, {x} \right) \sin ^2(\psi )-{y}^2
\left(1-2 \sqrt{2} \,{y} \right)\\
f'(x)&=\left(2x -6 \sqrt{2}\, x^2 \right) \sin ^2(\psi )
\end{align*}
Since $f$ is concave down the maximum value occurs at $x_0=\frac{1}{3 \sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$ This is a problem that I tried to solve and didn't come up with any ideas
.?$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n}}.$$
All I get is $\frac{1}{\sqrt{2n}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+2... | This proof true when $\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}.$
$$\frac { 1 }{ 2 } \cdot \frac { 3 }{ 4 } \cdots \frac { 2n-1 }{ 2n } \frac { 2n+1 }{ 2n+2 } <\frac { 1 }{ \sqrt { 2n+1 } } .\frac { 2n+1 }{ 2n+2 } =\frac { 1 }{ \sqrt { 2n+3 } } .\frac { \sqrt { 2n+3 } }{ \sqrt { 2n+1 } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting
$$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$
but to no avail. Could someone point me in the right direction?
| Intro: this is the business about taking the gcd of a polynomial and its derivative, in order to detect a repeated factor. In this case, the repeated factor just squares to give the original, so this is one of the simplest types. For this answer yesterday, How can one find the factorization $a^4 + 2a^3 + 3a^2 + 2a + 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 2
} |
How to determine that the solution to the following problem is unique? The question is:
[BMO2 2000 Q3] Find positive integers a and b such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20\sqrt[3]{6}$$
It suffices to find one solution to gain full marks, which I did by expanding and assuming that neither $a$ nor $b... | I apologize, this solution is a bit hand-wavy and (it may in fact be wrong) as my skills/knowledge in this topic is rather weak. Perhaps someone else can later provide a more rigorous version of this argument if it is correct.
We have,
$$\sqrt[3]{a^2}+\sqrt[3]{b^2}-2(\sqrt[3]{a}+\sqrt[3]{b}) = 20\sqrt[3]{6}.$$
Consider... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Solve $\frac{xdx-ydy}{xdy-ydx}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$
Solve $\dfrac{xdx-ydy}{xdy-ydx}=\sqrt{\dfrac{1+x^2-y^2}{x^2-y^2}}$
I tried doing this-
$\sqrt{x^2-y^2}d(x^2-y^2)=x^2\sqrt{1+x^2-y^2}d(\frac{x}{y})$ but it does not get better from here.
I even tried putting $\sqrt{x^2-y^2}$ as $u$ but that does not simp... | I would suggest to you a more intricate change of variables:
\begin{align}
x &= z \cosh{s} = z \,\left(\frac{e^{s} + e^{-s}}{2}\right)\\
y &= z \sinh{s} = z \left(\frac{e^{s} - e^{-s}}{2}\right)
\end{align}
Then
\begin{align}
dx = \cosh(s) \, dz + z\,\sinh(s)\, ds \\
dy = \sinh(s) \, dz + z\,\cosh(s)\, ds
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
From $\lim_{n\to\infty}n\sum_{k=1}^\infty\int_0^1\frac{dx}{(1+n^2x^2)(k+x)^3}$ to a multiple of $\zeta(3)$ This afternoon I tried do the specialisation of another problem due to Furdui. THat is PROBLEMA 103, La Gaceta de la RSME, Volumen 12, número 2 (2009), see the first identity in page 317 of the solution (in spanis... | Using partial fraction decomposition $$\frac{1}{(1+n^2x^2)(k+x)^3}$$ write $$\frac{n^2 \left(3 k^2 n^2-1\right)}{\left(k^2 n^2+1\right)^3 (k+x)}+\frac{2 k
n^2}{\left(k^2 n^2+1\right)^2 (k+x)^2}+\frac{1}{\left(k^2 n^2+1\right)
(k+x)^3}+\frac{\left(k^3 n^6-3 k n^4\right)+x \left(n^4-3 k^2 n^6\right)}{\left(k^2 n^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
In a triangle $ABC$, $AB = a-b$ and $BC = 2\sqrt{ab}$, then find $\angle B$? Is this question solvable?
In $\Delta ABC$, $AB = a-b$ and $BC = 2\sqrt{ab}$, then $\angle B$ is
(a) $\: 60^{\circ}$
(b) $\: 30^{\circ}$
(c) $\: 90^{\circ}$
(d) $\: 45^{\circ}$
| HINT:
What do we know from $\triangle\text{ABC}$:
*
*$$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ=\pi\space\text{radians}$$
*For the length $\text{AB}$:
$$\text{AB}=\text{a}-\text{b}$$
*For the length $\text{BC}$:
$$\text{BC}=2\cdot\sqrt{\text{a}\cdot\text{b}}$$
Using the law of sinus and the law of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Secondary school level mathematical induction
*
*It is given that
$$1^3+2^3+3^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$$
Then, how to find the value of
$2^3+4^3+\cdots+30^3$?
Which direction should I aim at?
*Prove by mathematical induction, that $5^n-4^n$ is divisible by 9 for all positive even numbers $n$.
$$5^n-... | 1) $2^3 + 4^3 + 6^3 + .... +30^3 = 2^3*1^3 + 2^3*2^3 + 2^3*3^3 + ... + 2^3*15^3 = 2^3(1^3 + 2^3 + .... + 25^3)$
2) Perfect. You did great. (Better than I did when I didn't realize it was only true for even numbers.)
3) If $n$ is odd then you don't want $a^{2n+1} + b^{2n+1}$ for the inductive step but either: $a^{n+2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Intuition for $\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3$ Trying to get some intuition behind why: $$ \lim_{x\to\infty}\sqrt{x^6-9x^3}-x^3=-\frac{9}{2}. $$
First off, how would one calculate this? I tried maybe factoring out an $x^3$ from the inside of the square root, but the remainder is not factorable to make anything ... | Hmmm, no one has pointed out the obvious:
$\lim \sqrt{x^6 - 9x^3} - x^3 = \lim \sqrt{x^6 - 9x^3 + 36/4} - x^3$
$ = \lim \sqrt{(x^3 - 9/2)^2} - x^3 = \lim x^3 - 9/2 - x^3 = -9/2$
Intuition.... hmm .... I guess realizing the $x^3$ from $\sqrt {x^6 + stuff}$ was going to cancel the $-x^3$. So I want some $\lim \sqrt {Y_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 2
} |
Reduce a fraction Given the function
$f(x)=\frac{x^2-5}{x+\sqrt{5}}$
If I draw this function in maple, I will get a line. How can that be true? I should expect a line except in area of $x = -\sqrt{5}$, where $f(x) \rightarrow \infty$ or $f(x) \rightarrow -\infty $.
Of course Maple has factorized the numerator and red... | Note that $x^2 - 5 = x^2 - \sqrt 5^2 = (x-\sqrt5)(x+\sqrt 5)$, so we have
$$
f(x) = \frac{x^2 - 5}{x + \sqrt 5} = \frac{(x-\sqrt 5)(x + \sqrt 5)}{x+\sqrt 5} = \frac{x-\sqrt 5}{1} = x-\sqrt 5
$$
However, this manipulation does not suddenly make $f$ defined when $x = -\sqrt 5$. The only thing it does is that it simplifie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to factorise this expression $ x^2-y^2-x+y$ This part can be factorised as $x^2-y^2=(x+y)(x-y)$, How would the rest of the expression be factorised ?
:)
| Another method to factor $${ x }^{ 2 }-x+\frac { 1 }{ 4 } -{ y }^{ 2 }+y-\frac { 1 }{ 4 } ={ \left( x-\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( y-\frac { 1 }{ 2 } \right) }^{ 2 }=\left( x-\frac { 1 }{ 2 } +y-\frac { 1 }{ 2 } \right) \left( x-\frac { 1 }{ 2 } -y+\frac { 1 }{ 2 } \right) =\left( x-y \right) \left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Set of values of $a$ for which function always increases If the set of all values of the parameter $a$ for which the function $$f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$$ increases for all $x \in R$ and has no critical point for all $x \in R$ is $(- \infty, m- \sqrt n) \cup (\sqrt n, \infty)$, then find the value of ... | Taking the derivative
$$f'(x) = 2\cos(2x) - 8(a+1)\cos x + (4a^2 + 8a -14)$$
Now both $\cos (2x)$ and $\cos x$ oscillate between $-1$ to $1$. Thus assuming $a \ge -1$,
$$-2 - 8(a+1) + 4a^2 + 8a - 14 \le f'(x) \le 2 + 8(a+1) + 4a^2 + 8a - 14$$
$$4a^2 - 24 \le f'(x) \le 4a^2 + 16a - 4$$
$$4(a^2 - 6) \le f'(x) \le 4(a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Absolute value inequality with variable on both sides I am trying to solve the following inequality:
$$|3-5x| \le x$$
I am not familiar with inequalities including one absolute value with variables on both sides. I tried to solve it as follows:
$$-x \le 3-5x \le x$$
Then I solved for each side separately,as follows... | You need to break up the absolute value into its intervals:
$$
|x| = \begin{cases}
x & x > 0 \\
-x & x < 0 \\
0 & x = 0
\end{cases}
$$
Therefore for $|3-5x|$ you need to find the interval when it's less than zero and when it's greater than zero:
$$
|3-5x| = \begin{cases}
3 - 5x & 3 - 5x > 0 \rightarrow 3 > 5x \rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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.